NCERT Exemplar Solutions for Class 10 Maths Chapter 7 Co-Ordinate Geometry

Class 10 Maths Chapter 7 Co-Ordinate Geometry

(1) Vertices of the given triangle are

Co-Ordinate Geometry

  • Introduction: Co-ordinate geometry is the branch of mathematics in which geometry is studied algebraically, i.e., In which geometrical figures (as points, lines etc.) are studied using equations.
  • Rene Descartes (1596-1665), a French mathematician was the first who introduced the Co-ordinate Geometry or Analytical Geometry or Cartesian Geometry.
  • In class IX, we have studied how to locate the position of a point on a plane. Now, we will study to find the distance between two points, the section formula and the area of a triangle.

NCERT Exemplar Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Distance Between Two Points

Let X’OX and Y’OY be coordinate axes. P and Q are two points in this cartesian plane with coordinates (X1, Y1) and (X2, Y2) respectively.

Co Ordinate Geometry Distance Between Two Objects

PL and QM are perpendiculars from P and Q respectively to the X-axis. PN is perpendicular from P to QM.

Now, O L =x1, P L=y1

O M =x2, Q M=y1

P N =L M

=O M-O L=x2-x1

P N=L M

and Q N =Q M-M N=Q M-P L

=y2 – y1

⇒ \(\triangle P Q N\) is a right-angled triangle.

⇒ \(P Q^2=P N^2+Q N^2\)

⇒ \(P Q^2=\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2\)

⇒ \(P Q=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

Distance Between Origin And Point \(x_1, y_1\)

Distance between origin and point \(\left(x_1, y_1\right)=\sqrt{\left(x_1-0\right)^2+\left(y_1-0\right)^2}=\sqrt{x_1^2+y_1^2}\)

Condition Of Collinear Points On The Basis Of Distance

If the sum of any two distances is equal to the third distance, then the three points will be collinear.

Solved Examples

Example 1. Find the distance between the following points :

  1. (3, 4) and (5, 2)
  2. (0, 2) and (4, – 1)
  3. (a, 2a) and (- a, – 2a)
  4. (4, – 3) and (- 6, 5)

Solution.

(1) Distance between the points (3,4) and (5,2)

=\(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{(5-3)^2+(2-4)^2}\)

=\(\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}\) units

(2) Distance between the points (0,2) and (4,-1)

=\(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{(4-0)^2+(-1-2)^2}\)

=\(\sqrt{(4)^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}\)=5 units

(3) Distance between the points (n, 2a) and (-r,-2 a)

=\(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{(-a-a)^2+(-2 a-2 a)^2}\)

=\(\sqrt{(-2 a)^2+(-4 a)^2}=\sqrt{4 a^2+16 a^2}=\sqrt{20 a^2}=2 \sqrt{5} a \)units

(4) Distance between the points (4,-3) and (-6,5)

=\(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{(-6-4)^2+(5+3)^2}\)

=\(\sqrt{(-10)^2+(8)^2}=\sqrt{100+64}=\sqrt{164}=2 \sqrt{41}\)units

Example 2. Find the distance between the points (5, 8) and (- 3, 2).

Solution:

Distance between the points (5, 8) and (- 3, 2).

=\(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{(-3-5)^2+(2-8)^2}\)

=\(\sqrt{(-8)^2+(-6)^2}=\sqrt{64+36}=\sqrt{100}\)=10 units

The distance between the points =10 units

Example 3. Find the distance of the point (a cos \(\theta\), a sin \(\theta\)) from the origin.

Solution:

Distance between the points {a cos \(\theta\), a sin θ) and origin (0, 0)

= \(\sqrt{(a \cos \theta-0)^2+(a \sin \theta-0)^2}\)

= \(\sqrt{a^2 \cos ^2 \theta+a^2 \sin ^2 \theta}=\sqrt{a^2\left(\cos ^2 \theta+\sin ^2 \theta\right)}\)

= \(\sqrt{a^2} (\cos ^2 \theta+\sin ^2 \theta=1)\)

= a units

The distance of the point = a units

Example 4. Find the distance of the point (3, 4) from the origin.

Solution:

Distance of the point (3, 4) to the origin

= \(\sqrt{(3-0)^2+(4-0)^2}=\sqrt{9+16}=\sqrt{25}\)=5 units

The distance of the point (3, 4) from the origin=5 units

Example 5. If the distance between the points (x, 2) and (6, 5) is 5 units, find the value of x.

Solution:

Distance between the points (x, 2) and (6, 5)

=\(\sqrt{(6-x)^2+(5-2)^2}=\sqrt{x^2-12 x+36+9}=\sqrt{x^2-12 x+45}\)

Given that, \(\sqrt{x^2-12 x+45}\)=5

⇒ \(x^2-12 x+45=25\)

⇒ \(x^2-12 x+20=0\)

⇒ \(x^2-2 x-10 x+20=0\)

x(x-2)-10(x-2)=0

(x-2)(x-10)=0

x-2=0 or x-10=0

x = 2 or x = 10

The value of x 2 or 10

Example 6. If the distances of P(x,y) from A(5, 1) and B(-1,5) are equal, then prove that 3x = 2y.

Solution:

Since P(x,y) is equidistant from A(5, 1) and B(-1, 5).

PA = PB

⇒ \(\sqrt{(x-5)^2+(y-1)^2}=\sqrt{(x+1)^2+(y-5)^2}\) (by using distance formula)

Squaring both sides, we get

⇒ \((x-5)^2+(y-1)^2=(x+1)^2+(y-5)^2\)

⇒ \(x^2-10 x+25+y^2-2 y+1=x^2+2 x+1+y^2-10 y+25\)

-10 x-2 y+26=2 x-10 y+26

-10 x-2 x=-10 y+2 y

12 x=8 y

3 x = 2 y

Hence Proved.

Example 7. Prove that the points (5, -2), (-4, 3) and (10, 7) are the vertices of an isosceles right-angled triangle.

Solution:

Given

(5, -2), (-4, 3) and (10, 7)

Let the points are A (5, – 2), B (- 4, 3) and C (10, 7).

Therefore,\(A B^2=(-4-5)^2+(3+2)^2=(-9)^2+(5)^2=81+25=106\)

⇒ \(B C^2=(10+4)^2+(7-3)^2=(14)^2+(4)^2\)=196+16=212

\(A C^2=(10-5)^2+(7+2)^2=(5)^2+(9)^2\)=25+81=106

A B=A C=\(\sqrt{106}\)

and \(A B^2+A C^2=B C^2\)

⇒ \(\triangle A B C\) is an isosceles right-angled triangle.

Hence Proved.

Example 8. Prove that the points (a, a),(-a,-a) and \((-a \sqrt{3}, a \sqrt{3})\) are the vertices of an equilateral triangle.

Solution:

Given

(a, a),(-a,-a) and \((-a \sqrt{3}, a \sqrt{3})\)

Let the points are A(a, a), B(-a,-a) and \(C(-a \sqrt{3}, a \sqrt{3})\).

A B=\(\sqrt{(-a-a)^2+(-a-a)^2}=\sqrt{(-2 a)^2+(-2 a)^2}\)

= \(\sqrt{4 a^2+4 a^2}=\sqrt{8 a^2}=2 \sqrt{2} a\)

B C=\(\sqrt{(-a \sqrt{3}+a)^2+(a \sqrt{3}+a)^2}\)

=\(\sqrt{3 a^2+a^2-2 \sqrt{3} a^2+3 a^2+a^2+2 \sqrt{3} a^2}=\sqrt{8 a^2}=2 \sqrt{2} a\)

C A=\(\sqrt{(-a \sqrt{3}-a)^2+(a \sqrt{3}-a)^2}\)

=\(\sqrt{3 a^2+a^2+2 \sqrt{3} a^2+3 a^2+a^2-2 \sqrt{3} a^2}\)

=\(\sqrt{8 a^2}=2 \sqrt{2} a\)

A B =B C = C A

∴ \(\triangle A B C\) is an equilateral triangle.

Hence Proved.

Example 9. Prove that the points (2, – 1), (4, 1), (2, 3) and (0, 1) are the vertices of a square.

Solution:

Given

(2, – 1), (4, 1), (2, 3) and (0, 1)

Let the points are A(2, – 1), B(4, 1), C(2, 3) and D(0, 1).

⇒ \(A B^2 =(4-2)^2+(1+1)^2=4+4=8\)

A B =\(\sqrt{8}=2 \sqrt{2}\)

⇒ \(B C^2 =(2-4)^2+(3-1)^2=(-2)^2+(2)^2\)=4+4=8

B C =\(\sqrt{8}=2 \sqrt{2}\)

⇒ \(C D^2 =(0-2)^2+(1-3)^2=(-2)^2+(-2)^2\)=4+4=8

C D =\(\sqrt{8}=2 \sqrt{2}\)

⇒ \(D A^2 =(2-0)^2+(-1-1)^2=(2)^2+(-2)^2\)=4+4=8

D A =\(\sqrt{8}=2 \sqrt{2}\)

Co-ordinate Geometry

⇒ \(A C^2 =(2-2)^2+(3+1)^2\)=0+16=16

A C =\(\sqrt{16}=4\)

and \(B D^2 =(0-4)^2+(1-1)^2=16+0=16\)

B D =\(\sqrt{16}=4\)

Now, A B=B C=C D=D A and A C=B D

⇒  A B C D is a square.

Example 10. Show that the points A(- 3, 3), B(7, – 2) and C(l, 1) are collinear.

Solution:

Given

A(- 3, 3), B(7, – 2) and C(l, 1)

A B =\(\sqrt{(7+3)^2+(-2-3)^2}=\sqrt{(10)^2+(-5)^2}\)

=\(\sqrt{100+25}=\sqrt{125}=5 \sqrt{5}\)

B C =\(\sqrt{(1-7)^2+(1+2)^2}=\sqrt{(-6)^2+(3)^2}\)

= \(\sqrt{36+9}=\sqrt{45}=3 \sqrt{5}\)

and \(A C =\sqrt{(1+3)^2+(1-3)^2}=\sqrt{(4)^2+(-2)^2}\)

= \(\sqrt{16+4}=\sqrt{20}=2 \sqrt{5}\)

Now, A C+B C=2 \(\sqrt{5}+3 \sqrt{5}=5 \sqrt{5}=A B\)

Points A, B and C are collinear.

Example 11. Show that the points (9, – 2), (- 5, 12) and (- 7, 10) lie on that circle whose centre is the point (1,4).

Solution:

Given

(9, – 2), (- 5, 12) and (- 7, 10)

Let the given points are A (9, -2), B (-5, 12) and C (- 7, 10),

If point‘O’is (1, 4), then

O A =\(\sqrt{(1-9)^2+(4+2)^2}=\sqrt{(-8)^2+(6)^2}\)

=\(\sqrt{64+36}=\sqrt{100}=10\)

O B =\(\sqrt{(1+5)^2+(4-12)^2}=\sqrt{(6)^2+(-8)^2}\)

= \(\sqrt{36+64}=\sqrt{100}=10\)

O C =\(\sqrt{(1+7)^2+(4-10)^2}=\sqrt{(8)^2+(-6)^2}\)

= \(\sqrt{64+36}=\sqrt{100}=10\)

OA = OB = OC

Point ‘O’ is equidistant from the points A, B and C.

Point (1,4) is the centre of that circle at which the points (9, -2), (-5, 12) and (- 7, 10) lie.

Hence Proved.

Example 12. In the given figure, \(\triangle\)ABC is an equilateral triangle of side 3 units. Find the coordinates of the other two vertices.

Solution:

Given

In the given figure, \(\triangle\)ABC is an equilateral triangle of side 3 units.

Since B is at a distance of 3 units from A on the X-axis in the positive direction, so B will be 5 units away from the origin.

Co Ordinate Geometry The Coordinates Of The Other Two Vertices

So, B = (5, 0)

Let M be the mid-point of AB

⇒ \(A M=\frac{1}{2} A B=\frac{3}{2}\) units

AC = 3 Units

In right \(\triangle C M A\), by Pythagoras theorem

Co Ordinate Geometry The Coordinates Of The C By Using Pythagoras Theroem

⇒ \(C M^2 =A C^2-\Lambda M^2\)

= \(9-\frac{9}{4}=\frac{27}{4}\)

⇒  \(C M =\sqrt{\frac{27}{4}}=\frac{3 \sqrt{3}}{2}\)

So, the coordinates of C are (O M, M C)

C=(O A+A M, M C) .

C=\(\left(2+\frac{3}{2}, \frac{3 \sqrt{3}}{2}\right)=\left(\frac{7}{2}, \frac{3 \sqrt{3}}{2}\right)\)

Example 13. The coordinates of two vertices of an equilateral triangle are (0, 0) and (3,\(\sqrt{3}\)). Find the coordinates of the third vertex of the triangle.

Solution:

Given

The coordinates of two vertices of an equilateral triangle are (0, 0) and (3,\(\sqrt{3}\)).

Let the \(\triangle\) ABC be an equilateral triangle in which the coordinates of points B and C are (0, 0) and (3,3, \(\sqrt{3})\) respectively.

Let the coordinates of the third vertex A be (x,y).

Co Ordinate Geometry The Co Ordinates Of The Third Vertex Of The Triangle

In equilateral \(\triangle\)ABC

A B=A C=B C

⇒ \(A B^2=A C^2=B C^2\)

⇒ \(A B^2=A C^2\)

⇒ \((x-0)^2+(y-0)^2=(x-3)^2+(y-\sqrt{3})^2 \quad B(0,0)\)

⇒ \(x^2+y^2=x^2-6 x+9+y^2-2 \sqrt{3} y+3\)

⇒ \(6 x+2 \sqrt{3} y=12 \quad \Rightarrow \quad 3 x+\sqrt{3} y=6 \)

⇒ \(\sqrt{3} x+y=2 \sqrt{3} \quad y=2 \sqrt{3}-\sqrt{3} x\)

and \(A B^2=B C^2\)

⇒ \((x-0)^2+(y-0)^2=(3-0)^2+(\sqrt{3}-0)^2\)

⇒ \(x^2+y^2=9+3\)

⇒ \(x^2+(2 \sqrt{3}-\sqrt{3} x)^2=12\)

⇒ \(x^2+12+3 x^2-12 x=12\)

∴ \(4 x^2-12 x\)=0

4 x(x-3)=0

x=0 or x-3=0

x=0 or x=3

Put these values in eq. (1)

x=0, then y=2\( \sqrt{3}-0=2 \sqrt{3}\)

x=3 , y=2\( \sqrt{3}-3 \sqrt{3}=-\sqrt{3}\)

Co-ordinates of third vertex =(0,2 \(\sqrt{3}\)) or (\(3,-\sqrt{3})\)

Example 14. What point on the X-axis is equidistant from (7, 6) and (-3, 4)?

Solution:

Given

(7, 6) and (-3, 4)

We know that the y-co-ordinate of a point on the X-axis is always 0. So, let a point on the X-axis be P(x, 0) and let two given points be A(7, 6) and B(-3, 4).

According to the condition,

P A=P B

⇒ \( \sqrt{(x-7)^2+(0-6)^2}=\sqrt{(x+3)^2+(0-4)^2}\)

Squaring both sides, we have

⇒ \(x^2-14 x+49+36 =x^2+6 x+9+16\)

Required point is (3,0) 20 x =60 \(\Rightarrow x=3\)

Example 15. Find the equation of the set of all points which are twice as far from (3, 2) as from (1, 1).

Solution:

Given

(3, 2) and (1, 1)

Let P(x,y) be a point and let A(3, 2) and B = (1, 1) be two other points on the plane, such that

P A=2 P B

⇒ \(\sqrt{(x-3)^2+(y-2)^2}=2 \sqrt{(x-1)^2+(y-1)^2}\)

Squaring both sides, we have

⇒ \(x^2-6 x+9+y^2-4 y+4=4\left(x^2-2 x+1+y^2-2 y+1\right)\)

∴ \(3 x^2+3 y^2-2 x-4 y-5=0\) Which is the required equation.

Example 16. If A(2, 2), B(-2, -2), C(-2\(\sqrt{3}\),2\(\sqrt{3}\)) and D(-4 – 2\(\sqrt{3}\),4 + 2\(\sqrt{3}\)) are the co-ordinates of 4 points. What can be said about these four points?

Solution:

Given

(2, 2), B(-2, -2), C(-2\(\sqrt{3}\),2\(\sqrt{3}\)) and D(-4 – 2\(\sqrt{3}\),4 + 2\(\sqrt{3}\)) are the co-ordinates of 4 points.

A B=\(\sqrt{(-2-2)^2+(-2-2)^2}=4 \sqrt{2}\) units .

B C=\(\sqrt{(-2+2 \sqrt{3})^2+(-2-2 \sqrt{3})^2}\)

=\(\sqrt{4+12-8 \sqrt{3}+4+12+8 \sqrt{3}}=4 \sqrt{2}\) units

C D=\(\sqrt{(-2 \sqrt{3}+4+2 \sqrt{3})^2+(2 \sqrt{3}-4-2 \sqrt{3})^2}\)

= \(\sqrt{16+16}=4 \sqrt{2}\) units

A C=\(\sqrt{(2+2 \sqrt{3})^2+(2-2 \sqrt{3})^2}\)

=\(\sqrt{4+12+8 \sqrt{3}+4+12-8 \sqrt{3}}=4 \sqrt{2}\) units

A D=\(\sqrt{(2+4+2 \sqrt{3})^2+(2-4-2 \sqrt{3})^2}\)

=\(\sqrt{36+12+24 \sqrt{3}+12+4+8 \sqrt{3}}=\sqrt{64+32 \sqrt{3}}\) units

B D=\(\sqrt{(-2+4+2 \sqrt{3})^2+(-2-4-2 \sqrt{3})^2}\)

=\(\sqrt{4+12+8 \sqrt{3}+36+12+24 \sqrt{3}}=\sqrt{64+32 \sqrt{3}}\) units

Here, AB = BC = CD = AC and also, AD = BD

So, in first view, it seems to be the vertices of a square.

(But),Here, AB, BC, CD and DA are not equal, (the order of A, B, C and D must be cyclic in the case of the square).

Also, AD and BD are equal but they cannot be diagonals. So, they do not form a square. Actually, A, B and D lie on a circle with C as the centre (as CA = CB = CD i.e., C is equidistant from A, B and D).

To Divide A Line Segment In a Given Ratio

Internal Division

The coordinates of a point which divides the line segment joining the points A(x1,y1) and B(x2,y2) in the ratio m: n internally are

⇒ \(\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}\right)\)

Co Ordinate Geometry To Divide A Line Segment In Given Ratio By Internal Division

Proof: Let X’OX and Y’OYare the coordinate axes and A (x1,y1) and B (x2,y2) be any two points in this cartesian plane. Let P(x,y) be any point on line segment AB which divides AB in the ratio m: n.

⇒ \(\frac{A P}{B P}=\frac{m}{n}\)

AL, PN and BM are perpendiculars from, P and B respectively on X-axis.

AH and PIC are perpendiculars from and P respectively to PN and BM.

Now, OL = x1, OM = x2, ON = x

AL = y1, BM = y2> PN =y

AH = LN = ON -OL=x-x

PH = PN – NH = PN – AL =y -y1

Pk = NM = OM – ON =x2-x

and BIC = BM – ICM = BM – PN =y2 -y

Here, \(\triangle\)AHP and \(\triangle\)PKB are similar.

⇒  \(\frac{A H}{P K}=\frac{P H}{B K} =\frac{A P}{P B}\)

⇒  \(\frac{x-x_1}{x_2-x}=\frac{y-y_1}{y_2-y}=\frac{m}{n}\)

⇒  \(\frac{x-x_1}{x_2-x} =\frac{m}{n}\) and \(\frac{y-y_1}{y_2-y}=\frac{m}{n}\)

⇒  \(\frac{x-x_1}{x_2-x} =\frac{m}{n} \quad \Rightarrow \quad n x-n x_1=m x_2-m x\)

⇒  \(m x+n x =m x_2+n x_1 \quad \Rightarrow \quad(m+n) x=m x_2+n x_1\)

x =\(\frac{m x_2+n x_1}{m+n}\)

Similarly,\(\frac{y-y_1}{y_2-y}=\frac{m}{n} \quad \Rightarrow \quad n y-n y_1=m y_2-m y\)

⇒  \(m y+n y=m y_2+n y_1 \quad \Rightarrow \quad(m+n) y=m y_2+n y_1\)

y=\(\frac{m y_2+n y_1}{m+n}\)

Therefore, the co-ordinates of point P=\(\left(\frac{m x_2+m x_1}{m+n}, \frac{m y_2+m y_1}{m+n}\right)\)

Co-ordinates of the Mid-point of a Line Segment

The co-ordinates of the mid-point of the line segment joining the points A(x1,y1) and (x2,y2) are \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)

Co Ordinate Geometry Co Ordinate Midline Of A Segment

Proof: Let P (x, y) be the mid-point of the line joining the points A (x1,y1) and B (x2, y2)

Ap : PB = 1:1

x =\(\frac{1\left(x_2\right)+1\left(x_1\right)}{1+1}=\frac{x_1+x_2}{2}\)

y =\(\frac{1\left(y_2\right)+1\left(y_1\right)}{1+1}=\frac{y_1+y_2}{2}\)

Co-ordinates of point P =\(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)

Solved Examples

Example 1. Find the coordinates of a point which divides the line segment joining the points (5, 1) and (-10, 11) in the ratio 2 : 3 internally.

Solution:

Given

(5, 1) and (-10, 11)

Let the coordinates of the required point he (x,y).

Here, \(\quad\left(x_1, y_1\right)=(5,1)\) and \(\left(x_2, y_2\right)=(-10,11)\)

and \(\left(x_1, y_1\right)\) =(5,1) and \(\quad\left(x_2, y_2\right)=(-10,11)\)

m: n = 2: 3

x =\(\frac{m x_2+n x_1}{m+n}=\frac{2(-10)+3(5)}{2+3}=\frac{-20+15}{5}=\frac{-5}{5}=-1\)

y =\(\frac{m y_2+n y_1}{m+n}=\frac{2(11)+3(1)}{2+3}=\frac{22+3}{5}=\frac{25}{5}=5\)

Co-ordinates of required point =(-1,5)

Example 2. Suppose point P lies on the line segment joining points A{-3, 4) and B(- 2, – 6) such that 2AP=3BP then, find the coordinates of point P.

Solution:

A{-3, 4) and B(- 2, – 6) such that 2AP=3BP

Given That, 2 AP = 3 BP

⇒  \(\frac{A P}{B P}=\frac{3}{2}\)

m : n = 3 : 2

(x1, y1) = (- 3, 4) and (x2,y2) = (-2,- 6)

Now, let the coordinates of point P are (x,y)

x=\(\frac{3(-2)+2(-3)}{3+2}=\frac{-12}{5}\)

and y=\(\frac{3(-6)+2(4)}{3+2}=-2\)

Co-ordinates of point P=\(\left(\frac{-12}{5},-2\right)\)

Example 3. Find the co-ordinates of the mid-point of the line segment joining the points A(3, -5) and B(1, 1 )

Solution:

Co-ordinates of the mid-point of AB = \((-1 + 3 5 + (-1)^ \left(\frac{3+1}{2}, \frac{-5+1}{2}\right)=(2,-2)\)

Example 4. The coordinates of the endpoints of a diameter are (-1, 5) and (3, -1). Find the coordinates of the centre and the radius of the circle.

Solution:

Given

The coordinates of the endpoints of a diameter are (-1, 5) and (3, -1).

Let the coordinates of the end points of diameter AB be A (-1, 5) and .8 (3, -1).

Co-ordinates of the centre P = Co-ordinates of mid-point of AB

= \(\left(\frac{-1+3}{2}, \frac{5+(-1)}{2}\right)=(1,2)\)

and radius of circle = length of P A

= \(\sqrt{(1+1)^2+(2-5)^2}=\sqrt{4+9}=\sqrt{13}\) units

Example 5. The coordinates of the mid-point of the line joining the points A and B are (2, -3). If the coordinates of point A are (-3, 4), then find the coordinates of point B.

Solution:

Given

The coordinates of the mid-point of the line joining the points A and B are (2, -3). If the coordinates of point A are (-3, 4),

Let the co-ordinates of point B=\(\left(x_2, y_2\right)\)

Given that,\(\frac{-3+x_2}{2}=2 \quad \Rightarrow \quad-3+x_2=4\)

⇒ \(x_2=7\) and \(\frac{4+y_2}{2}=-3\)

4+y_2=\(-6 \quad \Rightarrow \quad y_2=-10\)

Co-ordinates of point B=(7,-10)

Example 6. Find the ratio in which the X-axis divides the line segment joining the points (8, 5) and (-3,-7).

Solution:

Let the X-axis divide the join of points (8, 5) and (- 3, – 7) in the ratio k: 1.

We know that at X-axis

y = 0

⇒ \(\frac{k \cdot y_2+1 \cdot y_1}{k+1} =0\)

Co Ordinate Geometry The Ratio Of X Axis Divides The Line Segment Joining The Points

⇒ \(\frac{k(-7)+1(5)}{k+1}\) =0

-7 k+5 =0

k = \(\frac{5}{7}\)

Required ratio =5: 7

Example 7. In what ratio does the point \(\left(\frac{24}{11}, y\right)\) divide the line segment joining the points P(2,-2) and Q(3, 7)? Also, find the value of y.

Solution:

Let \(M\left(\frac{24}{11}, y\right)\) divide the line segment joining the points.

P(2,-2) and Q(3,7) in the ratio k: 1.

⇒ \(\frac{24}{11}=\frac{k(3)+1(2)}{k+1}\) (by using section formula)

⇒ \(11(3 k+2)=24(k+1) \quad \Rightarrow \quad 33 k+22=24 k+24\)

33 k-24 k=24-22 \(\quad \Rightarrow \quad 9 k=2\)

⇒ \(k=\frac{2}{9}\)

Required ratio =k: 1

i.e.,\(\frac{2}{9}: 1\)

i.e.,2: 9 internally.

Required ratio = 2:9

Example 8. The co-ordinates of the vertices of \(\triangle\)ABC are A(3, 2), B( 1, 4) and C(-1, 0). Find the length of the median drawn from point A.

Solution:

Given

The co-ordinates of the vertices of \(\triangle\)ABC are A(3, 2), B( 1, 4) and C(-1, 0).

Let AP be the median drawn from vertex A.

The midpoint of BC is P.

Co Ordinate Geometry The Length Of Median Drawn From The Point A

Now, the coordinates of P

=\(\left(\frac{1+(-1)}{2}, \frac{4+0}{2}\right)=(0,2)\)

A P =\(\sqrt{(3-0)^2+(2-2)^2}\)

=\(\sqrt{9+0}=3 units\)

Example 9. Find the coordinates of the points of trisection of the line joining the points (3, -2) and (-3, -4).

Solution:

Let P (a, b) and Q (c, d) trisect the line joining the points A(3, -2) and B(-3, -4).

Co Ordinate Geometry The Coordinates Of The Points Of Trisection Of The Line Segment

Now, point P (a, b) divides the line AB in the ratio 1:2

a =\(\frac{1(-3)+2(3)}{1+2}=\frac{3}{3}=1\)

b =\(\frac{1(-4)+2(-2)}{1+2}=-\frac{8}{3}\)

Therefore, coordinates of point P=\(\left(1,-\frac{8}{3}\right)\)

Q(c, d) divides the line A B in the ratio 2: 1.

c=\(\frac{2(-3)+1(3)}{2+1}=-1\)

d=\(\frac{2(-4)+1(-2)}{2+1}=\frac{-10}{3}\)

Therefore, co-ordinates of Q=\(\left(-1, \frac{-10}{3}\right)\)

Co-ordinates of points of trisection of A B=\(\left(1, \frac{-8}{3}\right) and \left(-1, \frac{-10}{3}\right)\)

Example 10. If two adjacent vertices of a parallelogram are (3, 2) and (-1, 0) and telagona intersect at (2, -5), then find the coordinates of the other two vertices.

Solution:

Given

Two adjacent vertices of a parallelogram are (3, 2) and (-1, 0) and telagona intersect at (2, -5)

Let two adjacent vertices of a parallelogram be A = (3, 2) and B(-1,0).

Co Ordinate Geometry Two adjacent Vertices Of A Parallelogram And The Diagonals Intersect

Let the coordinates of the other two vertices be C(X1, y1) and D(x2, y2)

We know that diagonals of a parallelogram bisect each other.

Mid-point of AC and the mid-point of BD are the same i.e., point O(2, -5).

⇒ \(\frac{3+x_1}{2}=2 and \frac{2+y_1}{2}\)=-5

⇒ \(x_1=1 and y_1=-12 \Rightarrow C \equiv\left(x_1, y_1\right) \equiv(1,-12)\)

Also, \(\frac{x_2-1}{2}=2 and \frac{y_2+0}{2}=-5\)

x2=5 and y2=-10 ⇒ D = (5,-10)

Hence, the remaining vertices are (1,-12) and (5,-10).

Example 11. The coordinates of three consecutive vertices of a parallelogram are (-1,0), (3, 1) and (2, 2). Find the coordinates of the fourth vertex of the parallelogram.

Solution:

Given

The coordinates of three consecutive vertices of a parallelogram are (-1,0), (3, 1) and (2, 2).

Let A(-1, 0), B (3, 1), C (2, 2) and D (x,y) be the vertices of a parallelogram ABCD.

We know that the diagonals of a parallelogram bisect each other.

Co-ordinates of the mid-point of AC = Co-ordinates of the mid-point of BD.

⇒  \(\left(\frac{-1+2}{2}, \frac{0+2}{2}\right)=\left(\frac{3+x}{2}, \frac{1+y}{2}\right)\)

⇒ \(\left(\frac{1}{2}, 1\right)=\left(\frac{3+x}{2}, \frac{1+y}{2}\right)\)

⇒ \(\frac{1}{2}=\frac{3+x}{2} and \quad 1=\frac{1+y}{2}\)

1=3+x and 2=1+y

x=-2 and y=1

Co-ordinates of fourth vertex =(-2,1)

Example 12. In which ratio the line y-x+2 = 0 Divides the line segment joining the points (3, -1) and (8,9)

Solution:

Let the ratio the line y-x+2 = 0 Divides the line segment joining the points (3, -1) and (8,9) in the ratio k: 1

Co-ordinates of P \(\equiv\left(\frac{8 k+3}{k+1}, \frac{9 k-1}{k+1}\right)\)

Co Ordinate Geometry Ratio Of The Line Divides The Line Segment Joining The Points

but this point P lies on the line y-x+2=0

⇒\(\frac{9 k-1}{k+1}-\frac{8 k+3}{k+1}+2\) =0

9 k-1-8 k-3+2 k+2 =0

3 k =2

k =\(\frac{2}{3}\)

Required ratio =\(\frac{2}{3}: 1=2: 3\)

Example 13. Find a point on the line through A{5, – 4) and B(-3, 2), that is, twice as far from A as from B.

Solution:

Given

A{5, – 4) and B(-3, 2)

Let P(x,y) be a point on AB such that

Co Ordinate Geometry A Point Of The Line Through Twice As Far From A As B

PA = 2PB

P A=2 P B

⇒  \(\frac{P A}{P B}=\frac{2}{1} \quad \Rightarrow \quad P A: P B=2: 1\)

So, by using the section formula for internal division,

x=\(\frac{m x_2+n x_1}{m+n}, y=\frac{m y_2+n y_1}{m+n}\)

x=\(\frac{2(-3)+1(5)}{2+1} \quad \Rightarrow \quad x=-\frac{1}{3}\)

and y=\(\frac{2(2)+1(-4)}{2+1} \Rightarrow \quad y=0\)

Required point =\(\left(-\frac{1}{3}, 0\right)\)

But, This is not the end of this question.

Think: Is it not possible that P(x, y) divides AB externally in the ratio 2: 1?

So, by using the section formula for external division,

x=\(\frac{2(-3)-1(5)}{2-1} \quad \Rightarrow x=-\frac{11}{1} \quad \Rightarrow \quad x=-11\)

and \( y=\frac{2(2)-1(-4)}{2-1} \Rightarrow y=\frac{8}{1} \quad \Rightarrow \quad y=8\)

So, the coordinates of p are (-11,8) also.

Hence, required points are \(\left(-\frac{1}{3}, 0\right)\) and (-11,8).

Example 14. Find the centroid of the triangle whose vertices are A(-1, 0), B{5, -2) and C(8, 2).

Solution:

Given

A(-1, 0), B{5, -2) and C(8, 2)

Centroid, the point where the medians of a triangle intersect, divides each median in the ratio 2: 1. Let AD be the median and G{x,y) be the centroid of \(\triangle\)ABC.

Co Ordinate Geometry The Centroid Of The Triangle Of The Vertices

D is the mid-point of BC

D =\(\left(\frac{5+8}{2}, \frac{-2+2}{0}\right)\) (mid-point formula)

=\(\left(\frac{13}{2}, 0\right)\)

Now, G(x, y) divides the line segment joining A(-1,0) and \(D(\frac{13}{2}\), 0) internally in the ratio 2: 1.

So, by using section formula,

x=\(\frac{2\left(\frac{13}{2}\right)+1(-1)}{2+1} \quad \Rightarrow \quad x=4\)

y=\(\frac{2(0)+1(0)}{2+1} \quad \Rightarrow \quad y=0\)

and Centroid of \(\triangle A B C\)=(4,0)

Example 15. A line intersects the Y-axis and X-axis at points P and Q respectively. If (2, -5) is the mid-point of PQ, then find the coordinates of P and Q.

Solution:

Given

A line intersects the Y-axis and X-axis at points P and Q respectively. If (2, -5) is the mid-point of PQ

We know that at Y-axis, the x-coordinate is zero and at X-axis, y-coordinate is zero.

Co Ordinate Geometry A Line Intersect The Y Axis And X Axis At The Points

So, let P = (0,y)

and Q = (x, 0)

Since M(2, -5) is the mid-point of PQ.

By using the mid-point formula,

\(\frac{0+x}{2} =2, \quad \frac{y+0}{2}\)=-5

x =4, y=-10

P = (0,-10) and Q=(4,0)

Example 16. Point P(h, k) divides a line segment between the axes in the ratio 1: 2. Find the lengths (intercepts) on the axes made by this line segment. Also, find the area of the triangle formed by the line segment and the axes.

Solution:

Given

Point P(h, k) divides a line segment between the axes in the ratio 1: 2.

Let AB be the line segment joining A (0,y) and B(x, 0) between the axes.

Co Ordinate Geometry The area Of The Triangle Formed By The Line Segment And The Axes

P(h, k) divides the line segment in the ratio 1: 2.

Now, question arises that whether PA : PB = 1 : 2 or PB : PA = 1:2

The answer is that we always take the former part of the ratio towards the X-axis and the latter part of the ratio towards the Y-axis.

So, here we will take PB: PA = 1: 2

By using the section formula,

h=\(\frac{1(0)+2(x)}{1+2} \Rightarrow h=\frac{2 x}{3} \quad \Rightarrow \quad x=\frac{3 h}{2}\)

and k=\(\frac{1(y)+2(0)}{1+2} \Rightarrow k=\frac{y}{3} \Rightarrow y\)=3 k

So, length of intercept on the X-axis =O B=x=\(\frac{3 h t}{2}\)

and the length of intercept on the Y-axis =O A=y=3 k.

Area of \(\triangle A O B=\frac{1}{2} \times O B \times O A=\frac{1}{2} \times \frac{3 h}{2} \times 3 k=\frac{9}{4}\) hk square units

Area Of Triangle:

If the coordinates of the vertices of a triangle are [x1,y1), (x2,y2) and (x3,y3), then the area of a triangle (A) is given by

⇒ \(\Delta=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)

Co Ordinate Geometry The Co Ordinates Of The Vertices Of The Triangle

or \(\Delta=\frac{1}{2}\left[\left(x_1 y_2+x_2 y_3+x_3 y_1\right)-\left(x_2 y_1+x_3 y_2+x_1 y_3\right)\right]\)

Proof: Let X’OX and Y’OY be co-ordinate axes.

Let the co-ordinates of the vertices of \(\triangle\)ABC are A (x1,y1), B (x2,y2) and C (x3,y3). Draw the 1 perpendicular AM, BN and CL from A, B and C. Y-axis.

□ ALMLC, □ CLNB and □ AMNB are trapezium.

Now, the Area of AABC

= Area of □ AMLC + Area of □ CLNB – Area of □ AMNB

= \(\frac{1}{2}(A M+L C) \times M L+\frac{1}{2}(C L+B N) \times L N-\frac{1}{2}(A M+B N) \times M N\)

=\(\frac{1}{2}\left(y_1+y_3\right)\left(x_3-x_1\right)+\frac{1}{2}\left(y_3+y_2\right)\left(x_2-x_3\right)-\frac{1}{2}\left(y_1+y_2\right)\left(x_2-x_1\right)\)

= \(\frac{1}{2}\left[\left(y_1+y_3\right)\left(x_3-x_1\right)+\left(y_3+y_2\right)\left(x_2-x_3\right)-\left(y_1+y_2\right)\left(x_2-x_1\right)\right]\)

=\(\frac{1}{2}\left[y_1 x_3-y_1 x_1+y_3 x_3-y_3 x_1+y_3 x_2-y_3 x_3+y_2 x_2-y_2 x_3-y_1 x_2+y_1 x_1-y_2 x_2+y_2 x_1\right] \)

=\(\frac{1}{2}\left[x_1 y_2-x_1 y_3+x_2 y_3-x_2 y_1+x_3 y_1-x_3 y_2\right]\)

=\(\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)

=\(\frac{1}{2}\left[\left(x_1 y_2+x_2 y_3+x_3 y_1\right)-\left(x_2 y_1+x_3 y_2+x_1 y_3\right)\right]\)

If the area of a triangle is negative then we neglect the negative sign, because the area of a triangle is always positive.

If it is given that the area of a triangle is 10 then it will be taken ± 10 for calculations to evaluate the value (s) of unknown terms.

Collinear Points

Three points A (x1, y1 )B (x2,y2) and C (x3, y3) are collinear if Area of \(\triangle\)ABC = 0

Note: If the order of description of the boundary is anticlockwise, then the area is considered to be positive, but if the order of description is clockwise, then the area is considered to be negative.

Co Ordinate Geometry Collinear Points In Anticlockwise And Clockwise Direction

Remark: To move from A to 6 and then from B to C, we are moving in a clockwise direction. So the above area comes out to be negative.

Co Ordinate Geometry Collinear Points In Clockwise And Anticlockwise Direction

Now, if we take the points in an anticlockwise direction, as A(2,1), C(6,3) and B(4,5) then

area =\(\frac{1}{2}[2(3-5)+6(5-1)+4(1-3)]\)

= \(\frac{1}{2}(-4+24-8)\)=+6 square units.

Solved Examples

Example 1. Find the area of the triangle, whose vertices are (2,1), (4,5) and (6, 3).

Solution:

Given

Vertices are (2,1), (4,5) and (6, 3)

Area of triangle = \(\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)

= \(\frac{1}{2}[2(5-3)+4(3-1)+6(1-5)]=\frac{1}{2}(4+8-24)\)=-6

But the area of a triangle cannot be negative

Area of triangle = 6 square units

Example 2. Find the area of a triangle, whose vertices are (2, 3), (7, 5) and (-7, -5).

Solution:

Given

Vertices are (2, 3), (7, 5) and (-7, -5)

  • Area of triangle = =\(\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\)]
  • = \(\frac{1}{2}[2(5+5)+7(-5-3)-7(3-5)]=\frac{1}{2}(20-56+14)\)=-11
  • But the area of a triangle cannot be negative
  • Area of triangle = 11 square units

Example 3. Find the area of the triangle, whose vertices are (a, c +a), (a,c) and (-a, c, -b).

Solution:

Given

Vertices are (a, c +a), (a,c) and (-a, c, -b)

  • Area of triangle =\(\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)
  • = \(\frac{1}{2}[a(c-c+a)+a(c-a-c-a)-a(c+a-c)]\)
  • = \(\frac{1}{n}\left[a^2-2 a^2-a^2\right]=-a^2\)
  • But the area of a triangle cannot be negative
  • Area of triangle = \(a^2\) square units

Example 4. Prove that the points (6, 4) (4, 5) and (2, 6) are collinear.

Solution:

Given

(6, 4) (4, 5) and (2, 6)

Area of triangle = \(\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)

⇒  \(\frac{1}{2}[6(5-6)+4(6-4)+2(4-5)]=\frac{1}{2}[-6+8-2]=0\)

Therefore, the given points are collinear.

Example 5. If the points A{x,y), B( 1, 4) and C(-2, 5) are collinear, then show that x + 3y = 13.

Solution:

Given points are collinear

Area of triangle = 0

⇒ \(\frac{1}{2}[x(4-5)+1(5-y)-2(y-4)]\) =0

-x+5-y-2 y+8 =0

x+3 y =13

Example 6. For what value of T, the points (k, 1), (1,-1) and (11,4) are collinear?

Solution:

The given points will be collinear if the area of the triangle = 0

⇒ \(\frac{1}{2}[k(-1-4)+1(4-1)+11(1+1)]\) =0

-5 k+3+22 =0

5 k =25

k =5

Example 7. If a \(\neq\) b \(\neq 0\), prove that the points \(\left(a, a^2\right),\left(b, b^2\right)\),(0,0) will not be collinear.

Solution:

Let the 3 points A=\(\left(a, a^2\right), B=\left(b, b^2\right)\) and C=(0,0) form a triangle ABC.

Area of triangle A B C = \(\frac{1}{2}\left|a\left(b^2-0\right)+b\left(0-a^2\right)+0\left(a^2-b^2\right)\right|\)

= \(\frac{1}{2}\left[a b^2-a^2 b\right]=\frac{1}{2} a b(b-a) \neq 0\)

Since a\((\triangle A B C) \neq 0\).

So, \(\triangle A B C\) will be formed.

Therefore 3 points A, B and C will not be collinear.

Hence Proved.

Example 8. If (x,y) is any point on the line segment joining the points (a, 0) and (0, b), then prove that \(\frac{x}{a}+\frac{y}{b}\)=1

Solution:

Given three points are collinear

Area of \(\triangle\)=0

⇒ \(\frac{1}{2}|x(0-b)+u(b-y)+0(y-0)|\)=0

-bx + ab – ay=0

Divide both sides by ab

–\(\frac{x}{a}+1-\frac{y}{b}=0 \quad \Rightarrow \quad \frac{x}{a}+\frac{y}{b}\)=1

Example 9. If P be a point equidistant from points A(3, 4) and B( 5, -2) and area \(\triangle\)PAB is 10 square units, then find the coordinates of point P.

Solution:

Let the coordinates of point P be (x,y).

Now, PA = PB

⇒ \(\sqrt{(x-3)^2+(y-4)^2}=\sqrt{(x-5)^2+(y+2)^2}\)

⇒ \(x^2+9-6 x+y^2+16-8 y=x^2+25-10 x+y^2+4+4 y^{\prime}\)

4 x-12 y=4

x-3 y=1  → Equation 1

and area of Δ=10

⇒ \(\frac{1}{2}[x(4+2)+3(-2-y)+5(y-4)]= \pm 10\)

⇒ \(6 r-6-3 y+5 y-20= \pm 20\)

6x + 2y – 26= \(\pm\) 20

6 x+2 y-26=20 or 6 x+2 y-26=-20

3 r+y=23  → Equation 2

or 3 r+y=3  →  Equation 3

From eqs. (1) and (2) From eqs. (1) and (3)

x=7, y=2 , x=1, y=0

Required co-ordinates =(7,2) or (1,0)

Example 10. The area of a triangle is 5 sq. units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is \(\left(\frac{7}{2}, y\right)\), find the value of y.

Solution :

Areat of a \(\triangle A B C\)=5 sq units (given)

Co Ordinate Geometry The Value Of Y By Using The Vertices

⇒ \(\frac{1}{2}\left|2(-2-y)+3(y-1)+\frac{7}{2}(1+2)\right|=5\)

⇒ \(\left|-4-2 y+3 y-3+\frac{21}{2}\right|=10\)

⇒ \(y+\frac{7}{2}= \pm 10\)

⇒ \(y+\frac{7}{2}=10\)

y=\(10-\frac{1}{7}\)or \(y=-10-\frac{1}{7}\)

y=\(\frac{13}{2}\) or y=\(-\frac{27}{2}\)

Hence, the value of y can be \(\frac{13}{2}\) and \(-\frac{27}{2}\).

Co-Ordinate Geometry Miscellaneous Examples

Example 1. If the point P(k – 1, 2) is equidistant from the points A(3, K) and B(k, 5), find the values of k.

Solution:

Given that. P A=P B

⇒ \((k-1-3)^2+(2-k)^2=(k-1-k)^2+(2-5)^2\)

⇒ \((k-4)^2+(2-k)^2=(-1)^2+(-3)^2\)

⇒ \(k^2-8 k+16+4+k^2-4 k=1+9\)

⇒ \(2 k^2-12 k+10=0\)

⇒ \(k^2-6 k+5=0\)

⇒ \(k^2-5 k-k+5=0\)

⇒ \(k(k-5)-1(k-5)=0\)

(k-5)(k-1)=0

k-5=0 or k-1=0

k=5 or k=1

The values of k is 5 or 1.

Example 2. Find the point on X-axis which is equidistant from the points (5, -2) and (-3, 2).

Solution:

Let the required point on the X-axis be P(x, 0) and the given points be (5, -2) and B(—3, 2).

Now, given that

P A =P B

⇒ \(P A^2=P B^2\)

⇒ \((x-5)^2+(0+2)^2 =(x+3)^2+(0-2)^2\)

⇒ \(x^2-10 x+25+4 =x^2+6 x+9+4\)

-16 x =-\(16 \quad \Rightarrow x\)=1

The required point is (1, 0).

Example 3. Points and B(5, 7) lie on a circle uadi centre 0(2, -3y). Find the values of y. Hence, find the die radius of the die circle.

Solution:

Given

Points and B(5, 7) lie on a circle uadi centre 0(2, -3y).

Here, OA = OB (radii of a circle)

O A =O B

O A² =O B²

Co Ordinate Geometry The Radius Of The Circle By Using The Value Y

⇒ \((2+1)^2+(-3 y-y)^2=(2-5)^2+(-3 y-7)^2\)

⇒ \(9+16 y^2=9+9 y^2+42 y+49\)

⇒ \(7 y^2-42 y-49=0\)

⇒ \(y^2-6 y-7=0\)

⇒ \(y^2-7 y+y-7=0\)

⇒ y(y-7)+1(y-7)=0

⇒ y-7=0 or y+1=0

y=7 or y=-1

Now, the co-ordinates of centre O are either (2, -21) or (2, 3) when the centre is 0(2, -21), then radius = OB

= \(\sqrt{(2-5)^2+(-21-7)^2}\)

= \(\sqrt{9+784}=\sqrt{793}\) units

When centre is 0(2, 3), then

radius = OB

= \(\sqrt{(2-5)^2+(3-7)^2}\)

= \(\sqrt{9+16}=\sqrt{25}\)=5 units

The die radius of the die circle =5 units

Example 4. The points A(-1, 7), B{p, 3) and C(7, 3) are the vertices of a right triangle, right-angled at B. Find the value of p.

Solution:

Given

The points A(-1, 7), B{p, 3) and C(7, 3) are the vertices of a right triangle, right-angled at B.

In \(\triangle\)ABC

⇒ \(\angle\)B = 90°

⇒ \(A B^2+B C^2=A C^2\)

⇒ \((p-4)^2+(3-7)^2+(p-7)^2+(3-3)^2=(7-4)^2+(3-7)^2\)

⇒ \(p^2-8 p+16+p^2-14 p+49+0=9\)

⇒ \(2 p^2-22 p+56=0\)

⇒ \(p^2-11 p+28=0\)

⇒ \(p^2-7 p-4 p+28=0\)

⇒ \(p(p-7)-4(p-7)=0\)

⇒ \((p-7)(p-4)=0\)

p-7=0 or p-4=0

p=7 or p=4

when p=7, then the points B and C coincide and so no triangle is formed.

⇒ \(p \neq 7\)

Hence, p=4

Example 5. Find the coordinates of the point of trisection of the line segment joining the points A(-5, 6) and B(4, -3).

Solution:

Let P and Q be the points of trisection of AB, and then P divides AB in the ratio 1:2.

Co-ordinates of point P \(\equiv\left(\frac{-5 \times 2+4 \times 1}{1+2}, \frac{6 \times 2+1 \times-3}{1+2}\right) \equiv(-2,3)\) and point Q divides A B in the ratio 2: 1.

Co-ordinates of point Q=\(\left(\frac{-5 \times 1+4 \times 2}{2+1}, \frac{6 \times 1+2 \times-3}{1+2}\right)\)=(1,0)

The coordinates of the point of trisection are (-2,3) and (1,0).

Example 6. Find the ratio in which the point P(x, 2) divides the line segment joining the points. A(12, 5) and B(4, -3). Also, find the value of. v.

Solution:

Let point P(x, 2) divide AB in the ratio k: 1.

x=\(\frac{12 \times 1+k \times 4}{k+1}\) and 2=\(\frac{5 \times 1+k \times-3}{k+1}\)

Now, 2=\(\frac{5-3 k}{k+1}\)

2 k+2=\(-3 k+5 \quad \Rightarrow \quad 5 k=3\)

k=\(\frac{3}{5}\)

Required ratio =3: 5

and x=\(\frac{4 k+12}{k+1}=\frac{4 \times \frac{3}{5}+12}{\frac{3}{5}+1}=\frac{12+60}{3+5}=9\)

x=9

Example 7. Find The lengths of the medians AD and BE of \(\triangle\)ABC whose vertices are A( 7, 3),B(5, 3) and C(3, -1 ).

Solution:

Co-ordinates of mid-point D of BC

= \((\frac{3+5}{2}\), \(\frac{-1+3}{2})=(4,1)\)

Co Ordinate Geometry The Length Of The Medians And Their Vertices Of Triangle ABC

Ciomodinates of mid-point E of AC :

= \(\left(\frac{3+7}{2}, \frac{-1-3}{2}\right)=(5,-2)\)

A D=\(\sqrt{(7-4)^2+(-3-1)^2}=\sqrt{9+16}=\sqrt{25}\)=5 units

and \(B E=\sqrt{(5-5)^2+(3+2)^2}=\sqrt{0+25}=\sqrt{25}=5 units\)

Example 8. If the points A(-1, -4), B(b,c) and C(5, -1) are collinear and 2b + c = 4, find the values of b and c.

Solution:

Points A(- 1, -4), B(b, c) and C(5, -1) are collinear

area of \(\triangle\)ABC = 0

-1(c + 1)+ b(-1 -2 + 4) + 5(-4-c) = 0

– c – 1 + 3b – 20 -5c = 0

3b – 6c = 21

b -2c – 7 …(1)

Given that, 2b + c = 4 (2)

from eqs. (1) and (2), we get

b = 3 and c = -2

Co-Ordinate Geometry Exercise 7.1

Question 1. Find the distance between the following pairs of points :

  1. (2, 3), (4,1)
  2. (-5, 7), (-1,3)
  3. (a, b), {-a, -b)

Solution :

(1) Distance between the points (2, 3) and (4, 1)

= \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

= \(\sqrt{(4-2)^2+(1-3)^2}\)

= \(\sqrt{(2)^2+(-2)^2}\)

= \(\sqrt{(4+4)}=\sqrt{8}=2 \sqrt{2}\) units

(2) Distance between the points (-5, 7) and (-1, 3)

= \(\sqrt{(-1+5)^2+(3-7)^2}\)

= \(\sqrt{(4)^2+(-4)^2}\)

= \(\sqrt{16+16}=\sqrt{32}=4 \sqrt{2}\) units

(3) Distance between the points (a, b) and {-a, -b).

= \(\sqrt{(-a-a)^2+(-b-b)^2}\)

= \(\sqrt{(-2 a)^2+(-2 b)^2}\)

= \(\sqrt{4 a^2+4 b^2}=\sqrt{4\left(a^2+b^2\right)}\)

=2 \(\sqrt{a^2+b^2}\) units

Question 2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B where town A is located 36 km east and 15 km north of town

Solution :

Distance between the points (0, 0) and (36, 15)

= \(\sqrt{(36-0)^2+(15-0)^2}\)

= \(\sqrt{1296+225}\)

=\(\sqrt{1521}\)=39 units

We can find the distance between two cities. The coordinates of the given cities in the cartesian coordinate system are A = (0,0) and B (36, 15).

The distance between these points is AB = 39 1cm.

Question 3. Determine if the points (1, 5), (2, 3) and (- 2, -11) are collinear.

Solution :

Let the given points are A(l, 5), B{2, 3) and C(-2,-ll).

A B =\(\sqrt{(2-1)^2+(3-5)^2}\)

=\(\sqrt{1+4}=\sqrt{5}\) units

B C =\(\sqrt{(-2-2)^2+(-11-3)^2}\)

=\(\sqrt{16+196}=\sqrt{212}\) units

C A =\(\sqrt{(-2-1)^2+(-11-5)^2}\)

=\(\sqrt{9+256}=\sqrt{265}\) units

Now, AB+B C \(\neq\) C A

Given points are not collinear.

Question 4. Check whether (5, -2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Solution :

Let the given points are A(5, -2), B{6, 4) and C(7, -2).

Now, A B =\(\sqrt{(6-5)^2+(4+2)^2}\)

=\(\sqrt{1+36}=\sqrt{37}\) units

B C =\(\sqrt{(7-6)^2+(-2-4)^2}\)

=\(\sqrt{1+36}=\sqrt{37}\) units

In a \(\triangle\)ABC,AB = BC

⇒ \(\triangle\)ABC is an isosceles triangle.

Therefore, the given points are the vertices of an isosceles triangle.

Question 5. In a classroom, 4 friends are seated at points A, B, C and D as shown in the figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using the distance formula, find which of them is correct

Co Ordinate Geometry Square Graph

Solution :

In the figure, the coordinates of A, B, C and D are (3, 4), (6, 7), (9, 4) and (6, 1) respectively.

A B =\(\sqrt{(6-3)^2+(7-4)^2}\)

= \(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2} units\)

B C =\(\sqrt{(9-6)^2+(4-7)^2}\)

=\(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)units

C D =\(\sqrt{(6-9)^2+(1-4)^2}\)

=\(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\) units

D A =\(\sqrt{(3-6)^2+(4-1)^2}\)

=\(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\) units

A C =\(\sqrt{(9-3)^2+(4-4)^2}\)

=\(\sqrt{36+0}=\sqrt{36}=6\)units

B D =\(\sqrt{(6-6)^2+(1-7)^2}\)

=\(\sqrt{0+36}=\sqrt{36}=6 units\)

Now, AB =BC = CD = DA and AC = BD

ABCD is a square.

So, Champa is correct.

Question 6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:

  1. (-1,-2). (1,0), (-1,2), (-3,0)
  2. (-3,5). (3, 1), (0,3), (-1,-4)
  3. (4, 5). (7. 6), (4, 3), (1,2)

Solution:

(1) Let points be A (-1, -2), (1,0), C (-1,2) and D (-3,0).

A B =\sqrt{(1+1)^2+(0+2)^2}

=\(\sqrt{4+4}=\sqrt{8}=2 \sqrt{2} units\)

B C =\(\sqrt{(-1-1)^2+(2-0)^2}\)

=\(\sqrt{4+4}=\sqrt{8}=2 \sqrt{2} units\)

C D =\(\sqrt{(-3+1)^2+(0-2)^2}\)

=\(\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}\)units

D A =\(\sqrt{(-1+3)^2+(-2-0)^2}\)

=\(\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}\) units

A C =\(\sqrt{(-1+1)^2+(2+2)^2}\)

=\(\sqrt{0+16}=\sqrt{16}\)=4 units

B D=\(\sqrt{(-3-1)^2+(0-0)^2}\)

=\(\sqrt{16+0}=\sqrt{16}\)=4 units

AB = BC = CD = DA and AC = BD

ABCD is a square.

Therefore, given points are the vertices of a square.

(2) Let given points are A(-3, 5), B(3, 1), C(0, 3) and D(-l, -4).

A B =\(\sqrt{(3+3)^2+(1-5)^2}\)

=\(\sqrt{36+16}=\sqrt{52}\)

=\(2 \sqrt{13}\)units

B C =\(\sqrt{(0-3)^2+(3-1)^2}\)

=\(\sqrt{9+4}=\sqrt{13} units\)

A C =\(\sqrt{(0+3)^2+(3-5)^2}\)

=\(\sqrt{9+4}=\sqrt{13}\) units

Now, BC + A C =\(\sqrt{13}+\sqrt{13}\)

=2 \(\sqrt{13}\)=A B

A, B and C are collinear.

Therefore, no quadrilateral will be formed from the given points.

(3) Let given points are A(4, 5), B(7, 6), C(4, 3) and D( 1,2).

A B=\(\sqrt{(7-4)^2+(6-5)^2}\)

=\(\sqrt{9+1}=\sqrt{10}\) units

B c =\(\sqrt{(4-7)^2+(3-6)^2}\)

=\(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\) units

C D =\(\sqrt{(1-4)^2+(2-3)^2}\)

=\(\sqrt{9+1}=\sqrt{10}\) units

=\(\sqrt{(4-1)^2+(5-2)^2}\)

=\(\sqrt{9+9}=\sqrt{18}\)

=3 \(\sqrt{2}\) units

A C =\(\sqrt{(4-4)^2+(3-5)^2}\)

=\(\sqrt{0+4}=\sqrt{4}\)=2 units

B D =\(\sqrt{(1-7)^2+(2-6)^2}\)

=\(\sqrt{36+16}=\sqrt{52}\)

=2 \(\sqrt{13}units\)

AB = CD, AD = BC and AC \(\neq\) BD

ABCD is a parallelogram.

Therefore, given points are the vertices of a parallelogram.

Question 7. Find the point on the A-axis which is equidistant from (2, -5) and (-2, 9).

Solution :

Let any point of the A-axis be P(x, 0). The distance of this point from the points A(2, -5) and B(-2, 9) are equal.

PA = PB

P A^2 =P B^2

⇒ \((x-2)^2+(0+5)^2 =(x+2)^2+(0-9)^2\)

⇒ \(x^2-4 x+4+25 =x^2+4 x+4+81\)

⇒ \(x^2-4 x-x^2-4 x =4+81-4-25\)

-8x = 56

x = -7

Coordinates of required point = (-7, 0).

Question 8. Find the values of y for which the distance between the points P(2, – 3) and Q(10,jV is 10 units.

Solution :

According to the problem, PQ = 10

⇒ \(P^2=100 \)

⇒ \((10-2)^2+(y+3)^2=100\)

⇒ \(64+(y+3)^2=100\)

⇒ \((y+3)^2=100-64\)

⇒ \((v+3)^2=36\)

⇒ \((y+3)= \pm 6\)

y+3=6 or y+3=-6

y=3 or y=-9

Required values of y = 3, – 9

Question 9. If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of .r. Also, find the distances QR and PR.

Solution :

Given that, Q is equidistant from P and R.

QP = QR

Q P = Q R

⇒ \(Q P^2=Q R^2\)

⇒ \((5-0)^2+(-3-1)^2=(x-0)^2+(6-1)^2\)

⇒ \(25+16=x^2+25\)

⇒ \(x^2=6 \Rightarrow x= \pm 4\)

⇒ \(Q R=\sqrt{( \pm 4-0)^2+(6-1)^2}\)

= \(\sqrt{16+25}=\sqrt{41}\) units

and P R =\(\sqrt{(5 \pm 4)^2+(-3-6)^2}\)

= \(\sqrt{(5+4)^2+81}\) or \(\sqrt{(5-4)^2+81}\)

= \(\sqrt{81+81}\) or \(\sqrt{1+81}\)

= \(\sqrt{162}\) or \(\sqrt{82}\)

P R =9 \(\sqrt{2} units\) or \(\sqrt{82} units\).

Question 10. Find a relation between x and; such that the point (x,y) is equidistant from the point (3, 6) and (- 3, 4).

Solution :

Given that, the distance of point {x,y) to (3, 6) = distance of point {x,y) to (-3, 4).

⇒ \(\sqrt{(x-3)^2+(y-6)^2}=\sqrt{(x+3)^2+(y-4)^2}\)

⇒ \((x-3)^2+(y-6)^2\)

= \((x+3)^2+(y-4)^2\)

⇒ \(x^2-6 x+9+y^2-12 y+36\)

=\(x^2+6 x+9+y^2-8 y+16\)

⇒ \(0=x^2+6 x+9+y^2-8 y+16-x^2 +6 x-9-y^2+12 y-36\)

12x- + 4y – 20 = 0

3x + y – 5 = 0

3x + y = 5

Co-Ordinate Geometry Exercise 7.2

Question 1. Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2:3.

Solution:

Let A=(-1,7),B=(4,-3)

Let p(x, y) divide AB in the ratio 2: 3.

A(-1,7), P(x, y),B(4,-3)

⇒ \((x_1, y_1)=(-1,7),(x_2, y_2)=(4,-3)\)

and m: n=2: 3

Now, x=\(\frac{m x_2+m x_1}{m+n}\)

x=\(\frac{2 \times 4+3 \times(-1)}{2+3}=\frac{8-3}{5}=\frac{5}{5}\)=1

and y=\(\frac{m y_2+n y_1}{m+n}\)

y=\(\frac{2 \times(-3)+3 \times 7}{2+3}=\frac{-6+21}{5}=\frac{15}{5}\)=3

Coordinates of required point =(1,3).

Question 2. Find the coordinates of the points of trisection of the line segment joining (4,-1) and (-2,-3).

Solution :

Given (4,-1) and (-2,-3)

Let A = (4, – 1) and B = (-2, – 3).

Let P and Q trisect the line segment AB.

Co Ordinate Geometry The Coordinates Of The Points Of Trisection Of The Line Segment

AP:PB= 1:2 and

AQ:QB = 2 : 1

For point P, x=\(\frac{m x_2+n x_1}{m+n}\)

x=\(\frac{1(-2)+2(4)}{1+2}=\frac{-2+8}{3}=\frac{6}{3}=2\)

and y=\(\frac{m r_2+n y_1}{m+n}\)

⇒ \(y=\frac{1(-3)+2(-1)}{1+2}=\frac{-3-2}{3}=\frac{-5}{3}\)

Coordinates of point P \(\equiv\left(2, \frac{-5}{3}\right)\)

For point Q, \(x^{\prime}=\frac{m_1 x_2+n_1 x_1}{m_1+n_1}\)

⇒ \(x^{\prime}=\frac{2(-2)+1(4)}{2+1}=\frac{-4+4}{3}=0\)

and \(y^{\prime}=\frac{m_1 y_2+n_1 y_1}{m_1+n_1}\)

⇒ \(y^{\prime}=\frac{2(-3)+1(-1)}{2+1}=\frac{-6-1}{3}=\frac{-7}{3}\)

Coordinates of point \(Q=\left(0, \frac{-7}{3}\right)\)

Question 3. To conduct Sports Day activities, in your rectangular-shaped school ground ABCD, lines have been drawn with chalk powder at a distance of lm each. 100 flower pots have been placed at a distance of lm from each other along AD, as shown in the figure. Niharika runs \(\frac{1}{4}\)th the distance AD on the 2nd line and posts a green flag.

Preet runs \(\frac{1}{5}\) through the distance AD on the eighth line and posts a red flag. What is the distance between both flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Co Ordinate Geometry The Distance Beyween The Two Persons From The Rectangular Shaped Ground

Solution :

100 flower pots are placed on side AD at a distance of 1 m from each other.

AD = 100 m

Position of Niharika’s flag = distance of \(\frac{1}{4}\)th part of AD in second row

⇒ \(\left(2,100 \times \frac{1}{4}\right)=(2,25)\)

Position of Preet’s flag = distance of \(\left(\frac{1}{5}\right)\) th part of AD in eighth row

= \(\left(8,100 \times \frac{1}{5}\right)=(8,20)\)

Distance between flags

= \(\sqrt{(8-2)^2+(20-25)^2}\)

= \(\sqrt{36+25}=\sqrt{61} \mathrm{~m}\)

Coordinates of the mid-point of (2,25) and (8,20)

= \(\left(\frac{2+8}{2}, \frac{25+20}{2}\right)=\left(5, \frac{45}{2}\right)\)

So, Rashmi should post her flag in the 5th row along AD at a distance of \(\frac{45}{2} \mathrm{~m}\).

Question 4. Find the ratio in which the line segment joining the points (-3, 10) and (6, – 8) is divided by (- 1, 6).

Solution :

Given

The line segment joining the points (-3, 10) and (6, – 8) is divided by (- 1, 6)

Let the points (-1, 6) divide the line segment joining the points (-3, 10) and (6, -8) in the ratio m: 1.

⇒ \(-1=\frac{m(6)+1(-3)}{m+1}\)

and \(6=\frac{m(-8)+1(10)}{m+1}\)

⇒ \(-m-1=6 m-3 and 6 m+6=-8 m+10\)

-7 m=-2 and 14 m=4

m=\(\frac{2}{7}\) and m=\(\frac{4}{14}=\frac{2}{7}\)

Required ratio = 2: 7.

Question 5. Find the ratio in which the line segment joining A (1, -5) and (-4, 5) is divided by the Ar-axis. Also, find the coordinates of the point of division.

Solution:

Given

(1, -5) and (-4, 5)

A=\((1,-5)=\left(x_1, y_1\right)\)

and \(B=(-4,5)=\left(x_2, y_2\right)\)

Let the x-axis divide the line segment A B at point (x, 0) in the ratio m: 1.

Now, 0=\(\frac{m \cdot y_2+1 \cdot y_1}{m+1}\)

⇒ \(0=m(5)+1(-5) \Rightarrow m=1\)

m: 1=1: 1

Required ratio =1: 1

Now, \(x=\frac{m x_2+1 \cdot x_1}{m+1}=\frac{1(-4)+1(1)}{1+1}\)

= \(\frac{-4+1}{2}=\frac{-3}{2}\)

Coordinates of point of division =\(\left(-\frac{3}{2}, 0\right)\)

Question 6. If (1, 2), (4,y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y-0

Solution:

Given

(1, 2), (4,y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order

Let the vertices of the parallelogram ABCD are A = (1, 2), B = (4, y), C = (x, 6) and D = (3, 5).

We know that the diagonals of a parallelogram bisect each other.

Coordinates of mid-point of AC = Coordinates of mid-point of BD

⇒ \(\left(\frac{1+x}{2}, \frac{2+6}{2}\right)=\left(\frac{4+3}{2}, \frac{y+5}{2}\right)\)

⇒ \(\frac{1+x}{2}=\frac{4+3}{2} and \frac{2+6}{2}=\frac{y+5}{2}\)

1+x=7 and 8=y+5

x=6 and y=3

Question 7. Find the coordinates of point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).

Solution :

Given

AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4)

Coordinates of center O = (2, -3)

Coordinates of point B = (1,4)

Let coordinates of point A = (h, k)

Now, coordinates of the mid-point of AB = coordinates of O

⇒ \(\left(\frac{h+1}{2}, \frac{k+4}{2}\right)=(2,-3)\)

⇒ \(\frac{h+1}{2}=2\) and \(\frac{k+4}{2}=-3\)

h+1=4 and k+4=-6

h=3 and k=-10

Coordinates of point A=(3,-10) .

 Question 8. If A and B are (- 2, -2) and (2, -4) respectively, find the coordinates of P such that AP = \(\frac{3}{7}\) AB and P lies on the line segment AB.

Solution :

Given

A and B are (- 2, -2) and (2, -4) respectively

A P=\(\frac{3}{7} A B\)

7 A P =3 A B\( \quad \Rightarrow \quad 7 A P=3(A P+B P)\)

7 A P =3 A P+3 BP

4 A P =3 B P \(\quad \Rightarrow \quad \frac{A P}{B P}=\frac{3}{4}\)

⇒ \(A B: B P=3: 4 \quad \Rightarrow \quad m: n=3: 4\)

and \(A A =(-2,-2)=\left(x_1, y_1\right)\)

B =\((2,-4)=\left(x_2, y_2\right)\)

Let coordinates of point P=(x, y).

x=\(\frac{m x_2+n x_1}{m+n}\)

x=\(\frac{3 \times 2+4 \times(-2)}{3+4}=\frac{6-8}{7}=\frac{-2}{7}\)

and y=\(\frac{m y_2+n y_1}{m+n}\)

⇒ \(y=\frac{3 \times(-4)+4 \times(-2)}{3+4}\)

=\(\frac{-12-8}{7}=\frac{-20}{7}\)

Coordinates of point P=\(\left(\frac{-2}{7}, \frac{-20}{7}\right)\)

Question 9. Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.

Solution:

Given

A(-2, 2) and B(2, 8)

Co Ordinate Geometry The Coordinates Of The Points Divide The Line segment Joining Into Four Equal Parts

Coordinates of mid-point Q of AB

=\(\left(\frac{-2+2}{2}, \frac{2+8}{2}\right)=(0,5)\)

Coordinates of mid-point P of AQ

=\(\left(\frac{-2+0}{2}, \frac{2+5}{2}\right)=\left(-1, \frac{7}{2}\right)\)

Coordinates of mid-point R of QB

=\(\left(\frac{0+2}{2}, \frac{5+8}{2}\right)=\left(1, \frac{13}{2}\right)\)

Coordinates of the points dividing A and B into equal parts are

∴ \(\left(-1, \frac{7}{2}\right),(0,5),\left(1, \frac{13}{2}\right)\)

Question 10. Find the area of a rhombus in its vertices as (3, 0), (4, 5), (— 1, 4) and diagonals (-2.-1) taken in order. [Hint : Area of a rhombus \(\frac{1}{2}\)(product of its diagonals)]

Solution :

Given

(3, 0), (4, 5), (— 1, 4) and diagonals (-2.-1)

Let the vertices of a rhombus be in the following order :

A =(3,0), B=(4,5),

C =(-1,4), D=(-2,-1)

A C=\(\sqrt{(-1-3)^2+(4-0)^2}\)

= \(\sqrt{16+16}=\sqrt{32}=4 \sqrt{2}\) units

B D =\(\sqrt{(-2-4)^2+(-1-5)^2}\)

= \(\sqrt{36+36}=\sqrt{72}=6 \sqrt{2}\) units

Now, area of rhombus =\(\frac{1}{2} \times\) product of diagonals

= \(\frac{1}{2} \times A C \times B D\)

= \(\frac{1}{2} \times 4 \sqrt{2} \times 6 \sqrt{2}\)

=24 sq. units

The area of a rhombus =24 sq. units

Co-Ordinate Geometry Exercise 7.3

Question 1. Find the area of the triangle whose vertices are :

  1. (2, 3), (-1,0), (2,-4)
  2. (-5-1), (3,-5), (5, 2)

Solution :

(1) Vertices of the given triangle are

⇒ \(\left(x_1, y_1\right)=(2,3)\),

⇒ \(\left(x_2, y_2\right)=(-1,0)\),

⇒ \(\left(x_3, y_3\right)=(2,-4)\)

Area of triangle =\(\frac{1}{2}[(x_1 y_2+x_2 y_3+x_3 y_1)\)\(-(x_2 y_1+x_3 y_2+x_1 y_3)]\)

= \(\frac{1}{2}[\{2 \times 0+(-1) \times(-4)+2 \times 3\}\)-{(-1) \times 3+2 \times 0+2 \times(-4)\}][/latex]

= \(\frac{1}{2}[(0+4+6)-(-3+0-8)]\)

= \(\frac{1}{2}(10+11)=\frac{21}{2}\)

Area of triangle =\(\frac{21}{2}\) sq. units

(2) Vertices of triangle are

⇒ \(\left(x_1, y_1\right)=(-5,-1),\left(x_2, y_2\right)=(3,-5),\left(x_3, y_3\right)=(5,2)\)

=\(\frac{1}{2}[(x_1 y_2+x_2 y_3+x_3 y_1\) \(-(x_2 y_1+x_3 y_2+x_1 y_3)]\)

=\(\frac{1}{2}[\{(-5) \times(-5)+3 \times 2+5 \times(-1)\}-\{3 \times(-1)+5 \times(-5)+(-5) \times 2\}]\)

=\(\frac{1}{2}[(25+6-5)-(-3-25-10)]\)

=\(\frac{1}{2}(26+38)=32\)

Area of triangle =32 sq. units

Question 2. In each of the following find the value of ‘k’, for which the points are collinear.

  1. (7,-2), (5, 1), (3, k)
  2. (8,1),(k,-4),(2,-5)

Solution :

(1) Let the given points are as follows:

A=\(\left(x_1, y_1\right)=(7,-2), \quad B=\left(x_2, y_2\right)=(5,1)C=\left(x_3, y_3\right)=(3, k)\)

A, B and C are collinear.

Area of \(\triangle A B C\)=0

⇒ \(\frac{1}{2}[(x_1 y_2+x_2 y_3+x_3 y_1) – (x_2 y_1+x_3 y_2+x_1 y_3)]\)=0

⇒ \((x_1 y_2+x_2 y_3+x_3 y_1)\)

⇒ \(-(x_2 v_1+x_2 v_2+x_1 y_3)\) =0

⇒ \({7 \times 1+5 \times k+3 \times(-2)\}-\{5 \times(-2)+3 \times 1+7 \times k}\) =0

(7+5 k-6)-(-10+3+7 k) =0

(1+5 k)-(-7+7 k) =0

1+5 k+7-7 k =0

8-2 k=0 \(\Rightarrow k =4\)

(2) Let the given points are as follows:

A \(\equiv\left(x_1, y_1\right)=(8,1), B=\left(x_2, y_2\right)=(k,-4)\),

C=\(\left(x_3, y_3\right)=(2,-5)\)

A, B and C are collinear

Area of \(\triangle A B C\)=0

⇒ \(\frac{1}{2}[(x_1 y_2+x_2 y_3+x_3 y_1)\).

⇒ \(\left.-\left(x_2 y_1+x_3 y_2+x_1 y_3\right)\right]\)=0

⇒ \(\left(x_1 y_2+x_2 y_3+x_3 y_1\right)\)

– \(\left(x_2 y_1+x_3 y_2+x_1 y_3\right)\)=0

⇒ \({8 \times(-4)+k \times(-5)+2 \times 1}\)

⇒ \({k \times 1+2 \times(-4)+8 \times(-5)}\)=0

⇒ \((-32-5 k+2)-(k-8-40)\)=0

(-30-5 k)-(k-48)=0

-30-5 k-k+48=0

-6 k+18=0 \(\Rightarrow k=3\)

Question 3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

Solution :

Given

(0, -1), (2, 1) and (0, 3)

Let vertices of \(\triangle\)ABC are

A=\(\left(x_1, y_1\right)=(0,-1), B=\left(x_2, y_2\right)=(2,1)\)

and C=\(\left(x_3, y_3\right)=(0,3)\).

Area of \(\triangle A B C\)

= \(\frac{1}{2}[(x_1 y_2+x_2 y_3+x_3 y_1\)

⇒ \(\left.-\left(x_2 y_1+x_3 y_2+x_1 y_3\right)\right\}\)

=\(\frac{1}{2}[\{0 \times 1+2 \times 3+0 \times(-1)\}\)

⇒ \(-\{2 \times(-1)+0 \times 1+0 \times 3\}]\)

=\(\frac{1}{2}[(0+6+0)-(-2+0+0)]\)

=\(\frac{1}{2}(6+2)\)=4 sq. units

Coordinates of mid-point P of AB

= \(\left(\frac{0+2}{2}, \frac{-1+1}{2}\right)=(1,0)\)

Coordinates of mid-point Q of BC

=\(\left(\frac{2+0}{2}, \frac{1+3}{2}\right)=(1,2)\)

Coordinates of mid-point R of CA

=\(\left(\frac{0+0}{2}, \frac{3-1}{2}\right)=(0,1)\)

Area of \(\triangle P Q R\)

= \(\frac{1}{2}[(1 \times 2+1 \times 1+0 \times 0) -(1 \times 0+0 \times 2+1 \times 1)]\)

= \(\frac{1}{2}[(2+1+0)-(0+0+1)\)

= \(\frac{1}{2}(3-1)\)=1 sq. unit

Now,\(\frac{\text { area of } \triangle P Q R}{\text { area of } \triangle A B C}=\frac{1}{4}\)=1: 4

Question 4. Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3,-2) and (2,3).

Solution:

Given

(-4, -2), (-3, -5), (3,-2) and (2,3)

Let the vertices of □ ABCD in order are as follows :

A=\(\left(x_1, y_1\right)=(-4,-2)\)

B=\(\left(x_2, y_2\right)=(-3,-5)\)

C=\(\left(x_3, y_3\right)=(3,-2)\)

D=\(\left(x_4, y_4\right)=(2,3)\)

Area of  Square A B C D

= \(\frac{1}{2}\left[\left(x_1 y_2+x_2 y_3+x_3 y_4+x_4 y_1\right)\right.\)

–\((x_2 y_1+x_3 y_2+x_4 y_3+x_1 y_4)]\)

= \(\frac{1}{2}[\{(-4) \times(-5)+(-3) \times(-2)+3 \times 3 +2 \times(-2)\}-\{(-3) \times(-2)+3 \times(-5)\) +\(2 \times(-2)+(-4) \times 3\}]\)

= \(\frac{1}{2}[(20+6+9-4)-(6-15-4-12)]\)

= \(\frac{1}{2}(31+25)\)=28 sq. units

Question 5. The median of a triangle divides it into two triangles of equal areas. Verify this result for \(\triangle\)ABC whose vertices are A(4, -6), B(3, -2) and C(5, 2).

Solution:

Conlinates of mid-point D of BC’

=\(\left(\frac{3+5}{2}, \frac{-2+2}{2}\right)=(1,0)\)

Area of \(\triangle ABD\)

=\(|\begin{array}{c} \frac{1}{2}\,\{4 \times(-2)+3 \times(1)+4 \times(-6)\} -\{3 \times(-6)+4 \times(-2)+4 \times(0) \mid \end{array}\|\)

=\(\left|\frac{1}{2}\right|(-8+0-2.4)-(-18-8+(0)||\)

=\(\left|\frac{1}{2}(-32+20)\right|=|-3|\)=3 sq. units

Area of \(\triangle ACD\) е:

\( =\begin{array}{r}
\frac{1}{2}, 1(4 \times 0+4 \times 2+5 \times(-6) \mid \\
-14 \times(-6)+5 \times 0+4 \times 2) \mid
\end{array}|\)

=  \(|\frac{1}{2}(0+8-3(0)-(-24+0+8)|\)

=\( \frac{1}{2}(-22+|(0)|=|-3|\)=3sq. units

⇒ I\(\triangle ABD\)I= \(\triangle ACD\)

Therefore, the median AD divides it into two triangles of equal areas. Hence Proved.

Co-Ordinate Geometry Exercise 7.4

Question 1. Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

Solution :

Let the line 2x + y – 4 =0 divides the line segment joining the points A (2, -2) and P (3, 7) in the ratio m: 1 and the coordinates of the point of division are (x,y).

x=\(\frac{m(3)+1(2)}{m+1}=\frac{3 m+2}{m+1}\)

and y=\(\frac{m(7)+1(-2)}{m+1}=\frac{7 m-2}{m+1}\)

Point (x, y), lies on the line 2 x+y-4=0

⇒ \(2\left(\frac{3 m+2}{m+1}\right)+\frac{7 m-2}{m+1}-4=0\)

6 m+4+7 m-2-4(m+1)=0

13 m+2-4 m-4=0

9 m=2 \(\quad \Rightarrow \quad m=\frac{2}{9}\)

Required ratio =2: 9

Question 2. Find a relation between x and y if the points (x, y), (1,2) and (7, 0) are collinear.

Solution :

Given

Points (x, y), (1,2) and (7, 0) are collinear.

Area of \(\Delta\)=0

⇒ \(\frac{1}{2}[\{x \cdot 2+1 \cdot 0+7 \cdot y\}\)

–\(\{1 \cdot y+7 \cdot 2+0 \cdot x\}]\)=0

(2 x+7 y)-(y+14)=0

2 x+7 y-y-14=0

x+3 y-7=0

which is the required relation between x and y.

Question 3. Find the centre of a circle passing through the points (6, -6), (3, -7) and (3, 3).

Solution :

Given

(6, -6), (3, -7) and (3, 3)

Let the points A(6, -6), B(3, -7) and C(3, 3) lie on the circumference of a circle whose centre is P(h,k)

P A=P B=P C (radii of the circle )

⇒ \(P A^2=P B^2=P C^2\)

Now, vP A^2=P B^2[/latex]

Now, \(P A^2=P B^2\)

⇒ \((h-6)^2+(k+6)^2 =(h-3)^2+(k+7)^2\)

⇒ \(h^2-12 h+36+k^2 +12 k+36\)

= \(h^2-6 h+9+k^2+14 k+49\)

-12 h+6 h+12 k-14 k+72-58=0

-6 h-2 k+14=0

3 h+k=7

and \(P B^2=P^{\prime} C^2\)

⇒ \((h-3)^2+(k+7)^2=(h-3)^2+(k-3)^2\)

⇒ \((k+7)^2=(k-3)^2\)

⇒ \(k^2+14 k+49=k^2-6 k+9\)

14 k+6 k=9-49

20 k=-40

k=-2

Put the value of k in equation (1),

3 h-2 =7

h =3

Coordinates Of the centre of the circle = (3,2)

Question 4. The two opposite vertices of a MJU.UV ;uv (-1, 2) aiui (3, 2). Find the coordinates of the other two vertices.

Solution :

Given

The two opposite vertices of a MJU.UV ;uv (-1, 2) aiui (3, 2).

Let two opposite vertices of square ABCD be A(-1,2) and C(3, 2) which are known.

Let the coordinates of vertex B be (h, k).

Now, AB = BC (sides of the square)

⇒\(A B^2=B C^2\)

⇒ \((h+1)^2+(k-2)^2=(h-3)^2+(k-2)^2\)

⇒ \((h+1)^2=(h-3)^2\)

⇒ \(h^2+2 h+1=h^2-6 h+9\)

2 h+6 h=9-

8 h=8

h=1

⇒ \(Again, \quad A B^2+B C^2=A C^2 \quad\left( \angle B=90^{\circ}\right)\)

⇒ \((h+1)^2+(k-2)^2+(h-3)^2+(k-2)^2\)

=\((3+1)^2+(2-2)^2\)

⇒ \((1+1)^2+k^2-4 k+4+(1-3)^2+k^2-4 k+4\)

=16+0

⇒ \(4+2 k^2-8 k+8+4=16\)

⇒ \(2 k^2-8 k=0 \quad \Rightarrow 2 k(k-4)=0\)

k=0 or k=4

Therefore, the remaining two vertices of the square = (1,0) or (1,4)

Question 5. The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.

Co Ordinate Geometry The Students Are To Sow Seeds Of Flowering Plants On The Remaining Area Of The Plot

  1. What will be the coordinates of the vertices of A BQR if C is the origin?
  2. Also, calculate the areas of the triangles in these cases. What do you observe?

Solution :

(1) If A is the origin, then

P = (4. 6), Q=(3, 2), R=(6, 5)

Area of \(\triangle P Q R\)

=\(\frac{1}{2}[(4.2+3.5+6.6)-(3.6+6.2+4.5)] \)

=\(\frac{1}{2}[(8+15+36)-(18+12+20)]\)

=\(\frac{1}{2}(59-50)=\frac{9}{2}\) sq. units

(2) If C is the origin, then

P =(-12,-2), Q=(-13,-6),

R =(-10,-3)

Area of \(\triangle P Q R\)

= \(\frac{1}{2}[\{(-12)(-6)+(-13)(-3)+(-10)(-2)\}\)

–\(\{(-13)(-2)+(-10)(-6)+(-12)(-3)\}]\)

= \(\frac{1}{2}[(72+39+20)-(26+60+36)]\)

= \(\frac{1}{2}(131-122)=\frac{9}{2}\) sq. units

Area Of the Triangle is both in some cases

Question 6. The vertices of a \(\triangle A B C\) are A (4,6), B(I, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that \(\frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{4}\).

Calculate the area of the \(\triangle A D E\) AB AC 4 and compare it with the area of \(\triangle A B C\).

Solution:

Given

The vertices of a \(\triangle A B C\) are A (4,6), B(I, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively,

A=(4,6), B=(1,5), C=(7,2)

Co Ordinate Geometry The Area Of Triangle ADE And Compare With The Area Of Triangle ABC

Area of \(\triangle A B C\)

= \(\frac{1}{2}[\{4.5+1.2+7.6\}\)

–\(\{1.6+7.5+4.2\}]\)

= \(\frac{1}{2}[(20+2+42) \quad \overbrace{B(1.5)}^{A(4,6)} -(6+35+8)] \)

= \(\frac{1}{2}(64-49)=\frac{15}{2}\) sq. units

Given, \(\quad \frac{A D}{A B}=\frac{1}{4} \quad \Rightarrow \quad \frac{A D}{A D+B D}=\frac{1}{4}\)

4 A D=A D+B D

3 A D=B D \(\quad \Rightarrow \quad \frac{A D}{B D}=\frac{1}{3}\)=1: 3

Point d divides B in the ratio 1 :  3. Therefore, the coordinates of point D

=\(\left(\frac{1 \times 1+3 \times 4}{1+3}, \frac{1 \times 5+3 \times 6}{1+3}\right)\)

=\(\left(\frac{1+12}{4}, \frac{5+18}{4}\right)=\left(\frac{13}{4}, \frac{23}{4}\right)\)

Again,\(\frac{A E}{A C} =\frac{1}{4} \quad \Rightarrow \quad \frac{A E}{A E+E C}=\frac{1}{4}\)

4 A E =A E+E C \(\Rightarrow 3 A E=E C\)

⇒ \(\frac{A E}{E C} =\frac{1}{3}=1: 3\)

Point E divides A-C in the ratio 1: 3. Therefore, the coordinates of point E

= \(\left(\frac{1 \times 7+3 \times 4}{1+3}, \frac{1 \times 2+3 \times 6}{1+3}\right)\)

= \(\left(\frac{7+12}{4}, \frac{2+18}{4}\right)=\left(\frac{19}{4}, 5\right)\)

Now, A = (4,6), D=\((\frac{13}{4}, \frac{23}{4}), E \equiv(\frac{19}{4}, 5\)

Area of \(\triangle A D E\)

= \(\frac{1}{2} {\left[\left\{4 \cdot \frac{23}{4}+\frac{13}{4} \cdot 5+\frac{19}{4} \cdot 6\right\}\right.}\)

–\({\frac{13}{4} \cdot 6+\frac{19}{4} \cdot \frac{23}{4}+4.5}]\)

=\(\frac{1}{2}[(23+\frac{65}{4}+\frac{114}{4})-(\frac{78}{4}+\frac{437}{16}+20)]\)

⇒ \(\frac{1}{2}\left[\frac{92+65+114}{4}-\frac{312+437+320}{16}\right] \)

= \(\frac{1}{2}\left[\frac{271}{4}-\frac{1069}{16}\right]=\frac{1}{2}\left(\frac{1084-1069}{16}\right)\)

= \(\frac{15}{32}\) square units

Now, \(\frac{\text { Area of } \triangle A D E}{\text { Area of } \triangle A B C}=\frac{15 / 32}{15 / 2}=\frac{1}{16}\)=1: 16

Co-Ordinates Multiple Choice Questions

Question 1. The distance of the point (3,4) from the X-axis is :

  1. 3 units
  2. 4 units
  3. 5 units
  4. 1 unit

Answer:

Question 2. The distance of the point (8, -6) from the origin is :

  1. 10 units
  2. 6 units
  3. 8 units
  4. 14 units

Answer:

Question 3.  The perimeter of a triangle with vertices (0, 0), (-3, 0) and (0, ‘1) is:

  1. 14 units
  2. 7 units
  3. 1 unit
  4. 12 Units

Answer:

Question 4. The points (-1. 0), (-1, 0) and (0, It) are the vertices of:

  1. a right-angled triangle
  2. an isosceles triangle
  3. an equilateral triangle
  4. a scalene triangle

Question 5. A(7, 0), B(4, 0) and C(S, 4) are the vertices of \(\triangle ABC\). The area of this triangle is :

  1. 14
  2. 28
  3. 8
  4. 6

Answer:

Question 6. (A-2, 3), B(6, 7) and C(8, 3) are three vertices of a parallelogram ABCD. The coordinates of vertex D are :

  1. (0,1)
  2. (0,-1)
  3. (c) (-1, 0)
  4. D(1,0)

Answer:

Question 7. The perpendicular bisector of the line segment joining the point (1,5) and (4, 6) intersects the Y-axis at point:

(0,13)

  1. (0,-13)
  2. (0, 12)
  3. (13, 0)

Answer:

Question 8. The mid-point of line segment joining the points A(-4, 2) and B( 5, 6) is \(P\left(\frac{a}{8}, 4\right)\). Then the value of A is

  1. -8
  2. -4
  3. 2
  4. 4

Question 9. The distance of the point (-3, 5) from the Y-axis is :

  1. -3
  2. 2
  3. 5
  4. None of these

Answer:

Question 10. The distance between two points (2, 3) and (4, 1) will be :

  1. 2
  2. 2 \(\sqrt{3}\)
  3. 2\( \sqrt{2}\)
  4. 3

Answer:

Question 11. The distance between the points P(2, -3) and (3(10, y) is 10 units. The value of y will be

  1. -3, 9
  2. -9, 3
  3. 9, 3
  4. -9, 2

Answer:

Question 12. A point on X-axis, equidistant from the points A(2, -5) and 2, 9) will be :

  1. (-7, 0)
  2. (-6, 0)
  3. (-2, 0)
  4. (2, 0)

Question 13. The distance between the points (-1, -3) and (5, 2) is :

  1. \(\sqrt{61}\) units
  2. \(\sqrt{37} units\)
  3. \(\sqrt{17}\) units
  4. 3 units

Answer:

NCERT Exemplar Solutions For Class 10 Maths Chapter 10 Circles

Circles

  • Circle: A circle is a collection of all points in a plane which are at the same
  • Centre: The fixed point from which all points in a plane are at the same constant distance is called the centre.
  • Radius: The distance between the centre and circumference of a circle is called the radius.
  • Chord: A straight line segment joining two points on a circle is called a chord of the circle.
  • Secant: A straight line which intersects a circle in two distinct points is called a secant to the circle.
  • Tangent: A straight line meeting a circle only at one point is called a tangent to the circle at that point.
  • Point Of Contact: The point where the straight line touches (or meets at only one point) the circle is called its point of contact.
  • Concentric Circles: Circles having the same centre are said to be concentric circles.

NCERT Exemplar Solutions For Class 10 Maths Chapter 10 Circles

Theorem 1:

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Given: A circle with centre O and a tangent AB at a point P of the circle.

To Prove: OP \(\perp\) AB

Circles The Tangent At Any Point Of A Circle Is Perpendicular To The Radius

Construction: Take a point Q other than P on AB. Join OQ.

Proof: Q is a point on the tangent AB, other than the point of contact P.

Q lies outside the circle.

Let OQ interest the circle at R.

Then, OR < OQ

But, OP = OR

Therefore, OP < OQ

Thus, OP is shorter than any other line segment joining O to any point of AB.

It means OP is the shortest distance among all the lines drawn from O to. the point on the tangent other than the point of contact.

Also, we know that perpendicular distance is the shortest distance So, \(\perp\)

i.e., the radius through the point of contact is perpendicular to the tangent.

Theorem 2:

A line drawn through the end of a ratlins and perpendicular to it is a tangent to the circle.

Given: A circle with centre O in which OP is a radius and AB is a line through P such that OP \(\perp\) AB.

Circles A Line Drawn Through The End Of A Radius And Perpendicular

To Prove: AB is a tangent to the circle at the point P.

Construction: Take a point Q different from P, on AB. Join OQ.

Proof: We know that the perpendicular distance from a point to a line is the shortest distance between them.

  • OP \(\perp\) AB
  • OP is the shortest distance from O to AB.
  • OP < OQ
  • Q lies outside the circle.
  • Thus, every point on AB other than P, lies outside the circle.
  • AB meets the circle at point P only. Hence, AB is the tangent to the circle at the point P.

An Important Result of the Above Theorem

If two circles touch internally or externally, the point of contact lies in a straight line through their centres.

Given: Two circles with centre O and O’ which touch each other at P.

To Prove : P lies on the straight line 00′ i.e., the line joining the centres.

Circles Touch Externally And Internally

Construction: Join OP, and O’P and draw a common tangent PT to the two circles at point P.

Proof: When Circles Touch Externally :

  • ∠1 = 90° …(1) (radius through the point of contact is perpendicular to the tangent)
  • ∠2 = 90° (same reason) …(2)
  • ∠1 + ∠2 = 90° + 90° [from (1) and (2)]
  • ∠l+∠2=180°
  • OPO’ is a straight line. (L.P.A.) …(3)
  • When Circles Touch Internally :
  • ∠OPT = ∠O’PT = 90° (radius through the point of contact is perpendicular to the tangent)
  • O’OP is a straight line. (O’P, OP are both 1 to PT at the same point P and only one

∴ \(\perp\) can be drawn to a line through one point on it) …(4) From both (3) and (4), we conclude that P lies on the straight line OO’ i.e., P lies on the straight line joining the centre of the circles.

Theorem 3:

The lengths of tangents drawn from an external point to a circle are equal.

Given: Two tangents AP and AQ are drawn from point A to a circle with centre O.

Circles The Lengths Of Tangents Drawn From An External Point To A Circle

To Prove: AP = AQ

Construction: Join OP, OQ and OA.

Proof: AP is a tangent at P and OP is the radius through P.

OP\(\perp\) AP

Similarly, OQ \(\perp\) AQ

In the right \(\triangle\) OPA and \(\triangle\) OQA

OP = OQ

OA = OA

OPA = ∠OQA

⇒ \(\triangle\) OPA = \(\triangle\) OQA

Hence, AP = AQ

Corollary : (A) If two tangents are drawn from an external point then

  • They subtend equal angles at the centre, and
  •  they are equally inclined to the line segment joining the centre to that point, (or, tangents are equally inclined at the centre).

Given: A circle with centre O and a point A outside it. Also, AP and AQ are the two tangents to the circle.

To Prove : \(\angle\)AOP = \(\angle\) AOQ and \(\angle\) OAP = \(\angle\) OAQ

Circles A Circle With Centre O And A Point A

Proof : In \(\triangle\)AOP and \(\triangle\)AOQ, we have

AP = AQ (tangents from an external point)

OP = OQ (radii of the same circle)

OA = OA (common)

⇒ \(\triangle\)AOP = \(\triangle\)AOQ (by SSS congruence)

Hence, \(\angle\)AOP = \(\angle\)AOQ and \(\angle\)OAP = \(\angle\)OAQ

(2) Prove that the tangents drawn at the endpoints of a chord of a circle make equal angles with the chord.

Given: Let AB be a chord of the given circle. PA and PB are the tangents at endpoints A and B.

To Prove : \(\angle\)5 = \(\angle\)6

Proof: Since OA and OB are the radii of a circle.

Circles The Tangent Drawn At The End Points Of A Chord Of A Circle

⇒ \(\angle\)1 = \(\angle\)2 …(1 )(each 90°, as radius through point of contact is 1 to the tangent)

Also in \(\triangle\)OAB,

Since OA = OB

⇒ \(\angle\)3 = \(\angle\)4 (angles opposite to equal sides are equal)

Subtracting equation (2) from equation (1), we get

⇒ \(\angle\)1 – \(\angle\)3 = \(\angle\)2 – \(\angle\)4

⇒ \(\angle\)5 = \(\angle\)6

Hence, tangents drawn at the endpoints of a chord of a circle make equal angles with the chord.

Solved Examples

Example 1. Find the length of tangent drawn to a circle of radius 6 cm, from a point at a distance of 10 cm from the centre.

Solution:

Since the tangent is perpendicular to the radius through the point of contact.

Circles The Length Of Tangent Drawn To Circle Of Radius

  • \(\angle O T P=90^{\circ}\)
  • In the right triangle OTP, we have
  • \(O P^2 =O T^2+P T^2\)
  • \(10^2 =6^2+P T^2\)
  • \(P T^2 =100-36=64\)
  • P T =8 cm
  • Hence, the length of the tangent is 8 cm.

Question 2. AP is tangent to circle O at point P. What is the length of OP?

Solution:

Let the radius of the given circle be r.

Circles Radius Through Point Of Contact Is Perpendicular To The Tangent

OP = OB = r

OA = 2 + r, OP = r, AP =4

⇒ \(\angle OPA\) = 90° (radius through the point of contact is perpendicular to the tangent)

In right \(\triangle O P A\),

⇒ \(O A^2 =O P^2+A P^2\)

⇒ \((2+r)^2 =r^2+(4)^2\)

⇒ \(4+r^2+4 r =r^2+16\)

⇒ \(4 r =12 \quad \Rightarrow \quad r=3\)

O P = 3 cm.

Example 3. If the angle between two tangents drawn from an external point P to a circle of radius V and centre O, is 60°, then find the length of OP.

Solution:

PA and PB are two tangents from an external point P such that

Circles The Angle Between Two Tangents Drawn From An External Point

⇒ \(\angle\)APB = 60°

⇒ \(\angle\)OPA = \(\angle\)OPB = 30° (tangents are equally inclined at the centre)

Also, \(\angle\)OAP = 90° (radius through the point of contact is perpendicular to the tangent)

Now in right \(\triangle\)OAP,

⇒ \(\sin 30^{\circ} =\frac{O A}{O P}\)

⇒ \(\frac{1}{2} =\frac{a}{O P} \quad \Rightarrow \quad O P\)=2 a units

Example 4. In the adjoining figure, PQ is a chord of a circle and is the tangent atP such that \(\angle\)QPT = 60°. Find \(\angle\)PRQ.

Solution:

Circles In The Adjoining The Chord Of A Circle And Tangent

Join OP and OQ. Take any point S on the circumference in the alternate segment. Join SP and SQ.

Since OP \(\perp\) PT (radius through the point of contact is 1 to the tangent

⇒ \(\angle 2+\angle 1=90^{\circ}\)

⇒ \(\angle 2+60^{\circ}=90^{\circ}\)

⇒ \(\angle 2=90^{\circ}-60^{\circ}=30^{\circ}\) (given)

But O P=O Q (each radii)

⇒ \(\angle 2=\angle 3\) (angles opposite to equal sides are equal)

But O P=O Q

⇒ \(\angle 2=\angle 3\) (angles opposite to equal sides are equal)

Now in \(\triangle P O Q\),

⇒ \(\angle 2+\angle 3+\angle 4 =180^{\circ}\)

⇒ \(30^{\circ}+30^{\circ}+\angle 4 =180^{\circ}\)

⇒ \(\angle 4 =120^{\circ}\)

⇒ \(\angle 5 =\frac{1}{2} \times \angle 4\) (angle sum property) [from (1) and (2)]

(The degree measure of an area is twice the angle subtended by it in an alternate segment)

⇒ \(\angle 5=\frac{1}{2} \times 120^{\circ}\)

⇒ \(\angle 5 =60^{\circ}\)

∴ Also,\(\angle 5+\angle 6 =180^{\circ}\)

⇒ \(60^{\circ}+\angle 6 =180^{\circ}\)

∴ \(\angle 6 =\angle P R Q=120^{\circ}\)

Example 5. In the given figure two circles touch each other at point C. Prove that the common tangent to the circle at C bisects the common tangent at P and Q.

Solution:

In the given figure, PR and CR are both tangents drawn to c the same circle from an external point R

Circles Two Circles Touch Each Other At A Point

PR = CR …(1)

Also, QR and CR are both tangents drawn to the same circle from an external point R.

QR = CR -(2)

From ( 1 ) and (2) we get

PR = QR

R is the mid-point of PQ

i.e., the common tangent at C bisects the common tangents at P and Q

Example 6. Two circles of unequal radii neither touch nor intersect each other. Are the common tangents AB and CD always equal? If no, then give an explanation of it and if your answer is yes, then prove it.

Circles Two Circles OF Unequal Radii Neither Touch Nor Intersect Each Other

Solution:

Let the two tangents AB and CD on producing meet at P.

Since PA and PC are tangents from an external point P to the circle with centre O

PA=PC

Also, PB and PD are tangents from an external point P to the circle with centre O’.

Circles The Tangents From A External Point To The Circle With Centre

PB = PD …(2)

Subtracting (2) from (I), we get

PA – PB = PC- PD

AB = CD

So, the direct common tangents are of equal length

Example 7. In the adjoining figure, common tangents AB and CD to two circles intersect at P. Prove that AB = CD.

Solution:

Circles In Adjoining Common Tangents Of Two Circles Intersect

  • Since PA and PC are two tangents to a circle with centre O from an external point P.
  • PA =PC
  • Also, since PB and PD are two tangents to a circle with centre O’ from an external point P.
  • PB = PD
  • Adding (2) and ( 1 ), we get
  • PA + PB = PC + PD
  • AB = CD (AB and CD are two straight lines)
  • Hence Proved.

Example 8. In the given diagram, PQ and RS arc common tangents to the two circles with centres C and D. Find the length of PQ and hence the area of trapezium RSDC.

Solution:

Circles The Common Tangents To The Two Circles With Centers

Draw CM//PS so that DCMSP becomes a rectangle.

Now, we have CR = 2 cm, DS = 7cm and CD = 13 cm

DM = DS- MS

= DS- CR

= 7-2 = 5 cm

In right ADMC, by Pythagoras theorem,

Circle The Length Of A Common Tangents To Two Circles Are Always Same

⇒ \(C M^2 =C D^2-D M^2\)

=\((13)^2-(5)^2=(12)^2\)

RS = 12 cm (opposite sides of the rectangle are equal)

PQ = 12 cm (length of common tangents to two circles are always same)

Now, ar(Trapezium RSDC)

= \(\frac{1}{2} \times\) h sum of parallel sides)

= \(\frac{1}{2} \times C M(C R+D S)\)

= \(\frac{1}{2} \times 12(2+7)=54 \mathrm{~cm}^2\)

Example 9. AB is the diameter of a circle with centre O. AH and BK are perpendiculars from A and B to the tangent at a point P on the circle. Prove that AD + BK = AB.

Solution:

Let AH = x, BK=y and BM = z

Circles AB Is A Diameter Of A Circle With Centre

⇒ \(\triangle M B K \sim \triangle M A H\) (AA corollary)

⇒ \(\frac{B K}{A H}=\frac{B M}{A M} \quad \Rightarrow \quad \frac{y}{x}=\frac{z}{2 r+z}\)

⇒ \(2 r y+y z=x z \quad \Rightarrow \quad z(x-y)=2 r y\)

z=\(\frac{2 r y}{x-y}\)

Similarly, \(\triangle M B K \sim \triangle M O P\)

⇒ \(\frac{B K}{O P}=\frac{B M}{O M} \quad \Rightarrow \quad \frac{y}{r}=\frac{z}{z+r}\)

y z+y r=z r

z(r-y)=y r

z=\(\frac{y r}{r-y}\)

From (1) and (2), we get

⇒ \(\frac{2 y}{x-y} =\frac{y r}{r-y} \quad \Rightarrow \quad \frac{2}{x-y}=\frac{1}{r-y}\)

2 r-2 y =x-y

x+y 2 r

A H+B K =A B

Example 10. In the given figure, if AB =AC, prove that BE = EC.

Solution:

We know that lengths of tangents from an external point are equal.

Circles Length Of Tangents From An External Point Are Equal

AD =AF  → Equation 1

DB=BE    → Equation 2

EC =FC  →  Equation 3

Now, it is given that

AB=AC

AD + DB =AF + FC

AD + DB = AB + E C From 1 And 3

D B = E C

B E = E C [From (2), DB = BE]

∴ Hence proved

Example 11. In the given figure ABC is a right-angled triangle with AB = 6 K cm, and BC = 8 cm. A circle with a centre O has been inscribed inside the triangle. Find the radius of the circle.

Solution:

Circles A Circle With Centre Has Been Inscribed Inside The Triangle

Let x be the radius of the circle. In the right triangle ABC

⇒ \(A C^2 =A B^2+B C^2\) (by Pythagoras Th.)

⇒ \(A C^2 =6^2+8^2\)

⇒ \(A C^2\) =36+64

⇒ \(A C^2\) =100

A C =10

Now in quadrilateral OPBR

⇒ \(\angle B =\angle P=\angle R=90^{\circ}\) each

⇒ \(\angle R O P =90^{\circ}\)sum of all angles of a quadrilateral is 360°) (each radii)

and also OP = OR (each radii)

Hence, OPBR is a square with each side x cm.

Therefore, CR = (8 -x) and PA = (6 -x)

BP = RB = x cm

Since the tangents from an external point to a circle are equal in length

AQ = AP = (6-x) and CQ = CR = (8 -x)

Now, AC = AQ + CQ

10 = 6 -x + 8 -x

10 = 14 – 2

2x = 4 = 2 cm

Example 12. A circle is touching the side BC of a \(\triangle\) ABC at the point and touching/IB and AC produced at Q and li respectively. Prove that AQ = \(\frac{1}{2}\) (perimeter of \(\triangle\) ABC).

Solution:

Given : \(\triangle ABC\) and a circle which touches BC, AB and AC in P, Q and R respectively.

Proof: Since the length of the two tangents drawn from an external
point to a circle are is equal, therefore,

Circles A Circle Is Touching The Side At A Point

AQ = AR

BQ = BP

CP = CR

Now, perimeter of \(\triangle ABC\) = AB + BC + AC

= AB +BP + PC +AC

= AB + BQ + CR + AC [from (2) and (3)]

= AQ+AR = 2AQ [from (1)]

Perimeter of \(\triangle ABC\) = \(\frac{1}{2} \times \text { (perimeter of } \triangle A B C)\)

Example 13. In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. IfPA = 12 cm, QC = QD = 3 cm, then find PC + PD.

Solution:

Circle The Length Of The Two Tangents Drawn From An External Point To A Circle

Since the lengths of the two tangents drawn from an external point to a circle are equal,

PA = PB  → Equation (l)

CA = CQ  → Equation (2)

DB = DQ  →  Equation (3)

Now, PA = 12

PC + CA= 12 (given)

PC + CQ = 12  From 2

PC + 3 = 12 ⇒ PC= 9 cm  →  Equation 4

PB=PA = 12

PD +DB= 12

PD +DQ= 12

PD + 3 = 12

PD = 9 cm  → Equation 5

PC + PD = (9 + 9) cm = 18 cm [From 4 and 5]

Example 14. O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects the circle at E. If AB is the tangent to the circle at E, find the length AB.

Circle O Is The Centre Of The Circle Of Radius

Solution:

Since, \(\angle O P T=90^{\circ}\) (radius through point of contact is \(\perp\) to the tangent)

In right \(\triangle O P T\),

⇒ \(OP^2+P T^2=O T^2\)

⇒ \(P T^2=O T^2-O P^2\)

⇒ \(P T^2=(13)^2-(5)^2=(12)^2\)

⇒ \(P T=12 \mathrm{~cm}\)

Let A P=x cm

At E=A P=x (lengths of tangents from an external point are equal)

A T=T P-A P=12-x

E T=O T-O E=13-5=8 cm

⇒ \(\angle A E T=90^{\circ}\)

Now; since \(\angle A E O=90^{\circ}\)(radius through point of contact is \(\perp\) to the tangent)

In right \(\triangle A E T\), by Pythagoras theorem,

⇒ \(A E^2+E T^2=A T^2\)

⇒ \(x^2+(8)^2=(12-x)^2\)

⇒ \(x^2+64=144+x^2-24 x \)

⇒ \(24 x=144-64=80\)

x=\(\frac{80}{24}=\frac{10}{3}\)

Similarly, B E=\(\frac{10}{3} \mathrm{~cm}\)

⇒ \(A B=A E+B E=\left(\frac{10}{3}+\frac{10}{3}\right) \mathrm{cm}=\frac{20}{3} \mathrm{~cm}\)

A B=\(\frac{20}{3} \mathrm{~cm}\)

Example 15. In the given figure, T is tangent to the circle with centre O such that O T=4 cm and \(\angle O T A=30^{\circ}\). Find the length of segment AT.

Circle A Tangent To The Circle With The Centre

Solution:

In right \(\triangle O A T\),

⇒ \(\cos 30^{\circ} =\frac{A T}{O T}\)

∴ \(\frac{\sqrt{3}}{2} =\frac{A T}{4} \quad A T=2 \sqrt{3} \mathrm{~cm}\)

Example 16. In the given figure, OP is equal to the diameter of the circle. Prove that \(\triangle ABP\) is an equilateral triangle.

Solution: 

Circles Op Is Equal To The Diameter Of The Circle Of An Equilateral Triangle

Let, \(\angle O P A=\angle O P B=\theta\)( tangents are equally inclined at the centre) and the radius of the circle be r.

Since, \(\angle 1=90^{\circ}\) (radius through point of contact is \(\perp\) to the tangent)

In right \(\triangle O A P\),

⇒ \(\sin \theta =\frac{O A}{O P}=\frac{r}{2 r}=\frac{1}{2}=\sin 30^{\circ}\)

⇒ \(\theta \Rightarrow \quad 30^{\circ} \quad \Rightarrow A P B=2 \theta=2 \times 30^{\circ}=60^{\circ}\)

Now, since PA = PB (length of tangents from an external point are equal)

⇒ \(\angle 2=\angle 3\) (angles opposite to equal sides are equal)

In \(\triangle M P B\).

⇒ \(\angle 2+\angle 3+\angle A P B=180^{\circ}\)

⇒ \(\angle 2+\angle 2+60^{\circ}=180^{\circ}\)

⇒ \(\angle 2=\angle 3=60^{\circ}\)

⇒ \(\angle 2+\angle 3+\angle A P B =180^{\circ}\) (angle sum property)

⇒ \(\angle 2+\angle 2+60^{\circ} =180^{\circ}\) [from (1) and (2)]

2 \(\angle 2 =120^{\circ} \quad \Rightarrow \quad \angle 2=60^{\circ}\)

⇒ \(\angle 2=\angle 3=60^{\circ}\)

So, all the angles of \(\triangle A P B are 60^{\circ}\).

∴ \(\triangle A P B\) is an equilateral triangle.

Example 17. If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that \(\angle D B C=120^{\circ}\), prove that B C+B D=B O.

Solution:

⇒ \(\angle 1+\angle 2 =120^{\circ}\)

But \(\angle 1 =\angle 2\)

Circle An External Point B Of A Circle With Centre O

⇒ \(\angle 1+\angle 1 =120^{\circ}\)

2 \(\angle 1 =120^{\circ}\)

⇒ \(\angle 1 =60^{\circ}\)

Also, \(\angle O C B =90^{\circ}\) (radius through point of contact is \(\perp\) to the tangent)

Now, in right \(\triangle O C B\),

⇒ \(\cos 60^{\circ}=\frac{B C}{O B} \quad \Rightarrow \quad \frac{1}{2}=\frac{B C}{O B}\)

O B=B C+B C

O B=2 B C \(\quad \Rightarrow \quad O B=B C+B C\)

O B=B C+B D(length of tangents from an external point are equal)

∴ Hence Proved.

Example 18. In the adjoining figure, AB is a chord of length 9.6 cm of a circle with centre O and radius 6 cm. The tangents at A and B intersect at P. Find the length of PA.

Solution:

Join OP, OA and OB.

Circle In The Adjoining Figure, AB Is a Chord Of Length

Let PA = x cm and PM = y cm

P A =P B (length of tangents from an external point are equal)

P M =P M (common)

⇒ \(\angle 1 =\angle 2\) (tangents are equally inclined at the centre)

⇒ \(\triangle A M P \cong \triangle B M P\) (SAS congruency)

Circle The Length Of A PA With A Centre O

A M=M B=\(\frac{9.6}{2}=4.8 \mathrm{~cm}\)

and \(\angle P M A=\angle P M B\)

But \(\angle P M A+\angle P M B=180^{\circ}\)

⇒ \(\angle P M A=\angle P M B=90^{\circ}\)

Now, in right \(\triangle A M P\),

⇒ \(x^2=y^2+(4.8)^2\) (by Pythagoras theorem)

Also, in right \(x^2=y^2+(4.8)^2\)\triangle A M O\(x^2=y^2+(4.8)^2\),

⇒ \((4.8)^2+O M^2 =(6)^2\)

⇒ \(O M^2 =36-23.04=12.96\)

⇒ \(O M =\sqrt{12.96}=3.6 \mathrm{~cm}\)

Now, \(\angle O A P=90^{\circ}\) (radius through point of contact is \perp to the tangent)

In right \(\triangle M O P\),

⇒ \(O P^2 =O A^2+A P^2 \Rightarrow(y+3.6)^2=36+x^2\)

⇒ \(y^2+12.96+7.2 y =36+y^2+(4.8)^2\)

⇒ \(7.2 y =36+23.04-12.96 \quad \Rightarrow \quad 7.2 y=46.08 \)

y =\(\frac{46.08}{7.2}=6.4 \mathrm{~cm}\)

Put this value of y in equation (1),

⇒ \(x^2 =(6.4)^2+(4.8)^2=40.96+23.04=64\)

x = 8 cm

Hence, x=8 cm and y=6.4 cm

Example 19. The radii of two concentric circles are 1 3 cm and 8 cm. AB is the diameter of the bigger circle. BD is a tangent to the smaller circle touching it at D. Find the length AD.

Solution:

Produce BD to E which cuts the circle at E. Join AE and OD.

Since AB is the diameter of the bigger circle.

⇒  \(\angle A E B=90^{\circ}\) (angle in a semicircle is right angle)

Also, \(\angle O D B=90^{\circ}\)(radius through point of contact is \(\perp\) to the tangent)

Circle The Radii Of Two concentric Circles

Now, in \(\triangle B O D\) and \(\triangle B A E\)

⇒ \(\angle B =\angle B\) (common)

⇒ \(\angle O D B =\angle A E B\) (each \(90^{\circ}\) )

⇒ \(\Delta B O D =\triangle B A E\)

⇒ \(\frac{O D}{A E} =\frac{O B}{A B}\)(corresponding sides of similar triangles are proportional)

⇒ \(\frac{8}{A E}=\frac{r}{2 r} \quad \Rightarrow A E=16 \mathrm{~cm}\)

Since, \(O D \perp E B\)

D E=D B (\(\perp\) drawn from the centre to the chord bisects the chord)

In right \(\triangle O D B\),

⇒ \(D B^2 =O B^2-O D^2=(13)^2-(8)^2=169-64=105\)

D B =\(\sqrt{105} \mathrm{~cm}=E D\)

Now, in right \(\triangle A E D\), by Pythagoras theorem

⇒ \(A D^2 =A E^2+E D^2\)

⇒ \(A D^2 =(16)^2+105=256+105=361\)

∴ \(A D =\sqrt{361}\) i.e.. 19 cm

Hence, A D =19 cm.

Circles Exercise 10.1

Question 1. How many tangents can a circle have?

Solution :

Infinitely many tangents can be drawn on a circle.

Question 2. Fill in the blanks :

  1. A tangent to a circle intersects it in point (s).
  2. A line intersecting a circle in two points is called a
  3. A circle can have parallel tangents at the most.
  4. The common point of a tangent to a circle and the circle is called

Answer :

  1. one
  2. secant
  3. two
  4.  point of contact

Question 3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm, Length PQ is :

Circle A Tangent At A Point P Of A Circle Of Radius

  • 12 cm
  • 13 cm
  • 8.5 cm
  • \(\sqrt{119} \mathrm{~cm}\)

Solution : 4. \(\sqrt{119} \mathrm{~cm}\)

Here OP = 5 cm,

OQ = 12 cm In \(\triangle\)POQ

In \(\triangle P O Q \)

⇒ \(P Q^2 =O Q^2-O P^2\)

=\(12^2-5^2=119\)

P Q =\(\sqrt{119}\) cm

Question 4. Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

Solution:

  1. Let the centre of the circle be O. The line AB lies outside the circle. Draw the perpendicular OM from O to AB.
  2. The perpendicular OM intersects the circle at P. Draw a line EPF from P parallel to AB.
  3. EF is the required tangent of the circle.
  4. Draw a line CD parallel to AB which intersects the circle at two points. It is the required secant of the circle.

Circle A Circle And Two Lines Parallel To A Given Line

Circles Exercise 10.2

Question 1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is :

  1. 7 cm
  2. 12 cm
  3. 15 cm
  4. 24.5 cm

Answer: 1.

Here, PQ = 24 cm and OQ = 25 cm

In \(\triangle O P Q\)

⇒ \(O P^2=O Q^2-P Q^2\)(from Pythagoras theorem)

Circle The Radius Of Circle From A Point

= \(25^2-24^2=625-576=49\)

OP = 7 cm

Radius of circle = 7 cm

Question 2. In the figure, if TP and TQ arc the Uvo tangents to a circle with centre O so that ZPOQ = 110°, then Z PTQ is equal to :

  1. 60°
  2. 70°
  3. 80°
  4. 90°

Answer: 2.

Here TP and TQ are the tangents to the circle.

Circle The Two Tangents To A Circle With The Centre

⇒ \(\angle\)OPT = \(\angle\)OQT = 90°

In □ OPTQ,

⇒ \(\angle\)OPT + \(\angle\)PTQ + \(\angle\)OQT + \(\angle\)POQ = 360°

90° + ZPTQ + 90° + 110° = 360°

⇒ \(\angle\)PTQ = 360° -290°

∴ \(\angle\)PTQ = 70°

Question 3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to :

  1. 50°
  2. 60°
  3. 70°
  4. 80°

Answer: 1. 50°

PA and PB are two tangents of the circle.

⇒ \(\angle\)PAO = \(\angle\)PBO = 90°

Circle The Tangents From A Point Of Circle With The Centre

Given : \(\angle\)APB =80°

Now, in □PAOB,

⇒ \(\angle\)AOB + \(\angle\)PAO + \(\angle\)APB + \(\angle\)PBO = 360°

⇒ \(\angle\)AOB + 90° + 80° + 90° = 360°

⇒ \(\angle\)AOB = 360° – 260° = 100°

Now, PO bisects \(\angle\)AOB.

⇒ \(\angle\)POA = – \(\angle\)AOB = – x 100° = 50°

Question 4. Prove that the tangents drawn at the ends of the diameter of a circle are parallel.

Solution :

Let AB be the diameter of a circle with centre O. PA and PB are the tangents to the circle at A points A and B respectively.

Circle The Tangents Drawn At The Ends Of A Diameter Of A Circle

Now Z\(\angle\)PAB = 90°

and \(\angle\)QBA = 90°

\(\angle\)PAB +QBA = 90° + 90° = 180°

PA || QB Hence Proved.

Question 5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Solution :

Given: A circle with centre O a tangent AQB and a perpendicular PQ is drawn from the point of contact Q to AB.

To Prove: The perpendicular PQ passes through the centre of the circle.

Proof: AQ is the tangent of the circle at point Q.

AQ will be the perpendicular to the radius of the circle.

PQ \(\perp\) AQ

The centre of the circle will lie on the line PQ.

Perpendicular PQ passes through the centre of the circle.

Hence Proved.

Question 6. The length of a tangent from point A at a distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Solution :

Let O be the centre of the circle and PQ is a tangent to the circle from point P.

Given that, PQ = 4 cm and OP = 5 cm

Circle The Length Of A Tangent From A Point At A Distance

Now,\(\angle O Q P=90^{\circ}\)

In \(\triangle O Q P\) ,

⇒  \(O Q^2=O P^2-P Q^2\)

=\(5^2-4^2\)

= 25-16 = 9

OQ = 3 cm

Radius of circle = 3 cm

Question 7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Solution :

Here, we draw two circles C1 and C2 with radii = 3 cm and r2 = 5 cm c2 respectively.

Circles The Length Of The Chord Of The Larger Circle Touches The Smaller Circle

Now, we draw a chord AB which touches the circle C1 at D.

O is the centre of concentric circles.

Now we draw the perpendicular from O to AB which bisects AB at D.

i.e., AD = BD

In right \(\triangle O B D\),

⇒ \(O B^2=O D^2+D B^2\) (from Pythagoras theorem)

⇒ \(5^2 =3^2+D B^2\)

⇒ \(D B^2\) =25-9=16

DB = 4 cm

Length of chord = AB = 2BD

= 2 x 4 = 8 cm

Question 8. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD=AD + BC.

Circle An Quadrilateral ABCD Is Drawn To The Circumscribe A Circle

Solution :

The sides of quadrilateral ABCD touch the circle at P, Q, R and S as shown in the figure. We know that the tangents drawn from an external point to the circle are equal.

AP = AS, BP = BQ,

CR = CQ, DR = DS

On adding, AP + BP + CR + DR

= AS + BQ + CQ + DS

⇒ AB + CD = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC

Hence Proved.

Question 9. In the figure, XY and X’ Y’ are two parallel tangents to a circle with centre O and another tangent AB with the point of contact C intersecting XY at A and X’ Y’ at B. Prove that \(\angle\)AOB = 90°.

Circle Two Parallel Tangents To A Circle With Centre O And Another Tangent

Solution :

The tangents from an external point to a circle are equal.

AP = AC

In \(\triangle A P O\) and \(\triangle A C O\),

AP = AC

AO = AO (common)

OP = OC (radii of a circle)

From S.S.S. congruency,

⇒ \(\triangle A P O \cong \triangle A C O\)

⇒ \(\angle P A O=\angle O A C\)

⇒ \(\angle P A C=2=\angle C A O\)

Similarly, we can prove that

⇒ \(\angle C B O =\angle O B Q\)

⇒ \(\angle C B Q =2 \angle C B O\)

⇒ \(Y \| X^{\prime} Y^{\prime}\)

⇒ \(\angle P A C+\angle Q B C=180^{\circ}\)

(sum of interior angles of the same side of a transversal is \(180^{\circ}\) )

2. \(\angle C A O+2 \angle C B O =180^{\circ}\)

⇒ \(\angle C A O+\angle C B O =90^{\circ}\)

In \(\triangle A O B\),

⇒  \(C A O+\angle C B O+\angle A O B=180^{\circ}\)

⇒ \(\angle C A O+\angle C B O=180^{\circ}-\angle A O B\) From equations (1) and (2)

⇒ \(180^{\circ}-\angle A O B=90^{\circ}\)

∴ \(\angle A O B=90^{\circ}\) Hence Proved.

Question 10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtends ey the line segment joining the points of contact at the centre.

Solution :

PA and PB are the tangents of the circle.

\(\angle O A P=\angle O B P=90^{\circ}\)

In O A P B

⇒  \(\Rightarrow \quad 90^{\circ}+\angle A P B+90^{\circ}+\angle A O B=360^{\circ}\)

⇒ \(\angle A P B+\angle A O B=180^{\circ}\)

⇒ \(\angle A P B\) and \(\angle A O B\) are supplementary.

Hence Proved.

Question 11. Prove that the parallelogram circumscribing a circle is a rhombus.

Circle The Parallelogram Circumscribing A Circle

Solution :

Let a parallelogram ABCD is given. Let the parallelogram touch the circle at points, P, Q, R and S.

AP and AS are the tangents drawn from an -external point A to the circle.

AP = AS …(1)

Similarly, BP = BQ _ (2)

CR = CQ (3)

DR = DS (4)

Adding equations (1), (2), (3) and (4),

AP + BP + CR+DR = AS + BQ + CQ + DS

(AP + BP) + (CR + DR)

= (AS+ DS) + (BQ + CQ)

AB + CD = AD+BC

AB + AB = AD + AD {CD = AB, BC = AD, opposite sides of a parallelogram)

2 AB = 2 AD

AB = AD

So, ABCD is a rhombus. (adjacent sides of a parallelogram are equal)

Hence Proved.

Question 12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.

Circle A Triangle ABC Is Drawn To The Circumscribe A Circle Of Radius

Solution :

Given : CD = 6 cm,

BD = 8 cm and radius = 4 cm

Join OC, OA and OB.

Circle The Tangents Drawn From An External Point To A Circle

We know that the tangents drawn from an external point to a circle are equal.

CD = CF = 6 cm and BD = BE = 8 cm

Let AF = AE = x cm c

In \(\triangle\)OCB,

area of triangle \(A_1 =\frac{1}{2} \times \text { base } \times \text { height }\)

= \(\frac{1}{2} \times C B \times O D\)

= \(\frac{1}{2} \times 14 \times 4=28 \mathrm{~cm}^2\)

In \(\triangle O C A\),

area of triangle

⇒ \(A_2 =\frac{1}{2} \times A C \times O F \)

=\(\frac{1}{2}(6+x) \times 4\)

=12+2 x

In \(\triangle O B A\).

area of triangle

⇒ \(A_3 =\frac{1}{2} \times A B \times O E\)

=\(\frac{1}{2}(8+x) \times 4\)

=16+2 x

Now, semiperimeter of triangle ABC,

s =\(\frac{1}{2}(A B+B C+C A)\)

s =\(\frac{1}{2}(x+6+14+8+x)\)

=14+x

Now, area of \(\triangle A B C\)

=\(A_1+A_2+A_3\)

=28+(12+2 x)+(16+2 x)

=56+4 x

From Heron’s formula,

Area of \(\triangle A B C\)

= \(\sqrt{s(s-a)(s-b)(s-c)}\)

= \(\sqrt{(14+x)(14+x-14)(14+x-x-6)} \quad(14+x-x-8)\)

= \(\sqrt{(14+x)(x)(8)(6)}\)

= \(\sqrt{(14+x) 48 x}\)

From equations (1) and (2),

⇒  \(\sqrt{(14+x) 48 x}=56+4 x\)

Squaring both sides,

⇒ \((14+x) 48 x =4^2(14+x)^2\)

3 x =14+x

⇒ \(2 x=14-\Rightarrow \quad x\) =7

A C=6+x=6+7 =13 cm

A B=8+x=8+7 =15 cm

Question 13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Solution :

We know that the tangents drawn R from an external point to a circle subtend equal angles at the centre.

⇒ \(\angle 1=\angle 2\) ,

⇒ \(\angle 3=\angle 4\) ,

⇒ \(\angle 5=\angle 6\)

and \(\angle 7=\angle 8\)

Circles The Opposite Sides Of A Quadrilateral Circumscribing A Circle

Now, \(\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8 =360^{\circ}\)

⇒ \(2 \angle 2+2 \angle 3+2 \angle 6+2 \angle 7 =360^{\circ} \)

⇒ \((\angle 2+\angle 3)+(\angle 6+\angle 7) =180^{\circ}\)

⇒ \(\angle A O B+\angle C O D =180^{\circ}\)

Similarly, \(\angle B O C+\angle A O D =180^{\circ}\)

Hence Proved.

Circles Multiple Choice Questions

Question 1. If the angle between two radii of a circle is 1 10° then the angle between the tangents drawn at the ends of these radii is :

  1. 110°
  2. 100°
  3. 90°
  4. 70°

Answer: 4. 70°

Question 2. If two tangents PA and PB drawn from point P are of equal length 4 cm, then the radius of the die circle is :

Circle The Two Tangents Are Drawn From A Point Are Equal Length

  1. 1 cm
  2. 2 cm
  3. 4 cm
  4. 3 cm

Answer: 3. 4 cm

Question 3. In the adjoining figure, PA and PB are tangents to a circle with centre 0 such that \(angle\)APB = 40°. \(angle\)OAB is :

Circle The Tangent To A Circle With The Centre O

  1. 20°
  2. 40°
  3. 30°
  4. 15°

Answer: 1. 20°

Question 4. If two tangents of a circle of radius 6 cm are drawn such that the angle between them is 60 then the length of each tangent is :

  1. 2 \(\sqrt{3} \mathrm{~cm}\)
  2. 6 \(\sqrt{3} \mathrm{~cm}\)
  3. 3 cm
  4. 6 cm

Answer: 2. 6 \(\sqrt{3} \mathrm{~cm}\)

Question 5. Two tangents PQ and PR are drawn to a circle of radius 5 cm where P is 13 cm away from centre O. The area of quadrilateral PQOR is :

  1. \(60 \mathrm{~cm}^2\)
  2. \(30 \mathrm{~cm}^2\)
  3. \(65 \mathrm{~cm}^2\)
  4. \(32.5 \mathrm{~cm}^2\)

Answer: 1. \(60 \mathrm{~cm}^2\)

Question 6. In the adjoining figure, O is the centre of the circle, PQ is the chord and the tangent PR drawn from points on the circle makes a 50° angle from chord PQ, \(\angle\) POQ is :

Circle The Chord And The Tangent Are Drawn From A Point

  1. 90°
  2. 80°
  3. 100°
  4. 75°

Answer: 3. 100°

Question 7. The radii of two circles are 3 cm and 4 cm and both circles touch each other externally. The distance between their centres is :

  1. 1 cm
  2. 3 cm
  3. 5 cm
  4. 7 cm

Answer: 4. 7 cm

NCERT Exemplar For Class 10 Maths Chapter 9 Some Applications Of Trigonometry

Some Applications Of Trigonometry

Practical Use Of Trigonometry

The main purpose of studying trigonometry is to determine the height of buildings, towers, telephone poles, trees, the width of the river, the distance of the ship from the lighthouse etc. Although it is not easy to measure them, we can determine these things by using knowledge of trigonometric ratios, before doing so, let us first discuss some necessary definitions.

NCERT Exemplar For Class 10 Maths Chapter 9 Some Applications Of Trigonometry

Line Of Sight: When an observer looks at an object then the line joining the observer’s eye to the object is called the line of sight.

Angle Of Elevation: When an observer sees an object situated in an upward direction, the angle formed by the line of sight with a horizontal line is called an angle of elevation.

Applications Of Trigonometry Angle Of Elevation

Angle Of Depression: When an observer sees an object .situated in a downward direction, the angle formed by the line of sight with a horizontal line is called an angle of depression.

Applications Of Trigonometry Angle Of Depression

In the adjoining above given, θ is the angle of depression of the object as seen from O.

Solved Examples

Example 1. The length of the shadow of a vertical pole is \(\frac{1}{\sqrt{3}}\) times its height. Show that the angle of elevation of the sun is 60°.

Solution:

Given:

The length of the shadow of a vertical pole is \(\frac{1}{\sqrt{3}}\) times its height.

Let PQ be a vertical pole whose height is \(\frac{h}{\sqrt{3}}\). Its shadow is OQ whose height is.

Let the angle of elevation of the sun be \(\angle\)POQ = 0

Applications Of Trigonometry The Angle Of Elevation Of Sun Is 60 Degrees

In \(\triangle\)POQ,

⇒ \(\tan \theta =\frac{P Q}{O Q}=\frac{h}{h / \sqrt{3}}=\sqrt{3}=\tan 60^{\circ}\)

⇒  \(\theta =60^{\circ}\)

The angle of elevation of the sun = 60°.

Example 2. If a tower 30 m high, casts a shadow 10 \(\sqrt{3}\) m long on the ground, then what is the angle of elevation of the sun?

Solution:

It is given that AB = 30 m be the height of the tower and BC = \(\sqrt{3}\) m its shadow on the ground.

Applications Of Trigonometry The Angle Of The Elevation Of Sun

Let \(\theta\) be the angle of elevation.

In a right triangle,

⇒  \(\tan \theta =\frac{A B}{B C}\)

= \(\frac{30}{10 \sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}\)

= \(\tan 60^{\circ}\)

⇒  \(\theta =60^{\circ}\)

Hence, the angle of elevation of the sun= 60°

Example 3. A ladder 15 metres long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, find the height of the wall.

Solution:

Given:

A ladder 15 metres long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall

Let PR be a ladder of length 15 m and QR, a wall of height h.

Applications Of Trigonometry The Ladder Makes An Angle Then The Height Of The Wall

Given that \(\angle\)PRQ = 60°

In \(\triangle P Q R\),

⇒  \(\cos 60^{\circ} =\frac{h}{P R} \quad \Rightarrow \quad \frac{1}{2}=\frac{h}{15}\)

⇒  \(\Rightarrow \quad h =\frac{15}{2} \mathrm{~m}\)

Height of the wall =\(\frac{15}{2} \mathrm{~m}=7.5 \mathrm{~m}\)

Question 4. The shadow of a tower standing on a level plane is found to be 50 m longer when the sun’s elevation is 30° than when it is 60°. Find the height of the tower.

Solution:

Given

The shadow of a tower standing on a level plane is found to be 50 m longer when the sun’s elevation is 30° than when it is 60°.

Let AB be a tower of height ‘h’ metres and BD and BC be its shadows when the angles of elevation of the sun are 30° and 60° respectively.

Applications Of Trigonometry The Shadow Of A Tower Standing On A Level Plane Then The Height Of The Tower

⇒  \(\angle A D B=30^{\circ}, \angle A C B=60^{\circ} \text { and } C D=50 \mathrm{~m}\)

Let BC = x metres.

In \(\triangle A B C\)

⇒  \(\tan 60^{\circ} =\frac{A B}{B C} \quad \Rightarrow \quad \sqrt{3}=\frac{h}{x}\)

⇒  \( x =\frac{h}{\sqrt{3}}\)

In \(\triangle A B D\)

⇒  \(\tan 30^{\circ}=\frac{A B}{B D} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+50}\)

⇒  \(\sqrt{3} h=x+50 \Rightarrow \sqrt{3} h=\frac{h}{\sqrt{3}}+50\)

⇒  \(h=h+50 \sqrt{3} \quad \Rightarrow \quad 2 h=50 \sqrt{3} \quad \Rightarrow \quad h=25 \sqrt{3}\)

Height of the tower =\(25 \sqrt{3} \mathrm{~m}\)

Question 5. The angle of elevation of the top of a tower from a point on the ground is 30°. After walking 40 m towards the tower, the angle of elevation becomes 60°. Find the height of the tower.

Solution:

Given

The angle of elevation of the top of a tower from a point on the ground is 30°. After walking 40 m towards the tower, the angle of elevation becomes 60°.

Let AB be a tower of height 7i’ metres. From points D and C on the ground, the angle of elevation of top A of the tower is 30° and 60° respectively.

Applications Of Trigonometry The Angle Of Elevation Of The Top Of A Tower From A Point

Given that CD = 40 m

Let BC = x metres

In \(\triangle A B C\)

⇒ \(\tan 60^{\circ} =\frac{A B}{B C} \quad \Rightarrow \quad \sqrt{3}=\frac{h}{x}\)

x =\(\frac{h}{\sqrt{3}}\)

In \(\triangle A B D\)

⇒ \(\tan 30^{\circ} =\frac{A B}{B D} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{40+x}\)

⇒ \(\sqrt{3} h =40+x \Rightarrow \sqrt{3} h=40+\frac{h}{\sqrt{3}}\)

⇒ \(3 h =40 \sqrt{3}+h \Rightarrow 2 h=40 \sqrt{3}\)

⇒ \(h =20 \sqrt{3}\)

Height of the tower =\(20 \sqrt{3} \mathrm{~m}\)

Question 6. The angle of elevation of the top of a tower from two points distant ‘s’ and ‘t’ from its foot are complementary. Prove that the height of the tower is \(\sqrt{st}\) .

Answer:

Given

The angle of elevation of the top of a tower from two points distant ‘s’ and ‘t’ from its foot are complementary.

Let BC be a tower of height ‘h’. Let AC = s and DC = t.

From points A and D, the angle of elevation of top B of the tower is complementary.

Applications Of Trigonometry The Angles Of Elevation Of The Top Of Tower From Two Points At A Distance

Let \(\angle B A C=\theta\)

⇒ \(\angle B D C=90^{\circ}-\theta\)

In \(\triangle B A C\)

⇒ \(\tan \theta=\frac{B C}{A C}=\frac{h}{s}\)

In \(\triangle B D C\)

⇒ \(\tan \left(90^{\circ}-\theta\right) =\frac{B C}{C D} \Rightarrow \cot \theta=\frac{h}{t}\)

⇒ \(\Rightarrow\frac{1}{\tan \theta} =\frac{h}{t} \quad \Rightarrow \frac{s}{h}=\frac{h}{t}\)

⇒ \(h^2 =s t \Rightarrow h=\sqrt{s t}\)

height of the tower =\(\sqrt{s t}\)

Example 7. A tree is broken by the wind. The top struck the ground at an angle of 30° and at a distance of 4 m from the root. Find the height of the tree before broken.

Solution:

Given

A tree is broken by the wind. The top struck the ground at an angle of 30° and at a distance of 4 m from the root.

Given that A B=4 m and \(\angle\) B A C=\(30^{\circ}\). Also C D=C A In \(\triangle \)A B C,

Applications Of Trigonometry A Tree Broken By The Wind Then The Height Of The Tree Before Broken

⇒  \(\tan 30^{\circ}=\frac{B C}{A B} \quad \Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{B C}{4}\)

B C=\(\frac{4}{\sqrt{3}} \mathrm{~m}\)

and \(\quad \cos 30^{\circ}=\frac{A B}{A C} \quad \Rightarrow \quad \frac{\sqrt{3}}{2}=\frac{4}{A C}\)

A C=\(\frac{8}{\sqrt{3}} \mathrm{~m} \Rightarrow C D=\frac{8}{\sqrt{3}} \mathrm{~m}( A C=C D)\)

Now the total height of the tree =B C+C D

= \(\frac{4}{\sqrt{3}}+\frac{8}{\sqrt{3}}=\frac{12}{\sqrt{3}}=4 \sqrt{3} \mathrm{~m}\)

The height of the tree before broken = \(4 \sqrt{3} \mathrm{~m}\)

Example 8. From the top of a tower 71 m high, the angles of depression of two objects, which are in line with the foot of the tower are \(\alpha\) and \(\beta(\beta>\alpha)\). Find the distance between the two objects.

Solution:

Given

From the top of a tower 71 m high, the angles of depression of two objects, which are in line with the foot of the tower are \(\alpha\) and \(\beta(\beta>\alpha)\).

Let AB be a tower of height ‘h’m. From the top A’ of the tower, the angle of depression of two objects D and C are ‘ \(\beta\) And \(\alpha\) respectively.

Applications Of Trigonometry The Top Of A Tower High And The Angles Of Depression Then The Distance Between The Two Objects

In \(\triangle A B C\)

⇒ \(\tan \beta=\frac{A B}{B C} =\frac{h}{B C}\)

B C =\(\frac{h}{\tan \beta}=h \cot \beta\)

In \(\triangle A B D\)

⇒ \(\tan \alpha =\frac{A B}{B D}=\frac{h}{B D}\)

B D =\(\frac{h}{\tan \alpha}=h \cot \alpha\)

Subtract equation (1) from (2), we get

⇒ \(B D-B C=h \cot \alpha-h \cot \beta\)

⇒ \(C D=h(\cot \alpha-\cot \beta)\)

⇒ \(C D=h(\cot \alpha-\cot \beta)\)

The distance between the objects =\(h(\cot \alpha-\cot \beta) \mathrm{m}\).

Example 9. Two men are on opposite sides of a tower. They measure the angles of elevation of the top of the tower as 30° and 60°. If the height of the tower is 150 m, find the distance between the two men.

Solution:

Given

Two men are on opposite sides of a tower. They measure the angles of elevation of the top of the tower as 30° and 60°. If the height of the tower is 150 m

Lot CD bo a tower of height 1 50 m. Two men A and B are on the opposite sides of the tower.

Applications Of Trigonometry The Distance Between The Two Men

Given that, \(\angle D A C=60^{\circ}\) and \(\angle D B C=30^{\circ}\)

In \(\triangle D A C\)

⇒ \(\tan 60^{\circ} =\frac{D C}{A C} \quad \Rightarrow \quad \sqrt{3}=\frac{150}{A C}\)

⇒ \(A C =\frac{150}{\sqrt{3}}=50 \sqrt{3} \mathrm{~m}\)

In \(\triangle B C D\)

⇒ \(\tan 30^{\circ} =\frac{D C}{B C} \Rightarrow \frac{1}{\sqrt{3}}=\frac{150}{B C} \Rightarrow B C=150 \sqrt{3} \mathrm{~m}\)

A B = A C+B C

= \((50 \sqrt{3}+150 \sqrt{3}) \mathrm{m}=200 \sqrt{3} \mathrm{~m}\)

Therefore the distance between two \(\mathrm{men}=200 \sqrt{3} \mathrm{~m}\)

Example 10. A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h, At a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are a and p respectively. Prove that the height of the tower is \(\left(\frac{h \tan \alpha}{\tan \beta-\tan \alpha}\right)\)

Solution:

Given:

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h, At a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are a and p respectively.

Let AB be a tower and BC be the flagstaff.

Let O be the observer.

Applications Of Trigonometry A Vertical Tower Stands On A Horizontal Plane

Now, \(\angle A O B=\alpha\), \(\angle A O C=\beta\) and B C=h

Let A B = H and O A = x

In \(\triangle O A B\)

\(\tan \alpha=\frac{A B}{O A}=\frac{H}{x} \quad \Rightarrow \quad x=\frac{H}{\tan \alpha}\) →  Equation 1

In \(\triangle O A C\)

⇒ \(\tan \beta=\frac{A C}{O A}=\frac{H+h}{x} \Rightarrow x=\frac{H+h}{\tan \beta}\)  → Equation 2

From equations (1) and (2), we get

⇒ \(\frac{H}{\tan \alpha} =\frac{H+h}{\tan \beta}\)

⇒ \(H \tan \beta =H \tan \alpha+h \tan \alpha\)

⇒ \(H(\tan \beta-\tan \alpha) =h \tan \alpha\)

H =\(\frac{h \tan \alpha}{\tan \beta-\tan \alpha}\)

Height of the tower =\(\frac{h \tan \alpha}{\tan \beta-\tan \alpha}\)

Example 11. The angle of elevation of the top of a vertical lower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 47′. Find the height of the tower.

Solution:

Given

The angle of elevation of the top of a vertical lower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 47′.

Let All he a vertical lower. From point C, the angle of elevation of the top of the tower is 60° and from point) as shown, the angle of elevation of the tower is 45°.

⇒ \(\angle A C B=60^{\circ}, \angle A D E=45^{\circ}\) and \(D C=10 \mathrm{~m}\)

In \(\triangle D E\),

Applications Of Trigonometry The Height Of The Tower From The Top

⇒ \(\tan 45^{\circ}=\frac{A E}{D E} \quad \Rightarrow \quad A E=D E \ldots(1)\left( \tan 45^{\circ}=1\right)\)

In \(\triangle A B C\),

⇒ \(\tan 60^{\circ} =\frac{A B}{B C} \quad \Rightarrow \quad \sqrt{3}=\frac{A B}{D E} \quad( B C=D E)\)

\(A B =\sqrt{3} D E\)

A E+B E =\(\sqrt{3} A E\)

⇒ \(B E =A E(\sqrt{3}-1)\)

A E =\(\frac{B E}{\sqrt{3}-1}=\frac{10}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}\)

⇒ \(\frac{10(1.732+1)}{3-1}=5 \times 2.732=13.66\)

Now, A B =A E+B E=A E+C D

= \((13.66+10) \mathrm{m}=23.66 \mathrm{~m}\)

Height of the tower =23.66 m

Example 12. The angle of elevation of the top Q of a vertical towerPQ from a point Y- on the ground is 60°. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of the tower is 45°. Find the height of the tower PQ and the distance PX. (Use \(\sqrt{3}\)=1.73)

Solution:

Given

The angle of elevation of the top Q of a vertical towerPQ from a point Y- on the ground is 60°. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of the tower is 45°.

Let height of tower PQ be h m and let P X = x m

Since, X Y=40=P Z

Q Z=P Q-P Z=h-40.

In right \(\triangle Q Z Y\),

Applications Of Trigonometry The Angle Of Elevation Of The Top Of A Vertical Tower

⇒ \(\tan 45^{\circ}=\frac{h-40}{x} \Rightarrow 1=\frac{h-40}{x}\)

⇒ \(h-40=x \Rightarrow x=h-40\)

In right \(\triangle Q P X\),

⇒ \(\tan 60^{\circ}=\frac{h}{x} \quad \Rightarrow \quad \sqrt{3}=\frac{h}{x}\)

x=\(\frac{h}{\sqrt{3}}\)

From equations (1) and (2), we get

h-40=\(\frac{h}{\sqrt{3}} \quad \Rightarrow h \sqrt{3}-40 \sqrt{3}=h\)

⇒ \(h(\sqrt{3}-1)=40 \sqrt{3} \Rightarrow h=\frac{40 \sqrt{3}}{\sqrt{3}-1}\)

h= \(\frac{40 \sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=20(3+\sqrt{3})\)

h = \(20(3+1.73)=20 \times 4.73=94.6 \mathrm{~m}\)

x =h-40

=94.6-40=54.6 m

Hence, the height of tower PQ is 94.6 m and the distance P X is 54.6 m.

Example 13. As observed from the top of a lighthouse, 100 m high above sea level, the angles of depression of a ship, sailing directly towards it, change from 30° to 60°. Find the distance travelled by the ship during the period of observation.(Use \(\sqrt{3}\)=1.73)

Solution:

Given

As observed from the top of a lighthouse, 100 m high above sea level, the angles of depression of a ship, sailing directly towards it, change from 30° to 60°.

Let the height of the lighthouse be A B=100 m and let D be the ship which is sailing towards it. Also, let D C=x m.

Applications Of Trigonometry The Distance Travelled By The Ship During The Period Of Observation

Here, \(\angle C A D =60^{\circ}-30^{\circ}=30^{\circ}\)

⇒ \(\angle C D A =\angle C A D\)

D C=A C=x (angles opposite to equal sides are equal)

Now, in right \(\triangle A B C\),

⇒ \(\sin 60^{\circ}=\frac{A B}{A C} \quad \Rightarrow \quad \frac{\sqrt{3}}{2}=\frac{100}{D C}\)

⇒ \(D C=\frac{200}{\sqrt{3}}=\frac{200 \sqrt{3}}{3}=\frac{200 \times 1.73}{3}\)

⇒ \(D C=\frac{346}{3}\)=115.3 m

Hence, the distance travelled by ship is 115.3 m.

Question 14. An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45° and 60° respectively. Find the width of the river. (Use \(\sqrt{3}\)=1.73)

Solution:

Given

An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45° and 60° respectively.

Let an aeroplane P fly at a height of 300 m above the x-ground

PM = 300m

Depression angles are \(\angle X P A=45^{\circ}\) and \(\angle Y P B=60^{\circ}\).

Applications Of Trigonometry An Aeroplane Is Flying At A Height Above The Ground

Now, in right \(\triangle P M A\),(alternate angles)

⇒ \(\tan 45^{\circ}=\frac{P M}{A M} \quad \Rightarrow \quad 1=\frac{300}{A M} . \quad \Rightarrow \quad A M=300 \mathrm{~m}\)

In right \(\triangle P M B\),

⇒ \(\tan 60^{\circ}=\frac{P M}{M B} \quad \Rightarrow \quad \sqrt{3}=\frac{300}{M B} \quad \Rightarrow \quad M B=\frac{300}{\sqrt{3}}=100 \sqrt{3}\)

Width of river =A M+M B

= \((300+100 \sqrt{3}) \mathrm{m}=(300+100 \times 1.732) \mathrm{m}\)

=(300+173.2) m=473.20 m

Hence, the width of the river = 473.20 m

Example 15. A man observes a car from the top of a tower, which is moving towards the tower at a uniform speed. If the angle of depression of the car changes from 30° to 45° in 12 minutes, find the time taken by the car now to reach the tower.

Solution:

Given

5. A man observes a car from the top of a tower, which is moving towards the tower at a uniform speed. If the angle of depression of the car changes from 30° to 45° in 12 minutes

Let AB = h m be the height of the tower.

Applications Of Trigonometry The Time Taken By The Car Now To Reach The Tower

Since the depression angle changes from 30° to 45° in 12 minutes, therefore, time taken from D to C = 12 min.

Let DC = x m and CB -y m

Now, in right \(\triangle A B C\),

⇒ \(\tan 45^{\circ}=\frac{h}{y} \Rightarrow 1=\frac{h}{y} \Rightarrow h=y\)

In right \(\triangle A B D\),

⇒ \(\tan 30^{\circ} =\frac{h}{x+y} \quad \Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{x+y}\)

⇒ \(x+y =h \sqrt{3}\)

x+y =y \(\sqrt{3}\)

⇒ \(y(\sqrt{3}-1) =x\)

Now time taken by car in moving \(x \mathrm{~m}=12 \mathrm{~min}\)

Time taken by car in moving \(y(\sqrt{3}-1) \mathrm{m}=12 \mathrm{~min}\)

Time taken by car in moving \(y \mathrm{~m}=\frac{12}{\sqrt{3}-1} \mathrm{~min}\)

= \(\frac{12(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} \min =\frac{12(1.732+1)}{2}\)

=6 \(\times 2.732=16.39 \mathrm{~min}\)

Hence, required time =16.39 minutes.

The time taken by the car now to reach the tower =16.39 minutes.

Example 16. A bird is sitting on the top of an 80 m-high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such a way that it remains at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30°. Find the speed of flying of the bird.(Take, \(\sqrt{3}\)=1.73)

Solution:

Given

A bird is sitting on the top of an 80 m-high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such a way that it remains at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30°.

Let P be the position of a bird at the height of 80 m with the angle of elevation 45° from A.

Let after 2 seconds, it reaches Q from where its elevation angle is 30°.

Applications Of Trigonometry The Speed Of Flying Of The Bird On The Top Of A High Tree

Now, in right \(\triangle P B A\),

⇒ \(\tan 45^{\circ} =\frac{P B}{A B} \quad \Rightarrow \quad 1=\frac{80}{A B}\)

A B = 80 m

In right \(\triangle Q C A\),

⇒ \(\tan 30^{\circ} =\frac{Q C}{A C} \Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{80}{A C}\)

⇒ \(A C =80 \sqrt{3} \mathrm{~m}\)

BC =A C-A B

=80 \(\sqrt{3}-80 \quad=80(\sqrt{3}-1)\)

= \(80(1.732-1)=80 \times 0.732=58.56 \mathrm{~m}\)

Now, speed of bird =\(\frac{\text { Distance }}{\text { time }}=\frac{58.56}{2} \mathrm{~m} / \mathrm{sec}=29.28 \mathrm{~m} / \mathrm{sec}\) .

The speed of flying of the bird =\( 29.28 \mathrm{~m} / \mathrm{sec}\) .

Example 17. The angle of elevation of a cloud from a point 7z’ metres above a lake is a and the angle of depression of its reflection in the lake is p. Prove that the distance of the cloud from the point of observation is \(\frac{2 h \sec \alpha}{\tan \beta-\tan \alpha}\)

Solution:

Given:

The angle of elevation of a cloud from a point 7z’ metres above a lake is a and the angle of depression of its reflection in the lake is p.

Let AB be a lake. The angle of elevation of cloud P at point A on height TP from the lake is a and the angle of depression of the reflection F of cloud is \(\beta\).

Applications Of Trigonometry The Distance Of The Cloud From The Point Of Observation

⇒ \(\angle P C E=\alpha and \angle F C E=\beta\)

Let B P=F B=d

P E=B P-E B

P E=B P-A C=d-h

and F E=F B+B E=F B+A C=d+h

Let C E = x

In \(\triangle C E F\),

⇒ \(\tan \beta=\frac{E F}{C E} \quad \Rightarrow \quad \tan \beta=\frac{d+h}{x}\)  → Equation 1

In \(\triangle P E C\),

⇒ \(\tan \alpha=\frac{P E}{C E} \quad \Rightarrow \quad \tan \alpha=\frac{d-h}{x}\)  → Equation 2

Subtract equation (2) from equation (1), we get

⇒ \(\tan \beta-\tan \alpha=\frac{d+h}{x}-\frac{d-h}{x}=\frac{2 h}{x} \quad \Rightarrow \quad x=\frac{2 h}{\tan \beta-\tan \alpha}\)

In \(\triangle P E C\),

⇒ \(\cos \alpha=\frac{C E}{P C}=\frac{x}{P C} \quad \Rightarrow \quad P C=\frac{x}{\cos \alpha}=x \sec \alpha\)  Equation 3

PC = \(\frac{2 h \sec \alpha}{\tan \beta-\tan \alpha}\)

Hence, the distance of the cloud from the point of observation =\(\frac{2 h \sec \alpha}{\tan \beta-\tan \alpha}\)

Example 18. A man standing on the deck of a ship which is 14 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Find the distance of the hill from the ship and the height of the hill.

Solution:

Given

A man standing on the deck of a ship which is 14 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°.

Let the height of the hill AB = h m and let the position of man at 14 m above the sea level is D.

Let B C=x \(\mathrm{~m} \Rightarrow D E=x \mathrm{~m}\)

In right \(\triangle D C B\),

Applications Of Trigonometry The Distance Of The Hill From The Ship And The Height Of The Hill

⇒ \(\tan 30^{\circ}=\frac{D C}{B C} \quad \Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{14}{x} \quad \Rightarrow \quad x=14 \sqrt{3} \mathrm{~m}\)

In right \(\triangle A E D\),

⇒ \(\tan 60^{\circ} =\frac{A E}{D E} \quad \Rightarrow \quad \sqrt{3}=\frac{h-14}{x}\)

⇒ \(x \sqrt{3} =h-14 \quad \Rightarrow \quad 14 \sqrt{3} \times \sqrt{3}\)=h-14 [ from (1) ]

h =42+14=56 m

Hence, the die distance of the hill from the ship is x i.e., \(14 \sqrt{3} m\) and the height of the hill is 56 m.

Some Applications Of Trigonometry Exercise 9.1

Question 1. A circus artist is climbing a 20 m-long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°.

Solution :

Given:

A circus artist is climbing a 20 m-long rope, which is tightly stretched and tied from the top of a vertical pole to the ground.

In \(\triangle A B C\),

Applications Of Trigonometry The Angle Made By The Rope With The Ground Level

⇒ \(\sin 30^{\circ} =\frac{A B}{A C}\)

⇒ \(\frac{1}{2}=\frac{A B}{20}\)

B =10

Height of pole = 10 m

Question 2. A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Solution :

Given

A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m.

Let the part CD of the tree BD break in the air and touch the ground at point A.

Applications Of Trigonometry A Tree Breaks Due To Storm And Broken Part And The Height Of The Tree

According to the problem,

AB = 8 m and \(\angle\) BAC = 30°

In \(\triangle A B C\),

⇒ \(\tan 30^{\circ} =\frac{B C}{A B} \quad \stackrel{A 0^{\circ}}{\longleftrightarrow} \mathrm{m}_8\)

⇒ \(\frac{1}{\sqrt{3}} =\frac{B C}{8}\)

⇒ \(B C =\frac{8}{\sqrt{3}} \mathrm{~m}\)

and \(\cos 30^{\circ} =\frac{A B}{A C} \Rightarrow \frac{\sqrt{3}}{2}=\frac{8}{A C}\)

⇒ \(A C =\frac{16}{\sqrt{3}} \mathrm{~m}\)

C D =\(\frac{16}{\sqrt{3}} \mathrm{~m}\) (A C=C D)

Now, the height of the tree = BC + CD

=\(\frac{8}{\sqrt{3}}+\frac{16}{\sqrt{3}}=\frac{24}{\sqrt{3}}=8 \sqrt{3} \mathrm{~m}\)

The height of the tree =\( 8 \sqrt{3} \mathrm{~m}\)

Question 3. A contractor plans to install two slides for the children to play in a park. For children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Solution :

Given

A contractor plans to install two slides for the children to play in a park. For children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground.

Let the slide for elder children be AC and for younger children be DE.

Applications Of Trigonometry The Length Of The Slide In Each Case

In \(\triangle A B C\),

⇒ \(\angle A B C=90^{\circ}\)

A B=3 m

⇒ \(\sin 60^{\circ}=\frac{A B}{A C} \Rightarrow \frac{\sqrt{3}}{2}=\frac{3}{A C}\)

A C=\(\frac{6}{\sqrt{3}}=2 \sqrt{3} \mathrm{~m}\)

In \(\triangle B D E\),

⇒ \(\angle D B E =90^{\circ}\)

B E=1.5 m

⇒ \(\sin 30^{\circ} =\frac{B D}{D E}\)

⇒ \(\frac{1}{2} =\frac{1.5}{D E} \Rightarrow D E\) =3 m

Length of slide for elder children

=2\( \sqrt{3} \mathrm{~m}\)

and length of slide for younger children =3 m

Question 4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Solution :

Let AB be the tower.

The angle of elevation of the top of the tower from point C, 30m away from A is 30°.

Applications Of Trigonometry The Angle Of Elevation Of The Top Of A Tower From A Point On The Ground

In \(\triangle B A C\),

⇒ \(\tan 30^{\circ}=\frac{A B}{A C}\)

⇒ \(\frac{1}{\sqrt{3}}=\frac{A B}{30}\)

⇒ \(A B=\frac{30}{\sqrt{3}}=10 \sqrt{3} \mathrm{~m}\)

Height of the tower =\(10 \sqrt{3} \mathrm{~m}\)

Question 5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Solution:

Given

A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°.

Let the height of kite A from the ground is 60 m and AC is the string.

Applications Of Trigonometry The Length OF The String, Assuming That There Is No Slack In The String

Given : \(\angle A C B=60^{\circ}\)

In \(\triangle A B C\),

⇒ \(\sin 60^{\circ}=\frac{60}{A C}\)

⇒ \(\frac{\sqrt{3}}{2}=\frac{60}{A C}\)

A C=\(\frac{120}{\sqrt{3}}=40 \sqrt{3} \mathrm{~m}\)

Length of string =40 \(\sqrt{3}\) m

Question 6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building

Solution:

Given

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building.

Let, the height of building AN = 30 m

Here, BM = height of boy = 1.5 m

DN = BM = 1.5 m

AD = AN -AD = 30- 1.5 = 28.5 m

Applications Of Trigonometry The Distance Walked Towards The Building

In \(\triangle A C D\),

⇒ \(\tan 60^{\circ} =\frac{A D}{C D} \quad \Rightarrow \quad \sqrt{3}=\frac{28.5}{C D}\)

⇒ \(C D =\frac{28.5}{\sqrt{3}}=9.5 \sqrt{3} \mathrm{~m}\)

In \(\triangle A B D\),

⇒ \(\tan 30^{\circ}=\frac{A D}{B D} \Rightarrow \frac{1}{\sqrt{3}}=\frac{28.5}{B D}\)

⇒ \(B D=28.5 \sqrt{3} \mathrm{~m}\)

Now, \(B C=B D-C D\)

=28.5 \(\sqrt{3}-9.5 \sqrt{3}=19 \sqrt{3} \mathrm{~m}\)

Distance walked by boy towards the building \(19 \sqrt{3} \mathrm{~m}\)

Question 7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Solution:

Given

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively.

Let, CD be the height of the transmission tower.

Here, the height of the building

Applications Of Trigonometry The Height Of The Tower From A Point On The Ground

BC = 20 m

In \(\triangle A B C\),

⇒ \(\tan 45^{\circ} =\frac{B C}{A B}\)

=\(\frac{20}{A B}\)

A B =20 m

In \(\triangle A B D\),

⇒ \(\tan 60^{\circ}=\frac{B D}{A B} \quad \Rightarrow \sqrt{3}=\frac{B D}{20}\)

B D=20 \(\sqrt{3} m\)

B C+C D=20 \(\sqrt{3}\)

20+C D=20 \(\sqrt{3}\)

C D=20\((\sqrt{3}-1) \mathrm{m}\)

Height of transmission tower =20\((\sqrt{3}-1) \mathrm{m}\)

Question 8. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Solution:

Given

A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°.

Let AB be the statue of height 1.6 m at the top of pedestal BC.

⇒ \(\angle B D C =45^{\circ}\)

and \(\angle A D C =60^{\circ}\)

Let C D =x

and BC =h

In \(\triangle B C D\),

Applications Of Trigonometry The Height Of The Pedestal Of The Statue

⇒ \(\tan 45^{\circ} =\frac{B C}{C D}\)

1 =\(\frac{h}{x}\)

h =x (1)

In ACD.

⇒ \(\tan 60^{\circ} =\frac{A C}{C D}\)

⇒ \(\sqrt{3} =\frac{h+1.6}{x}\)

⇒ \(\sqrt{3} =\frac{h+1.6}{h}\) [from eqn. (1)]

⇒ \(\sqrt{3} h\) =h+1.6

h\((\sqrt{3}-1)\) =1.6

h =\(\frac{1.6}{\sqrt{3}-1}=\frac{1.6 \times(\sqrt{3}+1)}{3-1}\)

=0.8(1.732+1)=2.1856

Height of pedestal \(\approx 2.18 \mathrm{~m}\)

Question 9. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building

Solution:

Given

The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high

Let AB be the tower and CD be the building.

Applications Of Trigonometry The Angle Of Elevation Of The Top Of A Building From The Foot Of The Tower

Here, the height of Tower AB is 50 m

⇒ \(\angle A C B=60^{\circ}\),

⇒ \(\angle D B C=30^{\circ}\)

In \(\triangle A B C\),

⇒ \(\sqrt{3}=\frac{50}{B C}\)

⇒ \(\tan 60^{\circ}=\frac{A B}{B C}\)

B C=\(\frac{50}{\sqrt{3}}\)

In \(\triangle B C D\),

⇒ \(\tan 30^{\circ}=\frac{C D}{B C} \quad \Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{C D}{B C}\)

C D=\(\frac{B C}{\sqrt{3}}=\frac{50}{\sqrt{3} \times \sqrt{3}}=\frac{50}{3}\)

C D=16.67 m

Height of building =16.67 m

Question 10. Two poles of equal height are standing opposite each other on either side of the road, which is 80 m wide. them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the points from the poles.

Solution:

Given:

Two poles of equal height are standing opposite each other on either side of the road, which is 80 m wide. them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively.

Let two poles AB and CD of equal heights ‘h’ be on either side of a road 80 m broad.

Applications Of Trigonometry Two Poles Of Equal Heights Are Standing Opposite Each Other On Either Side Of The Road

At point E,

Given : \(\angle C E D=60^{\circ}\) and \(\angle A E B=30^{\circ}\)

Let D E=x

B E = 80-x

In \(\triangle C D E\),

⇒ \(\tan 60^{\circ} =\frac{C D}{D E} \quad \Rightarrow \quad \sqrt{3}=\frac{h}{x}\)

h = \(x\sqrt{3}\) Equation (1)

In \(\triangle A B E\)

⇒ \(\tan 30^{\circ}=\frac{A B}{B E} \quad \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{80-x}\)

⇒ \(\frac{1}{\sqrt{3}}=\frac{x \sqrt{3}}{80-x}\) [from Equation (1)]

3 \(\mathrm{r}=80-x \Rightarrow 4 x=80\)

x = 20

80-x=80-20=60 and h=20 \(\sqrt{3}\)

Height of each pole =20 \(\sqrt{3} \mathrm{~m}\)

Distance of two poles from point E=20 m and 60 m.

Question 11. A TV tower stands vertically on the bank of an A canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point D 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.

Applications Of Trigonometry The Height Of The Tower And The Width Of The Canal

Solution:

Given

A TV tower stands vertically on the bank of an A canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point D 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°.

In \(\triangle B C\).

⇒ \(\tan 60^{\circ} =\frac{A B}{B C}\)

⇒ \(\sqrt{3} =\frac{A B}{B C}\)

⇒ \(A B =\sqrt{3} \cdot B C\)

In \(\triangle A B D\),

⇒ \(\tan 30^{\circ} =\frac{A B}{B D}\)

⇒ \(\frac{1}{\sqrt{3}} =\frac{\sqrt{3} B C}{B C+C D}\)

3 BC =B C+C D

2 BC =20

BC =10

Put in equation (1),

A B =\(10 \sqrt{3} \mathrm{~m}\)

Height of tower =10 \(\sqrt{3} \mathrm{~m}\)

and width of canal =10 m

Question 12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Solution :

Let CD be a building of height 7m and AB be a cable tower.

Given :

⇒ \(\angle A D E=60^{\circ}\) and \(\angle D B C=45^{\circ}\)

Applications Of Trigonometry The Height Of The Tower From The Top Of A High Building

In \(\triangle B C D\),

⇒ \(\tan 45^{\circ} =\frac{D C}{B C} \Rightarrow 1=\frac{7}{B C}\)

B C = 7 m

In \(\triangle A D E\),

⇒ \(\tan 60^{\circ}=\frac{A E}{D E} \Rightarrow \sqrt{3}=\frac{A E}{B C} \quad(D E=B C)\)

⇒ \(A E=\sqrt{3} B C=1.732 \times 7=12.124 \mathrm{~m}\)

⇒ \(A B=A E+B E=12.124+7=19.124 \mathrm{~m}\)

Height of tower =19.124 m

Question 13. As observed from the top of a 75 m high lighthouse from the sea level, the angles of depression of the two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships

Solution :

Given

As observed from the top of a 75 m high lighthouse from the sea level, the angles of depression of the two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse

Let AB be a lighthouse whose height is 75 m. The position of the two ships is at C and D.

Applications Of Trigonometry The Distance Between The Two Ships

In \(\triangle A B C\),

⇒ \(\tan 45^{\circ} =\frac{A B}{A C}\)

1 =\(\frac{75}{A C} \Rightarrow A C=75 \mathrm{~m}\)

In \(\triangle A B D\),

⇒ \(\tan 30^{\circ}=\frac{A B}{A D} \Rightarrow \frac{1}{\sqrt{3}}=\frac{75}{A D}\)

⇒ \(A D=75 \sqrt{3} \Rightarrow C D+A C=75 \sqrt{3}\)

⇒ \(C D=75 \sqrt{3}-75=75(\sqrt{3}-1)\)

Distance between two ships =\(75(\sqrt{3}-1)\) m

Question 14. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After elevation reduces to 30°. Find the distance travelled by the balloon during the interval.

Applications Of Trigonometry The Angle Of Elevation Of The Balloon From The Eyes OF The Girl

Solution:

Given

A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After elevation reduces to 30°.

Let C be the position of the girl. The two positions of the balloon are A and P.

PD = AB = 88.2 – 1.2 = 87 m

In right \(\triangle A B C\),

Applications Of Trigonometry The Distance Travelled By The Balloon During The Interval

⇒ \(\tan 60^{\circ} =\frac{A B}{B C}\)

⇒ \(\sqrt{3}=\frac{87}{B C} \quad \Rightarrow B C=\frac{87}{\sqrt{3}}\)  → Equation 1

In right \(\triangle PDC\),

⇒ \(\tan 30^{\circ}=\frac{P D}{C D}\)

⇒ \(\frac{1}{\sqrt{3}} =\frac{87}{C D} \quad \Rightarrow C D=87 \sqrt{3}\)

Now, BD = C D – B C = 87

⇒ \(\sqrt{3}-\frac{87}{\sqrt{3}}=87\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)\)

=87 \(\times \frac{3-1}{\sqrt{3}}=\frac{2 \times 87}{\sqrt{3}} \mathrm{~m}\)

= \(\frac{2 \times 87}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{2 \times 87 \times \sqrt{3}}{3}\)

= 58 \(\sqrt{3} \mathrm{~m}\)

Therefore, distance between two positions of balloons =58 \(\sqrt{3} m\)

Question 15. Question 15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Solution :

Given

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°.

Let the height of the tower be AB and the two positions of the car be C and D.

Applications Of Trigonometry The Time Taken By The Car To Reach The Foot Of The Tower

⇒ \(\angle D B C=\angle D B X-\angle C B X=60^{\circ}-30^{\circ}=30^{\circ}\)

In \(\triangle B D C\),

⇒ \(\angle D B C =\angle D C B\)

C D =B D

⇒ \(\left({cach} 30^{\circ}\right)\)

(the sides opposite to equal angles are equal) In right \(\triangle B A D\),

⇒ \(\cos 60^{\circ}=\frac{A D}{D B} \Rightarrow \frac{1}{2}=\frac{A D}{D B} \Rightarrow D B=2 A D\)

CD = 2 AD

Now, time taken to cover distance CD = 6 sec.

Time taken to cover distance 2AD = 6 sec.

Time is taken to cover distance AD = 3 sec.

Question 16. The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Solution :

Given

The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary.

Let AB be a tower of height h. Two points C and D are at 4 m and 9 m distances respectively from B.

Let, \(\angle A D C=\theta\)

Applications Of Trigonometry The Angles Of Elevation Of The Top Of Tower From Two Points At A Distance

⇒ \(\angle A C B=90^{\circ}-\theta\)

In \(\triangle A B D\),

⇒ \(\tan \theta=\frac{A B}{B D}=\frac{h}{9}\)  → Equation 1

In \(\triangle A B C\),

⇒ \(\tan \left(90^{\circ}-\theta\right)=\frac{A B}{B C}\)

⇒ \(\cot \theta=\frac{h}{4}\)

⇒ \(\frac{1}{\tan \theta}=\frac{h}{4} \quad \Rightarrow \frac{9}{h}=\frac{h}{4}\) [from equation 1 ]

⇒ \(h^2=36 \quad \Rightarrow h=6\)

Height of tower =6 m

Hence Proved.

NCERT Exemplar For Class 10 Maths Chapter 8 Introduction To Trigonometry

Introduction To Trigonometry

The word ‘trigonometry’ is derived from the Greek words: trigonon and metron. The word trigonon means a figure formed by three sides i.e., triangle, and metron means a measure. So, we can say that in trigonometry we solve the problems related to the sides and angles of a triangle.

NCERT Exemplar Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Identity

  1. An equation is a statement of equality between two expressions that is true for all values of the variable involved (taking into consideration the domain i.e., limitations of the variable), and is called an identity.
  2. An equation that involves trigonometric ratios of an angle and is true for all values of the angle is called a trigonometric identity.

Angle

An angle is considered as a figure obtained by rotating a given ray about its endpoint. The measure of an angle is the amount of rotation from its initial side to the terminal side. If the ray rotates in an anticlockwise direction then the angle will be positive. If the ray rotates in a clockwise direction then the angle will be negative.

Trigonometry An Angle Is Obtained A Given Ray

Trigonometric Ratios

The ratio of any two sides of a right-angled triangle is called trigonometric ratio. In the adjoining figure, \(\angle\)YAX = 0 is an acute angle. Consider a point C on ray AY. Draw perpendicular CB from C to AX.

⇒ \(\triangle\) ABC is a right-angled triangle in which

Trigonometry The Ratio Of Sides Of A Right Angle Triangle

⇒ \(\angle\)ABC = 90°

In \(\triangle\) ABC, let \(\angle\)BAC = 0

For \(\angle\)BAC = \(\theta\),

perpendicular P = BC

base B = AB

And hypotenuse H = AC

The ratio \(\frac{\text { perpendicular }}{\text { hypotenuse }}\) is called the sine of \(\theta\) and is written as \(\sin \theta\).

⇒ \(\sin \theta=\frac{P}{H}=\frac{B C}{A C}\)

The ratio \(\frac{\text { base }}{\text { hypotenuse }}\) is called the cosine of \(\theta\) and is written as \(\cos \theta\).

⇒ \(\cos \theta=\frac{B}{H}=\frac{A B}{A C}\)

The ratio \(\frac{\text { perpendicular }}{\text { base }}\) is called the tangent of \(\theta\) and is written as \(\tan \theta\).

∴ \(\tan \theta=\frac{P}{B}=\frac{B C}{A B}\)

Introduction to Trigonometry

The ratio \(\frac{\text { base }}{\text { perpendicular }}\) is called the cotangent of \(\theta\) and is written as \(\cot \theta\).

⇒ \(\cot \theta=\frac{B}{P}=\frac{A B}{B C}\)

The ratio \(\frac{\text { hypotenuse }}{\text { base }}\) is called the secant of \(\theta\) and is written as \(\sec \theta\).

⇒ \(\sec \theta=\frac{H}{B}=\frac{A C}{A B}\)

The ratio \(\frac{\text { hypotenuse }}{\text { perpendicular }}\) is called the cosecant of \(\theta\) and is written as \(cosec\theta\).

cosec \(\theta=\frac{H}{P}=\frac{A C}{B C}\)

Therefore, \(\sin \theta=\frac{1}{{cosec} \theta}=\frac{\text { perpendicular }}{\text { hypotenuse }}\), cosec \(\theta=\frac{1}{\sin \theta}=\frac{\text { hypotenuse }}{\text { perpendicular }}\)

⇒ \(\cos \theta=\frac{1}{\sec \theta}=\frac{\text { base }}{\text { hypotenuse }}\), \(\sec \theta=\frac{1}{\cos \theta}=\frac{\text { hypotenuse }}{\text { base }}\)

⇒ \(\tan \theta=\frac{1}{\cot \theta}=\frac{\sin \theta}{\cos \theta}=\frac{\text { perpendicular }}{\text { base }}\), \(\cot \theta=\frac{1}{\tan \theta}=\frac{\cos \theta}{\sin \theta}=\frac{\text { base }}{\text { perpendicular }}\)

Remember :

  •  \(\sin \theta \neq \sin \times \theta\)
  • \(\cos \theta \neq \cos \times \theta\)
  • \(\sin ^2 \theta =(\sin \theta)^2 \neq \sin ^2 \theta^2 \neq \sin ^2\)
  • \(cosec \theta =\frac{1}{\sin \theta}=(\sin \theta)^{-1} \neq \sin ^{-1} \theta\)

Perpendicular, Base, And Hypotenuse In A Right-Angled Triangle

See carefully the following right-angled triangles :

Trigonometry The Values Of Trigonometric Ratio In The Different Angles

Let us see the values of trigonometric ratios in the different figures in terms of AB, BC, and AC.

Remember:

  1. The side of the triangle at which the right angle (90°) and the given angle lie, is called the base.
  2.  The side opposite to the 90° angle is called the hypotenuse.
  3. The third remaining side is called the perpendicular.

Now,

In (1). \(\sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{A B}{B C}\)

In (2). \(\tan \theta=\frac{\text { perpendicular }}{\text { base }}=\frac{A C}{A B}\)

In (3). \(\sec \theta=\frac{\text { hypotenuse }}{\text { base }}=\frac{B C}{A C}\)

In (4). cosec \(\theta=\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{A C}{B C}\)

In (5). \(\cos \theta=\frac{\text { base }}{\text { hypotenuse }}=\frac{A C}{A B}\)

Solved Examples

Example 1. In \(\triangle\) ABC, \(\angle\) B = 90°, if AB = 5 cm, BC = 12 cm, then find the values of the following :

  1. sin A
  2. cos A
  3. cot A
  4. cosec C
  5. sec C
  6. tan C

Solution.

In \(\triangle\) A B C, from Pythagoras theorem

⇒ \(A C^2=A B^2+B C^2 =5^2+12^2\)

Trigonometry In Triangle ABC Form Pythagoras Theorem

=25+144=169

A C=13 cm

For \(\angle\) A, the base is AB, perpendicular is BC. For \(\angle C\), the base is BC, the perpendicular is AB, while the hypotenuse is the same i.e., AC for both angles.

  1. \(\sin A=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{B C}{A C}=\frac{12}{13}\)
  2. \(\cos A=\frac{\text { base }}{\text { hypotenuse }}=\frac{A B}{A C}=\frac{5}{13}\)
  3. \(\cot A=\frac{\text { base }}{\text { perpendicular }}=\frac{A B}{B C}=\frac{5}{12}\)
  4. cosec C=\(\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{A C}{A B}=\frac{13}{5}\)
  5. \(\sec C=\frac{\text { hypotenuse }}{\text { base }}=\frac{A C}{B C}=\frac{13}{12}\)
  6.  \(\tan C=\frac{\text { perpendicular }}{\text { base }}=\frac{A B}{B C}=\frac{5}{12}\)

Question 2. In \(\triangle\) ABC, \(\angle B=90^{\circ}\) and \(\sin A=\frac{4}{5}\), then find the values of all other trigonometric ratios for \(\angle\) A.

Solution:

We know that, \(\sin A=\frac{4}{5}=\frac{\text { perpendicular }}{\text { hypotenuse }}\)

Now construct a \(\triangle\) A B C in which

Trigonometry In Triangle ABC The Values Of All Other Trigonometric Ratio

⇒ \(\angle B=90^{\circ}\), B C=4 k and A C=5 k .

In \(\triangle\) A B C, from Pythagoras theorem

⇒ \(A B^2+B C^2=A C^2\)

⇒ \(A B^2=A C^2-B C^2=(5 k)^2-(4 k)^2\)

=25 \(k^2-16 k^2=9 k^2\)

AB = 3k

Now,\(\cos A=\frac{\text { base }}{\text { hypotenuse }}=\frac{A B}{A C}=\frac{3 k}{5 k}=\frac{3}{5}\)

⇒ \(\tan A=\frac{\text { perpendicular }}{\text { base }}=\frac{B C}{A B}=\frac{4 k}{3 k}=\frac{4}{3}\)

cosec A=\(\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{A C}{B C}=\frac{5 k}{4 k}=\frac{5}{4}\)

⇒ \(\sec A=\frac{\text { hypotenuse }}{\text { base }}=\frac{A C}{A B}=\frac{5 k}{3 k}=\frac{5}{3}\)

∴ \(\cot A=\frac{\text { base }}{\text { perpendicular }}=\frac{A B}{B C}=\frac{3 k}{4 k}=\frac{3}{4}\)

Example 3. In \(\triangle\) PQR, \(\angle R=90^{\circ}\) and \(\tan \theta=\frac{5}{12}\) where \(\angle\) QPR = \(\theta\). Find the values of all other trigonometric ratios for \(\angle \theta\).

Solution:

We know that,

⇒ \(\tan \theta=\frac{\text { perpendicular }}{\text { base }}=\frac{5}{12}\)

Trigonometry The Values Of Trigonometric Ratio For Theta

Now, construct a \(\triangle P Q R\) in which

P R=12 k, Q R=5 k and \(\angle Q R P=90^{=}\). }

Let \(\angle Q P R=\theta\)

In \(\triangle\) PQR,

From Pythagoras theorem

⇒ \(P Q^2=P R^2+Q R^2=(12 k)^2+(5 k)^2=144 k^2+25 k^2=169 k^2\)

PQ = 13k

Now,\(\sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{Q R}{P Q}=\frac{5 k}{13 k}=\frac{5}{13}\)

⇒ \(\cos \theta=\frac{\text { base }}{\text { hypotenuse }}=\frac{P R}{P Q}=\frac{12 k}{13 k}=\frac{12}{13}\)

⇒ \(cosec 0 =\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{P(2}{Q R}=\frac{13 k}{5 k}=\frac{13}{5}\)

⇒ \(\sec \theta =\frac{\text { hypotenuse }}{\text { base }}=\frac{P Q}{P R}=\frac{13 k}{12 k}=\frac{13}{12}\)

⇒ \(\cot \theta =\frac{\text { basse }}{\text { perpendicular }}=\frac{P R}{Q R}=\frac{12 k}{5 k}=\frac{12}{5}\)

Example 4. In \(\triangle A B C\), \(\angle C=90^{\circ}\) and cosec A=\(\frac{13}{12}\) find the values of all other trigonometric ratios for \(\angle A\).

Solution:

We know that,

cosec A=\(\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{13}{12}\)

Now, construct a \(\triangle A B C\) in which A B=13 k, B C=12 k and \(\angle A C B=90^{\circ}\).

Trigonometry The Value Of Other Trigonometric Ratio For Angle A

In \(\triangle A B C\), from Pythagoras theorem

⇒ \(A C^2+B C^2 =A B^2\)

⇒ \(A C^2 =A B^2-B C^2=(13 k)^2-(12 k)^2\)

=\(169 k^2-144 k^2=25 k^2\)

AC = 5 k

Now, \(\sin A=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{B C}{A B}=\frac{12 k}{13 k}=\frac{12}{13}\)

⇒ \(\cos A=\frac{\text { base }}{\text { hypotenuse }}=\frac{A C}{A B}=\frac{5 k}{13 k}=\frac{5}{13}\)

⇒ \(\tan A=\frac{\text { perpendicular }}{\text { base }}=\frac{B C}{A C}=\frac{12 k}{5 k}=\frac{12}{5}\)

⇒ \(\sec A=\frac{\text { hypotenuse }}{\text { base }}=\frac{A B}{A C}=\frac{13 k}{5 k}=\frac{13}{5}\)

∴ \(\cot A=\frac{\text { base }}{\text { perpendicular }}=\frac{A C}{B C}=\frac{5 k}{12 k}=\frac{5}{12}\)

Example 5. If cos A=\(\frac{1}{3}\), then find the values of sin A and tan A

Solution:

Construct a right-angled triangle ABC in which cos A=\(\frac{1}{3}=\frac{\text { base }}{\text { hypotenuse }}=\frac{A B}{A C}\)

Let AB = k and AC = 3 k

From Pythagoras theorem,

Trigonometry The Values Of Sin A And Tan A

From Pythagoras theorem, \(B C^2+A B^2 =A C^2\)

⇒ \(B C^2 =A C^2-A B^2=(3 k)^2-(k)^2\)

= \(9 k^2-k^2=8 k^2\)

BC = \(\sqrt{8 k^2}=2 \sqrt{2} k\)

The values of sin A and tan A are

⇒ \(\sin A=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{B C}{A C}=\frac{2 \sqrt{2} k}{3 k}=\frac{2 \sqrt{2}}{3}\)

and \(\tan A=\frac{\text { perpendicular }}{\text { base }}=\frac{B C}{A B}=\frac{2 \sqrt{2} k}{k}=2 \sqrt{2}\)

Example 6. In \(\triangle A B C\), \(\tan B=\sqrt{3}\) find the values of cosec B and cos B.

Solution:

Construct a right-angled triangle ABC in which

⇒ \(\tan B=\sqrt{3}=\frac{\sqrt{3}}{1}=\frac{\text { perpendicular }}{\text { base }}=\frac{A C}{A B}\)

Let \(A C=\sqrt{3}\) k and AB = k

From Pythagoras theorem

Trigonometry The Values Of Cosec B And Cos B

⇒ \(B C^2 =A B^2+A C^2\)

=\((k)^2+(\sqrt{3} k)^2\)

⇒ \(B C^2 =k^2+3 k^2=4 k^2\)

BC =2 k

The values of cosec B and cos B are

⇒ \(cosec B =\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{B C}{A C}=\frac{2 k}{\sqrt{3} k}=\frac{2}{\sqrt{3}}\)

and \(\cos B =\frac{\text { base }}{\text { hypotenuse }}=\frac{A B}{B C}=\frac{k}{2 k}=\frac{1}{2}\)

Example 7. If \(\cos \theta=\frac{4}{5}\), then find the value of \(\left(\sin \theta \cos \theta+\tan ^2 \theta\right) \)

Solution:

Given that, \(\cos \theta=\frac{4}{5}\)

Construct a right-angled \(\triangle A B C\) in which \(\angle B A C=90^{\circ}\), AC = 4 k and BC = 5k.

In \(\triangle ABC\),

Trigonometry Construction Of An Right Angle Triangle ABC

From Pythagoras theorem

⇒ \(A B^2+A C^2 =B C^2\)

⇒ \(A B^2 =B C^2-A C^2=(5 k)^2-(4 k)^2\)

= \(25 k^2-16 k^2=9 k^2\)

AB = 3k

and \(\sin \theta =\frac{A B}{B C}=\frac{3 k}{5 k}=\frac{3}{5}\)

⇒ \(\tan \theta =\frac{A B}{A C}=\frac{3 k}{4 k}=\frac{3}{4}\)

Now, \(\sin \theta \cos \theta+\tan ^2 \theta=\frac{3}{5} \cdot \frac{4}{5}+\left(\frac{3}{4}\right)^2=\frac{12}{25}+\frac{9}{16}\)

= \(\frac{192+225}{400}=\frac{417}{400}\)

The value of \(\left(\sin \theta \cos \theta+\tan ^2 \theta\right) \)= 417/400.

Example 8. If sec A=2, then find the value of \(\frac{1}{\cot A}+\frac{\cos A}{1+\sin A}\).

Solution:

Given that, \(\sec A=2=\frac{2}{1}\)

Construct a right-angled \(\triangle A B C\) in which

⇒ \(\angle A B C=90^{\circ}\), AB = k and AC = 2k .

In \(\triangle A B C\),

From Pythagoras theorem

Trigonometry Construction Of An Right Angle Triangle ABC From Pythagoras Theorem

⇒ \(A B^2+B C^2 = A C^2\)

⇒ \(B C^2 =A C^2-A B^2=(2 k)^2-(k)^2=4 k^2-k^2=3 k^2\)

⇒ \(B C=\sqrt{3} k\)

Now, \(\cot A=\frac{A B}{B C}=\frac{k}{\sqrt{3} k}=\frac{1}{\sqrt{3}}\)

⇒ \(\sin A=\frac{B C}{A C}=\frac{\sqrt{3} k}{2 k} =\frac{\sqrt{3}}{2}\) and \(\cos A=\frac{A B}{A C}=\frac{k}{2 k}=\frac{1}{2}\)

⇒ \(\frac{1}{\cot A}+\frac{\cos A}{1+\sin A}=\frac{1}{\frac{1}{\sqrt{3}}}+\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}=\sqrt{3}+\frac{\frac{1}{2}}{\frac{2+\sqrt{3}}{2}}\)

= \(\sqrt{3}+\frac{1 \times(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}=\sqrt{3}+\frac{2-\sqrt{3}}{4-3}\)

= \(\sqrt{3}+2-\sqrt{3}=2\)

The value of \(\frac{1}{\cot A}+\frac{\cos A}{1+\sin A}\) = 2

Example 9. If \(\cot A=\frac{b}{a}\), then prove that:

Solution:

⇒ \(\frac{a \sin A-b \cos A}{a \sin A+b \cos A}=\frac{a^2-b^2}{a^2+b^2}\)

We have,

L.H.S. =\(\frac{a \sin A-b \cos A}{a \sin A+b \cos A} =\frac{a \frac{\sin A}{\sin A}-b \frac{\cos A}{\sin A}}{a \frac{\sin A}{\sin A}+b \frac{\cos A}{\sin A}}\)

=\(\frac{a-b \cot A}{a+b \cot A}=\frac{a-b \times \frac{b}{a}}{a+b \times \frac{b}{a}}\) (dividing Nr. and Dr. by sin A ) (\(\cot A=\frac{b}{a}\))

=\(\frac{\frac{a^2-b^2}{a}}{\frac{a^2+b^2}{a}}=\frac{a^2-b^2}{a^2+b^2}\)= R.H.S

Hence Proved.

Example 10. In the adjoining figure, AM = BM and \(\angle B=90^{\circ}\). If \(\angle B C M=\theta\), then find the values of the following :

  1. sin θ
  2. tan θ
  3. sec θ

Solution:

In \(\triangle A B C\),

From Pythagoras theorem

Trigonometry In The Adjoining The Values Of Triangle ABC

⇒ \(B C^2 =A C^2-A B^2=b^2-(2 a)^2=b^2-4 a^2\)

⇒ \(B C =\sqrt{b^2-4 a^2}\)

Now, \(B M=\frac{A B}{2}=a\)

In \(\triangle B C M\),

From Pythagoras theorem

⇒ \(C M^2 =B C^2+B M^2=\left(b^2-4 a^2\right)+a^2=b^2-3 a^2\)

⇒ \(C M =\sqrt{b^2-3 a^2}\)

⇒ \(\sin \theta=\frac{B M}{C M}=\frac{a}{\sqrt{b^2-3 a^2}}\)

⇒ \(\tan \theta=\frac{B M}{B C}=\frac{a}{\sqrt{b^2-4 a^2}}\)

∴ \(\sec \theta=\frac{C M}{B C}=\frac{\sqrt{b^2-3 a^2}}{\sqrt{b^2-4 a^2}}\)

Example 11. In the adjoining figure, \(\angle B C D=\angle A D B(each 90^{\circ} )\). If B C=3 cm and the length of the side opposite \(\angle C\) in \(\triangle B C D\) is 5 cm, then find the square root of the length of the side opposite to \(\angle D\) in MDB.

Solution:

Draw DE \(\perp A B\)

In right \(\triangle B C D\), by Pythagoras theorem,

Trigonometry The Square Root Of Length Side Opposite

In right \(\triangle A E D\), by Pythagoras theorem,

⇒ \(A D^2=A E^2+D E^2\)

⇒ \(y^2=x^2+9\)

In right \(\triangle A D B\), by Pythagoras theorem,

Trigonometry In Right Triangle ADB By Pythagoras Theorem

⇒ \(A B^2=A D^2+D B^2\)

⇒ \((x+4)^2=y^2+25\)

⇒\(x^2+8 x+16=\left(x^2+9\right)+25\)

8 x=18

x=\(\frac{18}{8}=2.25 \mathrm{~cm}\)

⇒\(A B=x+4=2.25+4=6.25 \mathrm{~cm}\)

⇒\(\sqrt{A B}=\sqrt{6.25}=2.5 \mathrm{~cm}\)

Signs Of The Trigonometric Ratios

Let a rotating line rotate \(\angle XOA\) = \(\theta\) in an anticlockwise direction, starting from its initial position OX. Here, PM is perpendicular from P to OX where P is a point on side OA.

Trigonometry Signs Of The Trigonometric Ratios

In the first quadrant, 0 is the acute angle.

Here, OM > 0, PM > 0, OP > 0

Now,

⇒ \(\sin \theta=\frac{P M}{O P}>0\) ,\(\cos \theta=\frac{O M}{O P}>0\)

⇒ \(\tan \theta=\frac{P M}{O M}>0\) ,\(\cot \theta=\frac{O M}{P M}>0\)

⇒ \(\sec \theta=\frac{O P}{O M}>0 , cosec \theta=\frac{O P}{P M}>0\)

Therefore, all trigonometric ratios for all angles in the first quadrant are positive.

Trigonometric Ratios Of Specific Angles

1. Trigonometric Ratios for 30° and 60°

Trigonometry Trigonometric Ratios Of Specific Angles

⇒ \(\triangle ABC\) is an equilateral triangle whose side is ‘2a’.

⇒ \(\triangle A B C\) is an equilateral triangle whose side is ‘ 2 a ‘.

⇒ \(\angle A B C=\angle A C B=\angle B A C=60^{\circ}\)

AD is perpendicular from A to BC.

and \(\angle BAD =\angle C A D=30^{\circ}\)

B D = CD = a.

In \(\triangle A B D\), from Pythagoras theorem

⇒ \(A D^2+B D^2=A B^2\)

⇒ \(A D^2+a^2 =(2 n)^2\)

⇒ \(A D^2 =3 a^2\)

A D =a \(\sqrt{3}\)

For \(30^{\circ}\), in \(\triangle A B D\)

Base AD = a \(\sqrt{3}\), perpendicular B D= a and hypotenuse AB = 2a

⇒ \(\sin 30^{\circ}=\frac{B D}{A B}=\frac{a}{2 a}=\frac{1}{2}\)

⇒ \(cosec 30^{\circ}=\frac{A B}{B D}=\frac{2 a}{a}=2\)

⇒ \(\cos 30^{\circ}=\frac{A D}{A B}=\frac{a \sqrt{3}}{2 a}=\frac{\sqrt{3}}{2}\)

⇒ \(\sec 30^{\circ}=\frac{A B}{A D}=\frac{2 a}{a \sqrt{3}}=\frac{2}{\sqrt{3}}\)

⇒ \(\tan 30^{\circ}=\frac{B D}{A D}=\frac{a}{a \sqrt{3}}=\frac{1}{\sqrt{3}}\)

⇒ \(\cot 30^{\circ}=\frac{A D}{B D}=\frac{a \sqrt{3}}{a}=\sqrt{3}\)

For \(60^{\circ}\), in \(\triangle A B D\)

⇒ \(\sin 60^{\circ}=\frac{A D}{A B}=\frac{a \sqrt{3}}{2 a}=\frac{\sqrt{3}}{2}\)

⇒ \(\cos 60^{\circ}=\frac{B D}{A B}=\frac{a}{2 a}=\frac{1}{2}\)

⇒ \(\tan 60^{\circ}=\frac{A D}{B D}=\frac{a \sqrt{3}}{a}=\sqrt{3}\)

⇒ \(cosec 60^{\circ}=\frac{A B}{A D}=\frac{2 a}{a \sqrt{3}}=\frac{2}{\sqrt{3}}\)

⇒ \(\sec 60^{\circ}=\frac{A B}{B D}=\frac{2 a}{a}=2\)

⇒ \(\cot 60^{\circ}=\frac{B D}{A D}=\frac{a}{a \sqrt{3}}=\frac{1}{\sqrt{3}}\)

2. Trigonometric Ratios for 450

In \(\triangle\)ABC, \(\angle\)ABC = 90° and \(\angle\)BAC = 45°.

Trigonometry Ratios Of 45 Degrees

Therefore, \(\angle\)ACB = 45°

Let AB = BC = a

From Pythagoras theorem

In \(\triangle A B C\), \(\angle A B C=90^{\circ}\) and \(\angle B A C=45^{\circ}\). Therefore, \(\angle A C B=45^{\circ}\)

Let AB = BC = a

From Pythagoras theorem

⇒ \(A C^2 =A B^2+B C^2\)

=\( a^2+a^2=2 a^2\)

A C =a \(\sqrt{2}\)

For A=\(45^{\circ}\),

⇒ \(\sin 45^{\circ}=\frac{B C}{A C}=\frac{a}{a \sqrt{2}}=\frac{1}{\sqrt{2}}\)

⇒ \(cosec 45^{\circ}=\frac{A C}{B C}=\frac{a \sqrt{2}}{a}=\sqrt{2}\)

⇒ \(\cos 45^{\circ}=\frac{A B}{A C}=\frac{a}{a \sqrt{2}}=\frac{1}{\sqrt{2}}\)

⇒ \(\sec 45^{\circ}=\frac{A C}{A B}=\frac{a \sqrt{2}}{a}=\sqrt{2}\)

⇒ \(\tan 45^{\circ}=\frac{B C}{A B}=\frac{a}{a}\)=1

⇒ \(\cot 45^{\circ}=\frac{A B}{B C}=\frac{a}{a}=1\)

3. Trigonometric Ratios for 0°

Trigonometry Ratio For Zero

In \(\triangle \)ABC, \(\angle\)BAC = 0 and \(\angle\)ABC = 90°.

For angle \(\theta\), base = AB, perpendicular = BC and hypotenuse = AC.

In \(\triangle\)ABC, it is clear that as the value of ‘\(\theta\)’ decreases, the length of BC decreases, and for 0 = 0°, BC = 0 and AC = AB. Therefore,

Introduction to Trigonometry

⇒ \(\sin 0^{\circ}=\frac{B C}{A C}=\frac{0}{A C}=0\)

⇒ \(cosec 0^{\circ}=\frac{A C}{B C}=\frac{A C}{0}=\infty\)

⇒ \(\cos 0^{\circ}=\frac{A B}{A C}=\frac{A B}{A B}=1\)

⇒ \(\sec 0^{\circ}=\frac{A C}{A B}=\frac{A B}{A B}=1\)

⇒ \(\tan 0^{\circ}=\frac{B C}{A B}=\frac{0}{A B}=0\)

⇒ \(\cot 0^{\circ}=\frac{A B}{B C}=\frac{A B}{0}=\infty\)

4. Trigonometric Ratios for 90°

In \(\triangle\)ABC, it is clear that as the value of ‘0’ increases, the length of AB decreases, and for 0 = 90°, AB = 0 and AC = BC.

Trigonometry Ratio For 90 Degrees

Therefore, \(\sin 90^{\circ}=\frac{B C}{A C}=\frac{A C}{A C}=1\)

⇒ \(cosec 90^{\circ}=\frac{A C}{B C}=\frac{B C}{B C}=1\)

⇒ \(\cos 90^{\circ}=\frac{A B}{A C}=\frac{0}{A C}=0 \sec 90^{\circ}=\frac{A C}{A B}=\frac{A C}{0}=\infty\)

∴ \(\tan 90^{\circ}=\frac{B C}{A B}=\frac{B C}{0}=\infty \cot 90^{\circ}=\frac{A B}{B C}=\frac{0}{B C}=0\)

Trigonometry Values Of Trigonometric Ratios

Trigonometry Solved Examples

Example 1. Evaluate : \(\sin ^2 60^{\circ} \tan 45^{\circ}-\cos ^2 45^{\circ} \sec 60^{\circ}\)

Solution:

We know that,

⇒ \(\sin 60^{\circ}=\frac{\sqrt{3}}{2}, \tan 45^{\circ}=1, \cos 45^{\circ}=\frac{1}{\sqrt{2}}, \sec 60^{\circ}=2\)

⇒ \(\sin ^2 60^{\circ} \tan 45^{\circ} -\cos ^2 45^{\circ} \sec 60^{\circ} \)

= \(\left(\frac{\sqrt{3}}{2}\right)^2(1)-\left(\frac{1}{\sqrt{2}}\right)^2(2)=\frac{3}{4}-\frac{1}{2} \times 2=\frac{3}{4}-1=\frac{3-4}{4}=-\frac{1}{4}\)

\(\sin ^2 60^{\circ} \tan 45^{\circ}-\cos ^2 45^{\circ} \sec 60^{\circ}\) = -1/4.

Example 2. Evaluate : \(\cos 60^{\circ} \cos 30^{\circ}+\sin 60^{\circ} \sin 30^{\circ}\)

Solution:

We know that,

⇒ \(\cos 60^{\circ} =\frac{1}{2}, \cos 30^{\circ}=\frac{\sqrt{3}}{2}, \sin 60^{\circ}=\frac{\sqrt{3}}{2}, \sin 30^{\circ}=\frac{1}{2}\)

⇒ \(\cos 60^{\circ} \cos 30^{\circ}+ \sin 60^{\circ} \sin 30^{\circ}\)

= \(\frac{1}{2} \times \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} \times \frac{1}{2}=\frac{\sqrt{3}+\sqrt{3}}{4}=\frac{2 \sqrt{3}}{4}=\frac{\sqrt{3}}{2}\)

\(\cos 60^{\circ} \cos 30^{\circ}+\sin 60^{\circ} \sin 30^{\circ}\) = 1/2

Example 3. Show that : \(\cos 60^{\circ}=2 \cos ^2 30^{\circ}-1\)

Solution:

L.H.S. =\(\cos 60^{\circ}=\frac{1}{2}\)

R.H.S. = \(2 \cos ^230^{\circ}-1=2\left(\frac{\sqrt{3}}{2}\right)^2-1=2\left(\frac{3}{4}\right)-1=\frac{3}{2}-1\)

=\(\frac{3-2}{2}=\frac{1}{2}\)

L.H.S. = R.H.S. Hence Proved

Example 4. If \(A=15^{\circ}\), then find the value of \(\sec 2 A\).

Solution: 

A =\(15^{\circ} \Rightarrow 2 A=2 \times 15^{\circ}=30^{\circ} \)

⇒ \(\sec 2 A =\sec 30^{\circ}=\frac{2}{\sqrt{3}}\)

The value of \(\sec 2 A\) =\( \frac{2}{\sqrt{3}}\)

Example 5. If \(\sin x=1\), then find the value of \(\tan \frac{x}{3}\).

Solution:

⇒ \(\sin x =1\)

⇒ \(\sin x =\sin 90^{\circ} \Rightarrow x =90^{\circ}\)

⇒ \(\frac{x}{3} =\frac{90^{\circ}}{3}=30^{\circ}\)

∴ \(\tan \frac{x}{3}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}\)

The value of \(\tan \frac{x}{3}\) =[latec]\frac{1}{\sqrt{3}}[/latex]

Example 6. If \(\sin (A+B)=\frac{\sqrt{3}}{2}\) and \(\cos (A-B)=\frac{\sqrt{3}}{2}\), then find the values of A and B Given that, \(\sin (A+B)=\frac{\sqrt{3}}{2}\)

Solution:

⇒ \(\sin (A+B)=\sin 60^{\circ} \quad \Rightarrow \quad A+B=60^{\circ}\) Equation 1

and \(\cos (A-B)=\frac{\sqrt{3}}{2}\)

⇒ \(\cos (A-B)=\cos 30^{\circ} \quad \Rightarrow \quad A-B=30^{\circ}\) Equation 2

Adding equations (1) and (2)

⇒ \(A+B =60^{\circ}\)

2 A =\(90^{\circ} \quad\Rightarrow \quad A-B =30^{\circ} \)

A = \(45^{\circ}\)

Put A=\(60^{\circ}\) in equation (1)

⇒ \(45^{\circ}+B=60^{\circ} \quad \Rightarrow \quad B=60^{\circ}-45^{\circ}=15^{\circ}\)

The values of A and B are

⇒ \(A=45^{\circ}\) and \(B=15^{\circ}\)

Example 7.  In an acute-angled AABC, if tan (A + B – C) = 1 and see (B + C – A) = 2, then find the value of cos (45 – 3A).

Solution:

Solution. We have.

tan (A + B – C) = 1 = tan 45°

=* A + B- C = 45° …(1)

Also sec (B + C – A) = 2 = sec 60°

B + C – A = 60° …(2)

Adding equations (1) and (2), we get

25 = 105° ⇒ 5 = 52.5° …(3)

Subtracting equation (1) from equation (2), we get

2C – 2A = 15° ⇒ C-A = 7.5° …(4)

We know that A + 5 + C = 180°

⇒ A + C = 180° – 52.5° [from (3)]

⇒ A + C= 127.5° …(5)

⇒ -A + C = 7.5° …(4)

Adding equations (4) and (5), we get

2C = 135° ⇒ C = 67.5°

A = 127.5°-67.5° = 60°

cos (45 – 3A) = cos (4 x 52.5° – 3 x 60°) = cos (210° – 180°) = cos 30° = \(\frac{\sqrt{3}}{2}\)

The value of cos (45 – 3A) = \(\frac{\sqrt{3}}{2}\)

Example 8. If sin A= cos A, then evaluate \(\tan A+\sin ^2 A+1\).

Solution:

We know that, in A = cos A, then

Now, A =\(45^{\circ}\)

⇒ \(\tan A+\sin ^2 A+1 =\tan 45^{\circ}+\sin ^2 45^{\circ}+1\)

= \(1+\left(\frac{1}{\sqrt{2}}\right)^2+1=2+\frac{1}{2}=\frac{5}{2}\)

\(\tan A+\sin ^2 A+1\) = \(\frac{5}{2}\)

Example 9. Find the value of \(\left(\theta_1+\theta_2\right)\) if

Solution:

⇒ \(\tan \left(\theta_1+\theta_2\right)=\frac{\tan \theta_1+\tan \theta_2}{1-\tan \theta_1 \cdot \tan \theta_2}\)

where, \(\tan \theta_1=\frac{1}{2}\) and \(\tan \theta_2=\frac{1}{3}\).

⇒ \(\tan \left(\theta_1+\theta_2\right) =\frac{\tan \theta_1+\tan \theta_2}{1-\tan \theta_1 \cdot \tan \theta_2}=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}\)

= \(\frac{\frac{3+2}{6}}{1-\frac{1}{6}}=\frac{\frac{5}{6}}{\frac{5}{6}}=1=\tan 45^{\circ}\)

∴ \(\theta_1+\theta_2 =45^{\circ}\)

The value of \(\left(\theta_1+\theta_2\right)\) = 45°

Example 10. Evaluate : cos 1° cos 2° cos 3° … cos 179°

Solution:

cos 90° whose value is zero lies in between cos 1° cos 2° cos 3°.., cos 179°

cos 1° cos 2° cos 3°… cos 179°

= cos 1° cos 2° cos 3° …. cos 90° cos 179°

= cos 1° cos 2° cos 3° …. X 0 X cos 179°

= 0 (0 x finite number = 0)

cos 1° cos 2° cos 3° … cos 179° = 0

Question 11. In the adjoining figure, a right-angled triangle $A B C$ is shown in which AM = CM = 3 m. If \(\angle A C M=15^{\circ}\), then find A C:

Solution:

Here, AM = CM

⇒ \(\angle A C M = \angle C A M\) (opposite angles of equal sides)

Trigonometry In Adjoining A Right Angled Triangle ABC

⇒ \(\angle C A M=15^{\circ}\)

⇒ \(\angle A C B=90^{\circ}-15^{\circ}=75^{\circ}\)

and \(\angle B C M=\angle A C B-\angle A C M=75^{\circ}-15^{\circ}=60^{\circ}\)

In \(\triangle B C M\),

⇒ \(\cos (\angle B C M)=\frac{B C}{C M}\)

⇒ \(\cos 60^{\circ}=\frac{B C}{3}\)

⇒ \(\frac{1}{2}=\frac{B C}{3}\)

⇒ \(B C=\frac{3}{2} \mathrm{~m}\)

and \(\sin (\angle B C M)=\frac{B M}{C M}\)

⇒ \(\sin 60^{\circ}=\frac{B M}{3}\)

⇒ \(\frac{\sqrt{3}}{2}=\frac{B M}{3}\)

⇒ \(B M=\frac{3 \sqrt{3}}{2} \mathrm{~m}\)

⇒ \(A B=A M+B M=\left(3+\frac{3 \sqrt{3}}{2}\right) \mathrm{m}\)

Now, in \(\triangle A B C\), from Pythagoras theorem

⇒ \(A C^2 =A B^2+B C^2=\left(3+\frac{3 \sqrt{3}}{2}\right)^2+\left(\frac{3}{2}\right)^2=9+\frac{27}{4}+9 \sqrt{3}+\frac{9}{4}\)

= \(18+9 \sqrt{3}=9(2+\sqrt{3})=\frac{9}{2}(4+2 \sqrt{3})\)

AC =\(\sqrt{\frac{9(4+2 \sqrt{3})}{2}}=\frac{3(\sqrt{3}+1)}{\sqrt{2}} \mathrm{~m}\)

Relation Between Trigonometric Ratios

1. \(\sin \theta \times cosec\theta=1\)

  • \(\sin \theta=\frac{1} {cosec \theta}\)
  • \(cosec \theta=\frac{1}{\sin \theta}\)

2. \(\cos \theta \times \sec \theta=1\)

  •  \(\cos \theta=\frac{1}{\sec \theta}\)
  •  \(\sec \theta=\frac{1}{\cos \theta}\)

3. \(\tan \theta \times \cot \theta=1\)

  • \(\tan \theta=\frac{1}{\cot \theta}\)
  •  \(\cot \theta=\frac{1}{\tan \theta}\)

4. \(\tan \theta=\frac{\sin \theta}{\cos \theta}\)

5. \(\cot \theta=\frac{\cos \theta}{\sin \theta}\)

If the value of a trigonometric function is known, then we can find the values of other trigonometric functions. We can use Pythagoras’ theorem and the above results for it.

Trigonometric Identities

We know about the algebraic equations.

The algebraic equation satisfies a particular value of the variable but in trigonometry, the equation can satisfy all values of the variable, such equations are called trigonometric identities.

  1.  \(\sin ^2 \theta+\cos ^2 \theta=1\)
  2.  \(\sec ^2 \theta=1+\tan ^2 \theta\)
  3. \(cosec^2 \theta=1+\cot ^2 \theta\)

Identity 1. \(\sin ^2 \theta+\cos ^2 \theta\)=1

Proof : Let \(\triangle\)ABC is a right-angled triangle in which \(\angle\)ABC = 90°

Trigonometry Identities

Let \(\angle\)ACB = \(\theta\)

For this angle ‘\(\theta\)’

AB = perpendicular

BC = base

AC = hypotenuse

From Pythagoras theorem,

⇒ \(A B^2+B C^2=A C^2\)

Now,\(\sin \theta=\frac{A B}{A C}\) and \(\cos \theta=\frac{B C}{A C}\)

Divide each term in equation (1) by A C^2

⇒ \(\frac{A B^2}{A C^2}+\frac{B C^2}{A C^2}=\frac{A C^2}{A C^2}\)

⇒ \(\left(\frac{A B}{A C}\right)^2+\left(\frac{B C}{A C}\right)^2=1 \quad \Rightarrow \quad \sin ^2 \theta+\cos ^2 \theta\)=1

Identity 2. \(\sec ^2 \theta=1+\tan ^2 \theta\)

Proof: In \(\triangle M B C\),

⇒ \(A B^2+B C^2=A C^2\)

Now,\(\tan \theta=\frac{A B}{B C}\) and \(\sec \theta=\frac{A C}{B C}\)

Divide each term by \(B C^2\) in equation (1)

⇒ \(\frac{A B^2}{B C^2}+\frac{B C^2}{B C^2}=\frac{A C^2}{B C^2}\)

⇒ \(\left(\frac{A B}{B C}\right)^2+1=\left(\frac{A C}{B C}\right)^2 \Rightarrow \tan ^2 \theta+1=\sec ^2 \theta\)

Identity 3. \(cosec^2 \theta=1+\cot ^2 \theta\)

Proof : In \(\triangle A B C\),

⇒ \(A B^2+B C^2=A C^2\)

Now,\(\cot \theta=\frac{B C}{A B}\) and \(cosec \theta=\frac{A C}{A B}\)

Divide each term in equation (1) by \(A B^2\)

⇒ \(\frac{A B^2}{A B^2}+\frac{B C^2}{A B^2}=\frac{A C^2}{A B^2} \)

⇒ \(1+\left(\frac{B C}{A B}\right)^2=\left(\frac{A C}{A B}\right)^2 \Rightarrow 1+\cot ^2 \theta={cosec}^2 \theta\)

Alternate Proof: Identity (2) \(\sec ^2 \theta=1+\tan ^2 \theta\) and (3)\(cosec^2 \theta=1+\cot ^2 \theta\) can be proved with the help of identity (1) \(\sin ^2 \theta+\cos ^2 \theta\)=1.

Proof of \(\sec ^2 \theta=1+\tan ^2 \theta\) :

From identity (1)

⇒ \(\sin ^2 \theta+\cos ^2 \theta\)=1

Divide each term by \(\cos ^2 \theta\)

⇒ \(\frac{\sin ^2 \theta}{\cos ^2 \theta}+\frac{\cos ^2 \theta}{\cos ^2 \theta}=\frac{1}{\cos ^2 \theta}\)

⇒ \(left(\frac{\sin \theta}{\cos \theta})^2+1=\left(\frac{1}{\cos \theta}\right)^2\)

⇒ \(\tan ^2 \theta+1 =\sec ^2 \theta \Rightarrow \sec ^2 \theta=1+\tan ^2 \theta\)

Proof of \(cosec^2 \theta=1+\cot ^2 \theta\) :

From identity ( 1 ) \(\sin ^2 \theta+\cos ^2 \theta\)=1

Divide each term by \(\sin ^2 \theta\)

⇒ \(\frac{\sin ^2 \theta}{\sin ^2 \theta}+\frac{\cos ^2 \theta}{\sin ^2 \theta}=\frac{1}{\sin ^2 \theta}\)

⇒ \(1+\left(\frac{\cos \theta}{\sin \theta}\right)^2=\left(\frac{1}{\sin \theta}\right)^2\)

⇒ \(1+\cot ^2 \theta= cosec^2 \theta\)

∴ \(cosec^2 \theta=1+\cot ^2 \theta\)

Other Form of the Above Identities :

1. \(\sin ^2 \theta+\cos ^2 \theta=1\)

  • \(\cos ^2 \theta=1-\sin ^2 \theta\)
  •  \(\sin ^2 \theta=1-\cos ^2 \theta\)

2. \(\sec ^2 \theta=1+\tan ^2 \theta\)

  • \(\tan ^2 \theta=\sec ^2 \theta-1\)
  • \(\sec ^2 \theta-\tan ^2 \theta=1\)

3. \(cosec^2 \theta=1+\cot ^2 \theta\)

  •  \(\cot ^2 \theta= cosec^2 \theta-1\)
  •  \(cosec^2 \theta-\cot ^2 \theta\)=1

Trigonometric Solved Examples

Example 1. Simplify : \(\left(1+\tan ^2 \theta\right)(1-\sin \theta)(1+\sin \theta)\)

Solution:

⇒ \(\left(1+\tan ^2 \theta\right)(1-\sin \theta)(1+\sin \theta) \)

⇒ \(=\left(1+\tan ^2 \theta\right)\left[(1)^2-(\sin \theta)^2\right]=\left(1+\tan ^2 \theta\right)\left(1-\sin ^2 \theta\right)\)

= \(\sec ^2 \theta \cdot \cos ^2 \theta \quad \text { (using the identities } \sec ^2 \theta=1+\tan ^2 \theta \text { and } \sin ^2 \theta+\cos ^2 \theta=1 \text { ) }\)

= \(\frac{1}{\cos ^2 \theta} \cdot \cos ^2 \theta\)=1

\(\left(1+\tan ^2 \theta\right)(1-\sin \theta)(1+\sin \theta)\) = 1

Example 2. Prove that : \(\cos ^2 \theta \cdot cosec\theta+\sin \theta= cosec \theta\)

Solution:

L.H.S. =\(\cos ^2 \theta \cdot cosec \theta+\sin \theta=\cos ^2 \theta \cdot \frac{1}{\sin \theta}+\sin \theta\)

= \(\frac{\cos ^2 \theta}{\sin \theta}+\frac{\sin \theta}{1}=\frac{\cos ^2 \theta+\sin ^2 \theta}{\sin \theta}=\frac{1}{\sin \theta}\)

= cosec \(\theta\)= R.H.S.

\(\cos ^2 \theta \cdot cosec\theta+\sin \theta= cosec \theta\)

Question 3. Prove that : \(\sec ^4 \theta-\tan ^4 \theta=1+2 \tan ^2 \theta\)

Solution:

L.H.S. =\(\sec ^4 \theta-\tan ^4 \theta=\left(\sec ^2 \theta\right)^2-\left(\tan ^2 \theta\right)^2\)

=\(\left(\sec ^2 \theta+\tan ^2 \theta\right)\left(\sec ^2 \theta-\tan ^2 \theta\right)=\left(1+\tan ^2 \theta+\tan ^2 \theta\right)\left(1+\tan ^2 \theta-\tan ^2 \theta\right)\)

=\(\left(1+2 \tan ^2 \theta\right)(1)=1+2 \tan ^2 \theta= R.H.S\).

Hence Proved

\(\sec ^4 \theta-\tan ^4 \theta=1+2 \tan ^2 \theta\)

Question 4. Prove that : \(\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta+\sin \theta \cos \theta}=\cot \theta\)

Answer:

L.H.S. =\(\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta+\sin \theta \cos \theta}=\frac{\left(1-\sin ^2 \theta\right)+\cos \theta}{\sin \theta(1+\cos \theta)}=\frac{\cos ^2 \theta+\cos \theta}{\sin \theta(1+\cos \theta)}\)

=\(\frac{\cos \theta(\cos \theta+1)}{\sin \theta(1+\cos \theta)}=\frac{\cos \theta}{\sin \theta}=\cot \theta\)= R.H.S.

\(\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta+\sin \theta \cos \theta}=\cot \theta\)

Example 5. If \(\tan \theta=\frac{4}{3}\) , then find the value of \(\frac{3 \sin \theta-2 \cos \theta}{3 \sin \theta+5 \cos \theta}\) .

Solution:

⇒ \(\frac{3 \sin \theta-2 \cos \theta}{3 \sin \theta+5 \cos \theta}=\frac{3 \frac{\sin \theta}{\cos \theta}-2 \frac{\cos \theta}{\cos \theta}}{3 \frac{\sin \theta}{\cos \theta}+5 \frac{\cos \theta}{\cos \theta}}\)

=\(\frac{3 \tan \theta-2}{3 \tan \theta+5}=\frac{3 \times \frac{4}{3}-2}{3 \times \frac{4}{3}+5}=\frac{4-2}{4+5}=\frac{2}{9}\)

The value of \(\frac{3 \sin \theta-2 \cos \theta}{3 \sin \theta+5 \cos \theta}\) =\( \frac{2}{9}\)

Example 6. Prove that : \((\sec A+\tan A)(1-\sin A)=\cos A\)

Solution:

L.H.S. =\((\sec A+\tan A)(1-\sin A)\)

= \(\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)(1-\sin A)=\frac{(1+\sin A)}{\cos A}(1-\sin A)\)

= \(\frac{1-\sin ^2 A}{\cos A}=\frac{\cos ^2 A}{\cos A}=\cos A\) = R.H.S.

\((\sec A+\tan A)(1-\sin A)=\cos A\)

Example 7. Prove that : \({cosec A-\cot A=\frac{1}{cosec} A+\cot A}\)

Solution:

L.H.S. = \(cosec A-\cot A=(cosec A-\cot A) \cdot \frac{(cosec A+\cot A)}{(cosec A+\cot A)}\)

= \(\frac{cosec^2 A-\cot ^2 A}{cosec A+\cot A}=\frac{1}{cosec A+\cot A}\)= R.H.S.

Hence Proved.

\({cosec A-\cot A=\frac{1}{cosec} A+\cot A}\)

Example 8. Prove that: \(\frac{\sec A+1}{\tan A}=\frac{\tan A}{\sec A-1}\)

Solution:

L.H.S. =\(\frac{\sec A+1}{\tan A}=\frac{\sec A+1}{\tan A} \times \frac{\sec A-1}{\sec A-1}\)

[divide numerator and denominator by (sec A-1)]

= \(\frac{\sec ^2 A-1}{\tan A(\sec A-1)}=\frac{\tan ^2 A}{\tan A(\sec A-1)}\)

= \(\frac{\tan A}{\sec A-1}\)= R.H.S.

Hence Proved.

\(\frac{\sec A+1}{\tan A}=\frac{\tan A}{\sec A-1}\)

Example 9. Prove that : \(\sin \theta(1+\tan \theta)+\cos \theta(1+\cot \theta)= cosec \theta+\sec \theta\)

Solution:

L.H.S. =\(\sin \theta(1+\tan \theta)+\cos \theta(1+\cot \theta)\)

=\(\sin \theta(1+\tan \theta)+\cos \theta\left(1+\frac{1}{\tan \theta}\right)\)

= \(\sin \theta(1+\tan \theta)+\cos \theta\left(\frac{\tan \theta+1}{\tan \theta}\right)\)

=\((1+\tan \theta)\left[\sin \theta+\frac{\cos \theta}{\left(\frac{\sin \theta}{\cos \theta}\right)}\right]=(1+\tan \theta)\left(\sin \theta+\frac{\cos ^2 \theta}{\sin \theta}\right)\)

=\(\left(1+\frac{\sin \theta}{\cos \theta}\right)\left(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin \theta}\right)=\left(1+\frac{\sin \theta}{\cos \theta}\right)\left(\frac{1}{\sin \theta}\right)\)

⇒ \(=\frac{1}{\sin \theta}+\frac{\sin \theta}{\cos \theta} \times \frac{1}{\sin \theta}\)

= \(cosec \theta+\sec \theta\)= R.H.S.

\(\sin \theta(1+\tan \theta)+\cos \theta(1+\cot \theta)= cosec \theta+\sec \theta\)

Example 10. If \(\sec \theta+\tan \theta=p\),then prove that : \(\frac{p^2-1}{p^2+1}=\sin \theta\)

Solution:

L.H.S. =\(\frac{p^2-1}{p^2+1}=\frac{(\sec \theta+\tan \theta)^2-1}{(\sec \theta+\tan \theta)^2+1}=\frac{\sec ^2 \theta+\tan ^2 \theta+2 \sec \theta \tan \theta-1}{\sec ^2 \theta+\tan ^2 \theta+2 \sec \theta \tan \theta+1}\)

= \(\frac{\left(\sec ^2 \theta-1\right)+\tan ^2 \theta+2 \sec \theta \tan \theta}{\sec ^2 \theta+\left(\tan ^2 \theta+1\right)+2 \sec \theta \tan \theta}\)

= \(\frac{\tan ^2 \theta+\tan ^2 \theta+2 \sec \theta \tan \theta}{\sec ^2 \theta+\sec ^2 \theta+2 \sec \theta \tan \theta}\)

= \(\frac{2 \tan ^2 \theta+2 \sec \theta \tan \theta}{2 \sec ^2 \theta+2 \sec \theta \tan \theta}=\frac{2 \tan \theta(\tan \theta+\sec \theta)}{2 \sec \theta(\sec \theta+\tan \theta)}\)

= \(\frac{\tan \theta}{\sec \theta}=\frac{\sin \theta}{\cos \theta} \times \frac{\cos \theta}{1}\)

= \(\sin \theta\) = R.H.S.

Hence Proved

\(\frac{p^2-1}{p^2+1}=\sin \theta\)

Alternative Method : We have \(\sec \theta+\tan \theta=p\)

⇒ \(\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}=\frac{p}{1}\Rightarrow \quad \frac{1+\sin \theta}{\cos \theta}=\frac{p}{1}\)

Squaring both sides, we get

⇒ \(\frac{(1+\sin \theta)^2}{\cos ^2 \theta}=\frac{p^2}{1}\)

⇒ \(\frac{(1+\sin \theta)^2}{1-\sin ^2 \theta}=\frac{p^2}{1}\){ (using identity } \(\sin ^2 \theta+\cos ^2 \theta=1 \text { ) }\)

⇒ \(\frac{(1+\sin \theta)^2}{(1+\sin \theta)(1-\sin \theta)}=\frac{p^2}{1} \quad \Rightarrow \quad \frac{1+\sin \theta}{1-\sin \theta}=\frac{p^2}{1}\)

Applying componendo and dividends, we get

⇒ \(\frac{2}{2 \sin \theta}=\frac{p^2+1}{p^2-1}\)

∴ \(\frac{1}{\sin \theta}=\frac{p^2+1}{p^2-1} \Rightarrow \quad \sin \theta=\frac{p^2-1}{p^2+1}\)

Example 11. Prove that : \(\frac{\sec \theta+1-\tan \theta}{\tan \theta+1-\sec \theta}=\frac{\sin \theta}{1-\cos \theta}\)

Solution:

Note: In such type of questions, it is better to write \(\sec ^2 \theta-\tan ^2 \theta or cosec^2 \theta-\cot ^2 \theta\) in only numerator.

If in R.H.S. the single term in either numerator or denominator is \(sin \theta\) then convert the question in cosec \(\theta\) and cot \(\theta\) and if the single term is cos \(\theta\) then convert the question in see \(\theta\) and tan \(\theta\).

As in this question in R.H.S. single term \(sin \theta\) is in the numerator so we will use \(cosec^2 \theta-\cot ^2 \theta\) for 1.

(dividing Nr and Dr by \(\sin \theta\) to convert it in } \(cosec \theta and \cot \theta \text { ) }\)

= \(\frac{{cosec} \theta+\cot \theta-1}{1+\cot \theta-{cosec} \theta}=\frac{{cosec} \theta+\cot \theta-\left({cosec}^2 \theta-\cot ^2 \theta\right)}{1+\cot \theta-{cosec} \theta}\)

= \(\frac{({cosec} \theta+\cot \theta)-({cosec} \theta+\cot \theta)({cosec} \theta-\cot \theta)}{1+\cot \theta-{cosec} \theta} \)

= \(\frac{({cosec} \theta+\cot \theta)[1-{cosec} \theta+\cot \theta]}{1+\cot \theta-{cosec} \theta}={cosec} \theta+\cot \theta \)

= \(\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}=\frac{(1+\cos \theta)}{\sin \theta}=\frac{(1+\cos \theta)(1-\cos \theta)}{\sin \theta(1-\cos \theta)}\)

= \(\frac{1-\cos ^2 \theta}{\sin \theta(1-\cos \theta)}=\frac{\sin ^2 \theta}{\sin \theta(1-\cos \theta)}=\frac{\sin \theta}{1-\cos \theta}=\text { R.H.S. }\)

Hence Proved.

\(\frac{\sec \theta+1-\tan \theta}{\tan \theta+1-\sec \theta}=\frac{\sin \theta}{1-\cos \theta}\)

Example 12. If x=r \(\sin A \cos C\), y=r \(\sin A \sin C\) and z=r \(\cos A\), then prove that :

⇒ \(r^2=x^2+y^2+z^2\)

Solution:

Here, x=r sin A cos C, y=r sin A sin C and z=r cos A

Now, R.H.S. = \(x^2+y^2+z^2\)

= \((r \sin A \cos C)^2+(r \sin A \sin C)^2+(r \cos A)^2\)

= \(r^2 \sin ^2 A \cos ^2 C+r^2 \sin ^2 A \sin ^2 C+r^2 \cos ^2 A\)

= \(r^2 \sin ^2 A\left(\cos ^2 C+\sin ^2 C\right)+r^2 \cos ^2 A \quad\left(\cos ^2 C+\sin ^2 C=1\right)\)

= \(r^2 \sin ^2 A+r^2 \cos ^2 A\)

= \(r^2\left(\sin ^2 A+\cos ^2 A\right)\)

= \(r^2\) =L.H.S.

\(r^2=x^2+y^2+z^2\)

Hence Proved.

Identities And Equations

Identities are special + type of equations that are true for all values of the variable while equations are true for some particular values of the variable.

Solved Examples

Example 1. Check whether the equation \(\frac{\tan \phi+\sin \phi}{\tan \phi-\sin \phi}=\frac{\sec \phi+1}{\sec \phi-1}\) is an identity or not?

Solution:

L.H.S. = \(\frac{\tan \phi+\sin \phi}{\tan \phi-\sin \phi}=\frac{\frac{\sin \phi}{\cos \phi}+\sin \phi}{\frac{\sin \phi}{\cos \phi}-\sin \phi}\)

= \(\frac{\sin \phi \sec \phi+\sin \phi}{\sin \phi \sec \phi-\sin \phi}=\frac{\sin \phi(\sec \phi+1)}{\sin \phi(\sec \phi-1)}=\frac{\sec \phi+1}{\sec \phi-1}\)= R.H.S.

L.H.S. = R.H.S.

And this equation is true for all values of \phi.

The given equation is an identity.

Example 2. Check whether the following equation \(\tan ^4 \theta+\tan ^6 \theta=\tan ^3 \theta \sec ^2 \theta\) is an identity or not?

Solution:

Given equation,

⇒ \(\tan ^4 \theta+\tan ^6 \theta =\tan ^3 \theta \sec ^2 \theta\)

⇒ \(\tan ^4 \theta\left(1+\tan ^2 \theta\right) =\tan ^3 \theta \sec ^2 \theta\)

⇒ \(\tan ^4 \theta \sec ^2 \theta =\tan ^3 \theta \sec ^2 \theta\)

L.H.S. \(\neq\) R.H.S.

It is not an identity.

Again, \(\tan ^4 \theta \sec ^2 \theta-\tan ^3 \theta \sec ^2 \theta=0\)

⇒ \(\tan ^3 \theta \sec ^2 \theta(\tan \theta-1)\) =0

⇒ \(\tan ^3 \theta\)=0 or \(\sec ^2 \theta=0\) or \(\tan \theta-1\)=0

Therefore, given equation satisfies only for \(\theta=0^{\circ}\) and \(\theta=45^{\circ}\).

The given equation is not an identity.

Alternate Method : For \(\theta=60^{\circ}\)

L.H.S. =\(\tan ^4 60^{\circ}+\tan ^6 60^{\circ}\)

= \((\sqrt{3})^4+(\sqrt{3})^6=9+27=36\)

and R.H.S. =\(\tan ^3 60^{\circ} \cdot \sec ^2 60^{\circ}\)

= \((\sqrt{3})^3(2)^2=12 \sqrt{3}\)

L.H.S. \(\neq\) R.H.S.

The given equation is not an identity.

Example 3. Solve : \(2 \sin ^2 \theta=\frac{1}{2}, 0^{\circ}<\theta<90^{\circ}\)

Solution:

⇒ \(2 \sin ^2 \theta =\frac{1}{2}\)

⇒ \(\sin ^2 \theta =\frac{1}{2 \times 2}=\frac{1}{4}\)

⇒ \(\sin \theta =\sqrt{\frac{1}{4}}=\frac{1}{2}\)

⇒ \(\sin \theta =\sin 30^{\circ}\)

∴ \(\theta =30^{\circ}\)

Example 4. Find the value of \(\theta if 2 \cos 3 \theta=1\) and \(0^{\circ}<\theta<90^{\circ}\).

Solution:

2 \(\cos 3 \theta\) =1

⇒ \(\cos 3 \theta =\frac{1}{2}\)

⇒ \(\cos 3 \theta =\cos 60^{\circ}\)

3 \(\theta =60^{\circ}\)

∴ \(\theta =20^{\circ}\)

The Value of θ = 20°

Example 5. Find the value of \(\theta\) if \(\sec ^2 \theta+\tan ^2 \theta=\frac{5}{3}\) and θ lies in first quadrant.

Solution:

⇒ \(\sec ^2 \theta+\tan ^2 \theta =\frac{5}{3}\)

⇒ \(1+\tan ^2 \theta+\tan ^2 \theta =\frac{5}{3} \quad \Rightarrow \quad 2 \tan ^2 \theta=\frac{5}{3}-1=\frac{2}{3}\)

⇒ \(\tan ^2 \theta=\frac{1}{3} \quad \Rightarrow \quad \tan ^2 \theta=\left(\frac{1}{\sqrt{3}}\right)^2\)

⇒ \(\tan \theta=\frac{1}{\sqrt{3}}\) (talking positive sign only)

⇒ \(\tan \theta=\tan 30^{\circ} \Rightarrow \theta=30^{\circ}\)

The value of \(\theta\) if \(\sec ^2 \theta+\tan ^2 \theta=\frac{5}{3}\) = 30°

Example 6. If \(0^{\circ}<\alpha<90^{\circ}\), then solve the equation \(\frac{\sin \alpha}{1-\cos \alpha}+\frac{\sin \alpha}{1+\cos \alpha}=4\).

solution:

⇒ \(\frac{\sin \alpha}{1-\cos \alpha}+\frac{\sin \alpha}{1+\cos \alpha} =4\)

⇒ \(\frac{\sin \alpha(1+\cos \alpha)+\sin \alpha(1-\cos \alpha)}{(1-\cos \alpha)(1+\cos \alpha)}=4\)

⇒ \(\frac{\sin \alpha+\sin \alpha \cos \alpha+\sin \alpha-\sin \alpha \cos \alpha}{1-\cos ^2 \alpha}\) =4

⇒ \(\frac{2 \sin \alpha}{\sin ^2 \alpha}\)=4

⇒ \(\frac{2}{\sin \alpha}\)=4

⇒ \(\sin \alpha =\frac{1}{2}=\sin 30^{\circ}\)

⇒ \(\alpha=30^{\circ}\)

Example 7. If \(0^{\circ}<\theta<90^{\circ}\), then find the value of \(\theta\) from the equation \(\frac{\cos ^2 \theta}{\cot ^2 \theta-\cos ^2 \theta}=3\)

Solution:

⇒ \(\frac{\cos ^2 \theta}{\cot ^2 \theta-\cos ^2 \theta}=3\)

⇒ \(\frac{\cos ^2 \theta}{\frac{\cos ^2 \theta}{\sin ^2 \theta}-\cos ^2 \theta}=3 \Rightarrow \frac{\cos ^2 \theta}{\cos ^2 \theta\left(\frac{1}{\sin ^2 \theta}-1\right)}=3\)

⇒ \(\frac{1}{\frac{1}{\sin ^2 \theta}-1}=3 \quad \Rightarrow \quad \frac{1}{\frac{1-\sin ^2 \theta}{\sin ^2 \theta}}=3\)

⇒ \(\frac{\sin ^2 \theta}{1-\sin ^2 \theta}=3 \quad \Rightarrow \quad \frac{\sin ^2 \theta}{\cos ^2 \theta}=3\)

⇒ \(\tan ^2 \theta=(\sqrt{3})^2\)

⇒ \(\tan \theta=\sqrt{3}\) (taking positive sign only)

⇒ \(\tan \theta=\tan 60^{\circ}\)

∴ \(\theta=60^{\circ}\)

The Value of θ = 60°

Complementary Angles

Two angles are said to be complementary angles if their sum is 90°.

\(\theta\) and (90°- \(\theta\)) are complementary angles.

Trigonometric Ratios Of Complementary Angles

Let a rotating ray rotate 90° in an anticlockwise direction from the initial position OX and reach OY and after this, it rotates ‘9’ angle in a clockwise direction and reaches the OA position.

⇒ \(\angle\)XOA = 90° – \(\theta\).

Now, P is a point on side OA. PM and PN are perpendiculars from P to OX and OY respectively.

⇒ \(\sin \left(90^{\circ}-\theta\right)=\frac{P M}{O P}=\frac{O N}{O P}=\cos \theta\)

⇒ \(\cos \left(90^{\circ}-\theta\right)=\frac{O M}{O P}=\frac{P N}{O P}=\sin \theta\)

⇒ \(\tan \left(90^{\circ}-\theta\right)=\frac{P M}{O M}=\frac{O N}{P N}=\cot \theta\)

⇒ \(\cot \left(90^{\circ}-\theta\right)=\frac{O M}{P M}=\frac{P N}{O N}=\tan \theta\)

⇒ \(\sec \left(90^{\circ}-\theta\right)=\frac{O P}{O M}=\frac{O P}{P N}= cosec \theta\)

⇒ \({cosec}\left(90^{\circ}-\theta\right)=\frac{O P}{P M}=\frac{O P}{O N}=\sec \theta\)

⇒ \(\sin \left(90^{\circ}-\theta\right)=\cos \theta\)

⇒ \(\cos \left(90^{\circ}-\theta\right)=\sin \theta\)

⇒ \(\tan \left(90^{\circ}-\theta\right)=\cot \theta\)

⇒ \(\cot \left(90^{\circ}-\theta\right)=\tan \theta\)

⇒ \(\sec \left(90^{\circ}-\theta\right)={cosec} \theta\)

⇒ \({cosec}\left(90^{\circ}-\theta\right)=\sec \theta\)

Trigonometry Trigonometric Ratios Of Complementary Angles

Summary :

⇒ \(\sin \left(90^{\circ}-\theta\right)=\cos \theta\)

⇒ \(\cos \left(90^{\circ}-\theta\right)=\sin \theta\)

⇒ \(\tan \left(90^{\circ}-\theta\right)=\cot \theta\)

⇒ \(\cot \left(90^{\circ}-\theta\right)=\tan \theta\)

⇒ \(\sec \left(90^{\circ}-\theta\right)= cosec \theta\)

cosec\(\left(90^{\circ}-\theta\right)=\sec \theta\)

Solved Examples

Example 1. Evaluate the following :

  1. \(\frac{\sin 58^{\circ}}{\cos 32^{\circ}}\)
  2.  \(\frac{\sec 42^{\circ}}{{cosec} 48^{\circ}}\)

Solution:

  1.  \(\frac{\sin 58^{\circ}}{\cos 32^{\circ}}=\frac{\sin \left(90^{\circ}-32^{\circ}\right)}{\cos 32^{\circ}}=\frac{\cos 32^{\circ}}{\cos 32^{\circ}}\)=1
  2. \(\frac{\sec 42^{\circ}}{{cosec} 48^{\circ}}=\frac{\sec \left(90^{\circ}-48^{\circ}\right)}{{cosec} 48^{\circ}}=\frac{{cosec} 48^{\circ}}{{cosec} 48^{\circ}}\)=1

Example 2. Evaluate:

  1. \(\tan 42^{\circ}-\cot 48^{\circ}\)
  2. \(\sec 36^{\circ}- cosec 54^{\circ}\)

Solution:

(1)\(\tan 42^{\circ}-\cot 48^{\circ} =\tan 42^{\circ}-\cot \left(90^{\circ}-42^{\circ}\right)\)

=\(\tan 42^{\circ}-\tan 42^{\circ}\)=0

(2) \(\sec 36^{\circ}- cosec 54^{\circ}=\sec 36^{\circ}- cosec\left(90^{\circ}-36^{\circ}\right)\)

=\(\sec 36^{\circ}-\sec 36^{\circ}\)=0

Example 3. Prove that :

  1. \(\sin 42^{\circ} \cos 48^{\circ}+\sin 48^{\circ} \cos 42^{\circ}=1\)
  2.  \(\cos 70^{\circ} \cos 20^{\circ}-\sin 70^{\circ} \sin 20^{\circ}=0\)

Solution:

(1) L.H.S. =\(\sin 42^{\circ} \cos 48^{\circ}+\sin 48^{\circ} \cos 42^{\circ}\)

=\(\sin 42^{\circ} \cos \left(90^{\circ}-42^{\circ}\right)+\sin \left(90^{\circ}-42^{\circ}\right) \cos 42^{\circ}\)

= \(\sin 42^{\circ} \sin 42^{\circ}+\cos 42^{\circ} \cos 42^{\circ}\)

= \(\sin ^2 42^{\circ}+\cos ^2 42^{\circ}\)

= 1 = R.H.S.

Hence Proved.

(2) L.H.S. = cos 70° cos 20° – sin 70° sin 20°

= cos 70° cos 20° – sin (90° – 20°) sin(90° – 70°)

= cos 70° cos 20° – cos 20° cos 70°

= 0 = R.H.S.

Example 4. Without using trigonometric tables, evaluate :

⇒ \(\left(\frac{\tan 20^{\circ}}{{cosec} 70^{\circ}}\right)^2+\left(\frac{\cot 20^{\circ}}{\sec 70^{\circ}}\right)^2+2 \tan 15^{\circ} \tan 37^{\circ} \tan 53^{\circ} \tan 60^{\circ} \tan 75^{\circ}\)

Solutions:

⇒ \(\left(\frac{\tan 20^{\circ}}{{cosec} 70^{\circ}}\right)^2+\left(\frac{\cot 20^{\circ}}{\sec 70^{\circ}}\right)^2+2 \tan 15^{\circ} \tan 37^{\circ} \tan 53^{\circ} \tan 60^{\circ} \tan 75^{\circ}\)

= \(\left\{\frac{\tan 20^{\circ}}{{cosec}\left(90^{\circ}-20^{\circ}\right)}\right\}^2+\left\{\frac{\cot 20^{\circ}}{\sec \left(90^{\circ}-20^{\circ}\right)}\right\}^2\)

+2 \(\tan 15^{\circ} \tan 37^{\circ} \tan \left(90^{\circ}-37^{\circ}\right) \cdot(\sqrt{3}) \tan \left(90^{\circ}-15^{\circ}\right)\)

⇒ \(+2 \tan 15^{\circ} \tan 37^{\circ} \tan \left(90^{\circ}-37^{\circ}\right) \cdot(\sqrt{3}) \tan \left(90^{\circ}-15^{\circ}\right)\)

= \(\left(\frac{\tan 20^{\circ}}{\sec 20^{\circ}}\right)^2+\left(\frac{\cot 20^{\circ}}{{cosec} 20^{\circ}}\right)^2+2 \tan 15^{\circ} \tan 37^{\circ} \cot 37^{\circ} \cdot(\sqrt{3}) \cot 15^{\circ}\)

= \(\left(\frac{\sin 20^{\circ} / \cos 20^{\circ}}{1 / \cos 20^{\circ}}\right)^2+\left(\frac{\cos 20^{\circ} / \sin 20^{\circ}}{1 / \sin 20^{\circ}}\right)^2+2 \sqrt{3} \tan 15^{\circ} \tan 37^{\circ} \cdot \frac{1}{\tan 37^{\circ}} \cdot \frac{1}{\tan 15^{\circ}}\)

= \(\sin ^2 20^{\circ}+\cos ^2 20^{\circ}+2 \sqrt{3}=1+2 \sqrt{3}\)

Example 5. Without using trigonometric tables, evaluate the following :

⇒ \(\frac{{cosec}^2\left(90^{\circ}-\theta\right)-\tan ^2 \theta}{4\left(\cos ^2 48^{\circ}+\cos ^2 42^{\circ}\right)}-\frac{2 \tan ^2 30^{\circ} \sec ^2 52^{\circ} \sin ^2 38^{\circ}}{\left({cosec}^2 70^{\circ}-\tan ^2 20^{\circ}\right)}\)

Solution:

⇒ \(\frac{{cosec}^2\left(90^{\circ}-\theta\right)-\tan ^2 \theta}{4\left(\cos ^2 48^{\circ}+\cos ^2 42^{\circ}\right)}-\frac{2 \tan ^2 30^{\circ} \sec ^2 52^{\circ} \sin ^2 38^{\circ}}{\left({cosec}^2 70^{\circ}-\tan ^2 20^{\circ}\right)}\)

= \(\frac{\sec ^2 \theta-\tan ^2 \theta}{4\left\{\cos ^2\left(90^{\circ}-42^{\circ}\right)+\cos ^2 42^{\circ}\right)}-\frac{2\left(\frac{1}{\sqrt{3}}\right)^2 \sec ^2\left(90^{\circ}-38^{\circ}\right) \sin ^2 38^{\circ}}{{cosec}^2\left(90^{\circ}-20^{\circ}\right)-\tan ^2 20^{\circ}}\)

= \(\frac{1}{4\left(\sin ^2 42^{\circ}+\cos ^2 42^{\circ}\right)}-\frac{2{cosec}^2 38^{\circ} \cdot \sin ^2 38^{\circ}}{3\left(\sec ^2 20^{\circ}-\tan ^2 20^{\circ}\right)}\)

= \(\frac{1}{4}-\frac{2{cosec}^2 38^{\circ} \times \frac{1}{{cosec}^2 38^{\circ}}}{3}=\frac{1}{4}-\frac{2}{3}=\frac{-5}{12}\)

Example 6. Prove that :

  1. \(\sin \left(40^{\circ}-\theta\right)-\cos \left(50^{\circ}+\theta\right)\)=0
  2. \(\sec \left(65^{\circ}+\theta\right)-{cosec}\left(25^{\circ}-\theta\right)\)=0

Solution:

(1) L.H.S. =\(\sin \left(40^{\circ}-\theta\right)-\cos \left(50^{\circ}+\theta\right)\)

=\(\sin \left\{90^{\circ}-\left(50^{\circ}+\theta\right)\right\}-\cos \left(50^{\circ}+\theta\right)\)

=\(\cos \left(50^{\circ}+\theta\right)-\cos \left(50^{\circ}+\theta\right)\)=0= R.H.S.

(2) L.H.S. =\(\sec \left(65^{\circ}+0\right)- {cosec}\left(25^{\circ}-\theta\right)\)

=\(\sec \left\{90^{\circ}-\left(25^{\circ}-0\right)\right\}-{cosec}\left(25^{\circ}-\theta\right)\)

=\({cosec}\left(25^{\circ}-\theta\right)- {cosec}\left(25^{\circ}-\theta\right)\)=0= R.H.S.

Example 7. Express each of the following in terms of trigonometric ratios of angles between 0° and 45°.

  1. sin 70° + sec 70°
  2.  tan 65° + cosec 65°
  3. cos 81° + cot 80°

Solution:

  1. sin 70° + sec 70° = sin (90° – 20°) + sec (90° – 20°) = cos 20° + cosec 20°
  2. tan 65° + cosec 65° = tan (90° – 25°) + cosec (90° – 25°) = cot 25° + sec 25°
  3. cos 81° + cot 80° = cos (90° – 9°) + cot (90° -10°) = sin 9° + tan 10°

Example 8. If sin 3A = cos (A – 26°) where 3A is an acute angle, then find the value of A

Solution:

Given that,

sin 3A = cos (A – 26°).

cos (90° – 3A) = cos (A – 26°) ⇒ 90° – 3A =A – 26°

⇒ -471=-116° ⇒ A =29°

The value of A =29°

Example 9. If sin (\(\theta\)+ 24°) = cos \(\theta\) and \(\theta\) + 24° is an acute angle, then find the value of \(\theta\).

Solution:

Given that,

sin (\(\theta\)+ 24°) = cos \(\theta\)

⇒ sin (\(\theta\) + 24°) = sin (90° – \(\theta\))

⇒ \(\theta\) + 24° = 90° – \(\theta\)

⇒ 2\(\theta\) = 66°

∴ \(\theta\)= 33°

The value of θ = 33°

Example 10. If A, B, C are the angles of \(\triangle\) M B C, show that \(\sin \frac{B+C}{2}=\cos \frac{A}{2}\).

Solution:

In \(\triangle\) A B C,

A+B+C=\(180^{\circ}\)

B+C=\(180^{\circ}-A\)

⇒ \(\frac{B+C}{2}=90^{\circ}-\frac{A}{2}\)

⇒ \(\sin \frac{B+C}{2}=\sin \left(90^{\circ}-\frac{A}{2}\right)\)

⇒ \(\sin \frac{B+C}{2}=\cos \frac{A}{2}\)

Example 11. If \(\sin 36^{\circ}\)=p, then find \(\sin 54^{\circ}\) in terms of p.

Solution: 

We have, \(\sin 36^{\circ}\)=p

⇒ \(\sin ^2 36^{\circ}=p^2 \quad \Rightarrow \quad 1-\cos ^2 36^{\circ}=p^2\)

⇒ \(\cos ^2 36^{\circ}=1-p^2 \quad \Rightarrow \quad \cos ^2\left(90^{\circ}-54^{\circ}\right)=1-p^2\)

⇒ \(\sin ^2 54^{\circ}=1-p^2\)

∴ \(\sin 54^{\circ}=\sqrt{1-p^2}\)(taking only positive sign as \(54^{\circ}\) lies in 1 quadrant)

Example 12. If \(\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \tan 4^{\circ} \ldots \tan 89^{\circ}=x^2-8\), then find the value of x .

Solution:

⇒ \(x^2-8=\left(\tan 1^{\prime \prime} \tan 89^{\circ}\right)\left(\tan 2^{\prime \prime} \tan 88^{\circ}\right)\left(\tan 3^{\circ} \tan 87^{\circ}\right) \ldots\left(\tan 44^{\circ} \tan 46^{\circ}\right) \tan 45^{\circ}\)

⇒ \(\left.=\left[\tan 1^{\circ} \tan \left(90^{\circ}-1^{\circ}\right)|| \tan 2^{\circ \prime} \tan \left(90^{\circ}-2^{\circ}\right)\right] \mid \tan 3^{\circ} \tan \left(90^{\circ}-3^{\circ}\right)\right]\) … \(\left|\tan 44^{\circ} \tan \left(90^{\circ}-44^{\circ}\right)\right| \times 1\)

=\(\left(\tan 1^{\prime \prime} \cot 1^{\circ}\right)\left(\tan 2^{\circ} \cot 2^{\circ}\right)\left(\tan 3^{\circ} \cot 3^{\circ}\right) \ldots\left(\tan 44^{\circ} \cot 44^{\circ}\right)\)

=\(1 \times 1 \times 1 \times \ldots \times 1 \quad(\tan x \cdot \cot x=\tan x \cdot \frac{1}{\tan x}=1). \)

⇒ \(x^2-8\)=1

⇒ \(x^2=9 \quad \Rightarrow \quad x= \pm 3\)

The value of  \( \quad x= \pm 3\)

Introduction Of Trigonometry Exercise 8.1

Question 1. In \(\triangle A B C\), right-angled at B, AB = 24 cm, BC = 7 cm. Determine :

  1. sin A, cos A
  2. sin C, cos C

Solution :

In \(\triangle A B C\),

Trigonometry In Triangle ABC, Right Angled At B

AB = 24 cm, BC = 7 cm and \(\angle B=90^{\circ}\)

From Pythagoras theorem,

⇒ \(A C^2 =A B^2+B C^2=24^2+7^2\)

=576+49=625

AC = 25 cm

(1) \(\sin A=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{B C}{A C}=\frac{7}{25}\)

⇒ \(\cos A=\frac{\text { base }}{\text { hypotenuse }}=\frac{A B}{A C}=\frac{24}{25}\)

(2) \(\sin C=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{A B}{A C}=\frac{24}{25}\).

⇒ \(\cos C=\frac{\text { base }}{\text { hypotenuse }}=\frac{B C}{A C}=\frac{7}{25}\).

Question 2. In figure, find \(\tan P-\cot R\),

Solution :

In \(\triangle\) PQR,

Trigonometry The Value Of Tan P - Cot R

⇒ \(P Q^2+Q R^2 =P R^2\)

⇒ \(Q R^2 =P R^2-P Q^2\)

⇒ \(Q R^2 =(13)^2-(12)^2\)

=169-144=25 \(\Rightarrow Q R=5 \mathrm{~cm}\)

Now, \(\tan P=\frac{\text { perpendicular }}{\text { base }}=\frac{5}{12}\)

⇒ \(\cot R=\frac{\text { base }}{\text { perpendicular }}=\frac{5}{12}\)

∴ \(\tan P-\cot R=\frac{5}{12}-\frac{5}{12}=0\)

\(\tan P-\cot R\) = 0

Question 3. If \(\sin A=\frac{3}{4}\), calculate cos A and tan A.

Solution :

⇒ \(\sin A=\frac{3}{4}\)

In \(\triangle A B C\),

Trigonometry The Value Of Cos A And Tan A

Let BC = 3k

and AC = 4k

⇒ \(A B^2+B C^2 =A C^2\)

⇒ \(A B^2 =A C^2-B C^2\)

⇒ \(A C^2 =(4 k)^2-(3 k)^2\)

= \(16 k^2-9 k^2=7 k^2\)

A B = \(\sqrt{7} k\)

Now, \(\cos A=\frac{\text { base }}{\text { hypotenuse }}=\frac{\sqrt{7} k}{4 k}=\frac{\sqrt{7}}{4}\)and \(\tan A=\frac{\text { perpendicular }}{\text { base }}=\frac{3 k}{\sqrt{7} k}=\frac{3}{\sqrt{7}}\)

Question 4. Given 15 cotA = 8, find sin A and Sec A.

Solution :

⇒ \(15 \cot A=8 \Rightarrow \cot A=\frac{8}{15}\)

Trigonometry The Value Of Sin A And Sec A

Let base =8 k=A B

and Perpendicular =15 k=B C

In \(\triangle A B C\),

⇒ \(A C^2=A B^2+B C^2\)

= \((8 k)^2+(15 k)^2\)

= \(64 k^2+225 k^2=289 k_A^2\)

AC = 17 k

Now, \(\sin A=\frac{\text { perpendicular }}{\text { hypotenuse }}\)

= \(\frac{B C}{A C}=\frac{15 k}{17 k}=\frac{15}{17}\)

and \(\sec A=\frac{\text { hypotenuse }}{\text { base }}\)

= \(\frac{A C}{A B}=\frac{17 k}{8 k}=\frac{17}{8}\)

Question 5. Given \(\sec \theta=\frac{13}{12}\), calculate all other trigonometric ratios.

Solution :

⇒ \(\sec \theta=\frac{13}{12}\)

Introduction to Trigonometry

⇒ \(\sec \theta=\frac{\text { hypotenuse }}{\text { base }}=\frac{13}{12}\)

Let, in \(\triangle A B C\), \(\angle B=90^{\circ}\) and \(\angle A=\theta\)

Let, A C=13 k and A B=12 k

Now, \(A B^2+B C^2=A C^2\)

⇒ \(B C^2 =A C^2-A B^2=(13 k)^2-(12 k)^2\)

= \(169 k^2-144 k^2=25 k^2\)

BC = 5 k

Now, \(\sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{5 k}{13 k}=\frac{5}{13}\)

⇒ \(\cos \theta=\frac{\text { base }}{\text { hypotenuse }}=\frac{12 k}{13 k}=\frac{12}{13}\)

⇒ \(\tan \theta=\frac{\text { perpendicular }}{\text { base }}=\frac{5 k}{12 k}=\frac{5}{12}\)

cosec \(\theta=\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{13 k}{5 k}=\frac{13}{5}\)

∴ \(\cot \theta=\frac{\text { base }}{\text { perpendicular }}=\frac{12 k}{5 k}=\frac{12}{5}\)

Question 6. If \(\angle A\) and \(\angle B\) are acute angles such that \(\cos A=\cos B\), then show that \(\angle A=\angle B\).

Solution :

Let, in \(\triangle A B C, \angle C=90^{\circ}\)

Trigonometry The Acute Angles Of Triangle ABC

⇒ \(\angle A\) and \(\angle B\) are acute angles.

Given,\(\cos A=\cos B\)

Question 7. If \(\cot \theta=\frac{7}{8}\), evaluate :

  1.  \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\),
  2. \(\cot ^2 \theta\)

Solution :

⇒ \(\cot \theta=\frac{7}{8}\)

Trigonometry In Triangle ABC The Value Of Cot

⇒ \(\cot \theta=\frac{\text { base }}{\text { perpendicular }}=\frac{7}{8}\)

Let, in \(\triangle A B C\), \(\angle B=90^{\circ}\) and \(\angle A=\theta\)

Let base AB = 7k

and perpendicular BC = 8k

Now, \(A C^2=A B^2+B C^2\)

=\((7 k)^2+(8 k)^2\)

=\(49 k^2+64 k^2=113 k^2\)

A C =\(\sqrt{113} k\)

Now, \(\sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{B C}{A C}\)

=\(\frac{8 k}{\sqrt{113 k}}=\frac{8}{\sqrt{113}}\)

and \(\cos \theta=\frac{\text { base }}{\text { hypotenuse }}=\frac{A B}{A C}\)

= \(\frac{7 k}{\sqrt{113 k}}=\frac{7}{\sqrt{113}}\)

(1) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\)

= \(\frac{\left(1+\frac{8}{\sqrt{113}}\right)\left(1-\frac{8}{\sqrt{113}}\right)}{\left(1+\frac{7}{\sqrt{113}}\right)\left(1-\frac{7}{\sqrt{113}}\right)}\)

= \(\frac{(1)^2-\left(\frac{8}{\sqrt{113}}\right)^2}{(1)^2-\left(\frac{7}{\sqrt{113}}\right)^2}=\frac{1-\frac{64}{113}}{1-\frac{49}{113}}\)

= \(\frac{113-64}{113-49}=\frac{49}{64}\)

(2) \(\cot ^2 \theta=\left(\frac{7}{8}\right)^2=\frac{49}{64}\)

Question 8. If \(3 \cot A=4\), check whether \(\frac{\left(1-\tan ^2 A\right)}{\left(1+\tan ^2 A\right)}=\cos ^2 A-\sin ^2 A\) or not.

Solution :

3 cot A=4

⇒ \(\cot A=\frac{4}{3}\)

Now, \(\cot A=\frac{\text { base }}{\text { perpendicular }}=\frac{4}{3}\)

Trigonometry In Triangle ABC The Value Of Cos

In \(\triangle A B C, \angle B =90^{\circ}\)

base A B =4 k, perpendicular B C = 3 k

⇒ \(A C^2 =A B^2+B C^2=(4 k)^2+(3 k)^2\)

= \(16 k^2+9 k^2=25 k^2\)

A C = 5 k

Now, \(\tan A=\frac{\text { perpendicular }}{\text { base }}=\frac{B C}{A B}=\frac{3 k}{4 k}=\frac{3}{4}\)

⇒ \(\cos A=\frac{\text { base }}{\text { hypotenuse }}=\frac{A B}{A C}=\frac{4 k}{5 k}=\frac{4}{5}\)

⇒ \(\sin A=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{B C}{A C}=\frac{3 k}{5 k}=\frac{3}{5}\)

Now, \( \frac{1-\tan ^2 A}{1+\tan ^2 A}=\frac{1-\left(\frac{3}{4}\right)^2}{1+\left(\frac{3}{4}\right)^2}\)

= \(\frac{1-\frac{9}{16}}{1+\frac{9}{16}}=\frac{16-9}{16+9}=\frac{7}{25}\)

and \(\cos ^2 A-\sin ^2 A=\left(\frac{4}{5}\right)^2-\left(\frac{3}{5}\right)^2\)

=\(\frac{16}{25}-\frac{9}{25}=\frac{16-9}{25}=\frac{7}{25}\)

∴ \(\frac{1-\tan ^2 A}{1+\tan ^2 A}=\cos ^2 A-\sin ^2 A\)

Question 9. In triangle ABC, right-angled at B, if \(\tan A=\frac{1}{\sqrt{3}}\), find the value of:

  1. \(\sin A \cos C+\cos A \sin C\)
  2.  \(\cos A \cos C-\sin A \sin C\)

Solution :

In \(\triangle A B C, \angle B=90^{\circ}\)

Trigonometry In Triangle ABC, Right Angled At B Of Tan A

⇒ \(\tan A=\frac{1}{\sqrt{3}}\)

⇒ \(\tan A =\frac{\text { perpendicular }}{\text { base }}\)

= \(\frac{1}{\sqrt{3}}\)

Let perpendicular B C = k and base \(A B=k \sqrt{3}\)

Now, \(A C^2 =A B^2+B C^2=(k \sqrt{3})^2+k^2\)

= \(3 k^2+k^2=4 k^2\)

A C = 2 k

Now, \(\sin A=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{B C}{A C}=\frac{k}{2 k}=\frac{1}{2}\)

⇒ \(\cos A =\frac{\text { base }}{\text { hypotenuse }}=\frac{A B}{A C}=\frac{k \sqrt{3}}{2 k}=\frac{\sqrt{3}}{2}\)

⇒ \(\sin C =\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{A B}{A C}\)

= \(\frac{k \sqrt{3}}{2 k}=\frac{\sqrt{3}}{2}\)

⇒ \(\cos C=\frac{\text { base }}{\text { hypotenuse }}=\frac{B C}{A C}=\frac{k}{2 k}=\frac{1}{2}\)

(1) \(\sin A \cos C+\cos A \sin C =\frac{1}{2} \times \frac{1}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\)

= \(\frac{1}{4}+\frac{3}{4}=1\)

(2) \(\cos A \cos C-\sin A \sin C=\frac{\sqrt{3}}{2} \times \frac{1}{2}-\frac{1}{2} \times \frac{\sqrt{3}}{2}=0\)

Question 10. In \(\triangle\) PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P, and tan P.

Solution :

In \(\triangle P Q R\), \(\angle Q=90^{\circ}\)

and P Q =5 cm

P R+Q R =25  → Equation 1

Now, \(P R^2=P Q^2+Q R^2\)

Trigonometry The Values OF Sin P, Cos P, Tan P

⇒ \(P R^2-Q R^2=P Q^2\)

⇒ \((P R-Q R)(P R+Q R) =5^2\)

⇒ \((P R-Q R) \times 25 =2\)

P R-Q R =1  → Equation 2

Adding equations (1) and (2),

2 \(\cdot P R=26 \quad \Rightarrow \quad P R=13\)

From equation (1)

Now, Q R =25-P R=25-13=12

⇒ \(\sin P =\frac{Q R}{P R}=\frac{12}{13}\)

⇒ \(\cos P =\frac{P Q}{P R}=\frac{5}{13}\)

∴ \(\tan P =\frac{Q R}{P Q}=\frac{12}{5}\)

Question 11. State whether the following are true or false. Justify your answer.

  1. The value of tan A is always less than 1.
  2.  sec A = \(\frac{12}{5}\) for some value of angle A.
  3. cos A is the abbreviation used for the cosecant of angle A.
  4. cot A is the product of cot and A.
  5. \(\sin \theta=\frac{4}{3}\) for some angle \(\theta\).

Solution :

(1) False, \(\tan A=\frac{\text { perpendicular }}{\text { base }}\)

tan A < l is possible only when the perpendicular is smaller the base but it is not always necessary, hypotenuse

(2) True, sec A=\(\frac{\text { hypotenuse }}{\text { base }}\)

Hypotenuse is always greater than the base.

Therefore, sec A = \(\frac{12}{5}\), is true tor some angle A

(3) False, cos A, is the brief form of the cosine of \(\angle\)A

Introduction To Trigonometry Exercise 8.2

Question 1. Evaluate the following :

  1.  \(\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}\)
  2.  \(2 \tan ^2 45^{\circ}+\cos ^2 30^{\circ}-\sin ^2 60^{\circ}\)
  3. \(\frac{\cos 45^{\circ}}{\sec 30^{\circ}+{cosec} 30^{\circ}}\)
  4.  \(\frac{\sin 30^{\circ}+\tan 45^{\circ}-{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\)
  5. \(\frac{5 \cos ^2 60^{\circ}+4 \sec ^2 30^{\circ}-\tan ^2 45^{\circ}}{\sin ^2 30^{\circ}+\cos ^2 30^{\circ}}\)

Solution :

(1) \(\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}\)

=\(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{1}{2}\)

=\(\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1\)

(2) \(2 \tan ^2 45^{\circ}+\cos ^2 30^{\circ}-\sin ^2 60^{\circ}\)

=\(2(1)^2+\left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{\sqrt{3}}{2}\right)^2\)

=\(2+\frac{3}{4}-\frac{3}{4}=2\)

(3) \(\frac{\cos 45^{\circ}}{\sec 30^{\circ}+{cosec} 30^{\circ}}\)

=\(\frac{1 / \sqrt{2}}{\frac{2}{\sqrt{3}}+\frac{2}{1}}\)

⇒ \(=\frac{1}{\sqrt{2}\left(\frac{2+2 \sqrt{3}}{\sqrt{3}}\right)}\)

⇒ \(=\frac{\sqrt{3}}{2 \sqrt{2}(\sqrt{3}+1)}\)

= \(\frac{\sqrt{3} \cdot \sqrt{2}(\sqrt{3}-1)}{2 \sqrt{2}(\sqrt{3}+1) \cdot \sqrt{2}(\sqrt{3}-1)}\)

= \(\frac{\sqrt{2}(3-\sqrt{3})}{2 \cdot 2(3-1)}=\frac{3 \sqrt{2}-\sqrt{6}}{8}\)

(4) \(\frac{\sin 30^{\circ}+\tan 45^{\circ}-{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\)

= \(\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}\)

= \(\frac{\frac{\sqrt{3}+2 \sqrt{3}-4}{2 \sqrt{3}}}{\frac{4+\sqrt{3}+2 \sqrt{3}}{2 \sqrt{3}}}=\frac{3 \sqrt{3}-4}{3 \sqrt{3}+4}\)

= \(\frac{3 \sqrt{3}-4}{3 \sqrt{3}+4} \times \frac{3 \sqrt{3}-4}{3 \sqrt{3}-4}\)

= \(\frac{27+16-24 \sqrt{3}}{(3 \sqrt{3})^2-(4)^2}\)

= \(\frac{43-24 \sqrt{3}}{27-16}\)

= \(\frac{43-24 \sqrt{3}}{11}\)

(5) \(\frac{5 \cos ^2 60^{\circ}+4 \sec ^2 30^{\circ}-\tan ^2 45^{\circ}}{\sin ^2 30^{\circ}+\cos ^2 30^{\circ}}\)

=\(\frac{5\left(\frac{1}{2}\right)^2+4\left(\frac{2}{\sqrt{3}}\right)^2-(1)^2}{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\)

=\(\frac{\frac{5}{4}+4 \times \frac{4}{3}-1}{\frac{1}{4}+\frac{3}{4}}=\frac{\frac{15+64-12}{12}}{1}=\frac{67}{12}\)

Question 2. Choose the correct option and justify your choice :

(1) \(\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}\)= ?

  1. \(\sin 60^{\circ}\)
  2. \(\cos 60^{\circ}\)
  3. \(\tan 60^{\circ}\)
  4. \(\sin 30^{\circ}\)

(2) \(\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}=\) ?

  1. \(\tan 90^{\circ}\)
  2. 1
  3. \(\sin 45^{\circ}\)
  4. 0

(3) \(\sin 2 A=2 \sin A\) is true when A=

  1. \(0^{\circ}\)
  2. \(30^{\circ}\)
  3. \(45^{\circ}\)
  4. \(60^{\circ}\)

(4) \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\)=

  1. \(\cos 60^{\circ}\)
  2. \(\sin 60^{\circ}\)
  3. \(\tan 60^{\circ}\)
  4. \(\sin 30^{\circ}\)

Solution:

(1) 1

⇒ \(\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}=\frac{2 \times \frac{1}{\sqrt{3}}}{1+\left(\frac{1}{\sqrt{3}}\right)^2}\)

= \(\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}=\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}=\frac{2}{\sqrt{3}} \times \frac{3}{4}=\frac{\sqrt{3}}{2}=\sin 60^{\circ}\)

(2) 4

⇒ \(\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}=\frac{1-(1)^2}{1+(1)^2}=\frac{1-1}{1+1}=\frac{0}{2}=0\)

(3) 1

If A =\(0^{\circ}\)then \(2 A=0^{\circ}\)

⇒ \(\sin 2 A =\sin 0^{\circ}=0\)

and \(2 \sin A =2 \sin 0^{\circ}\)

=2 \(\times 0=0\)

So, for \(A=0^{\circ}, \sin 2 A=2 \sin A\)

(4) 3

⇒ \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}} =\frac{2 \times \frac{1}{\sqrt{3}}}{1-\left(\frac{1}{\sqrt{3}}\right)^2}=\frac{2 / \sqrt{3}}{1-\frac{1}{3}}=\frac{2 / \sqrt{3}}{2 / 3}\)

= \(\frac{2}{\sqrt{3}} \times \frac{3}{2}=\sqrt{3}=\tan 60^{\circ}\)

Question 3. If \(\tan (A+B)=\sqrt{3}\) and \(\tan (A-B)=\frac{1}{\sqrt{3}}\) ; \(0^{\circ}<A+B \leq 90^{\circ}\) ; A>B, find A and B.

Solution :

⇒ \(\tan (A+B)=\sqrt{3}\)

⇒ \(\tan (A+B)=\tan 60^{\circ}\)

⇒ \(A+B=60^{\circ}\)

and \(tan (A-B)=\frac{1}{\sqrt{3}}\)

⇒ \(\tan (A-B)=\tan 30^{\circ} \Rightarrow A-B=30^{\circ}\)

Adding equations (1) and (2)

⇒ \(A+B=60^{\circ}\)

⇒ \(A-B=30^{\circ}\)

⇒ \( 2 A=90^{\circ}\)

⇒ \(A \quad A=45^{\circ}\)

Put the value of A in equation (1),

⇒ \(45^{\circ}+B=60^{\circ} \Rightarrow \quad B\)

A=\(45^{\circ}\) and B=\(15^{\circ} \quad 45^{\circ}=15^{\circ}\)

Question 4. State whether the following are true or false. Justify your answer.

  1. \(\sin (A+B)=\sin A+\sin B\)
  2. The value of \(\sin \theta\) increases as \(\theta\) increases.
  3.  The value of \(\cos \theta\) increases as \(\theta\) increases.
  4. \(\sin \theta=\cos \theta\) for all values of θ.
  5.  cot A is not defined for A=\(0^{\circ}\)

Solution :

(1) False,

Let A=\(30^{\circ}\) and B=\(60^{\circ}\)

⇒ \(\sin (A+B)=\sin \left(30^{\circ}+60^{\circ}\right)=\sin 90^{\circ}=1\)

and \(\sin A+\sin B=\sin 30^{\circ}+\sin 60^{\circ}\)

= \(\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{\sqrt{3}+1}{2} \neq 1 \)

⇒ \(\sin (A+B) \neq \sin A+\sin B\)

(2) True, as the value of \(\theta\) varies from \(\theta^{\circ}\) to \(90^{\circ}\) then the value of \(\sin \theta\) varies from 0 to 1 .

(3) False, as the value of \(\theta\) varies front 0° to 90° then the value of cos \(\theta\) varies from 1 to 0,

i.e, the value of cos \(\theta\) decreases.

(4) False,

⇒ \(\theta\) = 0° then sin \(\theta\) = sin 0° = 0

and \(\cos 0=\cos \theta^{\circ}=1\)

⇒ \(\sin 0 \times \cos 0\), if \(\theta=0^{\circ}\)

(5) True,

⇒ \(A=0^{\circ}\) than \(\cot A=\cot 0^{\circ}\) which is not defined.

Exercise 8.3

Question 1. Evaluate :

  1. \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
  2. \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)
  3. \(\cos 48^{\circ}-\sin 42^{\circ}\)
  4. \(cosec 31^{\circ}-\sec 59^{\circ}\)

Solution :

(1) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}} =\frac{\sin \left(90^{\circ}-72^{\circ}\right)}{\cos 72^{\circ}}\)

=\(\frac{\cos 72^{\circ}}{\cos 72^{\circ}}=1\)

\(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\) = 1

(2)\(\frac{\tan 26^{\circ}}{\cot 64^{\circ}} =\frac{\tan \left(90^{\circ}-64^{\circ}\right)}{\cot 64^{\circ}}\)

=\(\frac{\cot 64^{\circ}}{\cot 64^{\circ}}=1\)

\(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\) =1

(3)\(\cos 48^{\circ}-\sin 42^{\circ} =\cos \left(90^{\circ}-42^{\circ}\right)-\sin 42^{\circ}\)

=\(\sin 42^{\circ}-\sin 42^{\circ}=0\)

\(\cos 48^{\circ}-\sin 42^{\circ}\) =0

(4) \({cosec} 31^{\circ}-\sec 59^{\circ}\)

= \({cosec}\left(90^{\circ}-59^{\circ}\right)-\sec 59^{\circ}\)

=\(\sec 59^{\circ}-\sec 59^{\circ}=0\)

\(cosec 31^{\circ}-\sec 59^{\circ}\)= 0

Question 2. Show that:

  1. \(\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}=1\)
  2. \(\cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin 52^{\circ}=0\)

Solution :

(1) L.H.S. =\(\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}\)

=\(\tan 48^{\circ} \cdot \tan 23^{\circ} \cdot \tan \left(90^{\circ}-48^{\circ}\right)\)

⇒ \(\cdot \tan \left(90^{\circ}-23^{\circ}\right)\)

= \(\tan 48^{\circ} \cdot \tan 23^{\circ} \cdot \cot 48^{\circ} \cdot \cot 23^{\circ}\)

= \(\tan 48^{\circ} \cdot \tan 23^{\circ} \cdot \frac{1}{\tan 48^{\circ}} \cdot \frac{1}{\tan 23^{\circ}}\)

= 1 = R.H.S.

Hence Proved.

(2) L.H.S.=\(\cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ}+\sin 52^{\circ}\)

=\(\cos \left(90^{\circ}-52^{\circ}\right) \cdot \cos \left(90^{\circ}-38^{\circ}\right)\)

⇒ \(-\sin 38^{\circ} \cdot \sin 52^{\circ}\)

=\(\sin 52^{\circ} \cdot \sin 38^{\circ}-\sin 38^{\circ} \cdot \sin 52^{\circ}\)

= 0 = R.H.S

Hence Proved.

Question 3. If \(\tan 2 A=\cot \left(A-18^{\circ}\right)\), where 2 A is an acute angle, find the value of A.

Solution:

⇒ \(\tan 2 A =\cot \left(A-18^{\circ}\right)\)

⇒ \(\cot \left(90^{\circ}-2 A\right) =\cot \left(A-18^{\circ}\right)\)

⇒ \(90^{\circ}-2 A =A-18^{\circ}\)

⇒ \(90^{\circ}+18^{\circ} =A+2 A\)

⇒ \(3 A =108^{\circ}\)

A =\(36^{\circ}\)

The value of A =\(36^{\circ}\)

Question 4. If \(\tan A=\cot B\), prove that

A+B=\(90^{\circ}\).

Solution :

⇒ \(\tan A=\cot B\)

⇒ \(\tan A=\tan \left(90^{\circ}-B\right)\)

⇒ \(A=90^{\circ}-B\)

⇒ \(A+B=90^{\circ}\)

Hence Proved.

Question 5. If \(\sec 4 A={cosec}\left(A-20^{\circ}\right)\), where 4 A is an acute angle, find the value of A.

Solution :

⇒ \(\sec 4 A = cosec \left(A-20^{\circ}\right)\)

cosec\(\left(90^{\circ}-4 A\right) = cosec\left(A-20^{\circ}\right)\)

⇒ \(90^{\circ}-4 A =A-20^{\circ}\)

⇒ \(90^{\circ}+20^{\circ} =A+4 A\)

⇒ \(5 A =110^{\circ}\)

A =\(22^{\circ}\)

The value of A =\(22^{\circ}\)

Question 6. If A, B and C are interior angles of a triangle A B C, then show that \(\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}\)

Solution :

In \(\triangle A B C\)

⇒ \(A+B+C=180^{\circ}\)

⇒ \(B+C=180^{\circ}-A\)

⇒ \(\frac{B+C}{2}=\frac{180^{\circ}-A}{2}\)

= \(\frac{180^{\circ}}{2}-\frac{A}{2}=90^{\circ}-\frac{A}{2}\)

⇒ \(\sin \left(\frac{B+C}{2}\right)=\sin \left(90^{\circ}-\frac{A}{2}\right)\)

⇒ \(\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}\)

Hence Proved.

Exercise 8.4

Question 1. Express the trigonometric ratios \(\sin A, \sec A\) and \(\tan A\) in terms of \(\cot A\).

Solution :

⇒ \(\sin A=\frac{1}{cosec^2 A}\)

⇒ \(\sin A=\frac{1}{\sqrt{{cosec}^2 A}}\)

⇒ \(\sin A=\frac{1}{\sqrt{1+\cot ^2 A}}\)

⇒ \(\sec A=\sqrt{\sec ^2 A}\)

⇒ \(\sec A=\sqrt{1+\tan ^2 A}\)

⇒ \(\sec A=\sqrt{1+\left(\frac{1}{\cot A}\right)^2}\)

⇒ \(\sec A=\sqrt{1+\frac{1}{\cot ^2 A}}=\sqrt{\frac{\cot ^2 A+1}{\cot ^2 A}}\)

and \(\tan A=\frac{1}{\cot A}\)

Question 2. Write all the other trigonometric ratios of \(\angle A\) in terms of \sec A.

Solution :

⇒ \(\sin A=\sqrt{\sin ^2 A} =\sqrt{1-\cos ^2 A}\)

= \(\sqrt{1-\frac{1}{\sec ^2 A}}=\sqrt{\frac{\sec ^2 A-1}{\sec ^2 A}}\)

⇒ \(\sin A =\frac{\sqrt{\sec ^2 A-1}}{\sec A}\)

cos A =\(\frac{1}{\sec A}\)

tan A =\(\sqrt{\tan ^2 A}\)

⇒ \(\tan A =\sqrt{\sec ^2 A-1}\)

⇒ \(\cot A =\frac{1}{\tan A}=\frac{1}{\sqrt{\tan ^2 A}}\)

⇒ \(\cot A=\frac{1}{\sqrt{\sec ^2 A-1}}\)

⇒ \(cosec A=\sqrt{{cosec}^2 A}=\sqrt{1+\cot ^2 A}\)

= \(\sqrt{1+\frac{1}{\tan ^2 A}}\)

= \(\sqrt{\frac{1+\tan ^2 A}{\tan ^2 A}}=\sqrt{\frac{\sec ^2 A}{\sec ^2 A-1}}\)

cosec A=\(\frac{\sec A}{\sqrt{\sec ^2 A-1}}\)

Question 3. Evaluate:

  1. \(\frac{\sin ^2 63^{\circ}+\sin ^2 27^{\circ}}{\cos ^2 17^{\circ}+\cos ^2 73^{\circ}}\)
  2. \(\sin 25^{\circ} \cos 65^{\circ}+\cos 25^{\circ} \sin 65^{\circ}\)

Solution:

(1) \(\frac{\sin ^2 63^{\circ}+\sin ^2 27^{\circ}}{\cos ^2 17^{\circ}+\cos ^2 73^{\circ}}\)

=\(\frac{\sin ^2\left(90^{\circ}-27^{\circ}\right)+\sin ^2 27^{\circ}}{\cos ^2\left(90^{\circ}-73^{\circ}\right)+\cos ^2 73^{\circ}}\)

=\(\frac{\cos ^2 27^{\circ}+\sin ^2 27^{\circ}}{\sin ^2 73^{\circ}+\cos ^2 73^{\circ}}=\frac{1}{1}=1\)

  1. \(\frac{\sin ^2 63^{\circ}+\sin ^2 27^{\circ}}{\cos ^2 17^{\circ}+\cos ^2 73^{\circ}}\)=1

(2) \(\sin 25^{\circ} \cdot \cos 65^{\circ}+\cos 25^{\circ} \sin 65^{\circ}\)

= \(\sin 25^{\circ} \cdot \cos \left(90^{\circ}-25^{\circ}\right)\)

⇒ \(+\cos 25 \sin \left(90^{\circ}-25^{\circ}\right)\)

= \(\sin 25^{\circ} \cdot \sin 25^{\circ}+\cos 25^{\circ} \cdot \cos 25^{\circ}\)

= \(\sin ^2 25^{\circ}+\cos ^2 25^{\circ}=1\)

\(\sin 25^{\circ} \cos 65^{\circ}+\cos 25^{\circ} \sin 65^{\circ}\) = 1

Question 4. Choose the correct option. Justify your choice.

(1) 9 \(\sec ^2 A-9 \tan ^2 A\)=

  1. 1
  2. 9
  3. 8
  4. 0

(2) \((1+\tan \theta+\sec \theta)\)

  1. \((1+\cot \theta- cosec \theta)\)=
  2. 0
  3. 1
  4. 2

(3) \((\sec A+\tan A)(1-\sin A)=\)

  1. sec A
  2. sin A
  3. cosec A
  4. cos A

(4) \(\frac{1+\tan ^2 A}{1+\cot ^2 A}\)=

  1. \(\sec ^2 A\)
  2. -1
  3. \(\cot ^2 A\)
  4. \(\tan ^2 A\)

Solution :

(1) Answer. (2)

⇒ \(9 \sec ^2 A-9 \tan ^2 A=9\left(\sec ^2 A-\tan ^2 A\right)\)

=9 \(\times \)1=9

9 \(\sec ^2 A-9 \tan ^2 A\)= 9

(2) Answer. (3)

⇒ \((1+\tan \theta+\sec \theta)(1+\cot \theta-{cosec} \theta)\)

=\(1+\cot \theta-{cosec} \theta+\tan \theta+\tan \theta \cdot \cot \theta\)

⇒ –\(\tan \theta \cdot {cosec} \theta+\sec \theta+\sec \theta \cdot \cot \theta-\sec \theta \cdot{cosec} \theta\)

= \(1+\cot \theta-{cosec} \theta+\tan \theta +\frac{\sin \theta}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta} \cdot \frac{1}{\sin \theta}+\sec \theta\)

⇒ \(+\frac{1}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta}-\sec \theta {cosec} \theta\)

=\(1+(\cot \theta+\tan \theta)-{cosec} \theta\)

⇒ \(+1-\sec \theta+\sec \theta+ cosec \theta\)

⇒ \(-\sec \theta cose \theta\)

= \(2+\left(\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}\right)-\sec \theta {cosec} \theta\)

= \(2+\frac{\cos { }^2 \theta+\sin ^2 \theta}{\sin \theta \cdot \cos \theta}-\sec \theta{cosec} \theta\)

= \(2+\frac{1}{\sin \theta \cos \theta}-\frac{1}{\cos \theta \sin \theta}=2\) .

\((1+\tan \theta+\sec \theta)\) =1

(3) Answer. (4)

⇒ \((\sec A+\tan A)(1-\sin A)\)

= \(\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)(1-\sin A)\)

= \(\left(\frac{1+\sin A}{\cos A}\right)(1-\sin A)\)

= \(\frac{1-\sin ^2 A}{\cos A}=\frac{\cos ^2 A}{\cos A}=\cos A \)

\((\sec A+\tan A)(1-\sin A)=\) =cos A

(4) Answer. (4)

⇒ \(\frac{1+\tan ^2 A}{1+\cot ^2 A} =\frac{1+\tan ^2 A}{1+\frac{1}{\tan ^2 A}}=\frac{1+\tan ^2 A}{\frac{\tan ^2 A+1}{\tan ^2 A}}\)

=\(\tan ^2 A\)

\(\frac{1+\tan ^2 A}{1+\cot ^2 A}\)= \(\tan ^2 A\)

Trigonometry Multiple-Choice Questions

Question 1. The value of \(\sin 45^{\circ}+\cos 45^{\circ}\) is :

  1. 2
  2. \(\sqrt{2}\)
  3. \(\frac{\sqrt{3}}{2}\)
  4. 1

Answer: 2. \(\sqrt{2}\)

The value of \(\sin 45^{\circ}+\cos 45^{\circ}\) is \(\sqrt{2}\)

Question 2. If \(\sin A=\frac{3}{2}\) then the value of tan A is:

  1. \(\frac{5}{3}\)
  2. \(\frac{4}{3}\)
  3. \(\frac{3}{4}\)
  4. \(\frac{3}{5}\)

Answer: 3. \(\frac{3}{4}\)

The value of tan A is \(\frac{3}{4}\)

Question 3. If \(\sin A+\sin ^2 A=1\) then the value of \(\left(\cos ^2 A+\cos ^4 A\right)\) is :

  1. 1
  2. \(\frac{1}{2}\)
  3. 2
  4. \(\frac{1}{3}\)

Answer: 1. 1

The value of \(\left(\cos ^2 A+\cos ^4 A\right)\) is 1.

Question 4. If \(\cos 8 \alpha=\sin \alpha\) and \(8 \alpha<90^{\circ}\) then the value of \(\tan 3 \alpha\) is :

  1. 0
  2. 1
  3. \(\sqrt{3}\)
  4. \(\frac{1}{\sqrt{3}}\)

Answer: 4. \(\frac{1}{\sqrt{3}}\)

The value of \(\tan 3 \alpha\) is  \(\frac{1}{\sqrt{3}}\)

Question 5. In \(\triangle A B C\), \(\angle C=90^{\circ}\). The value of \(\cos (A+B)\) is :

  1. 1
  2. 0
  3. -1
  4. \(\frac{1}{2}\)

Answer: 2.  0

The value of \(\cos (A+B)\) is 0

Question 6. If \(4 \tan \theta=3\) then the value of \(\frac{4 \sin \theta-3 \cos \theta}{\sin \theta+\cos \theta}\) is :

  1. \(-\frac{16}{25}\)
  2. \(\frac{16}{25}\)
  3. 0
  4. 4

Answer: 3.  0

Question 7. The value of \(\sin \left(60^{\circ}+\theta\right)-\cot \left(30^{\circ}-\theta\right)\) is :

  1. 0
  2. 2 \(\tan \theta\)
  3. \(2 \cot \theta\)
  4. \(2 \sqrt{3}\)

Answer: 1.  0

Question 8. If \(\cos \theta=\sin \theta, 0 \leq \theta<90^{\circ}\), then angle \(\theta\) is equal to :

  1. \(0^{\circ}\)
  2. \(30^{\circ}\)
  3. \(45^{\circ}\)
  4. \(60^{\circ}\)

Answer: 3. \(45^{\circ}\)

\(\theta\) is equal to \(45^{\circ}\)

Question 9. If \(\sin A=\frac{3}{5}\) then the value of \(\cos A\) is:

  1. \(\frac{5}{4}\)
  2. \(\frac{4}{5}\)
  3. \(\frac{5}{3}\)
  4. \(\frac{4}{3}\)

Answer: 2. \(\frac{4}{5}\)

The value of \(\cos A\) is \(\frac{4}{5}\)

Question 10. If \(\sec \theta=2\) then the value of \(\theta\) is:

  1. \(30^{\circ}\)
  2. \(45^{\circ}\)
  3. \(60^{\circ}\)
  4. \(90^{\circ}\)

Answer: 3. \(60^{\circ}\)

The value of \(\theta\) is \(60^{\circ}\)

Question 11. If \(\cos ^2 \theta=\frac{1}{2}\) then the value of \(\sin ^2 \theta\) is:

  1. \(\frac{1}{4}\)
  2. \(\frac{\sqrt{3}}{2}\)
  3. \(\frac{1}{\sqrt{2}}\)
  4. \(\frac{1}{2}\)

Answer: 4. \(\frac{1}{2}\)

The value of \(\sin ^2 \theta\) is \(\frac{1}{2}\)

Question 12. If \(\tan \theta=\frac{2 a b}{a^2-b^2}\) then the value of \(\cos \theta\) is:

  1. 1
  2. \(\frac{a^2-b^2}{a^2+b^2}\)
  3. \(\frac{a^2+b^2}{a^2-b^2}\)
  4. \(\frac{2 a b}{a^2+b^2}\)

Answer: 2.  \(\frac{a^2-b^2}{a^2+b^2}\)

The value of \(\cos \theta\) is \(\frac{a^2-b^2}{a^2+b^2}\)

Question 13. If cosec A= A, \(0^{\circ} \leq A \leq 90^{\circ}\) then \(\angle A\) is equal to :

  1. \(120^{\circ}\)
  2. \(60^{\circ}\)
  3. \(45^{\circ}\)
  4. \(30^{\circ}\)

Answer: 3. \(45^{\circ}\)

Question 14. If \(3 \cot A=4\) then the value of \(\sec A\) is :

  1. \(\frac{3}{4}\)
  2. \(\frac{5}{4}\)
  3. \(\frac{2}{3}\)
  4. \(\frac{3}{2}\)

Answer: 2. \(\frac{5}{4}\)

The value of \(\sec A\) is \(\frac{5}{4}\)