Class 11 Mathematics

# Trigonometric Functions

## NCERT Solution

### Exercise 1

Question 1: Find the radian measures corresponding to the following degree measures:

(i) 25° (ii) -47°30’ (iii) 240° (520°

Solution:We know, π radian = 180°

So, 1°=(π)/(180°)

25°=25xx(π)/(180°)

=(5π)/(36) radian

-47°30’=-(95)/2

So, (95)/2=(95)/(2)xx(π)/(180)

=-(19π)/(72) radian

240°=240xx(π)/(180)

=(4π)/3 radian

520°=520xx(π)/(180)

=(26π)/(9) radian

Question 2: Find the degree measures corresponding to the following radian measures ( Use π=22/7).

(i) (11)/(16) (ii) – 4 (iii) (5π)/3 (iv) (7π)/6

Solution: We know that π radian = 18°

Therefore, 1 radian =(180°)/(π)°

(i) (11)/(16) radian = (11)/(16)xx(180)/(π)

=(11)/(16)xx(180)/(22)xx7

=1/4xx(45)/(2)xx7=(315)/(8)°

=39°22’30’’

(ii) -4 radian =-(4xx(180°)/π)

=-4xx(180)/(22)xx7=-(2520)/(11)

=-229°5’27’’

(iii) (5π)/(3) radian =(5π)/(3)xx(180)/(π)

=5xx60=300°

(iv) (7π)/(6) radian =(7π)/(6)xx(180)/(π)

=7xx30=210°

Question 3: A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

Solution: We know that 1 minute = 60 second

Since, in 60 sec number of revolutions = 360

Hence, in 1 sec number of revolutions =(360)/(60)=6

In one revolution, the wheel makes angle of 360°

So, in 6 revolutions angle made by wheel =360°xx6

=2xx180°xx6=2xxπxx6 radian

=12 π radian

Question 4: Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm. (Use π = 22/7)

Solution: Given, length of arc, l = 22cm

Radius of the circle, r = 100 cm

We know that, l = r θ

Therefore, θ=l/r

Where, l = length of arc

ѳ = angle subtended at the centre.

Therefore,

θ=(22)/(100) radian

Or, θ=0.22 radian

Now, 0.22 radian =0.22xx((180)/(π))°

=(0.22xx(180xx7)/(22))°

=((180xx7)/(100))°

=12.6°=12°36’

Question 5: In a circle of diameter 40 cm, the length of cord is 20cm. Find the length of minor arc of the chord.

Solution:

Let ABC is a circle with centre O.

The diameter AB of the circle = 40cm

The BC is the chord of length = 20cm

The length of arc BD =?

In the given figure, the OC = 0B as they are radius of triangle, therefore it is a equilateral traiangle.

Hence, the angle COB = 600

=60xx(π)/(180) radian

=(π)/(3) radian

Now, we know that,

The length of arc, l = radius of the circle, r × angle subtended by the arc

=20xx(π)/(3) cm

=(20)/(3)xx(22)/(7) cm

=20(20)/(21) cm

This is the length of the arc

Question 6: If in two circles, arcs of the same length subtend angles 600 and 750 at the centre, find the ratio of their radii.

Solution: Let r1 and r2 are the radii of two given circles.

Given, The angle subtend by the arc in one circle = 600

The angle subtend by the arc in second circle = 750

The length of arcs is same.

θ_1=60°=60xx(π)/(180) radian

=(π)/(3) radian

θ_2=75°=75xx(π)/(180) radian

=(5π)/(12) radian

Now, we know that

l = rѲ

In the given circle

l_1=r_1\θ_1

l_2=r_2\θ_2

As the length of both the arcs are same.

Therefore,

r_1\θ_1= r_2\θ_2

Or, (r_1)/(r_2)=(θ_2)/(θ_1)

Or, (r_1)/(r_2)=((5π)/(12))/((π)/(3))

Or, (r_1)/(r_2)=5/4

Therefore, r_1:r_2=5:4

Question 7: Find the angle in radian through a pendulum swings if its length is 75cm and the tip describes an arc of length

(i) 10 cm (ii) 15cm (iii) 21cm

Solution:

(i) r = 75cm

length of arc, l = 10cm

The angle Ѳ =?

We know that,

θ=l/r

=(10)/(75) radian

=(2)/(15) radian

(ii) r = 75cm

length of arc, l = 15cm

The angle Ѳ =?

We know that,

θ=l/r

=(15)/(75) radian

=(1)/(5) radian

r = 75cm

length of arc, l = 21cm

The angle Ѳ =?

We know that,

θ=l/r

=(21)/(75) radian