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

r = radius of circle

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

Therefore, radius OB = 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 = 60^{0}

`=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 r_{1} and r_{2} are the radii of two given circles.

Given, The angle subtend by the arc in one circle = 60^{0}

The angle subtend by the arc in second circle = 75^{0}

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

`=(7)/(25)` radian