# Class 11 NCERT Solutions – Chapter 3 Trigonometric Function – Exercise 3.1

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

### (i) 25° (ii) -47°30′ (iii) 240° (iv) 520°

(i)As we know that180° = π radian

So, 1° = π/180° radian

Then, 25° = (π/180°) × 25°

= 5π/36 radians

Hence, 25° equals to 5π/36 radians.

(ii)As we know that180° = π radian

So, 1° = π/180°

And 60′ = 1°

30′ = (1/2)°

So, -47°30′ = -47 (1/2)°

-47(1/2)° = (π/180) × (-95/2) = (-19π/72) radian.

Hence, -47°30′ is equals to -19π/72 radian.

(iii)As we know that180° = π radian

1° = π/180° radian

So 240° = (π/180°) × 240°

= 4π/3 radians

Hence, 240° equals to 4π/3 radians.

(iv)As we know that180° = π radian

1° = π/180° radian

So 520° = (π/180°) × 520°

= 26 π/9 radians

Hence, 520° equals to 26 π/9 radians.

### 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

(i)11/16 radian = (11/16) (180°/π) {as 180° = π radian, then 1 radian = 180°/π}= (11/16) × (180° × 7/22)

= (11 × 180° × 7/16 × 22)

= 315/8°

= 39 (3/8)°

= 39(3/8)°

= 39° + (3/8)°

Again 1° = 60′

So (3/8)° = 60′ × (3/8)

= 22 (1/2)’

= 22 (1/2)’

= 22′ + 1/2′

Again 1′ = 60″

= (1/2)’ = 30″

So 39 (3/8)° = 39° 22′ 30″

Hence, 11/16 radian results to 39° 22′ 30″.

(ii)-4 radian = -4 × (180°/π) {as 180° = π radian, then 1 radian = 180°/π}.= -4 ×180° × 7/22

= -229° (1/11)

= -229 (1/11)°= -229° + (1/11)°

Again(1/11)° = (1/11) × 60′. {as 1° = 60′}

= 5(5/11)’

Also, 5 (5/11)’ = 5′ + (5/11)’

(5/11)’ = (5/11) × 60″ {as 1′ = 60″}

= 27″

So, -229(1/11) = -229° 5’27”

Hence, -4 radian results to -229° 5′ 27″.

(iii)5π/3 radian = (5 π/3) × (180/π) {as 180° = π radian, then 1 radian =180°/π}.= (5 × 180/3)°

= 300°

Hence, 5π/3 results to 300°.

(iv)7π/6 radian = (7π/6) × (180°/π) {as 180° = π radian, then 1 radian =180°/π}.= (7 × 180/6)°

= 210°

Hence, 7π/6 radian results to 210°.

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

**Solution:**

Given that

Total revolutions made by the wheel in one minute is 360.

1 second = 360/6 = 60

We know that

When a wheel revolves once it covers 2π radian of distance.

In one minute, it will turn an angle of 360 × 2π radian = 720 π radian

In one second, it will turn an angle of 720 π radian/60 = 12 π radian {as 1 minute = 60 seconds}

Hence, in one second, the wheel turns an angle of 12π 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 that

The radius of circle (r) = 100 cm.

Length of the arc (l) = 22 cm.

Let us consider the angle subtended by the arc is θ.

Also, we know that θ = l/r

The angle subtended (θ) = 22/100 radian

For finding the degree measure we have to multiply 180°/π with radian measure

So, θ = (22/100) × (180/π)

θ = (22/100) × (180 × 7/22)

θ = (22 × 180 × 7/22 × 100)

θ = 126/10 degree

θ = 12 (3/5) degree

We know that 1° = 60′

(3/5)° = 60′ × (3/5)

= 36′

So 12 (3/5)° = 12° 36′

Hence, the degree measure of the angle subtended at the Centre of a circle is 12° 36′

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

**Solution:**

Given that

Diameter of circle (d) = 40 cm

Radius (r) = d/2 = 40/2 = 20 cm

Let us consider AB as the chord of circle having length 20 cm, and Centre at O.

It forms a triangle OAB,

Having Radius = OA = OB = 20 cm

Also, chord AB = 20 cm

Hence, In ΔOAB OA = OB = AB. (equilateral triangle.)

So angle subtend = (π/3) radian

We know that θ = l/r (where θ = angle subtended by the arc

l = length of arc

r = radius)

Putting values of r and θ we get

π/3 = l/20

So.

l = 20 π/3

Hence, length of the arc is 20π/3 cm.

### Question 6. If in two circles, arcs of the same length subtend angles 60° and 75° at the Centre, find the ratio of their radii.

**Solution: **

Given that

Angle subtend by 1st arc (θ

_{1}) = 60Angle subtend by 2nd arc (θ

_{2}) = 75We know that θ = l/r

For 1st arc θ

_{1}= l_{1}/r_{1}For 2nd arc θ

_{2}= l_{2}/r_{2}θ

_{1}/θ_{2}= (l_{1}/r_{1})/(l_{2}/r_{2})θ

_{1}/θ_{2}= (l/r_{1})/(l/r_{2}) {here l_{1}= l_{2}= l}θ

_{1}/θ_{2}= r_{2}/r_{1}60/75 = r

_{2}/r_{1}r

_{2}/r_{1}= 4/5r

_{1}/r_{2}= 5/4Hence, ratio of their radius is 5:4.

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

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

**Solution:**

(i)Given thatLength of an arc (l) = 10 cm

Radius which represents length of pendulum(r) = 75

As We know that θ = l/r

So θ = 10/75 = 2/15 rad

Hence, θ = 2/15 rad

(ii)Given thatLength of an arc (l) = 15 cm

Radius which represents length of pendulum (r) = 75

As We know that θ = l/r

So θ = 15/75 = 1/5 rad

Hence, θ = 1/5 rad

(iii)Given thatLength of an arc (l) = 21 cm

Radius which represents length of pendulum(r) = 75

As We know that θ = l/r

So θ = 21/75 = 7/25 rad

Hence, θ = 7/25 radian