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

• Last Updated : 01 Dec, 2020

### (i) 25Â° (ii) -47Â°30â€² (iii) 240Â° (iv) 520Â°

(i) As we know that

180Â° = Ï€ radian

So, 1Â° = Ï€/180Â° radian

Then, 25Â° = (Ï€/180Â°) Ã— 25Â°

Hence, 25Â° equals to 5Ï€/36 radians.

(ii) As we know that

180Â° = Ï€ radian

So, 1Â° = Ï€/180Â°

And 60′ = 1Â°

30′ = (1/2)Â°

So, -47Â°30′ = -47 (1/2)Â°

-47(1/2)Â° = (Ï€/180) Ã— (-95/2) = (-19Ï€/72) radian.

Hence, -47Â°30′ is equals to -19Ï€/72 radian.

(iii) As we know that

180Â° = Ï€ radian

1Â° = Ï€/180Â° radian

So 240Â° = (Ï€/180Â°) Ã— 240Â°

Hence, 240Â° equals to 4Ï€/3 radians.

(iv) As we know that

180Â° = Ï€ radian

1Â° = Ï€/180Â° radian

So 520Â° = (Ï€/180Â°) Ã— 520Â°

= 26 Ï€/9 radians

Hence, 520Â° equals to 26 Ï€/9 radians.

### (i)11/16 (ii) -4 (iii) 5Ï€/3 (iv) 7Ï€/6

(i) 11/16 radian = (11/16) (180Â°/Ï€)  {as 180Â° = Ï€ radian, then 1 radian = 180Â°/Ï€}

= (11/16) Ã— (180Â° Ã— 7/22)

= (11 Ã— 180Â° Ã— 7/16 Ã— 22)

= 315/8Â°

= 39 (3/8)Â°

= 39(3/8)Â°

= 39Â° + (3/8)Â°

Again 1Â° = 60′

So (3/8)Â° = 60′ Ã— (3/8)

= 22 (1/2)’

= 22 (1/2)’

= 22′ + 1/2′

Again 1′ = 60″

= (1/2)’ = 30″

So 39 (3/8)Â° = 39Â° 22′ 30″

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

(ii) -4 radian = -4 Ã— (180Â°/Ï€) {as 180Â° = Ï€ radian, then 1 radian = 180Â°/Ï€}.

= -4 Ã—180Â° Ã— 7/22

= -229Â° (1/11)

= -229 (1/11)Â°= -229Â° + (1/11)Â°

Again(1/11)Â° = (1/11) Ã— 60′. {as 1Â° = 60′}

= 5(5/11)’

Also, 5 (5/11)’ = 5′ + (5/11)’

(5/11)’ = (5/11) Ã— 60″ {as 1′ = 60″}

= 27″

So, -229(1/11) = -229Â° 5’27”

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

(iii) 5Ï€/3 radian = (5 Ï€/3) Ã— (180/Ï€) {as 180Â° = Ï€ radian, then 1 radian =180Â°/Ï€}.

= (5 Ã— 180/3)Â°

= 300Â°

Hence, 5Ï€/3 results to 300Â°.

(iv) 7Ï€/6 radian = (7Ï€/6) Ã— (180Â°/Ï€)  {as 180Â° = Ï€ radian, then 1 radian =180Â°/Ï€}.

= (7 Ã— 180/6)Â°

= 210Â°

Hence, 7Ï€/6 radian results to 210Â°.

### 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  2Ï€ radian of distance.

In one minute, it will turn an angle of 360 Ã— 2Ï€ radian = 720 Ï€ radian

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

Hence, in one second, the wheel turns an angle of 12Ï€ 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  Ï€ = 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 Î¸.

Also, we know that Î¸ = l/r

The angle subtended (Î¸) = 22/100 radian

For finding the degree measure we have to multiply 180Â°/Ï€ with radian measure

So, Î¸ = (22/100) Ã— (180/Ï€)

Î¸ = (22/100) Ã— (180 Ã— 7/22)

Î¸ = (22 Ã— 180 Ã— 7/22 Ã— 100)

Î¸ = 126/10 degree

Î¸ = 12 (3/5) degree

We know that 1Â° = 60′

(3/5)Â° = 60′ Ã— (3/5)

= 36′

So 12 (3/5)Â° = 12Â° 36′

Hence, the degree measure of the angle subtended at the Centre of a circle is 12Â° 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 Î”OAB OA = OB = AB. (equilateral triangle.)

So angle subtend = (Ï€/3) radian

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

l = length of arc

Putting values of r and Î¸ we get

Ï€/3 = l/20

So.

l = 20 Ï€/3

Hence, length of the arc is 20Ï€/3 cm.

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

Solution:

Given that

Angle subtend by 1st arc (Î¸1) = 60

Angle subtend by 2nd arc (Î¸2) = 75

We know that Î¸ = l/r

For 1st arc Î¸1 = l1/r1

For 2nd arc Î¸2 = l2/r2

Î¸1/Î¸2 = (l1/r1)/(l2/r2

Î¸1/Î¸2 = (l/r1)/(l/r2)  {here l1 = l2 = l}

Î¸1/Î¸2 = r2/r1

60/75 = r2/r1

r2/r1 = 4/5

r1/r2 = 5/4

Hence, ratio of their radius is 5:4.

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

Solution:

(i) Given that

Length of an arc (l) = 10 cm

Radius which represents length of pendulum(r) = 75

As We know that Î¸ = l/r

So Î¸ = 10/75 = 2/15 rad

Hence, Î¸ = 2/15 rad

(ii) Given that

Length of an arc (l) = 15 cm

Radius which represents length of pendulum (r) = 75

As We know that Î¸ = l/r

So Î¸ = 15/75 = 1/5 rad

Hence, Î¸ = 1/5 rad

(iii) Given that

Length of an arc (l) = 21 cm

Radius which represents length of pendulum(r) = 75

As We know that Î¸ = l/r

So Î¸ = 21/75 = 7/25 rad

Hence, Î¸ = 7/25 radian

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