# Class 11 RD Sharma Solutions – Chapter 4 Measurement of Angles – Exercise 4.1 | Set 2

Last Updated : 10 May, 2021

### Question 11. A rail road curve is to be laid out on a circle. What radius should be used if the track is to change direction by 25o in a distance of 40 meters.

Solution:

Let AB denote the given rail-road.

We are given âˆ AOB = 25o. We know 180o = Ï€ radians = Ï€c or 1o = (Ï€/180)c

Hence 25o = 25 Ã— Ï€/180 = 5Ï€/36 radians

Also since, Î¸ = Arc/Radius â‡’ âˆ AOB = AB/OA â‡’ (5Ï€/36)c = 40/r â‡’ r = 288/Ï€ = 91.64 m

Hence, the radius of the track is 91.64 m.

### Question 12. Find the length which at a distance of 5280m will subtend an angle of 1â€™ at the eye.

Solution:

Let Î¸ = 1â€™ and the length of the arc that subtends Î¸ be l.

Radius = OA = OB = 5280m

We know, 1â€™ = 60o â‡’ 1â€™ = (1/60)o.  Since 180o = Ï€ radians = Ï€c or 1o = (Ï€/180)c,

â‡’ Î¸ = 1â€™ = (1/60 Ã— Ï€/180)

Also since, Î¸ = Arc/Radius â‡’ (1/60 Ã— Ï€/180)c = l/5280 â‡’ l = 1.5365 m

Hence, the length of the arc is 1.5365 m.

### Question 13. A wheel makes 360 revolutions per minute. Through how many radians does it turn in 1 second?

Solution:

Since the wheel makes 360 revolutions in 1 minute, number of revolutions made by it in 1 second = 360/60 = 6.

Angle made by the wheel in 1 revolution = 360o

Thus, angle made by the wheel in 6 revolutions = Angle made in 1 second = 360 Ã— 6 = 2160o

We know 180o = Ï€ radians = Ï€c or 1o = (Ï€/180)c

Hence, 2160o = (2160Ï€/180)c = 12Ï€ radians

Thus, the wheel turns 12Ï€ radians in 1 second.

### (i) 10 cm

Solution:

Let OA be the length of the pendulum. â‡’ OA = 75 cm = 0.75 m

Let the arc be denoted by AB. â‡’ AB = 10 cm = 0.1 m

Hence, the angle is 2/15 radians.

### (ii) 15 cm

Solution:

Let OA be the length of the pendulum. â‡’ OA = 75 cm = 0.75 m

Let the arc be denoted by AB. â‡’ AB = 15 cm = 0.15 m

Hence, the angle is 1/5 radians.

### (iii) 21 cm

Solution:

Let OA be the length of the pendulum. â‡’ OA = 75 cm = 0.75 m

Let the arc be denoted by AB. â‡’ AB = 15 cm = 0.21 m

Hence, the angle is 7/25 radians.

### Question 15. The radius of a circle is 30 cm. Find the length of the arc of the circle, if the length of the chord of the arc is 30 cm.

Solution:

Let OA = OB = Radius of the circle = 30 cm = 0.3 m, and chord AB = 30 cm = 0.3 m. Let l be the length of the arc AB.

Since, OA = OB = AB = 0.3 m, the triangle AOB is an equilateral triangle.

âˆ AOB = 60o. We know 180o = Ï€ radians = Ï€c or 1o = (Ï€/180)c

Hence, 60o = 60 Ã— Ï€/180 = Ï€/3 radians

Also since, Î¸ = Arc/Radius â‡’ (0.3Ï€/3) = 0.1 0.1Ï€ m = 10Ï€ cm

Hence, the length of the arc is 10Ï€ cm.

### Question 16. A railway train is travelling on a circular curve of 150 meters radius at the rate of 66 km/hr. Through what angle has it turned in 10 seconds?

Solution:

In the given circular track, OA = OB = r = 150 m

Let Î¸ denote the angle the train turns in 10 seconds.

We are given that speed = 66 km/hr = {66 Ã— 1000/60 Ã— 60} m/sec = 110/6 m/sec

Hence, the train will run 1100/6 m/sec in 10 seconds. â‡’ arc AB = 1100/6 m

Also since, Î¸ = Arc/Radius = 1100/6 Ã— 1500 = 11/90 radians

Hence, the angle is 11/90 radians.

### Question 17. Find the distance from the eye through which a coin of 2 cm diameter should be held so that the full moon, whose angular diameter is 31â€™ can be concealed?

Solution:

We are given Î¸ = 31â€™ and arc AB = 2 cm = 0.02 m

Since, 1â€™ = 60o â‡’ 1â€™ = (1/60)o.  Since 180o = Ï€ radians = Ï€c or 1o = (Ï€/180)c,

â‡’ Î¸ = 31â€™ = (31/60 Ã— Ï€/180)c

Also since, Î¸ = Arc/Radius â‡’ (31/60 Ã— Ï€/180) = 0.02/r â‡’ r = 2.217 m

Hence, the coin shall be placed at a distance of 2.217 m from the eye.

### Question 18. Find the diameter of the sun in kilometer supposing that it subtends an angle of 32â€™at the eye of the observer. Given that distance of the sun is 91 Ã— 106 km.

Solution:

We are given Î¸ = 31â€™ and r = 91 Ã— 106 km

Since, 1â€™ = 60o â‡’ 1â€™ = (1/60)o.  Since 180o = Ï€ radians = Ï€c or 1o = (Ï€/180)c,

â‡’ Î¸ = 32â€™ = (32/60 Ã— Ï€/180)c

Also since, Î¸ = Arc/Radius â‡’ (32/60 Ã— Ï€/180) = (AB/91 Ã— 106) km = 847407.4 km

Hence, the distance of sun is 847407.4 km

### Question 19. If the arcs of the same length in two circles subtend angles 65o and 110o at the centre, find the ratio of their radii.

Solution:

Let C1 and C2 denote the two given circles with the same arc length l.

Hence, AB = CD = l

Let Î¸1 and Î¸2 be the angles subtended, and OA = OB = r and OC = OD = R

Given, Î¸1 = 65o = (65Ï€/180)c and Î¸2 = 110o = (110Ï€/180)c

Also since, Î¸ = Arc/Radius â‡’ Î¸1 = AB/r = l/r or r= l/Î¸1     …….(1)

and, Î¸2 = CD/R = l/R or R = l/Î¸2        …..(2)

From equations (1) and (2), we get,

r/R = l/Î¸1 / l/Î¸2 = 110Ï€/180 / 65Ï€/180 = 22/13

Hence, the ratio of radii of both circles is 22:13.

### Question 20. Find the degree measure of the angle subtended at the centre of the circle of radius 100 cm by an arc of length 22 cm, using Ï€= 22/7.

Solution:

Let O denote the centre of the circle and AB denote the arc.

Hence, arc AB = 22 cm, and OA = OB = radius = 100cm

Let Î¸ be the angle subtended by the arc at the centre O by arc AB.