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

**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 25**^{o }in a distance of 40 meters.

^{o }in a distance of 40 meters.

**Solution:**

Let AB denote the given rail-road.

We are given ∠AOB = 25

^{o}. We know 180^{o}= π radians = π^{c}or 1^{o}= (π/180)^{c}Hence 25

^{o}= 25 × π/180 = 5π/36 radiansAlso since, θ = Arc/Radius ⇒ ∠AOB = AB/OA ⇒ (5π/36)

^{c }= 40/r ⇒ r = 288/π = 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 1’ at the eye.**

**Solution:**

Let θ = 1’ and the length of the arc that subtends θ be l.

Radius = OA = OB = 5280m

We know, 1’ = 60

^{o}⇒ 1’ = (1/60)^{o}.^{ }Since 180^{o}= π radians = π^{c}or 1^{o}= (π/180)^{c},⇒ θ = 1’ = (1/60 × π/180)

^{c }Also since, θ = Arc/Radius ⇒ (1/60 × π/180)

^{c }= l/5280 ⇒ 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 = 360

^{o}Thus, angle made by the wheel in 6 revolutions = Angle made in 1 second = 360 × 6 = 2160

^{o}We know 180

^{o}= π radians = π^{c}or 1^{o}= (π/180)^{c}Hence, 2160

^{o}= (2160π/180)^{c}= 12π radians

Thus, the wheel turns 12π radians in 1 second.

**Question 14. Find the angle in radians through which a pendulum swings if its length is 75 cm and the tip describes an arc of length:**

**(i) 10 cm**

**Solution:**

Let OA be the length of the pendulum. ⇒ OA = 75 cm = 0.75 m

Let the arc be denoted by AB. ⇒ AB = 10 cm = 0.1 m

Also since, θ = Arc/Radius = 0.1/0.75 = 2/15 radians

Hence, the angle is 2/15 radians.

**(ii) 15 cm**

**Solution:**

Let OA be the length of the pendulum. ⇒ OA = 75 cm = 0.75 m

Let the arc be denoted by AB. ⇒ AB = 15 cm = 0.15 m

Also since, θ = Arc/Radius = 0.15/0.75 = 1/5 radians

Hence, the angle is 1/5 radians.

**(iii) 21 cm**

**Solution:**

Let OA be the length of the pendulum. ⇒ OA = 75 cm = 0.75 m

Let the arc be denoted by AB. ⇒ AB = 15 cm = 0.21 m

Also since, θ = Arc/Radius = 0.21/0.75 = 7/25 radians

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.

∠AOB = 60

^{o}. We know 180^{o}= π radians = π^{c}or 1^{o}= (π/180)^{c}Hence, 60

^{o}= 60 × π/180 = π/3 radiansAlso since, θ = Arc/Radius ⇒ (0.3π/3) = 0.1 0.1π m = 10π cm

Hence, the length of the arc is 10π 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 θ denote the angle the train turns in 10 seconds.

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

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

Also since, θ = Arc/Radius = 1100/6 × 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 31**^{’} can be concealed?

^{’}can be concealed?

**Solution:**

We are given θ = 31’ and arc AB = 2 cm = 0.02 m

Since, 1’ = 60

^{o}⇒ 1’ = (1/60)^{o}. Since 180^{o}= π radians = π^{c}or 1^{o}= (π/180)^{c},⇒ θ = 31’ = (31/60 × π/180)

^{c}Also since, θ = Arc/Radius ⇒ (31/60 × π/180) = 0.02/r ⇒ 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 32’** **at the eye of the observer. Given that distance of the sun is 91 × 10**^{6} km.

^{6}km.

**Solution:**

We are given θ = 31’ and r = 91 × 10

^{6 }kmSince, 1’ = 60

^{o}⇒ 1’ = (1/60)^{o}. Since 180^{o}= π radians = π^{c}or 1^{o}= (π/180)^{c},⇒ θ = 32’ = (32/60 × π/180)

^{c}Also since, θ = Arc/Radius ⇒ (32/60 × π/180) = (AB/91 × 10

^{6}) 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 65**^{o} and 110^{o} at the centre, find the ratio of their radii.

^{o}and 110

^{o}at the centre, find the ratio of their radii.

**Solution:**

Let C

_{1 }and C_{2}denote the two given circles with the same arc length l.Hence, AB = CD = l

Let θ

_{1}and θ_{2}be the angles subtended, and OA = OB = r and OC = OD = RGiven, θ

_{1}= 65^{o}= (65π/180)^{c}and θ_{2}= 110^{o}= (110π/180)^{c}Also since, θ = Arc/Radius ⇒ θ

_{1}= AB/r = l/r or r= l/θ_{1}…….(1)and, θ

_{2}= CD/R = l/R or R = l/θ_{2}…..(2)From equations (1) and (2), we get,

r/R = l/θ

_{1}/ l/θ_{2 }= 110π/180 / 65π/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 **π**= 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 θ be the angle subtended by the arc at the centre O by arc AB.

We know, θ = Arc/Radius = 22/100 radians

Since π radians = 180

^{o}or 1 radian = 1^{c}= (180/π)^{o}Hence, 22/100 radians = (22/100 × 180/π)

^{o }= 12.6^{o}= 12^{o}36′

Thus, the angle subtended by the arc at the centre of the circle is 12^{o}36′.

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