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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. 

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We are given ∠AOB = 25o. We know 180o = π radians = πc or 1o = (π/180)c



Hence 25o = 25 × π/180 = 5π/36 radians

Also 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’ = 60o ⇒ 1’ = (1/60)o.  Since 180o = π radians = πc or 1o = (π/180)c,

⇒ θ = 1’ = (1/60 × π/180)



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 = 360o

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

We know 180o = π radians = πc or 1o = (π/180)c

Hence, 2160o = (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 = 60o. We know 180o = π radians = πc or 1o = (π/180)c

Hence, 60o = 60 × π/180 = π/3 radians

Also 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?

Solution:

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

Since, 1’ = 60o ⇒ 1’ = (1/60)o.  Since 180o = π radians = πc or 1o = (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 × 106 km.

Solution:

We are given θ = 31’ and r = 91 × 106 km

Since, 1’ = 60o ⇒ 1’ = (1/60)o.  Since 180o = π radians = πc or 1o = (π/180)c,

⇒ θ = 32’ = (32/60 × π/180)c

Also since, θ = Arc/Radius ⇒ (32/60 × π/180) = (AB/91 × 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 θ1 and θ2 be the angles subtended, and OA = OB = r and OC = OD = R

Given, θ1 = 65o = (65π/180)c and θ2 = 110o = (110π/180)c

Also since, θ = Arc/Radius ⇒ θ1 = AB/r = l/r or r= l/θ1     …….(1)

and, θ2 = CD/R = l/R or R = l/θ2        …..(2)

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

r/R = l/θ1 / l/θ2 = 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 = 180o or 1 radian = 1c = (180/π)o

Hence, 22/100 radians = (22/100 × 180/π)o = 12.6o = 12o36′

Thus, the angle subtended by the arc at the centre of the circle is 12o36′.




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