Revision and Self Assessment on Angles and Arc Lengths Class 11 OP Malhotra Exe-3B ISC Maths Solutions Ch-3. In this article you would learn to solve extra problems on Angles and Arc Lengths. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.

Angles and Arc Lengths Class 11 OP Malhotra Revision and Self Assessment ISC Maths Solutions Ch-3

Board	ISC
Publications	S Chand
Subject	Maths
Class	11th
Chapter-3	Angles and Arc Lengths
Writer	OP Malhotra
Exe-3(B)	For Revision and Self Assessment.

Revision and Self Assessment on Angles and Arc Lengths

OP Malhotra ISC Class 11 Maths Solutions

Que-1: Convert 40° 20′ into radian measure.

(a) 121π/540 rad
(b) 121π/3 rad
(c) 121π/180 rad
(d) None of these

Sol: (a) 121π/540 rad
Step 1: Convert minutes into degrees
20′ = 20/60 = 1/3°
Step 2: Total angle
40° 20′ = 40 + 1/3 = 121/3°
Step 3: Convert degree to radian
Radian = Degree × (π / 180)
⇒ (121/3) × (π/180)
= 121π / 540

Que-2: Find the degree measure corresponding to the radian measure 7π/6.

(a) 200°
(b) 210°
(c) 180°
(d) None of these

Sol: (b) 210°
We know:   π radians = 180°
7π/6 = 7 × (180° / 6)
= 7 × 30° = 210°

Que-3: Find the perimeter of the shape drawn here (π = 22/7).

(a) 280 cm
(b) 253 cm
(c) 340 cm
(d) 305 cm

Sol: (d) 305 cm
Perimeter of the shape = (Straight sides) + (Curved parts)
Sum of straight lengths = 176 cm
Sum of curved lengths = 104 cm
Total perimeter = 176 + 104 = 280 cm

Que-4: The radius of the circle whose arc of length 15 cm makes an angle of 3/4 radian at the centre is

(a) 10 cm
(b) 11 1/4 cm
(c) 20 cm
(d) 22 1/2 cm

Sol: (c) 20 cm
Formula: Arc Length (l) = r × θ
Given: l = 15 cm, θ = 3/4 radian
15 = r × (3/4)
r = 15 × 4 / 3
r = 20 cm

Que-5: The acute angle in radians between the minute and the hour hand of a clock when the time is 4:20 is

(a) π/18
(b) π/9
(c) π/6
(d) π/3

Sol: (a) π/18
Minute hand angle = 6 × 20 = 120°
Hour hand angle = 30 × 4 + 0.5 × 20 = 120 + 10 = 130°
Angle between them = |130 − 120| = 10°
Convert into radians:
10° = (10 × π) / 180
= π/18

Que-6: Find the angle in radians through which a pendulum swings if its length is 75 cm and the tip describes an arc of length 21 cm.

(a) 25/7
(b) 25π/7
(c) 7/25
(d) None of these

Sol:  (c) 7/25
We know that,
Arc length (s) = r × θ
Given:
s = 21 cm
r = 75 cm
θ = s / r = 21 / 75
= 7 / 25 radians

Que-7: If the arcs of the same lengths in two circles subtend angles 60° and 75° at the centre, find the ratio of their radii.

(a) 5 : 4
(b) 4 : 5
(c) 28 : 35
(d) None of these

Sol: (a) 5 : 4
Arc length formula:   l = rθ
Since arc lengths are equal:
r₁θ₁ = r₂θ₂
⇒ r₁ / r₂ = θ₂ / θ₁
⇒ r₁ / r₂ = 75° / 60° = 5 / 4

Que-8: A horse is tied to a post by a rope. If it moves along a circular path, keeping the rope tight and describes 132 m when it has traced out an angle of 108° at the centre, then the length of the rope is

(a) 54 m
(b) 66 m
(c) 70 m
(d) 81 m

Sol: (c) 70 m
Arc length = (θ/360) × 2πr
132 = (108/360) × 2πr
132 = (3/10) × 2πr
132 = (6πr)/10
132 = (3πr)/5
r = (132 × 5) / (3π)
r = 660 / (3π)
Using π = 22/7
r = 660 ÷ (66/7)
= 660 × 7 / 66
= 70 m

Que-9: A circular wire of radius 15 cm is cut and bent so as to lie along the circumference of a loop of radius 120 cm. The angle subtended by it at the centre is

(a) 30°
(b) 45°
(c) 60°
(d) None of these

Sol: (b) 45°
Length of small wire = circumference of circle of radius 15 cm
= 2πr = 2π × 15 = 30π cm
This length becomes arc of bigger circle (radius = 120 cm)
Angle (in radians) = Arc length / radius
= 30π / 120 = π/4
Convert into degrees:
π/4 = 180°/4 = 45°

Que-10: A train is travelling along a curve of radius 700 m at 21 km/hr. The angle in radians through which the train turns in 2 minutes is

(a) 2 rad
(b) 1 rad
(c) 1/2 rad
(d) None of these

Sol: (b) 1 rad
Speed = 21 km/hr = (21 × 1000) / 3600
= 35/6 m/s
Time = 2 minutes = 120 seconds
Distance travelled = speed × time
= (35/6) × 120 = 700 m
Radius (r) = 700 m
Angle in radians (θ) = distance / radius
θ = 700 / 700 = 1 radian

Que-11: The perimeter of a certain sector of a circle is equal to the length of the arc of the semicircle, then the angle at the centre of the sector in radians is

(a) π / 3
(b) 2π / 3
(c) π − 2
(d) π + 2

Sol: (c) π − 2
Let radius = r and angle of sector = θ radians
Perimeter of sector = 2r + rθ
Arc length of semicircle = πr
Given:
2r + rθ = πr
Divide by r:
2 + θ = π
θ = π − 2

–: End Angles and Arc Lengths Class 11 OP Malhotra Exe-3B ISC Math Ch-3 Solutions :–

Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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