Regular engagement with OP Malhotra Class 11 Solutions Chapter 3 Angles and Arc Lengths Ex 3 can boost students confidence in the subject.
S Chand Class 11 ICSE Maths Solutions Chapter 3 Angles and Arc Lengths Ex 3
(Take π = \(\frac { 22 }{ 7 }\))
Question 1.
Express the following angles in degrees :
\(\frac { 1 }{ 2 }\)
Solution:
We know that π rad = 180°
⇒ 1 rad = \(\frac { 180° }{ π }\)
∴ \(\frac{\pi}{6} \mathrm{rad}=\frac{180^{\circ}}{\pi} \times \frac{\pi}{6}\) = 30°
\(\frac{14 \pi}{15} \mathrm{rad}=\frac{180^{\circ}}{\pi} \times \frac{14 \pi}{15}\) = 168°
\(\frac{11 \pi}{18} \mathrm{rad}=\frac{180^{\circ}}{\pi} \times \frac{11 \pi}{18}\) = 110°
and \(\frac{7 \pi}{90} \mathrm{rad}=\frac{180^{\circ}}{\pi} \times \frac{7 \pi}{90}\) = 14°
Question 2.
Express the following angles in radians
(i) 1′
(ii) 20°
(iii) 135°.
Solution:
We know that n radian = 180°
Question 3.
Express in radians and also in degrees the angle of a regular polygon of (i) 40 sides, (ii) n sides.
Solution:
We know that exterior angle = \(\frac{360^{\circ}}{\text { no. of sides }}\)
(i) Here n = 40
∴ exterior angle = \(\frac { 360° }{ 40 }\) = 9° = \(\frac { π }{ 20 }\) rad [∵ π rad = 180°]
∴ interior angle of polygon = 180° – 9° = 171°
(In radians) interior angle of polygon
= π – \(\frac { π }{ 20 }\) = \(\frac { 19π }{ 20 }\)
(ii) each exterior angle = \(\frac { 360° }{ n }\) or \(\frac { 2π }{ n }\)
∴ each interior angle of polygon with sides
n = 180° – \(\frac { 360° }{ n }\) = \(\frac{(n-2)}{n} 180^{\circ}\)
or \(\pi-\frac{2 \pi}{n}=\left(\frac{n-2}{n}\right) \pi\) radians
Question 4.
The perimeter of a certain sector of a circle is equal to the length of the arc of the semi-circle having the same radius, express the angle of the sector in degrees, minutes and seconds.
Solution:
Given perimeter of sector of circle with radius r = 2r + s
Also, length of arc of semi-circle with radius r = πr
according to given condition, 2r + s = πr
⇒ 2r + rθ = πr [∵ θ = \(\frac { s }{ r }\) ⇒ s = rθ]
⇒ 2 + θ = π ⇒ θ = π – 2
⇒ θ = \(\frac{22}{7}-2=\frac{22-14}{7}=\frac{8}{7}\) [since π rad = 180°]
⇒ θ = \(\frac{8}{7} \times \frac{180^{\circ}}{\pi}=\frac{8}{7} \times \frac{180^{\circ}}{22} \times 7\)
= \(\frac { 720° }{ 11 }\) = 65°27’16”
Question 5.
The length of a pendulum is 8 m while the pendulum swings through 1.5 rad, find the length of the arc through which the tip of the pendulum passes.
Solution:
Given length of pendulum = r = 8m
Also, angle through pendulum swings θ = 1.5 rad
∴ required length of arc through which tip of the pendulum passes = s = rθ
= (8 x 1.5) m = 12m
Question 6.
The minute hand of a clock is 15 cm long. How far does the tip of the hand move during 40 minutes ? (Take π = 3.14)
Solution:
given length of minute hand of clock = r = 15 cm
since angle moved by minute hand in 60′ = 360°
∴ angle moved by minute hand in 40′ = \(\frac{360^{\circ}}{60}\) x 40 = 240°
∴ θ = 240°
We know that n radians = 180°
Thus the tip of the minute hand moving during 40 minutes be 62.85 cm.
Question 7.
A central angle of a circle of radius 50 cm intercepts an arc of 10 cm. Express the central angle 0 in radians and in degrees.
Solution:
Given radius of circle = r = 60 cm
length of arc = l = 10 cm
∴ required central angle of circle
θ = \(\frac{l}{r}=\frac{10}{50}=\frac{1}{5}\) rad
We know that, π radians = 180°
∴ θ = \(\frac{1}{5} \mathrm{rad}=\frac{180^{\circ}}{\pi} \times \frac{1}{5}\)
= \(\frac{180^{\circ}}{22} \times \frac{7}{5}=\frac{126^{\circ}}{11}=11^{\circ} 27^{\prime} 11^{\prime \prime}\)
Question 8.
The moon’s distance from the earth is 360000 km and its diameter subtends an angle of 31′ at the eye of the observer. Find the diameter of the moon.
Solution:
Here θ = 31′ = (\(\frac { 31 }{ 60 }\))° [∵ 1° = 60′ ⇒ 1′ = (\(\frac { 1 }{ 60 }\))°]
Also π rad = 180° ⇒ 1° = \(\frac { π }{ 180 }\) rad
∴ θ = (\(\frac { 31 }{ 60 }\) x \(\frac { π }{ 180 }\)) rad
Since the distance between the moon and earth is quite large so we consider diameter AB as arc AB
We know that, θ = \(\frac { l }{ r }\)
⇒ \(\frac{31}{60} \times \frac{\pi}{180}=\frac{\text { diameter }}{360000}\)
diameter of moon = \(\frac{31}{60} \times \frac{\pi}{180}\) x 360000
= \(\frac{31}{60} \times \frac{22}{7 \times 180}\) x 360000
= \(\frac { 68200 }{ 21 }\) km
Question 9.
A railway train is travelling on a curve of 750 m radius at the rate of 30 km/h, through what angle has it turned in 10 seconds ?
Solution:
Given radius of curve = r = 750 m
given distance travelled by train in 1 hour = 30 km = (30 x 1000) m
∴ distance covered by train in 1 second
= \(\frac { 30000 }{ 60×60 }\)
Thus distance covered by train in 10 seconds
= l = (\(\frac { 30000 }{ 3600 }\) x 10)
Therefore required angle has it turned in 10 seconds = θ = \(\frac{l}{r}=\frac{30000 \times 10}{3600 \times 750}=\frac{1}{9}\) rad.
Question 10.
A horse is tethered to a stake by a rope 810 cm long. If the horse moves along the circumferences of a circle always keeping the rope tight, find how far it will have gone when the rope has traced out an angle of 70° ?
Solution:
Let O denote the position of point and OP denote the length of rope in tight position.
Question 11.
The area of a sector of 6.024 cm² and its angle is 36°. Find the radius, (π = 3.14).
Solution:
Let O be the centre of circle with radius r
Then area of sector AOB with central angle θ
Hence, the required radius of sector be 4 cm.
Question 12.
Find the area of sector of a circle, radius 5 m bounded by an arc of length 8 in.
Solution:
Given radius of sector of circle = r = 5 m
given arc of length = l = 8 m
∴ Central angle θ = \(\frac { l }{ r }\) = \(\frac { 8 }{ 5 }\) rad
Thus, required area of sector of circle
= \(\frac{1}{2} r^2 \theta=\frac{1}{2} \times 5^2 \times \frac{8}{5}\) = 20 m²
Question 13.
The diagram shows a windscreen wiper cleaning a car windscreen.
(i) What is the length of the arc swept out?
(ii) What area of the windscreen is not cleaned?
Solution:
(i) Given θ = \(\frac { 8π }{ 9 }\)
∴ required length of arc swept out = l = rθ
= 45 x \(\frac { 8π }{ 9 }\) = 40π = \(\frac{40 \times 22}{7}\) cm
(ii) area of car windscreen = (55 x 100) cm²
= 5500 cm²
∴ area of windscreen is not cleaned = area of car windscreen – area of windscreen cleaned
= 5500 – \(\frac { 1 }{ 2 }\)r² θ
= (5500 – \(\frac { 1 }{ 2 }\) x 45 x 45 x \(\frac { 8 }{ 9 }\) x \(\frac { 22 }{ 7 }\)) cm²
= (5500 – 2828.57) cm² = 2671.42 cm²
Question 14.
Find the area of the shaded segment in given below figure.
Solution:
Question 15.
What is the ratio of the areas of the major sector in diagram A to the minor sector in a diagram B?
Solution:
