The availability of OP Malhotra Class 11 Solutions Chapter 3 Angles and Arc Lengths Chapter Test encourages students to tackle difficult exercises.
S Chand Class 11 ICSE Maths Solutions Chapter 3 Angles and Arc Lengths Chapter Test
Question 1.
Find the radian measure of
(i) 25°
(ii) 240°.
Solution:
We know that n radians = 180°
⇒ 1° = \(\frac { π }{ 180 }\) rad
(i) ∴ 25° = 25 x \(\frac { π }{ 180 }\) rad
(ii) 240° = 240 x \(\frac { π }{ 180 }\) = \(\frac { 4π }{ 3 }\)
Question 2.
Find the degree measure of (i) \(\frac { 5π }{ 3 }\) (ii) – 4.
Solution:
We know that, π rad = 180°
⇒ 1 rad = \(\frac { 180° }{ π }\)
(i) ∴ \(\frac { 5π }{ 3 }\) = \(\frac { 5π }{ 3 }\) x \(\frac { 180° }{ π }\) = 300°
(ii) – 4 = \(\frac { 180 }{ π }\) x (- 4) = \(\frac { -720 }{ 22 }\) x 7
= \(\frac{-360 \times 7}{11}\) = – 229°5’27”
Question 3.
If an angle measures D degrees or C radius, show that \(\frac{D}{90}=\frac{2 C}{\pi}\).
Solution:
We know that 180° = π rad
⇒ 180 D = πC ⇒ \(\frac{2 \mathrm{D}}{\pi}=\frac{\mathrm{C}}{90}\)
Question 4.
One angle of a triangle in 54° and another angle is \(\frac { π }{ 4 }\) radians. Find the third angle is centesimal unit.
Solution:
Since π radians = 180° ⇒ 1 rad = \(\frac { 180° }{ π }\)
⇒ \(\frac{\pi}{4} \mathrm{rad}=\frac{\pi}{4} \times \frac{180^{\circ}}{\pi}\) = 45°
given angles of triangle are 54° and 45°. since the sum of angles of triangle is 180°.
∴ 54° + 45° + θ = 180°
⇒ θ = 180° – 99° = 81°
Hence the required third angle of triangle be 81°.
Question 5.
Express in circular measure and also in degrees the angle of a regular octagon.
Solution:
Given no. of sides of regular octagon = n = 8
We know that, each interior angle of regular polygon = (\(\frac { n-2 }{ n }\)) x 180°
∴ each interior angle of regular octagon = (\(\frac { 8-2 }{ 8 }\)) x 180°
= \(\frac { 6 }{ 8 }\) x 180° = 135° [∵ π rad = 180°]
Thus each interior angle in radians
= 135 x \(\frac { π }{ 180 }\) = \(\frac { 3π }{ 4 }\)
Question 6.
If in two circles, arcs of the same length subtend angles of 60° and 75° at the centre, find the ratio of their radii.
Solution:
Let l be the arc length of both circles and let r1, r2 be the radii of two circles with central angles 60° and 75°
Here θ1 = 60° = \(\frac { π }{ 3 }\)
and θ1 = 75° = \(\frac { 75π }{ 180 }\) = \(\frac { 5π }{ 12 }\)
We know that, θ = \(\frac { 1 }{ r }\)
∴ θ1 = \(\frac{l}{r_1}\) ⇒ l = r1θ1
θ2 = \(\frac{l}{r_2}\) ⇒ l = r2θ2
⇒ r1θ1 = r2θ2
⇒ \(\frac{r_1}{r_2}=\frac{\theta_2}{\theta_2}=\frac{\frac{5 \pi}{12}}{\frac{\pi}{3}}=\frac{5 \pi}{12} \times \frac{3}{\pi}=\frac{5}{4}\)
Hence the required ratio of their radii
= r1 : r2 = 5 : 4
Question 7.
In a circle of diameter 60 cm the length of a chord is 30 cm. Find the length of the minor and major arcs of the chord.
Solution:
Since ∆OAB be an equilateral triangle
∴ θ = 60° = \(\frac { π }{ 3 }\)
∴ Minor arc = \(\widehat{\mathrm{AB}}\) = rθ = 30 x \(\frac { π }{ 3 }\) = 10 π cm
and Major arc = circumference of circle – minor arc
= (2πr – 10π) cm
= (2.71 x 30 – 10π) cm
= 50π cm
Question 8.
Find the angle in radian through which a pendulum swings and its length is 75 cm and the tip describes an arc of length 21 cm.
Solution:
Given length of pendulum = r = 75 cm
and length of arc described by tip of pendulum = l = 21 cm
We know that θ = \(\frac{l}{r}=\frac{21}{75}=\frac{7}{25}\) rad
Hence the required angle through which a pendulum swings be \(\frac { 7 }{ 25 }\) radians.
Question 9.
Find the area of the sector of a circle of radius 14 cm having central angle \(\frac { π }{ 4 }\).
(Take π = \(\frac { 22 }{ 7 }\)).
Solution:
Given radius of circle = r = 14 cm
∴ area of sector of circle with central angle
Question 10.
A horse is tied to a post by a rope. If the horse moves along a circular path always keeping the rope tight and describes 88 metres when it has traced out 72° at the centre, find the length of the rope.
Solution:
Let OP be the length of the rope in tight position = r m
Then θ = ∠POQ = 12° and arc PQ = 88 m
Hence the required length of the rope be 70 m.