ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.1

Well-structured ML Aggarwal Class 11 Solutions Chapter 3 Trigonometry Ex 3.1 facilitate a deeper understanding of mathematical principles.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.1

Very short answer/objective questions (1 to 5) :

Question 1.
Draw diagrams for the following angles:
(i) – 135°
(ii) 740°.
In which quadrant do they lie ?
(iii) Find another positive angle whose initial and final sides are same as that of – 135°, and indicate on the same diagram.
Solution:
(i) Let OX be the initial side and OP be the terminal side.

Clearly – 135° lies in III rd quadrant.

(ii) Let OX be the initial side and OP be the terminal side.

Clearly 740° lies in Ist quadrant.

(iii) Let OX be the initial side and OP be the terminal side.
Clearly – 135° lies in 3rd quadrant.

Clearly another positive angle whose initial and final sides are same as that of – 135° be
360° – 135° = 225°.

Question 2.
If θ lies in second quadrant, in which quadrant the following will lie?
(i) \(\frac{\theta}{2}\)
(ii) 2 θ
(iii) – θ
Solution:
Given θ lies in second quadrant.
∴ 90° ≤ θ ≤ 180°

(i) ∴ 45° ≤ \(\frac{\theta}{2}\) ≤ 90°
∴ \(\frac{\theta}{2}\) lies in first quadrant.

(ii) ∴ 180° ≤ 2θ ≤ 360°
∴ 2 θ lies in 3rd and 4th quadrant.

(iii) – 90° ≥ – θ ≥ – 180°
⇒ 180°≤ – θ ≤ – 90°
Clearly – θ lies in 3rd quadrant.

Question 3.
Express the following angles in radian measure :
(i) 240°
(ii) – 315°
(iiî) 570°
Solution:
We know that,
π radians = 180°
⇒ 1° = \(\frac{\pi}{180}\) rad

(i) 240° = 240 × \(\frac{\pi}{180}\) rad
= \(\frac{4 \pi}{3}\) rad

(ii) – 315° = – 315° × \(\frac{\pi}{180}\) rad
= – \(\frac{7 \pi}{4}\) rad

(iii) 570° = 570 × \(\frac{\pi}{180}\) rad
= \(\frac{19 \pi}{6}\) rad

Question 4.
Express the following angles in degree measure :
(i) \(\frac{5 \pi}{3}\)
(ii) \(\frac{13 \pi}{4}\)
(iii) – \(\frac{24 \pi}{5}\)
Solution:
We know that,
π radians = 180°
⇒ 1 rad = \(\frac{180^{\circ}}{\pi}\)

(i) \(\frac{5 \pi}{3}=\frac{5 \pi}{3} \times \frac{180^{\circ}}{\pi}\)
= 300°

(ii) \(\frac{13 \pi}{4}=\frac{13 \pi}{4} \times \frac{180^{\circ}}{\pi}\)
= 585°

(iii) \(-\frac{24 \pi}{5}=-\frac{24 \pi}{5} \times \frac{180^{\circ}}{\pi}\)
= – 864°

Question 5.
Express the following angles in radian measure :
(i) 35°
(ii) 520°
(iii) 40° 20’
(iv) – 37° 30’.
Solution:
We know that,
π radians = 180°
⇒ 1° = \(\frac{\pi}{180}\) rad

35° = 35 × \(\frac{\pi}{180}\)
= \(\frac{7 \pi}{36}\) rad

(ii) 520° = 520 × \(\frac{\pi}{180}\)
= \(\frac{26 \pi}{9}\) radians

(iii) 40° 20’ = 40° + 20’
= 40° + \(\left(\frac{20}{60}\right)^{\circ}\)
= 40° + \(\left(\frac{1}{3}\right)^{\circ}\)
= \(\left(\frac{121}{3}\right)^{\circ}=\left(\frac{121}{3} \times \frac{\pi}{180}\right)\) rad
= \(\frac{121 \pi}{540}\) rad

(iv) – 37° 30′ = – (37° + 30’)
= – \(\left[37^{\circ}+\left(\frac{30}{60}\right)^{\circ}\right]\)
[∵ 1° = 60’
1’ = \(\left(\frac{1}{60}\right)^0\)]
= \(\left[37^{\circ}+\left(\frac{1}{2}\right)^{\circ}\right]\)
= – \(\left(\frac{75}{2}\right)^{\circ}\)
= \(\left[\frac{75}{2} \times \frac{\pi}{180}\right]\)
= – \(\frac{5 \pi}{24}\) rad

Short answer questions (5 to 10) :

Question 6.
Find the degree measures corresponding to the following radian measures :
(i) 6
(ii) \(\frac{3}{4}\)
(iii) – 3
Solution:
We know that,
π radians = 180°

(i) 6 rad = 6 × \(\frac{180^{\circ}}{\pi}\)

(ii) \(\frac{3}{4}\) rad
= \(\left(\frac{3}{4} \times \frac{180^{\circ}}{\pi}\right)\)

(iii) – 3 rad

Question 7.
A wheel makes 360 revolutions in a minute. Through how many radians does it turn in one second ?
Solution:
Since the angle traced out by wheel in one revolution = 2π rad
∴ angle traced out by wheel in 360 revolutions in 1 minute = (2π × 360) rad
Thus, angle traced out by wheel in 360 revolutions in 1 second = \(\left(\frac{2 \pi \times 360}{60}\right)\) rad
= 12π radians

Question 8.
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
(ii) 15 cm.
Solution:
(i) Given radius of pendulum = r = 75 cm
arc length that tip of pendulum describes = s = 10 cm
We know that θ = \(\frac{s}{r}\)
where θ radians be the angle through which the pendulum swings.
θ = \(\frac{10}{75}\) rad
= \(\frac{2}{15}\) rad

(ii) Here s = 15 cm ; r = 75 cm
∴ θ = \(\frac{s}{r}\)
= \(\frac{15}{75}=\frac{1}{5}\) rad

Question 9.
Find the radius of the circle in which a central angle of 45° makes an arc of length 187 cm. (use π = \(\frac{22}{7}\)).
Solution:
Let r cm be the radius of circle
Here, arc length = s
= 187 cm
and θ = 45°
= 45 × \(\frac{\pi}{180}\)
= \(\frac{\pi}{4}\) rad
[∵ π radians = 180°]
We know that,
θ = \(\frac{s}{r}\)
⇒ r = \(\frac{s}{\theta}\)
= \(\left(\frac{187 \times 4}{\pi}\right)\) cm
⇒ r = \(\left(\frac{187 \times 4 \times 7}{22}\right)\)
= 238 cm

Question 10.
Find the length of an arc of a circle of diameter 20 cm which subtends an angle of 45 at the centre.
Solution:
Given diameter of circle = 20 cm
∴ radius of circle = r = 10 cm
Let s be the required arc length of a circle with radius 10 cm
and θ = 45° = \(\frac{\pi}{4}\) rad
We know that,
θ = \(\frac{s}{r}\)
⇒ s = 10 × \(\frac{\pi}{4}\)
= \(\left(\frac{10 \times 22}{7 \times 4}\right)\) cm
= \(\frac{5 \pi}{2}\) cm

Long answer questions (11 to 20) :

Question 11.
An engine is travelling along a circular railway track of radius 1500 metres with a specd of 60 km/h. Find the angle in degrees turned by the engine in 10 seconds.
Solution:
Given radius of circular railway track = 1500 m
given speed of engine = 60 km / h
Thus, the distance covered by engine in 1 hour = 60 km
∴ distance covered by engine in 1 second = \(\frac{60 \times 1000}{60 \times 60}\)
Thus disiance covered by engine in 10 seconds = s
= \(\frac{60 \times 1000}{60 \times 60}\)
= \(\frac{500}{3}\) m
Let θ in radians be the angle turned by engine in 10 seconds
Then θ = \(\frac{s}{r}\)
θ = \(\frac{500}{3 \times 1500}\)
= \(\frac{5}{45}=\frac{1}{9}\) rad
[∵ π rad = 180°]

Question 12.
If the arcs of the same length in two circles subtend angles of 65° and 110° at their respective centres, find the ratio of their radii.
Solution:
Let r₁ and r₂ be the radii of two given circles
and let their arcs of same length say s and subtends an angles 65° and 110° at respective centres.
Now 65° = 65 × \(\frac{\pi}{180}=\frac{13 \pi}{36}\)
[∵ π rad = 180°]
and 110° = 110 × \(\frac{\pi}{180}=\frac{11 \pi}{18}\)
We know that
θ = \(\frac{s}{r}\)
⇒ r = \(\frac{s}{\theta}\)
i.e. r₁ = \(\frac{s}{13 \pi}\) × 36 ………………..(1)
and r₂ = \(\frac{s}{11 \pi}\) × 18 …………………(2)
On dividing (1) and (2) ; we have
\(\frac{r_1}{r_2}=\frac{36}{13} \times \frac{11}{18}\)
= \(\frac{22}{13}\)
Thus,
r₁ : r₂ = 22 : 13

Question 13.
Large hand of a clock is 21 cm long. flow much distance does its extremity move in 20 minutes ?
Solution:
The angle traced out by large hand of clock in 60 minutes = 2 π rad
∴ angle traced out by large hand of clock in 20 minutes = θ
= \(\frac{2 \pi}{60}\) × 20
= \(\frac{2 \pi}{3}\) radians
given radius of clock = length of large hand = r = 21 cm
Let s be the required length of the arc moved by the tip of minute hand
Then s = rθ
= (21 × \(\frac{2 \pi}{3}\))
= 14 π
= (14 × \(\frac{22}{7}\)) cm
= 44 cm

Question 14.
Find the angles in degrees through which a pendulum swings if Its length is 50 cm and the tip describes an arc of length :
(i) 10 cm
(ii) 16 cm
(iii) 26 cm (use π = \(\frac{22}{7}\))
Solution:
(i) The pendulum describes a circle of radius 50 cm and its tip describes an arc of length 10 cm.
Let θ radians be the angle through which the pendulum swings.
Here r = 50 cm ;
s = 10 cm
Then θ = \(\frac{s}{r}\)
= \(\frac{10}{50}=\frac{1}{5}\) rad
= \(\frac{1}{5} \times \frac{180^{\circ}}{\pi}\)
= \(\left(\frac{1}{5} \times \frac{180}{22} \times 7\right)^{\circ}\)
= \(\left(\frac{126}{11}\right)^{\circ}=\left(11 \frac{5}{11}\right)^{\circ}\)
= 11° + \(\left(\frac{5}{11} \times 60\right)^{\prime}\)
= 11° + \(\left(27 \frac{3}{11}\right)^{\prime}\)
= 11° + 27’ + \(\left(\frac{3}{11} \times 60\right)^{\prime \prime}\)
= 11° + 27’ + 16″
= 11° 27’ 16″

(ii) The pendulum describes a circle of radius 50 cm and its tip describes an arc of length 16 cm.
Let θ (in rad) be the angle through which the pendulum swings.
Here r = 50 cm ;
s = 16 cm

(iii) The pendulum describes a circle of radius 50 cm and its tip describes an arc of length 26 cm.
Let θ (in radians) be the angle through which the pendulum swings.
Here r = 50 cm ;
s = 26 cm
Then θ = \(\frac{s}{r}\)
= \(\frac{26}{50}=\frac{13}{25}\) rad
= \(\left(\frac{13}{25} \times \frac{180}{22} \times 7\right)^{\circ}\)
θ = 29° + \(\left(\frac{1638}{55}\right)^{\circ}=\left(29 \frac{43}{55}\right)^{\circ}\)
= 29° + \(\left(\frac{43}{55} \times 60\right)^{\prime}\)
= 29° + \(\left(46 \frac{10}{11}\right)^{\prime}\)
= 29° + 46′ + \(\left(\frac{10}{11} \times 60\right)^{\prime \prime}\)
= 29° 46′ 55″

Question 15.
Find the length of an arc of a circle of radius 75 cm that spans a central angle of measure 126°. Take π = 3.1416.
Solution:
Let s be the length of arc of circle of radius 75 cm that subtends an angle of 126° at the centre.
Here, r = 75 cm ;
θ = 126°
= (126 × \(\frac{\pi}{180}\))
= \(\frac{126 \times 3.1416}{180}\)
[∵ π radians = 180°]
∴ θ = 2.19912 rad.
Then θ = \(\frac{s}{r}\)
⇒ s = rθ
= (75 × 2.19912) cm
= 164.934 cm

Question 16.
Find the angle ¡n radians between the hands of a clock at 3.30 A.M.
Solution:
Here, Given time be 3.30 A.M.
Hour hand and minute hand were together at 12 night.

Time of rotation for hour hand = 3 hour 30 minute
= (3 + \(\frac{30}{60}\)) hr
= \(\frac{7}{2}\) hr
angle trace out by hour hand in 12 hours = 360°
∴ angle trace out by hour hand in \(\frac{7}{2}\) hr
θ = \(\left(\frac{360}{12} \times \frac{7}{2}\right)^0\) = 105°
at 3.00 A.M., i.e. after 3 hours, the minute hand is at 12.
thus the angle trace out by minute hand at 3.00 A.M. be 0.
Now angle trace out by minute hand in 60 minutes = 360°
∴ angle trace out by minute hand in 30 minutes Φ = \(\left(\frac{360}{60} \times 30\right)^{\circ}\) = 180°
Thus, required angle between the hands of clock at 3.30 A.M. = Φ – θ
= 180° – 105° = 75°
= 75 × \(\frac{\pi}{180}\) rad
= \(\frac{5 \pi}{12}\) rad

Question 17.
The circular measures of two angles of a triangle are \(\frac{1}{2}\) and \(\frac{1}{3}\). Find the third angle in degree measure. Take π = \(\frac{22}{7}\).
Solution:
Since the sum of all angles of triangle = 180° or π rad
or Now \(\frac{1}{2}\) rad = \(\left(\frac{1}{2} \times \frac{180}{\pi}\right)^{\circ}\)
[∵ π rad = 180°]
∴ \(\frac{1}{2}\) rad = \(\left(\frac{90 \times 7}{22}\right)^{\circ}\)
= \(\left(\frac{315}{11}\right)^{\circ}\)

Question 18.
The difference between two acute angles of a right angled triangle is \(\frac{\pi}{5}\) in radian measure. Find these angles in degrees.
Solution:
Let the other two required angles in radians be θ and Φ.
Since the sum of angles of triangle be π rad.
∴ θ + Φ + 90° = 180°
θ + Φ = 90°
= \(\frac{\pi}{2}\) rad
also given,
0 – Φ = \(\frac{\pi}{5}\) rad
On adding (1) and (2) ; we have

Question 19.
The angles of a triangle are in A.P. and the greatest angle ¡s double the least. Find all the angles in circular measure.
Solution:
Let the required angles of triangle in A.P. be (a – d)°, a° and (a + d°).
Then, a – d + a + a + d = 180
⇒ a = 60
Now least angle = (a – d)° = (60 – d)°
and greatest anale = (a + d)° = (60 + d)°
according to question ; we have
(a + d)° = 2 (a – d)°
⇒ a + d = 2a – 2d
⇒ a = 3d
⇒ 3d = 60
⇒ d = 20
Thus the required angle are (60 – 20)°, 60°, (60 + 20)° i.e. 40°, 60°, 80°
since π rad = 180°
⇒ 1° = \(\frac{\pi}{180}\) rad
i.e. angles in radians are ;
(40 × \(\frac{\pi}{180}\)) ; 60 × \(\frac{\pi}{180}\), 8o × \(\frac{\pi}{180}\)
i.e. \(\frac{2 \pi}{9}, \frac{\pi}{3}, \frac{4 \pi}{9}\)

Question 20.
Estimate the diameter of the sun supposing that it subtends an angle of 32′ at the eye of an observer. Given that the distance of the sun is 91 × 10⁶ km.
Take π = \(\frac{22}{7}\).
Solution:
Let r km be the radius of Sun that subtends an angle of 32′ at the eye of an observer.
Given distance of sum from observer r = 91 × 10⁶ km
Here θ = 32’
= \(\left(\frac{32}{60}\right)^{\circ}=\left(\frac{8}{15}\right)^{\circ}\)
[∵ 1° = 60 rad
⇒ 1′ = \(\left(\frac{1}{60}\right)^{\circ}\)]
⇒ θ = \(\left(\frac{8}{15} \times \frac{\pi}{180}\right)\) rad
= \(\frac{2 \pi}{675}\) rad
[∵ π rad = 180°]
We know that,
θ = \(\frac{s}{r}\)
⇒ s = rθ
= (91 × 10⁶ × \(\frac{2 \pi}{675}\)) km
⇒ s = 847407.4 km

Hence the required diameter of the sun = arc length AB = 847407.4 km.