Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.2

Question 1.
Express each of the following in radian measure.
(i) 30°
(ii) 135°
(iii) -205°
(iv) 150°
(v) 330°
Answer:
(i) 30°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 1

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

(ii) 135°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 2
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 3

(iii) – 205°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 4

(iv) 150°
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Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

(v) 330°
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Question 2.
Find the degree measure corresponding to the following radian measures.
(i) \(\frac{\pi}{3}\)
(ii) \(\frac{\pi}{9}\)
(iii) \(\frac{2 \pi}{5}\)
(iv) \(\frac{7 \pi}{3}\)
(v) \(\frac{10 \pi}{9}\)
Answer:
(i) \(\frac{\pi}{3}\) radians
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Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 8

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

(ii) \(\frac{\pi}{9}\) radians
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(iii) \(\frac{2 \pi}{5}\) radians
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(iv) \(\frac{7 \pi}{3}\) radians
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(v) \(\frac{10 \pi}{9}\) radians
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Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

Question 3.
What must be the radius of a circular running path, around which an athlete must run 5 times in order to describe 1 km?
Answer:
Let the radius of the circular path be = r m.
Length of the circular path s = 1 k. m
s = 1000 m.
Athlete runs 5 times around the path to cover 1 k. m distance
∴ θ = 360° × 5
θ = 360° × 5 × \(\frac{\pi}{180}\) radians
θ = 10 π radians
s = r θ
1000 = r 10 π
r = \(\frac{1000}{10 \pi}\)
r = \(\frac{1000 \times 7}{10 \times 22}=\frac{350}{11}\)
r = 31 .818 meters
Radius of the circular path = 31 .82 meters

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

Question 4.
In a circle of diameter 40 cm a chord is of length 20 cm. Find the length of the minor arc of the chord.
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 13

Given Diameter AB = 40 cm
∴ Radius r = 20 cm
Chord CD = 20 cm
O – Centre of the circle
OC = OD = radius = 20 cm.
∴ Triangle OCD is an equilateral triangle.
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 14
To find the length of the minor arc CD.
Let s = minor arc CD.
The arc CD subtends 60° at the centre.
θ = 60°
θ = 60° × \(\frac{\pi}{180}\) radians.
θ = \(\frac{\pi}{3}\) radians
We have s = rθ
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Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

Question 5.
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.
Answer:
Given radius r = 100 cm.
Length of arc s = 22 cm.
Angle subtended by the arc at the centre = θ radians
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Question 6.
What is the length of the arc intercepted by a central angle of measure 41° in a circle of radius of 10 feet?
Answer:
Central angle subtended by the arc θ = 41°
θ = 41 × \(\frac{\pi}{180}\) Radians
The radius of the circle r = 10 feet
Length of the arc = s
s = rθ
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Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

Question 7.
If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.
Answer:
Let r1 and r2 be the radii of the two circles and l be the length of the arc.
Given central angle θ1 = 60°
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Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

Question 8.
The perimeter of a certain sector of a circle is equal to the length of the arc of a semi-circle having the same radius. Express the angle of the sector in degrees, minutes, and seconds.
Answer:
Let OAB be the sector of a circle of radius r.
The angle of the sector is θ.
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 19
Perimeter of the sector = OA + arc AB + OB arc AB = rθ
∴ Perimeter of the sector = r + r θ + r
= 2r + rθ
= r(2 + θ) ———- (1)
Length of the arc of the semi – circle of radius
l = nπ ——– (2)
Given that perimeter the circular sector = Length of the arc of the semi circle of radius r
From equations (1) and (2), we have
r(2 + θ) = πr
2 + θ = π
θ = π – 2
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Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 21

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

Question 9.
An airplane propeller rotates 1000 times per minute. Find the number of degrees that a point on the edge of the propeller will rotate in 1 second.
Answer:
Given An airplane, the propeller rotates 1000 times per minute.
∴ A point on the edge of the propeller also rotates 1000 times in 1 minute.
∴ In 1 minute the point describes 1000 × 2π radians angle at the centre.
In 60 seconds the point describes 1000 × 2π radians angle.
∴ In 1 second the angle described = \(\frac{1000 \times 2 \pi}{60}\) radians
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Question 10.
A train is moving on a circular track of a 1500 m radius at the rate of 66 km/hr. What angle will it turn in 20 seconds?
Answer:
Radius of the circular track r = 1500 m.
Speed of the train = 66 km/hr
Let θ be the angle made by the path of train at the centre in 20 seconds.
In 1 hr distance moved by train along the circular path = 66 km
In 60 × 60 seconds distance moved = 66 km
∴ In 20 seconds distance moved s = \(\frac{66}{60 \times 60}\) × 20
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Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

Question 11.
A circular metallic plate of radius 8 cm and thickness 6 nuns is melted and molded into a pie (a sector of the circle with thickness) of radius 16 cm and thickness 4 mm. Find the angle of the sector.
Answer:
Radius of the circular metallic plate r = 8 cm
Thickness of the plate h = 6 mm = \(\frac{6}{10}\)
Radius of the Pie l = 16 cm
Thickness of the Pie ( h) = 4mm = \(\frac{4}{10}\) cm
Given volume of the cylinder = Volume of the sector
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