Chapter 3 of the Class 11 NCERT Mathematics textbook, "Trigonometric Functions," explores the fundamental aspects of trigonometric functions, including their definitions, properties, and various applications. Exercise 3.1 focuses on solving basic problems related to these functions to build a solid foundation in trigonometry.
Chapter 3 Trigonometric Function - Exercise 3.1
This section provides detailed solutions for Exercise 3.1 from Chapter 3 of the Class 11 NCERT Mathematics textbook. The exercise includes problems designed to test and reinforce students' understanding of trigonometric functions, including their definitions, ranges, and graphical representations. Solutions are presented step-by-step to help students master the concepts effectively.
Class 11 NCERT Mathematics Solutions - Exercise 3.1
Question 1. Find the radian measures corresponding to the following degree measures:
(i) 25° (ii) -47°30′ (iii) 240° (iv) 520°
(i) As we know that
180° = π radian
So, 1° = π/180° radian
Then, 25° = (π/180°) × 25°
= 5π/36 radians
Hence, 25° equals to 5π/36 radians.
(ii) As we know that
180° = π radian
So, 1° = π/180°
And 60' = 1°
30' = (1/2)°
So, -47°30' = -47 (1/2)°
-47(1/2)° = (π/180) × (-95/2) = (-19π/72) radian.
Hence, -47°30' is equals to -19π/72 radian.
(iii) As we know that
180° = π radian
1° = π/180° radian
So 240° = (π/180°) × 240°
= 4π/3 radians
Hence, 240° equals to 4π/3 radians.
(iv) As we know that
180° = π radian
1° = π/180° radian
So 520° = (π/180°) × 520°
= 26 π/9 radians
Hence, 520° equals to 26 π/9 radians.
Question 2. Find the degree measures corresponding to the following radian measures(Use π = 22/7)
(i)11/16 (ii) -4 (iii) 5π/3 (iv) 7π/6
(i) 11/16 radian = (11/16) (180°/π) {as 180° = π radian, then 1 radian = 180°/π}
= (11/16) × (180° × 7/22)
= (11 × 180° × 7/16 × 22)
= 315/8°
= 39 (3/8)°
= 39(3/8)°
= 39° + (3/8)°
Again 1° = 60'
So (3/8)° = 60' × (3/8)
= 22 (1/2)'
= 22 (1/2)'
= 22' + 1/2'
Again 1' = 60"
= (1/2)' = 30"
So 39 (3/8)° = 39° 22' 30"
Hence, 11/16 radian results to 39° 22' 30".
(ii) -4 radian = -4 × (180°/π) {as 180° = π radian, then 1 radian = 180°/π}.
= -4 ×180° × 7/22
= -229° (1/11)
= -229 (1/11)°= -229° + (1/11)°
Again(1/11)° = (1/11) × 60'. {as 1° = 60'}
= 5(5/11)'
Also, 5 (5/11)' = 5' + (5/11)'
(5/11)' = (5/11) × 60" {as 1' = 60"}
= 27"
So, -229(1/11) = -229° 5'27"
Hence, -4 radian results to -229° 5' 27".
(iii) 5π/3 radian = (5 π/3) × (180/π) {as 180° = π radian, then 1 radian =180°/π}.
= (5 × 180/3)°
= 300°
Hence, 5π/3 results to 300°.
(iv) 7π/6 radian = (7π/6) × (180°/π) {as 180° = π radian, then 1 radian =180°/π}.
= (7 × 180/6)°
= 210°
Hence, 7π/6 radian results to 210°.
Question 3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Solution:
Given that
Total revolutions made by the wheel in one minute is 360.
1 second = 360/6 = 60
We know that
When a wheel revolves once it covers 2π radian of distance.
In one minute, it will turn an angle of 360 × 2π radian = 720 π radian
In one second, it will turn an angle of 720 π radian/60 = 12 π radian {as 1 minute = 60 seconds}
Hence, in one second, the wheel turns an angle of 12π radian.
Question 4. Find the degree measure of the angle subtended at the Centre of a circle of radius 100 cm by an arc of length 22 cm (Use π = 22/7)
Solution:
Given that
The radius of circle (r) = 100 cm.
Length of the arc (l) = 22 cm.
Let us consider the angle subtended by the arc is θ.
Also, we know that θ = l/r
The angle subtended (θ) = 22/100 radian
For finding the degree measure we have to multiply 180°/π with radian measure
So, θ = (22/100) × (180/π)
θ = (22/100) × (180 × 7/22)
θ = (22 × 180 × 7/22 × 100)
θ = 126/10 degree
θ = 12 (3/5) degree
We know that 1° = 60'
(3/5)° = 60' × (3/5)
= 36'
So 12 (3/5)° = 12° 36'
Hence, the degree measure of the angle subtended at the Centre of a circle is 12° 36'
Question 5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.
Solution:
Given that
Diameter of circle (d) = 40 cm
Radius (r) = d/2 = 40/2 = 20 cm
Let us consider AB as the chord of circle having length 20 cm, and Centre at O.
It forms a triangle OAB,
Having Radius = OA = OB = 20 cm
Also, chord AB = 20 cm
Hence, In ΔOAB OA = OB = AB. (equilateral triangle.)
So angle subtend = (π/3) radian
We know that θ = l/r (where θ = angle subtended by the arc
l = length of arc
r = radius)
Putting values of r and θ we get
π/3 = l/20
So.
l = 20 π/3
Hence, length of the arc is 20π/3 cm.
Question 6. If in two circles, arcs of the same length subtend angles 60° and 75° at the Centre, find the ratio of their radii.
Solution:
Given that
Angle subtend by 1st arc (θ1) = 60
Angle subtend by 2nd arc (θ2) = 75
We know that θ = l/r
For 1st arc θ1 = l1/r1
For 2nd arc θ2 = l2/r2
θ1/θ2 = (l1/r1)/(l2/r2)
θ1/θ2 = (l/r1)/(l/r2) {here l1 = l2 = l}
θ1/θ2 = r2/r1
60/75 = r2/r1
r2/r1 = 4/5
r1/r2 = 5/4
Hence, ratio of their radius is 5:4.
Question 7. Find the angle in radian 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 (iii) 21 cm
Solution:
(i) Given that
Length of an arc (l) = 10 cm
Radius which represents length of pendulum(r) = 75
As We know that θ = l/r
So θ = 10/75 = 2/15 rad
Hence, θ = 2/15 rad
(ii) Given that
Length of an arc (l) = 15 cm
Radius which represents length of pendulum (r) = 75
As We know that θ = l/r
So θ = 15/75 = 1/5 rad
Hence, θ = 1/5 rad
(iii) Given that
Length of an arc (l) = 21 cm
Radius which represents length of pendulum(r) = 75
As We know that θ = l/r
So θ = 21/75 = 7/25 rad
Hence, θ = 7/25 radian
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Conclusion
Chapter 3 of the Class 11 NCERT Mathematics textbook, "Trigonometric Functions," introduces basic concepts of trigonometry, including the definitions and properties of trigonometric functions. Exercise 3.1 focuses on fundamental problems to help students understand and apply these concepts. Detailed solutions are provided to support effective learning and mastery of trigonometric functions.