NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions Exercise 3.1

Chapter 3, Trigonometric Functions of Class 11 Maths, is categorised under the CBSE Syllabus for 2023-24. The chapter contains questions and concepts revolving around Trigonometric functions. Exercise 3.1 of NCERT Solutions for Class 11 Maths Chapter 3 – Trigonometric Functions is based on the following topics:

  1. Introduction
  2. Angles
    1. Degree measure
    2. Radian measure
    3. Relation between radian and real numbers
    4. Relation between degree and radian

These solutions are helpful for the students to get an idea of how to answer the questions from the board exam perspective. View online or download the NCERT Solutions for Class 11 to get a hold of all the concepts covered in the chapter.

NCERT Solutions for Class 11 Maths Chapter 3 – Trigonometric Functions Exercise 3.1

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Access other exercise solutions of Class 11 Maths Chapter 3 – Trigonometric Functions

To access the other exercise answers from NCERT Class 11 Maths Solutions Chapter 3, click on the links below.

Exercise 3.2 Solutions 10 Questions

Exercise 3.3 Solutions 25 Questions

Exercise 3.4 Solutions 9 Questions

Miscellaneous Exercise on Chapter 3 Solutions 10 Questions

Access Solutions for Class 11 Maths Chapter 3.1 exercise

1. Find the radian measures corresponding to the following degree measures:

(i) 25° (ii) – 47° 30′ (iii) 240° (iv) 520°

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 1

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 2

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 3

(iv) 520°

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 4

2. Find the degree measures corresponding to the following radian measures (Use π = 22/7).

(i) 11/16

(ii) -4

(iii) 5Ï€/3

(iv) 7Ï€/6

Solution:

(i) 11/16

Here, π radian = 180°

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 5

(ii) -4

Here, π radian = 180°

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 6

(iii) 5Ï€/3

Here, π radian = 180°

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 7

We get

= 300o

(iv) 7Ï€/6

Here, π radian = 180°

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 8

We get

= 210o

3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

Solution:

It is given that

No. of revolutions made by the wheel in

1 minute = 360

1 second = 360/60 = 6

We know that

The wheel turns an angle of 2Ï€ radian in one complete revolution.

In 6 complete revolutions, it will turn an angle of 6 × 2π radian = 12 π radian

Therefore, in one second, the wheel turns at an angle of 12Ï€ radian.

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:

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 9

5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of the minor arc of the chord.

Solution:

The dimensions of the circle are

Diameter = 40 cm

Radius = 40/2 = 20 cm

Consider AB as the chord of the circle, i.e., length = 20 cm

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 10

In ΔOAB,

Radius of circle = OA = OB = 20 cm

Similarly AB = 20 cm

Hence, ΔOAB is an equilateral triangle.

θ = 60° = π/3 radian

In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre,

We get θ = 1/r

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 11

Therefore, the length of the minor arc of the chord is 20Ï€/3 cm.

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:

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 12

7. Find the angle in the 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:

In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = 1/r

We know that r = 75 cm

(i) l = 10 cm

So we get

θ = 10/75 radian

By further simplification,

θ = 2/15 radian

(ii) l = 15 cm

So, we get

θ = 15/75 radian

By further simplification,

θ = 1/5 radian

(iii) l = 21 cm

So, we get

θ = 21/75 radian

By further simplification,

θ = 7/25 radian

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