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

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

Edited By Komal Miglani | Updated on Apr 22, 2025 05:37 PM IST

Trigonometry is one of the oldest branches of mathematics. It was developed initially to solve the problems based on triangles but now its applicability is far beyond algebra. Trigonometry was widely used by ancient navigators, astronomers, and captains mostly rely on it for navigation. Trigonometry helps in understanding the relationship between angles and the side lengths of the triangles. Exercise 3.1 of NCERT mainly focuses on measurements of angles in degrees and radians.

This Story also Contains
  1. NCERT Solutions Class 11 maths chapter 3 Exercise 3.1
  2. Topics covered in Chapter 3 Trigonometric Functions: Exercise 3.1
  3. NCERT Solutions of Class 11 Subject Wise
  4. Subject-Wise NCERT Exemplar Solutions

In this exercise, students will learn how to convert angles from degrees to radians and vice versa and how to measure angles. Solutions for Exercise 3.1 are designed systematically and comprehensively that helps students to understand concepts easily. Students can also check NCERT Solutions to get detailed solutions from Class 6 to Class 12 for Science and Maths

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NCERT Solutions Class 11 maths chapter 3 Exercise 3.1

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

(i) 25(ii)47 30 (iii) 240 (iv) 520

Answer:

It is solved using relation between degree and radian

(i) 25

We know that 180=π radian

So, 1=π180 radian
25=π180×25 radian

=5π36 radian

(ii) 4730

We know that4730=4712 degree =952

Now, we know that 180=π1=π180 radian

So, 952=π180×(952) radian

19π72 radian

(iii) 240

We know that
180=π1=π180 radian

So, 240=π180×2404π3 radian

(iv) 520

We know that
180=π1=π180 radian

So, 520π180×520 radian

26π9 radian


Question:2 Find the degree measures corresponding to the following radian measures. (Use π=227 )

(i)1116
(ii)4
(iii)5π3
(iv)7π6

Answer:

(1) 1116

We know that
π radian =1801radian=180πdegree

So, 1116radian=180π×1116degree (we need to take π=227 )

1116radian=180×722×1116degree3158degree

(we use 1=60 and 1' = 60'')

Here 1' represents 1 minute and 60" represents 60 seconds
Now,

3158degree=3938degree=39+3×608minutes39+22+12minutes39+22+303158degree=392230

(ii) -4
We know that

π radian =1801radian=180πdegree (we need to take π=227 )


So, -4 radian = 4×180π4×180×722252011degree


(we use 1=60 and 1' = 60'')

252011degree=229111degree=229+1×6011minutes229+5+511minutes=229+5+27252011=229527

(iii) 5π3

We know that
π radian =1801radian=180πdegree (we need to take π=227 )


So, 5π3radian=180π×5π3degree=300
(iv) 7π6

We know that
π radian =1801radian=180πdegree (we need to take π=227 )


So, 7π6radian=180π×7π6=210

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

Answer:
Number of revolutions made by the wheel in 1 minute = 360
Number of revolutions made by the wheel in 1 second = 36060=6
( 1 minute = 60 seconds)
In one revolutions wheel will cover 2π radian
So, in 6 revolutions it will cover = 6×2π=12π radian

In 1 the second wheel will turn 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=227 )

Answer:

We know that
l=rΘ ( where l is the length of the arc, r is the radius of the circle and Θ is the angle subtended)

here r = 100 cm
and l = 22 cm
Now,
Θ=lr=22100radian

We know that
πradian=180So,1radian=180πdegree22100radian=180π×22100degree180×722×22100=635degreeSo,635degree=1235degree=12+3×605minute=12+36635degree=1236
So,
Angle subtended at the centre of a circle Θ=1236

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.

Answer:

Given :- radius (r)of circle = Diameter2=40cm2=20cm
length of chord = 20 cm

We know that
θ=lr (r = 20cm , l = ? , θ = ?)

Now,
1655285292995 AB is the chord of length 20cm and OA and OB are radii of circle 20 cm each
The angle subtended by OA and OB at centre = θ
OA = OB = AB
Δ OAB is equilateral triangle
So, each angle equilateral is 60
θ=60 =π3radian
Now, we have θ and r
So,
l=rθ=20×π3=20π3
the length of the minor arc of the chord (l) = 20π3 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.

Answer:

Given:-
θ1=60θ2=75 and l1=l2

We need to find the ratio of their radii r1r2=?

We know that arc length l=rθ
So,
l1=r1θ1l2=r2θ2
Now,
l1l2=r1θ1r2θ2 ( l1=l2 )
So,
r1r2=θ2θ1=7560=54 is the ratio of their radii

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

Answer:

(i) We know that

l=rθ
Now,
r = 75cm
l = 10cm

So,
θ=lr=1075=215radian

(ii) We know that

l=rθ
Now,
r = 75cm
l = 15cm

So,
θ=lr=1575=15radian

(iii) We know that

l=rθ
Now,
r = 75cm
l = 21cm

So,
θ=lr=2175=725radian


Also read

Trigonometric Functions NCERT Exercise 3.2

Trigonometric Functions NCERT Exercise 3.3

Trigonometric Functions NCERT Miscellaneous Exercise

Topics covered in Chapter 3 Trigonometric Functions: Exercise 3.1

This Exercise introduces students to angles, measures of degrees and radians. In this exercise, students will find questions that primarily revolve around the following key topics:

Trigonometry: It deals with the relationship between angles and sides of triangles. Trigonometric ratios like sin, cos, tan, cot, sec, and cosec are used to relate angles to sides of triangles.

Angles: An Angle measures the rotation of a line about its initial point.


Degree measure: It is the amount of rotation from initial side to the terminal side of the angle. Degree measure is measured in degree.

Complete rotation = 3600

Right angle = 900

Straight angle = 1800

Radian measure: It is the measure of the central angle of a circle that subtends an area which is equal to the length of the radius of the circle. Angle subtended at the centre by an arc of length 1 unit in a unit circle is said to have a measure of 1 radian.

Also Read

NCERT Solutions for Class 11 Maths Chapter 13

NCERT Exemplar Solutions Class 11 Maths Chapter 13

NCERT Solutions of Class 11 Subject Wise


Subject-Wise NCERT Exemplar Solutions

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Frequently Asked Questions (FAQs)

1. Which topics are covered in Exercise 3.1 Class 11 Maths?

Application of trigonometric functions for finding the arc length, chord length etc. is discussed in this exercise.

2. IS this chapter related to Inverse trigonometric functions?

Yes, the basics of this chapter are used in the inverse trigonometric functions.

3. What is the weightage of this chapter in the CBSE exam ?

Approx 10 marks questions are asked in the final exams of 11 Class.

4. Is it necessary to memorise the values of trigonometric functions ?

It can help to save time in the exam but more focus should be on the concepts.

5. Can multiple choice questions be solved by a shortcut method?

Yes, but in the CBSE examination it is not recommended but it can be used in JEE Main and NEET to save time.

6. How much difficult is the questions asked from this chapter in CBSE board exam ?

Questions of easy, moderate and a few difficulty level are asked. Most of the questions are of moderate level.

7. How many solutions are given in the exercise 3.1 Class 11 Maths ?

Total 7 questions are discussed in this exercise.

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2.45×10−3 kg

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K/2\,

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