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 Sumit Saini | Updated on Jul 29, 2022 03:18 PM IST

NCERT solutions for class 11 maths chapter 3 exercise 3.1 is about some identities related to trignometric functions. In class 10 maths NCERT syllabus chapter 8 students have already familiarised the basic trigonometric functions, in this chapter applications of trigonometric functions are discussed which includes length of Arc and Chord of a circle etc. NCERT book Exercise 3.1 Class 11 Maths will be used in the class 12 chapter of Inverse Trigonometric functions. Solving NCERT Solutions for class 11 maths chapter 3 exercise 3.1 will be useful to make a strong base in Trigonometry. Many questions can be seen in the previous year question papers from class 11 maths chapter 3 exercise 3.1. Following exercise of NCERT can be explored for further information about upcoming exercises.

## NCERT solutions for class 11 maths chapter 3 trigonometric functions-Exercise: 3.1

(i) $25 \degree$(ii)$-47 \degree$ $30'$ (iii) $240\degree$ (iv) $520\degree$

It is solved using relation between degree and radian

(1) $\frac{11}{16}$

We know that
$\pi$ radian $= 180\degree \Rightarrow 1 radian = \frac{180}{\pi} degree$

So, $\frac{11}{16} radian = \frac{180}{\pi}\times \frac{11}{16}degree$ (we need to take $\pi = \frac{22}{7}$ )

$\frac{11}{16}radian = \frac{180\times 7}{22}\times \frac{11}{16}degree \Rightarrow \frac{315}{8}degree$

(we use $1\degree = 60'$ and 1' = 60'')

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

$\frac{315}{8}degree=39\frac{3}{8}degree\\ \\ =39\degree +\frac{3\times 60}{8}minutes \Rightarrow 39\degree +22' + \frac{1}{2}minutes \Rightarrow 39\degree +22' +30''\\ \\ \Rightarrow \frac{315}{8}degree = 39\degree22'30''$

(ii) -4
We know that

$\pi$ radian $= 180\degree \Rightarrow 1 radian = \frac{180}{\pi} degree$ (we need to take $\pi = \frac{22}{7}$ )

So, -4 radian = $\frac{-4\times 180}{\pi} \Rightarrow \frac{-4\times 180\times 7}{22} \Rightarrow \frac{-2520}{11}degree$

(we use $1\degree = 60'$ and 1' = 60'')

$\Rightarrow \frac{-2520}{11}degree = -229\frac{1}{11}degree =-229\degree + \frac{1\times 60}{11}minutes \\ \\ \Rightarrow -229\degree + 5' + \frac{5}{11}minutes = -229\degree +5' +27''\\ \\ -\frac{2520}{11} = -229\degree5'27''$

(iii) $\frac{5\pi}{3}$

We know that
$\pi$ radian $= 180\degree \Rightarrow 1 radian = \frac{180}{\pi} degree$ (we need to take $\pi = \frac{22}{7}$ )

So, $\frac{5\pi}{3}radian = \frac{180}{\pi}\times \frac{5\pi}{3}degree = 300\degree$
(iv) $\frac{7\pi}{6}$

We know that
$\pi$ radian $= 180\degree \Rightarrow 1 radian = \frac{180}{\pi} degree$ (we need to take $\pi = \frac{22}{7}$ )

So, $\frac{7\pi}{6}radian = \frac{180}{\pi}\times \frac{7\pi}{6} = 210\degree$

Number of revolutions made by the wheel in 1 minute = 360
$\therefore$ Number of revolutions made by the wheel in 1 second = $\frac{360}{60} = 6$
( $\because$ 1 minute = 60 seconds)
In one revolutions wheel will cover $2\pi$ radian
So, in 6 revolutions it will cover = $6\times 2\pi = 12\pi$ radian

$\therefore$ In 1 the second wheel will turn $12\pi$ radian

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

here r = 100 cm
and l = 22 cm
Now,
$\Theta = \frac{l}{r} = \frac{22}{100}radian$

We know that
$\pi radian = 180\degree\\ \\So, 1radian = \frac{180}{\pi}degree\\ \\ \therefore \frac{22}{100}radian = \frac{180}{\pi}\times\frac{22}{100}degree\Rightarrow \frac{180\times7}{22}\times\frac{22}{100} = \frac{63}{5}degree \\ \\ So, \\ \\\frac{63}{5}degree = 12\frac{3}{5}degree = 12\degree + \frac{3\times60}{5}minute = 12\degree + 36'\\ \\ \therefore \frac{63}{5}degree = 12 \degree36'$
So,
Angle subtended at the centre of a circle $\Theta = 12\degree36'$

Given :- radius (r)of circle = $\frac{Diameter}{2} = \frac{40cm}{2} = 20 cm$
length of chord = 20 cm

We know that
$\theta = \frac{l}{r}$ (r = 20cm , l = ? , $\theta$ = ?)

Now,
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 = $\theta$
$\because$ OA = OB = AB
$\therefore$ $\Delta$ OAB is equilateral triangle
So, each angle equilateral is $60\degree$
$\therefore$ $\theta = 60\degree$ $= \frac{\pi}{3}radian$
Now, we have $\theta$ and r
So,
$l = r\theta = 20\times\frac{\pi}{3}=\frac{20\pi}{3}$
$\therefore$ the length of the minor arc of the chord (l) = $\frac{20\pi}{3}$ cm

Given:-
$\theta_1 = 60\degree\\ \theta_2 = 75\degree\\$ and $l_1 = l_2$

We need to find the ratio of their radii $\frac{r_1}{r_2} = ?$

We know that arc length $l = r \theta$
So,
$l_1 = r_1 \theta_1\\l_2 = r_2\theta_2$
Now,
$\frac{l_1}{l_2}=\frac{ r_1 \theta_1}{ r_2\theta_2}$ ( $l_1 = l_2$ )
So,
$\frac{ r_1 }{ r_2}= \frac{\theta_2}{\theta_1} = \frac {75}{60} = \frac{5}{4}$ 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

(i) We know that

$l = r \theta$
Now,
r = 75cm
l = 10cm

So,
$\theta = \frac{l}{r} = \frac{10}{75} = \frac{2}{15}radian$

(ii) We know that

$l = r \theta$
Now,
r = 75cm
l = 15cm

So,
$\theta = \frac{l}{r} = \frac{15}{75} = \frac{1}{5}radian$

(iii) We know that

$l = r \theta$
Now,
r = 75cm
l = 21cm

So,
$\theta = \frac{l}{r} = \frac{21}{75} = \frac{7}{25}radian$

## More About NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.1

The NCERT class 11 maths chapter Trigonometric functions is one of the most important chapters as it has linkages with a lot of other chapters in Maths as well as physics. Exercise 3.1 Class 11 Maths Mainly discusses the application of various trigonometric functions to find out the arc length, chord etc. Hence NCERT Solutions for class 11 maths chapter 3 exercise 3.1 must be done seriously by the students.

## Benefits of NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.1

• The class 11th maths chapter 3 exercise is prepared by expert faculties.

• Exercise 3.1 Class 11 Maths can help one in a tremendous way as it has application in a lot of upcoming chapters of class 11 as well as class 12.

class 11 maths chapter 3 exercise 3.1 solutions provided here are one stop solution for theory as well as question practice.

Also see-

## NCERT Solutions of Class 11 Subject Wise

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## Subject Wise NCERT Exampler Solutions

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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