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

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

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

Answer:

It is solved using relation between degree and radian

(i) We know that = radianSo, radian
radian radian(ii) We know thatNow, we know that radianSo, radian radian(iii) We know that
radianSo, radian(iv) We know that
radianSo, radian radian

Question:2 Find the degree measures corresponding to the following radian measures. (Use $\small \pi =\frac{22}{7}$ )

$\small (i) \frac{11}{16}$
$\small (ii) -4$
$\small (iii) \frac{5\pi }{3}$
$\small (iv) \frac{7\pi }{6}$

Answer:

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

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

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 $1^\circ = 60'$ and 1' = 60'')

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

$\frac{315}{8}degree=39\frac{3}{8}degree\\ \\ =39^\circ +\frac{3\times 60}{8}minutes \Rightarrow 39^\circ +22' + \frac{1}{2}minutes \Rightarrow 39^\circ +22' +30''\\ \\ \Rightarrow \frac{315}{8}degree = 39^\circ22'30''$

(ii) -4
We know that

$\pi$ radian $= 180^\circ \Rightarrow 1 radian = \frac{180}{\pi} degree$ (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 $1^\circ = 60'$ and 1' = 60'')

$\Rightarrow \frac{-2520}{11}degree = -229\frac{1}{11}degree =-229^\circ + \frac{1\times 60}{11}minutes \\ \\ \Rightarrow -229^\circ + 5' + \frac{5}{11}minutes = -229^\circ +5' +27''\\ \\ -\frac{2520}{11} = -229^\circ5'27''$

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

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


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

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


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

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
$\therefore$ 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

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 $\small \pi =\frac{22}{7}$ )

Answer:

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
$\pi radian = 180^\circ\\ \\So, 1radian = \frac{180}{\pi}degree\\ \\ \therefore \frac{22}{100}radian = \frac{180}{\pi}\times\frac{22}{100}degree\Rightarrow \frac{180\times7}{22}\times\frac{22}{100} = \frac{63}{5}degree \\ \\ So, \\ \\\frac{63}{5}degree = 12\frac{3}{5}degree = 12^\circ + \frac{3\times60}{5}minute = 12^\circ + 36'\\ \\ \therefore \frac{63}{5}degree = 12 ^\circ36'$
So,
Angle subtended at the centre of a circle $\Theta = 12^\circ36'$

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 = $\frac{Diameter}{2} = \frac{40cm}{2} = 20 cm$
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 $60^\circ$
$\therefore$ $\theta = 60^\circ$ $= \frac{\pi}{3}radian$
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

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:-
$\theta_1 = 60^\circ\\ \theta_2 = 75^\circ\\$ 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

Answer:

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