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)$-{47}^{\circ }$ ${30}^{\prime }$ (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 $\pi =\frac{22}{7}$ )

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

Answer:

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

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

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 $1^{\circ }={60}^{\prime }$ and 1' = 60'')

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

$$\begin{gathered} \frac{315}{8}degree=39\frac{3}{8}degree \\ ={39}^{\circ }+\frac{3\times 60}{8}minutes\Rightarrow {39}^{\circ }+{22}^{\prime }+\frac{1}{2}minutes\Rightarrow {39}^{\circ }+{22}^{\prime }+{30}^{″} \\ \Rightarrow \frac{315}{8}degree={39}^{\circ }{22}^{\prime }{30}^{″} \end{gathered}$$

(ii) -4
We know that

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

$$\begin{gathered} \Rightarrow \frac{-2520}{11}degree=-229\frac{1}{11}degree=-{229}^{\circ }+\frac{1\times 60}{11}minutes \\ \Rightarrow -{229}^{\circ }+5^{\prime }+\frac{5}{11}minutes=-{229}^{\circ }+5^{\prime }+{27}^{″} \\ -\frac{2520}{11}=-{229}^{\circ }{5}^{\prime }{27}^{″} \end{gathered}$$

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

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

Answer:

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

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

We know that
$$\begin{gathered} \pi radian={180}^{\circ } \\ So,1radian=\frac{180}{\pi }degree \\ \therefore \frac{22}{100}radian=\frac{180}{\pi }\times \frac{22}{100}degree\Rightarrow \frac{180\times 7}{22}\times \frac{22}{100}=\frac{63}{5}degree \\ So, \\ \frac{63}{5}degree=12\frac{3}{5}degree={12}^{\circ }+\frac{3\times 60}{5}minute={12}^{\circ }+{36}^{\prime } \\ \therefore \frac{63}{5}degree={12}^{\circ }{36}^{\prime } \end{gathered}$$
So,
Angle subtended at the centre of a circle $\mathrm{\Theta }={12}^{\circ }{36}^{\prime }$

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 = $\frac{Diameter}{2}=\frac{40cm}{2}=20cm$
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
$\therefore$ $\mathrm{\Delta }$ 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:-
$$\begin{gathered} {\theta }_1={60}^{\circ } \\ {\theta }_2={75}^{\circ } \end{gathered}$$ 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,
$$\begin{gathered} l_1=r_1{\theta }_1 \\ {l}_2=r_2{\theta }_2 \end{gathered}$$
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,
$\theta =\frac{l}{r}=\frac{21}{75}=\frac{7}{25}radian$