Exercise 3.1 (Revised) - Chapter 3 - Trigonometric Functions - Ncert Solutions class 11 - Maths

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Chapter 3: Trigonometric Functions | NCERT Solutions for Class 11 Maths

Ex 3.1 Question 1.

Find the radian measures corresponding to the following degree measures:
(i) $25^{\circ}$
(ii) $-47030^{\prime}$
(iii) $240^{\circ}$
(iv) $520^{\circ}$

Answer.

(i) $25^{\circ}=\left(25 \times \frac{\pi}{180}\right)^c=\left(\frac{5 \pi}{36}\right)^c$
(ii) $-4730^{\prime}=-\left(47 \frac{30}{60}\right)^{\circ}=-\left(\frac{95}{2}\right)^{\circ}=-\left(\frac{95}{2} \times \frac{\pi}{180}\right)^c=-\left(\frac{19 \pi}{72}\right)^{\circ}$
(iii) $240^{\circ}=\left(240 \times \frac{\pi}{180}\right)^c=\left(\frac{4 \pi}{3}\right)^c$
(iv) $520^{\circ}=\left(520 \times \frac{\pi}{180}\right)^c=\left(\frac{26 \pi}{9}\right)^c$
Ex 3.1 Question 2.

Find the degree measures corresponding to the following radian measures $\left(\operatorname{Use} \pi=\frac{22}{7}\right)$

(i) $\frac{11}{16}$
(ii) -4

(iii) $\frac{5 \pi}{3}$
(iv) $\frac{7 \pi}{6}$

Answer.

(i) $\left(\frac{11}{16}\right)^c=\left(\frac{11}{16} \times \frac{180}{\pi}\right)^{\circ}=\left(\frac{11}{16} \times \frac{180 \times 7}{22}\right)^{\circ}=\left(\frac{315}{8}\right)^{\circ}=\left(39 \frac{3}{8}\right)^{\circ}=\left(39 \frac{3}{8} \times 60\right)^{\circ}$
$
=39^{\circ} 22 \frac{1}{2}=39^{\circ} 22^{\prime} \frac{1}{2} \times 60=39^{\circ} 22^{\prime} 30^{\prime \prime}
$
(ii)
$
\begin{aligned}
& (-4)^{\circ}=-\left(4 \times \frac{180}{\pi}\right)^{\circ}=-\left(4 \times \frac{180 \times 7}{22}\right)^{\circ}=-\left(\frac{2520}{11}\right)^{\circ}=-\left(229 \frac{1}{11}\right)^{\circ}=-229^{\circ}\left(\frac{1}{11} \times 60\right)^{\circ} \\
& =-229^{\circ} 5^{\prime}\left(\frac{5}{11} \times 60\right)^{\prime \prime}=-229^{\circ} 5^{\prime} 27 \prime
\end{aligned}
$
(iii) $\left(\frac{5 \pi}{3}\right)^c=\left(\frac{5 \pi}{3} \times \frac{180}{\pi}\right)^{\circ}=300^{\circ}$
(iv) $\left(\frac{7 \pi}{6}\right)^c=\left(\frac{7 \pi}{6} \times \frac{180}{\pi}\right)^{\circ}=210^{\circ}$

Ex 3.1 Question 3.

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

Answer.

Number of revolutions in 1 minute $=360$
$\therefore$ Number of revolution in 60 seconds $=360$
$\Rightarrow$ Number of revolutions in 1 second $=\frac{360}{60}=6$ revolutions
$\therefore$ Angle made by wheel in 6 revolutions $=360 \times 6=2160^{\circ}$

$
\Rightarrow 2160^{\circ}=\left(2160 \times \frac{\pi}{180}\right)^c=(12 \pi)^c
$
Ex 3.1 Question 4.

Find the degree measure of the angle subtended at the centre of a circle of radius 100 $\mathrm{cm}$ by an arc of length $22 \mathrm{~cm}\left(\right.$ Use $\left.\pi=\frac{22}{7}\right)$.

Answer.

Here $\gamma=100 \mathrm{~cm}$ and $l=22 \mathrm{~cm}$
$
\begin{aligned}
& \because \theta^c=\frac{l}{r} \\
& \therefore \theta^c=\frac{22}{100}=\left(\frac{11}{50}\right)^c \\
& \Rightarrow\left(\frac{11}{50}\right)^c=\left(\frac{11}{50} \times \frac{180^{\circ}}{\pi}\right)=\left(\frac{11}{50} \times \frac{180^{\circ} \times 7}{22}\right) \\
& =\left(\frac{63}{5}\right)^{\circ}=12^{\circ}\left(\frac{2}{5} \times 0\right)^{\prime}=12^{\circ} 36^{\prime}
\end{aligned}
$
Ex 3.1 Question 5.

In a circle of diameter $40 \mathrm{~cm}$, the length of a chord is $20 \mathrm{~cm}$. Find the length of minor arc of the chord.

Answer.

Given: Diameter $\mathrm{AB}=40 \mathrm{~cm}$, Radius $\mathrm{OA}=20 \mathrm{~cm}$ and Chord $\mathrm{AC}=20 \mathrm{~cm}$
$\therefore \triangle \mathrm{AOC}$ is an equilateral triangle.

$
\therefore \angle \mathrm{AOC}=60^{\circ}=\left(60^{\circ} \times \frac{\pi}{180}\right)^c=\left(\frac{\pi}{3}\right)^c
$

Now, $\theta^c=\frac{l}{\mu} \Rightarrow \frac{\pi}{3}=\frac{l}{20}$
$
\Rightarrow l=\frac{20 \pi}{3} \mathrm{~cm}
$
Ex 3.1 Question 6.

If in two circles, arcs of the same length subtend angles $60^{\circ}$ and $75^{\circ}$ at the centre, find the ratio of their radii.

Answer.

Let $r_1$ and $r_2$ be radii of two circles in which arcs of same length $l$ subtend angles $\theta_1=60^{\circ}$ and $\theta_2=75^{\circ}$ respectively.
$
\begin{aligned}
& \therefore \theta_1=\frac{l}{r_1} \Rightarrow\left(60 \times \frac{\pi}{180}\right)^c=\frac{l}{r_1} \\
& \Rightarrow r_1=\frac{3 l}{\pi}
\end{aligned}
$

And $\theta_2=\frac{l}{r_2} \Rightarrow\left(75 \times \frac{\pi}{180}\right)^c=\frac{l}{r_2}$

$
\begin{aligned}
& \Rightarrow r_2=\frac{12 l}{5 \pi} \\
& \therefore \frac{r_1}{r_2}=\frac{3 l / \pi}{12 l / 5 \pi}=\frac{5}{4} \\
& \Rightarrow r_1: r_2=5: 4
\end{aligned}
$
Ex 3.1 Question 7.

Find the angle in radians through which a pendulum swings if its length is $75 \mathrm{~cm}$ and the tip describes an arc of length:

(i) $10 \mathrm{~cm}$
(ii) $15 \mathrm{~cm}$
(iii) $21 \mathrm{~cm}$

Answer.

(i) Given: length of pendulum $(\gamma)=75 \mathrm{~cm}$ and length of arc $(l)=10 \mathrm{~cm}$
$
\therefore \theta^c=\frac{l}{\mu}=\frac{10}{75}=\left(\frac{2}{15}\right)^c
$
(ii) Given: length of pendulum $(\nu)=75 \mathrm{~cm}$ and length of arc $(l)=15 \mathrm{~cm}$
$
\therefore \theta^c=\frac{1}{\mu}=\frac{15}{75}=\left(\frac{1}{5}\right)^c
$
(iii) Given: length of pendulum $(\mu)=75 \mathrm{~cm}$ and length of arc $(l)=21 \mathrm{~cm}$
$
\therefore \theta^c=\frac{1}{\gamma}=\frac{21}{75}=\left(\frac{7}{25}\right)^c
$