Exercise 3.2 - Chapter 3 Trigonometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

$\mathbf{E} 3.2$
Question 1.
Express each of the following angles in radian measure:
(i) $30^{\circ}$
(ii) $135^{\circ}$
(iii) $-205^{\circ}$
(iv) $150^{\circ}$
(v) $330^{\circ}$
Solution:
(i) $30^{\circ}=30 \times \frac{\pi}{180}=\frac{\pi}{6}$ radians
(ii) $135^{\circ}=135 \times \frac{\pi}{180}=\frac{3 \pi}{4}$ radians
(iii) $-\mathbf{2 0 5}^{\circ}=-205 \times \frac{\pi}{180}=\frac{-41 \pi}{36}$ radians
(iv) $150^{\circ}=150 \times \frac{\pi}{180}=\frac{5 \pi}{6}$ radians
(v) $330^{\circ}=330 \times \frac{\pi}{180}=\frac{11 \pi}{6}$ radians
Question 2.
Find the degree measure corresponding to the following radian measures
(i) $\frac{\pi}{3}$
(ii) $\frac{\pi}{9}$
(iii) $\frac{2 \pi}{5}$
(iv) $\frac{7 \pi}{3}$
(v) $\frac{10 \pi}{9}$

Solution:
(i) $\frac{\pi}{3} \quad=\frac{\pi}{3} \times \frac{180}{\pi}=60^{\circ}$
(ii) $\frac{\pi}{9}=\frac{\pi}{9} \times \frac{180}{\pi}=20^{\circ}$
(iii) $\frac{2 \pi}{5}$. $=\frac{2 \pi}{5} \times \frac{180}{\pi}=72^{\circ}$
(iv) $\frac{7 \pi}{3}=\frac{7 \pi}{3} \times \frac{180}{\pi}=420^{\circ}$
(v) $\frac{10 \pi}{9}=\frac{10 \pi}{9} \times \frac{180}{\pi}=200^{\circ}$
Question 3.
What must be the radius of a circular running path, around which an athlete must run 5 times in order to describe $1 \mathrm{~km}$ ?
Solution:
Distance travelled in 5 rounds $=1 \mathrm{~km}=1000 \mathrm{~m}$
Distance travelled in 1 round $=\frac{1000}{5}=200 \mathrm{~m}$
Let the radius of the circular path be $r$ metre
So $2 \pi \mathrm{r}=200$
$
\text { (i.e) } \begin{aligned}
2 \times \frac{22}{7} \times r & =200 \\
\therefore r & =\frac{200 \times 7}{44}=31.82 \mathrm{~m}
\end{aligned}
$
Question 4.
In a circle of diameter $40 \mathrm{~cm}$, a chord is of length $20 \mathrm{~cm}$. Find the length of the minor arc of the chord.
Solution:
$\mathrm{O}=$ centre of the circle
$\mathrm{PQ}=$ diameter $=40 \mathrm{~cm}$
$\therefore \mathrm{OQ}=20 \mathrm{~cm}$
radius $=20 \mathrm{~cm}$
$\Rightarrow \mathrm{OA}=\mathrm{OB}=20 \mathrm{~cm}$
chord $\mathrm{AB}=20 \mathrm{~cm}$
$\mathrm{OC} \perp$ r $\mathrm{AB}$
$
\therefore \mathrm{AC}=\mathrm{CB}=10 \mathrm{~cm}
$
Now from the right angled triangle OCB

$
\begin{aligned}
& \sin \theta=\frac{10}{20}=\frac{1}{2} \Rightarrow \theta=\pi / 6 \\
& \text { So } \angle \mathrm{AOB}=2 \times \frac{\pi}{6}=\frac{\pi}{3}=60^{\circ} \\
& \text { Now } r=20 \mathrm{~cm}, \theta=\pi / 3 \\
& \text { So } \operatorname{arc} \mathrm{AB}=\frac{60}{360} 2 \pi r \\
& =\frac{60}{360} \times 2 \times \pi \times 20=\frac{20 \pi}{3} \mathrm{~cm} \text { (or) } 20.95 \mathrm{~cm} \text {. } \\
&
\end{aligned}
$
Question 5.
Find the degree measure of the angle subtended at the centre of circle of radius $100 \mathrm{~cm}$ by an arc of length $22 \mathrm{~cm}$
Solution:
$
\begin{aligned}
& \mathrm{r}=100 \mathrm{~cm} ; \text { arc length }=22 \mathrm{~cm} \\
& \text { Arc length }=\frac{\theta}{360} \times 2 \pi r \\
& \Rightarrow \frac{\theta}{360} \times 2 \times \frac{22}{7} \times 100=22 \quad \Rightarrow \theta=\frac{7 \times 22 \times 360}{44 \times 100} \quad \frac{126}{10}=12^{\circ} \quad \frac{6}{10} \times 60=12^{\circ} 36^{\prime}
\end{aligned}
$
Question 6.
What is the length of the arc intercepted by a central angle of measure $41^{\circ}$ in a circle of radius $10 \mathrm{ft}$ ?

Solution:

$
\begin{aligned}
& \theta=41^{\circ}, \mathrm{r}=10 \mathrm{ft} \\
& \text { Arc length }=\frac{\theta}{360} \times 2 \pi r \\
&=\frac{41}{360} \times 2 \times \frac{22}{7} \times 10 \\
&=\frac{451}{63}=7.16 \mathrm{ft}
\end{aligned}
$
Question 7.
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.
Solution:
Let the two radii be $r_1$ and $r_2$
The central angles arc $60^{\circ}$ and $75^{\circ}$
The arc lengths be $s_1$ and $s_2$
we are given $s_1=s_2$
$
\begin{array}{cl}
\Rightarrow \quad & \frac{60}{360} \times 2 \pi r_1=\frac{75}{360} \times 2 \pi r_2 \\
60 r_1=75 r_2 \\
\Rightarrow \quad \frac{r_1}{r_2}=\frac{75}{60}=\frac{5}{4} \\
r_1: r_2=5: 4
\end{array}
$
So their radii are in the ratio $5: 4$.
Question 8.
The perimeter of a certain sector of a circle is equal to the length of the arc of a semicircle having the same radius. Express the angle of the sector in degrees, minutes and seconds.
Solution:
Let $r$ be the radius and so perimeter of a sector $=1+2 \mathrm{r}$
Length of arc of the semicircle $=\pi \mathrm{r}$
we are given $1+2 r=\pi r$
(i.e) $1=\pi \mathrm{r}-2 \mathrm{r}$

$
\begin{aligned}
& \frac{\theta}{360} \times 2 \pi r=\pi r-2 r=r(\pi-2) \\
& \Rightarrow \frac{\theta}{360} \times 2 \pi=\pi-2 \\
& \theta==\frac{\pi-2}{\pi} \times 180=\frac{\frac{22}{7}-2}{\frac{22}{7}} \times 180 \\
& =\frac{8}{7} \times \frac{7}{22} \times 180 \\
& =\frac{8 \times 90}{11}=\frac{720}{11}=65 \frac{5}{11}=65^{\circ} \frac{5}{11} \times 60 \\
& =65^{\circ} \frac{300^{\prime}}{11}=65^{\circ} 27^{\prime} \frac{3}{11} \\
& =65^{\circ} 27^{\prime} \frac{3 \times 60}{11}=65^{\circ} 27^{\prime} 16^{\prime \prime}
\end{aligned}
$
Question 9.
An airplane propeller rotates 1000 times per minute. Find the number of degrees that a point on the edge of the propeller will rotate in 1 second.
Solution:
Number of rotations in $1 \mathrm{~min}=1000$
So Number of rotations in $1 \mathrm{sec}=\frac{1000}{60}=\frac{50}{3}$
The angle rotated in 1 rotation $=360^{\circ}$
So, the angle rotated in $\frac{50}{3}$ rotation $=\frac{50}{3} \times 360^{\circ}=6000^{\circ}$
Question 10 .
A train is moving on a circular track of $1500 \mathrm{~m}$ radius at the rate of $66 \mathrm{~km} / \mathrm{hr}$. What angle will it turn in 20 seconds?
Solution:

Speed of the train $=66 \mathrm{~km} / \mathrm{hr}$
$
\begin{aligned}
& =66 \times \frac{5}{18} \mathrm{~m} / \mathrm{sec} \\
& =\frac{55}{3} \mathrm{~m} / \mathrm{sec}
\end{aligned}
$
(i.e) Distance travelled in $\mathrm{sec}=\frac{55}{3} \mathrm{~m}$
So, distance travelled in $20 \mathrm{sec}=\frac{55}{3} \times 20=\frac{1100}{3} \mathrm{~m}$
Now radius of the circular track $=1500 \mathrm{~m}$
we are given $\frac{\theta}{360} \times 2 \times \frac{22}{7} \times 1500=\frac{1100}{3}$
$
\Rightarrow \theta=\frac{1100}{3} \times \frac{360 \times 7}{2 \times 22 \times 1500}=14^{\circ}
$
Question 11.
A circular metallic plate of radius $8 \mathrm{~cm}$ and thickness $6 \mathrm{~mm}$ is melted and molded into a pie (a sector of the circle with thickness) of radius $16 \mathrm{~cm}$ and thickness $4 \mathrm{~mm}$. Find the angle of the sector.
Solution:
Area of the circular plate melted

$
\begin{aligned}
=\pi\left(8^2\right)=64 \pi \mathrm{cm}^2 & \\
\text { Thickness }=6 \mathrm{~mm}=\frac{6}{10} \mathrm{~cm} & \\
\text { So, volume } & =\text { area } \times \text { thickness } \\
& =64 \pi \times \frac{6}{10}=\mathrm{V}_1 \\
\text { Area of the pie made } & =\frac{\theta}{360} \times \pi\left(16^2\right) \\
& =(256 \pi) \frac{\theta}{360} \\
\text { Thickness of the pie made } & =4 \mathrm{~mm} \\
\therefore \text { Volume of the pie made } & =\frac{4}{10} \mathrm{~cm} \\
& (256 \pi) \frac{\theta}{360} \times \frac{4}{10}=\mathrm{V}_2
\end{aligned}
$
We are given $\mathrm{V}_1=\mathrm{V}_2$
$
\begin{aligned}
\Rightarrow \quad 64 \pi \frac{6}{10} & =(256 \pi)\left(\frac{\theta}{360}\right) \times \frac{4}{10} \\
\theta & =64 \times \frac{6}{10} \times \frac{1}{256} \times 360 \times \frac{10}{4} \\
& =135^{\circ} \text { (i.e) } \theta=\frac{3 \pi}{4}
\end{aligned}
$