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

Updated On May 15, 2024

By SaraNextGen

Additional Questions
Question 1.
An athlete runs 4 times around a circular running track to describe $1760 \mathrm{~m}$. What is the (radius of the tract) degrees subtended at the centre of the circle, after he has run a distance of $308 \mathrm{~m}$ ?
Solution:

[Hint: Distance travelled in 4 rounds $=1760 \mathrm{~m}$ ]
Distance travelled in 1 round $=\frac{1760}{4}=440 \mathrm{~m}$
Now $2 \pi r=440 \Rightarrow r=\frac{440 \times 7}{2 \times 22}=70 \mathrm{~m}$
Here $l=308 \mathrm{~m}$. We know $l=r \theta \Rightarrow \theta=\frac{l}{r}$
(i.e.) $\theta=\frac{308}{70}=(4.4)^{\mathrm{C}}$ radians
$
=\left(\frac{44}{10} \times \frac{180}{\pi}\right)^{\circ}=\left(\frac{44}{10} \times \frac{180}{22} \times 7\right)^{\circ}=252^{\circ}
$
Question 2.
Find the angle 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) $19 \mathrm{~cm}$
Solution:

Here, $\mathrm{r}=75 \mathrm{~cm}$
(i)
$
\begin{aligned}
& l=10 \mathrm{~cm} \\
& \text { Now, } \quad \theta=\frac{l}{r}=\frac{10}{75} \text { radians } \\
& =\left(\frac{180}{\pi} \times \frac{10}{75}\right)^{\circ}=\left(\frac{180 \times 7}{22} \times \frac{10}{75}\right)^{\circ} \\
& =\left(\frac{12 \times 7}{11}\right)^{\circ}=\frac{84}{11} \text { degrees }=7^{\circ} 38^{\prime} 11^{\prime \prime} . \\
&
\end{aligned}
$

(ii) $\quad l=15 \mathrm{~cm}$
Now, $\quad \theta=\frac{15}{75}$ radians $=\frac{1}{5}$ radians $=\left(\frac{1}{5} \times \frac{180}{\pi}\right)$ degrees
$
=\left(\frac{36}{22} \times 7\right) \text { degree }=\frac{126}{11} \text { degrees }=11^{\circ} 27^{\prime} 16^{\prime \prime}
$
(iii)
Now,
$
\begin{aligned}
l & =19 \mathrm{~cm} \\
\theta & =\frac{19}{75} \text { radians }=\left(\frac{19}{75} \times \frac{180}{\pi}\right) \text { degrees } \\
& =\left(\frac{19}{15} \times \frac{36 \times 7}{22}\right) \text { degrees }=\frac{798}{55} \text { degrees }=14^{\circ} 30^{\prime} 33^{\prime \prime}
\end{aligned}
$
Question 3.
The angles of a quadrilateral are in A.P and the greatest angle is $120^{\circ}$. Express the other angles in radians.
Solution:
Let the angles of the quadrilateral be $a^{\circ},(a+d)^{\circ},(a+2 d)^{\circ}$ and $(a+3 d)^{\circ}$.
Since sum of all angles of a quadrilateral is $360^{\circ}$, we have .
$
\begin{aligned}
& \mathrm{a}^{\circ}+(\mathrm{a}+\mathrm{d})^{\circ}+(\mathrm{a}+2 \mathrm{~d})^{\circ}+(\mathrm{a}+3 \mathrm{~d})^{\circ}=360^{\circ} \Rightarrow 4 \mathrm{a}+6 \mathrm{~d}=360^{\circ} \\
& \Rightarrow 2 \mathrm{a}+3 \mathrm{~d}=180^{\circ} \ldots(1)
\end{aligned}
$
Now, greatest angle is $120^{\circ}$
So, $\mathrm{a}+3 \mathrm{~d}=120^{\circ} \ldots$ (2)
Solving (1) and (2), we have
$
\mathrm{a}=60^{\circ}, \mathrm{d}=20^{\circ}
$
Hence, the angles are $60^{\circ}, 80^{\circ}, 100^{\circ}, 120^{\circ}$.

$
\begin{aligned}
& \text { (i.e.,) } \Rightarrow 60^{\circ}=60 \times \frac{\pi}{180}=\frac{\pi}{3} \text { radians } \\
& 80^{\circ}=80 \times \frac{\pi}{180}=\frac{4}{9} \pi \text { radians } \\
& 100^{\circ}=100 \times \frac{\pi}{180}=\frac{5}{9} \pi \text { radians } \\
& 120^{\circ}=120 \times \frac{\pi}{180}=\frac{2}{3} \pi \text { radians } \\
&
\end{aligned}
$
Question 4.
A railroad curve is to be laid out on a circle. What radius should be used if the track is to change direction by $25^{\circ}$ in a distance of 40 metres?
Solution:
Let the radius of the circle on which the railroad curve is to be laid down be $\mathrm{x}$ metres and the angle subtended by it at the centre is
$
\begin{array}{rlr}
25^{\circ} \text { or }\left(25 \times \frac{\pi}{180}\right) \text { radians } & \\
x\left(\frac{25 \pi}{180}\right) & =40 & {[\therefore r \theta=l]} \\
x & =\frac{180 \times 40}{25 \pi}=\frac{180 \times 40}{25} \times \frac{7}{22} \\
& =\frac{720 \times 7}{55}=\frac{144 \times 7}{11}=\frac{1008}{11}=91.636 \text { metres }
\end{array}
$
Question 5.
A horse is tied to a post by a rope. If the horse moves along a circular path always keeping the rope tight and describes 88 metres when it has traced out $72^{\circ}$ at the centre, find the length of the rope.
Solution:
Let $O$ be the post.
Let $\mathrm{A}, \mathrm{B}$ be the two positions of the horse.
Here 1, the length of the arc $\mathrm{AB}=88$ metres
Angle subtended $=72^{\circ}$

$\begin{aligned}
\text { Angle subtended } & =72^{\circ} \\
& =\left(72 \times \frac{\pi}{180}\right) \text { radians }=\frac{2 \pi}{5} \text { radians } \\
\theta & =\frac{2 \pi}{5} \text { radians }
\end{aligned}$

If $r$ (=length of the rope) be the radius of the circle

$
\begin{aligned}
& \text { Then, } \quad \theta=\frac{l}{r} \text { gives } \frac{2 \pi}{5}=\frac{88 \mathrm{~m}}{r} \\
& \Rightarrow \quad r=88 \times \frac{5}{2 \pi} \mathrm{m}=\left(88 \times \frac{5}{2} \times \frac{7}{22}\right) \mathrm{m}=70 \text { metres } \\
&
\end{aligned}
$
Hence, the length of the rope $=70$ metres

Question 6.
A circular wire of radius $3 \mathrm{~cm}$ is cut and bent so as to lie along the circumference of a sector whose radius is $48 \mathrm{~cm}$. Find in degrees the angle which is subtended at the centre of the sector.
Solution:
Length of arc $=$ Circumference of wire of radius $=3 \mathrm{~cm}$ $1=2 \pi \mathrm{r}=2 \pi \times 3=6 \pi \mathrm{cm}$
The radius of the sector $(\mathrm{r})=48 \mathrm{~cm}$
$
\theta=\frac{l}{r}=\frac{6 \pi}{48}=\frac{\pi}{8} \text { radians }
$
$\therefore$ Angle in degrees which is subtended at the centre of the sector
$
=\left(\frac{\pi}{8} \times \frac{180}{\pi}\right)^{\circ}=\left(\frac{45}{2}\right)^{\circ}=22^{\circ} 30^{\prime}
$
Question 7.
Let $\alpha, \beta$ be such that $\pi<\alpha-\beta<3 \pi$. If $\sin \alpha+\sin \beta=-21 / 65$ and $\cos \alpha+\cos \beta=-27 / 65$, then find the value of $\cos \frac{\alpha-\beta}{2}$.

Solution:
$
\begin{aligned}
& (\sin \alpha+\sin \beta)^2+(\cos \alpha+\cos \beta)^2=\left(\frac{-21}{65}\right)^2+\left(\frac{-27}{65}\right)^2=\frac{1170}{4225}=\frac{18}{65} \\
& \Rightarrow\left(\sin ^2 \alpha+\sin ^2 \beta\right)+\left(\cos ^2 \alpha+\cos ^2 \beta\right)+2(\sin \alpha \sin \beta+\cos \alpha \cos \beta)=\frac{18}{65} \\
& \text { (i.e.) }\left(\sin ^2 \alpha+\cos ^2 \alpha\right)+\left(\sin ^2 \beta+\cos ^2 \beta\right)+2(\sin \alpha \sin \beta+\cos \alpha \cos \beta)=\frac{18}{65} \\
& \Rightarrow \quad 1+1+2 \cos (\alpha-\beta)=\frac{18}{65} \Rightarrow 2+2 \cos (\alpha-\beta)=\frac{18}{65} \Rightarrow 2(1+\cos (\alpha-\beta))=\frac{18}{65} \\
& \Rightarrow \quad 2\left(2 \cos ^2 \frac{\alpha-\beta}{2}\right)=\frac{18}{65} \Rightarrow 4 \cos ^2 \frac{\alpha-\beta}{2}=\frac{18}{65} \\
& \Rightarrow \quad \cos ^2 \frac{\alpha-\beta}{2}=\frac{9}{65 \times 2} \Rightarrow \cos \frac{\alpha-\beta}{2}= \pm \frac{3}{\sqrt{130}} \\
& \text { Now, } \pi<\alpha-\beta<3 \pi \text {, i.e. } \frac{\pi}{2}<\frac{\alpha-\beta}{2}<\frac{3 \pi}{2} \\
& \therefore \quad \cos \frac{\alpha-\beta}{2}= \pm \frac{3}{\sqrt{130}} \\
&
\end{aligned}
$