Exercise 3.2-Additional Questions - Chapter 3 - Trigonometry - 11th Maths Guide Samacheer Kalvi Solutions

You can Download the Exercise 3.2-Additional Questions - Chapter 3 - Trigonometry - 11th Maths Guide Samacheer Kalvi Solutions with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

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}
$