Unit Exercise 6 - Chapter 6 Trigonometry 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Unit Exercise 6
Question 1.
Prove that
(i) $\cot ^2 A\left(\frac{\sec A-1}{1+\sin A}\right)+\sec ^2 A\left(\frac{\sin A-1}{1+\sec A}\right)=0$
(ii) $\frac{\tan ^2 \theta-1}{\tan ^2 \theta+1}=1-2 \cos ^2 \theta$
Solution:
$
\begin{aligned}
& \text { (i)L.H.S. }=\cot ^2 \mathrm{~A}\left(\frac{\sec \mathrm{A}-1}{1+\sin \mathrm{A}}\right)+\sec ^2 \mathrm{~A}\left(\frac{\sin \mathrm{A}-1}{1+\sec \mathrm{A}}\right) \\
& \Rightarrow \text { L.H.S }= \\
& \frac{\cot ^2 \mathrm{~A}(\sec \mathrm{A}-1)(\sec \mathrm{A}+1)+\sec ^2 \mathrm{~A} \cdot(\sin \mathrm{A}-1)(1+\sin \mathrm{A})}{(1+\sin \mathrm{A})(1+\sec \mathrm{A})} \\
& =\frac{\cot ^2 \mathrm{~A}\left(\sec ^2 \mathrm{~A}-1\right)+\sec ^2 \mathrm{~A}\left(\sin ^2 \mathrm{~A}-1\right)}{(1+\sin \mathrm{A})\left(1+\sec ^{\mathrm{A}}\right)} \\
& =\frac{\cot ^2 \mathrm{~A} \tan ^2 \mathrm{~A}+\sec ^2 \mathrm{~A}\left(\sin ^2 \mathrm{~A}-1\right)}{(1+\sin \mathrm{A})(1+\sec \mathrm{A})}
\end{aligned}
$

$
\begin{aligned}
& =\frac{\cot ^2 \mathrm{~A} \cdot \tan ^2 \mathrm{~A}-\sec ^2 \mathrm{~A} \cdot\left(1-\sin ^2 \mathrm{~A}\right)}{(1+\sin \mathrm{A})(1+\sec \mathrm{A})} \\
& =\frac{\cot ^2 \mathrm{~A} \tan ^2 \mathrm{~A}-\sec ^2 \mathrm{~A} \cdot \cos ^2 \mathrm{~A}}{(1+\sin \mathrm{A})(1+\sec \mathrm{A})} \\
& =\frac{1-1}{(1+\sin \mathrm{A})(1-\sec \mathrm{A})}=0=\mathrm{RHS} .
\end{aligned}
$
$
\text { (ii) } \begin{aligned}
\text { L.H.S } & =\frac{\tan ^2 \theta-1}{\tan ^2 \theta+1}=\frac{\frac{\sin ^2 \theta}{\cos ^2 \theta}-1}{\frac{\sin ^2 \theta}{\cos ^2 \theta}+1} \\
& =\frac{\frac{\sin ^2 \theta-\cos ^2 \theta}{\cos ^2 \theta}}{\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos ^2 \theta}} \\
& =\frac{\sin ^2 \theta-\cos ^2 \theta}{1} \\
& =1-\cos ^2 \theta-\cos ^2 \theta \quad\left(\because \sin ^2 \theta=1-\cos ^2 \theta\right) \\
& =1-2 \cos ^2 \theta=\text { RHS. }
\end{aligned}
$
Question 2.
Prove that $\left(\frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}\right)^2=\frac{1-\cos \theta}{1+\cos \theta}$
Solution:

$
\begin{aligned}
& \text { LHS }=\left(\frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}\right)^2 \\
& =\frac{(1+\sin \theta)^2+\cos ^2 \theta-2(1+\sin \theta)(\cos \theta)}{(1+\sin \theta)^2+\cos ^2 \theta+2(1+\sin \theta)(\cos \theta)} \\
& =\frac{1+2 \sin \theta+\sin ^2 \theta+\cos ^2 \theta-2 \cos \theta-2 \sin \theta \cos \theta}{1+\sin ^2 \theta+2 \sin \theta+\cos ^2 \theta+2 \cos \theta+2 \sin \theta \cos \theta} \\
& =\frac{2+2 \sin \theta-2 \cos \theta-2 \sin \theta \cos \theta}{2+2 \sin \theta+2 \cos \theta+2 \sin \theta \cos \theta} \\
& =\frac{1+\sin \theta-\cos \theta-\sin \theta \cos \theta}{1+\sin \theta+\cos \theta+\sin \theta \cos \theta}(\because \text { dividing by } 2) \\
& =\frac{(1+\sin \theta)-\cos \theta(1+\sin \theta)}{(1+\sin \theta)+\cos \theta(1+\sin \theta)} \\
& =\frac{(1-\cos \theta)(1+\sin \theta)}{(1+\cos \theta)(1+\sin \theta)}=\frac{1-\cos \theta}{1+\cos \theta}=\text { RHS. }
\end{aligned}
$
Hence proved
Question 3.
If $x \sin ^3 \theta+y \cos ^3 \theta=\sin \theta \cos \theta$ and $x \sin \theta=$ $y \cos \theta$, then prove that $x^2+y^2=1$.
Solution:
$
\begin{aligned}
& \mathrm{x} \sin ^3 \theta+\mathrm{y} \cos ^3 \theta=\sin \theta \cos \theta ; x \sin \theta \mathrm{y} \cos \theta \text {. } \\
& \mathrm{x}(\sin \theta)\left[\sin ^2 \theta+\cos ^2 \theta\right]=\sin \theta \cos \theta \\
& \text { (as } x \sin \theta=y \cos \theta \text { ) } \\
& \therefore \quad x \sin \theta=\sin \theta \cos \theta \\
& \Rightarrow \quad x=\cos \theta \\
& y=\sin \theta \\
& \therefore \quad x^2+y^2=\sin ^2 \theta+\sin ^2 \theta=1 \text {. } \\
&
\end{aligned}
$

Question 4.
If $\mathrm{a} \cos \theta-\mathrm{b} \sin \theta=\mathrm{c}$, then prove that $(\mathrm{a} \sin \theta+\mathrm{b} \cos \theta)=\pm \sqrt{a^2+b^2-c^2}$
Solution:
If $a \cos \theta-b \sin \theta=c$
$a \sin \theta+b \cos \theta=\pm \sqrt{a^2+b^2-c^2}$
L.H.S. we have
$
\begin{aligned}
& (a \cos \theta-b \sin \theta)^2+(a \sin \theta+b \cos \theta)^2 \\
& =a^2 \cos ^2 \theta+b^2 \sin ^2 \theta-2 a b \cos \theta \sin \theta \\
& \quad+a^2 \sin ^2 \theta+b^2 \cos ^2 \theta+2 a b \cos \theta \sin \theta \\
& =a^2\left(\cos ^2 \theta+\sin ^2 \theta\right)+b^2\left(\sin ^2 \theta+\cos ^2 \theta\right) \\
& =a^2+b^2 \quad(\because a \cos \theta-b \cos \theta=c) \\
& \therefore c^2+(a \sin \theta+b \cos \theta)^2=a^2+b^2 \\
& \Rightarrow(a \sin \theta+b \cos \theta)^2=a^2+b^2-c^2 . \\
& a \sin \theta+b \cos \theta=\pm \sqrt{a^2+b^2-c^2} . \\
& \text { Hence Proved. }
\end{aligned}
$
Question 5.
A bird is sitting on the top of a $80 \mathrm{~m}$ high tree. From a point on the ground, the angle of elevation of the bird is $45^{\circ}$. The bird flies away horizontally in such away that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is $30^{\circ}$. Determine the speed at which the bird flies. $(\sqrt{3}=1.732)$.
Solution:
Let $s$ be the speed of the bird. In 2 seconds, the bird goes from $\mathrm{C}$ to $\mathrm{D}$, it covers a distance ' $\mathrm{d}$ '

$\begin{aligned}
& \therefore d=s \times t=2 \mathrm{~s} . \\
& \text { In } \triangle \mathrm{ABC}, \tan 45^{\circ}=\frac{\mathrm{BC}}{\mathrm{AB}}=1 \text {. } \\
& \therefore \quad \mathrm{BC}=\mathrm{AB} \text {, } \\
& \therefore \quad \mathrm{AB}=80 \mathrm{~m} \text {. } \\
& \text { In } \triangle \mathrm{ABE}, \tan 30^{\circ}=\frac{\mathrm{BE}}{\mathrm{AB}}=\frac{80-x}{80}=\frac{1}{\sqrt{3}} \\
& \Rightarrow 80 \sqrt{3}-\sqrt{3} x=80 \\
&
\end{aligned}$

$
\begin{aligned}
-\sqrt{3} x & =80-80 \sqrt{3} \\
\sqrt{3} x & =80(\sqrt{3}-1) \\
x & =\frac{80(\sqrt{3}-1)}{\sqrt{3}} \\
\text { In } \Delta \text { CDE, } \tan 30^{\circ} & =\frac{\mathrm{CE}}{\mathrm{CD}}=\frac{x}{d} . \\
\frac{1}{\sqrt{3}}=\frac{x}{d} & =\frac{80(\sqrt{3}-1)}{\sqrt{3} d} \\
\Rightarrow \quad d & =80(\sqrt{3}-1) \\
& =80(0.732) \\
& =58.56 \mathrm{~m} \\
\text { speed } & =\frac{\text { distance }}{\text { time }} \\
& =\frac{58.56 \mathrm{~m}}{2 \text { seconds }} \\
& =29.28 \mathrm{~m} / \mathrm{s.}
\end{aligned}
$
Question 6.
An aeroplane is flying parallel to the Earth's surface at a speed of $175 \mathrm{~m} / \mathrm{sec}$ and at a height of 600 $\mathrm{m}$. The angle of elevation of the aeroplane from a point on the Earth's surface is $37^{\circ}$ at a given point. After what period of time does the angle of elevation increase to $53^{\circ} ?\left(\tan 53^{\circ}=1.3270, \tan \right.$ $\left.37^{\circ}=0.7536\right)$
Solution:
Let Plane's initial position be A. Plane's final position = D Plane travels from $\mathrm{A} \rightarrow \mathrm{D}$.

In $\triangle \mathrm{ABC}$, $\tan 37^{\circ}=\frac{\mathrm{AC}}{\mathrm{BC}}=\mathrm{BC}=\frac{\mathrm{AC}}{\tan 37^{\circ}}=\frac{600}{\tan 37^{\circ}}$ $\mathrm{BC}=\mathrm{BE}+\mathrm{EC}$ $\mathrm{EC}=\mathrm{AD}=$ speed $\times$ time, speed $=175 \mathrm{~m} / \mathrm{sec}$ $\therefore \mathrm{EC}=175 \mathrm{~m} / \mathrm{sec}$. $\mathrm{BE}=\mathrm{BC}-\mathrm{EC}=\frac{600}{\tan 37^{\circ}}-175 t$
In $\triangle \mathrm{BED}$,
$
\begin{aligned}
& \tan 53^{\circ}=\frac{\mathrm{DE}}{\tan 53^{\circ}} \\
& \mathrm{BE}=\frac{\mathrm{DE}}{\tan 53^{\circ}}=\frac{600}{\tan 53^{\circ}}
\end{aligned}
$

Equating (1) and (2)
$
\begin{aligned}
& \frac{600}{\tan 37^{\circ}}-175 t=\frac{600}{\tan 53^{\circ}} \\
& \therefore \quad 175 t=600\left(\frac{1}{\tan 37}-\frac{1}{\tan 53}\right) \\
& =600 \times \frac{\tan 53-\tan 37}{\tan 37 \cdot \tan 53^{\circ}} \\
& \tan 37^{\circ}=\tan \left(90-53^{\circ}\right)=\cot 53^{\circ} \text {. } \\
& \therefore \quad 175 t=\frac{600(\tan 53-\tan 37)}{1} \\
& =600(1.327-0.753) \\
& 175 t=600 \times 0.574 \\
& t=1.96 \\
&
\end{aligned}
$
Question 7.
A bird is flying from A towards B at an angle of $35^{\circ}$, a point $30 \mathrm{~km}$ away from $\mathrm{A}$. At $\mathrm{B}$ it changes its course of flight and heads towards $\mathrm{C}$ on a bearing of $48^{\circ}$ and distance $32 \mathrm{~km}$ away.
(i) How far is B to the North of A?
(ii) How far is B to the West of A?
(iii) How far is $\mathrm{C}$ to the North of B?
(iv) How far is $\mathrm{C}$ to the East of $\mathrm{B}$ ?
$\left(\sin 55^{\circ}=0.8192, \cos 55^{\circ}=0.5736\right.$,
$\left.\sin 42^{\circ}=0.6691, \cos 42^{\circ}=0.7431\right)$
Solution:

(i) In $\triangle \mathrm{ABD}$,

$
\begin{aligned}
\frac{\mathrm{BD}}{\mathrm{AB}} & =\sin 35^{\circ} \\
\therefore \mathrm{BD} & =\mathrm{AB} \sin 35^{\circ}=30 \sin 35^{\circ} \\
& =30 \times 0.8192=24.58 \mathrm{~km} \text { (approx.) }
\end{aligned}
$
(ii)

$
\begin{aligned}
\frac{\mathrm{AD}}{\mathrm{AB}} & =\cos 35^{\circ} \\
\therefore \mathrm{AD} & =\mathrm{AB} \cos 35^{\circ}=30 \cos 35^{\circ} \\
& =30 \times 0.5736=17.21 \mathrm{~km} \text { (approx.) }
\end{aligned}
$
(iii)
$
\begin{aligned}
\frac{\mathrm{CE}}{\mathrm{BC}} & =\sin 48^{\circ} \\
\therefore \mathrm{CE} & =\mathrm{BC} \sin 48^{\circ}=32 \sin 48^{\circ} \\
& =32 \times 0.6691=21.41 \mathrm{~km} \text { (approx.) }
\end{aligned}
$
(iv)
$
\begin{aligned}
\frac{\mathrm{BE}}{\mathrm{BC}} & =\cos 48^{\circ} \\
\therefore \mathrm{BE} & =\mathrm{BC} \cos 48^{\circ}=32 \cos 48^{\circ} \\
& =32 \times 0.7431=23.78 \mathrm{~km} \text { (approx.) }
\end{aligned}
$

Question 8.
Two ships are sailing in the sea on either side of the lighthouse. The angles of depression of two ships as observed from the top of the lighthouse are $60^{\circ}$ and $45^{\circ}$ respectively. If the distance between the ships is $200\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)$ metres, find the height of the lighthouse.
Solution:
From the figure $\mathrm{AB}$-height of the light house $=\mathrm{h} \mathrm{CD}$ - Distance between the ships

$\tan 45^{\circ}=\frac{h}{x}=1$
$\Rightarrow h \quad=x$
$\tan 60^{\circ}=\frac{h}{200\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)-x}=\sqrt{3}$
$
\begin{aligned}
h= & \sqrt{3}\left(\frac{200(\sqrt{3}+1)}{\sqrt{3}}-x\right) \\
& =\frac{\sqrt{3}(200(\sqrt{3}+1)-\sqrt{3} x)}{\sqrt{3}} .
\end{aligned}
$

$
\begin{aligned}
h & =200 \sqrt{3}+200-\sqrt{3} x \\
h & =200 \sqrt{3}+200-\sqrt{3} h(\because h=x) \\
h+\sqrt{3} h & =200(\sqrt{3}+1) \\
h(1+\sqrt{3}) & =200(\sqrt{3}+1) \\
h & =200 \mathrm{~m} .
\end{aligned}
$
$\therefore$ The height of the light house is 200 metres.
Question 9.
A building and a statue are in opposite side of a street from each other $35 \mathrm{~m}$ apart. From a point on the roof of building the angle of elevation of the top of statue is $24^{\circ}$ and the angle of depression of base of the statue is $34^{\circ}$. Find the height of the statue.
$\left(\tan 24^{\circ}=0.4452, \tan 34^{\circ}=0.6745\right)$
Solution:
Let $\mathrm{AB}$ be the building \& $\mathrm{CD}$ be statue.
In $\triangle \mathrm{ACE}$,

$\begin{aligned}
\quad \tan 24^{\circ} & =\frac{\mathrm{CE}}{\mathrm{AE}} \\
\therefore \quad \mathrm{CE} & =\mathrm{AE} \tan 24^{\circ} \\
& =35 \tan 24^{\circ} \\
\text { In } \Delta \mathrm{AED}, \tan 34^{\circ} & =\frac{\mathrm{DE}}{\mathrm{AE}} \\
\therefore \quad \mathrm{DE} & =\mathrm{AE} \tan 34^{\circ} \\
& =35 \tan 34^{\circ} \\
\therefore \text { Height of statue } & =\mathrm{CE}+\mathrm{ED} \\
& =35 \tan 24^{\circ}+35 \tan 34^{\circ} \\
& =35\left(\tan 24^{\circ}+\tan 34^{\circ}\right)= \\
& =35(0.4452+0.6745) \\
& =35 \times 1.1197 \\
& =39.19 \text { metres. }
\end{aligned}$