A T.V. Tower stands vertically on a bank of a river. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is $ 60^{\circ} $. From a point $ 20 \mathrm{~m} $ away this point on the same bank, the angle of elevation of the top of the tower is $ 30^{\circ} $. Find the height of the tower and the width of the river.
Given:
A T.V. Tower stands vertically on a bank of a river. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is \( 60^{\circ} \). From a point \( 20 \mathrm{~m} \) away this point on the same bank, the angle of elevation of the top of the tower is \( 30^{\circ} \).
To do:
We have to find the height of the tower and the width of the river.
Solution:
Let $AB$ be the height of the tower and $BC$ be the width of the river.
Let point $C$ be the point of observation on the other bank and point $D$ be \( 20 \mathrm{~m} \) away from point $C$ on the same bank.
From the figure,
$\mathrm{CD}=20 \mathrm{~m}, \angle \mathrm{ADB}=30^{\circ}, \angle \mathrm{ACB}=60^{\circ}$
Let the height of the tower be $\mathrm{AB}=h \mathrm{~m}$ and the width of the river be $\mathrm{BC}=x \mathrm{~m}$.
This implies,
$\mathrm{DB}=20+x \mathrm{~m}$
We know that,
$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$
$=\frac{\text { AB }}{BC}$
$\Rightarrow \tan 60^{\circ}=\frac{h}{x}$
$\Rightarrow \sqrt3=\frac{h}{x}$
$\Rightarrow x(\sqrt3)=h \mathrm{~m}$
$\Rightarrow h=x\sqrt3 \mathrm{~m}$.........(i)
Similarly,
$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$
$=\frac{\text { AB }}{DB}$
$\Rightarrow \tan 30^{\circ}=\frac{h}{x+20}$
$\Rightarrow \frac{1}{\sqrt3}=\frac{x\sqrt3}{x+20}$ [From (i)]
$\Rightarrow x+20=x\sqrt3(\sqrt3) \mathrm{~m}$
$\Rightarrow 3x-x=20 \mathrm{~m}$
$\Rightarrow x=\frac{20}{2} \mathrm{~m}$
$\Rightarrow x=10 \mathrm{~m}$
$\Rightarrow h=10\sqrt3 \mathrm{~m}$
Therefore, the height of the tower is $10\sqrt3 \mathrm{~m}$ and the width of the river is $10 \mathrm{~m}$.
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