Question

A TV Tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is ${60}^o$. From another point $20m$ away this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is ${30}^o$ (see fig.). Find the height of the tower and the width of the canal.

• A. $h=10\sqrt{3}cm$
• B. $h=15\sqrt{2}cm$
• C. $h=10\sqrt{7}cm$
• D. None of these

Answer 1

The correct option is A $h=10\sqrt{3}cm$

Let $PQ=h$ meters be the height of the tower and $BQ=x$ meters be the width of the canal $\angle PBQ={60}^o$
Now, the angle of elevatio n of the top of the tower from the point $A={30}^o$
i.e., $\angle PAQ={30}^o$ where $AB=20$ meters
From $△PBQ$
$\frac{h}{x}=\tan {60}^o$ $\Rightarrow$ $\frac{h}{x}=\sqrt{3}=x\sqrt{3}......\left(1\right)$
From $△PAQ$,
$\frac{h}{20+x}=\tan {30}^o=\frac{1}{\sqrt{3}}$ $\Rightarrow h=\frac{20+x}{\sqrt{3}}.........\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$, we have $x\sqrt{3}=\frac{20+x}{\sqrt{3}}$
$\Rightarrow 3x=20+x$ or $2x-20$ $\Rightarrow x=10$
From $\left(1\right)$,$h=10\sqrt{3}m$