Question

There are two towers one on each bank of a river, just opposite to each other one tower is $50m$ high. From the top of this tower, the angles of depression of the top and foot of the other tower are ${30}^o$ and ${60}^o$ respectively. Find the width of the river and the height of the other tower.

Answer 1

Let $AB$ and $CD$ be the two tower and $BC$ be the width of the river. Then $\angle ADE={30}^{\circ }$ and $\angle ACB={60}^{\circ }$

Given: $AB=50m$

In $\Delta ABC,tan{60}^{\circ }=\frac{AB}{BC}$

$\Rightarrow \sqrt{3}=\frac{50}{BC}$

$\Rightarrow BC=\frac{50}{\sqrt{3}}=28.867m$

In $\Delta AED,\tan {30}^{\circ }=\frac{AE}{ED}=\frac{AB-BE}{BC}=\frac{AB-DC}{BC}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{50-DC}{\frac{50}{\sqrt{3}}}$

$\Rightarrow \frac{50}{3}=50-DC$

$\Rightarrow DC=50-\frac{50}{3}=\frac{100}{3}=33.33m$

Hence the width of the river is $28.867m$ and height of the other tower is $33.33m$