Question

There are two temples, one on each bank of a river, just opposite to each other. One temple is $50$ m high. From the top of this temple, the angle of depression of the top and the foot of the other temple are ${30}^0$ and ${60}^0$ respectively. Find the width of the river and the height of the other temple.

Answer 1

Let $AB$ and $CD$ be the two temples and $BC$ be the width of the river.

$AB=50m$

$\Rightarrow$ In $△ABC,\tan {60}^o=\frac{AB}{BC}$

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

$\Rightarrow$ $BC=\frac{50}{\sqrt{3}}=\frac{50\sqrt{3}}{3}=28.867m$ $(Take\sqrt{3}=1.73)$

$\Rightarrow$ In $△AED,\tan {30}^o=\frac{AE}{ED}$

$\Rightarrow$ $\frac{1}{\sqrt{3}}=\frac{AB-BE}{BC}=\frac{AB-DC}{BC}$

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

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

$\Rightarrow$ $DC=50-\frac{50}{3}=\frac{150-50}{3}=\frac{100}{3}=33.33m$

$\therefore$ The width of the river is $28.867m$ and height of the temple is $33.33m$