There are two temples, one on each bank of a river, just opposite to each other. One temple is 50 m high.
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 angles of depression of the top and the foot of the other temple are 30° and 60° respectively. Find the width of the river and the height of the other temple.

Solution:

Let  and  are two temples each at the bank of river. The top of the temple CE makes angle of depressions at the top and bottom of tower AB are 30° and 60°

Let $C E=50 \mathrm{~m}$ and $A B=H \mathrm{~m}$ and $\angle C B E=60^{\circ}, \angle D A E=30^{\circ}$

The corresponding figure is as follows

$\ln \triangle A D E$

$\Rightarrow \quad \tan 30^{\circ}=\frac{h}{x}$

$\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{x}$

$\Rightarrow \quad x=h \sqrt{3}$

Again in $\triangle B C E$,

$\Rightarrow \quad \tan 60^{\circ}=\frac{50}{x}$

$\Rightarrow \quad \sqrt{3}=\frac{50}{x}$

$\Rightarrow \quad 50=\sqrt{3} \times h \sqrt{3}$

$\Rightarrow \quad h=\frac{50}{3}$

Now the distance between the temples

$x=h \sqrt{3}$

$=\frac{50}{3} \times \sqrt{3}$

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

Therefore $H=50-\frac{50}{3}$

$\Rightarrow \quad H=33.33$

Hence distance between the temples is $\frac{50}{\sqrt{3}} \mathrm{~m}=28.83 \mathrm{~m} \mathrm{~m}$ and height of temple is $33.33 \mathrm{~m}$.