There are two temples, one on each bank of a river, just opposite to each other. One temple is 50 m high.

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 $33.33 \mathrm{~m}$.