Question

There are two temples one on each bank of a river just opposite to each other. One temple is 54 m high From the top of this temple, the angles 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 in m.

Answer 1

Let AB and CD be the two temples and AC be the river
Then AB=54 m
Let AC = X meters and CD = h meters.
$\angle ACB={60}^0,\angle EDB={30}^0$
$\frac{AB}{AC}=\tan {60}^0=\sqrt{3}$
$\Rightarrow AC=\frac{AB}{\sqrt{3}}=\frac{54}{\sqrt{3}}=(\frac{54}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}})=18\sqrt{3}$
$DE=AC=18\sqrt{3}m$
$\frac{BE}{DE}=\tan {30}^0=\frac{1}{\sqrt{3}}$
$\Rightarrow BE=(18\sqrt{3}\times \frac{1}{\sqrt{3}})=18m$
$\therefore CD=AE=AB-BE=(54-18)m=36m$
So Width of the river $=AC=18\sqrt{3}m$
$=(18\times 1.73)m=31.14m$
Height of the other temple $=CD=18m$