Question

A bridge across a valley is $800$ metres long. There is a temple in the valley directly below the bridge. The angle of depression of the top of the temple from the two ends of the bridge have measure 30${}^o$ and 60${}^o$.
Find the height of the bridge above the top of the temple.

Answer 1

Let $h$ be the height of the bridge above the top of the temple.
In the above figure, $BC$ represents the length of the bridge.
$AD=h$
Let $CD=x,BD=800-x$
In $\Delta ADC$,
$\tan {60}^o=\frac{AD}{CD}$
$\therefore \tan {60}^o=\frac{h}{x}$
$\therefore \sqrt{3}=\frac{h}{x}$
$\therefore x=\frac{h}{\sqrt{3}}$ ......... $\left(i\right)$
In $\Delta ADB$,
$\tan {30}^o=\frac{AD}{BD}$
$\therefore \tan {30}^o=\frac{h}{800-x}$
$\therefore \frac{1}{\sqrt{3}}=\frac{h}{800-x}$
$\therefore 800-x=h\sqrt{3}$
$\therefore x=800-h\sqrt{3}$ ....... $\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$,
$\therefore \frac{h}{\sqrt{3}}=800-h\sqrt{3}$
$\therefore h=800\sqrt{3}-3h$
$\therefore 4h=800\sqrt{3}$
$\therefore h=200\sqrt{3}$
$\therefore$ Height of the bridge above the temple $=200\sqrt{3}m$