Question

The angle of elevation of the top of a building from the foot of the tower is ${30}^{\circ }$ and the angle of elevation of the top of the tower from the foot of the building is ${60}^{\circ }$. If the tower is $50\text{m}$ high, find the height of the building.

Answer 1

Let $AB$ be the tower and $DC$ be the building.

In $△ABC$

$\tan {60}^o=\frac{AB}{BC}$

$BC=\frac{50}{\sqrt{3}}\text{m}$

In $△CBD$,

$\frac{CD}{BC}=\tan {30}^o$

$CD=\frac{50}{\sqrt{3}}\times \frac{1}{\sqrt{3}}=\frac{50}{3}\text{m}$