Question

The angle of elevation of the top of a building from the foot of the tower is ${30}^0$ and the angle of elevation of the top of the tower from the foot of the building is ${60}^0$ . If the tower is $50m$ high, find the height of the building .

Answer 1

Given height of tower $CD=50$m

Let the height of the building,$AB=h$m

In right angled $△BDC$,

$\tan {60}^{\circ }=\frac{CD}{BD}$

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

$\Rightarrow BD=\frac{50}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{50\sqrt{3}}{3}=28.86$m

$\therefore h=28.86$m

Thus the height of the building is $28.86$m