Question

The angles of depression of the top and bottom of a tower as seen from the top of a $60\sqrt{3}$-m-high cliff are ${45}^{\circ }and{60}^{\circ }$ respectively. Find the height of the tower.

Answer 1

Let the height of the tower be AB = h

Hence DE = h

Given height of the cliff, $CD=60\sqrt{3}$ m

Hence $CE=60\sqrt{3}–h$

In right triangle AEC,

$tan{45}^o=\frac{CE}{EA}$

$1=\frac{(60\sqrt{3}–h)}{EA}$

$EA=60\sqrt{3}–h$ ----(1)

In right triangle CDB

$tan{60}^o=\frac{CD}{DB}$

$\sqrt{3}=\frac{60\sqrt{3}}{DB}$

Hence DB = 60 m

That is EA = 60 m

Equation (1) becomes,

$60=60\sqrt{3}–h$

$h=60\sqrt{3}–60=60(\sqrt{3}–1)$ m

Thus the height of the tower is $60(\sqrt{3}–1)$ m