Question

From the top of a cliff, 60 metres high, the angles of depression of the top and bottom of a tower are observed to be ${30}^{\circ }and{60}^{\circ }$. Find the height of the tower.

Answer 1

Let the tower be AB and cliff be CD =60 m.
Join AC&CB.
Let E be a point on CD such that BDEA is a rectangle

$\angle ACE={60}^{0}and\angle BCD={30}^0$

$\tan {30}^0=\frac{BD}{CD}=\frac{BD}{CE+ED}CE=\frac{AE}{\tan {60}^0}so\tan {30}^0=\frac{BD}{\frac{AE}{\tan {60}^0}+AB}\to \frac{AB}{\sqrt{3}}=\frac{2}{3}BD\to AB=\frac{2}{\sqrt{3}}60AB=40\sqrt{3}$

So height of the tower = 40$\sqrt{3}$