Question

From top of a cliff 60m high ,angle of depression of top & bottom of a tower are observed as ${30}^{\circ }$&${60}^{\circ }$ degrees.Find height of the tower.

Answer 1

Let BC be the building and AD be the tower.
Let the height of tower, AD be hm.
Angles of depression of the top D and the bottom A of the tower CB are ${30}^{\circ }and{60}^{\circ }$ respectively.
$\therefore \angle CDE={30}^{\circ }$
$\angle CAB={60}^{\circ }$
Since, BC = 60 m.
$\therefore$ CE = (60 - h)m
Let AB = DE = xm
In $\Delta$DEC,
$tan{30}^{\circ }=\frac{CE}{DE}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{60-h}{x}$
$\Rightarrow x=\sqrt{3}(60-h)\cdots \left(1\right)$
$In\Delta CBA,$
$tan{60}^{\circ }=\frac{BC}{AB}$
$\Rightarrow \sqrt{3}=\frac{60}{x}$
$\Rightarrow x=\frac{60}{\sqrt{3}}\cdots \left(2\right)$
Equating equation (1) and (2),
$\frac{60}{\sqrt{3}}=\sqrt{3}(60-h)$
$\Rightarrow 3(60-h)=60$
$\Rightarrow 180-3h=60$
$\Rightarrow 180-60=3n$
$\Rightarrow 120=3h$
$\Rightarrow h=\frac{120}{3}$
$\Rightarrow h=40$
Thus, the height of the tower is 40m.