Question

A tower subtends and angle 60 at a point A in the plane of its base and the angle of depression of the foot of the tower at a point 10 meters just above A is 30. The height of the tower is

• A. $\frac{10}{\sqrt{3}}meters$
• B. 10 meters
• C. $10\sqrt{3}$
• D. 30 meters

Answer 1

The correct option is D 30 meters

REF.Image

Let DC be the height of the tower and AB = 10 cm.

In $\Delta BAC$,

$tan{30}^{\circ }=\frac{BA}{CA}$

$\frac{1}{\sqrt{3}}=\frac{10}{CA}$

$\Rightarrow CA=10\sqrt{3}m$ _________ (1)

In $\Delta ACD$,

$tan{60}^{\circ }=\frac{DC}{CA}$

$\sqrt{3}=\frac{DC}{10\sqrt{3}}$ [ from (1)]

$\Rightarrow DC=10\sqrt{3}.\sqrt{3}$

= 10.3

= 30 m.

Hence height of the tower = 30 m.