Question

The angle of depression of a car parked on the road from the top of a $150m$ high tower is ${30}^o$. The distance of the car from the tower (in meters) is

• A. $50\sqrt{3}$
• B. $150\sqrt{3}$
• C. $150\sqrt{2}$
• D. $75$

Answer 1

Answer: (B)

$150\sqrt{3}$

Given: Height of Tower $=150m$
Angle of depression = ${30}^{\circ }$
$\angle DCB={30}^{\circ }$
$\angle DCB=\angle ABC$ (Alternate angles)
Let, distance of car from tower AB be x metres
In $△ABC$,
$\tan {30}^{\circ }=\frac{AC}{AB}$

$\frac{1}{\sqrt{3}}=\frac{150}{x}$
$x=150\sqrt{3}m$