Question

The angle of depression of a car parked on the road from the top of a 150-m-high tower is ${30}^{\circ }$. The distance of the car from the tower is

(a) $50\sqrt{3}$ m

(b) $150\sqrt{3}$ m

(c) $150\sqrt{2}$ m

(d) 75 m

Answer 1

Answer:
We have a high tower whose height (AC)= 150m
Angle of depression = ${30}^{\circ }$

Here given angle of depression $\angle DCB={30}^{º}$​
We know ,
$\angle DCB=\angle ABC$ (Alternate angles)
Let ,
Distance of the car from tower AB = x m
In $△ABC$ we know that
$tan{30}^{\circ }=\frac{AC}{AB}$
$\frac{1}{\sqrt{3}}=150\times x$
⇒ $x=150\sqrt{3}$ m

(Ans) (b) is the correct option.