The angle of depression of a car parked on the road from the top of a 150 m high tower is 30°. The distance of the car from the tower (in metres) is

Answer: Option B

Let AB be the tower of height 150 m
C is car and angle of depression is 30°

Therefore, ∠ACB = 30° (alternate angle)
In right - angled triangle ABC,
$$\begin{array}{rl}& \frac{BC}{AB}=\cot 30^{\circ }\\ & \Rightarrow \frac{BC}{150}=\sqrt{3}\\ & \Rightarrow BC=150\sqrt{3}m\end{array}$$
That is, distance of the car from the tower is $$150\sqrt{3}m$$