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

Solution (By Examveda Team)

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,
$$\eqalign{ & \frac{{BC}}{{AB}} = \cot {30^ \circ } \cr & \Rightarrow \frac{{BC}}{{150}} = \sqrt 3 \cr & \Rightarrow BC = 150\sqrt 3 \,m \cr} $$
That is, distance of the car from the tower is $$150\sqrt 3 \,m$$