In the figure, AB is the tower and car is at the point C. Join A and B to C. So we get a right-angled triangle ABC with right angle at B. Also height of the tower is 150 m. So, AB = 150 m. Again, draw a line AD parallel to BC. The angle of depression of the car from the top of AB is 30°. So, ∠DAC = ∠ACB = 30°. We have to find the distance of the car from the tower, that is BC. For this, we will use the trigonometric ratio tan in ∆ABC.

So from ∆ABC,

or,

So, the correct option is (B).