Question

A man observes a car from the top of a tower, which is moving towards the tower with a uniform speed. If the angle of depression of the car changes from 30° to 45° in 12 minutes, find the time taken by the car now to reach the tower. [CBSE 2017]

Answer 1

Suppose AB be the tower of height h meters. Let C be the initial position of the car and let after 12 minutes the car be at D. It is given that the angles of depression at C and D are 30º and 45º respectively.
Let the speed of the car be v meter per minute. Then,
CD = distance travelled by the car in 12 minutes
CD = 12v meters



Suppose the car takes t minutes to reach the tower AB from D. Then DA = vt meters.
$$\begin{gathered} \mathrm{In}∆\mathrm{DAB},\mathrm{we}\mathrm{have} \\ \tan 45°=\frac{\mathrm{AB}}{\mathrm{AD}} \\ \Rightarrow 1=\frac{h}{vt} \\ \Rightarrow h=vt.....\left(i\right) \\ \mathrm{In}∆\mathrm{CAB},\mathrm{we}\mathrm{have} \\ \tan 30°=\frac{\mathrm{AB}}{\mathrm{AC}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{vt+12v} \\ \Rightarrow \sqrt{3}h=vt+12v.....\left(ii\right) \end{gathered}$$
Substituting the value of h from equation (i) in equation (ii), we get
$$\begin{gathered} \Rightarrow \sqrt{3}t=t+12 \\ \Rightarrow t=\frac{12}{\sqrt{3}-1}=\frac{12}{\left(\sqrt{3}-1\right)}\times \frac{\left(\sqrt{3}+1\right)}{\left(\sqrt{3}+1\right)}=6\left(\sqrt{3}+1\right)\min \end{gathered}$$