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 ${30}^{\circ }$ to ${45}^{\circ }$ in 12 minutes, find the time taken by the car now to reach the tower.

Answer 1

S

Let AB be the vertical tower. Suppose D and C be the positions of the car when the angle of depression from the top of the tower is 30° and 45° respectively.

Consider the uniform speed of car be V m/min.

Time taken for the angle of depression to change from 30° to 45° = 12 min

∠EAD = ∠ADB = 30° (Alternate angles)

∠EAC = ∠ACB = 45° (Alternate angles)

Suppose AB = h m and BC = x m.

CD = Distance covered by car in $12min=Vm/min\times 12=12Vmin$

In ΔABC,

$tan{45}^o=\frac{AB}{BC}1=\frac{h}{x}h=x----\left(1\right)$

In ΔADB,

$tan{30}^o=\frac{AB}{CD+BC}\frac{1}{\sqrt{3}}=\frac{h}{12V+x}\Rightarrow \sqrt{3}h=12V+x\Rightarrow \sqrt{3}x=12V+x\left[\text{Using eqn (1)}\right]\Rightarrow (\sqrt{3}-1)x=12V\Rightarrow \frac{x}{V}=\frac{12}{(\sqrt{3}-1)}min\Rightarrow \frac{x}{V}=\frac{12(\sqrt{3}-1)}{(\sqrt{3}-1)(\sqrt{3}+1)}min\Rightarrow \frac{x}{V}=\frac{12(\sqrt{3}-1)}{3-1}min\Rightarrow \frac{x}{V}=6(\sqrt{3}-1)min\Rightarrow \frac{x}{V}=983sec$

Time taken by car to reach the tower from $C=\frac{x}{V}$

∴ Time taken by car to reach the tower is 983 sec.