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

Let AB is a tower, car is at point D at ${30}^{\circ }$ and goes to C at ${45}^{\circ }$ in 12 minutes.

In $\Delta ABC,$
$\frac{AB}{BC}=tan{45}^{\circ }$
$\Rightarrow$ $\frac{h}{x}=1\Rightarrow h=x$ ...(i)
In $\Delta ABD,$
$\frac{AB}{BD}=tan{30}^{\circ }$
$\Rightarrow$ $\frac{h}{x+y}=\frac{1}{\sqrt{3}}$ $\Rightarrow h=\frac{x+y}{\sqrt{3}}$ ...(ii)
Comparing eq. (i) & (ii), we get
$x=\frac{x+y}{\sqrt{3}}\Rightarrow \sqrt{3}x=x+y$
$\Rightarrow (\sqrt{3}-1)x=y$
Car covers the distance y in time = 12 min
So $(\sqrt{3}-1)x$ distance covers in 12 min
Distance x covers in time $=\frac{12}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$=\frac{12(\sqrt{3}+1)}{3-1=2}$
$=6(\sqrt{3}+1)min$
$=6\times 2.732=16.9$
Now car reaches to tower in 16.39 minutes.