Question

A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it If it takes $12$ minutes for the angle of depression to change from ${30}^0$ to ${45}^0$ how soon after this will the car reach the observation tower?

• A. $14$ min. $35$ sec.
• B. 15 min. 49 sec.
• C. 16 min. 23 sec.
• D. 18 min. 5 sec.

Answer 1

Answer: (C)

16 min. 23 sec.

Let AB be the tower and C and D be the point of positions of two cars

Given $\angle ACB={45}^0$$\angle ADB={30}^0$

Let AB=h ,CD=x and AC=y

In $\Delta ABC$

$tan{45}^0=\frac{AB}{AD}$

$\Rightarrow 1=\frac{AB}{AD}$

$\Rightarrow AB=AD\Rightarrow y=h$

In $ABD$

$tan{30}^0=\frac{AB}{AD}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+y}$

$\Rightarrow x+y=\sqrt{3}h$

$\therefore x=(x+y)-y=\left(\sqrt{3}\right)h-h=h(\sqrt{3}-1)$

So car cover distance $h(\sqrt{3}-1)$ in 12 seconds

Then car cover distance h in=$\left[\frac{12\times h}{h(\sqrt{3}-1)}\right]=\frac{12}{0.73}=16min23sec$