Question

A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes $18$ min, for the angle of depression of the car to change from ${30}^o$ to ${45}^o$; then after this, the time taken(in min) by the car to reach the foot of the tower, is?

• A. $9(1+\sqrt{3})$
• B. $\frac{9}{2}(1\sqrt{3}-1)$
• C. $18(1+\sqrt{3})$
• D. $18(\sqrt{3}-1)$

Answer 1

The correct option is A $9(1+\sqrt{3})$

Given, $\angle HOA={30}^{\circ }$ and $\angle HOB={45}^{\circ }$. Let $OH=h$, then

$tan\left(\angle HOA\right)=\frac{HA}{h}$ $or$ $HA=h\times tan{30}^{\circ }$

$tan\left(\angle HOB\right)=\frac{HB}{h}$ $or$ $HB=h\times tan{45}^{\circ }$

It takes 18 mins to travel distance AB, hence speed $=\frac{HB-HA}{18}=\frac{h-\left(h/\sqrt{3}\right)}{18}$

Time taken to travel HA = $\frac{HA}{speed}=\frac{\frac{h}{\sqrt{3}}}{\frac{(h-\frac{h}{\sqrt{3}})}{18}}$

= $\frac{18\times \sqrt{3}}{\sqrt{3}(\sqrt{3}-1)}$

= $9(\sqrt{3}+1)$

Option A is correct.