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 12 min . for the angle of depression of the car to charge from $30$ degree to $45$ degree. 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. $18(1+\sqrt{3})$
• C. $18(\sqrt{3}-1)$
• D. $6(\sqrt{3}+1)$

Answer 1

The correct option is D $6(\sqrt{3}+1)$

Let $AB=h$ be the height of the tower.

$\angle AQB={45}^{\circ }$

Now, in $△ABQ,\tan {45}^{\circ }=\frac{AB}{BQ}$

$\Rightarrow 1=\frac{h}{BQ}$

$\Rightarrow BQ=h$

In $△APB,\tan {30}^{\circ }=\frac{AB}{PB}$

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

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

$\Rightarrow x=h(\sqrt{3}-1)$

Thus,speed$=\frac{h(\sqrt{3}-1)}{12}m/min$

Time for remaining distance,

$\frac{\frac{h}{h(\sqrt{3}-1)}}{12}=\frac{12(\sqrt{3}+1)}{3-1}$

$=\frac{12}{2}(\sqrt{3}+1)$

$=6(\sqrt{3}+1)$

Time$=6(\sqrt{3}+1)$ minutes