Question

A man from top of a 100 meters high towers sees a car moving towards the tower at an angle of depression of ${30}^0$. After some time, the angle of depression becomes ${60}^0$. The distance (in meters) travelled by the car during this time is

• A. $100\sqrt{3}$
• B. $\frac{200\sqrt{3}}{3}$
• C. $\frac{100\sqrt{3}}{3}$
• D. $200\sqrt{3}$

Answer 1

The correct option is B $\frac{200\sqrt{3}}{3}$

Let the distance traveled by x

In $\Delta ADE\Rightarrow tan{30}^o=\frac{ED}{AE}$

$\Rightarrow AE=ED\sqrt{3}=100\sqrt{3}$

In $\Delta AFC\Rightarrow tan{60}^o=\frac{CF}{AF}$

$\Rightarrow AF=\frac{100}{\sqrt{3}}$

$x=BD-BC$

$=AE-AF$

$=100\left(\frac{3-1}{\sqrt{3}}\right)=\frac{200}{\sqrt{3}}$

$x=\frac{200}{3}\sqrt{3}$