Question

A man from the top of a 50 metre high tower sees a car moving towards a tower at an angle of depression $30°$. After sometime, the angle of depression is $60°$. Then the distance travelled by the car during the time is

• A. $\frac{200}{\sqrt{3}}$ m
• B. $\frac{100}{\sqrt{3}}$ m
• C. $\frac{50}{\sqrt{3}}$ m
• D. $\frac{150}{\sqrt{3}}$ m

Answer 1

Answer: (B)

In $\Delta ABC,$ we have $\tan 30°=\frac{50}{x+y}\frac{1}{\sqrt{3}}=\frac{50}{x+y}$...(i)
In $\Delta ABD,$ we have $\tan 30°=\frac{50}{x+y}$
$\Rightarrow \sqrt{3}=\frac{50}{y}\Rightarrow y=\frac{50}{\sqrt{3}}\Rightarrow \frac{1}{\sqrt{3}}=\frac{50}{x+\frac{50}{\sqrt{3}}}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{50\sqrt{3}}{\sqrt{3}x+50}\Rightarrow \sqrt{3}x+50=50\times 3$
$\Rightarrow x=\frac{100}{\sqrt{3}}m$