Question

The base of a cliff is circular. From the extremities of a diameter of the base angles of elevation of the top of the cliff are $30^{\circ} $ and $ 60^{\circ}$ . If the height of the cliff be $500\, m$, then the diameter of the base of the cliff is

• A. $ \frac{2000}{\sqrt{3}}m $
• B. $ \frac{1000}{\sqrt{3}}m $
• C. $\frac{2000}{\sqrt{2}} m$
• D. $ 1000\sqrt{3}\,m $

Answer 1

Answer: (A)

In $\Delta AEC$
$\tan 60^{\circ}=\frac{500}{d_{1}}$
$\Rightarrow d_{1}=\frac{500}{\sqrt{3}} m$ ... (i)

and in $\Delta B E C, $
$\tan 30^{\circ}=\frac{500}{d_{2}}$
$\Rightarrow d_{2}=500 \sqrt{3} \,m$ ... (ii)
$\therefore $ Required diameter, $A B=d_{1}+d_{2}$
$=\frac{500}{\sqrt{3}}+500 \sqrt{3}=\frac{2000}{\sqrt{3}} \,m$