question_answer

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 \[\text{3}0{}^\circ \] and\[\text{6}0{}^\circ \]. If the height of the cliff be 500 m, then 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) \[1000\sqrt{3}m\]

Correct Answer: A

Solution :

In \[\Delta AEC,\] \[\tan {{60}^{o}}=\frac{500}{{{d}_{1}}}\] \[\Rightarrow \]               \[\sqrt{3}=\frac{500}{{{d}_{1}}}\] \[\Rightarrow \]               \[{{d}_{1}}=\frac{500}{\sqrt{3}}m\] and in \[\Delta BEC\], \[\tan {{30}^{o}}=\frac{500}{{{d}^{2}}}\]                 \[\Rightarrow \]               \[{{d}_{2}}=500\sqrt{3}m\]                 \[\therefore \] Required diameter \[={{d}_{1}}+{{d}_{2}}\] \[=\frac{500}{\sqrt{3}}+500\sqrt{3}=\frac{200}{\sqrt{3}}m\]