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 \[{{30}^{o}}\]and\[{{60}^{o}}\]. 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) \[\frac{2000}{\sqrt{3}}m\]          

D)         \[1000\sqrt{3}\,m\]

Correct Answer: A

Solution :

In\[\Delta AEC\], \[\tan {{60}^{o}}=\frac{500}{{{d}_{1}}}\] \[{{d}_{1}}=\frac{500}{\sqrt{3}}m\]                                                         ... (i) and in\[\Delta BEC\],                 \[\tan {{30}^{o}}=\frac{500}{{{d}_{2}}}\] \[\Rightarrow \]               \[{{d}_{2}}=500\sqrt{3}m\]                                          ? (ii) \[\therefore \]Required diameter,                 \[AB={{d}_{1}}+{{d}_{2}}\]                 \[=\frac{500}{\sqrt{3}}+500\sqrt{3}=\frac{2000}{\sqrt{3}}m\]