question_answer

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

A)  \[1000\sqrt{3}m\]

B)  \[\frac{2000}{\sqrt{3}}m\]

C)  \[\frac{1000}{\sqrt{3}}m\]               

D)  \[\frac{2000}{\sqrt{2}}m\]

Correct Answer: B

Solution :

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