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AB is a vertical pole with B at the ground level and A at the top. A man finds that the angle of elevation of the point A from a certain point C on the ground is 60. He moves away from the role along the line BC to a point D such that CD = 7m from D, the angle of elevation from point A is \[{{45}^{o}}\], then the height of the pole is

A)  \[\frac{7\sqrt{3}}{2}\left( \frac{1}{\sqrt{3}+1} \right)m\]            

B)  \[\frac{7\sqrt{3}}{2}\left( \frac{1}{\sqrt{3}-1} \right)m\]

C)  \[\frac{7\sqrt{3}}{2}(\sqrt{3}+1)m\]      

D)  \[\frac{7\sqrt{3}}{2}(\sqrt{3}-1)m\]

Correct Answer: C

Solution :

In \[\Delta ABC,\,\,\frac{h}{x}\,=\tan {{60}^{0}}\,\Rightarrow \,h=\sqrt{3}x\]  ?(1) In \[\Delta ABD,\,\frac{h}{x+7}\,=\tan {{45}^{0}}\,\Rightarrow \,h=x+7\] ?(2) \[\therefore \] From (1) and (2), we get \[h=\frac{7\sqrt{3}}{\sqrt{3}-1}\,=\frac{7\sqrt{3}}{2}\,(\sqrt{3}+1)m\]