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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 pole along the line BC to a point D such that CD = 7 m. From D the angle of elevation of the point A is \[{{45}^{o}}\]. Then the height of the pole is       AIEEE  Solved  Paper-2007

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

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

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

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

Correct Answer: D

Solution :

                \[\tan {{60}^{o}}=\frac{h}{BC}\]                                ?. (i) and \[\tan {{45}^{o}}=\frac{h}{7+BC}\]       ?.           (ii) tan 45 ?(ii)                 \[\Rightarrow 7+BC=h\Rightarrow BC=h-7\] From(i)\[\sqrt{3}=\frac{h}{h-7}\Rightarrow \sqrt{3}h-7\sqrt{3}=h\Rightarrow \sqrt{3}h-h=7\sqrt{3}\] \[h=\frac{7\sqrt{3}}{\sqrt{3}-1}\Rightarrow \frac{7\sqrt{3}\left( \sqrt{3}+1 \right)}{2}m\]