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$A B$ 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 $B C$ to a point $D$ such that $C D = 7 m$ . From $D$ the angle of elevation of the point $A$ is $45 ° .$ Then the height of the pole is

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$\frac{7 \sqrt{3}}{2} (\sqrt{3} + 1) m$

$\frac{7 \sqrt{3}}{2} (\sqrt{3} - 1) m$

$\frac{7 \sqrt{3}}{2} \frac{1}{\sqrt{3} + 1} m$

$\frac{7 \sqrt{3}}{2} \frac{1}{\sqrt{3} - 1} m$

answer is A.

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Detailed Solution

From the traingle $A B D$ , since the measure of angle $C$ is $45 °$ the length of the pole is equal to $(x + 7) m$

Hence, from the triangle $A B C$

$\begin{array}{rcl} \tan 60 & = & \frac{x + 7}{x} \\ x (\sqrt{3}) & = & x + 7 \\ x (\sqrt{3} - 1) & = & 7 \\ x & = & \frac{7}{\sqrt{3} - 1} \\ = & \frac{7 (\sqrt{3} + 1)}{2} \end{array}$

Therefore, the length of the pole is

$\begin{array}{rcl} x + 7 & = & \frac{7}{2} (\sqrt{3} + 1) + 7 \\ = & \frac{7 \sqrt{3}}{2} (\sqrt{3} + 1) \end{array}$

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