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A vertical tower stands on a declivity which is inclined at 15° to the horizon from the foot of the tower a man ascends the declivity from 80 feet and then finds that the tower subtends an angle of 30°. The height of the tower is

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$20 (\sqrt{6} - \sqrt{2})$

$10 (\sqrt{6} - \sqrt{2})$

$40 (\sqrt{6} - \sqrt{2})$

$40 (\sqrt{6} + \sqrt{2})$

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{BC}{\sin 75^{\circ}} = \frac{AB}{\sin 30^{\circ}} \\ 2 A B = B C (\frac{2 \sqrt{2}}{\sqrt{3} + 1}) \\ A B = 80 (\frac{\sqrt{2} (\sqrt{3} - 1)}{(\sqrt{3} + 1) (\sqrt{3} - 1)} \\ AB = 40 (\sqrt{6} - \sqrt{2}) \end{array}$

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