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A tower PQ stands on a horizontal ground with base Q on the ground. The point R divides the tower in two parts such that QR = 15 m. If from a point A on the ground the angle of elevation of R is 60° and the part PR of the tower subtends an angle of 15° at A, then the height of the tower is :

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$5 (\sqrt{3} + 3) m$

$10 (2 \sqrt{3} + 1) m$

$5 (2 \sqrt{3} + 3) m$

$10 (\sqrt{3} + 1) m$

answer is A.

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Detailed Solution

$\frac{15}{AQ} = \tan 60^{\circ}$ ......(1)

$\frac{15 + x}{A Q} = \tan 75^{\circ}$ .....(2)

$\frac{(1)}{(2)} \Rightarrow \frac{15}{15 + x} = \frac{\sqrt{3}}{2 + \sqrt{3}} \Rightarrow 30 + 15 \sqrt{3} = 15 \sqrt{3} + x \sqrt{3} x = 10 \sqrt{3} P Q = 15 + X = 15 + 10 \sqrt{3}$

So, $PQ = 5 (2 \sqrt{3} + 3) m$

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