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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 $Q R = 15 m$ . If from a point A on the ground the angle of elevation of R is $60^{0}$ and the part PR of the tower subtends an angle of $15^{0}$ at A then height of the tower is

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

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

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

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

answer is A.

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Detailed Solution

Let $P R = x$

$\tan 60^{0} = \frac{15}{A Q}$ ………… (1)

$\tan 75^{0} = \frac{x + 15}{A Q}$

On dividing (i) by (ii)

$\frac{\sqrt{3}}{2 + \sqrt{3}} = \frac{15}{x + 15} \Rightarrow x + 15 = \frac{15 \times (2 + \sqrt{3})}{\sqrt{3}} \Rightarrow x = 10 \sqrt{3}$

$P Q = 15 + 10 \sqrt{3} = 5 (3 + 2 \sqrt{3})$

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