A horizontal park is in the shape of a triangle OAB with AB = 16. A vertical lamp post OP is erected at the point O such that $∠$ PAO = $∠$ PBO = 15⁰and $∠$ PCO= 45⁰, where C is the midpoint of AB. Then (OP)² is equal to

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$\frac{16}{\sqrt{3}} (2 - \sqrt{3})$

$\frac{32}{\sqrt{3}} (2 - \sqrt{3})$

$\frac{16}{\sqrt{3}} (\sqrt{3} - 1)$

$\frac{32}{\sqrt{3}} (\sqrt{3} - 1)$

answer is B.

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Detailed Solution

$I n ∆ O A P, \frac{O P}{O A} = \tan 15^{0} \Rightarrow O A = O P c o t 15^{0} I n ∆ O C P \frac{O P}{O C} = \tan 45^{0} \Rightarrow O P = O C N o w, O P = \sqrt{O A^{2} - 8^{2}} \Rightarrow O P^{2} = {(O P)}^{2} c o t^{2} 15^{0} - 64 \Rightarrow O P^{2} = \frac{64}{c o t^{2} 15^{0} - 1} = \frac{64}{{(2 + \sqrt{3})}^{2} - 1} = \frac{64}{6 + 4 \sqrt{3}} = \frac{32}{\sqrt{3} (2 + \sqrt{3})} \Rightarrow O P^{2} = \frac{32}{\sqrt{3}} (2 - \sqrt{3})$