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Mathematics Question on Some Applications of Trigonometry

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=15∘ and ∠PCO=45∘, \begin{array}{l}\angle PAO = \angle PBO = 15^\circ ~\text{and}~ \angle PCO = 45^\circ,\end{array}where C is the midpoint of AB. Then (OP)2 is equal to

A

 323(3−1)\begin{array}{l} \ \frac{32}{\sqrt{3}}\left(\sqrt{3}-1\right) \end{array}

B

 323(2−3)\begin{array}{l}\ \frac{32}{\sqrt{3}}\left(2-\sqrt{3}\right) \end{array}

C

163(3−1)\begin{array}{l} \frac{16}{\sqrt{3}}\left(\sqrt{3}-1\right) \end{array}

D

163(2−3)\begin{array}{l}\frac{16}{\sqrt{3}}\left(2-\sqrt{3}\right) \end{array}

Answer

 323(2−3)\begin{array}{l}\ \frac{32}{\sqrt{3}}\left(2-\sqrt{3}\right) \end{array}

Explanation

Solution

Triangle OAB

From the above figure:

OP=OA tan 15=OB tan 15⋯(i) OP=OC tan 45⇒OP=OC⋯(ii)\begin{array}{l}OP = OA ~tan~15 = OB ~tan~ 15 \cdots \left(i\right)\\\ OP = OC ~tan~ 45 \Rightarrow OP = OC \cdots \left(ii\right) \end{array}

OA=OB⋯(iii) OC2+82=OA2\begin{array}{l}OA = OB \cdots \left(iii\right) \\\ OC^2 + 8^2 = OA^2\end{array}

OP2+64=OP2(3+13−1)2\begin{array}{l} OP^2+64=OP^2\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)^2\end{array}

64=OP2[(3+1)2−(3−1)2(3−1)2]\begin{array}{l} 64=OP^2\left[\frac{\left(\sqrt{3}+1\right)^2-\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}-1\right)^2}\right] \end{array}

=OP2(43(3−1)2)\begin{array}{l} =OP^2\left(\frac{4\sqrt{3}}{\left(\sqrt{3}-1\right)^2}\right)\end{array}

OP2=64(3−1)243=323(2−3)\begin{array}{l} OP^2=\frac{64\left(\sqrt{3}-1\right)^2}{4\sqrt{3}}=\frac{32}{\sqrt{3}}\left(2-\sqrt{3}\right)\end{array}