Question

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

• A. \(\begin{array}{l} \ \frac{32}{\sqrt{3}}\left(\sqrt{3}-1\right) \end{array}\)
• B. \(\begin{array}{l}\ \frac{32}{\sqrt{3}}\left(2-\sqrt{3}\right) \end{array}\)
• C. \(\begin{array}{l}  \frac{16}{\sqrt{3}}\left(\sqrt{3}-1\right) \end{array}\)
• D. \(\begin{array}{l}\frac{16}{\sqrt{3}}\left(2-\sqrt{3}\right) \end{array}\)

Answer 1

The Correct Option is B

From the above figure:

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

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

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

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

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

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