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

• A. $\frac{32}{\sqrt{3}}(\sqrt{3}-1)$
• B. $\frac{32}{\sqrt{3}}(2-\sqrt{3})$
• C. $\frac{16}{\sqrt{3}}(\sqrt{3}-1)$
• D. $\frac{16}{\sqrt{3}}(2-\sqrt{3})$

Answer 1

Answer: (B)

$\frac{OP}{OA}=\tan 15^{\circ }$
$\Rightarrow OA=OP\cot 15^{\circ }$
$\frac{OP}{OC}=\tan 45^{\circ }\Rightarrow OP=OC$
$\text{ Now, }OP=\sqrt{O{A}^2-8^2}$
$\Rightarrow O{P}^2=(OP{)}^2{\cot }^{2}15^{\circ }-64$
$\Rightarrow O{P}^2=\frac{32}{\sqrt{3}}(2-\sqrt{3})$