Question

A horizontal park is in the shape of a triangle $O A B$ with $A B=16$. A vertical lamp post $O P$ 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{ OA ^2-8^2} $
$\Rightarrow OP ^2=( OP )^2 \cot ^2 15^{\circ}-64 $
$ \Rightarrow OP ^2=\frac{32}{\sqrt{3}}(2-\sqrt{3})$