A horizontal park is in the shape of a triangle OAB with AB=16.

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:

Accepted Answer

$$OP = OA\tan 15 = OB\tan 15$$ ...... (i)

$$OP = OC\tan 45 \Rightarrow OP = OC$$ ...... (ii)

$OA = OB$ ....... (iii)

$O{C^2} + {8^2} = O{A^2}$

$$O{P^2} + 64 = O{P^2}{\left( {{{\sqrt 3  + 1} \over {\sqrt 3  - 1}}} \right)^2}$$

$$64 = O{P^2}\left[ {{{{{\left( {\sqrt 3  + 1} \right)}^2} - {{\left( {\sqrt {3 - 1} } \right)}^2}} \over {{{\left( {\sqrt 3  - 1} \right)}^2}}}} \right]$$

$$ = O{P^2}\left( {{{4\sqrt 3 } \over {{{\left( {\sqrt 3  - 1} \right)}^2}}}} \right)$$

$$O{P^2} = {{64{{\left( {\sqrt 3  - 1} \right)}^2}} \over {4\sqrt 3 }} = {{32} \over {\sqrt 3 }}\left( {2 - \sqrt 3 } \right)$$

Solutions · 1