JEE Main 2022 — Heights and Distances Question with Solution

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

Now plotting the graph of given value we get,

Now in $△PAO$ {given $\angle PAO=15°$}

So, $\frac{OP}{OA}=\tan 15°$  $\Rightarrow OA=OP\cot 15°$

And in $△PCO$ {given $\angle PCO=45°$ }

So, $\frac{OP}{OC}=\tan 45°\Rightarrow OP=OC$

Now, $OP=\sqrt{O{A}^2-8^2}$

$\Rightarrow O{P}^2={\left(OP\right)}^2{\cot }^{2}15°-64$

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

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

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

$\Rightarrow O{P}^2=\frac{32}{\sqrt{3}}\left(2-\sqrt{3}\right)$