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

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 $∠ PAO = ∠ PBO = 15 °$ and $∠ PCO = 45 °$, where $C$ is the midpoint of $AB$. Then ${\left(OP\right)}^{2} is equal to</p><p>Now plotting the graph of given value we get,</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/759e04d9-7653-4313-87ed-d37f1eb52200-image.png" style="width: 278px; height: 250px;" /></p><p>Now in$△ PAO${given ∠\text{ PAO } = 15 °}$

So, $\dfrac{OP}{OA} = \tan 15 °$ $⇒ OA = OPcot15 °$

And in $△ PCO$ {given $∠ PCO = 45 °$ }

So, $\dfrac{OP}{OC} = \tan 45 ° ⇒ OP = OC$

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

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

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

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

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

⇒ O{P}^{2} = \dfrac{32}{\sqrt{3}}\left(2 - \sqrt{3}\right)