Each side of a square subtends an angle of 60 ^ o at the top of a...

Let $ABCD$ be a square of each side of length $a$ meters.

It is given that $∠BPC = {60}^{o} .$
Let $M$ be the midpoint of $BC .$
Then, $∠BPM = ∠CPM = {30}^{o}$
In $∆ BMP,$ right angled at $M,$
We have
$\tan \left(∠BPM\right) = \dfrac{BM}{PM} ⇒ \tan \left(30 °\right) = \dfrac{a / 2}{PM} = \dfrac{1}{3}$

$⇒ PM = \dfrac{\sqrt{3}a}{2}$
In $∆ OPM$,

$PM^{2} = OM^{2} + OP^{2}$
$\dfrac{3a^{2}}{4} = \dfrac{a2}{4} + {5}^{2}$
$∴ a^{2} = 50 ⇒ a = 5\sqrt{2}m$