Question

Each side of a square subtends an angle of $60^\circ $ at the top of a tower $5$ meters high standing at the centre of the square. If $a$ meters is the length of each side of the square, then $a$ is equal to $m\sqrt{n},$ then $m^{2}+n^{2}=$

Answer 1

Let, $ABCD$ be a square with each side of length $a.$
It is given that $\angle BPC=60^\circ $
Let, $M$ be the midpoint of $BC.$
Then, $\angle BPM=\angle CPM=30^\circ $
In $\Delta BMP,$ right angled at $M,$
we have, $\tan\left(\angle B P M\right)=\frac{B M}{P M}$
In $\Delta OPM$
$\frac{3 a^{2}}{4}$ = $\frac{a^{2}}{4}+5^{2}$
$\therefore a^{2}=50\Rightarrow a=5\sqrt{2}$ meters
So, $m=5,n=2$
$\Rightarrow m^{2}+n^{2}=29$