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

Answer: 29

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, $\tan \left(\angle BPM\right)=\frac{BM}{PM}$
In $\Delta OPM$
$\frac{3a^2}{4}$ = $\frac{a^2}{4}+5^2$
$\therefore a^2=50\Rightarrow a=5\sqrt{2}$ meters
So, $m=5,n=2$
$\Rightarrow m^2+n^2=29$