Question

Each side of a square subtends an angle of $60^{o}$ at the top of a tower $5$ meters high standing in the center of the square. If $a$ meters is the length of each side of the square, then $a$ is equal to

• A. $3\sqrt{2}$
• B. $4\sqrt{2}$
• C. $5\sqrt{2}$
• D. $6\sqrt{2}$

Answer 1

Answer: (C)

Let $ABCD$ be a square of each side of length $a$ meters.
It is given that $\angle BPC=60^{o}.$
Let $M$ be the midpoint of $BC.$
Then, $\angle BPM=\angle CPM=30^{o}$
In $\Delta BMP,$ right angled at $M,$
We have
$tan \left(\angle B P M\right)=\frac{B M}{P M}\Rightarrow tan⁡\left(30 ^\circ \right)=\frac{a / 2}{P M}=\frac{1}{\sqrt{3}}$
$\Rightarrow PM=\frac{\sqrt{3} a}{2}$
In $\Delta OPM$ ,
$PM^{2}=OM^{2}+OP^{2}$
$\frac{3 a^{2}}{4}=\frac{a^{2}}{4}+5^{2}$
$\therefore a^{2}=50\Rightarrow a=5\sqrt{2}m$