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
$tan\left(\angle BPM\right)=\frac{BM}{PM}\Rightarrow tan\left(30^{\circ }\right)=\frac{a/2}{PM}=\frac{1}{\sqrt{3}}$
$\Rightarrow PM=\frac{\sqrt{3}a}{2}$
In $\Delta OPM$ ,
$P{M}^2=O{M}^2+O{P}^2$
$\frac{3a^2}{4}=\frac{a^2}{4}+5^2$
$\therefore a^2=50\Rightarrow a=5\sqrt{2}m$