Question

Two straight roads $OA$ and $OB$ intersect at $O$ . A tower is situated within the angle formed by them and subtends angles of $45^{o}$ and $30^{o}$ at the points $A$ and $B$ where the roads are nearest to it. If $OA=100$ meters and $OB=50$ meters, then the height of the tower is

• A. $25\sqrt{2}$ meters
• B. $50$ meters
• C. $25\sqrt{6}$ meters
• D. $25\sqrt{3}$ meters

Answer 1

Answer: (C)

Let, $PQ$ be the tower of height $h$ .
Let, $PA,PB$ be perpendiculars from $P$ upon $OA$ and $OB$ respectively.
Then, $\angle PAQ=45^{o}$ and $\angle PBQ=30^{o}$
$OA=100,OB=50$
$\frac{P A}{h}=cot 45^{o}=1\therefore PA=h$
$\frac{P B}{h}=cot 30^{o}=\sqrt{3}\therefore PB=\sqrt{3}h$
Now, $OP^{2}=PA^{2}+OA^{2}=PB^{2}+OB^{2}$
$\Rightarrow h^{2}+\left(100\right)^{2}=3h^{2}+\left(50\right)^{2}\Rightarrow 2h^{2}=\left(100\right)^{2}-\left(50\right)^{2}$
$\Rightarrow h=25\sqrt{6}$ meters