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 $\angle PBQ=30^o$
$OA=100,OB=50$
$\frac{PA}{h}=cot45^o=1\therefore PA=h$
$\frac{PB}{h}=cot30^o=\sqrt{3}\therefore PB=\sqrt{3}h$
Now, $O{P}^2=P{A}^2+O{A}^2=P{B}^2+O{B}^2$
$\Rightarrow h^2+{\left(100\right)}^2=3h^2+{\left(50\right)}^2\Rightarrow 2h^2={\left(100\right)}^2-{\left(50\right)}^2$
$\Rightarrow h=25\sqrt{6}$ meters