Two straight roads OA and OB intersect at O . A tower is situated...

Let,& $PQ$ be the tower of height $h$.

Let, $PA, PB$ be perpendiculars from $P$ upon $OA$ and $OB$ respectively.
Then, $∠PAQ = {45}^{o}$ and& $∠PBQ = {30}^{o}$
$OA = 100, OB = 50$
$\dfrac{PA}{h} = cot {45}^{o} = 1 ∴ PA = h$
$\dfrac{PB}{h} = cot {30}^{o} = \sqrt{3} ∴ PB = \sqrt{3}h$
Now, $OP^{2} = PA^{2} + OA^{2} = PB^{2} + OB^{2}$
$⇒ h^{2} + {\left(100\right)}^{2} = 3h^{2} + {\left(50\right)}^{2} ⇒ 2h^{2} = {\left(100\right)}^{2} - {\left(50\right)}^{2}$
$⇒ h = 25\sqrt{6}$ meters