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 the perpendiculars from $P$ upon $OA$ and $OB$ respectively.
Then, $∠PAQ = {45}^{o}$ and $∠PBQ = {30}^{o}$
$OA = 400, OB = 300$
$⇒ \dfrac{PA}{h} = cot {45}^{o} = 1 ∴ PA = h$
$⇒ \dfrac{PB}{h} = cot {30}^{o} = \sqrt{3} ∴ PB = \sqrt{3}h$
Now, $OP^{2} = PA^{2} + OA^{2} = PB^{2} + OB^{2}$
$⇒ h^{2} + {\left(400\right)}^{2} = 3h^{2} + {\left(300\right)}^{2} ⇒ 2h^{2} = {\left(400\right)}^{2} - {\left(300\right)}^{2}$
$⇒ h = \sqrt{\dfrac{{4}^{2} - {3}^{2}}{2}} \times 100 = 50\sqrt{14}$ meters