The angle of elevation of the top of a vertical tower from a point...

Let height of the tower be 'H'
Let $AP = x$
$BP = y$
In $∆ APQ ⇒ tan {45}^{o} = \dfrac{H}{x} ⇒ H = x$
In $∆ BPQ ⇒ tan {30}^{o} = \dfrac{H}{y} ⇒ y = \sqrt{3}H$

Since, $∆ APB$ is a right-angled triangle.
$⇒ {x}^{2} + {\left(54\sqrt{2}\right)}^{2} = {y}^{2}$ $($Pythagoras Theorem $)$
$⇒ {H}^{2} + {\left(54\sqrt{2}\right)}^{2} = 3{H}^{2}$
$⇒ {\left(54\sqrt{2}\right)}^{2} = 2{H}^{2}$
$⇒ 54\sqrt{2} = \sqrt{2}H$
$⇒ 54 = H$