The angle of elevation of the top of a vertical tower from a point A, due east of it is 45 o . The angle of…

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

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