top of the bottom of an …

Let h is height of big building,here as per the diagram.
AE = CD = 8 m (Given)
BE = AB-AE = (h - 8) m
Let  AC = DE = x
Also, {tex}\angle F B D = \angle B D E = 30 ^ { \circ }{/tex} 
{tex}\angle F B C = \angle B C A = 45 ^ { \circ }{/tex}

In {tex}\triangle {/tex}ACB, {tex}\angle A = 90 ^ { \circ }{/tex} 
{tex}\tan 45 ^ { \circ } = \frac { A B } { A C }{/tex} 
{tex}\Rightarrow {/tex} x = h, ...(i)
In {tex}\vartriangle {/tex}BDE, {tex}\angle E = 90 ^ { \circ }{/tex} 
{tex}\tan 30 ^ { \circ } = \frac { B E } { E D }{/tex} 
{tex}\Rightarrow \quad x = \sqrt { 3 } ( h - 8 ){/tex}  .(ii) 
From (i) and (ii), we get 
{tex}h = \sqrt { 3 } h - 8 \sqrt { 3 }{/tex}
h(√3 - 1) = 8√3
h = {tex}\frac{8\sqrt3}{\sqrt3-1}=\frac{8\sqrt3}{\sqrt3-1}×\frac{\sqrt3+1}{\sqrt3+1}{/tex}
= {tex}\frac{1}{2}×(24+8√3)=\frac{1}{2}×(24+13.84)=18.92 m{/tex}
Hence height of the multistory building is 18.92 m and the distance between two buildings is 18.92 m.