The angle of elevation of the …

Height of tower = 100 m
Let height of rock = h m = CD
Let distance between them = xm = AC
Using {tex}\triangle {/tex} ACD,{tex}\frac{x}{h} = \cot {45^ \circ }{/tex} 
{tex}\Rightarrow {/tex} {tex}\frac{x}{h} = 1 \Rightarrow x = h{/tex}.........(i)
In {tex}\triangle {/tex} BED,{tex}\frac{x}{{h - 100}} = \cot {30^ \circ }{/tex}  (since DE = DC - EC)
= DC - AB
= (h-100m)
{tex}\Rightarrow {/tex} {tex}\frac{x}{{h - 100}} = \sqrt 3 {/tex}
x = {tex}\sqrt 3 h - 100\sqrt 3 {/tex}
From (i) and (ii), we get

{tex}\sqrt 3 h - 100\sqrt 3 = h{/tex}
{tex}\Rightarrow {/tex} {tex}\sqrt 3 h - h = 100\sqrt 3 {/tex}
{tex}\Rightarrow {/tex} {tex}\left( {\sqrt 3 - 1} \right)h = 100\sqrt 3 {/tex}
{tex}\Rightarrow {/tex} {tex}h = \frac{{100\sqrt 3 }}{{\sqrt 3 - 1}} = \frac{{100\sqrt 3 }}{{\sqrt 3 - 1}} \times \frac{{\sqrt 3 + 1}}{{\sqrt 3 + 1}}{/tex}
= {tex}\frac{{300 + 100\sqrt 3 }}{{3 - 1}} = \frac{{300 + 100 \times 1.73}}{2}{/tex}
h = {tex}\frac{{300 + 173}}{2} = \frac{{473}}{2} = 236.5\,m{/tex}
Hence, the height of rock is 236.5 m.