In  $∆DAB, \mathrm{tan }\mathrm{\theta }=\frac{64}{d}$



$\Rightarrow d=\mathrm{64 }\mathrm{cot }\mathrm{\theta }$                   ...(i)

In  $∆CDE, \tan \left(90°-\mathrm{\theta }\right)=\frac{(100-64)}{d}$

$\Rightarrow d=36\mathrm{ tan }\mathrm{\theta }$                   ...(ii)

From Eqs. (i) and (ii), we get

$d^2=36\times 64 \Rightarrow d=48 \mathrm{m}$