Xavier Aptitude Test 2022 Paper

Given the distances are :
$\mathrm{AE}=4\mathrm{meters},\mathrm{EB}=8\mathrm{meters}$ and $\mathrm{EC}=16\mathrm{meters}$
Considering the length of $\mathrm{ED}=\mathrm{K}$
Given the $\angle \mathrm{DAE}$ and $\angle \mathrm{DCE}$ are complementary.
Hence the angles are $\mathrm{A}$ and $90-\mathrm{A}$.
$\mathrm{Tan}\left(90-\mathrm{A}\right)=\mathrm{Cot}\mathrm{A}$
$\mathrm{Tan}\mathrm{DAE}=\frac{\mathit{k}}{4}$ and $\mathrm{Tan}\mathrm{DCE}=\frac{1}{\mathrm{Tan}\mathrm{DAE}}=\frac{\mathit{k}}{16}$
Hence, $\frac{\mathit{k}}{16}=\frac{4}{\mathit{k}}$

$\mathit{k}=8\mathrm{meters}$
The $\angle \mathrm{DBE}$ is given by
$\mathrm{Tan}\mathrm{DBE}=\frac{\mathit{k}}{8}=1$
Hence the angle is equal to $45°$.