In $\Delta BCE$,
$\begin{array}{rcl}
\cos 40^\circ & = & \dfrac{BC}{BE} \\
BE &= & \dfrac{2}{\cos 46^\circ} \\
BE & = & 2.610\ 814\ 579 \text{ cm}
\end{array}$
Since $ABCD$ is a rectangle, $\angle ABC = \angle BCE = 90^\circ$.
By applying cosine law to $\Delta ABE$, we have
$\begin{array}{rcl}
AE^2 & = & AB^2 + BE^2 – 2 \times AB \times BE \times cos \angle ABE \\
AE^2 & = & 3^2 + 2.610\ 814\ 579^2 – 2 \times 3 \times 2.610\ 814\ 579 \cos(90^\circ – 40^\circ) \\
AE & = & 2.397\ 322\ 922\text{ cm}
\end{array}$
By applying sine law to $\Delta ABE$, we have
$\begin{array}{rcl}
\dfrac{AB}{\sin \angle AEB} & = & \dfrac{AE}{\sin \angle ABE} \\
\dfrac{3}{ \sin \angle AEB} & = & \dfrac{2.397\ 322\ 922}{\sin 50^\circ} \\
\sin \angle AEB & = & 0.958\ 624\ 851 \\
\angle AEB & = & 73.460\ 730\ 65^\circ
\end{array}$
In $\Delta BCE$,
$\begin{array}{rcll}
\angle BEC & = & 180^\circ – \angle CBE – \angle BCE & \text{($\angle$ sum of $\Delta$)} \\
\angle BEC & = & 180^\circ – 40^\circ – 90^\circ \\
\angle BEC & = & 50^\circ
\end{array}$
Hence, $\angle AED$
$\begin{array}{cll}
= & 180^\circ – \angle BEC – \angle AEB & \text{(adj. $\angle$s on a st. line)} \\
= & 180^\circ – 50^\circ – 73.460\ 730\ 65^\circ \\
= & 56.539\ 269\ 35^\circ \\
\approx & 57^\circ
\end{array}$