By symmetry $\angle B A E=\angle D E A=\left[180^{\circ} \times(5-2)-150^{\circ} \times 3\right] \div 2$
$=45^{\circ}(\angle$ sum of polygon $)$

Produce $A B$ and $E D$ to intersect at $F$.
$\angle A F E=180^{\circ}-45^{\circ}-45^{\circ}=90^{\circ}(\angle \mathrm{s}$ sum of $\triangle)$
By symmetry, $\angle B F C=\angle D F C=45^{\circ}$
$\angle B C F=\angle D C F=\left(360^{\circ}-150^{\circ}\right) \div 2=105^{\circ}(\angle \mathrm{s}$ at a pt. $)$
Let $A B=y=5 \sec 15^{\circ} \mathrm{cm}=C D=D E$, let $z=B F$.
Apply Sine rule on $\triangle A B C, \frac{z}{\sin 105^{\circ}}=\frac{y}{\sin 45^{\circ}}$
$z=\sqrt{2} \sin 105^{\circ} y$
$$
\begin{array}{l}
\text { In } \triangle A E F, x=(y+z) \sec 45^{\circ}=\sqrt{2}\left(y+\sqrt{2} \sin 105^{\circ} y\right) \\
\quad=y \sqrt{2}\left(1+\sqrt{2} \sin 105^{\circ}\right)
\end{array} \begin{array}{c}
=5 \sec 15^{\circ} \cdot 2\left(\frac{1}{\sqrt{2}}+\sin 105^{\circ}\right)=10 \sec 15^{\circ}\left(\sin 105^{\circ}+\sin 45^{\circ}\right) \\
=10 \sec 15^{\circ}\left(2 \sin 75^{\circ} \cos 30^{\circ}\right)=20 \sec 15^{\circ} \cdot \cos 15^{\circ} \frac{\sqrt{3}}{2}=10 \sqrt{3}
\end{array}
$$

Method 2
Join $A C, C E$. With similar working steps, $\angle B A E=\angle D E A=45^{\circ}$
$\angle B A C=\angle B C A=15^{\circ}=\angle D C E=\angle D E C(\angle \mathrm{s} \mathrm{sum}$ of isos. $\triangle)$
$\angle C A E=45^{\circ}-15^{\circ}=30^{\circ}=\angle C E A$
$A C=C E=2 y \cos 15^{\circ}=2 \times 5 \sec 15^{\circ} \times \cos 15^{\circ}=10$
$x=2 A C \cos 30^{\circ}=20 \times \frac{\sqrt{3}}{2}=10 \sqrt{3}$
G3.3 In Figure 2, an ant runs ahead straightly for $5 \mathrm{sec} 15^{\circ} \mathrm{cm}$ from point $A$ to point $B$. It then turns $30^{\circ}$ to the right and run $5 \mathrm{sec} 15^{\circ} \mathrm{cm}$ to point $C$. Again it repeatedly turns $30^{\circ}$ to the right and run $5 \mathrm{sec} 15^{\circ}$ twice to reach the points $D$ and $E$ respectively. If the distance of $A E$ is $x \mathrm{~cm}$, find the value of $x$.