Let the side of the pentagon be of unit length. Denote the angle $P E D$ by $\alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_02_93c3030777938584cbb8g-1.jpg?height=648&width=767&top_left_y=229&top_left_x=777)

One possible way to calculate $\angle P E D$ using the notation in the figure is as follows: By applying the Law of Sines in triangles $D E P$ and $P E A$:

$$
x=\frac{\sin 108^{\circ}}{\sin \left(72^{\circ}-\alpha\right)}, \quad \text { and } \quad x=\frac{\sin 58^{\circ}}{\sin \left(14^{\circ}+\alpha\right)}
$$

From this:

$$
\frac{\sin 108^{\circ}}{\sin \left(72^{\circ}-\alpha\right)}=\frac{\sin 58^{\circ}}{\sin \left(14^{\circ}+\alpha\right)}
$$

Using the known identities for $\sin (\alpha+\beta)$ and $\sin (\alpha-\beta)$, we get:

$$
\sin \alpha\left(\cos 14^{\circ} \sin 108^{\circ}+\cos 72^{\circ} \sin 58^{\circ}\right)=\cos \alpha\left(\sin 72^{\circ} \sin 58^{\circ}-\sin 108^{\circ} \sin 14^{\circ}\right)
$$

From this, $\operatorname{tg} \alpha=0.4865$, hence $\alpha=25^{\circ} 57^{\prime}$.
In a regular pentagon $A B C D E$, for a point $P$ on the side $C D$, $\angle P A B = 50^{\circ}$. What is the measure of $\angle P E D$?