What's the measure of the $\angle x$ in the figure below?

In a regular convex pentagon, diagonals are in golden ratio to its sides. If we have a regular pentagon $ABCDE$ with side length $1$, the diagonal will be $\frac{1 + \sqrt 5}{2}$. See $\triangle ABD$ where $AD = BD = \frac{1 + \sqrt 5}{2}$ and $\angle ADB = 36^\circ$.

Now in the given diagram, we extend $AD$ and draw $BE = AB$. Then $\angle ABE = 136^\circ \implies \angle CBE = 36^\circ$.

As $BE = AB = BC = \frac{1 + \sqrt 5}{2}, CE = 1$.

That leads to $\triangle CDE ~$ being isosceles.

As $\angle CED = 50^\circ$, $\angle x = \angle DCE - \angle BCE = 80^\circ - 72^\circ = 8^\circ$.