For reference: Na figura temos: $AB=BC=\frac{\sqrt5+1}{2}$, ${CD}=1$.
Calculate: $x$
(Answer:$8$)
My progress:
$\frac{\sqrt5+1}{2}$ = golden ratio
Draw ${BE}={BC}={BA}\implies $
$\triangle ACE$ (right) $\implies$ $\sin(50^{\circ}) = \frac{AC}{EC}$, therefore, ${AC}=\sin(50^{\circ}) \sqrt5 + 1$
$Draw {AC} \implies \angle {CAD}=18^{\circ}$ $\angle {ADC} = (122^{\circ}+x)$
Law of sines:
$\frac{AC}{sin122+x}=\frac{DC}{sin18}\implies x=8^o$
Is it possible to solve without trigonometry?
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 $\angle CED = 50^\circ$, $\angle x = \angle DCE - \angle BCE = 80^\circ - 72^\circ = 8^\circ$.