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?