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

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?

• @soupless I posted in the new image Nov 27, 2021 at 12:41

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