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For reference: Na figura temos: $AB=BC=\frac{\sqrt5+1}{2}$, ${CD}=1$. Calculate: $x$ (Answer:$8$) enter image description here

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?

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

1 Answer 1

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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 $\angle ABE = 136^\circ \implies \angle CBE = 36^\circ$.

enter image description here

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

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

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

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  • $\begingroup$ @petaarantes any further questions on this? $\endgroup$
    – Math Lover
    Nov 29, 2021 at 19:09
  • $\begingroup$ No..sorry..I hadn't confirmed the answer, $\endgroup$ Nov 29, 2021 at 19:23

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