<s>[INST] If $\sin(18^\circ)=\frac{a + \sqrt{b}}{c}$, then what is $a+b+c$? If $\sin(18)=\frac{a + \sqrt{b}}{c}$ in the simplest form, then what is $a+b+c$?
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Attempt:
$\sin(18)$ in a right triangle with sides $x$ (in front of corner with angle $18$ degrees), $y$, and hypotenuse $z$, is actually just $\frac{x}{z}$, then $x = a + \sqrt{b}, z = c$. We can find $y$ as
$$ y = \sqrt{c^{2}- (a + \sqrt{b})^{2}} $$
so we have 
$$ \cos(18) = \frac{y}{z} = \frac{\sqrt{c^{2}- (a + \sqrt{b})^{2}}}{c}$$
I also found out that
$$b = (c \sin(18) - a)^{2} = c^{2} \sin^{2}(18) - 2ac \sin(18) + a^{2}$$
I got no clue after this.

The solution says that $$ \sin(18) = \frac{-1 + \sqrt{5}}{4} $$
I gotta intuition that we must find $A,B,C$ such that
$$ A \sin(18)^{2} + B \sin(18) + C = 0 $$ 
then $\sin(18)$ is a root iof $Ax^{2} + Bx + C$, and $a = -B, b = B^{2} - 4AC, c = 2A$.

Totally different. This question is not asking to prove that $sin(18)=(-1+\sqrt{5})/4$, that is just part of the solution.
 [/INST] Hint This is a classic problem and probably a duplicate here, but I was unable to find another instance of this question as written. Edit Found one.
First, note that $\sin 18^\circ = \cos 72^\circ$; we find the latter.
Recall that (for $n \geq 2$) the sum of the $n$th roots of unity sum to zero, and taking real parts leaves $$\sum_{k = 0}^{n - 1} \cos \left(\frac{k}{n} \cdot 360^\circ\right) = 0 .$$
Setting $n = 5$ and using the symmetry of the cosine function gives
$$1 + 2 \cos 72^\circ + 2 \cos 144^\circ = 0 .$$
Using the double angle formula---in particular that $$\cos 144^\circ = 2 \cos^2 72^\circ - 1$$ ---and substituting in the previous display equation gives a quadratic expression in $\cos 72^\circ$:
$$4 \cos^2 72^\circ + 2 \cos 72^\circ - 1 = 0 .$$
By the quadratic equation,
$$\cos 72^\circ = \frac{-2 + \sqrt{(2)^2 - 4 (4) (-1)}}{2(4)} = \frac{-1 + \sqrt{5}}{4} .$$
(To resolve the $\pm$ ambiguity it's enough to know that $\cos 72^{\circ} > 0$.)
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