<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] Let $ABCDE$ be a regular pentagon.
In right triangle $GHC$ we see $\sin 18^{\circ} = {y\over 2x} $. Let $k=y/x$. 
From triangle similarity of $GFC$ and $ACE$ we have $${x\over y} = {2x+y\over a}$$
and from triangle similarity of $BFC$ and $EDC$ we have $${a\over x} = {2x+y\over a}$$
Eliminate $a = {x^2\over y}$ and we get $$x^3 = y^2(2x+y)\implies k^3+2k^2-1 =(k-1)(k^2+k-1)=0 $$
Clearly $k\ne -1$ so $$k_{2,3} = {-1\pm \sqrt{5}\over 2} $$
and so $$\sin 18^{\circ} = {-1+ \sqrt{5}\over 4} $$

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