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.
 A: Let $A = 18°$
$$5A = 90°$$
$$⇒ 2A + 3A = 90˚$$
Taking sine on both sides, we get 
$$\sin 2A = \sin (90˚ - 3A) = \cos 3A $$
$$⇒ 2 \sin A \cos A = 4 \cos^3 A - 3 \cos A$$
$$⇒ 2 \sin A \cos A - 4 \cos^3A + 3 \cos A = 0 $$
$$⇒ \cos A (2 \sin A - 4 \cos^2 A + 3) = 0 $$
Dividing both sides by cos A
$$⇒ 2 \sin A - 4 (1 - sin^2 A) + 3 = 0$$
$$⇒ 4 \sin^2 A + 2 \sin A - 1 = 0$$
After solving this quadratic
$$ \sin18°=\frac{-1+√5}{4}$$
A: Edit:  This is now obsolete - it applies to the original version of the question, without the condition "in simplest form":
It's clear that $a+b+c$ is not determined, since $$\frac{2a+\sqrt{4b}}{2c}=\frac{a+\sqrt b}c$$but $$2a+4b+2c\ne a+b+c.$$
A: 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 $$\sin 18^{\circ} = {-1+ \sqrt{5}\over 4} $$

A: 
In the figure, let $AB=AC=x$. Note that $AD=CD=2$. As $\triangle ABC\sim\triangle CDB$, $\dfrac x2=\dfrac 2{x-2}$
So, $x^2-2x-4=0$ and hence $x=1+\sqrt{5}$.
$\sin18^\circ=\dfrac 1x=\dfrac{\sqrt{5}-1}4$.
A: 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 $\cos 72^{\circ} > 0$.)
