If $\sin(18^\circ)=\frac{a + \sqrt{b}}{c}$, then what is $a+b+c$?

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 $$ \sin18°=\frac{-1+√5}{4}$$

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 $$2a+4b+2c\ne a+b+c.$$

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 $$\sin 18^{\circ} = {-1+ \sqrt{5}\over 4} $$