How to derive sides for 30-60-90 triangle? I first convert everything to radians about a unit circle.
$\frac{\pi}{6}$ angle from the center, $\frac\pi2$ right angle, $\frac\pi3$ angle remains. Hypotenuse is 1.
How do I figure the ratios so I can derive x or y with the Pythagorean Theorem? At first I thought the side opposite $\frac\pi3$ must be twice that of the side opposite $\frac\pi6$ but that was wrong.
 A: A $30-60-90$ is one of the must basic triangles known in geometry and you are expected to understand and grasp it very easily.
In an equilateral triangle, angles are equal.  As they add to $180$ then angles are are all $\frac {180}{3} = 60$.  And as the sides are equal all sides are equal. (see image)

So that is a $60-60-60$ triangle.
If we draw the perpendicular bisector of the the base you cut the triangle into two congruent triangles.  (see image)

$\triangle CAD$ (and $\triangle CBD$) are two congruent triangles that have angles $30- 60-90$.  As it is half of an equilateral triangle, the short leg, is $\frac 12$ of the side of the equilateral, while the hypotenuse is the full side of the equilateral.  SO $AD$ (and $DB$) $= \frac 12$ and $AC$ (and $BC$) $= 1$.
And because it is a right triangle $AD^2 + DC^2 = AC^2$ or $(\frac 12)^2+ DC^2 = 1$ so $DC^2 = 1 - (\frac 12)^2=1-\frac 14 = \frac 34$ so $DC = \sqrt {\frac 34} = \frac {\sqrt 3}{\sqrt 4} = \frac {\sqrt 3}2$.
And like all right triangles:
We can get that $AD = \cos 60^\circ *hypotenuse$ so $\cos 60^\circ = \frac{1}{2}$ and $DC = \sin 60^\circ*hypotenuse$ so $\sin 60^\circ = \frac{\sqrt{3}}{2}$.
 
You can always do this for any right triangle.  $AD = \cos \angle CAD*AC$ and $DC = \sin \angle CAD*AC$.  Always!
But this is one of the BASIC possible right triangles were the values of $\sin 60^{\circ}$ and $\cos 60^{\circ}$ should follow immediately and !!!VERY!!! easily directly from geometry.
A: In $\triangle ABC$ let $\angle ABC=\pi /2$ and $\angle BCA=\pi/3$ and $\angle CAB=\pi /6.$ 
Take $D\ne C$ on the line containing $C$ and $B,$ with $BC=BD.$
Then $\triangle   CBA $ is congruent to $\triangle DBA $ because $CB=DB,$ and $\angle CBA=\angle DBA=\pi /2,$ and they have side $BA$ in common. ("Side-angle-side"). 
So $\angle BCA=\angle BDA=\pi /3.$
(I). Therefore $\angle DCA=\angle BCA=\pi /3=\angle BDA=\angle CDA.$
Now  $\angle DAB=\angle CAB$ because $\triangle DBA$ is congruent to $\triangle CBA$, so $\angle DAB =\pi /6.$  
(Ii). Therefore $\angle CAD=\angle CAB+\angle DAB=\pi /6+\pi /6=\pi /3.$
Therefore each angle of $\triangle CDA$ is $\pi /3$   by (I) and (II).
Therefore the sides of $\triangle CAD$ are all equal, so $$AC=CD=BC+BD=2\cdot BC.$$ 
So if $AC=1$ then $BC=1/2$ and $BA=\sqrt {AC^2-BC^2}\;=\sqrt {1-1/4}=\sqrt 3\; /2.$
To grasp all this as a whole, it helps to follow it step-by-step on a diagram.  
A: You can use the law of sines .Let the sides of the triangle be $\,a,b,c$ where a is the hypotenuse  ,b is the side opposite to the $\frac\pi3$ angle  and c opposite to $\frac\pi6$.
Then, $\,\frac{1}{\sin(\frac\pi2)} = \frac{b}{\sin(\frac\pi3)} =\frac{c}{\sin(\frac\pi6)}$
$\implies b = \sqrt3c $ 
