Let a circle be drawn with a diameter of one (and thus a radius of one half). Then let a triangle with vertices A, B, and C be inscribed in the circle (i.e. points A, B, and C are arbitrary points on the circle).
Then a, the side of the triangle opposite angle A is equal to sin(A)
Likewise, b=sin(B) and c=sin(c). I have attempted to find or devise a proof of this, but I don't know where to start!
On
Although lab bhattacharjee has already said, we have to use the Law of Sines. If you aren't familiar with it or its proof, see the link. I will tell you how to proceed in a detailed manner.
Here we have our $\triangle ABC$ and its circumscribed circle with center $O$. We now construct a diameter $BOD$. So, $\angle BAC=\angle BDC$ and $\angle BCD=90^{\circ}$. Now,
$$\sin\angle A=\sin\angle BDC=\frac{a}{2r}$$
Where, $a=BC$ and $r$ is the radius. You can similarly draw conclusions for $\angle B$ and $\angle C$. This gives rise to what we call the, extended law of sines: $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=D$$
Where $D$ is the diameter of the circumradius. It is a very useful theorem, and applying to your triangle gives:
$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=1$$
Done! There is one caveat though, we did not prove the extended law of sines for right triangles [it should be obvious] and obtuse triangles. However, we can it do it similarly, and I leave the proof as an exercise for you.
As lab bhattacharjee pointed out, what you just described is the law of sines, and it applies to all triangles. Google "law of sines proof" for more info.