or

The Sine Law in terms of Circum-Radius R.
Statement: In any ΔABC, where R is the Circum-Radius of ΔABC.

Proof:
Consider the circum-circle of ΔABC with center O and radius R. Then we can have three possible figures.

Where the angle A is acute in fig i, or right angled in fig ii or obtuse in fig iii.
Let O be the center of the circum-circle of ΔABC and R be the circum-radius. Join BO and produce it to meet the circle at D.

Now in figure i,
BAC = BDC = A, BCD = 90º (angle at the semicircle).

= Sin(BDC)
= SinA
↣ 2R

In fig iii, 
1 = Sin90º = SinA
↣ SinA ( Since BC = BD )
↣ SinA 
↣ 2R

In fig iii, SinA 
↣ SinA
2R

Therefore in each figure it is found that 2R
Similarly we can prove that: 
Combining these we get, Proved.