or

The Sine Law

Statement: In any triangle the sides are proportional to the Sine of the opposite angles.
In other words, in any ΔABC

Proof:
Let ABC be a triangle with a = BC, b = CA and c = AB, then three cases are possible. The angle C is either acute or right or obtuse angle.

Draw AD perpendicular to BC ( produce BC if necessary )
In ΔABC, SinB =  for all figures
         
or, AD = cSinB 
→ (i)

Also in 
ΔACD,
{ SinC in fig (i), Sin90 = SinC in fig (ii) and Sin(π-C) = SinC in fig (iii) }

Therefore for all the figures,
SinC = 

AD = bSinC → (ii)

From (i) and (ii) we can write,

AD = bSinC = cSinB
↣ 
Similarly we can show,


Combining these we get,
Proved.