⁕ 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,
1 { 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.