The Sine Law

Statement: In any triangle the sides are proportional to the Sine of the opposite angles.

In other words, in any ΔABC,

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.