♪⚝Sanjiv❀Jaiswal❁⩸

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 inΔACD,
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.

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