In ABC, prove that a ( B - C )+ b ( C - A )+ c ( A - B )=0 - Vidyadip

Question

In $\triangle ABC$, prove that $a (\sin B -\sin C )+ b (\sin C -\sin A )+ c (\sin A -\sin B )=0$

✓

Answer

$\text { L.H.S. }= a (\sin B -\sin C )+ b (\sin C -\sin A )+ c (\sin A -\sin B )$
$=\quad a \sin B - a \sin C + b \sin C - b \sin A + c \sin A - c \sin B$
$=( a \sin B - b \sin A )+( b \sin C - c \sin B )+( c \sin A - a \sin C )$
$=0+0+0$
$=0=\text { R.H.S. }$

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