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Question: In \(\triangle A B C\) a \(\sin ( B - C ) + b \sin ( C - A ) + c \sin ( A - B ) =\)...

In △ABC\triangle A B C a sin⁡(B−C)+bsin⁡(C−A)+csin⁡(A−B)=\sin ( B - C ) + b \sin ( C - A ) + c \sin ( A - B ) =

A

0

B

a+b+ca + b + c

C

a2+b2+c2a ^ { 2 } + b ^ { 2 } + c ^ { 2 }

D

2(a2+b2+c2)2 \left( a ^ { 2 } + b ^ { 2 } + c ^ { 2 } \right)

Answer

0

Explanation

Solution

asin⁡(B−C)+bsin⁡(C−A)+csin⁡(A−B)a \sin ( B - C ) + b \sin ( C - A ) + c \sin ( A - B )

=k(Σsin⁡Asin⁡(B−C)= k ( \Sigma \sin A \sin ( B - C ) =k{Σsin⁡(B+C)sin⁡(B−C)}= k \{ \Sigma \sin ( B + C ) \sin ( B - C ) \}

=k{Σ12(cos⁡2C−cos⁡2B)}=0= k \left\{ \Sigma \frac { 1 } { 2 } ( \cos 2 C - \cos 2 B ) \right\} = 0.

Note: Students should note here that most of the expressions containing the cyclic factor associating with ‘–’ reduces to 0.