Question

If A+B+C=${180}^{\circ }$ then a sin (B-C)+b sin (C-A) + c sin (A-B) is :

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is A 0

Given, $A+B+C={180}^o$

$a\sin (B-C)=2R\sin A\sin (B-C)$

$=2R\sin (\pi -(B+C\left)\right)\sin (B-C)$

$=2R\sin (B+C)\sin (B-C)$

$=2R[{\sin }^{2}B-{\sin }^{2}C]$

Similarly

$b\sin (C-A)=2R[{\sin }^{2}C-{\sin }^{2}A]$

$C\sin (A-B)=2R[{\sin }^{2}A-{\sin }^{2}B]$

$\Rightarrow a\sin (B-C)+b\sin (C-A)+c\sin (A-B)$

$=2R[{\sin }^{2}B+{\sin }^{2}C+{\sin }^{2}A-{\sin }^{2}C-{\sin }^{2}A-{\sin }^{2}B]$

$=0$.