Question

In $△ABC,$ sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively. Then the value of $(b-c)\sin A+(c-a)\sin B+(a-b)\sin C=$

• A. $0$
• B. $a+b+c$
• C. $ab+bc+ca$
• D. $a^2+b^2+c^2$

Answer 1

The correct option is A $0$

We know,
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$
$\Rightarrow (b-c)\sin A+(c-a)\sin B+(a-b)\sin C$
$=\frac{1}{2R}\left[\right(b-c)a+(c-a)b+(a-b\left)c\right]$
$=0$