Question

In $\Delta ABC,(b-c)\sin A+(c-a)\sin B+(a-b)\sin C$ is equal to

• A. $ab+bc+ca$
• B. $a^2+b^2+c^2$
• C. $0$
• D. None of these

Answer 1

The correct option is C $0$

We know, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

Therefore

$(b-c)\sin A+(c-a)\sin B+(a-b)\sin C$
$=\frac{1}{k}\left[a\right(b-c)+b(c-a)+c(a-b\left)\right]$

$=\frac{1}{k}(ab-ac+bc-ab+ac-bc)$

$=0$.