Question

Consider the following statements
Statement I In any triangle $ABC$, the value of $a\sin (B-C)+b\sin (C-A)+c\sin (A-B)$ is 0 .
Statement II $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
Choose the correct option.

• A. Statement I is true; Statement II is true; Statement II is a correct explanation for Statement I
• B. Statement I is true; Statement II is true; Statement $\Vert$ is not a correct explanation for Statement I
• C. Statement I is true; Statement II is false
• D. Statement I is false; Statement II is true

Answer 1

Answer: (A)

Consider,
$a\sin (B-C)=a[\sin B\cos C-\cos B\sin C]\dots (i)$
Now, $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=K\text{ (say) }$
Therefore, $\sin A=aK,\sin B=bK,\sin C=cK$
Substituting the values of $\sin B$ and $\sin C$ in Eq. (i) and using cosine formulae, we get
$a\sin (B-C)=a\left[bK\left(\frac{a^2+b^2-c^2}{2ab}\right)-cK\left(\frac{c^2+a^2-b^2}{2ac}\right)\right]$
$=\frac{K}{2}\left(a^2+b^2-c^2-c^2-a^2+b^2\right)=K\left(b^2-c^2\right)$
Similarly, $b\sin (C-A)=K\left(c^2-a^2\right)$
and $c\sin (A-B)=K\left(a^2-b^2\right)$
Hence, $a\sin (B-C)+b\sin (C-A)+c\sin (A-B)$
$=K\left(b^2-c^2+c^2-a^2+a^2-b^2\right)$
$=0$