Question

Consider the following statements
Statement I In any triangle $A B C$, 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 $\|$ 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] \ldots (i)$
Now, $ \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=K \text { (say) }$
Therefore, $\sin A=a K, \sin B=b K, \sin C=c K$
Substituting the values of $\sin B$ and $\sin C$ in Eq. (i) and using cosine formulae, we get
$a \sin (B-C) =a\left[b K\left(\frac{a^2+b^2-c^2}{2 a b}\right)-c K\left(\frac{c^2+a^2-b^2}{2 a c}\right)\right]$
$ =\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, $a \sin (B-C)+b \sin (C-A)+c \sin (A-B)$
$=K\left(b^2-c^2+c^2-a^2+a^2-b^2\right)$
$=0$