Question

Mark the correct alternative in each of the following:
In any ∆ABC, \[\sum^{}_{} a^2 \left( \sin B - \sin C \right)\] = 

Options

• \[a^2 + b^2 + c^2\] 

• \[a^2\] 

• \[b^2\] 

•  0   

Answer 1

Using sine rule, we have \[\sum^{}_{} a^2 \left( \sin B - \sin C \right)\] 

\[= a^2 \left( \frac{b}{k} - \frac{c}{k} \right) + b^2 \left( \frac{c}{k} - \frac{a}{k} \right) + c^2 \left( \frac{a}{k} - \frac{b}{k} \right)\]
\[ = \frac{1}{k}\left( a^2 b - a^2 c + b^2 c - b^2 a + c^2 a - c^2 b \right)\] 

This expression cannot be simplified to match with any of the given options.  

However, if the quesion is "In any ∆ABC, 

\[\sum^{}_{} a^2 \left( \sin^2 B - \sin^2 C \right)\] = then the solution is as follows.
Using sine rule, we have \[\sum^{}_{}$ a^2 \left( \sin^2 B - \sin^2 C \right)\]

\[= a^2 \left( \frac{b^2}{k^2} - \frac{c^2}{k^2} \right) + b^2 \left( \frac{c^2}{k^2} - \frac{a^2}{k^2} \right) + c^2 \left( \frac{a^2}{k^2} - \frac{b^2}{k^2} \right)\]
\[ = \frac{1}{k^2}\left( a^2 b^2 - a^2 c^2 + b^2 c^2 - b^2 a^2 + c^2 a^2 - c^2 b^2 \right)\]
\[ = \frac{1}{k^2} \times 0\]
\[ = 0\] 

Hence, the correct answer is option (d).