Question

Mark the correct alternative in each of the following:

In any ∆ABC, $\sum _{}^{}{a}^2\left(\sin B-\sin C\right)=$

(a) $a^2+b^2+c^2$ (b) $a^2$ (c) $b^2$ (d) 0

Answer 1

Using sine rule, we have

$\sum _{}^{}{a}^2\left(\sin B-\sin C\right)$

$$\begin{gathered} =a^2\left(\frac{b}{k}-\frac{c}{k}\right)+b^2\left(\frac{c}{k}-\frac{a}{k}\right)+c^2\left(\frac{a}{k}-\frac{b}{k}\right) \\ =\frac{1}{k}\left(a^{2}b-a^{2}c+b^{2}c-b^{2}a+c^{2}a-c^{2}b\right) \end{gathered}$$

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

$\sum _{}^{}{a}^2\left({\sin }^{2}B-{\sin }^{2}C\right)$

$$\begin{gathered} =a^2\left(\frac{b^2}{k^2}-\frac{c^2}{k^2}\right)+b^2\left(\frac{c^2}{k^2}-\frac{a^2}{k^2}\right)+c^2\left(\frac{a^2}{k^2}-\frac{b^2}{k^2}\right) \\ =\frac{1}{k^2}\left(a^2{b}^2-a^2{c}^2+b^2{c}^2-b^2{a}^2+c^2{a}^2-c^2{b}^2\right) \\ =\frac{1}{k^2}\times 0 \\ =0 \end{gathered}$$

Hence, the correct answer is option (d).


Disclaimer: The question given in the book in incorrect or there is some printing mistake in the question.