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Mark the Correct Alternative in Each of the Following: in Any ∆Abc, ∑ a 2 ( Sin B − Sin C ) = - Mathematics


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


  • \[a^2 + b^2 + c^2\] 

  • \[a^2\] 

  • \[b^2\] 

  •  0   

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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).

Concept: Sine and Cosine Formulae and Their Applications
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RD Sharma Class 11 Mathematics Textbook
Chapter 10 Sine and cosine formulae and their applications
Q 1 | Page 26
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