Question

In any $\Delta ABC$, find the value of the following expression
$\sum {a\left( {\sin B - \sin C} \right) = } $
$
  {\text{A}}{\text{. 2s}} \\
  {\text{B}}{\text{. }}{{\text{a}}^2} + {b^2} + {c^2} \\
  {\text{C}}{\text{. 0}} \\
  {\text{D}}{\text{. None of these}} \\
$

Answer 1

Hint: To solve this question first we have to expand summation series and using property of triangle, $\dfrac{{\sin A}}{a} + \dfrac{{\sin B}}{b} + \dfrac{{\sin C}}{c} = k$. Using this property we can solve further very easily.

Complete step-by-step answer:

We are given,
$\sum {a\left( {\sin B - \sin C} \right)} $
Now on expanding that means on removing summation sign we get,
$a\left( {\sin B - \sin C} \right) + b\left( {\sin C - \sin A} \right) + c\left( {\sin A - \sin B} \right)$
Now as we know property of solution of triangle,
$\dfrac{{\sin A}}{a} + \dfrac{{\sin B}}{b} + \dfrac{{\sin C}}{c} = k$
$\therefore \sin A = ka,\sin B = kb,\sin C = kc$
Now putting all these value in question, we get
$
   \Rightarrow a\left( {kb - kc} \right) + b\left( {kc - ka} \right) + c\left( {ka - kb} \right) \\
   \Rightarrow kab - kac + kbc - kab + kac - kbc \\
   \Rightarrow 0 \\
 $
Hence option C. is the correct option.

Note: Whenever we get this type of question the key concept of solving is you have to first open the series and as it is written in question it is a triangle so to solve this type of question we have to think about properties which can suit that type of question.