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Question: In any \(\Delta ABC\), find the value of the following expression \(\sum {a\left( {\sin B - \sin ...

In any ΔABC\Delta ABC, find the value of the following expression
∑a(sin⁡B−sin⁡C)=\sum {a\left( {\sin B - \sin C} \right) = }
A. 2s B. a2+b2+c2 C. 0 D. None of these  {\text{A}}{\text{. 2s}} \\\ {\text{B}}{\text{. }}{{\text{a}}^2} + {b^2} + {c^2} \\\ {\text{C}}{\text{. 0}} \\\ {\text{D}}{\text{. None of these}} \\\

Explanation

Solution

Hint: To solve this question first we have to expand summation series and using property of triangle, sin⁡Aa+sin⁡Bb+sin⁡Cc=k\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,
∑a(sin⁡B−sin⁡C)\sum {a\left( {\sin B - \sin C} \right)}
Now on expanding that means on removing summation sign we get,
a(sin⁡B−sin⁡C)+b(sin⁡C−sin⁡A)+c(sin⁡A−sin⁡B)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,
sin⁡Aa+sin⁡Bb+sin⁡Cc=k\dfrac{{\sin A}}{a} + \dfrac{{\sin B}}{b} + \dfrac{{\sin C}}{c} = k
∴sin⁡A=ka,sin⁡B=kb,sin⁡C=kc\therefore \sin A = ka,\sin B = kb,\sin C = kc
Now putting all these value in question, we get
⇒a(kb−kc)+b(kc−ka)+c(ka−kb) ⇒kab−kac+kbc−kab+kac−kbc ⇒0  \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.