Question

In \[\Delta ABC\] find \[a\sin \left( {B - C} \right) + b\sin \left( {C - A} \right) + c\sin \left( {A - B} \right)\]
A. 0
B. \[a + b + c\]
C. \[{a^2} + {b^2} + {c^2}\]
D. \[2\left( {{a^2} + {b^2} + {c^2}} \right)\]

Answer 1

Hint: Using sine law, we will find the value \[\sin A\], \[\sin B\], and \[\sin C\]. Then using difference formula of sin we will find the value of of \[\sin \left( {B - C} \right)\], \[\sin \left( {C - A} \right)\], \[\sin \left( {A - B} \right)\] and substitute in the given expression. After simply the expression we will get the required solution.

Formula used:
Sine law:
\[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}\]
Difference of sine function formula:
\[\sin \left( {a - b} \right) = \sin a\cos b - \cos a\sin b\]

Complete step by step solution:
 Given expression is
\[a\sin \left( {B - C} \right) + b\sin \left( {C - A} \right) + c\sin \left( {A - B} \right)\]
We know that, \[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = k(say)\]
Now calculating the value of \[\sin A\], \[\sin B\], and \[\sin C\].
\[\sin A = ak\],\[\sin B = bk\], \[\sin C = ck\]
Then find \[\sin \left( {B - C} \right)\], \[\sin \left( {C - A} \right)\], \[\sin \left( {A - B} \right)\] using the formula \[\sin \left( {a - b} \right) = \sin a\cos b - \cos a\sin b\]
\[\sin \left( {B - C} \right) = \sin B\cos C - \cos B\sin C\]
\[ \Rightarrow \sin \left( {B - C} \right) = bk\cos C - ck\cos B\]
\[\sin \left( {C - A} \right) = \sin C\cos A - \cos C\sin A\]
\[ \Rightarrow \sin \left( {C - A} \right) = ck\cos A - ak\cos C\]
\[\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B\]
\[ \Rightarrow \sin \left( {A - B} \right) = ak\cos B - bk\cos A\]

Now putting the value of \[\sin \left( {B - C} \right)\], \[\sin \left( {C - A} \right)\], \[\sin \left( {A - B} \right)\]in the given expression
\[ = a\left( {bk\cos C - ck\cos B} \right) + b\left( {ck\cos A - ak\cos C} \right) + c\left( {ak\cos B - bk\cos A} \right)\]
Simplify the above equation
\[ = abk\cos C - ack\cos B + bck\cos A - abk\cos C + cak\cos B - bck\cos A\]
=0
Hence option A is the correct option

Note: If we calculate the value of a, b, c from the sine law and substitute it in the given expression, then we are unable to reach the correct answer. So from the sine law we will find the \[\sin A\], \[\sin B\], and \[\sin C\].