Question

In any triangle ABC, prove that a sin (B – C) + b sin (C – A) + c sin (A – B) = 0.

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Answer 1

Answer:

solution

asinB- asinC +bsinC- bsin A +csinA-csinb

=0

Answer 2

$\large\underline{\sf{Solution-}}$

We know, In triangle ABC,

$\boxed{ \rm{ \:A + B + C = \pi \: }}$

Now,

Consider LHS

$\rm \: a sin (B – C) + b sin (C – A) + c sin (A – B)$

We know, Sine law states that, In triangle ABC,

$\rm \: \dfrac{a}{sinA} = \dfrac{b}{sinB} = \dfrac{c}{sinC} = k$

So, from we get

$\rm \: a = k \: sinA$

$\rm \: b = k \: sinB$

$\rm \: c = k \: sinC$

So, above expression can be rewritten as

$\rm \: = ksinA sin (B – C) + ksinB sin (C – A) + ksinCsin (A – B)$

$\rm \: = k\bigg(sin[\pi - (B + C)] sin (B – C) + sin[\pi - (C + A)] sin (C – A) + sin[\pi - (A + B)]sin (A – B) \bigg)$

$\rm \: = k\bigg(sin(B + C)sin (B – C) + sin(C + A)sin (C – A) + sin(A + B)sin (A – B) \bigg)$

$\rm \: = k\bigg( {sin}^{2}A - {sin}^{2}B + {sin}^{2}B - {sin}^{2}C + {sin}^{2}C - {sin}^{2}A\bigg)$

$\rm \: = \: k \times 0$

$\rm \: = \: 0$

Hence,

$\rm\implies \:\boxed{ \rm{ \:a sin (B – C) + b sin (C – A) + c sin (A – B) = 0 \: \: }}$

$\rule{190pt}{2pt}$

Formulae Used :-

$\boxed{ \rm{ \:sin(\pi - x) = sinx \: }}$

$\boxed{ \rm{ \:sin(x + y)sin(x - y) = {sin}^{2}x - {sin}^{2}y \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\boxed{ \rm{ \:cosA = \frac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc} \: }}$

$\boxed{ \rm{ \:cosB = \frac{ {c}^{2} + {a}^{2} - {b}^{2} }{2ca} \: }}$

$\boxed{ \rm{ \:cosC = \frac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab} \: }}$

$\boxed{ \rm{ \:a \: = \: bcosC + ccosB \: }}$

$\boxed{ \rm{ \:b \: = \: acosC + ccosA \: }}$

$\boxed{ \rm{ \:c \: = \: acosB + bcosA \: }}$