asin(B-C) +bsin(C-A) +csin(A-B) =0
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yes by properties of triangle
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Answer:
Use sine formula ,
∴sinA = ak
sinB = bk
sinC = ck
Also sin(A - B) = sinA.cosB - cosA.sinB
= akcosB - cosA.bk
= K(acosB - bcosA}
Similarly, sin(B - C) = k(bcosC - ccosB)
sin(C - A) = k(ccosA - acosC)
LHS = asin(B- C) + bsin(C - A) + csin(A - B)
= ak(bcosC - ccosB) + bk(acosC - ccosA) + ck(acosB - bcosA)
= k(bccosA - bccosA) + k(accosB - accosB) + k(abcosC - abcosC)
= 0 + 0 + 0 = 0 = RHS
Explanation:
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