English, asked by sachin180, 1 year ago

asin(B-C) +bsin(C-A) +csin(A-B) =0

Answers

Answered by aditya159
1
yes by properties of triangle
Answered by kushangchowdhary
0

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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