Question

In any triangle ABC prove that a(SinB-SinC)+ b(SinC-SinA)+ c(SinA-SinB)=0

Answer 1

a(sinB-sinC)+b(sinC-sinA)+c(sinA - sinB) = 0
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sin A/a = sin B/b = sin C/c = k ........... sine rule triangle
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LHS = a(sinB-sinC)+b(sinC-sinA)+c(sinA - sinB)
= a (kb - ck)+b (ck - ak)+c (ak - bk)
= k[ab - ac + bc - ab + c a - bc]
= k[0]
= 0
= RHS

Answer 2

Answer:

Step-by-step explanation:

a(sinB-sinC)+b(sinC-sinA)+c(sinA - sinB) = 0

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sin A/a = sin B/b = sin C/c = k ........... sine rule triangle

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LHS = a(sinB-sinC)+b(sinC-sinA)+c(sinA - sinB)

= a (kb - ck)+b (ck - ak)+c (ak - bk)

= k[ab - ac + bc - ab + c a - bc]

= k[0]

= 0

= RHS

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