Question

已知在三角形ABC中,三内角A,B,C所对的边分别为a,b,c,且sin(A-B)/sin(A+B)=-(a+c)/c求

Answer 1

用正弦定理换：sin(A-B)/sin(A+B)=-(sinA+sinC)/sin(A+B)

所以sin(A-B)=-(sinA+sinC)=-(sinA+sin(A+B))

所以2sinAcosB=-sinA

cosB=-1/2

B=120°