已知三角形ABC中,sinA(sinB+cosB)-sinC=0,sinB-cos2C=0求角A、B、C分别多大?

sinA(sinB+cosB)-sinC=0 ----- sinB(sinA-cosA) = 0
sinA = cosA ,A = 45
sinB-cos2C= sinB - cos(2B + 90) = sinB - sin2B = sinB(1-2cosB) = 0
cosB = 1/2 ,B = 60
C = .