数学题`在三角形ABC中sinA(sinB+cosB)-sinC=0,sinB+cos2C=0,求A,B,C的大小

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数学题`在三角形ABC中sinA(sinB+cosB)-sinC=0,sinB+cos2C=0,求A,B,C的大小

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 = 75