题目内容：

已知在三角形abc中,sina(sinb cosb)-sinc=0,sinb cos2c=0,求角a.b.c大小

优质解答

sinA(sinB+ cosB)-sinC=0,∴sinAsinB+sinAcosB=sin(A+B)=sinAcosB+cosAsinB,∴sinAsinB=cosAsinB,sinB>0,∴tanA=1,A=45°.C=135°-B,sinB+ cos2C=0,sinB+cos(270°-2B)=0,sinB-sin2B=0,cosB=1/2,B=60°,∴C=75°....