已知三角形ABC中,角A=120度,AB=5,BC=7,则sinB/sinC是多少

A=120°BC/sinA=AB/sinC7/(√3/2)=5/sinCsinC=5*(√3/2)/7=5√3/14C是锐角（因为A是钝角）cosC=11/14sinB=sin(A+C)=sinAcosC+cosAsinC=(√3/2)*(11/14)+(-1/2)*(5√3/14)=3√3/14sinB/sinC=(3√3/14)/(5√3)/14=3/5...