Question

△ABC中，已知sinA=3/5,cosB=-5/13,求sinC ,cosC

Answer 1

解：∵ cosB=-5/13<0；A、B、C均大于0且小于180º
∴ B为钝角，A、C为锐角；
cosA=4/5；sinB=12/13
sinC=sin(A+B)
=sinAcosB+cosAsinB
=33/65
cosC=√[1-(sin²C)]
=56/65

Answer 2

∵cosB=-5/13∴B为钝角∴A、B为钝角
∴ cosA=4/5 sinB=12/13
sinC=sin(A+B)=sinAcosB+cosA sinB=3/5*(-5/13)+4/5*12/13=33/65
cosC=√[1-(sinC)^2]=56/65