Question

已知A.B.C是三角形ABC的内角.且cosA=4/5.cosB=5/13.求cosC的值?

Answer 1

cosA=4/5,cosB=5/13

A,B为第一象限角,sinA=3/5,sinB=12/13

cosC

=cos(派－(A+B))

=cos派cos(A+B)-sin派sin(A+B)

=-cos(A+B)

=-cosAcosB+sinAsinB

=-20/65+36/65

=16/65