在△ABC中,cosA=-4/5,sinB=5/13,求cosC
问题描述:
在△ABC中,cosA=-4/5,sinB=5/13,求cosC
答
cosA=-4/5 sinB=5/13
∴A>90°B< 90°
∴sinA=3/5 cosB=12/13
∴cosC
=cos[180°-(A+B)]
=-cos(A+B)
=-cosAcosB+sinAsinB
=-(-4/5)×12/13+3/5×5/13
=48/65+15/65
=63/65