Question

在三角形ABC中,若cosA=3/5,sinB=5/13求cosC的值

Answer 1

cosA=3/5,

sinA=√(1-cos^2A)=4/5,

sinB=5/13,

cosB=±√(1-sin^2B)=±12/13.

cosC=cos[180-(A+B)]=-cos(A+B)

=sinA*sinB-cosA*cosB.

当cosB=12/13时,

cosC=sinA*sinB-cosA*cosB=-16/65.

当cosB=-12/13时,

cosC=sinA*sinB-cosA*cosB=56/65.

则,cosC的值为:-16/65或56/65.