Question

三角形ABC中,已知cosA=5分之3,sinB=13分之5.,求sinC的值

Answer 1

三角形内角正弦大于0

sin²A+cos²A=1

所以sinA=4/5

sinB=5/13

cosB=±12/13

当cosB=12/13时

sinC=sin[180-(A+B)]

=sin(A+B)

=sinAcosB+cosAsinB

=4/5*12/13+3/5*5/13

=48/65+15/65

=63/65

当cosB=-12/13时

sinC=sin[180-(A+B)]

=sin(A+B)

=sinAcosB+cosAsinB

=4/5*（-12/13)+3/5*5/13

=-48/65+15/65

=-33/65