Question

在锐角三角形abc中,sinA=3/5,tan(A-B)=-1/3,求sinB,cosC的值

Answer 1

锐角三角形ABC,sinA=3/5,则有tanA=3/4,cosA=4/5.

tan(A-B)=(tanA-tanB)/(1+tanAtanB)=-1/3,

代入tanA=3/4,得tanB=13/9,则cosB=9√10/50,sinB=13√10/50.

cosC=cos(π-(A+B))=-cos(A+B)=-(cosAcosB-sinAsinB)=3√10/250.