若锐角三角形ABC中,sinA=3/5,tan(A-B)=-1/2,则sinB= ,cosC=
问题描述:
若锐角三角形ABC中,sinA=3/5,tan(A-B)=-1/2,则sinB= ,cosC=
答
sinA=3/5 cosA=√[1-(3/5)^2]=4/5 (锐角)
tan(A-B)=sin(A-B)/cos(A-B)=(sinAcosB-cosAsinB)/cosAcosB+sinAsinB=-1/2
tanB=2 sinB=2√5/5 cosB=√5/5
cosC=cos(180-A-B)=-cos(A+B)=-[cosAcosB-sinAsinB]=2/25