在三角形ABC中,已知B=派/3,cosA=4/5,b=根号3

解(1)：
∵cosA=4/5
∴ sinA=√(1-cos²A)
=√[1-(4/5)²]
=√(9/25)
=3/5
sinC=sin[180°-(A+B)]
=sin(A+B)
=sinAcosB+cosAsinB
=3/5×cosπ/3+4/5×sinπ/3
=3/5×1/2+4/5×(√3)/2
=(3+4√3)/10
(2)：
∵a/sinA=b/sinB
∴a=bsinA/sinB
=(√3×3/5)(/sinπ/3)
=(3√3/5)/(√3/2)
=6/5
三角形ABC面积=(1/2)absinC
=(1/2×6/5)×√3×(3+4√3)/10
=(9√3+36)/50
∵a/sina=b/sinB=c/sinC=2R (R为外接圆半径)
∴2R=a/sinA=(6/5)/(3/5)=2
∴ 外接圆面积=π×2²=4π