根据正弦定理得
c/sinC=a/sinA=b/sinB
即2/sin60°=a/sinA=b/sin(120°-A)
故a=4√3/3sinA,b=4√3/3sin(120°-A)
故a+b=4√3/3sinA+4√3/3sin(120°-A)
=4√3/3[sinA+sin(120°-A)]
=4√3/3[sinA+sin120°cosA-cos120°sinA]
=4√3/3[sinA+√3cosA/2+sinA/2]
=4√3/3[3sinA/2+√3cosA/2]
=4[√3sinA/2+cosA/2]
=4[sinAcos30°+cosAsin30°]
=4sin(A+30°)
在锐角三角形ABC中,
因0