根据正弦定理得
c/sinC=a/sinA=b/sinB
即√6/sin60=a/sinA=b/sin(120-A)
故a=2√2sinA,b=2√2sin(120-A)
故a+b=2√2sinA+2√2sin(120-A)
=2√2[sinA+sin(120-A)]
=2√2[sinA+sin120cosA-cos120sinA]
=2√2[sinA+√3cosA/2+sinA/2]
=2√2[3sinA/2+√3cosA/2]
=2√6[√3sinA/2+cosA/2]
=2√6[sinAcos30+cosAsin30]
=2√6sin(A+30)
在锐角三角形ABC中,
因0