由正弦定理得到a/sinA=b/sinB=c/sinC
因此,a+c=b(sinA+sinC)/sinB
=(sinA+sinC)/sinB
因?yàn)?B=A+C,A+B+C=180°
B=60°
A+C=120°
由于sinA+sinC=sinA+sin(120°-A)
=sinA+(√3/2)cosA+(1/2)sinA
=(3/2)sinA+(√3/2)cosA
=√[(3/2)^2+(√3/2)^2]sin{A+arctan[(√3/2)/(3/2)]}
=√3sin[A+arctan(√3/3)]
=√3sin（A+30°)
所以a+c=√3sin（A+30°)/sin60°=2sin(A+30°)
因?yàn)?°所以30°所以（1/2）因此1即a+c的取值范圍為（1,2]