∵m∥n
∴满足m=入n (入≠0)
∴sinB=入cos2B,√3=入[4cos²(B/2)-2]
∴cos2B/sinB=2*[2cos²(B/2)-1]/√3
即2sinB*[2cos²(B/2)-1]-√3cos2B=0
∴2sinB*cosB-√3cos2B=0
∴sin2B-√3cos2B=0
∴2sin(2B-60°)=0
又△ABC为锐角三角形
∴2B-60°=0
故B=30°
则A+C=150°
由正弦定理,有
b/sinB=a/sinA=c/sinC
(a+c)/(sinA+sinC)=b/sinB=1/sin30°=2
∴a+c=2(sinA+sinC)
=2[sinA+sin(150°-A)]
=4sin75°cos(75°-A)
∵sin75°=sin(30°+45°)
=sin30°cos45°+cos30°sin45°
=(√2+√6)/4
∴a+c=(√2+√6)*cos(75°-A)
∵0