（a+b+c)(a-b+c)=[(a+c)+b][(a+c)-b]=(a+c)^2-b^2=a^2+c^2-b^2+2ac=3ac
a^2+c^2-b^2-ac=0,cosB=(a^2+c^2-b^2)/(2ac)=1/2,B=π/3
tanA+tanC=3+√3=tanA+tan(120-A)=tanA-tan(A+60)=tanA-(tanA+√3)/(1-√3*tanA)
设tanA=x,3+√3=x-(x+√3)/(1-√3*x)=-(√3*x^2+√3)/(1-√3*x)=-√3*(x^2+1)/(1-√3*x)
(x^2+1)/(1-√3*x)=-(√3+1),(x^2+1)=-(√3+1)*(1-√3*x)=3*X^2+√3*x-√3-1
2*X^2+√3*x-(√3+2)=0,[2x+(√3+2)](x-1)=0,x=1或x＝-(√3+2)/2
若tanA=x＝-(√3+2)/2<0,tanC>4+3√3/2,则A>90,C>60,B=60,A+B+C>180,不符,舍
所以x=1,tanA=1,A=π/4,C=5π/12
A*cosB=4√3,cosB=√3/2,a=8,a/sinA=b/sinB,b=4√6,
sinC=sin(π/4+π/6)
sinC=(√6+√2)/4,也可得：c=4+4√3