S=根号3/2*AB*BC=根号3/2*|AB||BC|*cos(180-B)又有S=1/2|AB||BC|sinB故有sinB=-根号3cosBtanB=-根号3即角B=120度.b^2=(a^2+c^2-2accosB=a^2+c^2-2ac*(-1/2)4=(a+c)^2-ac由于有(a+c)^2>=4ac(a+c)^2>=4[(a+c)^2-4]3(a+...纠正：已知2SΔΑΒC=√3BA的向量乘以BC的向量S=根号3/2*BA*BC=根号3/2*|BA||BC|*cos(B)又有S=1/2|AB||BC|sinB故有sinB=根号3cosBtanB=根号3即角B=60度.b^2=(a^2+c^2-2accosB=a^2+c^2-2ac*(1/2)4=(a+c)^2-3ac由于有(a+c)^2>=4ac(a+c)^2>=4[(a+c)^2-4]/3(a+c)^2<=16a+c<=4又a+c>b=2即有范围是(2,4]