锐角△ABC,a=2根号3,△ABC面积S=根号3/4(b^2+c^2-a^2),周长L的取值范围第一问已将算出A=60°_数学解答

已知a=2√3,A=60°
根据正弦定理得
a/sinA=b/sinB=c/sinC
即2√3/sin60=b/sinB=c/sin(120-B)
故b=3sinB,C=3sin(120-B)
故b+c=3sinB+3sin(120-B)
=3[sinB+sin120cosB-cos120sinB]
=3[sinB+√3cosB/2+sinB/2]
=3[(3/2)sinB+(√3/2)cosB]
=3√3[√3/2sinB+(1/2)cosB]
=3√3sin(B+30)
在锐角三角形ABC中,
因0故30故1/2故(3√3)/2<3√3sin(B+30)<=3√3
即(3√3)/2则(7√3)/2