Question

数学谢谢。，

Answer 1

(1)根据正弦定理得：BC/sinA=2R
则2R=(2√3)/sin(π/春兆袜3)=4
∴AC=2RsinB=4sinx
AB=2RsinC=4sin(π-A-B)=4sin(2π/3 - x)
∴y=AB+BC+AC
=4sin(2π/3 - x) + 2√3 + 4sinx
=4[sin(2π/3)cosx - cos(2π/3)sinx] + 4sinx + 2√3
=2√3cosx + 2sinx + 4sinx + 2√3
=6sinx + 2√3cosx + 2√3
=6√3sin(x + π/6) + 2√3
∵B是△ABC的内角，且扒激A=π/3
∴定义域是0<x<2π/3
(2)∵0<x<2π/3
∴π/6<x + π/猜燃6<5π/6
则1/2<sin(x + π/6)≤1
∴y的最大值是6√3 + 2√3=8√3