Question

在△ABC中，已知内角A=π/3边BC=2根号3，设内角B=x周长为y，--人在

求函数y=f（x）的解析式和定义域

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Answer 1

利用正弦定理BC/sinA=AC/sinB=AB/sinC
BC/sinA=4=AC/sinx=AB/sin(2/3π-x)
f(x)=AB+BC+AC=2根号3+4sinx+4sin(2/3π-x)
定义域由x>0,2/3π-x>0可以算出 （三角形的每个内角都大于零）

Answer 2

角A=π/3，角B=x，则角C=π-x-π/3=2π/3-x,边BC=2根号3，
由正弦定理得BC/sinA=AB/sinC=AC/sinB,
AB=BCsinC/sinA=(2根号3)sin(2π/3-x)/sin(π/3)
=(2根号3)sin(x+π/3)/[(根号3)/2]
=4sin(x+π/3),
AC=BCsinB/sinA=(2根号3)sinx/sin(π/3)=(2根号3)sinx/[(根号3)/2]=4sinx
周长为y=AB+BC+AC=4sin(x+π/3)+2根号3+4sinx=4[sinxcos(π/3)+cosxsin(π/3)]+2根号3+4sinx
=4[sinx*(1/2)+cosx*(根号3)/2]+2根号3+4sinx=6sinx+2(根号3)cosx+2根号3
即f（x）=6sinx+2(根号3)cosx+2根号3
B=x>0，且角C=π-x-π/3=2π/3-x>0,所以0<x<2π/3,f（x）定义域为(0,2π/3) 望采纳

Answer 3

解：根据公式有，a/sinA=b/sinB=c/sinC

a=2√3,∠A=60°

=>b=4sinB,c=4sin(120°-B)

=>周长l=a+b+c

=2√3+4sinB+4sin(120°-B)

=2√3+4（sinB+√3/2cosB+1/2sinB)

=2√3+2(3sinB+√3cosB)

=2√3+2√3(√3sinB+cosB)

=2√3+4√3(√3/2sinB+1/2cosB)

=2√3+4√3sin(B+30°）

因为-1<=sina<=1且0<B<120°

1/2<sin(B+30°)<=1 此时B=60°，即现在是等边三角形了。

=>l周长=6√3

Answer 4

角A=π/3，角B=x，则角C=π-x-π/3=2π/3-x,边BC=2根号3，
由正弦定理得BC/sinA=AB/sinC=AC/sinB,
AB=BCsinC/sinA=(2根号3)sin(2π/3-x)/sin(π/3)
=(2根号3)sin(x+π/3)/[(根号3)/2]
=4sin(x+π/3),
AC=BCsinB/sinA=(2根号3)sinx/sin(π/3)=(2根号3)sinx/[(根号3)/2]=4sinx
周长为y=AB+BC+AC=4sin(x+π/3)+2根号3+4sinx=4[sinxcos(π/3)+cosxsin(π/3)]+2根号3+4sinx
=4[sinx*(1/2)+cosx*(根号3)/2]+2根号3+4sinx=6sinx+2(根号3)cosx+2根号3
即f（x）=6sinx+2(根号3)cosx+2根号3
B=x>0，且角C=π-x-π/3=2π/3-x>0,所以0<x<2π/3,f（x）定义域为(0,2π/3)