Question

已知三角形ABC中，sinA(sinB+根号3cosB)=根号3sinC

若BC=3，求三角形ABC周长的取值范围

Answer 1

sinA=(根号3)/2，cosA=1/2
BC/sinA=AB/sinC=AC/sinB
AB=BCsinC/sinA=3sinC/sin(π/3)=2根号3sinC
AC=BCsinB/sinA=3sinB/sin(π/3)=2根号3sinB
周长=AB+AC+BC=2根号3sinC+2根号3sinB+3
=2根号3sin(π-A-B)+2根号3sinB+3
=2根号3sin(A+B)+2根号3sinB+3
=2根号3(sinAcosB+cosAsinB)+2根号3sinB+3
=2根号3(sinAcosB+cosAsinB)+2根号3sinB+3
=2根号3(根号3/2cosB+1/2sinB)+2根号3sinB+3
=3cosB+根号3sinB+2根号3sinB+3
=3(cosB+根号3sinB)+3
=6(1/2cosB+根号3/2sinB)+3
=6sin(B+π/6)+3
0＜B＜2π/3
π/6
＜ B+π/6 ＜ 5π/6
sin(B+π/6) 值域（1/2,1】
周长值域（6,9】
即取值范围：大于6，并且小于等于9