Question

在三角形ABC中B=60°，AC=√3，则AB+2BC的最大值为？

Answer 1

三角形ABC中B=60°，AC=√3，则
AB=AC*sinC/sinB=2sinC
BC=AC*sinA/sinB=2sinA
∵C=180°-B-A=120°-A
∴AB+2BC=2sinC+4sinA=2sin(120°-A)+4sinA=√3cosA+5sinA
AB+2BC的最大值为√[(√3)²+5²]=√28

Answer 2

由余弦定理，3=a^+c^-ac,
a^-ac+c^-3=0,
△=c^-4(c^-3)=12-3c^>=0,c^<=4,c<=2,
取a=（c+√△)/2,设c=2sint,t∈(0,90°],
AB+2BC=c+2a=2c+√(12-3c^)=4sint+2√3cost=2√7sin(t+m),
其中m=arctan(√3/2),
当t=90°-m时sin(t+m)=1,AB+2BC取最大值2√7.

Answer 3

AB=AC*sinC/sinB=2sinC
BC=AC*sinA/sinB=2sinA
∵C=180°-B-A=120°-A
∴AB+2BC=2sinC+4sinA=2sin(120°-A)+4sinA=√3cosA+5sinAAB+2BC的最大值为√[(√3)²+5²]=√28