Question

在三角形ABC中,A=60度,a=3,求三角形ABC周长最大值,面积范围.？

Answer 1

/sinB=c/sin(120°-B)=a/sinA=3/(√3/2)=2√3
三角形ABC周长=a+b+c=3+2√3sinB+2√3sin(120°-B)
=3+2√3sinB+2√3*(√3/2cosB+1/2sinB)
=3+3√3sinB+3cosB
=3+6(√3/2sinB+1/2cosB)
=3+6sin(B+π/6)
当B=π/6时,即 A=B=C=π/6时,周长最大,为9
S=1/2*bcsinA
=√3/4*[2√3sinB]*[2√3sin(120°-B)]
=3√3sinBsin(120°-B)
=3√3sinB*(√3/2cosB+1/2sinB)
=9/4sin2B+3√3/2sin²B
=9/4sin2B+3√3/2*(1-cos2B)/2
=3√3/2+9/4sin2B-3√3/4cos2B
=3√3/2+3√3/2(√3/2sin2B-1/2cos2B)
=3√3/2+3√3/2sin(2B-π/6)
-1/2,5,上图
你说 a=3是哪一条边,3,