Question

1,在△ABC中,角A、B、C的对边为a,b,c,已知sin(B/2)=sin(A/2)*sin(C/2).

Answer 1

1.1

sin(B/2)=sin[(∏-A-C)/2]=cos(a/2+c/2)

=cos(a/2)cos(c/2)-sin(a/2)sin(c/2)

代入原式得:cos(a/2)cos(c/2)=2sin(a/2)sin(c/2)

故tan(A/2)*tan(C/2)=1/2.

1.2

利用正弦定理可得a=csinA/sinC ,b=csinB/sinC=csin(A+C)/sinC

要证明a+c=3b,即证明csinA/sinC + c=3csin(A+C)/sinC

简化后得sinA+sinC=3(sinAsinC+cosAcosC)

然后用万能公式代入此式,化简后将tan(A/2)*tan(C/2)=1/2代入,即可得等号两边是相等的,已验证可解,详细过程从略.

2.1

由方程可得:b+c=18 ,bc=60

由余弦定理得a^2=b^2+c^2-2bccosA=(b+c)^2-3bc=144 ,a=12

2.2

由正弦定理得sinB=bsinA/a ,sinC=csinA/a

sinB*sinC=bcsin^2A/a^2=(60×3)/(144×4)=5/16

2.3

过D点作两条分别垂直于b、c的辅助线

设AD长为 x ,则两条辅助线的长均为x/2

则x/(2sinB)+x/(2sinC)=12 ,可得

x(sinB＋sinc)/2sinBsinC=12

sinB+sinC=bsinA/a＋csinA/a=(b+c)sinA/a=3√3/4

将sinB+sinC与sinBsinC代入即得

x=40√3/3