tanA:tanB:tanC=1:2:3,求AC/AB

Question

tanA:tanB:tanC=1:2:3,求AC/AB

Answer 1

在三角形ABC中,设tanA＝a,则tanB＝2a,tanC＝3a.
又A+B+C＝180度,得tan（A+B+C）＝0,tan(A+B)+tanC=0
(tanA+tanB)/(1-tanAtanB)+tanC=0,tanA+tanB+tanC-tanAtanBtanC=0
将tanA＝a,tanB＝2a,tanC＝3a代入得tanA＝+-1.
当tanA＝-1时,tanB＝-2,tanC＝-3,三个角都大于90度,不成立.
故tanA＝1,A＝45度.
作辅助线：作BM垂直于AC交AC于M,作CN垂直于AB交AB于N.设AN=b,则
CN＝b,BN＝b/2,AB=3b/2,AM=BM=3b/2√2,CM=b/2√2,AC=2b/√2.
AC:AB =(2b/√2):(3b/2)=2√2/3.
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