Question

问题1 在△ABC中，∠B＝90度，点D在BC边上，∠BAD＝2∠C,AC＝12，CD＝8，求AB

Answer 1

设AB=x,BD=y,∠B=90°，CD=8,

tanC=AB/BC=x/(y+8),

∠BAD=2∠C,

所以tan2C=y/x=[2x/(y+8)]/[1-x^2/(y+8)^2]=2x(y+8)/[(y+8)^2-x^2],

去芦派分母得y[(y+8)^2-x^2]=2x^2*(y+8),

整理得y(y+8)^2=x^2*(3y+16),

x^2=y(y+8)^2/(3y+16),①

由勾股定理陪粗贺，AC^2=x^2+(y+8)^2=144.②

把①代入②*（3y+16）,得y(y+8)^2+(3y+16)(y+8)^2=144(3y+16),

整理得(y+4)(y+8)^2=36(3y+16),

(y+4)(y^2+16y+64)=108y+576,

y^3+16y^2+64y

........+4y^2+64y+256

..................-108y-576=0,

y^3+20y^2+20y-320=0,

解得y≈3.30142383，或y^2+23.30142 y+ 96.92788 ≈0，无正根。

代入①，x^2≈16.27781931，

AB=x≈4.03457796.

仅供参考。 凳运