已知AB是圆O的直径,AC切圆O于点A,连接CO延长叫圆O于点D、E
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2个回答

1

角C=C

角CEA=DAC

三角形AEC∽DAC

EA/AD=AC/CD

AD*AC=DC*EA

2

AC=nAB=2nOA

CO^2=AC^2+OA^2=[(2n)^2+1] OA^2

CO=√(4n^2+1)OA

CD=(√(4n^2+1)-1)OA

EA//BF,CF/AC=CD/CE

CF=AC*CD/(CD+DE)=2n*OA*[√(4n^2+1)-1]OA/[√(4n^2+1)+1]OA

=2n*OA*[√(4n^2+1)-1]/[√(4n^2+1)+1]

Scof=(1/2)CF*OA=(1/2)CO*h

h=CF*OA/CO=CF/√(4n^2+1)

过F作FG垂直CO于G

h/CG=OA/AC=1/2n

CG=2nh

OG=OC-CG=√(4n^2+1)OA-2n*CF/√(4n^2+1)

tanCOF=h/OG=[CF/√(4n^2+1)]/[√4n^2+1)OA-2nCF/√(4n^2+1)

=CF/[(4n^2+1)OA-2nCF]

=2n*[√(4n^2+1)-1]/(√(4n^2+1)+1] / [(4n^2+1)-2n(√(4n^2+1)-1)/(√(4n^2+1)+1)]

=2n[√(4n^2+1)-1]/[(4n^2+1)(√(4n^2+1)+1)-2n(√4n^2+1)-1))

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