已知三角形ABC中斜边AB=m高CD=n,E,F是AB边的两个三等分点,求＜ECF的大小
已知三角形ABC中斜边AB=m高CD=n,E,F是AB边的两个三等分点,求＜ECF的大小
三角形为直角三角形···
三角形为直角三角形···
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设AD=x,DB=y
则：x+y=AB=m,xy=CD^2=n^2
ED=x-(m/3),FD=y-(m/3)
tan角ECD=ED/CD=(x-(m/3))/n
tan角FCD=FD/CD=(y-(m/3))/n
tan角ECF=tan(角ECD+角FCD)=(tan角ECD+tan角FCD)/(1-tan角ECD*tan角FCD)
=n(x+y-(2m/3))/[n^2-(x-(m/3))(y-(m/3))]
=n(m-(2m/3))/[n^2-xy+(m/3)(x+y)-(m^3/9)]
=(1/3)mn/[(m^2/3)-(m^2/9)]
=3n/(2m)
角ECF=arctan[3n/(2m)]
则：x+y=AB=m,xy=CD^2=n^2
ED=x-(m/3),FD=y-(m/3)
tan角ECD=ED/CD=(x-(m/3))/n
tan角FCD=FD/CD=(y-(m/3))/n
tan角ECF=tan(角ECD+角FCD)=(tan角ECD+tan角FCD)/(1-tan角ECD*tan角FCD)
=n(x+y-(2m/3))/[n^2-(x-(m/3))(y-(m/3))]
=n(m-(2m/3))/[n^2-xy+(m/3)(x+y)-(m^3/9)]
=(1/3)mn/[(m^2/3)-(m^2/9)]
=3n/(2m)
角ECF=arctan[3n/(2m)]
1年前
10
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