Question

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

Answer 1

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)]