Question

如图,将正方形ABCD的边BC延长到点E,使CE=AC,AE与CD相交于点F.求∠EFC的正切值.

Answer 1

设正方形ABCD的边长 AB =a  ，则：

AC=a√2, CE=AC=a√2.

△ADF∽△FCE，AB/(DC-FC)=CE/FC，a/(a-FC)=a√2/FC,  FC=(a√2)/(1+√2).

∠EFC的正切值=CE/FC=(a√2)/[(a√2)/(1+√2)]=1+√2.