Question

如图,正方体abcd-a1b1c1d1的棱长为a,E,F分别是棱AB,BC的中点

Answer 1

说下思路,立体几何在知道中不好打出来

(1)取BB'中点G,连接：FG、EG,显有FG=EG=EF,∠GFE=60°,又∵A'D||B'C||FG,∴异面直线所成角是60°.

（2）A'D'||B'C',∠B'C'B=45°,∴异面直线所成角是45°

(3连接A'C',AC,A'B,∵A'C'=BC'=A'B,∴∠A'C'B=60°,又∵A'C'||AC||EF,∴异面直线所成角是60°