设F是C′D′的中点.有:MN‖BC′‖EF.
则∠DEF为MN和DE的夹角.看⊿DEF:设AB＝2.
则DE＝√5.EF＝BC′＝2√2.DF＝√5.
从余弦定理：DF²＝DE²＋EF²－2DE×EF×cos∠DEF
5＝5+8-2×√5×2√2×cos∠DEF
cos∠DEF＝√10／5.