已知平行六面体ABCD-A'B'C'D'各棱长都是1∠BAA'=∠DAA'=60°∠BAD=45°求AC'

Question

已知平行六面体ABCD-A'B'C'D'各棱长都是1 ∠BAA'=∠DAA'=60°∠BAD=45
°求AC'

Answer 1

从A'作A’H⊥平面ABCD,作A’M⊥AB,A‘N⊥AD,连结HM,HN,
AM=AA‘cos60°=1/2,〈A’AM=〈DAA‘=60度,H点在〈DAB的平分线上,而ABCD是菱形,故H在对角线AC上,AM/AH=cos30度,AH=（1/2）/（√3/2）=√3/3,
cos