正方体中ABCD-A'B'C'D',ABCD是正方形,其对角线AC,BD相互垂直平分,即 AC ⊥ 面 DBB'D'
设AC,BD相交于点O.在直角△AOD'中,
令AB=2,则AC=AD'=2√2,AO=√2,OD'=√6
sin∠AD'O = AO/AD'=√2 / 2√2 = 1/2
即∠AD'O = 30°