摘要:∴DE=1.AD=A1D=.tanA1ED==.故∠A1ED=60°为所求.(Ⅲ)作BF⊥AC.F为垂足.由面A1ACC1⊥面ABC.知BF⊥面A1ACC1.∵B1B∥面A1ACC1.∴BF的长是B1B和面A1ACC1的距离.

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