摘要:解:如图9―81.(Ⅰ)作A1D⊥AC.垂足为D.由面A1ACC1⊥面ABC.得A1D⊥面ABC∴∠A1AD为A1A与面ABC所成的角.∵AA1⊥A1C.AA1=A1C.∴∠A1AD=45°为所求.(Ⅱ)作DE⊥AB.垂足为E.连A1E.则由A1D⊥面ABC.得A1E⊥AB.∴∠A1ED是面A1ABB1与面ABC所成二面角的平面角.

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