取A1B1的中点E，连结C1E，AE，由正三棱柱性质得面A1B1C1⊥面A1B1BA，交线是A1B1．又C1E⊥A1B1，∴C1E⊥面A1B1BA．∴∠C1AE为所求．∵AB=a，C1C=2a，∴Rt△C1EA中，C1E=3a2，AE=32a．∴tan∠C1AE=C1EAE=33．∴∠C1AE=...