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1,∵ACd在AA1CC1面且在ABC面∴AC为二面交线又∵AA1=A1C,O为AC中点∴AO⊥AC又面AA1CC1⊥面ABC∴AO⊥面ABC2,过C1做AC的垂线交AC于M点连接BM,连接BO∵AA1=AC=A1C,AC∥ A1C1∴∠ C1=60°∴∠ C1MC=30°,CM=CC1/2=1∴C1M=√3∵AB⊥BC,AB=BC,且AC=2∴AB=BC=√2BM=√(BO^2+OM^2)=√(1+4)=√5直线BC1与面ABC所成角=∠MBC1sin∠MBC1=C1M/BM=√3/√5=(√15)/5... |
