摘要:∵FG⊥AB1. ∴DG⊥AB1. ∴∠FGD是二面角B―AB1―D的平面角 ∵A1A=AB=1.
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(1)设P是棱BB1的中点,证明:CP∥平面AEB1;
(2)求AB的长;
(3)求二面角B-AB1-E的余弦值.
| PF |
| PB |
| CG |
| CE |
(1)求证:FG∥平面PDC;
(2)求λ的值,使得二面角F-CD-G的平面角的正切值为
| 2 |
| 3 |
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| PF |
| PB |
| CG |
| CE |
| 2 |
| 3 |