已知斜三棱柱ABC-A1B1C1的底面为正三角形,侧面A1ACC1为菱形,∠A1AC=60°,且平面A1ACC1⊥平面ABC,M是C
已知斜三棱柱ABC-A1B1C1的底面为正三角形,侧面A1ACC1为菱形,∠A1AC=60°,且平面A1ACC1⊥平面ABC,M是C1C的中点.(1)求证:A1C⊥BM;...
已知斜三棱柱ABC-A1B1C1的底面为正三角形,侧面A1ACC1为菱形,∠A1AC=60°,且平面A1ACC1⊥平面ABC,M是C1C的中点.(1)求证:A1C⊥BM;(2)求二面角B-A1A-C的正切值.
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解答:(1)证明:取AC中点P,则BP⊥AC
∵平面A1ACC1⊥平面ABC,平面A1ACC1∩平面ABC=AC,
∴BP⊥平面A1ACC1,
∵A1C?平面A1ACC1,∴A1C⊥BP
∵A1C⊥AC1,AC1∥PM
∴A1C⊥PM
∵BP∩PM=P
∴A1C⊥面BPM
∵BM?面BPM
∴A1C⊥BM;
(2)解:作PQ⊥A1A于告仿雀Q,连接BQ
∵BP⊥袜早平面A1ACC1,∴A1A⊥BP
∵BP∩PQ=P,∴A1A⊥面BPQ
∵BQ?面BPQ,大扒∴A1A⊥BQ
∴∠BQP为二面角B-A1A-C的平面角
斜三棱柱ABC-A1B1C1的底面为正三角形,侧面A1ACC1为菱形,∠A1AC=60°,设AC=2,则BP=
,PQ=
∴tan∠BQP=
=2.
∵平面A1ACC1⊥平面ABC,平面A1ACC1∩平面ABC=AC,
∴BP⊥平面A1ACC1,
∵A1C?平面A1ACC1,∴A1C⊥BP
∵A1C⊥AC1,AC1∥PM
∴A1C⊥PM
∵BP∩PM=P
∴A1C⊥面BPM
∵BM?面BPM
∴A1C⊥BM;
(2)解:作PQ⊥A1A于告仿雀Q,连接BQ
∵BP⊥袜早平面A1ACC1,∴A1A⊥BP
∵BP∩PQ=P,∴A1A⊥面BPQ
∵BQ?面BPQ,大扒∴A1A⊥BQ
∴∠BQP为二面角B-A1A-C的平面角
斜三棱柱ABC-A1B1C1的底面为正三角形,侧面A1ACC1为菱形,∠A1AC=60°,设AC=2,则BP=
3 |
| ||
2 |
∴tan∠BQP=
BP |
PQ |
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