题目内容
(本题满分12分)
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在正三角形
中,
、
、
分别是
、
、
边上的点,满足AE:EB=CF:FA=CP:PB=1:2(如图1)。将△
沿
折起到
的位置,使二面角A1-EF-B成直二面角,连结A1B、A1P(如图2)
(Ⅰ)求证:A1E⊥平面BEP;
(Ⅱ)求直线A1E与平面A1BP所成角的大小;
(Ⅲ)求二面角B-A1P-F的大小(用反三角函数表示)
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在正三角形
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(Ⅰ)求证:A1E⊥平面BEP;
(Ⅱ)求直线A1E与平面A1BP所成角的大小;
(Ⅲ)求二面角B-A1P-F的大小(用反三角函数表示)
.解法一:不妨设正三角形ABC的边长为3
(1) 在图1中,取BE中点D,连结DF. AE:EB=CF:FA=1:2∴AF=AD=2而∠A=600 , ∴△ADF是正三角形,又AE="DE=1," ∴EF⊥AD在图2中,A1E⊥EF, BE⊥EF, ∴∠A1EB为二面角A1-EF-B的平面角。由题设条件知此二面角为直二面角,A1E⊥BE,又
∴A1E⊥平面BEF,即 A1E⊥平面BEP
(2) 在图2中,A1E不垂直A1B, ∴A1E是平面A1BP的垂线,又A1E⊥平面BEP,
∴A1E⊥BE.从而BP垂直于A1E在平面A1BP内的射影(三垂线定理的逆定理)设A1E在平面A1BP内的射影为A1Q,且A1Q交BP于点Q,则∠E1AQ就是A1E与平面A1BP所成的角,且BP⊥A1Q.在△EBP中, BE=EP=2而∠EBP=600 , ∴△EBP是等边三角形.又 A1E⊥平面BEP , ∴A1B=A1P, ∴Q为BP的中点,且
,又 A1E=1,在Rt△A1EQ中,
,∴∠EA1Q=60o,
∴直线A1E与平面A1BP所成的角为600
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(3)在图3中,过F作FM⊥ A1P与M,连结QM,QF,∵CP=CF=1,
∠C=600,∴△FCP是正三角形,∴PF=1.有
∴PF=PQ①,
∵A1E⊥平面BEP,
∴A1E=A1Q,
∴△A1FP≌△A1QP从而∠A1PF=∠A1PQ②,
由①②及MP为公共边知△FMP≌△QMP,
∴∠QMP=∠FMP=90o,且MF=MQ,
从而∠FMQ为二面角B-A1P-F的平面角.
在Rt△A1QP中,A1Q=A1F=2,PQ=1,又∴
. ∵ MQ⊥A1P∴
∴
在△FCQ中,FC="1,QC=2," ∠C=600,由余弦定理得
在△FMQ中,
∴二面角B-A1P-F的大小为
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(1) 在图1中,取BE中点D,连结DF. AE:EB=CF:FA=1:2∴AF=AD=2而∠A=600 , ∴△ADF是正三角形,又AE="DE=1," ∴EF⊥AD在图2中,A1E⊥EF, BE⊥EF, ∴∠A1EB为二面角A1-EF-B的平面角。由题设条件知此二面角为直二面角,A1E⊥BE,又
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(2) 在图2中,A1E不垂直A1B, ∴A1E是平面A1BP的垂线,又A1E⊥平面BEP,
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∴直线A1E与平面A1BP所成的角为600
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(3)在图3中,过F作FM⊥ A1P与M,连结QM,QF,∵CP=CF=1,
∠C=600,∴△FCP是正三角形,∴PF=1.有
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∴PF=PQ①,
∵A1E⊥平面BEP,
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∴△A1FP≌△A1QP从而∠A1PF=∠A1PQ②,
由①②及MP为公共边知△FMP≌△QMP,
∴∠QMP=∠FMP=90o,且MF=MQ,
从而∠FMQ为二面角B-A1P-F的平面角.
在Rt△A1QP中,A1Q=A1F=2,PQ=1,又∴
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在△FMQ中,
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∴二面角B-A1P-F的大小为
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略
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