如图,已知三棱柱ABC-A1B1C1中,侧棱A A1⊥底面ABC,AB⊥BC;
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如图,已知三棱柱ABC-A1B1C1中,侧棱A A1⊥底面ABC,AB⊥BC;
(Ⅰ)求证:平面A1BC⊥侧面A1ABB1.
(Ⅱ)若AA1=AC=a,直线AC与平面A1BC所成的角为
(Ⅰ)求证:平面A1BC⊥侧面A1ABB1.
(Ⅱ)若AA1=AC=a,直线AC与平面A1BC所成的角为
π |
6 |
(Ⅰ)证明:如图,
已知AA1⊥平面ABC,BC⊂面ABC,∴AA1⊥BC,
又已知AB⊥BC,且AB∩AA1=A,∴BC⊥平面AA1BB1,
而BC⊂面A1BC,∴平面A1BC⊥面A1ABB1;
(Ⅱ)过点A在平面AA1BB1内作AD⊥A1B,垂足是D,连结CD,
∵平面A1BC⊥面A1ABB1,且面A1BC∩面A1ABB1=A1B,
∴AD⊥平面A1BC,则CD为CA在平面A1BC内的射影,
∴∠ACD为直线AC与平面A1BC所成角.
即∠ACD=60°,
∵AC=a,∴AD=
a
2,
在Rt△A1AD内,A1A=a,AD=
a
2,
∴∠AA1B=
π
6,
在Rt△AA1B内,AB=atan
π
6=
3
3a.
已知AA1⊥平面ABC,BC⊂面ABC,∴AA1⊥BC,
又已知AB⊥BC,且AB∩AA1=A,∴BC⊥平面AA1BB1,
而BC⊂面A1BC,∴平面A1BC⊥面A1ABB1;
(Ⅱ)过点A在平面AA1BB1内作AD⊥A1B,垂足是D,连结CD,
∵平面A1BC⊥面A1ABB1,且面A1BC∩面A1ABB1=A1B,
∴AD⊥平面A1BC,则CD为CA在平面A1BC内的射影,
∴∠ACD为直线AC与平面A1BC所成角.
即∠ACD=60°,
∵AC=a,∴AD=
a
2,
在Rt△A1AD内,A1A=a,AD=
a
2,
∴∠AA1B=
π
6,
在Rt△AA1B内,AB=atan
π
6=
3
3a.
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