如图,在四面体ABCD中,平面ABC⊥ACD,AB⊥BC,AD=CD, ∠CAD=30° (Ⅰ)若AD=2,AB=2BC,求四面体AB...如图,在四面体ABCD中,平面ABC⊥ACD,AB⊥BC,AD=CD, ∠CAD=30°(Ⅰ)若AD=2,AB=2BC,求四面体ABCD的体积.(Ⅱ)若二面
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![如图,在四面体ABCD中,平面ABC⊥ACD,AB⊥BC,AD=CD, ∠CAD=30° (Ⅰ)若AD=2,AB=2BC,求四面体AB...如图,在四面体ABCD中,平面ABC⊥ACD,AB⊥BC,AD=CD, ∠CAD=30°(Ⅰ)若AD=2,AB=2BC,求四面体ABCD的体积.(Ⅱ)若二面](/uploads/image/z/2513134-46-4.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E5%9B%9B%E9%9D%A2%E4%BD%93ABCD%E4%B8%AD%2C%E5%B9%B3%E9%9D%A2ABC%E2%8A%A5ACD%2CAB%E2%8A%A5BC%2CAD%3DCD%2C+%E2%88%A0CAD%3D30%C2%B0+%EF%BC%88%E2%85%A0%EF%BC%89%E8%8B%A5AD%3D2%2CAB%3D2BC%2C%E6%B1%82%E5%9B%9B%E9%9D%A2%E4%BD%93AB...%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E5%9B%9B%E9%9D%A2%E4%BD%93ABCD%E4%B8%AD%2C%E5%B9%B3%E9%9D%A2ABC%E2%8A%A5ACD%2CAB%E2%8A%A5BC%2CAD%3DCD%2C+%E2%88%A0CAD%3D30%C2%B0%EF%BC%88%E2%85%A0%EF%BC%89%E8%8B%A5AD%3D2%2CAB%3D2BC%2C%E6%B1%82%E5%9B%9B%E9%9D%A2%E4%BD%93ABCD%E7%9A%84%E4%BD%93%E7%A7%AF%EF%BC%8E%EF%BC%88%E2%85%A1%EF%BC%89%E8%8B%A5%E4%BA%8C%E9%9D%A2)
如图,在四面体ABCD中,平面ABC⊥ACD,AB⊥BC,AD=CD, ∠CAD=30° (Ⅰ)若AD=2,AB=2BC,求四面体AB...如图,在四面体ABCD中,平面ABC⊥ACD,AB⊥BC,AD=CD, ∠CAD=30°(Ⅰ)若AD=2,AB=2BC,求四面体ABCD的体积.(Ⅱ)若二面
如图,在四面体ABCD中,平面ABC⊥ACD,AB⊥BC,AD=CD, ∠CAD=30° (Ⅰ)若AD=2,AB=2BC,求四面体AB...
如图,在四面体ABCD中,平面ABC⊥ACD,AB⊥BC,AD=CD, ∠CAD=30°
(Ⅰ)若AD=2,AB=2BC,求四面体ABCD的体积.
(Ⅱ)若二面角C-AB-D为60°,求异面直线AD与BC所成角的余弦值.
如图,在四面体ABCD中,平面ABC⊥ACD,AB⊥BC,AD=CD, ∠CAD=30° (Ⅰ)若AD=2,AB=2BC,求四面体AB...如图,在四面体ABCD中,平面ABC⊥ACD,AB⊥BC,AD=CD, ∠CAD=30°(Ⅰ)若AD=2,AB=2BC,求四面体ABCD的体积.(Ⅱ)若二面
(I)设F为AC的中点,由于AD=CD,
所以DF⊥AC.
故由平面ABC⊥平面ACD,
知DF⊥平面ABC,即DF是四面体ABCD的面ABC上的高,且DF=ADsin30°=1,
AF=ADcos30°=3,
在Rt△ABC中,因AC=2AF=23,AB=2BC,
由勾股定理易知BC=2155,AB=4155.
故四面体ABCD的体积V=13•S△ABC•DF=13×12×4155×2155=45.
(II)设G,H分别为边CD,BD的中点,则FG∥AD,GH∥BC,
从而∠FGH是异面直线AD与BC所成角或其补角.
设E为边AB的中点,则EF∥BC,由AB⊥BC,知EF⊥AB,
又由(I)有DF⊥平面ABC,故由三垂线定理知DE⊥AB,
所以∠DEF为二面角C-AB-D的平面角,由题设知∠DEF=60°.
设AD=a,则DF=AD•SsinCAD=a2,
在Rt△DEF中,EF=DF•cotDEF=a2•33=36a,
从而GH=12BC=EF=36a,因Rt△ADE≌Rt△BDE,
故在Rt△BDF中,FH=12BD=a2.
又FG=12AD=a2,从而在△FGH中,因FG=FH,
由余弦定理得cosFGH=(FG²+GH²—HF²)÷(2FG×GH)=六分之根号三
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