如图,已知四棱锥E-ABCD的底面为菱形,且∠ABC=60°,AB=EC=2√2,AE=BE=2. (I)求证:平面EAB⊥平面ABCD; (II)求二面角A-EC-D的余弦值.
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![如图,已知四棱锥E-ABCD的底面为菱形,且∠ABC=60°,AB=EC=2√2,AE=BE=2. (I)求证:平面EAB⊥平面ABCD; (II)求二面角A-EC-D的余弦值.](/uploads/image/z/5553048-48-8.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%B7%B2%E7%9F%A5%E5%9B%9B%E6%A3%B1%E9%94%A5E-ABCD%E7%9A%84%E5%BA%95%E9%9D%A2%E4%B8%BA%E8%8F%B1%E5%BD%A2%2C%E4%B8%94%E2%88%A0ABC%3D60%C2%B0%2CAB%3DEC%3D2%E2%88%9A2%2CAE%3DBE%3D2%EF%BC%8E+%EF%BC%88I%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9A%E5%B9%B3%E9%9D%A2EAB%E2%8A%A5%E5%B9%B3%E9%9D%A2ABCD%EF%BC%9B+%EF%BC%88II%EF%BC%89%E6%B1%82%E4%BA%8C%E9%9D%A2%E8%A7%92A-EC-D%E7%9A%84%E4%BD%99%E5%BC%A6%E5%80%BC%EF%BC%8E)
如图,已知四棱锥E-ABCD的底面为菱形,且∠ABC=60°,AB=EC=2√2,AE=BE=2. (I)求证:平面EAB⊥平面ABCD; (II)求二面角A-EC-D的余弦值.
如图,已知四棱锥E-ABCD的底面为菱形,且∠ABC=60°,AB=EC=2√2,AE=BE=2. (I)求证:平面EAB⊥平面ABCD; (II)求二面角A-EC-D的余弦值.
如图,已知四棱锥E-ABCD的底面为菱形,且∠ABC=60°,AB=EC=2√2,AE=BE=2. (I)求证:平面EAB⊥平面ABCD; (II)求二面角A-EC-D的余弦值.
过A作平面AGH⊥EC分别交EC、DE于G、H.
∵ABCD是菱形,∴AD=AB=2、AD∥BC,又∠ABC=60°,∴∠DAF=120°.
由余弦定理,有:DF^2=AB^2+AF^2-2AD×AFcos∠DAF=4+1-2×2×1×cos120°=7,
∴DF=√7.
∵EF⊥平面ABCD,∴EF⊥DF,
∴由勾股定理,有:DE=√(DF^2+EF^2)=√(7+1)=2√2.
∵△ABC是等边三角形,∴AC=AB=2.
由余弦定理,有:
cos∠AEG=(AE^2+EC^2-AC^2)/(2AE×EC)=(2+4-4)/(2×√2×2)=√2/4.
∵EG⊥平面AGH,∴EG⊥AG,∴EG/AE=cos∠AEG=√2/4,∴EG=(√2/4)AE=1/2,
∴由勾股定理,有:AG=√(AE^2-EG^2)=√(2-1/4)=√7/2.
∵ABCD是菱形,∴CD=AB=2,又EC=2、DE=2√2,∴DE^2=CD^2+EC^2,
∴由勾股定理的逆定理,有:CD⊥EC,而CD=EC,∴∠GEH=45°.
∵EG⊥平面AGH,∴EG⊥GH,又∠GEH=45°,∴GH=EG=1/2、EH=√2/2.
由余弦定理,有:
cos∠AED=(DE^2+AE^2-AD^2)/(2DE×AE)=(8+2-4)/(2×2√2×√2)=6/8=3/4.
由余弦定理,有:
AH^2=EH^2+AE^2-2EH×AEcos∠AED=1/2+2-2×(√2/2)×√2×(3/4)=1.
由余弦定理,有:
cos∠AGH=(AG^2+GH^2-AH^2)/(2AG×GH)=(7/4+1/4-1)/[2×(√7/2)×(1/2)]
=1/(√7/2)=2√7/7.
∵AG⊥CE、HG⊥CE、H∈DE,∴∠AGH=二面角A-CE-D的平面角,
∴二面角A-CE-D的余弦值为 2√7/7.