Question

如圖,在梯形ABCD中,AB∥CD,AD=DC=CB=a,∠ABC=60°,平面ACFE⊥平面ABCD,四邊形ACFE是矩形,AE=a,點(diǎn)M在線段EF上.

(1)求證:BC⊥平面ACFE;

(2)求二面角B-EF-D所成平面角的余弦值.

Answer 1

解:(1)在梯形ABCD中,

∵AB∥CD,AD=DC=CB=a,∠ABC=60°,

∴四邊形ABCD是等腰梯形,且∠DCA=DAC=30°,∠DCB=120°.

∴∠ACB=90°.∴AC⊥BC.

又∵平面ACFE⊥面ABCD,交線為AC,

∴BC⊥平面ACFE.

(2)取EF中點(diǎn)G,EB中點(diǎn)H,連DG、GH、DH,

∵DE=DF,∴DG⊥EF.

∵BC⊥平面ACFE,∴BC⊥EF.

又∵EF⊥FC,∴EF⊥FB.

又∵GH∥FB,∴EF⊥GH.

∴∠BGH是二面角B-EF-D的平面角.

在△BDE中,DE=a,DB=a,BE==a,

∴∠EDB=90°.∴DH=a.又DG=a,GH=a,

在△DGH中,由余弦定理得cos∠DGH=,

即二面角B-EF-D的余弦值為.