如图在几何体P-ABCD中,PB⊥底面ABCD,CD⊥PD,底面ABCD是直角梯形,AD//BC,AB⊥BC,AB=AD=PB=3,点E在棱PA上,PE=2EA1.求证:PC//平面EBD2.求平面PCD与平面PAB所成的角
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![如图在几何体P-ABCD中,PB⊥底面ABCD,CD⊥PD,底面ABCD是直角梯形,AD//BC,AB⊥BC,AB=AD=PB=3,点E在棱PA上,PE=2EA1.求证:PC//平面EBD2.求平面PCD与平面PAB所成的角](/uploads/image/z/8482174-70-4.jpg?t=%E5%A6%82%E5%9B%BE%E5%9C%A8%E5%87%A0%E4%BD%95%E4%BD%93P-ABCD%E4%B8%AD%2CPB%E2%8A%A5%E5%BA%95%E9%9D%A2ABCD%2CCD%E2%8A%A5PD%2C%E5%BA%95%E9%9D%A2ABCD%E6%98%AF%E7%9B%B4%E8%A7%92%E6%A2%AF%E5%BD%A2%2CAD%2F%2FBC%2CAB%E2%8A%A5BC%2CAB%3DAD%3DPB%3D3%2C%E7%82%B9E%E5%9C%A8%E6%A3%B1PA%E4%B8%8A%2CPE%3D2EA1.%E6%B1%82%E8%AF%81%EF%BC%9APC%2F%2F%E5%B9%B3%E9%9D%A2EBD2.%E6%B1%82%E5%B9%B3%E9%9D%A2PCD%E4%B8%8E%E5%B9%B3%E9%9D%A2PAB%E6%89%80%E6%88%90%E7%9A%84%E8%A7%92)
如图在几何体P-ABCD中,PB⊥底面ABCD,CD⊥PD,底面ABCD是直角梯形,AD//BC,AB⊥BC,AB=AD=PB=3,点E在棱PA上,PE=2EA1.求证:PC//平面EBD2.求平面PCD与平面PAB所成的角
如图在几何体P-ABCD中,PB⊥底面ABCD,CD⊥PD,底面ABCD是直角梯形,AD//BC,AB⊥BC,AB=AD=PB=3,点E在棱PA上,PE=2EA
1.求证:PC//平面EBD
2.求平面PCD与平面PAB所成的角
如图在几何体P-ABCD中,PB⊥底面ABCD,CD⊥PD,底面ABCD是直角梯形,AD//BC,AB⊥BC,AB=AD=PB=3,点E在棱PA上,PE=2EA1.求证:PC//平面EBD2.求平面PCD与平面PAB所成的角
1.
连续AC,BD交于F点
∵AB=AD,
∴Rt△ABD为等腰直角三角形,
∴∠ABD=45°
∠DBC=45°
CD⊥PD,PB⊥CD
∴CD⊥面PBD
∴CD⊥BD,∠DBC=45°
∴Rt△CBD为等腰直角三角形
CD=BD=3√2,BC=6,AC=3√5
△BFC∽△AFD
AF/CF=AD/BC=1/2=AE/PE
∴△ACP中,EF//PC
∴PC//平面EBD
2.
延长AB,CD交于M点,连续PM
过A作AN⊥PM交PM于N,连结DN
DA⊥面PAB,
∴DA⊥PM,
AN⊥PM
∴PM⊥面DAN
∴角DNA即为所求二面角
AB=AM=3,
PM=3√5,PA=3√2
sin∠PMB=3/3√5=√5/5=AN/AM
AN=3√5/5
MN=√(9-9/5)=√(36/5)=6/√5=6√5/5
DN²=DM²-MN²=18-36/5=54/5
AD=3
cos∠DNA=(AN²+DN²-AD²)/2AN*DN
=(9/5+54/5-9)/(6√54/5)
=18/6√54
=√6/6
∠DNA=arccos(√6/6)