(2013•烟台一模)如图,在梯形ABCD中,AB∥CD,AD=DC=CB=1,∠ABC=60°,四边形ACFE为矩形,平面ACFE⊥平面ABCD,CF=1. (Ⅰ)求证:BC⊥平面ACFE; (Ⅱ)点M在线段EF上运动,设平面MAB与平面
问题描述:
(2013•烟台一模)如图,在梯形ABCD中,AB∥CD,AD=DC=CB=1,∠ABC=60°,四边形ACFE为矩形,平面ACFE⊥平面ABCD,CF=1.
(Ⅰ)求证:BC⊥平面ACFE;
(Ⅱ)点M在线段EF上运动,设平面MAB与平面FCB所成二面角的平面角为θ(θ≤90°),试求cosθ的取值范围.
答
(I)证明:在梯形ABCD中,∵AB∥CD,AD=DC=CB=1,∠ABC=60°,∴AB=2∴AC2=AB2+BC2-2AB•BC•cos60°=3∴AB2=AC2+BC2∴BC⊥AC∵平面ACFE⊥平面ABCD,平面ACFE∩平面ABCD=AC,BC⊂平面ABCD∴BC⊥平面ACFE(II)由(I...