在直角三角形ABC中,已知AB=a,∠ACB=30°,∠B=90°,D为AC的中点,E为BD的中点,AE延长线交BC与F，将△ABD延BD

显示全部楼层 · 2010-12-14 21:05:24

起，二面角A'-BD-C的大小记为θ。θ为何值时，A'B⊥CD。此时，求C到平面A'BD的距离。
ATTENTION:“将△ABD沿BD折起”

千问 · 2010-12-14 21:05:24

当θ=90°时，A'B⊥CDCD=AD=AB=aC到平面A'BD的距离=CD*sin60°=3^0.5/2*a

千问 · 2010-12-14 21:05:24

当θ=90°时，A'B⊥CDCD=AD=AB=aC到平面A'BD的距离=CD*sin60°=3^0.5/2*a

千问 · 2010-12-14 21:05:24

to train ABD proof for the equilateral triangle, which can get BD an AE, BD an EF, according to the line surface vertical decision to teach BD AEF surface, then according to the as

千问 · 2010-12-14 21:05:24

当θ=90°时，A'B⊥CDCD=AD=AB=aC到平面A'BD的距离=CD*sin60°=3^0.5/2*a