...中,底面ABCD是一直角梯形,∠BAD=90°,AD∥BC,AB=BC=1,AD=2,PA⊥...
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发布时间：2024-10-21 19:38
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时间：2天前
（Ⅰ）证明：连接AE．
∵PA⊥底面ABCD，PD与底面成45°角，
∴∠PDA=45°，△PAD为等腰直角三角形．
∵点E是PD的中点∴AE⊥PD，
PA⊥底面ABCD，PA?面PAD，
∴面PAD⊥底面ABCD，
而面PAD∩底面ABCD=AD，∠BAD=90°，∴BA⊥AD，∴BA⊥面PAD，PD?面PAD，∴BA⊥PD，AE∩BA=A，∴PD⊥面ABE，
BE?面ABE，∴BE⊥PD．
（Ⅱ）解：
连接AC，∠PCA为二面角P-CD-A的平面角．
取AD中点F，连接CF，∠BAD=90°，AB=BC=1，四边形ABCF是正方形，∠ACF=45°，又AD=2，
∴FD=CF=1，∠FCD=45°，
∴∠ACD=90°，即AC⊥CD．又PA⊥CD，
∴CD⊥面PAC，
∴PC⊥CD，即∠PCA为二面角P-CD-A的平面角．
在RT△PAC中，AC=2，PA=AD=2，PC=AC2+PA2=6．cos∠PCA=ACPC=26=33．所以二面角P-CD-A的余弦值为已赞过已踩过你对这个回答的评价是？评论收起 ._1uevpeq{zoom:1;background-color:#fff;border:0;margin-bottom:10px;padding:30px 0 20px 42px;position:relative}._1uevpeq.ec-1841{padding:20px 0}._1uevpeq.ec-2246{padding:20px 0 10px}.ec-1841 .y7we4hu{font-size:16px;margin-bottom:-5px}.y7we4hu{color:#7a8f9a;height:25px;line-height:25px;overflow:hidden;position:relative}.y7we4hu h2{margin:0;padding:0}.y7we4hu:after{clear:both;content:" ";display:block;height:0;visibility:hidden}a.tycfu7u{color:#666;float:right;font-size:12px;margin-left:8px;text-decoration:none}.hhhv6ex{color:#666;font-size:13px;line-height:normal;line-height:20px;margin-top:10px}.vnsdjzp{margin-top:15px;position:relative}.vnsdjzp h3{font-weight:400;padding:0}.vnsdjzp a{text-decoration:none}.vnsdjzp em{color:#d81419;font-style:normal}.ec-2246 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