(I)证明:∵∠ABC=90°,AB=BC=1,∴AC=![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105127-21269.png) ∵四边形ABCD为直角梯形,AD=2,AB=BC=1 ∴CD= , ∴AC2+CD2=AD2,∴∠ACD=90° ∴DC⊥AC ∴平面PAC⊥平面ACD,平面PAC∩平面ACD=AC. ∴DC⊥平面APC; (II)建立如图所示的空间直角坐标系, 则A(0,0,0),B(1,0,0),D(0,2,0),P( ) ∴ , = ,![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105128-25272.png) 设平面APB的法向量为 , 平面APD的法向量为![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105128-20315.png) ∴ ,![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105129-64618.png) ∴![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105129-23413.png) ∴可取![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105129-33643.png) 同理![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105129-34989.png) ∴ =![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105130-43957.png) ∵二面角B﹣AP﹣D的平面角为钝二面角 ∴二面角B﹣AP﹣D的余弦值为 .
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105130-90371.png) |