题目内容
已知四棱锥P-ABCD的底面为直角梯形,AB∥DC,
底面ABCD,且PA=AD=DC=
AB=1,M是PB的中点。
(1)求直线AC与PB所成角的余弦值;
(2)求面AMC与面PMC所成锐二面角的大小的余弦值。![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231441407143466.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140652566.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140683225.gif)
(1)求直线AC与PB所成角的余弦值;
(2)求面AMC与面PMC所成锐二面角的大小的余弦值。
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231441407143466.gif)
(1)![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140730309.gif)
(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140761224.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140730309.gif)
(2)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140761224.gif)
因为PA⊥PD,PA⊥AB,AD⊥AB,以A为坐标原点AD为x轴,AB为y轴,AP为z轴,建立空间直角坐标系,不妨设AD=1,则各点坐标为A(0,0,0)B(0,2,0),
C(1,1,0),D(1,0,0),P(0,0,1),M(0,1,
…2分
(1)解:因![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140792759.gif)
…6分
(2)解:由题得:平面PMC的法向量为![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140823508.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140855930.gif)
所以
解得:
….9分
同理设平面AMC的法向量为![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140901521.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140933951.gif)
所以
解得:
….12分
故
, 即所求锐二面角的余弦值为
…..14分
注:几何法求解,相应分步给分。
C(1,1,0),D(1,0,0),P(0,0,1),M(0,1,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140777242.gif)
(1)解:因
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140792759.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231441408081748.gif)
(2)解:由题得:平面PMC的法向量为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140823508.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140855930.gif)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231441408701067.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140886494.gif)
同理设平面AMC的法向量为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140901521.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140933951.gif)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231441409481026.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140964502.gif)
故
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144141057928.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823144140761224.gif)
注:几何法求解,相应分步给分。
![](http://thumb2018.1010pic.com/images/loading.gif)
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