题目内容
已知从一点P引出三条射线PA、PB、PC,且两两成60°角,则二面角A-PB-C的余弦值是( )
A.
B.
C.
D.![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823130616939187.gif)
A.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823130616846170.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823130616846175.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823130616877185.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823130616939187.gif)
A
如图,AO⊥面BPC,由题意得:∠APB=60°,∠BPO="30°,"
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231306169555434.jpg)
由cos∠APB=cos∠BPO·cos∠APO,即cos60°=cos30°·cos∠APO,
得cos∠APO=
.作AE⊥PB,E为垂足,连接OE,则∠AEO就是二面角A-PB-C的平面角,不妨设PA=a,则AO=
,PO=
,又在Rt△OPE中,∠OPE=30°,所以OE=
,则在Rt△AEO中,tan∠AEO=
,则cos∠AEO=
.
所以二面角A-PB-C的余弦值为
,故选A.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231306169555434.jpg)
由cos∠APB=cos∠BPO·cos∠APO,即cos60°=cos30°·cos∠APO,
得cos∠APO=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823130616971204.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823130616986230.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823130617002225.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823130616986230.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823130617033547.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823130616846170.gif)
所以二面角A-PB-C的余弦值为
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823130616846170.gif)
![](http://thumb.zyjl.cn/images/loading.gif)
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