题目内容
已知正四棱锥P-ABCD,PA=2,AB=![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221406463175627/SYS201311012214064631756011_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221406463175627/SYS201311012214064631756011_ST/images1.png)
【答案】分析:先通过平移将两条异面直线平移到同一个起点M,得到的锐角或直角就是异面直线所成的角,在三角形中再利用余弦定理求出此角即可.
解答:
解:如图,连接AC,BD交与点O,
连接OM,∠OMB为异面直线PA与BM所成角
PA=2,OM=1,OB=1,BM=![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221406463175627/SYS201311012214064631756011_DA/0.png)
cos∠OMB=
,
故答案为![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221406463175627/SYS201311012214064631756011_DA/2.png)
点评:本小题主要考查异面直线所成的角,考查空间想象能力、运算能力和推理论证能力,属于基础题.
解答:
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221406463175627/SYS201311012214064631756011_DA/images0.png)
连接OM,∠OMB为异面直线PA与BM所成角
PA=2,OM=1,OB=1,BM=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221406463175627/SYS201311012214064631756011_DA/0.png)
cos∠OMB=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221406463175627/SYS201311012214064631756011_DA/1.png)
故答案为
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221406463175627/SYS201311012214064631756011_DA/2.png)
点评:本小题主要考查异面直线所成的角,考查空间想象能力、运算能力和推理论证能力,属于基础题.
![](http://thumb2018.1010pic.com/images/loading.gif)
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