试题分析:球心为点O,作AB中点D,连接OD,CD,说明SC是球的直径,利用余弦定理,三角形的面积公式求出S
△SCD,和棱锥的高AB,即可求出棱锥的体积。
设球心为点O,作AB中点D,连接OD,CD 因为线段SC是球的直径,所以它也是大圆的直径,则易得:∠SAC=∠SBC=90°所以在Rt△SAC中,SC=4,∠ASC="30°" 得:AC=2,SA=2
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003826490344.png)
又在Rt△SBC中,SC=4,∠BSC="30°" 得:BC=2,SB=2
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003826490344.png)
则:SA=SB,AC=BC
因为点D是AB的中点所以在等腰三角形ASB中,SD⊥AB且SD=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003826521988.png)
在等腰三角形CAB中,CD⊥AB且CD=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003826537495.png)
又SD交CD于点D 所以:AB⊥平面SCD 即:棱锥S-ABC的体积:V=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003826537327.png)
AB•S
△SCD,
因为:SD=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003826552523.png)
,CD=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003826537495.png)
,SC="4" 所以由余弦定理得:cos∠SDC=(SD
2+CD
2-SC
2)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003826583693.png)
则:sin∠SDC=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003826599540.png)
由三角形面积公式得△SCD的面积S=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003826615338.png)
SD•CD•sin∠SDC="=3"
所以:棱锥S-ABC的体积:V=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003826537327.png)
AB•S
△SCD=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003826412337.png)
,故选C
点评:本题是中档题,考查球的内接棱锥的体积的求法,考查空间想象能力,计算能力,有难度的题目,常考题型.