题目内容
((本小题满分12分)
如图,已知
,
,
,![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181452861343.gif)
,
.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231814529393921.gif)
(Ⅰ)求证:![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181452939423.gif)
;
(Ⅱ) 若
,求二面角
的余弦值.
如图,已知
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181452799526.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181452830467.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181452846372.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181452861343.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181452877432.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181452893389.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231814529393921.gif)
(Ⅰ)求证:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181452939423.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181452971254.gif)
(Ⅱ) 若
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181452986470.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453002311.gif)
证法一(Ⅰ):如图(1),取
的中点M,连接AM,FM,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231814530335005.jpg)
,
∴
.
,
∴
,∴AM∥BE
又∵
,
,
∴
.
∵CF="FD,DM=ME, " ∴MF∥CE,
又∵
,
,
∴
, 又∵
,
∴
,
∵
,
∴
.-------5分
证法二:如图(2),取CE的中点N,连接FN,BN,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231814535324370.jpg)
∵
,![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453080529.gif)
∴
,
∵CF=FD,CN="NE, " ∴
,
,
又
, ∴
,
,
∴
,
∴AF∥BN, 又∵
,
,
∴
.------5分
(Ⅱ)解法一:如图(3)过F作
交AD于点P,作PG⊥BE,连接FG.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231814540005086.jpg)
∵
,
,
∴![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454047647.gif)
∴
∴FG⊥BE(三垂线定理).
所以,∠PGF就是二面角
的平面角.
由
,
,知△
是正三角形,
在Rt△DPF中,
,
,∴PA=3,
∴![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454187730.gif)
,
∵
, ∴![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454250588.gif)
∴在Rt△PGF中,由勾股定理,得
,
∴
,即二面角
的余弦值为
.----12分
解法二:以A为原点,分别以AC,AB为
轴、
轴的正方向建立空间直角坐标系
,
如图(4)所示,则A(0,0,0),B(0,0,2),
,
,于是,有
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231814547964912.jpg)
,
,
,
设平面BEF的一个法向量为
,则
令
,可得,![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181455155590.gif)
设平面ABED的一个法向量为
,则
,可得,![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181455326514.gif)
∴![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181455513588.gif)
所以,所求的二面角
的余弦值为
.------12分
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453017240.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231814530335005.jpg)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453049541.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453080529.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453111429.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453127657.gif)
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453142683.gif)
又∵
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453158561.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453173414.gif)
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453189534.gif)
∵CF="FD,DM=ME, " ∴MF∥CE,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/2014082318145322065.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453236557.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453329408.gif)
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453345530.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453361544.gif)
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453376618.gif)
∵
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453392529.gif)
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453407504.gif)
证法二:如图(2),取CE的中点N,连接FN,BN,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231814535324370.jpg)
∵
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181452799526.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453080529.gif)
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453735412.gif)
∵CF=FD,CN="NE, " ∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453766410.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453782497.gif)
又
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453782495.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453813404.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453829398.gif)
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453922660.gif)
∴AF∥BN, 又∵
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453938461.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453953374.gif)
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453407504.gif)
(Ⅱ)解法一:如图(3)过F作
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453985409.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231814540005086.jpg)
∵
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181452799526.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454031540.gif)
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454047647.gif)
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454063449.gif)
所以,∠PGF就是二面角
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453002311.gif)
由
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454109403.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181452986470.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454141360.gif)
在Rt△DPF中,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454156463.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454172304.gif)
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454187730.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454203295.gif)
∵
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454234567.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454250588.gif)
∴在Rt△PGF中,由勾股定理,得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454265472.gif)
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454281547.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453002311.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454531227.gif)
解法二:以A为原点,分别以AC,AB为
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454702183.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454718128.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454733377.gif)
如图(4)所示,则A(0,0,0),B(0,0,2),
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454749503.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454765373.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231814547964912.jpg)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454811503.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454827554.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454905531.gif)
设平面BEF的一个法向量为
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454921486.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181455108943.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181455123274.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181455155590.gif)
设平面ABED的一个法向量为
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181455170509.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181455186796.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181455311380.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181455326514.gif)
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181455513588.gif)
所以,所求的二面角
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181453002311.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823181454531227.gif)
略
![](http://thumb.zyjl.cn/images/loading.gif)
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