【答案】
分析:解法一:(1)作SO⊥BC,垂足為O,連接AO,說明SO⊥底面ABCD.利用三垂線定理,得SA⊥BC.
(Ⅱ)由(Ⅰ)知SA⊥BC,設(shè)AD∥BC,連接SE.說明∠ESD為直線SD與平面SBC所成的角,通過
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/0.png)
,求出直線SD與平面SBC所成的角為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/1.png)
.
解法二:(Ⅰ)作SO⊥BC,垂足為O,連接AO,以O(shè)為坐標原點,OA為x軸正向,建立直角坐標系O-xyz,通過證明
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/2.png)
,推出SA⊥BC.
(Ⅱ).
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/3.png)
與
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/4.png)
的夾角記為α,SD與平面ABC所成的角記為β,因為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/5.png)
為平面SBC的法向量,利用α與β互余.通過
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/6.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/7.png)
,推出直線SD與平面SBC所成的角為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/8.png)
.
解答:![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/images9.png)
解法一:
(1)作SO⊥BC,垂足為O,連接AO,
由側(cè)面SBC⊥底面ABCD,得SO⊥底面ABCD.
因為SA=SB,所以AO=BO,
又∠ABC=45°,故△AOB為等腰直角三角形,AO⊥BO,
由三垂線定理,得SA⊥BC.
(Ⅱ)由(Ⅰ)知SA⊥BC,
依題設(shè)AD∥BC,
故SA⊥AD,由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/9.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/10.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/11.png)
.
又
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/12.png)
,作DE⊥BC,垂足為E,
則DE⊥平面SBC,連接SE.∠ESD為直線SD與平面SBC所成的角.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/13.png)
所以,直線SD與平面SBC所成的角為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/14.png)
.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/images16.png)
解法二:
(Ⅰ)作SO⊥BC,垂足為O,連接AO,
由側(cè)面SBC⊥底面ABCD,得SO⊥平面ABCD.
因為SA=SB,所以AO=BO.
又∠ABC=45°,△AOB為等腰直角三角形,AO⊥OB.
如圖,以O(shè)為坐標原點,OA為x軸正向,建立直角坐標系O-xyz,
因為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/15.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/16.png)
,
又
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/17.png)
,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/18.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/19.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/20.png)
.S(0,0,1),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/21.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/22.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/23.png)
,所以SA⊥BC.
(Ⅱ)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/24.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/25.png)
.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/26.png)
與
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/27.png)
的夾角記為α,SD與平面ABC所成的角記為β,因為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/28.png)
為平面SBC的法向量,所以α與β互余.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/29.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/30.png)
,
所以,直線SD與平面SBC所成的角為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074014_DA/31.png)
.
點評:本小題主要考查空間線面關(guān)系、直線與平面所成的角等知識,考查數(shù)形結(jié)合、化歸與轉(zhuǎn)化的數(shù)學(xué)思想方法,以及空間想象能力、推理論證能力和運算求解能力.