题目内容
在△ABC中,角A的对边长等于2,向量![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_ST/0.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_ST/1.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_ST/2.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_ST/3.png)
(1)求
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_ST/4.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_ST/5.png)
(2)在(1)的条件下,求△ABC面积的最大值.
【答案】分析:(1)根据平面向量的数量积的运算法则求出
•
,然后根据三角形的内角和定理,利用二倍角的余弦函数公式化简后进行配方得到
•
=-2
,由
为锐角,利用二次函数求最值得到
•
取最小值时sin
=
,根据特殊角的三角函数值求出A即可;
(2)由a=2,根据第一问求得cosA的值,利用余弦定理和基本不等式求出bc的最大值,根据S△ABC=
bcsinA=
bc,把bc的最大值代入到面积公式里得到面积的最大值.
解答:解:(1)
•
=2
-
.
因为A+B+C=π,所以B+C=π-A,
于是
•
=
+cosA=-2
=-2
.
因为
,所以当且仅当
=
,即A=
时,
•
取得最大值
.
故
•
取得最大值时的角A=
;
(2)设角、B、C所对的边长分别为a、b、c由余弦定理,得b2+c2-a2=2bccosA
即bc+4=b2+c2≥2bc,所以bc≤4,当且仅当b=c=2时取等号.
又S△ABC=
bcsinA=
bc≤
.当且仅当a=b=c=2时,△ABC的面积最大为
.
点评:考查学生会进行平面向量的数量积的运算,灵活运用二次函数求值的方法及灵活运用余弦定理化简求值.会利用基本不等式求最值.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/0.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/1.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/2.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/3.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/4.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/5.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/6.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/7.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/8.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/9.png)
(2)由a=2,根据第一问求得cosA的值,利用余弦定理和基本不等式求出bc的最大值,根据S△ABC=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/10.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/11.png)
解答:解:(1)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/12.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/13.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/14.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/15.png)
因为A+B+C=π,所以B+C=π-A,
于是
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/16.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/17.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/18.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/19.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/20.png)
因为
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/21.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/22.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/23.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/24.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/25.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/26.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/27.png)
故
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/28.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/29.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/30.png)
(2)设角、B、C所对的边长分别为a、b、c由余弦定理,得b2+c2-a2=2bccosA
即bc+4=b2+c2≥2bc,所以bc≤4,当且仅当b=c=2时取等号.
又S△ABC=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/31.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/32.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/33.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212443715814761/SYS201310232124437158147014_DA/34.png)
点评:考查学生会进行平面向量的数量积的运算,灵活运用二次函数求值的方法及灵活运用余弦定理化简求值.会利用基本不等式求最值.
![](http://thumb.zyjl.cn/images/loading.gif)
练习册系列答案
相关题目