【答案】
分析:△ABC中設(shè)AB=c,BC=a,AC=b,由sinB=cosA•sinC結(jié)合三角形的內(nèi)角和及和角的正弦公式化簡可求 cosC=0 即C=90°,再由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/0.png)
,S
△ABC=6可得bccosA=9,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/1.png)
可求得c=5,b=3,a=4,考慮建立以AC所在的直線為x軸,以BC所在的直線為y軸建立直角坐標(biāo)系,由P為線段AB上的一點(diǎn),則存在實數(shù)λ使得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/2.png)
=(3λ,4-4λ)(0≤λ≤1),設(shè)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/3.png)
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/4.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/5.png)
,由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/6.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/7.png)
=(x,0)+(0,y)=(x,y)可得x=3λ,y=4-4λ則4x+3y=12而
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/8.png)
,利用基本不等式求解最小值.
解答:![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/images9.png)
解:△ABC中設(shè)AB=c,BC=a,AC=b
∵sinB=cosA•sinC∴sin(A+C)=sinCcosnA
即sinAcosC+sinCcosA=sinCcosA
∴sinAcosC=0∵sinA≠0∴cosC=0 C=90°
∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/9.png)
,S
△ABC=6
∴bccosA=9,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/10.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/11.png)
,根據(jù)直角三角形可得sinA=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/12.png)
,cosA=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/13.png)
,bc=15
∴c=5,b=3,a=4
以AC所在的直線為x軸,以BC所在的直線為y軸建立直角坐標(biāo)系可得C(0,0)A(3,0)B(0,4)
P為線段AB上的一點(diǎn),則存在實數(shù)λ使得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/14.png)
=(3λ,4-4λ)(0≤λ≤1)
設(shè)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/15.png)
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/16.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/17.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/18.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/19.png)
=(x,0)+(0,y)=(x,y)
∴x=3λ,y=4-4λ則4x+3y=12
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/20.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/21.png)
故所求的最小值為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/22.png)
故選:C
點(diǎn)評:題是一道構(gòu)思非常巧妙的試題,綜合考查了三角形的內(nèi)角和定理、兩角和的正弦公式及基本不等式求解最值問題,解題的關(guān)鍵是理解把已知所給的
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/23.png)
是一個單位向量,從而可用x,y表示
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024914123163/SYS201311031030249141231009_DA/24.png)
,建立x,y與λ的關(guān)系,解決本題的第二個關(guān)鍵點(diǎn)在于由x=3λ,y=4-4λ發(fā)現(xiàn)4x+3y=12為定值,從而考慮利用基本不等式求解最小值