若△ABC的三边长a、b、c满足a2-a-2b-2c=0且a+2b-2c+3=0,则它的最大内角的度数是( )
A.150°
B.135°
C.120°
D.90°
【答案】
分析:联立已知的两等式,把a看作已知数解得b,c,显然c>b,假设c>a,列出关于a的不等式,求出不等式的解集得到a的范围,刚好符合题意,得到三角形最大边为c,由余弦定理表示出cosC,将表示出的c及b代入,整理后求得cosC的值,由C为三角形的内角,利用特殊角的三角函数值即可得到C的度数.
解答:解:把a
2-a-2b-2c=0和a+2b-2c+3=0联立可得,b=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101230831662881734/SYS201311012308316628817002_DA/0.png)
,c=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101230831662881734/SYS201311012308316628817002_DA/1.png)
,显然c>b.
接下来比较c与a的大小,
由b=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101230831662881734/SYS201311012308316628817002_DA/2.png)
>0,解得:a>3或a<-1(为负数,舍去),
假设c=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101230831662881734/SYS201311012308316628817002_DA/3.png)
>a,解得:a<1或a>3,其中a>3刚好符合,
∴c>a,即三角形最大边为c,
∴△ABC中C为最大角,
由余弦定理可得:c
2=a
2+b
2-2ab•cosC,
将b=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101230831662881734/SYS201311012308316628817002_DA/4.png)
,c=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101230831662881734/SYS201311012308316628817002_DA/5.png)
代入得:
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101230831662881734/SYS201311012308316628817002_DA/6.png)
=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101230831662881734/SYS201311012308316628817002_DA/7.png)
-2a•
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101230831662881734/SYS201311012308316628817002_DA/8.png)
•cosC,
解得:cosC=-
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101230831662881734/SYS201311012308316628817002_DA/9.png)
,又C为三角形的内角,
则C=120°.
故选C
点评:此题属于解三角形的题型,涉及的知识有:一元二次不等式的解法,余弦定理,以及特殊角的三角函数值,解题的思路为:根据已知的两等式用a表示出b与c,判断出c为最大边,C为最大角,然后利用余弦定理及特殊角的三角函数值来解决问题.