已知在锐角△ABC中，内角A，B，C的对边分别为a，b，c，若a＝1，2cosC＋c＝2 在锐角△ABC中，角A，B，C的对边分别为a，b，c．已知s...
\u5df2\u77e5\u9510\u89d2\u25b3ABC\u4e2d\uff0c\u5185\u89d2A\uff0cB\uff0cC\u7684\u5bf9\u8fb9\u5206\u522b\u4e3aa\uff0cb\uff0cc\uff0c\u4e14c=6\uff0c sin2C=- 3 cos2C \uff0c\uff081\uff09\u6c42\u89d2C\uff081\uff09\u2235sin2C=- 3 cos2C\uff0c\u5373tan2C=- 3 \uff0c\u53c8C\u4e3a\u9510\u89d2\uff0c\u22342C\u2208\uff080\uff0c\u03c0\uff09\uff0c\u22342C= 2\u03c0 3 \uff0c\u2234C= \u03c0 3 \uff1b\uff082\uff09\u2235\u5728\u9510\u89d2\u25b3ABC\u4e2d\uff0csinA= 1 3 \uff0csinC= 3 2 \uff0cc=6\uff0c\u2234\u6839\u636e\u6b63\u5f26\u5b9a\u7406\u5f97\uff1a c sinC = a sinA \uff0c\u5373a= csinA sinC = 6\u00d7 1 3 3 2 = 4 3 3 \uff0c\u2235\u53c8sinA= 1 3 \uff0c\u4e14A\u4e3a\u9510\u89d2\uff0c\u2234cosA= 1-si n 2 A = 2 2 3 \uff0c\u2234sinB=sin\uff08A+C\uff09=sinAcosC+cosAsinC= 1+2 6 6 \uff0c\u2234S \u25b3ABC = 1 2 acsinB= 2 3 +12 2 3 \uff0e
\uff08\u2160\uff09\u7531sin\uff08A-B\uff09=cosC\uff0c\u5f97sin\uff08A-B\uff09=sin\uff08\u03c02-C\uff09\uff0c\u2235\u25b3ABC\u662f\u9510\u89d2\u4e09\u89d2\u5f62\uff0c\u2234A-B=\u03c02-C\uff0c\u5373A-B+C=\u03c02\uff0c\u2460\u53c8A+B+C=\u03c0\uff0c\u2461\u7531\u2461-\u2460\uff0c\u5f97B=\u03c04\uff0c\u7531\u4f59\u5f26\u5b9a\u7406b2=c2+a2-2cacosB\uff0c\u5f97\uff0810\uff092=c2+\uff0832\uff092-2c\u00d732cos\u03c04\uff0c\u6574\u7406\u5f97\uff1ac2-6c+8=0\uff0c\u89e3\u5f97\uff1ac=2\uff0c\u6216c=4\uff0c\u5f53c=2\u65f6\uff0cb2+c2-a2=\uff0810\uff092+22-\uff0832\uff092=-4\uff1c0\uff0c\u2234b2+c2\uff1ca2\uff0c\u6b64\u65f6A\u4e3a\u949d\u89d2\uff0c\u4e0e\u5df2\u77e5\u77db\u76fe\uff0c\u6545c\u22602\uff0c\u5219c=4\uff1b\uff08\u2161\uff09\u7531\uff08\u2160\uff09\uff0c\u77e5B=\u03c04\uff0c\u2234A+C=3\u03c04\uff0c\u5373C=3\u03c04-A\uff0c\u2234\u5229\u7528\u6b63\u5f26\u5b9a\u7406\u5316\u7b80\u5f97\uff1aacosC?ccosAb=sinAcosC?cosAsinCsinB=<table cellpadding="-1" cellspacing="-1" style="margin-rig
解：∵a=1，2cosC+c=2b，
∴2acosC+c=2b，
2sinAcosC+sinC=2sinB
2sinAcosC+sinC=2sin（A+C）
2sinAcosC+sinC=2sinAcosC+2cosAsinC
sinC=2cosAsinC
2cosA=1
cosA=1/2
cosA=（b²+c²-a²）/2bc=（b²+c²-1）/2bc=1/2
b²+c²-1=bc
（b+c）²-1=3bc，
∵bc≤1/4（b+c）²
∴（b+c）²-1≤3/4（b+c）²，
∴（b+c）²≤4
∴b+c≤2，
∴a+b+c≤3，
∵b+c＞a（三角形两边之和大于第三边），
∴a+b+c＞2，
∴△ABC的周长取值范围（2，3]
（1）2acosc+c=2b，利用正弦定理2sinacosc+sinc=2sinb，
将sinb=sin（a+c）=sinacosc+cosasinc代入得sinc=2cosa
sinc，
即cosa＝
1
2
，a＝
π
3
（6分）
（2）由
b
sinb
＝
c
sinc
＝
a
sina
＝
2
3
得，l△abc＝
2
3
(sinb+sinc)+1，
将c＝
2π
3
?b代入化简得l△abc＝2sin(b+
π
6
)+1，因为
π
6
＜b+
π
6
＜
5π
6
所以周长的取值范围是（2，3]（12分）
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