在△ABC中,内角A.B.C的对边分别是a.b.c,其中b=(根号3)÷2,tanA+tanC+tan60°=tanAtanCtan60°. 在△ABC中,内角A,B,C的对边分别是a,b,c,已知a,...
\u5728\u4e09\u89d2\u5f62ABC\u4e2d\u5df2\u77e5\u5185\u89d2A\u3001B\u3001C\u6240\u5bf9\u7684\u8fb9\u5206\u522b\u4e3aa\u3001b\u3001c.\u5411\u91cfm=\uff08tanA+tanC\uff0c\u6839\u53f73\uff09\u2235m\u2225n,
\u2234(tanA+tanc)*1=\u221a3(tanAtanC-1)
(tanA+tanC)=-\u221a3(1-tanAtanC)
tanA+tanC
_____________=-\u221a3
1-tanAtanC
\u2234tan(A+C)=-\u221a3
\u2234 tanB= \u221a3
\u2220B=60\u00b0
a,b,c\u6210\u7b49\u6bd4\u6570\u5217,\u6240\u4ee5a*c=b^2
\u6839\u636e\u6b63\u5f26\u5b9a\u7406\uff0ca/sinA=b/sinB=c/sinC
\u6240\u4ee5sinA=a/b*sinB,sinC=c/b*sinC
\u6709cosB=3/5
sinB=\u6839\u53f71-9/25=4/5
cosA/sinA+cosC/sinc
=(cosA*sinC+sinA*cosC)/sinA*sinC
=sin(A+C)/[(a/b*sinB)*(c/b*sinB)]
=sinB/[(a/b*sinB)*(c/b*sinB)]
=1/sinB
=5/4
解：
（1）
由tanA+tanC+tan(π/3)=tanAtanCtan(π/3)
可以得出 tanA+tanC=-√3*（1-tanAtanC）
（tanA+tanC）/（1-tanAtanC）=tan(A+C)=-√3
在三角形中 tanB=-tan(A+C)=√3
∴B=π/3
（2）
正弦定理a/sinA=b/sinB=c/sinC
∴（a+c）/(sinA+sinC)=b/sinB=(√3/2)/(√3/2)=1
即a+c=sinA+sinC
∵B=π/3
∴C=2π/3-A
a+c=sinA+sin(2π/3-A) 展开再化简
a+c=√3sin(A+π/6)
∵A∈（0，2π/3)
∴A+π/6∈（π/6，5π/6)
∵1/2<3sin(A+π/6)≤1
∴a+c的取值范围是：（√3/2，√3]
a²-b²=√3bc
sinC=2√3sinB→2R*sinC=2R*2√3sinB→c=2√3b→c²=2√3bc
cosA=（b²+c²-a²）/（2bc）
=（c²-（a²-b²））/（2bc）
=（2√3bc-√3bc）/（2bc）
=√3/2
所以A=π/6
根据正弦定理，b/sinB=a/sinA,a=2√3，
A=π/3，B=x,b=4sinx,c/sinC=a/sinA,
c=2√3/(√3/2)*sinC=4sinC=4sin(A+B)
=4sin(π/3+x)=2√3cosx+2sinx
周长:y=a+b+c=2√3+4sinx+2√3cosx+2sinx
=2√3+6sinx+2√3cosx
0<x<2π/3,
y=2√3+2√3(√3sinx+cosx)
=2√3+4√3[sinx*cos(π/6)+cosx*sin(π/6)]
=2√3+4√3sin(x+π/6)
当sin(x+π/6)=1时函数有最大值，
y=6√3
希望能帮到你
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