在三角形ABC中,角A,B,C所对的边分别为abc,满足(a+c)/b=(sinA-sinB)/(sinA-sinC),求（a+b）/c的范围。 在三角形abc中角ABC的对边分别为abc且满足bcosA=...
\u5728\u4e09\u89d2\u5f62ABC\u4e2d,\u89d2A,B,C\u6240\u5bf9\u7684\u8fb9\u5206\u522b\u4e3aabc,\u6ee1\u8db3(a+c)/b=(sinA-sinB)/(sinA-sinC),\u6c42\uff08a+b\uff09/c\u7684\u8303\u56f4\u3002\u6839\u636e\u6b63\u5f26\u5b9a\u7406\uff1a
a/sinA=b/sinB=c/sinC=k
(a+c)/b=(sinA-sinB)/(sinA-sinC),
=(a/k-b/k)/(a/k-c/k)
=(a-b)/(a-c)
(a+c)(a-c)=(a-b)b
a²-c²=ab-b²
a²+b²=c²+ab
\u6839\u636e\u4f59\u5f26\u5b9a\u7406
a²+b²=c²+2*cosC*ab
\u6240\u4ee5\u5bf9\u6bd4\u5f97\u5230\uff1a2*cosC=1
\u6240\u4ee5cosC=1/2
C=60\u00b0
\u56e0\u4e3a\uff1aa/sinA = b/sinB = c/sinC = 2R \uff0cR \u4e3a\u25b3ABC \u5916\u63a5\u5706\u7684\u534a\u5f84\u3002\u6240\u4ee5\u6709\uff1a
a = 2RsinA, b = 2RsinB, c = 2RsinC
\u90a3\u4e48\uff0c\u4ee3\u5165\u8fd9\u4e2a\u6761\u4ef6\u5f0f\u4e2d\uff0c\u53ef\u4ee5\u5f97\u5230\uff1a
2RsinBcosA = (4RsinC+2RsinA)cos(A+C)
sinBcosA = (2sinC+sinA)cos(180\u00b0-B)
sinBcosA = (2sinC+sinA)(-cosB)=-2sinC*cosB - sinAcosB
\u79fb\u9879\uff0c
sinBcosA + cosBsinA = -2sinC*cosB
sin(A+B) = -2sinC * cosB \u6ce8\uff1asin(\u03b1+\u03b2)=sin\u03b1cos\u03b2 +cos\u03b1sin\u03b2
sin(180\u00b0-C)=-2sinC * cosB
sinC = -2sinC * cosB
\u6240\u4ee5\uff0ccosB = -1/2
\u56e0\u6b64\uff0cB = 120\u00b0
利用正弦定理化简已知等式得：
（a+c）/b=(a−b)/(a−c)，
化简得a^2+b^2-ab=c^2，
即a^2+b^2-c^2=ab，
∴cosC=(a^2+b^2−c^2)/2ab=1/2，
∵C为三角形的内角，
∴C=π/3
(a+b)/c
=(sinA+sinB)/sinC
=2/√3[sinA+sin（2π/3-A）]
=2sin（A+π/6），
∵A∈（0，2π/3），
∴A+π/6∈（π/6，5π/6），
∴sin（A+π/6）∈（1/2，1]，
则(a+b)/c的取值范围是（1，2]．
绛旓細锛1锛夊洜涓篵=1,c=鈭3,鈭燙=2/3蟺,鎵浠ョ敱姝ｅ鸡瀹氱悊寰楋細sinB=1/2锛孊=30掳锛屾墍浠osB=鈭3/2,(2)鍥犱负C=120掳锛孊=30掳锛屾墍浠=30掳锛屾墍浠=b=1
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