在三角形ABC中，角A.B.C所对的边分别为a.b.c，a（cosC+根号3sinC）=b. 在△ABC中，角A，B，C所对的边分别为a，b，c，a（co...

\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0c\u89d2A,B,C\u6240\u5bf9\u7684\u8fb9\u5206\u522b\u4e3aabc\uff0ca(cosC+\u6839\u53f73sinC)=b

a(cosC+\u6839\u53f73sinC)=b
\u7531\u6b63\u5f26\u5b9a\u7406\u5f97\u5230\uff1a
sinAcosC\uff0b\u221a3sinAsinC\uff0dsinB=0
sinAcosC\uff0b\u221a3sinAsinC\uff0dsin(A\uff0bC)=0
sinAcosC\uff0b\u221a3sinAsinC\uff0dsinAcosC\uff0dcosAsinC=0
\u221a3sinAsinC\uff0dcosAsinC=0
\u221a3sinA=cosA
\u56e0tan(A)=\u221a3/3
\u5f97\uff1aA=30\u00b0

(2)S=1/2bcsinA=1/2bc*1/2=\u6839\u53f73/2,\u5f97\u5230bc=2\u6839\u53f73
\u53c8\u6709a^2=b^2+c^2-2bccosA
1=(b+c)^2-2bc-2bc*\u6839\u53f73/2
\uff08b+c)^2=1+4\u6839\u53f73+\u6839\u53f73\uff1d1+5\u6839\u53f73
\u4e0ebc=2\u6839\u53f73\u8054\u7acb\u5c31\u53ef\u89e3\u5f97b\u548cc\u4e86\u3002

\uff08I\uff09\u7531\u6b63\u5f26\u5b9a\u7406\u5f97\uff1asinA\uff08cosC+3sinC\uff09=sinB\uff0c\u53c8sinB=sin\uff08A+C\uff09\uff0c\u5316\u7b80\u5f97\uff1a3sinAsinC=cosAsinC\uff0c\u2235sinC\u22600\uff0c\u22343sinA=cosA\uff0c\u5373tanA=33\uff0c\u2235A\u4e3a\u4e09\u89d2\u5f62\u7684\u5185\u89d2\uff0c\u2234A=\u03c06\uff1b \uff08II\uff09\u6839\u636e\u9898\u610f\u5f9712bcsinA\uff1d32b2+c2?2bccosA\uff1da2\uff0c\u628aA=\u03c06\uff0ca=1\u4ee3\u5165\u89e3\u5f97\uff1ab\uff1d2c\uff1d3\u6216b\uff1d3c\uff1d2\uff0e

解答：
利用正弦定理a/sinA=b/sinB=c/sinC
∵a(cosC+√3sinC)=b
∴sinA(cosC+√3sinC)=sinB=sin(A+C)
∴sinAcosC+√3sinAsinC=sinAcosC+cosAsinC
∴√3sinAsinC=cosAsinC
∴ tanA=√3/3
（1）∴ A=30°
（2）S=(1/2)bcsinA=√3/2，
∴bc=2√3 ①
利用余弦定理
a²=b²+c²-2bccosA
∴ 1=b²+c²*4√3*(√3/2)
∴ 1=b²+c²-6
∴ b²+c²=7 ②
解方程组①②
解得b=√3，c=2或b=2，c=√3

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