在三角形中内角ABC的对边分别为abc？ 在三角形ABC中，内角ABC对边分别为abc已知c方等于a方...
\u4e09\u89d2\u5f62abc\u5185\u89d2a\uff0cb\uff0cc\u5bf9\u8fb9\u5206\u522b\u4e3aa\uff0cb\uff0cc\u5df2\u77e5\u7b49\u5f0f\u4e24\u8fb9\u540c\u4e58\u4ee5a\uff0c\u5f97abcosC+accosB=2a
\u7531\u6b63\u5f26\u5b9a\u7406\u5f97a(sinBcosC+cosBsinC)=2sinA
asin(B+C)=2sinA
asinA=2sinA
a=2
cosA=2\u221a6/5
sinA=\u221a(1-cos²A)=\u221a[1-(2\u221a6/5)²]=1/5
\u7531\u6b63\u5f26\u5b9a\u7406\u5f97a/sinA=2R
R=a/(2sinA)=2/(2\u00b71/5)=5
S=\u03c0R²=\u03c0\u00b75²=25\u03c0
\u4e09\u89d2\u5f62ABC\u7684\u5916\u63a5\u5706\u9762\u79ef\u4e3a25\u03c0\u3002
\uff081\uff09\u3001
\u56e0\u4e3aa²+b²+ab=c²\uff0c\u5373a²+b²=c²-ab\uff0c
\u6240\u4ee5\u6839\u636e\u4f59\u5f26\u5b9a\u7406\u6709cos\u2220C=(a²+b²-c²)/2ab=(c²-ab-c²)/2ab=-ab/2ab=-1/2\uff0c
\u6613\u77e5\u2220C=120\u00b0\u3002
\uff082\uff09\u3001
\u56e0\u4e3a\u2220C=120\u00b0\uff0c\u6240\u4ee5sin\u2220C=\u221a3/2\uff0c\u53c8\u56e0\u4e3asin\u2220A=2/3\uff0ccos\u2220A=1-sin²\u2220A=\u221a5/3\uff0c
\u6240\u4ee5sin\u2220B=sin[180\u00b0-(\u2220A+\u2220C)]=sin(\u2220A+\u2220C)=sin\u2220Acos\u2220C+cos\u2220Asin\u2220C
=(2/3)(-1/2)+(\u221a5/3)(\u221a3/2)=(\u221a15-2)/6\uff0c
\u6839\u636e\u6b63\u5f26\u5b9a\u7406b/sin\u2220B=c/sin\u2220C\u53ef\u77e52/[(\u221a15-2)/6]=c/(\u221a3/2)\uff0c\u7b97\u5f97c=(18\u221a5+12\u221a3)/11\uff0c
\u6240\u4ee5\u25b3ABC\u7684\u9762\u79ef\u4e3a1/2\u00b7sin\u2220A\u00b7bc=(1/2)\u00d7(2/3)\u00d72\u00d7[(18\u221a5+12\u221a3)/11]=(12\u221a5+8\u221a3)/11\u3002
\uff08\u867d\u7136\u8fd9\u4e2a\u7ed3\u679c\u5f88\u957f\uff0c\u4f46\u662f\u6211\u5728\u51e0\u4f55\u753b\u677f\u9a8c\u8bc1\u8fc7\u4e86\uff0c\u5c31\u662f\u8fd9\u4e2a\u7b54\u6848\uff0c\u5982\u4e0b\u56fe\u6240\u793a\u3002\uff09
由正弦定理:a/sinA=b/sinB=c/sinC
则
2sinBcosC=2sinA+sinC
A+B+C=π
则A=π-(B+C)
带入：2sinBcosC=2sin(B+C)+sinC=2sinBcosC+2cosBsinC+sinC
则： 2cosBsinC+sinC=0
因为C≠0，则sinC≠0， 则
cosB=-1/2，则B=120°
~~~~~~~~~~~~~~~~~~
延长D至E,使得DE=BD,连接AE
因为BD=DE,AD=CD,∠BDC=∠ADE
则，可知△ADE≌△CDB
则 AE=BC
则∠BAE=∠B=120°
在△ABE中，由余弦定理:
c^2+a^2-2accosB=BE^2
即：c^2+4-2*2*c*(-1/2)=(2√3)^2
c=4 , c=-2(舍去）
2bcosC=2a+c， 2bcosC=(a^2+b^2-c^2)/a=2a+c, a^2+b^2c^2=2a^2+ac, a^2+c^2-b^2+ac=0, a^2+c^2-b^2=2accosB=ac, cosB=1/2, B=60度
AC=2√3, 2/sinA=2√3/sin60度， sinA=1/2, A=30度， C=90度
c²=b²+a²-2ab·cosC=b²+a²-a(2a+c)=b²-a²-ac==> ac=b²-a²-c²
b²=a²+c²-2ac·cosB==> cosB=(a²+c²-b²)/2ac=-1/2==> B=120°
如图，取BC中点E，连接中位线DE∥AB，由tan60°=√3，可知BD⊥CB，AB=2DE=4
tanC=√3/2，C=arctan√3/2=40.8934°。【估计印刷错误，题意应该是求c=AB=4】
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