在三角形ABC中,角A,B,C所对的边长分别是a,b,c.且满足c=bcosA(1)求角B的值
\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0c\u89d2A.B.C\u7684\u5bf9\u8fb9\u957f\u5206\u522b\u4e3aa.b.c\uff0c\u4e14\u6ee1\u8db3\uff082c\uff0da\uff09cosB\uff0dbcosA=0\uff081\uff09
\uff082c\uff0da\uff09cosB\uff0dbcosA=0
\u6b63\u5f26\u5b9a\u7406\uff1a
a=2RsinA,b=2RsinB,c=2RsinC
\u2234(2sinC-sinA)cosB-sinBcosA=0
\u22342sinCcosB-(sinAcosB+cosAsinB)=0
\u2235sinAcosB+cosAsinB=sin(A+B)=sinC>0
\u22342sinCcosB-sinC=0
cosB=1/2
\u2234B=60º
(2)
\u2235b=7,B=60º
\u4f59\u5f26\u5b9a\u7406\uff1a
b²=a²+c²-2accosB
\u2234a²+c²-ac=49 \u2460
\u2235a+c=13
\u2234a²+c²+2ac=169 \u2461
\u2460\u2461==>
ac=40
\u2234\u4e09\u89d2\u5f62ABC\u7684\u9762\u79ef
S=1/2accosB=1/2*40*\u221a3/2=10\u221a3
\u89e3\u6790\uff1a
\u7531\u6b63\u5f26\u5b9a\u7406\u6709\uff1aa/sinA=b/sinB=c/sinC
\u800c\u5df2\u77e5acosB\uff0dbcosA\uff1d(1/2)*c\uff0c\u90a3\u4e48\uff1a
sinAcosB\uff0dsinB*cosA\uff1d(1/2)*sinC
\u5373sin(A-B)=(1/2)*sinC
\u5219\u53ef\u77e5\uff1a-1/2 \u2264sin(A-B)\u22641/2\uff0c\u800c-\u03c0<A-B<\u03c0
\u53ef\u89e3\u5f97\uff1a5\u03c0/6\u2264A-B<\u03c0\u6216-\u03c0/6\u2264A-B\u2264\u03c0/6\u6216-5\u03c0/6\u2264A-B<-\u03c0
\u6240\u4ee5\uff1a\u5f53A-B=\u03c0/6\u5373sin(A-B)=(1/2)*sinC=1/2\u65f6\uff0ctan(A\uff0dB)\u6709\u6700\u5927\u503c\u4e3a(\u221a3)/3
\u6b64\u65f6sin(A-B)=(1/2)*sinC=1/2\uff0c\u5373sinC=1
\u89e3\u5f97\uff1a\u2220C=90\u00b0\u3002
则2c^2=c^2 + b^2 - a^2
则a^2+c^2 = b^2
勾股定理，B为直角
COS2分之A=5分之2倍根号5
则cosA=2(COS2分之A)^2 -1=3/5
则c=3，则b=5，得a=4，S=3*4/2=6
余弦定理和正玄定理，谢谢
因为c=bcosA，得出三角形ABC为直角三角形，角B为直角=90度。
cosA=2cos（A/2）^2-1=2*（2/5*√5）^2-1=3/5。b=c/cosA=3/（3/5）=5，a=√（5^2-3^2）=4
三角形ABC面积=1/2*4*3=6
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