在三角形ABC中,角A、B、C成等差数列,cosC=1/3,b=3√3,求三角形ABC的面积 在三角形ABC中,a=3根号2,b=2根号3,cosC=1/...
\u5728\u4e09\u89d2\u5f62abc\u4e2dabc\u6210\u7b49\u5dee\u6570\u5217b\u7b49\u4e8e\u4e09\u6839\u53f7\u4e09\uff0ccosA=1/3\uff0c\u6c42\u89d2B\uff0c\u6c42a\u53cacosC\u8fd0\u75282RsinA=a\uff0c2RsinB=b\uff0c2RsinC=c\uff0c\u6839\u636e2b=a+c\uff0c\u6c42\u51fa\u89d2B\uff0c\u5e94\u8be5\u4e0d\u96be\u3002
C\u662f\u4e09\u89d2\u5f62\u5185\u89d2
\u6240\u4ee5sinC>0
sin²C+cos²C=1
\u6240\u4ee5sinC=2\u221a2/3
\u6240\u4ee5S=1/2absinC=4\u221a3
所以,B=60°
而cosC=1/3,sinC=(2√2)/3
sinA=sin(180°-B-C)
=sin(120°-C)
=sin(120°)cosC-sinCcos(120°)
=(√3+2√2)/6
b=3√3
由正弦定理知道
b/sinB=c/sinC
所以,c=bsinC/sinB=4√2
所以,S三角形ABC
=(1/2)bcsinA
=4√3+3√2
是等差数列,所以2b=a+c,再根据余弦定理,c/2=a/2+b/2-2ab*cosc,将a=2b-c代入余弦定理,b和cosc都知道,把c求出来,然后a和c都求出来了,面积就好求了,根据cosc求出sinc,面积就为(a*b*sinc)/2
对不起啊,做错了,我的这个做法是边长a,b,c成等差数列,看到楼上才知道我错了,楼上的做法正确!
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