在△ABC中，内角A，B，C的对边分别为a，b，c，且满足cosA=3/5，向量AB*向量AC=3。（1）求三角形ABC的面积
\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0c\u89d2A B C\u6240\u5bf9\u7684\u8fb9\u5206\u522b\u4e3aa b c\uff0c\u4e14\u6ee1\u8db3cosA=3/5\uff0c\u5411\u91cfAB*\u5411\u91cfAC\uff081\uff09cosA=3/5
\u2234sinA=\u221a(1-cos²A)=4/5
\u2235\u5411\u91cfAB*\u5411\u91cfAC=3
\u2234|AB|*|AC|cosA=3
\u2234|AB||AC|=5
\u2234\u4e09\u89d2\u5f62ABC\u7684\u9762\u79ef
S=1/2|AB||AC|sinA=1/2*5*4/5=22
\uff082\uff09\u2235c=1 ,bc=5,b=5
\u6839\u636e\u4f59\u5f26\u5b9a\u7406 a²=b²+c²-2bccosA=25+1-2\u00d75\u00d73/5=20
\u5f97a=2\u221a5
向量AB*向量AC=|AB|*|AC|*cosA=3 所以|AB|*|AC|=5
cosA=3/5 所以sinA=4/5
三角形面积S=1/2*|AB|*|AC|*sinA=2
sinA=4/5
I向量ABI*I向量ACI=向量AB*向量AC/cosA=5
三角形ABC的面积=1/2x5X4/5=2
cosA=3/5则sinA=4/5
向量AB*向量AC=|AB|*|AC|cosA=3
则|AB|*|AC|=5
三角形ABC面积=1/2*|AB|*|AC|sinA=2
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