在△ABC中,a^+b^=mc^,若cotC/(cotA+cotB)=1002,求m的值
问题描述：
在△ABC中,a^+b^=mc^,若cotC/(cotA+cotB)=1002,求m的值
答
cos C=(a^2+b^2-c^2)/2ab=(m-1)/2abcotC/(cotA+cotB)=cos C*sin A*sin B/(sin Acos B+sin Bcos A)*sinC=cos C*sin A* sin B/sin^2C=(m-1)/2ab*sin A*sin B/sin^2C=1002sin A/a=sin B/b=sin C/c（m-1）/2=1002m=2005...