Question

在△ABC中,a^+b^=mc^,若cotC/(cotA+cotB)=1002,求m的值

Answer 1

cos C=(a^2+b^2-c^2)/2ab=(m-1)/2ab

cotC/(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=1002

sin A/a=sin B/b=sin C/c

（m-1）/2=1002

m=2005