Question

一道解三角形的题已知三角形ABC的三边长分别是2m+3,m*m+2m,m*m+3m+3(m>0),求最大内角的度数

Answer 1

三边长分别是2m+3,m*m+2m,m*m+3m+3,显然m^2+3m+3为最长边则其所对应的角是最大角.

根据余弦定理：

cosA=[(2m+3)^2+(m^2+2m)^2-(m^2+3m+3)^2]/2(2m+3)(m^2+2m)

=(4m^2+12m+9+m^4+4m^2+4m^3-m^4-9m^2-9-6m^3-6m^2-18m)/2(2m^3+4m^2+3m^2+6m)

=(-2m^3-7m^2-6m)/2(2m^3+7m^2+6m)

=-1/2

A=120