Question

在△abc,若(a^3+b^3+c^3)/(a+b+c)=c^2,且tanatanb=3,试判断△abc的形状

Answer 1

(a^3+b^3+c^3)/(a+b+c)=c^2

=> (a^3+b^3+c^3)=(a+b+c)*c^2

=> a^2+b^2-ab=c^2 又c^2=a^2+b^2-2abcosC

=> C=60

tan(A+B)=tan120=3^(1/2)=(tanA+tanB)/(1-tanAtanB) tanAtanB=3

=> tanA+tanB=-2*3^(1/2)

=> tanA=tanB=-3^(1/2)

所以是等边三角形