Question

在△ABC中,设tanA+tanB+根号3=根号3tanAtanB,则角C为多少度?

Answer 1

tanA+tanB+√3=√3tanAtanB

tanA+tanB＝√3(tanAtanB-1)

所以-√3=(tanA+tanB)/[1-tanAtanB]

tanC=tan(180-A-B)=-tan(A+B)=-(tanA+tanB)/[1-tanAtanB]

所以tanC＝√3,C＝60度,A