A+B=∏-C
tan(A+B)=tan(∏-C)
tan(A+B)=-tanC
tan =-tan(A+B)
And tana+tanb = tan (a+b) (1-tana× tanb).
So tanA+tan b+ tanC & gt;; 0
tan(A+B)( 1-tanA tanB)+(-tan(A+B))& gt; 0
tan(A+B)( 1-tanA tan B- 1)>0
tan(A+B)(-tanA tanB)>0
tanA tanB & lt0
When tan (a+b) >; O is 0 degrees; 0
This situation does not exist.
When tan(a+b)< 0 is 90 degrees.
And tanatanb >; O then angle a and angle b are acute angles, which is true.
So triangle ABC is an acute triangle.