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.