In ΔABC, if $\tan A + \tan B + \tan C = 3\sqrt{3}$, then the triangle is :

Equilateral

$\tan A + \tan B + \tan C = 3\sqrt{3};\frac{\tan A + \tan B + \tan C}{3}=\sqrt{3}=A.M.$

In a ΔABC, tan A + tan B + tan C = tan A . tan B . tan C $∴ \tan A . \tan B . \tan C = 3\sqrt{3}$

$(\tan A . \tan B . \tan C)^{1/3}=\sqrt{3}$ ∵ A.M. = G.M

∴ tan A = tan B = tan C ∴ A = B = C ⇒ Δ is equilateral