Question

In a $△ABC$ with usual notations, if $\tan A,\tan B$ satisfy the equation $\sqrt{3}{x}^2-4x+\sqrt{3}<0$ and $(a-b{)}^2>c^2-ab$, then number of such triangles possible is

Answer 1

Given, $(a-b{)}^2>c^2-ab...(1)$
$\sqrt{3}{x}^2-4x+\sqrt{3}<0\Rightarrow \sqrt{3}{x}^2-3x-x+\sqrt{3}<0\Rightarrow (\sqrt{3}x-1)(x-\sqrt{3})<0\Rightarrow \frac{1}{\sqrt{3}}<x<\sqrt{3}\Rightarrow \frac{1}{\sqrt{3}}<\tan A,\tan B<\sqrt{3}\Rightarrow \frac{\pi }{6}<A,B<\frac{\pi }{3}\Rightarrow \frac{\pi }{3}<C<\frac{2\pi }{3}\Rightarrow -\frac{1}{2}<\cos C<\frac{1}{2}\Rightarrow -\frac{1}{2}<\frac{a^2+b^2-c^2}{2ab}<\frac{1}{2}\Rightarrow -ab<a^2+b^2-c^2<ab\Rightarrow c^2-ab<(a-b{)}^2+2ab<c^2+ab\Rightarrow c^2-3ab<(a-b{)}^2<c^2-ab...\left(2\right)$

Since, eqn $\left(1\right)$ and $\left(2\right)$ are contradict each others.
Therefore, no such triangle is possible.