Question

In a $△ABC,\tan A$ and $\tan B$ satisfy the inequation $\sqrt{3}{x}^2-4x+\sqrt{3}<0,$ then

• A. $a^2+b^2+ab>c^2$
• B. $a^2+b^2-ab<c^2$
• C. $a^2+b^2>c^2$
• D. none of these

Answer 1

Answer: (A), (B)

$a^2+b^2+ab>c^2$
$a^2+b^2-ab<c^2$

$\sqrt{3}{x}^2-4x+\sqrt{3}<0$
The factors are $(x-\sqrt{3})(\sqrt{3}x-1)<0$
$\therefore \frac{1}{\sqrt{3}}<x\sqrt{3}$
$\frac{\pi }{6}<A<\frac{\pi }{3}$ or $\frac{\pi }{6}<B<\frac{\pi }{3}$
$\frac{\pi }{6}+\frac{\pi }{6}<A+B<\frac{\pi }{3}+\frac{\pi }{3}$
$\Rightarrow \frac{\pi }{3}<A+B<\frac{2\pi }{3}$
$\Rightarrow \frac{\pi }{3}<\pi -C<\frac{2\pi }{3}$
$\Rightarrow -\pi -\frac{\pi }{3}<-C<-\pi -\frac{2\pi }{3}$
$\Rightarrow \pi +\frac{\pi }{3}<C<\pi +\frac{2\pi }{3}$
$\Rightarrow C>\frac{\pi }{3}$
$\because C>\frac{\pi }{3}$
$\cos C<\frac{1}{2}\Rightarrow \frac{a^2+b^2-c^2}{2ab}<\frac{1}{2}$
$\Rightarrow a^2+b^2-c^2<ab$
$\Rightarrow a^2+b^2-ab<c^2$
and $C<\frac{2\pi }{3}$
$\therefore \cos C=\cos \frac{2\pi }{3}=\cos (\pi -\frac{\pi }{3})=-\cos \frac{\pi }{3}=-\frac{1}{2}$
$\therefore \cos C>\frac{-1}{2}$
$\Rightarrow \frac{a^2+b^2-c^2}{2ab}>-\frac{1}{2}$
$\Rightarrow a^2+b^2-c^2>-ab$
$\Rightarrow a^2+b^2+ab>c^2$