Question

In a triangle $ABC$ if $\tan C$ satisfy the inequation $\sqrt{3}{x}^2-4x+\sqrt{3}<0$ then

• A. $0^0<C<{30}^0$
• B. ${30}^0<C<{45}^0$
• C. ${30}^0<C<{60}^0$
• D. ${45}^0<C<{90}^0$

Answer 1

The correct option is C ${30}^0<C<{60}^0$

$(\sqrt{3}x-1)(x-\sqrt{3})<0$
$+1-1+$
$\frac{1}{\sqrt{3}}\sqrt{3}$
$\frac{1}{\sqrt{3}}<tanC<\sqrt{3}$
${30}^0<C<{60}^0$