Question

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

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

Answer 1

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

$\sqrt{3}{x}^2-3x-x+\sqrt{3}=0$

$\sqrt{3}x(x-\sqrt{3})-1(x-\sqrt{3})=0$

$x=\frac{1}{\sqrt{3}},x=\sqrt{3}$

$(x-\sqrt{3})(\sqrt{3}x-1)\angle 0$

$\therefore \frac{1}{\sqrt{3}}\angle x\angle \sqrt{3}$

$\frac{1}{\sqrt{3}}\angle \tan A\angle \sqrt{3}$

${30}^0\angle A\angle {60}^0$