Question

If $3ta{n}^2\theta -4\sqrt{3}tan\theta +3=0$ , find the acute angles $\theta$ .

• A. ${30}^{\circ }$ or ${60}^{\circ }$
• B. ${45}^{\circ }$ or ${45}^{\circ }$
• C. ${45}^{\circ }$ or ${30}^{\circ }$
• D. ${90}^{\circ }$ or $0^{\circ }$

Answer 1

The correct option is A ${30}^{\circ }$ or ${60}^{\circ }$

$3{\tan }^2\theta -4\sqrt{3}\tan \theta +3=0$
$3{\tan }^2\theta -3\sqrt{3}\tan \theta -\sqrt{3}\tan \theta +3=0$
$3\tan \theta (\tan \theta -\sqrt{3})-\sqrt{3}(\tan \theta -\sqrt{3})=0$
$(3\tan \theta -\sqrt{3})(\tan \theta -\sqrt{3})=0$
$\tan \theta =\frac{1}{\sqrt{3}}$ or $\tan \theta =\sqrt{3}$
$\theta ={30}^{\circ }$ or $\theta ={60}^{\circ }$