Question

If $\tan A,\tan B$ are the roots the quadratic equation $\sqrt{3}{x}^2-2x-\sqrt{3}=0$, $0<A+B<\pi$, then $A+B$ is equal to

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

Answer 1

The correct option is B ${30}^{\circ }$

$\because \tan A$ and $\tan B$ are the roots of the equation $\sqrt{3}{x}^2-2x-\sqrt{3}=0$
$\Rightarrow \tan A+\tan B=\frac{2}{\sqrt{3}}$ and $\tan A\cdot \tan B=-1$

From $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$
$=\frac{\frac{2}{\sqrt{3}}}{1-(-1)}=\frac{1}{\sqrt{3}}$
$\Rightarrow \tan (A+B)=\tan {30}^{\circ }$
$\Rightarrow A+B={30}^{\circ }$