Question

In a $\Delta ABC,\tan A$ 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)

The correct options are
A $a^2+b^2-ab<c^2$
B $a^2+b^2+ab>c^2$

From the given equation,

$\tan A+\tan B=\frac{4}{\sqrt{3}}$.

And $\tan A.\tan B=1$.

Hence, $\tan (A+B)=\infty$

$\therefore A+B=\frac{\pi }{2}$

$\therefore C=\frac{\pi }{2}$.

Hence,
$a^2+b^2=c^2$

$a^2+b^2-c^2=0$...(i)

Now the sides of a triangle are always positive.

Hence, $ab>0$
$a^2+b^2+ab>a^2+b^2$

$a^2+b^2+ab>c^2$.

Similarly, $ab<0$
$a^2+b^2-ab<c^2$.