Question

In a $\Delta ABC,\tan \frac{A}{2}$ and $\tan \frac{B}{2}$ satisfy $6x^2-5x+1=0.$ Then

• A. $a^2+b^2>c^2$
• B. $a^2-b^2=c^2$
• C. $a^2+b^2=c^2$
• D. none of these

Answer 1

The correct option is C $a^2+b^2=c^2$

$\tan \frac{A}{2}+\tan \frac{B}{2}=\frac{5}{6},\tan \frac{A}{2}.\tan \frac{B}{2}=\frac{1}{6}$
Now $\tan \left(\frac{A+B}{2}\right)=\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2}.\tan \frac{B}{2}}=\frac{5/6}{1-1/6}=1$
$\Rightarrow \frac{A+B}{2}={45}^0\Rightarrow A+B={90}^0\Rightarrow \angle C={90}^0$
$\therefore c^2=a^2+b^2$