Question

In $△ABC,\Sigma \cos 3A=1$, then:

• A. $△ABC$ is a right angled triangle
• B. $△ABC$ is an obtuse angled triangle
• C. $△ABC$ is an acute angled triangle
• D. In radius ${}^{\prime }{r}^{\prime }$ of $△ABC$ is either $\sqrt{3}(s+a)$ or $\sqrt{3}(s+b)$ or $\sqrt{3}(s+c)$

Answer 1

The correct option is B $△ABC$ is an obtuse angled triangle

$\cos 3A+\cos 3B+\cos 3C=1$
Or
$2\cos \left(\frac{3(A+B)}{2}\right)\cos \left(\frac{3(A-B)}{2}\right)=1-\cos 3C$
Or
$-2\sin \left(\frac{C}{2}\right).\cos \left(\frac{3(A-B)}{2}\right)=2{\sin }^2\frac{3C}{2}$
Or
$2\sin \left(\frac{3C}{2}\right)\left[\cos \right(\frac{3(A-B)}{2})+\sin (\frac{C}{2}\left)\right]=0$
Or
$\sin \frac{3C}{2}=0$
Or
$\frac{3C}{2}=\pi$
Or
$C=\frac{2\pi }{3}$

$={120}^0$.
Hence an obtuse angled triangle.