Question

If in a triangle ABC, $\cos 3A+\cos 3B+\cos 3C=1$, then

• A. $\Delta ABC$ is a right angled triangle.
• B. $\Delta ABC$ is an obtuse angle triangle
• C. $\Delta ABC$ is an acute angled triangle.
• D. Inradius 'r' of the triangle ABC is either $\sqrt{3}(s-a)$ or $\sqrt{3}(s-b)$ or $\sqrt{3}(s-c)$.

Answer 1

Answer: (B)

$\Delta ABC$ is an obtuse angle triangle

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