Question

If in $△ABC,\cos 3A+\cos 3B+\cos 3C=1$, then one angle must be exactly equal to

• A. ${90}^{\circ }$
• B. ${45}^{\circ }$
• C. ${120}^{\circ }$
• D. none of these

Answer 1

Answer: (C)

${120}^{\circ }$

Given, $\cos 3A+\cos 3B+\cos 3C=1$

$\therefore \cos 3A+\cos 3B=1-\cos 3C$

$2\cos \left(3\frac{A+B}{2}\right)\cos \left(\frac{3(A-B)}{2}\right)=2{\sin }^2\frac{3C}{2}$

$\cos \left(\frac{3\pi -3C}{2}\right)\cos \left(\frac{3(A-B)}{2}\right)={\sin }^2\frac{3C}{2}$

$-\sin \left(\frac{3C}{2}\right)\cos \left(\frac{3(A-B)}{2}\right)={\sin }^2\frac{3C}{2}$

${\sin }^2\left(\frac{3C}{2}\right)+\sin \left(\frac{3C}{2}\right)\cos \left(\frac{3(A-B)}{2}\right)=0$

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

$\sin \left(\frac{3\left(C\right)}{2}\right)=0$

$\frac{3C}{2}=\pi$

$C=\frac{2\pi }{3}$

Hence obtuse angled triangle.