Question

If in a $△ABC,\sum \cos 3A=1,$ then exactly one angle of $△$ is

• A. ${60}^0$
• B. ${30}^0$
• C. ${120}^0$
• D. ${150}^0$

Answer 1

The correct option is C ${120}^0$

$\sum \cos 3A=1\Rightarrow \cos 3A+\cos 3B+\cos 3C=1$

$\Rightarrow 2\cos \left(\frac{3A+3B}{2}\right)\cos \left(\frac{3A-3B}{2}\right)+\cos 3C=1$

$\Rightarrow 2\cos \frac{3}{2}(\pi -C)\cos (\frac{3A}{2}-\frac{3B}{2})+\cos 3C=1$

$\Rightarrow 2\cos (\frac{3\pi }{2}-\frac{3C}{2})\cos (\frac{3A}{2}-\frac{3B}{2})+1-2{\sin }^2\frac{3C}{2}=1$

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

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

$\Rightarrow -2\sin \frac{3C}{2}\times 2\sin \frac{3A}{2}\sin \frac{3B}{2}=0\Rightarrow \sin \frac{3A}{2}\sin \frac{3B}{2}\sin \frac{3C}{2}=0$

$\Rightarrow$ either $\frac{3A}{2}$ or $\frac{3B}{2}$ or $\frac{3C}{2}={180}^0$

$\Rightarrow A$ or $B$ or $C={120}^0$

But exactly one of $A,B$ and $C$ can be ${120}^0$