In a Delta ABC,if cos 3A + cos 3B + cos 3C = | vedclass

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(C) Given $\cos 3A + \cos 3B + \cos 3C = 1$.Using the identity for $\cos 3A + \cos 3B + \cos 3C$ in a triangle where $A+B+C = \pi$:$\cos 3A + \cos 3B

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$\Delta ABC$ is obtuse-angled

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(C) Given $\cos 3A + \cos 3B + \cos 3C = 1$.Using the identity for $\cos 3A + \cos 3B + \cos 3C$ in a triangle where $A+B+C = \pi$:$\cos 3A + \cos 3B + \cos 3C = 1 - 4 \sin \frac{3A}{2} \sin \frac{3B}{2} \sin \frac{3C}{2} = 1$.This implies $4 \sin \frac{3A}{2} \sin \frac{3B}{2} \sin \frac{3C}{2} = 0$.Thus,$\sin \frac{3A}{2} = 0$ or $\sin \frac{3B}{2} = 0$ or $\sin \frac{3C}{2} = 0$.For $\sin \frac{3A}{2} = 0$,we have $\frac{3A}{2} = \pi \Rightarrow A = \frac{2\pi}{3} = 120^\circ$.Since one angle is $120^\circ$,which is greater than $90^\circ$,the triangle is obtuse-angled.

In a $\Delta ABC$,if $\cos 3A + \cos 3B + \cos 3C = 1$,then:

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