If in a Delta ABC,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 $A+B+C = 180^o$,we have $\cos 3A

Vedclass · Accepted Answer

$120$

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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 $A+B+C = 180^o$,we have $\cos 3A + \cos 3B + \cos 3C = 1 - 4\sin \frac{3A}{2} \sin \frac{3B}{2} \sin \frac{3C}{2}$.Equating this to $1$,we get $1 - 4\sin \frac{3A}{2} \sin \frac{3B}{2} \sin \frac{3C}{2} = 1$.This implies $4\sin \frac{3A}{2} \sin \frac{3B}{2} \sin \frac{3C}{2} = 0$.Therefore,$\sin \frac{3A}{2} = 0$ or $\sin \frac{3B}{2} = 0$ or $\sin \frac{3C}{2} = 0$.This means $\frac{3A}{2} = 180^o$ or $\frac{3B}{2} = 180^o$ or $\frac{3C}{2} = 180^o$.Thus,$A = 120^o$ or $B = 120^o$ or $C = 120^o$.

If in a $\Delta ABC$,$\cos 3A + \cos 3B + \cos 3C = 1$,then one angle must be exactly equal to .......$^o$

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