The perimeter of Delta ABC is 6 times the A.M | vedclass

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(D) Given that the perimeter $a+b+c = 6 	imes \frac{\sin A + \sin B + \sin C}{3}$.Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \si

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$\frac{\pi}{6}$

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(D) Given that the perimeter $a+b+c = 6 	imes \frac{\sin A + \sin B + \sin C}{3}$.Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$,where $R$ is the circumradius.Substituting these into the equation: $2R(\sin A + \sin B + \sin C) = 2(\sin A + \sin B + \sin C)$.This implies $2R = 2$,so $R = 1$.Since $a = 2R \sin A$ and $a = 1$,we have $1 = 2(1) \sin A$.Therefore,$\sin A = \frac{1}{2}$,which gives $\angle A = \frac{\pi}{6}$ (or $30^{\circ}$).

The perimeter of $\Delta ABC$ is $6$ times the $A.M.$ of the sines of its angles. If $a = 1$,then $\angle A = $

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