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(A) Given that the side length $a = BC = 1$ unit. The perimeter of the triangle is $a + b + c$. The arithmetic mean of the sines of its angles is $\fr

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$\frac{\pi}{6}$

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(A) Given that the side length $a = BC = 1$ unit. The perimeter of the triangle is $a + b + c$. The arithmetic mean of the sines of its angles is $\frac{\sin A + \sin B + \sin C}{3}$.According to the problem,$a + b + c = 6 	imes \frac{\sin A + \sin B + \sin C}{3}$.This simplifies to $a + b + c = 2(\sin A + \sin B + \sin C)$.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: $2R(\sin A + \sin B + \sin C) = 2(\sin A + \sin B + \sin C)$.Since $\sin A + \sin B + \sin C 
eq 0$ for a triangle,we get $2R = 2$,which implies $R = 1$.From the sine rule,$\frac{a}{\sin A} = 2R$.Substituting $a = 1$ and $R = 1$,we get $\frac{1}{\sin A} = 2(1) = 2$.Therefore,$\sin A = \frac{1}{2}$,which gives $A = \frac{\pi}{6}$ (since $A$ is an angle of a triangle).

The perimeter of a $\triangle ABC$ is $6$ times the arithmetic mean of the sine of its angles. If its side $BC$ is of unit length,then $\angle A=$

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