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The perimeter of a $\triangle \mathrm{ABC}$ is 6 times the arithmetic mean of the values of the sine of its angles. If its side $\mathrm{BC}$ is of unit length, then $\angle \mathrm{A}=$
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$\frac{\pi}{6}$
$6\left(\frac{\sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C}}{3}\right)=a+b+c$
$\begin{aligned} & =2(\sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C})=2 \mathrm{R}(\sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C}) \\ & 2 \mathrm{R}=2 \\ & a=1 ; a=2 \mathrm{R} \sin \mathrm{A} \\ & 1=2 \sin \mathrm{A} \\ & \sin \mathrm{A}=\frac{1}{2} \\ & \mathrm{~A}=30^{\circ}\end{aligned}$
$\begin{aligned} & =2(\sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C})=2 \mathrm{R}(\sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C}) \\ & 2 \mathrm{R}=2 \\ & a=1 ; a=2 \mathrm{R} \sin \mathrm{A} \\ & 1=2 \sin \mathrm{A} \\ & \sin \mathrm{A}=\frac{1}{2} \\ & \mathrm{~A}=30^{\circ}\end{aligned}$
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