In a triangle ABC,2A + C = 300^{circ}. If the | vedclass

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(B) Given $2A + C = 300^{\circ}$ and $A + B + C = 180^{\circ}$.Subtracting the equations: $(2A + C) - (A + B + C) = 300^{\circ} - 180^{\circ} \implies

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$\frac{1}{4}$

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(B) Given $2A + C = 300^{\circ}$ and $A + B + C = 180^{\circ}$.Subtracting the equations: $(2A + C) - (A + B + C) = 300^{\circ} - 180^{\circ} \implies A - B = 120^{\circ}$.Also,$R = 8r$. The formula for the inradius is $r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.Substituting $R = 8r$,we get $r = 4(8r) \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \implies 32 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = 1$.Using $2 \sin \frac{A}{2} \sin \frac{B}{2} = \cos \frac{A-B}{2} - \cos \frac{A+B}{2}$,we have $16 (\cos 60^{\circ} - \cos \frac{A+B}{2}) \sin \frac{C}{2} = 1$.Since $A+B = 180^{\circ} - C$,$\cos \frac{A+B}{2} = \sin \frac{C}{2}$.$16 (\frac{1}{2} - \sin \frac{C}{2}) \sin \frac{C}{2} = 1 \implies 8 \sin \frac{C}{2} - 16 \sin^2 \frac{C}{2} = 1$.$16 \sin^2 \frac{C}{2} - 8 \sin \frac{C}{2} + 1 = 0 \implies (4 \sin \frac{C}{2} - 1)^2 = 0$.Therefore,$\sin \frac{C}{2} = \frac{1}{4}$.

In a $\triangle ABC$,$2A + C = 300^{\circ}$. If the circumradius $R$ of the $\triangle ABC$ is eight times its inradius $r$,then $\sin \frac{C}{2} = $

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