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(B) Let the center of the sector be $O$ and the radius of the sector be $r$. Let the center of the small circle be $C$ and its radius be $R$.The angle

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$\frac{r}{3}$

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(B) Let the center of the sector be $O$ and the radius of the sector be $r$. Let the center of the small circle be $C$ and its radius be $R$.The angle of the sector is $60^{\circ}$. The line $OC$ bisects this angle,so $\angle COP = 30^{\circ}$,where $P$ is the point of tangency on the radius of the sector.In the right-angled triangle formed by the center of the small circle $C$,the point of tangency $P$,and the center of the sector $O$,we have:$\sin 30^{\circ} = \frac{CP}{OC}$$\frac{1}{2} = \frac{R}{OC}$$OC = 2R$Also,the distance from the center of the sector $O$ to the arc along the bisector is the radius of the sector $r$. This distance is the sum of the distance $OC$ and the radius of the small circle $R$ (since the small circle is tangent to the arc).$r = OC + R$$r = 2R + R = 3R$$R = \frac{r}{3}$

$A$ circle is drawn in a sector of a larger circle of radius $r$,as shown in the figure. The smaller circle is tangent to the two bounding radii and the arc of the sector. The radius of the small circle is

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