1. **State the problem:** We have a sector AOB of a circle with center O and radius $3r$ cm. Inside this sector, a smaller circle with radius $r$ cm touches the two radii OA and OB and the arc AB. The angle $\theta$ at O is in radians, with $0 < \theta < \frac{\pi}{2}$. (a) Find the exact value of $\theta$. (b) Given the shaded area between the sector and the smaller circle is $8\pi$ cm², find $r$. --- 2. **Formula and important rules:** - Area of sector with radius $R$ and angle $\theta$ is $\frac{1}{2} R^2 \theta$. - The smaller circle is tangent to both radii and the arc, so its center lies on the angle bisector of $\theta$. - The distance from the center of the smaller circle to O is $3r - r = 2r$ (since the smaller circle touches the arc of radius $3r$). - The smaller circle radius is $r$. --- 3. **Find $\theta$:** The smaller circle touches both radii OA and OB, so its center lies on the angle bisector of $\theta$. The distance from the center of the smaller circle to each radius line is $r$ (the radius of the smaller circle). Since the center lies on the bisector, the distance from the center to each radius line is $r = d = \text{distance from center to radius line}$. The distance from the center to each radius line is $d = (2r) \sin(\frac{\theta}{2})$ because the center is at distance $2r$ from O along the bisector, and the perpendicular distance to each radius line is $2r \sin(\frac{\theta}{2})$. Set this equal to $r$: $$r = 2r \sin\left(\frac{\theta}{2}\right)$$ Divide both sides by $r$ (assuming $r \neq 0$): $$\cancel{r} = 2 \cancel{r} \sin\left(\frac{\theta}{2}\right) \Rightarrow 1 = 2 \sin\left(\frac{\theta}{2}\right)$$ So, $$\sin\left(\frac{\theta}{2}\right) = \frac{1}{2}$$ Therefore, $$\frac{\theta}{2} = \frac{\pi}{6} \Rightarrow \theta = \frac{\pi}{3}$$ --- 4. **Find $r$ given shaded area = $8\pi$ cm²:** The shaded area is the area of the sector minus the area of the smaller circle: $$\text{Shaded area} = \frac{1}{2} (3r)^2 \theta - \pi r^2 = 8\pi$$ Substitute $\theta = \frac{\pi}{3}$: $$\frac{1}{2} (9r^2) \frac{\pi}{3} - \pi r^2 = 8\pi$$ Simplify: $$\frac{9r^2 \pi}{6} - \pi r^2 = 8\pi$$ $$\frac{3}{2} r^2 \pi - \pi r^2 = 8\pi$$ Factor out $\pi r^2$: $$\pi r^2 \left(\frac{3}{2} - 1\right) = 8\pi$$ Simplify inside parentheses: $$\pi r^2 \left(\frac{1}{2}\right) = 8\pi$$ Divide both sides by $\pi$: $$r^2 \frac{1}{2} = 8$$ Multiply both sides by 2: $$r^2 = 16$$ Take positive root (radius positive): $$r = 4$$ --- **Final answers:** (a) $\theta = \frac{\pi}{3}$ radians (b) $r = 4$ cm