1. **State the problem:** We know the area of a sector of a circle, $A$, varies directly as the angle of the sector, $\theta$, and the square of the radius, $r$. Given $r=6$ cm, $\theta=70^\circ$, and $A=22$ cm². 2. **Express $A$ in terms of $\theta$ and $r$:** Since $A$ varies directly as $\theta$ and $r^2$, we write: $$A = k \theta r^2$$ where $k$ is the constant of proportionality. 3. **Find $k$ using the given values:** Substitute $A=22$, $\theta=70$, and $r=6$: $$22 = k \times 70 \times 6^2$$ $$22 = k \times 70 \times 36$$ $$22 = 2520k$$ Divide both sides by 2520: $$\frac{22}{\cancel{2520}} = k \times \cancel{2520}$$ $$k = \frac{22}{2520} = \frac{11}{1260}$$ 4. **Final formula for $A$:** $$A = \frac{11}{1260} \theta r^2$$ 5. **Calculate $\theta$ when $A$ doubles and $r=12$ cm:** The new area is $2 \times 22 = 44$ cm². Use the formula: $$44 = \frac{11}{1260} \theta \times 12^2$$ $$44 = \frac{11}{1260} \theta \times 144$$ Multiply both sides by 1260: $$44 \times 1260 = 11 \theta \times 144$$ $$55440 = 1584 \theta$$ Divide both sides by 1584: $$\frac{55440}{\cancel{1584}} = \theta \times \cancel{1584}$$ $$\theta = \frac{55440}{1584} = 35$$ **Answer:** (a) $A = \frac{11}{1260} \theta r^2$ (b) The angle of the sector when the area doubles and $r=12$ cm is $35^\circ$.