1. **State the problem:** We need to find the angle of rotation about the origin that maps triangle $\triangle XYZ$ with vertices $X(-4,2)$, $Y(-6,6)$, $Z(-2,6)$ to triangle $\triangle X'Y'Z'$ with vertices $X'(2,4)$, $Y'(6,6)$, $Z'(6,2)$. 2. **Recall the rotation formula:** A rotation about the origin by an angle $\theta$ transforms a point $(x,y)$ to $(x',y')$ using: $$ \begin{cases} x' = x\cos\theta - y\sin\theta \\ y' = x\sin\theta + y\cos\theta \end{cases} $$ 3. **Choose corresponding points to find $\theta$:** Use point $X(-4,2)$ and its image $X'(2,4)$. Substitute into the formulas: $$ 2 = -4\cos\theta - 2\sin\theta $$ $$ 4 = -4\sin\theta + 2\cos\theta $$ 4. **Rewrite the system:** $$ -4\cos\theta - 2\sin\theta = 2 $$ $$ -4\sin\theta + 2\cos\theta = 4 $$ 5. **Solve for $\cos\theta$ and $\sin\theta$:** Multiply the first equation by 2 and the second by 1 to align coefficients: $$ -8\cos\theta - 4\sin\theta = 4 $$ $$ -4\sin\theta + 2\cos\theta = 4 $$ 6. **Add the two equations:** $$ (-8\cos\theta + 2\cos\theta) + (-4\sin\theta - 4\sin\theta) = 4 + 4 $$ $$ -6\cos\theta - 8\sin\theta = 8 $$ 7. **Isolate $\cos\theta$:** $$ -6\cos\theta = 8 + 8\sin\theta $$ $$ \cos\theta = -\frac{8 + 8\sin\theta}{6} = -\frac{4}{3} - \frac{4}{3}\sin\theta $$ 8. **Use the Pythagorean identity:** $$ \sin^2\theta + \cos^2\theta = 1 $$ Substitute $\cos\theta$: $$ \sin^2\theta + \left(-\frac{4}{3} - \frac{4}{3}\sin\theta\right)^2 = 1 $$ 9. **Expand and simplify:** $$ \sin^2\theta + \left(\frac{16}{9} + \frac{32}{9}\sin\theta + \frac{16}{9}\sin^2\theta\right) = 1 $$ $$ \sin^2\theta + \frac{16}{9} + \frac{32}{9}\sin\theta + \frac{16}{9}\sin^2\theta = 1 $$ 10. **Combine like terms:** $$ \left(1 + \frac{16}{9}\right)\sin^2\theta + \frac{32}{9}\sin\theta + \frac{16}{9} = 1 $$ $$ \frac{25}{9}\sin^2\theta + \frac{32}{9}\sin\theta + \frac{16}{9} = 1 $$ 11. **Multiply both sides by 9 to clear denominators:** $$ 25\sin^2\theta + 32\sin\theta + 16 = 9 $$ 12. **Bring all terms to one side:** $$ 25\sin^2\theta + 32\sin\theta + 7 = 0 $$ 13. **Solve quadratic equation for $\sin\theta$:** $$ \sin\theta = \frac{-32 \pm \sqrt{32^2 - 4 \cdot 25 \cdot 7}}{2 \cdot 25} = \frac{-32 \pm \sqrt{1024 - 700}}{50} = \frac{-32 \pm \sqrt{324}}{50} $$ $$ = \frac{-32 \pm 18}{50} $$ 14. **Calculate both solutions:** $$ \sin\theta_1 = \frac{-32 + 18}{50} = \frac{-14}{50} = -0.28 $$ $$ \sin\theta_2 = \frac{-32 - 18}{50} = \frac{-50}{50} = -1 $$ 15. **Find corresponding $\cos\theta$ values:** For $\sin\theta = -0.28$: $$ \cos\theta = -\frac{4}{3} - \frac{4}{3}(-0.28) = -\frac{4}{3} + \frac{1.12}{3} = -1.333 + 0.373 = -0.96 $$ For $\sin\theta = -1$: $$ \cos\theta = -\frac{4}{3} - \frac{4}{3}(-1) = -\frac{4}{3} + \frac{4}{3} = 0 $$ 16. **Check which pair satisfies $\sin^2\theta + \cos^2\theta = 1$:** For $\sin\theta = -0.28$, $\cos\theta = -0.96$: $$ (-0.28)^2 + (-0.96)^2 = 0.0784 + 0.9216 = 1.0 $$ Valid. For $\sin\theta = -1$, $\cos\theta = 0$: $$ (-1)^2 + 0^2 = 1 + 0 = 1 $$ Also valid. 17. **Determine the angle $\theta$:** $$ \theta_1 = \arcsin(-0.28) \approx -16.26^\circ $$ $$ \theta_2 = \arcsin(-1) = -90^\circ $$ 18. **Check which angle maps $X$ to $X'$ correctly:** Using the rotation formula for $\theta = -90^\circ$: $$ x' = x\cos(-90^\circ) - y\sin(-90^\circ) = 0 + y = y $$ $$ y' = x\sin(-90^\circ) + y\cos(-90^\circ) = -x + 0 = -x $$ For $X(-4,2)$: $$ x' = 2, y' = 4 $$ Matches $X'(2,4)$. 19. **Final answer:** The angle of rotation about the origin is $\boxed{-90^\circ}$ (or equivalently $270^\circ$ counterclockwise).