1. **State the problem:** Given an acute angle $\theta$ in a right triangle with $\sin \theta = \frac{1}{3}$, find $\cos \theta$. 2. **Recall the Pythagorean identity:** $$\sin^2 \theta + \cos^2 \theta = 1$$ This identity relates sine and cosine of the same angle. 3. **Substitute the given sine value:** $$\left(\frac{1}{3}\right)^2 + \cos^2 \theta = 1$$ $$\frac{1}{9} + \cos^2 \theta = 1$$ 4. **Solve for $\cos^2 \theta$:** $$\cos^2 \theta = 1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9}$$ 5. **Take the square root to find $\cos \theta$:** Since $\theta$ is acute, $\cos \theta > 0$. $$\cos \theta = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}$$ **Final answer:** $\cos \theta = \frac{2\sqrt{2}}{3}$