1. **Problem statement:** Given $\cos \theta = \frac{12}{13}$, find $\sin \theta$. 2. **Formula used:** We use the Pythagorean identity: $$\sin^2 \theta + \cos^2 \theta = 1$$ This means: $$\sin^2 \theta = 1 - \cos^2 \theta$$ 3. **Substitute the given value:** $$\sin^2 \theta = 1 - \left(\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{169}{169} - \frac{144}{169} = \frac{25}{169}$$ 4. **Find $\sin \theta$ by taking the square root:** $$\sin \theta = \pm \sqrt{\frac{25}{169}} = \pm \frac{5}{13}$$ 5. **Determine the sign:** Without additional information about the quadrant of $\theta$, $\sin \theta$ can be either positive or negative $\frac{5}{13}$. Usually, if not specified, the positive value is taken. **Final answer:** $$\sin \theta = \frac{5}{13}$$