1. **State the problem:** Given a point $P = (8, -6)$ on the terminal side of angle $\theta$, find $\sin \theta$. 2. **Recall the formula:** For a point $(x, y)$ on the terminal side of $\theta$, the sine of the angle is given by: $$\sin \theta = \frac{y}{r}$$ where $r$ is the distance from the origin to the point, calculated by: $$r = \sqrt{x^2 + y^2}$$ 3. **Calculate $r$:** $$r = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$$ 4. **Calculate $\sin \theta$:** $$\sin \theta = \frac{y}{r} = \frac{-6}{10} = -\frac{3}{5}$$ 5. **Interpretation:** Since $y$ is negative and $r$ is positive, $\sin \theta$ is negative, which matches the value $-\frac{3}{5}$. **Final answer:** $$\sin \theta = -\frac{3}{5}$$