1. **State the problem:** Given that $\cos \theta < 0$ and $\sin \theta = -\frac{2}{3}$, find $\tan \theta$ and the coordinates of point $P = (x, y)$ on the unit circle. 2. **Recall the Pythagorean identity:** $$\sin^2 \theta + \cos^2 \theta = 1$$ This allows us to find $\cos \theta$ given $\sin \theta$. 3. **Calculate $\cos \theta$:** $$\sin^2 \theta = \left(-\frac{2}{3}\right)^2 = \frac{4}{9}$$ $$\cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{4}{9} = \frac{5}{9}$$ $$\cos \theta = \pm \sqrt{\frac{5}{9}} = \pm \frac{\sqrt{5}}{3}$$ Since $\cos \theta < 0$, we take the negative value: $$\cos \theta = -\frac{\sqrt{5}}{3}$$ 4. **Find $\tan \theta$ using the definition:** $$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{2}{3}}{-\frac{\sqrt{5}}{3}} = \frac{-2/3}{-\sqrt{5}/3}$$ Cancel the denominators: $$\tan \theta = \frac{\cancel{-2/3}}{\cancel{-\sqrt{5}/3}} = \frac{-2}{-\sqrt{5}} = \frac{2}{\sqrt{5}}$$ 5. **Coordinates of point $P$ on the unit circle:** Since $P = (\cos \theta, \sin \theta)$, $$P = \left(-\frac{\sqrt{5}}{3}, -\frac{2}{3}\right)$$ 6. **Check the options:** Option B states: $$\tan \theta = \frac{2}{\sqrt{5}}; \quad P = (-\sqrt{5}, -2)$$ The $y$-coordinate matches the sign and ratio but the $x$ and $y$ coordinates are scaled by 3 compared to the unit circle coordinates. Given the handwritten note $2^2 + b^2 = 32$ and $4 + b^2 = 9$ seems inconsistent, but the problem likely expects the unit circle values. **Final answer:** $$\tan \theta = \frac{2}{\sqrt{5}}$$ $$P = \left(-\frac{\sqrt{5}}{3}, -\frac{2}{3}\right)$$ This matches option B in tangent value and sign pattern, with coordinates scaled by 3.