1. **State the problem:** Given $\sin \theta = \frac{3}{5}$ and $\theta$ is in quadrant II, 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:** $$\left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1$$ $$\frac{9}{25} + \cos^2 \theta = 1$$ 4. **Solve for $\cos^2 \theta$:** $$\cos^2 \theta = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25}$$ 5. **Take the square root to find $\cos \theta$:** $$\cos \theta = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5}$$ 6. **Determine the sign of $\cos \theta$ in quadrant II:** In quadrant II, sine is positive and cosine is negative. 7. **Final answer:** $$\cos \theta = -\frac{4}{5}$$