If angle \( \theta \) lies in Quadrant II and \( \sin \theta=\frac{3}{\sqrt{45}} \). Determine possible coordinates for point P on the terminal arm of angle \( \theta \).

Given that angle \( \theta \) lies in Quadrant II and \( \sin \theta = \frac{3}{\sqrt{45}} \), we can determine the coordinates of a point \( P(x, y) \) on the terminal arm of \( \theta \). 1. **Simplify \( \sin \theta \):** \[ \sin \theta = \frac{3}{\sqrt{45}} = \frac{3}{3\sqrt{5}} = \frac{1}{\sqrt{5}} \] 2. **Using the Pythagorean identity:** \[ \sin^2 \theta + \cos^2 \theta = 1 \\ \left(\frac{1}{\sqrt{5}}\right)^2 + \cos^2 \theta = 1 \\ \frac{1}{5} + \cos^2 \theta = 1 \\ \cos^2 \theta = \frac{4}{5} \\ \cos \theta = \pm \frac{2}{\sqrt{5}} \] Since \( \theta \) is in Quadrant II, \( \cos \theta \) is negative: \[ \cos \theta = -\frac{2}{\sqrt{5}} \] 3. **Determine the coordinates \( (x, y) \):** - Let the hypotenuse (radius) \( r = \sqrt{45} \). - \( y = r \sin \theta = \sqrt{45} \times \frac{1}{\sqrt{5}} = \frac{3\sqrt{5}}{\sqrt{5}} = 3 \) - \( x = r \cos \theta = \sqrt{45} \times \left( -\frac{2}{\sqrt{5}} \right) = -\frac{6\sqrt{5}}{\sqrt{5}} = -6 \) Therefore, the coordinates of point \( P \) on the terminal arm of angle \( \theta \) are: \[ \boxed{\,(-6,\ 3)\,} \]