I want to learn how to complete questions like these but when I look at the equation all I think I can do is simplify $\sqrt{45}$ into $3\sqrt{5}$. What do I do next? It says quadrant $2$, so the sin must be positive and everything else negative...
$\sin\theta= \dfrac{y}{r}$, so I have one $y$-coordinate (3), but what about $x$?
One way to solve the problem is to use the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$.
\begin{align*} \sin^2\theta + \cos^2\theta & = 1\\ \cos^2\theta & = 1 - \sin^2\theta\\ |\cos\theta| & = \sqrt{1 - \sin^2\theta} \end{align*} Since $\theta$ is a second-quadrant angle, $\cos\theta < 0$. $$\cos\theta = -\sqrt{1 - \sin^2\theta}$$ Finding $\cos\theta$ will tell you the $x$-coordinate of the point where the terminal side of the angle intersects the unit circle. We know that the $y$-coordinate of that point is $\sin\theta$.
If you wish to find the $x$-coordinate of the point at which the terminal side of the angle intersects the circle with radius $\sqrt{45} = 3\sqrt{5}$, you will have to scale the $x$-coordinate you find by $3\sqrt{5}$.
Another way would be to draw a right triangle in the second quadrant with opposite side $3$ and hypotenuse $3\sqrt{5}$. You can then use the Pythagorean Theorem to find the magnitude of the $x$-coordinate. Just keep in mind that the $x$-coordinate must be negative.