Angles satisfying $(\cos\theta,\sin\theta)=(\frac{\sqrt a }{\sqrt{a+b}},\frac{\sqrt b}{\sqrt{a+b}})$ for integer $a$, $b$

If we list the commonly used angles in the first quadrant $\theta_0=0$, $\theta_1=\dfrac{\pi}{6}$, $\theta_2=\dfrac{\pi}{4}$, $\theta_3=\dfrac{\pi}{3}$, and $\theta_4=\dfrac{\pi}{2}$, then $$(\cos \theta_i, \sin \theta_i)=\left(\dfrac{\sqrt{4-i}}{2},\dfrac{\sqrt{i}}{2}\right) \qquad\mbox{for}\qquad i=0,\dots,4.$$

The angles $\dfrac{\pi}{6}, \dfrac{\pi}{4},\dfrac{\pi}{3} $ are obtained by bisecting or trisecting the first quadrant.

Question: Are there special constructions for angles $\theta$ such that $$(\cos \theta, \sin \theta)=\left(\dfrac{\sqrt{a}}{\sqrt{a+b}},\dfrac{\sqrt{b}}{\sqrt{a+b}}\right)$$ where $a,b$, and $\sqrt{a+b}\ $ are positive integers? Or specifically, is there a rational number $r=r(a,b)$ such that $\theta=r\pi$?

Note that if $a,b$ are nonnegative integers such that a+b=4, then we are describing the angles $\theta_0,\dots,\theta_4$.

Regarding the second question, there are no other possible values of $\theta\in[0,\pi/2]$. Squaring, we have $$\cos^2\theta=\frac{a}{a+b}\Leftrightarrow\frac{1+\cos 2\theta}{2}=\frac{a}{a+b}\Leftrightarrow\cos 2\theta=\frac{a-b}{a+b} $$ This means that $\cos2\theta$ is rational and $2\theta$ is a rational multiple of $\pi$ (since $\theta$ is a rational multiple of $\pi$). By Niven's theorem, $\cos2\theta=0,\pm\frac 12$ or $\pm 1$, which gives us $\theta\in\left\{0,\frac{\pi}{6}, \frac{\pi}{4},\frac{\pi}{3},\frac{\pi}{2}\right\}$.