Assume I want to compute one of the angles of a right triangle doing n...

There is a third option you did not list, which I think would be the most natural. You can average the measurements of the opposite side to get your best estimate of the opposite side and similarly for the adjacent side, then use those values to compute $\theta$. I will use subscripts to identify observations-they are more standard and easier to type. We would have
$$\begin{gathered} o=\frac{1}{n}\sum _i{o}_i \\ a=\frac{1}{n}\sum _i{a}_i \\ {\theta }_3=\arctan \left(\frac{o}{a}\right) \end{gathered}$$
Note that your second is really ${\theta }_2=\frac{1}{n}\sum _i{\theta }_i$ and let your first be ${\theta }_1$
To see which is more accurate, we can chase the impact of a single error through the computation. As the errors are small, we expect that they add linearly. For each calculation, we want to assess $\frac{d\theta }{d{o}_i}$ and $\frac{d\theta }{d{a}_i}$ So we have
$$\begin{gathered} \frac{d{\theta }_3}{d{o}_i}=\frac{1}{1+(\frac{o}{a}{)}^2}\frac{1}{n} \\ \frac{d{\theta }_3}{d{a}_i}=\frac{1}{1+(\frac{o}{a}{)}^2}\frac{-1}{a^2}\frac{1}{n} \\ \frac{d{\theta }_2}{d{o}_i}=\frac{1}{n}\frac{1}{1+(\frac{o_i}{a_i}{)}^2} \\ \frac{d{\theta }_2}{d{a}_i}=\frac{1}{1+(\frac{o_i}{a_i}{)}^2}\frac{-1}{a_i^2}\frac{1}{n} \end{gathered}$$
Which give just about the same result as long as your errors are small and symmetric. I believe you will get the same result with ${\theta }_1$, but haven't checked.