iN the 1780 textbook The Resolution of Triangles, by Hugh Worthington’s, the following approximation to solve the case of a right triangle when the angles and one side is given:
(1) "In any right angled triangle, as $ CAB$, let $ \alpha$ be the lesser angle in degrees. Then $\frac{57.3}{\alpha} + \frac{3\alpha}{1000} : 1 = AC : AB$, nearly"
where $AC$ and $AB$ are the hypotenuse and the opposite side to the angle $\alpha$, respectively. Also,
(2) $ 1 : 1 - \frac{1\frac{1}{2}\alpha^2}{10000} = AC : BC $
where $BC$ is the adjacent side.
The question is: how to derive the approximations (1) and (2)? The numerator of the first fraction in (1) is $57.3$. It should be related to $180/\pi \approx 57.29...$
Using $\frac{180}{\pi}$ for $57.3$, rewriting (1) so that $\alpha$ is in radians, and noting that $AC:AB = \csc \alpha$, one gets $$\frac{3 \left(\pi \left(60000+\frac{180 \alpha }{\pi }\right)^2\right)}{1000\pi\cdot\frac{ 180 \alpha }{\pi }} = \frac{1}{\alpha} + \frac{27}{50\pi}\alpha.$$ The MacLaurin series for $\csc\alpha$ is $$\frac{1}{\alpha }+\frac{\alpha }{6}+\frac{7 \alpha ^3}{360}+\frac{31 \alpha ^5}{15120}+\frac{127 \alpha ^7}{604800}+\frac{73 \alpha ^9}{3421440}+O\left(\alpha ^{11}\right),$$ and $\frac{1}{6} \approx 0.1666667$ while $\frac{27}{50\pi}\approx 0.171887$.
Note that just $\frac{1}{\alpha}$ is a pretty good approximation for $\alpha$ small. Since $\alpha$ is the smaller angle, it should be a decent approximation all by itself. In the graph below, $\csc\alpha$ is the blue line, the approximation in (1) is the orange line, $\frac{1}{\alpha}$ the green line, and $\frac{27}{50\pi}\alpha$ the red line.
The second equation is similar; rewriting for $\alpha$ in radians and looking at the Maclaurin series of the resulting LHS and of $\sec\alpha$ shows that the first two terms are pretty close to each other.