The approximate value of the angle in a right-angled triangle If we have a right-angled triangle $c^2$=$a^2$+$b^2$ and if we assume $a<b$ and    We have the formula
$ \frac {180}{pi}\cdot\frac{(\frac43\cdot(2\cdot\sqrt{\frac{(c-b)\cdot c}2}-a)+a)}{c}$~$\angle a $.
For example, if we have a right-angled triangle $10^2$=$5^2$+$8.660254...^2$ we will get
$ 57.295..\cdot\frac{(\frac43\cdot(2\cdot\sqrt{\frac{(10-8.660254..)\cdot 10}2}-5)+5)}{10} = $29.99.. .What is the approximate value of the angle  of 30 degrees  ($\angle a $).This formula is valid for any right-angled triangle (easy to check). So the question is whether my formula can be otherwise express, and if there is such a similar formula?
 A: Let $\theta$ be the angle opposite $a$.  We have $\theta\in[0,\frac\pi2]$.  Then
\begin{align*}
\frac{\frac43\left(2\sqrt{\frac{(c-b)c}{2}}-a\right)+a}{c}
&= \frac83\sqrt{\frac{1-\frac bc}{2}} -\frac13\frac ac \\
&= \frac83\sqrt{\frac{1-\cos\theta}{2}} - \frac13\sin\theta \\
&= \frac83\sin\frac\theta2 - \frac13\sin\theta \\
&= \frac23\sin\frac\theta2\left(4-\cos\frac\theta2\right)
\end{align*}
Wolfram Alpha shows that the approximation looks pretty good over the range we care about:

The Taylor expansion is
$$ \frac23\sin\frac\theta2\left(4-\cos\frac\theta2\right)
= \theta - \frac{\theta^5}{480} + O(\theta^7)
$$
The small coefficients on the powers $\theta^2$ through $\theta^6$ explains, in a sense, why the approximation remains so good even when $\theta$ is fairly far from $0$.  The even powers being zero can be explained by the oddness of the function; a person could easily cook up such a formula with the goal of getting zero in the $\theta^3$ term; but the small $\theta^5$ coefficient seems quite fortuitous.
A similar approximation, $\theta\approx\frac{3\sin\theta}{2+\cos\theta}$, is considered in the post "Revisiting Pythagoras Goes Linear" on the blog God Plays Dice (whose author posts here from time to time, if memory serves).
