A very little known approximation for the lesser angle of a triangle In the article A note on an Approximation in Trigonometry is proved a very interesting approximattion to the lesser angle of a triangle (in degrees):
$ (1)  A \approx \frac{344\Delta}{2s(s-a)+bc}$
Where $\Delta$ is the area of the triangle, $s$ is the semiperimeter, and $bc$ are the two adjacent sides of the angle $A$. For the case of a right triangle, the formula becomes simpler:
$A \approx \frac{172a}{b+2c}$
where $c$ is the hypotenuse.
This approximation seems to be very old, and have been appeared in a XVI century book on trigonometry. It is very little known, and have good accuracy.
The proof in the article is purely analytical. So, my question is: How to prove the approximate formula $(1)$ by means of plane geometry only?
 A: I am sure one of the very skilled geometers of the 16th century could deal with this in a much more geometrical way, but here is how I see it based on the article you linked to.
We base the derivation on the well known formulas $\Delta=\frac12bc\sin(A)$ and $\cos(A)=\frac{b^2+c^2-a^2}{2bc}$.
From these combined we derive
$$
\frac{2\Delta}{2s(s-a)+bc}=\frac{\sin(A)}{2+\cos(A)}
$$
where $2s=a+b+c$ so $s$ is the semiperimeter.

Now the rest relies on the approximation
$$
\frac{\sin(x)}{2+\cos(x)}\approx\frac x3
$$
where of course $x$ is in radian measure. This is a surprisingly good approximation for $x<\frac\pi3$. Here is a plot of the LHS and RHS above in the same coordinate system:

One way to establish how good this approximation actually is could be to consider the Taylor Expansions of the numerator and denominator of the LHS, which is what they did in the article, namely:
$$
\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...
$$
and
$$
2+\cos(x)=3-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...
$$
But a different approach could be to simply analyze the function
$$
f(x)=\frac{\sin(x)}{2+\cos(x)}
$$
which has a derivative that simplifies to
$$
f'(x)=\frac{2\cos(x)+1}{(2+\cos(x))^2}
$$
and since $\cos(x)$ is strictly decreasing for $x\in(0,\pi/3)$ we have
$$
f'(0)=\frac13<f'(x)<f'(\pi/3)=\frac{2}{6.25}
$$
so the slope of $f$ stays very close to $1/3$ throughout the interval $(0,\pi/3)$ giving an idea of the close resemblence to the RHS $g(x)=x/3$ on this interval.

It is important to note that the precision of the approximation
$$
A\approx\frac{6\Delta}{2s(s-a)+bc}\quad\text{(in radian measure)}
$$
relies solely of the size of the angle $A$ being approximated - not the sizes or proportions of the sides $a,b$ and $c$. The relative error will be exactly the relative error of the approximation
$$
\frac{\sin(x)}{2+\cos(x)}\approx\frac x3
$$

Finally, here is a plot of the relative error as a function of the angle size $x$ in radian measure:

And as you can see, the relative error stays below $1\%$ for angles less than $\pi/3$ in radian measure.

To convert to degrees we re-scale the formula by re-scaling the constant in the numerator as follows
$$
6\cdot\frac{180}{\pi}\approx343.774677
$$
which for convenience is rounded to $344$ corresponding to using $g(x)=x/3.00197$ which introduces a little more error for small angles and a little less error for larger angles:


Since the precision depends solely on the size of the angle involved, you may use the formula for any angle less than $\pi/3$ or correspondingly $60^\circ$ and it seems to have an accuracy of roughly 2-3 significant figures for angles in that range, for some angles the precision is considerably higher as you may infer from the relative error graphs.
If a triangle contains two small angles (and one obtuse), you may use the approximation for both of the small angles as long as they stay within $\pi/3$  or correspondingly $60^\circ$.
A: I have done a Monte Carlo simulation on $10^5$ random triangles (using Matlab), in order to have an idea of the quality of the proposed approximation.
Let (with @Semiclassical version):
$$F:=\dfrac{6 \Delta}{2s(s-a)+bc}$$
Approximation :
$$A\approx F$$ 
is always


*

*such that $A > F$ (always from under, which is understandable by considering the neglected terms in the proof given in the reference article).

*very accurate :  $1 < \dfrac{A}{F} < 1.005$ in about 94% cases. And another 4% between $1.005$ and $1.015$.
Here is the histogram of values of $\dfrac{A}{F}$:

