Parameterization of an ellipse by angle from the center is
$$
\gamma(\phi)=(\cos(\phi),\sin(\phi))\frac{ab}{\sqrt{a^2\sin^2(\phi)+b^2\cos^2(\phi)}}
$$
and
$$
\left|\gamma'(\phi)\right|=ab\sqrt{\frac{a^4\sin^2(\phi)+b^4\cos^2(\phi)}{\left(a^2\sin^2(\phi)+b^2\cos^2(\phi)\right)^3}}
$$
and integrating $\left|\gamma'(\phi)\right|$ gets extremely messy.
So instead, we use $\theta$, where
$$
b\tan(\theta)=a\tan(\phi)
$$
Then
$$
\gamma(\theta)=(a\cos(\theta),b\sin(\theta))
$$
and
$$
\left|\gamma'(\theta)\right|=\sqrt{a^2\sin^2(\theta)+b^2\cos^2(\theta)}
$$
Now, integrating $\left|\gamma'(\theta)\right|$ is a lot simpler.
$$
\int\left|\gamma'(\theta)\right|\,\mathrm{d}\theta
=b\,\mathrm{EllipticE}\left(\theta,\frac{b^2-a^2}{b^2}\right)
$$
However, to go from from arc length to angle, we still need to invert the Elliptic integral.
Solution of the problem given using Mathematica
Here we get the solution to 30 places.
In[]= Phi[a_,b_,s_,opts:OptionsPattern[]] := Block[ {t}, t+ArcTan[(b-a)Tan[t]/(a+b Tan[t]^2)] /. FindRoot[b EllipticE[t,(b^2-a^2)/b^2] == s, {t, 0}, opts]]
In[]= Phi[4, 2.9`32, 3.31`32, WorkingPrecision->30]
Out[]= 0.87052028743193111752524449959
Thus, $\phi\doteq0.87052028743193111752524449959\text{ radians}$.
Let me describe the algorithm a bit.
FindRoot[b EllipticE[t,(b^2-a^2)/b^2] == s, {t, 0}, opts]
inverts the elliptic integral to get $\theta$ from $a$, $b$, and $s$.
Now, we want to find $\phi$ so that $b\tan(\theta)=a\tan(\phi)$. However, simply using $\tan^{-1}\left(\frac ba\tan(\theta)\right)$ will only return a result in $\left(-\frac\pi2,\frac\pi2\right)$. To get the correct value, we use the relation
$$
\begin{align}
\tan(\phi-\theta)
&=\frac{\tan(\phi)-\tan(\theta)}{1+\tan(\phi)\tan(\theta)}\\
&=\frac{\frac ba\tan(\theta)-\tan(\theta)}{1+\frac ba\tan(\theta)\tan(\theta)}\\
&=\frac{b\tan(\theta)-a\tan(\theta)}{a+b\tan(\theta)\tan(\theta)}\\
\phi
&=\theta+\tan^{-1}\left(\frac{(b-a)\tan(\theta)}{a+b\tan^2(\theta)}\right)
\end{align}
$$
This is why we have
t+ArcTan[(b-a)Tan[t]/(a+b Tan[t]^2)]
Best Answer
Let $a=3.05,\ b=2.23.$ Then a parametric equation for the ellipse is $x=a\cos t,\ y=b \sin t.$ When $t=0$ the point is at $(a,0)=(3.05,0)$, the starting point of the arc on the ellipse whose length you seek. Now it's important to realize that the parameter $t$ is not the central angle, so you need to get the value of $t$ which corresponds to the top end of your arc. At that end you have $y/x=\tan 50$ (degrees). And in terms of $t$ you have $y/x=(b/a)\tan t$. Solving for $t$ then gives $$t=t_1=\arctan \left( \frac{a}{b}\tan 50 \right).$$
[note I'd suggest using radians here, replacing the $50$ by $5\pi/18.$]
For the arclength use the general formula of integrating $\sqrt{x'^2+y'^2}$ for $t$ in the desired range. In your case $x'=-a \sin t,\ y'=b \cos t$, so that you are integrating $$\sqrt{a^2 \sin^2t+b^2 \cos^2t}$$ with respect to $t$ from $0$ to the above $t_1$. There not being a simple closed form for the antiderivative (it's an "elliptic integral), the simplest approach now would be to do the integral numerically. This seems the more appropriate in your problem as you only know $a,b$ to two decimals, apparently.
* When I did this numerically on maple I got about $2.531419$ for the arclength.