Let \(\displaystyle{a}={3.05},{b}={2.23}\). Then a parametric equation for the ellipse is \(\displaystyle{x}={a}{\cos{{t}}},{y}={b}{\sin{}}\). When \(\displaystyle{t}={0}\) the point is at \(\displaystyle{\left({a},{0}\right)}={\left({3.05},{0}\right)}\), 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 \(\displaystyle{\frac{{{y}}}{{{x}}}}={\tan{{50}}}\) (degrees). And in terms of t you have \(\displaystyle{\frac{{{y}}}{{{x}}}}={\left({\frac{{{b}}}{{{a}}}}\right)}{\tan{}}\). Solving for t then gives

\(\displaystyle{t}={t}_{{{1}}}={a}{r}{c}{\tan{{\left({\frac{{{a}}}{{{b}}}}{\tan{{50}}}\right)}}}\).

[note I'd suggest using radians here, replacing the 50 by \(\displaystyle{\frac{{{5}\pi}}{{{18}}}}\).]

For the arclength use the general formula of integrating \(\displaystyle\sqrt{{{x}'^{{{2}}}+{y}'^{{{2}}}}}\) for t in the desired range. In your case \(\displaystyle{x}′=-{a}{\sin{{t}}},{y}′={b}{\cos{}}\), so that you are integrating

\(\displaystyle\sqrt{{{a}^{{{2}}}{{\sin}^{{{2}}}{t}}+{b}^{{{2}}}{{\cos}^{{{2}}}{t}}}}\)

with respect to t from 0 to the above \(\displaystyle{t}_{{{1}}}\). 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.

\(\displaystyle{t}={t}_{{{1}}}={a}{r}{c}{\tan{{\left({\frac{{{a}}}{{{b}}}}{\tan{{50}}}\right)}}}\).

[note I'd suggest using radians here, replacing the 50 by \(\displaystyle{\frac{{{5}\pi}}{{{18}}}}\).]

For the arclength use the general formula of integrating \(\displaystyle\sqrt{{{x}'^{{{2}}}+{y}'^{{{2}}}}}\) for t in the desired range. In your case \(\displaystyle{x}′=-{a}{\sin{{t}}},{y}′={b}{\cos{}}\), so that you are integrating

\(\displaystyle\sqrt{{{a}^{{{2}}}{{\sin}^{{{2}}}{t}}+{b}^{{{2}}}{{\cos}^{{{2}}}{t}}}}\)

with respect to t from 0 to the above \(\displaystyle{t}_{{{1}}}\). 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.