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Consider a translation surface $(X,\omega)$ and let $p_1\in X$ be a point of conical singularity and $p,p_2\in X$ non singular points.

Let $s_1$ and $s_2$ be two geodesics for the singular flat metric (often called saddle connections) respectively from $p$ two $p_1$ and from $p$ to $p_2$ such that $l(s_1)=d(p,p_1)$ and $l(s_2)=d(p,p_2)$ where the length and the distance are the ones induced by $\omega\overline{\omega}$.

Consider the triangle $T\subset \mathbb{R}^2$ with sides of length $d(p,p_1),d(p,p_2),d(p_1,p_2)$. Call $\alpha$ the angle at the vertex between sides of lengths $d(p,p_1)$ and $d(p,p_2)$.

Is $\alpha$ equal to the angle at $p$ in $(X,\omega)$ between $s_1$ and $s_2$?

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The angles are different (in general). Let $q$ be a singular point, $p_1$ another singular point. Then you can find regular points $p, p_2$ distinct from $q$ such that the angles between the three segments $qp_1, qp_2, qp$ are equal to $\pi, \pi$ and $2\pi$ respectively. Then the angle at $p$ between $pp_1, pp_2$ is $0$ (both segments pass through $q$), but the corresponding comparison angle $\alpha$ is $\ne 0$.

Note furthermore that if $T$ is complete and simply-connected, it is a CAT(0) space, hence by the hinge comparison theorem angles in $T$ are $\le$ the comparison angles in $E^2$. As a reference, take a look at Ballmann's book "Lectures on spaces of nonpositive curvature".

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    I'm confused: $\beta$ and $\alpha$ are the angles at the non singular point, so it can't be $\beta>\pi$. Also it would follow $\alpha\ge \beta>\pi$ which is impossible since $\alpha$ is an angle in an euclidean triangle. Maybe you mean that when the angle $\beta_1$ in $X$ at the singular point $p_1$ is $>\pi$ then the angle $\beta$ in $p$ is $<α$? Why should it be so? – 2017-01-13
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    please could you elaborate a little more? – 2017-01-13
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    @User29983: Oh, I see, I missed the assumption that $p$ is a nonsingular point. But an example with nonsingular $p$ is similar: You can have $\beta=0$ (in $X$) but $\alpha> 0$. The point is that geodesics in $X$ branch. – 2017-01-13
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    Thank you very much, but I still don't understand how can it be $\beta=0$ in $X$: I know that near a zero of $\omega$ the metric is isometric to a n-fold covering of $(\mathbb{C},|dz|^2)$, but I don't understand how this should imply the possibility of $\beta=0$ – 2017-01-13
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    If I understand correctly, you're saying that, calling $\beta_1$ the angle at the singular point $p_1$ and $\alpha_1$ the corresponding angle in the triangle in $\mathbb{R}^2$, it should also be $\beta_1\le \alpha_1$. This should have to do with the fact that there is a negative curvature concentrated in the zero of $\omega$ and so triangles are "thin". I tried to write down the metric and prove it directly but I failed, do you know a reference where this fact is explained with examples? – 2017-01-13
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    W.Ballmann "Lectures on spaces of nonpositive curvature"; also Bridson-Haefliger. – 2017-01-13
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    thank you, but I meant a reference for the fact that triangles in a neighborhood of a conical singularity are thin, not a reference for generic non positively curved spaces – 2017-01-13
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    Thank you, but excuse me if I ask one last question. It's clear to me that the angle at $p$ between $pp_1$ and $pp_2$ is zero. But if the angle at $q$ between $qp_1$ and $qp_2$ is $\pi$, isn't it greater than the angle between the corresponding segments in the triangle in $\mathbb{R}^2$ (since it will measure less than $\pi$)? This is unclear to me: if it's possible to have angles greater than $\pi$ between segments in $X$, how can the condition of thin triangles be satisfied (i.e. angles in $X$ are $\le$ than angles of corresponding triangles in $\mathbb{R}^2$)? – 2017-01-14
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    @User29983: Remember that you have to use the comparison triangle for comparing the angles. It is exactly the fact that the excess angles at vertices are $\ge 2\pi$ that ensures the CAT(0) condition. As before, my advice is to read Ballmann's book. – 2017-01-14
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    yes, but I can't understand what I'm getting wrong: in the example of your answer the angle in $X$ at $q$ between $qp_1$ and $qp_2$ is $\pi$, and if I consider the triangle in $\mathbb{R}^2$ with lengths $d(q,p1),d(q,p_2),d(p_1,p_2)$, the angle at $q$ is less than $\pi$. So the condition doesn't seem satisfied. – 2017-01-14
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    @User29983: In the comparison theorem one uses the "Alexandrov angle" to define angles in singular spaces, this angle is (in general) different) from the naive angle that you are using. However, at the "smooth" points (in your case, points where the metric is flat) the naive angle equals Alexandrov angle. – 2017-01-14