Metric angles in Riemannian manifolds of low regularity Given three points $a,b,c$ in a (geodesic) metric space $X$, one defines a comparison angle $\angle(a,b,c)$ by the cosine law:
$$
 \angle(a,b,c) = \arccos \frac{|ab|^2 + |ac|^2 - |bc|^2}{2\cdot|ab|\cdot|ac|}
$$
where $|ab|$, $|ac|$ and $|bc|$ are distances in $X$. In other words, $\angle(a,b,c)$ is the angle of a planar triangle with the same side lengths as the triangle $abc$ in $X$.
Given two paths $\xi,\eta:[0,\varepsilon)\to X$ starting at the same point $p=\xi(0)=\eta(0)$, define a metric angle $\angle(\xi,\eta)$ by
$$
 \angle(\xi,\eta) = \lim_{t,t'\to 0} \angle(p,\xi(t),\eta(t'))
$$
if the limit exists.
If $X$ is a smooth manifold with a $C^2$ Riemannian metric (which defines a geodesic distance in the usual way) and $\xi,\eta$ are $C^1$ regular curves, then a curvature comparison argument shows that the metric angle is well-defined and equals the Riemannian angle between the velocity vectors $\dot\xi(0)$ and $\dot\eta(0)$. But what if the Riemannian metric is only $C^1$, or even $C^0$? More precisely:


*

*Let $X$ be a smooth manifold with a $C^1$ Riemannian metric. Does the metric angle exist for every pair of $C^1$ regular curves $\xi,\eta$?

*If not, does every $C^1$ regular curve have a well-defined angle with itself? (Of course, this angle is zero if it exists).

*If the answer is affirmative, what about $C^0$ Riemannian metrics?
Remark.
The limit in the definition trivially exists (even in the $C^0$ case) if the ratio $t/t'$ is bounded away from zero and infinity as $t$ and $t'$ go to zero. The trouble is with very "thin" triangles $p,\xi(t),\eta(t')$ where $t\ll t'$ or vise versa.
 A: For $C^1$ metrics  the answer is yes. In fact, the answer is yes for any Hoelder-continuous Riemannian metric. The key observation is that geodesics are uniformly $C^{1,\alpha}$ in this case. From this one can draw the conclusions about the angles by using  a few traingel inequalities.
 For $C^0$ metrics there is a closely related question, to which I do not know the answer:  Is the angle between any pair of geodesics starting in a point well defined. 
I do believe  that the answer is nbo as well. But one cannot construct a counterexample  just by choosing  a wrong  parametrization of the manifold.
A: To Deane Yang,  concerning the smoothness of geodesics.  These issues have been studied by Hartman 
and his coauthors more than 40 years ago.  Together with Calabi he claimed to prove 
that in a $C^{\alpha}$ continuous metric all geodesics are uniformly $C^{1,\alpha}$.
(I do not have mathscinet access now, but I think the paper was called
"On the smoothness of isometries", or something like that).  The paper is not quite correct,
the computations prove that geodesics are only $C^{1,\alpha /2}$
 (this was observed at some point by  Reshetnyak).  The optimal regularity appears in my paper with Asli Yaman
"On Hoelder continuous Riemannian and Finsler metrics"  (I  apologize for citing my own work).
  The starting point of all these computations is a metric characterization  of $C^{1,\alpha}$
curves in metric terms (in terms of the distance of the midpoint of a pair of points on the curve and the image of the corresponding midpoint of the interval of the definition).   It is contained in the work of Hartman, but is probably much older.
To Sergei Ivanov:  The questions of extendibility of geodesics have also been studied by Hartman. I again do not remember the correct reference, but can check it on Monday. I think he constructs example of even $C^{1,\alpha}$ Riemannian metrics, for any $\alpha <1$, that do not have extendible geodesics.  (For $\alpha =1$, the metric has curvature bounded from both sides).
Now the main point. I am very sorry and have to apologize for my previous short posting.
 I slightly misunderstood your question. But I hope that my answer was still 
correct. Is the following more detailed  explanation correct?  


*

*We assume that $t'>>t$ and want to prove that 
$d(\xi (t), \eta (t') )=  
d(\eta (t'), p) - cos (\alpha ) t + o(t)$. Here $o(t) /t \to 0$ and $\alpha$ is 
the "usual Euclidean " angle between $\xi$ and $\eta$.


*

*The statement about the uniform "smoothness" of geodesics, show that it suffices to prove the claim in the case, when $\eta$ and $\xi$ are geodesics (since the "usual" angles
between $\eta$ and any  geodesic between $\eta (t')$ and $p$ goes to $0$).

*Going on the geodesic $\eta$ first to the point $\eta (rt)$, with large, but fixed $r$, and then to $\xi (t)$, one obtains the right upper bound for $d (\eta (t'),p)$.

*In a "general" metric space it would be now difficult to obtain the right lower bound 
of  $d(\eta (t'),p)$. One could do it, if one knew that the geodesic $\eta$ was extendible beyond $p$. Then one could use a traingle inequality and the distance from $\xi (t)$ and $\eta (-rt)$. I hope I understood your question correctly and this was what you had in mind.

*Here one can use another trick to obtain a lower bound. Observe namely, that from the
"smoothness" of geodesics the "usual" angle between $\xi$  and any geodesic from
$\eta (t') $ to $\xi (t)$ is very close to $\alpha$ (goes to $\alpha$ , with $t' \to 0$).
Now  we just look at the triangle $\xi (t),p, \eta (t')$ from $\xi (t)$ and not from $p$.
The inequality obtained in point 3. above gives us the right upper bound of $d(\eta (t'),p)$.
Just a side remark. Your question is very closely related to the question about the 
correctness of the first formula of variation. I have thought about that question
and proved that the formula holds true in Riemannian manifolds with Hoelder continuous metrics in "Differentiation in metric spaces". I just hope that the above proof is correct. Other wise the corresponding statement in my paper is wrong as well.
A: For $C^0$, the answer is "NO". 
Start with Euclidean plane and choose a function $\alpha(r)$ such that $\alpha(r)\to\infty$ as $r\to 0$,
but it does it really slow (so $r\cdot\alpha'(r)\to 0$ sufficiently fast).
Consider the induced metric by the map in polar coordinates $(\theta, r)\mapsto (\theta+\alpha(r),r)$;
it is continuous.
The geodesics from the origin in the new metric are formed by spirals which go around origin infinitely many times.
In particular, if $\xi$ is a smooth curve starting at the origin then $\angle (\xi,\xi)$ is undefined. 
