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.