Geodesics meeting with angle $0$ in ${\rm CAT}(0)$ space

Question

Consider two distinct geodesics $\gamma_1$ and $\gamma_2$ in a CAT($0$) space, issued from the same base point. A trivial example where we have $\angle(\gamma_1, \gamma_2)=0$ is when $\gamma_1(t) = \gamma_2(t)$ for $t$ smaller than some $\varepsilon > 0$. In this case, I say that they define the same germ.

My question is, is there examples of CAT(0) spaces with geodesics meeting with angle 0 and every geodesic is non-branching ?

The only class of examples I know of non-branching geodesics meeting with angle $0$ are constructed as follow: consider the following subset of the Euclidian plane, with induced length-metric, $X = \{ (x,y); 0 \le x \le 1, 0 \le y \le x^2 \}$. It is a CAT(0) space, the geodesic between $(0,0)$ and $(1,0)$ is the segment and the geodesic between $(0,0)$ and $(1,1)$ is the arc of parabola. These geodesics meet at $(0,0)$ with angle $0$ and they don't define the same germ.

However, in this space there exist branching geodesics, for example the geodesic between $(0,0)$ and $(1, t)$ for $t>0$ all define the same germ.

Edit: I add a description of these geodesics. The geodesic from $(0,0)$ to $(1,t)$ is an arc of parabola from $(0,0)$ to $P$ concatenated with the segment $[P, (1,t)]$ where $P$ is the point on the parabola closest to the origin such that the segment $[P, (1,t)]$ lies in $X$.

Answer 1

Consider an arc $x=0,\ (y-1)^2+z^2=1,\ 0<y\leq 1,\ z>0$. When we rotate it wrt $z$-zxis, then we have $X$, homeomorphic to a disc with a hole.

Note that it has a nonpositive curvature. A completion of universal cover of $X$ is homeomorphic to $\bigcup_{t\in \mathbb{R}}\ [oc(t)]$ where $c:\mathbb{R}\rightarrow \mathbb{S}^2$ is an immersion, $o$ is origin, and $[ab]$ is a line segment in $\mathbb{E}^3$.

Geodesics with Angle $0$ : Assume that $$p=(0,1-\cos\ \theta,\ast),\ q=(0,-1+\cos\ \theta ,\ast)\in X$$

Now consider a completion of $X$ : $|p-o|=\theta =|q-o|$. And $|p-q| < \pi(1-\cos\ \theta )$.