How to compute hyperbolic angle between a vertical geodesic and a circular geodesic? I am trying to compute the area of an ideal triangle.
I know I have to find the angles where the vertical geodesic meets the semi-circle but I am stuck trying to find the hyperbolic angle measure.
I have the circle: $(x-\frac{1}{2})^2+(y)^2=\frac{1}{4}$ (on the poincare model so $y>0$).

How do I go about computing the angle?
 A: As said in the comments, the angles $\theta$ and $\phi$ at the points $0$ and $1$ in your diagram are simply $\theta = \phi = 0$. One can see this using Euclidean geometry: at each of $0$ and $1$, the vertical line and the semicircle are tangent, meeting at a zero angle. There is also a third angle $\psi$, located at the third "ideal vertex" of the triangle, and its value is also $\psi = 0$, although to see this using Euclidean geometry you must first apply the coordinate change map $w = - \frac{1}{z}$.
However, there is something special going on here that you should be sharply aware of. The hyperbolic metric is defined only in the open upper half plane $\mathbb H = \{(x,y) \in \mathbb R^2 \mid y > 0\}$. Therefore it is not correct to appeal to conformality of this metric when measuring angles at points that are not in the open upper half plane. At points on "extended line at infinity" $\mathbb R \cup \{\infty\}$ such as the three ideal vertices $0$, $1$ and $\infty$ of the triangle depicted, angle measurement cannot be verified using conformality of the hyperbolic metric: you are stuck using Euclidean geometry for that verification.
Nonetheless, by some miracle, the area formula which you brought up in your comment still applies for angles at infinity whose angle measure is determined using Euclidean geometry. So the area can be calculated as
$$\pi - \theta - \phi - \psi = \pi - 0 - 0 - 0 = 0
$$
