= Solution
In the <Poincaré half-plane model>, the <Riemannian metric>
$$
ds^2=\frac{dx^2+dy^2}{y^2}
$$
assigns to a <smooth curve> $\gamma(t)=(x(t),y(t))$ the length
$$
L(\gamma)=\int
\frac{\sqrt{\dot x(t)^2+\dot y(t)^2}}{y(t)}\,dt.
$$
The metric is conformal to the Euclidean metric, so hyperbolic and Euclidean angles agree. Its area element is
$$
dA=\frac{dx\,dy}{y^2}.
$$
The two geodesics from $e^{i\alpha}$ and $e^{i\beta}$ to infinity are the vertical lines $x=\cos\alpha$ and $x=\cos\beta$, while the third side is the unit semicircle. The triangle lies above that semicircle, so its area is
$$
\begin{aligned}
\operatorname{Area}
&=\int_{\cos\beta}^{\cos\alpha}
\int_{\sqrt{1-x^2}}^\infty\frac{dy\,dx}{y^2}\\
&=\int_{\cos\beta}^{\cos\alpha}
\frac{dx}{\sqrt{1-x^2}}
=\beta-\alpha.
\end{aligned}
$$
Its interior angles are $\alpha$, $\pi-\beta$, and zero at the <ideal> vertex, and hence
$$
\operatorname{Area}
=\pi-\bigl(\alpha+(\pi-\beta)+0).
$$
Triangulating a geodesic polygon with $n$ sides into $n-2$ triangles gives
$$
\boxed{
\operatorname{Area}
=(n-2)\pi-\sum_{j=1}^n\theta_j
},
$$
the <area of a hyperbolic geodesic polygon> and the polygonal <Gauss-Bonnet theorem>.
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