= Solution
Because the half-plane metric is a <conformal rescaling of a Riemannian metric>[conformal rescaling] of the Euclidean metric, its angles are Euclidean angles. For each $y>0$, the <hyperbolic lines meeting the imaginary axis at a fixed angle> $\alpha$ at $iy$ are the Euclidean circles
$$
(x-y\cot\alpha)^2+Y^2=y^2\csc^2\alpha,
$$
where $Y$ denotes the vertical coordinate. These circles have centres on the real axis and are therefore hyperbolic lines; varying $y$ gives the required collection.
To construct a triangle, put one vertex at $i$ and a second at $it$ on $\ell_+$, with $t>1$. Draw from these vertices the lines making interior angles $\alpha$ and $\beta$ with $\ell_+$. Their Euclidean centres and radii are
$$
C_1=\cot\alpha,
\quad R_1=\csc\alpha,
\qquad
C_2=-t\cot\beta,
\quad R_2=t\csc\beta.
$$
If their other intersection has angle $\gamma$, the <law of cosines> in the Euclidean triangle formed by the two centres and that intersection gives
$$
\cos\gamma
=\frac{R_1^2+R_2^2-(C_1-C_2)^2}{2R_1R_2}
=\frac{\sin\alpha\sin\beta}{2}\left(t+\frac1t\right)-\cos\alpha\cos\beta.
$$
Equivalently,
$$
\frac12\left(t+\frac1t\right)
=\frac{\cos\gamma+\cos\alpha\cos\beta}{\sin\alpha\sin\beta}.
$$
The right side is greater than one exactly when $\alpha+\beta+\gamma<\pi$. Since $(t+t^{-1})/2$ ranges continuously and strictly increasingly from $1$ to infinity for $t>1$, such a $t$ exists. The three lines therefore bound a <hyperbolic triangle> with the prescribed angles.
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