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{\bf Question}

Let $T$ be a triangle with angles $\alpha$, $\beta$, and
$\frac{\pi}{2}$.  Let $a$ be the hyperbolic length of the side of
$T$ opposite the vertex with angle $\alpha$.  Prove that $\cosh(a)
\sin(\beta) = \cos(\alpha)$.

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{\bf Answer}

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\un{use lcII}:

\begin{eqnarray*} \cos(\alpha) &amp; = &amp;
-\cos(\beta)\cos(\frac{\pi}{2})+\sin(\beta)\sin(\frac{\pi}{2})\cosh(a)\\
&amp; = &amp; \sin(\beta)\cosh(a)\end{eqnarray*} as desired.

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