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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 $\alpha$, and let $b$ be the
hyperbolic length of the side of $T$ opposite the vertex with
angle $\beta$.  Prove that $\tanh(b) = \sinh(a)\tan(\beta)$, that
$\sinh(b) = \sinh(c) \sin(\beta)$, and that $\tanh(a) = \tanh(c)
\cos(\beta)$.
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{\bf Answer}

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\begin{itemize}
\item
\un{ls}
$\ds\frac{\sinh(a)}{\sin(\alpha)}=\ds\frac{\sinh(b)}{\sin(\beta)}$

\un{lcII}
$\cos(\beta)=-\cos(\alpha)\cos(\frac{\pi}{2})+\sin(\alpha)\sin(\frac{\pi}{2})\cosh(b)$

\un{from ls}:
$\sinh(b)=\ds\frac{\sinh(a)\sin(\beta)}{\sin(\alpha)}$

\un{use lcII}:
$\sinh(b)=\ds\frac{\sinh(a)\sin(\beta)}{{\cos(\beta)}/{\cosh(b)}}
=\sinh(a)\cosh(b)\tan(\beta)$

So $\tanh(b)=\sinh(a)\tan(\beta)$ as desired. $(\star)$

\item
$\sinh(b)=\sinh(c)\sin(\beta) =
\ds\frac{\sinh(c)}{\sin(\beta)}\sin(\beta)$ $\left({\rm{with}}\
\gamma=\ds\frac{\pi}{2}\right)$

immediately from ls.

\item
from above $(\star)$: $\tanh(b)=\sinh(a)\tan(\beta)$

\un{from ls and lcII}:

$\cosh(c)=\cot(\alpha)\cot(\beta)$

(with $\gamma=\ds\frac{\pi}{2}$ and using

$\cos(\gamma)=-\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)\cosh(c))$

$\sinh(c)=\ds\frac{\sinh(a)}{\sinh(\alpha)}$

\begin{eqnarray*}
\tanh(c) & = &\ds\frac{\sinh(a)}{\sin(\alpha)} \cdot
\tan(\alpha)\tan(\beta)\\ & = & \sinh(a) \cdot \tan(\beta) \cdot
\ds\frac{1}{\cos(\alpha)} \end{eqnarray*}

and so \begin{eqnarray*} \tanh(c)\cos(\alpha) & = &
\sinh(a)\tanh(\beta)\\ & = & \tanh(b) \end{eqnarray*} (from
$(\star)$) (as desired)
\end{itemize}
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