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I'm trying to prove that for a right triangle $\Delta ABC$ with right angle $B$, the angle $BAC \le \sin^{-1}(sech AB)$

I'm not really able to find a way to bring this proof together. I've tried converting $\sin^{-1}(sech AB)$ to $\int_{0}^{sec(AB)} \frac 1 {\sqrt{1-z^2}}dz$ and trying to relate that integral to the area of a hyperbolic triangle, but I'm having no luck.

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By the angle of parrallelism https://en.wikipedia.org/wiki/Angle_of_parallelism

When C is an ideal point $ AC = BC = \infty , \angle C = 0 $

Then by the angle of parallelism https://en.wikipedia.org/wiki/Angle_of_parallelism you get

$\sin \Pi(AB) = \sin \angle BAC = \frac {1}{\cosh (AB) } = sech(AB)$.

When $ \angle BAC$ is larger than $ \Pi(AB)$ there is no triangle (the "sides" won't meet)

and when $ \angle BAC$ is smaller than $ \Pi(AB)$ C is not an ideal point and you have a normal triangle.

To prove all of this you are in a difficult point , it all depends on the curvature (in the formula's above it is supposed to be $-1$)

If the curvature is not suppoost to be $-1$ the curvature first need to be calculated and then AB must be replaced by $ \frac{AB}{-K} $

so i guess you will need to find any proof of the angle of parallelism and tweak it a bit.

hopes this helps.

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