Another Hyperbolic triangle problem (all given except angle $\angle C$, and side $c$)

I thought that after asking How to solve an hyperbolic Angle Side Angle triangle? I could solve all hyperbolic triangles, but I am still stumped with SSA and AAS from both you can via the hyperbolic law of sinus calculate one extra value to get to AASS but then I got stuck again therefore:

If from an hyperbolic triangle $ \triangle ABC$ the angles $\angle A$, $\angle B$ and sides $a$ and $b$ are given. (so a combination of ASS or AAS )

How can I calculate the angle $\angle C$ or side $c$?

I thought with the two variations of the Hyperbolic law of cosines ( https://en.wikipedia.org/wiki/Hyperbolic_law_of_cosines )

or the hyperblic law of sinus ( https://en.wikipedia.org/wiki/Law_of_sines#Hyperbolic_case )

It must be easy, but neiter law can solve this problem.

Or am I overlooking something (obious) again?


Somehow the following formula's work:

$$ \tanh\left(\frac{c}{2}\right) = \tanh\left(\frac{1}{2} (a-b)\right) \frac{\sin{\left(\frac{1}{2}(A+B)\right)}}{\sin{\left(\frac{1}{2}(A-B)\right)}} $$


$$ \tan\left(\frac{C}{2}\right) = \frac{1}{\tan\left(\frac{1}{2} (A-B)\right)} \frac{\sinh\left(\frac{1}{2}(a-b)\right)}{\sinh\left(\frac{1}{2}(a+b)\right)} $$

They are hyperbolic alternatives to the spherical formulas mentioned at https://en.wikipedia.org/wiki/Solution_of_triangles#A_side.2C_one_adjacent_angle_and_the_opposite_angle_given

(maybe they can be simplified)

I don't know exactly why they work (and i won't accept this answer till i know)

user MvG advised me to investigate
so maybe after that I can explain them :)

proof based on the proof for spherical geometry by I Hothunter, section 52

(http://www.gutenberg.org/ebooks/19770 )

But then rewritten for hyperbolic geometry

The hyperbolic sinus rule ( https://en.wikipedia.org/wiki/Law_of_sines#Hyperbolic_case )

$$ \frac{\sin A}{sinh(a)} = \frac{\sin B}{sinh(b)}= \frac{\sin C}{sinh(c)} $$

from it follows that $ \frac{\sin A + \sin B}{sinh(a) + sinh(b)} = \frac{\sin C}{sinh(c)} $

and $ \frac{\sin A - \sin B}{sinh(a) - sinh(b)} = \frac{\sin C}{sinh(c)} $

In the following $ \frac{\sin C}{sinh(c)} $ or any of the other equivalents is $ SQ $

The hyperbolic cosinus rule ( https://en.wikipedia.org/wiki/Hyperbolic_law_of_cosines )

has two forms

(CR1) $ cosh(a) = cosh (b) cosh(c) - sinh(b) sinh(c) \cos A $


(CR2) $ \cos A = - \cos B \cos C + \sin B \sin C cosh(a) $

CR2 rewritten

(1) $ \sin B \sin C cosh(a) = \cos A + \cos B \cos C $

and for $ \angle B $

(2) $ \sin A \sin C cosh(b) = \cos B + \cos A \cos C $

1 and 2 involving SQ

(3) $ SQ \sin C cosh(a) sinh(b) = \cos A + \cos B \cos C $

(4) $ SQ \sin C cosh(b) sinh(a) = \cos B + \cos A \cos C $

adding 3 and 4 together

(5) $ SQ \sin C ( cosh(a) sinh(b) + cosh(b) sinh(a) ) = (\cos B + \cos A ) * (1 + \cos C )$

(6) $ sinh(x + y) = sinh (x) cosh (y) + cosh (x) sinh (y) $


(7) $ SQ \sin C sinh(a+b) = (\cos B + \cos A ) (1 + \cos C )$

(8) $ \frac {\sin X}{1 + \cos X } = \tan (\frac {X}{2}) $


(9) $ SQ \tan (\frac {C}{2}) sinh(a+b) = (\cos B + \cos A ) $

(10) $ SQ = (\frac {\sin A + \sin B }{ sinh(a) + sinh(b)} $


(11) $ (\sin A + \sin B ) \tan (\frac {C}{2}) sinh(a+b) = (\cos B + \cos A ) (sinh(a) + sinh(b))$

(12) $ \tan(\frac {1}{2} (A + B) ) = \frac {\sin A + \sin B}{\cos B + \cos A} $

(13) $ \tan(\frac {1}{2} (A + B) ) \tan (\frac {C}{2}) sinh(a+b) = (sinh(a) + sinh(b)) $

(14) $ \tan (\frac {C}{2}) = \frac {1}{\tan(\frac {1}{2} (A + B) ) } \frac {sinh(a) + sinh(b)}{sinh(a+b)} $

That should do :)


Suppose you know angles $\alpha$ and $\beta$ as well as edge lengths $a$ and $b$. As you said, one of these can be computed from the other three using the law of sines. So you need to find $c$ and $\gamma$.

Pick one of the law of cosines, e.g.

$$\cosh a=\cosh b\cosh c - \sinh b\sinh c\cos \alpha$$

Then combine that with the well-known identity $\cosh^2 c-\sinh^2 c=1$ and eliminate one variable, e.g. $\sinh c$ using e.g. a resultant. You end up with a quadratic equation for the remaining variable, in this case

$$(\sinh^2b\cos^2\alpha - \cosh^2b)\cosh^2c + (2\cosh a\cosh b)\cosh c - (\sinh^2b\cos^2\alpha + \cosh^2a)=0$$

As you can see, this is a quadratic equation in $\cosh c$. Compute its two solutions, and check which of them is the one you need. Perhaps one value is outside the range of $\cosh$? To find $\gamma$ you can again use the law of sines.

  • 1
    $\begingroup$ that doesn't seem to work, How do you get from $\cosh a=\cosh b\cosh c - \sinh b\sinh c\cos \alpha$ to $\cosh c = \frac{\cosh a}{\cosh b - \sinh b\cos\alpha}$ ? $\endgroup$ – Willemien Feb 3 '15 at 12:26
  • $\begingroup$ @Willemien: I had misread my own formula, hadn't noticed that one part was $\sinh$ not $\cosh$. It's more correct now, but also a lot less elegant. We'll see whether someone else comes up with a better solution. $\endgroup$ – MvG Feb 3 '15 at 13:00
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    $\begingroup$ found someting at en.wikipedia.org/wiki/… retweaking it gives tanh( \frac{c}{2}) = \cot(\frac{1}{2} (A-B) sqrt{\frac{\frac{cosh(a-b) -1}{cosh(a+b) -1}} or someting like that, it works but I don't understand why $\endgroup$ – Willemien Feb 3 '15 at 14:13
  • $\begingroup$ @Willemien: Well, I suggest you answer your own question providing both the link and the formula, and if you really want to know how this works and can't work it out even after looking at Napier's analogies (spherical) and perhaps also tangent half-angle formulas and this complex relationship, then you can post a separate question asking about that. $\endgroup$ – MvG Feb 3 '15 at 14:20

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