How to solve an Hyperbolic triangle when all is given except angle C and side c)

Question

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?

Answer 1

Somehow the following formula's work:

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

and

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

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
http://en.wikipedia.org/wiki/Spherical_trigonometry#Napier.27s_analogies
http://en.wikipedia.org/wiki/Tangent_half-angle_formula
and
http://en.wikipedia.org/wiki/Hyperbolic_function#Hyperbolic_functions_for_complex_numbers
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 $

and

(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) $

therefore

(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}) $

therefore

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

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

therefore

(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 :)

Answer 2

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.