In an Euclidean plane, we know that the area of a triangle is determined by the length of base and the height, i.e., $$ S_{\Delta}=\frac{1}{2}a.h, $$ where $a$ is the length of base and the $h$ is the height. Is there a similar formula in Spherical and hyperbolic spaces?
In Euclidean plane, we know that the cosine law says that (suppose $\alpha$, $\beta$, $\gamma$ are the angles, and $a, b, c$ are the lengths opposite $a, b, c$ respectively.) $$ \cos\gamma=\frac{a^2+b^2-c^2}{2ab}. $$ Then is there a analogue in spherical and hyperbolic geometry? I have noted that the First and Second Cosine law are not so clearly relevant with this formula.
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$\begingroup$ Posted even on MathStackexchange, here: math.stackexchange.com/questions/76084/… $\endgroup$– agtCommented Oct 26, 2011 at 16:00
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$\begingroup$ Is that illegal to a same question on two site? if it is, I owe an apologize. $\endgroup$– van abelCommented Oct 26, 2011 at 16:10
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3$\begingroup$ It’s certainly legal, but rather frowned upon, because it fragments the discussion and leads to duplication of effort. It makes sense to repost a question on the other site if you do not receive any answers, but you should wait a couple of days before doing so. In any case, you should crosslink the two questions to each other (which Giuseppe just did) so that people know. $\endgroup$– Emil JeřábekCommented Oct 26, 2011 at 16:17
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$\begingroup$ Thanks, That is reasonable, I will do that in my following questions. $\endgroup$– van abelCommented Oct 27, 2011 at 3:02
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1 Answer
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For hyperbolic geometry, you can find the answers to your questions on Wikipedia. An interesting thing is that there is an absolute upper bound on the area of a hyperbolic triangle, even though lengths are unbounded.
For spherical geometry, you can find the area of a spherical triangle here and corresponding trigonometric formulas here.
answered Oct 26, 2011 at 15:37
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$\begingroup$ Another interesting thing is that the area of a hyperbolic triangle is determined by its three angles. $\endgroup$ Commented Oct 26, 2011 at 16:00
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$\begingroup$ And the area of a spherical triangle likewise. $\endgroup$ Commented Oct 26, 2011 at 17:30
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$\begingroup$ And in both cases, a countably infinite set of circle-square pairs of equal area, both of which are constructible with compass and straightedge. But no algorithm that starts with either a square or circle of unknown area and produces the matching figure. I'm just saying. $\endgroup$ Commented Oct 26, 2011 at 18:46
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$\begingroup$ @Will. Good point. I guess I used the where I meant to use an. I did not intend to imply the uniqueness of interestingness. I have edited accordingly. $\endgroup$ Commented Oct 26, 2011 at 22:45