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## Is there a similar formula in spherical and hyperbolic geometry as Euclidean Geometry? [closed]

1. 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?

2. 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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Posted even on MathStackexchange, here: math.stackexchange.com/questions/76084/… – Giuseppe Tortorella Oct 26 2011 at 16:00
Is that illegal to a same question on two site? if it is, I owe an apologize. – van abel Oct 26 2011 at 16:10
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. – Emil Jeřábek Oct 26 2011 at 16:17
Thanks, That is reasonable, I will do that in my following questions. – van abel Oct 27 2011 at 3:02

## closed as off topic by Benoît Kloeckner, Bruce Westbury, Anton Petrunin, Ryan Budney, Gerry MyersonOct 26 2011 at 21:29

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.

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 Another interesting thing is that the area of a hyperbolic triangle is determined by its three angles. – Emil Jeřábek Oct 26 2011 at 16:00 And the area of a spherical triangle likewise. – John Stillwell Oct 26 2011 at 17:30 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. – Will Jagy Oct 26 2011 at 18:46 @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. – Tony Huynh Oct 26 2011 at 22:45