In an Euclidean plane, we know that the area of a triangle is determined by the length of base and the height, then is there a similar thing do happen 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.
Is there a similar formula in spherical and hyperbolic geometry as Euclidean Geometry?
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1For spherical trigonometry, the collection of formulas is much larger than in the Euclidean case. The [following](http://en.wikipedia.org/wiki/Spherical_trigonometry) gives a start. There is some reason to think that in fact spherical trigonometry, because of the needs of astronomy, is older than plane trigonometry. The area of a spherical triangle is determined by the "excess" of the sum of the angles, that is, the sum of the angles minus $180^\circ$. – 2011-10-26
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0thanks for your reply, however I am not so satisfied, since my first question is that: is there a formula as $$S_{\Delta}=\frac{1}{2}a.h $$ in spherical or hyperbolic geometry? – 2011-10-26
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0I didn't quite understand, your answer is negative for my first question, but the reasoning is not so convinced to me. As your use the scale (similarity) to get a contradiction, but did there a similar (geodesic) triangle in sphere? Since we know that the triangle is totally decided by its angles, if the scale is keeping the angles invariant, then how to define the concept of similarity? – 2011-10-26
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0There is no such thing as "similarity" for spherical polygons. See [this](https://secure.wikimedia.org/wikipedia/en/wiki/Spherical_trigonometry#Lines_and_angles_on_a_sphere). – 2011-10-26
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0I would remark that the same identical question has been posted even on Mathoverflow, here: http://mathoverflow.net/questions/79167/is-there-a-similar-formula-in-spherical-and-hyperbolic-geometry-as-euclidean-geom – 2011-10-26
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0I don't know the general case is true or not, but for a special case the formula (in first question) is validated. In a unit sphere, we know that the area of a triangle is $$ S=\alpha+\beta+\gamma-\pi. $$ Take a triangle with two right angles (eg. one side is on equator and the other two are perpendicular to it, then the vertex may be viewed as the north pole.) Suppose the angle make in north pole is $\alpha$, then the base is also $\alpha$. Thus, the area for this triangle is $$ S=\frac{\pi}{2}+\frac{\pi}{2}+\alpha-\pi=\alpha. $$ In which case it do double if you double the base. – 2011-10-26
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0Nice observation, you are right, works more generally. – 2011-10-26
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0For hyperbolic space, the area of a triangle is $\pi - (\alpha + \beta + \gamma)$. So a triangle consisting of three hyperparallel lines asymptotic to each other is $\pi$, and all other triangles are smaller. – 2011-10-26
1 Answers
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Yes,there are analogues in spherical and hyperbolic geometry.
Start with a right angled triangle $\gamma=\pi/2$.
$$\cos(c/R)=cos(a/R) cos(b/R) $$
can be expanded in power series:
$$( 1-(c/R)^2/2)\approx (1-(a/R)^2/2)*(1-(b/R)^2/2) $$
and let R go to infinity to derive Pythagoras theorem approaching from spherical side getting
$ c^2 = a^2 + b^2 $
Likewise,
$\cosh(c/R)=\cosh(a/R) \cosh(b/R) $, expand cosh, let R go to infinity to derive it approaching from hyperbolic side:
$ c^2 = a^2 + b^2 $
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0Gro-Tsen's article http://www.madore.org/~david/weblog/d.2013-12-17.2175.trigonometrie-triangle.html#d.2013-12-17.2175 expands on these analogs in more detail. – 2018-12-31