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  1. 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?

  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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For spherical trigonometry, the collection of formulas is much larger than in the Euclidean case. The following 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 $180^\circ$. –  André Nicolas Oct 26 '11 at 15:13
thanks 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? –  van abel Oct 26 '11 at 15:18
I 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? –  van abel Oct 26 '11 at 15:40
There is no such thing as "similarity" for spherical polygons. See this. –  Guess who it is. Oct 26 '11 at 15:43
I would remark that the same identical question has been posted even on Mathoverflow, here: mathoverflow.net/questions/79167/… –  Giuseppe Tortorella Oct 26 '11 at 15:58

1 Answer 1

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) $, expand cos, let R go to infinity to derive Pythagoras theorem approaching from spherical side. 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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