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If two spherical triangles have the same base $\theta$ and the same altitude $\phi$, do they have the same area. Initially I believed they would have by the same logic flat triangles do. However I'm starting to doubt myself that skewing a spherical triangle will no longer have its edges on great circles. I have not studied spherical geometry much at all.

user137794
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  • Are you aware that the area of a triangle is proportional to the sum of the angles, minus $\pi$ (or $180^\circ$ if you’re using degrees)? I don’t know what you mean by “skewing a triangle will no longer have...”, since the sides of a triangle are presumed always to be arcs of great circles. – Lubin Apr 09 '14 at 20:36
  • Well for flat triangles with a base on the x-axis, you can sort of skew it in the direction of the x-axis. It will have the same altitude and area. For a spherical triangle with a base on $\phi=\pi/2$ I was thinking of skewing the triangle about the axis perpendicular to the base. – user137794 Apr 09 '14 at 21:07
  • I still don’t know what you mean by “skew”. You seem to be thinking of Cartesian coordinates, but they make no sense here. – Lubin Apr 09 '14 at 22:16
  • I imagined cutting the triangle into latitude-discs and rotating them – user137794 Apr 09 '14 at 23:44
  • Two things go wrong: the usual construction depends on the cuts being parallel to each other. But there is no concept of parallelism in spherical geometry. And the usual construction depends on the cuts remaining the same length under the transformation you think you imagine. But this doesn’t happen either. – Lubin Apr 10 '14 at 02:06

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No, the area has nothing to do with the altitude. I have a thought experiment that would show it very clearly, if you believed it (which I’m not sure I do). But a direct computation will do the trick, remembering that the area is proportional to the “spherical excess”, that is $\angle_1+\angle_2+\angle_3-\pi$.

Take, first, a right triangle with legs of $30^\circ$ and $60^\circ$. I’m going to use the basic (“Napier”) rule for right triangles, that $\tan(\text{angle})=\tan(\text{opposite})/\sin(\text{adjacent})$. You can find Napier’s Rules for Spherical Triangles in any standard reference. In this case, the big angle has tangent equal to $\tan(60^\circ)/\sin(30^\circ)=\sqrt3/(1/2)=2\sqrt3$, and the small angle has tangent equal to $\tan(30^\circ)/\sin(60^\circ)=(1/\sqrt3)/(\sqrt3/2)=2/3$ So the spherical excess is $\arctan(2\sqrt3)+\arctan(2/3)-90=17.5880^\circ$.

The second triangle has base of $60^\circ$ and the same altitude of $30^\circ$, but has the altitude erected at the midpoint. Thus this triangle is the union of two right isosceles triangles with side $30^\circ$. The acute vertex angle has tangent equal to $\tan(30^\circ)/\sin(30^\circ)=1/\cos(30^\circ)=2/\sqrt3$. So in this case, the spherical excess is $4\arctan(2/\sqrt3)-180=16.4264^\circ$.

Lubin
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