Do spherical triangles with the same base and altitude have the same height? 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.
 A: 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 $4\arctan(2/\sqrt3)-180=16.4264^\circ$.
