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I've forgotten too much math to do this myself:

A guy in a general forum proposed a regular tetrahedron inscribed inside a sphere (specifically, the Earth), and then the borders of the sides are projected onto the sphere (making arcs of great circles) to entirely cover the sphere. His assertion is that for this case the sum of the angles of each triangle will be 360°. Is this correct, and either way why?

A further question for this is if this regular tetrahedron has its top vertex at the north pole, what will be the latitude of the remaining vertices?

Traildude
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  • Let us know your thoughts and what attempts you have made to solve the question. – FD_bfa Jan 03 '23 at 23:52
  • Questions to think about: Are all the angles the same? Why (or why not)? How many angles are there in total? How many circles of $360^\circ$ do they add up to? – Henry Jan 04 '23 at 00:12
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    The area of a spherical triangle is the sum of the angles minus $180$ ... the idea "Spherical excess" is discussed here ... https://en.wikipedia.org/wiki/Spherical_trigonometry#Area_and_spherical_excess – Donald Splutterwit Jan 04 '23 at 00:15
  • To cover the entire sphere these areas must add up to $720$ ... So a bit of linear algebra will get you to the value ($360$) mentioned. – Donald Splutterwit Jan 04 '23 at 00:19
  • It's very easy to find a triangle with internal angles all right-angles on the sphere. Just draw two great circles through one pole at right angles and then consider its intersection with the equator. – CyclotomicField Jan 04 '23 at 00:24

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If the tetrahedron is indeed regular, then after it's projected onto the sphere each triangle will have 120° angles at every vertex.

If a tetrahedron consists of four congruent trianular faces with any possible shape, such as by twisting a regular one about one of its $D_{2d}$ symmetry axes, then the angles of any triangular face match those meeting at any vertex. So in this more general case the angles in each face on the spherical projection still add up to 360°.

Oscar Lanzi
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