If we take a sphere of radius 1 and travel a quarter-circumference south from the north pole, turn 90 degrees, travel another quarter circumference, then return North, we form a triangle with an angle of 270$^\text{o}$. According to Mathworld, the area of such a triangle would be given by $$\Delta=R^{2}[(A+B+C)-\pi]$$ With A, B and C the angles of the triangle.

Now if we consider lines of latitude and longitude (I understand these always intersect at $90^\text{o}$), and shift the base of the triangle closer to the north pole but so that it remains on a line of latitude, this would suggest that the base of the triangle still meets the two longitudinal lines at 90$^\text{o}$. This would suggest that the area stays the same, which seems very wrong.

I am also unsure of the maximum sum of angles of a spherical triangle. On the same Mathworld page it states:

The sum of the angles of a spherical triangle is between $\pi$ and $3\pi$ radians (180 degrees and 540 degrees; Zwillinger 1995, p. 469)

However, considering a very small triangle (s.t. the sum of interior angles tends to 180$^\text{o}$, and taking this as the outside of the triangle (and the rest of the sphere as the inside) this would create a 'triangle' with angles summing 900$^\text{o}$. My guess would be that this might not count as a triangle because it perhaps cannot be mapped appropriately to Euclidaen space. So then, what would be an appropriate mapping to Euclidean space? Would it result in a 'triangle' with curved sides?


A spherical triangle must be formed by arcs of great circles, hence you cannot take a parallel as side, apart from the equator.

And of the two parts into which a sphere is divided by the sides of a spherical triangle, only the smaller (and convex) one is considered "the inner part" of the triangle. That explains the 540° limit.

  • $\begingroup$ Right you are. It’s a common error to think that parallels of latitude can be sides of a spherical triangle. $\endgroup$ – Lubin Dec 19 '18 at 18:14

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