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There's a problem that's bothering me lately: What's the area of a section of a unit sphere the shape of a non-euclidian triangle given that, when connecting the vertices of the triangle to the sphere's center, the angles formed by the three circular sectors of length $1$ are $\alpha$, $\beta$ and $\gamma$? My guess is to seperate the sphere in $\frac{360}{\alpha}$ parts so that each stripe has one circular sector of angle $\alpha$, then take the average of $\beta$ and $\gamma$ to make the triangle isosceles. Can anyone help me develop further or find a duplicate discussion? Calculus is fine. ThanksExample

As an example, here, $\alpha$, $\beta$ and $\gamma$ are $\angle BAC$, $\angle BAD$ and $\angle CAD$ (on their plane), and the area we are searching for is non-euclidian $BCD$.

TNT1288
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  • From "of length $1$", do you really intend that all three spherical triangle sides are length $1$? (If so, recall that on a unit sphere, the length of an (great circle) arc is equal to the angle it subtends at the center, so we would conclude $1 = \alpha = \beta = \gamma$.) – Eric Towers Mar 12 '24 at 03:31
  • Think about a triangular pyramid. Then, three sides of the pyramid would have length 1, and the common vertex those three sides would share would be the unit circle's center, and the three 'unit' sides would be three of its radii. The pyramid's base would be the non-euclidean triangle, so the three other triangles (with sides 1) are the three circular sectors, each with an angle. – TNT1288 Mar 12 '24 at 03:38
  • This might be a more similar duplicate, to find the area given the three angle / arc lengths $\alpha, \beta, \gamma$: Area of a spherical triangle – peterwhy Mar 12 '24 at 03:59
  • @peterwhy : Yes, the answer https://math.stackexchange.com/a/66731/123905 there, completely resolves this Question. – Eric Towers Mar 12 '24 at 04:05

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