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I am asked to calculate the area of a spherical triangle of points $(0,0,1),(\frac{1}{\sqrt2},0, \frac{1}{\sqrt2})$ and $(0,1,0)$. I know I will have to use Gauss Bonnet formula , after having found the internal angles but since we are given no formula to find them I managed to find by visual analysis just one angle of $90$ degrees. Help would be appreciated. Regards |
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Think of the $y$ axis as the north-south axis, with $(0,1,0)$ the north pole. Observe that both $(1,0,0)$ and $(1/\sqrt{2},0,1/\sqrt{2})$ lie on the equator (given by $y=0$ on the sphere). This means that the triangle's angle at the vertex $(0,1,0)$ is $45$ degrees. Since the great circle connecting $(0,1,0)$ and any point on the equator is perpendicular to the equator, the other two vertices of the triangle have angle $90$ degrees. So the triangle takes up an eighth of the northern hemisphere, thus a sixteenth of the area of the entire sphere: $\frac{4\pi}{16}$. |
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