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Interior angles of polygon on a sphere 
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#1
Feb2212, 05:25 PM

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Hi can anyone help me out with finding the interior angles of a pentagon on a sphere. I know two of the interior angles already and I know all the angles that correspond with the arc lengths of the sides of the pentagon. How do I find the other three interior angles?
Thanks 


#2
Feb2212, 07:36 PM

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P: 9,488

this is really interesting. on a sphere of radius R, we call K = 1/R^2 the curvature. The triangles on a sphere look moire like Euclidean ones the smaller they are. I.e. if I recall correctly, the formula for angle sum of a triangle on a sphere of radius R, and curvature K = 1/R^2, is something like angle sum = π + K.area.
Thus the smaller the area of the triangle, in comparison to the radius squared, the closer is the angle sum of a triangle to π. Of course if you know the a sum of a triangle you can also calculate that for a pentagon. 


#3
Feb2312, 01:59 PM

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P: 9,488

even if you require your polygons to lie in a given hemisphere, note that by taking the vertices closer and closer to being on the great circle bounding that hemisphere, you can make the interior angles of any polygon closer and closer to π (=180 degrees). Thus on a sphere of radius one, the sum of the interior angles of a pentagon lies between 3π and 5π, with the difference (angle sum  3π) equaling the area of the pentagon.



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