L'Huillier's theorem or calculation of spherical excess of "spherical triangle" formed between the unit vectors on unit sphere can find out the area, but how to explain this formula from purely plane trigonometry standpoint (i.e. without assuming any pre-requisite knowledge on spherical trigonometry)? The solid angle can be found by spherical trigonometry rules, and I am well aware of it. I want to introduce this problem to anyone with knowledge of plane trigonometry, but no knowledge of spherical trigonometry.
Dear Sumit, from my own experience, I can tell you that a number of people can't see in 3 dimensions, not even metric issues, like the one in your question. I would introduce them, gently, in 3D, and then go towards finer issues.
Dear Sumit, from my own experience, I can tell you that a number of people can't see in 3 dimensions, not even metric issues, like the one in your question. I would introduce them, gently, in 3D, and then go towards finer issues.
The the spherical angle over a spherical triangle T in the sphere $ r = 1$, is the integral
$\int_T \Omega, where \Omega = sin(\theta)d\theta d\phi
Now how do we compute this integral?
Suppose the triangle is spanned by e_0, e_1, e_2, on the unit sphere (obviously not in general orthonormal). Choose coordinates such that e_2 = NP = (0,0,1), e_0 = (x_0, 0, z_0) (ie. e_0, e_2 span the X-Z plane. Then in spherical coordinates e_0 is
r =1,
\phi_0 = 0,
cos(\theta_0}) = /||e_0||||e_2|| = z_0.
Like wise e_1 = (x_1, y_1, z_1) and in polar coordinates r = 1, (cos, sin)(\phi_1) = (x_1, y_1)/sqrt(x_1^2+ y_1^2), cos(\theta_1) = z_1. Thus in spherical coordinates the the spherical sides are given by
e_2 e_0 : \phi =0, 0
Dear Sumit Bhowmick
I hope you find the following discussion is useful
https://math.stackexchange.com/questions/202261/solid-angle-between-vectors-in-n-dimensional-space#:~:text=The%20formula%20of%203D%20solid,c)%7C%E2%86%92a%7C.
Best regards
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