Given a sphere with a specified radius, and two perpendicular arcs produced by angles, phi and theta.

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To be clear, phi and theta are the angles which give rise to the arcs which meet at a right angle. I placed the labels, phi and theta, on the actual arcs because it gets crowded near the center of the sphere. I am looking for the interior angle which gives rise to the oblique arc on the sphere's surface.

I read these links:



How do I measure distance on a globe?

relationship between a great circle arc and a latitude circle arc at a given latitude

but the derivation of an answer eludes me.

I realize that spending time to answer a question so fundamental and which has probably been answered clearly in some text is a waste of bandwidth. So maybe someone knows of a clear, simple exposition of this problem to which I can be referred?


2 Answers 2


The unit vectors in the directions of the sides of the unknown angle are (in a convenient basis) $$ v_1=(\cos\phi,0,\sin\phi),\qquad v_2=(\cos\theta,\sin\theta,0) $$ Therefore, the cosine of the angle between them is their dot product $$ \langle v_1,v_2\rangle=\cos\phi\cos\theta $$ That is, the angle is $$ \cos^{-1}(\cos\phi\cos\theta). $$

  • $\begingroup$ I do not understand the words that come after your 'therefore'. Where can I understand why 'the cosine of the angle between them is their dot product'. My understanding of dot product is that graphically it is the projection of the second vector onto the first vector. $\endgroup$ Nov 2, 2021 at 21:03
  • $\begingroup$ For any pair of vectors, the cosine of the angle between them is equal to $\langle u,v\rangle/||u||||v||$. This is because the dot product is $\langle u,v\rangle=||u||||v||\cos(u,v)$. If your vectors are unit vectors, you have $\langle u,v\rangle =\cos(u,v)$. $\endgroup$
    – GReyes
    Nov 2, 2021 at 22:06

The spherical right triangle in non-euclidean geometry this relation can be written as $$ \cos\gamma= \cos\phi\cdot \cos\theta $$

In series expansion upto second degree $$ 1- \gamma^2/2!\approx (1- \phi^2/2!)\cdot(1- \theta^2/2!) $$

It can be readily derived using spherical trig Law of Cosines $ \cos \ \alpha \approx 1- \alpha^2$ etc., it should be memorized. It is a sort of deeper non-linear hypotenuse.

Reducible in Euclidean geometry to

$$ \gamma^2\approx \phi^2+\theta^2\;;\;c^2=a^2+b^2.$$

[ Although not relevant here, tempted to mention in hyperbolic geometry its mirroring relation.. $ \cosh\gamma= \cosh\phi\cosh\theta $].

  • $\begingroup$ thank you narasimham. but I am looking for a derivation so I can UNDERSTAND it. My problem is so many authors like to demonstrate their erudition by making their texts dense and skipping many steps. $\endgroup$ Nov 2, 2021 at 21:00

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