# Spherical Triangle: Law of Sines with Clairaut's theorem

The spherical law of sines states that

On the sphere $$\mathbb{S}^2$$ we consider a triangle, i.e. three points connected by geodesics. We denote the side lengths with $$a, b, c < \pi$$ and the angles opposite to these sides with $$\alpha, \beta, \gamma \in [0, 2\pi]$$.

Then: $$\frac{\sin \alpha}{\sin a} = \frac{\sin \beta}{\sin b} = \frac{\sin \gamma}{\sin c}.$$

As part of my studies, I have been given the task of proving the above theorem. However, I need to somehow apply Clairaut's theorem, that is:

Let $$f$$ be a rotation surface, i.e. $$f(t, \varphi) := \left( r(t)\cos\varphi, r(t) \sin \varphi, h(t)\right)$$ and let $$\gamma(s) := \left( t(s), \varphi(s)\right)$$ be a regular curve and $$c := f \circ \gamma$$. Futher, we denote by $$\theta(s)$$ the angle between the curve $$c(s)$$ and the line of latitude through $$c(s)$$. That means: $$\cos \theta(s) = \frac{\langle c'(s), \partial_2f(\gamma(s)) \rangle}{\lvert c'(s) \rvert \lvert \partial_2f(\gamma(s)) \rvert}.$$

Then: $$s \mapsto r\left(t(s)\right) \cos \theta(s) \; \text{is constant}.$$

What I did: I proved Clairaut's theorem.

Remark: I know that many answers are already given here – however I think I need a little help to apply Clairaut's theorem.

IDEA: Maybe it could be benificial to put one corner of the triangle onto the north pole.