I know that the spherical law of sines is $\frac{\sin A}{\sin a}=\frac{\sin B}{\sin b}=\frac{\sin C}{\sin c}$ but how does one prove this rule?
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3No, it is an unsolved conjecture ... – hmakholm left over Monica Apr 10 '16 at 11:14
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@Henry There better be, otherwise is not a "law" but at best "A good shot" or a "What are the odd this time it won't work?" guess – DonAntonio Apr 10 '16 at 11:17
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You can see here: http://isites.harvard.edu/fs/docs/icb.topic246919.files/March-07.pdf – Emilio Novati Apr 10 '16 at 11:29
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Thanks @EmilioNovati, that's perfect – Henry Lock Apr 10 '16 at 12:00
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Choose a coordinate system so that the three vertices of the spherical triangle is located at $$(1,0,0),\quad (\cos a,\sin a,0)\quad\text{ and }\quad(\cos b, \sin b\cos C, \sin b\sin C)$$ The volume of the tetrahedron formed from these 3 vertices and the origin is $\frac16 \sin a\sin b\sin C$. Since this volume is invariant under cyclic relabeling of the sides and angles, we have $$\sin a\sin b\sin C = \sin b\sin c\sin A = \sin c \sin a \sin B \implies\frac{\sin A}{\sin a} = \frac{\sin B}{\sin b} = \frac{\sin C}{\sin c}$$
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