For Euclidean $\triangle ABC$, $\frac{a}{\sin A}$ is the circumdiameter. What is $\frac{\sin a}{\sin A}$ for spherical $\triangle ABC$? Given plane $\triangle ABC$, it is well known that the common value of the ratios appearing in the Law of Sines ...
$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$$
... is equal to the diameter of the circle which passes through the three vertices.

For a spherical $\triangle ABC$, is there a similar (or any) nice interpretation for the common ratios in the spherical Law of Sines?
$$\frac{\sin a}{\sin A}=\frac{\sin b}{\sin B}=\frac{\sin c}{\sin C}$$

Thanks
 A: In the Law of Sines for spherical triangles,
$$
\frac{\sin a}{\sin A} = \frac{\sin b}{\sin B} = \frac{\sin c}{\sin C}
= \frac{\sin a \sin b \sin c}
       {\sqrt{1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c}}.
$$
The formula on the far right bears some vague resemblance to a formula for the corresponding constant in plane geometry,
but I do not have a "nice" interpretation of it.
This notation assumes that the "lengths" of the sides are measured by the angle each side subtends at the center of the sphere. (If the sphere has radius $1$, this is the same as the length of the arc forming each side.)
The relationship between the circumradius of the triangle and the ratio $\frac{\sin a}{\sin A}$ is not going to be as simple as the corresponding relationship for plane triangles, as a simple example will demonstrate. Consider the triangles $\triangle ABC$ and $\triangle AB'C'$ with the following spherical coordinates:
$$
\begin{eqnarray}
\phi &=& 0 &\mbox{ at } A \\
(\phi.\theta) &=& \left(\frac\pi4,0\right) &\mbox{ at } B \\
(\phi.\theta) &=& \left(\frac\pi4,\frac\pi2\right) &\mbox{ at } C \\
(\phi.\theta) &=& \left(\frac{3\pi}4,0\right) &\mbox{ at } B' \\
(\phi.\theta) &=& \left(\frac{3\pi}4,\frac\pi2\right) &\mbox{ at } C'
\end{eqnarray}
$$
That is, $A$ is on the axis from which $\phi$ is measured; $B$ and $B'$ are on a common great circle through $A$; and $C$ and $C'$ are on a common great circle through $A$. Clearly both triangles have the same angle at $A$, and by symmetry the length of the side opposite $A$ is the same in both triangles; that is, if the arc from $B$ to $C$ is $a$ and from $B'$ to $C'$ is $a'$, then $a = a'$ and
$$\frac{\sin a}{\sin A} = \frac{\sin a'}{\sin A}.$$
But these triangles clearly do not have congruent circumcircles, 
so if $\frac{\sin a}{\sin A}$ equals some function of the circumradius of $\triangle ABC$, 
it follows that $\frac{\sin a'}{\sin A}$ does not equal the same function of the circumradius of $\triangle AB'C'$.
Some interesting facts about this are shown by Odani, Kenzi. "On a Relation between Sine Formula and Radii of Circumcircles for Spherical Triangles." 愛知教育大学研究報告. 自然科学編 59 (2010): 1-5.
https://aue.repo.nii.ac.jp/?action=pages_view_main&active_action=repository_view_main_item_detail&item_id=633&item_no=1&page_id=13&block_id=21
One of these facts is that if $R$ is the circumradius of $\triangle ABC$, then
$$
\left(\frac{\sin a}{\sin A}\right)^2
 = 4 (\tan^2 R) \left(\cos^2\frac a2\right) \left(\cos^2\frac b2\right) \left(\cos^2\frac c2\right).
$$
An immediate consequence of this is that for any non-degenerate spherical triangle $\triangle ABC$,
$$
\frac{\sin a}{\sin A} < 2 \tan R.
$$
But for "most" triangles, the ratio $\frac{\sin a}{\sin A}$ is not close to this limit.
If $\triangle ABC$ is a large triangle 
whose circumcenter lies within the triangle, 
the upper bound of $\frac{\sin a}{\sin A}$ is quite a bit smaller than this.
The reference above is quite explicit about what the upper (and lower) bounds of the
ratio are; there are four separate cases to consider, and in each case the upper and lower bounds are functions of the circumradius, but not always as simple as $2\tan R$.
A: There are two Spherical Laws of Cosines:
$$(1)\quad \cos c = \cos a \cos b + \sin a \sin b \cos C$$
$$(2)\quad \cos A = -\cos B \cos C + \sin B \sin C \cos a$$
where $a,b,c$ are lengths, and $A,B,C$ are the opposite angles.
