How should the Law of Cosines
$$\cos c = \cos a \cos b + \sin a \sin b \cos C$$
be modified if sides $(a,b,c) $ are not geodesics but are small circles with geodesic curvatures $k_a,k_b,k_c?$
It may be useful for navigation on the globe as airplanes and ships do not always take the shortest path.
EDIT1:
Parallelly could we consider using radii ( of curvatures instead of curvatures directly? ... like $(R_a,R_b,R_c) $ along with triangle sides because there seems to be a possible advantage for direct spherical trigonometric calculations and simplifications on each triangle side. A schematic with a rough hand sketch of small circles:
Set the stage on the unit sphere with a (proper) spherical triangle $\triangle ABC$ whose (great arc) sides have lengths $a$, $b$, $c$. For specificity, although I'll be doing any and all coordinate calculations off-the-page, we can take the vertices to be $$A = (1,0,0) \qquad B = (\cos c, \sin c, 0) \qquad C = ( \cos b, \sin b \cos A, \sin b \sin A) \tag{0}$$ Importantly, $a$, $b$, $c$ match central angles of the sphere (eg, $a=\angle BOC$), and $A$, $B$, $C$ are dihedral angles between faces angles of tetrahedron $OABC$.
To save a little visual clutter, I'll denote half-angles with a subscripted $2$; eg, $a_2 := a/2$.
In the "modified" figure with small-arc sides, let those arcs have lengths $a^\star$, $b^\star$, $c^\star$ and radii $\sin\alpha := 1/k_a$, $\sin\beta :=1/k_b$, $\sin\gamma :=1/k_c$. (Radii are more convenient here than curvature. And writing them as sines are helpful for simplifying frequent $\sqrt{1-\text{radius}^2}$ expressions.) Finally, let the "angles" —defined by tangent vectors to the "sides" at the vertices— be $A^\star$, $B^\star$, $C^\star$.
If $A'$ is the center of the circle through $B$ and $C$, then both $\angle BOC$ and $\angle BA'C$ subtend chord $BC$ of the sphere, and we have
$$2 \sin a_2 = |BC| = 2 \sin\alpha \sin\frac{a_2^\star}{\sin\alpha} \quad\to\quad \frac{\sin a_2}{\sin\alpha} = \sin\left(\frac{a_2^\star}{\sin\alpha}\right) \tag{1}$$
Note that in the great circle case, $\alpha=\pi/2$, so that $(1)$ reduces to $a^\star=a$, as expected.
Define $\overline{\alpha} := a_2^\star/\sin\alpha$ as the angle on the right-hand side of $(1)$ (likewise $\overline{\beta}$ and $\overline{\gamma}$) so that we can write $$\sin a_2 = \sin\alpha \sin\overline{\alpha} \tag{2}$$ The reader can verify that point $A'$ is given by $$A' = \frac{\cos\alpha \sin\alpha}{w\cos(a/2)} \left(\begin{array}{l} \phantom{+} A \sin a \cos\overline{\alpha} \\ + B \sin b \;( \cos\alpha \sin\overline{\alpha} \sin C - \cos\overline{\alpha} \cos C ) \\ + C \sin c \;( \cos\alpha \sin\overline{\alpha} \sin B - \cos\overline{\alpha} \cos B ) \end{array}\;\right) \tag{3}$$ where $w := \sin A\sin b \sin c=\sin a \sin B \sin c = \sin a \sin b \sin C$ (these equalities guaranteed by the spherical Law of Sines).
It should be noted that there's an ambiguity in these circles. While there is a unique great circle (of radius $1=\sin(\pi/2)$) through $B$ and $C$, for every radius $\sin\alpha$ in between there are two circles. (The two smallest circles coincide, but the semi-circular arc corresponding to $a^\star$ is ambiguous.) Equation $(3)$ conveniently resolves this ambiguity if we allow $\alpha$ to vary from $0$ to $\pi$; the positive-vs-negative values of $\cos\alpha$ for acute-vs-obtuse angles $\alpha$ distinguish the two positions of $A^\prime$.
Now, the vector at $C$ tangent to the "modified" side $CB$ is perpendicular the plane of $\bigcirc A'$ and perpendicular to $A'C$. Thus, $$\begin{align} t_A &:= \phantom{+}A' \times (C - A') \\ &\phantom{:}=- A \sin a \cos\alpha \sin\overline{\alpha} \\ &\phantom{:=}+ B \sin b \;( \cos\alpha \sin\overline{\alpha} \cos C + \phantom{\cos a} \cos\overline{\alpha} \sin C ) \\ &\phantom{:=}+ C \sin c \;( \cos\alpha \sin\overline{\alpha} \cos B - \cos a \cos\overline{\alpha} \sin B ) \\[4pt] |t_A| &= w \cos a_2 \end{align} \tag{4}$$ Swapping $A$- and $B$-related elements gives the vector $t_B$ tangent at $C$ to "modified" side $CA$. We find then that
$$\cos C^\star = \frac{t_A\cdot t_B}{|t_A||t_B|} = \frac{\left(\begin{array}{l} \phantom{+} ( \cos\overline{\alpha} \cos\overline{\beta} - \cos\alpha \sin\overline{\alpha} \cos\beta \sin\overline{\beta} ) \cos C \\ - ( \cos\alpha \sin\overline{\alpha}\cos\overline{\beta} + \cos\overline{\alpha}\cos\beta \sin\overline{\beta}) \sin C \end{array}\right)}{\cos a_2\cos b_2} \tag{5}$$
Again, in the great circle case, $\alpha=\beta=\pi/2$ and $\overline{\alpha}=a/2$ and $\overline{\beta}=b/2$, so that $(5)$ reduces to $\cos C^\star = \cos C$.
Risking notational overload, let us define $a':=\arctan(\cos\alpha\tan\overline{\alpha})$ (likewise $b'$ and $c'$), so that
$$\cos a' = \frac{\cos\overline{\alpha}}{\cos a_2} \qquad \sin a' = \frac{\cos\alpha \sin\overline{\alpha}}{\cos a_2} \tag{6}$$ Then $(5)$ happens to simplify to $$\cos C^\star = \cos(a'+b'+C)\tag{7}$$ which makes me think there's a much easier route to it. Nevertheless ...
From here, "all we need to do" is rewrite $(5)$ or $(7)$ without the elements of (proper) spherical triangle $\triangle ABC$. This can be done via $$\begin{align} \cos C &= \frac{\cos c - \cos a \cos b}{\sin a\sin b} \\[4pt] \sin C &= \frac{\sqrt{1+2\cos a\cos b \cos c-\cos^2 a-\cos^2 b-\cos^2 c}}{\sin a\sin b} \end{align} \tag{8}$$ (from a spherical Law of Cosines) and replacing the trigs of $a$, $b$, $c$ with expressions in terms of $\sin a_2$, $\sin b_2$, $\sin c_2$, and finally invoking $(2)$ to get a relation involving only arc-lengths $a^\star$, $b^\star$, $c^\star$, radii $\sin\alpha$, $\sin\beta$, $\sin\gamma$, and the "modified" angle $C^\star$.
The result is a mess that doesn't seem worth typing-up. If there's an elegant relation hiding in there, I haven't found it.