What is the relationship between the side length and the angle of a spherical lune? Is there a sine theorem? 
1，Assuming that the red side is a big arc and the blue side is a small arc, what is the relationship between spherical angle A or B and the side length? 
2，If both sides are small arcs, what is the relationship between sides and angles? Is there a sine theorem?
3，If all the sides of a spherical triangle are small arcs, what is the relationship between its angle and its side length? Is there a sine theorem?
Why raise this question? Because in the work, we have to solve the length of the small arc between two points on the sphere. The known conditions are the length of a large arc and the spherical angle A or B.
 A: As you may know, spherical trigonometry has a nice Law of Sines (and nice Laws of Cosines) relating the lengths of arcs and the angles bounded by them, but these laws only work for great arcs. Small arcs defy nice relations, as shown below.

First, an observation about the arc-length formula:
$$\text{arc length} = \text{radius}\cdot\text{subtended central angle} \tag{1}$$
We'll find expressions for $\text{radius}$ and $\cos(\text{subtended central angle})$ in terms of a couple of parameters. Unfortunately, because these parameters appear both inside and outside a trig function, we won't be able to solve for them analytically, dashing hopes of a clean, explicit relation between lune angle to arc-lengths. Any work translating between these measurements will require numerical methods (which are beyond the scope of this discussion).

Consider an origin-centered sphere of radius $r$, with points $A$ and $B$ on the $xy$-plane; we'll define $C$ as the midpoint of the (minor) great arc $\stackrel{\frown}{AB}$, $C^\prime$ as the antipode of $C$, and $M$ as the midpoint of $\overline{AB}$. Specifically, defining $\theta := \frac12 \angle AOB$, we can take
$$A = (r\cos\theta,r \sin\theta,0) \quad B = (r\cos\theta, -r\sin\theta,0) \qquad C = (r,0,0) \quad M = (r\cos\theta,0,0)$$
Also, by $(1)$, we hava
$$|\stackrel{\frown}{ACB}| = 2r\theta \tag{2}$$
Now, consider a plane through $\overline{AB}$ that makes an angle $\phi$ with  the $xy$-plane; write $P$ for the midpoint of (minor) $\stackrel{\frown}{AB}$ in that plane. Thus, $\phi = \angle CMP$. Let $O^\prime$ and $p$ be the center and radius of the circle of intersection with plane and sphere; and let $P^\prime$ be the antipode of $P$ in that circle.
Looking at the $xy$-plane "edge-on", with $A$ and $B$ coalescing with $M$, we have this:

Calculating the power of point $M$ with respect to circle in two ways gives
$$\begin{align}|MP||MP'| &= |MC||MC'|  \\
(p+r\cos\theta\cos\phi)(p-r\cos\theta\cos\phi) &= (r+r\cos\theta)(r-r\cos\theta) \\
p^2 - r^2 \cos^2\theta\cos^2\phi &= r^2 \sin^2\theta \\
p^2 &= r^2 \left(\sin^2\theta+\cos^2\theta\cos^2\phi\right) \tag{2} 
\end{align}$$ 
So, we have the $\text{radius}$ part of $(1)$. For the $\text{angle}$ part, let's change our view to $\bigcirc ABP$ ...

... where we see that
$$|\stackrel{\frown}{APB}| = |O'P|\;\angle AO'B =2 p \arcsin\frac{|MA|}{|O'A|}=2p\arcsin\frac{r\sin\theta}{p}  \tag{4}$$
That is, incorporating $(3)$,
$$|\stackrel{\frown}{APB}| = 2r\;\sqrt{\sin^2\theta+\cos^2\theta\cos^2\phi}\;\arcsin\frac{\sin\theta}{\sqrt{\sin^2\theta+\cos^2\theta\cos^2\phi}}\tag{5}$$
Writing in terms of arccosine will help accommodate major arcs:
$$|\stackrel{\frown}{APB}| = 2r\;\sqrt{\sin^2\theta+\cos^2\theta\cos^2\phi}\;\arccos\frac{\cos\theta\cos\phi}{\sqrt{\sin^2\theta+\cos^2\theta\cos^2\phi}}\tag{5'}$$
We're getting closer, but we haven't yet considered the "lune angle", by which we mean the angle made by $\stackrel{\frown}{ACB}$ and $\stackrel{\frown}{APB}$ on the surface of the sphere at $A$ (or $B$). Equivalently, this is the angle between tangent vectors to those arcs at $A$. Each such vector must be perpendicular to the normal of the arc's plane (because the vector lies in that plane), and to the vector to $A$ from the arc's center (because tangents are perpendicular to radii); thus, we can calculate those vectors as
$$\begin{align}
v &:= \text{(normal to $\triangle OAB$)}\times \overrightarrow{OA} = (0,0,1)\times A \\
& = (-r\sin\theta,r\cos\theta,0) \\
w &:= \text{(normal to $\triangle O'AB$)}\times \overrightarrow{O'A} = (-\sin\phi,0,\cos\phi)\times (A-O') \\
&= (-r \cos\phi \sin\theta, r \cos\phi \cos\theta, -r \sin\phi \sin\theta)
\end{align}$$
Then, writing $\Phi$ for the lune angle,
$$\cos\Phi = \frac{v\cdot w}{|v||w|}=\frac{r^2\cos\phi}{r^2\sqrt{\sin^2\theta+\cos^2\theta\cos^2\phi}}=\frac{\cos\phi}{\sqrt{\sin^2\theta+\cos^2\theta\cos^2\phi}} \tag{6}$$
We can re-write this as
$$\cos^2\phi = \frac{\sin^2\theta\cos^2\Phi}{1-\cos^2\theta\cos^2\Phi} \tag{6'}$$
which allows us to trade the parameter $\phi$ for $\Phi$ in $(5')$

$$|\stackrel{\frown}{APB}| = \frac{2r\sin\theta}{\sqrt{1-\cos^2\theta\cos^2\Phi}}\;\arccos\left(\cos\theta\cos\Phi\right) \tag{$\star$}$$

Herein, we encounter the problem described in the observation related to $(1)$: because $\Phi$ exists inside and outside of the arccosing in $(\star)$, the equation is transcendental, so we can't solve for the lune angle directly (without appealing to something like Lambert's $W$ function).
Now, $(\star)$ doesn't completely answer the question, as it only covers how the (small) arc $\stackrel{\frown}{APB}$ relates to the (great) arc $\stackrel{\frown}{ACB}$. If our lune is bounded by arbitrary arcs $\stackrel{\frown}{APB}$ (which makes lune angle $\Phi$ with the $xy$-plane) and, say, $\stackrel{\frown}{AQB}$ (which makes lune angle $\Psi$ with the $xy$-plane, which is related to the arc length by a counterpart of $(\star)$), then the total lune angle is $\Lambda := \Phi-\Psi$ (or $\Phi+\Psi$, depending upon how you want to assign a sign to $\Psi$). Due to the transcendental nature of $(\star)$, we cannot get a direct relation between $\Lambda$ and the lengths of the bounding arcs. Numerical methods will have to bridge the gap. (Triangles formed of small arcs doesn't get any easier.) $\square$

Addendum: Sanity checks for $(\star)$.


*

*If $\Phi=0$, so that $\stackrel{\frown}{APB}$ coincides with $\stackrel{\frown}{ACB}$, then $(\star)$ gives 
$$|\stackrel{\frown}{APB}| = \frac{2r\sin\theta}{\sqrt{1-\cos^2\theta}}\;\arccos\left(\cos\theta\right) = 2r\theta = |\stackrel{\frown}{ACB}| \quad\checkmark$$

*If $\Phi=\pi/2$, so that $\stackrel{\frown}{APB}$ is a semicircle of radius $\frac12|AB| = r\sin\theta$, perpendicular to the plane of $\triangle ABC$, then $(\star)$ gives
$$|\stackrel{\frown}{APB}| = \frac{2r\sin\theta}{\sqrt{1-0}}\;\arccos 0 = \pi r\sin\theta \quad\checkmark$$ 

*If $\theta = \pi/2$, so that $\overline{AB}$ is a diameter and any arc from $A$ to $B$ must be a great semicircle of radius $r$, then $(\star)$ gives
$$|\stackrel{\frown}{APB}| = \frac{2r}{\sqrt{1-0}}\;\arccos 0 = \pi r \quad\checkmark$$
A: I'm trying to answer, I don't know if it's right.

$AcB$ is a big arc, $ApB$ is a small arc, capitalized $O$ is the center of sphere, lowercase $o$ is the center of small arc, $∠OMA = π/2$, $∠oMA = π/2$, and the radius of sphere is $R$.
$∠AOB = \frac{ AcB }{ R}$
$ AM = Rsin(\frac{ AcB}{2R})$      (1)
$ OM=Rcos(\frac{ AcB}{2R})$
Let $φ=∠pMc$
$ Oo=OMsinφ= Rcos(\frac{ AcB}{2R}) sinφ$
Let the radius of a small circle be r.
$ R^2= r^2+( Rcos(\frac{ AcB}{2R}) sinφ)^2$
$ r=\sqrt {R^2-(Rcos(\frac{ AcB}{2R}) sinφ)^2}  $
$∠AoB= \frac{ ApB}{r} =\frac{ ApB/}{\sqrt {R^2-(Rcos(\frac{ AcB}{2R}) sinφ)^2}}   $
$ AM= \sqrt {R^2-({Rcos(\frac{ AcB}{2R}) sinφ}) ^2} sin(\frac{ ApB}{2 \sqrt{R^2-({Rcos(\frac{ AcB}{2R})sinφ}) ^2 } }) $        (2)
Formula (1) equals Formula (2)
$ Rsin(\frac{AcB}{2R})= \sqrt {R^2-({Rcos(\frac{ AcB}{2R}) sinφ}) ^2} sin(\frac{ ApB}{2 \sqrt{R^2-({Rcos(\frac{ AcB}{2R})sinφ}) ^2 } })  $
or
$Rsin(\frac{ AcB}{2R})= rsin(\frac{ ApB}{2r})$
