Example 3. Given two places $A$ and $B$ with a longitude difference of $\alpha^{\circ}$, both located on the same latitude line at $\beta$.

Find the shortest distance between $A$ and $B$ on the sphere.
Let $\angle A O B=\theta$, take the midpoint $M$ of $A B$, connect $O M$, $O_{1} M$, to form a tetrahedron $O-M B O_{1}$ enclosed by four right triangles, in which the following three conditions are known.
$$
\begin{array}{l}
\angle M O_{1} B=\frac{\alpha}{2}, \angle O B O_{1}=\beta \quad O B=R \\
=6370 \text { km. }
\end{array}
$$
$\therefore$ We can definitely solve for the unknown $\angle M O B$ in the tetrahedron $O-M B O_{1}$, i.e., find $\frac{\theta}{2}$. $B O_{1}=R \cos \beta, M B$
$$
\begin{array}{l}
=R \cos \beta \sin \frac{a}{2} \cdot \sin \frac{\theta}{2}=\frac{R \cos \beta \sin \frac{a}{2}}{B} \\
=\cos \beta \sin \frac{a}{2} . \\
\therefore \frac{\theta}{2}=\arcsin \left(\cos \beta \sin \frac{a}{2}\right), \\
\theta=2 \arcsin \left(\cos \beta \sin \frac{\alpha}{2}\right) .
\end{array}
$$

Let the shortest distance between $A$ and $B$ on the Earth be $\boldsymbol{S}$. Then
$$
\begin{aligned}
\boldsymbol{S} & =2 R \arcsin \left(\cos \beta \sin \frac{a}{2}\right) \\
& =2 \times 6370 \arcsin \left(\cos \beta \sin \frac{\alpha}{2}\right) \\
& =12740 \arcsin \left(\cos \beta \sin \frac{a}{2}\right)(km)
\end{aligned}
$$
$\therefore$ The shortest distance between $A$ and $B$ on the Earth is $12740 \arcsin \left(\cos \beta \sin \frac{\alpha}{2}\right)$ km.