= Solution
On the unit sphere, the <spherical excess formula> gives the area of a spherical triangle as
$$
\boxed{A=\alpha+\beta+\gamma-\pi}.
$$
For a sphere of radius $R$, the right side is multiplied by $R^2$.
Join an interior point of a convex spherical $n$-gon to its vertices by geodesics. The resulting $n$ spherical triangles contain all polygon angles and angles summing to $2\pi$ at the interior point. Adding their spherical excesses gives
$$
\boxed{A=\sum_{i=1}^n\alpha_i-(n-2)\pi}.
$$
For a regular polygon with interior angle $\alpha$, positivity of area requires $n\alpha>(n-2)\pi$, while strict convexity requires $\alpha<\pi$. Both limiting values can be approached by regular polygons, so
$$
\boxed{\frac{n-2}{n}\pi<\alpha<\pi}.
$$
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