Geometry of spherical triangle Using the formula for the area of a spherical triangle, find and prove a formula relating the angle sum of a spherical polygon to its area

Thought:
Area (spherical triangle) $=\mathbb{R}^2(\alpha+\beta+\gamma-\pi)$ where $\alpha, \beta, \gamma$ are the interior angles
Let $n$ be the number of vertices/sides of a polygon.
Consider a quadrilateral $n=4$ with interior angle $(\alpha,\beta,\gamma,\delta)$
Area (spherical quadrilateral) $=\mathbb{R}^2(\alpha+\beta+\gamma+\delta-2\pi)$ 
then how do I induce to infinite $n$???
 A: The natural thing seems to be the area is the sum of the angles minus $\pi$ times two less than the number of sides.  You should be able to prove this by cutting the polygon into triangles.This means that in Euclidean space all polygons have zero area.
A: A polygon with an infinite number of vertices/sides is a circle. So, what is the area of a circle in spherical geometry? It is the surface area covered by the region on $\mathbb{S}^2$ bounded by the curve resulting from intersecting $\mathbb{S}^2$ with a plane.
A: It is obvious that typically if the no. of sides, of a regular polygon having a finite circumscribed radius say $r$, tends to infinity i.e. $n \to \infty \space$ then the regular polygon becomes a circle with a radius $r$. 
Now, let $r$ be the radius of circle on a spherical surface with a radius $R$ then the solid angle ($\omega$) subtended by the circle at the center of the sphere is given as
$$\omega=2\pi \left(1-\sqrt{1-\left(\frac{r}{R}\right)^2}\right)$$
Hence, the area covered by a circle (i.e. a polygon having infinite no. of sides) on the sphere is given as
$$Area=(solid \space angle) \cdot ({(radius)}^2)=\omega R^2=2\pi R^2\left(1-\sqrt{1-\left(\frac{r}{R}\right)^2}\right)$$  
