We know that the area of an spherical triangle (in a unit sphere) is given by $A(\triangle) = \alpha + \beta + \gamma - \pi$, where $\alpha$, $\beta$, and $\gamma$ are the interior angles of the spherical triangle.
I would like to see how plane (Euclidean) geometry works as a limit when the radius of the sphere goes to infinity. Clearly the curvature of the sphere $1/r$ becomes zero and a sphere turns into a plane. What happens to the area of the triangle?
If we say that the area of the triangle is \begin{equation} A(\triangle) = r^2 [(\alpha + \beta + \gamma) - \pi] \end{equation} clearly $\alpha + \beta + \gamma - \pi$ go to zero, but not at the rate that $r^2$ goes to infinity. It seems that this limit is infinity.
There seems to me that we can not find something like $b h/2$ (base times height over two) from spherical geometry. Right?
Of course objects become amplified in area by $r^2$ or length by $r$ so we would need to have something to pull them back.
Thanks.
Update: One way to pull back is to think that the actual arc lengths of the stretched triangle segments are $a=r \alpha$, $b=r \beta$, and $c= r \gamma$, so we can pull one $r$ inside the formula above and have
\begin{equation} A(\triangle) = r [(a+b+c) - \pi r] \end{equation}
where now $a,b$, and $c$ are the actual lengths of the sides. Pulling $r$ inside again shows me the area of a circle and... it seems that we better point toward
and forget about base x height/2. Heron's formula is fine to me.