Area of a plane triangle as limit of a spherical triangle 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
Heron's formula
and forget about base x height/2. Heron's formula is fine to me.
 A: I found a connection here. We need to use 
Cagnoli's Theorem
That is, given excess $E=\alpha+\beta+\gamma-\pi= A(\triangle)$ 
Cagnoli's Theorem establishes that:
\begin{equation}
\sin \frac{E}{2} = \frac{\sqrt{ \sin s \sin (s-a) \sin (s-b) \sin(s-c)}}{
{2 \cos \frac{a}{2} \cos \frac{b}{2} \cos \frac{c}{2}}}
\end{equation}
Then as we write the trigonometrical functions as Taylor series:
\begin{eqnarray*}
   \sin x &=& x - \frac{x^3}{3} + H.O.T. \\
    \cos x &=& 1 - \frac{x^2}{2} + H. O. T.
\end{eqnarray*}
When $r \to \infty$ we get an asymptotic solution by retaining only
the leading order terms here. That is
\begin{equation}
\lim_{r \to \infty} \sin \frac{E}{2} =  \lim_{r \to \infty} \frac{E}{2}.
\end{equation}
(note that the radius $r$ is implicit on these equations, in addition
when $r \to \infty$, $E \to 0$ since the excess will be nothing once you get from a sphere to a plane.).
Similarly for the right hand side term, as $r \to \infty$ we find
\begin{equation}
\lim_{r \to \infty}
\frac{\sqrt{ \sin s \sin (s-a) \sin (s-b) \sin(s-c)}}{
{2 \cos \frac{a}{2} \cos \frac{b}{2} \cos \frac{c}{2}}} =
 \lim_{r \to \infty} \frac{\sqrt{s(s-a)(s-b)(s-c)}}{2}
\end{equation}
Please observe that for very large $r$ the interior angles
$a$, $b$, and $c$ become really small and so their semi-perimeter $s$.
That is, if $r=1$, then $a$, $b$, and $c$ are simultaneously lengths of
the triangle sides (arc segments) and central angles in the sphere.
If we want to increase $r \gg 1$, then, in order to preserve the
size of the segments, we need to shrink the angles by a factor of
$1/r$, so that the central angles $a$, $b$, and $c$ shrink to zero,
but the length of the segments $a$, $b$, and $c$ remain constant. Hence
there is a duality on the meaning of the symbols $a$, $b$, and $c$. 
As arguments of the sine and cosine functions, they are angles in radians, but as lengths they are the fixed lengths for $r=1$ and they should be 
preserved as the sphere explodes. What I call "pull back" in my question is this shrinking of $a$, $b$ , and $c$ as central angles of the sphere to keep the arc lengths of $a$, $b$, and $c$ constant.
From the previous two equations we find that
in the limit as $r \to \infty$:
\begin{equation}
E = \sqrt{s (s-a) (s-b) (s-c)}.
\end{equation}
which is Heron's formula. The conversion from here to base x altitude/2 should be a common problem solved elsewhere.
