Area of a plane triangle as limit of a spherical triangle

Question

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.

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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.

Answer 1

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.

Answer 2

We apply the Spherical Law of Cosines for Angles and render the planar area from the limiting form of this law.

Begin by rendering the arc lengths on a sphere of radius $R$* as $a,b,c$ and the respective opposite angles as $\alpha,\beta,\gamma$. The planar limit is to be obtained as $R\to\infty$ with $\alpha,\beta,c$ fixed.

(* -- We do not assume a unit radius, so "extra" factors of $R$ will appear in formulas below.)

From the spherical trigonometry law described above, we then have

$\cos\gamma=-\cos\alpha\cos\beta+\sin\alpha\sin\beta\cos(c/R)$

We render $\cos(c/R)$ as $1-[1-\cos(c/R)]$ and apply the formula for the cosine of a sum to obtain

$\cos\gamma+\cos(\alpha+\beta)=-\sin\alpha\sin\beta[1-\cos(c/R)]$

Next comes the sum-product relation for cosines:

$2\cos\left(\dfrac{\alpha+\beta+\gamma}2\right)\cos\left(\dfrac{\alpha+\beta-\gamma}2\right)=-\sin\alpha\sin\beta[1-\cos(c/R)]$

$2\sin\left(\dfrac{\Delta}{2R^2}\right)\cos\left(\dfrac{\alpha+\beta-\gamma}2\right)=\sin\alpha\sin\beta[1-\cos(c/R)]$ (***)

where $\Delta$ is the area of the triangle and $\Delta/R^2=\alpha+\beta+\gamma-\pi$ is the spherical excess in radians.

We now apply our approximations for the planar limit. First:

$\sin\left(\dfrac{\Delta}{2R^2}\right)\sim\dfrac{\Delta}{2R^2}$

$\cos\left(\dfrac{\alpha+\beta-\gamma}2\right)\sim\cos\left(\dfrac{\pi-2\gamma}2\right)=\sin\gamma$

$1-\cos(c/R)=\dfrac{1-\cos^2(c/R)}{1+\cos(c/R)}\sim\dfrac{\sin^2(c/R)}2$

So then (***) becomes

$\left(\dfrac{\Delta}{R^2}\right)\sim\left(\dfrac{\sin\alpha\sin\beta}{\sin\gamma}\right)\dfrac{\sin^2(c/R)}2$

We now plug in the Law of Sines to render

$\dfrac{\sin^2(c/R)}{\sin^2\gamma}=\left(\dfrac{\sin(a/R)}{\sin\alpha}\right)\left(\dfrac{\sin(b/R)}{\sin\beta}\right)$

and obtain

$\Delta\sim\frac12R^2\sin(a/R)\sin(b/R)\sin\gamma\sim\frac12ab\sin\gamma,$

which matches the planar formula.