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This Google Books search finds a page that says this:

74. Legendre's theorem

${}\qquad\qquad{}$ If each of the angles of a spherical triangle whose
sides are small when compared with the radius of the sphere
be diminished by one third of the spherical excess, the triangle
may be solved as a plane triangle whose sides are equal to
the sides of the spherical triangle, and whose angles are these
reduced angles.

"Spherical excess" I take to mean the amount by which the sum of the three angles exceeds a half-circle.

Similarly imprecise statements are found by additional googling. The sides are "small when compared to", etc., and the solution is approximate. So I'm thinking there should be something maybe with a Landau little-o, saying as the size goes to $0$, the error goes to $0$ like some power of the size. Maybe.

Does anyone know of a mathematically precise statement of this proposition?

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  • $\begingroup$ Your help is appreciated. But I've altered it further since then. $\endgroup$ – Michael Hardy Sep 6 '13 at 17:41
  • $\begingroup$ That last change looks much better! Software was outwitted in the end... $\endgroup$ – rschwieb Sep 6 '13 at 17:41
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There is a paper about Legendre's theorem by Zbyněk Nádeník, Legendre theorem on spherical triangles. The paper states and proves several versions of the theorem (with different error terms).

As I understand, here is what Legendre originally proved (see also Wikipedia for historical references):

Theorem [Legendre, stated in 1787, proved in 1798]. Consider a spherical triangle with sides $a$, $b$, and $c$ and angles $\alpha$, $\beta$, and $\gamma$. Then the plane triangle with sides $a$, $b$, and $c$ has angles $\alpha'$, $\beta'$ and $\gamma'$:

\begin{align*} \alpha' &= \alpha - \varepsilon/3 + O(\varepsilon \cdot M^2),\\ \beta' &= \beta - \varepsilon/3 + O(\varepsilon \cdot M^2),\\ \gamma' &= \gamma - \varepsilon/3 + O(\varepsilon \cdot M^2), \end{align*} where $\varepsilon = \alpha + \beta + \gamma - \pi$, $M = \max(a,b,c)/R$, and $R$ is the radius of the sphere. Note that $\varepsilon = O(M^2)$.

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