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We know area of a plane triangle $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$ where $s=\frac{a+b+c}{2}$.

I was just thinking: let we have a triangle with arc length $a,b,c$ on a sphere of radius $r$, do we have any similar kind of formula for that spherical triangle? when radius of $r\to \infty$ we get the plane, so do we have any estimate of area of spherical triangle when $r\to\infty$?

Any reference and article link are also welcome! Thank you.

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Do you mean this? – ABC Mar 21 '13 at 11:58
that is plane triangle right? spherical triangle is a triangle on the surface of a sphere. – La Belle Noiseuse Mar 21 '13 at 12:02
Now got the question. – ABC Mar 21 '13 at 12:04
Also posted at… – Joel Reyes Noche Mar 24 '13 at 13:09
@JoelReyesNoche The link seems to be broken. Gives a 404... – YatharthROCK Mar 26 '13 at 10:22
up vote 1 down vote accepted

Source: This Dr. Math Article

Novice here, so please excuse any mistakes. The ratio should be:

$$ \frac{ 180 \cdot \sqrt{ s(s-a)(s-b)(s-c) } } { 4 \cdot \pi \cdot R^2 \cdot \arctan \left( \sqrt{ \tan \left( \frac{s}{2} \right) \cdot \tan \left( \frac{s-a}{2} \right) \cdot \tan \left( \frac{s-b}{2} \right) \cdot \tan \left( \frac{s-c}{2} \right) } \right) } $$

Still needs some simplification, though...

Basically, I just placed Heron's formula for the area of planar $\Delta$s above the $\frac{ \pi \cdot R^2 \cdot E}{180}$ formula for the area of a spherical $\Delta$. Since I didn't have the angular measures required to calculate $E$, I used this formula:

$$ \tan \left( \frac{E}{4} \right) = \sqrt{ \tan \left( \frac{s}{2} \right) \cdot \tan \left( \frac{s-a}{2} \right) \cdot \tan \left( \frac{s-b}{2} \right) \cdot \tan \left( \frac{s-c}{2} \right) } $$

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