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I'm studying spherical geometry and was having some trouble solving an exercise problem. More specifically, this is in the section for the law of cosines. The exercise problem is a navigation/mapping application.

The Bermuda Triangle is a region in the Atlantic Ocean where, it is said, a large number of ships and planes have disappeared for mysterious reasons. The boundaries of the Bermuda Triangle are ill-defined but for the sake of argument we shall take them to be a spherical triangle with vertices at Miami, Florida (lat. $25°46'$N, long. $80°12'$W); San Juan, Puerto Rico (lat. $18°29'$N, long. $66°8'$W); and Hamilton, Bermuda (lat. $32° 18'$N, long. $64°47'$W). If a ship sinks at an unknown location in the Bermuda triangle, how many square miles must be searched to find the survivors?

My first instinct was to convert the latitude and longitude to degrees, and also to use the formula:

$$A + B + C = \pi + \dfrac{\text{area}(\triangle ABC)}{R^2}$$

where $A$, $B$, and $C$ each refer to the vertex angle of the triangle, and $R$ is the radius of the sphere it's in.

The converted degrees are:

$$ \begin{array}{|c|c|c|} \hline & \text{Latitude} & \text{Longitude} \\ \hline \text{Miami} (A) & 25.76° & 80.20°\\ \hline \text{Hamilton} (B) & 32.30° & 64.78° \\ \hline \text{San Juan} (C) & 18.48° & 66.13° \\ \hline \end{array} $$

My initial method was to set a point $N$ as the north pole of the sphere, and put $B$ and $C$ on the great circle passing by the north pole. If we say that $a$, $b$, and $c$ are each the arc length opposite of their capital-case vertices, I got the following results:

$$ \begin{align} a & = 32.3° - 18.48° = 13.82° \\ b & = 80.2° - 64.78° = 15.42° \\ C & = 80.2° - 66.13° = 14.07° \end{align} $$

since $a$ is the difference in latitude between $B$ and $C$, $b$ is the difference in longitude between $A$ and $C$, and $C$ is the difference in longitude between $A$ and $B$. I believe this is where the problems occur because I followed the steps of the previous exercise, but didn't fully comprehend how the differences in latitude and longitude were to be expressed as angles and arc lengths.

If we use the law of cosines for $c$, we eventually end up with $c \approx 106.2°$, but this can't be right because this means that arc $AB$ (i.e. the distance between Miami and Hamilton) would be

$$\dfrac{106.2°}{360°} \times 40,075\text{km} = 11,820.125\text{km}$$

Where exactly did I go wrong? I can't seem to wrap my head around what the proper steps would be. Thanks!

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  • $\begingroup$ I'm kind of surprised that there isn't a formula online for the area of a spherical triangle in terms of the spherical coordinates of its vertices, but if I were doing this from scratch, I would use the vertices to define planes through the center of the sphere. The vertex angle $ABC$ is the angle between the normal vectors to the planes containing $O$, $A$ and $B$, and $O$, $A$, and $C$ respectively. $\endgroup$ – Matthew Leingang Apr 10 at 17:38
  • $\begingroup$ Starting with "I got the following results", nothing you wrote makes any sense to me at all. What were "the steps of the previous exercise"? My guess is you tried to plug in the parameters of this exercise into the other's solution in a places where they don't belong. $\endgroup$ – David K Apr 10 at 20:10
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The area of the spherical triangle is given by $$ \Delta=ER^2, $$ where $R$ is the sphere radius and $E$ is so called spherical excess given by the l'Huilier's theorem as: $$E=4\arctan\left(\sqrt{\tan\left(\frac{s}{2}\right)\tan\left(\frac{s-a}{2}\right)\tan\left(\frac{s-b}{2}\right)\tan\left(\frac{s-c}{2}\right)}\right)\tag1 $$ where $a,b,c$ are the angular arcs of $BC,CA,AB$, respectively, and $s=\frac{a+b+c}2$.

It remains only to compute the arcs which is straightforward: $$ c=\arccos(\sin\theta_A\sin\theta_B+\cos\theta_A\cos\theta_B\cos(\phi_A-\phi_B)),\dots\tag2 $$ where $\theta$ and $\phi$ are the latitude and longitude, respectively.

Possibly one can simplify the expression (1) after substitution of (2) but I have not yet tried. If I get some nice result I will update.

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  • $\begingroup$ For your equation $(2)$ it would be better to use the Haversine formula instead. Read the WIkipedia article for details. $\endgroup$ – Somos Apr 10 at 19:58
  • $\begingroup$ @Somos Thank you very much for the reference. I never heard the name before. $\endgroup$ – user Apr 10 at 20:06

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