The numbers in today’s date—11, 28, and 23—make up the sides of a triangle. This doesn’t always happen; the two smaller numbers have to add up to more than the larger number.
We’ll look at triangles with sides 11, 23, and 28 in the plane, on a sphere, and on a hypersphere. Most of the post will be devoted to the middle case, a large triangle on the surface of the earth.
Solving a triangle in the plane
If we draw a triangle with sides 11, 23, and 28, we can find out the angles of the triangle using the law of cosines:
c² = a² + b² – 2ab cos C
where C is the angle opposite the side c. We can find each of the angles of the triangle by rotating which side we call c.
If we let c = 11, then C = arccos((23² + 28² − 11²)/(2 × 23 × 28)) = 22.26°.
If we let c = 23, then C = arccos((11² + 28² − 23²)/(2 × 11 × 28)) = 52.38°.
If we let c = 28, then C = arccos((11² + 23² − 28²)/(2 × 11 × 23)) = 105.36°.
Solving a triangle on a sphere
Now suppose we make our 11-23-28 triangle very large, drawing our triangle on the face of the earth. We pick our unit of measurement to be 100 miles, and we get a triangle very roughly the size and shape of Argentina.
We can still use the law of cosines, but it takes a different form, and the meaning of the terms changes. The law of cosines on a sphere is
cos(c) = cos(a) cos(b) + sin(a) sin(b) cos(C).
As before, a, b, and c are sides of the triangle, and the sides b and c intersect at an angle C. However, now the sides themselves are angles because they are arcs on a sphere. Now a, b, and c are measured in degrees or radians, not in miles.
If the length of an arc is x, the angular measure of the arc is 2πx/R where R is the radius of the sphere. The mean radius of the earth is 3959 miles, and we’ll assume the earth is a sphere with that radius.
We can solve for the angle opposite the longest side by using
C = arccos( (cos(c) – cos(a) cos(b)) / sin(a) sin(b) )
where
a = 2π × 1100 / 3959
b = 2π × 2300 / 3959
c = 2π × 2800 / 3959
It turns out that C = 149.8160°, and the other angles are 14.3977° and 29.4896°.
Importantly, the sum of these three angles is more than 180°. In fact it’s 193.7033°.
The sum of the vertex angles in a spherical triangle is always more than 180°, and the bigger the triangle, the more the sum exceeds 180°. The amount by which the sum exceeds 180° is called the spherical excess E and it is proportional to the area. In radians,
E = area / R².
In our example the excess is 13.7033° and so the area of our triangle is
13.7033° × (π radians / 180°) × 3959² miles² = 3,749,000 miles².
Now Argentina has an area of about a million square miles, so our triangle is bigger than Argentina, but smaller than South America (6.8 million square miles). Argentina is about 2300 miles from north to south, so one of the sides of our triangle matches Argentina well.
Note that there are no similar triangles on a sphere: if you change the lengths of the sides proportionately, you change the vertex angles.
Solving a triangle on a pseudosphere
In a hyperbolic space, such as the surface of a pseudosphere, a surface that looks sorta like the bells of two trombones joined together, the law of cosines becomes
cosh(c) = cosh(a) cosh(b) + κ sinh(a) sinh(b) cos(C)
where κ < 0 is the curvature of the space. Note that if we set κ = 1 and delete all the hs this would become the law of cosines on a sphere.
Just as the sum of the angles in a triangle add up to more than 180° on a sphere, and exactly 180° in a plane, they add up to less than 180° on a pseudosphere. I suppose you could call the difference between 180° and the sum of the vertex angles the spherical deficiency by analogy with spherical excess, but I don’t recall hearing that term used.
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