In navigation, for most purposes the surface of the Earth can be considered a sphere. The distance between points

For example, the distance from the North Pole to a point on the Equator is 1/4 of a great circle: 90

The upshot is that calculating the distance between two points amounts to calculating the central angle they determine on the great circle that passes through both of them. This is where spherical trigonometry becomes useful.

2. "In book XI of [Ptolemy's] *Almagest* the principles of spherical
trigonometry are stated in the form of a few simple and useful lemmas;
plane trigonometry does not receive systematic treatment..." [1]. This
is because of the great importance of spherical trigonometry for
astronomy. We have seen how Aryabhata (456-550) tabulated sines. The
next great progress came with Al-Battani (or, Albategnius; c.868-929),
the greatest Islamic mathematician and astronomer
of his time. He used the function *sinvers* defined by
sinvers(α) = 1 - sin(π/2 - α) essentially equivalent
to our cosine, in astronomical calculations; but despite insinuations to
the contrary (e.g. in [1]), he did not formulate the law of cosines
for a spherical triangle. The first formulation we know,
still in terms of sinvers, is due to
Regiomontanus (1464). Details from [3] are summarized in
The True History of the Law
of Cosines.

The the **spherical law of cosines**
states, for a spherical triangle with surface angles
*A*, *B*, *C* opposite sides corresponding to central angles
*a*, *b*, *c*, that

3. This law has the planar law of cosines as limit when
*a*, *b*, and *c* tend to zero. Use
the approximations cos*a* = 1 - (1/2)*a*^{2}, etc.,
and sin*b* = *b*, etc. valid for small *a, b, c*, and
discard the higher-order term (1/4)*b*^{2}*c*^{2}.

4. When we locate a point *X*
on the Earth by giving its latitude and longitude,
we are giving *angular* coordinates.
Latitude is the size of
the central angle, along a meridian (the great circle through *X*
and the North Pole *P*), between *X* and the equator. The latitude
is specified as North (N) for points above the equator, and South
(S) for points below, and runs from 0^{o} to 90^{o}.
New York City is at latitude 40^{o}47'N, Moscow is at
latitude 55^{o}45'N and
Buenos Aires is at latitude 34^{o}35'S. (Sometimes
latitude is counted negative for points in the Southern Hemisphere).
Longitude
is the size of the surface angle, at the North Pole, between the
"Prime Meridian" (the meridian through Greenwich, England) and the
meridian through *X*. It is measured East (E) or West (W) and
runs in each direction from 0^{o} to 180^{o}.
New York City is at longitude 73^{o}58'W, while Moscow
is at 37^{o}42'E and Beijing
is at 113^{o}20'E.

5. Suppose we want to calculate the air distance from New York
to Moscow. We know their coordinates.

New York: Latitude *l*_{1} = 40^{o}47'N,
Longitude *L*_{1} = 73^{o}58'W

Moscow: Latitude *l*_{2} = 55^{o}45'N,
Longitude *L*_{2} = 37^{o}42'E. We can mark
these coordinates on a globe:

New York (*N*), Moscow (*M*) and the North Pole (*P*)
are the vertices of a spherical triangle; we label the sides (central
angles) oppsite the vertices by *n, m, p *. We know side *m*:
it is
*m* = 90^{o} - *l*_{1} = 49^{o}13'.

Similarly side *n* is
*n* = 90^{o} - *l*_{2} = 34^{o}15'.

We also know the face angle at *P*: it is the *sum* of
the two longitudes, since one is East and the other is West:
*P* = *L*_{1} + *L*_{2} = 111^{o}40'.

In this context, the law of cosines gives:
cos*p* = cos*m* cos*n* + sin*m* sin*n* cos *P*.

Using a calculator, we find
cos*p* = 0.6532x0.8266 + 0.7572x0.5629x(-0.3692) = 0.3826, and
arccos(0.3826) = 67.51^{o}=4050.4'. This gives the great-circle
distance from
New York to Moscow as 4050 nautical miles, or 7501km.

Exercises:

- Calculate the great-circle distance from New York to London
(Latitude = 51
^{o}32'N, Longitude = 0^{o}5'W). Note that in this case the longitudes are both West and must be subtracted. - Calculate the great-circle distance from Anchorage, Alaska
(Latitude = 61
^{o}10'N, Longitude = 150^{o}1'W) to Moscow. - What is the maximum latitude along the New York-Moscow great-circle
route? (Hint: it occurs at the
*vertex V*of the route, the point where the route makes a right angle with a meridian). You will need to solve for surface angle*M*by a second application of the law of cosines; then use the**spherical law of sines**sin*A*/sin*a*= sin*B*/sin*b*= sin*C*/sin*c*to solve the rectangular spherical triangle*MPV*. - [2] Find the initial course and the distance for a voyage along a great
circle from Los Angeles (
*l*= 34^{o}03'N,*L*= 118^{o}15'W) to Aukland (*l*= 41^{o}18'S,*L*= 174^{o}51'E). [The initial course is "the angle of departure, measured from the north around through the east [i.e.*measured clockwise*] from 0^{o}to 360^{o}"[2].] Answer: Course = 224^{o}8' 48", distance = 5832 miles. - [2] Find the distance by great circle from New York
(
*l*_{1}= 40^{o}40'N,*L*_{1}= 4^{h}55^{m}54"W) to Cape of Good Hope (*l*_{2}= 33^{o}56'S,*L*_{2}= 1^{h}13^{m}55"E). [Longitude is given here in hours, minutes and seconds*of time*. Because of the rotation of the Earth with respect to the Sun, 1^{o}of longitude corresponds to a 4^{m}time difference.] Answer: 6779.9 miles.

[2] Kells, Kearn and Bland,

[3] Anton von Braunmühl,

*Corrected Dec 6, 2006.*