R
/2. Notice that
the lunar angle for each of these lunes is /4 radians, or 45
degrees.
R. A great circle
divides the sphere into two congruent hemispheres. Each of these will
have area 2R. Another great circle, which meets the first at
right angles, divides the sphere into four congruent lunes, each with
area R. We can continue this process by dividing each
of these four lunes into two by bisecting the angle. We then get
eight congruent lunes, each with area R /2. Notice that
the lunar angle for each of these lunes is /4 radians, or 45
degrees.
/q, and the area of
each lune is 2R/q. Now if we look at
the union of p of these lunes, we get one lune with
lunar angle p/q, and the area will be 2pR/q.
Thus if 
= p/q is the lunar angle, then the area of the
lune is 2R. Thus
(lunar angle). We
have proved this for angles of the form
= p/q, but from this
it follows for general angles by approximating the angle with rational
multiples of 
.
It is interesting to consider the formula for the area of a lune
when the angle is being measured in degrees instead of radians. Since
one radian is equal to 180/
degrees, and one degree is equal to /180 radians, the formula becomes
) = 2R2 
/180 = R2/90
.
Comparison of the formulas provides yet another reason for using radians.
A spherical triangle is defined just like a planar triangle. It
consists of three points called vertices, the arcs of great circles
that join the vertices, called the sides, and the area that is
inclosed therein. However as soon as that is said you realize that
there is a lot of ambiguity. First of all, there are two segments of
the great circle that join each pair of points. Then, once we have
decided which of the two segments to use, the resulting figure is the
boundary of two different regions --- what might be called the outside
as well as the inside. Since only one "long" side can be used without
getting extra points of intersection of the sides, we can count 8
regions that might be called triangles, all of which have these three
points as vertices. Of course the situation is even more complicated
if two of the points are antipodal.
To get away from this complexity, we will deal only with small triangles. Given the three vertices, no pair of which are antipodal, the small triangle has as sides the short segments of great circles that join the vertices. We notice that there is a hemisphere that contains the three points, and the region we choose is the region which is bounded by the sides and lies entirely in that hemisphere. Actually the results we will discover are true for "large" triangles, but we will leave the verification of that to the reader.
Exercise: Experimenting with spherical triangles.
Draw three spherical triangles of different sizes on a beach ball (Boyd Hemphill has a way to increase the precision with which this can be done). Measure the angles of the triangles and compare the sum of the angles with the relative size of the triangle. If there are others working on this, compare your results with theirs.Do you notice any pattern in the data? In particular how does the sum compare with the 180 degrees we expect in the plane? What happens to the sum of the angles as the size of the triangle gets larger?
Exercise: The area of a spherical triangle.
Can you find a formula that relates the area of a spherical triangle to the sum of its angles?This is a difficult problem, but you know everything you need to know to solve it. Think about the three sides of the triangle and the great circles of which they are a part. For each angle of the triangle, these great circles bound two congruent lunes, one of which contains the triangle and one which does not. If you can visualize the configuration of these six lunes on the sphere, you will be on the way to a solution.
The solution to this problem is Girard's Theorem, and it is proved in the next section.
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The next section contains a proof of Girard's Theorem. |
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The previous section discusses incidence on the sphere. |
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Table of Contents. |