(These notes are based on the first chapter of *Geometry from a Differential Viewpoint* by John McCleary, (Cambridge University Press, 1994).)

Whereas basic plane geometry is concerned with points and lines and their
*interactions*, most of the early
geometry of the Babylonians, Arabs, and Greeks was spherical geometry--the study of the
Earth, idealized as a sphere.
This early *science* was astronomy and the need to measure time accurately by the
sun.

We have to come to some agreement on what *lines* and *line segments* on a
sphere are going to be.

A **great circle**
on a sphere is the intersection of that sphere with a plane passing through the center
of the sphere.

Examples of great circles are the equator and the lines of constant longitude (such as the Greenwich Mean Time Line ). These are good choices to play the role of lines on our sphere. For example, given any two non-antipodal points there is a unique great circle joining those two points. This is easy to see when you remember that three non-collinear points determine a plane. Take these two points and the center of the sphere as the three non-collinear points. The intersection of the plane determined by these points and the sphere is the great circle joining the two given points.

If great circles are to be lines, then we can measure the angle between two intersecting great
circles as the angle formed by the intersection
of the two defining planes with the plane tangent to the sphere at the point of
intersection. See the figure below (Figure 1).
With this definition of angle, we can form triangles on the sphere whose interior angle sum is
greater than two right angles. In fact, we will show
that the interior angle sum of * all* triangles on the sphere is greater than
two right angles. We will be working with radian measure
for most of the semester (to make many of the calculations neater). Thus, we are claiming
that on a sphere the interior angle sums of any triangle
is greater than .

In order to discuss how to measure the angle sums, we will first discuss the concept of the area of a triangle on the sphere. We will discuss this more fully later, but for the present time we will accept the following four properties of area for polygonal regions on the sphere:

- The sphere of radius
*R*has area ; - the area of a union of
**non-overlapping**regions is the sum of their areas; - the areas of congruent regions are equal;
- the ratio of the area enclosed by two great circles to the area of the whole sphere is the same as the ratio of the angle between them to .

**Theorem 1.1** * On a sphere of radius R, a triangle with
interior angles , , and
has area given by:*

First, some terminology: a **lune** is one of the regions between two great circles and
the antipodal points where they cross. Antipodal points or figures are those that
correspond to opposite ends of a diameter of the sphere.
A lune is given by the angle at the center of the sphere between the two defining
planes. Note that this is also the angle
between the two great circles, **cf.** Figure 2.

**Proof**(due to Euler, 1781). From our assumptions above, Assumption (4) tells us

To make our proof easier, we may assume that the triangle lies in one hemisphere. If not you
can use Assumption (2) to break the triangle up into smaller triangles that do lie in one
hemisphere. You can then use the following argument on each piece and then add them
together.

A triangle is the intersection of three lunes: the lune given by , the lune given by and the lune given by . Each of these three lunes has an antipodal lune and the intersection of these three antipodal lunes is an antipodal triangle .

We will denote the lunes by the angle that includes them, *BCA*, *BAC*, and *ABC*.
Starting with the triangle and the lune *ABC*, add to them the lune
*ACB*. This covers the triangle three times. Now, add in the antipodal
lune to lune *CAB* and the antipodal triangle . This covers the
hemisphere determined by the line , counting the triangle three times. So we
get:

Thus, we get our desired formula:

We call the value of the **angle excess** of the spherical
triangle.

Note that since every triangle on the sphere has area, every triangle on the sphere has interior angle sum greater than . If the radius of the sphere is very large and the triangle very small in area, then the angle excess will be almost 0.

Let denote the sphere of radius *R* centered at the origin in .
In rectangular coordinates

and in spherical coordinates

The **length** of a path between two points along a great circle is found by measuring
the angle, in radians, made by the radii at the two endpoints and multiply it by the radius of
the sphere, *i.e.* . Note that the length of a segment is unchanged when we
rotate the sphere around some axis or reflect it through a great circle.

**Theorem 1.2:**[Spherical Pythagorean Theorem]
*For a right triangle
on a sphere of radius R with a right angle at vertex C and sides of length a, b, and
c, then
*

**Proof**:
Rotate the sphere so that the point *A* has coordinates (*R*,0,0) and the point *C* lies in
the *xy*-plane. This will make and the angles determining the point
*B*. Here is the central angle determined by side *AC*, is the angle
determined by *AC* and is the angle determined by *AB*. This gives us the
following spherical coordinates for the vertices of the triangle:

Using what we know about the dot product in , we can find the cosine of the
angle .

Now, in radian measure

and we are done.

Why is this called the spherical Pythagorean theorem? First, note that it does give a
relationship of the hypotenuse of a right triangle with the two sides of this triangle. To see
how it is related to the classical Pythagorean theorem, recall the Maclaurin series (Taylor
series at *a*=0) for the cosine function:

Using this in the formula from the spherical Pythagorean theorem gives us:

Now, as *R* gets arbitrarily large while *a*, *b*, and *c* stay small, we get the classical
Pythagorean theorem as a limiting case.

**Theorem 1.3:**[Spherical Sine Theorem]
*Let be a spherical triangle on a
sphere of radius R. Let a, b, and c denote the lengths of the sides in radians, and
let , and denote the interior angles at each vertex. Then
*

**Proof: **
We will first prove this for a right triangle and then use that result to prove the general case.
In addition we can restrict our attention to triangles that lie entirely within a quarter of a
hemisphere.

Let be a right triangle with right angle at *C*. Extend the radius *OA* to
*OA*', where *A*' is the on the intersection of the tangent plane to the sphere at *B* with
the line *OA*. Likewise, extend *OC* to *OC*'. This means that the triangles
and are right triangles with right angle at *B*.
Additionally .

Now, we have other right angles as well. First, planes *A*'*BC*' and *OBC* are perpendicular,
from above. Now, plane *OCB* is perpendiculare to plane *OCA* since is a
right angle. Look at the two planes *A*'*BC*' and *OAC*. They intersect in the line
. Since lies in the plane *OCA*, and lies in the
plane *OCB*, the two lines must be perpendicular. Thus, is a right triangle
with right angle at *C*'. The classical Pythagorean theorem gives:

Combining these equations gives us

and so is a triangle with right angle at *C*'. Let , and
denote the angles at the center of the sphere determined by the arcs *BC*,
*AC*, and *AB*, respectively. The definition of the sine function gives us:

and so

We can reverse the roles of *A* and *B* and this gives

and so

Altitudes

Now, if we have an arbitrary triangle, we can construct the spherical analogue of the altitude
to reduce the relation for two sides to the case of a right triangle, which we just finished.
For example in the following figure, inserting the altitudes from *A* and from *B*, we get

From the other altitude we have

and the theorem follows.

These theorems show that there are exactly six related pieces of information determining a spherical triangle -- the three sides and the three angles. Given any three of these pieces you can determine the other three pieces. This means that similar triangles must be congruent! This is not immediate. We need one other piece which we are able to prove, but will not do so now.

**Theorem 1.4:*** Suppose that is a right triangle on the sphere of radius
R with a right angle at C. Then
*

** **Homework

- Prove that two great circles intersect.
- Prove that two great circles bisect each other.
- The sphere of radius one can be considered as

In these two coordinate systems determine the distance along great circles between two arbitrary points on the sphere as a function of the coordinates. - A
*pole*of a great circle is one of the endpoints of a diameter of the sphere perpendicular to the plane of the great circle. For example, the North and South poles are both poles for the equator. Prove that through a given point, not on a given great circle and not the pole of the great circle, there is a unique great circle ghrough the given point perpendicular to the given great circle.