Analytic Spherical Geometry: Begin with a sphere of radius R, with center at the origin O. Measuring the length of a segment (arc) on a sphere. Let A and B be any two points on the sphere. We know that the measure of the arc is the measure of the central angle , which we will measure in radians. So,

the arc length is

AB AOB∠

22 2

m AOB m AOBC R R mππ π

∠ ∠⋅ = ⋅ ∠AOB⋅ = . That is

( )AB R AOB= ⋅ ∠ . (1.1)

O

A

B

Notice that this means when we discuss the length of an arc on the sphere, we will generally work with the angle from the center of the sphere. Measuring the area of polygons on the sphere.

I) The bi-angle (or lune) – a section of the sphere between two half great circles. The vertices of the lune are antipodal points.

Let the angle at each vertex be θ , and recall that the area of the surface of the sphere is . The portion of the sphere occupied by the lune, then, is

24 Rπ

24 22

Rθπ θ

π= 2R (1.2)

II) The triangle : Let be the radian angle measures. Extend the sides of the triangle

to make three lines (great circles). This creates three pairs of bi-angles. Notice the congruent triangle where are the antipodal points to A, B, C, respectively. Label the triangle on the figure below.

ABC∆

,A B

, ,A B C∠ ∠ ∠

′A B C′ ′ ′∆A B C′ ′ ′∆

,C′ ′

III) Exercise 1: Use the result in (I) to derive the formula for the area of a triangle. What is significant about the area of a spherical triangle versus the area of a Euclidean triangle.

Triangles on the sphere

I) Right triangles. Let be a right triangle with a right angle at ∠ and let be the lengths of the sides opposite , respectively. So,

ABC∆A B∠ ∠

C , ,a b c, , C∠

( ), ( ), (a BC b AC c AB= = = )

2

. (1.3)

Recall that a triangle in the sphere may have more than one right angle. Pythagorean Theorem: In Euclidean geometry we know that a b for a right triangle (Pythagorean Theorem). Let’s try to establish a Pythagorean Theorem for our right triangle on the sphere. To more easily follow the remaining discussion, on each of the right triangle pictures that follow,

2 2 c+ =

you should sketch in the x,y,z coordinate axes. As we are working on the sphere, let’s use spherical coordinates. Without loss of generality, assume that our sphere is centered at the origin O . Recall: (0,0,0)=

(1.4) cos sinsin sincos

x Ry Rz R

θ φθ φφ

===

where (same as polar coordinates in the xy-plane – i.e. measured from the positive x-axis toward the positive y-axis) and is measured from the positive z- axis toward the xy-plane.

0 2θ π≤ ≤0 φ π≤ ≤

WLOG assume that A is on the positive x- axis, C is in the xy-plane, and B is such that ∠ = is a right angle. Recalling that each point is on the sphere of radius R, express each in spherical coordinates. It is clear that

since they are both in the xy-plane. Now is a right angle so

the line (great circle) containing the arc on the sphere must contain the “North Pole” since C is on the “equator” (where the xy-plane intersects the sphere). Therefore, in the expressions for spherical coordinates in

(1.4), θ = and

C AC∠

C∠

(0,0, )R

B

( ,0,0), ( cos( ), sin( ),0)A R C R AOC R AOC= = ∠ ∠

BC

AOC∠2

BOCπ− ∠φ = for the point B(see the picture).

The spherical coordinates for B, then, are

cos( )sin( 2 ) cos( )cos( )sin( )sin( 2 ) sin( )cos( )cos( 2 ) sin( ).

x R AOC BOC R AOC BOCy R AOC BOC R AOC BOCz R BOC R BOC

ππ

π

= ∠ − ∠ = ∠ ∠= ∠ − ∠ = ∠ ∠= − ∠ = ∠

Now we just need to determine the angle ∠ , which will also give us the length of the arc . If we look at the vector representations of the points A, B, and C, which will be denoted , and , we have

AOB ABC,A B

( ,0,0)

( cos( )cos( ), sin( )cos( ), sin( ))

( cos( ), sin( ),0).

A R

B R AOC BOC R AOC BOC R BOC

C R AOC R AOC

=

= ∠ ∠ ∠ ∠ ∠

= ∠ ∠

(1.5)

Now the angle formed by the two vectors and is found from AOB∠ A B

|| || || || cos( )A B A B AOB= ∠i . (1.6) Calculating, we have or 2 cos( )cos( ) cos( )R AOC BOC R R AOB∠ ∠ = ⋅ ⋅ ∠

(1.7) cos( ) cos( )cos( )

orcos( / ) cos( / )cos( / )

AOB BOC AOC

c R a R b R

∠ = ∠ ∠

=

The last expression in (1.7) is referred to as the Pythagorean Theorem on the Sphere. Exercises:

2) Use the power series expansion of to show that for small values of and c (when the triangle is most like a Euclidean Triangle), that the Pythagorean Theorem on the sphere in (1.7, second expression) is equivalent to the Pythagorean Theorem in the plane. For simplicity, you may assume that R=1.

cos x , ,a b

3) Let (1,0,0), (1/ 4, 3 / 4, 3 / 2), (1/ 2, 3 / 2,0)C= = =A B be three points on the unit sphere.

Find the angles AOC, BOC, AOB and show that this is a right triangle by showing it satisfies the Pythagorean theorem. Find the length of each side of the triangle.

Right Triangle Trigonometry Often we will use vector representations of points to obtain information on the sphere. This was seen above already in the proof of the Pythagorean Theorem. We will need some more information here. Preliminaries: If u are any two vectors in then: ,v 3R

• , where θ is the angle between the two vectors. This is often used to find the angle between two vectors, when the two vectors are given in coordinates.

|| || || || cosu v u v θ=i

• , where θ is the angle between the two vectors. This can also be used to find the angle between two vectors. The cross product of two vectors is a vector perpendicular to both vectors, with direction determined by the right-hand rule.

|| || || || || || sinu v u v θ× =

• A plane in is completely determined by a point in the plane and a vector perpendicular to the plane (called the normal vector n ). The equation of a plane is where

and a point on the plane is ( . Example: The -plane is given by , so and is a point on the plane. (Note: could also use )

3R

))

( ) ( ) ( ) 0a x h b y k c z l− + − + − = yz x =

( 1,0,0)n = −

( , ,n a b c=

(0,0,0, , )h k l 0 (1,0,0)n =

• A plane is also completely determined by two non-parallel vectors . The cross product, , is perpendicular to both, so it is a normal vector to the plane.

3R ,u v u v×

• The angle between two planes is the angle between their normal vectors.

Theorem: Let be a right triangle with a right angle at and let be the lengths of the sides opposite ∠ ∠ , respectively. Again, we are on a sphere of radius R, with center at the origin O.

ABC∆,A B C∠

C∠ , ,a b c,

sin( )cos( ) sin( / )cos( / )cossin( ) sin( / )AOC BOC b R a RA

AOB c R∠ ∠

= =∠

(1.8)

Proof: First, notice that the second equality follows immediately from equations (1.1) and (1.3). The angle at is the angle between the -plane and the plane containing the vectors and . Now this is the same as the angle between their normal vectors, which are (see the bullets above). Using the vector representations in (1.5), and the definition of cross product,

A xy AA×

B

1 2(0,0,1) andn n= = B

C

. 2 22 (0, sin( ), sin( )cos( ))n R BOC R AOC BO= − ∠ ∠ ∠

To find the cosine of the angle A, then, we use the dot product; . This gives 1 2 1 2|| || || || cosn n n n θ=i

22sin( )cos( ) 1 || || cosR AOC BOC n∠ ∠ = ⋅ A , or

2

2

sin( )cos( )cos|| ||

R AOC BOAn

∠ ∠=

C .

But, || , and so 22 || || || || || || || sin( ) sin( )n A B A B AOB R AOB= × = ∠ = ∠

sin( )cos( )cossin( )AOC BOCA

AOB∠ ∠

=∠

(1.9)

Exercise 4: Use the cross-product technique as in the proof of the above theorem to show that

sin( / )sinsin( / )

a RAc R

= (1.10)

Exercise 5: Use the power series expansions of sine and cosine to show that the results of (1.8), second equality, and (1.10) gives the usual definition for and co if the lengths are very small. sin A s A , ,a b c Exercise 6: Let (1,0,0), (1/ 4, 3 / 4, 3 / 2), (1/ 2, 3 / 2,0)A B C= = = be three points on the unit sphere. a) Find the equations of the three planes – the plane containing each pair of vertices and the center of the sphere. b) Calculate the angles ∠ ∠ , by calculating the angles between the planes in (a). , ,ABC ACB BAC∠

Exercise 7: In the Euclidean plane the ratio of the sides of a 30-60-90 triangle are 1 3, ,2 2

1. Determine if

this is the case for 30-60-90 triangles on the sphere.