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Download 03 Spherical Geometry
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Transcript
Spherical Trigonometry—Preliminaries Students will learn about the polar triangle of a spherical triangle. Then other basic triangle properties will be explored. So we’re going to study spherical trigonometry which will be the relations among the sides and angles of a spherical triangle. The relationships should be similar to the plane trigonometry relationships, but correct for the curvature of the sphere. The setup: we will need a spherical triangle 4ABC. As usual we will denote the angles by the vertices where the sides of the angle meet, and the sides a, b, c will be opposite the correspondingly-named vertex. Recall that the lengths of the sides can be measured as angles because we are assuming a sphere of radius one. If you have a sphere of a different size, you can always scale the results after you find them! The first thing we will do is to create another triangle that is closely related to the original triangle. Given 4ABC we know that the great circle through B and C cuts the sphere into two hemispheres. One of those contains A. Let A0 be the pole of that hemisphere. That is, if the great circle through B and C is the equator and A is in the northern hemisphere, then A0 is the north pole. We can similarly find B 0 and C 0 . The triangle 4A0 B 0 C 0 is called the polar triangle of 4ABC. An important thing to notice is that the distance from A0 to either B or C is π/2 because A0 is the pole of the hemisphere the B and C on the equator. The same distances apply to B 0 from A and C, and to C 0 from A and B. 1 Theorem: if 4A0 B 0 C 0 is the polar triangle of 4ABC then 4ABC is the polar triangle of 4A0 B 0 C 0 . Proof: by the distances just noted, A is π/2 from B 0 and C 0 , so is the pole of one of the hemispheres determined by the great circle through B 0 and C 0 . But since A and A0 are in the same hemisphere determined by B and C, their distance from each other must be less than π/2. Since A0 can’t be more than π/2 from A, it must be in the same hemisphere cut by B 0 and C 0 as A, so A is the polar point of A0 . Similarly for the other points. The other thing about the polar triangle is that its sides are related to the angles of the original triangle, while its angles are related to the original triangle’s sides: Theorem: A = π − a0 , B = π − b0 , and C = π − c0 . (And, of course, since the polar triangle of the polar trinagle is the original, A0 = π − a and so forth.) Proof: If necessary, extend the great circle arcs of AB, AC and B 0 C 0 until the former two each cross the latter. Call these crossing points D and E. Now the spherical angle at A is the same as the measure of the arc DE. Also, notice that since D is on great circle AB we have the distance from D to C 0 is π/2. Similarly the distance from B 0 to E is π/2. But then a0 = B 0 C 0 = B 0 E + C 0 D − DE = π − DE = π − A. Right away, this tells us that the sum of the angles of a spherical triangle is less than 3π, just by adding up the three equations for the three angles—the sides a0 , b0 , and c0 can be made as small as you like but their total can’t be zero! Such a triangle can be realized by having three points all just north of the equator. On the other hand, the three sides of a spherical triangle can never total 2π or more. This is because the three sides are the same as the three angles where the radii from the center of the sphere to the vertices of the triangle meet. They must total less than 2π (to see this, note that the center of the sphere could be considered the vertex of a cone with the other three points on the cone somewhere; now just unroll the cone). If they total 2π exactly then they are coplanar, and the three points on the sphere would be on the same great circle, hence collinear, and would not form a spherical triangle. But then the total of the angles in a spherical triangle is A + B + C = π − a + π − b + π − c is always greater than π. The angles of a spherical triangle always total more than 180◦ . 2 Questions to consider: Is the spherical equivalent of the triangle inequality true? What about other theorems about the sides of triangles—are the base angles of an isosceles triangle equal? If two angles of a triangle are equal is the triangle isosceles? Is the longest side of a triangle always opposite the largest angle? The answers to all these questions is “yes.” The first is a consequence of the fact that the three sides are equivalent to the three angles at the center of the sphere, and since we’re not allowing angles greater than 180◦ the triangle inequality is true for those central angles. The second can be proven by working with the tangents instead of the great circles. Let sides a and b be equal. Extent the radius through point C and draw the tangents to great circles AC and BC in space. They will both hit this extended radius, (call this point S) and since they are two tangents to a sphere from a single point in space they will be equal in length. Similarly, the tangents from A and B toward each other will also meet at point T , is these two tangents are equal to each other. Since the distance ST equals itself, spatial triangles 4AST and 4BST are congruent. So angles ∠SAT and ∠SBT are equal—but these are the dihedral angles along the radii to A and B, hence they are equal to the spherical angles A and B. The third is a trivial consequence of the second plus using polar triangles! The last can be proven from the second and third statements nearly identically to the way it is proven in Euclidean geometry. All of these can probably be proven more easily once we have the trigonometric relationships for spherical triangles, so perhaps we should get to those! 3