posite angles (598), Bowditch's Rules (604), (605), Theorems for the solution of spherical oblique triangles, The sines of the sides are proportional to the sines of the up SECTION II. Solution of Spherical Oblique Triangles, when two sides and the included angle are known (615-620), when a side and the two adjacent angles are known (629 - 634), or (749), (758), when two sides and an angle opposite to one of them are when two angles and a side opposite to one of them are known when the three sides are known (674), or (686 - 688), or Page. 26 (731), (734), (836), (839), Theorems on the equality of triangles (861), (863), CHAPTER IV. Surfaces of Spherical Triangles, Definitions of degree of surface, lunary surface, Equality of surface in spherical triangles (876), polygon (903), (699 - 704), when the three angles are known (791), or (803–805), or rules for determining when the sides and angles are obtuse or 36 (713), 47 The sum of the sides and angles (713), (822), 32 35. 38 41 55 29 48 65 46 66 66 66 67 69 71 SPHERICAL TRIGONOMETRY. CHAPTER I. Definitions. 1. Spherical Trigonometry treats of the solution of spherical triangles. A Spherical Triangle is a portion of the surface of a sphere included between three arcs of great circles. In the present treatise those spherical triangles only are treated of, in which the sides and angles are less than 180°. 2. The angle, formed by two sides of a spherical triangle, is the same as the angle formed by their planes. 3. An isosceles spherical triangle is one which has two of its sides equal. An equilateral spherical triangle is one which has all its sides equal. 4. A spherical right triangle is one which has a right angle, all other spherical triangles are called oblique. We shall in spherical trigonometry, as we did in plane trigonometry, attend first to the solution of right triangles. CHAPTER II. Spherical Right Triangles. SECTION. I. Napier's Rules for the Solution of Spherical Right Triangles. 5. Problem. To investigate some relations between the sides and angles of a spherical right triangle. Solution. The importance of this problem is obvious; for, unless some relations were known between the sides and the angles, they could not be determined from each other, and there could be no such thing as the solution of a spherical triangle. Fig. 1 Let, then, ABC (fig. 1.) be a spherical right triangle, right angled at C. Call the hypothenuse AB, h; and call the legs BC and (427) AC, opposite the angles A and B, 0. respectively a and b. A Let O be the centre of the B B a sphere. Join OA, OB, OC. The angle A is, by art. 2, equal to the angle of the planes BOA and COA. The angle B is equal to the angle of the planes BOC and BOA. The angle of the planes BOC and AOC is equal to the angle (428) C, that is, to a right angle; these two planes are, therefore, perpendicular to each other. Moreover, the angle BOA, measured by BA, is equal to BA or h; BOC is equal to its measure BC (429) or a, and AOC is equal to its measure AC or b. Through any point A' of the line OA, suppose a plane to pass perpendicular to OA. Its intersections A'C' and A'B', with the planes COA and BOA must (430) be perpendicular to OA', because they are drawn through the foot of this perpendicular. As the plane B'A'C' is perpendicular to OA, it must be perpendicular to AOC; and its intersection B'C', with the plane BOC, which is also perpendicular to AOC, must likewise be perpendicular to AOC. Hence B'C' must be perpendicular to A'C' and OC' which pass through its foot in the plane AOC. All the triangles A'OB', A'OC', B'OC', and (431) A'B'C' are then right-angled; and the comparison of them leads to the desired equations, as follows: First. We have from triangle A'OB' by (5) and (429), OA' OB and from triangles A'OC' and B'OC' OA' cos. A'OC' =cos. b OC OC OB ; cos. B'OC'=cos. a= The product of the two last equations is cos, a cos, b = OA' OC OA' OB' OB (436) (438) (439) Secondly. From triangle A'B'C' we have by (5) (437) and (428), and the fact that the angle B'A'C' is equal to the inclination of the two planes BOC and BOA, cos. B'A'C' cos. A: = (440) (441) (442) (444) hence from the equality of the second members of equations, (432) and (435), cos. h =cos. a cos. b. tang. C'OA'=tang. b= A'C A'B' and, from triangles A'OC' and A'OB', by (5) and (429), cotan. h: = cotan. B'OA' = hence, by (438), A'C A'O' Α' Ο A'B Α' Ο A'C ; = cos. Atang. b. cotan. h. Thirdly. Corresponding to the preceding equation between the hypothenuse h, the angle A, and the adjacent side b, there must be a precisely similar equation between the hypothenuse h, the angle B, and the adjacent side a; which is (443) cos. B = tang. a cotan. h. Fourthly. From triangles B'OC', B'OA', and B'A'C', by (5), (429), and (437), sin. B'OC=sin. a= B'C OB |
