sphere having its centre either at the centre of the earth, or at the eye of the observer. PROBLEM. 15. To investigate properties and equations from which the solution of the several cases of spherical trigonometry may be deduced. In order to this let us recur to the spherical tetraedron oa Bc, where the angles A, B, C, of the spherical triangle are the diedral angles between each two of the three planes Aoc, AoE, &c. and the sides a, b, c, are the measures of the plane angles cob, coA, &c. Here it is 1st, evident that the three sides of a spherical triangle are together less than a circle, or, a + b + c < 360°. For the solid angle at o is contained by three plane angles, which (Euc. xi. 21) are together less than four right angles; therefore, the sides a, b, c, which measure those plane angles are together less than a circle. 16. Let the tetraedron oasc be cut by planes perpendicular to the three B edges; they will form an- - * other tetraedron of A'B'C'; their faces will be respectively perpendicular, two and two. But, in the quadrilateral AoEA' since the angles A and B are right angles, the plane angle o is the supplement of AA'B which measures the diedral angle AA'o'B. The same may be shown with respect to the other plane angles that meet at of as well as of the plane angles at o', in reference to the diedral angles of the tetraedron oa BC. Therefore, either of these tetraedrons, has each of its plane angles supplement to a diedral angle in the other: it is hence called the supplementary tetragdron. And if they become spherical tetraedrons referred to equal spheres, or to different parts of the same sphere, their bases will be spherical triangles respectively supplemental to each other. 17. It is obvious from this that the problems in spherical trigonometry become susceptible of reduction to half their number; since, if there are given, for example, the three angles A, B, C, and the three sides a, b, c, are required; let the triangle which has for its sides a', b', c', the supplements of the measures of A, B, and c, have its angles A, B', c', determined; their measures will be the supplements of the required sides a, b, and c. 18. On the surface of the sphere, the supplemental triangle is formed by the intersections of three great circles described from the angles of the primitive tri-angle as poles. Besides the supplemental triangle, three others are formed in each hemisphere by the mutual intersections of these three great circles; but it is the central triangle (of each hemisphere) that is supplementary. 19. Every angle between two planes being less than two right angles, it follows, that the sum of the angles of a spherical triangle is less than 3 times 2, or than 6 right angles. At the same time, it is #. than 2 right angles: for the sum a' + b + c of the sides of the supplemental triangle is less than 360° (art. 15 above): taking the supplements, we have 3 x 180°– (a' + b + c ) > 180°, or A + b + c > 180°. 20. To deduce the fundamental theorems, we may proceed thus, From any point A of the edge Ao of the tetraedron, let fall on the plane or face Boc the perpendicular AD : draw, also, in that plane, the lines DH, DC, perpendicular to ob, oc, respectively; and join AH, Ac; then will a H be perpendicular to ob, and Ac to oc. It is evident, therefore, that the angles AcD, AHD, measure the angles between the planes Aoc, coB, and Aob, coB, that is, the angles c and B of the spherical triangle Abc. It is also evident that the plane angles in o, are Aor = c, Aoc = b, Boc = a. This being premised, the triangles Aco, AcD, the former right angled in c, the latter, in D, give sin a T sin & T sing ' ' ' ' (1.) Hence, the sines of the angles of a spherical triangle are proportional to the sines of the opposite sides. 2i. Draw ce and DF, respectively perpendicular and parallel to ob; then will the angle DCF = Eoc = a. 180° — A', 180° — B", 180°– c', &c. and for the angles A, B, C, 180° — a', 180° — bo., &c. we shall have - cos A" = cos B' cos co — sin B'sin c' cos a'. Here again we have three symmetrical equations applying to any spherical triangle, viz. COSA F COS a Sln B Sin C – COS B COS C cos B = cos 5 sin A sin C — cos A cos : (3.) COS C = COS C SIn A Sin B – COSA COS B 23. Another important relation may be readily deduced. For, substituting for cos b in the third of the equations marked (2) its value in the second; substituting also for coso a its value 1 — sin” a (chap. i. 19), and striking out the common factor sin a, we shall have cos c sin a = sin c cos a cos B + sin b cos c. But, equa. (1) gives sin b = +. Therefore cot c sin a = cos a cos B + sin B cot c. Thus, again, we get three symmetrical equations, cot a sin b = cos b cos c + sin C cot A cot b sin c = cos c cos A + sin A cot § (4.) cot c sin a = cos a cos B + sin B cot C 24. The classes of equations marked 1, 2, 3, 4, comprehend the whole of spherical trigonometry: or, in truth, the equations (2), from which the others may be made to flow, may be regarded as comprehending the whole. They require, however, some modifications to fit them for logarithmic computations, and become simplified in their application to some kinds of triangles. We shall, therefore, now show the pupil * they be come transformed when they are applied to the principal cases which occur in practice. SECTION II. Resolution of Right angled Spherical Triangles. 25. Suppose, in the first diagram in this chapter, the angle A to be right, or the faces oAB, oac, to be perpendicular to each other. Then, since sin A = 1, the equations marked (1) become 1. sin B – sinc. Consequently, - - whence cos a = = cot B cotc.... (7.) sin n sin c Upon the same hypothesis, cot A becomes = o, so that the first of equa. (4) becomes cot a sin b = cos bcos c. Or, dividing by sin b, The two last of equa. (3) give also, upon the same hypothesis, cos B = sin c cos b7 9 coso- in ot . . . . (9) And, lastly, from equa. (4) we have cot B = cot b sinc 10 cot c= cot c sin b ( " " ' " (10.) From these equations by a few obvious transformations, the six usual cases of right angled spherical triangles may be o, as below. |