contained by arcs of the same circles, and having the same perpendiculars CD and CG. 1. Now, because CE is nearer to the arc CHG than CB is, CE is greater than CB, and therefore CE and CA are greater than CB and CA ; wherefore CB Ek and CA are less than a semicircle ; but because AD is, by supposition, of less than DB, AC is also less than CB (Sp. Ge. 13); and therefore in this case, namely, when the per- F pendicular falls within the triangle, and when the sum of the sides is less than a semicircle, the least segment is adjacent to the least side. 2. Again, in the triangle FCA the two sides FC and CA are less than a semicircle ; for, since AC is less than CB, AC and CF are less than BC and CF. Also, AC is less than CF, because it is more remote from CHG than CF is; therefore the least segment of the base AD is in this case also adjacent to the least side. 3. But in the triangle FCE the two sides FC and CE are greater than a semicircle; for, since FC is greater than CA, FC and CE are greater than AC and CE. And because AC is less than CB, EC is greater than CF, and EC is therefore nearer to the perpendicular CHG than CF is, wherefore EG is the least segment of the base, and is adjacent to the greater side. 4. In the triangle ECB the two sides EC, CB, are greater than a semicircle ; for, since by supposition CB is greater than CA, EC and CB are greater than EC and CA. Also, EC is greater than CB; wherefore in this case, also, the least segment of the base EG is adjacent to the greatest side of the triangle. PROPOSITION XVIII. Any two circles of the sphere, passing through the poles of two great circles, intercept equal arcs upon them. Let AFB, CFD, be the two great circles intersecting one another in F, and let F be the pole of the great circle ACBD, cutting them in the diameters AEB, CED. The circle ACBD passes through their poles ; let the diameter MN be perpendicular to AB, and PQ to CD; then M, N, are the poles of AFB, and P, Q, the poles of CFD. Let the small circle PGN pass through the poles P, N, and cut the circle ACBD in the line of common section PKLN; the arcs BH, DG, of the circles af AFB, CFD, intercepted by the circle passing through P, N, are equal. For, let "EH, HK, EG, GL, PG, NH, be drawn. In the triangles PEL, NEK, angle P=N, for PE=NE; also NEK, PEL, being right angles, are equal; hence (Pl. Ge. I. 26) PL = NK, and EL = EK. “Again, the quadrants PG, NH, are equal, and taking away HG, the arc PH = NG, and the chord PG=NH. Hence, in the triangles PLG, NKH, PLENK, PG=NH, and angles LPG, KNH, standing on equal arcs NG, PH, are equal; hence, LG=KH. Again, in the triangles EKH, ELG, having their sides respectively equal, angle HEK = GEL, and hence the arc HB = GĎ. SPHERICAL TRIGONOMETRY. Spherical Trigonometry treats of those relations between the sides and angles of spherical triangles, by which their numerical values may be computed. The trigonometrical lines defined in Plane Trigonometry are employed in reference to the sides and angles of spherical triangles. PROPOSITION I. In right-angled spherical triangles, the sine of either of the sides about the right angle, is to the radius of the sphere, as the tangent of the remaining side is to the tangent of the angle opposite to that side. Let ABC be a triangle, having the right angle at A ; and let AB be either of the sides, the sine of the side AB will be to the radius, as the tangent of the other side AC to the tangent of the angle ABC, opposite to AC. Let D be the centre of the sphere ; join AD, BD, CD, and let AF be drawn perpendicular to BD, which therefore will be the sine of the arc AB, and from the point F, let there be drawn in the DS plane BDC the straight line FE at right angles to BD, meeting DC in E, and let AE be joined. Since therefore the straight line DF is at right angles to both FA and FE, it will also be at right angles to the plane AEF (So. Ge. I. 4); wherefore the plane. ABD, which passes through DF, is perpendicular to the plane AEF (So. Ge. I. 17), and the plane AEF perpendicular to ABD; but the plane ACD or AED is also perpendicular to the same ABD; therefore the common section, namely, the straight line AE, is at right angles to the plane ABD (So. Ge. I. 18), and EAF, EAD, are right angles. Therefore, AE is the tangent of the arc AC; and in the rectilineal triangle AEF, having a right angle at A, AF is to the radius as AE to the tangent of the angle AFE (Pl. Tr. 2); but AF is the sine of the arc AB, and AE the tangent of the arc AC, and the angle AFE is the inclination of the planes CBD, ABD (So. Ge. I. Def. 7), or the sphe rical angle ABC; therefore the sine of the arc AB is to the radius as the tangent of the arc AC to the tangent of the opposite angle ABC. Cor.---And since by this proposition the sine of the side AB is to the radius, as the tangent of the other side AC to the tangent of the angle ABC opposite to that side ; and as the radius is to the cotangent of the angle ABC, so is the tangent of the same angle ABC to the radius (Pl. Tr. 4, Čor. to Def.); by equality, the sine of the side AB is to the cotangent of the angle ABC adjacent to it, as the tangent of the other side AÇ to the radius. PROPOSITION II. In right-angled spherical triangles the sine of the hypotenuse is to the radius, as the sine of either side is to the sine of the angle opposite to that side. Let the triangle ABC be right angled at A, and let AC be either of the sides; the sine of the hypotenuse BC will be to the radius as the sine of the arc AC is to the sine of the angle ABC. Let D be the centre of the sphere, and let CE be drawn perpendicular to DB, which will therefore be the sine of the hypotenuse BC; and from the point E let there be drawn in the plane ABD the straight line EF perpendicular to DB, and let CF be joined ; CF will be at right angles to the plane ABD, as was shown in the preceding proposition of the straight line EA; wherefore CFD, CFE, are right angles, and CF is the sine of the arc AC; and in the triangle CFE, having the right angle CFE, CE is to the radius, as CF to the sine of the angle CEF (Pl. Tr. 1). But, since CE, FE, are at right angles to DEB, which is the common section of the planes CBD, ABD, the angle CEF is equal to the inclination of these planes (So. Ge. I. Def. 4), that is, to the spherical angle ABC. The sine, therefore, of the hypotenuse CB is to the radius as the sine of the side AC is to the sine of the opposite angle ABC. Cor.--Of these three, namely, the hypotenuse, a side, and the angle opposite to that side, any two being given, the third may be found. PROPOSITION III. In right-angled spherical triangles, the cosine of the hypotenuse is to the radius as the cotangent of either of the angles is to the tangent of the remaining angle. Let ABC be a spherical triangle, having a right angle at A, the cosine of the hypotenuse BC will be to the radius E as the cotangent of the angle ABC to the tangent of the angle ACB. Describe the circle DE, of which B is the pole, and let it meet AC in F, and the circle BC in E; and since the circle BD passes through the pole B of the cirele DF, DF will pass through the pole of BD (Sp. Ge. I. Cor. 7). And since AC is perpendicular to BD, AC will also pass through the pole of BD; wherefore, the pole of the circle BD is in the point where the circles AC, DE, intersect, that is, in the points F. The arcs FA, FD, are therefore quadrants, and likewise the arcs BD, BE. In the triangle CEF, right angled at the point E, CE is the complement of BC, the hypotenuse of the triangle ABC; EF is the complement of the arc ED, which is the measure of the angle ABC; FC, the hypotenuse of the triangle CEF, is the complement of AC; and the arc AD, which is the measure of the angle CFE, is the complement of AB. But (Sp. Tr. 1) in the triangle CEF, the sine of the side CE is to the radius, as the tangent of the other side EF is to the tangent of the angle ECF opposite to it; that is, in the triangle ABC, the cosine of the hypotenuse BC is to the radius as the cotangent of the angle ABC is to the tangent of the angle ACB. Cor. 1.-Of these three, namely, the hypotenuse and the two angles, any two being given, the third will also be given. Cor. 2.-And since by this proposition the cosine of the hypotenuse BC is to the radius as the cotangent of the angle ABC to the tangent of the angle ACB, and since the radius is to the cotangent of ACB, as the tangent of ACB to the radius (Pl. Tr. 4, Cor. to Def.); therefore, by equality, the cosine of the hypotenuse BC is to the cotangent of the angle ACB, as the cotangent of the angle ABC to the radius. Schol.-The triangle CEF is called the complementary triangle. |