SPHERICAL TRIGONOMETRY. DEFINITIONS. 1. A great circle of a sphere is a section of the sphere by a plane passing through the centre. 2. The axis of a circle is a right line passing through the centre of the sphere, perpendicular to the plane of the circle and the two points, where the axis intersects the surface of the sphere, are called the poles of the circle. 3. A spherical angle is the inclination of two great circles. 4. A spherical triangle is a part of the surface of the sphere included by the arches of three great circles; which arches are called the sides of the triangle. 5. If through the poles A and F(fig. 11.) of two great circles DF and DA, standing at right angles, two other great circles ACE and FCB be conceived to pass, and thereby form two spherical triangles ABC and FCE, the latter of the triangles so formed is said to be the complement of the former; and vice versa. COROLLARIES. 1. It is manifest (from Def. 1.) that the section of two great circles (as it passes through the centre) will be a diameter of the sphere; and consequently, that their peripheries will always intersect each other in two points at the distance of a semicircle, or 180 degrees. 2. It also appears (from Def. 2.) that all great circles, passing through the pole of a given circle, cut that circle at right angles; because they pass through or coincide with the axis, which is perpendicular to it. 3. It follows morever, that the periphery of a great circle is everywhere 90 degrees distant from its pole; and that the measure of a spherical angle CAD*(fig. 12.) is an arch of a great circle intercepted by the two circles ACB, ADB forming that angle, and whose pole is the angular point A. For let the diameter AB be the intersection of the great circles ADB and ACB (see Corol.1.), and let the plane, or great circlé, DEC be conceived perpendicular to that diameter, intersecting the surface of the sphere in the arch CD; then it is manifest that AD BD = 90°, and AC = BC : (Corol. 1.), and that CD is the measure of the angle DEC (or CAD), the inclination of the two proposed circles. 90° 4. Hence it is also manifest, that the angles B and E, (fig. 11.) of the complemental triangles ABC and FCE, are both right angles; and that CE is the complement of AC, CF of *Note. Although a spherical angle is, properly, the inclination of two great circles, yet it is commonly expressed by the inclination of their peripheries at the point where they intersect each other. BC, BD for the angle F) of AB, and EF of ED (or the angle A). THEOREM I. In any right-angled spherical triangle it will be, as radius is to the sine of the angle at the base, so is the sine of the hypothenuse to the sine of the perpendicular; and as radius to the co-sine of the angle at the base, so is the tangent of the hypothenuse to the tangent of the base. DEMONSTRATION. Let ADL and AEL (fig. 13.) be two great circles of the sphere intersecting each other in the diameter AL, making an angle DOE, measured by the arch ED; the plane DOE being supposed perpendicular to the diameter AL, at the eentre O. Let AB be the base of the proposed triangle, BC the perpendicular, AC the hypothenuse, and BAC (or DAE = DE=DOE) the angle at the base : moreover, let CG be the sine of the hypothenuse, AK its tangent, AI the tangent of the base, CH the sine of the perpendicular, and EF the sine of the angle at the base; and let I, K and G, H be joined. Because CH is perpendicular to the plane of the base (or paper), it is evident, that the plane GHC will be perpendicular to the plane of the base, and likewise perpendicular to the diameter AL, because GC, being the sine of AC, is perpendicular to AL. Moreover, since both the planes OIK and AIK are perpendicular to the plane of the base (or paper), their intersection IK will also be perpendicular to it, and consequently the angle AIK a right angle. Therefore, seeing the angles OFE,GHC, and AIK are all right angles, and that the planes of the three triangles OFE, GHC, and AIK are all perpendicular to the diameter AL, we shall, by similar triangles, have that is, OE : EF :: GC : CH OE: OF :: AK: AI Š [Radius sine of EOF (or BAC): : sine of Radius : co-sine of EOF (or BAC) :: tang. COROLLARY I. Hence it follows, that the sines of the angles of any oblique · spherical triangles ADC are to one another, directly, as the sines of the opposite sides. by the form For let BC (fig. 14.) be perpendicular to AD; then radius: sine A:: sine AC : sine BC 2 since radius: sine D :: sine DC: sine BCS er part of the theorem; we shall have, sine A x sine AC (= radius x sine BC) = sine D x sine DC (by 16. 6.), and consequently sine A: sine D :: sine DC: sine AC; or sine A : sine DC: sine D : sie AC. COROLLARY 2. It follows, moreover, that, in right-angled spherical triangles ABC, DBC, having one leg BC common, the tangents of the hypothenuses are to each other, inversely, as the co-sines of the adjacent angles. For radius: co-sine ACB : : tan. AC : tan. BC since radius: co-sine DCB : : tan. DC: tan. BC by the latter part of the theorem; we shall (by arguing as above) have co-sine ACB: co-sine DCB:: tang. DC; tang. AC. THEOREM II. In any right-angled spherical triangle (ABC) (fig. 11.) it will be, as radius is to the co-sine of one leg, so is the cosine of the other leg to the co-sine of the hypothenuse. DEMONSTRATION. Let CEF be the complemental triangle to ABC, according to what has been already specified; then it will be (by Theor. 1. Case 1.) Radius sine F: : sine CF; sine CE; that is, : Radius: co-sine BA :: co-sine CB: co-sine AC (see Cor. 4. p. 27.) 2. E. D. COROLLARY. Hence, if two right-angled spherical triangles ABC, CBD (fig. 14.) have the same perpendicular BC, the co-sines of their hypothenuses will be to each other, directly, as the co-sines of their bases. For Srad: co-sin. BC:: co-sin. AB: co-sine AC, sincerad: co-sin. BC: : co-sin. DB : co-sine DC, therefore, by equality and permutation, co-sine AB: co-sine DB co-sine AC: co-sine DC. |