NAPIER'S RULES. PROPOSITION IX. Napier's Rules. If, in a right angled spherical triangle, we denominate CIRCULAR PARTS, 273 The side opposite the right angle with the including angles, and the complements of the other two sides, then Cos. Mid. Part = Product of Sines of Opp. Parts (568) observing that each of the five parts will be adjacent to two and opposite to two, no account being made of the right angle. In order to demonstrate this proposition, which is a most remarkable example of artificial memory, it is only necessary to make 90° and seek among the preceding forms, values for a, B, C ; 90° — b = b ̧‚* 90° — c = cc ; A = we find cosa = cosb cosc, or cos B cos C = sinB sin C cosa; = Cor. 1. Napier's Rules may be employed in solving (569) Quadrantal Triangles, by observing that the circular parts will be the supplement of the angle opposite the quadrant, with the sides adjacent, and the complements of the other two angles. and the triangle ABC, polar to ABC, being right angled at A, gives (568) = cosa sinb, sine, cotB cotC, .. (560) cos(180° - A) = sin[90° (180° - B)]sin(C-90) or, putting 180° - A = A., &c., and reducing, It will be observed that the quadrant, a, is regarded as not separating B, C. For illustration, let it be required to find B when a and c are given, and b = 90°. We have cos(180° - B) = cosB, cota cotc. Cor. 2. Napier's Rules may be employed in the solution (570) of Isosceles Spherical Triangles, by observing that the circular parts will be the equal side, equal angle, half unequal angle, and complement of half unequal side. For, let c = a; then joining B and the middle point of b, we shall have formed two right angled triangles, in each of which the parts will be (538), (568), Cor. 3. Napier's Rules may be employed in the solution (571) of Oblique Triangles, by dropping a perpendicular so that two of the known parts shall fall in one of the right angled triangles thus formed; there will result also, Bowditch's Rules. 1o. The cosines of the parts opposite the perpendicular, are proportional to the sines of those adjacent. 2o. The cosines of the parts adjacent to the perpendicular, are proportional to the cotangents of the parts opposite. For, let the angle, C, be separated into the parts, M, N, by the perpendicular, p, dividing the opposite side, c, into the corresponding parts, m, n, subjacent to a, b, and we find, m MN Fig. 1212. n CASES IN SPHERICAL TRIGONOMETRY. 275 Cases in Spherical Trigonometry. Given. Forms for Solution. I. Two sides and included angle. (535); (539); (566); (571). (535) and (384); (551), [?]. Scholium I. There will be a choice in forms, not only on account of logarithmic operations, but also for the purpose of avoiding the cosines of very small arcs, and the sines of those differing little from a quadrant. Scholium II. Problems pertaining to spherical trigonometry will generally find their easiest solution by constructing the triangle so that one of its angles shall be at the pole of the sphere. Cape Horn, S. America, Cincinnati (Fort Washington), Ohio, 39 5 54 55 58 41 Greenwich (Obs.), Eng., 10. Required the distance and direction from New York to Greenwich. [69'2 miles to 1°.] Polar distance of New York = 49° 17′ 20′′, = the problem, therefore, belongs to Cases I., III. Solve the same by (566), also by (571). 20. Required the distances between New York, Cincinnati and Washington, the angles of the spherical triangle thus constructed, and its area. 3. Required the distance and bearing of Chicago from Boston, reckoning the latitudes of both places at 42°. 4°. Required the direction and distance from Greenwich to Quito, the latter place being under the equator, and having 79° W. lon., nearly. 5°. Required the breadth of South America between Cape Blanco and Cape St. Roque, reckoning both to have 4° 30′ S. lat., and a difference of longitude = 46°. 6°. To reduce an angle to the horizon, Around the point of observation, O, imagine a sphere to be described with radius = 1; then will one of its arcs, a, be the measure of the inclined angle, SOS', and the other two, b, c, of the angles, S'OP, SOP, made with the vertical, OP; whence the spherical angle, A, opposite a, will be equal to the horizontal angle required. (572) -S' a Fig. 121 3. For example, let a = 36'7°, b = 110°, c = 75 25°; then will A be found 12° 0' 20". = 7°. Through any two given points and a third upon the surface of a sphere, which do not lie in the circumference of the same great circle, there may be made to pass two equal and parallel small circles; that is, one of them through the first two given points, and the other through the third given point; and every spherical arc which is terminated by these circles shall be bisected by the circumference of the great circle, to which they are parallel.* 8°. If there be two equal and parallel small circles, and if a great circle meets one of them in any point, it will meet the other in the opposite extremity of the diameter which passes through that point. 9°. If a great circle cuts one of two equal and parallel small circles, it will cut the other likewise; also, if it touches one of them, it will touch the other likewise. 10°. Lunular portions of surface, which are contained by equal spherical arcs with the arcs of equal small circles of the same sphere, are equal to one another; so also are the pyramidal solids, which have these portions for their bases, and their common vertex in the centre of the sphere. 11°. Spherical triangles, which stand upon the same base and between the same equal and parallel small circles, are equal to one another. 12°. If equal triangles, ABC, EBC, stand upon the same base, BC, and the same side of it, the points, A, E, and B, C, lie in the circumference of two equal and parallel small circles. 13°. Of equal spherical triangles upon the same base, the isosceles has the least perimeter. 14°. Of all triangles which are upon the same base, and have equal perimeters, the isosceles has the greatest area. 15°. If two spherical triangles have two sides of the one equal to the two sides of the other, each to each, and the angle which is contained by the two sides of the first equal to the sum of the other two angles of that triangle, but the angle which is contained by the two sides of the other not so; the first triangle shall be greater than the other. 16°. Two given finite spherical arcs, together with a third inde * This and the following exercises, drawn from the Library of Useful Knowledge, may be performed at the student's leisure. |