Hence, by substituting AB for AC, and c for B, we obtain (626) By denoting the three sides of the triangle by a, b, c, and their opposite angles by A, B, C, the following proportions are deduced, where c and b are constant quantities. 1. da da:: rad2: sin b. sin c. 2. da: da:: rad2: sin c. sin B. .. da: da:: cosec b: sin c:: cosec c: sin b, and da: da:: cosec c: sin B:: cosec B: sin c. 3. da: dB:: rad. sin a: -cos c. sin b. 4. da: dc:: rad. sin a: cos B. sin c. PROPOSITION VIII. (Plate IV. Fig. 6.) (627) In any oblique-angled spherical triangle ABC, suppose the two angles B and C to remain constant*, it is required to find the ratio of the differentials of the other parts. Take the supplemental triangle DEF (302), then DE and DF will be constant, therefore by Proposition VII. * La Lande's Astronomy, art. 4034 to 4045; Cagnoli, page 325, &c. 1. dD: der: cosec DF: sin F: cosec F: sin DF, 4. do: dr:: rad. sin EF: cos E. sin c. sin DE, and de : def::cot F: sin EF, (628) By denoting the three sides of the triangle by a, b, c, and their opposite angles by A, B, C, we derive the following proportions, wherein B and C are constant quantities. 1. da: da :: cosec c: sin b:: cosec b: sin c. 2. da da:: cosec B: sin c:: cosec c: sin B. (629) In all the preceding propositions if the sides of the triangle be diminished without limit, the triangle may be considered as plane, and instead of the sines and tangents of the sides, we may substitute the sides themselves (440). Hence the variations of plane triangles are readily deduced from those of spherical triangles, in every case where the differentials are proportional to the sines or tangents of the sides. Thus, by the 4th proportion (626) we have shown that da: dc':: rad. sin a cos B. sin c, that is (supposing the triangle plane), da: dc::a. rad: c. COS B. Again, by the 8th proportion (628) db: dc:: tan b : tan c, that is, when the triangle is rectilineal, db: dc::b: c, and in the same manner the rest may be deduced. The variations of rectilineal triangles may be deduced from the triangles themselves, without reference to spherical triangles, in a manner exactly similar to those deduced from the spherical triangles. Vide Traité de Trigonométrie, par M. Cagnoli, Chapitre X. THE APPLICATION OF THE DIFFERENTIAL ANALOGIES. PROPOSITION IX. * (630) To find when that part of the equation of time dependent on the obliquity of the ecliptic is the greatest possible.+ Here the sun's longitude will form the hypothenuse of a right-angled spherical triangle, his right ascension will be the base, and the obliquity of the ecliptic will be a constant angle. Let the hypothenuse be denoted by b, and the base by c, then db dc: sin 26: sin 2c (346). : Hence, when b-c is a maximum, db-de-0, or db=dc, and therefore sin 26 sin 2c; consequently 26 must be the supplement of 2c, or b+c=90o. The equation of time dependent on the obliquity of the ecliptic is therefore the greatest possible when the sun's longitude and his right ascension together are equal to 90° from the equinoctial points. This will happen, in 1840, on May 6 and November 8. PROPOSITION X. (631) Given the parallax in altitude of a planet, to find its parallax in latitude and longitude. A Let B represent the pole of the ecliptic, a the zenith, and c the place of the planet. * A variety of examples will be met with in the perusal of La Lande's Astronomy, vol. iii. + Simpson's Fluxions, page 550. Then do will represent the parallax in altitude, dв the parallax in longitude, and da the parallax in latitude. If EC be supposed to represent a part of the ecliptic, then BE and BC will be quadrants (290) and CEB and ECB will be right angles (291). In the right-angled triangle CEA, making EA the middle part, da sin EA sin b -db, and because the altitude of the nonagesimal degree of the ecliptic is an arc of a great circle comprehended between the zenith of any place of the pole of the ecliptic (550), we obtain the following proportion : sine of the zenith distance is to cosine of the altitude of the nonagesimal degree, as the parallax in altitude is to the parallax in latitude, viz. sin b: sin EA=cos c:: db: da. (632) Again, in the right-angled triangle CEA, making the ECA the middle part, we have rad . COS ECA=tan Ec. cot b; but cot b = rad 2 tan b' therefore by substitution dB = rad tan EC = (sin c=) supposed to be in or very near to the ecliptic, sin a=rad) hence, tangent of the planet's zenith distance is to the tangent of its longitude from the nonagesimal degree, as the parallax in altitude is to the parallax in longitude, viz. (633) Given the altitude of the nonagesimal degree of the ecliptic, the longitude of a planet from the nonagesimal degree, and its horizontal parallax, to find its parallax in latitude and longitude. It is shown in the preceding proposition, that da= and Bd= tan EC tan b rallax of any planet, its parallax in altitude db= db. Now, if H represent the horizontal pa H sin b nearly (185). By substituting this value of db in the above equations B tan EC. sin b and dB: = rad. tan b In the right-angled triangle CEA, making b the middle part, rad cos bcos EA . cos EC; but cos b = rad. sin b and tan b sin b tan b H COS EA. Sin EC H COS EA. sin EC tan EC. rad rad 2 hence tan EC substitution dв = but EC is the therefore dB = general rules. measure of the LB (474), and cos EA sin c, H sin c. sin B; hence are derived the following 1. Radius is to the cosine of the altitude of the nonagesimal degree of the ecliptic, as the horizontal parallax is to the parallax in latitude. 2. The square of the radius is to the rectangle of the sines of the altitude of the nonagesimal degree and the planet's longitude from thence, as the horizontal parallax is to the parallax in longitude.* PROPOSITION XII. (634) To determine the correction for finding the time of apparent noon, from equal altitudes of the sun. It is obvious that if the sun's declination was invariable, half the interval of time between equal altitudes would show the instant of noon; but by the variation in the sun's declin * Simpson's Fluxions, vol. ii. page 286. |