(C) In any oblique-angled spherical triangle ABC, suppose the two sides AB and AC to remain constant*, it is required to find the fluxions of the other parts. Let BC and AC change their positions to вn and an, An being equal to AC; produce AC and An to the points o and p, so that Ao and Ap may be quadrants, and join cn. Also, produce BC and вn to the points q and r, making вq and B quadrants, and through c draw cm parallel to qr. Then op will be the measure of the increment of the ▲ A, qr the measure of the decrement of the B, and mn will be the increment of the side BC. I. Sine Ap sine an:: sine op: sine cn, viz. rad sine AC :: Also, sine Bq sine BC:: sine viz. rad sine BC :: Z : sine cn= qr : sine cm, sine AC rad II. In the small straight-lined triangle cmn. 1. rad cn: sine mcn : mn. (Y. 33.) But sine c sine AB :: sine B: sine AC. (S. 194.) cx sine AC=sine BX sine AB, consequently sine BX sine AB mnc :: cn: cm. (Y. 33.) Again, by substituting AB for AC, and c for 4B, we obtain * La Lande's Astronomy, vol. III. art. 4015 to 4034; Cagnoli, page 321, &c. (D. 317.) (F) 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. and A : à :: cosec c: sine B : : cosec B: sine c. (G) In any oblique-angled spherical triangle ABC, suppose the two angles B and c to remain constant, it is required to find the fluxions of the other parts. Take the supplemental triangle DEF (U. 133.) then DE, and DF will be constant, therefore by Proposition vII. 1. D: EF: cosec DF: sine / F:: cosec 4F: sine DF, ZA Viz. BC: LA :: cosec Lc: sine AC :: cosec AC : sine c. 2. D: EF:: cosec DE: sine LE:: cosec E: sine DE, Viz. ic: A:: cosec B: sine AB:: cosec AB: sine ▲ B. * La Lande's Astronomy, art. 4034 to 4045; Cagnoli, page 325, &c. 3. 4D ZE rad x sine EF: cos FX sine FD, Viz. BC AB :: rad x sine ZA: COS ACX sine c. 4. 4D: 4F:: rad x sine EF: COS ZEX sine DE, Viz. BC: AC :: rad x sine A: COS AB X sine B. (H) By denoting the three sides of the triangle by a, b, and c, and their opposite angles by A, B, and c, we derive the following proportions, wherein B and c are constant quantities. 1. a: A cosec c: sine b:: cosec b: sine C. 2. a: A :: cosec B: sine c :: cosec c: sine B. (I) To find when that part of the equation of time depen dent 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. * A variety of examples will be met with in the perusal of La Lande's Astronomy, vol. III. + Simpson's Fluxions, page 550. Let Let the hypothenuse be denoted by b and the base by c. : :: when bc, then sine 2 b=sine 2 c; but when two arcs have equal sines, the one must be the supplement of the other (K. 31.). Consequently b+c=90°, therefore when b-c=0, that is when b-c is a maximum, b+c=90°. • 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°. The sun being in the first quadrant of the ecliptic. PROPOSITION X. (K) Given the parallax in altitude of a planet, to find its parallax in latitude and longitude. Let B represent the pole of the ecliptic, A the zenith, and c the place of the planet. Then will represent the parallax in altitude, B the parallax in longitude, and a the parallax in latitude. If Ec be supposed to represent a part of the ecliptic, then BE and BC will be quadrants (H. 130.) and CEB and ECB will be right angles (I. 130.). In the right-angled triangle CEA, making EA the middle part, we have a sine EA sine b × b, 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 and the pole of the ecliptic (R. 245.), we obtain the following proportion. (L) 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. sine b: sine EA=cos c :: b: a. Again, in the right-angled triangle CEA, making the ECA the the middle part, we have rad x cos ECA = tang ECX cot b; but (sine c) tang Ec = tang Ec xb (because the planet is sup posed to be in or very near to the ecliptic, sine a=rad) hence, (M) 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. tang b: tang Ec :: b: B. PROPOSITION XI. (N) 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 shewn in the preceding proposition that a=; sine b Now, if H represent the horizontal pa rallax of any planet, its parallax in altitude b= (T. 90.). By substituting this value of b in the above equations In the right-angled triangle CEA, making b the middle part, rad x cos bcos EAX COS EC; but cos b= rad x sine b tang b and sine b COS EA X sine EC tang b tang Ecx rad XH= tang ECX rad COS EA X sine Ec and by substitution B= COS EA X sine Ec rad2 × H, but Ec is the measure of the B (Z. 187.) |