To the log. sine of the longitude add the log. tangent of B, the sum, rejecting 10 ín the index, will be the log. tangent of the latitude of the same name as B. Remark. As the Tables of this collection are not marked above 180°, you must subtract 180° from the right ascension when it exceeds that quantity, and find the log. tangent and log. co-secant of the remainder; and then the arch, corresponding to the log. tangent of the longitude, is to be taken of the same affection as this remainder, and 180°, added thereto, the sum will be the longitude, unless B is greater than 90°, in which case the supplement of that sum to 360 is to be taken as observed above. EXAMPLE. By Table VIII. the right ascension of a Pegasi, July 16, 1808, was 22h. 55′ 14′′= 343° 48′ 30′, and its declination 14° 11′ N. the mean obliquity of the ecliptic 23° 27/ 47". Required its longitude and latitude? Declin. 14° 11' 0 N. tang. 9.40266 9.46295 sec. 10.15020 tang. The longitude and latitude of a celestial object being given, with the mean obliquity of the ecliptic E, to find the right ascension and declination. RULE. To the log. tangent of the latitude add the log. co-secant of the longitude, the sum, rejecting 10 in the index, will be the log. tangent of the arch A, which is to be called north or south as the latitude is. If the longitude is less than 1800, call the obliquity E north; if above 180°, south. If A and E are of the same name, take their sum, otherwise their difference, which call B, marking it with the same name as the greater number. Then add together the log. secant of A, the log. co-sine of B, and the log. tangent of the longitude, the sum, rejecting 20 in the index, will be the log. tangent of the right ascension in the same quadrant as the longitude, unless B be greater than 90°, in which case the quantity found in the same quadrant as the longitude, subtracted from 360°, will be the right ascension. To the log. sine of the right ascension add the log. tangent of B, the sum, rejecting 10 in the index, will be the log. tangent of the declination of the same name as B. Remark. If the longitude exceeds 180 you must subtract 180 from it, and find the log. tangent and log. co-secant of the remainder. The arch corresponding to the log. tangent of the right ascension is to be taken of the same affection as this remainder, and 180° added thereto will be the right ascension, unless B is greater than 90°, in which case the supplement of that sum to 360° is to be taken as was observed above. EXAMPLE. By Table XXXVII. the mean longitude of a Pegasi, July 16, 1808, was 350° 49′ 11′′, its latitude 19° 24′ 47′′ N. and the mean obliquity of the ecliptic 23° 27′ 47′′. Required its right ascension and declination? 9.54705 10.79712 targ. 9.20847 Lat. 19° 24′ 47 N. tang. tang. 9.40257 If the given longitude, latitude, and obliquity are the mean values, the resulting right ascension and declination will be the mean values, but if the proposed quantities are corrected for aberration and nutation, the resulting quantities will also be corrected. This remark is equally applicable to the preceding Problem. SPHERIC TRIGONOMETRY. Most of the rules given in the preceding Problems may be easily demonstrated by Spheric Trigonometry. As for example that of Problem XVII. may be investigated as follows. In Plate XII. Fig. 1, let A be the place of the moon, C that of the sun, CP an arch of the ecliptic, and AP a circle of latitude passing through the moon and cutting the ecliptic at right angles at P. Then the difference of longitude of the sun and moon is equal to the arch CP, and the moon's latitude is AP, whence the distance AC may be found by the rule of Napier, radius X co-s. AC = co-s. AP X co.s. CP. This in logarithms gives log. co-s. AC=log. co-s. AP+log. co-s. CP-log. radius, which is the formula made use of. Want of room prevents the insertion of the demonstrations of the methods of calculating the other Problems. The celebrated rules given by Lord Napier for solving the problems of Right-Angles Spheric Trigonometry being very easily remembered, are much made use of by mathematicians. In a paper communicated by the author of this work to the American Aczdemy of Arts and Sciences, and published in the third volume of the memoirs of that society, a method was given for the more easy application of those rules to oblique Spherie Trigonometry, and as the tables of this collection may sometimes be made use of in solving various problems of Spherics besides those given in the former part of this work, it was thought proper to insert this improved method, with the formulas most frequently made use of, to enable any person acquainted with Spheric Trigonometry to make use of the tables, without the trouble of referring to another work, for the rules. In every Right-angled Spheric triangle there are five circular parts; namely, the two legs, the complement of the hypotenuse, and the complements of the two oblique angles, which are named adjacent or opposite, according to their positions, with respect to each other. The right-angle is not included as one of the circular parts, neither is it supposed to separate the legs. In all cases of right-angled Spheric Trigonometry, two of these parts are given to find the third. If the three parts join, that which is in the middle is called the middle part; if they do not join, two of them must, and the other part which is separate, is called the middle part, and the other two opposite parts, as in Plate XII. fig. 1, 2. Then putting the radius equal to unity, the equations given by Napier will become = Sine of middle part Rectangle of the tangents of the adjacent parts. Rectangle of the co-sines of the opposite parts. The method of applying these solutions to the various cases of Right-angled Spheric Trigonometry is very simple, and is explained in several treatises. To apply the method to Oblique-angled Spheric Trigonometry, it is necessary to divide the triangle into two right-angled spheric triangles by means of a perpendicular AP (Plate XII. fig. 3, 4, 5, 14.) let fall from the point A upon the opposite side BC: the perpendicular being so chosen as to make two of the given things fall in one of the right-angled triangles, or in other words the perpendicular ought to be let fall from the end of a given side and opposite to a given angle. Each triangle thus found, contains, as above, five circular parts, the perpendicular being counted and bearing the same name in each of them: consequently the parts of each triangle similarly situated with respect to the perpendicular, must have the same name. In every case of Oblique-angled Spheric Trigonometry, there are three parts given to find a fourth, and in making use of the method of a solution by means of the perpendicular, there will in general be two of these parts in each of the triangles ACP, ABP, similarly situated with respect to each other. To each of these must be joined the perpendicular AP, and there will be three parts in each triangle, which are to be named midille, adjacent or opposite, according to the above directions. Then the equations for solving all the cases of Right-angled, and all except two cases of Oblique-angled Spheric Trigonometry are, Tangents of the adjacent parts.† α Co-sines of the opposite parts. These equations, when applied to right-angled spheric triangles, signify as before, that the sine of the middle part is equal to the rectangle of the tangents of the adjacent parts, or to the rectangle of the co-sines of the opposite parts; but when applied to an oblique-angled triangle, they signify, that the sines of the middle parts are proportional to the tangents of the adjacent parts; or that the sines of the middle parts are proportional to the co-sines of the opposite parts of the same triangle; observing that the perpendicular being common to both triangles APB, APC, and bearing the same name in each of them, must not be made use of in the analogies, nor counted as a middle part. This can produce no embarrassment, because the cases of Oblique Spheric Trigonometry may in general be solved in the shortest manner without calculating the perpendicular. The first case not included in the above rules, is where the question is between two sides and the opposite angles, which may be solved by the noted theorem, that the sines of the sides are proportional to the sines of the opposite angles, or as it may be expressed in an abridged form or more easy reference. 2. Sine side ∞ sine opp. angle. This, combined with the above improved formula, furnish a complete solution of the various cases of Spheric Trigonometry, except where three sides are given to find an angle, or (which is nearly the same thing, by taking the supplementary triangle) three angles to find a side. The above rules marked (1,) (2,) are simple in their When this can be done in two different ways (as in Cases II. IV.) it will generally produce the shortest solution to make use of that perpendicular which does not divide the required angle or side into segments. It will be of considerable assistance in remembering these rules to note that the second letters of the words tangent and co-rine are the same as the first letters of adjacent and opposite, form, and the first varies but little from that made use of by Napier, so that it is extremely easy to remember them. The case not included in these rules may be solved by one of the formulas of case V. or VI. which may be committed to memory with little trouble. To illustrate these rules, the following examples are given, which include all the cases of Oblique Spheric Trigonometry. CASE I. PLATE XII. Fig. 3, 4, 5, 14. Given AB, AC, and the opposite angle C, to find BC and the angles A, B. In the right-angled spheric triangle APC are given AC and C, and by marking it as in fig. 2, CP may be found by the rules sine mid.=tang. adj. which gives sine (co. C)= tang. CPXtang. (co. AC,) or tang. CP=co-s. CXtang. AC.* Then in the triangles ABP, ACP are given AB, AC and CP to find BP. If to these is joined the perpendicular AP it will be found that in the triangle ACP the complement of AC is the middle part (as in Fig. 3), and CP an opposite part. The triangle ABP is to be marked in a similar manner. Then the rule sine mid. ∞ co-s. opp. gives sine (co. AC): co-s. CP:: sine (co. AB): co-s. BP, and BC=BP+CP. By marking the segments as in Fig. 4, the rule sine mid. ∞ tang. adj. gives sine CP: tang. (co. C.): : sine BP : tang. (co. B.) Having found BC, the angle A may be found by the rule sine side. ∞ sine opp. angle which gives sine AB: sine C:: sine BC: sine A. Otherwise-If the side BC is not required, the angles A, B, may be found in the following manner. The rule sine mid.=tang. adj. gives by marking as in Fig. 1. sine (co. AC)=tang. (co. C) Xtang. (co. CAP) or cot. CAP. co-s. ACXtang. C, and by marking as in Fig. 5, the rule (sine mid. x tang. adj. or) tang. adj. ∞ sine mid. gives tang. (co. AC): sine (co. CAP) : : tang. (co. AB): sine (co. BAP,) then A= BAP+CAP. By marking the segments as in Fig. 14, the rule (sine mid. x co-s. S opp. or) co-s. opp. ∞ sin. zid. gives co-s. (co. CAP): sine (co. C): co-s. (co. BAP) : sine (co. B) or sine CAP: co-s. C:: sine BAP : co-s. B. Having A, C, and AB, BC may be found by the rule sine side ∞ sine opp. angle, which gives sine C sine AB: sine A: sine BC. : CASE II. Fig. 3, 4. Plate XII. Given AC, BC and the included angle C, to find AB, and the angles A, B. The rule sine mid.=tang. adj. gives as in Case I. tang. CP co-s. CXtang. AC, then BP-BC+CP and the rule co-s. opp. ∞ sine mid. gives by marking, as in Fig. 3. co-s. S CP: sine (co. AC): : co-s. BP: sine (co. AB,) and by marking as in Fig. 4, the rule sine mid. ∞ tang, adj. gives sine CP : tang. (co. C) : : sine BP : tang. (co. B.) Having found AB we may find A, by the rule sine side sine opp. angle, which gives If the angle A had been required and not B, it would have been shorter to let the perpendicular fall upon the point B, by which means the required angle A would not be divided into segments. In this case the side AB and the angle A might be found in a similar manner to that by which AB and B are found above. CASE III. Fig. 3, 4, 5, 14. Plate XII. Given the angles B, C, and the opposite side AC to find BC, AB, and the angle A. The rule sine mid. x tang. adj. gives as in Case I. tang. CP=co-s. C X tang. AC. Then the rule tang. adj. ∞ sine mid. gives, by marking as in Fig. 4, tang. (co. C.): sine CP tang. (co. B): sine B P, then BC=CP+BP. Again, the rule co-s. opp. cs sine mid. gives by marking as in Fig. 3, co-s. CP: sine (co. AC): co-s. B P: sine (co. AB.) Having found BC, the rule sine side o sine opp. angle, gives sine AC: sine B sine BC sine A. : Otherwise-The rule sine mid.=tang. adj. gives as in Case I. cot. CAP-co-8. ACX tang. C, and the rule sine mid. ∞ co-s. opp. gives by marking as in Fig. 14, sine (co. C.): co-s. (co. CAP) :: sine (co. B): co-s. (co. BAP) or co-s. C.: sine CAP. :: co-s. B: sine BAP, and A=CAP+BAP. Then the rule sine mid. c tang. adj. gives by marking as in Fig. 5, sine (co. CAP): tang. (co. AC): AB.) Having found A the rule, sine side o sine opp. :: sine A: sine BC. sine (co. BAP) : tang. (co. angle gives sine B: sine AC In putting this or any similar expression in logarithms, the radius must be neglected in the sum of the two logarithms of the second nuinber. CASE IV. Fig. 5, 14. Plate XII. C and the included side AC, to find AB, BC and the angle B. tang. adj. gives as in Case I. cot. CAP=co-s. ACXtang. The rule sine mid. ∞ tang, adj. gives by marking as in Given the angles A, S Fig. 5, sine (co. CAP): tang. (co. AC): : sine (co. BAP) : tang. co. (AB.) The rule cos. opp. ∞ sine mid. gives by marking as in Fig. 14, co-s. (co. ČAP). sine (co. C.):: co-s. (co. BAP) : sine (co. B) or sine CAP: co-s. C:: sine BAP : co-s. B. Having found B, the rule sine side ∞ sine opp. angle gives sine B: șine AC :: sine A: sine BC. If the side BC had been required and not AB, it would be shorter to let the perpendicular fall from the point C, by which means the required side BC would not be divided into segments. In this case the side BC and the angle B might be found in a similar manner to that by which AB and B are found above. CASE V. Fig. 3. Given AB, AC, and BC, to find either of the angles az A. Put S=} (AB+AC+BC,) then the angle A may be found by either of the following theorems, in which for brevity the words sine, co-sine, &c. are used for log, sine, log. co-sine, &c.. (3) Sine & A= Sine (SAB) + sine (S-AC)+co-sec. AB+co-sec. AC-20. (4) Co-s. A= 2 Sine Ssine (S-BC)+co-sec. AB + co-sec. AC-W. 2 CASE VI. Fig. 3. Given the angles A, B, C, to find either of the sides as BC. Put S= (A+B+C.) Then the side BC may be found by either of the following theorems, adapted to logarithms as in the last example. (5) Sine BC= (6) Co-sine BC= Co-sine S co-sine (S-A) + co-sec. B+co-sec. C-20. 2 Co-sine (S—B)+co-sine (SC)+co-sec. B+co-sec. C—N. 2 The above includes all the cases of Oblique Trigonometry. The 2d. and 4th. cases may be solved in a different manner by the following theorems, which on some occasions may be found very useful. Thus both the angles in Case II. may be found by the following theorems. (7) Sine (AC+BC) : sine § (BC ∞1⁄2 AC) : : cot. § C : tang. § (A—B). (8) Co-sine (AC+-BC): co-sine (BC AC): cot. C: tang. (A+B). (A-B) is less than 90° and § (A+B) is of the same affection as } (AC-BC), The sum and difference of the terms (A-B) and § (A+B) will give A and B. Both the sides in Case IV. may be found thus: (9) Sine (A+C): sine (A C:: tang. AC: tang. (BC AB). (10) Co-sine (A+C): co-sin. (AC) : : tang, AC: tang. (BC÷AB.) | (BC AB) is less than 90°, and 4 (BC+AB) is of the same affection as (A+C.) Then the sum and difference of § (BC 1⁄2 AB) and § (BC+-AB) give AB and BC. The improved rule for solving the cases of Oblique Spheric Trigonometry by the cir cular parts, may be easily deduced from those given by Lord Napier. For if we put M for the middle part, A for the adjacent part, and B for the opposite part of the triangle APC (Fig. 3, 4, 5, 14, Plate XII.) m, a, b, for the corresponding parts of the triangle APB; and P for the perpendicular AP. Then if P is an adjacent part, the rules of sine M sine m sine M sine m and tang. P tang. A, sequently sine M: tang. A:: sine m. tang. a. sine M sine m, and co-s. P co-s. b, Napier will give tang. P will give co-s. P—————— hence tan. ", con tang. a, tang. « If P is an opposite part, the same rule sine M sine m hence co-s. B co-s. b, -consequently co-s. B sine M: co-s. B:: sine m: co-s, b, which are the two rules to be demonstrated. APPENDIX TO THE SIXTH EDITION. ON FINDING THE LATITUDE BY TWO ALTITUDES. SINCE the part of this work for the finding the Latitude by two altitudes was in the press, the following Table XLVIII. has been computed, by means of which the correction of either one of the observed altitudes can be computed for the change of declina. tion of the observed object during the elaped time between the observations, and thus the Problems of double altitudes of the sun, moon, planet, or fixed star, can be reduced to the case of the declination, being invariably the same as at the time of the observation of the altitudes which is not corrected, and then the Problem comes under the first (or second) method of solution, which is much more simple and free from cases than the general solution by the third method. This process of correcting the altitude is somewhat similar to that before taught, for making allowance for the run of a ship during the time elapsed between the observations; and the same altitude, which is corrected for the run of the ship, can also be corrected for the change of declination. This method of correcting one of the altitudes is particularly applicable to the case where both observations are made on the same heavenly body, and the declination does not vary but few minutes, or in extreme cases more than one or two degrees; but the same process may be used when two different objects are observed, provided their declinations are nearly equal, or do not differ more than one or two degrees. As cither one of the altitudes may be corrected, the Problem admits of two different ways of solution. For the sake of precision, the altitude which is selected to be corrected, will be called the first altitude; and the corresponding declination, the first declination ; the other altitude, which is not corrected, will be called the second altitude, and the corresponding declination, the second declination. These terms, first and second, having no reference to the order in which these observations are taken, since the altitude here defined as the first altitude, may be actually observed either before or after the other obser vation The proposed table gives for various declinations, altitudes, and latitudes, the change of the first altitude, corresponding to a variation of 100" in the first declination. Thus, with the latitude 50° N. the sun's altitude 30°, and the declination 14° N. the Table gives 77" for the variation of that altitude arising from a change of 100" in the declination. If the actual change of declination is greater, or less than 100" the tabular number 77" must be increased or decreased in the same proportion. Thus, if the change of declination be 200′′, the change of altitude will be 200′′ × 77=154". If the change of declination be 60′′, the change of altitude will be 60 × 7746'. The correction of this first altitude having been found, it is to be applied to the first altitude, corrected as usual, for dip, refraction, semi-diameter and parallax, and the corrected first altitude will be obtained, such as it would have been, if the declination at the time of observing that altitude had been equal to the second declination. With this corrected first altitude, the second altitude and second declination without correction, and the observed elapsed time, or hour angle, the computation of the latitude may be made by the First Method, explained in page 133. This Table is calculated for every 2° of declination, from 0° to 26°. If the change of declination is not very great during the elapsed time, it will in general be sufficiently exact to enter the table with the nearest declination, and take proportional parts for the degrees of altitude and latitude. The latitude by account is to be used in finding the numbers from this table, it being sufficiently accurate, since an error of 1° of latitude rarely produces more than 2" change in the numbers of the Table. Suppose now, that the tabular number was required, when the latitude was 37° N. the first altitude 290,the tfirst declination 6° 25′ S. In this case, using the declination 6o, and the altitude 20°, he tabular numbers corresponding to the latitudes 30° S. and 40° S. are, respectively, 57′′ and 73′′, whose difference 16′′ corresponds to a change of 10o of latitude, and by TT Tab. |