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 TREATISE. 
 ON 
 
 PLANE AND SPHERICAL 
 
 TRIGONOMETRY: | 
 
 WITH THEIR MOST USEFUL PRACTICAL APPLICATIONS. 
 
 BY JOHN BONNYCASTLE, 
 
 PROFESSOR OF MATHEMATICS IN THE ROYAL MILITARY 
 ACADEMY, WOOLWICH. 
 
 THE SECOND EDITION, 
 
 > 
 
 CORRECTED AND IMPROVED, 
 
 LONDON: 
 
 PRINTED FOR C. LAW; CADELI AND DAVIES; B. AND R. CROSBY 
 AND CO.; JOHN RICHARDSON; CRADOCK AND Joy; 
 AND JOS. JOHNSON AND GO. 
 
 PS13." 
 
INTRODUCTION. 
 
 At what period Trigonometry first began to be culti- 
 vated, as a branch of the mathematical sciences, is ex- 
 tremely uncertain, no records having been left by the 
 ancients, which enable us to trace it to a higher age 
 than that of Hipparchus, who flourished about a cen- 
 tury and a half before Christ, and is reported by Theon, 
 in his Commentary on Ptolemy’s Almagest, to have 
 written a work, in twelve books, on the chords of cir- 
 cular arcs, which, from the nature of the title, must 
 evidently have been a treatise on Trigonometry. 
 
 But the earliest work on the subject, now extant, is 
 the Spherics of Theodosius, a native of Tripoli in 
 Bythinia, who, soon after the time above mentioned, 
 collected the scattered principles of the science, which 
 had been discovered by'his predecessors, and formed 
 them into a regular treatise, in three books, containing 
 a variety of the most necessary and useful propositions 
 relating to the sphere, arranged and demonstrated with 
 great perspicuity and elegance, after the manner of 
 Euclid’s Elements (a). 
 
 The next of the Greek writers, after Theodosius, 
 
 (a2) This work of Theodosius, which came to us through the 
 medium of an Arabic version, has been published both in Greek 
 and Latin by several writers; but the Latin edition of Dr. Barrow, 
 8vo. London, 1675, and that of Hunt, Svo. Oxford, 1709, are 
 
 2 , 
 
V4. 
 
 who has treated professedly on this subject, is Mene+ 
 laus, an astronomer and mathematician of considerable 
 eminence, who ‘lived‘about the middle of the first cen- 
 tury after Christ, and of whom we have three books 
 on Spherical Triangles, containing, besides the first 
 
 principles of the science, a number of propositions of a. 
 
 more difficult kind, which at that time were but little 
 known ; but the six books which he is said to have 
 written on the subtenses, or chords, of circular arcs, 
 being: probably a treatise on the ancient method of con- 
 structing trigonometrical ‘tables, has not been trans- 
 mitted to our times (6). | 
 
 This loss, however, has been, 1 in some measure, re- 
 paired by Ptolemy, who in the first book of ‘his Alma- 
 gest, published about the beginning of the second cen- 
 tury after Christ, has given us a table of arcs and their 
 chords to every half degree of the semicircle ; in the 
 forming of which it is observable, that he divides the 
 radius, and the arc whose chord is equal to the radius, 
 each into sixty equal parts, and then estimates all other 
 arcs by sixtieths of that arc, and the chords. by sixti- 
 
 reckoned the best. The third book, which is the most dificult, 
 has been commented upon and elucidated by Pappus, in his Ma- 
 thematical Collections. 
 
 (2) A translation of the Spherics of Menclaus had been under- 
 taken by. Regiomontanus, but was first published by Maurolycus 
 in Latin (Messane, 1558, fol.), together with the Spherics of 
 Theodosius and his own. Halley also prepared a new edition of 
 this work, corrected from a Hebrew manuscript, which was pub- 
 lished in Svc. 1758, without the preface which he had projected 
 for it, by Costard, the author of a History of Astronomy. 
 
 OE 
 
‘the Greeks in the sexagesimal division of the radius, 
 
 vil 
 
 ‘eths of that chord, or of the radius; being probably 
 the method used by Hipparchus and other ancient 
 writers on'this subject. He has here also proved, for 
 the firsttime that we Know of, that the rectangle of 
 ithe two diagonals of any quadiilateral inscribed ina 
 circle, as. equalito:the sumof the rectangles of its op- 
 ‘posite sides (c). 
 
 After the time of Ptolemy and his commentator 
 Theon, little more is known on this subject till about 
 the close of the eighth century after Christ, when the 
 ancient method of computing by the chords was changed 
 ‘for that of the sines, which were first introduced into 
 the science by the Arabians; to whom we are also in- 
 debted for the several axioms and theorems which are 
 at present considered as the foundation of our modern 
 Trigonometry, ‘as well as'for some other propositions 
 which such an alteration ‘in the system naturally re- 
 quired. 
 
 The Arabians, however, though they had been long 
 acquainted with the Indian, or decimal scale of arith- 
 metic, now used, do not appear'to have deviated from 
 
 (¢) Claudius Prolemeus was born at Prolemais in Egypt, and 
 taught astronomy at Alexandria, where he died in the year of 
 Christ 147, being the 78th year of his age. His Almagest, like 
 most of the celebrated works’ of antiquity, has had many editors 
 and commentators ; but a ‘good translation, both of this work 
 and the Commentary of Theon, is still much wanted ; the only 
 
 ‘tolerably complete ‘Latin edition (published at! Basil, 1551,) 
 ‘which we now possess, being that of George of Trebizonde, which 
 
 was so severely and justly criticised by Regiomontanus. 
 a2 | 
 
vill 
 
 which continued in use till about the middle of the 
 15th century, when an alteration was first made by 
 Purbach, a native of a small place of that name be- 
 tween Austria and Bavaria; who constructed a table 
 of sines to a division of the radius into 600000 equal 
 parts, and computed them for every ten minutes, or 
 sixth part of a degree, in parts of this radius, by the 
 decimal notation, | 
 
 This project of Purbach was also still further pro- 
 secuted by his disciple and friend John Muller, com- 
 monly called Regiomontanus, of the little town of 
 Mons Regius, or Konigsberg, in Franconia, who first 
 began his mathematical career by extending and im- 
 proving the tables of his master; but, afterwards, dis- 
 liking that plan, as evidently imperfect, he computed 
 a table of sines, for every minute of the quadrant, to 
 the radius 1000000. He also introduced the tangents 
 into this science, and enriched it with many theorems 
 and precepts, which, except for the use of logarithms, 
 renders the trigonometry of this author but little in- 
 ferior to that of our times (d). 
 
 Soon after the period here.mentioned, several other 
 mathematicians also contributed to the advancement of 
 this science, either by some useful alterations in the 
 
 (¢) The Treatise of Regiomontanus, on Plane and Spherical 
 Trigonometry, in five books, was written about the year 1464, — 
 and printed in folio at Nuremberg, 1533. In the fifth book, — 
 some of the problems relating to plane triangles are resolved by 
 means of Algebra, a proof that this science was known in Europe — 
 before the treatise of Lucas de Burgo appeared. : 
 
ix 
 
 form of the tables, or by other improvements; among 
 whom may be reckoned John Werner of Nuremburg, 
 and Nicholas Copernicus of Thorn, in Prussia, the ce- 
 lebrated restorer of the true system of the world, who 
 wrote a brief treatise on plane and spherical Trigono- 
 metry, with a description and construction of the ca- 
 non of chords, nearly in the manner of Ptolemy ; 
 which tract, together with a table of sines and their dif- 
 ferences, for every 10 minutes of the quadrant, to ra- 
 dius 100000, is inserted in the first book of his Revo- 
 lutiones orbium caelestium, published in folio, at Nu- 
 remburg, 1543. 
 
 To these cultivators and improvers of the science, 
 we may likewise add Erasmus Rheinold, professor of 
 mathematics in the academy of Wurtemburg, who 
 published his Canon feecundus, or Table of tangents, 
 in 1553; and Maurolycus, abbot of Messina, in Si- 
 cily, one of the most able geometers of the sixteenth 
 century, to whom we are indebted for the Tabula 
 Lenefica, or Canon of secants, which came out about 
 the same time. | 
 
 But the most complete work on the subject, which 
 had hitherto appeared, was a treatise, in two parts, by 
 Vieta, printed in folio at Paris, 1579, during the au- 
 thor’s lifetime, in the first part of which, entitled Ca- 
 non mathematicus seu ad triangula, cum appendicibus, 
 he has given a table of sines, tangents and secants for 
 every minute of the quadrant, to radius 100000, with 
 their differences ; and towards the end of the quadrant, 
 the tangents and secants are extended to 8 or 9 places 
 
x 
 
 of figures. ‘They are.also arranged like our tables at 
 present, increasing from the left-hand side to 45°, and 
 then returning backwards, from the right hand, to 90°; 
 so that each number and its complement stand on, the 
 same line. 
 
 The second part of the volume, which. is entitled; 
 Universalium inspectionum ad canonem mathemati- 
 cum liber singularis, contains, besides a regular ac- 
 count of the construction of the tables, a compendious. 
 treatise on plane and spherical ‘Trigonometry, with their 
 application to a variety of curious subjects im geome- 
 try, mensuration, and other branches of mathematics 5: 
 as also a number of particulars relating to the quadra- 
 ture of the circle, the duplication of the cube, and simi- 
 lar problems; which are all treated of ina manner 
 worthy the genius of the author (e). 
 
 Beside the performance above mentioned, there are, 
 likewise, several other smaller tracts on trizonometrical 
 subjects in the general. collection of Vieta’s works, pub- 
 lished at Leyden in 1646, by Schooten; among which 
 are the curious theorems, here first given by our au- 
 thor, relating to angular sections, or the multiples and 
 submultiples of arcs; as also general formule for the 
 chords, and consequently sines, of the sums and dif- 
 ferences of arcs, and such as are in arithmetical pro- 
 gression; which have since been so extensively and 
 
 (e) This curious performance, which was published sepa- 
 rately from the other works of Vieta, and without his name, 
 is extremely scarce, few copies of it having ever reached this 
 country. 
 
XI 
 
 usefully applied, both in this science, and in some of 
 the higher branches of the modern analysis Cry: 
 
 The next writer on this subject, whose labours de- 
 serve particular notice, is George Joachim Rheticus, 
 a pupil of Copernicus, and professor of mathematics 
 in the university of Wittemburg, who formed the de- 
 sign of computing the trigonometrical canon, for every 
 ten seconds of the quadrant, to 15 places of decimals ; 
 and though he was prevented, from executing the 
 whole of this laborious enterprise, he never theless ac- 
 complished that part of it relating to the sines and co- 
 sines; all of which he calculated according to his ori- 
 ginal plan, besides those of every single second of the 
 first and last degrees ; ; but was deterred from publish- 
 ing the work, on account of the expense attending 
 the impression. 
 
 Soon after his death, however, which happened in 
 1576, Valentine Otho, one of his disciples and friends, 
 engaged, according to the dying request of Rheiicus, 
 to finish this great undertaking ; and seNrarigrese iy: 
 a variety of difhculties and obstacles, which eiates 
 his labours, he at length gave it to the public, under 
 the title of Opus Palatinum de Triangulis (Hecidel- 
 bergze, folio, 1596): in which work, we have, for 
 the first time, an entire table of sines, tangents ‘and 
 
 ot f ) The demonstrations of most of the trigonometrical t theo- 
 rems in this work, relating to angular sections, were supplied by 
 Alexander Anderson, at that time professor of mathematics at 
 Paris, but a native of Aberdeen in Scotland, 
 
XU 
 
 secants, for every ten seconds of the quadrant, to ten 
 places of decimals, with their differences; the five last 
 figures of the former ‘Table being here omitted. 
 
 But as this performance, though highly valuable in 
 other respects, was afterwards found to contain a num- 
 ber of errors, particularly in the cotangents and co- 
 secants, which the sines that Otho had employed were 
 not sufficiently extensive to prevent, Bartholomew 
 Pitiscus, an able mathematician of that time, under-. 
 took the revision of it. And having procured, with. 
 some difficulty, the original manuscript of Rheticus, 
 he added to it an auxiliary table of sines of small arcs, 
 to 21 places of decimals, for the purpose of supplying 
 the defect above mentioned, and published the work, 
 with the augmentations he had made to it, under the 
 name of Thesaurus Mathematicus, &c. Gal 
 folio, 1613. ) | 
 
 This being done, he now recalculated, by means of 
 the materials he had thus prepared, the cotangents and 
 cosecants of the Opus Palatinum of Otho, to the end of 
 the first six degrees of the quadrant; and as this ren- 
 dered the work sufficiently exact for all astronomical 
 purposes, even to fractions of seconds, he published 
 the corrections in separate sheets, in 86 pages in folio, 
 for the purpose of replacing those of the former im- 
 pression: But the original work having been partly 
 sold off, and its purchasers neglecting to procure the 
 new sheets, these corrected copies are become so ex- 
 tremely rare, that few of them are to be found, either 
 
xi 
 
 in the possession of individuals or in the public libra- 
 ries (g). | 
 About the close of the 16th century, several other 
 persons also wrote on the subject of Trigonometry, and 
 the construction of the triangular canon; among whom 
 may be reckoned Philip Lansberg, a native of Zealand 
 in Holland, who in 1591 published his Geometria 
 Triangulorum, in four books, with the usual tables; 
 being the first compendium of this kind in which the 
 tangents and secants are continued to the end of the 
 last degree of the quadrant, to 7 places of decimals. 
 The Trigonometry of Pitiscus, also, which was pub- 
 lished at Francfort in 1599, is a very complete work, 
 having been long considered, both with respect to the 
 
 (zg) For a more detailed account of these valuable works, see 
 a paper by Prony, in the Mémoires de I’ Institut, vol. v., where 
 he says that he knows of only two of the corrected copies of the 
 Opus Palatinum that are now to be found; one being in the li- 
 brary of the Council of State at Paris, and the other that pur- 
 chased by himself of the bookseller Duprat; both of which 
 can be easily distinguished from the old work, by the difference 
 of the colour of, the paper and type, in aie, sheets that are 
 changed. 
 
 The title of the corrected copies is as follows: 
 
 Georgii Joachimi Rhetici Magnus Canon Doctrine trian- 
 gulorum ad decades secundorum scrupulorum, et ad partes 
 10000000000. _ 
 
 Recens emendatus 4 Bartholomeo Bitte silesio. Addita est 
 brevis commonefactio de fabrica et usu hujus canonis, &c. 
 
 Canon hic una cum brevi commonefactione de ejus fabrica et 
 usu, etiam separatim ab opere peatne venditur. In Bibliopoleio 
 harnischiano. ‘ 
 
X1V: 
 
 cortectness of the tables, and its numerous practical 
 applications, as the most commodious and useful trea- 
 tise'on the subject then, extant. 
 
 To these writers may likewise be added Christopher 
 Clavius, a German Jesuit, who, in the first volume of 
 his works, printed at Mentz, 1612, in 5 vols. fol., has 
 given us an ample and circumstantial treatise on Trigo- 
 nometry, with tables of sines, tangents and seconds, for 
 every minute of the quadrant, to 7 places of decimals, 
 and ina form continued forwards to the end of the 
 quadrant. The sines have also their differences set 
 down to every second, and the construction of the 
 tables is clearly explained, according to. the methods of 
 Ptolemy, Purbach, and Regiomontanus. 
 
 About the year 1600 Ludoph van Ceulen, a Dutch 
 mathematician, of considerable talents, also. published 
 his well-known treatise De Circulo et adscriptis, in 
 which he treats of the properties of lines drawn in and 
 about a circle, and especially of chords, or subtenses, 
 with the construction of the canon of sines. He here, 
 likewise, determined the ratio of the diameter ofa civ 
 cle to its circumference, to 36 places of figures; show- 
 ing that if the diameter be 1, the circumference will be 
 3.14159 26535 89793 23846 26433 83279 50288 
 &c.; which ratio, in imitation of the example of Ar- 
 chimedes, 1s said to have been engraved, by his order, 
 on his tombstone in the churciyard at Leyden. 
 
 This curious tract, with some other of Ceulen’s dis- 
 sertations on similar subjects, was translated into Latin, 
 and. published at Leyden, in 1619, by Willebrord 
 
xV 
 
 Snell; who has himself given, in his Doctrine trian.. 
 gzulorum canonice, the construction.of the sines, tan- 
 gents and secants, together with.a very. useful synopsis. 
 of the calculation of triangles, both plane and spherical. 
 
 Francis van Schooten also published, at Amsterdam, 
 in 1627, a table of sines,.tangents and secants, in a 
 small neat form, for every minute of the quadrant, to: 
 7 places of figures, which has a great character for 
 accuracy, being declared, by its author, to be without a 
 single error; though this must not be understood of | 
 the last figure of the numbers, which 1s sometimes. er- 
 roneous in excess, and sometimes in defect, by not 
 being always set down to the nearest unit. 
 
 These are the principal writers on.'Trigonometry, and 
 the tables of sines, tangents. and secants, before the 
 change that was made in the subject by the introduce 
 tion of the logarithmic calculus, which first began to 
 be employed in this science about the commencement 
 of the 17th century, by its celebrated inventor Baron 
 Napier, of Merchiston, in Scotland; who, in the year 
 1614, published his work entitled Mirificit logarith- 
 morum canonis descriptio (A), which contains the 
 
 (4) The principles of logarithms, and the method of comput- 
 ing the tables, are not given by Napier in this performance, but 
 were afterwards published by his son, Robert Napier, who in the 
 year 1619 gave a new edition of his father’s work, together with 
 the Logarithmorum canouis constructio, and other pieces; in which 
 performance it may be observed, that the geometrical method, 
 used by Napier in computing his logarithms, is similar to that 
 which was afterwards employed by Newton in the generation of 
 smagnitudes, in his doctrine of fluxions. 
 
XVI 
 
 logarithms of numbers, and the logarithmic sines, tan-° 
 gents and seconds, for every minute of the quadrant, 
 together with the description and use of the tables. 
 
 _ But as the text, or descriptive part of this work,, 
 was in the Latin language, it was soon afterwards trans- 
 lated into English by Mr. Edward Wright, the inven- 
 tor of the principles of what has been usually, though 
 erroneously, called Mercator’s Sailing; who having” 
 finished his manuscript, sent it to Edinburgh to be re- 
 vised and improved by the author; but on his dying a 
 short time after he had received it back, it was pub-. 
 lished, with a preface by Briggs, in the year 1616, by 
 his son Samuel Wright, together with the tables, but 
 each number to one figure less than in the original. 
 Soon after this, several other logarithmic tables, of, 
 a kind nearly similar to those of Napier, were publish-. 
 ed by Speidell, Ursinus, and Kepler ; but being. in ge- 
 neral very compendious, and formed upon principles 
 which have since been found incommodious in practice, 
 they are now chiefly curious on account of the ideas 
 and artifices displayed by their authors in their different 
 modes of computing them ; in which respect the per- 
 formance of Kepler, though frequently abstruse and 
 obscure, is particularly deserving of notice, both from 
 the originality of his plan, and the able manner in 
 which it is developed (7). 
 
 (1) The treatise of Kepler here mentioned, is entitled Chilias lo- 
 garithmorum ad totidem numeros rotundos, premissa demonstratione 
 legitima ortus logarithmorum eorumque usus, &c. (Marpurg, 1624.) 
 To which performance, the year following, he added a supple- 
 
XVii 
 
 ‘The person, however, to whom we are chiefly in- 
 debted for the new and more advantageous form which 
 this admirable. mode of computation has since assumed, 
 is Mr, Henry Briggs, at that time professor of geome- 
 try in Gresham College, London, and afterwards Sa- 
 vilian professor at Oxford; who, besides his eminent 
 talents as a mathematician, has the merit of having first 
 proposed, both to the public in his lectures, and to the 
 illustrious inventor of the doctrine himself, that happy 
 improvement in the system of these numbers, which 
 consists in making the radix of the system 10, instead 
 of 2.71828182845, &c. as was done by Napier; or, 
 which is the same thing, by changing them from what 
 are usually called hyperlolic, or Napierian logarithms, 
 to the present common or tabular logarithms. 
 
 It may here likewise be further remarked, that Briggs, 
 in addition to his other attainments, was also a most 
 indefatigable calculator, having laboured from the be- 
 ginning, with great zeal and diligence, at the compu- 
 tation of logarithms of this kind, of which he wasthe 
 inventor and promoter. And, as the first fruits of his 
 industry, he produced, in 1624, his Arithmetica Lo- 
 garithmica, a stupendous work for so short a time ; 
 which contains the logarithms of all numbers from 1 
 
 ment, containing the logarithms of integer numbers, and of such 
 of the natural sines as nearly coincide with them 
 
 It may also be observed, that the work of rithors above men- 
 tioned, entitled Trigonometria (Cologne, 1624), is not unworthy 
 of attention, as containing a table of natural sines and their loga- 
 rithms, of the Napierian form, to every 10 seconds of the qua- 
 drant; which the author had been at great pains in computin®? 
 
XVII 
 
 to 20000, and from ‘90000 to 100000 to 15 places 
 of figures, besides the index. " 
 
 The table, however, being imperfect, the ‘remaining 
 logarithms were ‘soon afterwards supplied by Adrian 
 Vlacq, of Gouda, in Holland, who completed the '70 
 intermediate chiliads, and republished the 4rithmetica 
 Logarithmica at that place, with these additional num- 
 bers, in 1627 and 1628, in which'state it contains the 
 logarithms of all numbers, from 1 to 100000, to 10 
 places of decimals, together with a table of logarithmic 
 sines, tangents and secants, to the same extent, for 
 every minute of the quadrant. 
 
 To this we may also add, that beside the work above 
 mentioned, Briggs lived to complete a table of loga- 
 rithmic sines and tangents for the hundredth part of 
 every degree, to 14 places of decimals, together with 
 a'table of natural sines for the same parts to 15 places, 
 and the tangents and secants of the same to 10 places, 
 with the’construction of the whole; which work was 
 likewise’printed at Gouda, by Vlacq,'in 1633 ; and on 
 his death, a preface to it was supplied by Mr. Henry 
 Gellibrand, at that time professor of astronomy in Gre. 
 sham College, who also added to it the application of 
 logarithms to ‘plane and spherical trigonometry, and 
 published it the same year, under the title of Trigono- 
 metria Britannica. 
 
 These’ two performances of Briggs also contain, be- 
 ‘sides the extensive tables above mentioned, and the me- 
 thod of constructing them, a variety of other matters 
 of.great utility and importance in this science; among 
 
xix 
 
 the most remarkable of which may be mentioned, the 
 method of interpolating logarithms by their differences, 
 as afterwards treated of by Cotes, in his Canonotechnia, 
 and the proof ‘he has given, in his chapter on angular 
 sections, of the curious property, that the sines of equi- 
 different arcs, with their 2d, 4th, 6th, &c. differences, 
 and the cosines of the mean arcs, with their Ist, 3d, 
 Sth, &c, differences, are in geometrical progression (A). 
 
 In thesame year, likewise, and during the time that he 
 was superintending the printing of the Zrigonometria 
 Britannica of Briggs, Vlacq published, at Gouda, his 
 own great work, entitled Trigonometria Artificialis, 
 which contains the logarithmic sines and tangents of 
 every 10 seconds of the quadrant, to 10 places of 
 figures, besides the index; and the logarithms of the 
 first Z0COO numbers, to the same number of : places, 
 with the differences of each; the whole being preceded 
 by an ample description of the tables, and the:applica- 
 tion of them to some of the principal problems in plane 
 and spherical ‘Trigonometry (/). 
 
 (%) Besides what relates more immediately to trigonemetrical 
 subjects, Briggs has shown, in his 7rigonometria Britannica, the 
 method of generating the coefficients of the terms of any integral 
 power of a binomial, successively from each other, independently 
 of any other power ; a property which Newton afterwards exhi- 
 
 ited in the form of a general theorem, algebraically expressed, 
 ‘and serving for all kinds of powers or roots, whethér integral or 
 fractional. 
 
 (/) A ‘new edition of the Trigonometria Artificialis of Viaeq, 
 “which has always been considered as'a work of great use to’ As- 
 ‘tronomers, has been lately published at Leipzig, by ‘Wega, under 
 the title of Thesaurus Mathematicws. 
 
AX 
 
 Several smaller tables of these logarithins were also 
 ‘published about the same time by Gunter, Wingate, 
 Roe, and others; the two latter of whom considerably 
 improved their form and disposition. Gunter is like- 
 wise deserving of notice, from his having first applied 
 the logarithms of numbers, together with those of the 
 sines and tangents, to a ruler, in the form of a two-foot 
 scale, that still goes by his name : by which proportions 
 in trigonometry, navigation, and other subjects, may 
 be performed, to a degree of accuracy sufficient for 
 many practical purposes, by the mere application of a 
 pair of compasses; being a method founded on the 
 well-known property, that the logarithms of the terms 
 of equal ratios are equidifferent (7). 
 
 But the common logarithmic canon was first reduced 
 to its most convenient form by John Newton, in his 
 Trigonometria Britannica, printed at London in 1658, 
 which work contains the logarithms of the first 100000 
 numbers, to 8 places of decimals, besides the index, 
 arranged in the same manner as they are in our best 
 tables at present ; as also the logarithmic sines and 
 tangents to the same extent, for every 100dth part of 
 a degree, with their differences, and for the 1000dth ° 
 part in the first three degrees, according to the decimal 
 division of Briggs. 
 lege at the time that Briggs was professor of geometry there, i is 
 also said to have first introduced the use of arithmetical comple- 
 
 ments into logarithmic computations, and to have been the in- 
 
 ventor, or at least to have first started the idea, of the logarithmic 
 curve. 
 
XXi 
 
 The greater part of these tables, however, have since 
 been, insome measure, superseded ‘by those of a later 
 date; among the most accurate and convenient of 
 which, for common use, may be reckoned the edition 
 of Vlacq’s small volume of tables, printed at Lyons 
 in 1670, and another work of this kind, printed at the 
 same place, in 1760; but more particularly by the edi- 
 tion of Sherwin’s Mathematical Tables, in 8vo. 1742, 
 as revised by Gardiner ; also Hutton’s Mathematical 
 Tables, in 8vo,, first printed in 1785; the Tables of 
 Vega, 2 vols. 8vo., printed at Leipzig in 1797; and 
 the 1st edition of the Tables Portatives de Logarithmes 
 of Callet, in small 8vo., printed at Paris 1783 (n); all 
 of which are adapted to the sexagesimal division of the 
 circle, used by Vlacq and most of the later compilers. 
 Besides these, several other tables, of a different kind . 
 have been lately published by the French; in which 
 the quadrant is divided, according to their new system 
 of measures, into 100 degrees, the degree into 100 
 minutes, and the minute into 100 seconds; the prin- 
 cipal of which are the 2d edition of the Tables Porta- 
 tives of Callet, beautifully printed in stereotype, at 
 Paris, by Didot, 8vo. 1795, with great additions and 
 improvements; the Trigonometrical Tables of Borda, 
 in 4to. an. ix, revised and enriched with various new 
 NS MEE Tees alll» & as oo ANAL UG a ee 
 (x) This neat portable work, which is now become extremely 
 scarce, contains all the tables in Gardiner’s 4to vol. hereafter men- 
 tioned, with several additions and improvements ; and is, by far, 
 
 the most useful and convenient performance of the kind that has 
 yet been offered to the public. 
 
 b 
 
xxii 
 
 precepts and formule by Delambre; and the tables 
 lately published at Berlin, by Hobert and Ideler, which 
 are also adapted to the decimal division of the circle, 
 and are highly praised for their accuracy by the French 
 computers. | | 
 
 Among the various tables, however, of the sexage- 
 nary kind, none have been more esteemed for their 
 usefulness and accuracy than those of Gardiner, print- 
 ed in 4to. at London, in 17423; which contain the lo- 
 garithms of all numbers from 1 to 102100, and the 
 logarithmic sines and tangents for every ten seconds of 
 the quadrant, to 7 places of decimals, with several other 
 necessary tables; a new edition of which work was also. 
 printed at Avignon, in France, in 1770, under the 
 care of Pezenas, who added to it the sines and tangents - 
 of every single second, for the first 4 degrees, and a 
 small table of hyperbolic ee taken from Simp- 
 son’s Fluxions. 
 
 But of all the trigonometrical tables hitherto pub-— 
 lished, the most extensive and best adapted for obtain- 
 ing accurate results, in many delicate astronomical and 
 geodetical observations, are those of ‘Taylor, printed 
 in large 4to. at London, 1792; which contain the lo- 
 garithms of the first common numbers from 1 to | 
 1260, to eight places of decimals ; the logarithms of all” 
 numbers from 1 to 101000, to 7 places; and the loga- 
 rithmic sines and tangents of every second in the qua- 
 drant, to 7 places; as also a preface, and various pre- 
 cepts for the explanation and use of the tables, which, 
 from the author’s dying before the last sheet of his 
 
} 
 f! 
 
 XXiil 
 
 work was printed off, were supplied by Dr. Maskelyne, 
 the astronomer royal. 
 
 It may here likewise be observed, that besides the 
 common tables hitherto mentioned, which contain the 
 logarithms of numbers in their usual order, others, of a 
 different kind, have been constructed, for the more rea- 
 dily finding the number corresponding to any given 
 logarithm ; of which the principal one, of any consi- 
 derable extent, is the Antilogarithmic Canon of Dod- 
 son, published at London, in 17423; which contains 
 the numbers corresponding to every logarithm, from 
 1 to 100000, to eleven places of figures, with their 
 differences and proportional parts; and, though little 
 used at present, is a performance of great labour and 
 merit (0). 
 
 In consequence also of the decimal division of the 
 circle, now generally used by the French mathemati- 
 cians, anumber of persons have been employed, for 
 several years past, at the Bureau de Cadastre, at Paris, 
 under the direction of Prony, in computing new tri- 
 gonometrical tables of this kind, to a far greater extent 
 than any that have hitherto been devised. But not- 
 withstanding the work appears to have been nearly 
 completed a considerable time since, it has not yet been 
 
 offered to the public : which is much to be regretted ; 
 
 (0) Dr. Wallis informs us, in the 2d vol. of his mathematical 
 works, that an antilogarithmic canon was begun by Harriot, the 
 algebraist (who died in 1621), and finished by Warner, the editor 
 of his works, about the year 1640; but which was lost for want 
 
 of encouragement to printit.. 
 
 bg 
 
XXIV 
 
 as these Tables, though too bulky and high priced to 
 be brought into common use, might be very advanta- 
 geously consulted, in many points of delicate calcula- 
 tion, which are otherwise not easily determinable. 
 Our common tables could likewise be corrected from 
 
 them, or new ones published under an abridged form; , 
 
 it is therefore to be hoped, that this great monument 
 of calculation will soon make its appearance, under 
 the auspices of a government which declares itself to 
 be ’ami des Arts et des Sciences (p). 
 
 To this brief account of the works of some of the 
 most celebrated writers on the subject here treated of, 
 and the tables which, at different times, have been 
 composed for facilitating its practical operations, it may 
 be proper to subjoin alight sketch of the improve- 
 ments which it has undergone in passing through 
 the hands of the later analysts, who, by means of a 
 more commodious algorithm, and the resources of a 
 ready and comprehensive calculus, unknown to their 
 predecessors, have greatly enlarged the boundaries of 
 the science, and simplified its rules and processes. 
 
 These advantages, and the consequent discoveries 
 
 which attended them, have chiefly arisen from thenew ~ 
 
 views of the subject that had been opened to mathe- 
 maticians, by the theorems, first given by Vieta, for the — 
 
 chords of the sums, differences, and multples of arcs | 
 
 (p) Fora detailed account of the contents of this great work, * 
 
 and the manner in which it was computed, see the Report of De- 
 lambre, Mémoires de nc inna vol, v, 
 
 qe 
 
XXV 
 
 and their supplements; which though left without de- 
 monstration, and, in the latter case, probably formed 
 by induction from the law of the terms and their co- 
 efficients, have, nevertheless, been the germ of most 
 of the numerous and elegant formule which have since 
 enriched this branch of the subject. | 
 
 Weare also, in this respect, no less indebted to Na- 
 pier, not only for his admirable discovery of logarithms, 
 but for the new and excellent analogies which he in- 
 troduced into that part of the science relating to the 
 solution of spherical triangles, which still go by his 
 name; as likewise for his other well known rules, 
 called the Five Circular Parts (¢) : which, though too 
 artificial and restricted in their application, to be em- 
 ployed by skilful computists, in the present advanced 
 state of the science, are sufficient proofs of the dexterity 
 and address with which he investigated every branch of 
 a subject so intimately connected with the invention 
 that has gained him such just celebrity. 
 
 The works of Briggs, already mentioned, also greatly 
 contributed to the advancement of this branch of the 
 
 (g) It isan exaggerated account of the excellence of the two 
 rules here mentioned, to assert, as some writers have done, that 
 the bare recollection of them, alone, is sufficient to enable a per- 
 son to carry the whole of spherical Trigonometry in his mind ; 
 for, besides their not being applicable to the two cases in which 
 the 3 sides, or the 3 Z*, ofa A are given, a number of other par- 
 ticulars, not easily remembered, are equally necessary to be 
 known, respecting the falling of the perpendiculars within or 
 without the 43, ‘and the affections of the sides and angles, be- 
 fore a complete solution of them can be obtained. 
 
KXVI 
 
 science, both by the ‘assistance which they afforded ta 
 the practical calculator, in many intricate and abstruse 
 computations, and by the numerous improvements and 
 discoveries of a higher kind, with which they abound: 
 The method, in particular, which he appears to have first 
 used in raising logarithms from their differences, and 
 
 his skilful application of analytical principles to several _ 
 
 other subjects of difficult investigation, erititle him to 
 rank with the first mathematicians of the age in which 
 he lived. 
 
 The logarithmic curve, likewise, and thdse of a si- 
 milar kind, which first began to be introduced about 
 this time, greatly facilitated the conception of these 
 numbers, by exhibiting some of their most remarkable 
 properties 1% a more perspicuous way than could be 
 done by the abstract methods of investigation employed 
 by Napier and others. And though the doctrine itself 
 has no-necessary connection with these or any other 
 
 geometrical figures, 1t was from this source that the 
 
 new and advantageous mode oi expressing logarithms 
 by series was first derived. 
 
 This happy improvement, which was inerodugelia into 
 the science about the year 1668, by Mercator and James 
 Gregory, who were led to the discovery of some of 
 the most simple forms of these series by contemplating 
 
 the nature of the hyperbola, was soon afterwards ex-" 
 tended to the trigonometrical part of the subject, or * 
 
 the arithmetic of sines, as it is now frequently called ; 
 
 which Newton, Leibnitz, the Bernoullis, and others, 
 enriched with similar formule ; and by this means as- 
 
 ii 
 ' 
 
<XVII 
 
 similated the principles of logarithms and trigonometry 
 with those of the new calculi, of which they were the 
 inventors and improvers. 
 
 The exponential formulz, also, for the sines and 
 cosines of arcs, which were first given by Demoivre; 
 have greatly contributed to the progress of the analy- 
 tical branch of this subject, by abridging its operations, 
 and shortening the labour of investigation; and though 
 
 some writers, who appear to be unacquainted with the 
 
 way in which imaginary symbols arise, have represented 
 all expressions of this kind as nugatory and absurd; 
 yet.their cormmodious form, and the ease and certainty 
 with which they can be applied in many intricate inqui- 
 ries, will always cause them to be regarded by the skil- 
 ful analyst as an important acquisition to the science (r). 
 
 Many other improvements, of more or less import- 
 ance, have since been made; both in the practical and. 
 theoretical branches of this subject, by later writers ; 
 but of these, none have proved of such general ad- 
 vantage to the science as the substitution of the analy- 
 tical mode of notation in the place of the geometrical ; 
 which useful change was first introduced by Euler ; 
 who, besides this simplification of the former methods, 
 has developed and extended, in his numerous works, al- 
 most every part of the trigonometrical analysis ; which, 
 
 (r) Lagrange, one of the most profound analysts of the pre- 
 sent age, regards the exponential formule of Demoivre, here 
 alluded to, for expressing the sines and cosines by means of cer= 
 tain imaginary functions of the arcs, as one of the greatest analy= 
 fical discoveries of the 17th century. 
 
 - ot, : oe . 
 ‘APIS Ba 
 
XXVH 
 
 under his masterly hand, assumed the form of a-new 
 science. | | 
 
 With respect to the projection of the sphere, which 
 is also.a branch of science intimately connected with 
 this subject, little more is known of the early part of 
 its history than what can be collected from the writings 
 of Ptolemy, who in his treatise on the planisphere, as 
 well as in his geography, has left us a number of pro- 
 positions relating to the stereographic method, as it is 
 now generally called, of representing the surface of 
 the sphere upon the plane of one of its great circles, 
 and its application to the construction of maps and 
 charts (s). | 
 
 It is evident, however, that Ptolemy, whose work 
 above mentioned is the oldest of the kind that we now 
 possess, was not the author of this mode of projection, 
 as he has commionly been thought to be; since it ap- 
 pears from a letter, still extant, addressed by Synesius; 
 
 a disciple of the celebrated Hypatia, and afterwards 
 
 bishop of Ptolemais in Libya, to Paonius, on his send- 
 ing him a silver astrolabe, that the method of exhi- 
 biting the surface of the sphere in this manner, upon 
 a plane, was known at an earlier period, and had been 
 particularly employed, for geographical and other pur- 
 
 (s) It may here be observed, that the term Stereographic, 
 which denotes stidity, or a projection of three dimensions, is an 
 improper appellation; atid though the word is: of Greek origin, 
 
 its application to this subject is modern, having been‘first pros: 
 posed: and employed by Aguilon, ity his Optics, printed at Anvers,’ 
 
 16125 before which time it had-the name off Planisphere. + 
 
 = 
 
XXIX 
 
 poses, by Hipparchus, to whom he expressly ascribes 
 the invention, 
 
 But though the foundation of this doctrine was esta- 
 blished by the ancients, they appear to have been un- 
 acquainted with some of its most important principles ;. 
 as it was not known to Ptolemy that all the circles of 
 the sphere, excepting those whose planes pass through 
 the eye, are, in the stereographic method, projected 
 into circles; or that any two projected great circles 
 make the same angle, by their intersection on the plane 
 of projection, which the original circles make on the 
 sphere. 
 
 The first of these remarkable properties was, in fact, 
 for along period within the reach of mathematicians, 
 without their availing themselves of it. For Apollo- 
 nius having demonstrated, in his Conics, that the sub- 
 contrary section of a cone Is a circle, the only thing 
 that remained to be done was to prove that the plane 
 of projection, in this method, forms a subcontrary sec- 
 tion in every cone, the vertex of which is the eye, and 
 the base a circle of the sphere; but, easy as this step 
 appears, it was not achieved tili fifteen hundred years 
 after the time of Hipparchus. 
 
 It is uncertain, indeed, by whom, or at what time, 
 these two useful propositions were first introduced into 
 this branch of the subject, as the latter is not to be 
 found in the large treatise of Clavius on the astrolabe ; 
 and though the former is distinctly mentioned by Jor- 
 danus in his Planisphere, printed at Bale in‘1536, the 
 first clear and rigorous demonstration of it, by means 
 
& 
 
 XXX 
 
 bf the subcontrary section, was given by Commandin; 
 in his Commentary on the Planisphere of Ptolemy, pub: 
 lished at Venice; 1558. The same obscurity also attends 
 whatever relates to the origin and progress of the ortho- 
 graphic and gnomonical projections, of which no ac- 
 count is to be found in any of our mathematical histories, 
 though the theory and practice of these methods have 
 been amply treated of by several writers, who have 
 neglected, with singular indifference, all inquiries con- 
 cerning the authors and improvers of these useful in- 
 ventions. 
 
ADVERTISEMENT 
 TO 
 
 THE SECOND EDITION. 
 
 Berorz the time of the first publication of the present 
 performance, no work of a similar kind had appeared, 
 in this country, which could be considered as partaking, 
 in any degree, of the great discoveries and improve- 
 ments which had been made, in the subject here treated 
 of, within the last fifty years. 
 
 The author was therefore induced to lay before the 
 public an additional treatise on this useful and interest- 
 ing branch of knowledge, that should be more con- 
 formable, both in matter and method, to the present 
 state of the science; and from the favourable reception 
 it has met with, he flatters himself, that the under- 
 taking has not been regarded as either useless or un- 
 necessary. 
 
 A careful revision, however, of the several parts of 
 the work, appeared to the author to be still wanting ; 
 and he has accordingly made such alterations and im- 
 provements in the present Edition of it, without mate- 
 tially departing from his original plan, as he trusts will 
 render it still more deserving the public favour, 
 
 Among other things, of this kind, that might be men- 
 tioned, nearly all the practical questions before given, 
 many of which had passed, in the same state, from one 
 
XXX 
 
 work to another, for more than a century past, are 
 here reproposed with new data; and in every case, 
 where it was judged necessary, both the examples and 
 the rules upon which they depend, have been carefully 
 attended to, and reduced to their most commodious 
 and approved forms. 
 
 The analytical part of the work, also, where the au- 
 thor first introduced to the notice of the English stu- 
 dent, the many elegant theorems and formule, with 
 which Euler, Delambre, Legendre, and other eminent 
 mathematicians have enriched the science, is here con- 
 siderably improved; and, though comprised within a 
 short compass, will, it is conceived, be found to contain 
 all the most useful.and essential principles of this 
 extensive and important branch of analysis. 
 
 J. BONNYCASTLE. 
 
 Royal Military Academy, 
 Oct. /7th, 1812. 
 
TABLE OF CONTENTS. 
 
 Tr ROT ee creer a ie css fee 8 
 
 Page. 
 
 lil 
 
 OCT RSI aI SSE a a-Si ee XXX! 
 
 PTR EG oe o0c) sd a io! phate nig 's 014.8 e000.) s9.0 
 Preperties Of Plane WiANPIES' .. <4 os cece es es ee en 
 General solutions of the cases of plane triangles .... 
 Table of solutions of all the cases of right-angled 
 
 PIAHEMTTGME Pepe aoe wile ab nin eines olewie Sols otis epee 
 
 a 
 1] 
 
 19 
 
 Table of solutions of all the cases of oblique-angied — 
 
 DOM GpETIAM PIES is,. eis 4 piss a-nle d9\s erpinia ain sib in, 3p 
 DER TAAC OES EM AUIDICH 4 tiv. 0's 0.0 wb o)é. esha.p ein, 0c0:9°a.9 
 Mensuration of heights and distances, &c. .....0.. 
 Miscellaneous examples in ditto ......eceseeccees 
 
 SSPHETICAN PREONGMICHY vec been te sheet os cue ees ee, 
 
 Definitions and axioms @eeeoet*#eseoevoeevpee ee oeoeee7 eee 
 General properties of spherical triangles........... 
 Ambiguous cases of spherical triangles ........... 
 
 Properties and affections of right-angled spherical tri- 
 
 RMR ca Sur aiila ee. wc kee hoe cabele WN. slate wal lytic aia ie ad 
 General solutions of the cases of right-angled sphe- 
 TOBANTEVIAUIRIER. she's Sein C site Has «ais sis siaie 6 a ages 
 Of quadrantal spherical triangles .........ceesces 
 Properties and affections of quadrantal spherical -tri- 
 WORE Ge iota" 5 itiMatotaceipt ora WG mH ope’ y'n!o.s epele tee’ ee» wie 
 General solutions of the cases of quadrantal spherical 
 RSAC ere recite aisalyuielalcets als ale Cilahe.eso'e p 
 General solutions of the cases of oblique-angled sphe- 
 OT EN ES SCIENCE 7." ER a en, 
 Table of solutions of all the cases of right-angled 
 Pereeraeal LITA GICG, . estes vss «$8 0 « cictere vids 
 Table of solutions of all the cases of quadrantal sphe- 
 MME EER’ 's V.e'n VRP EC F'0't-s, 0's 8% ule paw ot ee a 
 
 | 112 
 113 
 114 
 131 
 156 
 
 168 
 
KXX1V CONTENTS. 
 
 Table of solutions of all the cases of oblique-angled 
 spherical triangles js eeuly wyiateels «o's 5's «e+ seal 
 Miscellaneous examples for exercising the preceding 
 LADIES oppo. oy nis) oie Fishy SMR nt eatnslS 9 erie fialarae 
 Application of spherical trigonometry to the resolu- 
 tion of astronomical problems....,..eeeeeveeee 
 Various astronomical tables ........0gesecceoses 
 Miscellaneous astronomical DIObICINS 34," eet aaa 
 Of the signs of trigonometrical quantities ........ 
 Ofatniconometrical forma} ysis) sean o'shgis eek enna 
 Values of the sines, cosines, &c. of certain arcs, in 
 SUT NUMDETS sc siciccs ee os «94 fe'n.8 fin'y gies ne Te 
 Formule for the multiples and powers of the sines, 
 Gosines, coc. Of ‘Certain ares’ss : L's ¢ oy pee ee 
 Logarithmic fopmulee oso... dees sige n't 0 SOR aE ae 
 Demonstrations of the principal theorems in plane 
 {IBONGINELLY, é:acs;s 4:0 eo: tae Sy isle gue D RO ARE ti teg 
 Demonstrations of the principal theorems in spheri- 
 __ cal trigonometry, with other particulars ........ 
 Spherical theorems demonstrated geometrically ..... 
 Of the stereographic projection of the sphere ....., 
 Miscellaneous problems and theorems ......-.ese. 
 Solutions of all the cases of right-angled plane tri- 
 angles, independently of any CUVEE RRS, ine 
 Solutions of ail the cases of oblique- angled plane tri- 
 angles, independently of any tables........00+5 
 
 Of the increments and fluxions of the sines and tan- 
 
 gents of arcs, OF ABIES sees eee ee eeee eee ee ees 
 Solutions of quadratic equations by the tables of sines, 
 BCC. e's + MME Shs 0a. 6 5 +: 4 wcve 9 elem 
 Solutions of cubic equations by ditto ..........+. 
 Of the admeasurement of altitudes by the barometer 
 and thermometer ..4... sums Cte eeeeneeegnans 
 
 Page. 
 179 
 193 
 195 
 271 
 278 
 287 
 295 
 
 311 
 
 315 
 344 
 
 351 
 356 
 377 
 397 
 408 
 
 416 
 
 419 
 
} 
 
 PLANE TRIGONOMETRY. | 
 
 Prane Triconometry is the science which treats 
 of the analogies of plane triangles, and of the methods 
 of determining their sides and angles. 
 
 It also comprehends whatever relates to the proper- 
 ties and relations of certain right lines drawn in and 
 about a circle, called sines, tangents, &c. 
 
 -The sides of plane triangles are estimated in feet, 
 yards, chains, &c; or by abstract numbers, according 
 to the purpose intended. 
 
 A right lined angle is measured by an arc of a cir-_ 
 cle contained between its two legs, and having the 
 angular poiat for its centre (a). 
 
 Every circle is supposed to be divided into 360 
 equal parts, called degrees; each degree into 60 equal 
 parts, called minutes ; each minute into 60 equal parts, | 
 
 called seconds, &c. 
 
 So that a semicircle, or half the circumference, con- 
 
 (a) It is proved by Euclid, and ins. vi. prop. 24 of my Ge- 
 ometry, that angles at the centres of equal circles (and conse-_ 
 
 ' quently at the centre of the same circle) are proportional to the 
 _ arcs on which they stand ; and in p, viii. prop. 6, that the cir- 
 _ cumferences of different circles, or any similar parts of them, 
 _ are as the radii of those circles ; from which properties, the rea- 
 
 son for assuming arcs of circles as the measures of angles, will 
 
 _ be sufficiently evident. 
 
 B 
 
sae 
 
 tains 180 degrees; and a quadrant, or fourth part of 
 the circumference, 90 degrees: also.a sextant Is an 
 arc of 60 degrees, and an octant an arc of 45 degrees. 
 
 An angle is likewise said to be of as many degrees, 
 minutes, seconds, &c, as there are divisions of this 
 kind contained in the arc, or part of the circumference, 
 by which it is measured (0). 
 
 Hence, a right angle, being measured by a quarter 
 of the circumference, is 90 degrees; an obtuse angle 
 is greater than 90 degrees; and an acute angle less 
 than 90 degrees. ns, 
 
 Degrees are marked at the top of the figure, or 
 figures, by which the arc is denoted, by a small °, 
 minutes by ’, seconds by ”, &c. 
 
 Thus, 39° 28° is 39 degrees, 28 minutes; and 
 47° 12’ 19” is 47 degrees, 12 minutes, and 19 se- 
 conds. 
 
 The complement of an arc is*the difference be- 
 tween that arc and a quadrant, or 90°; and the sup- 
 plement of an arc is the difference between that arc 
 
 (4) From this method of estimation, it is plain that an angle | 
 may be of any magnitude less than 180 degrees; and that it is 
 the same thing whether the circle which is described from the 
 angular point be larger or smaller, as the arc contained between 
 the two legs is-always the same part of the whole circumference. 
 It may also be remarked, that it is not the angles of triangles 
 themselves that are employed in calculation, but certain lines 
 called the sines, tangents, &c of those angles; as the latter can — 
 always be compared with the sides of the triangle, being quar- 
 tities of the same kind, which is not the case withthe ay Lcoge or 
 ‘arcs by which they are measured. 
 
3 
 
 and a semicircle, or. 180°. Thus, 8 c is the comple- 
 ment of AB; and B a is its supplement (c). 
 
 c 
 a A 
 
 » In like manner, the complement of an angle is the 
 difference between that angle anda right angle, or 90°; 
 and the supplement of an angle is the difference be- 
 tween that angle and two right angles, or 180°. Thus 
 Bo c is the complement of ao B, and 804 is its 
 
 supplement. 
 
 The sine of an arc is a right line drawn from one 
 extremity of the arc perpendicular to the diameter 
 which passes through the other extremity. Thus Bp 
 is the sine of a B, or of B a. 
 
 bad 
 
 The cosine of an arc is the sine of the complement . 
 
 (c) Thé complement is common to two arcs, or angles, which 
 are the supplements of each other. Thus, p.cits the complement 
 of a B or of Ba; butin most \practical questions, it isvsually un- 
 derstood to be what any acute angle, or an arc, wants of 90°. 
 
 Bo 
 
4. 
 
 of that arc, or the part of the diameter which lies be- 
 tween the centre of the circle and the sine. Thus B E£, 
 or its equal o p, is the cosine uf a B orof B a, or 
 the sine of its complement B c. | 
 
 The tangent of an arc is a right line drawn perpen- 
 dicular to the diameter, at one extremity of the arc, 
 and terminated by a right line drawn from the centre 
 through the other extremity. Thus a £ is the tangent 
 of aB, orof Ba. 
 
 The cotangent of an. arc is the tangent of the com- 
 plement of that arc. Thus cz is the cotangent of a B, 
 or of B a, or the tangent of its complement B c. 
 
 ‘The secant of an arc is a right line drawn from the 
 
 centre, through one extremity of the arc, and termi- 
 nated by the tangent, or a line drawn perpendicular to, 
 
5 
 
 the diameter at the other extremity. Thus o £ is the 
 secant of a B, or of Ba. 
 
 The cosecant of an arc is the secant of the comple- 
 ment of that arc. Thus o £ is the cosecant of a B, 
 or of B a, or the secant of its complement B c. 
 
 The versed sine of an arc is that part of the diameter 
 which lies between the beginning of the arc and the 
 sine. Thus a D is the versed sine of a B, and D ais 
 the versed sine of its supplement B a. 
 
 A: 
 
 The coversed sign of an arc is the versed sine of the 
 complement of that arc. Thus cE is-the coversed sine 
 of a B, or the versed sine of its complement B c. 
 
6 
 
 The chord of an arc isa right line joining the two — 
 ‘extremities of that arc. Thus a Bis the chord of the ~ 
 arc A B, and B ais the chord of its supplement (d). 
 
 B 
 
 The cochord of an arc is the chord of the comple- 
 ment of that arc. Thus B c is the cochord of A B, or 
 the chord of its complement BC; ands p is the chord 
 of the supplement of c B. 
 
 The lines here described, belong equally to an angle 
 as to the arc by which it is measured; and, except the 
 chords and versed sines, they are all common to two 
 arcs or angles which are the supplements of each other. 
 
 So that if the sine, tangent, &c of any arc or angle 
 above 90° be required, it is the same thing as to find 
 
 a 
 ar ee 
 
 (d) The sine or cosine of any arc or angle can never exceed 4 
 the radius, and the secant and cosecant are never less than the 3 
 radius; but the tangent and cotangent admit of all possible de- | 
 grees of magnitude. . 
 
 It may likewise be remarked that the chord of 66°, the sine ~ 
 and versed sine of 90°, and the tangent of 45° are all equal té6 © 
 the radius, whatever may be the magnitude of the circle; also a 
 the sine of 30°, and the cosine, or versed sine, of 60°, are half 
 the radius, and the secant of 60° is double the radius. 
 
1 
 
 the sine, tangent, &c of its supplement, or what it 
 wants of 180°. 
 
 It may also be remarked, that these lines are called 
 the natural sines, tangents, &c of the arcs or angles to 
 which they belong; and the logarithms of the num- 
 bers by which they are represented, are the logarith- 
 mic sines, tangents, &c. 
 
 And as one or other of these make a part of every 
 trizonomietrical operation, they have been calculated to 
 a given radius, for every degree, minute, &c of the 
 quadrant, ‘and ranged in tables for use. 
 
 Whence, by the help of such a table, the sine, tan- 
 gent, &c of any arc or angle may be found by inspec- 
 tion; and, vice versa, the arc, or angle, to which any 
 sine, tangent, &c belongs. 
 
 Upon these tables, also, and the doctrine of similar 
 triangles, depends the solution of the several cases of 
 plane trigonometry, which may be performed either 
 by the natural or logarithmic sines, tangents, &c, as 
 occasion requires. 
 
 But the logarithmic sines, tangents, &c, are thi 
 - most commonly used ; as the calculations, in this case, 
 are all performed by adding and subtracting only, in- 
 stead of multiplying and dividing, as is required by 
 the natural sines, &c (e). 
 
 (c) The sine, tangent, &c of any arc or angle being of the same 
 magnitude as the sine, tangent, &c of its supplement, it is plain 
 that a table of these lines made for every degree, minute, &c of 
 the quadrant, or 96°, will serve for the whole circle. 
 
 It is also to be observed, that, in every such table, the natural 
 sines, tangents, &c are usually calculated to radius 1; but in 
 
3 
 
 In every plane triangle, three things must be given 
 
 to find the rest ; and.of these three one at least must 
 be a side, because the same angles are common to an 
 infinite number of triangles. 
 _ These circumstances having been stated, it may be 
 observed, that all the varieties that can possibly hap- 
 pen in the solution of plane triangles, are comprised 
 in the three following cases: viz. 
 
 1.’ When two of the three given things are a side 
 and its opposite angle. 
 
 2. When two sides and their included angle are 
 given. 
 
 3. When the three sides are given. 
 
 Each of which cases may be resolved, either by 
 geometrical construction, by arithmetical computation, 
 or instrumentally. 
 
 In the first of these methods, the triangle is con- 
 structed by laying down the sides from a scale of 
 equal parts, and the angles from a scale of chords, or 
 a protractor, and then measuring the unknown parts 
 
 order that the logarithmic sines, tangents, &c may be all positive, 
 they are calculated to radius 10000000000, or 1 with 10 ciphers, 
 
 the logarithm of which is 10, so that the latter are the logarithms . 
 
 of the former with 10 added to the index. | 
 
 and as the natural sines, tangents, &c of any angles, or arcs of 
 different circles, are proportional to the radii of those circles, 
 their values may be readily found, or made to correspond to any 
 radius whatever. ? 
 
 Such circumstances as relate to the state of the sine, tangent,’ 
 
 &c with respect to their being positive or negative, will be noticed 
 in another part of the work, as they do not interfere with the 
 common practice. 
 
 ee 4 Be 
 
9 
 
 by the same scale or instrument from which the others 
 were taken. 1 
 
 In the second method, having stated the proportion, 
 according to the proper rule, multiply the second and 
 third terms together, and divide the product by the 
 first, and the quotient will be the fourth term required, 
 for the natural numbers. Or, in working by loga- 
 rithms, add the logarithms of the second and third 
 terms together, and from the sum take the logarithm 
 of the first; then the number answering to the remain- 
 der will be the fourth term sought. 
 
 -In the third method, or instrumentally,. as suppose 
 by the logarithmic lines on one side of the common 
 two foot scales, extend the compasses from the first 
 term to the second or third, as they may happen to be 
 of the same kind; and that extent will reach from the 
 other term to the fourth term required, taking both 
 extents towards the same end of the scale. 
 
 The second of these methods, however, or that in 
 which the operation is performed, by logarithms, is the 
 one generally employed; the other two being chiefly 
 of use as checks on the calculations, or, in certain 
 simple cases, where a near approximate value! of the 
 quantities to be determined is thought sufficient (f). 
 
 _ But it may here be remarked, that when one or 
 more logarithms are to be subtracted, in any operation, 
 
 (f) In working any question by logarithms, it is not always , 
 necessary to make an accurate construction of the figure, as the 
 
 ‘learner should accustom himself to such as are readily formed 
 
 by the pen, and used only for the purpose of guiding him in the 
 calculation. i 
 
10 
 
 it will be better to write down their complements, or 
 what each of them wants of 10, instead of the loga- 
 rithms themselves, and then add them all together, 
 abating as many tens in the index of the sum as there 
 are logarithms to be subtracted. 
 
 Thus, if the logarithm to be subtracted be 3.4932758, 
 itis the same thing as to add its complement6.5067242, 
 and then abate 10 in the index of the sum; and if the 
 logarithm be 9.0743260, it is the same as to add its 
 complement 0.9256740, and abate 10 as before; ob- 
 serving that the readiest way of doing this, is to begin 
 at the left hand and subtract each figure of the loga- 
 rithm from 9, except the last significant figure on the 
 right, which must be subtracted from 10. 
 
 If the index of the logarithm, whose complement 
 isto be taken, be greater than 10, write down what 
 it wants of 19, and take the rest of the figures from 9 
 as before; observing, after the addition is made, to 
 abate 20 from the index of the sum. And if the lo- 
 garithm of a decimal, which has always a negative in- 
 dex, is to be subtracted, add the index, considered 
 as afirmative, to9, and then find the complement of 
 the rest of the figures as usual; and abate 10 in the 
 index uf the sum. | 
 
 Thus, the complement of the logarithm 12.4907327 
 is 75092673 ; and the complement of the logarithm 
 2.5972648 is 11.4027352 (g). 
 from adding the complements of the logarithms instead of sub- 
 tracting the logarithms themselves, may be rendered obvious as 
 
 follows: 
 When the index of the logarithm (1) is under 10, or between 
 
11 
 
 PROPERTIES OF PLANE TRIANGLES, REQUIRED IN 
 THE PRACTICAL PART OF THE SCIENCE. 
 
 1. The sum’ of all the three angles of any plane 
 triangle is equal to two right angles, or 180° (A). 
 
 2. The greater side of any plane triangle is opposite 
 to the greater angle; and the less side to the less angle. 
 
 8. The sum of any.two sides of a plane triangle is 
 greater than the third side; and the difference of any 
 two sides is less than the third side. 
 
 4. The triangle is equilateral, isosceles, or scalene, 
 according as its three angles are all equal, or only two 
 of them equal, or all three unequal. 
 
 5. The angles opposite to the two least sides of any 
 
 10 and 20, which is as far as any practical use of the tables 
 extends, it is plain that — 1 = (10 —L)—10; or —L = (20 — L) 
 —20, agreeably to the rule. 
 
 But if the index of the logarithm (1) to be subtracted be ne- 
 gative, let it be denoted by i, and the decimal part by d; then 
 —.=—(—i4d)=+i—d =(9+i)4+(1—d)—10, which is, 
 also, the same asthe rule. 
 
 The method of obtaining the complement, by beginning at 
 the left hand, and subtracting each figure of the logarithm, ex- 
 cept the last, from 9, is the same in effect as subtracting them 
 - from 10 in the usual way, and is merely employed upon this 
 occasion as being more commodious. 
 
 (4) Since the sum of all the three angles of any plane triangle is 
 180°, if one of the acute angles of a right-angled triangle be sub- 
 tracted from 90°, the remainder will be the other acute angle. 
 
 In like manner, if the sum of any two angles of a plane 
 triangle be taken from’ 186°, it will leave the third angle; and 
 if any one of the three angles be taken from 180°, it will leave 
 the sum of the other two. 
 
12 
 
 plane triangle are acute; and if there be an obtuse 
 angle, it will be opposite to the greatest side. 
 
 6. A perpendicular drawn from the opposite angle 
 of any plane triangle to the longest side, will fall with- 
 in the triangle; and the greater segment will lie next 
 the greater side, and the least segment next the Jeast 
 side (7) 
 
 . In any isosceles plane triangle, a perpendicular, 
 ais from the vertex, will bisect both the base and 
 the vertical angle. 
 
 8. Ina right-angled plane triangle, the hypothenuse 
 is equal to the square root of the sum of the squares 
 of the other two sides; and either of the sides is equal 
 
 to the square root of the difference of the squares of 
 
 the hypothenuse and the other side. 
 
 Note, also, that if the half difference of any two 
 quantities be added to their half sum, it will give the 
 greater of those quantities; and, if it be subtracted 
 from the half sum, it will give the less. 
 
 CASE I. 
 
 When two of the three given things are a side and 
 its opposite angle, to find the rest. 
 
 (1) Besides the case here mentioned, which is sufficient for all 
 practical purposes, it may be further observed, that if the angles 
 at the base be both acute, the perpendicular will fall within the 
 triangle; but, if one of them be obtuse, it will fall without the 
 triangle, on the side of the obtuse angle, And in either of these 
 cases, as in the former, the greatest segment will lie next the 
 greatest side, and the least segment next the least side. 
 
 — 
 ee ee ee 
 
13 
 
 ; RULE. 
 
 The sides of any plane triangle are to each other as 
 the sines of their opposite angles; and vice versa :— 
 That is, | | 
 
 As any side is to the sine of its opposite angle, so is 
 any other side to the sine of its opposite angle: or as 
 the sine of any angle is to its opposite side, so is the 
 sine of any other angle to its opposite side. 
 
 Heice, to find an angle, begin the proportion with 
 a side opposite to a given angle; and to find a side, 
 begin with an angle opposite to a given side. 
 
 Note. When two sides and. an angle opposite to 
 one of them are given, to find the rest, the question 
 Is sometimes ambiguous, or will admit of two diffe- 
 rent answers. | 
 
 Thus, if the given angle be opposite to the least of. 
 the two given sides, the angle to be found, by the 
 rule, may be either an acute angle or its supplement ; 
 but, if it be opposite to the greater side, the required 
 angle will be acute (A). 
 
 (4) The ambiguity in this case does not arise, as has com- 
 monly been thought, from the required angle being found by 
 means Of its sine, which isthe same both for an acute angle and 
 its supplement, but solely from circumstances attending the con- 
 struction of triangles from given data. For if;the side c a, in 
 the following. figure, be proposed of such a length that it shall 
 be less than c 8, and greater than a perpendicular drawn from 
 c to the opposite side, it must necessarily cut the indefinite line 
 BAa@ in two points; while, in all other cases, it will either fall 
 short of it, or touch it only in one point. Hence, when c ais 
 proposed less than cB x sin B, the triangle is impossible: when 
 it is greater than cs there can be only one triangle; and if it 
 
 \ 
 
14 
 
 EXAMPLE I. 
 In the plane triangle.a Bc, 
 AC 248 tae Required the 
 
 Given, BC 354 other parts. 
 ZB 41° 36 | 
 
 Ye Weer Wea ar al C 
 BY CONSTRUCTION. 
 
 1. Lay down the line B c = 354, from some con- 
 venient scale of equal parts; and make the 43 = 
 41° 36’ by ascale of chords, or other instrument. 
 
 2. With the centre c, and radius 248 (ac) taken 
 from the’same scale of equal parts, cross B A in A ora, 
 and joinc aor ca, andasBc, orasce, will be the 
 triangle required. | 
 
 Then the Z*c and a, measured by the scale of 
 chords, and the side B 4, or Ba, by the scale of equal 
 parts, will be found to be as follows, viz. 
 
 £0 292° | Za 712° | AaB 1854 
 or 67; |:,or,1082°.| sor $433 
 
 fall between these limits, the triangle is ambiguous, or admits of 
 two different answers. | 
 
 Thus, in the above figure, where the least side ac is op-, 
 posite to the given acute angle 8, it “is evident that either 
 43corasc may be the triangle sought. But when the given 
 angle is right or obtuse, it will be opposite to the greatest side s 
 and in this case there can be no ambiguity; for then neither’ 
 of the other angles can be obtuse, and the geometrical con- 
 struction will accordingly give only one triangle. 
 
15 
 
 BY CALCULATION. 
 As side ac --- 248- - - - 2.3944517 
 
 _ 1,6055483 
 Isto sine ZB --- 41° 36’ - 9.8221198 
 So is side BC - - - 354 --- 2.5490033 
 
 To sine Za 71° 23 or 108° 37’ 9.976671 4 
 4? 36°. 417: 36. 
 Sum 112° 59’ or 150° 13° 
 Subtract 180° O’ 180° 0 
 Leaves 67 Poor: 29° 477-220 
 
 Then, 
 Sate ee dd SF i, maemo! iets 9.8221198 
 0.1778802 
 PRS BCs a ee oh etm te te 2.39445 17 
 Sete Ch OF OA Pid em ae 9.6962012 
 = Side a BB T85368 Be eS 2.268532 1 
 Or, 
 © BEE LB chides 9.8221198 
 0Q.1778802 
 PST aS. 5D Lisa «eit el ol a 2.3944.517 
 Pied © GY biker lye 9,.964.0797 
 ft a Are a ee eee 2.5304116 
 
 —-- 
 
 INSTRUMENTALLY. 
 
 In the first proportion, extend the compasses from 
 + 248 (ac) to 354 (BC) upon the line of numbers, and « 
 that extent will reach, upon the sines, from 414° (2B) 
 to 714°, for the Za. 
 In the 2d proportion, extend from 412° (ZB) to 
 29° or 67° upon the sines, and that extent will 
 
16 
 
 reach, upon the line of numbers, from 248 to 1851, 
 or 3433. for the side a B, or a B. 
 
 EXAMPLE I. 
 In the plane triangle aBc, 
 AB 237 AC131.78 
 Given< Bc197 Ans.< 24 56°13’ 
 £c 90 2B 33° 47 
 Required the other parts. 
 
 EXAMPLE Ill. 
 In the plane triangle a Bc, 
 Bc 310 AC’ 2873S 
 Given Zp 62° 9’. Ans.< aB 209.95 
 ee Ormade LS LN a eee 
 Required the other parts. 
 EXAMPLE IV. 
 In the plane triangle a8 c, 
 AB 217 ZA 104° 47’ 
 Givens ac 305 Ans.< Zc 30° 12’ 
 B45 BC 417.03 
 Required the other parts. 
 
 EXAMPLE V. 
 
 On the plane triangle a Bc, 
 
 Ac 329 f AB 805.69 
 Givend BC 516 or 516.82 
 PA Val is pan? Ans. Ze 34° 8 
 
 ‘ or 71° 34/ 
 
 ZA 108° 43’ 
 
 Required the other parts. [| or 71°17’ 
 
17 
 
 ¢ 
 
 CASE Il. 
 
 When two sides and their included angle are given, 
 to find the rest. 
 
 RULE. 
 
 As the sum of any two sides of a plane triangle, is 
 to their difference, so is the tangent of half the sum of 
 their opposite Pant to the tangent of half their diffe- 
 rence. 
 
 Then the half difference of these angles, added to 
 their half sum, gives the greater ae and subtracted 
 from it gives the less. 
 
 And as all the angles are now known, the remain- 
 ing side may be found by Case 1. 
 
 Note. Half the sum of the two opposite, or .un- 
 known angles, is found by subtracting the given angle 
 from 180°, and dividing the remainder by 2 (/). 
 
 (/) Instead of the tangent of half the sum of the two un- 
 known angles, we may use the cotangent of half the given ari- 
 gle, or the tangent of half its supplement, which are all equal 
 . to each other, 
 
 It may also be further observed with respect to the general 
 solution above given, that when the triangle is isosceles, the 
 angles at the base are each equal to the complement of half the 
 given angle, or that at the vertex ; and consequently, in this case, 
 the third side may be found directly by Case 1. 
 
 Also, if the angle included between the two given sides of 
 the triangle be a right angle, or 90°, the third side, or hypothe. 
 nuse, may be found by extracting the square root of the sum of 
 the squares of the other two sides. 
 
 Cc 
 
18 
 EXAMPLE. 
 
 In the plane triangle ABC, | 
 Given : AB EES | feet Required the rest. 
 
 BC 2384 
 ZB 35° 46’ 
 
 A 
 Bz , Cc 
 
 BY CONSTRUCTION. 
 
 1. Take Bc = 2384 from a scale of equal parts; 
 and set off the 4 8=35° 46° by a scale of ahi or 
 other instrument. 
 
 2. Make az = 1082 by the same scale of equal 
 parts, as before, and join a, c, and aBc will be the 
 triangle required. 
 
 Then, the several parts being measured, we shall 
 have ZA=1211°, Zc = 223°, and side ac = 1630 
 feet. 
 
 BY CALCULATION. 
 
 > AB+ BC 3466 -.--- - 3.5398286 
 
 6.46017 14 
 
 : AOR TS08i a pines ST Ae 
 
 : Tan. = ~e ROI veiw ai ml 10.4032614 
 
 - Tan. ——— 49° 20’ ~~ ~'- 10.0660238 
 Sum 121° 27Za %. 
 
 Diff. 22° 47’Zc 
 
19 
 
 Then, ae Core 
 : Sine ZA 121° 27’ or 58° 33’ 9.9309978 
 | 0.0690022 
 : Sidepc- - -2384 ---* 3.38773063 
 ;oine £ Bp - - + 35° 46 DS a ).7667739 
 
 Sideac- -'- 1630.3 - + 3.215 3.21 30824 
 
 mae eee 
 
 INSTRUMENTALLY. 
 
 In the first proportion, extend from 3466(a B-+BC), 
 to 1302 (ABS 8) on the line of numbers, and that 
 extent will reach, on the tangents, from 72° 7’ (the 
 contrary way, because. the tangents are set back again 
 from 45°) to beyond 45°; which being set so far back 
 from 45°, falls upon 492°, the fourth term, 
 
 ‘ In the second proportion, extend from 581° (44) 
 to 352° ( ZB) on the sines, and that extent will reach, 
 
 on the numbers, from 2384 (B c) to 1620, the fourth 
 term, or side ac. 
 
 EXAMPLE Il. 
 
 In the plane triangle a Bc, 
 
 | AB 907 Zc, 96° 44/ 
 Given. < Bc 512 ATS .< L&. Ba oe 
 t Zp 49° 10° Ac 691.02 
 
 Required the other parts. 
 
 EXAMPLE Ill. 
 
 fn the plane triangle ABC, 
 
 ( das potas bo, La §2 9 
 Given BC 1004 Anse’ 2 0087° 51° 
 
 Required the other parts, 
 | £2 
 
20 
 
 EXAMPLE IV. 
 In the plane triangle a BC, 
 
 AB 530 ZA 80° 55’ 
 Given< Bec 830 Ans.< Zc 39° 5 
 ZB 60° AC 727.94 
 
 Required the other parts. 
 EXAMPLE V. 
 In the plane triangle a Bc, 
 AB 406 Lo 67° 45 
 Given< Bc 359 Ans.< ZA 54° 55’ 
 4.8 S{ 720° AC 369.29 
 Required the other parts. 
 
 CASE III. 
 When the three sides are given, to find the angles. 
 
 RULE, | 
 
 Make the longest side the base, and let fall a per- 
 pendicular upon it from the opposite angle. 
 
 Then, as the base, or sum of the sezments, made 
 by the perpendicular, is to the sum of the other two 
 sides, so is the difference of those sides to the diffe- 
 rence of the segments of the base. 
 
 And half this difference, added to half the base, 
 will give the greater segment, or that next the greater 
 side; and, subtracted from it, will give the less. 
 
 Whence, in each of the right-angled triangles, 
 formed by the perpendicular, two sides and an angle 
 opposite to one of them will be known; and conse- 
 quently the other angles may be found, by Case 1. 
 
 Or either of the angles may be found at one opera- 
 tion, by the logarithmic formula given in Case vi, of 
 the Table for oblique-angled plane triangles, p. 33. 
 
21 
 
 EXAMPLE I. 
 ” 
 
 In the plane triangle a Bc, 
 
 AB 467 
 Given) e 262 bart Required the 
 BC 698 angles. 
 
 fA 
 
 Be ee 
 
 D 
 BY CONSTRUCTION. 
 
 Take B c = 698, by a scale of equal parts, and with 
 the centres B, c, and radii 467 and 352, taken from 
 the same scale, describe arcs intersecting each other in 
 A; and join aB, ac, and aBc will be the triangle 
 required. | 
 
 Then, by measuring the angles with a protractor, 
 or by the scale of chords, they will be found to be 
 nearly as follows, viz. Za = 1164°, 43 = 27°, and 
 OLS So 87°: . 
 
 BY CALCULATION, 
 Having let fall the perpendicular A p, it will be 
 
 BC or BD+DcC 698 -- 2.8438554 
 7.1561446 
 > AB+ac -= 819 --= 2.9132839 
 SDA BSAC. ee 418 - - . 2.0606978 
 BDS pc = 18493 .-. 2.1301263 
 8+ 134.93 
 Hence ee 28 aide egiet iin 
 8— 134. 
 
 arti 2a Rs Ren 8 a 
 
29 
 
 Then, in the triangle a gp, right 4° at p, 
 
 - AB .& #.% = SOF ae: @ 1 4sp095169 
 
 > Bp --=-=s 416.46. -» 2.6195733 
 
 -: Sine ZD=-- Y90°.- - + - 10,0000000 
 
 : Sine ZpaD 63° 6 --~ 9,9502564 
 90° Of Hee. 
 26° 54° ZB 
 
 And, in the triangle ac p, right Z‘at p, 
 
 > AG ---+- 3852 --- 25465427 
 
 SD «ee 281.53 - = 9 2.4495247 
 
 2: Smee Zip -2- OG": ==) 2 GjQGGO000 
 
 2 OSin6L CAO) oS: Gon “9,9029820 
 90° 0 | 
 36° 54 Le 
 
 Also . 63° 6 ZBAD 
 And Sr O 7 CAL 
 Makes 116° 12) ZBaAc 
 
 Whence 4B = 26° 54, Zc = 36°54¥,andZBac 
 
 =< 116° 12’. 
 | INSTRUMENTALLY. 
 
 In the first proportion, extend from 698 (8c) to 
 819 (aB-+ ac) onthe line of numbers, and that ex- 
 tent willreach, on the same line, from. 115(aBo Ac) 
 to 135, the difference of the segments of the base 
 (BD o> DC). 
 
 In the second proportion, extend from 467. (4B) 
 to 416 (2p) on the numbers, and this extent will 
 reach, on the sines, from 90° ( 4p) to 63°, (2 BAD). 
 
 In the third proportion, extend from 35% (ac) to 
 2814.(p c) onthe numbers, and that extent will reach, 
 on the sines, from 90° (2p) to 53°2, (Ze aD). 
 
23 
 
 EXAMPLE II. 
 In the plane triangle a Bc, 
 
 AB 750 La OO 
 
 Given<e ac 340 / Ans.~< ZB 26° 7’ 
 
 Bec 592 Ze 103° 55’ 
 Required the angles. 
 
 ‘EXAMPLE Iil. 
 In the plane triangle a Bc, 
 | AB 500 Za 53° 8 
 civeny c 300 Aas ZB 36°.52° 
 Bc 400 Zc 90° 
 Required the angles. 
 
 EXAMPLE Iv. 
 In the plane triangle aBc, 
 
 ‘a B 760 Za 69° 3 
 
 Giveny ac 429 Ans.< ZB 33° 27’ 
 
 BC 727 4c 77° 30° 
 Required the angles. 
 
 EXAMPLE V. 
 In the plane triangle 4 3 c, 
 
 AB 1002 ZA 133° 46’ 
 Giveny ac 1960 Ans 2"25 30°69" 
 Bc 2750 LOOPS 15" 
 
 Required the angles. 
 
 EXAMPLE VI. 
 In the plane triangle a Bc, 
 AB 169 Za 90° 29’ 
 Given} Ac 169 Ans.< 4283 44° 46’ 
 Be. 240 £0 44° 46 
 Required the angles. 
 
24 
 
 These three problems include all the cases or vari- 
 eties of plane triangles, as well right-angled as oblique, 
 that can possibly happen ; but the following theorem, 
 for right-angled triangles, will sometimes be found 
 more convenient in practice, than the analogy which 
 is furnished by the general rule. 
 
 THEOREM, or CASE IV. 
 
 In any right-angled triangle, As radius is to the 
 tangent of either of the acute angles, so is the leg ad- 
 jacent to that angle to the leg opposite to it; and vice 
 versa, 
 
 Or, As radius is to the cotangent of either of the 
 acute angles, sois the leg opposite to that angle to the 
 Jeg adjacent ta it; and vice versa (m). 
 
 It may also be observed, that as the sine of either of 
 the acute angles of a right-angled triangle is equal 
 to the cosine of the other, the latter may be used in- 
 stead of the former, whenever it renders the operation 
 more simple. | 
 
 EXAMPLE I. 
 In the right-angled plane triangle az c, 
 
 BC 327 Required the other 
 4B S417 parts. 
 
 A 
 We 
 Mier ee See. 4 C 
 
 (m ‘To this we might also have added the following analogy : 
 As radiusis to the secant of either of the acute angles, so is the 
 
 Given 
 
25 
 
 . BY CONSTRUCTION. 
 
 Make nc = 327, and Zp -= 54° 17's then raise 
 the perpendicular ¢ a, meeting Ba in A; and aBc 
 will be the triangle required: in which a B is found 
 to measure 560, a c 4542, and 2A, which is the com. 
 plement of ZB, 35%". 
 
 é 
 
 BY CALCULATION. 
 
 Rad. or sine -- - 90° - = = 10,.0000000 
 Danny Uae Petey. 2e.4009432626 
 s +. Side Bait wists et B27 ~ © 2,51454783 
 
 >: Sideac ---- 454.79 - 2.6578104 
 
 ee eo 
 
 : SineZa or cos. ZB 54°17 . - 9.7662473 
 : Side Bc alae) dian deed. eohige’ » Wa ee aera ge 
 >: Rad.orsine Zc - 90° - - = - 10.0000000 
 
 ¢ Side as - - -.- 560.14. - <)  9.7483005 
 
 And 90 —54° 17’= 35° 43’ Za. 
 INSTRUMENTALLY. 
 
 Extend the compasses from 45° to 541° ( ZB) on 
 the tangents, and that extent will reach from 327 (Bc) 
 to 4542, on the line of numbers, for the side 4c. 
 And the extent from 541° (4a) to 90° on the 
 
 leg adjacent to that angle to the hypothenuse. Or, As radius is 
 to the cosecant of either of the acute angles, so is the leg oppo- 
 site to that angle to the hypothenuse. 
 
 But the rule for this case is as readily performed by the sines 
 and cosines, which are always to be found in the logarithmic 
 tables, where the secant is frequently omitted, 
 
26 
 
 sines, will reach from 327 (B c) to 560, on the line of 
 numbers, for the hypothenuse a 8B. 
 
 EXAMPLE Il. 
 
 In the right-angled triangle a Bc, 
 Given d 2 c 520 f AC 550.85 
 LA 45 21° Ans.< AB 757.52 
 ZB 46° 39° 
 
 To find the other sides and angle. 
 
 EXAMPLE III. 
 In the right-angled triangle 4 ze, 
 Given { Ac 807 { ZB 36° 6’ 
 Bc 421 Ans.< ZA 53° 54 
 AB $21.05 
 To find the other sides and angles. 
 
 EXAMPLE IV. 
 In the right-angled trianele a Bc, 
 Given AG 490 chia 1 B C 607.62 
 L RS SPOT Ans.< AB 780.58 
 A. 4 B.88° 53° 
 To find the other sides and angle. 
 
 To these rules it has been judged necessary to add 
 the following tables, which contain the solutions of all 
 the cases of plane trigonometry before given ; together 
 with such additional formule, for the tangents, as are 
 better adapted, in certain instances, to the producing 
 of accurate results than those derived from the sines 
 and cosines in the former analogies, 
 
 The reason of this deficiency of the sines and co- 
 sines, in the cases alluded to, is, that if an arc near 
 
97 
 
 90° is to be found by means of its sine; or a very 
 small arc, or one near 180°, by means of its cosine, 
 the variation of these lines is so small, that they will 
 not change in the tables for many seconds. 
 
 Thus, if the log-sine, or log-cosine, of a required 
 arc should come out 9.9999998, this number, in the 
 tables, is the sine of an arc from 89° 56’ 19” to 89° 
 57’ 8 or the cosine of an arc from 2’ 52” to 3’ 41”; 
 and consequently it is impossible to say what arc or 
 angle, between these limits, is to be taken, owing to 
 the tables not being continued to more than ‘seven 
 places of decimals. 
 
 An error also, of an opposite kind, will arise in 
 finding arcs near 90° by their tangents, or those near 
 0° by their cotangents, as the largeness of the loga- 
 rithmic differences, in this part of the quadrant, will 
 render the method of determining them by propor- 
 tional parts, incorrect, when they are to be found to 
 seconds, by the common tables, or to parts of seconds, 
 by Taylor’s tables. 
 
 But in all other cases, the magnitude and great va- 
 riation of the logarithmic differences render the use 
 of the tangents preferable to that of the sines or co- 
 sines ; besides which, the above error, in the use of 
 the common tables, may always be avoided, by sub- 
 tracting the log-tangent, or log-cotangent, from 20, 
 and then finding the corresponding arc, or angle, 
 in the first part of those tables, where it, is usually | 
 given to seconds for the first two or three degrees of 
 the quadrant. 
 
28 
 
 To this we may further add, that when a sine, cosine, 
 &c of the kind here mentioned enters into the calcula- 
 tion, as one of the data, it is rather favourable than 
 otherwise to the accuracy of the result, or the value 
 of the thing sought; as any small error in the given 
 arc, or angle, will not much affect the tabular value of 
 its sine or cosine. 
 
 Note. € u is used, in the following formulz, to 
 denote the co-log, or the complement of the common 
 tabular logarithm of the number answering to the let- 
 ter or expression to which it is prefixed. And the 
 sign % expresses the difference of the two quantities 
 between which it is placed, when it is not known which 
 of them is the greater. 
 
 SOLUTIONS OF ALL THE CASES OF RIGHT-ANGLED 
 PLANE TRIANGLES. 
 A 
 
 BZ 
 
 az 
 
 I. Given the hypothenuse and either of the oblique 
 angles, to find either of the legs. sci 
 
 RULE. 
 sin given Z its opp. leg 
 As rad : hyp:: or ‘ 
 cos given Z its adjt. leg 
 Or, 
 esin. A 6 COS. B csin.B  ¢ COS. A 
 = or a or —— 
 
 Tr r 
 
29 
 
 La =ic-+1 sin a (or L cos B)—10; rb= ret 
 L sin B (or Lcos A)—10. 
 c? sin. 2a c* sin 2B 
 
 4r gf 4r * 
 
 L area = 2Lc + L sin 2a —10.6020600 3 or 2Lc 
 +. 1 sin 28B—10.6020600. 
 
 Area A= 
 
 II. Given the hypothenuse and either of the legs, to 
 find either of the oblique angles. 
 
 RULE I. : 
 sin its opp. 4 
 As hyp: rad: : given leg : ' or 
 cos its adj'. Z 
 ee 
 rb 
 
 ° ra ° 
 Sin a (or cos B) = ——; sin B(orcosa) = —— 
 
 L sin A (or LcosB) = €Leo+L4a;3 LsinB(orn 
 COS A yice eC Herb. 
 
 If a be near 90°, find 4B, which is its complement, 
 and the contrary when 4B is near 90°. 
 
 RULE ll. 
 comb c—a 
 ae eee. _— *° 4 Ales aa e 
 Tania rv 7) tan +B an iene 
 €.L i tC oom +10 
 Ltan is Se ch 
 
 Where 4 a, or 1B, is always less than 45°. 
 Fe ne OL OB As et at 
 e Area A = = (ea) (c—2) ors (c+é) {c—b) « 
 Be L area == 4 $i (cu) +L (c—a)t +La— 
 - .8010300; or 4 $i (c4+8) +i (c—b)t +26 — 
 
 _ .3010800. 
 
30 
 
 Ill. Given the hypothenuse and either of the ES 
 to find the other leg. 
 
 RULE TI. 
 
 Find either of the oblique 2° by Case il ; and then 
 the required leg by Case 1. 
 
 RULE II. 
 CAS MV (c+) x (c—4) 3 b xe WV (c+a) x (ea) 
 Lie SE Grid ar AG ~~ L Ger meNGor 
 
 eg ert LN 
 
 IV. Given either of the legs and either of the ob- 
 lique angles, to find the other leg. 
 
 RULE. 
 tan Z adj‘. given leg 
 As rad; given leg: : or : req’. lec. 
 cot Zopp. givenleg = 
 Ce: 
 __ dbtan a b cot B atanB acota 
 or 
 
 ; b =——— or ‘ 
 
 ff r LF Tr 
 
 La = Lb + itan a (or. cot B)— 10 ; LU = La a+/ 
 L tan B (or L cot a) — 10. 
 Ares ‘Apenit’ tan Bs te a? tan B 
 Zr 2r 
 L area == 2ub + ttan a — 10. 3010800; or 2La 
 -+-+ ti tan B — 10.3010300. 
 
 V. Given either of the legs and either of the oblique 
 angles, to find the hypothenuse, 
 
31 
 
 RULE. 
 As sin Z opp. given leg 
 Or cos Z adj". given leg 
 
 Or, 
 
 hi given leg :; rad: hyp. 
 
 ra rb 
 
 ¢ = SS 
 sin A (or cos B) sin B (or cos A) 
 Le=€Lsin a (or €LcosB) + La= e€Lsins 
 (or €L cos A) + LO. 
 
 VI. Given the two legs, to find either of the oblique 
 angles. 
 
 RULE. 
 
 tan its opp. 4 
 As either leg : rad :: other leg : or 
 cot its adj. 2 
 
 Or, 
 
 Tan a (or cot B) =; tan B (or cota) = nf 
 
 Ltana(orLcots) = eLrb+ La; tans (or Lh 
 cota)= €La+ Lé. 
 
 If 2a be near 90°, find 48, which is its comple- 
 ment ; and the contrary when 438 is near 90°. 
 
 Area A =1ab3 L area La + Lb —.3010200. 
 
 Vil. Given the two legs, to find the hypothenuse. 
 
 RULE I. : 
 Find either of the oblique 4° by Case vi. and 
 then find the hypothenuse by Case v. 
 
32 
 RULE Il. 
 c= VepR =a (45) 20 7 (14-2) 
 
 which formulz do not admit of convenient logarithmic 
 expressions. 
 
 SOLUTIONS OF ALL THE CASES OF OBLIQUE-ANGLED 
 PLANE TRIANGLES. 
 
 A 
 
 Ahi. 
 
 I. Given a side and two angles, to find either of the 
 
 Cc 
 
 pee 
 a 
 
 other two sides. 
 
 RULE. 
 Find the third, or remaining 4, if necessary, by sub- 
 tracting the sum of the two given 2* from 180°. 
 ‘Then, 
 As sin Z opp. given side : given side :: sin 4 opp. 
 required side : required side. 
 
 Or, 
 
 bsina 
 
 sin B 
 La = €Lsng+uLusin at+1.b —10. 
 Either of the other sides may also be expressed in 
 the same form, by taking the side and angles which are 
 similarly situated with respect to the side whose value 
 
j 
 
 33 
 is sought. And the same may be observed in all the 
 
 other cases, where only one side or an angle is exhi. 
 bited by the formula. — 
 
 *cin A é 
 Area A = one aos (A “a Bye 
 Larea= €Lsinp-+4.Lisin(a+3)+uLtsnat2rb 
 
 — 20.8010300. 
 
 II. Given two sides and an angle opposite to one of © 
 them, to find the other angles. 
 RULE I. | 
 
 As side opp. given 4 : sin given 4 :: other side: 
 
 sin its opp. Z. 
 Which Z is acute, if it be opp. to the least of the 
 
 - two given sides; but, if it be opp. to the greater, it 
 
 may be either an acute 4 or its sup‘. or else a right 
 angle, 
 The 3d, or remaining Z = 180°— sum of the other 
 two Z£°%, | 
 Or, 
 asin B 
 
 b 
 Lsina=eLéi+‘1ra+ sin Bs — 10. 
 
 Sin A= 
 
 RULE II. 
 
 Lever b + 1a+.Lsing—10=L tan@. 
 : 10 an (45° 
 Then, 1 tan (45°o4ta)—= aia ee). 
 Where arc @ can never exceed 45°; and 4 a, which 
 is found by subtracting 2(45°=.1a) from 90°, is 
 subject to the same ambiguity as in rule 1. 
 D 
 
34 
 
 Note. In this and the following case, the given sides 
 and Z must be so taken, that the result, found by 
 rule 1, or the 1st part of mle 11, shall not be greater 
 than radius, otherwise the A is impossible. - 
 
 Ill. Given two sides and an angle opposite to one of 
 them, to find the remaining side. 
 
 RULE, 
 Find the other two 2* by case 11, “observing that 
 
 they will be equally subject to the ambiguity there — 
 
 mentioned ; and then find the remaining side by case 1. 
 
 Or, the side, and area, may be found by the follow- 
 
 ing formule : 
 es bcos A aa MW ¢* qt—l? sin? A 
 
 r 
 
 ee eae 
 
 aa 
 
 But neither of these admit of convenient logarith- 
 _ mic expressions. 
 
 If the triangle be isosceles, or have a = , the rule 
 will give c = *“cos a, orLceo=La vi LcosA — 
 
 9.6989700; and the area of the A = sin 2a, or 
 
 L area = 2na-+ sin 2a — 10.3010300, 
 
 IV. Given two sides and their included angle, to 
 find the other two angles. 
 
 RULE. ° ‘ 
 
 As sum of the two given sides : their dieeaen oc 
 
 cot + included 4: tan4 difference other two Z5. 
 Which 2 2 difference, addédito the complement of + 
 
 SS ets eS Oe Oe ee 
 
35 
 
 the given 2, gives the 4 opposite the greater of the 
 two given sides; and, subtracted from it, gives the Z 
 
 Gpposite the less, 
 mee 
 
 Tani (poc)= Fercot | A 
 
 L tan 2 (BC) =eLn(b+c)+ Ble L cot 
 LA 10. | 
 
 Then, (90°—1 a) + 4(85Cc) = Z opp. greater 
 side; and (90°—1 a) —+(Boc)= Zopp. less side. 
 
 If the values of b,c, in this case, should be given in 
 
 ‘logarithms, instead of the natural numbers, as is some- 
 
 times the case in astronomical calculations, the follow- 
 ing formulze will be found more convenient in practice 
 than that given above. 
 Let €L greater side + L less side = L tan @3 in 
 which case arc (®) is always less than 45°. 
 Then, 
 L tani (Bac) = xLcotya+ Ltan(45°— @)—10. 
 
 Where (90°—24) +i(B&c) = Z opp. greater 
 side, and (90°—ia)—1(Boc) = Z opp. less side, 
 as before. | 
 
 Or, either of the two 4° may be found by the fol- 
 lowing formule: : 
 
 rsin A ) rsin A 
 Tan Base a) CR ott De 
 Fe hae ? aig ey 
 ster (=) COSA ay cos A 
 bsin a csin A 
 
 ee doe Qbe we P tents SINS AS i482 cos a 
 rT r 
 
 Dg 
 
36 
 
 But these do not admit of convenient logarithmic 
 expressions, ° 
 
 or 
 L.area = Lb +uc+.xsin a—10.3010300. 
 
 V. Given two sides and their included angle, to 
 find the other side. 
 
 RULE. 
 Find the other two 4* by case Iv; eile then the 
 remaining side by case I. 
 
 Or, the side may be found by the following formule: 
 
 aN (ant) being 1a? Sintgc= BURGE REP Toe 
 Or, 
 c= VJ (a4) 2 costs = So 4 = 2 003 C. 
 
 But these do not admit ay convenient tegaxnnatic, 
 
 expressions. | | 
 If the A be isosceles, or have a= J, the rule will give 
 2asinic 
 ¢=—— , Or L¢ = La-+i sin; c —9,6989700. 
 
 VI. Given the three sides, to find either of the i Soa | 
 
 RULE I. ‘ 
 As the longest side, taken as a base: sum of the 
 other two sides : : difference of those sides : difference 
 of the segments of the base, made by a perpendicular 
 
 from the opposite 2. 
 
 Then, + this difference added to 1 the base, vee 
 
37 
 
 the greater serment, or that next the greater side; and 
 subtracted from it, gives the less. 
 
 _ And as the perpendicular divides the A into two 
 right-angled A‘, in each of which the hypothenuse and 
 a leg are known, the angles may be found by case II. 
 of right-angled triangles. 
 
 RULE Il. 
 Tan} a= raf (as—b) x ) x ($s—e), 
 4sx (45-4) 
 
 aoa eas i Vee, 
 L tan $a. gies sina Gla ata eae at cA” a) 
 
 Where s denotes the sum of the three sidesa+d+c; 
 and 1 Z a is always acute. 
 
 Or, the latter of these two formulz may be express- 
 ed in words at length, as follows : 
 
 Add together the log. of 3 the sum of the three 
 sides, and the log. of the difference between this } sum 
 and the side opposite the Z sought, and find the com- 
 
 _ plement of their sum, by subtracting the index from 
 
 19, and the rest of the figures from 9, as usual. 
 Then to this complement add the logarithms of the 
 
 differences between the said.4 sum and each of the: 
 
 other two sides, and the result, divided by 2, will give 
 
 _ the tangent of } the required angle. 
 
 Area A = 1 / (b4cf~—a? x a—(bocy; 
 Or, V3s5 x (Ls—a) XX (4 s—-b) x (4 5—c). 
 
 L area =ifuisu(ts—a) +1 (49-4) + 
 L(;s—c) ¢ 
 
38 
 
 To the formula above given, we ae also add the 
 following : 7 
 
 Singa= raf (4s—5) x G: dite) CcOSsA= rrf is (45—4), 
 bec ae Lea 
 Or, 
 
 ns €ub L L(t —~b 10 41 ee CY? 
 
 i Sie ke 
 1 €rb64+ €1¢e4+r.4s5 41 (4s—c) 
 ECGS 2 A) Se ar ee 
 
 The former of which, or that for the sine, must be 
 used when + a differs considerably from ane , and the 
 latter when it is near 90°. 
 
 It may likewise be observed, that when the A is ‘ ‘ 
 
 isosceles, or has 6 = c, we shall have 
 
 Sint A =a Coss A = ayM (21 +2)x(2b—a) 
 Or, usin} a=erb +14 a — .301300, and L cos 
 pasperab+ye fu (2b+a)+1(Qb—a? 
 And if the A be equilateral, having each of its sides 
 = a, we shall have area A = 1a’ 733 or area i 
 = 21a — .3634993. : 
 We may here also subjoin the two following formule: 
 
 ‘ ar Wily pf IROS © ee ae 
 Sina = raf f—Und" 5 cos A=r C ) 
 
 which have been un highly useful in many trigono- i 
 metrical investigations. | 
 
 MISCELLANEOUS EXAMPLES. 
 
 1. How -many inches does an angle of 1” subtend i 
 at the distance of six miles; supposing the object to — 
 be perpendicular to the horizon? e 
 3 Ans, 14 inch nearly, _ 
 
$9 
 
 2. The hypothenuse of a right-angled triangle being 
 13 feet 10 inches, and the base 9 feet 7 inches, it is 
 ee to find the perpendicular. 
 
 Ans. 9 feet 11-2, inches. 
 
 3. The hypothenuse of a right-angled triangle being 
 6395 feet, and one of the acute angles 39° 50’ 38”, it 
 is required to find the two legs. Cams 
 
 Ans. 4097.262 and 4910.035 
 
 4, The hypothenuse of a right-angled triangle being 
 _ 19630040, and one of the legs 19630000, it is re- 
 quired to find the two acute angles. 
 
 | Ans. 6° 56”2 and 89° 53° 3”3 
 
 5. If the base of an oblique-angled plane triangle be 
 60, and the other two sides 30 and 40, what is its 
 : perpendicular altitude ? | ' Ans. 17.77561 
 
 6. If the base of a plane triangle be 60, and the 
 other two sides 30 and 40, what are the lengths of the 
 segments of the base, made by a line bisecting the 
 vertical angle? 
 
 Ans. 34,28571 and 25.771428 
 
 7. Supposing one angle of a plane triangle to be 
 
 139° 54’, and the two sides about that angle in the 
 
 ratio of 5 to 9, it is required to find the other two 
 
 angles, - Ans. 26°.0' 10%2 and 14° 5’ 492 
 
 8. The sices of a plane triangle being 14373, 13172, 
 and 11845, it is fequured to find the three angles. 
 
 ) Ans .69° 54’ 10”, 59° 23’ 19”, and 50° 42’ 31” 
 9. If the three ates of a plane triangle be 104° 
 49! 15”, 49° 24’ 5”, and 25° 53’ 40”, and the ‘side 
 opposite the greatest angle 476.75 yards, what are the 
 other two sides? Ans. 215.2533 and 374.2469 
 
40 
 
 10. If the sides of a plane triangle be in proportion 
 
 to each other as 1,4, and 1, what are the angles? 
 Ans. 32° 44 39”, 44° 3) 41”, and 103° 11° 407 
 
 11. In a right-angled plane triangle, the three sides 
 are 3, 4, and 5, from which it is required to find the 
 angles, Ans. 36° 52’ 11°83, 53°'7' 48”4, and 90° 
 
 12. In an oblique. angled plane iriangle, the three 
 sides are 4, 5, and 6, what are the three angles? 
 
 Ans. 41° 24! 348, 55° 46’ 1648, and 82° 49° 9”2, 
 
 13. There are three towns, A, B, and c, the peer 
 of a from B is 5 miles, of B from c 9 miles, and of 
 c from a 7 miles, what are their respective bearings 
 from each other? Ans. Za 95° 44° 21", CIS I. 
 33’ 26”, 43 50° 49’ 13” 
 
 14. How must three trees, A, B, C, be planted, so 
 that the angle at a may be double that at B, and the 
 angle at B double that at c ; and that a line of 100 
 yatds may go just round them? 
 
 Ans. AB 19.80621, a c 35.68958, and Bc 44.50418 
 
 OF THE MENSURATION OF HEIGHTS AND 
 DISTANCES. 
 
 _The measuring of heights and distances depends 
 
 upon the use of certain instruments for taking angles, 
 
 and the rules of Plane Trigonometry, already delivered; 
 which being separately or jointly applied, as the case 
 may require, will resolve most questions of this kind * 
 
 * 
 
 that can occur in practice (p). 
 
 (p) Horizontal apd vertical angles are usu ally taken with a | 
 theodolite furnished with one or two telescop pes, and a vertical 
 
(> 
 
 41 
 
 In addition, however, to’the properties of plane tri- - 
 angles, given in page 11, it may be necessary to lay 
 down a few others relating to angles, parallel lines, &c. 
 which in several instances, will be found of great 
 use in facilitating both the constructions and calcula- 
 tions. » ; 
 
 ‘1. The two angles, which are made wd one right 
 
 line meeting another, are together equal to two right 
 
 angles, or 180°. om 
 
 2, If two right lines intersect each other, the verti- 
 
 cal or opposite angles will be equal. 
 
 3. lf a richt line intersects two parallel richt lines 
 5 p 8 2 
 
 it makes the alternate angles equal, and the outward 
 angle equal to the inward opposite angle, on the same 
 
 side. 
 
 4. If one side of a plane triangle be produced, the 
 
 arc ; and if the circles of the instrument are about 3} inches ra- 
 
 dius, the observed angles may be read off to half a minute. 
 But if the angles are oblique to the horizon, they must be taken 
 with a-sextant, or Hadley’s quadrant, which is to be held in 
 
 such a position that its plane may pass through both objects and 
 the eye of the observer ; and when altitudes are determined by 
 
 this instrament, it is done by reflecting the object from an artt- 
 ficial horizon. 
 
 Short bases, for temporary use only, are commonly measured 
 with rods, or Gunter’s chain, of 66 feet in length; but the 
 
 common 50 or 100 feet tapes are better adapted for expedition. 
 
 With these lines, when the ground is tolerably level, and the 
 
 ' direction, or ali, nement, of the base preity correct, the error in 
 
 distance will probably be about 3.inches in 50 feet, or 349 of the 
 whole measurement, as long as the tapes are kept dry. 
 
42, © 
 
 ¥ 
 
 outward angle will be equal to the sum of the two in- 
 ward opposite angles. ie Se Five 
 5. All angles in thé same sala of a circle, ok 
 which stand upon the same arc, are equal. » 
 6. An angle at the centre of a circle is double that 
 
 at the circumference, when they both stand on | the 
 game arc. =” | 
 7. An angle in a semicircie, or that which stand 
 ee half the circumference; is a right angle, or 90°. 
 . If a right line be drawn parallel to one of the 
 
 fia of a plane triangle, it will cut he other two sides. 
 proportionally. | 
 ' It may also be remarked, that some of the simplest 
 cases of heights and distances, may be resolved without 
 the assistance of trigonometry, or of any instrument 
 for taking angles, by one or other of the following 
 methods : ie. 
 1. By the property of similar triangles; from doiliichy 
 
 it is known that objects are in Proper sive to each other 
 as the saath of their shadows. 
 
 B 
 Thus, if the height of the pole ac be 10 feet, the 
 length of its shadow c b 8 feet, and the shadow cB, of 
 the object a c, 50 feet: 
 Then 8 (cb) : 10 (ac) :: 50 (c 2): 624 feet = 
 height ac. ! 
 
 . 
 
\ PSS 
 
 43 
 
 2. Another method is by means of two poles of un. 
 equal lengths, set up parallel to the object, so that the 
 observer may see the top of the object over the tops of 
 
 both the poles. 
 ‘i A. 
 
 K 
 AG ; 
 
 Thus, let the length af the pole de be 6 feet, that 
 of the pole fg 8 feet, their distance asunder e g 10 
 feet, and the distance ec, of the shorter pole from the 
 object, 190 feet. 
 
 Then the triangles dk f and dk a being similar, 
 dhoregz:hf::dk orec: KA; 0r10:8—6:: 
 190: 38 feett=ak. Henceak.+kKCmAK+ 
 de=38+6=>44= A. | 
 
 8. A third method, is by viewing the image of the 
 top of the object reflected from some smooth surface, 
 as a mirror placed horizontally, a vessel of water, &c. 
 
 Thus, let p be the reflecting surface, at the distance 
 
 of 94 feet from the bottom of the object a c; and let 
 
 a person at p, 6 feet from xs, with his eye 54 feet 
 above the ground, view the image of the top. of the 
 object at 8. 
 
 * 
 
44 
 
 Then, because, from the principles of optics, the i 
 angle of incidence A Bc, is equal to the angle of re- 
 
 ‘flection r BD, the triangles Bp F, Bc A, will be simi- 
 lar; and consequently BD: DF::BC;CA}3 oré6: 
 543294: 861 feet = ac, 
 
 4, A fourth method, is for the observer to fix a pole _ 
 
 | upright in the ground, by trials, so that having laid 
 
 himself on his back, with his feet against the bottom a 
 
 of it, he may see the top cf the pole and that of the 
 object in the same right line. | 
 
 2°" NS 
 < 
 A 
 « 
 
 In which case the distance rp from the foot of the 
 pole to the eye of the observer, will be in proportion 
 to the height of the pole c p, as the whole distance F B 
 is to the height of the object a B. 
 
 And if the height of the pole c p be equal in length 
 
 to the observer rp, the distance FB will be equal ta 
 
 the height of the object a B. | 
 
 PRACTICAL QUESTIONS. 
 
 1. Having measured a distance of 220 feet, in a direct if 
 horizontal line, from the bottom of a steeple, Ithen 
 
 found the angle of elevation of its top to be 46° 30’; 
 required the height of the steeple (q).. 
 
 (¢) In Mauduit’s Trigonometry (Crakelt’s Trans. p. 182.) it is 
 shown that the error of any altitude a ¢ is to the error committed 
 
 »} 
 ~ 
 
45 
 
 es 
 ‘ ry 
 * 
 As rad or sine = - - 90° - - - - 10,0000000 
 Istobce .----- 290 feet - -. 2.3424297 
 
 Soistan 4 abc - 46°30’ - - 10.0227500 
 
 Toheightac - - 231.83 feet- 2.3651'727 : 
 which added to the height of the instrument 8 J, will 
 give the whole height a c. 
 
 2. It is required to find the perpendicular Helene of 
 a hill, the angle of elevation of which, taken at the 
 _ bottom, was 46° 10’, and 110 yards further off, on a 
 level with the bottom, and in the same vertical plane 
 with the former, 29° 56’. 
 
 a 
 
 an 
 
 ‘in taking the Z adc, as double the height ac is to the sine of. 
 “double the observed Z abc. Whence the error that may arise 
 in taking the altitude of an object, by a sextant, or theodolite, 
 will be the least possible when the sine of double the observed £ 
 is the greatest possible ; which is when it is 45°. So that in 
 
 finding altitudes, the observed Z should be taken as near 45° as 
 
 can conveniently be done. At an éxact altitude of 45°, if an 
 error of 1’ be made in the determination of the observed Z, the 
 error in altitude will be -~ part of the whole; and if the ob- 
 served Z be greater or less than 45°, the error in altitude will be 
 increased in the ratio of ai to the sine of double the said Z. 
 
46 
 
 ZaBpc ----- 46 
 Z£ADC -*-85 31° 
 
 LDAB *+-++- 15° 
 , Assn ZDAaAB -- 16 14°. - - 9.4464591 
 Is to DB.) = 2a) = 110 ~~ - = 2.0413927 
 Soissn Zp --- 29°56 - - 9.6980936 
 STO (GAN BAG omc clekin dieh wc one eee 2.2929272 ie 
 Then, ! 
 As rad, or sin --- 90° «+ = 10.0000000 
 TSitQ: ABD mip asian ne ae ener eis 2.2929272 . 
 
 Soissin 4B --- 46°10 - 9.8581505 
 To heght/ace: t- <5) T41 GES - (.2, LolOg ee 
 
 8. It is required to find the perpendicular height of 
 
 we 
 
 a cloud, or other object, when its angles of elevation, 
 as taken by two observers at the same time, on the 
 ‘same side of it, and in the same vertical plane, were — 
 64° 10’ and 35° 25’; their distance asunder being half — 
 a mile, or 880 yards. . 
 
 L£DAB -- + =~ 28° 45" | 
 Assin Z pap -- 98°45’ -- 9.6891349 | 
 . | 0.3178651 
 Is to opp. sideD B - 880 ---- 29444897 
 
 Soissin Z apc - 35°25 -- 9.7630671 
 To opp. side a B= 1060.29 °-. 3,0254149 
 
47 
 
 | Then, 
 
 Asrad, or sin Zc - -90° - - - - 10.0000000 
 
 Is to opp. side AB - 1060.291°- 3.0254149 
 
 Soissin Z ABC = 64°10 -- 9.9542741 
 
 To height ac. --- 954.31 yds. 2.9796890 
 
 4, From the top of a tower, 125 feet high, which 
 lay in the same right line with two trees, I took the 
 angles of depression formed. by lines conceived to be . 
 drawn from my eye to the bottom of each tree, and 
 another line parallel to the horizon, and found them 
 to be 56°3 and 40}: what is the distance of the two 
 objects (r)? . : 
 
 As rad, or siny - --- 90° »- - - - 10.0000000 
 
 Istoac ---+-- 125 feet - - 2.0969100 
 Soiscot ZABC- 56°15 -- 9.82489296 
 To 8.0 ee 4 - ee BS b2oe — F109 T8088 
 a | Then, | 
 * As rad, orsin - -- 90° -%- 10,0000000 
 Istoac .--=--+- 125 --- 2.0969100 
 SoiscotZ apc - 40°30’ - ~ 10.0685011 
 BAO DON ieta fabian. 146.351 -- 9.1654111 
 Min 82.522 a 
 
 Dist. Dp B 62.829 
 
 TS 
 
 (r) An angle taken from the top of any object, or the one 
 usually called the angle of depression, is that which is made by 
 
48 
 
 5. Wanting to know the breadth of a river, I meas 
 sured a distance of 120 yards in a straight line by the 
 side of it, and at each end of this line, J found the 
 angles subtended by the other end and a tree, close by 
 the opposite side of the river, to be 51° 45° and 78° 
 50°: what is its perpendicular breadth ? 
 
 | a | : 
 pe ba ee 
 Mey & 
 
 LB Nee int ae] As 
 
 LG eva nln | TOMO: 
 
 130° 35° 
 
 180° O 
 
 Lipac (~~~. 499195" 
 Assn ZBAC «© 49°95’ - - - 9.8805052 
 0.1194948 
 IoWep Cc <- -8S oot: CME 8 ee 07913812 
 So is sin 4c «- - #78° 50° -' -*- 9.991699! 
 To ‘Alp iki. eee es I OOS ba 
 
 Then, 
 AS SIN ZT! a ete OO eee. oe 10.0000000 
 TS tO ANB ie nH ee ya ie 2S ne eS oe 
 
 Soissin 4B -- 51°45’ --- 9.8950450 
 Wo breadth ap, 121.74) °-'-  @ogsagg) 
 
 a right line passing from the eye to the object and another line 
 drawn parallel to the horizontal plane. Thus, £ axis the 2 of 
 depression of the object 8, which by the nature of parallel lines 
 is equal tothe 2 4Bc3 andits complement is the Z Bac. 
 
 U 
 
49 
 
 6. Wanting to find the height of an obelisk, stand- 
 ing on the top of a declivity, I measured from its bot- 
 tom a distance of 60 yards, and there found the angle 
 formed by the plane and an imaginary line drawn to 
 the top of the object to be 454°; and after measuring 
 on, in the same direction, 40 yards further, the angle, 
 formed as before, was only 24° 45’: what was the 
 height of the obelisk ? 
 
 . Then, in the triangle 8B a c, 
 Assin ZBAC -- 20°45’ .- 9.5493602 
 0.4506398 
 
 Is to opp. side Bc 40 yards - 1.6020600 
 Soissin ZaBC - 24°45° -- 9,6218612 
 To opp. side ac - 47.268 yards 1.6745610 
 
 ~_—— 
 
 In the triangle a c b, 
 As sum sides c A, cc D 107.268 - 2.0304378 
 nan, 7.9695622 
 Is to diff. sidesc a, cD 12.732 -~ 1.1048966 
 Soistantsum Z°a,D 67° 15 - 10.3774391 
 To tan + diff. 2° a, p.15°.48’. = 9.4518979 
 
 And 67° 15’ — 15° 48’ = 51°27" Zoap. 
 | E 
 
‘50 
 Lastly, in the same triangle a cD, 
 Assn Zc Abd - 51°27 - - 9.8932426 
 
 bala: 0.1067574 
 
 Isto opp. sidec D 60 yards - - 1.7781513 
 
 So.is.sin:Z.c...<. - =) 45°,.30% -\-. 918582421 
 . 54.721 yards, ® 
 
 Toheight ap - - 30 164.168 feate 1,7381508 
 
 7. Wanting to know the height of an inaccessible 
 object, I took its angle of elevation, at the least di- 
 stance I could from its bottom, which was found to be 
 572° ; and going 105 yards further, in a right line, the 
 angle was then found to be only 321°; required its 
 
 height, and my distance from it at the first station, the 
 
 instrument being 5 feet above the ground at each ob- 
 servation. . 
 
 Za D1 ~ «5745 
 LBC Vai) ~ 82°80) 
 
 4 
 
 ZA aves eS 
 
 Then in the triangle a B’e, 
 Assn ZBAC =- 25°15’ - - 9.6299890 
 0.8700110 
 Isto opp. sideBc - 105 -+- 2,0211893 
 Soissn ZB +--+ 82°30 -- 9.7302165 
 To opp. sde'ac tr -') - - - \ 121d Ge 
 
 ie 
 
§1 
 
 - And, in triangle a c p, 
 
 ‘Assin’ 2’ p {i225 90°. = -"-" 10.0000000 
 
 Istoopp.sideac - - - - - 21214168 
 
 Soissin Zc --- 57° 45° - = 9.9272306 
 
 To opp. side-aD -- 101.85. - 20486474 
 1.66 
 
 100.19 yds. height a ». 
 
 Lastly, in the same triangle a c D, 
 
 Assn Z.D ---- 90° - - - 10.0000000 
 Is toopp.sideac - - - - - 21214168 
 Soiscos4c --- 57°45 - 9.7272276 
 TosidecD ---- 70.574yds. 1.8486444 
 
 8. Wanting to know the height and distance of the 
 object o, on the top of a hill, I measured from the 
 station B, a base B c of 130 yards, up sloping ground, 
 directly from o ; and having set up a staff BE, of an 
 equal height to that of the theodolite ac, I found the 
 angle of elevationo ac, at the station c, to be 27° 
 10’, and the angle of depression c a £, of the top of 
 the staff 5, 8° 45’; and at the station B the angle of 
 elevation 0 EF, was 39° 40’: required the horizontal 
 distance B H, the height ou, and the height c p of 
 the station c above B. 
 
 Here, according to the question, we have the follow- 
 ~ Ing angles: 
 E 2 
 
 LIBRARY 
 
 — 
 
 IRIVERSITY OF Hy 
 
 ae 
 
 Nate 
 
34 
 
 at 
 
 5 
 
 LZoace 27°10. LAERPSAUT 15’ sup'ZG4r 
 Y GA Bis) 82456 ZOE Ress 89°40! 
 nia — SA. 55. Loh kid Bouma 
 LSARS5" x, ORM O, 
 S655... 167) so 
 167° 80. 12° BO". ZA OF 
 Then, 
 + Sin! ZA! O:Balee' vcard 2n80 ieee oe Oe OOS 
 0.6646632 
 3 Fy Se ee ay Best @ ume Meee IE NG OS bac 
 2 SinwZoaA EB) <= "85°. 55. - -y 917683480 
 OE Ci ibie Hier itn ek wae BASS 
 : Rad,orsinZeEFO 90° - - - - 10,0C00000 
 St) Ath eiloipaty. i= a. eeraete sy 2404586 
 > 3. Sin 20. E Fae =i 89> 40% =, ~ 9 9.8050385 
 a Yen elle ta 294,91 -- 2.3519931 
 : Rad,orsin - -- 90° - - - ~ 10.0000000 
 8 OUR fee aa ale. Se tlh a eH 5469 54.6 
 2: Cos ZOEF*- = 39° 40 y= - /9.8863616 
 EF, Or BH: 271.21; - = 24333162 
 > Rad,orsin ~- 90° - - - - 10.0000000 
 *" AEorcB --- 130 --- -. 2,1139434 
 >: SinZcBporcar 8°45 -@ 9.1821960 
 Bib ihe ean 19.776 -- 1.2961394 
 
 And if = B, or rH, the height of the theodolite, be 
 added to o F, it will give the height of o above the 
 horizontal line p #. 
 
 It may here be further remarked, that in certain 
 trigonometrical operations, when a base is measured © 
 
 ." 
 
53 
 
 on sloping ground, it is sometimes necessary to ree 
 duce it to the corresponding horizontal line; which 
 may be done thus. 
 
 o=* 
 ooh 
 . 
 °? 
 
 Pr) 
 eee? 
 oe? 
 Vl A Sua 
 -o* 
 oe 
 
 Let a xB be the measured base, o 8 a theodolite, and 
 AR a staff equal in height to the instrument: also, 
 suppose Ho R to be the angle of depression of the top 
 R of the staff, below the horizontal Ime Ho; then, if 
 co. be perpendicular to Ho, the line ac, which is 
 parallel to Ho, will be the horizontal base correspond- 
 ing to as. And, by case 1. of right-angled triangles : 
 
 As rad: AB:: coSHOR (or BAC): AC (s). 
 
 AB X COS HOR 
 
 Or cm 
 2.08 rad. 
 
 It also frequently happens that the angles subtended _ 
 by distant objects, lie in planes oblique to the horizon, 
 in which case they may be reduced to the se i emia 
 ing horizontal angles, as follows ; 
 
 &y é 
 (s) If asbe S00 yards, and the 2 of depression Ho R 5°, the 
 
 . a 
 horizontal line a c will be 298.9 yards, differing from the mea- 
 
54 
 
 Let ac be any two points in the horizontal plane 
 ABC, eB adistant spire, or other object, e ac, ec a, 
 the oblique angles, taken at a and c, and eas, ecs, 
 the angles of elevation. | 
 
 Then, 
 Cos Z elevatione AB: cosgiven Zeac:: rad. 
 or sin 90° : cos reduced 4 BAC. 
 
 And cos Z elevation ecB : cos given 4 eCA?: 
 rad. or sin 90° : cos reduced ZA cB (Z). 
 
 sured base a 8 by only 1+), yards; so that, except in cases where 
 great accuracy is required, a reduction of this kind seems un- 
 necessary, when the. measured base is inclined to the horizon in 
 a small angle. b 
 (t) For by right 24 A‘, en? = ea? — A B% = ec? — CBP; 
 hence, also, ¢ a® — ec? = a3* — c B*; and by adding a c® to 
 both sides of this equation, and then dividing each of them by 
 ea*+ac?—ec*?  ap*+ac*—c B? 
 
 2 asogrwe shall have See as. a re eae 
 SN o APC 
 
 But by the demonstration of Theorem 1v. in Plane Trigono- 
 AB*-+- A C?—Cc 3B? 
 DAO MATE URS 
 
 metry, Part 2, itis shown that cos BAC = r( 
 
 ea® -- ac*?—ec* 
 and cose ac = 7 ( + AC ). 
 A Oise A 
 
 eA 
 Hence AB COSBAC=3 ¢ACOSEAC} or COS B A C==—— COs 
 Ab 
 
 yea 
 
 eac, 
 But, by right 24 A*, ea: AaBi:r:cosea B; and conse- 
 eA r 
 quently _— = ; 
 AB COS €AB 
 
 ee rcoseac 
 Whence, by substitution, we have cos Bac =__ ”. 
 
  COSEAB 
 And in the same manner it may be shown that cos acs 
 
 ._ FCOS@éCa 
 
 cosecsB 
 
55 
 
 Also 4 A ec will be reduced to Z a 3c, by taking 
 £Bpac+ ZacB from180.. And if ac bea 
 known base, the horizontal distances a B, c B, may be 
 determined by case 1. of oblique-angled triangles. 
 
 But if it should be required to reduce the angle 
 A eC, taken at the top of any eminence e 8, to its cor- 
 - responding horizontal angle a Bc, by employing only 
 the observed angle and those of depression, it may 
 be done by the following rule: Sint ape = 
 
 Tgp SE ee Ane 3). 3) s sin J y (ae | (Aec+ecu—eas) 
 
 COS €A B COSECB 
 
 Where e a3, ec, are the angles of depression of 
 the two distant objects a, c, A ec the observed angle, 
 and r = rad, or sine 90° (w). 
 
 (uw) Inthe A asc, we have, by the demonstration of the last 
 
 problem, cosanc=r(———_-____}, or, Ac? = A B? 
 PA ALBe, SM uBeC 
 AE BCCOSABC : - 
 +Bce— os mul and “iniA cae do,’ cos Ae c 
 Tr 
 
 QeaxXeccos Aec, 
 
 Qea KX ec 
 
 ea®*+ec*—a c? 
 = 7 ( 28 )> or A C?=ea?+ec?— 
 
 r 
 Whence, by equality and transposition, e a? — a B® + ec? 
 
 Py) 9 
 é@A @CcosAéc Ze, OO Bre COS ABC 
 r r 
 
 cause both ¢ a? —, A B’, and ‘e c* —.p c* = es*) it will be 
 _ ¢4 X eccosaec ABX BCCOSABC 
 r r 
 
 But, by Plane Trigonometry, ¢e 3 = 
 
 eesinecs €ACOSEAB eccoseces 
 Ae = pane BoC Saree 
 r r r 
 
 or, 
 
 ————3 — 
 
 3 
 
 os @€AXKECSINCABSINECB e€AXECCOSAEC 
 whence, by substitution, £“* = 
 
 — 
 
 r? r 
 
 €AX @€CCOS€ABCOS€ECBCOSABC 
 roy CAE. A Or, rsineABsinecsB 
 
 r 
 
56 
 
 The same formula may also be applied to the redu- 
 cing’ an angle 2 c D, subtended by any two distant 
 objects EB, DA, when taken at a point c in the hori. 
 zontal plane a Bc, by using only the Pe angle 
 
 and those of elevation. ; 
 
 —- s¥coS AEC — COS€ ABCOS EC B COS ABC; and consequently, 
 eee re 
 
 ee 
 
 COs AB Cee ¥ ( 
 COS €A BCOSe€CB 
 
 But since, by Article 26, Part 2, cosane = 
 
 aos a es Ware 
 
 r 
 1 ‘9 a ’ 
 we shall have sin? tanec = —(r—cos apc) = — 
 %) ) 
 cP] 
 x (r bt mesa ea 2 Paes 
 COS €ABCOSECB 2 
 
 2 (ae eABCOSECB4+r7SNeABSiNEeCB—r*COSAEC 
 tT Ee Se ee mee re ieee) tS 
 COS €ABCOSECB 
 
 i a a nn ee eee 
 
 COS €A B COS€ CB 
 
 3 c - NRT 
 = (by Article 20) oa i cos (€AB—ecB) —? paseet 
 
 — se 
 
 2 
 
 Pa) Se (caB—ecs) ~ oe aeet = (by Article 
 
 COS €A BCOSé€C B 
 
 29) 7 resin > (eac+eaB—ecs)sind (Mili eC Bin e Am) 5 
 COS GAB COSECB’ 
 
 Whence, sin 3a Bc 
 
 me aE eae (AeC+ecB~—eaB) 
 cos €AB COSeECB 
 as by the rule. 
 
 And in the same manner it may be shown for the following 
 
 rcoSsECD=—SINECBSINDC 
 formula, thatcosBcAs=r (——__——___ . ) 
 
 COSECBCOSDCA 
 And the sin 3 BCA as it is there given. 
 
57 
 
 Thus, sin Bc as 
 
 sing (ECD+ECB—DCA) sin} (ecD4+DCA—ECB) 
 COSECBCOSDCA 
 
 Where £ cB, De 4, are the angles of elevation of the 
 two objects, EB, D A, and Ec D the observed angle. 
 
 Besides this reduction of angles to the plane of the 
 horizon, it is also sometimes necessary to attend to 
 what is usually called the depression or dip of the ho- 
 rizon, which makes the observed angle of elevation 
 greater than it would otherwise be. 
 
 Thus, if an observer, whose eye is at D, takes the 
 altitude of an object with the sextant, by bringing it 
 to the water’s edge at B, instead of to the horizon pz, 
 the altitude will, evidently, be too great by the angle 
 EDB, which angle may be found by the following rule: 
 
 rad X AB Reka Kh AB 
 
 os ZEDB= ——-____. = 
 C ac(oras)+pbec AD 
 
 Where 4B or-4c is the radius of the earth, which 
 is known to be 3979 miles, and pc the height of the 
 observer’s eye above the surface of the earth. 
 
 Another source of error, in taking angles of eleva- 
 tion, arises from the effect of refraction, which always. 
 makes objects appear more elevated than they really ° 
 are, on account of the rays of light, in their passage 
 
 » 
 
58 
 
 through the atmosphere, being continually bent down- 
 wards, and coming to the eye in the form of a curve, 
 instead of proceeding in a right line. _ 
 
 Thus, if = be the place of the observer’s eye, EH 
 the horizontal plane, and o any elevated distant object, 
 this object will not appear in its true place at o, but 
 will be seen in the direction £ T, which 1s a tangent to 
 the curve at the point £; and therefore the apparent. . 
 angle of elevation r x u will be greater than the true 
 angle of elevation o £ u, by the.angle T E 0, considering 
 Eo asa right line (v), 
 
 E H 
 
 Various trigonometrical problems may also be form- 
 ed from the different situations which objects may be 
 
 (v) This kind of refraction, which is called terrestrial, in order 
 to distinguish it from that which affects the altitudes of the hea- 
 venly bodies, is not constant at the same elevation and distance ; 
 but is found to vary with the density and other changes in the 
 atmosphere. At the distance of 8 or 10 miles, it is sometimes 
 no more than about 30”; but, in particular states of the air, it 
 has been found to amount to upwards of 2’. Maskelyne makes 
 it -1. of the intermediate arc c B (See fig. to dip of horizon) be- 
 tween the observer and the object: Bouguer 5, Legendre 1, 
 and Gen. Roy from 4 to 31, ; which allowances differ too much 
 from each other for any reliance to be placed on them. 
 
 * 
 
59 
 
 supposed to have with respect to each other; but, in 
 general, these are only applications of the preceding 
 rules. 
 
 As, for instance, the distances of the most remark- 
 able places in a town, or of several villages from each 
 other, the plan of a camp, or of a country, &c. may 
 be taken from what has been already explained. — 
 
 Thus, if a, c, D, E, B, H, G, F, &c. be several objects, 
 the situations of which are to be laid down ina map, or 
 plan; choose a convenient situation a B for a base, 
 from which you can see all the objects, and let it be 
 as long as possible, in proportion to the most distant 
 of them. ‘Then, from the extremity a, measure the 
 anglesE AB, DAB,CAB,&c. HAB, GAB, FAB, &c. 
 And from the other extremity B measure the angles 
 CBA, DBA, EBA,&Xc. FBA,GBA,HBA,&c. And as 
 the common base A B, and the several angles of all the 
 - triangles are now known, the sides ac, AD, AE, &c. 
 and consequently the points c, p, zr, &c. may be de- 
 termined by the first rule in plane trigonometry. 
 
 But, in order to insure the accuracy of the operation, 
 the objects c, p, E, &c. should be all intersected from 
 some third station 0, in the base a B; or otherwise the 
 figure may appear, in the plotting of it, to be right, 
 
60 
 
 when it is not so, and there will be no means of know- 
 ing whether the angles have been justly taken, without 
 - going over the work again. ) 
 
 A measurement may also be carried on, or the di- 
 stance of any two remote places may be found, by 
 means of a series of triangles, formed from a measured 
 base, in a manner similar to that generally practised ir 
 taking the trigonometrical survey of a country. 
 
 8 ° 
 ‘Thus, let a B be the measured base, and c, Dp, any 
 
 two objects that can be seen from the stations a,B; then 
 
 if the angles cA B, CBA, DAB, DBA, be taken with'a 
 theodolite, or other instrument, we can thence find the 
 sides CB,C a, DB, DA; and the angle BDA. 
 
 Also, knowing the angles DAB, CAB, we know 
 their difference D ac; from which, and the two sides 
 ¢ A, DA, we can find the side cp, and the angles 
 Abdc,acpb. Andif £,F, be any other two objects, 
 visible from ¢ p, we can determine the lengths of the 
 sides Dz, DF, and the angles at = and » in a similar 
 manner, | jae 
 
 And in the same way the operations may be con- 
 tinued from one base to another to any distance ; but, 
 in order to render the conclusion more accurate, the 
 
61 
 
 mensuration should be carried on from one base to the 
 next by different sets of triangles, all leading to the 
 same two objects, and then taking the mean of the re- 
 sults. . 
 
 - The distances from the first stations to the last, may 
 also be readily determined; for the two sides pF, Da, 
 and the included angle rp a being known, we can 
 thence find the side ar; and from the sides pg, pr, 
 and the included angle 8 p F, we can find the side x Fr. 
 
 It will be proper, however, in examples of this kind, 
 to find every angle of the triangles by observation, if 
 the situations will permit, as the difference of their sum 
 from 180° will enable us, in some measure, to judge 
 of the accuracy of the work. All the principal di- 
 stances should also be laid down from a scale of equal 
 parts; because a triangle can be more accurately con- 
 structed by means of its sides than from the angles (zw). 
 
 (w) After carrying on a series of triangles, in the manner here 
 described, to some distance, it is customary to find, by an ac- 
 tual measurement, the interval of two objects whose distance 
 has been previously obtained by calculation, in order to deter- 
 mine the error of the calculated distance: which line, so mea- 
 sured, is called the Base of Verification. 
 
 A survey of this kind is at present carrying on in this country, 
 from a base of 274061 feet, first measured on Hounslow Heath ; 
 from which it appears, that by continuing the measurement to 
 Salisbury Plain, through a tract of near 60 miles, the distance of 
 two objects was there found, by calculation, from the mean 
 result of several series of triangles, to. be 36574.% feet ; and, by 
 
62 
 
 - When this part of the work ts finished, the distance’ 
 AF may be reduced to the meridian Ns; and the 
 length of the corresponding are am, of that circle, be 
 determined, as follows : 
 
 Let o be the point of the horizon where the sun 
 sets, and take the angleo as; then supposing the 
 latitude of the place a to be known, the sun’s ampli- 
 tude o a nN can be readily found, and consequently the 
 difference of these angles B 4 N. | 
 
 Hence, drawing 5, Fm perpendicular to the me- 
 ridian, we have Ba, an, and the angle B an given; 
 from which the angle am, and the sides an anda 
 may be found; also, if from the angle asc there be 
 taken the angle ann, the remainder, added to c B F, 
 will give the angle rBu; from which, and the side FB, 
 we can find Fw, or its equal mm; and thence the 
 whole meridional distance, or length of the arc am. 
 
 This being done, the difference of latitude between 
 the two stations a and F may also be easily determined; 
 for by taking, with a proper instrument, the true ze- 
 nith distances of a star, at each of these places, their dif- 
 
 an actual measurement, it was found to be 36574.3, differing 
 but little more than an inch from the computed distance, | 
 
 The area, or content, of any extent of land, measured in this 
 way, may be found thus:—Area of the A anc =ianxBcx 
 sin ZA BC, radius being unity; areaofacp=1cRxXBD xX sin 
 Z£cBpD3 area of BDE=isDxX BEX sin Z EBD; area of 
 EBF=mLEBX BF X sinZEBFs; and thus we may proceed for 
 any number of triangles into which the whole is divided. 
 
63 
 
 ference will give the difference of latitude sought ; and 
 by proceeding in a similar way, the same may like- 
 wise be determined between a and either of the other 
 stations (x). 
 
 It also sometimes happens, in making a survey, that 
 the distance between two objects c, Dp, having been 
 determined, it is required to find the distance a B of 
 two eminences A, B, which are conveniently situated 
 for extending the series of triangles. 
 
 c 
 
 This is done by measuring the angles c ap, cB, 
 DBC, DBA: andas there are not sufficient data in any 
 of the triangles to compute the other parts, we must 
 assume a value for a B, and thence compute the value 
 of ¢ p, as in the last proposition. ‘Then, as the com- 
 puted value of cp is to its true value, so is the as- 
 sumed value of a B to its true value (y). 
 
 Military sketches, or small surveys, where much 
 accuracy is not required, may also be taken by means 
 
 ———. 
 
 (x) For a more ample detail of particulars respecting this 
 part of the subject, see Trigonométrie de Cagnoli, 2d Edit. and 
 the Traité de Géodésie and Traité de Topographie de Puissant. 
 
 (y) For, by changing the value of a8 while the angles at aand 
 remain the same, the whole figure will continue similar to itself ; 
 and consequently a 8 will vary in the same proportion as c p. 
 
64 
 
 of a pocket compass, fitted to the top of a staff, which 
 being stuck in the ground, so that the needle may play 
 freely, the angular distances, or bearings, must then 
 be taken from the magnetic meridian. 
 
 Thus, let n s represent the needle, or magnetic me- 
 ridian, n the true north point, and £, w, the east and 
 west points; then, if the sights of the compass be di- 
 rected to the object a, and the angle n o a, for example, 
 be 40°, the object is said to bear n. w. 40°; and if the 
 sights, when directed to the object 8, make the angle 
 NOB 110°, it is said to bear nN. E. 110°. 
 
 The compass will, likewise, be found useful in re- 
 connoitring a country with a map or plan, when the 
 direction of the meridian is laid down, and we know 
 the magnetic variation; and, in such cases, a distance 
 may be measured by pacing, in order to adapt a scale 
 to the plan or sketch. 
 
 Note. ‘The variation of the magnetic needle is, at 
 this time, between 23° and 24° westward, at London. 
 
 MISCELLANEOUS EXAMPLES. 
 
 1. At B, the top of a castle, which stood on a hill 
 near the sea shore, the 2 of depression 4 B s, of a ship 
 at anchor, was 5° 4, and at rx, the bottom of the 
 castle, its depression N R s was 4° 46%, required the ho- 
 
65 
 
 rizontal distance of the vessel, and the height of the- 
 building above the level of the sea, supposing the 
 castle itself to be 68 feet high. 
 
 Ans. A s 12891.61 feet, and AB 1142,986 feet. 
 
 2. Wanting to know the distance between two ob- 
 jects a, B, which could only be seen from a particular 
 place p, I set up two poles at c, E, and took the 4$ 
 ADc 89° 10", ADB 72° 35’, and BpE 54° 36’... I 
 then measured a distance px of 216 yards, and an- 
 other p c of 200, and took the Z° 3B ED 88° 34’, and 
 ~pca 50° 27’: required the distance a B. 
 
 Ans. AB 367.56 yards. 
 
 3. Wanting to know the distance between two in- 
 accessible objects A, B, | measured a base c p of 500 
 yards: at c the Z2Bc D was 48° 20’, and the ZacB 
 47° 48’; and at p the 2 cpa was 54° 19’, and Z 
 ADBA7° 15; required the distance a B. 
 
 Ans. AB 742.682 yards, 
 
4. Wanting to know the distance 4 c of a hill from 
 the station a, and also its height oc, we measured a 
 base A B of 376 yards, on ground nearly level, and at 
 the extremities a, B, observed the horizontal angles 
 BAO 43° 17, and ABO 78° 29’, and at 4, the angle 
 of elevation oc was 4° 41’: required the distance — 
 Ac, and height co. . 
 
 | Ans. ac 431.898, 0 ¢ 35.382 yards. 
 Oo 
 
 5. Ata mileeiane nN on theascending road ns, | 
 observed the angle s N w between the next mile-stone 
 s and the windmill w, on the top of a hill, and found 
 it to be 47° 36’, and the angle of elevation w Nn P was 
 8° 47’; also at the mile-stone s, the angle Ns w was 
 91° 5’: from which it is required to find the hori- 
 zontal distance n P, and the height p w. 
 
 Ans. N P 2659.492, and pw 175.8662 yds. 
 
 % 
 er 
 
67 
 
 6. Wanting to know the height of a castle cz, 
 standing upon a hill, and the ground not permitting 
 me to retreat from it in a right line, I measured a base 
 D A of 57 yards, and at p took the angles c DB 59°, 
 CDE 26°, and apc 72°20. Ata I also took the 
 angle c aD 65° 30’: from which it is required to find 
 the height of the castle c z, and that of the hill g g, 
 above the level of the first station p. 
 
 Ans. c E 40.38671, B E 25.8431 yards. 
 ot 
 
 D Lb 
 7. It is required to find how far the Peak of Tene- 
 riffe can be seen at sea, supposing its height a B to be 
 2 miles, and the radius of the earth pc 3979 miles. 
 Ans. dist. ACG 126.2036 miles. 
 
 8. From a window A, near the bottom of a house, 
 which seemed to be on a level with the bottom of a 
 church c p, I took the 4 of elevation c 4 p of the top 
 of the steeple equal to 40° 21’, and from another win- 
 dow B, 20 feet directly above the former, the Z of ele- 
 vation C B E was 37° 36°: from which it is required to 
 find the height and distance of the steeple. 
 
 Ans. C D 213.8357, AD 251.7008 feet. 
 E2 
 
68 
 
 9. Being at the station a, on an horizontal plane, 
 and. wanting to know the height of a tower c D, placed 
 on the top of an inaccessible hill, I took the angle of 
 elevation p a £, of the top of the hill, equal to 40° 15’, 
 and of- the top of the tower ca Ez equal to 51°13; 
 then, measuring on, in a direct line from it, to the 
 distance a B of 110 yards, I found the Z of elevation 
 of the top of the tower c BE to be 34° 45°: what then 
 is its height? Ans, height c p 55.1337 yds. 
 
 10. Suppose a and c to be two stations on sloping 
 ground, o an object on the top of a hill, and the 4° 
 OC A, OAC, measured with a sextant, to be 78° 23’ and 
 64° 11’ respectively: also, suppose the 4 of elevation 
 at Als 7° 26’, and at c 6° 22’; what are the horizontal 
 distances and height of the object, ac being 430 
 yards? Ans. AG 686.848, cB 632.891, 
 
 OB 70.61731, 06 91.64509 
 
69 
 
 11. Wanting to know the distance between two in- 
 accessible objects H, M, and also their heights mr, 
 fi P, we measured a base a B of 680 yards, on. ground 
 nearly horizontal, and at the extremities a, B, took the 
 following angles: 
 ph in 40° 26°N { aBH 42°26 = { MAR 7°58 
 < 
 From which it is required to find the heights and 
 distance. 
 
 ssasisee HM 1235.38 eas MR 202.7185,and HP69.668 
 
 se) <q 
 MAH 57° 30° x HBM 72° T HBP 3° 47° 
 
 12. Knowing the distance of two inaccessible ob- 
 "jects Cc, D, to be 2860 yards, and wishing to determine 
 the distance of two other objects a, B; I took the angles” 
 c AD 49°20’, c aB 88°99’, pBc 53°16’ andDBA 
 84° 39°; from which it is required, by means of aD 
 assumed distance, to find the true distance a B. 7 
 
 Ans. A B 3080.49 yards. 
 
 13, In the year 1784, a base B c being measured on 
 ’ Blackheath, of a mile in length, the angles of elevation 
 
70 
 
 of Lunardi’s balloon were taken, at the same time, by 
 observers placed at its two extremities and in the mid- 
 dle; the one at B being 46° 10’, that at p 55° 8’ and 
 that at c.54° 30’: required the height oa of the 
 balloon (7). Ans. 0 A 1.04298 miles. 
 
 14. Supposing it were possible to see a light-house 
 or other object a, in the horizon, at the distance of 
 180 miles, it is required to find its height a B, the 
 circumference of the earth B c p being 25000 miles. 
 
 Ans. 4.1, miles nearly, being about the height of 
 
 (y) As the height o a, in this question, cannot be determined 
 
 - by any of the direct rules before given, it will be proper to show 
 
 how the formula for the solution of it is obtained ; which may 
 
 OA COtOBA, 
 r 
 
 be done thus: By case 1v, of right 24 A’, an= 
 
 OACOLOPA OACOLOCA 
 Pes LO SAO And by 2a well-known 
 
 r r 
 prop. in Geometry, a B* + ac*= 2a Pp? + 98 P%, or A B® + 
 0 A? cot? OBA 
 
 r® 
 
 ac?—2ap*= 28 p*; hence, by substitution, 
 
 g 2 a 2 
 Oa* cot* OCA 20 A*COt*OP A 
 Ee ai ee ee A Pe == 2B p*, Or Oo A” 
 7 
 277 Bp? 
 
 aati ees aies ep reyes APLAR, ERR BUTTE and consequently 
 cot? oBA+ cot*oca—2 cot? oP A 
 
 B Pp? 
 i ay ll 
 
 cot? oBA+ cooca—2cot*ora 
 
71 
 Chimborazo, the highest mountain of the Andes, and, 
 probably, in the world. 
 
 15. Suppose 0 B to be an object standing on the lio: 
 rizontal plane aBc, and that ac is a measured base 
 of 260 yards, at the extremities of which a, c, the 
 following angles have been taken, o ac 54°46’,0CA 
 63° 54’, and the Z of elevation oaB 6° 42’; it is 
 required to find its height on, the reduced Z aBc, 
 and the horizontal distances AB, CB. 
 
 Ans. AB 264.2875, C B 240.0387, 0 B 31.0466 yds. 
 and Zaxc 61°50. 
 
 16. At the top of a tower c p 160 feet high, I took 
 the angle a cB, subtended by two distant objects a, B, 
 69° 37’, also the angles of depression c aD 1° 46’, 
 and c BD 2° 17’, from which it is required to find the 
 reduced £aDB, and the distances AB, BD, DA. 
 
 Ans. Z aDB 65° 43’, B D 4032.76, D A 5187.408, 
 AB 5342.652. 
 
17. At the point c in the horizontal plane az c, I 
 took the angle ‘ge c p 58° 20’ subtended by the tops of 
 two towers, whose heights D a, EB, were known to 
 
 be 169 and 213 feet respectively, and also the angles | 
 
 of elevation Dc A 3°18’, andECB 4°21’; ; from which 
 it is required to find the distances CA, CB, and AB. 
 Ans. CA 2930. 993, cB 2800, 125, AB 2801, 2579. 
 
 C A ta 
 18. Observing an object o a at a distance, I took “its 
 angle of elevation o B A at the place where I stood, 
 and found it to be 50° 23’; I then measured a distance 
 p B of 60 yards, in the most convenient direction the. 
 
 s) 
 
 ground afforded, and at this station found its elevation “* 
 
 o PA to be 40° 33’; after which I measured on, in the 
 same line, 50 yards further to c, and at this place 
 found its elevation o c A to be 30° 40’; from, which 
 it 1s required to determine the height of fi’ object, and 
 its distan¢e from each of the three stations B, P, C (z). 
 Ans. AB 65.417, A P 92. 368, eC 132. 573, aa 
 “‘heighto a. 79.029. 
 
 _ (#) The height o 4, in this questiori,,may be obtained in a 
 similar manner with that of the balloon, in quest. 14, as follows: 
 
 é 
 
 « 
 
 ™ 
 * 
 
 VR Poo 
 
19. Observing three objects a, B, c, whose distances 
 astinder are known to be as follows: a B 5307 yards, 
 BC,7022 yards, and ac $556 yards, I took their 
 angles of position.from the place where I stood p, 
 which was furthest from the object a, and found the 
 
 nels’ > 
 
 . OAcotoBA OACOtLOPA 
 Dy case wy right’ 27 Ae) 4 Bs oe AP = es 
 
 >? 
 
 r r 
 
 OACOLOCA , 
 By SOP ee ek aieade fede But, by case vi, of oblique 24 A’, 
 r » 
 
 BP? + a p? — A B? . 
 Me —, and cos c P A, or its equal 
 
 COS "BP A= Tf X 
 
 ee, 
 
 sare! QBPX PA 
 ye Pp _ BP*--A P*—A B* 
 —COSBPA=m==r xX crt Ea Hence, Boe ote iat. 
 Dick se? wd; * BP 
 
 cp?+ a r?— ac? ie + 
 ate a: Th “©; or, by transposition and addition, ¢ P 
 CP 
 # x AB?+ BPXAC*—(CP+BP) xX Be eat © P)XCPXBP. 
 
 oa® cot? OoBA 
 ‘ _And consequently, by substitution, OF Mri algpayth i at 
 7 
 rm x 0 A* cot*oca. o A# cot? OPA 
 
 %. 5 BOM il etched : erat) 
 r r 
 ef 
 
 Xx CPXBP; ; or, by reducing and simplifying the equation 
 
 % ~ ae m(ce+pp)xcPr x BP 
 ‘ AR PR SEA Leak De AR ee A | 
 . o CP COP OBA + BP co OCA—(CPHBF)COMOPA’ 
 ‘ Or,e A=r . (cep + BP) X CPX BP 
 a Mrs ¢ pcot?0B A+B P cot? oc A— GP-B iP color A: 
 ek 
 
 wae ba which formula the height o 4 may be readily computed. 
 
 : ; 
 ap 
 
14 
 
 Zane to be 17° 35, and BDA 26° 21’; required 
 my distance from each of the objects. 
 Ans. DB 8544.854, DA 11369.175, pc 9916. 558 yds. 
 
 ZK. 
 
 20. Supposing the objects a, B, c, as seen from the 
 point D, which is the nearest to a, to stand as im the 
 figure below; and that their distances are aB 5307, 
 Bc 7022, and ac 3556 yards, the angle Bp a being 
 27°13’, and c D A 29° 25’; it is required to determine 
 the distances DA, DB, DC. 
 
 Ans. D A 3494,.058, Dc 6158.059, DB 8167.893 
 
 yards. 
 E 
 
 D 
 
 21. Supposing the three objects a, B, c, as seen 
 
 from a point p, within the triangle, to stand as below; 
 and that their distances are AB 5307, Bc 7022, and .— 
 Ac 3556 yards, the angle B pc being 122° 37’, and | 
 
 ADC 131° 54’; required the distances D A, Dc, and 
 
 DB. Ans. D a 870.925, DB 5007.778, Dc 2914.783 
 
 yards, 
 
22, Supposing the objects a, B, c, as seen froma 
 point D, in the line a B produced, to stand as in the 
 figure below; and that their distances are a B 5307, 
 Bc 7022, and ac 3556 yards, the angle 8 pc being 
 18° 7’; it is required to find the distances p a, DC, DB. 
 Ans. DB 4481.997, Dc 11142.11, Da 9788.997 yds. 
 
 A 
 
 D Cc 
 
 23. Supposing the three objects a, B, C; as seen 
 from a point D, in the base Bc, to stand as below; 
 and that A Bis 5807, a c 3556, and Bc 7022 yards, 
 
 ~ the angle a pz being 108° 5’: required the distances 
 DA, DC, and DB. 
 
 Ans. D B 3761.025, Da 2754.544, and Dc 3260.975 
 
 | _ yards, | ; 
 
 j xf 
 
76 
 
 24, Supposing the three objects a, B, Cc, to bein a 
 right line, as below; and that their distances are.ac 
 1297, a B 7022, and Bc 5725 yards, the anglea pe 
 being 19°10 andppc 84° 139’: required the distances 
 DA, DC, and DB. | 
 
 Ans. D A 3604.086, D C 2873.22, DB §249.771. 
 E “i ss 
 
 D | t 
 25. At a certain place in the county of Kent, the 
 fort on Shooter’s Hill bore from men. gz. 444°; and 
 after going 15 miles in the direction n. w. 692°, I 
 perceived the fort again, which now bore n. £. 674°: 
 required my distance from it at each station. 
 Ans. Ac 26.5773 miles, and Bc 35.4352 miles. 
 
 > 
 
 26. Being ata certain place a, St. Paui’s church c, “4 
 at London, bore from me Nn. £. 151°, and after tra-_ 
 velling 20 miles further to B, in the direction N. w. © 
 421°, it bore N. E. 514°: required my distance from 
 it at the last place of observation. ° 4% 
 
 wie Ye 
 ‘Ma 
 ¥ 
 
 Ans. B Cc 28.6052 miles. — 
 
 4 
 % 
 
97. From a ship at sea A, I observed a point of land 
 c to bear N. E. 1024°, and after sailing 13 miles in the 
 direction N. E. 401° to B, it bore N.E. 1414°: required 
 the distance of the last place of observation from the 
 point of land. Ans. Bc 18.099 miles. 
 
 28. Coasting along shore, I observed two headlands, . 
 the Ist, B, bore n. w. 232°, and the 2d, c, N.E. 
 36° 56’; then steering 12 miles in the direction n. £. 
 17° 52’, the 1st headland bore n. w. 681°, and the 
 ad n. 3. 70° 45’: required the bearing and distance 
 
 _ of the two headlands from each other. 
 Ans. dist. Bc 17.178 miles, and bearing of c from 
 
 B, N.E. 95° 581. 
 
78 
 
 29. Let a,c, be any two points in the horizontal 
 plane a BC, distant from each other 627 feet, ande pa 
 distant spire, or other object ; then, having the oblique 
 
 Zeeac 57°12 ec A 49° 37’, taken at A and c, and 
 the Z$ of elevation e AB 15° 12’, ecB 13° 45’: it is 
 required to find the height of the object e8, its di- 
 stance from each station, and the reduced angles a cB, 
 pac. Ans. ZAacB 48° 9'49%, ZBac 55° 51’ 4” 
 AB 481.473, c B 534.811, height eB 130.867. 
 
 & - 
 
 80, Wanting to know my distance from an object o, 
 on the other side of a river, and having no instrument 
 for taking angles, I took two stations a, B, 400 yards, 
 asunder, and then measured a c, BD, each 100 yards, 
 in a direct line from the object : I also found the dia- 
 gonal a p to be 450 yards, and Bc 460: required the 
 distance of o from each of the stations A, B. 
 
 Ans. A 0 444.5657, BO 415.5711. 
 
 AO 
 
 Cc D 
 
 $1. The side az of a regular pentagon being 180 
 toises, the face of the bastion A ¢ 50, and the perpen- 
 
79 
 
 dicular kK L 30, it is required to find, by trigonometri- 
 cal calculation, all the other lines and angles of the 
 fortification, supposing the lines of defence Au, BG, 
 to be equal to the lines drawn from the salient angles 
 A, B, to the shoulders p, c. 
 
 These lines and angles, when found, may be com- 
 pared with those determined by construction in Mul- 
 ler’s Elements of Fortification ; or with a Table of the 
 various dimensions of such plans, given by Robertson, 
 at the end of his Navigation. 
 
 Note. A number of other questions might have been 
 given, in this part of the work, relating to the redu- 
 cing of angles, taken at a small distance from one of 
 the stations, in certain geodetical operations, to the 
 true angles at that station, which reduction becomes 
 necessary when the centre of the instrument cannot be 
 exactly placed in the vertical line occupied by the axis 
 of a signal, and in other similar instances: but as the 
 error arising from this cause, which seldom amounts 
 to more than a few seconds, is commonly neglected, 
 except in cases where the greatest accuracy is required, 
 the reader is referred for further particulars on this 
 head, to the works of Cagnoli and Puissant, before 
 mentioned, | 
 
SPHERICAL TRIGONOMETRY. 
 
 SPHERICAL Trigonometry is the science which. treats 
 of the properties and relations of spherical triangles, 
 and of the methods of determining their sides and 
 angles. eM 
 
 1. A sphere, or globe, is a solid contained under 
 ohe uniform round surface, which is every where 
 equally distant from a point within it, called its centre; 
 
 as ABCD. 
 ‘ 
 
 C 
 2. A diameter, or axis, of a sphere, is a right line 
 passing through the centre, and terminated on each 
 side by the convex surface; as A B. 
 
 3. A great circle of the sphere, is that which divides 
 the surface of it into two equal parts; and a small 
 circle is that which divides it into two unequal parts. 
 
Sl 
 
 Thus a Bc isa great circle of the sphere, and pura 
 small circle (¢). 
 
 4. Hence, also the plane of any great circle passes 
 through the centre of the sphere, and divides the solid 
 into two equal parts; as ABCD. 
 
 5. The poles of any circle, are the two extremities 
 of that diameter, or axis, of the sphere, which is per- 
 pendicular to the plane of that circle ; thus p, p are the 
 poles of the circle a Bc. 
 
 p 
 
 —— 
 
 A. ani 
 
 | 
 Mas C 
 ‘ “p 
 
 7 
 
 6. Hence, either pole of any circle is equidistant 
 from every part of its circumference ; and if it be a 
 
 (4) Small circles of the sphere are not used in trigonometrical 
 computations, on account of the diversity of their radii; but it 
 may be observed, that any arc of a great circle, is to an arc of 
 a small circle, of the same number of degrees, as the radius of the 
 sphere is taithe sine of the distance of the small circle from its pole. 
 Thus a3 : cp :: rad of the sphere : sine of pc. (fig. to def. €.) 
 
 G 
 
82 
 
 great circle, its pole is 90° from the circumference ; 
 thus if aB be an arc of a great circle, Pa ig equal to 
 PB, being each ac and in the small circle c p, 
 
 Pc is equal to PD. 
 ETN ste 
 
 (PRY 
 
 cy 
 
 7. A spherical angle, is the inclination, or opening, 
 of the arcs of two great circles of the sphere, which 
 meet each other in a point on its surface; as ABC. 
 
 A 
 
 B c 
 8. ‘The measure of a spherical angle, is the arc of a 
 great circle, drawn at the distance of 90° from its angu- 
 lar point, and intercepted by its two legs; thus if B a, 
 BC, be quadrants, ac will be the measure of the 
 angle ABC, A 
 
 reer 
 
 9. A spherical triangle, is a portion of the surface 
 of a sphere, contained by the arcs of three great cir- 
 cles; as A Bc. A 
 
83 
 
 i0, A right-angled spherical triangle, is that which 
 has a right angle, or one of 90°; as ABC. 
 A 
 
 B C 
 
 11. A quadrantal spherical triangle, is that which 
 has one of its sides a quadrant, or 90°; as ABC. 
 
 - An oblique-angled spherical triangle, is that 
 oe has each of its sides, or angles, greater or less 
 than 90°; as ABc. 
 
 / 
 F B Baie Ot 
 * 
 
 13. A great circle of the sphere is perpendicular to 
 another circle, when its plane is perpendicular to the 
 plane of that circle, and vice versa; thus the circle 
 PBP Is perpendicular toABC. 
 
84 
 
 14. Any two sides, or angles, of a spherical triangle 
 are said to be dike, or of the same kind, when they are 
 each equal to, each greater, or each less than 90°. 
 
 15. And if one of the sides, or angles, be equal to, 
 or greater than, 90°, and the other less, they are said 
 to be unlike, or of different hinds. 
 
 AXIOMS. 
 
 1. Every section of a sphere, by a plane passing 
 through it, is a circle. 
 
 2, The centre of a sphere is the centre of all its 
 great circles, and its axis is the common section of the 
 planes of all the great circles which pass through its 
 two extremities. | 
 
 3. A great circle can be drawn through any two 
 points on the surface of a sphere. 
 
 4, All parallel circles of the sphere have the same 
 poles; and no two great circles can have a common 
 pole. . 
 
 5. Any two great circles of the sphere cut each 
 other twice, at the distance of 180°, and make the 
 angles at each point of section equal. 
 
 6. A great circle, which passes through the poles of 
 any other circle, cuts it at right angles; and, if a great 
 circle cut any other circle at right angles, it will pass 
 through its poles (w). | 
 
 (2) Most of the principles, here laid down as axioms, will be 
 rendered sufficiently evident, by considering the position and na-: 
 ture of the circles usually drawn on a common globe. The 6th 
 in particular, which is, perhaps, not quite so obvious, may be 
 readily conceived, from observing that all the meridians pass 
 through the north and south poles, and are perpendicular to the 
 equator, and to all the parallels of latitude. 
 
85 
 
 GENERAL PROPERTIES OF SPHERICAL TRIANGLES. 
 
 1, Any side, or angle, of a spherical triangle, is less 
 than a semicircle, or 180°. 
 
 2. The greater side of any spherical triangle, is 
 opposite to the greater angle, and the less side to the 
 less angle. 
 
 3. The sum of any two sides of a spherical triangle, 
 is greater than the third. side ; 3 and their difference is 
 less than the third side. ; 
 
 4, The difference of any two sides of a spherical 
 triangle 1s less than a semicircle, or 180°; and the sum 
 of the three sides is less than 360°. 
 
 5. The sum of the three angles of any spherical 
 triangle, is greater than two right-angles, or 180°; and 
 less than six right-angles, or 540°. 
 
 6. The sum of any two anglesof a spherical triangle, 
 is greater than the supplement of the third angle. 
 
 7. If the three sides of a spherical triangle be equal 
 to each other, the three angles will also be equal; and 
 vice versa. 
 
 8. If any two sides of a spherical triangle be equal 
 to each other, their opposite angles will also be equal, 
 and vice versa. | 
 _ 9. If the sum of any two nae of a spherical tri- 
 angle be equal to 180°,-the sum of their opposite 
 angles will also be equal to 180°; and vice versa. 
 
 10. If the three angles of a spherical triangle be 
 all acute, or all right, or all obtuse, the three sides will 
 be, accordingly, all less than 90°, or all equal to 90°, 
 or all greater than 90°; and vice versa. 
 
86 
 
 11. Half the sum of any two sides of a spherical 
 triangle, is of the same kind as half the sum of their 
 opposite angles; or the sum of any two sides is of the 
 same kind, with respect to 180°, as the sum of their 
 opposite angles. 
 
 To these may also be dae the following properties 
 of the polar triangle; by which the data, in any case, 
 may be changed from sides to angles, and from angles 
 to sides. | . 
 
 If three arcs of great circles be described from the 
 angular points A, B, c, of any spherical triangle 4 Bc, 
 as poles, or at 90° distance from them, the sides and 
 angles of the new triangle p F £, so formed, wwill be 
 the supplements of the opposite angles and sides of the 
 other ; and vice versa. 
 
 Thus, peE=180°—c; EF=180°—Aa; FD=180°—B: 
 and p=180°—3Bc; E=180°—ac; F=180°—AB. 
 Also, AB=180°—F; BC=180°—p; ac=180°—E: 
 and A=180° FES B=180°—Fp; C=180°—DE 
 (). | 
 OF THE 
 AMBIGUOUS CASES OF SPHERICAL TRIANGLES. 
 
 Any three of the six parts of a spherical triangle be- _ 
 
 (v) The discovery of this useful property of the polar triangle — 
 is commonly attributed to Lansberg, who first gave it in his 
 Geometria Triangulorum, published in 1591. 
 
t+ 
 
 ditions of the question. 
 
 87 
 
 ing given, the rest may always be found; except that 
 in what are usually called the-two amliguous cases, the 
 data are sometimes insufficient for limiting the triangle. 
 
 These cases are, when two sides and an angle oppo- 
 site to one of them, or two angles and a side opposite 
 to one of them, are given, to find the rest; in which 
 instances there may be two triangles having the same 
 data; and, consequently, if there be no other restric- 
 tion or limitation, either of them will answer the con- 
 
 Thus, in the triangle a 8c, let there be given the 
 two sides AB, AC, and an opposite angle B ; then if 
 Ac can be drawn equal to a c, there will, evidently, 
 be two triangles, asc, and azBc, which have the 
 same given parts; whence the other angles may be 
 either acB, (=acC) or its supplement 4c B; or BAC, 
 
 or BAC; and the remaining side will be Bc or Bc. 
 
 A 
 
 C 
 CHER we 
 
 Also, in the triangle ac, let there be given the 
 two angles B and ac, and an opposite side Ac} 
 then if a c can be drawn equal to a c, and-B a, Bc, be 
 continued till they meet in p, there will be two tr- 
 
8&8 
 
 angles ABC, and apc, which have the same given 
 parts; whence the other sides may »be either a8, or 
 its supplement AD; or Bc or Dc; and the remaining 
 angle willbe BACorDAG 
 
 But as A c cannot, in all cases, be made equal to ac, 
 the question may be limited, or not, according to the 
 conditions of the data; which circumstance may be 
 readily known, by considering that half the sum of any 
 two sides of a spherical triangle is of the same kind as 
 half the sum of their opposite angles, and taking the 
 triangle accordingly. 
 
 Or, since the greater side of every spherical triangle 
 is Opposite to the greater angle, if only one of the va- 
 lues of the angle or side first found, or its supplement, 
 agrees with this theorem, the triangle is limited, being 
 that to which this value belongs. But if both the va- 
 lues are in conformity with the rule, the triangle is 
 ambiguous (w), 
 
 (w) Delambre, p. 469 Trig. de Cagnoli, 1st edit. and p. 100 
 Tablis des Log. de Borda, has given a theorem for determining, 
 a priori, whether the A be ambiguous or not; but it is easy to 
 show that the rule is not general, 
 
 Legendre also, p. 402 Elémnts de Géom. 4th edit. and ‘Lacroix, 
 p- 68, Elements de Tr'g. 2d edit. have pointed out all the cases 
 which furnish either one or two solutions; but they are too 
 
89 
 
 If the triangle a B.c be right-angled, or quadrantal, 
 there will be only one ambiguous case; which is when 
 a side a c and its opposite angle B or D are given, to 
 find the rest. For, if c be aright angle, it is plain 
 that the hypothenuse may be either 4 B, or its supple- 
 ment a Dp, and. the remaining side and angle either Bc 
 and Bac, or their supplements cp and c AD. 
 
 Also, if AB be a quadrant, it is evident that the 
 hypothenusal angle. may be either a cB, or its supple- 
 ment ac D, and the remaining side and angle either 
 Bc and Bac, or their supplements cp and cap. 
 And as the greater side in each of the triangles a Bc, 
 4 Dc, will be opposite to the greater angle, the ques- 
 tion, in either of these cases, will always admit of two 
 
 -solutions. 
 
 OF RIGHT-ANGLED SPHERICAL TRIANGLES. 
 
 The different cases or varieties that may happen in 
 the solution of right-angled spherical triangles, in which 
 two things, together with the right angle, are always 
 given, to find a fourth, are, in all, sixteen. But if 
 these be restricted to such as depend upon the same 
 principles, they may be reduced to six; or, when pro- 
 perly combined, to the three following general for- 
 mulz, which can be more easily remembered than if 
 they were expressed separately : 
 
 numerous to be remembered without reference to the table; and 
 are, indeed, of little practical use, as the ambiguity, when an 
 example is proposed in numbers, can always be readily disco- 
 vered from the known affections of the sides and angles. 
 
90 
 
 sin its opp. 4 X sin hyp. 
 1.rX sineith. leg = or | 
 cot its adj'Z X tan other leg. 
 _ (Cos its opp. leg & sin other Z 
 9,rx cos eith Z4= - or 
 . tan its adj'. lex x cot hyp. 
 cos one leg X cos other leg 
 8. ¥ 0 cos: Hyp/= | or 
 cot one 4 X cot other Z. 
 And, in order to apply these forms to every case of 
 right-angled spherical triangles, by converting them 
 into analogies, it will be sufficient to remark, that any 
 one of the terms on one side of the equation, is to 
 either of the terms on the other side, as the remaining 
 one of the latter is to the remaining one of the former ; 
 observ)», at the same time, that when a leg or an 
 angle is said to be opposite to another angle or leg, it 
 will be adjacent to the remaining one; and viceversa (2). 
 
 AFFECTIONS OF RIGHT-ANGLED SPHERICAL 
 TRIANGLES. 
 1. The legs are of the same kind as their opposite 
 angles; and conversely. 
 
 (x) In plane trigonometry, the knowledge of the three angles 
 is only sufficient for determining the ratio of the three sides, and 
 not their absolute values. But, in spherical trigonometry, where. 
 the sides are all arcs of great circles, they can be obtained from 
 the three angles, without any other data. 
 
 Another remarkable, difference between plane and spherical 
 trigonometry is, that in the former, the third angle may always 
 be determined from the other two; whereas in the latter, all the 
 three angles are independent of each other, and must, therefore, 
 be found separately. 
 
91 
 
 2. The hypothenuse is less or greater than 90°, ac- 
 cording as a leg and its adjacent angle, or the two legs, 
 or the two angles, are like or unlike. 
 
 3. A leg is less or greater than 90°, according as 
 its adjacent angle and the hypothenuse, or the other, 
 leg and the hypothenuse, are like or unlike. 
 
 4, An angle is acute or obtuse, according as its ad- 
 jacent leg and the hypothenuse, or the other angle 
 and the hypothenuse, are like or unlike. 
 
 hiss ay PROPERTIES OF RIGHT-ANGLED SPHERICAL 
 TRIANGLES. 
 
 1. If the hypothenuse be 90°, one of the legs and 
 its opposite angle will-be each 90°; and the other leg 
 and angle will be both measured by the same number 
 of degrees. 
 
 2, And if a leg, or an angle, be 90°, the opposite 
 angle, or leg, and the hypothenuse, will be each 90°; 
 andthe other leg and angle will be both measured by 
 the same number of degrees. 
 
 3. If a leg be less than the hypothenuse, their sum 
 will be less than 180°; and if it be greater than the 
 hypothenuse, their sum will be greater than 180° (y). 
 
 (y) Properties similar to this, and the following one, are given 
 by Cagnoli, p. 244, Traité de Trig., and by. Maskelyne, in his In- 
 troduction to Taylor’s Logarithms ; but they are so expressed, in 
 each of these works, that, if followed without any other restric- 
 tion, they would frequently lead to an impossible triangle. ‘The 
 same observation may also be applied,to the two corresponding 
 properties of quadrantal spherical triangles. 
 
92 
 
 4. If a leg be less than its opposite angle, their sum 
 will be less than 180°; and if it be greater than its op- 
 posite angle, their sum will be greater than 180°. 
 
 5. The difference of the two oblique angles is less 
 than 90°; and their sum is greater than 90°, and less 
 than 270°. 
 
 6. The three sides are either all equal to, or all less 
 than, 90°; ortwo of them are greater than 90°, and 
 the other less (z). 
 
 The six cases of right-angled spherical triangles, 
 before mentioned, may be ranged as follows: 
 
 [A leg and its opp. Z ) 
 A leg and its adj’. Z | 
 
 ay ay ‘he hyp. and a leg 
 nes The hyp. and an Z 
 The two legs | 
 
 (The two 28 J 
 
 ‘to find the other parts. 
 
 CASE I. 
 
 _ When a leg and its opposite angle are given, to find - 
 the rest, 
 
 1. To find the other leg. 
 As rad : tan giv. leg :: cot opp. or giv. 4 : sin 
 other leg. 
 Which leg may be either an arc less than 90°, or its 
 supplement. 
 
 {z) A right-angled spherical triangle may have either, 
 1. One right Z, and two acute, or two obtuse 28; 
 2. Or two right Z%, and one acute, or one obtuse 2; 
 3. Or all its three Z* may be right 2°. 
 
4S 
 
 2. To find the other Z. 
 As cos given leg : cos opp. or giv. 4 :: rad: sin 
 other Z. : 
 Which Z may be either an acute Z, or its -supple- 
 ment. 
 3. To find the hypothenuse. 
 As sin giv. Z : sin opp. or giv. leg :: rad : sin hyp. 
 Which hyp. may be either an arc less than 90°, or 
 its supplement. 
 
 , EXAMPLES. 
 1. In the right-angled spherical triangle a8 c, hav- 
 ing the leg Bc 42° 25’, and its opposite Za 46° 30’, 
 to find the rest. 
 
 BY CONSTRUCTION. 
 Pp A. 
 
 1. Describe the circle a D ad with the chord of 60°; 
 and draw the diameters aa, pd, at right angles to 
 each other. 
 
 2. Set the semitangent of the complement of the 
 angle a (43° 30°) from o to m, and through the three 
 points a, m, a, describe a circle. 
 
 3. From o as a centre, with the semitangent of the 
 complement of Bc (47° 35’) as a paar deserie an 
 arc, cutting the circle Am a in B. 
 
94: 
 
 4, Through the points o, B, draw the line o Bc; and 
 ABC, or aBc, will be the A required, each having the 
 same data; which shows this case to be ambiguous. 
 
 ‘To measure the required paris. 
 
 1. Set off the semitangent of the given Z a (46° 30’) 
 from o to p; and take c P equal to the chord of 90°. 
 
 2. Through the points B, p, draw the line pe BT, 
 cutting the circle in m and r. 
 
 3. Then pn, taken on the scale of brea wives the 
 ZB 68°48, ar, taken on the same scale, gives. AB 
 68° 25°, and ac on the same scale is GO° 6’. 
 
 BY CALCULATION. 
 
 suhohtad, ofisin’: "+. 2902 v24->>.-410.0000000 
 > Tanpe ----. 42°95’ --.- 9.9607842 
 se Ot LAT use On ome 9. OF 7250s 
 Sin ac 60° 6 55” or 119° 53’ 5” + -9.9380342 
 CGS BC = sae ig 05’ 9 8689088 
 sc Cos 4 A) -#etewee 46° 30’ 3c.) 98378190 
 ¢: Rad, or snore. YO". ~ =... 10.0000000 
 > Sin ZB 68°48'45’or111°11'15” 9.9696034 
 Sin Le A sien a\ oy 4630 Wwe OB BO SRO? 
 > SIN BG). -.- - ari 42° 95! 2 NO BORggaDG 
 >: Rad,orsin -- 90° --- = 10,.0000000 
 
 ‘> Sin aB 68° 25’ 3” 0r111°34' 57” 9.9684208 
 
 ee ee 
 
 INSTRUMENTALLY. 
 
 1. Extend the compasses from 46° 30’ to 42° 25’ on 
 the line-of tangents, and that extent will reach, on the 
 sines, from 90° to 60° 6’, the side a c. 
 
95 
 
 2, Extend from 47° 35’ (the comp’. of Bc) to 43° 
 30’ (the comp' of 4 a) on the sines, and that extent 
 will reach, on the same line, from 90° to 68° 48’ the 
 ZB. 
 
 3. Extend from 46° 30’ (2a) to 42° 25’ (Bc) on 
 the sines, and that extent will reach, on the same line, 
 from 90° to 68° 25’, the side a B (a). 
 
 2. In the right-angled spherical triangle a Bc, 
 
 Gn Thelegpci5° 9 Ac 36° 52! 44" or 143° 7/16" 
 “Ults opp. 2 4 94° 17’ Ans.<~ 23 70° 47' 40” or 109° 12’ 20" 
 Required the other parts. A B 39°27! 24” or 140° 32! 36” 
 
 3. In the right-angled spherical triangle a 3c, 
 
 n § The leg 8 c 97° 20’ Ac 45° 33’ 39” or 134° 26" 21" 
 "Ults opp. 2 4 84° 45' Ans,< 23 45° 47/46" or 134° 12! 14" 
 Required the other parts. wee 84° 52! 22" or, 969 7 Bol" 
 
 (az) In following the three general rules, which have been . 
 given for right-angled spherical triangles, the proportion used in 
 the calculation is not always that which is adapted to the instru- 
 mental solution; but the former may be easily reduced to the 
 latter by proper substitutions: Thus, since rad: tansc:: cot 
 
 72 
 
 A: sin Ac by the first formula, if be put in the place of 
 
 tan A 
 
 its equal, cot a, the proportion will become tan a: tanBe:: 
 rad: sinac, which is that applied to the instrument ;.and, if 
 thought necessary, will equally serve for the numeral solution. 
 
 It may also be observed, that, in the instrumental computation, 
 when the extent on the tangents reaches beyond the line, it must _ 
 be set as far back as it reaches over; the method of doing which 
 may be seen in the solution of case rv. following ; where it is 
 more fully described, 
 
 ey 
 
96 
 4. Inthe right-angled spherical triangle a B c, 
 
 Itsopp. 2B 42° 51’ £4. 76°.17' +3? er 109°. 497 577 
 Required the other parts. a B 67° 16! 28” or 112° 43’ 39” 
 
 Note. Several of the problems, given in this part of 
 the work, may often be more conveniently resolved by 
 means of some of the formule in the table of cases, 
 page 156 et seq., where the limits of the data, and other 
 circumstances, are more particularly pointed out. 
 
 The leg ac 36°51", { Bc 60°15" 40" or 119° 44" 20" 
 Ans 
 
 CASE I. 
 When a leg and its adjacent angle are given, to find 
 the rest. 
 1. To find the other leg. 
 As cot given Z: sin adjacent, or ibe leg 2: rad: 
 tan other leg. 
 Which leg is like its opposite  . 
 2 To find the other Z. 
 As rad : sin given Z :: cos sao or given leg : 
 cos other Z. 
 Which Z is like its opposite leg. 
 
 3. To Jind the hypothenuse. 
 
 _ As tan given leg : cos adjacent, or given Z:: rad : 
 
 cot hyp. 
 Which hyp. is less than 90° nF the given leg and Z 
 are like; but greater than 90° if they are unlike. i: 
 
 EXAMPLES. 
 i. In the right-angled spherical triangle, a B c, hav- 
 ing the leg ac 54° 46, and its adjacent angle 4 47° 
 50’, to find the rest. 
 @ 
 
BY CONSTRUCTION. 
 
 1. Describe the circle a p a d with the chord of 60°; 
 and draw the diameters aa, pd, at right angles to 
 each other. 
 
 2. Set the semitangent of the complement of the 
 
 angle a (42° 4') from o to m; and through the points 
 
 A, m, a, describe a circle. 
 
 3. Set off ac (54° 46’) by the scale of chords, and 
 draw c 0, cutting the circle a main B; then aBc will 
 be the triangle required. 3 
 
 Lo measure the required parts. 
 
 i. Set off the semitangent of the given Z a (47° 56’) 
 from o to p, and make c p equal to the chord of 90°. 
 
 2. Through the points B, p, draw the line np Br, 
 cutting the circle in r and n. 
 
 3. Then pn, taken on the scale of chords, gives the 
 43 64° 38’, ar, on the same scale, gives AB 64° 40’, 
 and o B, taken on the line of semitangents, and then 
 subtracted from 90°, gives Bc 42° 8’. 
 
 BY CALCULATION, 
 
 ca C,00, 2 AmEae ee 47 56) 3) FOS54535 
 oe OUYA Chae a ees = 54° 46° =~ - 9.9121207 
 >: Rad, or sin - - - 90° - - - - 10.0000000 
 
 : Tanbo ---- 49° 8/46" - 9.9566672 
 Which side is acute, being like its opposite Z a, 
 
 H 
 ’ 
 
 “«< 
 
98 
 
 Rad or sin --- 90° - - - = 10.0000000 
 Rin ZA. sue ~~ AY DO oO TUG Ine 
 MOS AG ah ee 54°40) Pel Ot Obs 
 “Gos ZB - 2 = - GRR) 81”) Ue benTa4e 
 
 Which Z is acute, being like its opposite leg a c. 
 
 eo ee es 
 
 Pan A) elec ae me 6. Oo OL ae 
 0 Ga Za) SS at Ae et as = | 9.8 260 gale 
 : : Rad, or sin_ ane OO 7 Onan anette 10.0000000 
 
 CMC Ot AB cloths 640-40 eae Ola? 505 70 
 
 Which side is less than 90°, because ac and ZA 
 are like. 
 
 INSTRUMENTALLY. 
 
 1. Extend the compasses from 90° to 54° 46’ (ac) 
 on the line of sines ; and this extent will reach, on the 
 tangents, from 47° 56’ (4 A) to 42° 8’, the leg Bc. 
 
 2. Extend from 90° to 47° 56’ ( 4.4) on the sines, 
 and this extent will reach, on the same line, from 35° 
 14’ (comp*. of ac) to 25° 22’, the complement of Z B. 
 
 8. Extend from 90° to 42° 4° (comp'. of ZA). on 
 the sines, and this extent will reach, on the tangents, 
 from 35° 14 (comp*. of ac) to 25° 20’, the comple- 
 ment of aB (0). | 
 
 Di 
 
 2 + 
 be substituged for cot a in the first stating of the 
 
 r 
 (2) If 
 tan A | 
 numeral calculation of this case, it will become rad: sin ac:: 
 2 
 . 7 
 tana: tansc;andif 
 
 be put for tan ac, in the third 
 cot ac ” 
 
 stating, it will become rad: cos a:: cot Ac: cot AB; which 
 
 are the proportions adapted to the instrumental solution. 
 
99 
 
 2. In the right-angled spherical triangle ax c, 
 Given J The leg ac 36° 52’ BOG 8 45" 
 Its adj’. 44 24°17 Ans.< £8 70°47! 28” 
 
 Required the other parts. AB 39° 26' 40° 
 
 8. In the right-angled spherical triangle a Bc, 
 
 ; The leg a c 76° 20° Be OL Ly Th, 
 Given.» Tes adits Z 4 91°18 ans ZB 76° 20'12” 
 
 Required the other parts. AB 90° 18’ 14” 
 
 4, In the right-angled spherical triangle a Bc, 
 
 Gi Theilée's ce 119%),'6y 0 = AB 111° 36’ 46” 
 iven fe Oats oO i 
 Itsadj. ZB 44°37 Ans.< 44109° 58’ 24 
 Required the other parts. Ac 40° 46’ 
 
 CASE II. 
 
 When the hypothenuse and a leg are given, to find 
 the rest. | 
 
 1. Tofind the opposite 4. 
 As sin hyp. : rad :: sin giv. leg : sin its opp. 4. 
 Which 2 is like its opposite leg. 
 
 2. To find the adjacent Z. 
 As rad : cot hyp. :: tan giv. leg. : cosits adj*. 2. 
 Which Z is acute, if the hyp. and given leg are like; 
 but obtuse, if they are unlike. 
 
 8. Lo find the other leg. | 
 As cos giv. leg : rad :: cos hyp. :: cos other leg. 
 Which leg is less than 90° if the hyp. and given leg 
 are like; but greater than 90° if they are unlike. 
 H 2 
 
100 
 
 EXAMPLES. 
 1. In the right-angled spherical triangle asc, having 
 the hypothenuse A B 65° 50’, and the leg BC 49° 20', 
 to find the rest. | 
 
 BY CONSTRUCTION. 
 
 1, With the chord of 60° describe the circle a D a d, 
 and draw the diameters a a, pd, at right angles to each 
 other. Tian 
 
 2. Set off the hyp. aB (65°50’), by a scale of 
 chords, each way, from a to m; and draw dm, cut- 
 ting Aq ins. 
 
 3. Through the points m, s, m, describe a circle ; 
 and from o, with the semitangent of the complement 
 of Bc (47° 40°) as a radius, intersect the former circle 
 in B. . 
 
 4. Describe a circle through the points a, B, a, and 
 draw oBc; then asc will be the triangle required. 
 
 To measure the required parts. 
 
 1. Measure the distance o 7, in degrees, on a scale 
 of semitangents, and set off its complement on the 
 same scale, from o to p. 
 
 2. Through the points 8, p, draw the line npn, cut- 
 ting the circle in; and take cp equal to the chord 
 of 90°. 
 
10} 
 
 3. Then px, on the scale of chords, gives Z 3 
 65° 52’, or, taken on the scale of semitangents, and 
 then subtracted from 90°, gives Z 4 47° 34, and ac, 
 on the scale of chords, is 56° 22’. ; 
 
 BY CALCULATION. 
 
 me Sin AR - <2 2) 65°50. = PICOLGSS 
 > Rad, orsin -- °90° -- -- 10.0000000 
 -:Singo »=--- 42°20 -- 9.8283006 
 
 Siva. hws AIP Bae 20" 9.8681351 
 
 Which Z is acute, being like its opposite leg B c. 
 Rad; orsin: - - 90° “== ~~" 10.0G00000 
 MotAB.- - -- .65.50 <=. 9.6519742 
 
 han. BC. le = =) 48 207.» -"- \ 9.9895156 
 
 GOS ILE bin, 9 =) O65 162 20° =)" 96124897 
 
 eo ee oe 
 
 Which Z is acute, because aB and ZB are like. 
 
 seo COS RC. ais +). "A220 Vee.” 98687851 
 ; Rad,orsm -- 90° ---  10.0000000 
 <9 GOS AB oe ata 6S a ee et  N6121S97 
 
 : Cosac ---- 56°25'20" 9,7433546 
 Which side is less than 90°, because 48 and Bc are like. 
 
 INSTRUMENTALLY. | 
 
 1. Extend the compasses from 65° 50’ (AB) to 90° 
 
 on the sines, and that extent will reach, on the same 
 line, from 42° 20’ (Bc) to 472°, the Za. 
 
 2. Extend from 65° 50’ (4.8) to 42° 20’ (pc), on 
 the tangents, and that extent will reach, on the sines, 
 from’ 90° to 24° 8’, the complement of Zs. 
 
 3. Extend from 47° 40 (comp. of Bc) to 90° on 
 the sines, and that extent will reach, on the same 
 
102 
 
 line, from 34° 10’ (comp*. of a B) to 24° 8’, the com. 
 plement of ac (c). 
 
 2. In the right-angled spherical triangle a Bc, 
 Given § Lhe hyp. a8 37° 25° LAT SS 18" 
 TINE" 7 The leg. Bc 15909 Ansi<, ZB 72° 18-477 
 To find the other parts. AC 35° 21" at 
 . In the right-angled spherical triangle.a B 4 
 Given J Phehyp.as 317% 1h £ &119°59' 29% 
 ‘i The leg Bc 148°27’ Ans.< 23144° 5’ 8” 
 
 Required the other parts. A €159°14' 49” 
 4. In the right-angled spherical triangle a Bc, 
 Thehyp. a B 69° 27’ {2 A 54° 6 48” 
 
 Cy abe les ac 57°24’ Ans,< 2B 64° 7 9” 
 
 Required the other parts. Bc 49° 20' 35” 
 CASE IV. 
 When the hypothenuse and an angle are given, to 
 find the rest. » 
 
 1. To jind the opposite leg. 
 
 As rad : sin hyp. :: sin giv. Z : sin its opp. leg. 
 
 Which leg is like its opposite Z . 
 
 2. To find the adjacent leg. | 
 
 As cot hyp. : rad':: cos giv. Z : tan its adj‘. leg. 
 
 _ Which leg is less than 90° when hyp. and given Z 
 are like; but greater than 90° when they are unlike. 
 
 s 
 
 8 a 
 (c) By SURE ES “__ for cot AB, in the second stating 
 B 
 
 of the numeral snes it will become tana B:tanac:3 rad : 
 cos 8 for the instrumental solution. ge 
 
103 
 
 3. To find the other Z. 
 As cot giv. Z : rad:: cos hyp.: cot other 2. 
 Which 4 is acute when hyp. and given Z are like ; 
 but obtuse when they are unlike. 
 
 EXAMPLES. 
 
 t 
 
 1. In the right-angled spherical triangle asc, hav- 
 ing the hypothenuse a B 65° 5’, and the angle a 48° 12’ 
 to find the rest. 
 
 BY CONSTRUCTION. 
 
 1. With the chord of 60° describe the circle ap ad, 
 and draw the diameters A a, Dd, at right angles to each 
 
 _ other. 
 2. Set off o m equal to the semitangent of the com- | 
 plement of the Z a (41° 48’), and through the points 
 A, m, a, describe a circle. 
 _ _ 8. Set off the semitangent of the Za (48° 12 ’) from 
 0 to p, and take a7, on the line of chords, equal to 
 _ AB (65° 5’). 
 ; 4. Through the points 7, p, draw the line rs pn, 
 be: the circle a ma in B, and the former circle in 
 len, through the points o, 8, draw the line oc, 
 af axe will be the triangle required. 
 
 il 
 
 d 
 
104 
 
 To measure the required pas: 
 
 1. Set off c P equal to the chord of 90°; then Pn, 
 taken on the scale of chords, will give Z’B (64° 46° sith 
 and a c, on the same scale, is 55° 7’. | 
 
 2. And if o8 be taken on the line of semitangents, 
 and then subtracted from 90°, it will give B c 42° 32’, 
 
 ‘BY CALCULATION. 
 Rad, orsin- -- 90° - - - - 10.0000000 
 Sin "AB. =! -<. OD tone —-o—" "ise TOOeg, 
 Pegi 2 a a el = 48° 19’ -. 9.8724337 
 - Sinpe ---- 49°39/19” 9.830003 
 
 ee, ey 
 
 Which side is acute, being like its opposite 2 a. 
 
 SeCOUA Bienes ef a GD 6) arial POCO Te LAGS 
 >: Rad,orsin -- 90° - - - - 10,0000000 
 “2 COSZA - ates AR? TO 4.5 OS OseelsS 
 ee Tan AVG ps fi ey os BD he? = 101567999 
 
 "Which side is less than 90°, because a B and Z a are like. 
 
 UME Gis MU SIN a ata Pees: a BS 248 
 
 -- 9.9513876 
 s*. Rad, or’ sin’ - -"- 90° ("= = + 10,0000000 
 
 2 COs Bis = es 65° 5" Vain eed SOTT 
 --GorZpo oo. 64° 46" 14” "9567390385 
 
 Which Z is acute, because az and Za are like, 
 
 INSTRUMENTALLY. 4 
 1. Extend the compasses from 90° to 65° 5’ (AB) q 
 on the sines, and that extent will reach, on the same ~ 
 line, from 48° 12’ ( Za) to 42° 39’, the leg BC. | 
 * 2. Extend from 90°, on the sines, to 41° 48’ (comp. | 
 of ZA); then apply | this extent from 45° on th > tan 
 gents, towards the left hand, and, keeping the ke tt 
 
“405 
 
 N 
 
 » point of the compasses fixed, turn the other leg, and 
 extend it till it reaches to 65° 5’ (a 8); which last ex- 
 tent will reach from 45° to 55° 7’, the leg ac. 
 
 3. Extend from 24° 55’ (comp*. of AB) to 90° On 
 the sines; then apply this extent from 45° on the tan- 
 gents, backwards, on the left hand, and keeping the 
 latter point of the compasses fixed, turn the other leg 
 till it reaches to 41° 48° (comp*. 2 Ae ; and this extent 
 will reach from 45° to 64° 46, the 23; the divisions, 
 in this case, falling as much under 45° as they would 
 have fallen beyond it (d). 
 
 2. In the right-angled spherical triangle a Bc, 
 
 Given £ The hyp. «8 39° a7 FBC 16° 53) 4” 
 The Za - - 27° 12’ Ans.< ac 36°11’ 58” 
 Required the other parts. ZB 6S" 21°17" 
 8. In the right-angled spherical triangle a B c, 
 Given The hyp. a B 79° 25° BC 41°29 14” 
 The Za - - 42°15 Ans.< ac 75° 50’ 
 ® " ie) / tt 
 Required the other parts. £880 31°45 
 4, In the right-angled spherical triangle a Bc, 
 Given J Lhe hyp. a B 98° 20° BC 105°20'°11” 
 | The 2 8 - - 57°43’ Ans.< ac 56°46’ 1g 
 é Required the other parts. Za l02*55°14" 
 : (d) By putting : Go for cot ate in the second numeral sta- 
 KP a B 
 . ae: it will become rad : cos a :; tan as : tan ac; and if 
 
 - be put for cot B in the third stating, it will become cos 4B : 
 
 ; tan B 
 “rad ; cot A: tan 8; which are the analogies adapted to the in- 
 tal solution. 
 
106 
 
 CASE V. ° 
 When the two legs are given, to find the rest. » 
 
 1. To find either of the Z°. 
 As tan one of the legs: rad :: sin other leg : cot 
 its adjacent Z. 
 Which Z is like its opposite leg. 
 2. To find the hypothenuse. 
 As rad : cos either leg : : cos other leg : cos hyp. 
 
 Which hyp. is less than 90° if the legs are like; but 
 greater than 90° if they are unlike. 
 
 EXAMPLES. 
 
 1. Inthe right-angled spherical triangle 4 B c, hav- 
 ing the leg ae 52° 13’, and the leg B c 42°17’, to find 
 the rest. 
 
 BY CONSTRUCTION. 
 
 i. From o, asa centre, with the chord of 60° de- 
 scribe a circle, and draw the diameter Ec. 
 
 2, Set off c a (52° 13’) from the scale of chords, - 
 and make o B equal to the semitangent of the comple- 
 ment of Bc (47° 43’). 
 
 3. Draw the diameters aa, pd, at right angles to 
 each other, and through the points a, B, a, describe a 
 circle; then Bp Bc willbe the triangle required. 53 
 
107 
 
 To measure the required parts. 
 
 1. Take the measure of o n in degrees, on the scale 
 of semitangents, and set its complement, on the same 
 scale, from o to p. 
 
 2. Through p, 8, draw the line mpsr, cutting the 
 circle in m andr; and set off c Pp equal to the chord 
 of 90°. | 
 
 3. Then Pm, taken on the line of chords, gives the ° 
 43 62° 27°; ar, on the same line, gives a B 63° 2°; 
 and on, taken on the line of semitangents, and then 
 subtracted from 90°, gives Z A 49°. 
 
 BY CALCULATION. 
 SWAT ASC Uae, me OE LSet. 105786 
 
 SSA. OF SI | <= a0”. =\'= 421 O.O000000 
 IR as = ee ae Ee es OD OOTSS4 8 
 S AGGZ BR kh oie OA EO 9 TL SORT 
 
 ee 
 
 Which Z is acute, being like its opposite leg ac. 
 
 > ‘Tampoc ----- 42°17) - -°. 9.9587542 
 SMe MOn ISIN. sj! 990, 0.'= 256 LO QOOO000 
 ss SMAC ---- 52°13 -- 9.8978103 
 MGot ZA 22 =. 149°.'00 25749. 9990561. 
 
 . Which Z is acute, being like its opposite leg 8 c. 
 
 weiad: Or sin ~ +, 90°. - = « «'16,0000000 
 Ree OSV A Oita — ham OO 1S. a ok GIR 79ST 7 
 Aton BC ao ate ae Yor, 8691 SOL 
 
 “: Cosas ---- 68° 245” . 9.6563618 
 
 Sex, : o e , 
 Which side is less than 90°, because ac and Bc are like, 
 ey F 4 
 
108 
 
 INSTRUMENTALLY. 
 
 1. Extend the compasses from 42° 17’ (3 c) to 90° 
 on the sines, and this extent will reach, on the tan- 
 gents, from 52° 13’ (A c) to 62° 27’, the ZB (e). 
 
 2. Extend from 52° 13’ (a c) to 90°, on the sines, 
 and this extent will reach, on the tangents, from 42° 117’ 
 (pc) to 49°, the 4 a. | 
 
 3. Extend from 90°, on the sines, to 37° 4'7’ (comp*. 
 of ac), and this extent will reach, on the same line, 
 from 47° 43’ (comp. of Bc) to 26° 58’, the comple- 
 ment of AB. 
 
 2. In the right-angled spherical triangle ac, 
 
 G; Uheleg ac 29° 17" “LA Cfo 1728 
 Wwe) ) The leg Bc 49° 27 Ans.d ZB 36° 25! 44° 
 
 Required the other parts. AB 55° 27 23° 
 
 3. In the right-angled spherical triangle a Bc, 
 Given § Theleg ac 76° 25° ZA 38° 18” His 
 The leg Bc 37° 31° Ans.< 48 81° 37° 46 
 Required the other parts. l AB 79° 15° 51" 
 
 4, In the right-angled spherical triangle a Bc, 
 Gj Theleg ac 97°29’ LALOR 80/2" 
 wWeD ) The leg Bc 104° 19! Ans 
 
 Required the other parts. AB vee 10° 10” 
 
 v2 
 
 (ce) By substituting for cot 4 8 in the first logarithmic 
 
 tan B 
 72 
 
 stating, and for cot A in the second, the two analogies be- 
 
 tan A . | 
 comesinepc:rad::tanac:tans,andsnac:rad::tanspc: 
 tan a; which are those used in the instrumental solution. 
 
 L' BLOT T5025 6 ae 
 
 ee ee 
 
109 
 
 CASE VI. 
 When the two oblique angles are given, to find the rest. 
 1. To find either of the legs. 
 As sin one of the given 2°: rad :: cos other 4 : 
 cos its opposite leg. 
 Which leg is like its opposite 2 . 
 2. To find the hypothenuse. 
 As rad: cot either Z :: cot other Z : cos hyp. 
 Which hyp. is less than 90° when the given Z$ are 
 like; but greater than 90° when they are unlike. 
 EXAMPLES. 
 1. In the right-angled spherical triangle a Bc, hav- 
 ing the angle a 48° 13’, and the angle g 64° 27’, to 
 find the rest. | 
 
 BY CONSTRUCTION. 
 
 1. Describe the circle z D ec with the chord of 60°; 
 and draw the diameters £ e, D c, at right angles to each 
 other. 
 
 2. From 5, with the chord of ZB (64° 27’), set off, 
 each way, rb, 24; and from o take on equal to the 
  semitangent of the complement of that angle. 7 
 
 3. Through the points b, n, b, describe a circle; 
 and from o, with the semitangent of 4 a (48° 13’) 
 describe the arc cc, cutting the former in p. 
 
110 
 
 4. Through po, draw the diameter p s, and another 
 
 Aa, at right angles to it. 
 
 5. Set off o m equal to the semitangent of the com- 
 plement of Z a (41° 47’); then through the points 
 A, m, @, describe’a circle, cutting Dc inB; and aBC 
 will be the triangle required. 
 
 To measure the required parts. 
 Through the points p, 8, draw the line p Br, cutting 
 the circle RDecin7; then a7, taken on the chords, 
 
 gives AB 64° 42°, ac on the same scale is 54° 39’, and 
 o 8, taken on the line of semitangents, and then sub- 
 tracted from 90°, gives Bc 42° 23’, 
 
 BY CALCULATION. 
 
 CMAN 6 1 A 
 ¢ Rad, or sin - 
 *i¢\ Cos (ZB. c=, 
 5 COS cA Ona 
 
 43° 13’ - - 9.8725466 
 90° - - - - 10.0000000 
 64° 27 -~ ~ 9.6347'780 
 
 54° 39’ 42” 9.7622314 
 
 Which side is less than 90°, being like its opp. ZB. 
 
 eZ Bil == 
 saad, or sin - 
 bos ZA: 2. 
 pROS IBC) 4 'e'- 
 
 64° 27’ - = 9,9553073 
 90° - - - - 10.0000000 
 48° 13’ ~ - 9.8236800 
 49° 23' 35” 9.8683727 
 
 Which side is less than 90°, being like its opp. Z a. 
 
 SERS OT) STD vide tele 
 PCIe Oe eat een AT 
 ‘GOL Bilvel ey «= 
 > GosaB----- 
 
 90° - - - - 10.0000C00 
 48°13’ - = 9,9511334 
 64° 27° . - 9.6794708 
 64° 42’ 42” 9.6306042 
 
 -_ 
 
 Which side is less than 90°, because Z° a and B are like. 
 
: iit 
 |. INSTRUMENTALLY. 
 
 1. Extend the compasses from 48° 13° ( 4 A) to 90’ 
 on the sines, and that extent will-reach, on the same 
 line, from 25° 33 (comp'. of 2B) to 35°21’, the 
 complement of ac. 
 
 2, Extend from 64° 27’ ( ZB) to 90° on the sines, 
 and that extent will reach, on the same line, from 41° 
 47’ (comp*. of Z a) to 47° 37’, the complement of Bc. 
 
 3. Extend from 48° 13’( Za) to 25° 33’ (comp. of 
 £8) on the tangents,and this extent will reach, on 
 the sines, from 90° to 25°1 8’, the complement of aB(/). 
 
 2. In the right-angled spherical triangle a Bc, 
 
 Gian The Z a 24° 30’ AC 36° 4 28” 
 The ZB 70° 25’ Soe BOE Pe 
 
 Required the other parts. A B 38° 40° 52° 
 
 3. In the right-angled spherical triangle a-B c, 
 Given J Lhe4 4 39° 12" {: C77 ilar” 
 TheZ gp 82° 9’ Ans.< Bc 38°31’ 51” 
 Required the other parts. AB 80°16" 2" 
 
 4, In the right-angled spherical triangle azo, 
 
 Given fbes a 105° 8" . AC 27°46 5” 
 he 3 31° 20). Ans.< B.¢ 12078) 5% 
 
 Required the other parts. AB 116° 22’ 25” 
 
 -(f) The analogy for obtaining the value of « 2 by the instru- 
 
 , os ~ . re ° 
 ment, is got by substituting for cot a, in the numeral solu- 
 
 . 
 « 
 
 tion, which then becomes tan a; cotz.:: rad: cos AB. 
 
112 
 
 OF QUADRANTAL SPHERICAL TRIANGLES, 
 
 The different cases, or varieties, that may. happen in 
 the solution of quadrantal spherical triangles, in which 
 two things, together with the quadrantal side, are al- 
 
 ways given, to find a third, are the same as in right- 
 
 angled spherical triangles. » 
 
 And since the sides and angles of any quadrantal 
 spherical triangle are the supplements of the opposite 
 ancles and sides of a right-angled spherical triangle, 
 described from its angular, points as poles, the three 
 general formula: which have been given for the latter, 
 may be readily converted into the following ones, 
 which are equally applicable to all the cases of ae 
 drantal spherical triangles ; 
 
 sin its opp. side X sin hyp’. 2, 
 
 lipae inch. 2={ or 
 : cot its adj‘. side x tan other Z. 
 
 cos its opp. 4 & sin other side, 
 2.rX<cos eith. side = or 
 —tan its adj‘. Z x cot hyp! 2, 
 
 —cos one Z Xcos other Z, 
 3.7 XiCOS DY Die Zoom OF ae 
 ) — cot one side X cot other side. 
 Where it is to be remarked, that the angle opposite 
 the quadrantal side is called the hypothenusal angle, 
 and the other parts simply the sides and angles. And 
 in applying these forms to practice, it is only necessary 
 
 to attend to the observations that have been already 
 
 made for right-angled spherical tr iangles. 
 
 =< ~~ 
 
' 113 
 
 AFFECTIONS OF 
 QUADRANTAL SPHERICAL TRIANGLES. 
 
 . The sides are of the same kind as their ina 
 ile ; and conversely. 
 
 2. The hypothenusal angle is greater or less than 
 90°, according as a side and its adjacent angle, or the 
 _ two sides, or the other two angles, are like or unlike. 
 
 8. An angle at the quadrant is obtuse or acute, ac- 
 cording as its adjacent side and the hypothenusal angle, 
 or the other angle and 2A alae angle, are like 
 or unlike. | 
 
 4, A side is greater or less than 90°, according as 
 its adjacent angle and the hypothenusal angle, or the 
 other side and the hypothenusal angle, are like or un- 
 like. ; 
 
 OTHER PROPERTIES OF 
 " QUADRANTAL SPHERICAL TRIANGLES. 
 
 1. If the hypothenusal angle be 90°, one of the 
 other angles and its opposite side will be each 90°, and 
 the other side and angle will be both measured by the 
 same number of degrees. 
 
 - 2. If an angle, or a side, be 90°, the opposite side, or 
 angle, and the hypothenusal Atle will be each, 90° ; 
 _and the other angle and side will be both measured i. 
 the same number of degrees. 
 
 3. If an angle at the quadrant be less than the hy- 
 pothenusal angle, their sum will be less than 180°; 
 and if it be greater than the hypothenusal angle, their 
 
 um will be greater than 180°. 
 
 I 
 
114 
 
 4. If a side be greater than its opposite angle, their 
 sum will be less than 180°; and if it be less than its 
 opposite angle, their sum will be greater than 180°. 
 
 5. The difference of the two sides is less than 90° ; 
 and their sum is greater than 90°, and less than 270°. 
 
 6. The three angles are either all equal to, or all 
 less than 90°; or two of them are greater than 90°, 
 and the other less. 
 
 The six cases of quadrantal spherical triangles al- 
 ready mentioned, may be ranged as follows: 
 
 A side and its opp.2 | 
 
 A side and its adj. 4 | 
 
 Givers The hyp Zand a side « to find the other 
 ) The hyp!. 4 and another 2 | parts. 
 
 The two sides 
 LThe two 25 
 
 CASE I, 
 When a side and its opposite angle are given, to 
 find the rest. 
 
 1. To find the other Z. 
 As rad: cot giv. side :: tan giv. 2: sin other 2. 
 Which 4 may be either an acute 4 or its sup’. 
 
 2. To find the other side. 
 As cos giv. 4 : cos opp. or giv, side : : rad : sin other side. 
 Which side may be an arc less than 90°, or its sup’, 
 
 8. To find the hypothenusal Z. 
 As sin giv. side : sin opp. or giv. 4:: rad: sin hyp. Z. 
 Which hypothenusal 2 may be either an acute Z, 
 or its supplement. 
 
115 
 
 EXAMPLES: 
 i. In the quadrantal spherical triangle a Bc, having 
 Ac 63° 58’, and its opposite pens B 54° 19’, to find 
 the rest. 
 
 BY CONSTRUCTIONe 
 
 1. Describe the circle a 4 a 8 with the chord of 60°, 
 and draw the diameters aa@,Bb, at right angles to 
 each other. 
 
 2. Set off ad, Ad, each way, with the chord of ac 
 (63° 58’), and take on equal to the semitangent of the 
 complement of that arc (26° 2’), and o m equal to the 
 semitangent of the complement of 4B (35° 41°). 
 
 8. Through the points B, m, b, and d, n, d, describe 
 two circles, cutting each other inc; then if a circle 
 be described through the points a, c, a, the triangle 
 ABC, or l ac, will be the one required, each having 
 the same data; which shows the case to be ambiguous. 
 
 To measure the required parts. 
 1. Take or on the scale of semitangents, and set 
 off its complement, on the same scale, from o to P. 
 2. Make op equal to the semitangent of theZ3 
 (54° 19'), and through the points cp, cp, draw he 
 lines ce, andc f 
 12 
 
116 
 
 8. Then bs, taken on the scale of chords, gives 
 BC 48° 48; ef, on the same scale, givésZ c 64° 41’; 
 and the complement of the degrees in 07, taken on the 
 semitangents, gives Z A 42° 51’. 
 
 BY CALCULATION. 
 
 & Os ZB it Uk RO el ni do OOS Ba OW 
 > Cosac---- 63°58 = -- 9,6423596 
 >: Rad, or sin’ -)+. 90°. + - = = 10.0000000 
 >; Singc 48°48’ 3” or 131° 1157" 9.8764639 
 
 Rad, or sin -- 90° ---  10.0000000 
 
 10 PAN A Barton arn wy LD aie DA oo 
 BOE ALC tal cm ute RT ek tee onl Cit eases are 
 : SinZ a 42°5)’ 25” or 137° 8 35” 9.8326185(g) 
 & Oma CS She, BUGS: be Oe aeeene 
 Sin: Zip lives. die wASS21:97 b ~ eis 1IQI909G015 
 & TRA, OLISIN | yes.) IOS | ~ ier: sree O.O00GQ00 
 
 oe 
 
 - Sin Zc 64° 41’ 6” or 115° 18’ 54” 9.0561546 
 
 * 
 
 INSTRUMENTALLY. 
 1. Extend the compasses from 35° 41’ (comp*. of | 
 £8) to 26° 2’ (comp*. of ac) on the sines, and this 
 extent will reach, on the same line, from 90° to 48° 
 48’, the side Bc. sig vd 
 2. Extend from 63° 58° (ac) to 54° 19 (2B) on © 
 the tangents, and this extent will reach, on the sines, 
 from 90° to 42° 51, the Za. 
 
 (g) Thisanalogy, by proper substitution, becomes tanac: 
 tan Bs: rad: sin A, which is that used in the instrumental 
 solution. 
 
117 
 
 8. Extend from 63° 58’ (ac) to 54° 19' (28) on 
 the sines, and’ this extent will reach, on the same line, 
 from 90° to 64° 41’, the Zc. 
 
 2. In the quadrantal spherical triangle ABC, 
 
 7 The side ac 112° 27’ BC §5°47' 39” or 124°12'21" 
 The opp. 23 117° 30’ Ans. { Ls ao 32 Yor 127° 27'S" 
 Required the other parts. 4c 73°41'17" or 106° 18/43" 
 
 3. In the quadrantal spherical triangle a B c, 
 
 Ge The side ncl2i° 9 ¢ 4c 6291235" or 117°47'25” 
 The opp. Za 54°13’ Ans. { £8 56°59! 24" or 123°, 0°35" 
 leita the other parts. » £4 071°25/ 30" or 108° 34’ 30" 
 
 , CASE II. 
 When aside and its adjacent angle are given, to 
 find the rest. 
 
 1. To find the other Z. 
 As cot given side : sin adjacent or given Z :: rad: 
 tan other Z. 
 Which Zis of the same kind as its opp. side. 
 
 2. To find the other side. 
 As rad : sin giv. side : : cos adj. or giv. Z : cos other side. 
 
 Which side is of the same kind as its opp. 2. 
 
 3. To find the hypothenusal Z. 
 As tan given 2: cos adjacent or given side :: rad : 
 - cot hypothenusal 2. 
 Which hypothenusal 2 i is greater than 90° when the 
 
 given — ‘side and 4 are like; but less than 90° when 
 they are unlike. 
 
* 
 AMES 
 
 , 118 
 
 EXAMPLES. 
 1. In the quadrantal spherical triangle a B C, having 
 given the side a c 63° 58’, and its adjacent angle a 
 42° 51’, to find the rest. 
 
 BY CONSTRUCTION, — 
 
 1. Describe the circle a 6 a 8 with the chord of 60°, 
 and draw the diameters a a, BJ, at right angles to each 
 other. 
 
 2. Set off ad, Ad, each way, with the chord of ac 
 (63° 58’), and take on equal to the semitangent of the 
 complement of that arc (26° 2’), and or equal to the 
 semitangent of the complement of Z a (47° 9’). 
 
 3. Through the points d,n,d, and a, 7, a, describe 
 
 circles cutting each other in c; then if a circle be de- | 
 
 scribed through the points 2, c, B, the triangle anc 
 will be the one required. : 
 
 To measure the required parts. 
 1. Set off the semitangent of the Z a (42° 51°) from 
 0 to P; also take om, on the same scale, and set cff 
 its complement from o to p. 1) , 
 2. Through the points c, p, and c, p, draw the lines 
 ¢e and sc; then 8s, taken on the scale of chords, 
 
 gives BC 48°47’; ef, on the same scale, gives 4 
 
 | 
 
 : 
 
m9 
 
 25° 19", (or 154° 41’ 13”); and the complement of 
 the degrees in om, taken on the semitacn tes gives 
 ZB 54° 18%, 
 
 BY CALCULATION. 
 
 - Rad, orsin--- 90° - = - - 10.0000000 
 - Dil es eel 63° 58 -- 9.9535369. 
 Se Meta ho se AO SL ie 4 oO, SOL eae 
 *. COSR EC = = = Hu, 40047 149°" °9'8187218 
 Which side is less than 90°, being like its opp. 4 a. 
 (A): 8 Cotta G4. 3° 2 86S7 BS! aie 115 96B88927 
 2G ORI. 4a SR EO ag ORB OE EOO 
 :: Rad, or sin - - - 90° - - - 10.0000000 
 
 ¢ )VanZ nid?-)S 4) + 64°.18°.47, +10)! 10.143 7382 
 
 Which 4 is acute, being like its opposite side a c. 
 
 bee fe ee 9.9673759 
 
 Cos A @) Sr S0s ngewe eh Ls pOlemesi5naS 
 2a Radvorrsin += 190? ~ 6-1/4 10, 0.0000000 
 COG eC? a ot ode, St OLe* 9.674983 a. 
 
 Which Z is obtuse, because Ac and Za are like. 
 
 INSTRUMENTALLY. 
 
 1, Extend the compasses from 90° to 63° 58’ (4c) 
 on the sines, and that extent will reach, on the same 
 line, from 47° 49° (comp. of 4a) to 41° 1%, the 
 complement of Bc. 
 
 2. Extend from 90° to 42° 51’ (Za) on the sines, 
 and this extent will reach, on the tangents, from 63° 
 58 ( c) to 54°18’, ZB. 
 
 (4) The two last of these analogies, by proper substitutions, 
 
 rad: sina::tanac:tans : 
 Bagong } trad > cosac::cot a: cot an phn arg. thane adgpted 
 
 to the use of the instrument. 
 
 ag 
 
‘120 
 
 3, Extend from 90° to 26°2’ (comp’. of a c) on the 
 sines, and this extent will reach on the tangents, from 
 47° 9’ (comp*. of Z A) to 25° 19°, the compt. of c. 
 
 2. In the quadrantal spherical triangle a Bc, 
 
 Civ The side ac 112°27’ BC117°4455" 
 ven 4 Theadjt. Z a 120°15’ Ans.< 28115°33'46" 
 
 ° / a” 
 Required the other parts. £C102°33'18 
 
 3. In the quadrantal spherical triangle anc, 
 Given J Thesidese 59° 16° AC1386°13" 3° 
 ven ) Theadjt. 28 147° 8’ Ans.< 2. 42°93'20" 
 ™ oO , se 
 Required the other parts. 40 513927 
 
 CASE III. 
 
 When the hypothenusal angle and either of the 
 other angles are given, to find the rest. 
 
 1. To find the side opposite to the given Z. 
 As sin hyp'. Z : rad :: sin given 2; sin opposite 
 or required side. 
 Which side is like its opposite 2. 
 
 2. To find the side adjacent to the given Z. 
 As rad: cot hyp’. 4: tan given 4 : : cos adjacent 
 or required side. 
 "Which side is greater than 90° when the given 4° 
 are like; but less than 90° when they are unlike, 
 
 8. To find the remaining Z. 
 As cos given 2: rad :: cos hyp!. 2: cos remaining — 
 or required 2. 
 Which 4 is obtuse when the given 4° are like; but 
 acute when they are unlike. 
 
12) 
 
 f EXAMPLES. 
 1. In the quadrantal spherical triangle a Bc, having 
 the angle a 42° 51’, and the hyp’. angle c 115° 19’, 
 to find the rest. 
 
 - 
 
 BY CONSTRUCTION. 
 
 1. Describe the circle a p ad with the chord of 60°, 
 and draw the diameters aa, Dd, at right angles to each 
 other. 3 
 
 2, Set off o B equal to the semitangent of the com- 
 plement of 4 a (47° 9°) and through the points 
 A, B, a, describe a circle. . 
 
 3. From o, 8 with the tangent and secant of Pe 
 (115° 19! or 64° 41°) describe arcs cutting each other 
 ino; then from the point 0, as a centre, with the 
 radius 0B, describe the circle cB c, and aB C gpl be 
 the triangle required. 
 
 To measure the required parts, bs 
 
 1. Inoo set off op equal to the semitangent of 
 Zc 115° 19 or 64° 41’), and through p, 8, draw the 
 line rs, cutting the circle in r and s. 
 
 2. Then rp, taken on the scale of chords, gives 28 
 54° 19’; cs, on the same aia gives C B 45° 47’; and 
 AC Is 6° 58, 
 
122 
 
 BY CALCULATION, 
 
 4 SinZo » -6 5 = 115° 19") noe 9.9561483 
 
 > Rad, orsin» -.- 90° ‘= - 10.0000000 — 
 Se Sin ZA se 3 = RSI (0 iee TERRACE OS 
 
 : Snape ---- 48°47'36" 9.3764126 
 
 Which side is less than 90°, being like its opp. Z a, 
 
 s Rad,orsin -- 90° - - - - 10,0000000 
 
 $ lan ZA.) = - -. 42d yet | 99673759 
 
 2: CotZc  - = + 115° 19" = - (9.6749105 
 
 > CosAc «=~ 63°58’ 17". 9,6422864 (i) 
 
 Which side is less than 90°, because the 4° a andc _ 
 are unlike. 
 
 Cos 4.4% sie, ey? 51s) = 55 98651849 
 Rad, or sin - - - 90° -- - - 10.0000000 
 "COS LO” - ein @ TAG. 1D) mn, OL GOG Re 
 Cos2 Bo j~ jo oo eee IS Boa Oe aaRo sy 
 
 s@ 6@ ee 
 
 —_ 
 
 Which 2 is acute, because the given 4 * are unlike. 
 
 INSTRUMENTALLY. 
 
 1. Extend the compasses from 47° 9" (compt. of 
 Z a) to 90° on the sines, and this extent’ will reach, 
 on the same line, from 25° 19" (compt. of Z c) to 
 955 41, the complement of Z s. | 
 
 2. Extend from 64° 41" (Zc) to 90° on the sines, 
 and this extent will reach, on the same line, from 
 42° 51’ (Za) to 48° 47’, the side Bc. 
 
 3. Extend from 64° 41° (Zc) to 42° 51’ (2 a) on 
 the tangents, and this extent will reach, on the sines, 
 from 90° to 26° 2’, the complement of ac. 
 
 (1) This analogy may be converted, as before, into tan c : 
 tan a:: rad: cos ac for the instrumental computation. 
 
123 
 
 2, In the quadrantal spherical triangle a Bc, 
 
 Gi Thehyp!. 2c 102° 9° £B119°31'29" 
 Ven) TheZa - 115°W7’ Ans.< Be112°20'35” 
 ° +f “P 
 
 Required the other parts. AG BME 6 59 
 
 3. In the quadrantal spherical triangle a Bc, 
 Given J Lhe hyp!. Zo 102" 5 { pei 6~ 14” 
 hese... STS Ans.< Bie TO 133 36 
 Required the other parts, Ac 59° 19°38" 
 
 CASE IV. 
 When the hypothenusal angle and a ag are gives, 
 io find the rest. 
 
 1. To find the Z opposite that side. 
 As rad : sin hyp!. 2: : sin given side : sin opposite 
 or required ve 
 Which 4 is like its opp. side. 
 
 2. To find the £ adjacent given side. 
 Ascot hyp!. 2: rad:: cos given side: tan its adj’. 2. 
 Which Z is obtuse when its adjacent side and the 
 hyp! Zare like; but acute when they are unlike. 
 
 3. Lo find the other side. 
 As cot given side: rad : cos hyp!. 4+: cot other side. 
 Which side is greater than 90°, when the other side 
 and the hyp!. Z are like; but less than 90° when they 
 are unlike, 
 
 EXAMPLES. 
 ie i the quadrantal spherical triangle a Bc, having 
 A C 63° 58’, and the hypothenusal LB VLE, 
 find the rest, 
 
124 
 
 BY CONSTRUCTION? 
 
 1. Describe the circle a p a d with the chord of 60°, 
 and draw the diameters aa, Dd, at right 4 § to each other. 
 2. Set off ac (63° 58’) from a scale é€ chords, and 
 draw the diameter c c, and another E e, at right angles 
 to it. is 
 3. Take o 7 equal to the semitangent of the com- 
 plement of the 4 c (25° 19’), and through the points 
 c, 2, c, describe a circle, cutting the diameter pd ins. 
 . Then if a circle be described through the points 
 A,B, @, the triangle a 8 c will be the one required. 
 To measure the required parts. 
 
 1. Make o p equal to the semitangent of the angle c 
 (115° 19’ or 64° 41°), and through p, 8, draw the line 
 r % giants the circle in 7 ands. di yeh ig age 
 
 Then os, taken on the semitangents, and sub- 
 eianrran 90°, gives 4 a 42° 51’; rp, on the chords, 
 gives Zp 54° 18°; ands, on the same scale, gives 
 CB 48°47’. ; 
 
 BY CONSTRUCTION. 
 
 Me Or sig. on’ 90? ie ase 6c OOOO 
 97 eID A Ce fi het TING HOT: | C9 ea OU aes 
 5:2 DIMVA Co tverage 63° 58.0 9.9 asp 569 
 
 Sin ZB -.--- 54°18 56". 9.9096852 
 
 — 
 
 Which Zis acute, being like its opposite side a c. 
 
125 
 
 > Cot Ze “Srh eheclns?yhOPieye 9.6749105 
 Rad, orsin -- 90° - - - - 10.0000000 
 2: Cosac ---- 63°58 = - -9.6423596 
 
 © Van Za enna  42°S5R 17.) 9,9674491 
 Which Z is acute, because a c and Z a are unlike. 
 
 > | GGA ose: 2h 6S%58" 3 2 O68eR2e7 
 
 2 Qosie: Yi=)— s>° 115° 19’ - - 9.6310589 
 >: Rad, orsin - - - 90° + - - 10,0000000 
 : CosBc ---- 48° 47°57" 9.9429362(h) 
 
 Which side is less than 90°, because ac and Z ¢ are 
 unlike, 
 
 INSTRUMENTALLY. 
 
 1, Extend the compasses from 90° to 64° 41’ (Zc) 
 on the sines, and this extent will reach, on the same 
 line, from 63° 58’ (a c) to 54° 18’, the ZB. 
 
 2. Extend from 90° to 26° 2” (comp*. of 4c) on the 
 sines, and that extent will reach, on the tangents 
 from 64° 41° ( Zc) to 42° 514, the Za. 
 
 3. Extend from 90° to 25° 19’ (comp. of Zc) on 
 the sines, and this extent will reach, on the tangents, 
 from 63° 58’ (Ac) to 41° 13’, the comp’. of Bc. 
 
 3 
 
 2. In the quadrantal spherical triangle a Bc, 
 
 Given J rhe side ac 114°17' BC 115° 0’ 33” 
 Thehyp!. Zc 102° 9’ Ans.< 43B116°59'15” 
 
 Required the other parts. 4411738758" 
 
 (£4) These twoanalogies are easily converted into the following: 
 , ‘rad ¢.cOs AC ?:tan ¢: tan A 
 rad/rcos:-Z2 @ 3 tart a'é 2 cot BC; 
 which are those used for the instrumental computation, 
 
126 
 3. In the quadrantal spherical triangle a Be, 
 Given 1 ‘The side Bc 79° 23" ‘ AC 41°32" 6" 
 Thehyp’. £4 c 102°13" Ans.< 48 40°23'44 
 } Required the other parts. LATZ°52 4" 
 oe CASE V. 
 When the two sides are given, to find the rest. 
 1. To find eiiher of the other 2°. 
 As sin either side : rad :: cos other side : cos-op- 
 posite or required 4 . 
 Which 4 is like its opposite side. 
 2. To find the hypothenusal Z . 
 As rad : cot either side: : cot other side : cos hyp'. 2. 
 Which Zis obtuse when the given sides are like; 
 but acute when they are unlike. 
 EXAMPLES. 
 1. In the quadrantal spherical triangle a Bc, having 
 the side a c 63° 58’, and the side B c 48° 48%, to find 
 
 the rest. 
 BY CONSTRUCTION. 
 
 1. Describe a circle with the chord of 60°, and draw 
 the diameters 4 a, p‘d, at right angles to oh other. 
 
 2. Take a c equal to the chord of 63° 58’, and set 
 off c b,c b, each way, with the chord of 48° 48’ (cB). 
 
 3. Make o n equal to the semitangent of the com- 
 plement of cB (41° 12’), and through the points 4, 2, 
 b, describe a circle, cutting pd in B. 
 
127 
 
 4, Draw the diameter c c, and through the points 
 ¢, B, c, describe a circle, and through a 8 a another ; 
 then agBc, will be the triangle required... 
 
 To measure the required parts. 
 
 1. Draw the diameter ze perpendicular to cc; and 
 having taken o m, in degrees, on the semitangents, set 
 off its complement from o to p. | 
 
 2. Through p, B, draw B pr, cutting the circlein r; 
 then 7 p, on the chords, gives 4 8 54° 18’; o 8, taken 
 on the semitangents, and substracted from 90°, gives 
 £A 42° 51’; and o m, taken on the same line, and then 
 substracted from 90°, gives Z c 64° 42° or 115° 18’ 
 
 BY CALCULATION. 
 
 > Sinpc ---- 48°48’ -- 9.8764574 
 : Rad, orsin - = - 90° - - - -.10,0000000 
 >; Cosac =--- 63°58 -- 9.6423596 
 > CosZp----- 54° 18° 58” 9.7659022 
 Which 4 is acute, being like its opposite side a c. 
 SIMA CO, < - le) == 63° 58’ - - 9,9535369 
 24 nad, OF si) “=~ “#902, os. °."<'10.0000000 
 >: CosBc --+ = 48°48’ -- 9,8186807 
 Sw COSLGA 5 am ay cathli AOM5 1 O99, 8651498 
 Which < is acute, being like its opposite side 8 c. 
 : Rad, orsin - - -' 90° --- - 10.0000000 
 >: Cotac ---- 63°58’ -- 9.6888927 
 >: CotBc --=-- 48°48’ -- 9.9499933 
 > CosZe --- - 115° 18’ 57" 9.6310460(2) 
 
 _—_——-— 
 
 Which Z is obtuse, because 4 c and Bc are like. 
 
 a 
 
 (4) This analogy becomes tan 4.¢: cot 3c ::rad:cose for 
 the instrumental computation. 
 
128 
 
 INSTRUMENTALLY. 
 
 1. Extend the compasses from 63° 58’ (ac) to 41° 
 12’ (compt. of Bc) on the tangents, and that extent 
 will reach, on the sines, from’ 90° to 25° 18’, the com- 
 eon oF ie. | 
 
 . Extend from 48° 48’ to 90° on the sines, and 
 a extent will reach, on the same line, from 26° 2’ 
 (comp'. of ac) to 35° 42’, the complement of Z B. 
 
 3. Extend from 63° 58’ (ac) to 90° on the sines, 
 and that extent will reach, on the same line, from 41° 
 12’ (comp*. of B c) to 47° 9’, the complement of Z a. 
 
 2, In the quadrantal spherical triangle a Bc, 
 
 Given The side A c 109°192’ Zo 98°11'26” 
 Thesidep c 112915’ Ans.< 2B 110°48' 48” 
 
 Required the other parts. 44 113°38'16" 
 
 3. In the quadrantal spherical triangle asc, — 
 Given J beside ac 145°177 LG Caen” 
 ven) Thesidesc 81°12 Ans 4B 1461651” 
 Required the other parts. Za 74825 4° 
 
 CASE VI. 
 When the two angles are given, to find the rest. « 
 1. To find either of the two sides. 
 As tan either given £4: rad :: sinother Z : cot its 
 adjacent side. | 
 Which side is like its opposite angle. 
 2. To find the hypothenusal angle. 
 As rad : cos eith. giv. 4°: : cos other Z : cos hyp’. 2 . 
 Which hyp’. 4 is obtuse when the given 4 § are like; 
 but acute when they are unlike. 
 
129 
 
 EXAMPLES 
 1. Inthe quadrantal spherical triangle a B c, having 
 the angle a 42°51’, and the angles 54° 19’, to find 
 the rest. 
 BY CONSTRUCTION. 
 
 ra 
 
 1. Describe a circle with the chord of 60°, and draw 
 the diameters a a, B 6 at right angles to each other. 
 
 2. Set off o m equal to the semitangent of the com- 
 plement of 2 a (47° 9"), and through the points A, m, @ 
 describe a circle. 
 
 3. Set off on equal to the semitangent of the com- 
 plement of / 8 (35° 41’), and through the points 
 b, n, B describe a circle, cutting the former inc; then 
 4 B C will be the triangle required. 
 
 * To measure the required parts. 
 
 Set off o P, op, equal to the semitangents of the 4° 
 A and sB, and through the point c draw the lines er 
 and fs: then rs, onthe chords, gives £4 c 115° 19’; 
 
 A e, on the same line, gives Ac 63° 58°; and Bf gives 
 B C 48° 47’ 
 
 ‘ 
 
 é 
 BY CALCULATION. 
 
 Meee ZA = 8 ns 4 Ls - 9,9673759 
 >: Rad, or sin - - - 90° - - - - 10.0000000 
 Pte Bw = 3 + 2 64° 19’. + 9.9096915 
 : ‘Cotpe --- 48°47’ 39”. 9,9423156 
 
 Which side is acute, being like its opposite 4 a: 
 K 
 
130 
 
 é Tan Zp weirs) “a salehotorcianbOs1487986 
 
 : Rad,orsin -- 90° ----- 10.0000000 
 4° Gin'd Abd Sweet Sah.” Sk-AOB825609 ui 
 > Cotesia > ~ -  6ares an" 9.6887651 
 
 ——— 
 
 Which side is acute, being like its opposite 2's. 
 
 Rad or sin <<) (9@u,". 5 <= 10,0000000 
 "Cos Za -=- 49°51’ - = = 918651849 
 ben BP LHLOR eae 19" . --- 9.7658957 
 : Cos Zc --+115°19’ 4”. 9.6310806(”) 
 
 Which Z is obtuse, because the 4° a and zB are like. 
 
 eo oo ee 
 
 | INSTRUMENTALLY. 
 
 1. Extend the compasses from 54°19’ ( Z B) to 90° 
 on the sines, and this extent will reach, on the tan- 
 gents, from 42° 51’( 2 A) to 48° 47’, the side Bc. 
 
 2. Extend from 42° 51’ ( Z a) to 90° on the sines, 
 and this extent will reach, on the tangents, from 54° 19° 
 ( 4B) to 63° 58, the side ac. 
 
 3. Extend from 90° to 47° 9’ (comp. of ZA) on. 
 the sines, and this extent will reach, on the same line, 
 from 35° 41’ (comp’. of ee to 25° 19’, the comple- 
 ment of Z c. 
 
 2. In the quadrantal spherical triangle az c, 
 
 Given TheZ a 82° 19’ Bic) 82° SS uz 
 The Z 8 73° 21° Ans.< ac 73°29" 29” 
 
 Required the other parts. LO SRATT 45% 
 
 (m) The first two of these analogies are convertible into the fol- 
 
 sin B:rad::tana:tansec 
 
 lowing ones: { ‘ 
 § sma:irad:;tane:tanac 
 
 } for the instrumen- 
 
 tal solution. 
 
13) 
 
 3. Inthe quadrantal spherical triangle a Bc, 
 
 CG; The Z a 104° 19’ BC 96230 :2'7” 
 WED The 8.163027 “Ans. AC .159%49.19” 
 
 ‘Required the other parts. Zo 102° 46° 47° 
 
 OBLIQUE-ANGLED SPHERICAL TRIANGLES. 
 
 The different cases, or varieties, that may happen 
 in the solution of oblique-angled spherical triangles, 
 where any three things are given to find a fourth, are, 
 in all, twelve ; but, by restricting them to such as de- 
 pend upon the same principles, they may be reduced 
 to six. And if a perpendicular be drawn from one of 
 the angles to the opposite side, each of these cases, 
 except the two where the three sides or the three an- 
 gles are given, may be resolved by means of the rules 
 already proposed for right-angled spherical triangles. 
 
 Or, all the cases of oblique-angled spherical triangles 
 may be resolved, without drawing a perpendicular, by 
 means of one or the other of the four following the- 
 orems; which are better adapted to practice, and more 
 easily retained in the memory, than the various parti- 
 culars which must be attended to in the former method, 
 with respect to the falling of the perpendicular, and 
 the species of the different parts of the triangle. 
 
 I: 
 
 Sin either side : sin its opp. 4 :: sin any other side 
 > sin its opp. 4; and conversely. 
 
 . II. 
 
 Sin ; sin ae 
 or >gsumany twosides: or +4 their diff. :: cot. 
 Cos cos 
 
 I 
 
 , I diff. 
 5 mcluded 2: tan4 or other two 4’. 
 $ sum 
 
 K 2 
 
132 
 Ill. 
 
 Sin l sin a 
 r ef sum any two4*: or pd their diff. :: tan d 
 Cos} Cos | 
 
 included side : any 
 
 : or of an 
 « Rect. sines of any two sides FS Aseahy: wePaalee 
 two Z 
 f sin 4"sum 3 sides X sin diff. this $ sum and 3d side 
 os or | 
 3 t cos4sum 34° cos diff. this $ sum and 3d 4 
 
 : cos’ y included Ps or sin’ 4 included side}. 
 
 To ied rules it is proper to add, that in the 2d _ 
 and 3d cases, the terms of the proportion may be taken - 
 either directly, inversely, or alternately ; and that, in 
 the 4th case, the first four terms of the two analogies, 
 contained in the rule, are to be used for finding an 
 angle, and the four latter for finding a side. 
 
 Tt may here also be observed, that the six cases of > 
 oblique-angled spherical triangles, already mentioned, 
 are as follows: 
 
 (‘Two sides and an Z opp. to one of them) + 
 
 Two 4*and a side opp. to one of them | x 
 
 ica. d ‘Two sides and their included 4 2a 
 Two 4* and their included side bi, . 
 
 The three sides | A= 
 
 The three angles 9 
 
 | CASEL in 
 When two sides and an angle opposite te one of 
 them are given to find the rest. 
 
133 
 
 1, To find the other opposite angle. 
 
 As sin side opp. given 2: sin that 2 :: sin other 
 given side : sin its opposite 2. 
 
 Which Z is either an acute Z or its supplement, ac- 
 cording as it makes the greater side opposite the greater 
 angle. And if each of them agree with this rule, the 
 triangle is ambiguous, or admits of two different s So- 
 lutions. 
 
 2. To find he angle contained by the given sides. 
 
 Find the Z opp. the other given side, by rule 1, 
 and note whether it be ambiguous or not. 
 
 ibis sin + dif. two given sides : sin 4 their sum 
 
 : tan 4 dif. theif opp. Z*: cot 4 inclu’. Ze 
 
 Hawvhias 4 4 Z is always acute ; iad if the angle found 
 by rule 1. be ambiguous, the required angle will be 
 ambiguous, otherwise not. 
 
 3. To find the third side. 
 
 Find the angle opp. the other given side, by rule 1, 
 and note whether it be ambiguous or not. 
 
 Then, Sin 4 dif. these 2°: sin + their sum: : tan 
 2 dif. given sides : tan + remaining zie | 
 
 Which 2 side its sieeeg less than 90°; and if the 
 angle ene by rule 1. be ambiguous, the required side 
 will be ambiguous, otherwise not. 
 
 EXAMPLES, 
 1. In the oblique-angled spherical triangle a-B oc, 
 having the side a B 68° 37’, the side a c 62° 40’, and 
 the angle 3 51° 28’, to find the rest. 
 
134 
 
 BY CONSTRUCTION. 
 
 1. Describe a circle with the chord of 60°, and draw 
 the diameters a a, pd at right angles to each other. 
 
 2. Set off the side a B (68° 37’) from a tos, by the 
 scale of chords, and draw the diameter 8 b, and an- 
 other Ee, at right angles to it. | | 
 
 3. Take on equal to the semitangent-of the com- 
 plement of ZB (38° 32’), and pnb the points 8, n, b 
 describe a circle. 
 
 4, Set off the side ac (62° 40’), by a scale of chords, 
 from a each way, toh, h; and make o m equal to the 
 semitangent of the complement of a c (27° 20’). 
 
 5. Through the points h, m, h describe a circle, cut- 
 ting the former Bzd-inc; then, if a circle be de- 
 scribed through the points a, c, a, the triangle azBc, 
 or ABC, will be the one required, each having the same 
 data; which shows the case to be ambiguous. 
 
 To measure the required parts. 
 
 1. Take o v in degrees, on the semitangents, and set | 
 ‘off its complement from o to Pp: also take o p equal to 
 the ie ia of Z'B.(51, 28), 
 
 . Through the points c, p and c, p heer the lines 
 or}: a st, cutting the circle in 7,s, and ¢; then7’s, © 
 on the chords, gives the Zc 55° 4’; ov, taken onthe | 
 line of semitangents, and then added to 90°, gives Lig | 
 122° 4°; and B ¢, on the chords, gives Bc 105° 45. 
 
135 
 
 And if the triangle a B c had been taken, the angle 
 8 Ac would have been found 8° 44’, angle B ac 24° 
 
 36’, and Bc 9° 55’. 
 
 BY CALCULATION. 
 
 : SinaG ---- 62°40 - += 99485842 
 0.0514158 
 LAP IA) Diet itm ie, ba Bike, wm ite DG 48 
 Se MBs min SOS ay = 9.9690252 
 > SinZc 55° 4 45”, or 124° 55'15” 9.9137843 
 Sin “== - - 9°58'30' - 8.7151692 
 an a 1.2848308 
 <:..Sm 5 - - 65°38 30” - 9.9595107 
 >: Tan ~-- 1°48'29” . 8.4987656 
 $ ZA Q 2407 
 pp ot ee GL 12 s  O.7aS1 071 
 
 . 2 
 122) 46922" LA 
 
 And had Zc 124° 55’ 15” been taken, 
 
 ot 
 
 the result, by 
 
 the same method, would have given Z a, or BAc, 8° 44’, 
 
 Sin satis cao 1° 48°22” - - 8.4985498 
 1.5014502 
 Soar Fe eas 53° 16’ 22” ~ - 9.9038989 
 MAGE ceo os O° B880" =. 89857549 
 - Tan --- - 52°53/18” - 10.1211040 
 
 Z 
 
 2 
 105° 46’ 26” Bc. 
 
 ——— 
 
 —as 
 
 And had 4 c 124° 55’ 15” been taken in this case, 
 
136 
 
 the result, by the same method, would have given Bc 
 9° 55° 44 (n). 
 INSTRUMENTALLY. 
 1. Extend the compasses from 62° 40° (4 c) to 
 51° 28'( 4B) on the sines, and this extent will reach, 
 on the same line, from 68° 37’ (aB) to 55° 4 Zc. 
 
 eo) to 65° 38" (PE) 
 
 on the sines, and this extent will reach, on the tangents, 
 
 2. Extend from 2° 58° ( 
 
 from 1° 48° (= “) to 28° 58’, the comp‘. of + Za. 
 
 3. Extend from 1° 48’ (Se) to 53°16 (ae on 
 
 the sines, and this extent will reach, on the tangents, 
 
 from 2° 58° (—3-) to 52° 53’, which is 4 sideBc. 
 2.-In the oblique-angled spherical triangle a Bc, 
 The side aB 56°19’ BC 87" (3 52" 
 Given< Theside ac 114° 12% Ans.< Zc 50°48’ 39” 
 The ZB .- 121° 50° | £468°28 7” 
 
 Required the other parts. 
 3. In the oblique-angled spherical triangle a Bc, 
 The side ac 62° 27’ "A BIG 47 22° 
 Given < The side Be 73°19’ Ane Z 8: AGT’ 32° 
 ( TheZa - 51°14 Zc 138%23° 50° 
 Required the other parts. oe 
 
 (#) This example affords an opportunity of remarking, that the 
 angle first found, or that opposite the other given side, in this 
 case, is always either an acute angle or its supplement ; but itis 
 
 evident, both from the construction and calculation, that this — 
 
 may not be the case with respect to the remaining side, or the fe. 
 
 maining angle; for the two values of Z 4 are 122° 4’ 22” and 8° 
 
 44’, and of Bc 105° 46! 26” and $° 5 55' 44)’, which are not sup- 
 plements of each other, 
 
137 
 
 4. In the oblique-angled spherical triangle a Bc, 
 The side a B 59° 19" Ac 43° 18’ 36” 
 Given< The sides c 48° 7 Ans.< 2a 59° 50°57” 
 TheZc - 87°15 ZB 52° 48' 58” 
 Required the other parts. 
 
 CASE II. 
 When two angles and a side opposite to one of them 
 are given, to find the rest. 
 
 1. Yo find the other opposite side. 
 
 Assin Z opp. given side ; sin that side : : sin other 
 given Z : sin its opposite side, 
 
 Which side is an arc less than 90°, or its supplement, 
 according as it makes the greater ‘side opposite the 
 greater angle. 
 
 And if each of them agree with this rule, the tri- 
 angle is ambiguous, or admits of two different solutions. 
 
 (2. To find the side included by the given 4°. 
 
 Find the side opp. the other given angle, by rule 1, 
 
 and note whether it be ambiguous or not. 
 ; Then, 
 
 Sin 4 dif. two given Z*: sin 4 their sum :: tan 4 
 dif. ther opposite sides ; tan + included side. 
 
 Which 4 side 1s always less see 90°; and if the side 
 found by file i. be ambiguous, the required side will 
 be ambiguous, otherwise not. 
 
 3. To find the remaining angle. 
 Find the side opp. the other given angle, by rule 1, 
 and note whether it be ambiguous or not, 
 
138 
 
 Then, 
 Sin 4 dif. two given sides : sin 4 their sum :; tan 4 
 dif. their opposite Z§ : cot 4 jadinded is 
 Which + Zis always see ; and if the side, found 
 by rule 1, be ambiguous, the required angle will be 
 ambiguous, otherwise not. 
 
 EXAMPLES. 
 
 1. In the oblique-angled spherical triangle a Bc, 
 having the angle c 51° 38’, the angle a 59° 27’, and 
 the side a B 64° 5’, to find the rest. 
 
 BY CONSTRUCTION. 
 
 1. Describe a circle with the chord of 60°, and draw 
 the diameters aa, D d at right angles to each other. 
 
 2. Take on equal to the semitangent of the com- 
 plement of Z a (30° $3’), and through the points 
 A, 7, a describe a circle. 
 
 8. Set off ar, ar, each equal to az (64° 5°) from 
 a scale of chords, and having made o m equal to the 
 semitangent of the complement of a B (25° 55’), de- 
 scribe the circle r mr, cutting a Ba in B. , 
 
 4, With the tangent of Zc (51° 38’), and o asa 
 centre, describe an arc; and with the secant of the 
 same angle, and B as a centre, cross it in 0. 
 
 5. From the centre oa, with radius o B, describe the 
 
139 
 
 circle b3c; and ABC, or ABC’, will be the triangle 
 required, each having the same data; which shows the 
 question to be ambiguous, : 
 
 To measure the other parts. 
 
 1. Set off the semitangent of the comp‘. of Z a 
 (30° 33’) from o to P,. and draw B Ps}; also, with the 
 semitangent of Zc (61° 38’), and o as a centre, cross 
 oain p, and draw vp Be. 
 
 2. Then s v, taken on the scale of chords, and sub- 
 tracted from 180°, gives ZAB Cc 132° 24’; cc, on the 
 4.3-and ac is 122° 6’, 
 
 And, if the triangle a 8 c’ had been taken, the angle 
 a Bc’ would have been found 154° 33’, the side Bc’ 
 98° 55’, and ac’ 150° 28’. 
 
 . BY CALCULATION. 
 
 same scale, givesBc 81° 
 
 23 Sir 4.G"- - 22 - §1° 38’ .- - 918943464 
 : 0.1056326 
 
 SinaBo --- 64° 5’ .~- .9.9539677 
 >: SnZa whe mi SOOT 46! 5 9,9350969 
 
 : Sinpc 81° 4’ 56” or 98° 55’ 4” = 9,9947182 
 
 Where it is to be observed, that, as each of these 
 values of Bc, agree with the rule, in making the 
 greater side opposite to the greater 4, the A, in this 
 case, is ambiguous. 7 | 
 
 - Cea A , Ww 
 ; Sin-3- ---- | 3° 54’ 30” - 8.8335321 
 | 1.1664679. 
 
 « C+A. / 
 ; Sin-S* ---- 55°39/ 30” - 9.9162106 
 > Tan = -- 8°29" 58” - 9.1744700 
 peta 2 OETA GLO SON TORT TELS 5 
 
 Ree 
 
 2 
 
 $22" Gs Dae 
 
, 140 
 
 And had 8 c’ (98° 55’ 4”) been taken, the result, 
 by the same method, would have given AC’ 150° 
 28 32” 
 
 Sin [== - -- 8° 29/58” - 9.1696739 
 0.8303261 
 
 : Sin <=Z"° . - . 72° 34’ 58” - 9.9796169 
 :: Tan3- ---- 9°54/30" - 8.8845423 
 Cot eee ee 66°12 1” - 9.6444853 
 
 2 
 132° 24° 2” Lape 
 And had B c’ (98° 55’ 4”) been taken, the result, by 
 the same method, in this case, would have given 
 £88 C154 33''32"-(0). 
 
 INSTRUMENTALLY. 
 
 1. Extend the compasses from 51° 38’ (Ze) to 
 64° 5° (a B) on the sines, and this extent will reach, 
 on the same line, from 59° 27’ (2) to 81° 4’, the 
 side BC. | 
 
 Cio A C+A 
 
 2, Extend fot 3° 54 5 ) to 55° 39’ ee ) on 
 
 es sines, and this extent wil reach, on the tangents, 
 
 from 8° 29’ (—— 
 
 } to 61° 3’, which is 2 ac. 
 
 (0) From this example it also appears that the side BC, first 
 found, must be either an are less than 90°, (81° 4’ 5 56") or its 
 supt. (98° 55’ 4”) ; but the two vaiues of the side ac (129° 6’ 2" 
 and 150° 28’ $2") are both obtuse, as are, also, the two values 
 
 of. 38 (132° 4! 3" and 154° 33’ 32”). 
 
141 
 
 3. Extend froni 8° 29’ (—) 10 '72° 34’ eases 
 
 on the sines, pe this extent will reach, on the tan- 
 
 Bes A 
 
 = ——} to 23° 49’, the complement 
 
 gents, from 3° 54 
 of as 
 
 2. In the oblique-angled spherical triangle a Bc, 
 TheZc--- 49° Fog aB 55° 53/91” 
 Given {he 24 hem ghd Qaeda And Ac 85°. 53' 27" 
 The side 5c 115° 6’ 28 65° 36! 58" 
 Required the other parts. 
 3. In the oblique-angled spherical triangle a Bc, 
 
 The zc - - 47°19 Bc 63°13! 22" or 116246! 39” 
 G {Theat - - 51°39 Aos.4 ac 97° 2'46” or 170° 2¥' 32" 
 
 The side as 56°57’ ZB 119°29' 34” or 171° 34’ 6" 
 
 Required the other parts. 
 
 CASE Ill. 
 
 When two sides and their included angle are given, 
 to find the rest. 
 
 3. Lo find either of the other two angles. 
 
 As cos 1. sum given sides : cos + their diff. : : cot 4 
 their ‘eli 4:tan 4. sum other two Zs, 
 
 And, As sin sum Bey sides : sin + their diff. : : 
 cot + their ers Z: tan + diff. other abs vas} 
 
 ‘Then, if the + diff. of ise two 4° be added to + 
 their sum, it will give the Z opp. the greater side; and, 
 if it be subtracted from the + sum, it will give the 
 angle opposite the less side. | 
 
 In which case, 2 the sum of the two 4 ‘is like 4 the 
 sum of their opposite sides, and 4 their differ ence is 
 always less than 90°. 
 
142 
 
 2. To find the remaining side. 
 Find the two remaining angles by the first part of 
 the rule. Then, — 
 Sin + diff. of these 2°: sin 4 their sum:: tant 
 dif. their opp. sides : tan 4 remaining side. 
 Which + side is always less than 90°. 
 EXAMPLES. 
 1. In the oblique-angled spherical triangle aBc, 
 having the side a B 76° 20’, the side Be 119° 17’, and 
 the included angle B 52° 5,, to find the rest. 
 
 BY CONSTRUCTION. 
 
 1. Describe the circle a p ad with the chord of 60°, 
 and draw the diameters a a, pd at right angles to each 
 other. 
 
 2. Set off az (76° 20’) froma scale of chords, and 
 draw the diameter 8 4, and another £ e, at right angles 
 to it. 
 
 3. Make on equal to the semitangent of the com- 
 plement of ZB (37° 55’), and through the points 
 B, n, 6 describe a circle. 
 
 4, Set off bs, bs each equal to the chord of the sup- 
 plement of Bc (60° 43’), and having made o r equal 
 to the semitangent of the complement of the same arc, 
 describe the circle srs, cutting the circle Bz in c. 
 
143 
 
 5. Through the points a, c, a@ describe a circle; 
 and a Bc will be the triangle required. _ 
 
 To measure the required parts. 
 
 1. Take o m, o n, in degrees on the line of semitan- 
 gents, and set their complements from o to p and p; 
 and through the points c, p, c, p, draw the lines xc, vw. 
 
 2. Then a w, measured on the scale of chords, gives 
 Ac 66° 5’; wv, onthe chords, gives 4c 56°58’; and 
 om, taken on the semitangents, and added to 90°, 
 gives ZA 131° 10’. 
 
 BY CALCULATION. 
 
 : Cos**t** . ~ .97°48'30" - 9.1330906 
 
 ' seh) 0.8669094 
 
 a¢€ ee = - 21° 98’ 30” - 9.9687595 
 ZB 
 
 >: Cot ---- 26° 230" 10,3110171 
 
 , & Z 
 5 Lan, om = $5° 55" 10” or 94° 4’ 50" 11.1466790 
 
 The latter of which 94° 4’ 50” must be taken, to ° 
 make the ¢ sum of the 4° agree with the 4 sum of 
 their opposite sides. 
 
 : Sin“*5=" - - 97°48'30" - 9.9959545 
 
 0.0040455 
 -~ = 21°28' 30° + 9.5635941 
 
 2: Cot- ---- 26° 2’30" 10.3110171 , 
 : Tan as Eg? 5) 50"... 0. Sugg rey 
 94° 4! 50” 
 
 131° 10’ 42” ZA 
 56° 58 58’ Zo 
 
l4a4 
 
 . LAL bs a 
 : Sin ““S-* -- 37° 5°52” - 97804444 
 | , 0.2195553 
 a tah Aiden Lic , wt : 
 : Sin "Ee 2. 94° 4’ 50” - 9.9988977 
 >; Tan "= -. 21°28'80” - 95948416 
 : Tan ---- 39° 9 48” . 9.8192946 
 2 
 66° 5 36" A Cs 
 INSTRUMENTALLY,. 
 1. Extend the compasses from 7° 48’ (comp*. of 
 
 ASB 
 
 ne as : ~) to 68° 32’ (comp'. of —~ 5) on the sines, 
 
 and ‘his extent will reach, on the es from 63° 58’ 
 
 comp’. of $28) to 85° 55, the }.sum of 2 a and:c, 
 
 2. Extend from 82° 112’ (supp*. of = ae) to 21° 
 
 9! C3 NAN —) on the sines, and the same extent will 
 
 reach, on the tangents, from 63°58" (comp'. of 3 Zs) 
 to 37° 5’, the 4 diff. of 4° a and c. 
 
 3. Extend from 37° 5(~5-~*) to 94° Pai ks 
 
 on the sines, and this extent will reach, on the tan- 
 gents, from OSES. (—; oes =") to 33° 2’, the 4 of ac. 
 2. In the oblique.angled spherical triangle a Bc, 
 The side aB 59° 13’ La (62? 859° 
 Given < Thesidepc 81°17’ Ans.< Zc 50°19/58” 
 : Theinc®. 4B 125° 36’ AC114°37'54" 
 Required the other parts. 
 
145 
 3: In the oblique-angled spherical triangle a Bc, 
 ‘The side a B 109° 19° 4c 108°44" 1” 
 Given< Theside pc 78°14 Ans.< 4a 79°14°37" 
 Theinc?. ZB 91° 9° AC 94°56 
 Required the other parts. t 
 
 CASE IV. 
 When two angles and their included side are given, 
 to find the rest. , 
 
 1. To find either of the other two sides. 
 
 As cos 4 sum given Z*: cos % their diff. :: tan 3 
 
 included side : tan 4 sunt other two sides. 
 And, | 
 
 As sin 4 sum given Z®: sin 4 their diff. :: tan 4 
 included side : tan 4 diff. other two sides. 
 
 And if § the diff. of the sides be added to 3 their 
 sum, it will give the side opposite the greater Z; and, 
 if it be subtracted from 4 the sum, it will give the side 
 opposite the less 4. 
 
 In which case, 3 the sum of the two sides is like 2 
 the sum of their opposite 4’, and 4 their diff. is less 
 than 90°. 
 
 2. To find the remaining angle. 
 Find the two remaining sides by the former part of 
 the rule, 
 Then, 
 Sin 3 diff. these sides : sin 2 their sum :: tan 2 dif, 
 given Z* : cot 4 remaining Z. _ 
 Which $ Zis always less than 90°, 
 iL 
 
146 
 
 EXAMPLES. 
 
 1. In the oblique-angled spherical triangle ABC, 
 having the angle a 130° 56’, the angle B 52° 5’, and 
 the included side a B 76° 48’, to find the rest. 
 
 BY CONSTRUCTION. 
 
 1. Describe the circle a p ad with the chord of 60°, 
 and draw the diameters aa, p d, at right angles to 
 each other. 
 
 2. Set off AB (76° 48’) froma scale of chords, and 
 draw the diameter 8 4, and another « e at right angles 
 to if. 
 
 3. Make o x equal to the semitangent of the com- 
 
 plement of the ZB (37° 55’), and through the points ~ 
 
 B, 2, 6 describe a circle. 
 
 4, Also, make o m equal to the semitangent of 40° 
 56° (the excess of Za above 90°); and through the 
 
 points a, m, a describe a circle, cutting the former in 
 
 c; and asc will be the triangle required. 
 
 To measure the required parts. 
 
 1. Set off the semitangent of 52° 5’ (48) from — 
 
147 
 
 oto p, and the semitangent of 49° 4 (sup*. of 2 a) 
 from o to P. 
 
 2. Through the points p, c and p, c draw the lines 
 rs,av: then as, onthe chords, gives ac 65° 44; 
 Bv, on the same line, gives Bc 119° 11’; andrais 
 57° 24, the Zc. 
 
 BY CALCULATION. 
 
 : Cos a ~ = 91°30’ 30” - 8.4203945 
 | 1.579675 
 : Cos44>4? .. 39°95’ 30” - 9.8878741 
 
 Z 
 
 : Tan “- wien i: BB? Bae. car Olg90Gas 
 
 : Tan ces 87° 32/17" or 92° 27/43” 11.3665983 
 
 Se eeeeneeeeeel 
 
 eo 
 
 The latter of which 92° 27’ 43” must be taken, to 
 make the 4 sum of the sides agree with the $ sum w 
 their sheets angles. 
 
 : Sin — ~- 91°30'30’ - 9.9998495 
 0.0001505 
 - Sin 44748 _ "39° 95’ 30” - 9.8028200 
 
 -embnee oC.  -.) 98°. 24 - 9.899048'7 
 
 Z otha te 
 > Tan SS ~ + 26°43’ 34” ~ 9,.7020192 
 99° 27’ 43” 
 119° 11'17’ Be 
 
 i: 65° 44° 9’ ac. 
 
 L 2 
 
148 
 
 ; Sin -. 26° 43’ 34” - 9.6529480 
 0.3470520 
 Sin S*= -- 92°97/43” - 9.9995989 
 
 Zz : yw | 
 2: Tan 424% _ 39° 25°30” - 9.9149459 
 Cot 5 Sb po eisiie’ *? 1Gig6 ede 
 
 ee 
 
 2 
 57° 24°12" iL, 
 
 INSTRUMENTALLY. 
 
 1. Extend the compasses from 1° 30’ (comp*. of | 
 
 A+B 
 Z 
 
 extent seal reach, on the tangents, from 38° 24° 
 
 (> ") to 87° 32’, which is aces, 
 
 2, Extend from 88° 30° (sup‘. of <=) to 39° 25’ 
 
 ASB 
 ee ) on the sines, and this extent will reach, on the 
 
 Cop 
 
 tangents, from 38°24 (> ~ ) to 26° 43’, which i is ~ <3. 
 
 & OSs 0) F 
 
 3. Extend from 26° 43’ (-S 
 A a 
 of 
 
 —— ) to 2° 27’ (comp*. 
 =) on the sines, and ms same extent will reach, 
 
 ZA LB 
 on - tangents, from 39° 25’ (——;— 
 
 ) to 61° 18’, 
 Y dy 2 
 the complement of --. 
 
 2. In the oblique-angled spherical triangle a B c, 
 
 Phen ar 71 cl 8. Bc 80°15'27” 
 G*.Z TheZ.3p =:-'-,130°.15' Ans.~ ac127°26’3) @ 
 
 Theinc’.sideaB 82° 9’ 
 
 : AC 72°11 tee 
 Required the other parts. 
 
 ) to 50° 35’ (comp'. of “> ) on the sines, and this. 
 
 ~ 
 Se Cae 
 
149 
 3. In the oblique-angled spherical triangle a 8 c, 
 The Za -- - 110°16 4B 83°21" 32” 
 ord Thee -+- 56°22’ ans} A BSG )32 GC" 
 Theine’.sideac 95°36’ BC 109°57'42” 
 Required the other parts. | 
 CASE V, 
 
 When the three sides are given, to find either of the 
 angles, 
 
 As rect. sines of any two sides : sin 4 sum of the 
 three sides X sin diff. this ¢ sum and 3d side :: rad’ 
 ; cos’ g their included Z. 
 
 Which 4 Z is always acute. . | 
 
 Or the same theorem may be expressed logarithmi- 
 cally thus :—-Add together the logarithmic sines of 
 the two sides about the required 2, and find the com- 
 plement of their sum, by subtracting the index from 
 19, and the other figures from 9, as usual. Then, to 
 this complement add the log sine of $ the sum of the 
 three sides, and the log sine of the diff. of this 4 sum 
 and the third side, and the result, divided by 2, will 
 give the log. cos. of the Z sought. 3 
 
 EXAMPLES. 
 
 1. In the oblique-angled spherical triangle asc, 
 having ag 79° 56’, Bc 119° 36’, andc a 64° 5’, to 
 find the angles. 3 
 
 BY CONSTRUCTION. 
 
150 
 
 1, Describe a circle with the chord of 60°, and draw 
 the diameters 4 a, Dd at right angles to each other, 
 
 2. Set off ac (64° 5’) from a scale of chords, and: 
 draw the diameter c c, and another £ e at right angles 
 to it. 
 
 3. Set off the side a B (79° 56’) by a scale of chords, 
 each way from a tor; also make o m equal to the 
 semitangent of the complement of a B (10° 4°); and 
 through the points r, m, r describe a circle. 
 
 4. Set off the supplement of c B (60° 24’) by a scale 
 of chords, each way from ¢ to s; also make on equal 
 to the semitangent of the complement of cB (29° 36’), 
 and through the points s, , s describe a circle, cutting 
 the former in 3. . 
 
 5. Through the point 8 draw the circles aB a, and 
 Cc Be, cutting pdinz, and Ee inz, and aBc will be 
 the triangle required. 
 
 To measure the angles. 
 
 i. Take ow, in degrees, on the semitangents, and 
 set off its complement from oto p; also take oz, 
 in the same way, and set off its complement from 
 © to P. 
 
 2. Through the points c, p, draw the line c pu; and 
 through B, p and g, P draw Bv and Bw. 
 
 3. Then v w, on the chords, gives Z B 52° 18°; cu, 
 on the same line, gives Zc 60° 1’; and oz, taken on 
 the line of semitangents, gives 40° 5’ for the excess of 
 
 A above 90°. : ’ 
 
151 
 
 BY CALCULATION. 
 Ake REIT 5G. 
 
 BiChe Min eh 19° “SB, 
 CAM me 64° 75" 
 
 -9|963° eT 
 Oa bod 48° 30° 30” - +sum 
 64° Be a = AC 
 67° 43’ 30” - Esum—ac. 
 Sis i eS) r9° 56’ - - 9.9932621 
 ERIC tee. elim 139°: 36" /-" «+ 9.9392671 
 19.9325292 
 0.0674708 
 
 SinZ sum - - - - 131° 48’ 30” 9.8723772 
 
 Sin (Zsum—ac) - 67° 43’ 30° 9.9663179 
 
 am 2|19.906 1659 
 
 Cos$4p ---- 26° 9'20° 9,9530829 
 2 
 
 52° 18 40’ ZB, 
 
 ae 
 
 131° 48’ 30” . - dsum 
 
 79° 56’ eh ACB 
 51° 59 30” - - $sum—AB. 
 sme «] 2S 2 199° 36H) =) \9.9392671 
 BINA CGH wlccos «642 5 -  -9.9539677 
 9.8932348 
 0.1067652 
 Sin sum - ~ - - - 131° 48 30° = 9.8723772 
 
 Sin ($ sum—as) - 51° 52’.30° riddle 
 
 —— RS 
 A ee ee 
 
 Cos$ic ----- 30° 0’ 53” “f 9374663 
 
 a ENG fe PES 
 
 60° 1. 46; Ze. 
 
152 
 
 131° 48’ 30° - - 4 sum 
 119% 36ig i= BS 
 
 12° 12 30” --4 sum— B Cy 
 Sin a’) wae - - = - 640 Br. 9.958967 
 SNAB -----= 79° 56’ -.-. 9.9932621 
 | -9,9472298 
 ~ 0,0527702 
 Sind sum --.- - 181° 48 90" 9.8723772 
 Sin (5 sum—sc) ~- 12° 12°30" 9.3252425 
 2|19.2503899 
 Cosi Za ---- 65° 247" 9.6251949 
 
 2 
 
 130.4534, Loa. 
 
 2. In the oblique-angled spherical triangle ABC, 
 
 {1 sideaB 59° 12’ { ZA 62°39! 49% 
 G". 
 
 The sideBc 81°17 Ans.< 2B 124° 50’ 50” 
 The sideacll4° 3’ Zc 50°31° 42” 
 Required the 2. 
 
 3. In the oblique-angled spherical triangle azc, 
 {ti side AB 112° 19° { Za 62°29’ 48” 
 G*. ! 
 
 The sideBc 85° 16 Ans.< 2B 49° 55 5A 
 The sideac 49° 56’ — Zc 124° 34’ 40” 
 
 Required the 4°. | 
 4. In the oblique-angled spherical triangle a xc, 
 The side aB 73° 13’ Za 44° 18% 
 Given< The sidepc 62°42’ Ans. < 48 136° 40 
 ) The side ac 119°: 5’ Zc 48° 48/ 
 Required the 2 *, 
 
155 
 
 CASE VI. 
 
 When the three angles are given, to find either of 
 the sides. ey 
 
 As rect. sines of any two Z$ : cos 4 sum of the three 
 Z° x cos diff. this } sum and $d Z:: rad? ; sin® 4 in- 
 cluded side. 
 
 Which side is always less than 90°. 
 
 Or the same theorem may be expressed logarithmi- 
 cally thus : | 
 
 Add together the logarithmic sines ef the two Z° 
 adjacent to the required side, and find the complement 
 of their sum by subtracting the index from 19, and 
 the rest of the figures from 9, as usual. 
 
 Then to this complement add the log cosine of 4 
 the sum of the three Z*, and the log cosine of the diff. 
 of this $ sum and the 3d 4, and the result, divided by 
 2, will give the log sine of $ the side sought. 
 
 EXAMPLES. 
 
 1. In the oblique-angled spherical triangle a Bc, 
 having the Z a 131° 35’, the ZB 63° 30’, and the 4c 
 59° 25°, to find the sides, 
 
 BY CONSTRUCTION, 
 
 1. Describe the circle a p ad with the chord of 60°, 
 and draw the diameters aa, Dd, at right angles to each 
 other. 
 
154 
 
 2, Make om equal to the semitangent of 41° 35” 
 (the excess of 4 4 above 90 °) and trotee the points 
 A, m, a describe a circle, 
 
 3. Set off o p equal to the semitangent of the com- 
 plement of the same arc (48° 25’), and draw a pr; 
 from which point p set off pw, Pw, each equal to the 
 chord of Zc (oy 25) | 
 
 4. Draw aw’, aw, cutting the diameter p din s, s, 
 and upon ss describe a semicircle: also, from the 
 point o, with the semitangent of ZB (63° 30°) as ra- 
 dius, describe an arc cutting the former in x. 
 
 5. Through the points 7, 0, draw the diameter E e, 
 on which take or equal to the semitangent of a 
 comp. of Z B-(26° 30°). 
 
 6. Make bb perpendicular to re, and through the 
 points B, 7, b describe a circle, cutting the circle ama 
 in c; and aBC will be the triangle required. 
 
 To measure the sides. 
 
 Draw pw through the points p c, and xv through 
 x, C3 then av, taken on the chords, gives a c 80° 17’; 
 Bw, onthe same line, gives Bc 124° $1’; and aA Bon 
 the same line, is 71° 28’. 
 
 BY CALCULATION, 
 LAW = peek BO 
 Bi we oe oe OOF, 
 WON Ly dink wae 
 2|254° 30° 
 127° 18’, = 6) 295itm 
 5G? 25’ :ai eae GS 
 67° 50’ - - (Esum—c). 
 
 eee ee ee 
 
155 
 
 Sin Z Ay Ses = 8 181% B35 > |) 9.8738965 
 Sin: LiBy serene, 2) OBS me 99517912 
 9.8256877 
 0.1743123 
 Cos}sum --- 127°15 = - 9.7819664 
 Cos (F sum < c) 67° 50° - - 9.5766892 
 2|19.5329679 
 Sndap --- 35°44'20" —9.7664839 
 2 
 71° 28’ 40” AB. 
 Vege tse nee suit 
 63° 30° -- ZB 
 6659/44'.9". 2 sums), 
 
 Civ is ew 2191! 35%s. ~~ 9.878896 
 Se a mnie, wie GO? Bo setae yO IS40497 
 19.8088442 
 ~0.1911558 
 Cos$sum ---+ 127°15 = - 9.7819664 
 Cos (4 sumo B) 63° 45° -- 9.6457058 
 
 9|19.6188280 
 SinZac .--- 40° 8 58” 9.8094140 
 2 
 80° 17°56" Ac. 
 151” 35 0- ee 
 
 127° 15° -- dsum 
 4a°zO «- (ftsumoa). 
 Sn Zp +«---- 637 807 ~'= 9.051 7912 
 
 SinZc --*+--- 59° 25° - - 9.93494'77 
 | 9.8867389 
 Cosisum --- 127°15' -- 9.7819664 
 Cos (4 sum & A) 4°20 -- 9.9987567 
 | 7 9)19.8939842 
 Sndpo ---- 62°15°49"  9.9469921 
 d iad 
 
 124° 31’ 39” Be. 
 
156 
 
 2. In the oblique-angled spherical triangle a Bc, 
 
 ZA 71° 49° AB 52° 50 44” 
 Givend ZB 125° 37’ Ans.< Bc 95° 56’ 10” 
 
 Ze 49° $2’ AC 121° 36 31” 
 
 Required the sides. 
 
 3. In the oblique-angled spherical triangle a Bc, 
 Za 49° 16° AB 118° 38’ 
 Given< ZB 63° 17’ Ans.xeBic.” 57° 45° 2" 
 oe Les oe Kees rol Soe 
 Required the sides. 
 4. In the oblique-angled spherical triangle a Bc, 
 PR LOL ABB) abe? te 
 Given, 2B 34° 33° Ans AC 40° 26’ 45” 
 eho Se BCT eae el 
 Required the sides. 
 
 \ 
 
 SOLUTIONS OF ALL THE CASES OF RIGHT-ANGLED 
 SPHERICAL TRIANGLES. 
 
 I. Given the hypothenuse and either of the oblique 
 angles, to find the other oblique angle. 
 
 RULE. 
 Asrad : cos hyp: : tan given 4: cot rem§, or req’. Z. 
 Which Z is acute when the hyp. and given 4 are 
 like; but obtuse (or the supplement of the former) 
 when they are unlike. | 
 Where it is to be observed, that this affection, as well 
 as all the rest in the following tables, except in the 
 
 iota: 
 
 NE 
 
157 
 
 ambiguous cases, may be readily deduced from the 
 algebraic formule, by attending to the signs of the 
 quantities which compose them (/). 
 
 Tm Or, 
 
 ‘ cos ¢ tan B cos ¢tan A 
 Cot a= ———; 3 cots =. 
 
 Lcota=Lcosc-+LtanB— 10; LcotB=L cos 
 c-+ iL tan a—10. 
 
 Note. In this, and the two folie cases, the hyp. 
 and given 4 may be each of any magnitude under 
 180°; but, if either of them be 90°, the required Z 
 or leg will be 90°, or else indeterminate. 
 
 When the hyp. is = given Z, cot a will be = sin 
 c, or sin B, and cot B = sin-c, or sin A; and when 
 the sum of the hyp. and given 4 = 180°, cot a will 
 be =—sinc, or sin B; and cot B =—sinc, or sin a. 
 
 II. Given the hypothenuse and either of the oblique 
 angles, to find the leg adjacent to that angle. 
 
 RULE. 
 As rad : tan hyp. :: cos giv. 2: tan adj‘. or req. leg. 
 Which leg is less than 90° when the hyp. and given 
 Z are like; but greater than 90° (or the supplement 
 of the former) when they are unlike. 
 
 (p) When a side or angle near 90°, in these or any of the 
 following cases, is to be determined by its tangent, subtract the 
 log tangent from 20, and it will give the log cotangent of the 
 same arc ; which can be found, by inspection, to seconds, in the 
 first part of the common tables; and if it should come out in a 
 cotangent, the log tangent may be found, in like manner, by sub- 
 tracting the log cotangent from 20, See the observations made 
 en this subject in page 27. 
 
158 
 Or, 
 
 tan ¢ COS B - tan ¢ccos A - 
 Tan a= —— ; tan b= ——_— 
 
 Ltan a=L tanc-+1L cosB—103; utan b= tan c+ 
 L cos A—10, 
 
 When the hyp. is = given Z, tan a will be = sin c, 
 or sin B; and tan b = sinc, or sin aA. 
 
 And when the sum of the hyp. and given Z = 180°, 
 tan a will be =— sin c, orsinB; andtan b =—sinc, 
 or sin A. 
 
 III. Given the hypothenuse and either of the ob- 
 lique angles, to find the leg opposite to that angle. 
 
 RULE I. 
 As rad : sin hyp :: sin giv. Z: sin opp. or req’. leg, 
 Which leg is less than 90° when the opp. or given 
 4 is acute; but greater than 90° (or the supplement 
 of the former) when it is obtuse. 
 
 Or, 
 : sincsinA  . sin ¢csin B 
 Sin. a= PEs dcaad On 
 
 Lsina=Lsine+Lsina—~10; rsnb=usinc+ 
 L sinB — 10. 
 
 RULE il. 
 
 Find the other oblique angle by case1; then, by 
 means of the hyp. and this angle, find the leg adjacent 
 to it by case 11; which will be the leg fete or that 
 opposite the given angle. 
 
 When the hyp. is = given Z, or the sum of the 
 
 sin? ¢ 1 
 r 2 or OF 
 
 hyp. and given Z = 180°; sina will be = 
 
 vers 2c. 
 
 A 
 
159 
 
 IV. Given the hypothenuse and either of the legs, 
 to find the angle adjacent to that leg. 
 
  RULEL | 
 As rad: cot hyp :: tan giv. leg : cos adj®. or req’. 2. 
 Which Z is acute when the hyp. and given leg are 
 
 like; but obtuse (or the supplement of the former) 
 when they are unlike. fi ; 
 
 Or, 
 
 tan Acote . __ tan acote 
 
 ae Oe Fe ee tS ar WE 
 A r 
 
 Cos A. = 
 
 LcosA=Ltan 6+1Lcotc—10; LcosB=L tana 
 -+- tL cotc— 10. 
 
 RULE II, 
 Poe sin (c—4) 
 en ae es (c-+5)" 
 
 __ €xusin (c+) + vsin (ec—b) + 16 
 See eons” Os a aR 
 
 The tan of 3 B may also be found by the same for- 
 mula, using the leg a instead of 4: and in each of 
 these cases the 4 4 is always acute. 
 
 Note. In this, and the two following cases, when 
 the given leg is less than the hypothenuse, their sum is 
 less than 180°; and when it is greater than the hypo- 
 thenuse, their sum is greater than 180°. If the hyp. 
 be equal to the leg, the triangle is indeterminate. 
 
 If the hyp. be = the leg, or the sum of the hyp. 
 and leg = 180°, the A is indeterminate. 
 
 ; 1 
 Ltang A 
 
 V. Given the hypothenuse and either of the legs, to 
 find the angle opposite to that leg. 
 
160 
 RULE I. 
 As sin hyp : rad :: sin given leg : sin opp. or req’. 2. 
 
 Which Z is acute when the opp. or given leg is less 
 than 90°; but obtuse (or the supplement of the former) 
 when it is greater than 90°. 
 
 Or, | “ 
 Sina = ibeit ing, sin B = “Mai 
 
 ; sinc sin ¢ 
 
 LSNA=€LsSinc+Lsna; LSNB=€ELsne+ 
 L sin 6, 
 
 RULE Il. 
 Tan (45° — 24) = +V tan 4 (c—a) cot 4 (ca) 
 
 Ltant poles +L ey (c +2) 
 L tan (45°ofa)= TDM id Mix 4 ek 
 
 The tan of (45°— £8) may also be determined by 
 the same form, using 6 instead of a: and the whole 4° 
 Aor B, which are found by subtracting 2 (45° 44a), 
 or 2 (45° £8) from,90°, will be acute or obtuse, 
 according to the rule given above. 
 
 VI. Given the hypothenuse and either of the legs, 
 
 to find the other leg. 
 | RULE I. 
 
 As cos giv. leg : rad :: cos hyp : cos rem&. or req’. leg. 
 
 Which leg is less than 90° when the-hyp. and given 
 leg are like; but greater than 90° (or the supplement 
 of the former) when they are unlike. 
 
 Or, 
 
 cosc¢ rcos¢ 
 5 Cobh ee eae 
 os b cosa 
 
 r 
 Cosa = 
 Cc 
 
 Lcosa=e€eLcosb+4nLcosc; tcosb=e€eLcsa 
 +L cos ¢. y 
 
161 
 
 RULE Il. 
 
 Tan hand he se) 
 
 The tangent of 3 b may also be éutie by the same 
 form, using the tai a instead of b: and in each of 
 these cases the 4 leg is always less than 90°. 
 
 VII. Given either of the legs and its adjacent angle, 
 
 to find the hypothenuse. 
 RULE. 
 
 As cos given Z : tan adjacent or given leg :: rad: 
 tan hyp. ! 
 
 Which hyp. ts less than 90° when the given leg and 
 angle are like; but greater than 90° (or the supple- 
 ment of the.former) when they are unlike. | 
 
 Or, . : 
 r tana r tanb 
 ‘Tan'c =) 
 COs B COS A 
 
 Ltanc==€LcosB+Ltana=€L cos a+Ltan 6, 
 Note. In this, and the two following cases, the given 
 leg and Z may be each of any magnitude under 180°. 
 
 rtana@ 
 When a= s, tanec = Se and ifa + 8B = 180°, 
 Sa 
 f rtana 
 tan Cc tien esa i 
 COS @ 
 
 VIII. Given either of the legs and its adjacent angle, 
 to find the other leg. 
 RULE. 
 As rad : sin given leg :: tan adjacent or given Z : 
 tan opposite or required leg. 
 M 
 
162 
 
 Which leg is less than 90° when the opp. or given 2 
 is acute; but greater than 90° (or the supplement of 
 the former) when it is obtuse. 
 
 ' Or, 
 
 __ sindtana sin a tan B 
 Tan a a teh tan b =- 
 
 z 
 Ltana=Lsin 6+1 tan a—103 Ltand=1 sna+ 
 
 LtanB—10. 
 sin 3 sn 
 
 When =a, tana = and if b+ a=180°, 
 
 sin 6 tan b 
 tang =— Fh TITRE ie 
 
 / 
 
 IX. Given either of the legs and its ae angle, 
 to find the other angle. 
 
 | RULE I. | 
 
 As rad : sin given Z :: cos adjacent or given leg : 
 cos opposite or required Z. 
 
 Which 4is acute when the opp. or given leg is less 
 than 90°; but obtuse (or the supplement of the for- 
 mer) when it is greater than 90°. | 
 
 Or, 
 
 cos @ sin Be cos ésina 
 
 Cos A = —— 3; cosB= . 
 r r 
 
 Lcos A=L cosa+LsinB—10; LE cosB= Lcosb 
 +L sin a—10. 
 
 RULE Il. 
 Find the hyp. by case vm; then, by means of the 
 hyp. and given 2, find the other Z by case 1. 
 When a= s, cos a==4 sin 2a; and if a+B=180°, 
 cos A=—sin 2a. 
 
163 
 X. Given the two legs, to find either of the oblique 2°. 
 
 RULE. | 
 As sin either leg : rad :: tan other leg: tan opp. 
 or required Age | 
 Which Z is acute when its opp. leg is less than 90° ; 
 but obtuse (or the supplement of the former) when it 
 is greater than 90°. | 
 | Or, 
 
 rtana rtan bd 
 Tana = Aen ty COLL Bs e 
 sin b sin 
 
 Ltana= exisind+ tana; LtanpB=eLsma 
 -+ ttan b.. 
 
 Note. In this, and the following case, the two given 
 legs may be each of any Pade las under 180°. 
 
 When a=), tan a = seca, OF earsis and if a+6= 
 
 2 
 
 180°, tan a —=—sec a, or— , 
 cos a 
 
 XI. Given the two legs, to find the hypothenuse. 
 
 RULE I. 
 As rad : cos either leg : : cos other leg : cos hyp. 
 Which hyp. is less than 90° when the legs are like; 
 but greater than 90° (or the supplement of the former 
 arc) when they are unlike. 
 | Ora 
 
 cos acos 6 
 
 L cos ¢ = Lcosa+1Lu cos 6—10. 
 
 RULE Il. 
 Find either of the oblique 2% by case x; then, by 
 M 2 
 
164 
 
 means of this angle and its adjacent leg, find the hy- 
 pothenuse by case vil. 
 
 2 
 When a= b, cose = —< ; and ifa+ b= 180°, 
 
 cos? a 
 
 cos ¢ =— 
 r 
 
 XII. Given the two oblique angles, to find the hy- 
 pothenuse. : 
 RULE I, 
 As rad : cot of eith. of giv. 2° :: cot other Z : cos hyp. 
 Which hyp. is less than 90° when the 2% are like; 
 but greater than 90° (or the supplement of the former 
 
 arc) when they are unlike. 
 Or, 
 cot A cot B 
 Cos ¢c = ————_. 
 T 
 -L cOS c = Lcot A+Lcot B—10. 
 
 RULE Il. 
 Ta ib —cos (A+B) 
 rng A xf cos (At B)* 
 
 Ltandc= €.cos (ac.B) i Lcos (A+B) + 10. 
 
 Where 3c is always less than 90°. 
 
 Noite. In this, and the following case, the two ob- 
 lique angles must be so taken, that their difference 
 shall be less than 90°, and their sum between 90° and 
 270. , 
 
 "cola ; 
 When a= 8, cos c = ——, and if a + B= 180°, 
 
 cot#Aa 
 cos ¢ = — ——,, 
 r 
 
 XIII. Given the two oblique alibi to find either 
 of the legs. 
 
165 
 
 RULE I. 
 As sin of either of given 2°: rad :: cos other Z : 
 cos opposite or required leg. 
 Which leg is less than 90° when its opp. 4 is acute; 
 but greater than 90° (or the supplement of the former 
 arc) when it 1s obtuse. 
 
 bei 
 at cos B 
 
 ree RAs 4 
 “sin 3 sin A 
 
 Lcosa=€L sin B+1L cos a; Lcosb=e€Lsin A+ 
 L COS B. - 
 
 : RULE II, 
 
 baal 
 
 Yan 3 ta=Vtan = — 45°) cot (25 =+-4.5°) 
 i 
 
 ‘The tan of 4 6 may also be found by the same form, 
 putiing a in the place of B, and zB in that of a: andin 
 each of these cases the 3 leg is always less than 90°. 
 
 -When-a= 8, cosa = cota; and if a+B = 180°, 
 cos A =— cot a. 
 
 L tan 1 (REA 459) +2 cot (pees 
 I, tan - 5 oom 2 \ 
 
 Z 
 
 XIV. Given either of the legs and its opposite angle, 
 to find the other leg. 
 
 RULE I. 
 
 As rad : cot given Z :: tan of opp. or given leg : 
 sin remaining or required leg. 
 
 Which leg is ambiguous: that is, it may be either 
 an arc less than 90°, or its supplement. 
 
Or, 
 
 A tan J cot B 4 tanacota 
 Sin a=—— 5 outt 0 am 
 
 ; r 
 Lsina=Ltanhb+ .Lcotp—10; tsnd =.Ltana 
 + Lcot a—10. 
 
 RULE Il. 
 nn EIS ORC Mor sin (B—5) 
 Yan (45°— $a) st TH 
 auth (45°S2a)= €usin (B44) ae (Bo5) + e 
 
 The tangent of (45°—4b6) may also be found by 
 the same form, using a and a instead of 8B and 6: and 
 a or b, which are found by subtracting 2 (45°44 a), 
 or 2 (45° 4b) from 90°, is subject to the same 
 ambiguity as in rule 1. 
 
 Note. In this, and the two following cases, if the 
 given leg be less than its opposite Z, their sum will be 
 less than 180°; and if it be greater than its opposite Z , 
 their sum will be greater than 180°. 
 
 Aiko, if the leg be = its opposite Z, or their sum 
 be = 180°, the other leg, its opposite Z, and the hyp. 
 will be each 90°, or else indeterminate. 
 
 oe weeertnenenceminns 
 
 XV. Given either of the legs and its opposite a 
 to find the other angle. 
 
 RULE I. 
 As cos given leg : cos opp. or givenZ :: rad : 
 siiY remaining or required Z. 
 Which 4 is ambiguous; that is, it may be either 
 an acute Z or its supplement. 
 
OS B ° el 7 COS A 
 Sin A —$ ae 
 a cos a , 
 Lsin A= € L cos pg cOSB; LsinB=€Lcosa+ 
 
 LCOS A. 
 
 RULE II. 
 
 Tan (45°—£ a) = ,/tan + (3-+-6) tan t (p—S) 
 
 i aH 
 L tan (45° 44a) = ee), 
 
 The tangent of (45° —4 8) may also be found by the 
 same form, using a and a instead of g and b: and a 
 or B, which is found by subtracting 2 (45°54), or 
 2 (45°28) from 90°, is subject to the same am- 
 biguity as in rule 1. 
 
 XVI. Given either of the legs and its opposite angle, 
 to find the hypothenuse. 
 
 RULE I. 
 As sin given Z : sin opp. or given leg :: rad : sin hyp. 
 Which hyp. is ambiguous; that is, it may be either 
 an arc less than 90°, or its supplement. 
 
 Or, 
 rsina __ rsinb 
 Sine a 
 sin A sin B 
 
 LSne=€Lsin a+ Lsna=exLsins-+ usin b. 
 
 RULE Il. 
 
 _ Tan (45°—3 c) =,/cot 4 (a+a) tang (a—a) 
 | Lcot4 (a-+-a) + vtint (ana) 
 MG eee STN DE wT 
 
 Ltan(45°43¢)= 3 
 
168 
 
 The same form may also be applied to the other side 
 and its opposite 2 , using B and Z instead of « and a: 
 and the hyp. ¢, Shick is found by subtracting 2 
 ( 45°44) from 90°, will be subject to the same am- 
 biguity as if found by rule 1. 
 
 SOLUTIONS OF ALL THE CASES OF QUADRANTAL 
 SPHERICAL TRIANGLES, 
 
 A 
 pee oN 
 ae 
 Ye é 
 ki 
 B ete, ls) Me 
 
 I. Given the hypothenusal angle and either of the 
 sides, to find the other side, _ 
 
 | RULE. 
 
 As rad : cos hyp!. Z :: tan given side : cot remain- 
 ing or required side. » 
 
 Which side is less than 90° when the given side and 
 hyp'. Zare unlike; but greater than 90° (or the sup- 
 plement of the former) when they are like. 
 
 —;cotb=— ee ee 
 
 Or, 
 tan 6 cos tan acosc 
 Cot a= — : & 
 
 Lcot@a =Ltanb+nLcosc—10; Lcoth= L tana 
 +1 cosc—10. 
 
 Note. In this, and the two following cases, the given 
 side and hyp’. Z may be each of any magnitude under 
 
169 
 
 180°; but, if either of them be 90°, the remaining side 
 and angle will be indeterminate. 
 
 When b=c, cota=—sins; andifbt+c= 180°, 
 cota = sind, 
 
 Ii. Given the hypothenusal angle and either of the 
 sides, to find the angle adjacent to that side. 
 
 RULE. 
 
 As rad : tan hyp!. Z :: cos given side : tan adjacent 
 or required Z. 
 
 Which Z is acute when the given side and hyp'. 2 
 are unlike; but obtuse (or the supplement of the for- 
 mer) when they are like. 
 
 Or, 
 
 s cos 6tanc cosatanec 
 Tana =— ——}; tanB = — —, 
 fs 
 
 rtana=tLcos$+itanc—10; LtanB=Lcosa 
 +41Ltanc—10. . 
 
 When b=c, tana=—sin 6 ; and if b-+c = 180°, 
 tan A=sin 0, | 
 
 * 
 
 III. Given the hypothenusal angle and either of the 
 sides, to find the angle opposite to that side. 
 
 » RULE I. 
 Asrad : sin hyp’. Z :: sin given side < sin opposite 
 or required Z. 
 Which 4 is acute when the opp. or given side is 
 less than 90°; but obtuse (or the supplement of the 
 former) when it is greater than 90°. 
 
; sinasinc . sindésine | 
 SitvA == NG SARE eee 
 
 LSinaA==Lsina+Lsnc—10; psnsap=usinb’ 
 +1 sinc— 10. 
 
 RULE Il. 
 
 Find the other side by case 1; then, by means of the 
 hyp'. 4 and this side, find the 2 adj‘. to it by case 11; 
 which will be the req’. 2, or that opp. the given side. 
 
 When a=c, or a+c=180°, sin a == 4 vers 2a, 
 
 sin? a 
 or 
 
 IV. Given the hypothenusal angle and either of the 
 other angles, to find the side adjacent to that angle. 
 
 RULE I. 
 
 As rad : cot hyp. Z :: tan given 2: cos adjacent or 
 required side. 
 
 Which side is less than 90° when the hyp’. Z and 
 given Z are unlike; but greater than 90° (or the sup- 
 plement of the former) when they are like. 
 
 Or, 
 
 tan Bcotc 
 
 tan acotc 
 Nos ae ee 
 
 r 
 
 § Cosa=— 
 
 Lcos@==LtanB-+L cotc—10; tcosb==.tana 
 +1 cot c—10. | / 
 
 RULE Il. 
 x Ps sin (c+8) 
 Tania =rJ — ee, 
 
 €usin (¢48) + isin (C48) + 10" 
 
171 
 
 The tan of 3 6 may also be expressed by the same 
 form, using 4 a instead of B; and, in either of these 
 cases, the 4 side will be always less than 90°. 
 
 Note. In this, and the two following cases, if the 
 given Z be less than the hypothenusal 4, their sum 
 will be less than 180°; and if it be greater than the 
 hypothenusal 4, their sum will be greater than 180°. 
 Ifthe two Z* be equal, the A is indeterminate. | 
 
 V. Given the hypothenusal angle and either of the 
 other angles, to find the side opposite to that angle. 
 RULE I. 
 As sin hyp'. Z: rad :: sin given Z : sin opposite 
 or required side. 
 Which side is less than 90° when the ‘opp. or given 
 4 is acute; but greater than 90° (or the supplement 
 
 of the former) when it is obtuse. 
 OFS 
 
 Sin a= si ; Sin 0 CRO 
 sin C 
 Lsna=€eLsnce+Lsna; Lsn =e. sinc+ 
 L sin B. 
 3 RULE Il, 
 Tan (45°—4a) = tan 4(c—a) cot$ (c+a) 
 ee Ge ss a)= Ltand (coa) + L cot (c+) 
 ws 2 
 The tan of (45°— 14) may also be expressed by 
 the same form, using 28 instead of a; and the whole 
 sides a or 4, which are found by subtracting 2 
 (45° — Za), or 2 (45° 4D) from 90°, will be less or 
 greater than 90°, according to the rule given above. 
 
 A 
 
172 
 
 VI. Given the hypothenusal angle and either of the 
 other angles, to find the remaining angle. 
 
 RULE I. 
 As cos given 4 : rad :: cos hyp’. 4 : cos remaining 
 or required Z. : 
 Which Z is acute if hyp'. Z and given Z are un- 
 like ; but obtuse (or the supplement of the former) if 
 
 they are like. 
 Cor. 
 
 T- COSC F COSC 
 
 Cos A ==— ; cos B == — ——,. 
 COS B COS A 
 
 LcOS AS=€ LCOS Br} L COSC; LCOSB==> €LCOSA 
 +- L cos Cc. 
 RULE II. 
 Cot £ a== Vtan 3 (c-+p) tan } (c—B)” 
 
 Ry 1 
 1, ~_ Ltan} (c+B) + i tan} (ces) 
 TiO op A aa aera 
 
 The cot of 3 8 may also be expressed by the same 
 form, using ve A instead of B3; and in each of these 
 cases the 4 Z will be always acute. If the two angles 
 be equal, the triangle will be indeterminate. 
 
 VII. Given either of the sides and its adjacent angle, 
 to find the hypothenusal angle. 
 
 RULE. 
 As cos given side : tan adj‘. or given Z 3: rad: tan 
 hypothenusal 2. 
 Which hyp'. Z is acute ae the given side and Z 
 are unlike; but obtuse (or the supplement of the for- 
 mer) when they are like. 
 
173 
 
 Or, 
 ; rtana rtansB 
 Pan Goes 20S an 
 cos b cos a 
 
 Ltanc=e€Lcos$+1L tan am€Lcosa-+L tan B. 
 Note. In this, and the two following cases, the given 
 side and angle may be each of any magnitude under 
 1 80°. x , ; 
 
 tz b ° 
 When J= a, tanc=— a and if b-- a= 
  rtand 
 180°, tan c —"—, 
 cos 6 
 
 Vill. Given either of the sides and its adjacent an- 
 gle, to find the other angle. 
 
 \ 
 
 RULE. 
 
 As rad : sin given Z : : tan adj‘. or given side : tan 
 remaining or required 2. 
 
 Which 4 is acute when opp. or given side is less 
 than 90°; but obtuse (or the supplement of the for- 
 mer) when it is greater than 90°. 
 
 Or, 
 
 tan asin B tandsina 
 r r 
 
 Tan a==— 
 
 Ltan A==1L tana+xLsins—10; LtanB==Ltanb 
 +L sin a—10. 
 
 tanasina : 
 ah and if a+p= 
 
 When a =, tan a= — 
 
 tan a sina 
 180°, tan a = 
 
 re 
 IX. Given either of the sides and its adjacent angle, 
 to find the other side. 
 
174: 
 
 RULE I. ; 
 
 As rad : sin given side :: cos adjacent or perven 4 
 cos remaining or required side. 
 
 Which side is less than 90° when opp. or given Z 
 is acute; but greater than 90° (or the supplement of 
 the former) when it 1s obtuse. 
 
 Or; 
 
 sin @COSA h sin @ COS B 
 
 — 
 
 > 
 
 Cosa = 
 
 rv 
 
 Lcosa==Lsinb + LcosA—103; Lcosb=L sina 
 +L cos B—10. 
 
 RULE Il. 
 
 Find the hyp!. 4 by case vir; then, by means of 
 this angle and the given side, find the remaining or 
 required side by case 1. 
 
 When b==a, cosa sin 26; and if b+ a= 
 180°, cosa ==— $sin2b. 
 
 X. Given the two angles, to find either of the sides. 
 _ RULE. 
 As sin either 4 : rad :: tan other Z : tan opposite 
 or required side. 
 Which side is less than 90° when opp. or given Z 
 Is acute; but greater than 90° {or the supplement of 
 the former) when it is obtuse. 
 
 rta 
 Tan a==——-; tan == —__. 
 sin B SIN A 
 
 Ltana=eLtsnse+Ltana; ttanb=e€eLtsna 
 -+Ltan zB. 
 
175 
 
 Note. In this, and the following case, the two given 
 angles may, be each of any magnitude under 180°. 
 
 Pid . 
 When a= B, tana =sec a, or —3; andif a+b 
 f res ae, COSMAS 
 
 = 186°, tan a=— sec A, or— ‘ 
 COSA 
 
 XI, Given the two angles, to find the hypothenusal 
 
 angle. | 
 | RULE I. 
 
 As rad : cos either Z :: cos other 4 : cos hyp!. Z. 
 
 Which hyp!. Z is acute when the given angles are 
 unlike; but obtuse (or the supplement of the former ) 
 when they are like. | 
 
 Or, 
 
 COS ACOSB 
 Cosc =— 
 
 r 
 
 Lcos C = Lcos A + LcosB—10. 
 
 RULE Il. 
 Find either of the sides by case x; then, by means 
 of this side and its adj‘. 4, find the hyp’. Z by case vi. 
 
 cos* A ; 
 When a=3, cos c=— ——; and if a+s=180°, 
 r 
 
 ~. COS* A ” 
 
 XII. Given the two sides, to find the hypothenusal 
 angle. | 
 RULE I. 
 As rad : cot either given sides :: cot other side : 
 cos hypothenusal 2 . 
 
176 
 
 Which hypothenusal Z is acute when the given sides 
 are unlike ; but obtuse (or the supplement of the for- 
 mer ) when they are like. 
 
 Or, 
 
 cota cot} 
 
 Cos C= aT RR 
 r 
 
 L cos Cc=Lcot a+ 1 cot b—10. 
 
 RULE II. 
 Tan Zc =f, Re find aa a 
 —cos (a+B) 
 _ €xucos (a+b) + icos (ab) + lo 
 
 Which 5 4c is always less than 90°. 
 
 Note. The two sides, in this and the following case, 
 must be so taken, that their difference shali be less 
 than 90°, and their sum between 90° and 270°. 
 
 t?a ; rm 
 ; and if a+3 = 180, 
 
 When a=4, cos c= — 
 
 bt Ryegate Na 
 08 atte 
 
 XIilI. Given the two sides, to find either of the angles. 
 
 RULE I. 
 As sin either given sides : rad :: cos other side : 
 cos opposite or required Z. | 
 Which Z is acute when its opposite side is less than 
 90°; but obtuse (or the supplement “of the former) 
 when it is greater than 90°. 
 cox, 
 
 \ r cosa rcosb 
 Cosa = ——-+ 5 cosB= ——. 
 sin @ 
 
7 
 
 L cos hag bop cha ee Be cosa; LCOSB=€LSina 
 Pe 
 
 RULE II. 
 
 Cot La= et lee — 45°) tan (5*+45°) 
 
 L cot (P44 _ 45°) +L tan (4 uo 45° ) 
 LcotZ a= _ 5 
 
 The cot of $ B may also be expressed by the same 
 form, putting a in the place of b, and 2 in that of a; 
 and the $ angle, in each of these cases, is always less 
 than 90°, | 7 
 
 When a=), cosa=cota; andifa + b6=—180°, 
 cos A =—cof A. 
 
 XIV. Given either of the sides and its opposite an- 
 
 gle, to find the other angle. 
 RULE I. 
 
 As rad : cot given side :: tan opp. or given "hee 
 sin remaining or required Z. 
 
 Which 4 is ambiguous; that is, it may be either 
 an acute angle, or its supplement. 
 
 Or, 
 
 : Scot & tab *% cotatan A 
 Sin A= ——- 3 Sin B= ; 
 
 £ 
 Lsna=Lcotb-+-Ltanp—10; LsingB=Lcota 
 | RULE Il. 
 
 ° Sit b— 
 _ Tan (46°—4 a) = easy 
 
 €. sin (+8) +1sin (d-48) + 10 
 2 ° 
 
 ,Ltan (45°. 4.4) = 
 
 N 
 
178 
 The tan of (45°—4 8) may also be expressed by the 
 same form, using a, A instead of b, 3B; and a ors, 
 which are found by subtracting 2 (45°44) or 2 
 
 (45°>28) from 90°, are subject to the same ambiguity 
 as In vile I. 
 
 Note. In this, and the two following cases, if the 
 given angle be less than its opposite side, their’sum 
 will be less than 180° and if it be greater than its opp. 
 side, their sum will be greater than 180°. If the side 
 and opp. angle be equal, the triangle is impossible. 
 
 XV. Given either of the sides and its so angle, 
 to find the other side. 
 
 | RULE. 
 As cos given 4 : cos opp. or given side :: rad : 
 sin remaining or required side. 
 
 Which side is ambiguous ; that is, it may be either 
 an arc less than 90°, or its supplement. 
 
 Or, 
 
 oi. OEE r cosa 
 Sin a Si b =n s 
 COSA 
 
 L Sin eee L eke cos 4; nsn 6 = ex cosa 
 + Lcosa. 
 
 RULE Il. ra 
 
 (oe ee 
 
 Tan (45°—$ a) = tan £(b-+B) tani (b—B) 
 L tan $(d+8) + + .tan 4 (28) 
 hehe 
 
 The tan of (45°—4 6) may also be expressed by the 
 same form, using @, A instead of b, 8; and aor 6, © 
 
 L tan (45° 43 a)= 
 
#19 
 
 which are found by subtracting 2 (45°—4 a), or 2 
 (45°— 6) from 90°, is subject to the same ambiguity 
 as in rule 1. 
 
 XVI. Given either of the sides and its opposite an- 
 gle, to find the hypothenusal angle. 
 
 RULE I. 
 As sin given side : sin opp. or given Z :: rad : 
 sin hypothenusal Z. 
 “Which hyp’. angle is ambiguous; that is, it may be 
 either an acute angle or its supplement. 
 
 Or, 
 . rsina r sin B 
 Sin l= a. 
 sin @ sin b 
 Lsnc=€.Lsina+usin a=e€x sin b—.sin 5b. 
 RULE. II. 
 
 Tan (45°—4 c) =“ coti (aa) tant (a—a). 
 
 L 
 L tan (45°41 c) ek (a) ae. 
 
 The same form may also be applied to the other 
 side and its opposite 2, using b, B instead of a, a; 
 and in each of these cases the hypothenusal Zc, which 
 is found by subtracting 2 (45°—1c) from 90°, will 
 be subject to the same ambiguity as in rule 1. 
 
 SOLUTIONS OF ALL THE CASES OF OBLIQUE-ANGLED 
 SPHERICAL TRIANGLES; 
 A 
 
180 
 
 I. Given two sides and an angle opposite to one of 
 them, to find the angle opposite the other. 
 
 RULE I. : 
 As sin side opp. given 2: sin given 4 :: sin other © 
 side : sin opposite or required 2. ‘ 
 Which is either an acute Z or its supplement, ac- 
 cording as it makes the greater 4 opposite the greater 
 side ; and if each of them agree with this rule, the A 
 is ambiguous, or admits of two different solutions, 
 sli: @ SiN B U 
 “sin } 
 Lsna=e€xsinb +r sina +- Lsin B—10. 
 
 Or, sin A = 
 
 RULE Il. 
 LeteLrsnd-+-Lsna+Lsins—10=.Ltan®@.. 
 10 + 1 tan (45°—@) | 
 aE 
 
 Where arc © can never be greater than 45°; and 
 Za, which is found by taking 2 (45° <4.) from 90°, 
 is subject to the same ambiguity as in rule 1. , 
 
 The 4 Borc may also be expressed by the same 
 form, using the sides‘and angle which are situated like 
 those above; and the same observation may, be ex- 
 tended to all the other cases of the table. 
 
 Note. In this, and the two following cases, the given 
 sides and angle must be so taken, that the result, found | 
 by the Ist rule, shall not be greater than radius ; other- 
 wise the triangle 1 is impossible. 
 
 When the two sides are equal, the req’. 2 will be 
 equal to the given 4; and if the sum of the two sides 
 be 180°, the req*. 2 will be the supp‘. of the given Z. 
 
 Then, i tan (45° 4fa)= 
 
181 
 
 II. Given two sides and an angle opposite to one of 
 them, to find the included angle. 
 
 RULE. 
 
 Find the angle opp. the other given side, by case 1, 
 and note whether it be ambiguous or not. 
 
 Then, 
 
 As sin 3 diff, given sides : sin 4 their sum :: tan 4 
 diff. their opp. 24% : cot 4 sat be i. 
 
 Which # Z is always tae ; and ifthe Z found by 
 the first part of the rule be ambiguous, the required 4 
 will be ambiguous, otherwise not. 
 
 Or, 
 
 Ppegite’ 3 (2+4) 
 Cot 3 = alte 
 
 L cotd c==exisind (a4b)+1 sing (atedvh-t 
 Ltan$ (AsB)—10. 
 
 tan3 (a cB). 
 
 Or Z the included Z may be found directly, by the 
 following formula : 
 
 Tan 6 om sin a cot A »/ sin? acot* a bul 7 sin (a—b) 
 2 © Pesin ab) hes sin®, (@o-b) sin (a+b) _ 
 t 
 When a=8, cot $c = 4" aes and ifa+b= 
 cos@atana 
 180°, tan 3 C= Pima 
 
 r 
 
 III. Ce two sides and an angle opposite to one of 
 them, to find the remaining side. 
 
 RULE. 
 Find the angle opp. the other given side by case J, 
 and note whether it be ambiguous or not. 
 
182 
 
 Then, 
 As sin 3 diff. these 2° : sin $ their sum :: tan 4 
 diff. given aie : tan 3 ee side. 
 Which # side is always less than 90°; and if the Z 
 found by the first part of the rule be ambiguous, the 
 required side will be ambiguous, otherwise not. 
 Or, 
 Tan 4-¢== ss (05) 
 Ltangc= €Lsing (AGB) +1 sing (A+B) + 
 L tang (a—b)—10. | 
 Or 3 the remaining side may be found directly, by 
 
 the following formula : 
 
 tan 3 (a~b). 
 
 sin bcos A sin?scost?a abet eee 
 cos a+cos B_— (cosa+ cos! cosh )* cosa+cosé 
 
 —;andifia+6= 
 
 Tan 3c= 
 
 tan @ COSA 
 
 Whena=J,tan$c= 
 
 tan a@cos A 
 
 180°, cot 4 aC Se aT ae 
 
 IV. Given two angles and a side opposite to one of 
 
 them, to find the side opposite the other. 
 RULE I. 
 
 As sin eith. given ZS: sin its opp. side :: sin other 
 given Z : sin its opp. side. 
 
 Which side is an arc less than 90° or its supplement, _ 
 according as it makes the greater side opp. the greater i 
 angle ; and if each of them agree with this rule, the A | 
 is ambiguous, or admits of two different solutions, 
 
 Or, 
 
 sin d sin a 
 
 Sin a eer 
 
 sin B 
 
 LSMma=eLusnea+usna-—+ rsinb—10. 
 
183 
 
 RULE It. 
 
 Let € psins+trsin a+4 sin b—10= 1 tan @. 
 
 Then, bt tan (45°oia) = ae eee. 
 
 Where arc @ is always less than 45°; and side a is 
 subject to the same ambiguity as in rule 1. 
 
 Note. In this, and the two following cases, the given 
 Z* and side must be so taken, that the result, found 
 by the first rule, shall not be greater than radius, other- 
 wise the triangle is impossible. 
 
 If the two 2° be equal, the req‘. side will be equal 
 to the given side; and if the sum of the two Z$ be 
 180°, the req’. side will be the sup‘. of the given side. 
 
 V. Given two angles and a side opposite to one of 
 them, to find the included side. 
 RULE. 
 Find the side opp. the other given angle by case tv, 
 and note whether it be ambiguous or not. 
 Then, 
 As sin 3% diff. given 2° : sing theirsum :: tan 3 
 diff. their opp. sides : tan 4 included side. | 
 Which § side is always ase than 90°; and if the side 
 found by the first part of the rule be ambiguous, the 
 required side will be ambiguous, otherwise not. 
 Or, 
 
 __ sin} (4+8) 
 Tang c= ie sing. (a8), 
 
 Ltan$Zc= €exsind (Acs) +L sing (a+s8) + 
 
 tang(a-—b). 
 
184 
 
 Or 4 the included side may be found directly, by 
 the Lien: formula : 
 
 cota sin Ay cot® a sin* a sin (A+B 
 Tani c= ———— + pee ti Pe ge nh tn 
 
 sin (AB) sin?(a—B) sin (a—B) 
 
 tan acos A ° 
 eda ; p APP ai ies e — 
 When A=B8, tan fc = ;andifa+t+sp= 
 » + 
 tan @ £08 A 
 
 ae 
 
 VI. Given two angles and a side opposite to one of 
 them, to find the remaining angle. 
 
 RULE. 
 
 - 
 
 Find the side opp. the other given angle by case 1v, 
 and note whether it be ambiguous or not. 
 
 Then, 
 As sin 3 diff. these sides : sin } theirsum :: tan $ 
 diff. lane Z* : cot 4 remaining 2. 
 Which 4 Z is Ate acute; and if the side found 
 by the first part of the rule be ambiguous, the required 
 angle will be ambiguous, otherwise not. 
 
 Or, 
 
 sing (0-48) 
 sin} (a%5) 
 Lootzc=exLsnz(aob) 4+ .sini(a+)h) + 
 
 L tand Oe areas | 
 
 Cots ¢ = tan 2 (A438). 
 
 Or } the remaining angle may be found directly, by 
 the following formula : 
 
 cosfsina cos? & sin? ; 
 Tash J Xigame b sin? a 72 ye COS A+ COS B 
 
 COSA==COSB—*/ (COSA—cosB)* | COS A—coOsB 
 
185 
 
 cos a tan A 
 When a =8, cot}c = 
 
 —; and if a+ B= 
 
 cosa tana 
 180°, tan 4 c= —-— . 
 
 r 
 
 VII. Given two sides and their included angle, 
 find either of the other angles. 
 
 ‘RULE. 
 As cos sum given sides : cos + their diff. 
 their included Z : tan % sum other two Z°, 
 | And, 
 As sin + sum given sides : sin 4 their diff. :: cot 
 
 ce Cae 
 
 ban © 
 
 their included 4 : tan 3 diff. other two 2°, 
 Where 4 the sum of the two 4° is like J the sum 
 of their binalite sides ; and 4 their difference is al ways 
 less than 9O°. 
 Then, if 5 the difference of these two 2 be added 
 
 o § their sum, it will give the 4 opposite the greater 
 side ; and if subtracted from the 3 sum, it will give 
 the Z opposite the less side. 
 
 to]H 
 
 | Or, 
 cos} (25 
 Pee (a+3)= = cosh (a+ non eae 
 sin 2 (ath 
 Tani (aoB)= na ey couric, 
 
 Ltant (a+sB)=€Lcos+ 1 (atb)+1Lcoss (ab) 
 -+- Lcotic—10. 
 
 Ltan}(aoB)=€L sin (a+b) +L sin 2 (ab) 
 +- Lcotic—10. 
 
 Or eithet angle may be fowhd Hideety by the fol- 
 lowing formula : 
 
 cot a sind’—cos bcosc 
 Cota = —— ; 
 sin Cc 
 
 — 
 
186 
 
 Note. In this, and the two following cases, the two 
 given sides and their included angle may be each of 
 
 any magnitude under 180°. 
 cosatanic 
 
 When a=0, cota = ———+—; andifa+b= 
 r 
 O° cos acot$c 
 _.180°, cot a = 4, 
 
 VIII. Given two sides and their included angle, to 
 
 find the remaining side. 
 RULE I. 
 
 As rad : cos given Z :: tan either given ai : 
 tan arc 2. 
 
 Then, 
 
 As cos arc @ : cos side above used : : cos diff. other 
 side and ® : cos remaining or required side. 
 
 Where arc @ is less than 90°, when the given angle 
 and side, used in the first part of the rule, are like; but 
 greater than 90° when they are unlike. 
 
 The required side is also less than 90°, when the 
 given angle and the diff. of the other side and arc © 
 are like; but greater than 90° when they are unlike. 
 
 Or, put Lt tand+1 cos c—10= 1 tan @. 
 Then, tcosc=€1L cos$+1 cosb+1 cos (a4 )—10. 
 | RULE II. 
 Find the two remaining angles by case vil. 
 
 Then, As sin 4 diff. these Z° : sin + their sum :: 
 tan 4 diff. their opp. sides : tan 4 remaining side. 
 
 Which 4 side is always Take than 90°. 
 Or, 
 
 sin 4 (A+3) 
 sin 3 (AcB) 
 
 Tanic= tani (a—b). 
 
187 
 
 Ltanic=e€Lsint (Acs) -+Lsint (a+sB)+ 
 L tan + (a4b)—10. 
 Or the remaining side may be found by the following 
 formula : 
 rcosacosé + sinasindcosc 
 
 Cos ¢ = — 
 ‘ r 
 
 , sinasinic p , 
 When a=J, sin 4 c= —-———;; and if a+b= 
 r 
 
 , sina cosic 
 
 7? 
 
 IX. Given two angles and their included side, to 
 find either of the opposite sides. 
 
 RULE. 
 
 As cos 3 sum given 2°: cos + their diff. : : tan 4 
 included side : tan + sum other two sides. 
 
 And, 
 
 As sin + sum given 4% : sin 2 their diff. :: tan 4 
 included side : tan + diff. other a sides. 
 
 Where +. the sum of the two sides is like + the sum 
 of their opp. 4°; and + their diff. is less hart 90°. 
 
 And if 4 the diff. of the sides be added to + their 
 sum, it will give the side opposite the greater 2, and 
 if it be subtracted from the + sum it will is the side 
 opposite the less 2. 
 
 Or, 
 site, \ . CO8s (ANB) 
 Jan3 (4-74) = cosd (ata) Aa Gs 
 Tan3(a—b)= 2 Ss Ls 
 
 sin} (a+B) 
 
188 
 
 L tang (a+b)=€1 cos £ (a+-B)+1 cos $ (acs) 
 +1 tani c—10. 
 
 Litand(aob)=exLsing (A+B)+tLsn3(acs) 
 | + Ltangc—10. 
 Or the side may be found directly, by the following 
 
 formula : 
 
 cota sinB -+cosccos B 
 
 Cota = 
 SIN c 
 
 Note. In this, and the following case, the two given 
 4 and their included side may be each of any mag- 
 
 nitude under 180°. 
 
 cot1¢ cosa 
 When a= ,, cot a “aaa and if A+B= 
 
 . tani ccosa 
 180°, cota = —+——. 
 
 X. Given two angles and their included side, to find 
 the remaining angle. 
 
 RULE I. 
 
 As rad : cos given side :: tan either given 2° : 
 cot arc @. | 
 
 TED, ious Hace 
 
 As sin arc @ : cos £ above used :: sin diff. other 
 4 and © : cos remaining or required Z. 
 
 Where arc @ is less than 90° when the given side 
 and angle, used in the first part of the rule, are like ; 
 but greater than 90° when they are unlike. 
 
 The required angle is also like the angle above men- 
 tioned, when arc @ is less than the other given angle ; 
 but unlike it when it Is greater. 
 
189 
 
 Or, 
 fan ACOS¢ COS A Sin Bes 
 == 'cot @;_ then cos c’= cosa sin (1g) ¢). 
 r sin p 
 Or, 
 
 Ltana + Lcosc—10 = L cot ®. 
 Then, t cosc=exsind+.sin(s0)+1Lcos a —10, 
 RULE II. 
 Find the two remaining sides by case 1x. 
 Then, 
 As sin 3 diff. these sides : sin 4 their sum :: tan $ 
 dif. ee 4° ; cotd remaining 2. 
 Which 4 Z is always less than 90°. 
 
 __sin}(a+4) 
 Cot 2 a ~ Bin (4) tan 4 a (A oH B). 
 
 Loot}c=exnsnd(a4b) + xusind(a+b)+ 
 L tang (A B)—10. 
 
 Or the remaining angle may be found directly, by 
 the following formula : 
 
 cos¢ sin A SIN Br cos ACOSB 
 
 Cosc a 
 
 e i 
 
 cosicsina : 
 
 When a= s, cos3.c = ———-— ; and if a+3 = 
 T 
 
 Siti cosicsina 
 90", sin 3 CHE. 
 
 XI. Given the three sides to find either of the angles. 
 
 RULE. 
 
 As rect. under the sine of 4 the sum of the three 
 sides and the sine of the differcnve between this 4 sum 
 and the side opp. the 2 sought, is to dhsg liar of 
 rad. 
 
190 
 
 So is rect. under the sines of the differences of the 
 same 2 sum and each of the other two sides, to the 
 square of the tan of 5 required 2. 
 
 Or, | 
 
 If s be made to denote the sum of the three sides, 
 c the side opp. the required 4 , and a, b the sides about 
 that 2. : 
 
 / 
 
 Then, 
 af sin (i s—a) sin (4s—5) 
 smissin(ds—c) ~ 
 €vsinds+€xsin($s—c)+ 1 sin (sa) +xsin (4:-5) 
 2 
 
 Fan 2 O'=r 
 Ltandc= 
 
 Which 4 Z is always acute. 
 
 Or this ace: formula may be expressed in 
 words, as follows : 
 
 Add together the log sine of 4 the sum of the three 
 sides, and the log sine of the difference between this 2 o 
 sum and the side opp. the angle sought, and find the 
 complement of their sum, by taking the index from 
 19, and the rest of the figures from 9, as usual. 
 
 Then to this complement add the log sines of the 
 differences between the same 4 sum and each of the 
 other two sides, and the result, divided by 2, will give 
 the log tan of 3 the required 2. 
 
 Or the shold angle may be found by the following 
 formula : 
 
 7 , cos acos$ 
 Cos C = SS 
 sin asind 
 
 Note. Inthis case the three sides must be so taken, 
 that the sum of any two of them may be greater than the 
 third, and the sum of all three of them less than 360°. 
 
191 
 
 When all the three sides of the A are equal, the 
 
 : cot atanta 
 cos of either 2, asc = ees Oh sn 2 c= 
 
 rsing a __ 
 
 % Sec A. 
 sin a 2 
 
 If two of the three sides only are equal, sinc = 
 T sin $¢ 
 
 2”; and when the sum of two of the sides is 180°, 
 rcostc 
 
 itc= 
 COS oe sin @ 
 
 XII. Given the three angles, to find either of the sides. 
 
 RULE. 
 
 As rect. under the cosine of 4 the sum of the three. 
 Z* and the cosine of the diff, sisceiicat this - sum and 
 the Z opp. the side sought, is to the square of rad. 
 
 So is the rect. under the cosines of the differences 
 of the same 4 sum and each of the other two 2°, to 
 the square of the cot + required side. 
 
 Or, 
 
 If s be made to denote the sum of the three Z i C 
 the Z opp. the required side, and a,B the 2§ adj‘ 
 that side. 
 
 Then, 
 Cot £ I om ra/ 008 (B8- Tema)eos | cos cos (4 ‘s—B). 
 —cos 4s cos scos (4. sc)” 
 1 corte €Y cos4s+€xcos(4 he (is—a) 4+ 1cos(is—z) 
 Which 2 side is always less than 90°. 
 
 Or this logarithmic formula may be expressed in 
 words, as follows : 
 
 Add together the log cosine of 4 the sum of the 
 
192 
 
 three Z* and the log cosine of the diff. between this 4 
 sum and the Z opp. the side sought, and find the com- 
 plement of their sum, by taking the index from 19, 
 and the rest of the figures from 9, as usual. 
 
 Then to this complement add the log cosines of the 
 differences between the same + sum and each of the 
 other two 28, and the result, divided by 2, will give 
 the log cot. of 3 the required side. 
 
 Or the whole side may be found by the following 
 formula : 
 
 si A Sin B 
 Jote. In this case the three angles must be so taken, 
 that the sum of any two of them may be greater than 
 the supplement of the third, and that the sum of all 
 three of them may fall between 180° and 540°, 
 When all the three 25 of the A are equal, the cos 
 
 : | cot ccotic r 
 of either side, as c cos = ———#_, or cos $c = 
 
 rcosic 
 
 Sapo == 2.cosec.3.c. 
 
 If two of the three 2% only are dy hth cos$c = 
 rcos3ic 
 
 ae eee and when the sum of two of the Z* is 180°, 
 
 aes § 4 fee 
 see Sa sin A CQ). 
 
 (g) As every isosceles spherical A is divided into two equal 
 right-angled spherical A’, by a perpendicular drawn from the 
 vertical Z to the base, it is evident that the solutions of all the 
 cases of the former, as given in the foregoing Tables, may be 
 derived from those of the latter, by referring them to the right- 
 angled a to which they belong. 
 
 Also, in the species of 4 which has the sum of two of its sides 
 
193 
 
 MISCELLANEOUS QUESTIONS FOR EXERCISING THE 
 RULES IN THE PRECEDING TABLES. 
 
 1. Ina right-angled spherical triangle a BC, one of 
 the oblique angles a is 60°, and the other B 45°; re- 
 sa the side Bc opposite the former. Ans. Bc 45°. 
 
 2. Ina right-angled isosceles spherical triangle ABC; 
 the two equal sides a c, cB are each 30°; required the 
 hypothenuse a B. Ans, AB 41° 24° 35”, 
 
 3. It is required to find the angles of an equilateral 
 
 spherical triangle aBc, each side of which is 60°.- 
 Ans. each Z 70° 31’ 44”. 
 
 4. In a quadrantal spherical triangle a B c, the hy- 
 pothenusal angle c is 874°, and one of the other an- 
 gles a 95° 13’; required the side a c adjacent to that 
 angle. Ans. ac 61° 25’ 53”. 
 
 5. In a night-angled spherical triangle azc, the 
 hypothenuse a B is 84° 23’ 20”, and one of the legs 
 AC 96° 36’ 22”; required the angle s adjacent to that 
 leg. Ans. 4 a 148° 1° 49% 
 
 6. In an isosceles spherical triangle a Bc, each of 
 the two equal sides aB, ac are 952°, and their in- 
 cluded angle 100°; required the base Bc. 
 
 Ans. B c 99° 22’ 24”, 
 
 7. In-an oblique-angled spherical triangle a Bc, one 
 of its sides a B is 96° 50’, the other ac 83° 10’, and 
 
 equal to 180°, the sum of their opposite £* will likewise be equal 
 to 180°; and, consequently, if the base and one of these sides 
 be produced to semicircles, or till they meet, the supplemental a, 
 thus formed, will be isosceles, and can, therefore, be resolved 
 by some of the cases of right-angled triangles, in a similar man- 
 ner with the former. 
 
 0 
 
194 
 
 their included angle a 120°; required the base B ¢, and 
 the other two angles B and c. 
 Ans. Bc 120° 28’ 10”, 23 86° 413”, 
 Lai Qs 55" 407% 
 
 8. In a right-angled spherical triangle a 8 c, the hy- 
 pothenuse AB is 61° 4’ 56”, and the leg B c 40° 30’ 20"; 
 required the other leg a c, and the angles a and B. 
 
 Ans, ac 50°30'29”, 2.4 47°54'21”, 23 61°50 28". 
 
 9. Inthe quadrantal spherical triangle a B c, the side 
 BC is 132° 5’ 40”, and the side ac 118°9' 31”; it is 
 required to find the hypothenusal angle c, and the 
 other two angles a and B. 
 
 Ans. c = 118° 55 4”, 24 139° 29’ 41”, 
 4B 129% 29° 307: 
 
 10. In the oblique-angled spherical triangle a Bc, 
 the side aB is 76°35 36”, ac 50° 10’ 30”, andsec 
 40° 0’ 10”; required the three angles a, Band c. 
 
 Ans. 4 A 34°15’ 3", 23 42° 15’ 132”, 
 4c 121° 36 20°, 
 
 11. At each of three objects, on the surface of the 
 earth, the angles subtended by the other two were 
 54° 29’ 36”, 93° 32° 59”, and 31°57’ 25” respectively, 
 from which it is required to find the distances of the 
 objects from each other. 
 
 Ans. 106213, 130224.4, and 69058 feet. 
 
 Note. A great variety of examples of this kind may 
 be found in that part of the Trigonometrical Survey of 
 England and Wales already published; in which, as 
 weil as in several other similar performances, these 
 kind of geodetical operations are amply detailed. 
 
195 
 APPLICATION OF SPHERICAL TRIGONOMETRY 
 TO THE 
 RESOLUTION OF ASTRONOMICAL PROBLEMS. 
 
 Astronomical Problems are such as chiefly relate to 
 the determination of the relative situations and positions 
 of the heavenly bodies with respect to each other, or 
 to certain imaginary points, lines, and circles of the 
 sphere, that have been fixed upon, by astronomers 
 and geographers, for this purpose ; each of which be- 
 long equally to the globe of the earth and the concave 
 sphere of the heavens that surrounds it, and are distin- 
 guished as follows: 
 
 The axis of the celestial sphere, is an imaginary line 
 passing through the centre of the earth, about which 
 the sun and stars appear to perform their daily revo- 
 lutions (7). 
 
 The north and south poles of the celestial sphere, are 
 the two extremities of its axis; and the points lying 
 directly under them, on the terrestrial sphere, are the 
 north and south poles of the earth (s). 
 
 (r) Although it is the diurnal rotation of the earth upon its 
 axis in 24 hours, and its annual revolution round the sun in 365 
 days, 5 hours, 48 minutes, 51.6 seconds, which occasions the 
 various phenomena of the heavens, yet as the places and ap- 
 pearances of the celestial bodies will be the same, whether the 
 earth moves round the sun, or the sun and the rest of the celes- 
 
 tial bedies move round the earth, astronomers, for the ease of 
 calculation, usually assume the earth as a point at rest in the 
 centre of the system, and ascribe to the heavenly bodies that 
 motion which they appear to have’ to a spectator on its surface. 
 
 (s) The star commonly called the north-pole star is not directly 
 
 o 2 
 
196 
 
 The equinoctial, or celestial equator, is a great cir- 
 cle which divides the northern half of the heavens 
 from the southern; and, when referred to the earth, it 
 is called simply the equator. 
 
 The ecliptic is a great circle which cuts the equator 
 in two opposite points, at an angle of 23° 28’ nearly, 
 being that in which the sun appears to perform his 
 annual revolution through the heavens (¢). 
 
 This circle, like all others, is divided into 360°, and 
 also into twelve equal parts, of 30 degrees each, called 
 signs, the names and characters of which are as fol- 
 
 lows: 
 
 ea Taurus Gemini Cancer Leo Virgo 
 ee 8 bat & SF mm 
 
 ea Scorpio Sagittarius Capricornus Aquarius Pisces 
 ana aha tf VS a x 
 
 The sun continues about a month in each of these 
 signs, and goes through nearly a degree in a day. 
 
 Meriaiansare great circles passing through the poles 
 of the world, and cutting the celestial equator at right 
 angles. 
 
 in the point which is the true north pole of the heavens ; its decli- 
 nation, for the beginning of the year 1790, having been found 
 to be 88° 11’ 8”, or 1° 48! 52” from the pole, and its annual va- 
 riation 192”, 
 
 (¢) It has been found, by comparing the observations of the 
 
 ancients with those of the moderns, as well as from physical 
 principles, that the obliquity of the ecliptic, or the angle which 
 it makes with the equator, is continually diminishing towards a 
 certain limit; and that it will, afterwards, return again; but the 
 variation is very small, and has been differently stated by diffe- 
 rent authors. © 
 
 ie 
 i Na 
 
" 
 197 
 
 They are also called hour circles, or circles of right 
 ascension ; and, when referred to the earth, circles of 
 terrestrial longitude. 
 
 The horizon is a circle which is distinguished by 
 several names, according to the sense in which it is 
 employed: thus, 
 
 The rational, or true horizon, is a great circle which 
 divides the upper half of the heavens from the lower ; 
 being formed bya plane passing through the centre of 
 the earth, at the distance of 90°, each way, from the 
 spectator. 
 
 The astronomical horizon, is an imaginary plane, 
 touching the earth in the point supposed to be the 
 centre of view of the spectator, and extending inde- 
 finitely on all sides; being that to which the risings 
 . and settings of the heavenly bodies are always referred. 
 
 Where it is to be observed, that this latter horizon, 
 with regard to the fixed stars, coincides, as to sense, 
 with the rational horizon; but for the sun, their di- 
 ‘stance from each other is about 9 seconds, and for 
 the moon about 33 minutes. 
 
 The sensible or visible horizon is a small circle of 
 the earth, parallel to the rational horizon, and ts the 
 _ boundary of the spectator’s view at sea or land ; being 
 more or less extended, according to the height from 
 which it is observed (w). 
 
 It may also be remarked, that the first of the three 
 
 (w) It may here be further observed, that the astronomical, as 
 well as the sensible horizon, is frequently referred to a parallel 
 plane passing through the earth’s centre; in which case, it is 
 then called the reduced horizon. 
 
198 
 
 circles here mentioned, 1s divided, by mariners, into $2 
 equal parts, of 11°15’ each, called the points of the 
 compass ; the principal of which, east, west, north and 
 south, are denominated the four cardinal poinis. 
 
 The zodiac is a space which extends about 8° on 
 each side of the ecliptic, like a belt or girdle, being 
 that in which the paths or orbits of the principal planets 
 are situated. 
 
 Circles of celestial longitude are great circles pass- 
 ing through the poles of the ecliptic, and cutting it at 
 right angles. 
 
 The tropics are two small circles, at the distance of 
 23° 28° from the equator; that in the northern hemi- 
 sphere being called the tropic of cancer, and the other, 
 in the southern hemisphere, the tropic of capricorn. 
 
 The polar circles are two small circles at the distance 
 of 23° 28" from the poles ; that in the northern hemi- 
 sphere being called the arctic circle, and the other, in 
 the southern hemisphere, the antarctic circle. 
 
 All circles of this kind, which are parallel to the 
 equator, are also called parallels of terrestrial latitude, 
 or of declination. 
 
 Parallels of celestial latitude are small circles pa- 
 rallel to the ecliptic ; and parallels of altitude, or alma- 
 canters, are small circles parallel to the horizon. 
 
 The equinoctial points are the two points where the 
 ecliptic cuts the equinoctial; and the solsticial poznts 
 are the two points where it touches the tropics (2). 
 
 (x } When the sun appears in either of the equinoctial points the 
 days and nights are equal, or 12 hours each, all over the world; 
 
199 
 
 The zenith and nadir are the two poles of the hori- 
 zon; the former being the point directly over our 
 heads, and the latter that under our feet. 
 
 The equinoctial colure is a meridian passing through 
 the equinoctial points; and the so/sticial colure is a 
 meridian passing through the solsticial points. 
 
 Azimuih, or vertical circles, are great circles pass- 
 ing through the zenith and nadir of any place, and 
 cutting the horizon at right angles. 
 
 The prime vertical, is the azimuth circle which 
 passes through the east and west points of the horizon ; 
 being that on which the sun rises and sets when the 
 days and nights are equal. 
 
 The six o’clock hour-circle, is that meridian which 
 cuts the 12 o’clock hour-circle, or meridian of the place, 
 at right angles; being that on which the sun is at six 
 o’clock, both in the morning and evening. 
 
 The latitude of any celestial object, is its distance 
 north or south from the ecliptic, as measured on a cir- 
 cle of celestial longitude passing through its centre. 
 
 The latitude of any place on the earth, is its distance 
 
 which happens about the 21st of March and the 22d of Septem- 
 ber; the former being called the vernal equinox, and the latter 
 the autumnal equinox, | 
 
 Also, when the sun comes to the northern solsticial point, or 
 tropic of cancer, it is the longest day ; and when he comes to the 
 southern solsticial point, or tropic of capricorn, it is the shortest 
 day ; the former of these happening about the 21st of June, and 
 the latter about the 22d of December. 
 
 The solsticial points cancer (og) and capricorn (¥§) are so 
 called, because when the sun isin this situation he seems to have 
 nearly the same altitude at noon for several days together, and 
 on that account appears to stop or stand still, 
 
200 
 
 north or south from the equator, as measured on a 
 meridian passing through that place. 
 
 The longitude of any celestial object, is the distance 
 of the circle of longitude, which passes through its 
 centre, from the first point of aries, as reckoned in de- 
 grees, minutes, &c. quite round the ecliptic (y). 
 
 The longitude of any place on the earth, is the di- 
 stance east or west of the meridian, which passes over 
 that place, from the first meridian, reckoned in degrees, 
 minutes, &c. on the equator. : 
 
 Note. The first meridian, in this country, is that 
 which passes over the Royal Observatory at Greenwich; 
 and for other countries, it is that of the capital of the 
 kingdom. sa 
 
 The co-latitude, or polar distance, of any place on 
 the earth, is an arc of a meridian, contained between 
 that place and the north or south pole of the world. 
 
 The co-latitude, or polar distance, of any celestial 
 object is an arc of a circleof celestial longitude con- 
 tained between the centre of that object and the north 
 or south pole of the ecliptic. 7 
 
 The dip of the horizon is the angle formed by an 
 horizontal line, passing through the eye of the ob- 
 
 (y) The sun being always situated in some part of the ecliptic, 
 has no latitude: and instead of his longitude, his distance in 
 signs, degrees, minutes, &c. from the first point of aries, is some- 
 times called his place in the ecliptic. 
 
 It may also be observed, that the latitude of any place is equal 
 to the height*of the pole above the horizon, or to the distance of 
 the zenith of that place from the celestial equator, as counted 
 upon the meridian of that place. 
 
201 
 
 server, and another line drawn so as to be a tangent - 
 to the earth, at some point on its surface (z). 
 
 Refraction is the difference between the real and 
 apparent place of an object, as occasioned by the rays 
 of light passing through the atmosphere in a curvili- 
 near instead of a right-lined direction. 
 
 Parallax is the difference between the places of any 
 celestial object as seen from the surface of the earth 
 and from its centre. 
 
 Horizontal parallax is the angle under which the 
 semidiameter of the earth would appear if seen directly 
 from the centre of the sun, or a planet. 
 
 Parallax in altitude is the angle under which the 
 semidiameter of the earth would appear if seen obliquely 
 from the centre of the sun, or a planet (a). 
 
 The altitude of any of the heavenly bodies, is an are 
 of an azimuth, or vertical, circle, contained between 
 the centre of the object and the horizon. 
 
 (z) Asthe Z of depression, or dip of the horizon, occasions ob- 
 jects to appear higher than they really are, it must be subtracted 
 from the observed altitude, in order to obtain the apparent 
 altitude. i 
 
 (a) The more elevated a planet is above the horizon, the less is 
 its parallax, its distance from the earth’s centre continuing the 
 same, When the’planet is in the zenith it has no parallax, and 
 when inthe horizon its parallax is the greatest. 
 
 Refraction and the dip of the horizon make all objects appear 
 higher than they really are; but parallax, having a contrary effect, 
 makes them appear lower, The fixed stars are at such immense 
 distances that they have no sensible parallax. 
 
202 
 
 The observed altitude is that which is taken simply 
 with a sextant, or other instrument, without being cor- 
 rected for the dip of the horizon, refraction, or parallax, 
 
 The apparent altitude is that which has been cor- 
 rected for the dip of. the horizon, without considering 
 the effect of refraction or parallax. 
 
 The true altitude is that which is found after making 
 all the corrections arising from the dip of the horizon, 
 refraction, and parallax. 
 
 The zenith distance is the complement of the alti- 
 tude, or the arc of a vertical circle, contained between 
 the centre of the object and the zenith. 
 
 The declination of the sun, moon, or stars, is an arc 
 of a meridian contained between the centre of that ob- 
 ject and the celestial equator. 
 
 The amplitude of any of the heavenly bodies, is an 
 arc of the horizon contained between the centre of the 
 object, when rising or setting, and the east or west 
 points of the horizon, 
 
 The azimuth of any of the heavenly bodies, is an 
 arc of the horizon contained between an azimuth cir. 
 cle, passing through the centre of the object, and the 
 north or south points of the horizon (0). 
 
 The righi ascension of any celestial body, is an arc 
 of the equinoctial contained between the first point of 
 
 (2) The amplitude of any of the heavenly bodies may also be 
 considered as its bearing, when in the horizon, from the true east 
 or west point of the compass ; and the azimuth, as its bearing from 
 one of these points, when it is at any height above the horizon. 
 
203 
 
 aries and a meridian passing through the centre of the 
 object (c). | 
 
 Or, it is that measure of time which shows how 
 much sooner, or later, than the first point of aries any 
 star comes to the meridian ; and consequently the dif- 
 ference of time between one star coming to the meridian 
 and another. 
 
 The oblique ascension, or descension, is the distance 
 ‘ of the first point of aries from the horizon when the 
 object is rising or setting, as reckoned on the equator ; 
 or it is that point of the celestial equator which rises or 
 sets with the object in an oblique sphere. 
 
 The ascensional difference is the difference between 
 the right and oblique ascension or descension ; or it is 
 the time that the sun rises before six o’clock in the 
 summer, or sets before six in the winter. 
 
 Aberration is an apparent change of place in the 
 stars, arising from the motion of the earth combined 
 with the motion of light. 
 
 The achronical rising or setting of a planet, or star, 
 is when it rises at sun-set, or sets at sun-rise; and the 
 cosmical rising or setting is when it rises or sets with 
 the sun. 
 
 The heliacal rising of a star, is when it appears 
 
 (c) The sun’s longitude and right ascension, are usually 
 reckoned from the first point of aries, quite round the globe ; 
 hence, although at equal distances from the equinoctial points, his 
 declination may be of the same name, n. or s., and the same 
 quantity, yet the longitudes and right ascensions will be mate- 
 tially different. 
 
204 
 
 above the horizon just before the sun in the morning 5 
 and its heliacal setting, is when 1t sinks below the ho- 
 rizon immediately after him in the evening. 
 
 Time is a portion of duration, which, in astrono- 
 mical computations, is divided into three kinds, viz. 
 apparent, true or mean time, and sidereal time. 
 
 Apparent time is that shown by the sun; being rec- 
 koned from the instant of his passing over any meridian, 
 in one day, to the instant of his passing over it the next. 
 
 Sidereal time is that which is measured by a well- 
 regulated clock or watch, so adjusted as to count 24 
 hours from the passage of a star over any meridian till 
 it returns to that meridian again. 
 
 True, or mean time, is that which ts measured by a 
 clock or watch, so adjusted as to count 24" 3™ 563§ 
 in a mean solar day, or 365% 55 48™ 51.65 in a mean 
 solar year. 
 
 True, or mean noon, is 12 o'clock, as shown by a 
 well-regulated chronometer, so adjusted as to go 24 
 hours in a mean solar day; and apparent noon, is 
 when the sun comes to the meridian of the place, as 
 shown by the same machine (@). 
 
 (d) It may also be remarked, that besides the divisions of time 
 above mentioned, there are four kinds of lunar months, and three 
 kinds of solar years, which are distinguished as follows; 
 
 The periodical month is the period in which the moon returns to 
 the first point of aries, consisting of 27474 43m 5s, 
 
 The sidereal month is the period in which the moon returns to 
 the same star, consisting of 27¢ 75 43™ 12s, 
 
 The synodical month is the period in which the moon returns to 
 the sun, consisting of 294 12" 44™ 3s, 
 
 _ 
 
205 
 
 The right ascension of the mid heaven is the distance 
 of the first point of aries from the meridian, at the time 
 and place of observation, as reckoned on the equator. 
 
 The nonagesimal degree, or medium ceeli, is the 
 90th degree of the ecliptic, reckoned from its inter- 
 section with the eastern part of the horizon at ay 
 given time. 
 
 Its altitude is equal to the distance between the ze- 
 nith and the pole of the ecliptic; or it is measured by 
 the angle which the ecliptic makes with the horizon, 
 at any elevation of the pole. 
 
 The crepusculum circle, is a small circle parallel to 
 the horizon, at the distance of about 18° below it, 
 where the twilight is supposed to begin and end. 
 
 The rising of any celestial object, is when its centre 
 appears in the eastern part of the horizon; and its 
 setting is when its centre comes to the western part of 
 the horizon. 
 
 The culminating of any celestial object, is the time 
 when it ¢ranszts, or comes to the meridian of any place. 
 
 The variation of the compass, is the deviation of the 
 magnetic needle from the true north or south point ; or 
 
 The anomal:stic month is the period in which the moon returns 
 to its apogee, consisting of 274 138 1§™ 34s, 
 
 The tropical year is the period in which the sun’ returns to the 
 same point in the ecliptic, consisting of 3659 54 48™ 5.68. 
 
 The sidereal year is the period in which the sun returns to the 
 same star, consisting of 3654 64 S™ 1(Cs, 
 
 The anomalistic year is the period in which the sun returns to 
 the same apsis, consisting of 3654 65 15™ 46%, 
 
306 
 
 the difference between the true and magnetic amplitude, 
 or azimuth, of any of the heavenly bodies. 
 
 The hour angle, is an angle at the pole of the equa- 
 tor, contained between the meridian of any place and 
 the meridian which passes through the sun or a star. 
 
 Diurnal and nocturnal ares; are such portions of 
 the parallels of declination, above and below the hori- 
 zon, as are described by any celestial body from the 
 time of its rising to its setting, and from its setting to 
 its rising (e). 
 
 The equation of time, is the difference between ap- 
 parent or solar time, as shown by a true sun-dial, and 
 true or mean time, as shown by a well-regulated clock 
 or watch (f). 
 
 Consequentia is the apparent motion of a planet east- 
 ward, or according to the order of the signs; and an- 
 tecedentia is its apparent motion westward, or iit 
 to the order of the signs. 
 
 (e) In places situated on the equator, the rational horizon cuts 
 all the parallels of declination into two equal parts; and, in this 
 case, the sun and stars are 12 hours above the horizon and 12 
 hours below it, Butin places lying between the equator and the 
 elevated pole, the parallels of declination are unequally divided, 
 
 the greater arc being above the horizon, and the less below it. | 
 
 And in all cases between the equator and the depressed pole, the 
 greater arc is below the horizon, and the less above it. 
 
 (f) The equation of time arises from three causes, viz. the obli- 
 quity of the ecliptic, the earth’s unequal motion in its orbit, and 
 the precession of the equinoxes ; which variation is always equal 
 to the difference between the sun’s apparent increase of right 
 ascension in 245 and $™ 561°, the mean increase in that time. 
 
 ae 
 
207 
 
 The conjunctions, oppositions, &c. of the planets, 
 and the characters by which they are denoted, are as 
 follows : 
 
 dS conjunction, is when two planets are referred to 
 the same point of the ecliptic. 
 
 ¥ sextile, is when they are 2 signs, or 60° distant 
 from each other. 
 
 O quartile, is when they are 3 signs, or 90° distant 
 from each other. 
 
 A trine, is when they are 4 signs, or 120° distant 
 from each other. 
 
 f opposition, is when they are 6 signs, or 180° di- 
 stant from each other. 
 
 The conjunctions and oppositions are also called the 
 syzyges, and the quartile aspects the quadratures, these 
 terms being chiefly applied to the moon. 
 
 The nodes are the two points in which the orbits of 
 the primary planets intersect the ecliptic ; or the points 
 in the orbits of the primary planets which are the inter- 
 sections of the orbits of their secondaries, or satellites. 
 
 That point, or node, where the planet ascends from 
 the south towards the north of the ecliptic, is called the 
 north, or ascending node ; and the other the south or 
 descending node ; the two points being marked thus : 
 
 8 Moon’s north, or ascending node. 
 & Moon’s south, or descending node. 
 
 The names and characters of the sun, and all the 
 primary planets, yet known, are as follows: 
 
208 
 
 © The sun, % mercury, ? venus, @earth, 
 3 mars, Y% jupiter, kh saturn, H uranus, 
 2 ceres, $ pallas, # juno, § vesta (g). 
 
 The nutation of the earth’s axis, is a periodical re- 
 volution of it, depending upon the place of the moon’s 
 ascending node; by which the situations of the stars 
 are apparently changed. 
 
 The precession of the equinoxes is an uniform retro- 
 grade motion of the equinoctial points in the plane of 
 the ecliptic, which affects the longitude, right ascension, 
 and declination of the stars (/). 
 
 A constellation is a collection of stars, supposed to 
 be circumscribed by the outlines of some assumed 
 figure, as a ram, a serpent, an hercules, &c. which 
 division was found necessary in order to direct an ob- 
 server to that part of the heavens where any particular 
 star is situated (7). 
 
 (z) The four last of these planets, which have been lately dis- 
 covered, lie between Mars and Jupiter, and are too small to be 
 seen with the naked eye. | 
 
 (4) By comparing together the places of the fixed stars, as de- 
 duced from observation, astronomers have found that their longi- 
 tudes increase about 501” annually ; which increase must necessa- 
 rily cause an irregular’motion in the same star with respect to the 
 equator: hence, the right ascensions and declinations of the stars 
 are constantly varying ; so that some of those which had formerly 
 north declination have now south declination, and the contrary. 
 Their latitudes are also subject to a small variation, from the 
 same cause. 
 
 (?) In order that the memory may not be overburthened by a 
 
 a 
 
209 
 
 PROBLEM I. | 
 To reduce the time, under any Known meridian, to 
 the corresponding time at Greenwich; and the converse, 
 
 ; RULE. | 
 
 Convert the longitude of the place into time, by 
 reckoning 15 degrees to an hour. : 
 
 Then, this time, added to the time from the preced- 
 ing noon, at the given place, if it be west, or subtracted 
 from it, if east, will give the Greenwich time. 
 
 Or, conversely, 
 
 To the time from the preceding noon at Greenwich, 
 add the longitude of the place in time, if east, or sub- 
 tract it, if west; and the sum or difference will be the 
 corresponding time under the meridian of that place(A). 
 
 multiplicity of names, astronomers mark the stars of each constel- 
 lation with a letter of the Greek alphabet, denoting those which 
 are the most conspicuous by a, the next by 6, and so on in succes- 
 sion; by which means they may be easily spoken of and referred 
 to, as occasion requires. Several of the brightest stars have also 
 proper names, as Arcturus, Orion, Aldebaran, Antares, &c. 
 
 (£) Since the earth makes one revolution on its axis, from west 
 to east, in 24 hours, the sun must apparently make one revolution 
 round the earth, from east to west, inthe same time. And as the 
 longitude of all places on the earth, is reckoned on the equator, 
 the whole of this circle, which is divided into360°, must pass the 
 sun in 24 hours; therefore, every 15 degrees of motion is one 
 hour in time, every degree 4 minutes, and every minute 4: se- 
 conds. Whence a place one degree eastward of Greenwich 
 will have noon, and every hour of the day, 4 minutes sooner than 
 at Greenwich ; and a place one degree westward of Greenwich, 
 will have noon, and every hour of the day, 4 minutes later, 
 
 It may also be observed, that the astronomical day is supposed 
 
 P Pe 
 
 Are 
 Ae 
 Ditiey 
 
 dips 
 
210 
 
 Example 1. _ 
 It is required to find the time at Greenwich, when it 
 is 5" 15’ p. m. ata place in longitude 74° 21’ w. 
 3\74° 21° 
 Kou cate 
 4b 5" 240 
 
 ee 
 
 Time at given place --- 5°15 
 Longitude in time --- 4° 57! 24” w. 
 Time at Greenwich - - 108 19’ 24” p,m. 
 
 Example 2. 
 What is the time in longitude 62° 9’ w. when it is 
 15511' 14’ at Greenwich ? 
 
 3|62° 9° 
 
 5}20° 43° 
 
 pee BBG. 
 Apparent time at Greenwich - 15° 11/14” 
 Longitude in time’ ----- 4° & 36" w. 
 
 Time at the given place --- 112 9’ 33” 
 
 Example 3. 
 It is required to find the time at Greenwich answer- 
 ing to 3° 45’ 39” of May Ist in lon. 110° 2’ 15” east. 
 Ans. April 30th at 205 25° 30”. 
 
 _ Example 4. 
 It ts required to find the time at a place in longitude 
 
 to begin at noon, or 12 hours later than the civil day of the same 
 denomination; and is counted up to 24 hours, or the succeeding 
 noon, when the next day begins: so that, for instance, January 
 10th, at 15 hours, is the same as January 11th at 3 in the morn- 
 ing, by the civil reckoning. 
 
211 
 
 107° 48’ east, which corresponds with 4° 51’ 30” 
 Greenwich time. Fee al ue A 
 PROBLEM Il, | 
 
 To reduce the declination of the sun, as given in the 
 Nautical Almanac for Greenwich, to any other meri- 
 dian, and to any given time of the day. 
 
 RULE. 
 
 Convert the longitude of the place (if different from 
 that of Greenwich) into time, as in the last problem. 
 
 Then, as 24 hours : change of the sun’s declination 
 in that time :: the time from noon : the proportional 
 part required. : 
 
 Which added to, or subtracted from, the declination 
 at noon, according as it is increasing or decreasing, will 
 give the declination at the time and place required. 
 
 Note. The sun’s change of longitude, or of right 
 ascension, may also-be reduced to any given time and 
 place, by a similar process (/). 
 
 Example 1. i 
 
 Required the sun’s declination August 9th 1808, 
 at 45 46’ a. m. in longitude 123° 7’ west. 
 
 128° 7’ == 8 12’ 28”, the time at which the inhabi- 
 tants of Greenwich have the sun Lefore those in longi- 
 tude 123°'7’ w. Hence, when it is 4° 46’ in the morn- 
 ing at the latter place, it is 12" 58’ 28” at Greenwich, 
 or 58’ 28” p. M. 
 
 (2) The sun’s longitude, right ascension in time, and declina- 
 tion, are given in the Nautical Almanac for every day in the 
 year, at noon. 
 
 P2 
 
i 212 
 
 Sun’s declin. at noon Aug. 9th 1808, 15° 51° 58” N. 
 Ditto- - - - - - 10th - - 15° 34 82° Nn. 
 Decrease of declination in 24 hours O. 17 268 
 
 As 24% + 17°26” 3: 58’ 28” : 42.46 
 
 Sun’s declin. at noon Aug. 9th 1808, 15° 51’ 58” 
 Decrease of declination in 58’ 28” - 4.2.46" 
 Sun’s true declin. at given place - - (16% SAG LSE” ita 
 
 ee 
 
 Example 2. 
 Required the sun’s declination on the 17th of May 
 1808, at 115 48’ Greenwich time. Ans. 19°27’ 59”. 
 
 Example 3. 
 Required the sun’s declination on Feb. 20th 1808, 
 at 2° 15’ p.m. inlon. 134° 56’ east. Ans. 11° 19'°4. 
 
 | Example 4. 
 What was the suh’s right ascension at Greenwich, 
 October 18th 1808, at 7° 40’p.m.? Ans. 1$°33°35”. 
 
 Example 5. 
 What was the sun’s right ascension at noon onthe 
 27th of April 1808, in longitude 86° 45’ west? 
 | Ans. 2°) 19°35". 
 Example 6. | 
 What was the sun’s longitude, January 19th 1808, 
 at 4" 35’ in longitude 21° 15’ east of Greenwich? 
 
 PROBLEM Ill. | 
 
 To reduce the declination of the moon, as given in 
 the Nautical Almanac, for Greenwich, to any given 
 time under ‘a known meridian. 
 
213 
 
 RULE. 
 
 Reduce the given time (if necessary) to the meridian 
 
 of Greenwich, by Prob. 1; and find the variation of 
 declination in 12 hours by the almanac. 
 
 Then, as 12 hours : this variation :: the interval 
 between the reduced time and the preceding noon or 
 midnight : the proportional part required. 
 
 Which part being added to, or subtracted from, the 
 moon’s declination at the preceding noon or midnight, 
 according as it is increasing or decreasin 
 
 g, will give 
 the declination at the time required. 
 
 5? 
 
 Note. The moon’s right ascension, semidiameter, 
 and horizontal parallax, may also be reduced to any 
 time and place by a similar process (m). See Prob. xxi. 
 
 Example 1. : 
 Required the declination of the moon at Greenwich 
 on the 13th of August 1808, at 8" 15° 58” P.M. ap- 
 parent time. 
 Moon’s declin. at noon 13th Aug. 1808, 15° 37’ N. 
 
 Dittojat midnight. = os. 44 05 ge ase ¥ 6° 59 
 
 Increase of declination in 12 hours - - i° 16’ 
 
 12>) e120? 1G tisys SRS ye Get BOL 
 Atnoon. = -.- ./1GRs7 Ws 
 Increase - - - 52° 204" 
 
 Declin. required 16° 29° 203” 
 
 (m) ‘The moon’s right ascension, declination, semidiameter, ho- 
 rizontal parallax, and time of passing the meridian at Greenwich, 
 _ are given in the Nautical Almanac for every day in the year, at 
 
 « noon and midnight. 
 
214 
 
 Example 2. | 
 ~ Required the moon’s declination Sept. 20th 1808, at 
 6" 13’ apparent time, under the meridian of Green- 
 wich, Ans, 4° 80 10" s, 
 
 Example 3. 
 Required the moon’s declination Dec. 17th 1808, at 
 15° 9’ apparent time, in longitude 64° 14° w. 
 Ans, 19° 1° 59”, 
 
 Example 4, 
 Required the moon’s horizontal parallax. and semi- 
 diameter, December 7th 1808, at 10" 15’, in longi- 
 tude 39° 40° east, 
 
 ne (S$ Parallax - - 56’ 6.7” 
 ° ) €8 Semidiam. - 15’ 17” 
 
 PROBLEM IV. 
 To find the culminating of the stars, or the times sok 
 their coming to the meridian.. 
 
 RULE. 
 
 Subtract the sun’s right ascension for the given day 
 from the right ascension of the star (increased by 24 
 hours, if necessary ), and the remainder will he the time 
 of the star’s culminating nearly. 
 
 Then, as 24 hours, added to the increase or decrease 
 of the sun’s right ascension in that time : 24 hours : 
 the time of the star’s culminating nearly : the true time 
 of culminating, at Greenwich. 
 
 And i the time of culminating be required for any 
 other meridian than that of Greenwich, add the longi- 
 tude in time to the true time of culminating at Green- » 
 
 ‘he 
 
218 
 ° ° 4 i . : ] ® WH if 
 wich, if the place be west, or take their cifierence, 1 
 
 it be east, and the result will give the time of culmi- 
 nating at the given place (7). 
 
 Example 1. 
 
 At what time did the star Arcturus come to the 
 meridian of Greenwich on the 19th of Dec. 1803? 
 Arcturus’s right ascen. (1808) + 24" - 38" 67 54° 
 Sun’s right ascension ------+--- 175 49’ 6” 
 Time of star’s culminating nearly - - - 20" 17’ 48” 
 
 ee eee 
 
 Sun’s right ascen. atnoon Dec. 19, 1808, 17° 49" 6” 
 
 Pitte s oe ~~ 2 Dec. 20, 1808) FT" sacmer 
 Sun’s increase of right ascen.in 24 hours 4° 26.7" 
 
 hen, 240 4696 bo 248 22 20 17) 48" 5. 907 144.3" 
 true time’of Arcturus’s culminating at Greenwich, or 
 gh 14’ 3” December 20th 1808, 
 
 » Example 2. 
 At what time did Aldebaran culminate at Greenwich 
 on the 23d of November 1808, his right ascension be- 
 ing 4° 24" 54.8"? Ans. 125 26’ 573” at night. 
 
 es Example 3, 
 
 At what time did Regulus culminate at Greenwich 
 on the 9th of February 1808, his right ascension being 
 TSS) as Ans, 27’ 48” past 12 at night. 
 
 (x) If to any given time there be added the sun’s right ascen- ~ 
 sion for that time, the sum (rejecting 24, if necessary) is the 
 right ascension of the mid heaven, which, being sought for in a 
 table of the right ascensions, will show what star is on, or near, 
 
 * the meridian at that time. 
 
216 
 
 Example 4. 
 ~ At what time, on the 21st of February 1808, was 
 Sirius, or the dog-star, on the meridian of a place whose 
 longitude is 166° 30’ east ? Ans. 9° 12’ 142”, 
 PROBLEM V. 
 To find the time when the moon, or a planet, will 
 culminate, or pass the meridian. 
 
 RULE. 
 
 From the planet’s motion in right ascension in 24 
 hours, (increased by 24 hours, if necessary, ) take that 
 of the sun, if the planet be progressive; or add them 
 together, if it be retrograde. : 
 
 Then, as 24 hours, diminished by this sum or diffe- 
 rence, when the planet’s motion is greater than the 
 sun’s, or increased by it when the sun’s apparent mo- 
 tion is the greater : 24 hours :: the planet’s right as- 
 cension at noon, ( + 24-hours, if necessary, ) diminished 
 by the sun’s : the time of its transit at Greenwich. 
 
 And if the time of culminating be required for any 
 other meridian than that of Greenwich, it may be ob- 
 tained by the following rule: 
 
 As 24 hours, added to the daily change of the moon 
 or planet in right ascension : the longitude of the place, 
 converted into time :: the daily change above men- 
 tioned : the reduction required. Which added to, or 
 subtracted from, the time of the moon’s passing the 
 meridian of Greenwich, according as the longitude is 
 west or east, will give the apparent time of the transit 
 at the required place. 
 
217 
 
 And if the longitude in time be added to, or sub- , 
 tracted from, this, it will give the time at Greenwich (0). 
 
 Example 1. 
 Required the time of the moon’s culminating at 
 Greenwich, lat. 51° 28’ 40” n. on the 17th Aug. 1808. 
 
 Sun’s right ascen.at noon 17th Aug. 1808, 9b 4.6’ 30.40 
 Dittot aos Ge o5eld + seh sae o 4 ORSO! Taare 
 Increase of motionin 24 hours - - - 3’ 43.8” 
 
 Moon’s right ascen. at noon 17th Aug. 1808, 89° 20° 
 Ditto - - ~ = -jertybsthy .- (+.9-), 102745" 
 Increase of the moon’s motion in 24 hours 13° 25° 
 
 From 13° 25’ = 53’ 40” of time 
 take 2 43.8¢ 
 leaves 49° 56.2” 
 
 Which is the excess of the moon’s motion above the 
 sun’s in 24 hours. 
 
 Moon’s right ascension 89° 20’ =. 5" 57’ 20” 
 | 94h 
 | | | 308 By" 20" 
 Sun’s right ascension - - - - - 9 46° 30.4” 
 Mpurercice C=. ee = = 201d 405" 
 
 Then, as 24 hours — 49° 56.2” (23" 10 3.8”) : 24. 
 hours :: 20° 10° 49” : 205 54’ 19” the true time of 
 the moon’s passing the meridian of Greenwich, or 
 8" 54’ 19.4" a.m. Aug. 18th 1808, - 
 
 : / 
 (o) The right ascensions of the planets are not given in the 
 Nautical Almanac; but they may be readily computed from 
 their geocentric longitudes and latitudes (which are to be found 
 in that work), and the obliquity of the ecliptic. 
 
218 
 
 | Example 2 
 It is required to find the time of the moon’s passing 
 the meridian of Greenwich on the 21st of July 1808. 
 Ans, 1}° 3! 1.6" a. m. July 22d)1808. 
 
 Example 3. 
 
 Required the time at Greenwich when the moon’s 
 centre passed the meridian of a place in longitude 21° 
 west, on the 14h of November 1808. 
 
 Ans. 10° 54” 50,9” a. mM. Noy, 15th 1898. 
 
 Example 4. 
 It is required to find the time when the planet Mer- 
 cury passed the meridian at Greenwich on the 22d of 
 December 1808, his right ascension being 19" 44’. 
 
 PROBLEM VI. 
 
 The sun’s declination, and the obliquity of the eclip- 
 tic, being given, to find his longitude and right ascen- 
 sion, ts 
 
 Example 1. 
 
 On the 23d of April 1808, the sun’s declination was 
 12° 39’ 58”, and the obliquity of the ecliptic 23° 27’ 42”; 
 required his longitude and right ascension (/). 
 
 aia 
 
 (p) The construction of the figures in all the following examples 
 is left for the exercise of the learner. 
 
219 
 
 1. To find ©’s longitude. 
 -Assin Z@ YA - - 23° 27/ 42” 9.6000308 
 Isto sindec.@ A >- 12° 33°58" 9.3375910 
 So is rad, or sin - - 90° - + = 10.0000000 
 Tosin@’slon.r © 63° 7 28" Y.7375602 
 
 2. To fad ©’s right ascension. 
 As rad, orsin - - 90° - -- - 10.0000000 
 Isto tan dec. OA’ - 12° 33’ 58” = 9.3481208- 
 SoiscotZ@YA - 23° 27°49" 10.36249296 
 To sinrightasc. ra 30° 54’ 10.8” 9.7106134 
 
 And if this be converted into time, at the rate of 15° 
 to an hour, the right ascension will be 2" 3’ 36.7” 
 
 Note. As the sun’s declination is the same at equal 
 distances from the equinoctial points, it is necessary to 
 know the time of the year, or what part of the ecliptic 
 he was in when an observation was made, in order to 
 determine his longitude and right ascension, which are 
 reckoned from the first point of aries quite round the 
 globe. 
 - Thus, while the sun is moving from r towardss, 
 or is in the 1st quadrant of the ecliptic, the longitude 
 is 7 ©, the declin. © a, and the right ascen. ¥ a. 
 
 When he has passed the solstice 3, and is descend- 
 ing towards +, he is then in the 2d quadrant, and his 
 longitude, or distance from 1, must be taken from 
 180°, in which>case the remainder + © becomes the 
 hypothenuse, and the declination is still north; but 
 the arc a is the supplement of the right ascension, 
 and must, therefore, be taken from 180°. 
 
 When he has passed the point £, and is descending 
 e f 
 
220 
 
 towards vv, he has got into the 3d quadrant; in which 
 case the excess above 180°, or his distance from =, 
 will be the hypothenuse 2 ©; the declination will be 
 south, and the are: 2. must be added to 180°, for the 
 right ascension, estimated from r. 
 
 When he has passed the solstice vw, and is ascend- 
 ing towards 7, he is then in the 4th quadrant; in 
 which case the longitude must be taken from 360° to 
 give the hypothenuse = ©; the declination is south, 
 and the arc 2 a must be taken from 360° to give the 
 right ascension from 1. 
 
 Example 2 
 On the Ist of Jan. 1808, the obliquity of the ecliptic 
 was 23° 27’ 42”, and the sun’s declination 23° 5’ 8” s 
 and increasing ; required his longitude and right as- 
 
 cension. | 
 Maisie long. - - 279° 59° QL” 
 ©’s right asc. 280° 51 502” 
 Example 3. 
 
 On the Ist July 1808, the obliquity of the ecliptic 
 was 23° 277’ 40%; and the sun’s right ascension in 
 time 6° 40’ 48.3”; required his longitude and decli- 
 
 nation. 
 ’s long. 359° 22° 24.7” 
 rahe pe ©’sdecl. 23° 7 46.6’ nN. 
 Any two of the four things mentioned in this pro- 
 blem being given, the rest may be found by one of 
 
 the cases for right-angled spherical triangles. 
 
 PROBLEM VII. f 
 The latitude of the place, and the sun’s declination, 
 
221 
 
 being given, to find his amplitude, ascensional diffe- 
 rence, and the time of his rising and setting. 
 
 Example 1. 
 
 Given the latitude of Greenwich 51° 28’ 40” Nn., 
 and the sun’s declination on the 2!st of June 1808, 
 
 being the longest day, 
 
 23° 27° Al” w.3 required his 
 
 amplitude, ascensional difference, and the time of his 
 
 rising and setting. 
 
 s 
 
 ° 
 e. 
 
 1. To find the sun’s amplitude. 
 Sin colat. 24 © 38° 31 20” 9.7943612 
 Sin declin. 4@ - - 23° 27’ 41” 9.6000259 
 
 ydead, Orin 212.902) < 2 TO, 0000000 
 
 Sin.©’sampl.r © 39° 44' 6.9” 9.8056647 
 
 Which is the amplitude from the east or west point 
 
 j of the horizon ; 
 
 and its complement 50° 15’ 53.1 
 
 4/ 
 
 ‘shows how far has the north the sun rises or sets on 
 
 the longest day, at Greenwich. 
 
 Og ‘ 
 
 2. To find the sun’s ascensional di ifference. 
 Rad, OFSIN (pe 3) = 90° - - - - 10,0000000 
 Tan lat. Z4N 7 @ 51° 28’ 40” 10.0990491 
 
 > Lan declin Oia 14293°.27' 41” 9. 6375010 | 
 Sin asc. dif. ra 33° 2° 16.2” 9.7365501 
 
 Where if 33° 
 
 —_——_—, 
 
 16.2” be converted into time, at 
 
> ad 
 
 the rate of 15° to an hour, it gives 2" 12” 9” for the 
 
 time the sun rises before, and sets after 6 o’clock, on 
 
 the longest day. ) 
 
 Hence 6»—<" 12" 9°= 38» 47’ 51” time of sun rising : 
 and 65-2" 1¥’ 9°= 8" 12’ 9” time of sun setting. 
 Also, if the latter of these be doubled, it gives 
 
 16" 24’ 18” for the length of the longest day at Green- 
 
 wich ; and if the former be doubled, it gives 7° 35° 4.2” 
 
 for the length of the night. 
 
 But on the shortest day at Greenwich, whitch is 
 when the sun has 23° 27’ 41” of s. declination, the 
 lengths of the days and nights change place, the day 
 being then 7 35’ 42”, and the night 16" 24° 18”. 
 
 It may also be remarked, that if r s be a parallel of 
 declination as far to the south ‘as 2 m is to the north, 
 the hour-circle NB s, passing through ©, the place of 
 the sun, at its rising or setting, will forma A T @B= 
 Ar ©a, where the amplitude + © is to the south- 
 ward of the east and west points. 
 
 From which it is evident, that when the latitude and 
 declination have the same name, the sun rises before 
 and sets after 6 , but when they are of contrary names, 
 the sun rises ae and sets before 6. 
 
 ‘When the sun’s declination is equal to, or greater 
 than, the co-latitude of the place, (which can only 
 happen to places upon or within the polar circles) the 
 parallel of declination nm will not cut the horizon H 0, 
 and consequently the sun will neither rise nor set at 
 these times. And the same will hold with respect to 
 those stars whose co-declination, or polar distance N ©, 
 
is equal to, or less than, the latitude of the place, or 
 
 the elevation of the pole No, and in the same hemi- 
 
 sphere (q). | 
 Example 2. 
 
 Given the sun’s amplitude 39° 44’ 6.9”, and his de-__ 
 clination 23° 27’ 41” N., to find the latitude of the 
 place, and the time of the sun’s rising and setting. 
 
 latitude *51° 28’ 40” n. 
 Ans.< © rises 3" 47" 51” 
 @Q S6tsi; Ske il 
 Example 3. 
 
 Given the latitude of the place 51° 28’ 40” n., and 
 the sun’s amplitude 39° 44’ 6.9" N. of the east, re- 
 quired his declination, ascensional difference, and time 
 of rising and setting. 
 
 : declin. - 23° 27’ 41” n, 
 asc. diff. 33° 2 16,2” 
 Ans. . sb oy 4 vr 
 © rises-+ 3° 47 $1 
 \ © sets - | 84 197° 9” 
 
 Note. Any two of the five things mentioned in this 
 
 problem being given, the rest may be found by some 
 
 of the cases in right-angled spherical triangles. 
 
 PROBLEM VIII. . 
 
 The latitude of the place being given, and the decli- 
 nation of a star, to find its amplitude, ascensional dif- 
 ference, and the time of its rising and setting, 
 
 (g) When the latitude and declination have the same name, 
 the difference between the right ascension and the ascensional dif. 
 ference is the oblique ascension ; and their sum is the oblique 
 descension, But when they are of contrary names, their sum is 
 the oblique ascension, and their difference the oblique descension. 
 
224 
 
 Example 1. 
 ~ Required the amplitude, ascensional difference, and 
 theitime of rising and setting of Arcturus, at Green- 
 wich, lat. 51° 28’ 40° w. on December 19th 1808, its 
 __ declination being 20° 11’ 21.8”, 
 
 r; 
 
 1. To find the amplitude T %. 
 Sinco-lat. Z aT % 38° 31'20" 9.7943612 
 RAC OF SSID = {omy OO) mea OU 
 2 pom deéclin. av <. 20° 11’ 21.8” 9.5379757 
 : Sinamp. r% - 33°39’ 3.8” 9.7436145 
 
 —_—— 
 
 - Which amplitude is always of the same name as the 
 _ declination ; and since the variation of a star’s declina- 
 tion is extremely small, the same star may be consi- 
 dered as having constantly the same amplitude during 
 the whole year, in the same latitude. 
 
 2. To find the ascensional difference ¥ A. 
 Rad, orsin - - - 90° - +--+  10.0000000 
 : Cot co-lat. ar %° 38° 31’ 20” 10.0990491 
 : : Yan declin.a-% - 20° 11’.21.8” 9.5655151 © 
 Sin asc. dif. ar 27° 30’ 392” 9.6645642 
 
 _ Which, converted into time, gives 15 50’ 2.6” 
 And 6'-+ 1h 50’ 2.6”=='7" 50’ 2.6”==arc a £, or half 
 - the time of the star’s continuance above the horizon. 
 
From time of %’s culmin’. prob. 1v. 8° 14° 3.4” a.m. 
 Rakeitvie’ pps aa, AOR es 4 7S 6 
 Time of %’s risingin morn’.Dec.20th 24’ ys (nea 
 
 ~ 
 
 To time of %’s np olga =~ ete On fae ee 
 Add. wae okie 5 he ee oe 7. 50’ 2.6" 
 Time of the %’s setting - - - - - - "16" a 6". 
 
 Or 46 4’ 6” in the afternoon Dec. 20th 1808. 
 
 Where it may be observed, that on account of the* 
 
 small change in the declination of the stars, the same 
 star, in any latitude, may be considered as having the 
 same ascensional difference throughout the year. And. 
 as the diurnal difference of the same star’s rising, cul- 
 minating, and setting, in the same latitude, is nearly 
 equa! to the diurnal difference of the sun’s right ascen- 
 sion, which is 3° 563”, this may be taken for the daily 
 difference of the rising, southing, and setting of any 
 fixed star in the same latitude (7). 
 
 Example 2. 
 
 It is required to find at what time Sirus, or the dog- 
 star, rose and set at a place in lat. 51° 28’ 40” n., and 
 long. 166° 30’ £., on the 21st of February 1808, its 
 declination being 16° 27’ 29.9” s. 
 
 Pa Sirius rises at # 29’ 222” in the morning. 
 Ray cow. oo 'séts ated? 46’, ° 6 Vin the afternoon. 
 
 (7) The mode of solution made’use of in this and the preceding 
 problem may be also applied to the rising or setting of the moon, 
 or a planet. But when great exactness is required, the declination 
 of the sun or planet must be calculated as near to the time of rising 
 and setting as possible, especially for the moon, on account of her 
 
 Q 
 
eee 
 
 226 
 
 Example 3. 
 
 It is required to find the amplitude, and time of 
 rising and setting of Aldebaran at Greenwich, on the 
 23d of Nov. 1808, its declination being 16° 6’ 47” N. 
 
 Amplitude 26° 27’ 45” N. 
 Ans. < Rises at.5h 1’ 51” evening. 
 Sets at - 72 52 4” next morning. 
 
 Note. Any two of the five things, mentioned in this 
 ‘problem, being given, the rest may be found, asin the 
 former. . 
 
 PROBLEM IX. 
 
 The latitude of the place, and the sun’s declination, 
 
 being given, to find his altitude and azimuth at 6 o’clock. 
 Example 1. ¥ 
 
 At Greenwich in latitude 51° 28’ 40” n. on the 
 longest day, when the sun’s declination is 23°27’ 4aTn., 
 it is required to find his altitude and azimuth at 6 
 
 o'clock in the morning, or evening. 
 
 . To find the eens Oa, 
 : Rad,orsin - - - 90° - - - - 10.0000000 
 : Sin declin.y.© - 25° 2741” 9.6000259 
 >: Sin lat.Z.© ra - 51° 2840"  9.8934103 
 > Sin alt.Oa - - - 18°''8’ 55.3” 9.4934362 
 
 ——a 
 
 _ swift andirregular motion. Also, the declination of the sun near 
 the equinoxes changes considerably in the compass of an hour. 
 
 se 
 
227 
 
 2. To find the azimuth a o. 
 
 : Rad,orsin -- - 90° - - - - 10.0000000 
 > Cos lat. O ra - - 51° 28’ 40° 9.7943612 
 :: Tan declin. r © - 23° 27/41” 9.63'7750!0 
 : Cotazim. ao - - 74° $2 25.9” 9.4318622 
 
 ~~_—__-- — 
 . 
 
 On the shortest day, at London, the parallel of s. 
 declination rs cuts the 6 o’clock hour-circle n s be- 
 low the horizon; and as the AS © ra, © ra, are 
 equal in all their parts, the depression a@© below the 
 horizon, on the shortest day, at 6 o’clock, will be 
 equal to the altitude © a, at the same hour on the 
 longest day ; and the azimuth will also be equal to it, if 
 estimated from the south. So that on the 21st of June, 
 at Greenwich, the sun will bear n. 74° 52’ 25.9” E. at 
 6 o’clock in the morning, and n. 74° 52° 25.9" w. at 
 6 in the evening; but on the 2ist of December, at 
 the same hours, it will bear s. 74° 52° 25.9” r., and 
 B74 52 25.9" w. (s). 
 
 Example ‘2. ‘ 
 it 
 
 Given the sun’s declination 23° 27’ 41” N. and his 
 
 (s) From a due consideration of this problem, it is evident, 
 that as the declination increases, the altitude increases and the 
 azimuth lessens ; and the contrary, when the declination is di- 
 minishing. So that on the days of the equinoxes, on which the 
 
 sun has no declination, his altitude at 6 o’clock will be nothing, 
 - or he will then be in the horizon ; and the azimuth, in this case, 
 being 90° from the north, the sun will be due east in the morn- 
 ing, and west in the evening; that is, on the days of the equi- 
 noxes, the sun rises and sets at 6 o’clock, in the east and west 
 points of the horizon. 
 
 Q2 
 
228 
 
 altitude at 6 o’clock in the morning 18° 8’ 55.3”, re- 
 quired his azimuth and the latitude of the place. 
 ie Azimuth - 74° 52 25.9” from N. 
 
 Lat. of place 51° 28° 40” N, 
 
 Example 3. 
 
 Given the sun’s declination 25° 27’ 41” n., and his 
 azimuth at 6 o’clock in the morning 74° 52° 26” from 
 the N., required his altitude and the latitude of the 
 
 place. 
 Ans, 5 Altitude - - 18° 8" S55 
 Lat. of place 51° 28’ 40” N. 
 
 Any two of the four things, mentioned in this pro- 
 blem, being given, the rest may be found as before. 
 
 PROBLEM X. 
 
 The latitude of the place, and the dasliaauion ofa 
 star being given, to find at what time it will be upon 
 the 6 o’clock hour-circle, and its altitude and azimuth 
 at that time. 
 
 Example 1. 
 
 At what time, on the 19th of Dec. 1808, did the 
 star Arcturus appear upon’the 6 o’clock hour-circle, 
 at Greenwich, lat. 51° 28’ 40” n.; and what was its 
 altitude and azimuth at that time, its declin. being 
 a Wi Fs 
 
 oR 
 
229 
 
 2. The time of the star’s passing the meridian, as 
 found by problem rv, willbe 20" 14° 3”; and conse- 
 quently it will be upon the 6 o’clock hour-circle at 
 (205 14/ 3”—6») 14" 14’ g” in the eastern hemisphere ; 
 and at (20° 14’ 3”-++6") 26 14’ 3”, in the western 
 
 hemisphere. 
 2, To find the altitude KA. | 
 or Wad, Or cin “= -. = 9Ow-s% = 241, TOMOGOO0O 
 sae dechn. re". QO? VV 218% 955879757 
 copa latseria. Y's 61928" 4007» > D489 84108 
 >; Sinalt.%a - - - 15° 39’ 54” =9.4318860 
 
 8. To find the azimuth ao. 
 : Rad,orsinm .- - - 90° - - - - 10.0000000 
 -'Van declin.’- <'-' 20°11’ 31.8" 925655151 
 >: Cos lat.% ra -- 51° 2840” 9.7943612 
 
 Cot azim. ao - 77° 61.7" 9.3598763 
 
 In which case, it may be observed, that on account 
 of the small change in the right ascension and declina- 
 tion of astar, it may, without material error, be said 
 to have the same altitude and azimuth every time it 
 arrives at the six o’clock hour-circle; and the difference 
 of the times it arrives there may be considered as equal 
 to the diurnal difference of the sun’s right ascension. 
 
 | Example 2. 
 
 At what time did Aldebaran appear upon the 6 
 o'clock hour-circle at Greenwich, lat. 51° 28’ 40” N., 
 on the 23d of November 1808, and what was its alti- 
 tude and azimuth at that time, its declination being 
 16° 6 47” N.? 
 
 ~ In the evening at 6» 26’ 573” 
 Ans.< %’saltitude - - 12° 32’ 26.8” 
 ¥%’sazimuth - - 79° 48’ 1” from Nn. 
 
230 
 Example 3. 
 
 At what time did Regulus appear upon the 6 o’clock 
 hour-circle at Greenwich, lat. 51° 28' 40” n., on the 
 9th of February 1808, and what was its altitude and 
 azimuth at that time, its declination being 12° 54’ 7”? 
 
 In the evening at 6° 27’ 48” 
 
 A %’s altitude - - 10° 3° 38” 
 ¥%’s azimuth - - 81° 52’ 50".N. 
 Noite. Any two of the five things, mentioned in this 
 problem, being given, the rest may be found as before. 
 
 PROBLEM XI. 
 The latitude of the place, and the sun’s declination, 
 being given, to find his altitude and the time when he 
 will be due east or west. 
 
 Example 1. 
 
 In latitude 51° 28’ 40” n., when the sun’s declina- 
 tion is 19° 47’ 27”, it is required to find his altitude, 
 and the time when he will appear upon the prime ver- 
 tical, or due east or west. 
 
 1. To find the altitude vr ©. 
 Sin lat ar © - - 51° 28’ 40" 9.8934103 
 Sin declin.@ a - - 19° 47° 27" 9.5296707 i 
 :: Rad, or sin .- - #°90°:> ~'-.-°-40.0000000 
 Sin alt.r © -- - 25°38 37" | 9. 6362604 
 
231 
 
 2. To find the hour from 6, Y a. 
 : Rad, orsin - - - 90° -.-.- - 10,0000000 
 : Tan declin.@a = 19947 27” 9.5561111 
 51°28 40” 9.9009509 
 : Sin va +4 = - > Tores’ 46”  9.4570620. 
 
 Which converted into time, wives 15 6’ 35” for the 
 time from 6 o’clock. 
 
 Hence, the sun will be exactly east at 7° 6’ 35” in the 
 morning, or west at 4° 53’ 25” in the afternoon (¢). 
 
 Example 2. 
 
 The sun’s declination being 19° 47’ 27” n. and his 
 altitude, when upon the prime vertical, 25° 38’ 37”, it 
 is required to find the latitude of the place, and the 
 
 ‘hour of the day. 
 Lat. 51° 28’ 40” n. 
 a J 
 
 Ans.< Time 7 6/35 morning 
 Ox :48. 58.256 afternoon. 
 
 Example 3. 
 In latitude 51° 28’ 40” n. the sun’s altitude, when 
 
 (¢) From this problem it appears, that when the latitude of 
 the place and the sun’s declination have the same name, the alti- 
 
 . tude and time from six o’clock increase as the latitude.and decli- 
 
 nation increase; and having contrary names, the same thing 
 happens, with this difference, that in the former case the days 
 lengthen, on account of the increase of the latitude and declina- 
 _ tion, whereas, in the latter they shorten on that account. When 
 the latitude of the place is less than the declination, the sun never 
 appears on the prime vertical. 
 If this problem be worked, in the A ¥ © 4, for the longest day, 
 at Greenwich, it will give 4> 39’ 16” for the time before and after 
 “noon when the sun is due east and west; andinthea 7 O4, it 
 will give the time for the shortest day. 
 
 + 
 
232 
 
 on the pritne vertical, was 25° 38’ $7.3”, required his 
 declination and the hour of the day. 
 Declin. 19° 47’ 27” we 
 Ans «| Mine - 7 6 35" ’ morning . 
 Or - 4°53’ 25” afternoon. 
 
 Note. Any two of the four things, mentioned in this 
 problem, being given, the rest may be found as before. 
 
 PROBLEM XII. 
 The latitude of the place being given, and the de- 
 clination of a star, to find its altitude, and the time 
 when it will be due east, or west. 
 
 Example 1 
 At what time did Arcturus appear due east or west 
 from Greenwich, on the 19th of December 1808; and 
 what was its altitude at that time, its right ascension 
 being 14° 6° 25”, and declination 20° 11’ 21.8” N.? 
 
 1. The star culminates at 20° 14’ 34”, as found by 
 prob. iv. 
 
 1. To find the altitude + *%. 
 >: Sinlat. ar *% - - 51°28’ 40" 9.8934103 
 3+ Wace OL SIN 5-4. 90 ip - — — LO,OO00U0 
 : : Sin declin. %a, ,-..20° 11’, 21.8” 9.5: 5379757 ” 
 
 Sin alts: rt wo oie 262106830 eo 9.64456. 5 de 
 
253 
 
 2. To find the hour from 6, ¥ A. 
 Rad, or sin - - - 90° - - ~ = 10,0000000 
 Tan declin, Xa - 20° 13/21 8” 9.5655151 
 >: Cot latZ% ra - 51° 2840"  9.9009509 
 : Sinra----- 17° 117.2” 9.4664660 
 
 And if 72° 58’ 42.8” (the complement of 7 a) be 
 converted into time, it gives 4" 51’ 54.8”, which sub- 
 
 tracted from the time of the star’s southing, leaves 
 15° 22 82” or 3° 22’ 84” the next morning, (Decem- 
 ber 20th,) when the star appeared due east, and added, 
 gives 25° 5’ 58”, or 1" 5’ 58” p. m. (December 20th,) 
 when the star appeared due west (w). 
 
 Example 2. 
 
 At what time did Aldebaran appear due east or west 
 at»Greenwich, lat. 51° 28’ 40” n., on the 23d Novem- 
 ber 1808, and what was its altitude at that time, its 
 declination being 16° 6’ 47” n. ? , 
 
 / Altitude 20° 46’ 382” 
 Ans. < East at - 7, 20' 8a” evening 
 West at 5°33’ agi" next morning. 
 
 Example 3. 
 At what time did Regulus appear due east or west 
 at Greenwich, lat. 51° 28’ 40” n., on the 9th of Fe- 
 
 (u) The height of the same star upon the prime vertical, in any 
 place, is always nearly the same, for the reasons already ass! igned ; 
 and the difference ofthe times of its coming to the prime vertical 
 will be equal to the difference of the times of its culminating, 
 
 which i is nearly equal to the divrnal difference of the sun’s right 
 ascension. 
 
234 
 
 bruary 1808; and what was its altitude at that time, 
 its declination being 12° 54 7” w.? 
 Altitude 16° 34’ 57” 
 Ans. < East - - 7° 9° 492” evening © 
 West - 5 45°46” next morning. 
 Note. Any two of the things, mentioned in this pro- 
 blem, being given, the rest may be found as before. 
 
 PROBLEM XIII. 
 The sun’s declination, and the latitude of the place, 
 being given, to find the time of day-break in the morn- 
 ing, and the end of twilight in the evening. 
 
 Example 1. 
 
 Given the latitude of the place 51° 28’ 40” w., and 
 the sun’s declination 9° 48’ 20” n., to find the time 
 _of day-break in the morning, and the end of twilight » 
 in the evening. 
 
 The crepusculum circle rs being 18° below the hori- 
 zon, we shall have ? 
 _ Sun’s polar dist.o n = 80° 11’ 40” 
 -Sun’s zenith dist. o z = 108° 
 Compt. of lat. - wz = 38° 31 20” 
 2|226° 43°  =sum 
 118° 21’ 30”=Fsum 
 
235 
 
 113° 21’ 30” 113° 21’ 30” 113° 21’ 30” 
 
 108° 80° 11’ 40” "> 38° 31220" 
 
 5291.30", BS De 50" 74° 50! 10” 
 Log sine - - - - = - 113° 21’ 30” 9.9628631 
 Log sine - - - - - - 5° 21' 30" 8.9702738 
 Sum of log SIMeS es ot Raat tiie 18.9331369 
 Its complement - - - - - -  1.0668631 
 Log sine - - - - - - 33° 9’ 50” 9.7380157 
 oe sme -+--- - 74° 50° 10” 9.9846090 
 
 2\20.7894878 
 Tan + rw ZNQ@ ---=-. 68° 3/10” 10.3947439 
 
 136° 6 20”ZzNO. 
 
 And if this be converted into time, it gives 9" 4’ 25” 
 _ ==time from noon when the ©is 18° below the horizon. 
 
 Hence the day breaks at 2° 55° 342” in the morn, 
 ing, and twilight ends at 9" 4’ 25” in the evening, sup- 
 posing the sun’s declination to undergo no cheese 
 during that time (2). 
 
 (x) When the declination becomes greater than the difference 
 between the co-latitude and 18°, the parallel of declination n 
 will not cut the parallel rs, and consequently there will then be 
 ngunight at that place, the twilight continuing from sun-setting 
 to sun-rising ; which takes place at Tpuag from 22d May to 
 about the 21st of July. 
 
 _ It may also be observed, that as the sun sets more obliquely at 
 some times of the year than at others, it follows that he will be 
 longer in ascending or descending through an are of 18° below 
 the horizon at one season than at another. When he is on the. 
 same side of the equator as the visible pole, the duration of twi- 
 light will constantly increase till he enters the tropic, at which 
 
236 
 
 Example 2. 
 
 Given the latitude of the place 51° 28’ 40” n., and 
 the sun’s declination 9° 48° 20” s., to find the time 
 of day-break in the morning, and the end of twilight 
 in the evening. 
 
 ie h 1” 
 yin Day breaks at 4: 53° 255" morning. 
 
 Twilight ends at z 6 34” evening. 
 Example 3. 
 
 Given the latitude of the place 51° 28’ 40” n. and 
 the sun’s declination on the shortest day 23° 27’ 36” s., 
 to find the time of day-break in the morning, and the 
 end of twilight in the evening. 
 
 Weis Day breaks at - - 6 O 59” " morning 
 ’ . Twilight ends at 5°59’ 1” evening. 
 
 PROBLEM XIV. 
 The latitude of the place, and the sun’s declination 
 
 time it will be the longest. It will then decrease till some time 
 after he passes the equinox, but will increase again before he 
 enters the other tropic; whence, there must be some point be- 
 tween the tropics, where the duration of twilight is the shortest ; 
 which point may be found by the following analogy : 
 
 Asrad: tan 9° (half the distance of the crepusculum circle 
 from the horizon) :: sin lat. of the place : sin sun’s declina- 
 tion, when the twilight 1s the shortest. 
 
 Which declination is always of a contrary name with the lati- 
 tude. 
 
 Note. At Greenwich, lat. 51° 28’ 40" w. the time of shortest 
 twilight is when the sun has 7° 7’ 25" s. declination, answering 
 to March 2d and October 11th; between which days it increases, 
 while from the latter to the former it decreases; its whole dura- 
 tion being 15 56’32". For a demonstration of the above analogy, 
 see Emerson’s Miscellanies, p. 492, and Vince’s Astronomy, vol. i. 
 
 a ' 
 Lad 
 
237 
 
 and altitude, being given, to find his azimuth and the 
 hour of the day. 
 Example 1. 
 
 In latitude 51° 28’ 40” n., the sun’s true altitude 
 ‘was found to be 46° 23’ when his declination was 
 23° 97’ 41” ~.3 what was his azimuth, and the hour of 
 day, when the observation was made ? 
 
 Co-lat. - - zN = 38° 31’ 20” 
 Co-alt? <'- ‘2 ©: as" 3'7’ 
 Co-declin. NO© = 66°32’ 19” 
 9\148°. 40’ 39° = sum 
 _ 14° 20 192” = 2 sum. 
 74° 20° 192”. 74° 20° Wis She 20' 194” 
 38° 31’ 20” 66° 32 19” 43° 37" 
 85° 48° 591” 7° 48 08”. 30° 48" 192” 
 Logsine- - - - 74°20'192” 9.9835697 
 ; en Pi Aim lll coh 0.5” _9.1326373 
 S 
 
 oflog sines - - - - - 19.1 1162070 
 Complement - - - -  0,8837930 
 Logsine- - - - 35° 48" 592" 9.7672981 
 Log sine- - -~ - 30°43 (1p 9.7083140 
 
 2/20.3594051 
 Tan$ZNZOQ - - 56°31’ 46” 10.1797025 
 : sll st 
 
 oe et Le 3’ 32" azim. from n. 
 
 ¥, 
 
aa0: °° 
 
 Log sine - -. - 74°20’ 193” - - 9.9835697 
 Log sing - - + 30°43°193" -- 9.7083140 . 
 
 Sum of log smes -> = - - - 19.6918837 
 
 Complement - = - - » - 0.3081163 
 
 Logsine- - - 7 48 0.5" - 9.1826373 
 
 Log sine - - = 35° 48 593” - 9.7672981 
 
 | 2|19.2080517 
 
 Tani Z@nz - 21° 53/272" - 9.6040258 
 2 
 
 "43° 46’ 55”=hour Z from noon. 
 
 Which converted into time, gives 2" 55° 72”. 
 
 Hence, the observation was made either at 95 4’ 52” 
 in the morning, or at 2° 55" 7y in the afternoon (y). 
 
 raiinple 9. 
 Given the latitude of the place, 51° 28’ 40” n. the 
 
 sun’s declination, 19° 34’ 26” n. and the altitude of his: 
 
 centre, 38° 20°; required the azimuth and the hour 
 from noon. 
 
 ‘ uf sid 
 sid . Azimuth $79" Se OSs E. 
 
 Hour from noon 3 29’ 334”. 
 
 (y) If the declination and latitude are of contrary names, the 
 things required may be found by the same mode of operation, 
 except that the side n © being, in this case, greater than 90°, the 
 declination must be added to 90°, instead of subtracted from it, 
 as in the above example. . 
 
 Note. The observed altitude of the sun’s upper or-lower limb, 
 must be corrected for refraction, parallax, and the dip of the ho- 
 rizon, in order to obtain the trve altitude of his centre, which is 
 that to be used. See problem xxileg | 
 
 ie 
 
Example 3. 
 In latitude 48° 52’s., the sun’s declination was 
 19° 34’ 26” n., and his true altitude 20° 30’, required 
 the hour of the day. Ans. 0° 54’ 124” from noon. 
 
 Example 4. 
 
 At Greenwich, in latitude 51° 28’ 40’ N., in the 
 afternoon, when the sun’s declination was 9° 39’ 40”, 
 and the altitude of his centre 24° 59° 50”, what was 
 his azimuth from the north, and the hour of the day? 
 
 ‘Ane | Azimuth 152° 0° 183” 
 Hour -- 1° 42° 154" p. M. 
 Example 5. 
 
 In latitude 39° 55’ n., longitude 35° 50° w. the alti- 
 tude of the sun’s lower limb, on the 7th of May 1808, 
 at 5° 30’ 35” Pp. M., per watch was 15% 40° 53”; how 
 much was the watch too fast or too slow (z) ? 
 
 _ Ans. Watch too slow 1’ 54.7”. 
 
 Note. Any three of the five things, mentioned in 
 _ this problem, being given, the rest may be found by 
 some of the cases of oblique-angled triangles. 
 
 PROBLEM XV. 
 Given the time of the year, the latitude of the place, 
 
 (z) In the practical application of problems of this kind, it will 
 be proper to take several altitudes of the sun or star; and the 
 corresponding times, per watch, within one or two minutes of 
 time of, each other, and to use the mean of these observations 
 instead of any one of them, taken singly, as the errors arising 
 from the imperfection of the instrument, &c. will, in this case, 
 be rendered almost insensible. 
 
240 
 
 and the altitude of a known fixed star, to find the hour 
 of the night when the observation was made. 
 
 Example: by:! by 
 Some time in the night on the 19th of Dibeinbdi 
 1808, in the latitude of Greenwich, 51° 28’ 40” n., 
 the altitude of Arcturus was observed to be 26° 10’ 30”; 
 required the hour of the night. | 
 
 Colt. = 's @ me 6 31 20" 
 
 Co-alt. - - 2% 63° 49’ 30” 
 
 Co-declins w% 69° 48’ 398” 
 
 9|172° 9 28” 
 86° 4°44" 86° 4! 44” 86° 4° 44” 
 TORRES Wf 63° 49’ 30” 69° 48° 38” 
 47° 33°94" | OP 15 1a” S16 oe 
 
 —_—_—- 
 
 ren ee 
 
 Logsin - - - - 86° 4 44° 9,9989822 
 Log sin, =. - -' =)" 29°95 14” 9.5783088 
 Sum of logsines “- *- =) - = -19.5772910 
 @oriptement 2s ees eo eeer eee 
 Log sin - - - = 47°33 24” 9,8680242 
 Log sin. - .- - + 16°16, 6" 9.4473686 . 
 
 | 2|19.7381018 
 Tani Zzn% - - 36° 29'34” 98690509 
 
 2 
 
 72° 59’ 8”hour from noon. 
 
241 
 
 Which converted into time, gives 4° 51’ 562” for 
 the time before the star came to the meridian. 
 
 e s 9 M4 4 he ey 
 Time of %’s culmin’. at Green } 90% 14’ 34” Dec. 19th. 
 
 wich (prob. Iv.) - - - = 
 Time before % passes the meridian 4°51’ 565” 
 Hour of the night ee ae yk Ee . 
 
 Or 3) 22 372” a.m. Dec. 20th 1808. 
 
 : Example 2. 
 
 In latitude 48° 55’ 40” n. longitude 65° 59’ 20” w. on 
 the 14th of April 1808, the altitude of Aldebaran, 
 when west of the meridian, was 22° 20’ 20”; required 
 the apparent time of observation. 
 
 Ans. Hour of the night, 12° 14’ 45”. 
 
 Example 3, 
 
 In latitude 10° 26's. and longitude 166° 30° E. on 
 the 2ist of February 1808, the altitude of Sirius, when 
 east of the meridian, was observed to be 21° 10° 50”; 
 required the true time of the night when the observa- 
 tion was made. 
 
 Ans. February 21st at 45 28" 592”. 
 
 Example 4. 
 
 On the 30th of January 1808, in latitude 53° 24’ N. 
 and longitude 25° 18° w. the mean of several altitudes 
 of Procyon to the west, and of Aiphacca, or a Co- 
 rong, to the east of the meridian, were observed se- 
 parately, by two persons, at the same instant of time, 
 the mean of which was 16° 38’ 38”, the mean altitude 
 
 B 
 
242 
 
 of Procyon 19° 51°18”, and the mean altitude of 
 « Coron 42° 8’; from which it is required to find 
 
 the error of the riod 
 Hour at night for Procyon 16" 36’ 37” 
 Ans. < Ditto -  - - for Alphacca 16° 35’ 27” 
 Watch too fat - - - - - 235° 
 
 PROBLEM XVI. 
 
 The right ascension and declination of :a fixed star, 
 or planet, being given, to find its latitude and longitude. 
 Example 1. 
 
 Required the right ascension and declination of Al- 
 debaran in Taurus, its latitude being 5° 28’ s. its lon. 
 gitude ¢° 6°56’, and the obliquity .of the ecliptic 
 23°28" (ap, 
 
 1. Zo find the co-declination s%. ~’ 
 J M+. == 23°28’ the obliquity of ecliptic 
 Heres 2 %ms=156° 56’ star’s isa 
 q MH - - = 84°32’ compt. of star’s latitude. 
 
 (a) A method of resolving this problem, which is better 
 adapted for obtaining accurate results, in certain cases, is given — 
 by Maskelyne, in his Introduction to Taylor’s Logarithms. 
 The problem itself may be varied so as toadmit of several cases ; 
 but except that given above, and its converse, none of these are 
 of any great practical use. ei 
 
243 
 
 Lie: by case vill. of oblique Z* spherical A’, 
 
 Rad, or sin - - - 90° 
 
 CosZ ¥%ms - - 156° 56° 
 
 > Tan m% + - 
 
 Tanarc® --- 
 
 Cosarc® --.- 
 
 >, Gosm MK), vever- 
 >: Cos(smoQ) - 
 
 Cos s-% 
 
 Declination -- . 
 
 PI Be mu ee TE OT, 
 
 : ‘Sind Sms  ~S 156 56. 
 
 Sm geo sca Sarre 
 
 Sin Z Ses ogo gah 6 
 90° 
 
 ' The right ascension 65° 54° 
 
 Example 2. 
 
 84° 32° 
 
 95° 57’. 
 
 95° 57’ 
 
 84° 32’ 
 72° 29° 
 106° 57’ 
 90° 
 
 16° 57's. 
 
 oo 
 
 - —10,0000000 
 - 9,9638112 
 - 11.0190794 
 - 10.9828906 
 ~. °9.0156135 
 0.9343865 
 
 - 8.9789408 
 £9.4785425 
 - 9.4418696 
 
 ae 
 
 2. To find the co-right ascension % 8m. 
 
 9.9807120 
 0.0392880. 
 -  9.5930666 
 -  9.9980202 
 
 - 9.6103748 
 
 oe 
 
 Required the latitude and longitude of Spica, or 
 Virginis, its right ascension being 13 15’ 52”, its de- 
 clination 10° 9’ 74” s. and the obliquity of the eclip- 
 tic 23° 27 42”, 
 
 Ans. 
 
 Example 3. 
 
 Latitude «.-- 2° 2’ i7” s. 
 Longitude 621° 9° 41” 
 
 The datitude of the moon being 5° 6 37” n. her 
 
 R 2 
 
244 
 longitude 10° 21° 363°; and the obliquity of the eclip- 
 tic 23° 27' 42”; it is required to find her right ascen- 
 sion and declination. 
 
 Ans. 
 
 Right ascen, 322° 19° 4.2” 
 Declination - 9° 2x’ 11” 
 Example 4. 
 
 Required the right ascension of the planet Mercury 
 in time, on the 3lst of January 1808, its geocentric la- 
 titude being 2°s. and its geocentric longitude 10°4°57’. 
 | Ans. 208 81’ 17". 
 
 PROBLEM XVII. 
 
 The right ascensions and declinations of two stars, 
 or their latitudes and longitudes, being given, to find | 
 their distance. 
 
 Example 1. 
 
 It is required to find the distance between Sirius in 
 Canis Major, and Procyon in Canis Minor, the right 
 ascension of the former being 6" 36’ 41”, and its de- © 
 _ Clination 16° 27° 30” s. and the right ascension of the 
 latter 75 29° 14”, and its declination 5° 42’ 34” wn. 
 
 $ % = 73°32'30" comp‘. of Sirius’s declin. 
 
 £%sp==13° 815” thediff. of right ascen. %° 
 Here 
 § P= 95°42’ 34” Procyon’s declin.+ 90°. 
 
245 
 
 Whence, by case vin. of oblique-angled spherical A’, 
 : Rad, Or sin = +s + 90° = == +) 10,0000000 
 : “GésZ ¥ sp 92+ .03°°18'15" © 9.9894820 
 ¢: Tans % 92 « + 4°.78°'82" 30" 1015295562 | 
 
 Tah are@ = <= 5793 27 1015 180982 
 
 ee ne ee 
 
 : CosarcQ ---- M3? 9 OF” . G9 4698447 
 
 —_——~- 
 
 0.5371552 
 ot GiiS shae Tv STEERS? BO" 80" .” 94599749 
 ¢: Cos(sps@) - - 24°35 6” 9.9653471 
 
 ae ee 
 
 > Cos %Pp' - - ) 4:25° 43° 45” | 9.9547765 
 
 ne 
 
 Where P is the distance of the two stars required. 
 Example 2, 
 
 It is required to find the distance between Rigel in 
 Orion, and Procyon in Canis Major, the right ascen- 
 sion of the former being 5" 55’ 19”, and its declination 
 8° 25° 49” s.; and the right ascension of the latter 
 7 ay’ 14”, and its declination 5° 42’ 34” n. 
 
 Ans. dist. 27° 21’ 32”, 
 Example 3. ! 
 
 Required the distance between Sirius and Spica 
 Virginis, the latitude of the former being 39° 33’ 22” s. 
 and its longitude 3° 11° 46’ 10”; and the latitude of 
 the latter 2° 2’ 17” s, and its longitude 6° 21° 9’ 49”. 
 
 Ans. dist. 96° 10° 18”. 
 
 PROBLEM XVIII. | 
 The places of two stars being given, and their di- 
 stances from a third star, or comet, to find the place 
 of the third object. 
 
246 
 
 Example 1. 
 
 Suppose the distance of a comet, or mew Star, as 
 measured by a sextant, to be 65° 47’ from Sirius, 
 «whose latitude is 39° 33’ s. and longitude 3° 11° 137, 
 and 51° 6’ from Procyon, whose latitude is 15° 58’ s. 
 and longitude 3° 22° 55’; it is required to find the lati- 
 tude and longitude of this comet or star. 
 
 1. In the Asn p, we have given 
 
 52 == 129° 33’ dist. of Sirius from pole of eclip. 
 pn = 105° 58’ do. of Procyon from same pole. 
 
 4£snp== 11°42’ diff. longitude of the two stars. 
 Hence, by case vii. of oblique 2% spherical A’, 
 
 Rad, or sin, =.= 90°, - - - , 10,0000008 
 Cos.2snP. <=) ULL 44 ci) Ope uOb ey L 
 : Tan's? ‘= » « -/1 29 So "me modem foo 
 Tan arc® --- 130° 8’ - 10.0739526 
 
 : Cosarc® --- 130° 8’ = 9.809269) 
 
 -0.1907309 
 
 Cossn --- + 129° 33’ -  9.8039699 
 : Cos (Pro) - 24 10° - 9.9601655 | 
 Coss pws pe -) 959.40’ .-- ., 9,9688663 
 
 e 
 
 i Sin SP coos oid gil lal owe MM A, SA 
 
 | 0.3628516 
 Z£SnP -+-+- 41°44 -  9,3082590 
 
 >: Sinpn . - «= = 105° 58 ~ . 9.98299140 
 
 SinZnspP 
 
 26° 48’ - 9.6540246 
 
247 
 
 2. Inthe Acs p, we have given 
 s P = 25° 4%’ dist. of Sirius and Procyon 
 Cc s== 65° 47’ dist. of comet and Sirius 
 cp== 51° 6' dist. of comet and Procyon. 
 Hence, by case x1. of oblique 4 ¢ spherical ge i 
 
 25° 4.37 
 65° 47’ 
 5s. 6’ 
 
 2\142° 35 
 
 a 
 
 24 86. 
 2 
 
 49° 12’ Zcosp 
 26° 48’ ZnsP 
 
 » 22° 24’ Zasn 
 
 —_— 
 
 oe 
 
 fea 
 25° 42° 
 A5° 35 
 
 9,.9764036 
 9.5378508 
 
 19,5142544 
 
 a a ree ee 
 
 0.4857456 
 8.9815729 
 9.8538619 
 
 2|19.3211804 
 
 ees 
 
 - - 9.6605902 
 
 —— 
 
 3. In the Ansc, we have given 
 sn = 129° 33’ Sirius’s dist. from pole of eclip. 
 sc = 65°47’ dist. of Sirius and comet 
 
 Zcosn= 22° 24’ found as above. 
 
248 
 
 Hence, by case vitt. of oblique 2¢ spherical A’, 
 > Rad, orsin - = - 90° - - - - 10 0000000 
 
 - CosZeosnm -- = 29° 24 - 9,9659285 
 ©: Tange os 4°37 2765? 47" 2 10,5870119 
 : Tanared ---~+- 64° 3 - 10.8129404 
 : Cosarc® ----+ 64 3° - 9.640640 
 
 0.3589360 
 
 ¢ C68 's'c oa PCR 4ST) Oo QEne9633 
 >: Cos(sueQ) +--+ 65°30' - 9.6177270 
 9 AER Ota tenn to = ee 67° 8’ - 95896463 
 
 29° 52 lat. of comet. 
 
 : Sincn -+--- 67° 8 - 9.9644537 
 
 0.0355463 
 > SnAZcsn---- 22°24 - 9.5810052 
 >: Sinsc ---+-- 65°47 - 9,9599952 
 ; SinZcns - -- 22°10 - 9.5765467 
 
 nn 
 
 101° 13’ long. of Sirius 
 22° 10° diff. long. 
 79° 3° long. of comet. 
 
 Note. The same things may also be determined from 
 the right ascensions and declinations of the two stars, by 
 referring them to the equator instead of the ecliptic. 
 
 Example 2. 
 
 Suppose an unknown star was found to be 66°10’ 12” 
 distant from Capella, whose declination is 45° 47’ 15° N. 
 and its right ascension 5" 2’ 31”; anid that its distance 
 was 25° 15’ 18” from Procyon, whose declination is 
 5° 42’ 34” n. and right ascension '7° 29° 14” ; required 
 
249 
 
 the declination and right ascension of the unknown 
 star, 
 
 ‘nyt Declination - - 12° 6 23” 
 
 *'. Right ascension 9° 8’ 14” 
 
 Example 3. 
 
 Suppose the distance of an unknown star, or planet, 
 was observed to be 67° 45 24” from Sirius, whose right 
 ascension is 99° 10’ 15”, and declination 16° 27’ 30” s. 
 and to be 51° 4’ §3” distant from Procyon, whose right 
 ascension is 112° 18’ 30”, and declination 5° 42’ 34’N.; 
 it isrequired to find the right ascension and declina- 
 tion of this unknown star or planet (£). 
 
 hee Declination at: 49° LT" 36” N. 
 Right ascension 80° 43° 15”. 
 
 PROBLEM XIX. 
 
 The day of the month, the sun’s declination, and 
 the latitude of the place, being given, to find the ap- 
 parent time of his centre appearing in the horizon. 
 
 Example 1. 
 Given the sun’s declination, at noon, on the 2!st of 
 June 1808, 23° 27' 41”, and the latitude of the place 
 
 (4) By the assistance of the above problem, and a knowledge 
 of the true situation of any particular star, the places of all the 
 rest may be determined; it being chiefy by this means that 
 astronomers have rectified the places of the fixed stars, and 
 thence, by a similar mode of proceeding, found the true places 
 of the planets. In the same manner, we may also find thie dis 
 stance between any two places on the surface of the earth. — 
 
250 
 
 51° @8’ 40” nN; required the apparent time of his centre 
 appearing in the horizon (c). 
 
 Here, the example being given for the longest day, 
 the sun’s. declination may be considered the same at 
 his rising as at noon; the variation not being more 
 than 5” in 24 hours at this time, though near the equi- 
 noxes it varies above 1’ in an hour. | 
 
 Hence, by the tables, his horizontal refraction being 
 == 33’, his parallax 9”, and his semidiameter 15’ 47”, 
 if s be the point where the upper limb of the sun rises, 
 and b that of its apparent rising; we shall have 3374. 
 15’ 47” —9”= b © the distance of his centre below the 
 horizon ; and consequently, 
 
 7, © ==90° 48’ 36.8” app. dist. of ©’s centre from zen. | 
 N © =66° 32’ 19” dist. of ©’s centre from N. pole 
 ZN ==38° 31’ 20” the complement of the latitude. 
 
 (c) The apparent time of the rising and setting of the heavenly 
 bodies always differs from the true time, on account of their be- 
 ing elevated by refraction, and depressed by parallax. The sun’s 
 horizontal parallax is about 9”; and therefore his upper limb 
 will appear in the horizon when it is 32’ 51” below it ; and as his 
 semidiameter is then 15’ 47 
 zon when it is 48’ 38” below it: but a star, having no sensible 
 parallax, will appear in the horizon when it is 33’ below it, or 
 
 90° 33’ from the zenith. 
 
 ’, his centre will appear in the hori- 
 
251 
 
 / 
 Hence, by case x1. of oblique-angled spherical’ A’, 
 90° 48’ 36.8” 
 ee ts, 
 38° 31’ 2 
 21) 195° 52 15.8 yt 
 
 97° 56’ 7.9% 97° 56’ 9” 97° 5G. 7.9” 
 
 66° 32’ 19” 90° 4836.8" 38" 31° 20” 
 Ba AMET pul wit: Cp 59° 24! 4.7.9" 
 Logsin - - - - 97 56’ 7.9" 9,9958912 
 Logsin ai & oo 4 oh? 78B1 990935608 
 7 19,0893820 
 ~0.9106179 | 
 
 Logsin - - - - 31° 93' 48.9"  9,7168075 
 
 Log sin - - - + 59° 24 47.9" 9,9349396 
 2|20.5623581 
 * Log tan -. - - -,62° 22° 23.2” 10,2811790 
 
 een + 
 
 2 
 124° 44’ 462” Zz2No. 
 
 Which, converted into time, gives 8" 18’ 59”= the 
 true time from noon, when the sun’s centre appears in 
 the horizon. 
 
 Hence, his apparent central rising is 3" 41’ 1”, and 
 his setting 8" 18’ 59” (d). 
 
 The true time of the sun’s rising and setting on this 
 day, has before been shown to be 3" 47’ 51”, and 
 
 (2) As refraction is the means of occasioning an error in the 
 rising and setting of all the celestial objects, the same cause will 
 necessarily produce an error in their arnplitudes, as may be seen 
 by comparing the triangles 7 as and 34; but this may he 
 avoided, by making the altitude of ©’s lower limb = 16’+ di; 
 of the horizon. 
 
558 3 
 
 8° 19’ 9”, hence the apparent day ig 13” 40” longer 
 than the astronomical day. 
 | Example 2. 
 
 Required the apparent rising and setting of the sun’s 
 centre, in latitude 51° 28’ 40” w. supposing his decli- | 
 nation to be 23° 27° 36” s. 
 
 Ans. 8" 5° 22” and 3° 54/ 38”, 
 Example 3.. 
 
 Required the apparent time of the central ri ing and 
 setting of the sun, in latitude 51° 28’ 40” n, when his 
 declination is 19° 47’ 20” N. supposing it to undergo 
 no variation from sun-rise to sun-set. 
 
 Ans, 4" 6’ 15” and 7° 53° 442”. 
 PROBLEM XX. 
 
 The latitude and longitude of a place, and the day 
 of the month, being given, to find the time of the 
 moon’s rising and setting at that place. . 
 
 Example 1. 
 
 Required the time of the rising and setting of the 
 moon at Greenwich, latitude 51° 28° 40° N. on the 
 18th of August 1808 (e). 
 
 (¢) The horizontal parallax of the moon, being from 53’ to 62’, 
 always exceeds the horizort: i] refraction ; therefore, when the 
 moon’s upper limb appears :n the horizon she is really above it, by 
 a quantity equal to the horizontal parallax, minus the refraction. 
 
268 
 
 Let n be the place of the moon, when in the hori- 
 zon, m the point where she becomes visible, and e her 
 place on the meridian; in which case, de will repre - 
 sent her change, of declination from the time of her 
 rising to that of her transit. , 
 
 Then, by calculating as in problem v. the true time 
 of her passing over the meridian will be 8° 54’ 19” in 
 the morning. And by prob. 111. her declination, when 
 on the meridian, will be found to be 18° 56’ 5124” n., 
 and her horizontal parallax 56 31.3”; also, by the 
 tables, the horizontal refraction = 33’. 
 
 Henee, in the A zn m, we have given 
 
 Z m==90°-+33'—56 31.3” = 89° 36’ 28.6” 
 
 { Ma IONS BERIT A = 71 S82" 
 
 ZN=90°—51° 28° 40° 38° 31’ 20” 
 Therefore, by case x1. of oblique-angled spherical A’, 
 Hal iG ace 
 89° 36° 28.6” 
 38° Bi" Or 
 2/199° 1 10’ 56.9” 
 
 oe 
 
 99° 35’ 28.4" 99° 35’ 98.4" 99° 35' 28.4” 
 pela ag. 2” 88° 81""90" 89° 36’ 28.6” 
 98° 3% 20.2” 6ie 4’ 8.47 ak Ba! 80. 3” 
 
 — = -——-——— 
 
 Log sin - -.- 99° 35' 28.4” 99938863 
 
 Loesin” - - =) 9° 58° 59.8" 9.2389495 
 | 19.2328288 
 “0.7671 711 
 
 Log sins == (98°.32° 20.2” 9.6792062 
 Log sin - ~ = 61° 4 84° 9.9421088 
 
 2|20.388486L 
 Logtan - - + 57° 2221.9" 10.1942439 
 
 9 an eect ne 
 
 114° 49° 43.8" £Z 
 
 (Cameos 
 
254 
 
 * 
 
 ‘ 
 See ee 
 
 Which converted into time, gives 75 99’ 14,9” == 
 nearly the time the moon rises before she comes to the. 
 neridian. Whence 8" 54° 19.”4 — 75 39’ 14.9 = 
 18 15! 4,4’s= estimated time of the moon’s rising, 
 
 Then, by calculating as in problem 111. the moon’s 
 declination at nearly the time of her rising, will be 
 found’ 2°19" 26° 12.27 Ney and her horizontal paral- 
 lax 36° 17.2". I 
 
 Hence, in the A zN_m, we have, again, given 
 
 Zz m=90° +33 —56' 17.2" = 89° 36’ 42.7” 
 . m==90°— 19° 26°.12.2" =.70° 339’ 47.7". 
 Z N=90°— 51° 28' 40” SR el ee 
 89° 36’ 42.7” 
 10° 33°°47.7” 
 
 38° 31' 20” 
 9/198° 41’ 50.4” 
 
 | 
 
 99° 20’ 55.2” 99° 20' 55.2” 99° 20’ 55.2” 
 70°.83) 47.27) 1895 36.42% 38° 31’ 20” 
 > BR 107 A 9° 4.4'.19.5" , 60° 49° 35.2” 
 
 ~_ 
 
 Log sn - - - 99° 20'°°55.2” 9.9941930 
 Log'sin’.=!(¢)- ong? 4a! a5” 0 Queggoona). 
 | 19.2223950 
 
 0.7776049 
 
 Log. sin 2-5 =) [982 477.47 9,6896237 
 Logsin - = ~ 60°49’ 35.2”  9.9410874 
 | 2\20.4013162 
 
 Log tan - - = 57° 47’ 20.8” 10,200658k. 
 
 115° 34’ 40.7” ZZN Mm. 
 
 had 
 ‘Which, converted into time, gives 7 4.2’ 18.7” for 
 , : a { 
 the time the moon rises before she comes to the me- 
 ridian. 
 
255 
 
 Whence, 8" 54’ 19.4”— 75 42° 18.7” = 15 12’0.6” 
 the time from noon when the moon rises ; which dif- 
 fers only 3’ 3.7” from the time found aboVe. | 
 
 And if still greater exactness be required, the moon’s 
 declination may be found for 1° 12° 0.6”, and the ope- 
 ration repeated as before. 
 
 It is also evident that 8" 54’ 19.4”--b-'75 492’ 18.7”= 
 .16" 36’ 38.1” = nearly the time of her setting ; but in 
 order to obtain this time more rigorously, the variation 
 declination of must be allowed for as in the preceding 
 part of the problem. 
 
 Example 2. 
 Required the time of the moon’s rising at Green- 
 wich, latitude 51° 28° 40” wn. on the 22d of July 1808. 
 Ans. 3° 24’ 342” apparent time in the morning. 
 
 Example 3. 
 Required the time of the moon’s rising and setting 
 at Paris, latitude 48° 50’ 14” n. and longitude 2° 20’ r., 
 on the 25th of September 1808. 
 Ans. Rises 32° 342” afternoon, apparent time at 
 Paris ; and sets at 9® 10° 5534” in the evening. 
 
 PROBLEM XXI1. 
 _ The sun’s declination, two altitudes, and the time 
 between the observations, being given, to find the la- 
 titude of the place. 
 Example 1. 
 
 Ata place in the northern hemisphere, the sun’s de- 
 clination being 19° 39’ 12” n., the true altitude of his 
 centre, in the forenoon, was found to be 38° 20’ 30”, 
 
286 .. 
 
 and at the end of an hour and a half afterwards 
 50° 26° 10”; required the latitude of the place. 
 
 Here, s being the place of the sun at the time of the 
 Ist observation, and B his place at the time of the 2d, 
 the polar distances N a, NB will be equal, supposing 
 the declination to remain the same for the whole of the 
 interval, 
 
 Hence, in the isosceles A N AB, there is given 
 
 N A = 70° 20’ 48” the sun’s polar dist, at 1st obser". 
 N B = 70° 20 48’ ditto. - - «+ - at 2d obser". 
 LZ ANB = 22° 30 the measure of elapsed time. 
 
 1. To find the side ap. - 
 Rad, or sin - -- 90° - - - = 10.0000000 
 Sin NA or NB »- 70° 20'48" 9.9739332 
 >: Sn dZanep -- 11° 15+ =: 9.2902357 
 : Sindap  --+ - 10° 36'12.5” 9.9641689 
 
 -_— ree ee 
 
 2 
 
 oe 
 
 ve 
 
 _———— + 
 
 21° 10° 25" AaB. 
 
 —_—_-—— 
 
 2, To find the angle ABN. 
 
 Coti Zane -- 11°19 -- 107018882 
 Rad, orsin --- 90° - - -. 10.000G000 
 ; CoSNAOorNB - 70° 20°48" 9.5267634 | 
 
 ae) 
 
 U nna 
 
 : Cot ABN ~-+- 6° 10 21.4” 88254252 
 
 a9 
 
 so 
 e 
 
 —_——_——- 
 
257 
 
 XL. In the oblique-angled A a8 Z, there is given 
 
 A Z=51° 39’ 30” complement of 1st altitude 
 B Z=39° 33’ 50” complement of 2d altitude 
 A B=21° 10 25” as before found. 
 Whence, by case x1. of oblique- 4 4 spherical A’, 
 §1° 39° 30” 
 39° 33° 50” 
 21° 10° 25” 
 91112° 29’ 45” | 
 é poe see, Lee ee 
 56° 11° 52” 56° 11’ 52” 56° 11’ 52” 
 
 39° 33’ 50” 51° 39° 30” OTF Oe, 
 16° 38° 9” 4° 39/ 29" Soc ede 
 Logsin - «= - + 56°11' 52”. .9,9195823 
 Logsin - - + - 4° 32/22” 8.8984402 
 18.8180225 
 
 , 861819774 
 
 Log sin - + += + 16°38 2” 9.4567571 
 Log sin - . - - 35° 127” 9,7588541 
 | | 2120.3975887 » 
 Tan -) ¢ + = « 87° 40' 47” ~10,1987943 
 
 2 
 
 115°, 317.95" Lanz 
 ‘86°10 21” Z ABN 
 
 99° 11’ 14° ZN BZ, 
 
 II. In the triangle 8 zn, there is given 
 _ (NB=70° 20’ 48” sun’s polar distance 
 " £ Z = 39° 33’ 50” complement of the 2d alt, 
 4NBZ= 29° 11’ 14” included Z. 
 Whence, by case vit. of oblique-angled spherical A3, 
 « Rad, or sin..- ~ - 90° - - -.-..10.0000Q00 
 Cos4nBz -- 29°11'14" 9.9410899° 
 > Tanpz -- - ~ 39° 33’ 50” 9.9170909 
 Tanarc@ = - ~~ 35° 4811” 9.5581201 
 
 ee 8e@ ee 
 
 oe 
 
 s 
 
258 
 
 > Cosarc@ =~ +.35° 48°11" ' 9.90903%6 
 0.0909623 
 > (Cos BZ. =\+ -)=.89%33''50" °9.8870064 
 :: Cos(@onsB) - 34° 82°36"  9.9157669 
 2 \COSNZ ‘++ n= <138% 98.52". , 9.8037857 
 90° 
 51°31’ 54” lat. required (f). 
 
 Example 2. 
 
 Ona day, in the northern hemisphere, when the 
 sun’s declination was 20° 13’ 30” Nn. his true altitude 
 in the forenoon was observed to be ai 32’ 5", and 
 three hours afterwards it was 39° 58’ 12”; from itich 
 it is required to find the latitude of the bhace 
 
 Ans. 60° 10° 24” N. 
 Example 3. 
 
 - When the sun’s declination was 22° 37’ N. his cor- 
 rect altitude at 108 55’ 13” in the forenoon was 
 53° 28’ 9”, and at I" 17’ 16” in the afternoon it was 
 52° 47’ 29”; required the latitude of the place, sup- 
 posing it to be north. Ans. 57° 3’ 422” wn. 
 Example 4, — 
 
 At a place in the northern hemisphere, when the 
 sun’s declination was 23° 28’ 36” n. his corrected alti- 
 tude, at 8° 53’ 48” inthe forenoon, was 48° 42’ 20”, 
 
 and at 9° 46’ 9” it was 55° 47/ 48” ; required the true™ 
 
 latitude. ie Ans. 49° 59 183” N. 
 
 eee 
 
 (f) The principal use to which this problem is applied,. is in 
 
 questions of a similar nature to that given-above; but several » 
 other things may be determined from the same data: such as: - 
 
 the hour from noon when each altitude was’taken, the azimuth 
 ‘at each observation, &c. 
 
259 
 
 PROBLEM XXII. 
 
 The apparent distance of the moon from the sun, 
 or a star, and their apparent altitudes, or zenith di- 
 stances, being given, to find their true distances, as 
 seen from the earth’s centre (¢). | 
 
 N 
 
 Let zm be the observed zenith distance of the moon, 
 and zm her true zenith distance, mm being the dif- 
 ference between the moon’s refraction and her parallax 
 in altitude. 
 
 Also let zs be the observed zenith distance of the 
 sun, or a star, and Zs itstrue zenith distance; ss 
 
 (¢) Since the observed altitude of any celestial object is affected 
 by refraction and parallax, the effects of which are always pro- 
 duced in a vertical direction, it is obvious that the observed di- 
 stance between any two bodies will also be affected by the same 
 causes, With regard to the fixed stars, the parallax vanishes, so 
 that their places are only changed by refraction ; but in obser- 
 vations of the moon particularly, the effect of parallax is very 
 sensible, on account of her proximity to the earth: for which 
 reasons, the true distance between the moon and any celestial 
 object is, for the most part, considerably different from the ob- 
 
 served distance. 
 
 It may also be remarked, that since the refraction of the sun, at 
 the same altitude, is always greater than his parallax, his true 
 place will be lower than his apparent place; and because the 
 moon’s parallax, at any given altitude, is always greater than 
 the refraction at that altitude, her true place will be higher than 
 her apparent place. 
 
 Ss 2 
 
260 
 
 being the difference between the sun’s refraction and 
 parallax, or the refraction of a star. 
 
 Then, since in the triangle zs, the three sides 
 ZS, ZM, and s M, are given, the vertical angle sz mM 
 may be found, by case x1. of oblique-angled spherical 
 triangles. 
 
 And, because in the triangle zsm, the two sides 
 zs, zm, and the included angle s z m, are also known, 
 the true distance sm may be found by case vit. of 
 oblique-angled spherical triangles. 
 
 But as this method, though direct and obvious, re- 
 quires three separate statings, or analogies, for obtain- 
 ing the true distance, it may be rendered more com- 
 modious in practice by incorporating the analytical for- 
 mulze for finding the angle z and the side sm into a 
 single expression; which, when converted into loga- 
 rithms, gives the following rule, using the altitudes 
 
 instead of the zenith distances. 
 
 RULE. 
 
 1. Take the difference of the apparent altitudes of 
 the moon and star, or moonand sun, and half the dif- 
 ference of their true altitudes. 
 
 2. Also, take half the sum and half the difference 
 of the apparent distance and the difference of the appa- 
 rent altitudes. 
 
 3. To the log sines of this half sum and half diffe- 
 rence, add the log cosines of the true altitudes, and the 
 
 complements of the log cosines of the apparent alti- . 
 
 tudes, and take half the sum. 
 
261 
 
 4, From this half sum take the log sine of half the 
 difference of the true altitudes, and look for the re- 
 mainder among the log tangents ; which being found, 
 take out the corresponding log cosine, without taking 
 out the arc, which Is unnecessary. 
 
 5. Subtract this log cosine from the log sine of half 
 the difference of the true altitudes, above found, in- 
 
 creased by 10.in the index; and the remainder will 
 _be the log sine of half the true distance (/). 
 
 Example 1 
 
 ( Apparent dist. of ) and % 51° 28’ 30” 
 
 | Apparent alt. of ) ’scentre 12° 30’ 4.” 
 Given< Apparent alt. of % - - - 24° 48717” 
 | True altitude of » *s centre 13° 20’ 40” 
 
 UT rue altitude of % --- 24° 46 14” 
 
 Required the true distance of the moon and star. 
 
 24° 48° 177” 51, 28.20, 
 
 12° 30° 4” 12°18) 13" 
 
 12° 18°13” - 9|63° 46’ 43” 
 
 94° 46) 14” 31° 53° 212” - I sum 
 
 13° 20° 40" 2|39° 10° Ton”, | 
 2|11° 25° 34” 19° 35 85° Bers 2 diff. 
 
 5° 4D 47” 
 
 (4) The method of reducing the apparent to the true distance, 
 or of clearing it of the effects of refraction and parallax, being 
 the most tedious part of the calculation for ascertaining the lon- 
 gitude, by the common spherical analogies, various compendious 
 formule have been proposed, at different times, for the more easy 
 solution of this problem; among the best of which are those 
 of Borda, la Caille, de Lambre, Messrs. Dunthorn, Lyons, 
 Maskelyne, Witchel, and Mendoza, 
 
262 
 
 Log sin - - - 31° 53214" -~ 9.°7228640 
 Log sin -)- - 19°35’ 88” -~ 9.5253252 
 
 Log cos = - - 13°20'40" - 9.9881129 
  Logceos - - - 24° 46'14" «+ 9.9580824 
 € Logcos - - ~- 12°30 4” ~- 0.0104203 
 
 eLogcos - - - 24° 48°17" + 0.0420371 — 
 ; 2139.2468421 
 
 ee 
 
 19.6234210 
 
 Log sin - - - 5°42'47" - 8:9980259 
 
 Log tanofanarc - - = = - 10.6258951 
 
 Corresponding log cosine - - - “9.3627458 
 
 Logsin - - - 25°34 54" - 9.6352801 
 Z 
 
 51° 9 48” true distance. 
 
 Example 2. 
 
 (Apparent dist. © and )’s centres 90° 21°17” — 
 
 | Apparent alt. of )’scentre--- 5°17’ 9” | 
 ele Apparent alt. of Q's centre - - - 84° 7° 20” © 
 True altitude of )’s centre - - - 6° 9' 14” 
 
 LTnie algae af ©’s centre - - - 84° 7° 15” | 
 
 Required the true distance of the sun and moon’s 
 centres. | Ans. 89° 29° 18”, 
 Example 3. 
 ( Apparent dist. © and )’s centres 38° 45" 38" @ 
 | Apparent alt. of ) ’scentre - - - 29° 31° 10" © 
 Givens Apparent alt. of ©’s centre - + - 35°43" 6° 
 
 | True altitude of > ’s centre - - - 30°.19° 43” 
 UTrue | altitude of ©’ scentre - - - 35° 41° 540 ‘i 
 
 Required the true distance of the sun and moon’s | 
 centresi7 gam. Ans. 38° 28’ 20", 
 
263 
 
 PROBLEM XXIII. 
 The observed altitudes of the sun and moon, or of 
 the moon and a star (z), and their apparent distance, 
 
 (z) The principal stars used in finding the longitude are the fol- 
 lowing ; which have been chosen on account of their lying near 
 the moon’s path : 
 
 1. @ Arietis, a small star lying without the zodiac, about 22° 
 - to the right hand of the Pleiades. 
 
 2. Aldebaran, in the Bull’s eye, a large conspicuous star, lying 
 about half way between the Pleiades and the star which forms 
 the western shoulder of Orion. 
 
 3. a Pegasi, a star lying something less than 45° to the right 
 of a Arietis, being nearly in a line with this latter star and the 
 Pleiades. ' 
 
 4. Pollux, a star a little to the northward of Aldebaran ; being 
 the left-hand one of two stars lying near together, called Castor 
 and Pollux. 
 
 5. Regulus, a star about 38° s. £. of Pollux ; being the southern- 
 most of four bright stars to the n. £. of Aldebaran, forming a 
 zignzag line. 
 
 6. Spica, or a& Virginis, a white sparkling star, about 54° s. £. 
 of Regulus, 
 
 7. Antares, a star lying to the right-hand of aon and about 
 
 45° from Spica Virginis. 
 
 8. Fomahault, a star lying about 45° to the south of & Pegasi. 
 
 9. a Aquila, a star about 47° to the westward, or to the right- 
 hand of a Pegasi, 
 
 These stars may be readily known, by finding them on a com- 
 mon celestial globe ; or by means. of their calculated distances, 
 which are given in the Nautical Almanac, for every 3 hours of 
 apparent time at Greenwich. For, the sextant being fixed to 
 the distance betweerf the moon and the star which ought to be 
 observed, and the moon found upon the horizope pales) | itis only 
 necessary to look to the east or west of the moon, “according as 
 the distance corresponds to the 8th and oth or 10th and 11th 
 pages of the Nautical Almanac, guiding the sextant in a line 
 with the moon's shortest axis. 
 
264 
 
 being given, together with the time, and the longitude 
 by account, to find the true altitude. 
 
 RULE. | 
 
 I. Add the longitude by account, converted into 
 time, to the time at the given place, if west, or sub- 
 tract it from that time, if east, and the result will give 
 the time at Greenwich nearly ; which call the reduced 
 time. # 
 
 II. ‘Take the moon’s semidiameter, and horizontal 
 parallax, from the Nautical Almanac, for the noon 
 and midnight between which the reduced time falls, 
 and find their differences. 
 
 Then, as 12 hours : the difference of the semi- 
 diameters at those times :: the reduced time : a fourth 
 number; which being added to, or subtracted from, 
 the preceding semidiameter, according as the tables 
 are increasing or decreasing, will give a result, to which 
 if the augmentation of the semidiameter (tab. 1v.) be 
 added, the sum will be the moon’s true semidiameter, 
 at reduced time. 
 
 Also, As 12 hours ; the difference of the horizon- 
 tal parallaxes at those times-:: the reduced time : a 
 fourth number; which being added to, or subtracted 
 from, the preceding parallax, according as the tables 
 are increasing or decreasing, will give the moon’s ho- 
 rizontal parallax, at reduced time. a 
 
 Ill. Add the difference between the moon’s semi- 
 diameter, at reduced time, and the dip of the horizon, 
 
 to the observed altitude of her lower mb, and the sum 
 
 will give the apparent altitude of ihe moon’s centre. 
 Then, to the cosine of the altitude, thus found, add 
 
 R 
 
265 
 
 the logarithm of the horizontal parallax in seconds, at 
 reduced time, and the sum, abating 10 in the index, 
 will be the logarithm of the moon’s parallax in altitude, 
 in seconds ; from which take the refraction in altitude, 
 and the result, added to the apparent altitude of the 
 centre, will give the true altitude of the moon’s centre. 
 
 LVEF the sun be used, add the difference between 
 his semidiameter on the given day (Naut. Alm.) and 
 the dip of the horizon (tab. 11.) to the observed alti- 
 tude of his lower limb, and the sum will give the ap- 
 parent altitude of his centre ; from which take the 
 difference between his refraction in altitude and his 
 parallax (tab. 11. and 1v.), and the remainder will be 
 the true altitude of the sun’s centre. 
 
 Or, if a star be used, take the dip of the horizon 
 (tab. 11.) from its observed altitude, and the remainder 
 will be its apparent altitude ; from which subtract the 
 refraction, and the result will be the star’s true alti- 
 tude. ve 
 
 V. To the observed distance of the sun and moon’s 
 nearest limbs add their semidiameters, at reduced time, 
 or subtract them for the furthest limbs, and the re- 
 sult will be the apparent distance of the sun and 
 moon's centres, 
 
 Or, to the observed distance of a star from the 
 moon’s nearest limb add her semidiameter, at reduced 
 time, or subtract it for the furthest limb, and the re- 
 sult will be the apparent distance of the star from the 
 moon’s centre. : 
 
 Vi. With the apparent altitudes, the true altitudes, 
 and the apparent distance, find the true distance, by 
 
266 
 
 problem xx11. And if the watch has not been pre. 
 viously regulated, the true time must now be found 
 from the altitude of the sun’s centre, or a star, and the 
 latitude of the place, as in problem xv. observing to 
 proportion the sun’s declination to the reduced time. 
 
 VII. Look in the Nautical Almanac, on the given 
 month and day, for the computed distance between 
 the moon and sun, or star, and if it be found exactly, 
 the time at Greenwich will stand at the top of the 
 column; but if not, find the nearest distances to it, 
 both less and greater, and take their difference, and 
 also the difference between the computed distance and 
 the earliest Ephemeris distance. 
 
 Then, as the. first difference : 3 hours : : the second 
 difference : a fourth number; which being added to 
 the time standing over the earliest Ephemeris distance, 
 will give the true lime at Greenwich. 
 
 And, if the difference between the time at the given 
 place and the time at Greenwich, be converted into 
 degrees, it will give the longitude required ; which will 
 be east or west, according as the time at the given place 
 is greater or less than the time at Greenwich. | 
 
 Example I. 
 
 On the 30th of January 1796, in longitude 10° 46’ 5. 
 by account, at 10" 15’ p.m. the distance of the moon’s 
 furthest limb from the star Regulus was. 63° 50° 20”, 
 the altitude of the moon’s lower limb 24° 18’ 40”, and 
 the altitude of the star 45° 13’ 15”, the eye being 18 
 feet above the plane of the horizon ; required the true 
 longitude of the place. 
 
267 
 
 i. 
 
 Time per watch - Me. 10h 15! 0" Pe. 
 
 Longitude 10°46’ east - - - 
 
 Reduced time - - - 
 
 49°40" 
 gh 31' 90" P.M. 
 
 —_——» 
 
 ‘ II. 
 
 ) ’s semid*. at noon - - 15’ 3”} Horizontal parallax - 55’ 14" 
 Do. at midnight - - -(14' 59") Dittlo- - - - - - 54 59” 
 
 First diff.- - Of 4” Second diff. - - 0" 15” 
 {oh 241s) -OB 31’ 20" O13" 1.12R 15": + OB 31’ 207s) O12" 
 )’s semid*®. at noon - - 15! 3” Horizontal par. at noon 55! 14” 
 -) ’ssemid®. at red. time 15/0” Ditto at reduced time 55’ 2" 
 )’saugmentation -- 0/7” 60 
 
 -) ’s true semid!, 
 time - - 
 
 as 15! 7” Inseconds - =. -.'- 3302 — 
 
 Ii. 
 
 >’ s observed alt. - 24°18’ 40"; Cos )’s app. alt. 
 
 Semidiam. 15’ rig fay rr 
 Di eeey Jy Cll’ 4 
 App.alt.of ) ’scent. 24° 29’ 44” 
 Correction 48% 1! 
 
 Truealt.of p’scen. 25° 17! 45” 
 
 IV, 
 *’s observed alt. - 45°13’ 15” 
 
 - 9.9590383 
 Hor. paral. 3302 log 3.5187771 
 
 Par. in alt, 3005 log 3. 3.477815 
 
 3005 sec. 50'S” 
 y’srefrac. 2’ 4 
 
 Correction 48/1" 
 
 Vv. 
 Obs. dist. pand * 63°50’ 20" 
 
 Dip of horizon - - 4! 3!! )?s semd'. red. time aie ad 
 *’s apparent alt. - 45° 9/12”| App. dist. ) and * 65° 35’13" 
 Refraction “7 470% 57? 
 
 *’struealt. - - - 48° 8’ 15” 
 
 Vi. 
 
 #*’s apparent alt. - 45° 9’ 12” %’s true altitude - 45° 8' 15" 
 ) ’s apparent alt. - 24° 29’ 44”! )’s true altitude 25° 17’ 45” 
 Difference - - - 20° $9’ 28’! Difference - - 19° 50’ 30” 
 
 - = 63° 35’ 13" 
 ~ = 84° 147 41” 
 = - 42° 55! 45" 
 
 App. dist. - 
 Bum =) '« ‘= 
 
 Difference - 
 
 4 Difference -~ 9° 5515" 
 
 49° 7 204" - - ¢sum 
 Q\° 97! 524" - = 3 diff. 
 
268 
 
 Logsin - - - - 42 7° 203” 9.8265388 
 Logsin - - - - 21° 27° 523" 9.5633933 
 Log cosine - - - 25°17°45"  9,9562230 
 Log cosine - - - 45° 815"  9,8484403 
 eLog cos - - - - 24° 29/44"  0.0409617 
 
 eLogcos - - - - 45° 9 12”  0,1516804 
 2\39.8872375 
 
 19.6936187 
 Logsin - - - =) 9°55 15" 9,.2362231 
 Log tan ofan. arc sy praia) Sill 10.4573956 
 Corresponding log cos - - - - 9.5176693 
 Log'sin *- + = - 91782’ 16%” 9.'7185538 
 
 9 
 True distance - - - 63° “4 33” 
 
 —— 
 
 VII. 
 
 . Dist. at 9% - 62°49’ 15” | True dist. - - 63° 4/35” 
 Dist. at 125 - 64° 19’ 56” | Ear. Eph. dist. 62° 49/15” 
 First diff. - - 1° 30'41” | Second diff. - 0° 15'18” 
 
 ee 
 
 ASA BOAT 0s Ons 21S 8 SOT 80/28" 
 Earliest Eph. distance - - -- 9 
 
 True time at Greenwich aoe er Gus oe 4 
 Time at the given place - - - 10515’ 0” 
 Difference of time - - - -- 0% 44’ 38” 
 
 ‘Which converted into degrees, at the rate of 15° to 
 an hour, gives 11° 9’ 30”, the longitude of the place 
 east; the time at the place being greater than that at 
 Greenwich. , 
 
 Example 2. | 
 On November 8th 1804, in longitude 24° w. by 
 account, at 3" 50° 10” p.m. the observed distance’ be- 
 
269 
 
 tween the néarest limbs of the sun and moon wag 
 
 67° 48’ 29”, the observed altitude of the moon’s lower 
 
 limb 31° 10’, and that of the sun’s 14° 46’, the height 
 
 of the eye being 12 feet ; required the true longitude 
 
 of the place. Ans. 24° 29’ 30” w. 
 Example 3. 
 
 On October 10th 1804, in latitude 15° 15’N. and 
 longitude 68° £. by account, at about 6 4’ p.m, the 
 watch not being well regulated, the distance of the 
 moon’s furthest limb from Fomahault was 60° 37’ 35”, 
 the observed altitude of the moon’s upper limb 46° 30’, 
 and of the star 21° 24’, the height of the eye being 
 14 feet; required the true longitude of the place. 
 
 Ans. 68° 1’ 45”, 
 Example 4. 
 On the 13th of June 1796, in longitude 45° w. by 
 account, the watch being well reguiated, the following 
 observations were taken: 
 
 : Alt. ©’s Alt: p 7s Dist. nearest 
 dachrhdelctrtantgeag eee rsty upper limb. limbs, ~ 
 
 3h 6'40” 45°54’ 0” PG° SB I4.6)" 106° 16’ 45” 
 3°13 14 45 45 O 19 5a. 0 106 17 45 
 
 5 ls Ye AS?) 45 18 45 Zorn ite Oe 106 18 30 
 
 3 22 34 45 4 O 20 17 30 106 18 45 
 
 5 26 46 44,48 30 20 34 O 106 19 15 
 
 5\16 26 40 5/226 50 15 ~ 5{100 20 80 5/581 31 0 
 mean3 1720 45 22 3 90 4 6 106 18 12 
 Errors of the quadrants 5S 12:0 eae th 
 True mean « 45° 21' 5” ~ 90° 3’ 6" 106°15’ 35" 
 
 fips -_—_—— 
 
 — 
 
 Required the true longitude, the eye being 21 feet 
 above the sea. Ans. 42° 52’ 15” w. 
 
 n 
 
 Example 3. 
 On the Ist of April 1796, in latitude 65° 36’ north, 
 
270 
 
 and longitude by account, 2° 15’ west, the watch not 
 regulated, the following observations were taken, the 
 eye being 18 fect above the level of the sea : 
 
 ‘Times Pp. M. 
 
 0h 47" 140" 
 
 20 50 11 
 
 20 55 26 
 
 $[62 32 51 
 mean 26" 40’ 57" 
 
 Alt. @’s 
 
 lower limb. 
 
 wu ol 
 
 oo Fa" 
 
 Alt y’s Dist. nearest 
 lower limb. limbs. 
 86° 18’ 65° 36’ 
 80 36. 69 37 
 $1.9 69 38 
 m2 8 208 51 
 80° 41 65° 37° 
 
 —s 
 
 Required the true longitude of the place. 
 Ans. 2° 4:7” 15° w Ww, 
 
 Times P.M. 
 
 12236" 14” 
 12 39 5 
 12 41 4) 
 12 43 45 
 12 45.53 
 563 2 95 38 
 
 12 41 19 
 
 Error of quadrant 
 
 Example 6. 
 
 On the 15th of May 1796, in longitude 1° 30’ x. 
 by account, the watch being well regulated, the fol- 
 lowing observations were taken : | 
 
 Alt. p’s 
 
 lower limb. 
 
 igo 6” Of 
 18 21°0 
 18- 39 30 
 18 55 O 
 19'¢ “BRAD 
 
 93 10 30 
 
 18.38 6 
 7 30+ Error of quadrant 
 
 18 45 36 
 
 Alt. Spica Mg. 
 
 19° 50/ 80” 
 20042." 
 20.15 0 
 
 90 29 0 
 
 20 40 O 
 
 101 16 30° 
 
 ~90 18 18 
 
 Dist. )’s 
 furth. limb. 
 312 30% 4.5" 
 31 31 30 
 31°33 O 
 81 34 O 
 35 35 45 
 157 45 0 
 8) .83.-0 
 +, 24 
 31 33 24 
 
 —_———. 
 
 Required the true longitude, the eye being 21 feet 
 
 above the level of the sea. 
 
 Ans. Moon’s true semidiameter 15’ 
 
 31”, horizontal 
 
 parallax 56’ 35”, apparent alt. D ’s centre 18° 56’ 45”, 
 correction ).’s altitude 50’ 46”, apparent altitude +¢’s 
 centre 20° 10’ 56”, correction %’s altitude 2’ 35”, ap- 
 parent distance of their centres 31° 17’ 53”, true di- 
 stance 31° 11° 43”, true longitude 15’ west. 
 
271 
 TABLE I. 
 
 Of the Sun’s Right Ascension and Declination for 1808. 
 
 January. . February. Maren. 
 Right Declin. Right | Declin. Right Declin. 
 ascen. $. ascen. Ss. ascen. 3. 
 
 — 
 
 Be Ns Ss Gat MM. Sael He uM) S.5) D. My 7 8.1.Ha Mes Sole D Mie 
 18.43.28 | 23. 5. 8| 20.56. 2117.20.12! 22.49. 8} 7.31.28 
 18.47.53 | 23. 0.18/21. G. 7/17. 3.12] 22.52.5217. 8.37 
 18.52.18 | 22.55. 1| 21. 4.11 | 16.45.55 | 22.56.36| 6.45.39 
 
 18.56.42 | 22.49.17] 21. 8.14 | 16.28.20} 23. 0.19| 6 22.36 
 
 eee Se Pd . 
 
 19; 4..6 | 22.43,°.6 ) 21.12.16}: '6:10281 23. 4. 2 mee 
 6| 19. 5.30.| 22.36.27 | 21.16.17 | 15.5220] 23. 7.441)'5 ne. 
 7110. 0.53 | 22.290.22.|.21,29.18 | 15.33.56] 23.11.20) 5.12.57! 
 8 | 19.14.16 22.2) 50 | 21.24.18 | 15.15.15| 23.15. 7 MGy4 
 9 | 19.18.38 | 22.13.51 | 21.28.16 | 14.56.19| 23.18 4g | 4.26.12! 
 10!) 19.22 590 |22: 5.27} 21.32.15 (14.37. 8|23.22.20| 4. 2.44 
 
 — 
 
 14.17.43 | 23.26.10 | 3.39.12} 
 13.58. 3 ee 30} 
 13.38.10 180 | 2.52, 3] 
 13.18. 4\ 23.37, Q | 2.28.25) 
 
 12.57.44) 23.40.48 2. 4.46 
 
 12,| 19.31.41 | 21.47.21 | 21-40. 8 
 13 | 19.36. 1 | 21.37.40 | 21.44. 4 
 19.40.20 | 21.27.33 | 21.47.59 
 19.44 38 | 21.17. 2| 21.51.53 
 
 : 4927220 1.21550 .37 |-21.30,12 
 
 — 
 
 19.48 56 | 21. 6. 7 | 21°55.47 
 
 3.44.97 LAL. 6 
 
 19.53.13 | 20.54.47 | 21.59.40 tO-2 SAS! iO} dt aeo 
 10.57 .30.1'20:43.'3 |.22..3.32 1 11.55.31123.51 451} 0.53.43 
 207140: )20:30555 4) 22.2723 "1 ¥.34.94'123 55:25 0.30, |) 
 20, °6.°1)/-20:18.25 + 22.31.4497 9F 137% 5123 '50, 210. 0:20 
 } North 
 
 20.10,15 | 20. §.31-}22.15: 4 
 20.14.29 | 19.52.15 | 22 18.54 
 20.18.42 | 19.38.36 | 22.22.42 
 20.22.54 | 19.24.36 | 22 20.31 
 20.27. 5 | 19.10.14 | 22.30.18 
 
 20.31.16] 18.55.31 | 22.34. 5 
 20.35.26 | 18.40.27 | 22.37.52 
 20.39.34 | 18.25. 3 | 22.41.38 
 20.43 43} 18. 9.19 | 22.45.33 | 
 20.47.50 | 17.53.15 | ——— I 
 
 20.51.56 ! 17.36.53 | ——~—— | ———~-} 0 39. 1] 4.12.17] 
 
 OS a a re ee Fee 
 
 eS) ‘ 
 
 ow GO 
 
 a5 
 
 Nene Se 
 
 bo bo 
 
 G2 
 
 in Gr ok A 
 
 CO +e 
 
 bo 
 
 roe Lee owe dt & 
 ee eee EIDE EEE 
 
 a cunnnmnnannetinenmnenr an eeemaremas 
 OWN wed & bo Ww 1 
 SESSNSIL GARGS 
 
 LO ARN A RN CRE et EE el ek ee a 
 
“ 
 
 A RN RR A a EF I NR A I 
 
 OO Te een 
 
 Gy feniaa06. tm, SABCT 
 
 te 
 
 yg 
 
 TABLE I. 
 
 Of the Sun’s Right Ascension and Declination for 1808: 
 
 | 
 
 April. 
 
 ~— 
 
 Right | Declin. 
 
 ascen. 
 
 2.14.54 
 2.18.40 
 pe Mey, 
 2.20.15 
 D, 3 Ohara 
 
 er ee 
 
 || 10. 9 39 
 
 N. 
 De jves (Bis 
 4.35.26 
 4.58.31 
 5.21.30 
 5 44.24 
 Gie7 12 
 6.29.53 
 O.52/27 
 714.54 
 7.37.14 
 7.59.26 
 8.21.30 
 8.43.26 
 Q. 5.13 
 9.26.51 
 9.48 20 
 
 10 30 49 
 10 51.48 
 11.12.36 
 11.33.14 
 11.53 41 
 12 13 56 
 12.33.58 
 12.53 49 
 13.13 27 
 13 32.5 
 13.52. 
 
 We Sh eal Ph 
 
 14 29.46 
 14.48.15 
 
 os 
 
 2 
 3 
 ] 
 
 Right 
 ascen. 
 
 FEANE SS. 
 2.33.52 
 2.3741 
 2.41.31 
 2.45.21 
 2.49.11 
 
 253.3 
 2.56.55 
 3. 0.47 
 3. 4.40 
 3. 8.33 
 
 3.12.28 
 3.10.29 
 3.20.18 
 3.24.13 
 3.28.10 
 
 lay. 
 
 oe re ee 
 
 Declia. 
 N. 
 
 DES Ss 
 15. 6,29 
 15.24.29 
 15.42.14 
 15.59.43 
 16.16.56 
 
 16.33.53 
 16.50 33 
 17. 6.56 
 17.23. 3 
 17.38.52 
 17.54.23 
 18. 9 37 
 18.24.32 
 18.39. 9 
 18.53.27 
 
 10..'7.26; 
 
 16.21.60 
 19.34.26 
 19.47.27 
 20. 0. 7 
 
 20.12.27 
 20.24.27 
 20.36. 5 
 20.47.23 
 20.58.19 
 
 424. 8.53 
 21.19. 6 
 
 21.28.56 
 21.38.24 
 21.47.30 
 21.56.13 
 
 SE er a SE as ee Ee 
 BRS SOS te a el St Ee See Se 
 5 . 
 
 Light 
 ascen. 
 
 aS Se eee eee 
 ac a 
 
 June, 
 
 Declin, 
 eas 
 
 22.40.24 
 22.46.23, 
 22.51.59 
 22.57.10 
 23. 1.58 
 23. 0.21 
 23.10.19 
 93.13.54 
 23.17: 3 
 23.19.49 
 
 23.22. 9 
 93.24. 5 
 23.25.36 
 23.26.42 
 23.27.24 
 
 23.27.41 
 23:27 32 
 23.20.59 
 23.90, 2 
 23.24.39 
 
 29.22.52 
 
 | 
 | 
 | 
 
 oe 
 a mn ee 
 
 . et ee 2 
 
July. 
 
 Right 
 ascen. 
 
 H. M.S. 
 
 oO 
 
 jet) 
 
 e 
 
 ] 
 
 2 | 6.44.56 
 3 649. 4 
 4 
 5 
 6 
 7 
 
 6.53.12 
 
 2532 
 
 11 | 7.21.53 
 12} 7.25.58 
 13/730) 2 
 141734. 5 
 15 | 7.38. 8 
 
 —| 
 
 | 
 
 eee ew ee 
 
 Declin. 
 N. 
 
 D. M. S. 
 
 | 6.40.48 | 23. 7.47 
 
 23..3.33 
 22.58.54 
 22.53.52 
 
 | 6.57.19 (22.48.26 
 
 ee «ee 
 
 | 
 
 \ 
 | 
 
 | 
 i 
 { 
 
 | 
 
 17. 1.25 | 22.42.36 
 
 22.36.23 
 22.20.46 
 22.22.45 
 224522 
 
 22. 7.36 
 21.59.26 
 21.50.54 
 21.42. O 
 21.32.43 
 
 ee 
 
 — 
 
 16 7.42.11 | 21.23. 4 
 17 740.13 21.13. 4 
 18 7.50.14 21. 2.41 
 19 | 7.54.15 20.51.58 
 20} 7.58.16 20.40.53 
 
 — 
 
 -{21) 8. 2.16 
 22 | 8. 0.15 
 
 23.| 8.10.14 
 
 24 | 8.14.12/ 19.53. 6 | 
 
 25 | 8.18.10 
 26 |-8.22. 7 
 27 | 8.26. 4 
 28 | 8.29.59 
 29 | 8.33.55 
 30 | 8.37.49 
 }31 | 8-41.43 
 
 ! 
 ' 
 
 | 
 
 | 
 
 \ 
 
 20.29.27 
 20.17.40 
 
 2040.33 
 
 19.40.19 
 19.27.12 
 19.13.46 
 1, O-r I 
 18.45.57 
 18.34.35 
 18.16.54 
 
 273 
 
 TABLE I. 
 Of he Sun’s Right Ascension and Declination for 1808. 
 
 \ August. 
 
 Right 
 ascen. 
 
 H. M. S&S. 
 8.45.37 
 8.49.29 
 $,53:22 
 8.57.13 
 9.1.4 
 Q. 4.54 
 Q. 8.44 
 9.12.43 
 9.10.22 
 9.20.10 
 
 0.23.57 
 9.27.44 
 9.31.30 
 9.35.16 
 
 9.39. 1 
 
 0.42.46 
 9.40.30 
 9.50.14 
 9.53.58 
 9:57.40 
 
 10. 1.23 
 10.5. 5 
 
 10. 8.46 
 10.12.27 
 10.16. 8 
 10.19.48 
 10.23.28 
 10.27. 7 
 10.30.46 
 1034.24 
 
 —_——s 
 
 Declin. 
 
 10.42.45 
 16.26. 5 
 16. 9.10 
 15.51.58 
 15.34.32 
 
 15.16.50 
 14.58.53 
 14.40.42 
 14.22.17 
 14. 3.38 
 13.44.45 
 13.25.40 
 13. 6.21 
 12.46.50 
 12.27. 6 
 os oN ge! 
 11.47, 4 
 
 11.26.46 
 LVA6.1 
 10.45.37 
 10.24.47 
 10. 3.47 | 
 9.42 38 
 9.21.20 
 8.59.52 
 
 September. 
 
 Right 
 
 ascen. 
 
 bs RI HET 
 10.41.41 
 10.45.18 
 10.48.56 
 10.52.33 
 10.56. 9 
 10.59.46 
 bling 3.28 
 11. 6.59 
 11.10.35 
 11.14.11 
 11.17.46 
 11.21.22 
 11.24.58 
 11.28.33 
 1¥.32,°9 
 11.35.44 
 11.39.20 
 11.42.55 
 11.40.31 
 11.50...7 
 11.53.42 
 11.57.18 
 
 12. 0.54 
 12. 4.30 
 12. 8. 6 
 12.11.42 
 12.15.1G 
 12:18.55 
 12.22.32 
 12.26. 9 
 
 BO Rahal Sache L/S) me 
 
 by 
 
 Se eeeetial 
 
 Declin. 
 N. 
 
 D.M. 8s, 
 8.16.33 
 7 54.4) 
 73241 
 7.10.34 
 0.48.20 
 6.26. O 
 0. 3.33 
 5.41. O 
 5.18.2) 
 4.55.30 
 4.32.47 
 4. 9.52 
 3.40.53 
 3.23.50 
 3. 0.43 
 2.37.33 
 2.14.19 
 1.51. 2 
 1.27.43 
 1y.422 
 0.40.50 
 0.17.35 
 South. 
 0.5.51 
 0.29.17 
 0.52.43 | 
 1.16.. 9 
 1.39.34 
 2.2.50 
 2.26.23 
 2.40.45 
 
274 
 
 TABLE I. 
 Of the Sun's Right Ascension and Declination for 1808. 
 
 bee ee — —" : ie € 
 ahkoawe ! Qoob Oryolagpere she 
 
 16 
 
 Oo Om 
 
 bo 
 co) 
 
 S| 
 
 23 
 24 
 25 
 26 
 27 
 28 
 29 
 30 
 
 Ociober. 
 Right Declin. 
 ascen. Ss. 
 
 H. M. .S.[° D2 BM. ..S; 
 b220.47. |°3.13.; 6 
 12.33.24 | 3.36.24 
 12.37. 2} 3.59.39 
 12.4040} 4.22.52 
 12.44.19 |"4.46. 2 
 12.47:58 15500. 8 
 12'51.37 |" 5.32.10 
 LBS: 17 445-55, 6 
 12.58.57 | 6.18. 2 
 13. 2.38 | 640.51 
 rae ete PY oS 
 SIO. PP F 26.29 
 13.13.43 | 7.48.44 
 13.17.26 | 8.11.10 
 1S.3T. Ot 833,20 
 13.24.53 | 8.55.40 
 13.28.38] 9.17.44 
 | 13.32.23 | 9.39.40 
 13.36. 8} 10. 1.28 
 13:39.55 | 10.23. 7 
 13.43 42 | 10.44.36 
 13.47.29 | 11. 5.56 
 13.51.18 | 11.27. 6 
 13.55. 7111.48. 6 
 13.58.56 | 12. §.54 
 14. 2.47 | 12.20.31 
 114. 6.38 | 12.49.57 
 14.10.30 | 13.10.10 
 (14.14.23 | 13.30.11 
 (14.18.16 | 13.49.59 
 
 $31] 14.22.10 
 
 14. 9.34 
 
 November. 
 
 Right 
 ascen. 
 
 —— 
 
 FY BE. Sa 
 14.26. 5 
 14.30. 1 
 14.33.57 
 14.37.54 
 14.41.53 
 14.45.52 
 14.49.51 
 14.53.52 
 14.57.54 
 TS "1.56 
 15,559 
 15/10.:4 
 15.14. 9 
 15.18.14 
 15°39 Dt 
 
 15.26.29 
 15.30.37 
 15.34.47 
 15.38.57 
 15.43. 8 
 15.47.20 
 15 51.32 
 15.55.46 
 16. 0. O 
 16. 4.14 
 16. 830 
 10.12.46 
 IGAZ. 3 
 16.21.21 
 10.25.39 
 
 Declin. 
 s. 
 
 Dy M. 'S: 
 14.28.54 
 14.48. ] 
 15. 6.54 
 15.25.82 
 15.43.54 
 16. 2. 1 
 16.19.52 
 16.37.26 
 16.54.43 
 17.11.44 
 17.28.27 
 17.44.51 
 18. 0.58 
 18.16.45 
 1832.13 
 18.47.22 
 19. 2.11 
 19.16.39 
 19.30.40 
 19.44.32 
 19.57.56 
 20.10.59 
 20.23.39 
 20.35.56 
 20.47.50 
 
 20.59.21 
 21.10.28 
 21.21.11 
 21.31.30 
 21.41.24 
 
 December. 
 
 Right 
 ascen, 
 HH, Me? Ss, 
 16.29.58 
 
 16.34.18 | 
 
 16.38.28 
 16.42.59 
 10.47.20 
 16.51.42 
 16.56. 5 
 17. 0.28 
 17.04.51 
 vgs Aly 
 17.13.40 
 17.18. 4 
 17.22.30 
 17.26.55 
 7.37.21 
 17.35.47 
 t7.40.13 
 17.44.39 
 17.49. 6 
 7 Sa.a8 
 17.57.59 
 18. 2.00 
 18. 6.53 
 18.11.20 
 18.15.46 
 18.20.13 
 18.24.39 
 18.29. 5 
 18.33.31 
 
 18.37.57. 
 
 18.42.22 
 
 Declin. 
 s. 
 D. M. S. 
 21.50.53 
 
 22:31.56 
 22.38.51 
 92.45.18 
 A251 FO 
 29.56.53 
 WB. \\2>6 
 23. 6.39 
 23.1051 
 23.14.35 
 23.37.51 
 23.20.39 
 23.22.59 
 23.24.5] 
 23.20.14 
 23.27.10 
 
 23.27.36 
 23.27.35 
 23.27..5 
 23.26. 7 
 23.24.41 
 23.22.46 
 23.20.24 
 23.17.33 
 23.14.14} © 
 23.10.27 
 23. 6.12) 
 
975 
 
 TABLE I. 
 
 Refraction in Altitude and Dip of the Horizon. 
 
 | 
 
 Apparent 
 altitude. 
 Refraction. 
 Apparent 
 altitude. 
 Refraction. 
 
 31.22 
 30.35 
 
 2.55\14.52 
 28.22/13. 0114.36 
 27 A1ll3. 5 
 27. O||3.10}; 
 26.20)'3.15 
 
 50 25.42/|3.20 
 55|25. 5||\3.25 
 24.29| 3.30 
 23.541/3.40 
 23.20)/3.50 
 
 Seat A & fb EE eee 
 
 22.47)|4. O 
 1.20129.15/ (4.10 
 1.25}21.44!/4.20 
 4i.30!21.15/ 4.30! 
 1.35|20.46| 4.40 
 
 —_—_—_—— 
 
 0 
 
 12.40 
 12.15 
 11.51 
 11.29 
 11. 8 
 10.48 
 10.29 
 
 ee = —~----—— 
 
 1.40/20.18 14.50|10.11 
 1.45|19.51|'5. O| 9.54 
 1.50/19.25||5.10] 9.38 
 1.55/19. 01/5.20| 9.23 
 . 0118.35/15.30] 9. 8 
 
 oe 
 
 —— = 
 
 . 5|18.11]|5.40} 8.54 
 .10}17.48||5 50] 8.41 
 2.15117.26)6. 0} 8.28 
 20117. 416.10} 8.15 
 2.25}16.44}|6.20| 8. 3 
 
 33. O|/2.30/16.24 
 32.10)/2.35)16. 4 
 2.40)15.45' 
 2.45|15.27)| 7. O 
 
 14.20 
 
 13.34 
 13.20. 
 13. 6 
 
 latitude. 
 
 Apparent 
 Refraction. 
 
 6.30/7.51 
 6.40)7.40 
 6.50 
 
 . 0 
 Q.10)5.42 
 9.20)5.30 
 9.30/5.31 
 9.4015.25 
 
 9.50 
 
 (5.20 
 10. O/5.15 
 10.15|5. 7 
 10.30/5. O 
 10.45/4.53 
 
 —- .— | -——. —_— 
 
 11. 014.47 
 11.1514.40 
 11.30/4.34 
 11.45/4.29 
 
 112. 0/4.23 
 
 altitude, 
 
 Apparent 
 Refraction 
 
 19.30|2.39 
 20. 012.35 
 20.30/23 | 
 
 et | ee 
 
 21.30)2.2 4 
 22. 0/2.20 
 23. O/2.14 
 
 24, O12. FE 
 
 25. Ol2. 2 
 26. O}1.56 
 27. O|L.SL 
 28. OH.A7 
 
 29. 011.42 
 
 44 
 
 9 
 
 asd 
 
 Refraction. 
 
 L2i 
 1.18 
 1.16 
 38/1.13 
 1.10 
 
 40|1. 8 
 41j1. s 
 42/1. 3 
 43li. 1 
 440.59 
 
 45 
 46 
 
 97 
 
 1179 
 
 Refraction 
 f the hor. 
 at sea. 
 
 i Height ot eye. 
 
 z Dip o 
 | 
 
 63/0.29 
 64,0.28 
 65'0.26| 
 66'0.25|| 6/2.20 
 67\0.24| 7/231 
 68|0.23!| 8|2.42 
 69|0.22,| 9}2.52 
 
 70\0.21//10}3. 1 
 7110.19|}t 1}3.10 
 17 2|0.18]12/3.18 
 7310.17) 13 3.26 
 7410.16 |14|3.34 
 
 Mey 
 
 —— ee 
 
 7510.15 |15|3.42 
 76,0.14/}10)3.49 
 77/0.13||17|3.56 
 78)0.12)18]4. 3 
 19|4.10 
 20/4.16 
 21/4.22 
 2/4.28 
 23)/4.34 
 244.40 
 
 80 
 81 
 82 
 83 
 84 
 85 
 \86 
 87 
 83 
 
 26|4.52 
 28/5. 3 
 305.14 
 
 ( 
 
 58 
 \59 
 
 35|5.30 
 140.6. 2 
 
276 
 
 TABLE Il. 
 
 Right Ascension and Declination of 36 principal Stars. 
 
 4 Ann. 
 Mean R.A. in ; 4 Annual 
 rity. Mag. Sidereal ‘Tinie. Biba Mean. UseHnaitt| gariation. 
 hm 5. s. + oct ieee S. 
 
 y Pegasi....2] O 3 15,40 | 3,069 | 14 6 24,90 Nn. | +20,20 
 a Arietis, .2.3 1 50 15,60 | 3,347 | 22 32 24,908 N. | 417,47 
 aw Cethy: ie ee 2 '}482) 52 8,88) fa, 115 . 19 22,40n. | +14.75 
 
 Aldebaran 1 | 4 24 48,00 (3,420 0 3i1,40N.] + : 
 Capella...1 15 222,62 [4,415 vy 47 5,88n.{ +4,57 
 Rigel... 1 | 5 55 13,11 | 2,876 | 8 25 59,32 s. | — 4,92 
 )6 Tauri....2| 5 14 2,17 | 3,781 | 28 25 52,50 N.| + 3,91 
 ra, Orionis...1 | 544° 40;234 3,243 | 7 2b°38,10N. 1° 4 1;49 
 Sirs ta a | 6 36 36,00 vt 6534 16. 27 ZY 54 7h 4 0} 
 Ne OAs ton. isis 2 | 7 22 11,92 | 3,853 | 32 18 3,76n.| — 7,06 
 TaeProcyonisk2 |. 7.29) 8,17 3,142 | 5 42 51,48 wn. | — 8,53 
 SPoll@xs: 2.92 | 7 33 25,43 | 3,688} 28 29 2.58 N.| — 7,03 
 ka Hydre...2| 9 18 8,08 | 2,940), 7 49 19,70 s. |, +15,10} , 
 Regulus...1 | 9 58 1,65 ie 212 | 12 54 41,74N. ] —17,19) © 
 ; 6 Leonis®: he | 11 15 30 25,24N. | —20.04 
 | 8 Virginis . of 18 i 35,2 2743, 126 2 51 32,42 nN. | —20,22 
 ia Virginis..1 | 13 LOY 53 147°) 10 8 29,80 s. | + 18,80 
 Arcturus . ] | 14 6 48,$ se | 2,723 | 20 11 59.41 N. | —18,79 
 "a Libra....2 | 14 89 58,60 | 3,296 | 15 10 49,06 s. | | 475,19 
 4. Libra, i's2-2 14 Oy Gag 13207 {15 12 26784: SoA 495 21 
 a Corona 2.3 | 15°26 28,63 23845 1.27 22 34,54N. | —12,49 
 a Serpentis.2 | 15 34 43,17 [2,945] 7 48,60 n.| —11,70 
 Antares. 1 }:16 17 32,06| 3,658 |25 59 4,92's.|.+ 8,43 
 a Hercules2.3}17 5 18,33. QAs1 | 14-37 .26, 48n.| — 448 
 ‘a@Ophiuchi2 | 17 25 55,91 | 2,776 | 12 42 47,88 n.| — 3,03 
 am tyre tt 18 30 22,08 | 2,027 | 38 30 36.34N. |] + 2,91 
 a Aquile..3 19 3 1,91 Tae 10} 0 8 HAD we: ot 8,38) 8 
 a Aquile 1.2 | 19 41 18,83 | 2,925 [ 8 22 2,64n.] + 9,11 
 B Aguile 3.4] 19 45 46,85 /2,944[{ 5 56 1,28N.| + 8,57 
 'y Capricorni4 | 20 6 53.03 That 113 5 39,70 8. | — 10,80 
 *¢ Capricorni3 ; 20 7 10,83} 3,339] 13 7 58,30 s.| —10,81 
 a Cygni.. 1.2 | 20 34 409,06 | 2,038 | 44 35 33,S4N.[ +!2,56 
 a Aquarii.. 3 | 21 55 48,75 | 3,081 1 15.15,66 s.|.—17,36 
 
 Komalhaut 1.2 | 22 46 54,18 
 | @ Pegasi...21:22 55° 6,12 
 aAndromedx2 | 23 58 22,89 
 
 3,343 | 30 38 26,30 s. | —19,10 
 2,973 9 59,32 nN. | +10,43 
 
 3.070 | 27 58 $4,24.N. | +19,99 
 
pod 
 i 
 
 TABLE VI. 
 
 To convert Time into Longitude. 
 BH M.{D.'M.{|\M. 
 
 —_—- | —~—-—_——. 
 
 4 
 151225 ||/15): 
 16 (240 16/4 
 17 255)\17 4 
 18/}270 :18.4 
 19 | 28519 /4 
 20|300/|20 5 
 21 315/101 |5 
 22 |330}|22'5 
 
 23 (345123 '5 
 
 241300 |24'6 Oo 
 
 ) 
 
 aa 45 || 57 
 28,7 0|/58) 
 129.7 15.159. 
 
 TABLE IV. 
 Angmentation of the Moon's 
 Semidiameter. 
 
 D’s alt, Augm | D’s alt. | Augm. 
 
 25 6 15 
 |26 6 30 
 
 12 
 
 14 
 
 30/7 30'|60/15 
 
 | 3 
 
 OC — 
 Oa ono Go 
 
 Seon nawmoOON 
 
 TABLE V. 
 The Sun's Parallax in 
 ' “Altitude. 
 G)’s alt |Parallax.| ©)’s alt.jParallax.( 
 
 — 
 
 D. S: D. Ss 
 
 8) 9 60 4 
 10 9 65 4, 
 BOT yd 7U 3 
 30 8 75 Q 
 4K (fs 80 2 
 50 6 85 1 
 55 5 90 0) 
 
 TABLE VIL. 
 L 
 
 To convert Longitude into Time. 
 
 D. | H. M. 
 
 D. { H. Me a a 
 
 M.|M.. 8.|| M.} M, 8. | Bi 2 a 
 Sob. Toll. StS. ted. Coley ae 
 1;0 4131|;2 4! 70} 4 40 
 2/0 8|32/2 8|| 80} 5 20 
 3/0 12|/33/2 12]) gol 6 o 
 4/0 16/134|/2 16] 100 6 40 
 5|0 20/135|2 20] 110] 7 20 
 6\0 24136|2 24/120) 8 oO 
 710 28|13712 231/130) 8 40 
 810 321/38} 2 321/140} 9 20 
 9/0 36||39|2 36])150)10 O 
 10|0 40|'40|2 401) 160/10 40 
 11|0 441/41]}2 44]/17011 20 
 12/0 481142}2 48]18012 O 
 1310 52|!43]2 52]|190|12 40 
 14/0 56]|44|2 56|]20013 20 
 1511 0145/3 O},210)14 O 
 16/1 4! 46/3 4/1220114 40 
 17/1 8147/3 8|/230])15 20 
 18/1 12|!48}3 12]/24016 O 
 19/1 16/49 3 16|/250/16 40 
 20) 1 20]5013 20] 260,17 20 
 2111 24115113 24| 270118 0 
 22/1 28|52/3 28), 280|18 40 
 2311 32| 53/3/32'290|!9 20 
 2411 36] 54/3 361/300|20 O 
 25\1 40] 55\3 40 |) 310)20 40 
 26)1 44|'56/3 44'1320121 20 
 27|1 4857/3 481/33/|22 0 
 28|1°52/ 5813 52||340\22 40 
 29/1 50/'59|3 56/1\350\23 20 
 3012 0] 60/4 O||360\24 0 
 
278 
 
 ‘MISCELLANEOUS ASTRONOMICAL PROBLEMS. 
 
 1. In what latitude north, will the ‘shortest day be 
 Ans. Lat. 41° 23’ '7”, 
 2. At what time of the day, in the month of May, 
 
 just 2 of the longest? 
 
 when the sun’s declination is 20° 16’ 32”, will the. 
 
 shadow of a perpendicular object be just equal to its 
 length? Ans, 9° 13’ 16” a.M. 
 3. In what latitude, on the first of June, will the 
 sun’s altitude, when due east, be double his altitude 
 at 6 o’clock? Ans. Lat. 49° 6’ s. 
 4, At what time in the afternoon, on the 9th of 
 June, is the sun’s altitude exactly the same in the la- 
 titudes of 50° and 60° north? = Ans. 4" 49° 11” p. a. 
 5. On the 24th uf May, in latitude 50° 12’ Nn. re- 
 quired the time it will take for the body of the sun to 
 rise out of the horizon. Ans. 3’ 58”. 
 6. On July ist 1792, in latitude 57° 9’ n. and lon- 
 gitude 2° 8’ w. the stars Vega and Altair were observed 
 on the same vertical circle, at 10" 9’ per watch; re- 
 quired the apparent time of observation, and the error 
 of the watch. sik dy 
 _ Ans, Apparent time 105 8’ 14”, watch too fast 46”. 
 7. On October 25th 1792, in longitude 21° z. by 
 account, the interval in mean time, between the rising 
 of Aldebaran and Rigel, was 3" 17’ 30”; required the 
 apparent time of the rising of Aldebaran. 
 : Ans. 6° 36’ 4”, 
 8. On July 4th 1804, in latitude 35° 48’ s.. and 
 longitude 23° 26’ E. the mean times per watch, of 
 
279 
 
 several equal altitudes of Altair, were 8° 21° 152” and 
 14° 3,’ 4124"; required the error of the. watclys for 
 apparent time. Ans. Watch too fast 53”. 
 9. On the 13th of December 1804, in latitude 
 $7° 46° Nn. longitude 21° 15° £. a certain phenomenon 
 was observed at the same instant that the altitude of 
 Arcturus, when east of the meridian, was found to be 
 84° 62’; required the apparent time of observation, 
 the height of the eye being 10 feet. 
 _ Ans. Apparent time 16" 33’ 34”, 
 10. On February 14th 1792, in latitude 43° 26’ N, 
 and longitude 54° 16’ w. the altitude of Jupiter, at 
 12" 23’ 5” per watch, was found to be 16° 9.7’; re- 
 quired the error of the watch, the height of the eye 
 being 14 feet. Ans. Watch exact. 
 11. In latitude 51° 31’Nn. having found the azimuth 
 of an object to be s. 48° 10’ E. ; it is required to find 
 the error in the apparent time of observation, corre- 
 sponding to a supposed error of 10’ in the altitude, 
 Ans. 1° 26”. 
 12. In latitude 54° 42’ by account, having found 
 the azimuth of an object to be 26° 17’; it is required 
 to find the error in the apparent time, answering to a 
 supposed error of 10’ in the latitude. Ans. 2° 20”. 
 13. At 9° 51’ 58” a. m. per watch, the correct alti- 
 tude of the sun’s centre was 21° 11’, at 10° 48’ 54” 
 it was 24° 40’, and at 112 29’ 42” it was 26°; required 
 the apparent time when the greatest altitude was ob- 
 served, and the error of the watch. _ | 
 Ans. Apparent time 115 29’ 16”; watch too fast 26”. 
 
280 
 * 
 
 14. On November 21st 1792, at 10 hours, reduced 
 time, the true distance between « Orionis and the 
 
 ‘moon’s centre was found to be 105° 26' 14”; from 
 
 which it is required to find the moon’s longitude. 
 Ans. 118 10° 5’ 50". 
 15. On December 30th 1792, the moon’s eastern 
 limb was observed to pass the meridian-at 13553’33.875 
 trom which it is required to find-the longitude of the 
 place of observation. | Ans. 78° 13°. 
 16. On Nov. 15th 1804, in latitude 45° 35’ 25” N. 
 and longitude by account 20° w. the meridian altitude 
 of the moon’s lower limb was found to be 48° 33’ 40°; 
 required the longitude of the place of observation, the 
 height of the eye being 15 feet. Ans, 19° 45° w. 
 17. On October 2d 1800, the beginning of a lunar 
 eclipse was observed at 6" 14’ p. m. per watch, and the 
 end at 125 52’ p.m.; required the longitude of the 
 place of observation, the watch, for apparent time, 
 being 131’ too slow. Ans, 41° 30’ w. 
 18. Required the altitude and longitude of the no- 
 nagesimal degree, in latitude 57° 8.9’ N. and longi- 
 tude in time 8’ 40° w. on November 26th 1787, at 
 
 11" 18’ 8” apparent time. . 
 
 Ans. Altitude 53° 22’, longitude 65° 24.1’, 
 19. On June $d 1788, in latitude 57° 9’ n. and 
 longitude, by account, 8’ 32” in time, the beginning 
 of a solar eclipse was observed at 19" 33’ 19” apparent 
 time, and the end at 20" 49’ 29”; required the lon- 
 gitude of the place of observation. Ans. 2° 9 w. 
 ‘20. On November 26th 1787, in latitude 57° 9’ n. 
 
28) 
 
 longitude by account 8’ 32” w. in time, the Immersion 
 of 7 II, in an occultation of that star by the moon, was 
 observed at 11° 18’ 8” apparent time, and the emer- 
 sion at 12° 43’ 12”; required the longitude of the 
 place of observation. 
 Ans. Long. in time 7’ 25° w. 
 21. On December 9th 1803, an emersion of the 
 first. satellite of Jupiter was observed at 16" 58’ 35” 
 per’ watch. which was 3° 58” too slow for apparent 
 time; required the longitude of the place of observa- 
 tion. | Ans. 12°'35' w. 
 22, On November 6th 1804, in longitude 158° w. 
 the meridian altitude of the sun’s lower limb was 
 87° 37'N.3 required the latitude of the place, the 
 height of the eye being 12 feet. Ans. 18° 194° s. 
 23, On December 25th 1804, the meridian altitude 
 of Saturn was found to be 68° 42’ N.; required the 
 latitude of the place, the height of the eye being 15 
 feet. Ans! 25°14 s. 
 24. On December 14th 1804, in longitude 30° w. 
 the meridian altitude of the moon’s lower limb was 
 found to be 81° 15° N.3; required the latitude of the 
 place, the height of the eye being 16 feet. | 
 | Ans. 15° 17x, 
 25. In north latitude, at 11° 10’ and at 12" 40’ per 
 watch, the altitude of the sun’s lower limb, when cor- 
 rected, was 26° 55, and his declination 5° 17’s.3. re- 
 quired the latitude of the place. Ans. 57° 9! w. 
 26. At 9° 23° 20° a.m. apparent time, the true alti- 
 
282 
 
 tude of the sun’s centre was 34° 29°, and. at 11» 9’ 39" 
 it was 42°19’; required the latitude and declination. 
 Ans. Lat. 7°7’ n. Dec. 108. 2 ¥onte 
 27. On June 4th 1804, in latitude by account 37/n. 
 at 10" 29’ in the forenoon, per watch, the correct alti- 
 tude of the sun’s centre was 65° 24’, and at 125 31’ 
 it was 74° 83, required the true latitude. | 
 Ans. 86° 57'N. 
 28. On January Ist 1805, in north latitude, the true 
 altitude of Capella was 69° 23’, and, at the same in- 
 stant, the true altitude of Sirius was 16° 19’; required 
 the true latitude. Ans. 57° 8’ x. 
 29. On the 12th of December 1804, being in 
 north latitude, and in longitude 24° w. by account, at 
 5° 24 p.m. per watch, the altitude of the moon’s 
 lower limb was 41° 33’, and at 7> 12’ it was 52° 563 
 required the true latitude of the place, the height of 
 the eye being 20 feet. Ans. 48° 52’. 
 30. On the 15th of May 1804, in latitude 33° 10’ n. 
 and longitude 18° w. about 5 o’clock a.m. the sun 
 was observed to rise x. by N.; required the variation 
 of the compass. Ans. 11° 26 w. 
 31. On November 19th 1806, in latitude 50° 22’ n. 
 and longitude 24° 30’ w. about three quarters past 
 8 o’clock a.m. the altitude of the sun’s lower limb 
 was 8° 10’, and his bearing per compass s. 21°18'z.3 . 
 required the variation, the height of the eye being 20 
 feet. Ans. 24° 12’ w. 
 32. On August 19th 1810, in latitude 41° 46’ n. 
 
283 
 
 longitude 144° 20’ w. at 9 13°44” A. M. apparent time, 
 the magneticazimuth of the sun’s centre was s. 80° 20’ w. 
 required the variation. Ans. 16° 42’ g. 
 33. On August 26th 1809, in the forenoon, the 
 sun’s magnetic azimuth was s. 22° 41’ 2. and his cor- 
 rect central altitude 33° 145 and some time afterwards, 
 the magnetic azimuth was s. 14° 53’ w. and the true 
 altitude 42° 36’; required the latitude and variation. 
 Ans. Lat. 57° 82’ n. variation 29° 31’ w. 
 34. Required the latitude and longitude of a star, its 
 right ascension being 16° 14’, its declination 25° 5I n., 
 and ‘sis obliquity of the ecliptic 23° 28’. 
 Ans. Lat. 46° 64’ n. Longitude 234° 36’, 
 35. In latitude 20° x. the gnomon of an horizontal 
 dial being perpendicular to the plane of the horizon, 
 it is required to find at what hour in the afternoon, on 
 the longest day, the shadow of the gnomon will stand 
 still, and how many degrees it will run back. 
 Ans. Stands still at 25 12’ 8”, and runs back 32° 32’. 
 36. At London, on the 10th of December 1780, at 
 what time of the night were the stars Aldebaran and 
 Rigel on the same azimuth circle! 
 Ans. 9" 32’ 23” evening. 
 $7. At what time in the evening were the stars 
 Betelguese and Pollux on the same almacanter, or 
 of one common altitude above the horizon of Lon- 
 don, on the 10th of December 1780? 
 Ans. 95 12° 24", 
 38. Being at sea, in an unknown latitude, | ob- 
 
254 
 
 served the star Schedar in Cassiopeia, and Almaach in 
 Andromeda, to have the samé azimuth, when the alti- 
 . tude of Schedar was 37° 157; required the latitudé of 
 the place. | Ans. 50° 44, 
 39. Being at sea, in an unknown place, the star 
 Aldebaran was observed to rise 3515’ later than the 
 pie star in Aries ; required the latitude of the place. 
 Ans. 54° 38’. 
 40. Some time in the month of May 1780, ata place 
 in the Western ocean, the sun’s meridian altitude was 
 observed to be 62°, and J 48’ 14” after it was found 
 to be 54° 30’; required the day of the month and the 
 latitude of the place. 
 Ans. 19th May, latitude 48°.0' 192” v. 
 41. Some time in July, in wn. latitude, the sun was 
 observed to rise at 4 24’ 36” a.m. and his altitude, at 
 noon, in the same place, was 62°; required the day 
 of the month and the latitude of the place. 
 Ans. July 23d, latitude 48° 0" i10°N. 
 42. Ata place in the Western ocean, in the month 
 ,of May, the sun’s altitude, at 6" a.m. was 14° 432, 
 and at 8° 8’ 11” it was 36°; required the latitude of 
 the place, and the day of the month. 
 Ans. Latitude 48° x. May 19th. 
 43, At a place in the Western ocean, the sun was 
 observed, at rising, to be 59° 15’ 40” from the true 
 north point of the horizon, and his altitude at 65 a. m. 
 was 14° 432’; required the latitude and declination. 
 Ans. Latitude 48°.n. Declination 20°. 
 
235 
 
 ~ 44, Ata place in the Western ocean, some time in 
 May 1763, the altitude of the sun’s centre, at 85.8’ 11” 
 A. M. was 36°, and at 9°10’ 51” it was 46°; required 
 the latitude and declination. 
 Ans; Latitude 48° n. De pee i 19° 59’. 
 45. Some time in July 1763, three descending alti- 
 tudes of the sun were found to be 545°, 46°, and 36°, 
 and the intervals of time between them 60’ 55” and 
 62’ 40"; required the times when the observations were 
 made. | Ans. ‘Time 2d obser. 2" 48! 59%, 
 Time $d obser. gh 5139”, 
 46. It is required to find the declination of a plane, 
 upon which the sun, on June 10th, in latitude 51°52’ n, 
 will continue 9" 50’. Ans. 60° 31’, 
 47. To determine the latitude of the place, where 
 the shadow of the gnomon, on an horizontal dial, will 
 move, -between 3 and 4 o’clock, with the greatest ve- 
 locity possible. % Ans, Latitude 49° 29" 43”, 
 48. It is required to find the greatest error of an 
 horizontal dial, made for a place in latitude 51° 32’ Nn, 
 
 Os 
 
 but placed in latitude 54°. ei A ns! 4.27”, 
 
 » 49. It is required to find, in latitude 66° ne when 
 the continuance of the morning and evening twilight 
 is iis equal to the length of the day. 
 Ans. January 28th and November 12th. 
 50. In what latitude, on the 21st of June, will the 
 sun be due east, when he has run half his course be- 
 tween the time of his rising and noon? 
 poe ae ? Ans. Latitude 64° $5’ 48” N, 
 
 » we 
 
286 
 
 51. At what time, on the 10th of June, in latitude 
 12° 30’ n. will the sun’s azimuth be the greatest pos- 
 sible ? Ans. 8° 5’ A.M. or 3° 55’ P.M. 
 
 52. In what points of the ecliptic, between r and s 
 does the sun’s longitude exceed his right ascension. the 
 greatest possible f Ans. In 16° 14 16° of T aurus. 
 
 53. On what day of the year, in latitude 51° 32" N. 
 does the sun’s azimuth increase the most possible in 2 
 hours after rising! Ans. Nov. 19th or Jan. 22d. 
 
 54. On what day of the year, in latitude 51°32’ w. 
 does the length of the afternoon. exceed that of the 
 forenoon the most possible, reckoning the day to begin 
 at sun-rise and to end at sunset ? 
 
 Ans. 19th April, when sun’s long. is 29° 82’. 
 
 55. eon has an horizontal dial, made for the 
 latitude of 51° 32’ w. how must he set it so that it shall 
 go true in latitude 53° 15° aaa. NR eer ae 
 
 Ans. Directly nN. and s. inclining 38° 17 fiom 
 
 the zenith, 
 56. To determine a place on the earth, bee a 
 
 degree of the meridian is equal to a degree of the equa- A 
 tor; supposing the ratio of the two axes to beas 229 
 230. Ans. From lat.54°17' 38to lat.55°17! 38”. 
 pee The latitude of London being 31° 30 , the lati. 
 tude and longitude of Moscow 55° 45’ and 38° 0 and 
 those of Constantinople 41° 30’ and 29° ae respec- 
 tively, itis required to find the latitude and. longitude | 
 of aplace that shall be equidistant from all the three, - 
 Ans. Lat. of the place 51° 17’ and long. 19° 13”. 
 
 $ at 
 og . gh 
 
287 
 
 OF THE SIGNS OF 
 TRIGONOMETRICAL QUANTITIES. 
 
 As no account has been given, in the former part of 
 this work, of the change of signs which the various 
 trigonometrical lines are liable to undergo, according 
 to the magnitude of the arc to which they belong, it 
 will be here proper to enter into some explanation of 
 this part of the subject, which requires to be particu- 
 larly attended to in many analytical investigations. 
 
 1. For this purpose, let asc B p be a circle, having 
 the sines, tangents, &c. represented as in the figure ; 
 and suppose one of the extremities a, of an arc a M to 
 remain fixed, while the other extremity m passes suc- 
 
 cessively over the circumference of the circle, from a 
 through c, B, D to A again. 
 vy c 
 
 > 
 7 
 
 op) 
 
 » Then’as the sine m P continually recedes from a, till 
 the point M arrives at B, and afterwards approaches to- 
 wards a, on the other side of the diameter a B, till it 
 is united with it again, it is plain that the sines of all 
 arcs, in the first semicircle A CB, areaffirmative or ++, 
 and that the sines of all arcs, in the second semicircle 
 BDA, are negative, or —. 
 
 Tr is also evident, that the sine m P increases from o, 
 during the 1st quadrant a c, till, at the end of it, it be- 
 
288 
 
 comes equal to the radius; and that it decreases during 
 the 2d quadrant cB, ull it again becomes o. After 
 this, it passes to the other side of the diameter, and 
 increases negatively during the 3d quadrant B p, till it 
 becomes equal to —radius; and then decreases nega- 
 tively during the 4th quadrant p a, till it becomes o, 
 as before. | , | e 
 
 It appears, likewise, from an inspection of the figure, 
 that the sine  pP increases faster in the first part a M, 
 of the quadrant ac, than when it approaches near the 
 end of it at c ; and, on the contrary, that 1t decreases 
 more slowly in the first part c m,of the 2d quadrant 
 c B, than when it arrives near the end of it at B ; owing 
 to the convexity of the circle being more or less fa- 
 vourable to this variation. . | 
 
 2. In like manner, the cosine o Pp, heii referred to 
 the centre o, will become negative as often as it passes 
 that point; and as this takes place both when the arc 
 A m becomes greater than a c, and when, by its further 
 increase, it is greater than ACBD, it 1s evident that the 
 cosines of all arcs in the Ist and 4th quadrants ac, DA 
 wil! be positive, or 4+-, and that the cosines of all arcs 
 in the 2d and 3d quadrants CB, BD) will be negative, 
 or —. i i Ml | 
 It is also plain, that the cosine 0 P is puts to ra- 
 dius when the arc aM is 0, and that it continually 
 decreases during the Ist quadrant ac, till, at the end 
 of it, itis o. After this, it increases negatively during 
 the 2d quadrant cB, at the end of which it is equal to 
 
 —radius. It then decreases negatively during the 3d . 
 
 f 
 
oy 289 
 
 " . 
 
 quadrant Bp, at the end of which it is 0; and in the 
 
 4th quadrant pa, it again becomes positive, and in- 
 — creases till it is equal to radius, as before. 
 
 And since the cosine of the arc a m is equal to the 
 sine of its complement m c, it follows, reversedly, from 
 what has been said respecting the variation of the sines, 
 that the cosine o P increases slower in the first part a M 
 of the quadrant Ac, than when it approaches near the 
 end of it, at c; and, on the contrary, that it decreases 
 faster in the first part c m’ of the 2d quadrant c B, than 
 when it arrives near the end of it, at B. , 
 
 3. The tangent a r becomes negative as often as it 
 meets the radius o m produced, on the opposite side of 
 the point a, or diameter a B, from that in which it is first 
 drawn ; and as this takes place both when the arc a m 
 
 ~ becomes greater than ac, and when, by its further in- 
 
 * 
 
 crease, it. is greater than ac BD, it follows that the 
 tangents of all arcs in the ist and 3d quadrants ac, BD 
 are positive, or +, and that the tangents of all arcs in 
 
 -the 2d and.4th quadrants c B, D A are negative, or — 
 
 a 
 
 Tn the ist quadrant Ac, the tangent a T increases 
 _ from o till it becomes infinite, or greater than any given’ 
 quantity 5 and during the 2d quadrant CB, it decreases 
 “negatively, from an infinite quantity too. After this, it 
 is again afhir mative in the 3d quadrant B b, and increases 
 from o to infinity, as in ‘the Ist quadrant ; ; and in the 
 4th quadrant p a, it decreases from an infinitel y nega- 
 tive quantity to 0, as in the 2d quadrant. 
 
 {It is also apparent, from the nature of the figure, 
 
 U 
 
290 3 ¥: 
 
 ‘ ‘ 
 that the tangent a T increases more slowly about the 
 middle of the quadrant ac than in any other part of 
 it; and that it augments with great rapidity as the point 
 Mm approaches near c; having no limit to its increase, 
 as the sine has, but admitting of all possible degrees 
 of magnitude, from o to infinity. 
 
 4. Again, as the cotangent cs is computed from the 
 point c, in the same manner as the tangent a T is com- 
 puted from 4, it will evidently vary in its direction and 
 length like the latter, being ++ in the Ist and 3d qua- 
 drants ac, BD, and—in the 2d and 4th cB, pa. In 
 the Ist quadrant ac, it decreases from infinity to 0 ; 
 and in the 2d quadrant cB, it increases negatively 
 from o to infinity. After this, it becomes affirmative 
 in the 3d quadrant B p, and-decreases from infinity to 
 o; and in the 4th quadrant pb a, it increases negatively © 
 from o to infinity, as in the 2d quadrant. 
 
 5. In like manner, the secant becomes — as often 
 as the revolving radius 0 m passes the centre o, and 
 changes with the cosine, being + in the Ist and 4th 
 quadrants ac, Da, and — in the 2d and 3d c B, 
 BD. Inthe ist quadrant a c it increases from radius 
 to infinity ; and in the 2d quadrant c B, it decreases, 
 negatively, from an infinite quantity to radius. After 
 this, it increases negatively in the 3d quadrant 8 p from 
 radius to infinity; and in the 4th quadrant pb 4 it is 
 again afirmative, and decreases from infinity to radius, . 
 as in the 1st quadrant. 
 
 6. In the same way it may also be shown, that the 
 
291 
 
 cosecant o T changes with the sine, being +- in the Ist 
 and 2d quadrants ac, cB, and — in the 3d and 4th 
 DB, DA. In the first quadrant a c, it decreases from 
 infinity to radius, and in the 2d quadrant c B, it in- 
 creases from radius to infinity. After this it decreases 
 negatively in the 3d quadrant.B D, from infinity to ra- _ 
 dius; and in the 4th quadrant pa, it again increases 
 negatively till it becomes infinite. 
 
 7. From an inspection of the figure, it likewise ap- 
 pears, that the versed sine a P increases from o during 
 the 1st semicircle ac B, till it becomes equal to the di- 
 ameter a B, which is its utmost limit. It then decreases 
 during the 2d semicircle 8 pa, till it becomes 0; but 
 being always computed in the same direction, from a 
 towards B, it is positive; or +, in every part of the cir- 
 cumference. The same also takes place with respect 
 to the chords Am, mM, &c. each of them being com- 
 mon to two arcs, which are, together, equal to the 
 whole circle, 
 
 8. It may, also, be further observed, by way of eluci- 
 dation, that all these lines change their directions as 
 often as they become either infinite or nothing. When 
 they become infinite, their increase is at its utmost 
 limit ; after which they take a contrary direction, and 
 decrease. When they become o, their decrease is at 
 Its utmost limit; after which they again increase in a 
 contrary direction ; this changing their algebraic signs 
 whenever they pass through a state of infinity ora 
 state of nothingness. 
 
 U2 
 
’ 
 
 292 
 
 9, These changes, in the signs of the several trigo- 
 nometrical lines, may be commodiously exhibited, at 
 one view, as in the following table: » 
 
 | Istquad. 2d quad. 3dquad. 4th quad. 
 Sin and cosec -+ se oa 
 Cosandsec + £— — of 
 Tan and cot + _ -|- -~ 
 
 10. The value of these lines, at the termination of 
 each quadrant of the circle, may also be exhibited in 
 
 a similar manner, as below : | ; 
 
 Qo 90° 180° 270° 36" 
 Sin O r On: don O 
 Cos 7 O —r O r~ 
 ji Tan 0 oo O wre 0. 
 ee: Cot er) O co O or) 
 | Sec 7 co ome oo r 
 Cosec © r co —r ors) 
 
 11. Beside the limits here mentioned, it is common, 
 
 in many analytical processes, to employ, indifferently, 
 arcs of all magnitudes, whether negative or positive, or 
 
 greater or less than 360°; in which cases their sines, 
 
 cosines, &c. may be derived from the above figure, in 
 nearly the same way with the former, — | 
 Thus, if to any arc a M, there be added one or more 
 circumferences of the circle, it is evident that they will 
 terminate again exactly in the point m, and that the arc, 
 so augmented, will have the samne positive or negative 
 sine, cosine, &c. with the arc am. Whence, if c de- 
 note an entire circumference, or 360°, we shall have 
 
 _—— 
 ~ 
 
293 
 
 Sin w=sin (ctw) =sin (2c-+a2) =sin (3c+2) &c. 
 cos v=cos (c-++x) = cos (2c-+4)=cos(3c+ 2) &c. 
 tanv=tan(c+a) =tan (2c-+2x) =tan (3c+~2) &c. 
 cotx=cot (c+) =cot(2c+2) =cot (38c+2) &c. 
 SGC. | 
 
 2. Also, if two equal arcs, AM, A m be taken in 
 ic directions on the Gt ouitifevames of the circle 
 ACBD, one being considered as positive, and the other - 
 as negative, their sines, in this case, will be equal, but” 
 affected with contrary signs, while the cosines will be 
 the same for each. Whence — 
 
 Sin (—a)=—sina; cos (—a)=-+cos a 
 
 Tan (—a) =—tana; cot (—a)=—cota 
 
 Sec (—a) =+ seca; cosec (—a) =—cosec @ 
 &e. 
 
 13. Also, if « be made to denote the semicircum- 
 ference of a circle, the radius of which is r, and n be 
 either 0, or any whole number, the above table may be 
 rendered general for an arc of any magnitude what- 
 ever. ‘Thus, 
 
 . snag = O C0829 ws aT 
 ee a | ” 4, ] 
 Sin 2-7 = or Cos ass Oo 
 ti 2 
 ._ An+t? Avcag abs 2, 
 Bin ee Cos Ap ~—=—? 
 ys 9 
 ~ 4A 8 4. 3 
 Sin bs tay fd Sea Cos mh Guine ge ee o 
 2 y 
 3s A. 4: 4 
 _ Sin I" 0 rete ikea T 
 2 9 
 > 4n — ] 4n—I] : 4 
 Sin —p- w==—r | Cos —7—7= 0 (2). 
 
 (z) This table is formed by adding the circumference of the 
 circle continually to the values of the quadrants in the former 
 table, } 
 
 * 
 
% 
 
 294 
 
 And sin a=sin (7 — a) = sin (27 — a) =sin 
 (37—a) &c. 
 
 14. It is likewise evident, from what has been said, 
 that if a singlé trigonometrical line only be given, it 
 
 » can always be placed in two different quarters of the 
 circle, with its proper sign ; but if two of these lines 
 
 be given, which are not the reciprocals of each other, 
 
 there will be only one quadrant in which both their 
 ms signs will agree. 
 
 Thus, if the tangent of any arc or angle be expressed 
 by the form — , the numerator a may be considered as 
 
 representing a sine, and the denominator 0 a cosine ; 
 and the union of the two signs determines the quadrant 
 in ei the are or angle must be placed: as, for in- 
 
 stance, as Reet to the 1st quadrant, — —, to the 2d 
 
 quadrant, — ; to the 3d quadrant, and — i: = to the 4th 
 quadrant. 
 
 15. In addition to these observations, it may also be 
 
 ‘remarked, that the sine, tangent, &c. of any arc being 
 
 of the same magnitude as the cosine, cotangent, &c. of 
 its complement, and vice versa, the values of these lines 
 
 _ may be expressed in terms of each other, as follows: 
 
 Sin a = cos (90°—a), cos a = sin (90°—a) 
 tan a = cot (90°—a), cota = tan (90°—a) 
 , &e. 
 
 16. Moreover, as the sine, cosine, &c. of any arc, 
 is of the same magnitude as the sine, cosine, &c. of its 
 supplement, the same lines may be expressed, in a 
 similar‘manner, with their proper signs, thus: 
 
 rie; 
 
295 
 
 Sin a = sin (180°—a), cos a =—cos (180°—a) 
 tan @ =— tan (180°—a), cot a= —cot (180°—a) 
 &ec. . 
 
 17. Or, by substituting 90°—a for a, in each of the 
 latter forms, it will appear that the sine, cosine, &c. of 
 any arc or angle below 90°, is equal to the sine, cosine, 
 &c. of an arc or angle as much above 90° as the other 
 is less. Thus, 
 
 Sin (90°—a) =,sin (90° a); cos (90°— ay =— 
 cos (90°-+-a). 
 
 Tan (90° —a)=— tan (90°); cot (90°-+a) = 
 — cot (90°—a) 
 
 &e. 
 
 In each of these forms, however, regard must be had 
 to the change of signs, when the arc or angle a is 
 greater than 90°, which may be easily done: for since 
 cos a = sin (90°—a), it is plain, that knowing how to 
 value the sine in all possible cases, we shall be able to 
 value the cosine, and thence all the rest of the trigo- 
 nometrical lines. 
 
 OF TRIGONOMETRICAL FORMULE. 
 
 18. From the figure above given, it will be easy, by 
 means of right-angled and similar triangles, to deduce 
 the following formula, for the sine, cosine, tangent, _ 
 
 &c. of any arc or angle in terms of the rest. Thus, 
 g 
 
 ° re ee AS rtana f 
 -§in a= WV r*+—cos?a — Sa AES Fane AO 
 VW r?-+tan? a V r* + cot? a 
 ra/ secta — r@ r cosatana rcosa 
 eo ae tae cosec 4 r cot 4 
 cos asec a rtana tana cota 
 
 ———— 
 
 cosec a seca cosece 
 
296 
 F . 
 aq . r £ Gots. 
 Cas a= / rt — sir? a == Se ee ee” 
 VW +2 + tan?a . Wr? + cotta 
 Be PW cosec® ze it sin a __ sina cota 
 
 tan acota sin a cosec a 
 
 Sec @ sec a 
 
 rsina Ea "pita ti ts 
 2 FM ¥ —cos* a = / se0u—r 
 
 | 
 
 Tana = ee 
 co.d V 72—sin® a cos a 
 
 — 
 
 rv? rsina +? COS a cos aseca 
 
 AN Ebehes yiace® cos a sin a coi? a cot a 
 
 rsec a sin a coseca a 
 
 a 
 
 cosec a cota 
 Cotta Sh ee ee ee ee ee 
 tan a sin @ V/ r+—costa MW sec? a —r* 
 ry ¢ SORBET, sina _ cos asec a2 
 
 =.  cosec*ta—r? = += - = > oa 
 | sin a cos @ tan’? a tan @ 
 
 r coseca sina cosec a 
 
 | 
 
 "seca Mt tang 
 
 r cosec a r3 +2 te 
 Sec (Foo Tae ee SNE aN noe ge VW r?-Ltan® a 
 VW coseta—r? V p—sinta COSA 
 
 TY P4cotra—_Trtand _ cotatana __ sinacoseca 
 
 aaa ee 
 
 coud si @ cos a cos @ 
 ee r __ rcosec@ _ tanacoseca 
  sinacota . cota ) a r 5 
 cgi se ad OS 25 eae 
 V secta—r? SNA A 2? —COS* a tana 
 bl rd rcota tan acota r 
 = fj? + cota = —— = -—____. 
 cosa sin @ cos @ tana 
 (s $8C a." Cota $eC a0 i Cosa sec a 
 Si a Rea aee ohh r Poe SO a 
 
 19, The versed sines and chords, being seldom used, 
 are omitted in the above table; but if, in any case, they 
 should be wanted, they may be readily expressed in 
 terms of the rest, by substituting the particular values 
 of. them, given below, in any of those forms. 
 
 pee wa | — 
 
297 
 
 Vers a = r—cos a, covers @ =r— sin a, supvers a 
 = 71 -f €0s aa | 
 
 Cha= W2r(r—cosa), coch a= 27 (r— sina) 
 supch a = 27 (r +c08 a). 
 
 In all of which forms regard must be had to the 
 ’ change that takes place in the signs, when arc a is 
 greater than 90°. : 
 
 Having thus exhibited the various,expressions of the 
 sine, tangent, &c. of a single arc, it will now be pro- _ 
 _ per to show the method of obtaining the sine and co- 
 sine of the sum and difference of any two arcs, upon 
 which the greater part of the most useful properties, 
 both of these and the other trigonometrical lines, en- 
 tirely depend. 
 
 20. For this purpose, let ig eit greater of two 
 proposed ares, be denoted by a, and Bc, the less, by 4: 
 also, make B D equal to Bc; and having joined cD, oB, 
 let fall the perpendiculars c c, mn, B¥, DE, and draw 
 m H, DK parallel to the radius o a. 
 
 Then, because the chord c p is bisected in m, HC 
 will be equalto HK, ormr, and prto?rk: also, mH 
 is equal to Gn, orn E, andmn to GH. 
 
 And since c m0, Hmn areright angles, if the angle 
 
298 
 m0, which is common, be taken away, the remiain- 
 ing angle c mu will be equal to o mn, or OBF. 
 
 Hence, the triangles c H m, on m, and oF B being 
 
 similar, we shall have 
 sin a cos 4 
 
 OB 20 22 BF! wn We mao 
 r 
 ord 
 . sin 6 cos a4 
 OB308 % cm : HC Ho, = ‘ 
 But-mn+H c=Gc.H+HC=CG, which is the sine 
 
 of ac, or ofa+d,; and mm—H c=¢ H—HK=KG 
 ==peE, which is the sine of a p, or of a—b. Whence 
 
 sinacos 6 + sindcosa. 
 
 Sin (a—L) = 
 
 sin acos 6 — sindbcosa 
 fett 
 
 r 
 cos acosé 
 
 O88. > OF SS meee e — . 
 __ sinasin } 
 Oe Ba tea ae ace, all be red ti ee 
 
 r 
 
 But on—mH=Oon—GnN=OG, wacieh is the cos 
 sine of Ac, orofa+; andon+mH=on+en 
 —on+neE= ok, which is the cosine of ap, or of 
 a—b. Whence, also 
 
 Cos (a+b) = . 
 Cos (a—L) = cos ary thee ©. BE, Ls 
 
 21. From these expressions for the sine and cosine of 
 the sum and difference of any two arcs, and the values 
 of the simple arcs, given in the preceding table, the 
 formule for their tangents, cotangents, &c. may be 
 readily obtained, as follows : 
 
 cos acosé—sinasinb 
 
299 
 Tan (a+b) = 
 Cot (a+b) 
 
 r* (fan a + tan d) 
 r? + tana tan b- 
 
 cot acot & + 7? 
 
 cot d + cota 
 
 rsecasecé 
 
 Sec (ata) 
 Cosec(a-b)= 
 
 r? + tanatand 
 cosec acosec & 
 cots + cota 
 
 The formulze for the chords and versed sines are, 
 
 also, as below: 
 
 Ch (at b)= 
 
 cha supch b+ chésupcha 
 
 Zr 
 
 Vers (a+b) —< i( VA vers a supvers br WV versb supvers a)" 
 
 AR 5s 
 
 22. Hence, if 7 be made to represent the semicir- 
 
 cumference of a circle, as before, and 3 7, 7, $ 
 
 WT, &C. 
 
 be substituted for a, and a for b, in the expressions 
 above given, for the sine and cosine of the sum and 
 
 difference of any two arcs, we shall have 
 
 Sin (7-+a) =-+ cos a 
 Cos (4 7-+a)=— sin a 
 
 Sin (z + a)=— sina 
 
 Cos(7-+ a) =— cos a 
 Sin ($%7-+a) =— cos a 
 Cos (27+a) —-+- sin a@ 
 Sin (27-+a) =+ sina 
 Cos(27-+-a) =-+ cos a 
 
 23. Or, if n be zero; 
 whole number whatever, 
 
 or 
 
 Sin (47—a)=-+ cos a 
 
 Cos (4 e—a)=-+ sin a 
 
 Sin (a — a) =+ sina 
 
 Cos (7 — a) =— cosa 
 Sin ($7—a) =— cos a 
 Cos(37—a) =— sina 
 Sin (2 7—a) =— sina 
 
 Cos (2 r—a) =-bcos a 
 
 any positive or negative 
 
 the same formule may be 
 
 rendered more general, as follows : 
 
300 
 
 in (- fet or tog) cos @ | Sin (Hea) = +cos @ 
 EES: sIn a Gés (rita a)=-+ sin a 
 Sin (- tr ha)=— sin @ Sin (“tp a = + sin a 
 Cos (3 bik a-+a)=—cos a|Cos (7S pa) =—cosa 
 Sin (Sr ea) <= —cos a/Sin (Ae ea) =—cos @ 
 Cos (ta) = 4 sin a Cos (tz —a ia a | 
 Sin gamer + a)=-+ sin @{Sin eaeauiace =— sina 3 
 Cos (te ta) = = -++-cos a| Cos (ae ae Ol eos Oe 3 
 Sin lis +e ¥ cos a | Cos (- ee ta) 4 sing 
 
 24. }'rom the expressions for the sine, cosine, &c. 
 of the sum of two arcs, we may also easily deduce the 
 following formule for the sine, cosine, &c. of the 
 double arc, by barely taking 6 equal to a, and then 
 substituting the values of the lines, thus obtained, in 
 terms of the rest. ‘Thus, 2 
 
 \ Q2sinacosa 2 sin? a 2 cos? a 2r*tana 
 Sin 2a = = ha oie a 
 r tana cot a r+ tan"@ 
 21° reo 4 2rsiva » -2rtana 
  tana+cot a. “egecot*a  aeseca 9 ae ceuems 
 2 rt cota 
 UT HLL x ch 4a 
 cosec? a 2 
 cos? a — sin? a r— 9 sin* a 2co%a— 7% | 
 Cos2 a= ——_——_- = en 
 r r . tf 
 r(r?—tan*a) __ r (cot? a—7*) r (cot a —- tan a) 
 r?+tan? a coat Je # cot @-+tanae 
 r(2r?—sec*a) rr (cosec?a—2Zr*) a 
 = -—__—_+ = + ~_—_ =4 supch4a@ % 
 sec* a cosec? a ~ bi 
 
 ww R 
 jay 
 
 ae 
 
301 
 
 Tan 2a: 2 rt tana’ ‘ 27% Qrcosasina frsinacosa 
 “> 7@—tan*a —cota—tana r’—2sin’a 2 cov ar 
 
 2r? cota 2r* tana Qrcot a 
 
 9° 
 
 = er: fa eo) 2a oe 
 col? a—r?” 9 rt?—sec’a cosec* a— 2 r* 
 
 cota? “r'—tan*a cota—tana r({r°—2sin?a) 
 Cot2a = , all 
 
 9cota i.2tana ~ 2 — Qsin acosa 
 r (2cos*a—r’) Qre—seca __ cosec?a—Qr? 
 ~~ 2$in a cos @ : Ztan g 2 cota 
 7 “a 42 sg 3 
 5 r sec’ a r cosec’ a r 
 ec 2a= 2 goers a 2 Th ie et Slat care 
 2r— seca cosecta — Zr ro—Z sina 
 ri r{r?+ tan? a) r (cot? a + r*) 
 ats re ee eee 
 Qcosa—r 8 #—tan’?a cova—?r 
 
 r (cota + tana) 
 (hw eota—tana 
 
 7 P? 4+ tag@a cot a + tan a 
 COSec Bapent ei ace . = 
 j 2 sina.cos@ Z tana Z 
 “7 4+ cot? a Sec a cosec? 4 sec* a 
 5 Teopia i) \ nian a Beota' . 2 tana 
 sec a@.cosec a r cosec a 
  aenel Ds Fe CE EES brrze ————_— , & 
 ne or 2cos a 
 
 25. ‘The versed sines and chords of the double arcs 
 may also be expressed in terms of the rest, by substi, 
 tuting the following particular values of them in any of 
 the above forms. | 
 
 2sin* a r?— 2sinacosa 
 
 Vers 2a = —, covers 2 a= Rare yer' pment e ai 
 7 
 __ 2sin? (45°—a) -2 cos? a 
 = er UpVels 2g ————. 
 r 
 
 Ch 2a=2sina, coch 2a=2 sin (45°—a), supch 
 2a= 2 cos a. 
 
 26. In like manner, if 4 a be substituted for a, in 
 each of the above expressions, for the sine, cosine, &c. 
 of the double arc, we shall obtain the following for- 
 
302 
 
 mulz for the sine, cosine, &c. of the single arc, in 
 terms of the sine, cosine, &c. of the half arc. 
 
 Sin Qsindacosta_ Zsin*ta Qcostta  2r*tan 34 
 ne= —— = well Rice 
 r tae COW a one 4+ tan*ia 
 A a 2r* cotta @rsinga  2r*tanta 
 ae 74 ee. ok en 
 tanda+cotha r@+cot*ia secza sec*ia4 
 2] $1 
 _ 2@rcot4a_ Btanacos oo Qrcosta_ ha 
 aieiend Ce AE hele AN —Sc,| oo =2 6 a. 
 cosec*4a r - cosec da 
 Chan cos*ta—sin?ta . r*—2sin®ta 2cos*4 a—?r 
 a r i wt sl r 
 
 r(r°—tan*4a) __r(eot?Za—r*) _r(cot} a—tanga) 
 
 r’?+tan? ta cot ta+r?  cotia+tanta 
 r (2r 4a) __r(cosec*ta—2r° , me Loy Be 4 
 scOhi a. MMGOSe tae P 
 NS 2rtania Q 2 _ Qrsintacosha 
 Tana == 2S ee 
 r—tan’ta cot }a—tanta@ r?=-2 sin? ¢ a 
 Zrsing acosta 2r>cotia 2r tanga 
 Zco*ita—r — co®?La—r Qr—set sa 
 . Zr cotia 
 >), cosec?. i ga—Dir?” 
 2I7 2a 2s Ue cay Bases i 
 Cc ie i cov 44 ce (OM 3 tan, 5 2p Cot Sa tan 3 @ 
 t= 3 ae cine 
 Zcotia Ztanta . 2 
 r(r°—2 sin? $a) fl #2 cos" Aigner 2aimegec? Sa 
 .. Qsindacosda Me @snigcmiea ” tanta 
 cosec? t a—2r? 
 2 cot 5 me es 
 rsecta ' rcosec? t re 
 Se a= 53 Se oe So 
 2r—sec? sa cosec? 3 a—Zr" v—2snrta 
 ? r(r’?+tan? ta) a (cot? $ a+r) 
 =. Danek dubs? ie Eee ee cei ae 
 2 cos t+ a—r r—-tan’ ga cot? 4 ar" 
 
 r (cot 2 ap a0 ae 
 
 cot 3 .a—tan $ aa 
 
 a 
 
308 
 
 Cosec a == e _ rt+tan?}a cotza+tanie 
 QsinJacosta 2tanja ) 
 
 r+cotia  rsecka  cosec?ia sec* 
 
 I I 
 ae ee pata 2° 
 Sled fe eeesin + 2Zcotha 2tan la 
 
 Sec = sec 3ac a EOSet 4 GA. r cose’ =a 
 
 me eee, 2cos £ a 
 . 2 sin? ia 2". 
 Also, versa = ——=-, covers a==—sin’ (45°— La) 
 
 _2cos* Za 
 Ch ae sin a, coch a=2 sin (45°—4 a), supch a 
 2 cos + 
 
 eet: And ‘ finding the values of the sine, cosine, &c. 
 of 3 ain the most commodious of these latter equa- 
 tions, we shall obtain the sine, cosine, &c. of the half 
 arc in terms of the sine, cosine, &c. of the whole arc. 
 
 Thus, 
 : a Pee aus, r— COS a4 
 Sin e a=) 4 WV +r sina 5 ‘Por sive TA eS 
 sec a—r r vers a 
 = peed. ea | 
 r ey en ae 
 BS secs ag 2 sec a § 2 2 
 
 as cos a 
 
 Gos at Ver sina y Vr—rsina = 7 f— 
 
 sec d-+-r r supv a 1 
 =r f/f —-—_—- = — =s5 SU ch a. 
 vs 2 sec a yf 2 P 
 rsina r?—r cosa r— cosa 
 Tania= —— = —————- = TV 
 ; r+ cosa sin a r+ cos @ 
 seCa—r vers @ 
 =f 4 TH = CSCC A—COTt A. 
 VE a+r v 2r—versa 
 -rsinag r+rcos a r-+ cos 4 
 ms r—cosa sina r— COS @ 
 seca-+r 2r—vers a 
 =f Se 7, — = cosec a-cot a. 
 a=? MG vers @ 3 
 Qtan a 2 sec a 
 Sec 4a=r =Prff———_ =r V¥—— 
 ima Parag V Tees o 
 Qr J 
 
 isi VEE ayhie 
 
 versa s supva 
 
; aj. 
 
 Dr | : 2tana 10 She 
 a nt al em all Ys — sr —— - 
 om sa pi COS " tan a—sina ol sec g—f' 
 or 
 = PER Tr ae ie ° 
 Va a. 2r—supv a 
 
 ae vers 4 @ =r—cos}a, coversfa=r— 
 sin 4 a, supvers $ @ = r — cos 
 
 ena Vv 2r(r—cos $a), coch 3 
 supch 4 = WV 2r(r-+-cos £ a). 
 
 2 = tole 
 
 = Wee (Peron nia)s 
 
 28. The formules given in art. 20 for the sine, co- 
 sine, &c. of the sum or difference of any two arcs, also 
 furnish a number of other useful expressions ; among 
 which the following may serve to change the product 
 of two or more sines, cosines, &c. into their sums and 
 differences, or vice versa. 
 
 Nes fie i sa : : 
 
 Sin a sin 6==37r cos (a—b) — 3 r cos (a+) 
 
 . (bynes | » % ¥ 
 Cosacosb = $rcos(a—b) + 4 rcos (a+b) 
 ie 
 2 
 ae 
 2! 
 
 Sin a cost = 
 
 rsin(a+-b) + +r sin (a—b) 
 * sin (a+ 6) —'trsin(a—b), 
 
 Sin 6 cosa = 
 
 To these we may also subjoin the additional forms, 
 9 608 (ats)) —cos (a+b) 
 cos (a+5) +00: ( 5 (ath) 
 
 cos (a+b) -+cos (ab) 
 cos (a2) —cos (a+6) 
 , sin (a+4)+:in (a—b) 
 
 Tan a cot b — 22 sin (4-4) —sin (a—J) 
 
 ‘an 0 tan b= #7 
 
 Cot acot 6 = 7r2 
 
 sin (a+4)—sin (a—b) 
 sin sin (a+) +sin (a—d) 
 cos’ (a—b)—cos? (a+b) 
 
 Sina? Si) Uae a 
 4 cos a cos 6 *: 
 
 Tan J cot a= 7? 
 
 Tan a tan b= ua Pycosila—4) —e0s (244) 
 _2cosacos d Js 
 
305 ‘ 
 
 sin? 4 (a-+b)—sin? $ (a—b) 
 Sin a sin’ = ' or 
 
 cos® 3. (a4 b)—cos? 1 (a43) 
 cos® $ (a4 b)—sin? $ (a+b) 
 Cos @ cos b = ' 
 
 or 
 
 cos? 4 (a+b)—sin? £ (a—b) 
 29. And if in these formule: there be substituted 
 5 (a+) for a, and 4 (a—b) for b, we shall obtain 
 the following ones, which are often employed in trigo= 
 nometrical computations for reducing the sum or diffe. 
 rence of two factors to their product. 
 
 Sin at: sin 6 = = sin Z (a-+6) cos $ (a—b) 
 Sin a—sin b = “sin  (a—b) cos } (a+b). 
 Cos a-+-cos b == cos: 5 (a+b) cos 4 (a—b) 
 
 Cos b—cosa ==sin eae (a—b) 
 
 Fin tn Be 4 a Plt 
 tm en bain Fa _ le 
 Cot a--cot bias was a ae ums Suk] 
 Cathie ey oat = le 
 Cnt wan ba it 4 Ek eed 
 coramend ist tt tele 
 
 And if 96°—4 be substituted for b, we shall have 
 the following formule, which will be found applicable 
 x 
 
306 
 
 to somite particular cases of spherical trigonometry, 
 where the common rules would require two analogies. 
 
 a+s 
 pa 45" ) 
 
 2. a 
 Sin a+cos b = ~ sin. (45°--25- 
 
 2 a—b a+é 
 = —cos (45° —5=) cos (45° = ——) 
 
 ng: Ra d—b, . ak . 
 Sin a—cos 6 = ey (Aes sin ( ee 45°) 
 . b 
 =" sin (45°— ey; sin (A irl ! 
 
 2 
 Sin b--cosa = rate OS PT °) cos (44 145°) 
 aS 
 
 mics Ce Nempests 
 
 2. jamb ve 
 Sin b—cos a =~ sin (“ 3 +4.5°) sin eee, mh gc 
 
 b b 
 =* cos (A5° “S~) cos (45° oe cm ) 
 
 pent ‘ sin a tan a 
 30. Also, since = 
 cos a 
 
 — —"_| we shall ob- 
 Lf a 
 
 tain, by division, the following expressions, which will 
 be found convenient in many logarithmic computations. 
 
 Sina+sin 6 . sin xy (2+d) cos $(a—b) _ tanZ 3 (a+b) 
 sin a—sin bia: COS es i 1 (a+b) sin 4 (apy! tan 2 L (a—b) 
 Cosatcosh _ cos (a+4) cos$(a—b) _— cat ¥ (a+d) 
 cos b—cosa sin % (a+) sin$g (a—b) ~~ tan I (a—d) 
 Sina+sin b __ cosé—cosa __ tan $ (a+4) 
 cosa+cosé, sn a—sinb’ a 
 "Sin a+sinb  cosa-+cosd cot £ cot $ (a— —4) 
 cos b—cos a sing—sin b r 
 - Sin a—sin b __. 608, 4—cos a __ tani (a—b) 
 7 cosa+cosé sin a+sin d bea 
 
 Sina—sinb __ cosa+cosé _ cot 5 cot 3 (a5) 
 cs Po tsa = | sin sin a+ sin yim r 
 
"Tan Tan @-+tan b 
 tanag—tanb 
 Tan a+-tan b 
 
 cota —cotb 
 
 Tan a+tcot b 
 cota+tanb 
 
 Tan a+tan 6 
 
 cota—tand 
 
 Cot a+cotb 
 
 cota—tange 
 Tan a—tan b 
 cota +tanb” 
 
 ‘Coth— cota 
 tan 2 + coia~ 
 
 Sin a—cos b 
 
 Sin a+cosé 
 ie tan (45 
 
 __ cot a+cot b 
 
 tan (BBP aye 
 
 307 
 
 feat __ sn (a+b) 
 cotb—cota sin (a—b) 
 ___ tan tan a—tan 6 _. tana tan b 
 cot b—cota r? 
 cotb—tana __ tan a cot d 
 cota—tan b r 
 __ tang tan (a+) de tan (a+4) 
 7? Ton Ceti 
 _. cothtan(a +4) __ tan (¢+4) 
 ait tan d 
 tanatan (a—d) __ tan (u—3) 
 r? mega 
 cot é tan (a —b) fia tan (a—5) 
 7 tan 5 
 a—b 
 cot (—5 — 45°) 
 D 
 ee cot (2 45°) 
 
 To these may also be subjoined the additional forms, 
 
 i / © } 3 (4—- 
 Sin (44-2) cos $ (a+b £), or si ascent l(a 6)sin(a+) 
 sinag+4-sind cos} (a— cos 4 (a-s) 
 
 ; ol Leg ees 
 Daa (a5) sin (a tee oie ie sty Slt (4 ~2)sin(a 4 5) 
 sina—sin 6 sin $ (a—b) sin} (a+) 
 
 31. And if one of the expressions in art. 29, be 
 multiplied by the other, we shall obtain, after some 
 simple substitutions for the two last cases, the follow- 
 ing formulz for the difference of the squares of the 
 sines, tangents, &c. of any two arcs. 
 sin (a+6) sin (a—6) 
 sin (a+b) sin (a—b) 
 cos (a+b) cos (a—b) 
 
 r* sin (a+4) sin (2-8) | 
 cos* 4 cos” } 
 x 2 
 
 — 
 
 Sin? a= sin? b 
 
 —— 
 — 
 
 Cos’ £ — cos? a 
 
 —— 
 —— 
 ¥ 
 
 Cos? Ga sin” b 
 
 Tan* @—. tan* b= 
 
308 
 
 __rtsin (a+) sin (ab) 
 sint?a sinh 
 
 r*cos (a+4) cos (a—é 
 Cot? a — cot? b = re ye pub iss 
 sin? a cos? é 
 cos (a%5) sec asec } 
 
 r 
 
 Cot? b — cot?a 
 
 yr ttanatand = 
 
 1% sail taal tat iiee— cos (arent aroes 
 
 $2. Also, if 45° be put for a, and a for b, in the 
 
 formule for the sine, tangent, &c, of the sum and 
 difference of any two arcs, (art. 20, 21,) we shall have, 
 after a few simple substitutions and reductions, the fol- 
 
 lowing expressions for the sine, tangent, &c. of 45° 
 
 + 
 
 a, 45° +4a, and 30° +a. 
 ; cosa -& sing 
 PF Ibi 
 cosa— sina 
 a2 
 r (r+sina) 
 ai ticles 
 
 Sin (45°-+a) = cos (45°—a) = 
 
 Sin (45°—a) = cos (45°--+a)= 
 
 Sin (45°-+3 a) = cos (45°—-d.a) = 
 
 A oh (r—sin 2) 
 
 Sin (45°—4a) = cos (45°+ia)= 
 
 y+sina 
 r—sina 
 
 Tan (45°-+-2 a) = cot (45°—2 a)= rv 
 
 r—sina 
 r+sin 2 
 
 Tan (45°—$a) = cot (45°42 aj=ry 
 
 Tan (45° +4 aye ete. reosa__rcote 
 
 cos a r—sina@ coseca—r 
 r(r—sina rcosa rcota 
 Tan (45° —t 4 a)-= = Ars ne. Se ee 
 cosa 7¥+sing cosec a+ 7 
 4 r(r--tan a) r-esin2a 
 Tach (ae eee ey Se ae 
 ( ok ) r—tan a r—sin 2a 
 O r(r—tan a) r—sin 2a 
 Tan (45° a) = SS = Vee 
 r+-iana r+-sin 2: 
 r(r—tana) __ r(cota—r) 
 
 Cot (45 iin. = 
 
 geetang r+cote 
 
309 
 
 Cot (45°—a) = r(r+tana) _ r(cota+r) 
 
 r—tana@ = cota—r 
 
 . ° a 6 __ cosa + sin a,f3 
 Sin (30°+-a) = cos (60°—a) = Sa a 
 
 Sin (30°—a) = cos (60°-++-a) = Sk Ete he 
 
 33. From art. 38 and 40, we may also readily ob- 
 tain the formule given below, which have been found 
 of considerable use in the computation of the common 
 
 trizonometrical tables. 
 
 Bin (n +1)a me (el Sibi (n—1) @ 
 
 eae ey att Pacnaactgs aheoeio dD 2 
 
 r 
 Or, 
 rsin(n—2)a+2sinacos(z—1)a 
 r 
 
 Sin na= 
 
 r cos (n—2) a—2 sinasin (n—1) a 
 aA Ot EG oe ang oO 2 
 
 Cosna = 
 
 84. The following expressions may likewise be ob- 
 tained in nearly a similar way with the former; but, 
 being of less use in their application, they are here 
 given separately. : 
 
 Sin(45°+ 2) —sin(45°—a) _ cos (45° —a)—cos (45°+ a) 
 f2 V2 
 
 i ek Sin(30°+ a) —sin(30°—2a) __cos(60%-a)—cos (60°) 
 
 Sin 2. 
 
 J3 vv? 
 pcs Rt gM A cn ei Se 
 r?-f tan? 45° — 40): tan(4i°4+2a) 4% an( sha) 
 
 (4) Here, if a be 1°, andz be put = 2, 3, 4, &c. successively, 
 we shall have 
 Sin 2°= sin 6°-+-2 sin 1° cos 1° 
 Sin 3°= sin 1°+2 sin 1° cos 2° 
 Sin 4°= sin 2°42 sin 1° cos 3° &c. 
 And the cosines of 2°, 3°, 4°, &c. may be expressed in a similar 
 manner, 
 
} i Qr* Det s 
 Cosas yj 1g) -+tan(45°—La) — cot(45° 34) + (cots 4a) 30) 
 
 Qcos(45°4 4a)cos(45°-1a)  2sin(45°—}h.a)sin(45° +4) 2) 
 a= OS a 
 
 r ) r 
 tan(45°-4-4a)-tan( 45° ta) cot(45°~3a)}—cot(45°+4 + Ja) 
 2attess Pe ek ona ne err eM Te! 
 
 35. Finally, to these may be added the equations 
 given below, which have been found of great service 
 in the computation of the sines, tangents, &c. of the 
 common trigonometrical tables. 
 
 Sin (30°-+-a)=cos(60°—a)= cos a—sin (30°—a) 
 
 Cos (30°-+-a)=sin (60°—a) = cos ( 30° —a)—sina 
 
 Sin (60°+a)=cos(30°—a)= sin (60°—a)+sin a 
 
 Cos (60°--a)=sin (30°—a)=cos a—cos (60—a) 
 Z 
 
 cot (30°4+-44 )—tan(30°4 4a) 
 2 
 
 Tan (80°--a)=cot (60°—a)= 
 
 Tan(30°—a@)=cot (60°+-a)= 
 
 Tan (45°7-+ a) = tan (45°—a) + 2 tan 2 a=cot 
 '(45°+a) + 2tan 2a. 
 
 Sec @ = tan Coats tan (45° — iqa)=tan.(45°+4 a) 
 — tana, 
 
 86. (J) Having thus given a variety of the most 
 simple and useful expressions for the sines, tangents, 
 &c. of the sum or difference of any two arcs, it may 
 not be improper to subjoin the following formule for 
 
 (7) Formule of this kind may be, obviously, multiplied with- 
 outend ; butit is conceived that the collection here given will be 
 found to be more complete and methodical than any which has 
 hitherto appeared. ’ : 
 
311 
 
 the arcs themselves, i in which radius i is supposed to be 
 unity, or 1. 
 
 Arc tan # +- arc tan y = arc tan = . 
 mes 
 
 re tan v — arc tan ate:tan | 
 ig 3 arc tan y = rere 
 
 xy—1 
 Arc cot x arc cot y = are cot ~—— 
 ig J w+y 
 
 1 
 Arc cot # — arccot y = arc cot pa! 
 Mr a 
 2 1—x* 
 Arc tan = 4 arc tan = = 2 arc cot —— 
 2 1—x < 2 x 
 
 5 oes 
 
 1 r 
 : 1 2 — 1 Pecans 
 Arc cot # = gare cot->— = g arc kant 31 
 
 1 ae 
 Arc sec 7 = arc cos - 5 arc cosec ® = arc sin ‘i 
 - x 
 
 : x See 
 - Arc versv=2arc sin / 53 are covers x= 90° + Zare 
 
 sin V 55 arc chord v=2 are sin =, 
 Also, 
 Arc to tani + arc to tan 4 = arc 45° 
 Arc totan + + Qarc totan2 = arc 45°. 
 37. It may also be remarked, that, by means of the 
 formulze here given, and the known values of the sines 
 of 30°, 45°, 60° and 90°, it is easy to obtain the values 
 of the sines, cosines, &c. of a great variety of other 
 arcs, in surd numbers; the most simple and commo- 
 dious of which are the following : 
 SiO. =" tos 90" v==.0 
 
 s ; - Views t 
 Sin TE — cos 82i° niga! Gisela v6 
 
 Sin, 9°. (ia cpg Sd ince. (4g es os WAG lines 
 
 Sin 1jt° = .COs 782° aed 
 
g12 
 ‘ . ; rd 
 
 Sin 15° = cos75° = ‘ (/6— 2) 
 Sin 18° = cos 72° == 5 (—14+¥5) 
 
 ° ¢ _—_—--— 
 Sin 223° = cos 672° = 5 VIP 
 
 : a: Mt Me he ik ad ———- 
 sin 27° = cos 63° = q V54f5-Z V3—W/8 
 
 Sin 30°. == cos 60° = 
 
 Sin 333° == cos 564° VW 9. pig, 
 
 Sin 36° = cos 54° = — / i0—2,75 
 
 44 /2— V6 
 
 Sin, 874° == cos 522° == 5 
 
 Sin 45° == cos 45° =< 
 
 x 
 bo 
 
 Ate /6— 4/2 
 v tw 
 
 (14+¥5) 
 
 Sin 523° = cos 374° = 
 Sin 54° = cos 26° = 
 
 Sin 564° = cos 333° => / 24 / 2,2 
 
 a 
 0 
 
 Sin 60° cos 30° = 
 
 I 
 
 Sin 63° = cos 27° = 5 V5¢f8-+ TV 3— 5 
 
 Sin 672° = cos 222° = = W/ 94,72 
 
 Sin 72° -=.cos 18° = — /1042,/5 
 
 S BLY tops ls cols wo] x wl eofs eOofs cola Bf s wopr ofr 
 
 Sin 75° == cos 15° = | (V6+ sf 2) 
 
 . Bitar Al MRP et 
 Sin 78} = cos lle = > 704/24 72 
 
313. 
 
 P dani’ 0 of yee tp 
 Sin 81° = cos ef = ZVv3+/5 +3 VY 5— 45 
 
 ‘ ee oT A MI+ VS 
 Sin 825° = cos 73° = ZG V¥— yp 
 
 Sin 90° = cosh 0. =r (mes 
 
 Similar surd expressions may also be readily found 
 for the tangents, secants, &c. of the same arcs, by 
 
 : rsin r cos 
 means of the equations, tan = » cot = ——} 
 cos sim 
 
 r r 
 sec = —, and cosec = --. 
 
 COS Sith 
 
 88. Again, if the arc b be taken equal to 2a, 3 a, 
 
 4a, &c. successively, the sine of any multiple of a sin- 
 gle arc may be readily determined, by means of the 
 known formula for the'sine of the sum of two arcs, 
 given in art. 20. 
 
 (m) Since V3 4 75 =i V104+1 /2,andV73— y5=4 
 of 10-1 4/2, it is plain, from the above table, that, considering 
 the ,/2, ./5, and ,/10, as known quantities, there will be only 
 four extractions of the square root required to obtain the values 
 of the sines and cosines of all the arcs which are multiples of 9°. 
 
 It also appears, from the same table, that sin (60 + 7) = sin 
 
 (60 — x) + sin x; sin 54° — sin 18° = sin 90° — sin 30° = oi 
 and sin 81° + sin 27° + sin 9° = sin 45° +4 sin63°. Or, more 
 
 generally, 
 Sin (18°+2z) + sin (18°—x) = 2Zsin 18% cos ¢ 
 Sin (64°42) +4 sin (54°—r) = 2 sin 54° cos # 
 
 And because sin §4°—sin 16° = 4, and cos #= sin (90°—z) 
 we shall have, sin (54°42) 4 sin (54° — x) — sin (18° + 2)— 
 sin (18°—z) = sin (90°—z). Which formule may serve as very 
 useful checks in the calculation of trigonometrical tables. 
 
314 
 
 (n) Thus, if ==2 a, and radius =1, we shall have 
 sin 3a = sina cos2a-+sin 2acosa; but'sna= 
 
 del 
 
 — (art. 24),and cos 2a = 2cos' a — 13 whence 
 
 sin 
 26 se a— 1) =sin 2a cosa— 
 sin @ cos oS oom Pe ) = sin 2a cos 
 
 sin2a 
 2cosa 
 
 substituted in the first equation, gives 
 
 = sin 9 acosa— sina; which value, being 
 
 Sin 3a = 2cosasin 2a — sinu. 
 
 And, by following the same mode of investigation, 
 we shall readily obtain the sines of the arcs 4a, 5a, 
 &c. in terms of the sines and cosines of the inferior 
 arcs, as below: 
 
 Sity aus sin @ 
 Sin 2a = 2cosasina 
  Sin3a = Z2cosasinZa — sing 
 Sin 4a = 2cosasin3Sa —sin2a 
 Sin Sa = 2cosasn4a — sn 3a, &c. 
 
 Which series, it is obvious, may be continued at 
 pleasure, without any new calculation, by multiplying 
 twice cos a by the preceding sine, and then subtracting 
 the sine preceding the last, for the next following one. 
 
 (z) When the radius, in any ‘trigonometrical expression, is 
 denoted by 1, instead of r, the latter symbol may be readily in- 
 troduced into the equation, by joining either the simple letter r, 
 or its powers, to such of the factors as will render them all ho- 
 mogeneous, or of the same dimensions. Thus, if radius = 1, 
 
 and 4:sin’ a == 3 sin?a — cosa + 2, in which the highest term is 
 of 3 dimensions, the equation, by introducing r, will become 
 4 sin? a= 3.r sin? a—r? cosa +27°. 
 
815 
 
 : Thus, 
 Sinn a = 2cos asin (n—1) a—sin (n—2) a 
 Or, 
 
 Sin 2 a = sin(n—2) a+2 sin a cos (n—1) a 
 
 And by substituting the values of cos a, sin 2a, sin 
 3a, &c. as taken in terms of the sine and cosine of the 
 single arc, the same series may be easily varied, as 
 follows: — 
 
 Sin @= sina 
 Sin 2a = 2 sin aW1—sin’a 
 Sin 3a = 3sina — 4 sin’ a@ | 
 Sin 4a = (4sina — 8 sin’ a) Y1—sin’a 
 Sn 5a = 5sin a — 20sini a + 16 sin’ a 
 
 Sin 6@ = (6 sina—32 sin'a +32sin* a) /1—sin'a 
 Sn 7a = 7sina — 56sin}a +112 sin’ a—64sin’ g 
 
 &c. 
 Or, 
 
 Sm @==V 1—cos’a 
 Sin 2a==2 cos @ /1—cos?a 
 Sin 3 a==(4 cos’ a—1) fT—cos* a 
 Sin 4. a==(8 cos? a—4 cos a) f 1—cosa 
 Sin 5 a=(16 cos’ a— 12 cos’ a+1)V I—cos’a 
 Sin 6 a==(32 cos’ a—32 cos’ a+6 cos a) / 1—costa 
 Sin 7 peal G4 cos’a4— 80.cos*a-+- 24cos’a—1) 4/ Icosta 
 
 &c. 
 
 39. Hence, observing the law of the coefficients 
 and exponents of the several terms in the above ex- 
 pressions, we may readily obtain the following general 
 formula, for the sine of any multiple of an arc, in 
 terms of the sine of the simple arc. ‘Thus, 
 
316 
 
 . 2 2] 
 Sin na=n sin a — po ; ) sin’ a Si 
 
 n(1? —1) (n°—S*) — 5%) (n*® ir 8 ti 
 9,3.4.5°6.7 7 
 
 a (n*—} ) (it 32) 
 2.3.4.5 14> 
 n—~l . nel 
 sin a 
 
 5 
 
 sin ad— -2 
 
 Be 
 
 nem nm 
 Sinna==n sina — 555 =n) sin’a— > = (8) 5 sin*a 
 
 7? — 5? ° 4 Se =) <2 
 ee (c)sin’a— +, (D) sin’a, &c. 
 
 Where a, 8, c, &c. are the preceding terms with 
 their proper signs; and if 2 be an odd number, the 
 series will terminate; but otherwise not. 
 
 Again, if these two formule: be divided by the ex- 
 panded value of Nee ee fe ye ae ~ sits &c. 
 and then multiplied by 4/;2—sin? a, we shall have 
 n( 1? —2*)(n? —~#) 
 
 : bi gral Uae n( n°. nll ‘ 
 Sinna = {—sina— ear a Wap ee 
 
 “2.3 5 
 
 n(n®@—2*) (7?—4°) (n?—6%) . 
 TR RSET A Borde oe 
 
 n—l eee Abie Werte 
 n at VW 7? — sin®s 
 
 sin’ a— 
 
 Or, 
 
 Rime GZ 
 
 © . id 
 Sinn a= a2 aS sa ER sin'a— = _(B)sin°a 
 
 yt — 5? i} 
 
 6.7 72 ‘(c) Si a, &. } V resin? ae 
 
 In which case, iat series will terminate when nis an 
 
 even number. 
 And if - be substituted for x,-in each of the two 
 
 formulz in the first part of this article, we shall have 
 the following series for the sine of any part or sub- 
 multiple of i arc a. 
 
317 
 
 Pe m(m—n) m(m? — Re ate) 
 Sl tne he a Po sin a+ 2345. a re 
 m(m°—2") (m°—3" n°) (m'—5* n*) 
 nS in’ a, 
 sin’ a— ea ie sin’ a, &c. 
 ath 
 Sina = ste Fae tithes TL 4) sin? @ + ———— Btn! — mi 
 vee < Aye 
 
 (B) sin’ my iy saint gee : ™ (0) sin?a, &c. (0) 
 
 40. Moreover, if we still suppose the arc 6 succes- 
 sively equal to a, 2a, 3a, 4a, &c. and proceed in a 
 similar manner with the formula for the cosine of the 
 sum of two arcs, given in art. 20, we shall obtain the 
 following expressions: 
 
 Cosa = cosa 
 
 Cos 2a= 2cosacosa— 1 
 
 Cos 3 a = 2cos a cos 2a — cosa 
 Cos 4 a@ = 2cosacos3a—cos2a 
 
 Cos 5 a = 2cosacos 4a — cos 3a, &c. 
 
 Where it is plain, as*in the former table of sines, 
 that multiplying any cosine by 2 cosa, and then sub- 
 tracting the next preceding cosine, we shall obtain the 
 next following one. Thus, 
 
 Cos na = 2cosacos (n—1) @ — cos (n—2) a. 
 
 Or 
 
 Cos n a = cos(n—2) a—2 sin asin (n—1) a. 
 
 And by substituting the preceding values of cos 2a, 
 cos 3a, &c. in each of the succeeding forms of the 
 above table, the same series may be varied as follows: 
 
 (o) The formule here given, for the sine and cosine of any are 
 or its multiple, are equally applicable to the chord of the arc, and 
 the chord of its supplement, using the diameter, or Yr, instead of. 
 
Cos a= cosa 
 
 Cos 2a = 2cos a—1 
 
 Cos 3a = 4cos’ a—3 cos a 
 
 Cos 4 @ = 8 cos* a—8 cos? a1 
 
 Cos 5 a = 16 cos’ a—20 cos’ a+5 cos a 
 
 Cos 6a = 82 cos’ a—48 cos‘ a+18 cos’ a—1 
 
 Cos 7 a = 64 cos’ a—112 cos’ a+56 cos’? a—7 cos tt, 
 &c. Or, | 
 
 Cos ya 4/1 4sin2z 
 
 Cos 2 a=1—2sin? a 
 
 Cos 3 a=(1—4 sin’ a) / 1—sin* a 
 
 Cos 4 a=1—8 sin’ a+8 sin’ a 4 
 
 Cos 5 a=(1—12 sin* a+ 16 sin’ a) /1—sin?a 
 
 Cos 6 a=1—18 sin* a+48 sin* a—32 sim’ a 
 
 Cos 7a=(1—24sin’a-+80sin® a—64sin® a) / 1—sita 
 &c. | 
 41. Hence, observing the law of the coeffictents and 
 
 exponents of the several terms of these expressions, 
 
 we shall obtain the following general formule for the 
 
 cosine of any multiple of an arc, in terms of the co 
 
 “ 
 
 sine of the simple arc. ‘Thus, 
 
 Ba! gat & . n(n—~3) 
 Cos n a= — {cos dar ACOs ¥y1 4a ai 
 4 nh li (i— 2) 6 n(n—5){n—6)\(n—7) 
 rcoS a reso 249 eos" ne gluta Na? 5 AMOR OWI 0. P 
 “+ 
 cos Way) Bec. 
 Or, 
 n ng 2 oy a 
 Cos alia a car see pm ago Beh y (n—1) (n-5) r> 
 Ox? * COs*a 2.240874 3.22 n= costa 
 
 He ati cota D, &¢. 
 
 4.2% (n—4) cos* 2 
 
319 
 
 e ° j a 1 ? 
 Which series must be continued to"s— terms, when 
 
 nis an odd number, and to —— * when i itis even; but 
 
 a 
 the same series will be ahi true when 2 is any 
 
 fractional number. 
 
 x r tan a tan2a 
 42. In like manner, since tan 3a = samt Fa 
 l—tanatanZe@ 
 
 2 tan im a 
 
 2g = = 
 (art. 21), and tan 2a=7——, 
 
 - (art. 24), if this latter 
 
 expression be substituted in the sas equation, and 
 the whole be reduced to its most simple terms, we 
 shall have | 
 
 3 tan a—tan® a a 
 1—3tan?a 
 
 And by following the same mode of investigation, 
 we shall readily obtain the tangents of the arcs 4a, 
 5 a, &c. in terms of the tangent of the single arc, as 
 
 given below. — | 
 
 Tan 3 Ga 
 
 tan @ 
 
 Tan az 
 a» 
 9 tana 
 Tan 2a= ——; 
 l—tan* 2 
 ie Stan a—tan? a 
 As he nog eee en gare 
 1] 2Sixand a 
 4tan a—4tan? a 
 -Tan4a= 
 
 ul tan? a+ tan? a 
 ' 5 tana—10tan8a+tan' a 
 FAD Dy Gort. F Tagan ap Seaame 1S 
 43. Hence, also, by observing the law of the co- 
 efficients and exponents of these last expressions, it 
 will be easy to deduce the following general formula 
 
 for the tangent of any multiple of a. Thus, tanna= 
 
 © a(n m-2) a(n—1) (n—2) (n—S) (n-4) 
 natana SO ara ae Vat Tee athe ——_——tan' a, &c. 
 n(n—1) 5 n(n—}) (1—2) (n—3) 2 
 ] oe tan*@ + on a Th MF res bays a, Kes 
 
320 
 Op, Tan git ie, | 
 m tana wdten Mars 2) (a) tan?a+ Goria (8) tan? g &e, 
 1 — iN a tan? a+ ee (4’) tanta, &c. 
 
 Where m may be any number, either whole or frac- 
 tional; but the series can only terminate when x is a 
 whole number. 
 
 44, The cotangents of the multiple arcs 2a, 3a, 
 4a, &c. may also be readily expressed, by substituting 
 
 tn in the place of the tangent of a, in the last table; 
 
 which being done, we shall have 
 
 Cot a= cota 
 
 cova—1 
 nelle Tees 
 cot? a3 cota 
 O22 ee 
 Cot 3 3 cot? a—I1 
 
 cot*a—6 cot?a+] 
 Cot 4a = OO er ae 
 
 4 co a—4 cota 
 cot? a— 10 cot® Sa+5 cota cot 2 
 
 Cot 5 a = 5 coita—10 cot a+l , Bee, i 
 
 45, Hence, also, it will be easy, either from the 
 above expressions, or from the former series for the 
 tangent, to deduce the following formula for the co- 
 tangent of any multiple arc. Thus, cot 2 a= 
 
 n(2~-1) bate -2 ine a aE ) (n=) 
 
 nm 
 y" cot” a, &c. 
 
 | 3 Rie CRD Ne = 
 coi"a 5 7*cot 2.5.4. 
 
 n—1  n(n—l)(m—2) , 1-3 n(n-1)(n-Z)(n-3 (3-4) n— 
 ncot. @——— ect ss a rect 
 2.3 2.3.4.5 
 
 ‘ Or, Cot na= 
 eT (n—1)r? (n=2) (2-3)? 4 (7-4) (1-5) °° Pee 
 cot"*g— ————— ~— Bee +, > 
 * 2cota 2.4 cotta 5.6 cot a , 
 ami (WL i(n—z)r (1=3) (m-4r? | (n- 5 mG jr” 
 n cot PA Sac ee a! ees or oe —c', &c. 
 2.3 cos 4.5 cov a 6.7 cot’ 2 
 
321 
 46. In like manner, the secants of the same multi- 
 ple arcs, ‘ a, 3a, 4a, &c. may be nie by sub- 
 stituting — for cosine of a, in the former table of 
 
 those ae which will give 
 Sec a=seca 
 
 Q 
 sec’ a 
 Sec UN ee Speen ta 
 Z— seca 
 sect a 
 Sec Sq. py 
 , 4imm 3 sec a 
 sect a 
 Sec 4q = ———-———_ 
 8—S sec? a+secta 
 sec? & 
 Sec §a== — &c. 
 
 16—20 sec? a+5 secta 
 Or, generally, sec na = 
 
 sec" aq 
 nf 5 eee 
 gif) n—l ar secta+ 2 n(n—3)r sectg— 2("—4) (n—5)r 
 "9,94 933,99 
 &c. Also, 
 
 n—] n — 
 Vers na == n' vers a— paar cA 2 atin ae 7'(3) 
 
 2 — 
 i 
 47, Also, s means of art 40, it will be easy to find 
 the powers of the sine and cosine of the simple arc, in 
 terms of the sine and cosine of certain multiples of that 
 arc. For, since 2sin® a=1—cos2a; 4sin'a=3sina 
 —sin 3a; and 8 sin*a=cos4a+8sin°a—Il=cos 44 
 —4 cos Za+3, &c. we shall have | 
 Sin a=sin a 
 
 if a 
 vers a — = (c) Vero ee (anal Be vers a, &c. 
 
 2 sin’ a==1—cos 2a 
 4. sin’ a==3 sina—sin 3a 
 8 sint a=3—4 cos 2 a-{- COS 4a 
 16 sin’ a=10sina— 5 sin 3 a--sin 5a 
 $2 sin°@=10—15 cos2a+6 cos 4a—cos 6a 
 64 sin’ a= 35 sin a—21 sin 3a+7sin 5a—sin 7a, &c. 
 ¥ 
 
322 
 
 Where, beginning at the last term, the Jaw of the 
 coefficients is manifest, being the same with that of the 
 coefficients of a binomial, raised successively to its seve- 
 _ ral powers, except that the numbers 1, 3, 10, &c. stand- 
 ing by themselves, are only half the coefficients of the 
 corresponding terms in the like powers of the binomial. 
 
 48. Hence, we shall have sin" a = 
 
 me +sin n azn sin (n—2) a+ ‘ Y sin(n—4)a, 
 ot 
 &c. er, Sie 
 n(n—1) 
 at Lcosnat nos (n—2)a +- ae il tee (n—4) a 
 
 In the first of which series, the upper sign must be 
 used when 7 is an odd number, and equal to 4m-+1, 
 and the lower sign when z is equal to 4m—1, m being 
 any number whatever. | 
 
 In the second series, the upper sign must be used when 
 nis equal to 4 times any number m, and the lower 
 signs when n is equal to twice any odd number m. 
 
 49, Similar formule may also be found for the suc- 
 cessive powers of the cosine of any simple arc, by ap- 
 plying the expressions in art. 40 in the same manner 
 as above. ‘Thus, 
 
 Cos a=cos a 
 2 cos? a=1-+cos 2 a 
 4.c0s3 a=3 cos a-+-cos 3a 
 8 cos’ a=3-+4 cos 2a+cos 44 
 16 cos a=10cos a+5 cos 3a-+cos 5a 
 32 cos’ a=10+15 cos 2a-+6 cos 4a-+cos 6 a 
 64 cos’a=35 cosa+21 cos3a+7 cos §5a-+cos 7 a, 
 &c.. &c. 
 
323 
 
 Where the coefficients observe the same law as in the 
 former table of the sines; and if we regard the last 
 terms of these equations as the first, or take them all 
 backwards, the following general formula will be rea- 
 dily obtained: Cos” 
 
 1 
 wat : cosna-—-ncos(n— 2)a aE 
 
 oi 
 
 ae 
 
 2 (n—1)(n— _ sn(n—1)(n—2) &e. 
 se heos(n—6) a REE cipal TWN PRIA 
 
 cos (n—n) a, or cos a, according as n is an even or an 
 odd number. 
 
 50. From these latter series, expressions may like- 
 wise be derived for determining the value of the sine, 
 cosine, &c, in terms of the arc, and vice versa; but 
 as the mode of deduction, commonly used for this pur- 
 pose, is not so clear and satisfactory as could be wished, 
 it will be better to employ the doctrine of fluxions, 
 from which. they may be easily obtained, as follows: 
 
 Let z == the length of the arc, and # = sine; then, 
 
 by a known formula, 
 
 ia yhgai(E+it+ at B54 
 5.7 x8 Sate . 3.5a%x 3.5.7 08 o 
 ey bc.) = a si ce et eee 
 x 3 2 
 
 c.3 the fluent of which is z= a 
 &c.; the fluent of which is z Da iat nde at, 
 
 S627 SH Te é 
 ape + a6) _& And, by reverting the 
 : hall have «=z — =~ - 
 Series, we snail! have x : 2. 2.3.4.5 rt 
 
 2” 
 
 2.3.4.5.6.7 r9 + &c. 
 
324 
 
 Whence a = 
 
 be a sink a 3 sin’ a 3.5 sin? a 3.5.7 si? a 
 
 [SP or oS T ons  Taer + THEA OC 
 < Or, 
 O% an? be i142 m2 atnd 
 | ni? ee o sherk oe sia Ds ee, e 
 Ising + 55 + 4.5 r* Bor 6.7 7* Chappe &e. 
 Sing = 
 
 [ as ae a’ ao & 
 2347 2345 2.54.5.67 oti aa a 
 < Or, 
 
 | o— set Loading 6 oleae oe VS 
 wor 4.5 r® 6.7 \7* 3.9 r? 10.11 i r* 
 
 And the same series will equally apply to the chord 
 of any arc a, substituting the chord for the sine, and 
 ae 2 aah d, or 2r, instead of r. | 
 
 Also, if 90° — a be substituted for a, we shall 
 sin’ —— ~~ ge. 
 
 have 90°—a=sin (90°—a) + 
 
 a——90°— cos a— ——. &c. Andcos a=r—-—sin?+i a 
 2.3 r? r 2 
 
 3 
 
 a ae a 
 a Gare giao ee 
 
 at a2 
 ies T54 Teaser Tt & 
 Whence a = 
 
 C%s% cos? a 3 cos? a 3.5 costa, 
 (70 Sem See ER Page eae 
 < Or, 
 r-cosa r=—cossa., 3 Pecan 3.5. ,ri—cos’ @ & 
 eeigh (RTOS. PR -) 24.6 7 ae 
 
 | (Cos aes 
 i. a® at ae a® 
 i on gags Taser T ag. 7 ep ee 
 < Or, 
 a® a 2 2 2 
 BS An es, A ei) eae a f 
 LT GAA gan8— Baal Zan DO 9.10 BSC 
 
325 
 
 52. Again, let z = length of the arc, as before, and 
 t= tangent; then, by a well known ke — 
 
 rt ORAL 8 4 
 Baril 4 Stemi 94%, be 
 
 7° 
 
 the fluent of which is 2=2— 2.4 4, fee. And by 
 
 Sr’ 
 
 ‘ A 9x5 
 reverting the series we shall have =z 2 b= ~~ by oe ae &c. 
 
 T 
 
 Whence 
 tan3a@, tanta tan’ a tan?a tang 
 @ = tana— ——-- —— — TC ee. 
 
 3? 39 5% 7 7 9 78 AoE 
 
 Tana=a-+ ga she - =f ink i Ay Cmte ld 
 
 15 + r* 315 7 76 2335 7? 
 1382 a! 21844 gl te 9295690" 4 
 155925 © © 6081075 r® 636519875 74 
 ye 2 
 
 53. And since tan a= -—, or cota= , we 
 co tana 
 
 shall readily obtain, by means of the above series, the 
 following expressions for the cotangent. 
 
 72 yr? 7 6 +8 pie 
 
 . 4 cot 2 3cova ! 5coa 7 cot’ a + 9 cot? a 
 
 72 ay 
 11 cotta cps 
 yr? a a° 2 a° a’ 2 a? 
 ee wre ar a Nha OLS ECT LAME Teor 
 a 3 45 r 945 r 4725 r 93555 r 
 1382 all 4 a8 
 = a oe eee a tee &c, 
 638512875 r'9 18245225 r 
 re r2 
 54. Also, because sec @ = ——, orcosa = ‘ 
 COS @ sec @ 
 
 the series before given for the cosine, may be easily 
 converted into the following expressions for the secant : 
 7" - $46 a & 
 Sea Ys ee RT Pe 
 seca 2.3sec3a~ 2.4.5 sec? a” 2.4.6.7 sec? a? 
 Or, 
 
 seca-r seca-i* 3 (secPa—*). 3.. 5{ sec7'a— 7) 
 C. 
 
 a= 90° — 
 
 a=7T y —— 
 sec a 1 12.3 sec a +54 4,5 sec? a @ "34.6.7 sec a 
 
326 
 
 61 a 77 a® 50521 ats 
 Sec. = ae ot oe 24 3 ene 720 T S0Gn7? 1 S6u880079- 
 540553 ae 
 +E ass00sa0 71 C+ 
 % 
 55. In like manner, because cosec a = “ ,orsin @ 
 7 : : 
 == ——, we may obtain, by means of the series be- 
 cosec a 
 fore given for the sine, the following expressions for 
 
 the cosecant.: 
 
 tee yr? rt 3 78 3.5 r° Rie 
 “~ éosee a TS Geto a +245 cosec> a+ 34.6.7 cosec’ a re" 
 4 7a 31.25). 127 a’ 
 Cosec a ==-~ +. 4% Meira See soar 
 a x 6 att 360 +r? Be 15120 77 1 oss00 8 
 
 73 a? 1414477 a 
 s121440 8 +} SSssc7isi000 r= © 
 
 56. Also, because vers a = r — cos a, we may ob- 
 tain, by means of the forrnule for the cosine, the fol- 
 lowing series for the versed sine; observing that d in 
 the second is equal to the diameter, or 2 r. 
 
 versa 3vers*a 3.5 versa 
 
 ee ae ris Are PY at genet ee) 
 
 Or: 
 vers a 52versa ®versa@ 
 se ray 4 Se | 
 Wa. rerye (14+ 557[4 + Ce a as mao (c) ke.) 
 Versa = 
 
 a’ 
 
 ed aa (4) = a) = (c)— SUE (pb) &c. 
 
 57. And since covers a = r— sina, and supv a= 
 r---cos a, we may obtain, in like manner, the follow- 
 ing expressions for those lines. 
 
 at 2 a 
 
 Qiart 2.3.4.57% + 33: 23.4.5. 6.7 ee 
 
 Covers a = r—a-+- 
 
 a? 
 a EC 
 9.3.4460.6.7.8.9 55 y 
 
327 
 Supvers ee rr | 
 a® a* aus . 
 Pod smmew neg Clee Bde 6c) = 7678 (D) &c. 
 58. The sine, cosine, &c. of any arc may also be 
 expressed by meanis of certain factors, which are found 
 by taking such values of the arc that each side of the 
 
 equation shall vanish when the sine, or cosine, &c, of 
 the arc, so taken, becomes zero. 
 
 Thus, sin + n x being=o (art. 13), if sin z ; sg x 
 
 (arse + se: Say) t be supposed = z Xz Xz” 
 2.3 2.3.4.5 } 
 
 &c. it is plain that sin z will be o, whenever z has any 
 
 of the values that can be taken from the expression 
 
 i eer an AS ees will also be = 0, when +2 
 
 is substituted for z, if this be taken asa general factor, 
 independently of the first z, we shall have 
 
 Sinz =z xX (I— =) x (1+ —) x a — sik 
 (+ 2)x V- )x Gt -&) &e. 
 
 i= fart. 13), 
 
 , : 4 nl 
 59. In like manner, since cos = 
 
 if cos e(=1—-% -- 5 ay ec.) be supposed = 2’ X 
 
 2’ x« 2” &c. it is evident that cos z will be o, whenever 
 z has any of the values that can be taken row the ex- 
 
 ° 42+ | 
 pression 2 =4—5~ 7. And as 1} eg a is also 
 4n+1 . : . 
 —=0, when + aes a is substituted for z, if this be 
 taken as a general factor, we shall readily obtain the 
 
 following formula for the cosine of .z3 viz. 
 
828 
 
 22 
 
 22, 22 
 EE or) es) 
 
 Da ts 22, 
 Nee Dat “Pa eee 
 
 Cose= (1) x4 
 
 « fu % e : ° 
 And, if a be substituted for z, in each of the above 
 
 expressions, we shall have the following formule for 
 the sine or cosine of any part of the quadrant or semi- 
 
 circumference. 
 q 7 ons. Qn-+-n 4 — 12 
 sn Ee (SS) KGS) eS) x 
 4in tn 6 1— 6 n+ 
 peasy ci Cay oem 
 —n i 3n— 3n4- 
 cos = (52) x (42) x CE KOE) 
 5n— 5 ? 
 (SEY x EEN) Be 
 60. Or, since sin (em) — cog a, and cos cine 4 
 Lu i 
 
 _ mr | : 3 
 = sin a2 if n—7 be put in the place of m, we shall 
 
 have (p) 
 
 (#) Ifthe first of these expressions; for the sine of os be divided 
 £ 1 
 
 mri 9 25 49 
 by the second, we shall have l= erg ears 
 
 2 = Sih pe 3h Sal 
 i 3 aes & = 2 (In) x (ay) x 
 (1—-35) x (l—_), &c. which is the same as the series given by 
 Wallis in his Arithmetic of Infiaites, for the value of } of the cir- 
 cumference of the circle. 
 
 &c.) and con- 
 
 T 
 sequently. a= 
 
 The first of these formulz for the sine also gives 7 ee ite ale 
 : , jane tate, Qn 
 
 2 2g 4. ; 4. 6 
 + Cr) . ise x oan) * ae) * oe &c. 
 
4 $n—1 9 sehin 
 Gos 2 = (ESE) x (AE) x (HE) x (2) 
 
 x (Samm at 
 
 4n On ! 
 Sin ae CEM) x Ft) x (HE) x 
 
 An Sn—m 
 ( ——") x \— : —} &e. 
 61. Also, if He first of these expressions for the 
 sine be divided by the. second for the cosine, and the 
 terms of the series, thus obtained, be afterwards in- 
 
 verted, we shall have | : 
 me m 2n—m Qn +m 44 — 
 tant = 52, (SSE) x GED x GRD 
 
 Zn n— nm 3n—m 3m + m 
 x e =) &c. 
 oa Samael ane OE eae eR eg 
 x (=) &c. 
 
 E hase 
 which, on the supposition of — = =1, is the same as the former. 
 
 : , we shall have, (by taking — —=) wily 
 n 
 
 V2 Z 
 
 ‘ ae 4 
 Or, since sin — = 
 4: 
 
 f2 16 64 144 256 
 “1 (15 63 143 255 
 And if this be combined with the series of Wallis it will give 
 2* x 6? x 10? x 147 x 18? &c. : 
 AP eiscobae sit x9 mlx 18 | Teg Bel ER 
 more remarkable, as it cannot be obtained by any other method. 
 Alsouw = 1441 (l1—$)+ L(1— sy) +1 (1—ay) + 
 (1—j,) &c. whatéver may be the kind of log’, tabular or hyper- 
 bolic. And as these log*., when expanded, form series, of which 
 the terms, taken vertically, can be summed, we shall readily ob- 
 tain hyp. v ¢ = 1.14472985884940017414342, and tab. pr = 
 0.49714987269413385435126. . 
 
830 
 
 62. In like manner, the following expressions for 
 the secants and cosecants may be obtained. 
 mar n n Sn Sn 
 Bec = (2) (gettin gota) 
 
 n—m\n-- nt 3n—m 3n+m 
 
 ) &c. 
 
 on 
 sei or | 
 Covet 95 = (Faw Kaif As ) x ( hs (x 
 
 2n—m 
 
 5n 
 ips) x ( n—m &c. 
 
 63. And if the former expressions for the sine and 
 cosine be combined with the others, we shall have 
 WG sue m 1(2n—m) ©, 3(2n+m) 
 ‘Tan Ga. 8 x n——m 2(n+m) 2(32—-m) 
 3(41—m) ; 
 
 4(32+m) 
 ; ma. 1 3{2n— 
 
 3 
 ae 
 
 SE ape Mong te zig Rig Ming ae 
 
 4. 
 
 Ket &c. 
 
 Cosee = = = x = x 5 x px 
 x at &c. 
 
 64. Again, if 2 be put for m, and the sine and co- 
 sine of the angle * — = be found in the same way as the 
 
 former, we shall vk by dividing the first Fil egg 
 
 by the latter, the following formule: : 
 . men 
 Sin=— 
 
 Zn m Za—m Qn+-m 4n—-m 4n+tm 
 Fae i ppmay Ouek’ gay etal ‘ere 
 1 er 
 
 Zn 
 
831 
 mT , ; me ast 
 Yn m Qn—m 2nt+-m 4Anu—m 4n+m 
 
 in nk n4k * Sn—k™ Snph™ bak 
 
 Zn : 
 mm 
 Cos— 
 Ln nm—m n+t+m 3n—m col 5n—m rmdir’ 
 oat eee =X c. 
 ke nuk n+h Snes Suk 5n—k- 
 cos —— 
 Zn 
 mar 
 Cos — i 
 22, n—m n+-m 3n—m 3n+-m 5n—m 
 
 "peer ek MG aes Znt+k “s 4n—k x 4An-+k ac. 
 
 Whence, taking an angle im of which the sine and 
 
 Yn 
 
 cosine are given, we can readily find the sine and co- 
 : : nT 
 sine of any other angle oe 
 
 65. From these and the former expressions, the na- 
 
 tural and logarithmic sines, cosines, &c. may be obtain- 
 
 ed much more readily than by the methods employed 
 by the first calculators of our present tables. For since 
 
 : ne x a 
 Sin f=2— 2.3 ia Q345 234.5.6.7 Bc. 
 x? at x 
 Cos r=1— Ca -t- O34 — 05.4.5.6 &c. 
 if “(2 ‘ =) be substituted for x, we shall have 
 . om, m 1 sm a3 1 m 
 
 Sin x (5) =F 7 a(S 3) cy, 934.5 Hees 2 ) — &e. 
 
 my Ty (a ? ] m 
 CRieMeal at a4 ca). t oad yaa hoo 
 
 And by taking 7=3.1415926535897932, and ca- 
 culating the coefficients to 16 decimals, the followity 
 formulz will be readily obtained. (q) : 
 
 (g) See the Analysis Infinitorum of Euler, or the French tran. 
 lation of that work, with notes, by Labey: where ‘these aid 
 various other series, of a similar kind, are investigated. 
 
332 
 
 Sin po 70°= 
 } a 
 1.5707963267948966 ~ 
 3 
 — 0.6459640975062468-,- 
 m®> 
 -+- 0.07969262624616 0—- 
 7 
 - 0.0046817541853187—, 
 $ 
 3 m 
 +-0 00016 4411847874 
 ll 
 — (,0000035988432352 “sf 
 ms 
 -4- 0.000000056921 7292—, 
 15 
 os 0.0000000006688035-—, 
 4 (i 
 
 ” mit 
 “+ 0." 000000000060669-—, 
 
 19 
 ate 0.0000000000000438—, 
 
 | ml 
 __ +0.0000000000000003—, 
 
 ° mer 
 66. Also, because sin roe el 
 
 Cog 90° = 
 n° 
 1.0000000000000000. 
 2 
 —~ 1.2837005501361698— 
 amt 
 + 0.2536695079010450 — 
 ’ é 
 — 0.0208684807633530—> 
 7 
 § 
 -+-0.0009192602748391 = 
 2 
 10 
 — 0.0000252020423731—, 
 
 12 
 “+0 000. 002710874779" 
 
 1¢ 
 
 — 0,0000000063866031—. 
 rp A 
 
 16 
 
 -+-0.000000000065 $596, 
 2 
 
 1s 
 
 — 0.00000000000052047— 
 iL 
 
 20 
 -+ 0.0000000000000034-—, 
 71 
 
 mer ( 
 
 ey 
 
 aay (akew 
 
 | on an 2n 
 x (Gatien pate) &c. and cos a coe —* 
 ty Y eee) x ( sib &c. if these factors 
 
 De multiplied two by two, we shall have, by taking 
 
 their logarithms 
 L sin 
 +L(1— 
 | L cos =—= 
 
 tia ee i i me 
 
 mT nim ! 
 
 m? mm 
 EE ey Sen 
 my ont 
 
 = pele ae 
 ap! Natty &c. 
 
 Fee 
 
 m1? 
 +L 0- Ten 
 
 ) 
 
 a 
 
333 
 
 67. And by reserving the logarithms of a few of the 
 first terms, in order to render the series more rapid, 
 and calculating the coefficients of the expanded terms 
 of the others, the following formule for the tabular 
 log sines and cosines of any arc may be readily obtained. 
 
 L sin — ~ YO? == Cos = 90° = 
 
 ntti ath L(Q2n+-m)——3in L(n—m)+L(n-+m)—~2Lx 
 +- 9.5940598857021903 | ++ 10,0000000000000000 
 
 = 0.0700228268059019"— oe 0.1014948598418929"— 
 
 — 0.001 172664416618" — 0.003187294065451 a 
 
 — 0.0000392291 464599", — 0.0002094858000174", 
 
 — 0,0000017292707984—~ Ts 0,0000163488485983".— 
 —o, ooovoosss629863", pol 0,000001801 989860", 
 a 0.00000000854871.55 a+ 0.0000001865022729" 
 —0 .0000000002819312"" — 0,00000001 29817147 
 —0. soon00d0001286912 — 0.000000001261471 pee 
 ~ 0.0000000000007027"., ab 0,00000000012456 17 
 ee 0.0000000000000395"", a 0.0000000000124559".- 
 
 — 0, eaoo000o0000022"" — 0. cen0o00000012881 
 
 68. iene: as the sines and cosines of arcs, 6h 
 - zero to 45°, comprehend the sines and cosines of arcs 
 
 from 45° to 90°, = in these formule may always be 
 
 taken less than 4; and as the series are thus rendered 
 very convergent, it will only be necessary, in many 
 
S34 
 
 cases, to calculate a few of their terms, to obtain the 
 sine or cosine of the arc required. 
 
 69. Several other formulz for expressing the sine, 
 cosine, &c. in functions of the arc, may also be ob- 
 tained by means of certain imaginary factors, which 
 have been found of considerable use in physical astro- — 
 nomy, and various other branches of the modern 
 analysis. | 
 Thus, radius being supposed = 1, we shall have 
 cos? x-+sin® x=1, of which the first member of the 
 equation is the product of the two imaginary factors 
 cos a-+sin « f—1 and cos r—sin & / —1. And if 
 any two similar factors cos x+sin # 4/—1 and cos y 
 + sin y / —1 be multiplied together, their product will 
 be = cos x cosy—sin x sin y + (sin w cos y + sin y 
 cos ) f—1= cos (x+y) 4 sin (x+y) /—1. 
 
 In like manner, (cosx + sin x /—1) X (cosy + 
 sin y / —1)X(cos z + sinz f —1)=cos (v-+y+2z) 
 4 sin (v-+y+z) /—1, &c. each of which are like 
 the simple factors, and are produced in a similar way 
 with logarithms, by barely adding the arcs. 
 
 And if the arcs x, y, z, &c. be supposed equal to 
 each other, we shall have (cos 7 + sin x »/ —1)%= cos 
 2x 4sin 2x  —1,and for the three factors (cos x 
 + snz /—1)=cos3r4sn3r/—1. Conse- 
 quently, in general, 
 
 '(Cosr4sine /—1)*=cosnx +snner f/—l. 
 
 Hence, by transposition and division, we shall obtain 
 the two following equations for the sine and cosine of 
 any multiple of the arc x, in terms of the arc. 
 
835 
 
 __ (cos « + sinx 4/—1)" + (cos —sin x ,7—])» 
 Cos 1x== Ta. Se ake SOE RE Co 
 
 (cos # + sin x,/—1)® — (cos x — sinx,/—1)s 
 27-1 
 
 70. The sine, cosine, &c. of any arc, or multiple of 
 
 that arc, may also be readily derived from the well- 
 
 Sin A i a 
 
 known iti expression e? = 1 ++ = + ose — 
 23 
 1.2.3. + jase T2840? 
 the hyperbolic logarithm is 1. 
 For, if in this formula, x ” —1 and — *./—1 be 
 successively substituted for x, we shall have 
 
 &c. where e is the number of which 
 
 Lf—1__ xf —l x 2 /—1 xt 
 
 be Pe abie ss oe FE gee a 
 LB f—!l 
 Seta ont hee 
 2.3.4.5 
 
 —2Jf—l__ xf —1 eee w/t xs 
 
 e eh 1 2 I ERE Ke 
 2 f—l 
 Ce. aah &c. 
 
 And if these be added to and subtracted from each 
 other, the results, after being divided by 2, and 2./ —1, 
 will give 
 
 2 a a Se Ea xt 28 re 
 NUE cz 2 2.3.4: Taso th : 
 
 Saiki dale agra aaa 
 CEPT Sy os. 2.5 Tb 234.5 ~ 234567 ce 
 
 Of which series the second members are the known 
 values, as before found, of the cosine of « and sine 
 of x. . Whence 
 
 ell at} gv lel 
 
 Cos x=% , ; sinxv=— 
 
 2,f—l 
 
$36 
 x i a 3 
 a (— ME OR) 
 ov) rt/-l Safa hy. gota al 
 > Sayan aN hee aT oi Poet ford 
 Or, if each of the terms of the numerator and de- 
 nominator of the second members of the two last equa- 
 
 k iM say | aun | 
 tions be multiplied by & Y~, they will become 
 2x f-1 
 
 e 
 Tan <== a ay y ae im 
 And, if mx be ei for x, In each of these 
 formulze, we shall obtain the sine, cosine, &c. of any 
 
 1 
 
 9 
 Tae cat #==4/ —1 mary 
 
 multiple of those arcs: thus, Cos m* = 
 Mell ma /—) Aa ae ak mx af—l_ ,—mx f-l 
 SEAL pine’ $54 a Ske CRRA cae cao SRR Sins 
 Tan mx = —— x 
 a | 
 
 te 2 as 
 maf i _ por me f eal 
 prmxf—i_, 
 
 1 
 eee Sh cot mem f —1( 
 
 71. The same formule also give pa, ioe hose - 
 
 ie bie mess), Ayah AA : 
 sin * / —1, and e aud = cos¥ — sNnx f/f —1;5 
 
 whence, by division, we shall any wa meee 
 
 cosx + sine f—1 1 + tane /f— 
 cos x — sin x f/a-1l 1—tanz /— 
 
 asi and by taking the 
 
 1+tanz,/—1 
 an? 
 But log (+= 2) = 2(z+a242 z+ +2 &e.); 
 
 therefore, putting tan # »/—1 in the place of z, and 
 dividing each member of the cua by 4f — 1, we. 
 shall obtain 
 
 * = tanw — J tan’x + ttan’x — 1+ tan’ x &c, 
 
 logarithms of each member, 244/ —1=log 4 
 
 tee * 
 
337 
 
 Which is the known series for the value of any arc 
 in terms of its tangent. 
 72. It may here also be shown, that any trigonome- 
 
 m Sin & 
 
 trical expression, of the form tan z = ———-—. can be 
 1+ mcosz 
 
 converted into a series of the various multiples of the 
 sine of z, taken progressively. 
 
 For, if in this equation, there be put instead of the 
 tangent of w and the sine and cosine of z, their values 
 in exponentials (art. '70), we shall obtain 
 
 x Vv—liy Pee 2 ( af aril. ade r/—!l) 
 eal 0 
 And, consequently, by reduction, 
 
 ered fi Aare Fis FL ick in 
 
 v—1 
 
 ee Oia ey og 
 Whence, by taking the logarithm of each member 
 of the equation, and converting the second into a series, 
 according to the form log (1 4+ z) = 2-327 +4123 
 — 12° &c. we shall have2 2 f/—1 = 
 Lof—1 : reat ieee enn, ca ie de 
 MN fe Plantae Zt 
 But @V—! — eT ?V—! = asing f'— Wwek® vod 
 — e *V—1—o9sin Qa —1 &c. (art. 70); whence, 
 dividing each side of the equation by 2 «/ —1, we shall 
 have | 
 
 : ‘mo. m or 
 r=msinae— > sin2e->—z sin3x——{~sin4x + &c. 
 Or, 
 
 ve m2 ° m> ° om ° 
 t==mM sin * — 3 sin 2e— <> Sl S + i sin 4 &c, 
 
 - 
 
 z 
 
338 
 
 The former of which series answers to the case tan x 
 
 oe oo and the latter to tan, y= 
 1 + mcos — mM COS % 
 
 73. Several other expressions aoe hs sine, cosine, 
 . &c. of multiple arcs, may be derived from one or 
 other of the preceding formule ; the most curious and 
 useful of which are the following : 
 
 Sin a Qe ee Sinise 
 
 m Sin % 
 pepe 
 
 Sn27= 2sin z sin (> —z) 
 Sin 3 z= 4sin z sin (4 —z) sin(+-+ 2) 
 
 . ‘ é 
 Sin 4 2% = 8 sin 'z sin (4 —x) sin (| + z) sin (F- z) 
 
 ° ° e 2 ‘ 2 
 ins z= 16 sin x sin (2 —z) sin (2+ 2) sin ($2). 
 
 “ee ay 
 aes) vn OBCe 
 » Or, generally, 
 . —TI. ° oy ae . 
 Sinng = 2" sin z sin (=-— z) sin (~ + x) sin 
 Qn np OO 2 om 3 
 (= — z) sin(= “+ +z) sin (—- — z) sin (— + 2) &e. 
 Where the series must be continued to as many fac~ 
 ‘tors as there are units in 7. 
 
 74. Cosz = sin( > — ay 
 Cos. 2% = 2 sin (= — z)sin(— + 2) 
 
 (r) The simple series 2 = sin z —~$sin2%2+4sn32—4 
 sin 42, &c. of which that given above is a more general form, was 
 first discovered by Euler; as was, also, the series x = sin x sec Za 
 sec } gw sec i vr, &c. 
 
 For the niechcal solutions of several bf the cases of spherical 
 triangles, according to this form, see Tables ‘Trigonométriques dé 
 Borda; wherethe following theorem is likewise given for the decimal 
 
 MSIN SG —_ Em sin Zz tnd sin 32 
 
 division of the circle, r= ——=-. 2—— +. 2_—_ + F+ &e. 
 4 sind”. sin 1” sin 1” 
 
 ax 
 
339 
 bse a ae sur Sim Mgouoet | 
 Cos 3 z =4 sin a — z)sin(— +z) sin eo *) 
 : . wv . fie ge a. eae ° 
 Cos 42 = 8 sin (> — x) sin (+ 2)sin (G —2) sin 
 ont 
 Cg 7 2) 
 yd _ 8 0) Sem 
 Cos 5z= 16 sin (75 — 2) sin (7 + 2) sin (Gp = 2) 
 . OR o OW | 
 sin (75 + 2)sin (55 — 2) 
 &c. Or, generally, 
 : nm]. oo tte onl) Se 
 Cosnz=2 “sin(,- —z) sin (=~ + z)sin ([- — 2) 
 s | Ser Oe be 7 
 sin (= + 2) sin(3* — %) sin (= +2) &c. to x factors. 
 7865,Cos 2 =="cosg 
 . 7 
 Cos 2% = 2 cos (— + 2) cos(7 — z) 
 € (93 
 Cos 3z = 4 cos (= +z) cos(= — z) COs z 
 C ps on on w 
 os 4% = 8 cos(= + z%) cos( — 2) cos (> +2) 
 T 
 cos(z — 2) 
 4g 4a Qn 
 Cos 5% = 16 cos (75 +%) cos(=5 —2) cos (55 + 2) 
 
 2 
 cos (= — 2) cos Z. 
 
 &c. _ Or, generally, 
 
 1 
 
 _ —1 —1 
 Cosnx= 2" cos (— aw -+- 2) cos (= 7 — 2) 
 
 COS Cx 7 + 2) cos ae 7 2%) Cos om a +2)cos 
 
 (ome 
 27. 
 WOs, OCC, B= Sec. & 
 
 7. — %) cos (eau aw -+ z) &c. to n factors. 
 
 3 Sec 3 % = sec (= +2) +sec(F — 2) — sec (=) 
 
 3. 
 Z2 
 
340 
 
 5 Sec 5.z = sec (= + $a) pe0e(— eS ma) 
 
 — sec (+ — z) + secz 
 
 Tec Tz = see (F ~e z)+sec(2 — 2) —see( -+-z) 
 
 Qa ; 
 —sec (+ —z) +sec (+ +z) +-sec(=—z)—secz 
 
 &c. | 
 Or, generally, putting n = 2m +1 
 
 n SOO7 Z = sec (7 $2) + sec (= 7 — %) — sec 
 Le —2 
 (== +2) —sec (“= 7 — Z) + sec Oo ne a) 
 —2 
 + sec (= 3 — 2) — = ee 4 SEC R 
 77. Cosec z= cosec z% , 
 3 Cosec 3 = = cosec x -4- cosec (S—2)— cosec (= +2) 
 5 Cosec 5 % = cosec z + cosec (= — z)—cosec G — %) 
 g g 
 — cosec = — z) + cosec (<= + 2). 
 &c. Or, generally, puttng n = 2m-+ 1 
 7 cosec 7 & = cosec z-+-cosec (= —z)—cos (= +2) 
 2 Qn 3 
 ~- CoOsec (= — 2) --cosec (= ++ z) + cosec (= — 2) 
 = COSEC a +2)—---F cosec om — %) + cosec 
 
 +2) 2 
 
 (“2 
 
 Where the upper signs take place when m is an even 
 
 number, and the lower when it 1s an odd eo 
 "8. Tanz <=. tan ‘z 
 
 8 Tan 3 z= tan z + tan (+ -- %) + tan (< + 2) 
 
 ae SS aes 
 
341 
 
 5 tan 5 z = tan 2 -+ tan (4 + z) + tan (% +2) + 
 tan (<2 + 2) + tan (2 + 2) 
 &c. 
 Or, since tan v= — tan (7—v), we shall have 
 Tan z = tan z | 
 3 tan3z = tan z—tan (4 — z)-+ tan G+ z) 
 
 Stans z= tan 3 — tan(-> —z) + tan G+2)— tan 
 
 29 La 
 (a —_ z) + tan (= + Z) 
 &c. | 
 Or, generally, putting n = 2m -+ 1 
 
 T T 
 ntann z= tanz— tan(- —2)-+ tan (- +2) — tan 
 
 (Seep tan (Se aan Rhee ee 
 — tan (“2 —2z)+ tan (~—" + 2). 
 ion Also, 
 
 Tan 2 s=etan z 
 a Ossi tan z tan (7 — 2) tan (G + 2) 
 ey : a . Qn | 
 Tan 5 z == tan x tan(=—2) tan Ge -- 2) tan Ca) 
 2% 
 tan (— + 2). 
 Or, generally, putting » = 2 m + 1, as before, 
 ) 
 Tann z= tan ztan(" —z)tan(= + z)tan(— — z) 
 74 72 1/3 
 or 37 m WF 
 tan (— +} 2) tan ea) es ~ x tan te 2) 
 mT 
 tan o* + 2). 
 79.mCot z= cot 2 
 2 Cot 2 z==cotz + cot (5 +2) 
 
342 
 
 8 uf ¢==cot z+ cot (= % +2) -+- cot (== 7 +2). 
 
 4 Cot 4z = cot + cot (= +2)+ cot (“p+h2) beet 
 (J : + 2) 
 
 5 Cot 5 z= cot s + cot (Z+2)+cot (= -+-2)-+ cot 
 (“FE +2)+4cot (=% +2). 
 
 &c. 
 Or, since cot v = — cot (7—v), we shall have 
 
 Cot z = cot 2 
 2 Cot 2z = cot z — cot (~ —2) 
 
 3 Cot 3 z = cot z — cot (4 2) 
 
 4 Cot 4 2 = cot 3 — cot (F a ube nevi Ne 
 
 (Ea) 
 5 Cot 5 z = cot z — cot (= — 2)-- cot (= -+- 2) cot 
 2 2 
 (F — 2)+cot (+ 2). 
 &e. : Or, generally, j 
 
 n cot nz = cot z — cot (= —2z)-+cot (= +2) — 
 
 g 8 * 3 
 cot (— —2%,-+-cot on -+z)—cot(—™ — 2) + cot — 
 
 +2) — &c. to n terms. 
 fea 
 
 — 2 cot2z = tan + tan( 
 
 + 2) . : 
 
 eee enn | 
 tan fess +- 2) 
 
 — 6 cot 6z = tan % + tan (= +2) -+- tan ( (7 42) a 
 
 tan qs * 42) -4tan(-Z -+z) -+-tan Cz +z) 
 &c. 
 
S43 
 
 | Ors 
 2 Cot 2 = —tan x + tan(- —2z)- 
 
 . 4Cot4z2 = —tanz + tan (= —#) —tan(Z +2)+ 
 tan (53 — z) 
 
 6 Cot 6z = — tanz + tan(= —2)—tan(= +2) + 
 tan tt —z)—tan Ge -+2)-Ftan (~~ — 2) 
 ic. Wes. Or, generally, 
 ncotnz = — tang + tan ( —2)—tan (= +2) + 
 
 m w : 
 33 
 
 tan (2 —2)—tan(== + +-z)+-----+ tan 2B er 
 80. The sum of the sines of any number ‘of arcs 
 in arithmetical progression, may be also exhibited as 
 follows : ; 
 Sina + sin (a+b) +sin (a+26) -+sin (a+3 6} 
 
 cos (a—Z 5) 
 
 + sin (a+42) &c. ad infinitum = aaa ee 
 Also, 
 Sin a+ sin (a-+6)+sin(a+2 b)-+sin (a+3 b)-+sin 
 fats 6) 4 in (anea ee ple 1(a+4 oa (n$1)b 
 
 81. And the sum of the cosines of any number of 
 
 arcs, similarly taken, is as below: 
 Cos a + cos (a+46) +cos(a+2 b) ats cos (a+3 b) 
 -- cos (a+4 b), &c. ad infinitum = — ——.-,?—. 
 - Also, 
 Cos a--cos (a+ B)-peos(a-+21)-feas( a+3b) + co 
 (at4b)&e. -- +cos(a+nb)= Coste spa sinattiatad )9 
 
 “sin ZO 
 
344 
 
 82. As some of the common logarithmic formule, | 
 for numbers, are often required inthe investigations of 
 trigonometrical expressions, I shall here subjoin such 
 of them as are the most useful and necessary, for the © 
 sake of reference. Thus, 
 
 Log (I-+p)== (p —3p'+$2'—4 pt +4 p— &c.) 
 1 
 
 Log p=; = xp (PEE PUES DEEP Fa pte p' &e.) 
 Loe Po 244 +P +s p° ++p'+4p ty p'&c. ) 
 
 ‘ Or, Loz a = 
 2{(a=1)—$ (a1) + 5 (@=1)' 4 (a1) Bee, b 
 7 ‘96 = 
 1 fa—l ened ‘1 
 at Sate) a4 +37 Kb Bee } 
 into 
 2 fa-—l eo waa! ys : 
 ™ la+l in spear ib 5 Moa? ++ eo) +éc.b 
 Also, Log 5 = 
 J ob arb atsb at st 
 “tie ies Ey + 4 (= 4 sp + tc. } 
 ings 
 1 Cand bent 2 b, ss be, ; 
 feet tg eee Cay +(S2y $ 8c 
 acsb } renh aes a eh a 
 (5+ ar ia oe UME P + Yt &e.b 
 Or, Log a= 
 
 Log (ot) + eb et tte be} 
 
345 
 
 Loga = log (a — 1) + 
 lef tebe at. 1} b-, 1 | 
 ae oF 2(i—1)? bu 3(a—1)8 ig uaniy &e. } 
 Log a = log (a — 1) pa | 
 2s 1 LA ] 
 a dani * Sap t Baye ae agate t & + 
 Where m == 1 for hyperbolic logarithms, or = 
 2.302585093 for the common tabular logarithms; 
 which number is the hyperbolic logarithm of iO, or 
 what is usually called the modulus of the system. And 
 
 if its reciprocal be used, it becomes — == 494294482 
 for a multiplier. 
 
 83. To these formulz may also be added the follow- 
 
 ing, which will be found useful upon particular occa- 
 sions (¢). 
 
 oie nase \—i(a?—a yes. (a—a- a 
 | 4. {a3 Te te )i&e, t 
 
 (@) For an investigation of the doctrine of lo: ele watt from 3 
 most simple algebraical principles, see article Logarithms, given 
 by the Author, in the Supplement to Hutton’s, Mathematical 
 Dictionary, or Lagrange, ‘Théorie des Fonctions Analytiques. 
 
 Te the Jogarithmic formule, given above, may also be added 
 those of Borda and Haros, cited by Lacroix, in his Legons d’Al- 
 gebre, which are as follows : 
 
 Log (n+4+2) = log (n—2Z) + 2 bog (n4+1) — 2 log (n—1) + 
 2 2 Z 2 
 ee) se 3 5 
 M be: RP foe a ¢ n® 7) se. f 
 Lag(n+-5)=log(x—3) +log(x+3 {ih co Abe) 
 
 Bx “™Y 72 
 =. ea mana aes {ot 1052 +72 “be (SI 5x 
 
 5 Be ber wir re bs of Bo, be 
 
 Lad 
 — 250° +72 
 
 —25 nt + 2479 
 
S46 
 Log (a + z) = 
 “Loge +E -$ES+4 (D4) + Bed 
 — Log (a—z) = 
 Lega Stet ay +4 (lb HB he} 
 Log (a+z) = 
 Cee Lae tilage) fa Bo) ak ma ya. § 
 Log a == — af vV (a—1)—3(Ya—1)°+4(Va—1) 
 —1(Va—1)* &c.} mq ORF : 
 
 Also, aa 
 SEY eS) tis Geta eee Ba 
 =1+ 
 ue a 1 log @ loga@ log a 
 +1 4 08 ee ) fe. 
 
 “a5 
 meh | ae stigtaat 23.45 &¢ 
 
 Where e = beste whose hyperbolic logarithm is 
 1, or the radix of the system of the common tabular 
 logarithms, which is known to be 2.7182818284. 
 
 84. After these formulse, which will be found con- 
 venient upon many occasions, it will be proper to show | 
 the method of obtaining the sine of the sum or differ- 
 ence of any two arcs, in the form of a series ; sehigh 
 ‘may be done thus: 
 
 sin @ cos z%-- cos asin z 
 Sinteie |) =e 
 
 cos a cosz + sin a Sl sin 2% Z, 
 
 ‘ , and cos(atz)= 
 3 
 
 (art. 20); oe os aK +. 
 
 vo 
 
 oe i 
 xT ok slips a and cos =7 — = | ot =—> &c.; whence, 
 
 by substitution, we shall have 
 
347 
 
 Sin (a+) = sin’ 4- 
 
 0s @ sin @ cos @ sin @ cos @ 
 
 eters ORT ge Fs eR tal of Boon &e: 
 7 oA CSP Oba A Gade 
 : Sin (a—z) = sina — 
 cosa sin a cosa sin a cos a . 
 —— 2 — — gS 98 SE Ay aceon 3 ae OE Oy 
 s Oya 92,3" Tagan" —95457 
 Cos (az) = cosa — 
 sin @ cosa sina cosa sin @ 
 
 . z— —. 2? EL Bae a et  _, F&O} 
 r 97? + 2.373 na 9.3.4r% 2.3.4.575 
 Cos (a—z) = cosa + 
 
 sin @ COS a sina cos a sin @ 
 : in 238" * 230A T.5 3 i458 
 
 In each of which series, if ~ —, =" be substituted for z, 
 
 the arc will be expressed in i bees instead of by its 
 leneth; 7 being = radius, and r° = number of de- 
 grees in an arc of equal length with the radius; which 
 is 57°.2957795 or 206264”.8. 
 
 85. The logarithmic sines, cosines, tangents, &c. of 
 any arc, may also be found from the expressions already 
 given for their natural sines, sitgl ii as follows: 
 
 : a? a 
 sina==a 2.37% + 294.574 2.3.4.5 6 a z5 Be. ( -50) 
 
 a? at ae & } 
 ah eee 7" P9464 O8AG.TR TS 
 “a Whence 
 a* at 
 Log sin a= log a + log (1— 5-34 + 5355 &c.) 
 
 2 
 
 Or, by putting p= — Grek! son —— &c.) we shall 
 have, log sin a = loga + log (!—p) =loga — 
 {pre ptt P &e.¢, as is evident by changing -++ p 
 
 to — p, in the first farm of logarithms, art. 82. 
 
348 
 
 And, by restoring the value of p, it will become 
 
 Log sina =loga~ +9". + =“ et, Beit 
 
 Of SING = 108 Ow ase F 5a S57 
 . at 
 
 Or "93473 
 
 &c. (art. 51) and ial: has log ‘cos a = 
 
 86. In like manner, cosa =r — 
 
 a 
 2.3.4.5.67° 
 ed Geet oi hich 
 
 og tf log oa itegs 2 ve é:c.) whic 
 being treated as in the former article, gives 
 
 lea aé a® 
 Log cos a = log r——1 ss ++ a8 -- ir ae 
 
 1 
 87. Again, tana = af Sp 22 ra wag Be 
 
 (art. 52); whence log tana = log a + log Q+5 
 
 +e — te if = &c.); and by finding the logarithm 
 
 of the series, according to the form log (1-++p), we 
 shall have 5 
 Log tana =loga-+ 3 ft -+- eh + oe ic. ? 
 From which three formulz we can also readily ob- 
 tain, by addition and subtraction, the expressions for 
 the logarithmic secants, cosecants, and cotangents. _ 
 88. The logarithmic sines, cosines, &c. of the sum 
 or difference of any two arcs may likewise be readily 
 found, in nearly a similar manner with those of the 
 
 345.77 7r® 
 
 simple arcs, as follows: 
 
 Log sin (a+ z) = log Ssin a cos %-+ cos a sin zt 
 
 ° . cos 2 sin 
 
 men Gs sin PEON NET tenet | 
 
 ‘(1 + tan 2 cot 2 do log sin @ -+ 1 08. or 9B, Ut 
 tan 2. cot :a):. 
 
 Whence, log sin (@-+-2) = log si sin a--log cos z+- 
 
 ~ {tang cota — } tans cot a+-+tan’z cota— &c, t 
 
 bees log }sin & COS & 
 
349 
 
 And log sin (a — z) = log sin @ ++ log cos = ~ 
 tan z cot a+-itan’* z cot” a-+--+ tan’ z cotta cit 
 89. In like manner, L cos (az) = log} COS @ COS Z 
 
 sin asinz 
 +sin @ sin ab = — log Pee cme } = log 
 COS @COS% 
 
 cos a + log cos s + log (1 } tana tan z). 
 Whence log cos (a+2) = L cos a + log cos s + 
 
 — } tan atan s—+ tan’ atan’2 + 3 tan’ atan’z &c. } 
 And log cos (a — 2) = log cos a + log cos s — 
 : + }tan a tan s-+4 tan’ a tan? 2 -+ Ltan3atan’ z &c. } 
 ” 90, One of the most commodious forms for the tan. 
 
 tan (a+2z)—tena 
 
 gent may be found from the formula — Prince 
 
 sin (a+%—a) | _ sin x 
 sin (a+z+a) sin (2a+z) (art. 30), which, by reduc. 
 ty sin 2 
 . ‘ seu sin (Za+z) 
 tion, gives tan (a-++2) =tan a ——— | 
 sin 3 a+z) 
 24 sins, 
 
 Whence log tan (a-++-2)=log tane+ — 
 
 sin % sin % 
 “1 (gers rh et sin (ss pay) &c. 
 
 From which expressions the formule for the loga- 
 
 m (sin (2a+2%) 
 
 rithmic secants, cosecants, and cotangents may be rea- 
 dily obtained. 
 
 91. The logarithmic sines, cosines, &c. of the sum 
 or difference of any two arcs may also be otherwise 
 expressed, as follows: | 
 
 ! ae sin (@-+-2) = log sin a + 
 
 i jon em 2, cosa (as EEE Ae, me ge 2 
 
 m (sina QZsin’a 3 sin? a 1% siu* a “§ 
 Log sin (a—z) = log sina — . 
 
 1 (cos a 1 COS a’ P+-2 cos? a 
 
 mM @sin a "ese 3 sin? @ a LZ sin* a be. 5 
 
350 
 
 hs cos (a+z)= log cos a— 
 2 fsna sin @ . 1+2sin* a sin? a 
 ; “Pinta thameen: {KD 
 . aa ocate PP gh “12 c08% a 24 &e, | 
 
 m |lcosa 3 cos* a 
 ‘Log cos riage cos a+ 
 1s fs tiay 1 32 sna | _ 1+2:sin’ a 
 M { cos a. 2cosa.” oe 3 cova 12 cos* a a4 &e ' 
 
 92. It is also evident, from 2 algebraic expressions 
 for the natural sine, cosine, &c. of any arc, given in 
 art. 18, that their logarithmic sines, cosines, &c. will 
 be as below. 
 
 Log sin a= 20 — log cosec a = log cos a+log tan a 
 —10= 10 -+ log tan a — log sec a= 10 + log 
 cos a — log cot a. r 
 
 Log cos a=20—log sec a=10+log sin a—log tan a 
 = 10 + log cot a — log cosec a = log sin a+-log 
 cot a — 10. 
 
 Log tan a = 20 — log cota = 10+ log sin a — log 
 cos a@ = log sin a—log cot a-+-log cosec a = log 
 cos a + log sec a — log cota. 
 
 Log cot a = 20 —log tan a= 10+ log cos a — log 
 
 sina = log cos a + log sec a — log tan a = log 
 sin a + log cosec a — log tana. 
 
 Log sec a = 20 — log cos a=10 + log tan a — log 
 sina = 30 — log sina — log cota = 10 + log 
 cosec a — log cot a. 
 
 Log cosec a = 20 — logsina = 10 + log cot a — 
 log cos a = 30 — log cos a — log tan a = 10+- 
 log sec a — log tan a. 
 
 Log versa = log 2+2 log sin} a — 10= 2 log sin 
 4 a — 9.69899700. Be 
 
 Log supvers a = log 2 + 2 log cos$ a—10=2 log 
 cos 7 &@—~ 9.69899700. 
 
$51 
 
 DEMONSTRATIONS OF THE ‘PRINCIPAL THEOREMS 
 IN PLANE TRIGONOMETRY, 
 
 Having treated, pretty fully, in the first part of this 
 work, of the practical part of Plane Trigonometry, and 
 its most useful applications in the determination of 
 heights, distances, &c. it will be here proper to give 
 the investigations of the principal theorems from 
 whence those calculations are derived; which are the 
 three following : 
 
 THEOREM I. 
 
 93. The sides of any plane triangle a Bc, are to 
 each other as the sines of their opposite angles; and 
 conversely. 
 
 For let the triangle be circumscribed by a circle, and 
 from the centre o, with the radius of the tables, de- 
 scribe the circlea bc; and having joined oa, 0 B, oc, 
 draw the chords a b, bc, ca. 
 
 Then, because the angles ao B, BOC, COA at the 
 centre, are double the angles ac B, BAC, ABC at the 
 circumference, and that the chords ad, le, ca are twice 
 the sines of half the former of these angles, or of their 
 equals ao b, boc, coa, they will be twice the sines 
 of the whole angles ACB, BAC, ABC. 
 
 And since 0 a, 0 L, oc are equal to each other, being 
 radii of the circle a bc, as also o A, 08, OC, the lines 
 
 \ 
 
252 
 
 a b,-b c, ca will be respectively parallel to the Eerste 
 sides of the triangle AB, BC, CA. 
 
 ~ Hence the corresponding triangleso a B, oah, OAc, 
 xr similar,oA:0a::AB: a6, and oa :0@ 
 
 PAC 2aC} “indi bi also, by equality, AB: a b: 
 AC: ac; oraB:i Rey ee I . 
 
 But 3 al and 4 ac have been shown to be the sines 
 of the angles c andB; whence ap:sin £4 c::ac: 
 sin: ZB; or'sin 2°C YArB: 7 sin 2 BDA Gy 0.4. D: 
 
 Cor. Since aB:a6::a0:a@0,0raB:2sin Ze 
 
 : ao: ao by similar triangles, it follows that each 
 side of the triangle 4 Bc Is to twice the sine of its op- 
 posite angle, as the radius of the circumscribing circle 
 is to the radius of the tables. 
 
 THEOREM Il. 
 
 94. The sum of any two sides of a plane triangle 
 ABC, is to their difference, as the tangent of half the 
 sum of their opposite angles is to the tangent of half 
 their difference. 
 
 For about one of the angular points 3, of the tri- 
 angle, and with the greater side B a as a radius, describe 
 a circle, meeting BC, produced, in x and F, and ac in 
 p: also join DB, E A, F A, and draw Fc parallel to ac. 
 
 Then, because B a is equal to B £, itis plain that zc 
 is the sum of the two sides a B, BG, and c F their dif- 
 
 ference. | vite 
 
 — - 
 
353 
 
 And because the outward angle a B £ is equal to the 
 sum of the inward angles B Ac, Bc A, and the angle 
 AFE at the circumference is half the angle a BE at 
 the centre, the angle  F # will be half the sum of the 
 angles BAC, BC A. 
 
 Also, since the angle a cB is equal to the sum of 
 the angles cB D, CDB, or CBD, CAB, the angle FBD 
 will be the difference of the angles BA c, BC A; and 
 the angle F a p at the circumference, or its equal AFG, 
 will be the half difference of those angles. 
 
 But the angle £ a F, in the semicircle, being a right 
 angle, ac, AE will be the tangents of the angles a Fc, 
 A Fe to the radius F a. 
 
 Hence, since ac, GF are parallel, Ec istoc Fas EA 
 isto AG; thatis, the sum of the sides AB, BC is to their 
 difference, as the tangent of half the sum of their op- 
 posite angles B Ac, Bc Ais to the tangent of half their 
 difference. OE, De 
 
 THEOREM III. 
 
 95. The base of any plane triangle a Bc, is to the 
 sum of the sides, as the difference of the sides is to the 
 difference of the segments of the base. 
 
 For, about one of the angular points a, of the tri- 
 angle, as a centre, and with the greater side ac asa 
 radius, describe a circle, meeting a xB produced in £ 
 
 ZA 
 
354 ‘ 
 
 and ¥, and the base c 8 in G : also draw the perpendi- 
 
 cular A D. 
 Then, because az, AF are each equal to ac, it is 
 manifest that B x is the sum of the sides a B, AC, and 
 B F their difference. | 
 And because the perpendicular ab, from the centre, 
 bisects CG IN D, it is also plain that 3 c is the difference 
 of the segments of the base cD, DB, or DG, DB. 
 
 But since the lines cc, EF in the circle cut each 
 
 other in B, the rectangle of BE, B F is equal to the rect- 
 
 angle of cB, BG. | 
 
 Hence, cB is toBE as BF isto BG; that is, the base 
 Bc is to the sum of the sides a B, ac, as the difference 
 of those sides is to the difference of the segments of 
 the base c D, DB. OE Dis 
 
 ScHoLium. When the perpendicular Ap falls without 
 the triangle, the segments of the base must, be both 
 reckoned the same way, from the angle c or B, to the 
 foot of that perpendicular. 
 
 To these three propositions, which furnish solutions 
 to all the problems that can occur in Plane Trigono- 
 metry, it may be proper to add the following, which, 
 when applied to each of the angles, is alone sufficient 
 for that purpose. 
 
 THEOREM Iv. 
 
 96. As twice the rectangle of any two sides of a 
 -plane triangle: radius: : sum of the squares of the 
 
 same two sides—the square of the other side : cosine 
 
 -of the angle included by those sides. 
 
$55 
 
 : AD oa OE : 
 For let a p be perpendicular to the base 8 ¢, falling 
 either within or without the triangle, as in the figures. 
 Then, in the case in which the perpendicular falls 
 within the triangle, we shall have a c* = a B’-++B C'— 
 
 A B® ++ BC#— Ac? 
 2BC y 
 
 But 4 8p being a right-angled triangle, it will be as 
 
 po BD, or's D-= 
 
 ° ABCOS B 
 rad : AB:: SMBAD OrcosB: BD; orBD=-———-. 
 
 And if this value. be substituted in the first equation, 
 
 eae. CALE COS R A B® -+- BC?—a C? 
 it will give — = ——>~--——— 3; orcosB=7rxX 
 r Zee 
 
 AB? -+BCc?— ac? 
 
 ZABX BC ' 
 Again, if the perpendicular a p falls without the tri- 
 
 angle, we shall have, ac’ = aB’ + Bo*?-+ 2BCXBD, 
 
 bata Sak AC* — ABY—— BC# 
 The Qnt . 
 
 And since Azp isa right-angled triangle, rad: axB:: 
 
 . AB COS ABD 
 sm BAD OrcosaABD:BD; Or BD= Per aeetia ha 
 
 But asp being the supplement of anc, cos aBD = 
 ABCOS ABC 
 
 — cos ABC; whence B D =— —_—-—-. 
 
 . And if this value be substituted in the first equation, 
 
 Ac* — Ap* — 3c* ABCOSABC 
 
 we shall have Ben ee aT CeO 
 
 ZA 
 
' 356 
 Sox ae aha Ts whence, if these equations be 
 turned into analogies, we shall have 2 AB x1 B ci : rad 
 >:AaB? +Be?— ac’: cos Z ABC. 
 ScHo.ium. If the three angles of the triangle be 
 denoted by a, 8, c, and their opposite or correspond- 
 ing sides by a, 4, c, the four theorems here demon- 
 
 strated may be exhibited in general terms, as follows : 
 
 1. Sina Redes 
 ; bh 
 i es bese A 
 9, Tan ‘i rar cot z 
 ae 
 3. BD & fete y ons add AD 
 Bt Co mm Q* 
 LiMn tw anita ben r( pea 
 
 DEMONSTRATIONS OF THE PRINCIPAL THEOREMS 
 IN SPHERICAL TRIGONOMETRY. 
 
 ‘In treating of the practical part of Plane Trigono- 
 metry, no distinction was. made between right-angled 
 and oblique-angled triangles, on account of the three 
 rules, which are there given, being sufficient to solve 
 every problem that can occur in this branch of the sub- 
 ject, whatever may be the species of the triangle. 
 ‘But in Spherical Trigonometry, where, from the na- 
 ture of the subject, the rules of practice become more 
 niimerous, it was judged proper to class the various 
 species of these kind of triangles under the three heads 
 
 of right-angled, quadrantal, and oblique-angled trian- . 
 
 - gles, and to give the rule for each case separately. 
 The common rules, however, for the various cases in 
 
 this branch of the science, as well asm the former, may 
 
357 
 
 be all derived from the three following theorems, which 
 have the same generality with those mentioned above ; 
 and are here demonstrated, from geometrical princi- 
 ples, in a manner equally simple and perspicuous. 
 
 THEOREM I. 
 x97, Jn any right-angled spherical triangle a Bc, the 
 sine of either of the legsis to radius, as the tangent of 
 the other leg is to the tangent of its opposite angle. 
 
 ? 
 
 E 
 A 
 
 7g 
 
 For let o be the centre of the sphere, and having 
 joined oa, oc, oB, draw C E perpendicuiar to oB: also, 
 in the plane a o 8, draw FE perpendicular to the same 
 line oB, meeting o a, produced, in F; and join Fc. 
 
 Then, since 0 £ is at right angles to both Ec and EF, 
 it will be at right angles to the plane E Fc. 
 
 And because the plane c oB passes through o£, it will 
 also be perpendicular to the plane r Fc; or, which is 
 the same thing, the plane EFc will be perpendicular 
 to the plane cos. 
 
 But the angle acB being a right angle (by hyp.) 
 the plane co a or cor will be perpendicular to the 
 
 same plane.co B. | | 
 
 Hence the planes EFc, cor iin each perpendicu- 
 lar to the plane c 0B, their common section Fc will, 
 also, be perpendicular to co B. 
 
3858 
 
 And since ¥ ¢, EF, which lie in the planes coz, Aon, 
 are each perpendicular to oB, the angle Fz c will be 
 the measure of their inclination, or of the spherical 
 angle CBA. 
 
 Also, Fc 0, CE 0, being right angles, rc will be the 
 tangent of the arc ac, and c £ the sine of the arc ca, 
 to the radius of the sphere oc. 
 
 Hence, rcs being a right-angled plane triangle, right- 
 angled at c, weshallhaverc: rad::cr:tan 2FeC; 
 or sinv¢ 'B*s rad 2**'tan APCs tans ZA Bie: Q.E.D. 
 
 ScHoLium. .By using the same notation as in the 
 former théorem the present one becomes 7 tan. = 
 
 2 
 sina tan B; andif 3e8 be put for tan B, we shall have 
 ‘Sin a= tan bicou.Re 
 which is the. formula for the second case of right- 
 angled spherical triangles. 
 : THEOREM Il. 
 98. In any spherical triangle a 8 c, whether right- 
 
 angled or oblique-angled, the sines of the sides are as 
 the sines of their opposite angles; and conversely. 
 
 B 
 
 For, let o be the centre of the sphere, and having 
 joined 0A, OC, 0B, draw a D perpendicular to the plane 
 ©BC3 also make p & perpendicular to oB, and DF to 
 
 oc; and join ag, AF. 
 
“i 
 
 359 
 
 ¥ aly 
 
 - Then, because aD is perpendicular to the plane ozc, 
 each of the planes ADE, ar p3swhich pass through ap, 
 will also be perpendicular to that plane. 
 
 And since E D is perpendicular to oB, and the plane 
 ADE to the plane oB c, the line a 5, which lies in the 
 plane aps, and is drawn from the same point &, is also 
 perpendicular to os. 
 
 In like manner, because F D is perpendicular to oc, 
 and the plane arp to the plane ozc, the line Fa, which 
 lies-in the plane a FD, and is drawn from the same 
 point F, is perpendicular to oc, | 
 
 Hence the angles agp, arp, which measure the in- 
 clinations of the planes ao B, Aoc, will measure the 
 angles cB A, BCA of the spherical triangle az c. 
 
 Also, a F, being perpendicular to oc, is the sine of 
 the angle aor, or of the arc ac; and ag, whichis 
 perpendicular to o B, is the sine of the angle a o B, or 
 of the arc a B. 
 
 But a DE, AFD, being right-angled plane triangles, 
 right-angled at p, we shall have ap = az sin Z aED, 
 and aD= AFSIN 4 AFD.. 
 
 Whence, by equality, az sin 4 AED =aFsin 4 
 AFD; and consequently, Az: sin 4 aFD:: AF: sin 
 AED; Or SiN AB: sin Popes 4 Oy2sin 74 c)ssin 
 opposite Z B. 0. ED. 
 
 Scuorium. If the three angles of the triangle arc 
 
 be. denoted by a, B, c, and their corresponding oppo. . ° 
 
 site sides by a, 4, c, the proportion obtained above may 
 be represented by the following equation : 
 Sin asin B = sin 0 sin a, 
 
360 
 
 And if asc be aright-angled triangle, of which c is ~ 
 
 the right angle, and c the hypothenuse, we shall have _ 
 rsin@ = sinc sin A, 
 which is the formula for the first case of right-angled 
 spherical triangles. | 
 | THEOREM HI. 
 
 99. As the rectangle of the sines of any two sides of 
 a spherical triangle : radius : : rectangle of radius and 
 the cosine of the other side — the rectangle of the co- 
 sines of the same two sides : cosine of the angle in- 
 cluded by those sides. | 
 
 For having joined 04, 0 B, oc, draw FD in the plane 
 ope, and pe in the plane 08, each perpendicular to 
 their common section 0 B, and join EF. 
 
 Then, because the angle EDF is the measure of the 
 inclination of the planes 0 BC, O48, it is also the mea- 
 sure of the spherical angle a Bc or B. 
 
 r(DE? + pF? — EF” 
 And because cos EDF = seas il ) 
 
 ; ,and cos EOF 
 ZDEX DF 
 
 2 OF X OF COS EOF 
 , OF EF*= OE? -+ oF? ——__-_______, 
 ; ie; 
 
 __ r(on*or EF’) 
 
 ZOE X OF 
 
 if this be substituted in the first equation, we shall have 
 
 r(DE* 4+ DF’ — OZ’ — oF?) +2 of x OF COs EOF 
 2 DE X DF 
 
 But o:?— Ep* and oF*— pF’ are each equal to op*; 
 
 cos EDF = 
 
361 
 
 ; i : 
 whence, cos EDF, or its equal cos 4 B= 
 OE X OF COSEOF ~7 X OD? _ OE X OF COS AC—7 X OD® 
 DEX DF A DE X DF : 
 s OE of ) r OF ¥ 
 Or, since — = ———— == >= - = 
 DE sin DOE sin AB? DF sin DOF 
 r OD COSDOE © COSAB OD | COSDOF __ cos BC 
 sin BC? DE SINDOE SIN AB? DF sinDOF sin BC? 
 
 if these values be substituted in the former equation, 
 
 r? cOSAC —7 COS AB COSBC 
 
 we shall have cos B = , and con. 
 
 sin AB sin BC 
 sequently, sin AB X SIN BC :7r::7 COS AC — COS AB 
 x cos BC :.cos:'Z. Be. OB ab. 
 ScHOLIUM. By using the same letters for the sides 
 and angles of the triangle, as in the two former theo- 
 -rems, the above formula becomes 
 
 r-cos6—rcosacose 
 
 —. 
 
 sin a sinc 
 
 Cos 8B = 
 
 Which principle being applied successively to all the 
 three angles, furnishes three equations, which are suf- 
 ficient for resolving all the problems of spherical trigo- 
 nometry ; having the same generality with respect to 
 spherical triangles, that the theorem given in art. 96 
 has with respect to plane triangles. 
 
 A similar formula may also be readily obtained for 
 the cosine of either of the sides in terms of the sines and 
 cosines of the three angles. Jor since any spherical 
 _ triangle, whose sides are A, B, C, and opposite angles 
 a, 5, c, answers to the polar triangle, whose sides are 
 180°— a, 180°—8, and 180°—c, and the angles 180° 
 —a, 180°—b, 180°—c, we shall have : 
 Pi tds es r’cos( 180°) — rcos(180°-8)cos (180°-c) 
 
 vate) 
 Cos (1 BP er sin (180° — 8) sin oi — Cc) 
 
362 
 
 And because cos (180° — a) = — cos a, cos (180° 
 —a)=—cosa, &c. this equation, when reduced, 
 becomes 
 
 ° ¢ 
 rcOS A + 7 COS BCOS ¢€ 
 sin B sinc 
 
 Which latter formula resolves immediately the case 
 in which it is required to find a side by means of the 
 three angles, as the former does when it is required 
 to find an angle by means of the three sides. 
 
 100. Having obtained by art. 97 the proper formula 
 for thesolution of the first two cases of ri ght-angled sphe- 
 rical triangles, the rest may be easily derived from them, | 
 by means of the complemental triangle, as follows : 
 
 Cosa = 
 
 Let asc be a right-angled spherical triangle, and 
 D A E the complemental triangle, formed by producing 
 the sides B A, C A, If necessary, to quadrants. 
 
 Then, because the triangle pa & is also right-angled 
 at E, it follows, from theorem 2, that 7 sin a’ = sin e 
 sin A; and since sin a’== cos EF = cos Z B, andsine 
 = cos b, if these be substituted for a’ and e, we shall - 
 have, for the 3d case, 
 
 rcosB = cosOsin a. 
 
 Also, since r sin a’ = tan d cot p, by Ist theorem; 
 and sin a’=cos E F = cos B, tan d= cot c, andcot D 
 
863 
 
 = cot c F = tan a, we shall have, by substitution, for 
 the 4th case, 
 rcos B = tana cote. 
 
 In like manner, because 7 sin d==sin z sin p, by 
 ‘the 1st theorem, and sin d =cosc, sine = cos J, 
 and sin D = sin c F = cosa, we shall have, by a simi- 
 lar substitution, for the 5th case, 
 
 r cos c = cosa cos b. 
 
 Finally, because r sin d = tan a’ cot a, by the 1st 
 theorem, and sin d = cos ¢, tan a’=cot EF = cot ZB, 
 we shall have, by a like substitution, for the 6th case, 
 
 rcos c = cot a cot B. 
 
 Hence, ail the cases of right-angled spherical tri- 
 angles being collected together, may be commodiously 
 exhibited at one view, as follows : 
 
 rsina = sine sina = tan b cots 
 rcosB =cosbsin A = tan acotc 
 T€08 ¢ == COS @ COSdiss={COLrA cots. 
 
 The same forms will also hold in any case, by tak- 
 ing such other sides and angles as are similarly situated 
 with respect to each other. 
 
 101. The various cases of quadrantal spherical tri- 
 angles may also be derived, in a similar way, from the 
 principles above demonstrated, by means of the com- 
 
 _plemental triangle, as follows : | 
 
364. 
 
 ee. 
 
 Let a’x’c’ be a quadrantal spherical triangle, and — 
 
 A’c’D, or BC’F the Boplemental triangle, formed by 
 producing the sides a’c’, B’c’, if necessary, to quadrants. 
 Then, since by theorem 2 the sines of the sides of 
 
 any spherical triangle are as the sines of their oppan 
 
 site angles, it follows for the first case, that sin 4’ B’ 
 (sin 90°, or pehih) ysis MAC GA SING Sie By OF 
 
 r sin B’ = sin UO’ sinc’, 
 
 Also, because the triangle a’c’p is right-angled at | 
 
 D, 7 sin c = tana cotc’a’p, by Ist theorem; but sin 
 c= sin Z B, tan a = cota, and cot c’a’D = tan 
 pa‘c’, ortan 4 a’ (BA’D being aright 2); whence, 
 by substitution, we shall have, for the 2d case, 
 
 rsin B = cot a’ tan a’. 
 
 In like manner, because r sin a = sin U’ sin o’a’b, 
 and sin a = cosa’, sinc’a’D = cos BAC, or cos ZA’ 
 (s’a’D being a right 4), we shall Hotes by substitu- 
 tion, for the 3d case, 
 
 7.COS Giz= sin 6 COS A’. 
 
 Again, because 7 sin a = tan c cot c’, by the Ist the- 
 orem, and sin a = cos a’, tanc = tan a’B’D, or tanB, 
 we shall have, by substitution, for the 4th case, 
 
 r cos a’ = tan B’ cot c’. 
 
 Also, because r cos c’ = cos c sin c’a’D, as has been 
 shown for case 3 of right-angled triangles, and cos ¢ 
 =3'c6s"\Z By and sinc’ a’ p is='cos' B ACC) or cos #2 A’ 
 (’a’p being a right 2), we shall have, by substitu- 
 tion, for the 5t) case, 
 
 r cos Cc’ == cos A’ COS B. 
 
365 
 
 Lastly, because r cos c’ = tan a cot U’, as has been 
 shown for the 4th case of right-angled triangles, and 
 tan a = cot a’, we shall have, by substitution, for the 
 6th case, 
 *. r cos c’ = cot a’ cot Ll’, 
 
 Hence, also, by taking, for the sake of uniformity, 
 _ such sides and angles as correspond with those of the 
 former triangle, all the cases of quadrantal spherical . 
 triangles may be exhibited as below: 
 
 7S A == Sige sinc == cot O° tan B’ 
 
 r'cos b’ = sifa’ cos 8B = tan A’ cot ‘Cc’ 
 
 r COSC’ = cus A’ cos B = cot a’ cot b’. 
 
 Where, by comparing together the similar forms of 
 the two tables, it appears (as well as from the polar 
 triangle, def. 11) that if the sides and angles of any 
 guadrantal spherical triangle be considered, reversedly, 
 as the angles and sides of a right-angled one, the rules 
 for the latter will equally apply to all the cases of the 
 former; observing to change the terms /ife and unlike | 
 for each other, when the hypothenusal angle is con- 
 cerned. 
 
 102. Next, in order to apply the theorems, above 
 demonstrated, to the solution of the remaining cases 
 of oblique-angled spherical triangles, it will be proper, 
 ~ for the sake of convenience, to present the formule 
 already obtained, under all the varieties of which they 
 are susceptible; which, for the case of the sines, in 
 theorem 11, are as follows: 
 
 sina@asinB , sinasinc 
 
 sin 6 sin ¢ 
 
 Sin a= 
 
Sin Ls fees 
 Sin c s 
 Sin'g |= 
 Sin Ga 
 
 Sinc = 
 
 366 
 
 sin 6 sin A 
 
 sin a 
 
 sinc sin A 
 
 sin @ 
 sin csi A 
 
 sin ¢ 
 
 sinc sin B 
 
 sin c 
 
 sin asin c 
 
 sin A 
 
 | 
 
 sin &sin ¢ 
 sin ¢ 
 
 sinc sin B 
 
 von eee 
 
 sin bsin A 
 sin B 
 
 sin @.sin B 
 
 sin A 
 sin &6sinc 
 
 sin B 
 
 > 
 
 The three first of which equations furnish the means 
 of determining an angle of the triangle, by means of 
 
 two of the sides, and the angle opposite to one of / 
 them; and the three latter determine a side by means 
 of two of the angles and the side opposite to one of © 
 
 them. 
 
 103. In like manner, the general expression obtained 
 
 by theorem 111, when applied to all the sides and angles 
 
 of the triangle, admits of the following permutations :_ 
 
 Cos A 
 Cos B 
 Cos c 
 
 Ts Plomen 
 Cosh * == 
 
 ; Cos C ‘ieee 
 The three. first 
 
 —— 
 a 
 
 —— 
 
 r*cosa—rcoshcose 
 sin é sinc 
 
 . b 
 
 r cos6—rcosacose 
 
 sinasine 
 
 rz “ 
 
 rcos acos b+ sina sin écosc 
 i TR aa Te 
 of which equations give the angles 
 
 r?cos¢—rcosacos 
 
 sin asin b | 
 rcosécosc¢c + sin dsinccos a 
 i ad 
 
 rcosacosc + sin asince cos B 
 
 by means of the sides; and the three latter give either 
 of the sides, by means of the other two sides and their 
 
 contained angle. 
 
367 
 
 104, Also, the second general expression, obtained 
 from the same theorem, when applied as at gives 
 the following formule : 
 
 rv cos aA + rcosBcos¢c 
 Cos a = LIL Ah nl 
 sin B sin C 
 2 
 r*cosB + 7rcosacosc 
 Cos 8 = ————. —————_— 
 sin A sinc 
 rT? cos C + 7 CoS A COSB 
 Cos tae ee ee 
 sin A sin B 
 cOS a sinB sinc —r7r cos BCcOosSCc 
 (Costa ra paiee se 
 r 
 cos4 sin asin c — rcos Acosc 
 Cos B= 
 ay 
 -COS¢ SIN A SiN B — 7 COS A COS B 
 Cos C= OO > 
 
 7 
 
 The three first of which equations give the sides by 
 means of the angles; and the three latter give either 
 of the angles, by means of the two other angles and 
 their included side. 
 
 105. Again, if the value of the cosine of c in the 6th 
 of the 2d set of equations, be substituted for the cosine 
 of cin the 1st of the same set, we shall have r cos A 
 sinc =rcos a sinl —cosc sina cost; and by sub- 
 stituting in this equation the value of the sine of c, as 
 given in the 6th of the Ist set of the same formule, it 
 will become, after reducing it to its most simple form, 
 
 7 COS A rcosasin é’— sinacosécosc 
 ——— = cota = ——_—— : a eel 
 sin A sin a sinc 
 
 106. And if this expression be applied to each of the 
 three angles of the triangle, by using all the permu- 
 tations of which it is capable, we shall have the six 
 following formule : 
 
 r cos asin’ — sinacos fcosc 
 
 Cota == 
 
 sin @ sin C 
 
368 
 
 . rsinacos§é—cosasindcosc — 
 Cot 3 = 
 
 sin 6 sin c 
 rcOs asin ¢ — sina cose cos B 
 Cot A= Se eee 
 sin @ Sin B. 
 rsin @cos ¢ — cos @ sine COS B 
 sin ¢ sin B 
 rceos ésinc — sin &6cosccosa 
 Cot 8 = — 
 
 sin &sin A 
 
 r sin.b cos ¢ — cos db sinccos 4 i 
 sin¢ sin A 
 
 Which equations determine any two angles of the 
 triangle, when we know the third angle and the two 
 sides by which it is contained. | 
 
 107. The cotangent of either of the sides may also be © 
 determined, in a similar manner, by means of the 1st and 
 3d set of equations; or more readily by the 1st of the 3d 
 set and the polar triangle. For since cot (180°— @) = 
 
 rcos( 180°A sin (180° ) —sin( 180°—a }cos{ 180° Te Jeost 180" c)F 
 sin (180° — a) sin (180° — ¢) 
 . cos A sin B in A COS BCOS¢ 
 we shall have cot a = 7S 480 BF 
 SIn A SIN ¢ 
 And this, being applied to each of the three sides, 
 by using all the permutations of which the letters are 
 
 capable, will give the six following formule: . 
 
 : rcos A Sin B + coSc¢sin 4 COSB 
 
 Cot t= 
 sinc Siti A 
 
 r sin ACOS B+ .coS¢COSASINE 
 
 CO) a saa ee ee ee 
 . sin ¢ sin B 
 
 rcosa since -+cosdsinacos ec 
 sin 6 sin A 
 
 rsin aA cos c + cos dcos Asin c 
 
 Coble eh 
 
 sin ’sin ¢ 
 
 r cos BSIN C + COS@sin BCOSC 
 
 Cot 6 = 
 sin @ sin B 
 
 rsinBcosc 4+ cosacosBsin¢ 
 
 Cot. Eee 
 
 sin @ sim Cc 
 
369 
 
 Which equations serve to determine any two sides of 
 the triangle, when we know the third side and the two 
 angles between which it is contained (w). 
 
 108. Of the five classes.of formule here given, 
 which are sufficient for resolving directly all the cases 
 of spherical trigonometry, the four last deserve parti- 
 cular notice, not only on account of their elegance, but 
 from their possessing the property of showing whether 
 
 an arc or an angle be greater or less than a quadrant, 
 or 90°. , 1 
 
 For the cosine and cotangent of any are being — in 
 the first quadrant, and -+- in the second, which is the 
 limit of the sides and angles. of every triangle, if care 
 be taken to give to the known quantities, which enter 
 into any of these formule; their proper signs, the sign 
 of the result will show, the species of the arc or angle 
 sought. 
 
 But this cannot be known from the expression of the 
 sine of an arc or angle, as its value and sign are the 
 same both for the arc and its supplement. — 
 
 - 109. From these formule, which, except those 
 of the first class, are not adapted to logarithmic com- 
 
 (uw) These formule are equally applicable to every species of 
 spherical triangles, whether right-angled, quadrantal, or oblique- 
 angled ; and in the two former cases they may be easily resolved 
 into the simple forms before given for the solution of those trian- 
 gles. The whole doctrine of spherical trigonometry might, there- 
 fore, have been deduced from the 2d and 3d theorems above given, 
 or from the 3d alone; but for the sake of the learner, it was 
 judged proper, in the case of right-angled and quadrantal tri- 
 angles, to follow a mode of investigation which appears some- 
 ' thing more easy and natural. 
 
 223 
 
370 
 
 putation, we can readily deduce the four elegant the- 
 orems commonly known under the name of the Ana- 
 logies of Napier, which are of great use in facilitating 
 the solution of several cases of spherical.triangles. 
 
 Thus, by a combination of the values of cos a and 
 cos C, given in art. 103, we shall have the two follow- 
 ing equations : 3 
 
 rcos A sinc = rcosa sin b — cosc sina cosh 
 
 rcosBsinc=rcosbsina — cosc sin b cosa 
 And by adding these together, and reducing them, 
 there arises | 
 Sin'c (cos a-+cos B) = (r—cos c) sin(a+é). 
 ae ee —— = oe we shall have | 
 Sin c (sin a-++sin B) = sin c (sin a-+sin b) ! 
 Sin c (sin A—sin B) = sin c (sin a—sin b) 
 Which two equations, when divided by the preceding 
 y 
 
 But since 
 
 \ -.. Sina + sins sin C sin a + sin & 
 
 one, give ————— ——" S>=*K 
 cos A + COSB r—cos C sin (a + d) 
 sin A — sin B smc ., sina — sind 
 
 : 
 COs.A.-b'COS'B.) ig COSC. eine A PAY | 
 f 
 
 And reducing these, by means of the formule given 
 in art. 27 and 30, they become Y 
 cos £ (a—S) 
 Tan 3 (a+3) = — B(a48) cote | | ; 
 
 Tan 3 (a—B) = he Bg ae cot'4 c. 
 
 Which. equations 5 ena any two angles of a 
 spherical triangle by means of the two opposite sides 
 and their included angle; and are the same as the 
 practical rule given in the former part of the work. 
 
 And if the formule, thus obtained, be applied to 
 the polar triangle, by substituting 180°—a, 180°—B,, 
 
—STl 
 
 180°— a, 180°— 6, and 180° —c in the place of a, b 
 A, B, © respectively, the result will give the two fol. 
 
 lowing analogies, 
 cos ${A—B) 
 
 Tan 3 (a6) = — nary panies tan 7c 
 Tan 3 (a—b) = eee tan £ c. 
 
 Which equations determine any two sides of a sphe- 
 rical triangle, by means of the opposite angles and 
 their included side; and are the same as the practical 
 rule already given for this purpose. 
 
 110. The logarithmic formula for determining either 
 of the angles of a spherical triangle when the three 
 sides are known, may be also obtained from the prin- 
 ciples already laid down, as follows : 
 
 If the value of the cosine of c (art. 103) be substi- 
 tuted in the formula 2 sin*4c=r*—rcosc (art. 27), 
 we shall have | 
 
 2 si Eh cos Cc cos acos 6 + sina sinb — 7 cose 
 
 —= Sor Aan ra a 
 —s <= e 
 
 ° We 
 
 7? r | sin a sin} 
 Or, since cos a cos 6 + sinasinb =r cos(a—b), 
 (art. 20) the former expression will become 
 
 2sin? 5c _ rcos (a—b) — rcose 
 
 Tea 
 
 But by art. 29, r cos (a — 6) — rcose = 2 sin $ 
 (c+b—a) sins str alec ; whence 
 Sin? J 2C __ sin (c+s—a) sin 4 (c+a—6) 
 
 ae 
 
 sin asin 8 
 
 sin @ si j 
 
 figs. cht) Arcee if sin E (c+4—a) sint (cab) 
 
 7 
 
 sin @ sin ; 
 In like manner, if ate value of the cosine of c be 
 substituted in the formula 2 cos* 4c = 7° ++ rcos ¢ 
 
 (art. 27), we shall have 
 2B 2 
 
 8 
 
Zcos*ic cosc r cose —cosacosh + sin asing 
 — =1-+ ext Sa ge 
 rt r | sin @ sin } 
 Or since — (cos a cos b — sin a sinb) = — r cos 
 (a+b) (art. 20) the former expression will become 
 2co*$c — rcosc—rcos (a+ d) 
 beh sm @ sin é : 
 
 But by art. 29, rcosc — r cos(at+b) = 2 sin 5 
 (a+b—c)sni(atbteo); whence 
 
 Cos? I ee) sin $(a+é-+c) sin J (a+3—0) 
 ye oe ' sin-a ae 
 sin 5 (a+b+c) sin 3 (a+i—c) 
 
 Or, cosic =r 7 
 ? os 4 sin a sin } . 
 
 sin 2 | 
 Or, because = ; : == tan 4c, if the former of these 
 ‘ 2 s 
 expressions be divided by the latter, we shall have 
 1 sin 4 (c+b— }b—a) s sin £ 5 (ct+a—b) 
 Tan ZI sk Nae a ie L(a+b+c) : sin 4 (at+b—c)’ 
 Where either of thes three fabediii will deter- 
 
 mine an angle, when the three sides of the triangle are 
 
 given. 
 111. By following the same mode of investigation, 
 the logarithmic expression for determining either of 
 the sides of a spherical triangle, in terms of the three 
 angles, may be obtained as follows : 
 If the value of the cosine of a (art. 104), be substi- 
 
 ‘ 2 sin?la cos a 
 tuted in the formula ——,*- = 1 — -—— (art. 27), we 
 shall have 
 sin* ia ik sin B Siz C — COS B COSC — rcosAa 
 7 aA RW 2 sin B sinc ; 
 
 Or, since sin B sin Cc — cos B cosc — rcosA = — 
 r cos (B-+ c) — rcos a, the former expression will 
 become 
 
B'S 
 
 Sint fa —rcos(B+c)—rcosa 
 Bo Vianna FR TSULee tis 
 But, by art. 29,r cos (B-+ c) +r cosa==2 cos 4 
 (A+B+c) cos }(p+c—a); whence 
 Sint5a_ —cosh(a+s-+4c) cos i (n-+-c—a) 
 
 7 sin B sin c 
 
 : ro Coa A c)cos i (p+c—a) 
 (x) Or sind a sr fst tele 2 (B+ ) 
 
 sin B sin ¢ 
 In like manner, if the value ne the i of a, be 
 
 2 cos? Sa 
 > we shall 
 
 substituted in the formula 
 
 have 
 
 Cos? fa sin B sinc + cos Bcosc ++ 7COS A 
 Oe Sn EEeeEeeeem0000wwme 
 
 fF Zsin Bsinc 
 Or, since sin B sinc ++ cosB cos c =r cos (B—C), 
 the former expression will become 
 
 Cola rcos (B—c) aL, COR Ay 
 eo eon 2 sin B sin c 
 
 But, by art. 29, 7 cos (B—c) + rcosa = 2 cosi 
 (A+B—c) cos $ (Aa-++c—B), whence 
 
 Cos? La cos £ (a-t3—c) cos 3 (a+e—3)” 
 7 PH sin B sin Cc 
 
 cos 2 (a+s—c)cosZ ici ta 
 
 n oe 
 And, COS 5 a= 7 / sin B SIN C 
 
 22°) tan “3g; if the former of 
 COS 3 a 
 
 these expressions ie divided by the latter, we shall have 
 
 (2) It may here be observed, that the second member of this 
 equation, though under a negative form, is always affirmative ; for 
 4 (a+8-+ 0) being greater that 90°, its cosine will be negative; 
 and consequently the expression — cos} (a+ 3+ ¢c) will become 
 positive. And since 8 +c—a ean never surpass 180°, or t 
 (s+c—a) be greater than 90°, it is plain that cos 3 (B+ c—a) 
 must also be positive. 
 
B74: 
 oC ae tet — cos (B+c-+4a) cos £(34+c—a) 
 Tan ga=rv cos 4 (a+B—c) cost (a+c—B) * 
 Either of which formule will determine a side when 
 _ the three angles of the triangle are given. 
 
 112. It may be still further observed, that by taking 
 all the varieties of which these last formule for the 
 tangents are susceptible, we shall have 
 
 sin 4 (a+4—c) sind (a+c—4) 
 
 sind (+-c—a) sin i (a+b+c) 
 
 sin § (64+c—a) sind (a+4—c) 
 
 sin (a+¢—b) sind (ato+c) ° 
 Teh sin + (a+c—b) sin § (b4+c—a) 
 
 Yan 2 aarti t Vite 4 (a+b—c) sin H (a+b+c) 
 
 i ae Ve cos 3( (p+c—a)cos 4 (a+s8+c) 
 
 Tan SAS ray 
 
 Tan. 2°38 9 / ies 
 
 Tan 
 
 “cos £ (a+4+-8—c)icos (a4+c—3) 
 feahe SERRE — cos % (4 +c—s) cos 3 (a+8+c) 
 Ten dalieeiny cos 4 (B+c—a) cos 2(a4+3—C) 
 
 — cos a 4 
 Tan 2 ea Rome Hel c) cos4 ) (ase Bc) 
 
 cos i (a4+c—Bs) cosd(B+c—a) | 
 113. In like manner, by taking all the varieties of » — 
 
 which the preceding formule for the tangents are sus- 
 ceptible, we shall have 
 
 j— ind (p—a) 
 Tan ae ee ga 
 2 sin $ (BA) ans 
 Tag Lato nh hae 
 ae cost (B+) % 
 
 ee he ee ee ee = 
 
 —b sind (c—B) 
 Tan - onal Sekine aS 
 v7 sin 3 (c+B) ape hs 
 +5 cos (c—s) . | 
 et 4 (C43) tan 3 @ 
 te arate esa mien ' 
 Zz sin} ( 4 (atc) = b 
 
 a+ce cos 4 (A—c ' 
 Tan 23 ime mye ) Lh 
 
 —s “—_/ tan 1 
 ie? cos+ (a+c) ‘aa 
 
Tan 
 
 Tan 
 
 Tan 
 
 Tan 
 
 ‘lan —— == 
 
 A+c 
 Z 
 
 Tan 
 
 ($75 
 
 sin u (s—a) 
 sin 4 (+a) 
 cos + (4—a) 
 cos £ (b+a +a) 
 sin} (c— (c—b) 
 sin 4 sind (c+) 
 cos 4 (¢—b) 
 cos } oe 
 sin } (a— 
 
 sin a am 
 oat (a—c) 
 
 a cot 4 
 cos 4 (a+c) 
 
 cot 4 C 
 
 cot 4c 
 
 cot 4 A 
 
 E 
 
 cots a 
 cot 4 1s 
 
 114. And from these last 12 formule may be de. 
 duced the following, which serve to find the third side 
 or the third angle, when two sides and the angle oppo- 
 
 site to one of them are known. 
 Tan bc = ees} tan 4 (b—a) 
 cJanic= aaa tan 5 (b-+a) 
 Tet ar th ae es tan 3 (c—C) 
 Tan fas at a tan £ (c+) 
 Tan3.b = a ATO tan 3 an 3 (a—c) 
 Tan id= ea (hee) tan 5 (a+c) 
 Cots Cis = Ws tan 2 (B—a) 
 Cot 3 o = oa ras tan 2 (B++4) 
 Cotta = anes tan 3 (c—B) 
 Cotia= ey tan 4 (c+28) 
 
 cos 4 (cm 
 
% 
 
 376 
 
 sin % (a-+¢) 
 24 oS an + : 
 Cot is = Ste ba fan $ (Aa—c) 
 1 Oi ee ea 
 Cot $3 = — (gue) 0 2 (a+c). 
 
 Which formule, joined to those of art. 102, for de- 
 
 . termining a side or angle, when two sides and an angle 
 
 opposite to one of them, or two angles and a side op- 
 posite to one of them, are given, are sufficient for re- 
 solving, logarithmically,, every case of spherical tri- 
 angles. 
 
 It may here also be observed, that these equations 
 furnish the means of determining the affections of the 
 sides and angles of spherical triangles, in all the cases 
 which are not necessarily ambiguous, by barely at- 
 tending to the signs of the quantities of which they 
 are composed. 
 
 Thus, in the equation r cos a = cos b cos, for 
 right-angled spherical triangles, the 3 sides must be all 
 equal to 90°, or all less, or two of them greater and 
 the third less; as no other combination of them can ren- 
 der the sign of cos 6 cos clike that of cos a, as the 
 equation requires. | 
 
 Also, in the last analogy for oblique-angled spheri- 
 cal triangles, art. 113, as cot 4 B and cos 3 (@“c) are 
 both positive, tan 1 (a-+-c) and cos} (a+c) must 
 have the same sign; hence, half the sum of any two 
 sides is of the same kind as half the sum of their op- 
 posite angles: which consideration will sometimes take 
 away an ambiguity that might otherwise arise, in cases 
 where the quantity sought is to be determined by means 
 of a sine. my 
 
377 
 
 SPHERICAL THEOREMS. 
 
 THEOREM I. 
 115. If two arcs of circles meet each other, they | 
 make two angles, which are, together, equal to two 
 
 right angles, or 180°. 
 
 Let the are aB meet the,arc cp in the point 8; then 
 will the two 4° aBc, ABD be equal to two right angles, 
 
 For, suppose the arc EB to be perpendicular to cp, 
 then the ZS ERC, EBD are right angles. 
 
 And since the Z EzBDis equal tothe Z2* EBA, 
 ABD, thethree Z° EBC, EBA, ABD are equal to.two 
 right 4%. 
 
 But the two 4/* EBC, EBA are equal tothe Z Azc, 
 whence the two 4$ aBc, ABD are also equal to two 
 right angles. Q. BD. 
 
 THEOREM II. 
 
 116. If two arcs of circles intersect each other, the 
 
 vertical or opposite angles will be equal. 
 
 oy 
 Let the two arcs A B, c D intersect each other in &, 
 
 then will the 4 aAxEc be equalto DEB, and AED to 
 i | 
 GC. Be i 
 
378 
 
 For since the arc AE meets the arc cD, the 25 axc, 
 AED are, together, equal to two right 2% (theo. 1). 
 And because the arc D E meets the are a B, the Z§ 
 DEB, DEA are also equal to two right 2%. 
 Whence the sum of the 4* AEC, AED is equal to 
 the sum of the Z' pEB, DEA... ™ 
 And if the Z agp, which is common, be taken . 
 away, the remaining 2 axEc will be equal to the re- 
 maining 4 DEB. And in the same manner it may be 
 shown, that the Z a ED is equal to CEB. Q. B.D. 
 Cor. If two arcs of circles intersect each other, the 
 £* about the point of intersection are, together, equal 
 to four right 2°, 
 THEOREM It. (y) 
 117. Anangle made by any two great circles of the 
 sphere is equal to the angle of inclination of the planes 
 of those circles. 
 
 AVN 
 
 Let BAE be spherical angle, made by the two 
 great circles ¢ BA, CEA; then will this angle be equal 
 to the angle of inclination of the planes of those circles, — 
 
 For take the arcs AB, a E each equal to 90°, and 
 through the points B, E draw the arc of a great circle BE, 
 and from pb, the centre of the sphere, draw DB, DE. 
 
 (y) Any two great circles of the sphere anc, asec, which pass 
 through the poles a, ¢ of another great circle B E, cut all the pa- 
 rallels HG, Be into similar arcs, or such as contain the same 
 number of degrees. v 
 
oro - 
 
 » Then because a B, AE are quadrants, the lines p gz, 
 DE are each perpendicular to the common section ac ; 
 and consequently B px is the 4 of inclination of the 
 planes cBA, CEA. 
 
 But since DB, DE are equal, being radii of the sphere, 
 the Z BbeE, which is measured by the arc Bk, is equal 
 to the Z Bak, which is measured by the same arc. 
 
 And if rH be drawn in the plane cBa, and Fe inthe 
 plane c £ A, each perpendicular to the common section 
 Ac, the Z arc, which is equal to Z gpg, will also 
 be equal to the Z Bak. Q. Ben 
 
 Cor. The 4 B AE made by two great circles of the 
 sphere Ba, EA, is equal to the 4 7 am, formed by two 
 tangents drawn from the angular point a, one in each 
 plane, these tangents being each perpendicular to the 
 diameter Ac. 
 
 ScHotium. As the 2* Bag, BCE, formed by the 
 intersections of two great circles of the sphere, are 
 equal, so it may be easily proved that if the arcs acz, 
 ADB of any two circles, whether great or small, inter- 
 sect each other, either in a plane, or on the surface of 
 a sphere, the opposite Z4* of the lunule Bab, agp 
 
 will be equal. 
 
 For draw az, af touching the arcs AD, AC in a, 
 and BE, BF touching the arcs BD, BC in B, and meet- 
 ing each other in £ and F: also join gr. 
 
380 
 
 Then since BA, EB are tangents to the circle aps, 
 and meet in the point z, they are equal. 
 
 And, because F A, FB are also tangents to the circle 
 Acs, and meetin the point F, they are equal. 
 
 Hence the sides aE, aF of the A are being equal 
 to the sides BE, BF of the A BF &£, and the base EF 
 common, the Z EAF will be = the Z exsF. 
 
 But these 4° are equal to the curvilinear 4° BaD, 
 ABD (cor.); whence the latter are also equal. 
 
 THEOREM IV. 
 
 118. The distance of the poles of any two great 
 circles of the sphere is equal to the angle of inclination 
 of the planes of those circles. 
 
 Let AEB, CED be two great circles, and Pp, p their 
 poles ; then will the arc p p be equal to the angle of 
 their inclination Aoc, or BOD. | 
 
 For since p is the pole of the circle ar B, and p of 
 CED, thearc Pp A will be equal to p c, being each qua- 
 drants, or 90°. 
 
 And if p c, which is common to each, be taken away, 
 the remaining are P p, which is the distance of the two 
 poles, is equal to c a, which is the measure of the 2 
 of inclination Aoc. Q. E. D. 
 
381 
 
 THEOREM V. 
 
 119. If arcs be described from the three angular 
 points of any spherical triangle a3Bc, as poles, the sides 
 and angles of the triangle p EF, formed by their inter- 
 section, will be the pasa. of the angles and sides 
 of the former” and vice versa. 
 
 For let the sides of the A asc be produced, if ne- 
 cessary, till they meet the sides of the A pe F, in the 
 points G, H, I, K, L, M3 and draw the ares Ea, EC. 
 
 Then, since the points a, c are the poles of the arcs 
 EF, ED, the arcs cz, AE will be quadrants, or each 
 90°; and consequently & is the pole of ac. 
 
 In like manner, it may also be shown that F is the 
 pole of a B, and p the pole of Bc. 
 
 Since, therefore, £ is the pole of a u, and F the pole 
 of ac, the arcs Eu, Fe will-be each quadrants, and 
 their sum EH +FGOrEF+GH =180°. 
 
 But a being the pole of zF, the arcs AG, AH are 
 also each quadrants; and consequently cH is the mea- 
 sure of the Z BAC or A, 
 
 Whence, EF+cGH being = = 180°, zF-+ Za is also 
 
 = 180°; or EF = 180°— Z a. 
 
 And in the same manner it may be shown, that F pD 
 
 = 180° — 4 3B, andpE=180°— Zc. 
 
882 
 
 Again, c being the pole of mp, and B the pole of 
 1p, the arcs cM, BI are each quadrants; and their sum 
 cm-+ BlormMi+Bc = 180°, 
 
 But the arc m1 having the point p for its pole, is the 
 measure of the Z EDF or D3; whence Z D+ BCis 
 also = 180°; or Z D = 180°— Be. 
 
 And in the same way it may be shown, that Z ut 
 = 180° — ac, and Z ¥= 180° — AB. ; 
 
 Hence, also, reciprocally, 4 a = 180°— EF, ZB 
 — 180°— Fp, Zc = 180° — DE; andpc = 180° 
 —Zd,Ac=180°-— ZF, AB=180°— ZF. Q.E.D. 
 
 THEOREM VI. 
 
 190. If the three sides of one spherical triangle be 
 equal to the three sides of another, each to each, the 
 angles which are. opposite to the equal sides will be 
 
 equal. A " 
 
 | Spies 0 Ninn 
 
 Let aBc, D£EF be two spherical triangles, having 
 the side AB = DE, AC = DF, andBc==£F; ‘then 
 BL Le 9 Av seme Mey) AS ci eiieranicl c= ZF. 
 
 For take any two equal distances oc, OL, on the 
 equal radii o a, oD; and in the planes oac, oAB, draw 
 G K, GH each perpendicular too a; and in the planes 
 DOF, DOE draw LN, L M each perpendicular to op. 
 
 Then, because the side oc of the A oH G=oL of 
 the A oim, the 4 cox = Z Lon (being measured by 
 
283 
 
 the equal arcs ac, pF), and 4 ocK = 4 oLn (being 
 right 4°), the side cx will be = to Ln, and ok to on. 
 
 In like manner, because the side oc of the Aone 
 =o of the*A omu, the Z cou = ZLoM ( being 
 measured by the equal arcs aB, DE), andthe 4 ocH 
 == Z om (being right 2°), the side cu will be = to 
 LM, and ou to om. 
 
 Since, therefore, the sides oK, on of the A OHK, 
 are = the sideson, om of the A omn, and the Z HoK 
 = £ mon (being measured by the equal arcs Bc, EF), 
 the side x H will also be = to NM. 
 
 Hence, the three sides of the A cu kK being equal 
 to the three sides of the A umn, the Z2°KGH, NLM, 
 which are opposite to the equal sides KH, NM will also 
 be equal. ‘ 
 
 But the 4 xcu, which is the inclination of the two 
 planes aoc, 04B is = the spherical Z Bac, andthe 
 £ nuLM, which is the inclination of the two planes Dor, 
 DOE, Is = the spherical 4 EDF; whence the spherical 
 £* BAC, EDF are also equal. 
 
 And, if two equal distances be taken in the radii og, 
 OE, Or OC, OF, and perpendiculars be drawn from their 
 extremities in the other planes, as before, it may be 
 shown, inva similar manner, that the Z agscis= Z 
 D Eitye and, 4) A.C. B= Z DE, BA(z). Ceuta DD. 
 
 (x) Most of the trigonometrical writers have attempted to prove 
 this, and several other propositions in Spherics, by means of laying 
 one triangle upon the other, asin plane geometry. But this me- 
 thod, in several cases, is not exact, as there may be two spherical 
 triangles, as well as two solids, or two solid angles, which are 
 
384 
 
 THEOREM VI. 
 121. If the three angles of one spherical triangle be 
 equal to the three angles of another, each to each, the 
 
 sides which are opposite to the equal angles will be 
 L 
 
 equal, 
 
 cial oh M 
 tick ABC, DEF be two spherical A’, having 24 a = 
 
 Zp, 4B 4 8 and2 che" Z Fe ther? ah pe ee 
 
 Se FY and CuA == "Rp. 
 
 For, about the angular points of the two AS describe 
 the supplemental or polar A’ Gux, LMN. 
 
 Then, because the 4* a, B, c are, respectively, = 
 tothe Z*np, £, F, the sides HK, KG, GH which are the 
 supplements of the former, will be = sides MN, NL, 
 uM which are the supplements of the latter. 
 
 And since the three sides of the A G HK are equal 
 to the three sides of the A L MN, each to each, the 45 
 G, H, K will be respectively equal to the Z* L, M,N. 
 
 But the 2° c, Hu, K being the supplements of the 
 sides Bc, CA, AB, andthe 4’ 1, m, wn the supple- 
 ments of EF, FD, D£, the side an willbe = pz, BC 
 chee land Canaria yg Q. E. D. 
 
 equal in all their constituent parts, and yet are not superposable, or 
 equal by coincidence; asisshown by Legendre, prop. 11. b. vii. 
 of his Geometry; where he also observes (note 1.) that Dr. Simp- 
 son, in objecting to the demonstration of prop. 28, b. xi. of Euclid, 
 has himself fallen into this mistake, by founding his demonstra- 
 tion upon a coincidence that does not exist. 
 
385 
 
 THEOREM VIII. | 
 122. If two sides and the included angle of one 
 spherical triangle be equal to two sides and the in- 
 . cluded angle of another, each to each, the remaining 
 sides and angles will be equal. 
 
 Let apc, DEF be two spherical A® having the side 
 AB= Ds, AC=—=pF, and Z BAC Z EDF; then 
 will the sidesc = EF, Z asc = Z DEF,and Z acs 
 == (Li a F. 
 
 For take any two equal distances o Gc, ox on the 
 equal radiioa, op; and in the planes oac, oaB draw 
 G K, G H each perpendicular to oa ; and in the planes 
 DOF, DO E, draw tN, LM each perpendicular to op. 
 
 Then, because the side oc of the A oke is = to 
 oL of the A ons, the Z cox = Z LON (being mea- 
 sured by the equal arcs ac, pF), and 2 ocK = 4 OLN 
 (being right 2°), the side c x will be = to LN, and 
 OK to ON. i : 
 
 In like manner, because the side oc, of the A onc, 
 is = OL, of the A om, the 4 coH = 4 Lom (being 
 measured by the equal arcs aB, DE), and the 4 ocH 
 
 = £ 0O.L™ (being right ZS), the side cH will be = 
 ai. and oH tooM. 
 
 Since, therefore, the sides cx, Gu, of the AGHK, 
 are equal to LN, LM of the ALmMN,and 4 KcH = 
 
 2c 
 
386 
 
 Z wum (being the inclinations of the planes which 
 form the equal 4° Bnicy EDF), the side kK H will also 
 be equal to N M. | 
 
 Hence the three sides of the A u K o being equal to 
 the three sides of the Aomwn, the Z°HOK, MON, 
 which are opposite to the equal sides KH, NM, are | 
 also equal. 
 
 But these 4* being at the centre of the sphere, are 
 measured by the arcs Bc, EF at the circumference, which 
 subtend them ; whence the side B c is equal tor F. 
 
 And since the three sides of the A a Bsc are equal 
 to the three sides of the A pe F, it follows, from the 
 last proposition, thatthe 2 aBc = 4 peEF,and 4 
 AOR LOD Ue. | QUAESD: 
 
 THEOREM IX. 
 
 123. If a side and the two adjacent angles of one 
 spherical triangle be equal to a side and the two adja- 
 cent angles of another, each to each, their remaining 
 sides and angles will be equal. 
 
 Let anc, DEF be two spherical A‘, having the side 
 BiCiges Eh Bie ul) g, ‘andi Zi ciseuZieeehenewill 
 side aB = DE, ac=pF,and Za= Zp. 
 
 For, about the angular points of the two AS, de- 
 scribe the supplemental or polar A* GHK, LMN. 
 
387 
 
 Then, because the side pc = EF, the 4’ G, L, which 
 are their supplements, are also equal. 
 
 And because the 2°B,c are== 4° 8, F, thesides cx, 
 co H, which are the supplements of the former, are = 
 sides LM, LN, which are the supplements of the latter. 
 
 But since the sides cu, Gx and the included Z c, 
 of the A GHK, are = sides L M, MN and the included 
 4 1, of the A tn, the side nx will be= mn, ZH 
 = 2°Myand) 2 (Kk. == 2 N, 
 
 Hence, also, 4 a = Z pb, being the supplements of 
 HK, MN, and aB, Ac, being supplements of the 25 
 K, H, are = DE, DF, which are the supplements of £5 
 N, M. Oo Be Dy. 
 
 THEOREM X. 
 125. The angles at the base of an isosceles spheri- 
 cal triangle are equal; and if the angles at the base are 
 
 equal, their opposite sides are equal. 
 A 
 
 B Cc 
 
 D 
 
 Let asc be an isosceles A, having the side as = 
 to ac; then willthe 2 apc beequalto Z acs. 
 
 For, bisect the base Bc in p, and through the points 
 A, D draw the arc a D. 
 
 Then, because the two sides an, BD of the A aps, 
 are equal to ac, cp of the A apc, and the side ap is 
 common to each, the 4 axzc willbeequalto 4 acs. 
 
 2c2 
 
388 
 
 Again, if 4 apc be equalto 4 acs, the sideas 
 will be equal to ac. a 
 
 For if not, let a 8 be the greater, and take BE equal 
 to Ac, and draw the arc c E. 
 
 Then, because the two sides-EB, Bc, and the in- 
 cluded 4 EBC, of the A BEC, are equal to ac, cB, 
 and the included 4 acs,of the A pac,the 4 ECB 
 will be equal to 4 ase. 
 
 But Z ascis equal to Z acs (by hyp.); whence 
 4 £CB is alsoequal to Z acs, the less to the greater ; 
 which is impossible. | | 3 
 
 - The side aB ts, therefore, not greater than ac; and 
 in the same manner it may be shown that it is not less ; 
 whence they are equal. ey rs by 
 
 Cor. A perpendicular drawn from the vertex of an 
 isosceles spherical A to the base, bisects both the base 
 and the vertical angle, except when the two equal sides 
 are quadrants; in which case there are an. indefinite 
 number of perpendiculars. 
 
 | THEOREM XI. 
 
 125. The sum of any two sides of a spherical tri- 
 angle is greater than the third side; and the difference 
 of any two sidesis less than the third side. 
 
 A 
 
 BS Caen 
 Let asc bea spherical triangle; then will the sum 
 
389 
 
 of any two sides AB, Ac be greater than Bc; and the 
 difference of as, ac less than Bc. - | 
 
 For draw the chords as, Ac, Bc, which will fall 
 within the sphere. 
 
 Then, since these chords form a plane A, the sum 
 of any two of its sides will be greater than the third 
 side. | | 
 
 And because, in the same circle, the greater chord 
 subtends the greater arc, the sum of any two sides aB 
 ‘+ ac, of the spasrce A ABC, is greater than the 
 third side 8 c. 
 
 Also, since the side a 8B is less than pc + Ac, if Ac 
 be taken from each of them, the difference of a B and 
 AC is lessthan Bc. Q.E.D 
 
 Cor. The shortest distance between any two points 
 on the surface of a sphere is the arc which passes 
 through those points. 
 
 THEOREM XII. _ 
 
 126. The greater side of any spherical triangle is 
 opposite to the greater angle, and the least side to the 
 least angle; and conversely. 
 
 A 
 D 
 
 B CG 
 Let apc be aspherical A; then if 2 c be greater 
 than Z 3, AB will be greater than ac; and if aB be 
 greater than ac, 4 c willbe greater than 2 B. 
 
390 
 
 For through the point c draw the arc ¢ p, making 
 
 4 BcD equal to 4 DBC. 
 
 ‘Then, in the A apc, the sum of the sides aD “f 
 DC is greater than a c (theo. 11). 
 
 But 4:3cp being equal to Z pBo (by const.) the 
 side pc is equal to pB (theo. 10); whence, also, aD 
 -++ DB or A Bis greater than ac. 
 
 Again, if AB be greater than ac, 4 4c B will be 
 greater than Z ABC. | 
 
 For, if not, it must be either equal or less. 
 
 But angle ac B cannot be equalto 4 a Bc; for, in 
 this case, aB (theo. 10) would be equal to a c, which 
 it is not. 
 
 Neither can 4 acBbe less than 4 apc; for,-in 
 that case, AB (by Ist part prop.) would be less than 
 Ac, which it is not. 
 
 Whence 4 acs being neither equal to nor less than 
 4 ABC, must be greater than it. 
 
 And in a similar manner it may be shown that the 
 least side is opposite to the least 4, and the least 2 
 to the least side, Q. E. D. 
 
 | THEOREM XIII. 
 
 127. The sum of the three sides of any spherical 
 triangle is less than the circumference of a circle, or 
 
 860°; and the difference of ny two sides is less than 
 
 180°. 
 A 
 
 a eS 
 
391 
 
 Let Bc bea spherical A; then will the sum of its 
 three sides an -+3Be-+-ca be less than 360°; and the 
 difference of azn, ac less than 180°. 
 
 For, produce the sides B A, Bc till they meet in the 
 opposite point of the sphere at p. 
 
 Then, since the arcs BAD, BCD are ccfihical at 
 the sum of the arcs Ba + BC+ DA + Dc is equal to 
 a circle, or 360°. 
 
 But the sum of the two sides pa + Dc is greater 
 than ac (theo. 11); whence the sum of the three sides 
 Ba+sec-+ ac is less than a circle, or 360°. 
 
 Also, the difference of azn, ac being less than Bc 
 - (theo. 11), and Bc less than 180°, aB ~ Ac must be 
 less than 180°. 3 Q. E. D. 
 
 , THEOREM XIV. 
 
 128. The sum of the three angles of every spherical 
 triangle is greater than two right angles, or 180°, and 
 _ Jess than six, or 540°. 
 
 Let apc bea spherical A; then will the sum of its 
 £*a-+n-+c be greater than 180°, and less than 540°. 
 
 For, about the angular points a, B,c describe the 
 supplemental or polar A DEF. 
 
 Then, since 4 A==180° — EF, 4 B=180° — Fp, 
 and Zc =180°—pz,tharsum Z2a+Z3+ic 
 will be = 540° — (gF-+Frp+pk), 
 
392 
 
 - But the sum of the sides EF + FD Aap being less 
 than 360° (theo. 13), if this be taken from 540°, the 
 remainder will be greater than 180°. Ry 
 
 Whence, also, 4 a-+' 4 B+ 4 ¢, which is equal 
 to this difference, will be greater than 180°. 
 
 And because each Z of the A is less than 180°. 
 their sum A-+ B--c must be less than 540°. Q. E.D. 
 
 Cor. The sum of any two 4* of a spherical A, is 
 greater than the supplement of the third angle. 
 
 For Z a-+Z38-+ Zc being greater than two right 
 4’,orthan Zacsep-+ Z ace, if Z acsore be 
 taken away, thesum of the remaining 2‘ a + B will 
 be greater than Z acc (a). | 
 
 | THEOREM XV. 
 
 129. If the sum of any two sides of a spherical tri- 
 angle be equal to, greater, or less than a semicircle, 
 the sum’ of their opposite angies will, accordingly, be 
 equal to, greater, or less than two right angles; and 
 conversely. A 
 
 C 
 
 (2) Mr. Vince, in his Treatise of Trigonometry (p. 112, prop.17), 
 has endeavoured to prove that the sumof any two Z8 of a spherical 
 A is greater than the third. But this is not true, except in right- 
 angled A, as may be easily shown, either by partial examples, or 
 by a general investigation. Asan instance of the former kind, let 
 ABC (fig. to the prop.) be an isosceles A, having its equal Za 
 and 8 each less than 45°, then will their sum be less than 90°; 
 and consequently theremaining Z c must be greater than 90°; or 
 
393 
 
 Let apc bea spherical A; then if aB--ac be equal 
 ‘to, greater, “if less than. a semicircle, or 180°, the 
 sum of the 4*3-++c will be equal to, greater, or 
 less than 180°. 
 
 For produce the sides B a, Bc till they meet in the 
 opposite point of the sphere at p: 
 
 Then, if aB-+ ac be equal to the semicircle BAD, 
 the side ac willbe =ap,andthe Z acp= ZporB 
 (theo. 10).’ 
 
 But Z acp-+ Z AcB = two cn Zor 180°; 
 whence, also, 4 acB-+ 24 B= 180° 
 
 Again, if AaB + ac be greater than the semicircle 
 
 » sab, the side ac will be greater than ap; and 4 p 
 or B greater than 4 acD (theo. 12). 
 
 But Z acb+4 acB==two right 25, as before ; 
 whence, also, 4 acB-+ 4 B is greater than 180°. 
 
 Lastly, if az + ac be less than the semicircle 3 ap, 
 the side ac will be less than ap, and 4 p or B less 
 than Z acD (theo. 12). : 
 
 But 4 acp-+ Z acs being = two night 2°, or 
 180°, the sum of the Z* acs -++ B is less than 180°. 
 And, ina similar manner, it may be shown, that if 
 the sum of the two 2*B-+ c be equal to, greater, or 
 less than 180°, the sum of their opposite sides a B -+ 
 Ac, will also be equal to, greater, or less than 180°. 
 
 7 in aa 0. E.D. 
 
 Cor. 1. If each side,of a spherical A be equal to, 
 
 otherwise the sum of the three 2* of the A would be less than 
 180°; which being contrary to the above proposition, is a suffi- 
 cient proof that the principle is erroneous. 
 
594 
 
 greater, or less than 180°, each of the £5 will, accord- 
 ingly, be right, obtuse, or acute; and conversely. i 
 
 Cor. 2. Half the sum of any two sides of a spheri- 
 cal A is.of the same kind as half the sum of their op- 
 posite angles. | 
 
 THEOREM XVI. 
 
 130. In any right-angled or quadrantal spherical tri- 
 angle, the legs, or sides, are of the same kind as their 
 opposite angles ; and conversely. | 
 
 AAS 
 
 Let apc, a’Bc, or aBc bea right-angled spherical 
 A, of which c is the right 4; then will theleg ac, 
 A’c, or ac be like its opposite 2. 
 
 For let a’c be equal to a quadrant, ac less than a 
 quadrant, and ac greater; and through the points a, 
 A’, a, and B draw the circles a B, A’B, andas. 
 
 Then, because a’ is the pole of the circle cap, the 
 Z,a’Bc is a right 4, or 90° (def.); and, conse- 
 quently, the 4 aBc is less than 90°, andthe Z aBc 
 greater, agreeing with the opposite leg a’c, ac, or ac. 
 
 On the contrary, if a’Bc be aright 4, a’ will be 
 the pole of cB p, and a’c will be a quadrant; whence, 
 also, if the 4 asc be less than a right Z, and the 
 4 asc greater, the opposite leg ac will be less than 
 the quadrant a’c, and a c greater. ‘QV Be D. 
 
 The same will also hold if the A be quadrantal ; 
 for its sides and 2% being the supplements of the 4° 
 
- 
 
 895 
 and legs of the polar A, which, in this case, is right 
 £4, the similarity will be the same as before. 
 THEOREM XVII. ’ 
 131. In any right-angled spherical A the hypothe- 
 nuse is less or greater than 90°, according as the two 
 legs, or the two angles, or a leg and its adjacent ane 
 
 are like or unlike. 4 P 
 a 
 
 c D 
 
 ‘@ 6 
 
 ist. Ifthe A axsc, right-angled at c, have its legs 
 c A, cB each less than 90°, the hypothenuse 4B will 
 be less than 90°. 
 
 For make c P equal toa quadrant, and through the 
 points p, B draw the arc of a great circle PB. 
 
 Then, because P is the pole of the great circle c BD, 
 the arc Pp Bisa quadrant, or 90° (def.) 
 
 And since, in the right 24 A’ pcs, app, the leg 
 cB is less than 90°, and ps greater, the 2 cPB or APB 
 is also less than 90°, and the Z DAB, or PAB greater 
 than 90° (theo. 16), 
 
 But the less side of every A being opposite to the 
 less £, the hypothenuse a B is less than 90°, or than 
 the quadrant PB. 
 
 2dly. If the A acb haveits legs ca, cb each great- 
 er than 90°, the hypothenuse a b will, in this case also, 
 be less than 90°, 
 
 For produce ca, c 6 till they meet at p, which will 
 be aright angle, and through the points Pp, b draw 
 the quadrant p b. 
 
$96 
 
 - Then, since the legs pa, pb are each less than 90°, 
 it may be shown, as before, that the hypothenuse a b, 
 which is common to both the A*® ac 43 ap4, is less than 
 90°, or than the quadrant p 4. 7 
 
 3dly. If the A acs have one leg cB less than 90°, 
 and the other ca greaier, the hypothenuse a8 will be 
 greater than 90°. 
 
 For, since in the right 2¢ AS acs, Ppp, the leg cB 
 is less than 90°, and pz greater, the Z caB, or P@B, 1s 
 less than 90°, and the Z pps, or aps greater (theo. 12); 
 whence, also, a8 is greater than 90°, or than the qua- 
 drant PB. 
 
 Again, the 4 Sin either of the AS asc, abc, or aBc, 
 being of the same kind as their opposite legs (theo. 16), 
 it follows, that the hypothenuse az, a 6, or aB is less 
 or greater than 90°, according as the two oblique 4°, 
 of the A to which it belongs, are like or unlike. 
 
 And because a leg andan Z in each of these A‘ are 
 of the same kind as the two legs (the other leg being 
 like its opp. 4 ), it is plain that the hypothenuse a 8, 
 a b, or 2B is also less or greater than 90°, according as 
 either leg and its adjacent Z are like or unlike. Q.£.D. 
 
 Cor. It follows, reversedly, from this proposition, 
 that in any right-angled spherical A, either leg is less 
 or greater than 90°, according as its adjacent 4 and 
 the hypothenuse, or the other leg and the hypothe- 
 nuse, are like or unlike. 
 
 Also, that either of the oblique angles is acute or 
 obtuse, according as its adjacent leg and the hypothe- 
 nuse, or the other 4 and the hypothenuse, are like 
 or unlike. 
 
397 
 
 Scuotium. This proposition and its corollaries will 
 also hold for any quadrantal spherical A,, observing to 
 substitute the hypothenusal 4 for the hypothenuse, 
 and the terms greater or less for less or greater. 
 
 F 
 
 For the sides and angles of the quadrantal A age 
 are, evidently, like or unlike, according as the angles 
 or legs-of the right 24 polar A per, which are their 
 supplements, are like or unlike. 
 
 But the hypothenusal 4 c, being the supplement of 
 the hypothenuse pb £, will consequently be greater than 
 90° when pe is less, and less than 90° when pk is 
 greater; whichis, therefore, the only change that takes 
 pisce in the proposition. 
 
 OF THE 
 
 STEREOGRAPHIC PROJECTION OF THE SPHERE. 
 
 The stereographic projection of the sphere, is sucha 
 representation of the various parts of its surface, on the 
 plane of one of its great circles, as would be formed 
 by lines drawn from the pole of that circle to every 
 point of the figure to be delineated. | 
 
 Or, if taken in an optical sense, it is a view of the 
 points and circles of the sphere, as they would appear 
 on a transparent plane, passing through the centre, to 
 
398 
 an eye placed at one of the extremities of a ie ge 
 drawn perpendicular to that plane. | 
 
 The place of the eye, iscalled the projecting point, and 
 the plane, on which the points and circles of the sphere 
 are to be represented, is called the plane of projection. 
 
 The primitive circle, is that which lies in the plane 
 of projection; being the one to which all the other 
 circles and points of the sphere are referred. 
 
 A right circle, is that which, passing through the 
 eye, has its plane perpendicular to the plane of the pri- 
 mitive ; and, being seen edgewise, is Dain into a 
 _ right line. 
 
 A parallel circle, is that which is Metal to the 
 primitive; and an oblique circle is that igs Is seen 
 obliquely by the eye. | 
 
 It is also to be observed, that the projection of any 
 point of the sphere, is that point in the plane of projec- 
 tion, which is cut by a right line drawn from the origi- 
 nal point to the eye. And that lines flowing to the 
 projecting point, or place of the eye, from every point 
 of the circumference of a circle, form the convex sur- 
 face of a cone. 
 
 | LEMMA. 
 
 132. If a cone be cut by a plane parallel to its base, 
 
 the section will be a circle. 
 
399 
 
 Let ancp bea cone, either right or oblique, and era 
 a section parallel to its base Bcp; then will e Fe bea 
 circle. | 
 
 For let the planes acm, apm pass through the 
 axis a.m of the cone, meeting the section in the points 
 F, G57} 
 
 Then, because the section EFG is parallel to the base 
 BCD, and the planes cn, Dm meet them, nF will be 
 parallel to mc, and nc to mp. 
 
 And, because the A‘, formed by these lines, are si- 
 milar, Am:Ani:mcinForasmD:inG. — 
 
 But mc is equal to m b, being radii of the same cir- 
 cle; whence, also, x Fis equal to nc. 
 
 And the same may be shown for any other lines 
 drawn from the point x to the circumference of the 
 section EFG, which is therefore acircle. § Q. B.D. 
 
 THEOREM 1. 
 
 133. Every circle of the sphere, which does not pass 
 through the poles of the primitive, is projected into a 
 circle, : 
 
 Let ams be a circle to be projected on the plane tm, 
 which passes through the centre of the sphere, at right 
 4*to aradius drawn from the eye at £; then will its 
 representation a7 6, on that plane, bea circle. 
 
 For through 7, the centre of the circle a mB, draw 
 the plane rmc parallel to mM, and join gr, rm; the 
 
400 
 
 former of which will be the axis of the cone of rays 
 flowing to 5, and the latter the common ‘section of 
 the two planes AmB, FMG. - 
 
 Then, because the 4 E 4a is measured by half the 
 sum of the arcs EH, KB, or half E KB, it is equal to 
 the Z EAB, which is also measured by half eK B. 
 
 Also, because FG is parallel to ab, the Z EGF Is 
 equal to Eba, or EAB; and consequently, by similar 
 AM AT DTG 201E RS TB eOhiRT MiGs Ae Oo SER 
 
 But ar, 7B, rm being radii of the same circle ams, 
 the rectangle rr X re will be = rm’*; whence rme 
 is a circle; asis also the section anb, which is parallel 
 to it (Lem.). 
 
 Or the same thing may be shown independently of 
 the Lemma. 
 
 For Fe being parallel to ab, We shall have Er: ES 
 $2 7Fisda,.and Er: ES::.re:s63 whence, com- 
 poundedly, er* : Es*::Fr xX re (orrm’):sa es b. 
 
 But the plane F mc being parallel to anb, and the 
 plane nr cutting them, sz will be parallel to r Re and 
 consequently E7*: Es? i: 7rm*:sn* 
 
 Hence, also, : equality, ron? sa.% 80 sc" me < 
 sn; and, therefore, sa X sb being = s 7", the section 
 anb isa circle. - Q.\ EDs 
 
 Cor. The centres and poles of all circles of the 
 sphere, parallel to the plane of projection, will fall in 
 the centre of the primitive. 
 
 THEOREM Il. 
 
 134, The angle formed by two great circles on the 
 surface of the sphere, is equal to the angle formed by 
 their representatives on the plane of projection. 
 
Let & be the projecting point, or place of the eye, 
 uM the plane of projection, and c BD, c K D two great 
 circles of the sphere, meeting each other in c; then 
 will the projected Z be equal to the spherical Z BcK. 
 
 For, if to the arcs cB, cK there be drawn the tan- 
 gents CF, cc, meeting the plane tm in F and, the 
 former of these c F will be projected into cr, and the 
 latter cc into cc. 
 
 And because the AS z ce, Eocareright Z*at cand 
 o, and have the 4 exc common, the remaining Z Eec 
 will be == the 4 ocs, orits opposite Z ccr. 
 
 But the 4 EcF, or ccF, formed by the chord cE 
 and the tangent c F, being = the Z Eec in the alter- 
 nate segment, the Z ccF will be = the Z ccr, and 
 the side cFto cF. 
 
 In like manner, it may also be shown, that cc is = 
 eG; whence the two sides cr, cc of the A cer being 
 
 = to the two sides c F, cc of the A cer, and the base 
 GF common, the Z rce willbe = Z Fcc. 
 
 But since the Z made by the intersection of any two 
 arcs is equal to the 4 made by the tangents of those 
 arcs, drawn from the point of section, the projected 
 4 of which cr, cc are the tangents, is equal to the 
 ‘spherical 4 BCK. 
 
 2D 
 
402 
 
 Cor. The tangent and secant, of any arc of a great 
 circle of the sphere, are represented, on the plane of 
 projection, by right lines equal in length to the ‘tee, 
 
 THEOREM Ill. 
 
 135. The distance of any projected point of the 
 sphere, from the centre of the primitive, is equal to 
 the semitangent of the arc intercepted between the ori- 
 ginal point and the pole opposite to the eye. 
 
 : 
 
 Let ars be the primitive circle, lying in the plane of 
 projection L M, E the place of the eye, and ¢ any point 
 on the sphere; then will oc, the distance of the pro- 
 jected point c from the centre 0, be = the semitangent 
 of ec. | 
 
 For, having joined oc, the Z ec, oro£c at the 
 circumference of the circle eA EB, is half the Z eoc 
 at the centre. 
 
 And since the latter of these 4* éoc, is measured by 
 the arc ec, the former o£ c, will be measured by half 
 that arc. 
 
 But oc is the tangent of the Z o Ec, to the radius 
 of the sphere Eo; whence it is also the tangent of 
 half the Z eoc, or the semitangent of ec. 
 
 And if any other point p be taken on the opposite 
 side of the pole e, it may be shown, in like manner, that 
 o d is the semitangent of e pb. | Q.E. D, 
 
 | 
 
403 
 
 Cor. 1. Any arcec of aright circle, commencing 
 at the pole opposite to the eye, is projected 1 into oc, the 
 semitangent of that arc. _ 
 
 2. As the poles and extremities of the diameter of 
 any great or small circle are points on the surface of 
 the sphere, their projected distances from the centre 
 of the primitive will be the semitangents of their great- 
 est and least original distances from the pole opposite 
 
 to the eye. 
 THEOREM IV. 
 
 136. The distances of the projected poles of any 
 oblique great circle from the centre of the primitive, 
 are equal to the tangent and cotangent of half the 
 angle which the two circles make on the sphere; and 
 the distance of their centres is equal to the tangent of 
 the whole of that angle. 
 
 Let x be the place of the eye, ad the plane of pro- 
 jection, and c p the diameter of a great circle, of which 
 Pp, p’ are its projected poles, and g, the middle of the 
 projected diameter cd, its centre; then will op, op’ be 
 == the tangent and cotangent of 4 the 2 which this 
 circle and the primitive aB en on the sphere, and 
 0Q = the tangent of the whole of that angle. 
 
 For, P, e being the poles of cp, as, the arcs PC, eA 
 are quadrants; and, consequently, if ec, which is com- 
 mon, be taken away, the remainder e Pp will be = ac. 
 
 2D2 
 
4.04: 
 
 But op is the tangent of the Z oz/, or of } the 
 arc ep, to radius Eo; whence it is also the tangent of 
 ZAC, which is the measure of g the inclination of the 
 planes of the two circles aB, c D, or of 5 the 4 which 
 they make on the sphere. 
 
 In like manner, because the Z p’EP or p’Episaright 
 Z,the Z p’£o will be the complement of the Z o£; 
 and consequently o p’ is the cotangent of oz Por 3 eOP 
 to radius Eo, or of 4 the 4 which the two circles 
 make on the sphere. | | 
 
 Again, because the lines gE, Qd are equal, being 
 radii of the same circle ec E d, the outward Z oQ&, of 
 the A ged, will be double the inward opposite 2 gde. 
 
 Also, since the Z$cred, coer of the A’ Ecd, Eoc 
 are right 2%’, and the 4 Eco is common, the remain- 
 ing Z ceo will be=Qdes. | 
 
 And because an Z at the centre of a circle is dou- 
 ble that at the circumference, the Z coe is double the 
 Z cEO, or its equal gdz. 
 
 Hence, the 4*0Q#, Coe, being each double the 2 
 gdz, are equal; and consequently the 2* o£Q, coa, 
 which are their complements, are also equal. 
 
 But the Z co 4, being the inclination of the planes 
 of the two circles cp, AB, is the measure of the Z 
 which they : make on the sphere; hence 0g, which is 
 the tangent of the 4 o£ Q, to the radius £0, is also 
 the tangent of the 4 formed by those circles. 9. E. D. 
 
 Cor. It is also evident, from the figure, that the 
 radius EQ, Qc, or od is equal to the secant of the angle 
 which the two circles make on the sphere. 
 
405 
 
 THEOREM V. 
 
 187. The projected radius of any small circle of the 
 sphere, perpendicular to the primitive, is equal to the 
 tangent of its distance from its nearest pole; and the 
 line joing the centres of the two circles is the secant 
 of that distance. _é | 
 
 Let £ be the place of the eye, ad the plane of pro- 
 jection, and c p the diameter of a small circle, perpen- 
 dicular to the primitive aB; then if its projected di- 
 ameter cd be bisected in Q, the radius Qc will be = 
 the tangent of the arc Bc, and 0 Q will be its secant. 
 
 For, since the ASrce, doe are right Z“at c and 
 o, and have the Z ¢ common, the remaining Z e Ec, 
 or oc willbe = Z edoorcdg. 
 
 But gc being = gd, and oc to o£, the Z cdgqis 
 =Zdcg,and Z ozcto Z ocE; whence, also, dcg 
 po 2 OC B OF. OC Cc. 
 
 And if to each of these equals there be added the 2 
 ccQ, the whole Z ocq will be = the whole Z ccd. 
 
 But the 4 ccd being a right 2, the 4 ocgisalso 
 aright 4 ; and consequently gc isthe tangent of Bc, 
 and 0 Q its secant. Q. E.D. 
 
 Cor. 1. The distance oc of the circumference of the 
 projected circle from the centre of the primitive, is = 
 the seiitangent of the complement of the arc Bc. 
 
“4.06 
 
 2. It also appears from theorem 3, that the radius 
 of any projected great or small circle is = 4 the sum 
 or % the difference of the semitangents of its least and 
 greatest distances from the pole opposite to the eye, 
 according as this point is within or without the given 
 
 circle. 
 THEOREM VI. 
 
 138. Any projected arc of a great circle of the 
 sphere is measured by that arc of the primitive which 
 is cut off by right lines drawn from the projected pole 
 through the two extremities of the given arc. - 
 
 “ 
 
 A. - 
 
 i 
 
 Let 4cB be the primitive circle, lying in the pro- 
 jecting plane ad, x the place of the eye, and cfd the 
 projection of the great cirele cr p; then if right lines 
 pd; pf be drawn through its pole p, the arc fd will 
 be measured by cB. 
 
 For, since p d lies in both the planes ac B, APBE, 
 it is their common section ; and, consequently, will pass 
 through the point s. | 
 
 In like manner, because pc or pf is the common 
 section of the planes ac B, PGE, it will pass through 
 the point c. , 
 
 Hence the points.F, D being projected into f, d, it is 
 plain that the are F p will be projected into the similar 
 arc fd. 
 
407 
 
 But since PD, EB, PF and£G are each quadrants, 
 if-BD, GF, which are common, be taken away, the 
 remainder, or side pB will be = ED, and Pc to EF, 
 
 And because the opposite 4° BrcG, DEF, whichare 
 included by those sides, are equal, the base Bc will 
 be==pF. | 
 
 .Whencee, F D having been shown to be similar to, or 
 the measure. of fd, its equal Bc will, also, be the 
 measure of fd. Q. E.D. 
 
 THEOREM VII. 
 
 129. Any projected spherical angle, formed by the 
 representatives of two great circles of the sphere, is 
 measured by that arc of the primitive which is cut off 
 by right lines drawn from the angular point through 
 the projected poles of those circles (1). 
 
 Let GK c be any projected 2, and p, p’ the poles 
 of the arcs Kc, Kc by which it is formed; then, if 
 the lines K pL, Kk pm be drawn from the angular point 
 K, to meet the primitive cars, the intercepted arc LM 
 will be the measure of the Z Gkec. 
 
 (4) For a brief, but neat, treatise on the Stereographic Projection 
 of the Sphere, see an article by Delambre in the Mémoires de 
 VInstitut National, tom. v. also Mémoires de Académie de 
 Berlin, for 1779, where this subject is treated analytically with 
 great clearness and elegance. 
 
408 
 
 For, since the angular point of the original arcs, of 
 which Ke, Kc are the projections, is common to each 
 of them, and 90° distant from their poles, it will be 
 the pole of a great circle of the sphere which passes 
 through the two former poles. 
 
 And because the original Z on the sphere is mea- 
 sured by that arc of the abovementioned great circle 
 which lies between these two poles, the projected 4 
 Gkc, which is = the spherical one, will also be mea- 
 sured by the same, or an equivalent arc. 
 
 But p, p’ being the projection of this arc, and k its 
 projected pole, it is plain, from the last proposition, 
 that the arc tm of the primitive, which is cut off by 
 the lines kK pL, Kp’ will be the measure of the Z 
 GKC. Q.E.D. 
 
 MISCELLANEOUS PROBLEMS AND THEOREMS. 
 
 PROBLEM I. 
 
 140. As four right angles, or 360°, is to the angle 
 BAC, formed by two great circles of the sphere, or to 
 its measure Bm, so is the surface of the sphere to the 
 lunar area ABC A, 
 
 For, let the circle B m D, of which a 1s the pole, be 
 divided into any number of equal parts Bm, mm, &c. 
 and draw the circles amc, amc, &c. 
 
409 
 
 Then will these circles divide the surface of the 
 sphere into the same number of equal parts which the 
 circle B m pb is divided into. 
 
 And since the bases 8 m, mm, &c. of the A*. Ba m, 
 mam, &c. as well asthe sides AB, am, Am, &c. are 
 equal, the A‘ themselves, or their doubles, the lunule 
 ABC A,amca, &c. will be equal. 
 
 Hence, the sum of the parts B m, m m, &c. being equal 
 to the whole circumference of the circle BmpB, the 
 sum of the lunule apca, amca, &c. will also be 
 equal to the whole surface of the sphere. 
 
 And, consequently, the circumference of the circle 
 Bm DB, or 360° : the part, or arc, Bm :: surface of 
 the sphere : the area of the lune aBca. Q. E.D. 
 
 Cor. Since the surface of the sphere is equal to 
 four times the area of one of its great circles, if d be 
 put = diameter, c = circumference, and a@ = length 
 of the arc Bm, by the last mentioned proposition, the 
 area of the lune aBca will be = ad, and the area 
 of the whole surface of the sphere = c d. 
 
 PROBLEM II. 
 
 The area, or surface, of any spherical triangle, aBc, 
 is equal to the difference between the sum of its three 
 angles and two right angles, multiplied by the radius 
 
 of the sphere. Pk 
 
 Ve? 
 
410 
 
 For having completed the great circle aBaba, pro- 
 duce the sides ac, Bc till they meet in the other hemi- 
 sphere at c’. . | 
 
 _. Then, because c’lc is equal tobcs, and c’ac to 
 ac A (being each semicircles), if c b, ca, which are 
 common, be taken away, the remainders, or sides c’ 6, 
 c’ a will be equal to cB, ca. 
 
 And since the opposite 2$ c, c’ of the lune cac’be, 
 which are included by those sides, are also equal, the 
 triangle 4B c will be equal to the triangle alc’; or P 
 equal to p. ! i 
 
 Hence, by the last proposition, 
 180°: 4 a::Z surface of the sphere: p+e 
 180°: 4 8B::4 surface of the sphere: p+s 
 180°: 4 c:: 4 surface of the sphere: p-+rR (p+R) 
 Or, by composition, 
 “7BO7 = ZUR ER eZ guns surface of the sphere : 
 3pPp+totrR+s. 
 And, by division, 
 180°: Za+ 2438+ 4 c— 180°:: + surface of the 
 
 sphere: 3P+Q+R-+5— § surface, or 2 P 
 
 Hence, by multiplying the means and extremes of 
 this proportion, there will arise 
 180° X 2 p = E surface of the sphere X(4A-++ ZB 
 + 4 c— 180°) 
 
 Or, putting 180° = 4c, and the surface of the 
 sphere = cd, as in the Cor. to the last proposition, 
 we shall have p, or 
 
 Area A ABC =£(La+Lu4+Zco—180)=r 
 (Za+ 23+ 42¢— 180°) (c). 
 
 (c) This curious theorem, which has been of late years much 
 used, in various geodatical operations, for correcting the angles 
 
=- 
 
 411 
 
 Cor. Area of the A in square degrees = R° (a°-- 
 B°-+ c°— 180°) where r° = 57°.2957795 the degrees 
 in an arc of equal length with the radius. 
 
 Or, if the excess of the three angles of any spherical 
 triangle above two right angles, be required, # may 
 be obtained by the following practical rule: 
 
 From the logarithm of the area of the triangle, taken 
 as a plane one, in feet, subtract the constant logarithm 
 9.3267737, and the remainder will be the logarithm 
 of the excess of the 3 angles of the A above 180° in 
 seconds, nearly (d). 3 | 
 
 Scuoxium. Ifa, b, c be made to denote the 3 sides 
 of any spherical A a Bc; 4, B, c their opposite angles, 
 and z the semicircumference of a great circle of the 
 sphere, of which the radius is r, we shall have, by one 
 of the analogies of Napier, 
 
 cos 3 {a—5) 
 Tan ¢ (A-++B) = a: Sa) 
 
 From which the following formula may be easily 
 deduced. 
 
 Cot $ (a+s+c—z) a 
 
 pe 
 cot; Cc 
 
 cotZacotib+rcosc 
 sin C 
 
 Which may serve for determining the area of the A, 
 
 of observation, taken on the surface of the earth, is commonly 
 ascribed to Albert Girard, a Flemish Analyst of the sixth cen- 
 
 tury, who appears to have first given it in nearly the same form 
 
 as above, in his work entitled /uvention Nouvelle en Algébre, pub- 
 lished in 1629; where he has also treated of the measures of so- 
 
 lid angles. See Montucla Hist. Math. vol. ii. p. 8. 
 
 (d) For an investigation of this rule, see Trigonometrical Sur- 
 
 ’ vey of England and Wales, vol. i. 
 
412 
 
 or the spherical excess, when two sides and their in- 
 cluded angle are known. And by following the mode 
 of substitution used by Legendre (Elém. de Géom. 
 note 10.) this expression will become 
 
 Tan + Jseideathie. thine Pik , 
 
 b 2 hes poss b we 
 ~ ytan tt tan I= tan “FS tan a $ 
 
 Which elegant theorem was first given by Lhuillier, 
 of Geneva. Legendre, p. 418 of the above work, has 
 also given the following rule for reducing spherical 
 triangles, whose sides are small with respect to the ra- 
 dius of the sphere, to such as are rectilineal. 
 
 Any small spherical triangle, whose sides are a, 8, c, 
 and their opposite angles a, B, c, always answers to a 
 rectilinear triangle, of which the sides are equal in 
 length to the former, and whose opposite angles are 
 A—te B—t¢and c—+<¢, < being the excess of the 
 sum of the 3 angles of the spherical triangle above 
 2 right angles. 
 
 PROBLEM III. 
 
 142. Two sides and the included angle, of, any 
 spherical triangle 4B c, being given, to find the angle 
 included by the chords of those sides. 
 
 B Cc 
 a 
 By spherical trigonometry, cos a = sin b sin ¢ cos 
 sph.4 a+ cosccosh; orl —2smga=sinbsine 
 
418 
 cossph. Z a-+(1—2sin’ic) x (1—2sin? ib); and 
 consequently 2 sin*4c-+-2 sin? 3 b—2 sin? ta=sin c 
 sin b cossph. 4 a-+ 4sin’3 c sin’ J 0. 
 
 Again, by plane cig duit: a* = $+ e*#— 2h’ 
 cos rect. Z a; and because a’ = 2 sini a, b'=2sin 
 <b, and c’ = 2 sin 4c, we shall have, by substitution, 
 4sin’Sa=4sin?$c+4sin?$b— 8snicsindd 
 Cos Tect ZA, 
 
 And this value, being substituted in the former Bg 
 tion, will give, sincsin 4 cossph. Z a ++ 4 sin*4 csin? 
 4b=4sin $c sin} bcos rect. Z a. 
 
 From which may be readily obtained the following 
 simple expression for the angle contained By the 
 chords, viz. 
 
 Cos rect. Z a=sin $ bsin$ c-+cos$bcosic 
 cos sph. Z a. 
 
 Or, by restoring the value of the radius 7, 
 
 Cos rect!, Za = = (r sin 3 bsind c+ cos bcos 
 %ccossph. Z a), 
 ~ Cor. From the formula here given, it appears that 
 a right or obtuse spherical angle is always greater than 
 the corresponding rectilineal angle. 
 
 And that an acute spherical angle is less than its cor- 
 
 responding rectilineal angle, when its cosine is greater 
 ih rsinidsindc 
 
 r?— cost écosic’ 
 PROBLEM Iv. 
 143. Given the three sides of any spherical triangle 
 
 ABC, to find the angles contained by the chords of those 
 sides. 
 
b% +c — a’ 
 QW ¢ ee > 
 but a’ = 2 sin} a, b= 2 sin 3 b, and ¢ = 2 sinc. 
 OF. id ena l Oy 2r(r—cosa), 0 = = / 27 (r—cosb), and og 
 = VW 2r a cosc); Whence, by substitution and reduc- 
 tion, we shall readily obtain . 
 1+cos a—cos b—cosc¢ ? 
 ———} which, by 
 restoring the value of radius, becomes | 
 
 r-+cosa—cos4—cose 
 Cos rect. Z A=? et aa —-—-—.. Also cos 
 4 sin 6 sin 4 Cc 
 
 By plane trigonometry, cos rect! Z a = r(—_— 
 
 Cosirect))Z) a= oh One AEN 
 4 sind dsinic 
 
 r® cos SES Orie 
 
 sph. 4A = 
 
 sin 5 sine , 
 Cor. 1. When the three sides of the spherical tri- 
 angle are equal, their chords are also equal; in which 
 
 r—cos a 1 ia 
 
 case, cos rect. Z A= 7r?-—__- = — = cos 60°, as 
 a 4 sin? 5a Z 
 
 it ought. | 
 
 ~ 2, Also, since, in the same case, cos sph. Z a = 
 
 rtania r?—tan®ia . P ‘ 
 rie le 2—, if a = 60°, either of the 4 § will 
 
 are 
 
 tana Zr 
 be that of which the cosine is + or .3333, &c. that is 
 70° 31’ 43”; and as the angles are all equal, and their 
 sum greater than 180°, the spherical Z a must be 
 
 180° : on ee: oe 
 ereater than —> or 60°, which is the rectilineal Z a. 
 
 € 
 
 PROBLEM VY. 
 144, Given the three angles of any spherical triangle 
 
415. 
 
 ABC, to find the angles contained by the chords of the 
 sides. 
 
 A 
 
 The sine of $ any arc being equal to $ the chord of 
 the whole arc, we shall have, by spherical Pegncay, 
 
 —cos 5s cos (5 s—-A) 
 
 —cos 4 4S COS (4 S—B ) 
 a=2r ,¢=2r wind BiB 
 Ap toigie’ = sin B sin C 1 omen sin A sinc ? 
 —cosiscos (£s—c 
 COR _——- (3 ) where A, B, c denote the 
 sin A sin B 
 
 three Z° of the spherical A, and s their sum. 
 
 Also, by plane trigonometry, cos rect! Z a= r 
 b? +c? — q’? 
 eager +s whence, if the former values of a’, l’, c’, 
 be substituted in this equation, we shall obtain, after 
 proper reductions, the following expressions : 
 Cos rect'. Vig a 
 a1 sin B cos (is—s) + sinccos(is—c)—sinacos(4s—A) } 
 
 W sin B sin c cos (£s—B) cos (b s—c) 
 
 which, when a, B, c are equal, becomes cos rect’. 2 a 
 ’ * 
 = > as it ought. 
 
 And by observing the order of the letters, the other 
 two angles may be each expressed by a similar form (e). 
 
 (e) The angles formed by the chords, in all the other cases of 
 spherics, might be easily obtained in a similar way; but, as they 
 are less useful, and more complicated in their forms, it was thought 
 propertoomit them. From 
 
416 
 
 145. SOLUTIONS OF ALL THE CASES OF RIGHT-AN- 
 -GLED PLANE TRIANGLES, INDEPENDENTLY OF 
 ANY TABLES. 
 
 A 
 g b 
 B a ¢ 
 
 i. Given the hypothenuse and either of the oblique 
 angles, to find either of the legs. 
 
 _ From the expressions above given, we might, also, readily pass 
 to the solutions of the inverse cases of these problems, 
 
 Thus, in prob. 111, art. 142, since cos 5 45 coskcz= WV 72 —sin? 2 £0 
 x fr—sin tc; also sn $4 x sindc= Zh x 1y aa bic at 
 these values be substituted in the equation cos rect!. 2 a=sin 34 
 sin 3.¢+-cos2dcos%ccos sph, 4 A, the radius being considered as 1, 
 
 _cosrect. £A—j Hc 
 we shall have, cos sph. £ a= Wie ; Toast in the cal- 
 
 culation of which form, the sides 4’ c’ must be so taken that they 
 may be the chords of a circle, having 1 for its radius; which may 
 be easily done, by dividing their numerical values in feet, yards, 
 
 » &c. by such a power of 10, that neither of them shall exceed 2, 
 which is the greatest chord in the circle, 
 
 If the A be isosceles, and of consequence b= ¢, and d' =<’, 
 these two formule may be reduced to the following: sin J rect!. 
 ZAzccosidsinisph. Z a,andsinisph. Z a= mi shal 
 
 aa: 2 2 al ica i Wd oT V1 =F Bie * 
 And ifsph. Z 4 be 90°, the same forms will give cos rect!. Z a= 
 36'c’.. From the first of which expressions it appears, that the 
 vertical 4 of any isosceles spherical A is always greater than its 
 corresponding rectilineal angle. 
 
417 
 
 csin A c cos B 
 a= aoeel yarn 
 r r 
 De Rib ea eer Sere 1 ae 
 Se aphaigaay Gaia ke, 
 | x 2.3 (5 ) +3356 ' 2 2.5.4.5.6.7 (Fo) hag 
 a= C4 Or, 
 ° ; ° 
 
 Gita ei ee et Rae ait 
 ( a (oo) + a5. Ge) 2366 (xe! ne 
 
 And if 8° be put in the place of a° in the Ist of 
 these series, and a° in the place of B° in the 2d, they 
 will express the value of &. 
 
 Note. a° or B° denote the number of degrees and 
 decimal parts in the length of the arc which measures 
 the Z a or B; and R° = 57°.2957795 is the number 
 of degrees, &c. in an arc equal in length to the radius. 
 Also the length of any arc a, in parts of the radius, 
 
 =7 (5) =r(5) =1 (5). 
 
 IJ. Given the hypothenuse and either of the legs, 
 to find either of the oblique angles. 
 
 ' : b 
 
 Sina = —, sin = — 
 
 te Ried re 3.5 ,a@\, 
 = Rat tag (Olt ogg Slt ones Gl + xe. b 
 b bees Sik Bi et Otc al 
 =e itz " “+545 G + Fae + &e.f 
 
 III. Given the hypothenuse and either of the legs, 
 to find the other leg. _ 
 a= VP-P, b= fe—a 
 
 _— 
 
 pets 54. 3 7& 
 c— b1EC 3 =) taal "+ bag (> 4 sae (=) & ct 
 QE 
 
IV. Given either of the legs and either of the ob- 
 lique angles, to find the hypothenuse. 
 
 ne b 
 c = —sec B = —SeCA 
 Ris? r 
 i Bo v2 a Mh y 
 cma q1th(2s)+ 2 Br Say a R° o)' + S068 (x2 y tae b 
 
 BOE fi ey Aa, Ay Gh iyo ty CORT tng 
 cmb 1 48(Sy4 F(A wag (go)? + 8064: 30) ih ‘ 
 
 V. Given either of the legs and either of the ob- 
 lique angles, to find the other les. 
 
 __ btana _ bcots 
 | Tg wee 2 oe 
 Pe ire ey ee 17 62) Bs 
 1° wot 3 x veg 2 (ea 2” eee ene 
 
 Or, 
 
 a a 
 Be yacias B yas t huig Bont 
 i bf Ss —5( aii 3) ~ 945 (Ge) F795 5° a 
 And if a be put in the place of , B° in the place of 
 a° in the first of these series, and a° in the place of B” 
 in the second, they will express the values of 0. 
 
 Vi. Given the two legs to find either of the pi Me 
 angles. 
 Tans ‘= a seigies 3h | 
 1 1 Ring fe 1 
 ak —gOGht g(Gi 9 pig (gre 
 
 ao Or, ae 
 wie fms 1 oh Lye ae 
 pRe{ = — 5 iy uae as Ay (2 + +5 (poke. 
 
419 
 
 And if 6 be put in the place of a, and a in that of 
 b, these series will express the value of 8”. 
 
 VII. Given the two legs, to find the hypothenuse. 
 c= Vase =a v(1 +D=bvat+s 
 
 b J ae 7 
 pre Ong oA C+ xg sae (Gl &e: t 
 c= 4 7 Or, 
 bie: Vite De Mi By Say i 
 Li+2 245) — were 2.4.6 '5) — ade (5)" &e 5 
 Note. The following formula, which belongs equally 
 to both the tables, may here also be subjoined, as it 
 will be found useful upon particular occasions : 
 at+b+c 1] ;8&+P+2 8 ,A'+b+0 3.5 
 r hag la) obs cers ) + ae 
 
 q’ 7 
 6 7 +") &c. = 3.14159265358979, &c. = the cir- 
 
 cumference of a circle whose diameter is 1, a, 0, c be- 
 ‘ing the halves of the 3 sides of any triangle, and r the 
 radius of the circumscribing circle. 
 
 146. SOLUTIONS OF ALL THE CASES OF OBLIQUE=~ 
 ANGLED PLANE TRIANGLES, INDEPENDENTLY OF 
 ANY TABLES, 
 
 I. Given a side and two angles, to find either of the 
 other sides. 
 
420 
 
 h = Hee Boa e i (90°, 8) 
 
 sina 
 
 cos (90°) 
 Mere aaa 7 
 5 3 : 
 nas let a3a5 Gl er ne 
 baa coisa 4, a ye 
 nasal) apa ae 
 , o 0 
 lsd score a a aaa great) 
 
 In which case, the numerator of either of these series 
 may be placed over the denominator of the other, when 
 such a combination is found to render them more 
 convergent. 
 
 The side a or c may also be expressed in the same 
 manner, using the angles which are similarly situated 
 with respect to these sides, instead of those in the above 
 formula; and the same may be observed in all the 
 other cases when only one side or angle is exhibited. 
 
 II. Given two sides and an angle opposite to one 
 of them, to find the remaining side. 
 
 b j Oe Bs lho es! 
 c= - ; ©OS Axt- ~ Mat —58 sin? a 
 
 Fanaa f(y Ste —telte) ny ay } 
 —< Or, 
 ane eee Baws 4" (Fa) ee 2 
 [ 2.3-4 a3 (S <) 
 
 The first of which series only takes place when a is 
 equal to, or greater than, J; but if a be less than 4, 
 either of them will hold, as the question, in this case, 
 admits of two answers. 
 
421 
 
 III. Given two sides and an angle opposite to one 
 of them, to find either of the other angles. 
 Sin A = 5 sin B 
 
 a Pay «Ue ae [-a*)?-S 8) 40 
 Armee Lo aah (3 aoe ~& a) Ke f 
 
 3 sin A7b 1 PD Batata ASEH 
 Sin c = — [* cos A tk = WF a*—2 sina) 
 
 ° 
 
 c= 
 fb+a_a° _ 3h +4ab+a% Ao 80 a? —16 a&b+at—15 b* 
 6. pet ak 1% 2.8 af (3) 2.3.4.5 at 
 
 (=) &c, 
 
 ry 
 bea s° 3 4 abta® a° 15 6* + 16 a®4—30 a®h*§~— at 
 PC io roa ue Cu + 2.3.4.5 at 
 
 A° 
 —)§ & 
 (=) &c. 
 
 L 
 The first of which series, for c°, only takes place 
 
 when a is equal to, or greater than, 4; but if a be 
 less than b, either of them will hold, as the question, 
 In this case, admits of two answers. 
 
 IV. Given two sides and the included angle, to find 
 the remaining side, 
 
 c= Me cia Bia OP cost c 
 6 = VJ as) ad ooleees sac “+ sags Ss)? & cf 
 
 itm on, i 
 Ce af a j2— = cos? c 
 Be . 
 
 c=(a+b)x 
 070 a %_ co 
 jim, ab 180°—c oda AC Seah aS Cs Be, ‘ 
 2(a+6)? R° 2.3.4 (a+b) 
 
422 
 
 ‘The first of which series must be used when c° is 
 less than 90°, and the latter when it is greater than 90°. 
 
 V. Given two sides and the included angle, to find 
 either of the other angles. ! 
 
 Tan 4 a hee) =i cot 4 
 
 ‘ a+b ill 
 : b 2 R wi a! 8 ] 
 mr ee . Rd ARP Ens aD ie, 2. vee ye 
 A =90" c ' gem © ~ 23Rr° 8.45 
 
 Cc J.3 Ro 
 
 a ee oe — = (5) — &.t 
 
 (SP ke,} = et seit Bie Rete) Se) 
 
 T Sade 8.45, 
 Or, 
 b 90e—1co (at + 4%) 
 ae Bh a ic ° + at ° eee Pees BA 
 sa Pas i { x° 3 (a + bp 
 90° —Zc° 1 faa +40 an I 
 ar a, - ae 5 (a+b)t 9 (; y &c. 
 
 in sie of ain series the upper sign ++ must be 
 taken when a is greater than 4, and the lower sign — 
 when a is less than 4; observing, in either case, to take 
 the series which is found to be the most convergent. 
 
 VI. Given the three sides, to find either of the angles. 
 
 at ie 
 ol? a b* — c% 
 | | Cos c = 5 bing is | 
 ° o J ee —<° bat tel a ne l 
 — RWa bases 4) Tope 1.5.6 
 CYR 1 ah Ut Ooal cab) 88466 
 
 Which series will always converge, whatever may be 
 the values of the sides a, 0, c. 
 Note. If the angle c, in the 4th or 5th case, be very 
 
 “Ne 
 
423 
 
 obtuse, or near 180°, and, of consequence, the re- 
 maining angles a and B small, the side c, and either 
 of these angles, may be found, to a considerable degree 
 of exactness, by means of the Bereming formule. : 
 180° — c° 
 c=a+b— seh apy | 
 
 o__4 ieee par) 180° =<" 
 
 a+b 6 (a+b)? 
 * The angle and side, in this case, may a be ex- 
 pressed, in series, by means of the formula given in 
 
 page 337, as follows: 
 
 a=Ssino-+ Ssin 204 “sin 3c-+ “sin 4c &e. 
 es 
 B=<sine-+548in2c+ 5, ssin3 c+ >, = sin 40 &e. 
 i Crs log a— 
 =i: cos o + 5, cos20 + x "cos 830+ &e. 
 147. To the tables last given, there may likewise be 
 added the following solutions of some particular cases 
 
 of right-angled plain triangles, when one part is given, 
 and the sum or difference of two others. 
 
 A 
 7 b 
 B a Cc 
 } B 6? 
 i CP a) 5 a ae 
 (c—b)r ra 
 
 r i dts —_ 
 Gt 2. Tanga = ——— 
 
4.24 
 
 : c—b b 
 A Singasrvy>-; cos >. AF Vee 
 
 bed ih ae We oe 24 
 4 c--a =~, col 7B; C—~a=—~ tan gB 
 
 §. atb= oe cot (45° —3)=* °43) 
 
 6. aap — oe tan (45° — B) = ae (45°-++B) 
 : ° b : Lb 
 
 7, Sin (45°--B) = es sin (45 a 
 
 8 c—a==b tani sp, c—b=atanga. 
 
 148. The following additional fant for right- 
 angled plane triangles may also be subjoined to those 
 given above. | 
 
 . coe 9rab 
 
 1. Sin(B—a) = sodas fet ae ; cos(B—A) = ~ 
 = r(b-+-a) x (ba) | r(a°—b*) 
 2. Tan(p—a)= a aE ;tan(a—B)= Ee 
 
 res i Zr ab Sriab 0 2ab 
 
 5. 512 Aci. 2 Baa ae eS 
 | __ r(—a) | + (at), __ r(a—l?) 
 4. Cos 24 = PY alrncy PS WE che sR ned eh 
 
 Ar ab (b*—a*) 
 
 ct 
 
 4rab(a’—b*) 
 arent hae 
 
 5. Sn4a= ; sn4 a= 
 
 From which latter formula, it appears, that 4 a is 
 greater than 180°, when a is greater than b. 
 
 149. To these may also be subjoined the following 
 formulee, which serve for the solutions of some particu- 
 lar cases of oblique-angled plane triangles, when the 
 sum or differences of certain parts are given, instead of 
 the parts themselves. A 
 
425 
 
 a La I 
 1, Sin 5 (Bsc) = —— cosh a 
 
 b _ 
 2. Cos3 (Bac) = sinka 
 
 I (2+l)t+e. 
 3. SPL oS rae - tan > B 
 See ed unk oa Le 
 4, Tanga eed) tan > B 
 b . 
 5. Tanga = 7 cot 4 (BSC) 
 
 6b act 1 a tani (pac) 
 
 r: bte= —~ cot i a cot 4 (B4c) 
 
 150. Similar elie may likewise be given for the 
 solutions of some particular cases of right-angled sphe- 
 rical triangles; the most useful of which are the fol- 
 lowing : A 
 
 B Cc 
 
 1, r? Tan £ (c—a) = tan? $b cot $ (c-+-a} 
 Or, 7? Tan 4 (c-++-a) = tan’ 5 L b cot 1 1 (c—a) 
 Ohar? Bi (osha = sin b cosatan$ Bs. 
 Or, r? Sin (c—a) = tan b cosc tan 5B 
 3. r Cot 4 (45°41) = tan $b cot $ (c-+a) 
 4, Cos tava = 2 cos c — cos (a+) 
 5. r Tan (45°-+2 a) = tan 4 b cot $ (c—a) 
 6. 7? Sin (c—a) = tan’ £ B sin (ete) 
 7, 7 Sin (ca) = cot’; B sin (c—a) 
 
426 
 
 8. 7? Cos (A&B) = — cos (A+B) cot? dc 
 9. r* Cos (A+B) = — cos (AB) tan? ic 
 10. r* Tan4 (ca) =cot* 5 (45°-4 i SN 
 11. r Tan 4 + b= cot (45° ne A) tan} (c+a) 
 12. 7? Tan (e+a) = tan’ (45° $a) tan 5 (c—a) 
 13. 7 Tan 5 pies tan (45°-+2 a) tan 2.(c—a) 
 14, 7? Tan 5 (c+a)= tan’ 4 b cot $ 1. tien) 
 15 Gas C fee waits (a-}-b) + 5 cos (deb) 
 16. Sin b = cos (cuB) — 4 cos (c+B) 
 17. Sin a= 4 cos (cua) — 4 cos (ca) 
 ES Os A a sin (a-++B) — 4 sin (a—B) 
 19. Cos B =4sin (b4+a) —4 = (La) 
 Among tie formule it is evident, that the five 
 last, being formed by addition and subtraction only, 
 may be calculated by means of the natural sines and 
 cosines. | 
 To these, we may likewise add the four following 
 forms, which are applicable to any spherical triangle 
 
 whatever. 
 Sin 5 (b—c) _ sing Lox), Sin 5 (O-+<) __ 0983 (3p—c) 
 sniao= ~~ sin La 7 aM gia isin Bi a sin 5 a 
 
 Cos £ (4—c) sin F (B+c) | Cos 5 (b+c) apes a Ee) 
 aay. i > 
 
 cos 5a Cos 4a cos £a_ ara sin 3 A 
 
 151. OF THE INCREMENTS AND FLUXIONS OF THE 
 SINES AND TANGENTS OF ARCS, OR ANGLES. | 
 
 As formule of this kind are frequently employed in 
 astronomy and the higher branches of mathematics, in 
 ‘order to. show the changes that take place in the sides 
 and angles of triangles, from.small variations of some 
 of their parts, I shall here subjoin such of them as will 
 
42'7 
 
 be found sufficient to answer most of the purposes ‘to 
 which they are usually applied. 
 
 Thus, if z be made to represent the increment of 
 the arc z, and sin’ x the increment of its sine, &c. we 
 shall readily obtain the following expressions; which are 
 only the formule in article 26 under another form. 
 
 l. 5 msi, 2.== 2 pms 2’ cos. (z-+4 2’) 
 2. = r.cos Z = 2 sin 2 z sin, (z ---2 2%) 
 
 7? sin x 
 we Tan’ 2 gpl ear ae tran 
 ¢ cOs % Cos (2-+2') 
 4. Cb? . 7? sin 2! 
 oo y 
 
 De 4 Hh, Ses SIN Zein ( re pn’) 
 
 6. — Cos* z= sin z sin (2z2+4+2’) 
 
 . sin 2 sin (2 %-42') 
 
 cos? 2 cos® (2 4-2’) 
 
 8 — Cote = BER Gat) 
 
 _ Which expressions are rigorously exact, whatever 
 may be the magnitude of the quantity 7; and if the 
 
 second member of either of these equations is to be 
 
 : Tan? z= 
 
 employed with a negative sign, we must substitute 
 (z—z) instead of (z+2’), (z—$2z) for (c+32’), 
 and (2z—z’) for (2z+2’). 
 
 152. Itis also evident, that if z’, in these expressions, 
 be taken indefinitely small, or within less than any 
 assignable limit of O, its sine may be considered as 
 equal to the arc, and its cosine to the radius : whence, 
 by proper substitutions, we shall readily obtain the 
 following formule, for the fluxions of the sine, tan- 
 gent, &c, of any arc or angle ; being the same as those 
 usually given by the writers on this subject. 
 
Sin z= 2cosz = ZW Posin®z 
 v 
 
 9, — Cosz =z sin 2 = 2 / A cose 
 
 3 T ° z 2 % sec? x 
 2 anz= re a I 
 cos*z rt— sin? x" 7 
 ° % z % COSeC? 
 ag Cotizit= Uk se ar Bake a “spk ios oe mk 
 sin?z 7? — COS? x r 
 
 Sin?s = 2 Zz sin z cos 3 ==Z sin 2 z. 
 
 Tove han 2 2tanztanz = 2tanz x echt 
 
 8, — Cot? s = 2cot zcotz = 2cotz X —— 
 
 Which formulz are only to be employed when the 
 variation of the are is extremely small, as the result, 
 when its increase or decrease is considerable, is often 
 very erroneous (d). 
 
 153. From these latter expressions we may likewise 
 readily obtain the following, in which the fluxion of the 
 arc is expressed in terms of its sine, tangent, &c, 
 
 r sing 
 
 Lag Say eens see 
 fr? — sin? 2 
 
 7 COS B&B 
 
 /12—cos? % 
 
 9 
 Ne 
 | 
 
 e 
 r? tan % 
 7? 4+ tan? x 
 
 - 
 
 e 
 r* cot 
 r’+cot® % 
 
 s 
 4.2==— 
 
 (f) For a more copious collection of formule of this kind, 
 both incremental and fluxional, with their applications, see Traité 
 de Trigonométrie de Cagnoli, 2d edition, where this subject is very 
 ably and fully discussed. | 
 
429 © 
 
 o 
 < r* sec % 
 NAS Memes rmgemecees rn | 
 . sec % ,/sec® 4-7" 
 . r vers % 
 6. gas satniaes 
 
 Y2r vers z—vers’x. 
 
 And if these formule are expanded, and the fluents 
 of each term be taken, they will give the usual series 
 for the arc, in terms of its sine, tangent, &c. 
 
 154. Some of the trigonometrical formule, given in 
 the former articles of this work, may also be applied to 
 the solution of the various cases of quadratic and cubic 
 equations, as follows : | 
 
 SOLUTIONS OF QUADRATIC EQUATIONS. 
 loa + pe — 9 = 0. 
 Z 
 Puts V gq = tan 2 
 
 +vqXtan 3 z, 
 Then x = or; 
 me / 9X COl 
 
 Or, putting 
 10 + log 2 + 3 log q — log p =Iog tanz. 
 + $ log gq + log tan i z — 10, 
 Then log x. = or, 
 
 — 4 log g — log cot $x — 10, 
 2. = pct —¢q = 0; 
 
 ita 
 = 
 
 Put — WG == D2, 
 
 +Wq7X cotZ2, 
 Then x = or; 
 
 —V 9X tant z. 
 Or, putting 
 10 + log 2 + 4 log g — log p = log tan z. 
 
al — 
 
 430 
 
 “or, 
 
 + 4 log ¢ + log cot 4 x — 10, 
 Then log = =4 | 
 _— % log g — log tan $ z — 10. 
 do Rt ih act sf 
 Put V9 = sin. 
 
 —/ 7X tan} 2, 
 ‘Then w = or, 
 —VWqX coZ2 
 
 Qr, putting 
 10-+log2+4 log g.— log p = log sin x, 
 4 log q — log tan $ z— 10, 
 Then log x =\_, or, ii 
 — 3 log g — log cot 4 x — 10. 
 4, 2°—px+t+q=0. 
 Put — /q=sin x. 
 
 | +YqxX tandz, 
 Then v = | or, 
 +/9¢X cots 2. 
 
 Or, putting 
 10 + log 2 -++ 4 log g — log p = log sin z. 
 (+ 4logg + log tang z — 10, 
 Then log x =| Or, 
 + 3 log g + log cot 32 — 10. 
 In the two latter of which cases, if = vq be 
 
 greater than 1, or 10 + log 2+ Slog g — log p comes 
 out greater than 10, the two roots, or values of #, will 
 be impossible. 
 
431 
 
 155. SOLUTIONS OF CUBIC EQUATIONS. 
 1. a? -++-pr—g=0. 
 Put = 4 £(=)'= tan z, and ¥/tan (46°—2 2) 2) == tan uw; 
 MST Ren ¥ 212-9 C X cot 2% 
 Or, putting 
 Log —.--+.10,— br Po 1c tan z 
 8 9 POP) Ip wy Ciao eee 
 And 
 ; (log tan 45°F -}+ 20) = log tan w, 
 _ Then log x = $ log =? +- log cot 2 u — 10. 
 2xe-+pr+tq=0. 
 Put %(2)? = tanz, and Vim G5—2s) = tanu, 
 Then x=— 2 ve xX cot 2 u. 
 a ae 
 Log 2 +10——; = log 4 3 == log tan x, 
 ene | 
 2 (log tan 45°—2 x -+ 20) = log tan u, 
 4p 
 Then log « = 10 — 4 log -3- — log cot 2u. 
 3. x” 2 C= q tid og 
 This form has 2 cases, according as — = (f )? is less 
 
 or greater than 1. 
 In the 1st (fis 
 n the; Ist case, put — (ies = Cos % 
 
 And 7 tan 45°—)2z) = tan w; 
 
 Then 2 ==.2/ a X cosec 2 u. 
 
432 
 
 Or, dea 
 10 3 = log 4 aU RNa ‘ = log cos z, 
 And 
 ; (log tan 45° = + 20) = log tan uw; 
 Then log « = 10 ++ log < — log sin 2 u. 
 
 In the 2d case, put 2 ey = cos z, and x will have 
 the 3 following values: 
 
 x= +274 x cos= 
 
 $ 
 ' z 
 “==— 2 v ix cos (60° — =) 
 x=—2 Vv © x cos (60° + a) 
 g 3 
 Or, 
 
 Log xz = + log 2 + log cos — 10, 
 Log x = 4 log nf log cos (60° — 5) — 10 
 
 Logie = } log =2 ~£ + log cos (60° ++ 5) — 10, 
 
 Taking the value of «, answering to log x, positive 
 in the first, and negative in the two latter. 
 
 4.027 —pxr+-q=o0. 
 This form, like the one above, has also two cases, 
 
 t: WS Pe te | 
 according as — Ge): is less or greater than 1. 
 
 In be Ist case, put’ = oF ei = COS 2, 
 And ./tan (45°—i2) = tan u, as before; 
 
 Then « = ave “ cosec 2 2. 
 
433 
 
 Or, putting 
 10 + 5 log & — log z = log cos z, 
 ” And 
 ate: (tan 45°—3 z) + 20\ = log tan uw; 
 Then, — log x= 10+ log ~£ — log sin 2 u. 
 
 In the 2d case, put 5 ( ay = cos x, and w will have 
 
 the 3 following values : 
 
 x=—@2 vex cos = 
 
 x=+2 v£ x cos (60° — =) 
 
 x ot ke 2 vex cos (60° +- =) 
 Or; 
 L I 4 p 3% 
 og «= Ft log at tog cos; — 10, 
 
 Log x = 4 log ah log cos (60° — 5) — 10, 
 
 _ Log x = 3 log ara log cos (60° + 5) — 10, 
 
 Taking the value of x, answering to log x, negative 
 in the first, and positive in the two latter. 
 
 As an example of this mode of solution, in what is 
 usually called the Jrreducible Case of Cubic Equa- 
 tions, let x? — 3 x1, to find its 3 roots. 
 
 Here 4 (Got A ge ae & ee "5-=3 cos GO" sea} 
 
 j2= 2/8 x cos =z 2 cos 20° == 1.8793852 
 =| 
 Bid e=— 2.4/5 x cos (60 —=) = — 2c05 40 = — 1.532088 
 Q 
 @ 
 | ex2v8 x cos (60° + 5) = — 2cos 80° = — .3472964. 
 
 2F 
 
t 
 dIUOTT 
 
 454: 
 
 Also, let #3 — 3 ¥=— 1, to find its three roots. 
 Here, as before, 7 ae = 15, = COR OO Va a 
 
 4 
 Pe =— ay x cos = =— 2cos2° = — 1.8793852 
 a o ; 
 
 ii 5 Nae x cos (60° — # = 2 cos 40° == 1.5320888 
 [== —2 VE x cos (60 ‘)= 2 cos 80° = 0.3472964. 
 
 Where the roots are the negatives of those of the © 
 first case (¢). 
 
 OF THE ADMEASUREMENT OF ALTITUDES BY THE 
 BAROMETER AND THERMOMETER. . 
 
 156. Having treated, pretty fully, in the former part 
 of this work, of the methods of measuring elevations 
 and depressions geometrically, I shall here subjain one 
 of the most easy practical rules for determining the same 
 thing by means of the barometer and thermometer, 
 which is a mode frequently used at present; and 
 though not founded upon such sure and well-establish- 
 ed principles as the former, is nevertheless susceptible, 
 
 when performed by a skilful observer, with good in- 
 
 struments, of a considerable degree of accuracy. 
 
 For this purpose, the person who undertakes these 
 observations should be provided with two portable ba- 
 rometers, of the best construction (both filled with 
 mercury of the same specific gravity), on which, by 
 
 zn ) ‘ 
 .. (g) For the mode of investigating these kinds of formule, see 
 
 Cagnoli Traiié de Trig. and article Irreducible Case, given by the 
 author, in the Supplement to Hutton’s Mathematical Dictionary, 
 
435 
 
 7 
 
 means of a nonius, properly adapted to the scale, he 
 may read off the heights of the mercurial columns to the 
 200dth partof aninch. Each of the barometers must, 
 also, have an attached thermometer, set in the wooden 
 frame, in the same manner as the barometer is, and 
 having their balls of nearly the same diameter as that of 
 the barometer tube: besides which, there must be two 
 other thermometers detached from the barometers. 
 
 Then, one of these barometers, with its attached 
 and detached thermometers, is to be placed in the shade 
 at the top of the eminence whose height is required, 
 while the other remains below; and when they have 
 continued in their places a sufficient time for the des 
 tached thermometers to acquire the temperature of the 
 air, or till the fluid is stationary, the observer on the 
 eminence must note down the height of the mercurial 
 column in the barometer, as well as the temperatures 
 exhibited by the attached and detached thermometers ; 
 and, at the same time, another observer must make 
 the like observations on the instruments below. 
 
 This being done, the altitude of the object, at the 
 top and bottom of which the instruments were placed, 
 may be ascertained by observing the following pre- 
 cepts, or the practical rule which is deduced from 
 them. 
 
 (1. The height through which we must rise to pro- 
 duce a fall of mercury in the barometer, is inversely. 
 proportional to the density of the air, or to the height 
 of the mercury in the barometer. | : 
 
 2. When the barometer stands at 30 inches, and the 
 
436 
 
 air and quicksilver are of the temperature 32°, we 
 must rise through 87 feet to produce a depression of 
 lsth of an inch. 
 
 3. But if the air be of a different temperature, this 
 87 feet must be increased, or diminished, by about — 
 *'; of a foot for every degree of difference of tem- 
 perature above or below 32°. 
 
 4, Every degree of difference in the temperatures 
 of the mercury at the two stations makes a change of 
 2.383 feet, or 2 feet 10 inches in the elevation, 
 
 Hence the following Rule. 
 
 1. Take the difference of the barometric heights, 
 in tenths of an inch, and multiply the result by 30. 
 
 2. Multiply the difference between 32° and the mean 
 temperature of the air, by .21, and find the sum or 
 difference of this product and 87 feet, according as the 
 temperature is above or below 32°. 
 
 3. Multiply this sum, or difference, by the former pro- 
 duct, and the result,’ divided by the mean of the baro- 
 metric heights, will give the approximated elevation. 
 
 4, Multiply the difference of the mercurial tempera- 
 tures by 2.833 feet, and add this product to the ap- 
 proximated elevation, if that of the upper barometer 
 be the greatest, or otherwise subtract it, and the result 
 will be the corrected elevation. 
 
 Or, the same rule may be expressed algebraically 
 30 p (87 + 0.21 d) 
 
 thus: A 
 . in 
 
 2.833 5 
 
 Where dis the difference between 32° and the mean 
 temperattre of the air, p the difference of the baro- 
 
437 
 
 metric heights in tenths of an inch, m the mean baro- 
 metric height, 5 the difference between the mercurial 
 temperatures, and a the corrected altitude. 
 
 For example, suppose the mercury in the barome- 
 ter, at the lower station, was at 29.4 inches, its tem- 
 perature 50° of Fahrenheit’s thermometer, and the tem- 
 perature of the air 45°: also the height of the mer- 
 cury, at the upper station, 25.19 inches, its tempera- 
 ture 46°; and the temperature of the air 39°. 
 
 ( 7 — (45°—32°) + (39°— 32°) 
 
 Theng = (29.4 — 25.19) X 10 = 42.1 
 2 To eae Ye See = 4, 
 
 Lm = 2 (29.4 + 25,19) - -- == 27.295 
 
 ee 30 x 42.1 x (87 +. .21 x 10) 
 And Ayo a OF BORG ee oa 4 xX 2.833 
 
 = 4111.91 feet, or 685.32 fathoms, the correct 
 altitude. 
 
 And if two or three sets of observations be made at 
 each station, after short intervals of time, and the mean 
 of the results be taken, the probability of error will 
 be much diminished. 
 
 It may here be added, that the method of measuring 
 altitudes by means of the barometer, was first distinctly 
 pointed out by Halley, in No, 181 of the Philosophi- 
 cal Transactions ; but it was not till long after his time 
 that the method was turned to any real use. The 
 chief improvements are due to M. de Luc, who pub- 
 lished, at Geneva, the result of his experiments and 
 
 Inquiries, made on the high mountains of Switzerland, 
 
in‘a treatise on the barometer and-thermometer ; and’ 
 also in the Philosophical Transactions. Other valuable‘ 
 papers on this subject have likewise been given by Mas-’ 
 kelyne, Horsle y Sr George’Shuckburgh, and Gene- — 
 ral Roy, in the heron tyrone’ of these Transac-' 
 tions. | : 
 
 But the most complete ritacssitt that has yet appears 
 ed, of this useful and interesting part of the subject, 
 may be found in the additions to chap. 15, vol. ili, of 
 the Traité Elémentaire d’ Astronomie Physique, by 
 Biot, who has there entered into all the details that can 
 be wished for, on this head, both with respect to the 
 demonstrations of the formule, and their several appli- . 
 cations. 
 
 FINIS. 
 
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