a eae ae, Baty Paty ODT Te ICR NE SE CR Se ae Oe ehh Ain AM Sei Beg Pm Hin 5 Re Berd eR a ate Talia g intel eee Paint P Aca het oath tate Farclagtts BaP Bahn uae io RM BMatet 4, Gene he 2p Fate: Tad - Meg a aa Aste ste Pit eee Ne AO fi a0 teh O-appenten Fore tether ute peed, ee re re 1 raphe Pe Te al 2 ‘ : : Fn a hag “eh PEE: aS Sr mee: Fa Bein Mgt ae Me ; . r ath alt aa as Peete ALS “iit te ce, : : ‘ Hedy Sorby) ‘ “ ‘ * Sg ap ett ear : Nga gur th ate Ati Seat Fig B= 8: 9 8 Sn a tee a sain, ag Pig Pate BAA LS See BE a ae. ratte Si anensbere 0, ee ees OH Z 2 Named a ate’ 2, rae iatoee “ites pe thg hd in FFs gt eta PAA FIN atin ta r " oa. " SMe Saale ae eget at Aes ee iy We : tig hatha ng eh E ightn tae Me om RO nets? Bh Ate Nek Te Fe” A Movras a 4.4 ¥ Ae - : ? igtatetn An: The atctet st gS A als! Baa ee fe BG He Ae omy the BaP as Sates “ ete ne Sa seenaie® 42 MATHEMATICS i? 7. Digitized by the Internet Archive in 2022 with funding from University of Illinois Urbana-Champaign https://archive.org/details/elementsofplanesOOgreg NOTICE OF THE EDGES OF THIS a > : WA i. 4 Af - re ae ae ca : As aie tr - MAGAZINE HAS BEEN LEFT UNTRIMMED, BECAUSE OF AN EXTREMELY NARROW MARGIN HERTZBERG-NEW METHOD INC _facetecs Cocomere, ee fe Cr teste Ma 7, Pe be 7 Ses rte, ; : AMeemerveed, ' ni; A tafe tee Oxehecleecsre4 4 i Rita i We Re Ri f hoa g. Vivi Le | «Mearns tenet sta Sie COL! Lele’ ee é Vb x ha. Meee : | Decexvitt ve ae a Mes th WM, y LTHORAWS , | ; § ‘ f ' ELEMENTS PLANE AND SPHERICAL TRIGONOMETRY: WITH THEIR APPLICATIONS TO HEIGHTS AND DISTANCES, PROJECTIONS OF THE SPHERE, DIALLING, ASTRONOMY, THE SOLUTION OF EQUATIONS, AND GEODESIC OPERATIONS. BY OLINTHUS GREGORY, LL. D. Of the Royal Military Academy, _ WICE-PRESIDENT OF THE LONDON PHILOSOPHICAL SOCIETY, AND HONORARY MEMBER OF THE LITERARY AND PHILOSOPHICAL, AND OF THE ANTIQUARIAN, SOCIETIES AT NEWCASTLE UPON TYNE, 4 LONDON: PRINTED FOR BALDWIN, CRADOCK, AND JOY, 47, PATERNOSTER-ROW ; W. BLACKWOOD, EDINBURGH; AND Js CUMMING, & DUBLIN. 2 ered 1816. ee ae SS Ps See ee eee eee we “ST ov aL Nees 7 A ange aun MYATT Co “; VATAMOKO ODINT: aon TROLL ity SSM'S WD Fy, pot a ee a ens ~ t ok a Rabie Xt af enor TLOAT Soma tarc ‘aA arKo tie y ira TE, tii Pit FLARE cans hed A a en 3 fe te ‘ , J Pp it ’ = oie i “Rs ta r ’ 7 : sie ; ahs ’ ~ * ‘ ; Sake Ma owe , ‘ee 7 od et trons aUHTVLIO. _gysiabpaty ABV Sayed MAO” iy * vb ca Toe Xt ont 2agiH4 YOuUar mAy RQ made’ eR ¥. . na GORY HARNTH BAT 2 AGAMA Ts aofOn p, ee ee oe ty a PREAH AS TE bah sel 22 SWAESTATIORT rer 30. x + : ¢ , 4 hk at = by as °- 3 y it . ' Wee : we ; ‘ ay uy ere i te 2VOGMOL | (uh a aa 4 Set TOU-7 hi AD onh 10. MUTEIAT KOT RTH, 4 pa 2 WO-MATHOK BETAS Rtg id ig a ee. pWOLTAAIGS © oowaiee y MMU Fis ane ted by C.Maldwin, ey er Beart London, — a alee a S44 5/024 e ¢ §G Z 2. MATHEMATICS 72 ee? LIBRARY OF OP ebe ne « 633 515 ve PREFACE. fr who wishes to acquire any other reputation than that of being useful, will not devote his time to the pre- paration ofan elementary work. Such, however, isthe -humble honour after which I here aspire: and if the small volume I now lay before the public shall contri- bute, in any degree, to increase the knowledge and _ improve the taste of the mathematical student in the. ieee stages of his progress, I shall have attained my object. ) ~ Of late years the authors of elementary treatises on edifferent branches of mathematics, and especially Dr. . Hutton and Mr. Bonnycastle, in their compendious ma- » nuals for popular use, have shown that it is possible, by judicious arrangement and selection, to compress much hinteresting, valuable, and scientific matter, within the ‘compass of a small duodecimo volume, There can, I /think, be no doubt that to the extensive circulation of this class of books, together with the stimulus furnished by the annual problems in the Ladies’ and Gentleman’s ' Diaries, and those proposed in the Senate House at Cambridge, must principally be ascribed the circum- stance that mathematical knowledge, to a certain extent, is more widely diffused over the middle classes of so- ciety in this country, than in any other part of Eu- : rope. “ A HTS: a av PREFACE, What, however, has been so successfully effected in other departments of mathematics has not yet been attempted with regard to Trigonometry. We have some excellent works on this subject, whose value it would ill become me to depreciate. But such of them as go extensively into the business of Trigonometry and its applications are too large and expensive for general circulation; while others, being confined almost en- tirely to the elements, exclusive of the applications, must of necessity be restricted, both in point of circu- lation and utility. There is one treatise, that of Emerson, which is a most copious store-house of curious and ele- gant theorems: but they are so obscured by a defective notation, that the perusal of greater part of the book must, to a mathematical student, be as perplexing as the solution of a. perpetual string of enigmas. It has been my.aim to steer into a middle course, between that, in which is presented.a mere common- place book of principles and theorems, and that which, by leading far into'the detail of multifarious processes and methods, precludes the study of the science, except by, the sacrifice of much time.and expense. In order to this I have endeavoured to be select in my materials, and. have, for the most part, observed unity of method. A pook of three times the size and price might have been drawn up with far less intellectual labour (for the. fa- ‘tigue of selecting from a fund of valuable materials, and casting the result into one mould, is not slight); but such a work would not have tended to accomplish the purpose I havein view. By adopting a small type and a full page, and confining myself chiefly to the analyti- cal mode of investigation, I have been able to introduce, and I shall rejoice if it be thought I have treated perspi- cuously, a greater variety of the applications of plane and — spherical Trigonometry than are to be found in any other work on the subject with which L.am, acquainted. In the first three chapters I have exhibited the theory of PREFACE, Vv Plane Trigonometry geometrically, and have shown the application of that theory to the logarithmic solutions of the usual cases into which this portion of the subject is conveniently distributed... I have endeavoured to . conduct these introductory inquiries with the utmost perspicuity ; that the student, by obtaining a thorough comprehension of the principal topics of research, and by seeing a little of: their utility, may enter with the greater relish upon the subsequent investigations ; and by tracing the correspondence of these results with such as will afterwards appear in the analytical theory of Plane Trigonometry, may be prepared to lean with full confidence upon the analytical formule that are in other: places to be laid before him. The deductions from theory in the succeeding chapters are usually obtained by analytical processes ; and their utility is shown in the logarithmic and trigonometric solution of a great. number of problems, classified under the particular heads of the several applications, as specified in the ta- ble of contents, The whole, except what relates to the _ minute variations of the sides and angles of triangles, aud the differential analogies which apply to them, may, I am persuaded, be readily comprehended by any per son who is tolerably conversant with the elements of. Geometry and Algebra. | In the composition of the work I have freely availed. myself of all such matter as was likely.to answer my pure pose, especially in the productions of foreign mathema- ticians. The plan and method are of course my own: the materials have been collected, almest of necessity,. from all quarters. In addition to the acknowledgments: which will occur in different parts-of this little volume,. it would be unjust not to say here,.that the theory. of Projections, the general:problem in reference to Dial- ling, and the comprehensive table of Differential Equa-- tions for the variations of triangles, are taken, simply. with such alterations as fitted them better for general: AS. vi PREFACE. usefulness, from the. Chevalier Delamére’s admirable treatise on Astronomy, in three volumes quarto, The — transferring such curious and valuable matter ‘from an expensive treatise in a foreign language, into'a’ cheap volume in our/own, will mot I hope be regarded as per- forming a trifling service to the English student. - Fam aware that there/are some persons, into whose hands this work may fall, who will not approve it as they would have done had the demonstrations ‘been. exclu- sively geometrical. This is in consequence of a preju- dice against the analytical processes, most singularly . cherished in a country where the modern analysis ‘has received some of its most valuable improvements : ‘a prejudice which, though it is rapidly weakening, still ré< tains its hold upon the minds of severalbrespectable ma- thematicians; and on account of which it may be expe- dient to assign some of the reasons that have induced me to appropriate so largea portion of the following vox lume as I have done to thé analytical or algebraicaé mode of deducing properties and theorems. 10 oie 1. It is more concise, and thereforeallows of the in- troduction of a much greater quantity and variety of matter, in any proposed space,’ than ceuld possibly be exhibited and demonstrated according to the geometri- cal method of the antients. | 2, This method is more’ wniform than the other, ‘as well as more general and comprehensive. In the geome= © trical method as it is usually conducted, however con- vincing and elegant, the demonstrationsof one property or theorem may not have the remotest analogy to that _which will serveto establish-the truth of another, The demonstrations of a series of propositions such as are ob- _ viously connected in the logical arrangement of a trea- tise, may probably have nothing common in their ap- pearance, except that they are all geometrical; nor shall the manner of demonstrating one proposition sug- gest necessarily a single hint that may apply to the de- PREFACE i¥ monstration of the very next.° The separate chairs ‘of demonstraticniof the two: ‘propositions may be as dis- tinct (if I may be pardoned so familiaran allusion) as thé processes by which a sword and a/needle are ma- nufactured, ‘In the one:case both ‘are: geometrical, qn thei other both are ‘mechanical; but neither of the two,. whether ‘geometrical or mechanical, although’ beauti- fully adapted to their purpose, need be at’all alike. It ig not thus with regard: to the analytical method: the processes haveall:more ‘or less of resemblance, they are all conducted by the same general rules; and they com- monly lead to universal results, from which particular corollaries ave’ deducible at pleasure. The analytical method is at the same’ time much hey most comprehen= sive. ‘There are several curious and useful theorems'to be found in the analytical treatises on’ trigonometry, which Have not yet, ‘to my knowledge, been demon- strated’ in any other way; and: not a few which I am persuaded do not admit: of any other kind of proof. 8: This'methed is also much the easiest. The’ pro- eessés themselves are, ‘in the main, conducted with the sreatest possible simplicity; the substitutions and trans- formations are generally natural and obvious! in ‘truth, so.much so, that a'student no sooner ‘attains a ‘compe- tent acquaintance with the manner of conducting his in- vestigation, than he willbe enabled to develope prac- tical theorems nearly as fast as he can write them down. Nor is the mode of inquiry such as heed encumber the _ Memory; the operations being general, the requisite first principles few. This is a great recommendation; because every unnecessary load upon the memory tends more or less to weaken our mental elasticity, and “impede the intellectual operations. 1 am happy to for- tify my opinion on this point by an observation of the most profound mathematician and natural philosopher now living, Lar.ace. \‘ Préférez (says he) dans Pen- “ seignement les méthodes générales, attachez-vous & ‘3 | Vili PREFACE. ‘“‘ les présenter de la maniére, la plus'simple, et vous ‘* verrez en méme tems qu’elles sont presque toujours “‘ les plus faciles.”’ 4, The analytical method of establishing the prin- ciples, and deducing the formule of trigonometry, has this farther advantage, that it connects it more. inti- mately with the principal-topics of mixed mathematics, and causes it.to become a portal to the higher mechanics and the celestial physics... Any person who has looked, however cursorily, into. the best treatises on statics, dynamics, and physical astronomy, especially those which have been published on the continent, must have observed that they abound with trigonometrical for- mule. And they who have gone a little below the surface, know that several of the most striking results of physical astronomy turn.upon some obvious trigono- metrical truth. Thus, to select only one class of in- stances, our countryman Sznpson, in his researches into that part of the celestial physics which relates to the moon (Miscellaneous Tracts, p. 179), having shown that no terms enter the equation of the orbit but what are expressible by the cosine of an arc, or the cosines of its multiples, and, therefore, that no terms enter that equation but what. by a regular increase and decrease return to their former values; immediately infers that the moon’s ‘‘ mean motion, and the greatest quantities of the several equations, undergo no change from gra- vity.”’ Frist advanced still farther in the same line of induction. And farther yet Lagrange and Laplace; who have demonstrated that no term of the form A x nT, or A tan nT, or A Ccosec NT, or ~ =. (r denoting the time) can enter the analytical expression forany of the inequalities of the planetary motions, or those of their satellites: and have thus proved that the-system is stable, all its irregularities being confined within certain limits; just as all the modifications in the magnitude and posi- - a eae ee PREFACE. ‘ix tion of the sines.and cosines of arcs in the same circle ave confined within limits, such as the theory of trigo- nometry assigns them., This consideration stamps a value upon,the researches in this departnient of science which they would not otherwise possess; and in order that the mathematical student may fully avail himself of it, it is requisite that he understand the analytical method. Lastly, this method is preferable to the geometrical, because it tends to communicate to thestudent the habit of investigation, which that does not. It is one thing to be able to demonstrate, or to be able to understand by means of a demonstration, that a proposition is true or false: it is a totally distinct one to be able to znves- tigate propositions which shall inevitably be true. In this point of view I have often been struck with what I cannot but regard as a singular defect in the manner of teaching geometry, which prevails in most mathema- tical seminaries. If a student so apply himself to the admirable ELEMENTS of Euclid, or to those of Legendre, or others which need not be specified, as to understand and feel the force of each demonstration, and trace the exquisite concatenation and mutual dependence of the parts, his logical habits of arrangement, and the classi- fication of his thoughts in reasoning, must be improved; _ and superadded to this there may be a fondness for geo- _ metrical pursuits. But this latter consequence does not _ necessarily follow; for which the principal reason is, that he has not been taught the use of his instruments. _ That method of teaching geometry is essentially defect- _ive which does not include geometrical analysis: and yet, axiomatic as this would seem to be, Euclid’s Ele- ments, or books for a similar purpose, are almost univer- sally studied; while Euclid’s Data are almost as univer- sally neglected, For like reasons, every department of mathematics should be so taught as to enable the stu« dent, nay to stzmulate him, to pursue his researches: and 6 x PREFACE. this, as in every other region of abstract science, so in trigonometry, must, if the pursuit is: intended to be at — all extensive, be conducted, for the most part, (I de not say, exclusively) according to the principles of analysis. | | OLINTHUS GREGORY, Royal Military Academ Y; Woolwich, Feb, 1816, CONTENTS. Pa ge Cuar. 1. Preliminary definitionsand principles . . II, General properties and mutual relations of the lines and angles of circles and plane triangles Shed IIT. Solutions of the Se cases se bites tri- angles) -/i<% 4'¥« iyi {V. Plane trigonometry econadee analytically V. Application’ of plane*trigonometry to the _ determination of heights and‘distances. . VI, Spherical trigonometry, in six sections © 298/809 “Sect, 1. Fundamental principles, general pro- perties, and formulz .... . Sect. 2. Right angled spherical triangles . Sect. 3. Oblique angled ditto... .. | Sect. 4. Analogies of Napier . . . . ies bocnod “_- ‘Br Anerinonida of: spherical trigonometry id at wee Se 6. “Teas of spherical. ‘triangles and po- init h *, lygons, and measure of'solid aigles ° : vil Iomtennic computation ‘of! ih tri- | BOPACE coe ie. 5) amy: hes VIII, Projections of the pana ge in three sections Sect. 1. Astronomical definitions . . . Sect. 2. Orthographic projection . . . Sect, 3, Stereographic projection 1X. Principles of dialling . . 2... X. Astronomical problems . . ... . XI. Minute variations of triangles, with a table of differentialequations . .... . XII. Miscellaneous problems, in two sections Sect. 1. Problems with solutions . . . . Sect. 2. Problems without solutions . .. . 4 ] 79 i RAG: 3 2 Page 8, line4; for’ ” tt 2 ipked ap! —-- , , So Sie 14, Leta line 5 aeaed om C to the Sine of contact A, in the first figure. 29, line 8, for answer read answers 37, line 2, for and is read it is: WoT 4}, line 3, for ® read =* 156, ee 1, rae An, ee Crinmrnisoeried des. ems, pour Van 18{7, there is given a table by) M, Burckhardt to” shorten the computation of this problem, ‘167, line 9 from bottom, for limited, read limits 212, last line but one, for + 2.4/ Lay néak FP 24/ Zps sper le et } TRIGONOMETRY. CHAPTER I. Preliminary Definitions and Principles. - 1. THE word Trigonometry signifies the measure of triangles. But in an enlarged sense we comprehend under this name, the science by which we are enabled to determine the positions and dimensions of the different parts of space, by means of the previous knowledge of some of those parts. 2. If we conceive any different points whatever, posited in space, to be joined one to another by right lines, there will be presented to our consideration three things: Ist. The lengths of those lines. 2dly. The angles they respectively form. 3dly. The angles formed_re- spectively, by planes in which those lines are, or may be imagined to be, comprehended. On the comparison _of these three objects depends the solution of the various questions which can be proposed, as to the measure of extension, and of its parts. | _ §. The intersections of three or more lines in one and “the same plane, constitute angles limited by right lines, _and plane or rectilinear triangles, or polygons suscep- tible of being resolved into triangles. And the inter- sections of three or more planes, form plane angles and triedral or polyedral surfaces, the determination of the magnitudes and relations of which is facilitated by re- B 9g Definitions and Principles. ference to the surface of asphere. Hence mathemati- cians have been led to attempt the solution of two general problems. I. Knowing three of the six things, whether angles or sides, which enter the constitution of a rectilinear tri- angle, to determine the other three; when it is possible. — Il. Knowing three of the six things which compose a — triangle formed on the surface of a sphere, by the inter- sections of three planes, which also meet in the centre of that sphere, to determine the other three, when pos- sible. The resolution of the first of these general problems appertains to plane or rectilinear trigonometry: that of the second, to spherical trigonometry. 4. Lines and angles, being magnitudes of different kinds, do not admit of comparison. It becomes neces- — sary, therefore, to have recourse to quantities of an inter= — mediate kind, akin to the one, yet having an obvious — dependence upon the other, and serving as a common ~ vinculum, Such are the lineo-angular quantities deno- — minated sines, tangents, &c. which we are about to de- — fine. They are dines, but lines which admit of being © measured only by parts of an assigned line, the radius — of a certain circle; and lines which at the same time depend altogether, for their value, upon arcs of that — circle, which arcs are, themselves, adequate measures ef the angles included between the radii which limit such arcs. ~ | 5. By means of this happy invention of intermediate quantities, the business of trigonometry is greatly facili- tated. For, by imagining a perpendicular let fall from the vertical angle of an oblique angled plane triangle upon the base, or base prolonged, it will at once be manifest that the resolution of triangles generally, may — be referred to that of right angled triangles. Thus, as- suming for aterm of comparison, the hypothenuse of a right angled triangle, equal to unity, for example, com- puting the bases and perpendiculars of all possible right Plane Trigonometry. 3 angled triangles having the assigned hypothenuse, and arranging them in different columns of a table, the mag- nitudes of the different parts of any proposed triangle, would become determinable upon the known principles of similar triangles. Such a table as this, would, as will soon be seen, be no other than a table of natural sines. PLANE TRIGONOMETRY. 6. Plane trigonometry is that branch of mathematics, by which we learn how to determine or compute three of the six parts of a plane, or rectilinear triangle, from the other three ; when that is possible. _. This limitation is necessary, although there is only one case in which it can occur, namely, that in which the three angles of a rectilinear triangle are given. For, it is plain from Euc. vi. 4, that while the three angles of a triangle remain the same, the sides, though retaining the same mutual relation, may be greater or less, in all conceivable proportions, 7. Lemmal. Jet acs be a rectilinear angle: if, about the point c as a centre, and with any distance, or radius, cA, a circle be described, intersecting ca, cx, the right lines that include the angle AcB, in Aand 8; the angle ace will be D to four right angles, as the arc az to the whole circumference of the circle ADFE, F A Produce Ac to meet the circle in . F, and through the centre c draw pE, another diameter, to meet the circle E in D, E. } . Then, Euc. vi. 33, ang. acs : right ang. AcD :: are AB pee sarc AD, and, Euc. v. 4, cor. quadrupling the consequents, we have, angle Acs : 4 right angles ::; arc AB: 4AD, that is, to the whole circumference. BZ 4, Plane 1 rigonometry. 8. Lemma lI. Let acs be a rectilinear angle: if, about c as a centre with any two distances ca’, cA, two circles be described, meeting cA, cB, in A’, B’, A, B; the arc AB will be to the whole circumference of which it is a part, as the arc a’B’ to the whole circumference of which it is a part. By, lem. 1, arc aB: whole circum. :: angle acB: 4 right angles, and, arc A’B’:its whole circum. ::angle a’cs’:4rightangles. __ Therefore, arc AB: whole circum. :: arc A’B’: whole circum, of its respective circle. Definitions. 9. Let acs be a rectilinear angle, if about c as a centre, with any radius cA, a circle be described, inter- secting CA, CB, in A, B, the are AB is called the measure - of the angle acs. 10. The circumference of a circle is supposed to be divided, or to be divisible, into 360 equal parts, called degrees ; each degree into 60 equal parts, called minutes ; each of these into 60 equal parts, called seconds ; and so on, to the minutest possible subdivisions. Of these, the first is indicated by a small circle, the second by a single accent, the third by a double accent, &c. Thus, 47° 18’ 34” 45°”, denotes 47 degrees, 18 minutes, 34 seconds, and 45 thirds. So many degrees, minutes, seconds, &c, as are contained in any arc, of so many degrees, minutes, seconds, &c. is the angle of which that arc is the measure said to be. Thus, since a qua- drant, or quarter of a circle, contains 90 degrees, and a guadrantal arc is the measure of a right angle, a right angle is said to be one of 90 degrees. 11. The complement of an arc is its difference from a quadrant; and the complement of an angle is its differ- ence from a right angle. ead ¢ Definitions. 5 12. The supplement of an are is its difference from a semicircle ; and the supplement of an angle is its differ- ence from two right eee: 13. The sine of an arc is a perpendicular let fall from one extremity upon a diameter passing through the other. 14, The versed sine of an arc is that part of the dia- meter which is intercepted between the foot of the sine and the arc. | 15. The tangent of an arc is aright line which touches it in one extremity, and is limited by a right line drawn from the centre of the circle through the other extre- mity. 16. The secant of an arc is the sloping line which thus limits the tangent. 17. These are also, by way of accommodation, said to be the sine, tangent, &c, of the angle measured by the aforesaid arc, to its determinate radius. 18. The cosine of an arc.or angle, is the sine of the complement of that arc or angle: the cotangent of an are or angle is the tangent of the complement of that arc or angle. The co-versed sine, and co-secant are de- fined similarly. To exemplify these definitions by the annexed dia- gram: let AB be an assumed are of a circle described with the radius Ac, and let aE be a qua- drantal arc; let pp be demitted perpendicularly from the extre- mity B upon the diameter Aa’; parallel to it let at be drawn and limited by cr: let cp and £M be drawn parallel to aa’, the latter being limited by cr or cr produced. Then BE is the complement of BA, and angle scx the comple- ment of angle BCA; BEA’ is the supplement of BA, and angle sca’ the supplement of Bca; BD is the sine, DA the versed sine, AT the tangent, cr the secant, Ge the 6 Plane Trigonometry. cosine, GE the coversed sine, EM the cotangent, and cm the cosecant, of the arc AB, or, by convention, of the angle ACB. Note. ‘These terms are indicated by obvious con- tractions : : Thus, for sine of the arc AB we use sin AB, tangent... ditto’... tan AB, secant ....ditto...... sec AB, versed sine ditto..,... versin AB, cosine,....,. ditto,.,... COS AB, cotangent .. ditto,..... cot AB, cosecant.... ditto....4.. CoSsec AB, coversed sine ditto...... Ccoversin AB. Corollaries from the Definitions. 19. (a). Of any arc, less than a quadrant, the are is less than its corresponding tangent; and of any arc whatever, the chord is less than the arc, and the sine Jess than the chord. For, in the preceding diagram, the circular sector cAB is less than the triangle cat, the former being con- tained within the latter.. That is, by the rules for men- suration of surfaces, 3cA x arc AB is less than $cA x tan AT; whence, dividing by Jca, there results arc Az Jess than tangent AT. In a similar way it may be seen that chord a3, is less than arc. AB, and sine Bp, less than chord as. The same is also evident from the consideration that a right line. AB is less than any curve line terminated by the same points, and the perpendicular sp less than the hypothenuse a8, of a right angled triangle ADB. (3). The sine sp of an arc az, is half the chord br of the double arc BAF. ) ; (c). An are and its supplement have the same sine, tangent, and secant. (The two latter, however, are affected by different signs, + or —, according as they | appertain to arcs less or greater than a quadrant; the - *, Definitions. 7 reasons of this will be explained in a subsequent _ chapter.) (p). When the arc is evanescent, the sine, tangent, and versed sine, are evanescent also, and the secant be- comes equal to the radius, being its minimum limit. As the arc increases from this state, the sines, tangents, secants, and versed sines, increase; thus they continue, till the arc becomes equal to a quadrant Az, and then the sine is in its maximum state, being equal to radius, thence called the sine total; the versed sine is also then equal to the radius; and the secant and tangent become ing incapable of mutually limiting each other, are re- garded as infinite. (zc). The versed sine of an arc, together with its co- sine are equal to the radius. Thus, ap + Bc = ab +pc=ac, (This is not restricted to arcs less than a quadrant, as will be seen in the chapter on analytical plane trigonometry.) (F). The radius, tangent, and secant, constitute a right angled triangle cav. The cosine, sine, and ra- dius, constitute another right angled triangle cps, simi- lar to the former. So again, the cotangent, radius, and cosecant, constitute a third right angled triangle mec, similar to both the preceding. Hence, when the sine and radius are known, the cosine is determined by Euc. 1,47, The same may be said of the determination of the secant, from the tangent and radius, &c. &c. &c. (c). Further, since cb: pB:: cA: AT, we see that the tangent is a fourth proportional to the cosine, sine, and radius. | Also, cb: 0B::CA:cT; that is, the secant is a third proportional to the cosine and radius. Again, CG: GB::CE: 45M; that is, the cotangent is a fourth proportional to the sine, cosine, and radius. And, Bb: BC:: CE: cM; that is, the cosecant is athird proportional to the sine and radius. (u). Thus, employing the usual abbreviations; we should have ait 225 cio eal 8 Plane Trigonometry. 1 1. cos = ./(rad*— sin), 2. tan = 4/(sect—rad*), — 3. sec = 4/(rad? + tan?), 4. cosec = 4/(rad?-+ cot?). — rad x sin rad x cos 5. tan = ————. 6. cot = —_-—_- cos sin rad? rad? I 7. sec = . 8. cosec = ; cos sin ‘These, when unity is regarded as the radius of the — circle, become | 1. cos = 4/(1— sin’). 2. tan =: 4/(sec? — 1). Si seC == 4/ {1 4+ tan? ); 4, cosec = 4/(1 + Ccot?). 5. tan = Pa 6.icot— rab 7. sec = ze 8. cosec == Me . cos sin : cos sin 20. From these and other properties and theorems, some of which will be demonstrated as we proceed, ma- thematicians have computed the lengths of the sines, tangents, secants, and versed sines, to an assumed ra- dius, that correspond to arcs from 1 second of a degree, through all the gradations of magnitude, up to a quar drant, or 90°. ‘The results of the computations are are ranged in tables called Trigonometrical Tables for use, The arrangement is generally appropriated to two dis- tinct kinds of these artificial numbers, classed in their regular order upon pages that face each other. On the left hand pages are placed the sines, tangents, secants, &c, adapted at least to every degree, and minute, in the quadrant, computed to the radius 1, and expressed de- cimally. On the right hand pages are placed in suc- cession the corresponding logarithms of the numbers that denote the several sines, tangents, &c. on the res- pective opposite pages. Only, that the necessity of using negative indices in the logarithms may be pre- cluded, they are supposed to be the logarithms of sines, tangents, secants, &c. computed to the radius 10000000000. The numbers thus computed and placed on the successive right hand pages are called logarithmic gines, tangents, &c. The numbers of which these are the logarithms, and which are arranged on the left hand General Properties of Lines and Angles. % pages, are called natural sines, tangents, &c. In the small pocket tables these are usually omitted, and the logarithmic alone retained, as the most useful and expe- ditious in operation. Of the various trigonometrical tables which have been published at different times, those which deserve the warmest recommendation, as most accurate and best fitted for general use are ** The Mathematical Tables” of Dr. Hutton, in one vol. royal 8vo.; and the stereo- typed * Tadles Portatives de Logarithmes, par Francois Callet,”’ printed also in royal 8vo. in 1795. Dr. Hut- ton’s work contains a copious and valuable introduction, - comprizing the history, nature, construction and use of logarithmic and trigonometrical tables. The introduc- tion to Callet’s tables, likewise exhibits directions for their use, and some of the best formulae employed in their construction.* Of small tables for the pocket the best with which I am acquainted are those of Mr. Whiting, and the stereotyped tables of the Rev. F. A. Barker. CHAPTER IT. General Properties, and Mutual Relations of the Lines and Angles of Circles and Plane Triangles, Prop. I. ‘ : s : 1. UHE chord of any arc is 2 mean proportional be- tween the versed sine of that arc and the diameter of the circle. * The nature and use of logarithms being fully explained in Dr. Hutton’s valuable work, and, indeed, in every collection of logarithmic tables which a student ought to possess, I think it entirely unnecessary to occupy any portion of this *itroducticr, by an elucidation ef the properties of those useful numbers. BO 10 Plane Trigonometry. In the marginal figure, ax is the chord, and An is the versed sine of the arc AB. BL being B joined the angle LBA in a semicircle is x a right angle, and is therefore equal to angle BDA, BD being perpendicular to 7 LA. Hence, the triangles pba, tea, L CD A are similar, and we have AD: AB :: AB: AL. Prop. II. 2. As radius : cosine of any arc :: twice the sine of that are : the sine of double the arc. In the preceding figure cn is the cosine of the are BN, AB is twice the sine BH of that arc, and Bp is the sine of AB the double arc. From the similar triangles ACH, ABD, we have AC:CH:: AB: BD. Prop. III. S. The secant of any arc is equal to the sum of its tangent, and the tangent of half its complement. In the annexed diagram, where Aap is the proposed arc, let the tangent TA be produced downwards till TA + AR = cT the secant. Then, since angle R is the complement of acr, and of arc, it fol- j + lows that acR = arc, that is = 4 comp. LA AcT. But ar is the tangent of acr, or Cc A of the arc Ac; whence the proposition is ed manifest. tO Prop. IV. » 4 The sum of the tangent and secant of any are, is equal to the tangent of an are exceeding that by half its complement. . ; Produce ar, the tangent of the:assumed are, till the __ prolongation Ts becomes equal to cr the secant, and join Cs, intersecting the quadrantal arc Ax inp. Then, because Ts = re, angle rcs = Tsc = TCE, by reason of - BN chek ia VF fi Cooly ve General Properties of Lines and Angles. 11 the parallels rs, cx. Hence np = DE = 3 comp. AB, and As = tan AB + sec AB = tan AD = tan (AB + 3 comp AB). Prop. V. 5. The chord of 60° is equal to the radius of the cir- cle; the versed sine, and cosine of 60° are each equal to half the radius; and the secant of 60° is equal to double the radius. Let az be an arc of 60°, AB its chord, cp 4y its cosine, AD its versed sine, cr its secant. Then, 1. Since acs = 60°, andcsB = cAa,A=B = 1(180° — 60°) = 60°. Thatis,thethree ¥% angles of the triangle anc are equal, and therefore the triangle is equilateral. Conseq. AB = radius. , Cc D A 2, cD = DA = 3cA, because in an isosceles or equi- lateral triangle the perp. bisects the base. 3. BAT = comp. 60° = Bra; therefore BT = BA; and TB + Bc = Tc = 286 = 2 radius. Cor. If with the same radius an are were described from centre B, then cp would become the sine of 30°, which is consequently half the radius. Prop. VI. 6. The tangent of an are of 45° ts equal to the radius. Suppose aB in the last figure but one were an arc of 45°; then would acs be half a right angle, and conse- quently its complement atc. The sides ar, and ac, opposite to those angles, would then be equal; that is, tan 45° = radius. Cor. From this and prop. 5, it is evident that the sine of 30°, tangent of 45°, and secant of 60°, are in the ratio of the numbers 1, 2, and 4; or that tan 457 is a mean proportional between sin 30° and'sec 60°. 12 Plane Trigonometry. Prop. VII, 7. The square of the sine of half any arc or angle is equal to a rectangle under half the radius and the versed sine of the whole; and the square of its cosine, equal to a rectangle under half the radius and the versed sine of the supplement of the whole arc or angle. This will be at once obvious from the definitions, and the diagram to prop. 1, of this chapter. For 4an? = AB* = AL,AD = 2AcC.AD. Whence AH? = fac » AD, Also, 4cH? = LB? = AL. LD = 2ac, Lp. Whence @H? = ZAC. LD. Prop. VIII. 8. The rectangle under the radius and the sine of the sum or of the difference of two arcs, is equal to the sum or the difference of therectangles under their alternate sines and cosines. ey it Let as and sp be two unequal arcs, of the circle whose radius is ac; and let Bp’ = BD. YS ee: Then is ap the sum, and Ap’ the differ- ence, of the arcs AB and Bp. Also, BH == sine, CH = cosine, of arc AB; DF = sine, CF = cosine, of arc BD; DK y = sine, CK = cosine, of AD=AB+ ED; © - DK’ = sine, CK’ = cosine, of AD’ = AB — BD. The triangles cBH, crc, being similar, we have, cz >BH :: CF: FG = EK, whence CB, EK.= BH. CF, Again, the triangles cBH, DFE, are equiangular ; for each has a right angle, u and Fj and angle czhH = DFE, the former being equal to crc, which is the complement to ere, and the latter being also a complement to the game angle, : Hence, cB:CH:: DF:Dk, and conseq.CB.DE= CH. D¥.- But, since as above, cB. EK = BH. CF, Ao | and cB, DE = CH. DF, General Properises of Lines and Angles, 1% we have, by addition, cB, DK = BH.CF + CH.DF, which is the first part of the proposition. | _ Again, since the triangles DFE, D‘FE’, are equal as well as equiangular, ED = E’p: the preceding 1 rectangles may, therefore, be expressed thus, CB.EK’ =BH.CF, and CB.D’E = CH.DF, of which the difference is cn... D‘K’ = BH.CF — CH. DF, which is the second ‘part of the proposition. Cor. If the two arcs become equal, then we have for the sum, rad x sin 2AB = sin AB X 2 Cos AB, agreeing with prop: - a this chapter. inde oH : Prop. IX. Oy ‘The: rectangle under the radius and the cosine of the sum or the difference of two arcs, is equal to the difference or the sum of the rectangles under their res- pective cosines and sines, Recurring to the same diagram, we have from the similar triangles cBH, CFG, CBS GH :: CE 3CG; whence CB.CG = CH.CF: and the similar triangles cpu, and DEF, give CB: BH:: DF: EF or KG; whence CB. KG = BH. DF. The difference of these rectangles is, cB.CK =CH.CF — BH.DF; which is the first part of the proposition. The equivalent rectangles to the preceding, are CB.CG = CH.CF, . , and cB. K’G =BH.DF; + the sum of which gives cB. cK’ = CH’ CF + BH.DF, which is the second part of the proposition, Cor. When as and sc are equal, we have from this proposition rad x cos 2AB = cos?7AB — sin 7AB. Remark, The preceding figure is adapted to the case .where not only az, and sn, but their sum aD is less thana quadrant. But the properties enunciated in these two propositions are equally true, let the magnitudes of 14 Plane Trigonometry. the arcs, of their sum, and their difference, be what they may. ‘This might be rendered evident by a suitable modification of the diagram ; and will still farther ap- pear in the fourth chapter of this work. Prov. X. 10. As the difference or sum of the square of the radius and the rectangle under the tangents of two arcs, is to the square of the radius; so is the sum or differ- ence of their tangents, to the tangent of the sum or dif- ference of the arcs. Let AB, AD, be the two arcs; AT, AR, their tangents; also let Bs in the first figure P 1 be the tangent of the sum of those arcs, and Bs in the second figure the tangent of their difference: and from 7 R the point R let Ru, in both 3B figures, be drawn parallel pr to BS, or perpendicular to the radius es. Then, because of the similar triangles TAc, diet we shall have, TC:CAi:TR:RH, whence TC. RH = CA.TR, TC3TA?: TR: TH, whence TC. TH = TA.TR. Each of the last equal rectangles being taken from the square of tc, there will remain, Tc? — Te. TH, or Cc TC.(TC— TH), or TC.CH = TC? —TA.TR. Now, TC .CH (or TC? — TA,TR):TC.RH (Or CA. TR)::: CH : RH ?: CB (Or CA): BS:: CA*: CA. BS; whence, alter- Mango, £C2 —-. TA. TR? CA* St GA. TRSCAs BEA: TREBRL Of this proportion, the first term rc? — TA. TR, be- cause Tc? = CA? + AT?, may also be expressed by ca? + AT? —TA.TR, or by CA? + AT? — TA (TATE AR), or lastly, by cA? == AT. AC}; whence the proposition is. manifest. sumed circle, describe an arc dée, General Properties of Lines and Angles. 15 Prop. XI. 11. As the sum of the sines of two unequal arcs, 18 to their difference, so is the tangent of half the sum of those two arcs, to the tangent of half their difference. Let Ax and AB be two unequal arcs, of which EK and BG, are the sines; and let ex be produced to cut the circle in p, and sr be drawn parallel to the diameter A&A’. Draw 1p, 1£, from the cen- tre 1 with the radius ca of the as- and through 4 draw 1dM parallel to pE.. Then, it is evident that nr = Kr + RBG, is the sum of the sines of AE and As, and ND = KE — BG = KD — KN, is their difference. Also, since Biz (at the circumference) = $8CE (at the centre); and BID = 4BeD (Euc. iii, 20); 6m is equal to the tangent of $(Az+ AB), and dn = to the tangent of (azn — as), that is, of (ap — AB.) But, by reason of the parallels pr, Lm, we have EN: DN:: 6M: 01; which is evidently the theorem enun- ciated above. Cor. The sum of the cosines of two arcs, is to their difference, as the cotangent of half the sum of those two arcs, is to the cotangent of half their difference. For, the cosines being the sines of the complements, it follows from the proposition that the sum of the co- sines, is to their difference, as the tangent of half the sum of the complements, is to the tangent of half their difference. But half the sum of the complements of to arcs is the complement of half the sum of those two arcs, and half the difference of the complements is the same as the complement of half the difference; whence . the truth of the corollary. Prov. XII. 12, Of any three equidifferent arcs, it will be, as 1 16 Plane Trigonometry. radius, to the cosine of their common difference, so is the sine of the mean arc, to half the sum of the sines of the extremes; and, as radius, to the sine of the com- mon difference, so is the cosine of the mean arc to half. the difference of the sines of the two extremes. Let av’, AB, AD, (in the figure to prop. 8, of this chapter), be the three equidifferent arcs. Then pF = DF, is the sine of their common difference, and cr its cosine. Also Fe, being an arithmetical mean be-— tween the sines DK, D’k’, of the two extreme arcs, is - equal to half their sum, and DE equal to half their dif- ference. By reason of the similar triangles cBH, Cre, DFE, we have, CB: CF::BH:FG, and CB? DF ::CH:DE; which are the analogies in. the proposition. . Cor. 1. From the “Scien proportions, we have DF. CE D K)) = ndp be ‘ FG=5(DK+D'K’) = m Bae dpE= }(DK pK’ cd pp. Bete BH 7, SDE. CH Therefore DK + D’k = —.— and pk — D’K’= a Cor. 2. Hence, if the mean arc AB be one of 60°, its cosine CH, will (prop. 5) be equal to cs, and pK — b’K’ == DF: consequently Dk will in that case equal DF + 0D’x’. From this conjointly with the preceding corol- lary result these two theorems: ) (A). Ifthe sine of the mean of three equidifferent arcs (radius being unity) be multiplied into twice the cosine of the common difference, and the sine of either extreme be deducted from the product, the remainder will be the sine of the other extreme. (s): The sine of any arc above 60°, is equal to the sine of another arc as much below 60°, together with the sine of its excess above 60°. _ Remark. From this latter corollary, the sines below 60° being known, those of arcs above 60° are determin- able by addition only, : General Properties of Lines and Angles. 17 Thus, sin 60° 1’ = sin 59° 59’ + sin 1’, sin 60° 2’ = sin 59° 58’ + sin 2’, sin 60° 3’ = sin 59° 57’ + sin 34, sin 64° O’ = sin 56° 0’ + sin 4°, &e, &C. &C. Prop. XIII. 13. In any right angled triangle, the hypothenuse is to one of the legs, as the radius to the sine of the angle opposite to that leg; and one of the legs is tu the other, as the radius to the tangent of the angle opposite to the latter. Let anc be a triangle, right-angled at C B, and let AR on the leg as, be the ra- dius of the tables. With centre a and uy radius AR, describe an arc tocut the hy- yp pothenuse in p, and draw pH, TR, per- pendicular to as. Then pu is the sine, rr the tangent, and aT the secant, ofthe A HR B arc DR, or angle a: and the similar tri- angles ABC, ART, AHD, give . AC:CB:: AD: DH::rad:sin A; and AB: BC:: AR: RT::rad: tana. Cor. To the hypothenuse as a radius, each leg is the sine of its opposite angle; and to one of the legs as a radius, the other leg is the tangent of its opposite angle, and the hypothenuse is the secant of the same angle. Prop. XIV. | 14. In any plane triangle, as one of the sides, is to another, so is the sine of the angle opposite to the former, to the sine of the angle opposite to the latter. Let asc be a plane triangle,. C in which it is required to deter- mine the relation which subsists S between the sides ac and se. 4 With the angular point AasaGf AdH 6 RB 18 Plane Trigonometry. centre, and the distance ac, equal to the radius of the tables, describe the semicircle Gecu. From c draw ch parallel to cs, and cd perpendicular to az: draw also eA parallel to cs, and from the point of intersection e (of that line with the circle) demit the perpendicular ef on BA produced. Then cd is the sine of the angle caz, to the radius ac, and ef is the sine of the angle eac = angle cBa, to the same radius. But ac: Bc :: ac: dc, because of the parallels zc and dc, :: Ae: dc, because Ae = Ac, P 3: ef: cd, because of the similar triangles ef, bcd. That is, AC: BC::sin B: sin A, In a similar manner it may be shewn that AC: AB:;:8in B:sinc, and AB: BC::sin C:sin A. And by drawing a figure for each case, it will be seen that the circumstance of any one of the angles being obtuse will make no difference in the demonstration. Otherwise. From c the vertical angle of the C triangle let fall the perpendicular cp upon AB, or AB produced, according as the angles a and 8 are both acute, | x or one obtuse. - Cia. D Then (prop. 13) ac: ep::rad: sin A; also cD : CB:: sin B: rad; .. EX quo pertur. AC + CB isin Bi sin A, Prop. XY. 15. In any plane triangle it will be, as the sum of the sides about the vertical angle, is to their difference, so is the tangent of half the sum of the angles at the base, to the tangent of half their difference. By the preceding prop. Ac: BC::sinB: sin A, .*, by comp. and div. ac + Bc: Ac BCiSsiNB + sin A °SIN B= §ln A, Pe EES General Properties of Lines and Angles. 19 | “But, (prop. 11) sins + sina::sinB — sin A:: tang P (B+ A):tant(B—A), AC + BC: AC — BC::tand (B + A): tan 4 (B — A). 3 Pror. XVI. 16. In any plane triangle it will be, as the base, to the sum of the two other sides, so is the difference of those sides, to the difference of the segments of the base made by a perpendicular let fall from the vertical angle. From the centre c (namely, the vertical angle of the triangle azc) with the distance of the greater side Ac describe the circumference of a circle, meeting AB, cB, produced, in the points z, r, Gc. Then, it is obvi- ous that cB is equal to the sum of the sides, AC, CB, and FB equal to their difference. And because cD is per- pendicular to AE, AD is equal to nx, (Euc. iii.3). Wherefore, of the two AB, BE, one is the sum of the seg- ments of the base pa, pxB, and the other their difference. But, (uc. lil. 35) the rectangle under ex, and BF, is equal to the rectangle under aBand Br. Conse- quently (Euc. vi. 16) AB: GB:: BF: BE; which agrees with the proposition. Prop. XVII. 17. In any plane triangle it will be, as twice the rect- angle under any two sides, is to the difference of the sum of the squares of those two sides and the square of the base, so is the radius to the cosine of the angle con- ‘tained by the two sides. *Let agc be a plane triangle. From a, © one extremity of the base aB, draw AD per- D endicular to the opposite side sc: then { Euc. ii, 12, 13) the difference of the sum of the squares of ac, cB, and the square of 4 B 20 Plane Trigonometry. the base AB, is equal to twice the rect- jy angle nc.cp, But twice the rectangle c BC .CA, is to twice the rectangle Bc. cp, Sank that is, to the difference of the sum of the squares of AC, BC, and the square of az, * as. cA to cD; that is, (prop. 13) as radius to the sine of CAD, or, as radius to the cosine of Acs. b Cor. When unity is assumed as radius, we have! Ac? + BC? — AB? cos ¢ = ———____—___, ‘ . 2CB.CA gl Prop. XVIII. 18. As the sum of the tangents of any two unequal. angles, is to their difference, so is the sine of the sum of those angles, to the sine of their difference. Let Bca, acp, be the two proposed Cc angles, BA and Aap their tangents to the radius cA. Take 14 = aD, join cI, and draw wk, de, IK, 2k, perpendi- cular to Bc. Then, it is evident, since 1A = AD, andca perpendicular to gp, that ck = cD, ICA = pcA, and there- i fore BCI = BCA — ACD: also that ed is the sine of the sum of the given angles, and zé the sine of their differs | @ ce,to the assumed radius ac. % Now, by reason of the similar triangles ppr, BIK, it willbe, ’ : BD (= BA + AD): BI(= BA — AD) :: DE: IK. } Also, because the triangles cpr, ode, are similar, a8 well as the triangles c1x, cik; we have a CD:cd:: DE: de, and ci (= cp).:.e7 (= cd).:: 1K tigen. therefore DE: 1K :: de: th. 3 And consequently, Bp: B1I:: de: ik; sf that is, tan BCA + tanDcA:tanBcA —tanDCcA ::sin (BCA + DCA) :sin(BCA — DCA). Cor. Hence it follows, that the base of a plane tri-” angle cD, is to the difference of its two segments BA, General Properties of Lines and Angles. 21 AD, as the sine of the whole angle at the vertex, to the sine of the difference of the angles at the base. Prov. XIX. 19. As the sine of the difference of any two unequal angles, is to the difference of their sines, so is the sum of those sines, to the sine of the sum of the angles. Let a and B be the two unequal an- C : gles; both being bounded on one side by - the line AB, and on the other by lines, which, produced, meet at c, and form a rectilinear triangle acz. Demit cp per- pendicularly from ¢ upon AB; make pa = pA,andjoinca. Then weshall have ap = pB—DAy, Ca = CA, CaD = CAD, CaB = supp. CAB; and conseq. (ch. i, 19 c) sin cas = sin cas. Also (by the same) sinc = sin (A +B): andacB = Adc —Ba=A —B. Hence, in A Ack, sinc, or sin (A + B):sinA:: AB: CB. Also, in A acs, sinc, orsin (A — B):SiN Ai: GB: CB. Therefore, aB: AB::sin (A — B):sin(A +B). Again (prop. 14) cB: cA:: sin A: sin B. comp. and div.cs —cA:cCB+4CA::sinA— sin B:sinA + sin B. But, by the inversion of prop. 16, aB:CB = CA::CB + CA: AB. _ Therefore, comparing the corresponding antecedents and consequents, in the $d and Sth analogies, there re- sults, ‘sin (A —8):sin A — sinB::sin A + sinB:sin(A +8). Cor. If A and B be to each other as x + 1 ton, then ‘the last proportion will become ‘sin a:sin (xn + 1) A —sinva::sin(n +1) A+ sin7a ae :sin (224+ 1).A. | A Dia 6B ROE ScHOLIUM. | _ 90. Of the propositions in this chapter, some-are aoe il inthe solutions of plane triangles, and will find ir applications in the next chapter; others are useful ve is i bins Ma i; 22 Plane Trigonometry: in the construction of tables of natural sines, tangents, &c. To glance, for a moment, at this application, in- passing, suppose it were required to find the natural sines to each of the 10 first minutes of the quadrant. The radius of a circle being 1, the semi-circumference is known to be 3° 14:159265358979. This being divided successively by 180 and 60, or at once by 10800, gives -0002908882086657, for the arc of 1 minute. Of so” small an are the sine, chord, and are, differ almost im-_ perceptibly from the ratio of equality; so that the first ten of the preceding figures, that is, ‘0002908882 may ~ be regarded as the sine of 1’; and in fact the sine given in the tables which run to seven places of figures is” ‘0002909. By chap. i. art. 19, we have, for any arc,” cos = ,/(1 — sin*). This theorem gives, in the pre-" sent case, cos 1’ = ‘9999999577. Then, by prop. 12, cor. 2, (A), of this chapter, we shall have 2cosl’ x sin l’ — sin 0’ = sin 2’ = :0005817764 2cos 1’ x sin 2’ — sin ]’ = sin 3’ = :0008726646 2cosl’ x sin3’ —.sin 2’ = sin 4’ = -0011635526 2cos 1’ x sin if — sin 3’ =sin 5 = ‘0014544407 2cos 1’ x sin 5’ — sin 4’ = sin 6’ = -0017453284 &C. &c. &c. Thus may the work be continued to any extent, the whole diificulty consisting in the multiplication of each successive result by the quantity 2 cos 1’=1+99999991 54. Or, the sines of 1’ and 2’ being determined, the work might be continued by the last proposition, thus: sin 1’: sin 2’ — sin 1’:: sin 2’ + sin 1’: sin 3’ sin 2’: sin 3’ — sin 1’:: sin 3’ + sin 1’: sin 4” sin 3’: sin 4’ — sin 1’:: sin 4’ + sin 1’: sin 5’ sin 4’: sin 5’ — sin ]’:: sin 5’ + sin 1’: sin6’ &c. &c, &c. In like manner, the computer might proceed for the sines Seon ca ah &c. thus: sin 1°: sin 2° — sin 1°:: sin 2° + sin 1°: sin 3°? sin 2°: sin . — sin 1°%: sin 3° + sin 1°: sin 4° sin 3° :sin4° — sin 1° :: sin 4° + sin 1°: sin 5° &C. &e, &e. General Properties of Lines and Angles. 23 To check and verify operations like these, the pro- portions should be changed at certain stages. Thus, sin 1°: sin 3° — sin 2°::sin 8° + sin2°: sin 5° sin 1°: sin 4° — sin 3°:: sin 4° + sin 3°:sin 7° sin 4°: sin 7° — sin 3°:: sin 7° + sin 3°: sin 10°. The coincidence of the results thus obtained, with the analogous results in the preceding operations, will ma- nifestly establish the correctness of both. The sines and cosines of the degrees and minutes up to 30°, being determined by these or other processes (some of which will be indicated in the 4th chapter), they may be continued thus: sin 30° 1’ = cos 1’ — sin 29° 59’ sin 30° 2’ = cos 2’ — sin 29° 58’ sin 30° 3’ = cos 3’ — sin 29° 57’. | And these being continued to 60°, the cosines also become known to 60°; because cos 30° 1’ = sin 59° 59’ cos 30°’ = sin 59° 58%. The sines and cosines from 60° to 90°, are deduced from those between 0° and 30°. For sin 60° 1° ='cos'29° 59’: - sin 60° 2’ = cos 29° 58’ SG." Sr.) % Ke. 21. The sines and cosines being found, the versed sines are determined by subtracting the cosines from radius in arcs less than 90°, and by adding the cosines to radius in arcs greater than 90°. — 22, The tangents may be found from the sines and , . sin ° cosines. For since tan = —, (chap. i. 19), cos sin I’ ; we have tan |’ = aT = cot 89° 59 sin 2! i SF COS &C. &c. &c. Above 45° the process may be considerably simplified 24 Plane Trigonometry. by the theorem for the tangents of the sums and differ- ences of arcs. For, when the radius is unity, the tan-— gent of 45° is also unity, and tau (A + 8B) will be de- noted thus: + tan B tan (45° + B) = — tan 3 And this, again, may be me further simplified in practice. 23. The secants may readily be found from the tan- gents by addition, For (prop. 3) sec A = tan a + tan $comp A. Or, for the odd minutes of the quadrant the » I secants may be found from the expression sec = samt / 8 Other methods for all the trigonometrical lines are q deduced from the expressions for the sines, tangents, &c. of multiple arcs; but this is not the place to explain them, even if it were requisite to introduce them at. large into a cursory outline, 5 CHAPTER III. Solutions of the several Cases of Plane Triangles. 1. THERE being in every plane triangle, six things, namely, three sides and three angles, ‘of which some three are given to determine the other three; and the combinations that can be formed ae of six quantities, . O25 taken three and three being = rar first sight be imagined that 20 distinct rules would be required in this branch of trigonometry. But though the varieties of data are in truth thus numerous, it will — * or 20; it might at | ‘f * * 9 P Solution ofits Three Cases. 25 soon appear, that the number of cases which require distinct rules, are very few. Let a, B, c, denote the angles of a plane triangle, and a, 0, ¢, the sides respec- _ tively opposite; then the twenty varieties of data will be these, viz. | (3) abc... (1) aba... (1) abs... (2) adc L).aca 2)... (2iaen., w« (A)vdce 24 bcA ..- \ OCB .., tH bcc (4) apc... (1) ABa...(1) abd... (*) anc 1 MOO «din, (oA CD ac 1} ACC (*) Bea... ti} BCA... tH BCC. Here the varieties which are marked by the figure J, have this in common, that a side and its opposite angle are two of the given parts; and, if it be considered that when two angles of a plane triangle are known, the third is, in fact, given, because it is the supplement of their sum, it will appear that those varieties which are marked with an asterisk fall under this first case: so that the varieties comprehended,in this case are fifteen. Of the remaining five, three, indicated by the number 2, have this in common, that two sedes and an included angle are given; they, therefore, constitute a second case. ‘The example marked 3, has three sides given, but no angle ; this makes a third ease: and the remaining variety in which the three angles are given, but no side, would make a fourth case, were it not (see chap. i. 6) that for want of aside among the data, the problem thus ex- pressed is unlimited. Our twenty varieties, thereiore, only furnish three distinct cases, to the solution of which we shall now proceed. Case I, 2. When a side and its opposite angle are among the given parts. Here the solution may be obtained by means of chap. “ii. prop. 14, where it is demonstrated that the sides of plane triangles are respectively proportional to the sines _ of their opposite angles. oe | c 26 Plane Lrigonomeiry. In practice if a side be required, begin the proportion with a sine, and say, As the sine of the given angle, To its opposite side ; So is the sine of either of the other angles, To its opposite side. If an angle be required, begin the proportion with a side, and say, ' As one of the given sides, Is to the sine of its opposite angle ; So is the other given side, To the sine of its opposite angle. The third angle becomes known, by taking the sum of the two former from 180°. Note 1. Since sines are lines, there can be no impro- priety in comparing them with the sides of triangles; and the rule is better remembered by young mathema- ticians, than when the sines and sides are compared each to each. , Note 2. It is usually, though not always, best to work the proportions in trigonometry by means of the loga- rithms, taking the logarithm of the first term from the gum of the logarithms of the second and third, to obtain the logarithm of the fourth term. Or, adding the arith- metical complement of the logarithm of the first term to the logarithms of the other two, to ebtain that of the fourth. Note 3. It is an excellent plan to accustom the pupil | ’ to draw (previously to his commencing the computa-~ tions in plane trigonometry) not a rough, but a neat and accurate, sketch of the triangle proposed, from the given data, by means ofscale, protractor, and compasses. Such construction will enable him at once to trace the peculiarities of the problem, and to detect its ambi- guities, ifthere be any. It will also, if accurately per- _ | formed, upon a scale of moderate size, give the sides to within their 200th part, and the angles true to within half a degree. A ey BY, ee pe Solution of its Three Cases. 27 Note 4. The triangle being carefully constructed, and marked by suitable letters of reference, as A and B fur example, at the extremities of the base, o at the vertex, then, according to the nature of the problem, write down the requisite proportion in four distinct lines, with the letters of reference to each term, and the given numbers to the three first: against these numbers place their respective logarithms; find the legarithm to the fourth term by the directions in note 2, and ascer- tain its angular or lineal value, by means of the tables. *,* The last three notes are not restricted to the present case, but extend in their application to the usual practice of plane trigonometry. \ Example I. 2. Ina plane triangle are given two angles equal to 58° 7’, and 22° 37’, respectively, and the side between them 408 yards. Required the remaining angle and sides. Construction. On an indefinite right ¢ line, set off, from a convenient diagonal scale, the distance an = 408. Fromthe ; point a draw a right line ac, to make with AB an angle ef 58° 7’; and from 4 § the point B another line, turned towards the former, and to make with Ba an angle of 22°37’, The intersection c, of these two lines determines the triangle; and the sides AC, BC, measured upon the scale of equal parts from which as was laid down, are found to be 159 and 351 respectively, Computation. Two of the angles being known, their sum 60° 44’ taken from 180°, the sum of the angles in a plane triangle, leaves 99° 16’ for the third angle This, being an obtuse angle, its sine is to be found in the table by taking that of its supplement 80° 44’, which (chap. i, 19) is the same. Hence, A Rr gibt 1 Diem ae (frat, A, es 28 Plane T. rigonometry. : Logs. Logs. As sin c 99° 16’ 9:9942950| As sin c 99°16’ 9:9942950 To AB. ..408... 2°6106602/To aB...408... 2°6106602 Soissina58° 7’ 9°9289718|Soissin B 22° 37’ 9°5849685 To Bc..351:02 2°5453370|To ac..158°98 2°2013337 ac and Bc, therefore, are 351:02 yards, and 15898 yards, respectively. 4. In the preceding operation, instead of adding to- gether the logs. of the second and third terms, and sub- tracting that of the first from their sum, the work has been performed thus:—The right hand figure of the upper line was taken from 10, and each of the other figures from 9, and their several remainders added to the numbers below them in the respective columns. This is easily effected in practice by making those re- mainders emphatical in adding downwards. ‘Thus, in the operation for sc, we begin at the right hand, and adding downwards say, fen and 2 are 12, and 8 are 20; set down 0: carry 2, and four are 6, and 1 make 7: nothing added to 6 and 7, gives 13; set down 3: carry 1, and seven are 8, and 6 are 14, and 9 are 23; set down 3: carry 2, and jive are 7, and 8 are 15: and so on, to the left hand column, which added in the same way amounts to 12; of which the 2 are put down, and the 10 rejected, to compensate for what has been borrowed in the process by the arithmetical complement. This me- thod is very easy in practice, and is found less liable to produce error than that in which the arith. comp. is put down at once from the tables. Example I. 5. Ina plane triangle anc are given Ac = 216; cB == 117; the angle a = 22° 37’; to find the rest. Construction. Draw an indefinite right line ann’, tS S Solution of its Three Cases. from an assumed point a in which c draw a line to make the angle a = 99° 37’, Make ac = 216, and PN from c as a centre with a radius Be B’ = 117 describe an arc BB’, it'will cut the line AB in two points, from each of which draw- ing lines Bc, Bc, there will be formed two triangles azc, AB‘c, each of which answer the conditions of the ques- tion. ‘The required lines and angles being measured, give AB==117,...AcB = 221°..., ane = 19430" AB = 282 ....ACB’= 112° .....ABC= 45}° Computation. To find the angle s. AS Bes AT Oe Ys SUGEISS9 Tosin A. . 22° 37’.... 9°584.9685 So is Ac ..216...... 23344538 TosinB,..... 45° 14’ or 134° 46’,, 9°8512364 Add to each Z A 22° 37’... 22°377 ——————» Take the sums 67° 51’ .. 157° 23’ Promos oes 7 280" 005, 180% 0 Remain .... 112° 9’ or 22°37’, being acp’ and Acs, Here since acB = BAC, we have AB = Bc = 117, But to find as, reverse the first two terms of the former proportion, and say, As sin A..+..- 22° 37’... 9°5849685 TO Be) we see RR A RS 2:0681859 So is sin AcB’,.112°- 9’ .... 9:9667048 TOBE ass eka Por Lip ares 24499222 6. Remark. The ambiguity in this and similar ex- amples, does not, as has been often affirmed, depend upon the circumstance that an angle and its supplement have the same sine, but solely upon this, that in con- ! 80 Plane Trigonometry. structing triangles from analogous data, when the side cB has its length between certain limits, that is, be- tween the length of the side ac, and that of the per- pendicular.cd from c on the third side azn, it must ne- eessarily eat the indefinite right line ane’ in-two points. Beyond those limits there is no ambiguity; for when cs is proposed to be less than cd, the problem is impos- sible; and when cB exceeds ca, the angle a being ail along suppesed given, the line azz’ can only be cut in one point. Hence the practical maxim may be thus ex- pressed:—when cp is proposed Jess than ac sin a, the data are erroneous ; when it is given greater-than ac | there can be only one triangle; between those limits the problem is ambiguous. a. 3. Given two angles of a plane triangle 22° 37’ and 134° 46’, and the side between them 351. To find the remaining angle and sides, Ans. Angle 22° 37’; sides 351 and 648. ix, 4, Given two sides of a plane triangle 50 and 40 respectively, and the angle opposite to the latter equal to 32°; to determine the triangle. Ans. If the angle opposite to the side 50 be acute, then is it = 41° 28’, the third angle 106° 32’, and the remaining side 72°36. If the angle opposite to side 50 be obtuse, it is = 158° 32’, and the other angle and side 9° 28’ and 12°415 respectively. Case Il. | 7. When two sides and the included angle are given. The solution is effected by means of chap. il: prop. 15. Take the given angle from 180°, the remainder will be the sum of the other two angles. Then say,-—As the sum of the given sides, Is to their difference; So is the tangent of half the sum of ‘the remaining angles, To the tangent of half their difference. ee ee Solution of tts Three Cases. ro Half the difference added to half the sum of those angles, gives the greater of them; and taken from half the sum, leaves the less *. All the angles becoming known by this process, the third side is found by the rule in case 1. Example I. 8. Let there be: given in a plane triangle anc, Ac = 450, Bc = 540, and the included angle c = 80°; to find the remaining angles and side. Leaving the construction to be effected by the pupil, { shall proceed to the computation. Bc + ac = 990, Bc — Ac = 90, 180° — c = 100°. As BC + AC .o..0% 990... 2°9956352 * : To Bc —AC...... 90... 19542425 : So istan$(a +B) 50°..10 0761865 To tan (A — B)..6° 11’ .9:0347938 ee eee Hence 50° + 6° 11/2 56° 11’ = a; 50°—6°1 = 45°49 =. Then, to find the third side az, say, As sin B.... 43° 49’. . 9°8403276 To Ac ......450....2°6532125 Soissine ..80° 0’ ..9°9933515 To aB...... 640°08 . . 28062364 = Example I. 9. The two sides of a triangle are 40 and 32, and the included angle 90°, required the other angles and side. In this example the operation may be considerably shortened by working without, instead of with, the loga- * Leta+b=s, and a—b =d: then, by addition 2a=s + d, or ¢@ = 4s + 3d; and, by subtraction 2b = s — d, or b = $s —- $d. 32 «Plane Trigonometry. rithms. For, since the given angle is 90°, the half'sum — of the remaining angles is 45°, whose natural tangent is - unity; and the sum and the difference of the given sides are 72 and 8 respectively. Hence 8 ! 1 As 72:8::1 Sas iy os "1111111 = nat. tan. 6° 20’. Therefore 45° + 6°20’ = 51°20’, and-45° — 6° 20’ == 38° 40’, are the remaining angles. And the third side (Euc. i. 47) = 40? + 322 = 7 2624 = 8 4/41 = 51-225. ; *.* Other methods, still shorter, of solving right an- gied triangles, will be given! before we terminate the present chapter. Ez. 3. Given two sides of a plane triangle 1686, and 960, and their included angle 128° 4’; to find the rest, Ans. Angles 33° 35’, 18° 21’, side 2400, Casz III, 10. When the three sides of a plane triangle are given, to find the angles. Ist Method. Assume the longest of the three sides as base, then say, conformably with chap. ii. prop. 16. As the base, To the sum of the two other sides; So is the difference of those sides, To the difference of the segments of the base. Half the base added to the said difference, gives the greater segment, and made less by it gives the less; and thus, by means of the perpendicular from the ver- tical angle, divides the original triangle into two, each of which falls under the first case. 2d Method. Find any one of the angles by means of prop. 17, of the preceding chapter ; and the remaining angles either by a repetition of the same rule, or by the relation of sides to the sines of their opposite angles. ‘ 4 q Solution of its Three Cases. BS Example. 11. The three sides of a triangle are 40, 34, and 23- Required the angles. By Rule I. As AB: AC + BC?: AC—BC: AD — BD; C that is, 40:59::9: ara ay hore as b alin Then ae e sd 266370 == ADs A D&B gid ee 199698 yn, 2 STA S fre oi BE ower 1:5314789| As Ca ema ae 95.... 1°39T9400 Tosinnv.... 90° .. 10°0000000|To sinp... 96° ..10°0000000 So is aD... 26°6375 1°4254935)So is pp.. 13°3625 1:°1258878 Tosin acp.. 51935’ 9°8940146\Tosin scp ..32°18’ 9°T279478 a Hence 90° — 51° 35’ = 38°25’ = a; 90° — 32° 19’ =O = 1. 42 = and 51° 35’ + 32° 18’ = 83° 53’ = Acs. By Rule II. AC? = 1156, Bc? = 625, AB? = 1600, cB.cA = 850. From log ac* + Bc? — aB* = log 181 = 2:2576786 Take log 2cB .. ca = log 1700 = 3°2304489 Rem. -+- 10(intheindex) =log cos c=83°53’=9:0272297 The other two angles may be found by case 1. fiz. 2, When the sides of a plane triangle are 408, 351, and 159, what are the angles? Ans. 99° 16’, 58° 7’, and 22° 37’. Ex. 3. When the sides are 4, 5, and 6, what are the angles? Ans, 41° 24’ 35”, &5° 46’ 16”, and 82° 49’ 9” cé 84 Plane Trigonometry. RIGHT ANGLED PLANE TRIANGLES, 12. 1. Right angled triangles may, as well as others, be solved by means of the rule to the respective case under which any specified example falls: and it will then be found, since a right angle is always one of the data, that the rule usually becomes simplified in its ap- plication; as appeared in the solution of the second ex- ig to case 2. 2. When two of the sides are given, the third may be found by means of the property demonstrated in Euc. i. 47. Thus, Hypoth. = ,/ (base? + perp.?.) Base = 7/ (hyp.* — perp.) = W (hyp. + perp.). (hyp. — perp.) Perp.= (hyp.” — base*) = »/(hyp, + base) . (hyp. — base)s ‘3. There is another methed for right angled tr iangles, “situa by the phrase making any side radius; which is this. “To find a side. Call any one of the sides radius, and write upon it the word radius; observe whether the other sides become sines, tangents, or secants, and write those words upon them accordingly. Call the word “written upon each side the name of each side: then say, ** As the name of the given side, ‘* Is to the given side ; ‘¢ So is the name of the required side, ‘< To the required side? ¥ “To find an angle. Call either of the g7ven sides radius, and write upon it the word radius; observe whe- | ther the other sides become sines, tangents, or secants, and write those words on them accordingly. Call the word written upon each side the name of that side, Then say, «* As the side cade radius, ‘“* Is to radius; ‘*« So is the other given side, ‘To the name of that side: which determines the opposite angle.’? ; ,: Right angled Plane Triangles. 35 13. When the numbers which measure the sides of the triangle, are either under 12, or resolvable into factors which are each less than 12, the solution may be obtained, conformably with this rule, easier without logarithms than with them, For, Let azc be a right angled triangle, in which AB the base is assumed to be radius; BC is the tangent of a, and ac its secant, to that radius; or, dividing each of these ~@ : by the base, we shall have the tangent and A o secant of A, respectively, to radius 1. Tracing in like manner, the consequences of assuming Bc, and ac, each for radius, we shall readily obtain these expressions. ef erp.” Peay ie y, PEP — tan angle at base. pd gp GOor Fins * base I base 2. = tan angle at vertex, Pye perp. * at aif OL hyp. | a oe : “Ss == sec angle at base. | base a ae 4, —** = sec angle at vertex. perp. erp. * 5. : P. = sin angle.at base, 1Yp. base ; 6. = sin angle at vertex. hyp. > Example I. 14, In aright angled triangle are given, the hypo- thenuse and the base, 25 and 24 respectively ; to find the rest. | Perp. = ./(hyp.?—base?) = VY (25424). (25 +24) =7, pepe ot WL: ea Ppa heg i Seg) Tegbar arkay 2916666 = tan 16° 15’ 37”, angle at the base; whence 90° — 16° 15’ 37” = 73° 44° 23”, angle at vertex. | e 36 Plane Trigonometry. bas 9A oe a “= 34285714 tan 73°44'23”, ang. at vert. *perp. Or, thus, by the secants. =>s, = —— = 1:0416666 = sec 16° 15’ 37”, = 3'5714285 = sec 73° 44’ 23”. © PRP. 1 ee a8 Ez. 2. In a right angled triangle, given a leg and its opposite angle 384, and 53° 8’ respectively; to find the other leg and the hypothenuse. Ans. Leg 280, hypothenuse 480. Ex. 8. Tn a right angled triangle are given the base 195, its adjacent angle 47° 55’; to find the rest. Ans. Perp, 216, hyp. 291, vert. ang. 42° 5’. CHAPTER IV. Plane Trigonometry considered Analytically. 1. IN the preceding chapters the investigation of tri- gonometrical properties has been conducted geometri- cally ; the various relations of the sines, cosines, tan- gents, &c. of arcs and angles, whether depending upon triangles or not, being deduced immediately from the figures to which the several enquiries were referred. This method carries conviction at every step; and by keeping the objects of enquiry constantly before the Analytical Plane Trigonometry. 37 eye of the student, serves admirably to guard him against the admission of error; and is, therefore, well fitted for adoption at the commencement of this branch of science. But of late years, another method, first m- troduced by Euler, has been generally employed by the continental mathematicians, and very frequently by those of England. It is analytical. The nature and mutual relation of the lineo-angular quantities, sines, tangents, &c, being defined by a few obvious and simple equa- tions, every other theorem and formula that is likely to be of use, is deduced with great facility by the mere re- duction and transformation of the original equations. This method serves greatly to shorten almost all trigo- nometrical investigations, except a few which lie at the foundation of the science: and admitting of an exten- sion to which the geometrical method cannot lay claim, at the same time that it proceeds to. some of its most important results with great rapidity ; no treatise on tri- gonometry can be complete that does not assign to this manner of handling the subject a considerable place. 2. Previously to the student’s entrance upon this de- partment of enquiry, it will be expedient for him to trace the mutations which the principal intermediary or lineo-angular quantities undergo, when they relate to arcs found in different quadrants of the same circle. To this end let him draw afresh and lay before him, the diagram given at chap. i. art. 18, and first trace the mutations of the sines and the cosines; keeping in mind this general principle, that every variable algebraic quan- tity changes its sign after it becomes 0. 3. Let the arc be supposed to commence at 4, and to increase in the direction aBEA’B. As the arc augments through the first quadrant Az, the sine augments till it becomes equal to cE, the radius; passing from 90° to 180° through the quadrant £4’, the sine diminishes, but is to be regarded as positive till the arc becomes AEA’ er 180°. In that state of the arc the value of the sine 38 Plane Trigonometry. is obviously 0. Passing on to the third quadrant, as when the arc is AEA’B’, the sine D’s’ is directed contra- rily to what it was in the first two quadrants, and is then to be regarded as negative. In this state it continues to increase “through the third quadrant, at the end x’ of which it is agam equal to radius. From thence the ne- gative value ‘of the sine diminishes, till at the end a of the fourth arc, it again passes through zero, and the - sine becomes positive in the fifth quilleait ; as it obvi- ously ought to do, since the filth quadrant is coincident with the first. In the Ist and 2d quadrants, then, the sines are +, In the 3d and 4th, they are —. If the arc were taken negatively, its sine would, for like reasons, be negative in the first half, positive in the second half, of the circle. . 4. The cosine of an arc is equal to radius at the point A, where the arc is evanescent. From thence it dimi- nishes while the arc increases through the first quadrant; at = the cosine is 0. Then it changes its sign and con- tinues negative through the arc £a’r’, that is, from 90° to 270°. At’ the cosine is again 0; and, of course, from 5’ to a, through the fourth quadrant, it is again positive. The rule is the same for a negative are. 5. With regard to the signs of the tangents, it is evi- dent, since tan = ; (chap. i, 19), that when the signs of the sine and sikcs are both alike, whether positive or negative, the sign of the tangent is positive; and when those’signs are unlike, the sign of the tangent is nega- tive. The tangent, therefore, is positive in the first quadrant, the sine and cosine being then both positive : it is negative in the second quadrant, the cosine being then negative though the sine remains positive; it be- comes posttive again in the third quadrant, the sine and cosine being then both negative: finally, it is negative é , 2 ca A l 1 wtan <= —, and cot = —: cot wy Analytical Investigation of Properties. 39 in the fourth quadrant, since the sine is then negative and the cosine positive. The tangent, therefore, changes its sign at the end of every quadrant, where it passes alternately through nothing and infinity; these being, indeed, the algebraic indications of the changes of sign. If the arc is regarded as negative, the rule for the tangents becomes inverted; the tangent being then ne- gative in the first and third, posiitve in the second and fourth arcs. 6. For the cotangents the rule of the signs is the ‘same as for the tangents: this is evident, because tan 7. So again, the rule for the secants is the same as for P 1 the cosines; because sec = a. 3s 8. The rule for the cosecants is, also, the same as for * H the sines; because cosec = a8 9. Proceeding in this way, a second, a third, &c. time over the circumference of the circle, like muta- tions would occur. The results of the whole may be thus tabulated :— sin cos tan cot Ist..5th.. 9th..13th 2 hat vals telah ~2d..6th..10th,..14th{t s + —-—-— — pod... 7th. 11th’. 15th (73 —- — + + 4th... 8th..12th.. 16th} B—- + - = 10. Let it be remembered that when passes through infinity, the secant does, Sec COseC ++ m hey + putea! the tangent and for the same reason, viz. because they in that case could only limit each other at an infinite distance. ‘The principal changes, then, in point of magnitude, may be easily. traced, and tabulated as below; where the sign oo de- notes enfinity, 1 oe apts - 40 Plane Trigonometry. CPt GOMES TRO. ene tas COG OL ate Oly he Res ae” Sermons CET RO a 88 Re ef eee SEO SARA POO eee a bg s =~ OOne be GOS ie Teta ere nats remaere ine k OM, Wie ts COP ie DO oO el ore eet 8 COSeC SO RR ke OR S$8xnnoOo 11. These particulars premised, the properties and © relations of the sines, tangents, &c. of combined and — multiple arcs, as well as rules for the solutions of tri- — angles, may be investigated analytically, Let axc, in the annexed figure, be c any plane triangle, c the vertical angle, cp a perpendicular let fall from it upon z a the base or base produced, and let a, 4, and c, denote the sides respectively oppo- site to the angles a, 8, and c. i . Then, since ac = ; AD is the cosine of a to that radius ; consequently, ‘when radius is unity we have — Ap =6cosa. In like manner sp =acosz. Theres fore, AD + BD = AB=c=acosp+écosa. Ifone | of the angles, as a,-were obtuse, the result would, not- withstanding, be the same; because, while on the one | hand cos a would be negative, ap, lying on the contrary side of a to what it does in the figure, must be deducted from BD to leave aB, and a negative quantity subtracted, (B.) 4 +i td 4 3 a is equivalent to a positive quantity added. By letting q fall perpendiculars from the angles a and 8, upon the opposite sides, or their continuations, precisely analo- gous results will be obtained. They may be ape to- gether, thus :— a=bcosc+ccosB &b=acosc + com aba c= acosB + dcosa 12. If Bc, or a, be regarded as radius, cp will be the sine of the angle B to that radius; therefore, to the ra- dius unity, cp willbe = a sins. So again, for a like | | 4 2 j : Analytical Investigation of Properties. 41 reason, cb = ésin a. Consequently, asin & = dsin a, : a sin A 3 ‘ or ~ = ——. In a similar manner, we may obtain b sin B a sina b _ sing : ’ _ =--——, and - = ——. Or, changing the denomina- c sinc c sin ¢ tors, the relations of all the six quantities may be thus _ expressed, sinA sinB _ sinc Goa eelng Unban BI IY PY, These equations are, manifestly, of similar import with chap. ii. prop. 14. 13. From the last equations we have, sina sins a = C Says = C¢ 3 sinc sin ¢ Substituting these values of a and 6 for them in the equation c = acosB + 4cosa, and multiplying the ‘ MEO. equation so transformed, by —, it will become sinc = sina cos B + sin B COs A, Now, since in every plane triangle, the sum of the three angles is equal to two right angles, a + B = sup- plement of c; and, since an angle and its supplement have the same sine, it follows that sin (a + B) = sinc; whence | sin (A + B) = sina cos B + Sin B Cos A. 14. If in this equation B be regarded as subtractive, then will sin B obviously be subtractive also; but cos B will not change its sign, because it will still continue to be estimated in the same direction on the same radius The equation will, therefore, become sin (A — B) = sin AcosB— sinB cos A, 15. Conceive B’ to be the complement of 8B, andi O to be the quarter of the circumference, or the measure of a right angle: then will B’ = 4O —8, sin B’ = cos B, and cos B = sins. But, by the preceding article, sin (a—B’) =sinacosB’ —sinB’ cosa. Substituting for at Sa! Sy od - - , ge Ne ky ee. 42, Plane Trigonometry. sin B’, cos B’, their values, there will result sin (A — BY) = sina sin B—cosacoss. Hence, because B’ = ¢ © — 8B, we have sin (4 — 3’) = sin (a+B—i0) =sin[(a+B)—-310] =—sin(}O—-(a+B) == -~ COS(A + B)]. This value of sin (a — B’) being sub- stituted for it in the equation above, it becomes cos (A + B) = cos a cosB — sinasin 3. : 16. If 3 in this latter equation be made subtractive, | sin B will beceme — sin 8, while cos 8 will not change (art. 3,4). ‘The equation will consequently be trans-_ formed imto this: viz. a bot ” of oo © 5 a 19>) et joe Oo ss) oo ie) (G) / sindSa = 5s — 2053 + 1685 > sin 6a = (6s — 3253 + 3255) /(1 — 8°) J Be, &e. &e. &e. yy cos A= CC ~ ~ “eos2a = 2c7— I - cos3a= 403— 3c * < cos4a = Sct 807+ 1 eos 5a = 16c5 — 2003 + Se > * cos Ga = 320° — 48c4 + 180? — 1 a . } wail Bee , J " os Plane Trigonomei ry. 20. Other useful expressions for the sines and cosine of multiple arcs, may be found thus: , Take the sum of the expanded expressions for sin’ (3 + A) and sin (B -— a); that is, 9 add,.......sin(B + A) =sinBCcosA + CosBSiINA ‘ fO.........8in(B— A) =sinBcosa —cosBsinA — there results, sin (B + a) + sin (B— A) = 2cos Asin B, So again, the sum of the expressions for cos (B + A). and cos (B — A), is ; cos (B+ A) + cos.(B — A) = 2 cos A cos Be Whence, cos (B + a) = 2cos AcosB — cos (B — A). | Substituting in these expressions for the sine and co-) sine of'B + A, the successive values of a, 2a, 34, &c. instead of B, we shall have, | 4 sin2A =2cosAsin A { sin 3A = 2cosasin2a —sin A | sn4A=2cosasin3A—sin2a (3) } sinnA = 2cos asin (n —1)a—sin(n—2)A cos2A = 2cosacos A—cos 0 (=1) COSSA = 2cosacos2A —cos A | cos 4A = 2-cos A cos 3A — Cos2a (K.) 4 cosnA = 2cosacos(m—1)A—cos(n—2)a Jy 21, 1f the cosine of a be represented by a particular binomial, the formula (x) will be transformed into a_ class of very elegant and curious theorems: thus ‘ 7 7 make 2 cos A = 2 + - y then 2 cos 2a = 2 (2cos?a = 1) “#2 [3 (: + ~\ = 174 4 ] a | rts ee | A like substitution in the forms for 2 cos 9A,<2 cos 4A, i &c. will give ss : Analytical Investigation of Properties. 45 ® 1 2cos 3a = 23 + =| H a 4 2cos 4A 2 cos 5A = 25 as ot, + 2 cosna = 2" 22. By an inverted process, the sine affd cosine of a single arc may be-inferred from those of a double arc; er, which comes to the same, those of a half arc from those of a whole arc. Letx-z= yv(l— cos 2A), or x7 — xz 4+ 22 = 1 — cos 2A, and assume x? + 27 = 1; then 2rz = cos 2a; cos? ZA 1. ; Ax? 3 Hence x4 — z? + } =4(1 — cos?2a) = 4 Sin? 2a, eee ese ct g sin 2a, and + = 3 2 /(1 = sin 24) and z? = 4 = 3 sin 2a, and z = a7 == sin ZA). Hence, sina = 4 whl ange anak and, exterminating z, 2? + cosA= 3 /(1 + sin2a) +3 /(1 —sin2a - Or, for the half arc, sing A= /(1+sina) — (COS RAS 5 A/(1 hsm A) . But, since a is supposed greater than J, this angle will be greater than half a right angle, or will be measured by an arc greater. than 45°. tan « — tan 45° 1 + tanatan45°? and because tan 45° = r = 1, we have tan (@— 45°) = (SI) ery oe Consequently, from equa. (3) a—6_ tang(a—r)_ tang(a—an Hence, because tan (# — 45°) = a—b a+b ) : whence A+.6 tand(A +B) cot g¢ tan} (A — B) = cot dctan(«—45°).. - (10. ye Hence results the following enesienl rule:— Subtract the less from the greater of the given logs, the remainder will be the log. t tangent of an angle: take 45° from that angle, and to the log. tangent of the re- mainder add the Jog. cotangent of half the given angle, the sum will be the log. tangent of half the difference _ of the other two angles of the plane triangle. $8. Let us next return to the equation QQ) i in art, Tl of this chapter; namely, a a= b6cosc +ccosB i J b= acosc + ccosa ¢ =acosB + dcosa , ‘The first of these equations being multi iplied ‘by @, the second by 4, the third by c, and each of the equa- tions thus obtained, being taken from ie sum of the ¢ ia other meee there will arise “ Loe, | fe* ef) Cees (#. Sy, Fades ” 4 Plane Trigonometry. Gr 3 ? g b? + c?— a? b? + c?—a? = Ziccos A, whence cos A = a Wisbay Zhe a? +c? —}? @? +. c?— 5*= 2accos B, cos B = ————_ 711} 2ac a? +- 6%? —¢2 a? + 62—c? = 2abecosc 1. Woh -—ypnnltelii us ? ea ub =O These equations are, obviously, equivalent to prop. 17, chap. 1. They may be expressed in a form rather better suited to computation, thus: (64+ c+ a)(b+c—a) eNRUENET PORRMLARDIart) skips 2 Ds Where, since the expressions are perfectly symmetrical, those for cos B, cos c, need not be put down. ' 39. Substituting in equa. (12), for cos a its equiva- lent 2 cos? 34 — 1, we have 1 rl (a oe col ga, J ee ee ee cos A = If, in equa. (11) we substitute 1 —2 sin? $a for cos A, we shall have a—h—c? — (a+b—c) (a+e—b) 2aINthA Slap open SR eg os Thevetore 4 (A(a+b+c)—c].[4lat+b —b din 2A =, een eS Tf, again, equa. (14) be divided by equa. (13), there will result, Lhd Bovic) (chap, i. 19 1), we have Cos é 1 id Q Be 30 e080) = ays 8 YO 1 i ne A sec 60 cos 608 ri These may obviously be extended to other arcs, by means of formule (P), (@), (R), of this chapter, and ‘prop. 3 of chap. ii. 42. Operations of this kind ought not to be carried far without being subjected to checks and proofs. For this purpose, after the sines and cosines are found, the tangents, secants, &c. are easily \ verified by their mitial relations. ‘The sines and cosines’ themselves, are exa- mined by means of some of the following “ formule of verification.”” Qo sin (18° + a) + sin (18° — a) = 28m 18° cosa re _ sin (54° + A) + sin (54° — a) = 2sin 54° cos A | ce sin (54° + A) + sin (64° — a) — (18° mt _ Sy 4 (18° — 4). = sin (90° — a) Shane 2 FIL \ x sin A -+ sih {36° — A) -+ sin (72° + a) == sin (0° 3) \ | + sin (72°— a). Se Here a may have any assigned value less than the So least specified Are. Thus, if a were 9°, we should have : ‘ sin 9° + sin 27° + sin 81° = sin 45° + sin 68°, . ois and if from any table the sines thus taken make equal . sums, it is highly probable they are all passa ‘Faking: hese from Dr. Hutton’s tables, we hav a sin 9° = 1564345 1393 sin 27° = 4539905 sin 45° == « *7071068 - sin 81° =. -9876883 ~ gin 68° = “8910065 - 1°5981133 .... equal sums . . 15981138 ek . 43. Of the formule investigated in ie Rea hose which have /etters of reference (a), (B), (c), &c, relate to the sums, differences, multiples, &c, of sines, tan- zee, 26 see Lge ‘. ‘ 2 at ‘ke nth ere uote bo Gateohg 58 Plane Trigonometry. gents, &c. while those which have jigures of reference, (1), (2), (3), &c. will be employed in the solutions of plane triangles. Other kinds will find their application in subsequent parts of this introduction ; and the student will do well, after he has gone through their investiga- tion, to arrange them in separate tables for use. We shall terminate the present chapter by subjoining three examples, Example I. Given the three sides of a plane triangle 40, 34, and 25, respectively, to find the largest angle, by formula (15). Here 4 2 (a+ 6+c)—~c=15°5..log = 1:1903317 i (a+ b—c) —b = 245 .. log = 13891661 Their prod. shown by the sum of the logs ,. 2°5794978 + Also; 3 (a+ 5-—c)-—a= 95.., log =0-9777236 . eo (a+ b—c) = 49°5.. log = 16946052 Their prod. shown by the sum of the logs.. 2°6723288 The latter swm taken from the former borrowing 20 for radius squared, gives 19:9071690. The latter log. -- by 2 for the 4/, gives 9:9535845. This, by the tables, is nearly tan 41° 563’, and by the fitaiila it is tan 34; therefore 34 = 41°562’, and the required angle = 83° 53’ nearly. Note. In chap. iii. case 3, the same result is obtained by a difierent process. Example Vi. What are the angles of that plane triangle whose na- tural tangents are integers? It is evident from equa. (4) that the sum of the three p cogeuts F must be equal to their Cia ies gpa we ihe sashes H- ae ee sor (1 BL Off Led / By Ee pa toa ‘ Applications of the Formule. 59 the only three integers which possess this property, are 1, 2, and 3; which are, therefore, the tangents required. The angles of which they are the tangents, are 45°, 63° 26’ 6”, and 71° 39’ 54’; whose sum makes 180°, as it ought to do. Example II, There is a plane triangle whose sides are three con- secutive terms in the natural series of integer numbers, and whose largest angle is just double the smallest. Re- guired its sides and angles. That the student may. compare the two methods, we shall present both a geometrical.and an algebraical solu-- tion to this problem, Ist. Geometrically. Analysis. Suppose the triangle aze to be that whose sides CB, BA, AG, are respectively as three C consecutive terms in the increasing series . . of integer numbers, and the greater angle BE ABC equal to twice the less angle Bac; From c upon AB let fall the perpendicular cD, make pb) = ps, and join sc. Then, 9 bDB. _ because the angle cbs (= Asc) is equal _ : to 2caB, the points a and care in the circumference of a circle whose radius is ba, or 6c, and centre 0. But AB: AC + cB (= 2aB) :: AC — cB (=.2): AD. — DB. ag = 4) = cB; hence cp is given; ‘and be- cause AB’= cB + 1 = ac — 1 (by hypothesis), the three sides are given, to construct the triangle. Construction. Let the right: line AB be made equal to 5, on any scale of equal parts; from centres A and B with radii 6 and 4 respectively, describe arcs to intersect each other in c; draw ac, Bc, and agc is the required triangle. Demonstration. The sides cB, BA, AC, are three con-- secutive terms +in the series of natural. numbers (by. ey. “as oof 138-A sey: Spee Ags I.64f 20o/1T4Y 5,94 S 9 x p04 £4 0h af PYRE pple €0 Plane Trigonometry. const.) From c upon as demit the perp. cn, set off pb = DB, and draw dc. Then, ap (== 5): Ac + cB (=6 +4) ::ac —can(=6—4):aAD— pB = Ab= 4. Therefore ad = 5c = éc: and hence the points a and c are in the cireumference of the circle whose centre is 6 and radius A, or dc; and consequently (Euc. iii. 20) angle CUB = CBA = 2 angle cas. Odly. Algebraically. _ Denote sc by x—1, Ba by a, and ac byx +1; then, (Ee being perpendicular to ac the longest side) we have AC: AB + BC:: AB —3BC: AE — EC, that iS, 2: —] 7 2-1: Qe — D1: — 7° Half this added to half pe ee | xr+l x4 Ax ; AC, that MPs ec. @ “b Par EQNS re AE. But AB: rad :: AE: COS A, or, 2: 1::>—— :———— = cosa. Whence sin a = 4/(1 — cos 2A) = tA Sein 27 + 2 ] : , x+1 7/(38x? — 12} al ‘4 wie . setvor > Se a enaeee | Again, BC: sin A:: Ac SN Be oe ao and (by the quest.) this angle = 2a. But, equa. (1), e+ 4 (322 — 12) r+l° @r42 By equating these two values 6f sin B, we have r+ rta Tig gas nence a + 2c +] = xe? + 32 — 4, sn2A =2cosAsina — = sin B, andz= 5. The sides, therefore, of the triangle are 4, r e 7 3 5,and6. The cosines of the angles are, cos A = ra sie Thee ; } CO AG ORR re e sines are, sin A = 7 4/7, Heights and Distances. 61. And the angles are, A = 41° 24° 34” 34” / ¢ = 55° 46’ 16” 18” deme 82° 4.9’ 9” g” CHAPTER V. Application of Plane Trigonometry to the determination of Heights and Distances. k ONE of the most familiar, and at the same time, useful, applications of plane trigonometry, is to the de~ - termination of the altitudes and distances of remote ob- jects, the former usually designated by the term adti- metry, the latter by that of longimetry. It is not in- tended to treat them separately here, nor to treat them jointly much in detail; but simply to present and solve a selection of such problems as are most likely to occur in practice, and as are best calculated to suggest the modes of procedure in’ other cases. 2. The instruments employed to measure angles are ~ quadrants, sextants, and other circular instruments, For military men, perhaps, the best instrument is a pocket sextant, which, if accurately constructed by a skilful artist, will enable a careful observer to take an- gles to within a minute of the truth. But for general purposes the most proper instrument for the measuring of horizontal and vertical angles, is a theodolite fur- nished with a compass and level, one or two telescopes, and a vertical are. Such an instrument, when each circle is 6 or 7 inches diameter, and has a nonius adapted to it, will enable the observer to ascertain an- gles to half a minute, 62 Plane Trigonometry. 3, The space would be improperly occupied in giving particular descriptions of these instruments, and the manner of adjusting them for use. These matters will be much better learnt from an examination of the in- struments, and a few explanatory remarks from a judi- cious tutor. Nor shall we here embarrass the compu- tation and diagrams, by showing how to allow for the height of the instrument, for that is a matter which re- quires only a single hint from the person who teaches the use of the instrument employed. So again, in re- ference to chains, tapes, and other contrivances for measuring lines, descriptions are suppressed, It ought, however, to be observed, that whenever a base, or dis- tance between two stations, is measured on sloping ground, it must be reduced to the corresponding hori- zontal line, before it is employed in the general com- putation, if horizontal angles are taken at its extremities with a theodolite. This is in all cases easily effected : for, if the sloping or hypothenusal line be regarded as radius, the corresponding horizontal line will be the co- sine of the inclination of the plane on which the sloping line was measured. It is, therefore, simply requisite to multiply the natural cosine of the angle of inclination, by the length of the line measured, to obtain the true horizontal line. ' Suppose that on a plane inclined to the horizon in an. angle of 44 degrees, a distance of 400 yards were mea- sured, what would be the corresponding horizontal dis-. tance? Multiply nat. cos 43° = = 9969175 IBY: vin 0 old the 2'eld'e he en 400 Product gives hor. dist. = 398°7669200 Here the difference is not 1} yard, or not a 320th. part of the measured line. As this is not a greater de-. viation from accuracy than will occur in the usual pro- eesses for measuring distances with a chain or a tape,. Heights and Distances. 68 the reduction from the sloping to the horizontal line, may, in common cases, be disregarded, except the an- gle of inclination exceed 43° or 5°. 4. There are several simple methods of approximating to the heights of both accessible and inaccessible ob- jects, by means of shadows, mirrors, unequal vertical stafis, &c. But as these depend solely upon the prin- ciple of similar triangles, and do not require the theo- rems and formule of trigonometry, properly so called, they are not described here.. The student may, with- out being detained by farther observations, proceed to the solution of the following problems. ExaMpue |. In order to ascertain my distance from a tower, which was inaccessible by reason of an intervening river, I measured on a horizontal plane a base line of 600 yards, and_at each end took the angle included between the other end and the tower, finding them to be 57° 35’ and 64° 51’ respectively. Required the ery distance of the tower from each end of the measured base. In the annexed figure are given AB = 600 CAB = 57° 35" Wak tau CBA = 64° 5Y fos adi, eonseq. c= 180° ~ (A +8) = 57°34. P 4 E ogs Hence, sin c ....57° 34 0) 9°9263.507 Toas.., 600 .,,..2°7781513 So is sin A 57° 35’... ...9'9264310 To Bc... 600711... . 2°7782316 and, Sois sins 64° 5¥ .... 9°9567437 To ac... 643°49.... 28085443. Remark. These distances might have been deter- mined without the aid of an instrument to measure angles, Thus, suppose in the continuations of the res- 64 Plane Trigonometry. pective lines cA, cB, two distances, AD, BE were mea- sured, each = 120; and suppose, on measuring the dis- tances sD, AE, the former is found to be 660, the latter 672 yards; from these measures the required angles may be determined. For, in the triangle ane thus formed, we have (Euc. ii. 12) AE? = AB? + BE? + 2EB.BP, Whence, by trans- AE? — AB? — BE? 2EB the cosine of the angle asc, to the radius aB; se that, dividing the preceding expression by AB, we shall have the cosine of that angle to radius 1. A hike process will give the cosine of CAB. fae Au? — AB? — BE? 6722, — 600? — 120? Thus, cos ABC = Ii |r Ae" 7 ella TNS TRC = °425 = cos 64° 51’ BD? — AB? — Ap? 6602 — 600? — 120? Sid, COR. BAC = Tomi an om 240 . 600 = *536 = cos 57° 35. Hence, the angles anc, and Bac, being determined, the distances are found as before. position and division, BP = - But BP is EXxAMpte II. In order to find the distance between two trees A and g, which could not be directly measured because of a pool which occupied much of the intermediate space, I measured the distance of each of them from a third ob- ject c, viz. Ac = 588, Bc = 672, and then at the point c took the angle acz between the two trees = 55° 40". Required their distance. This is an example to case 2 of plane triangles, in which two sides and the included angle are given. The work, therefore, is left to exercise the student: the an- swer is 593°8 ; Examp te III. Wanting to know the distance between two inacces- sible objects, which lay in a direct line from the bottom Heights and Distances. 65 of a tower on whose top I stood, I took the angles of depression of the two objects, viz. of the most remote 251°, of the nearest 57°. What is the distance between them, the height of the tower being 120 feet? A The figure being constructed, as in the margin, AB = 120 feet, the altitude of the tower, and au the horizontal line drawn through its © top; there are given, WAD = 25° 30’, hence BAD = BAH — HAD = 64° 30’ HAC = 57° 0’, hence BAC = BAH — HAC = 33° OQ, Hence the following calculation. C D In right angled A Asc. In right angled A AED. AgraAnie es)... 10°0000000|As rad ....... 10:0000000 To AB..120.. 2:0791812!To AB ..120..° 2:0791812 Soistansac 33° 9°8125174|SoistanBap643°10°3215039 To sc... 77'929 1:8916986|Topp,..251'585 24006851 “= ———— - | Conseq. pp — BC = 173°656 feet, the distance required, Such is the way in which this problem is usually solved: it may, however, be done more easily and con- cisely, by means of the natural tangents. For, if az be regarded as radius, Bp and sc will be the tangents of the respective angles BAD, BAC, and cp the difference of those tangents. It is, therefore, equal to the product of the difference of the natural tangents of those angles into the height As. Thus, nat. tan 643° = 2°0965436 nat, tan 33° = 0°6494076 difference......+++. 1°4471260 multiplied by height . +120. gives distance cD... ,. 1736563200 Be RS ee ee ee! Satie por . ; 66 - «Plane Trigonometry. EXAMPLE IV, From the top of a hill I observed two mile-stones on a horizontal road, which ran straight from its bottom, and took their respective angles of depression below the horizontal plane passing through the place of my eye; that of the nearer mile-stone was 14° 3’, that of the far- ther was 3° 56’.. Required the height of the hill. The figure being drawn, it will be found analogous to that in exam. 3, to which, therefore, we shall refer. There are given cb = 1760 yards, the distance between the two mile-stones; ADB = HAD = 8° 56’; ACB = HAC == 14° 3’, This admits of three distinct modes of de- termination, as below. Ist Method. The angle Acp is equal to the sum of the angles cAD. and cpa (Luc. i. 32): therefore cAaD = ACB — ADB = 10° 7’. Then, in the triangle acp, it will be, as sin CAD: CD::sinCDA:cA. .Aind, in triangle acs, it will be, as rad. or sin B: AC:>sin ©: AB = 166°85; and so, if it be required, is sin cAB: cB = 666°75. According to this method the logarithmic process will require e7ght lines. | 2d Method. Since cAD is the difference of acz and apc, we have, cD.sinD i —D):cD::sinD: AC = -—-———-:_ Also, rad: ac sin (C —D):cp::sinD: AC = — rata Also, cD sin D ‘ cp sin ¢ sin D ; (= =—~———-) :: sin. c: AB-== ———-—.— But -(ch. i. sin (C—D) sin (C—D) 1 19) cosec (c — D) = ————-__ Hence aB = cb sin (C—D) : sin C sin D cosec (c — D), Or, making the terms homogeneous (ch. iv. 17), AB.rad3 = cp sin e sin p cosec (Cc — D). The logarithmic formula is, therefore, this: a, ‘Heights and Distances, 7 log AB = logen + log sin c + logsin» + logcosec(c =p) — 30 [in the index of the log]. Thus, log cp ...... -,. 1760 .... 3°2455127 log smc ...... 14° 3’... 9°3851924 log sinD,..... 3° 56’ .. 88362969 log cosec (C—D) 10° 7’ ..10°7553442 ES UY The sum — 30, is log aw 166°857 feet 2:2293462 This method is, obviously, applicable to all similar examples. 3d Method. If aB were radius, cp would evidently be the differ> ence of the tangents of BAD and BAC, or, of the cotan- ' cD gentsof BpbAandsca. Hence AaB = ———-~——* €0t D — cotc Thus, nat cot p =14°543833 nat cote = 3995922 10°547911)1760-000( 166-857ft. as bef. tereee 1054791 105209 - 632878 79334 63287 9047 8438 609 527 82 . 14 —y 68 Plane Trigonometry. From the comparison of these three methods, it will appear that the second ought to have the preference to the first: and that, considering the time employed in looking out the logarithms, the third method is prefer- able to the second. -ExAMPLE V. Wanting to know the height of a church steeple, to the bottom of which I could not measure on account of # high wall between me and the church, I fixed upon two stations at the distance of 93 feet from each other, on a horizontal line from the bottom of the steeple, and at each of them took the angle of elevation of the top of the steeple, that is, at the nearest station 55° 54’, at the other 33° 20°,. Required the height of the steeple. This is similar to exam, 4, and being worked by the 2d method, the height of the steeple is found to be 110°27 feet. : ExAmpte VI, ~ Wishing to know the height ofan obelisk standing at the top of a regularly sloping hill, I first measured from its bottom a distance of 36 feet, and there found the angle formed by the inclined plane and a line from the centre of the instrument to the top of the obelisk 41°; but after measuring on downward in the same sloping direction 54 feet farther, I found the angle formed in like manner to be only 23° 45’. What was the height of the obelisk, and what the angle made by the sloping ground with the horizon? 3 The figure being constructed, as in - the margin, there are given in the tri- angle acs, all the angles and the side AB, tafind gc. It will be obtained by this proportion, as sine (= 17°15’ = B— A): AB (= 54)::sin A (= 23°45/) : BC = 73°3392. ‘Then, in the triangle DBC are known BC as above, BD = 36, Heights and Distances. - 69 cBp = 41; to find the other angles, and the side cp. Fas first, as CB + BD: cB —BD::tang(p+c)= £ (139°) : tan 3 (Dp — c) = 42° 243’. Hence 69° 30’ be 42° 247’ = 112° 545° = cos, Bnd 69° 30’ — 42° 24.3’ =e ZOn 51) = = pep. Then, sin pcb: BD:: sin cBD: cD = 51 86, height of the obelisk. . The angle of inclination DAE = HDA = cpB — 90° = 22° 544% Remark. If the line pp cannot be measured, then the angle DAE of the sloping ground must be taken, as well as thd angles caB, and cpp. In that case DAE + 90° will be equal to cps: so that after cs is found from the triangle AcB, cD may be found in the triangle CBD, by means of the relation between sides and the sines of their opposite angles. ExAMPLeE VII, Being on a horizontal plane, and wanting to ascertain the height of a tower standing on the top of an inacces- sible hill, I took the angle of elevation of the top of the hill 40°, and of the top of the tower 51°, then mea- suring in a direct line 180 feet farther from the hill, I took in the same vertical plane the angle of elevation of the top of the tower 33°45’. Required from hence the height of the tower. The figure being: constructed, as in the margin, there are given, AB = 180, CAB = 33° 45’ ACB = CBE — CAE = 17° 15 cBD = 11°, Bpc = 180° — (90° — DBE) = 130°. And cp may CZs be found by two proportions, viz, 5 # Ist. As sin A€B: AB::sinCAB:cB, and @dly, as sin p 2CB::sincBD:cp, This process would require eight lines. But the operation may be shortened; for, by the principles of method 2, exam, 4, we shall have cp rad* = AB sin A SiN CBD Cosec ACB See DBE, 70 Plane Trigonometry. In logarithms thus: log aAB......180.... 22552725 SIN A we eee . 33° 45’ 2. 9°7447390 ‘ sinw ......11° 0’ .. 92805988 cosec ACB ..17° 15’ .. 10°5279144 - 8eC DBE,...40° 0’ ..10°1157460 The sum —-40, is log ep . .83'9983 feet .. 1°9242707 | eetemetiaioee t Exameie VIII, -At the top of a castle which stood on a hill near the sea-shore, the angle of depression of a ship at anchor was observed to be 4° 52’; at the bottom of the castle the angle of depression was 4° 2’, Required the hori- zontal distance of the vessel, and the height of the hill on which the castle stands above the level of the sea, the castle itself being 60 feet high, In the annexed diagram, where HT, OB, are parallel, and ar per- pendicular to the horizontal line AS, are given BT = 60 feet, uts = 4° 52’, consequently ars = 4 85° 8’, and oBs = 4° 2’, whence _ABS = 85° 58’; to find as and as. Here method 2, exam. 4, is obviously applicable: so that we have AS.rad} = BT sin ATS sin ABs cosec (ABS — ATS AB .rad3 = BT sin ATS cos ABS cosec (ABS — ATS), The logarithmic operation will stand thus: lo Sess. 5: 60 1°7781513|log TB..... 60 1°7781518 sin ATS 85° 8’ 9-9984315|sin ars 85° 8’ 9:9984315 sin ABS 85°58’ 9:9989230!cos ABS 85°58’ 8-84:71827 cosec (B— T)50' 11°8371392}cosec (B— 1) 50’ 11°8373192 Cae log As 4100-4 ft. 3-6128250llog an 289:12 ft. 24610847 RTE Z Heights and Distances. 7 ExampLeE IX. Wanting to know the distance of an object at p from two others A and 8 in the same horizontal plane, as well as the distance between a and B, a pole was set up at c ina right line with as, and the angle . AcD was found to be 57°. The dis- tance cp being measured was found to be 549°36 yards; and at p the angles cpDA and Aps were taken; the first = 14°, the latter = 41°30’. Required the above specified distances, Here, the sum of the angles c and cpa taken from 180°, leaves the angle cap, and the sum of the angles c and.cps taken from 180°, leaves the angle cup. ‘Then, it will be as sinepp:cpD::sinc: DB, As sincAD: cD t:sine:pA. And sin ABD: AD::sinApB: AB, These operations will give pp = 498'68, pA = 487°27, AaB = $49°52 yards. EXAMPLE X. ‘\ Wanting to know the horizontal distance between two inaccessible objects A and B, and not finding any station from which I could see them , RB both, I chose two points c and D dis- tant 200 yards, from the former of which a could be seen, from the latter B, and at each of the points c and pa flag-staff was set up. I then mea- “< sured rc = 200, pE = 200, and, hav- 7 E ing set up flag-staves at F and =, took the following angles, viz. arc = 83°, acF = 54° 3Y, acp = 53° 30’, Bpc = 156° 25’, BDE = 54° 30’, and BED = 88°30’. Required as. | Here, in the triangle arc, all the angles are given, and the side rc, to find ac. Then, in the triangle acp are given AC, CD, and the contained angle, to find the ether angles and the side ap, Next, in the triangle 72 Plane Trigonometry. BDE are given DE and all the angles to find pz. Lastly, in the triangle apg are known AD, bB, the included angle BDA = BDC — ADC, to find aB = 345°5 yards. EXAMPLE XI, In order to determine the distance between two inac- cessible objects & and w, on a horizontal plane, we mea- sured a convenient base AB of 536 yards, and at the ex tremities a and 8 took the following angles, viz, BAW == 40°16’, WAE = 57° 40’, ABE = 42°29, EBW = 71°77. Required the distance Ew. First, in the triangle ABE are given £ Ww all the angles, and the side As, to find BE. So again, in the triangle ABw, are given all the angles and Ax to find Bw, Lastly, in the triangle BEw are given the two sides EB, BW, and the included angle., EBW, to find ew = 939°52 yards. ft 5 Remark. In like manner the distances taken two and two, between any number of remote objects posited» around a convenient station line, may be ascertained. ExAmMeP_LeE XII. Suppose that in carrying on an extensive survey, the distance between two spires A and B has been found equal to 6594 yards, and that c and p are two eminences conveniently situated for extending the b : triangles, but not admitting of the de- B termination of their distance by actual admeasurement: to ascertain it, there- °¢ fore, we took at c and p the following angles, viz. | ACB = 85° 46’ Apc = 31°48’ ¥ = Ze = 93° 56’ ee = 68° 2’ , Required cp from these data. oe Jn order to solve this problem, construct a similar quadrilateral scdb, bua ea cd equal to 1, 10, or any Heights and Distances. 73 ether convenient number: compute Ad from the given angles, according to the method of the preceding ex- ample. Then, since the quadrilaterals acdb, acpB, are ~ similar, it will be, as Ad: cd's: AB: cD; and cp is found = 4694 yards. Bcastein XII. If the height of the mountain called the Pike of Te- neriffe be 3 miles, and the angle formed at its top be- tween the vertical or plumb line and a right line con- ceived to touch the earth in the horizon, or at the far- - thest visible point, be 87° 46’ 33”; it is required from hence to determine the diameter of the earth, supposing _it to be a perfect sphere. : __,Let.c, in the marginal diagram, be the centre of the earth, the-circle BTG a yertical sec- > Sap tion passing through the centre, aB the height of the Pike of Teneriffe, and at the tangential line drawn to the visible horizon: let, also, BE, a tangent to the earth’s surface at 8, meet the other tangent ar in E. Then, ‘in the triangle aBE, right an-. gled at 8, and having the angle BAr = 87° 46, 33”, it is, as rad: AB:: tan. A: BE, and *: sec A:AE. But it is evident, from Euc. iii. 37, that pe = ®T; therefore, AE + EB = AE + ey = AT. In the. triangle arc right angled at 1, we have, asrad: AT:: tan A.: Tc, the radius of the earth. The operation performed as above described occupies but small compass; it may, however, be shortened by means of’ch. ii. prop. 4. For, since tan A + sec A = tan (A + 3 comp. A), we shall, by incorporating the etek from which af, Be, and cr are deduced, lave eae cT. rad? = AB tan (A + 3 comp. A) tana, Or, log cr = log axB + log tan (A + 3 comp. a) + logtan aA — 20, in the index, E ~*~. 74 Plane Trigonometry. Thus, log AB... 0. eos SSA Re cee) OWE hark log tan (A + Lcomp. A) 88° 58’ 163” 11:7119309 log tan A @eaeeooveeoee Si 4.6/ 33” oe 11°4107381 log OF oe. eee ee 3979'15.... 35997903 The double of this, or 7958'3 miles, is the diameter of the earth, | If at be required, we have only to take radius (10 from the sum of the first two lines, the remainder 91890522, is the log of 154°54 the distance sought. Note 1. The log tangents are found in this example by the method taught at p. 153 of the Introduction to Dr. Hutton’s Logarithm Tables. Note 2. This method of determining the earth’s ra- dius, though elegant in theory, is rendered useless in practice, by the extreme irregularity of the horizontal refractions. EXAMPLE XIV. Given the angles of elevation of any distant object, taken at three places in a horizontal right line, which does not pass through the point directly below the ob- ject; and the respective distances between the stations; to find the height of the object, and its distance from either station. Let arc be the horizontal plane, rE the perpendicular height of the object above that plane, A, B,C, the three places of observation, FAE, EBE, FCE, the angles of elevation, and AB, BC, the given distances. Then, since the triangles AEF, BEF, CEF, are all right angled at £, the distances AE, BE, CE, will manifestly be as the cotangents of the angles of elevation at A, B, and c. Put az = bv, sc =d, EF = a, and then express alge- braically the following theorem, demonstrated at p. 128, Simpson’s Select Exercises: viz. Heights and Distances. 75 AE? .BC + CE*,AB = BE*, AC + AC. AB. BC, the line Es being drawn from the vertex © of the tri- angle ACE, to any point B in the base. The equation thence resulting is, dx* cot 2a + px? cot?7c = (p + d)x* cot?B + (D+) nd. Hence, transposing all the unknown terms to one side of the equation, dividing by the sum of the coeffi- cients, and extracting the square root, we shall have : : (p + d)pd pane RT ne eB dcot?a + pcot °c — (D + d) cot *B Thus EF becoming known, the distances AE, BE, CE, are found, by multiplying the cotangents of A, B, and c, respectively, by EF. Cor. When p = d, or p + d= 2p = 2d, the exprese sion becomes z=d-—+ ./(scot?a + 3 cot?c — cot zB), which is pretty well suited for logarithmic computation. The rule may, in that case, be thus expressed.— Double the log. cotangents of the angles of elevation of the extreme stations, find the natural numbers an- swering thereto, and take half their sun; from which subtract the natural number answering to twice the log. cotangent of the middle angle of elevation: then half the log. of this remainder subtracted from the log, of the measured distance between the Ist and 2d, or the 2d and 3d stations, will be the log, of the height of the object.* : : To illustrate these general methods, two particular examples are subjoined. EXAMPLE XV. Let the three angles of elevation be 36° 50’, 21° 24’, * For geometrical constructions of this general problem the atudent may consult Hutton’s Course, vol. iii. p. 128, Leybourn’s Ladies’ Diaries, vol. i. p. 397, or the Ladies’ Diary for 1748, E2 ; 76 Plane Trigonometry. and 14°, and the two equal measured distances 84 feet. Required the height of the object. Ans. 53°964: feet. _Examrere XVI. ) Let the distances be an = 100 yards, zc = 400 yards, and the angles, at a = 5° 24’, at B = 6° 273’, at c = 8°36’. What is the height of the object ? Ans. 44°46 yards. ExAmpeLe XVII. From a convenient station p, where could be seen three objects A, B, and c, whose distances from each other were known (viz, AB = 800, Ac = 600, Be = 400 yards), I took the horizontal angles apc = 33° 45, Bpc = 22° 30’. It is hence required to determine the re- spective distances of my station from each object. Here it will be necessary, as preparatory to the com- putation, to describe the manner of Construction. Draw the given triangle anc from any convenient scale. From the point a aah: draw a line Ap to make with aB an angle equal to 22° 30° and from B a line Bp to make an angle pa = 33°45". 4¢ Let a circle be described to pass — through their intersection p, and through the points a ands. Through c and p draw a right line to meet the circle again in P: so shall p be the point required. For, drawing PA, PB, the angle Arp is evidently = axp, since it stands on the same arc AD: and for a like reason BPD = BAD. So that Pp is the point where the angles have the assigned value. Manner of computation. In the triangle asc where the sides are known, find the angles. In the triangle ABD, where all the angles are known, and the side AB, Heights and Distances. ae find one of the other sides Aap. Take pAb from BAC, the remainder, DAc is the angle included between two known sides AD, Ac; from which the angles anc and Aacp may be found, by chap. iii. case 2. The angle CAP = 180°— (Arc + AacpD). Also, BCP = BCA — ACD; and pBc = ABC + PBA = ABC + sup. ADc. Hence, the three required distances are found by these propor- tions. As sin APC: AC:: SiN PAC: PC, and:: sin PCA: PA; and lastly, as sin Bec: BC::sinBcP: BP. ‘The results of the computation are, pA = 709°33, pc = 1042-66, PB = 934 yards. ' *,* The computation of problems of this kind, .how- ever, may be a little shortened by means of the fol- ‘lowing General Investigation. Put Ac =a, Bc = 6, apc = P, BPC = P’, ACB and let there be taken for unknown quantities pac PpBc =y. The triangles Pac and Pzgc give sin APC: SiN CAP?:: AC: CP and sin BPC : SIN CBP :: BC: CP; asin el & that is, sin P:sinw::a: 3 = CP sin P bsiny and sin P’: siny::6: —— = cp. sin P asineg bsin Hence — = seb ) which reduces to sinP sinP asin P’ sing — dsinrsiny =0. In the quadrilateral Acar, we have CBP = 360° — apc — BeC — ACB — CAP or y = 360° — Pp — Po —co— es. Make 360° — p — r’ —c = R, then we shall have y = R — 2, and consequently, asin p’ sina — dsinp (sink cos — cos Rsin zc) = 0. Dividing by sin « there results, 3 P ‘ cos x asin P’ — dsinP (sin R —— — cosRk) =0. SIN & 78 Plane Trigonometry. Whence we have cosx asinp’ + bsinPcosr —— = cor = sin x bsin Psink® This expression being separated intotwo parts, we have asin P’ cos R cots = Te FS bsin P sin R sink ae ( asin Pp +1); e r ] or, cotz = siu R \OsinPCOSR Base} ue asin P’ or, lastly, cotx = cot R ——- + 1) ; Hence, x being thus determined, we get y from the equation y = R — z, and cp from either of the expres- sions above given. Note. It will be a useful exercise for the student, to — work out the computation by both these methods. The comparison of the results will serve to give him confi- dence in the deductions from the analytical investi- gations. ExAMPLE XVIII. It is required to find the distances from Edystone light-house to Plymouth, Start Point, and the Lizard, respectively, from the following data: : Plymouth to Lizard. ., . 60 The distance from } Ea to Start Point ..7 of miles. Start Point to Plymouth 20 Plymouth .... North. Lizard 0... bears from Edystone rock } wav. Start Point... | E. by N. Ans, From Edystone to Lizard .....53°04 miles to Plymouth ., 14°333 ditto to Start Point.. 17-36 ditto. Remark. The general problem of which the last two exemples are particular cases, was originally proposed by Richard Townley, Esq. and solved by Mr. John Collens, in the Philosophical Transactions, N° 69, A. D. 1671. See New Abridgement, vol. i. p. 563. Py General Properties, &c. 12 The six cases there considered are, 1. When the sta- tion is out of the triangle made by the objects, but in one of its sides produced. 2, When the station is in one of the sides of the triangle. 3. When the three objects lie in a right line. 4. When the station is not within the triangle formed by the objects. 5. When the station 7s _ within that triangle. 6. When, by reason of the rela-_ tion of the sides and angles, the points c and p (see the preceding diagram) fall so near together as to make the continuation to p of doubtful accuracy. ‘To these, later writers on trigonometry have added another case, viz. 7. When the point c falls between the line az and the _ station P. It will be advantageous to the student, to modify the construction and computation to suit all these cases. CHAPTER VI. Spherical Trigonometry. 4 Section f. Fundamental Principles, General Properties, and . Formule. 1. THREE planes, AOC, AOB, BOC, all of which pass through the centre o of a sphere, in- e tersect the surface of that sphere in @ portions of great circles which form a INA spherical triangle anc. Thus also is Ath B constituted the spherical pyramid or tetraedron which has for its base the triangle ane, and for its vertex the centre o of the sphere. The angle a 86 Spherical Trigonometry. of the triangle, is the same as the diedral angle, be- tween the two planes BAO, CAO; it is, also, the angle formed by the: tangents to’the two arcs Ac, AB. The like may be said of the other angles, The sides are ma- nifestly the measures of dependent plane angles, viz. a the measure of the angle cos, b the measure of coa, c the measure of Aos. _ 2. A right angled spherical triangle has one right angle; the sides about the right angle are called legs; the side opposite to the right angle is called the hynpo- thenuse. ) : 3. A quadrantal spherical triangle has one side equal to 90°, or is a quarter of a great circle. 4. An isosceles, or an equilateral spherical triangle, has respectively two sides or three sides equal. 5. When the sides of a triangle are each 90°, it is not only an equilateral, but a guadrantal, and a right angled triangle. All its angles as well as its sides are equal; . and these sides may any of them be regarded as an hy- pothenuse, any of them as legs. Such is the case with the triangle that would be formed on a Celestial or terrestrial globe, by the horizon, the brazen meridian, and a quadrant of altitude, fixed at the zenith, and pass- ing through the east or west point. 6. Iwo arcs or angles, when compared together, are said to be alike, or of the same affection, when both are less, or both greater than 90°. They are said to be unlike, or of different affections, when one is greater and the other less than 90°. | 7. Every spherical triangle has three sides and three angles; of which, if any three be given, the remaining three may be found, 8. In plane trigonometry, the knowledge of the three angles is not sufficient for ascertaining the sides (chap. i, 6): but, in spherical trigonometry, the sides may always be determined when the angles are known. In plane triangles, again, two angles always determine the third; in spherical triangles they never do. So, farther, General Properties, &c. $1 the surface of a plane triangle cannot be determined from its angles merely; that of a spherical triangle always can. 9, A line perpendicular to the plane of a great circle, passing through the centre of the sphere, and termi- nated by two points diametrically opposite, at its surface, _ is called the axis of such circle; and the extremities of the axis, or the points where it meets the surface, are the poles of that circle. If we conceive any number of less circles, each pa- rallel to the said great circle, this axis will be perpendi- cular to them likewise; and the poles of the great circle will be their poles. : 10. Hence, each pole of a great circle is 90° distant from every point in its circumference; and all the arcs drawn from either pole of a little circle to its circum- ference, are equal to each other. 11. It likewise follows that all the arcs of great circles, drawn through the poles of another great circle, are perpendicular to it; for, since they are great circles by the supposition, they all pass through the centre of the sphere, and consequently through the axis of the said circle. The same thing may be affirmed in reference to small circles. 12. Hence, in order to find the poles of any circle, it is merely necessary to describe, upon the surface of the sphere, two great circles perpendicular to the plane of the former, the points where these circles intersect each other will be the poles required. : | 13. All great circles desect each other. For, as they have a common centre, their common section will be a diameter; and that manifestly bisects them. 14. The small circles of the sphere do not fall under consideration in spherical trigonometry; but such only as have the same centre with the sphere itself.. Hence appears the reason why spherical trigonometry is of such great use in practical astronomy, the apparent heavens being regarded as in the shape of a concave ES 82 : Spherical Trigonometry. sphere having its centre either at the centre of the earth, or at the eye of the observer. PrRogLEeM. 15. To investigate properties’ and equations from which the solution of the several cases of spherical tri- gonometry may be deduced. In order to this let us recur to the spherical tetrae- dron oAxzc, where the angles a, 3, c, of the spherical triangle are the diedral angles between each two of the three planes aoc, Aon, &c. and the sides a, 6, c, are the measures of the plane angles cos, coa, &c. Here it is Ist, evident that the three sides of a spherical triangle are. together less than a circle, or, a + 6+ c¢ < 360°. For the solid angle at o is contained by three plane angles, which (Euc. xi. 21) are together less than four , right angles; therefore, the sides a, 4, c, which measure those plane angles are together less than a circle. 16. Let the tetraedron oasc be cut by planes per- pendicular to the three edges; they will form an- other tetraedron 0’a’B‘C’; 5 iN their faces will be respec- O< igs yp O’ tively perpendicular, two i! GB and two: But, in the quadrilateral aoa’ since _ the angles a and B are right angles, the plane angle o is the supplement of aa’s which measures the diedral angle aA’o’s. The same may be shown with respect to the other plane angles that meet at 0; as well as of the plane angles at 0’, in reference to the diedral angles of the tetraedron oasc. Therefore, either of these tetrae- drons, has each of its plane angles supplement to a diedral angle in the other : itis hence called the supplementary tetraedron. And if they become spherical tetraedrons referred to equal spheres, or to different parts of the same sphere, their bases will be spherical triangles re~ spectively supplemental to each other. a ee oot a i war nA S&S Fundamental Theorems, &c. 88 17. It is obvious from this that the problems in sphe- rical trigonometry become susceptible of reduction to half their number; since, if there are given, for exam- ple, the three angles a, 8, c, and the three sides a, 4, c, are required ; let the, triangle which has for its sides a’, b’, c’, the supplements of the measures of a, 8B, and c, have its angles a’, B’, c’, determined; their measures will be the supplements of the required sides a, 6, and c. 18, On the surface of the sphere, the supplemental triangle is formed by the intersections of three great circles described from the angles of the primitive tri- angle as poles. Besides the supplemental triangle, three others are formed in each hemisphere by the mu- tual intersections of these three great circles; but it is the central triangle (of each hemisphere) that is sup- plementary. 19. Every angle between two planes being less than two right angles, it follows, that the sum of the angles of a spherical triangle is less than 8 times 2, or than 6 right angles. At the same time, it is greater than 2 right angles: for the sum a’ + & + c’ of the sides of the sup- plemental triangle is Jess than 360° (art. 15 above): taking the supplements, we have ; 3 x 180° — (a + & +c’) > 180°, ora + 3B +c > 180°. 20. To deduce the fundamental theorems, we may proceed thus. From any point a of the edge Ao of the tétraedron, let fall on the plane or face soc the perpen- dicular ap: draw, also, in that plane, the lines DH, pc, perpendicular to on, oc, respectively; and join AH, ac; then will au be perpendicular to op, and ac to oc. It is evident, therefore, that the angles acD, AHD, measure the angles between the planes aoc, cos, and AoB, CoB, that is. the angles c and B of the spherical triangle asc, It is also evident that the plane angles in 0, are Aoz =c, aoc =b,noc=a. This being premised, the tri- angles aco, AcpD, the former right angled in ¢, the latter IM D, give > 84 Spherical Trigonometry. AC = AOSin Aoc = Aosind AD = ACSIN ACD = AO sin JSsinc. In like manner, the triangles aou, ADH, right engled in H and p, give AH = AOSiN AOB = AOsine AD = AHSIN AHD = AOSINCSINB. Making these two values of A AD equal, we have AO SiN BSINc = AO Sin sin J, And therefore, by division, sinB sinc sind sing In a similar manner it may be shown that : sinA -sinB -—— == —-: therefore, sina sin & sin A sin B sinc Panes : . eecee (1.) sin c sna. sind’ / Hence, the sines of the angles of a spherical triangle are proportional to the sines of the opposite sides. 21. Draw ce and pr, respectively perpendicular and parallel to oB; then will the angle DCF = EOC = a, But the right angled triangles aoe, acp, FcD, give AC = AO Sin 6, pc = ac cosc = AO sin cosc and FD = DC sina = Aosinasinb cosc. Now oH = OF + EH = OF +. FD, er AO COSC = OC cosa + FD = Ao cosa cosh + Fp = AO cos acos 6 -+- Aosinasind cosc, Therefore, dividing by ao, we have cos c = cos acos 6 + sina siné cosc, Similar relations are deducible for the other sides a and 4: hence, generally cesa= cos d4cosc¢ + sin ésinc cosa cos 6 = cosa cos c + sinasinc cos B + (2.) cos¢ = cosacos6 + sinasiné cos c 22, ‘These equations apply equally to the supple- mental triangle, Thus, putting for the sides a, b, c, « ‘ , Fundamental Equations. 85 180° — A’, 180° — 8’, 180° — c’, &c. and for the angles A, B, C, 180° — a’, 180° — U’, &e. we shall have — cos A’ = cos B’ cos ¢’ — sin B’ sin c’ cosa’, Here again we have three symmetrical equations ap- plying to any spherical triangle, viz. COSA = COS@ SiN B sin C — COS B COSC cos B = cos bsin asin c — cos A cos ¢ (3.) COS C = COS C SIN ASINB — COS A COSB 23. Another important relation may be readily de- duced. For, ‘substituting for cos 6 in the third of the equations marked (2) its value in the second; substi- tuting also for cosa its value 1 — sin? a (chap. i. 19), and striking out the common factor sin a, we shall have cos c sina == sinc cosa.cosB + sind cosc. sinB sine > ives sin 6 = —— But, equa. (1) gives sin 6 = ——- Hence, by substitution, | sin B cos C sinc cos ¢ sina = sinc cOS a COSB + sinc Dividing by sin c we have, cosc , sin B cosc —— sina = cosa cosB -+- ————" sinc sinc But < = tan (chap. i. 19). Therefure cot c sina = cosa cos B + sin BCotc, Thus, again, we get three symmetrical equations, cotasin 6 = cosbcosc + sinc cota cot bsinc = cosc cos A + sina cot B + (4.) cot csing = cosa cosB + sinBcotc 24. The classes of equations marked 1, 2, 3, 4, com- prehend the whole of spherical trigonometry: or, in truth, the equations (2), from which the others may be made to flow, may be regarded as comprehending the whole. They require, however, some modifications to fit them for logarithmic computations, and become sim- plified in their application to some kinds of triangles. We shall, therefore, new show the pupil oe they be- 86 Spherical Trigonometry. come transformed when they are applied to the prin- cipal cases which eccur in practice. a Section II. Resolution of Right angled Spherical Triangles. 25. Suppose, in the first diagram in this chapter, the angle A to be right, or, the faces oAB, OAc, to be per- pendicular to each other. ‘Then, since sin a = 1, the equations marked (1) become aay | sinB SEG sind 1 Bia. 2. Sin oo Consequently, : __ sind , __ sine sinB = ——.,.sinc = ——... te sin 6 = sinBsina...sinc = sinc sina Also, since cos Ais then = 0, we have from equa. (2) cosa = cosbcosc.... (6. For the same reason, the first of equa. (3) gives COs @ Sin B sin C = COS B CosC; cos B COSC whence cos a = = cots cotc.:..(7.) sin B sinc Upon the same hypothesis, cot A becomes = 0, so that the first of equa. (4) becomes cotasin6 = cosécosc. Or, dividing by sin 4, cos b cota = —— cosc = cot dcosc.... (8.) sin & The two last of equa. (3) give also, upon the same hypothesis, cos B = sin C cos 6 cosC =sinBcosc} **"* (9.) -And, lastly, from equa. (4) we hare cot B = cot édsinc cot c = cot csinb¢ °*°+ (10) From these equations by a few obvious transforma- tions, the szx usual cases of right angled spherical tri- angles may be solved, as below. | 6 Right angled Triangles, «gt Case I. 26. Given the hypothenuse a Cc and an angle B ; to find the rest ; hs \ sin §= sinasinB; B A or, sin side req. = sin opp. angle. C X sin hypoth. tanc = tana cosB; or, tan side req. = tan hypoth. x cos included angle, cot.€' == cos a tan B;.'** or, cot angle req. = cos hypoth. x tan given angle. In this case there can be nothing ambiguous; for, in applying the first form, it is known that the angle and the opposite side are always of the same affection ; and in the two latter the rules for the changes of sines in the different quadrants (chap. iv. 9), will determine to which the result belongs. ise Il. 27. Given the hypothenuse a, and one of the sides 6; to find the rest. A sin b 3 __. sin given side sin B = ——; or, sin angle req. = ome pant cos hypoth. : _ cosa 4 cos¢ = ; or, cos side req. = cos b” cos c = tan 6 cota; or, cos angle req. = tan given side x cot hypoth, cos given side Case III. 28. Given the two sides including the right angle, namely, 6 and c; to find the rest. cosa = cos 6 cosc; or, cos hypoth. = rectangle cosines of the sides. tan b __ tanec. tan B = me ane a waht tan opp. side tan adj. skle va or,.tan angle req. = 88 Spherical Trigonometry. Case IV. 29. Given a side 4, and its opposite angle 8; to find the rest. 4 sin b A sin given side sin 4 = ——-; or, sin hypoth. = isso obs | iF f sin B sin opp. angle sinc = tan) cot B; or, sin side req. = tan given side x cot opp. angle. cos given angle cosB , sinc = — |: or, sin req. angle = os b cos given side Case V. - 30. Given a side c, and its adjacent angle 8; to find the rest. tan 6 = tan B sinc; or, tan side req. = tan opp. angle x sin given side, tana = =<; or, tan hypoth. = sea hil eh cos B cos given angle or, cota = cos Bcotc; . that is, cot hypoth. = cos given angle x tan given side. cos C = cosc sin B; or, cos angle req. = cos opp. side x sin given angle. Case VI. 31. Given the two oblique angles B and c; to find the rest. cos c = cot BCotc; or, cos hypoth. = rectangle cot’s given angles, py j cosB . SOE Or gin ae cos opp. angle cos req. side = ———-—-—> cos C sin adj, angle COSC = -= sin B Note 1. Here the rule of the signs (chap, iy. 9) serves all along to determine the kind or affection of the un- known parts, oa a cal i gal oem ie Oblique Spherical Triangles. 89 Note 2. In working by the logarithms, the student must observe, that when the resulting log. is the log. of a quotient, 10 must be added to the index: when it is the log. of a product, 10 must be subtracted from the index. This is done in conformity with the rule (chap. iv. 17), to make the terms homogeneous by multiplying or dividing by the powers of radius, Section III. Resolution of Oblique angled Spherical Triangles. 32. This may be effected by means of four general eases; each comprehending two or more problems. Case I. Given three of these four things, viz. two sides 4, c, and their opposite angles B, c; to find the fourth. This case comprehends two problems, in one of which the unknown quantity is an angle, in the other a side. They are both solved by means of equa. (1) of this chapter; from which we have sid csinB sin sind sinc = ———-,,.. sinc = — sin } * sin B Case Ul. 33. Of these four things, viz. the three sides a, 6, cy and an angle, any three being given, to find the fourth. This case comprises three problems. . 1. When the three sides are given, to find an angle. Here from equa. (2) we have cosa —cosbcosc } coSA = * = sin bsince cos 6 — cosacose COS} =. sin asince cose — cosacosd oe 2 Ce Ce nein a J 90 Spherical Trigonometry. But these are not fitted for logarithmic computation. Recurring, therefore, to (chap. iv. equa. v’), we have 1 + cos A = 2 cos* 2A, and 1 — cos a = Q sin? 3a. Hence, i sin dsine cosa —cosbcose 2 cos ZA MARCIE, Ei sg ee ae sin & sinc sin} sinc cos a— cos (b + c) neers sindsince Hence also, 2 sin? Za = Sa cea ela ‘gin ésinc The latter of these divided by the former, gives, cos (b —c)— cosa cos a — cos (b + c) Whence, from the 4th of equa, (uv) chap. iv. sing (a+ 6—c)sind(a + c—8) sing (a + b + e)sing (b + ¢ — a) Hence we have, for the tangents of the half angles, these three symmetrical equations: _ sing(a+b—c)sing(a+e¢—d)) sin$(a + 6 + c)sing(b + c—a) sin $(a + b—c)sin$ (6 + c—a) tan 3B ae \/ ii +6+0c)sing(@+c— bd) e (11) sing(a+c—b)sing(6+ ¢e—a) Io — ‘Hela a Ss Dosis A ht dicta AN A sing (a +6 + c)sing(a + 5b—c) tan? sa = _tan?4Za = tan 3a = The expressions for the sizes of the half angles might be deduced with equal facility. As they are symame- trical we shall put down but one, viz. rope sin $ (a + 6—c) sing (a + ¢ —b) gb aay sin 6 sine Expressions for the cosines and cotangents or the _ half angles, may be readily found from the above, by n COs h fi m = phir — ® the forms cos 7s cot = Oe ne ge Oblique Spherical Triangles. 91 Cor. When two of the sides, as 6 and c, become equal, | sin $a sind the expression for sin $A, becomes sin}a = Cor.2. If a=b=c = 90°, then sin 4A = — as £,/2 == sin 45°; and a= B= 6 =190°, Leaving other corollaries to’'be deduced by the stu- dent, let us proceed to the next problem in this case. 2. To find the side c opposite to the given angle ¢€; that is, given two sides and the included angle, to find the third side. Find from the data a dependent angle ¢, such that tan @ = cosc tand ....(12.) | Substitute for cos c in the third of equa. (2) its value in this, it will become cos c= cosacosé + sinacosdtan¢ cos acos > + sinasin¢g = cosb (eee) COs cos 6 cos (a — >) oe. (13.) cos > ‘Note. The equa. (12) obviously reduces to , or, COSc = i e . J ——, COSC = » or cot ¢ cos c = cot 8; which is tan » tan b analogous to equa. (8). So that 4 is the hypothenuse and ? one leg of a right angled triangle. ‘The above transformation, therefore, is equivalent to the division of the proposed triangle into two, by an arc from the | vertical angle a falling perpendicularly upon the oppo- site side a. 3. To find the side a, not opposite to the given angle; 5, c, and c, being given. _ Here find 9, as before, by equa. (12): then from equa. (13) we have | osc cos? cos(a—¢) = = see (14) cos & Hence a is known by adding ¢. 92 Spherical Trigonometry. Case III. 34. Of these four things, viz. two sides a and c, and two angles B and c, one opposite, the other adjacent; three being given, to find the fourth. This case presents four problems. 1. Given a, c, B; to find c. - Determine an arc ¢’ by this condition, cotc pales ? ——_ = f cot c = cot ¢’ cos B, or coent = Ct PH...» C15.) Substitute this value of cot ¢ for it in the third of equa. (4); it will become sin a cot $ cos B = cosa cos B + sin B Cotc; (cot ¢’ sina — cosa)cosB sin B _ = (cot ?’ sin a — cos a) cot B;. cot B sin (a — +’) Ser laar MACON PO ea €6 sé (16.) sing Note. The equa. (15) is akin to equa. (8); showing that the operation here performed is equivalent to let- ting fall a perpendicular arc from the angle a to the base a; the subsidiary arc ¢’ being the segment adja- cent to the angle B. 7 2. Given B, c, c; to find a. | Here 9?’ must be found by equa. (15), and then from equa. (16) we have | ees whence cot c = or, COL C\= cot csin o! abel sated ol caer ve ee (17) Whence a becomes known. 3. Given B, C, a; to find c. Find a dependent angle ¢” by making cotc cot c = cosatan ?”, or —— = tan @”.... (18.) Substitute this value of cot c for it in the third of equa. (4), and it will reduce to cot acos (B — 9”) COU: C = Ty ORT maaan PEUER, ~ - o*h j e ~ Oblique Spherical Triangles. 93 #. Given a, c, c; to find s. Determine ¢” from equa. (18); then cos (B — ?”) = eh ls . (20.) From which g becomes known. cota - Case IV. ‘ 35. Of these four things, viz. the three angles and a side, suppose c, any three being given, to find the fourth. This comprises three problems. 1. Given the three angles; to find a side. Suppose the first of equa. (11) to be applied to the solution of the supplemental triangle, by changing a, 6, ec, and c, into a’, ’, c’, andc’, Then, to bring it back to the triangle proposed, Jet there be substituted for a’, b’, c’, and c’, the corresponding values 180° — A, 180° — B, 180° — c, 180° — c. Those equations will then ‘be transformed into the following, applicable to the pre- sent problem. cos$ (A+ B—C)cos$(A—B+ C) cotje= 4 /S ee eee eo] —cos $(B + 6 + A)cos$ (B+ C— A) cos$(B +c—a)cos$(A + B—C) 1,= PLESSIS Ae NAN ES EASY PEE cot 36 yf Sa ere eee I ) ‘ cos (A + C— B)cos$(B + C— A) vide Sen —cosi (A+B + c)cos$(A+B—Cc)J The following are the expressions for oe sines of the half angles, viz. cos3(A + B+ c)cos}(B + C—A)) Y sin 3a 13 — sinBsinc cost (A + B+ c)cos3(A + c —B) Be 8h me gfe eet eer (O88) 2 — sin ASIN C ‘ cos} (A + B + C)cos$ apne earn ehh Tn saa iden Matives Ry ia DES a a, be — sin ASINE $4. Spherical Trigonometry. Note. In these expressions, although the denomi- nators are negative, the whole fractions under the radi- cal are always positive. The expressions for the tangents and cosines are omitted, to save room, 2. Given A, B, c; to find ce. Here, by applying in like manner, the equations (12), (13), to the supplemental triangle, we shall have cot ? = cosctanB,... (23. from which the subsidiary angle ¢ may be determined; and thence cosBsin (A — ¢) cosc = sides (ated sin > 3. Given B, Cc, c: to find A. Find @ from equa. (23); then from equa. (24) there results, cosc sing sin (A — 4) = aes eps (25.) from which A is known. Section IY. On the Analogies of Napier. 36. These are four simple and elegant formule dis- covered by the celebrated inventor of logarithms, of which two serve to determine any two angles of a sphe- rical triangle, by means of the two opposite sides and their included angle; while the other two serve to de- termine any two sides, by means of the opposite angles and*their contained side. Thus, therefore, they toge- ther With equa. (1), will serve for the solution of all the cases of oblique spherical triangles. The investigation of these analogies may be given, as below. | If from the first of equa. (2), cos c be exterminated, there will result, after a little reduction, - Napier’s Analogies. 5 and by a simple permutation of letters, : cos Bsin¢ = cos dsina ~— coscsinb cosa: adding these equations together, and reducing, we have sin ¢- (cos A + cosB) = (1 — cosc) sin(a ts 0). Now we have from equa. (1) “sing” sind sinc sinA sinB pei c Freeing these equations from their denominators, and respectively adding and subtracting them, there results, sinc (sin A+ sins) =sinc (sin a+ sin), and sinc (sin A — sinB) = sinc (sina — sind). Dividing each of these two equations by the preced-~ ing, ESE will be obtained, sin A +sinB sinc sina + sin’ cosa +cosB” J1—cose” sin(a +4) ‘sin A—sinB sinc sina — sin & oe REE owe eee teeta Ps ameemeneen e cosA + COSsB l—cosc sin(a@+ d) Comparing these with the equations in arts. 24, Se 26, chap. iv. we shall have 1 ja 5 Ber 4 (a — db) tan3(A +B) =cotic. rE CLE 0) 1 Wid a yaa 1, sin 4 (a — b) tan 3 (A — 8B) = cotic. ~~ hot) (26.) The above equations expressed as analogies, are cos § (@ + b): cosk (a — 8):: cot ge: tan § (A + B) sin 3 (@ + 4):sin 3 j (a — b):: cot 4c: tan $(A — B). These analogies being applied to the a ips triangle, by putting 180° — A, 180° — B, &c. fora, 5 &c, we have cos $(A + B): cos 3 (A — B): : tan gor tang (a + 8) sin §(A + 8B):sin £(a — B):: tan $c: tan 3 (@ — 6). 37. From a due consideration of the four analogies it results, Ist. That 3 (A—B) < 90°, or that the difference of two anples of a shane: triangle i is less than 180°. A Ie- hte ore 86 Seherical Trigonometry. Qdly. That 3 (a + 4) and 4 (a + 8) are always of the same affection. 3dly. ‘That the difference of two sides is always less than 180°. 4thly. That (@ — 6)and (A — 8) ors always the same sign ; whence it follows, that the greatest angle is opposite to the greatest side, and reciprocally. To these it may be added, 5thly. That the least angle is opposite to the least side, and the mean angle, to the mean side. One or other of these observations will serve to re- move the ambiguity in the doubtful cases, where either a, b, and 8B, or A, B, and 4, are given. 38. We may now collect the mdst commodious theo- rems, and present in one place all that will be usually required 1 in the solution of oblique angled spherical tri- angles. sind sinB sinc ‘f sina sinb sin ¢ sind(a+b—c)sm$(at+e—B 2: tanga = 4 / eo en ee oe ; sing (6 + c—a)sing (a+ 6+) sing(6+c—a)sing(a+b—c 3, tan $B = RAS poms AE ech sing (a +c—b)sing(a+b+c} sing(a +c—b) sing (@+c— a Potala, See a sind (a+ b—c)sing(a+5 + c) : —cos4(B + c—A)cos$(A + B+C) §. tan ; PRS ER Y MES i ECO cos (A + B— 6) cos$(A + C—B) 6.-tan ag fae cconentins ns e 2 = cos #(B + C — A) Cosg(A + B— C) ”, tan } | She G ESD A gies ee cos$ (A + C— B)cos$ (B+ C— A). ee a 4a sin $ (B— 4) 8. tan == tan $c eee RE: b+ a 1/.00S$ (B + 4). ch 9. tan —j— = tan eT 4 (Gh A) , aw a aS . = eo “ts ‘ SA « er 10. 4; 12. 13. 1A. 15. 16. 17. 18. ate 90. tan $c = tan 1(b — a) “21. tan dc = tan 3(b + a) ——— 22. tana = tans(c — d 23. tan ga = tan }(c + 5) Summary of Formule. tan —— tan tan tan tan tan tan tan tan tan b 2 2 a-e Cc = tan 3a b = tan $a = tan 3 = cot dc Bis I = cotta = coti gh = = cot 3B = cot is 94. tan 4b = tan}(a — % sin £(c —B) sin $(c + B) €0s $ (Cc — B) cos $(¢ + B) sin (A —c) sin % (A + ©) +, cos$ (A — cc) cos $ (4 + 6) sin 3 (5 — a) sin (5 + a) cos $(b — a) cos $ (6 + a) sin 3 (c — d) sin §(c + 6) cos ¢ (c — 4) cos £ (6 + 5) sin § (a — c) sin § (a + c) cos $ (a — c) cos § (@ + c) sin $(B + a) sin $(B ~— A) cos §(B + A) COs. 3 (B — A) sin $(c + B) sin $ (Cc —B) cos § (c + B) cos § (Cc — B) ‘sin $ (A +6) ©) sin §(A — ©) 25. tan $5 == tan $(a + c) cos $(A +) cos $ (A — C} $8 Spherical Trigonometry. thas 194") sin $ (6+ a) 26. cot ic = tan 3(B— A) > Va) cos 4 (b + a) cos 4 (b — a) sin 2(c + b) sin 4(c — d) 27. cot ic = tan3(B+ A) 28. cot}a = tan}(¢ — B) cos $(c + 5) cos $ (c'— b) sin $ (a +0) cos § (a + c) 29. cotsa = tan 3(c + B) 30 cot 3B = tan $(A-— C) 31. cot 3B = tan $(a sac) - cos 3 (a — ¢) -SecTion Ve Mnemonics of Spherical Trigonometry. 39, As it is, obviously, difficult to retain in recollee- tion the necessary rules and formule in spherical trigo- nometry, attempts have been made by different mathe- maticians, to assist the student by contrivances akin to those, which occur in repositories of artificial memory. Napier, to whom this department of science is so much indebted, at the end of his Mirifict Canonis Con- structio, obscurely suggested a simple and comprehen- sive method, characterised by the name of- Napier’s Rules for the Circular Parts ; which apply, Ist, to right angled spherical triangles; 2dly, by means of the polar triangle, to guadrantal triangles; and, 3dly, by means of a perpendicular from the vertical angle, to oblique angled spherical triangles. ‘These rules were developed much more perspicuously by Gellibrand; and have, since his time, been explained by almost every writer on spherical trigonometry. 40. Ina spherical triangle anc, right angled at A, we have, as was shown in arts. 25—31, of this chapter, * - Napier’s Circular Parts. 98 sill c == sina sinc = tan 6 cots c sin 86 = sina sin B = tanec cotc 4 ; cosa = cosc cos 6 = cot Bcotc B 7 cos B = cosdsin c = tanc cota eat’ - os o J —_* gf. \ > > 4 — cos C = cosc sin B = cota tan b : These; equations include only five of the six parts of the triangle, the sixth part, viz. the angle a being con- stant. Now, Napier’s circular parts are, the comple- ‘ments of the angles B and c, the complement of the hypothenuse a, and the other two sides 6 and ¢: that is, 90° — a, 90° — B, 90° — c, d, and.c, are the five circular parts. Any one of these five may be called a middle. part; then the two parts next adjacent, one on the - right, the other on the left, [not including the right angle] are called adjacent parts; and the next two, each separated from the middle part by an adjacent part, are called opposite parts. . This premised, the rules are, 1. The rectangle of the radius and the sine of the. middle part = rectangle of the tangents of the adjacent parts. 9.,The rectangle of the radius and the sine of the middle part = rectangle of the cosines of the opposite parts, ' These rules may be comprehended in one expression, extremely easy to remember, simply remarking that the vowels in the contractions 572, tan, cos, are respectively the sdme.as those in the first syllable of middle, adja- cent, opposite: for then, regarding unity as radius, we shall have GER Re PEI LT BE “sin mid = rect tan adja =: rect-cosop. . - . A rule which, by taking a, 6, c, &c. successively for the middle part, will be found to comprise the five expres- sions given above. | 5D aU @ ve 41.-By the circular parts of an oblique spherical tri- angle are meant its three sides, and the supplements of its three angles... Any of these six being assumed as a middle part, the opposite parts are those two that are : . 9 | Lah: | 100 Spherical Trigonometry. of the same denomination with it; ‘that is, if the middle part is one of the sides, the opposite parts are the other two; and if the.middle part is the supplement of one of the anglés, the opposite parts are the supplements of tlie other two. en ee | 42. Mr. Walter Fisher has also given, in ‘the Trans- ‘actions of the Royal Society of Edinburgh, rules of easy ‘recollection which will serve for the solution of all the cases of plane and spherical triangles. ‘They are in- ‘cluded in four theorems, which may be applied to plane triangles by taking, instead of the sine or tangent of a side, the side itself. Theor. 1, Given two parts, and an opposite one. S+A3S.0':5 24:8. 0, | Theor. 2. An included part given, or sought. Aa. Ata Oo-0O M 2 ed MET PP ain eee gy = AND A Ata o— 0 oO + ( e Ve ° ad baa ar ° re Ne Et Theor B47 Taal 5 ay oe Theor. 4. Given the three ‘sides or angles of .an,ob- lique angled triangle. , : (Ata) +l 4 (At+a)—T om S.,AX S.a4:R73268, 2 i 2 Ta ae) Here m denotes the middle part of the triangle, and must always be assumed defween two given parts. It is either a side or the supplement of an angle; and is sometimes given, sometimes not. | A and a are the two parts adjacent to the middle, and of a different denomination from it. o and o denote the two parts opposite to the adjacent parts, and of the same denomination with the middle art. A 1 is the ast, or most distant part, and of a different denomination from the middle part. die That these four theorems may be called to mind with the greater facility, the following four words formed by - Spherical Eacess, 10] abbreviating-the terms of the respective analogies should, be committed to memory, viz. Sao, satom, tao, sarsalm. Szctrion VI. On the. Areas of Spherical Triangles and Polygons, and the, Measures of Solid Angles, Theorem I. 43. In every spherical triangle, the following propor- tion obtains, viz. as four right angles, (or 360°), to the surface of. a hemisphere; or, as two, right angles (or 180°), to a great circle of the sphere; so is the excess of the three angles of the. triangle, above two right angles, to the area of the triangle, Let axzc be the spherical triangle. Complete one,of its sides, as BC, into the circle BCEF, __ aie Se which,may-be supposed to bound the upper hemisphere. Prolong also, at hoth ends, the two sides aB, Ae, un- til they form semicircles estimated from each angle, that is, until BAE = ABD =CAF= acp = 180° Then will opr = 180%= Bre; and conse- quently the triangle arr, on the an- 7 terior hemisphere, will be equal tothe triangle Bcp on the opposite hemisphere, Putting‘, mm’, to represent the surface of these triangles, » for that of the triangle BAF, g for that of cax, and a for that of the proposed triangle anc. Then a and m’ together (or: their equal a andm together) make up the surface of a spheric lune comprehended. between the, two semicircles ACD, ABD, inclined in the angle a: @ and p together make up the lune included between the semicircles CAF, CBF, making the angle c: a and g together make up the spheric lune included between the semicircles BcE, BAE, making the angle B. And the surface of each of these lunes, is to 102 Spherical Trigonometry. that of the hemisphere, as the angle made. by the eom- prehending semicircles, to two right angles. ‘Therefore, putting 4s for the surface of the hemisphere, we have 180°: Az: 4s:a+ m. 180° 9Bi:358 2a q. 180°: ci:48:a +p. | Whence, 180°:a+B4c::$s:3a-+m+ p+ 9 =2a+ 9s; and consequently, by division of proportion, as 180°:a +B+cC—180°::48:2a + $s — fs = 2a; a A+B+c—180° Ve Te: O 66 e cD | ' or, 180 cA + B | C= 180 A ce a = 98° 360° : QF.-D.,* Cor. 1. Hence the excess of the three angles of any spherical triangle above two right angles, termed tech- nically the spherical eacess, furnishes a correct measure of the surface of that triangle. . Cor. 2. If # = 3°141593, and d the diameter of the gaa RN Ee Boe 3 Fale sphere; themis wa ; Se enone + =' the area of the spherical triangle, SA Cor. 3. Since the'length of the radius, in any circle, is equal to the length of 57:2957795 degrees, measured onthe circumference of that circle; if the spherical ex- cess, be multiplied by 57'2957795, the-product. will, ex- press the surface of the triangle in square degrees. — ~ - Cor, 4.| When a = 0, then A+ 8 + c = 180°: and when a = $s, then A.+.B.+ Cc = 540°. Consequently the sum of the three angles of a spherical triangle, 1s always between 2-and 6:right angles: which is another confirmation of art. 19, p.83. On Ao! x Pais determination of the area of a spherical triangle is due’ to Albert Girard (who died about 1633), — But the’ demonstration now commonly given of therule was first published by Dr, Wallis.’ It was considered asa mere speculative, truth, until General Roy, in 1787, employed it very judiciously in the great Trigonome- trical Survey, to correct the errors of spherical angles, See Phil, Trans. vol, $0, and chap. xii, of this volume, Spherical Excess. 108 Cor. 5. When two of the angles of a spherical tri- angle are right angles, the surface of the triangle varies with ‘its othird angle. -And when a spherical triangle has three right angles its surface is one-eighth of the surface of the sphere. |‘ Remark, The mode of finding the spherical excess, and thence the area when the three angles of a spherical triangle are-given, is obvious enough; but it is often requisite’ to ascertain it by means of other data, as, when two sides and:the included angle are given, or when all the three sides are given. Inthe former case, let a and 6 be the tio sides, c the included angle, and x the sphe- « When the three sides a, 4, c, are given, the spherical excess may be found by the following very elegant theorem, discovered by Simon-Lhuillier: re : cot a. cot $b + cosc rical excess: then is cot $2 = —*—————— Si ei rar a+b—e a—-b+e Y E= I aie carta a Gee ORS RRO Cee tan } / (tan —> tan — tan ——| —at+b+e The investigation of these theoroms would eecupy more space than can be allotted to them ‘in the present volume. Theorem Il, ie 44, In every spherical polygon, or surface included by any number of intersecting great circles, the sub- joined proportion obtains, viz. as four right angles, or 360°, to the surface of a hemisphere ; or, as two right angles, or 180°, to a great circle of the sphere; so is the excess of the sum of th xe ‘angles above the product of 180° and two less than the number ot’ angles Berne spherical polygon, to its area For, if the polygon be supposed to be divided into as many triangles as it has sides, by great circles drawn from all the angles through any point within it, forming: 104 Spherical Trigonometry. at that point the vertical angles of all the triangles. Then, by theor, I, it will be as 860°: 45:3: A+ B+ 0 — 180°: its area. Therefore, putting P for the sum of all the angles of the polygon, » for their number, and v for the sum of all the vertical angles of its constituent triangles, it will be, by composition, as 360°: $s :: Pp + v— 180°. : surface of the polygon. . But vis manifestly equal to 360° or 180°x 2. Therefore, —(n—2 as 360°: Is::P — (n — 2) 180°: 4s. the. 360° area of the polygon. Q.E.D Cor.1. If and d represent the same quantities as in theor. 1, cor. 2, then the surface of the polygon will be ws wey ie) expressed by zd?. —— Cor. 2. If R° = 572957795, then will the surface of the polyg on in square degrees be = n°. [Pp — (n — 2) 180° “Cor. 3. When the surface of the polygon is 0, then p= (m — 2) 180°; and when it is a maximum, that j ig. e when it is equal to the surface of the hemisphere, then Pp = (n — 2) 180° + 360° = n ..180°: consequently v, the sum of all the angles of any spheric polygon, is always less than 2n right angles, but greater than (2n — 4) right angles, n denoting the number of angles of the ‘polygon, Nature and Measure of Solid asap - 45, A solid angle is defined by Euclid, that which is made by the meeting of more than two plane angles, which are net in the same plane, in one point. Others define it the angular space comprised between several planes meeting in one point. It may be defined still more generally, the angular space included between several plane surfaces or one or more curved surfaces, meeting in the point which forme | the summit of the angle. Solid Angles. 105 According to this definition, solid angles bear just the same relation to the surfaces’ which comprise them, as plane angles do to the lines by which they are included: so that, as in the latter, it is not the magnitude of the lines, but their mutual inclination, which determines the angle; just so, in the former, it is not the magnitude of the planes, but:¢hezr mutual inclinations, which de- termine the angles. And hence all those geometers, from the time of Euclid down to the present period, who have confined their attention principally to the magnitude of the plane angles, stead of their relative positions, have néver been able to develope the pro- perties of this class of geometrical quantities; but have affirmed that no solid angle can be said to be the half or the double of another, and have spoken of the bisection and trisection of solid angles, even in the simplest cases, as impossible problems. But. all this supposed difficulty vanishes, and the doc- trine of solid angles becomes simple, satisfactory, and universal in its application, by assuming spherical sur- faces for their measure; just as circular arcs are as- sumed for the measures of plane angles.* Imagine, that from the summit of a solid angle (formed by the mecting of three planes) as a centre, ¢ any sphere be de- * This disquisition on solid angles was first published in the 3d volume of Dr. Hutton’s Course of Mathematics, in the year 1811. At that time I thought the notion of measuring this class of geometrical magnitudes by means of spherical triangles and polygons was, though extremely obvious and natural, perfectly new. I have since found; by consulting Montucla’s ‘History of Mathematics, yol. ii, p. 8, that Albert Girard, in his Invention nouvelleen Algébre, advanced an analogous theory. "While I am adverting to the third volume of Dr. Hutton’s Course, I beg to'mention,’ in order'to account for any instances of close correspondence which: may be found between parts of this volume and of that (though they, I believe, will occur quite o6 seldom as can well be expected when the same person is treat- of the same subjects), that the 2d, 3d, 4th, 5th, 6th, Tth, Sth, oth, and 11th chapiers of that volume were composed by me; the remainder by my excellent friend the author of that Course. ox FS 106 Spherical. Trigonometry. scribed, and that those planes are produced till they’cut the surface of the sphere; then will the surface of the spherical triangle, included between those planes, be a proper measure of the solid angle niade by the planes at their common point of meeting: for no change can be conccived in the relative position of those planes, that is, in the magnitude of the solid angle, without a corresponding and proportional mutation in the surface of the spherical triangle. If, in like manner, the three or more surfaces, which by thei meeting constitute another solid angle, be produced till they cut the sur- face of the same or an equal sphere, whose centre coin- cides with the summit of the angle; the surface of the spheric triangle or polygon, included between the planes which determine the angle, will be a correct measure of that angle. And the ratio which subsists between the areas of the spheric triangles, polygons, or other sur- faces thus formed, will be accurately the ratio which subsists between the solid angles, constituted by the meeting of the several planes or surfaces, at the centre of the sphere, sf Hence, the comparison of solid angles becomes a matter of great ease and simplicity: for, since the areas of spherical triangles are measured by, the excess of the sums of their angles each above two right angles (theor. 1); and the areas of spherical polygons of n sides, by the excess of the sum of their angles above 2n — 4 right angles (theor. 2); it follows, that the magnitude of a trilateral solid angle, will be measured by the excess of the sum of the three angles, made respectively by its bounding planes, above two right angles; and the mag- nitudes of solid angles formed by bounding planes, by the excess of the sum of the angles of inclination of the several planes above 2n — 4 right-angles. . As to solid angles limited by curve surfaces, such: as the angles at the vertices of cones; they will manifestly be measured by the spheric surfaces cut off by the pro-_ Jongation of their bounding surfaces, in the same man- Solid Angles. 107 ner as angles determined by planes are measured by the triangles or polygons, they mark out upon the same, or an equal sphere. In all cases, the maximum limit of, solid angles, will be the plane towards which the various planes determining such angles approach, as they diverge further from each other about the same summit; just i as a right line is the maximum limit of plane angles, being formes by the two boundi ling lines when they make an angle of 180°. The maximum limit of solid angles is measured by the surface of a hemisphere, in like man- ner as the maximum limit of plane angles is measured by the arc of a semicircle. The solid right. angle (either angle, for example, of a cube) is 4. (= 47) of the maximum solid angle: while the plane right angle is half the maximum plane angle. The analogy between plane and solid angles pene thus traced, we may proceed to exemplify this {1 1eory by a few instances; assuming 1000 as the numeral mea- sure of the maximum solid angle = 4 times 90° solid ='360° solid. 1. The solid angles of right prisms are compared with great facility. For, of the three angles made by the three planes which, by their meeting, constitute every such solid angle, two are right angles; and the third is-~ the same as the corresponding plane angle of the poly- gonal base; on which, therefore, the measure’ of the solid angle depends. Thus, with respect to the right prism with an equilateral triangular base, each solid angle is formed by planes which respectively make angles of 90°, 90°, and 60°.. Consequently 90° + 90° + 60° — 180° = 60°, is the measure of such angle, compared with 360° the maximum angle. It is, there- fore, one-sixth of the maximum angle. A right prism with a square base has, in like manner, each solid angle measured by 90° + 90° + 90°,— 180° = 90°, which ia 4 of, the maximum angle. Aha thus it may be found, that each solid angle of a right prism, with an equi- ~ lateral 108 Spherical Trigonomeiry. triangular base is 1 max. angle = 4 . 1000; square base ig py base = 2. 1000. pentagonal baseis: ........... = $5. 1000. hexagonal: igh... .0 0... 4.2000 heptagonal BSIG8 Syl N ae Oe octagonal IBS Ee Tyas Be 41000: nonagonal WS), AY Clee, 10007 decagonal RYO U8 ee 8 1000 undecagonal’ )is) ie. = 9%. 1000. duodecagonal is3%......... 242. 1000. m gonal ey Tee ns 1000. Hence it may be deduced, that each solid angle of a regular prism, with triangular base, is half each solid angle.of a prism with a regular hexagonal base. Each with regular square base = § of each, with regular octagonal base, pentagonal = 2 .......ce0s0---- decagonal, hexagonal = 3 ................ Guodecagonal, m—4 : 1. peo ¢ 5m gonal Sot ttt eeeeeese m gonal base. Hence again we may infer, that the sum of all the solid angles of any prism of triangular base, whether that base be regular or irregular, is half the sum of the. solid angles of a prism of quadrangular base, regular or irregular, And, the sum of the solid angles of any prism of tetragonal baseis = }sum of ang. in prism of pentag. base, pentagonal ii). eRe de Ue Bea. OG, hexagonal, hexagonal »)."). 004. Bi eit se ati heptagonal, — m— rs gonal eer re © © Tisai ar it erveee er ee Oe (m+ 1)gonal. 2. Let us compare the solid angles of the five regular podies. In these bodies, if m be the number of sides of each face; » the number of planes which meet at each solid angle; 1Q = half the circumference or 180°; and a the plane angle made by two adjacent faces; > Solid Angles... 109 cos on O ee | | sin a5; O for the plane angle formed by every two contiguous faces of the tetraédron, 70° 31’ 42”; of the hexaédron, 90°; of the octaédron, 109° 28’ 18”; of the dodecaé- dron, 116° 33’ 54’; of the icosaédron, 138° 11’ 23”, But, in these polyedra, the number of faces meeting about each solid angle, are 3, 3, 4, 3, 5, respectively. Consequently the several solid angles will be determined by the subjoied proportions : then we have sin 44a = + This theorem gives ;3 SiVes, Solid Angle. 360° :3.70°31’42” —180°::1000: 87°73611 tetraédron. 360°: 3.90° — 180°:: 1000:250: hexaédron. 360°: 4.109°28'1 8” —360°:: 1000: 216°35185 octaédron. 860°: 3.116°33’54” —180°:: 1000:4.74:395 " dodecaédron. 360°:5 .138°1 1°23” — 540° :: 1000:419°30169 icosaédron. 3. The solid angles at the vertiees of cones, will be determined by means of the spheric segments cut off at the bases of those cones; that is, if right cones, instead of having plane bases, had bases formed of the segments of equal spheres, whose centres were the vertices of the cones, the surfaces of those segments would be measures of the solid angles at the respective vertices. Now, the surfaces of spheric segments, are to the surface of the hemisphere, as their altitudes, to the radius of the - «sphere; and, therefore, the solid angles at the vertices of right cones will be to the maximum solid angle, as the excess of the slant side above the axis of the cone, to the slant side of the cone. Thus, if we wish to as- certain the solid angles at the vertices of the equilateral and the right angled cones; the axis of the former is $4/3, of the latter, 34/2, the slant side of each being unity. Hence, eg hi, 27 Angle at vertex, Lt) — $,/3 :: 1000: 13307464, equilateral cone, ~ = Val — 3.4/2: 1000 : 29289322, right angled cone, 194. Voie pp MoCo AE said Gat F WE e (fas de aes $4 us a fae furs p pre Lee 8 ee RE 110 Spherical Trigonometry. 4, From what has been said, the mode of determining the solid angles at the vertices of pyramids will be suffi- ciently obvious. If the pyramids be regular ones, if N be the number of faces meeting about the vertical angle in one, and A the angle of inclination of each two of its plane faces; if m be the number of planes meeting about the vertex of the other, anda the angle of inclination of each two ofits faces: then will the vertical angle of the former, be to. the vertical angle of the latter pyramid, as NA — (N — 2) 180°, to na — (n — 2) 180°. If acube be cut by diagonal planes, into six equal pyramids with square bases, their vertices all meeting at the centre of the circumscribing sphere; then each of the solid angles, made by the four.planes meeting at each vertex, will be } of the maximum solid angle; and ~ each of the solid angles at the bases:of the pyramids, K will be .{, of the maximum solid angle. ‘Therefore, each solid angle at the base of such pyramid, is one- fourth of the solid angle at its vertex: and, if the angle at the vertex be bisected, as described below, either of the solid angles arising from the bisection, will be dou- ble of either solid angle at the base. Hence also, and from the first subdivision of this inquiry, each solid angle of a prism, with equilateral triangular base, will be half each vertical angle of these pyramids, and doudle . each solid angle at their bases. ; | | The angles made, by one plane with another, must be ascertained, either by measurement or by computation, according to circumstances. But, the general theory being thus explained and illustrated, the further appli- cation of it is left to the skiil and, ingenuity of geo- meters; the following simple examples, merely, being added here... ©. A i Ex. Let the solid angle at the vertex of a, square pyramid be bisected. Pisin Ist. Let a plane be drawn through the vertex and any two opposite angles of the base, that plane will bisect the solid angle at the vertex; forming two -trila- ; G 4 ip eee 12 2 © Oh AE SS et Pa ks U pA page? {! Pit, Sata nM he of «athe "or. Seman 5 Baten ae in ss iM Fe ee ee ve ora yal KEE Mg6. at Ef OF sts Ax PLb a 20os Z a we Lg te A — ¥ SO Sek OE a Be ~ above ZOL = % va eee a ; Dien we eee eee e Fle cate Loby, ” Be ts ui ee 7 f e r a a : eh fe Logarithmic Comtputations. Lit teral angles, each equal to half the original quadrilateral angle, . Qdly. Bisect either diagonal of the base, and draw any plane to pass through the point of bisection’and the vertex of the pyramid ; ‘such plane, if it do not coincide with the former, will divide the quadrilateral solid angle into two equal quadrilateral solid angles. For this plane, produced, will bisect the great circle diagonal of the spherical parallelogram cut “off by the base ‘of the pyra- mid; and any great circle bisecting such diagonal is known to bisect the spherical parallelogram, or square ; the plane, therefore, bisects the solid angle. ; Cor. Hence an indefinite number of planes may be drawn, each to bisect a given quadrilateral solid angle. Ex. 2. Determine the solid angles of a regular pyra- mid with hexagonal base, the altitude of the eet being to each side of the base, as 2 to 1. Ans. Plane angle between each two la+ teral taces* /S2 aS. 22 195° 29" SEY between the base and PAU LACE « cer oes oe a 66° 35° 12”. Solid angle at the vertex 89°60648 7 The max. angle . Each ditto at the base .. 218-19367 being 1000. CHAPTER ‘VII. Logarithmic Computation of Sphericat Triangles. 1. KOR the putposes of.exemplifying the rules and formule in the preceding chapter, and of assisting the student in deducing the logarithmic computations from the analytical expressions, a few ie silts are here added, ; ia ay | 9, *4¢ Lhd a: V4 te i 45 | a mie ee Oth. ioe Yensene {12 + Spherical Triangies. | Example I. _ In the right angled spherical a ABC, right an- gled at A, given the hypothenuse a = 64° 40’, and one leg b = 42°12’, Required the rest. This example falls under case 2, section 2, i the PERSIE, chapter, where sin given side | Ss oe == in B, a rise ay gle: sin hypoth, cos hypoth, 4 cos ¢, the side req. = cos C, a req. angle = tan given side x cot hypoth. Hence the following logarithmic operation, From log sin 42° 12’. . 9-8271887] From log cos 64° 40’. .9:6313258 Take log sin 64° 40’, .9°9560886| Take log cos42° 12’, 98697037 cos given side —_— Pear en ee Rem, sin p. 48° 0/..9°8711001|/Rem. cos c. .54° 43’. .9'7616221 | 3 To log tan 49° 12", ... 9°9574850 Add log cot 64° 40’ .... 9°6752372 a The sum is log cos c 64° 35’... 96327 222 Here 10 are added to the index of the remainder, and taken from the index of the sum; conformably with note 2, art. 31 of the preceding chapter, odly. To compute the same by means of Napier’s ctr- cular parts. Here, if the leg 4 be aicumed as the middle term, (90° — a), and (90° — %) are the opposite parts. and sin mid, = rect cos op. becomes, ‘sin 6 = sin-a sin B: sind ___ sin given fe . sina sin shypoth.< this agrees wilh the foregoing process, Again, if (80° ¢) be the middle part, then (90° — a) and’ are adjacentyparts, and — ” sin mid, = rect tan adja. becomes Spo X45 OD N« ton x BA on = ge : pn pan = LAE a =. it, scale ee ge? ei A YE aod og ’ a ym pid Logarithmic Computations. us cos c = cota tan 6 = cot hypoth. x tan given side; this also agrees with the preceding. 3dly. If (90° — a) be made the middle part, then 4 and c are the opposite parts, and sin mid. = rect cos oppos. becomes COS & cos a = cos 6 cosc; whence cos c = ——, cos 6 cos hypoth. that is, cos c = ; which also agrees with the cos given leg preceding. Hence the logarithmic computation need not be repeated, : f Example Vi. Given in a spherical triangle, the quadrantal side cA = 90°, an adjacent angle c = 42° 12’, and the op- posite angle B = 115° 20’; to find the other angle and sides. Suppose the side cs produced until cp = ca = 90°: then, it is evident from chap. vi. art. 33, | that both the angles, cap, and_p, are = 90°, and consequently that ap is the measure of the angle c, and therefore = 42°19’, It is also evident that aBn, as well as acp, is a right angled tri- angle, and that 7 ABp = 180° — aBc = 64° 40’.. Hence, to find as, &c. we make-use of the triangle ABD, of which we determine the hypothenuse fas oblique angles by case 4, of right angled triangles, hus, , sin hypoth, = sin given side | . cos given angle er een. * SID FOUL Ae eee et sin op, angle cos given side From log sin 42° 12’, .9'8271887| From log cos 64° 40. .9°6313258 Take log sin 64° 40’. .9°9560886|Take log cos 42° 12’. .9°8697037 Rem. sin as .. 489 ., 9°8711001|Rem. sin BAD 35°17’ 0 9.» % ~| or.coscaB 54°43’ pene Sin sin side req. == tan given side x cot op..angle Z 5 pepe patty Cre fin PF , bea Phebe» CY ot at Te ¥ le ff UMS aahille og Ore i “4 1t4 ; Spherical 1 riangles. To ‘log.tan'49°'19’ =. |, 9:9574850 Add log ‘cot 64° 40°, ).9°6752372 The sum is sin pp 25° 25’ vor cos BC 64°. 96/0 72» 26327222 2 Example IXK. Ina richt angled spherical triangle given the hypo- thenuse = 64° 404 andsan adjacent angle =. 64° 35’; to-find the rest. f pod Example lV. Given one leg = 42° 12’, and its opposite angle = * 48°; to find the rest. ° sats ‘Example V.- . Given,a leg = 54°43’, and its adjacent angle = 48°; to find the rest. 22-30B by. uy | Example VI, Given the two legs = 54° 13; and 42° 12, reapoete wely;\to find the rest... Example VU. Given the two oblique angles ='48° and 64° 35’ re- spectively; to find the rest, | PG Example VIII. “Given-a quadrantal side, one of the other sides = 115°:9',,and the angle comprehended between them = 115° 55‘; to find the rest. ere | 4 Ans.-Angles 101° 4’ and 117°34%, side 118° 18’, a scons Example 1X. | Given in an oblique angled spherical triangle, the Logarithmic Compilations. 115 side @-= 44° 19°45”, 6 = 84° 14729”, and their in cluded angle c = 36° 45’ 28”; te find the rest. ae example corresporids with case 2, prob. 2, of oblique angled spherical trian- gles; and may first be solved by means of the subsidiary arc, in the manner there ex- plained. Thus, first find ¢,,so0 that tan ¢ = cos € tan 6, To log*cos:c¢ = cos'36°45=28" x: ..) 9:9037261 Add tan 0 = tan 84° 14’ 29”... 10°9963395.. tan p= tan 82° 49 33”... 10:9000656 ! —— This arc 9 éxceeds a, therefore the perpendicular AD from the vertical’ angle falls’ on ‘the base produced: hence the ad expression ‘becomes © °’ | cos bcos (aw ¢) cos we ety fut eol i “Hence, to log cos 84° 14’ 99” .... soteena : add cos 36° pit 487 0.4. 918929604 fare the § SUM ....02000+ 18°3944236 take. cos 82°.49'.33”.... °9:0965132 Rem. cos = cos 51° 6 11”.... 9°7979104 To find the remaining parts use the known propor- tion of the sines of ‘sides to the sines of their opposite angles ; thus AS sine. ..20..0 SI? 6” 11”... 98911340 SOSIPOS Siccee ss 56°. 45°25" 5. OEE OLoS So-is‘sin a. 0 ).. 44°18° 45 5.2. 9°8435629 PPOPRI Ao ave 82° 267° 7 5. 5 9°T 2944.47 ‘And so issin-6.. 84°14" 29”... .9:9978028 aT Sin B20 12, 130° 8 91". 24. 9°8836846 Here the logarithmic sine 9° 8836846 answers either to 49° 54’ 39” or to its supplement 130° 5’ 21”; the 116 _ Sphericat Triangles. former of which is the exterior angle ABD, the latter the angle B of the triangle. 2d Method, by Napier’s. Analogies. Taking the 14th and 15th formule at the end of sect. 4, of the preceding chapter, we ny 2 (b — a) sin 2 (6 + a) ? cos 3 (b — a) cos 3 (6 a) The log. computation will therefore stand thus: tan 3 (B— ae = cot hc = and tan 3(B+ ‘<“ = cot he To log cot 3c....... 18° 29’ 44.7... . 10°4:785395 Add log sin 3 (6 — a) 20° 0'92”,... 9°5341789 ‘From the sum Shihan s erdhictin hod ae 0127184 Take log sin J (6 + a) 64°14" 77 9:9545255, Rem. log tan 3 (BA) 48° 49’ 38” .... 10°0581929 Also, to log cot ZC .. 18° 29’ 44”... 10°4785395 Add log cos $ (2 - _ a) LOCAL Do yon DOL AI0G0 From the enn pins Asa’ es Node Ol Gs Berd oon 30°4:515085 Take log cos 4 (6 a. a) 64° i 7” .... 96381663 Rem.log tan} (p+ a) 81° 15’ 44”... 10°8133422 Hence 81° 15’ 44” 4: 48° 49/38" = 130° 5/.99”"=.», and 81° 15’ 44” — 48°49’ 38” = 32°96" 67 = A; agreeing aes Fs with the: result, of the former compu- tation, 7 © © % Hest it al seem, oe a B siunad of .. me- thods, that the first is rather quickest in operation, while the last is probably the easiest to, remember, and pro- vides best against the eccasions of ambiguity. ; " Astronomical Definitions. 117 te : ; Example X. _ In an oblique angled spherical triangle azc, given the ‘side c = 114° 30’, side a = 56° 40’, and the angle c op- posite the first side = 125° 20’: to find the rest. 4 Ans. A = 48° 30’, B = 62° 54’, 6 = 83° 12’. Example XI. | Given a = 48° 30’, c = 125° 20’, ¢ = 114° 30’; to find ‘the rest. Example XII. Given @ = 56° 40’, c = 114° 30’, B = 62°. 54; to find the rest. | | | Example XUII. Given A = 48° 30’, c = 125° 20, 6 = 83° 12; to ‘find the rest. | | Example XIV. . Given a == 56°40, 6 = 83°12’, c = 114° 30’; ‘to find the rest. Example XV. ‘ “Given a ‘= 48°30, 5 = 62° 54’,'c == 125° 20’; to find the rest. *,* For more examples see chap. x. CHAPTER VIII. ‘On Projections of the Sphere. Section I, Astronomical Definitions. - 1 SINCE the figure of the earth differs but little from that of a sphere, it is usual in the greater part of the 1138 Projections of the Sphere. inquiries and computations of astronomers, to proceed as though it were a sphere in reality; and since, to an observer on the earth, the heavens appear, as a very large concave sphere, every part of which is equidistant from him, it has been found expedient to imagine vart- ous lines and circles to be described upon the earth, and the planes of several-of them to be extended every way until they mark other similar lines and circles upon the imaginary concave sphere of the. heavens. Some of these it now becomes necessary to explain, 2...Uhe axis of the earth is an imaginary right line passing through the centre, about which-ine it is sup- _ posed to turn uniformly once in a natural day... 4 3. The extremities ofthis axis are called the poles of the.earth. .. < _ go nar 4. That great circle of the earth, the poles of which are the poles of the earth, is called the equator. = 5. If the axis of the earth be supposed produced both ways to the concave heavens, it is then called the aais of the heavens ; its extremities are called the poles of the heavens; and the circumference formed by extend- ing the plane of the equator to the celestial concavity is called the celestzal equator, or thie eguincctial. 6. A secondary to the equator drawn through any place on the earth, and passing through the poles, is called the merzdzan of that place. 7. The latitude of any place upon, the surface of the earth, is its distance from the equator measured on an arc of the meridian passing through it. A less circle passing through anyplace. parallel to the equator is called a parallel of latitude. Places that lie between the equator and the north pole have north latitude; if they lie between the equator and the south pole they have south latitude. 8. All places that lie under the same meridian have the same /ongitude; and those places which lie under different meridians have different longitudes. The dif Jerence of longitude between any two places, is the dis- Asironomical Definitions. 119 tance of their meridians measured in degrees, &c. upos the equator. 9. The sensible horizon is a circle, the plane of which is supposed to touch the spherical surface of the earth, in the place of the spectator whose horizon it is. The rational horizon is a circle whose. plane passes through the centre of the earth, parallel to the plane of the sen- sible horizon. The radius of the earth being exceed- ingly minute compared with that of the celestial sphere, the sensible and rational horizon may, in many astrono- mical inquiries, be supposed, without error, to coincide. 10. Great, circles.which are drawn as secondaries to _the rational horizon, are. called vertical circles ; they serve to measure the altitude.or the depression of -any celestial object. 11. The two points in which all the vertical paohee that can be drawn to any rational horizon meet, are called, the one above the spectator the zenzth, and that which is below him the nadir, 12. Admucantars, or parallels of altitude, are circles parallel to the horizon, or whose poles are the zenith -and nadir. All the points of any one almucantar.are at equal altitudes above the horizon. 13. The real motion of the earth about the sun once in a year, gives rise to an apparent motion of the sun about the earth in the same ihterval of time... The circle in which the sun appears to move is called the ecliptic; the angle in which it crosses the equinoctial the obliquity of the ecliptic ;*. and the two points where it intersects that circle, the eguinoxes. | * The obliquity of the ecliptic is.a variadle quantity, oscil- lating between certain limits which it never passes, According to the profound investigations of Laplace in physical astronomy, the obliquity may always be-determined very nearly by this for- mula, viz. 23° 28/ 23-05 — 119F"2184 [1 — cos (¢ porters — 3347-0496 sin (¢ 32’°11575) 7 where ¢ denotes the number of years run over from 17503 it ts 120 Projections of the Sphere.. 14. The distance of the sun, or of any of the hea- venly bodies, from the equator, measured -on an are of the meridian, is called the declination ; north or south, accerding as the body is situated north or south of the equator. . 15. Secondaries to the celestial equator are called circles of declination ; of these, twenty-four, whieh di- vide the equator into equal parts of 15° each, are called | hour circles ; because the sun in his apparent diurnal motion passing over 360° of a circle parallel to the equator, goes through 54th of them, or 15°, in an hour. 16. The right ascension of a-celestial body is an arc of the equinoctial, intercepted between one of the equi- moxes, and a declination circle passing through the body; it is measured according to the order of the sun’s apparent motion through the twelve signs. 17. The longitude of a heavenly body is an arc of the ecliptic, contained between the Ist point of Aries (that is, one of the equinoctial points), and a secondary to the ecliptic, or a circle of latitude passing through the body. 13. The latitude of a body is its distance from the ecliptic measured upon a secondary to that circle. And the angle formed at the body by two great circles, one passing through the pole of the equator, the other through the pole of the ecliptic, is called the angle of — position. 19. The tropics are two circles parallel to the equi- noctial, and touching the ecliptic at the two points where it is most remote from the equator ; that is tosay, the first points of Cancer and of Capricorn; the former is denominated the tropic of Cancer, the latter the tropic of Capricorn. negative before, positive after, that epoch. This theorem is. found to answer very well up to the time of Pytheas, 350 years before the Christian era, The obliquity at the beginning of 1816 is 23° Q7/ 49/2, ‘Astronomical Definitions. 121 20. The points where the tropics touch the ecliptic are called solstices, because the sun when in either of them appears to be at a stand with regard to his decli- nation. 21. Colures are two secondaries to the equinoctial : one, passing through the equinoctial points, is ealled the equinocttal colure; the other, passing through the sol- stices,is called the solstitial colure. 22. Small circles. drawn at the distance of 23° 28’ (or correctly 23° 27’ 49”) from the north and south poles of the equator, are called polar circles ; the former the arctic, the latter the antarctic circle. 23. That vertical circle which intersects the meridian of any place“at right angles, is called the prime vertical : the points where it cuts the horizon are the east and west points; atthe distance of 90° from each of these on the horizon are the north and south; all four being called cardinal points. 24. The distance on the horizon of a vertical circle that passes through any body from the north or south . points is the azimuth of that body; the distance of the same circle from the east or west points is the ampli- tude. 25. In order to represent on a plane the celestial sphere with all its circles great and small, the ancients invented two kinds of projection. The first, named by them Analemma, has since received the name of Oriho- graphic Projection. The second, originally denoted by the generic term Planisphere, received from the Jesuit Aguiton the name of Stereographic Projection. The adjective orthographic is given to the former, because it is produced by lines which fail at r7ght angles upon the plane which represents the sphere, That of stereogra- phic was given to the other, because it results from the intersection of two sclids, a sphere and a cone. e7 122 Projections of the Sphere. SECTION II. Orthographic Projection. 96. In this projection the eye is supposed. at an infi- nite distance: in which case a great circle BCDF is ap- parently reduced to a right lime F equal to its diameter BD, the eye being imagined indefinitely dis- tant in the direction’ sc. Then, ‘also, every arc cA which has its B D origin at the apparent centre, has ‘for its projection a right line sa viys orga equal to the sine of that arc. The quadrant cB or cp will be pro- jected into its sine, or radius. An‘arc as AE = CE: cA, will be projected into: ae, = since — sin CAi= (by equa. u, chap. iv.) 2sin § (cz — Ca) cos 3.(CE + CA) = 2sin } Arcos (cA + $Ak).... (1s) 27. Every arc, as DA, from the edge of the disc to- wards the centre, is projected into its versed sine Da. The quadrant ne, therefore, is projected into the radius ps the versed sine of 90°; and the semicircle pcB into the diameter pz, the versed sine of 180°. 28. What is here remarked of the circle Bcp¥F applies equally to all great circles which intersect at s and form the visible hemisphere: each of these semicircles-is re- duced to its diameter, and the hemisphere is reduced to a disc. 29. In this projection every circle, great or small, whose plane prolonged does not. pass through the eye, will be seen obliquely, and under an elliptical form: for an oblique circle making throughout the same angle with the plane of projection, its several parallel ordi- . “nates are all reduced in a constant ratio ;: therefore, the projected ordinates are all in a constant ratio to the corresponding ordinates of the circle of equal diameter oN Orthographic Projection. _ 123 on the plane of projection, and together constitute an ellipse. (Hutton on Ellipse, prop. 3, cor. 1). } 30. In general, in the orthographic projection every great circle perpendicular to the plane of projection is represented by its diameter ; and every circle perpendi- cular to that plane is represented by a chord of the pri- mitive circle equal to its diameter. Thus if, by way of showing the use of this projection, we assume the me- Tidian for the plane of projection, the horizon will be re- presented by its di- ameter HO: the rime vertical by its iameter zZv, which cuts the furmer per- pendicularly: the 31x o’clock hour cir- cle will be repre- sented by its dia- meter which is the axis PP’, making with the horizon : the angle pco = the height of the pole = the latitude = 1: the equator will be projected into its diameter eg, making with the horizon an angle Hce = 90° — L: the parallels to the equator will be represented by chords, such as AB parallel to the diameter of the equator: the almucantars are projected into chords, such as RS, pa- rallel to the horizontal diameter Ho. 31. aB being the projection of a certain parallel, suppose that the star which in its apparent motion de- scribes that parallel, has its inferior transit of the meri=« dian at B. ‘The point 8 which is its place on the sphere, is also its place then in the projection. The star being in the horizon at 1, of will be the versed sine of its aziz muth, and cz the sine of its amplitude. Also, refer- G2 ; {24 Projections of the Sphere. ring the point 7 to the chord Ba, Br will be the versed gine of the arc described from the inferior transit of the star to its rising, or the semi-nocturnal arc; TG'will be the sine of the are which, remains to be déscribed before it reaches the 6 o’clock hour circle, or te will bé the sine of the arc which deducted from 90° will leave the semi-nocturnal arc; or TG will be the sine of the arc which added to 90°, equivalent to. ac, will give the semi-diurnal arc represented by AT. aw On ag as a diameter describe the semi-circle ADB ; from any points ¥ and £ answering tothe position of the star at different instants, erect the perpendiculars FF’, EE’; produce cp to D; then it is evident that if the _ semi-circle ADB were elevated perpendicularly on the plane of the meridian, F, G, and £, would be the respec- tive projections of the points F’, D, ale Sali Satay 39, To find the value of the arc DE’ projected into the rectilinear portion Gr, we have 5 rad: sin DE’ :: AG: GE; whence eExrad GE. GE GE. GE, | ) sin DE’ = = rotK AG GA sinPA cosAe . C@SD_ . Hence, to know the are which answers to GE,.GF, ct, &c. we must divide each of those lines estimated from the middle of the chord by the sine of the chord’s distance from its pole... . ey Again, from the triangle cet we have GT = cc tanecr = sinptan4.... (3.) pn being the declination of the parallel ; GT sin D : é . Re = tany =tanptanu = sin pr’..,. (4.) cos D cos D : ‘From the same triangle we have, also, . ‘ CG sin D. CT = sin ampl. = cos azim. = ce see ae Ge) é cosGcr — CoOsL 83. Through the point z, a projected place of a star, draw the chord Rs parallel to the horizon, Qe will be to the radius ar the éosine of the azimuth, and EL per- pendicular to the horizon will be the sine of the alti- yf Stereographic Projection. 125- tude. - To determine this from the projection, draw ca parallel to the horizon, then SIN.A = EL.= La 4-4. ab. ==.GM +. ak : CG sin. GGM + EG Sin EGA ti. sin DSINL + AG SIN DE’ Cos Geo sin D SiN L -F Cos D cos L cos hour angle sin D sin L + cos Dp cosLcosHu....(6.) Put 4: = 90° — p, & = 90° — L, and z = 90° — a; then will the last equation become cos Z = cos A cos E + sin Asin Ecos H, ; which accords’precisely with the fundamental equation (2) of spherical trigonometry; and the equa. (1 and 4:) may readily be deduced from the same diagram. HWW Wit Section III. Stereographic Projection. 34. In this projection, which appears to have been invented by Hipparchus, all circles, whether great or small, are represented by circles; and a second pro-° perty, equally general and more curious, is that in the projection all the circles make. respectively the same angles as on the sphere. EF cxig. , . 35, Let azop be a great circle of the sphere, o the place of the eye: the dia- : | meter OCA being drawn, and..the diameter Bcp perpendicular to it, BcD will Be the orthographic projection of a great cir- cle perpendicular to the . visual ray oA, or of the circle one of whose poles is O; its on the plane of. this circle that it is proposed to describe all the circles of the sphere, as they would appear at the ry \ 126 Projections of the Sphere. point 0. The circle, then, whose diameter is Ep is the plane of projection; the point c its centre; a and o its poles; and the point c is evidently the projection of the point A. : 36. Let p be any point assumed on the circumference OBAD: take pz = py, and draw the chord Er, it will be the orthographic projection, or the diameter of a small circle whose pole isp. Draw £0, Fo; to cut BD in s and T, sT will be the projection of the chord rF on Bp; and we propose to demonstrate that sv is the diameter of a circle which will be the projection of the circle described on EF. Now it is evident that rays from all points of the circumference of the circle whose diameter is uF to meet at o will form the surface of an oblique cone whose vertex will be o, and circle about EF its base; of which all sections parallel to that base will ( Hutton’s Geometry, theor. 113) be circles. In order to deter- mine the section of this cone whose orthographic pro- jection is sT, we may proceed thus: meas, of # FEO is Fo = Jon + Apr = 45° + IpF; meas. of Z sTo is 0B + 3D¥ = 45° + dpF: therefore Fso = sTo; and consequently, EFO = 180° — FEO — FOE = 180° — sto — TOS = OST. The triangles Ero, Tso, then, are similar; yet the. lines EF, ST, are not parallel, but are what is techni- cally denominated anti-parallel, or sub-contrary. Sup- pose, however, the cone.zoF to be turned half round upon the axis po, then (since both slant sides oF, or, make equal angles with op) oT would become oT’, and os would become os’; in that case 1’s’ would be parallel to the original chord EF, and the section of the cone (which can inno respect of magnitude or shape differ from the section projected into st) would evidently be acircle. st is, therefore, the diameter of a circulat section, Thus every circle, whether oblique or not to the visual ray directed to its pole, will be represented ow r & the projection by a circle. Stereographic. Projection. 127. 37: Draw from, the centre to the pole. Py the radius. cp, and through the point p the tangent x’Pr’G, to meet in G the plane of projection Bp prolonged. Then, meas. of Z GPK-== 4PD0 = 45° + 4PpD; meas, of Z PKG = 430 + PD = 45° + 4Pv; therefore GPK = GKP, and PG. = GK: that is, the tangent.eG is projected, into a. line. KG equat tout, 38. Since pox and Por are equal, the line oP will not bisect sv; but ks <. xr. Bisect st in m, then ms = mT = radius of circle of projection; and cs, cm, CT, will be in arithmetical progression. Hence, cm = cr + $cs = 3 tan Zar + gtantaAE= sin 3(AF + AE) __ sin AP 2 cos $AF COs SAE = Beos $ (AP + PE) cos $(4P — PE). Let d = cm distance of the centre m from the centre c of the plane of projection, 4 = pz = polar distance of the circle EF, D = ap = distance of the two poles, r= ms = mT: then sin AP sin D ‘ d= ——___—_- = CORRELL © ae aL (7.) cOSAP + CO3PE cospd+cosA eo Ase = 4 (ct — cs) = 4 (tan jar — tan 3Az) _ sin § (AF — AE) sin. A = 2cos $AR cos gan —~ cosp + cosa **** (8.) [See formule (u), &c. chap. iv.] Consequently, d:r::sinp:sinA.... (9) From these three theorems the whole doctrine of stereographic projection may be deduced, by tracing the’ mutations of p and A. The chief maxims and principles of construction may also be developed: geo- metrically, thus: 39. Beginning with great circles, let pz = 90°, and PF = PE; then will mF be a diameter. Draw the right lines orF, o&s, bisect Ts in vr; then7s = rt, willbe the radius of the circle into which the great circle whose diameter is EF will be projected. Through o and 7, draw. ornri, ‘The. circle described on st will pass 128 Projections of the Sphere. through 0, because FOr is a right angle. In like man- wer, it will pass through A, because cs is perpendicular to the middle of . Ao. Thereforero = rsizlpas con- seq. SO7 = OS7, and cro’= 2 oSr. cor = 90° — cro == 906° — 2 osr = 90° — (DO — BE) == 90°— 90° + BE = Be, Thatis; cor = inclination ~ of the plane of the great circle to the plane of projection. » Hence, or is manifestly = sec. inclination, and cr = tan inclination, to rad co. Thus, with the radius = sec p, and at the distance from the centre = tan p, it will be easy to describe the eircle. Again, take Ar = 28n = 2ap = 2p, and draw ol, the point of intersection r with ps will be the centre, and vo the radius. — . Also, since meas. of cor = BE, meas. of ocR = OF = 90° — BE; we have cor’+ ocr = 90°, and conseq. cro = 90°. Hence, by. drawing ory perpendicular’ to FE, we find the centre r. and the radius ro: also cr = sin BE = sin D, OR = cos D. 40. or, therefore, will always make with oc an equal equal to the inclination or. distance of the two poles. Let there be, then, a second circle whose inclination shall be cor’; the radius of its projection will be ro, and the radii ro, r’o, will make at their points of inter- section an angle ro?” which will be their difference of inclination, or the mutual inclination of the two circles, This is a particular case of the general theorem, 4]. Suppose now that ap = 90°, Pp will coincide with B, and the pole of the circle will be upon the limit of Tweanay “st ‘great circle which has its ‘origin at any point whatever M” ‘tangent of pp; ct will be the Stereographic Projection. 129 the projection.. Let su) = be = A polar distance of the circle to be projected, the chord ux will be per- pendicular to ps. , Draw ou and ‘ors, Gs will be the diameter sought. Bisect cs in n, x will be the centre, nE = NH = nS = nG, will be the radius of the projected circle. Meas. of enc = 2nse = 0D, BE= 90° — BE = 90°— BCE CnNE ++ nce = 90°... ‘cEn\= 90° \. En\== tan BE‘ stam A, andicn = sec BE = sec A, These values serve for all circles which have. their pole on the circumference of the circle of projection. If these are great circles, then A = 90°, and tan A, see 4 are infinite: consequently, the centres of the projec- tions falling at an infinite distance from c, the projec~ tions themselves, will be right dines passing through o, and“intefsecting under the: angles which such circles, tukéttwo and- two, form on the sphere. 22.-Eét 0 be the point of observation, or place of the eye, A the pole of the pro- jection, specs the plane of projection, Pp an arc.of a of ‘the circle: OBPE, Pt the \B secant, and s the projection of Pp. Dray'st: then from the rectilineal triangle sct we shall have, Si? = cS? + ci? — Zes. ek cos scé = tan? ap -+ sec? pp — 2 tan SAP sec PD COS DE = tan? dap 4- tan? pp + 1—2Ztan SAP SeC PDCOSDE = sec? JAP + tan? PD — 2 tan ZAP SeC PD COS DE. But the “spherical triangle Por right angled in & gives, (chap. vi. equa, 6), | ; cos PD = COS PE. COS DE = Sin AP COS DE; MD ye a ; cosPp. sin APCOSDE . Substituting this in the last value of s¢?, we have |. G5 - and, therefore, sec pp = 130 Projections of the Sphere. 2 tan ZAP cos DE sin AP COsSDE sf? = sec? Jap 4+ tan? PD — = sec? 3aP + tan? PD — ai Bs 2sin SAP COs SAP = sec? 3Ap + tan* pp — sec? 4p = tan? PD, Therefore st = tan pp = Pt. Consequently, the tangent Pt of an are of a great circle terminated at the plane of projection, is projected into a right line equal to it. Let p/, and rd’, two such tangents, be connected by the right line 颒 which will be in the plane of projec- tion. Let st, st’, be the projections of those tan- gents; the triangles ¢p?’, ist’, are (from the .above) equal in all respects: there- fore the angles opposite to the common side #¢’ will be the same in both: conse- quently, the tangents of any two arts terminated at the plane of projection, are projected into lines which are re- spectively equal to them, and which form an equal angle. Hence, two circles which intersect in P on the sphere, form on the projection an angle equal to that which they make on the sphere ; because, at the point of in- tersection the elements of the arcs coincide with those of their tangents. Therefore, all great circles intersect mutually on the plane of projection under the same angle as on the sphere; so also do little circles which intersect at the same points, and have, by consequence, common tangents. 43. By way of showing the application of these prin- ciples, let us suppose that the eye is at the south pole of the equator. The plane of projection will then be the equator itself; the centre of the equator will repre- sent the north pole; ap (fig. to art. 35) will be = 0; the projections of the parallels to the equator will all Stereographic Projection. 131 have for a common centre that of the projection; and the radii of those circles will be the tangents of the halves of the polar distances. Thus, for the polar circle ....7== tan ( 23° 98’) =tan11°44" tropic of cancer . r =tan3}( 66°32’) =tan33°16 tropicofcapricornr = tan 3 (113°28’) =tan 56°44" for the antarctic circle r = tan $ (156° 32’) = tan 78° 16 for any latitude L .. .. r= tan} (90°—L) = tan45°— 34 or, if the lat.be south 7 = tan (45° + gL). As for the meri- dians, whose planes all pass throughthe place of the eye, they all become diameters which divide the equator in its several de- grees, and form at the centre of the projection angles equal to the dif- ferences of longi- tude. For these circles d = oo, and 7 = oc(art. 38). This kind of pro- jection, the easiest of all to describe, serves very conveniently for eclipses © of the sun. The meridians and the parallels are herein divided mutually into degrees: those of the parallels are equal; those of the meridians unequal ; for the exe pression for one of their degrees is, d= tan$ (4+ 1°) —tanjA= “tye sin 30’ ™ cos $4 (cos £4 cos 30’ — sin $A sin 30’) tan 30’ tan 30’ sec? 3.4 nn i cos $4 — tan 30’ sin $A cos $A 1 — tan30' tan ga sin 30! cos $A cos$(d + 1°) 132 Projections of the Sphere. 44, On a planisphere of this kind, the stars are placed according to their right ascensions ‘and declinations. Then the ecliptic is traced, as well as its poles, the circles of latitude all intersecting mutually at those two poles, and then the parallels to the ecliptic. Thus, on the radius marked 270°, at a distance cr from the centre = tan 11° 44’, we mark the north pole # of the ecliptic. From that point,.with an opening of the compasses = cosec 23° 28’ we mark the point x, or, which amounts to the same, we make cr = cot 23° 28°, Through the point x we draw the indefinite perpendicular vex, which: is evidently the locus of the centres of all the circles of latitude intersecting mutu- ally in Hand 7: E will be the centre of that circle of latitude which passes through 0° and 180°, or of the equinoctial colure, which will be the circle vax. sEH will be the solstitial colure. In general, make ev’ = longitudé: from the centre G with the radius c’s we describe a circle, it will be the eircle of latitude which answers to the longitude sup- posed. Repeating the same operation on the other side of the line sz, we shall have the circles of latitude of the other hemisphere. For the circles parallel to the ecliptic we employ the formule tan$.(p + 4) + tan$(p — a) | ¢ =— Sara ; tan$(p + 4) —tan$(p— A) 3 : p being = 23° 27’ 49”, or nearly 23° 28’, the distance " between the poles of the ecliptic and the equator; and A the polar distance of the parallel.. Wheu A >‘p, the sign of the second term will be changed. 45. The principakdefect of this projection is the little resemblance and proportion between the arcs of the sphere and their projections. Thus, the arcs 7A and All represent arcs of 90°; 7p represents an are of 113° 23’, and pz, though greater, only represents 66° 32’; the arcs au, az, aH, &c. are of 90°; DE-represents 23° : i ~—— Siereographic Projection. 133 28’ = DAN = inclination of the ecliptic Ane to the equator Ang. It is true, however, that the greatest in- equalities are out of the circle ap@s which is properly the projection. If we regard the circle sv0xz as a map of half the terrestrial globe, then zv, zy, 7G, 7£, will represent arcs of 90°, though zv will be the only one of those four which is actually a quadrant. 46. Another inconvenience of this projection, is the difficulty of finding the true distance of two points of which we have the projections. Yet, let m and win the diagram to art. 42, be two such points: produce cm, CN, to m and » respectively; the arc mn will give us MCN which is the same as on the sphere. cM and cw are the tangents of the half distances from the pole of the projection. The spherical triangle will give (see chap. vi. equa, 2), COS MN = COS CM COs CN + Sin €M SIN CN cos Cc __ (1 — tan? $car) (1 — tan? gon) + 4 tan $cm tan Sen cose¢ a (1 + tan? 3cm) (1 + tan? gen) __ (i — om?) (1 — cn?) + 4cm . cn cos c oy (1 + om?y(L + cn?) ; Take cm’ = cm and draw om’m”, am” will be equal to the arc represented by cm. Proceed similarly for en. Then cos MN may be computed from the above. The third member of the equation’is obtained from the second. by substituting for cos cm, sin cm, their values in tan cm, &c. deduced by means of equa. rR, chap, iv. - 47, The projections ‘here treated serving for the usual purposes of astronomy, we need not enter upon the explication of the other kinds of projection, devised by geometers for different purposes. The principal of these is the gnomonic projection, in which the eye is supposed at the centre of the sphere, and the plane of projection a tangent plane to the sphere at any assumed point. All the points within adequate limits have their projections at the extremities of the tangents of their 2 : é 134 Dialling. distances from the point of contact; and those tangents form respectively the same angles as the arcs that mea sure the several distances from the principal point. In this projection too, a less circle will evidently be pro- jected into an ellipse, a parabola, or a hyperbola, acs cording as the distance of its most remote point is less, equal to, or greater than 90°, from the centre of the plane of projection. . But for a farther developement of these properties, and for the geometrical constructions derived from them, such as want to enter more minutely into this subject may consult Emerson’s Projection of the Sphere, the treatise in Bishop Horsley’s Elementary Treatises on Practical Mathematics, or that in the Traité de To- pographie, par Puissant. ; CHAPTER IX. On the Principles of Dialling, 1. DIA LLING, or gnomonics, is the art of drawing on the surface of any given body, whether plane, an- gular, or curved, a sun-dial, that is, a figure, the dif- ferent lines of which, when the sun shines, indicate by the shadow of a style or gnomon the time of the day. 2. The general principles which serve as a basis to the theory of dialling, cannot be more aptly illustrated than they have been by Ozanam and Ferguson, in the following contrivance. . Suppose a hollow transparent sphere prgp, of glass, to represent the earth as transparent, and its equator divided into 24 equal parts by so many meridian semi- circles a, 0, c, d, e, &c. one of which is the geographical . 4 Dialling. 355 meridian of any given place, as London, which it is supposed is at the point a; then if the hour of 12 were marked at the equator, both upon that meridian and the oppo- site one, and allthe rest | (// of the hours in order on the other meridians, those meridians would be the hour circles of London. Because, as the sun appears to move round the earth, which is in the centre of the visible heavens, in 24 hours, he will pass from one meridian to another inanhour. Then, if the sphere had an opaque axis, aS PEp, terminating in the poles e and g, the sha- dow of the axis, which is in the same plane with the sun and with each meridian, would fall upon every par- ticular meridian and hour when the sun came to the plane of the opposite meridian, and would consequently show the time at London, and at all other places on the same meridian. If this sphere were cut through the - middle by a solid plane axscp in the rational horizon of London, one half of the axis er would be above the plane, and the other half below it ; and if straight lines were drawn from the centre of the plane to those points where its circumference is cut by the hour circles of the sphere, such lines would be the hour lines of an ho- rizontal dial for London: for the shadow of the axis would fall upon each particular hour line of the dial, when it fell upon the like hour circle of the sphere. If the plane which intersects the sphere be imagined upright, and at the same time to face the meridian of the assumed place, the intersections of the several hour circles with the plane in this position, would give the hour lines of an erect, direct, south, or north dial. And proceeding, in like manner, to contemplate 136. Dialling. planes, or other surfaces, any way posited in this hollow terrestrial sphere, we should by the several intersections . of these surfaces with the hour circles, obtain the hour lines for every variety of dial. And since the earth it- self compared with its distance from the sun, may, in reference to this branch of inquiry, be regarded as a point, if a small sphere of glass, or a small sphere con- stituted of a wire axis and wire hour circles, be placed upon any part of the earth’s surface, so that its axis be parallel to the axis of the earth, and the sphere have lines upon it and. planes within it, such as those above described, it will indicate the time of the day as accu- rately as if it were placed at the earth’s centre, and the earth itself were as transparent as glass, | These general notions being premised, it will be “proper to annex a few definitions. 7 3. The plane erected perpendicularly to the face:of the dial, and the upper edge of which marks and bounds the shadow, is called the gxomon; the superior edge of the said plane is called the style of the dial, and it is always parallel to the earth’s axis. fe 4, The line in which the plane of the gnomon inter sects the plane of the dial is denominated the sudstyle.: 5. The angle included between the style and the sub- style is called the elevation of the style; in the followimg formulae it will be denoted by x. ~ In a horizontal dial this is, evidently, equal to the latitude of the place, or B= L. 6. While those dials whose planes are parallel to the plane of the horizon are called horizontal dials; such as have their planes perpendicular to the horizon are'called erect dials. 7. Those erect dials whose planes are either parallel or perpendicular to the plane of the meridian, are called ‘direct erect dials; they face one or other of the four -eardinal points, : y mah. 8. All other erect dials are called declining dials, _ 9. Those dials whose planes are neither parallel nox o General Problem. 137 perpendicular to the plane of the horizon, are called rnclining or reclining dials. They may, at the same time, be either direct.or declining, according as they present a sloping face. to the cardinal points, or not. 10. The arch of the horizon which is intercepted be- tween any given plane, and that of the prime vertical, is called the declination of the plane. It will be denoted by p, and will be regarded as posztive when the decli- nation is towards the north, negative when it is towards the south. 1]. The mnclination, 1, of a plane is the angle whiely it makes with a vertical plane. 12. The intersection of the plane of the dial and that of the meridian passing through the style is called the meridian of the dial, or the hour line of 32. 13. Those meridians passing through the style, which make angles of 15°, 30°, 45°, &c. with the meridian of the place (marking the hour line of 12) are called hour circles, and their intersections with the plane of the diak hour lines. 14, The angle for med by the substyle and the meri- dian is called the horary angle of me substyle: it will be denoted by m. / 15. The angle included between the substyle and the Becta a is termed the inclination of the subst. aE & it will be sanared by s. | ‘GreNERAL ‘PROBLEM: 16. Yo determine the requisites in a diak of any pro posed tnclination to @ vertical plane, ang declination from the prime vertical. Let Hor, in the mar- ginal diagram, represent . the horizon, 1zr the meri- dian’ of the place, z the zenith, zo the prime ver- tical, p the elevated pole, PO, : ey portions of hour circles, zpb, zpb’, corre- 138 Dialling. sponding hour angles, and\let the plane on which ‘it is proposed to draw the dial be coincident with the plane of the great circle mér. Also, let za perpendicular to mat be drawn and produced'to1. Then pr = 1, the latitude: of the place; za = 1, the inclination of the plane; of = Hi = D, its‘declination; ps drawn: per- pendicular to MT will be the: position of the substyle; and mé.is the inclination of the-hour line to the meri- dian. Now, we may regard the dial whose plane coincides: with MaT, as.a'vertical or erect dial atthe place whose zenith is mM, where M and z are on thersame meridian, and, of course, reckon the hours alike. Let mz=J, and p’ = comp. of zma. Then pz = 90° — i would become PM = PZ +: zM = 90°—1L +1=90°— (L—/). The right angled triangle mza, gives: tan Ma = sin'zatan Mza = sinitanD....(1.) Here ma is the’angle between the meridian and ver- tical, and is obviously evanescent when either 1 or Dare. When 1 = 90°, Ma becomes = HI = D. tan Za tant cosmMza comp’ '*" (2.) cos ZMa = Sin D’=sin Dcos1.... (3) sinPs = sin E=sin zMa sin PM=cos D’cos (L —/),. (4.) where x isthe height of the style, or elevation of the. pole above the plane. Then, Pd being.any hour circle whatever, the sphe- rical triangle PbM, gives us, from equa. (4) of spherical trigonometry [chap. vi. 23]. cot Md sin pM’ = cos PM cos PMb'+ sin PMO cot FP. But, cos pMd = sin D’, and sin pMd = cos D’; Also, tan zm = tan / = also, —— =cot. Therefore, dividing by sin PM, : foto . cop) cot Md = sim p’ tan (n— 1) + Se fon ¥ 55) tan MT = tan (Ma + 90°) = — cot Ma, where mt is the inclination ef the horizontal plane ta the meridian. General Problem. — 189: The above expression (5) is general, and by no means’ complex. It may, however, be rendered more conve- nient for further deductions and corollaries by extermi- nating / and pv’, Thus, for the first term, tani — tan? sin p’ tan (L — 7) = sin p'cos 1 —————— 1 + tant tan£ tan tan I [sin D cos I (tan'L' — a) [i+ cop tan L} sin DcosttanL —sinrtanp ee, 1 + tanisecp tan And, for the second term, cos D’ sinPMb sin? cos(L—2) cos(u— 1) _ sinZ cost.— 1) sink. hint ie sindcosicosu + sin®sinu Poa (tan /cosL + tan sin z) ee sin (1 + tan 2f ) ™ tanlcosi + tan lsin& ‘ tan % tant cos u ia Sas Ne i. : ° cOsD . 3 cot Mb = tan z sin D ++ ~— cotr.... (9) In this hypothesis equa. (7) becomes > |: Sotah MS == COLL sihtD = cots... (10). By which the angle between the substyle and the hori gomis determined. ; ~~~ 23 >, same acronis ~ And equa. (8) becomes sin £ =.cos\L. cos) .:..(11.) We have also, when m and z coincide, cot zps’ == sin L cotp.... (12.) Thus we obtain the herary angle of the substyle. 18. If p as well as:1 become = 0, then - ~ - mig “33 Pend ste coup ; vie. ROL MS =.= SS. COL.R SEC L; ‘ F COS L : J Cay aus . consequently pee, = eee OF, tam md = tan PcosL...> (13.) \ General Problem. In this hypothesis also, we have shire = cos LS EP opts = 0" ay The three last formula evidently serve for. erect a rect south or north dials; in which equa. (15) shows that the substyle is perpendicular to the horizon, i9. If while 1 = 0, p become = 90°; then will equa. (10) be transformed to cot § cot L (16), and equa. (11) to sine =.0...,(17.)) « This is the case of the erect direct bait on west dial, in which it appears that the style is parallel tothe plane, and the substyle inclined to the horizon in an aagle equal to the latitude. Here, since the style is parallel to the plane, the dial has no centre; all the hour lines, therefore, are parallel to the substylar line, which is the hour line of 6 o’clock: the respective distances of the hour lines from the sub- stylar are measured ‘by a cot i. on a perpendicular to * on a vertical line, the 6 o’clock hour line; or by — This gives an infinite distance ang een the 6 o’clock and 12 o’clock hour 'lines ; ; as there manifestly ought to be, because at noon.the solar rays will he perpendicular: to the plane of the dial. 20. If when, p ='0, we have tics = 90°; the equations (6) and. (8) become : ; * cotP ] sin L - cot md = ———, whence —— = ——; bin L °® cotmb. cote or, tan Md = sin L tanP. . (18. and sin E = sin L . (19. These theorems obviously apply to the hor izontal dial. 21. If, in the last: is wae L become equal to i * we shall haves tan. Md = tan P*...,.sin £& = radius. This would apply to a- horizontal dial at the poles, or ' a dial in any. latitude with its face posited parallel to the equator. Here the formule show that the style would become a pin placed perpendicularly to the centre of ; 142 Dialling. the dial, round which the. hour lines would. be, radii drawn from the foot of the style,to make angles of 15°; 30°, 45°,,&c..with the meridian, or hourjline of 12. ProBLem iI. 22. To describe a horizontal dial for any proposed la- ¥ : titude. This is the simplest dial to, draw next to the polar dial, just_ mentioned ; ,and it is the. most ,useful, be- cause, if it be posited where the sun’s rays meet with no obstructions, that luminary will shine upon it from his rising to his setting. The theorems from which the construction is to be deduced are, — . -tan M6.= sin L tan p, and sin. E = sin L. Here mé.is the measure of the ‘angle H, between the _12 o’clock and any other hour Jine on the dial, and Pp the hour angle from the meridian, as it varies at the poles, ofthe heavens; the latter, therefore, varies uni- formly, while the former only varies uniformly in cer- ‘tain limited cases; as, for example, in the horizontal dial at the latitude of 90°. In any other latitude, mak- ing the terms of the equation homogeneous, we have Sin 90° tan H = sin. tan P. | Hence, if two radii be assumed in the ratio of sin 90° to sin L, tan H referred to the former radius will always be equal to tan 1 referred to the latter. From this con- sideration the following construction is deduced: On the proposed P plane assume the right Jine 12 us for the me- ridian, or 12 o’clock hour line, parallel to § which draw, the line 12hs at a distance _ equal to the proposed ¢ _ thickness of the style. Perpendicularly to .5 these draw 6H6, for Horizontal Dial. 14g the east and west line of the dial, or the 6 o’clock hour line, Make the angle 12H¥ = the latitude of the place, and from 12 let fall the perpendicular 12F upon nF. Make 12p, upon. 412:prolonged, equal to 1gr. From P draw lines Pl, P2, P32, &c. (to terminate in the line 12-5, perpendicular to» 12H).and to make angles 12p1, 12P2, 12P3, &c. equal to 15°, 30°, 45°, &c. Then from the centre H draw H1, H2, 43, .H4, 45, for the hour lines of 1, 2, 3,:4,-and 5, in the afternoon. Take, on the other side of the: substylar line 12-11, 12-10, 12-9, &c. respectively equal to 12-1, 12-2, 12-3, &c. and from A draw the lines 47, 48, 29; &c. Produce the lines 4H, 5u, for the lines of 4 and 5 inthe morning ;. and pro- duce the lines 7h, 8h, for the hour .lines:of 7.and 8 o’clock in the afternoon. The truth of this construction is. manifest from the remarks which. precede it. For 12p. = 12F is. evidently the sine of the latitude to the radius H12. . And while 12-1, 12-2, 12-3, &c..are the tangents of 15°, 30°, 45°, &c. respectively, to the radius P12 ; ithe same lines:are tangents to the angles 12H1, 12H2, 12n3,:.&c. | Conse- quently, while the former, are ‘hour angles at the pole, .the latter are the correspanding hour angles. at the cen- tre of the dial. | _ The quarter and half hour lines are drawn by setting off angles of 3°45’, 7°30’, 11° 25’, &c, from, the. meri- dian line: but they are omitted in the diagram .to_pre- vent confusion. The angle 12uF is also.equal;to the elevation of the style; fork = L. As to the gnomon, it should be a-metallic triangle of the thickness hu, and: having one angle = t the latitude. It must be fixed: perpendicularly to the plane, on the space left for it:in the figure from hu towards 12, and having its angle nat 4H: then will the style of the dial be parallel to the earth’s:axis. be Bee) Dialling. : PROBLEM Ill. 23. To describe an erect direct south dial for any pro- posed north latitude; or an erect. direct north dial, for any proposed south latitude. 6) -° . - Here the formule are those numbered 13, 14, and ‘15,.0f which. the.two first are tan M).= cosL tan p, and sin rE = cos L. Substituting nm for md in the former of these, and ‘making the terms homogeneous, we have * This equation is similar to the equation of the hour -angles in-the horizontal dial; and the construction will employed instead of sin x. Qn the proposed plane draw the right line 6H6, for In the middle of this line set off hu ‘om $ : 7 posed thickness of Ss the gnomon, and $ at lines perpendicular AN ‘to 6u6 to terminate rallel ‘to 6H6 at a - ; convenient distance. Draw ur to make the angle dial is made, and let fall upon ur the perpendicular 127. Make 12p = ur, the cosine of the angle 12uF angles of 15°, 80°, &c. with p12: draw lines from 1 to meet these in the line 10-5; set off corresponding lines pleted. The demonstration is the same as in prob. 2. The angle made between the style and substyle is sin 90° tan H = cos LL tan P. ‘therefore be similar, except that cos L is here to be ‘the east and west line of the dial, or the hour line of 6. ‘equal to the pro- -through h, 4, draw 9 0, 1.72 -in the line 10-5, pa- P Z ‘12u¥ equal to the latitude of the place for which the to radius 112: draw from Pp lines Pl, P2, &c. to make on the left side of the dial, and the construction is com- here equal to the complement of the latitude. When East Dial. 145 the dial is placed vertically to face the proper cardinal point, the line 6H6 will be horizontal, and the style sloping downwards from at an angle of 90° — x will again-be parallel to the earth’s axis, as it ought to be. Note. An erect north dial for a place in north lati- tude is constructed in exactly the same manner as’ an erect south dial; but the position of the dial must be reversed. In the case of the north dial for north lati- tude the line 646, instead of being the top will become the bottom of the dial. The same may be observed in reference to an erect south dial for a south latitude. Prosyiem IV, 24. To describe un erect direct east dial, for any pro- posed north latitude, or an erect direct west dial for any south latitude. It appears from art. 19, that the substyle in this dial’ will make an angle with the horizon equal to the lati- tude, that the hour lines will be all parallel to the sub- style, and at the distances indicated by a cot P, a being. the height of the gnomon, and p the hour angle from the meridian at the pole of the sphere. Let az be assumed as the horizontal line on the pro- posed dial. From the corner 8 draw sp to: make an. angele ABD = the com- a plement of the latitude, Phe Bias Be BO TNS, 8 and about the middle, 1 H, of that line draw per- pee atly to it the ‘line 6H6 for the 6 0’clock hour line: this will, also, * be the substylar line, and will evidently make .4-— marae a with the hovimoa ef de ee SB gle 26x = the latitude of the place. Assume any point, as that marked 11, for the point where the 11 o’clock : ce | i 146" Dialling. hour line is to intersect the line Bu; and draw 11p.to make the angle u11p = 15°: so shall px be the height of the gnomon. Set off angles HP7, HP8, HP9, &c, respectively equal to 15°, 30°, 45°, &c. and through the points where the lines P7, P8, &c. intersect the line BD, draw lines parallel to pH: also, set oft, from H-towards D, HS = H7, H4 = HS, Ke. and, through the points 5, 4, 3, &c. draw other lines parallel to pH, these shall be the hour lines required. The truth of the construction is evident from what has preceded. A rectangular gnomon of altitude = pH; being set up perpendicularly to the plane of the dial, will, when the dial itself is posited vertically to face the east, have its upper edge parallel to the earth’s axis. It is here supposed that the, upper edge of the. gnomon is reduced to a mere line: if it have, any measureable thickness, allowance must be made for it in the construction, asin, the preceding problems. , Note. An erect direct west dial for any place in north latitude, may be constructed exactly in the same man- ner as is just taught, except that, instead of beginning the construction from the right hand, B8,. of the plane, the operation must commence at the left: hand, a, and, that the figures expressing the hours of 1, 2, 3, &c, in the afternoon, be placed between, and the hour line of 6. The like, may be observed with regard to. an erect direct east dial for any south latitude. PROBLEM V. 25. To describe an. erect. south dial for any proposed. north latitude, to have a given declination from the west 3 or an erect north dial for a south latitude, to decline JSrom the east. - The formula in the general problem which are appli- cable to the present, are those numbered (9), (10), and (11), viz. ct ce rate Sen Ft. Declining Dial. 147 . cospD cot M4, or, cot H = tani sinD + sory, Cot Ps for the determination of the hour lines ; cot Ss = cot L sin D, for that of the angle between substyle and horizon; and sin E = cos L cos D, for the elevation of the style. From these and a table of natural sines and tangents, the hour lines, &c. may easily be drawn, thus: Let azcp be the plane on which the dial is to be constructed.. Assume u for the centre of the hour lines, and the vertical line HM for the meridian, or 12 o’clock hour line. Now, 4 since the only variable quan- 6 tity in the preceding expres- sion for cot H, the cotangent of the angle between the me- ridian of the plane and any hour line, is cot p, the cotan- of « A at gent of the corresponding D 7 up“4, Op C&R hour angle at the pole, we may, by taking that = 0, get a fixed hour line to which the others may be referred. Thus, the expression, when the hour angle at the pole is 90° = 6 x 18°, becomes cot MH6 = tan AH6 = tani sinp; an equation by which the position ‘of the 6 o’clock hour line may be readily determined. For, if um be taken between and A, equal to the unit of any measure, as one inch, one foot, &c, we shall have mn = tan L sin pD, in the same measure; the product being at once determinable by means of # table of sines and tangents. Through u and 7 draw a line both ways, to meet ap and the prolongation of EB; then that portion of this right line which falls on the left hand side of the plane will be the hour line of 6 in the morning, while the other portion to the right of H, HZ 148 Dialling. and above the plane, would be the hour line of 6 in the afternoon: that, however, is useless in this dial. To ascertain the position of the other hour lines, we | have only to estimate the remaining term of the equation by the several corresponding values of cot P for the dif- ferent hours from 12 or 6, and set those values from ” to n',n ton”, n ton”, &c. on the line mnn’n” perpene dicular to az. These values are successively cosD cot 75° = cos D sec L tan 1 5° cCosD b —— cot 60° = cos pv sec L tan 30° cos L cosD © ° ae cot 45° = cos Dp sec L tan 45° = cos D sec L €0s pD Be cot 30° = cos p sec z tan 60° rm Oe ~~~ cot 15° = cos D sec L tan 75° (see chap. iv. 41.) cos L These being computed and set off, the several right lines un"7, Hn”8, Hn’’9, &c. drawn from u through the points x’, n”, #””, &c. will be the morning hour lines required. The afternoon hour lines may be computed and set off by the same formule. Indeed, taking the other vertical line mn‘n’n’”, &c. between H and B, (Hm == Hm) the several values of mn, nn’, nn”, &c. will be numerically the same, because the corresponding values of cot P are so; the only change being in the s7gn from +. to —, on account of passing the meridian, —__ Hence, since the values of nn’, nn’, nn’, &c. on the two parallel vertical lines are respectively equal, lines joining their extremities will form a series of parallelo- grams: and hence, the morning hour lines being drawn, the evening hour lines equally distant from 6 o’clock, may be determined by drawing n’n’, n“n", nn", &e. parallel to 66, and then drawing through 7’, n”, n’”, te the right, the lines Hm’5, Hn’4, Hn, &C. Declining Dial. 149 Thus, it appears, as has been observed by Ozanam, Emerson, and Delambre, that in all plane dials, one half of the dial being traced, determines the other by the several distances from the hour line of 6, measured on two vertical lines equidistant from the meridian. Thus these distances are equal between the lines from 6 to 7 in the morning and from 5 to 6 in the evening, PCO vies ¢ OPTI A otaece tate 4to6..... evening, &c. &c. &e, The position of the substyle, in this instance between HM and.HA, is determined by the preceding equation for cot s, and the elevation of the style by the preced- ing theorem for sin E. | 26. But this kind of dial, as well as horizontal and direct dials, may be constructed independent of coin- putation. Thus: G A H B oS C8 924 4 POs LE 12M 1 D On the proposed plane Ascp assume Hu for the centre from which the hour lines shall diverge, and draw the vertical line HM for the hour line of 12. Produce ma to G, so that HA shall be to aG, in the ratio of sin 90° to sin D; draw G6 to make the angle ac6 = 1, the lati- tude of the place, and join “6, which will be the 6 150. Dialling. o’clock hour line, From » draw ue parallel to 66 to meet a line o& parallel to an in &, From o the inter- section of o£ and HM, draw oF to make the angle EoR = complement of the given declination; make or = oF, draw rs perpendicular to o£, make os = Os, and join HS which will be the substylar lme. Perpendicular to Hs draw the right line 5s1, intersecting the meridian in 12, and the 6 o’clock hour line in 6. On 6-12 as a dia- meter describe a semicircle to cut Hs produced in P. With centre p and radius ps describe a semicircle, which divide into arcs of 15° each, both ways from the line P6, or, which amounts to the same, set off angles 6P7, 7e8, 8P9, 9P10, &c. each = 15°. Through the ‘points 5, 6,7, 8, &c. where the lines bounding these angles cut the lines 5sI, draw from Hu. the lines n44, H55, H66, H77, &c. they will be the hour lines required. For, first, since the expression for the 6 o’clock hour line, when the terms are rendered homogeneous, is sin 90° tan AH6 = sin D tan L, we shall have tan 1 to the radius.sin p equal to tan An6& to the radius sin 90°; which is obviously the case with regard to the triangles GA6, HA6,. Consequently, the 6 o’clock hour line is rightly determined. Again, since Eu is parallel to 6c, the angle orn is equal to the latitude, and ruo the co-latitude; therefore OE = oF, is the cotangent of the latitude to the radius oH; and since EoF is equal.to the co-declination, os = os is the sine of the declination to the radius oF = OF; or OS = 0S = sin DcotL. But os = tanous = cots (substylar angle with horizon) to radius Ho: therefore cot s = sinp cot L, as.it ought to be; and the substyle is rightly determined. But the substylar line is evidently a portion of the right line between the centre of the dial and the appa- rent pole: therefore, the apparent pole, p, lies in the prolongation of us. At the apparent pole, Pp, also the horary angle between 6 o’clock and 12, is 6 x 15°, or a right angle; and the angle in a semicircle is a right ay Astronomical Problems. 151 angle: therefore, the semicircle described upon 6812 as a diameter, will intersect the prolongation of ‘Hs in P, the apparent pole. . The truth of the remainder of the construction is manifest *. 27. For astronomical methods of determining the meridian, and the declination of any vertical plane, the student may turn to prob. 2 of the next chapter, ex~ amples 3, 4, and 5. 3 CHAPTER xX. Astronomical Problems. d. SINCE the science of astronomy has given birth to spherical trigonometry, it is to be expected that at least some elementary problems in astronomy may be ad- mitted into an introduction like the present. ‘To deter- mine the position of points in the apparent heavens, as- tronomers first referred to two planes, the horzzon and the meridian (see chap. viii. § 1), which are fixed in reference to any one place on the surface of the earth. But the necessity of comparing observations at different * The problems given in this chapter will suffice to show the application of the principles of dialling to the most useful cases, They who wish to pursue either the theory or the mechanical part of dialling, through all its modifications, may consult Leadbetter’s Mechanic Dialling, Emerson’s Dialling, the treatise on dialling in the 5th vol, of Gzanam’s Course of Mathematics, and that in the 3a. vol, of Dr. Hutton’s edition of Ozanam’s Recreations, There is also-an elegant essay on dialling by M. de Parcieux, at the end Of his Trigonometrie Rectiligne et Spherique; and a neat and sim- ple deduction of the practice of dialling from the principles of perspective, at the end of S’Gravesande’s Essay on Perspective. 152 Astronomical Problems, places, has led to the introduction of other planes and circles into the science, independent of the position of the observers, and even of the figure of the earth. Thus, when the situation of the celestial equator, and the manner of valuing the angles between meridians by the measure of time, became known, they were em= ployed to determine the position of the heavenly bodies by means of their right ascension and their observed declination. Afterwards, as it was found that a consi- derable portion of the celestial phenomena relative to the planetary system, occur in the plane of the ecliptic, or in planes but little inclined to it, it was found expe- dient to refer the position of the stars to the same plane, that is, to determine their latitude and longitude (chap. viii. art. 17, 18). These, and many other branches of astronomical in- quiry, which we shall not here be able to touch, de- pending upon the mutual relations and intersections of different czrcles of the sphere, fall necessarily within the department of trigonometry. A few only will here be selected. Prosuem I, 2. Given the obliquity of the ecliptic, and either the right ascension and declination of a star, or its latitude and longitude, to find the other two, and the angle of ‘position. Let rc in the annexed figure be a portion of the ecliptic, EQ a portion of the equa- tor, the two circles intersecting in © the first point of Aries, in an angle, 1, of 23°27 49”. Let P’be the elevated pole of the ecliptic, p that of the equator, rr’ a portion of the solstitial colure. Fromp’ y and p let quadrants p’sL, Psr, of EA great circles, be drawn through s, the place of the star. Then Ex Astronomical Problems. 153 will be the right ascension, a, of the star, sR its decli- nation, d, or Ps, its co-declination, EL its longitude, /, SL its latitude, L, or Ps its co-latitude, psp’ = p, the angle of position; and pr’ is given = 23° 27’ 49”, the present measure of the obliquity of the ecliptic. It is farther evident that sp’p is the complement of the lon- gitude, and p’ps = P’PE + EPR = 90° + right ascen- sion; as indicated at the poles of their respective circles. Now, if they are the right ascension and declination which are supposed known, in addition to the obliquity, we shall have, from the triangle spr’, (see chap. vi, equa. 2 and 4), ; cos P’S = sin PP’ sin SP COS P’PS + COS P’P COS SP, cot sP sin Pe’ — cos P’ps cos P/P cot PPs = —* sin PPS Adopting the. preceding literal representatives of these sides and angles, and remembering that cos P’Ps == cos (90° -+ a) = — sina, these become sinL = —sinicosdsina + cosisind....(1.) tandsini + sinacos? tan 2 = ———_—__—_——_ | (2.) cos a = tan dsinzseca + tana cos2 These two formulz may be accommodated to loga-. rithmic computation, by taking a subsidiary. angle $ such that sina tand © for then, exterminating sin a from the former, and tan #@ from the latter, we shall have . sind cos(p + 2 - sin L — Meidicostaict #) © @e-0 (So) cos > tan ¢ = tan l — Laan ets ae 2° (4): sin > 3. If, on the contrary, the latitude and longitude of the star are given, we shall have the declination and. HS 154 Astronomical Problems. right ascension by the following theorems similarly deduced, viz. COS PS = sin P’P Sin P’s cos PP’s + Cos PP’ COS P’S, cot P’s sin P’p — cos pp’s cos P/P : cot P’ps = Se sin PPS where again, introducing the literal values and observing that cot p’ps = — tana, we have sind = sini cosL sind + cosisiny....(5.) _. —tanusinz + sinZcosz tana = AT TTT ee ene T ett fie = —tanLsinzsecl + tan/ cosz These equations are obviously analogous to those we have just deduced for the latitude and longitude, the only difference being that here the obliquity enters the formule negatively. Hence by taking another subsi- diary angle ¢!, so that sin J tan’ tan ? = and eliminating as before, there will result, sin L cos (¢” — 2) Sarma te. tan / sin (4 —7Z : tan a = pincer ZAR (8.). sing sin d = 4, As for the angle of position psp’ = 9, it is easily _ found from the relation between the sines of angles and the sines of their opposite sides ; for from hence we have AY ie sinicosa f sinicosl sin p = ~~ , and sin p = aera. (9.) osSL 5, We have also, trom the same consideration, cos a cosd = cost cos/....(10.) And when 1 = 0, as is always the case with the sun, we have cos 2 ae d, eee ° cos a = ——~ = cost sec (11.) tan a = tan/ cosz....(12.) cos/ = cosacosd,.,. (13) Astronomical Problems, 155 sind = 6m / sini... (14) sin 2 = sind cosec? ...(15.) The preceding formule have been deduced upon the supposition that the heavenly body has not gone beyond the first quadrant of right ascension from the vernal equinox. But they are applicable to all positions by simply regarding the mutations of signs in the several sines, cosines, tangents, &c.- according to the arcs to which they refer. The right ascensions and longitudes, being reckoned from the first point of Aries through their respective circles, will at once indicate by their attendant signs, ++ or —, to which quadrant they be- long.’ And as to the declinations and latitudes, they being regarded as. positive when toward the clevated pole, will be régarded as negative. when towards the. contrary pole. Thus, with us, north latitudes and de- clinations will be positive, south latitudes and declina- tions, negative. Example i. The right ascension of Aldebaran being 67° 40’ 30”, its declination 16° 8’ 20” x. Required the longitude, latitude, and angle of position. sin a Here, taking ¢-so that tan 9 = are and substitut- mn f oi cos” sing? and cosd ing sec $, cosec 9, and sec d, for. —— in equa. 3, 4, and 9, they become sin L = sin d cos (@ + 7) sec Q, tan / = tana sin (9 + 7) cosec 9, and sin p = sin? cos / sec d the log. operations corresponding to which will be -as follows : From log sina . 67° 40’ 30". ... 9:9661625 Take ... tand..16° 8 20".... 9°4614544 Remains tan 9,.72° 37’ 46”... ,. 10'5047081 156 Astronomical Problems, Then, to find 1 the latitude, add together, bay (es Ae 16° 8’ 20” .... 9°4439927 cos ($+ i)..96° 5° 35”.... 90258936 SCC D....4272° 37 46”..,. 105249894 a rs ee The sum sini ...... 5° 40°18" s... 89948687 Here, because cos (¢ + 7) being in the second qua- drant is negative (chap. iv. art. 4), and the other terms: are positive, the product is negative, and therefore the. latitude is south. Next, to find 7 the longitude, add together, Log tana...... 67° 40’ 30”... . 10°3865391 sin(?+7)..96° 5’ 35”.... 9°9975396 cosec @ ,...72° 87’ 46”....10°0203305 The sum tan 7,..... 68° 29’ 28”.... 10°4044092 Here all the terms being positive, their product is positive; conseq. the longitude is in the first quad. Lastly, for p the angle of position, add together, Log sinz ....23° 27° 49” .... 9°6000647 cost .,..68° 29’ 28”.... 9°5642090. secd..,.16° 8 20”....10°0174616 ee The sumsinp..., 8° 44’ 96”,... 9°1817353 Lzxample UH. What are the latitude and longitude of the moon, when her right ascension is 304° 21’, and declination 22° ST. 8? ) Ans, Longitude 10° 1° 21’ 54”, latitude 3° 8’ 46” S,. Example IIT, When the longitude of the moon is 1* 7°. 41’ 23”, and’ ~ the latitude 3° 49’ 57”S,, what are the right ascension end declination? | Ans, Right ascension 36° 36’, declination 10° 28’ N. Astronomical Problems, 157 | Example IV. When the sun’s declination is 19° 18’ 20” N, what are’ his right ascension and longitude ? First, to find the longitude, employ *q4z, io. that is, sin? = sind cosec?. To logsind....19° 8’ 20”.... 9°5156873 Add log cosec?, . 23° 27’ 49”... .. 10°3999353 The sum islogsin?.,.55° 25’ 43”.... 9°9156226 Here since the declination is north, or positive, sin Z is positive, and lies, therefore, either in the first or second quadrant, / is, therefore, either 55° 25’ 43” (that is, 1° 25° 25’ 43”) as above expressed, or its supple- ‘ment 124° 34’ 17”, that is, 4° 4° 34’ 177. To find the right ascension, take equa. 11, that is, cos a = cos/ sec d. To log cos? .. 55° 25’ 43” ..,.. 9°7539143 Addlogsecd..19° 8’ 20”....10°0246938 The sum islog cosa ..53° 5’ 6”.... 9°7786081 This reduced to time at the rate of 15° to an hour, gives 3° 32™ 20°, for the right ascension corresponding to the longitude 55° 25’ 43”. If the other longitude had been taken, the resulting right ascension would have been 8° 27™ 40°, the supplement of the former to 12 hours. The days that correspond to these in the Nautical Almanac for 1816, are May 16 and July 27. Example V. Given the sun’s longitude 6° 8° 9’ 36”, to find the righ ascension, declination, and time in 1816. Ans. 12°29” 58%, right ascension, 3°14’ 24” S, declin. time Oct, 1, 1816, at noon. ) . 4, 158 Astronomical Problems, CAMB SA Il. 6. To investigate formule that shall be applicable to inquiries in’ reference to the times of risings and settings of the heavenly bodies, the azimuths, the duration of twilight, &e. Let uzr be the’ theridian of the place of observation, uvRk the horizon, z the zenith, Pe, : . S Zs Pp the elévated pole, 's the place ey of the sun, a fixed star, or other eet heavenly body, s’ss” part of the circle it appears to describe about the pole pe dusing the H earth’s diurnal’ rotation, zsv part of the vertical circle on which the body is at any proposed time, and Ps a portion of a great circle passing through the pole and the place of the body. Then ps is = 90° — d the co-declination; PZ = 90° — 1, the co-latitude; zs = n, thezenith distance, or = 90° — a the co-altitude; zps = Pp the horary or polar angle between the two meridians ZP, SP; and szp = z, the azimuth, measured also by the arc vr. Then the formula we may first employ in the present investi- gation, is equa, 2 of chap.vi. which suited to the case before us, is grew ws COS ZS = COS PS COs PZ + sin PS sin PZ cos P, .. (1*)- or adopting the characters just specified. : cos n = sind sind + cos d cos! cosP...... (1.) Suppose P = Q, or the. body in the meridian, then cos p = rad = 1, and the fundamental equation becomes, COS ZS = COS PS COs PZ + sin PS sin PZ = cos (PS — Pz) whence zs = + (ps ~pz) = + Pps > pz; ps =pz+ zs; and pr or 180° — pR = Rs”. + Ps” or HS’ + Ps’, ( ' (2. = mer. alt. from north or south + declin.. ' = sum. of mer. alts, in a circumpolar star. \ 7. Ifthe horary angler = pas itis when the body Jetetln iE gers Oh Fg LE eperecty geetlicrs SIME CaN LAI G0 ? Ba iANG= 50 YH 2 BE é iB i . i et NEE AEE re care er ee a ee Astronomical Problems. 159 is 6 hours from the meridian, then its cosine will vanish, and equa. 1, becomes 7 cos n or sina = sind sini... eee (Bu) from which the altitude, a, becomes canta In the same circumstances theorem 10, of right angled spherical triangles, gives | cot szp = cot ps sin PZ = tan d cos/.,.. (4) ‘ by which the azimuth,in that case may be determined. 8. When the body $ is in the horizon, or s and v co» incide, we have zs = 90°, or cos n =0; hence 0 = sin d sin / + cos d cos/ cos P, sin d sin pe or, cosP = — —tandtanl,. (5.) cos dcosl Here it is evident that when the declination and lati- tude are both ofthe same kind, cos P is negative, or P is greater than 90°; that is, the time occupied by the hea- venly body in passing from the horizon to the meridian, or fromthe meridian to the horizon, exceeds 6 hours. If the declination and latitude are of different kinds, cos P will be positive, and the time of passage from the horizon. to the meridian, or from the meridian to the ho- rizon, less than 6 hours. Thus, this theorem will serve to determine the times of the risings and settings of the sun, orother heavenly bodies, disregarding the changes of declination, and the effects of parallax and refrac- tion. To find the azimuth when s is in the horizon, we have “from the principles of quadrantal triangles, * cosPs ‘sind ° seks) acne 9... sac Bn Sees, : sin PZ cos — ; 9. If the body s be upon the prime vertical, then pzs = 90°, and the formule for right angled triangles - give cos P = tan zP pa: SEES FERC (7.) cos PS te Leg iM cos 82 = sing = —— = sin d cosec /,.,. (8) ; ER ips 8! eas GOD AS: fio? cor OF oth of Don t- ffi t- AY) - [Siam fa bp: Ab 1+ AB: 1S Gen Rae 6 tisha AC A> (od lip AK /_ 2L.20 s. 206 bo 4 o4 cal ae Ve, i60 Astronomical Problems. 10. When the parallel circle s’s s” which a heavenly body describes in consequence of the diurnal rotation can have a vertical zsv drawn to touch it, which it may while the declination of that body exceeds the latitude of the place, then the body when at the place of contact of the two circles will move vertically, and with the greatest apparent rapidity. In that case the spherical triangle Psz will be right angled at s, and we shall have from the formule for right angled triangles in chap. vi. cotd cos P = —~ = cotd tanl.... (9.) sin “Zs oo" = cos d secl.... (10.) cos l 2 sin J ; sin a = ~~ =sin/ cosecd.. (11.) 11. Returning to the first equation _ COS ZS = COS PS COS PZ + sin PS sin PZ Cos P it may evidently be employed to determine the altitude, or the zenith distance, when the latitude, declination, and horary angle, are known. The same may also be readily effected by an auxiliary angle, as explained in case 2 of oblique spherical triangles, equa. 12 and 13. If we wish to find the hour from the meridian, by the observed zenith distance, we have from the above equa- tion COS.ZS — COS PS COS. PZ cos P= : : SID PS sin PZ Or, from equa. 2, art. 38, spherical trigonometry, ne \/ sin 3 (Pz — ps + sz) sin} (sz + Ps — Pz) (12.) | sin 3 (PZ — sz + ps) sind (Ps +. sz. + PZ) a theorem fitted to logarithmic use. ; A like theorem will obviously serve for the determi- nation of the azimuths, viz. sin §(zP + ps — Sz) sing (ps + zs — ZP) (13 ) Lama, wii / sing (zp + Zs —Ps)sin$(Ps + Z8 + ZP) Astronomical Problems. 163. Or, the angles z, s, and the side zs, may be easily found by Napier’s analogies, thus : cot $psix 4 (ps — Pz) sin 3 (PS + PZ) cot 4p cos $ (Ps — Pz) tan} (z—s)= ig Amide egiReaT TORO ee tan 3 (pz + Ps) cos $(z + 8) I nics 2 pos sd “y cos $(z—3) 12. Since, according to the determinations of astro- nomers, the crepusculum or twilight commences and terminates when the sun is 18° below the horizon, either previously to his rising, or subsequently to his setting, it follows that equa. 12 will serve to. compute the time from noon ta the beginning or end of twilight, if zs the zenith distance be assumed equal to 90° + 18°, or 108°. Or, to determine the duration of twilight, regarding the effect of horizontal refraction, we have these for« mule: cos Pp’ = — sin 18° sec/ secd — tan / tan d, for the hour angle for the commencement of the morn- ing, or the end of the evening twilight. cos P= — sin 34’ sec /sec d — tan / tan d, for the hour angle from noon to sunrise or sunset, ine- cluding refraction. And 4, (er! — p) = duration of twilight .... (15.) These theorems may be employed in the solution of a variety of questions: a few are subjoined, Example I. At Cambridge, in north latitude 52°12! 35”, when does the sun rise and set, on the 4th of May, 1816, and what is the azimuth at rising and setting? ‘The sun’s declination on the proposed day at noon, is 16° 0’ 46” north; and the formula to be employed are. the 5th and 6th. For the first, viz. iF cos P= — tan d tan /, we have, (ote A 162 Astronomical Problems. log tand.. 16° 0’ 46”.... 9°4578618 tand. .. 52°12’ 85”... 10°1104697 The sum is log cos PB .. 111° 49’ 21” .... 95683315 Heresince the latitude and declination are both of the same kind, cos p is negative, and belongs to an arc be- tween 90° and 180°. This hour angle converted into time by-dividing by 15 or multiplying by 5s gives 7* 26" 534° from noon, for the time required. In cases where great accuracy is needed, the change of declination in the interval must be regarded. To find the azimuth, take the theorem : ‘cos Z = sin d sec, Log sind ..16° 0'46”.... 9:4406756 sec J ,, 52°12’85” ..,. 10°2127004 ‘Thesum-cos: z .. 63° 14447 _ azimuth from the north {** | 9°6533760 Example It, - -Required.the time of the sun’s rising and setting at the ’ equator, on the 14th of October, 1816. It is‘not necessary in the solution of this example to dook: out the: declination: for, since / the latitude is 0, tan / is 0, and consequently cos p = 0: therefore P = 90°, the hour angle of 6; that is, the sun rises and sets at 6 o’clock. Cor. Hence it appears that at the equator all the hea- venly bodies are 12 hours above and 12 hours below the horizon; neglecting the effects of refraction. Example It, Required the times when the sun is cast or west, at _ fe fob ate fhe Bech seth siti) LO x eee OV oe ee co genet 0 a Astronomical Problems. 168 Cambridge, on May 4th, 1816, and his altitude at those times, 4 " The theorems applicable to this example are 7 and 8, viz. ) } cos P = cot 1 tand, and sin a = sind cosec J, Thus, for the hour angle Log cot 1.. 52° 12’ 35” .... 9'8895736 tand.. 16° 0 46” ....9°4578618 The sum.cosP.. 77° $8 27” | aie hour angle from noon t » 93474354 Then, for the altitude ; Logsind ..16° 0'46”.... 9:4406756 cosec 7.. 62°.12' 35”.... 10°1022306 et The sumsina.. 20° 25’ 48”.... 9°5429062 Note. By means of the equations to this example, the solar time when the sun is due east or west may be deter- mined: and, therefore, in the summer half year, when the sun is above the horizon, when he is in the prime ver- tical, he will at those moments cast the shadows of opake objects in the east and west direction, and a horizontal perpendicular to such shadows will be a meridian line. Example IV. Given the latitude of the place 52° 12’ 35” N. the sun’s declination 15° 54’ 25” N. and the sun’s altitude 40°; to find his azimuth and the time from noon. Here the time and azimuth may be found by means of equa. 12and13. Or, the azimuth being found by means of equa. 13, the time may be found by the proportion between ‘the sines. of the sides and opposite angles of spherical triangles, when we shall have : sin Z sin ZS F sin Pp == ——-—————. = Sin Z cos a secd; sinsP 164 Astronomical Problems. The operation, which gives 119° 53’ 8” for the azimuth from the north, and 2" 55™ for the time from noon, is left for the learner’s exercise. Note. From such an example as this it will be easy to find the meridian, by simply drawing a line to make with the direction in which the sun is at the moment of observation, an angle equal to the azimuth found by this process. And this method, provided the sun’s altitude be corrected for refraction and parallax, will be found © very commodious in the determination of a meridian, as the changes in declination, which prevail in the method by corresponding altitudes, are in this avoided entirely. The same method serves evidently to determine the va- riation of the compass. Example V. In the datitude of 52°12’ 35” N. when the sun’s de- clination is 15° 54°25” N. what is the sun’s azimuth at 9» 5™ in the morning ? The azimuth, which is 119° 53’ 8” from the north, may be readily found, as well as the angle at the sun and the zenith distance, by the equations marked 14 in the pre- ceding investigation, The work is left for the pupil’s exercise. | : Note. An operation, like that required in this exam- ple, will serve to find the declination of any vertical plane, by knowing when the sun begins to shine uponit, or quits it. Example VI. At Peferborough, in latitude 52° $2’ N., on February 21, 1801, at what time from that star’s being on the me- ridian, was the apparent motion of Dubhe or « Ursa majoris vertical ; and what were the altitude and azimuth of the star at that time? On the year and day specified, the declination, ¢, of the star was 62° 50’. The logarithmic opperation coz responding to equa. 9, viz. cos P= cot d tan J, Astronomical Problems. 165 will give 47° 57’ 46”, equivalent to 3°11" 51%, in time, for the hour angle from the meridian. Also, equa. 10, or sin z = cos d sec J, will give 48° 38’ 28” azimuth from the north : And equa. 11, or sin a = sin / cosec d, will give 63° 8’ 29”, for the altitude of the star. The operation is again left for the learner’s exercise. Note. Allied to the principle of this example is the method of finding, not merely the meridian, but the la- titude, simply by observing two azimuths of the same known fixed star with the azimuth compass. If the azi- muth of a circumpolar star be taken on each side of the meridian when at the greatest from the elevated pole, that is, when the apparent motion of the star is vertical ; then it is evident that half the sum of these two azimuths will be the true azimuth from the north or south, ac- cording as the latitude of the place is north or south, and that half the difference will be the variation of the needle. Also, recurring to the diagram at page 158, there are given in the triangle pzs right angled at s, the star’s co- declination ps, and the azimuth pPzs, whence the co- latitude zp becomes known by means of Case 4 of right angled triangles; from which we have cos d cos f == ——— = cos d cosec Z sin Z@ Example Vit. At what time on the 9th of August, 1817, will the twilight begin and end at Cambridge ? Here the sun’s declination for the proposed time is 15° 55’ 25” N. and the latitude 52° 12’ 35” N. Putting 108° for zs, 37° 47’ 25” for pz, and. 74° 4’ 35” for ps, in equa. 12, the resulting hour angle reduced_to time will give 10" 122™ for the time from noon when twilight bes gins and ends. | | 166 Astronomical Problems. Proxswiem III. 13. Given the latitude of a place to find the time and the duration of shortest twilight. Let p be the elevated pole of the world, z the zenith of the observer, s the sun 18° below the horizon, zs = 90° + 18°, or more generally zs = 90° + 2a. The observer, whose zenith is z, will then see the commencement of. the morning twilight. By reason of the diurnal ro- tation, the declination circle Ps turning about the axis will bring the sun from s to s‘ in the horizon, if it be the heavens which turn about the earth: on, the contrary hypothesis of the earth turning on its axis, the zenith z will approach nearer to s by de- scribing about r the little circle zmba, such that the dis- tances PZ, pm, Pd, PQ, are all equal. The same consequences as to this problem may be drawn from both hypotheses; but the latter is a little the most convenient. When, therefore, the zenith shall have descended from z to any point m, such. that-ms = 90° the:sun will ap- pear at 90° from the zenith, the day will commence and the twilight terminate; and the arc zm of the little circle will be the measure of the angle zpm, and consequently of the duration of the twilight. ‘To determine this angle draw the are zpm of a great circle, and. to its middle s the perpendicular arc pz; then will wis weet sing zm sind zm sin 5 ZPm = Sin. ZPB = ——_ = —_——_: sin PZ cos L Now, in the spherical: triangle zms, sm + mz > zs, or 90° + mz > 90° + 2a. Therefore mz > 2a, and ynz >a. Let jmz—a-+ ax: then ne (sina+ a) — sinacos2# + cosasin x as cosL cos L Astronomical Prodlems. 167 sin a-+ cos.asin x — 2sina sin hz —o |e RR ae | ee SD we cos L {because 1 — cos. = 2sin *}a. ch. iv. art. 23.] _ sina | 2sin $x (cos acos dx —sinasin$x.)- Shinto) hehe cos L sina . 2sin3xcos(a+ $x) p2TesS a | cosL cosh Now 2a is essentially positive, and a and. $x small ans gles such that a + 3x < 90°; therefore sin. $zpm > : cos L It is farther evident that the twilight-wilkbe longer as zx is greater, and shorter as x is less ; and that it will be the shortest possible if x canbe nothing ; because in that case.the second term of the second member of the equa- tion will vanish. But this.is.what, occurs when. the, tri- angle zms is reduced to: the are.zs, that is, when the point m falls on 4, or, when the distance Ps: is such that the part zb of the vertical zs lying within. the little circle, zm is equal to 2a, andthe exterior part bs equal to 90% It is also manifest that if ps increases the opposite, angle pzs will become enlarged, and that on the contrary the said angle will diminish if ps diminishes. . In these varia-, tions the point m will approach-to’ or recede from,z, the intercepted part zd will vary, and it may vary between thelimited Oand z@ = 2pz. Thus the,intercepted part © may. have all: values between O'and 2(90° — L) = 180°, — 2x, and-consequently may have the.value 2a. Hences in the case where zb = 2a, and 6s =.90°, the shortest twilight will obtain; and its semiduration in degrees will be found by the equation. . ‘ sin @ . sin zPB = ——=sinasecL.. (1.) cos L. sin. a duration in time = -2,.ZPB .... (Z.) 14, Other theorems may be deduced from the resulting : 7 168 Astronomical Problems. construction. On thearc zb = 2a and with 7 the complement zp of the latitude, consti- tute the isosceles triangle zpd, and let-fall the perpendicular pm; zprd will be the an- gle which measures the duration. Prolong ® zd till bs = 90°, and draw the arc ps, which will be the sun’s polar distance for the day of the shortest twilight. T COS PZ cos “"(chap. vi. 27.) i Now, cos mP = ee therefore cos Zm: Cos mS :: Cos PZ: COS PS; or, cos a: cos (90° + a) :: sin L:sin D; or, COS @:— sin a@::sin Li: sinD; sin a whence sin D = — —— sink = — tan asin x (3.) Cos @ 15. Farther, the right angled triangle pzm, gives ‘tan zm = cos ztan pz (chap. vi. 26.) and cos z = tan Zm cot Pz = tana tan (4:.) Now z is the sun’s azimuth at the commencement of the twilight, and pz) = phz = 180° — pds. But vés is the sun’s azimuth at the instant when his centre is at the true horizon: therefore the sun’s azimuths at the begin- ning and end of the crepusculum are supplements to each other on the day of the shortest twilight. Hence, since cos pzs = tan @ tan L we have cos pbs = — tan atant.. (5.) 16. zps and bps are the hour angles at the beginning and end of the twilight; let the former be denoted by P’, the latter by p: then zps — bps = Pp’ —p = zpd, the. angle which measures the duration of twilight. Hence we have from what has been done above, : ; _ sina sin 3(P’ = P) Ae se (6.) Also mes = bps + mph = p+ }(pP’ —P) = 3(P’ + P) sinsm _ sin(90° + a) __ cos a. and sin mps sin PS CoeB CB” Astronomical Problems. 168 heir (2s) Hence from equa, 6 and 7, we may determine P’ and P, the hour angles at the beginning and end of the twi- light, p the sun’s declination being determined by — equa. 3. cos a _thatis, sin $(2’ +P) = es, sin pee (8.) ‘ cos D 17. Let st = 2a = 2b; and draw pt, then is zt = 90°, and T-is a point in the horizon for the moment when the zenith was in z. The triangle pzr gives (chap. vi. art. 21.) COS PT = COS Z SiN PZ Sin ZT + COS PZ COS ZT = COSZ SIN PZ = tan a tan Lcos L = tana sin L = sin D, Therefore 90° — pr = p, or PT = 90° — D. But ps = 90° +- p; therefore pT + ps = 180°... (9.) This is another remarkable symptom of the shortest twilight.* We have also cos s = Example. Required the day on which the twilight is shortest, at Woolwich, N. lat. 51° 28’ 40”, in the year 1816, with its duration and the time of its beginning and end. First, for the declination, sin Dp = — tan asin & SOW Cel gg tf 77s MRR» aD ai . 91997125 Add log sin u .. 51° 287’ .... 98934104 Thesum,logsinp., — 7° 7 5” .. 90931229 _Thus the declination is south, and answers to March gd, and October 11th. : * Many other curious particulars connected with this problem are given in Astronomie Théorique ct Pratique, par M, Delambre, chap. 14; whence the above investigation has been extracted. Another elegant investigation may be seen in Connaissance det Teme, pour Van 1818, pp. 382404, rt _ - 170 Astronomical Problems. Next, for the duration, sin $ (2° — Pp) = —— From log sina, S199. «- 91943324 Take log cos L.. 51° 283’.... 9°7943613 Rem. log sin.... 14° 32’41”.. 9:3999711 Hence pv’ — Pp = 29° 5’ 22”, or in time 1" 56™ 213°. ° y cos a Farther, sin }(P’ + P) = Hence, from log cosa... 9°.........- 9:9946199 takelogcosp .. 7° 7 5”.... 9°9966399 Rem. log sin,. ..95° 31’ 19”.. .. 9°9979800 This value of 1 (P’ + P) reduced to time is equivalent to 6" 22" 51°, while 4 (P’ — Pp) is 58" 103. The sum of these is p’ = 7" 20" 16}, the end of the evening twilight; their difference is p= 5" 23" 543°, the time of the sun setting or the beginning of evening twilight. ‘These re- spectively taken from 12, leave the times of the begin- ing and end of the morning twilight. PrRoBLEM LV. 18. Assuming the error in taking an altitude, it is re- quired to determine the corresponding error in time, in the case of example 4, prob. ii. Let £Qq in the marginal figure be the equator, Pp the elevated pole, ad the parallel of de- clination in which the sun or other 5 heavenly body is on the day of ob- servation; and let 7 be the real and s the apparent place of the body, or rs the error. in altitude. Draw ms parallel to the horizon, from z the zenith draw zm, zr, and from -P the pole, the meridians, pmp, Prq, to Astronomical Problems. 171 pass through m, 7, and cut the equator in the points, g. Thenm and r will be the apparent and real places of the body on the parallel of declination, and the arc mr on that circle, or yg on the equator, will measure the angle mpr, the corresponding error in time. The triangle msr being, of course, exceedingly small, may be regard- ed as a rectilinear triangle right angled at s. Hence, sr: rm::sinsmr: rad. and mr: pq:: cos gr: rad. Conseq. multiplying the corresponding terms we have Sr: pgs: sin smr cos gr rad? rad Whence pq = oF al par cong viele fb) But zrp = smr, srm being the comp. of each; and sin zrp (= sin smr) sin PZ:: sin rze: sin pr; Conseq. sin smr sin pr = sin smr cos gr = sin pz sin rzp. Therefore, by substituting for the denominator in equa. 1, its equivalent in the last, we have rs. rad? rs AT ail Meal rER 7 6 Ce Siherncs 7 (2.) Cor. Hence, since sin azimuth is. a mazimiim when the body is on the prime vertical, the error in.time will then be a minimum, the refraction-being accurately allowed for. Wherefore, Cor. 2. It is best to deduce the time from an altitude taken when the sun or other body is on or near the prime vertical, [See, on kindred topics, chap, xi.] Example. Suppose that in example 4, prob. 2. the error in alti- tude was 1’, what would be the corresponding error in time? Gas! Seam 1 i, 1 e % - Pd “Cos 59° 12 35" sin 119 53/8" = “198 x GOTO +882 This in time is = 7-528 seconds, 12 172- Astronomical Problems. PROBLEM V. 19. The right ascensions and declinations of two fixed stars, observed on the same vertical given in position, together with their distance, or the arc of that vertical included between them, being known; it is required to determine the latitude of the place of observation. If s,s’, inthe annexed diagram, be the places of the two stars in the vertical zso, HR Z the horizon, P the elevated pole, and EQ the equator ; then it is evi- dent that the three sides of the tri- angle pss’ are given, as well as the angle at p. For ps, ps’, are the H codeclinations of the stars (known O Q by hypothesis), ss’ their distance on the vertical given in position, and the angle sps’ the measure of the difference of their right-ascensions. Hence, by the proportiona- lity of the sines of opposite sides and angles, we shall have, sin SS’: SIN SPS:: Sin 8’ P:sin PSS’ = sin PSZ sins’P ,. = ——— sin SPs’. sInss Thus, two angles and a side opposite to one of them, will become known in the triangle pzs, namely, the azi- muth, pzs, the angle psz, and the side Ps; whence we shali have sin PZS: SiN PSZ:: Sin PS:sin PZ = cosL= sin Ps sin Psz _ sin Ps sin ps’ sin sps‘ sinPzs ~ sin ss’.sin Pzs = sin PS sin PS’ sin SPS’ COSEC $8’ COSEC PZS. 20. Cor. 1. It is evident that the hour angle zps may be readily found from the same data; and therefore if the sun’s place and his right ascension be known, the difference between it and that of the star, s, will become _ Astronomical Problems. 173 known; and if from this difference the arc ED be taken, the remainder will give the arc of the equator contained betwixt the meridian of the place and that which passes through the sun, whence the moment of observation may at once be deduced. It is equally obvious that, by pursuing the computation, the altitude of each of the stars at the moment of observation may be found, 21. Cor. 2. If, instead of the azimuth of the two stars, their altitudes were given, we might readily determine the elevation of the pole; for, in that case we should know the three sides of the triangle pss’ with the angle p; and thence the angle psz between the given sides Ps, zs, of the triangle psz, 22. Cor. 3. So again, if besides the declinations and right ascensions ofthe stars observed on the same verti- tical, the hour angle zrs were given, the latitude might be ascertained; for then, in the triangle psz, the angles ZPS, ZSP, and the side sp, would be known, and zp would be determinable by case 3, prob. 3, of oblique spherica) triangles. 23. Cor. 4. Hence, from this problem and corollaries the latitude of a place may be determined with great fa- cility ; a plumb line serving to show when two stars are in the same vertical, and the altitude being susceptible of being taken with sufficient accuracy for common pur- poses without employing large instruments. But correct tables of the places of the stars are indispensable. Example. Two stars, the right ascension of one of which s is 78° 94’, and its declination 27° 25’, the right ascension of the other, s’, is 104° 52’, and its declination 12°18’, of the same kind with the former, are distant from each other 28° 30’, on a vertical whose azimuth Ezo is 73° 36’. Required the latitude of the place ? Here rs = 90° — 27° 25’ = 62° 35, ps’ = 90° — 12°18’ = 77° 42’, 174 Astronomical Problems. 53° = 28° 30’, sps’ = 104° 52’ — 78° 94’ = 26° 28’, and PzZS = 180° — 73° 36’ = 106° 24’. The logarithmic computation from the formula is Log sin ps .... 62°35'...... 9°9482572 SBS Ne Ll OMS eo eu hahe 9:9899148 CON OO RS iS: RE 9'6490203 DORE SE. ic, oh ee ook tec, diate 10°3213371 cosec PZS .. 106° 24’ ...... 100180392 Logcosi;..... 32°23’ 18” .. 9:9265686 Hence the latitude is 32° 23’ 18”, of the same kind — with the declinations of the stars, ProsieM VI. 24. Given the apparent altitudes of the moon and sun, or of the moon and a fixed star, and their apparent distance to determine their true distance. This is an important problem in navigation, being of essential use in one of the best methods of determining the longitude at sea. The true distance of the moon and the sun, or a star at any given instant, being ascer- tained, we are enabled by means of the ‘ Nautical Al- manac,”’ and the Ephemerides of different countries, to find the time at the first meridian, which corresponds with the moment of observation : the difference between these times, reduced to degrees at the rate of 15 to an hour, shows the longitude required. Besides the error arising from the imperfection of in- struments, the observed altitudes of the heavenly bodies are affected by three causes, the depressionof the horizon, the refraction of rays of light in passing through the air, and the parallaa, or the inclination of two visual rays gassing from the celestial body, one to the earth’s centre, the other to the potnt where the observer ts placed on tts surface. The principal works on nautical astronomy contain tables by which the requisite allowances may be Astronomical Problems. 175 made for these at any place and for any heavenly body: and the purpose of making such allowances is, to reduce the observed or ‘apparent altitude of any body at or above the surface of the earth, to the real altitude at which it would appear from the earth’s centre if light were transmitted in right lines. 25. The apparent altitude of the sun always surpasses the true; for the refraction elevates the sun more than the parallax depresses that luminary, With regard to a fixed star the parallax Zz -is evanescent. Hence, the true place of the sun, or’of a fixed star, will always be below the apparent’ place. The moon, on the contrary, is more depressed by parallax than it is elevated by refrac- M s tion. Hence the true place of the moon ™ S will be’always above its apparent place, 26. Let, then,’s be the apparent place of the sun orstar, s its true place, on'the same vertical zs, m the apparent place of the moon, M its true place, on the same verti- cal.zm,,ms the. apparent distance of the two bodies. Then, there are known H = 90 — Zs, apparent altitude of the sun or star; Hu’ = 90° — zm, apparent altitude of the moon; D = ms, their apparent distance: also, h = H—r +p = 90° — zs, true altitude of the’sun or star; and h’ = wh’ — 7 + p= 90° — zM, true altitude of the moon; rand p, 7 and p’, being the refraction and pa- rallax corresponding respectively to the apparent alti- tude of the bodies. Consequently, in the spherical tri- angle zms are given all the sides to find the angle z: and then in the triangle zMs are given two sides 2M, zs, and their included angle z, to findthethirdside ms = d the true distance required. The solution of the problem thus-conducted.is, obviously as simple and natural as can well be wished, to men acquainted with theory and ac- customed to computation: yet it has been found em- barrassing to mariners; on which account most writers on the subject. of navigation have investigated other 176 Astronomical Problems. rules, several of which (especially those which : depend upon subsidiary tables) are more direct and expeditious in operation than the original rules deduced without mo- dification from the two triangles zms, zMs. 27. The second general theorem of spherical trigono- metry, when applied to these two triangles, gives cos sm — COS Zm COS Zs =" cos SM — COS ZM €os ZS ° cos Z = sin Zm sin Zs ™-<- sin zm sin zs Or, employing the preceding notation, } cos D — sin Hsin H’ cos d — sin hsin hk’ cos Z = SS cos It cos 1! cas h.cos hi - But since cos (H + H’) = cos H-cos H’ — sinjH sin\ H's we have sin H sin H’ = cos H Cos H’ —. Cos (1a +H’ wo and, intike manner sin A sinh = cosh cosh — cos (h +.h’)3: cos D + cos (1 + B’) — CosM COSAS [fy whence, cos 2 = cosH cos H! ml, cos d + cos (h + h') cos h cos hi ie. aaa nn TT r. cos h cosh Taking from the second member of this equation cOs HCos H’ . , coshcos hi’ ~ ———_———— = ], and fromthe third —————,, = 1; we COS HCOS H cosh cos h ete have for the result. te i cos D + cos (H + BH’) cos d + cos (h +h! ~ : fo rte Gee a Pit oe tay cos H COs H © cosh cos h Now it is evident from the formuls in chap. iv. that cos D + cos (uH +H’) = 2 cos 3 (H +H’ +p)! « cos $(H+H’—D), cos (h + h’) =2 cos? (h+h’)— 1, cos d = 1 — QZsin® id. These values being respectively substituted for the se- veral quantities in se ak A, it will be'‘transformed into 2 cos$(H + Hw’ — vd) cos$(H + H’ + D) cos Hos H! 1 — 2sin?3d + 2cos?h(h +h’) —1 cosh cos h’ 7 = —— Astronomieal Problems. 177 Multiplying by cos / cos h’, and reducing, we have cos h cos hh’ D+H+ FH D+nH +H’ cos H Cos H’ ( 2 Eh = — sin? 4d + cos*4(h +h’).... (B) This, ifs be put = 4 (D + H + H’) becomes, cos A cos hh’ 3 = — sin? 4d + cos?3(h +7’). —»v) cos cos cos (S —D) —— Hence 2 cos s cos (s — D) cos kh cos h’ sin? 1d = cos? 3 (h + h’) — nS 8 OWA BD) COSA a cos H COS H = cost (t+ h)(1 — cos scos (Ss —D) cos h cos a cos H cos A’ cos? $ (hk + h’) : cos s cos (s — D) cos hcos h’ sec wsec B’ (c) cos? $ (h + h’) iy Then the preceding equation becomes sin? 3d = cos?4$(h +h’) (1 —sin? F) = cos? 5 (A +h’) cos’F, whence sin 3d = cos 3 (h + h’) cos¥.... (D) This is the formula of Borda: it has the advantages of requiring only the usual tables, and of being free from ambiguity. ) 28. The legarithmic formule deduced from equations c and D, are 2 log sin ¥ = log cos-s + log cos (sD) + log cos h + log cos h’ + log sec u + log sec H’ + 2logsec3 (h+h’) —60.... (£) log sin 3d = log cos 3 (h + h) + log cos¥ — 10.. (£) The secants being introduced into the numerator for the cosines in the denominator, render the logarithmic computation entirely additive. Put sin? F = Example. The apparent distance of the moon’s centre from the star Regulus being 63° 35’ 13”, when the apparent alti- tude of the moon’s centre was 28° 29’ 44”, the apparent altitude of the star 45° 9’ 12;” the moon’s correction, or the difference between the refraction and the parallax in 15 178 Astronomical Problems. altitude 48’ 1”; the star’s correction, or the refraction 57”. Required the true distance ? Ist Method, directly from the triangles zms, ZMs. In the triangle zms (last figure) are given zm = 90° — 28° 29’ 44” = 61° 30’ 16”, zs = 90° — 45° 9°12” = 44° 50’ 48” sm = 63° 3513”: to find the angle z. This may be effected by chsvi. art. 38, equa. 2. which, ° suited to the present case, becomes rc a \/ as (am + ms — sz) He i (ms + Zs — zm) sin $ (zm + Zs — ms) sin $(ms + Zs + Zm) and this, when performing the logarithmic computation, may be best accomplished by adding the log cosecants of the terms in the denominator, instead of suhstracting their log sines, and deducting 10 from the index, at last. Thus, sin § (zm + ms — sz).. 40° 7° 20”.. 9.8091692 sin 4 (ms + Zs— zm).. 23° 27’ 53”.. 9°6000842 cosec 4 (zm-+ zs —ms).. 21° 22’ 38”. . 10 4382946 cosec § (ms + zs + zm)., 84°58’ 8”. . 10°0016765 2 | 39°849224.5 The halfsum — 10 is tan $z.. 40°3’ 7” .. 9:9246122 Pra Consequently z = 80° 6’ 14.” Then there are given, in the triangle zms, ZM = 90° — (28° 29! 44” + 4871”) = 60° 42° 15” Zs = 90°.— (45° 9’ 12” — 57’) = 44°51 45” and the included angle z = 80° 6’ 14”. The third side may be obtained as in case 2, prob. 2, by finding a subsidiary angle @ such that tan 9 = cos 80° 6 14” tan 44° 51’ 45” In logs...... cos 80° 6’ 14” .,... 9°2351804 tan 44° 51’ 45” .... 9°9979155 The sum tan? = tan 9° 42’ 21” .... 9°2330959 ~ PoLbBasma Astronomical Problems. 179: cos MS = 44° 51’ 45” cos (60° 42’ 15” — P) sec Oe In logs...... cos 44° 51’ 45” .... 9°8506086 cos 50° 59’ 54” .... 97938874 sec 9° 42’ 21” ....10°0062613 The sum cos Ms = cos 63° 5’ 11”.... 96557573 Thus the true distance between the moon and star is found to be 63° 5’ 11”. (Sa aE Qdly. By Borda’s Theorem. Here we have # = 45° 9’ 12”, h = 45° 8’ 15” WW’ = 28° 99’ 44, h’ = 29° 17°45", h + Ko 742 26 s = 68°37’ 45,"; g iD S671 ST S- The logarithmic computation for r, therefore, is. POSTS ot Sak 68° 87°44", 1. SO S6L 7995S cos(s—D).. 5° 1° 51%”...- 9°9983236 COSA. eb: 45° 8°15” .... 9°8484403 CORR: gine t 4. 99° 17 45” .... 9°9405687 SOC TE Ste. Xe 45° 9°12” ....10°1516804 BEC cree 98° 2Y 44” .,..10°056083Z2 Qsech(h +h’) 37°13’ 0” ....20°1977876 The sum — GO =2sin F.... seas 19°7546833 Its half issin¥.... 48°56’ 1” .... 9°8773416 Then to find 3d, and thence d, we have COS BoM hehe 48° 56° 1” .... 98175211 cos $ (kh + h’)..87°13’ O”.... 99011062 The sum is sin $d......31° 32’ 34”... 9°7186273 Consequently d is 63° 5’ 8”. | Note. Here the difference in the results is only 3 se- conds ; which is quite as little as can be expected when tables are employed in which the computer has to pro- portion for the seconds. Had the value of r been takem 180 Astronomical Problems. 48° 56’ instead of 48° 56’ 1” (and it is evident from the common tables that it lies between the two), then d would have been found about 63° 515”, The mean be- tween these two results is 63° 5’ 114”. Example II. Given the apparent distance between the centres of the sun and moon 83° 57’ 33”, the apparent altitude of the sun’s centre 48° 27’ 32”, its true altitude 48° 26’ 49”, the apparent altitude of the moon’s centre 27° 34’ 5”, its true altitude 28° 20’ 48”. Required the true distance of the centres of the two luminaries, igse 83°20 55% Example UT. The apparent distance of the moon’s centre and a star was 2° 20’, when the apparent altitude of the star’s cen- tre was 11° 14’, and that of the moon’s 9° 39’; the moon’s correction was 51’ 30”, the star’s 4°40”. Required the true distance of their centres, Ans, 1° 49, Nate 2. In examples of this kind the sum of the appa- rent zenith distances must always be greater than the apparent distance; for if not they cannot form a spheri- cal triangle. Thus it will be found that example 4, pa. 31, of the Requisite Tables, 2d edition, relates to an im- possible case. The third side 103°29' 27” is greater than 89° 50’ 22” the sum of the other two. Note 3, ‘Tlr3 being an important problem in nautical astronomy, we shall here refer to other works where more compendious rules, founded principally upon sub- sidiary tables, are given. Such are, the Requisite Ta- bles to be used with the Nautical Almanac; Mackay on the Longitude; Mendoza’s Tables for Navigation and Nautical Astronomy 3 Myers’s Translation of Rossel on the Longitude; Andrew’s Astronomical and Nautical Lables ; Kelly’s Spherics. See also, Mr. Sanderson’s e 5 Astronomical Problems. 181 rules in the Ladies’ Diary for 1787, or.in Leydourn’s cole lection of the Diaries, vol. iii. Dr. Brinkley’s in the Irish Transactions for 1808, and various others in De~« lambre’s Astronomy, vol, iii. chap. 36. Prosiem VII, 29, Given the longitudes and latitudes of two places upon the earth, regarded as a globe, to find their itine- rary distance, that is, the arc of the great circle com- prehended between them. Let a and B be the two places on the surface of the terraqueous globe, of which Pp is one of the poles, and conceive a spherical triangle pas to be described, such that PA, PB, shall be the respective distances of the two places from the pole p, or their respective co-latitudes, and the angle aps the difference of longitude of those two places. Hence, t and x’ being the respective lati- tudes, and Pp the difference of longitude, there are given in the triangle pas, two sides, viz. pA = 90° = 1, PB = 90° = L’, and P, to find the third side an. The problem, therefore, belongs to case 2, prob. 2, of oblique spherical triangles, the appropriate formule for which, suited to the present case, become tan? = cosptan PB.... (1.) for the subsidiary angle: and cos AB = cos PB sec ?cos (PA — @).... (2.) Example I. Given the latitude of the observatory of Paris, 48° 50’ 14” N. that of the observatory of Pekin, 39° 54’ 13” N. and their difference of longitude 114° 7’ 30”: to find their distance. | ' : Here. since the latitudes are both of the same kind, we have a | PA = 90° — L = 90° — 48° 50’ 14” = 41° 9’ 46” PB = 90° — L’ = 90° — 39° 54°13” = 50° 5’ 477 andp = diff. of long. — UD ARN BE Nie 182 Astronomical Problems. Hence to log cosp.... 114° 7’ 30”.. 96114352 add tan PB,.... 50° 5’ 47”, . 10:0776707 The sum — 10 = Jog tan 9—26° 2’ 53”,. 9°6891059 Now, since p exceeds 90° its cosine is negative, and consequently @ must be taken negatively. Hence PA — 9 = 41° 9 46” + 26° 2’ 53” = 67° 12’ 39”, and the work indicated by equa. (2) will stand thus: Log cos PB’...... 50°) 5 4°7.", .,..,.9°8071953 cos (PA — $) 67° 12’ 39%.... 9°5880938 SAO Ox ssh oe 26° 2 53”,...10°0465177 The sum — 20, cos AB 73° 56’40”.... 9°4418068 Thus the distance required is 73° 56’ 40” of a great circle, or in English miles, reckoning 693 to a degree, it is 5139 nearly. Example Ii. The latitude of St. Helena is 15° 55’ S. its longitude 5° 49’ W.; the latitude of the Bermudas 32° 35’ N, lon- gitude 63° 32’W. Required their distance ? Here since the latitudes are of different kinds, we shall have PA = 90° + L = 90° 4+ 15° 55’ = 105° 55’ PB = 90° — L’ = 90° — 32° 35’ = 57° 25’ and P = 63° 32 — 5° 49’ = 57° 43’. The preceding formule applied to these data, will give 73° 26’ for the distance required. Example Il, What is the distance on a great circle between St. Mary’s, latitude 37° N., longitude 25° W., and Cape Henry on the same parallel of latitude, but in longitude 76° 23’ W.? Ans. 40° 31, which is about 36 miles less than the distance measured upon the parallel itself. 7 Astronomical Problems. 183 Exampie IV. Required the distance on a great circle between the island of St. Thomas, latitude 0°, longitude 1° E., and Port St. Julian, in latitude 48° 51’S., and longitude 65° 10’ W. Ans, 74° 35’. Note. In the 3d example the operation becomes shortened because the triangle PAB is isosceles; in the 4th it is shortened because the triangle is guadrantal. Prosiem VIII. 30. To determine the elongation of a planet from the sun, at the time when it appears stationary; on the sup- position of concentric circular orbits, in one and the same plane. As thus restricted the problem may be solved by means of the principles of plane trigonometry, combined with those of central forces in dynamics. Let two concentric circles Ee1, Ppa, be drawn, the former to represent the orbit of the earth, the latter that of the planet, the common centre s of both circles being imagined the place of the sun. While the earth moves from £ to e in its orbit, let the planet meve from P to pin its orbit; then, when the planet appears sta- tionary the right lines ep, ep, drawn from the earth to the planet will be parallel. In that state of things draw the radii sz, se, sp, sp; from £ and Pp draw tangents to their respective circles to meet each other in T; and join sr. So will the triangles srr, spr, be right angled at E and Pp; and the parallel lines ep, ep, will be so near each other, that the intercepted arcs Ee, pp, of the orbits, may be regarded as coinciding with the corre- sponding portions of their respective tangents. Let v be the velocity of the earth in its orbit, v’ that of the planet. Then, since the arcs Ee, Pp, are de- scribed in the same interval of time, they will be to each other as those velocities, that is, | Ee: Pps Vi: V 184 Astronomical Problems. Also, since the arcs ze, pp, are confounded with the portions of their respective tangents which meet at rT, and the parallels ep, ep, divide TE, TP, proportionally, we have ET: pT:: Ee: Pp; and consequently, EPTPT ee VV. Put se = a, sp = 8, and the required angle sep of elongation = E. Then, since SP: SE::sin E: Sin EPS, we shall have e a. sin EPS = COS EPT = — sin E, Whence, since sin? EPT = rad? — cos? EPT, we have - Me sin EPT = »/(7? — = sin S503 Farther, because cos SEP = sin PET, we have sin PET = ,/(7? —sinE). But sin EPT: SiN PET :: ET: PT:: Viv. 2 Therefore, v: Vv’ :: ./(77 — = sin?E):4/(r? — 8in7E). 2 Whence v? (7? — sin 2B) = v? (7? — = sin 2B); tp sfirtsan 2 ee Ty Le sin? E = 770? 3 Vv and sinE = + rb aie ao a rie alle) 31. A still simpler and more convenient expression for sin E may be obtained by introducing the radii of the respective orbits for the velocities of the bodies. Thus, by the theory of central forces (Gregory’s Me- chanics, book ii, art. 282) v: v’:: /6: /@; — y2 eV us tre J we have 6:a::r? — Gr sin? BE: 7? — sin *E, Hence (43 — ab?) r+ = (63 — a3) sin *z, 2g b3 — ab?) r? or, sin* E = cor: maak Supposing r and a to be each unity, we shall have _ Astronomical Problems. 135 “4 B tee oe b? 27 = SS Se SS? 03 “Selon hae Le Be b I = sb ie rink sin E = TP tbt (2.) : b+ 1 Alse, cos E = A (1 — sin *E) = / wearmiet C) sin R b? and tan R= = eral ox (4s) Here the positive sign obtains in the case of the infe- rior planets, the negative sign in that of the superior planets. | Example. The mean distance of Jupiter from the sun being 52028 times that of the earth; required the elongation of the plans from the sun at the time of its apparent station : ; , The logarithmic operation for tan E indicated by equa. (4) isas follows: bi “From 2 log &.,.. 5:°2028.... 1°4324742 Take log (6 + 1) 6°2028.... 0°7925878 SDH 20 oe cs ee isc acess 206398864 Pees ©: °ts-half is tan B ..115° 35’... . 10°3199232 Here, the tangent being negative, it belongs to the second quadrant (chap. iv. art. 5). | Note. By like operations the mean elongations of the other planets at the time of their apparent stations may be found as below. DICTCORYS © a. e's 18° 12’ “Venus iid. a: AeS? 51" Marsan e0-T 2.0% <1 S6%a &: Cerene es LOG *C 76 Saturm (36.0%. ax 108° 4:7’ UTANUSias mien « fee LOS te * 186 Astronomical Problems. SCHOLIUM, _ $2. The inversion of our 4th equa. furnishes in cer- tain cases a convenient expression by which to approxi- mate to the mean distance of a planet from the sun. }2 For from tan —E = aye we find 6= i tan*z + tang (1 + } tan2n)?: an equation which gives the radius of the planet's orbit in terms of that of the earth, from the elongation at the instant when the motion of the planet became direct after having been retrograde. . Thus, in the case of Ceres it was found by observa- tion that # at the time of apparent station was 122° 37’ 40”; whence 4 was found = 32018. The slight dif- ference between this elongation and the one just given, arises from this,—that the latter was actually observed, the planet moving in an elliptical orbit, while the former is what would be observed if the orbit were circular. The following is the logarithmic operation, for find ing the mean value of % in. Ceres: Dinas Log 0:28) 22 oii ticssceces 13979400 Plogtan E ,.. 2.20. sek sy tee ,20:3873552 Sum — 20 islog 3 tan?= ..0-60995.%!,~ 11°7852952 Add .. ; @eeeses+re?8e8e ee Log (1 + }tan? £) = log .. 1:60995'... . 20:2068125 Its halfgys Savy. io, gtieantt as, 10°1034062 TOG LANE oes 6.0 Be esa ate 00 oe LO TOSETTE! Sum — 20........log + 1:9819.....,. 0°2970838 § tan? © oy. e+. 12199 — The sum='6== 3:2018 SE er ee) Minute Variations of Triangles. 187 The negative value of 1:9819 cannot here obtain, because 6 would in that case be negative, which is im- possible, CHAPTER XI. On the Investigation of Differential Equations for esti- mating the minute Variations in the Sides and Angles of Triangles. : 1. By reason of the imperfection of instruments, the unavoidable though generally slight inaccuracies of ob- servers, the effects of parallax, refraction, the preces- sion of the equinoxes, the varying obliquity of the ecliptic (see-note, p. 119), and other causes, the sides and angles of either plane or spherical triangles, whe- ther observed on the earth or in the heavens, can never be taken with perfect exactness, It, therefore, be- comes necessary in cases where great accuracy is re- quired, to strike out some means of estimating the ex- tent of error which may be occasioned in certain sides and angles of triangles, by any assignable or supposable errors in the other parts. This is an interesting depart- ment of research, in which the celebrated Cotes in his treatise De c@stimatione errorum in mixté mathest, and many subsequent mathematicians have laboured with considerabie ingenuity and success. 2. The inquiry betore us is one (of very few in my estimation) in which the contemplation of magnitudes as augmenting and diminishing by differences leads to @ more natural and satisfactory explication, than that in which magnitudes are regarded as varying in conse- 188 Differential Equations. quence of motion. Hence I shall in this chapter em- ploy the notation of finite and enfinitesimal differences, mstead of that of flucions: although I am fully per- suaded that in a great majority of mathematiéal in- quiries the fluxional notation and metaphysics are pre- ferable to those of differentials. 3. When variable quantities augment or diminish by portions which are finite or susceptible of mensuration, the portions which constitute the augmentation or dimi- nution are called differences. If the variations, instead of being finite, are indefinitely small, they are called differentials. The former are aptly denoted by the ca- pital Greek letter A placed before the letter which re- presents the variable quantity, as Ac, Ay, Az, &c. the latter by the small or lower case Greek letter dy as on, dy, dz. The processes of differentiation and integration being similar to those of finding fusions and fluents, as taught in our standard works*, need not here be ex- plained otherwise than in connexion with the present investigation. 4. To determine the difference, or finite variation of the sie of an arc or angle, let us take from formula (vu), chap. iv. the equation, sin A — sinB = 2sin 3 (A — B) cos (A 4 B). Here, supposing A greater than B we may denote by Az, the augmentation which a must receive to become equal to B; and, in like manner, by 4 sin B, the differ- ence which must be added to sin B to make it equal to sin A. Thus, we shall have a = B + Az, and sin a = sinB + Asinp, Substituting these values in-the pre- ceding equation, it becomes AsinB = 2sin 3AB cos (B + LAB).... (1.) 5. To obtain the difference of the cosine, take from the same class of formule the equation, cosB— cosa = 2sin} (A —B)sin 3 (a +B); * See the treatises on Fluxions by Maclaurin, Simpson, and Dealtry. Minute Variations of Triangles. 188 and, performing an operation similar to the preceding, there will result, | — AcosB = 2sin 3Apsin(B + $AB)....(2.) 6. To find the difference of the tangent of an arc or angle, assume SIN A sIn B cosA cos B tanA — tans = sin A COSB — SINBCOSA aed —e COSA COS B sin (A — B) = COS A COsB ; and, operating as before, there will result, Atan B= eid eau i Oe a cOs B COs (B + AB) 7. Taking, in like manner, COs B cos A sia (A — B) cotB—cota = —— —- —— = ————_: sInB sin A Sin A SIL B we shall obtain for the difference of the cotangent sin AB DASE Ga, lanka eae Bilsinig test, 8. For the difference of the secant, take first, ] 1 __ seca — seen OEE aes C TEAS FeamD ay bec AY Mae wines a > == (sec A — sec B) COSB COS A: whence, form. v, chap. iv. 2sing (A —B)sind(a + 8B) gw Gow AGW HARE ot Kir at SeCA— secB = Making in this equation substitutions analogous to the preceding, we shall have 2sin $\Bsin (2B + JAR) " COSRCOS(R+ DR) 9. So, again, since ] CQs€EC A Ry cosec B A sec R= are PY sn A — SIN B= » we may by introdu- 190 Differential Equations. cing the requisite substitutions in the value of sn A — sin B, form. vu, chap. iv. obtain 2sin SAB Cos (B + SAB) = A osec = e eeee ° : x sin B sin (B + SB) (6 ) 10. These six equations are rigorously correct what- ever the magnitude of 4B may be. Let us trace the modifications they will undergo when the variation be- comes indefinitely minute, or Ag becomes ds. Return- ing to the first equation, we shall, by expanding cos (B + 348) according to form. vu, chap. iv. have Asin B = 2 sin AB (cos B cos $AB — sinB sin} JAB) = 2 sin $4B cossB cos {AB — 2sin? AB sinB = sin AB cosB — sinB (1 — cos AB). Now, if 4s be indefinitely small, so as to approximate very nearly to evanescence, sin Az will: also be indefi- nitely small, or practically evanescent, while cos As will differ indefinitely little from radius. In-that case the differential equation, will become dsin B = sin dB cos B — sinB (1 — cosdB); which, since sin dB = op, and coséB = 1, reduces to dsinB = OB cosB.... (7.) 11. Proceeding similarly with equa. (2), wevobtain — 9 COS B = OB SIN B, Or dcosB = —OBSINB.... 8.) 12. In like manner from equa. 3 and 4, we obtain % ¢ 'B —~....(9.) oB oB otan B= And so on, for the differentials of the secant and the cosecant. 13. Or, having found the differentials of the sine and cosine of 8, others may be deduced thus: Since versin B = ] — cos 8, we have éversinB = — dcoss = dB sinB.,,, (11.) ‘ sin B Also, since tan B = ——, we have Coss 4 Minute Variations of Triangles, 191 cos B)sins — sins) cosEe étanB = : cos? 8B (cos? B + sin?B) 3p a SS OE cos?B But cos? 8 + sin? s = 1: therefore éB dtan B = ——, as before. Cos? B 1 3 tans myers tans —ds tB cotB = — —— = —__— = _ __, tan? B tan? B Cos?B sin?B because tan B cos B = sinB, So again, since cotn = , e 1 Farther, since sec 8B = ——, we have COosB cos B SBSINB dB tan B Cos B cos? B cos? B cos?B $e tans = dp tan BsecB,... (12.) cosB e 1 Lastly, since cosec 8B = ——, we have $1n B é cosec B = —- —— = — eBcotB sinB 14. The differences and differentials of the principal lineo-angular quantities (chap. i. art. 4) being thus de- termined, we may now proceed to trace the minute varia- tions of the six parts of triangles. In order to this, the general method consists in determining the relation of any two differentials, To determine this we must dif- ferentiate the formula which expresses the general rela- tion of the quantities under consideration: the rule is very simple and well known to all who have studied the modern analysis. Let the formula be « = ayz + b, a and 6 being con- stant quantities, : = — dB cots cosecB ..,, (13.) 19% Differential Equations. Instead of x in the first member put x. ; In the second, put dy instead of y, and you will have > az, a there is a second variable quantity, put also ¢z for z, and you have oz . ay. | a and 6, being constant quantities, furnish no va- riation. Hence results the differential equation, dc = dy.az +02. ay; If the values of dy and dz are known, we thence know dx’ = dy.az, and dx” = dz .ay: so that the total varia- tion of x is constituted of two parts, the one depending on dy, the other on dz. 15. In the solutions of problems relating to the varia- tions of triangles, we have only to substitute one by one the differentials of the sine, cosine, &c. in the appro- priate formula of the problem, and proceed agreeably to the above directions. Thus, suppose that in the spherical triangle whose angles are A, B, C, and sides respectively opposite a, 6, c, the sides 6 and c were constant. ‘Then, from equa. (2) chap. vi. we have cosa = cosAsindsinc + cos 0 cos c. This becomes by differentiation, — dasna = — asin Asin dsinc. The formula has only two terms, because it contains ai more than two variable quantities. It results from it that ; da sin Asin bsine sin B sin 6 sinc sin csin & sine — aes Ey sina sin 6 sinc Or, finally, that . a ° ° Ld = en = sin B sinc = sin Ccsin 0. 16, Upon these principles a complete series of dif- ferential equations for all possible variations of spherical 193 and plane triangles may be investigated. But as the inquiry would occupy more space than can be devoted to it in this work, it must be omitted. We shall here, however, lay before the reader, with a few alterations, corrections, and additions, the valuable summary of trigonometrical differential equations given by Delambre in his Astronomie Théorique et Pratique. Minute Variations of Triangles. OBLIQUE ANGLED SPHERICAL TRIANGLES, I, The stdes 6 and c constant. da 4 , ‘ i — = SINBSINC = sinc sin} oA 'B cotc cesc da” sina sinAsine tC cotB cos B ~ Mace kein a. -SiD A Sin B cB tan B —_-= = tan Bcotc éc tanc SB sin B cosc sin 6Bcosc fac” sin A sin a cA sin A sin A — ee oO 6c cos B sinc cos B SING II. 3A ta 3b ca dc ry dc 06 &b ta Sc ta The angles B and c constant. sin dsinc = sincsing cotc SIN A cot b sin A tanc tan b sin BCcosc¢c sin A sinc cos B sin a K cosc sin Csin a cos b sin @sinB = tanec cot bd sin & cos ¢ ee sin «a sin C cos b See ae SIR A 194° Differential Equations. III. Angle s and opp. side a constant. de = cose 773 ose ob __ tand 3B tans Sc __. lane Qc tance éc __. cose ~~ 3B cosbd éc sin c cot c oe tan B cosc SMT: Tease ain © GAeeNe éc tan 6 cos c sin ccotc bie Baia see oy wales SOON IV. Angle a and adj. side c constant. Sa — = COSC cb cc tanec —— = = tanc cot-a oa tana cc = —- = COS @ oB ob sin a sin? a Lisl sin a sin b he 2a sinc sinAsine sin Bsine BY] tan @ =———- SF = tan da cosec.c oc sinc oa sina ‘ —_- = = sinacotec. 4 oB tane : Ricutr ANGLED SPHERICAL TRIANGLES. 1. The sides b and c constant. This case needs no formule: for two constant sides with a right angle render the whele constant. II. Angle 8 = 90°, angle c constant. cA ale or cota Sa cot a = cot A tana Minute Variations of Triangles. 195 Sb 2 cotb cosb ca sin 2a ne sin a cosa ee cotc _-cosb __ cotb cose dt. fata “bing “Sina tana se tanec jp = COSA = ay = tance cot b & sin 2b cose da sina ~~ sina Se __ sin 2¢ Sar Stata. II. Reght angle a and hypoth. a constant. &c cote ee aa cot c tan} BY : tan } oB Sahara Gah tanB “Se : tan c tc ain Oi tanc eee cons = sin 2c cosec 2B oB sin 2B ec cosc cos C ~ %% = cospsina cosBsind — = = a sine = 4 tan B sin 2c, IV. Right angle a and adj. side c constant. oa Ob éc Sa dc 3B. 3b ez = cosc = sin B cosSc = cotatan db = tanc cot a = cot B coseca = cos a = cos 6 cos c = cot B cute sin a sin 2b sin © sin 28 KE? 196 Differential Equations. 8b tana _ 2tand |. 2 She” cae oa sin 2a SB ~=6cotB = sina cotc. QUADRANTAL TRIANGLES. I. Side b = 90°, c also constant. ca cot a ee ee = cotatana oA cota SB 2cotB da ~ sin 2a tc cotc — = = cotc cota da tana $B tanB —_-= = tan Bcotc oc tanc cB cos c "Sa is ; 3A 2tana sc sin 2c Il. Two angles B and c constant. Here there are no formule, because there are neces- sarily three constant quantities in the complementary - rectangled triangle. III, Szde a = 90°, opp. angle a constant. &c sin 2c rsh gin 26 tb tand —= = tan 6 cot B 3B tans ce tance —_= = tanccotc 2, $C tan c tc cotc } = — <= cotc tans tB cots IV Minute Variations of Triangles. 197 dc sin 2B paz Siw 2cotdb dc _—« « Bcotc a a sin2p Side c = 90°, adj. angle a constant. fa “cota TE tig Teor SRR fad &e tanc __ tans One A tana sina 3c cotc a Oe ase ose cotB Yi) sin 2b oe ~ sin 2p yi) 2cot bd tan a de ~ ‘sin@c 7” sinc * #a* gin 2a Mi sina cB. QtanB ~ tance PLANE TRIANGLES. I. Two sides b and c constant. oA a sin A cK. bcosc sin Bcuos¢ fA a sin A te ae ee “ oC cCOsB Sin Cc COSB cA 1 ] da ~—é‘ SN csinB oB tanB —_ = = tan B cot¢ éc tanc . ‘B cot c cos C i ca a asinc RY cots cosB are BLD abs inte FR | la a @sing 198 Differential Equations. Il. Two angles 8 and © constant. Sa te ta 06 ze) A is of consequence constant. III. Angle a and opp. side @ constant, tanc . iB te $e de cae commeal COs C Sec B IV. Angle a and adj, side ¢ constant. fa 0D tc ta de 3B ib op 06 tc ca TB COS € tane a Cosec © ad COsec © a. cot-c, = Minute Variations of Triangles. 199 17. In the preceding table the differential relations are exhibited in equations, both because in that form they occupy comparatively small space, and because they are in the majority of cases more commodious in application than when presented after the manner of analogies. They may, however, at once be read and regarded as analogies, by taking the numerator and de- nominator of the fraction in each member of the equa- tion, as the antecedent and consequent of a ratio. Thus, Ma s sz = cos Cc, becomes da: 0b:: cos c:rad Cc tc tanc ‘ — + =~, becomes oc: —da::tanc:a, and similarly of others. 18. It must be borne in mind that the equations in this table are merely approximative, and that the results they furnish are more accurate in proportion as the va- riations are small. In cases where the greatest possible correctness is required, the theorems for the finite di/- Jerences of sines, tangents, &c. must be introduced into the appropriate formula for the individual problem, in- stead of those for the differentials; and the results em- ployed, instead of those tabulated in the preceding pages. We here present a few examples of the use of the formulze above given, Example I. Let an object whose height is Ac, be measured. by taking its angle of elevation anc, at a given horizontal distance Bc from its base. It is required to ascertain what error may be committed with respect to the height AC, in consequence of any supposed error in the ob- served angle B? 7 | _ The fourth equation, class 4 of plane triangles, is ap- plicable to this example : 200 Differential Equations. ay 1) a SAC BC BC vlZ. ——- = ——, or —— = —— = —_*, oB sin c oB sinc COsB But (chap. ii, 2, iv. 18), sin 2B = 2sin B cosB; J QsinB whence —— = — ’ cosB sin 2B RSG oAC 2Bcsin B 2ac and, by substitution, —- = ———— = - : Ae: 7. oR sin 2B sin 2n tly bac = 2 ihe consequen y AC = CB ain’ on This thrown into a proportion becomes, As the sine of twice the observed angle, To double the computed height ; So is the error of the observed angle, To the corresponding error in height: Corollary. Hence dac, the error in the height will be a minimum, when sin 2B is a maximum, or is equal to radius. That is, the error in altitude, ceteris paribus, is the least when the angle of observation B is 45°. Remark, If an error of a minute is made in taking the angle B, supposed of 45°, the corresponding error in the altitude will be its 1719th part: for in that case a AC e noaine 9) — Ac = °0002909 x Zac = AT AC 7719” nearly. Example I. Given the inclination of the plane of a theodolite to the horizon; required the greatest error that can pos- sibly happen in. determining the magnitude of an angle. The face of the theodolite when posited horizontally, and when inclined in any assigned angle, may be re- garded as two great circles of a sphere whose centre is the centre of the face of the theodolite. Also, if the * The figure, as in many other cases, is left designedly to be sketched by the student, Minute Variations of Triangles. 201 index of the theodolite is horizontal when it passes through the points 0° and 180°; when it is turned round to measure any horizontal angle upon the in- clined plane, a vertical circle passing through the index in that position, would have cut the theodolite, if po- sited horizontally, in the true arc. Hence (sketching a right angled spherical triangle like that at p. 87) if B be the angle of inclination of the theodolite, Bc or a the arc measuring the angle upon the inclined face of the instrument, then will Ba or c be the arc that measures the true angle; and it is required to determine in what circumstances a — c is a maximum. Now equa. 5, class 2, of right angled spherical tri- angles, is applicable to the present case, and when adapted to our figure becomes fa ___ sin 2a te sin 2c Hence when da — dc = 0, which is the case when a -- c is a maximum, that is, when ea = 2v, sin 2a must = 2c, and consequently Za must be the supplement of 2c, or a +c must = 90°. Now, (chap. vi. art. 26) tanc = tana cos B; whence, rad: cosB:: tana: tanc comp. , * eh Stat. rad + cos B:rad —‘cosB:: tana + tance :tana@ — tanc and (chap. ii. art. 18) :: sin (a@ + c):sin (a —c): that is, rad + cosp:rad — cosB::rad:sin (a — c). Fron this proportion, since 2 is given, the measures a, c, and their difference, may be determined. But the expression admits of farther simplification, For, transformed into an equation, it becomes sin (a — c) ie rad — co3B rad sad + cos ‘ 1 —cosB or —c) = ————= cae iv. sin (4 —c) = Jory = tan? 48 (chap. iv. rn), that is, the tangent of half the angle of inclination is a K § ‘ 202 Differential Equations. mean proportional between radius, and the sine of the maximum error on the theodolite. j Suppose that the face of the theodolite, instead of being horizontal, was inclined in an angle of 5 degrees. Then log tan $B ....2° 30’... . 8°6400931 Multiplied by.......... 2 The prod, — 10, sin error 6’ 33” 16”. . '7-°2801862 ee Fee are tee Example Ill. To find when that part of the equation of time which depends on the obliquity of the ecliptic, is the greatest possible, Here the sun’s longitude will form the hypothenuse of a right angled spherical triangle, his right ascension will be the base, and the obliquity of the ecliptic is sup- posed constant, It is required to find when hyp. — base is a maximum; which is, evidently, as in the preceding example, when hyp. + base = 90°, that is, when sun’s long. + sun’s right ascension = 90°, from the equinoc- tial points. This happens in the year 1816, about May 7 and November 8. Example IV. To ascertain the error that may be committed in the observation of zenith distances, by any conceivable de- viation of the instrument from the vertical plane. Draw a figure in which Ho is the intersection of the horizon with the vertical plane zo, and Hz’so the posi- tion of the plane of the instrument, z’ being the appa- rent zenith upon that plane, and s the place of the hea- venly body when it has arrived at the plane of the in- strument. The are zz’ will measure the inclination 1 of the plane at H or o, and the base z’s = z of the right angled spherical triangle zz‘s will be the apparent zenith~ distance of the body, while the hypothenuse zs = z + « will be the true distance, 5 Minute Variations of Triangles. 203 Now from equa. 6, art. 25, chap. vi. we have cos ZS = cos zz’ cos zs: that is, costcosz=cos(Z+a) = cos Z cosa — sin ZsIN&. Whence sin z sin = cos z (cost — Cos 1) = cos z (1 —YZsin?2# — 1 + 2sin? 31) = 2 cos z (sin? 41 — sin? gr). Dividing by sin z, there results sin « = 2 cot z (sin 31 — sin? §z). Here since « is by hypothesis small, sin* $x is ex- tremely minute, and may be rejected: hence sin « = 2 cot Z sin? 41. Corol. As the zenith distance z increases, the error x diminishes, the inclination 1 remaining the same. Supposing the deviation from the vertical 5’, and the apparent zenith distance of a star 37°, the error would be ~2,9, of a second. Note. If z, z’, be the observed zenith distances of a circumpolar. star (corrected for refraction ) at the supe- rior and inferior transits of the meridian; z, 2’, the true zenith distances; x, 2’, the corrections due to the devi- ation of the circle from the meridian: then will v= 2-+2 2 PS Zo as therefore the distance of the pole from the zenith, or, (2 —z) =H §(4%4+e)—F(24+ 2). But in the inferior transit 2’ is imperceptible; there- fore : Jat. = 90° — 3 (2’ —z) = 90° — 4 (av — z) + ba. Consequently, in the case of employing a circumpolar star, the latitude is only affected by half the error pro- duced by the deviation of the instrument from the ver- tical plane. Example V. It is required to determine, at a given time and lati- tude, how long an interval is taken by the body of the sun to rise from the horizon, 204: Differential Equations. Here the triangle which is supposed to undergo a minute variation is oblique, one of the sides being pz, the co-latitude, or distance from the pole to the zenith, another being ps, the co-declination of the luminary at the time proposed, and the other zs = 90° the distance from the zenith to the horizon. It is required to ascer- tain the variation in the hour angle p which corresponds to any assigned variation in the opposite side zs. The differential equation which applies to this in- quiry is, obviously, the first of class 1, oblique angled spherical triangles, which, when accommodated to the ‘present notation, becomes ozs Z $zs —— = sin S sin PS; or, dp = ——___- oP sin S$ sin Ps Now, from the 2d fundamental theorem of spherical triangles, (chap, vi. p. 84), we have COS PZ = COS ZS Cos PS + SiN ZS sin PS COS S. But, in the example before us, cos zs = 0, and s COs PZ sin Zs == 1. Whence coss = igo 5 cos? PZ Consequently sin s = J (1 -— Ne - sin? ps $zs 3zs sinssinPs —_4/(sin? ps — cos? Pz) ft may be readily shown, by multiplying together the _ corresponding values of cos (A + B) and cos (A — 8), (equa. C, p. 42), and substituting 1 — cos? for sin?, that sin? PS — cos? PZ = cos (PZ + PS) cos (PZ — PS). Hence it follows that ozs ite Sarpaeermyemeneent primes From this we have sun’s diam, : 15 . 7 (cos (lat, + dec.) cos (lat, — dec.)} Let it be proposed to ascertain the time occupied by . time sun ris, = Minute Variations of Triangles. 205 the sun in rising from the horizon on the 25th of May, 1816, in latitude 50° 12’ N. It appears from the Nautical Almanac, that the sun’s declination on the given day is 20° 59’ N., while its ap- parent semidiameter is 15’ 483”. Hence, diam, — 15'= 2-1078. Log cos (L + D)..71°11’.... 9°5085850 cos (L — D)..29° 13’..., 9:9409048 Se mmemeetd SuiMe-t 1B, NUD tes 19°4494898 Qotienty. Lorre wg cose 97247449 Taken from log 2:1078...... -- 0°3238294 Rem. log 3°9727-+iw.'. eee. 0°5990845 So that the time required is 39727 minutes, or 3" 58°. Note. By a similar theorem we may find the time which the sun’s rising or setting is affected by hori- zontal refraction. Example V1. To determine the annual variation of the declination and right ascension of a fixed star, on account of the precession of the equinoxes. Here, if A (in the spherical triangle aBc) be the pole of the ecliptic, B the pole of the equinoctial, and c the place of the star; the sides aB or c, and ac or 0, must be regarded as invariable. The differential equa- tions applicable to this question are the ist and 2d of | class 1, oblique spherical triangles; from which the fol- lowing are at once deduced. 1, Var. dec, = preces. equinox x sin obliq. eclip, x sin right ascen. from solstitial colure. var. dec. x cot ang. of posit. 2, Var. right ascen. = — cos dec. 206 Miscellaneous Problems. Example VII. To determine the variation in right ascension and declination, occasioned by any variation in the obliquity of the ecliptic. In this example, the hypothenuse and the opposite angle of a right angled spherical triangle are assumed as constant. The 2d and 6th equations, class 3, right angled spherical triangles, give 1. var. dec. = var. obliq. x sin right ascen. = var, obliq. x cot obliq. x tan dec. 2, var. right ascen. = — var. obliq. % 4 tan obliq. x sin 2 right ascen.* CHAPTER XII. Miscellaneous Problems. Section I. rroblems with Solutions. ProsieM I. REQUIRED the are whose logarithmic tangent is 7°1644598. First, by rule 4, p. 55. * For more on this curious subject the reader may consult €agnoli’s Trigonometry, chap. xix, and, xxi., and Lalande’s As- tronomy, vol. iii. pp, 588—604, Miscellaneous Problems. 207 LO: ood ele wes 01644398 add........ 5°3144251 eee Brony ti cura wie. 5 3 12°4'788649 Take 2 ar. com. log cos 3 CR cee Rem, — 10, is log 301:2067.. 2°4788646 eae Conseq. arc = 301-2067 = 5’ 12067. 2. By Hutton’s tables. 3. By Borda’s tables. Log tan 5’ 2” ,.7°1655821|Log tan 9’ 30”. . 7°1646031 Log tan 5’ 1” ..7°1641417|Log tan 9’ 20”. .'7-1599080 Difference.... 14404] Difference..,. 46951 Given log tan . . ’7°1644398]Given log tan . .'7°1644398 Log tan 5’ 1” ,,.'7'1641417|Log tan 9 20”, . 7°1599080 Difference. ... 2981; Difference.... 45318 998] ; 453180 =. mite a fr 7.0) a ? ? 4, are he i4404= 1”:2069\arc = decim. 9’ 20 ene = 9 29’"652 = sexiges..5. 1°-2082: Hence it appears that in this part of the tables Hut- ton’s has the advantage of Borda’s in point of accuracy. Borda, however, gives a rule to approximate more nearly to the truth; while in other parts of his tables the decimal division supplies great facilities in the use of proportional parts. ProsiemM If. It is required to demonstrate that if a, 6, and c, re- present the three sides of a plane triangle, then will a? + &* = c?, indicate a right angled triangle, a* + ab + 6? = c?, one whose angle c is 120°, a? — ab + 6% =”, one whose angle c is 60°. 208 Miscellaneous Problems. It appears from equa. 11, chap. iv. that a? + 5? Oe c? 2db Substituting, then, for c? in this equation, its value in each of the three former, there will result, respec- tively, cos C = In the 1st case, cos c = 0 = cos 90° In the 2d cos c = — 5 = cos 120° In the 3d cos ¢ = + 4= cos 60°. Corollary. In like manner it may be shown, that when a? dab + a =c*, csc=t4,andc= ae: a 3ab-+ 6? = c*, cosc = t1, andc = } es a> fab + 6? = c*, cosc = £1, andc = } ates a? + ab + 6» = c*, csc= +-1,,andc= ae arch ~ab +. b* = c*, cos C = +; and the two va- lues of c supplements to each other. Prosiem III, Required a commodious logarithmic method of find- ing the hypothenuse of a right angled plane triangle, when the base and perpendicular are given in large numbers. . B denoting the base, Pp the perpendicular, and u the hypothenuse ; Find n so that 2 log p — log 8 = logn . and make 8B + N =™M. Then, § (log M + log 8B) = log u. 2 For, from the nature of logarithms, —= N; i p? also, > +8= = M; whence Trigonometrical Solution of Quadratics. 209 3 (log m + log 8) = $log Mp = 3 log (—t* x ) = log / (Pp? + B*) = log H; as it ought to be. Scholium. The following formule, the first three for right an- gled triangles, will often be of use, and may be easily demonstrated by the student. Let a and d be any two quantities, of which a is the ; b ‘ greater. Find 2; x, &c. so, that tan x =. ,/ —, sin’ 2 == a b & . b . J —-, sec y = = tan u =~ andsin¢# = -: then will a b a a log ./ (a? — 6+) = loga + logsin y= log é + log tan y log ./ (a? — 67) = 3 [log (a + b) + log (a — 6)]. log ./ (a? + 57) = loga + log secu =logd + log cosecu. log / (a + 5) 3 loga + logseca = $ loga + 3 log2 + log cos Ly. log / (a — 6) =3 loga+logcosz = Lloga + tlog2 + log sin dy. log (a +b)" = = [log a + log cost + log tan (45° + it)]. PROBLEM LV. To investigate a method of resolving quadratic equa- tions, by means of trigonometrical tables. 1. Leta? + px = q*be the equation «A proposed to be resolved. Suppose AB, and Bc, the perpendicu- lar and base of a right angled triangle : ABC, to be respectively equal to g and a4 Ee 3p: then CA = CD = CE= 4/ (AB? + BC?) = Vf (q? + tp?) = J (2? + px + 4p?) =x t bp. 210 Miscellaneous Problems. Consequently « = ca = 3p = vz or BE, 2 cee Now — = — = tan c = tan 2x (Euc. iii. 20.) BC p Also, DB: BA:: rad: tan D, or cot, or cot Je and EB: BA::rad:tan E or tan 4c. Therefore pB = = BA tan 3c = g tan 4c cot $c EB = ——~- = BA cotic =¢qcot 3c tan de i BY Fd 5 Tiers Which are trigonometrical values of the two roots. 2. Let ct: pa = «1 = ¢? be the proposed equation, Then making AB = g, AC = 3p, it may be shown, si- milarly to the above, that cA = re = neand pp, are the two roots: whence the appropriate trigonometrical ex- pressions may be readily deduced. The precepts for all the cases may hence be laid down thus :— . J, If the equation be of the form x? + pa = q?: 2 F Make tan c = ee then will the two roots be x==+qtan 3c.... c= — gq cot jc. 2. For quadratics of the form 2* — px = q?. 2 ; Make, as before, tanc = = then will cm —gitanjgc....“2£= +4 cot jc. 3. For quadratics of the form 2? + px =, — (9). 2 E Make sinc = a then will go —g tan 3c.... c= — g cot 4c. 4. For quadratics of the form x? — px = — (q*). ‘ 2 : Make sin c = —: then will e=-+gtanjc....c¢= + qcot ke. wate : : She In the last two cases, if fe exceed unity, sin c is ima- nary, and consequently the values of «. Solution of Quadratics. 214 The logarithmic application of these formule is very simple, as will be manifest from the following Example, 7 1693 Find the roots of the equation 2? 4. —2 = 19716? tables of cae and tangents. 1695 ‘ ot. Here p=; a 7? = Fa77q and the equation agrees with the ist form. Alsotanc = = / at —_ 4 in logarithms thus: | Log 1695 = 3:2291697 Arith.com.log12716=5 5:8956495 sum + 10 = 19'1248192 half sum = 9°5624096 = log g log 88 = 1°9444827 Arith, com. log 7 = 9:1549020 sum ~10=logtanc = 10°6617943 =log tan 77°42'312” log tan 3c = 9:9061115=logtan38°51'152" log g, as above = 9'5624096 sum — 10=logr = — 1:4685211 = log 2941176. 5 5 This value of x, viz. *2941176, is nearly equal to 77 To find if this be the exact root, take the arithmetical compliment of the last logarithm, viz. 0°5314379, and consider it as the logarithm of the denominator of a frac- tion whose naerakot is unity; thus is the fraction found to be =, exactly; which is manifestly equal to 212 Miscellaneous Problems. o Dhe other root of th jon is equal to — > a er root of the equation is equaito — 12716 TeAO te 339 BB per MCE ScHOLIUM. Cubic equations, as weil as quadratics, are readily solved by means of trigonometrical tables. The for- mulz demonstrated and given by Cagnoli and Borda for this purpose, are as follow: 1. For cubics of the form 23 + pr Eq = 0. 1 . Make tan w= =". 2 / Sp ..e. tan A = Stan $B. Then « = = cot2a.2 / 4p. 2. For cubics of the form 23 — pr Eq = 0. 4 Make sin p =" .2 J/ ip... tana=s & tan $s. Then « = = cosec 2A .2 4/ 5p. ; Here, if the value of sin B should exceed unity, B would be imaginary, and the equation would fall in what is called the irreducible case of cubics. In that case we 4 must make cosec 34a = a .2./%p: and then the three roots would be ge maode sin A 2 4/ 3p: x= +sin (60° > A).2 / 3p. x = = sin (60° + A) .2 / 3p. Ifthe value of sin B were 1, we should have 8 = 90°, tan A = 1; therefore a = 45°, anda = = 2.,/%p-. But this would not be the only root. ‘The second selution Oo. would give cosec 34 =1: therefore a = say and then gp aes + sin 30°.2 /3ip = VV p. go == sin 30°.2 / 5p = £ Vo Hp. _ & = sin 90°.2 f sp = + 2A Bp. Here it is obvious that the first two roots are equal, that Series for Circular Arc. 213 their sum is equal to the third with a contrary sign, and that this third is the one which is produced from the first solution. {n these solutions, the double signs in the value of 2, telate to the double signs in the value of qs ProsueMm Y, To investigate series in which any circular arc shall be expressed in terms of its sine, or of its tangent. There are various methods of solving this problem : one of the simplest is by means of the expressions for the differentials of the sine and tangent of an are, given in the preceding chapter. Thus, from the expression ? sins = 28 cos B, we have, esing S sins —Z heat ee ene dieses) 2 2. cos B V4 (Ll — sin *B) Se (i per B) Expanding the factor (1 — sin? B) by the binomial theorem, the whole expression becomes icy 6 Dg sin® B A —— 7 Ig; eae 2B = Asin B(1 + isin? + oq Sint B+ + &c.) The integration of this expression is very easy: for the integral of ¢B is B, that of dsin B x 1 is sin B, that ; ‘ . sineB of 3 sin? B sin B is 5-93 andsoon. Thus we have, at length, ° sin3B 63 sindB 3'5 sin’? B OLS Chr satath vataie 2467 3°S°T sind B 2°4°6°8°9 +, Xe. Suppose B to be = 30°, then because sin:30° = 3 (ch. il. pr. 5.) it results that 1 3 3°5 Br\O te ire Re ——— Te are 30° = 3 + 2-328 + 2°4°5,25 + 2°4°6°7 ,27 ss ‘ 214: Miscellaneous Problems. S57 2°4°6'8'9'29 circumference of a circle to the radius |. Again, the differential expression for the tangent is + &c. = °523598775 = a of the oB étanB=———; whence cos* B d,tan B } tan B op = tans cos? 8B = ——_ = —— = sec? B 1 + tan?s Stans (1+ tan?B) — > Here, expanding the factor (1 + tan? B)~ 1 by the binomial theorem, and integrating each term of the result, there will arise B= tanB — } tan3p +1tanSs — +tan7 B+ &ec. Thus if 8 = 45°, then (ch. ii. pr. 6.) tan B = 1; con- sequently, substituting this value of tan B, AGRI! Ata ty eid ote ns arc 45 =1-% 1s Atha HES Ey Note. This series obviously converges very slowly: Euler (in his Analysis Infinitorum, vol. i.), and Dr. Hutton (in his 8vo. Tracts, vol.i.) have transformed it into others of rapid convergence. &c. ScHOLIUM. Many curious, and indeed useful, results are deduci- ble from series of this kind, and of the opposite class, in which the sine, cosine, &c. is expressed in terms of the arc itself (ch. iv.equa. y.) As a specimen we select. the following, first discovered by John Bernoulli. The expression x8 x a7 3348 Bea5e7 + EC: when sin « = 0, becomes, after dividing by a, 1 1 1 hae Er et a ae 3 0=1—55% + spa5t sapeq? tT XS nz=in- — $in x ag 7 , Geodesic Operations. 215 q i Or, making + = Py ] 1 1 LE Aelia 2°322 7‘ Q°3°4°bzt 2-3-4567 28 + &e. Multiplying by 2™ 1 Seiden. | epee are =e in 4 2n—4 ee ee 53° + o345 2 &ec, Now, it is shown by writers on algebra, that the sum ofthe roots of every equation of this form is equal to the co-efficient of the 2d term with its sign changed ; . . ] e that is in the present case = g: it is known also (chap. iv. art. 3.) that the values of 2 which answer to the case of sin < = 0, are 7, 27, 37, 47, &c. 7 denoting the semi- circumference to the radius 1. These successive values 1 ; ’ ° of x or of — being substituted, their results 1 1 1 1 i ber wt ep Ppa t pb ees & tel be} OE 1 I n 2 Tg tat &e.= Prosiem VI. In Geodesic operations, it is required to deduce from angles measured out of one.of the stations, but near to it, the true angles at the station, : When the centre of the instrument with which hori- zontal angles are taken, cannot be placed in the vertical line occupied by the axis of the signal, the angles ob- served must undergo a reduction, according to circum- stances. 1. Let c be the centre of the sta- B A tion, p the place of the centre of (the instrument, or the summit of ‘the observed angle aps: it is re- quired to find c, the measure of ACB, supposing there to be known C.F, APB = P, BPC = p, cP = d, BO == L, AC = R, 216 Miscellaneous Problems. Since the exterior angle of a triangle is equal to the sum of the two interior opposite angles, we have, with respect to the triangle 1Ap, AIB = Pp + 1AP; and with regard to the triangle Bic, Ais = c + cpp. Making these two values of A1B equal, and transposing 1AP, there results C= P + IAP — CBP. But the triangles cap, CBP, give : ‘ cP , dsin (» + p) sin CAP = sin IAP = — sin APC = ——~~——: dsinp : cP ° sin CBP = — .81In BPC = BC And, as the angles cA?, cBP, are, by the hypothesis of the problem, always very small, their sines may be sub- stituted for their arcs or measures: therefore dsin (Pp + p) dsinp Cm P= L Or, to have the reduction in seconds, d sin(P + sin in | R L The use of this formula cannot in any case be embar- rassing, provided the signs of sin p, and sin (P + p) be attended to. Thus, the first term of the correction will be positive, if the angle (P + p) is comprised between 0 and 180°; and it will become negative, if that angle sur- pass 180°. The contrary will obtain in the same cir- cumstances with regard to the second term, which an- swers to the angle of direction p. The letter r denotes the distance of the object a to the right, 7 the distance of the object B situated to the left, and p the angle at the place of observation, between the centre of the sta- tion and the object to the left. 2. An approximate reduction to the centre may in- deed be obtained by asingle term; but it is not quite so correct as the form above. For, by reducing the two fractions in the second member of the last equation but one to a common denowinator, the correction becomes ~J Geodesic Operations. 91 dusin (Pp + — drsin © eee P ——— dusin (e+ p) — adRsin p LR RsinA RsinA But the triangle anc gives L = ——— = ——_—___ sin B sin (A + C) And because Pp is always very nearly equal to c, the sine of a + Pp will differ extremely little from sin (a + c), and may therefore be substituted for it, making L = RSINA sin (A + P) Hence we manifestly have dsinasin{e + p)— dsinpsin(a + P) sad F hig Aa a Cancer As cC—P= : -RsmA Which, by taking the expanded expressions for (p + p}, and sin (A + P), and reducing to seconds, gives d sin p sin (A — p) RsinA 3. When either of the distances r, 1, becomes infi- nite, with respect to d, the corresponding term in the expression art. 1 of this problem, vanishes, and we have accordingly d sin dsin (p + cet se See Bric pee gas Cais Lsin | R sin 1” The first of these will apply when the object a is a hea- venly body, the second when B is one. When both a and B are such, then c — P= 0. But without supposing either a or 8 infinite, we may have c — yp =O, or GC = P in innumerable instances: that is, in all cases in which the centre p of the instru- ment is placed in the circumference of the circle that passes through the three points a, B,c; or when the angle BPc_is equal to the angle Bac, or toBAc + 180°. Whence, though c should be inaccessible, the angle AGR may commonly be obtained by observation, without ‘any computation, It may further be observed, that when P falls in the circumference of the circle passing < 1b e ; | 218 me Trigonometrical, Surveying. through the three points A,.B,,c, the angles A, B, G, may be determined solely by measuring: the angles APB and prc. For, the opposite angles asc, Arc, of the quad- rangle inscribed i in a circle, are-—= 180°. Consequently, ABC = 180° — apc, and BAc = 180° .— (ABC + ACB) = 180° — (apc + arb), 4, If one of the. objects, viewed from a farther sta- tion, be a vane or staff in the centre of a steeple, it will frequently happen that such object, when the observer comes near it, is both invisible and inaccessible. Still there are various methods of finding the exact angle at c. Suppose, for example, the signal-siaff be in the cen- tre of a circular tower, and that the angle aps was taken at Pp near its base. Inet. the tangents pr, pr’, be marked, and on them two equal and. arbitrary, distances pm, pm’, be measured. Bisect mm’ at the pomt 2; and, placing there @ signal-staff, measure the- angle ups, which, (since px prolonged ‘obviously passes through Cc the centre, ) will be the angle p of the preeeding inves- tigation. Also, the distance ps added to the radius es 5 of the tower, will give pe‘=.d in the former investi- gation. . If the:cireumference of. the tower cannot, be mea- sured, and the radius thence inferred, proceed ‘thus: Measure the angles Brn, ‘Ber’, then will BeG = $/(BPT -++ BPT’) = D3 andopr <= Bet— Bec: Measure Pt, then pC = PB.secceT =d.. With-the values of p and d,- thus obtained, proceed as before. 5. If the base of the tower be polygonal and regular, as most pafteaanay happens; assume Pp in the point of in- tersection of two of the sides prolonged, and Bec’ = § (npr + BPT’) as. before, Pp = the distance from P to the middle of one of: the sides whose prolongation, passes through p; and hence rc is found, as above, If the figure be ar egular: hexagon, then, the tr langle Pmm’ is equilateral, and're = m’m/ 3. — Geodesic Operations. 919 Prostem VE. To reduce angles measured in a-plane- inclined to the horizon, to the corresponding angles in the horizontal plane. ET G8 Let nea be an angle measured in a plane inclined to the horizon, and let B’ca’ be the corresponding angle in the horizontal plane. Let d and d’ be the zenith dis- tances, or the complements of the angles of elevation . ACA’, BoB’. Then from z the zenith of the observer, or of the angle c, draw the arcs xa, xb, of vertical circles, mea- 7/2. in suring he zenith distances | ALIS ae Hi ed ; d, ad’, and draw the arc ab ip dale Seal of another great circle tomea- | pigsty :, sure the angle ¢. It follows Cues NFA from this construction, that the angle z, of the spherical triangle zad, is equal to the horizontal angle a’c’3; and that, to find it, the three sides za = d, xb = a) G0'=='Ct are given. Call the sum of these s; then the corres- ponding formula of ch. vi. pa, 90, applied to the present instanee, becomes sin 32 = sindc= VES SS rad 3 AU ia sin dsimd If h and h’ represent the angles of altitude aca’, Bos’, the preceding expression will become sing(c + h —h')sin3(c + A’~ hk) cos A cos h’ | Or, in logarithms, log sin 3c = 3 (20 + logsind (c +h —h’) + log sin 2(C + h’—h) — log cosh — log cosh’). sin SACB sin 36 = / Cor.1. Ifh = kh’, then is sin fo = PPR dedi fog sin La’ce’ = 10 + log sin acs — log cosh, Cor. 2. If the angles k and 4’ be very small, and nearly equal; then, since the cosines of gmall angles Lg 220 Trigonometrical Surveying. vary extremely slowly, we may, without sensible error, take log sin }a’cB’ = 10 4+ log sin acs — log cos4 (h + h’). Cor. 3. In this case the correction « = A’cB’.— ACB, may be found by the expression d +d’ d—d’ )? — cot dc ( 5 )). And in this formula, as well as the first given for sin 3c, d and d’ may be either one or both greater or less than a quadrant; that is, the equations will obtain whether aca’ and scp’ be each an elevation or a depression. _ Scholium. By means of this problem, if the altitude ef a hill be found barometrically, according to one of the methods described in Gregory’s Mechanics, vol. i. book 5, or geometrically, according to some of those described in heights and distances, (chap. v.); then, finding the angles formed at the place of observation, by any objects in the country below, and their respec- tive angles of depression, their horizontal angles, and thence their distances, may be found, and their relative places fixed in a map of the country; taking care to have a sufficient number of angles between intersecting lines, to verify the operations. x= sin 1” (tangc (40 — Prosiem VIII. In any spherical triangle, knowing two sides and the included angle; it is required to find the angle compre- hended by the chords of those two sides. Let the angles of the spherical triangle be a, B, Cc, the corresponding angles included rt by the chords a’, B’, c’; the sphe- eres rical sides opposite the former, a, b, c; the chords respectively oppo- site the latter, «, 6, y; then, there are given 6, c, and A, to find A’. Geodesic Operations. 92) Here, from equa. 2, chap. vi. we have cosa = sin d’sinc cos A + cosécosc. But cos c = cos ($¢ + 4c) = cos* 3c — sin* 4c (by art, 23, chap. iv.) = (1 — sin? 4c) — sin? 3c = 1 —2 sin? 3c. And in like manner cosa = 1 — 2 sin? fa, and cos = 1 — 2sin?36. Therefore the preceding equa- tion becomes 1 —2sin* 4a = 4-sin 1d cos $b sin $c cos 4c cos A + (1 —2sin? 46) (1 — 2sin* ic). But sin ia = La, sinihb = 44, sin4c = 47: which values substituted in the equation, we obtain, after a little re- duction, 2 2. 42 é oie te = fy cos 4b cos 3c cos A +} By? 62 5% — 42 23/ Now, (equa. 11, chap. iv), cos a’ = : There- fore, by substitution, : By cos A’ = &y cos thcos $c cos A + 452y?; whence, dividing by #y, there results cos A’ = cos 34 cos 3c ces A + $8 37; or, lastly, by restoring the values of $2, 5v, we have cos A’ = cos 44 cos $c cos A + sin $b sin $c.... (1.) Cor. 1. It follows evidently from this formula, that when the spherical angle is right or obtuse, it is always greater than the corresponding angle of the chords. Cor. 2. The spherical angle, if acute, is /ess than the corresponding angle of the chords, when we have cos A sin 45 sin 3c reater than ———_——_ E " 1 —cosgacos ge ProgpureM IX, Knowing two sides and the inciuded angle of a recti- linear triangle, it is required to find the spherical angle of the two arcs of which those two sides are the chords. Here 4, y, and the angle a’ are given, to find a. Now, since in all cases, cos = 4/(1 — sin*), we have cos $0 cos 3c = /[ (1 — sin? 44) (1 —sin? 3c)]; we have also, as above, sin 3 = 3, and sin $c = dy. 222 Trigonometrical Surveying. Substituting these values in the.equation ].of the pre- ceding problem, there will result, by reduction, sine eos A’ — 437 C08 AEP aey le a el bey gS To compute by this formula, the values of the sides , y, maust be reduced to the corresponding values of the chords of a circle whose radius is unity. This is easily effected by dividing the values of the sides given in feet, or toises, &c. by such a power of 10, that neither of the sides shall exceed 2, the value of the greatest chord, when radius is equal to unity. ' From this investigation, and that of the preceding problem, the following corollaries may be drawn. Cor.1. Ifc = b, and of consequence vy = 8, then will cos A’ = cos A cos* gc + sin? $c; and thence 1 —2sin* 3a’ = (1—2 sin? Za) cos? $¢ + (1 — cos? $c): from which may be deduced . sin $a’ = sin $a ‘cos 3c.... (3. Cor. 2. Also, sinee cos $¢ = 4/(1 — sin* Ec) = v/f1 — 447), equa. 2 will, in this case, reduce to sin 3a’ Cor. 3. From the equa. 3, it appears that the vertical angle of an iseseeles spherical triangle, is always greater than the corresponding angle of the chords. , Cor. 4. If A = 90°, the fornaule 1, 2, give cos A’ = sin 30 sin 3c = d6y.... (5.) These five formule are strict and rigorous, whatever be the magnitude of the triangle. But if the triangles be small, the arcs may be put instead of the sines in equa. 5, then 7 Cor. 5. As cos a’ = sin (90° — a’) = in this case, 90° — a’; the smallexcess of the spherical right angle aver the corresponding rectilinear angle, will, supposing the arcs 4, c, taken in seconds, be given in seconds by the following expression : ‘gin ak =: b b 90° — a’ = 2 = TO...) Geodeste Operations. 223 The error in this formula will not amount to & second, when 6 + cis less than 10°, or than 700 miles measured on the earth’s surface. | Cor. 6. If the hypothenuse does not exceed 11°, we may substitute @ sin c instead of c, and a cos c instead of 6; this will give dc = a* sinc cosc = $a? sin 2 (90° — B) = ja’ sin 28: whence asin@e a sin Qn a (903 ADF geet ee CTD Ifa = 13°, and B = c ='45° neafly; then will 90° — a’ ex 177. Cor, 7. Retaining the same hypothesis of a = 90°, and a = or < 14°, we have i b? cot B bo RN ae aa pee Ray ee Also'e — ¢C a Cnt ‘e eo (9.): Cor..8, Comparing formule .8, 9, with 6, we have B- Bo=¢ — Cc’ = 3 (90° — a’). Whence it appears ‘that the sum of the two.excesses of the oblique spherical ‘angles, over the corresponding angles of the chords, in a small right. angled triangle, is equal to the excess of the right angle over the corresponding angle of the chords. So that. either of the formule 6, 7, 8, 9, will suffice to determine the difference of each of the three angles of a small right angled spherical triangle, from the corresponding angles of the chords, And hence this method may be applied to the measuring an arc of the meridian by means of a series of triangles. Prostem X. In a spherical triangle anc, right angled in a, know- ing the hypothenuse se (less than 4°) and the angle 3p, it is required to find the error e committed through _ finding by plane trigonometry, the opposite side ac. 4 j Referring still to the diagram of prob. 8, where we 224 Trigonometrical Surveying. now suppose the spherical angle a to be right, we have (equa. 5, chap. vi) sind = sinasing. But it has been proved (chap. iv. equa. y) that the sine of any arc A is equal to the sum of the following series: i A3 AS Al & VECO A ee ee ee ee RIES ol eS eR ee e ’ 230 1 254bew, 2B AEOA ORS : A3 AS A? or, Sin A =A — —— + 120 Pitney aes And, in the present inquiry, all the terms after the second may be neglected, because the 5th power of an arc of 4° divided by 120, gives a quotient not exceeding 0”-01. Consequently, we may assume sin 6 = 6 — 16, sina = a — 4a}; and thus the preceding equation will become 6 — 153 = sin B (a — 143) or, b= asin B — 1 (a3 sin B — 43), Now, if the triangle were considered as rectilinear, we sheuld have 6 = asinB; a theorem which manifestly gives the side 6 or ac too great by 1 (a3 sinB — 63). But, neglecting quantities of the fifth order, for the reason already assigned, the last equation but one gives 63 = a3sin3s. Therefore, by substitution, e = — a3 sin B (1 —sin?): or, to have this error in seconds, take rn” = the radius expressed in seconds, so shal! a cos? B 3 6RR” : Cor. 1. If a = 4°, and B = 35° 16’, in which case the value of sin B cos? B is a maximum, we shall find = — 45". Cor. 2. If, with the same data, the correction be ap- plied, to find the side ¢ adjacent to the given angle, we should have e=—asins, a a? sin? B- é€ =—=acossB “origi: So that this error exists in a contrary sense to the other; the one being subtractive, the other additive. e Spherical Excess. 995 Cor. 3. The data being the same, if we have to find the angle c, the error to be corrected will be sin 2B dn” As to the excess of the arc over its chord, it is easy to find it correctly from the expressions in prob. 8: but for arcs that are very small, compared with the radius, a near approximation to that excess will be found in the same measures as the radius of the earth, by taking ,', of the quotient of the cube of the length of the arc di- vided by the square of the radius. ws @-=> a+ ProspLtem XI. It is required to investigate a theorem, by means of which, spherical triangles, whose sides are small com- pared with the radius, may be solved by the rules for plane trigonometry, without considering the chords of the respective arcs or sides. The sum of the angles of a spherical triangle exceeds 180°, by a quantity © called (chap. vi. p. 102) the sphe- raced excess, which is a correct measure of the area of the triangle. The sum of the angles of a egnce triangle ABc, exceeds 180° by 277. 3ac.ABsinA = 377AB. AC SIN A = surface of the triangle ; r being the radius of the sphere. Or, when r= iS we have: A+B+c=180°+ SAB. ACSINA == 180° + FAR .CB SIN B = 180° + Jac. Bcsinc — ISU tas. ACSINA + 1AB.CBSINB + lac. Be sin c. Distributing this excess equally, by taking its third part from each angle, we shall have A’ =A — 1AB.acsSin A, B= B— Tas. CB Sin B, =c—1AC.zBCSiNC: and the flideias steps may readily be traced by i LO 226 _ Trigonometrical Surveying. who have carefully studied chapters iv. and vi. of this work :— __ Besin BY __-BCsin(B — 1an.cr sinB) ot sina’ dima — 1AB. AC sill A) __ BCsin (B — 48) sin (A — 38) BC sin B Cos $E — BC Cos B sin +B. sin A cost as cos Asin 4 gh BCOsinB ~*~ Be COs Bsin ie Sf CW Wiecrreemrpert tex guar . +- (1— cota tan}r) snA Sin A COSsE %; BCsinB 1—cotBtaniE sina © 1 —cota taniz BC SIN B , ‘ Ere (1 + tan Ze cot A —.tan $E cot B) = — (1 + $aB.ACsin A cota—taAB. RC sin B cot B) = SP" (1 {LAB WAC COSA — 1AB..BC C08 B) sin A s 6 = (1 + taB.AD — 1aB. BD) sina 6 6 [ AD and sp being segments of the base as by a per- pendicular cp from cl PERS ‘sin B — LAB (AD — BD Taina, Dh cahBA ) BC SinB = ——. ,AD* — {BD* sin A t+ Ai ths ] __ BesinB (1 + ac? — inc?) sina : - BCSINB BC.AC*sinB BC3 sinB in & 6sin A 6sina . ne BC sin B Ac3 sin A Bc3 sin B aera a or prensa lhentiaa- er saree: sin A 6 sin A 6sins BCSIN'B Bc3 sin B = —— + | SOHGLIUM. 272° SU CHTSR A bn We have already given (in sect. 6, chap, vi.), expres- sions for finding the spherical excess, “A few additional rules may with propriety-be presented here. 1. The spherical excess x, may be found in seconds, f* dW 4 , ‘ - er » £ puSH Lomtg ; R%s BIS PGURT Aas “ayy x oby the expression, z = vi where sis the. surface gf the'triangle (whose sidésare-a, 6) ¢; and angles A, By) $e AP esinho we od hice 04 Bene hee a ee = gbesImA c= dadisinee = fo sin B= 3a aura vae ois the radius:of the-earth, in the ‘same measutes.as a,’ b,. and ¢c, and r” = 206264’-8, the seconds in an arc equal ih Tength'to the'tradiag, 9° ogc! ol 12 EE ONO Af this formula be ‘applied Jogarithmieally, ‘then dg 108 * This curions theorem ywas first announced by M. Legendre,, va the Memoirs ‘of the Paris Academy, for’ 787, “hie investiga. ; tion here given is by M. Delambre, ~"— ‘228 Trigonometrical Surveying. 2. From the logarithm of the area of the triangle, taken as a plane one, in feet, subtract the constant log 9:3267737, then the remainder is the logarithm of the excess above 180°, in seconds nearly.* 3. Since s = 3dc sin A, we shall manifestly have E = R’ } 5 a OC sin A. Hence, if from the vertical angle B we demit the perpendicular Bp upon the base ac, dividin it into two segments , By we shall have b.= «+ 8, wd R"” uM a ence E = 55¢ (4+ &) sn A =o] ve. Sin J + 55 Fc. sina. But the two right angled triangles aBD, cBD, being nearly rectilinear, give «= acosc, ands=¢ cos A; whence we have ~ u ‘ rR” } . — i, f Bae ; Eo 57 2¢ Sin A: COSC + 5 2c SIMA COSA. Z yb tn yA Pe U ey Ds wane In like manner, the triangle axzc, which itself is so small as to differ but little from a plane triangle, gives c sin A =asinc. Also, sin A cos a = 3sin2a, and sin ccos¢ = 4sin2c (equa. 1, chap.iv). Therefore, finally, R! ; Rp! ’ A — 72 — 72 J co B= 730" sv 20 + a ° gin 2A: ; ' From this theorem a table may be formed, from ‘which the spherical excess may be fotind ; entering the table with each of the’sides above the base and, its adja- cent angle, as arguments. 4. If the baseband height h, of the triangle are given, ‘ p : ; rR” then we have evidently: = 304 = Hence results the following simple logarithmic rule: add.the logarithm of the base of the triangle, taken in feet, to the logarithm “of the perpendicular, taken in the same measure; de- duct from the sum the-logarithm 9°6278037; the re- jo itt his,is Mr. Dalby’s rule given by General Roy in the Phile- sophical Transactions, for 1790, p. 171. Se ae ae 4 GED Figure of the Earth: 229 mainder will be the common logarithm of the spherical excess in seconds and decimals. 5. Lastly, when the three’ sides of the triangle are given in feet; add to the logarithm of half their sum, the logs. of the three differences of those sides and that half sum, divide the total of these 4. logs. by 2, and from the quotient subtract the log 9:3267737; the re- mainder will be the logarithm of the spherical excess in seconds, &c. as before. One or other of these rules will apply to all cases in which the spherical excess will be required, * Prosiem XII. To determine the ratio of the earth’s axes from the measures of convenient portions of a meridian in any two given latitudes; the earth being supposed a sphe- roid generated by the rotation of an ellipse upon its minor axis. The most accurate way of solving this problem, will be to compare, not merely single degrees measured on different parts of the meridian, but large portions of 5, 6, or 7 degrees, the most extensive that have been cor- rectly measured; according to the method proposed by Professor Playfair (Edinburgh Transactions, vol. v.), which is as below. . 7 Let the ellipse prepa represent a terrestrial meridian passing through the poles p, », and cutting the equator in F,@, Let cbe the centre of the earth, cg the radius of the equator = a, and rc, half the polar axis = 4. * The intelligent student who wishes to go more minutely into the subject of geodesic operations, especially in reference to the determination of the figure and magnitude of the earth, may con- sult the chapter on that subject in the 3d vol. of Dr. Hutton’s Course, Colonel Mudge's “ Account of the Trigonometrical Survey of England and Wales,” M. Puissant’s works, entitled “* Geodésie”’ and ** Traité de Topographie, d’Arpentage, &c.” Mechain and Delambre, ‘‘ Base du Systéme Métrique Décimal,” and chap, xxxv, of Delambre’s quarto * Astronomie,” 230 Problems with Solutions. Let as be any small are of the meridian, having. its centre of curvature in H; join HA, HB, intersecting cq in K and L. Let @ be the nreasure of the latitude of a, or the measure of the angle QKA, expressed in decimals of the radius 1; not in degrees and minutes. Then, the excess of the angle aus above aKa, that is, the angle Lux or BHA will be = 29. Also, if the elliptic arc @a = 2, then will AB = cz = 7) X AH. Now it is shown by the writers-on conic sections, that the radius of curvature at A, that is, Arh? — = ab? (a? — a*sin? ® 4 An = — - (a? —a? sin?) + ‘b2 sin? YF pa Or ain? Yig Fey o. Af, therefore, c be the compression at the poles, or: =‘ ab. we shall have & = a? — Qac + c*;0r, rejecting the powers of c higher than the first,, because.c 1s very mall in comparison to a.(that is, about the 300th part), it will be 4? = a* — 2aé, whence, by substituuon, AH = a3 (a — 2c) (a? — a? sin? @ +e? sin? — 2ae, apy . ; . Oat POP sin? @)” ? = a3 (a — 2c) (a? — Zac sin® 2h ey pel 2c = sin? ?) z “44 But, (a?—Qacsin? 9) 7 * =a (l= le Weg. aba d as mts "(1 + = sin*¢) nearly ; #ejecting, as before, the terms that involve ¢?, ¢3,;&C..) Hence AH =(a— 2c) (1+ = sin? ¢). = a — 2¢ + 3csin??. Hence, also, Og = 00 x AH Ee (a — 2+ Sein? ) aia ‘ae . , = (a — 2c) do + Se sin? Gos bug’ : ag Se eS tae . - 231”, obtained —ae0lad for the resulting compression. (See my Collection of Dissertations and Letters relating to the Trigonometrical Survey of England and Wales, pa.47.) There is great reason to conclude that the true compression lies between these limits. i SEcTION I. es L0G yf Problems without Solutions. ” 1. Demonstrate the truth of the following analogy, viz. As the-sine of half the difference of two arcs which together make 60° or 90° respectively, is to the differ- ence of their sines; so is 1 to ./ 3 or »/ 2, respec- tively. _ vues %. Demonstrate that 4 times the rectangle of the " _.. sines of two arcs, is equal to the difference of the squares fA of the chords of the sum and difference of those arcs. 3. Demonstrate that of any arc a, ite eS ] ‘yer Ye Aten ~ A (1+ cot?a) 4 (1 + tan?a) 4, Also, that cota cosSA = ; / AW (1 + tan? a) ion Va + colzA). 5. And that | NV tana= We Cen xs 1) = _ 2 tanga Y Be A _ |—tan*ga 2 2cot Za = Cot ga — tanga ~ cot? ga — L 6. Demonstrate that cos? Problems without Solutions, 233 Pr ar ake bane ae sin (A — ®) sin (A + B) f cos? A COs? B Pe se 7. Also, that cot? B — cot7A 8. Demonstrate that log sin (A +B) = log sin a + log sin B + log (cot a + cots) — 20. sin (A — B)sin(A +8) = SO sin? A sin? B 9, Demonstrate that ot sec? Aa Reo 2 — sec? a’ vi sec3 A BHPeS Shim 4A—3sec?Aa ; Sh sect A c sec 44 = 8 — Ssecta + secta? sec ‘SAt=s ett “16 — 20 sec? a+ Ssecta” 10. Demonstrate that arc to tan 4 + arc to tan } = arc 45° arc to tan + + 2 arc to tan } = are 45° mH. retiodeate that the Laneeat of the sum a of any Bamber of ares will be represented by A—c+E—G Ke, 1—p+o—Fd&e, the sum ofall the tangents of the separate a ares being de- noted by a, the sum of all their rectangles by B, the. sum of all their solids by c, &c. i 12. Find the are whose secant and co- agent shall be equal. _ Ans. Arc whose sine is 3 4/ 5 — 3, or 38° 10’. 13. Find the are whose sine ance to its cosine shall be equal to a; and show the limits ai ED Ans, a* must never exceed 2. ae 14. What arc is that whose mibeht and “co tangent shall together be equal to four times s radius? dns. Arc of 75° or of 15°. . ee er ncmnentee Cellar, : : ie Sa ] 24 V4 te . . “* f hy} f \/ - - £e7 1 a ~ isi - G= eee y teamins’ E Z * y : or aah, af 4 Efe tA | A v ¥ “0 i A ‘ ye. : O34 | Problems without Solutions. > 15. Given the two segments of the base of a plane tri- angle made by a perpendicular from the vertical angle, equal to 9 and I respectively, and the vertical angle = 55° 46°16": required the other angles and sides, by means of cor. to prop. 18. ch. ii. | Ans, Angles, 40° 24’ 343” and $2°49' 9"; _ Sides, 8 and 12, } 16. In a plane triangle are given, the perimeter = 1502, and the sines of the angles. respectively as 232, 174, and $45: required the sides and angles. Ans. Sides, 464, 348, 690: — Angles; 27° 3’, 37°20’; 115° 37’. 17, Ifone angle of a plane triangle be 129° 34’, and the two sides about that angle in the ratio of 4 to 7; what are the other angles? 298 Ans. 32° 41’ 7” and 17° 44’ 53”, 18. At what distance on the horizontal plane passing through the foot of a steeple of 50 yards in height must an observer stand, that the said steeple may give an ap- parent elevation of 80 degrees? | — Ans. 863 yards, ae Cai 2 19. A statue of 12 feet high stands at the top of a co- umn of 48 feet high. At what distance from the base of the column, on the same horizontal plane, will the statue appear under the greatest possible angle, and what will that angle be? , , Ans. Distance 53-6656 feet, angle 6° 22’ 46”. [fu poof. 20. In aright angled triangle axc, right angled at B, ne Dy there are known a right line ax bisecting the angle a ”) } and falling on the opposite side Be, = 4, and a right line ‘ ep bisecting the angle c and terminating in AB, = 2: required the three sides of the triangle? is Ban ae, Ans. AB = 3:9193, Bc = 1'6683, ac = 42596. [j- ““""}, 2. The respective distances from a given point P to j jase, tthe three nearest angles of a square garden AxBcD, are —4/a 4 \PA = 70 yards, pB = 40, and pc = 60. Required the “Tt 4,04) length of aside of that garden. ‘ bethl 1a § PGE ~ Anse 81,4885 yards Y era shal LEA f . / Problems without Solutions. 235 « (22) A gentleman travelling towards York, discovered, by means ofa telescope, the top of the’tower of the ca- thedral, when just:in the horizon; and on travelling 20 miles nearer, he found the top of the tower elevated one degree. What is its height, that of the observer’s eye being 5 feet 6 inches? Ans. 2602 feet. 7 _ 23. There are four trees, a B,C, D, standing in’ @ “straight hedge row; -their distances are an = 60 yards, “Bc = 20, cp =40. .Where must I'stand to observe them, so that the three intervals may appear equal? Ans. At 12 yards from 8 towards c,-erect to the line ABCD a perpendicular mp of 24 yards; then will the an= _gles APB, BPC, CPD, be each = 45°; | 24, Having at a certain unknown distance taken the angle of elevation of a steeple, I advanced 60 yards “nearer (upon level ground), and then observed the ele- vation to be the complement of the former: advancing still 20 yards nearer, the elevation was found to be dou- ble the first. Hence the height of the steeple is required? » Ans. 7416198 yards. : _ 25. The excess of the three angles of a triangle, mea- _sured on the earth’s surface, above two right angles, is 1 second: what is its area, taking, the earth’s diameter at 79573 miles? 7 a Ans. '76:75299 square miles. ; e 26. The three sides of a triangle, measured on the -earth’s surface, and reduced to the level of the sea, are 17,18, and 10 miles, What is the spherical excess? _ 27. The angles subtended by two distant objects at a ‘third object is 66° 30’ 39”: one of those objects appear- -ed under an elevation of 25’ 47”; the other under a — depression of 1”. Required the reduced horizontal - angle? \ +. Ans. 66° 30°37”. “a 28. In two right angled spherical triangles axe, ADE{ . having one angle a ean ik there be given thetwe A y aes (ah i” y fe f ct > as Pde fe Kish 236 Problems without Solutions. perpendiculars Bc, DE, and the sum or difference of the hypothenuses AC, AE; then it will be, tan 3 (DE + BC) : tan 3 3 (DE — BC) :: tan t (AE + AC) , :tan} (AE — AC). Aya Loft Required a demonstration. /y 29. If, besides the two perpendiculars, there be given " dhe sum or difference of the bases, AB, ap; then it will be, sin (DE + BC):sin (DE — Bc)::tan.3 (AD + AB) tan § (AD — AB). Required a demonstration. 30. Given the hypothenuse ac, aid the sum aB- + BO, or difference AB — Bc of the legs of aright angled spherical triangle: then it will be, . 2 cos AC = cos (AB + BC) + cos (AB — BC) whence the legs become known, Required a demon- stration. 31. If, in a plane triangle asc there be given, the sides Ac, Bc, and the line cp drawn to bisect the verti- eal angle and terminating in the base: then we have AC. AC. BC — CD? — cp? AB = (AC + BC) ateaeors (Ac + BC) cD and cos ACD = cos BeD = ———_—— « 2ac.BC Required a demonstration, : 32. If the triangle azc be spherical, and the same parts be given; then sin (Ac + Bc) tanep cos ACD = Cos BCD = teen ECRRET ATE Required a demonstration. 33. Given the north latitude of three places, viz. Moscow 55° 30’, Vienna 48° 12’, Gibraltar 35° 30’, all lying directly in the same arc of a great circle. The ‘dif. ference of longitude between Vienna (situated between the other two) and Moscow, easterly, is equal to that | between Vienna and Gibraltar, westerly. It i s re uired ACO 4 Gaated CLE BED tre &: D Ane BE = f7 BUCY Hews & ov ES of os f Problems without Solutions. | 237 to find the true bearing and distance of each place from the other, as well as their difference of longitude. Ans. Difference of longitude between Moscow and Vienna 14° 13’ 27”, between Moscow and Gibraltar 28° 26° 54”. . Vienna and Gibraltar bear from Moscow. S, 56° 4’ W. ‘Moscow from Vienna........... operas NL 44°50’ E; Gibraltar from Moscow, 27° 59’ = 1937 miles. y 34. Two sides of a plane triangle are 80 and 50, and 9 ‘the line bisecting the verticle angle and terminating in 'the base is 52. Required the angles of the triangle, ‘without previously finding the base. _ Ans. 37° 38’ 16”, 64° 39’ 22”, and 77° 42’ 29”, 35. ‘Two sides of a plane triangle are 12 and 9, and T ‘the right line drawn from the vertical angle to the mid- dle of the third side is 8. Required the angles of the ‘triangle, Ans. 39° 45’ 10", 58° 29’ 55”, and 81°44’ 55”. 36. When all the three sides of any plane triangle are given, an angle a may be found by the following ‘theorem: : " 2(% perim — ap) (4 perim — ac) AB,AC ‘ Required a demonstration. 37. In any spherical triangle Axc, versin A = : versin cB — versin {CA — BA) versin A = a sags pf otpat r $8. Also, sin ACSIN AB : 3 [versince — versin(ca — BA ¥ si AC sin AB o o9. And ft versin cB — versin(caA — BA) LAT ET MALE eR MT, a are versin (CA + BA) — versin cp i Required the demonstration, 42 Qe 97 ee e lyhe+ bls lkb= kb : Gale plig he 2 tan? 3A = » ys | Sees OK ee Sh a — - 238 Problems without Solutions. 40. On a certain horizontal dial the sum of the ans gular distances of the hour lines of 3 and 4 in the after= noon from the 12 o’clock hour line, is 82° 48’. For what latitude was the dial madé?. Ans. The latitude of 41° 48’. 7 41. On another horizontal dial the distance between. the hour lines of 4 and.5 o’clock in the afternoon was 20 degrees. For. what latitude was this dial made? Ans. The latitude of 35° 53’. 42. In what latitude is the angle included between the hour lines of 12 and 1, on a horizontal dial, double ‘oa of that included between the same hour lings on a verti- \ ge 8% Cal-south dial? ; 4. Ans. Latitude 63°45’ 2Q”. i ' 48. Suppose a horizontal dial made for the latitude of 51° 32’, to be fixed horizontally in the latitude of 54°; % what will be the greatest error, and at what time of the “e day? 3 Ea The greatest error will be 4" 27° of time, and it will occur at. 6" 15" 27%, true time from noon. _ 44, On what day of the year, in the latitude of Lon- don, will the length of the afternoon exceed that of the morning by the greatest difference possible; conceiving | the day te begin at sun-rising and to end at sun-sets ting? ae When the sun’s longitude is 29° $2’ from the’ first pomt of Aries, that is, on the 19th day of April; the afternoon of that day exceeding the morning by 1” 445, | 45. In latitude 53° N. stands a tower, the shadow of whose summit-on June 10th, 1816, described on the ho- rizontal plane a hyperbolic curve whose transverse. axis. was 150 yards. What was the height of tliat tower? Ans, 43°688 yards. sie 46. Required the’ latitude of the place and the decli- nation of the sun when the length of the day is to that. of the night as $ to 2, and-the sun’s mid-day altitude to. . his midnight depression as 2 to 1? | "Ans, Lat.61° 57’ N. Declination 9°21’ N, LA Lage Ad by ca bbe AS Problems without Solutions, 239 . _ 47. In the north latitude of 51° 30’ I observed the sun to rise on a certain point of the compass; and eleven hours after the sun had arisen, I perceived my shadow to be projected towards the same point. What was the sun’s declination? | Ans. Declination 9° 54’ 41” N. 48. To what height must a person be raised above the _ city of London, on June 2Zlst, at midnight, so that he may just see the sun’s upper limb? : Ans. 155°4283 miles, the radius° of the earth being - supposed 3980. 49. In the latitudes of 36° and 50° north, on the same meridian, on the 2Ist of June in the morning, it is: required to determine the exact instant; when the sun’si altitude will be equal, if observed at both’ places; alsa, the latitude at that time? ee , Ans. Time 3°55" 16° from noon; altitude 37° 37’ 17”, 50. At what time. of the year is' the night’ (exclusive _ of twilight) longer at York (N. lat. 54°) than itis either at London (N. lat. 51°32’) or at Edinburgh (N, ‘lat. 56° 7’)? Ne ed Ans. Wiilst the sun’s declination south is. between: 14° 14° 46” and 14° 40° 34”; that is, on February 10th; and, November Isti — ; rei Oe tai 51. In what north latitude may an erect south dedlin- . ing dial be fixed, to have the declination: of. the plane, a the distance of the substyle from the: meridian, and the! height of the style, all equal:? Ans, The latitude of 38° 2’; and the plane’s. deéclina- tion will be 38° 13’. | 62, In the spring, quarter last. year, day broke at) 3:0’clock, and the sun’s: altitude, that morning when) due east, was 32° 42°. Where and: when did. this happen? | SY0UL _. Ans. Lat. 38°39’ 20” N.: Dec. 19°49’ 18” N. answer- Ing to May the 19th, _ 58. At a certain place [observed thesun to rise at 10 | _mubrutes past 4 o’clock, and! his altitude at noon to be Lois L > = Tinks UAL foe A uA Ue sage ae » 940 Problems without Solutions: 58°40’. What were the latitude and day of observa- tion? Ans. Lat. 51° 80 Ni; Dec. 20° 10'N.,, eae rCrine ye to May 21st and July 224. 54. On the longest day last year it was observed at a certain place, that the sun’s altitude when due east was 14°46’ more than it was at 6 o’clock the same morning ; what were those altitades, and what was the latitude of the place? Ans. Altitudes 32° 8’ 5” and 17° 22’ 5”, Latitude 48° 30’ 49” N. 55. What is the northern latitude, time of the year, and time of the day, in 1816, when the sun’s altitude, his azimuth from the east, the arc from noon, and the co-latitude of the place are equal to each other? Ans. Lat. 51° 28’ 53”, time of year Apr17, or Aug. 26, time of day 9°25" 56° A.M. 56. In what north latitude is the shortest day equal te 48, of the longest at London? Ans, Vatitude 43° 28’. 57. In’what latitude and time of the year does the sun rise at half past 5 o’clock, and appear due east at 10? Ans. Lat. 21°13’, Dec. 18° 35’, both of the same kind, 58. Where is the sun’s altitude at 6 o’clock, on the longest day, equal to the co-latitude? Ans, N. latitude 68° 17’ 12”. 59. To find the declination of that star whose change in azimuth is a maximum or minimum in a given time, reckoned from the time that it transits a given almu- cantar in a given latitude. Suppose the latitude of London, the: time one hour, and the almucantar 15° above the horizon. - Ans. Star’s declination 20° 25’ south. 60. What arc of a circle is equal to its tangent? Ans. Arcs of 257° 27’ 12”, 442° 37’ 28”, 624° 45° 38”, 805° 56’ 1”, 986° 40° 36”, 1167° 11’ 23251 347° 33° 55%, Problems without Solutions, 4b 1527° 51’ 9”, 1'708° 5’ 44”, or 1888° 16’ 12”, answer the conditions of the question; and these ares fall in. the 2d, 4th, 6thy 8th, 10th, 12th, &c, quadrants, running ¢con- tinually round the same cirele, 61. What are is that whose sine shall be equal to the sine of triple‘the same arc ? 62. What arc is that whose sine is half the sine of its triple arc? 63. What arc is that whose sine is the’ th part of the sine of its triple arc? Ans. The arc whose sine is } 4/ (3. — n). 64. What arc is that whose sine is 8 equal to sthe sine of its quadruple arc? 65. What are is that whose sine is equal to the sine of its quintuple are? 66. What arc is that whose sine is equal to 7 timesiits cosine ? n? Ans. The are whose sine is Vere . 67. What arc is that whose secant is equal to ” times its tangent? Ans, The arc whose tangent is ne — ) 68. In a right angled triangle the right lines drawn from the acute angles to the middle points of the oppo= site sides are equal to a and 6 respectively: required the acute angles. Ans. The acute angles are those whose tangents are ileesass, fuze. Thus if a = 4, 6 = 3, then the tangents are ~-and \/ an and the angles are 58° 54’ 32” and 31° 5’ 28”. 69. Demonstrate the truth of Dr. Maskelyne’s four. M 942 Problems without Solutions. rules for determining the log. sines and tangents of small arcs, given at pa. 54 of this volume. 70. Supposing the latitude of London to be 51° 30’ N., the latitude and longitude of Moscow 55° 45’ N. and 38° E,, and the latitude and longitude of Constantinople 41° 30’ N. and 29°15’ E. It is required to determine the Jatitude and longitude of a place which shall be equidistant from the former three. Ans. Lat. 519.17’ N., long. 19° 13’ E. 71. Three stars a, B, and c, were all observed to be in the arc of a great circle; the distance of a and B was found to be-10°, of Band c 20°; the difference of the ‘azimuths of a’and c was found to be 90°; and the mid- dlemost 8 was the least distance possible from the ze- nith. Required the altitudes of the three stars? Ans. Alt. of a, 72°18’ 14”, of B, 75° 19’ 32”, and of €}'657'22' 38" 0 | 72. In what north latitude will the sun appear due east, on the longest day, at the mid-time between sun- rise and noon?.~ | Ans. InN. lat. 64° 3.5’ 48.” ; 3 . BC 3 73, Let anc he a plane triangle of which —- = m;4. it is required to demonstrate, that the value of the angle A is €xpressed in seconds by the first or second of the following series, according as the perpendicular from Bon Ac falls within or out-of the triangle : viz. _ msinc m* sin 2c m3 sin 3c mt sin 4c & wtih gin 1” "sin Q” sin 3 sin 4” eA can sin c m* sin 2c m3sin 3c mi sin 4c = sink” sin? a sinS’ \. sin4”. +-&e., 74.’ Demonstrate also that Mm COsB msin2sn micos3B mésin4z - sin 1” gin 2!” sin 3” sin 4’ mcosB m* sin 2B m3 cos 3z mt sin At ee SS gee SOR ee sin 1” sin 2” ' gin 3’ sin 4” + &c. Problems without Solutions. 243 75. If in a plane triangle tan B = 7 tan a, then de- ‘monstrate that : eo ee —-- l—n_ sin@A 1—n)\2. sin4a (A— 3B) = e Ll+n sin 1” ltn sin 2’ 1— Ay sin 64 + (; + n/ Lana’ e oc and that if z = cos c, then : 3) tan? gc sin 2a tan4 £csin 4a Sire as sin 1” ny sin 2” tan® Scsin 6a . —= — &c, sin 3” 76. Let © be the spherical excess in seconds of a spherical triangle ano, then it is to be demonstrated that - gE = tan gb tan desin a _ tan? 3b tan? de sin 2a sin ]// sin 2! tan3 44 tan3 4c sin 3a sory Eitan de heh sin3 77, Demonstrate also, that cot 45 cot de | gin A and find u, when a = 6 =c¢ = 179°, Ans. % = 857° 59’ 58”, | 78. Given the latitude of the place, and the position _ of two hour circles, with respect to the meridian; to de- termine the declination of that star whose change in al- titude shall be the greatest possible in passing over the interval between those hour circles, Ans, Let h’ be the greatest, and h the least hour an- gle from the meridian, 1 the latitude, and p the decli- nation; then cot 4E= cota + sind (h’—h) * sin § (hk + h’) 79, Ifa person could approach so near to the moon as to see one third of her convex surface, what angle fan p\= fan Z DAA Problems without Solutions. at his eye would the diameter of her apparent dise sub- tend? : Ans. Twice the angle whose natural cosecant is thrice radius, that is, twice 19° 28’ 16” or 38° 56’ 32”. 80. In aplane triangle azc are given cot A cot B= 4, cot B cot © = 5} to find the angles a, B; and ¢. Ans. 45°, 63° 26’ 6”, and 71° 33° 54”. 81. Ina certain north latitude it was observed that on the morning of the longest day the body of the sun occupied precisely 8 minutes of time in rising out of the horizon. What was the latitude? Ans. 61° 29’ 64” disregarding refraction and parallax: or, 62° 28" 54”, taking the horizontal refraction at 33’; and the parallax at 9”, 3 THE ENR ©, Baldwin, Printer; New Bridge-street, London. _ Publications by the Author of this Work. reas 1, LETTERS TO A FRIEND, on the Evidences, Doc- trines, and Duties of the CHRISTIAN RELIGION, | Third Edition, in two vols, 8vo.; price 14s. boards, 2. LESSONS ASTRONOMICAL AND PHILOSOPHICAL, for the Amusement and Instruction of British Youth, 12mo, Fourth Edition, price 5s. boards, 3. 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