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A TREATISE 
 
 ON 
 
 PLANE AND SPHERICAL 
 
 TRIGONOMETRY; 
 
 NCLUDING THE 
 
 CONSTRUCTION OF THE AUXILIARY TABLES; 
 
 A CONCISE 
 
 TRACT ON THE CONIC SECTIONS, 
 
 AND THE 
 
 PRINCIPLES OF SPHERICAL PROJECTION. 
 
 BY ENOCH LEWIS. 
 
 PHILADELPHIA: 
 URIAH HUNT & SON, 
 
 No. 62 NORTH FOUXITII STltEUT. 
 
 1860. 
 
 r 
 
Entered, according to the Act ot Congress, in the year 1844, by 
 
 HECTOR ORR, 
 
 in the clerk's office of the District Court of the United States in and for the 
 Eastern District of Pennsylvania. 
 
 J. Pagan, Stereotyper. 
 

 PREFACE. 
 
 It will probably appear, to some, a work of super- 
 erogation to add another tract on Plane and Spherical 
 Trigonometry to the great number already before the 
 public ; especially as some of those treatises are the 
 productions of men whose talents and attainments 
 were unquestionably of the highest order. Still, it has 
 appeared to me that, however valuable many of those 
 works must be considered, there are none of them 
 exactly suited to the use of schools in which this 
 branch of mathematics is traced to its principles. It 
 is, indeed, no unusual thing to find young men who 
 have studied this science in the way it is commonly 
 taught, who are very imperfectly acquainted with the 
 nature, and almost entirely ignorant of the construc- 
 tion, of the tables which they are continually using. 
 And it must be admitted, that, when the nature and 
 construction of logarithms, and of sines and tangents, 
 are explained by Algebra and common Geometry, the 
 processes are generally either so obscure, or so prolix, 
 as to discourage the majority of students. The Diffe- 
 rential Calculus is well known to furnish the most 
 direct, if not the only direct, and simple method of 
 
 A* (v) 
 
vi PREFACE. 
 
 investigating the formulae by which those tables are 
 most expeditiously computed. But that calculus itself, 
 as commonly exhibited, presents so many refined 
 speculations, that very few, except those who have a 
 taste for mathematical studies, can avail themselves of 
 its advantages. 
 
 As it is evidently unscientific to erect a system, 
 either in theory or practice, upon unknown principles, 
 it has been my object, in the following work, to trace 
 every process which is required to be adopted, to 
 principles which are supposed to be previously under- 
 stood. The student is supposed to be already 
 acquainted with Algebra and Geometry. If the 
 student is master of the first six, and the eleventh 
 books of Euclid ; or, which is nearly the same thing, 
 of the first six, and the second supplementary book 
 of Playfair's Geometry ; and of as much algebra as is 
 contained in the first ninety pages of my treatise ; he 
 may proceed with confidence to the study of the 
 following tract. 
 
 This work was intended to include as much only 
 of the Differential Calculus, as the elucidation of the 
 science of Trigonometry required. I have therefore 
 confined myself to differentials of the first order; and, 
 by the use of proper expedients, have deduced the 
 requisite formulae from those differentials. Some of 
 
PREFACE. vii 
 
 the methods used in this work are supposed to be new ; 
 and, if so, they may be considered as improvements 
 upon the labours of my predecessors. Of this cha- 
 racter are the investigations of Gregory's theorems 
 for computing an arc in terms of its tangent, and for 
 computing logarithms. 
 
 The treatises on Spherical Trigonometry with which 
 our schools are supplied, are nearly all of them desti- 
 tute of anything on the subject of Spherical Projec- 
 tions. This appears to me an important defect. A 
 small tract on that subject, which I added, more than 
 thirty years ago, to a Philadelphia edition of Thomas 
 Simpson's Plane and Spherical Trigonometry, is the 
 only one, so far as I know, which is to be found in 
 our schools ; unless we consider Davies's Descriptive 
 Geometry as one. Simpson's work being now out of 
 print, and the work of Davies, notwithstanding its 
 merits, not appearing calculated to supply the place of 
 the appendix, I have revised, or rather written anew, 
 that part of my early labours, and subjoined it to this 
 work. 
 
 The present treatise being designed as an introduc- 
 tion or preliminary to Astronomy, a concise tract on 
 the Conic Sections, including all the properties of the 
 ellipse and parabola which are usually cited by writers 
 on that science, has been introduced. This appeared 
 
viii PREFACE. 
 
 requisite, because some of those properties were 
 unavoidably referred to in the tract on Spherical 
 Projections ; and, among the treatises on that subject 
 already in print, it was not easy to fix upon any one 
 to which I could refer, that would be generally known 
 to my readers. Besides, it appeared no difficult matter 
 to include in this work all the information which the 
 astronomic inquirer would need in the prosecution of 
 his studies. The Conic Sections being, as the name 
 implies, derived from the cutting of a cone, I thought 
 it more direct to deduce the primary propositions from 
 the section of the cone, than to lay down first a plane 
 figure, derived in a different way; and, after demon- 
 strating most of its properties, to prove at last its 
 identity with a conic section. 
 
 In the practical examples, some astronomical terms 
 are used which are not defined, because it was sup- 
 posed that few young persons would study this work 
 without some previous acquaintance with terms so 
 generally understood. 
 
 The references to the properties of geometrical 
 figures are made to Playfair's Euclid ; but they will 
 generally apply to Simson's translation of Euclid's 
 Elements. This oldest work on Geometry, with the 
 few improvements introduced by the Scotch geometer, 
 is, in my opinion, better calculated to lead the attentive 
 
PREFACE. ix 
 
 student into an accurate acquaintance with this noble 
 science, than any modern treatise which has fallen into 
 my way. Among those improvements, however, the 
 substitution of the language of Algebra, in the fifth 
 book, in place of that which Euclid made use of, Can 
 hardly be reckoned as one. Play fair's exposition of 
 the fifth definition is certainly a good one; but, in 
 other respects, I consider Simson's translation of that 
 book greatly preferable to the form in which Playfair 
 has left it. 
 
 Philadelphia, 10 Mo. 1844. 
 
PLANE TRIGONOMETRY. 
 
 INTRODUCTION. 
 
 In the practice of Trigonometry there are several tables 
 generally used, the construction and uses of which constitute 
 an essential part of the science. But when the construction 
 of these tables is deduced from Geometry and common Alge- 
 bra, the subject is certainly presented to the student in a very 
 discouraging shape. The rules by which these tables are 
 most readily computed, are easily derived from the Differen- 
 tial Calculus ; but that branch of science, when pursued to 
 any considerable extent, involves many refined and difficult 
 speculations. It therefore generally happens that the stu- 
 dents of Trigonometry acquire a practical acquaintance with 
 the auxiliary tables, but understand neither their construc- 
 tion nor nature. 
 
 As those parts of the Differential Calculus which the con- 
 struction of all the tables commonly used in trigonometrical 
 calculations absolutely demands, lie within a narrow compass, 
 and involve no very difficult inquiries, I shall employ a few 
 pages in explaining the elements of this science, so as to 
 enable the student to understand the nature and origin of the 
 formulae which are commonly used in the computation of the 
 trigonometrical tables 
 
 CM) 
 
12 PLANE TRIGONOMETRY. 
 
 Article 1. The Differential Calculus is founded essentially 
 upon the relation which variable and dependent quantities, 
 considered as decreasing till they vanish, bear to each other 
 in their evanescent or vanishing state. This rati-o is called 
 their ultimate ratio. Newton observes that the ultimate ratio 
 of evanescent quantities is not the ratio which they have 
 before they vanish, nor afterwards ; but the ratio with which 
 they vanish.* As the ultimate ratio of vanishing quantities 
 is not the ratio which they have before they vanish, but at 
 the instant when they vanish, it is most convenient to deter- 
 mine that ratio by supposing the evanescent quantities to 
 have actually vanished. This may often be done in a way 
 which leaves no room for error or doubt. One or two exam- 
 ples in common Algebra will render this obvious : 
 
 xr 
 
 — —a+x 
 
 fi — x? 
 
 — — = a iJ rax J r x~. 
 
 Now, these results being strictly correct, whatever value 
 may be assigned to a or x< let x be supposed at first less than 
 a, and to increase till it becomes equal to a ; d 1 — x 2 and 
 a — x, are thus rendered decreasing and evanescent quanti- 
 ties, whose ultimate ratio is the ratio which they have, not 
 before x becomes equal to a, but at the instant when that 
 
 equality is attained. But when x=a, a+x=2a=-y ; and 
 
 ar-\-ax-\-x 2 =3a 2 =-^-. Hence the ultimate ratio of a — x to 
 
 a- — or, when x approximates and ultimately equals a, is the 
 ratio of 1 : 2a; and the ultimate ratio of a — x to a a — x 3 , 
 under like circumstances, is the ratio of 1 : 3a\ 
 
 Principia, Scholium, Book I., Art. 1. 
 
INTRODUCTION. 13 
 
 Art. 2. In the investigations connected with the Differen- 
 tial Calculus, quantities are considered under two very dif- 
 rent aspects: constant and variable. 
 
 Constant quantities are such as remain unchanged, or 
 retain the same value throughout the investigation. Variable 
 quantities are those which change their value in the course 
 of the solution or demonstration. Constant quantities are 
 usually denoted by the initial letters, a, b, c, &c. ; variable 
 ones by the final letters, x, y, z, &c. 
 
 Art. 3. When two variable quantities enter into an equa- 
 tion, so that the value of one depends upon the value of the 
 other, and one of them is increased by any quantity or incre- 
 ment ; the quantity which must be added to the other to 
 preserve the equation, is called the corresponding increment. 
 Thus, if y=ax 2 , and we increase x by the increment h, so that 
 x becomes x-\-h, the new value of y will be 
 a(x+hy=ax 2 + 2axh + ah 2 ', 
 hence the corresponding increments of x and y are // and 
 2axh -f all 2 . 
 
 The variable quantity from which a differential is deduced 
 is called an integral. 
 
 Differentials of dependent variables may be considered as 
 the corresponding increments of those variables, in the ulti- 
 mate state of the increments, when they are diminished till 
 they vanish. Or differentials may be defined to be quantities 
 having to each other the ratio which is the ultimate ratio 
 of the corresponding increments, those increments being sup- 
 posed to decrease till they vanish. 
 
 Hence it follows that one differential cannot enter into 
 an equation without another. We never inquire into the 
 absolute, but merely into the relative values of differentials. 
 
 Art. 4. The differential of a variable quantity is denoted 
 by prefixing the letter d ; thus dx signifies the differential of 
 x. When the integral is a compound quantity, it is enclosed 
 in brackets, or a point is introduced between it and the letter 
 d : thus the differential of x n is expressed by d(x n ), or d.x n ; 
 
 B 
 
14 PLANE TRIGONOMETRY. 
 
 whereas dx n signifies the n power of dx, the same as {dx) n . 
 The differential of x+y is also expressed by d(x+y). In these 
 cases the letter d is not an algebraic quantity, but a sign, or 
 an abridgement of the words differential of. 
 
 Art. 5. Let y = x + z, and let x and z be increased by the 
 increments h and k respectively. Denote the new value of y 
 by y' ; so that 
 
 y = x + h -\- z + h ; 
 
 wherefore y — y = h + k; 
 
 or the increment of y = the sum of the increments of x and z. 
 Now, this being true, whether the increments are taken in 
 their vanishing state or any other, we evidently have 
 
 dy = dx + dz. 
 
 Again : Let y = ax, 
 
 and y' = a(x-\-h) = ax + ah ; 
 
 then y' — y — ah; 
 
 which is also true, whether the increment h is taken in its 
 vanishing or finite state : consequently, 
 
 dy = adx. 
 Art. 6. Let y — x 2 , 
 
 and y' = (x + h)" = *' + % xh + &; 
 
 whence y' — y = 2a:A + A 2 , 
 
 and y , * =%x -{■ h = — , — : 
 
 n 1 
 
 consequently, % ; : y — y : : 1 : 2# + A. 
 
 But the ultimate ratio of 1 : 2x + h is the ratio of 1 : 2x ; 
 
 wherefore, dx : dy : : 1 : 2a: ; 
 
 or . . . . dy = 2o%£r ; 
 
 that is, da: 2 = 2^^r. If the equation dy = 2xdx be put into 
 this form, 
 
INTRODUCTION. 15 
 
 dx ~ ** 
 the last member 2x is called the differential coefficient. 
 Art. 7. To find the differential of xz. 
 Put y = : x + z ; 
 
 then y 2 = (a? + z) 2 = af + 2xz + z 2 ; 
 
 and (Art. 5,) dy = dx + dz; 
 
 also c?.y 2 = d.(x + z) 2 = 
 
 (Art. 6) 2(x + z) (c?x + dz) = 2xdx + 2xdz + 2zdx + 2z<7z. 
 But ' d.y 1 = d(x> + 2xz + z 2 ) =* 
 
 (Arts. 5 & 6) 2.^ + 2^(xz) + 2zdz. 
 
 Comparing these values of d.y 2 , we have 
 
 2d.(xz) = 2xdz + 2zdx ; 
 or d.xz — xdz + zdx. 
 
 Hence, d.xyz = xd.yz + yzdx == xz/^z + rrzdy + yzcfo. 
 
 Again : 
 
 d.xyz _ dz dy dx 
 xyz z y x 
 
 and the same thing may be proved, whatever be the number 
 of variables. 
 
 Art. 8. To find the differential of x n , n being a whole 
 positive number. 
 
 It is obvious that x n = x.x.x.x to n terms. Hence, by Art. 7, 
 
 d.x n dx dx ndx 
 — — = h — + &c. to n terms = : 
 
 wherefore, by clearing of fractions, 
 
 d.x n p nx xl ~ ] dx. 
 
 x 
 Art. 9. To find the differential of — , both numerator and 
 
 denominator being variable. 
 
16 PLANE TRIGONOMETRY. 
 
 Put z = - ; 
 
 y 
 
 wlience, y% — x, 
 
 and, (Art. 7), ydz + zdy = dx : 
 
 consequently, 
 
 dx—^ 
 
 d% ^ ^—Z^M = & = ydx — x dy 
 
 y y y l ' , J 
 
 that is, d.'~ — 
 
 ydx — xdy 
 
 y y 
 
 m 
 
 Art. 10. To find d.x", m and n being positive integers. 
 
 m 
 
 Put y = a? ; 
 
 then, . . . y n — x m ; 
 
 and, (Art. 8) ny n ~hly = mx m - } dx; 
 
 whence, 
 
 mx m ~ } dx m x m ~ l m —~\ 
 
 Art. 11. To find the differential of a? _n = — . 
 
 x n 
 
 then yx n = 1. 
 
 But 1 being invariable, its differential is 0; therefore, 
 
 d.yx n = 0. 
 
 Or, yd.x n -f x n dy = ; 
 
 that is, nyx n ~~ ] dx + x n dy = : 
 
 whence, 
 
 j — nyx n ~ x dx x n ~ ] dx — ndx 
 
 d y = "Ji = — n -^ir =^+r=— nx — ^dx; 
 
 or, <7.ar~ n = — nx~' n ~~ ] dx. 
 
 From this article, and Arts. 8 and 10, it is evident that 
 
INTRODUCTION. 17 
 
 d.x n = nx n ' x dx, 
 whether n is integral or fractional, positive or negative. 
 
 Art. 12. Let the two ascending series, Ax* + Bx b + Cx c + 
 Dx d + &c, and Mx m + Nx n + Px p + Qx* + &c, be always 
 equal ; so that whatever value may be assigned to x, we shall 
 still have, 
 
 Ax a +Bx ,b + Gr s + Da: d + &c. = Mx m + Na: n + Pa: p + Q^ + &c. ; 
 then, a = m, b = n, c = p, &c. ; 
 
 and A - M, B = N, C = P, &c. 
 
 For, if possible, let a be less than m, and divide the equation 
 by # a ; then 
 
 A + Bo*" 11 + Gc c " a + Da " 8 + &c. = 
 Mx m - & + Nx"-* + Px p - a + Qx- q - a + &c. 
 But as the series are both ascending ones, b, c, d> m, n, p, q, 
 &c. are all greater than a. Hence, if x = 0, all the terms of 
 this equation, except the first, will vanish ; hence in that case 
 A = 0, which is evidently absurd. Therefore, a is not less 
 than m ; and in a similar way it may be proved that m is not 
 less than a;, therefore a = m, and the above equation be- 
 comes 
 
 A + Bx b ~* + O c - a + Dx da + &c. = 
 
 M + Ntf n - m + P^ p - m + Qx*~ m + &c. 
 And making x = 0, A = M. Consequently, A# a = Mx m ; 
 wherefore, 
 
 Bx h + Cx* + Dx A + &c. = Nx n + Pa: p + Q^ + &c. 
 Then, by the same process of reasoning, we find b = n; and 
 B = N. Hence the proposition is manifest. 
 
 Art. 13. An important application of the property just 
 announced may be exhibited in the demonstration of New- 
 ton's binomial theorem. 
 
 It is evident that in the general development of (1 -f- x) n , 
 the first term must be 1 ; for when x = 0, (1 + x) n = l n = 1. 
 We may therefore assume 
 
 (1 + x) n = 1 + Ax p + Bx« + Cx T + Da? 8 + &c. 
 
 3 B* 
 
18 PLANE TRIGONOMETRY. 
 
 in which A, B, &c., are unknown, but determinate coeffi- 
 cients; and p, q, r, &c, unknown exponents, integral or 
 fractional, positive or negative. Suppose x a variable quan- 
 tity. Then, differentiating both sides of this equation, and 
 dividing by dx, we have, % 
 
 ' «.(1 -f xf" 1 = pAx^ + qBx-*- 1 + rCx T ~ l + sT>x*~ l -f &c. 
 Multiplying bj I + x, 
 
 rc.(l + ;r) n = pAx»~ l + qBx*" 1 + rCx r ~ l + sDx 8 - 1 + &c. 
 pAx» + qBx* + rCx T + s&x 3 + &c. 
 Then, from first equation, multiplying by n, 
 
 n{\ + x) n = n -f nAx p + nBx q + nCx r + &c. 
 These equations being identical, we have by transposition, 
 
 pAx'- ] + qBx*- 1 + rCx T " 1 + sDx" 1 + &c. = n + 
 (n — p) Ax p + (n — q) Bx Q + (n — r) Cx T + (n — s) ~Dx 9 + &c. ; 
 and by comparing, first the exponents, and then the coeffi- 
 cients, (Art. 12,) we have 
 
 p — 1 = Q ; q — 1 = p; r — I — q ; &c. ; 
 or, p = 1 ; q = 2 ; r = 3 ; s = 4; &c. : 
 
 and 
 
 a t> n — 1 a n — l n n — %v> n— In— 2 
 
 A=w;B=— g— A=n. — g- ;C= — ^— B=n.—^-.—^-; &c. 
 
 Consequently, 
 
 w — 1 c w — In — 2 , 
 (1 + x) n = 1 + n# + n. — Q—ar+ w -~o — • — q~~ ^ + & c - 
 
 From this we readily obtain the development of (a + by. 
 For, 
 
 (a+by = a«(l + -)"; 
 
 in which we have — instead of x in the preceding. Conse- 
 quently, 
 , 7V ( , b n — \ b* n— In— 2 P . ) 
 
INTRODUCTION. 19 
 
 = a n + na n - } b + n. —^~a n - 2 b 2 + n. —^— . — ^— a n ~W &c. 
 
 As the equation d.x n — nx n ~~~^dx, on which this demonstration 
 is founded, is equally correct whether n is integral or frac- 
 tional, positive or negative, it is evident that the preceding 
 development of (a + b) n is also correct, whatever may be the 
 value of n. 
 
 Of Logarithms. 
 
 The calculations which are connected with Trigonometry 
 are much facilitated by the use of logarithms ; it will there- 
 fore be proper, in a treatise on that science, to explain their 
 nature and use. 
 
 Art. 14. If we take a series of numbers in geometrical 
 proportion, beginning with a unit, as 1, a, a-, a 3 , a 4 , a 5 , a fi , a 7 , 
 &c, it is manifest that the product of any two of these terms 
 is a term whose exponent is the sum of the exponents of the 
 factors. Thus, 
 
 a\a? =a a 7 ; a m .a n = a m +\ 
 If then A = a m , and B = a", , 
 AB = a m + n : 
 A 2 = A.A = a m + m = a 2ra : 
 
 On the other hand 
 
 A 3 = a m + m + m 
 
 a= a 3m , &c. 
 
 id, 
 
 a 1 . A 
 ^* ; B 
 
 a m 
 = — = a m - n ; 
 
 1 m 
 
 A' 2 = a 2 ; 
 
 1 m 
 
 A 3 = a 3 . 
 
 Hence it appears that a, being assumed equal to any num- 
 ber at pleasure, if we can find such values of m, n, &c, that 
 a m = A, a n = B, &c, A, B, being given numbers, then calling 
 m the logarithm of A, n the logarithm of B, &c. ; the loga- 
 rithm of AB will be the sum of the logarithms of A and B. 
 
20 PLANE TRIGONOMETRY. 
 
 The logarithm of ~ will be the logarithm of A diminished by 
 
 the logarithm of B. In other words, the business of multiply- 
 ing and dividing by given numbers may be effected by the 
 addition and subtraction of their logarithms. 
 
 As a m = A, a given number ; we readily perceive that, by 
 assuming different values of a, we shall change the value of 
 m ; that is, we shall have different numbers to denote the 
 logarithm of a given number A, by varying the value of a. 
 Thus it appears there may be an indefinite variety of systems, 
 according to the various values which may be taken for a. 
 This quantity a is called the radix or base of the system. 
 
 Art. 15. To investigate a formula by which the logarithm 
 of any given number may be computed, we may assume 
 a x = y ; y being any given number whatever ; then x = loga- 
 rithm of y : and the object in view is to find a general expres- 
 sion for x in terms of y. If we suppose x to be variable, it 
 is manifest that y — a x must also be variable. 
 
 In the first place, if x — 0, then y '= a — 1, whatever value 
 may be assigned to a ; it is therefore evident that the loga- 
 rithm of 1 is in every system. 
 
 Now, let y' = a x + h = a x .a h = ya b : 
 
 and, to reduce this second member to a more manageable 
 form, put 1 + b = a ; 
 
 then, (Art. 13,) 
 
 ^=y(l+5) fc f=y+y | hb+h.^bHh! 1 -^. — P + &c. | 
 
 Therefore, 
 
 y- - y = y \ hb + h~b- + h!^. ~V + &c. j ; 
 
 and, consequently, 
 
 y' — y c. h — 1_„ h— I h — 2, 
 
 h 
 
 ( 1 it — i 7C , n — i it — ^., e ) 
 
INTRODUCTION. 21 
 
 Now, when h = 0, the series in the second member of this 
 equation becomes 
 
 b — W + W — W + l^ 5 — &c.; 
 and this is the value to which this series approximates as h 
 is diminished, and to which it arrives only at the instant 
 when h = 0. Put, then, 
 
 b - W + ±V-\V + #J. &c. = ~; 
 
 and it will be V_Zll = 1-. 
 
 h m 
 
 Hence the ultimate ratio of h to y —~ y is the ratio of m to y, 
 
 dy y 
 
 consequently, j- =s — ; 
 
 or, =4i'i& (A) 
 
 In this equation, the value of m depends upon the value of a ; 
 and as a may be assumed at pleasure, we may assign any 
 value we please to m. This is more convenient than to as- 
 sume a value of a, and from that assumption to find the value 
 of m. It is usual to call m the modulus of the system. When 
 m is taken = 1, the logarithms thence deduced are called 
 hyperbolic logarithms, because they correspond with certain 
 areas contained between the curve and asymptotes of an equi- 
 lateral hyperbola. In Briggs', or the common logarithms, 
 the radix a is assumed = 10 ; but m is computed by a method 
 hereafter explained. 
 
 Art. 16. As no general method has been discovered by 
 which to express the logarithm of a number in finite terms 
 of the number itself, we are obliged to have recourse to infi- 
 nite series. When numbers are to be computed by means of 
 such series, it is of importance to have the series constructed 
 in such manner that the successive terms shall become smaller 
 and smaller ; so that, a limited number of terms being intro- 
 
22 PLANE TRIGONOMETRY. 
 
 duced into the computation, the rest of the Series may be 
 rejected without sensible error. 
 
 To find the logarithm of a + *,. a being constant, and z 
 variable. 
 
 By Art. 15, if x — log. of?/, 
 
 mdxi 
 dx= — -. 
 
 y 
 
 Assume, then, log. of a + z = log. a + Az n + Bz p -f Cz q +Dz r + 
 &c. ; in which the exponents to, p, q, &c, as well as the co- 
 efficients A, B, C, &c, are indeterminate. In this equation, 
 if z = 0, we have log. a = log. a, as it evidently ought to be ; 
 and the quantities to, p, q, A, B, C, &c, being susceptible of 
 any value, positive or negative, integral or fractional, the 
 above equation must express the log. of a + z, if it can be 
 expressed at all in terms of z. 
 
 Differentiating this equation, and dividing by dz, we have 
 
 7YI 
 
 ■ — — = nAz n ~ l + pBzP" 1 + qCz*- x + rDz'" 1 + &c. 
 
 a+z i ■ 
 
 By multiplication and transposition, 
 
 naAz n -~ l + poBz"™ 1 + qaCz«-~ l + raDzH + &c. ) _ Q 
 — m-\- nkz n + pBz p + qCz q + &c. I ~ 
 
 Equating the exponents and the coefficients respectively of 
 the corresponding terms, 
 
 to — 1 = 0, jo — 1 = to, q — I =p, r — I = q, &c. 
 togA — m=0, paB+TOA=0, qaC + pB=0, rdD + ^C=0, &c. 
 Hence, to = 1, p = 2, q = 3, &c. ; 
 
 and A = -,B= — -—0=^-^,0 = — -<— ;, &c. 
 
 a 2ar Sa* 4a { 
 
 Consequently, 
 
 *•:■■> , TOZZ TO2Z 2 TO2Z 3 7TOZ 4 
 
 log. (« + ,) = log. a + -- 2^ + 3^-3^+ &c. 
 Putting — z for -f z, 
 
INTRODUCTION. 23 
 
 , • x , mz mz 1 mz' ] mz i £ 
 log. (a — z) == log. a -— , — 5-3 j-j — &c. 
 
 Therefore, log. -- = 2m i — + ~ + - r — ( + &c. 
 
 ° a — z la 3(1* 5a' J 
 
 When a = 1, 
 
 io g .;_±H m j 2+ ! + ! + f + &c. 
 
 To find a number in terms of its hyperbolic log. 
 In this case, if x = log, y, 
 
 ? = </• 
 
 dx 
 Put y = 1 + Ax n + Bx p + O 1 -{- D,x r + &c. ; where y = 1, 
 when a? = 0, as it ought to be. 
 
 Differentiate and divide by dx ; and 
 
 J? (= y ) = nAx n ~ l + pBx*- 1 + qCx*~ l + rDaH + &c. 
 
 Comparing these values of y, and equating the coefficients and 
 exponents respectively, we have 
 
 Art. 17. From the formulae contained in the last article, 
 the logarithms of small numbers? are readily computed ; and 
 those of large ones are easily deduced from the smaller. A 
 few examples are subjoined. 
 
 Required the logarithm of 2. 
 
 Here * = ~j = J. 
 
 z = § = .3333333333 
 ^W|s =.0370370370 
 2 s = j^=. 0041 152263 
 
 z 7 = i z s = .0004572474 
 z 9 = .0000508053 
 
24 
 
 And 
 
 ►LANE 
 
 TRIGONOMETRY. 
 
 z" = 
 
 .0000056450 
 
 z 13 = 
 
 6272 
 
 Zl5 = 
 
 697 
 
 Z 17 = 
 
 77 
 
 % a 
 
 3333333333 
 
 ^ = 
 
 123456790 
 
 1^5 — 
 
 8230453 
 
 4% 7 = 
 
 653211 
 
 iz 9 = 
 
 56450 
 
 1 1 
 
 5132 
 
 1 ~13_ 
 
 483 
 
 _1 -15 = 
 
 1 5 Z — 
 
 46 
 
 1 ~17_ 
 
 5 
 
 .3465735903 
 Therefore, the log. of 2 = .6931471806 m; and log. of 8 = 
 2.0794415418 m. 
 
 Next, let the log. of f be required. Here, 
 
 This number being substituted for z in the foregoing equa- 
 tion, produces 
 
 log. off = . 2231435514m. 
 Then the sum of these logarithms is the 
 
 log. of 8*x I = 10. 
 Therefore, the log. of 10 = 2.3025850932 m. 
 
 If now we desire to compute the common logarithm of any 
 number, the radix of that system being 10, the log. of 10 is 
 = 1 : putting, therefore, the log. of 10 just found equal 1, 
 we find, 
 
 m = OTS0S33 = • 4342 9 44819 " 
 
 The coefficients of m, in the log. of 2 first found, being multi- 
 plied by this number, the product is the common log. of 2. 
 
INTRODUCTION. 25 
 
 Hence, the common log. of 2 = .3010299956. 
 
 From this log. we may find the logarithm of any power of 2 
 
 by multiplication alone. 
 
 Art. 18. The common logarithm of any prime number 
 may be readily computed when that of the next inferior 
 integral number is known. 
 
 Let p = the number whose log. is required ; q = the pre- 
 ceding whole number ; R = 2m = .8685889638 ; and putting 
 
 p _l + z 
 
 n j v— <1 * 
 
 we find, z = — : — = — ,— . 
 
 p + q p + q 
 
 As an example, let the log. of 9 be required ; that of 8, the 
 third power of 2, being known. 
 
 In this case, p = 9, q = 8, and z = fo Hence, the 
 
 log. of 1 = y 7 + Jpp + ,_!_ + >. Jl^ = .0511525224 ; 
 
 in which A is the preceding term ; B, C the preceding terms 
 without the divisors, 3, 5. 
 
 To the log. of f add the log. of 8, or three times the log. of 
 2 ; the sum .9542425093 = log. of 9 ; 
 
 and its half, or .4771212546 == log. of 3. 
 
 From these logarithms, the logarithms of all the powers of 
 3, and of all the products of 2 and 3, and of all the products 
 of their powers, may be obtained by multiplication and 
 addition. 
 
 As a second example, let the log. of 49 or T 2 be required, 
 the log. of 48 being known from those of 2 and 3. Here, 
 
 1 + z 
 p = 49 ; q = 48 ; , = || ; and z = ^. Hence, 
 
 R 
 
 97 ' *'(97V a 
 4 c 
 
 log. of i% \ = — + J.T^r = .0089548426 ; 
 
26 PLANE TRIGONOMETRY. 
 
 to this add the log. of 48 ; and the sum = 1.6901960797 is 
 the log. of 49, and its half = .8450980398 is the log. of 7. 
 
 Art. 19. Although the methods already explained are 
 sufficient to enable the student to compute the logarithm of 
 any given number, yet there are other expedients for abridg- 
 ing the labour of such computations; one of which is the 
 following : 
 
 Let «, b, c, be three equi- different numbers, whose common 
 difference is 1 ; so that a — b — 1, and c — b + 1 ; then ac =* 
 b 2 — 1, and ac + 1 = 6 2 ; consequently, 
 
 ac+l _ b* 
 
 ac ~~ ac' 
 
 If now we put the first member of this equation in place of?/ 
 1 + 
 1 — 
 
 1 + z 
 or ^ in the general equation, (Art. 16,) we shall have 
 
 1 
 
 z = 
 
 2ac + V 
 
 a quantity which will converge more rapidly, the greater a 
 and c are. Finding, then, the log. of 
 
 ac + l 
 
 we have the log. of 
 
 ac 
 
 l l 
 ac 
 
 If, now, the logarithms of any two of these numbers a, b, c, 
 are known, the log. of the third is immediately determined. 
 For, put 
 
 A=log. of a ; B=log. of b ; C==log. of c ; and S=log. of ; 
 
 ac 
 
 then, since 
 
 ac+l _b* ^ 
 ac ac 1 
 
INTRODUCTION. 27 
 
 it follows that S = 2B — A — C ; 
 
 whence either A, B, or C being required, is immediately de- 
 termined by means of the others. 
 
 As the series for computing the log. of 
 ac+l 
 
 converges more rapidly, when a, &c. are large numbers, than 
 when they are small ones, the labour is frequently abridged 
 by computing the log. of a power or multiple of the number 
 whose log. is required. 
 
 Let the log. of 1 1 be required, those above computed being 
 considered as known. 
 
 Here we may take a — 98, b — 99, and c = 100 ; whence 
 1 1 
 
 Z ~2ac+l~19601 ; 
 
 and S = log. of ^±^ = j^. =.0000443135, 
 
 ° ac 19601 
 
 the other terms being rejected, because they do not affect the 
 result short of the fourteenth decimal. Now, the log. of 98 
 is known from those of 49 and 2, and the log. of 100 = 2, the 
 log. of (10) 2 . Consequently, in the equation 
 
 S = 2B — A — C; 
 the terms are all known except B. Therefore, 
 
 B = i (A + C + S). 
 
 A _ c .3010299956 . 
 1 1.6901960797 . 
 
 . log. of 2. 
 . log. of 49. 
 
 C= 2. 
 
 . log. of 100. 
 
 S = .0000443135 
 
 
 2)3.9912703888 
 
 
 1.9956351944 - . 
 
 . log. of 99. 
 
 But .9542425093 . 
 
 . log. of 9. 
 
 wherefore, 1.0413926851 . 
 
 . log. of 11. 
 
28 PLANE TRIGONOMETRY. 
 
 Art. 20. As the radix of the common system is 10, the 
 log. of 10 = 1, the log. of 100 = 2, the log. of 1000 = 3, &c. ; 
 hence it follows that the log.of any number less than 10 con- 
 sists wholly of decimals ; the log. of a number which is more 
 than 10, but less than 100, is more than 1, but less than 2 ; 
 the log. of a number which is more than 100, but less than 
 1000, is more than 2, but less than 3, &c. ; that is, if the 
 number is between 1 and 10, the integral part of the log. is 
 ; if the number is between 10 and 100, the integral part of 
 the log. is 1 ; if the number is 100 or more, but less than 1000, 
 the integral part of the log. is 2. This integral part is usually 
 termed the index of the logarithm. 
 
 As a number, when multiplied or divided by any power 
 of 10, is still indicated by the same significant figures, the 
 position of the decimal point only being changed by the pro- 
 cess ; so the logarithm of a number being increased or .dimi- 
 nished by adding or subtracting the log. of any power of 10, 
 suffers no change except in the index or integral part. 
 Hence we readily perceive that the index of the log. will be 
 0, 1. 2, or 3, according as the first left-hand significant figure 
 of the corresponding natural number denotes units, tens, 
 hundreds, or thousands. 
 
 The log. of 1 being 0, the log. of a proper fraction must be 
 negative ; yet, as a decimal number is equivalent to an inte- 
 gral one divided by some power of 10, the log. of a decimal 
 number differs in nothing but the index from the log. of a 
 whole number which is indicated by the same significant 
 figures. The relation between the logarithmic index and the 
 power of 10 denoted by the left-hand digit of the correspond- 
 ing natural number, may be illustrated by arranging the 
 integral logarithms and their corresponding natural numbers 
 in adjacent lines, as follows : 
 
 3 2 10—1 —2 —3 —4 Log. 
 
 ioooioo io i T V=.iTk=.oi T oU-.oooi T 4oo=-oooirj§! 
 
 Here it is evident that if a natural number falls between 1 
 
INTRODUCTION. 29 
 
 and 10, its log. will fall between 0. and 1., or it will consist 
 wholly of decimals. If the number is between .1 and L, the 
 log. will be between — 1 and ; that is, the index of the log 
 will be — 1., while the decimal part of it will be positive. In 
 like manner, when the natural number lies between .01 and 
 .1, the index of the log. must be — 2, and the decimal part 
 of it a positive quantity. Hence we observe that when the 
 natural number consists wholly of decimals, the logarithmic 
 index will be — 1., — 2., — 3., &c, according as the left-hand 
 significant figure of the natural number denotes tenths, hun- 
 dredths, thousandths, &c. 
 
 In printing tables of logarithms, it is usual to omit the 
 index, leaving it to be supplied in practice upon the prin- 
 ciples above explained. 
 
SECTION I. 
 
 PLANE TEIGONOMETRY. 
 
 The object of Plane Trigonometry is, when of the sides 
 and angles of a plane triangle we have enough given to limit 
 it, to determine the parts which are not given. 
 
 As every oblique angled triangle may be divided into two 
 right angled ones, it is found most expedient to commence 
 the subject by examining the relations and properties of tri- 
 angles of the latter kind. The terms of the science are 
 therefore adapted chiefly to right angled triangles. 
 
 Definitions. 
 
 Article 21. Definition 1. An 
 arc of a circle is any part of the 
 circumference, usually taken less 
 than the whole. As AB, or BHD. 
 
 2. The chord of an arc is a 
 right line drawn from one end of 
 the arc tQ the other. Thus, BE 
 is the chord of the arc BAE, or 
 BDE. 
 
 3. The sine of an arc is a 
 straight line drawn from one ex- 
 
 p tremity of tKe arc, at right angles 
 
 to the diameter, which passes through the other extremity. 
 
 Thus, AD being a diameter to the circle, the line BF, at right 
 
 angles to it, is the sine, or right sine, of the arc AB or DHB. 
 
 4. The tangent of an arc is the right line which touches 
 
 (31) 
 
 
32 PLANE TRIGONOMETRY. 
 
 the circle at one extremity of the arc, and extends till it 
 meets another right line, which is drawn from the centre 
 through the other extremity. Thus AG, which touches the 
 circle at A, is the tangent of AB. 
 
 5. The secant of an arc is the right line intercepted be- 
 tween the centre of the circle and the extremity of the 
 tangent. Thus CG is the secant of the arc AB. 
 
 6. The versed sine of an arc is the part of the diameter 
 intercepted between one end of the arc, and the sine which 
 passes through the other end. Thus AF is the versed sine 
 of AB, and DF is the versed sine of DHB. 
 
 7. The part by which an arc differs, in excess or defect, 
 from a quadrant, or fourth part of the circumference, is called 
 its complement. Thus, the arc ABH being a quadrant, HB 
 is the complement of AB or of DHB. 
 
 8. The cosine, cotangent or cosecant of an arc, is the sine, 
 tangent or secant of the complement of that arc. Thus BI, 
 HK and CK, the sine, tangent and secant of HB, are termed 
 the cosine, cotangent and cosecant of AB. 
 
 9. What an arc lacks of a semicircle, is called its supple- 
 ment. Thus BUD is the supplement of AB. 
 
 10. The circumference of every circle is supposed to be 
 divided into 360 equal parts, called degrees ; each degree 
 into 60 equal parts, called minutes', each minute into 60 
 equal parts, called seconds, &c. Degrees, minutes and 
 seconds are designated thus, °, ', ". 
 
 11. As angles at the centre of a circle have to each other 
 the same ratio as the arcs on which they stand (33.X3) ; the 
 latter are usually termed the measures of the former. Hence 
 an angle at the centre of a circle is said to contain as many 
 degrees, minutes and seconds, as the arc which subtends it. 
 The sine, tangent,* &c, of an arc, is also called the sine, tan- 
 gent, &c. of the angle which is measured by the arc. Thus 
 BF, the sine of AB, is called the sine of the angle ACB. 
 
SECTION I. 
 
 General Properties and Relations of Arcs, Sines, Tan- 
 gents, $$c. 
 
 Art. 22. If the arcs AH and DH are quadrants, and 
 therefore equal, it follows (33.6) that the angles ACH and 
 DCH are equal, and therefore are right angles. Hence the 
 angle at the centre of a circle, subtended by a quadrant, is 
 always a right angle. 
 
 Art. 23. The lines AG and HK, 'which touch the circle 
 at A and H, are respectively at right angles to CA and CH 
 (18.3); hence CAG and KHC are right angled triangles. Now 
 the lines AG and CH, being at right angles to AC, are pa- 
 rallel to each other (28.1) ; consequently, the alternate angles 
 IICK and CGA are equal ; wherefore the triangles CHK and 
 GAC are similar. The triangle CFB is also evidently similar 
 to CAG; and CIB to CHK; therefore those four triangles 
 are similar to each other. Also, the figure CFBI being a 
 parallelogram, CI = FB, and CF = BI. From these trian- 
 gles we have of course the following analogies : 
 
 1. As CF : FB : : CA : AG, or as cosine : sine : : radius : tang. 
 
 2. As CF : CB : : CA : CG, or as cosine : rad. : : radius : sec't. 
 
 3. As CI : CB : : CH : CK, or as sine : radius : : radius : cosec. 
 
 4. As AG : CA : : CH : HK, or as tang. : radius : : radius : cotan. 
 
 5. As CG : AG : : CB : BF, or as secant : tang. : : radius : sine. 
 
 In the algebraic formula? used to express the relations of 
 sines, tangents, &c, it is most convenient to assume the ra- 
 dius = 1. Making, therefore, this assumption, we may con- 
 vert the foregoing analogies into the following equations. 
 
 , sine ■ sine 
 
 1. — : — = tangent ; sine = cosine. tangent ; cosine = . 
 
 cosme, - *=*? & tan 
 
 2. — = secant ; cosine.secant = 1 ; cos= — - = — - ■ 
 
 cosine • sec't Vl + tan 2 
 
 5 
 
34 
 
 PLANE TRIGONOMETRY. 
 
 3. - — = cosecant ; sine.cosecant == 1. 
 sine 
 
 1 
 
 _ 1 cosine 
 
 4. — = cotan ; tan.cotan = 1 ; cotan = - — = — : . 
 
 tan „* tan sine 
 
 Hence, taking P and Q any arcs, tan P. cotan P = tan Q. 
 cotan Q; consequently (16.6), 
 
 tan P : tan Q : : cotan Q : cotan P. 
 
 5. sine == 
 
 tangent tangent 
 
 secant V 1 + tan 2 
 
 Art. 24. It is sometimes necessary to attend to the alge- 
 braic signs of these quantities, particularly when they are 
 reduced to general formulae. 
 
 An arc estimated in one di- 
 rection is considered as posi- 
 tive, and in the opposite direc- 
 tion as negative. The same 
 may be said of the sines, tan- 
 gents, &c. Thus the arc AB, 
 its sine FB, cosine CF, tan- 
 gent AG, and secant CG, are 
 considered as positive ; but, 
 when estimated in the oppo- 
 site direction, they are consi- 
 dered as negative. Now, we 
 readily perceive that when 
 the arc is less than a quadrant, as AB is, the sine, tan- 
 gent, &c, are all positive. But if we take the arc more 
 than a quadrant, but less than a semicircle, as AL, the sine 
 LM is still positive, but the cosine CM is negative, being 
 measured from C in a direction opposite to CF. The tangent 
 AP and secant CP are also negative ; the former being drawn 
 in a direction opposite to AG, and the latter not produced 
 from C through L, the extremity of the arc, but in the oppo- 
 
SECTION I. 35 
 
 site direction. If we take the arc more than a semicircle, 
 but less than three quadrants, as AHDN ; the sine MN be- 
 comes negative, the cosine CM also negative, the tangent AG 
 positive ; but the secant CG, not being produced through N, 
 but in the opposite direction, is negative. If we take the arc 
 more than three quadrants, but less than four, as AHDE ; 
 the sine EF is still negative, but the cosine CF and the 
 secant CP are positive ; the secant being produced from the 
 centre, through the extremity of the arc, till it meets the 
 tangent ; but the tangent AP is negative. 
 
 These signs, when prefixed to the several quantities in the 
 preceding equations, are found to be conformable to the 
 algebraic rules for the adaptation of signs. In the first 
 quadrant, 
 
 + sine 11 1 
 
 ~ .— = + tan; — = + sec;- — — = + cosec; — — = -f cot. 
 
 + cosin + cos -f sine +tan 
 
 In the second quadrant, 
 
 + sine 11 1 
 
 — == — tan; = — sec; - — : — = -f cosec; — - — = — cot. 
 
 — cosin — cos +sine — tan 
 
 In the third quadrant, 
 
 — sine 11 1 
 
 — = + tan; = — sec; — : — == — cosec; — - — =+cot. 
 
 — cosin — cos — sine -ftan 
 
 In the fourth quadrant, 
 
 — sine 11 1 
 
 — tan;— =+sec; : — = — cosec; — — = — cot. 
 
 + cosin -fcos — sine — tan 
 
 Art. 25. It is easily perceived that the sine, tangent, &c, 
 of a given arc are limited, being dependent upon the length of 
 the arc ; but the sine, tangent, &c, of an angle, being the 
 sine, tangent, &c, of the measuring arc, whatever may be 
 the radius with which that arc is described, evidently admit 
 various values. Thus EC, HI, MN, which are the sines of 
 
16 
 
 PLANE TRIGONOMETRY. 
 
 BC, FI, KN, respectively, are also the sines of the angle at 
 A. The lines BL, FO, KP, which are the tangents of the 
 same arcs, are likewise the tangents of the angle at A. 
 
 Art. 26. It appears from cor. to 15.4, that the side of a 
 regular hexagon, inscribed in a circle, is equal to the radius 
 of the circle. But the side of a regular hexagon, inscribed 
 in a circle, subtends an arc of 60°; hence the chord of 60° is 
 equal to the radius of the circle. Again, since a quadrant 
 subtends a right angle at the centre of the circle (Art. 22), 
 it is evident that the sine of a quadrant, or 90°, is the radius 
 of the circle (see Fig. p. 34) ; thus HC the sine of AH, or 
 ACH is the radius of the circle. Further, if we suppose CG 
 to bisect the right angle ACH, we shall have CGA (which = 
 HCG, by 29.1) = ACG ; whence AG = CA ; that is, the tan- 
 gent of 45°, or half a right angle, = the radius. Thus it 
 appears that the chord of 60°, the sine of 90°, and the tangent 
 of 45°, are respectively equal to the radius of the circle. 
 
 Trigonometrical Propositions. 
 
 Art. 27. The sines of two angles adapted to any radius 
 have to each other the same ratio as the sines of the same 
 angles adapted to any other radius. 
 
SECTION I. 
 
 37 
 
 Let BAC and BAD be 
 two angles, whose sines 
 adapted to the radius 
 AC or AB, are EC and 
 FD; while the sines of 
 the same angles adapted 
 to the radius AG or AH, 
 are KH and LI. 
 
 Since the angles at E, 
 F, K and L, are right 
 ones, it is evident that 
 the triangles AEC and AK.H are similar ; as are also AFD 
 and ALL Consequently, 
 
 As AC : CE : : AH : HK ; 
 and alternately, 
 
 AC : AH : : CE : HK. 
 In like manner, 
 
 As AD : AI : : DF : IL ; 
 wherefore, CE : HK : : DF : IL; 
 
 and again alternately, 
 
 CE : DF : : HK : IL. 
 
 Q. E. D. 
 
 Cor. If we substitute the word tangent or secant in place 
 of sine, the proposition will still be true ; and the demonstra- 
 tion will be made out in the same manner by drawing tan- 
 gents to the circles at B and G, and using those tangents, or 
 their secants, instead of the sines. 
 
 Art. 28. In any right angled plane triangle, as the hy- 
 pothenuse is to the perpendicular, so is radius to the sine of 
 the angle at the base ; as the hypothenuse is to the base, so 
 is radius to the cosine of the angle at the base ; and as the 
 base is to the perpendicular, so is radius to the tangent of 
 the angle at the base. 
 
PLANE TRIGONOMETRY. 
 
 Let ABC be a 
 triangle, right an- 
 gled at B ; from A, 
 with the radius 
 AD, describe the 
 arc DE, measur- 
 ing the angle A ; 
 through E and D, 
 draw the sine EH 
 and tangent DF. 
 Then the triangles 
 
 ABC, ADF and AHE, being similar, 
 
 As AC : BC : : AE : HE ; 
 As AC : AB :: AE : AH; 
 AB : BC :: AD : DF; 
 As AC : BC : : radius 
 As AC : AB : : radius 
 As AB : BC : : radius 
 
 and 
 
 that is, 
 
 and 
 
 the sine of A ; 
 the cosine of A ; 
 the tangent of A. 
 
 Q. E. D. 
 
 Art. 29. In any right lined triangle, the sides have to each 
 other the same ratio as the sines of the opposite angles. 
 
 Let ABC be a trian- 
 gle ; make AE = BC ; 
 from the centres B and 
 A, with the radii BC 
 and AE, describe the 
 arcs CG and EH ; from 
 C and E, let fall on AB 
 c (produced if necessary) 
 the perpendiculars CD 
 and EF ; these perpen- 
 diculars are the sines 
 of CG and EH, or of 
 the angles B and A to 
 
SECTION I. 
 
 39 
 
 the radius BC or AE. Now, from the similar triangles 
 ACD,AEF; 
 
 As AC : CD :: AE : EF (4.6); 
 
 and alternately, 
 
 AC : AE : : CD : EF, : : sine of B : sine of A ; 
 
 these sines being suited to anv radius whatever (Art. 27). 
 
 Q. E. D. 
 
 Art. 30. In any right lined triangle, the sum of any two 
 sides is, to their difference, as the tangent of half the sum of 
 the angles, opposite to those sides, to the tangent of half their 
 difference. 
 
 Let ABC be the tri- 
 angle; AC, AB, the 
 sides. From the centre 
 A, with the distance 
 AC, describe the circle 
 DCEF; meeting AB, 
 produced in D and E ; 
 and CB, produced in 
 F ; join AF, DC ; and 
 through E draw EG 
 parallel to BC, meeting 
 DC produced in G. 
 Then it is evident that DB is the sum, and BE the difference, 
 of AC and AB. The outward angle CAD of the triangle 
 ABC, is equal to the two inward and opposite angles, ABC 
 and ACB (32.1). But AEC, at the circumference, is equal 
 to half the angle CAD at the centre (20.3) ; that is, AEC = 
 half the sum of ABC and ACB. Again, AC = AF ; there- 
 fore, AFB = ACB (5.1). But, 
 
 ABC = AFB + BAF (32.1) = ACB + BAF ; 
 
40 
 
 PLANE TRIGONOMETRY. 
 
 consequently, BAF = the difference between ABC and ACB ; 
 and therefore ECF = half that difference (20.3). But EG 
 being parallel to BC, the angle CEG = ECF. Furthermore, 
 the angle DCE in a semicircle being a right one (31.3), ECG 
 is also a right angle. Now, because BC is parallel to EG ; 
 
 DB : BE : : DC : CG (2.6). 
 
 But CD is the tangent of CED, and CG the tangent of CEG, 
 suited to the radius EC ; and these tangents have to each 
 other the same ratio as the tangents of the same angles 
 adapted to any other radius (Art. 27). Hence, 
 
 AC + AB : AC — AB : : tang of } (ABC + ACB) : tang of 
 J (ABC — ACB). 
 
 Q. E. D. 
 
 Art. 31. In any right lined triangle, having two unequal 
 sides ; as the less of those sides is to the greater, so is radius 
 to the tangent of an angle ; and as radius is to the tangent 
 of the excess of that angle above half a right angle, so is the 
 tangent of half the sum of the angles opposite to those sides, 
 to the tangent of half their difference. 
 
 B Let ABC be the trian- 
 
 gle ; AB the less, and AC 
 the greater side. Draw 
 AD at right angles to AC, 
 and equal to AB ; cut off 
 AE, also = AB ; and join 
 DE and DC. Then, DAC 
 being a right angle, 
 
 DA : AC : : rad : tangent 
 of ADC (Art. 28). 
 
 Now, because AE = AD, the angle ADE = AED ; hence 
 each of those angles is half a right angle. Since the triangles 
 
SECTION I. 41 
 
 ADE and ADC have the angle at A common, the angles 
 
 ADC + ACD = ADE + AED = 2ADE. 
 
 Again, since 
 
 ADC = ADE + EDC = AED + EDC ; 
 
 and AED = ACD + EDC (32.1) : 
 
 it follows that ADC — ACD = 2EDC ; 
 
 that is, ADE is half the sum, and EDC half the difference, 
 of ADC and ACD. Hence (Art. 30), 
 
 As tan of ADE : tan of EDC :: AC + AD : AC — AD :: tan 
 { (ABC + ACB) : tan J (ABC — ACB). 
 
 But the tangent of ADE = radius (Art. 26); hence the above 
 analogies are the same as those announced at the beginning 
 of this article. A Q. E. D. 
 
 Art. 32. In any plane triangle, as the base is to the sum 
 of the sides, so is the difference of the sides to twice the dis- 
 tance between the middle of the base and the perpendicular 
 falling upon it from the vertex of the triangle. 
 
 Let ABC be a trian- 
 gle, whose base is AB. 
 From the vertex C, with 
 the greater side AC, de- 
 scribe the circle AEGF, 
 cutting BC produced in 
 E and F, and AB pro- 
 duced in G; join AE, 
 FG; bisect AB in H, 
 and draw CD at right 
 angles to AB. Then, 
 since CD, which passes 
 through the centre of the circle, cuts AG at right angles, 
 
 6 D* 
 
PLANE TRIGONOMETRY. 
 
 that is, AB : AC + BC 
 
 AD = DG (3.3) ; 
 or AG = 2AD ; 
 and AB = 2AH; 
 
 therefore, 
 
 BG fe 2HD. 
 
 Now, the angle BAE — 
 BFG ; and AEB = BGP 
 (21.3); consequently, the 
 triangles ABE, FBG, are 
 similar ; wherefore, 
 AB : BE :: BF : BG; 
 
 AC — BC : 2HD. 
 
 Q. E. D. 
 
 Art. 33. If the half difference of two unequal magnitudes 
 be added to the half sum, the result is the greater magnitude ; 
 but if the half difference be subtracted from the half sum, 
 
 the result is the less magnitude. 
 
 E 
 
 B 
 
 1 Let AB and BC de- 
 
 c note any two unequal 
 magnitudes, whose sum is AC and half the sum AE or EC ; 
 make AD = BC ; then DE = EB, the half difference. Now, 
 AB = AE + EB ; and BC = EC — EB. Q. E. D. 
 
 Art. 34. When the sides of a triangle are given, we have 
 the three following proportions for finding either of the 
 angles. 
 
 Find half the sum of the three sides, and from that half 
 sum subtract the sides severally. Then, 
 
 1. As the rectangle of the half sum, and the excess thereof 
 above the side opposite the proposed angle, is to the rectan- 
 gle of the other two remainders ; so is the square of radius 
 to the square of the tangent of half the angle. 
 
 2. As the rectangle of the sides containing the required 
 angle is to the rectangle of the excesses of the half sum 
 
SECTION I. 
 
 above those sides respectively ; so is the square of radius to 
 the square of the sine of half the angle. 
 
 3. As the rectangle of the sides containing the required 
 angle is to the rectangle of the half sum, and the excess 
 thereof above the side opposite to the proposed angle ; so is 
 the square of radius to the square of the cosine of half the 
 angle. 
 
 Let ABC be the triangle ; produce AB, AC, to II and L. 
 Bisect the angles BAC, ABC and HBC, by the lines AG, BG 
 and BK, respectively; and let AG, BG, meet in G. Now, 
 since the angle CBH is greater than BAC (16.1), it is obvious 
 that HBK is greater than BAG; and, therefore, (13.1,) BAG 
 -f ABK is less than two right angles; consequently (cor. 29.1), 
 BK and AG produced will meet. Let them meet in K; and 
 A draw KH, KM, EL, 
 
 and GD, GF, GE, re- 
 spectively, perpendicu- 
 lar to AB, BC and AC. 
 Then it is obvious (4.4) 
 that DG, FG and EG 
 are equal ; as well as 
 KH, KM and KL. 
 
 Now, in the triangles 
 ADG, AEG, the side 
 AG being common, and 
 the perpendiculars DG, 
 EG, equal, we have 
 (47.1) AD = AE. For a like reason, BD = BF, CE = CF, 
 BH = BM, CL = CM, and AH = AL. As BH = BM, and 
 CM = CL, it follows that 
 
 AH + AL = BC + AB + AC. 
 
 Hence AH or AL is equal to half the sum of the sides. But 
 that half sum = AD + BD + CF=AD + BC=BD + AC. 
 Hence, AH — BC=AD; 
 
44 PLANE TRIGONOMETRY. 
 
 AH — AC = BD; 
 
 AH — AB = BH. 
 
 Now, the angles ABC 
 and CBH, being together 
 equal to two right angles 
 (13.1), DBG + HBK = 
 one right angle = DBG + 
 L BGD. Consequently, the 
 triangles DBG, HBK, are 
 equiangular. Also, the 
 triangles ADG and AHK 
 are equiangular. 
 
 Hence (4.6) 
 
 BD : DG :: HK : HB; 
 wherefore (16.6) BD.HB = HK.DG. 
 Also, AD : DG :: AH : HK; 
 
 consequently (23.6), 
 
 AD 2 : DG 2 : : AH.AD : HK.DG, or BD.HB. 
 But (Art. 28 and 22.6), 
 
 AD 2 : DG 2 : : rad 2 : tan 2 DAG or JBAC. 
 Therefore, 
 
 AH.AD : BD.HB : : rad 2 : tan 2 JBAC. 
 This is the first proportion. 
 
 Again, it has been proved that CF = CE, and GF = GE, 
 CG being common; hence (8.1) the angle GCF = GCE; 
 wherefore, 
 
 GCA+GAC(=CGK)=JBCA+pAC=JHBC(32.1)=HBK. 
 
 Consequently (13.1), AGC — ABK; these angles being the 
 supplements of CGK, HBK. Also the angle GAC = BAK. 
 Therefore, the triangles AGC and ABK are equiangular; 
 whence (4.6), 
 
SECTION I. 45 
 
 AB : AK :: AG : AC .-. AK.AG = AB.AC. 
 Also (4.6), AG : GD : : AK : HK; 
 
 wherefore (23.6), 
 
 AG 2 : GD 2 : : AK.AG : HK.DG : : AB.AC : BD.IIB. 
 But (Art. 28 and 22.6) 
 
 AG 2 : GD 2 : : rad 2 : sin 2 DAG or sin 2 ^BAC. 
 Hence, AB.AC : BD.HB : rad 2 : sin 2 JBAC ; 
 which is proportion 2d. 
 Further : 
 
 AG : AD : : AK : AH .-. (23.6) AG 2 : AD 2 : : AK.AG : 
 AH. AD :: AB.AC : AH. AD. 
 
 But (Art. 28 and 22.6), 
 
 AG 2 : AD 2 : : rad 2 : cos 2 DAG or cos 2 * BAC. 
 
 Consequently, 
 
 AB.AC : AH.AD : : rad 2 : cos 2 JBAC ; 
 
 which is the third proportion. 
 
46 
 
 PLANE TRIGONOMETRY. 
 
 SECTION II. 
 
 The properties of plane triangles, which are explained in 
 the preceding section, are sufficient, with the aid of the usual 
 auxiliary tables, to enable the student to solve all the com- 
 mon cases in Plane Trigonometry. But for the solution of 
 more complex problems, and particularly for the purpose of 
 understanding the manner in which the trigonometrical 
 tables are computed, it is necessary to investigate other 
 theorems. This is most readily effected by the analytical 
 method. In what follows, the radius to which the sines, 
 tangents, &c, are adjusted, is always taken = 1. But it 
 may be observed that whenever it is required to apply the 
 results, here obtained, to the case where the radius is denoted 
 by any other number, nothing more is necessary than to 
 change all the trigonometrical lines in the same ratio in 
 which the radius is changed. 
 
 Article 35. Let 
 AB, AC, AD, be 
 
 three equidifferent 
 arcs, whose common 
 difference is BC or 
 CD. From the cen- 
 tre O, draw OA, 
 OC ; from B, C, D, 
 draw BE, CF, DG, 
 at right angles to 
 OA ; join BD, meet- 
 ing OC inn; through 
 B, n, draw BH, nm, parallel to AO ; and np parallel to CF. 
 Then, since the arc BC = CD, if we suppose BO and DO 
 joined, those arcs will subtend equal angles at O (27.3). 
 
SECTION II. 47 
 
 Hence (4.1) Bn = Dn ; and BnO = DnO; consequently, DnO 
 is a right angle. Hence, BE = sin AB ; CF — sin AC ; DG 
 = sin AD ; Dn = sin CD or BC ; OE = cos AB ; OF = cos 
 AC ; OG == cos AD ; On = cos CD or BC. Since np is parallel 
 to CF, and nm to BH, it is obvious that the triangle Onp is 
 similar to OCF; and Dnm to DBH; and as DB = '2Dn, it 
 follows that BH = 2nm ; and DH = 2Dm. Since DnO = 
 mnp, both being right angles, Dnm = Onp ; and the angles at 
 m and p are right ones ; therefore Dnm, Onp, are similar tri- 
 angles. Of course, the three Dnm, Onp, OCF, are similar. 
 Hence, we have the following proportions : 
 
 As OC : CF : : On : np. 
 As OC : OF : : On : Op. 
 As OC : OF : : Dn : Dm. 
 As OC : CF : : Dn : nm. 
 
 Taking now OC = 1, and substituting for CF, OF, &c, 
 sin AC, cos AC, &c, these proportions furnish the following 
 equations : 
 
 np = sin AC. cos CD. (A) 
 
 Op ps cos AC. cos CD. (B) 
 
 Dm a cos AC. sin CD. (C) 
 
 nm == sin AC. sin CD. (D) 
 
 From equations A and C, 
 
 sin AD(=np + Dm)=sin AC. cos CD + cos AC. sin CD. (1) 
 and 
 
 sin AB(=np — Dm)=sin AC. cos CD — cos AC. sin CD. (2) 
 From equations B and D, 
 
 cos AD(=Op — nm) = cos AC. cos CD — sin AC. sin CD. (3) 
 and 
 cos AB(=Op+w»i)=cos AC. cos CD + sin AC. sin CD. (4) 
 
48 PLANE TRIGONOMETRY. 
 
 By adding equations 1, 2*; 
 
 sin AD + sin AB = 2 sin AC. cos CD. (5) 
 By subtracting, 
 
 sin AD — sin AB * 2 cos AC. sin CD. (6) 
 B j adding equations 3, 4 ; 
 
 cos AB + cos AD = 2 cos AC. cos CD. (7) 
 By subtracting, 
 
 cos AB — cos AD = 2 sin AC. sin CD. (8) 
 
 Art. 36. From 
 these equations, a 
 number of others 
 may be deduced. 
 
 Suppose AC = 
 CD ; and put AC= 
 a ; then AD = 2a, 
 and AB = 0; and 
 Fp G U 
 
 equation 1 becomes sin 2(1=2 sin a. cos a. (1) 
 
 equation 3, cos 2a = cos 2 a — sin 2 a. (2) 
 
 equation 4, cos 0=1 = cos 2 a + sin 2 a, 
 
 which corresponds with 47.1. 
 
 From the equations for sin 2a and cos 2a, it is manifest 
 that sin a = 2 sin \d. cos \a. (3) 
 
 E 
 
 cos a = cos 3 \t 
 
 sin 2 \a. 
 
 and 
 
 But cos 3 \a = 1 — sin 3 \a. .-. cos a = 1 
 
 From the last, 2 sin 3 \a=\ — cos a. 
 
 (4) 
 
 -2sin*ia. (5) 
 
 (6) 
 
SECTION II. 49 
 
 In equation 4, substitute for sin 3 \a its equal 1 — cos 2 \a, 
 and the equation becomes 
 
 cos a = 2 cos 3 %a — 1. (7) 
 
 From this equation, 
 
 2 cos 3 \a = 1 + cos a. (8) 
 
 By Art. 23.1, 
 
 sin \a 2 sin \a. cos \a . J L. 
 
 tan * a = ^ = — ai^T 1 =< e * 3 ' 8 > 
 
 (9) 
 
 1 -f cos a " 
 By Art. 23.4, 
 
 cos \a 2cosAa. sinia . _ ' sin a ,_-. 
 
 cotan 4a= -^-f- = o^Vi ' = ( e q- 3 > 6 )l • ( 10 ) 
 
 J sin fa 2sm~$a vn 7 1 — cos a x ' 
 
 sin 3 ia 2sin 2 \a ' „ _1 — cos a .,"- 
 
 tan " a ^s?fe = S5s?fcr< , » 6 ' 8 >tt^w < n > 
 
 cotanHa= ^.^L = |£2!;i_« =(eq . 6) 8)i±- c ^. ( i2) 
 
 2 sin 3 ia 2 sin 2 £a v n ' 1 — cos a v ' 
 
 Art. 37. Take now AC = a, CD = b ; whence 
 
 AD = a + & ; 
 and AB = a — b ; 
 
 and equations 1 and 2, Art. 35, become 
 
 sin (a ± 5) = sin a. cos b =fc cos a. sin 6. (1) 
 3 and 4 become, 
 
 cos (a =fc i) = cos a. cos Z> qp sin a. sin 5. (2) 
 Now, 
 
 . , .. sin(a ± Z/) sin a. cos b ± cos a. sin b ■ 
 
 tan (a =fc &) = — 7--^ = , 1 ■*- = (di 
 
 v ' cos(a± 0) cos a. cos zp sin a. sin v 
 
50 PLANE TRIGONOMETRY. 
 
 viding numerator and denominator by cos a. cos b, and using 
 
 . sin k tan a =b tan b 
 
 tan for ) ^ — -. (3) 
 
 cos ' 1 qp tan a. tan o x ' 
 
 Equations 5, 6, 7, 8, also become 
 
 sin (a + b) + sin (a — b) = 2 sin a. cos 6. (4) 
 
 sin (a + ft) — sin (a — 5) = 2 cos a. sin &. (5) 
 
 cos (a — b) + cos (a + b) = 2 cos a. cos b. (6) 
 
 cos (a — b) — cos (a + b) = 2 sin a. sin 5. (7) 
 
 By Art. 23.1, 
 
 sin a sin b sin a. cos 5 ± cos a. sin & 
 
 tan a ± tan & = ± , = t = 
 
 cos a cos o cos a. cos 6. 
 
 (eq .i)!l^±l) (8) 
 x n 'cos a. cos 6 x ' 
 
 By Art. 23.4, 
 
 cos b cos a sin a. cos b ± cos a. sin # 
 
 COt ft ± COt a =— r ± n = : : 7 ' = 
 
 sin b sin a sin a. sin b 
 
 sin a. sin b w 
 
 By changing our notation, other equations may be de- 
 duced. 
 
 Let AD = a ; AB = b ; then AC = $($ + &), and DC or 
 BC = ^(a — b). With this notation, equations 5, 6, 7, 8, Art. 
 35, become 
 
 sin a + sin b = 2 sin J(<z + 5). cos ^(a — b). (10) 
 
 sin a — sin b = 2 cos J(a + £). sin J(« — b). (11) 
 
 cos 6 + cos a = 2 cos ^(a + 6). cos J(a — &). (12) 
 
 cos b — cos a = 2 sin £(a + 5). sin |(a — &). (13) 
 
SECTION IL 51 
 
 Also, 
 
 U jl h\ - sin -a( a + & ) _ 2 sin *(a + 6). cos £( a — h ) _. 
 tan 3 (a + 6) - cog |^j + ft j - 2 cos " i (a + ^ cos |p _ ^ 
 
 / in io\ sin a + sin 6 
 
 (eq. 10, 12) T . (14) 
 
 v _■ ' cos a -f cos 6 
 
 sin |(« — ft) _ 2 sin *(a — ft), cos *(a + ft)_ 
 ~V ' ~~ cos J(fl — ft) — 2 cos i(a — b). cos l(a + b)~ 
 
 / n mx sin a — sin ft 
 
 (eq. 11, 12) ; r. (15) 
 
 v T ' cos a + cos b 
 
 cos i(<2 + ft) 2 cos Ua + ft), sin {{a — b) 
 
 cotan Ua + b) = — — 77- ' , , x = « — : — 77 — — tt — 5 — 77 r\ 
 
 ZK ' sin \{a + b) 2 sin |(a + b). sin £(« — b) 
 
 sin a— sin ft 
 
 = (eq. 11, 13) 7- . (16) 
 
 v ^ ' cos — cos a N ' 
 
 _ cos J (a — ft) 2 cos |(a — ft), sin *( a + ft) 
 cotan j(n — ft) =-2HT( fl -.fc) ~ 2 s i n ^(a — ft), sin J(a + 6) 
 
 ,^ 4«» sin a + sin ft ; 4 __ 
 
 = (eq. 10, 13) - — -t . (17) 
 
 v n ' cos — cos a x ' 
 
 tan Ua — ft) . , A , rx sin a — sin ft 
 
 f) — r r( = (eq. 14, 15)- .j. . , , (18) 
 
 tan £(a + ft) v ^ 'sin a + sin ft x ' 
 
 tan|(« — &) , 1K 1m cosft — cosa 
 
 From equations 15 and 16, 
 
 cot \{a -f ft) cos a + cos ft 
 tan J(a — b) — cos ft — cos a' 
 
 (20) 
 
 The following is an analytical investigation of the rules, 
 already given in Art. 34, for finding an angle of a plane tri- 
 angle., when the sides are known. 
 
 X 
 
52 PLANE TRIGONOMETRY. 
 
 Let ABC be the triangle ; and put the angle ABC = B ; 
 the side AB = c; AC = b; BC = a; BD = d; the line CD 
 being at right angles to AB. Then (12, 13.2), 
 
 c 2 + a 2 = & 2 ±2a7; 
 
 the sign + being used when B is acute, as in Fig. 1 ; and 
 the sign — when B is obtuse, as in Pig. 2. 
 
 Fig. 2. 
 
 Hence, 
 
 db<* = 
 
 c 3 + a' — fr 
 
 2c 
 
 But (Art. 28) 
 
 — = cos B ; 
 a 
 
 the cos B being positive or negative, according as the angle 
 is acute or obtuse (Art. 24). Consequently, 
 
 cos B = 
 
 c 9 + a 2 — b» 
 2ac 
 
 and 
 
 1 + cos B = 1 + 
 
 4 (cor. to 5.2) 
 
 c s-f a 3_&* __ c a +2ac+a s — ft* _(c+a) 3 -& 3 
 2ac 2ac 2ac 
 
 (c + a + b) (c + a— b) 
 2ac 
 
SECTION II. 53 
 
 Now, Art. 36, Form. 8, 
 
 1 -f cos B = 2 cos 3 JB ; 
 wherefore, 
 
 (c + a + b) (c + a—b) |(c>6>fe). {(c + a— b) 
 
 cos 3 AB = 
 
 4ac ac 
 
 c + a + b\ s.(s — b) 
 
 = (puUing , = — 2— )— ; ■— ; (A) 
 which is Rule 3, Art. 34. 
 
 c 3 + a* —b* , 
 
 Again, from the equation cos B = ~ » we nave 
 
 i ; c» + a s — 6 3 2ac+6 2 — c 2 — a s 
 1 — cos B = 1 - 
 
 2ac 2ac 
 
 b*—(c s —2 ac + a z ) _ b»—(c—a) a 
 Qac 2ac 
 
 But, Art. 36, Form. 6, 
 
 1 — cos B = 2 sin a JB ; 
 
 consequently, 
 
 6 3 — (c— a)* l(b + c—a). {(b + a—c) 
 
 sin 3 ^B = 
 
 iac ac 
 
 (B) 
 
 ac 
 which is Rule 2, Art. 34. 
 
 Since tan = -^ (Art. 23) ; tan 3 B = — -^ = (by eq. 
 cosine v ' cos 8 B v J ' 
 
 , -r^ ( s — a) .(s — c) 
 A and B) J / * ■ . x - ' ; 
 
 ' s . (s — b) 
 
 which is Rule 1, Art. 34. 
 
 E* 
 
PLANE TRIGONOMETRY. 
 
 Art. 38. Let ABC 
 
 be a semicircle ; ADC 
 its diameter; D the 
 centre ; AB the chord 
 of an arc; AE, BG, 
 lines touching the cir- 
 Iji ~^c cle in A and B, and 
 
 meeting in G. Join DG, DB, CB ; and produce DB, CB, to 
 
 meet AE in E and F. 
 
 Now, the angle ABC in a semicircle being a right one 
 (31.3), the adjacent angle ABF is also a right one (13.1). 
 Again, since AG and BG touch the circle, each of the angles 
 GAB, GBA == the angle ACB in the alternate segment (32.3) ; 
 hence GAB = GBA, and GB == GA (6.1). Furthermore, 
 since AFB + FAB = ABC (32.1) = ABF (31.3 and 13.1) =? 
 GBF + GBA ; it follows that GFB = GBF ; whence GB = 
 GF ; and AG + GB = AF. But the triangles ADG, BDG, 
 being evidently equal, the line AG is the tangent of half 
 the arc intercepted between A and B ; hence AF =s twice 
 the tangent of that half arc. The chord AB is also twice 
 the sine of the same half arc. Now, the triangles ACB, 
 FAB, being right angled at B, and having the angle ACB = 
 FAB, are similar ; whence AC : CB : : AF : AB : : AG : 
 JAB : : tan of ^ arc : sine of the same half arc. 
 
 Let, now, the point B move along the arc towards A ; the 
 lines which pass through B moving with it : then, as the 
 angle at C diminishes, the line CB must approximate to AC, 
 and ultimately become equal to it. Consequently, the ratio 
 of AF to AB, or of the tangent to the sine of half the arc 
 between A and B, is ultimately a ratio of equality. 
 
 Again. The angle ADG = JADB = ACB (21.3) ; con- 
 "sequently, DG is parallel to CF (28.1) ; and therefore, 
 
 EB : BD : : EF : FG (2.6) ; 
 
SECTION II. 
 
 or, doubling the consequents, 
 
 EB : AC : : EF : FA ; 
 whence, by composition (18.5), 
 
 EB + AC : AC : : AE : AF. 
 But AC : CB : : AF : AB (4.6) ; 
 
 therefore, ex equali, 
 
 AC + BE : CB : : AE : AB (22.5). 
 
 But as the point B approaches A, the line BE decreases 
 and ultimately vanishes. The angle at C ultimately vanish- 
 ing, 4he line CB becomes finally equal to AC. Hence the 
 ratio of AC + BE to CB, and consequently of the tangent 
 AE to the chord AB, becomes ultimately a ratio of equality. 
 
 Hence it is manifest that the ultimate ratio of the tangent 
 of an evanescent arc to its sine, or to its chord, is a ratio of 
 equality. 
 
 B Let AEB be a circular arc, 
 
 whose centre is C ;• AD, BD, two 
 right lines touching the circle in 
 A and B. Join CD, AB; and let 
 CD cut the arc in E, and the 
 chord AB in F. Through E, 
 draw GH parallel to AB ; and 
 join AE, BE. Then, from what 
 is above proved, AD = BD ; the 
 c angle ACF = BCF ; and conse- 
 quently (4.1), AFC = BFC; whence GIT, being parallel to 
 AB, and therefore at right angles to CE, touches the circle 
 in E. Also, AD is the tangent and AF the sine of the 
 arc AE. 
 
 Since AFE is a right angle, the angles AEF and ADF are 
 each less than a right angle (17.1). But AED = AFE + 
 EAF (32.1) is greater than a right angle. Hence, AD is 
 greater than AE, and AE than AF (19.1). That is, the tan- 
 
56 
 
 PLANE TRIGONOMETRY. 
 
 gent is greater than the chord, and the chord than the sine. 
 Again, since GD + DH are greater than GH (20.1), it is 
 obvious that AD + DB must be greater than AG + GH + 
 HB. But AG + GE being greater than AE (20.1), and EH 
 + HB than EB ; AG + GH -f HB must be greater than AE 
 + EB. Also AE + EB are greater than AB. 
 
 If we were to join CG, CH, and draw the tangents and 
 chords to the intercepted arcs, we might demonstrate, in the 
 same manner, that the sum of the tangents thus drawn would 
 be less than AG + GH + HB, and the sum of the chords 
 greater than AE + EB. By continued bisections, we thus 
 find the sum of the tangents continually decreasing, and the 
 sum of the chords always increasing. But the tangents and 
 chords are, by this process, brought to approximate still 
 more and more nearly to the circular arc which lies between 
 them. Hence we infer that when, by the evanescence of the 
 arc, the ratio of the tangent to the chord or sine becomes a 
 ratio of equality, the ratio of the arc itself to the tangent, 
 chord or sine, is a ratio of equality. 
 
 D 
 
 E Fp 
 
 tial Calculus. 
 Let AB = 
 OC being = 1. 
 
 Art. 39. As near- 
 ly all trigonometri- 
 cal calculations are 
 usually performed 
 by means of auxili- 
 ary tables, it be- 
 comes necessary to 
 explain the nature 
 and origin of those 
 tables. This is most 
 expeditiously effect- 
 ed by the Differen- 
 
 BE = y; OE = x; BCD = h; the radius 
 Then, as proved in Art. 35, equation C, 
 Dm = cos (z + jjk). sin \h ; 
 whence, sin (z + h) = sin % + 2 cos (z + ^).sin \h ; 
 
SECTION II. 57 
 
 consequently, 
 
 sin (z + h) — sin z _ 2 cos (z + *A).sin J A cos(z-f-|A).sin|Zt 
 A h \h 
 
 But (Art. 38) when h becomes evanescent, the ultimate ratio 
 of }Ji to the sin \h is a ratio of equality. Also the ultimate 
 ratio of cos (z + \K) to cos z is a ratio of equality. Hence, 
 
 e/.sinz /dy\ 
 
 —fa =te-) = cosz = *; 
 
 consequently, dy = xdz. 
 
 Again, x 2 -f y z = 1 ; whence, 2;«£c -f 2*/c?!/ = 0; or 
 
 — ydy — yxdz 
 
 dx = —2-2- = — - =—ydz. 
 
 x x J 
 
 Now, let t = tan z =— ; then 
 
 x 
 
 y xdy — ydx ' x l dz + y*d% 
 
 x x 4 
 
 „^_£~ m S=-«f ^ (since f + fim 1) 
 
 Z = (] + *2) <fe, because - 2 = 1 + t 2 (Art. 23, 2.) 
 
 x- 
 
 Art. 40. To find the length of an arc z in terms of its tan- 
 gent, t. No formula has been discovered for expressing an 
 arc in finite terms of its tangent ; recourse must therefore be 
 had to infinite series. 
 
 Let, then, z = At n + B* p + Cf + Df, &c, in which the 
 exponents and coefficients are unknow r n. As the tangent of 
 an arc becomes nothing at the same time the arc itself 
 vanishes, the series here assumed must express the arc z, if it 
 can be expressed at all in terms of t. 
 
 From this equation, 
 
 ■^=n At*- 1 + pBt^" 1 + qCt*- 1 + rBv- 1 + &c. 
 8 
 
58 PLANE TRIGONOMETRY. 
 
 But (Art. 39), 
 
 dz 
 Comparing these values of 77; n — 1=0, p — 1 = 2, q — 1 
 
 = 4, r — 1 = 6, &c. ; and nA. = 1 ; pB = — 1 ; qC = 1 ; ?*D 
 = — 1, &c. ; from which n = 1; p = 3; q = 5, &c. ; A = 1 ; 
 B = — ^ ; C = I; D = — \, &c. ; which values, substituted 
 in the primitive equation, give 
 
 z = t — jf + j** — *f + jf + &c. 
 
 The length of the arc being thus found in terms of the 
 tangent, the next object is to find a value of t which will 
 converge rapidly in this series, and at the same time corre- 
 spond to a known part of the circumference of a circle. 
 One of the most convenient methods yet discovered of effect- 
 ing this object, is the following : 
 
 Let a and b denote two circular arcs ; such that tan a=\, 
 and tan b = J; then (Art. 37, Form. 3), 
 
 
 
 tan {a + b) = 
 
 
 = 1. 
 
 
 
 ow (Art. 26) 1 
 
 the tangent of 45 
 
 ' = i. 
 
 Hence, a + b = 
 
 = 45°. 
 
 Now a - 
 
 1 
 = 2 
 
 11 11 
 
 1 1 
 
 &c.= 
 
 :. 463647609001 
 
 and b = -£ — i*-^ + y. 35— yyp &c.=.321750554397 
 
 a + b = (45°) = 785398163398 
 
 From which we find the semicircumference==3.141592653592; 
 and this is the whole circumference very nearly true to twelve 
 decimals, when the diameter is = 1. 
 
 Art. 41. The length of an arc of 45° being thus deter- 
 mined, the length of an arc of 1°, or of any fraction of it, is 
 
SECTION II. 59 
 
 readily found. But, from this datum, to find a general for- 
 mula for the sine and cosine, requires the aid of the Differen- 
 tial Calculus. And no method has been discovered to denote 
 the sine or cosine in terms of the arc, without recourse to 
 infinite series. 
 
 Let z = an arc ; y = sin z ; x = cos z ; and assume 
 
 x = 1 + Az n + Bz p + Cz q + Dz r + &c. ; 
 
 in which the first term of the series is taken = 1 ; because, 
 when z = 0, x = 1. From this equation, 
 
 dx 
 
 -^ = nAz n ~ l + pBz»- 1 + qCz*- 1 + rT>z r ~ l -f &c. 
 
 But (Art. 39), 
 
 dx 
 
 Therefore, 
 
 y = — nAz"" 1 — pBzP" 1 — qCz*- 1 — rDz'" 1 — &c. ; 
 
 and 
 
 du 
 
 f z = — ( n — 1).» Az"- 2 — (p — l).pBzP- 2 — (7 — l).?Cz<-» 
 
 — (r— l).rDz r ~ 2 — &c. 
 
 But (Art. 39), 
 
 -| = x = 1 + Az n + Bz p + Cz q + &c. 
 
 Comparing the corresponding terms of these series, 
 
 n — 2 = 0; p — 2=ti; q — 2 = p; r— -2 = q,&e. 
 (to — l).nA = — 1; (/> — l).pB = — A ; &c. 
 
 Whence n = 2; p = 4; ? = 6; r = 8; &c. 
 
 A- _Lr_J_ p_ 1 n_ X * 
 
 A ~ 2' *~ 2.3.4' U 2^Z^6 ; U_ 2.3.4.5.6.7.8 &c * 
 
 Consequently, 
 
60 PLANE TRIGONOMETRY. 
 
 x = l ~ 1 + 2^4-3X47^6 + &C - (A) 
 
 * = Z ~is + 2^475-2.3.416.7 + &C> (B) 
 
 To exemplify these series, let the sine and cosine of 1° be 
 required. In that case, 
 
 *=.017453292520;^= .000152308710 ; ^=.000000886080; 
 m .000000003866; *&-== .000000000013. 
 
 2.3.4 — ^ vyw ' 2.3.4.5 
 
 Substitute these values in series A and B. Whence x = 
 .999847695156; and y = .017452406453. 
 
 z* z 3 
 If, instead of these values of z,-^, s~q> & c -> we substitute 
 
 in the series A and B, \ of the first, \ of the second, J of the 
 third, &c, we shall obtain the cosine and sine of 30' ; and 
 from these results we may, by a similar process, find the 
 cosine and sine of 15'. 
 
 Thus, cos of 
 
 30'= 1— .000038077177 + .000000000242= .999961923095 ; 
 and sin of 
 
 30'=.00872664626O— .000000 11 0760=. 008726535500; 
 Also, cos of 
 
 15'= 1— .000009519299 + .000000000015= .999990480728 ; 
 and sin of 
 
 15'=. 004363323130— .000000013845=. 004363309285. 
 
 z 2 
 If, instead of the values of z, -~-, &c, first found, we take 
 
 e *3 of the first, gg 1 ^ of the second, &c, and substitute them 
 in equations A and B, we shall have the cosine and sine 
 
 of r. 
 
SECTION II. 01 
 
 Thus, cos of r=l.— .000000042308 =.999999957692; 
 and sin of 
 
 1'= .000290888209— .000000000004= .000290888205. 
 
 When the sines and cosines of two arcs are known, the 
 sine and cosine of their sum, or difference, are readily found 
 from equations 1 and 2, Art. 37. Or the sine and cosine of 
 any arc, v, being found, the sine and cosine of 2u, 3v, 4v, may 
 be determined in the following manner : 
 
 From Art. 37, Form. 4 and 7, we have, by transposition, 
 
 sin (a + b) = 2 sin a. cos b — sin (a — b) (C) 
 
 cos (a -f b) = cos (a — b) — 2 sin a. sin b (D) 
 
 Taking, then, b = any arc v ; and a successively = v 3 2v, 
 3v, &c. ; these equations become 
 
 sin 2v = 2 sin v. cos v ; - V C*-> 
 
 cos 2v = 1 — 2 sin 2 v ; 
 
 sin 3u = 2 sin 2v. cos v — sin v ; 
 
 cos 3v = cos v — 2 sin 2p. sin i; ; 
 
 sin 4u = 2 sin 3d. cos u — sin 2u ; 
 
 cos 4u == cos 2v — 2 sin 3v. sin u. 
 
 Hence, the sine and cosine of v being known, the sine and 
 cosine of any multiple of v may be found. 
 
 Art. 42. The sine of an arc, being half the chord of twice 
 the arc, and the chord of 60° = 1, (Art. 26,) the sine of 30° 
 = I ; consequently, if we take a = 30°, in the equations C 
 and D of the last article, we shall hnve 
 
 sin (30° + b) = cos b — sin (30° — b) ; 
 
 cos (30° + b) = cos (30°— b) — sin b. 
 
 If, then, the sine and cosine of every degree and minute, 
 as far as 30°, were computed by the preceding methods, 
 
 F 
 
 t 
 
 lA^s 
 
62 PLANE TRIGONOMETRY. 
 
 these equations furnisn a mode of computing the sines and 
 cosines of the remaining arcs by subtraction only. 
 
 To find the sine and cosine of 31°, 
 
 cos 29° = cos 30°. cos 1°+ sin 30°. sin 1° = .874619707108 
 
 sin 1° = .017452406453 
 
 cos 31°= .857167300655 
 
 cos 1°= .999847695156 
 sin 29° = sin 30°. cos 1° — cos 30°. sin 1° = .484809620238 
 
 cos. 31° = .515038074918 
 
 The sines, computed as above explained, and arranged in 
 a table, constitute a table of natural sines. Those sines, as 
 put down in the tables, are seldom extended to more than 
 seven decimals, and frequently not even so far ; but, in com- 
 puting such tables, it is necessary to extend the sines of the 
 primary arcs considerably further than the number of deci- 
 mals intended to be retained, in order to render the numerous 
 deductions from them sufficiently correct. 
 
 The tangents may be found from the sines and cosines, by 
 
 sine 
 
 simple division; for tan = — (Art. 23). The secants are 
 
 1 cosine v ' 
 
 also deduced from the cosines ; for secant = 
 
 cosine 
 
 Art. 43. The tables of sines, tangents, &c, which are 
 commonly used in trigonometrical calculations, are loga- 
 rithmic, and are easily deduced from a table of logarithms 
 and of natural sines. But it may be observed that the sine? 
 computed to a radius 1, are all decimals except the sine of 
 90°. Hence, the logarithms of those sines, if the common 
 logarithms are used, must all have negative indices, excep 
 the sine 90°, whe.se logarithm is 0. To avoid this inconve 
 nience the decimal point in each of the sines is removed tec 
 
SECTION II. 63 
 
 places towards the right, which is equivalent to finding the 
 sines to a radius of 10000000000. The logarithm of this 
 number is 10; and the sine of 1" computed to this radius is 
 48481.37, whose log. =4.6855749. From which it appears 
 that all arcs or angles which can occur in practice, have 
 their logarithmic sines positive. 
 
 The sines computed according to the preceding articles 
 have the decimal point, in each case, removed ten places to 
 the right ; the logarithms of the results are then taken from 
 a table of logarithms, and arranged in a table. This com- 
 poses a table of logarithmic or artificial sines. Then, since 
 
 cos : sine : ; rad : tang (Art. 23) ; 
 
 the index of the logarithmic sine being increased by 10, and 
 the logarithmic cosine subtracted, the remainder will be the 
 logarithmic tangent. And, since 
 
 cos : rad : : rad : secant (Art. 23) ; 
 
 if we subtract the logarithmic cosine from 20 (twice the log. 
 of radius), the remainder will be the secant. Again, 
 
 tang : rad :: rad : cotan; 
 
 consequently, the logarithmic tangent of an arc, being sub- 
 tracted from 20, will leave the logarithmic cotangent. 
 
 In trigonometrical calculations where logarithms are used, 
 it is most convenient to take the arithmetical complements 
 of the logarithms which are to be subtracted, (that is, what 
 those logarithms want of 10 or 20,) and add them with the 
 other additive logarithms, rejecting as many tens or twen- 
 ties from the result as there are complements used. When 
 the subtractive numbers are logarithmic sines, tangents or 
 secants, the arithmetical complements can be taken imme- 
 diately from the table ; for the cosecant is the arithmetical 
 complement of the sine ; the cotangent of the tangent ; and 
 the cosine of the secant. All this is manifest from the na- 
 are of logarithms, and the analogies in Art. 23. 
 
64 
 
 PLANE TRIGONOMETRY. 
 
 Art. 44. A few trigonometrical problems will now be 
 given to exercise the preceding rules. 
 
 1. Given, AB 35, AC 30, 
 BC 25, the three sides of a 
 triangle ; to find ihe distances 
 from the several angles to a 
 point E within the triangle, 
 such that the angles AEB, 
 AEC and BEC shall be equal 
 to each other. 
 
 Construction. On AB, one 
 of the sides, describe the 
 equilateral triangle ABD ; and 
 about that triangle describe a 
 circle; join DC, cutting the 
 circle in E ; then is E the point required. 
 
 Join AE, BE ; then, since the angles of the triangle ABD 
 are all equal (cor. to 5.1), each of them contains 60°, or ^ of 
 two.right angles (32.1). But AED= ABD ; and BED=BAB 
 (21.3) : therefore AED = 60°, and AEC = 120°. Also BE Y J 
 = 60°, and BEC = 120°. 
 
 Calculation. With the three sides, the angle BAC may Vf 
 found by Art. 34, Rule 2. 
 
 AB 35 AC 8.4559320 
 AC 30 AC 8.5228787 
 BC 25 
 
 sum 
 
 90 
 
 J sum 45 
 J sum— AB 10 log. 1. 
 i sum — AC 15 " 1.1760913 
 
 2)19.1549020 
 
 sinpAC 22°12J' 9.5774510 
 
SECTION II. 65 
 
 BAG 44° 25' 
 BAD 60° 
 
 DAC 104° 25 
 
 £DAC 52°12J' 
 ACD + ADC"^ 4 , 
 
 36 
 
 Then, by Art. 30, 
 
 As 
 is to 
 
 So is tan 
 to tan 
 
 AD + AC 
 
 65 AC 84870866 
 
 AD — AC 
 
 5 .6989700 
 
 ACD + ADC 
 
 2 
 
 37° 47£' 9.8895519 
 
 ACD — ADC 
 
 2 
 
 3° 24J' 8.7756085 
 
 ACD 
 
 41° 12' 
 
 ADC) 
 ABES 
 
 34° 23' 
 
 CAE 
 
 18° 48= AED— -ACE. 
 
 EAB 
 
 25° 37= CAB — CAE. 
 
 From Art. 29, 
 
 As sinAEB 120° AC .0624694 
 
 is to sin ABE 34° 23' 9.7518385 
 
 So is AB 35 1.5440680 
 
 to AE 22.823 1.3583759 
 
 As sinAEB 120° AC .0624694 
 
 istosinBAE 25° 37' 9.6358335 
 
 So is AB 35 1.5440680 
 
 to BE 17.473 1.2423709 
 
 9 F* 
 
G8 PLANE TRIGONOMETRY. 
 
 As sin AEC 120° AC .0624694 
 is to sin CAE 18° 48' 9.5082141 
 
 So is AC 30 1.4771213 
 
 to 
 
 CE 11.164 
 
 1.0478048 
 
 2. Given, the vertical angle 
 ACB 70°, the segments into 
 which the base is divided by 
 the line CD bisecting the ver- 
 tical angle, viz. AD 30, and DB 
 20, to determine the angles 
 and sides of the triangle, and 
 the line which bisects the ver- 
 tical angle. 
 
 Construction. Bisect the base AB in E ; through E draw 
 FEG at right angles to AB ; make the angle EAF = com- 
 plement of the vertical angle, above AB when that angle is 
 acute, but below when it is obtuse. From the centre F, 
 where the line AF meets the perpendicular, describe a circle 
 passing through A, and cutting FEG in G; join GD, and 
 produce it to cut the circle in C ; join CA, CB ; and ACB 
 will be the triangle required. 
 
 Since AE = BE, and the angles at E are right ones, the 
 line AF = BF (4.1) ; consequently, the circle must pass 
 through B. The angle AFE being equal to BFE, the arc 
 AG = BG (26.3) ; consequently, ACG = BCG (27.3). Also, 
 the angle AFE = twice ACG (20.3) = ACB ; therefore, ACB 
 is the complement of EAF. 
 
 Calculation. In the right angled triangle EAF we have, 
 besides the right angle, the side AE = 25, and the angle 
 EAF = 20° ; from which we find, by Art. 28, AF = 26.604, 
 
SECTION II. 
 
 67 
 
 and EF = 9.099 ; whence EG = 17.505 ; then, in the right 
 angled triangle GED, we have EG ; and ED = 5 ; from 
 which we find the angle EDG or CDB = 74° 3' ; then 
 
 CAD = CDB — ACD (32.1) = 39° 3'. 
 
 In the triangle ADC we then have AD = 30, and all the 
 angles, from which, by Art. 29, we find AC = 50.29, and 
 DC = 32.951. Lastly, from 3.6, we have 
 
 AD : DB : \ AC : BC = 33.527. 
 
 3. Given, the base AB 70; 
 the vertical angle ACB 75°; 
 and the ratio of the sides, viz., 
 AC : BC : : 4 : 3, to deter- 
 mine the rest. 
 
 Divide AB in D, so that 
 AD : DB : : 4:3 (10.6) ; then 
 the construction will be the 
 same as in the last example. 
 
 The calculation will like- 
 G wise be similar to the last. 
 
 The results are, AF = 36,235 ; EF = 9.378 ; BDC = 79° 
 27'; CAB = 41° 57'; ABC = 63° 3' ; AC = 64.596; BC = 
 48.447; DC = 43.927. 
 
 4. Given, the base AB 500 ; the difference of the sides 100 ; 
 and the vertical angle ACB 72°, to determine the rest. 
 
 O 
 
 Construction. On the base AB 
 describe the segment of a circle 
 containing an angle equal 90° + ^ 
 the vertical angle (33.3) ; place in 
 this circle the right line AD = 
 the difference of the sides ; pro- 
 
68 
 
 FLANE TRIGONOMETRY. 
 
 duce AD ; join DB ; and make the angle DBC = BDC ; then 
 is ABC the triangle proposed. 
 
 Draw CE at right angles to DB ; then, CD being = CB 
 (because the angle CBD = CDE), the angle DCB is evidently 
 bisected ; and the angle 
 
 ADB = DEC + DCE = 90° + JACB ; 
 also, AD = AC — BC : hence the construction is manifest. 
 
 Calculation. In the triangle ABD, AB, AD, and the an- 
 gle ADB, are known; whence the angle BAD may be 
 found ; from which and the given angle ACB, the angle ABC 
 becomes known. Then AB and all the angles of the triangle 
 being known, AC and BC are determined. The results are 
 BAC = 44° 41', ABC = 63° 19', AC = 469.74; and BC 
 = 369.74. 
 
 5. Given, the base AB 465, 
 the vertical angle ACB 75° ; 
 and the sum of the sides 760, 
 to determine the rest. 
 
 Construction. On AB de- 
 scribe a segment of a circle 
 containing half the vertical 
 angle ; from A, place the line 
 AD = the sum of the sides, 
 in this segment ; join DB ; 
 
 and make the angle DBC = BDC. Then will ACB be the 
 
 triangle proposed. 
 
 Because the angle DBC m BDC ; DC = BC ; and the exte 
 rior angle ACB = twice ADB ; also AC + BC = AD. 
 
 Calculation. With the sides AB, AD, and angle ADB ; the 
 angles ABD and BAD may be found ; whence ABC becomes 
 known. Then AC and BC are determined. Results : ABC 
 
SECTION II. 69 
 
 = 46° 45', or 58° 15' ; AC = 350.64, or 409.36 ; BC = 409.36 
 or 350.64 
 
 6. Given, the base AB 50 ; the line DC, drawn from the 
 middle of the base to the vertex 40 ; and the ratio of the 
 sides, AC : BC : : 3 : % to determine the sides 
 
 Construction. Divide AB in E, so that AE : EB in the pro- 
 posed ratio of AC : BC ; produce AB to F, so that BF shall 
 be a third proportional to AE — - EB and EB ; from the cen- 
 tre F, through E, describe the arc EC ; from the centre D, 
 with the given distance DC, describe an arc, cutting the 
 former in C ; join AC, DC and BC ; then ABC is the triangle 
 proposed. 
 
 From the proportion AE — EB : EB : : EB : BF, we 
 
 have (18.5), AE : EB : : EF : BF; 
 
 therefore (12.5) AE : EB : : AF : EF ; 
 
 consequently (19.5), 
 
 AF : EF :: EF : BF; 
 
 whence (F. 6), AC : CB : : AE : EB. 
 
 Calculation. In the triangle CDF, we have all the sides to 
 find ^ the angle FDC, which is the -J sum of DAC and DCA ; 
 then, with that J sum and the sides AD, DC, the angle DAC 
 and side AC may be found. 
 
 Result: AC = 55.50: BC = 37.00. 
 
70 
 
 PLANE TRIGONOMETRY. 
 
 7. Given* the sides of the 
 triangle ABC, viz., AB 90, AC 
 80, BC 70, to determine the dis- 
 tances AD, CD and BD, to a 
 point D, which is so situated 
 that the angles ADB and ADC 
 are 70° and 40° respectively. 
 
 Construction. On AB, and 
 on the side opposite to C, de- 
 scribe the segment of a circle 
 containing an angle of 70° ; 
 complete the circle; at the 
 point B make the angle ABE 
 = 40°; from C, through E, 
 where BE cuts the circle, draw CE, and produce it till it 
 cuts the circle again in D, the point required ; join DA, DB, 
 and the work is done. 
 
 The angle ADE == ABE (21.3) = 40 6 \ whence the con- 
 struction is manifest. 
 
 Calculation. Join AE ; then the angles ABE, BAE, and 
 the side AB> are known ; whence AE may be found. The 
 angle CAB may also be determined from the three sides; 
 hence CA, AE, and the contained angle, become known ; 
 from which ACE and AEC, and consequently AED, may be 
 found. But AED = ABD (21.3) ; therefore the angles of the 
 triangle ABD become known, as well as those of ADC ; from 
 which and the given sides, AD, BD and CD may be found. 
 
 Result: AD = 82.915; DB = 73.406; DC = 123.178; 
 
 and DAB = 50° 2' 6". 
 
 8. Given, the base AB 50 ; radius of the circumscribing 
 circle 30 ; and ratio of the sides, AC : BC : : 3 : 2, to find 
 the sides. 
 

 II 
 
 
 
 ' I 
 
 
 S C 
 
 A^ 
 
 P / \ 
 
 \ 
 
 33 
 
 1 
 
 
 (; 
 
 SECTION II. 71 
 
 Construction. Divide AB in 
 D so that AD : DB :: 3:2; 
 bisect AB in E ; draw EF at 
 light angles to AB ; from the 
 centre A, with the radius of 
 the circumscribing circle, cut 
 EF in F ; from the centre F, 
 with the radius FA, describe 
 a circle cutting FE produced 
 in G; join GD, and produce 
 it to meet the circle in C ; join 
 AC, BC; then ABC will be the triangle required. The 
 circle will pass through B (1.3) ; and the arc AGB is bisected 
 in G ; consequently, the angle ACB is bisected by the line 
 CD (27.3) ; wherefore (3.6), 
 
 AC : BC : : AD : DB. 
 
 Calculation. In the right angled triangle AEF, AE and 
 AF are given, from which EF and the angle AFE are deter- 
 mined ; but AFE = 2ACD = ACB. In the triangle EDG ; 
 EG and ED being known, the angle EDG or BDC is deter- 
 mined ; whence the angles of the triangle ABC are known, 
 and the sides AC, BC determined. 
 
 Result: AC = 59.447 ; BC *= 39.632. 
 
 9. Given, one angle of a triangle 50° ; the sum of the three 
 sides 120; and the radius of the inscribed circle 10, to 
 determine the sides of the triangle. 
 
 Construction. Make AB = \ the sum of the sides, 60 ; the 
 angle BAN = the given angle 50° ; bisect that angle by the 
 line AF, meeting BF, which is drawn at right angles to AB ; 
 make BC = the given radius of the inscribed circle, 10; 
 through C draw CD parallel to BA, meeting AF in D ; draw 
 DE, DG at right angles to AB, AN respectively ; from the 
 centre D, with the radius DE, describe the circle : that circle 
 
PLANE TRIGONOMETRY. 
 
 N 
 
 will touch the lines AB, AN, in E and G (4.4). On the 
 diameter DF, describe the semicircle DHLF, cutting AB in 
 H and L ; from either of these points L draw LK, touching 
 the circle GEK in K, and cutting AN in M ; then ALM is 
 the triangle proposed. 
 
 Join DL, DK, DM ; LF, MF ; and draw FN, FP at right 
 angles to AN, LK respectively. Now, since DE = DK, the 
 angles at E and K are right ones, and DL is common to the 
 triangles DEL and DKL, it follows that the angle DLE = 
 DLK. But DLF is a right angle (31.3) ; hence, DLE + FLB 
 =DLF (13.1) ; from these equals, take the equals DLE and 
 DLK ; and we have BLF == KLF. Thence, the angles at 
 B and P being right ones, it is obvious that LB = LP, and 
 FB = FP. Again, in the triangles FAB and FAN, we have 
 AF common, and the angles of the one respectively equal to 
 those of the other ; hence AN == AB, and FN = FB = FP ; 
 consequently (47.1), MN = MP. 
 
 Now, it has been proved that LP = LB ; consequently, 
 ML = MN + LB; 
 and, therefore, 
 
 AM + AL + ML = AB + AN= 2AB 
 
SECTION IT. 73 
 
 Calculation. In the right angled triangles AED, ABF, we 
 have the angle at A, and the lines ED and AB given; from which 
 AE, EB and BF are determined. Then the angle DCF being 
 a right one, the semicircle on DF must pass through C (con- 
 verse 31.3) ; consequently, CB.BF = LB.BH (cor. 36.3). If, 
 now, we suppose a line drawn from the centre of the semi- 
 circle, cutting EB at right angles in I, it is manifest (2.6 and 
 3.3) that EB and HL are both bisected in I ; whence EL = 
 HB ; and EL.LB = CB.BF, a known rectangle. Hence IL 2 
 {= IB- — EL.LB (5.2)) becomes also known. 
 
 Result: AL=50.306; AM=31.139; LM=38.555. 
 
 10. Given, the base 50; difference of the other sides 10; 
 and radius of the inscribed circle 12 ; to determine the sides. 
 
 Construction. Make AB= 
 the base 50, and bisect it in 
 C; lay down CD = J the 
 difference of the sides ; at D 
 erect a perpendicular DE = 
 radius of the inscribed circle 
 A~~ c D B 12; from E, with the dis- 
 
 tance ED, describe a circle ; through A and B draw the lines 
 AH, BH, touching the circle in F and G ; ABH is the triangle 
 required. 
 
 Join EF, EG; then (47.1) AD = AF ; BD = BG; FH = 
 GH; consequently, 
 
 AH — BH = AD — BD - 2CD. 
 
 Calculation. With AD, DE, and BD, DE, find the angles 
 BAE, ABE, from which BAH and ABH are known ; and 
 thence the sides AH and BH. 
 
 Result: AH == 45.79; BH = 35.79. 
 
 11. Given, the perimeter of a triangle 120; radius of the 
 inscribed circle 10 ; and vertical angle 70°, to determine the 
 sides. 
 
 10 G 
 
74 
 
 PLANE TRIGONOMETRY. 
 
 M 
 
 l\/?& 
 
 ^2B^>^ 
 
 A H 
 
 
 E /J) 
 
 K 
 
 Construction. Make AB= J the perimeter 60 ; at B erect the 
 perpendicular BC=radius of the inscribed circle 10; through 
 C, draw CG parallel to AB ; make the angle BCD = comple- 
 ment of half the vertical angle, 55° ; bisect AD in E ; draw 
 EF at right angles to AD, meeting CD produced in F ; from 
 F, as a centre, through A or D, describe the arc AGB, cut- 
 ting the line GC ; from one of the intersections G, draw GH 
 at right angles to AD; from G, with the radius GH, describe 
 the circle HLM ; through A and D draw AI, DI, touching 
 the circle in L and M ; then ADI is the triangle required. 
 
 Join AG, DG and IG; then it is obvious that those lines 
 bisect the angles DAI, ADI and AID (26.1); consequently, 
 
 DAG + ADG + AIG = a right angle (32.1) ; 
 
 that is, AIG is the complement of DAG + ADG. Now, since 
 EFD = the angle at the circumference, which stands on 
 AGD ; it follows that 
 
 EFD + AGD = 2 right angles (22.3) = GAD + GDA + AGD 
 
 (32.1); wherefore. 
 
 EFD = GAD + GDA; 
 
 their complements are therefore equal; that is, EDF or CDB 
 — LIG. Consequently, the whole angle AID is twice the 
 
SECTION II. 79 
 
 complement of DCB. Also, the angles LIG and ILG, being 
 respectively equal to BDC and DBC, and LG = BC, the side 
 IL = DB (26.1); hence the semiperimeter of the triangle 
 ADI = AB. 
 
 Calculation. Draw FK parallel to AB, meeting GH pro- 
 duced in K, and join FG ; then, in the triangle DBC, having 
 BC and all the angles, BD is found ; whence ED becomes 
 known ; from which, and the angles, DF and EF are found ; 
 then, in the right angled triangle FGIv, FG and GK are 
 known ; whence FK or EH becomes known ; whence AH 
 and BH are known.* 
 
 Result: AD=45.719 ; AI=27.02; DI=47.261. 
 
 12. Given, the sides of the triangle ABC> viz. : AB 4G4, 
 AC 418, and BC 385 ; it is required to find a point D within 
 the triangle, such that AD, BD and CD shall be to each other 
 in the ratio of 7, 6 and 5 respectively. 
 
 Construction. Divide AB in the point E, and AC in G, so 
 that AE : EB : : 7 : 6 ; and AG : GC : : 7 : 5 ; produce AB 
 and AC to F and H, so that BF and CH shall be third pro- 
 portionals to AE — EB and EB, and to AG — GC and GC 
 respectively ; from the centres F and H, with the distance? 
 FE and HG, describe arcs cutting each other in D ; join AD 
 
 * Examples 9 and 11 are essentially of the same nature, and might have 
 Seen solved by the same method ; the two solutions furnish a little variety. 
 
f 70 PLANE TRIGONOMETRY. 
 
 BD and CD ; and the figure is constructed. For, as was 
 proved in the 6th example, 
 
 AD : BD : : AE : EB : : 7 : 6 ; 
 
 and AD : CD :: AG : GC : : 7 : 5; 
 
 whence, BD : CD : : 6 : 5. 
 
 Calculation. Join DF, DH and FH ; then, in the triangle 
 ABC, we have all the sides to find the angle BAC ; then, in 
 the triangle AFH, we have the sides AF, AH, and the in- 
 cluded angle, to find the angle AFH and side FH ; in the 
 triangle FDH, the three sides are then known to find the 
 angle DFH ; whence the angle AFD becomes known : then, 
 in the triangle AFD, we have the sides AF, FD, and the 
 contained angle, to find the angle FAD and the side AD ; from 
 which BD and CD are found from the given ratios. 
 
 Results: BAD = 25° 59' 8"; CAD = 25° 27' 15"; AD = 
 283.688 ; BD = 243.161 ; CD = 202.635. 
 
 [The following ingenious construction of this problem, which 
 admits of a simpler calculation than that already given, has 
 been kindly furnished the author by Samuel Alsop, Principal 
 of Friends' Select School, Philadelphia.] 
 
 Construction. Make FE = t 
 6; and on it describe FEG 
 similar to CBA, the given 
 triangle, making 
 
 FE:EG:FG::BC:BA:AC. 
 
 s On FG describe the triangle 
 FGA, making FA = 7, and 
 AG a fourth proportional to 
 BC, AB and 5. Upon AE, on 
 the same or any other scale, lay down AB = 464, and com- 
 ulete the triangle ABC. Draw BD parallel to EF, cutting 
 VF in D, which will be the point required. For, join CD; 
 
SECTION II. 77 
 
 draw EH parallel to BC ; arid join FH. Then, since AEH 
 is similar to FEG, being both similar to BAC ; 
 
 AE : EG :: EH : EP ; 
 
 therefore (6.6), AEG and HEF are similar ; and 
 
 AB : BC : : AE : EH : : AG : FH. 
 
 But AB : BC : : AG : 5 ; 
 
 therefore, FH == 5. Consequently, 
 
 AD : BD : CD : : AF : FE : FH : : 7 : 6 : 5. 
 
 Calculation. In the triangle ABC, with the given sides, find 
 the angle BAC = FGE ; also find AG, GF and GE. From 
 the three sides of the triangle AGF, find the angle AGF; 
 whence AGE becomes known. In the triangle AGE, find 
 AE; then 
 
 AE : EF : : AB : BD ; 
 
 from which AD and CD are found from the given ratios. 
 
 13. In a right-angled isosceles triangle, the hypothenuse is 
 30 yards longer than one of the sides ; what are the sides ? 
 
 Ans. : hypoth. 102.4264 : sides 72.4264. 
 
 14. The hypothenuse of a right-angled triangle is 75, and 
 the sum of the sides is 105; what are the sides? 
 
 Ans. : 60 and 45. 
 
 15. The sides of a triangle are in the ratio of 4, 6, and 7 ; 
 and the line bisecting the greatest angle is 20 ; required the 
 sides. Result : 22.87, 34.31, 40.02. 
 
 16. Given the perimeter of a right-angled triangle 120, 
 and the radius of the inscribed circle 10 ; required the sides 
 of the triangle? Ans. : 50, 40, and 30. 
 
 a* 
 
78* PLANE TRIGONOMETRY. 
 
 17. From a position in a horizontal plane, I observe the 
 angle of elevation of a tower, which is 100 feet high, to be 
 60° ; how far must I measure back, to obtain a position from 
 which the elevation shall be 30°? Ans. 115.47 feet. 
 
 18. A person on the top of a tower which is 50 feet in 
 height, observes the angles of depression of two objects on 
 the horizontal plane, which are in the same straight line with 
 the bottom of the tower, to be 30° and 45°. Determine their 
 distance from each other and from the observer. 
 
 Ans. Distance from each other 36.60 feet. 
 From the observer 70.71, and 100 feet. 
 
 19. From the top of a tower, whose height is 108 feet, the 
 angles of depression of the top and bottom of a vertical 
 column, standing in the horizontal plane, are found to be 30° 
 and 60° respectively. Required, the height of the column. 
 
 Ans. 72 feet. 
 
 20. Suppose the angle of elevation of the top of a steeple 
 to be 40°, when the observer's eye is level with the bottom, 
 and that from a window 18 feet directly above the first sta- 
 tion the angle of elevation is found to be 37° 30'. Required, 
 the height and distance of the steeple. 
 
 Ans. Height, 210.44 feet. 
 Distance, 250.79 feet. 
 
 21. Two columns, 80 and 100 feet in height, standing on 
 a horizontal plane, are distant from each other 220 feet ; it is 
 required to find a point in the line joining their bases, from 
 which the angles of elevation of the two columns shall be 
 equal. Ans. The point is 122| feet from the higher column. 
 
 22. The altitude of a cloud was observed to be 34° 20', and 
 that of the sun in the same direction 50° ; also the distance 
 of the shadow of the cloud from the station of the observer 
 measured 375 yards. Determine the height of the cloud. 
 
 Ans. 600 yards. 
 
SECTION II. 79* 
 
 23. In a plane triangle there are given, the base 60, an 
 adjacent angle 55° 30', and the ratio of the side opposite the 
 given angle to the other unknown side 6 to 5 ; to determine 
 these sides. Ans. 50.047, and 41.706. 
 
 24. From a station in a horizontal plane, I observed the 
 angle of altitude of the summit of. a cliff which bore exactly 
 north to be 47° 30'. I then measured N. 87 W. 283 feet, and 
 again taking the angle of altitude, found it to be 40° 12'. 
 What was the height of the cliff? Ans. 354.53 feet. 
 
 Remark. In the solution of the preceding problem, it will 
 assist the pupil if he will observe, that when two right angled 
 triangles have the same perpendicular, their bases are to each 
 other as the cotangents of the angles at the base. 
 
 25. Three ships sailed from the same place to different 
 ports in the same parallel of latitude; the first sailed directly 
 south 55 leagues, when she arrived at the desired port ; the 
 other two sailed upon different courses, between the south 
 and west, till they arrived at their destined ports, which were 
 57 leagues asunder, and the angle included by their courses 
 at the port sailed from was 38°. Required, the course and 
 distance run by each of the two latter vessels. 
 
 Ans. S. 52° 12', W. 89.75 leagues; 
 and S. 14° 12', W. 56.73 leagues. 
 
 26. Walking on shore, I was surprised by the flash of a 
 gun, at sea, bearing S. 56° 15' E. ; seven seconds after the 
 flash I heard the report, and four seconds after that I heard 
 the echo from a castle bearing from me S. 56° 15' W. Re- 
 quired, the distance of the gun and castle; sound being esti- 
 mated to pass over 1142 feet in one second of time. 
 
 Ans. Distance of gun, 7994 feet; of castle, 3005.51 feet. 
 
 27. In a right angled triangle there are given, one of the 
 legs 94, and the segment of the hypothenuse adjacent to the 
 other leg, made by a perpendicular from the right angle, 66, 
 to determine the triangle. 
 
 Ans. The other leg is 93.56, and the hypothenuse, 132.62. 
 
80* PLANE TRIGONOMETRY. 
 
 28. Having given two sides of a triangle, 40 and 50, and 
 the line drawn from the included angle to the middle of the 
 third side, 34; to determine the third side. Ans. 59.80. 
 
 Construction. — Form the triangle BAE, making AB 40, 
 AE 50, and BE 68 ; complete the parallelogram ABCE, draw 
 the diagonal AC, and ABC will be the required triangle. 
 
 29. At three points in the same horizontal straight line, the 
 angles of elevation of an object were found to be 36° 50', 21° 
 24', and 14°, the middle station being 84 feet from each of the 
 others. Required, the height of the object. 
 
 Ans. 53.96 feet. 
 
 30. Tn a level garden there are two lofty firs, having their 
 tops ornamented with gilt balls : one is 100 feet high, the other 
 80, and they are 120 feet distant at the bottom. Now, the 
 owner wants to place a fountain in a right line between the 
 trees, to be equally distant from the top of each, and to make 
 a walk or path from the fountain, in every point of which he 
 shall be equally distant from each of the balfe ; also, at the 
 end of the walk he would fix a pleasure-house, which should 
 be at the same distance from each ball, as the two balls are 
 from each other. How must this be done? 
 
 Ans. From bottom of taller tree to fountain, 45 feet. 
 From ball to ball, 121.655 " 
 
 Length of the walk, 52.678 " 
 
 From bottom of taller tree to house, 69.282 " 
 
 31. Three objects, A, B, and C, are situated in the same 
 straight line, and are distant from D, 312, 150, and 123 yards ; 
 also, the distance of A from B is to the distance of C from 
 B as 22 to 13. How far is B from A and C ? 
 
 Ans. From A 198 yards, from C 117 yards 
 
 Construction. — Make AD = 312, and divide it in E so that 
 
 AE : ED : : 22 : 13. On DE form the triangle DEB, making 
 
 DB 150, and EB = §§ DC. Join AB, produce it till it meets 
 
 DC drawn parallel to BE in C, and the figure is constructed. 
 
SECTION II 81* 
 
 32. Given, the angles of elevation of an object taken at 
 three positions, A, B, and C, in the same horizontal straight 
 line, 17° 46', 33° 41', and 39° 6', respectively ; also, from A to 
 B is 264 feet, and from B to C 156 feet. Required, the height 
 of the object. Ans. 133.33 feet. 
 
 33. There are three towns, A, B, and C, whose distances 
 apart are as follows: from A to B 6 miles; from A to C 
 22 miles ; and from B to C 20 miles. A messenger is des- 
 patched from B to A, and has to call at a town D in a direct 
 line between A and C. Now, in travelling from B to D, he 
 walks uniformly at the rate of 4 miles an hour, and from D 
 to A at the rate of 3 miles an hour. Supposing him to per- 
 form his journey in three hours, it is required to determine 
 the position of the town D. 
 
 Ans. The distance of D from A is 4.72 miles. 
 
 In the above example, we have J AD + £ BD = 3, or 
 AD + |BD = 9. On AC lay off AE = 9, and join BE; 
 then in the triangle BDE the side BE and the angle BED 
 become known, and ED : DB : : 3 : 4. Hence the point D is 
 readily determined. 
 
 34. The lengths of three lines drawn from a given point to 
 three angles of a square are, 35, 46, and 50 yards ; to deter- 
 mine a side of the square. Ans. 59.95 yards. 
 
 35. Wishing to ascertain the length of a tree which leaned 
 in the plane of the meridian, I measured from the foot of the 
 tree north 85 feet, when I found the angle of elevation of the 
 top to be 35°. I then took a second station 50 feet east of 
 the former, at which the elevation was 30°. Required, the 
 length of the tree. Ans. 52.44 feet. 
 
 11 
 
SECTION III. 
 
 SPHERICAL TRIGONOMETRY. 
 
 Article 45. The business of Spherical Trigonometry is, 
 to investigate the properties of triangles formed on the sur- 
 face of a sphere, by the arcs of circles whose planes pass 
 through the centre. 
 
 As the diameter of a circle is the greatest straight line in 
 it (15.3), so the diameter of a sphere is necessarily the great- 
 est straight line in it. Hence, when a plane passes through 
 the centre of the sphere, the diameter of the circle which is 
 formed by the section of this plane and the sperical surface, 
 is greater than any other line in the sphere which is not a 
 diameter. 
 
 A plane cutting the sphere, but not passing through its 
 centre, forms, by its section with the spherical surface, a 
 circle whose diameter is less than the diameter of the sphere. 
 That the section is a circle, is readily inferred from 14.3 ; 
 and that the diameter of that circle is less than the diameter 
 of the sphere, is plain from 15.3. 
 
 Definition 1. Those circles whose planes pass through the 
 centre of the sphere, are called great circles ; but circles 
 whose planes do not pass through the centre of the sphere, 
 are called less circles. 
 
 Corollary 1. The diameter of every great circle is also a 
 diameter of the sphere. 
 
 Cor. 2. The common section of the planes of two great 
 circles, is a diameter to each of those circles. 
 
 (78) 
 
SECTION III. 
 
 79 
 
 Cor. 3. Every great circle in the sphere divides every 
 other great circle into two equal parts. 
 
 Def. 2. The axis of a circle is the right line which passes 
 through its centre, and is at right angles to the plane of the 
 circle ; and the poles of a circle are the points where its axis 
 meets the surface of the sphere. 
 
 Def. 3. A spherical angle, or the angle formed by two 
 great circles, is the inclination of their planes. 
 
 Cor. When two great circles are at right angles to each 
 other, each of them passes through the poles of the other ; 
 and if they pass through the poles of each other, they are at 
 right angles. Also, when the plane of a great circle is at 
 right angles to the plane of a less one, the former circle 
 passes through the poles of the latter. For the axis of every 
 circle passes through the centre of the sphere, and is at right 
 angles to the plane of its own circle. 
 
 Def. 4. A spherical triangle is formed by the arcs of three 
 great circles, each of which cuts the other two, but in such 
 manner that each of the arcs composing the triangle is less 
 than a semicircle. 
 
 Def. 5> If AD and 
 DF, two quadrants of 
 great circles, are placed 
 at right angles to each 
 other ; and through the 
 points A, F, two other 
 great circles, AE, FB ; 
 are described, cutting 
 each other in C; the 
 triangles ABC, FEC 
 are called complementat 
 triangles. 
 
 Art. 46. The arc of a great circle, intercepted betweer 
 another great circle and its pole, is a quadrant. 
 
80 
 
 SPHERICAL TRIGONOMETRY. 
 D 
 
 Let AEBF be a great circle, whose centre is C, and axis 
 DCG, its poles being D and G ; DAGB another great circle 
 passing through the axis DCG; these great circles are at 
 right angles to each other (17.2 sup.), and CA their common 
 section at right angles to CD ; hence the arcs AD, BD, AG 
 and BG, are quadrants. 
 
 Art. 47. The angle made by two great circles is measured 
 by the arc intercepted between them, at the distance of 90° 
 from the angular point. 
 
 Let ACB, ADB, be 
 two semicircles, whose 
 common section passes 
 through E, the centre 
 of the sphere ; from E, 
 draw EC, ED, at right 
 angles to AB, one in the 
 plane ACB, the other in the plane ADB ; and let the plane 
 
SECTION III. 
 
 81 
 
 passing through EC, ED, cut the surface of the sphere in the 
 arc CD ; CD is part of a great circle (Def. 1), and AE is at 
 right angles to its plane (4.2 sup.) ; consequently, AC and AD 
 are quadrants; and A, B are the poles of CD. Also, the 
 inclination of the planes ACB, ADB is the angle CED (Def. 
 4.2 sup.) ; and that angle is measured by the arc CD. 
 
 Cor. Since the plane of CED is at right angles to AB, ana 
 consequently to each of the planes ADB, ACB (17.2 sup.), it 
 must pass through the axes of those planes ; and therefore 
 the circle DC, continued, must pass through the poles of ADB 
 and ACB. Those poles being 90 degrees from their respect- 
 ive circles, the arc intercepted between them is manifestly 
 equal to CD, the measure of the spherical angle CAD. 
 
 Art. 48. In the complemental triangles ABC, FCE ; AC 
 is the complement of CE ; BC is the complement of FC ; AB 
 of the angle at F ; and the angle at A of the side FE. 
 
 For, since FD and 
 AD are quadrants at 
 right angles to each 
 other, F is the pole of 
 AD, and A is the pole 
 ofFD (Art. 46); hence 
 FB, AE are also quad- 
 rants; consequently, BD 
 is the measure of the 
 angle at F, and DE of 
 the angle at A ; whence 
 the proposition is ob- 
 vious. 
 
 Art. 49. In isosceles spherical triangles, the angles oppo- 
 site the equal sides are equal. 
 
 Let ABC be a spherical triangle, whose sides AB and AC 
 are equal ; it is to be proved that the angles ABC and ACE 
 are also equal. 
 
82 SPHERICAL TRIGONOMETRY. 
 
 Take D the centre of the sphere, 
 and join DA, DB and DC ; and 
 in the plane of ADB, draw AE 
 at right angles to DB. In like 
 manner, in the plane ADC, draw 
 AF at right angles to DC. Then, 
 since the arc AB is equal to AC, 
 "j!j the angle ADB at the centre of 
 
 the sphere is equal to the angle ADC ; therefore the triangles 
 ADE, ADF, right angled at E and F, having two angles of 
 the one respectively equal to two angles of the other, and 
 the side AD, opposite the right angle in each, common to 
 both ; have the sides AE, AF, adjacent to the right angles, 
 also equal (26.1). 
 
 Again, in the plane BDC, draw EG and FG at right angles 
 to DB and DC respectively, and let them meet in G. Then, 
 because AE and EG are both at right angles to DB, the line 
 DB is at right angles to the plane which passes through AE 
 and EG (4.2 sup.) ; and therefore the plane DBC is at right 
 angles to the plane AEG (17.2 sup.). In like manner, the 
 plane DBC is proved to be at right angles to the plane AFG; 
 consequently, the line AG, the common section of the planes 
 AEG, AFG, is at right angles to the plane DBC (18.2 sup.) ; 
 wherefore the angles AGE, AGF are right angles (1 Def. 2 
 sup.) 
 
 Now, the right angled triangles AGE, AGF, having the 
 perpendicular AG common, and the hypothenuse AE equal 
 the hypothenuse AF, must have their bases EG, FG, also 
 equal (47.1) ; and therefore the angles AEG, AFG, likewise 
 equal (8.1) ; that is, the spherical angles ABC, ACB, are 
 equal. Q. E. D. 
 
 Art. 50. If two angles of a spherical triangle are equal, 
 the sides opposite to them are also equal. 
 
 Let the spherical angles ABC, ACB be equal ; then the 
 sides AB, AC shall be also equal. ^ 
 
SECTION III. 83 
 
 Making the same construction as in the last article, w 
 have, as before, the angles AGE, AGF, both right angles s 
 also the angles AEG and AFG, which are the same as the 
 spherical angles ABC and ACB, likewise equal, and the side 
 AG common to the triangles AGE, AGF ; therefore AE is 
 equal to AF (26.1). Then, in the right angled triangles 
 ADE, ADF, we have the perpendiculars AE, AF equal, and 
 the hypothenuse AD common ; wherefore DE is equal to DF 
 (47.1), and consequently the angle ADE equal to ADF (8.1) : 
 whence AB is equal to AC (26.3). 
 
 Q. E. D. 
 
 Art. 51. Any two sides of a spherical triangle are together 
 greater than the third. 
 
 Let ABC be a spherical 
 triangle; any two of its 
 sides taken together are 
 greater than the third. 
 
 Take D the centre of the 
 sphere, and join DA, DB 
 and DC. Then the solid 
 angle at D is contained by 
 the three plane angles 
 ADB, ADC and BDC, of which any two taken together are 
 greater than the third (20.2 sup.) ; therefore any two of the 
 arcs which measure those angles are likewise together greater 
 than the third. Q. E. D. 
 
 Art. 52. The three sides of a spherical triangle are toge- 
 ther less than the circumference of a circle. 
 
 Let ABC be a spherical 
 triangle; the sides AB + 
 AC + BC are less than 
 360°. Continue two of 
 those sides AC, AB, till 
 
84 SPHERICAL TRIGONOMETRY. 
 
 they meet in D ; then ACD and ABD are semicircles (Art. 
 45, Cor. 3, Def. 1). But BD + CD are greater than BC. If 
 to these unequal quantities we add AB + AC, we have ABD 
 + ACD greater than BC + AB + AC ; that is, AB + AC + 
 BC are less than two semicircles, or 360°. 
 
 Q. E. D. 
 
 Art. 53. In any spherical triangle having unequal angles, 
 the greater angle has the greater side opposite to it. 
 
 In the spherical triangle ABC, let the angle ABC be greater 
 than ACB ; and take CBD = BCA ; then (Art. 50) BD=CD ; 
 consequently, AC = BD + AD ; but (Art. 50) BD + AD are 
 greater than AB ; that is, AC is greater than AB. 
 
 Q. E. D. 
 
 Conversely : If the side 
 AC is greater than AB, the 
 angle ABC is greater than 
 ACB. For if it is not 
 greater, it is equal or less. 
 If the angles were equal, 
 the opposite sides would also be equal (Art. 49) ; and if ACB 
 was greater than ABC, the side AB would be greater than 
 AC. 
 
 Art. 54. If the angular points of a spherical triangle are 
 made the poles of three great circles, these three circles, by 
 their intersections, will form a triangle, which is said to be 
 supplemental to the former ; and the two triangles are such, 
 that the sides of the one are the supplements of the arcs 
 which measure the angles of the other. 
 
 Let A, B, C, the angular points of the triangle ABC, be 
 the poles of the great circles FE, DE, DF, which form the 
 triangle FED ; and let the sides of the former triangle be 
 produced till they meet those of the latter. Now, since A 
 and C are the poles of EF and DF respectively, the distances 
 
SECTION III. 
 
 85 
 
 from F to A and from F 
 to C are quadrants (Art. 
 46) ; hence F is the pole 
 of K ACL. In the same 
 manner it is proved that 
 D and E are the poles of 
 NBCH and GABM ; con- 
 sequently, EM + LF = 
 180°; thatis,EFandLM 
 are supplements to each 
 other. In like manner it 
 may be proved that DE 
 and GH; DF and KN; 
 AC and KL; AB ar.d 
 GM; BC and NH, are 
 respectively supplements to each other. But ML, GH and 
 KN are the measures of the angles A, B and C ; also KL, 
 NH and GM are the measures of the angles F, D and E (Art. 
 47). Hence the proposition is manifest? Q. E. D. 
 
 Cor. Since the sides FE, FD and DE, together with the 
 measures of the angles A, B and C, are equal to three semi- 
 circles, or 540° ; and the three sides of any spherical triangle 
 are together less than two semicircles, or 360° (Art. 52), it 
 follows that the three angles of the triangle ABC are more 
 than 180°, but less than 540°. 
 
 Aet. 55. Let AEB and 
 
 AHB be semicircles, whose 
 
 planes are at right angles to 
 
 each other ; and AB the 
 
 common section of those 
 
 planes, a diameter to the 
 
 sphere; AH, HB, quad- 
 
 P e rants; and C any other 
 
 point than H in the semicircle AHB; then CD, CE, CF, 
 
 being arcs of great circles, intercepted between the point C 
 
 12 
 
86 SPHERICAL TRIGONOMETRY. 
 
 and the semicircle AEB ; the arc CA, which passes through 
 H, is greater, and CB, the remaining part of the semicircle, 
 is less, than any other arc contained between C and AEB ; 
 also those nearer to CHA are greater than those which are 
 more remote. 
 
 Draw CG and HI at right angles to AB ; then is CG at 
 right angles to the plane AEB (Def. 2, 2 sup.) ; hence, GD, 
 GE, GF, being drawn, the angles CGD, CGE, CGF, are all 
 right ones. And AH, BH, being quadrants, H is evidently 
 the pole, and I the centre, of AEB ; consequently, GA is the 
 greatest, and GB the least, of all the straight lines drawn 
 from G to the circumference ; and GD is greater than GE ; 
 and GE than GF (7.3). Now, 
 
 AO-CGM-GA 8 ; 
 and . DC 3 = CG 3 + GD 3 ; 
 
 of which GA 3 is greater than GD 3 ; wherefore AC 2 is greater 
 than DC 3 , and AC greater than DC. But the arcs AC and 
 DC are each less than a semicircle ; and, therefore, the greater 
 chord subtends the greater arc ; that is, the arc AC is greater 
 than DC. In the same manner it may be proved that the arc 
 DC is greater than EC, that EC is greater than FC, and 
 FC greater than BC. Q. E. I). 
 
 Art. 56. In a right angled spherical triangle, the sides 
 containing the right angle are of the same affection as the 
 angles opposite to them.* 
 
 Let ACB be a spherical triangle, right angled at A ; and 
 let AC, AB be continued till they meet in D ; and bisect ABD 
 in E ; then ACD, ABD are semicircles (Art. 45, Cor. 3 to 
 Def. 1) ; and AE is a quadrant. But the angle at A being a 
 right one, AEB passes through the pole of AC (Art. 45, Cor. 
 
 * Sides are said to be of the same affection when they are both less or 
 both greater than quadrants ; the same is said of angles when they are 
 both less or both greater than right angles. A side and an angle are also 
 of the same affection when the former is less or greater than a quadrant, 
 and the latter less or greater than a right angle. 
 
SECTION III. 87 
 
 to Def. 3). Consequently E is the pole of AC (Art. 46) ; CE 
 is a quadrant, and ACE a right angle. 
 
 Now, AC being taken 
 less than a quadrant, the 
 angle ACB will be less or 
 greater than ACE, ac- 
 cording as B lies be- 
 tween A and E, or between 
 E and D; that is, when AB 
 is less than a quadrant, the angle ACB is less than a right 
 angle ; and when AB is more than a quadrant, ACB is more 
 than a right angle. And if we suppose ACB to be less than 
 a right angle, it is manifest that AB is less than a quadrant ; 
 and, if greater, greater. 
 
 Again, in the right angled triangle DCB, in which DC is 
 greater than a quadrant, it is manifest that the angle DCB is 
 greater or less than a right angle, according as DB is greater 
 or less than a quadrant ; and vice versa. Q. E. D. 
 
 Art. 57. When the sides of a right angled spherical tri- 
 angle, about the right angle, are of the same affection, the 
 hypothenuse is less than a quadrant ; but when those sides 
 are of different affections, the hypothenuse is more than a 
 quadrant. 
 
 Retaining the construction used in the last article, and 
 bisecting ACD in G, we have G the pole of ABD, and CE a 
 quadrant as before. But CB is either greater or less than 
 CE, according as it is nearer to or farther from CGD than 
 CE is (Art. 55) ; that is, the hypothenuse is less than a quad- 
 rant when the sides are both less or both greater than a 
 quadrant ; but the hypothenuse is greater than a quadrant 
 when one side is less, and the other greater, than a quadrant. 
 
 Q. E. D. 
 
 Cor. 1. Conversely, when the hypothenuse of a right angled 
 spherical triangle is less than a quadrant, the sides are of the 
 same affection ; but, when the hypothenuse is greater than a 
 quadrant, the sides are of different affections. 
 
bd 
 
 SPHERICAL TRIGONOMETRY. 
 
 Cor. 2. Since the oblique angles of a right angled spherical 
 triangle are of the same affection as the opposite sides (Art. 
 56) ; therefore, according as the hypothenuse is greater or 
 less than a quadrant, the oblique angles will be of different, 
 or the same affection. 
 
 Cor. 3. Because the sides are of the same affection as their 
 opposite angles ; therefore, when an angle and the side adja- 
 cent are of the same affection, the hypothenuse is les-s than a 
 quadrant, and vice versa. 
 
 Art. 58. In any right angled spherical triangle, as radius 
 is to the sine of an oblique angle, so is the sine of the hy- 
 pothenuse to the sine of the opposite side. 
 
 Let ABC be the triangle, 
 right angled at B ; take D 
 the centre of the sphere ; 
 join DA, DB, and DC ; in 
 the plane ADC, draw CE 
 at right angles to DA ; 
 from E, draw in the plane 
 ADB, the line EP at right 
 angles to DA, meeting DB 
 in F ; and join CF. 
 
 Then, the lines EC and 
 EF being both at right angles to DA, the plane CEF is at 
 right angles to DA (4.2 sup.) ; consequently, the planes ADB 
 and CEF are at right angles to each other (17.2 sup.). But 
 the plane DBC is, by hypothesis, at right angles to DAB ; 
 hence the planes CEF and DBC, being both at right angles 
 to DAB, their common section FC is also at right angles to 
 the same plane (18.2 sup.) ; wherefore DFC and EFC are 
 right angles. Hence CF is the sine of CB ; also CE is the 
 sine of CA ; and the angle CEF is the inclination of the 
 planes CDA and BDA ; that is, Cl^F = the spherical angle 
 CAB. Now (Art. 28), 
 
 As radius : sine of CEF : : CE : CF ; 
 
SECTION III. 89 
 
 that i?, 
 
 As radius : sine of CAB : : sine of AC : sine of BC. 
 
 Q. E. D. 
 
 Art. 59. In any oblique angled spherical triangle, the sines 
 of the angles are to each other as the sines of the opposite 
 sides. 
 
 Let ABC be the trian- 
 gle ; and through C de- 
 scribe the arc CD of a 
 B great circle, at right an- 
 gles to AB ; then, by last 
 article, 
 
 As sin of A : radius : : sin DC : sin AC ; ' 
 
 radius : sin B : : sin BC : sin DC ; 
 
 therefore (23.5), 
 
 sin A : sin B : : sin BC : sin AC. 
 
 Q. E. D. 
 
 Art. 60. In any. right 
 angled spherical triangle, 
 as radius is to the sine of 
 one of the sides, so is the 
 tangent of the adjacent 
 angle to the tangent of 
 the opposite side. 
 
 Let ABC be the trian- 
 gle, right angled at B; 
 take D the centre of the 
 sphere, and join DA, DB, 
 DC; in the plane ADB, 
 draw BE at right angles 
 
90 
 
 SPHERICAL TRIGONOMETRY. 
 
 to DA ; from E, draw EF, in the plane ADC, at right angles 
 to AD, and meeting DC produced in F ; and join FB. Then 
 EB and EF being at right angles to DE, the plane FEB is 
 at right angles to DE (4.2 sup.) ; consequently, the plane 
 ADB, which passes through DE, is at right angles to the 
 plane FEB (17.2 sup.) ; therefore the common section BF of 
 the planes EBF and DBC, is at right angles to the plane 
 DAB (18.2 sup.) ; whence EBF and DBF are right angles ; 
 and, consequently, BF is the tangent of BC ; BE is also the 
 sine of AB ; and the angle BEF the same as the spherical 
 angle BAC. Now (Art. 28), 
 
 that is, 
 
 As radius : tan BEF : : BE : BF ; 
 
 As radius : tan BAC : : sin AB : tan BC ; 
 
 and alternately (16.5), 
 
 As radius : sin AB 
 
 tan BAC : tan BC. 
 
 Q.E.D. 
 
 Art. 61. If two right angled spherical triangles have the 
 same perpendicular, the sines of the bases are to each other 
 ^ reciprocally as the tan- 
 
 gents of the adjacent 
 angles. 
 
 Let ADC and BDC be 
 t the triangles ; DC the 
 common perpendicular; 
 then (Art. 60), 
 
 As sin AD : radius : : tan DC : tan A ; 
 
 and As radius : sin BD : : tan B : tan DC ; 
 
 whence (23.5), 
 
 AD : sin BD : : tan B : tan A. 
 
 Q. E. D. 
 
SECTION III. 
 
 91 
 
 Art. 62. In any right angled spherical triangle ; as radius 
 is to the cosine of the angle at the base, so is the tangent of 
 the hypothenuse to the tangent of the base. 
 
 p . Let ABC be the tri- 
 
 angle ; B the right, an- 
 gle ; and FCE the com- 
 plemental triangle. 
 Then (Art. 60), 
 
 As radius : sin FE : : 
 tan F : tan CE ; 
 
 that is (Art. 48), 
 
 As radius : cos A : : 
 cotan AB : cotan AC. 
 
 B But (Art. 23.4), 
 
 As tan P : tan Q : : cotan Q : cotan P ; 
 
 consequently, 
 
 As radius : cos A : : tan AC : tan AB. 
 
 Q. E. D. 
 
 Art. 63. If two right angled spherical triangles have the 
 same perpendicular, the cosines of the vertical angles are to 
 each other reciprocally as the tangents of the hypothenuses. 
 
 Let ADC, BDC (see fig. on opposite page), be the triangles 
 right angled at D ; then (Art. 62), 
 
 As cos ACD : radius : : tan DC : tan AC ; 
 
 and As radius : cos BCD : : tan BC : tan DC ; 
 
 therefore (23.5), 
 
 As cos ACD : cos BCD : : tan BC : tan AC. 
 
 Q.KD 
 
92 
 
 SPHERICAL TRIGONOMETRY. 
 
 Art. 64. In any right angled spherical triangle ; as radius 
 is to the cosine of the hypothenuse, so is the tangent of either 
 angle to the cotangent of the remaining angle. 
 
 In the triangle FCE (see fig. on opposite page), we have 
 (Art. 60), 
 
 As radius : sin CE : : tan FCE : tan FE ; 
 that is (Art. 48), 
 
 As radius : cos AC : : tan ACB : cot BAC : : (Art. 23.4) 
 tan BAC : cot ACB. 
 
 Q. E. D. 
 
 Art. 65. In any right angled spherical triangle ; as radius 
 is to the cosine of one of the sides, so is the cosine of the 
 other side to the cosine of the hypothenuse. 
 
 In the right angled triangle FCE, we have (Art. 57), 
 
 As radius : sin F : : sin FC : sin CE ; 
 
 that is (Art. 48), 
 
 As radius : cos AB : : cos BC : cos AC. 
 
 Q. E. D. 
 
 Art. 66. If two right angled spherical triangles have the 
 C same perpendicular, the 
 
 cosines of the hypothe- 
 nuses are to each other 
 as the cosines of the 
 * bases. 
 
 Taking ADC and BDC 
 as the triangles right 
 angled at D, we have (Art. 65), 
 
 As radius : cos DC : : cos AD : cos AC ; 
 
 and As radius : cos DC : : cos BD : cos BC ; 
 
 whence (16.5), 
 
SECTION III. 
 cos AD : cos BD : : cos AC : cos BC. 
 
 Q. E. D. 
 
 Art. 67. In any- 
 right angled spherical 
 triangle; as radius is 
 to the sine of either 
 oblique angle, so is the 
 cosine of the adjacent 
 side to the cosine of the 
 opposite angle. 
 
 In the right angled 
 triangle FCE, we have 
 b (Art. 58), 
 
 As radius : sin FCE : : sin CF : sin FE ; 
 
 that is (Art. 48), 
 
 As radius : sin ACB : : cos BC : cos BAC. 
 
 Q. E. D. 
 
 Art. 68. In two right angled spherical triangles, A CD, 
 BCD (fig. p. 92), having the same perpendicular CD, the co- 
 sines of the angles at the base are to each other as the sines 
 of the vertical angles. 
 
 By Art. 67 and A. 5, 
 
 As cos DAC : cos DC : : sin ACD : radius ; 
 and, by same article, 
 
 As cos DC : cos DBC : : radius : sin BCD ; 
 
 consequently (22.5), 
 
 As cos DAC : cos DBC : : sin ACD : sin BCD. 
 
 Q. E. D. 
 13 
 
94 SPHERICAL TRIGONOMETRY. 
 
 Art. 69. The same things being supposed as in the last 
 article, the tangents of the bases are to each other as the 
 tangents of the vertical angles. 
 
 By Art. 60, 
 
 As radius : sin CD : : tan ACD : tan AD ; 
 
 and As radius : sin CD : : tan BCD : tan BD ; 
 
 consequently (11 and 16.5), 
 
 tan ACD : tan BCD : : tan AD : tan BD. 
 
 Q. E. D. 
 
 Art. 70. In two 
 right angled spheri- 
 cal triangles ABC, 
 ADC, having the 
 same hypothenuse 
 AC, the cosines of the 
 bases are to each 
 other reciprocally as 
 the cosines of the per- 
 pendiculars. 
 For (Art. 65), 
 
 As radius : cos AB : : cos BC : cos AC ; 
 and from the same article inverted (A 5), 
 
 As cos AD : radius : : cos AC : cos DC ; 
 hence (23.5), 
 
 As cos AD : cos AB : : cos BC : cos DC. 
 
 Q. E. D. 
 
 Art. 71. The same things being supposed as in the last 
 article, the tangents of the bases are to each other as the 
 cosines of the adjacent angles. For (Art. 62), 
 
 As radius : cos CAB : : tan AC : tan AB ; 
 
SECTION III. r 5 
 
 and, by inversion, 
 
 As cos CAD : radius : : tan AD : tan AC ; 
 
 therefore (22.5), 
 
 As cos CAD : cos CAB : i tan AD : tan AB. 
 
 Q. E. D. 
 
 A 
 
 Art. 72. As the sum of the sines of any two unequal arcs 
 is to their difference, so is the tangent of half the sum of 
 those arcs, to the tangent of half their difference. 
 
 Let AB, AC be the arcs ; 
 L the centre of the circle ; 
 and AH the diameter passing 
 through A ; make AF=AB; 
 join BF, and let BF cut AH 
 in D ; draw CE parallel to 
 BF,andCG to AH; let CG 
 meet BF in I ; join GB, GF 
 and CF. Then, since AF 
 = AB, FD = BD; and 
 HDF = HDB ; hence BD is the sine of AB ; and CE, which 
 is parallel to BD, is the sine of AC ; therefore, FI is the sum, 
 and BI the difference, of the sines of AB and AC. 
 
 Again, the arc CF is the sum, and CB the difference, of 
 AB and AC ; therefore the angle FGC is measured by half 
 the sum, and BGC by half the difference, of AB and AC 
 (20.3). But the angle GIF = HDF (29.1), and is therefore 
 a right angle ; consequently, IF is the tangent of CGF, and 
 IB the tangent of CGB, to the radius GI ; therefore, for any 
 other radius (Art. 27, Cor.), 
 
 
 
 
 
 c 
 
 
 \ £ 
 
 I 
 7) 
 
 
 
 
 \ A 
 
 ^r 
 
 /E 1 
 
 
 
 
 Ug 
 
 
 As IF : IB : : tan CGF : tan CGB 
 i(AB-AC). 
 
 tan i(AB + AC) : tan 
 Q. E. D. 
 
 Art. 73. The sum of the cosines of two unequal arcs is, 
 
(£ SPHERICAL TRIGONOMETRY. 
 
 to their difference, as the cotangent of half their sum is to 
 the tangent of half their difference. 
 
 Retaining the construction of the last article, it is easily 
 perceived that GI is the sum, and IC the difference, of the 
 cosines of AB and AC ; also the angle GFI is the complement 
 of CGF; and IFC - BGC; hence, 
 
 cos AB + cos AC : cos AC — cos AB :: cotan ^(AB + AC) : 
 tan J(AB — AC). Q. E. D. 
 
 Art. 74. In any oblique angled spherical triangle, a per- 
 pendicular being let fall from the vertex on the base, i,t will 
 be, as the tangent of half the base is to the tangent of half 
 the sum of the sides, so is the tangent of half the difference 
 of those sides to the tangent of the distance between the 
 c perpendicular and mid- 
 
 dle of the base. 
 
 Let ABC be the trian- 
 gle, CD the perpendicu- 
 and E the middle of 
 base. Then (Art. 
 
 As cos AC : cos BC : : cos AD : cos BD ; 
 
 whence (E 5), 
 
 As cos AC + cos BC : cos BC — cos AC : : cos AD + cos BD ; 
 cos BD — cos AD ; 
 
 consequently (Art. 73), 
 
 As cotan l(AC + BC) : tan J(AC — BC) : : cotan A(AD + 
 DB) : tanl(AD — DB) :: cotan AE : tan ED ; 
 
 and, alternately, 
 
 As cotan i( AC + BC) : cotan AE :: tan £(AC— CB) : tan ED. 
 But (Art. 23.1), 
 cotan i(AC + BC) : cotan AE : : tan AE : tan J(AC+BC); 
 
SECTION Jill 97 
 
 therefore, 
 
 As tan AE : tan £(AC + BC) :: tan i(AC — BC) : tan ED. 
 
 q: E. D. 
 
 Art. 75. In any oblique angled- spherical triangle, a per- 
 pendicular being let fall from the vertex on the base, and an 
 arc described bisecting the vertical angle ; it will be, as the 
 cotangent of half the sum of the angles at the base is to the 
 tangent of half their difference, so is the tangent of half the 
 vertical angle to the tangent of the angle formed by the per- 
 pendicular and the arc bisecting the vertical angle. 
 
 Let ABC be the triangle, CD the perpendicular, and CF 
 the arc bisecting the vertical angle ; then (Art. 68), 
 
 As cos A : cos B : : sin ACD : sin BCD ; 
 
 hence (E 5) 
 
 cos A+cos B . cos A — cos B :: sin ACD -f sin BCD : sin 
 ACD — sin BCD ; 
 
 therefore (Arts. 72, 73) 
 
 As cotan J(A + B) : tan i(B — A) :: tan $(ACD + BCD) 
 :tan|(ACD — BCD) :: tan ACF : tan DCF. 
 
 Q. E. D. 
 
 Scholium. From the analogies demonstrated in Articles 60, 
 62, 63, 65, 66 and 67, we may frequently determine the affec- 
 tions of the sides and angles of the triangles, by adverting 
 to the signs of the terms, as explained in Art. 24. Thus, in 
 Art. 60, the base AB being always less than a semicircle, the 
 sin AB is positive; hence the tan BAC and tan BC are both 
 positive or both negative; consequently, BAC and BC are 
 both less or both more than 90°. In Art. 62, when the angle 
 at the base is acute, its cosine is positive ; consequently, the 
 tangents of the hypothenuse and base will be both positive or 
 both negative ; therefore the arcs themselves will be both more 
 or both less than 90° ; that is, they will be of the same affec- 
 13 i 
 
08 SPHERICAL TRIGONOMETRY. 
 
 tion. But when the angle at the base is obtuse, its cosine 
 will be negative ; and therefore the tangent of the hypothe- 
 nuse and that of the base will be one positive, and the other 
 negative ; consequently, the arcs themselves will be of dif- 
 ferent affections. In Art. 63, when the vertical angles are 
 of the same affection, their cosines have the same sign ; con- 
 sequently, the tangents of the adjacent sides will have the same 
 sign, and will therefore be of the same affection. The same 
 principles are applicable to the other cases. The conclusions 
 thus obtained are consonant to those obtained in a different 
 manner in Arts. 56, 57. 
 
 The analogies above demonstrated are sufficient to enable 
 the student to calculate the sides and angles of spherical tri- 
 angles from the usual data; yet there are various useful 
 forms, hereafter demonstrated, which are applicable to par- 
 ticular cases. We have also two concise rules, discovered 
 by Baron Napier, the celebrated inventor of logarithms, by 
 which the cases in right angled spherical triangles are con- 
 veniently solved ; and, being easily remembered, they are 
 frequently used in practice. 
 
 Art. 76. In right angled spherical triangles, there are five 
 parts which may have different values assigned to them 
 without changing the right angle, viz. : the hypothenuse, the 
 two sides, and the two oblique angles. Now, the sides, and 
 the complements of the hypothenuse and oblique angles, are 
 called the five circular parts ; one of which being assumed as 
 the middle part, the two which lie contiguous to this middle 
 part are called the adjacent extremes ; and the other two 
 are termed the opposite extremes. Then Napier's rules are : 
 
 1. The rectangle of radius and the sine of the middle part 
 is equal to the rectangle of the tangents of the adjacent 
 extremes. 
 
 3. The rectangle of radius and the sine of the middle 
 
SECTION III. 
 
 99 
 
 part is equal to the rectangle of the cosines of the opposite 
 extremes. 
 
 These rules may be explained and demonstrated in the 
 following manner : 
 
 c Let ABC be the tri- 
 
 angle, right angled 
 at B. Then, assum- 
 ing AB as the mid- 
 dle part, the side BC 
 and complement of 
 BAC are the adja- 
 cent extremes ; and 
 the complements of 
 AC and ACB are the 
 opposite extremes. Now (Art. 60), 
 
 As radius : sin AB : : tan BAC : tan BC ; 
 and, alternately, 
 
 As radius : tan BAC : : sin AB : tan BC. 
 But (Art. 23), 
 
 As radius : tan BAC : : cotan BAC : radius. 
 Hence, 
 
 cotan BAC : radius : : sin AB : tan BC ; 
 therefore (16.6), 
 
 radius . sin AB = cotan BAC . tan BC ; 
 which is Napier's first rule. 
 
 Again (Art. 58), 
 
 As radius : sin ACB : : sin AC : sin AB ; 
 whence (16.6), 
 
 radius . sin AB = sin ACB . sin AC ; 
 which is Napier's second rule. 
 
100 SPHERICAL TRIGONOMETRY. 
 
 If we assume BC the middle part, AB and the complement 
 of ACB become the adjacent extremes; and the complements 
 of BAC and AC, the opposite extremes. Napier's rules may- 
 then be demonstrated in that case exactly as before. 
 
 Assuming next the complement of BAC as the middle part, 
 AB and the complement of AC become adjacent extremes ; 
 and BC and the complement of BCA opposite extremes. 
 Then (Art. 62), 
 
 As radius : cos BAC : : tan AC : tan AB ; 
 
 alternately, 
 
 As radius : tan AC : : cos BAC : tan AB. 
 
 Hence (Art. 23), 
 
 As cotan AC : radius : : cos BAC : tan AB ; 
 
 consequently (16.6), 
 
 radius . cos BAC == cotan AC . tan AB ; 
 
 which is rule first. 
 
 Again (Art. 67), 
 
 As radius : sin BCA : cos BC : cos BAC ; 
 whence, 
 
 radius . cos BAC = sin BCA . cos BC ; 
 which is rule second. 
 
 In the same manner, the rule is demonstrated, when the 
 complement of ACB is taken as the middle part. 
 
 Lastly, assuming the complement of AC as the middle part, 
 the complements of BAC and BCA are the adjacent extremes ; 
 and AB, BC, the opposite extremes. Then (Art. 64), 
 
 As radius : cos AC : : tan ACB : cotan BAC ; 
 
 alternately, 
 
SECTION III. 101 
 
 As radius : tan ACB : : cos AC : cotan BAC ; 
 wherefore (Art. 23), 
 
 As cotan ACB : radius : : cos AC : cotan BAC ; 
 consequently, 
 
 radius . cos AC = cotan ACB . cotan BAC ; 
 which is rule first. 
 
 And (Art. 65), 
 
 As radius : cos AB : : cos BC : cos AC ; 
 whence, 
 
 radius . cos AC = cos AB . cos BC ; 
 which is rule second. 
 
 The following table exhibits the different cases, and the 
 equations arising from Napier's rules : 
 
 o 
 
 S 
 ■ 
 
 1 
 
 Middle 
 part. 
 
 Adjacent Ex- 
 tremes. 
 
 Opposite Ex- 
 tremes. 
 
 Equations. 
 
 AB 
 
 Comp BAC 
 BC 
 
 Comp ACB 
 Comp AC 
 
 a ■ kt> CcotBAC.tan BC 
 rad.sinAB=£ 8inACBt8inAC 
 
 •2 
 
 BC 
 
 CompBCA 
 AB 
 
 CompBAC 
 Comp AC 
 
 j • on Ccot BCA.tan AB 
 
 rad.sinBC= -? . t> a r< „: a n 
 
 £ sin fcsAC.sin AC 
 
 3 
 
 Comp BAC iCom PAC 
 
 Comp ACB 
 BC 
 
 j -d a n C cot AC. tan AB 
 rad.cosBAC=^ 8inACBcoBBC 
 
 4 
 
 CompBCA; 001 ^^ 
 
 Comp BAC 
 AB 
 
 j t)/-. a Ccot AC. tan BC 
 rad.cosBCA=^ sinBACcosAB 
 
 5 
 
 n a /-i ICompBAC 
 Comp AC ! Com £ BCA 
 
 AB 
 BC 
 
 . , n Ccot BAC.cotBCA 
 rad.cosAC=-? AT} m m ^^ 
 £ cos AB.cos BC 
 
 When any two of these circular parts are given to find a 
 third, we must assume such one to be the middle part as will 
 make the other two either both adjacent or both opposite 
 extremes. 
 
 The following practical examples will serve to exercise the 
 preceding theory : 
 
 14 
 
102 
 
 SPHERICAL TRIGONOMETRY. 
 
 Example 1. In the 
 spherical triangle 
 ABC, right angled 
 at B, given the side 
 AC 52° 15', and the 
 angle A 23° 28', to 
 find the other sides 
 and the remaining 
 angle. 
 
 The perpendicular 
 BC may be found by Art. 58 ; the side AB, by Art. 62 ; and 
 the angle C, by Art. 64. Or, using Napier's circular parts, 
 we find BC by the second equation, case 2, in the foregoing 
 table ; AB by the first equation, case 3 ; and the angle C by 
 the first equation, case 5. The results are, BC 18° 21' 9" ; 
 AB 49° 49' 57"; C 75° 6' 58". 
 
 Ex. 2. Given, the base AB 61° 25', and the adjacent angle 
 A 32° 45', to determine the rest. 
 
 The hypothenuse AC may be found by Art. 62 ; the per- 
 pendicular BC by Art. 60 ; and the angle C by Art. 67 ; or 
 by cases 3, 1, 4 of the circular parts. 
 
 The results are, AC 65° 22' 52" ; BC 29° 27' 32" ; and C 
 
 75°. 
 
 Ex. 3. Given, the base AB 75° 28', and the perpendicular 
 BC 41° 15', to find the rest. 
 
 The results are, AC 79° 7' 30"; A 42° 10' 32"; C 80° 
 18' 1". 
 
 Ex. 4. Given, the angle A 23° 28' 30", and angle C 75° 22', 
 to find the rest. 
 
 The results are, *$ 53° 2' 36"; AB 50° 38' 22"; BC 18° 
 33' 40". 
 
SECTION III. 103 
 
 Ex. 5. In the oblique 
 angled triangle ABC, 
 given the sides AB 70°, 
 AC 58°, and the angle 
 CAB 52° 30', to find the 
 rest. 
 
 Suppose the arc CD of 
 a great circle at right angles to AB to pass through C ; then 
 the given triangle will be divided into two right angled ones, 
 ADC and BDC. 
 
 In the triangle ADC, the side AD and angle ACD may be 
 computed by Arts. 62 and 64. Hence BD is known. Then 
 the angle B, the side BC, and the angle BCD, may be found 
 by Arts. 61, 66 and 69. 
 
 Results : B 64° 28' ; BC 48° 12' 46" ; ACB 91° 0' 21". 
 
 Ex. 6. In the spherical triangle ABC, given the angle BAC 
 50° 15', ACB 92°, and side AC 57° 30', to find the rest. 
 
 The arc CD being made perpendicular to AB, the side AD 
 and angle ACD may be found as in the last example ; whence 
 BD, BC, and the angle at B, may be computed by Arts. 69, 
 63 and 68. 
 
 Results: BA 69° 25* 2"; BC 46° 4' 16"; ABC 64° 12' 16". 
 
 Ex. 7. In the triangle ABC, given AB 71° 30' ; AC 59° 
 20'; BC 50° 10' ; to find the angles. 
 
 Drawing CD at right angles to AB, and taking AE = ■JAB, 
 the arc ED is found by Art. 74 ; from which AD and BD 
 become known ; and thence the angles may be found by Arts. 
 62 and 59. 
 
 Results: BAC 54° 3' 51"; ABC 65° 5' 4"; ACB 90° 48' 
 47". 
 
 Ex. 8. In the triangle ABC, given the angle BAC 51° ; 
 ABC 58°; and ACB 110°; to find the sides. 
 
104 
 
 SPHERICAL TRIGONOMETRY. 
 
 Using the construction of the last example, and describing 
 CF so as to divide the vertical angle into two equal angles, 
 the angle FCD may be found by Art. 75 ; whence the angles 
 ACD and BCD become known ; and thence the sides AC, BC 
 may be determined by Art. 64 ; and AD, BD, by Art. 67. 
 
 Results : AC 64° 28' 31" ; BC 55° 47' 13" ; AB 90° 44' 26". 
 
 Art. 77. It has been already mentioned that there are 
 various useful forms which are applicable to particular cases. 
 By means of these, the necessity of dividing an oblique 
 angled triangle into two right angled ones, is always 
 obviated. Of these forms, the following are the most im- 
 portant. They are investigated most conveniently by 
 algebra. 
 
 c Let ABC be a spherical 
 
 triangle ; CD a perpendi- 
 cular upon AB; and, to 
 accommodate the expres- 
 t sions to the language of 
 algebra, let the capital 
 letters A, B, C denote the 
 angles, and the small letters a, b, c, the opposite sides ; the 
 segments AD, BD, being represented by d, e, and the oppo- 
 site angles ACD and BCD by D and E respectively. In 
 these investigations, the radius is taken = 1. 
 
 Now (Art. 62), 
 
 1 : cos A 
 
 tan b : tan d. 
 
 But (Art. 23), 
 
 sin sin d sin b 
 
 ■= tan .-. r = cos A. 
 
 cos d 
 
 cos b 
 
 cosin 
 Again (Art. 66). 
 
 cos b : cos a : : cos d : cos (c — d) ; 
 wherefore, 
 
SECTION III. 105 
 
 cos a cos (c — d) ,- n „ _ ~,cos c.cos a 7 + sine. sin d 
 
 r = -- r J -= (Art. 37, Form. 2) j 
 
 cos o cos a x ' cos a 
 
 sin d . . sin b 
 — cos c + sine.-- — ; = cos c + cos A.sinc. r ; by putting 
 
 cos a cos o J l ° 
 
 . sin b . sin J 
 
 cos A. 1 instead ot .. 
 
 cos b cos a 
 
 Clearing this equation of fractions, 
 
 cos a = cos c.cos b + cos A.sin c.sin b, (1) 
 
 By Art. 64, 
 
 _ _ ; •'■- " •■' i^ sin B cos E 
 
 As 1 : cos a : : tan B : cot E : : (Art. 23) ~ : ~ — =cr •"• 
 
 v ' cos B sin E 
 
 cos E . sin B 
 
 COSG. 
 
 sin E 'cos B' 
 
 By Art. 68, 
 
 As cos B : cos A : : sin E : sin (C — E) .-. 
 
 cos A sin (C — E) , k _,_ _, fc% sin C.cos E — cos C.sin E 
 
 ^ = r^— =(Art. 37, F. 1) .— -^ 
 
 cos B sin E v ' sin E 
 
 . cos E sin B 
 
 = sin C- — ts — cos C. — cos a.sin C. ^ — cos G ; substi- 
 
 sin E cos B 
 
 cos E 
 tuting for- — ^. Then, clearing of fractions, 
 6 sin E 6 ■ 
 
 cos A = cos a. sin C.sin B — cos C.cos B. (2) 
 
 By Art. 62, 
 
 As 1 : cos B : : tan a : tan e : : (Art. 23, 4) cotan e : cotan a 
 
 _ _ cos e 
 
 .*. cotan a = cos B.cotan e => cos B.— ; 
 
 sin e 
 
 . . cotan a cos e 
 
 wherefore, =r- = — « • 
 
 cos B sin e 
 
 14 
 
100 SPHERICAL TRIGONOMETRY. 
 
 But (Art. 61), 
 
 As tan A : tan B : : sin e : sin (c — e) ; 
 
 therefore, 
 
 tan B sin (c — e) /k -V _ , x sin c.cos e. — cosc.sine 
 
 ; T = ^ = (Art. 37, F. 1) : 
 
 tan A sin e v ' sin e 
 
 cos e . cotan a 
 
 = sin c- cos c = sin c. ^ cos Ct 
 
 sin e cos J3 
 
 Consequently, by clearing of fractions, 
 
 cos B.tan B = cotan a.sin c.tan A — cos c.cos B.tan A. 
 
 Now, 
 
 cos B.tan B=sin B (Art. 23) and r = cotan A. 
 
 tan A. 
 
 sin B.cotan A+cos c.cos B ._. 
 
 Hence cotan a = : . (o) 
 
 sine . 
 
 By Art. 64, 
 
 As 1 : cos b : : tan D : cotan A=cos fr.tan D=cos b. ^ 
 
 cos D 
 
 By Art. 63, 
 
 As tan a : tan b : : cos D : cos (C — D) ; 
 
 tan b cos (C — D) ow _ x 
 
 .-. ; = * — ^r--= (Art. 37, Form. 2) cos C 
 
 tan a cos D v [ 
 
 . sin D sin D tan ft.cotan a — cos C 
 
 + sin C. ^ .-. =p: = ; — ~ ' 
 
 cos D cos D sm U 
 
 Consequently, 
 
 sin fr.cotan a — cos C.cos & 
 
 cotan A == : — ~ . (4) 
 
 sm C v ' 
 
 These forms are not suited to logarithmic computations ; 
 but they are useful in the investigation of othor equations, to 
 which logarithms are conveniently applied. 
 
SECTION III. 107 
 
 From Form. 1, above given, we find, by transposition and 
 division, 
 
 . cos a — cos c.cos b 
 
 cos A = - 1 — ; 
 
 sin c.sin b 
 
 hence, 
 
 cos a — cos c.cos b 
 
 1 — cos A = 1 
 
 sin c.sin b 
 
 cos c.cos b + sin c.sin b — cos a ,:, „ „ 
 
 1 r-t = (Art. 37, Form. 2) 
 
 sin c.sin b v ' 
 
 cos (c — b) — cos a 
 sin c.sin b 
 
 But (Art. 36, Form. 6), 
 
 1 — cos A = 2 sin 9 -JA ; 
 and (Art. 37, Form. 13), 
 
 cos (c — b) — cos a = 2 sin %(a + c — b).sm ±(a + b — c) 
 
 .-. sin 5 
 
 u _sin i(a + c — 6) .sin £(a + b — c) 
 
 ■syA — r ; r^ 
 
 sin c.sin b 
 
 (5) 
 
 . . c ,, .. . cos a — cos c.cos b 
 
 Again, trom the equation cos A = . we 
 
 sm c.sin b 
 
 have 
 
 cos a — cos c.cos b 
 
 1 + cos A = 1 +■ 
 
 sin c.sin b 
 
 cos a — cos c.cos b + sin c.sin b 
 
 "Iran"*- "= (Art. 37, Form. 2) 
 
 cos a — cos (c + b) 
 sin c.sin b 
 
 But (Art. 36, Form. 8) 
 
 1 + cos A = 2 cos 8 £A ; 
 
108 SPHERICAL TRIGONOMETRY. 
 
 * sin c.sin b v ' 
 
 Again : 
 
 tan* i A - sin3 ^ A - si" l(a+c—b) sin j(a +b— c) 
 
 * cos s |A sin i(c + b + a) sin i(c + b— a)' ™ 
 
 Equations 5, 6 and 7 furnish convenient expressions for 
 finding an angle, when the three sides are given. 
 From Form. 2, 
 
 cos A + cos C.cos B 
 
 cos a = . — ~r--. — s ; 
 
 sin C.sin B 
 
 wherefore, 
 
 _ cos A+cos C.cos B 
 
 1 COS G= 1 . — 77-: — =i = 
 
 sin C.sin B 
 
 sin C.sin B — cos C.cos B — cos A 
 sin C.sin B 
 
 — (cos C.cos B — sin C.sin B) — cos A _ — cos (B + C) — cos A 
 sin C.sin B sin C.sin B 
 
 cos (B + C) + cos A 
 
 = • n • p = ( Art - 37, Form. 12) 
 
 sin C.sin B v ' 
 
 — 2 cos |(B + C + A).cos fr(B + C — A ) 
 sin B.sin C 
 
 But (Art. 36, Form. 6), 
 
 1 — cos a = 2 sin 3 \a ; 
 
 Wherefore, 
 
 «*£■ -cosKB-fC+Aj^KB+C-A), 
 
 J sin B.sin C v ' 
 
 * Since the three angles of a spherical triangle are together greater than 
 180°, but less than 540° (Art. 54, Cor.), cos i(B+C + A) will always be 
 a negative quantity, and consequently — cosine a positive quantity. 
 
SECTION III. • 109 
 
 Again : 
 
 _ _ cos A+cos C.cosB 
 
 1 -f cos a= H -. — 7^—: — ts = 
 
 sin C.sin B 
 
 cos C.cos B -f sin C.sin B + cos A _ cos (C — B)+c os A 
 sin C.sin B sin C.sin B 
 
 (Art. 37, Form. 2). 
 
 But (Art. 36, Form. 8), 
 
 1 + cos a = 2 cos 3 \a ; 
 
 and (Art. 37, Form. 12) 
 
 cos (C — B)+cos A=2 cos£(A + C — B).cos J(A+B — C); 
 
 whence, 
 
 cos * i fl cos|(A+C-B).cosKA+B~C) 
 
 cos ?a- sin C.sin B * W 
 
 Further : 
 ♦ ,i sin 2 \a -cosKB+C+A).cosj(B+C--A) 
 tan ^ a ~~cos s \a ~ cos i(A + C — B).cos |(A+B— C)' ^ 1U > 
 
 Equations 8, 9 and 10 may be conveniently used for de- 
 termining the sides, when all the angles are given. 
 
 By Art. 59, 
 
 As sin b : sin a : : sin B : sin A; 
 wherefore (E. 5), 
 
 sin b + sin a : sin b — sin a :: sin B + sin A : sin B — sin A; 
 consequently, by Art. 72, 
 
 As tan J(6+a) : tan l(b— a) : : tan £(B + A) : tan £(B— A). (N) 
 wherefore, 
 
 ta„ K B-A)=ta„KB + A).^||=§ (P) 
 
 15 
 
110 SPHERICAL TRIGONOMETRY. 
 
 Again (Art. 68), 
 
 As cos B : cos A : : sin E : sin D ; 
 hence (E. 5), 
 
 As cos B + cos A : cos A — cos B : : sin E + sin D : sin D — sin E ; 
 consequently (Arts. 73.72), 
 
 As cotan J(B+ A) : tan J(B— A) : : tan JC : tan i(D— E) ; 
 whence 
 
 tan . ( D-E)=ta„ .C.-^i|^= (Art. 23.4) 
 
 tan JC.tan * (B — A).tan \ (B + A). 
 
 By Art. 63, 
 
 As tan b : tan a : : cos E : cos D ; 
 
 wherefore (E. 5), 
 
 As tan J + tan a : tan b — tan a : : cos E + cos D : cos E — cos D. 
 
 But (Art. 37, equation 8), 
 
 , , sin (b ± a) 
 
 tan b db tan a = 2 r ; 
 
 cos a.cos o 
 
 consequently, 
 
 sin (a + b) : sin (b — a) : : cotan JC : tan J(D — E) ; 
 
 wherefore, 
 
 tan i(D — E)=cotan^C. S ? n , ( f "7^ = ( Art - 36 > e q- 3 ) 
 
 JX ' sin (b + a) v ' ' 
 
 2 cos j( b — a). sin ±(b — a) 
 cotan nC-2-—-T( a + b).sm ±(a + b)' 
 
 Equating these values of tan £(D — E), 
 
 tan JC.tan J(B — A).tan i(B + A) = 
 
 cos h (b — q).sin | (b — a) 
 cotan u^' cos i ( a + ^. s j n J( a + &)" 
 
SECTION III. Ill 
 
 In this equation, substitute for tan J(B — A), its value given 
 in equation P ; then, 
 
 tan JCton' >(B + A) ' a "^ ~ a |= 
 2V '. tan l(b + a) 
 
 cos l(b — a). sin Mb — a) 
 C0tan i U cos i(a + b).s'm J(a + b)' 
 
 , i/p ,. A \ _ cotan 2 C »cos *(&— oQ.sin j(&— a).tan j (&+ a ) 
 tan 3 ^b+a;- tan i C . CO s |(«+^).sin |(a+6).tan |(^— a) * 
 
 Now f Art. 22.1V 
 
 sin 1 _ tan 
 tan ' * cos sin 
 
 Hence, 
 tn«i(J 
 
 Now (Art. 23.1), 
 
 and (Art. 23.4), 
 
 1 cotan 
 
 cotan = - — .*. — = cotan 3 . 
 
 tan tan 
 
 Our equation is therefore reducible to this : 
 
 tan' i(B+A)=cotan' i C . cos3 ^~ « > ■ 
 ^ X ' z cos- 8 l(a+b) 
 
 consequently, 
 
 tan*(B + A)=cotan^g=|. (11) 
 
 From equation P, 
 
 i/t> a\ in cos K^ — a). ten Mb — a) 
 
 tan |(B — A)=cotan-lC. 177— ; — (-7 — -tj—, — C = 
 
 2V y 2 cos l(b + a). tan i(o -f a) 
 
 ,>_ sin l(& — a) 
 cotan i fi, A ^ v (12) 
 
 sin i(b + a) K ' 
 
 From proportion N, 
 As tan h(B + A) : tan i(B— A) : : tan i(b + a) : tan £(&— a) 
 
112 SPHERICAL TRIGONOMETRY. 
 
 x tani(B — A) ^ 
 
 ... tan £(&-a)=tan W + fl)-^ h \ B + A j- (0) 
 
 By Art. 66, 
 
 As cos a : cos 5 : : cos e : cos d ; 
 therefore (E. 5 and Art. 73), 
 
 As cot z(b+a) : tan i(b — a) : : cot £c : tan i(d — e) ; 
 whence, 
 
 tan Ud — e)=cot ic.-^^r-^T = ( Art - 23, ec l- 4 ) 
 v ' cotan i(6 + a) v 
 
 cot ic.tan 1(6 -f a). tan £(& — a). 
 
 Again (Art. 61), 
 
 As tan B : tan A : : sin d : sin e ; 
 
 therefore (E. 5 and Art. 72), 
 
 As tan B + tan A : tan B — tan A : : tan \c : tan h(d — e) ; 
 
 . tan B — tan A , . .■ ow _ v 
 
 .-. tan Ud — e)=tan £c. s — - r=(Art. 37, eq. 8) 
 
 v ' tan B + tan A v 
 
 , sin (B — A) , . nrt _. 
 tan ^-sinlB— Al = ( Art - 36 ^ e ^ 3) 
 
 1 sini(B — A).cosi(B — A ) 
 tan 3jc. gin ^ R + A ^ cog i(B + A y 
 
 Equating these values of tan h(d — e), and substituting 
 for tan h(b — a) its value in equation Q; 
 
 , , „ tan l(B — A) 
 cot icm* ,(b + a) -J^g-j^ = 
 
 sini(B — A) cos KB — A) , /7 r 
 
 tan he - — .^ ' V n • — nfopr , Ai •*• tan * 5(6 + «)= 
 sin i(B -f A) cos A(B -f A) v ' 
 
SECTION III. 113 
 
 tan Ic.sin |(B — A).cos i(B — A).tan j(B+A) 
 cotk.sin i(B + A).cos £(B + A).tan i(B — A) ~ 
 
 cos» KB — A) 
 tan 9 \c- 
 
 cos 3 £(B + A) * 
 consequently, 
 
 „ i i cosi(B — A) 
 
 tanK^ + «)=tanic.^g T - A -J. (13) 
 
 From equation Q, 
 
 ta„ i ( J -a)=ta„ ic . S 4^|^. (14) 
 
 Equations 11 and 12 may be used when two sides and the 
 included angle are given to find the other angles ; and equa- 
 tions 13 and 14 when two angles and the side between them 
 are given to find the other sides.* 
 
 From these four last equations, the following are derived 
 by a very simple process : 
 
 cota„|C=tani(B + A).£|^. (15) 
 
 cotaniC=tan|(B-A).^||^. (16) 
 
 tanic=tan^+a).^|||^^. (17) 
 
 tan ic=tan J(A - a).^ l (B _S^y (18) 
 
 A few examples are given to exercise these equations. 
 
 * The discovery of these four equations is attributed to Baron Napier. 
 15 K * 
 
114 
 
 SPHERICAL TRIGONOMETRY. 
 Z 
 
 Ex. 1. In the spheri- 
 cal triangle ABZ, given 
 AZ=54° 10', BZ m 39° 
 25', AB = 72° 36', in 
 ZA produced Aa = 2', 
 and in ZB, B6 = 35', to 
 find ab. 
 
 First, with the three sides, find the angle Z, by equa- 
 tion 5. 
 
 ZA 54° 10' 
 
 cosec .0911273 
 
 ZB 39° 25' 
 
 cosec .1972569 
 
 AB 72° 36' 
 
 
 2)166° 11' 
 
 
 S 83° 5' 30" 
 
 
 S— ZA 28° 55' 30" 
 
 sine 9.6845440 
 
 S— ZB 43° 40' 30" 
 
 sine 9.8392057 
 
 
 2)19.8121339 
 
 \Z 53° 39' 32" 
 
 sine 9.9060669 
 
 BZA 107° 19' 4" 
 
 
 Then, by equations 11 and 12, 
 l(Za + Zb) 46° 31' sec .1623210 cosec .1393179 
 
 |(Za— Zb) 
 
 7° 41' 
 
 cos 9.9960834 sin 
 
 9.1261246 
 
 PZA 
 
 53° 39' 32" 
 46° 39' 15" 
 
 cot 9.8666880 cot 
 
 9.8666880 
 
 h(b+a) 
 
 tan 10.0250924 
 
 
 i(b~a) 
 
 7° 43' 12" 
 
 
 9.1321305 
 
 Zba 
 
 54° 22' 27" 
 
 

 SECTION 
 
 III. 
 
 By Art. 58, 
 
 
 
 Zba 
 
 54° 22' 27" 
 
 cosec .0899959 
 
 bZa 
 
 107° 19' 4" 
 
 sin 9.9798526 
 
 Za 
 
 54° 12' 
 
 72° 17' 8" 
 
 sin 9.9090550 
 
 ab 
 
 sin 9.9789035 
 
 Or the side ab 
 
 may be found by equation 18. 
 
 i(b-a) 
 
 7° 43' 12" 
 
 cosec .8718203 
 
 i(b + a) 
 
 46° 39' 15" 
 
 sine 9.8616681 
 
 l(Za—Zb) 
 
 7° 41' 
 36° 8' 33" 
 
 tan 9.1300413 
 
 \ab 
 
 tan 9.8635297 
 
 ab 
 
 72° 17' 6" 
 
 
 115 
 
 This is the direct solution of the celebrated problem of 
 clearing the observed distance between the moon and the 
 sun, or a star, from the effect of parallax and refraction. 
 
 c Ex. 2. In the spherical tri- 
 
 angle ABC, given AC 46° 
 18', AB 100° 26', and the 
 angle A 39° 50', to find the 
 rest. Result: ACB 136° 0' 
 54"; ABC 30° 41' 54"- 
 BC 65° 6' 34". 
 
 Ex. 3. In the triangle ABC, given AB 112° 56', BAC 40° 
 16', ABC 54° 20', to find the rest. 
 
 Result: AC 79° 44' 58"; BC 51° 31' 30"; ACB 130° 30' 
 20". 
 
 Ex. 4. Given, the side AB 96° 12', AC 57° 16', BC 49° 8', 
 to find the angles. 
 
116 SPHERICAL TRIGONOMETRY. 
 
 Result : BAC 31° 32' 42" ; ABC 35° 35' 15" ; ACB 136° 
 
 32' 48". 
 
 Ex. 5. Given, the angle BAC 50°, ABC 60°, ACB 85°, to 
 find the sides. 
 
 Result: AB 51° 59' 16"; AC 43° 13' 48"; BC 37° 17 
 26". 
 
SECTION IV. 
 
 CONIC SECTIONS. 
 
 Article 78. Definition 1. If, from a point in the circum- 
 ference of a circle, a right line be drawn to pass through a 
 fixed point which is not in the plane of that circle, and then 
 caused to revolve round that fixed point so as to describe the 
 whole circumference of the circle; the curve surface, de- 
 scribed by this revolving line, is called a conical surface; 
 and the solid included between this curve surface and the 
 generating circle, is called a cone. 
 
 Def. 2. The circle described by the revolving line is called 
 the base, and the fixed point the vertex, of the cone. 
 
 Def. 3. The straight line drawn from the vertex to the 
 centre of the base, is called the axis of the cone. 
 
 Def 4. When the axis is at right angles to the plane of the 
 base, the cone is called a right cone ; but when the axis i3 
 oblique to that plane, the solid is termed a scalene cone. 
 
 As the line which, by its revolution, describes the conical 
 surface, maybe indefinitely extended, two cones having a 
 common vertex, and equal solid angles at the vertex, may be 
 generated by the same revolution. 
 
 Art. 79. Let the cone ABCD be cut by a plane which 
 
 passes through its vertex A, and cuts the base in the right 
 
 line BC ; the common section of this plane with the surface 
 
 (117; 
 16 
 
118 
 
 SECTION IV. 
 
 of the cone, will be a triangle. The 
 common section of the base and cutting 
 plane is a right line (3.2 sup.) ; and the 
 right lines drawn from B and C to the 
 vertex, are in the cutting plane (2.2 
 sup.) ; and those lines correspond to the 
 position of the revolving line when it 
 passes through B and C ; they are there- 
 fore in the conical surface. 
 
 Q. E. D. 
 
 Art. 80. Let the cone ABC be cut by a plane which is 
 parallel to the plane of the base ; then the section of this 
 cutting plane with the conical surface, is a circle whose 
 centre is in the axis of the cone. 
 
 Let AF be the axis of the cone ; 
 DLE the cutting plane. In the cir- 
 cumference of the base take any point 
 K ; join FK ; and through AF, FK, 
 suppose a plane to pass, cutting the 
 conical surface in AK, and the cut- 
 ting plane in GH ; then (Art. 79, and 
 3.2 sup.) AK and GH are right lines. 
 Let also another plane ABC pass 
 through the axis ; its section with the 
 base will be a diameter,.because F is the centre of the circle ; 
 and the section of this plane with the conical surface is a tri- 
 angle (Art. 79). Take BC and DE, the sections of this plane 
 with the parallel planes BCK and DLE ; then (14.2 sup.) DE 
 and GH are respectively parallel to BC, FK ; consequently, 
 
 As AF : AG : : BF : DG : : FC : GE : : FK : GH. 
 
 But BF, FC and FK are all equal; therefore, DG, GE and GH 
 are also equal ; consequently (9.3), DLEH is a circle whose 
 centre is G. Q. E. D. 
 
CONIC SECTIONS. 
 
 11<> 
 
 Art. 81. Let AB, DE, two lines at right angles to each 
 
 other, such that AD.DB =± DE 2 ; 
 then a semicircle described on 
 the diameter AB will pass 
 through the point E. 
 
 Bisect AB in C ; then, since 
 AD.DB = DE 3 , 
 
 AD.DB + CD 3 = DE 3 + CD 2 ; 
 
 that is (5.2 and 47.1), CB 3 = CE a .-. CB £ CE. Conse- 
 quently, a circle described from the centre C, with the radius 
 CB, will pass through E. Q. E. D. 
 
 M ,~+ 
 
 Art. 82. Let ABLC be 
 a scalene cone; ABC the 
 triangle formed by the sec- 
 tion of the conical surface 
 with a plane which passes 
 through the axis, and stands 
 at right angles to the plane 
 of the base ; and let another 
 plane GHK, at right angles 
 to the plane ABC, cut that 
 plane in GK, making the 
 angle AGK = ACB, and 
 AKG = ABC;* then the 
 plane GHK cuts the coni- 
 cal surface in a circle. 
 
 In the section of the cut- 
 ting plane and conical sur- 
 face, take any point H; 
 through H let a plane 
 DHE pass, parallel to 
 
 the base of the cone, cutting the planes GHK and ABC in 
 the lines HF and DE respectively. Then, since the plane 
 
 * This is called a sub-contrary section. 
 
120 
 
 SECTION IV. 
 
 DHE is parallel to the plane of the base, it is at right angles 
 to the plane ABC (15.2 sup.). But the plane GHK is at 
 right angles to the same plane ; therefore the common section 
 HF is at right angles to the plane ABC (18.2 sup.), and con- 
 sequently to the lines DE and GK in that plane (Def. 1, 2 
 sup.) 
 
 Now, the angle GDP being = FKE ; and DFG = KFE 
 (15.1) ; the triangles GDF, EKF, are similar ; therefore, 
 
 As DF : FG : : FK : FE (4.6) 
 
 consequently, DF.FE=GF.FK (16.6). But DHE is a circle 
 (Art. 80) ; therefore, DF.FE = HF 3 (35.3). Hence, GF.FK 
 == HF 2 ; and consequently GHK is a circle (Art. 81), whose 
 diameter is GH. Q. E. D. 
 
 Art. 83. Let ABC be 
 a triangle formed by the 
 section of a cone with 
 a plane passing through 
 its axis at right angles 
 to the plane of its base ; 
 and let another plane 
 DFE, cutting the cone, 
 be at right angles to the 
 plane of the triangle, 
 and so situated that 
 FG, the common section 
 of these planes, shall be 
 parallel to AC, the op- 
 posite side of the trian- 
 gle ; then the common section of the plane DFE with the 
 conical surface, is a curve called a •q ^gboly, ; the general 
 property of which this article is intended to exhibit. 
 
 In this curve take any point H, and through II let a plane, 
 parallel to the base of the cone, be passed ; and let this plane 
 cut the plane of ABC in the line LM, and the plane DFE in 
 
CONIC SECTIONS. 
 
 121 
 
 HK. Then, because the planes LHM and DFE are at right 
 angles to ABC, their common section HK is at right angles 
 to the same plane (18.2 sup.) ; it is therefore at right angles 
 to FG and LM. Now, the section of the plane LHM with 
 the conical surface, is a circle (Art. 80) ; wherefore, LK.KM 
 = HK 3 (35.3). In like manner, BG.GC == DG 3 . The plane 
 LHM being parallel to the base, LK is parallel to BG (14.2 
 sup.) ; therefore, 
 As FG : FK : : BG : LK :: (1.6) BG.GC : LK.KM : : 
 
 DG 3 : HK 3 . 
 Def. 5. The line FG is called the axis, and F the vertex, 
 of the parabola ; any segment of the axis FK, reckoned from 
 the vertex, is called an abscissa ; and a perpendicular KH, on 
 the axis, is called an ordinate. 
 
 The demonstration in this article, therefore, shows that 
 any two abscissas are to each other as the squares of the 
 corresponding ordinates. 
 
 A Art. 84. Let ABC be a trian- 
 
 gle, formed by the common sec- 
 tion of a cone and a plane through 
 its axis, at right angles to the 
 plane of the ^ase ; DIEF the com- 
 mon section of the conical sur- 
 face, and a plane which is at 
 right angles to the plane of the 
 triangle ABC, passing through 
 its opposite sides, but neither pa- 
 rallel to the base, nor sub-con- 
 trarily situated ; the curve DIEF 
 is called an ellipsp : the general 
 property of which this article is 
 designed to explain. 
 
 Bisect DF, the common sec- 
 tion of DIEF and ABC, in L ; 
 take any other point K in DF ; 
 and through L and K let the 
 16 l 
 
 D 
 
 k O 
 
 M ( 
 
 ~~\iA 
 
 \v\N 
 
 iXLk^ 
 
 I\ 
 
 
 P I 
 
 
 "^7^° 
 
 
 
 
 VT 
 
 
 K vy 
 
 
 
 
 
 
 
122 SECTION IV. 
 
 planes MIN and PEO pass parallel to tho base of the cone 
 cutting ABC in MN, and PO, and DIEF in LI and KE. 
 Through D and F draw the lines DG and HF parallel to BC. 
 Now, since (15.2 sup.) the planes MIN and PEO are at 
 right angles to the plane ABC; LI and KE, the common 
 intersections of these planes and the plane DIEF, are at right 
 angles to the plane ABC (18.2 sup.); and consequently to the 
 lines MN,PO and DF in that plane (Def. 1.2 sup). Also the 
 common sections of the planes MIN and PEO, and the conical 
 surface, are circles, of which MN and PO are diameters (Art. 
 80). Therefore, ML.LN = LP, and PK.KO = KE 2 . Now, 
 the lines MN and PO are parallel to BC (14.2 sup). Hence, 
 by similar triangles, 
 
 
 
 AsDL 
 
 : DK :: 
 
 ML 
 
 :PK; 
 
 and 
 
 
 AsLF 
 
 : KF :: 
 
 : LN 
 
 : KO; 
 
 therefore 
 
 (23.6), 
 
 
 
 
 As DL.LF 
 
 : DK.KF 
 
 : : ML.LN : ] 
 
 PK.KO 
 
 As the line DL=LF, it is obvious that ML = ^HF, and LN 
 = JDG; therefore ML.LN = iDG.HF. Hence IL is a mean 
 proportional between JDG and ^HF. As DL a= LF, DL.LF 
 s= LF 2 . Consequently, the above analogy is, 
 As L^ 2 : LP : : DK.KF : KE 2 . 
 
 Def. 6. The lines LF, LI, are called the first and second 
 semi-axes ; DK, KF, the abscissas ; and KE an ordinate. 
 
 The property of the ellipse, demonstrated in this article, 
 therefore, is this : 
 
 As the square of the first semi-axis is to the square of the 
 second, so is the rectangle of the two abscissas to the square 
 of the ordinater 
 
 It is observable that if the plane DIEF is parallel to the 
 base, or sub-contrarily situated, all that is demonstrated in 
 this article continues to be true ; but in either of those cases 
 the curve becomes a circle (Arts. 80, 82) : and therefore LI 
 = LF, and DK.KF = KE 2 . 
 
CONIC SECTIONS. 
 
 123 
 
 Art. 85. Let ABC 
 be a triangle, formed 
 by the section of a 
 cone and a plane pass- 
 ing through its axis, 
 at right angles to the 
 plane of its base ; and 
 let DFE be a plane at 
 right angles to the 
 plane of the triangle, 
 so situated that GF, 
 the common section of 
 these planes, being 
 produced, will meet 
 CA, the opposite side 
 of the triangle also 
 produced, beyond the 
 vertex A ; then the 
 curve which is the 
 common section of the conical surface and the plane DFE, is 
 called an hyperbol a ; the general property of which is to be 
 shown. 
 
 In this section, take any point N; through which let a 
 plane pass parallel to the plane of the base ; and let ILK be 
 the common section of this plane with the plane of the tri- 
 angle, and NL its section with the plane DFE. 
 
 Now (18.2 sup.), DG, the section of DFE and the base of 
 the cone, and NL, are both at right angles to the plane of the 
 triangle. Also (Art. 80), the common section of the conical 
 surface, and the plane which passes through NL, is a circle ; 
 consequently, IL.LK^LN 2 , and BG.GC=GD 2 (35.3). Since 
 IK is parallel to BC (14.2 sup.), by similar triangles, 
 
 As FG : FL : : BG : IL ; 
 
 and 
 
 As HG : HL : : GC : LK ; 
 
124 SECTION IV. 
 
 consequently (23.6), 
 As FG.HG : FL.HL : : BG.GC : IL.LK : : GD 2 : NIA 
 
 Def. 7. The line HG is called the axis of the hyperbola ; 
 HL, FL, as likewise HG, FG, corresponding abscissas ; and 
 DG, NL, the ordinates. 
 
 The property of the hyperbola, proved in this article, there- 
 fore, is this : The rectangles of corresponding abscissas are 
 to each other as the squares of their ordinates. 
 
 From what has been demonstrated in the last six articles, 
 it appears that tkere are five different figures which may be 
 formed by the section of a plane and the surface of a cone, 
 viz., the triangle, circle, parabola, ellipse and hyperbola. 
 The properties of the triangle and circle being explained in 
 common Geometry, the remaining three are usually denomi- 
 nated the Conic Sections. A few of the most useful proper- 
 ties of these figures, deduced from the general relations 
 above demonstrated, are subjoined. 
 
 Of the Parabola. 
 
 Art. 86. Let BAC be a parabola ; AD, part of the axis, an 
 
 DC' 2 
 abscissa ; DC an ordinate ; AE = AF == rrfv EG perpendi- 
 cular, and CG parallel to ED ; then, FC being joined, FC 
 shall be equal to CG. 
 
 Since EF is bisected in A, 
 
 4AF.AD + DF 3 = ED 3 (8.2) = CG 3 . 
 But (47.1), 
 
 FC 3 = DC 3 + DF 3 = (by construction) 4AF.AD + DF 3 . 
 Therefore, CG 3 = CF* ; and CG = CF. Q. E. D. 
 
 Def. 8. The line EG is called the directrix ; 4AE, the latus 
 rectum; and the point F, the focus. 
 
CONIC SECTIONS. 125 
 
 E I P G R 
 
 
 
 A 
 
 
 
 
 TC / 
 
 
 r X 
 
 H 
 
 
 
 
 X. 
 
 \ \ 
 
 
 
 M\ 
 
 
 
 
 H^ 
 
 
 
 
 3 
 
 ) 
 
 
 c 
 
 \J 
 
 The proposition demonstrated in this article, therefore, may 
 be enunciated : If from any point in the parabola, two straight 
 lines be drawn, the one to the focus and the other at right 
 angles to the directrix, they will be equal to each other. 
 
 Art. 87. A right line KFH, drawn through the focus 
 parallel to the directrix, bounded at both ends by the para- 
 bola, is equal to the latus rectum; and the rectangle of the 
 latus rectum and abscissa is_equal to the square of the corre- 
 sponding ordinate. 
 
 It is evident from Art. 83, that KF = FH ; and (Art. 86), 
 FH = HI = 2AF ; therefore, KH = 4AE. 
 
 Draw any other ordinate LM ; then (Art. 83), 
 
 As DC 2 : LM* : : AD : AL :: (1.6) AD.KII : AL.KH. 
 
 But (by construction of Art. 86), 
 
 4AD.AF = AD.KH = DC 2 .-. AL.KH = LM 2 . 
 
 Q. E. D. 
 
 Art. 88. If a point be taken either within or without a 
 parabola, and from it a straight line be drawn to the focus, 
 and another at right angles to the directrix ; the former of 
 
 17 
 
126 SECTION IV. 
 
 these lines will be less or greater than the latter, according 
 as the point is within or without the parabola. 
 
 First, let N be taken within the parabola ; join FN, and 
 produce it till it meets the parabola in C ; let NP and CG be 
 at right angles to the directrix ; and join CP. Then CP is 
 greater than CG (17.1 and 19.1) ; but CN-HNP>CP. (20.1) 
 >CG;andCG = CF = CN + NF.-. CN + NP>CN + NP; 
 and therefore NP>NF. 
 
 Next, let O be without the parabola ; then, a similar con- 
 struction being used, OG > OR ; but OC + CG > OG. Also 
 OC + CG = OC + CF (Art. 86) = OF ; therefore, OF > OR. 
 
 Q. E. D. 
 
 Cor. Hence, a point is either in, within or without a para- 
 bola, according as the line drawn from it to the focus is equal 
 to, less or greater than the perpendicular falling from it upon 
 the directrix. 
 
 Art. 89. Let D be a point in the parabola ; DF the line to 
 the focus ; DB the perpendicular to the directrix ; and DG 
 a line bisecting the angle FDB ; then DG touches the para- 
 bola. 
 
 In DG, take any other point I ; and join IF, IB ; then the 
 angle FDB being bisected, we have (Art. 86) the sides BD, 
 DI, and the contained angle BDI, severally equal to FD,DI, 
 and the contained angle FDI; consequently, BI = FI (4.1). 
 But BI is evidently greater than IL, the perpendicular from 
 I to the directrix (19.1) ; hence the point I is without the 
 parabola (Art. 88, Cor.) ; and therefore DG touches the para- 
 bola. Q. E. D. 
 
 Cor. 1. A right line through the vertex, at right angles to 
 the axis, is a tangent to the parabola. For AM being drawn 
 through the vertex A, at right angles to EF, it is evident that 
 every point in AM, except the point A, is farther from F 
 than from the directrix. 
 
CONIC SECTIONS. 
 Or 
 
 127 
 
 L B 
 
 Cor. 2. If FB be drawn from the focus to the point where 
 the line through D parallel to the axis meets the directrix, it 
 is manifest that FB is bisected and cut at right angles by the 
 tangent DG. For the triangles BDH and FDH are in every 
 respect equal (4.1). 
 
 Art. 90. Let DG touch the parabola in D, and meet the 
 axis in G ; then DF being drawn to the focus, and DN at 
 right angles to the axis, FG = DF, and AG = AN. 
 
 Because DB is parallel to FG, the angle FGD = BDG 
 (29.1) = GDF (Art. 89). Hence FG "m FD (6.1). 
 
 Again, since FD = DB (Art. 86) = EN ; and EA = AF, 
 AG = AN. Q. E. D. 
 
 Def. 9. The line NG is called the subtangent ; the second 
 part of this article, therefore, shows that the subtangent is 
 double the abscissa ; or GN = SAN. 
 
 Def. 10. The line DP, drawn from the curve to the axis, 
 
128 
 
 SECTION IV. 
 
 perpendicular to the tangent, is called the normal ; and NP, 
 the segment of the axis between the ordinate and the normal, 
 is called the subnormal. 
 
 Art. 91. The subnormal is equal to half the latus rectum. 
 
 The angle GDP being a right angle, is equal to DGF + 
 DPF (32.1) ; and GDF = DGF, as shown in the last article; 
 the remainder FDP = DPF ; .-. FP = FD (6.1) fe DB (Art. 
 86) == NE. Taking FN from each, NP = EF = £ the latus 
 rectum. Q. E. D. 
 
 Cor. From this demonstration, we have DF = FP. 
 
 Abt. 92. Let DG touch the 
 parabola in D, and FI be the 
 perpendicular from the focus F 
 to the tangent ; A, the vertex ; 
 then shall FI be a mean pro- 
 portional between DF and 
 FA. 
 
 Join AI, and draw through 
 D the ordinate DN. Then, 
 since the angle DIF = GIF ; 
 and DF = GF (Art. 90) ; DI 
 = IG. Also, AN = AG con- 
 sequently, AI is parallel to DN 
 (2.6) ; and therefore the angles 
 at A are right angles ; where- 
 fore the triangle IAF is similar to GIF or DIF (8.6). Hence, 
 
 As DF : FI : : FI : FA. 
 
 Q. E. D. 
 Cor. As DF : FA : : DF 2 : FI 2 (cor. 2.20.6). 
 
 Art. 93. Let UK touch the parabola in H, and HR be 
 parallel to the axis AM ; from any point Q in the parabola, 
 let QV be drawn parallel to HK ; then QV, produced, will 
 
CONIC SECTIONS. 
 
 129 
 
 meet the parabola in another point ; and the line between its 
 points of section with the curve will be bisected by RH. 
 
 From Q, draw QS at right angles to the directrix ; and 
 from the centre Q, with the distance QS, describe a circle ; 
 this circle will evidently pass through the focus F (Art. 86), 
 and touch the directrix in S (cor. 16.3). Join FR, and let 
 QV cut FR in Y ; then, since HK is at right angles to FR, 
 and also bisects it (Art. 89, cor. 2) ; the angles at Y are right 
 angles ; and the point X, where the circle cuts FY the second 
 time, lies between F and R. Also, FY = YX (3.3). Make 
 RT = RS; draw TP parallel to AM, meeting QV in P ; and 
 through the points F, T, X, describe a circle. Then, since 
 FR cuts the circle FXS, and RS touches it, FR.RX = RS 2 
 (36.3) = RT 2 ; consequently, RT touches the circle FTX in 
 17 
 
130 
 
 SECTION IV. 
 
 T (37.3). Hence the centre of that circle is in TP, which is 
 at right angles to RT (19.3) ; it is also in VY, which bisects 
 FX at right angles (cor. 1.3) ; it is therefore in P; the point 
 P is of course in the parabola (Art. 86). 
 
 Now, PT, HR and QS being parallel, and TR = RS, it 
 follows (2.6) that PV = VQ. Q. E. D. 
 
 Def. 11. Any line OR, parallel to the axis, is called a dia- 
 meter of the parabola ; the point H, where the diameter meets 
 the curve, is called its vertex ; 4HR is called the latus rectum 
 of that diameter ; the line PV or VQ, parallel to the tangent 
 HK, is called an ordinate and HV an abscissa to the diameter 
 OH. 
 
 Art. 94. Let PZ be drawn at right angles to the diameter 
 OR, and PV parallel to the tangent HK ; then PZ 2 =4AF.VH. 
 
CONIC SECTIONS. 131 
 
 Retaining the construction of the last article, let the tan- 
 gent HK meet FR in a, and the diameter TP in c ; draw en 
 parallel to FR, and join Aa. Then, since FR is bisected in 
 a (Art. 89, cor. 2) ; and FX in Y (3.3) ; Ya or nc = ^XR : 
 also, Pc = VH (341). But (as was proved in Art. 92), the 
 triangle FAa is similar to FaK or Yen. Consequently, 
 
 As Fa : FA : : Pc : en ; 
 
 whence FA.Pc (or FA.VH) = Fa.nc (16.6). Hence, 
 
 4FA.VH = FR.RX = TR 2 = PZ 2 
 
 Q. E. D. 
 
 Art. 95. If, from two points P, A, ordinates, PV, AO, be 
 drawn to any diameter OR, the squares of those ordinates 
 shall be to each other in the same ratio as their abscissas ; 
 that is, 
 
 As PV 3 : AO 3 : : VH : OH. 
 
 Draw PZ, AW, at right angles to OR ; then (Art. 94) PZ 3 = 
 4FA.VH; and AW* = 4FA.OH; consequently (1.6), 
 
 PZ 3 : AW a : : VH : OH. 
 
 But the triangles PVZ, AOW, being similar, 
 
 PV 8 : AO 3 : : PZ 3 : AW 3 : : VII : OH. 
 
 Q. K D. 
 
 Art. 96. The square of any ordinate is equal to the rec- 
 tangle of its abscissa, and the latus rectum of the diameter. 
 
 Let PV be an ordinate to the diameter OH; from the 
 vertex of the axis let AO be drawn to the same diameter, 
 parallel to PV ; from H, draw HM at right angles to the 
 axis. Then, since AOHK is a parallelogram, 
 
 OH = AK - (Art. 90) AM = HW .-. OW = 20H. 
 
 Now, 
 
132 SECTION IV. 
 
 AO - AW 3 + OW 2 (47.1) = 4AF.0H + 40H 9 (Art. 94, and 
 cor. 2.8.2) = 4RW.OH + 4WH.OH = 4RH.OH. 
 
 But (Art. 95), 
 
 As AO- : PV 2 :: OH : VH :: (1.6) 4RH.OH : 4RH.VH. 
 
 Hence, PV 2 = 4RH.VH. Q. E. D. 
 
 Art. 97. A double ordinate passing through the focus of a 
 parabola, is equal to the latus rectum of the diameter to 
 which that ordinate is applied. 
 
 Let F be the focus ; VH a diameter ; EG the directrix ; 
 HD a tangent to the parabola ; PFO, the line through the 
 focus parallel to DH. Join FH ; then, PV being parallel to 
 DH, the angle DHG = FVH; and DHF =* HFV (29.1). 
 But DHG = DHF (Art. 89) ; therefore, FVH = HFV; and 
 HV = FH (6.1) - HG (Art. 86). Now, PV = VO (Art. 93), 
 and PV 2 = 4GH.HV (Art. 96) = 4HG 2 . Hence PV = 2HG 
 (cor. 2.8.2). Therefore, PO = 4HG. 
 
 The case of the double ordinate applied to the axis, is 
 proved in Art. 87. Q. E. D. 
 
CONIC SECTIONS. 
 
 133 
 
 H^^ 
 
 
 
 Xi 
 
 
 
 N 
 
 
 
 F 
 
 
 E 
 
 M 
 
 
 
 
 
 G j 
 
 X 
 
 
 
 
 
 1> 
 
 Of the Ellipse. 
 
 Def. 12. Let ACBD be an 
 ellipse; AE, CE, the semi- 
 axes ; and from C, the extre- 
 mity of the less, with a dis- 
 tance equal to the greater, 
 let an arc be described cut- 
 ting AB in F and G ; each 
 of those points, F, G, is 
 called the focus of the ellipse; and the line HI, passing 
 through either focus at right angles to AE, meeting the 
 ellipse in H and I, is called the latus rectum of the ellipse. 
 
 Art. 98. The rectangle of the abscissas AF.FB, into which 
 the axis AB is divided by the focus, is equal to the square of 
 the semi-axis CE. 
 
 Since AB is bisected in E, and divided unequally in F ; 
 
 AF.FB + FE 2 = AE 2 (5.2) = FC 2 = FE 2 + CE 2 (47.1) ; 
 therefore, AF.FB = EC 2 . Q. E. D. 
 
 Cor. Hence, HF is a third proportional to AE, EC. For 
 (Art. 84), 
 
 As AE 2 : EC 2 :: AF.FB (EC 2 ) : FH 2 ; 
 whence AE : EC : : EC : FH. 
 
 Art. 99. If from L, any point in the ellipse, a line LN be 
 drawn at right angles to the second or minor axis CD ; then, 
 
 As EC 2 : EB 2 : : CN.ND : NL 2 . 
 
 Draw LM at right angles to AB; then (Art. 84), 
 
 As EB 2 : EC 2 : : AM.MB (EB 2 — EM 2 ) : ML 2 or EN 2 . 
 
 Therefore (19.5), 
 
 As EB 2 : EC 2 : : EM 2 : EC 2 — EN 2 or CN.ND (5.2) ; 
 
 18 
 
1U 
 
 SECTION IV. 
 
 hence, by inversion, 
 
 As EC 2 : EB 2 : : CN.ND : EM 2 or NL 2 . 
 
 It is therefore manifest that the property demonstrated in 
 Art. 84 is equally true, whichever axis is divided by an 
 ordinate. 
 
 Cor. Since CN.ND is less than CE 2 (27.6), it follows that 
 NL 2 is always less than EB 2 , and consequently NL less than 
 EB. 
 
 Art. 100. If, on either axis of an ellipse, a circle be 
 described; and from any point in the ellipse a perpendi- 
 cular be drawn to that 
 axis, meeting the circle ; it 
 will be, as the axis on 
 which the circle is de- 
 scribed is to the other 
 i 
 
 axis, so is the ordinate to 
 the circle, to the ordinate 
 to the ellipse. 
 
 Let ACBD be the ellipse, 
 and first let the circle be 
 described on the greater axis AB, and let LM be the ordi- 
 nate ; then (Art. 84), 
 
 As EB 2 : EC 2 : : AM.MB : ML 2 . 
 But AM.MB = MP 2 (35.3). Hence (22.6), 
 As EB : EC : : MP : ML. 
 
 Again, let the circle be described on CD, and the ordinate 
 LN meet the circle in I ; then (Art. 99), 
 
 As CE 2 : EB 2 : : CN.ND : NL 2 . 
 
 But CN.ND = NI 2 (35.3) ; therefore (22.6), 
 
 CE : EB : : NI : NL. 
 
 Now, CE : EB : : CD : AB. 
 
 Hence the proposition is manifest. 
 
CONIC SECTIONS. 135 
 
 Cor. Two ordinates, ML, TR, being drawn to the same 
 axis, ML : TR : : MP : TS. 
 
 Art. 101. The sum of the lines FH, GH, drawn from any 
 point in the ellipse to the two foci, is equal to the greatei 
 axis AB. 
 
 Take E the middle of 
 AB,* and through it draw 
 the perpendicular CD ; this 
 will be the less axis. From 
 H draw the ordinate HI, 
 and take EL a fourth pro- 
 portional to EB, EG and EI. 
 Then (22.6), 
 
 As EB 2 : EG 2 :: EI 2 : EL 2 . 
 
 Consequently (19.5), 
 
 EB 2 : EG 2 : : EB 2 — EI 2 : EG 2 — EL 2 (that is, 5.2) : : AI.IB 
 : FL.LG. 
 
 Hence (17.5 and Def. 12), 
 
 EB 2 : EC 2 :: AI.IB : AI.IB — FL.LG. 
 
 Therefore (Art. 84), 
 
 AI.IB — FL.LG = IH 2 . 
 
 Again, BE 2 f EL 2 = 2BE.EL + BL 2 (7.2) ; 
 
 also, GE 2 + EI 2 = 2GE.EI + GI 2 . 
 
 Taking the latter of these equations from the former, and 
 remembering that BE 2 — EI 2 = AI.IB (5.2) ; that GE 2 — EL 2 
 = FL.LG ; and that BE.EL == GE.EI (16.6) ; we have 
 
 AI.IB — FL.LG = BL 2 — IG 2 ; 
 
 that is, IH 2 = BL 2 — IG 2 . Hence, IH 2 + IG 2 = BL 2 , Con- 
 sequently (47.1), GH 2 = BL 2 ; and GH = BL. 
 
 * The middle of the axis is usually called the centre of the ellipse. 
 
136 
 
 SECTION IV. 
 
 Taking AE 2 + EL 2 + 2AE.EL = AL 2 (4.2), and FE 2 + 
 EP + 2FE.EI = FI 2 , and proceeding as before, we have 
 
 IH 2 = AL 2 — FP .-. FP + IH 2 = AL 2 ; 
 
 whence (47.1), FH 2 = AL 2 ; therefore, FH = AL. Conse- 
 quently, 
 
 FH + GH = AL + BL = AB. 
 
 Q. E. D. 
 
 Cor. 1. Since FH = AL,and GH = BL, it is manifest that 
 
 EL = FH — AE = AE — GH. 
 
 Hence, FH — GH ==± 2EL ; or EL = $(FH — GH). 
 
 Cor. 2. If, from a point without the ellipse, two right lines 
 be drawn to the foci, their sum will be greater than the 
 greater axis of the ellipse ; but if a point be taken within the 
 ellipse, the sum of the lines drawn from it to the foci will be 
 less than the greater axis. This is evident, from what is 
 above proved and 21.1. 
 
 Cor. 3. Conversely : A point is either in, without or within 
 an ellipse, according as the sum of the lines drawn from it to 
 the foci is equal to, greater, or less than the greater axis. 
 
 Art. 102. If, from any point P of an ellipse, a straight line 
 PR = AE, half the greater axis, be applied to the less axis 
 
 K JT CD, cutting the greater axis 
 
 in S ; then shall SP = EC, 
 half the less axis. 
 
 From P draw PT at right 
 angles to AB ; then, because 
 of the similar triangles TSP, 
 ESR: 
 
 As PR: PS :: ET:ST; 
 hence (22.6 and 19.5), 
 
CONIC SECTIONS. • 137 
 
 As PR 2 ( = AE 2 ) : PS 2 :: AE 2 — ET 2 : PS 2 — ST 2 :: (5.2 
 
 and 47.1) AT.TB : TP 2 :: (Art. 84) AE 2 : EC 2 ; 
 therefore, PS = EC. Q. E. D. 
 
 The ellipsograph, or instrument for describing an ellipse, 
 is founded upon this property. 
 
 Art. 103. From a point H in an ellipse, two lines HF, 
 HG, being drawn to the foci, and one of them FH produced, 
 the line HV, which bisects the exterior angle, is a tangent to 
 the ellipse. 
 
 Make HW, in FH produced, = HG ; join GW, cutting the 
 bisecting line in V ; in VH, take any point K ; and join KG, 
 KW and KF. Then the triangles GHV, WHV, have GH= 
 WH, and the angle GHV=WHV; hence (4.1), GV=WV, 
 and the angle GVH = WVH. Consequently, GK=WK; 
 and, therefore, 
 
 FK + GK = FK + WK. 
 But FK +WK>FW (20.1), and FW=FH+HG=AB (Art. 
 101); therefore FK + GK>AB; and (Cor. 3, Art. 101) the point 
 K is without the ellipse. This being true of every point in 
 HV, except H, the line HV is a tangent to the ellipse. 
 
 Q. E. D. 
 
 Cor. 1. From this demonstration it is obvious that the line 
 HV, bisecting the exterior angle GHW, also bisects GW, and 
 cuts it at right angles in V. 
 
 Cor. 2. Hence the angles FHK, GHV, which the lines from 
 the foci to the point of contact make with the tangent, are 
 equal. For WHV = KHF (15.1). 
 
 Art. 104. Aright line through the vertex of either axis, 
 parallel to the other axis, is a tangent to the ellipse. 
 
 The line DM parallel to AB, makes the angle GDM=DGF, 
 and MDN = GFD (29.1), and DGF = GFD (5.1) ; conse- 
 quently, DM bisects the angle GDN, and is therefore a tan- 
 gent to the ellipse (Art. 103). 
 
 18 at* 
 
138 
 
 SECTION IV. 
 
 Next, take BY parallel to CD, and through any point X 
 in CD draw XZ parallel to BE, meeting the ellipse in Z ; 
 then XZ is less than EB, (Cor., Art. 99); that is, Z lies between 
 DC and BY ; therefore, the line BY is without the ellipse. 
 
 Art. 105. Let the right 
 lines FW and GW, of which 
 FW=AB, the greater axis, 
 be drawn from the foci to 
 meet in W ; and let VH, bi- 
 secting GW at right angles 
 in V, cut FW in H ; then 
 HV touches the ellipse in 
 H. Join HG ; then it is evi- 
 dent (4.1) that GH =WH, 
 and the angle GHV=WHV; hence the angle GHW is bisect- 
 ed by the line HV ; and since HG = HW, 
 
 FH + HG = FW = AB; 
 
 therefore (Art. 101, Cor. 3), the point H is in the ellipse; 
 and (Art. 103) HV is a tangent to the ellipse. 
 
 Cor. A line which cuts GW at right angles, but does not 
 bisect it, is not a tangent to the ellipse. For this line is 
 parallel to HV, and, if it cuts GW between V and W, it does 
 not meet the ellipse ; but if it falls between G and V, it must 
 cut the ellipse. 
 
 Art. 106. A straight line which meets an ellipse, but does 
 not bisect the exterior angle formed by the lines drawn to 
 the foci, is not a tangent to the ellipse. 
 
 Let HR meet the ellipse in H, but not bisect the exterior 
 angle GHW. On HR let fall the perpendicular GN; and 
 produce it to M, making NM = GN; join FM, meeting HR 
 in I ; join also GI* GH and HM. Then (4.1), GI = IM, GH 
 =HM, and the angle GHN=MHN. Hence the angle GHM 
 
CONIC SECTIONS. 
 
 139 
 
 is bisected by the line HR, but by supposition the angle 
 GHW is not ; consequently, HM does not coincide with 
 HW ; and therefore FHM is a triangle, of which FM is less 
 than FH + HM (20.1). But 
 
 FH + HM = FH + HG = AB (Art. 101). 
 
 Also FM = FI + IG. Therefore, FI + IG are less than AB, 
 and the point I is within the ellipse (Art. 101, Cor. 3). Hence 
 HR is not a tangent. 
 
 Cor. From this and Art. 103, it is evident that the tangent 
 must bisect the exterior angle formed at the point of contact 
 by right lines drawn to the foci ; and that only one right line 
 can touch the ellipse at a given point. 
 
 Art. 107. Let HM touch the ellipse in H, and meet the 
 greater axis AB produced in M ; and from H let HI be drawn 
 at right angles to AB ; then, E being the centre, EI, EB and 
 EM shall be proportionals. 
 
 Take EL a fourth proportional to EB, EG and EI ; this 
 line is half the difference between FH and GH (Art. 101, 
 
140 
 
 SECTION IV. 
 
 Cor. 1). Now, the tangent HM bisects the exterior angle of 
 the triangle FHG (Cor. 2, Art. 103), and, meets the base pro- 
 duced; therefore (A. 6), 
 
 AsFH : HG :: FM : MG; 
 
 consequently (E. 5 and Art. 101), 
 
 As 2EB : 2EL :: FM+MG (2EM) : FG (2EG) ; 
 
 wherefore (15.5), 
 
 AsEB : EL :: EM : EG? 
 
 consequently (1.6), 
 
 As EB 3 : EB.EL : : EM.EI : EG.EI. 
 
 But EB.EL = EG.EI (16.6) ; therefore, EB 3 = EM.EI ; and 
 
 (17.6), EI : EB : : EB : EM. 
 
 Q. E. D. 
 
 Cor. If the tangent MH meet the less axis EC produced 
 in P, and HN be drawn at right angles to EC ; then shall 
 EN, EC and EP be proportionals. 
 
CONIC SECTIONS. 141 
 
 By similar triangles, 
 
 As PE : PN :: EM : NH or EI :: (1.6) EM.EI : EI 2 :: 
 
 (by this article) EB 2 : EP. 
 Hence (D. 5), 
 As PE : EN : : EB 3 : EB 3 — EP : : (5.2, Art. 84 and 16.5) 
 
 EC 3 : EN 2 . 
 
 Again <1.6), As PE : EN : : PE.EN : EN 2 . 
 Consequently, PE.EN = EC*; or EN : EC : : EC : EP. 
 
 Art. 108. Let MHP touch the ellipse in H, and meet the 
 axes in M and P ; then if, on the axes, circles be described, 
 cutting the ordin^tes HI and HN in R and O, the lines MR 
 and PO shall touch the circles. 
 
 Join ER ; then, since ER = EB, we have (Art. 107), 
 
 As EI : ER : : ER : EM. 
 
 Hence (6.6), the triangles EIR, ERM, are similar; conse- 
 quently, ERM is a right angle ; wherefore (cor. 16.3), RM 
 touches the circle. In the same manner it may be proved 
 that OP touches the circle. 
 
 Art. 109. Let PT be a right line touching the ellipse in 
 O ; FP, GT, perpendiculars falling upon it from the foci F 
 and G ; then shall FP.GT = EC 2 , the square of the less semi- 
 axis. 
 
 Join FO, and let FO, 
 GT, produced, meet in W; 
 and join OG, ET. Then, 
 since the angle GOT = 
 WOT;andGTO=WTO; 
 the side OG = OW, and 
 GT=WT(26.1). Hence 
 FW = FO + OG = (Art. 
 101) AB. Now, in the 
 triangles FGW, EGT, FG 
 
142 
 
 SECTION IV. 
 
 = 2EG, GW == 2GT, and the angle at G common ; therefore 
 (6.6), FW = 2ET, or EB = ET. Consequently, the circle 
 described on AB will pass through T. In the same manner 
 it may be proved that it will pass through P. 
 
 Next, produce TE till it meets the circle in S, and join 
 FS ; then, because ES = ET, EF = EG, and the angle SEF 
 == TEG (15.1) ; the side FS must = GT, and the angle EFS 
 = EGT (4.1); consequently, FS is parallel to GT (27.1)- 
 that is, FS is in the same straight line with PF. For GT 
 and FP are at right angles to the same line, and are therefore 
 parallel. 
 
 Now, FP.FS == AF.FB (35.3) = EC 2 (Art. 98) ; 
 
 therefore, FP.GT == EC 2 . 
 
 Q. E. D. 
 
 Art. 110. Let RK be a tangent to the ellipse in H; the 
 li-ne HN a perpendicular to RK, meeting the greater axis in 
 M, and the less in N; HI, HP, ordinates to the axes AB, CD ; 
 then it will be, 
 
 AsEB 2 : EC 2 : : EI : IM; 
 
 and 
 
 As EC 2 : EB 2 : : EP : PN. 
 
 Join FH, GH ; and pro- 
 duce FH to L, making HL 
 = HG ; join GL, and draw 
 EO parallel to HM. Then 
 GL is at right angles to 
 HK (Cor., Art. 103) ; and 
 therefore parallel to HM or 
 OE. Hence EO, which bi- 
 sects FG, also bisects FL 
 (2.6); wherefore FO=EB. 
 Then, HL being = HG, and OL = FO; HO = half the dif- 
 ference of FH and HG. Consequently (Art. 32), 
 
 AsEB : EG :: EI : OH; 
 
CONIC SECTIONS. 143 
 
 also (2.6), 
 
 As FO (EB) : FE (EG) : : OH : EM; 
 
 hence (23.6), 
 
 AsEB 2 : EG 2 :: EI : EM; 
 
 therefore (D. 5), 
 
 As EB 2 : EB 2 — EG 2 : : EI : IM. 
 But EB 2 — EG 2 = EC 2 (see def. 12). Consequently, 
 As EB 2 : EC 2 : : EI : IM. 
 
 Again, fr.om similar triangles, 
 
 As EI (PH) : IM :: NP : HI (PE); 
 and by inversion, 
 
 IM : EI : : PE : PN. 
 Hence, EC 2 : EB 2 : : PE : PN. Q. E. D. 
 
 
 
 
 c 7 
 
 
 
 
 
 
 
 — -yjH\ 
 
 
 M 
 
 II 
 
 
 in 
 
 
 i 
 
 aI ] 
 
 f- 
 
 E 
 
 Xi . 
 
 I Or 
 
 B 
 
 Art. 111. Let 
 
 FK be the ordinate 
 through the focus 
 F; KR a tangent 
 to the ellipse at K 
 cutting the greater 
 axis AB produced 
 in R; arnd RM 
 a perpendicular to 
 AB ; then, if from any point H in the ellipse, HF be drawn 
 to the focus, and HM at right angles to RM, it will be, 
 
 As EA : EF : : HM : HF : : AR : AF. 
 
 From H draw HI, at right angles to AB ; and take EL a 
 fourth proportional to AE, FE and EI. Then (Art. 101), 
 FH = AL ; also (Art. 107), 
 
 As ER : EA : : EA : EF : : (by construction) EI : EL ; 
 
M4 SECTION IV. 
 
 therefore (12.5), 
 
 As ER : EA : : ER + EI : EA + EL. 
 
 But ER + EI = HM ; and EA + EL = FH. From the 
 analogy ER : EA : : EA : EF, we have (19.5), 
 
 ER : EA : : AR : AF. 
 
 Hence, EA : EF : : HM : HF : : AR : AF. 
 
 Q. E. D. 
 
 Def. 13. If we 
 make EN = ER, 
 and draw NP pa- 
 rallel to RM, each 
 of the lines RM, 
 NP, is called a di- 
 rectrix to the el- 
 lipse; and if HP, 
 HG, are drawn, the 
 
 latter to the focus And the former at right angles to the 
 
 directrix, it may be proved as before that 
 
 HP : HG : : BN : BG. 
 Cor. Because ER : EA : : EA : EF; 
 (16.6) EA 2 = ER.EF = EF 2 + EF.FR ; 
 
 therefore, AE 2 — EF 2 = EF.FR; that is, EC 2 = EF.FR. 
 
 M 
 
 
 y< C 
 
 1 i 
 
 T 
 
 
 
 U; 
 
 
 
 ii 
 
 AV 3 
 
 p E 
 
 X. . 
 
 t Cr J'JBj 
 
 i) 
 
 Art. 112. Retaining the construction of the last article, 
 it will be 
 
 FH = 
 
 EC 2 
 
 EG 2 
 
 AE — EF.cos BFH t AE + EF.cos AFH 
 From Art. 28, 
 
 As 1 : cos BFH : : FH : FI = FH.cos BFH. 
 Now, 
 
 MH | RF + FI = RF + FH.cos BFH; 
 
CONIC SECTIONS. 143 
 
 therefore (Art. Ill), 
 
 As AE : FE : : RF + FH.cos BFH : FII ; 
 consequently (16.6), 
 
 FH.AE i RF.FE + FH.FE.cos BFH; 
 whence, 
 FH.AE— FH.FE.cos BFH=RF.FE=(Cor., Art. Ill) EC 2 . 
 
 Consequently, 
 
 EC 2 EC 2 
 
 FH - AE — FE.cos BFH " AE + FE.cos AFH 
 
 For, cos AFH * —cos BFH. Q. E. D 
 
 Cor. If IH be produced to meet the semicircle on AB in 
 T, and ET be joined, then shall 
 
 FH = AE + EF.cos BET = AE — EF.cos AET. 
 By Art. 28, 
 As 1 : cos BET :: ET (=AE) : EI : : EF : EL=EF.cos BET. 
 But 
 FH=AL (Art. 101)=AE + EF.cos BET=AE— EF.cos AET. 
 
 Art. 113. Every diameter* to an ellipse is bisected in the 
 centre. 
 
 Let HI, passing through the centre E, meet the ellipse in 
 H and I ; then is EI = EH. For if it is not, take EP=EH 5 
 and from H, I, P, draw lines to the foci F and G. Then we 
 shall have EH = EP ; EF = EG ; and the angle HEF = 
 PEG (15.1) ; therefore (4.1), FH = PG. In like manner, GH 
 == FP ; therefore, 
 
 FH + HG = FP + PG. 
 
 But, FH + HG = FI + IG (Art. 101) ; 
 
 consequently, FP + PG = FI + IG ; 
 
 which is absurd (21.1). 
 
 * Any right line passing- through the centre, limited at both extremities 
 by the ellipse, is called a diameter. 
 
 19 N 
 
143 
 
 SECTION IV. 
 
 Art. H4. The tangents to an ellipse passing through the 
 extremities of any diameter, are parallel to each other. 
 
 Let HM and IN touch the ellipse in H and I, the extremi- 
 ties of the diameter HI ; and produce FH, GI, the lines from 
 the foci, to K and L ; then the angles GHK, FIL are bisect- 
 ed by the tangents HM, IN (Cor. 1, Art. 106). And since 
 GE, EH are respectively equal to FE,EI (Art. 112), and the 
 angle GEH = FEI (15.1), the angle EHG = EIF, and the 
 side HG = FI (4.1). In the same manner, FHE = GIE; 
 consequently, the whole angle FHG = FIG ; and, therefore, 
 (13.1) GHK = FIL; whence GHM *a FIN. But EHG = 
 EIF; therefore, EHM = EIN; whence (27.1) HM is parallel 
 to IN. 
 
 Cor. Hence, if tangents be drawn through the extremities 
 of any two diameters, they will form a parallelogram. 
 
 Def. 14. If the diameter OT be drawn parallel to the tan- 
 gents through the extremities of IH, then OT is said to be 
 conjugate to IH. 
 
 Art. 115. Let OT, which is conjugate to IH, cut the 
 
 radius vector, or line from the focus to the curve, FH in V; 
 then is HV = AE, half the greater axis. 
 
 Through the other focus G, draw GW parallel to OT, cut- 
 ting FH in W. Then, because OT, and consequently WG, 
 is parallel to the tangent HM, the angle HWG = KHM ; 
 
CONIC SECTIONS. 
 
 147 
 
 and HGW^GHM (29.1) ; therefore (Cor. % Art. 103), HWG 
 =HGW; and consequently HG = HW (6.1). Since EV is 
 parallel to GW, and FE = EG ; therefore (2.6), FV = VW. 
 Hence, VH=FV+HG; and therefore (Art. 101) VH=AE. 
 
 Q. E. D. 
 
 Art. 116. If the diameter OT is parallel to HM, the tan- 
 gent at H, thei#the diameter HI shall be parallel to TN, the 
 tangent at T. 
 
 
 s 
 
 
 Q 
 
 
 
 
 ^-"~c 
 
 
 
 
 //$) 
 
 
 
 
 
 
 \^ 
 
 / K 
 
 Ii \ 
 
 JN ^^ 
 
 
 
 ^\\T 
 
 
 
 ^c 
 
 D 
 
 
 t 
 
 
 
 11 
 
 
 
 
 
 Let HM, TN, meet the greater axis AB produced in M 
 and N ; through H and T draw the ordinates HK, TL, to 
 that axis ; and let them meet the circle described on AB, in 
 Q and P; join QM, QE, PN and PE ; then QM and PN are 
 tangents to the circle (Art. 108). Now, ET being parallel 
 to HM, the angle TEL « HMK (29.1), and ELT - MKH; 
 therefore (4.6), 
 
 As EL : LT :: MK : KH; 
 
 and, alternately, 
 
 EL : MK : : LT : KH : : (Cor., Art. 100) LP : KQ. 
 
 Hence, As EL : LP : : MK : KQ; 
 
 and the angles at L and K are equal ; therefore (6.6), the 
 angle LEP = KMQ ; consequently, EP is parallel to QM 
 
148 
 
 SECTION IV. 
 
 (27.1). But EQM is a right angle (18.3) ; therefore, QEP is 
 also a right angle (29.1) ; and EPN is a right angle (18.3) ; 
 therefore, PN is parallel to EQ (28.1). Consequently, the 
 angle LNP = KEQ (29.1) ; and as NLP = EKQ, we have, 
 
 As LN : EK : : LP : QK : : (Cor., Art. 100) LT : KH. 
 
 Hence, LN : LT : : EK : KH; 
 
 consequently, the triangles LNT, KEH are similar (6.6), and 
 the angle LNT = KEH; therefore (27.1), TN is parallel to 
 EH. Q.E.D. 
 
 Cor. 1. Hence, OT being conjugate to IH, IH is also con- 
 jugate to OT. 
 
 Cor. 2. Hence, also, if through the extremities H and T 
 of two conjugate diameters, ordihates, HK, TL, be drawn to 
 the greater axis, meeting the circle described on that axis in 
 Q and P; the tangent QM to the circle is parallel to the 
 radius EP, and the tangent PN to the radius EQ. 
 
 Art. 117. The sum of the squares of any two semi-conju- 
 gate diameters, is equal to the sum of the squares of the 
 semi-axes. 
 
 
 t""li" 
 
 
 Q 
 
 
 
 
 
 
 
 
 
 
 //£> 
 
 
 
 
 
 
 ^S 
 
 / K 
 
 Ii \ 
 
 IU "^^ 
 
 
 /x 
 
 X>w 
 
 
 
 
 
 \r 
 
 D 
 
 ___^^y/v 
 
 
 
 K 
 
 
 
 
 m 
 
 Let OT, IH, be conjugate diameters ; then, retaining the 
 construction of the last article, the triangle EMH is similar 
 
CONIC SECTIONS. 149 
 
 to NET, because EH is parallel to TN, and HM to ET. 
 The triangle HMK is likewise similar to TEL. Conse- 
 quently, 
 
 As EM : MH :: EN : ET; 
 
 and AsMH : MK :: ET : EL; 
 
 therefore (22.5), 
 
 As EM : MK : : EN : EL : : (Art. 107, and Cor. 2 to 20.6) 
 EB 2 : EL 2 . 
 
 Also, As EM : EK : : EB 2 : EK 2 . 
 
 Consequently (24.5), 
 
 EM : MK + EK : : EB 2 : EL 2 + EK 2 ; 
 
 wherefore, EB 2 = EL 2 + EK 2 . 
 
 If the tangents MH, NT, produced, meet the other axis 
 CD produced, in S and R ; and the ordinates HW and TX 
 be drawn to that axis ; we have, in like manner, the triangles 
 ESH, SHW, respectively similar to RET and ETX ; whence, 
 as before, 
 
 AsES : SW :: ED 2 : EX 2 ; 
 
 and As ES : EW : : EC 2 : EW 2 . 
 
 Consequently, 
 
 EC 2 = EX 2 + EW 2 = LT 2 + KH 2 . ' 
 
 Wherefore, 
 
 EB 2 + EC 2 - EL 2 + LT 2 + EK 2 -f KH 2 = (47.1) ET 2 +EH 2 . 
 
 Q. E. D. 
 
 Cor. Since EK 2 +EL 2 =EB 2 =EQ 2 =EP 2 , it follows (47.1) 
 that EL == KQ, and EK = LP. Hence, 
 
 As EB : EC : : EL : KH : : EK : TL. 
 
 In like manner, 
 
 As EC : EB : : EX : WH : : EW : XT. 
 20 
 
150 
 
 SECTION IV. 
 
 Art. 118. The parallelogram formed by the tangents 
 through the extremities of any two conjugate diameters, is 
 equal to the rectangle of the axes. 
 
 / c 
 
 ^<? 
 
 
 0/ 
 
 r s 
 
 
 
 
 
 ^T 
 
 Q 
 
 D 
 
 Let ET, EH be two conjugate semidiameters ; TM, HM, 
 tangents to the ellipse passing through their extremities ; Q, 
 the intersection of HM with the greater axis produced. 
 Draw HS, TR at right angles to AB, the greater axis ; and 
 EX at right angles to HM. Then (Cor., Art. 117), 
 
 EC : EB :: TR : ES; 
 
 and (Art. 107), EB : EQ : : ES : EB ; 
 
 therefore (22.5), EC : EQ : : TR : EB. 
 
 But in the similar triangles EQX, TER ; 
 
 AsEQ : EX :: ET : TR; 
 
 therefore (23.5), EC : EX : : ET : EB; 
 
 consequently (16.6), 
 
 EB.EC == ET.EX = the parallelogram ETMH. 
 
 Now, this parallelogram being one-fourth of that which cir- 
 cumscribes the ellipse, and EB.EC one-fourth of the rectan- 
 gle of the axis, it is obvious that the circumscribing paral- 
 lelogram is equal to the rectangle AB.CD. 
 
CONIC SECTIONS. 
 
 151 
 
 n 
 
 / c 
 
 o// 
 
 \X 
 
 H 
 
 jfy/'k 
 
 N^N 
 
 1^- — 
 
 p\ 
 
 
 12 
 
 Art. 119. Let OT, IH be conjugate diameters; MN, a 
 tangent at H ; FH, GH, lines from the foci to the point of 
 
 contact ; then shall the rect- 
 angle FH.HG-ET 2 . Draw 
 FM, GN, HP and EX at 
 right angles to MN or OT 
 and let OT cut FH in W 
 thenisWH=AE(Art.H5) 
 and OT being parallel to 
 MN, the angle HWE = 
 WHM (29.1). ButWHM 
 = GHN (Cor. 2, Art. 103); 
 consequently, the triangles FHM, GHN and HWP, being 
 right angled at M, N and P, are similar; therefore (4.6), 
 
 As HP (or EX) : HW (or EB) : : FM : FH : : GN : GH; 
 
 consequently (22.6), 
 
 As EX 2 : EB 2 : : FM.GN : FH.GH :: (Art. 109) EC* : FH.GH. 
 
 But (Art. 118 and 22.6), 
 
 As EX 2 : EB 2 :: EC 2 : ET 2 ; 
 
 therefore, FH.GH - ET 2 . 
 
 Cor. From the similar triangles FHM, GHN ; we have, 
 FH : GH : : FM : GN : : (1.6) FM e : FM.GN or EC 2 . 
 
 Art. 120. Let OT, IH be conjugate diameters; HP, a 
 perpendicular to OT, cutting the greater axis in V; then 
 shall HP.HV == EC 2 , the square of the less semi-axis. 
 
 Let the tangent at H meet the less axis produced in S ; 
 and draw HK, HL at right angles to the axes ; and EX 
 parallel to PH. Then the line ES being parallel to KH, and 
 EX to VH, the angle SEX = VHK (29.1), rfnd the angles at 
 X and K are right ones ; therefore, 
 
 SE : EX (or PH) : : HV : HK or EL. 
 
152 SECTION IV. 
 
 Therefore, 
 
 PH.HV = SE.EL = EC 2 (Cor., Art. 107). 
 
 Art. 121. Let CD, FG be conjugate diameters; DL, a 
 tangent to the ellipse ; HI, a chord parallel to DL, cutting CD 
 in K ; then shall HI be bisected in K ; and 
 
 As DE 2 : EG 2 : : DK.KC : HK 2 . 
 
 Let DL, HI meet the greater axis AB produced in L and 
 V; through H, D, I and G, draw HM, DO, IN and GW at 
 right angles to AB, meeting the circle on AB, in R, P, S and 
 W; join PL, RS, SV, EP and EW; and let RS cut EP in 
 T ; and join TK. Now, since IV is parallel to DL, the angle 
 NVI = OLD (29.1), and VNI = LOD, both being right an- 
 gles; therefore (4.6), 
 
 As VN : NI :: LO : OD. 
 
 But (Cor., Art. 100), 
 
 AsNI : NS :: OD : OP; 
 therefore (22.5), 
 
 As VN : NS :: LO : OP; 
 
 consequently (6.6), the angle NVS = OLP. Hence (28.1), 
 VS is parallel to LP. 
 
CONIC SECTIONS. 153 
 
 Again, from similar triangles, 
 
 As VM : MH :: VN : NI; 
 
 and (Cor., Art. 100), 
 
 MH : MR :: NI : NS; 
 
 therefore (22.5), 
 
 As VM : MR : : VN : NS. 
 
 If, therefore, we suppose VR joined, the angle MVR = NVS 
 (6.6) ; consequently, VS and SR are in the same straight line. 
 Now, as DL touches the ellipse in D, PL touches the circle 
 in P (Art. 108). Hence, EPL is a right angle (18.3) ; and 
 RS, which is parallel to PL, is bisected in T (3.3). Since 
 HV is parallel to DL, and TV to PL ; we have (2.6), 
 
 As EK : KD :: EV i VL : : ET : TP; 
 
 hence TK is parallel to PD (2.6), and consequently to SI and 
 RH; wherefore, 
 
 ST : TR : : IK : KH. 
 
 But ST = TR; therefore, IK = KH. 
 
 Again, since EW is parallel to PL (Cor. 2, Art. 116), and 
 therefore to RV ; and EG to DL or HV ; the angle EVR = 
 VEW, and EVH = VEG ; hence, HVR = GEW. Also, HR 
 being parallel to GW, the angle VHR — EGW ; for each of 
 them is equal to GtiR (29.1). Hence, 
 
 EW : EG : : VR : VH : : RT : HK. 
 
 Now, from similar triangles, 
 
 As ED : EP :: EK : ET; 
 
 and, therefore (22.6), 
 
 As ED 2 : EP 2 : : EK 2 : ET 2 : : (19.5) 
 
 ED 2 — EK 2 : EP 2 — ET 2 :: (5.2 and 47.1) DK.KC : RT 2 . 
 
 But EW 2 or EP 2 : EG 2 : : RT 2 : HK 2 : 
 
 20 
 
154 SECTION IV. 
 
 therefore (22.5), 
 
 As ED 2 : EG 2 : : DK.KC : HE 
 
 Q. E. D. 
 
 Of the Hyperbola. 
 
 Art. 122. Let ABC be a triangle 
 formed by the section of a cone and 
 a plane passing through its axis, at 
 right angles to the plane of its base ; 
 and let a plane at right angles to 
 the plane of the triangle cut the 
 cone in DFE and the opposite 
 cone, made by the extension of the 
 sides of the given cone on the other 
 side of A, in OHR ; then (Art. 85) 
 the curves DFE and OHR form 
 opposite hyperbolas. 
 
 The two branches of either hy- 
 perbola, as likewise the two opposite 
 hyperbolas, are like figures, and 
 equal to each other. 
 Let BDC be the base of the cone ; DE, the intersection of 
 BDC and the plane DFE ; and GFHP, the intersection of the 
 cutting plane and the plane of ABC. 
 
 In GF take any point L, make HP = FL, and through L 
 and P let planes pass parallel to the base of the cone ; then 
 the sections KNI, MRT, of these planes and the conical sur- 
 face, are circles, whose centres are in KI, MT (Art. 80), the 
 intersections of these planes with the plane of ABC, which 
 passes through the axis of the cone. But (18.2 sup.) SLN 
 and OPR are at right angles to the plane of ABC, and there- 
 fore (Def. 1.2 sup.) to KI and MT in that plane; therefore 
 (3.3), SL = LN, and OP = PR. Hence, the two branches 
 of either hyperbola are equal to each other. 
 
CONIC SECTIONS. 155 
 
 Again, since the planes MOT and KSI are parallel, MT is 
 parallel to KI (14.2 sup). Hence the triangles FLI and MFP 
 are similar, as likewise the triangles KHL and PHT ; there- 
 fore, 
 
 AsFL : FP :: LI : MP; 
 
 and AsLH : HP :: KL : PT; 
 
 consequently, 
 
 As FL.LH : FP.HP : : LI.KL : MP.PT : : (35.3) LN 2 : PR 2 . 
 
 But, by construction, FL === HP ; therefore, LH ±= FP, and 
 FL.LH = FP.HP; .-. LN 2 = PR 2 , or LN = PR. Conse- 
 quently, equal ordinates corresponding to equal abscissas, the 
 opposite hyperbolas must be like figures and equal to each 
 other. Q. E. D. 
 
 Art. 123. Let MAN, PBQ (see fig. on page 156) be two 
 hyperbolas, formed by the sections of a cone and a plane, as 
 described in Art. 122 ; AB, the interval between the points 
 where the cutting plane cuts the opposite cones, correspond- 
 ing to HF in the figure on page 154. 
 
 Bisect AB in E ; draw DEC at right angles to AB ; and, 
 drawing the ordinate HI from any point H in the curve, to 
 meet BA produced in I, make ED and EC such that 
 
 AI.IB : IH 2 :: EA 2 : ED 2 or EC 2 ; 
 
 then AB is called the first, and DC the second, axis of the 
 hyperbola. The points A, B, are called the vertices, and E 
 the cenfre, of the hyperbola. 
 
 From this construction and Art. 85, it is obvious that AE 2 
 is to ED 2 as the rectangle of any corresponding abscissas is 
 to the square of their ordinate. 
 
 Art. 124. Def. 15. Join AD ; and make EF, EG, each 
 equal to AD ; then F and G are called the foci of the hyper- 
 
156 
 
 SECTION IV. 
 
 M. 
 
 bolas; and the double ordinate RFS, passing through the 
 focus, is called the latus rectum. 
 
 The latus rectum is a third proportional to the first and 
 second axes. For BA is bisected in E ; therefore, 
 
 BF.FA + AE 2 = EF 2 (6.2) = AD 2 = AE 2 * ED 2 (47.1) ; 
 
 consequently, BF.FA = ED 2 ; hence (Art. 123), 
 
 AE 2 : ED 2 :: ED 2 : RF 2 or FS 2 ; 
 
 wherefore (22.6), 
 
 AE : ED :: ED : RF or FS; 
 and AB : DC : : DC : RS. 
 
 Cor. From this demonstration, it is obvious that 
 ED 2 = AF.FB = BG.GA. 
 
CONIC SECTIONS. 157 
 
 Art. 125. If from any point H in the hyperbola, a perpen- 
 dicular HK be let fall on the second axis, it will be, 
 
 As ED 2 : AE 2 : : ED 2 + EK 2 : HK 2 . 
 
 For (Art. 123), 
 
 As AE 2 : ED 2 : : AI.IB : IH 2 : : (12.5) 
 AE 2 + AI.IB : ED 2 + IH 2 or ED 2 + EK 2 . 
 
 But (6.2), AE 2 + AI.IB = EI 2 . 
 
 Hence, by inversion, 
 
 ED 2 : AE 2 : : ED 2 + EK 2 : EI 2 or HK 2 . 
 
 Q. E. D. 
 
 Art. 126. The difference of two right lines, HG, HF, 
 drawn from any point. H in the hyperbola to the foci, is equal 
 to AB, the first axis. 
 
 Take EL a fourth proportional to EA, EF and Ei ; HI 
 being an ordinate to the point H. Then (22.6), 
 
 AsEA 2 : EF 2 :: EI 2 : EL 2 ; 
 therefore (17.5), 
 AsEA 2 : EF 2 — EA 2 (ED 2 ) :: EI 2 : EL 2 — EI 2 :: (19.5) 
 EI 2 — E A 2 : EL 2 — EI 2 — ED 2 . 
 
 But (6.2), EI 2 — EA» = ALBI; 
 
 and (Art. 123), 
 
 AsEA 2 : ED 2 :: AI.IB : III 2 ; 
 
 therefore, EL 2 — EF — ED 2 =* IH 2 . 
 
 Again, EA 2 + EL 2 == 2AE.EL + AL 2 (7.2) ; 
 
 and EF 2 + EI 2 = 2EF.EI + FI 2 . 
 
 Taking the difference of these equations, and remembering 
 that EF 2 — EA 2 = ED 2 ; and AE.EL = EF.EI, because 
 AE : EF : : EI : EL ; we have, 
 
 EL 2 — EI 2 — ED 2 = AL 2 — FI 2 . 
 21 
 
158 
 
 SECTION IV. 
 
 Hence, AL 2 — FP=IH 2 ; or AL 2 = IH 2 + FP = (47.1) FH 2 ; 
 or AL = FH. 
 
 Further, EL 2 + EB 2 + 2BE.EL = BL 2 (4.2) ; 
 also, EI 2 + EG 2 + 2IE.EG £ IG 2 . 
 
 Taking the difference as before, 
 
 EL 2 — EI 2 — ED 2 = BL 2 — IG 2 . 
 But EL 2 — EI 2 — ED 2 = IH 2 ; 
 
 therefore, BL 2 — IG 2 =IH 2 ; 
 
 and, therefore, 
 
 BL 2 = IG 2 + IH 2 = (47.1) GH 2 ; or BL = GH. 
 Consequently, 
 
 BA=(BL — AL=)(>H — FH. 
 
 Q. E. D. 
 
CONIC SECTIONS. 159 
 
 Art. 127. If, from a point Z within an hyperbola, two 
 right lines, FZ, GZ, be drawn to the foci, the difference of 
 these lines is greater than AB, the first axis ; but if from a 
 point b without the hyperbola, two right lines, bG, bF, be 
 drawn to the foci, the difference of these lines will be less 
 than the axis AB. 
 
 First, let ZG meet the hyperbola in a, and join Fa ; then, 
 since FZ is less than Fa + aZ, the difference between GZ 
 and FZ is greater than between GZ and Fa + aZ ; that is, 
 than Ga — Fa. But (Art. 126) Ga — Fa = AB; therefore, 
 GZ — FZ is greater than AB. 
 
 Next, let Fb meet the hyperbola in d, and suppose Gd 
 joined; then, because Gd is greater than Gb — bd, Gd — dF; 
 that is, AB is greater than Gb — bF, or Gb — bF is less than 
 AB. Q. E. D. 
 
 Cor. Hence a point is either in, within, or without an hy- 
 perbola, according as the difference of the lines drawn from 
 it to the foci is equal to, greater, or less than the first axis. 
 
 Art. 128. If, from any point H (see fig. on page 160) in 
 the hyperbola, a right line HM be drawn bisecting the angle 
 FHG, made by Ikies to the foci F, G, the line HM will be a 
 tangent to the hyperbola. 
 
 Take on HG, the line HL =5 HF ; and take in MH any 
 other point P, and join PL, PF ; then (4.1), FP=LP. Now, 
 since HF == HL, LG must be equal to AB (Art. 126) ; hence, 
 
 AB = PL + LG — FP. 
 But PL + LG are greater than the right line joining P and 
 G ; hence the excess of that line above PF is less than AB ; 
 consequently, the point P is without the hyperbola (Cor., Art. 
 127). And this being true of every point in PM except the 
 point H, that line must be a tangent to the hyperbola. 
 
 Q. E. D. 
 
 Cor. 1. From this we may infer that the tangent must 
 bisect the angle FHG ; for if it was possible to draw a tan- 
 
160 
 
 SECTION IV. 
 
 gent through H which did not bisect the angle, we might 
 have two right lines touching the same curve in the same 
 point, and yet not coinciding with each other. 
 
 Cor. 2. The line ka through the vertex of the hyperbola, 
 at right angles to GF, is a tangent ; for the angles GAa, FAa, 
 are equal. 
 
 Art. 129. Let FN, GI be perpendiculars falling from the 
 foci F, G, upon a tangent HI ; then shall FN.GI = ED 2 , the 
 square of the second semi-axis. 
 
 Take, as in the last article, HL == HF ; join LN, NE ; and 
 produce NE to meet GI in K. Then, since HF = HL ; and 
 the angle FHN = LHN (Cor. 1, Art. 128) ; the angle HNF 
 must be equal to HNL (4.1) : consequently, HNL = a right 
 angle, and therefore FNL is a right line (14.1). Now, in the 
 triangles NFE, LFG, we have the angle at F common, and 
 the sides NF, FE, the halves of LF, FG, respectively; 
 whence (6.6) the angle FNE = FLG, and NE = half LG; 
 
CONIC SECTIONS. 1G1 
 
 consequently (28.1), NK is parallel to LG. But LN is pa- 
 rallel to GK; hence (34.1), LN = GK, and NK = LG = AB 
 (Art. 126). Hence, 
 
 EN = EK = JAB = EA. 
 
 Consequently, a circle described from the centre E, at the 
 distance EA, will pass through N and K ; it will also pass 
 through I, because KIN is a right angle (converse of 31.3). 
 Therefore (cor. 36.3), AG.GB = IG.GK; that is, (Cor., Art. 
 124), ED 2 = FN.GL Q. E. D. 
 
 Art. 130. Let HM touch the hyperbola in H, and meet 
 the first axis AB in M, and HS be an ordinate to that axis ; 
 then it shall be, 
 
 As ES : EA : : EA : EM. 
 
 Take ER a fourth proportional to EA, EF and ES ; then, as 
 proved in Art. 126, AR = FH, and BR = GH. Then, the 
 vertical angle of the triangle GHF being bisected by the line 
 HM (Cor. 1, Art. 128), 
 
 AsGH : HF :: GM : MF (3.6) ; 
 
 hence (E. 5) 
 
 GH+ HF : GH — HF :: GM + MF : GM — MF; 
 
 then, taking the halves of these quantities, 
 
 ER : EA :: EF : EM; 
 
 and, alternately (16.5), 
 
 ER : EF : : EA : EM. 
 
 But ER : EF :: ES : EA; 
 
 therefore (11.5), ES : EA : : EA : EM. Q. E. D. 
 
 Art. 131. Let FO, Go, be the ordinates through the foci ; 
 OT, ot, tangents to the hyperbolas at O and o, cutting the 
 first axis in T and t ; QTU and qtu, perpendiculars to AB ; 
 21 o* 
 
102 
 
 SECTION IV. 
 
 then, taking any point H in the hyperbola, and drawing HQ</ 
 parallel to the axis AB, it will be, 
 
 As FH : HQ :: FA : AT : : GH : Bq :: GA : BT. 
 
 Draw HS at right angles to the axis; and take ER a 
 fourth proportional to EA, EF, ES ; then (Art. 126) AR = 
 FH, and BR = GH. Now, by construction, 
 
 ER : ES : : EF : EA : : (Art. 130) EA : ET : : (19.5) 
 ER — EA : ES — ET :: EF— EA : EA — ET; 
 for the last four terms substituting their equals, 
 
 FH : HQ :: AF : AT. 
 Again (12.5), 
 
 ER : ES :: ER+EA : ES+ET :: EF+EA : EA+ET. 
 Hence, GH : Uq : : FB (GA): BT. 
 
 And these ratios in both cases are the same as EF : EA. 
 
CONIC SECTIONS. 
 
 163 
 
 Each of the lines QU and qu is called the directrix of the 
 hyperbola. 
 
 This article being compared with articles 86 and 111, it is 
 manifest that from any point in either of the three conic 
 sections, two straight lines being drawn, one of them to the 
 focus and the other at right angles to the directrix, they will 
 have to each other a constant ratio. In the parabola, the 
 perpendicular upon the directrix is equal to, in the ellipsis 
 it is greater, and in the hyperbola it is less, than the radius, 
 vector, or line front the focus to the curve. 
 
 Art. 132. Let AB, DC be the axes of the hyperbolas; AH, 
 AI, at right angles to AB, and each equal to half DC ; then, 
 right lines being drawn through E, the middle of AB, and 
 the points H and I, and indefinitely extended, they are called 
 the asymptotes ; the property of which, to be demonstrated 
 hereafter, is, that they continually approach the hyperbola, 
 but do not meet it. 
 
 The asymptotes do not meet the hyperbola. 
 
164 SECTION IV. 
 
 From any point N in the hyperbola, draw NM at right 
 angles to the axis, and let it meet the asymptote in X ; then 
 (from similar triangles and 22.6), 
 
 As EA 2 : AH* : : EM 2 : MK 2 . 
 
 But (Art. 123), 
 
 As EA 2 : ED 2 (AH 2 ) :: AM.MB : MN 2 ; 
 
 and EM 2 is greater than AM.MB (6.2) ; therefore, MK 2 is 
 greater than MN 2 , and MK greater than MN. 
 
 Art. 133. Retaining the construction in the last article, 
 produce KM to meet the asymptote EQ in L; then shall 
 KN.NL = ED 2 . 
 
 As in last article, we have, 
 
 EA 2 : ED 2 : : EM 2 : MK 2 : : AM.MB : MN 2 : : (19.5) 
 EM 2 — AM.MB : MK 2 — MN 2 : : (6.2 and 5.2) AE 2 . KN.NL. 
 Hence, KN.NL = ED 2 = HA.AI. 
 
 Cor. Hence, OPQ being drawn parallel to KL, the rect- 
 angle OP.PQ = KN.NL; and, therefore (16.6), 
 
 As KN : OP : : PQ : NL. 
 
 Art. 134. The asymptotes continually approach to the 
 hyperbola. 
 
 Taking KL and OQ as in the last article, it is evident that, 
 ER being greater than EM, PQ is greater than NL ; but, 
 
 KN : OP :: PQ : NL; 
 
 hence, OP is less than KN. 
 
 Art. 135. Through the vertex A, and any other point N 
 of the hyperbola, let the lines AS, NT be drawn parallel 
 to one of the asymptotes EQ, meeting the other in S and 
 T; then. 
 
 As ES : ET : : TN : SA. 
 
CONIC SECTIONS. 
 
 165 
 
 Draw AW, NV parallel to EO; then the triangles IAW, 
 LNV, are similar ; as are also HAS, KNT ; hence the follow- 
 ing analogies : 
 
 As AW : NV : : AI : NL : : (Art. 133) KN : HA : : TN : AS. 
 
 Therefore, As ES : ET : : TN : SA. 
 
 Cor. If PU be drawn parallel to EQ, then 
 
 EU : ET : : TN : UP. 
 
 Scholium. From the property demonstrated in Art. 135 is 
 deduced a relation between logarithms and the areas con- 
 tained between the hyperbolic curve and its asymptote. Let 
 EA =s ED, and consequently ES = SA, and SEW a right 
 angle. The hyperbola is then called an equilateral or rect- 
 angular one. If in that case we assume ES = 1 ; and of 
 course the square SW also = 1 ; then ET being estimated in 
 units of ES, and the area ASTN in units of SW, it is proved 
 by writers on differentials that ASTN is the logarithm of 
 
 22 
 
166 SECTION IV. 
 
 ET, provided the modulus (Art. 15) m = 1. Hence, those 
 logarithms are termed hyperbolic. 
 
 It is, however, observable, that these hyperbolic areas may 
 be made to express logarithms of other kinds. For, if the 
 relation between the axes is such that, ES being = 1, the 
 area of the parallelogram SW shall be expressed by the 
 modulus, the area of ASTN will be the logarithm of ET, 
 according to the system to which that value of m belongs. 
 But the demonstration of these properties would lead further 
 into the differential and integral calculus, than the design of 
 this work admits. 
 
SECTION V. 
 
 SPHERICAL PROJECTIONS. 
 
 Article 136. The business of Spherical Projections is to 
 represent by lines, drawn or described on a plane given in 
 position, the circles which are described on the surface of a 
 sphere. The lines thus drawn or described on the plane, are 
 called the projections of the circles which they represent ; 
 and are so framed that, to an eye properly located, every 
 circle on the sphere will appear coincident with its repre- 
 sentative. 
 
 Def 1. The plane on which the circles of the sphere are 
 represented, is called the plane of projection ; and the point 
 where the eye is supposed to be located, the projecting point. 
 A right line drawn from the projecting point to any point on 
 the sphere, and extended to meet the plane of projection, is 
 termed a projecting line. 
 
 Def. 2. Every circle on the sphere is called an original 
 circle; and the figure which represents it on the plane of 
 projection, a projected circle. 
 
 Def. 3. In the orthographic and stereographic projections, 
 the plane of projection is supposed to pass through the centre 
 of the sphere. Then the common section of this plane and the 
 spherical surface is a circle, w T hich is called the primitive 
 circle. This circle is evidently a great one (Art. 45) ; and, 
 being both on the sphere and plane of projection, may be 
 considered as an original circle, projected into itself. 
 
 Def. 4. In the orthographic projection, the projecting point 
 is in the axis of the primitive circle ; but so remote, that all the 
 
 (167) 
 
168 
 
 SECTION V. 
 
 projecting lines drawn to the different points of the sphere 
 may be considered as parallel. 
 
 Def 5. In the stereographic projection, the projecting point 
 is at one of the poles of the primitive circle. 
 
 Def. 6. The line of measures of any circle which is to be pro- 
 jected, is the common section of the plane of projection, and 
 another plane which passes through the axes, both of the 
 primitive circle and of the circle to be projected. 
 
 Def. 7. The semitangent of an arc is the tangent of half the 
 arc ; not half the tangent of the arc. 
 
 Of the Orthographic Projection. 
 
 Art. 137. If a right line AB be projected orthographically 
 upon a plane, the projection will be a right line ; and the 
 original line will be to its projection, as radius to the cosine 
 of the inclination of the original to the plane of projection. 
 
 Let EF be the plane 
 of projection, seen 
 edgewise ; Aa, Bb, 
 the projecting lines, 
 through the extremi- 
 
 ties of AB, meeting 
 
 the plane of projection 
 in a, b. Conceive a plane to pass through Aa, Bb ; this plane 
 will include AB, and be at right angles to the plane of pro- 
 jection (def. 4 and 17.2 sup). The common section of these 
 planes will evidently be the projection of AB ; but this sec- 
 tion is a straight line (3.2 sup.) contained between Aa and 
 Bb ; that is, it is the line ab. 
 
 Through A draw AD parallel to ab; then is DAB the 
 inclination of AB to the plane of projection. And (Art. 28) 
 
 radius : cosine DAB : : AB : AD or ab. 
 
 E 
 
SPHERICAL PROJECTIONS. 
 
 169 
 
 Cor. 1. When a line is parallel to the plane of projection, 
 its projection is equal to the original line. 
 
 Cor. 2. When two lines which make an angle with each 
 other are parallel to the plane of projection, their projections 
 make an equal angle with each other. This is obvious from 
 9.2 sup. 
 
 Cor. 3. Any figure which is delineated on a plane parallel 
 to the plane of projection, is projected into a figure similar 
 and equal to itself. 
 
 Art. ,138. A circle whose plane is at right angles to the 
 plane of projection, is projected into a right line equal to its 
 diameter. 
 
 Let EF be the 
 plane of projec- 
 tion, seen edgewise ; 
 ABDC, the circle; 
 Gil, IK, right lines 
 perpendicular to the 
 plane of projection, 
 
 •^ ^*^- : ^ ^ touching the circle 
 
 in A and D ; then the plane GHKI will evidently include the 
 circle ; and its intersection HK with the plane of projection, 
 must be the projection of the circle : but HK = AD, the 
 diameter of the circle. 
 
 Cor. Hence any figure which is delineated on a plane at 
 right angles to the plane of projection, is projected into a 
 right line. 
 
 Art. 139. A circle of the sphere, whose plane is parallel 
 to the plane of projection, is projected into a circle equal 
 to itself, and concentric with the primitive circle. 
 
 For every circle of the sphere which is parallel to the 
 plane of projection, has the same axis with the primitive 
 circle ; and that axis being at right angles to the plane of 
 22 p 
 
 Or 
 
 B 
 
 T 
 
 A 
 
 ( \ 
 
 D 
 
 V / 
 
170 
 
 SECTION V. 
 
 projection, is a projecting line (def. 4). Consequently, the 
 centre of the original circle must be projected into the centre 
 of the primitive. And every radius of that circle is projected 
 into a line equal to itself (Cor. 1, Art. 137). 
 
 Cor. As the radius of any circle on the sphere is the sine 
 of its distance from its own pole ; the radius of a projected 
 circle whose original is parallel to the primitive, is the sine 
 of the distance of that original circle from its pole. 
 
 Art. 140. A circle whose plane is inclined to the plane of 
 the primitive, is projected into an ellipse, whose major axis 
 is equal to the diameter of the original circle, and whose 
 minor axis is to the major, as the cosine of the inclination of 
 the planes is to radius. 
 
 Let AGBH be the ori- 
 ginal circle ; P, its centre ; 
 GH, its diameter parallel 
 to the plane of projection ; 
 AB, the diameter at right 
 angles to GH ; ABba, a 
 plane at right angles both 
 to the plane of projection 
 
 - — if. "" 
 
 g 
 
 and to the plane of AGBH. From any point D in the origi- 
 nal circle, let DE be drawn perpendicular, and DQ parallel 
 to BA ; and let the given circle be projected into the figure 
 
 agbh; the lines Aa, Gg, Bfr, Wi, P/?, Q?, Del, Ee, being the 
 projecting lines, which (def. 4) are necessarily at right angles 
 to the plane of projection. Then, since GH and DE are 
 parallel to the plane of projection, their projections are equal 
 to the lines themselves (Cor. 1, Art. 137) ; that is, gp = GP; 
 ph = PH ; de = (qp=) DE = QP. Hence, 
 
 gq.qh = GQ.QH = (35.3) DQ 2 . 
 
 But Pp, Ee, and Bb, in the plane aABb, being parallel, 
 
 BP : EP :: bp : ep (2.6) ; 
 
SPHERICAL PROJECTIONS. 171 
 
 consequently (22.6), 
 
 BP 2 : EF (or DQ 3 ) :: bp 2 : ef or dq 2 ; 
 
 that is, gp 2 : gq.qh : : bp 2 : dq 2 ; 
 
 wherefore agbh is an ellipse, whose major axis is gh = GH, 
 the diameter of the original circle (Art. 84). Also, gh 
 ( as AB) : ab : : radius : cosine of the inclination of AB to 
 the plane of projection (Art. 137). Now, the plane AGBH 
 and the plane of projection are both at right angles to the 
 plane aABb ; hence, if those planes were extended so as to 
 meet, their common section would be at right angles to the 
 same plane (18.2 sup.), and therefore at right angles to AB. 
 Hence, the inclination of AB to the plane of projection is the 
 inclination of the original circle AGBH to the same plane. 
 
 Cor. 1. The major axis of the ellipse into which a circle is 
 projected, is twice the sine of the arc of a great circle inter- 
 cepted between the original circle and its own pole. 
 
 Cor. 2. The minor axis of the same ellipse, or that axis 
 produced, passes through the centre of the primitive circle. 
 For the plane ABba, being perpendicular to the circle AGBH, 
 and passing through its centre, must pass through its axis ; 
 and that axis passes through the centre of the primitive 
 circle. 
 
 Cor. 3. The distances of the extremities of the minor axis 
 from the centre of the primitive circle, are the sines of the 
 greatest and least distance of the original circle from the 
 pole of the primitive. 
 
SECTION V. 
 
 Of the Stereographic Projection. 
 
 Art. 141. To explain the nature of this projection, let 
 ABED be a circle formed by the section of a spherical sur- 
 face and a plane which 
 passes through the centre 
 of the sphere; this plane 
 is here represented by the 
 plane of the paper; take 
 C the centre of this circle, 
 then C is also the centre 
 of the sphere; draw the 
 diameter ACE; and sup- 
 pose another plane, per- 
 pendicular to AE, to pass through C, the centre of the 
 sphere ; this plane will be perpendicular to the plane ABED 
 (17.2 sup.), and its common section with the spherical surface 
 will be a great circle ; which circle, seen edgewise, may be 
 represented by BD, a diameter to the circle ABED at right 
 angles to AE. If, then, A be taken as the projecting point, 
 the circle represented by BD will be the primitive circle, 
 whose centre is C ; and as the pole E, opposite to the pro- 
 jecting point, is projected in C, the pole of the primitive 
 circle is projected in its centre. 
 
 Art. 142. Any point on the sphere is projected into a point 
 distant from the centre of the primitive, the semitangent of 
 the arc of a great circle intercepted between the given point 
 and the pole of the primitive, opposite the projecting point. 
 
 Retaining the construction of the last article, let F and G 
 be two points to be projected ; join AF, AG ; and let AF, 
 AG, produced if necessary, cut the plane of the primitive in 
 I and H ; these points are evidently the projections of F and 
 G. But IC is the tangent of CAI, and CH the tangent of 
 
SPHERICAL PROJECTIONS. 173 
 
 CAH; that is, CI is the semitangent of ECF (20.3) or of EF ; 
 and CH the semitangent of ECG, or of EG. Q. E. D. 
 
 Cor. Any point in the hemisphere BED, opposite to the 
 projecting point, will be projected within the primitive cir- 
 cle; but a point on the hemisphere BAD, adjacent to the 
 projecting point, will be projected without the primitive. 
 For the distance of any point in the first hemisphere from 
 the pole E, is less than a quadrant ; but in the second it is 
 greater ; and the semitangent of 90° = tangent of 45° = 
 radius. 
 
 Art. 143. Every circle of the sphere which passes through 
 the projecting point, is projected into a right line at right 
 angles to its line of measures.* 
 
 The original circle, being in the same plane as the project- 
 ing point, cannot be projected out of that plane; it will, 
 therefore, be projected into the common section of that plane 
 and the plane of projection : but that section is a right line 
 (3.2 sup.) ; therefore, the projection of the circle is a right 
 line. And as the plane of the circle projected, and the plane 
 of projection, are at right angles to the plane which forms 
 the line of measures (Art. 136, Def. 6), their common section 
 is at right angles to that plane (18.2 sup.); and, therefore, 
 the line of measures is at right angles to the projection of the 
 given circle. 
 
 Cor. 1. A circle which passes through the poles of the 
 primitive, is projected into a right line which passes through 
 the centre of the primitive, at right angles to its line of 
 measures. 
 
 Cor. 2. Every circle which passes through the poles of the 
 primitive is projected into a line of semitangents. This is 
 evident from Art. 142. 
 
 * The projection of a circle which passes through the poles of the primi- 
 tive, is usually called a right ciiaie. £v*> 
 
 23 
 
174 
 
 SECTION V. 
 
 Art. 144. Every circle of the sphere which does not pass 
 through the projecting point, is projected into a. circle. 
 
 Case 1. When the plane of the original circle is parallel to 
 the plane of projection. 
 
 Let A be the projecting point ; 
 E, the opposite pole; BCD, the 
 plane of projection at right angles 
 to AE ; C, the centre of the pri- 
 mitive circle ; FHGI, the circle to 
 be projected; LMNP, its projec- 
 tion. If, now, while the point A 
 remains fixed, we suppose the line 
 AF carried round with a conical 
 motion, so as to describe the circle FHGI ; the common sec^ 
 tion of the conical surface and the plane of projection will 
 be LMNP, the projection of the circle FHGI. But the plane 
 of that section, being parallel to the plane of the base, is a 
 circle (Art. 80). 
 
 Cor. The radius, CN or CL, of the projected circle is the 
 semitangent of EG or EF, the distance of the original circle 
 from the pole opposite to the projecting point. 
 
 Case 2. When the circle to be projected is not parallel to 
 the plane of projection. 
 
 Let A be the projecting 
 point; E, the opposite pole; 
 FHGIj the circle to be project- 
 ed ; LMNP, its projection ; 
 ABED, the common section 
 of the spherical surface and a 
 plane which passes through 
 the axes both of the primitive 
 and of the circle FHGI, and 
 therefore at right angles to both these planes. Then BN, the 
 
SPHERICAL PROJECTIONS. 175 
 
 common section of this plane and the plane of projection, is 
 the line of measures for the circle FHGI (Art. 136, Def. 6). 
 
 Supposing, as before, the line AF to be carried round the 
 circle FHGI, it will describe a conical surface, whose com- 
 mon section with the plane of projection will be LMNP, the 
 projection of FHGI. 
 
 Because the plane ABED passes through the axis of 
 FHGI, it must pass through its centre and the axis of the 
 cone; therefore the line FG, the common section of this plane 
 and the plane of the circle, is a diameter, which is projected 
 into the line LN. 
 
 Draw GK parallel to NB. Then the angle LNA = KGA 
 (29.1) = AFG (21.3), because AK=AG; hence the triangles 
 AFG, ANL, which have the angle at A common, are equi- 
 angular to each other ; and the section LMNP is a subcon- 
 trary section, and therefore (Art. 82) is a circle, whose 
 diameter is LN. 
 
 Cor. The projected pole and centre of the projected circle 
 are both in the line of measures. 
 
 Art. 145. The centre of a projected less circle, at right 
 angles to the primitive, is in the line of measures, distant 
 from the centre of the primitive the secant of the circle's 
 distance from its own pole ; and the radius of the projected 
 circle is the tangent of the same distance. 
 
 Let A (see fig. on p. 176) be the projecting point ; ABED, as 
 before, the common section of the spherical surface and a plane 
 which passes through the centre of the sphere, and is at right 
 angles both to the plane of projection and the plane of the cir- 
 cle to be projected ; BCDN, the plane of projection, seen edge- 
 wise ; C, the centre of the sphere ; FG, the common section of 
 the circle to be projected and the plane ABED ; FG will then 
 represent that circle, seen edgewise. Join AF, AG, CG and 
 EG ; and let AF, AG meet BN in L and N ; these points 
 will then be the projections of F and G ; and, consequently, 
 
176 
 
 SECTION V. 
 E 
 
 the line LN will be the projection of FG, the diameter of the 
 circle to be projected. As the circle to be projected is per- 
 pendicular to the plane of projection, that plane must pass 
 through its poles ; hence the point D, where that plane cuts 
 the circle ABED, is one of the poles ; and, therefore, FD = 
 DG ; also, BN is the line of measures. Draw GP touching 
 the circle ABED in G, and cutting BN in P; then, as proved 
 in Art. 144, the angle ANL = AFG. And since GP touches 
 the circle AGDE, and GA cuts it, the angle PGN = GFA 
 (32.3) = GNP; consequently, PN - PG (6.1). 
 
 Again, since ACD = ECD, being both right angles ; and 
 CAF=CEG; it is plain (26.1) that EG cuts CD in L; then, 
 since CG=CE, the angle CGL=CI^L. Taking these equals 
 from the right angles CGP, LCA; the angle LGP=CLE= 
 PLG; hence, LP=PG. Consequently, PG — the radius of 
 the circle described on the diameter LN; but GP is the 
 tangent, and CP the secant of GD. 
 
 If now we suppose the figure to revolve on BN until the 
 plane of ABED becomes perpendicular to CA, the circle 
 ABED will be the primitive circle ; and the points L, D, P, 
 N, will remain unchanged : consequently, the circle described 
 from the centre P with the radius PL = PG, will be the pro- 
 jected circle proposed. 
 
SPHERICAL PROJECTIONS. 
 
 it- 
 
 Art. 146. Any oblique* great circle will be projected into 
 a circle whose centre is in the line of measures, distant from 
 the centre of the primitive, the tangent of its inclination to 
 the primitive, and the radius of the projected circle is the 
 secant of that inclination. 
 
 Let, as before, A be the projecting point ; ABED, the great 
 circle at right angles to the primitive, and to the circle to be 
 projected ; BN and FG, the common sections of the plane of 
 this circle w r ith the plane of projection, and with the plane 
 of the circle to be projected, respectively. Then BN will 
 represent the plane of projection, and FG the circle to be 
 projected, both seen edgewise ; the line BN will also be the 
 line of measures. Join AF, AG, meeting BN in L and N ; 
 these will be the projections of F and G, and LN the dia- 
 meter of the projected circle. Now, the plane of the primi- 
 tive circle, and of the circle to be projected, being both at 
 right angles to the plane of ABED, their common section, 
 which passes through C, the centre of the sphere (Art. 45), is at 
 right angles to that plane (18.2 sup.) ; hence CB and CF are at 
 right angles to that common section ; consequently, the angle 
 FCB is the inclination of the circle FG to the primitive (def. 
 
 * A circle whose plane makes an oblique angle with the plane of prnjpc 
 tion, is called an oblique circle. 
 
 23 
 
173 
 
 SECTION V. 
 
 4.2 sup). Draw AI, making the angle CAI == FCB. To 
 these equals add CAL = CFL, and LAI=FLB (32.1) = ILA 
 (15.1). Consequently, LI - AI (6.1). 
 
 Again, the angle FAG in a semicircle being a right angle 
 (31.3), is equal to BCF+CAF+CFA (32.1) = LAI + CFA. 
 Hence, IAN = CFA. But ANI^AFC, by subcontrary sec- 
 tion (Art. 144) ; wherefore, ANI= IAN, and AI=IN. Hence 
 the radius of the circle described on LN, that is, the projec- 
 tion of FG, is equal to AI. But AI is the secant, and CI the 
 tangent, of CAI or BCF, the inclination of the circle FG to 
 the primitive. 
 
 If, then, as before, we suppose the figure to revolve on BN 
 until the plane of ABED becomes perpendicular to AC, the 
 circle ABED will be the primitive ; and the circle described 
 from the centre I, with the radius IA or IL, will be the pro- 
 jection proposed. 
 
 Cor. 1. Hence an oblique great circle being projected on 
 the plane of the primitive; and, from the point where the 
 projected circle cuts the primitive, two right lines being 
 drawn to the centre of the primitive and of the projected 
 circle ; the inclination of those lines is the same as the incli- 
 nation of the original circle to the primitive. 
 
SPHERICAL PROJECTIONS. 
 
 179 
 
 Cor. 2. Of all projected great circles, the primitive is the 
 least ; for the secant of any arc is greater than the radius. 
 
 Def. 8. The angle made by two circles, whether on the 
 same or different planes, is the angle made by their tangents 
 passing through the point of intersection. 
 
 When the circles are both great circles, the tangents are 
 at right angles to the diameter of the sphere, passing through 
 the point of intersection, which is the common section of the 
 planes of these circles ; and, consequently, the angle made by 
 the tangents measures the inclination of the planes. 
 
 The definition contained in Art. 45 is therefore but a par- 
 ticular application of the more general one now given. 
 
 Cor. The angle made by the radii (drawn to the point of 
 intersection) of two circles on the same plane, is equal to the 
 angle made by the circles. 
 
 Art. 147. The angle made by two great circles on the 
 surface of the sphere, is equal to the angle made by their 
 representatives on the plane of projection. 
 
 Let A be the place 
 of the eye or projecting 
 point; B, the opposite 
 pole; O, the centre of 
 the sphere ; HCLG, the 
 primitive circle ; I and 
 K, the poles of the pro- 
 posed great circles ; 
 ACIB, a great circle 
 passing through A, B 
 and I ; BKLA, another 
 great circle passing 
 through A, B and K. 
 
 Draw AOB, the axis of the primitive ; and 01, OK, the axes 
 of the proposed circles. Let the plane ACIB cut the plane 
 of projection in the line OC ; and BKLA cut it in OL ; then 
 
180 SECTION V. 
 
 (def. 6), OC and OL will be the lines of measures of the cir- 
 cles, whose poles are I and K respectively. Also (Cor., Art. 
 47), the angles IOB, KOB and IOK are respectively equal to 
 the angles which the proposed circles make, on the surface 
 of the sphere, with the primitive and with each other. In 
 the plane ACIB, suppose the line AM to be drawn parallel 
 to OJ, meeting OC in M. Then, since the angle OAM=BOI 
 (29.1), the angle which the great circle, whose pole is I, 
 makes with the primitive ; and M is in the line of measures 
 of that circle ; it follows (Art. 146) that M is the centre, and 
 MA the radius, of the representative of that circle on the 
 plane of projection. In like manner, suppose AN, in the 
 plane BKLA, drawn parallel to the axis OK, and meeting the 
 line OL in N ; then N will be the centre, and NA the radius, 
 of the circle on the plane of projection which represents the 
 original circle, whose pole is K. Now, since AM is parallel 
 to 01, and AN to OK, it follows (9.2 sup.), that the angle 
 MAN == IOK, the inclination of the proposed circles. 
 
 Lastly, the points M, N, being in the plane of projection, 
 let the triangle MAN revolve on MN, till the point A falls 
 into the same plane, and its position will evidently be the 
 point where the projected circles intersect each other ; and 
 as the radii of those circles make, at the point of intersection, 
 the same angle as the original circles on the surface of the 
 sphere, the projected circles themselves make the same angle 
 (cor., def. 8). Q. E. D. 
 
 Art. 148. If a tangent to an original circle be projected, 
 the projected tangent will be a tangent to the projected cir- 
 cle, provided the original circle does not pass through the 
 projecting point. 
 
 A projected circle is the intersection of the plane of the 
 primitive and a cone, whose vertex is the projecting point, 
 and base the original circle. If a plane be supposed to pass 
 both through the projecting point and a tangent to the origi- 
 nal circle, this plane will evidently touch the surface of the 
 
SPHERICAL PROJECTIONS. 
 
 181 
 
 cone : and the intersection of this plane and the plane of pro- 
 jection will be a tangent to the intersection of the cone and 
 the plane of projection. But the former of these intersections 
 is the projected tangent, and the latter the projected circle. 
 
 Q. E. D. 
 
 Cor. If two original circles have a common tangent, the 
 projections of these circles will have their radii drawn to the 
 point of contact, in the same straight line. 
 
 For the radii of both the projected circles is in a line 
 drawn through the point of contact at right angles to the 
 projected tangent. 
 
 Art. 149. The angle made by any two circles on the sur- 
 face of the sphere, is equal to the angle made by their repre- 
 sentatives on the plane of projection. 
 
 Let DIBC be a great 
 circle of the sphere ; BT, 
 a straight line touching 
 it at the point B ; through 
 BT let another plane 
 pass, cutting the sphere 
 in the circle BGF; this cir- 
 cle is, by Def. 1, Art. 45, a 
 less circle. As BT is a 
 tangent to the great cir- 
 cle, it is a tangent to the 
 sphere, and therefore to 
 the circle BGF. Conse- 
 quently, if these circles are projected, their projections will 
 have their radii, which are drawn to the point of contact, in 
 the same straight line (Cor., Art. 148). If, then, through the 
 point B, another great circle and a less one, having a common 
 tangent, be supposed to pass, these circles, when delineated 
 on the plane of projection, will have their radii, which are 
 drawn to the point of contact, also in the same straight line. 
 
 24 
 
182 SECTION V. 
 
 But (Art. 147) the angle made by two great circles on the 
 surface of the sphere, is equal to the angle made by their 
 representatives on the plane of projection, or by the radii of 
 those representatives drawn to the point of intersection. And 
 (def. 8) the angle made by two circles is the angle made by 
 their tangents passing through the point of intersection; 
 hence it is obvious that the angle made by the two less cir- 
 cles, or by either of them, with a great circle touching the 
 other at the point of intersection, is equal to the angle made 
 by the two great circles. 
 
 Therefore the truth of the proposition is manifest. 
 
 Q. E. D. 
 
 Art. 150. The extremities of the diameter of a projected 
 circle are in the line of measures, distant from the centre of 
 the primitive, the semitangents of the least and greatest dis- 
 tances of the original circle from the pole of the primitive 
 opposite to the projecting point. . 
 
 Let A be the projecting 
 point; ABED, the great cir- 
 cle whose plane is at right 
 angles to the plane of the 
 primitive and of the circle to 
 be projected ; BD the primi- 
 tive, and FG the circle to be 
 projected, both seen edgewise. 
 Then the right line BD, which 
 is the common section of the 
 plane ABED and the plane of the primitive, is the line of 
 measures of the circle FG (Art. 136, Def. 6). As the extre- 
 mities F and G are projected into H and I, the line IH, which 
 is in the line of measures BD, is evidently the diameter of 
 the projected circle. Also, E being the pole opposite to the 
 projecting point, EF and EG are the least and greatest dis- 
 
SPHERICAL PROJECTIONS. 
 
 183 
 
 tances of the circle FG from that pole; and CH, CI are the 
 semitangents of EF, EG (Art. 142). 
 
 Cor. 1. The points where a projected oblique great circle 
 cuts the line of measures, within and without the primitive, 
 are distant from the centre of the primitive the tangent and 
 cotangent of half the complement of the inclination of the 
 original circle to the plane of the primitive. 
 
 The angle BCF is the inclination of the original circle FG 
 to the plane of the primitive (see Art. 146) ; and therefore 
 FCE is the complement of that inclination. But CL is the 
 tangent of LAC = tangent of half FCE (20.3). Also, since 
 FAG is a right angle (31.3), CN, the tangent of CAG, is the 
 cotangent of half FCE. 
 
 Cor. 2. The centre of a projected circle is in the line of 
 measures, distant from the centre of the primitive, half the 
 difference of the semitangents of the greatest and least dis- 
 tance from the pole opposite to the projecting point, when 
 the circle encompasses that pole; but half the sum of the 
 semitangents, when the circle lies wholly on one side of the 
 pole. • 
 
 Art. 151. Any circle and its poles being projected on the 
 plane of the primitive, the segments of the diameter inter- 
 
184 
 
 SECTION V. 
 
 cepted between its extremities and one projected pole, have 
 to each other the same ratio as the segments between the 
 same extremities and the other pole. 
 F E 
 
 Let A be the projecting point ; ABED, as before, the great 
 circle at right angles to the plane of projection and of the 
 circle to be projected ; BD, the line of measures ; FG, the 
 common section of the plane ABED and the plane of the 
 circle to be projected ; P, Q, the poles of the same circle. 
 Then PQ is a diameter to ABED at right angles to FG (Art. 
 45, Cor. to Def. 3) ; and the arc PF is equal to PG. Hence 
 F, P, G, being projected to H, p, I, the angle HAp = IAp 
 (21.1) ; consequently, 
 
 \p : H/> : : IA : HA (3.6). 
 
 Again, producing GA to K, and joining QF, the angle QAK 
 *9 QFG (22.3 and 13.1) = QAF (21.3), because QG = QF. 
 Hence (A. 6), 
 
 lq : qU :: IA : AH; 
 
 consequently, Ip : Up : : lq : ^H. 
 
 Cor. Hence, of the two segments into which the diameter 
 is divided by the projected pole, the greater is that which 
 is more remote from the centre of the primitive circle ; 
 
SPHERICAL PROJECTIONS. 
 
 185 
 
 and therefore the centre of the projected circle is furthei 
 from the centre of the primitive, than the projected pole. 
 
 Art. 152. The projected poles of any circle are in the line 
 of measures, within and without the primitive ; and distant 
 from its centre, the tangent and cotangent of half its inclina- 
 tion to the primitive. 
 
 Retaining the construction of the last article, the angle 
 ECP = the inclination of the primitive to the circle, whose 
 poles are P and Q (Art. 47, Cor.) ; and these poles being in 
 the circle ABED, are projected to p and q in the line of mea- 
 sures BD. But 
 
 Cp = tan CA^ = tan ^ECP (20.3) ; 
 and pAq being a right angle (31.3), 
 
 Cq == cotan Ckp = cotan JECP. 
 
 Cor. The projected pole of the primitive is its centre ; and 
 the projected pole of a right circle lies in the primitive. 
 
 Art. 153. If two planes cut the sphere, and also intersect 
 each other, and from the points where their common section 
 meets the spherical surface, taken as poles, two circles be 
 described at equal distances from those poles ; the arcs of 
 these circles, intercepted between the cutting planes, on the 
 same side of the common section, are equal to each other 
 
 Case 1. When the cutting planes both pass through the 
 centre of the sphere. 
 
 Let ABDPLG, ACEP, be the 
 cutting planes; AP, their com- 
 mon section ; BC, DE, the inter- 
 • cepted arcs ; BF, CF, DH, EH, 
 the common sections of their 
 planes and the cutting planes. 
 It is to be proved that BC=DE. 
 Because the cutting planes pass 
 24 q* 
 
186 
 
 SECTION V. 
 
 through the centre of the sphere, their common section AP is 
 a diameter to each of the circles formed by the spherical 
 surface and the cutting planes. And since A, P, are the 
 poles of the circles BC and DE, the line AP is perpendicular 
 to the planes of those circles (Art. 45, Def. 2) ; therefore, 
 AFB, AFC, PHD, PHE are right angles (def. 1.2 sup.) ; 
 whence FB is parallel to HD, and FC to HE (28.1) ; conse- 
 quently, the angle BFC = DHE (9.2 sup). The lines BF, 
 CF, DH, EH, are also equal, because they are sines of equal 
 arcs ; wherefore, BC = DE (26.3). 
 
 Case 2. When one of the cutting planes passes through 
 the centre of the sphere, and the other does not. 
 
 Let ABDP be the common 
 section of the spherical surface 
 and the cutting plane, which 
 passes through the centre; 
 ACEP, the common section of 
 the same surface and the other 
 plane ; AP, the common sec- 
 tion of the planes; BC and DE, 
 as before, the intercepted arcs 
 of the circles described from A and P ; O, the centre of the 
 sphere, and consequently of ABDP ; BF, DH, the common 
 sections of the planes of the circles BC, DE, and the plane 
 of ABDP ; N, I, the intersections of BF, DH, with the line 
 AP. Join AO, PO, FC, NC, HE, IE, AC, PE; then, as in 
 the first case, BF, CF, DH, EH, being sines of equal arcs, 
 are equal to each other ; and AFB, PHD, right angles : AF, 
 PH, are also equal, being versed sines of equal arcs. Now, 
 in the triangle AOP, the side AO = OP ; wherefore, OAP = 
 OPA; then, in the triangles AFN, PHI, we have AF=PH; 
 the angle FAN=HPI, and AFN- PHI ; whence (26.1), AN 
 = PI, and FN=HI. Then, in the triangles ANC, PIE, we 
 have AC = PE (29.3), AN = PI, and the angle NAC = IPE 
 
SPHERICAL PROJECTIONS. IB* 
 
 (21.3) ; consequently, NC = IE (4.1). Lastly, in the trian- 
 gles CFN, EHI, the sides are respectively equal ; wherefore 
 the angle NFC = IHE (8.1), and consequently the arc BC — 
 DE (26.3). 
 
 Case 3. When neither of the cutting planes passes through 
 the centre of the sphere. 
 
 Through the common section of these planes and the cen- 
 tre of the sphere, let a third plane pass ; then, by the last 
 case, the arcs of one of those equidistant circles, intercepted 
 between the third plane and each of the others, are respect- 
 ively equal to the corresponding arcs of the other similarly 
 intercepted : and, therefore, their sums or differences are also 
 equal. But, when the third plane passes between the other 
 two, the sum of the arcs contained between it and the other 
 planes is the arc in question. When it passes on the same 
 side of them, the difference is the arc proposed. 
 
 Art. 154. Let EFGH, efgh, be the projections of two 
 equal circles, of which EFGH is as far from its pole P as 
 efgh is from the projecting point ; then any two right lines 
 EP, FP, drawn through P, will intercept the representatives 
 of equal arcs of those circles; on the same side, if P falls within 
 the circles ; but on the contrary side, if it falls without ; that 
 is, EF = e/, and GH =£•/?. 
 
188 
 
 SECTION V. 
 
 For (Art. 153) two planes passing through the projecting 
 point and the pole of the original circle, which is represented 
 by EFGH, will cut off equal arcs from those circles. And 
 those planes will (Art. 143) be projected into right lines, 
 which will evidently pass through the projected pole P. 
 
 Cor. 1. Let a circle be project- 
 ed into a right line EF at right 
 angles to the line of measures 
 I EG; and from C, the centre of 
 the primitive, let a circle be de- 
 scribed through P, the projected 
 pole of EF ; then any two lines 
 PE, PF, will cut off from the circle an arc, ef, containing 
 the same number of degrees as the arc which is represented 
 by EF. And the arc, intercepted between PE and PF, of 
 any other circle which passes through P, will contain the 
 same number of degrees. 
 
 For any circle which is projected into a right line, must 
 pass through the projecting point (Arts. 143, 144) ; and, 
 therefore, the distance of that circle from its pole is the same 
 as the distance of the pole from the projecting point. Con- 
 sequently, the projected circle through P represents an origi- 
 nal circle, as far from the projecting point as the circle which 
 is projected into EF is from its own pole. Hence, EF and 
 ef represent equal arcs. The latter part of the corollary is 
 evident from 26.3. 
 
 Cor. 2. If two right lines be drawn through the projected 
 
SPHERICAL PROJECTIONS. 189 
 
 pole of a great circle, the intercepted arc of that circle will 
 contain the same number of degrees as the intercepted arc 
 of the primitive. 
 
 For any great circle is distant 90° from its pole ; and the 
 primitive is 90° from the projecting point. 
 
 Cor. 3. If, from the point where two projected great cir- 
 cles cut each other, two right lines are drawn through the 
 projected poles of those circles, the intercepted arc of the 
 primitive circle will measure the spherical angle made by 
 those circles at the point of their inter sec ti<*i. 
 
 For the arc of a great circle, contained between the poles 
 of two other great circles, is the measure of the angle which 
 the axes of these circles make with each other ; and that 
 angle is the same as the inclination of the planes of those 
 circles (Cor., Art. 47). 
 
 Scholium. If the circles of the sphere were to be projected 
 on a plane parallel to the plane of the primitive, the projec- 
 tions would be similar to those on the plane of the primitive 
 itself; for the projecting line, when carried round on the cir- 
 cumference of a circle which does not pass through the 
 projecting point, forms a conical surface ; and that surface 
 being cut by the plane of the primitive, and by any other 
 p-lane parallel thereto, the sections are similar, but of greater 
 or less dimensions. Thus projecting on the plane of a less 
 circle parallel to the primitive, instead of projecting on the 
 primitive itself, would be only changing the scale. In the 
 subsequent parts, however, of this section, the plane of the 
 primitive will be used. 
 
 In the following problems, the primitive circle is always 
 supposed to be described with the chord of 60° ; and the 
 secant, tangent, and semitangent referred to, are such as 
 correspond to the scale used for the primitive. These differ- 
 ent lines are frequently marked by the side of the scale of 
 chords, on the small scales introduced into boxes of mathe- 
 
 25 
 
190 
 
 SECTION V. 
 
 matical instruments ; but more frequently on the foot or two 
 feet scales, which contain Gunter's lines. 
 
 Art. 155. Problem. To describe a circle parallel to the 
 primitive, at a given number of degrees from its pole. 
 
 E 
 
 From the centre of the 
 primitive, with a radius equal 
 to the semitangent of the 
 given distance of the circle 
 from its own pole, describe 
 the circle required. Or draw 
 the diameters AB, DE at 
 right angles to each other; 
 and from the extremity E 
 of one of them, lay off the 
 proposed number of degrees on the primitive as EF ; join 
 DF, cutting AB in G; from the centre C, at the distance 
 CG, describe the circle required. 
 
 The radius CG is the tangent of CDG, or semitangent of 
 EF, as it ought to be (Cor., Case 1, Art. 144). 
 
SPHERICAL PROJECTIONS. 
 
 191 
 
 Art. 156. Prob. To describe a less circle at right angles 
 to the primitive, and at a given distance from its own pole. 
 
 Let B be the pole of the circle proposed ; through C, the 
 centre of the primitive, and the pole B, draw the right line 
 CD ; from C to D, lay down the secant of the given distance ; 
 from the centre D, with a radius equal to the tangent of the 
 same distance, describe the circle proposed. Or, from B, lay- 
 down BE on the primitive ; join CE ; draw ED touching the 
 circle; from D, with the distance DE, describe the circle 
 proposed. 
 
 It is obvious that DE is the tangent, and CD the secant, 
 of BE, as they ought to be, Art. 145. 
 
 Art. 157. Trob. To describe an oblique circle at a given 
 distance from a given pole. 
 
 hetp be the projected 
 pole ; through p draw 
 the line of measures 
 ApB ; apply C/>, the dis- 
 tance of the given pole 
 from the centre of the 
 primitive, to the line 
 Qf semitangents ; and, 
 having found the num- 
 ber of degrees, thus 
 measured, in Cp, take the sum and difference of this number 
 and the distance of the proposed circle from its own pole ; 
 lay down these results taken from the semitangents, on the 
 line of measures, from C to g and/; on the diameter fg de- 
 scribe the circle required. Or, having drawn the line of 
 measures, draw the diameter DCE at right angles to it; 
 draw Dp to meet the primitive in P ; from P, lay down on 
 the primitive PF, PG, each equal to the given distance of the 
 circle from its pole ; draw DF, DG, cutting the line of mea- 
 sures in /and g; on fg describe a circle ; it will be the circle 
 required. 
 
192 SECTION V. 
 
 This construction follows from Art. 150. 
 
 Scholium. This method is applicable to great circles as 
 well as less ; but the former cases are conveniently managed 
 by other methods hereafter given. 
 
 Art. 158. Prob. To describe a great circle, the projected 
 pole of which is given in position. 
 
 Case 1. When the given pole is in the primitive circle. 
 
 Through the given pole draw the line of measures ; and at 
 right angles thereto draw a diameter to the primitive circle ; 
 this diameter will represent the circle proposed. 
 
 Because the pole is in the primitive, the original circle is 
 at right angles to the primitive (Art. 45, Cor.. Def. 3) ; and, 
 being a great circle, it must pass through the poles of the 
 primitive. Consequently (Art. 143, Cor. 1), it is represented 
 by a right line through the centre of the primitive, at right 
 angles to the line of measures. 
 
 Case 2. When 
 circle. 
 
 the given pole is within the primitive 
 
 Let p be the projected pole ; 
 through p draw the line of 
 measures CpG; apply Cp to 
 the line of semitangents ; take 
 CG equal to the tangent of the 
 number of degrees in Cp; 
 from G, as a centre, with the 
 seGant of the same number 
 of degrees, describe the circle 
 D DHE, which will be the one 
 
 required. 
 
 Or, draw the diameter DCE at right angles to the line of 
 measures ; join Dp, and produce it to the circumference in 
 P ; make PF = PE ; draw DF cutting the line of measures 
 in G ; from the centre G, with the radius GD, describe the 
 
SPHERICAL PROJECTIONS. 
 
 193 
 
 circle DHE, the circle proposed. The line Cp is the projec- 
 tion of an arc of a great circle, intercepted between the pole 
 of the primitive and the pole of the circle proposed ; and that 
 arc measures the inclination of those circles (Art. 47, Cor). 
 The arc is also projected into a line of semitangents (Art. 
 143, Cor. 2) ; hence Cp, measured on the semitangents, or the 
 arc PE, indicates the same inclination. But, by the con- 
 struction, CG is the tangent and GD the secant of PE ; con- 
 sequently (Art. 146), DHE is the circle, whose pole is j>. 
 
 Art. 159. Prob. Through two given points, to describe a 
 great circle. 
 
 Let A, B be the 
 given points ; through 
 the centre of the pri- 
 mitive and one of the 
 given points draw the 
 right line ACF. If 
 that line passes through 
 B, the business is done ; 
 for ACF is the projec- 
 tion of a great circle 
 at right angles to the primitive (Art. 143, Cor. 1). But if 
 ACF does not pass through B, apply CA to the line of semi- 
 tangents, and make CF the semitangent of the supplement 
 of CA ; through ABF describe a circle, and the thing is done. 
 Or, draw the diameter DCE at right angles to ACF ; draw 
 DAG cutting the primitive in G ; draw the diameter GCH ; 
 join DH ; and let DH, produced if necessary, cut AF in F ; 
 and through ABF describe the circle ABF, as before. 
 
 From the construction, it is obvious that AF is the projec- 
 tion of a semicircle ; consequently, any circle which passes 
 through A and F must be a great one (converse to Cor. 3, 
 Def. 1, Art. 45). 
 25 
 
194 
 
 SECTION V. 
 
 Art. 160. Prob. About a given pole, to describe a circle 
 through a given point. 
 
 Let p be the given pole 
 and B the given point ; 
 through p and B describe 
 a great circle (Art. 159) ; 
 and draw BF touching it 
 in B (17.3) ; draw, from 
 the centre of the primi- 
 tive, the right line C/)F, 
 meeting the tangent in F ; 
 from the centre F, at the 
 distance FB, describe the circle BGH ; and the work is done. 
 The centre of a circle whose pole is p is in the line of mea- 
 sures QoF (Cor. to Case 2, Art. 144). The circle />B, passing 
 through the pole p of the proposed circle, is at right angles 
 to it (Cor. to Def. 3, Art. 45) ; hence the radii of their pro- 
 jections, drawn to the point of their intersection, must also 
 be at right angles to each other (Art. 149) ; consequently, the 
 centre of the required circle is in the tangent BF (18.3) ; it 
 is, therefore, at the intersection of Cp and BF. 
 
 Art. 161. Prob. To find the poles of a given projected 
 circle FNG. 
 
 E 
 
 Through the centre of the given circle and centre C of the 
 primitive, draw the right line FCGP, cutting the given circle 
 
SPHERICAL PROJECTIONS. 
 
 135 
 
 in F and G. Measure CF and CG on the line of semitan- 
 gents ; take the half sum or half difference of these measures, 
 according as F and G are on the same or opposite sides of C, 
 and lay its semitangent from C to p ; then is p one of the 
 poles required : observing, however, that p must be on the 
 same side of C as the centre of the given circle. Lay down 
 CP from the semitangents equal to the supplement of Cp, and 
 on the opposite side of C ; then is P the other pole required. 
 Or, draw the diameter ACE at right angles to FG ; draw 
 also AG, AF, cutting the primitive in L and M ; bisect LM 
 in I ; join AI, cutting FG in p ; draw the diameter IH to the 
 primitive circle ; draw AH, produced if necessary, to meet 
 FG produced in P ; then p and P are the poles required. 
 
 When the circle is a great one, as AGE, the poles may be 
 found with more facility in a different manner. 
 
 Draw FC from the centre of the given circle to the centre 
 of the primitive, and produce it. Measure CF on the line 
 of tangents: take half the number of degrees thus found, and 
 lay down the tangent of this result from C towards F to p ; 
 and its cotangent in the opposite direction to P ; then will p 
 and P be the poles required. Or, draw the diameter ACE 
 at right angles to FC ; join AF ; bisect the angle CAF by the 
 line Apl, cutting FC in p and the primitive circle in I ; draw 
 the diameter IH and the line AHP as before ; then p and P 
 are the poles required. 
 
196 
 
 SECTION V. 
 E 
 
 In the case of the less circle, CG is the semitangent of 
 EL, and CF the semitangent of EM; consequently (i\.rt. 
 150) EL and EM are equal to the least and greatest dis- 
 tances of the original circle represented by FNG, from the 
 pole of the primitive opposite the projecting point. Hence 
 EI, half the difference of EM and EL, when F and G are on 
 opposite sides of C (as in the figure), or half the sum when 
 F and G are on the same side, is the distance of the pole of 
 the original circle from the pole of the primitive. Cp, the 
 semitangent of EI, is therefore the distance of the projected 
 pole of FNG from the centre of the primitive (Art. 142). 
 And the projected pole lies in the line which joins the centre 
 of the projected circle and the centre of the primitive (Art. 
 144, Cor. to Case 2). Also, p and P are on the same project- 
 ed great circle, at the distance of 180°; hence, P is the pole 
 opposite to p. 
 
 In the case of the great circle, CF is the tangent of CAF, 
 the inclination of AGE to the primitive (Art. 146, Cor. 1). 
 But, by the construction, ECI = CAF ; consequently, the arc 
 EI measures the distance of the pole of the original circle, 
 represented by AGE, from the pole of the primitive (Cor. to 
 Art. 47). Hence Cp, the semitangent of EI, is equal to the 
 distance of the projected pole of AGE from the centre of the 
 primitive (Art. 142). 
 
SPHERICAL PROJECTIONS. 
 
 E 
 
 197 
 
 Art. 162. Prob. To describe a great circle making a given 
 angle with the primitive at a given point A. 
 
 Draw through the given point the diameter ACE ; and 
 through C, the centre of the primitive circle, draw a line CF 
 at right angles to AE ; make the angle CAF = the angle 
 proposed ; and from F, the intersection of AF, and CF, de- 
 scribe the circle AGE ; which will be the circle required. 
 Or, from the centre of the primitive, with the tangent of the 
 given angle, describe an arc; from the point A, with the 
 secant of the same angle, describe an arc cutting the former • 
 and from the point of intersection describe, through the point 
 A, the circle AGE. 
 
 This construction is obvious from Art. 146. 
 
 When the given angle is a right one, the lines AF and CF 
 are parallel ; hence, in that case, the centre of the required 
 circle is at an infinite distance ; consequently, the circle be- 
 comes a right line passing through the centre of the primitive. 
 See Cor. 1, Art. 143. 
 
 Art. 163. Prob. Through a given point P, to describe a 
 great circle, making a given angle with the primitive. 
 
 From the centre of the primitive, with the tangent of the 
 given angle, describe an arc; from the given point, with the 
 secant of the same angle, describe an arc cutting the former 
 in F ; from the centre F, through P, describe the circle APB ; 
 this will be the circle required. 
 
 26 
 
198 
 
 SECTION V. 
 
 Or, through P draw the 
 diameter DE ; at the cen- 
 tre C erect a perpendicular 
 CG ; make the angle CDG 
 = the given angle; from 
 C, through G, describe an 
 arc ; and from P, with the 
 distance DG, describe an- 
 other, cutting the former 
 in F ; then F is the centre 
 of the required circle. 
 
 This, like the last, de- 
 pends upon Art. 146. 
 
 N. B. If the circles described from C and P do not meet, 
 the problem is then impossible ; and this case occurs when 
 the required angle is less than that which would be measured 
 by PD, taken on the scale of semitangents from 90° towards 
 the beginning of the scale. 
 
 Art. 164. Prob. To describe a great circle, making at a 
 given point P a given angle with a given great circle APB. 
 
 Through the given point P draw the diameter DE, meeting 
 the circle again in H; find F the centre of the given circle; 
 draw FI at right angles to DE ; join PF ; make the angle 
 FPL = the given angle ; then, from L, describe the circle 
 MPN ; this is the circle required. 
 
 The line FI, being at right angles to PH, bisects it (3.3) ; 
 consequently, every circle whose centre is in FI, and which 
 passes through P, will also pass through H. Now, as APB 
 and DCE are the projections of great circles, PBH is the 
 projection of a semicircle (Cor. to Def. 3, Art. 45) ; hence, 
 any other circle passing through P and H must be a great 
 circle. MPN is therefore a great circle ; also (Art. 147) the 
 angle BPN = FPL. 
 
SPHERICAL PROJECTIONS. 199 
 
 Art. 165. Prob. Through a given point P, to describe a great 
 circle making a given angle with a given great circle DE. 
 
 Find the pole of the given 
 circle (Art. 161); about that 
 pole describe a circle GFH 
 at a distance equal to the 
 measure of the given angle 
 (Arts. 155-6-7); about the 
 point P as a pole, describe a 
 great circle IFK, cutting GFH 
 in F and K (Art. 158) ; about 
 D one of those points F as a 
 
 pole, describe the great circle LPM; and the work is done. 
 
 Since P is the pole of IFK, every point in that circle is 90° 
 from P ; consequently, the great circle whose pole F is in 
 IFK, must pass through P. And as the distance of the poles 
 of two great circles is the measure of the angle which those 
 circles make with each other, the construction is manifest. 
 
 If the given angle is a right one, the circle must be de- 
 scribed through P and the pole of DE, by Art. 159. 
 
 Art. 166. Prob. To describe a great circle making given 
 angles with two given great circles AB and CD. See fig. on 
 page 200. 
 
 Find the poles r, s of the given circles (Art. 161) ; describe 
 about r and ft less circles, at distances respectively equal to 
 the measures of the given angles (Art. 157) ; from the inter- 
 section E of these circles, as a pole, describe the great circle 
 FG (Art. 158) ; that circle is the one required. 
 
 This construction evidently depends upon the principle, 
 that the distance between the poles of two great circles is 
 the measure of their inclination (Cor. to Art. 47). 
 
 If the circle to be described is to be at right angles to each 
 of those which are given, it must be described through their 
 poles, by Art. 159. 
 
in 
 
 SECTION V. 
 
 Art. 167. Prob. To describe a right circle (that is, a great 
 circle at right angles to the primitive) making a given angle 
 with a given great circle CD. 
 
 Find s, the pole of CD ; about s describe a circle at a dis- 
 tance equal to the measure of the given angle ; from the point 
 H, where this circle cuts the primitive, lay down HI, on the 
 primitive, = 90° ; through I draw the diameter IL, the right 
 circle required. 
 
 Every great circle at right angles to the primitive is pro- 
 jected into a right line through its centre (Art. 143, Cor. 1). 
 
 [Note. There is a limit in this and the last article. If, in 
 Art. 166, the circles about r and s do not meet ; or, in this 
 article, if the circle about s does not meet the primitive ; the 
 problem is impossible.] 
 
 Scholium. When the proposed angle is a right one, lay down 
 90' from C on the primitive circle ; and through the point 
 thus found draw a diameter for the right circle required. 
 
 Art. 168. Prob. Through a given point Z, to describe a 
 great circle which shall touch a given less circle ABC. 
 From Z, as a pole, describe the great circle DGE (Art 
 
SPHERICAL PROJECTIONS. 
 
 201 
 
 158) ; find P the internal pole of ABC (Art. 161), and Q the 
 
 opposite pole ; about Q de- 
 scribe a circle FGH, at a dis- 
 tance from Q equal to the 
 complement of PB, the dis- 
 tance of ABC from its own 
 pole (Arts. 155-6-7); from 
 the point G, where these cir- 
 cles cut each other, taken as 
 a pole, describe the great cir- 
 cle ZIL (Art. 158) ; this cir- 
 cle will touch the given circle 
 ABC. Through PGQ de- 
 scribe a circle cutting ABC in I ; this will be a great circle, 
 because P and Q are opposite poles, and PI 4- QG are by- 
 construction = 90° ; hence GI === 90° : and G being the pole 
 of ZIL, that circle must pass through I, and make the angle 
 ZIP a right angle. Hence (Art. 55) PI is less than any other 
 arc of a great circle contained between P and ZIL ; there- 
 fore, those circles touch each other at I. 
 
 Art. 169. Prob. To lay down a given number of degrees 
 on a given great circle, or to measure an arc of it. 
 
 Case 1. When the given circle is the primitive, lay down 
 or measure the arc by the scale of chords. 
 
 For the primitive is an original circle, and is therefore 
 measured as in common Geometry. 
 
 Case 2. When the given circle is a right one, that is, one 
 passing through the centre of the primitive, lay down or 
 measure the arc on the scale of semitangents (Art. 143, Cor. 
 2), observing that an arc beginning at the centre of the pri- 
 mitive must be measured from the beginning of the scale ; 
 but one beginning at the primitive must be measured from 
 the 90th degree on the scale towards the beginning or end 
 of the scale, according as the arc extends towards or from 
 the centre of the primitive. 
 26 
 
202 
 
 SECTION V. 
 
 Or, let ACB be the right cir- 
 cle; draw the diameter DCE 
 at right angles to AB ; then, to 
 lay down any proposed number 
 of degrees from A or C, lay 
 them on the primitive from A 
 or E to F ; join DF, cutting 
 AB in G ; or, to measure AG 
 or CG, draw DG to cut the 
 primitive in F ; then AF is the 
 measure of AG, and EF of CG. 
 
 Suppose the figure to revolve on AB till CD becomes per- 
 pendicular to the plane of projection; then is D the project- 
 ing point, and AEB the semicircle passing through the pole 
 of the primitive ; consequently, AG is the projection of AF, 
 and CG of EF. 
 
 Case 3. When the given circle is an oblique one. 
 
 Let DUE be the circle ; find its internal pole I (Art. 161) ; 
 then, to lay down an arc HL or EL, lay the proposed num- 
 ber of degrees on the primitive from A or E to F, and join IF, 
 cutting the given circle in L, the point required ; or, to mea- 
 sure HL or EL, join IL, and produce it to F in the primitive ; 
 then AF is the measure of HL, and EF of EL. 
 
 The primitive circle is as far from the projecting point, as 
 DHE is from its pole ; therefore (Art. 154), the right lines 
 IA, IF, cut off corresponding arcs AF, HL. 
 
 Art. 170. Prob. To lay down any proposed number of 
 degrees on a less circle, or to measure a given arc of it. 
 
 Case 1. When the given less circle is parallel to the 
 primitive. 
 
 Lay the proposed number of degrees on the primitive 
 circle; and through the extremities of the arc draw right 
 lines to the centre ; the intercepted arc of the less circle is 
 
SPHERICAL PROJECTIONS. 
 
 203 
 
 that proposed ; or, to measure the arc, draw right lines from 
 
 the centre, through its extremi- 
 ties, to the primitive, and mea- 
 sure the intercepted arc of the 
 latter. Thus BH is the measure 
 of IL, and EH of ML ; for the 
 B projected less circle parallel to 
 the primitive is formed by cut- 
 ting the conical surface by a 
 plane parallel to the base ; con- 
 sequently, the projected circle 
 differs from its original in nothing 
 but its dimensions. 
 
 Case 2. When the circle is not parallel to the primitive. 
 
 Let ABC be the less cir- 
 cle; find its pole D (Art. 
 161) ; describe a circle 
 FGH, as far from the pro- 
 jecting point as ABC is 
 from its pole (Art. 155) ; 
 then any arc of ABC may 
 be laid down or measured 
 by the aid of FGH as an 
 arc of a great circle is, by 
 means of the primitive in Art. 169, Case 3 ; observing that 
 an arc of FGH is laid down or measured as directed in Case 
 1. Thus, I being the centre of the primitive, ML, a part of 
 its circumference, is the measure of FG ; and FG the mea- 
 sure of AE (Art. 154). 
 
 Art. 171. To measure the angle made by two great circles 
 whose position is given. 
 
 Find the poles of the circles (Art. 161) ; from the angular 
 point, through those poles, draw two right lines; and the 
 intercepted arc of the primitive is the measure required. 
 
204 
 
 SECTION V. 
 
 Let ACB, ECF, be the circles ; 
 G, I, their poles ; then the arc of 
 a great circle, contained between 
 I and G, would measure the angle 
 ACF (Art. 154, Cor. 3). Or, draw 
 lines from the point of intersection 
 C to the centres of the circles ACB 
 and ECF; then the angle con- 
 tained between these lines will 
 measure the angle ACF (Cor., 
 Def.8). 
 
 Art. 172. Prob. To form a general projection of the sphere 
 on the plane of a meridian.* 
 
 Let ZONH denote the me- 
 ridian ; Z, the zenith ; N, the 
 nadir; P, S, the north and 
 south poles; EQ, the equator; 
 HO, the horizon ; then ZE = 
 OP, the latitude of the place. 
 Then, circles being described 
 through P and S, making suc- 
 cessively angles of 15°, 30°, 
 45°, 60°, 75° and 90°, with 
 the primitive; these will be 
 the meridians, or hour circles, for the different hours. 
 
 A few examples will now be given to exercise the student 
 in Spherical Projections and Calculations. 
 
 Of Rectangular Spherical Triangles. 
 
 1. Given, the hypothenuse 70° 15', and the adjacent angle 
 30° 30', to find the rest. 
 
 * See Definitions, page 217. 
 
SPHERICAL PROJECTIONS. 
 
 205 
 
 Construction. Describe 
 the primitive circle ABC, 
 and the oblique great 
 circle ADC, making the 
 given angle with the pri- 
 mitive at the point A 
 (Art. 162); on AD lay 
 AE equal to the given 
 hypothenuse (Art. 169) ; 
 and through F, the pole 
 of the primitive, and the 
 point E, describe the 
 great circle FEG, cutting 
 
 the primitive circle in G; then AGE is the triangle proposed, 
 
 of which G is the right angle. 
 
 Calculation. 
 
 As rad : cos A : : tan AE : tan AG (Art. 62) = 67° 23'. 
 
 As rad : sin A : : sin AE : sin EG (Art. 58) F 28° 32'. 
 
 As rad : cos AE : : tan A : cot AEG (Art. 64) = 78° 45'. 
 
 It is obvious, from the construction, that there is no ambi- 
 guity in this problem ; for the points E and F being given, 
 the great circle passing through them can have but one 
 position. 
 
 Again, the side EG is of the same affection as the angle A 
 (Art. 56) ; also, AG is of the same affection with EG, or a 
 different one, according as AE is less or greater than a quad- 
 rant (Art. 57) ; and the angle AEG is of the same affection 
 as AG (Art? 56). 
 
 2. Given, the hypothenuse 125° 25', and one leg 37° 40', to 
 find the rest. 
 
 Construction. Having described the primitive ^ : icle ABC, 
 lay AG on it equal the given leg; through G and the 
 
 27 
 
206 
 
 SECTION V. 
 
 pole F, draw the right 
 circle GFE ; from the 
 point A, as a pole, at a 
 distance equal to the hy- 
 pothenuse (or from the 
 opposite pole C with its 
 supplement), describe a 
 less circle DEH (by Art. 
 156), cutting GFE in E ; 
 through A and E describe 
 the great circle AEC 
 (Art. 159) ; then AGE is 
 the triangle proposed. 
 
 Calculation. 
 
 rad : cos EAG (Art. 62) = 123° 18'. 
 rad : sin AEG (Art. 58)- 48° 34'. 
 rad : cos EG (Art. 65) = 137° 4'. 
 In this problem there is no ambiguity ; for the angle AEG 
 is of the same affection with the side AG (Art. 56) ; EG is 
 of the same affection with AG, when*AE is less than a quad- 
 rant, and of a different one when AE is greater (Art. 57) ; 
 and the angle EAG is of the same affection as EG (Art. 56). 
 
 As tan AE 
 As sin AE 
 As cos AG 
 
 tan AG 
 sin AG : 
 : cos AE 
 
 3. Given, one leg 75° 
 26', and the adjacent an- 
 gle 40° 10', to find the 
 rest. 
 
 Construction. Describe 
 the primitive circle ABC, 
 and the oblique great cir- 
 cle AEC, making BAC 
 equal to the given angle 
 (by Art. 162); on the 
 primitive lay AG equal 
 the given leg; and through 
 
SPHERICAL PROJECTIONS. 
 
 207 
 
 G and the pole of the primitive, draw the right circle GEF, 
 cutting AC in E ; then is AGE the triangle in question. 
 
 Calculation. 
 As rad : sin AG : : tan A : tan EG (Art. 60) = 39° 15'. 
 As cos A : rad : : tan AG : tan AE (Art. 62) = 78° 46'. 
 As rad : sin A : : cos AG : cos AEG (Art. 67) = 80° 40'. 
 
 This problem includes no ambiguous case ; for the side EG 
 and the angle at E are respectively of the same affection 
 with the angle A and the side AG (Art. 56) ; and the hypo- 
 thenuse is less or greater than a quadrant, according as AG 
 and GE are of the same or different affections (Art. 57). 
 
 4. Given, one leg 36° 45', and the opposite angle 42° 16', 
 to find the rest. 
 
 Construction. Describe 
 the primitive circle ABC, 
 and the oblique great cir- 
 cle AC, making at the 
 point A an angle equal to 
 the given one (Art. 162) ; 
 i about the pole F of the 
 primitive, at a distance 
 equal to the complement 
 of the given leg, describe 
 a less circle, cutting the 
 oblique circle AC in E 
 (Art. 155); through E 
 and the pole F describe the great circle GEF (Art. 159), 
 cutting the primitive in G ; then AGE is the triangle pro- 
 posed. 
 
 Calculation. 
 
 As sin A : sin EG : : rad : sin AE (Art. 58) = \ 62 ° 49 '' 
 
 ( 117° 11'., 
 
208 
 
 SECTION V. 
 
 As tan A : tan EG : : rad : sin AG (Art. 60) = j 
 As cos EG : cos A : : rad : sin AEG <Art. 67) = j 
 
 55° 15'. 
 124° 45'. 
 
 67° 27'. 
 112° 33'. 
 
 The cases contained in this problem are ambiguous, as is 
 obvious from the construction ; the unknown sides and angle 
 being susceptible of two values, which are supplemental to 
 each other. 
 
 5. Given, the two legs, 70° 29' and 30° 16', to find the 
 rest. 
 
 Construction. On the 
 primitive circle lay down 
 AG equal to one of the 
 legs ; and through G and 
 the pole of the primitive 
 describe the right circle 
 GF ; on which lay down 
 GE equal to the other leg 
 (Art. 169) ; through A and 
 E describe (by Art. 159) 
 the great circle AEC; 
 and AGE is the triangle 
 which was to be con- 
 structed. 
 
 Calculation. 
 
 As sin AG : rad : : tan EG : tan A (Art. 60) = 31° 46'. 
 
 As sin EG : rad : : tan AG : tan E (Art. 60) = 79° 52'. 
 
 As rad : cos AG : : cos EG : cos AE (Art. 65) == 73° 14'. 
 
 This problem contains no ambiguity ; for the angles at A 
 and E are of the same affections as the opposite sides (Art. 
 56) ; and AE is less than a quadrant, when AG and GE are 
 of the same affection (Art. 57). 
 
SPHERICAL PROJECTIONS. 
 
 209 
 
 6. Given, the two oblique angles, 28° 19' and 75° 15', to 
 find the sides. 
 
 Construction. Describe the oblique great circle AC, making 
 
 with the primitive at A 
 an angle equal to one of 
 those given (Art. 162) ; 
 about the pole P of this 
 circle, at a distance equal 
 to - the measure of the 
 other given angle, or of 
 its supplement if obtuse, 
 describe a less circle cut- 
 ting the primitive in H 
 and I (Art. 157); from H 
 nearest to A, when the 
 second angle is acute, or 
 from I the more remote 
 point of intersection,when 
 the angle is obtuse, lay down HG on the primitive equal a 
 quadrant; through G describe a great circle at right angles 
 to AG, cutting AC in E ; then AGE is the triangle proposed. 
 For, H is evidently the pole of GEF ; and the distance HP 
 is the measure of the angle AEG. 
 
 Calculation. 
 As tan A : cotan E : : rad : cos AE (Art. 64) = 60° 45'. 
 As sin A : rad : : cos E : cos AG (Art. 67) = 57° 32'. 
 As sin E : rad : : cos A : cos EG (Art. 67) = 24° 27'. 
 
 This problem contains no ambiguity; for the sides AG and 
 EG are of the same affection as the opposite angles (Art. 
 56) ; and when those sides-, or their opposite angles, are of 
 the same affection, the hypothenuse is less than a quadrant 
 (Art. 57). 
 
 27 
 
u 
 
 210 SECTION V. 
 
 Of Oblique Angled Spherical Triangles. 
 
 1. Given, two sides, AC 45° 30', BC 30° 30', and the an^ 
 A opposite one of them, 36° 45', to find the rest. 
 
 Construction. On the primitive circle lay down AC, one 
 of the given sides ; about the pole C describe a less circle, at 
 the distance BC of the other given side ; describe a great 
 circle, making, at the point A, with the primitive, an angle 
 equal to the given one (Art. 162), intersecting the less circle 
 in B ; through C and B describe a great circle (Art. 159) ; 
 and ABC is the triangle to be made. 
 
 When BC, the side opposite the given angle, is the less of 
 the two, there may be two points of intersection, and, conse- 
 quently, two positions of B. Hence, in that case, the problem 
 is ambiguous. 
 
 Calculation: Describe the great circle CD through the 
 pole of AB; then ADC, BDC are rectangular triangles. 
 Hence, 
 
 As rad : cos A : : tan AC : tan AD (Art. 62) = 39° 12'. 
 
 As cos AC : cos BC : : cos AD : cos BD (Art. 66)- 17° 41'. 
 
SPHERICAL PROJECTIONS. 211 
 
 As rad : cos AC : : tan A : cotan ACD (Art. 64)=62° 22'. 
 
 As tan BC : tan AC : : cos ACD : cos BCD (Art. 63)=36° 46'. 
 
 As sin BC : sin AC : : sin A : sin ABC (Art. 59) =57° 14' 
 or 122° 46'. 
 
 Hence, AB = 56° 53', or 21° 31' ; and ACB = 99° 8', or 
 25° 36'. 
 
 Or, without a perpendicular. 
 Find ABC as above. Then, 
 
 Ascosi(AC — BC) : cos i(AC + BC) :: tan i(ABC + BAC) 
 : cotan JACB (Art. 77, eq. 15)=49° 34', or 12° 48'. 
 
 Hence, ACB = 99° 8', or 25° 36'. 
 
 And, 
 As cos i(ABC — BAC) : cos i(ABC + BAC) :: tan -J(AC + 
 
 BC) : tan £AB (Art. 77, eq. 17) =28° 26^', or 10° 451'. 
 Whence, AB 56° 53', or 21° 31'. 
 
 2. Given, two sides, 75° 20' and 60° 16', and the included 
 angle 40° 18', to find the rest. 
 
 Construction. Describe 
 a great circle, making at 
 the given point A an 
 angle with the primitive 
 equal to the given one 
 (Art. 162). Make AB, 
 AC, respectively, equal 
 to the given sides (Art. 
 169) ; and describe a 
 great circle through B 
 and C (Art. 159); then 
 ABC is the triangle pro- 
 posed. 
 
 In this problem, there is no ambiguity. 
 
212 SECTION V. 
 
 Calculation. Draw CD at right angles to AB. Then, 
 
 As rad : cos A :: tan AC : tan AD (Art. 62) = 53° 10'. 
 As sin BD : sin AD : : tan A : tan B (Art. 61) = 60° 56'. 
 As cos AD : cos BD : : cos AC : cos BC (Art.66)=39° 59'. 
 As rad : cos AC : : tan A : cotan A CD (Art. 64)= 67° 11'. 
 As tan AD : tan BD :: tan ACD : tan BCD (Art. 69) =35° 57'. 
 Hence, ACB = 103° 8'. 
 
 Or, without a perpendicular, 
 
 As cos J(AB + AC) : cos £(AB — AC) :: cot £BAC 
 : tan i(ACB + ABC) (Art. 77, eq. 11) = 82° 2'. 
 
 As sin i(AB + AC) : sin |(AB — AC) : : cotan £BAC 
 : tan £(ACB— ■ ABC) (Art. 77, eq. 12) == 21° 6'. 
 
 Whence, ACB = 103° 8', and ABC = 60° 56'. 
 
 As sin ABC : sin BAC : : sin AC : sin BC (Art. 59)=39° 59'. 
 
 3. Given, one side 80° 44', and the two adjacent angles, 
 40° 50' and 70° 12', to find the rest. 
 
 Construction. On the 
 primitive circle, lay down 
 AB equal to the given 
 side; and describe two 
 great circles, making an- 
 gles with the primitive, 
 at the points A and B, 
 equal to the given ones 
 (Art. 162), and let those 
 circles cut each other in 
 C; ABC is the triangle 
 proposed. 
 
SPHERICAL PROJECTIONS. 213 
 
 Calculation. Through B, the extremity of a given side, 
 and P the pole of AC, describe a great circle, cutting AC 
 (produced, if necessary) in D. Then, 
 
 As rad : cos A : : tan AB : tan AD (Art. 62)= 77° 50'. 
 As rad : cos AB : : tan A : cotan ABD (Art. 64)= 82° 5'. 
 As cos CBD : cos ABD :: tan AB : tan CB (Art. 63)=40° 49'. 
 As tan ABD : tan CBD :: tan AD : tan CD (Art. 69)= 7° 44'. 
 As sin ABD : sin CBD :: cos A : cos BCD (Art. 68) =80° 57'. 
 Hence, AC = 70° 6', and ACB == 99° 3'. 
 Or, without a perpendicular, 
 
 As cos J (B + A) : cos i(B — A) : : tan £AB : tan £(AC + 
 BC) (Art. 77, eq. 13) = 55° 27'. 
 
 As sin i(B + A) : sin J(B — A) : : tan ^AB : tan \(A& — 
 BC) (eq. 14) = 14° 39'. 
 
 As cos !(AC — BC) : cos i(AC + BC) : : tan J(B + A) 
 
 : cot JACB (eq. 15) = 49° 31'. 
 Whence, AC = 70° 6', BC = 40° 48', and ACB = 99° 2'. 
 
 4. Given, two angles, 50° 16' and 60° 36', and a side 42° 
 34', opposite one of them, to find the rest. 
 
 Construction. Describe 
 (Art. 162) a great circle, 
 making with the primi- 
 tive, at the point A, an 
 angle equal to the given 
 one, to which the given 
 side is adjacent. On that 
 circle lay down AC equal 
 to the given side (Art. 
 169) ; through C describe 
 (by Art. 163) a great 
 circle, making with the 
 primitive an angle equal 
 28 
 
21-1 
 
 SECTION V. 
 
 to the other given one ; and let that r.ircle cut the primitive 
 in B : then ABC is the triangle. 
 
 Calculation. Draw CD at right angles to AB. Then, 
 As rad : cos A : : tan AC : tan AD (Art. 62)= 30° 25'. 
 As tan B : tan A : : sin AD : sin BD (Art. 61)= 20° 4' 21". 
 As rad : cos AC : : tan A : cotan ACD (Art. 64)=48° 27' 26". 
 As cos A : cos B : : sin ACD : sin BCD (Art. 68) =35° 5' 9". 
 As sin B : sin A : : sin AC : sin BC (Art. 59)= 36° 39' 46". 
 Wherefore, AB = 50° 29' 21", and ACB = 83° 32' 35". 
 
 Or, without a perpendicular. 
 Fin»l BC as above. Then, 
 
 As cos ^(B — A) : cos J(B + A) : : tan J(AC + BC) 
 : tan £AB (Art. 77, eq. 17) = 25° 14' 42". 
 
 As cos i r (AC — BC) : cos ^(AC + BC) : : tan £(B + A) : 
 cotan JACB (eq. 15) = 41° 46' 17". 
 
 Wherefore, AB = 50° 29' 24", and ACB = 83° 32' 34". 
 
 5. Given, the three sides, 80 c 
 find the angles. 
 
 16', 60° 44', and 50° 20', to 
 
 Construction. On the 
 primitive, lay down AB 
 equal to one of the given 
 sides ; from A and B, as 
 poles, at distances equal 
 to the other sides respec- 
 tively, describe (Art. 156) 
 two circles, ICK and 
 GCH, cutting each other 
 in C ; through A, C, and B, 
 C, describe (Art. 159) two 
 great circles; then ABC 
 is the triangle proposed. 
 
SPHERICAL PROJECTIONS. , 215 
 
 Calculation. Through C, describe the great circle CD at 
 right angles to AB, and bisect AB in E , then, 
 
 As tan AE : tan ^(AC + BC) : : tan J(AC — BC) : tan ED 
 
 (Art. 74) = 8° 56' 14". 
 Whence, AD = 49° 4' 14", and BD = 31° 11' 46". 
 As tan AC : tan AD : : rad : cos A (Art. 62)=49° 44' 18' . 
 As tan BC : tan BD : : rad : cos B = 59° 51' 33". 
 
 As sin AC : sin AD : : rad : sin ACD (Art. 58) - 60° 0' 17". 
 As sin BC : sin BD : : rad : sin BCD = 42° 17' 25" 
 
 Wherefore, ACB = 102° 17' 42". 
 
 Or, without a perpendicular. 
 
 Take P = half the sum of the sides ; then, 
 
 As sin P.sin (P— BC) : sin (P — AC).sin (P — AB):: rad 2 
 
 : tan 2 £A (Art. 77, eq. 7) « 24° 52' 9". 
 Wherefore, A = 49° 44' 18". 
 And (Art. 59), B = 59° 51' 34", and C = 102° 17' 42". 
 
 6. Given, three angles, 60° 36', 66° 20', and 99° 50', to find 
 the sides. 
 
 Construction. At the centre of the primitive, make an angle 
 C equal to the greatest given angle ; and describe a great 
 
216 SECTION V. 
 
 circle, making, with the right circles including that angle, 
 two angles respectively equal to the other two given ones 
 (Art. 166) ; and let A and B be those angles : then ABC is 
 the triangle proposed. 
 
 Calculation. Through C let two great circles, CD and CE, 
 pass ; the former at right angles to AB, and the latter bisect- 
 ing the angle ACB. Then, ^ 
 
 As cotan £(B + A) : tan K B "— A ) : : tan ACE : tan ECD 
 (Art. 75) = 6° 47' 44". 
 
 Whence, ACD = 56° 42' 44", and BCD = 43° 7' 16". 
 
 As |an A : cotan ACD : : rad : cos AC === 68° 17' 13". 
 
 As tan B : cotan BCD : : rad : cos BC (Art. 64) =62° 5' 42". 
 
 As sin A : rad : : cos ACD : cos AD (Art. 67)= 50° 57' 5". 
 
 As sin B : rad : : cos BCD : cos BD= 37° 9' 41". 
 
 AB = 88° 6' 46". 
 Or, without a perpendicular, 
 
 As sin B.sin C : cos |(A + C — B).cos $(A + B — C) : : rad 2 
 : cos 2 ^BC (Art. 77, eq. 9) ; 
 
 whence, BC = 62° 5' 58". And (Art. 59), AB = 8«8° 6' 46", 
 AC = 68° 17' 12". 
 
 Otherwise. From A and B, as poles, describe the great 
 circles FG, FH, cutting the primitive in G and H ; then the 
 triangle FGH is supplemental to ABC (Art. 54). That is, 
 GH = 180° — ACB ; FH = 180° — ABC ; FG = 180° — 
 BAC ; whence the sides of FGH are known. Then, 
 
 As sin FH.sinFG : sin i(HG + FH — FG).sin i(HG + FG 
 — FH) :: rad 2 : sin 2 JHFG or cos 2 JAB ; 
 
 whence, AB = 88° 6' 46", as before. 
 
SPHERICAL PROJECTIONS. 
 
 Promiscuous Examples. 
 
 The following astronomical terms being occasionally used 
 in the succeeding examples, it is deemed advisable to insert 
 their definitions. 
 
 The axis of the earth is the line through the centre on 
 which it revolves; the points where the axis meets the 
 surface of the earth are the poles of the earth ; and the points 
 where the axis produced meets the concave surface of the 
 visible heavens, are the celestial poles. 
 
 The common section of the earth's surface and a plane 
 passing through its centre at right angles to its axis, is 
 termed the terrestrial equator; and the section of the same 
 plane and the concave surface of the visible heavens, is called 
 the celestial equator, or equinoctial circle. 
 
 Meridians are great circles passing through the poles, and, 
 consequently, cutting the equator at right angles. 
 
 A right line drawn from the centre of the earth, through 
 the place of an observer, and continued till it meets the celes- 
 tial sphere, cuts it in a point which is termed the zenith of 
 the place. The point where the same line, extended beyond 
 the centre, meets the celestial sphere, is termed the nadir. 
 
 The great circle of which the zenith and nadir are the 
 poles, is termed the horizon. 
 
 A meridian passing through the zenith of any place, is 
 called the meridian of that place. 
 
 The angle formed by the meridian of a place, and a great 
 circle passing through the zenith, and a celestial object, is 
 called the azimuth. 
 
 The distance, reckoned in degrees, minutes, &c, on the 
 meridian, between the equator and a place on the earth's 
 surface, is termed the latitude of that place. 
 
 The angle contained between the meridian of a place, and 
 28 ■ t 
 
213 SECTION V 
 
 some other assumed as a first meridian, is termed the longi- 
 tude of the place. The longitude is usually reckoned east- 
 ward or westward as far as 180°. 
 
 The common section of the plane of the earth's orbit and 
 the celestial sphere, is called the ecliptic. This circle is the 
 sun's apparent annual path. 
 
 The points where the ecliptic cuts the celestial equator are 
 termed the equinoxes. The point in which the sun appears 
 when passing from the southern to the northern side of the 
 equator, is termed the vernal equinox. 
 
 The arc of the celestial equator, reckoning eastward, 
 between the vernal equinox and the point where a meridian 
 through any celestial object cuts the equator, is called the 
 right ascension of that object ; and the arc of the meridian 
 between the object and the equator, is termed the decli- 
 nation. 
 
 The arc of the ecliptic, reckoned eastward, between the 
 vernal equinox and the point where a great circle, passing 
 through the pole of the ecliptic and any celestial object, cuts 
 the ecliptic, is termed the longitude of that object ; and the 
 arc of that great circle, betw r een the object and the ecliptic, 
 is termed its latitude. 
 
 The angle formed by the ecliptic and the celestial equator 
 is termed the obliquity of the ecliptic. 
 
 The right ascension of the point in the celestial equator 
 which is cut by the meridian of a place, is called the right 
 ascension of the mid-heaven. 
 
 The point of the ecliptic which is cut by a great circle 
 passing through its pole and the zenith of a place, is called 
 the nonagesima degree. 
 
 Ex. 1. Required, the distance on a great circle of the earth 
 between Point Venus in Otaheite, and Edinburgh ; also, the 
 direction of each from the other ; the latitude of the former 
 berns 17° 29' South, and longitude 149° 29' West; and the 
 
SPHERICAL PROJECTIONS. 
 
 219 
 
 atitude of the latter 55° 57' North, and longitude 3° 11' 
 West. 
 
 Cons/ruction. Assume the 
 primitive circle as the meri- 
 dian of Point Venus ; and 
 take P as the north pole; 
 from P lay down PA = the 
 polar distance, 107° 29', of 
 the place ; through P describe 
 a great circle, making, with 
 the primitive at P, an angle 
 146° 18', equal to the differ- 
 ence of longitudes (Art. 162) ; on this circle lay down PB 
 34° 3' = the polar distance of Edinburgh (Art. 169) ; through 
 A and B describe a great circle. Then AB is the arc, and 
 PAB, PBA the angles required. 
 
 Computation. The sides AP, BP, and the angle APB, being 
 given, the angles at A and B are found (Art. 77, eq. 11, 12), 
 viz. : PAB = 25° 32', PBA = 47° 15'. Then (eq. 18), AB ** 
 133° 53'. Consequently, Edinburgh bears from Point Venus, 
 N. 25° 32' E. ; and Point Venus bears from Edinburgh, N. 
 47° 15' W. ; distance 133° 53'. 
 
 Ex. 2. Required, the bearing and distance on a great circle 
 of the Observatory at Greenwich from the Capitol at Wash- 
 ington: the latitude of the former being 51° 28' 40" N. ; the 
 latitude of the latter 3S° 53' N., and longitude 77° 2' W. from 
 the meridian of Greenwich. 
 
 Ans. N. 49° 20' E. ; dist. 53° 8'. 
 
 Ex. 3. When the sun's declination is 23° 28' N., how long 
 after midnight does it rise in latitude 39° 57' N., and how far 
 from the northern point of the horizon ? 
 
 The construction of this problem is readily understood 
 
220 
 
 SECTION V. 
 
 from the figure ; taking HO 
 for the horizon; OP=39° 57', 
 the latitude of the place ; de- 
 scribing a circle about the 
 pole P, at the distance 66° 
 32', the sun's polar distance ; 
 supposing this circle to cut 
 HO in n ; then, the great cir- 
 cle PwS being described, the 
 triangle POrc, right angled at 
 O, will contain the elements 
 required ; the angle ?vPO, con- 
 verted into time at the rate of one hour to 15°, or four min- 
 utes to 1°, will be the time required ; and nO, the required 
 distance from the northern part of the horizon. 
 
 Result: On 58° 42', 0?n 68° 41'. 
 
 Ex. 4. Given, the latitude of the place, 39° 56' N. ; decli- 
 nation of the sun, 23° 28' N. ; and zenith distance, 60° 30' ; 
 to find the azimuth and polar angle. 
 
 To construct this on the 
 plane of the meridian, take 
 the primitive circle HZPO 
 for the meridian; HO, the 
 horizon ; Z, the zenith ; P, 
 the north pole ; and, there- 
 fore, PO = the latitude of 
 the place. About P describe 
 a circle at a distance = 66° 
 32', the sun's polar distance 
 (Art. 156) ; and about Z, at 
 F a distance = 60° 30', the 
 
 sun's zenith distance, de- 
 scribe another circle, cutting the former in N ; then describe 
 great circles through NZ and NP (Art. 159) ; and in the 
 triangle NPZ, the angle PZN is the azimuth, and ZPN the 
 
SPHERICAL PROJECTIONS. 
 
 221 
 
 polar angle required. Those angles may be computed by- 
 Art. 77, eq. 5, 6, or 7. 
 
 Result: NZP 82° 56', ZPN 70° 20'. 
 
 Ex. 5. In the beginning of 1842, the right ascension of 
 Aldebaran (the bull's eye) was 66° 42' 50", and the declina- 
 tion 16° 8' 36" N. Required, the longitude and latitude at 
 that time ; the obliquity of the ecliptic being 23° 27' 40". 
 
 To project this on the 
 plane of the celestial equa- 
 tor, take the primitive 
 ABCD to denote that cir- 
 cle; A and C being the 
 equinoxes, and P the north 
 pole : through A, C, describe 
 the circle AGCH, making 
 an angle of 23° 27' 40" with 
 the primitive ; then AGCH 
 will represent the ecliptic ; 
 take Q the pole of AGCH ; 
 make AE = the star's right 
 ascension; through E, P, 
 draw the line EP ; this will represent the meridian passing 
 through the star. On EP lay down ES (Art. 169) = the 
 star's declination ; then S is the place of the star. Through 
 SQ describe (Art. 159) a great circle cutting the circle AGC 
 in F ; then AF is the longitude, and FS the latitude required. 
 Through AS describe a great circle; then the angle EAS 
 may be computed by Art. 60, and thence AF and FS by 
 Arts. 71 and 70. 
 
 Result : long. 67° 34' 23" ; lat. 5° 31' 18" S. 
 
 Ex. 6. Given, the latitude of the place, 40° north ; the 
 obliquity of the ecliptic, 23° 28' ; and the right ascension of 
 the mid-heaven, 60° ; to find the longitude and altitude of the 
 nonagesima degree. 
 
 29 
 
222 
 
 SECTION V. 
 
 To project this example on the plane of the meridian, let 
 the primitive circle HZMO denote the meridian; HO, the 
 horizon; Z, the zenith, or pole of HO. On the primitive 
 circle lay down HP, ZM, each equal to the latitude of the 
 place, or 40° ; then P will be the pole of the equator. About 
 P, at the distance 23° 28', the obliquity of the ecliptic, de- 
 scribe a less circle; through M and C (the centre of the 
 primitive), draw the right circle CMA ; this will represent 
 the equator. Make MA = 60°, the right ascension of the 
 mid-heaven; the point A will be the vernal equinox. Make 
 CL = 30° ; then L will be the autumnal equinox, for CM = 
 90°. About the pole L describe the great circle QPS, cut- 
 ting the less circle in Q ; then Q is the pole of the ecliptic. 
 Through L describe a great circle having Q for its pole ; this 
 circle will denote the ecliptic, and pass through A. Lastly, 
 through Q and Z describe a great circle QZNR, cutting the 
 
SPHERICAL PROJECTIONS. 
 
 223 
 
 ecliptic in N, and the horizon in R ; N is the nonagesima, 
 whose longitude is AN, and altitude NR or QZ. 
 
 Calculation, Let QP cut the equator in E ; then AE — AM 
 =EM=ZPE» Hence, QPZ becomes known = 150°; then, 
 in the triangle ZPQ, ZP == 50°, PQ = 23° 28' ; with which 
 and the contained angle, we may find the angle ZQP = 23° 
 53' 44", and side QZ == NR = 71° 0' 18". Consequently, the 
 longitude of the point N = 66° 6' 16". 
 
 Ex. 7. Given, the latitude of the place 41° 36' north, anc 4 
 the sun's declination 22° 10' north, to find the time from noon, 
 or the polar angle, when the sun is on the vertical circle 
 which passes through the east and west points of the horizon, 
 and the sun's altitude at the same time. 
 
 Result: Polar angle 62° 41', or time from noon 4 hours 11 
 minutes ; altitude 34° 38'. 
 
 Ex. 8. On the first day of the year 1836, when it was noon 
 at Greenwich, the right ascension of Jupiter was 101° 55', 
 and declination 23° 4' 2" north ; at the same time, the right 
 ascension of Saturn was 212° 17', and the declination 10° 30' 
 50" south. What was their distance on the arc of a great 
 circle 1 
 
 Ans. 112° 44' 59'. 
 
 Ex. 9. In the beginning 
 of 1842, the declination of 
 the polar star was 88° 28' 
 north. What was its azi- 
 muth, when its elongation 
 from the meridian was the 
 greatest, the observer being 
 in latitude 40° north ? 
 
 Ans. 2° 0' 6". 
 
224 
 
 SECTION V. 
 
 Ex. 10. In latitude 40° north, required the duration of 
 twilight at the several times when the sun's declination is 
 23° 28' south, 5° 50' south, and 23° 28' north ; the twilight 
 being supposed to begin when the sun's centre is 49' below 
 the horizon,* and to end when it is 18° below. 
 
 Ans. 1 h. 35 m. ; 1 h. 29 m. ;f and 2 h. 4 m. 
 
 Ex. 11. Given, two zenith distances of the sun's centre, 
 65° 20' and 60° 18', taken at the same place, both being in 
 the forenoon ; the interval between the observations, mea- 
 sured by a good time-piece, 1 hour 32 minutes ; the sun's 
 declination 20° south, and the approximate latitude of the 
 place 40° 15' north ; to find the time of the last observation, 
 and the correct latitude of the place. 
 
 Let HO be the horizon; 
 HZO, the meridian ; Z, the 
 zenith; P, the north pole; 
 BZ and AZ, the given zenith 
 distances. Then PZ is the 
 colatitude ; and PB, PA, the 
 sun's polar distance. 
 
 Now (by Art. 77, eq. 1), 
 
 cos AZ = cos ZPA. sin AP 
 sin ZP + cos AP.cos ZP ; 
 
 and 
 
 cos BZ = cos ZPB.sin BP.sin ZP + cos BP.cos ZP. 
 Hence (AP being = BP), 
 cos AZ — cos BZ = (cos ZPA — cos ZPB) sin AP.sin ZP ; 
 
 * The refraction of the sun's light, when in the horizon, is 33', and 
 apparent semi-diameter about 16'; hence his upper limb is just visible 
 when the centre is 49' below the horizon. 
 
 j- This is the shortest twilight in latitude 40° N.,and occurs twice in the 
 year, viz., in spring and »itumn. when the sun's declination is 5° 50' S. 
 
SPHERICAL PROJECTIONS. 225 
 
 consequently (Art. 37, eq. 13), 
 
 sin J(BZ + AZ).sin J(BZ — AZ) = sin J(BPZ + APZ). 
 sin i(BPZ — APZ) sin AP.sin ZP ; 
 
 of which, BPZ — APZ is given from the elapsed time. Hence, 
 
 • imp 7 m AP7^ Bin « BZ + AZ).sini(BZ- AZ) 
 sin i(BPZ + APZ) = ^jxjfcye^ ^BPZ^-A JZ) = 
 
 15° 51' 11". 
 
 Whence APZ - 4° 21' 11", and (by Art. 59) AZP == 175° 
 17' 24" ; from which we find (by Art. 77, eq. 18), ZP ==- 49° 
 50' ; and therefore PO, the corrected latitude, 40° 10' ;» and 
 time from noon, when the least zenith distance was taken, 17 
 minutes 24 seconds. 
 
 Ex.12. Given, the approximate latitude, 39° 26' N.; the 
 sun's declination, 20° 41' N. ; sun's corrected zenith distance 
 at 11 h. 30' 15", by watch, 21° 30' ; and at 12 h. 26' 28", by 
 watch, 18° 52'. Required, the corrected latitude, and error 
 of the watch. 
 
 Ans. Lat. 39° 29' ; watch too fast, 18 min. 57 sec. 
 
 Ex. 13. At the time when Sirius and Aldebaran were in 
 the same vertical circle, the true zenith distance of the latter 
 was found to be 30° 16' ; the stars being on the east of the 
 meridian. Required, the latitude of the place, and right 
 ascension of the mid-heaven ; the right ascension of Sirius 
 being 99° 33' 30", and its declination 16° 30' 18" south ; the 
 right ascension of Aldebaran 66° 43' 45", and its declination 
 16° 11' 12" north. 
 
 Ans. Lat. 35° 8' 42" N. ; right ascension of mid-heaven 
 39° 17' 23". 
 
 * When the latitude thus found differs considerably from the approxi- 
 mate latitude, the computation ought to be repeated, with the result first 
 obtained substituted for the approximate latitude. 
 
 29 
 
226 
 
 SECTION V. 
 
 Examples of a Mixed Character. 
 
 Ex. 1. From the top of a cliff near a river, two buoys at 
 anchor being observed, whose distance from each other was 
 known to be 300 yards, their angles of depression below the 
 plane of the observer were found to be 30° and 40° respec- 
 tively; and the angle at the eye, subtended by the line joining 
 them, was 37°. Required, the distance of each buoy from 
 the observer, and the altitude of the cliff above the level of 
 the water. 
 
 The observed depressions, each increased by 90°, form two 
 sides ; and the angular distance, the base of a spherical tri- 
 angle, with which the angle opposite the base is found =44°. 
 This is the horizontal angle, subtended by the line joining the 
 buoys. 
 
 Drawing then a vertical line through the position of the 
 observer to meet the plane of the water ; and, from the point 
 where it meets that plane, drawing lines to the buoys ; those 
 lines will be to each other as the cotangents of the given 
 angles of depression ; and the angle which they make with 
 each other will be 44°, as above found. The construction is 
 this : 
 
 Take AB=300, the given dis- 
 tance ; on AB describe a segment 
 of a circle ACB, containing an 
 angle of 44° ; complete the cir- 
 cle, and bisect the arc AEB in 
 E; make the angles ABF and 
 BAF = 30° and 40° respective- 
 ly ; draw FG at right angles to 
 AB; join EG, and produce it to 
 meet the circle in C; join CA, 
 CB ; then, since the angle ACB is bisected by the line CG, 
 
 As AC : CB : : AG : BG (3.6) : : cot BAF : cot ABF. 
 
SPHERICAL PROJECTIONS. 227 
 
 Consequently, C is the point in the plane of the water which 
 is cut by a vertical line passing through the place of the 
 observer. 
 
 The calculation is easily made. For (Art. 28 and Art. 37, 
 eq. 8), 
 As sin (BAF+ ABF) : sin (BAF — ABF) : : AB : BG— AG. 
 
 Hence AG is known. If we join AE, BE, the side AB, and 
 all the angles of the triangle ABE, are given ; whence AE 
 becomes known. Then, in the triangle AGE, the sides AE, 
 AG, and the contained angle, are known; from which the 
 angle AEG = ABC is found. In the triangle ABC, we then 
 have the base AB and all the angles, to find AC and BC. 
 Then, from either of these and the angle of depression, the 
 altitude of the cliff may be found. Lastly, with the distances 
 AC, BC, and the angles of depression, the distances from A 
 and B to the place of the observer are determined 
 
 Result : Altitude of cliff, 249 ; distances, 388 and 49S 
 
 Ex. 2. Suppose an observer on a frozen lake takes the 
 altitudes and angular distance of two cliffs on the shore, as 
 follows : altitude of first, 50° ; of second, 55° 30' ; angular 
 distance, 25° 20'. Then, advancing on the ice 500 yards, in 
 the vertical plane which passes through the first cliff, the 
 altitudes are 57° and 59° respectively. Required, the dis- 
 tance of the cliffs from <Mkcfi utfier, and their respective 
 altitudes above the surface of the lake. 
 
 Ans. Distance of cliffs, 2347 or 2850 yards ; altitude of 
 first, 2636 yards ; of second, 4069 or 552 yards. 
 
 Ex. 3. The crew of a vessel at sea discovering a light in 
 the horizon, which they suppose to be a vessel on fire, sail 
 directly towards it, over 1° of a great circle, when they per- 
 ceive that it is a fire on a mountain, which is then 1° 30' 
 above the horizon. Required, the distance of the light when 
 first seen, and the height of the mountain above the level of 
 
228 SECTION V. 
 
 the sea ; the earth being considered as a spnere wnose radius 
 is 3968 miles. 
 
 Answer. Distance, 138.5 miles ; height, 2.4 miles. 
 
 Ex. 4. Given, AB, a horizontal line, 1785 yards in length, 
 running exactly north ; D, C, two elevated peaks, eastward 
 from AB, such that the elevation of C above the plane of the 
 horizon, seen from A, is 16° 30', and the elevation of D 20° 
 40'. But, seen from B, the elevation of C is 14° 25', and the 
 elevation of D 13° 15'. Also, the angle BAG, taken in the 
 oblique plane passing through AB and C, is 38° 16' ; the 
 angle CAD, taken in the plane which passes through A, C 
 and D, 87° 20'. Required, the distance and bearing of DC 
 when reduced to the horizontal plane on which AB lies. 
 
 Result : DC, N. 29° W. ; 2,556 yards. 
 
 TUB END 
 
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