hyM.UkliULixiiiULili.itit»iiiiIiiii]yi:iiliih.i TREATISE OF A. PLANE TRIGONOMETRY. TO WHICH IS PREFIXED a summary view of the nature and use or jLOG.IRITHMS: RZING THE SECOND PART OP A COURSE OF MATHEMATICS. ADArTED TO THE METHOD OF INSTRUCTION TT THE AMERICAN COLLEGES. BY JEREMIAH DAY, D.D. LL.D Prtiidenl of Yale College. THE SECOND EDITION, \V'ITH ADDITIONS AND ALTERATION' NEW-HAVEN : PUBLISHED BY HOWE & SPALDING 8. coNVKRSE, Prin^tr 1824. ^% » ♦ • ' • DISTRICT OF CONNECTICUT, sb. Be it remembered, That on the twenty-ninth day ot ^July, in the forty-ninth year of the Independence of the Uni- ted States of America, Jeremiah Day, of the said District, hath deposited in this Office the title of a Book, the right whereof he claims as Author, in the words following — to wit : " A treati&e of Flanc Trigonometry ; to which is prefixed a summary "view of the nature and use of I^ogarithms : being the second part of a *' course of Matliematics, adapted to the method of instruction in the Ameri- *< can Colleges. By Jeremiah Day, D. D. LL. D., President of Yale College. " The second edition, with additions and alterations." In conformity to the Act of the Cono^ress of the United States, entitled " Au Act for the encouragement of learning, by securing the copies of Maps, Charts, and Books, to the Authors and Proprietors of such copies, during the times therein mentioned." CHAS. A. INGERSOLL, Clerk of the District of Connecticut. A true copy of Record, examined and sealed by rne, CHAS. A. INGERSOLL, Clerk of the District of Connecticut. r^ Mathemat'ci^I nPHE plan upon which this work was originally commen- -*■ ced, is continued in this second part of the course. As the single object is to provide for a class in college, such matter as is not embraced by this design is excluded. The mode of treating the subjects, for the reason*, mentioned in the preface to Algebra, is, in a considerable degree, diffuse. It was thought better to err on this extreme, than on the other, especially in the early part of the course. The section on right angled triangles will probably be con- sidered as needlessly minute. The solutions might, in all cases, be effected by the theorems which are given for ob- lique angled triangles. But the applications of rectangular trigonometry are so numerous, in navigation, surveying, as- tronomy. &c. that it was deemed important, to render famil- iar the various methods of stating the relations of the sides and angles; and especially to bring distinctly into view the principle on which most trigonometrical calculations are founded, the proportion between the parts of the given tri- angle, and a similar one formed from the sines, tangents, fac. in the tables. CONTENTS LOGARITHMS. Page. Section I. Nature of Logarithms ----- f II. Directions for taking logarithms and their numbers from the tables - - 10 III. Methods of calculating by logarithms Multiplication .,---. 17 Division -.------21 Involution ------- 22 Evolution -------- 25 Proportion - -* 27 Arithmetical Complement - - - 28 Compound Proportion - - - - 30 Compound Interest ----- 32 Increase of Population - - . - 35 Exponential Equations - - - - 39 JV. Different Systems of Logarithms - - 42 Computation of Logarithms - - 45 TRIGONOMETRY. Section I. Sines, Tangents, Secants, &ic. - - - 49 II. Explanation of the Trigonometrical Tables 58 III. Solutions of Right angled Triangles - 66 IV. Solutions of Oblique angled Triangles 80 V. Geometrical Construction of Triangles 91 VI. Description and use of Gunter's Scale 97 Vn. Trigonometrical Analysis - - - - 105 VIII. Computation of the Canon - - - 123 IX. Particular Solutions of Triangles - - 127 Notes 137 Table of Natural Sines and Tangents 147 LOGARITHMS. SECTION I NATURE OF LOGARITHMS.* 'T'HE operations of Multiplication and Division. \RT. 1. J. ^yjjgj^ ^^Qy g^pg ^Q be often repeated, become so laborious, that it is an object of importance to substitute, in their stead, more simple methods of calculation, such as Addition and Subtraction. If these can be made to perform, in an expeditious manner, the otBce of multiplication and division, a great portion of the time and labour which the latter processes require, may be saved. Now it has been shown, (Algebra, 233, 237,) that powers may be multiplied, by adding their exponents, and divided, by subtracting their exponents. In the same manner, roots may be multiplied and divided, by adding and subtracting their fractional exponents. (Alg. 280, 286.) When these ex- ponents are arranged in tables, and applied to the general purposes of calculation, they are called Logarithms. 2. LOGARITHMS, thkn, are the EXPONENTS of a -CRIES OF POWERS AND ROOTS. f In forming a system of logarithms, some particular num- ber is tixed upon, as the base, radix^ or first power, whose log- arithm is always 1. From this, a series of powers is raised, and the exponents of these are arranged in tables for use. To explain this, let the number which is chosen for the first *= Maskal3ne's Preface to Taylor's Logarithms. Introduction to Ilutton's Tahlcs. Keil on Logarithms. Maseres Scriptores Logarithmici. Briggs' Logarithms. Dodson's Anti-logarithmic Canon. Kuler's Algfl'nt. ■J- See notn A. S2 NATURE OF power, be represented by a. Then taking a series ol' pow- ers, both direct and reciprocal, as in Alg. 207 ; a/^f «^, «-, a\ «% «-', a~-, a~^, a "•, &c. The logarithm of a^ is 3, and the logarithm of a~' is —1, of a' is 1, of a~2 ig _2^ offlO is 0, of a-3 is — 3,&c, Universally, the logarithm of a"" is x. 3. In the system of logarithms in common use, called Brig^s"^ logarithms, the number which is taken for the radix or base is 10. The above series then, by substituting 10 for a. becomes lOS 10^ 10% 10% 10% 10-% 10-% 10-%&;c. Or 10000, 1000, 100, 10, 1, ^V, tU, toVo, &ic. Whose logarithms are 4, 3, £!, 1, 0, —1, -2, -3, &c. -1. The fractional exponents of roots^ and of powers of roots, are converted into decimals, before they are inserted in the logarithmic tables. See Alg. 255. The logarithm of a^, or a^'^ ^2% is 0.3333, of a^, or «'-^^'^% is 0.6666, of cr, or «^-^-«% is 0.4285, of « 2'^ or^3.0666js 3.6666, kc. These decimals are carried to a greater or less number of places, according to the degree of accuracy required. 5. In forming a system of logarithms, it is necessary to obtain the logarithm of each of the numbers in the natural series 1, 2, 3, 4, 5, &;c.: so that the logarithm of any number may be found in the tables. For this purpose, the radix of the system must first be determined upon ; and then every other number may be considered as some power or root of ihis. If the radix is 10, as in the common system, every other number is to be considered as some power of 10. That a power or root of 10 may be found, which shall be equal to any other number whatever, or, at least, a very near approximation to it, is evident from this, that the exponent rnay be endlessly varied; and if this be increased or dimin- j-hed. the pov:er will be increased or diminished. LOGARITHM.^. -^ If tke exponent is a fraction, and the numerator be increas- ed, the power will be increased : but if the denominator be increased the power will be diminished. 6. To obtain then the logarithm of any number, accor- ding to Briggs' system, we have to tind a power or root of 10 which shall be equal to the proposed number. The expo- nent of that power or root is the logarithm required. Thus 7=100.845 1^ fof 7 is 0.8451 20=10'-'°»° I therefore the ,' of 20 is 1.3010 30=10^*^' '* J logarithm ] of 30 is 1.4771 400=10='^*'2°J Lof 400 is 2.6020, &€, 7. A logarithm generally consists of two parts, an integer and a decimal. Thus the logarithm 2.60206, or, as it is some- times written, 2-f .60206, consists of the integer 2, and the decimal .60206. The integral part is called the characteris- tic or index* of the logarithm ; and is frequently omitted, in the common tables, because it can be easily supplied, when- ever the logarithm is to be used in calculation. By art. 3d, the logarithms of 10000, 1000, 100, 10, 1, .1, .01, .001, &c. are 4, 3, 2, 1, 0, -1, —2, -3, &c. As the logarithms of 1 and of 10 are and 1, it is evident, that, if any given number be between 1 and 10, its logarithm will be between and 1 , that is, it will be greater than 0, but less than 1. It will therefore have for its index, with a decimal annexed. Thus the logarithm of 5 is 0.69897. For the same reason, if the given number be between 10 and 100, ) ' the log. Cl and 2, i. e. 1+the dec. part 100 and 1000, > ■ will be <2 and 3, 2-t-the dec. part 000 and 10000,^ , between (3 and 4, 3-l-the dec. part We have, therefore, when the logarithm of an integer or mixed number is to be found, this general rule. * The term index, as it is used here, may possibly lead to some confusion m the mmd of the learner. For the logarithm itself is the index or exponent of a power. The characteristic, therefore, is the index of an index. A NATURE OF 8. Tlic index of the logarithm is ahvays one less, than the number of integral Jigurts, in the natural number whose loga- rithm is sought : or, ihe index shows how far the first figure of the natural number is removed from the place of unit?. Thus the logarithm of 37 is 1.5G820. Here, the number of figures being two^ the index of the logarithm is I. The logarithm of 253 is 2.40312. Here, the proposed number 253 consists oi three figures, the first of which is in the second place from the unit figure. The index of the logarithm is therefore 2. The logarithm of 62.8 is 1.79796. Here it is evident that the mixed number 62.8 is between 10 and 100. The index of its logarithm must, therefore, be 1. 9. As the logarithm of 1 is 0, the logarithm of a number less than 1, that is, of any p rope r/?ac/ion, must be negative. Thus by art. 3d The logarithm of yV or .1 is —1, of T^o or .01 is - 2, of ToVo or -001 is —3, &c. 10. If the proposed number is betzoeen y|o and y^* its logarithm must be between —2 and —3. To obtain the logjarithm, therefore, we must either subtract a certain frac- tional part from — 2, or add a fractional part to — 3 ; that is, we must either annex a negative decimal to —2, or a positive one to —3. Thus ihe logarithm of .008 is either -2 -.09691, or — 3 + .90309.* The latter is generally most convenient in practice, and is more commonly written 3.90309. The line over the index * That these two expressions are of the same value will be evident, if we subtract the same quantity, +.90309 from each. The remainders will be •qual, and therefore the quantities from which the subtraction is made must be equal. See note B. LOGARITHMS. b ilcnote?, that thai is negnlivc, while the decimal pari of the logarithm is positive. of 0.3, isT47712, The logarithm r oi u.:5, IS 1.4/ / 1:;, < of 0.00, is ^^77815, ^ of 0.009, is 3795424. And universally, 1 1. The negative index of a log-irithm shows hozv far the first significant fg}ire of the natural numher. is removed from the place of units, on the right ^ in the same manner asa;?os- itive index shows how far the first figure of the natural num- ber is removed from the place of units, on tlie left. (Art. 8.) Thus in the examples in the last article, The decimal 3 is in the/r5^ place from that of units. 6 is in the second place, 9 is in the third place ; And the indices of the logarithms are 1, 2, and 3. 12. It is often more convenient, however, to make the in- dex of the logarithm positive, as well as the decimal part. This is done by adding 10 to the index. Thus, for —1, 9 is written; for —2, 8, &;c. Because — 1-f 10 = 9, -2+10 = 8, 8.65963, .004567, 3.65963, 7.65963. This property, which is peculiar to Briggs' system, is of great use in abridging tlie logarithmic tables. For when we have the logarithm of any number, we have only to change the index, to obtain the logarithni of every other number, whether integral, fractional, or mixed, consisting of the same significant iigures. The decimal part of the logarithm of a fraction found in this way, is always yoo^/Z/r^. For it is the same as the decimal part of the logarithm of a whole nu!n- ber. 15. in a series effractions continually decreasing^ the neg- ative indices of the logarithms continually increase. Thus In the series 1, .1, .01, .001, .0001, .00001, &£c. The logarithms are 0, -1, -2, -3, -4, —5, Sic. If the progression be continued, till the fraction is reduced to 0, the negative logarithm will become greater than any as- signable quantity. The logarithm of 0, therefore, is infinite and negative, (Alg. 447.) 16. h is evident also, that all negative logarithms belong to fraclions which are between 1 and 0: \y\i\\e positive loija- J.OGARITHMS. 'i rithms belong to natural numbers which are greater than 1. A? the whole range of nunnberg, both positive and negative, is thus exhausted in supplying the loorarithms of integral and fractional positive quantities; there can be no other numbers to furnish logarithms for negative quantities. On this ac- count the logarithm of a negative quantity is, by some wri- ters, considered as impossible. But as (here is no difference in the multiplication, division, involution &lc. of positive and negative quantities, except in applying the signs ; they may be considered as all positive, while these operations are per- forming by means of logarithms ; and the proper signs may be afterwards affixed. 17. If a series of numbers be in geometrical progression^ their logarithms will be in aiuthmetical j?rogTes5io/i. For, in a geometrical series ascending, the quantities increase by a common multiplier ; (Alg. 436.) that is, each succeeding term is the product of the preceding term into the ratio. But the logarithm of this product is (he sum of the logarithms of the preceding term and the ratio ; that is, the logarithms increase by a common addition, and are, therefore, in arith- metical progression. (Alg. 422.) In a geometrical progres- sion descending, the terms decrease by a common divisor and their logarithms, by a common difference. Thus the numbers 1, 10, 100, 1000, 10000, &ic. are in ge- ometrical progression. And their logarithms 0, I, 2, 3, 4, &£c are in arith- metical progression. T^niversally, if in any geometrical series, « = the least term, r=the ratio, L = its logarithm, /=its logarithm ; Then the logarithm oi ar is L+/, (Art. 1.) of ar^ is L-{-2/, oi ar^ is L-^U, &c. Here, the quantities «, ar^, ar^, ar^, &LC, are iugeo- tnetrical progression. (Alg. 436.) And their logarithms L, L-\-l, L-\-^l, L-i-Sl, he. are in arithmetical progression. (Alg. 423.) * See note C. LOGARITHMIC THE LOGARITHMIC CURVE. 19. The relations of logarithms, and their corresponding numbers, may be represented by the abscissas and ordinates of a curve. Let the line AC (Fig. 1.) be taken for unity. Let AF be divided into portions, each equal to AC, by the points 1, 2, 3, &c. Let the line a represent the radix of a given system of logarithms, suppose it to be 1.3 ; and let a^, a^, &c. correspond, in length, with the different powers of a. Then the distances from A to 1, 2, 3, &c. will represent the logarithms of «, a- , a^, he. (Art. 2.) The line CH is called the logarithmic curve, because its abscissas are proportioned to the logarithms of numbers represented by its ordinates. (Alg.527,) 20. As the abscissas are the distances from AC, on the line AF, it is evident, that the abscissa of the point C is 0, which is the logarithm of 1=AC. (Art. 2.) The distance from A to 1 is the logarithm of the ordinate «, which is the radix of the system. For Briggs' loijarithms. this ought to be ten times AC. The distance from A to 2 is the logarithm of the ordinate a^ ; from A to 3 is the logarithm of a^, &ic. 21. The logarithms of numbers less than a unit are nega- tive. (Art. 9.) These may be represented by portions of the line AN, on the opposite side of AC. (Alg. 507.) The or- dinates rt~S ^~^? «~^i <^c. are less than AC, which is taken for unity -, and the abscissas, which are the distances from A to -1, -2, -3, he. are negative. 22. If the curve be continued ever so far, it will never meet the axis AN, For, as the ordinates are in geometrical progression decreasing, each is a certain portion of the pre- ceding one. They will be diminished more and more, the farther they are carried, but can never be reduced absolutely to nothing. The axis AN is, therefore, an asymptote of the curve. (Alg. 545.) As the ordinate decreases, the abscissa increases; so ihat, when one becomes infinitely small, the other becomes infinitely great. This corresponds with what has been stated, (Art. 15.) that the logarithm ofO hinfinite and nesrative. CURVE. 9 23. To find the equation of this curve, Let a=the radix of the system, x=any one of the abscissas, y=the corresponding ordinate. Then, by the nature of the curve, (Art. 19.) the ordinate to any point, is that power of a whose exponent is equal to the abscissa of the same point ; that is (Alg. 528.) * For other properties of the logarithmic curve, see Fluxions. SECTION II. )lRi:CTIOiNS FOR TAKING LOGARITHiMS AND THEIR NUMBERS FROM THE TABLES.* \ ^'^ 1 rP^^E purpose which logarithms are intended to ans\yer, is to enable us to perform arithmetical operations w'lih greater expedition, than by the common meth- ods. Before anyone can avail himself of this advantage, he must become so familiar with the tables, that he can readily find the logarithm of any number ; and, on the other hand, ihe number to vvliich any logarithm belongs. In the common tables, the indices to the logarithms of the first 100 numbers, are inserted. But, for all other numbers, the decimal part only of the logarithm is given ; while the index is left to be supplied, according to the principles in arts. 8 and 11. 25. To find the logrrithm of any 7iumber betzoeen 1 and ] 00 •, Look for the proposed number, on the left ; and against it. in the next column, will be the logarithm, with its index. Thus Tne log. of 18 h 1.25527. The log. of 73 is 1.86332. 96. To find the loganthm of any number between 100 and 1000 ; or of any number consisting of not more than three significant figures, with ciphers annexed. In the smaller tables, the three first tigures of each num- ber, are generally placed in the left hand column ; and the fourth tigure is placed at the head of the other columns. Any number, therefore, between 100 and 1000, may be found on the left hand ; and directly opposite, in the next column, is the decimal part of its logarithm. To this the index must be prefixed, according to the rule in art. 8. * The be?t En-lish Tables are Hutton's in 8vo. and Taylor's in 4to. h these, the logarithms are carried to seven places of decimals, and proportionsi parts are placed in the margin. The smaller tables are numerous : ai; .' Avheu accurately printed, aresuHicient for common calculation?. THE LOGARITHMIC TABLES. Jl The log. of 458 is 2.66087, The log. of 935 is 2.97081, of 796 2.90091, of 086 2.58659. If there arc ciphers annexed to I'hc significant fii^ures, the logarithm may be found in a similar manner. For, by art. 14, the decimal p;irt of the logarithm of any number is the same, as that of the number muhiplied into 10, 100, &c. All the difference will he in the index : and this may be supplied by the same general rule. The log. of 4580 is 3.66087, The log. of 326000 is 5.5 1 322, 'of 79600 4.90091, of 8010000 6.90363. 27. To find the logarithm of any number consisting o/four figures^ either zvithj or without, ciphers annexed. Look for the three first figures, on the left hand, and for the fourth figure, at the head of one of the columns. The logarithm will bo found, opposite the three first figures, and in the column which, at the head, is marked with the fourth figure.* The log. of 6234 is 3.79477, The log. of 783400 is 5.89398, of5231 3.71858, ofG281000 6.79803. 28. To find the logarithm of a number containing more than FOUR significant figures. By turning to the tables, it will be seen, that if the dij^er- ences between several numbers be small, in comparison with the numbers themselves ; the diflfbrences of the logarithms will be nearly proportioned to the differences of the nym- hers. Thus The log. of 1000 is 3.00000, ^rmni Q nr\r\io Here the di.Tereuces in the of 1 00 1 3.00043, ,^^^^j,^,^^ ^^^^ 1 , 2, 3, 4, &c. and of 1002 3.00087, the correspondiDg diflerences in of 1003 3.00130, thelo-arithmsare43, 87, 13(', of 1004 3.00173, &c. ^'^'^''• Now 43 is nearly half of 87, one third of 130, one fourth ofl73, s ^66362, or 8.66362. of_JL- 3.04376, or 7.04376. 32. If the logarithm of a mixed number \s required, reduc* it to an improper fraction, and then proceed as before. The logarithm of 3^=- V is 0. 57724. 33. To find the natural number belonging to an-^ loga- rithm. In computing by logarithms, it is necessary, in the first place, to take from the tables the logarithms of the numbers which enlcrinto the calculation ; and, on the other hand, al: the close of the operation, to find the number belonging to rABLES. i., the loganihm obtained in the result. This is evidently done, hy reversing the methods in the preceding articles. Where great accuracy is not required, look in the tables for the logarithm which is nearest to the given one ; and directly opposite, on the left hand, will be found the three first fig- ures, and at the top, over the logarithm, the fourth figure, of the number required. This number, by pointing off decim- als, or by adding ciphers, if necessary, must be made to cor- respond with the index of the given logarithm, according to arts. 8 and 11. The natural number belonging to 3.86493 is 7327, to 1.62672 is 42.24, to2.90141_ 796.9, to^89115 0.07783. in the last example, the index requires that the first signifi- cant figure should be in the second place from units, and therefore a cipher must be prefixed. In other instances, it is necessary to anjiex ciphers on the right, so as to make the number of figures exceed the index by 1. The natural number belonirinti to 6.71 5i3-Z is 5196000, to T.6.5677 is 0.004537, to 4.67062 46840, to "iTo 9 302 0.0003963. 34. When great accuracy is required, and the given loga- rithm is not exactly, or very nearly, found in the tables, it will be necessary to reverse the rule in art. 28. Take from the tables two logarithms, one the next greater, the other the next less than the given logarithm. Find the difference of the two logarithms, and the difference of their natural numbers; also the difference between the least of the two logarithms, and the given logarithm. Then say. As the difference of the two logarithms, To the difference of their numbers ; So is the difference between the given logarithm and the least of the other two. To the proportional part to be added to the least of the two numbers. 16 THE LOGARITHMIC TABLES. Required the number belonging to the logarithm 2.67325. Nextgreatlog. 2.67330. Its numb. 471.3. Given log. 2.67325. Next less 2.67321. Its numb. 471.2. Next less 2.67321. Differences 9 0.1 Then 9 : .1 ! :4 : 0.044, which is to be added to the number 471.2 The number required is 471.244. The natural number belonging to 4.37627 is 23783.45, to J_.73698 is 54.57357, to 3.69479 4952.08, toT7o9214is 0.123635. 35. Correction of the tables. The tables of logarithm> have been so carefully and so repeatedly calculated, by the ablest computers, that there is no room left to question their general correctness. They are not, however, exempt from the common imperfections of the press. But an errour of this kind is easily corrected, by comparing the logarithm with any two others to whose sum or difference it ought to be equal. (Art. 1.) Thus 48=24X2 = 16X3 = 12X4=:8X6. Therefore, the logarithm of 48 is equal to the sum of the logarithms of 24 and 2, of 16 and 3, &ic. And3 = |=:V = ¥ = '/ = V> <^c. Therefore, the loga- rithm of 3 is equal to the difference of the logarithms of 6 and 2, of 12 and 4, &c. SECTION III. METHODS OF CALCULATING BY LOGARITHMS. . - ^PHE arithmetical operations for which loga- ^^* * rithms were originally contrived, and on which theirgreat utility depends, are chiefly multiplication, division, involution, evolution, and finding the term required in single and compound proportion. The principle on which all these calculations are conducted, is this; If the logarithms of two numbers be added, the sum will be the logarithm of the product of the numbers ; and If the logarithm of one number be subtracted from that of an- other, the iJiFFERENCE will be the logarithm of the quotient of one of the numbers divided by the other. In proof of this, we have only to call to mind, that loga- rithms are the exponents of a series of powers and roots. (Arts. 2, 5.) And it has been shown, that powers and roots are multiplied by adding their exponents; and divided, by subtracting their exponents. (Alg. 233, 237, 280, 286.) MULTIPLICATION BY LOGARITHMS. 37. ADD the logarithms of the FACTORS : the SUM WILL BE THE LOGARITHM OF THE PRODUCT. In making the addition, 1 is to be carried, for every 10, from the decimal part of the logarithm, to the index. (Art. 7.) Numbers. Logarithms, Numbers. Logarithms, Mult, 36.2 (Art. 30.) 1.55871. Mult. 640 2.80618 Into 7.84 0.89432. Into 2.316 0.36474 Prod. 283.8 2.45305 Prod. 1482 3.17092 The logarithms of the two factors are taken from the ta- bles. The product is obtained, by finding, in the tables, the natural number belonging to the sum. (Art. 33.) 4 18 MULTIPLICATION BY Mult. 89.24 1.95056 Mult. 134. 2.12710 Into 3.687 0.56667 Into 25.6 1.40824 Prod. 329. 2.51723 Prod. 3430 3.53534 38. When any or all of the indices of the logarithms are negative, they are to be added according to the rules for the addition of positive and negative quantities in algebra. But it must be kept in mind, that the decimal part of the loga- rithm is ;)05?7tre. (Art. 10.) Therefore, that which is car- ried from the decimal part to the index, must be considered positive also. Mult. 62.84 1.79824 Mult. 0.0294 2^46835 Into 0.682 T.83378 Into 0.8372 1.92283 Prod. 42 86 1.63202 Prod. 0.0246 2.39118 In each of these examples, +1 is to be tarried from the decimal part of the logarithm. This added to — 1, the lower index, makes it ; so that there is nothing to be added to the upper index. If any perplexity is occasioned, by the addition of positive and negative quantities, it may be avoided, by borrowing 10 to the index. (Art. 12.) Mult. 62.84 Into 0.682 1.-9824 9.83378 Mult. Into Prod. 0.0294 0.8372 8.46835 9.92283 Prod. 42.86 1.63202 0.0246 8.39118 Here 10 is added to the negative indices, and afterwards rejected from the index of the sum of the logarithms. Multiply 2G.83 J^.42862 1.42862 Into 0.00069 4.83885 or 6.83885 Product 0.0185 2.26747 8.26741 LOGARITHMS. 10 Here -f-l carried to —4 makes it — 3, which added to the upper index +1, gives —2 for the index of the sum. Multiply .00845 3.92686 or 7.92686 Into 1068. 3.02857 3 02857 Product 9.0246 0.95543 0.95543 The product of 0.0362 into 25.38 is 0.9188 of 0.00467 into 348.1 is 1.625 of 0.0861 into 0.00843 is 0.0007258 39. /7«y number of factors may be multiplied together by adding their logarithms. If there are several positive^ and several negative indices, these are to be reduced to one, as in algebra, by takiiig the difference between the sum of those which are negative, and the sum of those which are positive, increased by what is carried from the decimal part of the logarithms. (Alg. 78.) Multiply 6832 3.83455 3.83455 Into 0.00C63 X^93601 or 7.93601 And 0.651 _K81358 9.81358 And 0.0231 "2.36361 or 8.36361 And 62.87 1.79344 1.79844 Prod. 55.74 1.74619 ^.74619 Ex. 2. The prod, of 36.4X7.82X68.91X0.3846 is 7544] 3. The prod, of 0.00629 X2. 647 X0.082 X 278.8 X0.00063 is 0.0002398. 40. Negative quantities are multiplied, by means of loga- rithms, in the same manner as those which are positive. (Art. 16.) But, after the operation is ended, the proper sign must be applied to the natural number expressing the product, ac- cording to the rules for the multiplication of positive and negative quantities in algebra. The negative index o(di log- 20 MULTIPLICATION BY arithm, must not be confounded with the sign which denotes that the natural number is negative. Th:it which the index of the logarithm is intended to show, is not whether the nat- ural number is positive or negative^ but whether it is greater or less than a imit. (Art. 16.) Mult, -f-36.42 1.56134 Mult. —2.681 0.42330 Into —67.31 1.82808 Into +37.24 1.57101 Prod. —2451 3.38942 Prod. —99.84 1.99931 In these examples, the logarithms are taken from the ta- bles, and added, in the same manner, as if both factors were positive. But after the product is found, the negative sign is prefixed to it, because + is multiplied into — . (Alg. 105.) Mult. 0.263 j741996 Mult. 0.065 "^81291 Into 0.00894 T.95134 Into 0.693 T84073 Prod. 0.002351 3.37130 Prod. 0.04^)04 2.5364 Here, the indices of the logarithms are negative, but the product is positive, because the factors are both positive. Mult. —62.59 1.79650 Mult. —68.3 183442 Into —0.00863 "3'.93601 Into --0.0096 "3.98227 Prod. -1-0.5402 1.73251 Prod. -f 0.6557 1.81669 LOGARITHMi^. 21 Division nv Logarithms, 41. From THE logarithm of the DIVIDEND, SUB- TRACT THE LOGARITHM OF THE DIN ISOR ; THE DIF- FERENCE WILL BE THE LOGARITHM OF THE QUOTIENT. (Art. 36.) Numbers. Logarithms. Numbers. Logarithms. Divide 6238 3.79505 Divide 896.3 2.y5245 Bv 2982 3.47451 By 9.847 0.89330 Quot. 2.092 0.32054 Quot. 91.02 1.959 42. The decimal part of the logarithm may be subtracted as in common arithmetic. But for the indices, when either of them is negative, or the lower one is greater than the upper one, it will be necessary to make use of the general rule for subtraction in algebra ; that is, to chanjie the signs of the subtrahend, and then proceed a- in addition. (Alg. 82,) When 1 is earned i'rom the decimil part, this is to be considered affirmative, and applied to the index, before the sign is changed. Divide 0.8697 7793937 or 9.93937 By 98.65 1.99410 1.99410 Quot. 0.008816 3.94527 7.94527 In this example, the upper logarithm being less than the lower one, it is necessary to borrow 10, as in other cases of subtraction ; and therefore to carry 1 to the lower index, which then becomes +2. This changed to — 2, and added to — 1 above it, makes the index of the difference of the log- arithms — 3. Divide 29.76 1.47.363 1.47363 By 6254 3.79616 3.7Q616 Quot. 0.00476 3.67747 or 7.67747 22 INVOLUTION BY Here, 1 carried to the lower index, makes it -{-4. This changed to — 4, and added to 1 above it, gives — 3 for the index of the difference of the logarithms. Divide 6.832 0-83455 Divide 0.00634 3.80209 By .0362 2.55871 By 62.18 1.79365 Quot. 188.73 2.27584 Quot. 0.000102 4^0844 The quotient of 0.0985 divided by 0.007241, is 13.6. The quotient of 0.0621 divided by 3.68, is 0.01687. 43. To divide negative qnnntities, proceed in the same manner as if they were positive, (Art. 40 ) and prefix to the quotient, the sign which is required by the rules for division in algebra. Divide +3642 3.56134 Divide —0.657 r81757 By —23.68 1.37438 By -f0.0793 2789927 Quot. —153.8 2.18696 Quot. —0.285 0.91830 In these examples, the sign of the divisor being different from that of the dividend, the sign of the quotient must be negative. (Alg. 123.) Divide— 0.364 T.56110 Divide —68 5 ^83569 By —2.56 0.40824 By -f 0.094 Y.913]3 Q„ot. 4-0.1422 T 15286 Quot. —728.7 2.86256 Involution by Logarithms. 44. Involving a quantity is multiplying it into itself. By means of logarithms, multiplication is performed by addition If, thon. the logarithm of any quantity be added to itself, the LOGARITHMS. 23 lofjanthm o{ a power of that quantity will be obtained. But adding a logarithm, or any other quantity, to itself, is multi- plication. The involution of quantities, by means of loga- rithms, is therefore perfornied, by multiplying the logarithms. Thus the logarithm of 100 is 2 of 100X100, that is, of 1002 is 2+2 =2X2. of 100X100X100, 2^3 is 24-2 + 2 =2X3. of 100X100X100X100, 100* is 2 + 2 + 2 + 2 =2X4. On the same principle, the logarithm of lOO" is 2Xn. And the logarithm of x" , is (log. x)Xn. Hence, 45. To involve a quantity by logarithms. MULTIPLY THE LOGARITHM OF THE QUANTITY, BY THE INDEX OF THE I'OWER REQUIRED. The reason of the rule is also evident, from the considera- tion, that logarithms are the exponents of powers and roots, and a power or root is involved, by multiplying its index into the index of the power required. (Alg. 220,288.) Ex. 1. What is the cube of 6.296? Root 6.296, its log. 0.79906 Index of the power 3 Power 249.6 2 39718 2. Required the 4th power of 21.32 Root 21.32 log. 1.32879 Index 4 Power 206614 5.31516 3. Required the 6th power of 1.689 Root 1.689 log. 0,2276 Index G Power 23.215 1.36578 I 24 INVOLUTION BY 4. Required the I44th power of 1.003 Root 1.003 log. 0.00130 Index 144 Power 1.539 0.18720 46. It must be observed, as in the case of multiplication, (Art. 38.) that what is carried from the decimal part of the logarithm \s positive, whether the index itself is positive or negative. Or, if 10 be added to a negative index, to render it positive, (Art. 12.) this will be multiplied, as well as the other figures, so that the logarithm of the square, will be 20 too great ; of the cube, 30 too great, he. Ex. 1. Required the cube of 0.0649 Root 0.0649 log. "2.81224 Index Power 0.0002733 4.43672 or 8.81224 3 6.43672 2. Required the 4th power_of 0.1234 Root 0.1234 log. 1.09132 or 9.09132 Index 4 4 Power 0.0002319 4.36528 6.36528 3. Required the 6th power of 0.9977 Root 0.9977 ]oicr the logarithm of the given quantity into the index of the power, (Art. 45.) and then dividing the product by the num- ber expressing the root. (Art. 47.) 1. What is the value of (53)'', that is, the 6th power of the 7th root of 53 ? Given number 53 log. 1.7242S Multiplying l)y 6 Dividing by 7)10.34568 Power required 30.06 1.47795 2. What is the 8th power of the 9th root of 654 .' Proportio.v by Logarithms. 52. In a proportion, when three terms are given, tlie fourth is found, in common arithmetic, by multiplying together the second and third, and dividing by the first. But when lo^a- arithms are used, addition takes the place o "multiplication, and subtraction, of division. To find then, by locrarithms, the fourth term in a propor- tion, ADD THK LOGARITHMS OF THE SECOND AND THIRD TERMS. AND from the .^7ir/* SUBTRACT the LOGARiTHnr of 28 ARITHMETICAL THE FIRST TER3I. Tlic remainder will be the logarithm oi the term required. Ex. 1. Find a fourth proportional to 7964, 378, and 27960. Numbers. Logarithms. Second term 378 2.57749 Third term 27960 4.44654 7.02403 First term 7964 3.90113 Fourth term 1327 3.12290 2. Find a 4th proportional to 768, 381 and 9780. Second term 381 2.58092 Third term 9780 3.99034 6.57126 First term 7G8 2.88536 Fourth term 4852 3.68590 Arithmetical Complement. 53. When one number is to be subtracted from another, it is often convenient, first to subtract it from 10, then to add the difference to the other number, and afterwards to reject the 10. Thus, instead o( a — 6, we may put 10 — h-\-a — 10. In the first of these expressions, b is subtracted from a. In the other, b is subtracted from 10, the difference is added to f/, and 10 is afterwards taken from the sum. The two ex- pressions arc equivalent, because they consist of the same terms, with the addition, in one of them, of 10— 10=0. The alteration is, in fact, nothin;jj more than borrowing 10, for the sake of convenience, and then rejecting it in the result. Instead of 10, we may borrow, as occasion requires, 100, 1000, &c. Thus a— fc = 100— /; + a- 100 = 1 000— i + «— 1000, &c. 54. The DIFFERENCE between a given number and 10, or 100, or 1000, 4^c. is called the arithmetical complement of that number. COMPLEMENT. J9 The arithmetical complement of a number consisting: of Gjie integral figure, either with or without decimals, is found, by subtracting the number from 10. If there are izvo inte- gral figures, they are subtracted from 100 ; if three, from 1000, &c. Thus the arithmetical compl't of 3.46 is 10 — 3.46=6.54 of 34.6 is 100—34.6=65.4 of 346. is 1000— 346. =654. &c. According to the rule for subtraction in arithmetic, any number is subtracted from 10, 100, iOOO. &;c. by beginning on the right hand, and taking each figure from 10, after ///- rreasiiicr all except the first, by carrying I. Thus, if from 10.00000 We subtract 7.63125 The difference, orarith"! comp't is 2.3G875, which is ob- tained, by taking 5 from 10, 3 from 10, 2 from 10, 4 from 10, 7 from 10, and 8 from 10. But, instead of taking each fig- ure, increased by 1, from 10; we may take it rvithoiU being increased, from 9. Thus 2 from 9 is the same as 3 from 10, 3 from 9, the same as 4 from 10, kc. Hence, 55, To obtain the arithmetical complemext of a num- ber^ subtract the right hand significant Jigurc from ]0, arid each of the other figures from d. If, however, there are ciphers on the right hand of all the significant figures, they are to be set down without alteration. In taking the arithmetical complement of a logarithm, if the index is negative^ it must be added to 9 ; for adding a negative quantity is the same as subtracting a positive one. (Alg. 81.) The difference between — 3 and + 9. i.-? not 6, but 12. The arithmetical complement of 6.24897 is 3.75103 of2. 70649 is 11.29351 of 2.98643 7.01357 of3.G4200 6.35800 of 0.62430 9.37570 of 9.35001 0.64999 30 COMPOUND PROPORTION. 56. The principal use of the nrilhmeticr.l complement, is in working proportions by logiuithms; where some of the terms are to he added, and one or more to be subtracted. In the Rule of 'Pliree or simple proportion, two terms are to be added, and from the sum, the first term is to be subtracted. But if, instead of the logarithm of the tirst term, we substi- tute its arithmetical complement, this may be added to the sum of the other two, or more simply, all three may be ad- ded together, by one operation. After the index is diminish- ed by 10, the result will be the same as by the common meth- od. For subtracting a number is the same, as adilinji; its arithmetical complement, and then rejecting 10, 100, or 1000, from the sum. (Art. 53.) It'willgenerally be expedient, to place the terms in the same order, in which they are arranged in the statement of the pro- portion. I. As C'i73 «. c. 6.20252 2. As 253 a, c 7.59688 Is to 769.4 2.88615 Isto672.5 2.82769 So is 37.61 1.67530 So is 497 2.69636 To 4.613 0.G6397 To 1321.1 3.12093 As 46.34 a. c. 8.33404 4 As 9.85 a. c. 9.00656 Is to 892.1 2 95041 Is to 643 2.80821 So is 7.633 0.88298 So is 76.3 1 .88252 To 147 2. 1 674:3 To 4981 3.69729 Compound Piioi'ORTioN 57, In compound, as in single proportion, th<^ term !<'- quired may be found by logarithms, if we substitute addition for multiplication, and subtraction for division. Ex. 1. If the interest of $365, for 3 years and 9 months, be g82.13 ; what will be the interest of\$8940, for 2 years and 6 months ? In common arithmetic, the statement of the question is made in this manner, * See VVebl.ier's Arithmetic. C03irOUND PROPORTION. 31 .365 dollars } . o-, ^ ^ a m - ■ S 8940 dollars ; ^^. } : 82.13 dollars., s r^ t, t ' 3. /o years ) ( 2.5 years ) And the method of calculation is, to divide the product ol" the third, fourth, and tilth terms, by the product of the two first.* This, if lo^^arithms are used, will be to subtract the ^um of the logarithms of the two first terms, from the sum of the logarithms of the other three. .p .. , . ( 365 log. 2.56229 Iwo hrst terms ^ ^ -- n --?^no ( 3, to O.o7403 Sum of the logarithms 3.13632 j Third term 82.13 1.91450 r ,, 1 ^A, , ^ 8940 3.95134 lourthand htth terms < ^ r /% oo-o , 2.5 0.39/94 Sum of the logs, oflhr 3d, 4th, and 5th G. 26378 Do. ' 1st and 2d 3.13632 Term required 1341 3.12746 58. The calculation will be more e-i.nple, if, instead of .subtracting the logarithms of the two first terms, we add their arithmetica I, complements. But it must be observed, that eac/i arithmetical complement increases the index of (he logarithm by 10. If the arithmetical complement be introduced into tioo of the terms, the index of the sum of the logarithms will be 20 too great ; if it be in three terms, the index will be 30 too great, &ic. ^ 365 ^3.75 ,,, ^ , . J ..^.. a. c. /.43771 Iwo hrst terms < . -,_ o . ^^n-; ^o a. c. 9.42o97 Third term 82 13 1.91450 V tu 1 ^-Ai ^ S^^^^ 3.95134 b ourth and hfth terms < ^.5 0.39794 Term required 1341 23.12746 The result is the sjme as before, except that tiie index of the logarithm is 20 too great. 5i; COMPOUND INTEREST. Ex. 2, lilhe wages of 53 men for 42. days be 2200 dol- lars ; what will be the wages of 87 men for 34 days ? 53 men > . com-' J ^"^ ^^^l 42 days 5 ' --^^-- ^ 34 days J rp r , . (^ 53 a, c. S.27572 1 wo First terms •; ,^ „ c,-,^-,- (42 a, c, 8.37G70 Third term 2200 3.34242 terms < Fourth and fifth terms ^ i\^14r Term required 2923.5 3.46589 59. In the same manner, if the product of any number oi quantities, is to be divided, by theproduct of several others; we may add together the logarithms of the quantities to be divided, and the arithmetical complements of the logarithms of the divisors. Ex. If 29.G7X 346.2 be divided by G9.24 X7.862 X497 ; what will be the quotient ? Numbers to be divided Divisors 1.47232 2.53933 ( 29.67 ( 346.2 C 69.24 a. c, 8.i5964 ^ 7.862 a. c. 9.10447 ( 497 a, c. 7.30364 Quotient 0.03797 8.5794 In this way, the calculations in Conjoined Proportion may be expeditiously performed. CoiiPouxD Interlst. GO. In calculating compound interest, the amount for the first year, is made the principal for the second year ; the amount for the second year, the principal for the third year, what will be the number, at the end of the century? Ans. 132,730,000. 3. If the population of a country, at the end of a centur}^, is found to be 45,800,000; and if the yearly rate of increase has been j^o ; what was the population, at the commence- ment of the century ? Ans. 20 millions. 4. The population of the United States in 1810 was 7,240,000; in 1820, 9,625,000. What was the annual rate of increase between these two periods, supposing the in- crease each year to be proportioned to the population at the beginning of the year? IN'CREASE OF POPULATION. 37 loir. 9,625,000 — locr. 7.240,000 Here log. «=— ^- rr: Therefore, «=1.029; and yff^, or 2.9 percent, is the rale of increase. 5. In how many years, will the population of a i.ountry advance from two millions to five millions; supposing the yearly rate of inciease to be 3I0 ? Ans. 47| years. 6. If the population of a country, at a given time, be seven n^.illions ; and if the yearly rate of increase be ^\{h ; what will be the population at the end of 35 years ? 7. The population of the United States in 1800 wa> 5,306,000. VVhat was it in 1780, supposing the yearly rate of increase to be ^V ' 8. In what time, will the populalioii of a country advance from four millions to seven millions, if the latio of increase be -^- ? 9. What must be the rate of increase, that the population of a place may change from nine thousand to fifteen thou- sand, in 12 years ? If the population of a country is not aiPected by immigra- tion or emigration, the rate of increase will be equal to the ditference between the ratio of the ^irM^, and the ratio of the deaths, when compared with the whole population. Ex. 10- If the population of a country, at any given time, he ten millions ; and the ratio of the armual number of births to the whole population be ^\, and the ratio of deaths jj, what will hi the number of inhabitants^, at the end of 60 years ? Here in? yearly rate of increase = VT — t =T^n. And the population, at the end of 60 year?= 31,750,000. The rate of increase or decrease from immig ration or em- igration, will be equal to the ditrorence between the ratio of immigration and the ratio of emigration ; and if this differ- r?«>Or' 38 INCREASE OF POPULATION. ence be added to, or subtracted from, the difference between the ratio of the births and that of the deaths, the whole rata of increase will be obtained. Ex. 11. If in a country, the ratio of births be 3 01 the ratio of deaths ^^j a oi _J_. G 01 the ratio of immigration j\ the ratio of emigration and if the population this year be 10 millions, what will it be 20 years hence ? The rate of the natural increase =-^^__i_ = _i_ j That of increase from immigration = 5V~ 6'o = 3io ; The sum of the two is -- 6 0? And the population at the end of 20 years, is 12,611,000. 1 2. If the ratio of the births be -^'oi of the deaths, 3V, of immigration, j\, of emigration, ~, in what time will three millions increase to four and a half millions ? If the period in which the population will double be given; the numbers for several successive periods, will evidently be in a geometrical progression, of which the ratio is 2; and as the number of periods will be one less than the number of terms: If P=the first term, A=thc last term, n=the number of periods ; Then will A = PX2", (Alg. 439.) Or log. A = log. P. = log. P-{-nXlog. 2. Ex. 1. If the descendants of a single pair double once in 25 years, what will be their number, at the end of one thou- sand years ? The number of periods here is 40. And A = 2X2^ " = 2.199.200.000,000. EXPONENTIAL EHUATIONS. ^9 ■2. If the descendants of Noah, beginning with his three sons and their wives, doubled once in 20 years for 300 years ; what was their number, at the end of this time ? Ans. 196,608. 3. The population of the United States in 1820 being 9,625,000 ; what must it be in the year 2020, supposing it to double once in 25 years ? Ans. 2,464,000,000. 4. Supposing the descendants of the first human pair to double once in 50 years, for 1650 years, to the time of the deluge, what was the population of the world, at that time ? Exponential Equations. 62. An Exponential equation is one in which the letter expressing the unknown quantity is an exponent. Thus cf=^b, and 3c'=bc, are exponential equations. These are most easily solved by logarithms. As the two members of an equation are equal, their logarithms must also be equal. If the logarithm of each side be taken, the equation may then be reduced, by the rules given in algebra. Ex, What is the value of x in the equation 3^=243? Taking the logarithms of both sides log. (3*)=log. 243 But the logarithm of a power is equal to the logarithm of the root, multiplied into the index of the power. (Art. 45.) Therefore (log.3)Xi=log. 243 ; and dividing by log. 3, log. 243 2.38561 ^=l^i7r=o:^i2 = 5- So that 3^ =243. 63. The preceding is an exponential equation of the sim- plest form. Other cases, after the logarithm of each side is taken, may be solved by Trial and Err our ^ in the same man- ner as affected equations. (Alg. 503.) For this purpose, make two suppositions of the value of the unknown quantity, and find their errours : then say, 40 EXPONENTIAL As the dilFerence of the crroiirs, to the dif- ference of the assumed numbers ; So is the least errour, to the correction required in the corresponding assumed number. Ex. 1. Find the value ofx in the equation x" =256 Taking the logarithms of both sides (log. a[;)Xa; = log. 256 Let X be supposed equal to 3.5, or 3.6. By the first supposition. liy the second supposition . .x = 3.5, and log. a; = 0.54407 a=3.6, and log. a; = 0.55630 Multiplying by 3.5 Multiplying by 3.6 (log. ?;)Xx=l. 90424 (log. a^) X.r ==2.00268 log. 25G = 2.40ry24 log. 256=2.40824 Errour —0.50400 Errour —0.40556 Difference of the errours 0.09844 Then 0.09844 : 0.1 : 10.40556 : 0.41 19, the correction. This added to 3.6, the second assumed number, makes the value of x=4.0119. To ct)rrect this farther, suppose a = 4.01 !, or 4.012. By the first supposition. By the second supposition. .i=4.0n,andlog. a: = 0.60325 a;=4.012,and log. a; = 0.60336 Multiplying by 4.011 Multiplying by 4.012 (log. x)Xx=2.41963 (log. a:) X a; =2.42068 log. 256=2.40824 log. 256=2.40824 Errour +0.01139 Errour +0.01244 Difference of the errours 0.00105 Then 0.00105 : 0.001 : ! 0.01 139 : 0.011 very nearly. Subtracting this correction from the first assumed number 4.011, we have the value of x=4, which satisfies the condi- tions of the proposed equation ; for 4* =256, 2. Reduce the equation 4a:* =100a:^. Ans. x = 5. 3. Reduce the t-quaiion a - =9,t. EQUATIONS. 41 04. The exponent of a power may be itself a power, as in the equation a^" =6. where x is the exponent of the power nf , which is the expo- nent of the power a"'"" . Ex. 4. Find the ralue of x, in the equation 9^ =1000. 3^ X(log. 9) = log. 1000. Therefore 3' =^^1^:9 =3.14 log. 9 Then as 3 =3.14. x (log. 3) = log. (3. 14.) Therefore x=^'^^^d^Ll^-^:\^\%\^ = \.OA. log. 3. In cases like this, wliere the factors, divisors, &c. are loga- rithms, the calculation may be facihtated, by taking the /oo-- arithms of the logarithms. Thus the value of the fraction •HttIt! 's most easily found, by subtracting the logarithm of the logarithm which constitutes the denominator, from the logarithm of that which forms the numerator. Find the value of x, in the equation ^^"^ + are, in any circle, each equal to And the tangent of Ab the radiusj and therefore equal to each other. Demonstration* 1 In the quadrant ACH, (Fig. 5.) the arc AH is 90<*. The sine of this, according to the definition, (Art. 82.) is CH, the radius of the circle. 2. Let AS be an arc of 60°. Then the angle ACS, be- ing measured by this arc, will also contain 60° ; (Art. 75.) and the triangle ACS will be equilateral. For the sum of the three angles is equal to 180°. (Art. 76.) From this taking the angle ACS, which is 60°, the sum of the remain' ing two is 120°. But these two are equal, because they are subtended by the equal sides, CA and CS, both radii of the circle. Each, therefore, is equal to half 120", that is to 60°. * Sines is here put for the square of the sine, cos 2 for the square ol the cosine &c. o6 TRIGONOMETRY. All the angles being equal, the sides are equal, and therefore AS, the chord of 60°, is equal to CS the radius. 3. Let AR be an arc of 45°. AD will be its tangent, and the angle ACD subtended by the arc, will contain 45°. The angle CAD is a right angle, because the tangent is, by defi- nition, perpendicular to the radius AC. (Art. 84.) Subtract- ing ACD, which is 45°, from 90°, (Art. 77.) the other acute angle ADC will be 45° also. Therefore the two legs of the triangle ACD are equal, because they are subtended by equal angles; (Euc. 6. 1.) that is, AD the tangent of 45°, is equal to AC the radius. Cor. The cotangent of 45° is also equal to radius. For the complement of 45° is itself 45°. Thus HD, the cotan- gent of ACD, (Fig 5. is equal to AC the radius. 98. The sine of 30° is equal to half radius. For the sine of 30° is equal to half the chord of 60°. (Art. 82. cor.) But by the preceding article, the chord of 60° is equal to radius. Its half, therefore, which is the sine of 30° is equal to lialf radius. Cor. 1. The cosine of 60° is equal to half radius. For the cosine of 60° is the sine of 30° (Art. 89.) Cor. 2. The cosine of 30° = |v^3. For Co5230°=R— sm230=l -\ = h Therefore, Cos 30°=v/| = i\/3. 96. 6. The sine of 45°=-77;. For R-=l=5m2 45°-f-co52 45°=2 sm^ 45° J_ 72 Therefore, Sin 45° = v'^ = 97. The chord of any arc is a mean proportional^ between the diameter of the circle, and the versed sine of the arc. Let ADB CFig. 6.) be an arc, of which AB is the chord, BF the sine, and AF the versed sine. The angle ABH is a right angle, (Euc. 31. 3.) and the triangles ABH and ABF are similar. (Euc. 8. 6.) Therefore, AH : AB::AB : AF. . SINES, TANGENTS, kc, 57 That is, the diameter is to the chord, as the chord to the versea sine. In Fig. 6th, let the arc AD=a, and ADB=2a. Draw BF perpendicular to AH. This will divide the right angled tri- angle ABH into two similar triangles. (Euc. 8. 6.) The an- gles ACD and AHB are equal. (Euc. 20. 3.) Therefore the four triangles ACG, AHB, FHB, and FAB are similar; and the line BH is twice CG, because BH : CO * : HA : Ca! The sides of the four triangles are AG=5m cr, CG =cos a, HF =vers. sup. Sa, AB = 2sin a, BH = 2co5 a, AC=the radius, BF=sin2a, AF=vers 2a, AH=the diameter. A variety of proportions may be stated, between the ho- mologous sides of these triangles : For instance, By comparing the triangles ACG and ABF, AC : AG: :AB : AF, that is, R : sin a::2s{na : vers 2a AC : CG: : AB : BF R : cos a: \'2sin a : sin 2a AG : CG : : AF : BF Sin a : cos a '. '.vers 2a t sin 2a Therefore, RXvers 2a=2sin^a R X sin 2a =2sm a X cos a Sin aXsin 2a=vers 2a X cos a By comparing the triangles ACG and BFH, AC : CG: :BH : HF, that is, R : cos a: \2cosa : vers, sup, 2a AG : CG : : BF : HF Sin a l cos a : '.sin 2a *, vers, sup, 2a Therefore, R X vers. sup. 2a=2cos^ a Sin aXvers. sup, 2a-=-cos aXsin 2a rrond. 62 THE TRIGONOMETRICAL 60" is 48 ; the difference for 10" must be a proportional pari of 48. That is, 60" : 10": :48 : 8 the correction to be ad- ded to the sine of 14° 43'. Therefore the sine of 14^ 43' 10" is 9.40498 ?. What is the logarithmic cosine of 32° 16' 45" ? The cosine of 32° 16' is 9.92715 of 32° 17' 9.92707 Difference 8 Then 60" : 45":: 8 : 6 the correction to be subtracted from the cosine of 32° 16'. Therefore the cosine of 32° 16' 45" is 9.92709 The tangent of 24° 15' 18'' is 9.65376 Thecotangentof 31° 50' 5" is 10.20700 The sine of 58° 14' 32" is 9.92956 The cosine of 55° 10' 26" is 9.75670 If the given number of seconds be any even part of 60, as ^, i, i, k : c/«::CD : Cd, TRIANGLES. 69 That IS, as the sine in one circle, to the sine in the otlier: so is the tangent in one, to the tangent in the olhei ; and so is the secant in one, to the secant in the other. This is a general principle, which may be applied to most trigonometrical calculations. If one of the side.^ of the pro- posed triangle be made radius, each of the other sides will be the sine, tangent, or secant, of an arc described by this ra- dius. Proportions are then stated, between these lines, and the tabular radius, sine, tangent, &ic. 120. A line is said to be made radius, when a circle is de- scribed, or supposed to be described, whose semidiameter is equal to the line, and whose centre is at one end of it. 121. In any right angled triangle, if the hypothenuse be made radius, one of the legs will be a sine of its opposite an- gle, and the other leg a cosine of the same angle. Thus, if to the triangle ABC (Fig. 14.) a circle be applied, whose radius is AC, and whose centre is A, then BC will be the sine, and BA the cosine, of the angle at A. (Art. 82, 89.) If, while the same line is radius, the other end C be made the centre, then BA will be the sine, and BC the cosine, of the angle at C. 122. If either of the legs he made radius, the other leg will be a TANGENT of its opposite an^le, and the hypothenuse will be a SECANT of the same angle; that is, of the angle between the secant and the radius. Thus, if the base A.^ (Fig. 15.) be made radius, the centre being at A, BC will be the tangent, and AC the secant, of the angle at A. (Art. 84, 85.) But, if the perpendicular BC (Fig. 16.) be made radius, with the centre at C, then AB will be the tangent, and AC the secant, of the angle at C. 123. As the side which is the sine, tangent, or secant of one of the acute angles, is the cosine, cotangent, or cosecant of the other; (Art. 89.) the perpendicular BC (Fig. 14.) is the sine of the angle A, and the cosine of the angle C ; while the base AB is the sine of the angle C, and the cosine of the angle A. If the base is made radius, as in Fig. 15, the perpendicular BC is the tangent of the angle A, and the cotangent of the an- gle C ; while the hypothenuse is the secant of the angle A, and the cosecant of the angle C. If the perpendicular is made radius, as in Fig. 16, the base AB is the tangent of the angle C, and the cotangent of the 7u RIGHT ANGLED angle A; while the hypoihenuse is the secant of the angle C, and the co^eca?it ol' the angle A. 124. Whenever a right angled triangle is proposed, whose sides or angles are required ; a similar triangle may be formed, from the sines, tangents, &ic. of the tables, (Art. 117, 118.) The parts required are then found, by stating proportions be- tween the similar sides of the two triangles. If the triangle proposed be ABC, (Fig. 17.J another abc may be formed, having the same angles with the first, but differing from it in the length of its sides, so as to correspond with the numbers in the tables. If similar sides be made radius in both, the re- maining similar sides will be lines o^ the same name ^ that is, if the perpendicular in one of the triangles be a sine, the per- pendicular in the other will be a sine ; if the base in one be a cosine, the base in the oiber will be a cosine, &;c. If the hypothenuse in each trian^^le be made radius, as in Fig. 14, the perpendicular be will be the tabular sine of the angle at a ; and the perpendicular BC will be a sine of the equal angle A, in a circle of which AC is radius. If the base in each triangle be made radius, as in Fig. 15, then the perpendicular be will be the tabular tangent of the angle at a ; and BC will be a tangent of the equal angle A, in a circle of which AB is radius, he. 125. From the relations of the similar sides of these trian- gles, are derived the two following theorems, which are suf- ficient for calculating the parts of any right angled triangle whatever, when theVequisite data are furnished. One is used, when a side is to be found ; the other, when an angle is to be found. Theorem I. 126. When a side is required ; As THE TABULAR SINK, TANGENT, &tC. OF THE SAME NAME WITH THE GIVEN SIDE, To THE GIVEN SIDE; So IS THE TABULAR SINE, TANGENT, &;C. OF SAME NAME WITH THE REQUIRED SIDE, To THE REQ,UIRED SIDE. It will be readily seen, that this is nothing more than a statement, in general terms, of the proportions between the TRIANGLES. 71 similar sides of two triangles, one proposed for solution, and the other formed from the numbers in the tables. Thus if the hypothenusfe be given, and the base or perpen- dicular be requiml ; then, in Fig. 14, where ac is the tabular radius, be the tabular sine of «, or its equal A, and ah the tab- ular sine of C; (Art. 124.) ac : AC: :6c : BC, that is, R : AC::Sin A : BC. ac : kCwah : AB, R : AC::Sin C : AB. In Fig. 16, where ah is the tabular radius, ac the tabular secant of A, and be the tabular tangent of A ; ac : AC: :5c : BC, that is Sec A : AC: :Tan A : BC. ac :AC::ah : AB, Sec A : AC::R : AB. In Fig. 16, where be is the tabular radius, ac the tabular secant of C, and ab the tabular tangent of C ; ac : AC ::bc : BC, that is, Sec C : AC::R :BC. ac : ACy.ab: AB, Sec C : AC : :Tan C : AB. Theorem II. 127. When an angle is required ; As THE GIVEN SIDE MADE RADIUS, To THE TABULAR RADIUS ; So 15 ANOTHER GIVEN SIDE, To THE TABWLAR SINE, TANGENT, &IC. OF THK SAME NAME. Thus if the side made radius, and one other side be given, then, in Fig. 14, AC : ac: :BC : he, that is, AC : R: :BC : Sin A. AC : ac:: AB : ah AC :R::AB : Sin C In Fig, 15, AB : a6::BC : be, that is, AB : R::BC : Tan A. AB : abr.AC :ac AB : R: :AC : Sec A. In Fig. 16. BC : he:: AB : a6,thatis, BC :R::AB : Tan C. BC : hc::AC : er BC : R::AC : Sec C. 7i RIGHT ANGLED It will be observed, that in these theorems, angles are not introduced, though they are among the quantities which are either s'ven or required, in the calculation of triangles. But the tabular sines, tangents, &£c. may be considered the repre- sentatives of angles, as one may be found from the other, by merely turning to the tables. 128. In the theorem for finding a side, the first term of the proportion is a tabular number. But, in the theorem for finding an angle, the fir^t term is a side. Hence, in applying the proportions to particular cases, this rule is to be observed, To find a side, begin xvith a tabular number, To find an ANfJLE, begin with a side. Radius is to be reckoned among the tabular numbers. 129. In the theorem for finding an angle, the first term is a side made radius. As in every proportion, the three first terms must be given, to enable us to find the fourth, it is evi- dent, that where this theorem is applied, the side made radi- us must be a given one. But, in the theorem for finding a side, it is not necessary that either of the terms should be ra- dius. Hence, 130. To find a side, any side may be made radius. To find an angle, a given side must be made radius. It will generally be expedient, in both cases, to make radi- us one of the terms in the proportion ; because, in the tables of natural sines, tangents, iic. radius is 1, and in the logarith- mic tables it is 10. (Art. 103.) 131. The proportions in Trigonometry are of the same nature as other simple proportions. The fourth term is found, therefore, as in the Rule of Three in Arithmetic, by multiply- ing together the second and third terms, and dividing their pro- duct by the first term. This is the mode of calculation, when the tables o^ natural sines, tangents, Sic. are used. But the operation by logarithms is so much more expeditious, that it has almost eniirfcly superseded the other method. In loga- rithmic calcut^iioas, addition takes the place of multiplica- tion ; and subtraction the place of division. The logarithms expressing the lengths of the sides of a tri- an^!: f^re to be taken from the tables of common logarithms. Tiv: logarithms of the sines, tangents, he. are found in the ta- bles of artificial sines, &c. The calculation is then made by TRIANGLES, 73 adding the f^econd and third terms^ and subtracting the first, (Art. b%) 132. The logarithmic radius 10, or, as it is written in the tables, 10.00000, is so easily added and tJubtracted, that the three terms of which it is one, may be considered as, in ef- fect, reduced to two. Thus, if Ihf* tabular radius is in the first term, we have only to add the other two terms, and then take 10 from the index ; for this is subtracting the first term. If radius occurs in the second term, the first is to be subtracted from the third, after its index is increased by 10. In the same manner, if radius is in the third term, the first is to be subtracted from the second. 133. Every species of right angled triangles maybe be sol- ved upon the principle, that the sides of similar triangles are proportional, according to the two theorems mentioned above. There will be some advantages, however, in giving the exam- ples in distinct classes. There must be given, in a right angled triangle, tzoo of the parts, besides the right angle. (Art. 1 16.) These may be ; 1. The hypothenuse and an angle ; or 2. The hypothenuse and a leg ; or 3. A leg and an angle ; or 4. The two less. Case I. io>. r- S The hypothenuse, } . r^ i S The base and 134. Given < * j .. i^ > to fand < ^ i;^ i I And an angle : ^ ( Perpendicular. Ex. 1. If the hypothenuse AC (Fig. 17.*) be 45 miles, and the angle at A 32^ 20', what is the length of the base AB, and the perpendicular BC ? In this case, as sides only are required, ani/ side may be made radius. (Art. 130.) If the hypothenuse be made radius, as in Fig. 14, BC will be the sine of A, and AB the sine of C, or the cosine of A. (Art. 121.) And if abc be a similar triangle, whose hypoth- enuse is equal to the tabular radius, be will be the tabular sine of A, and ab the tabular sine of C. (Art. 124.) * The parts which are given are distinguished by a mark across the line, or at the opening of the angle, and the parts required^ by a cipher, 1! 74 RIGHT ANGLED To find the perpendicular, then, by Theorem I, wc have this proportion ; ac : AC: :6c : BC. Or R : AC::SinA : BC. Whenever the terms Radius, Sine, Tangent, &c. occur in a proportion like this, the /a^w/ar Radius, &c. is to be under- stood, as in arts. 126, 127. The numerical calculation, to find the length of BC, may be made, either by natural sines, or by logarithms. See art. 131. By natural Sines. \ : 45::0.53484 : 24.068=BC. Computation by Logarithms, As Radius 10.00000 To the hypothenuse 45 1.65321 So is the Sine of A 32° 20' 9.72823 To the perpendicular 24.068 1.38144 Here, the logarithms of the second and third terms are ad- ded, and from the sum, the first term 10 is subtracted. (Art. 132.) The remainder is the logarithm of 24.068 = BC. Subtracting the angle at A from 90°, we have the angle at C==57° 40'. (Art. 1 16.) Then, to find the base AB ; ac I AC: °.ab t AB Or R : AC::SinC : AB=38.023. Both the sides required are now found, by making the hy- pothenuse radius. The results here obtained may be verifi- ed, by making either of the other sides radius. If the base be made radius, as in Fig. 15, the perpendicu- lar will be the tangent, and the hypothenuse the stcant of the angle at A. (Art. 122.) Then, Sec A : AC::R: AB R: AB::Tan A : BC TRIANGLES. 76 By making the arithmetical calculations, in these two pro- portions, the values of AB and BC will be found the same as before. U the perpendicular be made radius, as in Fig. iO, AB will be tbe tangent, and AC the secant of the angle at C. Then, Sec C : AC::R : BC R : BC::TanC : AB Ex. 2. If the hypothenuse of a right angled triangle be 250 rods, and the angle at the base 46^ 30' ; what is the length of the base and perpendicular ? Ans. The base is 172.1 rods, and the perpendic. 181.35. Case II. lOK n^.r^^ ^ The hypothenuse, > , ^ , ( The aneles and 135. Given ^ ^^^ ^^^ ^^^ ^ ,« find ^ ^^^ ^^^^^ 1^^^ Ex. 1. If the hypothenuse (Fig;. 18.) be 35 leagues, and the base 26 ; what is the length of the perpendicular, and the quantity of each of the acute angles ? To find the angles it is necessary that one of the given sides be made radius. (Art. 130.) If the hypothenuse he radius, the hsise and perpendicular will be sines of their opposite angles. Then, AC : R::AB : Sin C=47° 58'i And to find the perpendicular by Theorem I ; R : AC::Sin A : BC=23.43 If the base be radius, the perpendicular will be tangent, and the hypothenuse secant of the angle at A. Then, AB : R::AC : Sec A R : AB::TanA : BC In this example, where the hypothenuse and base arc giv- en, the angles can not be found by making the perpendicular radius. For to find an angle, a gzr^/i side must be made ra- dius. (Art. 130.) 76 RIGHT ANGLED 136. Ex. 2. If the hypothenuse (Fig.iy.) be 64 milesj. and the perpendicular 48 miles, what are the angles, and the base ? Making the hypothenust radius. AC : R::BC : Sin A R: AC::SinC : AB The numerical calculation will give A=62° 44', and AB =24.74. Making the perpendicular radius, BC :R::AC : Sec C R: BC::TanC : AB The angles can not be found by making the base radius, when its length is not given. Case III. 137. Given 11^^""^^^ ?to find f 1^",^^"^"' 5 A"d one leg ) (. And the other leg. Ex. 1. If the base (Fig. 20,) be 60, and the angle at the base 47° 12', what is the length of the hypothenuse and the perpendicular ? In this case, as sides only are required, any side may be ra- dius. Making the hypothenuse radius. SinC : AB::R : AC = 88.31 R : AC::Sin A : BC=64.8 Making the base radius. R : AB::SecA : AC R : AB::Tan A : BC Making the perpendicular radius^ TanC : AB::R:BC R:BC::SecC : AC TRIANGLES. 77 138. Ex. 2. If the perpendicular (Fig. 21.) be 74, and the angle C 61° 27', what is the length of the base and the hjpothenuse ? Making the hypothenuse radius. Sin A : BC::R : AC R : AC::SinC : AR Making the base radius. Tan A : BC::R: AB R: AB::SecA : AC Making ihe perpendicular radius. R : BC::SecC : AC R : BC::TanC : AB The hypothenuse is 154.83 and the base 136. Case IV. 139. Given ( The base, and ? x ^ j J Tlie hypothenuse, I Perpendicular 5 ^^ """^ { And the angles. Ex. 1. If the base (Fig. 22, be 284, and the perpendicular 192, what are the angles, and the hypothenuse ? In this case, one of the legs nnust be made radius, to find an angle; because the h)'pothenuse is not given. Making the base radius. AB: R:.:BC : Tan A = 34M' R : AB: :Sec A : AC = 342.84 Making the perpendicular radius. BC : R::AB: Tan C R:BC::SecC : AC Ex. 2. If the base be G40, and the perpendicular 480, what are the angles and hypothenuse ? Ans. The hypothenuse is 800, and the angle at the base 36° 52' 12". 78 KIGHT ANGLED Examples for practice, 1. Given the hypothenuse 68, and the angle at the base 39° 17' ; to find the base and perpendicular. 2. Given the hypothenuse 850, and the base 594, to find the angles, and the perpendicular. 3. Given the hypothej-ase 78 and perpendicular 57, to find the base, and the angles. 4. Given the base 723, and the angle at the base 64° 18', to find the hypothenuse and perpendicular. 5. Given the perpendicular 632, and the angle at the base 81° 36', to find the hypothenuse and the base. 6. Given the base 32, and the perpendicular 24, to find the hypothenuse, and the angles. 140. The preceding solutions are all effected, by means of the tabular sines, tangents, and secants. But, when any two sides of a right angled triangle are given, the third side may be found, without the aid of the trigonometrical tables, by the proposition, that the square of the hypothenuse is equal to the sum of the squares of the two perpendicular sides (Euc. 47. 1.) If the legs be given, extracting the square root of the sum of their squares, will give the hypothenuse. Or, if the hy- pothenuse and one leg be given, extracting the square root of the difference of the squares, will give the other leg. Let A=the hypothenuse ^ ^=the perpendicular > of a right angled triangle. 6=the base S Then h^=^b'--\-p\ or (Alg. 296.) h=^b^+p^ By trans. b^=h^-p'', or b=\ /h^ -p"" And p^-=h^-b\ or p = ^h^-b^ Ex. 1. If the base is 32, and the perpendicular 24, what is the hypothenuse ? Ans. 40. 2. If the hypothenuse is 100, and the base 80, what is the perpendicular? Ans. 60. 3. If the hypothenuse is 300, and the perpendicular 220, what is the base ? Ans. 300 —220=4160, the root of which is 204 nearly. TRIANGLES. 79 141. It is generally most convenient to find the difference of the squares by logarithms. But this is not to be done by subtraction. For subtraction^ in logarithms, performs the of- fice of division. (Art. 41.) If we subtract the logarithm of b^ from the logarithm of /i^, we shall have the logarithm, not of the difference of the squares, but of their quotient. There is, however, an indirect, though very simple method, by which the difference of the squares may be obtained by logarithms. It depends on the principle, that the dfference of the squares of two quantities is equal to the product of tlw. sum and difference of the quantities. (Alg. 235.) Thus t h^~-b^-={h-\-b)X{h—b) as will be seen at once, by performing the multiplication. The two factors may be multiplied by adding their loga- rithms. Hence, 142. To obtain the difference of the squares of tzoo quanti- ties, add the logarithm of the sum of the quantities, to the log- arithm of their dfference. After the logarithm of the differ- ence of the squares is found ; the square root of this differ- ence is obtained, by dividing the logarithm by 2. (Art. 47.) Ex, 1. If the hypothenuse be 75 inches, and the base 45, what is the length of the perpendicular? Sum of the given sides 120 log. 2,07918 Difference of do. 30 1.47712 Dividing by 2)3.55630 Side required 60 1.77015 2. if the hypothenuse is 135, and the perpendicular 108- what is the length of the base ? Ans. 81. SECTION IV. SOLUTIONS OF OBLIQUE ANGLED TRIANGLES. A 14^ T^HE sides and angles of oblique angled trian- ^^' * -*- gles may be calculated by the following the- orems. Theorem I. In any plane triangle, the sines of the angles are as THEIR OPPOSITE SIDES. Let the angles be denoted by the letters A, B, C, and their opposite sides by cr, 6, c, as in Fig. 23 and 24. From one of the angles, let the line p be drawn perpendicular to the op- posite side. This will fail either within or without the tri- angle. 1. Let it fall within as in Fig. 23. Then, in the right an- gled triangles ACD and BCD, according to art. 126, R : 6: :Sin A : p R : a::SinB : p Here, the two extremes are the same in both proportions. The other four terms are, therefore, reciprocally proportion- al: (Alg. 387.*) that is, a : i::Sin A : SinB. 2. Let the perpendicular p fall ivithoiit the triangle, as in Fig. 24. Then, in the right angled triangles ACD and BCD : R : 6: :Sin Alp R : a: :Sin B : p Therefore as before, a : 6: : Sin A J SinB. * Euclid 23. 5 OBLIQUE ANGLED TRIANGLES 81 Sin A is here put both for the sine of DAC, and for that of BAG. For, as one of these angles is the supplement of the other, they have the same sine. (Art. 90.) The sines which are mentioned here, and which are used in calculation, are tabular sines. But the proportion will be the same, if the sines be adapted to any other radius. (Art. 119.) Theorem !!• 3 44. In a plane triangle, As THE SUM OF ANY TWO OF THE SIDES, To THEIR DIFFERENCE ; So IS THE TANGENT OF HALF THE SUM OF THE OPPOSITE ANGLES ; To THE TANGENT OF HALF THEIR DIFFERENCE. Thus the sum of AB and AC (Fig. 25) is to their differ- ence ; as the tangent of half the sum of the angles ACB and ABC, to the tangent of half their difference. Demonstration. Extend CA to G, making AG equal to AB ; then CG is Uie sum of the two sides AB and AC. On AB, set off AD equal to AC ; then BD is the difference of the sides ABand AC. The sum of the two angles ACB and ABC, is equal to the sum of ACD and ADC ; because each of these sums is the supplement of CAD. (Art. 79.) But, as AC = AD by con- struction, the angle ADC = ACD. (Euc. 5. 1.) Therefore ACD is half the sum of ACB and ABC. As AB=AG, the angle AGB = ABG or DBE. Also GCE or ACD=ADC=: BDE. (Euc. 15. 1.) Therefore, in the triangles GCE and DBE, the two remaining angles DEB and CEG are equal : (Art. 79.) So that CE is perpendicular to BG. (Euc. Def. 10. 1.) If then CE is made radius, GE is the tangent of GCE, (Art, 84.) that is, the tangent of half the sum of the angles op- posite to AB and AC. If from the greater of the (wo angles ACB and ABC, there be taken ACD their half sum ; the remaining angle ECB will be their half difference. ( Alg. 34 1 .) The tangent of this angle, CE being radius, is EB, that is, the tangent of half the. dffcrence oj the angles opposite to AB and AC. We have then, I? Sii OBLiqUE ANGLED CG=tiie iiim of the sides AB and AC; T)B=t]ieir difference ; GE=the tangent of half the sum of the opposite angles 5 f'iB=the tangentof half their difference. But, by similar triangles, CG :DB::GE : EB. Q. E. D. Theorem III. 145. If upon the longest side of a triangle, a perpcndicu] Jar be drav/n from the opposite angle ; As THE LONGEST SIDE, To THE SUM OF THE TWO OTHERS ; So IS THE DIFFERENCE OF THE LATTER. To THE DIFFERENCE OF THE SEGMENTS MADE BY THE PERPENDICULAR. in the triangle ABC, (Fig. 26.) if a perpendicular be drawn from C upon AB; AB : CB+CA::CB -CA : BP-PA.* Demonstration* Describe a circle on the centre C, and with the radius BC. Through A and C, draw the diameter LD, and extend BA 10 H. Then by Euc. 35, 3, ABXAH=ALXAD Therefore, AB : AD::AL : AH J^ut AD=CD4-CA=CB-FCA And AL=:CL-CA=CB-CA AndAII = HP-PA = BP-PA (Euc. 3. 3.) If then, for the three last terms in the proportion, we sub* atitute their equals, we have, AB : CB+CA::CB-CA : PB-PA 146. It is to be observed, that the greater segment is next the areater side. If BC is greater than AC, (Fig. 26.) PB is * See note F. TRIANGLES. b:^ greater than AP. With the radius AC, describe the arc AN. The segment NP=AP. (Euc. 3. 3.) But BP is greater than NP. 147. The two segments are to each other, as the tangents of the opposite angles, or the cotangents of the adjacent an- gles. For, in the right angled triangles ACP and BCP. (Fig. 26.) if CP be made radius, (Art. 126.) R:PC::Tan ACP: AP R:PC::TanBCP : BP Therefore, by equality of ratios, (Alg. 384.*) Tan ACP : AP::TanBCP : BP That is, the segments are as the tangents of the opposite angles. And the tangents of these are the cotangents of the angles A and B. (Art. 89.) Cor. The greater segment is opposite to the greater an- gle. And of the angles at the base, the less is next the greater side. If BP is greater than AP, the angle BCP i^ greater than ACP : and B is less than A. (Art. 77.) 148. To enable us to find the sides and angles of an ob- lique angled triangle, three of them must be given. (Art. 1 1 4.) These may be, cither 1. Two angles and a side, or 2. Two sides and an angle opposite one of them, oi- 3. Two sides and the included angle, or 4. The three sides. The two first of these cases are solved by theorem 1, (An. 143.) the third by theorem II, (Art. 144.) and the fourth bv theorem III, (Art. 145.) 149. In making the calculations, it must be kept in mind, that the greater side is always opposite to the greater angle, (Euc. 18, 19. I.) that there can be only one obtuse angle in a triangle, (Art. 76.) and therefore, that the angles opposite to the two lea«t sides must be acute. oBJJQUE ANGLKI Case I. 150. Given. Two ancrl<^?-^rind f , r i S -t^'c remaining angle, and A side. 5 ^ ( The other two sides. The third angle is found, by merely subtracting the sura of the two which are given from 180^ (Art. 79 ) The sides are found, by stating, according to theorem I, the following proportion ; As tiie sine of the angle opposite the given side, To the length of the given side ; So is the sine of the angle opposite the required side, To the length of the required side. As a side is to be found, it is necessary to begin with a tab' ular number, Ex. 1. In the triangle ABC (Fig. 27.) the side h is given 32 rods, the angle A 56° 20', and the angle C 49° 10', to find the angle B, and the sides a and c. The sum of the two given angles 56° 20'-}-49° 10'= 105* 30'; which subtracted from 180°, leaves 74° 30' the angle B. Then, c- -D 1 . . ( Sin A : a .smB:6.. ^ gin C : ,: Calculation by logarithms. As the Sine of B 74° 30' a. c. 0.01609 To the side ^^ 32 1.50515 So is the Sine of A 56° 20' 9.92027 To the side a 27.64 74° 30^ «. 32 49° 10* 25.13 1.44151 As the Sine of B To the side b So is the Sine of C c. 0.01609 1.50515 9.87887 To the side c 1.40011 TRIANGLES. 06 The arithmetical compUment used in the tirst term here, maybe found, in the usual way, or by taking out the cosecant of the given angle, and rejecting 10 from the index. (Art. 113.) Ex. 2. Given the side h 71. the angle A 107° 6' and the angle C 27° 40 ; to find the angle B, and the sides a and c. The angle B is 45° 14'. Then Sin B : h (Sin (Sin A : a=95.58 C :c = 46.43 When one of the given angle? is obtuse, as in this example, the sine of its supplement is to be taken from the tables. (Art. 99.) Case 11. 151. Given Two sides, and ^ t fi 1 5 "^^^ remaining side, and An opposite angle, ) ^ \ The other two angles. One of the required angles is found, by beginning with a -ide, and, according to theorem I, stating the proportion. As the side opposite the given angle, To the sine of that angle ; So is the side opposite the required angle, To the sine of that angle. The third angle is found, by subtracting the sum of the oth- er two from 180"^ ; and the remaining side is found, by the proportion in the preceding article. 152. In this second case, if the side opposite to the given angle be shorter than the other given side, the solution will be ambiguous. Two different triangles may be formed, each -of which will satisfy the conditions of the problem. Let the side 6, (Fig. 28.) the angle A, and the length of the side opposite this angle be given. With the latter for ra- dius, (if it be shorter than 6,) describe an arc, cutting the line AH in the points B and B'. The lines BC and B'C will be equal. So that, with the same data, there may be formed two different triangles. ABC and AB'C. 6u UBLiqUK ANGLED There will be the same ambiguity in the numerical calcu- lation. The answer found by the proportion will be the sine of an an^le. But this may be the sine, cither of the acute angle ABC, or of the obtuse angle ABC. For, BC being equal to B'C, the angle CB'B is equal to CBB'. Therefore ABC, which is the supplement of CBB' is also thesupplement of CB'B. But the sine of an angle is the same, as the sine of its supplement. (Art. 90.) The result of the calculation will, therefore be ambiguous. In practice however, there will generally be some circumstances which will determine whether the angle required is acute or obtuse. If the side opposite the given angle be longer than the other given side, the angle which is subtended by the latter, will necessarily be acute. For there can be but one obtuse angle in a triangle, and this is always subtended by the long- est side. (Art. 149.) If the given angle be obtuse, the other two will, of course, be acute. There can, therefore, be no ambiguity in the so- lution. Ex. 1. Given the angle A, (Fig. 28.) 35° 20', the oppo- site side a 50, and the side 6 79 ; to tind the remaining side, and the other two angles. To find the angle opposite to b, (Art. 151.) c : Sin A: :6 : Sin B The calculation here gives the acute angle AB'C 54°3' 50''. and the obtuse angle ABC 125° 56' 10''. If the latter be added to the angle at A 35° 20', the sum will be 1 6 1 ° 1 6' 1 0", the supplement of which 18** 43' 50" is the angle ACB. Then in the triangle ABC, to find the side c=AB, Sin A : a: tSin C : c=27.7G If the acute angle AB'C 54° 3' 50' be added to the angle at A 3o° 20', the sum will be 89° 23' 50", the supplement of which 90° 36' 10'' is the angle ACB'. Then, in the triangle ABC, Sin A : CB': :Sin C : AB'=86.45 Ex. 2. Given the angle at A 63° 35' (Fig. 29.) the side b 64, and the side a 72 : to find the side c. and the angles B' and C TRIANGLES. 87 « : Sin A: :6 : Sin B=52° 43' 25" Sin A : a::SinC : c = 72.05 The sum of the angles A and B is 116° 20' 25', the sup- plement of which 63° 39' 35" is the angle C. In this example the solution is 7iot ambi^uous^ hecause the side opposite the given angle is longer than the other given side. Ex. 3. In a triangle of which the angles are A, B, and C, and the opposite sides a, 6, and c, as before ; if the angle A be 121° 40', the opposite side a 68 rods, and the side 6 47 rods ; what are the angles B and C, and what is the length of the side c '? Ans. B is 36° 2' 4", C 22° 17' .56", and c 30.3. In this example also, the solution is not ambiguous, because Case III. the gi-Dtn angle is obtuse. \o3. Given Two sides, and ^ f fi 1 ^ '^^^ remaining side, and The included angle, ) X The other two angles. In this case, the angles are found by theorem II. (Art. 144.) The required side may be found by theorem I. In making the solutions, it will be necessary to observe, that by subtracting the given angle from 180°, the sum of the other two angles is found ; (Art. 79.) and, that adding half the difference of tzvo quantities to their half sum gives the great' er quantity, and subtracting the half difference from the half sum gives the less, (Alg. 341.) The latter proposition may be geometrically demonstrated thus ; Let AE (Fig. 32.) be the greater of two magnitudes, and BE the less. Bisect AB in D, and make AC equal to BE. Then AB is the sum of the tv/o magnitudes : CE their difference ; DA or DB half their su7n ; IJE or DC Juilf their difference ; But DA-}-DE=AE the greater magnitude ; And DB - DE=BE the less, Ex. 1. ]n the triangle ABC (Fig. 30.) the angle A is giv- 88 OBLIQUE ANGLED en 26° 14' the side b 39, and the side c 53 ; to find the an- gles B and C, and the side a. The sum of the sides b and c is 53-{-39=92, And ihe'ir differetice 53 — 39 = 14. The sum of the angles B and C = 180°-26° 14' = 153° 46' And half the sum of B and C is 76° 53' Then, by theorem If, (b+c) : (6-c)::Tani(B4-C) : Tan K^^-C) 'I o and from the lialf sum 76° 53' Adding and subtracting the half difference 33 8 50 We have the greater angle HO 1 50 And the less angle 43 44 10 As the greater of the two given sides is c, the greater an- gle is C, and the less angle B. (Art. 149.) To find the side a, by theorem 1, Sin B : ^>::Sin A : a=24.94 Ex. 2. Given the angle A 101° 30' the side b 76, and the side c 109 ; to find the angles B and C, and the side «. B is 30° 57f, C 47° 32^, and a 144.8. Case IV^ 154. Given the three sides, to find the angles. In this case, the solutions may be made, by drawing a per- pendicular to the longest side, from the opposite angle. This will divide the given triangle into two rii^ht angled inanities. The two segments may be found by theorem 111. (Art. 145.) There will then be {Zjiven, in each of the right angled trian- gles, the hypothenuse and one of the legs, from which the angles may be determined, by rectangular trigonometry. (Art. 135.) Ex. 1. Jn the triangle ABC (Fig. 31.) the side AB is 39, A.C 35, and BC 27. What are the angles ? Let a perpendicular be drawn from C. dividing the long- TRIANGLES. 89 est side AB into the two segments A? and BP. Then by theorem III, AB : AC-|-BC::AC-BC : AP-BP As the longest side 39 a. c. 8.40894 To the sum of the two others 62 1 .79239 So is the difference of the latter 8 0.90309 To the difference of the segments 1 2.72 1 . 1 0442 The greater of the two segments is AP, because it is next the side AC, which is greater than BC. (Art. 146.) To and from half the sum of the segments 19.5 Adding and subtracting half their difference, (Art.153.) 6.36 We have the greater segment AP 25.86 And the less BP 13.14 7'hen, in each of the right angled triangles APC and BPC, we have given the hypothenuse and base ; and by art. 135, AC : R: :AP : Cos A=42° 21' 57' BC : R::BP: Cos B = 60° 52' 42" And subtracting the sum of the angles A and B from 180°, we have the remaining angle ACB = 7G° 45' 21". Ex. 2. If the three sides of a triangle are 73, 96, and 104 ; what are the angles ? Ans. 45° 41' 48 ", 61° 43' 27", and 72° 34' 45". Examples for Practice* 1. Given the angle A 54° 30', the angle B 63° 10', and the side a 164 rods ; to find the angle C, and the sides b and c. 2. Given the angle A 45° 6' the opposite side a 93, and the side b 108 ; to find the angles B and C, and the side c. 3. Given the angle A6T 24', the opposite side a 62, and the side h 46 ] to find the angles B and C, and the side c. 4. Given the angle A 127° 42', the opposite side a 381, and the side b 1 84 : to find the angles B and C, and the side c. 13 90 OBLIQUE ANGLED TRIANGLES. 5. Given the side b 58, the side c 67, and the included an- gle A=36° ; to tind the angles B and C, and the side a, 6. Given the three sides,631 ,2G8,and 546-, to find the angles. 155. The three theorems demonstrated in this section, have been here applied to oblique ansrled triangles only. But they are equally applicable to risrhl a/i^/erZ triangles. Thus, in the triangle ABC, (Fig. 17.) according to theo- rem I, (Art. 143.) SinB: AC::Sin A:BC This is the same proportion as one stated in art. 134, ex- ceptthat, in the first term here, the sine of Bis substituted for radius. But, as B is a right angle, its sine is equal to radius, (Art. 95.) Again, in the triangle ABC, (Fig. 21.) by the same theorem ; Sin A :BC::SinC : AB This is also one of the proportions in rectangular trigo- nometry, when the hypothenuse is made radius. The other two theorems might be applied to the solution of right angled triangles. But, when one of the angles is known to be a right angle, the methods explained in the pre- ceding section, are much more simple in practice.* * For the application of Trigonometry to the Mensuration of Heights and Distances, see Navigation and Surveying. SECTION V. GEOMETRICAL CONSTRUCTION OF TRIANGLES, BY THE PLANE SCALE. . . P ^^O facilitate the construction of geometrical -*- figures, a nunnber of graduated lines are put upon the common two feet scale ; one side of which is call- ed the Plane Scale, and the other side, Gunter^s Scale. The most important of these are the scabies of equal parts, and the line o( chords. In forming a given triangle, or any other right lined figure, the parts which must be made to agree with the conditions proposed, are the lines, and the angles. For the former, a scale of equal parts is used ; for the latter, a line of chords. 157. The line on the upper side of the plane scale, is di- vided into inches and tenths of an inch. Beneath this, on the left hand, are two diagonal scai\es of equal parts,* divided into inches and half inches, by perpendicular lines. On the lar- ger scale, one of the inches is divided into tenths, by lines which pass ohliqueli^ across, so as to intersect the parallel lines which run from right to left. The use of the oblique lines is to measure hundredths of an inch, by inclining more and more to the right, as they cross each of the parallels. To take off. for instance, an extent of 3 inches, 4 tenths, and 6 hundredths ; Place one foot of the compasses at the intersection of the perpendicular line marked 3 with the parallel line marked 6, and the other foot at the intersection of the latter with the oblique line marked 4, The other diagonal scale is of the same nature. The di- visions are smaller, and are numbered from left to right. 158. In geometrical constructions, what is often required, is to make a figure, not equal to a given one, but only similar* Now figures are similar which have equal angles, and the * These lines are not represeated in the plate, as the learner is supposed to have the scale before him. 92 GEOxMETRICAL CONSTRUCTION * sides about the equal angles proportioned, (Euc. Dei. 1. 6.; Thus a hind surveyor, in plotting a field, makes the several lines in his plan to have the same proportion to each other, as the sides ol' the field. For this purpose, a scale of equal parts may he used, of any dimensions whatever. If the sides ofthc field are 2, 5, 7, and 10 rods, and the lines in the plan are 2, 5, 7, and 10 inches^ and if the angles are the same in each, tlie figures are similar. One is a copy of the other, upon a smaller scale. So any two right lined figures are similar, if the angles are the same in both, and if the number of smaller parts in each side of one, is equal to the number of larger parts in the cor- responding sides of the other. The several divisions on the scale of equal parts may, therefore, be considered as repre- senting any measures of length, as feet, rods, miles, &;c. All that is necessary is, that the scale be not changed, in the con- struction of the same figure ; and that the several divisions and subdivisions be proptrly proportioned to each other. If the larger divisions, on the diagonal scale, are units, the smal- ler ones are tenths and hundredths. If the larger are tens, the smaller are units and tenths. 159. In laying down an angle, of a given number of de- grees, it is necessary to measure it. Now the proper measure of an angle is an arc of a circle. (Art. 74.) And the measure of an arc, where the radius is given, is hs chord. For the chord is the distance, In a straight line, from one end of the arc to the other. Thus the chord AB (Fig. 33.) is a meas- ure of the arc .ADB, and of the angle ACB. To form the line of chords, a circle is described, and the lengths of its chords determined for every degree of the quad- rant. These measures are put on the plane scale, on the line marked CHO. 160. The chord of 60° is equal to radius. (Art, 95.) In hiying down or measuring an angle, therefore, an arc must be draun, with a radius which is equal to the extent from to 60 on the line of chords. There are generally on the scale, two lines of chords. Either of these may be used ; but the angle must be measured by the same line from»which the radius is taken. 161. To make an angle, then, of a given number of de- grees ; From one end of a straight line as a centre, and with a radius equal to the chord of 60° on the line of chords, de- scribe an arc of a circle cuttina; the straight line. From the OF TRIANGLES. poiot of intersection, extend the chord of the given number of degrees, applying the other extremity to the arc ; and through the place of meeting, draw the other line from the angular point. Jf the given angle is obtuse, take from the scale the chord of Art//* the number of degrees, and apply it twice to the arc. Or make use of the chords of any two arcs whose sum is equal to the given number of degrees. A right angle may be constructed, by drawing a perpen- dicular without using the line of chords. Ex. I. To make an angle of 32 degrees. (Fig. 33.) With the point C, in the line CH, for a centre, and with the chord of 60° for radius, describe the arc ADF. Extend the chord of32'' from A to B ; and through B, draw the line BC. Then is ACB an an^le of 32 degrees. 2. To make an angle of 140 degrees. (Fig. 34.) On the line CH, with the chord of 60°, describe the arc ADF ; and extend the chord of 70° from A to D, and from D to B. The arc ADB=70°x2 = 140.° On the other hand, 162. To measure an angle ; On the angular point as a cen- tre, and with the chord of 60° for radius, describe an arc to cut the two lines which include the angle. The distance be- tween the points of intersection, applied to the line of chords, will give the measure of the angle in degrees. If the angle be obtuse, divide the arc into two parts. Ex. 1. To measure the angle ACB. (Fig: 33.) Describe the arc ADF cutting the lines CH and CB. The distance AB will extend 32° on the line of chords. 2. To measure the angle ACB. (Fig. 34.) Divide the arc ADB into two parts, either equal or unequal, and meas- ure each part, by applying its chord to the scale. The sum of the two will be 140°. 163. Besides the lines of chords,and of equal parts, on the plane scale ; there are also lines of natural sines, tangents, and secants, marked Sin. Tan. and Sec. oi semitangents, marked S. T. o( longitude, marked Lon. or M. L. of rhumbs, marked Rhu. or Rum. he. These are not necessary in trigonomet- rical constructions. Some of them are used in Navigation ; and some of them, in the projections of the Sphere. 164. In Navigation, the quadrant, instead of being gradua- ted in the usual manner, is divided into eight portions, called 94 GEOMETRICAL CONSTRUCTION Rhumbs, The Rhumb line^ on the scale, is a line ol chords, divided into rhumbs and quarter-rhumbs, instead of degrees. 165. The line o{ Longitude is intended to show the num-= ber ot" geographical miles in a degree of longitude, at different distances from the equator. It is placed over the line of chords, with the numbers in an inverted order : so that the fig- ure above shows the length of a degree of longitude, in any latitude denoted by the figure below.* Thus at the equator, where the latitude isO, a degree of longitude is GO geographi- cal miles. In latitude 40, it is 46 miles; in latitude 60, 30 miles, &ic. 166. The graduation on the line of secants begins where the line of sines ends For the greatest sine is only equal to radius; but the secant of the least arc is greater than radius. 167. The semitangents are the tangents of half the given arcs. Thus the semitangent of 20° is the tangent of 10°. The line of semitangents is used in one of the projections of the sphere. 168. In the construction of triangles^ the sides and angles which are given, are laid down according to the directions in arts. 158, 161. The pans r eg iiired ure then measured, ac- cording to arts. 158, 162. The following problems corres- pond with the four cases of oblique angled triangles; (Art. 148.) but are equally adapted to right angled triangles. 169. Pkob. 1. The angles and one side of a triangle being given ; to find, by construction, the other two sides. Draw the given side. From the ends of it, lay off two of the given angles. Extend the other sides till they intersect j and then measure their lengths on a scale of equal parts. Ex. 1. Given the side 6 32 rods, (Fig. 27.) the angle A 56° 20', and the angle C 49° 10 ; to construct the triangle, and find the length.:3 of the sides a and r. Their lengths will be 25 and 27i. 2. In a right angled triangle, (Fig. 17.) given the hypoth- enuse 90, and the angle A 32° 20', to find the base and per- pendicular. The length of AB will be 76, and of BC 48. * Sometimes the line of loDgitude is placed tt/Lder the line of chord?. OF TRIANGLES. 95 3. Given the side AC 68, the angle A 124°, and the angle C 37° : to construct the triangle. 170 Prob. II. Two sides and an opposite angle being giv- en, to find the remaining side, and the other two angles. Draw one of the given sides ; from one end of it, lay off the given angle ; and extend a line indefinitely for the required side. From the other end of the first side, with the remain- ing given side for radius, describe an arc cutting the indefinite line. The point of intersection will be the end of the requir- ed side. If the side opposite the given angle be less than the other given S'de, the case will be ambiguous (Art. 152.) Ex. 1. Given the angle A 63° 35' (Fig. 29.) the side b 32, and the side a 36. The side AB will be 36 i^.early, the angle B 52° 45V and C 63° 391' 2. Given the angle A (Fig. 26.) 35° 20'; the opposite side a 25. and the side h 35. Draw the side b 35, make the angle A 35° 20', and extend AH indefinitely. From C wi».h radius 25, describe an arc cutting AH in B and B'. Draw CB and CB', and two trian- gles will be formed, ABC and AB'C, each corresponding with the conditions of the problem. 3. Given the angle A 1 16°, the opposite side a 38, and the side b 26 ; to construct the triangle. 171. Prob. Hi. Two sides and the included angle being given ; to find the other side and angles. Draw one of the given sides. From one end of it lay off the given angle, and draw the other given side. Then con- nect the extremities of this and the first line. Ex. i: Given the angle A (Fig. 30.) 26° 14', the side // 78, and the side c 106; to find B, C. and a. The side a will be 50. the angle B 43° 44', and C 110° 2. 2. Given A 86°, b 65, and c 83 ; to find B, C, and a. 172. Prob. IV. The three sides being given; to find the angles. Draw one of the sides, and from one end of it, with an ex- tent equal to the second side, describe an arc. From the other end, with an extent equal to the third side, describe a second arc cutting the first; and from the point of intersec- tion draw the two sides. (Euc. 22. 1.) Ex. 1. Given AB (Fig. 31.) 78, AC 70, and BC, 54 ; to find the angles. 3C GEOMETRICAL CONSTRUCTION OF TRIANGLES. The angles will be A 42° 22', B 60^^ 52|', and C 76° 451'. 2. Given, the three sides 68, 39, and 40 ; to find the an- gles. 173. Any right lined figure whatever, whose sides and an- gles are given, may be constructed, by laying down the sides from a scale of equal parts, and the angles from a line of chords. Ex. Given the sides AB (Fig. 35.) =20, BC=22, CD= 30, DE = 12; and the angles B = 102°, C = 130% D = 108°, to construct the figure. Draw the side AB = 20, make the angle B — 102°, draw BC=22, make C = 130», draw CD=30, make D=108% draw DE = 12, and connect E and A. The last line EA may be measured on the scale of equal parts; and the angles E and A, by a line of chords. SECTION VI. DESCRIPTION AND USE OF GUNTER'S SCALE. \ I "4 A^ expeditious method of solving the prob- ^ RT. . -^!^|gjj^g in trigonometry, and making other loga- rithmic calculations, in a mechanical way, has been contrived by Mr. Edmund Gunter. The logarithms of numbers, of sines, tangents, &:c. are represented by lines. By means of these, multiplication, division, the rule of three, involution, evolution, &;c. may be performed much more rapidly, than in the usual method by figures. The logarithmic lines are generally placed on one side only of the scale in common use. They are, A line of artificial Sines divided into Rhumbs, and mark- ed S. R. A line of artificial Tangents, do. T, R. A line of the logarithms o( numbers, Num. A line of artificial Sines, to every degree. SIN. A line of artificial Tangents, do. TAN. A line of Versed Sines, V. S- To these are added a hue of equal parts, and a line o( Me- ridional Parts, which are not logarithmic. The latter is used in Navigation. The Line of JVumbers. 175. Portions of the line oi Numbers, are uitended to rep- resent the logarithms of the natural series of numbers 2, 3, 4, 5, &c. The logarithms of 10, 100, 1000, &c. are 1, 2, 3, &c. (Art. 3.) Tf then, the log. of 10 be represented by a line of 1 foot; the log. of 100 will be repres'd by one of 2 feet ; the log. of 1000 by one of S feet; the lengths of the several lines being proportional to the cor- responding logarithms in the tables. Portions of a foot will represent the logarithms of numbers between 1 and 10; 14 98 rillGONOMETRV- and poitions of a line 2 feet long, the logarithmb of numbers between 1 and 100. On Gunter's scale, the line of the logarithms of numbers begins at a brass pin on the left, and the divisions are num- bered 1, 2, 3, (foe. to another pin near the middle. From this the numbers are repeated, 2, 3, 4, &;c. which may be read ^0, 30, 40, &IC. The logarithms of numbers between 1 and 10 are represented by portions of the first half of the line ; and the logarithms of numbers between 10 and 100, by por- tions greater than half the line, and less than the whole. 17C. The logarithm of 1, which is 0, is denoted, not by any extent of line, but by a point under 1, at the commence- ment of the scale. The distances from this point to differ- ent parts of the line, represent other logarithms, of which the, figures placed over the several divisions are the natural /lumbers. For the intervening logarithms, the intervals be- tween the ligures, are divided into tenths, and sometimes in- to smaller portions. On the right hand half of the scale, as the divisions which are numbered are tens, the subdivisions are units. Ex. 1. To take from the scale the logarithm of 3.6 ; set one foot of the compasses under 1 at the beginning of the scale, and extend the other to the 6th division after the first figure 3. 2. For the logarithm of 47 ; extend from 1 at the begin- ning, to the 7th subdivision after the second figure 4.* 177. It will be observed, that the divisions and subdivi- sions decrease, from left to right; as in the tables of loga- rithms, the differences decrease. The difference between the logarithms of 10 and 100 is no greater, than the differ- ence between the logarithms of 1 and 10. 178. The line of numbers, as it has been here explained, fijrnishes the logarithms of all numbers between 1 and 100. And if the indices of the logarithms be neglected, the same scale may answer for all numbers whatever. For the deci- mal part of the logarithm of any number is the same, as that of the number multiplied or divided by 10, 100, 8zc» (Art. 14.) In logarithmic calculations, the use of the indices is to determine the distance of the several figures of the natural numbers from the place of units. (Art. 11.) But in those cases in which the logarithmic line is commonly used, it will * If the compasses will not reach the distance required ; first open them so as to take afffijalff or any part of the distance, and then the remaining part. GUNTER-S SCALE. d9 not generally be difficult to determine the local value of the figures in the result. 179. We may, therefore, consider the point under 1 at the left hand, as representing the logarithm of 1, or 10, or 100 ; or y*;j, or y^oi <^c. for the decimal part of the logarithm of each of these is 0. But if the first 1 is reckoned 10, all the succeeding numbers must also be increased in a tenfold ra- tio ; so as to read, on the first half of the line, 20, 30, 40, &.c. and on the other naif, 200, 300, kc. The whole extent of the logarithmic line, is from 1 to ICO, or from 0.1 to 10. or from 10 to 1000, or from 0.01 to 1, or from 100 to 10000, kc. or from O.OOI to 0.1, &c. Different values may, on diflerent occasions, be assigned to the several numbers and subdivisions marked on this line. But for any one calculation, the value must remain the same. Ex. Take from the scale 3C5. As this number is between 10 and 1000, let the 1 at the beginning of the scale, be reckoned 10. Then, from this point to the second 3 is 300 ; to the 6th dividing stroke is 60 ; and half way trom this to the next stroke is o. ISO. Multiplication, division, (Sic. are performed by the line of numbers, on the same principle, as by common loga- rithms. Thus, To multiplt/ by this line, add the logarithms of the two factors; (Art. 37.) that is, takeoff, with the compasses, that length of line which represents the logarithm of one of the factors, and apply this so as to extend forward from the end of that which represents the logarithm of the other factor. The sum of the two will reach to the end of the line reprc- senting the logarithm of the product. Ex. Multiply 9 into 8. The extent from 1 to 8, added to that from 1 to 0, will be equal to the extent from 1 to 72 the product. 181. To divide by the logarithmic line, subtract the loga- rithm of the divisor from that of the dividend; (Art. 41.) that is, take off the logarithm of the divisor, and this extent set back from the end of the logarithm of the dividend, will reach to the logarithm of the quotient. Ex. Divide 42 by 7. The extent from 1 to 7, set back from 42, will reach to 6 the quotient. 182. Involution is performed in logarithms, by multiplying • , • rn RXsin K : cos! :tan : sin, that is, ran= ' ' cos The sign of the quotient is determined from the signs of the divisor and dividend. (Alg. 123.) The radius is consid- ered as always positive. If then the sine and cosine be both positive, or both negative, the tangent will be positive. But if one of these be positive, while the other is negative, the tangent will be negative. Now, by the preceding article. In the 2d quadrant, the sine is positive, and the cosine ne- gative. The tangent must therefore be negative. In the 3d quadrant, the sine and cosine are both negative. The tangent must therefore be positive. In the 4th quadrant, the sine is negative, and the cosine positive. The tangent must therefore be negative. 196. By the 9th, 3d, and 6th proportions in art. 93. 1. Tail : R: :R : cot, that is, Cot= — • ANALYSIS. 107 Therefore, as radius is uniformly positive, the cotangent must have the same sign as the tangent. 2. Cos : R: :R : sec. that is, Sec= — • The secanf, therefore, must have the same sign as the cosine. R2 3. Sin : R: :R : cosec, that is, Cosec = — r- • sm The cosecant, therefore, must have the same sign as the sine. The versed sine, as it is measured from A, in one direction only, is invariably positive. 197. The tmgent AT (Fig. 36.) increases, as the arc ex- tends from A towards B. See also Fig. 11. Near B the in- crease is very rapid ; and when the difference between the arc and 90^, is less than any assignable quantity, the tangent is ^rea/er than any assignable quantity, and is said to he infi- nite, (Alg. 447.) If the arc is exactly 90 degrees, it lias, strictly speaking, no tangent. For a tangent is a line, drawn perpendicular to the diameter which passes through one end of the arc, and extended till it meets a line proceeding from the centre, through the other end. (Art. 84.) But if the arc is 90 degrees, as AB, (Fig. 36.) the angle ACB is a right an- gle, and therefore AT is parallel to CB; so that, if these lines be extended ever so far, they can never meet. Still, as an arc infinitely near to 90° has a tangent infinitely great, it is frequently said; in concise terms, that the tangent of 90° is infinite. In the second quadrant, the tangent is, at first, infinitely great, and gradually diminishes, till at D it is reduced to nothing. In the third quadrant it increases again, becomes infinite near H, and is reduced to nothing at A. The cotangent is inversely as the tangent. It is therefore nothing at B and H, (Fig. 36.) and infinite near A and D. 198. The secant increases with the tangent, through the first quadrant, and becomes infinite near B ; it then dimin- ishes, in the second quadrant, till at D it is equal to the radius CD. In the third quadrant it increases again, becomes infi- nite near H, after which it diminishes, till it becomes equal to radius. The cosecant decreases, as the secant increases, and v. v. It is therefore equal to radius ;»t B and H, and infinite near A and D. 108 TRIGONOMETRICAL 199. The sine increases through the first quadrant, till at B (Fig. 36.) it is equal to radius. See also Fig. 13. It then diminishes, and is reduced to nothing at D. In the third quad- rant, it increases again, becomes equal to radius at H, and is reduced to nothing at A. The cositie decreases through the first quadrant, and is re- duced to nothing at B. In the second quadrant, it increases, till it becomes equal to radius at D. It then diminishes again, is reduced to nothing at H, and afterwards increases till it be- comes equal to radius at A. In all these cases, the arc is supposed to begin at A, and to extend round in the direction of BDH. 200. The sine and cosine vary from nothing to radius, which they never exceed. The secant and cosecant are never less than radius, but may be greater than any given length. The tangent and cotangent have every value from nothing to infinity. Each of these lines, after reaching its greatest limit, begins to decrease ; and as soon as it arrives at its least limit, begins to increase. Thus the sine begins to decrease, after becoming equal to radius, which is its greatest limit. But the secant begins to increase after becoming equal to radius, which is its least limit. 201 V The substance of several of the preceding articles is comprised in the following tables. The first shows the signs of the trigonometrical lines, in each of the quadrants of the circle. The other gives the values of these lines, at the ex- tremity of each quadrant. Quadrant 1st 2d 3d 4th Sine and cosecant + + — — Cosine and secant + — — 4- Tangent and cotangent + — -f 0° 90° 180° 270° 360° Sine r r Cosine r r r Tangent CD CD Cotangent CD CD CO Secant r CO r CO r Cosecant CO r CO r op Here ris put for radius, and co for infinite. 202. By comparing these two tables, it will be seen, that each of the trigonometrical lines changes from positive to ne- gative, or from negative to positive, in that part of the circle ANALYSIS. lOD m which the line is either nothing or infinite. Thus the tan- gent changes from positive to negative, in passing from the first quadrant to the second, through the place where it is infinite. It becomes positive again, in passing from the second quadrant to the third, through the point in which it is nothing. 203. There can be no more than 360 degrees in any cir- cle. But a body may have a number of successive revolu- lioBS in the same circle ; as the earth moves round tlie suuj nearly in the same orbit, year after year. In astronomical calculations, it is frequently necessary to add together parts of different revolutions. The sum may be more than 360*^ But a body which has made more than a complete revolution in a circle, is only brought back to a point which it had pass- ed over before. So the sine, tangent, &c. of an arc great- er than SGO"^, is the same as the sine, tangent, &;c. of some arc less than 360'. If an entire circumference, or a number of circumferences be added to any arc, it will terminate in the same point as before. So that, if C be put for a whole circumference, or 360°, and x be any arc whatever: sin x=sin (C + x) = sin (2C-f--i)=sin (3C + a:), &;c. tan x = tan(C-h.T)=tan (2C+T) = tan (3C + x), &c. 204. It is evident also, that, in a number of successive rev olutions, in the same circle; The first quadrant must coincide with the 5ih, 9th, 13th, 17th. The second, with the 6th, 10th, 14th, 18th, &;c. The third, with the 7th, 11th, 15th, 19th, &:c. The fourth, with the Sth, 12ih, 16th, 20th, &c. 205. If an arc extending in a certain direction from a given point, be considered /?05tYj're; an arc extending from the same point, in an opposite direction, is to be considered negative. (Alg. 507.) Thus, if the arc extending from A to S (Fig. 36.) be positive; an arc extending from A to S'" will be negative. The latter will not terminate in the same quadrant as the oth- er; and the sines of the tabular lines must be accommodated to this circumstance. Thus the sine of AS will be positive, while that of AS'" will be negative. (Art. 194.) When a greater arc is subtracted from a less, if the latter be positive* ihe remainder must be negative. (Alg. 58, 9.) Trigonometrical Formula:. 206. From the view which has here been taken of the changes in the trigonometrical lines, it will be easy to see, in 110 TR1G0N03IETKICAL what parts ofthe circle each of them increases or decreases. But this does not determine their exact values, except at the extremities of the several quadrants. In the analytical in- vestigations which are carried on by means of these lines, it is necessary to calculate the chanf;es produced in them, by a given increase or diminution of the arcs to which they be- long, 'n this there would be no difficulty, if the sines, tan- gents, &1C. were proportioned to their arcs. But this is far from ooing the case. If an arc is doubled, its sine is not ex- actly doubled. Neither is its tangent or secant. We have to inquire, then, in what manner, the sine, tangent, &ic. of one arc may be obtained, from those of other arcs already known. The problem on which almost (he whole of this branch of analysis depends, consists in deriving, from the sines and co- bines of two given arcs, expressions for the sine and cosine of t'leir swn and difference. For, by addition and subtrac- tion, a few arcs may be so combined and varied, as to pro- duce others of almost every dimension. And the expres- sions for the tangents and secants may be deduced from those of the sines and cosines. Expressions for the sine and cosine of the sum o/jc/ differ- ence of arcs, 207. Let rt = AH, the greater of the given arcs, And 6 = HL = MD, the less. (Fig. 37.) I'hen «-f i = AU-|-HL=AL, the swn of the two arcs, And a -b=^All--HD=^ADythe\Y difference. Draw the chord DL, and the radius CII, which may be represented by R. As DM is, by construction, equal to HL; DQ is equal to QL, and therefore DL is perpendicular to CH. (Euc. ^. 3.) Draw DO, [IN, QP, and LM. each per- pendicular lo AC ; and DS and QB parallel to AC. From the definitions of the sine and cosine, (Art. 82, 0,) >J is evident, that fof AH, that is, .9m a=HN, '""^ "''^, ofAL,i.e.sin(a-f6) = LM, lof AD,i.o.?in(«-^)x=DO. ANALYSIS. 1 1 1 fof AH, that is, cos a=CN, r,„ . ] of HL, that is, cos 6=CQ, J he cos.ne <;^ of AL, i.e.cosr^+M^CM, |^ofAD,i.e.cos(a-6)=CO. The triangle CHN is obviously similar to CQP ; and it i^ also similar to BLQ, because the sides of the one are per- pendicular to those of the other, each to each. We have, then, 1. CH : CQ: :IJN : QP, that is, R : cos b: [sin a : QP. 2. CH : QL: :CN : BL, R : sin bwcos a : BL, 3. CH :CQ::CN : CP, R x cos bWcos a :CP. 4. CH : QL: :HN : QB, R : sin br.sin a : QB, Converting each of these proportions into an equation ; si7i a cos b* cos a cos b I- Qp= R 3. cp= — n — sin b cos a sin a sin b 2. BL= R 4. QB= ^ QP4-BL Then adding the first and second. sin a cos b~^sin b cos a QP-BL= R Subtracting the second from the first. sin a cos b — sin b cos a CP-QB CP4-QB: R Subtracting the fourth from the third. cos a cos b — sin a sin b R Adding the third and fourth. cos a cos b-\'sin a sin h put R In these formulae, the sign of multiplication is omitted ; stn a cos b bejin» for sin a vi cos b, that i? the product of the sine of a into thp rr>smp nf '. 1 12 TRIGONOMETRICAL But it will be seen, from the figure, that QP+BL=BM-f BL=LM=5m (a+b) QP-BL=QP-QS = D0=5UJ (a-b) CP-QB=CP-PM==CM=c-^s (a+b) CP+QB=CP+SD=CO=co5 (a-b) 208. If then, for the first member of each of the four equa- tions above, we substitute its value, we shall have, I. o^^i^+h)- '"'' ^ cosb+sinb cos a H. / i-x sin a cos b — 5m b cos a . sin {a—o)=' III. co^(a+6) = R cos a cos b — sin a sin b R TIT / IS cos a cos b-^siyi a sinb ly, cos (a — b)= ^ R Or, multipljing both sides by R, Rsin {a-\-b)=sin a cos b+sin b cos a^ R sin {a — b)=sin a cos b — sin b cos a R {cos {a-]rb)=cos a cos b —sin a sin b R cos (a — b) — cos a cos b-\-sina sin b That is, the product of radius and the sine of the mm of two arcs, is equal to the product of the sine of the first arc into the cosine of the second -f the product of the sine of the second into the cosine of the first. The product of radius and the sine of the difference of two arcs, is equal to the product of the sine of the first arc into the cosine of the second — the product of the sine of the second into the cosine of the first. The product of radius and the cosine of the sum of two arcs, is equal to the product of the cosines of the arcs — the product of their sines. The product of radius and the cosine of the difference of two arcs, is equal to the product of the cosines of the arcs -f the product of their sines. These four equations may be considered as fundamental propositions, in what is called the Arithmetic of Sines and Co- sines, or Trigonometrical Analysis* ANALYSIS. 113 Expressions for the sine and cosine of a double arc* 209. When the sine and cosine of any arc are given, it is easy to derive from the equations in the preceding article, ex- pressions for the sine and cosine of double that arc. As the two arcs a and b may be of any dimensions, they may be sup- posed to be equal. Substituting, then, a for its equal 6, the tirst and the third of the four preceding equations will be- come, R sin (a-{-a)=5m a cos a-^sia a cos a R cos (a-f-a)=c05 a cos a — sin a sin a That is, writing sin- a for the square of the sine of «, and ^'os'^a for the square of the cosine of «, I. R sin 2a = 2s ill a cos a II. R cos 2a=cos^a — sin^a* Expressions for the sine and cosine of half a given arc* 210. The arc in the preceding equations, not being neces- sarily limited to any particular value, may be half a, as well as a. Substituting then |a for a, we have, R sin a:=2sin -i-o; cos -^a R cos a=cos^ ^a — sin^ ^u Putting the sum of the squares of the sine and cosine equal 10 the square of radius, (Art. 94.) and inverting the members of the last equation. cos 2ia-|-5m-|a=R* co5^Ja — sm^ Ja=R cos a If we subtract one of these from the other, the terras con- taining cos^^a will disappear ; and if we add them, the terms containing siti^^a will disappear : therefore, 2sin^^a=K^ — -R cos a 3cos2|«=R2+Rco5G TI4 TRIGONOMETRICAL Dividing by J, and extracting the root of both sides, U.cos^a= ^/Ul'^-h^RXcos a Expressions for the sines and cosines o/'multiple ai'cs, 211. In the same manner, as expressions for the sine and cosine of a double arc, are derived from the equations in art. 208 ; expressions for the sines and cosines of other muUiple arcs ma}^ be obtained, by substituting successively 2a, 3a, Sic. for h, or for h and a both. Thus, r R^m 3rt=R sin{a+2a): -? Ksin 4ff=R sinla-\-^)- (^ R.fi?i 5a=R sin{a-\-4a) 'Rsin 3rt=R sin(a+2a)==sin a cos 2a-{-sm 2a cos a i=sin a cos 3fl-{-sm 3a cos a ■sin a cos ^a-^-sin 4a cos a &LC. CR cos 3a =R co5(a-i-2a)=co5 a cos 2a — sin a sin 2a IT. -f-tan a "^ cot c+tan b Dividing the numerator and denominator of the expres- sion for the tangent of a, (Art. 210.) by tan \a, we have tana ^ot 4«-tan ^a These formulas may be multiplied almost indefinitely, by •mbining the expression? for the sines, tangents, &c. The 120 TRIGONOMETRICAL following are put down without demonstrations, tor the exer cise of the student. 1 —cos a tan ^a=cot|(z — 2 cot a. tan -5-a~ sin a sin a I —cos a tan ia=7-r— "~ tan"|a=7-; l-t-cos« 2 i-j-cosff 2 tan |fl ^ 1— tan ^J« 5in C5 — i4.tan2|«: cos a — i-^tanH^ cot |a — tan \a - 2 cos a cot ifl+tan »« sm a —cot |a+tan ^« 1 1 sin a—cot |« - cot a sin a cot «-|-tan la Expression for the area of a triangle, in terms of the sides. 221. Let the sides of the triangle ABC (Fig. 23.) be ex- pressed by a, b, and c, the perpendicular CD by p, the seg- ment AD by dy and the area by S. Thena2=62-f.c=-2ctZ, (Euc. 13. 2.) Transposing and dividing by 2c d= 6^-fc^-Q" (6M-c2--a3)2 (Alg. 2c • Therefore rf^= iT^ • [223.) By Euc. 47. 1 , p2 =^2 _ ^2 =^3 _:^ _ — l- Reducingthe fraction, (Alg. 150.) and extracting the root o^ both sides. ANALYSIS. 121 3 This gives the length of the perpendicular, in terms of the sides of the triangle. But the area is equal to the -product of the base into half the perpendicular height. (Alg. 518.) that is, . ^=^\cp = W\b'c^ - {b' -\'C'-a^y Here we have an expression for the area, in terms of the sides. But this may be reduced to a form much better adaj^ed to arithmetical computation. It will be seen, that the quantities 4h-c-, and {b- -\-c'' -a^Y are both squares ; and that the whole expression under the radical sign is the difference of these squares. But the difference of two squares is equal to the product of the sum and difference of their roots. (Alg. 233.) Therefore 46-c=^ -(6^ -fc' -a')' maybe resolved into the two factors, ( 26c4-(^' 4-c2 -a2) which is equal to (^-hc)^ -a» ( 26c — (6=^-1- c^" -a^) which is equal toa^—^^^ — c)^ Each of these also, as will be seen in the expressions on the right, is the difference of two squares ; and may, on the same principle, he resolved into factors, so that, Ub-\-cY-a^-=ib'\-c-\-a)X{b-\-c-a) la^-{b- c)' =(a-f-6-c) X(«— 6+c) Substituting, then, these four factors, in the place of the quantity which has been resolved into them, we have, S = iv/(6-{-c-f a) X (6+c-a) X(a+6-c) X(a-6+c) * The expression for the perpendicular is the same, when one of the angples 13 obtuse, as in Fig. 24. Let AD=:d. Then a^ =6^ -f c^ +2crf. (Euc. 12. 2.) And£?= Yc (-^2 -c2-j-a2)2 (^2^c2_a2)2 4c2 = ^ Therefore (^^ = -'2 = 73 ^rAl.219,) And/;— ^7 as above 17 122 TRIGONOMETRICAL ANALYSIS. Here it will be observed, that all the three sides, a, 6, and c, are in each of these factors. Let h=l{a-{-b'\'c) half the sum of the sides. Then S=s/hy.{h'-a)X{h-b)x{h''c) 222. For finding the area of a triangle, then, when the three sides are given, we have this general rule ; From half the sum of the sides, subtract each side severally / multiply together the half sum and the three remainders ; and extract the square root of the product. H SEcnoN vm COMPUTATION OF THE CANON. Art 2^3 HPHE trigonometrical canon is a set of tables -*- containing the sines, cosines, tangents, fee- to every degree and minute of the quadrant. In the com- putation of these tables, it is common to find, in the first place, the sine and cosine of one minute ; and then, by suc- cessive additions and multiplications, the sines, cosines, Sic* of the larger arcs. For this purpose, it will be proper to begin with an arc, whose sine or cosine is a known portion of the radius. The cosine of 60° is equal to half radius. (Art. 96. Cor.) A formula has been given, (Art. 210.) by which, when the cosine of an arc is known, the cosine of half ihdii arc may be obtained. By successive bisect! ions of 60°, we have the arcv 30° 0° 28' 7' 30'" 15° 14 3 45 T" 30' 7 1 52 30 3° 45' 3 30 56 15 1° 52' 30' 1 45 28 7 30 0°56' 15" 0' 52" 44'" 3"" 45"'" By formula II, art. 210, cos ia = v^iR2-fiRxco5 a If the radius be 1, and if a=60°, 6=30*^, c=15°, &c. ; then cos6=cos |a=v/i-f-iX 1=0.8660254 cos c=cos ^b = y/^-\-lcos 6=0.9659258 cos d=cos ^c=\/i-f-icos c=0.99 14449 cos e=co5 ^i(29° 59') (30° 2')=cos 2' - sin(29° 58') (30° 3')=cos 3' — sin{29° 51') &IC. &;c. If the 517165 be calculated from 30° to 60°, the cosinea wjii also be obtained. For the sines of arcs between 30° and 45°, are the cosines of arcs between 45° and 60*. And the sines of arcs between 45° and 60°, are the cosines of arcs between 30° and 45°.* (Art. 96. 227. By the methods which have here been explained, the natural sines and cosines are found. The logarithms of these, 10 being in each instance added to the index, will be the artificial sines and cosines, by which trigonometrical calculations are commonly made, (Art. 102, 3.) 228. The tangents, cotangents, secants, and cosecants, art easily derived from the sines and cosines. By art. 93, * See note \. !6 COMPUTATION OF THE CANON R t cos ! I tan * sin cos t Rl'.R * sec R : sin : '. cot : cos sin t R '.< R I cosec Therefore, R X sin , , R- The lanG;ent=— The secant= ° cos cos R X cos ^ R* The cotangent=— ^— The cobecant=^^-^ Or, if the computations are made by logarithms, The tangent= lO-f 5m — coi, The secant = 20— co5, The cotangent=5ilO-{-co.9 — 5iw, The cosecant =20— 5m. SECTION IX. PARTICULAR SOLUTIONS OF TRIANGLES.* A 9*^1 \^^ triangle whatever may be solved, by ART. ^oi, x\. ^j^g theorems in sections IIL IV. But there are other methods, by which, in certain circumstances, the calculations are rendered more expeditious, or more accu- rate results are obtained. The differences in the sines of angles near 90°, and in the cosines of angles near 0^, are so small as to leave an uncertain- ty of several seconds in the result. The solutions should be varied, so as to avoid finding a very small angle by its cosine, or one near 90° by its sine. The differences in the logarithmic tangents and cotangents are least at 45°, and increase towards each extremity of the quadrant. In no part of it, however, are they very small. In the tables which are carried to 7 places of decimals, the least difference for one second is 42. Any angle may be found within one second, by its tangent, if tables are used which are calculated to seconds. But the differences in the logarithmic sines and tangents, within a few minutes of the beginning of the quadrant, and in cosines and tangents within a few minutes of 90°, though they are very large, are too unequal to allow of an exact determin- ation of their corresponding angles, by taking proportional parts of the differences. Very small angles may be accurate- ly found, from their sines and tangents, by the rules given in a note at the end. t 232. The following formulae may be applied to right angled triangles, to obtain accurate resuUs, by finding the sine or tan- gent of Aa//*an arc, instead of the whole. In the triangle ABC (Fig. 20, PI. II.) making AC radius, AC : AB::i : Cos A. By conversion, (Alg. 369, 5.) AC: AC-AB::l : l-CosA. * Simson's, Woodhouse'?. and Caj^oli's Tngonometry. t See Note K. Ua PARTICULAR SOLUTIONIS TherefoFe, AC-AB ^^ =1— Cos A=2sin^iA. (Art. 210.; Sin ^ A=v^ 2AC Again, from the first proportion, adding and subtracting lerms, (Al§. ,iS9, 7.) AC+AB : AC-AB::l-l-CosA : l~Cos A. Therefore, AC-AB 1-CosA , „ AC + AB^TTC^^^^^^ ^^- ^^'^'' ^'^-^ Or, AC-AB TaniA=v'-^_p^ 233. Sometimes, instead of having two parts of a right an- gled triangle given, in addition to the right angle; we have only one of the parts, and the sum or difference of two others. In such cases, solutions may be obtained by the following proportions. By the preceding formulae, and Art. 140, 141, AC-AB 1. Tan^ iA=^:cTAB 2. BC2=(AC-AB)(AC-fAB) Multiplying these together, and extracting the root, we hare, TaniAxBC=AC-AB Therefore, I. Tani A : i::AC-AB : BC That IS, the tangent of half of one of the acute angles, is to 1, as the difference between the hypothenuse and the side at the angle, to the other side. If, instead of multiplying, we divide the first equation abovxj by the second, we have TanfA 1___ BC ~AC+AB OF TRIANGLES. 129 Therefore, II. 1 :tan ^A::AC-f AB : BC Again, in the triangle ABC, Fig. 20, AB : BC : l::tan A Therefore, AB+BC : AB-BC::l + tan A : 1-tan A Or, 1 fori A AB-f BC : AB-BC : : 1 : t-; t l-f tan A By art. 218, one of the arcs being A, and the other 45°, the tangent of which is equal to radius, we have, _ 1 — tan A Tan i'''-^)=rH^^ Therefore, III. 1 : tan (45°-A)::AB4-BC : AB-BC. That is, unity is to the tangent of the difference between 45° and one of the acute angles ; as the sum of ihe perpendic- ular sides is to their difference. Ex. 1. In a right angled triangle, if the difference of the hypothenuse and base be 64 feet, and the angle at the base 33°f, what is the length of the perpendicular.^ Ans. 211. 2. If the sum of the hypothenuse and base be 185.3, and the angle at the base 37°; what is thje perpendicular.'* Ans. 620. 3. Given the sum of the base and perpendicular 128.4, and the angle at the base 41°i, to find the sides. 1 : tan(45°—4Pi):: 128.4 : 8.4, the difference of the base and perpendicular. Half the difference added to, and subtracted from, the half sum, gives the base 68.4, and the perpendicular 60. 4. Given the sum of the hypothenuse and perpendicular 83, and the angle at the perpendicular 40°, to find the base. 5. Given the difference of the hypothenuse and perpendic- ular 16.5, and the angle at the perpendicular 37^ j, to find the base. 6. Given the difference of the base and perpendicular 35, and the ang^le at the perpendicular 27°|, to find the sides. 18 130 PARTICULAR SOLUTIONS 234. The following solutions may be applied to the third 'And fourth cases o( oblique angled triangles ; in one of which, two sides and the included angle are given, and in the other, the three sides. See pages 87 and 88. Case IU. In astronomical calculations, it :s frequently the case, that two sides of a triangle are given by their logarithms. By the following proposition, the necessity of finding the corres- ponding natural numbers is avoided. l^HEOitEM A. In any plane triangle, oj the two sides which include a given angle, the less is to the greater ; as radius to the tangent of an angle greater than 45° : And radius is to the tangent of the excess of this angle above 45® ; as the tangent of half the sum of the opposite angles, to the tangent of half their difference In the triangle ABC, (Fig. 39.) let the sides AC and AB, and the angle A be given. Through A draw DH perpendic- ular to AC. Make AD and AF each equal to AC, and AH equal to AB. And let HG be perpendicular to a line drawn from C through F. Then AC : AB::R : Tan ACH And R : Tan(ACH - 45°) : : Tani( ACB-f B):Tani(ACB - B) Demonstration, In the right angled triangle ACD, as the acute angles are subtended by the equal sides AC and AD, each is 45°. For the !^ame reason, the acute angles in the triangle CAF are each 45°. Therefore, the angle DCF is a right angle, the angles GFH and GHF are each 45°, and the line GH is equal to GF and parallel to DC. In the triangle ACH, if AC be radius, AH which is equal to AB will be the tangt nt of ACH. Therefore, AC : AB::R : Tan ACH. In the triangle CGIJ, if CG be radius, GH which is equal to FG will be the tangent of HCG. Therefore, R : Tan (ACH-45°): iCG : FG. OF TRIANGLES. 131 And, as GH and DC are parallel, (Euc. -2. 6.) CG : FG::DH : FH. But DH is, by construction, equal to the sum, and FH to the difference of AC and AB And by theorem II, [Art. 144.] the sum of the sides is to their difference ; as the tangent of half the sum of the opposite angles, to the tangent of half their difference. Therefore, R : Tan (ACH-45°): iTan ^(ACB-f B) : Tan ^(ACB^B) Ex. In the triangle ABC, (Fig. 30,) given the angle A = 26° 14', the side AC = 39, and the side AB=53. AC 39 1.5910646 R 10. AB 53 1.7242759 Tan 3° 39' 9" 9.1823381 R 10. Tan ^(B4-C)76«'53' 10.632^181 Tan 53° 39' 9" 10. 1332 11 3 Tani(B^C)33°8'50" 9.8149562 The same result is obtained here, as by theorem II, p. 75. To find the required side in this third case, by the theo- rems in section IV, it is necessary to find, in the first place, an aiigle opposite one of the given sides. But the required side may be obtained, in a different way, by the following proposition. Theorem B. Iu a plane triangle, twice the product of any two sides, is to the differeiice between the sum of the squares oj those sides, and the square of the third side, as radius to the §osine of the angle included between the two sides. Id the triangle ABC, (Fig. 23.) whose sides are a, b, and c, 2bc : 62-f-c2-«2;:R ; Cos A For in the right angled triangle ACD, b : d'.'.R : Cos A Multiplying by 2c, 26c : 2rfc: :R : Cos A But, by Euclid 13. 2, 2dc = b^ -{-c^ -a'' Therefore, 26c : b'' -^-c^ -a^ ::R : Cos A. The demonstration is the same, when the angle A is obtuse, as in the triangle ABC, (Fig. 24.) except that a^ is greater 132 PARTICULAR SOLUTIONS than b--{-c' ; ''Eiic. 12. 2.) so that the cosine of A is nega- tive. See art 194. From this theorem are derived expressions, both for the sides of a tr'angle, and for the cosines of the angles. Con- verting the last proportion into an equation, and proceeding in the same mannei with the other sides and angles, we have the following expressions; For the angles. Cos A=R X Cos B=R X Cos C=Rx 26c 2ac 2^6 For the sides. 26c Cos A\ a=v/(62-f-c2 -^ \ 2ac Cos B 6 = v/ a^-fc-~ Uos ti\ -rT-) c=v/(a2-|-62 2a6 Cos C uosu\ R— ; These formulae are useful, in many trigonometrical investi- gations; but are not well adapted to logarithmic computa- tion. Case IV. When the three sides of a triangle are given, the angles may be found, by either of the following theorems; in which a. S, and c are the sides, A, B, and C, the opposite angles, and A=halfthe sum of the sides. 2R J Sin A= -T-x/h{h-a)(h-b)(h-c) be 2R Theorem C. i Sin B= ^Z h{h~-a){h-b){h-c 2R [Sm C= -^ x/ h{h-a){h-b){h-c) The quantities under the radical sign are the same in all the equations. In the triangle ACD, (Fig. 23.) R : i::Sin A :p. Therefore, Sin Ax6=Rxp. ^^^ _ ^^b^c^ - {b^^+c'~^^\ (Art. 221. p. 105.) •'^ 2c OF TRIANGLES. 1'>r'^ This, by the reductions in page 106, becomes V2hx2{h - a)X2{h-b)x2{h-c) P= 2c ^ Substituting this value of/?, and reducing, 2R Sin A=-7 — \/h{h — a){h — b)(h — c) The arithmetical calculations may be made, by adding the logarithms of the factors under the radical sign, dividing the sura by 2, and to the quotient, adding the losiarithms of radi- us and 2, and the arithmetical complements of the logarithms of 6 andc. (Arts. 39, 47, 59.) Ex. Given a = l34, 6 = 108, and c=80, to find A, B, and C. For the angle A. A 161 log. 2.2068259 /i-a 27 log. 1.4313638 h-b 53 log. 1.7242759 fi-c 81 lo-. 1.9084850 RX2 2)7.2709506 * 3.6354753 loz. 10.3010300 108 a. 80 a. 13.9365053 c. 7.9665762 c. 8.0969100 For the a 134 a c 80 a angle B. 13.9365053 c. 7.8728952 .c. 8.0969100 SiaB. 9.9063105 B^53°42'9" For the angle C- 13.9365053 a 134 a. c. 7.87289.52 b 108 a. c. 7.9665762 Sin A. 9.9999915 Sin C. 9.7759767 0=36° 39' 20" Si„jA=Rv/^-^*)^ Theorem D. <( Sin iB= Rv/ Sin iC= Rv/ {h-a){ h-c) ac ah By art. 210, 2 Sin2iA=R» -Rxcos A. Substituting for cos A, its value, as given in page 132, 624-c2-a2 2SinHA=R2-R2X 26c * This is the logarithm of the area of the triangle. (Art. 222.^ 134 PARTICULAR SOLUTIONS Therefore 2Sin2iA=R2 x ^^ Bui2bc-\'a^-b''-c^=a^ -(h-cy={a+h-c){a-h-{-c) (Alg. 2o5.) Putting then A=|(a+6-|-c), reducing, and extracting, Sin i A=Rv/ Y^ Ex. Given a, 5, and r, as before, to find A and B. For the angle A. For the angle B. h-b 53 }.l2iTio9 h-a 27 1.4313638 h—c 81 1.^)084850 h-c 81 1.9084850 b 108 a. c. *;.:•.« »762 a 134 a. c. 7,872,8952 c 80 a. c. 8.<:.-r9!00 c 80 a. c. 8.0969100 2)19.(;962471 2)19.8096540 SiniA 9.8481235 Sin^B 9 654H270 A=::89''38' 31" B=53''42'9'' [cosiA=R/-(^^ TH.OH.M E. i (,^^ .B^R^^CLJO icosiC=Rv/'^^ By art. 210, 2Cos^iA=R2+Rxcos A, Substituting and reducing, as in the demonstration of the last theorem, 25c4-&-+c2-«2 (b-^-c+aYb-^c-a) 2Cos-A=R^ X -,,-^ =R^X^- i- -^ Pulling h=l{a-{-b-\-c) reducing and extracting, h(h-c) Cos jA=Rv/ ^ , * be Ex. Given the sides 134, 108, 80; to find B and C. OF TRIANGLES. l3;-» For the angle B. k 161 2.2068259 h-b 53 1.7242759 a 134 a. c. 7.8728952 c 80 a. c. 8.0969100 2)19.9009070 Cos iB 9.9504535 B==53*42'9" For the angle C. h 161 2.2068259 A— c 81 1.9084850 a 134 a. c. 7 8728952 b 108 a. c. 7.9665762 2)19.9547823 CosiC 9.9773911 0=36" 39' 20" TuEO REM F. I Tan .A=R^-^^^ \ Tan iB=Rv/ I lTan^C=Rv/ {h-a){h-c) h{h-b) (^k-a){h-b) h(h-c) The tangent is equal to the product of radius and the sine, divided by the cosines. (Art. 216.) By the last two theorems, then, Ta„xA=^''"^^ COS RV (h-b){h-c) be ~Rv/ h(h — a) be Thatis,Tan^A=Rv^ (h-bXh-c) h{h-a) Ex. Given the sides as before, to find A and C For the angle A. k-b 53 1.7242759 k-c 81 1.9084850 h-a 27 a. c. 8.5686-362 h 161 a. c. 7.7931741 For the angle B. h—a 27 1.4313638 h-b 53 17242759 h—c 81 a. c. 8.0915150 k 161 B.C. 7.7931741 Tan iA 2)19.9945712 9.9972856 A=89°38'3r Tan iC C=36°39' 20" 2)19.0403288 9.5201644 The three last theorems give the angle required, tvithout ambiguity. For the half of any angle must be less than 90.° Of these different methods of solution, each has its advan- tages in particular cases. It is expedient to find an angle, sometimes by its sine, sometimes by its cosine, and some- times by its tangent. By the first of the four preceding theorems marked C, D, E, and F, the calculation is made for the sine of the whole an- gle ; by the others, for the sine, eosine, or tangent, of hnjfihf' 136 PARTICULAR SOLUTIONS, &c. angle. For finding an angle near 90°, each of the three last theorems is preferable to the first. In the example above, A would have been uncertain to several seconds, by theorem C, if the other two angles had not been determined also. But for a very small angle, the first method has an advan- tage over the others. The third, by which the calculation is made for the cosine of half the required angle, is in this case the most defective of the four. The second will not answer well for an angle which is almost 180°. For the Aa//*of this is almost 90° ; and near 90°, the differences of the sines are very small. notes- Note A. Page 1 rpHE 7iame Loo:arithm is from >^oyog ratio, and a^i&ixos num- ber. Considering the ratio of a to 1 as a simple ratio, that of a^ to 1 is a duplicate ratio, of a^ to.l a triplicate ra- tio, &:c. (A!g. 354.) Here the exponents or logarithms 2, 3, 4, &ic, show how many times the simple ratio is repeated as a factor, to form the compound ratio. Thus the ratio of 100 to 1, is the square of the ratio of 10 to 1 ; the ratio of 1000 to 1 , is the cube of the ratio of 1 to 1 , &tc. On this account, logarithms are called the iiieasurts of ratios ; that is of the ratios which different numbers bear to unity. See the In- troduction to Hutton's Tables, and Mercator*s Logarithmo- Technia, in Masseres' Scriptores Logarithmici. Note B. p. 4. If 1 be added to -.09691, it becomes 1 —-09691, which is equal to -f-'90309. The decimal is here rendered positive, by subtracting the figures from 1. But it is made 1 too great. This is compensated, by adding - 1 to the intes:ral part oY the logarithm. So that — 2— .09691 = — 3+.90309. In the same manner, the decimal part of any logarithm which is wholly negative, may be rendered positive, by sub- tracting it from l,and adding —1 to the index. The sub- traction is most easily performed, by taking the right hand significant figure from 10, and each of the other figures from 9. (Art. 55.) On the other hand, if the index of a logarithm be negative, while the decimal part is positive ; the whole may be render- ed negative, by subtracting the decimal part from 1, and ta- king — 1 from the index. to 138 TRIGONOMETRY Note C. p. 7. It is common to define logarithms to be a series of num- bers in arithmetical progression, corresponding with another series in geometrical progression. This is calculated to per- plex the learner, when, upon opening the tables, he finds that the natural numbers, as thej stand there, instead of being in geometrical^ are in arithmetical progression ; and that the log- arithms are not in arithmetical progression. It is true, that a geometrical series may be obtained, by taking out, here and there, a few of the natural numbers ; and that the logarithms of these will form an arithmetical se- ries. But the definition is not applicable to the whole of the numbers and logarithms, as they stand in the tables. The supposition that positive and negative numbers have the same series of logarithms, (p. 7.) is attended with some theoretical difficulties. But these do not affect the practical rules for calculating by logarithms. Note D. p. 43. To revert a series, of the form X =:^un-\-bn^ -\-cn^ ■\-dn'^ -\-en^ -]- &c. that is, to find the value of??, in terms of ac, assume a series, with indeterminate co-efficients, (Alg. 490. 6.) Letn=Aa: + Bx2-fCa:34-Dx^-i-Ex^-f-&c. Finding the powers of this value of ?i, by multiplying the series into itself, and arranging the several terms according to the powers of x ; we have n^=A2a:2 4-2AB.x^+2AC> -I-2BC > 5 .&. ^3= A^r34-3A-Bx^ +3A*C> , , « J > X -f-KC. n' = + 3AB' A^r* -f 4A»Ba;5-|-&c. A^T^-f&C. NOTES. 139 Substiluting these values, for ?i and its powers, in the first series above, we have „l faA x+aB } ,-\-aC ) + «D^ + aE I +iA2 5 ^ +26AB> x^+ '2b AC | +26BC + cAM + ^B^ )>x'-\-2bAD -f.BcA^B I -\-3cA'C [x' -f- dA' J -{-3cAB^ -f4JA3B -h eA^ J Transposing x, and making the co-efficients of the several powers of x each equal to 0, we have aA - 1 =0, «B+6A^=0, uC-\-nAB-\-cA^=0, aD4-26AC + 6B2-h3cA2B + rfA*=0, a.E-\-2bBC-\'2bAD-\-3cA^C + 3cAB^ -\-^dA^B-^eA' =:0. And reducing the equations, A = i B=- C= D=-- 26^ —ac 5b^—5abc+a''d E=b l^b'—^^ab^'c-^'Sa^c^+ea^bd—a^e These are the values of the co-efficients A, B, C, &c. in ihe assumed series n=Ax-fBx2+Cx +Dx^-|-Ex5+&c. Applying these results to the logarithmic series. (Art. €C. p. 43.) 140 TRIGONOMETRY- in which we have, in the inverted series A=^=l D=— L- 2.3.4 B= -6 = 1 2.3 2.3.4.5 Therefore 2 ^2.3 ^2.3.4 ^2.3.4.5 ^^^* Note E. p. 50. According to the scheme lately introduced into France, of dividing the denominations of weights, measures, &c. into tenths, hundredths, &;c. the fourth part of a circle is divided into 100 degrees, a degree into 100 minutes, a minute into 100 seconds, &c. The whole circle contains 400 of these degrees ; a plane triangle 200. If a right angle be taken for the measuring unit ; degrees, minutes, and seconds, may be written as decimal fractions. Thus 36° 5' 49'' is 0.360549. C 10°=9° ^ According to the French division < 100' =54' > English. / 1000 ' = 324" S Note F. p. 82. If the perpendicular be drawn from the angle opposite the longest side, it will always fall within the triangle ; because the 05 her two angles must, of course, be acute. But if one of the angles at the base be obtuse^ the perpendicular will fall without the triangle, as CP, (Fig. 38.) In this case, the side on which the perpendicular falls, is to the sum of the other two ; as the difference of the latter. to i\iQ Sim of the segments made by the perpendicular. NOTES. 141 The demonstration is the same, as in the other case, ex- cept that AH = BP+PA, instead of BP-PA. Thus in the circle BDHL(Fig. 38.) of which C is the cen- tre, ABxAH=ALxAD; therefore AB : AD::AL : AH. But AD = CD-f-CA=CB+CA And AL=CL-CA=CB-CA And AH=HP4-PA = RP+PA Therefore AB :CB-}-CA::CB-CA : BP+PA. When the three sides are given, it may be known whether one of the angles is obtuse. For any angle of a triangle is obtuse or acute, according as the square of the sine subtend- ing the angle is greater, or less, thsin the sum of the squares of the sides containing the angle. (Euc. 12, 13. 2.) Note G. p. 104. Gunter's Sliding Rule is constructed upon the same prin- ciple as his scale, with the addition of a slider, which is so contrived as to answer the purpose of a pair of compasses, in working proportions, multiplying, dividing, &z;c. The lines on the fixed part are the same as on the scale. The slider con- tains two lines of numbers, a line of logarithmic sines, and a line of logarithmic tangents. To multiply by this, bring 1 on the slider, against one of the factors on the fixed part ; and against the other factor on the slider, will be the product on the fixed part. To divide, bring the divisor on the slider, against the dividend on the fixed part ; and against 1 on the slider, will be the quotient on the fixed part. To work ?l proportion, bring the first term on the slider, against one of the middle terms on the fixed part ; and against the other middle term on the slider, will be the fourth term on the fixed part. Or the first term may be taken on the fixed part ; and then the fourth term will be found on the slider. Another instrument frequently used in trigonometrical con- ftructionsj i< 142 TRIGONOMETRY. The Sector. This consists of two equal scales, moveable about a point as a centre. The lines which are drawn on it are of two kinds ; some being parallel to the sides of the instrument, and others diverging from the central point, like the radii of a circle. The latter are called the double lines, as each is repeated upon the two scales. The single lines are of the same nature, and have the same use, as tiiose which are put upon the common scale ; as the lines of equal parts, of chords, of latitude, (Sic. on one face ; and the logarithmic lines of numbers, of sines, and of tangents, on the other. The double lines are A line o( Lines, or equal parts, marked Lin. or L, A line of Chords, Cho. or C. A line of natural Smc5, Sin. or S. A line of natural Tangents to 45^. Tan. or T. A line of tangents above 45°. Tan. or T. A line of natural Secants^ Sec. or S. A line o( Polygons, Pol. or P. The double lines o( Chords, o( sines, and oi tangents to 45°, are all of the same radius ; beginning at the central point, and terminating near the other extremity of each scale ; the chords at G0% the sines at 90°, and the tangents at 45°. (See art. 95.) The line of lines is also of the same length, containing ten equal parts which are numbered, and which are again subdivided. The radius of the lines of secants, and of tangents above 45°, is about one fourth of the length of the other lines. From the end of the radius, which for the secants is at 0, and for the tangents at 45°, these lines extend to between 70° and 80°. The line of polygons is numbered 4, 5, 6, &;c. from the extremity of each scale, to- wards the centre. The simple principle on which the utility of fhese several pairs of lines depends is this, that the sides of similar triangles are proportional. (Euc. 4. 6.) So that sines, tangents, &;c. are furnished to any radius^ within the extent of the opening of the two scales. Let AC and AC (Fig. 40.) be any pair of lines on the sector, and AB and AB' equal portions of these lines. As AC and AC are equal, the triangle ACC is isos- ceks, and similar to ABB'. Therefore, AB: AC::BB':CC'. NOTES. 143 Distances measured from ihe centre on either scale, as AB and AC, are called lateral distances. And the distances between corresponding points of the two scales, as BB' and CC are called transverse distances. Let AC and CC be radii of two circles. Then if AB be the chord, sine, tangent, or secant, of any number of degreei? in one ; BB' will be the chord, sine, tangent, or secant, of the same number of degrees in the other. (Art. 119.) Thus, to find the chord of 30°, to a radius of four inches, open the sector so as to make the transverse distance from 60 to 60, on the lines of chords, four inches ; and the distance from 30 to 30, on the same lines, will be the chord required. To find the sine of 23°, make the distance from 90 to 90, on the lines of sines, equal to radius ; and the distance from 28 to 28 will be the sine. To find the tangent of 37"^, make the dis- tance from 45 to 45, on the lines of tangents, equal to radius ; and the distance from 37 to 37 will be the tangent. In find- ing secants, the distance from to mu?t be made radius. (Art. 201.) To lay down an angle of 34°, describe a circle, of any con- venient radius, open the sector, so that the distance from GO to GO on the lines of chords shall be equal to this radius, and to the circle apply a chord equal to the distance from 34 to 34. (Art. 161.) For an angle above 60°, the chord of Ac//' the number of degrees may be taken, and applied twice on the arc, as in art. 161. The line o{ polygons contains the chords of arcs of a circle which is divided into equal portions. Thus the distances from the centre of the sector to 4, 5, 6, and 7, are th ji h ^"^ 4 '^^ ^ circle, The distance 6 is the radius. (Art. 95.) This line is used to make a regular polygon, or to inscribe one in a given circle. Thus, to make a pentagon with the transverse distance from 6 to G for radius, describe a circle, and the distance from 5 to 5 will ha the length of one of the sides of a pentagon inscribed in that circle. The line of lines is used to divide a line into equal or pro- portional parts, to find fourth proportionals, &,c. Thus, t© divide a line into 7 equal parts, make the length of the given line the transverse distance from 7 to 7, and the distance from 1 to 1 will be one of the parts. To find f of a line, make the transverse distance from 5 to 5 equal to the given line ; and the distance from 3 to 3 will be f of it. In working the proportions in trigonometry on the sector, the lengths of the sides of triangles are tfiken from the linf 144 TRIGONOMETRY. of lines, and the degrees and minutes from the lines of sines, tangents, or secants. Thus in art. 135j ex. 1, 35 : R::26 : Sin 48°. To find the fourth term of this proportion by the sector, make the lateral distance 35 on the line of lines, a transverse distance from 90 to 90 on the lines of sines ; then the lateral distance 26 on the line of lines, will be the transverse dis- tance from 48 to 48 on the lines of sines. For a more particular account of the construction and uses of the Sector, see Stone's edition of Bion on Mathematical Instruments, Button's Dictionary, and Robertson's Treatise on Mathematical Instruments. Note H. p. 124. The errour in supposing that arcs less than 1 minute are proportional to their sines, can not affect the tirst ten places of decimals. Let AB and AB' (Fig. 41.) each equal 1 min- ute. The tangents of these arcs BT and B'T are equal, as are also the sines BS and B'S. The arc BAB' is greater than BS + B'S, but less than BT + BT. Therefore BA is greater than BS, but less than BT : that is, the difference be- tween the sine and the arc is less than the difference between the sine and the tangent, Now the sine of 1 minute is 0.000290888216 And the tangent of 1 minute is 0.000290888204 The difference is 0.00000000001 2 The difference between the sine and the arc of 1 minute is less than this •, and the errour in supposing that the sines of 1', and of 0' 52'' 44'" 3''" 45"'" are proportional to their arcs, as in art. 223, is still less. Note I. p. 125. There are various ways in which sinas and cosines may be more expeditiously calculated, than by the method which is NOTES. 146 given here. But as we are already supplied with accurate trigonometrical tables, the computation of the canon is, to the great body of our students, a subject of speculation, rath- er than of practical utility. Those who wish to enter into a minute examination of it, will of course consult the treatises in which it is particularly considered. There are also numerous formulas of veriftcatioji, which are used to detect the errours with which any part of the cal- culation is liable to be affected. For these, see Legendre's and Woodhouse's Trigonometry, Lacroix's Differential Cal- culus, and particularly Euler's Analysis of infinites. Note K. p. 127. The following rules for finding the sine or tangent of a very small arc, and, on the other hand, for finding the arc from its sine or tangent, are taken from Dr. IViaskelyne's Introduction to Taylor's Logarithms. To find the logarithmic sine of a very small arc. From the sum of the constant quantity 4.6855749, and the logarithm of the given arc reduced to seconds and decimals, subtract one third of the arithmetical complement of the log- arithmic cosine. To find the logarithmic tangent of a very small arc. To the sum of the constant quantity 4.6855749, and the logarithm of the given arc reduced to seconds and decimals, add two thirds of the arithmetical complement of the logarith- mic cosine. To find a small arc from its logarithmic sine. To the sum of the constant quantity 5.3144251, and the given logarithmic sine, add one third of the arithmetical com- plement of the logarithmic cosine. The remainder diminish- ed by 10, will be the logarithm of the number of seconds in the arc. To find a small arc from its logarithmic tangent. From the sum of the constant quantity 5.3144251, and the given logarithmic tangent, subtract two thirds of the arithmet- ical complement of the logarithmic cosine. The remainder diminished by 10, will be the logarithm of the number of see- onds in the arc. For the demonstration of these rules, see Woodhouse's Trigonometry, p. 189. 20 A TABLE OF NATURAL SINES AND TANGENTS; TO EVERY TEN MINUTES OF A DEGREE, IF the given angle is less than 45°, look for the title of the column, at the top of the page ; and for the de- grees and minutes, on the left. But if the angle is be- tween 45° and 90°, look for the title of the column, at the bottom ; and for the degrees and minutes, on the right. 148 NATURAL SINES AND TANGENTS D. M. 0° 0' 10 20 30 40 0° 50' Sine O.OOOOOOO 0029089 0058177 00S7265 0116353 0145439 r 0' 0.0174524 10 0203608 20 0232690 30 0^01769 40 0290847 r 50' 0319922 go 0' 0.0348995 10 0378065 20 0407131 30 • 0436194 40 0465253 QO 50' 0494308 '^° 0' 0.0523360 10 0552406 20 0581448 30 0610485 40 0639517 3° 50' 0668544 40 0' 00697565 10 0726580 20 0755589 30 0784591 40 0813587 4° 50' 0842576 5° 0' 0.0871557 10 0900532 20 0929499 30 0958458 40 0987408 5° 50' 10'6351 D. Cosine rHn 0029089 0058178 0087269 0116361 0146464 0.0174551 0203650 0232753 0261859 0290970 0320^86 0.0349208 0378335 0407469 0436609 0465757 0494913 0.0524078 0553251 0582434 0611626 0640829 0670043 0.0699268 0728605 0757755 0787017 0816293 0848583 0.0874887 0904206 0933340 0962890 0992257 10216 41 (.'otansent Infinite 343.77371 171.88540 114.58865 85.939791 68.750087 57.289962 49.103881 42.964077 38.188459 34.367771 31.241577 28.636253 26.431600 24.541758 22.903766 21.470401 20.205553 19.081137 18.074977 17.169337 16.349855 15 604784 14.924417 14.300666 13.726738 13.196883 12.706205 12.250505 11.826167 11.430052 11.059431 10.711913 16.385397 10.078031 9.7881732 Cosii Tangent 1.0000000 0.9999958 9999831 9999619 9999323 9998942 0.9998477 9997927 9997292 9996573 9995770 9994881 0.9993908 9992851 9991*709 9990482 9989171 9987775 0.9986295 9984731 9983082 9981348 9979630 9977627 0.9975641 9973669 9971413 9969173 9966849 9964440 D. M. 90° 0' 50 40 30 20 89° 10 89° 60 40 30 20 88° 10' 88° 0' 50 40 30 20 87° 10' 87° 0' 50 40 30 20 86° 10' 86° 0' 50 40 30 20 85° 10' 0.9961947 85° 9950370 50 9956708 40 9953962 30 9951132 20 9948217 84° 10 Sine D. M. NATURAL SINES AND TANGENTS. 149 ; D. 51 . 10 20 30 40 i 6o 50' I 7° i 10 20 , 30 40 I 7° 5o' i 8° 0' ! 10 20 30 40 j 8° 50' f 9° 0' 10 20 30 I 40 9° 50' 10' 10= 11 0' 10 20 30 40 50' Sine 0' 10 20 30 40 11° 50' D. M. 0.1045285 1074210 1103126 1132032 1160929 1189816 0.1218693 1247560 1276416 1305262 1334096 1362919 0.1391731 1420531 14493 19 1478094 1506857 1535667 0.1664345 1593069 1621779 1650476 1679159 1707828 0.1736482 1765121 1793746 1822355 1850949 1879528 O.J 908090 1936636 1965166 1993679 2022176 2050655 Cosine Tangent. 0.1051042 1080462 1109899 1139356 1168832 1198329 0.1227846 1257384 1286943 1316525 1346129 1375757 0.1404085 1435084 1464784 1494510 1524262 1554040 0.1583844 1613677 1643537 Cotangent 9 5143645 9.2553035 9.0098261 8.7768874 8.5555468 8.3449558 8.1443464 7.9530224 7.7703506 7.5957541 7.4287064 7.2687255 7.1153697 6.9682335 6.8269437 6.6911562 6.5605538 6.4348428 6.3137515 6-1970279 6.0844381 1703344 I 5.8708042 1733292 5.7693688 0.1763270 1793279 1823319 1853390 1883495 1913632 0.1943803 1974008 2004248 2034523 2064834 2095181 5.6712818 5.5763786 5.4845052 5.3955172 5.3092793 5.2256647 5.1445540 5.0658352 4.9894027 4.9151570 4.8430045 4.7728568 Cotangent | Tangent Cosine D. 84° M. 0' 0.9945219 9.942136 50 9938969 40 9935719 30 9932384 20 9928065 830 10' 0.9925462 83° 0' 9921874 50 9918204 40 9914449 » 30 9910610 20 9906687 82° 10' 0.9902681 82° 0' 9898590 50 9894416 40 9890159 30 9885817 20 9881392 81° 10' 0.9876883 81° 0' 9872291 50 9867615 40 9862856 30 9858013 20 9853087 80° 10' 0.9848078 80° 0' 9842985 50 9837808 40 9832549 30 9827206 20 9821781 79° 10' 0.9816272 79° 0' 9810680 50 9805005 40 9799247 30 9793406 20 9787483 78° 10' Sine « D. M. 150 NATURAL SINES AND TANGENTS. D. M . 12° 0~ 10 20 30 40 12° 60' 13° O 10 20 30 40 13° 50' 14° 0' 10 20 30 40 14° 50' 15° 10 20 30 40 60' 0' 10 20 30 40 50' 0' 10 20 30 40 60' Sine 16= 16= 17' -17' D. M. 0.2079117 2107561 2135988 2164396 2192786 2221168 0.2249511 2277844 2306159 2334454 2362729 2390984 0.2419219 2447433 2475627 2503800 2531952 2560082 0.2588190 2616277 2644342 2672384 2700403 2728400 0.2756374 2784324 2812251 2840153 2868032! 2895887 Tangent 0.2923717 2951522 2979303 3007058 3034788 3062492 Cosine 0.2125566 2165988 2186448 2216947 2^247485 2278063 0.2308682 2339342 2370044 2400788 2431575 2462405 0.2493280 2524200 2555165 2586176 2617234 2648339 0.2679492 2710694 2741945 2773245 2804597 2835999 0.2867454 2898961 2930521 2962135 2993803 3026527 0.3057307 30S9143 3121036 3152988 3184998 3217067 Cotangent Cotnn?ent 4.7046301 4.6382457 4.5736287 4.5107085 4.4494181 4.3896940 4.3314759 4.2747066 4.2193318 4.1652998 4.1125614 4.0610700 4.0107809 3.9616518 3.9136420 3.8667131 3.8208281 3.7759519 3.7320508 3.6890927 3.6470467 3.6058835 3.5655749 3.5260938 3.4874144 3.4495120 3.4123626 3.3759434 3.3402326 3.3052091 3.2708526 3.2371438 3.2040638 3.1715948 3.1397194 3.1084210 Cosine 1). IVi 09781476 9775387 9769215 9762960 9756623 9750203 0.9743701 9737116 9730449 9723699 9716867 9709963 0.9702957 9695879 9688719 9681476 9674152' 9666746 0.9659258 9651689 9644037 9636305 9628490 9620594 0.9612617 9604558 9596418 95^8197 9579895 9571612 0.9563048 9554502 9545876 9537170 9528382 9519514 18' IT 0' 50 40 30 20 10 77° 0' 60 40 30 20 76° 10' 76° 0' 50 40 30 20 75° 10' 50 40 30 20 10 74= 74° O' 50 40 30 20 75° 10' Tangent i^ine 73' 72' 0' 50 40 30 20 10' D. M NATURAL SINES AND TANGENTS. 151 D. M. 1 bine Tangent Cotangent Cosine 72° 0' 18° 0' 0.3090170 0.3249197 3.0776835 0.9510565 10 3117822 3281387 3.0474915 9501536 50 20 3145448 3313639 3.0178301 9492426 40 30 3173047 3345953 2.9886850 94i>3237 30 40 3200619 3378330 2.9600422 9473966 20 18° 50' 3228164 3410771 2.9318885 9464616 71° 10' 19° 0' 0.3255682 0.3443276 2.9042109 0.9155186 71° 0' 10 3283172 3475846 2.8769970 9445675 50 20 3310634 3508483 2.8502349 9436085 40 30 3338069 3541186 2.8239129 9426415 SO 40 3365475 3573956 2.7980198 9416665 20 19° 50' 3392852 3606795 2.7725448 9406835 70° 10' 20° 0' 0.3420201 0.3639702 2.7474774 0.93969261 70° oj 10 3447521 3672680 2.7228076 9386938 50 20 3474812 3705728 2.6985254 9376869 40 30 3502074 3738847 2.67462 i 5 9366722 30 40 3529306 3772038 2.6510867 9356495 20 20° 50' 3556508 3805302 2.6279121 9346189 69° 10' 21° 0' 0.3583679 0.3838640 2.6050891 0.9335804 69° 0' 10 3610821 3872053 2.5826094 9325340 50 20 3637932 3905541 2.5604649 9314797 40 30 3665012 3939105 2.5386479 9304176 30 40 3692061 3972746 2.5171507 9293475 20* 21° 50 3719079 4006465 2.4959661 9282696 68° 10' 22° 0' 0.3746066 0.4040262 2.4750869 0.9271839 68° 0' 10 3773021 4074139 2.4545061 9260902 50 20 3799944 4108097 2.4342172 9249888 40 30 3826834 4142136 2.4142136 9238795 30 40 3853693 4176257 2.3944889 9227624 20 22° 50 3880518 4210460 2.3750372 9216375 67° 10' 23° 0' 0.3907311 0.4244748 2 3558524 0.9205049 67° 0' 10 3934071 4279121 2.3369287 9193644 50 20 3960798 4313579 2.3182606 9182161 40 30 3987491 4348124 2.2998425 9170601 30 40 4014150 4382756 2.2816693 9158963 20 23° 50 4040776 4417477 2.2637357 9147247 66° 10/ D. M. D. M. Cosine Cotangent Tangent j iSine 152 NATURAL SINES AND TANGENTS. D. M. 24° . 0' 10 20 30 40 24° 50' 25° 0' 10 20 30 40 25° 50' 26° 0' 10 20 30 40 26° 50' 27° 0' 10 20 30 40 27° 50' 28° 0' 10 20 30 40 280 50' 29° 0' 10 20 30 40 29° 50' D. M. Sine 0.4067366 4093923 4120445 4146932 4173335 4199801 0.4226183 4252528 4278838 4305111 4331348 4357548 0.4383711 4409838 4435927 4461978 4487992 4513967 0.4539905 4565805 4591665 4617486 4643269 4669012 0.4694716 4720380 4740004 4771588 4797131 4822634 0.4848096 4873517 4898897 4924236 4949532 4974787 Cosine Tangent 0.4452287 4487187 4522179 4557263 4592439 4627710 0.4663077 4698539 4734098 4769755 4805512 4841368 0.4877326 4913386 4949549 4985816 5022189 5058668 0.5095254 5131950 5168755 5205671 5242698 5279839 0.6317094 5354465 5391953 6429557 5467281 5605126 0.5543091 5581179 5619391 5657728 5696191 5734783 Cotangent Cotangent 2.2460368 2.2285676 2.2113234 2.1942997 2.1774920 2.1608958 2.1445069 2.1283213 2.1123348 2.0965436 2.0809438 2.0665318 2.0503038 2.0352565 2.0203862 2.0056897 1.9911637 1.9768050 1.9626106 1.9485772 1.9347020 1.920982; 1.9074147 1.8939971 1.8807266 1 8676003 1.8546159 1.8417709 1.8290628 1.8164892 1.8040478 1.7917362 1.7795524 1.7674940 1.7555690 1.7437463 Tangent Cosine D. M. 0.9135455 66' 0' 9123584 50 9111637 40 9099613 30 9087511 20 9075333 65o 10' 0.9063078 65° 0' 9050746 50 9038338 40 9025853 30 9013292 20 9000654 64° 10' 0.8987940 64° 0' 8975151 50 8962285 40 8949344 30 8936326 20 8923234 63° 10' 0.8910065 6.3° 0' 8896822 50 8 38 3603 40 8870108 30 8856639 20 8843095 62° 10' 0.8829476 62° 0' 8815782 60 8802014 40 8788171 30 8774254 20 8760263 61° 10' 0.8746197 6I0 0' 8732068 60 8717844 40 8703657 30 8689196 20 8674762 600 10' Sine D. M. NATURAL SINES AND TANGENTS 153 ^dTIl Sine Tangent CotHiii^ent Cosine / D. M. 30° 0.500000n 0.5773503 1.7o2U508 0.8660254 60^ 0' 10 5025170 5812353 1.7204736 8645673 50; 20 5050298 5851335 1.70.90116 8631019 40 30 50763S4 6890450 1.6976631 8616292 30 40 5100426 5929699 1.6864261 8601491 20 l30° 60' i 6125425 5969084 1.6752988 8586619 59° 10' 1 131^ 0' 0.5150381 0.6008606 1.6642795 0.8571673 59° 0' 10 5175293 604826C 1.6533663 8556655 50 20 6200161 6088067 1.6425576 8541564 40 30 6224986 6128008 1.6318517 8526402 30 40 5249766 6168092 1.6212469 8511167 20 31° 50' 5274602 6208320 1.6107417 8495860 58° 10' 32'> 0' 0.5299193 0.6248694 1.6003345 0.8480481 58° C 10 5323839 62892 '4 1.5900238 84650.30 50 20 5348440 6329883 1.5798079 8449508 40 30 6372996 6370703 1.5696856 843 3914 30 40 6397507 6411673 1.5596552 8418249 20 3^° 50' 5421971 6452797 1.5197155 8402513 67° 10' 33^^ 0' 0.5446390 0.6494076 1.5398650 0.8386706 57° 0' 10 6470763 633651 1 1.5301023 8>.70827 50 20 6495090 6577103 1.5204261 8354878 40 30 5519370 6618S56 1.5108352 8338858 30 40 5543603 6660169 1.501.3282 8322768 20 33° 50' 5567790 6702843 1.4919039 8306607 56° 10' 34^ 0' 0.5591929 0.6745035 1.4825610 0.329037G 56° 0' 10 6616021 6737492 1.4732933 8274074 50 20 5640066 6830066 1.4641147 8257703 40 30 5664062 6872310 1.4550090 8241262 30 40 668801 1 6915725 1.4459801 8224751 20 34° 50' 6711912 G958813 1.4370268 8208170 55'' 10' 1 35° 0' 0.5735764 0. '700207 5 1.4281480 0.8191520 55° 0' 10 6759568 7045515 1.4193427 8171801 50 20 6783323 | 7089133 1.4106098 8158013 40 30 5807030 7132931 1.4019483 8141155 30 40 5830687 7176911 1.3933571 8124220 20 35° 50' 5854294 7221075 Cotangent 1.3848353 8107234 54° 10 D. -VI ^ D. M. Cosine Tangent . Sine ^1 I FA NATURAL SINES AND TANGENTS. D. m: 36° 0' Sine Tnngent. 0.7265425 .Cotanger^t 1.3763819 Cosine U.M. 0.5877853 0.8090170 54° 0' 1 ]0 5901361 730f>963 1.3679959 8073038 50 t 20 5924819 7354691 1.3596764 8055837 40 30 5948228 7399611 1.3514224 8038569 30 40' 5971586 7444724 1.3432331 8021232 20 360 50 5094893 7490033 1.3351075 8003827 630 10' =37° 0' 0.6U18150 0.7535541 1.3270448 0.7986355 53° C 1 10 6041356 7581248 1.3190441 7968815 50 ; 20 6064511 7627157 1.3111046 7951208 40 1 30 6087614 7673270 1.3032254 7933533 30' 40 6110666 7719589 1.2954057 7915792 20 37° 50' 6133666 7766118 1.2876447 78979&3 52° 10' 38° 0' 0.6156615 0.7812856 1.2799416 .0.7880108 52° 0' 10 6179511 7859808 1.2722957 7862165 50 20 6202355 7906975 1.2647062 7844157 40 30 6225146 7954359 1.2571723 7826082 30 40 6247885 8001963 1.2496933 7807940 20 38° 50' 6270571 8049790 1.2422685 7789733 61° 10' i39° 0' 06293204 0.8097840 1.2348972 0.7771460 51° 0' 1 10 6315784 8146118 1.2275786 7753121 50 20 6338310 8194625 1.2203121 7734716 40 30 6360782 8243364 1.2130970 7716246 30 40 6383201 8292337 1.2059327 7697710 20 39° 50 6405566 8341547 1.1.988184 7679110 50° 10' 40° 0' 0.6427876 0.8390996 1.1917536 0.7660444 50° 0' 10 6450132 8440688 1.1847370 7641714 50 20 6472334 8490624 1.1777698 7622919 40 30 6494480 8540807 1.170S496 7604060 30 40 6516572 8591240 1.1639763 7585136 20 40° 50' 6538609 8641926 1.1571495 7566148 49° 10' 41° 6560590 0.8692867 1.1503684 0.7547096 49° 0' 10 6582516 8744067 1.1436326 7527980 50 20 66043'66 8795528 1.1369414 750f.800 40 30 6626200 8847253 1.1302944 7489557 30 40 6647959 8899244 1.1236909 7470251 20 1 41° 50' 666~J66\ 8951506 Cotangent 1.1171305 7450881 48° lO'j D. M. ' D. M. Cosine Tangent Sine NATURAL SINES AND TANGENTS. 155 D. M. Sine Tanjjent Cotangent Cosine D. M. 42° 0' 0.6691306 0.9004040 1.1106125 0.7431448 48° 0' 10 6712895 9056851 1.1041365 7411953 60 20 6734427 9109940 1.0977020 7392394 40 30 6755902 91b3312 1.0913085 7372773 30 ^ 40 6777320 9216969 1.0849554 7353090 20 42° 50' 6798681 9270914 1.0786423 7333345 47° 10 43° 0' 0.6819984 0.932-5151 1.0723687 0.7313537 47° 0' 10 6841229 9379683 1.0661341 7293668 50 20 6862416 9434513 1.0599381 7273736 , 40 30 6883546 9489646 1.0537801 7253744 30 40 6904617 9545083 1.0476598 7233690 20 43° 50' 6925630 9600829 1.0415767 7213574 46° 10' 44° 0' 0.6946584 0.9656888 1.0355303 0.7193398 46° 0' 10 6967479 9713262 1.0295203 7173161 50 20 6988315 9769956 1.0235461 7152863 40 30 7009093 9826973 1.0176074 7132504 30 40 7029811 9884316 1.0117038 7112086 20 44° 50' 7050469 9941991 1.0058348 7091607 45° 10' 45° 0' 0.7071068 1.0000000 Cotangent 1.0000000 0.70710G8 45® 0' a M. 1 D. M. ' Cosine Tangent ' Sine The Secants and Cosecants^ which are not inserted in this table, may be easily supplied. If 1 be divided by the cosine of an arc, the quotient will be the secant of that arc. (Art. 228.) And if 1 be divided by the «ine. the quotient will be fhe cosecant. PI. I. y.^cflin ScM.ff. THICOWOMETHY. TRIGOXOMETRir, ri.ii. ,/.» Si.J^H. TRIOOITOSI GTR V. ^JuA I „->^ ^\ T«IGO:^©>tBTltY, i^m. TBICOyO>tKT»Y. r^ A PRACTICAL APPUCATION OP THE PRINCIPLES OF GEOMETRY TO THE OF SUPERFICIES AND SOLIDS Buiva THE THIRD PART OF A COURSE OF MATHEMATICS, ADAPTED TO THE METHOD OF INSTRUCTION IN THS AMERICAN COLLEGES. By JEREMIAH DAY, D. D. LL. D. president of Yale College, THE SECONB EDITION. NEW-HAVEN : PRINTED AND PUBLISHED BT S. CONTERSE' 1B25. District of Connecticuti ss. BE IT REMEMBERED, That on the second day of March, in tht fortieth year of the Independence of the United States of America, JER- EMIAH DAY, of the said District, hath tleposited in this Office the title of a look, the right whereof he claims as Author, in the words following, to wit :— .. T'Z^ " A practical Application of the Principles of Geometry to the Mensu* ration of Superficies and Solids : Being the Third Part of a Course of Mathematics, adapted to the method of instruction in the American Colleges. By Jeremiah Day, Professor of Mathematics and ITatunl Philosophy in Yale College." In conformity to the act of the Congress of the United States, enUtled " An act for the encouragement of learning, by securing the copi«R of Maps, Charts, and Books, to the authers »nd proprietors of such copies, during the times therein mentioned " HENRY W EDWARDS, Clerk of tht IHHrict of Conneetieut A true copy of Record, examined by ifte, HENRY W. EDWARDS, Cltrk of th$ Dittrict o/ Connecticut Thk following short Treatise contains little more than an application of the principles of Geometry, to the numerical calculation of the superficial and solid contents of such figures as are treated of in the Elements of Euclid. As the plan proposed for the.work of which this number is a part, does not admit of introducing rules and propositions which are not demonstrated ; tlie particular consideration of the areas of the Conic Sections and other curves, with the contents of solids produced by their revolution, is reserved for succeed- ing parts of the course. The student would be Httle profited by applying arithmetical calculation, in a mechanical way, to figures of which he has not yet learned even the definitions. But as this number may fall into the hands of some who will not read those which are to follow, the principal rules for conic areas and solids, and for the gauging of casks, are given, without demonstrations, in the appendix. Those who wish to take a complete view of Mensuration, in all its parts, are re- ferred to the valuable treatise of Dr. Hutton on the subject. CONTENTS, P»ge. Section I. Areas of figures bounded by right lines, 1 II. The Quadrature of the circle and its parts, 14 Promiscuous examples of Areas, - 25 HI. Solids bounded by plane surfaces, - 28 IV. The Cylinder, Cone, and Sphere, - 43 Promiscuous examples of Solids, - 59 V. Isoperimetry, - - - - 61 APPENDIX.— Part I. Mensuration of the Conic Sections, and other figures, 72 Part II. Gauging of Casks, ... - 80 Notes, -..--- 86 Table of Circular Segments, - - - 93 SECTION I. AREAS OF FIGURES BOUNDED BY RIGHT LINES. - J T^HE following definitions, which are nearly the ^ ' ' -^ same as in Euclid, are inserted here for the con- venience of reference. 1. Four-sided figures have different names, according to the relative position and length of the sides. A parallelogram has its opposite sides equal and parallel ; as ABCD. (Fig. 2.) A rectangle, or right parallelogram, has its opposite sides equal, and all its angles right angles; as AC. (Fig 1.) A square has all its sides equal, and all its angles right angles ; as AB GH. (Fig 3 ) A rhombus has all its sides equal, and its an- gles oblique: as ABCU. (Fig. 3.) A rhomboid has its op- posite sides equal, and its angles oblique ; as ABCD. (Fig. 2.) A trapezoid has only two of its sides parallel ; as ABLD. (Fig. 4 ) Any other four sided figure is called a trapezium. II. A figure which has more than four sides is called a polygon. A regular polygon has all its sides equal, and all its angles equal. III. The height of a triangle is the length of a perpendicu- lar, drawn from one of the angles to the opposite side; as CP. (Fig. 5.) The height oi a four-sided figure is the per- pendicular distance between two of its parallel sides; as CP. (Fig. 4.) IV. The area or superficial contents of a figure is the space contained within the line or lines by which the figure is bounded. 2. In calculating areas, some particular portion of surface is fixed upon, as the measuring unit, with which the given figure is to be compared. This is commonly a square; as a square inch, a square foot, a square rod, he. For this rea- son, determining the quantity of surface in a iigure is called squaring it, or finding its quadrature ; that is, finding a square or number of squares to which it is equal. 2 ^ MENSURATION OF 3. The superficial unit has generally the same name, as the linear un'\t which forms the side of the square. The side of a square inch is a linear inch ; of a square foot, a linear foot ; of a square rod, a linear rod, &;c. There are some superficial measures, however, which have no corresponding denominations of length. The acre, for instance, is not a square which has a line of the same name for its side. The following tables contain the linear measures in com- mon use, with their corresponding square measures. Linear Measures, Square Measures. 12 Inches = 1 Foot. 144 Inches = 1 Foot. 3 Feet = 1 Yard. 9 Feet = 1 Yard- 6 Feet = 1 Fathom. 36 Feet = 1 Fathom. 161 Feet = 1 Rod. 2721 Feet = 1 Rod. 51 Yards = 1 Rod. 301 Yards = 1 Rod. 4 Rods = 1 Chain. 16 Rods = 1 Chain. 40 Rods = 1 Furlong. 1600 Rods = 1 Furlong. 320 Rods = 1 Mile. 102400 Rods = 1 Mile. An acre contains 160 square rods, or 10 square chains. By reducing the denominations of square measure, it will be seen that 1 sq. mile=640 ;icre?=102400 rods=27a7n-100 ft.=401 4409600 inche?. 1 acre=10 chains=l60 rods=43560 feet^627^2640 inches. The fundamental problem in the mensuration of superfi- cies is the very simple one of determining the area of a right parallelogram. The contents of other figures, particularly those which are rectilinear, may be obtained by finding par- allelograms which are equal to them, according to the princi- ples laid down in Euclid. PROBLEM I. To find the area of a parali.elogram, square, rhombus, or rhomboid, 4. MULTIPLY THE LENGTH BY THE PERPENDICULAR HEIGHT OR BREADTH. It is evident that the number of square inches in the par- allelogram AC (Fig. 1 .) is equal to the number of linear inches in the length AB, repealed as many times as there are PLANE SURFACES. 3 inches in the breadth BC. For a more particular illustration of this, see Alg. 511 — 514. The oblique parallelogram or rhomboid ABCD (Fig. 2.) is equal to the right parallelogram GHCD. (Euc. 36. 1.) The area, therefore, is equal to the length AB multiplied into the perpendicular height HC. And the rhombus ABCD {Fig. 3.) is equal to the parallelogram ABGH. As the sides of a square are all equal, its area is found, by mulliplying out of the sides into itself, Ex. 1 . How many square feet are there in a floor 23^ feet long, and 18 feet broad ? Ans. 23^X18=423. 2. What are the contents of a piece of ground which is 66 feet square ? Ans. 4356 sq. feet=16 sq. rods. 3. How many square feet are there in the four sides of a room which is 22 feet long, 17 feet broad, and 11 feet high ? Ans. 858. Art. 5. If the sides and angles of a parallelogram are given, the perpendicular height may be easily found by trigonome- try. Thus CH (Fig. 2.) is the perpendicular of a right angled triangle, of which BC is the hypothenuse. Then (Trig. 134.) R : BC::SinB :CH. The area is obtained by multiplying CH thus found, into the length AB. Or, to reduce the two operations to one, As radius, To the sine of any angle of a parallelogram ; So is the product of the sides including that angle, To the area of the parallelogram. For the arca=AB XCH (Fig. 2.) But CH=JP^^'"-g R Therefore, The area = ^^^^^"^^'"^ OrR : SinB::AB X BC : the area. Ex. If the side AB be 58 rods, BC 42 rods, and the angle B 63*, what is the area of the parallelogram ? MENSURATION OF As radius 10.00000 To the sine of B 63» 9.94988 So is the product of AB 58 1.76343 Into BC (Trig. 39,) 42 1.62325 To the area 2170.5 sq. rods 3.3365G 2. If the side of a rhombus is 67 feet, and one of the angles 73^ what is the area ? Ans. 4292.7 feet. 6. When the dimensions are given in feet and inches, the multiplication may be conveniently performed by the arith- metical T[i\e o( Duodecimals; in which each inferior denom- ination is one twelfth of the next higher. Considering a foot as the measuring unit^ a prime is the twelfth part of a foot ; a second, the twelfth part of a prime, &c. It is to be ob- served, that, in measures of /en^//i, inches ^re primes; but in superficial measure they are seconds* In both, a prime is jj of a foot. But j\ of a square foot is a parallelogram a foot long and an inch broad. The twelfth part of this is a square inch, which is j^j of a square foot. Ex. 1. What is the surface of a board 9 feet 5 inches, by 2 feet 7 inches. F 9 5' 2 7 18 10 5 5 11 24 3 ir 11", or 24 feet 47 inches. 2. How many feet of glass are there in a window 4 feet 11 inches high, and 3 feet 5 inches broad ? Ans. 16F. 9' 7", or 16 feet 115 inches. 7. If the area and one side of a parallelogram be given, the other side may be found by dividing the area by the given side. And if the area of a square be given, the side may be found by extracting t'e square root of the area. This is merely re- versing the rule in art. 4. See Alg. 520, 521. Ex. 1. What is the breadth of a piece of cloth which is 36 yds. long, and which contains 63 square yds. Ans. If yds. PLANE SURFACES. 5 2. What is the side of a square piece of land containing 289 square rods ? 3. How many yards of carpeting \\ yard wide, will cover a floor 30 feet long and 221 broad ? Ans. 30X22|feet = 10x7i=75yds. And 75-f.li=60. 4. What is the side of a square which is equal to a paral- lelogram 936 feet long and 104 broad ? 5. How many panes ®f 8 by 10 glass are there, in a win- dow 5 feet high, and 2 feet 8 inches broad ? PROBLEM 11. To find the area of a t&ianole. 8. Rule I. Multiply one side nv half the perpen- dicular FROM THE OPPOSITE ANGLE. Op, multiply half the side by the perpendicular. Or, multiply the whole side by the perpendicular, and lake half the product. The area of the triangle ABC (Fig. 5.) is equal to | PCx AB, because a parallelogram of the same base and height is equal to PCxAB, (Art. 4.) and by Euc. 41. 1, the triangle is half the parallelogram. Ex. 1. If AB (Fig. 5.) be 65 ie^t, and PC 31.2, what is the area of the triangle ? Ans. 1014 square ieeU 2. What is the surface of a triangular board, whose base is 3 (eti 2 inches, and perpendicular height 2 feet 9 inches? Ans. 4F. 4' 3'', or 4 feet 51 inches. 9. If two sides of a triangle and the included angle, are given, the perpendicular on one of these sides may be easily found by rectangular trigonometry. And the area may be calculated in the same manner as the area of a parallelogram in art. 5. In the triangle ABC (Fig. 2 ) R:BC::SinB :CH And because the triangle is half the parallelogram of the same base and height, As radius, To the sine of any angle of a triangle ; So is the product of the sides including that angle, To twice the area of the triangle. ( Art. 5.) Ex. If AC (Fig. 5.) be 39 feet, AB G5 feet, and the angle at A 53° 7' 48", what is the area of the triangle ? Ans. 1014 square feet. 6 MENSURATION OF 9. b. \ione side and the angles are given ; then As the product of radius and the sine of the angle opposite the given side, To the product of the sines of the two other angles ; So is the square of the given side, To twice the area of the triangle. If PC (Fig. 5.) be perpendicular to AB. R : Sin B::BC :CP Sin ACB : Sin A::AB : BC Therefore (Alg. 390, 382.) R X Sin ACB : Sin A X Sin B : : AB X BC : CPxBC:: AB^ : ABxCP= twice the area of the triangle. Ex If one side of a triangle be 57 feet, and the angles at the endsof this side 50° and 60°, what is the area ? Ans. 1 147 sq. feet. 10. If the sides only of a triangle are given, an angle may be found, by oblique trigonometry, Case IV, and then the perpendicular and the area may be calculated. But the area may be more directly obtained, by the following method. Rule II. When the three sides are given, from half their sum subtract each side severally ^multiply together the half sum and the three remainders, and extract the square root of the product. If the sides of the triangle are a, h, and c, and if A=half their sum, then The area=\/hX{h'-a)X{h — b)X{h-c) For the demonstration of this rule, see Trigonometry, Art. 221. If the calculation be made by logarithms, add the loga- rithms of the several factors, and half their sum will be the logarithm of the area. (Trig. 39, 47.) Ex. I. In the triangle ABC (Fig. 5.) given the sides a 52 feet, b 39, and c 65 ; to find the side of a square which has the same area as the triangle. ^(a+b-\'c)^h=7Q A- 6 = 39 h'-a=2G A — c=13 Then the area=v'78 X26 X39 X 13 = 1014 square feet. PLANE SURFACES. By logarithms. The half sum =78 1.89209 First remainder =26 1.41497 Second do. =39 1.69106 Third do. =13 1.11394 2)6.01206 The area required =1014 2)3.00603 Side of the square =31.843 (Trig. 47) 1.60301 2. If the sides of a triangle are 134, 108, and 80 rods, what is the area ? Ans. 4319. 3. What is the area of a triangle whose sides are 37 J , 264, and 225 feet ? 11. In an equilateral triangle, one of whose sides is a, the expression for the area becomes ^hx{h-a)x{h—a)X(h-a) But as h = ^a, and h — a = ^a — a = laf the area is y/laX^aX^aXia = y/j\a'=laW3 (Alg. 271.) That is, the area of an equilateral triangle is equal to i the square of one of its sides, multiplied into the square root of 3, which is 1.732. Ex. 1. What is the area of a triangle whose sides are each 34 feet.'* Ans. 500^ feet. 2. If the sidesof a triangular field are each 100 rods, how manf acres does it contain .'' PROBLEM III. To find the area of a trapezoid. 21. Multiply half the sum of the parallel sides into their perpendicular distance. The area of the trapezoid ABCD (Fig. 4.) is equal to half the sum of the sides AB and CD, multiplied into the perpen- dicular distance PC or AH. For the whole figure is made up of the two triangles ABC and ADC ; the area of the first of which is equal to the product of half the base AB into the perpendicular PC, (Art, 8.) and the area of the other is equal to the product of half the base DC into the perpendic- ular AH or PC. B MENSURATION OF Ex. If AB (Fig. 4.) be 46 feet, BC 31, DC 38, and the angle B 70®, what is the area of the trapezoid ? R : BC::SinB : PC=29.13. And 42x29.13=1223^. 2. What are the contents of a field which has two parallel sides 65 and 38 rods, distant from each other 27 rods P PROBLEM IT. To find the area of a trapezium, or of an irregular polygon. 13. Divide the whole figure into triangles, by drawing diagonals, and find the sum of the areas of these triangles. (Alg.519.) If the perpendiculars in two triangles fall upon the same di- agonal, the area of the trapezium formed of the two trian- gles, is equal to half the product of the diagonal into the sum of the perpendiculars. Thus the area of the trapezium ABCH (Fig. 6.) is iBHxAL-f-iBHxCM=iBHx(AL4-CM) Ex. In the irregular poloygon ABCDH (Fig. 6.) CBH = S6 I ALi=5.3 if the diagonals < r^rj oc>' and the perpendiculars < CM=9.4 ^CH = 3J, r (DN=7.3 The area = 18x 14.6 + 16x7.3=379.6 14. If the diagonals of a trapezium^re given, the area may be found, nearly in the same manner as the area of a paral- lelogram in Art. 5, and the area of a triangle in Art. 9. In the trapezium ABCD (Fig. 8.) the sines of the four an- gles at N, the point of intersection of the diagonals, are all equal. For the two acute angles are supplements of the other two, and therefore have the same sme. (Trig. 90.) Put- ting, then, Sin N for the sine of each of these anglps, the areas of the four triangles ol which the trapezium is com- posed, are given by the following proportions; (Art. 9.) fBNxAN : 2 area ABN n.Q- r OP the versed sine, and BO the chord \ °*^^«iA ^he arc AOB. And in the right angled triangle CBP, m Vt ^ BP * ^'n BCP or BO ^D . IV.. ^^p ^ (^^gBCPoi BO Ex. 1. If the radius CO (Fig. 9.) =25. and the chord AB = 43 3 ; what is the length of the arc AOB ? CB : R: :BP : Sin BCP or BO=60° very nearly. The circumference of the circle =3.14159x50=157.08. And 360° : 60°::1 57.08 : 26. 1 8 =OB.Therefore AOB =52.36. 2. What is the length of an arc whose chord is 216^ in a circle whose radius is 125? Ans. 261.8. THE CIRCLE. 19 29. If only the chord and the height of an arc be given, the radius of the circie may be found, and then the length of the arc. If BA (Fig 9.) be the chord, andPO the height of the arc AOB, then (Euc. 35.3.) DP=-^ • AndD0=0P+DP=0P+-^5p- • That is, the diameter is equal to the height of the arc, -f- the square of half the chord divided by the height. The dianiL'ter being found, the length of the arc may be calculated by the two preceding articles. Ex. 1 If the chord of an arc be 173.2, and the height 50, what is the length of the arc 1 _ 86^6' The diameter =50-|--^ =200. The arc contains 120°; (Art. 28.) and its length is 209.44. (Art. 26.) 2. What is the length of an arc whose chord is 120, and height 45.? Ans. 160.8.* PROBLEM IV. To find the area of a circle. 30. Multiply the square of the diameter by the decimals .7854. Or, Multiply half the diameter into half the circum- ference. Or, multiply the whole diameter into the whole circumference, and take \ of the product. The area of a circle is equal to the product of half the diameter into half the circumference; (Sup. Euc. 5, 1.) or which is the same thing, { the product of the diameter and circumference. If the diameter be 1, the circum/erence is 3.14159; (Art. 23.) one fourth of which is 0.7S54 nearly. But the areas of different circles are to each other, as the squares of their diameters, (Sup. Euc 7, l.f) The area of any circle, therefore, is equal to the product of the square *> See note C. tEuclid2, 12. 20 MENSURATION OF of its diameter into 0.7854, which is the area of a circle whose diameter is 1. Ex. 1. What is the area of a circle whose diameter is 623 feet? Ans. 304836 square feet. 2. How many acres are there in a circular island whose diameter is 124 rods ? Ans. 76 acres, and 76 rods. 3. If the diameter of a circle he 113, and the circumfer- ence .355, what is the area? Ans. 10029. 4. How many square yards are there in a circle whose diameter is 7 feet ? 31. If the circumference of a circle he given, the area may be obtained, by first finding the diameter ; or, without finding the diameter, by multiplying the square of the circumference by .07958. For, if the circumference of a circle be 1, the diameter = 1-7-3.14159=0.31831 ; and i the product of this into the circumference is .07956 the area. But the areas of differ- ent circles, being as the squares of iheir diameters, are also as the squares of their circumferences. (Sup. Euc. 8. 1.) Ex. 1- If the circumference of a circle be 136 feet, what is the area ? Ans. 1472 feet. 2. What is the surface of a circular fish-pond, which is 10 rods in circumference .'' 32. If the area of a circle he given, the diameter may be found, by dividing the area by .7854, and extracting the square root of the quotient. This is reversing the rule in art. 30. Ex. 1. What is vhe diameter of a circle whose area is 380.1336 feet.^Ans.380.1336~.7854=484.AndV484=22. 2. What is the diameter of a circle whose area is 19.635.^ 33. The area of a circle, is to the area of the circum- scribed square ; as .7854 to 1 ; and to that of the inscribed square as .7854 to \. Let ABDF (Fig. 10.) be the inscribed square and LMNO the circumscribed square, of the circle ABDF. The area of the circle is equal to AD ^ X.7854. (Art. 30) Butthearea of the circumscribed square (Art. 4.) is equal to 0N^= ^D*. And the smaller square is half of the larger one. For the THE CIRCLE. ^1 laiter contains 8 equal triangles, of which the former contains only 4. Ex. VVhat is the area of a square inscribed in a circle whose area is 159.? Ans. .7854 : -^::159 : 101.22. PROBLKAI V. To find the area of a sector of a circle. 34. Multiply the radius into half the length of the arc. Or, As 360, TO THE NUMBER OF DEGREES IN THE ARC ; So IS THE AREA OF THE CIRCLE, TO THE AREA OF THE SECTOR. It is evident, that the area of the sector has the same ratio to the area of the circle, which the length of the arc has to the length of the whole circumference ; or which the number q{ degrees m the arc has to the number of degrees in the cir- cumference. Ex. 1. If the arc AOB (Fig. 9.) be 120^ and the diam- eter of the circle 226 ; what is the area of the sector AOBC? The area of the whole circle is 401 15. (Art. 30.) And 360=^ : 120°: : 401 15 : 1337 If, the area of tiio sector. 2. What is the area of a quadrant whose radius is 621 .'' 3. What is the area of a semi-circle whose diameter is 328 ^ 4. What is the area of a sector which is less than a semi- circle, if the radius be 15, and the chord of its arc 12 .? Half the chord is the sine of 23° 34'f nearly. (Art. 23.) The whole arc, tl'.en. is 47° 9'i The area of the circle is 706 86 And 360° : 47° 9'^: : 706.86 : 92.6 the area of the sector. 5. If the arc ADB (Fig. 9.) be 240 degrees, and the radiuj^ of the circle 113, what is the area of the sector ADBC ^ PROBLEM VI. To find the area of a segment of a circle, 35. Find the area of the sector which has the same arc, AND also THE AREA OF THE TRIANGLE FORMED BV THE CHORD OF THE SEGMENT AND THE RADH OF THE SECTOR. «2 MENSURATION OF Then, if the segment be less than a semi-circle, subtract the area of thf. triangle from the area of the sector. but, if the segment be greater than a semi-circle, add th«<: area of the triangle to the area of the sector. If the triangle ABC (Fig. 9.) be taken from the sector AOBC, it IS evident the difference will be the segment AOBP, less than a semi-circle. And if the same tri-dnii;le be added to the sector ADBC, the sum will be liie segment ADBP, greater than a semi-circle. The area of the triangle (Art 8.) is equal to the product of half the chord AB into CP which is the difference be- tween the radius and PO the height of the segment. Or CP is the cosine of half the arc BOA. If this cosine, and the chord of the segment are not given, they may be found from the arc and the radius. Ex. 1. If the arc AOB (Fig. 9.) be 120°, and the radius of the circle be 113 feet, what is the area of the segment AOBP ? In the right angled triangle BCP, R : BC::Sin BCO : BP=r.97.86, half the chord. (Art 28.) The cosine PC = iCO (Trig. 96, Cor.) =56 5 The area of the sector AOBC ^Art. 34.) = 1 337 J. 67 The area of the triangle ABC = BP xPC = £528.97 The area of the segment, therefore, = 7842.7 2, If the base of a segment, less than a semi-circle, be 10 feet, and the radius of the circle 12 feet, what is the area of the segment ? The arc of the segment contains 49] degrees. (Art. 28.) The area of the sector =6 1 .89 (Art. 34.) The area of the triansie =54.54 And the area of the segment = 7.35 square feet. 3. What is the area of a circular segQient, whose height is 19.2 and base 10? Ans. 947.86. THE CIRCLE. 23 4. What is the area of the segment ADBP, (Fig. 9.) if the base AB be 195.7, and the height PD 169.5 ? Ans. 32272.* 36. The area of any figure which is bounded partly by arcs of circles, and partly by right hnes, may be calculated, by rinding the areas of the segments under the arcs, and then the nrea of the rectilinear space between the chords of the arcs and the other right lines. Tims the Gothic arch ACB, (Fig. 11.) contains the two segments ACH, BCD, and the plane triangle ABC. Ex If AB (Fig. 11.) be 110, each of the lines AC and B( 00. and the iieight of each of the segments ACH, BCD 10.435 ; what is the area of the whole figure f The areas of the two sefijments are 1404 The area of the triangle ABC is 4593.4 And the whole figure is 5997.4 FROBLEM Vn. To find the area nf a circular zone. 37. From the area of the whole circle, subtract the two SEGMENTS ON THE SIDES OF THE ZONE. If from the whole circle (Fig. 12.) there be taken the two segments ABC and DFH, there will remain the zone ACDH. Or, the area of the zone may be found, by subtracting the segment ABC from the segment HBD : Or, by adding the two small segments GAH and VDC to the trapezoid ACDH. See art. 36. The latter method is rather the most expeditious in prac- tice, as the two segments at the end of the zone are equal, Ex. 1. ^Vhat is the area of the zone ACDH, (Fig. 12.) if AC is 7.7«, DH 6.93, and the diameter of the circle 8 ^ * For the method of finding the areas of segments by a tahle^ ?ee note P 24 MENSURATION OF The area ofthewln/io circle is 50.26 of the segment ABC 17.32 of the segment DFH 9.82 of the zone AC l)H 23.12 2. What is the area of a zone, one side of which is 23.25, and the other side 20.8, in a circle whose diameter is 24 ? Ans. 208. 38 If the diameter of the circle is not given, it may be found from the sides and the breadth of the zone. Let the centre of the circle be at O. (Fig. 12.) Draw ON perpend'cular to AH, NM perpendicular to LR, and HP perpenJicular to AL. Then AN=iAH, (Euc. 3. 3.) MN=i(LA4-RH) LM=iLK, (Euc. 2. 6.) PA=:LA-RH. The triangles APH and OMN are similar, because the sides of one are perpendicular to tliose of the other, each to each. Therefore PH:PA::MN:MO MO being found, we have ML-MO=OL. And the radius CO^V'OL^+CL^ (Euc. 47. 1.) Ex. If the breadth of the zone ACDH (Fig. 12.) be 6.4, $nd the sides 6.8 and 6 ; what is the radius of the circle ? PA=3.4-3=0.4. And MN = i(3.4-f o) = 3.2. Then 6.4 : 0.4::3.2 : 2=M0. And 3.2-0.2=3=OL And the radius C0='/3"^ +(3.4)3 =4.534. PROBLEM VIII. To find the area of a lhne or crescent, 39. Find the difference of the two segments which are be- tween THE ARCS OF THE CRESCENT AND ITS CHORD. If the segment ABC (Fig 14.) be taken from the seg- ment ABD; there will remain the lune or crescent ACBD. Ex. If the chord AB be 88, the height CH 20, and the height DH 40 ; what is the area of the crescent ACBD ? THE CIRCLE. 25 The area of the segment ABD is 2698 of the segment ABC 1220 of the crescent ACBD 1478 PROBLEM IX. To find the area of a ring, included between the peripheries of two concentric circles. 40. Find the difference of the areas of the two circles. Or, Muhiply the product of the sum and difference of the two diameters by .7854. The area of the ring (Fig. 13.) is evidently equal to the difference between the areas of the two circles AB and A B'. But the area of each circle is equal to the square of its diameter multiplied into .7854. (Art. o9.) And the differ- ence of these squares is equal to the product of the sum and difference of the diameters. (Alg. 235.) Therefore the area of the ring is equal to the product of the sum and differ- ence of the two diameters multiplied by .7854. Ex. 1. If AB (Fig. 13.) be 221, and A'B' 106, what is the area of the ring .** Ans. ("22^ X. 7854) -( foe 'x. 7854) = 29535. 2. If the diameters of Saturn's larger ring be 205,000 and 190.000 miles, how many square miles are there on one side of the ring ^ Ans. 395000 X 1 5000 X .7854=4,653,495,000. PROMISCUOUS EXAMPLES OF AREAS. Ex. 1. What is the expense of paving a street 20 rods long, and 2 rods wide, at 5 cents for a square foot ^ Ans. 544^ dollars. 26 MENSURATION OF 2. If an equilateral triangle contains as many square feet as there are inches in one of its sides ; what is the area of the triangle ? Let x= the number of square feet in the area. X Then 7^= the number of linear feet in one of the sides. 476 Reducing the equation, a?= -^ = 332.55 the area. 3. What is the side of a square whose area is equal to that of a circle 452 feet in diameter ? Ans. \/(452)2 X. 7854=400.574. (Art. 30 and 7.) 4. What is the diameter of a circle which is equal to a square whose side is 36 feet ? Ans. v/(36)2-i-0.7854=40.6217. (Art. 4. and 32.) 5. What is the area of a square inscribed in a circle whose diameter is 132 feet.^ Ans. 8712 square feet. (Art. 33.) 6. How much carpeting, a yard wide, will be necessary to cover the floor of a room which is a regular octagon, the sides being 8 feet each f Ans. 34| yards. 7. If the diagonal of a square be 16 feet, what is the area ? Ans. 128 feet. (Art. 14.) 8. If a carriage wheel four feet in diameter revolve 300 times, in going round a circular green ; what is the area of the green f Ans. 41541 sq. rods, or 25 acres, 3 qrs. and 34^ rods. 9. What will be the expense of papering the sides of a room, at 10 cents a square yard ; if the room be 21 feet long, 18 feet broad, and 12 feet high ; and if there be deducted 3 windows, each 5 feet by 3, two doors B feet by 41, and one fire-place 6 feet by 4i ? Ans. 8 dollars 80 cents. 10. If ^ circular pond of water 10 rods in diameter be sur- rounded by a gravelled walk 8} feet wide ; what is thci area of the walk ? Ans. 16i sq. rods. (Art. 40.) SUPERFICIES. 27 11. If CD (Fig. 17,) the base of the isosceles triangle VCD, be 60 feet, and the area 1200 feel ; and if there be cut off, by the line LG parallel to CD, the triangle VLG, whose area is 432 feet ; what are the sides of the latter triangle f Ans. 30, 30, and 36 feet. 12. What is the area of an equilateral triangle inscribed in a circle whose diameter is 52 feet ? Ans. 878.15 sq. feet. 13. If a circular piece of land is enclosed by a fence, in which 10 rails make a rod in length ; and if the field contains as many square rods, as there are rails in the fence; what is the value of the land at 120 dollars an acre ? Ans. 942 48 dollars. 14. If the area of the equilateral triangle ABD (Fig. 9.) be 219.5375 feet ; what is the area of the circle OBDA, in which the trian^jle is inscribed .'* The sides of the triangle are each 22.5167. (Art. 11.) And the area of the circle is 530.93 15. If G concentric circles are so drawn, that the space between the least or 1st, and the 2d is 21.2058, between the 2d and 3d 35. 343, between the 3d and 4th 49.4802, between the 4lh and 5th 63 6174, between the oth and 6th 77.7546; what are the several diameters, supposing the longest to be equal to 6 times the shortest.^ Ans. 3,6, 9, 12,15, and 18. SECTION III. SOLIDS BOUNDED BY PLANE SURFACES. ^, T^ ,4 PRISM is a solid bounded Art. 41. Definition L A ^^ pj^„^ ^^^^^^ ^^ ^^^^3^ ^^^ of which are parallel, similar, and equal; and,the others are parallelograms. II. The parallel planes are sometimes called the bases or ends ; and the other figures, the sides of the prism. The lat- ter taken together constitute the lateral surface. III. A prism is right or oblique^ according as the sides are perdendicular or oblique to the bases. IV. The height of a prism is the perpendicular distance between the planes of the bases. In a right prism, there- fore, the height is equal to the length of one of the sides. V. A Farallelopiptd is a prism whose bases are parallelo- grams, VI. A Cube is a solid bounded by six equal squares. Jt is a ric^ht prism whose sides and bases are all equal. VII. A Pyramid is a solid bounded by a plane figure call- ed the base, and several triangular planes, proceeding from the sides of the base, and all terminating in a single point. These triangles taken together constitute the lateral surface. VIII. A pyramid is regular, if its base is a regular poly- gon. and if a line from the centre of the base to the vertex of the pyramid is perpendicular to the base. This line is called the axis of the pyramid. IX. The height of a pyramid is the perpendicular distance from the summit to the plane of the base. In a regular pyra- mid, it is the length of the axis. X. The slont-height of a regular pyramid, is the distance from the summit to the middle of one of the sides of the base. XL A frustum or trunk of a pyramid is a portion of the solid next the base, cut off by a plane parallel to the base. The height of the frustum is the perpendicular distance of tbe two parallel planes. The slant-height of a frustum of a MENSURATION OF SOLIDS. 29 regular pyramid, is the distance from the middle of one of the sides of the base, to the middle of the corresponding side in the plane above. It is a line passing on the surface of the frustum, through the middle of one of its sides. XII. A Wedge is a solid of five sides, viz. a rectangular base, two rhomboidal sides meeting in an edge, and two tri- angular ends ; as ABHG. (Fig 20.) The base is ABCD, the sides ate ABHG and DCHG, meeting in the edge GH, and the ends are BCH and ADG. The height of the wedge is a perpendicular drawn from any point in the edge, to the plane of the base, as GP. XIII. A Prismoid is a solid whose ends or bases are par- allel, but not similar, and whose sides are quadrilateral. It differs from a prism or a frustum of a pyramid, in having its ends dissimilar. It is a rectaiigular prismoid, when its ends are right parallelograms. XIV. A linear side or edge of a solid is the line of inter- section of two of the planes which form the surface. 42. The common measuring vnit of solids is a cuhe^ whose sides are squares of the same name. The sides of a cubic inch are square inches ; of a cubic foot, square feet, &c. Finding the capacity, solidity * or solid contents of a body, is finding the number of cubic measures, of some given denom- ination contained in the body. In solid measure. 1728 cubic inches =1 cubic foot, 27 cubic feel =1 cubic yard, 4492| cubic feet =1 cubic rod, 32768000 cubic rods =1 cubic mile, 282 cubic inches =1 ale gallon, 231 cubic inches =1 wine gallon, 21i0.42 cubic inches =1 bushel, 1 cubic foot of pure water weighs 1000 Avoirdupois ounces, or 62^ pounds. * See note E. 3Q MENSURATION OF PROBLEM I. To find the solidity of a prism. 43. Multiply the area of the base by the height. This is a general rule, applicable to parallelepipeds whether right or nblique, cubes, triangular prisms, (fee. As surfaces are measured, by comparing them with a right paraVelogram (Art. 3.) ; so solids are measured, by compar- ing them with a r'l^hi par allelopiped. If ABCD (Fig. 1.) be the base of a right parallelopiped, as a stick of timber standing erect, it is evident that the number of cubic feet contained in one foot of the height, is eq-.ial to the number of square feet in the area of the base. And if the solid be of any other height, instead of one foot, the contents must have the same ratio. For parallelopipeds of the same base are to each other as their heights. (Sup. Euc. 9. 3.) The solidity of a right parallelopiped, there- fore, is equal to the product of its length, breadth^ and thick- ness. See Alg. 523. And an oblique parallelopiped being equal to a right one of ihe same base and altitude, (Sup. Euc. 7, 3.) is equal to the area of the base multiplied into the perpendicular height. This is true also of prisms, whatever be the form of their bases. (Sup. Euc. 2. Cor. to 8. 3.) 44. As the sides of a cube are all equal, the solidity is found by cubing one of its edges. On the other hand, if the solid contents be given, the length of the edges may be found, by extracting the cube root, 45. When solid measure is cast by Duodecimals, it is to be observed that inches are not primes of feet, but thirds. If the unit is a cubic foot, a solid which is an inch thick and a foot square is a prime ; a parallelopiped a foot long, an inch broad, and an inch thick is a second, or the twelfth part of a prime ; and a cubic inch is a third, or a twelfth part of a second. A linear inch is ^3 of a foot, a square inch yi^ of a foot, and a cubic inch jj\^ of a foot, Ex. 1. What are the solid contents of a stick of timber which is 31 feet long, 1 foot 3 inches broad, and 9 inches thick ? Ans. 29 feet 9", or 29 feet 108 inches. SOLIDS. 31 2. What is the solidity of a wall which is 22 feet long, 12 feet high, and 2 feet 6 inches thick f Ans. 660 cubic feet, 3. What is the capacity of a cubical vessel which is 2 feet 3 inches deep f Ans. llF. 4' 8" 3'", or 11 feet 675 inches. 4. If the base of a prism be 108 square inches, and the height 36 feet, what are the solid contents ? Ans. 27 cubic feet. 5. If the height of a square prism be 2\ feet, and each side of the base 10^ feet, what is the solidity f The area of the base =10i X lO^X 106| sq. feet. And the solid contents =106| X2i = 240i cubic feet. 6. If the height of a prism be 23 feet, and its base a regu- lar pentagon, whose perimeter is 18 feet, what is the soliiity ? Ans. 512.84 cubic feet. 46. The number o( gallons or bushels which a vessel will contain may be found, by calculating the capacity in inches^ and then dividing by the number of inches in 1 gallon or bushel. The weight of water in a vessel of given ditnensions is easily calculated ; as it is found by experiment, that a cubic foot of pure water weighs 1000 ounces Avoirdupois. For the weight in ounces, then, multiply the cubic feet by 1000 5 or for the weight in pounds, multiply by 62i. Ex. 1. How many ale gallons are there in a cistern which is 1 1 feet 9 inches deep, and whose base is 4 feet 2 inches square/ The cistern contains 352500 cubic inches ; And 352500—282=1259. 2. How many wine gallons will fill a ditch 3 feet 11 inches wide, 3 feet deep, and 462 feet long ? Ans. 4060S. 3. What weight of water can be put into a cubical vessel 4 feet deep ? Ans. 4000 Ib^. a^ MENSURATION OF PROBLEM II. To^ndMc LATERAL SURFACE of a RIGHT PRISM. 47. Multiply the length into the perimeter of the base. Each of the sides of the prism is a right parallelogram, whose area is the product of its length and breadth. But the breadth is one side of the base ; and therefore, the sum of the breadths is equal to the perimeter of the base, Ex. 1. If the base of a right prism be a regular hexagon whose sides are each 2 feet 3 inches, and if the height be 16 feet, what is the lateral surface ^ Ans. 216 square feet. If the areas of the two ends be added to the lateral sur- face, the sum will be the whole surface of the prism. And the superficies of any solid bounded by planes, is evidently equal to the areas of all its sides. Ex. 2. If the base of a prism be an equilateral triangle whose perimeter is 6 feet, and if the height be 17 feet, what ^s the surface .^ The area of the triangle is 1.732. (Art. 11.) And the whole surface is 105.464. PROBLEM III. To find the solidity of a pyramid. 48. Multiply the area of the base into ^ of the height. The solidity of a prism is equal to the product of the area of the base into the height. (Art. 43.) And a pyramid is ^ of a prism of the same base and altitude. (Sup. Euc. 15. 3. Cor. 1.) Therefore the solidity of a pyramid whether right or oblique, is equal to the product of the base into \ of the perpendicular height. Ex. 1. What is the solidity of a triangular pyramid, whose height is 60, and each side of whose base is 4 .'* The area of the base is 6.928 And the solidity is 138.56. 2. Let ABC (Fig. 16.) be one side of an oblique pyramid whose base is 6 feet square; letBC be 20 feet, and make an angle of 70 degrees with the plane of the base ; and let CP be perpendicular to this plane. What is the solidity of the pyramid .'* SOLIDS. 33 la the right angled triangle BCP, {Trig. 134.) R : BC.lSinB : PC = 18.79. And the sohdity of the pyramid is 225.48 feet. 3. What is the solidity of a pyramid whose perpendicular height is 72, and the sides of whose base are 67, 54, and 40 f Ans. 25920. PROBLEM IV. To find the lateral surface of a regular pyramid. 49. Multiply half the slant-height into the perimeter of THE BASE. Let the triangle ABC (Fig. 18.) be one of the sides of a regular pyramid. As the sides AC and BC are equal, the angles A and B are equal. Therefore a line drawn from the vertex C to the middle of AB \s perpendicular to AB. The area of the triangle is equal to the product of half this per- pendicular into AB. (Art. 8.) The perimeter of the base is the sum of its sides, each of which is equal to AB. And the areas of all the equal triangles which constitute the lateral surface of the pyramid, are together equal to the product of the perimeter into half the slant-height CP. The slant-height is the hypothenuse of a right angled tri- angle, whose legs are the axis of the pyramid, and the dis- tance from the centre of the base to the middle of one of the sides. See Def. 10. Ex. 1. What is the lateral surface of a regular hexagonal pyramid, whose axis is 20 feet, and the sides of whose base are each 8 feet f The square of the distance from the centre of the base to one of the sides (Art. 16.) =48. The slant-height (Euc. 47. 1.) =\/4S-f- 20^ =21.16. And the lateral surface =21.16 X 4x6 = 507.84 sq. {qqU 2. What is the whole surface of a regular triangular pyra- mid whose axis is 8, and the sides of whose base are each 20.78 .? The lateral surface is 312 The area of the base is 187 And the whole surface!^' 499 M MENSURATION OF 3. What is the lateral surface of a regular pyramid whose axis is 12 feet, and whose base is 18 feet square ? Ans. 540 square feet. The lateral surface of an oblique pyramid may be found, by taking the sum of the areas of the unequal triangles which form its sides.^ PROBLEM V. To find the solidity of a frustum of a pyramid, 50. Add together the areas of the two ends, and the square root of the product of these areas ; and multiply tfi» sum by a of the perpendicular height of the soltd. Let CDGL (Fig. 17.) be a vertical section, through the middle of a frustum of a right pyramid CDV whose base is a square, LetCD=nent or zone of a sphere is equal to the pro- duct of the height of the segment or zone into the circum- ference of the sphere. The surface of the zone produced by the revolution of the arc AB about ON, is equal ta ON xc/rc CP. And the surface of the segment produced by the rerolution of BD about DN is equal to DN XaVc CP, Cor. 2. The surface of a sphere is equal to four times the area of a circle of the same diameter; and therefore, the convex surface of a hemisphere is equal to twice the area of its base. For the area of a circle is equal to the product of half the diameter into half the circumference ; (Art. 30.) that is, to J the product of the diameter and circumference. Cor. 3. The surface of a sphere, or the convex surface of any spherical segment or zone, is equal to that of the cir- cumscribing cylinder. A hemisphere described by the rev- olution of the arc DBP, is circumscribed by a cylinder pro- duced by the revolution of the parallelogram D^CP, The convex surface of the cylinder is equaf to its height multi- plied by its circu.jiference. (Art. 62.) And this is also, the surface of the hemisphere. THE SPHERE. 51 So the surface produced by the revolution of AB is equal to that produced by the revolu'ion of ab. And the surface produced by BD is equal to that produced by hd, Ex. 1. Considering the earth as a sphere 7930 miles in diameter, how manv square miles are there on its surface ? Ans. 197,558,500. 2. If the circumference of the sun be 2.800,000 miles, what is his surface i* Ans. 2,495,547,600,000 sq. miles. 3. How many square feet of lead will it require, to cover a hemispherical dome whose base is 13 feet across ^ Ans. 2651. PROBLEM VIII. To find the solidity of a sphsrb. 70. I4 Multiply the cube op the diameter by .O^ot>« Or, 2. Multiply the squ.ire of the diameter by J- of the ctR- CCMFERENCE. Of, 3. Multiply the surface by | of the diameter. 1. A sphere is two thirds of its circumscribing cylinder. (Sup. Euc. 21. 3.) The height and diameter of the cylinder are each equal to the diameter of the sphere. The solidity of the cylinder is equal to its height multiplied into the area of its base, (Art. 64.) that is putting D for the diameter, DXD2X.7854 or D3X.7854. And the solidity of the sphere, being | of this, is D3X.5236. 2. The base of the circumscribing cylinder is equal to hal the circumference multiplied into half the diameter, (Art. 30.) that is, if C be put for the circumference, iC xD, and the solidity is iC xD^. Therefore the solidity of the sphere is fof iCxD2=D2xia •b^ ftlENSURATION OF 3. In the last expression, which is the same as C X D X ^Vr we may substitute S, the surface, for C xD. (Art. 69.) We then have the solidity of the sphere equal to SxiD. Or, the sphere may be supposed to be filled with small pyramids, standing on the surface of the sphere, and having their common vertex in the centre. The number of these may be such, that the difference between their sum and the sphere shall be less than any given quantity. The sotidity of each pyramid is equal to the product of its base into ^ of its height. (Art. 48.) The solidity of the whale, therefore, h equal to the product of the surface of the sphere into ^ of its radius, or i of its diameter. 71. The numbers 3.14159, .7854, .5236, should be made perfectly familiar. The first expresses the ratio of the ctrcvm/ercnce.of a circle to thcf/iame^er; (Art. 23.) the sec- ond, the ratio of the area of a circle to the square of the diam- eter (Art. 30.) ; and the third, the ratio of the solidity of a sphere to the cube of the diameter. The second is | of the first, and the third is | of the first. As these numbers are frequently occurring in mathematical investigations, it is common to represent the first of them by the Greek letter 11. According to this notation, *=3.14159, i*=.7854. i*=.5236. If D=the diameter, and R=the radius of any circle or sphere; Then D=2R D2=4R« D='=8R3. And-rrD) . i l-ffDa >= the area of I irD') , Or2^R^ =thepen;,A. J^j^, ^ ^^^ ^^^^^ orl.R3l =^»^^ solidity of the sphere, Ex. 1- What is the solidity of the earth, if it be a sphere 7930 miles in diameter ? Ans. 261,107,000,000 cubic miles. 2. How many wine gallons will fill a hollow sphere 4 feet in diameter ? Ans. The capacity is 33.5104 feet=250| gallons. 3. If the diameter of the moon be 2180 miles, what is its 4iOlidity ? Ans. 5,424,600,000 miles* THE SPHERE. 53 72. If the solidity of a sphere he given, the diameter may be found b}' reversing the first rule in the preceding article; that is, dividing by .5236 and extracting the cube root of the quotient. Ex 1. Wh^t is the diameter of a sphere whose solidity- is 65.45 cubic feet? Ans. 5 feet. 2. What must be the diameter of a globe to contain 16755 pounds of water? Ans. S feet. PROBLEM IX. To find the convex surface q/* a segment or zone of a ^ sphere* 73. Multiply the height of the segment or zonk into the cir- cumference OF THE sphere. For the demonstration of this rule, see art. 69. Ex. 1. If the earth be considered a perfect sphere 7930 miles in diameter, and if the polar circle be 23*' 28' from the pole, how many square miles are there in one of the frigid zones ? If PQOE (Fig. 15.) be a meridian on the earth, ADB one of the polar circles, and P the pole; then the frigid zone is a spherical segment described by the revolution of the arc APB about PD. The angle ACD subtended by the arc AP is 23° 28'. And in the right angled triangle ACD, R : AC: : Cos ACD : CD=3637. Then CP~CD=3965-3637=328=PD the height of the segment. And 328X7930X3.14159 = 8171400 the surface. 2. If the diameter of the earth be 7930 miles, what is the surface of the torrid zone, extending 23° 28' on each side of the equator ? If EQ (Fig. 15.) be the equator, and GH one of the tro- pics, then th*' angle ECG is 23° 28'. And in the right angled triangle GCM,- 54 MENSURATION OF R : CG: :Sin ECG : GM=CN = 1678.9 the height of half the zone. The surface of the whole zone is 78669700. 3. What is the surface of each of the temperate zones ? The height DN=CP-CN-PD=^0o8.1 And the surface of the zone is 51273000. The surface of the two temperate zones is 102,540,000 of the two frigid zones 16,342,000 of the torrid zone 78,669,700 of the whole globe 1 97, 558,500 PROBLEM X. To find the solidity of a spherical sector, 74. Multiply the spherical surface by ^ of the radius of THE sphere. The spherical sector, (Fig. 24.) produced by the revolu- tion of ACBD about CD, may be supposed to be filled with small pyramids^ standing on the spherical surface ADB, and terminating in the point C Their number may be so great, that the heightofeach shall differ less than by any given length from the radius CD, and the sum of their bases shall differ less than by any given quantity from the surface ABD. The tsolidity of each is equal to the product of its base into i of the radius CD. (Art. 48.) Therefore, the solidity of all of them, that is, of the sector ADBC, is equal to the product of the spherical surface into j of the radius. Ex. Supposing the earth to be a sphere 7930 miles in di- ameter, and the polar circle ADB (Fig. 15.) to be 23° 28' from the pole ; what is the solidity of the spherical sector ACBP? Ans. 10,799,867,000 miles THE SPHERE. 55 PROBLEM XI, To find the solidity of a spherical segment. 75. Multiply half the height of the segment into the area OF the base, and the cube of the height into .5236 ; and add the two products. As the circular sector AOBC (Fig. 9.) consists of two parts, the segment AOBP and the triangle ABC; (Art. 35.) so the spherical sector produced by the revolution of AOC about OC consists of two parts, the segment produced by the revolution of AOP, and the cone produced by the revolution of ACP. If then the cone be subtracted from the sector, the remainder will be the segment. Let CO=R the radius of the sphere, PB = r the radius of the base of the segment, PO=A the height of the segment, Then PC=R — A the axis of the cone. The sector=2'rrR xhx^R (Arts. 71, 73, 74.) = f -^r/tRs The cone='«'r3 xi(R-A) (Arts. 71,66.) = icrr5^R~^<;r/i;.2. Subtracting the one from the other, The segment =t'' Xi(A-R)=i^Ar« -i«-r«R Adding them together, we have as before, The segment= |^AR3 - J^r^ R-f \^hr^ , Cor. The solidity of a spherical segment is equal to half a cylinder of the same base and height -f-a sphere whose di- ameter is the height of the segment. For a cyHnder is equal to its height multiplied into the area of its base ; and a sphere is equal to the cube of its diameter multiplied by .6236. Thus ifOy (Fig. 15.) be half Ox, the spherical segment produced by the revolution of Oxt is equal to the cylinder produced by tvyx -f the sphere produced by Oyxz; sup- posing each to revolve on the line Ox. Ex. 1 . If the height of a spherical segment be 8 feet, and the diameter of its base 25 h^i ; what is the solidity ? Ans. 25'x.7854X4 + 8='X.5236=225l.58feet. 2. If the earth be a sphere 7930 miles in diameter, and the polar circle 23° 28' from the pole, what is the solidity of one of the frigid zones ? Ans. I,303,000,.0p0 miles. THE SPHERE. 57 PROBLEM XII. To find the solidity of a spherical zose or frusiunu 76. From the solidity of the whole sphere, subtract the two segments on the sides of the zone. Or, Jdd together the squares of the radii of the two ends^ and 1 the square of their distance ; and multiply the sum by three, times this distance^ and the product by .5236. If from the whole sphere, (Fig. 15.) there be taken the two segments ABP and GHO, there will remain the zone or frustum ABGH. Or, the zone ABGH is equal to the difference between the segments GHP and ABP. LetNP=H> ,. , . , . , r)p__ 7 > the heights ot the two segments. AQ~ > the radii of their bases. DN=c?=H — A the distcktice of the two bases, or the height of the zone. Then the larger segment=iTHR2-|-|'?rH=' \ , k^ -. \ And the smaller segment=^'TAr2 +iBD'. (Euc. 20,1.) Therefore, AD>BD'; BD>AD', (Euc. 47, 1.) and lBD>iAD'. But iBD, or BH, is the height of the isosceles triangle; (Art. 1.) and |AD' or AH', the height of the scalene triangle ; and the areas of two triangles which have the same base are as their heights. (Art. 8.) There- fore the area of ABC is greater than that of ABC. Among all triangles, then, of a given perimeter, and upon a given base, the isosceles triangle is a maximum. Cor. The isosceles triangle has a less perimeter than any scalene triangle of the same base and area. The triangle ABC' being less than ABC, it is evident the perimeter of the former must be enlarged, to make its area equal to the area of the latter. PROPOSITION II. 80. A triangle in which two given sides make a right ANGLE, has a greater area than any triangle in which the same sides make an oblique angle. If BC, BC, and BC" (Fig. 27.) be equal, and if BC be perpendicular to AB ; then the right angled triangle ABC, ISOPERIMETRY. 63 has a greater area than the acute angled triangle ABC', or the oblique angled triangle ABC". Let PC and PC" be perpendicular to AP. Then, as the three triangles have the sam^ base AB, their areas are as their heights ; that is, as the perpendiculars BC, PC', and PC". But BC is equal to BC, and therefore greater than P'C (Euc. 47, 1.) BC is also equal to BC", and therefore greater than PC". PROPOSITION III. 81. If all the sides except one of a polygon be given^ the area will be the greatest^ when the given sides are so disposed^ that the figure may be inscribed in a semicircle, of which the undetfrmined side is the diameter. If the sides AB, BC, CD, DE, (Fig. 28.) be given, and if their position be such that the area, included between these and another side whose length is not determined, is a maxi- mum ; the figure maj be inscribed in a semicircle, of which the undetermined side AE is the diameter. Draw the lines AD, AC, EB, EC. Py varying the angle at D, the triangle ADE may be enlarged or diminished, without affecting the area of the other parts of the figure. The whole area, therefore, cannot be a maximum^ unless this triangle be a maximum, while the sides AD and ED are given. But if the triangle ADE be a jnaximumy under these conditions, the angle ADE is a right single ; (Art. 80.) and therefore the point D is in the circumference of a circle, of which AE is the diameter. (Euc. 31, 3.) In the same man- ner it may be proved, that the angles ACE and ABE arc right angles, and therefore that the points C and B are in the cir- cumference of the same circle. The term polygon is used in this section to include iriaii- glesj and four-sided figures, as well as other right-lined figures. 82. The area of a polygon, inscribed in a semi-circle, in the manner stated above, will not be altered by varying the order of the given sides. The sides AB, BC, CD, DE, (Fig. 28.) are the chords of so many arcs. The sum of these arcs, in whatever order they are arranged, will evidently be equal to the semicir- ciimference. And the segments between the given sides and 64 MENSURATION. the arcs will be the same, in whatever part of the circle they are situated. But the area of the polygon is equal to the area of the semicircle, diminished by the sum of these seg- ments. 83. If a polygon, of which all the sides except one are given, be inscribed in a semicircle whose diameter is the un- determined side ; a polygon having the same given sides, cannot be inscribed in any other semicircle which is either greater or less than this, and whose diameter is the undeter- mined side. The given sides AB, BC, CD, DE, (Fig. 28.) are the chords of arcs whose sum is 180 degrees. But in a larger circle, each would be the chord of a less number of degrees, and therefore the sum of the arcs would be less than 180^ : and in a smaller circle, each would be the chord of a greater number of degrees, and the sum of the arcs would be greater than J80^ PROPOSITION IV. 84. A polygon inscriijed in a circle has a greater area, than any polygon of equal perimeter, and the same number oj" sides, which cannot be inscribed in a circle. If in the circle ACHF, (Fig. 30.) there be inscribed a po- lygon ABCDEFG ; and if another polygon abcdefg (Fig. 31.) be formed of sides which are the same in number and length, but which are so disposed, that the figure cannot be inscribed in a circle ; the area of the former polygon is greater than that of the latter. Draw the diameter AH, and the chords DH and EH. Upon de make the triangle deh equal and similar to DEH, and join ah. The line ah divides the figure abcdhefg into two parts, of which one at least cannot, by supposition, be inscri- bed in a semicircle of which the diameter is AH, nor in any other semicircle of which the diameter is the undetermined side. (Art. 83.) It is therefore less than the corresponding part of the figure ABCDHEFG. (Art. 81.) And the other part of abcdhefg is not greater than the corresponding part of ABCDHEFG. Therefore the whole figure ABCDHEFG is greater than the whole figure abcdhefg. If from these there be taken the equal triangles DEH and deh, there will remain the polygon ABCDEFG greater than the polygon abcdefg. ISOPERIMETRY. So 85. A polygon of which all the sides are given in number and length, can not be inscribed in circles of different diam- eters. (Art. 83,) And the area of the polygon will not be altered, by changing ti^e order of the sides. (Art. 82.) PROPOSITION V. 86. When a polygon has a greater area than any other, of the same number of sides, and of equal perimeter^ the sides are EQUAL. The polygon ABCDF (Fig. 29.) cannot be a maximum^ among all polygons of the same number of sides, and of equal perimeters unless it be equilateral. For if any two of the sides, as CD and FD, are unequal, let CH and FH be equal, and their sum the same as the sum of CD and FD. The isosceles triangle CHF is greater than the scalene trian- gle CDF (Art. 79.) ; and therefore the polygon ABCHF is greater than the polygon ABCDF ; so that the latter is not a maximum* PROPOSITION VI. 07. Jl REGULAR POLYGON has fl greater area than any other polygon of equal perimeter, and of the same number of sides. For, by the preceding article, the polygon which is a max- imum among others of equal perimeters, and the same num- ber of sides is equilateral, and by art. 84, it may be inscribed in a circle. But if a polygon inscribed in a circle is equilat- eral, as ABDFGH (Fig. 7.) it is also equiangular. For the sides of the polygon are the bases of so many isosceles trian- gles, whose common vertex is the centre C. The angles at these bases are all equal ; and two of them, as AHC and GHC, are equal to AHG one of the angles of the polygon. The polygon, then, being equiangular, as well as equilateral, is 2i regular polygon. (Art. 1. Def. 2.) Thus an equilateral triangle has a greater area, than any other triangle of equal perimeter. And a square has a greater area, than any other four-sided figure of equal perimeter. 10 ee MENSURATION. Cor. A regular polygon has a less perimeter than any other polygon of equal area, and the same number of sides. For if, with a given perimeter, the regular polygon is greater than one which is not regular; it is evident the perime- ter of the former must be diminished, to make its area equal to that of the latter. PROPOSITION VII. 88. If a polygon be described about a. circle, the areai of the two figures are as their perimeters* Let ST (Fig. 32.) be one of the sides of a polygon, either regular or not, which is described about the circle LNR. Join OS and OT, and to the point of contact M draw the radius OM, which will be perpendicular to ST. (Euc. 18, 3.) The triangle OST is equal to half the base ST raultiphed into the radius OM. (Art. 8.) And if lines be drawn, in the same manner, from the centre of the circle, to the extremi- tfes of the several sides of the circumscribed polygon, each of the triangles thus formed will be equal to half its base mul- tiplied into the radius of the circle. Therefore the area of the whole polygon is equal to half its perimeter multiplied into the radius : and the area of the circle is equal to half its circumference multiplied into the radius. (Art. 30.) So that the two areas are to each other as their perimeters. Cor. I. If different polygons are described about the same circle, their areas are to each other as their perimeters. For the area of each is equal to half its perimeter, multiplied into the radius of the inscribed circle. Cor. 2. The tangent of an arc is always greater than the arc itself. The triangle OMT (Fig. 32.) is to OMN, as MT to MN. But OMT is greater than OMN, because the for- mer includes the latter* Therefore the tangent MT is greater than the arc MN. ISOPERIMETRY. ^7 PROFOSITION VIII, 89. Jl CIRCLE has a greater area than any polygon of equal perimeter. If a circle and a regular polygon have the same centre, and equal perimeters; each of the sides of the polygon must fall partly within the circle. For the area of a circumscribing polygon is greater than the area of the circle, as the one in- cludes the other: and therefore, by the preceding article, the perimeter of the former is greater than that of the latter. Let AD then (Fig. 32.) be one si< (rf:5+ 3X7+^+ fxrr +^^-) C6 = the base or double ordinate, Fn which To find the solidity of a circular spindle, produced by the revo- lution of a circular segment about its base or chord as an axis. 115. From J of the cube of half the axis, subtract the product of the central distance into half the revolving cir- cular segment, and multiply the remainder bv four times ^.14159. If a=the area of the revolving circular segment, /=:half the length or axis of the spindle, e=the distance of this axis from the centre of the circle to which the revolving segment belongs ; The solidity=(|/3-iac)X4X3.14159. Ex. Let a circular spindle be produced by the revolution of the segment ABO (Fig. 9.) about AB. If the axis AB be 140, and OP half the middle diameter of the spindle be 38.4 ; what is the solidity ? The area of the revolving segment is 3791 The central distance PC 44.6 The solidity of the spindle 374402 problem XVI. To find the solidity of the middle frustum of a circular spindlt. 1 16. From the square of half the axis of the whole spindle, subtract ^ of the square of half the length of the frustum ; multiply the remainder by this half length ; from the product subtract the product of the revolving area into the central distance; and multiply the remainder by twice 3.14169. If L=half the length or axis of the whole spindle, /=half the length of the middle frustum, c=the distance of the axis from the centre of the circle, rt=the area of the figure which, by revolving, pro duces the frustum ; The solidity =(l7^rp7x/--ac)X2X 3.1 41 59. APPENDIX. 79 Ex. If the diameter of each end of a frustum of a circular spindle be 21.6, the middle diameter 60, and the length 70; what is the solidity i* The length of the whole spindle is 79.75 The central distance 11.5 The revolving area 1703.8 The solidity 136751.5 PROBLEM XVII. To find the solidity of a parabolic spindle, produced by the revo- lution of a parabola about a double ordinate or base. 117. Multiply the square of the middle diameter by f*^ of the axis, and the product by .7854. Ex. If the axis of a parabolic spindle be 30, and the mid- dle diameter 17, what is the solidity ? Ans. 3631.7. PROBLEM XVIII. To find the solidity of the middle frustum of a parabolic spindle. 118. Add together the square of the end diameter, and twice the square of the middle diameter, from the sum sub- tract I of the square of the difference of the diameters; and multiply the remainder by | of the length, and the product by .7854. If D and d = the two diameters, and / = the length; The solidity =(2D2+rf»-|(D~rf)2)Xi/X.7854. Ex. If the end diameters of a frustum of a parabolic spin- dle be each ]2inches, the middle diameter 16, and the length 30 ; what is the solidity ? Ans. 5102 inches. APPENDIX.— PART 11. GAUGING OF CASKS. At 11Q ^ AUGING is a practical art, whicli does not * ^^ admit of being treated in a very scientific manner. Casks are not commonly constructed in exact con- formity with any regular mathematical figure. By most wri- ters on the subject, however, they are considered as nearly coinciding with one of the following forms ; ' i The middle frustum < ^ u r ' • ji 2.5 ^ of a parabolic spmdie. 3. ) rp 1 r . ( of a paraboloid, - > Iwo equal irustums < c ^ ' 4. ) ^ ( ot a cone. The second of these varieties agrees more nearly, than any of the others, with the forms of casks, as they are commonly made. The first is too much curved, the third too little, and the fourth not at all, from the head to the bung. 120. Rules have already been given, for finding the capa- city of each of the four varieties of casks. (Arts. 68, 110, 112, 118.) As the dimensions are taken in Inches, these rules will give the contents in cubic inches. To abridge the computation, and adapt it to the particular measures used in gauging, the factor .7854 is divided by 282 or 231 ; and the quotient is used instead of .7854, for finding the capacity in ale gallons or wine gallons. .7854 Now-2g2-'=.002785,or .0028 nearly; . ,.■7854 And ^^=.0034. If then .0028 and ,0034 be substituted for .7854, in the rules referred to above ; the contents of the cask will be giv- en in ale gallons and wine gallons. These numbers are to each other nearly as 9 to 11. GAUGING. gi PROBLEM I. To calculate iht mmtcnis of a cask^ in the form of the middle frus- tum of a SPHEROID. 121. Add together the square of the head diameter, and twice the square of the bung diameter ; multiply the sum by I of the length, and the product by .0028 for ale gallons, or by 0034 for wine gallons. If D and ^^^ that the ratio of the pe- rimeter of a circujnscribf'd pol'.gon of 192 sides, to the diam- ter, is less than 3^^ : 1, that is, than 22 : 7. JShtins gave the ratio of 355 : 115, which is more accurate than any other expressed in small numbers. This was con- firn)ed by Vieta, who by inscribed and circumscribed poly- gons of 393il6 sides, carried the approximation to ten places of figures, viz. 3.141592653. T^an Ceulen of Leyden afterwards extended it, by the la- borious process of repeated bisections of an arc, to 36 places. This calculation was deemed of so much consequence at the time, that the numbers are said to have been put upon his lomb. But since the invention of ^Mjzon5, methods much more expeditious have been devised, for approximating to the re- quired ratio. These principally consist in finding the sum of a series, in which the length of an arc is expressed in terms of its tangent. If ^=the tangent of an arc, the radius being 1, t^ t' r t^ The arc =^--o--|-y — -=--{- -Q-- &c. See Fluxions. MENSURATION. §7 This series is in itself very simple. Nothing more is ne- cessary to make it answer the purpose in practice, than that the arc be smallj so as to render the series sufficiently coii- veraing, and that the tangent be expressed in such siniple numbers, as can easily be raised in the several powers. The given series will be expressed in the most simple numbers, when the arc is 45°, whose tangent is equal to radius, [f the radius be 1, The arcof 45° = l-i-fi-4-hi-&c. And this multi- plied by 8 gives the length of the whole circumference. But a series in which the tangent is smaller, though it be less simple than this, is to be preferred, for the rapidity with which it converges. As the tangent of 30'' = v/i, if the ra- dius be ], The arc of 30^=^^iX{l-^^+T^-~+^-&c. And this multiplied into 12 will give the whole circumference. This was the series used by Dr. Halley. By this also, Mr Ahrnham Sharp of Yorkshire computed the circumfer- ence to 72 places of figurt-s, Mr. John jMachin, Professor of Astronomy in Gresham college, to lOO places, and M, De Lagny to 128 place?. Several expedients have been devised, by Machin, Euler, Dr. Hutton, and others, to reduce the la- bour of summing the terms of the series. See Euler's Anal- ysis of Infinites, Hutton's Mensuration, Appendix to Maseres on the Negative Sign, and Lond. Phil. Trans, for 1776. For a demonstration that the diameter and the circumference of a circle are incommensurable, see Legendre's Georaety, Note. IV. The circumference of a circle whose diameter is 1, is 3. U 15926535, 8979323846, 2643383279, 5028841971, 6939937510, 5820974944, 5923078164, 0628620899, 8628034825, 3421170679, 8214808661, 3272306647, 09384i6 -f or 7 - . 8S NOTES. Note B. p. 17. The following multipliers may frequently be useful ; l=the side of an equal square. the side of an ins'bed sq're- the side of an inscribed [equilateral triangle. The diam'rof a circle i X.8862= < X.707 =1 { X.866 =1 C X. 2821=1 f.^ X. 2251 = 1 ( X .2756=1 X.2821=the side of an equal square. The circumf. <{ X.2251 = thfi side of an inscribed square. ithe side of an ins'bed eq'lat. trian. The side of a sq.-< X l.l28=the diam. of an equ'l circle. X3.545=the circ. of an equal circle. X 1.414=the dia. of the circumsc. circle. X 4.443= the cir. 6f the circumsc. circle. Note C. p. 19. The following approximating rules may be used for finding the arc of i circle. 1. The arc of a circle is nearly equal to | of the differ- encp between the chord of the whole arc, and 8 times the chord of half the arc. 2. If h^ihe height o( Sin arc, and c?= the diameter of the circle ; The ^rc=2d\/,jj^ Or, h 3h^ S5h^ 3. The zrc^2^/ dh X (1 -j-^^g^-f 2X5^^+ 2.4]b.7^^^-)^'' / 5h 4. The arc=(5d| \/ 5^33^ +'*v^^^) very nearly. 5, If «=the sine of an arc, aud r=the radius of the circle; Thearc=5X(l-h 2:3;:7+ 5 2.4r^ +7.2.4.6r"« ^''• See Button's Mensuration. NOTES. 89 Note D. p. 23. To expedite the calculation of the areas of circular seg-? ments, a table is provided, which contains the areas of seg- ments in a circle whose diameter is one. See the table at the end of the book, in which the diameter is supposed to be di- vided into 1000 equal parts. By this may be found the areas of segments of other circles. For the heights of similar segments of different circles are as the diameters. If then the height of any given segment be divided by the diameter of the circle, the quotient will be the height of a similar seg- ment in a circle whose diameter is 1. The area of the lat- ter is found in the table ; and from the properties of similar figures, the two segments are to each other, as the squares of the diameters of the circles. We have then the following rule : To find the area of a circular segment hy the table. Divide the height of the segment hy the diameter of the circle ; look for the quotient in the column of heights in the toAle ; takeout the corresponding number in the column of areas; and multiply it by the square of the diameter. It is to be observed, that the figures in each of the columns in the table are decimals. If accuracy is required, and the quotient of the height di- vided by the diameter, is between two numbers in the column of heights; allowance may be made for a proportional part of the difference of the corresponding numbers in the column ef areas; in the same manner, as in taking out logarithms. Segments greater than a semicircle are not contained in the table. If the area of such a segment is required, as ABD (Fig. 9.), find the area of the segment ABO, and subtract his from the area of the whole circle. Divide the height of the given segment by the diameter, subtract the quotient from 1, find the remainder in the column of heights, subtract the corresponding area from .7854 and multiply this remainder by the square of the diameter. 13 90 MENSURATION. Ex. I. What is the area of a segment whose height is 16, the diameter of the circle being 48 ? Ans. 526. 2. What is the area of a segment whose height is 32, the diameter being 48? Ans. 1281.55. The following rules may also be used for a circular seg- ment. 1. To the chord of the whole arc, add | of the chord of half the arc, and multiply the sum by | of the height. If C and c=the two chords, and A=the height ; The segment =(C+|c)fA nearly. 2. U h= the height of the segment, and d={he diameter of the circle ; h h' h' Thesegment=2A>/^Ax(f-^-285^-f2^3 ^c.) Note E. p. 29. The term solidity is used here in the customary sense, to express the magnitude of any geometrical quantity of three dimensions, length, breadth, and thickness ; whether it be a solid body, or a fluid, or even a portion of empty space. This use of the word, however, is not altogether free from objection. The same term is applied to one of the general properties of matter ; and also to that peculiar quality by which certain substances are distinguished (rom fluids. There seems to be an impropriety in speaking of the solidity of a body of water, or of a vessel which is empty. Some writers have therefore substituted the word volume for solidity. But the latter term, if it be properly defined, may be retained without danger of leading to mistake. Note F. p. 35. The geometrical demonstration of the rule for finding the solidity of a frustum of a pyramid, depends on the following proposition : NOTES. 91 ^ifntstum of a triangular pyramid is equal to three pyramids ; the greatest and least of which are equal in height to the frustum^ and have the fji'o ends of the frustum for their bases ; and the third is u mean proportional betuieen the other tn;o. Let ABCDFG (Fig. 34.) be a frustum of a triangular py- ramid. If a plane be supposed to pass through the points AFC, it will cut off the pyramid ABCF. The height of this is evidently equal to the height of the frustum, and its base is ACB, the greater end of the frustum. Let another plane pass through the points AFD. This will divide the remaining part of the figure into two triangulai? pyramids AFDG and AFDC. The height of the formel^ is equal to the height of the frustum, and its base is 13FG, the smaller end of the frustum. /^ To find the magnitude of the third pyramid AFCD, let F be now considered as the vertex of this, and of the second pyramid AFDG. Their bases will then be the triangles ADC and ADG. As these are in the same plane, the two pyramids have the same altitude, and are to each other as their bases. But these triangular bases, being between the same parallels, are as the lines AC and DG. Therefore the pyramid AFDC is to the pyramid AFDG as AC to DG ; and AFCD' : AFDG' : : AC' : IJG'. (Alg. 39L) But the pyramids ABCF and AFDG, having the same altitude, are as their bases ABC and DFG, that is, as AC^ and DG^. (Euc. 19, 60 We have then AFDC' : AFDG'^jAC'^G" ABCF: AFDG:: AC' : DG' Therefore AF DC" : AFDG\:ABCF : AFDG. And AFDC' =AFDGx ABCF. That is, the pyramid AFDC is a mean proportional be- tween AFDG and ABCF. Hence, the solidity of a frustum of a triangular pyramid is equal to ^ of the height, multiplied into the sum of the areas of the two ends and the square root of the product of these areas. This is true also of a frustum of any other pyramid. (Sup. Euc. 12, 3. Cor. 2.) If the smaller end of a frustum of a pyramid be enlarged, till it is made equal to the other end ; the frustum will become a prism, which may be divided into three equal pyramids. (Sup. Euc. 15, 3.) 4e MENSURATION. Note G. p. 59. The following simple rule for the solidity of round timber, or of any cylinder, is nearly exact : Multiply the length into twice the square of ^ of the circumference. If C=- the circumference of a cylinder; The area of the base=4^ = Y2:5gg-But 2^^^^ =j^^ It is common to measure hewn timber, by multiplying the length into the square of the quarter-girt. This gives exact- ly the solidity of a parallelepiped, if the ends are squares. But if the ends are parallelograms, the area of each is less than the square of the quarter-girt. (Euc. 27, 6.) Timber which is tapering may be exactly measured by the rule for the frustum of a pyramid or cone (Art. 50, 68.) ; or, if the ends are not similar figures, by the rule for a pris- moid. (Art. 55,) But for common purposes, it will be suffi- cient to multiply the length by the area of a section in the middle between the two ends. A TABLE 'DF THE SEGMENTS OF A CIRCLE, WHOSE DIAMETER IS 1, AND IS SUP- POSED TO BE DIVIDED INTO 1000 EQUAL PARTS. Height. Area Seg. Height. Area Seg I Height Area Seg. .001 .000042 .034 .008273 .067 .022652 002 000119 035 008638 068 023154 003 000219 036 009008 069 023659 004 000337 037 009383 070 024168 005 000471 038 009763 071 024680 006 000618 039 010148 072 025195 007 000779 040 010537 073 025714 008 000952 041 010932 074 026236 009 001135 042 011331 075 026761 010 001329 043 011734 076 027289 Oil 001533 044 012142 077 027821 012 001746 045 012554 078 028356 013 001968 046 012971 079 028894 014 002199 047 013392 080 029435 015 002438 048 013818 081 029979 016 002685 049 014247 082 030526 017 002940 050 014681 083 031076 018 003202 051 015119 084 031629 019 003472 052 015561 085 032186 020 003748 053 016007 086 032745 021 004032 054 016457 087 033307 022 004322 055 016911 088 033872 023 004618 056 017369 089 034441 024 004921 057 017831 090 035011 025 005231 058 018296 091 035585 026 005546 059 018766 092 036162 027 005867 060 019239 093 036741 028 006194 061 019716 094 037323 029 006527 062 020206 095 037909 030 006865 063 020690 096 038496 031 007209 064 021178 097 039087 032 007558 065 021659 098 039680 .033 .007913 .066 .022154 .099 .040276 1 94 TABLE OF CIRCULAR SEGMENTS. 1 Height. Area Seg. Height. Area Seg. Height. Area Seg. .100 .040875 .144 .069625 .188 .102334 101 041476 145 070328 189 103116 102 042080 146 071033 190 103900 103 042687 147 071741 191 104685 104 043296 148 072450 192 105472 105 043908 149 073161 193 106261 106 044522 150 073874 194 107051 107 045139 151 074589 195 107842 108 045759 152 075306 196 108636 109 046381 153 076026 197 109430 110 047005 154 076747 198 110226 111 047632 155 077469 199 111024 112 048262 156 078194 200 111823 113 048894 157 078921 201 112624 114 049528 158 079649 202 113426 115 050165 159 080380 203 114230 116 050804 160 081112 204 115035 117 051446 161 081846 205 115842 118 052090 162 082582 206 116650 119 052736 163 083320 207 117460 120 053385 164 084059 208 118271 121 054036 165 084801 209 119083 122 054689 166 085544 210 119897 123 055345 167 086289 211 120712 124 056003 168 087036 212 121529 125 056663 169 087785 213 122347 126 057326 170 088535 214 123167 127 057991 171 089287 215 123988 128 058658 172 090041 216 124810 129 059327 173 090797 217 125634 130 059999 174 091554 218 126459 131 060672 175 092313 219 127285 132 061348 176 093074 220 128113 133 062026 177 093836 221 128942 134 062707 178 094601 222 129773 135 063389 179 095366 223 130605 136 064074 180 096134 224 131438 137 064760 181 096903 225 132272 138 065449 182 097674 226 133108 139 066140 183 098447 227 133945 140 066833 184 099221 228 , 134784 141 067528 185 099997 229 135624 142 068225 186 100774 230 136465 .143 .068924 ;}.'' . .101553 1 .231 ,.137307 1 TABLE OF CIRCULAR SEGMENTS. 95 1 Height Area Seg. Height Area Seg Height Aaea. Seg. 232 .138150 .277 .177330 .322 .218533 233 138995 278 178225 323 219468 234 139841 279 179122 324 220404 235 140688 280 180019 325 221340 236 141537 281 180918 326 222277 237 142387 282 181817 327 223215 238 143238 283 182718 328 224154 239 144091 284 183619 329 225093 240 144944 285 184521 330 226033 241 145799 286 185425 331 226974 242 146655 287 186329 332 227915 243 147512 288 187234 333 228858 244 143371 289 188140 334 229801 245 1492.30 290 189047 335 230745 246 150091 291 189955 336 231689 247 150953 292 190864 337 232634 248 151816 293 191775 338 233580 249 152680 294 192684 . 339 234526 250 153546 295 193596 340 235473 251 154412 296 194509 341 236421 252 155280 297 195422 342 237369 253 156149 298 196337 343 238318 254 157019 299 197252 314 239268 255 157890 300 198168 345 240218 256 158762 301 199085 346 241169 257 159636 302 200003 347 242121 258 160510 303 200922 348 243074 259 161386 304 201841 349 244026 260 162263 305 202761 350 244980 261 163140 306 203683 351 245934 262 164019 307 204605 352 246889 263 164899 308 205527 353 247845 264 165780 309 206451 354 248801 265 166663 310 207376 355 249757 266 167546 311 208301 356 250715 267 168430 312 209227 357 251673 268 169315 313 210154 358 252631 269 170202 314 211082 359 253590 270 171089 315 212011 360 254550 271 171978 316 212940 361 255510 272 172867 317 213871 362 256471 273 173758 318 214802 363 257433 274 174649 319 215733 364 258395 275 175542 320 216666 365 259357 .276 .176435 .321 .217599 .366 , .260320 96 TABLE OF CIRCULAR SEGMENTS. Height ^ rea Seg. Heigiit Area &.„ Height. Area 5r"eg. .367 .261284 .412 .305155 .457 .349752 368 262248 413 306140 458 350748 369 263213 414' 307125 459 351745 370 264178 415 308110 460 352742 371 265144 416 309095 461 353739 372 266111 417 310081 462 354736 373 267078 418 311068 463 355732-' 374 268045 419 312054 464 356730 375 269013 420 313041 465 357727 376 269982 421 314029 466 358725 377 270951 422 315016 467 359723 378 271920 423 316004 468 360721 379 272890 424 316992 469 361719 380 273861 425 317981 470 362717 381 274832 426 318970 471 363715 282 275803 427 319959 472 364713 383 276775 428 320948 473 365712 384 277748 429 321938 474 .366710 385 278721 430 322928 475 367709 386 279694 431 323918 476 368708 387 280668 432 324909 477 369707 388 281642 433 325900 478 370706 389 282617 434 326892 479 371705 390 283592 435 327882 480 372704 391 284568 436 328874 481 373703 392 285544 437 329866 482 374702 393 286521 438 330858 483 375702 394 287498 439 331850 484 376702 395 288476 440 332843 485 377701 396 289454 441 333836 486 378701 397 290432 442 334829 487 379700 398 291411 443 335822 488 380700 399 292390 444 336816 489 381699 400 293369 445 337810 490 382699 401 294349 446 338804 491 383699 402 295330 447 339798 492 384699 403 296311 448 340793 493 385699 404 297292 449 341787 494 386699 405 298273 450 342782 495 387699 406 299255 451 343777 496 388699 407 300238 452 344772 497 389699 408 301220 453 345768 498 390699 409 302203 454 346764 499 391699 410 303187 455 347759 .500 .392699 .411 .304171 .456 .348755 _ >iBir suit A T J O^" I^l.I. M ;c / :H \ A dl,' AT B T ,' •R \ \" /■ &\ / h\ \ \ / a \ p c N P I> A n lf-.Ji>tt7in Ays. v^ »imtH\TKA.TI py. i jME:ysiTRATio:s'» rLU. -M Jiraf&n JV. »IEWST'RAT10jr. THE MATHExMATICAL PRINCIPLES NAVIGATION AND SURVEYING, WITH THE MENSURATION OF ; HEIGHTS AND DISTANCES. BEING THB rOU&TB PART OF .4 COURSE OP MATHEMATICS. ADAPTED TO THE METHOD OF INSTRUCTION IN THE AMERICAN COLLEGES. BY JEREMIAH U VY. 1). /. LL.D. President of Yale College. THE SECOND EDITION. NEWHAVE^T: miNTED AND PUBLISHED BY S. CONVERSE. 1824. DISTRICT OF CO^rjVECTICUT, ss. Be it remembered ; That on the twentieth day olWIarch, in the forty-first year of the Independence ol (he United States of America, Jeremiah Day, of the said District, h^ith deposited in this Office, the title of a book, the right whereof he claims as Author, in the words following, to wit : "The Mathematical Principles of Navigation and Surveying, with the Mensuration of Heights and Distances. Being the fourth part of a Course of Mathematics, adapted to the method of instruction in the American Colleges. By Jeremiah Day, Professor of Mathematics and Natural Philosophy, in Yale College." In conformity to the Act of the Congress of the United State?, entitled " An Act for the encouragement of learning, by securing the copies of Maps, Charts and Books, to the authors and proprietors of such copies, during the times therein mentioned." HENRY W. EDWARDS, Clerk of the District of Connecticut. A true copy of Record, examined and sealed by me, H. W. EDWARDS, Clerk of the District of Connecticut. As the following treatise has been prepared for the use of a class in College, it does not contain all the details which would be requisite for a practical navigator or surveyor. The object of a scientific education is rather to teach princi- pies, than the minute rules which are called for in profession- al practice.' The principles should indeed be accompanied with such illustrations and examples as will render it easy for the student to make the applications for himself when- ever occasion shall require. But a collection of rules merely, would be learned only to be forgotten, except by a few who might have use for them in the course of their bu- siness. There are many things belonging to the art of navi- gation, which are not comprehended in the mathematical part of the subject. Seamen will of course make use of the valuable system of Mackay, or the still more complete work ofBowditch. The student is supposed to be familiar with the principles of Geometry and Trigonometry, before he enters upon the present number, which contains little more than the applica- tion of those principles to some of the most simple problems in heights and distances, navigation, and surveving. CONTENTS. Page. Heiohts and Distances - - . _ . j NAVIGATION. Section I. Plane Sailing ---..--_ 15 11. Parallel and Middle Latitude Sailing - 25 III. Mercator's Sailing 32 IV. Traverse Sailing -----.-. 40 V. Miscellaneous Articles The Plane Chart - 45 Mercator's Chart 46 Oblique Sailing - - 47 Current Sailing ------- 49 Hadle)''s Quadrant ------ 51 SURVEYING. Section f. Surveying a field by measuring round it - 57 II Oilier methods of surveying - - - - 73 III. Dividing and laying out lands - - - - 62 IV. Levelling 86 V. The Magnetic Needle 92 Notes 95 Tables 109 HEIGHTS AND DISTANCES. Art. 1. rWlHE most direct and obvious method of determin- A ing the distance or height of any object, is to ap- ply to it some known measure of length, as a foot, a yard, or a rod. In this manner, the height of a room is found, by a joiner's rule ; or the side of a tield by a surveyors chain. Eut in many instances, the object, or a part, at least of the line which is to be measured is inaccessible. We may wish to determine the breadth of a river, the height of a cloud, or the distances of the heavenly bodies. In such cases it is ne- cessary to measure some other line ; from which the requir- ed line may be obtained, by geometrical construction, or more exactly, by trigonometrical calculation. The line first measured is frequently called a base line. 2. In measuring angles, some instrument is used which contains a portion of a graduated circle divided into degrees and minutes. For the proper measure of an angle is an arc of a circle, whose centre is the angular point. (Trig. 74.) The instruments used for this purpose arc made in different forms, and with various appendages. The essential parts are a graduated circle, and an index with sight-holee, for tak- ing the directions of the lines which include the angles. - 3. Angles of elevation, and of depression, are in a plane perpf^ndicular to the horizon, which is called a vertical plane. An anele of elevation is contained between a parallel to the horizon, and an ascending line, as BAG. (Fig. 2.) An angle o( depression is contained between a parallel to the horizon, and a descending line, as DCA. The complement of this is the angle ACB. 4. The instrument by which angles of elevation, and of depression, are commonly measured, is called a Quadrant. In its most simple form, it is a portion of a circular board 2 MENSURATION OF ABC, (Fig. 1.) on which is a graduated arc of 90 degree?, AB, a plumb line CP, suspended from the central point C, and two sight-holes D and E, for taking the direction of the object. To measure an angle o( elevation with this, hold the plane of the instrument perpendicular to the horizon, bring the cen- tre C to the angular point, and direct the edge AC in such a mattner, that the object G may be seen through the two sight-holes. Then the arc BO measures the angle BCO, which is equal to the angle of elevation FCG. For as the plumb-line is perpendicular to the horizon, the angle FCO is a right angle, and therefore equal to BCG. Taking from these the common angle BCF, there will remain the angle BCO=FCG. In taking an angle o^ depression, as HCIj (Fig. 1.) the eye is placed at C, so as to view the object at L, through the sight-holes D and E. 5. In treating of the mensuration of heights and distances, no new principles are to be brought into view. We have only to mak^ an application of the rules for the solution of triangles, to the particular circumstances in which the obser- ver may be placed, with respect to the line to be measured. These are so numerous, that the subject may be divided inta a great number of distinct cases. But as they are all solved upon the same general principles, it will not be necessary to give examples under each. The following problems may serve as a specimen of those which most frequently occur in practice Problem I. To FIND THE PERPENDICULAR HEIGHT OF AN ACCESSIBLE OB- JECT STANDING ON A HORIZONTAL PLANE. 6. MEASURE FROM THE OBJECT TO A CONVENIENT STATION, AND THERE TAKE THE ANGLE OF ELEVATION SUBTENDED BY THE OBJECT. If the distance AB (Fig. 2.) be measured, and the angle of elevation BAC; there will be given in the right angled triangle ABC, the base and the angles, to find the perpen- dicular. (Trig. 137.) As the instrument by which the angle at A is measured, is commonly raised a few feet above the ground ; a point B must be taken in the object, so that AB shall be parallel to HEIGHTS AND DISTANCES. 3 the horizon. The part BP, may afterwards be added to the heia:ht EC, found by trigonometrical calculation. Ex. 1. What is the height of a tower BC, (Fig. 2.) if the distance AB, on a horizontal plane, be 9S feet; and the an- gle BAC 35| degrees ? Makinor the hypothenuse radius, (Trig. 121.) Cos BAC : AB: :Sin BAC : BC=69.9 feet. For the geometrical constructio7i of the problem, see Trig. 169. 2. What is the height of the perpendicular sheet of water at the falls ofNiagara. if it subtends an angle of 40 degrees, at the distance of 163 feet from the bottom, measured on a horizontal plane? Ans. 136 J feet. 7. If the height of the object be known, its distance may be found by the angle of elevation. In this case the angles, and the perpendicular of the triangle are given, to find the base. Ex. A person on shore, taking an observation of a ship's mast which is known to be 99 feet high, finds the angle of elevation 3^ degrees. What is the distance of the ship from the observer ? Ans. 98 rods. 8. If the observer be stationed at the top of the perpen- dicular BC, (Fig. 2.) whose height is known ; he may find the length of the base line AB, by measuring the angle of depression ACD, which is equal to BAC Ex. A seaman at the top of a mast G6 feet high, looking at another ship, finds the angle of depression 1 degrees. What is the distance of the two vessels from each other ? Ans. 22f rods. We may find the distance between tzoo objects which are in the same vertical plane with the perpendicular, by calcu- lating the distance of each from the perpendicular. Thus AG (Fig. 2.) is equal to the difference between AB and GB. Problem. H. to find the height of an accessible object standing dn an inclined plane. 9. Measure the distance from the object to a con- venient STATION, AND TAKE THE ANGLES WHICH THIS BASE MAKES WITH LINES DRAWN FROM ITS TWO ENDS TO THE TOP OF THE OBJECT. 4 MENSURATION OF If the base AB ''Fig. 3.) be measured, and the angles BAG and ABC ; there will be given, in the oblique angled triangle ABC, the side AB and the angles, to find BC. (Trig. 150.) Or the height BC maj be found by measuring the distan- ces BA, AD, and taking the angles BAC and BDC. There will then be given in the triangle ADC, the angles and the side AD, to find AC; and consequently, in the triangle ABC, the sides AB and AC with the angle BAC, to find BC. Ex. If AB (Fig. 3.) be 70 (eet, the angle B 101° 25' and the angle A 44® 42' ; what is the heiglit of the tree BC? SinC : AB::SinA : BC=95.0feet. For the geometrical construction of the problem, see Trigc 169. 10. The following are some of the methods by which the height of an object may be found, without measuring the an- gle of elevation. 1. Bi^ shadows. Let the staff be (Fig. 4.) be parallel to an object BC whose height is required. If the shadow of BC extend to A, and that of 6c to a ; the rays of light CA and ca coming from the sun may be considered parallel ; and therefore the triangles ABC and aba are similar ; so that ab : bc'.AB : BC. Ex. \(ah be 3 feet, be 5 feet, and AB G9 Cect, what is the height of BC ? Ans. 1 1 5 (eet. 2. By parallel rods. If two poles am and en (Fig. 5.) be placed parallel to the object BC, and at such distances as to bring the points C, c, in a line, and \( ab be made parallel to AB; the triangles ABC, and abe will be similar ; and we shall have ab :k::AB: BC. One pole will be sufficient, if the observer can place his eye at the point A, so as to bring A, a, and C in a line. 3. By a mirror. Let the smooth surface of a body of water at A, (Fig. G.) or any plane mirror parallel to the hori- zon, be so situated, that the eye of the observer at c may view the top of the object C reflected from the mirror. By a law of Optics, the angle BAC is equal to bAe ; aod if be HEIGHTS AND DISTANCES. f. be made parallel to BC, the triangle bAc will be similar to BAG : ?o that A6 : bc::AB : BC. Problem HI. to ri.vd the height of an inaccessible object above a horizontal plane. 11. Take two stations in a vertical plane passing through the top of the object, measure the distance from one station to the other, and the angle of eleva- tion at each. If the base AR (Fig. 7.) be measured with the angles CBP and CAB ; as ARC is the supplement of CRP, there will be given, in the oblique angled triangle ARC, the side AB and the angles, to fmd BC; and then, in the right angled triangle BCP, the hypothenuse and the angles, to find the perpendic- ular CP. Ex. I. If C (Fig. 7.) be the top of a spire, the horizontal base line AB 100* feet, the angle of elevation BAC 40°, and the angle PBC 60° ; what is the perpendicular height oj; the spire ? The difference between the angles PBC and BAC is equal to ACB. (Euc. 32. 1) Then Sin ACB : AR::Sin BAC : BC = 187.9 And R : BC::Sin PBC : CP=162| feeU 2. If two persons 120 rods from each other, are standing on a horizontal plane, and also in a vertical plane passing through a cloud, both being on the same side of the cloud : and if they find the angles of elevation at the two stations to be G8= and 76° ; what is the height of the cloud ? Ans. 2 miles 135.7 rods. 12. The preceding problems are useful in particular ca- ses. But the following is 2i general rule, which may be used for finding the height of any object whatever, within mode- rate distances. 6 BIENSURATION OF Problem IV. to find tht: height of any object^ bv observations at two stations. 13. Measure the base line between the two stations, the angles between this ease and lines drawn from each of the stations to each end of the object, and the an- gle subtended by the object, at one of the stations. IfBC(Fig. 8.) be the object whose height is required, and if the distance between the stations A and D be meas- ured, with the angles ADC, DAC, ADB, DAB, and BAG; there will be given, in the triangle ADC, the side AD and the angles, to find AC ; in the triangle ADB, the side AD and the angles, to find AB; and then, in the triangle BAG, the sides AB and AC with the included angle, to find the re- quired height BC. If the two stations A and D be in the same plane with BC, the angle BAC will be equal to the ditTerence between BAD and CAD. In this case it will not be necessary to measure BAC. Ex. If AD=83 feet, (Fig. 8.) ( ADB=33°, ( ADC=51° (DAB=121° iDAC = 95° BAC=26^ What is the height of the object BC ? SinACD : AD:: ADC : AC = n5.3 Sin ABD : AD: : ADB : AB=103.1 (AC+AB) : (AC-AB)::Tan i(ABC-l-ACB) : Tani(ABC -ACB) = 13°38' Sin ACB : AB: :Sin BAC t BC=50.57 i^Qi. If the object BC be perpendicular to the horizon, its height, after obtaining AB and BC as before, may be found by taking the angles o( elevation BAP and CAP. The differ- ence of the perpendiculars in the right angled triangles ABP and ACP, will be the height required. Problem V. to find the distance of an inaccessible object. 14. Measure a base line between two stations, and the angles between this and lines drawn from each of the stations to the object. HEIGHTS AND DISTANCES. 7 If C (Fig, 9.) be the object, and if the distance between the stations A and B be measured, with the angles at B and A; there will be given, in the oblique angled triangle ABC, the side AB and the angles, to find AC and BC, the distan- ces of the object from the two stations. For the geometrical construction, see Trig. 169. Ex. 1. What are the distances of the two stations A and B (Fig. 9.) from the house C, on the opposite side of a river; if AB be 26.G rods, B 92^ 46', and A ^8" 40' .? The angle C = 180-(A-hB)=48^ 34'. Then Q- r- . ATJ.. ^Sin A :BC=22.17 Sm CAB,. (sinB:AC=35.44 2. Two ships in a harbour, wishins: to ascertain how far they are from a fort on shore, find that their mutual distance is 90 rods, and that the angles formed between a line from one to the other, and lines drawn from each to the fort are 4^° and 56° 15'. What are their respective distances from the fort.'* Ans. 76.3 and 64.9 rods. 15. The perpendicular distance of the object from the line joining the two stations may be easily found, after the dis- tance from one of the stations is obtained. The perpendicu- lar distance PC (Fig 9 J is one of the sides of the right an> gled triangle BCP. Therefore R:BC::SinB : PC Problem VI. to find the distance between two objects, when the passage fuom one to the other, in a straight line, is obstructf-d. 16. Measure the right lines from one station to each of the objects, and the angle included between these lines. If A and B (Fig. 10.) be the two objects, and if the dis- tances BC and AC be measured, with the angle at C; there will be given, in the oblique angled triangle ABC. two sides and the included angle to find the other two angles, and the remaining side. (Trig. 153.) Ex. The pai^sage between the two objec tsA and B (Fig 10.) being obstructed by a morass, the line BC was measured and found to be 109 rods, the line AC 76 rods, and the angle at C lOl'^ 30'. What is the di^:tance AB ^ Ans. 144.7 rods. MENSURATION OF Problem VII. to find the distance between two inaccessible objects. 17. Measure a base line between two stations and THE angles between THIS BASE AND LINES DRAWN FROM EACH OF THE STATIONS, TO EACH OF THE OBJECTS. If A and BfFig. 11.) be the two objects, and if the distance between the stations C and D be measured, with the angles BDC, BCD, ADC, and ACD ; the lines AC and BC may be found as in Problem V, and then the distance AB as in Problem VI. This rule is substantially the same as that in art. 13. The two stations are supposed to be in the same plane with the objects. If they are not, it will be necessary to measure the angle ACB. 18. The same process by which we obtain the distance of two objects from each other, will enable us to find the dis- tance between one of these and a third, between that and a fourth, and so on, till a connection is formed between a great number of remote points. This is the plan of the great Trig- onometrical Surveys which have been lately carried on, with surprising exactness, particularly in England and France. See Surveying, Section II. 19. In the preceding problems for determining altitudes, the objects are supposed to be at such moderate distances, that the observations are not sensibly affected by the spherical figure of the earth. The height of an object is measured from an horizontal plane, passing through the station at which the angle of elevation is taken. But in an extent of several miles, the figure of the earth ought to be taken into account. Let AB (Fig. 12) be a portion of the earth's surface, H an object above it, and AT a tangent at the point A, or a horizontal line passing through A. Then HT, the oblique height of the object above the horizon of A, is only a part of the height above the surface of the earth, or the level of the ocean. To obtain the true altitude, it is necessary to add BT to the height HT found by observation. The height BT may be calculated, if tiie diameter of the earth and the distance AT be previously known. Or if the height BT be first determined from observation, with the distance AT ; the diameter of the earth mav be thence deduced. HEIGHTS AND DISTANCES, PUOBLEM VIH. to find the diameter of the earth, from the known height of a distant mountain, whose summit is just vis- ible in the horizon. 20. From the square of the distance divided by the height, subtract the height. IfBT (Fig. 12) be a mountain whosn height is known, with the distance AT; and if the sanimit T be just visible in the horizon at A ; then AT is a tangent at the point A. Let 2BC = D, the diameter of the earth, AT = c^, (he distance of the mountain, BT=/i, its height. Then considering AT as a straight line, and the earth as a sphere, we have (Euc. 36. 3.) (2BC + BT)xBT = AT' ; Ihat is, (D+/0 X/^=£/^ and reducing the equation, D=^-A A Ex» The highest point of the Andes is about 4 miles above ihe level of the ocean. If a straight line from this touch the surface of the water at the distance of 178} miles; what is the diameter of the earth ? Ans. 7940 miles. 21. If the distance AT (Fig. 12.) be unknowny it may be found by measuring with a quadrant the angle ATC. Draw BG perpendicular to BC ; and join CG. The triangles ACG and BCG are equal, because each has a right angle, the sides AC and BC are equal, and the hypothenuse CG is common. Therefore BG and AG are equal. In the right angled trian- gle BGT, the angle BTG is given, and the perpendicular BT. From these may be found BG and TG, whose sum is equal to AT, the distance required.* 22. In the common measurement of angles, the light is supposed to come from the object to the eye in a straight line. But this is not strictly true. The direction of ihc light is atfected by the refraction of the atmosphere. If the object be near, the deviation is very inconsiderable. But in an cx- • This method of determining the diameter of the earth is not as accurate aa that by measuring; a de^^ree of Latitude, See Surveying, Sec. I J. 10 MENSURATION OF tent of several miles, and particularly in such nice observa- tions as determining the height of distant mountains, and the diameter of the earth, it is necessary to make allowance for the refraction.* Problem IX. To FIND THE GREATEST DISTANCE AT WHICH A GIVEN OB- JECT CAN BE SEEN ON THE SURFACE OF THE EARTH. 23. To THE PRODUCT OF THE HEIGHT OF THE OBJECT INTO THE DIAMETER OF THE EARTH, ADD THE SQ,UARE OF THE HEIGHT ; AND EXTRACT THE SQUARE ROOT OF THE SUM. Let 2BC = D, the diameter of the earth, (Fig. 12.) BT=/f, the height of the object, AT = f/, the distance required. Then (D+^)XA=.i'o'. mns, m'n's', Ctn and C't'n' are all similar to^A'B'C. The differ- ence of latitude is AB = Ao-\-ms-{-nt = A'o'-hm's' + n't' = A'B'. And the departure is om-{'sn+lC = o'm''\-s'n''{-f'C'=B'C', 20 NAVIGATION. 43. In plane sailing, then, the process of calculation is as accurate.,* and as simple, as if the surface of the ocean were a plane. Let NS (Fig 20.) be a meridian line. If a ship sails from A to C, and if BC is perpendicular to NS; then The Course, is the angle at A; and tlie complement of the course, the angle at C ; The Distance is the hypothenuse AC ; The Depar ure is the base BC, which is always opposite to the course; and The Difference of Latitude is the perpendicular AB, which is opposite to the complement of the course. Of these four quantities, any two heing given, the others may be found by rectangular trigonometry. (Trig. 116.) The parts given may be 1. The course and distance; or 2. The course and departure; or 3. The course and difference of latitude; or 4. The distance and departure; 5. The distance and difference of latitude; or 6. The departure and difference of latitude. The solutions may be made by arithmetical computation, by Gunter's scale or sliding rule^ or by geometrical construc- tion. (Trig Sec. Ill, V, VI.) The first method is by far the most accurate. As the student is supposed to be already familiar with trigonometry, these operations will not be repeat- ed here. In the geometrical construction, it will be proper to consider the upper side of the paper as north, and the low- er side south. The right hand will then be east, and the left hand west. Case I. ^^ ^. S The course, >^ ^ j ( The departure and 44. G>ven | ^^^ dis,a„ce; ] '° ""^ { Difference of latitude. Here we have the hypothenuse and angles given, to find the base and perpendicular. (Trig. 134.) Making then the distance radius, •D J . T^- . . . ^ Sin Course : Departure Rad.Dist.. I j.^^ Course :Diff.Lat. Example, I . A ship sails from A (Fig. 20.) SW by S, 38 miles to C= Required her departure and difference of latitude ?. ♦See Note B. PLANE SAILING. 21 The course is 3 point*;, or 33° 45 (Art. 35.) R . OQ. . 5 Sin 33° 4a' : 21.1 =Drpjrl. ^*^^' I Cos. 33° 45' : 31.6=Diff. Lat. Example 2. A ship sails S 20° E, 34 leagues. Her departure and dif- ference of latitude are required. Ans. 16.5 and 29.7 leagues. The proportions in this and the following cases may be va- ried, by making different sides radius, as in Trigonometrv Sec. in. Case II. Ar r*' ( The course, ") /. , ( The distance, and ^^•^^'^"Und departure; i '° ^°^ ^ Difference of latitude. Making the distance radius, Trig. 137.) c r> . T\ . ^ Rad : Distance Sin Course : Depart. . ^ f,^^ bourse : Diff. Lat. Example 1 . A ship leaving a port in latitude 42° N, has sailed S 3?° W, till she finds her departure 62 miles. What distance ha? she run, and in what latitude has she arrived .'' c- Q-70 » acy ^ Rad : 103=Distance. bm J7 . bJ. . ^ ^^g 3^, . 82.3==Diff. Lat The difference of latitude is S2.3 miles, or 1° I2'.3. (Art. 41.^ This is to be subtracted from the original latitude of the ship, because her course was towards the equator. The remainder is 40° 47'. 7, the latitude on which she has ar- rived. Example 2. A ship leaves a port in latitude 63° S, and runs N 54° E, till she makes a harbour where her departure is found to be 74 miles ; how great is the distance of the two places, and what is the latitude of the latter.'* The distance is91i miles; and the latitude of the latter place is 62° 06'.2. Case III. 4fi r* ^ ^^^ course, and ^ ♦ ^ j ^ The distance, ( Diff. of latitude ; J (. And departure. Making the distance radius, Cos Course : Diff. Lat: : ^ ^."V ^'Tf ,r, « ( Sm Course : Departure. 22 NAVIGATION. Example. A ship sails S 50^ E, from latitude T" N, to latitude 4^ S. Required her distance and departure. As the two latitudes are on different sides of the equator, the distance of the parallels is evidently equal to the sum of the given latitudes. This is 1 1° or 660 miles. The distance is 1026.8 miles, and the departure 786|. Case IV. A'f rirr « ^ The distance, 1 ^ ^ a S The course, and ^^•^'^^"^ And departure; ] ^° ^"^ 1 Diff. of latitude. Making the distance radius, (Trig. 135.) Dist : Rad: : Depart J Sin Course, Rad : Dist: :Cos Course : Diff. Lat. Example. A ship having left a port in Lat. 3° N, and sailing between S and E 400 miles, finds her departure ISO miles. What course has she steered, and what is her latitude? Her latitude is 2° 57'i S, and her course S 26° 44'J E. Case V. AQ n- ^ The distance, and > ^ n , ^ The course, 48. Given. < ri/r r i .-. j t to find w , , / ( Ditt. ol latitude; ) ^ And departure. Making the distance radius, Dist : Rad: : Diff. Lat : Cos Course, Rad : Dist: :Sin Course : Departure. Example, A vessel sails between N and E 66 miles, from Lat. 34° 50 to Lat. 35° 40'. Required her course and departure. The course is N 40° 45' E, and the departure 43.08 miles. Case VL 49, Given. \ ^^^ ^7^!-^^' '"^ \ '^ ^"d \ f ^^^ ^.^"^^^' ^ Diff. of latitude ; 5 ^ And distance. Making the difference of latitude radius, (Trig, 139.) Diff. Lat : Rad: :Depart : Tan Course, Rad : Diff. Lat: :Sec Course : Distance. Example. A ship sails from the equator between S and W, till her TRAVERSE TABLE. 23 latitude is 5^ 52', and her departure 264 miles. Required her course and distance. The course is S 36° 52'i W, and the distance is 440 miles. Examples for practice, 1. Given a ship's course S 46° E, and departure 59 miles; 10 find the distance and difference of latitude. 2. Given the distance 68 miles, and departure 47; to find the course and difference of latitude. 3. Given the course SSE, and the distance 57 leagues; to find the departure and difference of latitude. 4. Given the course NW by N, and the difference of lati- tude 2° 36' ; to find the distance and departure. 5. Given the departure 92, and the difference of latitude 86 : to find the course and distance. 6. Given the distance 123, and the difference of latitude 07 ; to find the course and departure. THE TRAVERSE TABLE. 50. To save the labour of calculation, tables have been prepared, in which are given the departure and difference of latitude, for every degree of the quadrant, or for every quar- ter of a degree. These are called Traverse tables, or tables oi Departure and Latitude. The distance is placed in the left hand column, the departure and difference of latitude di- rectly opposite, and the degrees if above 45° or 4 points, at the top of the page, but if under 45°, at the bottom. The titles at the top of the columns correspond to the courses at the top ; and the titles at the bottom, to the courses at the bot- tom ; the difference of latitude for a course ^rea/er than 45°, being the same as the departure for one which is as much /c55 than 45°. See Trig. 104. If the given distance is greater than any contained in the table, it may be divided into parts, and the departure and dif- ference of latitude found for each of the parts. The sums of the numbers thus found will be the numbers required. 24 NAVIGATION. The departure and difference of latitude for decimal parts may be found in the same manner as for whole numbers, by supposing the decimal point in each of the columns to be moved to the left, as the case requires. With the aid of a traverse table, all the cases of plane sail- ing may be easily solved by inspection. Ex. 1. Given the course 33° 45'; and the distance 38 miles; to find the departure and difference of latitude. Under SJ*^!, and opposite 38, will be found the difference of latitude 31.6, and the departure 21.11 ; the same as in page 21. 2. Given the course 57°, and the distance 163. The departure and diff. of lat. for 100 are 83.87 and 54.46 for 63 52.84 34.31 for 163 136.71 88.77 3. Given the course 39°, and the distance 18.23. The departure and diff. of lat. for 18. are 11.33 and 13.99 for .23 0.14 0.18 for 18.23 11.47 14.17 4. Given the course 41° 15', and the departure 60. Under 41°^, and against the departure 60. will be found the difference of latitude 68.42 and the distance 91. 5. Given the distance 63, and the departure 56. Opposite the distance 63, find the departure 56 ; in the adjoining column will be the latitude 28.85, and at the bot- tom, the course 62°|. 6. Given the departure 72, and the difference of latitude 37. Opposite these numbers in the columns of latitude and de- parture, will be found the distance 81, and at the foot of the columns, the course 62°f . 51. The traverse table is useful, not only for taking out departure and difference of latitude; but for finding by in- spection the sides and angles of any right-angled triangle whatever In plane sailing, the distance is the hypothenuse, (see Fig. 20.) the difference of latitude is the perpendicular, the departure is the base, and the course is the acute angle PARALLEL SAILING. 25 at the perpendicular. If then the hypotlienusc of any right- afigled triangle whatever, be found in the column of distances, in the traverse table; the perpendicular will be opposite in the latitude column, and the base in the departure column; the angle at the perpendicular, being at the top or bottom of the page. Ex. 1. Given the hypothenuse 24, and the angle at the perpendicular 54°A ; to find the base and perpendicular by inspection. Opposite 24 in the distance column, and over 54°| will be found the base 19.54 in the departure column, and the per- pendicular 13.94 in the latitude column. 2. Given the angle at the perpendicular 37°|, and the base 46 ; to find the hypothenuse and perpendicular. Under 37°i. look for 46 in the departure column ; and opposite this will be found the perpendicular 60.5 in the lati- tude column, and the hypothenuse 76 in the distance col- umn. 3. Given the perpendicular 36, and the base 30.21 ; to find the hypothenuse and angles. Look in the columns of latitude and departure, till the num- bers 38 and 30.21 are found opposite each other ; these will give the hypothenuse 47, and the angle at the perpendicular 40°. SECTION II. PARALLEL AND MIDDLE LATITUDE SAILING. 52. By the methods of calculation in plane sailing, a ship's course, distance, departure, and difference of latitude are found. There is one other particular which it is very important to determine, the difference of longitude. The departure gives the distance between two meridians in miles. But the situa- tions of places on the earth, are known from their latitudes and longitudes; and these are measured in degrees. The lines of longitude, as they are drawn on the globe, are farthest 5 26 NAVIGATION. from each other at the equator, and gradually converge to- wards the poles. A ship, in making a hundred miles of departure, may change her latitude in one case 2 degrees, in another iO, and in another 20. It is important, then, to be able to convert departure into difference of longitude ; that is, to determine how many degrees of longitude answer to any given number of miles, on any parallel of latitude. This is easily done by the following Theokem. 53. As THE cosine of latitude, To radius ; So IS THK DEPARTURE, To THE DIFFERENCE OF LONGITUDE. By this is to be understod. that the cosine of the latitude is to radius; as the distance between two meridians measured on the given parallel, to the distance between the same me- ridians measured on the equator. Let P (Fig. 21.) be the pole of the earth, A a point at the equator, L a place whose latitude is given, and LO a line per- pendicular to PC. Then CL or CA is a semi-diameter of the earth, which may be assumed as the radius of the tables; PL is the complement of the latitude, and OL the sine ofPL, that is, the cosine of the latitude. If the whole be now supposed to revolve about PC as an axis, the radius CA will describe the equator, and OL the given parallel of laiitude. The circumferences of these cir- cles are as their semi diameters OL and CA, (Sup Euc. 8. 1.) And this is the ratio which any portion of one circum- ference has to a like portion of the other. Therefore OL is to CA, that is, the cosine of latitude is to radius, as the distance between two meridians measured on the given par- allel, to the distance between the same meridians measured on the equator. Cor. 1. Like portions of different parallels of latitude are to each other, as the cosines of the latitudes. Cor. 2. A degree of longitude is commonly measured on the equator. But if it be considered as measured on a paral- lel of latitude, the lc?igth of the degree will he as the cosine of the latitude. PARALLEL SAILING. The following table cojitains the length of a decree oj longi tilde fur each degree of latitude. D.t -UiUs I) 1, Alil^ J) 1- .•Vlile, D.L .Miles D 1, Milts j.i. .MileK. 1 \ l'59.99j |16 57.67 |31i51.43!46 41.68 ;6li29.09| 76 14.52; ! 2 5jI.96 Il7 57.3S 32 50.88; 47 40.92 62 28.17 77 13.50| 3 59.92 18 57.06 ; 33 50.32' 48 40.15 ,63 27,24 78 12.47; 4 59.85 1^^,56 73||34|49.74| 49 39.36 !64 26 30 79 U.45i i^ 59.77 20 56.38 35 36 49.15 48.54 bO 38.57 j65 25.36 24.40 SO 8I1 10.42J 1 9 39i 6 59.67; 21 56 Oi 51 37.76 l66 1 7 50 55 22 55-63 37 47 92 5'2|36.94|';67 23.44 82 8.35| ' 6 59 42 23 55.23 38 47.28 oi 36.11 68 22.45 !83| 7.31 9 59.26] 24 54.81 39 46.63 54 .35.27 69 2/50 84i 6.27 10 59.08 25 54 38 1" 41 45.96 3.5 34.41 1 70 20.52 85 5.23J 11 58.S9I26 53.93 43.28 06 33.55 71 19.53 861! 4.19 12 58.68 27 53.4rJ 42 44.59||57 32.68 72 18 54 :87 3.141 13i58.46 2Si52.97| 43 43.88 58 31.80 73 17.54 '88; 2.091 14jo6.22 29 52 47! 44 43.16 59 30-90 174 16.54 75I15.53 89 i 1.05 15157 95!'30 51.96; 45 42.43 60 30.00 90 1 0,00; The length of a degree of longitude in different parallels is also shown by the Line of Longitude, placed over or under the line of chords, on the plane scale. See Trig. (165.) The sailing of a ship on a parallel of latitude* is called Par- allel Sailing. In this case, the departure is equal to the dis- tance. The difference of longitude may be found by the preceding theorem; or if the difference of longitude be giv- en, the departure may be found by inverting the terms of the proportion. (Alg. 380. 3.) 55. The Geometrical Construction is very simpld. Make CBD (Fig. 22.) a right angle, draw BC equal to the depar- ture in miles, lay off the angle at C equal to the latitude in degrees, and draw the hypothenuse CD for the difference of longitude. The angles C, and the sides BC and CD, of this triangle, have the same relations to each other, as the latitude, departure, and difference of longitude. For Cos C : BC: :R : CD (Trig. 121.) And Cos Lat : Depart: :R ; Diff. Lon. (Art. 53. ?ee Note C^ 28 NAVIGATION. 5Q. The parls of the triangle may be found by inspection in the traverse table. (Art. 51.) The angle opposite the departure is D the complement ol the latitude, and the differ- ence of longitude is the hypothenuse CD. If then the de- parture be found in the departure column under or over the given number of degrees in the co-latitude, the difference of longitude will be opposite in the distance column. Example L A ship leaving a port in Lat. 38° N. Lon. 1G° E. sails west on a parallel of latitude 117 miles in 24 hours. What is her longitude at the end of this time .'* Cos38° : Rad::ll7 : 148'^=2° 28'^ the difference of longitude. This subtracted from 1G° leaves 13° 31'^ the longitude re- quired. Example 11. What is the distance of two places in Lat. 46° N. if the longitude of the one is 2° 13' W. and that of the other 1° 17' E.? As the two places are on opposite sides of the first meridian, the difference of longitude is 2° 13' + 1° 17' = 3° 30', or 210 minutes. Then Rad : Cos 46°: :210 : 145.88 miles, the departure, or the distance between the two places. Example IIL A ship having sailed on a parallel of latitude 138 miles, finds her difference of longitude 4° 3' or 243 minutes. What is ht r latitude? Diff. Lon. 243 : Dep. 138! : Rad : Cos Lat. 55° 23'f. Example IV. On what part of the earth are the degrees of longitude half a's long as ftie equator ? AnSr In latitude 60. MIDDLE LATITUDE SAILLXG. 09 Middle Latitude Sailing. o7. By the method just explained, is calculated the dilTer- ence of longitude of a ship sailing on a parallel of latitude. But instances of this mode of sailing are comparatively few. It is necessary then to be able to calculate the longitude when the course is oblique. If a ship sail from A to C, (Fig. lo.) the departure is ecjual to om-\-sn-\-tC, But the sum of these small lines is /tS5 than BC, and greater than AD. (Art. 40.) The departure, then, is the meridian distance measur- ed not on the parallel from which the ship sailed, nor on that upon which she has arrived, but upon one which is be- tween the two. If the exact situation of this intermediate parallel could be determined, by a process sufficiently sim- ple for common practice, the difference of longitude would be easily obtained. The parallel usually taken for this pur- pose, is an arithmetical mean between the two extreme lati- tudes. This is called the Middle Latitude, The meridian distance on this parallel is not exactly equal to the depar- ture. But for small distances, the errour is notmateiial, ex- cept in high latitudes. The middle latitude is equal to half the sum of the two ex- treme latitudes, if they are both north or both south : but to half their difference. y if one is north and the other south. 58. In middle latitude sailing, all the calculations are made in the same way as in plane sailing, excepting the pro- portions in which the difference of longitude is one of the terms. The departure is derived from the difference of lon- gitude, and the difference of longitude from the departure, in the same manner as in parallel sailing, (Arts. 53, 54.) only substituting in the theorem the term middle latitude for lat. itude. Theorem I. As THE COSINE OF MIDDLE LATITUDE^ To RADIUS ; So IS THE DEPARTURE, To THE DIFFERENCE OF LONGITUDE. 59. The learner will be very much assisted in stating the proportions, by keeping the geometrical construction steadi- ly in his mind. In Fig. 20 we have the lines and angles in plane sailing, and in Fig. 22, those in parallel sailing. By 30 NAVIGATION.. briiif^'ing these together, as in Fig. 23, we have all the parts in middle latitude sailing. The two right angled triangles, being united at the common side BC, which is thci departure, Ibrm the oblique angled triangle ACD. 60. The angle at D is the complement of the middle lati- tude. (Art. 55.) Then in the triangle ACD, (Trig. 143.) Sin D : AC: :Sin A : DC ; that is, Theorem II. As THE COSINE OF MIDDLE LATITUDE, To THE DISTANCE ; So IS THE SINE OF THE COURSE, To THE DIFFERENCE OF LONGITUDE. Gl. The two preceding theorems, with the proportions in plane sailing, are sufficient for solving all the ci^ses in middle latitude sailing. A third may be added, for the sake of re- ducing two proportions to one. In the triangle BCD (Fig. 23.) Cos BCD : R: :BC : CD And in the triangle ABC, AB : R : : BC : Tan A. The means being the same in these two proportions, the extremes are reciprocally proportional. (Alg. 387.) We have then Cos BCD : AB: :Tan A : CD; that is, Theorem III. As the cosine of middle latitude. To the difference of latitude ; So is the tangent of the course, To the difference of longitude. Among the other data in middle latitude sailing, one of the extreme latitudes must always be given. Example L At what distance, and in what direction, is Montock Point iVom Martha's Vineyard ; the former being in Lat. 4l° Ol'N. Lon. 72° W., and the latter in Lat. 41° 17' N. Lon. 70° 48' W.? Here are given the two latitudes and longitudes, to find the course and distance. The difference of longitude is 72' The difference of latitude 13' The middle latitude 41° 10'\ lUlDDLE LATITUDE SAILING. 3i Beginning with the triangle in which there are two parts given, by theorem I, R : Diff. Lon: :Cos Mid. Lat ; Depart.=54.2 And by plane sailing, Case Vf, DiflT. Lat : Rad: ".Depart : Tan Course = 7G° 30'| Or to tind the course at a single statement, by theorem ni, Diff. Lat : Cos Mid. Lat: :Diff. Lon : Tan Course = 7G° 30'| To find the distance by plane sailing. Case III, Cos Course : Diff. Lat: :Rad : Dist. = 55.73 Example IL A ship leaving New-York light-house in Lat 40^^ 28' N. and Lon. 74° 08' W. sails S. E. 67 miles in 24 hours. Re- quired her latitude and longitude at the end of that time. By plane sailing, Rad : Dist: :Cos Course : Diff. Lat.=47'.4 The latitude required, therefore, is 39° 40'. 6, and the mid- dle latitude 40° 04'.3. Then by theorem II, Cos Mid. Lat : Dist: :Sin Course : Diff. Lon.=61'.9 Or by theorem III, Cos Mid. Lat : Diff. Lat: : Tan Course : Diff. Lon.=Gl'.9 The longitude required is 73° 06'. 1. Example III, A ship leaving a port in Lat 49° 57' N. Lon. 5° 14' W. sails S. 39° W. till her latitude is 45° 31'. Required her longitude and distance. Ans. 10° 34'.3 W. and 342.3 miles. Example IV, A ship sailing from Lat. 49° 57' N. and Lon. 5° 14' W. steers west of south, till her longitude is 23° 43', and her de- parture 789 miles. Required her course, distance, and lat- itude. Course 51° 5' W. Latitude 39° 20' N. Distance 10(4 miles. SECTION III. MERCATOR'S SAILING. A T 62 nn^^ calculations in middle latitude sailing are ' ' JL. simple, and sufficiently accurate for short dis- tances, particularly near the equator. But they become quite erroneous, when applied to great distance, and to high latitudes. The only method in common use, which is strict- ly accurate, is that called Mcrcator''s Sailings or Wright's Sailioi/. This is founded on the construction of a chart, pub- lished in 1556 by Gerard Mercator. About forty years after, Mr. Edwnrd Wright gave demonstrations of the principles of this chart, and applied them to the solution of problems in navigation. 63. In the construction of Mercator's chart, the earth is supposed to be a sphere. Yet the meridians, instead of con- verging towards the poles, as they do on the globe, are drawn parallel to each other. The distance of the meridians, there- fore, is every where too great, except at the equator. To compensate this» the degrees of latitude are proportionally enlarged. On :he artificial globe, the parallels of latitude are drawn at equal distances. But on Mercator's chart, the distances of the parallels increase from the equator to the poles, so as every where to have the same ratio to the dis- tances of the meridians, which they have on the globe. Thus in latitude 60% where the distance of the meridians must be doubled, to make it the same as at the equator, a de- gree of latitude is also made twice as great as at the equa- tor. The dimensions of places are extended in the projec- tion, in proportion as they are nearer the poles. The diam- eter of an island in latitude 60° would be represented twice as great as if it were on the equator, and its area four times as great. ♦ Robertson's Navigation, London Phil Trans, for 1666 and 1696, Hut- ton's Dictionary, Introduction to Hutton's Mathematical Tables, Bowditch's Practical Navigator, Emerson'? an^ M'Laurin's Fluxions, M'Kay'? Naviga- tion, Emerson's Pria. Navig, Barrow's Navigation. MERCATOR'S SAILING. 33 64. Table of Meridional Parts, If a meridian on a sphere be divided into degrees or minutes, the portions are all equal. But in Mercator's projection, they are extended more and more as they are farther from the equator. To facilitate the calculations in navigation, tables have been prepared, which contain the length of any number of degrees and minutes on this extended meridian, or the distance of any point of the projection from the equator. These are called tables of Me- ridional Parts. The common method of computing them is derived from the following proposition. 65. Any MixuTf: portion of a parallel of latitude, Is to a like portion of the meridian ; As radius, To THE SECANT OF THE LATITUDE. For. by the theorem in parallel sailing, (Art. 53.) the co- sine of latitude is to radius, as the departure to the difference of longitude measured on the equator ; that is, as a part of the parallel of latitude, to a like part of the equator. But on a sphere, the equator and meridian are equal. Therefore Cos Lat : Rad: '.a part of the parallel: a like part of the meridian. But Cos Lat : Rad: : Rad : Sec Lat. (Trig. 93. 3.) By equality of ratios then, (Alg. 384.) Apart oj the parallel : a like part of the merid: ! Rad : Sec Lat. By like parts of the parallel of latitude and the merid- ian are here meant minutes, seconds or other portions of a degree. The proposition is true when applied either to the circles on a sphere, or to the lines in Mercator's projection. For the parts of the latter have the same ratio to each other, as the parts of the former. (Art. 63.) The divisions of Mercator's meridian, however, should be made very small; for the measure of each part is supposed to be taken at the parallel of latitude, and not at a distance from it. In the com- mon tables, the meridian is divided into minutes. 66. Suppose then that the length of each minute of a de- gree of Mercator's meridian is required. By the proposition in the last article, 1' of the parallel I I' of the meridian.: iRad : Sec. Lat. But in this projection, the parallels of latitude are all equal. 6 M NAVIGATION. (Art. 63.) Wliaiever be tne latitude, then, the first term of the proportion is equal to a minute at the equator, or a geo- graphical mile; and if this is assumed as the radius of the trigonometrical tables ( Irig. 100.) the first and third terms are equal, and therefore the second and fourth must be equal also. (Alg. 695,) That is, the length of ayiij one minute of jMercatur's mertdmn is equal to the natural secant of the lati- tndi of that part of the meridian, Vu^-^"' , ? minme of ihe meridian is ( f ^ '"i""'e. 1 he second \ , ,^ ,,,^ ^^^^^^ ^^ / u>« mmutes, Ihe third ) ^ (^ mrec mmutes, he. &ic. The table of meridional parts is formed by adding together the several minutes thus found.* Beginning from the equa- tor, an arc of the meridian o( two minutes=st'C \'-\-sec 2', of three n\\nu\es= sec 1 -\-sec 2'-{-sec 3', of four minutes=5ec I'-j-^ec 2'-^sec 3'-f sec 4', he. &ic. See the table at the end of this number. To find from the table the length of any given number of degrees and minutes, look for the degrees at the top of the page, and the minutes on the side ; then against the minutes, and under the degrees, will be the length of the arc in nau- tical miles. 67. Meridional Difference of Latitude, An arc of Mer- cator's meridian contained between tzoo parallels of latitude^ is called meridional difference of latitude. It is found by subtractini^ the meridional parts for the less latitude from the meridional parts for the greater, if both are north or south ; or by adding them, if one latitude is north and the other south. Thus the lat. of Boston is 42° 23' Merid. parts 2S13 Baltimore 39 23 Merid. parts 2575 Proper difFerence of lat. 3° Merid. diff. of lat. 238 68. If one latitude and the meridional difFerence of lati- tude be given^ the proper difFerence of latitude is found by reversing this process. * See Note D. MERCATOR'S SAILING. 35 When the two latitudes are on the same side of the equator, subiiaciiiig the niendional difference of latitude from the meridional parts for the greater, will give the me- ridional parts for the less; or adding the meridional ditier- ence to the parts for the less latitude, will give the parts for the greater But il the two latitudes are on opposite sides of the equator, subtracting the parts lor the one lat* rtude from the meridional difference, will give the parts for the other. Thus the meridional difference of latitude between New- York and New-Orleans is 793 The lat. of N. Orleans is 29° 57 Merid. parts 1SS5 The lat of New-York 40 42 Merid. parts 267S 69. Sulutio7is 171 Mercator'*s Sailing. The Solutions in Mercator's sailing are founded on t/ie swiUaritu of tic o right- angled triangles, in one of 7chich the perpendicular sides arc the proper difference of I at dude and the departure ; and in the other the r eridional dfference of latitude and the difference of longitude. According to the principle of Mercator's projection, the enlargement of each minute portion of the meridian is pro- portioned to the enlargement of the parallel of latitude which crosses it. (Art. 63.) Any part of the meridian before it is enlarged, is proper dfference of latitude j and after it is en- larged, is meridional difference of latitude, A part of the parallel, before it is enlarged, is departure^ and after it is en- larged, is equal to the corresponding difference of longitude ; because in this projection, the distance of the meridians is the same on any parallel, as at the equator, where longitude is reckoned. If then we take a small portion of the distance which a ship has sailed, as A?/i, (Fig. 18.) Prop. Dif Lat. Ao ; Depart, om: : Merid. Dif Lat i Dif. Lon. In the triangle ABC, (Fig. 24.) let the angle at A = the course oA//?, (Fig. 18.) AB=the proper difference of latitude. AC = Am4-'"'i-h"C the distance, and BC = om-{-sn-rtC the departure. Then as the triangles Aoniy msn, 7itC are each similar to the trian2;!e ABC, (Fig, 24.) the difference of latitude for any one of the small dis- tances as Am, is to the corresponding departure ; as the whole difference of latitude AB to the zrholc departure BC. Therefore. 3t> NAVIGATION. ( for Am : i for Am, P. Dif. Lat. AB : Dep. BC : : Mer. Dif. Lat. ) for mn : Dif. Lon. ) for mn^ ( for nC : (for nC. But the whole meridional difference of latitude for the dis- tance AC, is equal to the sum of the differences for A??;, mn, and 7iC; and the whole difference of longitude is equal to the sum of the differences for Am, mn, and nC. Therefore, (Alg. 388. Cor. 1.) Prop. Dif. Lat. AB : Dep. BC: :Merid. Dif. Lat : Dif. Lon. Extend AB, (Fig. 24.) making Al equal to the meridional difference of latitude corresponding to the proper difference of latitude AB ; from L draw a line parallel to BC, and ex- tend AC to intersect this in D Then is DL the differenct of longitude. For it has been shown that the difference of longitude is ^fourth prdportionul to thej proper difference of latitude, the departure, and the meridional difference of lati* tude ; and by similar triangles, AB : BC::AL : LD. 70. To solve all the cases, then, in Mercator's sailing, we have only to represent the several quantities by the parts of two similar right-angled triangles, as ABC and ALD, (Fig. 24.) and to find their sides and angles. In the smaller trian- gle ABC the parts are the same as in plane sailing, and the calculations are made in the same manner. The sides AL and DL are added for finding the difference of longitude; or when the difference of longitude is given, to derive from it one of the other quantities. The course is common to both the triangles, and the complement of the course is either ACB or ADL. The hypothenuse AD is not one of the quantities which are given or required in navigation. 71. In the similar triangles ABC, ALD, (Fig. 24.) AB : AL::BC : LD; that is, Theorem I. As THE PROPER DIFFERENCE OF LATlTtJDE, To THE MERIDIONAL DIFFERENCE OF LATITUDE ; So IS THE DEPARTURE, To THE DIFFERENCE OF LONGITUDE. 72. In the triangle ALD, if AL be made radius. Bad : Tan A::AL : DL; that is, MERCATOR'S SAILING. 37 Theorem II. As RADIUS, To THE TANGENT OF THE COURSE ; So IS THE 51KRIDIONAL DIFFEUINCE OF LATITUDE, To THE DIFFERENCE OF LONGITUDE. By this theorem, the difference of longitude may be calcu- lated, without previously finding the departure. 73. In Mercator's as well as in middle latitude sailing, one latitude mi st always be given. This is requisite in con- verting proper difference of latitude and meridional difference of latitude into each other. (Arts. 67, 68.) 74. When the difference of latitude is veiy sniall^ the dif- ference of longitude will be more correctly found by middle latitude sailing, than by Mercator's sailing; unless a table is used in which the meridional parts are given to decwials. Mercator's sailing is strictly correct in theory. But the com- mon tables are not carried to a degree of exactness, sufficient to mark very minute differences. On the other hand, the errour of middle latitude sailing is diminished, as the differ- ence of latitude is lessened. c 41' The latitudesofMontock and Martha's Vineyard are < .. Example I. ' °4'N. 17 N. Their longitudes < ^^ ^g, y^' Required the course and distance from one to the other. Lat. of Martha's Vin, 41° 17' Merid. parts 2724 of Montock 41 04 Merid. parts 2707 Proper Diff. of Lat. 13' Mer. Dif. Lat. 17 (Art. 67.) The difference of longitude is 1* 12'=72 miles. To find the course by theorem II, (Fig 24.) Merid. Diff. Lat : Diff. Lon::Rad : Tan Course=76° 43.' To find the distance by plane sailing, Cos Course : Prop. Diff. Lat.: :Rad : Dist.=56.58. The results by middle latitude sailing, page 31, are a littfe different, as that method is not perfectly accurate. 38 NAVIGATION. Example IL A ship sailing from the Lizard in Lat. 49® 57' N. Lon. b'^ 15' W. proceeds S. 39° W till her latitude is found by obser- vation to be 45° 31' N. What is then her longitude, and what distance has she run ? Here are given the difference of latitude and the course, to find the distance and tlie difference of longitude. The proper difference of latitude is 4° 26'=266' The meridional difference of latitude 396 Then by plane sailing, Cos Course : Prop. Diff. Lat.: : Had : Dist.=342.3. And by theorem II, Had : Tan Course ::M.Dif. Lat : Uif Lon.=320'.7=5°2(y.T This added to the longitude of the Lizard 5° 14' gives the longitude of the ship 10° 34'.7 W. Example IIL A ship sailing from Lat. 49° 57' N. and Lon. 5° 14' W. steers west of south, till her latitude is 39° 20' N. and her departure 789 miles. Required her course, distance, and longitude. The proper difference of latitude is 10° 37'=637' The meridional difference of latitude 899 Then by theorem I, (Fig. 24 ) P. Dif. Lat. : M. Diff. Lat::Dep : Diff. Lon.=:1113'.5^ 18°33'.5 The longitude of the ship is therefore 23° 47'^ And by plane sailing, Prop Diff. Lat : Rad::Depart : Tan Course=51° 5' Rad : Prop. Diff. Lat. : :Sec Course : Distance=1014 miles. Example IV, A ship sailing from a port in Lat. 14° 45' N. Lon. 17° 33' W. steers S. 28° 7'i VV. till her longitude is found by ob- servation to be 29° 26' W. Required her distance and latitude. The difference of longitude is 11° 53'=713'. By theorem II, Tan Course : Rad: : Dif. Lon : M Dif Lat. = 1334 S. Lat. of the port 14° 45' N. Merid. parts 895 N. of the ship 7 18 S. Merid. parts 439 S. (Art. 68.) Diff. of Lat. 22° 3' = 1323' MERCATOR'S SAILING. 39 By plane sailing, Cos Course : Diff. Lat: :Rad : Distance =1500 miles. Example V, A ship sails 300 miles between north and west, from Lat. 37° N. to 41° N. What is her course and difference of longi- tude ? The course is N. 36° 52' W., and the difference of longi- tude 3° 52!. Example VI. A ship sails S. 67° 30' E. from Lat 50° 10' S till her de- parture is 957 miles. What is her distance, difference of latitude, and difference of longitude ? The distance is 1030 miles. The difference of latitude G° 36'.4 The difference of longitude 26° 53' Example VII. A ship sailing from Lat. 26° 13' N. proceeds S. 27° W. 231 miles. What is her difference of latitude and difference of longitude? Example III, A ship sailing from Lat. 14° S. 260 miles, between south and west, makes her departure 173 miles. What is her cburse, difference of latitude, and difference of longitude r* ♦ See Note E- SECTION IV. TRAVERSE SAILING. TJY the methods in the preceding sections, are Art. 7j. J-J found the difference of latitude, departure, &;c. for a single course. But it is not often the fact that a ship proceeds from one port to another in a direct line. Varia- ble and contrary winds frequently render a change of direc- tion necessary every few hours. The irregular path of the ship sailing in this manner, is called a traverse. Resolving a traverse is reducing the compound course to a single one. This is commonly done at sea every noon. From the several courses and distances in the log-book, the departure, difference of latitude, &c. are determined for the whole 24 hours. In the same manner, the courses of several successive days are reduced to one, so as to ascertain, at any lime, the situation of the ship. The following methods by construction and by calculation, are sufficiently accurate for short distances, at least near the equator. 76. Geometrical construction of a traverse. To construct a traverse, draw a meridian line and lay down the first course and distance ; from the end of this, lay down the second course and distance ; from the end of that, a third course, &ic. Then draw a line connecting the extremities of the first and last of these, to show the whole distance, and the direc- tion of the ship from the point of starting. This will be easily understood by an example. Example L A ship sails from a port in Lat. 32° N., and in 24 hours makes the following courses ; 1. N. 25° E. 16 miles, 2. S. 54° E. 11, 3. N. 13° W 7, 4. N. 61° E. 5, 5. N. 38° W. 18, It is required to find the departure, difference of latitude, distance, and course, for the whole traverse. TRAVERSE SAILING. 41 On A as a centre (Fig. 25.) describe a circle and draw the meridian NAS. Then considering the upper part as north, the right hand east, and the left hand west, draw the hnes Al, A2, A J, A4, and A5, to correspond with the several eourses; that is, make the angle NA1=25°, SA2=54°; NA^3=13°, NA4=61°, and NA5=38^ Make AlB=16, BC = 11 and parallel to A2, CD=7 and parallel to A3, DF = 5 and parallel to A4, FG=18 and parallel to A5 ; join AG, and draw GP perpendicular loNS. Then if the surface of the ocean be considered as a plane, G is the place of the ship at the end of the 24 hours, AG the distance from port, PG the departure, AP the difference of laliiufh, and GAP the course. The angles may be measured by a line of chords, and the distances taken from a scale of equal parts.^ (^Trig. 148, 161, 2.) The distance is 32.3 miles. The departure 7.38 The difference of lat. 31 45 The course 13^ 12' 77. Resolving a traverse, by Calculation or Inspection, When a ship sails on different courses for a short time, the difference of latitude, at the end of that time, is equal to the difference between the sum of the northings and the sum of ihe southings, and the departure is neaily equal to'the differ- ence between the sum of the eastings and the sum of the westings. See Arts. 78 79, If then the difference of lati- tude and the departure for each course be found by calcula- tion or inspection, and placed in separate columns in a table; the difference of latitude for the whole time mav be obtained exactly, and the departure nearly, by addition and subtrac- tion ; and the corresponding distance and course may be de- termined by tri^^onometrical calculation or inspection, as in the last case of plane sailing. (Art. 49.) The following; table contains the courses, distances, depart- ure, and difference of latitude in the preceding example. See Fig. 25, •^ See Note F. 7 42 NAVIGATION. Traverse Table. Courses. Distances. w- Lai, Departure. 1 N. S. . .. . 6.47 E. 6.76 8.90 4.37 1 1 w. I N. 25° E. 2. S. 54° E. 3 N. 13° W 4. N. 61° E 5. N. 38° W. N 13° 12^ E. AB 16 BC 11 CD 7 DF 5 FG 18 AG. 32.3 14.50 6.82 2.42 14 18 1.57 11.08 37.92 6.47 6.47 20 03 12.65 12.65 1 31.45 7.38 The sum of the northings is 37.92. Subtracting from this the southing 6.47, we have the difference of latitude AP 31.45 N. The sum of the eastings is 20.03. Subtracting from this the sum ot the westings 12.65 we have the departure GP 7.38 E. Then (Art. 49.) Diff. Lat : Rad::Depart : Tan Course NAG=13° 12^1 Rad : Diff. Lat: : Sec. Course : Distance AG =32.3 The latitude of the port is The difference of latitude TheJatitude of the ship The meridional difference of lat. 32° N. 0° 31'.45 N. 32° 31'.45 N. 37.5 Then by Mercator's sailing, Rad : Tan Course: iMerid. Diff. Lat : Diff. Lon.=8'.8 Example IL A ship sailing from a port in Lat. 42° N. makes the follow- ing courses and distances. 1. S, 13° E. 21 miles, 2. S. 18° W. 16, 3. N. 84° E. 9, 4. S. 67° E. 12, 5. N. 78° E. 14, 6. S. 12° W. 35. TRAVERSE SAILING. 43 The difference of latitude, departure, &ic. are required. The departure is 26'. 19 E. The difF. of latitude, P lO'f S. The diff of longitude, 35'.07 The direct course, S. 20° 18'| E. The distance, 75^ miles. Accurate method of resolving a traverse, 78. The preceding method of resolving a t'-averse is fre- quently used at sea, because it is simple, and in most cases is sufficiently accurate for a run of 24 hours. But it is found- ed on the assumption, that when a ship sails from one place to another by several courses, she makes the sarne departure^ as if she had proceeded by :i single course to the same place. This is not strictly true. Suppose a vessel, instead of sail- ing directly from A to C, (Fig i8.) proceeds by one course from A to H. and then by a different course from H to C. In the compound course, the whole departure, is 6f/+^H + tC ; (Art. 40.) which on account of the obliquity of the me- ridians, is less than om-\-sn-\~tC, the departure on the single course. If the compound course had been on the other side of the single one, nearer the equator, the departure would have been greater, 71). But the difference of latitude is the same, whether the ship proceeds from one place to the other, on a single course, or on several The difference of latitude AB (Fig. 18.) = \o-\-ms-\-nt==Kb-\-dg-\-Y{t, The difference of longitude is also the same, whether the course is single or compound. For the difference of longitude is the distance between the meridians of the two places measured on the equator. If then the difference of latitude and difference of longitude be calculated for each part of the compound course ; the whole difference of latitude and difference of longitude icill^be found by addition and subtraction ; and from these may be deter- mined the direct course and distance. The difference of longitude for each course may be obtained independently of the departure, by theorem II. of Mercator's sailing. It will facilitate the calculation of the longitude, to place in the traverse table, the latitudes at the beginning and end of each of the courses, the corresponding meridional parts, and the meridional differences of latitude. 44 NAVIGATION. In the followlnpj example, the courses and distances are the sanne as in Art. 7G. Ex. 1. The port from which the ship is supposed to sail, is in latitude 32° N. 1 Traverse Table. Utg Lut. Latitudes - ^. (n.a. Mend. Diff. Long.i Courses. Dist Parts. Dtf. Lai. N. 1 S. 32° ;12 14 '.50 2028 2045.5 17.5 7.5 E 1 W. I. N. 25° E 16 14.50 6.47 8.16 2. S. 54° E. 11 J2 8 03 2038 7.8 10.32 3. N 13° W. / 6.82 32 14.85 2045.8 2 5 1.80 4 N. 61° E 5 2.42 32 17.27 2048.3 17.2 4.51 5. N.38° W. 18 14 1!; r37.92 6.47 647 32 3145 2065.5 22.99 15.24 13.44 15.24 11° 40' 37" 32.12i 31.45 ' 7.75 The difference of longitude is here found to be 7.75, and in Art. 77, 8'. 8 the errour there bein^ 1.05. To find the direct course and distance from the port to the place of the ship. Merid. Dif. I.at : Dif. Lon::Rad : Tan Conrse = ll°40' 37" Rad : Prop. Dif. LatllSec Course : Distance=32.12. By comparing the results here with those in Art. 77, it will be seen that a small errour was introduced there, both into the course and the distance^ by making them dependent on the departure ; which being obtained from ihe several cour- ses, is not the same as for a single course. (Art. 78.) Ex. 2 A ship sailing from a port in latiude 78° 15' N. fnakes the following courses and distances. 1. N. 67° 3(y w. 154 miles. 2. S. 45 W. 96 3. N. 50 371 W. 89 4. N. 11 15 E. 110 5. N. 36 33 J W. 56 6. S. 19 4U E. 78 Required the difference of latitude, the difference of lon- gitude, and the distance the ship must have sailed, to reach the same place on a single course. The difference of latitude is 2* 7' The difference of longitude 22° 29' The direct course N. 63° 1' W. The distance . 279.9 miles. SECTION V. MISCELLANEOUS ARTICLES. I. The Plane Chart, -^ f SlHE Charts commonly used in navigation are eitber J_ Plane Charts, or Mercator's Charts, The latter are generally to be preferred. But plane charts will answer for short distances, such as the extent of a harbour or small bay. In the construction of the plane chart, that part of the sur- face of the globe which is represented on it, is supposed to be a plane. The meridians are drawn parallel ; and the lines of latitude at equal distances. Islands, coasts, &ic. are delineated upon it, by laying down the several parts accord- ing to their known latitudes and longitudes. 81. On a chart extending a small distance, each side of the equator, the meridians ought to be at the same distance from each other, as the parallels of latitude. A similar construction is frequently applied to different parts of the globe. But this renders the chart much more incorrect than is riecessary. A circular island in latitude 60 would, by such a construction, be thrown into a figure whose length from east to west would be twice as great, as from north to south ; the comparative distance of the meridians being made twice as great as it ought to be. (Art. 53. Trig. 96. cor.) But when the chart extends only a few degrees, if the distance of the meridians is proportioned to the distance of the parallels of latitude, as the cosine of the mean latitude to radius; (Art, 53.) the representation will not be mate- rially incorrect. The meridian distance in the middle of the chart will be exact. On one side, it will be a little too great; and on the other, a little too small. 82. To construct a Plane Chart, then, on one side of the paper draw a scale of equal parts, which are to be counted as degrees or minutes of latitude, according to the proposed x'xtent of the chart. Through the several divisions, draw tht 46 THE PLANE CHART. parallels of latitude, and at right angles to these, draw the meridians in such a manner, that their distance from each other shall be to the distance of the parallels of latitude, as the cosine of the latitude of the middle of the chart, to ra- dius. After the lines on all the sides are graduated, the po- sitions of the several places which are to be laid down, may be determined, by applying the edge of a rule or strip of paper, to the divisions for the given degree of longitude on each side, and another to the divisions for the degree of latitude. In the intersection of these, will be the point required. The distance which a ship must sail, in going from one place to another, on a single course, may be nearly found, by applying the measure of the interval between the two places, to the scale of miles, of latitude on the side of the chart.* H. Construction of Mercator's Chart. 83. In Mercator's chart, the meridians are drewn at equal distances, and the parallels of latitude at unequal distances, proportioned to the meridional differences of latitude. (Arts. 63, 67.) To construct this chart, then, make a scale of equal parts on one side of the paper, for the lowest parallel of lati- tude which is to be laid down, and divide it into degrees and minutes. Perpendicular to this, and through the dividing points for degrees, draw the lines of longitude. For the se- cond proposed parallel of latitude, find from the table, (Art. 67.) the meridional difference of latitude between that and the parallel first laid down, and take this number of minutes from the scale on the chart, for the interval between the two parallels. Iti the ssme manner, find the interval between the Second and third parallels, between the third and fourth, &ic. till the projection is carried to a sufficient extent. Places whose latitudes and longitudes are known, may be laid down in the same manner as on the plane chart, by the intersection of the meridians and lines of latitude passing through them. If the chart is upon a small scale, the least divisions on the graduated lines may be decrees instead of minutes ; and the meridians and parallels may be drawn for every fifth or every tenth degree. But in this case, it will be necessary to di- ^See Note G. MERCATOR'S CHART. 47 vide the meridional differences of latitude by 60, to reduce them from minutes to degrees. 84. The Line of Meridional Parts on Gunler'^s scale is di- vided in the same manner as Mercator's Meridian, and cor- responds with the table of meridional parts ; except that the numbers in the latter are minutes^ while the divisions on the other are degrees. Directly beneath the line of meridional parts, is placed a line of equal parts. The divisions of the latter being considered as degrees of longitude, the divisions of the former will be degrees of latitude adapted to the same scale. The meridional difference of latitude is found, by ex- tending the compasses from one latitude to the other. A chart may be constructed from the scale, by using the line of equal parts for the degrees of longitude, and the line of meridional parts for the intervals between the parallels of latitude. 85. It is an important property of Mercator's chart, that all the rhumb-lines projected on it are straight lines. This renders ii, in several respects, more useful to navigators, than even the artificial globe. By Mercator's sailing, theorem II, (Art. 72.) Merid. DifF. Lat : Diff. Lon! :Rad : Tan Course So that, while the course remains the same, the ratio of the meridional difference of latitude to the difference of lon- gitude is constant. If A, C, C', and C" (Fig. 26.) be several points in a rhumb-line, AB AB', and AB", the correspond- ing meridional differences of latitude, and BC, B'C, B' C", the differences of longitude j then AB : BC::AB' : BC::AB" : B'Q'. Therefore ABC, AB'C, and AB"C". are similar triangles^ and ACC'C" is a right line. (Euc. 32. 6.) III. Obl:q,ue Sailing. 86. The application of oblique angled trigonometry to the solution of certain problems in navigation, is called oblique sailing. It is principally used in bays and harbours, to de- termine the bearings of objects on shore, with their distances from the ship and from each other. A few examples will be sufficient here, in addition to those already given under heights and distances. ly NAVIGATION. One of the cases which most frequently occurs, is that in which the distance of a ship from land is to be determined, when leaving a harbour to proceed to sea. This is necessa- ry, that her difference of latitude and departure may be reck- oned from a fixed point, whose latitude and longitude are known. The distance from land is found, by taking the bearing of an object from the ship, then running a certain distance, and taking the bearing again. The course being observed, there will then be given the angles and one side of a triangle, to find either of the remaining sides. Example L The point of land C (Fig. 27.) is observed to bear N ^1^ 30' W. from A. The ship then sails S. 67^ 30' W. 9 miles from A to B; and the direction of the point from B is found to be N. 11° 15' E. At what distance from land was th« ship at A ^ Let NS and N'S' be meridians passing through A and B. Then subtracting CAN and BAS each 67°i from 180, we have the angle "CAB=45°. And subtracting CBN' ll°i from BAS or its equal ABN', we have ABC=56°i. The angle at C is therefore 78° 43'. And Sin C : AB; :Sin B : AC =7.63 miles. Example IL New-York iight-house on Sandy-Point is in Lat. 40° 2S' N. Lon. 74° 8' W. A ship observes this to bear N 76° 16' W., and after sailing S 35° 10' W. 8 miles, finds the bearing to be N. 17° 13' W. Required the latitude and longitude of the ship, at ll^ first observation. The latitude is 40° 26'J The longitude 73 58} In this example, as the difference of latitude is small, the difference of longitude is best calculated by middle latitude sailing. (Art. 74.) Example III, A merchant ship sails from a certain port S. 51° E. at the rate of 8 miles an liour. A privateer leaving another port 7 mi'es N E of the first, sails at the rate ot 10 miles an hour. What must be the course of the privateer, to meet the tihip. without a change of direction in either ? Ans. S. 7° 43' E. CURRENT SAILING. 49 Example IV, Two light-houses are observed from a ship sailing S. 38° W. at the rate of 6 miles an hour. The first bears N. 21° W., the other N. 47^ VV. At the end of two hours, the first is found to bear N. 5° E., the other N. 13° VV, Wliat is the distance of the hght-houses from each other f Ans. G miles and 30 rods. IV. Current Sailing. 87. When the measure given by the log-line is taken as the rate of the ship's progress, the water is supposed to be at rest. But if there is a tide or current, the log being thrown upon the water, and left at liberty, will move with it, in the same direction, and with the same velocity. The rate of sailing, as measured by tlie log, is the motion through the water. If the ship is steered in the direction of the current, her whole motion is equal to the rate given by the log, added to the rate of the current. But if the ship is iteered in opposi- tion to the current, her absolute motion is equal to the differ- ence between the current, and the rate given by the log. In all other cases, the current will not only affect the velocity of the ship, but will change its direction. Suppose that a river runs directly south, and that a boat in crossing it is steered before the wind, from west to east. It will be carried down the stream as fast, as if it were mere- ly floating on the water in a calm. And it will reach the op- posite side as soon, as if the surface of the river were at rest. But it will arrive at a different point of the shore. Let AB (Fig. 26.) be the direction in which the boat is steered, and AD the distance which the stream runs, while the boat is crossing. If DC be parallel to AB, and BC paral- lel to AD ; then will C be the point at which the boat proceeding from A will strike the opposite shore, and AC will be the distance. For it is driven across by the wind, to the side BC, in the same time that it is carried down by the current, to the line DC In the same manner, if Awt be zny part of AB, and mn be the corresponding progress of the stream, the distance sailed will be An. And if the velocity of the ship and of the stream continue uniforn), Am is to mn, as A B to BC^ so that AnC % 60 NAVIGATION. is a straight line, (Euc 32. 6.) The lines AB, BC, and AC*. form the tliree sidi s ol a triangle. Hence, 88. If the direction and rate of a ship's motion through the water, be represented by the position and length of one side of a triangle, and the direction and rale of the current, by a second sidr ; the absolute direction and distance will be shown by the third side. Example I. If the breadth of a river running south (Fig. 28.) be 300 yards, and a boat steers S. 75 E. at tht rate of 10 yards in a minute, while the progress of the stream is 24 yards in a minute ; wl.at is the actual course, and what distance must the boat go in crossing ? Cos BaP : AP: :R : AB=3io.6 And 10 : 24: : AB : BC = 745.44 Then in the triangle ABC, (BC+AB) : (BC-AB)::Tan i (BAC + BCA) : Tan -I- (BAG - BCA) = i7° 33' 50" The angle BAC is 55'' 3' 50" Then Sin BAC : BC::Sin ABC : AC' = 879the distance. AndDAC=BCA=19° 56 iO" the course. Example IL A boat moving through the water at the rate of five miles an hour, is endeavouring to make a certain point lying S, 22i° W. while the tide is running S. 78f° E. three miles an hour. In what direction must the boat be steered, to reach the point by a single course .'' Ans. S. 58° 33' W. 89. But the most simple method of making the calculation for the effect of a current, in common cases, especially in re- solving a traverse, is to consider the direction and rate of the current as an additional separate course and distance; and to find the corresponding departure and difference of latitude. A boat sailing from A (Fig. 28.) by the united action of the wind and current, will arrive at the same point, as if it were first carried by the wind alone from A to B, and then by the current alone from B to C. Example I. A ship sails S. 17° B. for 2 hours, at the rate of S miles HADLEY'S QUADRANT. 51 an hour ; then S. 18° VV. for 4 hours, at the rate of 7 miles an hour ; and during the whole time, a current sets N, 76° W. at the rate of 2 miles an hour. Required the direct course and distance. First Course ;Second do. Current S. 17° E. S. 18° \V. N. 76° W. Di,t, N. S. E. 4.68 W. 16 28 1 2 2.9 \b.3 26.6 8.65 11.64 20.29 4.68 41.9 2.9 Dif. Lat .39. D,p. 15.61 The course is 2*° 48' 50", and the distance 42 miles. Example IL A ship sails S. E. at the rate of 10 miles an hour by the lo^, in a current setting E. N. E. at the rate of 5 miles an hour. What, is her true course ? and what will be her dis- tance at the end of two hours ? The course is 66° IJ', and the distance 25.56 miles. V. HaDLLY'S QuADltANT. 90. In the preceding sections, has been particularly ex- plained the process of determining the place of a ship from her course and distance, as given by the compass and the log. But this is subject to so many sources cf errour, from variable winds, irregular currents, lee-way, uncertainty of the magnetic needle, &:c. that it ought not to be depended on, except for short distances, and in circumstances which forbid the use of more unerring methods. The mariner who hopes to cross the ocean with safety, must place his chief reliance, for a knowledge of his true situation from time to time, oa observatiofjs of the heavenly bodies. By these the latitude and longitude may be generally ascertained, with a sufficient degree of exactness. It belongs to astronomy to explain the methods of making the calculations. The subject will not be anticipated in this place, any farther than to give a de- scription of the quadrant of refl.xion. commonly called Had- hifs Quadrant ^^ by which the altitudes of the heaveniy bo- *See Note H. b2 NAVIGATION. dies, and their distances from each other, are usually measur-* ed at sea. The superiority of this, over most other astrono- mical instruments, for the purposes of navigation, is owing to the fact, that the observations which are made with it, are Jiot materially affected by the motion of the vessel, 91. In explaining the construction and use of this quadrant, it will be necessary to take for granted the following simple principles of Optics. 1. The progress of light when it is not obstructed, or turn- ed from its natural course by the influence of some contigu- ous body, is in n^A^ ^m€5. Hence a minute portion of light called a ray, may be properly represented by a line. 2. Any object appears in the direction in which the light from that object strikes the eye. If the light is not made to deviate from a right line, the object appears in the direction in which it really is. But if the light is reflected, as by a common mirror, the object appears not in its true situation, but in the direction of the glass, from which the light comes to the eye. 3. TTie angle of reflection is equal to the angle of incidence; that is, the angles which the reflected and the incident rays make with the surface of the mirror, are equal ; as are also the angles which they make with a perpendicular to the mir- ror. 92. From these principles is derived the following propo- sition ; li^cn light is reflected by two mirrors successively, the angle which the last reflpcted ray makes with the incident ray, is DOUBLE the angle between the mirrors. If C and D(Fig. 29.) be the two mirrors, a ray of light coming from A to C, will be reflected so as to make the an- gle DCM = ACB ; and will be again reflected at D, making HDM=CDE. Continue EC and ED to H, draw DG pa- rallel to BH, and continue AC to P. Then is CPM the an- gle which the last reflected ray DP makes with the incident ray AC ; and DHM is the angle between the mirrors. By the preceding article, with Euc. 29. I and 15. 1, GDC=DCM=ACB=PCM And HDM=EDC=EDG-hGDC=DH]M-fPCM But by Euc. 32. I and 15. 1, rPM+PCM=DHM-fHDM=2DHM-{-PCM Therefore CPM=2DHM HADLEY'S QUADRANT. 53 Cor. 1. If the two mirrors make an angle of a certain num- ber of degrees, the apparent direction of the object will be changed twice as many degrees. The object at A, seen by the eye at P, without any mirror, would appear in the direc- tion PA. But after reflection from the two mirrors, the light comes to the eye in the direction DP, and the apparent place of the object is changed from A to R. Cor. 2. If the two mirrors be parallel^ they will make no alteration m the appirent p'act? of the object. 93. The principal parts of Hadley's Quadrant an; the fol- lowing ; 1. A graduated arc kB {V\g. 17.) connected with the ra- dii AC and BC. 2. An index CD, one end of which is fixed at the centre C, while the other end moves over the graduated arc. 3. A plane mirror called the index glass^ attached to the index at C. Its plane passes through the centre of motion C, and is perpendicular to the plane of the instrument ; that is, to the plane which passes through the graduated arc, and its centre C. 4. Two other plane mirrors at E and M, called horizon glasses. Each of these is also perpendicular to the plane of the instrument. The one at E, called the/ore horizon glass, is placed parallel to the index glass when the index is at 0. The other, called the back horizon glass, is perpendicularto the first and to the index at 0. This is only used occasional- ly, when circumstances render it difficult to take a good ob- servation with the other. A parf of each of these glasses is covered with quicksilver, so as to act as a mirror ; while another part is left transpar-' cnt, through which objects may be seen in their true situation' 5. Two sight vanes at G and L, standing perpendicular to the plane of the instrument. At one of these, the eye is placed to view the object, by looking on the opposite hori- zon glass. In the fore sight vane at G, there are two perfo- rations, one directly opposite the transparent part of the fore horizon glass, the other opposite the silvered part. The back sight vane at L has only one perforation, which is opposite the centre of the transparent part of the back horizon glass. 6. Coloured glasses to prevent the eye from being injurc^i by the dazzling light of the sun. These are placed at H, be- 54 NAVIGATION. tween the index mirror and the fore horizon glass. They may be taken out when necessary, and placed at N between the index mirror and the back horizon glass. 94. This instrument, which is in form an octant, is called a quadrant, because the graduation extends to 90 degrees, al- though the arc on which these degrees are marked is only the eighth part of a circle. The light coming from the ob- ject is first reflected by the index glass C, (Fig. J 7.) and thrown upon the horizon glass E, by which it is reflected to the eye at G. if the index be biought to 0, so as to make the index glass and the horizon glass parallel ; the object will appear in its true situation. (Art. 92. Cor. 2.) But if the index glass be turned, so as to make with the horizon glass an angle of a certain number of degrees ; the apparent direction of the object will be changed twice as many degrees* Now the graduation is adapted to the apparent change in the situation of the object, and not to the motion of the in- dex. If the index move over 45 degrees, it will alter the ap- parent place of the object 90 degrees. The arc is common- ly graduated a short distance on the other side of towards P. This partis called the arc of excess, 95. The quadrant is used at sea, to measure the angular distances of the heavenly bodies from each other, and their elevations above the horizon. One of the objects is seen in its true situation, by looking throuph the transparent part of the horizon glass. The other is seen by reflection, by look- ing on the silvered part of the same glass. By turning the index, the apparent place of the latler may be changed, till it is brought in contact with tiie other. The motion of the index whicliis necessary to produce this change, determines the distance of the two objects.* 9G. To find the distance of the moon from a star. Hold the quadrant so that its plane shall pass through the two objects. Look at the star through the transparent part of the horizon glass, and then turn the index till the nearest edge of th ' im- age of the moon is brought in contact with the star. This will measure the distance between the star and one edcre of tlie moon. By adding the semi-diameter of the moon, we shall have the distance of its centre from the star- * For the a-f/Ui/z/icn/s of the quadrant, see V'ince's Practical Astronomy, Maokay's Navigation, or Bowditch's Practical Navigator. HADLEY'S QUADRANT. 56 The distance of the sun from the moon, or the distance of two stars fronn each other, nnay be measured in a similar man- ner. 97. To measure the altitude of the sun above the horizon. Hold the instrument so that its plane shall pass through the sun. ai d be perpendicular to the horizon Then move the index till the lower edge of the image of the sun is brought in contact with the horizon, as seen through the transparent part of the glass. The altitude of any other heavenly body may be taken in the same manner. 98. To measure altitudes by the back observation. When the index stands at o. the index glass is at right angles with the back horizon glass. (Art. 93.) The apparent place of the object, as seen by reflection from this glass, must therefore be changed ISO degrees; (Art. 92 Cor. 1.) that is. it must ap- pear in the opposite point of the heavens. In taking altitudes by the back observation, if the object is in the east, the ob- server faces the west ; or if it be in the south, he faces the north ; and moves the index, till the image formed by reflex- ion is brought down to the horizon. This method is resorted to, when the view of the horizoR in the direction of the object is obstructed by fog, hills, &ic. 99. Dip or Depression of the Horizon. In taking the alti- tude of a heavenly bodyat sea, with Hadley's Quadrant, the reflected image of the object is made to coincide with the most distant visible part of the surface of the ocean. A plane passing through the eye of the observer, and thus touching the ocean, is called {\\q marine horizon o{ i\\e place of obser- vation If BAB' (Fig. 13.) be the surface of the ocean, and the observation be made at T, the marine horizon is TA. But this is different from the true horizon at T, because the eye is elevated above the surface. Considering the earth as a sphere, of which C is the centre, the true horizon is TH perpendicular to TC The marine horizon TA falls below this. The angle ATH is called the dip or depression of the horizon. This varies with the height of the eye above the surface. Allowance must be made for it, in observations for determining the altitude of a heavenly body above ihe true horizon. In the right angled triangle ATC, the angle ACT is equal to the angle of depression ATH ; for each is the complement h6 NAVIGATION. of ATC. The side AC is the semi-diameter of the earth, and the hypothenuse CT is equal to the same semi-diameter added to BT the height of the eye. Then AC : R: :TC : Sec ACT=ATH the depression.* 100. Artificial Horizon. Hadley's Quadrant is particular- ly adapted to measurinji; altitudes at sea. But it may be made to answer the same purpose on land, by means of what is called an artificial horizon. This is the level surface of some fluid which can be kept perfectly smooth. Water will answer, if it can be protected from the action of the wind, by a covering of thin glass or talc which will not sensibly change the direction of the rays of light. But quicksilver, Barbadoes tar, or clear molasses, will not be so liable to be disturbed by the wind. A small vessel containing one of these substances, is placed in such a situation that the object whose altitude is to betaken may be reflected from the sur- hce. As this surface is in the plane of the horizon, and as the angles of incidence and reflection are equal, (Art. 91.) the image seen in the fluid must appear as far below the hori- zon, as the object is above. The distance of the two will, therefore, be double the altitude of the latter. This distance may be measured with the quadrant, by turning the index so as to bring the image formed by the instrument to coincide with that formed by the artificial horizon. 101. 77ie Sextant IS SL more perfect instrument than the quadrant, though constructed upon the same principle. Its arc is the sixth part of a circle, and is graduated to 120 de- grees. In the place of the sight vane, there is a small tele- scope for viewing the image. There is also a magnifying glass, for reading off* the degrees and minutes. It is common- ly made with more exactness than the quadrant, and is bet- ter fitted for nice observations, particularly for determining longitude, by the angular distances of the heavenly bodies. A still more accurate instrument for the purpose is the Circle of Reflexion. For a description of this, see Borda on the Circle of Reflexion, Rees' Cyclopedia, and Bowditch's Bractical Navigator. * See Note I, and Table II. SURVEYING. SECTION I. SURVEYING A FIELD BY MEASURING ROUND IT. \rt. 105. T^^ "^^^^ common method of surveying a fieid is to measure the length of each of the sides, and the angles which they make with the meridian. The lines are usually measured with a chain, and the angles with a compass. 106. The Compass, The essential parts of a Surveyor's Compass are a graduated circle, a magnetic needle, and sight- lioles for taking the direction of any object. There are fre- quently added a spirit level, a small telescope, and other ap- pendages. The instrument is called a Theodolite, Circum- lercntor, &zc. according to the particular construction, and the uses to which it is applied. For measuring the angles which the sides of a field make with each other, a graduated circle with sights would be suf- ficient. But a needle is commonly used for determining the position of the several lines with respect to the meridian. This is important in running boundaries, drawing deeds, (fee It is true, the needle does not often point directly north or south. But allowance may be made for the variation, when this has been determined by observation. See Sec. V, 107. The Chain, The Surveyor's or Gunter's chain is four rods long, and is divided into 100 li?iks. Sometimes a half chain is used, containing 50 links. A rod, pole, or perch, is 161 feet. Hence 1 Link =7.92 inches = |ofa foot nearly. 1 Rod = 25 links =16-'- feet. 1 Chain = 100 links =06 feet. 58 SURVEYING. 108. The measuring unit for the^rra of a field Is \hc acre, which coiitain? 160 square rods. If then (he contents iu square rod? be divided by 160, the quotient will he (he num- ber of acres. P/Ut it is commonly most convenient to make the computation for the area in square chains or links, which are decimals of an acre. For a square chain =4X4 = 16 squnre rods, which is the tenth part of an acre. And a square link =T^oXTio = Toiroe of » ^q^^re chain =-j-^^u__ of an acre. Or thus, 625 links, or 272{ feet = 1 square rod, 10000 4356 = 1 chain or 16 rods, 250C0 10890 = 1 rood or 40 rods, 100000 43560 = 1 acre or 160 rods. 109. The contents, then, being calculated in chains and links ; if/oM/ places of decimals be cut off, the remaining fig- ures will be square chains ; or \i five places be cut off, the remaining figures will be acres. Thus the square of 16.32 chains, or 1632 links, is 2663424 square links, or 266.3424 square chains, or 26 63424 acres. If the contents be con- sidered as square chains and decimals, removing the decimal point one place to the left will give the acres. 1 10. In surveying a piece of land, and calculating its con- tents, it is necessary, in all common cases, to suppose it to be reduced to a horizontal level. If a hill, or any uneven piece of ground, is bought and sold ; the quantity is compu- ted, not from the irregular surface, but from the level base on which the whole may be considered as resting. In run- ning the lines, therefore, it is necessary to reduce (hem to a level. Unless this is done, a correct plan of the survey can never be exhibited on paper. If a line be measured upon an ascent which is a regular plane, though oblique to the horizon ; the length of the cor- responding level base may be found, by taking the angle of elevation. Let AB (Fig. 30.) be parallel to the horizon, BC perpen- dicular to AB, and AC a line measured on the side of a hill. Then, the angle of elevation at A being taken with a quad- rant, (Art. 4.) R : Cos A::AC : AB, that is, .^.'^ radius, to the cosine of the angle of elevation ; So is the oblique line measured} to the corresponding hori- .zontnl base. SURVEYING. 59 if (he chain, instead of being carried parallel to the surface of the ground, be kept constantly parallel to the horizon ; the line thus measured will be the base line required. The line AB (Fig. 50.) is evidently equal to the sum of the parallel lines abf cd, and eC. Plotting a Survey. 111. When the sides of a field are measured, and their bearings taken, it is easy to lay down a plan of it on paper. A north and south line is drawn, and with a line of chords, a protractor, or a sector, an ancjle is laid off, equal to the an- gle which the first ?ide of the field makes with the meridian, and the length of the side is taken from a scale of equal oarts. (Trig. 15G — ir.l.) Through the extremity of this, a serood meridian is drawn parallel to the first, and another side i;- laid down ; from the end of this, a third side, &,c. till the plan is completed. Or the plot may be constructed in the sane manner as a traverse in navigation. (Art. 7o.) If the field is correctly surveyed and plotted, it is evident the extremity of the last side must coincide with the beginnmg of the first. Example I. Draw a plan ( of a field, from the following courses and dis fances, as noted in the field-book ; Ch. Links. 1. N. 7G° E. 2 46 2. S. 1G° W. 3 54 3. N. 8.3° W. 2 72 4. N. 12° E. 2 13 5. N. 60i°E. 95 Let A (Fig. 31 )be the first cornerof the field. Thro' A, drawthemerid N-S, make B AN = 78^ &: \B = 2.4G Thro' B,draw N'S' par. to NS,make S'BC=16°,&; BC = .S.54 Thro' C,drawN"S''par. toNS, makeDCN = 83°,ii CD=2.72 &c. &c. 112. To avoid the inconvenience of drawing parallel lines, the sides of a field may be laid down from the ans^les v.hich they make ivith each other, instead of the angles which they make with the meridian. The position of the line BC (Fig. 31.) is determined by the angle ABC, as well as by the an- gle S'BC. When the several courses are given, the angles 60 :^LKV EYING. which any two conliguous sides make with each other, may be known by the following rules. 1. Ifone course is North and the other South, one East and the other West ; subtract the less from the greater, 2. Ifone is North and the other South, but both East or West ; add them together, 3. If both are North or South, but one East and the other West ^subtract their sum from ISO degrees, 4. If both are North or South, and both East or West ; add together 90 degrees, the less course^ and the complement of the greater. The reason of these rules will be evident by applying thenn to the preceding example. (Fi^. 31.) The first course is BAN, which is equal to ABS'. (Euc. 29.1.) If from this the second course CBS' be subtracted, there will remain the an^jle ABC. If the second course CBS', or its equal BCN", be added to the third course DCN ; the sum will be the angle BCD. The sum of the angles CDS, NDE, and CDE, is 180 de- grces. (Euc. 13.1.) If then the two first be subtracted from 180 degrees, the remainder will be the angle CDE. Lastly, let EP be perpendicular to NS. Then the sum of the angles DES, PES, and AEP the complement of AEN, is equal to the angle DEA. We have then the angle ABC=G2°, DEA = 131i°, BCD=&9°, EAB = 162i°. CDE=85°, With these angles, the field may be plotted without draw- ing parallels, as in Trig. 173. Finding the Contents of a Field. 113. There are in common use two methods of finding the contents of a piece of land, one by dividing the plot into tri- angles, the other by calculating the departure and dijfer^nce of latitude for each of the sides. When a survey is plotted, the whole figure may be divided into triangles, by drawing diagonals from the different an- gles. The lengths of the diagonals, and of the perpendicu- lars on the bases of the triangles, may be measured on the same scale of equal parts from which the sides of the field were laid down. The area of each of the triangles is equal SURVEYING. 6.1 (o half the product of its base and perpendicular; and their sum is the area of the whole figure. (Mens. 13.) Example I, Let the plan Fig. 32 be the same as Fig. 31, the sides of which, with their bearing?, are given in art. 111. Then the triangle ABC = BCx^AP =3.84 sq. chains. ACE = ACxiEP'=1.53 DCE = ECxiDP"=2.89 The contents of the whole = 8.26 1 1 4. This method cannot be relied on, where great accur- acy is required, if the lines are measured by a scale and com- passes only. But the parts of the several triangles may be found by trigonometrical calculation^ independently of the projectiorj ; and then the area of each may be computed, either from two sides and the included angle, or from the three sides. (Mens. 9, 10.) The sides of the iield and their bearings being given by the survey, the angles of the original (igure may all be known. (Art. 112.) Tiicn in the triangle ABC (Fig. 32.) we have the sides AB and BC, with the angle ABC, to find the other parts. (Trig. 153.) And in the triangle CDE, we have the sides DC and DE, with the angle CDE. Subtracting the angle BAC from BAE, we shall have CAE ; and subtracting DEC from DEA, we shall have CEA. There will then be given, in the triangle ACE, the side EA and the angles. (Trig. 150.) The sides and bearings, as given in art. Ill, are 1. AB N. 70° E. 2.4G chains. 2. BC S. 16 VV.3.54 3. CD N 03 W. 2.72 4. DE N. 12 E. 2.13 5. EA N. GO}yE, 0.95 Then by Mensuration, art. 9, R : Sin ABC::ABXBC : 2 are« ABC = 7.69 sq. chains R : Sin AEC::AExEC : 2 area AEC=3.0G R : Sin CDE: :CDxDE : 2 area CDE =5.77 2)16.52 Contents of the whole field, 8.26 62 SURVEYING. Or the areas of tlie several triant^les may be found by the rule in Mensuration, art. 10, viz. If a, b, and c, be the sides of any triangle, and /t=ha!f their sum ; The area =\/hX{h — u)X{h- b)X{h — c) Example 11. Courses. Ch. Links. 1. E. 2G 34 2. S. 10° 30' E. 32 26 3. N.42 W. 13 35 4. S. 58 W. 23 52 5. N. 30 55 Contents of the field, 69.735 acres. The method which has been explained, of ascertaining the contents of a piece of land by dividing it into triangles, is of use in cases which do not require a greater degree of accura- cy, than can be obtained by the scale and compasses. But if the areas of the triangles are to be found by trigonometri- cal calculation, the process becomes too laborious for com- mon practice. The following method is often to be preferred. Finding the area of a field by departure and differ- ence OF latitude. 115. Let ABCDE (Fig. 33.) be the boundary of a field. At a given distance from A, draw the meridian line NS. Parallel to this draw L'R', AG, BH, and DK. These may be considered as portions of meridians passing through the points A, B, D, and E. For all the meridians which cross a field of moderate dimensions, may be supposed to be paral- lel, without sensible errour. At right angles to ISSdraw the parallels AL, BM, CO, EP,and DR. These will divide the figure LABCDR into the three trapezoids ABML, BCOM, and CDRO ; and the figure LAEDR into the two trapezoids DEPR and EALP. The area of the field is evidently equal to the difference between these two figures. The sum of the parallel sides of a trapezoid, multiplied in- to their distance, is equal to twice the area. (Mens. 12.) Thus (AL-\-BM)x AG =2 area ABML. Now AL is a given distance, and BM=AL-f BG. But BG is the departure^ and AQ the difference of latitude, cor- SURVEYING. 63 responding to AB one of the sides of the field. (Arts. 39, 40.) And bj art 44, r> 1 T^- J. A t> . . ^ Sin BAG : Depart. BG Rad : D.st. AB. . | ^^^ ^^q . jj.^ Lat. AG Or the departure and difference of latitude may be taken from the Traverse Table, as in Navigation. (Art. 50.) In the same manner, from the sides BC, CD, DE, and EA, may be found the deparlnre CH, CK. DR', AL'. and the dif- ferences of latitude BH, DK, ER', and EL'. We shall then have the parallel sides of each of the trapezoids, or the dis- tances of the several corners of the field from the meridian NS. For BM = AL4-BG, DR=CO-CK, CO=BM+CH, EP=DR-DR'. If the field be measured in the direction ABCDE, the dif- ferences of latitude AG, BH, and DK, will be SouthingSj v/hile R'E and E'L will be Korihings, The former are the breadths of the three trapezoids which form the figure LABCDR ; and the latter are the breadths of the two trape- zoids which form the figure LAEDR The difference, then between the sum of the products of the northings into the cor- responding meridian distances, and the sum of the products of the southings into the corresponding meridian distances, is twice the area of the field. It will very much facilitate the calculation, to place in a fable the several courses, distances, northings, southings, &c We have, then, the following Rule. IIG. Find the northing or southing, and the easting or west- i.ng, for each side of the field, and place them in distinct col" urnns in a table. To these add a column of Meridian Distan- ces,for the distance of one end of each side of the field from a given meridian ; a column of Multipliers to contain the pairs of meridian distances for the two ends of each of the sides ^ and columns for the north and south Areas, See Fig. 23, and the table for example 1. Suppose a meridian line to be drawn without the field, at any given distance from the first station ; and place the assumed distance at the head of the column of Meridian Distances, To this add the first departure^ if both be east or both io.cst ; but ,'^ifbtract, if one be east and the other west ; and place the sum 64 SUKVEYING. or difference in ike column of Meridian Distances^ against the first course. To or from the last numhcr^ add or subtract the second departure^ ^c. 4'c. For the column of Multipliers^ add together the first and 5f- cond numbers in the column of Meridian Distances ; the se- cond and thirds the third and fourth j ^c. placing the sums op- posite the several courses. Mullipli^ each 7iumber in the column of Multipliers into its corresponding northing or southing, aiid place the product in the column oj north or south areas. The difference between the sum of the north areas, and the sum of the south areas, zoill he twice the area of the field. This method of finding the contents of a field, as it depends on departure and difference of latitude, which are calculated by right-angled trigonometry, is sometimes called Rectangular Surveying. 117. If the assumed meridian pass through the eastern or western extiemity of the field, as L'ER' (Fig. 33.) the dis- tance EP will be reduced to nothing, and the figures AEL' and EDR' will be triangles instead of trapezoids. If the sur- vey be made to begin at the point E, cipher is to be placed at the head of the column of meridian distances, and the first number in the column of multiplier? will be the same, as the first in the column of meridian distances. See example II. 1 18. When there is a re-entering angle in a field, situated with respect to the meridian as CDE ; (Fig. 34.) the area EDM, being included in the tigure BCRA, will be repeated in the column of south areas. But, as it is also included in the figure DCRM, it will be contained in the column of north areas. Therefore the difference between the north areas and the south areas, will be twice the area of the field, in this case, as well as in others. 119. If any side is directly east or loest, there will be no difference of latitude, and consequently no number to be placed against this course, in the columns of north and south areas. See example 11. Course 1. AB (Fig. 34.) The number in the columns of areas will be wanting also, when any side of the field coincides with the assumed roe rrdian. See example II. Conne 5. EA (Fig. 34.) SURVEYING. 65 120. In finding the departure and difference of latitude from the traverse table, the numbers for the links may be looked out separately; care bcin^ taken to remove the deci- mal point two places to the left, because a link is the 100th part of a chain. Thus if the course be 29°, and the distance 23.46 chains ; The dif. of lat. &;depart. for 23 chains are 20. 1 2 and 11.15 for 46 links .40 .22 for 23.46 20.52 11.3: Example I, See Fig. 33. Covrus. Ditt. Dif. Lot Departure. M D. A L .0 E. Mult l^ ^tat o . 1 N. i S. E : w. it. BAG S. 64° E. AB 30 cA. I AG 1 13 15 , GB 26. 9f. BM 46.96 ALtBM 66.96 .Ai«.UL 880 5346 2. CBH S. 14° K BC 10 1 BH i 9.70 HC 2.42 CO 493-< BM+CO 96 34 2B.M0C 934 4980 3. CDK S. 35° W CD 30 KD 24.57 1 CK 17 21 DR 82.(7 COfDR 8i 55 ;;CDRO 2003 •:83i( 4. KDE N. 65° W. DE R'E 1 8.45 1 DR' 18.13 EP 14.04 DR+EP 2 DEPR 390.4715 ¥ 5. L'EA N. 8«42'E EA 39.42 EL' ; 38.97 : LA 5. 96 AL JO EP+AL 34.04 1 2EPLA 1 1326 5388 i7^2 47 42 35.34 35JJ 1 J|717.ll3:i 3818.7055 Twice the figure ABCDRL is 3818.7055 square chains ; Twice the figure AEDRL 1717.1133 The difference 2101.5922 The contents of the field 1050.7961 sq. ch. or 105.0796 acres. (Art. 109.) 10 o6 SURVEViNG. Example IL See Fig. S-i, Couries. /)«/. Dijf. Lai. N. 1 S. Depaiiurt. E. 1 VV. M. Dist. 00 Mult. N.Areas. S. Areas 1. E. AC 26.34 00 13.G4 30.55 44,19 1 00 31.72 12.47 44.19 26.34 5.8r 00 1 12.27 19.95 00 AB 26.34 CR 32.22 DM 19.95 00 00 00 00 00 2.S.10rE. BC 32.26 AB+CR 58.56 CR+PM 52.17 2ABCR 1857.5232 3. N.42^VV. CD 18.3.3 2CD.\^R 711.598» 4.S.58"\V. DE 23.52 DM 19.95 2DME 248.7765 5 i\. EA 30.55 00 1 1 711.5988 2106.29971 The contents of the field =^(2106.3-71 1.6)=:697..35 sq. ch. Or 69.735 acres. In this example, the meridian distance of the first station A being nothing, cipher is placed at the head of the column of meridian distances. (Art. 117.) The first side AB being directly east and west, has no difference of latitude, and therefore the numbcrin the column of areas against this course is wanting, as it is against the fifth course, which is directly north. (Art. 119.) The number against the fourth course, in the column of multipliers, is only the length of the line DM; the figure DME being a triangle^ instead of a trape- zoid. Example III, Find the contents of a field bounded by the following lines; 1. N. SS'' 30' E. 15 ch. 50Hnks. 2. N. 72 45 E. 18 70 3. S. 70 45 E. 18 70 4. S. 53 W. 12 45 5. S. 83 15 E. 24 10 6. S. 31 15 W.15 20 7. S. 62 45 W. 22 60 8. N. 73 30 VV.27 30 9. N. 17 25 W. 14 50 The area is 145^ acres. S.URVEYING. ,37 \2l. When a field is correctly surveyed, and the depar- tures and differences of latitude accurately calculated; it ij^ evident the sum of the northings must be equal to the sum of the southings, and the sum of the eastings equal to the sum of the westings. Jf upon adding up the numbers in the de- parture and latitude columns, the northings are not found to agree nearly with the southings, ;.nd the eastings with the westings, there must be an crrour, either in the survey or in the calculation, which requires that one or both should be revised. But if the difference be small, and if there be no particular reason for supposing it to be occasioned by one part of the survey rather than another ' it may be apportion- ed among the several departures or differences of latitude, according to the different lengths of the sides of the tield, by the following rule ; As the whole perimeter of the field, To the whole errour in departure or latitude ; So is the length of one of the sides, To the correction in the corresponding departure or latitude, This correction, if applied to the column in whieh the sum of the numbers is too small, is to be added ^ but if to the oth> er column, it is to be subtracted,* See the example on the next page. *Seo the fourth Number of the \x)H]yst publi?hpd of Phflaflelphin. ■—^•1- —i B •n - i -vj Oi Or ^ 03 ♦O — tz; fy3 ^ CA^ c/D c/3 !2; •0 to 03 ►t*' to C;t 0*.|«*|- 0,0- *.i- 00 00 j^ 5=1 < ?i < w p Courses. ; 1^ CO to !^ ^ lu ^ Ml- b»|- M|- 2 2. b OS c -t 03 Ci to — p 03 p CO Cr to GO Oi OJ Oi CD *t^ 6.70 8.43 13.96 5.85 -1 o c to CO to CO •— to ^ b CO ^ ►;^ -v) CO to ■i p «o JO ;;-J to -vl ot iii. b b OS CO -J i 1 i CO I -h 1 1 1 1 j t b b b b b b b •4 00 »f^ Cji fc^ Cjt 05 rp r* r' 1 1 1 4-1 ff c> '0 '0 h (D '0 c> CO fi^ o- cr. >f^ --■* 00 bo 0:1 to — p 03 p Id in 03 ? ^ b CO b 03 b to — CO Cj' K3 CO Id ^ CO CO J:^ -v> , 1 CO CO 7.03 11.95 7.45 2.48 KJ ^0 to — Q JO CO -- p ^ rfi- I^- CO CO <> CO CO CO 0. CO CO H- ^ »-* oo Oi' ku t^i. — to to 1— »0 CO fO ^^ ii. 4:^. CO b 6^ b CO CO — — C5 CO CO -J 1 ♦* Oj b to 153.46 112.61 51.85 r^ C/0 b 00 o» to Cjt 4:1. to ^ CO CO JO 03 ^ 00 to ^ b c> 03 -J to 6* SURVEYING. ^9 In this example, the whole perimeter of the field is IOO4. chains, the whole errour in latitude .34, the whole errour in departure .42. and the length of the first side 18. To find ;he corresponding errours, lOOi 'IS' * \'*^^ ' '^^ ^^^® errour in latitude, 2 * ' ' 1 .42 : .08 the errour in departure. The errour in latitude is to be added to 10.26 making: it 10.32, as in the column of corrected northings; and the er- rour in departura is to be added to 14.79 making it 14.87, as in the column of corrected eastings. After the corrections are made for each of the courses, the remaining part of the calculation is the same as in the precedmg examples. 122. If the length and direction of each of the sides of a field except one be given, the remaining side may be easily found by calculation. For the difference between the sum of the northings and the sum of the southing? of the ?iven sides, is evidently equal to the northing or southing of the re- maining side ; and the difference between the sum of the eastings and the sum of the westings of the given sides, is equal to the easting or westing of the remaining side. Hav- ing then the difference of latitude and departure lor the side required, its length and direction may be found, in the same manner as in the sixth case of plane sailing. (Art. 49.) Example V, What is the area of a field of six sides, of which five are <^iven, viz. 1. S. 56° E. 4,18 chains 2. N. 21 E. 480 3. N. 56 W. 3.06 4. S. 21 W. 0.13 5. 6. Th( N. I ai 66«i W. 1.44 ea is two acres. Exampl e F7. 1. N 38° W. 17.21 chains 2. N. 13 E. 21.16 3. N 72 E. 24.11 4. S. 41 E. 19 26 5. S. 11 W. 2435 6. 70 SURVEYIN«. 123. Plotting hy departure and difference of Latitude. A survey may be easily plotted from the northings and south- ings, eastings and westings. For this purpose, the column of Meridian Distances is used. It will be convenient to add al- so another column, containing the distance of each station from a given parallel of latitude, and /ormed by adding the northings and subtracting the southings, or adding the south- ings and subtracting the northings. Let AT (Fig. 33.) be a parallel of latitude passing through the first station of the field. Then the southing TB or LM is the distance of B, the second station, from the given parallel. To this adding the southing BH, we have LO the distance of CO from LT. Proceeding in this manner for each of the sides of the field, and copying the 7th column in the table, p. 65, we have the following differences of latitude and meridian distances. Diff. Lot, Merid, BisU AL 20 1. LM 13.15 BM 46.9G 2. LO 22.85 CO 49 38 3. LR 47 42 DR 32.17 4. LP 38.97 EP 14.04 To plot the field, draw the meridian NS, and perpendicu- lar to this, ibe parallel of latitude LT. From L set off the differences of latitude LM, LO, LR, and LP. Through L, M, O, R, and P, draw lines parallel to LT; and set off the meridian distances AL, BM, CO, DR, and EP. The points A, B, C, D, and E, will then be given. 124. When a field is a regular figure, cs a parallelogram, triangle, circle, &ic. the contents may be found by the rules in Mensuration, Sec. L and IL 125. The area of a field which has been plotted, is some- times found by reducing the whole to a TiiiANfiLB of the same area. This is done by changing the figure in such a manner as, at each step, to make the number of sides one less, till they are reduced to three. Let the side AB (Fig. 35.) be extended indefinitely both ways. To reduce the two sides BC and CD to one, draw a line from D to B, and another parallel to this from C, to in- tersect AB continued. Draw also a line from D to the point of intersection G. Then the triangles DBC and DBG are SURVEYING. 7i equal. (Euc. 37. 1.) Taking from cacli the common part DBH, there remains BGH equal to DCH. If then the triangle DCH be thrown out of the plot, and BGH be added, we shall have the five-sided figure AGDEF equal to the six- sided figure ABCDEF. In the same manner, the line EL may be substituted for the two sides AF and EF ; and then DM, for EL and ED This will reduce the whole to the triangle MGD, which is equal to the original figure. The area of the trian- gle may then be found by multiplying its base into half its height; and this will be the contents of the field. In practice it will not be necessary actually to draw the parallel lines BD, GC, he. It will be sufficient to lay the edge of a rule on C, so as to be parallel to a line supposed to pass through B and D, and to mark the point of intersec- tion G. 126. If after a field has been surreyed, and the area com- puted, the chain is found to be too long or too short ; the true contents may be found, upon the principle that similar figures are to each other as the squares of their homologous sides. (Euc. 20.6.) The proportion may be stated thus ; As the square of the true chain, to the square of that by which the survey was made ; So is the computed area of the field, to the true area. Ex If the area of a field measured by a chain 66.4 feet Jong, be computed to be 32.6036 acres ; what is the area as measured by the true chain 66 leet long.'* Ans. 33 acres. 127. A plot of a field may be changed to a different scale, that is, it may be enlarged or diminished in any given ratio, by drawing lines parallel to each of the sides of the original plan. To enlarge the perimeter of the figure ABCDE (Fig. 36.) in the ratio of aG to AG ; draw lines from G through each of the angular points. Then beginning at n, drawai parallel to AB, &c parallel to BC, &c. 72 SURVEYING. It is evident that the Gtigles are the same in the enlarged figure, as in the original one. And by similar triangles, AG :aG::BG : 6G::CG : cG::&c. And AG : aG::AB : ab::BC :bc::hc. Therefore ABCDE and abode are similar figures. (Euc. Def 1 6.) In the same manner, the smaller figure a'b'dd'e' may be drawn, so as to have its perimeter proportioned to ABCDE as a'G to AG. SECTION n. Methods of Surveying in particular cases, A i^ft T^MEASURING round a field, in the manner IT JL explained in the preceding section, is by far the most common method of surveying. The following problems are sometimes useful. They may serve to verify or correct the surveys which are made by the usual method. Problem I. To survey a field from two stations. 129. Find the distance of the two stations, and THEIR bearings FROM EACH OTHER ; THEN TAKE THE BEAR INGS OF THE SEVERAL CORNERS OF THE FIELD FROM EACH OF THE STATIONS. In the field ABCDE, (Fig. 37.) let the distance of the two Stations S and T be given, and their bearings from each other. By taking the bearing of A from S and T, or the angles AST and ATS, we have the direction of the lines drawn from the two stations to one of the corners of the field. The point A is determined by the intersection of these lines. In the same manner, the point B is determined, by the inter- section of SB and TB; the point C, by the intersection of SC and TC ; &c. &tc. The sides of the field are then laid down, by connecting the points ABCD, &£c. The area is obtained, by finding the areas of the several triangles into which the field is divided by lines drawn from one of the stations. Thus tlie area of AbCDE (Fig. 37.) is equal to ABT-fBCT+CDT+DET-fEAT or to ABS+BCS-fCDS + DES+EAS Now wc have the base line ST given and the angles, in the triangle AST, to find AS and AT j in the triangle BST, to find BS and BT, &ic. After these are found, we have two n 74 .^L'KVKVINCI. sides and the included angle in the triangles ABT, BCT, &:o. from which the areas may be calculated. (Mens. 9.) Example. Let the station T (Fig. 37.) be N. 80^ E. from S, the dis- tance ST 27 chains, and the bearings of the several corners of the field from S and T as follows ; TA N. 30° W. SA N. 17° E. TB N. 15 E. SB N 55 E. TC S. 53 E. SC S. 73 E. TD S. 65 W. SD S. 24 W. TE N. 70 W. SE N. 26 W. Tiiese will give the following angles j ATS= 70° AST= 63° ATB= 46* BTS = 115 BST= 25 BTC = U2 CTS = 133 CST= 27 CTD=I08 DTS= 25 DST = 124 DTE= 55 ETS=: 30 EST =106 ETA= 40 From which, with the base line ST, are calculated the fol- lowing lines and areas. AT=32.89 chains. ABT = 206.45 sq. chains BT= 17.75 BCT = 294.95 CT = 35.84 CDT = 740.7 DT=43.46 DET=665 1 ET = 37.3G EAT = 395. Contents of the field, =230.22 acres. The course and length of each of the sides of the field may be found, if necessary. After the parts mentioned above are calculated, there will be given two sides and the included an- gle, in the triangle ATB, to find AB, in BTC to find BC he. If the base line between the two stations be too short, com- pared with the sides of the field and their distances, the sur- vey will be liable to inaccuracy. It should not generally be less than one tenth of the longest straight line which can be drawn on the ground to be measured. 130. It is not necessary that the base line, from the extre- mities of which the bearings are taken, should be within the field. It may be one of the sides, or it may bo entirely with- out the field. SURVEYING. 75 Let S and T fFig. 38.) be two stations from wiilch all the corners of a field ABODE may be seen. If the direction and length of the base line be measured, and the bearings of the points A, B, C, D, and E, be taken at each of the sta- tions ; the areas of the several triangles may be found. The 6gure ABCTDE is equal to DET+EAT+ABT+BCT From this subtracting DCT, we have the area of the field ABCDE. In this manner, a piece of ground may be measured which, from natural or artificial obstructions, is inaccessible. Thus an island may be measured from the opposite bank, or an enemy's camp, from a neighbouring eminence. 131. The method of surveying by making observations from two stations, is particularly adapted to the measurement of a bay or harbour. The survey may be made on the water, by anchoring two vessels at a distance from each other, and observing from each the bearings of the several remarkable objects near the shore. Or the observations may be made from such eleva- ted situations on the land as are favourable for viewing the figure of the harbour. If all the parts of the shore cannot be seen from two stations, three or more may be taken. In this case, the direction and distance of each from one of the oth- ers should be measured. Proble :*i. To survey a field by measuring from one station. 132. Take the bearings of the several corners of THE field, and MEASIJRE THE DISTANCE OF EACH FROM THE GIVEN STATION. If the length and direction of the several lines AT, BT, CT, DT, and ET, (Fig. 37.) be ascertnined ; there will be given two sides and the included angle of each of the trian- gles ABT, BCT, CDT, DET, and EAT ; from which their areas may be calculated, (Mens. 9.) and the sum of thesQ will be the contents of the whole figure. 76 SURVEYINC:;. The station may be taken in one of the sides or angles of the field, as at C. (Fig. 32.) The lines CD, CE, CA, CB, and the angles DCE, EGA, ACB, being given, the areas of the triangles nnay be found. ^ Problem III. To survey a field by the chain alone. 133. Measure the SIDES of the field, and the DI- AGONALS BY WHICH IT IS DIVIDED INTO TRIANGLES. By measuring the sides (Fig. 32.) AB, BC, CD, DE, EA, and the diagonals CA and CE, we have the three sides of each of the triangles into which the whole figure is divided. They may therefore be constructed, (Trig. 172.) and theii areas calculated. (Mens. lO.) 134. To measure an angle with the chain, set off equal distances on the two lines which include the angle, as AB, AC, (Fig. 39.) and measure the distance from B to C. There will then be given the three sides of the isosceles triangle ABC, to firrd the angle at A by construction or calculation. The chain may be thus substituted for the compass, in sur- veying a field by going round it, according to the method explained in the preceding section ; or by measuring from one or two stations, as in problems I. and II, Problem IV. To survey an irregular boundary by means of offsets. 135. Run a straight line in ant convenient direc- tion, AND MEASURE THE PERPENDICULAR DISTANCE OF EACH ANGULAR point OF THE BOUNDARY FROM THIS LINE. The irregular field (Fig. 40.) may be surveyed, by taking the bearing and length of each of the four lines AE, EF, FI, lA, and measuring the perpendicular distances BB', CC, DD', GGr', HH', KK'. These perpendiculars are cal- led offsets, h is necessary to note in a field book the parts PURVEYING 77 into which the line that is measured is divided by the offsets^ as in the following example. (See Fig. 40.) Offsets on the left 1 Courses ctnd Distances. \ Offsetts on the right. ChaiDB. AEN.85°E. 12.74 ch. BB' 2.18 AB' 3.25 CC 2.18 BC 2.13 DI> 1.23 CD' 1.12 DE 6.24 1 EF S. 24° E. 7.23 Fl N. 87 W. 13.34 GG' 2.66 FG 3.84 HH' 1.48 G'H' 2.22 H'l 7.28 — ■ lA N. 26 VV. 5.32 IK' KK' 2.94 K'A As the offsets are perpendicular to the lines surveyed, the little spaces ABB', BB'CC, CC'DD' he. are either right an- gled triangles, parallelograms, or trapezoids. To find the contents of the field, calculate in the first place the area be- tween the lines surveyed, as the trapezium AEFIA, (Fig. 40.) and then add the spaces between the offsets, if they fall with- in the boundary line ; or subtract them, if they fall without, as AIK. When any part of a side of a field is inaccessible, equal off- sets may be made at each end, and a line run parallel to the boundary. Problem V, To measure the distance between any two points oh the sur- face of the earth, by means of a series of triangles extending from one to the other. 136. Measure a side of one of the triangles for a BASE LINE, TAKE THE BEARING of this or some oth- HR SIDE, AND MEASURE THE ANGLES IN EACH OF THE TRIANGLES. If it be required to find the distance between the two points A and I, (Fig. 41.) so situated that the measure can- not be taken in a direct line from one to the other; let a se- 78 SURVEYING. ries of triangles be arranged in such a manner between them, that one side shall be common to the first and second, as BC, to the second and ihird as CD, to the third and fourth, &tc. Then measure the length of BC for a base line, take the bearing oi" the side AB, and measure the angles of each of the triangles. These data are sufficient to determine the length and bear- ing of each of the sides, and the distance and bearing of I from A. For in the two first triangles ABC and BCD. the angles are given and the side BC, to find the other sides. When CD is found, there are given, in the third triangle CDE. one side and the angles, to find the other side. In the same manner, the calculation may be carried from one triangle to anoiher, till all the sides are found. The bpiirifigs of the sides, that is, the angles which they make with the meridian, may be determined from the bear- ing of the first side, and the angles in the several triangles. Thus if NS be parallel to AM, the angle BAP, or its equal ABN subtracted from ABD leaves NBD; and this taken from 180 degrees leaves SBD. From the bearing and length of AB may be found the southing AP, and the easliiigr PB. In the same manner are found the several southings PP, P'P", P"P"', P"'M. The sum of the southings is the line AM. And if the distance is so smfill, that the several meridians may be considered paral- lel, the difference between the sum of the eastings and the sum of the westings, is the perpendicular IM. We have then, in the right-angled triangle AMI, the sides AM and MI, to find the distance and bearing of I from A. 137. This problem is introduced here, for the purpose of giving the general outlines of those important operations which have been carried on of late years, with such admira- ble precision, under the name of Trigonometrical Surveying* Any explanation of the subject, however, which can be made in this part of the course, must bo very imperfect. In the demonstration of the problem, the several triangles are supposed to be in the same plane, and the distances of the meridians so small, that they may be considered parallel. But in practice, the ground upon which the measurement is to be made is very irregular. The stations selected for the angular points of the triangles, are such elevated parts of the country as are visible to a considerable distance. They should SURVEYING. 7;. be so situated, that a signal staff, tower, or other conspicuous object in any one of the angles, may be seen from tl)e other two angles in the same triangle. It will rarely be the case that any two of the triangles* will be in thi> same plane, or any one of them parallel to the horizon. Reductions will therefore be necessary to bring them to a common level. But even this level is not a plane In the cases in which this kind of surveying is commonly practised, the n]easnrement is carried over an extent of country of many miles. The se- veral points, when reduced to the same distance from the centre of the earth, are to be considered as belonging to a spherical surface. To make the calculations then, if the line to be measured is of any considerable extent, and if nice ex- actness is required, a knowledge of Spherical Trigonometry is necessary. 13S. The decided superiority of this method of surveying, in point of accuracy, over all others which have hitherto been tried, particularly where the extent of ground is great, is ow- ing partly to the fact that almost all the quantities measurecd in France, and glass tubes or steel chains in England. The results have in many instances been extreme- ly exact. A base was measured, on Hounslow Heath, by General Roy, with glass rods. Several years after, it was re- 80 SURVEYING. measured by Colonel Mudge, wilh a steel chain of very nice construciion. The dilFerence in the two measurements was 5css than three inches in more than five miles. Two parties measured a base in Peru ot 62T5i toises, or more than seven miles; and the difference in their results did not exceed two inches. Exact as these measurements are, the exquisite construc- tion of the instruments which have been used for taking the angles, has given to that part of the process a still higher de- gree of perfection. The amount of the errours in the angles of each of the triangles, measured by Ramsden's The- odolite, did not exceed three seconds. In the great surveys in France, the angles were taicen with nearly the same cor- rectness. 140. One of the most important applications of trigono- metrical surveying, is in meat-uring ores of the meridian, or of parallels of latitude, particularly the former. This is neces- sary in determining itie figure of the earth, a very essential problem in Geography and Astronomy. A degree of zeal has been dispiu)ed on this subject, proportioned to its prac- tical importance. Arcs of the meridian have been measured at great expense, in England, France, Lapland, Peru, &ic. Men of distinguished science have engaged in the under- taking. A meridian line has been measured, under the direction of General Roy and Colonel Mudge, from the Isle of Wight, to Clifton in the north of England, a distance of about 200 miles. Several years were occupied in this survey. An- other arc passing near Paris, has been carried quite through France, and even across a part of Spain to Barcelona. In measuring this, several distinguished mathematicians and as- tronomers were engaged for a number of years. These two arcs have been connected by a system of triangles running across the English Channel, the particular object of which was to determine the exact difference of longitude between the observatories of Greenwich and Paris. Besides the me- ridian arcs, other lines intersecting them in various directions have been measured, both in England and France. With these, the most remarkable objects over the face of the country have been so connected, that the geography of the various parts of the two kingdoms is settled, wilh a precision which could not be expected from any other method. SURVEYING. 81 141. The exactness of the surveys will be seen from a comparison of the lines of verification as actually measured, with the lengths of the same lines as determined by calcula- tion. These would be affected by the amount of all the er- rours in measuring the base lines, in taking the angles, in computing the sides of the triangles, and in making the ne- cessary reductions for the irregularities of surface. A base of verification measured on Romney Marsh in England, was found to differ but about two feet from the length of the same line, as deduced from a series of triangles extending more than 60 miles. A base of verification connected with the meridian passing through France, was found not to differ one foot from the result of a calculation which depended on the measurement of a base 400 miles distant A line of verifi- cation of more than 7 miles, on Salisbury Plain, differed scarcely an inch from the length, as computed from a system of triangles exiending to a base on Hounslow Heath.* ♦ S«« Note K. IS SECTION 111. LAYING OUT AND DIVIDING LANDS. A ' 142 nr^ ^^^°^^ ^^^° ^^^ familiar with the principles of ^ ' ' geometry, it will be unnecessary to give par- ticular rules, for all the various methods of dividing and lay- ing out lands. The following problems may serve as a spe- cimen of the manner in which the business may be conduct- ed in practice. Problem I. To lay out a given number of acres m the form of a square. 143. Reduce the number of acres to s(^uare rods ob CHAINS, and EXTllACT THE SQ,UARE ROOT. This will givC OnC side of the required field. (Mens. 7.) Ex. 1. What is the side of a square piece of land con- taining 1241 acres. Ans. 141 rods. 2. What is the side of a square field which contains 58f acres ? Problem IL To lay out a field in the form of a parallelogram, when one side and the contents are given, 144. Divide THE number of square rods or chains by THK LE^GTH OF THE GIVEN SIDE. The quotient will be a side perpendicular to the given side. (Mens. 7.) Ex. What is the width of a piece of land which is 286 rods long, and which contains 77 acres ? Ans. 44 rods. Cor. As a triangle is half a parallelogram of the same base and heiglit, a field may be laid out in the form of a triangle whose area and base are given, by dividing twice the area by the base. The quotient will be the perpendicular from the opposite angle. (Mens. 8.J SURVEYING. 83 Problem III. To lay out a p^iece of land in the form of a parallelograjn, the length of which shall be to the breadth in a given ratio. 145, As the length of the parallelogram to its breadth ; So is the area, to the area of a square of the same breadth. The side of the square may then be found by problem I, and the length of the parallelogram by problem II. If BCNM (Fig. 42.) is a square in the right parallelogram ABCD, or in the obh'que parallelogram ABC'D', it is evi- dent that AB is to MB or its equal BC, as the area of the parallelogram to that of the square. Ex. If the length of a parallelogram is to its breadth as 7 to 3, and the contents are 52^ acres, what is the length and breadth. Problem IV. TTie area of a parallelogram being given, to lay it out in such a form, that the length shall exceed the breadth by a given DIFFERENCE. 146. Let x=BC the breadth of the parallelogram ABCD (Fig. 42.) and the side of the square BCNM. J=AM the difference between the length and breadth. a=the area of the parallelogram. Then a = (x-{-d)Xx=x^-i'dx. (xMens. 4.) Reducing this equation, we have \^a^fd^- '^d=x. That is, to the area of the parallelogram, add one fourth of the square of the difference between the length and the breadth, and from the square root of the sum, subtract half the difference of the sides ; the remainder will be the breadth of the parallelogram. Ex. If four acres of land be laid out in the form of a parallelogram, the difference of whose sides is 12 rods, what is the breadth ? 84 SURVEYING. Problem V, To lay out a triangle whose area and angles are given, 147. Calculate the area of any supposed triangle WHICH has the same ANGLES. THEN As THE AREA OF THE ASSUMED TRIATJGLE, To THE AREA OF THAT WHICH IS REQUIRED ; So IS THE SQUARE OF ANY SIDE OF THE FORMER, To THE SQUARE OF THE CORRESPONDING SIDE OF THE LATTER. If the triangles B'CC and BCA (Fig. 43.) have equal an- gles, they are similar figures, and therefore their areas are as he squanjs of their like sides, for instance, as AC * CC • (Euc. 19. 6.) The square of CC being found, extracting the Square root will give the line itself To lay out a triangle of which one side and the area are given, divide twice the area by the given side ; the quotient will be the length of a perpendicular on this side from the opposite angle. (Mens. 8.) Thus twice the area of ABC (Fig. 45.) divided by the side AB, gives the length of the perpendicular CP. 148. This problem furnishes the means of cutting off, or laying out, a given quantity of land in various forms. Thus- from the triangle ABC, (Fig. 43.) a smaller triangle of a given area may be cut off, by a line parallel to AB. The line CC being found by the problem, the point C will be given, from which the parallel line is to be drawn. 1 49. If the directions of the lines AE and BD, (Fig. 44.) and the length and direction of AB be given ; and if it be required to lay off a given area, by a line parallel to AB ; let the lines AE and RD be continued to C. The angles of the triangle ABC with the side AB being given, the area may be found. From this subtracting the area of the given trapezoid, the remainder will be the area of the triangle DCE ; from which may be found, as before, the point E through which the parallel is to be drawn. If the trapezoid is to be laid off on the other side of AB, its area must be added to ABC, lo give the triangle D'CE'. 150. If a piece of land is to be laid off from AB, (Fig. 45. by a line in a given direction as DE, 7iot parallel to AB ; SURVEYING. 85 let AC parallel to DE be drawn through one end of AB The required trapezium consists of two parts, the triangf ABC, and the trapezoid ACED. As the angles and one side of the former are given, its area may be found. Sub- stracting this from the given area, we have the area of the trapezoid, from which the distance AD may be found by the preceding article. 151. If a given area is to be laid off from AB, (Fig. 46.) by a line proceeding from a s^ivrn point D ; first lay off the trapezoid ABCD. If this be too small, add the triangle DCE ; but if the trapezoid be too large, subtract the triangle DCE'. Problem VI. To divide the area of a triangle into parts having given ratios to each other, by lines drawn from one of the angles to the opposite base. 152. Divide the base in the same proportion as the PARTS required. If the triangle ABC (Fig. 47.) be divided by the Hues CH and CD ; the small triangles, having the same height, are to each other as the bases BH, DD, and AD. (Euc. 1.6,) Problem VII. To divide an irregular piece of land into any two given parts. 153. Run a line at a venture^ near to the true division line required, and find the area of one of the parts. If this be too large or too small, add or subtract, by the preceding articles, a triangle, a trapezoid, or a trapezium, as the case may require. A field may sometimes be conveniently divided by redu- cing it to a triangle, as in Art. 125, (Fig. 35.) and then divi- ding the triangle by problem VI. SECTION IV. LEVELLLVG. A 134 \^ *^ frequently necessary to ascertain how much one spot of ground is higher than anoth- er. The practicabihty of supplying a town with water from a neighbouring fountain, will depend on the comparative ele- vation of the two places above a common level. The di- rection of the current in a canal will be determined by the height of the several parts with respect to each other. The art of levelling has a primary reference to the level surface of water. The surface of the ocean, a lake, or a river, is said to be level when it is at rest. If the fluid parts of the earth were perfectly ^pAenca/, every point in a level surface would be at the same distance from the centre. The difference in the heights of two places above the ocean would be the same, as the difference in their distances from the centre of the earth. It is well known that the earth, though nearly spherical, is not perfectly so. It is not necessary, however, that the difference between its true figure and that of a sphere should be brought into account, in the compara- tively small distances to which the art of levelling is com- monly applied. But it is important to distinguish between the true and the apparent level. 155. The. TRUE LEVEL is a CURVE which either coincides Tcith, or is parallel to, the surface of water at rest. The APPARENT LEVEL is a STRAIGHT LINE which is fl TAN- GENT to the true level j at the point where the observation is made, Thusif ED (Fig. 48.) be the surface of the ocean, and AB a concentric curve, B is on a true level with A. But if AT be a tangent to AB, at the point A, the apparent level, as observed at A, passes through T. 156. When levelling instruments are used, the level is de- termined either by a/mrf, or a plumb-line. The surface of the former h parallel to the horizon. The latter h perpen- LEVELLING. 87 dicular. One of the most convenient instrunoents for the purpose is the spirit level, A glass tube is nearly filled with spirit, a small space being left for a bubble of air. The tube is so formed, that when it is horizontal, the air bubble will be in the middle between the two ends. To the glass is at- tached an index with sight vanes ; and sometimes a small telescope, for viewing a distant object distinctly. The sur- veyor should also be provided with a pair of levelling rods, which are to be set up perpendicularly, at convenient dis- tances, for the purpose of measuring the height from the sur- face of the ground to the horizontal line which passes through the spirit level. If strict accuracy is aimed at, the spirit level should be in the middle between the two rods. Considering D'ED"D as the spherical surface of the earth, and B'AB"B as a concen- tric curve ; a horizontal line passing through A is a tangent to this curve. If therefore AT' and AT" are equal, the points T' and T" are equally distant from the level of the ocean. But if the two rods are at T and T', while the spirit level is at A, the height TD is greater than T'D'. The dif- ference however will be trifling, if the distance of the stations T and T' be small. 157. With these simple instruments, the spirit level and the rods, the comparative heights of any two places can be ascertained by a series of observations, without measuring their distance, and however irregular may be the ground be- tween them. But when one of the stations is visible from the other, and their distance is known ; the difference of their heights may be found by a single observation, provided allowance be made for atmospheric refraction, and for the difference between the true and the apparent level. Problem I. To find the difference in the heights of two places by levelling rods. 1 53. Set up the levelling rods perpendicular to the horizon, and at equal distances from the spirit level ^ observe the points where the line of level strikes the rods before and behind, and measure the heights of these points above the ground ; level in the same manner from the second station to the third, from iht 88 LEVELLING. third to the fourth, <^c. The difftrenct helxveen the sum of the heights at the back stationSf and ai the fonoard stations, will be the difference betiveen the height of the first station and the last If the descent from H to H"(Fig. 49.) be required, let the spirit level be placed at A, equally distant from the stations H and H' ; observe where the line of level BF cuts the rods which are at H and li', and measure the heights BH and FH'. The difference is evidently the descent from the first station to the second. In the same manner, by placing the spirit level at A', the descent from the second station to the third may be found. The back heights, as observed at A and A', are BH and B H' ; the forward heights are FH' and F'H''. NowFH'-BH =the descent from H to H', And F'H' - B'H'=the descent from H' to H" ; Therefore, by addition, (FH'-f F'H'O -(BH-|-B'H')=the whole descent from H to H". 159. It is to be observed, that this method gives the true level, and not the apparent level. The lines BF and B'F' are not parallel to each other ; but one is parallel to a tan- gent to the horizon at N, the other to a tangent at N'. So that the points B and F are equally distant from the horizon, as are also the points B' and F'. The spirit level may be placed at unequal distances from the two station rods, if a correction is made for the difference between the true and the apparent level, by problem IL J 60. If the stations are numerous, it will be expedient to place the back and the forward heights in separate columns ill a table, as in the following example. Back heights. Fore heights. Feet. In. Feet. In. 1st. Observation 3 7 2 8 2. 2 5 3 1 3. 6 3 5 7 4. 4 2 3 2 5. 5 9 4 10 22 2 19 4 Difference 19 4 2 10 LEVELLING. 83 If the sum of the forward heights is less than the sum of ihe back heights, it is evident that the last station must bp higher than the first. Problem II. 161. To find the difference between the true and the ap- parent level, for any given distance. If C (Fig. 12.) be the centre of the earth considered as a sphere, AB a portion of its surface, and T a point on an ap- parent level with A ; then BT is the difference between the true and the apparent level, for the distance AT. Let 2BC = D, the diameter of the earth, AT=c/, the distance of T, in a right line, from A, BT=A, the height of T, or the difference between the true and the apparent level. Then by Euc. 36. 3, (2BC + BT)XBT=AT'; that is, {D+h)xh==d^ and reducing the equation, Therefore, to find h the difference between the true and the apparent level, add together one fourth of the square of the earth's diameter, and the square of the distance, extract the square root of the sum, and subtract the semi-diameter of the earth. 162. This rule is exact. But there is a more simple one, which is sutficiently near the truth for the common purposes of levelling. The height BT is so small, compared with the diameter of the earth, that D may be substituted for D-f^, without any considerable errour. The original equation above will then become Dxh=dK Therefore h=^— D That is, the difference between the true and the apparent level, is nearly equal to the square of the distance divided by *he diameter of the earth. Ex. I. What is the difference between the true and the app. nt level, for a distance of one English mile, supposing the earth to be 7940 miles in diameter ? Ans. 7.98 inches, or 8 inches nearly* 13 90 SURVEYING. , In the equation ^=7^-* as D is a constant quantity, it is evident that Aw varies as c?*. According to the last rule then, the differenee hetween the true and the apparent level varies as the square of the distance. The difference for 1 mile being nearly 8 inches, In. Feet. In For 2 miles, it is 8X22 = 32=2 8 nearly. For 3 miles, 8X3^= G For 4 miles, 8X43= jq g &c. &c. See Table IV. Ex. 2. An observation is made to determine whether wa- ter can be brought into a town from a spring on a neighbour- ing hill. At a particular spot in the town, the spring, which is 2i miles distant, is observed to be apparently on a level. What is the descent from the spring to this spot ? The descent is nearly 4 feet 2 inches for the whole dis- tance, or 20 inches in a mile; which is more than sufficient for the water to run freely. Ex. 3. A tangent to a certain point on the ocean, strikes the top of a mountain 23 miles distant. What is the height of the mountain ? Ans. 352 (eet, 1G3. One place may be below the apparent level of an- other, and yet above the true level. The difference between the true and the apparent level for 3 miles is 6 feet. If one spot, then, be only two feet below the apparent level of an- other 3 miles distant, it will really be 4 feet higher. If two places are on the same true level, it is evident that each is below the apparent level of the other. Problem III. To Jind the difference in the heights of two places whose dis- tance is known, 1G4. From the angle of elevation or depression, calculate how far one of tie places is above or below the apparent level of the other ; and then make allowance for the difference be- tween the apparent and the true level. By taking, with a quadrant, the elevation of the object whose distance is given, we have one side and the angles of LEVELLING. 91 a fight angled triangle, to find the perpendicular height above a horizontal plane. (Art. 6.) Adding this to the difference between the true and the apparent level, we have the height of the object above the true level of the place of observation. Whi'n an angle o( depression is taken, it will be necessary to subtract instead of a ^^ ^^^ ^^^^^ Mecham, 5 3. In Austria, by Liesganig, 48 43 57,086 60,835 6. In England, by Roy and Mudge, 52 2 57,074 60,827 10. In Lapland, by Maupertius, 66 20 57,422 61,184 Do. by Swanberg, 57,192 60,952 On a comparison of all the measurements which have been made, it is found that a degree of latitude is greater near the poles, than in the middle latitudes ; and greater in ttie middle latitudes, than near the equator. The earth is NOTES. 103 therefore coBipressed at the poles, and extended at the equa- tor. But it does not appear that it is an exact spheroid, or a solid of revolution of any kind. If arcs of the meridian which are near to each other and of moderate length be compared, they will not be found to increase regularly from a lower to a higher latitude. On the southern part of the line which was measured in France, the degrees increase ve- ry slowly ; towards the middle, very rapidly ; and near the northern extremity, very slowly again. Similar irregulari- ties are found in that part of the meridian which passes through England. These irregularities are too great to be ascribed to errours in the surveys. It is concluded, there- fore, that the direction of the plumb line, which is used in determining the latitude, is affected by local inequalities in the action of gravitation, owing probably to the different den- sities of the substances of which the earth is composed. These inequalities must also have an influence upon the fig- ure of the fluid parts of the globe, so that the surface ought not to be considered as exactly spheroidical. See Col. Mudge's account of the Trigonometrical Survey in England. Gregory's Dissertations, &c. on the Trigono- metrical Survey. Rees' Cyclopedia, Art. Degree. Playfair's Astronomy. Philosophical Transactions of London for 17G8, 1785,1787,1790,1791,1795,1797, 1800. Asiatic Research- es, vol. viii. Puissant. " Traite de Geodesic." Maupertius. " Degre du Meridian entre Paris et Amiens." Do. " La- Figure de la Terre.'' Cassini. " Expose des Operations, &ic." Delambre. *' Bases du systeme m^trique.'' Swanberg. " Exposition des Operations faites en Lapponie." Laplace. " Trait* de Mecanique Celeste." Note L. p. 94. One of the most simple methods of determining when the pole star is on the meridian, is from the situations of two other stars, Alioth and y Cassiopeiae, both which come on to the meridian a few minutes before the pole star, the one above and the other below the pole. Alioth, which is the star marked sin the Great Bear, is on the same side of the pole with the pole star, and about 30 degrees distant. The star 7 in the constellation Cassiopeiae, is nearly as far on the 104 SURVEYING. opposite side of the pole. The right ascension ot* the latter in 1810 was Oh. 45m. 24s., increasing about 3| seconds an- nually. The right ascension of Alioth was 12h. 45m. 36s., increasing about 2i seconds annually. These two stars, therefore, come on to the meridian nearly at the same time^ This time may be known by observing when the same verti- cal line passes through them both. The right ascension of the pole star in January 1810, was Oh. 54m. 36s., and in- creases 13 or 14 seconds in a year. So that this star comes to the meridian about 9 or 10 minutes after / Cassiopeiae. In very nice observations, it will be necessary to make allow- ance for nutation, aberration, and the annual variation in right ascension. About 10 minutes after aline drawn from Alioth to y Cas- siopeiae is parallel to the horizon, the pole star is at its great- est distance from the meridian. As this is the case only once in 12 hours, the two limits on the east side, and on the west side, cannot both be observed the same night, except at cer- tain seasons of the year. But on any clear night, one obser- vation may be made ; and this is sufficient for finding a me- ridian line, if the distance of the star from the pole, and the latitude of the place be given. The angle between the me- ridian and a vertical plane passing through a star, or an arc of the horizon contained between these two planes, is called the azunuth of the star. And by spherioal trigonometry, when the star is at its greatest elongation east or west, As the cosine of the latitude, To radius; So is the sine of the polar distance, To the sine of the azimuth. The distance of the pole star from the pole in 1810, was 1° 42' 19".6, and decreases 19|^ seconds annually. To observe the direction of the pole star when its azimuth is the greatest, suspend a plumb line 15 or 20 feet long from a fixed point, with the weight swinging in a vessel of wa- ter, to protect it from the action of the wind. At the distance of 12 or 15 feel south, fix a board horizontal on the top of a firm post. On the board, place a sight vane in such a man- ner that it can slide a short distance to the east or west. A little before the time when the star is at its greatest elonga- tion, let an assistant hold a lighted candle so as to illuminate (he plumb line. Then move the sight vane, till the star seen NOTES. 105 through it is in the direction of the line. Continue to follow the motion of the star, till it appears to be stationary at its greatest elongation. Then fasten the sight vane, and fix a candle or some other object in the direction of the plumb line, at some distance beyond it. As the declination of the needle is continually varying, the courses given by the compass in old surveys, are not found to agree with the bearings of the same lines at the present time. To prevent the disputes which arise from this source, the declination should always be ascertained, and the courses stated according to the angles which the lines make with the astronomical meridian. It must be admitted, after all, that the magnetic compass is but an imperfect instrument. It is not used in the accu- rate surveys in England. In the wild lands in the United States, the lines can be run with more expedition by the compass, than in any other way. And in most of the com- mon surveys, it answers the purpose tolerably well. But in proportion as the value of land is increased, it becomes im- portant that the boundaries should be settled with precision, and that all the lines should be referred to a permanent me- ridian. The angles of a field may be accurately taken with a graduated circle furnished with two indexes. The bear- ings of the sides will then be given, if a true meridian line be drawn through any point of the perimeter. EXPLANATION OF THE TABLES. Table I contains the parts of Mercator's meridian, to eve- ry other minuie. The parts for any odd niiniite maybe found with sufficient exactness, by takinji, the arithmetical mean between the next greater and the next less. For the uses of this table, see Navigation, Sec. IIL Table II gives the depression or dip of the horizon at sea for different heights. Thu,^ if the eye of the observer is 20 feet above the level of the ocean, the angle of depression is 4' 24". See Art. 99. This table is calculated according to the rule in note I, which gives the depression 59" for one foot in altitude ; allowance being made for the mean terres- trial refraction. In Table III. is contained the depression for different heights and different distances, when the view of the ocean is more or less obstructed by land. Thus if the height of the eye is 30 feet, and the distance of the land 2\ miles, the de- pression is 8'. See Note I. Table IV contains the curvature of the earth, or the differ- ence between the true and the apparent level ♦or different distances, according to the rule in art. 161. Thus for a dis- tance of 17 English miles, the curvature is 192 feei. Table V contains the distances at which objects of differ- ent heights may be seen from the surface of the ocean, in the mean state of the atmosphere. This is calculated by first finding the distance at which a given object might be seen, if there were no refraction, and then increasing this distance in the ratio of \/l : v/6. See Note A. 108 Explanation of the Tables. Table VI contains the polar distance, and the right ascen- sion in time, of the pole star, from IBOO to 1820. From this it will be seen, that the right ascension is increasing at the rate of about 14 seconds a year, and that the north polar dis- tance is decreasing at the rate of 19^ seconds a year. From the latitude of the place, and the polar distance of the star, its azimuth may be calculated, when it is at its greatest dis- tance from the meridian. The time when it passes the me- ridian may be ascertained by finding the difference between the right ascension of th« star and that of the sun. See Note L, TABLE I. MERIDIONAL PARTS. M. 0°| l'^ 2° 3° 4° 5° 6° 7° 8° 9° lOo 11° 12° M.; l_ 1 60 120 180 240 300 361 421 482 642 603 664 725 2 2i 62 122 182 242 302 363 423 484 544 60a 666 727 2| 4 4' 64 124 184 244 304 365 425 486 546 607 668 729 4i 6 6 66 126 186 246 306 367 427 488 5481609 670 731 6 8 8 68 128 188 248 308 369 429 490 550 611 672 734 8 10 10 70 130 190 250 310 371 431 492 552 613 674 736 10 12 12 72 132 192 252 312 373 433 494 554 615 676 738 12 14 14 74 134 194 254 314 375 435 496 556 617 678 740 14 16 16 76 136 196 256 316 377 437 498 558 619 680 742 16 18 18 78 138 198 258 318 379 139 500 560 621 682 744 18 20 20 80 140 200 260 320 381 441 502 562 623 684 746 20 22 22 82 142 202 262 322 383 443 504 565 625 687 748 22 24 24 84 144 204 264 324 385 445 506 567 627 689 750 24 26 26 86 146 206 266 326 387 447 508 569 629 691 752 26 28 28 88 148 208 268 328 389 449 510 571 632 693 754 28 30 30 90 150 210 270 331 391 451 512 573 634 695 756 30 32 32 92 152 212 272 333 393 453 514 575 636 697 758 32 34 34 94 154 214 274 335 395 455 516 577 638 699 76034 36 36 96 156 216 276 337 397 457 518 579 640 701 762 36 38 38 98 158 218 278 339 399 459 520 581 642 703 764 381 40 40 100 160 220 280 341 401 461 522 583 644 705 766 40 42 42 102 162 222 282 343 403 463 524 585 646 707 768;42 44 44 104 164 224 284 345 405 465 526 587 648 709 770 44 46 46 106 166 226 286 347 407 467 528 589 650 711 772 46 48 48 108 168 228 288 349 409 469 530 591 652 713 774 48 50 50 110 170 230 290 351 411 471 532 593 654 715 777 '50 52 52 112 172 232 292 353 413 473 534 595 656 717 779 52 54 54 114 174 234 294 355 415 476 536 597 658 719 781 ;54 56 56 116 176 236 296 357 417 478 538 599 660 721 783 ,56 58 68 118 178 238 298 359 419 480 540 601 662 723 785 :58 1 M. 0° r 20 3° 4° 5° 6° r 8° 9°|lO° 11° 12° |M. TABLE I. MERIDIONAL PARTS. M. 13° 14° 150 16° 17° 18° j 19° 200 21° 22° VI. (; 787 848 010, 973 1036|1098 - 1161 1225 1289 1354 2 789 85i (»13, 975 I037|1I00 1164 1227 1291 1366 2 4 791 bo.s 9!..'> 977 1039 1102 1166 1229 1293 1358 4 c 793 'cbo 917 979 1042 1105 1168 1232 1296 1360 6 8 79:-> 857 919 981 1044 1107 1170 1234 1298 1362 « 10 797 85'' 921 983J1046 1109 1172 1236 1300 1364 10 12 799 661 923 985 1048 nil 1174 1238 1302 1367 12 14]801 8bo '32 987J1050 1113 1176 124011304 1369 14 16803 ;)65 927 9891 1052 1115 178 I242il306 1371 16 18S05 86-; 929 991 1054 1117 1181 1244 1308 1373 18 20 807 SG9 031 994 1056 1119 1183 1246 1311 1376 20 22809 871 933 996|l058!l]21 1185 1249 1313 1377 0.2 24:811 87,i 935 998!1000!ll23 1187 1251 1316 1380 24 26 813 87j 937; 1000; 1063 1126 1189 1263 1317 1382 26 28 816 877 939 1002 1065 -1128 1191 1265 1319 1384 28 30;818 87P 942 1004 1067 1130 1193 1267 1321 1386 30 321820 882 944 1006 1009 1132 1195 1269! 1324 1388 32 34 822 >i^-l 946; 1008 1071 113411198 1261 1326 1 390 34 36 824 8.6 948il010 1073 113e5 1200 1264 1328 1393 36 38 826 88Ji 950 1012 1075 1138 1202 1266 13.'.0 1395 38 40 828 890 952 1014 1077 1140 1204 1268 1332 1397 40 42830 892 9rA 1016 1079 11 42' 1206 1270 1334 1399 42 441832 894 966 1019 IO81 1145 1208 1272 1 336 1401 44 46834 896 968 1021 10S4 1147 1210 1274 1339 1403 46 48 836 898 960 1023 1086 1149 1212 1276 1341 1406 48 60|838 f'OO 962 1025 1088 1151 1215 1278 i343 1408 50 52|840 9U2 964 1027 1 090 1163 ]2|7 1281 1346 1410 52 54 842 901 966; 10291 1092 1166 1219 128.: 1347 1412 64 56 844 906 969; 1 031 i 1094 1167 1221 1286 1349 1414 56 t 846 908 971 1 1033 1096 1159 1223 1287 1362 1416 22° 58 1 L JM. 13° 140 15°| 16° 17° 18° 1 9° 20° TABLE I. MERIDIONAL PARTS, 23° 24° 25° 26° 27° 1 28° 1 29° 1 30° 31° 32° 31. |l419 1484 1550 1616 1684 jl751 1819 1888 1950 2028 2 Il421 1486 1552 1619 16.6 11753 1822|'i8'^*l 1960 2031 2 4 14L'3 1488 i554 1621 1688 1756 1824 1893J1963 2033 4 6 1423 1491 1557 1623 1690 1758 1826 1895 1965 2035 6 8 1427 1493 1559 1625 1693 1760 1829 1898 1967 2038 8 10 1430 1495 1561 1628 1695 1762 1831 1900 1970 2040 10 1 '' 143i 1497 1563 163{j 1697 1765 1833 1902 1972 2043 12 14 1434 1499 1565 I G3- 1699 1767 1835 1905 1974 2045 14 16 1435 150211568 1634 J701 1769 1838 1907 1977 2047 ]e IG 1438 1504 1570 1637 1704 1772 1840 1909 1979 2050 18 20 1440 1506 1572 1639 1706 1774 1842 1912 1981 2052120 22 1443 1 508 1574 1''41 1 ro8 1776 1845 1914 1984 2054122 24 1415 1510 1577 1643 1711 1778 1847 1916 1986 2057:24 26 1447 151.S 1579 1645 1713 1781 1849 1918 1988 2059126 28 1449 1515 1581 1648 1715 1783 1852 1921 1991 2061 !28 30 1451 1517il583 1650 1717 1785 1854 1923 1993 2064 30 3-> 1453 1519 \oSo iu5^ 1720 1787 1856 1925 1995 2066 32 34 1456 lo2i I588|i654l 172J 1790 1858 1928 1998 2069 34 [30 145" 152 ^ 1590 1657 1724 *1792 1861 1930 2000 2071 ,'36 36 1460 1526 1592 1659 1726 1794 1863 1932 2002 2073 38 40 1462 1528 1594 1661 1729 1797jl865 1935 2005 2076 40 42 14^4 •530 1596 1663 1731 1799 1868 1937 2007 2078 42 44 1467 1532 15i?9 1666 1733 1801 1870 1939 2010 2080 44 4d 1469 1535 J601 1668 1735 1:03 1872 1942 2012 2083 46 48 1471 1637 1603 1670 1738 1806 1875 1944 2014 2085 48 50 1473 1539|1605 1672 1740 1808 1877 1946 2017 2088 50 52 1475 154i|1608 1675 1742 1810 1879 1949 2019 2090 52 54 1477 1513I1610 1677 1 7441 1813 1881 1951 2021 2092 54 56 1480 1546, 1612'! t>79j 17471 1815! 1884, 1953 2024 2095 d6 58 1482 154e 161411681 1749| 1817 28° 1836 1956 2026 2097 58 M. 23° 24° 1 25° 26° 27° 1 29° 30° 31° 32° >!.! TABLE 1. MERIDIONAL PARTS. 33° 34" 36" 36" 37° 38° 39° \ 40° f 41" 42° JVl. 2100 2171 2244 2318 2393 2468 2546 2623 2702 2782 2 2102 2174 2247 2320 2396 2471 2648 2625 2704 2784 2 4 2104 217C 2249 2323 2398 2473 2660 2628 2707 2787 4 6 2107 2179 2262 2326 2400 2476 26631 263 ij 27 10 2790 6 8 2109 2181 2264 2328 2403 2478 2665 2633 2712 2792 8 10 2111 2184 2251 2330 2405 2481 2568 2636 2716 2795 10 12 2114 2186 2269 2333 2408 2484 2560 2638 2718 2798 12 142116 2188 2261 2336 2410 2486 2663 264 i 2720 2801 14 16 2119 2191 2261 2338 2413 2489 2566 2644 2723 2803 16 18 2121 2193 2266 2340 2415 249J 2668 2646 2726 2806 18 20 2123 2196 2269 2343 2418 2494 2571 2649 2728 2809 20 22 2126 2198 2271 2345 2420 2496 2573 2661 2731 2811 22 24 2128 2200 2274 2348 2423 2499 267c (2654 2733 2814 24 26 2131 2203 2276 2350 2425 2501 2678 2657 2736 2817 26| 28 2133 2206 2279 2353 2428 2504 2581 2669 2739 2820 28 30 2136 2208 2281 2355 2430 2506 2584 2662 2742 2822 30 32 2138 2210 2283 2358 2433 2509 2586!2665 2744 2826 32 34 2140 2213 2286 2360 2435 251212589 2667 2747 2828,34i 36 2143 2216 2288 2363 2438 2514|2591 2670 2760 2830 36 38 2145 2217 2291 2365 2440 2617 2694 2673 2762 2833 38 40 2147 2220 2293 2368 2443 2619 2697 2675 2755 2836 40 42 2150 2222 2296 2370 2445 2622 2599 2678 2768 2839 42 44 2162 2226 2298 2373 2448 2524 2602 2680 2760 2841 44 46 2165 2227 2301 2375 2461 2627 2604 2683 2763 2844 46 48 2167 2230 2303 2378 2453 2630 2607 2686 2766 2847 48 50 2169 2232 2306 2380 2468 2532 2610 2688 2768 2849 50 62 2162 2236 2308 2383 2458 2535 2612 2691 2771 2862 62 64 2164 2237 2311 2385 2461 2537 2616 2694 2774 2865 64 6612167 22.39 2313 2388 2463 2540 2617 2696 2776 2858 56 68 2169 ) 2242 2316 2390 2466 2542 2620 2699 2779 2860 68 JM. 33° 34» 36° 36° 37° 38° 39° 40° 41° 42° M. TABLE 1. MERIDIONAL PARTS. 10 12 14 16 18 20 22 24 26 28 30 32 34 '36 38 40 42 44 46 43° 44° 45° 46° 4?° 48° 49 2863 2866 2869 2871 2874 2877 2880 2882 2885 2888 2891 2893 2896 2899 2946 2949 2951 2954 2957 3030 3033 3036 3038 3041 2960 3044 2963 3047 2902 2985 2904 2907 2910 2913 2915 2918 2921 2924 2926 48 2929 50 52 54 56 58 2965 2968 2971 2974 2976 2979 2982 2988 2991 2993 2996 2999 3002 3005 3007 3010 3013 2932 3016 3101 3050 3053 3055 3058 3061 3064 3067 3070 3073 3075 3078 3081 3084 3116 3118 3121 3124 3127 3130 3133 3136 3139 3142 3144 3147 3150 3153 3156 3159 3162 3165 3168 3171 3087 3173 3090 3176 30933179 3095 3098 2935|3019 2937 2940 2943 M. 43° 3021 3024 3027 3104 3107 3110 3113 3182 3185 3188 3191 3194 3197 3200 44° 45° 460 47° 3203 3206 3209 3212 3214 3217 3220 3223 3226 3229 3232 3235 3238 3241 3244 3247 3250 3253 3256 3259 3262 3265 3268 3271 3274 327r 3280 3283 3286 3289 3292 3295 3298 3301 3303 3382 3385 3388 50^ 3474 3478 3481 3391 3484 3394 3487 3306 3397 3309 3400 3312 3316 3319 3322 3325 3328 3331 3334 3337 3340 3343 3346 3349 3352 3355 3358 3361 3364 3367 3370 3373 3376 3379 48^ 3403 3407 3410 3413 3416 3419 3422 3425 3428 3431 3434 3437 3440 3443 3447 3450 3453 3456 5r 3569 3572 3575 3578 3582 3585 3588 3591 3490 3493 3496 349913594 350313598 350GI3601 3509 3512 3515 3518 3521 3525 3528 3531 3534 3537 3540 3543 3547 3550 3459 3553 3462 3465 3468 3471 49^ 3556 3559 3562 3566 50° 51 3604 3607 3610 3614 3617 3620 3623 3626 3630 3633 3636 3639 3643 3646 3649 3652 3655 3659 3662 52° M. 3665 3668 3672 3675 3678 3681 3685 3688 3691 J695 IS 3698 3701 3704 3708 3711 3714 3717 3721 3724 3727 3731 3734 3737 3741 3744 3747 3750 3754 3757 3760 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 52° 16 TABLE I. MERIDIONAL PARTS. >1. 53= 54^ 55" 56^ 57^ 5&0 59= 60' 61 62" 10 12 14 16 18 120 |22 24 i26 3764 3767 3770 3774 3777 3780 3784 3787 3790 3794 3797 3800 3804 3807 386o 3868 3871 3875 3878 3882 3885 3889 3892 3895 3899 3902 3906 3909 3968 3971 3975 3978 3982 3985 3989 3992 3996 4074 4077 4081 4085 4088 4092 4095 4099 4103 3999 4106 4003 4006 4010 4014 3811 3913 30 32 34 36 38 40 42 44 46 48 I 3814 3817 3821 3824 3827 3831 3834 3838 3841 3844 3848 3851 3854 3858 4110 4113 4117 4121 4017|4124 50 52 54 56 58 3861 3916 3919 3923 3926 3930 3933 3937 3940 3944 3947 3951 3954 3958 3961 3964 4021 4024 4028 4031 4128 4132 4135 4139 4183 4186 4190 4194 4197 4201 4205 4208 4212 4216 4220 4223 4227 4231 4234 4238 4242 4294 4298 4302 4306 4309 4313 4317 4321 4325 4409 4413 4417 4421 4425 4429 4433 4436 4440 4328 4444 4332 4336 4340 4344 4347 4351 4355 4448 4452 4456 4460 4464 4468 4472 4476 4480 4035 4142 4038 4042 4045 4049 4052 4056 4060 4063 4146 4150 4153 4157 4161 4164 4168 4172 424614359 424914363 4253 436714484 4257 4260 4264 4268 4272 4067 4175 4070 4179 4370 4488 4374 4378 4382 4386 4275 4390 4279 4283 4287 4291 4394 4398 4401 4405 4527 4531 4535 4539 4543 4547 4551 4555 4559 4564 4568 4572 4576 4580 4584 4588 4592 4596 4600 4604 4608 4492 4612 4495 4499 4503 4507 4511 4515 4519 4523 4616 4620 4625 4629 4633 4637 4641 4645 4649 4653 4657 4662 4666 4670 4674 4678 4682 4687 4691 4695 4699 4703 4707 4712 4716 4720 4724 4728 4733 4737 4741 4745 4750 4754 4758 4762 4766 4771 4775 4779 4784 4788 4792 4796 4801 4805 4809 4814 4818 4822 4826 4831 4835 4839 4844 4848 4852 4857 4861 4865 4870 4874 4879 4883 4887 4892 4896 4901 58 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 53° 54« 55° 56° 57° 58°|59*> 60o|61<» 62° M TABLE r MERIDIONAL PARTS. M 63° 64° 65 10 12 14 16 1 20 22 24 f26 28 3014972 32 34 36 38 4905 4909 4914 4918 4923 4927 4931 4936 4940 4945 4949 4934 4958 4963 4967 4976 4981 4985 4990 40 4994 42 4999 5003 5008 5012 5017 5021 5026 5030 5035 5039 5044 5049 5053 5058 5062 5067 5071 5076 5081 5085 5090 5095 5099 5104 5108 5113 5118 5122 5127 5132 5136 5141 5146 5151 5155 5160 5165 5/69 5174 5179 5184 5188 5193 5198 5203 5207 5212 5217 66= 67° 680 69° 70° 71° 72° 5324 5474 53285479 5333 5484 5338 5343 5348 5353 5358 5363 5222 5368 5226 5231 5236 5241 5246 5250 5255 5260 5265 5270 5275 5280 5284 5289 5294 5299 5304 5309 5314 5319 M. 63° 64° 65° 66° 67° 5489 5495 5500 5505 5510 5515 5631 5636 5795 5966 5800 5972 5642]5806|5978 5647 581 1|5984 565215817 5989 5658 5663 5668 5674 55205679 5373 5378 5383 5388 5393 5398 5403 5408 5413 5418 5423 j428 5433 5438 5443 5448 5454 5459 5464 5469 5526 5531 5536 5685 5690 5695 5541 5701 5546 5706 5552 5712 55575717 5562I5723 5567:5728 557315734 5573 5739 5583 5745 5588(5750 5594 5756 5599 5761 5823 5828 5834 5839 5995 6001 6OO7 6013 5845 6019 5851 5856 5862 5868 :;874 6025 6031 6037 6043 6049 587916055 5604 5610 5615 5620 5625 5767 5772 5778 5885 5891 5896 5902 59O8 5914 5919 5925 5931 5937 5943 5948 5783 5954 5789 5960 6061 6067 6O73 6146 6152 6158 6164 6 170 6177 6183 6335 6341 6348 6354 6361 6367 6374 618916380 6195 6387 6201 6394 6208 6214 6220 6226 623vi 6239 6245 6252 6258 6079 6264 6085 6091 6097 6103 6109 6115 6121 6127 6133 6140 6400 6407 6413 6420 6427 6433 6440 6447 6453 6460 6271 6467 6277 6473 680 690 jQo 7-1 o j2 6283 6290 6296 6303 6309 6315 6322 6328 6480 6487 6494 6500 6507 9614 6521 6528 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 • TABLE I. MERIDIONAL PARTS. M. 73° 74^^ 75^' 76" I 77° 78 6534 6541 6548 6555 6562 6569 6576 6583 6590 6597 6603 6610 6617 6624 6631 6639 6646 6653 6660 6667 6674 6681 6688 6695 6702 6710 6717 6724 6731 6738 6746 6753 6760 6768 6775 6782 6790 6797 6804 6812 6819 6826 6834 6841 6849 6970 6978 6986 6994 7001 7009 7017 7025 7033 7041 7048 7056 7064 7072 7210 7218 7227 7235 7467 7476 7485 7494 7243J7503 7252|7512 72607521 72687530 7277 7285 7294 7302 7311 7319 7080 7328 6850 7088 6864j7096 6871i7104 6879;7112 6886 7120 6894 6901 6909 6917 6924 7128 7136 7145 7153 7161 7336 7345 7353 7362 7371 7379 7388 7397 7406 7414 6932 7169 7423 6939|7177 69477185 6955 6963 M.| 73° 74° 75- 7194 7202 7432 7441 7449 7458 76' 7539 7548 7557 7566 7576 7585 7594 7603 7612 7622 7631 7640 7745 7754 7764 7774 7783 7793 7803 7813 7822 7832 79^ 8046 8056 8067 8077, 80- 81° 82" M. 8375 8739 8387 8752 8398 8410 8088;8422 I 8099 8109 8120 8131 8141 7842 8152 8492 8433 8445 8457 8469 8480 7852 7862 7872 7882 81638504 817418516 8185 8528 8196 8540 7892|8207 79028218 791218229 7922|8240 7932|8251 76507942 7659,7953 76687963 76787973 76877983 76977994 770618004 7716^8014 7725 7735 8025 8765 8778 8791 8804 8817 8830 8843 8856 8869 8883 8896 8909 8923 8936 8950 8963 8977 8991 9005 9018 9032 9046 9060 8318 8676 9074 832918688 9088 8552 8565 8577 8589 8601 8262 8614 8273 8284 8626 8638 8295;8651 830718663 9145 9160 9174 9189 9203 921 9233 9248 9262 9277 9292 9307 9322 9337 9353 9368 9383 9399 9414 9430 38 9445 40 10 12 14J 16 18 20 22] 24 26 28 30 32 34 36 9461 9477 9493 9509 42 44 4Q\ 48 9341 8352 8035 8364 8701 8714 8726 77° 78° 79° 80° 81° 82" M. 9525 50 9541 52 54 91039557 91179573 913l|9589 TABLE II. TABLE HI. Depression of the Ho- rizon of the Sea. Dip oi the Sea, at different Distances from the Observer. Depres- sion. 0* 39" 1 24 1 42 1 58 2 12 2 25 3 36 2 47 2 57 3 7 3 16 3 25 3 33 3 41 3 48 3 56 4 3 4 10 4 17 4 24 4 37 4 49 5 1 5 13 5 23 5 49 6 14 6 36 6 57 37 14 8 48 9 20 9 50 10 47 11 39 12 27 13 12 13 55 Dist. ot land m Height of the Eye above the &ea, in (eet. mile/. 5 |I0 |15 |20 |25 |30 |35 |40 i 11' 22' 34'145';56"6B'|79';90} i 6 11 17 122 128 ,34 39 45 a 4 8 12 15 19 23 27 30 1 4 6 9 12 15 17 20 23 I i 3 5 7 9 12 114 16 19 1 ; 3 4 6 S 10 11 14 15 2 2 3 5 6 8 lO 11 12 2 ^ 2 3 5 6 7 8 9 10 3 2 3 4 5 6 7 8 8 3 i 2 3 4 5 6 6 7 7 4 2 3 4 4 5 6 7 7 5 2 3 4 4 5 5 6 6 6 2 3 4 4 5 5 6 6 TABLE IV, Curvature of the Earth. Dist. in tJ eight. Dist. id Height. ;i 1 [oches. Feet. 1 4 1 2 15 149 i 2 16 170 1 8 i 17 192 Feet. ! 18 215 2 2.6 i ^^ 240 3 6. ! 20 266 4 10.6 ' 25 4i5 5 16.6 30 599 6 23.9 j 35 814 7 32.5 : 40 1064 8 42.5 : 45 1346 9 53.8 j 66.4 I 50 1662 10 60 2394 11 80.2 1 70 3258 12 95.4 1 80 4255 13 112. 90 5386 14 130 100 6649 TABLE V. Distances at which Objects can be seen at Sea. Height ir, Distance in Height ii ) Distance in leet. Eng. miles. feet. Eng. miles, 1 1.3 60 10.2 2 1.9 70 11.1 3 2.3 80 11.8 4 2.6 90 12.5 5 2.9 100 132 6 32 200 18.7 7 3.5 300 22.9 8 3.7 400 26.5 9 4. 500 29.6 10 42 600 32.4 12 4.6 700 35. 14 49 800 374 16 53 900 39.7 18 5.6 1000 41.8 20 59 2000 592 25 66 3000 72-5 30 7.3 4000 83-7 35 7.8 5000 93-5 40 8.4 10000 133- 45 8.0 15000 163- 50 9.4 20000 188- TABLE VI. The Polar Distance and Right Ascension af the Pole Star. Polar Distance. 1800 io 45' 35" 1801 45 15 1802 44 56 1803 44 36 1804 44 17 1S05 43 57 1806 1 43 33 1807 43 18 1808 4i 58 1809 42 39 1810 42 19 1811 42 18l£ 41 40 1813 41 21 1814 41 1 18)5 40 42 1816 40 23 1817 40 04 1818 39 45 1819 39 25 1820 39 05 Ann. Var. --19".5 I— 19".4 Right Ascension. | Ann. Var. \ h. m. 8. s. 52 24 + 12.9 52 37 52 50 53 3 53 16 53 20 53 42 53 55 54 9 54 22 s. 54 m 4-136 54 50 55 04 55 18 55 33 55 47 56 02 56 17 56 32 56 46 s. 57 01 + 14.3 S2=«ilj£RVErijrG yv.y/ « A^ ^ -^C' 5 » ^rt B c .3^ V.^.iT.j.^^^^^ .,.^ ^'m N' and SV^RVBYIXG. /^m. ^ B JJ _D JTAVIBATIOX and SrBVBYilVC. f^m AN ELEMENTARY TREATISE CONIC SECTIONS, SPHERICAL GEOMETRY. AJfS SPHERICAL TRIGONOMETRY. BY MATTHEW R. BUTTON, PROFESSOR 01 MATHEMATICS AND NATURAL PHIL080PHT IN YALE COLLEGE. BEING THE FIFTH AND SIXTH PARTS 0» A COURSE OF MATHEMATICS ADAPTED TO THE METHOD OE INSTRUCTION IN AMERICAN COLLEGES NEW-HAVEN. PUBLISHED BY HOWE & SPALDING. S. Converse, printer. 1824. DISTRICT OF CONNECTICUT, ss. BE IT REMEMBERED, That 00 the seventeenth day of March, in the forty-ei§:th year of the InJependence of th United States of America, Matthew R. Dutton, of the said District, hath de- posited in this Otfice the title of a book, the right whereof he clainjfc as Author, in the words following, to wit : — •' An Elementary Treatise on Conic Sections, Spherical Geometry, and Spher- <*ical Trigonometry. By Matthew R. Dutton, Profnssor of Mathematics and Nat- ♦* ural Philosophy in Yale College. Being the Fifth and Sixth Part? of a course of "Mathematics adapted to the method of instruction in American Colleges." In conformity to the act of the Congress of the United States, entitled, " An Act for the encouragement of learning, by securing the copies of Maps, Charts and Books, to the authors and proprietors of such copies, during the times therein men- tioned." CHAS. A. INGERSOLL, Clerk of the District of Connecticut. A true copy of Record, examined and sealed by me, CHAS. A. INGERSOLL, Clerk of the District of Connecticut, ADVERTISEMENT. In preparing the following Treatise on Conic Sections, it was ietermined to adopt the general plan and arrangement of propo- iitions in Button's Conic Sections ; the deviations from that plan however have become so numerous and considerable as to leave but a slight resemblance to it. The authors whose works tiave been consulted, and in some cases freely used, are Appollo- vius, Archimedes, Hamilton, Simpson, West, Vince, Emerson and Bridge ; especially the two authors last mentioned. This general acknowledgment is made, as it is inconvenient or im- possible to specify more particularly what is due to each. The authors whose works have supplied the principal materi- als for the Spherical Geometry and Trigonometry, are Theodo- sius, Simpson, Platfair, and Cagnoli ; the Projections and Examples are taken with slight alterations from the Mathematics of Webuer, for, in the following work, there has been no af- fectation of originality, when there appeared no room for improve- ment. ^ M. R. D. CONTENTS, CONIC SECTIONS. Definitions, ------ Parabola, ------. Ellipse, ------- HyPERBOLA, ------- Curvature of Conic Sections, - - - APPENDIX TO CONIC SECTIONS, 1. Similar and Sub-Contrary Sections, 2. Comparison of Conic Sections, - SPHERICAL GEOMETRY. Definitions and Sections of the Sphere, Spherical Triangles, - - - - - IWTErtSECTIONS <5lz:C. - - - - - APPENDIX TO SPHERICAL GEOMETRY, Projection, ----- Problems in Stereographic Projection, SPHERICAL TRIGONOMETRY. Geometrical Principles of - - - Arithmetical Calculation „ - - - Notes, .-.--,• CONIC SECTIONS. DEFINITIONS. 1. CONIC SECTIONS are the figures made by the mutual intersection of a cone and plane. 2. According to the different positions of the plane, five different figures or sections, are produced ; viz, a Triangle^ a Circle^ a Parabola^ an Ellipse^ and a Hyperbola. 3. If the plane pass through the vertex of the cone, and any part of the base, the section will be a triangle. Case 1. Let it pass through the vertex, perpendicular to the base, and coinciding with the axis of the cone; it will then coincide with the triangle by the revolution of which the cone is generated. The section therefore willgup.s.ii. be an isoceles triangle VCD, whose angle at the vertex Fig. i will be double that of the generating triangle. Case 2. Let the plane now cut the base obliquely. This section also will be a triangle, for every straight line drawn from the vertex of a cone to any point of the circumference of its base, is in the surface of the cone ; since it coincides with the Hypothenuse of the revolving triangle b) which the cone is generated. 4. If the plane cut the cone parallel to its base, the sec- Fig;. 2. tioH will be a circle, as ABD — for this section wll coin- cide with the revolution of a line perpendicular to the axis of the generating triangle, and will therefore be a circle, whose radius is equal to that perpendicular. 5. If the cone be cut by a plane parallel to one of its sides, the section is called a Pt/ra6o/a; as ACD. Fig. 3. As the plane in this case cuts one !«ide i4 the cone parallel to its opposite side, it is evident that it can never meet the latter; the section therefore may be ( ontinued i. lo indefinitely, if the cone is supposed to be indefinitely ex- pended. 2 ■ CONIC SECTIONS. G. If the plane cut the cone obliquelj, so as to make an &ngle with the base less than that made by the side of the ^ 'e- 4. cone, the section is called an Ellipse : as ABC. As the cutting plane in this case, is not parallel with the side of the cone V K, it will meet that side — that is it will intersect the surface of the cone on every side. This sec- tion therefore is complete and definite for every position of the plane. 7. If the cutting plane makes with the base of the cone Fie. 5. ^" angle greater than that made by the side of the cone, the section is called an Hyperbola ; as ACD, The cutting plane in this position will never meet the opposite side of the cone in the direction of V K, but it will meet it produced, on the other side of V, and if all the sides of the cone be supposed to be produced through the vertex V, forming an opposite and similar cone, and if the plane also be produced cutting this cone in the section Bed, the two sections ACD and Bed are called Opposite Hyperbolas, Tig. 6. 8. If there be four cones, the angles at whose vertices are together equal to four right angles ; and if their axes ].^ 13. be all placed in one plane and their vertices all meet in a given point V, the sides of the cones will touch each oth- er in the right lines LVF, BVH. If a plane per- pendicular to that in which the axes of the cone are pla- ced, and parallel to O P, the axis of the two cones BVL FVH,cut these cones in the opposite hyperbolas AD Bd ; if another plane, also perpendicular to that in which the axes of the cone are placed, pass through A or B par- allel to KVM, the axes of the remaining cones, cutting opposite hyperbolas in these cones, these two pairs of op- posite hyperbolas are mutually conjugate to each other. The two sections are commonly supposed to be placed in the same plane, the lines AB and ab bisecting each oth- er, as in fig. 7. If the angle at the vertex of each of the cones, (fig, 6) be a right angle, the cones will all be equal and similar, and the sections, being at the same distance from the ver- tex, will be equal and similar, and AB will be equal to ab ; in this case they are called Equilateral Hyperbolas, and may be all cut by one plane, parallel to that in which th« axes of the cones are placed ; as represented in fig. 7. Cor. CONIC SECTIONS. 7 t). As the properties of the triangle snd circle are de- monstrated in the Elements of Euclid, and are investiga- ted without any reference to the sections of a cone, they are not usually reckoned among the conic sections ; — this term being appropriately applied to the three remaining sections, — viz. the Parabola, Ellipse and Hyperbola. 10. The Vertices of a conic section, are the points where the cutting plane meets the opposite sides of the cone ; or the sides of a vertical triangular section through the axis of the cone, and perpendicular to the plane of the given section. In the Parabola, the plane does not meet the opposite ^^s- 3. side of the cone VH, (3) * This section therefore has but one vertex, as A . The Ellipse has two vertices, and opposite Hyperbolas have two ; as AB. ^^'^'^'^' 1 1. The Transverse axis, is the straight line which con- nects the vertices ; as AB. This definition is applicable to the Ellipse and opposite Hyperbolas, but not to the Parabola which has but one vertex. The axis of the Parabola, may be defined, the intersection of its plane with that of the vertical triangu- lar section to which it is perpendicular. This definition Sup. 2. 3. is applicable to all the conic sections, since it evidently coincides with that before given for the Ellipse and Hy- perbola. The line of their mutual intersection passes through the vertex of the Parabola and is of indetermi- nate length ; since the section itself may be indefinitely continued. 12. The Centre of a conic section is the middle of the transverse axis. Hence the Parabola has no centre ; that of the Ellipse is within the section, and that of the Hyperbola without it. between the opposite Hyperbolas. 13. The Conjugate axis of a conic section, is a line drawn through the centre, at right angles to the transverse axis. * This figure and others similar, refer to the articles numbered in tljis ^ieries of definifions. J CONIC SECTIONS. Hence the Parabola has no conjugate axis. In the El- lipse it is within the section and terminated by the curve. The conjugate axis of the Hyperbola is without the sec- tion and is bounded by the curves of the conjugate Hy- perbolas. Or more correctly, the conjugate axis is the transverse axis of the conjugate Hyperbolas. (8) Hence the two axes are mutually conjugates to each other. The preceding definition, of the conjugate axis of a Hyperbola, is correct, in all cases in which it is applicable. It is limited however in its application, to those cases in which the Hyperbola, is cut by a plane parallel to the Ax- is of the cone. The following limitation of the conjugate axis of a Hy- perbola is general. It is a mean proportional between the diameters of the circular sections, which pass through the vertices of the transverse axis of the Hyperbola. Thus the conjugate axis of A B, is equal to a mean pro- *'»§;• 5. portional between A Q and B R. 14. A Diameter of any conic section, is a straight line, drawn through the centre, and terminated in both direc- tions by the curve. This definition of a diameter evidently includes the axis. This definition is inapplicable to the Parabola which has no centre. (12). A diameter of a Parabola, may be defined, any line drawn parallel to its axis^ and terminated, like the axis, in one direction by the curve, while the oth^ er may be indefinitely extended, as Cd. 15. The extremities of a diameter, or its intersections with the curve are called its Vertices, 16. The Conjugate to any diameter, is a line drawn through the centre of the section, and parallel to the tangent * at the vertex of the diameter. 7^hus D E is the Pi^ 8. conjugate of the diameter C H. This definition evidently includes the conjugate axis (13.) If it is admitted, or assumed, that the tangent at the vertex of the transverse axis, is perpendicular to it. * Euclid's definition of a line touching a circle, may be applied to a tan- gent of a Conic Section. It is a line which meets the curve, but does not cut it. Fi-. 9. CONIC SECTIONS. 17. An Ordinate to any diameter is a right line, parallel to the tangent at its vertex, and terminated in one direc- tion by the curve, and in the other, by the diameter as K G. Figures 8 and 9. 1 8. The Jlbsciss to any ordinate is the part of the diam- eter, contained between its vertex and that ordinate, as ^^S- ^^ C G and G H. "^ ^• Hence in the Elhpse and Hyperbola, every ordinate has two determinate abscisses. In the Parabola, there is but one. the other being supposed indefinitely extended. (ll.)asGJ. Fig. 9. 19. The Parameter of any diameter, sometimes called the Lotus Rectum, is a third proportional to that diameter and its conjugate. This detinition is not applicable to the Parabola, (13.) but its Param(iier is a third proportional to any absciss and its ordinate. 20. The Focus of a conic section, is the point in the axis where the ordinate is equal to half the parameter. As F whore C F is equal to the semi-parameter of the fj?- 7. section. !i.&9. The Ellipse and Hyperbola have each two Foci as F, /, the Parbola has but one. 21. If tangents be drawn to the four vertices of the curves, in Conjugate Hyperbolas, forming an inscribed rectangle, thediagonals of this rectangle, are called the Assymptotes of the Hyperbolas. As L F and K H. p.^ -, 22. Similar conic sections, are those of the same kind, whose transverse and conjugate axes have to each other the same ratio. As this definition is not applicable to the Parabola, Sijn- ilar Parabolas^ are those whose abscisses and ordinates are to each other in the same ratio. 23. When a magnitude, which is supposed to vary ac- cording to a certain law, approaches continually towards equality with another given magnitude, so that ultimately the former differs from the latter less than by any assign- i^ CONIC SECTIONS able magnitude, then the latter is said to limit, or to ht the limit of (he former, and their ultimate or limiting ra- tio, is said to be a ratio of eqnahty. Thus Euclid, (Book XI 1 Prop. 2.) dennonstrates that i a regular polygon iriscribt;d in a circle, will, as the num- ber of its sides is supposed to be increased, approach con- tinuall) to equality with the circle, so as ultimately to dif- fer from the circle less than by any assignable magnitude. I By this truth, he demonstrates, that the areas of circles are as the squares of their diameters. In this example, the circle is called the limit of the pol- ygons. He also demonstrates that in this sr*.nse, the circle is also the limit of polygons circumscribed about it. *' The consideration of limits more or less dis^juised, must unavoidably enter into every investigation wh»ch has for its object, the mensuration of the circle." Leslie, "This principle is general, and is the only one by which we can possibly compare curvilineal with rectilineal spa- ces, or the length of curve lines with the length of straight lines, whether we follow the methods of the ancient or of the modern geometers." Playfair. General Scholium, The cutting plane being parallel to the base, forming a circular section, if it be supposed to revolve about A, at Fig. 10. the commencement of its motion, when it deviates in the least pos'^ible degree from parallelism with the base of the cone, the section will be an Ellipse, which in the language of mathematicians is infinitely near to a coincidence with the circular section. That is, it may differ from the circle less than by any assignable difference. If now the plane b*- supposed to revolve from B' to- t J-. 10. ^aj.(jg B''-in passinjr through B, where it is parallel to the base, the circle will be the section which forms the transition from the Ellipse immediately on • and AI . . DB : DE=» : : AD . DB : DI . DB ; .:. DE2=D1.DB. ScHOL. — The square of the ordinate DE, therefore (which equals DIFB) is less than the rectangle of the ab- sciss DB into the Parameter, or DHGB, by the rectangle IHGF similar and similarly situated to the whole rectan- gle PB It was on account of this deficiency in the square of the ordinate, compared with the rectangle of the ab- sciss into the Parameter, that Apollonius named this sec- tion the Ellipse, The rectangle ABGP or AB . P was called the Figure o( the section ; AB and BG its latera, and the perpendicular BG the Latus rectum. PROPOSITION III. As the square of the Conjugate axis, To the square of the Transverse, So is the rectangle of the abscisses of the Conjugate, To the square of their ordinate. That is fl6» : AB^ : :ad . db : dE^. Draw ED an ordinate to the Transverse AB, Fig. 2. Then (IICor. 1) AC^ : aC^r.-CA^-CD^ : DE^ ; But CD^ =t/E% and DE^ =C(/^ Therefore AC^ : aC^ wCA'—dE^ J CJS and AC^ : CA2 - dE^ : laC^ : Cd^, „ AC2 : c/E2 : :aC2 . ^q^ -Cd\ 5. E. „ AC^ XaC^y.dE^ XaC^-Cd\ aC^ : kC^WaC^-Cd^ : dE' ; But flC^ - C(/^ =ad . db, 2. s.Cor. Therefore aC^ : AC'l'.ad^ db t dE''. Q. £. D. 36 OP THE ELLIPSE. Cor. I. — From this Proposition it appears thaUhe Con- jugate axis, and the rectangles of its abscisses, have the same relations to their orditiates and to the Transverse, which is its conjugate (16), as those which have been de- monstrated concerning the Transverse axis. Hence 1st the Conjugate axis, also, bisects all its doub- le ordinates. 2. Ordinates at equal distances from its vertices, or from the centre, are eijual. 3. The Conjugate axis divides the Ellipse into two equal and similar parts. 4. The tangents'at its vertices are perpendicular to it. 5. Ordinates from the opposite extremities of any di- ameter are equal and equally distant from the centre. 6. The square of the Conjugate axis, is to the square of the Transverse, as the difference between the squares of the semi-conjugate, and the distance from the centre to any ordinate, is to the square of that ordinate. 7. If two Ellipses have the same Conjugate axis, the corresponding ordinates to this axis, will be to each other as their Transverse axes. 8. The Conjugate axis is to its Parameter, as the rect- angle of its abscisses, to the square of their ordinate. 9. The square of any ordinate is less than the rectan- gle of its absciss ana the Parameter of the Conjugate, by a rectangle similar and similarly situated to the rectangle of the Conjugate and its Parameter. (See Prop, i and II.) 10. The squares of any two ordinates areas the rec- tangles of their abscisses ; or in other words, the rectangles of the segments of its double ordinates are as the rectan- gles of the segments into which they divide the Conjugate axis. OF THE ELLIPSE. 37 ScHOL. From Prop. Ill compared with Prop. II, it is evident, that if two unequal straight lines bisect each oth- er at right angles, and either of them move parallel to it- self on the other, varying in its length contii»ually, so that the rectangles of its parts have always the same ra- tio to the rectangles of the parts into which it divides the other line, — the figure formed by the moving line will be an Ellipse. Cor 2. — If two circles be described on the two axes of an Ellipse as diameters, the one being inscribed within the Ellipse, and the other circumscribed about it, then any ordinate in either circle is to the corresponding ordinate in the Ellipse as the axis of this ordinate is to the other axis. That is, CA : Ca::DG : DE. and CaiCA:: dg : dE. For AD . DB =DGS Therefore (II) CA^ : Ca^ : iDG^ : DE^ ; Or CA :Ca::DG :DE. In the same manner. Ca I CA'.'.dg t dE, DGtDE::dE::dg. Or the corresponding ordinates, in the circlei and El lipse are reciprocally proportional. PROPOSITION IV. The square of either axis, is the square of the other, as the rectangle of the segments of a line in the Ellipse paral- lel to the former, is to the rectangle of the corresponding segments of a line parallel to the latter. Fig. 3. That is eh . hg : Eh . AG::AB2 lab^ Or Eh , hG : eh . hg'.'.ab^ : AB\ 38 OF THE ELLIPSE 5. E. For (I) ED» : eN^ : : AD . DB : AN . NB, and ED^ : ED^— eN^ : lAD.DB : AD.DB-AN.NB. But ED^-e]^^=ED'-hD'=-Eh . AG, And AD.DB-AN.NB=BD.DN+NA-BD+DN.NA; 2. 1. And since, BD . DN-hNA = BD . DN-f-BD . NA, And BD -f- DN . NA=BD . NA+DN . NA, Therefore, AD.DB-- AN . NB=DN . BD-NA . DN = Dw . DN = % . eh ; ED= : Eh . AG:: AD . DB::eh . hg, And ED^ : AD. DB::EA , hG : eh . hg. But (II.) ED2 : AD . DB::«62 : AB^, 11. .:. Eh . hG t eh , hgWab'- : AB^. Q. E, D. Cor. 1 . — Hence by equality of ratios, eh . hg : ek ,kg'.\ Eh . hG : M' . kl. Or, universally, the rectangles of the correspondiijg segments, of lines parallel to the two axes, mutually intersecting each other in the Ellipse, are to each other always in the same ratio; namely, the ratio of the squares of the two axes. ScHOL. — As the square of either semi-axis, or of an or- dinate, is the same as the rectangle of the two halves of the axis or double-ordinate. This proposition evidently in- cludes Prop. II and III, as it demonstrates, that the rela- tions there shown to exist between the two axes, extends to all lines in the section parallel to them. For AC2 : ac^:: AD . DB : DE^ is the same as AC . CB : aC . C6: : AD . DB : ED . DG which may be considered as included in this propo- sition. If the cutting plane revolve until the section becomes a| circle, we should infer, that the rectangles of lines, mu- tually intersecting each other at right angles, in the cir- cle, have always the same ratio. This is demonstrated by Euclid, (HI Book, 35 prop. See Schol. to Prop, I.) and also, that it is a ratio of eiquality. OF THE ELLIPSE. 39 If the plane be supposed to revolve in the opposite di- rection until the section becomes a Parabola, and the lon- ger segments of the lines parallel to the transverse axis be considered equal. Then the Prop. AD . DB : eh . ^^^: :ED . DG : EA . /iG becomes AD : eA : : ED . DG : E/i . AG The same as is demonstrated in the Parabola, Prop. II. Cor. 1. PROPOSITION. V. If lines parallel to the two axes, intersect each other, without the Ellipse, the rectangU:s of their corresponding segments are to each other as the squares of the axes to which they are parallel. Fie. 4. That is, EH , . UG : eU , Hgy.ab'' : ABS or eU . Hg : EH . hG::AB= : ab*. For (I,) e^^ : ED^ : : AN . NB : AD . DB, And eN= : eN^-ED2::AN.NB : AN.NB-AD.DB. But .N2-ED-^=HD— ED2=EH .HG, Also AN . NB = BN . AD-fND=BN . AD-f BN . DN, And AD . DB=AD . B +ND=BN . AD-f AD .DN, Therefore AN . NB - AD . DB=BN . DN - AD . DN, =DN . BN-AD=DN . BN-Br/ =DN. ^d=DN . Dn. Therefore (eN2::=)HD2 : EH . HG: :AN . NB : (DN . Dn = )eU . H^, And HD2 : AN . NB::EH . HG : eH . H^; But (II.) HD2(=eN0 : AN . i\B:: b^ : AB-', EH . HG : eU . Eg Wah^ : AB^ Q. E, D. 40 OF THE ELLIPSE. Cor. 1. — Since this proposition is equally applicable to all lines parallel to the axes, it follows by equality of ratios^ that, the rectangles of the segments of all lines parallel to the axes, intersecting each other wtthout the Ellipse, are to each other always in the same ratio ; viz. that of the squares of the two axes. That is EH . HG : eH . H^::KI . IL : el . ]g. CoR. 2. — If DH be supposed to move towards AP, the rectangle EH . HG approaches continually to an equali- ty with the square of AP, as its limitf and when D coin- cides with A, AP- ma) be substituted for EH . HG. In like manner, aS^ may be subitituted for eH . HG. That is, the rectangles of the segments of lines parallel to the axes, are to each other, as the squares of the inter- cepted parts of the tangents to the vertices of the axes. ^'?- 3. Cor. 3. -Since (IV,) Eh . hG : eh . hg y.ab' : AB-% ^'^' '^' and from this EH . HG : eH . H^ : ab^ : AB^, Therefore Eh . hG I eh , hg: :EH . HG : eH . H^. ScHOL. — When the Ellipse becomes a circle, the same relations exist betwen the rectangles of the corresponding segments, without the figure, and also between these rectangles and the squares of the tangents, as is demonstra- ted in Euclid 3 Book 36 and 37, but it is then a ratio of equality. Also if the plane revolve in the contrary direction, un- til the section becomes a Parabola, and the longer seg- ments are considered equal, then EH . HG : eH.H^y.Kl . IL : el . I^. becomes EH . HG : eH: :KI . IL : el ; and EH . HG : Kl . IL:>H .el. As demonstrated of the Parabola II, Cor. 1. The five preceding; propositions may all be embraced id ' the (o\\o\\\n^, gentrnl proposition. If lines parallel to one axis rf the Ellipse iutersect the other axis, or lines parallel to it, tithtr withiji or without the section ; the rectangles of OF THE ELLIPSE. 41 ihe corresponding segments will always have to each other, the same ratio. PROPOSITION VI. If a tane^ent and ordinate be drawn to any point of an Ellipse, meeting the transverse axis produced, the »emi- traiisverse is a mean proportional between the distances of the two intersections from the centre, That is, or CM . CT=CA2 ; CM : CA : CT, are continued proportionals. Fig. 5. For from the point T draw any other line TEH to cut the curve in two points E and H ; from which let fall the perpendiculars ED and HG , bisect DG in K, Then (I) AD . DB : AG .GB::DE2:GH% And bysim. tri. TD^ : TG^* : iDE^ : GH^ ; Therefore AD . DB : AG . GB ::TD-^ : TG^ But DB=CB4-CD=AC+CD = AG+DC-CG= iCK + AG, AndGB=CB-CG=AC-CG=AD+DC-CG=2CK + AD. .-. AD . 2CK-f AD . AG : AG . 2CK+AG . AD:: TD2 : TG% and DG . 2CK : (TG^ -TD^or) DG . 2TK: : AD . 2CK -f-AD. AG :TD% ^^- Vl^^ or 2CK : 2TK: : AD . 2CK-f AD . AG : TD=^ * ^ orAD.2CK : AD 2TK: :AD .2CK-f-AD . AG : TD^; 5. 19. .-. AD.2CK : AD.2TK::AD. AG : TD2-AD.2TK, and CK : TK : : AD . AG : TD^ - AD . 2TK, 5. a & is. and CK:TC::AD . AG : TD^-AD . TD-f-TA ; or CK : CT::AD .AG : AT^. But the limit of AD . AG, when the line TH comes in- to the position ofTL, is AM^ (See Parab. Prop. IV,) and K, then coincides with M. The proposition therefore becomes. CM : CT: : AM^ : AT^ : 42 Oi^ THE ELLIPSE. That is CM : CT: :CA-CM=' : CT-CA«, 5.19. or CM : CT::CIVP + CA^ : CA2+CT^ and CM :MT::CM2+CA2 : CT-^ -CM% or CM : MT : : CM^ +CA3 : CT+CM . MT, or CM^ : CM . MT : : CM^ ^-CA^ : CM . MT+ CT . MT ; Hence CM^ : CA-^ : :CM . MT : CT . MT, i^onr^ 3'^^ CM2 : CA^XCM : CT, 2 coa-'' •'• CM : CA : : CA : CT. versely. Q, E. D. The converse of this poposition, may be very concisely demonstrated in the following manner, which as it does not depend on the preceding demonstration, may be made the principal demonstration, and the preceding inferred as the converse of it. fig. 5. On AB describe the circle AGB, produce ML to G, draw a tangent to the circle at G, intersecting the axis produced in T, join CG, then CGT is a right angle, 6. 8. and (CG=)CA, is a mean proportional between CM and CT. Connect T and L, and the line TL is a tangent to the Ellipse at L. If not, it must cut it. Let it be supposed then to cut it at L and e, through c draw the ord.nate ed, and produce it to cut the circle in g, and the tangent in r. Then, sim. triang. TM : Td.'.ML : del and " " TM : Tdr.MG : dr, Mh:MG::de:dr, but (II, Cor. 3.) ML :MG::de: dg (Cfl. : CA.), dr=dg — ihe less equals the greater, which is impossible. Therefore TL does not cut the Ellipse in L; but it meets it ; it is therefore a tangent to the curve in L, and CT is a third proportional to CM and CA. That is CM : CA : CT, are continued proportion- als. Q. E. D. OF THE ELLIPSE. 43 Cor. 1.— CM : CT::A]VP : ATS that is, the distances from the centre to the two intersections of the tangent and ordinate with the axis, are as the squares of the distances of the same intersections from the vertex. CoR. 2. — If a tangent and ordinate be drawn from anj point in the Ellipse, cutting the conjugate axif; produced, the semi-conjugate is a mean proportional between the distances of the two intersections from the centre. For the properties of the transverse, from which this proposition has been deduced, are the same in the con- '^' jugate. See Prop. II and III. The preceding demonstrations therefore, are applicable in every respect to the semi-conjugate axis. That is Cd *, Ca : C/, are continued proportionals. Fig. 6. CoR. 3. — As the demonstrations are applicable to a tan- Fig;. 7. gent to the curve at either extremity of the double ordin- ate, the two tangents drawn to the two extremities of any double ordinate to the axis, meet the axis in the same point, and at equal angles. And, conversely, a line drawn through the intersection of two such tangents, if it bisect the angle, will also bisect the double ordinate ; and if it bisect the double ordinate, it will also bisect the angle, and in both cases it is the axis. Cor. 4. — Also KA:, the tangent at the vertex is bisect- ed by the axis, For TD:TA::DE : AK. and TD : TA::DH : A^, therefore DE : AK : : DH : AAr ; but DE=DH.-.AK=AA:. For a similar reason, NF=NG, And as NI =NM -.FI = MG. 44 OF THE ELLIPSE. Cor. 3. — If any number of Ellipses have the same transverse axis, and an ordinate in one be continued ta intersect the curves of all, the tangents from all the points of intersection, will meet the curve in the same point; for, since CD and CA, the two first of the continued pro- portionals, remain the same in all, the third CT will be the same. It continues the same, if the Ellipse becomes a circle, as is evident from the first steps of the demonstration. CoR. 6. — Two tangents to the curve at the extremities of any double ordinate, to the conjugate axis, cut it in the same point. CoR. 7. — Tf any number of Ellipses have the same conjugate axis, the tangents drawn to each at the inter- sections of the same double ordinate produced, will cut th( conjug e in the same point. The same is true also of tangents to a circle described upon the conjugate axis as its diameter. ^^S- 6. Cor. 8. — The following list of proportionals is de- rived directly from this proposition, and its second Corol- lary, viz. that CT:CA::CA :CD. And Ct I Ca y.Ca : Cd. Fig. 6. (1.) Then CT-f PA : CT-CA: :CA+CD : CA-CD that is BT : AT::BP : AD. or BT is harmonically divided. In the same manner bt : at'.'.bd * ad, (2.) Also CT : CT-hCA::CA : CA+CD. that is CT : BT::CA : BD. And Ct :bt::Ca I bd. (3.) Again (2.) CT-CA : BT-BD::CT : BT, That is AT : DT::CT: BT. And, at : di'.'.Ct : bt. OF THE ELLIPSE. 4^ (4.) Therefore sim. tri. AK : DE'.'.Ct : BH, and " " TK : TE'.'.Tt : TH. also ak : dE'.'.CT : bh, and tk : tE'.uT : th. PROPOSITION VII. The rectangle of the Focal distances from the vertices, is equal to the rectangle of one fourth the parameter, and the transverse axis. That is AF . FB==;iP . AB, Tig. 8. For (II Cor. -2.) AF . FB : FE-^ : : AB : P, therefore AF . FB : FE^ : : AB . AF : P . AF, and AF . FB: AB . AFiiFE^ : P . AF ; But (19) FE = iP.-.FE2=iPS AF . FB : AF . AB: :^P=^ : P . AF, g.s.Cor.s. FB : AB::iP : AF, and FB . FA=iP . AB, Q. E. A Cor. 1.— FB : AB::iP : AF. ScHOL. — If, as in the Parabola, FB and AB, be consid- ered equal, then :[P=AF as demonstrated in the V Prop, of Parabola. CoR. 2.-SinceFB.FA=iP.AB,and FB is less than AB, FA is greater than {P In the Parabola, FA = iP. CoR. 3.— AF . FB=Ca\ or the rectangle of the Focal distances, equals the square of the semi-conju* gate. For since (19.) P . AB=ab^--'.i? . AB=aC2=:AF . FB. or AF :aC ::aC : FB. 46 OF THE ELLIPSE. Cor. 4.— AF.FB(=aC^) = ia6^ or iAB.iP=AC.FE ; or AF : FE::AC :FB. Cor. 3.— (VI Cor. 1.) aC = =C/ . Cd=Ct . DE. but (V(. Cor. 7, 4.)C/ . DE=AK . BH, .-. AK . BH=aC2=CA . iP=AF . FB. CoR. 6. — Therefore HFK, is a right angle, for (Cor. 5) AK : AF::FB : Bfl, .'. the triangles AKF and BHF -e -infi'' r; -^ .-. - angle BHF=AFK and BFH = AKF, .*. also HFK is a right angle. The same may be proved, in referenre to the other Focus, therefore a circle described on HK, as a diameter, will pass through the two Foci. PROPOSITION VIII. The square of the distance of the Focus from the cen- tre, is equal to the dilFerence of the squares of the semi-axes. Fig. 8. That is CF2=CA2-Ca^ For (111 Cor. 1.) CA^ : Ca^ y.CA^ -CF^- : FKS 6 '»2 and (19) CA^ : Ca^y.Ca^ : FE2,(=|P2,) Therefore CA2-CF2=Ca2, and CF-^=CA2^Cfl2. Q. £. D. Fig. 9. Cor. 1.— Hence FF2=AB2—a69=AB+fl6-AB-a6, or F/is a mean proportional, between the sum, and dif- ference of the two axes. Cor. '2. — The two semi-axes, and the Focal distance fi'om the centre, are the sides of a right angled triangle, in which the hypotenuse, or distance of the Focus from the vertex of the conjugate axis, is equal to the semi- transverse. _. _ For Fa=-Ca2=CF2, ^^ • and CA2-Ca2=CF%.-.Fa=CA. OF THE ERLIPSE. 47 Cor. 3. — The square of the Focal distance from the centre, is equal to the rectangle of the semi-transverse, and the difference of the semi-transverse and the semi- parameter. That is FC^=CA . CA-FE, For FC2=CA^"^C^and aC^=(19) AC . FE, Therefore FC^ ^CA^ -AC . FE=AC rAC"^^FE, 2. i. and.'. AC:FC::FC : AC-FE. 6. i7. PROPOSITION IX. The sum of two lines drawn from the Foci to meet at any point in the curve, is equal to the transverse axis. That is /E4-/E=AB. Fi-. 9. For, draw AG parallel and equal to Ca the semi-conju- gate; and join CG meeting the ordinate DE in H ; also take CI a fourth proportional to CA, CF, CD : Then(Il,CoR.1.)CA2 : AG^ : iCA^ -CD^ : DE-% 6.22.4 And, (sim. tri.) CA2 : AG^ ; iCA^ -CD^ lAG^-DHS ^- ^• consequently DE^r^AG^ -DH2=Ca2 -DH^. Also FD, is the difference between CF. and CD, And FD=^=CF=^--2CF .CD-fCD2, FE-=FD2+DES Therefore FE-^=CF2+C«=^ -2CF . CD-f CD^^ -DH^ But (VIII, ) CF2+Ca2=CA2; and (Hyp.) 2CF . CD = 2CA . CI ; Therefore FE2=CA=^-2CA . CI-f■CD2—DH^ 2. 4. 48 OF THE ELLIPSE. Again (Hjp.) CA^ : CD^ : :CF^(or CA^^AG^^) : CI^ ; And CA3 :CD«::CA2-AG2 :CD-"-DH^ CP=CD2-DH2, Consequentl}.F'E-=CA2-2CA . Cl-f CI^. ^ And the side of this square is FE = CA — CI=AI ; .. 4. Cor. jjj ^^^ ^^^^ manner it is found that,/E=CA + Cl=BI ; FE+/E==AI-fBI=:=AB. Q. E. D. Cor. 1 .—The difference between the semi-transverse, and the distance from the Focus to any point in the curve, is a fourth proportional to the semi-transverse, the distance from the centre to the Focus, and the distance from the centre to the ordinate to that point. That is CA : CF::CD : CI. (or CA-FE), Cor. 2.— And /E-FE=2CI. Oi* the difference be- tween two hues, drawn from the Focus to any point in the curve is equal to twice the fourth proportional, to CA, CF, and CD. Therefore CA : CF: :2CD :/E-FE. !oR. 3.— Hence CA . /E - FE=t2CD . CF =(2CF or)/F . CD. Cor. 4 Also/E.FE=CA-fCl.CA-Ci=CA2-CP. Cor. 5. — From this proposition is derived the com- mon method of describing this curve mechanically, by points, or with a thread thus : Fig. Id. In the Transverse take the Foci F,/and any point L Then w^ith the radii AT, BI and centres F, j\ describe arcs intersecting in r„ which will be a point in the curve. In like manner, assuming other points I, as many other points will be fouiid in the curve. Then with a steady OF T^E ELLIPSE. 49 iiand, the curve line may be drawn through all the points of intersection e. Or take a thread of the length AB of the Transverse axis, and tix its two ends in the Foci F, / by two pms. Then carry a pen or pencil round by the thread, keeping it always stretched and its point will trace out the curve line. PROPOSITION X. If there be any tangent, and two lines drawn from (he Foci to the point of contact ; these two lines will make equal angles with the tangent. For draw the ordinate DE, andy*e' ; parallel to FE, Fig. i2i Then(lXCor.l)CA : CD::CF : CA-FE ; p, °^^ and (VI) CA : CD: :CT : CA, 3 therefore CT : CFmCA : CA-FE, and TF : T/::FE : 2CA-FE,(or/E)-(IX); 5 e' But (sim. tri.) TF : TF: :FE :/e', therefore /E=/e' and the angle f' = theangleyEe' ; but because FE is parallel tofe', the angle «' = the an- gle FET ; therefore The angle FET=the angle /Ee'. Q. E. D. ScHOL. — As opticians find that the ailgle of incidence is equal to the angle of reflection, it appears from this proposition, that rays of light issuing from one focus, and meeting the curve in every point, will be reflected into lines drawn from those points to the other focus, so the ray/E is reflected into FE, and this is the reason why the points F,/, are called ihe foci or burning points. Appollonius called the Foci, of the Ellipse and Hyper- bola Puncta ex comparatione facta, but he does not mention the Focus of the Parobala. We may hence conclude, per- haps, as well as from the term Latus Rectum, (see Prop. fK Schol.) that the latter was first fixed on, and that the 50 OF THE ELLIPSE. Foci, were discovered, or determined, from the relation of the urdin .tes to the Latus Rectum. N«^wton and others called the Foci Umbilici, (Apollo- tiii Conica — HamiUon's Conic Sections 102. Newton's Princjpia passim.) Cor. 1. — Hence the Normal^ EO, or the line drawn Fig. 9. perpendicular to the tangent, from the point of contact, bisects the angle made by the two lines, drawn from the Foci. For since the J^ormal, makes equal angles with the tan- ax 3 gent, OEP=OEF, or FE/is bisected by EO. 6. 3. Cor. 2.-— Consequently FE :/E: :F0 :/0 ; ^■^' and FE+/E :/E - FE: :F0+/0 :/0- FO ; That is (IX) 2CA : 2C1: :F/ : 2C0, therefore CA : C1::CF:C0. CoR. 3.-Therefore CA^ : CA . CI: :CF- : CF . CO, and CA^ : CF^ : :(CA . CI=) CF . CD : CF . CO, .-. CA^ : CF2::CD : CO. CoR. 4.-Therefore CA^ -CF^ : CF^ : :CD-CO : CO. thatis(Vlll) aC- :CF2::D0:C0. CoR. 5.— CA« : CA2-CF-::CD3 : CD-CO. that is CA2 : Ca2::CD : DO. 6.20. CoR. 6. -Hence (19) AB:P::CD:DO, ^''''^' and AC :iP::CD:DO, AC. DO=iP. CD. And if, as in the Parabola, AC and CD, be supposed equal, then DO, the subnormal^ equals half the Parame- ter. CoR. 7.-2CA : 2CI::/F: 2C0::CF : CO, 2CA : 2CA-2CI: :CF : CF-CO. thatis(IX) 2CA: 2FE ::CF:OF; or CA:FE ::CF:OF. OF THK ELLIPSE. 51 PROPOSITION XI. If a line be drawn from either Focus, perpendicular to a tangent to any point of the curve, the distance of their intersections from the centre will be equal to the Semi- transverse axis. That is CP and Cp each = CA or CB. Fig. 12, For throtigh, the point of contact E draw FE, and/E meeting FP produced in G, then the angle GEP = dngle FEP, being each equal to the angle/e/?, and the angles at P being right and the side PET being common, the two triangles GEP. FEP, are equal in all respects, and so GE = FE, and GP = FP. Therefore, since FP = iFG, and FC = |F/, and the angle at F common, the side CP will be ^iFG or lAB that is CP = CA or CB. And in the same manner C 2?=CA or CB. Q. E. D, CoR.l. — A circle described on the Transverse axis, will pass through the points of intersection P and p. CoR. 2. — If from the intersections of a tangent with a circle, circumscribing an Ellipse, perpendiculars be drawn they will pass through the Foci. Cor. 3. —The distances of the Foci from the point of contact, are to each other as the perpendiculars. For the triangles EPE,yyoE are similar ; Therefore FE : /E: :FP : //?. Cor. 4.-If PF and pC be produced, they will meet in the circumference of the circle in K, of which /?K will be the semi -diameter, for/PK is a right angle, also FKC, C;?/*are similar and equal triangles ; Therefore PF . pf=?V . FK = AF . FB=Co*. (Vlf. Cor. 5.) 5^ OF THE ELLIPSE. PROPOSITION XII. The distance of the Focus from any point in the curve, is equal to the ordinate to that point, continued until it meets the Focal tangent, Fig. n. that is Fa==C^ FA=AL, FB=Bz, and generally FM=» DG, for(19) FE = iP, and (VII Cor. 5) FE . C/=iPAC=FE . AC, therefore C^=AC=Fa. (VIII) Again (VI) CF : AC: : AC : CT, and AC : AF::CT : AT, Sim. tri. AT : TC::AL : C<, = CA ; AF : AC::AL : AC, therefore AF=AL. That is, the distance from the vertex to the Focus, is equal to the tangent lo the vertex, intercepted by the Focal tangent. (VI Cor. 7,) AF : FB: :AT : TB, .-.Sim tri. AT : TB: : AL (=AF) : BZ, therefore AF : FB: :AF : BZ, BZ=.BF. (VI.) AC^ =CF . CT=CF : CF+FT=CF2 -|-CF . FT, .♦. AC2-CF~^(=aC^)=CF.FT, Sim. tri. TF : FE: :TC : C/. = AC, but TC : AC: :TC . CF : AC . CF, - . TC.CF(=AC2) : AC .CF::AC:CF, that is no TF : FE : : (AC : CF) : : TA : AL : : TD :DG ; but (IX) CT : AC::CD :FM-CA, and CT+CD : AC + FM-CA: :CT : AC, that is TD: FM::CT: AC; .-, TD : DG: :TD : FM.-.DG=FM. Q. E. Z). CoR.l — If through T, a line be drawn perpendicular to TA, it is called the directrix. OF THE ELLIPSE. 53 The distance of any point in the curve, from the direc- trix as My IS equal to TD which has a constant ratio to DG oiF-M, the distance of the point from the Focus. In the Parabola it is a ratio of equality. PROPOSITION XIII. If tangents and ordinates be drawn from the extremities of any two conjugate diameters, meeting the Transverse axis produced, the distances of these intersections from the center are reciprocally proportional* That is CD :Cd:\Ct:CT, Pi?- J 4. For(V[) CD :CA::CA :CT, and ,, Cd t CA; :CA J C/, CD :Cd::Ct : CT. Q. E. D. Cor. 1. — Hence the distance from the center, to the ordinate, at the extremityfof one diameter, is a mean pro- portional between the distances of the intersections of the tangent and ordinate drawn from the extremity of the other. For since CD : Cdy.Ct : CT, and bysim. tri. Cd : DT: :C^ : CT, CD :Cd::Cd: DT. In like manner Cd : CD: :CD : dt. Cor. 2 — The ordinates, also, are reciprocally as their distances from the center. Forsim. tri. DE : de.'.Cd: CD. CoR. 3. — Therefore the ordinates are, as the distan- ces of the intersections of their tangents from the center, or (Cor. 1,) DE ideWCT : Ct . 8 54 OF THE ELLIPSE. Cor. 4. -Also by (Cor. 2,) The rectangles DE . CD therefore, the triangle CDE=-Crff. PROPOSITION XIV. If ordinates to the Transverse axis, be drawn from the extremities of any two conjugate diameters, the sum of their squares is equal to the square of the Semi-Conjugate axis, and the sum of the squares of two ordinates, drawn from the same extremities to the Conjugate axis, is equal to the square of the Semi-Transverse. Fig. 14. That is, I>e^-\-de^=CaK And D'e'-{-d'e^=CA\ For (VI) CD : CA::CA:CT, CD :CA::AD : AT ; And CD:DB::AD:DT, . . (XIII. 2) CD . DT (=Crf2)=AD . DB=CA2 ~CD^ CJ2=CA=^-CD2; And CD2 4-C(/2=CA2; That is D'E^-+d'e'=CAK In the same manner DE=-|-c?'«2=Ca2. q. E, D. CoR. 1.— Hence AD .DB=C(;% and AJ. c/B=CD^ Cor. 2.— Hence CD^ +0^^ =CA^ ; and CD'2 4-C(i'^=Ca2; Cor. 3.— Since CX' :Ca' '..{AD .DB=) Cd^:DE\ .'. CA:Ca::Crf:DE. and CA:Ca::CD:de. OF THE ELLIPSE. ^$ PROPOSITION XV. If from the extremity of any diameter a perpendicular be drawn to its conjugate, the rectangle of this perpendic- ular, and the part of it intercepted by the Transverse Axis, is equal to the square of the Semi-Conjugate axis. Fig:. 14. That is EP.EI=Ca^ Draw C^ parallel to EP, the triangle EDI, and Cyt are similar : Therefore Ct : Cyr.El : ED, That is Cr : EP::EI:C(/', Therefore EP . EI=C/ . Cd=^Ca'. Q. E. Z>. PROPOSITION XVI. All parallelograms circumscribed about an Ellipse whose sides are parallel to two conjugate diameters, are equal to each other, being each equal to the rectangle of the two axes. That is KQNS=RYZX=AB . ab. Fig. 13. For (II h XIV) CA^ : CaA ! (AD . DB or) C^/^ : DE^; Therefore C A iCay.Cd: DE, In like manner CA : Ca ! :CD : de ; Or CA:CD::Ca:de. But (VI) CA:CD::CT:CA, Ca : del :CT : Ca, Also, siui. tri. Ce : de: ;CT : Cy, Therefore Ca iCe'.'.Cy : CA ; And Ca . CA=C« . Cy or G^ . EP. 4Ca . CA, or AB . ab=iCe . EP, or QKSN. Q. E. D, 55 OF THE ELLIPSE. PROPOsrrioN xvii. The sum of the squares of every pair of conjugate di- am ers, equal to the same constant quantity, viz. the sum of th- squares ol the two xes. That is, AB -{-ab^ =EG' -{-egK Fig. 14. For CE2+Ce2=-CD2+CJ2-fDE2-fDe2; But XIV, D +Cri^=AC3;andDE^-fr/e^=Cfl2; therefore CE -|-Cc ^AC^+Cr-', and EG''+eg^=AB^-]-ab\ Q, £. A PROPOSITION XVIII. If the extremities of the transverse and conjugate axes be connected, the diameters which are parallel to the con- necting line, are conjugate and equal to each oiher. Fig. 16. That is, EL=GK, and they are conjugate to each other. For Aa=«B=B =6A.*.AaB6, is a parallelogram, also the angles GCB=a\B=hAB = LCB, and since AB=2CB. .Ba=2BR, aB is a double ordinate to GK, and EL parallel to it, is conjugate to GK. Also CG=CL or KG=EL. Q, E. D. Corollary. — These are the only two congugate di- ameters in the Ellipse, which are equal. This Corollary and the last step in the demonstration may easily be demonstrated by reductio ad absurdum. — See Emerson's Conic Sections — Ellipse — Prop. 3.5, 36. 37. OF THE ELLIPSE. PROPOSITION XIX. 57 If from any point in the Ellipse, a line be drawn to the conjugate axis, equal to the semi-transvere axis, the part of the line intercepted between that point, and the trans- verse axis, is equal to the semi-conjugate. That is, if MG=AC,"then MI=aC. Fi^. s. For, sim. tri. MG : MR: :MI : ID, and MG2 :MR ::MI-^ : IDS that is AC2 : ( :D»::MP : ID^, and AC^ : AC^-CD-.iMP ; MI'-ID^, AC : AD.DB:: MI^iMD, AC-- :MI^::AD.DB:MD2 ; but (II) AC- : «C2 : : AD . DB : MD% MV=aC^ and MI=aC. Q. E. a CoR.— Conversely— If ]MI=aC then MG = AC:. PROPOSITOIN XX. If a tangent and ordinate be drawn from the extremi- ties of the transverse axis, and of any other diameter, and be produced to meet that diameter and transverse axis, the triangles formed respectively by these lines, are equal. That is. CAE=CMT and AGT=MGE, Fig. 17. also CHN=CDiM and AHE=MDT. 1. For CDM and CAE, also CHA and GMT are similar. .;. CH : CM::CA : CT, ::CD : CA, ::CM : CE; -^ therefore DH, AM and TE are parallel; AME = AMT, and CAE = CMT. 2. Also taking CMGA from each, AGT=MGE. 6. 15. 3. * adding MDA, to each, xMDT = AHE. 4. therefore CDM=CHA. Q. E. D. d» OF THE ELLIPSE. CoR.l.— AME+AMD=MDAE=MDT ; AMH4-AMT= AHMT=AEH ; MDAE=AHMT. Cor. 2. QIR=AEFI. Forsim.tri.CA;AE::CD:DM::CD-fCA:DM-fAE; and CA : AE::CI : IF::Cl-|-CA : IF+AE; Again AI . IB=AI . IC+CA ; and AD.DB=AD.DC-f CA; therefore AI. IB :AD.DB::AI.IC4-CA: AD. DC+CA, :AI.AE+1F:AD.AE+DM, .2AEFI: 2AEMD, : AEFl : AEMD. Butsirn.tri. QIR : MDTliQi^ : MDS ::AI . IB : AD . DB, ::AEFI : AEMD; But (Cor. 1.) MDT=AEMD . . QIR=AEFI. Cor. 3. QLF=LMTR. For take CLR from CAE and from CMT, then LRTM=LRAE=LRIF+AEF1=LRIF-|-RQI =L(iF. PROPOSITION XXI. The square of any diameter, Is to the square of its conjugate, As the rectangle of its abscisses, Is to the square of their ordinate. Fig. 17. That is NM^ : VK^iiNL . LM : LQ^ For (XX. Cor. 3.) LQF=LRTM ; and as LRQ moves parallel to itself, towards CK, LRTM approaches to equality with CTM ; They may therefore be considered as ultimately equal, and CKO=CMT. OF THE ELLIPSE. 59 But Sim. tri. CMT : CRLr.CINP : CL^ 6. 20. and CMT : CM r-CKL: :C]VP : CM^— CL*, that is CMT : LRTMxCM^ : NL . LM ; Butsim. tri. CK-' : LQ^iiCKO : LQF, ::CMT: LRTM, ::CM' : NL.LM, therefore CK= CM^ : iLQ^ : NL . LM ; and NM^ : VK^::NL.LM : LQ^ Q. E. D, ScHOL. — The preceding proposition shows, that any di- ameter of an Ellipse, and its rectangles, have the same re- lation to its conjugate, and ordinates, as have been demon- strated in the case of the Axes. All the properties therefore, which have been deduced from this relation, in the former case in reference to lines intersecting each other, parallel to the axes, and to tangents intersecting these lines and the curve, may be applied to any conjugate diameters. The first six propositions, therefore, of this chapter, with their corollaries, are particular truths which may safely be gen- eralized by applying them to any two conjugate diameters, as well as to the two axes. The general propositions, however, will be stated, and the corresponding figures have been drawn and lettered in a similar manner, so that the demonstrations already given, for the axes, are appli- fjable, with slight variations, to those propositions also, which will be arranged like the corresponding propositions, concerning the axes. CoR. L — The squares of the ordinates o{ any diameter are to each other, as the rectangles of their abscisses. That is CK^ : LQ-^ : :NC . CM : NL . LM. (See I Prop.) Fig. 17. Cor. 2. — The corresponding ordinates on each side of any diameter are equal. CoR. 3. — Ordinates at equal distances from the center, or from the vertices of any diameter are equal ; and, con- versely, equal ordinates are at equal distances irom the center and from the vertices. 60 OF THE ELLIPSE. Cor. 4. — The whole Ellipse is divided by every diam- eter into two equal parts, which may be placed so as to coincide in every respect. Cor. 3. — Therefore the parts of the tangents, at the ex- tremities of the diameter, would coincide, and the two tan- gents are parallel, as demonstrated, Prop. 1 and III. Cor. 6. — Any diameter is to its parameter, as th e rec- tangle of its abscisses, is to the square of their ordinates. Cor. 7. — The rectangle of the absciss of any diameter and its parameter, exceeds (he square of the ordinate, by a rectangle similar, and similarly situated to the rectangle of the diameter and its parameter. That is DB . BG>DE-^ by FIHG. CoR. 8. — If a circle be described on any diameter, The ordinate in the circle. Is to the corresponding ordinate in the Ellipse. As the given diameter, To its conjugate. Fig. 18. That is CA : Ca: :DG : DE. PROPOSITION XXll. The square of any diameter, is to the square of its con- jugate, as tile rectangle of the segments of a line in the Ellipse, parallel to the former, is o the corresponding rec- tangle of the segments of a line parallel to the latter. 19. That is, CA=^ : Ca2 or AB^ : ah^ :\eh , hg-\-Eh . ^G. SeeProp. IV, &ic. Cor. 1. — Therefore by equality of ratios : The corresponding rectangles of the segments of lines intersecting each other, parallel to any diameter and its conjugate, have to each other, always the same ratio, viz. that of the squares of their diameters. That is, eh.hgiek, kg: :Eh . ^G : ik . kl. OF THE ELLIPSE. gj ScHOL. — If the Ellipse become a circle, then the same proposition holds true, as is demonstrated by Euclid, but the ratio is then, by alternating the terms, a ratio of equal- ity. Ill Book, 35. l( hgf and kg, are considered equal, as in the Parabola, then the proposition \s eh : ekwE^h . AG : ik . kl. As demonstrated of the Par. Prop. II. PROPOSITION XXIII. If lines in the Ellipse parallel to two conjugate axes, in- tersect each other without the Ellipse, the rectangles of corresponding segments, are to each other as the squares of the diameters to which they are parallel. That is, ; B^ : ab^ : :eH .HglEH . HG. See Prop.V, &;c. CoR. I. — Hence the rectangles of the segments of all lines, parallel to two conjugate diameters, and intersecting each other without the Ellipse, have to each other always the same ratio, being the ratio of the squares of those two diameters. That is, EH . HG : eH . H^: :K1 . IL : EI . I/. Cor. 2.— Also EH . HG : ES . SG: :eH . H^ : Sa^ €oR. 3.— Also ek.kniEk . kl: :EH . HG : cH . Bg. PROPOSITION XXIV. The squares of any two diameters, are to each other, as the rectangles of the segments of one of those diameters, or of lines parallel to it, are to the corresponding rectangles of the segments of lines parallel to the other. That is AB3 : EG^ : : AD . DB : KD . DL. Fig. 9 62 OP THE ELLIPSE. There is of course no corresponding proposition, in the caseof the two axes, as they have a definite position in regard to each other, but this more general proposition may be deduced from the preceding propositions by the ordinary changes of geometrical ratios. CoR.—ed .dg:Kd, dL::CB^ : CG' : : AB^ : EG», and ed . dg : eh . hg: '.Kd . dLi : Eh . AG. That is, the rectangles of the corresponding segments of all parallel lines, intersecting each other in the EllipsCy have to each other the same ratio. PROPOSITION XXV. If any two lines, cutting the Ellipse, intersect each oth- er, without it, the rectangles of their segments, are to each other, as the squares of the diameters, or semi-diameters to which thevare parallel. Fig. 23. That is EH . HG : eH . H^: :CG'-^ : C^'^ This is deduced in a manner similar to the last, CoR. 1. — The squares of tangents intersecting each oth- er, are to each other, as the squares of the diameters, or semi-diameters, to which they are parallel ; or the tangents, are as the parallel diameters. Fig. 22. That is HxM-^ : HN2 yCm^ ♦ Cn^ ; or HM : HN::Cm : Cw. rig. 22 Cor. 2.— IQ . QS : ED . DL: :T)N' : QM% andlQ.QS :ED . DL: :IP . PSiEP.PL: iCG^ :CK-. Cor. .3.— HN : HM : : TF : TM : : CG : CK, and LR.RE-.RF^I'.ED. DLiDN^iCK^ : CG^. ScHOL. — In the two preceding propositions, with their Corollaries, the principle involved in the preceding gene- ral propositions, (and of course in the particular prop osi- lions relative ro tlr? axis,) is seen in its most general form. It may be expressed as follows. If parallel lines in the OF THE ELLIPSE. G3 Ellipse intersect another line, or intersect other parallel lines, either within or without the Ellipse, the rectangles of their corresponding segments, ore always in the same ratio, and if any of the lines touch the Ellipse, the squares of the intercepted parts of those lines may be considered as rectangles, and are as the squares of diameters parallel to them. It will be seen by trial, (reccollecling that the square of a semi-diameter, or of an ordinate is equal to the rectangle of the equal parts of the whole diameter, or double ordi- nate) that this general proposition as now stated, is appli- cable to all the cases alluded to, and includes propositions I, II, III, IV, V, XXI, XXII, XXIII, XXIV, XXV, From it also may be very easily deduced, all the general propositions which are most useful, concerning diameters, tangents, he. PROPOSITION XXVI. If a tangent and ordinate be drawn to any point in the curve, meeting any diameter, the half of that diameter is a mean proportional between the distances of the two in- tersections from the centre. That is CD.CT=CA2, or CD : CA : CT, are continued proportionals. {This may be demonstrated in the same manner as in Prop. VI but it is immediately deduced from the Gene- ral Ppoposition just stated, as it is illustrated in Prop. XXV, Cor. 3.) For HB:AK::EH :EK::BD :DA::BT: AT; Fig. 24. andBD-DA (or 2 CD) : BD: :BT - AT (or BA) : BT, CD :BD::BC ;BT, and CD :BD-CD::BC:BT-BC; That is CD : AC:: AC : CT. q. E. D. CoR. 1.— The two angents, ET, ET drawn at the two ^'^S- 25. extremities of the same double ordinate, intersect the diam- eter in the same point. And, conversely, if a diameter pass through the point of intersection, it will bisect the line which connects the points of contact. 64 OF THE ELLIPSE. Fig. 24. Cor. 2.— HK and FG are bisected. Hence IF=MG. Cor. 3. — DT is harmonically divided, or DT : AT::BD : AD. Cor. 4.— TA : TD: :TC : TB. also TK : TE::T^ : TH. and AK: DE::C/-fBH. ScHOL. — As the Transverse Axis, passes through the Foci, it has particular relations to those points, which no other diameter can have. In the relations which other di- ameters have to lines drawn through the Foci, there are however some analogies to those of the axis, as will be seen in the Notes. OF THE HYPERBOLA. PROPOSITION I. The squares of any two ordinates to the axis, are to each other, as the rectangles of their abscisses. Let MVN be a triangular section through the axis of the ^^n- *• cone (3), AGIH, another section perpendicular to the for- mer, forming an Hyperbola; AH, the mutual intersection of the two, will be the transverse axis of the Hyperbola (11); let MIN, KGL be circular sections parallel to the base (4) ; FG, HI, the mutual intersections of these planes with that of the Hyperbola, will be ordinates, both in the Hyperbola, and in the circles, being perpendicular, both to MN and KL, and to AB. Then FG^ : HP : : AF . FB : AH . HB. For Sim. tri. AF : AH : : FL : HN, And FB:HB::KF :MH; Therect. .:. AF . FB : AH . HB::KF. FL : MH . HN, 6. C. But KF . FL=FG^ and MH . HN=HIS _ . _ Therefore AF . FB : AH . HB: :FG=» : HP. Q. E. D. 5. 7. ScHOL. — If then a straight line, placed perpendicularly \ipon another finite straight line produced, were to move parallel to itself, and to vary in its length continually, 60 that its squares should always have the same ratio to each other, as the rectangles of the parts between it and the extremities of the other line, the figure formed by the moving line, would be the portion of an Hyperbola between the axis and carve. 6S OP THE HYPERBOLA. Cor. 1. — As the proposition and demonstration, are equally applicable to ordinates on opposite sides of the axis, having the same abscisses, those ordinates are equal or GS, is double of FG. Hence it is called a double or- dinate. Cor. 2. — The squares of the ordinates are the same as the rectangles of the equal segments of the double or- dinates, therefore, Ihe proposition may be thus expressed. The rectangles of the segments of the double ordinates^ are to each other as the rectangles of the abscisses, of the tranS' verse produced. Cor. 3. — Ordinates at equal distances from either ver- tex, are equal ; since the abscisses and consequently the rectangles of the abscisses are equal. And conversely, if the ordinates are equal, their distances from the vertices are equal, CoR. 4. — Hence the Foci are equally distant from eith- er vertex, for their ordinates are equal, being each equal to the semi-parameter, (20), their distances from the cen- tre are therefore equal. CoR. 5. — Hence also every diameter is bisected in the centre. For since at equal distances from the centre, the T\%. 13. ordinates are equal, the third sides and the remaining an- 1. 14. ^^^ °^ *^^ T\g\\\. angled triangjes, CDE, Cd'G are equal, therefore ECG, is a right line bisected in the centre. And conversely, the ordinates from the ends of any diam- eter upon the transverse axis are equally distant from the centre. CoR. 6. — Hence also, the whole Hyperbola, is by the transverse axis, divided into two equal parts, which if pla- ced one upon the other will coincide in every respect. For if they should not coincide in any part, the or- dinate in one, at that point, would be unequal to the cor- responding ordinate in the other. CoR. 7. — For similar reasons, the two parts of the tan- gent at the vertex, would coincide ; that is, the tangent at the vertex makes equal angles with the axis, on each side of it. It is therefore perpendicular to the axis. OF THE HYPERBOLA. (57 Cor. 8. — The two opposite sections of a Hyperbola, are equal and similar, and if placed upon each other would coincide in every respect ; for. ag the ordinates at equal dis- tances from either vertex are equal, the points in the curve which limit these ordinates would coincide. Cor. 9. — The two diameters, drawn from the extrem- ities of any double ordinate to the axis, are equal. Cor. 1 0. — If a circular section as POC, pass through the ^'•0- i- centre of the Hyperbola, the semi-conjugate Ca, is a mean proportional between OC and CP. For the whole conjugate axis, is a mean proportional between BR and AQ , the diameters of the circular sec- tions, passing through the two vertices of the Hyperbola. (13.) or BR : aby.ab: AQ, but AB=2AC,.-.BR=2CP, and QA=20C, therefore OC : Ca'.'.Ca : CP. 5, 15. SciiOL. — It has been already remarked, that if the cut- ting plane be supposed to revolve about A, the axis AB will become more extended, the abscisses FB and HB, will ap- proach to a ratio of equality, and the section itself will ap- proach indefinitely near to a Parabola ; and when it be- comes parallel to the side of the cone, it will be a Parabo- la (5). If then the indelinite abscisses are considered Fig. 2. equal, the precedintj propositiofi will become, AF : AH::FG*::HP e. i. the same as the first proposition of the Parabola. Again if the cutting plane, move parallel to itself, on the base MN, towards M, A and B will approach continually towards V, FB and HB will also approach to equality with FA and HA. When A and B coincide in V, the section will become a triangle (3) and FB, HB will be equal to FA, HA, and the rectangles AF . FB, AH . HB, will be equal respectively to AF^, and AH^, then the proposition becomes AF2 : AH^::FG2 : Hi% AF : AH : :FG : HI. (See Euclid, 6, 1.) 68 OF THE HYPERBOLA. PROPOSITION II. As the square of the Transverse Axis, Is to the square of the Conjugate Axis, So is the rectangle of the abscisses of the Transverse, To the square of their ordinate. Fig 1- That is AB^ : a6» : : AD . DB : DE^. For sim. tri. BC : CO: :BF : FK, and „ „ AC:CP::AF:FL, AC . CB : CO . CP: :AF . FB : KF . FL ; that is AC2 : Ca^ : : AF . FB : FG=» ; so AC^ :Ca2::AD.DB: DES 5. 15. or AB2 : ah^ : : AD . DB : DE^ q. E. D. 2. 6. Cor. 1.— Since the rectangle AD . DB:=CD2 -CA% Therefore AC^ : aC^ : :CD» -CA^ : DE^; or AB^ : ab^ ::CD»— CA^ : DE^. CoR. 2.— (19)AB :ah lab : Parameter=P, cof'2 ^^^^ ^^ * ^ ::AB2 : a6-% 5.'ii: •• AB :P ::AD.DB:DES or The Transverse axis, Is to its Parameter, As the rectangle of its abscisses, To the square of their ordinate. Cor. 3.-If different Hyperbolas have the same Trans' verse axis the corresponding ordinates in each, will be to each other as their Conjugate axes. For as the first and third terms in the proposition, will then be the same, the second will vary as the fourth. Fig. 2. Cor. 4. — Let APGB be the rectangle of the Transverse axis, and Parameter BG, then D'HGB is the rectangle of the absciss DB into the Parameter, and D'IFB or the rec- tangle D'l . D'B is equal to the square of DE. For (Cor. 2.) AB : BG::AD' .D'B : DES 6. 1 and Sim. tri. AB : BG : : AD' : D'l : : AD' . D'B : D'l' . D'B, and .-. AD' . D'B : D'E^ : : AD' . D'B : D'P . D'B' ; D'E2=D'I . D'B. OF THE HYPERBOLA. 69 ScHOL. 1. — The square of the ordinate D'E, therefore (whichequalsD'IFB)is greater than the rectantjle of the ab- sciss D'B into the Parameter, or D'HGB, by the rectangle IHGF similar and similarly situated to the whole rectan- gle PB. It was on account of this excess in the square of the ordinate, compared with the rectangle of the ab- sciss into the Parameter, that Apollonius named this sec- tion the Hyperbola. ScHOL. 2. — If the two axes are equal, that is if the sec- tion is an Equilateral Hi/perbola, then the rectangles of the abscisses are equal to the square of the ordinate, the Equilateral Hyperbola, having the same relation to a cir* cle, (which may be called ^n Equilateral Ellipse,) 2iS any other Hyperbola, has to an E^llipse. ScHOL. 3. — From the manner in which the Hyperbolic section is made, it is evident, that as the cones are more or less obtuse, the Conjugate axis, will be encreased or di- minished, while the transverse remains the same ; and it appears from the 3d Cor. that the curve will also become more or less obtuse. A section therefore may have its Transverse axis, equal to the Conjugate of another sec- tion, and its Conjugate may be equal to the Transverse of the other ; and in this case they are Conjugate Hyper- bolas, such are all those considered in the remaining prop- ositions of this section, each pair of opposite Hyperbolas having the same properties. PROPOSITION HI. As the square of the Transverse axis, Is to the square of the Conjugate, So is the sum of the squares of the semi-transverse, and of the distance from the centre to any ordinate, To the square of that ordinate produced to the Conjugate Hyperbola. ^ 10 70 OF THE HYPERBOLA. Thatis, AB^ lab^y.CA^+CD' iDer-, For (II Cor. 3 and Schol.) Ca'- But, CX^::Cd'-Ca' :de^', and 5> or C«2 =Je2 and De^=Cd''. :De-::CA2 :CA-+CD% :CAMDe2 CA^+CDS CA-' :Ca2::CA3+CD^ : Dd^; AB2 :a62::CA2+CD2 iDe^. In like manner AB^ : ah^^ : CA^+CD'^ : DV^ also ab^ : AH-^ : Cfl^+C^i^ -. dE'\ Q. E. I), PROPOSITION IV. The square of either axis, is to the square of the other, as the rectangle of the segmentsof a line in theH^'perbola par- allel to the former, is to the rectangfe of the corresponding segments of a line parallel to the latter. f'iS 4. That is eh . hg I Eh , AG: : AB^' : ab\ Or Eh . hG I eh , hg'.'.ab'' : AB^. , ^ For (1) ED2 : eN^::AD . DB : AN . NB, and ED' : ED^-cN^ : :AD.DB : AD.DB-AN .NB. Bat ED2-eN==ED2-AD2=EA . AG, And AD.DB-AN.NB=BD.DN4-NA-BD-DN.NA o. 1. And since, BD . DN+NA = BD . DN-fBD . NA, And BD— DN . NA=BD . NA~DN . NA, Therefore, AD.DB — AN . NB=DN . BD+NA . DN = Dtt ,DN = hg '. ehf ED^ :Eh. AG:: AD . DBr.eh . hg, And ED^ : AD. DB::EA , hG l eh . hg. 5.11. But (II.) EDM AD. DB::a62 : AB% EA . AG : eA . hg'.'.ab^ : AB^. Q. E. D. OF THE HYPERBOLA. 71 Cor. 1 . — Hence by equality of ratios, the rectangles of the corresponding segments, of lines parallel to the two axes, mutually intersecting each other in the Hyperbola, are to each other always in the same ratio; namely, the ratio of the squares of the two axes. ScHOL. — As the square of either semi-axis, or of an or- dinate, is the same as the rectangle of the two halves of the axis or double-ordinate ; this proposition evidently in- cludes Prop. H and HI, as it demonstrates, that the rela- tions there shown (o exist between the two axes, extend to all lines in the section parallel to them. For AC^ : flc^ : : AD . DB : DE^ is the same as AC . CB : aC . C6: :AD . DB : ED . DG which may be considered as included in this propo- sition. If the cutting plane move parallel to itself, until the section becomes a triangle, the preceding proposition should be applicable to the opposite triangles ; which would also follow from similar triangles. If the plane be supposed to revolve until the section be- comes a Parabola, and the longersegments of the lines par- allel to the transverse axis be considered equal. Then the Prop. AD . DB : eh . A^: :ED . DG : Eh . AG, becomes AD : eA : : ED . DG : EA . AG, The same as is demonstrated in the Parabola, Prop. II, Cor. 1. PROPOSITION. V. If lines parallel to the two axes, intersect each other, without the Hyperbola, the rectangles of their correspond- ing segments are to each other as the squares of the axes to which they are parallel. 72 OF THE HYPERBOLA. Fi?. 5. That is, EH . HG : eH . Hgy.ab' : ABS or ell . Hg : EH . HG::AB=^ : abK For (T,) cN2 : ED^ : : AN . NB : AD . DB, And eN^ : cN2-ED^•:AN.NB : AN.NB— AD.DR. But eN='-ED2=HD2 -ED-^=EH .HG, Also AN . NB=BN . AD-hND=BN . AD-f BN . DN, And AD .DB=AD . BN - ND=BN . AD - AD . DN, Therefore AN . NB-AD . DB=-BN . DN + AD . DN, =DN . BN+AD=DN . BN+Brf =DN . Nf/=DN . D/i. Therefore (eN2=)HD2 : EH . HG: : AN . NB : (DN. I>n=)eH . Hg, And HD2 : AN . NB: :EH . HG : eH . Hg; But (II,) HD='(=eN-^) : AN . NBy.ab' : ABs EH . HG : eU . Hg :\ab' : AB^. Q. £. D. CoK. ]. — Since this proposition is equally applicable to all lines parallel to the axes, it follows by equality of ratios, that, the rectangles of the segments of all lines parallel to the axes, intersecting each other without the Ellipse, are to each other always in the same ratio ; viz. that of the squares of the two axes. Cor. 2. — If DH be supposed to move towards AP, the rectangle EH . HG approaches continually to an equali- ty with the square of AF, as its limit^ and when D coin- cides with A, AP2 may be substituted for EH . HG. That is, the rectangles of the segments of lines parallel to the axes, are to each other, as the squares of the inter- cepted parts of the tangents to the vertices of the axes. f> -' Cor. 3. -Since (IV,) Eh . hG : eh . hg Wab^ : AB% ^'^- ^' and from this EH . HG : eH . Hg : ab^ : AB% Therefore VJi . hG : eh . hgWF.R . HG : f H . H^. OF THE HYPERBOLA. 73 ScHOL. . If the Hyperbolas are Equilateral^ the ratio will in all cases be that of equality. That is EH . HG=eH . Hg. Also if the plane revolve until the section becomes a Parabola, and the longer segments are considered equal, then the rectangles of the double ordinates produced, are as the external parts of the diameters. That is EH . HG is as eU. As demonstrated of the Parabola II, Cor. 1. The live preceding propositions may all be embraced in the (oWow'm^ general proposition. If lines parallel to one axis of the Hijperhola intersect the other axis^or lines parallel to it^ either within or without the section ; the rectanf^les of the corresponding segments will ahuays have to each other, the same ratio. PROPOSITION VI. If a tang;ent and ordinate be drawn to any point of an Hyperbola, meeting the transverse. axis produced, the semi- tratisverse is a mean proportional between the distances of the two intersections from the centre, That is, or CM . CT^CA^ ; CM : CA : CT, are continued proportionals. Fig For from the point T draw any other line TEH to cut the curve in two points E and H ; from which let fall the perpendiculars ED and HG, bisect DG in K, Then (1) AD . DB : AG .GB.lDEMGH^, And bysim. tri. TD- : TG^ ; iDE^ ; GH^ ; Therefore AD . DB : AG . GB ::TD-^ : TG\ But DB=CB-|-CD=AC+CD=CG-|-DC-AG = 2CK -AG, And GB=CB + CG=AC+CG=CG-f DC - AD=2CK -AD. 74 OF THE HYPERBOLA. .!. AD . 2CK-AD . AG : AG . 2CK-AG . AD:: TD3 : TG% and DG . 2CK : (TG^ -TD^or) DG . 2TK: : AD . 2CK 5- 17- 16 -AD. AG :TDS or 2CK : 2TK: : AD . 2CK- AD . AG : TD^ 5. 19. or AD.2CK : AD. 2TK: :AD.2CK-AD. AG : TD^j .'. AD.2CK : AD.2TK::AD.AG: AD.2TK-.TD% 5.A&18. and CK : TK::AD . AG : AD . 2TK-TDS 2.2. &3. and CK:TC::AD. AG:TD2-AD.TJD+TA; or CK :CT::AD. AG . AT^. But the limit of AD . AG, when the line TH comes in- to the position of TL, is AM* (See Parab. Prop. IV,) and K, then coincides with M. The proposition therefore becomes, CM : CTltAM^ : AT^ ; ThatisCM : CT::CM-CA : CA - CT , 5 19. or CM :CT::CM2 + CA2 :CA2+CT^ and CM :MT::CM2-1-CA2 : CM^ -CT% or CM : MT: iCM^ -f CA^ : CT-f-CM . MT, or CM^ : CM . MT : iCM^ ^CA^ : CM . MT4- CT . MT; Hence CM^ : CA^ : :CM . MT : CT . MT, ^^^ and CM^ :CA-:CM:CT, ^^on • •'• CM:CA ::CA:CT. v«rsely. Q. E. D» Cor. l.—CM : CT::AM2 : AT^, that is, the distances from the centre to the two intersections of the tangent and ordinate with the axis, are as the squares of the distances of the same intersections from the vertex. CoR. 2. — If a tangent and ordinate be drawn from any point in the Hyperbola, and the tangent be produced to cut the conjugate axia the semi-conjugate is a mean proportion- al between the distances of the two intersections from the centre. OF THE HYPERBOLA. 75 For CD:CL::AC :CT, .:. CD:CT::€D^ : CA^ .!. (II, Cor.)DT : CD: :CD^ -CA^ : CA» : :ED^ : aC' ^ But Sim. tri. DT : CT: :ED : Ct, ED :Q::ED2 : C^S That is Cd : Ci::Cd^ ! Ca% Therefore, Cd : Ca'.'.Ca : Ct, 6.19. Cor. That is Cd : Ca : Ct, are continued proportionals. Cor. 3. — As the demonstration is applicable to a tan- ^^?- '^• gent to the curve at either extremity of the double ordin- ate, the two tangents drawn to the two extremities of any double ordinate to the axis, meet the axis in the same point, and at equal angles. And, conversely, a line drawn through the intersection of two such tangents, if it bisect the angle, will also bisect the double ordinate ; and if it bisect the double ordinate, it will also bisect the angle, and in both cases it is the axis. Cor. 4. — Also Kk, the tangent at the vertex is bisect- ed by the axis, For TD:TA::DE : AK, and TD : TA.iDH : Ak, therefore DE : AK: :DH : AA;; but DE=DH.-.AK=AA:. CoR. 5* — If any number of Hyperbola* have the same transverse axis, and an ordinate in one be continued to intersect the curves of all, the tangents from all the points of intersection, will meet the curve in the same point; for since CD, and CA, the two first of the continued pro- portionals, remain the same in all, the third CT will be the same. CoR. 6. — Two tangents to the curve at the extremities of any double ordinate, cut the conjugate axis, at equal dis- tances from the centre. 76 OF THE HYPERBOLA. Fig. 8. Cor. 7. — If a perpendicular from either extremity of the transverse axis, be produced to intersect the Conju- gate Hyperbola, a tangent to the curve at the point of in- tersection, will pass through the other extremity of the transverse axis. Fig. 8. Cor. 8. — The following list of proportionals is de- rived directly from this proposition, and its second Corol- lary, viz. That CT:CA::CA:CD. And Ct : Ca : :C« : Cd. (1.) Then CT-f CA : CA-CT: I'^/A+CD : CD-CA that is BT : AT::BD : AD. In the same manner ht : at'.'.bd I ad. (2.) Also CT : CT-|-CA::CA : CA-j-CD. that is CT : BT::CA: BD. And Ct:bi::aC : bd. (3.) Again (2.) CA-CT : BD-BT::CT : BT, That is AT:DT::CT:BT. And, at : dt::Ct : bt. (4.) Therefore sim. tri. AA: : DE: :C^ : BH, and " " Tk: TE::Tt : TH, Cor. 9.— When CD is indefinitely great, CT is indefi- nitely small, or the centre is the point to which the tan- gent approaches as a limit when the absciss recedes from it. OF THE HYPERBOLA. 77 PROPOSITJON VII. The rectangle of the Focal distances from the vertices, is equal to the rectangle of one fourth the parameter, and the transverse axis. That is AF . FB=iP . AB. Fig. 8. For (II Cor. 2.) AF . FB : FE^ : : AB : P, therefore AF . FB : FE^ : : AB . AF : P . AF, and AF . FB : AB . AF.iFE^ : P . AF ; But (19) FE = iP.-.FE-"=iPS AF . FB : AF . AB : {P^ : :P . AF, FB : AB::{P : AF, and FB . FA^^P : AB. Q. E. D. Cor. 1.— FB : ABii^P : AF. ScHOL. — If, as in the Parabola, FB and AB, be consid- ered equal, then ^P=AF as demonstrated in the V. Prop, ef Parabola. CoR. 2.— Since FB . FA=iP . AB, and FB is greater than AB, FA is less than {P. In the Parabola, FA=iP, in the Ellipse FA>iP. CoR. 3.— AF . FB=CaS or the rectangle of the Focal distances, equals the square of the semi-conju- gate. For since (19.) P. AB=a62.-,iP . AB=aC2=AF. FB. or AF : aC: : aC : FB. Cor. 4.— AF.FB(=flC2) = |a^2. or iAB.iP=AC.FE ; f'^g- 10. or AF: FE::AC :FB. 11 4> 70 OF THE HYPERBOLA. ^iS' 8. Cor. 5.--(VI Cor. 1.) aC^=C< . C^=C/ . DK, but (VI. Cor. 7, 4.)C< . DE = AA: . BH, ^ .-. A^ . BH=flC-^=CA . |P=AF. FB. Cor. 6. — Therefore H/A:, is a right an^le. for (Cor. 5) AA; : AF::FB : BH, .*. the triangles A^and BHF are similar ; .-. - angle BH/=A/A: and B/H = A^/; .»*. also Hfk is a right angle. The same may be proved, in reference to the other Focus, therefore a circle described on HA:, as a diameter, will pass through the two Foci. PROPOSITION VIII. The square of the distance of the Focus from the cen- tre, is equal to the sum of the squares of the semi- axes. **•§:• »• That is CF^ =CA2+C. OF THE CURVATURE OF CONIC SECTIONS. iQl PROPOSITION VI. The diameter of curvature (CO) is a third proportional to twice CP and HK. That is 2CP : HK : CO, are con- tinued proportionals, or 2CP . CO=HK2. For, sim. tri. CO : CL : : Cc : OP : : 2Cc : 2CP, therefore (V) CO . 2CP=(2Cc) or CG . CL=HK% Q. E. D. PROPOSITION VII. The chord of curvature which passes through the Focus, is a third proportional to the transverse axis, and the diam- eter conjugate to that vi-hich passes through the point of contact That is AB : HK : CV, are continued proportionals. For, Sim. tri. CV : CO: :CP : CN: :2CP : 2CN, But (XI) CN=Ac, and 2CN=AB, CV:C0::2CP: AB, and (VI) CV . AB=CO . 2CP = HK-% or AB:HK:: HK:CV. Q. E. D. CoR. 1. — When the point of contact, is the vertex, the diameter of curvature, the chord which passes through the centre, and the chord which passes through the Focus, all coincide with each other, since they all coincide with the axis ; they are therefore equal, and the diameter of Curva- ture is therefore a third proportional to the Transverse and Conjugate axes, that is, it is equal to the Parameter. (See Prop. II. Schol.) Cor. 2. — When the point of contact is the extremity of the Conjugate axis, the Transverse axis is then conju- gate to that which passes through the point of contact ; and then the equation becomes CV . AB = ABS and CV=AB. That is, the chord of curvature passing through the Fo- cus, then equals the Transverse axis. 14 10S OF THE CURVATURE OF CONIC SECTIONS. CoR> 3. — The chord of curvature which passes through the centre, is to that which passes through the Focus, as the Transverse axis, is to the diameter which passes through the point of coi^tact. For (demonstration) C V : CO : : 2CP : AB, and (VI) CL: CO::2CP:CG, CL: CV::AB:CG. Cor. 4. — The diameter of curvature, the chord which passes through the centre, and the chord which passes through the Focus, are inversely as the parts cut off from each, by the diameter conjugate to that which passes through the point of contact. For (Cor. 2) CL : CV: :AB : CG: :CN : Cc, and (VI) CO : CL: :CG : 2CP: :Cc : CP, 5.20. .:. ( CO : CL : CV, are as I CN : Cc : CP. CoR. 5. — The diameter of Curvature, varies as the cube of the diameter, conjugate to that which passes through the point of contact. For (Ell.XVl) 2CP . HK=AB . ab, but (VI) 2CP . CO=HK», 2CP . HK : 2CP . CO : : AB . a6 : HK% or HK ; CO : : AB . ah : HK^, ' . .-. the parallelopiped CO . AB . ab=EK\ But the base AB . ab is constant, CO varies as HK^. Consequently the curvature itself (II) at any point is in- versely as the diameter conjugate to that which passes through that point, it is iherefore least at the extremity of the Conjugate axis, and greatest at that of the Transverse. Cor. 6. — The diameter of Curvature, at the intermedi- ate points, varies as the cube of the normal, CI. For (Ell. XV) CI . CP=Ca=» .-. CP varies as CI, and (Ell. XVI) CP . HC=Ac . ca .'. CP varies as HC, CI varies as HC, and CP asHC, asHK% .;. CO which varies (Cor. 5) as HK^, varies as CP. OF THE CURVATURE OF CONIC SECTIONS. PROPOSITION VII L In the hyperbola, as in the Ellipse, the chord of cur- vature, which passes through the centre, is equal to the parameter of the diameter which passes through the point of contact. That is CG : HK : CL, are continued proport's. For Ci .tG :Er«::CG2 : HK% whichisulti. ED . CG : ED . CL::CG» : HK^ and CG : CL::CG2 : HK^ CL:CG::HK2 : CG% HK.CL : HK.CG::HK» : CG% HK . CL : HK» : : HK . CG : CG% CL : HK::HK :CG. But (19) P' : HK::HK : CG. .-. P'=CL, and CG : HK : CL, contin'd. prop'Is. Q. E. D, PROPOSITION IX. The Diameter of Curvature, is a third proportional to 2CP and HK ; or 2CP : HK : Co, are continued prop's. Forsim. tri. CO: CL: :Cc : CP: :2Cc : 2CP, CO . 2CP=CL . CG=HK% Q. E. D, PROPOSITION X. The chord of curvature which passes through the Focus, is a third proportional to the transverse-axis, and the diameter conjugate to that which passes through the point of contact. That is AB : HK : CV, are continued proportionals, Insim. tri. CV : COl.CP : CN::2CP : 2CN==AB .'. CV.AB=C0.2CP=HK^ i»r AB : HK : CV, are coatinued proportionals. Q. E. D, 103 104 ^^ THE CURVATURE OF CONIC SECTIONS. Cor. 1. — At the vertex of the Hyperbolas, the diame- ter of curvature, coincides with the chord which passes through the centre, and the chord which passes through the Focus, since they all coincide with the axis; it is there- fore equal to the Parameter. (See Schol. Prop. II.) The diameter of curvature varies as the cube of the normal (CI), and therefore the curvature itself is inversely asCP. Schol.— The analogy between the curvature of the Ellipse and Hyperbola is complete, since the corresponding prop- ositions are capable of being expressed in the same words. The terms, however, as was remarked in reference to the general properties of the two sections, have somewhat dif- ferent significations ; and in particular the normal^ the cube of which is proportional to the radius of curvature, in- creases more rapidly in the Hyperbola than in the Ellipse, and never decreases again, after reaching a maximum^ as it does in the Ellipse. The curvatures of the two sections therefore are widely different. That of the Hyperbola, con- stantly decreases, and the curve never returns into itself. When the two conjugate axes, are unequal, the curvature is greatest at the vertex of the greater axis, as in the Ellipse, and if the two axes of a Hyperbola are equal to the two axes of an Ellipse, each to each, the curvature at their vertices, in one section is the same as in the other. If the two axes of a Hyperbola are equal to each other, (that is if the Hyperbolas are Equilateral,) in which case the section corresponds to the circle into which the El- lipse passes as its two axes become equal, the curvature at the fourvertices of the Conjugate Equilateral Hyperbolas is the same, and is equal to that of a circle, whose diame- ter is equal to the axis ; and the diameter of curvature to any other point, is as the distance of that point from the centre. The curvature itself therefore of Equilateral Hy- perbolas is the same, in any point, as that of a circle, whose circumference is at an equal distance from its cen- tre. In other words, in both figures, the curvature is al- ways inversely as the distance of the curve at that point, from the centre. In this respect, therefore, there is a striking analogy between an Equilateral Hyperbola, and the circle, which may be considered as an Equilateral El- lipse, The difference however, between the two curves OF THE CURVATURE OF CONIC SECTIONS. 105 is great, arising from the fact, that in one the curve is convex towards the centre, and in the other it is con- cave. In a general connparison of the curvature of the circle, with that of the other Conic Sections, the distinguishing peculiarity of the former is, that its curvature is uniform^ remaining the same in every point of its circumference, while that of the three other sections changes continually, and is never the same at two successive points. In all of them, the curvature is greatest at the vertex of the sec- tions, and decreases according to a given law, as any point is removed from the vertex. This continual Variation of Curvature ^ as it is termed, is the reason why no definite portion of any Conic Sec- tion, coincides with its Circle of Curvature. If the curva- ture of the section, remained the same, orwas uniform for any distance, that portion would be a part of a circle, and would consequently coincide with its Circle of Curvature, since it has been demonstrated that the latter has the sanip. curvature as the curve of the section at the point of con- tact. Since however the curvature of a Conic Section, as has been shown, never continues the same, in two suc- cessive points, but varies continually, the curve immedi- ately deviates from the tangent more or less than the Cir- cle of Curvature does, and therefore passes in the former case within, and in the latter without the circle. As the cur- vature in all the sections, is greater at the vertex, than in any other point, the whole section passes without the Circle of Curvature to that point ; fora similar reason, the Ellipse is wholly within the Circle of Curvature, to the curve at the ex- tremity of its Conjugate axis: and in all the sections, that fj^ 7. part of the curve which is towards the vertex from any point, being more curved than at the point, passes within the Cir- cle of Curvature to that point, while the part more remote, being less curved, passes without it. APPENDIX TO CONIC SECTIONS. J. Similar and Sub-contrary Sections. Def. 1. — If a point be taken above the plane of a cir^ clet and one extremity of a straight line, remain in that point, while the line moves around in the circumference of the circle, the figure described is called an Oblique PI. VII. ^onc. Fi?. 8. as VLK. 2. — If either of the Conic Sections be supposed to re- volve around its axis, which remains fixed, the solid thus generated, may be called a Conoid,* Conoids receive specific names, from the diflferent fig- ures by which they are described ; as Paraboloids, Ellip- soids and Hvperboloids. f^'^;^^!' ' As GAH, AHBG, and GAHO. PROPOSITION I. All the sections of a cone, nvide by parallel planes, are similar figures. tlZ. Parabolic sections, AD : ad: :DG : d^ Fig. 2. Ellipses, AB : ab:: A-B' : a'b'. , * Perhaps this use of the term is not authorized : but some general name was needed in the following propositions, and it seemed most easy and proper to extend the application of a term already in use. Archimedes gives the name Conoids to Paraboloids and Hyperholoids, (probably be- cause they obscurely resemble cones,) and for a similar reasofn to Elif^ soids, he o;ives the name of Spheroids. APPENDIX TO CONiC SECTIONS. lOT Hyperbolic Sections, AB : conj. of AB y.ab : conj. of Fig. 3. For, the conj.of AB=a mean proportional between AQ and BR, and the conjugate of ff6=a mean proportional between ag and hr. therefore AB : its conjugate Wnb \ its conjugate. Cor. 1. — If a plane perpendicular to thr triangular section, pass through the vertex of the cone, cutting the base, it will cut off similar segments, from similar parallel sections ; that is, in the Parabola, AD : DG: \ad : dg, F,g ,^ J5 5» Ellipse, AD \T>E:\Mdxde. " ?. and „ BD: DE::B'^: (/e. AD: A'(/::AB:A'B'. DE : de'.'.ab : ah', ., Hyperbola AD \ DGWad : dg, ?< 3 BD:DG::W:rf^, AD : a(/::AB : «^, DG : dg: : conjugate of y^B ! conjugate o( ab. Cor. 2. — Similar segments may be cut from all Parab- olas, that is, all Parabolas are similar sections. y^i'% For, if there be taken DG-^ : DG'^^AD : AH, ' ^^ then AD :DG::AH:HK. Cor. 3. — Similar Polygons may be inscribed in similar segments of similar sections. PROPOSITION H. If any Conoid, be cut by a plane parallel to its base, 01' to the plane of its revolving axis, the section will be a cir ofle. Fig. 4.6. Thus MZN and HS^^G are circular sections. and 6. ]Qg APPENDIX TO CO?JlC SEGT10>'3. PROPOSITION III. If any Conoid, be cut by a plane, coinciding with its ax- is, the section will be a figure, equal and similar to that, by the revolution of which the Conoid was generated. Kig. 4.5. For it coincides with the revolving figure, in one of its positions. . Thus GAHO, may be considered a section of the Co- noids. PROPOSITION IV. If a Conoid be cut by a plane parallel to its fixed axis, the section will be of the same kind with that by the revolution of which, the Conoid was generated, and simi- I a?' to it. ^'•g- 4 Suppose a plane GAHO to coincide with the axis produ- ^•*" " cing the section by which the conoid was described, and cutting the given plane at right angles. Let there be two circular sections MZN and HS^G perpendicular to the fixed axis, and therefore at right angles to each of the two other planes. The mutual intersection of these planes RZ and DG will be ordinates, both in the circular sections, and in the first sections, and it will be easy to prove, fi?- 4. 1. In the Parabola, that the squares of the ordinates are as their abscisses (See Par. Prop. 1.) That is, aR : ad.'.KZ'' : dg^. And that the section, is similar to that which coincides with the axis, or to that by which the Conoid was genera- ted. That is, ad : dg: .AD : DG. Fig. 5. 3. In the Ellipse, that the squares of the ordinates are as the rectangles of theirabscisses (Ellipse Prop. I.) That is aR.RB:ad.db::RZ^:dg'. and that AD : DG '.'.ad : dg, and AB :ab::HG iSldg. APPENDIX TO CONIC SECTIONS 1(J9 u ill the H}perbola, if a plane parallel to that in which the axes of four Conjugate Hyperboloids are placed, cat them, it may, in a similar manner, be proved, that the two opposite sections, are Hyperbolas, and in every respect equal and similar to each other, and that the four sections are Conjugate Hyperbolas, similar to the four by whicli the Hyperboloids were generated. That is cR' -ca^ : cd' -ca^ : .'RZ^ : dg', ''.f ^^-^ and AB : abr.A'B' la'b'. Cor, — If a plane cut two opposite Hyperboloids, be- ing perpendicular to the plane in which their axes are pla- ced, and pasjifii; through their centre (and therefore coin- ciding with a diameter of the Hyperbolas, by which the Hyperboloids were generated,^ the sections will be two opposite H\perbo|as equal to each other in every respect, but not similar to the generating Hyperbolas. Thati^OIQ is a Hyperbola, orim . mH : I5 . sH iini^ : sv", and the opposite section will be equal and similar. If then another plane cut the two Conjugate Hyperbo- loids in the same manner, but coinciding with KB" the di- ameter which is conjugate to the other, this section will al- so form two opposite Hyperbolas which will be conjugate to the other two. and III- : KB"- : 'Am . mH : mi-. Fig- 7. PROPOSITION V. if in any Conoid, a plane pass so as to intersect all sides of it, the section in every case, will be an Ellipse. p. In addition to the pianos supposed in the last figure, lets, aidi, a plaiie coincide with RZ (the. mutual intersection of the circular section MZN, and the section parallel to the axis, ttZgd) and let it cut the axis of the Conoid and all its sides ; it can easily be shewn that this oblique section has in every case the properties of an Ellipse, (See Ellipse, Prop. 1.) That is XR . RO : XP . PO: :RZ^ : PS^. CoR. — All sections parallel to any oblique section are Ellipses an^j are similar. 15 no APPENDIX TO CONIC SECTIONS PROPOSITION VI. Every section of an oblique cone, which is parallel to its base, is a circle. Every section which is sub-contrary to the base is a circle. Every other section is an Ellipse, and all the Elliptic sections which 2iVQ between the sections parallel and sub- contrary to the base, may be called Ellipses sub-contrary to all other Elliptic sections of the same cone, that is their Transverse and Conjugate axis, are placed in contrary po- sitions. 1. Sections parallel to the base, are proved to be cir- cles, by shewing that all lines drawn in them, from the ax- is of the cone to its circumference are equal. Fig. 8. that is cB=BA. 2 If a plane, cut the cone so as to make the same an- gles with the axis and with the opposite sides of the cone, that are made by a section parallel to its base, but in a contrary order, it is called a sub-contrary section, and is proved to be a circle, as in case first. That is, if VBA=VAB and cd be the mutual intersec- tion of the circular section, parallel to the base B'cR', and the given sub-contrary section, BA, then Be . c\=cd^. 3. That all other sections are Ellipses, is proved as by Prop. I Ellipse. 4. Of any section between the parallel and sub-contrary sections as AB', AB' is the conjugate, or smaller axis. ', ^ ScHOL. — Those circular sections which are parallel to the base, and those which are sub-contrary, in an oblique cone, are limits between sub-contrary Elliptic sections, and hold the place of transition from one of those Elliptic sections to the other. Thus, the sections AB, AB are limits hetvfeen AB' and AB' , two Elliptic sections, one on each side, but having their axes in a contrary position. APPENDIX TO CONIC SECTIONS Hi PROPOSITION VII. Every section of a Cylinder, parallel to its base is a cir- cle, every other section is an Ellipse. The first case is evident. — The second is proved in a manner similar to Prop. I, of the Ellipse. Eig That is, BD . DA : BS . SA: iDE^' : SZ«. ScHOL. — The demonstrations of the preceding proposi- tions have not been given in full, on the supposition that to persons well acquainted with the propositions on Con- ic Sections, the truth of them will be almost immediately evident ; to others, it may afford a useful and agreeable ex- ercise, to make out the demonstrations, by deducing them from the properties of the curves previously demonstrated. APPENDIX TO CONIC SECTIONS, n. COMPARISON OF CONIC SECTIONS, ^i$§i PROPOSITION I. • A Parabola is equal to four thirds of a Triangle, having the same base and altitude. Or it is two thirds of a circum PI VIII scribmg parallelogrann. Fig. 10.' That is Par BAC=| Tri. BAG. Let the base of the Parabola BAG, be bisected contin- ually, dividing it into the equal parts, BR, RG, GS, SD, DX, XE, EY, YC, and let lines be drawn through the intersections parallel to the axis AD, meeting the curve. Let GA be drawn parallel to the axis, and made of such length: that Ca : DA::4 : 3. Connect aB also AB, AH, HB, AG, AF, FG, form- ing the inscribed polygon BHAFG. Then RP pass- ing throng]^ that division of the base, which is next to B, and meeting Ba in d and BH, in / ; and GH, meeting oQ^ AB, in c and L ; HQ also being drawn parallel to BG, and Vq parallel to BA, and DO, HO perpendicular to AB, Then tri. BaG:BAC::Gfl : DA::4 : 3, and sim.tri.BaG=4B6D, or B6D = iBaG, BflG:6aGD::4: 3, therefore 6aGD=BAG. Again tri. BHA : BD A : : HO' : DO : : HL : AD, BHA-i-AFC : BAG: :HL ( = AQ) : kD, ::HQ»:BDs :: 1 : 4. And DGc6: GD6a(=BAC) :: 1 : 4. Therefore BHA+AEC=DGcJ. APPENDIX TO CONIC SECTIONS. 113 In like manner BPH-f-HMA + ANF+FNC : BHA+AFC, ::P/: HL, P72 : AL ::RG=: BG% :: 1 : 4, BuiGRr^c : Gc6D( =BH A + AFC) : ! 1 : 4 ; Therefore BPH + HMA + ANF+FVC = GRrfc. In the like manner if the parts of the base be bisected continually, and the sides of the inscribed polygon multipli- ed in the same proportion, the polygon will in all cases, be equal to the trapezium cut from the triangle BaC, by the parallel to the axis, nearest to B. But the polygon, thus approaches continually and indefinitely towards equality with the Parabola, which is its limit; also the trapezium approaches indefinitely towards the triangle BaC, which is evidently its limit; and since the polygon is always equal to the trapezium, the lijnit of xhe polygon, that is, the Parabo- la is equal to the limit of the trapezium, or the triangle BaC. If this conclusion is denied, it can be shown that the de- nial leads to absurdity ; for if the triangle BaC. for instance were supposed to be greater than Parabola, then the pro- cess could be carried on, until the trapezium taken from the triangle, should be also greater than the Parabok ; but the trapezium is always equal to the polygon inscribed in the Parabola, and therefore always less than the Parabola. A like absurdity follows the supposition that the Parabo- la is greater than the triangle. », Therefore the triangle BaC. is equal to the Parabola. But the triangle BAC = 5 BaC, .*. also the inscribed triangle BAC = f Parabola, Or the Parabola =^ Triangle BAC. Q. E. D. CoR. 1. — Hence if two Parabolas have equal altitudes, they are to each other as their bases. For this is true of the circumscribing parallelograms, or the inscribed trianglei. ^ sf 15 CoR. 2. — And if two Parabolas have equal bases, thejr are to each other as their altitudes. 6. I. 114 AFPEXDIX TO CONIC SECTIONS. Cor. 3. — Any two Parabolas are to each other, in the ratio compounded of their bases and altitudes. Cor. 4. — Hence similar Parabolic Sections have to each other, the duplicate ratio of their bases, or altitudes. PROPOSITION II. The Ellipse has to the circle which is described upon Its Transverse Axis, the same ratio, which the Coujugate axis has to the Transverse ; — and to the circle described upon its Conjugate axis, the same ratio which the Trans- verse has to the Conjugate. PI. VIII. That is Ellip. Aa Bb : Cir. AGE: [ab : AB. Fig. 12.* And Ellip. Aa Bb : Cir. agb: :AB : ab. Let AGB. be a semi-circle, described upon AB, the Transverse Axis, of the Ellipse AKB, let NHF, MKG be any two ordinates in the Ellipse, produced until they meet the circle, join GF and KH, Then (Ell. Ill Cor. 2.) MK: MG: iNH'-NF ; Therefore GF. KH, produced will meet in R, in MN produced. Also, il GF be bisected in D, and the ordinate DEI be drawn, then the tangents to the Circle and Ellipse at the points D and E, will meet CA produced in the same point T. (Ellip. VI. Conversely.) 6. 1. Then. tri. MGR : MKR : : MG : MK, Also " NFR : NHR: :NP : NH: :MG : MK, 5 ^9 .'. trap MGFN : MKHN: :MG : MK. also tri. NEA:NHA::NF :NH::MG:MK, and " MGB : MKB : : MG : MK, .-. PolAFDGB : aHEKB::MG: MK::CA:Ca. 6. 4. But, DT is parallel to OR, .-. ER is parallel to KR and the triangles GDF, KEH, are each greater than half their segments GDF, KEH, therefore the inscribed polygons, as the sides are multi- plied, will approach indefinitely to equality with the semi- circle and the semi-ellipse, as their limits. .-.Cir : Ellipse : : Pol. AGB : Pol. AKB: :CA : Ca. PI in In like manner Ellipse A«B : Cir. Dab: :CA : Co. Fig. 2. Q. E. D. APPENDIX TO COMC SECTFOJ^S. US Cor. 1. — Any Ellipse is a mean proportional, between ihe circles described upon its two axes. That is, Cir. AGB : Ell. AaBb : : Ell. AaBb : Cir. agb. Cor. 2. — An Ellipse is equal to a circle whose diameter fs a mean proportional between its two axes. For let Z be such a circle, ^'=* ^^ Then cir. AGB : Z : : AB =^ : A B' ^ Also, cir. AGB : Ellip.: : AC : Ca, ::AC-^ : AC.Ca, ::AB= :2AC . 2Ca=A'B ^ CoR. 3. — Hence any circle is to the square of its diam- eter, as the inscribed Ellipse, is to the rectangle of two axes. CoR. 4. — The areas of any two Ellipses, are to each, ether as the rectangles of their Transverse and Conjugate axes. That is,Ellip. A'aB'6 : Ellip. Aa'B6': :A'B' . ab : AB a'b', Fi-. 14. ScHOL. — The two preceding propositions are demon- strated after the manner of the ancient Geometers. See Archimedes " De Quadraturae Parabolae" and ''De Con- ©idibus et Spheroidibus." It is called the method of Exhaustions, and is much ad- mired for its ingenuity and rigorous exactness. " Though few things, says Playfair, more ingenious than this method have been devised, and though nothing could be more conclusive than the demonstrations resulting from it, yet it laboured under two very considerable defects. In the first place the process by which the demonstrations were obtained was long and difficult, and in the second place it was indirect, giving no insight into the principle on which the investigation was founded." This method has been much abridged by the moderns, especially by Cavaleri, in the method of Indivisibles, a method not only more concise, than that of the ancients, but much more easily and extensivly applicable to a great variety of propositions, and at the same time no less rigo- rously exact, when properly managed. f 116 APPENDIX TO COiMC SECTIONS. An application of it to the second of the preceding prop- ositions will illustrate its nature, and prepare the student for the demonstrations which follow. f''?- 11- Any trapezium as KMGD, is to the corresponding trapezium in the Ellipse KLED, as the sum of the parallel sides KM and DG, is to the sum of KL, and DE, for each trapezium may be divided into two triangles, which will be to each other, respectively as these lines. Consequently, all the trapeziums taken together, or the polygon in the semi-circle, will be to the polygon in the Ellipse, as the sum of the ordinates in the circle, is to the sum of the ordinates in the Ellipse, that is (Euclid V. 12, as any one ordinate in the circle, to the corresponding ordi :; te in the Ellipse; in other words, as the Transverse to the Cc n- jugate axes. But by increasing the number of or .;jiates, and consequently the number of sides of the polygons, the latter approach indefinitely to equality with the circle and Ellipse; therefore the circle is to the Ellipse, in the same ratio as the polygon in the circle, to the polygon in the El- lipse, that is, as the sum of the ordinates m one, to the sum of the corresponding ordinate in the other, or as the Trans- verse :o the Conjugate axes. This method of Indivisibles, or of supposing ordinates to be encreased indefinitely in number while their distances are diminished in the same proportion, often appears unsat- isfactory, not only on account of the abridged mode of ex-' pression, by which important steps in the demonstration are omitted, but because it is common to say, that the spaces compared are made up of an infinite number of parallel or- dinates, whereas it is evident from the definitions of Geom- etry, that no possible number of lines can " make up" a space, or form any thing but a line; not to mention the in consistency of an '* infinite number" of lines The truth is, as has just been shown, that the rectilineal spaces compared, have to each other the same ratio which the sums of corresponding ordinates have, and curvilineal spaces, which are their limits, have exactly the same ratio. The exact truth of this Proposition, may be rigorously de- monstrated in all cases by a reductio ad absurdum. APPENDIX TO CONIC SECTIONS. ||7 The method of Indivisibles then, when properly consid- ered and applied, is an abridged form of the method of Exhaustions; the ungeometrical language, so often used, and before alluded to, will be avoided in the following propositions. PROPOSITION III. Two Ellipses having the same, or equal Transverse ax- es, are to each other as their Conjugate axes. That is Ellip. Aa Bh : Ellip. Aa Bb'y.ab I a'h\ For any ordinate DE : De : :CA : Ca, P|; Vm. Therefore bv indivis. CAa t CAa': :Ca : Ca' ; ^'^' ^^• Or Ellipse AaBh : Aa'Bb' : :ab X 'ah'. Q. E. IX CoR. 1. — This reasoning is equally applicable to two Ellipses, having the same or equal conjugate axes. They are to each other as their transverse axes. That is, Ellip. A'aB'b : AaB^>: :A'B' : AB. Cor. ^. — Any Ellipses have to each other the ratio compounded of the ratios of their two axes. For Ellipse Aa'Bb' : Ellip. AaBb: \a'b' : ab, and Ellip. AaBb : Ellip. A'aB'b'.'.A'B' : AB, .'. Ellip. Aa'Bb' : Ellip. A'aB'b in a ratio com- pounded of the ratios of a'b' to ab, and of A'B' to AB. 5. C* CoR. 3. — Similar Ellipses have to each other the du- plicate ratio which either of their axes have. That is Ellipses Aa'Bb' : A'aB'br.AB^ I A'B'^ : '.a'b'^:ab\ For(22)AB : a'b'WA'B' : ab, and AB2 : a'b' . AB: iA'B'^ : ab . A'B', .:. AB^ : A'B'= : : AB . a'b' : AB' .a6: :EHip. Aa'B6'Ellip. A'aB'6. *- Simpson's Euclid.— See also Book VI, Prop. XXIIT. 16 118 APPENDIX TO CONIC SECTIONS. PROPOSITION IV. Two Hyperbolas, having the same or equal Transvene axes and equal abscisses, are to each other as their Conju- gate axes. r>g. 15. That is, ADE : ADE': :Ca : Ca'. This is evident from the method of indivisibles, (See Hyp. II, Cor. 3.) PROPOSITION V. Similar segments of similar Conic Sections, are to each other in the ratio compounded of the ratios of their ab- PI. VII. scisses and ordinates. Fig:. 1. ADG : adg: :AD , DG : ad . dg. Fig. 2: ADE : A'Je: : AD . DE : Arf . de, Fig. 3. ADG : adg: \ AD . DG : a^/ . dg. For their abscisses, being divided into anequal number of parts, these parts will be to each otherasthe whole abscisses, also the corresponding trapeziums in the inscribed polygons, will be to each other in the ratio compounded of their bases and altitudes ; and the whole polygons, and consequently the Conic Sections which are their limits, will be to each other as the sums of the bases and altitudes, or as the ab- scisses and ordinates. CoR. 1. — Similar segments are to ^ach other as the rectangles of the Transverse and Conjugate axes of their sections. CoR. ^.-Similar segments, are to each other, in the du- plicate ratio of either of their axes. ScHOL. — This proposition and corollaries, follow from Prop. X^ Cor. 3, Append. I. APPENDIX TO CONIC SECTIONS. H9 PROPOSITION VI. Any diameter of any Conic Section, bisects any seg- ment included between a double ordinate or the conju- jrate diameter, and the curve. 1. In the Parabola HA and HV, bisect the segments ^'s- ^2. AlandKVI. '""^ '^' Or KAH=HAI,andKVH=HVI. Fig. 3. 2. In the Ellipses AC, bisects aAb and EAG, Or ACa=AC6, and ADE=ADG. also CG^ = CE^=CEe = CGe. The same reasoning evidently applies to the circle. Fig. 13. In the Hyperbola AD bisects the segment EAG, or ^»g- ^' ADE-=ADG. Also in a triangle, whether isosceles or scalene, a line which bisects the base, bisects the triangle. For in each case the diameter bisects the double ordinate which limits the segment, and all double ordin- ates within the segment, therefore by the method of in- divisibles, it bisects the space in which they are drawn. Cor. 1. — The spaces included between the curve and two tangents drawn from the extremities of any double or- dinate, are bisected by the diameter of that double ordin- ate. That is CAT=TAD, TNC=TNL and TAE=TAH. Fiff. i^ Also the space between the curve, the tangent and tP. 7* double ordinates produced are equal. ^ of EU That is COP = LiR, FEI = GHM, also AKE=AK/i. & Hyp.' CoR. 2. — The spaces between the curve of an Ellipse, and a circumscribing parallelogram, whose sides are par- allel to two coujugate axes, are all equal. That is, EKG=s^SG=EQe=Gfe. ^'^S- 13. This is equally applicable to the circle. 120 APPENDIX TO CONIC SECTIONS. Cor. 3. — The spaces between the curve of a Hyper- bola and its asymptotes, and bounded by a double or- dinate, are bisected by the diameter of that double or- dinate. That is CAEH=CAE^. PROPOSITION VIII. Similar Conoids, are to each other, in the ratio com- pounded of the ratios of their bases and altitudes, or, of the squares of the diameters of their bases and altitudes. Similar Segments, being supposed to revolve on their axes, generate similar Conoids, and if the inscribed poly- gons, of which they are limits, be supposed to revolve with them, it will generate a solid, of which the Conoid is the limit. Hence, by Indivisibles or Exhaustions, the proposi- tion may be deduced. Cor. 1. — Similar Conoids are to each other, in the ra- tio compounded of the ratios of the squares of the diame- ters of their bases and their altitudes, or of the squares of their revolving axes or ordinates, and their fixed axes or abscisses. CoR. 2. — Similar Conoids are to each other, in the triplicate ratio of their altitudes, or the diameters of their bases. PROPOSITION IX. 1. An OblateSpheroid, or one described about the Con- jugate axis, is to a Prolate Spheroid, or one about the Transverse of the same Ellipse, as the Transverse axis is to the Conjugate axis. 2. The circumscribing sphere, is to the oblate El- lipsoid, as the Transverse axis is to the Conjugate. 3. The Prolate Ellipsoid is to the inscribed sphere, as the Transverse axis, is to the Conjugate. APPENDIX TO CONIC SECTIONS. 121 4. The circumscribing sphere, is to the oblate Ellipsoid, as the prolate Ellipsoid is to the inscribed sphere. That is, the four solids are Geometrical Proportionals. 5. They are continued proportionals, each having to the following, the ratio of the Transverse axis to the Conju- gate 5 therefore the circumscribing sphere has to the in- scribed sphere, the triplicate ratio of the Transverse axis to the Conjugate, as is evident also from Euclid, Sup. Ill, n-21. PROPOSITION X. Every Paraboloid, is equal to one half its circumscri- bing cylinder. This is proved, in a manner very similar to that in which the corresponding proposition concerning the sphere is demonstrated. (Plate XVI. 9) PROPOSITION XL Every Ellipsoid, Prolate or Oblate, is f of its cirt:um- scribing cylinder. The Ellipsoid and the sphere, have evidently the same pj yjjj relations to the circumscribing cylinders. Fig. 12! SPHERICAL GEOMETRY. DEFINITIONS AND SECTIONS OF THE SPHERE. Def. 1. — A Sphere is a solid, generated by the revolu- g^"*^^ tion of a semi-circle about its diameter, which remains ^^{[^^ fixed. 2. — The Axis of the sphere, is the fixed diameter about ^ which the semi-circle is supposed to revolve. Def. 9. 3. — The Centre of a sphere is the same as that of the semi-circle, by which it is described. 4. — The Radius of the sphere, is any straight line, drawn from the centre to the circumference of the sphere. Cor. 1. — The Radii of the sphere are all equal; each being equal to the radius of the generating semi-circle. In other words, every point on the surface of the sphere is equally distant from its centre. CoR. 2. — Hence if two portions of the surface of the same sphere or of equal spheres, be placed upon each other they will coincide in one superfices. Def. 5.— The Z)iawc/er of the sphere, is any straight line which passes through the centre, and is terminated g^ .^ both ways by the surface of the sphere. Def/**ia 124 SPHERICAL GEOMETRY. Cor.— The diameters of a sphere are all equal ; being each equal to twice the radius, or to the diameter of the generating semi-circle. Any diameter, therefore, may be assumed, as the axis of the sphere. 6.— *The Poles of the sphere, are the extremities of iti axis. Cor. — The extremities of any diameter may be assum- ed as the poles of the sphere. (Def. 5, Cor.) 7. — A plane touches the sphere when it meets the sur^ face of the sphere, but doei not cut it. 8. — A straight line touches the sphere when it meets the surface, but does not cut it. PROPOSITION L If a plane cut the sphere passing through its centre, the mutual intersection of the sphere and plane, will be a circle, which is equal to that by which the sphere was described. This is evident because every point of the surface of the •phere, (Def. 4, Cor. 1.) is equally distant from the cen- tre ; and this distance is equal to the radius of the genera- 1 Def.i2.ting semi-circle. CoR. 1. — All circles on the sphere whose planes pass through the centre, are equal to each other. Scholium. — Circles on the sphere whose planes pass through the centre, are called Great Circles. Such upon the Armillary sphere, are the Equator, the Horizon, the Ecliptic and the Meridians. Cor. 2. — A great circle may pass through any two points in the surface of the sphere. SPHERICAL GEOMETRY. 125 For a plane may be supposed to pass through arty three ^ii-en points. As the plane of a great circle passes through the centre of the sphere, it may also pass through any two points in its surface. Cor. 3. — The straight line which connects any two points, in the surface of the sphere, falls wholly within the sphere. For it is a chord of the great circle which passes through these two points. Cor. 4. — Any straight line which touches the sphere, touches the great circle whose plane coincides with the line. (Dcf. 8.) 3 Dei. h Cor. 5. — The straight line which passes through the point of contact and the centre of the sphere, is perpen- dicular to a line touching it. In other words, any diameter of the sphere, is perpen- dicular to all lines, which meet its extremity and touch the sphere. 3, ig, CoR. 6. — An indefinite number of great circles may pass through any given point in the surface of the sphere. For planes may pass through the centre, through a giv- en point in the surface, and through any other point in the surface. And the intersection of each of them with the sphere will be a great circle. (Cor. 2.) ScHOL. - Thus there are an indefinite number of Me- ridians, all passing through the poles of the terrestial sphere. Cor. 7.— The mutual intersections of a plane which touches a sphere, and the planes of any number of great circles passing through the point of contact, will be straight lines touching the sphere, (Def. 7.) and touching those great circles respectively. (Cor. 4.) 17 Sup. 2. Dei; I. 126 SPHERICAL GEOxMETRY. ScHOL. — Thus if a plane touch the globe at one of its poles, the mutual intersections of this plane with the planes of the Meridians, will all be lines touching those Meridians, and also touching the sphere. Cor. 8. — The radius or diameter of the sphere, which passes through the point of contact, is perpendicular to \\\ii plane touching the sphere. For it is perpendicular to every straight line which it meets in that plane. (Cor. 5, and 7.) ScHOL. — Thus the axis of the Earth, is perpendicular to planes touching it, at its poles. CoR. 9. — The mutual intersection of two great circles, is a diameter of the sphere. For as the plane of each great circle passes through the centre of the sphere, the line of their mutual intersection ihcrtfore passes through the centre of the sphere. It is therefore a diameter. (Def. 5.) CoR. 1 0. — Hence all great circles mutually bisect each other : that is, they divide each other into two equal parts, or semi-circles. ScHOL.— Thus the Ecliptic and the Equator, bisect each other. All the Meridians bisect each other, &c. PROPOSITION II. If any plane cut a sphere, but do not pass through the centre, the section will be a circle, whose radius is equal to that ordinate in the generating semi-circle which coin- cides with the plane. For in describing the sphere, the revolutions of that ordinate will describe the circle. CoR. 1. — The circles formed by the section of planes which do not pass through (he centre of the sphere, since they are to each other as the squares of their diameters or radii, — that is, the squares of the ordinates in the semi- SPHERICAL GEOMETRY. 127 circle, are as the rectangles of the segments into which their planes divide the axis. That is, the cir. EcQ : HDI: :PC . C/> : PD . Dp. Fig;, i- ScHOL. — As these circles are smaller than the genera- ting circle of the sphere, and therefore smaller than the circular sections which pass through the centre, they are called Small Circles of the sphere. Such are all the allels of latitude on the terreatial sphere. uli- CoR. 2. — Small circles whose planes are equally distant from the centre, are equal. •' And conversely. — If they are equal, they are equally distant from the centre. For then, the corresponding rectangles, of the axis, will be equal. (Cor. 1.) Cor. 3. — If two small circles are unequal, the less is more distant from the centre, and conversely. ScHOL. — Thus the two Tropics, or the Polar Circles, are respectively equal, and equally distant from the cen- tre ; but the Polar Circles are more distant from the centre than the Tropics, and smaller. Cor. 4. — A small circle may pass through any three points, in the surface of the sphere, if those points are not all in the same great circle. For a plane may pass through ani/ three points, and if it do not pass also through the centre of the sphere, the sec- tion will be a small circle. Cor. 5. — If any two circles meet, but do not cut each other, on the surface of the sphere, the straight line which touches one of the circles, touches the other also. For it meets the other circle, but does not cut it. If the line did cut the second circle, it would fall wholly within the sphere (I, Cor. 2.) and therefore would not touch the first circle, contrary to the suppo- sition. 128 SPHERICAL GEOMETRY. 1 And conversely, Circles which touch the same straight line touch each other. Cor. 6. — Hence the mutual intersection of the planes of two circles which touch each other on the sphere, is a straight line which touches both of them. For the line is in the planes of both the circles, and it touches each of them. ScHOL. — As illustrations : The Tropics touch the Ecliptic. Each of them therefore, and the Ecliptic touch the same straight line, at the point of contact, and that line is the common intersection of their planes. Def. 9. — The extremities of the axis, are the poles of all circles which are perpendicular to the axis. Cor. 1. — Hence two great circles cannot have the same poles, not having a common axis. Cor, 2. — There may be an indefinite number of small circles, all parallel to a great circle, and having the same poles with it, being all perpendicular to the same axis. ScHOL. — Thus all the parallels of latitude have the same poles as the Equator. CoR. 3. — Circles which have the same poles are par- allel to each other, and the axis passes through the cen- tre of each of them. The axis therefore passes through the poles of any cir- cle, through its centre, and through the centre of the sphere. CoR. 4. — The plane of any great circle, is at right an- "^' ' gles to all the circles through whose poles it passes. For it coincides with the axis, which (Def. 9.) is perpendicular to all their planes. ScHOL. — Thus the Meridians are at right angles to the Equator and to all its parallels. SPHERICAL GEOMETRY. 129 Cor. 5. — And conversely, All the great circles, which are at right angles to a given great circle, intersect each other in its poles. For they all pass through its poles, and must therefore intersect each other there. ScHOL. — For example, all the Meridians are at right angles to the Equator, and they all intersect each other in its poles. CoR. 6. — So also, if two great circles are at right angles to each other, the poles of each are in the circumference of the other : and if any number of great circles pass through the poles of another great circle, the poles of all the former, are in the circumference of the latter. ScHOL. — Thus the poles of all Meridians, are in the E- quator. Cor. 7. — If three great circles are at right angles to each other, the pole of each is in the circumference of both the others, and therefore in the point of their inter- section. CoR. 8. — A circle whose plane coincides with the axis, and therefore passes through the poles, bisects all the cir- cles which are perpendicular to the axis. For it passes through their centres. (Cor. 3.) CoR. 9. — The angle made at the centre of the sphere, by the axes of two great circles, is equal to the inclinatioi of their planes. That is PCP=QCL, For PCQ=P'CL= a right angle, therefore taking away ^»S- 2 the common part KCQ, and PCP'=QCL. CoR. 10. — If two circles touch each other, on the sur- face of the sphere, the piano which passes through the point of contact, and is perpendicular to the line, touching both of them. (11 Cor. 5.) that is, to the line of their mu- ]3() SPHERICAL GEOMETRY. Sup.2.i7.tual intersection, (II Cor. 6.) is perpendicular also to their planes. It therefore passes through their poles, (Cor. 5.) and through the centre of the sphere, (Cor. 3.) Conversely, The plane of a great circle which passes Sup.2.l8.through the poles of two other circles which touch each other, passes also through the point of contact and is per- pendicular to the line of their mutual intersection. ScHOL. — For example. The solstitial Colure. passes through the points of contact, of the Ecliptic and Tropics, and through their poles. It is perpendicular also to their planes, and to the lines of their muttil intersection, each of which lines touches both the Ecliptic and Tropic in the points of their contact. CoR. 11. — Any circle on the sphere cuts off equal arcs from all the great circles which pass through its Fig. 2. poles. That is, PH=PF and PE=PG. For in the description of the sphere, the arc of the gen- erating semi-circle between the poles, and any plane per- pendicular to the axis, will coincide successively with the arcs of all great circles passing through the pole. ScHOL.— Thus the Equator, or any of its parallels, cuts off equal arcs from all the Meridians. Cor. 12. — Hence any two parallel circles, intercept equal arcs of all great circles passing through their poles. That is EH=GF. This is inferred from the last, by taking equal arcs from those which are equal. SPHERICAL GEOMETRY. 13^ ScHOL. — Thus arcs of Meridians, intercepted between two parallels of Latitude, or between any parallel and the Equator, are equal. PROPOSITION III. If two circles on the sphere, pass through the remotest poles of two great circles, they will cut oif equal arcs from those circles. That is Ke=Gg, Fig. 3. Forthe arcPE=;}G; .'. PG=;oE, Therefore the chord PG =pE, and PD=;?D. 3. 3C. A^ain the chord Fe=pg^ .', arc Fe=^pg. Therefore the arc Pg=pe and chord, Fg=pe, and Vd^=pd^ Therefore, the triangle DPc/=D/)c/ in all respects; i- 8- .'. also GPg = E/?e, and chord gG=Ee; and arc Gg^=Ee. Q. E. D. PROPOSITION IV. If from any point which is not the pole of a great cir- cle, there be described arcs of great circles to that circle, the greatest is that which passes through its pole. And the other arc of the same circle is the least; and of the others, that which is nearer to the pole, is greater than any other which is more remote. Let the common section of the planes of the great cir- Fij. 7. cles ACB, ad B at right angles to each other, be AB; and from C, draw CG perpendicular to AB, which will ^jj; ^ also be perpendicular to the plane ADB, Jom GD, GE, GF, CD, CE, CF, CA, CB. Of all the straight lines drawn from G, to the cir- cumference ADB, GA is the greatest and GB the least ; and GD which is nearer to GA is greater than GE, which is more remote. The triangles CGA, CGD are right angled at G, and they have the common side CG ; therefore the squares of CG, GA together, that is the square of CA, is greater than the squares of CG, CD to- gether, that is. the square of CD : and CA is greater than 3. 7 ^32 SPHERICAL GEOMETRY. a 28. CD, and therefore the arc CA is greater than CD. In the same manner, since GD is greater than GE, and GE than GF, &;c. it is shown that CD is greater than CE, and CE than CF he, and consequently, the arc CD, greater than the arc CE, and the arc CE greater than the arc CF, &c. And since GA is the greatest, and GB the least of all the straight lines drawn from G to the circumference ADB, it is manifest that CA is the greatest, and CB the least of all the straight lines drawn from C to the circumference : and therefore the arc CA is the greatest, and CB the least of all the circles drawn through C, meeting ADB. Q. E. D, SPHERICAL GEOMETRY. II. OF SPHERICAL, TRMJVGLES. Def. i— a spherical angle is the angle made by two circles which intersect each other, on the surface of the sphere. Or it is the angle made by the straight lines 7vhich touch those circles at the point of their intersection* ScHOL. — The first of the preceding definitions, is one frequently given by writers on spherics. '* The angle made by tiuo circles^^^ however is two indefinite an expres- sion. CagnoH endeavors to render it more precise, by adding that ?7 is the angle made by the arcs, " considered in the points immediately contiguous to that in which they meet each other**'' (consideree dans les points immediatement contigiis a celui dans lequel Us se rencontrent^ (279.) and this angle he infers is the same as that made by the lines touching them in that point. But not to mention the want of mathematical precision in the phrase "points imme- diately contiguous to that AC, and AB-f AC>BC, and BC+AC Fig. 8. >AB. For the planes of the three circles all pass through D, the centre of the sphere, forming a solid angle at D, any two angles of which are greater than the third ; therefore the Sup. 2. same is true of the arcs AB, BC and AC which have to 20. each other the same ratios as the angles. 6. 33. Q. E. D, PROPOSITION \n. The three sides of a spherical triangle, are together less than a circle. For any two great circles, as ACD, ABD, bisect each Fig. other, therefore ACD and ABC, are together equal to a circle. l^et the arc CB, intersect both these circles. In the triangle CAB the two sides CA and AB, are to- other greater than the third side CB (VI,) therefore CD DB and BC, are together less than CD, DB, BA and AC, that is less than a circle. Q. E. D. CoR. -Each bf the three sides of a spherical triangle may be greater than a quadrant. Also each of them, evidently, may be less than a quad- rant. 10. 140 SPHERICAL GEOMETRY. PROPOSITION Vlll. In any spherical triangle, the greatest side is opposite the greatest angle ; and conversely. Fig. 10. That is, if the angle ACE is greater than AEC then al- so AE is greater than AC. For, let the angle ACB be made equal to BAC, then BA=BC (III) and therefore CB+BE=AE ; but CB-h BE>CE, therefore AE>CE, that is the greater side, AE in the triangle ACE, is opposite to the greater angle C. Q. E. D. PROPOSITION IX. If the three angles of any spherical triangle are made the poles of three great circles, the intersection of those circles will be a triangle, whose sides will be supplemen- tal to the measures of the angles of the given triangle, and the measures of its angles will be supplemental to the sides of the given triangle. Fig 9. That is DF, FE, and ED are supplemental of the an- gles C, B and A, and the angles at E, D and F, are sup- plemental of the arcs DF, FE and ED. Since A is the pole of FE and the circle AC passes through A, EF will pass through the pole of AC (9, Cor. 6.) and since C, is the pole of FD, FD also will pass through the pole of AC ; therefore the pole of AC is in the point F, in which the arc DF, EF intersect each other. In the same manner D is the pole of BC, and E the pole of AB. And since F, E, are the poles of AL and AM, FL and EM are quadrants, and FL, EM together, that is FE and ML together, are equal to a semi-circle. But since A is the pole of ML, ML is the measure of the angle BAC. (II, I Cor. 8.) In the same manner, ED, DF are the supplements of the measures of the angles ABC, BCA. SPHERICAL GEOMETRY. 141 Since likewise CN, BH are quadrants, CN, BH togeth- er, that is, NH, BC together are equal to a semi-circle ; and since D is the pole of NH; NH is the measure of the angle FDE ; therefore the measure of the angle FDE is the supplement of the side BC. In the same man- ner, it is shown that the measures of the angles DEF, EFD are the supplements of the sides AB, AC, in the tri- angle ABC. Q. E. D. ScHOL. — The triangle DEF, is called the Supplemental or Polar triangle of ACB. PROPOSITION X. If two triangles have the three angles of one equal to the three angles of the other each to each, the two trian- gles shall be equal in every respect. For the sides of their Polar Triangles are then equal, each to each (IX,) therefore the two Polar Triangles shall be equal to each other in every respect,{V.) Therefore the sides of the give7i triangles, which are supplemental to the angles, respectively, of i\\e polar triangles, shall be equal to each other, each to each, therefore the two given triangles are equal in all respects. (V.) Q. E. D, PROPOSITION XI. The three angles of any spherical triangle are together greater than two right angles, and less than six right angles. That is A-j-B-{-C> 2 right angles and < 6 right angles. Fig. 9. First they are greater than two right angles, for the sides of its polar triangle DP, FE and ED, which are. sup- plemental to their angles, are together less than a circle (VII) which is the measure of four right angles, therefore their three supplemental angles C, A, B are greater than two right angles. 19 ^42 SPHERICAL GEOMETRY. Second, As any spherical angle (II Def. 1, Cor. 3.) is less than two right angles, the three angles of any spherical triangle, are together less than six right angles. Q. E. jD. ScHOL. — In the property now demonstrated, there is H striking difference between spherical and plane triangles, (see Euclid I, 32.) which is the foundation of correspond- ing differences in all the relations deducible from these dif- ferent properties. Cor. 1, — If a side of a spherical triangle be produced, the exterior angle, is less than the sum of the two interior and opposite angles. Fig. 10. That is, CED is less than ECA + EAC. For this exterior angle CED together with its adjacent interior angle CEA is equal to two right angles, but the angle CEA together with the two opposite interior angles, is greater than two right angles. Cor. 2. — Also if the sides of a spherical triangle be produced, the exterior angles are together less than four right angles. CoR. 3. — The three angles of a spherical triangle, may be each greater than a right angle. PROPOSITION XII. In any spherical triangle, if the sum of the tivo sides, be equal to a semi-circle, the external angle at the base, will be equal to the interior and opposite angle ; and therefore the sum of the two angles at the base, will be equal to two Fi-. 10. ri?*^^ angles. That is, if AC4-CE = AD,then CED=CAE. and CAE-f CEA=two right angles. For ihen CE=CD.-.(IV) CED=CDE=CAE. And CAE-f CEA=CEA4-CED=rtwo right angles. Q. E. D. SPHERICAL GEOMETRY. j43 Cor. 1. — Conversely, if the angles at the base, are equal to two right angles, the sides are together equal to a semi- circle; and if the angles are unequal, that opposite the greater angle, is greater than a quadrant, (Vl[I)and that op- posite the less, is less than the quadrant : and conversely. But if the angles are equal, the angles at the base are right angles. That is, if CAEH-CEA=2 right angles, then AC4-CE = AD. And if CAE = CEA, then CE and CA, each equal a quadrant. If CAE>CEA " CE is greater, and CA less than a quadrant. CoR. 2. — If the two sides are together greater than a semi-circle, the interior angle will be greater than the ex- terior and opposite angle, and the sum of the angles at the base, will be greater than two right angles, and the great- er side will be greater than a quadrant. Conversely, if the interior is greater than the exterior angle, then the two interior angles are together greater than two right angles, and the two opposite sides are together greater than a semi- circle. Thatis, IfCA-f-CECEA, and CAE+CEA >2 right angles, and conversely, if CAE>CED, then CAE-f CEA> 2 right angles, and CA + CE > AD. Also if the two sides be less than a semi-circle, the inte- rior angle will be less than the exterior and opposite angle, and the two angles at the base, will be less than two right angles. Also the less side will be less than a quadrant. That is. ifCA-fCE'ig- 1^- 2. Let CD fall without the triangle ; then the angle B is of the same affection with (.'D; and the angle CAD is of the same affection with CD ; therefore the angle CAD and B are of the same affection, and the angle CAB and B are therefore of different affections. Q. jE. D. (^OR. — Conversely, if the angles at the base of any triangle, be of the same affection, the perpendicular will fall within the triangle, but if the angles be of different af- fection, the perpendicular will fall without the triangle. SPHERICAL GEOMETRY. l47 PROPOSITION XVII. A perpendicular being drawn to the base of a spherical triangle, if the sum of the sides is less than a semi-circle, then the least segment of the base is adjacent to the least side of the triangle; but if the sum of the sides be greater than a semi-circle, the least segment is adjacent to the greatest side. Let ABEF be a great circle of a sphere, H its pole, and GHDany circle passing through H, which therefore is per- pendicular to the circle ABEF. Let A and B be two points in the circle ABEF, on opposite sides of the point D, and let D be nearer to A, than to B, and let C be any point in the circle GHD between H and D. Through the points A and C, P. and C, let the arcs AG and BC be drawn, and let them be produced, till they meet the circle CBEF in the points E and F, then the arcs ACE, BCF are semi- circles. Also, ACB, ACF, CFE, ECB are four spherical tri- angles, contained by arcs of the same circles, and having the same perpendiculars CD and CG. 1. Now because CE is nearer to the arc CHG than CB is, CE is greater than CB, and therefore ('E and CA are greater than CB and CA; wherefore CB and CA are less than a semi-circle; but because AD is by supposition less than DB, AC is also less than CB. (I. Prop. IV.) and therefore in this case, viz. when the perpendicular falls within the triangle, and when the sum of the sides is less than a semi-circle, the least segment is adjacent to the leastside. 2. Again, in the triangle FCA the two sides FC and CA are less than a semi-circle; for since AC is less than CB, AC and CF are less than BC and OF. Fi-. 11. 148 SPHERICAL GEOMETRY. Also AC is less than CF, because it is more remote from CHGthan CF is, therefore in this case, also. viz, when the perpendicular falls without the triangle, and when the sum of their sides is less than a semi-circle, the least seg- ment of the base AD is adjacent to the least side. 3. But in the triangle FCE, the two sides FC and CE are greater than a semi-circle; for since FC is greater than CA, FC and CE, are greater than AC and CE. And because AC is less than CB, EC is greater than CF, and EC is therefore nearer to the perpendicular CHG than CF is, wherefore EG is the least segment of the base, and is adjacent to the greatest side. 4. In the triangle ECB, the two sides EC, CB are great- er than a semi-circle, for since by supposition CB is threat- greater than CA, EC and BC are greater than EC and CA. Also, EC is greater than CB; wherefore in this case also, the least segment of the base FG is adjacent to the greatest side of the triangle. Wherefore, when the sum of the sides is greater than a semi-circle, the least segment of the base is adjacent to the greatest side, whether the per- pendicular fall within or without the triangle : and it has been shewn, that when the sum of the sides is less than a semi-circle, the least segment of the base is adjacent to the least of the sides, whether the perpendicular fall within or without the triangle.* Q. E, D. PROPOSITION XVIII. The suras of the opposite angles of a quadrilateral figure, PI xvi.^" ^^® sphere, which can be inscribed in a small circle, are equal Fig. 7 That is A+C=B+D. * When the perpendiculars fall without the triangle, it is that whirh is noarest to the triftr.^le to which the demonstration is applicable. SPHERICAL GEOMETRY. 140 For if arcs of great circles pass through the Pole of the small circle and each angle of the quadrilateral, they will divide it into isosceles triangles, (Def. 9. Cor 7.) whose an- gles at the base in each will be equal, (IV.) and conse- quently the sums of equals will be equal. That is PBA+PBC-f PDA+PDC=PAB4-PAD+ PCD-fPCB. Or ABC + BDC=BAD4-BCD. q. E. D. PROPOSITION XIX. If two spherical triangles, have two right angles, in other words, If their sides intersect each other in the Pole of their base, then the two triangles are to each other as their bases. p.^ If a great circle should pass through P, and bisect the ^*' " base, EG, of the triangi- PEG, it would also bisect the triangle, ( V ) for the two triangles into which it would divide PEG would be equal. Also if any multiple of the base should be taken, the triangle formed by a great cir- cle passing through its extremity and the Pole (P ) would be the same multiple of the triangle PEG. Therefore the demonstration of Euclid (VI, 33.) con- cerning the angles at the centre of a circle, and the circu- lar sectors, is applicable in every respect, to the angles at the Pole of a great circle on the sphere, and to the tri- angles formed between it, as a vertex, and the arcs of that circle as bases. That is, the angles at the Pole, and also the triangles themselves, have to each other respectively, the same ra- tio, which their bases have. q. E. D. Cor. 1 . — In such triangles, the area, varies as the base, or as the angle at the Pole, of which the base is a meas- ure ; that is the base is a measure of the area of the trian- gle. CoR. 2. — Hence, since the triangle E;7G, equals EPG in every respect, therefore, the arc EG which is the meas- 20 150 SPHERICAL GEOMETRY. ure of the angle at the Pole, is also the measure of the lune FEpG ; which is composed of the two triangles. Thus, when the lune is right angled, or EG= a quadrant, it is one half of a hemisphere, and as EG approaches to a semi circle, the lune approaches to a Hemisphere. Also a spherical triangle which has its three angles right, being the half of a right-angled lune, is equal to one fourth of a Hemisphere. PROPOSITION XX. As Four right angles, is to the excess of the angles of any spherical triangle above two right angles ; called the Spherical Excess ; so is the area of the Hemisphere, to the area of the triangle. Let ABC be a spherical triangle, BCEF, be the cir- Pl. XVI. cle, of which BC is an arc ; let ABDCA, be the lune for- ^»g- 6. med by the two semi-circles, of which AB and AC are arcs, and the parts CAF, BAE two semi-circles, meeting the circle BCEF, (I,Prop. I Cor. 1 0,) then the triangle EFA =BDC on the opposite Hemisphere (II) therefore the lune ABDC, which equals ABC+DBC, equalsalso ABC-f AFE. Also ABC-f ABF= lune CAFB, and ABC4-ACE=luneBAEC, and ABC + AFE + ABF-f ACE= the Hemisphere. Now these lunes are to each other respectively as their angles (XIX, Cor 2.), that is, as the angles of the spher- ical triangle, and each of them, is to the Hemisphere as its angle, is to two right angles, That is, 2. right angles : angle A: :Hem. : ABC-j-AFE, and 2. „ „ : „ B::Hem. : ABC + ACE, 2. „ „ : „ C: : Hem. : ABC-f ABF, two right angles are to the angles at A, B,C, taken 5. 12. together, as a Hemisphere is to 3 ABC+AJE + ACE-f ABF ; therefore also two right angles, are to the Excess and^D ^^ A-fB-f-C, above two right angles, as a Hemisphere is SPHERICAL GEOMETRY. 151 to the excess of 3ABC-f AFE-f-ACE-fABF, above a Hemisphere, that is to 2 ABC. Or 4 right angles : Spherical Excess .''IS Hera. : 2ABC, :: Hem.: ABC. Q. E. D. Cou. - As the first and third terms are constant, there- fore, by equality of ratios the Spherical Excess is the measure of the area, of a spherical triangle. III. INTERSECTIONS IN THE PLANE OF A GREAT CIRCLE OF THE SPHERE. PROPOSITION L If a line be drawn from any point of the Sphere, to the pole of a great circle, the distance from the centre of that circle, to the point where the line meets its plane, is equal to the tangent * of half the distance of the given point, from the other pole of the circle. Fig- 18. That is Cf=pi, where /?0=|joE. For the triangles CP/=/)C^, are similar and equal in all respects. C- JS. D. PROPOSITION II. If one extremity of a straight line, remain in the pole of a great circle, while the line is carried around in the cir- cumference of any parallel circle, it will describe the con- vex surface of a right cone ; and the intersection of this surface, with the plane of the great circle, is a circle whose centre will coincide with that of the Primitive circle. t * The word tangent is here used, a3 in Trigonometry, for that part of a line touching a circle, which is intercepted between the point of contact, and any diameter of the circle produced. Also, distance on the sphere, is used for the arc of a great circle^ which passes through any two points ; for this arc on the sphere, like a straight line, in a plane, is the least line which can be drawn between two points. t The term Primitive is here applied to that great circle, in whose plane the intersections are supposed to be made. SPHERICAL GEOMETRY. I53 1. The Figure FPI described by the motion of the line, fig. 17, is evidently a right cone. (See EucHd, Sup. Ill, def. 11.) 2. Its intersection with the plane of the Primitive is a circle. (Con. Sect. def. 4.) 3. The centre will coincide with that of the Primitive, for the axis of the cone, PH, is the axis of the sphere and of the great circle EQ, through the centre of which, and of all circles parallel to it, it passes. (Def. 9, Cor. 3.) Q. E. D. Cor. — If the vertex of the cane, or P, be the more dis- tant pole, the circular intersection will be within the Primitive, as Jhi, and consequently less than the Primi- tive. But if the vertex be the nearer pole, the intersec- tion will be without the Primitive, and larger asft* PROPOSITION HI. The same being supposed as in the last proposition, but the circle oblique to the Primitive, the figure described will be an oblique cone, and its intersection with the Prim- itive will, be a circle. 1 . The figure described FPI is an oblique cone, (Con. Fig, le. Sect. Appendix Def. 1.) 2. Its intersection with the Primitive, fhi^ is a circle, i- 29, for it.is a sub-contrary section, since IPx =11/, and also=c ^' ^^* IFF. (Con. Sec. App. VI.) Q. E, D. Con. 1 . — If F be between E and P, a part of the in- tersection will be in the plane of the Primitive, and the re- mainder without it, and if the oblique circle is a Great Circle, the corresponding circular intersection, will cut the primitive in two points diametrically opposite. (Fugf. I. Cor. 10.) The preceding proposition and demonstration, are ap- plicable to any small circle which is at right angles to the 154 SPHERICAL GEOMETRY. Fig^ 19. primitive as EFI, whose corresponding circular intersec- tion isfhi. PROPOSITION IV. The centres of the circular intersections in the plane of the Primitive arc, in all cases, are in the line of mutual in- tersection of the Primitive and a plane passing throug:h the centre of the sphere, and through the poles of the Primi- tive, and of the given circle, upon its surface. Fig, 16. That is, the centre oi fhi ism QE, which is the inter- 17. & l9gection of the Primitive, and of I^E, or a great circle pass- ing through the poles of the Primitive, and of the circle FHI. For the circle I/^DE, is perpenrl'cular to the Primitive, and to the circle FHI, {Dei, 9. Cor. 4.) and coincides with the axes of both, and consequently passes through the centres of both, bisecting both thecircle FHI, on the sphere, and the corresponding circularintersectjon in the plane of the prim- itive. Q. £. D. Cor. 1. — If the circle on the sphere be an oblique great circle, the centre of the corresponding circular intersec- tion will be in that diameter of the Primitive produced, which is perpendicular to its intersection with the oblique Fig. 23. circle. 3. 1. That is, in QEt, which is perpendicular to FI. CoR. 2. — Hence if any number of oblique great circles, cut the Primitive in the points F and I, the centres of all their circular intersections, will be in the same line. Viz. in QEt produced. Fig. 20. CoR. 3. — In like manner if a small circle be at right angles at the Primitive, the centre of the corresponding circular intersection, will be that diameter of the Primitive, which is perpendicular to its intersection with the small circle. And the centres of all such circles, will be in the same straight line, Ei produced. SPHERICAL GEOMETRY. 155 Cor. 4. — If any number of great circles oblique to the ^»g- 21. Primitive, intersect each other m one point on the Sphere, their corresponding circular sections, will intersect each oth- er in one point in the plane of the Primitive, and their cen- tres will be in a straight line KL, perpendicular to the di- ameter of the Primitive which passes through the point of their mutual intersection, and passes through, h, the centre of the circular lection whose chord, at its intersections with t-he Primitive, is parallel to KL. PROPOSITION V. The distance between the centres of the Primitive and of any circular intersection of a great circle in its plane, is equal to the tangent of the arc which measures the inclina- tion of the circles on the sphere, I. Prop. Cor. I. That isPA=Q^ Eig. 23. For APQ, being the inclination of the circles on the sphere, and Q^, the tangent of QA, which measures their inclination. Then QPA=:FPK; h FPK-hFPA=right angle. ^. Also AIP+PU4-HIi*=PlA-fPAI = right angle, ^\\'^ .:. AlF'hhli=Phl=h[i+hil=2hli; .:. A\P=hli ; h 2AIP, or APF=2hIz=PM, .:. APF-fPIA=FPK4-APF, .-. PIA=FPK=APQ^ hPh=qt. Q. E. D, CoR. 1. — Hence \h the radius of the circular intersec- tion F/Ii, is equal to P^, the secant of the arc QA, which measures the inclination of the cicles. CoR. 2. — PA:, the distance from the centre of the Prim- itive to the point, where a line connecting the pole of the oblique circle with that of the Primitive, meets the plane of the latter, is equal to the tangent of half the the angle of their inclination. F0rPIA:=iFPK=iAPQ. n^ SPHERICAL GEOMETRY PROPOSITION VI. If a small circle be at right angles to the Primitive, the radius of its circular intersection, will be the tangent of its distance from the nearer pole. Fig. 20. That is, Fh is the tangent of FE. For xDF=DBF=PFB, And x'Di=hiF=hFi, 3. 31. .-. PFB=AFi, .-. PF/i-:/Fi=right angle. .'. Fh is the tangent of FE. Cor. — Fh the distance of the centre of the circle F/ li, from that of the Primitive, is the Secant of FE, the distance of the small circle from its nearest pole. PROPOSITION VII. The angle made by two circles which cut each oth- er on the sphere, is equal to the angle made by the corres- ponding circular intersections in the plane of the Primtive. Fig. 22. That is, DFK,ar D FK'=ci/A; or DfK', For suppose planes to coincide with FK and FD, the tangents to the circles, at the point of their intersection, and to pass through P ; these planes will touch the con- vex surfaces of the cones, whose bases are the two circles. Also let another plane pass through P, and cut those tan- gents, in K and D, forming a pyramid PKDG, which is cut by a plane KDG, parallel to the primitive. Let FI be parallel to EQ. ^ 32. Then PIF=PFK' = GFK, also PIF=PFI=FGK, Therefore FK=KG. In like manner FD=DG. SPHERICAL GEOMETRY. m Therefore KFD=KGD=^/J, in a plane parallel to ^ '^^ KGD. But since the planes PFK, PFD touch the cones, their intersection with the primitive, will be straight lines touching the circular intersections, in the same plane. And the angle made by two circles which cut each other, in a plane, is here considered as equal, or is the same as the angles made by lines touching them at the point of intersection. In most cases, the demonstration will be much more con- cise and simple, by taking the vertical angle of the tangents, viz: D'FK', and proving its equality with D/ K' in the plane of the Primitive. For DTP=FIP=DyF, therefore D'F=Dy. In like manner KT=K/, and therefore, D FK' D'/K'. Q. E. b. PROPOSITION VIII. If on the sphere, any number of small circles pass through one pole of the Primitive, and the farther pole of any ob- lique great circle, their circular intersections will all pass through the centre of the Primitive, and also through the point, where a line drawn from the other pole of the Prim- itive through the pole of the oblique circle, meets the plancp^'' ^j.^' of the Primitive. That is, the circular intersections, will all pass through P andy. This is evident from the demonstrations of the preceding propositions. PI. XIV. Cor. 1. — MO, ON are respectively equal to the arcs wo, on, of the oblique circle on the sphere. CoR. 2. — ^The intersections of the planes of the small Fig. 4. circles, with the Primitive, are straight lines which intersect each other, in the point (y) where aline drawn from the given pole of the oblique circle, to that of the Primitive, meets the plans of the latter. 21 tSB SPHERICAL GEOMETRY. Pj- XL PB, PN, PX, are the intersections of the pimitive, and of *^* ' the planes of small circles, passing through its pole, and the farther pole of the oblique circle CRSOD. Cor. 3.— RS and SO correspond to arcs of the oblique circle on the sphere, respectively equal to XN and NB. ivr APPENDIX TO SPHERICAL GEOMETRY 13 PROJECTIONS. Def. 1. — To Project oin object^ is to represent every point of it, on a plane, as it appears to the eye in a certain position. 2. — The Plane of Projection, is that on which the object represented. 3. — The Projecting Pointy is that point where the eye is supposed to be placed. 4, —The Orthographic Piojection of the sphere is thai in which a great circle is assumed as the plane of projection, and a point at an infinite distance in the axis produced, as the projecting point. 5. — The Stereographic Projection of the sphere is thai, in which a great circle is assumed as the plane of projec- tion, and one of its poles, as the projecting point. 6. — The Gnomonic Projection of the sphere, is that in which the plane of projection touches the sphere, and the centre is the projecting point. ScHOL. — In the Theory of Projections, the rays of light are supposed to move in strai^iht lines, from any point of an object to the eye. and the projection of each point, is the intersection of such a line, and the plane of projection. 7. — A direct circle is parallel to the plane of projection. 160 APPENDIX TO SPHERICAL GEOMETRY. 8. — ^n oblique circle is oblique to the plane of projec- tion. 9. — M right circle, is that whose plane coincides with the axis of the eye. Cor. — A Great Circle which is right, is perpendicular to the plane of projection. The following results or Laws of Projection will follow as Corollaries, from the preceding definitions and Scholi- um. I. Of Orthographic Projection » N .)( 1. — The Rays of light being supposed to come from an indefinite distance, 'ttiay be considered as parallel to each oilier, and perptndicular to the plane of projection, 2. — A straight line, perpendicular to the axis of the eye, is projected into a point. 3. — A straight line parallel to the plane of Projection, is projected into a line equal to itself. 4. — A straight line oblique to the plane of Projection is projected into a line less than itself, in the ratio of the sine of the angle which it makes with any ray of light to radius, 6. — So a plane, perpendicular to the plane of Projection, is projected into a straight line, and a plane parallel to the plane of Projection, is projected into a plane equal and similar to itself. n. VIII. 7. — A circle oblique to the plane of Projection is pro- =' ^ ' jected into an Ellipse. For let AMB represent the circle to be projected, and ALB the figure into which it is projected. Any ordinates in the circle will be projected into lines less than them- sel^fcs, in the same ratio. (4.) APPENDIX TO SPHERICAL GEOMETRY. 161 Thatis,KM:DG::KL:DE; which is the property of an Ellipse. As Orthographic Projection, is not necessarily hmited m its application, to the sphere, having no particular relation to it, and is principally used in a noore general application to mathematical figures, it is sufficient here, to give a mere sketch of its general properties. 11. Stereographic Projection, is confined to the sphere, and on many accounts, is the most convenient method of representing on a plane, figures on the surface of the sphere. By comparing the preceding definitions with the propo- sitions, in the third part of Spherical Geometry, it will be evident, 1. — That all circles, on the surface of the sphere, are projected into straight lines, or into circles. This renders the practical operations in this method of projection very easy. 2. — That these projected circles, in all cases, make the same angle, on the plane of Projectioii, which the circles to be projected make on the sphere. 3. — That the centres of projected circles and their pro- jected poles are accurately and easily found, by geometri- cal operations. III. Gnomonic Projection, has a particular reference to the sphere, but is of a very limited application ; being used to explain the theory of Dialing, or the Geometrical construc- tion of Dials. If the Earth were supposed transparent, and its axis ca- pable ofcasting a shadow, this shadow from the Earth's ro- tation would coincide successively with the planes of dif- ferent Meridians ; and as the rotation of the Earth is sup- posed to be uniform, the shadow, would in equal times, co- incide with Meridians at equal distances from each other, or which make Vvith each other equal angles. 162 APPENDIX TO SPHERICAL GEOMETRY. If successive meridians were taken, making angles of 15^ with each other, the shadow of the axis would move from one to the other is one hour, or j\ part of an entire revolu- tion. Such Meridians are called hour circles. If the planes of these Meridians or hour circles, weve produced to intersect a plane touching the sphere, their mutual intersections would be straight lines, which to an eye placed at the centre of the sphere, would coincide with the planes of the Meridians. They would therefore be the Projections of those planes, or of their circumjerences on the surface of the Sphere. Hence in every kind of Di- al, the edge or part of the Gnomon^ which casts the shad- ow must be placed parallel to, or coinciding with the Earth's axis, and supposing a sphere to surround it, the hour lines are the intersections of the hour circles with the dial plate, which is the plane of projection. PI. XV. Thus in the common Horizontal Dial, CAB, is a triangu- p-^ 1. ^^^ plate, of which the edge, which casts the shadow, is CA, and the angle CaB being made equal to the latitude of the place, i. e. to the elevation of the pole, CA will be parallel to the axis of the earth. Suppose a sphere to sur- round CA, as its axis, whose centre is D, then DF per- pendicular to CA, will be the intersection of its equator with the plane CAB, andHFG will be the intersection with the Dial plate. Making Fl and^ each equal to FD, the quadrants, FK, fk will represent half the Equator, or a circle concentric with it, equally divided by Im, \m he. m, in, &c. which represent meridians, or hour circles; Ym, F/w,, &c. also//i, fn, he. are the intersections of these circles, with the line HG, in the plane of the Dial. Hour lines drawn from A and a, through m, m, he. n, n,hc. will intersect the circumference of the Dial plate, in the divisions which mark the hours. The same principles, will serve for the construction and explanation of all Dials. APPENDIX TO SPHERICAL GEOMETRY- PROBLEMS IN STEREOGRAPHIC PROJECTIO]^. PROBLEM L To project a great Circle, of zchich a given point is thr projected Pole* 1 . TTie given point being the centre of the Primitive. Then the Primitive itself is the circle required. 2, The given point being in the circumference of the Primi- tive. Let the given point be A. Draw through the centre of Fig. i the Primitive the diameter ACE, and a perpendicular di- ameter BCD, which is the projected circle, whose poles are A and E. Also B and D are the poles of the projected circle ACE. 3. TTie given point, being in the plane of the Primitive, Let the given point be p. Draw the diameter BpD, i-'i?- - and its perpendicular AC. From A reduce^ to G, in the circumference of the Primitive ; make GF equal to a quad- rant, project F in E ; a circle through A, E and C, is the projected circle, whose pole is p. 164 appelndix to spherical geometry. Cor. — By reversing the process in each case, the pro jected pole of a given projected circle may be found. PROBLEM IL To Project a Small Circle, parallel to a given projected Great Circle, and at a given distance Jrom it, 1. The given Great Circle being the Pnmitive* Fig. 15. Draw any diameters as BCD, and its perpendicular. From B set the given distance to O; project O in X, then the circle XFE, is the circle required. 2. The given Circle being a right Circle, Fig. 16. Let the given projected circle, be BD. Set the given distance from B to F, project F in G on the perpendic- ular diameter AY ; a tangent to the primitive, at F, will intersect AY produced in (K) the centre of the required circle, which will pass through F and G. 3. The given Circle being oblique. Fig. n. Let the given oblique circle, be AED. Find its Pole (Prob. L Cor.) P ; reduce P to N, and from N set off on each side, the complement of the given distance of the small circle, to R and Q ; project R and Q in X and Y, XY is the diameter of the small circle (BXY) required. CoR. — By reversing the process, the pole of a given small circle, parallel to a large circle, may be found, anij also the distance of the small circle from its large circle. APPENDIX TO SPHERICAL GEOMETRY. 1^5 ** PROBLEM III. To project a great circle^ through two given points in the Primitive, I . One of the points being in the centre of the Primitive, Let the given points be C and A, or Cand R. Then Fig. the diameter ACE or YRB will be the circle required. 2« One of the points being in the circumference of the Primi- tive, Let the given points be C and K. Draw the diameter F'?- 8« CXD, the circle required will pass through C, K and D. 3. Keither of the points being in the centre or circumference of the Primitive* Let the given points be P and Y. Draw the diameter Fig. 9. BPA and its perpendicular CD, reduce P to G ; the re- quired circle will pass through, YP and G. PROBLEM IV. Through any given point, to project two circles luhich shall make a given angle. 1. When the angle is at the centre of the Primitive. Draw any diameter as ACE, set the measure of the giv- Fig. 1, en angle from A to H, draw HC, then AC and HC are pro- jected circles making the required angle at C. 22 166 APPENDIX TO SPHERICAL GEOMETRY. 2. When the angular point is in the circumference of the- Primitive. fiS' 3. Let A be the angular point. Draw tbe diameter AC, and the perpendicular DB. Set the measure of the given angle from D to E; project E to F, AFC, is a projected circle which makes the given angle with the Primitive at A. 3. When the angular point is in the plane of the Primitive, Fig- 5. Let O be the given point. Draw the perpendicular di- ameters, aOB and (JD. From C reduce O to I • make DK equal to Al ; project K in P, CPD is the projected circle, of which O is the projected pole. (Prob. 1.) Set the 2;iven angle from C to L, let a line from L to O (which will be the projection of a small right circle) intersect the circle CPD in Q, from Q as a pole, project the great circle EOF, (Pkob. I.) which makes the given angle EOA, with the right circle AOB. 4. When the circles required are both oblique. Fig. 14. Let E be the given point. Through E project any ob- lique sreat circle, as AED. Find its pole- P. Let the straight line EP intersect the Primitive in 1'; set off TV the measure of the given angle ; draw the diameter VY, and its perpendicular FN, NEA, is the required angle, NEF, -iED, the required circles. Cor. — By reversing the process, in each case the angle made by two given projected great circles can be measur- ed. PROBLEM V. Through a given point to project a great circle perpen- rlirylar to a given great circle. APPENDIX TO SPHERICAL GEOMETRY. 1^7 1. When the given circle is the Primitive. Let Y or X be the given point. The diameter YCB, or ^ AXCE, will be the circle required. '^" 2. When the given circle is righty and the given point the centre of the Primitive* Let AB be the given projected circle, the perpendicular ^'^s- 8- CXD is the circle required. 3. IVhen the given circle is right., and the given point not in the centre of the Primitive, Let the given projected circle be AB, and K, in it, the given point. Draw the perpendicular diameter CD, then the required circle will pass through C, K, and D. 4. When the given circle is oblique, and the given point in the middle. Let DCK be the given oblique circle, and K the given Eig. 8. point. The diameter AKB, is the required projected cir- cle. 5. When the giveyi circle is oblique, and the given point not in the middle of it. Let Y be the given point in CYD. Find the poles P, f^'g- ^■ p of CYD, and FYp, is the circle required. PROBLEM VI. To project a great circle, through a given point which sjiali make a given angle roith the Primitive, Let O be the given point. Describe the circle CPD, ^^o' 5. whose pole is O, as in Prob. IV. 3. Make BL equal to 168 APPENDIX TO SPHERICAL GEOMETRY. th6 given angle, project L in M ; and describe the small circle EQN, which is parallel to the Primitive and whose distance from the centre is equal to the given angle, inter- secting CPD, in Q. From Q as a projected pole, describe the circle EOF, which makes the required angle OEA. CoR. — Conversely BL is the measure of the angle, AEO. PROBLEM VII. To project a great circle, which shall make given angles with two given circles. 1 . When one of the given angles is a right-angle. Fig- 3. Let the Primitive and the right circle DFB, be the giv- en circles. It is required to project a great circle which shall make right angles with DB, and any given angle with the Primitive. Draw AC perpendicular to DB. Set off the given angle from D to E, project E in F, and CFA, i? the circle required, F and A the required angles. 2. When neither of the angles are right. Fig. 3'2. Let the Primitive and ABd be the given circles. Find q the pole of ABt/, around it describe a small circle at the distance of the angle B. (Prob. 11.) Set the angle C from d to 6, project b in a, around the centre or pole of the Primitive, describe a small circle at the distance of a, intersecting the other small circle in s, which is the pole af the circle required, viz, CBe which describe by Prob. I- PROBLEM VIII. ^'^To set off any number of degrees on a projected circle,^^ Or, to cut off from a projected circle an arc. v^hich corres- ponds to an arc of the circle to be projected equal to a given arc. APPENDIX TO SPHERICAL GEOMETRY. Igg i . fVhen the projected circle is the Primitive, or one of its parallels. Make DG, equal to the given arc, it is the arc required, ^^o- 15. EF on a parallel circle, is an arc of the same " number of degrees,'' or it is the same portion of the circle EFX, that DG is of DAB. 2. When the projected circle is a right circle. Let the projected circle be ACB ; make AE or DH, Fig. lO RN or RX, equal to the given arc; then AF or CG, OH ^ ^^• or OG will be the required arcs. If TU be made equal to the given arc, then OB will be Fig. 18. the required arc, on the great circle, and XY, a proportion- al arc, on the parallel small circle MN. 3. When the projected circle is oblique. Let the projected circle be CSD. Find its pole P. Fig. 12- (Prob. L Cor.) Draw PSN, from N, cut ofFNB or NX equal to the given arc ; draw PB and PX, intersecting CSD in R and O ; SR or SO, are the arcs required. Here PX and PB, represent a small circle passing through the remotest poles of the Primitive, and of the oblique cir- cle CSD. (Sp. Geom. HL Prop. VHL Cor. 2.) If CPA, represent the given oblique circle, and NL the Fig. 19. given arc, then OQ will be the required arc on the great circle, and XY the corresponding arc on a small circle par- allel to it. CoR. — Conversely, if the arc of a projected circle be given, the corresponding arc, on the circle to be projected mav be found and measured. thus AE and DH, RN and RX are the measures of AF Fig. m and CG, OH and OG. ^ ^^• Also TL^, measures OB and XY, and LN measures is A- ly OQandXY. 170 APPENDIX TO SPHERICAL GEOMETRY. PROBLEM IX. To project a Hemisphere on the plane of (he Equator VL X\V. Let the Primitive ENWS, represent the Equator. Draw the perpendicular diameters EW, NS. Divide each quadrant into nine equal parts, then the lines drawn from their division to the centre, will be the projection of Meridians. Project parallels of Latitude, corresponding to the divi- sions of the Primitive, (Prob. II.) also the Tropic ami Po- hr circles, at the proper distances from the centre or pole of the Primitive ; The Ecliptic may be projected by Proh. IV. PROBLEM X. ^'To project a Hemisphere on the plane of the Meridian. Describe the Primitive and divide it as before. Let it Represent the solstitial colure. WE is the projected Equator. Project parallels to it, with the Tropics and Polar circles, by Prob. II. Tangents to the Primitive, at the several divisions, will intersect NS produced in the centres of their parallel cir- cles, as in y, y. The meridians may be projected by Prob. IV. (iheii centres are all in WE produced.) The projection of the Ecliptic is obvious. SPFiERICAL TRIGONOMETRY, PART I. Geometrical Principles of Spherical Trigonome- try ; or the mutual relations of the Trigonomerriral lines, corresponding to the arcs and angles of Spherical Triangles. PROPOSITION I. In right angled spherical triangles, the sine of either ol the sides about the right angle, is to the rauius of the sphere, as the tangent of the remaining side is to the tan- gent of the angle opposite to that side. Let ABC be a triangle having the right angle at A. PI- X. Then, sine of AB : rad: ".tang. AC : tang. ABC. Y\%. 24. Lei D be the centre of the sphere; join AB, AD, AC, and let AF be drawn perpendicular to BD, which there- fore will be the sine of the arc AB ; and from the point F, let there be drawn in the plane BDC the straight line FE at right angles to BD, meeting DC in E, aad let AE be joined. Since therefore the straight line DF is at right an- gles to both FA and FE, it will also be at riiiht angles to the plane AEF, wherefore the plane ABD, which passes^up. ^2. 4 through DF is perpendicular to the plane AEF, and the " "^ plane AEF, is perpendicular to ABD : But the plane ACD orAED. is also perpendicular to the same ABD be- cause the spherical angle BAC is a right an^le : There- fore AE, the common section of the planes y^ED, AEF is P' * ' it at right angles to the plane ABD and EAF, EAD are right angles. i72 SPHERICAL TRIGONOMETRY. Therefore AE is the tangent of the arc AC and in the rectilineal triangle AFE having aright angle at A, AF : rad.::AE : tang. AFE, but AF=sine ofAB, and AE=tang. of AC, and AFE= [ABC. .'. sin AB : rad.: :tang. AC : ABC. Q. E. D. Cor. — Since sin AB : rad: :tang. AC : tang. ABC, and (PI. Trig. 93.) R : cot. ABC: : tang. ABC : rad, .-. Sin. AB : cot. ABC: : tang. AC : rad. PROPOSITION II. In right angled Spherical Triangles, the sine of the Hy- potenuse is to the radius, as the sine of either side is to the sine of the angle opposite to that side. Fig. 25. That is, sin BC : rad :; sin AC : sin ABC. Let D be the centre of the sphere, and let CE be drawn perpendicular to DB, which will therefore be the sine of the Hypotenuse BC : and from the point E let there be drawn in the plane ABD the straight line EF, perpendicu- lar to BD, and let CF be joined : then CF will be at right Sup.2.i8.angles with the plane ABD. Wherefore CFD, CFE are right angles, and CF is the sine of the arc AC : and in the triangle CFE, having the right angle CFE, CE :rad::CF : sin CEF, ButCEF=ABC, (Spher. Geom. II Def. l.cor. 5.) .'. sin. BC : rad: : sin AC ; sin ABC. Q. £. D, SPHERICAL TRIGONOMETRY. 173 PROPOSITION HI. In right-angled spherical triangles, the cosine of the Hypotenuse is to the radius, as the cotangent of either of the angles is to the tangent of the remaining angle. That is, cos. BC : rad.: :cot. ABC : tan. ACB. Fig. 56. Describe the circle DE, of which B is the pole, and let it meet AC in F, and the circle BC in E ; and since the circle BD passes through the pole B, of the circle DF, DF must pass through the pole of BD. And since AC is perpendicular toBD, (Sp. Geom. Def, 9, cor. 5.) therefore AC must also pass through the pole of BAD, wherefore the pole of the circle BAD is in the point F, where the circles AC, DE intersect. The arcs FA, FD are therefore quadrants, and likewise the arcs BD, BE. Therefore, in the triangle CEF right- angled at the point E, CE is the complement of BC, the Hypotenuse of the triangle ABC ; EF is the complement of the arc ED, the measure of the angle ABC; and FC, the Hypotenuse of the triangle CEF, is the complement of AC; and the arc AD, whicli is the measure of the angle CFE, is the complement of AB. But (I)Sin.CE : R::tan. EF : tan. ECF, That is, COS. BC : R : : cot. ABC : tan. ACB. q. JE. D. CoR.— Since cot. ACB : R: :R : tan. ACB, i^i- Trig. .-. cot. ACB : COS. BC: :R : cot. ABC. ^^• PROPOSITION IV, In right-angled spherical triangles, the cosine of an an- gle is to the radius, as the tangent of the side adjacent to that angle, is to the tangent of the Hypotenuse. That is, cos. ABC : rad. : itang. AB : tang. BC. Fig. 26. 2.3 174 SPHERICAL TRIGONOMETRY. For (I) sine. FE : rad.: itang. CE : tan. CFE, But sine EF = cos. ABC, tang. CE = cot. BC, and tang. CFE = cot. AB. .- . COS. ABC : rad. : ! cot. BC : cot. AB. Again cot. BC : rad.: irad. : tang. BC, And cot. AB : rad.: irad. : tang. AB, .'. cot. BC : cot. AB: : tang. AB. : t; Therefore, cos. ABC : rad. : : tang. AB : 1 : cot. AB: :tang. AB. : tang BC; ABC : rad. : : tang. AB : tang. BC. q. E. D. CoR. 1. —From the demonstration it is manifest, that the tangents of any two arcs AB, BC are reciprocally pro- portional to their cotangents. CoR. 2 —Cos. ABC : cot. BC: :tan. AB : rad. PROPOSITION V. In right-angled spherical triangles, tne cosine of either of the sides is to the radius, as the cosine of the Hypote- nuse, is to the cosine of the other side* That is, cos. CA : rad. : cos.BC : cos. AB. For (II) Sin. CF : rad. : :sin. CE : sin. CFE, ButSin.CF=cos.CA, sin.CE=cos.BC,and sin.CFE = [cos AB. .'. cos. A : rad. : :cos. BC : cos. AB. Q. E. D. PROPOSITION VI. In right-angled spherical triangles, the cosine of either of the sides is to the radius, as the cosine of the angle op- posite to that side, is to the sine of the other angle. That is, cosine CA : rad : : cosine ABC : sin. BCA. SPHERICAL TRIGONOMETRY. J75. For (11) sin. CF t rad.::sin. EF : sin. ECF, But '' sin. CF=cos. CA, sin. EF=cos. ABC, and sin. ECF=sin BCA. ^. COS. CA : rad.: .cos. ABC : sin. BCA. Q. E. D, PROPOSITION VII. In spherical triangles, whether right-angled, or oblique- angled, the sines of their sides are proportional to the sines of the angles opposite to them. That is, sin. AC : sin. B: isin. AB : sin.C . Fi?. 13. 1. For (II) sin. BC : rad. (=sin. A.): isin. AC : sin. B, Also *' sin. BC : sin. A: :sin. AB : sin. C, .:. " sin. AC : sin. B: :sin. AB : sin. C. 2. Then, sin. BC : sin AC: Isin A: .'sin B, Fig. 14. Through C, draw a perpendicular CD to the opposite ^ ^^' ^ide, then (II) sin. BC : rad.: :sin. CD : sin. B, " sin. AC : rad.: :sin. CD : sin. A. .:. sin. BC : sin. AC: :sin. A : sin. B. In like manner, sin. BC : sin. AB: :sin. A : sin. C. Q. E, D, PROPOSITION VIII. In oblique angled spherical triangles, a perpendicular arc being drawn from any of the angles on the opposite side, the cosine of the angles at the base, are proportion- Fi^. u. al to the sines of the segments of the vertical angle. Let ABC be a triangle, and the arc CD perpendicular to the base BA Then, cos. B : cos. A: :sin. BCD : sin. ACD. 176 SPHERICAL TRIGONOMETRY. For (VI) COS. CD : rad.: :cos. B : sin. DCB, and " COS. CD : rad.! :cos. A : sin ACD, COS. B : COS. A: :sin. DCB : sin ACD. Q, E. D. PROPOSITION IX. The same things remaining, the cosines of the sides are proportional to the cosines of the segments of the base* rig. 14. That is, COS. BC : cos. AC : :cos. BD : cos. AD. For (V) cos. BC : COS. BD: :cos. DC : rad. And " COS. AC : cos. AD: :cos. DC : rad. COS. BC : cos. BD::cos. AC : COS. AD, And COS. BC : cos. AC:: cos. BD : cos. AD. Q. E, D, PROPOSITION X. The same construction remaining, the sines of the seg- ments of the base are reciprocally proportional to the tan- gents of the angles at the base. That is, sin. BD : sin. AD: : tan. A : tan. B. For (I) sin. BD : rad.: : tan. DC : tan. B, and " sin. AD : rad.: :tan. BC : tan. A, .*. " sin. BD : sin. AD : : tan. A : tan B. Q. E. D. PROPOSITION XI. The same construction remaining, the cosines of the segments of the vertical angle, are reciprocally propor- tional to the tangents of the sides. SPHERICAL TRIGONOMETRY. 1 77 That is, COS. BCD : cos. ACD: :tan. AC : tan. BC. For (IV) COS. BCD : R: :tan. CD : tan. BC, And " COS. ACD ; R: :tan. CD : tan. AC, COS. BCD : COS. ACD: :tan. AC : tan. BC. Q. E. 2>. PROPOSITION XII. The tangent of half the sum of the segments of the base, is to the tangent of half the sum of the sides, as the tangent of half the difference of the sides, is to the tangent of half the difference of the segments of the base. Then, tan. i BD + aD : tan. ^BC-fAC : : tan. i BC- AC : tan. iBD-AD. For (IX.) COS. BC : cos. AC: :cas. BD-f cos. AD, *. cos BC+cosAC : COS. BC— cos. AC::cos. BD-f-cos. AD : COS. BD-cos. AD. But cos. BC+cos. AC : COS. BC~cos. AC::cot. | BC-fAC : tan. | BC-AC,* Also, COS. BD-f COS. AD : cos. BD-cos. AD::cot. | BD-f AD : tan. i BD-AD. cot. iBB-fAC Man. i BC - AC: :cot. 1 BD-f AD : tan. iBD-AD. But, tan. i BC-fAC X cot. i BC-fAC : tan. 1 BC-fAC Xtan. 1 BC-AC::tan. i BD-f ADXcot. i BD-f AD : tan. 1 BD-f ADxtan. i BD-AD, But, tan. I BC-fAC Xcot. 1 BC-f AC=rad.=tan. I BD-f AD cot. 1 BD+AD, tan. i BC-f AC Xtan. i BC-AC=tan. i BD-f AD Xtan. i BD-AD, * Plaue Trigf. 178 SPHERICAL TRIGONOMETRV. Or tan. i BD + AD : tan. i BC4-AC: :tan. IBC-AC : tan. iBD-AD. Q. E. D, ScHOL. — The preceding proposition will be easily re- membered, from its resemblance to the corresponding proposition in Plane Trigonometry. PROPOSITION XIII. DAT. tfthe ratios of the radius of any circle, to the trigonomet- rical lines corresponding to any given arc, were known, and conversely; then if any two of the parts of a ri^ht-angled spherical triangle were given, the remaining parts would also be given, i, e. they might be found. I. If the hypotenuse and either of the adjoining angles were given, the sides and the remaining angle, would be PI. XII. given also. Fig. 21. That is, if AC, and A are given, AB, BC, and C may be found. 1. Because the radius, the sine of AC, and the sine of A are given, the sine of BC and consequently the arc BC, may be found. For (II) R : sine AC:: sin. A : sin. BC ; and a fourth 6. 12. proportional to any three straight lines may be found. Therefore the sine of BC may be found. But as the radius, and the sine of BC are given, their ratio is given, and therefore by supposition the arc BC is given also. It is to be remarked that any sine may be the sine of two arcs which are supplemental to each other, but as A is given, it is known of what affection it is, i. e. whether it is greater or less than a right-angle, and BC is of the same affection, (Sp. Geom. XIV.) therefore BC is given without ambiguity. SPHERICAL TRIGONOMETRY. I79 2. In a similar manner AB, may be found. For (IV) R : COS. A: :tang. AC : tang. AB. If AC be less than a quadrant, AB is of the same affec- tion as A which is given. If AC be greater than a quad- rant, AB, and A, are of different affections, in either case AB is given without ambiguity. (Sp. Geom. XIV. XV.) 3. Also, C may be found. For (III) R : cos. AC: :tang. A : cot. C. If AC is less than a quadrant, C is of the same affectioa as A; otherwise they are of different affections. In either case C is given unambiguously. In the same manner, if AC and C are given, BC, AB, and B may be found. II. If the hypotenuse and either of the sides are given the angles and the remaining sides are given. That is, if AC, BC, are given, AB, A and C may be found. 1 . The side AB may he found. For (V.) cos. BC : cos. AC: :R : cos. AB. If AC is less than a quadrant, AB is of the same affec- tion as BC, if not, they are of different affections. 2. The angle A may be found. For (11) sine AC : sine BC : : R : sin. A. The angle A is of the same affection as BC. 3. The angle C may be found. For (IV) tangent AC : tan. BC: :R : cos. C. If AC is less than a quadrant, BC and C, are of the same affection, otherwise they are of different affections. In the same manner, if AC and AB are given, BC, A and ^ may be found. 180 SPHERICAL TRIGONOMETRY. III. Jf one side and an adjacent angle are given, the hy- potenuse and the remaining side and angle, are given. That i», if BC and C are given, AC, AB and A may be found. 1. The hypotenuse JlC may be found* For (IV) COS. C : R: :tan BC : tan. AC. If BC and C, (both of which are given,) be of the same affection, AC is less than a quadrant, otherwise it is greater. 2. The side AB may he found. For(I.)R : sin. BC::tan. C : tan AB. AB is of the same affection as C. 3. The ansrle A may he found. For (VI) k : cos. 'BC: :sin. C : cos. A, The angle A is of the same affection as BC. In like manner, if AB and A are given, AC, BC and C may be found. IV. If one side, and the angle opposite are given, the hypotenuse and the remaining side and angle are given. That is, if BC and A are given, AC, AB and B may be found. 1. The hypotenuse AC may be found. For (II) sin. A : sin. BC : : R : sin. AC. The hypotenuse may be either the arc AC, or its sup- plement Cd^ for both have the same sine. The two triangles ABC, and BCc/ have each the parts given in this case. The hypotenuse therefore is given ambiguously. It is one of two given arcs. If the affec- tion of either of the remaining parts of the triangle, should be given, the ambiguity would be removed. 2. The side AB may he Jound, For(l) tan. A : tan. BC::R : sin AB. The arc is ambiguous, being either AB or its supple- ment B(f. The sine of both being the same. SPHERICAL TRIGONOMETRY. 3. The angle C may he found. For, (VI) COS. BC : cos. A: :R : sin. C, which is ambi- guous, being either BCA, or BCJ, which have the same sine. In like manner if AB and C are given, AC, BC, and A may be found. V. If the two sides are given, the hypotenuse and an- gles are given. That is, if AB and BC are given, AC, A and C may be found. 1 . The hypotenuse AC may he found. For, (V) R : cos. AB::cos. BC : cos. AC, which is less than a quadrant, when AB and BC, are of the same affection, but greater than a quadrant when they are of different affections. (Sp. Geom. XV.) 2. The angle A may he found. For, (I) sin. AB : R: :tan. BC : tan. A ; which is of the same affection as BC. 3. In the same manner B may be found. For, (I) sin.BC : R::tang. AB : tan. C ; which is of the same affection as AB. VI. If the two angles are given, the Hypotenuse and the opposite sides are also given. That is, if A and C are given, AC, AB, and BC may be, found. 1. The hypotenuse AC may be found. For, (HI) tan. A : cot. C: :R : cos. AC. When B and C are of the same affection, AC is less than a quadrant; but when they are of diffi^reiU affections, AC is greater than a quadrant. (Sp. Geom. XIV. XV.) 2. The side AB may be found. For (VI) sin. A : cos. C: :R : cos.AB ; which is of the same affection as C. 24 181 182 SPHERICAL TRIGONOMETRY. 3. In the same manner BC may he found. For, (VI) sin. C : cos. A: :R : COS. BC, which is of th« same affection as A. PROPOSITION XIV. DAT. The same being supposed as before, if any three of the parts of an obhque angled triangle be given, the other three are also given, ^'o 32. I, jf ^^Q sides artd the included angle afe given, the re- maining side and angles are also given. That is, if AC and CB and C, are given, AB and A and B ; may be found. 1. The angle B, may be found. For let fall the perpendicular AP, from the other un- known angle A, on BC, Then, (IV) R : cos. C: :tan. AC : tan. CP ; which is therefore given. And (X) sin. BP : sin. CP: :tan. C : tan B ; which is of the same affection as A when the perpendicular falls within the triangle, but if it falls without, B and A are of differ- ent affections. In like manner, the angle A may be found ; or each of its parts CAP, and PAB, may be found ; as in case HI, Prop. XIII Dat. 2. The side AB may be found. For (IV) R : COS. C: :tan. AC : tan. CP, And (IX) COS. CP : COS. BP: :cos. AC : COS. AB ; which is of the same affection as AC, when AP falls within the triangle, but different when it falls without. (Sp. Geom. XVI.) Or AB, may be found, after B is found, by Prop. VII. II. If two angles and the included line are given, the re-" maining sides and angle are given. That is, if A, C, and AC, are given^ AB; CB and B may be found. SPHERICAL TRIGONOMETRY. 183 I. If AB may he found. For, letting fall the perpendicular as before, Then (III) R : COS. AC: :tan. C : cot. CAP, Therefore BAP may be found (=CAB-CAP,) which is of the same affect as C, when AC is less than a quadrant. and (XI) COS. BAP : cos. CAP : :tan. AC : tan. AB. AB, is of the same affection as AC, when the perpen- dicular falls within the triangle, i, e, when CAPC6,C>AB6, But,, AB-f-AC2 right-angles AC -1-AB>C^ and„C-hB<2 „ „ AC+ABCe,C-}-B>2 right-angles, and the greater side, is opposite to the greater angle. VI. If the three angles of a spherical triangle are given, the three sides are given. The supplements of the angles, are the sides of the sup- plemental triangle, (Sp. Geom. X.) which are there- fore given, therefore its angles are given by the last ; but the supplements of those angles are the sides of the sup- posed triangle which therefore are given. ScHOL. If the sides of a spherical triangle are reduced indefinitely, they approach continually to equality with their sines and tangents, and to coincidence with them. i. e. the limit of the ratio of the sides to their sines, or tan- gents, is a ratio of equality. If they be considered as actually coinciding in their evanescent state, the triangle, becomes a plane triangle, and as the properties of the spherical triangle belong to it, whatever be the length of its sides, the propositions which have been demonstrated concerning spherical trian- gles, may be transferred to plane triangles, by substituting the sides of such plane triangles in the places of the sines or tangents, of the sides of the spherical triangle. It will be seen, that these results conform to the Propositions in Plane Trigonometry. NAPIER'S RULES OF THE CIRCULAR PARTS. The rule of the Circular PartSy invented by Napier, is of use in spherical trigonometry, by reducing all the the- orems employed in the solution of right-angled triangles to two. These two are not new propositions, but are mere- ly enunciations, which, by help of a particular arrange- ment and classification of the parts of a triangle, include the first six propositions, with their corollaries, which have been demonstrated above. They are perhaps the happi- est examples of artificial memory that is known. Definition 1. — If in a spherical triangle, we set aside the right angle, and consider only the five remaining parts of the triangle, viz, the three sides and the two oblique angles, then the two sides which contain the right-angle, and the complements of the other three, namely, of the two angles and the Hypotenuse, are called the Circular Parts . Fig. 13. Thus in the triangle ABC right-angled at A, the circular parts are AC, /iB with the complements of B, BC, and C. These parts are called circular ; because, when they are named in the natural order of their succession, they go round the triangle. Def. 2. — When, of the five circular parts, any one is taken, for the middle part, then of the remaining four, the two which are immediately adjacent to it, on the right and left, are called the adjacent parts ^ and the other two, each of which is separated from the middle by an adjacent part, are called opposite parts, * This account of Napier's Circular Parts, is taken principally from Play- fair's Appendix to Spherical Trigonometry. NAPIER'S RULES OP THE CIRCULAR PARTS. Thus, if AC, be reckoned the middle part, then AB and the complement of C, which are contiguous to it on differ- ent sides, are called adjacent parts ; and the complements of B and BC are the opposite parts. In like manner if AB be taken for the middle part, AC and the complement of B are the adjacent parts, and the complements of BC and C the opposite. If the complement of BC be the middle part, the complements of B and C are adjacent, AC and AB opposite parts. PROPOSITION. In a right-angled spherical triangle, the rectangle of the radius and the sine o( the middle part, is equal to the rec- tangle of the tangents of the adjacent parts ; or to the rec- tangle of the cosines of the opposite parts. The truth of the two theorems included in this enunci- ation may be easily proved, by taking each of the live cir- cular parts in succession for the middle part, when the gen- eral proposition will be found to coincide with some one of the analogies already given for the resolution of the cases of right-angled spherical triangles. If a spherical triangle have one side a quadrant, the sup- plemental triangle will be right-angled ; and as the sines, tangents, &£c of an arc are the same as those of its supple- ment, Quadrantal, or Rectilateral spherical triangles, may evidently be solved like Rectangular Triangles. The same proposition of Napier, is applicable to both, if the following are taken as the Circular Parts in the former. The Quadrant is in the place of the right-angle, and is not supposed to separate the circular parts, which are, the two angles adjacent to the Quadrant, the complements of the other two sides, and of the remaining angle. SPHERICAL TRIGONOMETRY, PART 11. Calculation of the sides and Angles of Spheri- cal Triangles. It has been saidbyPlayfair, that, "Trigonometry is the apphcation of numbers to express the relations of the sides and angles of triangles to each other.'' This is true of Trigonometry, as applied to the actual calculation of the sides and angles of triangles. The nature and use of this application of numbers will be evident, by remarking that the Data, Prop. XIII, and XIV. are founded on the supposition that the ratios of radius to the Trigono- metrical lines belonging to any given arc, are given. These ratios however are not given, and cannot, except in a very few cases, be expressed by numbers.* If however the circumference of the circle be divided into any number of equal parts, as 360, the ratios of radius to the trigonometrical lines, belonging to these arcs, may be calculated to any required degree of exactness. {Days Trigonometry, " Computation of the canon''\) This is done, and numbers expressing the relations of their parts to radius, considered as unity, are arranged in tables. By means of these, the sides and angles of spherical, as of plane triangles can be easily computed, from the pre- ceding propositions. *^Ca5noli,Chap. VI, I. RIGHT ANGLED SPHERICAL TRIANGLES. CASE I. Given the Hypotenuse (=55° 8',) and one side Fig. 21. (=32° 12',) it is required to find the other parts. Project the Triangle,* a( the circumference of the Primitive, making {Ap = ) A' =55° 8' and {fy=fz = ) BC = 32° 12'. (Prob. VIII.), on circles described through ACrf, and B y. Attht centre^ make (m n=AP=) AC=Hyp. and (xy=) BC = side, on the obhque circle described through m and Fig. 22. C. In the plane, from the point A, make (y2=ym=) AC= a 33, Hyp. on the oblique great circle described through y A and C, and BC, (on the oblique circle described through 3cCrf)=given side. 1, To find the other side, AB. By Circular Pa ts; R . cos. AC=BC . cos. AB. Or, Prop. XIII.2 . 1 . ; cos EG : R: :cos. AC : cos. AB. Cos. BC = 32° 12' av . co. 0.0725305. R 10 Cos AC=55° G' 9.7571444. Cos. AB=47° 30' 4' 9.8296749. * The Projections are not given in full. To a person familiar with the Problems in Stereographic Projections, a mere reference to the steps of the process, it is presumed, will be sufficient. 25 190 SPHERICAL TRIGONOMETRY. 'Tlie side AB is less than a quadrant, like BC, because the Hypotenuse is less than a quadrant. 2. To find the angle C. Cir. Paris; R . cos . C=tan. BC . cot. AC, Or prop.(Xni. 2. 3.) tan. AC : tan.BC:: R:cos.C.=^ 68° 58' 30". C < 90°, like BC. 3. To find the angle A, Cir. Parts; R . sin. BC=sin. AC . sin. A. Or Prop. (XIII. 2. 2.) sin. AC : R: :sin BC : sin. A, 40^30 5". A likeBC< 90°. CASE II. Given the Hypotenuse (=55° 8') and one angle (=40** 30' 5 'to find the remaining angle, and the sides. Project the Triangle, making A=40° 30' 5" and AC= 55° 8'. Fi«', 21. The projections at the circumference and centre are evi- 24. dent. From a point, as A, in the plane, project the circle, -^- making A =40° 30' 5'', (Prob. IV.) From A cut off AC=Hypotenuse (VIII) project the circle jt?Co, through C ; BC, AB are the required sides; 1. Cir. Parts; R . cos. 55° 8'=cot. 40° 30* 5" . cot. Q. C<90°, like A, because AC<90°. 2. Sin. A . sin. AC=^R . cos. BC. BC<90°. hke A . because AC<90°. 3. R . cos. A=cot. AC . tan. AB. AB<90 . like C; because AC < 90. SPHERICAL TRIGONOMETRY. 191 CASE III. Fis:. ^. Given one side (=47° 30' 4") and its opposite angle (=63° 58 30") to find the other parts. Project the triangle, at the circumference. Make AB=47° 30' 4"=Draw the perpendicular diam- eters Br, yz. Make ym=^n =63° 58' 30' and draw the small parallel circle, (Prob. II.) from /, as a pole, pro- ject the oblique circle AC^R. ACB, or RCx is the tri- angle required. Each of them contain the given parts. Jit the centre. Cut ofF AB = given side (Prob. VHI.) Fig. 27. Project the circle dBe (Prob. III.); around x, its pole, describe the small circle, at the distance of the given angle. (Prob. II.) From h, where it intersects the primitive, as a pole, project the right circle fad; then ACB, or its supplement is the triangle. From a point in the plane, as .^, cut off AB, on a right ^^g- "^S. circle, equal to the given side. (Prob. V^III..) Project the oblique circle arBz, (Prob. III.) find its pole m, and around it, describe the small circle, at the distance of the given angle; r, its intersection with the circle xvZj whose pole is A, is the pole of a circle passing through A, making the required angle atC. 1. R . COS. C=cos. AB (= Rx) . sin. A or CRa? its Fig.26. supplement. For the angle is ambiguous. 2. Tan. . AB (or Rx) . cot. C=R sin. BC (or Ca:.) The side is ambiguous. BC or its sup' Ca. 3. R . sin. AB=sin. C . sin. AC, or CR . which is am- biguous. ^g^ SPHERICAL TRIGONOMETRY. CASE IV. Given one side (=47° 30' 4"), and the adjacent angle (40° 30' 5') to find the other parts. Fig 21. Project the Triangle, mMngAB^^I"^ 30' 4", andA= 2^; 40° 30' 5". (Prob.IV.) 1. R . sin. AB=cot. A . tang. BC. EC is of the same affection as A. 2. Cos. AB . sin. A=R . Cos. C. C is of the same affection as CB. 3. R . cos. A = tang. AB . cot. AC. The Hypotenu^^e is less then a quadrant, because AB and A are of the same affection. CASE V. Given two sides (=47° 30' 4" and 32° 12') to find the other parts. Pig. 21. Project the Triangle, making AB and BC equal to the 22. 23. given sides. 1. Cos. AB . cos. BC=R . cos. AC. AC< 90, because AB and BC are alike. 2. R . sin. AB=tang. BC . cot. A. A<90, likeBC. 3. R . sin. BC=tan. AB . cot. C. C<90,likeAB. SPHERICAL TRIGONOMETRY* 193 CASE VI. Given the two angles (=40° 30' 5" and 63° 58' 30") to iind the other parts. Project the Triangle, Fig. 39. At the circumference, make A =40 30' 5". Find y the pole of ACx, and around it describe the small circle, at the distance of 63° 58' 30", its intersection with the primitive in S, is the pole of the right circle BC, which makes the given angle at C, with ACx. At the centre, make A=one of the given angles about x the *' 30. pole of CArf, describe the small circle yz at the distance of the other given angle from x, their intersection is the pole of 5BC/ making the given angle atC. From K, a point in the plane of the Primitive ; make " 31, A= given angle, about m the pole of ACx, describe a small circle, at the distance of the other angle, intersecting AB^? in p which is the pole of dBCf making C=the other ^iven angle. 1. Cot. A . cot. C=R. cos. AC. AC<90, because A and C are alike. 2. R . cos. C=sin. A . cos. AB. AB<90, like opposite angle C. 2. R . COS. A=sin. C . cos. BC. BC<90 1ikeA. OBLIQUE-ANGLED TRIANGLES. CASE I. Given two sides (=58°, and 79° 17' 14",) and the angle opposite one of them (=62°' 34 6",) it is required to find Fig. 32. the other parts. Project the Triangle, making AC = 58° and describe the circle CBe, making C=62° 34' G", and the small circle kg, at the distance of 79° 17' 14" from A, ABrf will com- plete the triangle. Through, q^ the pole of CBc project the circle Apqd. perpendicular to Be. 1. Sin. AB : sin. C: Isin. AC : sin. B. i. c. sin. ABC. or ABe. This case is ambiguous, 2. R . cos. C=cot. AC . tan. PC. PC<90, because C,and AC< 90. 3. R . COS. B=cot. AB . tan. BP. PB<90, as before, and PC+PB=CB. 4. Sin. AC : sin. B: isin. BC : sin. A, Which is ambiguous. The projection gives B acute, and A obtuse, but with the same things given, another triangle might be project- ed, containing angles equal to the supplements of these. SPHERICAL TRIGONOMETRY. 1^5 That two spherical triangles may have two sides, and the angle opposite to one of them in each equal, while the remaining side and angles are unequal may be easily shown, as in fig. 6, where the triangles APB, APB', have AP, AB. or AB', and the angle P, in each equal. The projection of these two triangles is given PI. XVI, fig. 2, where the ambiguity of A, B and BC is exhibited. (BC = PC^PB,) and A, B, are either acuta or obtuse. CASE 2. Given two sides ( = 58°. and 110°.) and the included an- gle, (62°. 34' 6") to find the other parts. Project the Triangle ; making AC = 58° C=62°. 34' 6" Fig,-33. and CB = nO°, by drawing the small circle pBr^, at the distance of 180°. — 1 10°. = 70°. from e as its pole. Describe A Br?, through B, and APrf through the pole of Ce. 1. R. cos.C=cot. AC . tan. PC ; <90°. ThenBC-PC=PB. 2. (X; Sin. PB : sin. PC: :tan. C : tan. B. B< 90°. like C, because AP is within the triangle. .3. Sin. B : sin. AC:: sin. C : sin. AB<90^ Because B and BP are both acute. 4. Sin. AC : sin. B::sin. BC : sin. A .>90°. ForBC-fBA>180°5.:.(S. G. XII)B Ae90°. ForAB+BC>180°.-.A>BC(/. CASE 4. Given two angles (=50° and 62° 34' 6") and the side FiM 34 between them ( = 110°) to find the other parts. Project the Triangle ; making CB = 110°. C=62° 34' 6'', and B=50° ; also draw BPti through the pole of CAe. 1 . R . cos. BC =cot. C . cot PBC > 90° ThenPBC-ABC=PBA. 2. Sin. BC.sin. C=R . sin. PB<90^ 3. R. cos. PBA=tan. PB . cot. BA<90«. 4. R. cos. BA=cot. PBA . cot. PAB<90. Then 100°-PAB = BAC>90. 5. R . COS. BA=cos. PB . cos. PA<90^. ThenPC-PA=AC<90. CASE 5. Given the three sides (=79*' 17' 14" , 110° and 58*) to find the angles. Fig. 33. Project the Triangle ; making AC=58°, AB=79° 17' 14" and BC = 110°, by describing small circles rg and pq, at the distance of 79° 17' 14" and 70° from A and e as poles SPHERICAL TRIGONOMETRY. 197 and then drawing ABJ, CBe through the point of their in- t€rsection. 1. T an. ^BC : tan, iAC+AB::tan. |AB - AC : tan. igp PC= 'D. Then iBC+^iD=BP, and iBC - iD=PC. For the least segment, is adjacent to the least side. (Sp. Geom. II Prop. XVII.) 2. The other parts are easily found as before. CASE 6. Given the three angles ( = 121°54' 5G'\ SO^' and 62* 34' 6'') to find the angles. The supplements of these angles, are the sides of the supplemental triangle, the angles of which are found, as in Case 5. Then the supplements of those angles are the sides of the given triangle. Spherical Trigonometry is extensively applicable to the lolution of questions connected with Geography, and espe- cially in Trigometrical Surveying, and Geodesic opera- tions.* Its use and importance in astronomical investigations, is indicated by the declaration of M. de La Lande, himself an eminent Astronomer, '' La Trigonometrie Spherique est la veritable Science de VAstronome.^'' The following questions will illustrate its application to each of these sciences. 1. Given the Latitude of St. Petersburg, 59^ 66' N. and its Longitude, 27° 59' 30" E. from Paris, also the Lati- tude and Longitude of Conception in South America, 36° 42' 53' S. arid 75° W; required the distance of the two places, as measured on the arc of a great circle.f In the triangle NAB two sides and the included angle P'-.^^'- are given ; its solution therefore is according to Case 2, of '*' * Oblique angled Triangles. * Hutton'3 Math. Vol. II. I Cagr.oli rilSO.^, 26 - ' 198 SPHERICAL TRIGONOMETRY. 2. Given the Longitude and Latitude of two stars, and the distance of a third star from each of them, required the Longitude and Latitude of the third Star. I'V 4. Let EL represent the Ecliptic *PF, PH and PL arcs of great circles perpendicular to it. A and B the two stars whose places are given, and C the place ofthe third star. Then in the Triangle APB, two sides and the included angle are given ; therefore the remaining side and angles may be found. Then in the triangle ABC, the three sides are given, whence the angles may be found. Then in the Triangle PA(% two sides (PA and AC) are given, and the angle included by them. Therefore the remaining side PC, (which is the complement of the Latitude of C) and APC, (the difference of Longitude belwean A and C) may be found. If the Declination and Right Ascension of each of the stars A and B were given, then EL would represent the Equator; and the solution would give the Declination dm^ Right Ascension ofthe third Star. * The Longitude of aheaveuly body is measured by an arc of the Eclip* tic and its Latitude is its disianct from the same circle. NOTES TO CONIC SECTIONS. Definitions. Conic Sections arc treated differently by different au- thors ; some defining them by the sections of a cone, and deriving their properties directly from the intersection ot* the cone and plane, while others define them as figures de- scribed in a plane, and derive their properties from their mechanical description. Each method has its advantages. If the latter, in some cases, renders the demonstrations more simple and easy; the former, which is adopted in this treatise, has the important advantage, of deriving the fig- ures from solids whose properties have been demonstrated, and by operations which have become familiar in the Ele- ments of Geometry. " Thus far,'' says Newton, •' I think 1 have expounded the construction of solid Problems by operations whose manual practice is most simple and expeditious. So the Ancients, after they had obtained a method of solving these Problems by a composition of solid places, thinking the Constructions by the Conick Sections as useless, by reason of the Difficulty of describing them, sought easier Constructions by the Conchoid, Cissoid, the Extension of 'threads, and by any Mechanic Application of Figures.— " If the Ancients had rather construct Problems by Fig- ures not received into Geometry at that Time, how much more ought these Figures now to be preferred which are received by many into Geometry as well as the Conick Sections. However I do not agree to this new sort of Geometri- cians who receive all Figures into Geometry. -In ray judg- ment, no Lines ought to be admitted into plain Geometry besides the right Line and the Circle, unless some Distinc- 200 NOT£:s TO CONIC SECTIONS. tion of Lines might be first invented, by which a circular Line might be joined with a right Line, and separated from all the rest. But truly plain Geometry is not then to be augmented by the number of Lines. For all figures are plain that are admitted into plain Geometry, that is, those which the Geometers postulate to be described in planoJ^"* — " All these descriptions of the Conicks in plano^ which the Moderns are so fond of, are foreign to Geome- try. Nevertheless, the Conick Sections ought not to be flung out of Geometry. They indeed are not described Geometrically mp/ffno, but are generated in the plane Superficies of a Geometrical Solid. A Cone is constituted geometrically, and cut by a Geometrical Plane. Such a segment of a Cone is a Geometrical Figure, and has the same place in solid Geometry, as the Segment of a Circle has in Plane, and for this reason its base, which they call a Conick Section, is a Geometrical Figure. Therefore ft Conick Section hath a place in Geometry, so far as it is the Superficies of a Geometrical Solid ; ELLIPSE PROPOSITION XXVI. The properties referred to, are the following, PROPOSITION A. If any line in the Ellipse pass through either the of th« Foci, and a tangent be drawn through one of its extremities, a line drawn from the centre, parallel to the line passing i\OTES TO CONIC SECTIONS. 201 through the Focus and intercepted by the tangent, is equal to the semi-transverse axis. That is, C^=CA. Fig. 26. Bor, (IX Cor. 1.^ CF . CD=CA2-CA . EF, CA.EF=CA2-CF.CD; but, (VI; CA»=CD . CT ; ..CA . EF=CD|. CT [-CD.CF=CD.TF, :TF. sim. tri. TC : C/: :TF : FE: :TF . CD (=CA . FE) [FE . CD, ::CA:CD, therefore, Ct , CA=TC . CD^CA^. Ct=CA. Q. E. D, CoR. 1.— Hence TF . CD=CA . FE. 3rf step ofdem. Cor. 2. — AB : kz : EH are continued proportionals. For, draw the diameter LCIK, bisecting EH, and draw Ed parallel to it. Then, (XXVI.) Cd: Ck : C^,are continued proportionals, That is IE:CA::CA, '' HE:zA::BA, " and zA^^HE.BA. PROPOSITION B. The rectangle of the segments of any line in the Ellipse, passing through the Focus, is equal to the rectangle of one fcurth of the Parameter, into the same line. That is, EF . FH = iP . EH Fig. §6 502 NOTES TO CONIC SECTIONS. For, (XXIV,) Ck^ : CA« : :EF . FH : AF .FB ; But (VII. Cor. 3 ) AF FB = (Ca^=) CA . iP, Therefore, C^^rEF.FH.-rCA^' : CA . iP," But, CA : \P\ :CA . EI : iP . EI, And,CA2 :CA. iP: :CA . EI : i P . EI. .-. Ck^ :EF.FH::CA.EI:iP. EI. But, fXXVII. Cor. 2 ) CA:-^ =EI. CA. EF . FH=iP . [EI, or lEH . iP, ... EF.FH=iP.EH. q. E. d: CoR. 1. — As the proposition is applicable to all lines passing through the Focus, AF . FB = iP . AB, as demonstrated. Prop. VII. AndEF. FH = iP.EH, Therefore, AB : EH: :AF . FB : EF . FH. (See Pa- rab. XIV. Cor. 7.) Cor. 2. — When EH, is perpendicular to the Transverse Axis, it is a double ordinate to it, and since the rectangle of the abscisses at the Focus equals the square of the semi- conjugate, and EH then equals the parameter, and the rectangle EF . FH is the same as the square of EF, the proposition becomes As the Transverse Axis Is to the Parameter, So is the square of the Semi-Conjugate To the square of the Focal-Ordinate. PROPOSITION C. The rectangle of the two lines drawn from the Foci, to any point in the curve, is equal to the square of half the di- ameter conjugate to that which passes through the point. Fig. 12. ThatisFE./E=CpS CURVATURE OF CONIC SECTIONS. 2()3 Let ER be drawn perpendicular on eh, Then, sim. triang. FE : FP: :EI : ER, And •« '' yE :/)?::EI:ER, Therefore FE . /"E : FP .^: EP : ER^ But, (IX. Cor. 4.) FP - j5?=Ca2 ; and EP =€?«=( VII) Therefore, FE ./E : Ca^ ; iCA^ : ER^ ; But, (XVI.) Ce . ER=AC . Ca, .:. Ce:Ca::CA : ER, And, Ce^ iCa^iiCA^ : ER% Therefore, FE./E=Ce^ Q. £. Z>. CURVATURE OF CONIC SECTIONS. The application of Conic Sections to Physical Astrono- my, which is one of their most important applications, re- quires an acquaintance with their Curvatures, and espe- cially with the method of reasoning employed in treating of this part of Conic Sections. {See Kewtons Principia Lib* I Sect. 1. Cavalh^s Philosophy, Introduction, Lem- mas, Enfield'' s Phil, Central Forces, Lem.) As the doc- trine of the Curvature of Conic Sections, gives the stu- dent new and interesting views of their general properties, and is attended with no peculiar difficulties, a few elemen- tary propositions are introduced in this part of the course* NOTES TO SPHERICAL TRIGONOMETRY, PROPOSITION V. •■•The angles at the base of an isosceles spherical triangle are Symttrkal magnitudes, not admitting of being laid on one another, nor of coinciding, notwithstanding their equality. It might be considered as a sufficient proof that they are equal, to observe that they are each determined to be of a certain magnitude rather than any other, by conditions which are precisely the same, so that there is no reason why one of them should be greater than anoth- er. For the sake of those to whom this reasoning may not prove satisfactory, the demonstration below is given, which is strictly geometrical." Playfair^s Sph, Ti-ig. . „y. Let ABC be a spherical triangle, havmg the side AB Fig. 1. equal to the side AC ; the spherical angles ABC and ACB are equal. Let D be the centre of the sphere ; join DB, DC, DA, and from A on the straight lines DB, DC, draw the per- pendiculars AE, AF ; and from the points E and F draw in the plane DBC the straight lines EG, FG perpendicu- lar to DB and DC, meeting one another in G ; Join AG. Because DE is at right angles to each of the straight lines AE. EG, it is at right-angles to the plane AEG, which NOTES TO SPHERICAL TRIGONOMETRY. 40" passes through AE, EG (4.2. Sup.) ; and therefore, every plane that passes throu^ DE is at right-angles to the plane AEG (17. 2. Sup.) ; wherefore, the plane DBC is at right- angles to the plane AEG. For the same reason, the plane DBC is at right-angles to the plane AFG, and therefore AG, the common section of the planes AFG, AEG is at right-angles (18. 2. Sup.) to the plane DBC, and the an- gles AGE, AGF are conseqnently right-angles But since the arch AB is equal to the arch AC, the an- gle ADB is equal to the angle ADC. Therefore the tri- angles ADE, ADF, have the angles EDA, FDA equal, as also the angles AED, AFD, which are right-angles ; and they have the side AD common, therefore the other sides are equal, viz. AE to AF, (26. 1.), and DE to DF. Again, because the angles AGE, AGF are right-angles, tho squares on AG and GE are equal to the square of AE ; and the squares of AG and GF to the square of AF. But the squares of AE and AF are equal, therefore the squares of AG and GE are equal to the squares of AG and GF, and taking away the common square of AG, the remain- ing squares of GE and GF are equal, and GE is therefore equal to GF. Wherefore, in the triangles AFG, AEG, the side GF is equal to the side GE, and AF has been proved to be equal to AE, and the base AG is common, therefore, the angle AFG is equal to the angle AEG (8. 1.). But the angle AFG is the angle which the plane ADC makes with the plane DBC (4. def. 2. Sup.) because FA and FG, which are drawn in these planes, are at right-an- gles to DF, the common section of the planes. The an- gle AFG (3. def) is therefore equal to the spherical angle ACB ; and, for the same reason, the angle AEG is equal to the spherical angle ABC. But the angles AFG, AEG are equal. Therefore the spherical angles ACB ABC are also equal. Q. E. D. The converse of this proposition is thus demonstrated by PI ay fair, • Let ABC be a spherical triangle having the angles ABC ACB equal to one another ; the sides AC and AB are al- so equal. 27 408 NOTES TO SPHERICAL TRIGONOMETRY. Let D be the centre of the sphere ; join DB, DA, DC, and from A on the straight hnes DB, DC, draw the per- pendiculars AE, AF ; and from the points E and F, draw in the plane DBC the straight lines EG, FG perpendicu- lar to DB and DC, meeting one another in G ; join AG. Then, it may be proved, as was done in the last propo- sition, that AG is at right angles to the plane BCD, and that therefore the angles AGF, AGE are right-angles, and also that the angles AFG, AEG are equal to the an- gles which the planes DAC, DAB make with the plane DBC. But because the spherical angles ACB, ABC are equal, the angles which the planes DAC, DAB make with the plane DBC are equal, (3. def.) and therefore the an- gles AFG, AEG are also equal. The triangles AGE, AGF have therefore two angles of the one equal to the two angles of the other, and they have also the side AG com mon, wherefore they are equal, and the side AF is equal to the side AE. Again, because the triangles ADF, ADE are right-an- gled at F and E, the squares of DF and FA are equal to the square of DA, that is, to the squares of DE and DA , now, the square of AF is equal to the square of AE, therefore the * square of DF is equal to the square of DE, and the side DF to the side DE. Therefore, in the triangles DAF, DAE, because DF is equal to DE, and DA common, and also AF equal to AE, the angle ADF is equal to the an- gle ADE ; therefore also the arches AC and AB, which are the measures of the angles ADF and ADE, are equal to one another ; and the triangle ABC is isosceles. Q. £. D. ERRATA. The following errors of the press should be corrected with a pen or pencil, before reading the book. Page 10 line 4 fr. top, for XII Prop. 2, read Sup. I, Prop. 47 Prop. IX „ /E4-/E „ FE+/E, 51 line 13 fr. top ., iFG „ ^/G, 52 „ 7 „ „ „ PAC „ P . AC^ „ „ 21 „ „ „ CF : CF + FT „ CF . CF+FT, 54 Prop: XIV „ De^ and D e^ „ DE^ and D E^ 55 line 5 fr. hot. „ Ca, „ CA, .. " 2 „ „ ,, Ggr „C^, 57 „ 10 „ „ „ GMT, „ CMT, 178,, 2fr. top, „ iBD-AD, „ iBD-AD, 185 „ 2 „ „ „ lAC +AB and j AC -ABrea d lAC+ABandiAC— AB. 204 for Prop. V, read Prop. IV. ERRATA. [Of the following Errata, some were obviously owing to the press, and many to inaccura- cies of tbe manuscript, which had been transcribeil, and was written originally in haste, and not very legibly A considerable number of them are unimportant, but it was thought zulviseable to make tbe list as complete as possible.] 12 margin, - - . for - - 66 - - read - - 6.8. 15 24th line from top - " - - Fg3 - (( - /«2. 17 10 {( «k bottom, insert that o/j after equal to. ~ 20 9 (( (( top, for AF read AF.2 _ 12 « h do. '* 2AF it -2AF. 23 last line, - « (VII) (( (VI). 27 5 t4 from top, " circle - {( curve. _ 27 (( " do. " II Cor. - n (XI Cor. 4.) 29 11 (1 (( bottom, " (II Cor. 1) (( (XI Cor. 4.) - 2 n do. " (Viii; - « (XI; 30 6 u top, " 2CIVI:TP u 2CM.TP. 34 - margin, « Fig. 6. Fl. VII (( PI. VIII. Fig. 14. 35 1&2 a bottom, " ad.ab <; ad.db. 38 14 n top, " eLkl <{ ik.kt. 46 18 « do. « (III. « (11. _ 22 i( do. '^ FF2 t( F/3. 47 49 16 do. *« /E±/-E - do. " TF tt FE+FE. T/. 50 12 (( do. " OEP ii OE/. 51 8 (( do. " fep, ti /Ef - 13 (( do. ♦* iFG u i/£. . 4 u bottom, " semi-diameter a diameter. - - n '' /PK . i( »PK. 52 7 21 <( u top, « PAC do. " CF:CF-I-FT (4 P.AC- — CF.CF-I-FT. 54 lO&ll « n do. " De2D'c2 (( DE2 D'Ea . 55 - - margin, " Fig. 14. - i< Fig. 13. - 5 « u bottom, " Ca u CA. - 2 ti u do. « G^ i( Cg- 56 5 it (( top, " De2 <( de2 . 57 10 (t ti bottom, ** GMT « CMT. 60 8 (( (i do. « + (C 64 - - top, « HK (( kK. - 2&3 <( (( do. «' DT (( BT. - - - margm, '« Fig2 4. - '4 Fig. 25. - 6 U « top, « -f. £( g. •««*. 75 - - do. " CL t( CA. - 3 a n do. " CD <( CT. 77 - - margin, " Fig. 10. - u Fig. 9. 78 6 15 top, « BHF do. " (III. it BH/. 2 ERKATAi Page 78 I7th line from top, for CA2 - CP2 read CF8 - CAa , - 2 jeet%on. 166 14 ** " do. insert and H /o P after reduce P^ to G. - 15 « « do. for Y, P & G read P, Y & p. 166 9 '« " bottom, ft TV - ft TH, and after VY insert through X w/ierc EH tnter- sects the great circle, whose pole is E. (Cor. 3.) 171 10 ft ft top, " AB, AD, AC, read DB.DA, & DC. 172 5 ft « do. "ABC - ft tang. ABC 174 5 '• ft bottom,** Cos. A. - ** Cos. CA ERRATA. 176 5th line from bottom, for tang. BC read tang. DC. 178 2 n (I top. (( iBD— AD t( iBD— AD 177 11 il (( do (( + u ^ 15 i( l( do. (( I T (•' i ^ 48 (( t( do. (4 BB (i BC. ^ 4 (( u bottom. rad n radi . _ 5 i( u do. i» cot a xoU 181 20 u (( top, u B u C. 185 2 {( (( do. (( iAC+AB & iAC— AB, read iACtAB & iAC- AB. 189 7 (( (( do. for AP read Ap. 11 (( (( do. (( y\C (( ^A/. „ 15 « ATi: IV. 10\ a ADnoUUU Sc p^ rt V / b. V+'K PLATE \I. PLATE ^n^ «. G B ^ [. •" R ^t- •■, K^ PLATE ym. 'C^ — ■"'^i^ ;--.._ l"' / i . ■■■V'N h- .-// •■^-^.^ PLATE K. PLATE XI, 3. A \ iB D /G ^- .J A DooUUU sc. v^xif r yc 3:\--\p\^ A-v, \ /y 24- jr I n S/ 'h^ ^M' V''' \; \ 32. ef~^- 3/7. ^i! s/UL \ ■;;;;; /^j//r\ \ / />