A TREATISE ON S U R V E Y I N G, CONTAINING THE THEORY AND PRACTICE: TO WHICH IS PREFIXED A PERSPICUOUS SYSTEM OF PLANE TRIGONOMETRY. EXN WHOLE CLEARLY DEMONSTRATED AND ILLUSTRATED BY A LARGE NUMBEX OF APPROPRIATE EXAMPLES, PARTICULARLY ADAPTED TO THE USE OF SCHOOLS. BY JOHN- GUMMERE, A.., FELLOW OF THE AMERICAN PHILOSOPHICAL SOCIETY, AND CORRESPONDING MEMBER OF THI ACADEMY OF NATURAL SCIENCES, PHILAJELPHIA. SEVENTEENTH EDITION, CAREFULLY REVINSED, A ENLARGED BY THE ADDITION OF ARTICLES ON THE THEODOLITE, LEVELLING, AND TOPOGRAPHY. ALSO, HINTS TO YOUNG SURVEYORS, AND RULES FOR SURVEYING THE PUBLIC LANDS OF THE UNITED STATES. BY GEORGE H. HOLLIDAY, M. A. PHILADELPHIA: URIAH HUNT & SON. CINCINNATI: APPLEGATE & CO. ENTERED, according to Act of Congress, in the year 1853, by URIAH HUNT, EI the Clerk's Offico of the District Court of the United States in and for the Eastern District of Pennsylvania. PRINTED BY SMITH & PETERS, Franklin Buildings, Sixth Street, below Arch Philadelphia CONTENTS. PAGN Logarithms....... 9 Geometrical Definitions. 25 Geometrical Problems..... 29 Plane Trigonometry...... 37 Application of Plane Trigonometry to the Mensuration of Distances and Heights...... 70 Practical Questions. 80 Dimensions of a Survey...... 83 Supplying Omissions in the Dimensions of a Survey.. 109 Problems for finding the Content of Land... 120 Laying out and Dividing Land..... 165 Variation of the Compass..... 205 Miscellaneous Questions...... 215 Theodolite...... 219 Levelling........ 239 Topography....... 257 Hints to Young Surveyors.... 282 PREFACE. THE following compilation originated in the belief that our schools are in want of a Treatise on Surveying, adapted to the methods practised in this country, and freed from the defects of the systems now in use. Notwithstanding the importance of the science, and the large number that make it an object of study. it is believed we are not in possession of a treatise on this subject, suited to the wants of the student. The works of Gibson and Jess are the only ones at present in general use; the former, though much the better of the two is deficient in many respects. It may be sufficient here, merely to advert to its want of exampies, which renders it entirely unsuitable for a school book. From the latter, the student would in vain expect to become acquaint. ed with the principles of the science, or the rationale of any of the rules, necessary in performing the various calculations.* In order to understand the principles of surveying, a previous knowledge of Geometry is absolutely necessary; and this knowledge will be best acquired frlom a regular treatise on the subject. In the demonstrations, therefore, throughout this work, the student is supposed to be acquainted with the elemnents of that science. The references are adapted to Playfair's Geometry but they will in general apply equally well to Simson's translation of Euclid's Elements. As there are many who wish to obtain a practical knowledge of Surveying, whose leisure may be too limited to admit of their Each of these works has lately gone through a new edition, in which considerable additions are stated to have been made. On exanlination, however, it does not appear, that those additions are such as to supplly the deficiencies. The additions made to Gibson, consist principally of some nautical problems quite foreign to a treatise on Surveying. Those made to.Jess, consist of a few extracts fromn Gibson, in one of which the Pennsylvanlia method of calculation is introdllced, as being quite different fromn that given!by Jess; whereas it is well krown to hbe the method given bly that author, atd used, as well in the preceding, as is the subsequent part of his work. 5 Vi PREFACE going through a course of Geometry, the author has adapted his work to this class, by introducing the necessary geometrical definitions and problems, and by giving plain and concise rules, en. tirely detached from the demonstrations; the latter being placed in the form of notes at the bottom of the page. Each rule is exemplified by one wrought example; and the most of them by several unwrought examples, with the answers annexed. In the laying out and dividing of land, which forms the most difficult part of surveying, a variety of problems is introduced, adapted to the cases most likely to occur in practice. This part of the subject, however, presents such a great variety of cases, that we should in vain attempt to give rules that would apply to all of them. It cannot therefore be too strongly recommended to every one, who has the opportunity, to make himself well acquainted with Geometry, and also with Algebra, previous to entering on the study of Surveying. Furnished with these useful auxiliaries, and acquainted with the principles of the science, the practitioner will be able to perform, with ease, any thing likely to occur in his practice. The compiler thinks proper to acknowledge, that in the arrangement of the work, he availed himself of the advice of his learned preceptor and friend, E. Lewis of New-Garden; and that several of the demonstrations were furnished by him. ADVERTISEMENT TO THE FOURTH EDITION In preparing this edition for the press, several alterations have been made, which, it is believed, will be found to be real improvements. A number of new Problems has been introduced, and a more methodical arrangement of the whole has been adopted. Instead of three different rules for calculating the content of a Survey, one general rule, including these, is now given. It may be further added, that the rules for solving several of the problems in Division of Land, have been considerably simplified. The Mathematical Tables have been stereotyped, after carefully revising them and comparing them with the most correct European Editions. THE AUTHOR. ADVERTISEMENT TO THE FOURTEENTH EDITION. To meet the wants of the Student of Civil Engineering, this edition has been enlarged by the addition of several chapters, in whichl the Theodolite and Levelling Instrument are described, the methods of adjusting and using them are given, and the principles and practice of Levelling and Topography are explained and illustrated.-The whole has been carefully revised and the few typographical errors existing in former editions have been corrected. J. G. 7 CONTENTS. PAC Legarithms,... 9 —......,,, 9 Geometrical Definitions, - - - - - - - 25 Geometrical Problems, - - - - - - - - - - - - - - - 29 Plane Trigonometry, - - - - - - - - - - - - 35 Application of Plane Trigonometry to the Mensuration of Distances and Heights, - - - - - - - - - - - - - - - - - - 68 Practical Questions, - - - - - - - - - - - - - - - - - 78 Dimensions of a Survey, - - - - - - - - - - - - - - - 81 Supplying Omissions in the Dimensions of a Survey, - -.. - - - 107 Problems for finding the Content of Land, - - - - - - - - - 118 Laying out and Dividing Land, - -. -..-. - 163 Variation of the Compass, - - - - - - - - - - - - - - - 202 Miscellaneous Questions, - - - - - - - - - - - - - - - 213 Theodolite, - - - - - - - - - - - - - - - - - 217 Levelling, - - - - - - - - -. - - - - - - 237 Topography, - - -. - - - - - - - - 254 EXPLANATION OF THE CHARACTERS USED IN THIS WORK. + signifies phts, or addition. minuzs, or subtraction. X multiplication. division.:::: proportion. = e aC(llality. / squllS(;tre root. diltbircrlce between two quantities *lvecll it is not known which is tJt greater. A TREATISE ON SURVEYING. OF LOGARITHMS. LOGARITHMS are a series of numbers so contrived, that by them the work of multiplication is performed by addition, and that of division by subtraction. If a series of numbers in arithmetical progression be placed as indices, or exponants, to a series of numbers in geometrical progression, the sum or difference of any two of the former, will answer to the product or quotient of the two corresponding terms of the latter. Thus, 0. 1. 2. 3. 4. 5. 6. 7. &c., arith. series, or indices. 1. 2. 4. 8. 16. 32. 64. 128. &c., geom. series. Now 2+3=5. also 7-3=4. And 4X8-=32. and 128~. 8=16. Therefore the arithmetical series, or indices, have the same properties as logarithms; and these properties hold true, whatever may be the ratio of the geometrical series. There may, therefore, be as many different systems of logarithms, as there can be taken different geometrical series having unity for the first term. But the most B 9 10 OF LOGARITHMS. convenient system is that in which the ratio of the geomotrical series is 10; and this is the one in common use. Thus, 0. 1. 2. 3. 4. 5. &c. indices or logar. 1. 10. 100. 1000. 10000. 100000. &c. natural numbers In this system the log. of 1. is 0,* the log. of 10 is 1 the log. of 100, is 2, &c. Hence it is plain that the log of any number between 1 and 10, will be expressed by a decimal, the log. of any number between 10, and 100, by 1 and a decimal, -the log. of any number between 100 and 1000, by 2 and a decimal, &c. The numbers, 0, 1, 2, 3, &c. that stand before the decimal part of logarithms, are called indices and are always less by unity, than the number of figures in the integral part of the corresponding natural number. The index of tie logarithm of a number, consisting in whole, or in part of integers, is affirmative; but if the number be a decimal, the index is negative, and is marked by a negative sign (-) placed either before or above it. If the first significant figure of the decimal be adjacent to the decimal point, the index is, -1, or 1; if there be one cipher between them, the index is -2, or 2; if there be two ciphers between them, the index is -3 or 3, &c. The decimal parts of the logarithms of numbers, consisting of the same figures and in the same order, are the same, whether the number be integral, fractional, or mixed. This is illustrated as follows: * In every system the logarithm of 1 is 0. OF LOGARITHMS. 11 Number 18960 Logarithm 4.27784 1896 3.27784 189.6 2.27784 18.96 1.27784 1.896 0.27784.1896 -1.27784.01896 2.27781.001896 -3.27784.0001896 -4.27784 The method of finding logarithms in' the tables, and of multiplying, dividing, &c. by them, is contained in the following problems. PROBLEM I. To find the Logarithm of a given number. If the given number consists of one or two figures only, find it in the column marked No. in the first page of the table, and against it, in the next column, marked log. is the logarithm. Thus the log. of 7 will be found 0.84510, and the log. of 85 will be found 1.92942. But if the given number be either wholly or in part decimal, the index must be changed accordingly. Observing that the index must alwavs be one less than the number of figures in the integral part of the given number; also, when the given number is wholly a decimal, the index is negative, and must be one more than the number of the ciphers between the decimal point and first significant figure on the left hand. Thus the log. of.7 is -1.84510, and the log. of.0085 is -3.92942. If the given number consists of three figures, find it in one of the other pages of the table, in the column marked No. and against it, in the next column, is the decimal 12 OF LOGARITHMS. part of the logarithm. The index must be placed before it agreeably to the above observation. Thus the log. of 421 is 2.62428, the log. of 4.21 is 0.62428, and the log. of.0421 is -2.62428. If the given number consists of four figures, find the three left hand figures in the column marked No. as before, and the remaining, or right hand figure at the top of the table; in the column under this figure, and against the other three, is the decimal part of the logarithm. Thus the log. of 5163 is 3.71290, and the log. of.6387 is -1.80530. If the given number consist of five or six figures, find tile logarithm of the four left hand figures as before; then take the difference between this logarithm and the next greater in the table. Multiply this difference by the remaining figure or figures of the given number, and cut off one, or two figures to the right hand of the product, according as the multiplier consists of one, or two figures; then add the remaining figure or figures of the product to the logarithm first taken out of the table, and the sum will be the logarithm required. Thus, let it be required to find the logarithm of 59686; then, Logarithm of 5968 is - - 77583 The next greater log. is - 77590 Difference - - - - 7 Remaining figure - - - 6 Product - - - - - 4,2 To - - - - 77583 Add - - 4 Decimal part of the log. - - 77587 LOGARITHMS. 13 Tile natural number consisting of five integers, the in. dex must be 4; therefore the log. of 59686 is 4.77587. Again, let it be required to find the log. of.0131755; then, Logarithm of 1317 is - 11959 The next greater log. is - 11991 Difference - - 32 Remaining figures - - - 55 Product - - - - 17 60 To - - - 11959 Add - - - - 18* Decimal part of the log. - - 11977 As the given number is a decimal, and has one cipher between the decimal point and first significant figure, the index must be — 2; therefore the log. of.0131755 is -2.11977. EXAMPLES. 1. Required the log. of 4.3 Ans. 0.63347 2. Required the log. of 7986 Ans. 3.90233 3. Required the log. of.3754 Ans. -1.57449 4. Required the log. of 596.87 Ans. 2.77588 5. Required the log. of 785925 Ans. 5.89538 6. Required the log. of 6543900 Ans. 6.81583 7. Required the log. of.0027863 Ans. -3.44503 * Because 17.6 is nearer IS than 17. 2 14 LOGARITHMS. PROBLEM II. To find the natural number corresponding to a given logarithnm. If our figures only be required in the answer, look in the table for the decimal part of the given logarithm, and if it cannot be found exactly, take the one nearest to it, whether greater or less; then the three figures in the first column, marked No. which are in a line with the logarithm found, together with the figure at the top of the table directly above it, will form the number required. Observing, that when the index of the given logarithm is affirmative, the integers in the number found, must be one more than the number expressed by the index; but when the index of the given logarithm is negative, the number found will be wholly a decimal, and must have one cypher less, placed between the decimal point and first significant figure on the left hand, thain the number expressed by the index. Thus the natural number corresponding to the logarithm 2.90233 is 798.6, the natural number corresponding to the logarithm 3.77055 is 5896, and the natural number corresponding to the logarithm -3.36361 is.00231. If the exact logarithm be found in the table, and the figures in the number corresponding do not exceed the index by one, annex ciphers to the right hand till they do. Thus the natural number corresponding to the logarithm 6.64068 is 4372000. If five or six figures be required in the answer, find, in the table, the logarithm next less than the given one, and take out the four figures answering to it as before. Subtract this logarithm from the next greater in the table, and also from the given logarithm; to the latter lifference, annex one or two ciphers, according as five LOGARITHMS. 15 or six figures are required, and divide the number thus produced, by the former difference; annex the quotient to the-right hand of the four figures already found, and it will give the natural number required. Thus let it be required to find the natural number corresponding to the logarithm 2.53899 true to fi.t figures; then, Given logarithm -.53899 Next less - - -.53895 the natural number corresponding is 3459 Diff. with one cipher annexed 40 Next less log. - -.53895 Next greater - -.53908 Difference 13 Divide 40 by 13 and the quotient will be 3, which, annexed to the right hand of 3459, the four figures already found?, makes 34593; therefore as the index is 2, the required natural number is 345.93. Again let it be required to find the natural number corresponding to the logarithm 4.59859, true to six figures; then, Given logarithm - -.59859 Next less - - - -.59857, the natural number answering to it is 3968. Diff. with two ciphers annexed 200 Next less log. - - - 59857 Next greater - - - 59868 Difference - - - - 11 Divide 200 by 11, and the quotient will be 18, which annexed to the right hand of.3968 the four figures al 16 LOGARITHMS. ready found, malkes 396818; therefore as the index is 4, the required natural number is 39681.8. EXAMPLES. 1. Required the natural numnber answering to the logarithm 1.88030. Ans. 75.91. 2. Required the natural number answering to the logarithm 5.37081. Ans. 234861. 3. Required the natural number answering to the logarithm 3.11977. Ans. 1317.56. 4. Required the natural number answering to the logarithxn-2.97435. Ans..094265. PROBLEM III. To multiply numbers by means of logarithms. Case 1.-When all the factors are whole or mixed numbers. RULE. Add together the logarithms of the factors, and the sum will be the logarithm of the product. EXAMPLES. 1. Required the product of 84 by 56. Logarithm of 84 is 1.92428 Do. of 56 is 1.74819 Product 4704 Sum 3.67247 2. Requircd the continued product of 17.3, 1.907 and 34. Logarithm of 17.3 is 1.23805 Do. 1.907 is 0.28035 Do. 34. is 1.53148 Product 1121.71 Sum 3.04988 LOGARITHMS. 17 3. Find by logarithms the product' of 76.5 by 5.5 Ans. 420.75. 4. Find by logarithms the continued product of 42.35, 1.7364, and 1.76. Ans. 129.424. CASE 2. —When some or all of the factors are decimal numoers. RULH. Add the decimal parts of the logarithms as before, and if there be any to carry from the decimal part, add it to the affirmative index or indices, or else subtract it front the negative. Then add the indices together, when they are all of' the same kind; that is, all affirmative or all negative; but when they are of different kinds, take the difference be. tween the sums of the affirmative and negative ones, and irefix the sign of the greater. Note.-When the index is affirmative, it is not necessary to place any sign before it; but when it is negative, the sign must not be omitted. EXAMPLES. 1. Required the continued product of 349.17, 25.43, 93521 and.00576. Logarithm of 349.17 is 2.54303 Do. 25.43 is 1.40535 Do..93521 is -1.97090 Do..00576 is -3.76042 Product 47.83 Sum 1.67970 *~~ C 18 LOGARITHMS. In this example there is 2 to carry from the decimal part of the logarithms, which added to 3, the sum of the affirmative indices, makes 5; from this taking 4, the sum of the negative indices, the remainder is 1, which is the index of the sum of the logarithms, and is affirmative, because the sum of the affirmative indices, together with the number carried, exceed the sum of the negative indices. 2. Required the continued product of.0839,.7536, and.003179. Logarithm of.0839 is -2.92376 Do..7536 is -1.87714 Do..003179 is -3.50229 Product.000201 Sum -4.30319 In this example there is 2 to carry from the decimal part of the logarithms, which subtracted from 6, the sum of the negative indices, leaves 4, which is the index of the sum of the logarithms, and is negative, because the sum of the negative indices is the greater. 3. Required the continued product of 13.19,.3765, and.00415. Ans..02061. 4. Required the continued product of 343, 1.794, 5.41, and.019. Ans. 63.25. PROBLEM IV. To divide numbers by means of Logarithms. CASE 1.-When the dividend and divisor are both whole or mixed numbers LOGARITHMS. 19 RULE. From the logarithm of the dividend, subtract the logarithm of the divisor, the remainder will be the logarithm of the quotient. Note.-When the divisor exceeds the dividend, the question must be wrought by the rule given in the next case. EXAMPLES. 1. Required the quotient of 3450 divided by 23. Logarithm of 3450 is 3.53782 Do. 23 is 1.36173 Quotient 150 Remainder 2.17609 2. Required the quotient of 420.75 divided by 76.5. Ans. 5.5. 3. Required the quotient of 37.1542 divided by 1.73958. Ans. 21.3585. CASE 2.-When the dividend or divisor, or both of them, are decimal numbers. RULE. Subtract the decimal parts of the logarithms as before, and if 1 be borrowed in the left hand place of the decimal part, add it to the index of the divisor when that index is affirmative, but subtract it when negative. Then conceive the sign of the index of the divisor changed from affirmative to negative, or from negative to affirmative; and if, when changed, it be of the same name with that of the dividend, add the indices togethel 2 '20 LOGARITHMS. but if it be of a different name, take the difference of the indices, and prefix the sign of the greater. EXAMPLES. 1. Required the quotient of.7591 divided by 32.147 Logarithm of.7591 is -1.88030 Do. 32.147 is 1.50714 Quotient.02361 Remain. -2.37316 In this example, the index of the divisor, with its sign changed, is -1, which added to -1, the index of the dividend, makes -2, for the index of the quotient. 2. Required the quotient of.63153 divided by.00917 Logarithm of.63153 is -1.80039 Do..00917 is -3.96237 Quotient 68.8683 Remain. 1.83802. In this example there is 1 to carry from the decimal part of the logarithm, which subtracted from -3, the index of the divisor, leaves — 2; this, with its sign changed, is +2; from which subtracting 1, the index of the dividend, the remainder is 1, and is affirmative, be. cause the affirmative index is the greater. 3. Required the quotient of 13.921 divided by 7965.13. Logarithm of 13.921 is 1.14-367 Do. 7965.13 is 3.90125 Quotient.001748 Remain. -3.24242 LOGARITHMS. 21 In this example there is 1 to carry from the decimal part of the logarithm, which added to 3, the index of the divisor, makes 4; this, with its sign changed, is, -4; from which subtracting 1, the index of the dividend, the remainder is -3. 4. Required the quotient of 79.35 divided by.05178. Ans. 1532.46. 5. Required the quotient of.5903 divided by.931. Ans..63404. PROBLEM V. To involve a number to any power, that is, to find the square, cube, 4fc. of a number, logarithmically. RULE. Multiply the logarithm of the given number by the index of the power, viz. by 2 for the square, by 3 for the cube, &c. and the product will be the logarithm of the power. Note.-When the index of the logarithm is negative, if there be any to carry from the decimal part, instead of adding it to the piroduct of the index and multiplier, subtract it, and the remainder will be the index of the logarithm of the power, and will always be negative. EXAMPLES. 1. Required the square of 317. Logarithm of 317 is 2.50106 2 Square 100489 5.00212 22 LOGARITHMS. 2. Required the 5th power of 1.735. Logarithm of 1.735 is 0.23930 5 5th power 15.7218 1.19650 3. Required the cube of.08761. Logarithm of.08761 is -2.94255 3 Cube.0006724 -4.82765 4. Required the cube of 7.503. Ans. 422.37. 5. Required the 7th power of.32513. Ans..0003841 PROBLEM VI. To extract any root of a number logarithmically. RULE. Divide the logarithm of the given number by the index of the root, that is, by 2 for the square root, by 3 for the cube root, &c. and the quotient will be the logarithm of the required root. Note. —When the index of the logarithm is negative, and does not exactly contain the divisor, increase the index by a number just sufficient to make it exactly divisible by it, and carry the units borrowed, as so many tens, to the left hand figure of the decimal part; then proceed with the division in the usual manner. LOGARITHMS. 23 EXAMPLES. 1. Required the cube root of 391.27. 3) Logarithm of 391.27 is 2.59248 Cube root 7.314 0.86416 2. Required the square root of.08593. 2) Logarithm of.08593 is -2.93414 Square root.29314 -1 46707 3. Required the cube root of.7596. 3) Logarithm of.7596 is -1.88058 Cube root.9124 -1.96019 4. Reauired the cube root of.0000613. 3) Logarithm of.0000613 is -5.78746 Cube root.03943 -2.59582 5. Required the square root of 365. Ans. 19.105. 6. Required the 5th root of.9563. Ans..9911. 7. Required the 4th root of.00079. Ans..16765 Of the Arithmetical Complements of Logarithms. When it is required to subtract several logarithms from others, it will be more convenient to convert the subtraction into an addition, by writing down, instead of the logarithms to be subtracted, what each of them wants of 10.00000, which may readily be done, by writ 24 LOGARITHMS. ing down what the first figure, on the right hand, wants of 10, and what every other figure wants of 9; this remainder is called the Arithmetical Complement. Thus, if the logarithm be 2.53061, its arithmetical complement will be 7.46939. If one or more figures to the right hand be ciphers, write ciphers in their place, and take the first significant figure from 10, and the remaining figures from 9. Thus, if the logarithm be 4.61300, its arithmetical complement will be 5.38700. In any operation, where the arithmetical complements of logarithms are added to other logarithms, there must be as many tens subtracted from the sum, as there are arithmetical complements used. As an example, let it be required to divide the product Df 76.4 and 35.84, by the product of 473.9 and 4.76. 473.9 - - Ar. Co. 7.32431 4.76 - - Ar. Co. 9.32239 35.84 - - log. 1.55437 76.4 - - log. 1.88309 Quotient 1.214 0.08416 GEOMETRY. DEFINITIONS. 1. GEOMETRY is that science wherein the properties of magnitude are considered. 2. A point is that which has position, but not magnitude. 3. A line has length but not breadth. 4. A straight, or right line, is the shortest line that can be drawn between any two points. 5. A superficies or surface is that which has length and breadth, but not thickness. 6. A plane superficies is that in which any two points being taken, the straight line which joins them lies wholly in that superficies. Fig. 1. 7. A plane rectilineal angle is the incli- / nation of two straight lines to one another, which meet together, but are not in the same straight line, as A, Fig. 1. A. Note.-When several angles are a Fig. 2. D formed about the same point, as at B, Fig. 2, each particular angle is expressed by three letters, whereof the middle letter shows the angular point, and the other two the A B lines that form the angle; thus, CBD or DBC signifies the angle formed by the lines CB and DB. DP 25 26 GEOMETRY. D Fig. 8. A 8. The magnitude of an angle depends on the inclination which the lines that form it have to each other, and not on the length --- E a of those lines. Thus the angle DBE is greater than the angle ABC, Fig. 3. 9. When a straight line stands on another straight Fig. 4. line so as to incline to neither side, but makes the angles on each side equal, then each of those angles is called a right angle, and the line which stands on the other is said to be perpendicular A ) B to it. Thus ADC and BDC are right angles, and the line CD is perpendicular to AB, Fig. 4. 10. An acute angle is that which is less than a right angle, as BDE, Fig. 4. 11. An obtuse angle is that which is greater than a right angle, as ADE, Fig. 4. Fig. 5. 12. Parallel straight lines are those cwhich are in the same plane, and A Bwhich, being produced ever so far both ways, do not meet, as AB, CD, Fig. 5. 13. A figure is a space bounded by one or more lines. Fig. 6. a 14. A plane triangle is a figure bounded by three straight lines, as ABC, Fig. 6. Fig. 7. Fig. 8. 15. An equilateral triangle has its three sides equal to each other, as A, Fig. 7. 16. An isosceles triangle has only two of its sides equal, as B, Fig. 8. GEOMETRY. 27 17. A scdlene triangle has three unequal sides, as ABC, Fig. 6. Fig. 9. 0 18. A right angled triangle has one right angle, as ABC, Fig. 9; in which the side AC, opposite to the right angle, is called / the hypothenuse. A B 19. An obtuse angled triangle has Fig. 10. one obtuse angle, as C, Fig. 10. 20. An acute angled triangle has all its angles acute, as ABC, Fig. 6. 21. Acute and obtuse angled triangles are called oblique angled triangles. 22. Any plane figure bounded by four right lines, is called a quadrilateral. 23. Any quadrilateral, whose oppo- Fig. 11. site sides are parallel, is called a paral- / D lelogram, as D, Fig. 11. / 24. A parallelogram, whose angles are all Fig. 12. right angles, is called a rectangle, as E, Fig. 12. 25. A parallelogram whose sides are all Fig. 13. equal, and angles right, is called a square, as F, Fig. 13. 26. A rhomboides is a parallelogram, whose opposite sides are equal, and angles oblique, as D, Fig. 11. 27. A rkombus is a parallelogram, Fig. 14. whose sides are all equal and angles oblique, as G, Fig. 14. G 28 GEOMETRY. 28. Any quadrilateral figure that is not a parallelogram, is called a trapezium. 29. A trapezium that has two parallel sides is called a trapezoid. 30. A right line joining any two opposite angles of a quadrilateral figure, is called a diagonal. 31. That side upon which any parallelogram, or triangle is supposed to stand, is called the base; and the perpendicular falling thereon from the opposite c Fig. 15. rE angle is called the altitude of the parallelogram, or triangle. Thus AD is the base of the parallelogram ABEC, or triangle ABC, and CD is A JD B the altitude, Fig. 15. 32. All plane figures contained under more than four sides, are called polygons; of which those having five sides, are called pentagons; those having six sides, hexagons, and so on. 33. A regular polygon is one whose angles, as well as sides, are all equal. Fig. 16. 34. A circle is a plane figure, bounded / B ~ by one curve line called the circumfer/ t: ence or periphery, every part of which is equally distant from a certain point within the circle; and this point is called the centre, Fig. 16. 35. The radius of a circle is a straight line drawn from the centre to the circumference, as CB, Fig. 17. 36. The diameter of a circle is a straight line drawn through the centre. and terminated both ways by the GEOMETRY. 29 circumference, as AE, Fig. 17. It di- Fig. 17. vides the circle into two equal parts, called semicircles. 37. A quadrant is one quarter of a circle, as ACB, Fig. 17. Note. —The fourth part of the circumference of a circle is also called a quadrant. 38. A segment of a circle is the figure contained by a right line, and the part of the circumference it cuts off: thus AEBA and AEDA are segments of the circle ABED, Fig. 16. 39. An arc of a circle is any part of the circumfer ence; as AD or DE, Fig. 17. 40. Ratio is a mutual relation between two quantities of the same kind with respect to magnitude. N~ote. —A ratio is generally expressed, either by two numbers or by two right lines. 41. When two quantities have the same ratio as two other quantities, the four quantities taken in order are called proportionals; and the last is said to be a fourth proportional to the other three. 42. When three quantities of the same kind are such that the first has to the second the same ratio which the second has to the third, the third is called a third proportional to the first and second, and the second is called a mean prcrportional between the first and third. 3* 30 GEOMETRICAL PROBLEMS. GEOMETRICAL PROBLEMS. PROBLEM I. To bisect a right line, AB, Fig. 18. Fig. 18. Open the dividers to any dis-.-' tance more than half the line AB, and with one foot in A, describe the arc CFD; with the -ellL, F B same opening, and one foot in, I B, describe the arc CGD, meeting the first arc in C and D; from C to D draw the right line CD, cutting AB in E, which will be equally distant from A and B. PROBLEM II. At a given point A, in a ~right line EF, to erect a perpendicular, Fig. 19. Fig. 19. IT,~B From the point A, lay off on each side, the equal distances AC, AD; from C and D, as centres, EC D with any radius greater than AC A or AD, describe two arcs intersecting each other in B; from A to B, draw the line AB, which will be the perpendicular required. GEOMETRICAL PROBLEMS. 31 PROBLEM III. To raise a perpendicular on the end B of a rig/lt line AB, Fig. 20. Take any point D not in the Fig. 20. line AB, and with the distance from D to B, describe a circle cutting AB in E; from E D" through D draw the right line EDC, cutting the periphery in A'- C, and join CB, which will be perpendicular to AB. PROBLEM IV. To let fall a perpendicular upon a given line BC, from a given point A, witthout it, Fig. 21. In the line BC take any Fig. 21. point D, and with it as a centre and distance DA describe an arc AGE, cutting BC in G; with G as centre, and distance B GA, describe an arc cutting! AGE in E, and from A to E draw the line AFE; then AF will be perpendicular to AB. PROBLEM V. Through a given point A to draw a Fig. 22. a right line AB, parallel to a given right line CD, Fig. 22. i From the point A to any point c F, in the line CD, draw the right 32 ) GEOMETRICAL PROBLEMS. line AF; with F as a centre and distance FA, describe the arc AE, and with the same distance and centre A describe the arc FB; make FB equal to AE, and through A and B draw the line AB, and it will be parallel to CD. PROBLEM VI. At a given point B, in a given right line LG, to make an angle equal to a given angle A, Fig. 23. Ffg. 29. F With the centre A D/'">-' and any distance AE, / describe the arc DE, A E: -L B G and with the same distance and centre B describe the arc FG; make HG equal to DE, and through B and H draw the line BH; then will the angle HBG be equal to the angle A. PROBLEM VII. To bisect any right lined angle BAC, Fig. 24. Fig. 24. c In the lines AB and AC, from the point A, set off equal distances, AD!/~. and AE; with the centres D and E and any distance more than half A B,, DE, describe two arcs cutting each other in F; from A through F draw the line AG, and it will bisect the angle BAC. GEOMETRICAL PROBLEMS. 33 PROBLEM VIII. To describe a triangle that shall have its sides respectively eoual to three right lines, D, E, and F, of which any two must be together greater than the third, Fig. 2p. Make AB equal to D; Fig. 25. a with the centre A and E distance equal to E, de-, -B describe an arc, and with the centre B and distance equal to F describe another arc, cutting the former in C; draw AC and BC, and ABC is the triangle required. PROBLEM IX. Upon a given line AB to describe a square, Fig. 26. At the end B of the line AB, by Problem Fig. 26. III. erect the perpendicular BC, and make it equal to AB; with A and C as centres, mnd distance AB or BC, describe two arcs cutting each other in D; draw AD, and CD, then will ABCD be the square required. PROBLEM X. To describe a circle that shall pass through the an-gular points A, B, and C, of a triangle ABC, Fig. 27. By Problem I. bisect any two of c Fig. 27. p the sides, as AC, BC, by the perpen- D diculars DE and FG; the point H where they intersect each other will A be the centre of the circle: with a/ this centre, and the distance from 34 GEOMETRICAL PROBLEMS. it to either of the points.A, B, or C, describe the circle. PROBLEM XI. To divide a given right line AB into any number of equal parts, Fig. 28. Fig. 28., Draw the indefinite right line AP, making an angle with AB, AXE F\_B also draw BQ, parallel to AP, in each of which, take as nmany equal n.~ L ~/o parts AM, MN, &c. Bo, on, &c., as a-'/" the line AB is to be divided into; then draw Mm, Nn, &c., intersecting AB in.E, F, &c., which will divide the line as required. PROBLEM XII. To make a plane diagonal scale, Fig. 29. Draw eleven lines parallel to, and equidistant from each other; cut them at right angles by the equidistant lines BC; EF; 1, 9; 2, 7; &c. then will BC, &c. be divided into ten equal parts; divide the lines EB, and FC, each into ten equal parts; and from the points of division on the line EB, draw diagonals to the points of division on the line FC: thus join E and the first division on FC, the first division on EB, and the second on FC, &c. Fig. 29. 6I: 5 4 3 2 1 E2 4 6 8 k_ - t_[5 JI I = ~ D —-i) -- 1 ~ b 7 9 F Note.- Diagonal scales serve to take off dimensions or numbers of three figures. If the first large divisions be units, the second set of divisions, along EB, will be GEOMETRICAL PROBLEMS. 35 10th parts, and the divisions in the altitude, along BC, will be 100th parts. If IIE be tens, EB, will be units, and1 BC will be tenth parts. If HIE be hundreds, BE will be tens, and BC units. And so on, each set of divisio:ns being tenth parts of the former ones. For example, suppose it were required to take off' 242 from the scale. Extend the dividers from E to 2 towards IH; and with one leg fixed in the point 2, extend the other till it reaches 4 in the line EB; move one leg of the dividers along the line 2, 7, and the other along the line 4, till they come to the line'marked 2, in the line BC, and that will give the extent required. PROBLEM XIII. To find a third proportional to two given rzg'ht lhnes, A and B. Draw two right lines, CD, CE B containing any angle; make CF AD —equal A, and CG, CH, each equal F B; join FG and draw HL parallel to it: then will CL be the third proportional required. C CG PROBLEM XIV. To find a fourth proportional to three given right line, s A, B and C. A. Draw two right lines, DE, DFB _3 containing any angle; make DG C - equal A, DH equal B, and DL equal C; join GHI and draw LM G parallel to it: DM will be a fourth proportional to A, B, / iid C. D HM 36 CGEOMETRICAL PROBLEMS. PROBLEM XV. foJind a mean proportional between two given right lines A and B. Draw any right line CE and in it take CD equal A, and DE equal A B; bisect CE in F, and with the B centre F and radius FC or FE describe the semicircle CGE; draw DG perpendicular to CE: then DG will be a mean proportional between A and B. C ID F E PROBLEM XVI. To divide a given right line AB into two parts that slall have the same ratio to each other as two given lines C and D. Draw AE making an angle _c with AB; in AE take AF equal ID E C and FE equal D: join EB and draw FG parallel to it; then AG will have to GB the same ratio A that C has to D. PROBLEM XVII. To divide a given right line AB in two parts in the point D, so that AD may be to DB in the ratio of two given numbers m and n. For example, let m=3, and n=4. Draw AC making any angle with AB; take c the number m from any convenient scale of equal parts, and lay it on AC, from A to E; and take the number n from the same scale, Eg and lay it from E to C; join CB and draw ED parallel to it; then AB will be divided as required. A 1) B PLANE TRIGONOMETRY. DEFINITIONS. 1. PLANE TRIGONOMETRY is the art by which, when any three parts of a plane triangle, except the three angles, are given, the others are determined. 2. The periphery of every circle is supposed to be divided into 360 equal parts, called degrees; each degree into 60 equal parts, called minutes; and each minute into 60 equal parts, called seconds, &c. 3. The measure of an angle is the Fig. 30. are of a circle, contained between the D two lines that form the angle, the an-. /j A gular point being the centre; thus the angle ABC, Fig. 30, is measured by the arc DE, and contains the same number of degrees that the arc does. The measure of a right angle is therefore 90 degrees; for DIH, Fig. 31, which measures the right angle DCII, is one-fourth part of the circumference, or 90 degrees. Note.-The degrees, minutes, seconds, &c., contained in any are, or angle, are written in this manner, 500 18' 35"; which signifies that the given arc or angle contains 50 degrees, 18 minutes, and 35 seconds. 4. The complement of an arc, or of an angle, is what it wants of 90~; and the supplement of an arc, or of an angle, is what it wants of 1800. 5. The chord of an arc, is a right line drawn kfom one extremity of the arc to the other: thus the line BE is the chord of the are BAE or BDE, Fig. 31. 37 38 PLANE TRIGONOMETRY. Fig. 31. G 6. The s;ine of an arc, is a right II i line drawn from one cxtremity of X I /the arc perpendicular to the diameter which passes through the F — (,/ A other extremity: thus BF is the sine of the arc AB or BD, Fig. 31. 7. The cosine of an arc, is that part of the diameter which is intercepted between the sine and the centre: thus CF is the cosine of the arc AB, and is equal to BI, the sine of its complement HB, Fig. 31 8. The versed sine of an arc, is that part of the diameter which is intercepted between the sine and the arc: thus AF is the versed sine of AB; and DF of DB, Fig. 31. 9. The tangent of an arc, is a right line touching the circle in one end of the are, being perpendicular to the diameter which passes through that end, and is terminated by a right line drawn from the centre through the other end: thus AG is the tangent of the arc AB, Fig. 31. 10. The secant of an are, is the right line drawn from the centre and terminating the tangent; thus CG is the secant of AB, Fig. 31. 11. The cotangent of an arc, is the tangent of the complement of that arc; thus HK is the cotangent of AB, Fig. 31.. 12. The cosecant of an are, is the secant of the complement of that are; thus CK is the cosecant of AB, Fig. 31. 13. The sine, cosine, &c., of an angle is the same as the sine, cosine, &c., of the arc that measures the angle. PLANE TRIGONOMETRY. 39 PROBLEM I. Fig. 82. F To construct the lines of cords, inles, ta;ngents, and secants, to any / radius. Fig. 32. Describe a semicircle with any convenient radius CB; from the 70. centre C draw CD perpendicular to AB, and produce it to F; draw / / BE parallel to CF and join AD. m Divide the arc AD into nine 6o ---- / equal parts, as A 10; 10, 20, &c., and with one foot of the dividers 50 | | / in A, transfer the distances A, 10; A, 20, &c., to the right line AD;.0'- -' then will AD be a: line of chords con- 4 structed to every, -. 4 5o.\\ ten degrees. 30- 31 Divide BD into / 7 I I0 nine equal parts, 0 /5 and from the points C of division, 10, 20, 30, &c., draw lines parallel to CB,* and meeting CD in 10, 20, 30, &c., and CD will be a line of sines. From the centre C, through the divisions of the are BD, draw lines meeting BE, in 10, 20, 30, &c., and BE will be a line of tangents. With one foot of the dividers in C transfer the distances from C to 10, 20, &c., in the line BE to the line CF, which will then be a line of secants. * To avoid confusion, these lines are not drawn in the figure. 40 PLANE TRIGONOMETRY. By dividing the arcs AD and BD each into 90 equal parts, and proceeding as above, the lines of chords, sines, &c., may be construed to every degree of the quadrant. PROBLEM II. At a given point A in a given right line AB, to make an angle of any proposed number of degrees, suppose 38 degrees. Fig. 33. Fig. 33. e With the centre A, and a radius equal to 60 degrees, taken from a scale of chords, describe an are, cutting AB in m; from the same scale of chords, A, take 38 degrees and apply it to the are from m to n, and from A through n draw the line AC; then will the angle A contain 38 degrees. Note. —Angles of more than 90 degrees are taken off at twice. PROBLEM III. To measure a given angle A. Fig. 34. Fig. 34. Describe the are mn with the chord of 60 degrees, as in the last problem. Take the arc mn between the dividers, and that extent applied to the scale of chords will show the degrees in the given angle. N3ote. —When the distance mn exceeds 90 degrees, it must be taken off at twice, as before. PLANE TRIGONOME'rtRY. 41 OF THE TABLE OF LOGARITHMIC OR ARTIFICIAL SINES, TANGENTS, &c. THIS table contains the logarithms of the sine, tangent, &c. to every degree and minute of the quadrant, the radius being 10000000000, and consequently its logarithm 10. Let the radius CB, Fig. 32, be supposed to consist of 10000000000 equal parts as above, and let the quadrant DB be divided into 5400 equal arcs, each of these will therefore contain 1'; and if from the several points of division in the quadrant, right lines be drawn perpendicular to CB, the sine of every minute of the quadrant to the radius CB will be exhibited. The lengths of these lines being computed and arranged in a table, constitute what is usually termed a table of natural sines. The logarithms of those numbers taken from a table of logarithms, and properly arranged, form the table of logarithmic or artificial sines. In like manner the logarithmic tangents and secants are to be understood. The method by which the sines are computed is too abstruse to be explained in this work, but a familiar exposition of this subject, as well as of the construction of logarithms may be seen in Simpson's trigonometry. To find, by the table, the sine, tangent, L4c. of an arc or angle. If the degrees in the given angle be less than 45, look for them at the top of the table, and for the minutes, in 4* F 42 PLANE'rRIGONOMETRY. the left hand column; then in the column marked at the top of the table, sine, tangent, &c. and against the minutes, is the sine, tangent, &c. required. If the degrees are more than 45, look for them at the bottom of the table, and for the minutes, in the right hand column; then in the column marked at the bottom of the table, sine, tangent, &c. and against the minutes, is the sine, tangent, &c. required. Note. —The sine of an angle and of its supplement being the same, if the given number of degrees be above 90, subtract them from 180~, and find the sine of the re mainder. EXAMPLES. 1. Required the sine of 32~ 27' Ans. 9.72962. 2. Required the tangent of 57~ 39' Ans. 10.19832. 3. WVhat is the secant of 890 31' Ans. 12.07388. 4. What is the sine of 1570 43' Ans. 9.57885, bo find the degrees and minutes corresponding to a given sine, tangent, &c. Find, in the table, the nearest logarithm to the given one, and the degrees answering to it will be found at the top of the table, if the name be there, and the minutes on the left hand; but if the name be at the bottom of the table, the degrees must be found at the bottom, and the minutes at the right hand. EXAMPLES. 1, Required the degrees and minutes in the angle whose sine is 9.64390. Ans. 260~ 8. PLANE TRIGONOMETRY. 43 2. Required the degrees and minutes in the angle whose tangent is 10.47464. Ans. 71~ 28'. ON GUNTER'S SCALE. GUNTER'S scale is an instrument by which, with a pair of dividers, the different cases in trigonometry, and many other problems, may be approximately solved. It has on one side, a diagonal scale, and also the lines of chords, sines, tangents, and secants, with several others. On the other side there are several logarithmic lines as follow: The line of numbers marked Num., is numbered front the left hand of the scale towards the right, with 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, which stands in the middle of the scale; the numbers then go on 2, 3, 4, 5, 6, 7, 8, 9, 10, which stands at the right hand end of the scale. These two equal parts of the scale are similarly divided, the distances between the first 1, and the numbers 2, 3, 4, &c. being equal to the distances between the middle 1, and the numbers 2, 3, 4, &c. which follow it. The subdivisions of the two parts of this line are likewise similar, each primary division being divided into ten parts, distinguished by lines of about half the length of the primary divisions. The primary divisions on the second part of the scale, are estimated according to the value set upon the unit on the left hand of the scale. If the first I be considered as a unit, then the first 1, 2, 3, &c. stand for 1.2, 3, &c the middle 1 is 10, and the 2, 3, 4, &c. follcwi.ng stand 44 PLANE TRIGONOMETRY. for 20, 30, 40, &c. and the ten at the right hand for 100. If the first 1 stand for 10, the first 2, 3, 4, &c. must be counted 20, 30, 40, &c. the middle 1 will be 100, the second 2,3, 4, &c. will stand for 200, 300, 400, &c. and the 10 at the right hand for 1000. If the first 1 be considered as l-lo of a unit, the 2, 3, 4, &c. following will be -i, 3,, &c. and the middle 1, and the 2, 3, 4, &c. following, will stand for 1, 2, 3, 4, &c. The intermediate small divisions must be estimated according to the value set upon the primary divisions. Sinles.-The line of sines, marked Sin., is numbered from the left hand of the scale towards the right, 1, 2, 3, 4, &c. to 10, then 20, 30, 40, &c. to 90, where it terminates just opposite 10 on the line of numbers. Tangents.-The line of tangents, marked Tan., begins at the left hand, and is numbered 1, 2, 3, &c. to 10, then 20, 30, 40, 45, where there is a brass pin, just under 90 in the line of sines; because the sine of 90~ is equal to the tangent of 45~. From 45 it is numbered towards the left hand 50, 60, 70, 80, &c. The tangent arcs of above 45~ are therefore counted backward on the line, and are found at the same points of the line as the tangents of their complements. There are several other lines on this side of the scale. as Sine Rhlumbs, Tangent Rlzumbs, Versed Sines, &c.; but those described are sufficient for solving all the problems in plane trigonometry. Remarks on Angles, Triangles, cc. I. If from a point D in a right line AB, one or more right lines be drawn on the same side of it, the angles thus formed at the point D will be together equal to two PLANE TRIGONOMETRY. 45 right angles, or 180~; thus ADE Fig. 35. C +EDB= two right angles, or 180: also ADC + CDE + EDB - two / right angles, or 1800. Fig. 35. 2. Since the angles thus formed A D B at the point D, on the other side of AB, would also be equal to two right angles, the sum of all the angles formed about a point is equal to four right angles, or 360~. 3. If two right lines cut one ano- Fig. 36. ther, the opposite angles will be equal: thus AEC- BED, and AED \E =CEB, Fig. 36. 4. The sum of the three angles of D a plane triangle is equal to two rightangles, or 180~. 5. If the sum of two angles of a triangle be sub tracted from 1800, the remainder will be the third angle. 6. If one angle of a triangle be subtracted from 180~, the remainder will be the sum of the other two angles. 7. In right-angled triangles, if one of the acute angles be subtracted from 90~, the remainder will be the other acute angle. 8. The angles at the base of an isosceles triangle are equal to one another. 9. If one side of a triangle be pro- Fig. 37. duced, the external angle will be equal to the sum of the two internal and opposite angles: thus the A B external angle CBD, of the triangle ABC, is equal to the sum of the internal and opposite angles A and C, Fig. 37. 46 PLANE TRIGONOMETRY. Fig. 38. 10. The angle at the centre of a A circle is double of the angle at the circumference, upon the same base, /E \ \ that is upon the same part of the circumference: thus the angle BEC B is double of the angle BAC, Fig. 38. 11. The angles in the same segFig. 39.-E ment of a circle are equal to one -/A another: thus the angle BAD is equal to the angle BED; also the angle BCD is equal to the angle HB BFD, Fig. 39. 12. The angle in a semicircle is a right angle: thus the angle ECF, Fig. 45, is a right angle. 13. This mark' placed on the sides or in the angles of a triangle, indicates that they are given; and this mark 0 placed in the same way, indicates that they are required. PRACTICAL RULES FOR SOLVING ALL TIIE CASES OF PLANE TRIGONOMETRY. CASE 1. The angles and one side of any plane triangle being givzen, to find the other sides. RULE. As the sine of the angle opposite the given side, Is to the sine of the angle opposite the required side, So is the given side, To the required side.* * DEMONSTRATION. Let ABC, Fig 40, be any plane triangle; take BF== AC, and upon AB let fall the perpendiculars CD and FE, which will be PLANE TRIGONOMETRY. 47 Note 1. —The proportions in trigonometry are worked by logarithms: thus, from the sum of the logarithm of' the second and third terms, subtract the logarithms of' the first term, and the remainder will be the logarithm of the fourth term. 2. The logarithmic sine of a right angle or 900 is 10.00000, being the same as the logarithm of the radius. EXAMPLES. 1. In the triangle ABC, there are given the angle A =32~ 15', the angle B =1140 24', and consequently the angle C= 33~ 21', and the sides AB =98;* required the sides AC and BC. By Construction, Fig. 41. Make AB equal to 98 by a scale of Fig. 41. equal parts, and draw AC, making the angle A= 320 15'; also make the angle B=_1140 24', and produce BC, A AC, till they meet in C, then is ABC B the triangle required; and AC measured by the same scale of equal parts, is 162, and BC is 95. the sines of the angles A and B to the equal c Fig. 40. radii AC and BF. Now the triangles BDC and BEF being similar, we have CD: FE:: BC: BF: or AC; that is sin. A: sin. B:: BC: AC. In like manner it is proved, that A - sin. A: sin. C:: BC: AB. WVhen one of the angles is obtuse, the demonstration is the same. Hence it appears, that in any plane triangle, the sides are to one another as the sines of their opposite angles. * This 98 may express so many feet, or yards, &c., and the other sides will be of the same denomination as the given. 48 PLANE TRIGONOMETRY. By Calculation. As sine of the angle C 33~ 21' - - 9.74017 Is to sine of the angle B 114~ 24' - - 9.95937 So is AB 98 - - 1.99123 11.95060 9.74017 To AC 162.3 - - 2.21043 As sine of C 33 21' - - - - 9.74017 Is to sine of A 32~ 15' - - - - 9.72723 So is AB 98 - 1.99123 11.71846 9.74017 To BC 95.12 - - 1.97829 By Gunter's Scale. Extend the compasses, on the line of sines, front 33~ 21' to 650 36' the supplement of the angle B; that extent will reach, on the line of numbers, from 98 to 162, the side AC. Extend the compasses from 33' 21' to 320 15' on the line of sines; that extent will rea( h, on the line of num bers, from 98 to 95, the side BC. 2. In the right-angled triangle ABC, are given the hypothenuse AC=480, and the angle A=_53~ 8'. To find the base AB and perpendicular BC. PLANE TRIGONOMETRY. 49 From 90~ subtract the angle A= 530 8'; the remain. der 360 52' will be the angle C. The angle B, being a right angle, is 90~. By Cornstruction, Fig. 42. This may be constructed as in the preced- Fi-. 42. ing example, or otherwise thus, Draw the line AB of any length, and draw AC, making the angle A- 53~ 8'; make AC -480 by a scale of equal partLand from C diraw CB perpendicular to AB, then ABC is A B the triangle required. AB measured by the same scale of equal parts, will be 288, and BC will be 384. By Calculation. As sine of B 90 - 10.00000 Is to sine of A 530 8' - 9.90311 So is AC 480 -2.68124 12.58435 To BC 384 - 2.58435 As sine of B 900 - - 10.00000 Is to sine of C 36~ 52' - - - - 9.77812 So is AC 480 - 2.68124 12.45936 To AB 2S8 - - 2.45936 G 650 PLANE TRIGONOMETRY. By Gunter's Scale. Extend the compasses, on the line of sines, from 90~ to 53~ 8', that extent will reach, on the line of numbers, from 480 to 384, the perpendicular BC. Extend the compasses, on the line of sines, from 900 to 360 52', the complement of the angle A; that extent will reach, on the line of numbers, from 480 to 288, the base AB. 3. In the triangle A3C, are given the angle A-790 23', the angle B=540 22', and the side BC=125; required AC and AB. Ans. AC-103.4, and AB-=91.87. 4. In a right-angled triangle, there are given the angle A=560 48', and the base AB=-53.66, to find the perpendicular BC and hypothenuse AC. Ans. BC —82 and AC-9 8. 5. In the right-angled triangle ABC, are given the angle A=390 10', and the perpendicular BC=407.37, to find the base AB and hypothenuse AC. Ans. AB=500.1, and AC=645. CASE 2. Two sides and an angle opposite one of them being given, to find the other angles and side. RULE. As the side opposite the given angle, Is to the other given side, So is the sine of the angle opposite the former, To the sine of the angle opposite the latter.* * This is evident from the demonstration of the rule in the preceding case. PLANE TRIGONOMETRY. 51 Add the angle thus found to the given angle, and subtract their sum from 1800, the remainder will be the third angle. After finding the angles, the other side may be found by Case 1. Note.-The angle found by this rule is sometimes ambiguous; for the operation only gives the sine of the angle, not the angle itself; and the sine of every angle is also the sine of its supplement. When the side opposite the given angle is equal to, or greater than the. other given side, then the angle opposite that other given side is always acute; but when this is not the case, that angle may be either acute or obtuse, and is consequently ambiguous. EXAMPLES. 1. In the triangle ABC, are given the angle C=330 21', the side AB=.98 and the side BC=.7912; required the angles A and B, and the side BC. By Construction, Fig. 43. Make BC=.7912 by a scale Fig. 43. of equal parts, and draw CA, making the angle C=33~ 21'; / with the side AB=.98, in the compasses, taken from the same scale of equal parts, and B as Bj b a centre, describe the arc ab, cutting AC in the point A, and join BA; then is ABC the triangle required: the side AC, measured by the scale of equal parts, will be 1.54, and the angles A and B, measured by a scale of chords, will be 26~ 21' and 120~ 18'. Here the arc ab cuts AC in one point only, because AB is greater than BC; therefore the angle A is acute, and not ambiguous 62 PLANE TRIGONOMETRY. By Calculation. As AB,.98 — 1.99123 Is to BC,.7912 - - - — 1.89829 So is sine of C, 33~ 21' - 9.74017 9.63846 To sine of A, 26~ 21' - 9.64723 To the ang e C=33~ 21' add the angle A=26~ 21' arid the sum is 59~42' which subtracted from 180~ leaves the angle B -- 120 18' As sine of C, 33~ 21' - 9.74017 Is to sine of B, 120~ 18. 9.93621 So is AB,.98 - - — 1.99123 9.92744 To AC, 1.539 - - - - 0.18727 By Gunter's Scale. Extend the compasses from.98 to.79 on the line of numbers, that extent will reach from 33~ 21' to 260 21', the angle A, on the line of sines. Add the angle A-26~ 21' to the angle C=-33 21', and the sum will be 59~ 42'; then extend the comnpasses from 33~ 21' to 590 42', on the line of sines, that extent will reach fron.98 to 1.54, the side AC, on the line of numbers. 2. In the triangle ABC, are given the angle C-33~ 21' the side BC=95.12 and the side AB-60, to find tho angles A and B, and tile side AC. PLANE TRIGONOMETRY. 53 By Construction, Fig. 44. This is constructed in the same Fig. 44. manner as the preceding exam- a A ple; only, AB being shorter than'r... BC, the arc ab cuts AC in two A points on the same side of BC; hence the angle A may be either acute or obtuse. The side required has also two values, as AC and AC. By Calculation. As AB, 60 1.77815 Is to BC, 95.12 - - - - - 1.97827 So is sine C, 330 21' - - - - 9.74017 11.71844 To sine of A 119 22 obtuse 9.94029 The sum of the angles C and A subtracted from 180O leaves the angle B- 860 1' if A be acute, or 270 17' if A be obtuse. To find the side AC answering to the acute value of the angle A. As sine of C, 330 21' - - - - 9.74017 Is to sine of B, 860 1' - - - - 9.99895 So is AB, 60 - - - - - 1.77815 11.77710 To AC, 108.9- - - - - - 2.03693 To find the side AC, answering to the obtuse value of the angle A. 54 PLANE TRIGONOMETRY. As sine of C, 33~ 21' - - - 9.74017 Is to sine of B, 27~ 17 - - 9.66124 So is AB,60 - - - - 1.77815 11.43939 To AC, 50.03 - - - - 1.69922 3. In a triangle ABC, the side AB is 274, AC 306. and the angle B 78~ 13'; required the angles A and C. and the side BC. Ans. A=40~ 33', C=61~ 14', and BC-203.2. 4. In a right-angled triangle, there are given the hypothenuse AC= 272, and the baseAB-= 232; to find the angles A and C, and the perpendicular BC. Ans. A=31~ 28' C-58~ 32' and BC —142. 5. In a right-angled triangle ABC, the hypothenuse AC is 150, and one side BC 69; required the angles and other side. Ans. C —62~ 37', A= 27' 23', and AB 133.2. CASE 3. Two sides and the included angle being given, tofind the other angles and side. RULE. Subtract the given angle from 180~, and the remainder will be the sum of the two unknown angles. Theit, As the sum of the two given sides, Is to their difference; PLANE TRIGONOMETRY. 55 So is the tangent of half the sum of the two unknown angles, To the tangent of half their difference.* This half difference of the two unknown angles, added to their half sum, will give the angle opposite the greater of the two given sides, and being subtracted * DEMONSTRATION. Let ABC, Fig. 45, be the Fig. 45. G proposed triangle, having the two given sides C AB, AC, including the given angle A. About A as a centre, with AC the greater of the given sides for a radius, describe a circle meeting AB F produced in E and F, and BC in D; join DA, A EC, and FC, and draw FG parallel to BC, meeting EC produced in G. The angle EAC (32.1.) is equal to the sum of the unknown angles ABC, ACB; and the angle EFC at the circumference, is equal to the half of EAC at the centre (20.3.;) therefore EFC is half the sum of the unknown angles; but (32.1.) the angle ABC is equal to the sum of the angles BAD and ADB, or BAD and ACB; therefore FAD is the difference of the unknown angles ABC, ACB; and FCD, at the circumference is the half of that difference; but because of the parallels DC, FG, the angles GFC, FCD are equal; therefore GFC is equal to half the difference of the unknown angles ABC, ACB; but since the angle ECF in a semicircle, is a right angle, EG is perpendicular to CF, and therefore CF being radius, EC, CG are the tangents of the angles EFC, CFG; it is also evident that EB is the sum of the sides BA, AC, and that BF is the difference; therefore since BC, FG are parallel, EB: BF:: EC: CG (2.6.:) that is, the sum of the sides AC, AB, is to their difference, as the tangent of half the sum of the angles ABC, ACB, is to the tangent of half their difference. To demonstrate the latter part of the Fig. 46. rule, let AC and AB, Fig. 46, represent C F 3 any two magnitudes whatever; in AB produced, take BD equal to AC the less, and bisect AD in E. Then because AE is equal to ED, and AC to BD, CE is equal to EB; therefore AE or ED is half the sum of the given magnitudes AB, AC, and CE, or EB is half their difference; but AB the greater is equal to AE, EB, that is to half the sum added to half the difference; and AC the less, is equal to the excess of AE, half the sum, above CE, half the dif. ference. 56 PLANE TRIGONOMETRY. from the half sum will give the angle opposite the less given side. After finding the angles, the other side may be found by Case 1. EXAMPLES. 1. In the triangle ABC, there are given AB= 128, AC-= 90, and the angle A =480 12', to find the angles B and C, and the side BC. By Construction, Fig. 47. Fig. 47. Draw AB-=128, and make the x~\% angle A=480 12'; draw AC-90, and join BC. The angle B will A~ 4. X measure 440 37', the angle C 87~ 11', and the side BC 95.5. By Calculation. AB 128 1800 0' AC 90 Angle A - - - - 48 12 Sum 218 Sum of the angles B and C 131 48 Difference 38 Half sum do. 65 54 As the sum of the sides, AB, AC, 218 - - 2.33846 Is to their difference, 38 - - 1.57978 So is the tangent of half the sum of the 10.34938 angles B and C, 650 54' 34938 11.92916 To tang. of half their difference, 210 17' - - 9.59070 Half sum of the angles B and C - - 650 54' Add and subtract half their difference - 21 17 Angle C - - - - - - - - -87 11 Angle B - - 44 37 PLANE TRIGONOMETRY. 57 To find the side BC. As sine of B, 44~ 37' - - - 9.84656 Is to sine of A, 48 12 - - - 9.87243 So is AC, 90 - - - 1.95424 11.82667 To BC, 95.52 1-.98011 By Gunter's Scale. Extend the compasses from 218, the sum of the sides, to 38, their difference, on the line of numbers, and apply this extent to the line of tangents from 45~ to the left hand; then keeping the left leg of the compasses fixed, move the other leg to 650 54' the half sum of the angles; that distance will reach from 45~ on the same line, to 210 17', the half difference of the required angles. Whence the angles are obtained as before. To extend the second proportion, proceed as directed in Case 1st. 2. In a triangle ABC, are given AB= 109, BC=76, and the contained angle B- 101~ 30', to find the other angles and side. i Ans. The angle A=300 57', C=47~ 33', and tlhe side AC= 144.8. 3. Given, in a right-angled triangle, the base AB=890 and the perpendicular BC —787, to find the angles and hypothenuse. Ans. The angle A —41~ 29', C=480 31', and the hypothenuse AC= 1188. 58 PLANE TRIGONOMETRY. CASE 4. Given the three sides, tofind the angles. RULE 1. Consider the longest side of the triangle as the base, and on it let fall a perpendicular from the opposite angle. This perpendicular will divide the base into two parts, called segments, and the whole triangle into two right-angled triangles. Then, As the base, or sum of the segments, Is to the sum of the other two sides; So is the difference of those sides, To the difference of the segments of the base.* Fig. 48. * DEMONSTRATION. Let ABC, Fig. 48, be a triangle, and CD be perpendicular upon AB / About C as a centre, with the less side BC for a radius, describe a circle, meeting AC proC duced, in G and E, and AB in F. Then it is evident that AE is equal to the sum of the sides AC, BC, and that AG is equal to their rA D B difference; also because CD bisects FB (3.3,) it is plain that AF is the difference of the segments of the base; but AB xAF = AEx AG (36.3. cor.;) therefore AB: AE:: AG: AF )16.6;) that is, the base, is to the sum of the sides, as the difference of the sides, is to the difference of the segments of the base. Cor. If AF be considered the base of the triangle AFC, then will CD be a perpendicular on the base produced; AE will be equal to the sum of the sides AC, FC, and AG will be equal to their difference; also AB will be equal to the sum of the segments AD, FD. But by the preceding demonstration, and (16.5,) AF: AE:: AG: AB; hence, when the perpendicular falls without the triangle, the base is to the sum of the sides, as the difference of the sides is to the sum of the segments of the base. A rule might, therefore, be given, making either side of a triangle the base; and such a rule would be rather more convenient, in some cases, than the one above: but then, on account of the perpendicular, sometimes falling within and sometimes without the triangle, it would require twc cases, and consequently would be less simple. PLANE TRIGONOMETRY. *59 To half the base, add half the difference of the segInents, and the sum will be the greater segment; also from half the base, subtract half the difference of the segments, and the remainder will be the less segment. Then, in each of the two right-angled triangles, there will be known two sides, and an angle opposite to one of them; consequently the other angles may be found by Case 2. 1. In the triangle ABC, are given AB-=426, AC = 365, and BC= 230; required the angles. By Construction, Fig. 49. Draw AB = 426; with AC = 365 in Fig. 49. the dividers, and one foot in A, describe an arc, and with BC =230, and one foot in B describe another arc, / B A D B cutting the former in C; join AC, BC, and ABC will be the triangle required. The angles measured by a scale of chords, will be A- 32~ 39', B = 580 56', and C = 88 25'. By Calculation. AC - 365 BC - - - - - - 230 Sum - - 595 Difference - 135 60 PLANE TRIGONOMETRY. As the base AB -426 2.62941 Is to the sum of the sides AC, BC - 595 2.77452 So is the diff. of the sides AC, BC - 135 2.13033 4.90485 To the diff. of the segments AD, DB 188.6 2.27544 Half diff. of the segments - - - 94.3 Half base - 213. Segment AD - - - - 307.3 Segment BD - - -. —118.7 As AC - 365 2.56229 Is to AD - - - - 307.3 2.48756 So is sine of ADC - - 90~ 10.00000 12.48756 To sine of ACD - - - - 570 21' 9.92527 90 00 Angle A - - - 32 39 As BC - - - 230 2.36173 Is to BD - - - - - - 118.7 2.07445 So is sine of BDC - - - - 90 10.00000 12.07445 To sine of BCD 310 4' 9.71272 90 0 Angle B - - - - - 58 56 From 180~ subtract the sum of the angles A, and B, 91~ 35', and the remainder 88~ 25' is the angle C. PLANE TRIGONOMETRY. 61 By Gunter's Scale. Extend the compasses from 426 to 595 on the line of numbers, that extent will reach on the same line from 135 to 188.6 the difference of the segments of the base. Whence the segments of the base are found as before. To extend the other proportions, proceed as dir cted in Case 2. 2. In a triangle ABC, there are given AB —64, AC= 47. and BC=34; required the angles. Ans. Angle A -31l 9', B —45~ 38', and C=103~ 13'. 3. In a triangle ABC, are given AC=88, AB=-108, and BC= —54, to find the angles. Ans. Angle A=-29~ 49', B=540 7', and C=96~ 4'. RULE 2. Add together the arithmetical complements of the logarithnls of the two sides containing the required angle, the logarithm of the half sum of the three sides, and the logarithm of the difference between the half sum and the side opposite the required angle. Then half the sum of these four logarithms will be the logarithmic cosine of half the required angle.* DEMONSTRATION. Let ABC, be a triangle of which the side AB, i greater than AC: make AD=AC, join DC, bisect it in E, and join AE:z draw EH parallel and equal to CB; join IIB and produce it to meet AE produced in G. Now in the triangles AED, AEC, all the sides of the one are equal to the sides of the other, each to each; therefore (8.1) the angle EAD /N =E.A, and A El )= AEC; consequent A't. ly AED is a right angle. ~ ~]n~~~I 62 PLANE TRIGONOMETRY. EXAMPLES. 1. In the triangle ABC, are given AB=426, AC=365 and BC-230; required the angle A. By Calculation. BC 230 AC 365 Ar. Co. 7.43771 AB 426 Ar. Co. 7.37059 2)1021 Half sum 510.5 log. 2.70800 Difference 280.5 log. 2.44793 2)19.96423 Cos. i A 16~ 20) 9.98211 2 Angle A 32 40 Because EH is equal and parallel to BC, BIT is also equal and parallel to EC (33.1;) now in the triangles EDF and HBF, the angle EFD —BFH, the angle FED=FHB (29.1) and ED=EC=BH; therefore (26.1) EF=FH, and FD=FB. Again, the angle HGE=DEA=a right angle; if therefore with the centre F, and radius FE=FH, a circle be described, it will past through the point G (31.3.) Now 2AF-2 AD + 2 DF AD ~ AD ~ DBAD 4- ABAC AB; therefore AF-AC+ ~AB; also FK=-IIK=-E-==-BC; hence, by adding equals to equals, AF+FK= —AC+ 1AB+IBC, or AK=- (AC+ AB+ BC;) agaiL, AI AK-IKI —-(AC+ AB+ BC)-BC. But (Dem. to rule, case 1st.) AD: AE:: sin. AED: sin. ADE: rad.: cos. EAD (cos. ~BAC.) Also, AB: AG:: sin. AGB: sin. ABG:: rad.: cos. BAG (cos. -BAC.) IHence (c. 6) AB X>< AD:: AG X AE:: rad.2: (cos. JBAC)2. But AB X AD=AB x AC, and (cor. 36.3) AG X AE=AK X AT=I(AC+ AB+ BC) x ['(AC+ AB+ BC)-BC;] therefore ABx AC: ~(AC+ AB+- BC)x [I(ACAB+BC)-BC]:: rad.2: (cos. ~BAC)2. Now it is evident, that in working this proportion by logarithms PLANE TRIGONOMETRY. 6 3 If the other angles are required, they may be found by Case 1. 2. In a triangle ABC, are given AB 64, AC =47, and BC- 34, to find the angle B. Ans. Angle B450 38'. 3. In a triangle ABC, are given AC= 88, AB -108, and BC =-54, to find the angle C. Ans. C-=960 4'. The preceding rules solve all the cases of plane triangles, both right-angled and oblique. There are, however, other rules, suited to right-angled triangles, which are sometimes more convenient than the general ones. Previous to giving these rules, it will be necessary to make the following Remarks on right-angled triangles. 1. ABC, Fig. 50, being a right-angled Fig. 50. triangle, make one leg AB radius, that is, r with the centre A, and distance AB, describe an arc BF. Then it is evident that A G the other leg BC represents the tangent of the arc BF, or of the angle A, and the hypothenuse AC the secant of it. 2. In like manner, if the leg BC, Fig. 51, Fig. 51. be made radius; then the other leg AB will represent the tangent of the arc BG, or angle C, and the hypothenuse AC the secant of it. A/ Tang. (C and taking the arithmetical complements of the logarithms of the first term, viz. of the two sides, including the required angle, if we omit the logarithm of the square of radius, which is 20, it is just equivalent to rejecting 20 from the sum of the logarithms, which would otherwise have to be done. 64 PLANE TRIGONOMETRY. Fig. 52. 3. But if the hypothenuse be made C radius, then each leg will represent the sine of its opposite angle; namely, the \ leg AB, Fig. 52, the sine of the are AE A. Sin. - or angle C, and the legt BC the sine of the arc CD, or angle A. The angles and one side of a right-angled triangle being given, to filnd the other sides. RULE. Call any one of the sides radius, and write upon it the word radius; observe whether the other sides become sines, tangents, or secants,.and write these words on them accordingly. Call the word written upon each side the name of that side. Then, As the name of the side given, Is to the name of the side required; So is the side given, To the side required.* Two sides of a right-angled triangle being given, to find the angles and other side. RULE. Call either of the given sides radius, and write on them as before. Then, Fig. 53. * DEMONSTRA.TION. Let ABC, Fig. 53, be a rightangled triangle; then it is evident that BC is the tangent, and AC the secant of the angle A, to the radius AB. Let AD represent the radius of the tables, and draw DE perpendicular to AD, meeting AC produced A 1; in E; then DE is the tangent, and AE the secant of the angle A, to the radius AD. But because of the similar triangles ADE, ABC, AD: DE:: AB: BC; that is, the tabular radius: tabular tangent:: AB: BC. Also AD: AE:: AB: AC; that is, the tabular radius: tabular secant:: AB: AC. These proportions correspond with the rule. When either of the other sides is made radius, the demonstration will be similar. PLANE TRIGONOMETRY. 65 As the side made radius, Is to the other given side; So is radius, To the name of that other side.* After finding the angle, the other side is found as ill the preceding rule. EXAMPLES. 1. In a right-angled triangle ABC, are given the base AB —208, and the angle A=35~ 16', to find the hypo thenuse AC and perpendicular BC. By Calculation. The hypothenuse AC being radius. As the sine of C, 54~ 44' - - 9.91194 Is to radius - 10.00000 So is AB 208 - 2.31806 12.31806 To AC 254.8 - 2.40612 As the sine of C 54 44' - - - 9.91194 Is to the sine of A, 35 16 - - - 9.76146 So is AB 208 - --- 2.31806)G 12.0795' To BC, 147.1 -2.1G675 Thl:is is the converse of the preceding rule. 6* I 6 6 PLANE TRIGONOMETRY. The base AB being radius. As radius - 10.00000 Is to the secant of A, 35~ 16' - - 10.08806 So is AB 208 - 2.31806 12.40612 To AC 254.8 2.40612 As radius - 10.00000 Is to tangent of A, 350 16' - - 9.84952 So is AB, 208 - 2.31806 12.16758 To BC 147.1 - - 2.16758 The perpendicular BC being radius. As tangent of C 54 44' - - - - 10.15048 Is to secant of C, 54 44 - - - - 10.23854 So is AB, 208 - - 2.31806 12.55660 To AC 254.8 - 2.40612 As tangent of C, 54 44' - - - - 10.15048 Is to radius, - 10.00000 So is AB 208 2.31806 12.31806 To BC, 147.1 - - 2.16758 2. In a right-angled triangle ABC, there are given the lhypothenuse AC= 272, and the base AB= 9232; required the angles A and C, an(d the perpendicular [C. PLANE TRIGONOMETRY. 6 By Calculation. The hypothenuse AC being radius. As AC, 272 - 2.43457 Is to AB, 232 - -2.36549 So is radius - 10.00000 12.36549 To sine of C, 58~ 3Z - - - - 9.93092 As radius - -. 10.00000 Is to sine of A, 31~ 28' - - - - 9.71767 So is AC, 272 - 2.43457 12.15224 To BC, 142 2.15224 The base AB being radius. As AB, 232 - - - - - - - 2.6549 is to AC, 272 — - - - 2.43457 So is radius - - - 10.00000 12.43457 To secant of A, 31~ 28' - - - 10.06908 As radius - 10.00000 Is to tangent of A, 310 28' - - - 9.78675 So is AB, 232 - 2.36549 12.15224 To BC,142 2.15224 PLANE TRIGO.NOhMETRY. 3. In a right-angled triangle, are given the hypothe-,use AC-36.57, and the angle A=270 46', to find the base AB, and perpendicular BC. Ans. Base AB-=32.36, and perpendicular BC-=17.04 4. In a right-angled triangle, there are given, the perpendicular=193.6, and the angle opposite the base 47' 51'; required the hypothenuse and base. Ans. Hypothenuse= —288.5, and base=-213.9. 5. Required the angles and hypothenuse of a rightangled triangle, the base of which is 46.72, and perpen~ diicular 57.9. Ans. Angle opposite the base 38~ 54', ancle opposite ( the perpendicular 510 6', and hypothenuse 74.4. When two sides of a right-angled triangle are given, the other side may be found by the following rules, without first finding the angles. 1. When the hypothenuse and one leg are given, to find the other leg. RULE. Subtract the square of the given leg from the square of the hypothenuse; the square root of the rclmaiader will be the leg required.* Or by logarithms thus, To the logarithm of the slnm of the hypothenuse and given side, add the logaritllll of' tleir diflerenlce; half this sumn will be tthe logaritlln of the leg re(cuired. * DEMONSTRATTON. The s(quare (,f tle hypothenluse of a rithlt-anled tri. ungle is equal to the surl (f the squares of' ti.e sities (47.1.)'Tlieretobre the truth of thle first part of each of tlhe rules is evidenLt PLANE TRIGONOMETRY. 69 2. When tihe two legs are given to find the hypotlienuse. RULE. Add together the squares of the two given legs; the square root of the sum will be the hypothenuse.* Or by logarithms thus, From twice the logarithmn of the perpendicular, subtract the logarithm of the base, and add the corresponding natural number to the base; then, half the sum of the logarithms of this sum, and of the base, will be the logarithm of the hypothenuse. EXAMPLES, 1. The hypothenuse of a right-angled triangle is 272. and the base 232; required the perpendicular. Calculation by logarithms. Hypothenuse - - 272 Base 232 Sum - - - - 504 log. 2.70243 Difference - - - 40 1.60206 2)4.30449 Perpendicular - -142 2.15224 * Put h=the hypothenuse, b=the base, and p=the perpendicular, then (47.1) p2-h2 —b2=(5.2 cor.) hi-bxh —b, or p=Vh+-bxh —b; whence, from the nature of logarithms, the latter part of the first rule is evident. Also (47.1) h2=b2+p2=bX b+p or h=/(bx+p which, solved by logarithms, will correspond with the latter part of tsh second rule. t0 APPLICATION OF 2. Given the base 186, and the perpendicular 152, to find the hypothenuse. Calculation &y Logarithms. Perpendicular 152 log. 2.18184 2 4.36368 Base - - - 186 2.26951 2.26951 124.2 2.09417 310.2 log. 2.49164 2)4.76115 Hypothenuse 240.2 log. 2.38057 3. The hypothenuse being given equal 403, and one leg 321; required the other leg. Ans. 243.7. 4. What is the hypothenuse of a right-angled triangle, the base of which is 31.04 and perpendicular 27.2? Ans. 41.27. The following examples, in which trigonometry is applied to the mensuration of inaccessible distances and heights, will serve to render the student expert in solving the different cases, and also to elucidate its use. Fig. 64. THE APPLICATION OF PLANE TRIGONOMETRY TO THE MENSURATION OF DIS-,. — TANCES AND *HEIGHTS. EXAMPrLE 1, Fig. 54. A B Being on one side of a river and wanting to know the distance to a house on the other side, PLANE TRIGONOMETRY. 71 I measured 500 yards along the side of the river in a right line AB, and found the two angles* between this line and the object to be CAB-= 71~ 14' and CBA=490 23'. Required the distance between each station and the object. Calculation. The sum of the angles CAB and CBA is 1230 37', which subtracted from 1800 leaves the angle ACB= 56~ 23'. Then by Case 1; s. ACB s. CBA:: AB AC 560 23' 490 23' 500 455.8 and s. ACB s. CAB BA BC 56~ 23' 740~ 14' 500 577.8 EXAMPLE 2, Fig. 55. Suppose I want to know the dis- Fig. 65. tance between two places, A and B, H accessible at both ends of the line AB, A and that I measured AC -735 yards, and BC= 840; also the angle ACB= 550 40'. What is the distance between A and B? Calculation. The angle ACB-550 40', being subtracted from 180~, leaves 1240 20'; the half of which is 620 10'. Then by Case 3, * The angles may be taken with a common surveyor's compass; or more accurately with an instrument called a theodolite. 72 APPLICATION OF CAB +CBA. CAB-CBA BC+AC: BC-AC:: tangent 2' tangent 2 1575 105 620 10' 70 12' CAB+CBA CAB-CBA To and from 2 =62~ 10' add and sub. 2 =-7 12', and we shall have CAB 69~ 22', and CBA= 54~ 58'. Then, s. ABC: s. ACB:: AC: AB 540 58' 550 40' 735 741.2 EXAMPLE 3, Fig. 56. Fit,. 56. Wanting to know the distance between two inaccessible objects A and B, I measured a base line CD=-300 yards: at C the angle D i BIBCD was 580 20' and ACD 950 i~~ 20'; at D the angle CDA was 530 30' and CDB 980 45'. Required the distance AB. Calculation. 1. In the triangle ACD, are given the angles ACD 950 20', ADC — 530 30', and the side CD -- 300, to find AC — 46 5.98. 2. In the triangle BCD, are given the angle BCD — 680 20', BDC=-980 45', and side CD= 300, to find BC 761.47. 3. In the triangle ACB we have now given the angle ACB=-ACD-BCD=-37~, the side AC-465.98 and BC-761.47, to find ABI-479.8 yards, the distance required. Fig. 57. EXAMLPIE 4, Fig. 57. 2 r Being'on one side of a river.N and observing three objects, A. B and C stand on the others side, whose distances apart I knew to be, AB-= 3 miles, AC PLANE TRIGONOMETRY. 73 =2, and BC =1.8, I took a station D, in a straight line with the objects A and C, being nearer the former, and found the angle ADB =170 47'. Required my distance from each of the objects. Construction. With the three given distances, describe the triangle ABC; from B, draw BE parallel to CA, and draw BD making the angle EBD- 170 47' (the given angle ADB) and meeting CA produced, in D: then AD, CD and BD will be the distances required.* Calculation. 1. In the triangle ABC we have all the sides given, to find the angle C 1040 8'. 2. Subtract the sum of the angles D and C from 1800, the remainder 58~ 5' will be the angle DBC; then in the triangle BCD we know all the angles and the side BC to find DC 5.002 and BD=5.715; therefore DA=DC-AC = 3.002. EXAMPLE 5, Fig. 58. From a station at D, I perceived Fig. 68. three objects, A, B and C, whose distances from each other I knew to be as follows: AB — 12 miles, BC % A =7.2 miles, and AC=8 miles; at D, I took the angle CDB=250 and ADC = 19~. Hence it is required / to find my distance from each of the objects. * DEMONSTRATION. By construction, the distances AB, BC and AC are tqual to the giver. distances; also the angle (29.1) BDC=the angle DBE -the given angle. K 74 APPLICATION OF Construction. With the given distances describe the triangle A 3C; at B, make the angle EBA4- 19~-the given angle ADC, and at A, make the angle EAB=25~ the given angle BDC; draw AE, and BE meeting in E, and (by prob. 10,) describe a circle that shall pass through the points A, E and B: join CE and produce it to meet the circle in D, and join AD, BD, then will AD, CD, and BD be the distances required.* Calculation. 1. In the triangle ABC, all the sides are given, to find the angle BAC=35~ 35'. 2. In the triangle AEB, are given all the angles, viz. EAB-25~, EBA= 19~, and AEB — 136~, and the side NAB=12, to find AE-5.624. DEMONSTRATION. The angle ADC standing on the same are with the angle ABE is equal to it (21.3.) For the same reason the angle BDC is equal to the angle BAE; but by construction the angles ABE and BAE are equal to the given angles; therefore the angles ADC and BDC are equal to the given angles. Note. —When the given angles ADC, BDC are respectively equal to the angles ABC, BAC, the point E will fall on the point C, the circle will pass through the points A, C, and B, and the point D may be any where in the arc ADB; consequently, in this case, the situation of the point D, or its distance from each of the objects A, B, C, cannot be determined from the data given. It may not be improper also to observe, that even when the angle ADB, which is the sum of the given angles, is equal to the sum of the angles ABC, BAC, on which is the same thing, is the supplement of the angle ACB, the circle passes through the points A, C, B; but then the angles ADC, BDC, unless they have been erroneously taken, will be respectively equal to the angles ABC, BAC. PLANE TRIGONOMETRY. 75 3. In the triangle CAE we have given the side AC — 8, AE= 5.624, and the angle CAE =BAC-EAB= 100 35', to find the angle ACE-=220 41'. 4. In the triangle DAC, all the angles are given, viz. ADC=19~, ACD=-220 41' and CAD=1800~-the sum of the angles ADC and ACD=-138~ 19', and the side AC=8, to find AD 9.47 miles, and CD-16.34. In the triangle ABD, we have the angle ADB ADC +BDC —440, the angle BAD= CAD-BAC- 102~0 44', and the side AB13=12, to find BD-16.85 miles. EXAMPLE 6, Fig. 59. A person having a triangular field, Fig. 69. the sides of which measure AB 50 - perches, AC=46 perches, and BC= 40 perches, wishes to have a well, dug in it, that shall be equally distant from the corners A, B and C. What must be its distance from each corner,, and by what angle from the corner A, may its place be found? Construction. With the given sides construct the triangle ABC, and (by Prob. 10.) describe a circle that shall pass through the points A, B, and C; then the centre E of this circle is the required place of the well.* Calculation. 1. In the triangle ABC, all the sides are given, to find the angle ABC=600 16'. 2. Join CE and produce it to meet the circumference in D; also join AE and AD; then the angles ADC, ABC * The demonstration of this is plain (1.3 cor.) 76 APPLICATION OF being angles in the same segment, are equal; also the angle DAC being an angle in a semicircle, is a, right-angle: therefore in the right-angled triangle DAC, we have the angle ADC-=ABC =600 16', and the side AC, to find CD —52.98 perches. The half of CD is=26.49 perches ==CE- the distance of the well from each corner. 3. The angle ACD=900 —ADC=-290 44; but because AEC is an isosceles triangle, the angle CAEACE-=290 44' the angle required. EXAMPLE 7, Fig. 61. Fig. 61. Wishing to know the height of o a steeple situated on a horizontal aip plane, I measured 100 feet in a ~/ i riglht line from its base, and then took the angle of elevation* of the |!i top, which I found to be 470 30', the centre of the quadrant being 5._ | feet above the ground: required the height of the steeple. Fig. 60. * Angles of elevation, or of depression, are usually taken with an instrument called a quadrant, the arc of which is divided into 90 equal parts or degrees, and those, when the instrument is sufficiently large, may be B 0 subdivided into halves, quarters, &c. From o the centre a plummet is suspended by a fine silk thread. Fig. 60 is a representation of this instrument. To take an angle of elevation, hold the quadrant in a vertical position, and the degrees being numbered from B towards C, with the eye at C, look along the side CA, moving the quadrant till the top of the object is seen in a range with this side; then the angle BAD made by the plummet'with the side BA, will be the angle of elevation required. Angles of depression are taken in the same manner, except that then the eye is applied to the centre of the quadrant. NXote. —In finding the height of an object, it is best to take such a position that the observed angle of altitude may be about 45~; for when the observed angle is 450, a small error committed in taking it, makes the least error in the computed height of the object. PLANE TRIGONOMETRY. 77 Calculation. In the righllt-angled triangle DEC, we have the angle CDE-47~ 30', and the base DE=AB=100 feet, toc find CE — 109.13 feet; to CE add EB13=DA —5 feet, the height of the quadrant, and it will give BC:= 114.13 feet, the required height of the steeple. EXAMPLE 8, Fig. 62. Wishino to knowqthe height of Fig. 62. c a tree situated in a bog, at a station D, which appeared to be on a level with the bottom of the tree, I took the angle of eleva- A l) tion BDC = 51~ 30'; I then measured DA -75 feet in a direct line from the tree, and at A, took the angle of elevation BAC —26~ 30'. Required the height of the tree. Calculation. 1. Because the exterior angle of a triangle is equal to the sum of the two interior and opposite ones, the angle BDC DAC + ACD; therefore ACD =BDCDAC -- 25~: now in the triangle ADC we have DAC26~ 30', ACD-=25~, and AD= 75, to find DC —79.18. 2. In the right-angled triangle DBC are given DC= 79.18, and the angle BDC= 510 30'to find BC 61.97 feet, the required height of the tree. EXAM.TPLE 9, Fig. 63. Wanting to know the Fi heigllt of a tower EC, which stood upon a hill, at A, I took the angle of, elevation CABI=44~; I then mleasured AD 134 / yards, on level ground, in D, straight line towards the 78 APPLICATION OF tower; at D the angle CDB was 670 50' and EDB 510 Required the height of the tower and also of the hill. Calculation. 1. In the triangle ADC we have the angle DAC440, the angle ACD-BDC-DAC =230 50', and the side AD, to find DC=230.4. 2. In the triangle DEC all the angles are given, viz. CDE=BDC-BDE=160 50', DCE=900 —BDC =22 10', DEC= 1800=the sum of the angles CDE and DCE =141~, and CD=230.4, to find CE=106 yards, the height of the tower. 3. In the right-angled triangle DBC, we have the angle BDC=670 50', and the side DC=230.4, to find BC= 213.4; then BE BC-CE-213.4 —106 =107.4 yards, the height of the hill. EXAMPLE 10, Fig. 64. Figr. 64. An obelisk AD standing I- t on the top of a declivity, I measured from its bottom a distance AB=40 feet, and then took the angle ABD=410; going on in /zc the same direction 60 feet IC / ~~// farther to C, I took the angle ACD=230 45': what was the height of the obelisk? Calculation. 1. In the triangle BCD, we have given the angle BCD-=230 45', the angle BDC-ABD-BCD= 170 15', and side BC =60, to find BD= 81.49. 2. In the triangle ABD are given the side AB=40, BD=81.49, and the angle ABD=41~, to find AD= 57.64 feet, the height of thc obelisk. PLANE TRIGONOMETRY. 79 EXAMPLE 11, Fig. 65. Wanting to know the height Fig. 65. of an object on the other side of - a. a river, but which appeared to D b)e on a level with the place where I stood, close by the side M of the river; and not having NA room to go backward on the same plane, on account of the immediate rise of the bank, I placed a mark where I stood, and measured in a direct line from the object, up the hill, whose ascent was so regular that I might account it a right line, to the distance of 132 yards, where I perceived that I was above the level of the top of the object; I there took the angle of depression of the mark by the river's side equal 42~, of the bottom of the object equal 27~, and of its top equal 190: required the height of the object. Calculation. 1. In the triangle ACD, are given the angle CADEDA-27~, ACD-=1800-CDE (FCD)=-1380, and the side CD-=132, to find AD=194.55 yards. 2. In the triangle ABD, we have given ADB = ADEBDE=8~, ABD= BED+BDE 109~ and AD= 194.55, to find AB 28.64 yards, the required height of the object. EXAMPLE 12, Fig. 66. Fig. 66. A May-pole whose height was 100 feet standing on a horizontal plane, was broken by a blast of wind, and the extremity of the top D:.. F part struck the ground at the distance of 34 feet from the bottom of the pole: required the length of each part. A D APPLICATION OF Construction. Draw AB=34, and perpendicular to it, make BC= 100; join AC and bisect it in D, and draw DE perpendicular to AC, meeting BC in E; then AE=-CE- the part broken off.* Calculation. 1. In the right-angled triangle ABC, we have AB=34 and BC —100, to find the angle C-18~ 47'. 2. In the right-an(led triangle ABE, we have AEB= ACEI-+CAE=2ACE=37~ 34', and AB=34, to find AE: =55.77 feet, one of the parts; and 100-55.77=44.23 feet the other part. PRACTICAL QUESTIONS. * 1. At 85 feet distance from the bottom of a tower, the angle of its elevation was found to be 52~ 30': required the altitude of the tower. Ans. 110.8 feet.'2. To find the distance of an inaccessible object, 1 measured a line of 73 yards, and at each end of it took * DEMONSTRATION. In the triangles AED, DEC, the angle ADEF-CDE, the side AD=CD, and DE is commnon to the two triangles, therefore (4.1) AE —(E. iVote. —'rlThis qlestion may be neatly solved in the fjllowinr manner without fin lingr eitlier of' the angrles.'Thus, draw DIp perpendicalar to BC, th;en, (:31.3 and cor. 8.6) FC: C:: DC: CE; conseI)C' ACZ ABW+ 13C2 quently CE-=FC; but DC' - i-= —'_ and FC= —BC; thereAB'+ BC-';34" 11001 1156+- 10000 11156 fore rCE -- -~ — ~__ --- 55.7 9, the f =re C 2BC 00 =- 2 20U 200 sane as before nearly. PLANE TRIGONOMET RY. 81 the angle of position of the object and the other end, and found the one to be 90~, and the other 61~ 45': required the dista'nce of the object from each station. Ans. 135.9 yards from one, and 154.2 from the other. 3. Wishing to know the distance between two trees C and D, standing in a bog, I measured a base line AB=339 feet; at A the angle BAD was 100~ and BAC 36~ 30'; at B the angle ABC-was 121~ and ABD 49~: required the distance between the trees. Ans. 6974 feet. 4.; Observing three steeples, A, B and C, in a town at a distance, whose distances asunder are known to be as follows, viz. AB=213, AC=404, and BC=262 yards, I took their angles of position from the place D where I stood, which was nearest the steeple B, and found the angle ADB=13~ 30'; and the angle BDC=29~ 50'. Required my distance from each of the three steeples. Ans. AD==571 yards, BD=389 yards, and CD=514 yards. 5. A May-pole, whose top was broken off by a blast of wind, struck the ground at 15 feet distance from the lbot of the pole: what was the height of the whole Maypole, supposing the length of the broken piece to be 39 feet? Ans. 75 feet. 6. At a certain place the angle of elevation of an inaccessible tower was 26~ 30; but mneasuring 75 fet its a direct line towards it, the angle was then found to be 51~ 30': required the height of the tower and its distance from the last station. Ans. Height 62 feet, distance 49. 7. From the top of a tower by the sea side, of 143 fcet hilgh, I observed that the angle of depression of a ship3' L 82 APPLICATION, &C. bottom, then at anchor, was 35~; what was its distance fronm the bottom of the wall? Ans. 204.2 feet. 8. There are two columns left standing upright in the ruins of Persepolis; the one is 64 feet above the plane, and the other 50; in a right line between these stands a.n ancient statue, the head of which is 97 feet from the summit of the higher, and 86 from that of the lower column; and the distance between the lower column and the centre of the statue's base is 76 feet: required the distance between the tops of the columns. Ans. 157 feet. SURVEYIN -G. CHAPTER I. ON THIE DIMENSIONS OF A SURVEY. 1. SURVEYING is the art of measuring, laying out and dividing land. 2. A Four-Pole Chain is an instrument used for measuring the boundaries of a survey. It is, as its name imports, 4 poles or 66 feet in length, and is divided into 100 equal parts or links. The length of a link is therefore 7.92 inches. Note.-A Four-pole Chain is frequently called simply a chain. 3. A Two-pole Chain is 2 poles or 33 feet in length, and is usually divided into 50 equal parts or links. ~When it is thus divided, the links are of the same length as in a four-pole chain; and the measures taken with it are reduced to four-pole chains previous to using theni in calculation. Sometimes the two-pole chain is divided into 40 links; in which case, each two links is the one-tenth of a perch. Measures taken with a two-pole chain, thus divided, are uisually expressed in perches and tenths. 814 DIMENSIONS OF A SURVEY. [CHAP. I. 4. The Distance of a line in surveying, is its length, estimated in a horizontal direction. It is generally expressed either in chains and links, or in perches and tenths. 5. A Mleridian or Meridian Line is any line that runs due north or south. Note.-All the meridians passing through any survey of moderate extent may be considered as straight lines, parallel to one another.* 6. The Bearing or Course of a line, is the angle which it makes, with a meridian passing through one end; and it is reckoned from the North or South Points of the horizon towards the East or West Points. Fig. 77. Thus, supposing the line NS, Fig. 1 77, to be a meridian, and the angle A..-' SAB to be 500; then the bearing of AB from the point A, is 500 to the east of south; which is usually expressed thus: S. 500 E, and read, -._._B south, fifty degrees east. 7. The Reverse Bearing of a line is the bearing taken from the other end of the line. Note.-The bearing and the reverse bearing of a line, are angles of the same niagnitude,t but lying between * The meridians are, in reality, curve lines which meet in the north and south poles of the earth. No two of them are therefore exactly parallel; but in usual surveys their deviation from parallelism is so very small, that there is no sensible error in considering them so. t As the meridians are not exactly parallel, this is not strictly true, except in a few cases; but the difference is too small to be observed in practice. In CHAP. I.] DIMIENSIONS OF A SURVEY. 85 directly opposite points. Thus, if the bearing of A B, from the end A, is S. 50~ E., the bearing of the same line from the end B, is N. 500 W. 8. A Circumferentor or Surveyor's Compass, is an ilstrument used to take the bearings of lines. The circumference of its face is divided into degrees, and in some of the larger ones into half degrees, in such manner that two opposite points may be exactly in the direction of the sights with which the instrument is fiurnished. These points are the north and south points of' the instrument. Midway between them, on the circumference, are tile east and west points. The degrees are numbered from 0~ to 90~, each way from the north and south points to the east and west ones. In the centre of the face is a pin, finely pointed, which supports a Magnetic Needle, moving freely within the instrument. The instrument, when used, is placed on a staff, having a pointed iron at the bottom, and a ball and socket at the top. The Chain and Compass are the instruments with which the dimensions of surveys in this country are generally taken. It is important to have them accurately made. In the selection of a compass, particular attention should be directed to the settling of the needle. If, when the needle has been moved out of its natural position, it settles very soon, it is defective; either its magnetic virtue is weak, or it does not move with sufficient freedom on the pin. 9. The Difjerence of Latitude, or the Northing or the latitude of Philadelphia the greatest difference between the bearing and reverse bearing of a line, a mile in length, is only 44". In higher latitudes the difference is greater. 8 06 DIMENSIONS OF A SURIVEY. LCHAP. I. Southing of a line, is the distance that one end is further north or south than the other end; or it is the distance which is intercepted on a meridian passing through one end, between this end and a perpendicular to the merilian, from the other end. Thus, if NS, Figr. 77, be a meridian passing through the end A, of the line AB, and Bb be perpendicular to NS, then is Ab the difference of latitude or southing of AB. 10. The Departure or the Easting or Itesting of a line is the distance that one end is further east or west than the other end; or it is the distance from one end, perpendicular to a meridian passing through the other end. Thus Bb, Fig. 77, is the departure or casting of the line AB. But if ns be a meridian, and AC perpendicular to it, and if the bearing of the line be taken from B to A, ther is BC the difference of latitude or northing, and AC thl departure or westing, of the line AB. Note. —It is evident from the definitions, that the Distance, Difference of Latitude, and the Departure form the sides of a right-angled triangle; in which, considering the departure as the base, the perpendicular is the difference of latitude, the hypothenuse is the distance, and the angle at the perpendicular is the bearing. 11. The Meridian Distance of any station, is its distance from a meridian passing through the first station of the survey, or any other assumed point. 12. The Traverse Table, is a table containing the dif CHAP. I.] DIMENSIONS OF A SURVEY. 87 Ferences of latitude and the departures, computed to difIbrent courses and distances. 13. The Area or Content of a tract of land is the horiZontal surface included within its boundaries, expressed ill known measures, as Acles, Roods, and Perches. 14. In going round a tract of land and returning to the place of beginning, it is evident that the whole northing which has been made, must be equal to the southing, and the easting to the westing; or in other words, that the sum of all the nortlhings must be equal to that of the southings, and the sum of the eastings, to that of the westings. This principle enables us to judge of the accuracy of a survey, when the bearings and distances of all the sides have been taken. If the sums of the computed northings and southings are equal, and also those of the eastings and westings; or, if, though not exactly equal, they are very nearly so, we may conclude that the survey has been correctly made; as very small differences in these sums may be imputed to little, unavoidable errors in taking the bearings and measuring the distances. But when the sum of the northings differs considerably from that of the southings, or that of the eastings from that of the westings, we must infer that an error has been made, too great to be admitted. In this case a re-survey should be taken. It is a rule with some of our best practical surveyors, that when the difference between the sums of the northings and southings, called the error in latitude, or that between the sums of the eastings and westings, called the error in departure, exceeds one link for every five chains in the sun of the distances, a re-survey ought to} be made. DIMENSIONS OF A SURVEY. [CHAP. I. \Whlen the errors in latitude and departure fall withir. tile limits just mentioned, they should be properly apportioned among the several latitudes* and departures; we shall thus obtain what are called the corrected latitudes and departures. The method of doing this will be given in one of the foliowing problems. PROBLEi I. To reduce two-pole chains and links to four-pole chains and links. RULE. 1. If the number of chains is even, divide it by 2, and to the quotient annex the given number of links. 2. If the number of chains is odd, divide by 2 as before, for the chains; and for the I that is off, add 50 to the given number of links. EXAMPLES. 1. In 16 two-pole chains and 37 links, how many fourpole chains and links? Ans. 8 ch. 37 links, or 8.37 ch. 2. How many four-pole chains and links are there in 17 two-pole chains and 42 links? Ans. 8.92 ch. 3. How many four-pole chains and links are there in 22 two-pole chains and 7 links? Ans. 11.07 ch. * In order to conciseness of expression, difference of latitude is frequently called simpiy, latitude. CAP. 1.] DIMENSIONS OF A SURVEY. 89 PROBLEM II. To reduce two-pole chains and links to perches and hundredths of a perch. RULE. Multiply the links by 4, for the hundredths, and the chains by 2, for the perches. If the hundredths exceed 100, set down the excess, and add 1 to the perches. Note.-This rule supposes the two-pole chain to be divided into 50 links. EXAMPLES. 1. Reduce 17 two-pole chains and 21 links to perches and hundredths. Ans. 34.84 per. 2. Reduce 15 two-pole chains and 38 links to perches and hundredths. Ans. 31.52 per. 3. Reduce 57 two-pole chains and 49 links to perches and hundredths. Ans. 115.96 per. PROBLEM III. To reduce square four-pole chains to acres. RULE. Divide by 10, and the quotient will be the acres. If there is a decimal in the quotient, multiply it by 4, for the roods; and the decimal of these py 40, for the perches. 8* M 90 DIMENSIONS OF A SURVEY. LCHAP. I. EXAMPLES. 1. Reduce 523.2791 square chains to acres. 10)523.2791 52.32791 4 1.31164 40 12.46560 Ans. 52 ac. 1 r. 12p. 2. Reduce 41.9682 square chains to acres. Ans. 4 ac. Or. 31p. 3. Reduce 132.925 square chains to acres. Ans. 13 ac. 1 r. 6.8p. PROBLEM IV. To reduce acres, roods and perches to square chains. RULE. Divide the perches by 40 and prefix the roods; divide tlhe result by 4 and prefix the acres; then this latter result, multiplied by 10, will give the square chains. Or reduce the given quantity to perches and divide 16. EXAMPLES. 1. Reduce 13 ac. 1 r. 10p. to square chains. 40)10 4)1.25 13.3125 A-133.125 sq. ch. CHAP. 1.] DIMENSIONS OF A SURVEY. 91 2. Reduce 127 ac. 3r. 23p. to square chains. Ans. 1278.9375 sq. ch. 3. Reduce 35 ac. 0 r. 20 p. to square chains. Ans. 351.25 sq. ch. PROBLEM V. To Jfnd the bearing of a line. 1. Let a stake of six or eight feet in length be set up perpendicularly, at the far end of the line. Set up the compass staff perpendicularly, at the beginning of the line, and placing the compass on the staff, adjust it to a horizontal position; the ball and socket admitting a motion for that purpose. This position can be deterruined with sufficient accuracy, by observing whether, when the compass is turned round, the ends of the needle remain at the same height above the face of the instrument. 2. Turn the compass round so as to bring the south end of it towards the stake at the far end of the line Then applying the eye to the sight at the north end, move the compass gently round till the stake can be seen through the fine slits in both sights, and let it remain in this position. 3.'When the needle has settled, observe the number of degrees and parts of a degree, that are intercepted between the south end of the needle and the north or south point of the compass, to whichever it is nearest; which will be the bearing, reckoning it from that point, towrards the east if the south end of the needle is to the right hand, but towards the west if it is to the left hand Note 1.-The bearing thus obtained may be, and should be, verified by going to the far end of the line, 92 DIMENSIONS OF A SURVEY. [CHAP. I and from thence taking the bearing to the first end; which, if both bearings are correct will be the reverse of the former. Note 2.-When there is a fence on the side, or other obstacle in the way, preventing the stake at one end from being seen through the compass sights at the other,end, the bearing may be obtained by setting up the compass and stake at small equal distances to the right or left, so that the line joining them may be parallel to the side. Note 3.-The method of obtaining the bearing between two stations when there are obstacles in the way, which also prevent a parallel bearing being readily taken, or when the stations are too distant to be seen from each other, will be noticed in the next chapter. PROBLEM VI. To measure the distance of a line. For convenience in marking the termination of the chain in measuring, ten iron pins should be provided, about a foot in length, and terminated at top by a small ring, to which a piece of red flannel or other conspicuous substance should be tied, in order that the pins may be readily found, when set up among high grass or in other situations where they would not otherwise be easily discovered. Let the person who is to go foremost in carrying the chain, take nine of the pins in his left hand, and one end of the chain and the other pin in his right hand; then he moving on in the direction of the line, let another person take the other end of the chain and hold it at the beginning of the line. When the leader has moved on till the chain is stretched tight, he must set down the pin, per CHAP. I.] DIMENSIONS OF A SURVEY. 93 pendicularly, exactly at the end of the chain, the hinder chain-man taking care that the chain is in the direction of the line; which is readily determined by observirg whether it is in a range with a stake previously set up at the far end of the line. When the leader has not his end of the chain in the direction of the line, the hinder chain-man can direct him which way to move, by a motion of his left hand. When the distance of one chain or half chain* has been thus determined, the carriers. taking hold of the two ends of the chain, move on till the hinder one comes to the pin which was set up by the other; then the chain buing stretched, the person at the fore end of it sets up another pin as before; the hinder chain-man then taking up the pin at his end, they proceed to a third distance of the chain; and so on. When the person at the fore end of the chain has set up all his pins, he still moves on another length of the chain, and then setting his foot on it to keep it in place, lihe cries " out." The hinder chain man then comes forward, and counts to him the ten pins; and he setting up one of them at the end of the chain, again moves on, dragging the chain after him, till he is checked by the hinder chain-man, who, getting the hind end of the chain, applies it as before to the pin set up. The number of outs should be carefully noticed; each out being ten chains, when a four-pole chain is used, but only five, when the measuring is done, with a two-pole chain. When arrived at the end of the line, the number of pins, which the one at the fore end of the chain has set up since the last out, and the number of links from the last pin to the end of the line, must be carefully noted. From these, and * When a two-pole chain is used, one length of it may properly be called,' alf chain. 94 DIMENSIONS OF A SURVEY. [CHAP. I. the number of outs, the distance measured is readily determined. All slant or inclined surfaces, as the sides of a hill, should be measured horizontally, and not on the plane or surface of the hill. To effect this, the hind end of the chain, in ascending a hill, should be raised from the ground till it is on a level with the fore end, and, by means of a plummet and line, or when the hill is not very steep, by estimation, should be held perpendicularly above the termination of the preceding chain.'In descending a hill, the fore end of the chain should be raised in the same manner, and the plummet being suspended from it will show the commencement of the succeeding chain. PROBLEM VII. Toprotract a Survey, having the bearings and distances of the sides given. The method of doing this will be best understood by an example. Thus, Suppose the following field notes to be given, it is required to protract the survey. Fig. 75. Ch. 7 S X 1. N. 50~ E. 9.60 /i ~ 2. S. 320 E. 16.38 3. S. 41Y W. 6.30 4. West 8.43 "1 b":" /' \ 6. N. 50 E. 11.25 7. s. 830 E. 6.48 I'Id/ M.ethod 1st. Fn./ IDraw NS, Fig. 75, to represent a meridian line; then N standing for the north and cRAP. I.] DIMENSIONS OF A SURVEY. 95 S for the south, the east will be to the right hand, and the west to the left. In NS take any convenient point as A for the place of beginning, and apply the straight edge of the protractor to the line, with the centre to the point A, and the arch turned towards the east, because the first bearing is easterly; then holding the protractor in this position, prick off 500 the first bearing, from the north end, because the bearing is from the north; through this point and the point A, draw the line AB on which lay 9.60 chains, the first distance from A to B. Now apply the centre of the protractor to the point B, with the arch turned toward the east, because the second bearing is easterly, and move it till the line AB produced, cuts the first bearing 500; the straight edge of the protractor will then be parallel to the meridian NS; hold it in this position, and from the south end prick off the second bearing 32~; draw BC and on it lay the second distance 16.38 chains. Proceed in the same manner at each station, observing always, previous to pricking off the succeeding bearing, to have the arch of the protractor turned easterly or westerly, according to that bearing, and to have its straight edge parallel to the meridian; this last may always be done by applying the centre to the station point, and making the preceding distance line produced if necessary, cut the degrees of the preceding bearing. It may also be done by drawing a straight line through each station, parallel to the first meridian. When the survey is correct, and the protraction accurately performed, the end of the last distance will fall on the place of beginning. Method 2d. With the chord of 60~ describe the circle NESW 96 DIMENSIONS OF A SURVEY. [CHAP. I. Fig. 76, and draw the diameter NS. Take the several bearings from the line of chords, and lay them off on the circumference from N or S according as the bearing is northerly or southerly, and towards E or W according as it is easterly or westerly, and number them 1, 2, 3, 4, &c., as in the figure. From A, the centre of the circle, to I draw A 1, on which lay the first distance from A to B; parallel to A 2 draw BC, on which lay the second distance from B to C; parallel to A 3 draw CD, on which lay the third distance from C to D; proceed in the same manner with the other bearings and distances. Fig. 76. EXAMPLE 2. The following field notes are given, to "B' - 1 protract the survey. i,/"'\: i'~ Ch.' a. /',E 1. N. 150 00' E. 20 2. N. 370 30' E. 10 3. East 7.50 4. S. 110 00' E. 12.50 5. South 13.50 6. West 10. 7. S. 360 30' W. 10. 8. N. 380 15'W. 8.50 PROBLEM VIII. T'te bearing of two lines from the same station being given, to find the angle contained between them. RULE. When they run from the same point of the compass, towards the same point, subtract the less from the treater. CHAP. I.] DIMENSIONS OF A SURVEY. 97 When they run from the same point, towards diferent points, add them together. When they run from diferent points, towards the same point, add them together, and take the supplemnent of the sum. When they run from different points, towards different points, subtrhct the less from the greater, and take the sutpplement of the remainder. Note. —When the bearing of one of the lines is given towards the station, instead of from it, take the reverse bearingf of such line; the angle may then be found by the above rule. EXAMPLES. Fig. 67. 1. Given the bearing of the line AB, Fig. 67, N. 340 E., and AD, N. 580 E.; required the angle A. AD, N. 58~ E. / AB, N. 340 E. Angle A =24 o 2. Given the bearing of BA, Fig. 57, S. 34~ W., and' BC, S. 350 E.; required the angle B. Ans. B=690. 3. Given the bearing of BC, Fig. 67, S. 35~0 E., and CD, S. 87~ W.; required the angle C. Ans. 58~. 4. Given the bearing of DC, Fig. 67, N. 87~ E., aIld DA, S. 58~ W.; required the angle D. Ans. 1510~. 9 N 98 DIMENSIONS OF A SURVEY. LCH4P. I. PROBLEM IX. To change the bearings of the sides of a survey in a corresponding manner, so that any pxtrticular one of them may become a Meridian. RULE. Subtract the bearing of the side that is to be made a imeridian, trom those bearings that are between the same points that it is, and also from those that are between points directly opposite to them. If it is greater than either of the bearings from which it is to be subtracted, t.lke the difference, and change E. to t., or W. to E. Add the bearing of the side which is to be made a meridian, to those bearings which are neither between the same points that it is, nor between the points that tare directly opposite to them. If either of the sums exceeds 90~, take the sup'plement and change N. to S., or S. to N.* Note. —W hen the bearings of soame, or all, of the sides of a survey have been thus changed, and by calculation the changed bearing of another side or line has been * The changing of the bearings so as to make a given side become a meridian, may be illustrated by means of a protracted survey. If a protracted survey or plot is held horizontally, with the meridian in a north and south direction, the north end being towards the north, the bearings of the sides of the plot will then correspond with the bearings of the sides of the survey. If then, keeping the paper horizontal, it be turned round till any particular side of the plot has a north and south direction, or becomes a meridian, the bearings of all the other sides of the plot will have been changed by a like quantity. But it is evident, that neither the Relation of the different parts of the. plot to one another, the area nor the lengths of the sides will have been altered by this change. It magy be here observed, that some calculations in surveying are considerably shortened by changing the bearings so as to make a certain side become a meridian. The method was communicated to me by Robert Patterson, late Professor of Mlathemnatics and Nalur'tal _Philosophy in the t-nicersity of Pennsylvania. CHAP. I.] DIMENSIONS OF A SURVEY. 99 found, its true bearing will be obtained by applying to the changed bearing, the bearing of the side which was made a meridian, in a contrary manner to what is directed in the rule; that is, by adding in the case in which the rule directs to subtract, and by subtracting in the case in which it directs to add. EXAMPLES. 1. Given the bearings of the sides of a survey as follow; 1st. S. 45~0 W.; 2d. N. 50~ W.; 3d. North; 4th. N. 850 E.; 5th. S. 47~ E.; 6th. S. 20i~ W.; and 7th. N. 514~ W. Required the changed bearings, so that the 5th side may be a meridian. 1st. S. 45i~ dW. 47 92i J~ncUCI;L'i~~ 180 chang. bear. N. 87]~ W. 2d. N. 500 WV. 47 chang. bear. N. 30 W. 3d. N. 0o E. 47 chang. bear. N. 470 E. 4th. N. 85~ E. 47 132 180 chang. bear. S. 480 E. 100 DIMENSIONS OF A SUItVEY. [CPHM. I. 5th. side, changed bearing, South. 6th. S. 204~ W. 47 chang. bear. S. 674 7th. N. 510 W. 47 chang bear. N. 44 W. 2. Given the following bearings of the sides of a survey; 1st. S. 40~0 E.; 2d. N. 540 E.; 3d. N. 29~~ E.; 4tll. N. 280~ E.; 5th. N. 57~ W.; and 6th. S. 47~ W.; to find the changed bearings so that the 2d. side may be a meridian. Ans. 1st. N. 8540 E.; 2d. North; 3d. N! 44]\WV.; 4th. N. 25i~ W.; 5th. S. 690 W.; 6th. S. 7~ E. 3. Given the bearings as in the 1st. example; viz. 1st. S. 452~ W.; 2d. N. 50~ W.; 3d. North; 4th. N. 85~ E.; 5th. S. 47~ E.; 6th. S. 20~0 V; 7th. N. 51~~ W.; to find the changed bearings so that the 6th side may be a meridian. Ans. 1st. S. 25~ XV.; 2d. N. 700~ WV.; 3d. N. 920~ XV.; 4th N. 64~~ E.; 5th. S. 67~0 E.; 6th. South 7th. N. 71 i W. PROBLEM X. Of the bearing, Distance, Difference of Latitude and Departure, any two being given, tofind the other two. RULE. IWFen the bearing and distcance are given. As Rad.: cos. of bearing: distance: dif. of latitude. 1tad. aila. of bearing- distance;,leparture. CHAP. I.] DIMENSIONS OF A SURVEY. 101 Whlen thle bearing and cifference of latitude are given. As Rad.: sec. of bearing:: diff. lat.: distance. Rad.: tang. of bearing:: diff. lat.: departure. When the bearing and departure are given. As Rad.: cosec. of bearing:: departure: distance. Rad.: cotang. of bearing:: departure: diff. lat. lWhen the direrwen of latitude and the departure are given. As diff. lat.: departure:: rad.: tang. of bearing. Rad.: sec. of bearing:: diff. lat.: distance. When the distance and dfference of latitude are given. As Diff. lat.: distance:: rad.: see. of bearing. Rad.: tang. of bearing:: diff. lat.: departure. When the distance and departure are given. As Distance: departure:: rad.: sin. of bearing. tad.: cos. of bearing:: distance: diff. lat. Note.-It is evident the above proportions are the solutions of a right-angled triangle, having for its sides the distance, difference of latitude, and departure. EXAMPLES. I Given the bearing of a line, N. 53~ 20( E., distance 13.25 ch.; to find the difference of latitude and the departure. Ans. Diff. lat. 7.91 N.: dcp. 10.63 E. 9 * 102 DIMENSIONS OF A SURVEY. [CHAP I 2. Given the bearing of a line, S. 32~ 30' E., and the departure 10.96 ch. to find the distance and difference of latitude. Ans. Dist. 20.40 ch.; diff. lat. 17.20 S. 3. Given the distance of a line, running between the north and east, 44 ch. and its difference of latitude 34.43 clh.; to find the bearing and departure. Ans. Bearing, N. 38~ 30' E.; dep. 27.39 ch. E. 4. The bearing of a line S. 320 30' E., and the difference of latitude 17.21 ch. being given, to find the distance and departure. Ans. Dist. 20.41 ch.; dep. 10.96 E. 5. Given the difference of latitude of a line 27.92 N., hald the departure 5.32 E.; to find the bearing and distance. Ans. Bearing, N. 10~ 47' E.; dist. 28.42. 6. The distance of a line, running between the north and west, is 35.35 ch., and its departure 15.08 ch., required the bearing and difference of latitude. Ans. Bearing N. 25~ 15' W.; diff. lat. 31.97 N. PROBLEM Xl. Tofind the difference of latitude and departure correspond. ing to any given bearing and distance, by means of the Traverse >able. RULE. When the distance is any number of whole chains or perches,?not exceeding 100. Find the given bearing at the top or lottom of the table, according as it is less or more than 45~. Then against cHAP. I.J DIMENSIONS OF A SURVEY. 103 the given distance, found in the column of distances at tile side of the table, and under or over the given bearing, is the difference of latitude and departure; which must be taken as marked at the top of the table, when the bearing is at the top; but as marked at the bottom, vllhen the bearing is at the bottom..Wihen the distance is a number of whole chains or perches, exceeding 100. Separate the distance into parts that shall not exceed 100 each; and find, as before, the difference of latitude and departure, corresponding to the given bearing and to each of those parts; the sums of these will be the dif frence of latitude and departure required. WYhen the distance is expressed by chains or perches arnd the decinwl of a chain or perch. Find, as above, the difference of latitude and departure corresponding to the given bearing and to the whole chain or perches. Then considering the decimal part as a whole number, find the difference of latitude and departure corresponding to it, and remove the decimal point in each of them, two figures to the left hand if thcre gare two decimal figures in the distance, or one figure to the left if there is but one; then these added to the former will give the difference of latitude and departure required. Note. When the number of whole chains or perches is less than 10, and the second decimal figure is a cipher,,he difference of latitude and departure may be taken out at once, by considering the mixed number, rejecting 104 DIMENSIONS OF A SURVEY. [CHAP. I. the cipher, as a whole number. The difference of latitude and departure thus found, must have the decimal I)oint in each, removed one figure to the left hand. EXAMPLES. 1. Given the bearing of a line S. 354~ E., dist. 79 ch.; required the difference of latitude and departure by the traverse table. Ans. Diff. lat. 64.51 S.; dep. 45.59 E. 2. A line bears N. 202~ E., 117 ch.; required the difference of latitude and the departure. Dist. 100, gives diff. lat. 93.67 and dep. 35.02 17 15.92 5.95 Whole dist. 117 diff. lat. 109.59 N. dep. 40.97 E. 3. Required the difference of latitude and the departure of a line which bears, S. 414~ W., 57.36 ch. Dist. 57.00 gives diff. lat. 42.53 and dep. 37.96 36.27.24 Whole dist. 57.36 diff. lat. 42.80 S. dep. 38.20 W. 4 Required the difference of latitude and departure of a line which bears, N. 72~ W., 124.37 ch. Dist. 100.0 gives diff. lat. 30.90 and dep. 95.11 24.00 7.42 22.83.37.11.35 Whole dist. 124.37 diff. lat. 38.43 N. dep. 118.29 W. CHAP. I.] DImZENSlONS OF A SURVEY. 105 5. Given the bearing and distance of a line, N. 39i~ WV. 15.20 ch., to find its difference of latitude and departure. Ans. Diff. lat. 11.72 N., and dep. 9.67 W. 6. The bearing and distance of a line are N. 46~ E., 27.25 ch.; required its difference of latitude and departure. Ans. Diff. lat. 18.93 N. and dep. 19.60 E. 7. The bearing and distance of a line are S. 37~~ W., 137.50 ch.; required its difference of latitude and departure. Ans. Diff. lat. 109.45 S., and dep. 83.23 W. 8. Required the difference of latitude and departure of a line, whose bearing and distance are S. 6.~ E., 5.60 ch. Ans. Diff. lat. 5.56 S., and dep. 0.63 E. ROBLEM XII. Given the bearengs and distances of all the sides of a tract of land to obtain the corrected latitudes and departures. RULE. 1. Rule a table as in the annexed example, in the first vertical column of which, place the letters designating the sides, or the numbers denoting the stations at the beginning of each side; in the second column, place the )bearings; and, in the third, the distances. 2. Find, by the last problem, the difference of latitude and the departure, corresponding to each side, and place them in the next four columns, under their proper heads of N. or S., E. or W. Add up the nqrthings and southings; and if the sums are not equal, find their difference 0 106 DIMKENSIONS OF A SURVE Y. [CHAP. I. which will be the error of the survey ill difference of latitude; which call by the same name as the least sum. Proceed in the same manner with the eastings and Nvest. aigs, and find the error in departure. Also add up the column of distances. Then it will be, As the sum of the distances, Is to any particular distance, So is the error in lati ude or departure To the correction of latitude or departure, correspondingr to that distance. 3. Find, by the above proportion, the corrections of latitude and departure corresponding to all the sides calculating them to the nearest two decimal figures, and place them in the next two columns, heading them with the same names as the errors in latitude and departure. If the sums of these corrections, are not respectively equal to the errors in latitude and departure, which, in consequence of the fractions neglected, will sometimes be the case, alter some of them by a unit in the second decimal figure, to make them so. 4. Apply these corrections to their corresponding differences of latitude and departures, by adding when of the same name, but by subtracting when of different names, and the corrected differences of latitude and departures will be obtained; which may be placed in the four succeeding columns. In these the sums of the northin(gs and southings will be equal, and also those of the eastings and westings.* * The directions given in the rule, for correcting the errors in difference of latitude and departure, are deduced from the rule given and demonstrated in No. 4, of a periodical work, called the Analyst, by Nathaniel Bowditch, A. M., and also by the editor, Professor Adrain. The demonstration is too long, and not of a nature for insertion here. CHAP. 1.] DIMENSIONS OF A SURVEY. 107 Note 1.-In the proportion for finding the correction of the latitude or departure, the decimal parts of the sum of the distances and of the particular distance may be omitted, taking, in each case, the nearest number of whole chains. 2. The corrections may be frequently estimated with sufficient accuracy without the trouble of working out the proportions. 3. When one or two of the sides are hilly, or when there are other difficulties in the way of obtaining their bearing or distances with accuracy, it is better to allow a. considerable part of the errors, on the latitudes and departures corresponding to them, and afterwards to apportion the remaining part among the others. EXAMPLES. 1. Given the bearings and distances of the sides of (a tract of land as follow: 1 st. S. 401 E. 31.80 ch.; 2nd. N. 54~ E. 2.08 ch.; 3rd. N. 294~ E. 2.21 ch.; 4th. N. S`28 E. 35.35 ch.; 5th. N. 570 W. 21.10 ch.; 6tll. S. 47~ WI. 31.30 ch. Required the corrected differences of latitude and departures. Sta. Cour. E. 5.3L.1E.D. W. D. r. D. W.D.1 Corses.| Ch. N. L. S. L. BE. IN. L.S. L.E. D1WD. 1S. 402 E. 31.80 24.18 20.65.03.05 24.2120.70 t2 N. 54 E. 2.08 1. 23 1.68.00.00 1.23 1.68 i3 N. 291 E. 2.21 1.92 1.08.00.00 1.92 1.08 4 N. 283 E. 35.35 31.00 17.00.04.05 30.96 11 7.05 5 N. 57 W. 21.10 11.49 17.69.02.03 11.4_71 17.66 16 S.47. W. 31.30 21.34 22.89.03.04 21.371 22.85 123.845.6445.52 40.41 40.5812 17 45.58 45.5S.4().51 40.51 45.52 40.41 12 Er. S..17 Er. E 108 DIMENSIONS OF A SURVEY. [CHAP. I. As 124: 32::.12:.03 As 124: 32::.17:.04 or.05* 124: 2::.12:.00 124: 2::.17:.00 24: 35::.12:.03or.04 124: 35::.17:.05 124: 21::.12:.02 124: 21:.17:.03 124: 31::.12:.03 124: 31::.17:.04 2. Given the bearings and distances of the sides of a tract of land as follow: 1st. N. 75~ E. 13.70 ch.; 2d. N. 202 E. 10.30 ch.; 3d. East 16.20 ch.; 4th. S. 33i~ W. 35.30 chl.; 5th. S. 76 W. 16 ch.; 6th. North 9 ch.; 7th. S. 840 W. 11.60 ch.; 8th. N. 5314~ W. 11.60 ch.; 9th. N. 36~0 E. 19.36 ch.; 10th. N. 221~ E. 14 ch.; l1th. S. 76~0 E. 12 ch.; 12th. S. 15~ W. 10.85 ch.; 13th. S. 18~ W. 1'0.62 ch.; to the place of beginning. Required the corrected latitudes and departures. Ans. 1st. 3.56 N. 13.26 E.; 2d. 9.66 N. 3.62 E.; 3d. 0.02 N. 16.22 E.; 4th. 29.39 S. 19.44 W.; 5th. 3.85 S. 15.50 W.; 6th. 9.01 N. 0.01 E.; 7th. 1.19 S. 11.52 W.; 8th. 6.96 N. 9.27 XV.; 9th. 15.54 N. 11.61 E.; 10tlh. 12.95 N. 5.38 E.; 11th. 2.73 S. 11.70 E.; 12th. 10.46 S. 2.80 W.; 13th. 10.08 S. 3.27 W. * When, as in this case, the correction is found to be nearly midway be. tween two numbers, it is best to note them both. Then, if in using the one that is nearest to the true value, the sum of the corrections does not equal the whole error, the other should be taken. CHAP. II.] SUPPLYING OMISSIONS. 109 CHAPTER II. On supplying omissions in the dimensions of a survey. When the bearings and distances of all the sides of a survey are known, except one bearing and one distance, or two bearings, or two distances, these can be obtained by calculation, provided those that are known can be depended on, as sufficiently accurate. This may sometimes be necessary when there are obstacles in the way of obtaining one or two of the bearings or distances; or when, after they have all been taken on the ground, the notes of one or two of them happen to be obliterated. As, however a bearing, or distance thus. obtained, must be affected by any error or errors that may have been made in taking the otthers, it is better, when practicable, to have the bearings and distances of all the sides, as taken on the ground. PROBLEM I. The bearings and distances of all the sides o a tract of land, except the bearing and distance of one side, being given, toJind these. RULE. Find by prob. 11, of the preceding chapter, the differences of latitude and the departures for the sides whose bearings and distances are given, and place them in their proper columns in a table ruled for the purpose: tdd up the northings and soutllins, and taking the dif10 110 SUPPLYING OMISSIONS. VCHAP. I1. ference of their sums, place it opposite the unknown side, in the column whose sum is the least. The sums of the two columns will then be equal. This is called balancing the latitudes. Do the same with the eastings and westings. The two numbers inserted to make the latitudes and the departures balance, will be the diffeience of latitude and the departure of the unknown side: with which its bearing and distance may be found, by prob. 10, of the preceding chapter. Note 1.- By the application of this rule, the bearing and distance of a line joining two corners or stations, may be found, when there are obstacles in the way which prevent our going directly from one corner to the other, or when one cannot be seen from the other. To do this, let one or two, or more stations, if necessary, be taken out of the line, and take the bearing and distance from the first corner to the first assumed station; from this station to the second; and so on, to the second corner. Then considering these bearings and distances, as the bearings and distances of the sides of a survey, the required bearing and distance of the line may be found by the above rule. The bearing thus found must be reversed, in order to have the bearing from the first corner to the second. 2. In the same way the bearing and distance of a straight road to run between two given places, may be found, by taking the several bearings and distances of the old road if there is one; or of lines joining assumed stations and extending from one of the places to the other. EXAMPLES. 1. The bearings and distances of the side of a tract of' land, except the bearing and distance of one side whichl CHAP. 1I.] SUPPLYING OMI.SSIONS. 111 aire not known, are as in the following field-notes; required the unknown bearing and distance. Chains. Chains. 1. S. 450A~ W. 15.16 5. - 2. N. 500 W. 22.10 6. S. 20 MWro. 23.80 3. North 18.83 7 N. 514 W~. 26.47 4. N. 850 E. 35.65 Sta. Bearings. Dist. N. S. E. W. 1 S. 452 W. 15.16 10.5.162 10.81 2 N. 50 W. 22.10 14.20 16.93 3 North 18.83 18.83 4 N. 85 E. 35.65 3.11 35.52 5 (19.79) (21.20) 6 S. 20 W. 23.80 22.29 8.33 7 N. 511 W.126.47 16.56 20.65 52.70 52.70 56.72 56.72 As diff. of lat. 19.79 S. Ar. Co. 8.70355: dep. 21.20 E. - - - 1.32674 rad. - 10.00000: tang. bear. S. 470 E. - - -10.03029 As rad. - - -10.00000: sec. bearing 47- - - - - 10.16622::diff. lat. 19.79 - - - - 1.29645: dist. - - 29.02 - - - - 1.46267 Ans. S. 47~ E. 29.02 ch. 1 12 SUPPLYING OMISSIONS. LCHAP. II. 2. Given tile bearings and distances of the sides of a tract of land, as follow: 1st. N. 154 W 9.40 ch.; 2d N. 63* E. 10.43 ch.; 3d. S. 490 E. 8.12 ch.; 4th. S. 132 E. 8.45 ch.; 5th. S. 163 E. 6.44 ch.; 6th. Unknown; 7th. N. (0~ W. 9.72 ch.; 8th. N. 174 E. 7.65 ch.; required the bearing and distance of the 6th. side. Ans. S. 600 8' WV. 12.27 ch. 3. One side of a tract of land of which a survey is to be taken, passes through a pond. Two stations are therefore taken on one side of the pond as represented in Fig. 80. The bearings and distances from the first end of the side to the first station, from that to the second, and thence to the other end of the side are; 1st. S. 52~ W. 10.70 ch.; 2d. S. 7~~ W. 13.92 ch.; and 3d. -S. 341 E. 9 ch. Required the bearing and distance of the side. Ans. S. 10~ 33' W. 28.31 ch. 4. Given the bearings and distances of an old road, running between two places, as follow; 1st. S. 10~ E. 92.20 ch.; 2d. S. 15~ W. 120.50 ch.; 3d. S. 185 W. 205. ch.; 4th. S. 711 E. 68 ch. Required the bearing and distance of a straight road, that shall connect the two places. Ans. S. 20 8' W. 423.47 ch. PROBLEM II. rGiven all the bearings and distances of the sides of a survey, except the distances of two sides, to find these. RULE. By prob. 9, of the preceding chapter, change all the riven bearings, in a corresponding manner, so that one of the sides whose bearings only are given, may become a meridian. With the changed bearings and given distances find the corresponding differences of latitude an(1 the departures. Add tlp the eastings and westinls, and take the difference of their sums, which will be thl} CHAP. i i.] SUPPLY.INL, OMISSIONS. II 3 departure of that unknown side, which is not made a meridian. With this departure and the changed bearing, find by prob. 10, of the preceding chapter, the distance and difference of latitude of this side, which place ii; their proper columns. Now add up the northings and southings, and take their difference, which will be tile distance of the side made a meridian.* EXAMPLES. Given tile following bearings and distances of the sides of a survey; 1st. S. 45~ W. 15.16 ch.; 2d. N. 50~ W. 22.10 ch.; 3d. North 18.83 ch.; 4th. N. 850 E. 35.65 ch.; 5th. S. 470 E. dist. unknown; 6th. S. 201 W. dist. munknown- 7th. N. 51~ W. 26.47 ch. to the place of be& ginning. Required the unknown distances. _ _ 3. JDist. N. S. E.___ S. 45'0W. N. 87O NV. 15.16 0.66 15.1.) 2 N. 50 W. N 3 W. 3W 22.10. 1.16 3 North N. 47. 18.83 12.85 13.77 4 N. 85. S. 48 E. 35.65 23.85 26.49 5 S. 47 E. South (29.02) (29.02) 6 S. 20 W. S. 67 W. (23.80) (9.11) (21.99) N. 51k W. N. 41 W. 26.47 26.401 1.96 61.98 61.98j 40.26 40.26 * The reason of the rule is obvious. For as the side made a meridian h]as no departure, the difibrence of the sums of the departures, nmust be the departure of the other unknown side. And when the difibrence of latitude of this side has been found and placed in its proper situ-ition, the difference o!' the sumins of the latitudes must evidently be the difference of latitude of' tl,, side made a mneridian; or which, in this case, is the same thing, its distance. 10' P' 114 SUPPLYING OMISSIONS. [CHAP. H. As rad. - 10.00000: cosec. chang. bearing 67W~ - - 10.03438:: dep. - - - - - 21.99 1.34223: Dist. 6th side - - 23.80 - 1.37661 As rad. -10.00000: cotang. chang. bearing 671 - - 9.61722:: dep. - - - - - 21.99 - 1.34223: diff. lat. 6th. side - - 9.11 - 0.95945 Ans. 5th. side 29.02 ch. and 6th. side 23.80 ch. 2. Given the bearings and distances of a tract of land as follow: I st. S. 401 E. 31.80 ch.; 2d. N. 540 E. dist. unknown; 3d. N. 29' E. 2.21 ch.; 4th. N. 283 E. 35.35 ch.; 5th. N. 57~ W. dist. unknown; 6th. S. 470 WV. 31.30 ch.; to the place of beginning. Required the distances of the 2d. and 5th. sides. Ans. 2d. side. 2.08,ch. and 5th. side 20.90 ch. PROBLEM III. Given the bearings and distances of all the sides of a survey except two; one of which has only its bearing given, and the other, the distance and the points of the compass between which it runs; tofind the unknown bearing and distance. RULE. As in the last problem, change all the given bearings, so that the side whose bearing only is given, may become a meridian. Find the differences of latitude and the departures, corresponding to the changed bearings and the given distances. Take the difference of the sums of the eastings and westings, which will be the de CHAP. II.] SUPPLYING OMISSIONS. 115' parture of the side whose bearing is not given With the given distance and this departure, find by chap. 1. prob. 10. the changed bearing and difference of latitude. and place them in their proper columns. From the changed bearing, the true bearing may be readily found by note to prob. 9. chap. 1. Lastly, take the difference of the sums of the northings and southings, and it will be the distance of the side, changed to a meridian. Note.-The changed bearing as found by the rule, must be reckoned from the north, or the south point of the compass, according as the one, or the other, will render the true bearing when found from it, conformable to the given points. The point from which the changed bearing must be reckoned determines also the column in which the difference of latitude must be placed. Sometimes the changed bearing when reckoned from either north or south, will render the true bearing conformable to the given points. In such cases, there are two different bearings and distances that will answer the conditions of the problem; and we can only know which of them is the right one by previously knowing the required bearing nearly. EXAMPLES. 1. Given the bearings and distances of a survey as follow: 1st. S. unknown W. 15.16 ch.; 2d. N. 50~ CT. 22.10 ch.; 3d. N. 18.83 ch.; 4th. N. 85~ E. 35.65 ch.; 5th. S. 470 E. 29.02 ch.; 6th. S. 201 W. dist. unknown; 7th. N. 51*~ W. 26.47 ch. Required the unknown bearing and distance. 116' SUPPLYING OMISSIONS. [CHAP. 11. Chan ged Sta. Bearings. bearings. Dist. N. S. E. W. 1 S. (45~30 36') W. (S. 25~ 6' W.) 15.16 I 13.73 (6.43) 2 N. 50 W. N. 70 W. 22.10 7.37 20.831 North N. 201 W. 18.83 17.641 6.59 N. 85 E. N. 641 E. 35.6 15.35 32.18 | 5 S. 47 E. S. 671 E. 29.02 11.11 26.81 6 |S. 201 W. South (23.81) (23.81) _ N._51W. N.. 71 W. 2647 8.29-5 48.65 48.65 58.99 1 58.9)9 As dist. 1st. side 15.16 Ar. Co. 8.81930: dep. do. 6.43 0.80821 r: rad. - 10.00000: sin. chan. bear. 25 6' - - - 9.62751 As rad. - - 10.00000: cos. chang. bearing 250 6' - - 9.95692: dist. - 15.1 6 - - 1.18070: diff. lat. - - - - 13.73 - - 1.13762 Ans. 1st. S. 450 36' VW.; 6th. 23.81 ch. 2. Given the following bearings and distances of a survey: 1lst. S. 401~ E. 31.80 ch.; 2d. N. 540 E. dist. unknown 3rd. N. 29~I E. 2.21 ch.; 4th. N. unknown E. 35.35 ch.; 5th. N. 57~ WV. 20.90 ch.; 6th. S. 47~ vW. 31.30 chi; to place of becinIling. Required the bearing Cf the 4th. side and distance of the 2d. side. Ansu. Bearing, of 4th. side N. 28{~ E.; dist. of 2d. sidle, 2.09 ch. NCAP. I.] SUPPLYING OMISSIONS. 117 PROBLEM IV. Given all the bearings and distances of the sides of a tract. of land, except the bearings of two sides, to fjid these bearings. RULE. 1. Find the difference of latitude and the departure of each side, whose bearing and distance are both given. Take the difference of the sums of the northings and southings of these sides, and also the difference of the sums of the eastings and westings. These differences will be the difference of latitude, and the departure of a line, which, with those sides, would form a closing survey; and which may therefore be called a closing line. 2. With the difference of latitude and departure of the closing line, find, by prob. 10. chap. 1, its bearing and distance. Take the closing line and the two sides whose bearings are not given, for the three sides of a triangle, and calculate the angles. 3. To the bearing of the closing line, apply, by addition or subtraction, as the case may require, the angle contained between it, and the sid(e which is the one coming first in the order of the survey; and it will give the bearing of that side. Then to the reverse bearing of that side, apply in a proper marnner, the angle contained between the two sides which are sides of the survey, and it will give the bearing of the second of those sides.* * It is easy to see the reason of the r.ule, by consideri-ng that the two sides wipse bearings are not given, bcing made to ftrm xwit~h thle c-losing line, thle i 18 SUPPLYING OMISSIONS. [CHAP. II. EXAMPLES. 1. Given the bearings and distances of the sides of a tract of land as follow: 1st. S. unknown V. 15.16 ch.; 2d. N. 500 W. 22.10ch.; 3d. North 18.83 ch.; 4th. N. 850 E. 35.65 ch.; 5th. S. unknown E. 29.02 ch.; 6th S 20o~ W. 23.80 clh.; and 7th. N. 51 WV. 26.47. ch. Re quired the unknown bearings. Sta. Bearings. Dist. N. S. E. W. 1 S. W. 15.16 2 N. 500 W. 22.10 14.20 16.93 3 North 18.83 18.83 4 N.85 E. 35.65 3.11 35.52. ~ ~..Ei. _35.52 5 S. E. 29.02 6 S. 20 W. 23.80 122.29 8.33 7 N. 514 W.26.47 16.56 20.65 52.70 22.29 35.52 45.91 22.29 35.52 30.41 S. 10.39 E sides of a triangle, the sum or difference of their ditIerences of latitude, wil! necessarily be equal to the difference of latitude of the closing line; and that, therefore, their-differences of latitude will be such as to make the sums of the northings and southings of the whole survey equal; and the same for the departures. CHAP. II.] SUPPLYING OMISSIONS. 119 As diff. lat. - - - 30.41 S. Ar. Co. 8.51698: dep. - - - - 10.39 E. 1.01662:: rad. - - 10.00000 ~ tang. of bear. of clos. line, S. 18~ 52' 9.53360 As rad. 10.00000 diff. lat. - - 30.41 - - 1.48302:: sec. of bear. of cldos. line - 180 52' - 10.02398 dist. of clos. line - - - 32.14 1.50700 Fig. 78. Let DE, Fig. 78, represent the closing line, DF, the 1st side of the survey, and FE, the 5th side. Then, DE 32.14 DF 15.16 Ar. Co. 8.81930 FE 29.02 8.53730 2)76.32 Half sum 38.16 log. 1.58161 E$ tRem. 6.02 - 0.77960 Cos.' F 430 44' 2)19.71781 F 87 28 9.85890 As DE 32.14 Ar. Co. 8.49295 FE 29.02 1.46270 sin. F 87 28' 9.99958 sin. D 640 26' 9.95523 DE, S. 180 52' E. FD, N. 450 34' E. Angle D -64 26 Angle F 87 28 1st side, S. 45 34 W. 133 2 180 00 5th side, S. 46 58 E. 120 CONTEN-,T OF LAND. [cHAP. III. 2. Given the bealrings and distances of the sides of a tract of land as follow: 1st. S. ukn7ceown E. 31.80 ch.; 2d. N. 540~ 1. 2.08 ch.; 3d. N. 291 E. 2.21 ch.; 4th. N. 284~ E. 35.35 ch.; 5th. N. 57~ W. 20.90 chl.; and 6th. S unklnown W. 31.30 ch. to the place of beginning. Required the unknown bearings'. Ans. 1st. S. 40~ 29' E.; and 6th. S. 470 W. CIIAPTER III. Problems for finding the Conteit of Land. When the sides of a survey are right lines, and all the? bearings and distances are given, the area may be found by a problem that will be given in this chapter. If one or two of the bearings or distances are not known, they may be found by the problems in the last chapter. Although the problem alluded to, is general, and may be applied whatever number of sides there may be, yet there are some particular rules for finding the areas of triangles and quadrilaterals, which are often useful. These, and also rules for finding the areas of circles and ellipses, are given in the first part of the chapter. When a part of the boundary of a tract of land, is irregular, as is frequently the case, if one or more of the sides are bounded by water, it is sometimes very troublesome to take all the bearings and distances requisite to obtain the area with accuracy. In these cases, it is usual to run one or more straight lines, called stationary, lines, near to such boundary, and so as to connect thle straight sides of the survey. In measuring these sttationary lines, perpendicular distances are measured fromn them, to each bend in the irregular boundary. These perpendicular distances are called ob-sets. The lengths cHAP. Ill.j CONTENT OF LAND. 121 of the off-sets, and the distance of the foot of each, from the commencement of the stationary line, should be carefully noted in the field book; observing also that slch a number of off-sets should be taken, that the part of the irregular boundary intercepted between each adjacent two, may, withoutmaterial error, be considered a straight line. From these notes, the area or areas of the land contained between the stationary line or lines, and the irregular boundary, may readily be calculated. This area added to the area enclosed by the stationary lines, and straight sides of the survey, when they are on the outside of the stationary lines, or subtracted from it, when oil the inside, will give the area of the survey. In those cases in which water is a boundary of a tract of land, if that water is a brook or rivulet, it is usual to consider a line running through its middle as the true boundary; and the off-sets must be measured accordingly. oWhen tide water is the boundary, the land is considered as extending to the line of low water mark. If the bearings of all the corners of a tract of land from two stations, taken either within or out of the tract are given, and also the bearing and distance of these stations from each other, the area may be calculated. It is however necessary, that the two stations should be so taken that they shall not be in a straight line, or very nearly in a straight line, with either of the corners of the land. This method of obtaining the area, though not practically so accurate as where the bearings and (distances of the sides are correctly given, nlay sometimes be found useful. Some surveyors, in order to calculate the area of a survey, first protract it; then dividing the plot into tri11 Q 122 CONTENT OF LAND. [CHAP. III. angles and trapeziums by lines joining opposite corners, they measure with the scale and dividers the lengths of such lines and perpendiculars as are requisite for calculating the areas of these. The sum of the areas thus obtained, is the area of the survey. When the survey is carefully protracted, and proper attention is given to take the measures with the utmost precision, this method serves to give a near value of the content; but is by no means to be depended on as equally accurate with the general problems mentioned above. The area of a field or small tract of land, the corners of which can be seen from one another, may readily be found by means of the chain only. To do this, the lengths of the sides must be measured, and also the length of diagonals joining opposite corners, so as to divide the field into triangles. Or instead of the diagonals, the distances from some assumed point within the field, to the several corners, may be used. Having then all the sides of the several triangles, the area of each may be found; and the sum of these areas will be the area of the tract. PROBLEM I. To find the area of a Parallelogram, whether it be a Square, a Rectangle, a Rhombus, or a Rhomboides. RULE. Multiply the length by the height or perpendicular breadth, and the product will be the area.* Fig. 68. * DEMONSTRATION. Let ABCD (Fig. 68) be a rectangle; and let its length AB and CD, L _1 and its breadth AD and BC, be each divided into as many equal parts, as are expressed by the number of times they contain the A I I l I l l 8 lineal measuring unit; and let all the opposite points of division be connected by right CHAP. III.] CONTENT OF LAND. 123 Note.-Because the length of a square is equal to its height, its area will be found by multiplying the side by itself. EXAMPLES. 1. Required the area of a square field, a side of which measures 7.29 four-pole chains. 7.29 Ch. 7.29 6561 1458 5103 10)53.1441 Area 5A. 1 R. 10P. 5.31441 4 1.25764 40 10.30560 2. Required the area of a rectangular field whose length is 13.75 chains, and breadth 9.5 chains. lines. Then, it is evident that these lines divide the rectangle into a number of squares, each equal to the superficial measuring unit; and that the number of these squares is equal to the number of lineal measuring units in the length, as often repeated as there are lineal measuring units in tlhe breadth, or height; that is, equal to the length multiplied by the breadth. But the area is equal to the number of squares or superficial measuring units; and therefore the area of a rectangle is equal to the product of the length and breadth. Again, a rectangle is equal to any oblique parallelogram (f an equal length and perpendicular height (36.1;) therefore the area of every parallelogram is equal to the product of its length and height. 124 CONTENT OF LAND. [CHAP. IIL 13.75 Ch. 9.5 6875 12375 10)130.625 Area 13 A. 0 R. o10 P 13.0625 4.2500 40 10.000 3. Required the area of a field, in the form of a rhomboides, whose length AB is 42.5 perches, and perpendicular breadth CD is 32 perches. Fig. 15. 42.5 P. 32 850 1275 410)13610.0 4)34 8A. 2R. 4. What is the area of a square tract of land whose side measures 176.4 perches? Ans. 194 A. 1 R. 36.96 P. 5. What is the area of a rectangular plantation whose length is 52.25 chains, and breadth 38.24 chains? Ans. 199 A. 3 R. 8.6 P. 6. The length of a field, in the form of a rhombus, measures 16.54 chains, and the perpendicular breadth 12.37 chains: required the area. Ans. 20 A. 1 R. 33.6 P. CHAP. 111.] CONTENT OF LAND. 125 7. Required the area of a field in the fornl of a rhomboides, whose length is 21.16 chains, and perpendicular ureadth 11.32 chains. Ans. 23 A. 3 R. 32.5 P. PROBLEM II. bTofind the area of a triangle when the base and perpendicular height are given. RULE. Multiply the base by the perpendicular height, and half the product will be the area.* EXAMPLES. 1. Thle base AB of a triangular piece of ground, measures 12.38 chains, and the perpendicular CD 6.78 chains: required the area. Fig. 49. 12.38 Ch. 6.78 9904 F666 7428 2)83.9364 10)41.9682 Area, 4 A. 0 R. 31 P. 4.19682 4.78728 40 3 1.491 20 * DEMONSTRATION. A triangle is half a parallelogram of the same base and altitude, (41.1) and therer)re the truth of the rule is evident. 11* 126 CONTENT OF LAND. [CHAP. II!. 2. Required the area of a triangular field, one side of which measures 18.37 chains, and the distance from this side to the opposite angle, 13.44 chains. Ans. 12A. IR. 15P. 3. What is the area of a triangle whose base is 4(9 perches and height 34 perches? Ans. 5 A. 0 R. 33 P PROBLEM III. Tofind the area of a triangle when two sides and thezr included angle are given. RULE. As radius, Is to the sine of the included angle; So is the rectangle of the given sides, To double the area.* EXAMPLES. 1. In a triangular lot of ground ABC, the side AB measures 64 perches, the side AC 40.5 perches, and their contained angle CAB 30~: required the area. Fig. 49. * DEMONSTRATION. In the triangle ABC, Fig. 49, let AB and AC be the given sides, including the given angle A, and let CD be perpendicular on AB. Then by trig. rad.: sin. A:: AC: CD;bllt (cor. 1.6) AC: CD:: AC XAB; CD x AB; therefore (11..5) rad.: sin. A:: AC x AB: CD x AB; but CD x AB is equal to twice the area of the triangle: hence the truth of the rule is evident. CHAP. III.] CONTENT OF LAND. 127 As radius -10.00000 Is to sin. A, 30~ - 9.69897 S64- - 1.80618 So is AB, AC 640 - 1. 60746 40.5 1.60746 13.11261 To double the area 1296 perches 3.11261 410)6418 4)16 8 4A. OR. 8P. 2. Whllat is the area of a triangle, two sides of which measure 15.36 chains and 11.46 chains respectively, and their included angle 47~ 30'? Ans. 6 4. 1 R. 38 P. 3. One side of a triangular, field bears N. 12~ E. dis tance 18.23 chains, and at the same station the other adjacent side bears N. 78~ 30' E. distance 13.84 chains: required the area. Ans. 11 A. 2 R. I1 P. 4. Required the area of a triangular piece of ground, one side of which bears N. 82~ 30' W. dist. 19.74 chains, and at the same station the other adjacent side S. 24~ 1,5 E. dist. 17.34 chains. Ans. 14 A. 2 R. 8 P. PROBLEM IV. To find the area of a triangle when one side and the two adjacent angles are given. RULE. Subtract the sum of the two given angles from 180~; the remainder will be tile angle opposite the given side. Then, 128 CONTENT OF LAND. [CHAP. II1. As the rectangle of radius and the sine of the angle opposite the given side, Is to the rectangle of the sines of the other angles, So is the square of the given side, T's double the area.* EXAMPLES. 1. In a triangular field ABC, the side AB measures 76 perches, the angle A 60~, and the angle B 50~: required the area. Fig. 47. The angle ACB-=180~ —the sum of the angles A and B,=70~. A rad. Ar. Co. 0.0000t As rad.x sin. C, }sin. C. 700 Ar. Co. 0.02701 sin. A. 60~ 9.93753 ~ sin. AX sin. B, i sin. B. 50~ 9.8842*: AB2- AB X ABD, S AB 76 1.88081 AB 76 1.88081:double area in perches 4078 3.61041 40)2039 4)50 39 12A. 2R. 39P. * DEMONSTRATION. Let AB, Fig. 49, be the given side of the triangle ABC, and A and B the given angles; also let CD be perpendicular on AB: Then by trig. sin. ACB: sin. B.:: AB: AC rad. l sin. A:: AC: CD. Therefore (23.6) rad. x sin. ACB: sin A x sin. B:: AB x AC: CD X AC: (Cor. 1.6) AB: CD:: AB 2: AB x CD; but AB x CD is equal to double tho area of the triangle ABC; therefore (11.5) rad. x sin. ACB: sin. A X sin.,: AB: double the area of the triangle ABC. CHAP. III.] CONTENT OF LAND. 129 2. One side of a triangle measures 24.32 chains, and the adjacent angles are 63~ and 74~: required the area. Ans. 37 A. O R. 22 P. 3. What is the area of a triangular field, one side of which is 17.36 chains, and the adjacent angles 370 30', and 480 15'? Ans. 6 A. 3 R. 18 P. PROBLEMS V. To find the area of a triangle when the three sides are given. RULE. From half the sum of the three sides subtract each side severally; multiply the half sum and the three remainders continually together, and the square root of the last product will be the area.* * DEMONSTRATION. Let ABC, Fig. 69, Fig. 69. be the triangle. Bisect any two of the angles, BAC, ABC, by the straight lines D AG, BG, meeting in G; let fall on the G three sides of the triangle, the perpen- \ diculars GD, GF, GE, and join GC; also produce AB, AC, and bisect one of the exterior angles, JIBC, by the line BK, meeting AG* produced in K, join KC, H\ and let fall tile perpendiculars KII, KM, and KL. Then (26.1) AD is equal to AE and DG to GE; also BD is equal to BF, and DG to GF; hence GF and GE are equal, and consequently (47.1) CF is equal to CE. In like manner it mnay be proved that AII is equal to AL, BII to BM, and CM to CL; as likewise that KIH, KM, and KL are equal to each other. Now since BII is equal to BMO and CL to UM, it is manifest that AII and AL together, are equal to the sun):f the thrce sides AB, AC, and BC; hence AII or AL is equal to the semiperimeterof the triangle ABC. But since twice AD, twice BD, and twice CF are, together equal to the sum of the sides of the triangle, or twice All, it is Gb* The nngle BAC is less than the angle IIBC (16.1;) consequently BAG is less than 11BK, andti AG, KBA, are together lees than 11tBK, KBA; but II3K, KBA, are togretbcr equal to two right angles; hence BAG, KBA, are less than two right angles; therefore (cor. 29.1) the line BK will meet the line AG produced. 130 CONTENT OF LAND. ICHAP. III EXAMPLES. 1. Required the area of a triangular tract of land whose three sides are 49.00, 50.25 and 25.69 chains. vious that AD, BD and CF together, are equal to AH; consequently CF is equal to BH or BM; hence CM or CL is equal to BF or BD; and therefore 1)DH and BC are equal. HIence, if from the semiperimeter AH, the three sides AB, AC and BC be severally taken, the remainders will be BHI, CL, (or BD) and AD respectively. Again, since the angles DBF and DGF are together equal to two right angles, as likewise DBF and FBH together equal to two right angles, it is manifest that the angle DGF is equal to the angle HBF; and the angle DGB to the angle HBK; the triangles DBG and HKB are therefore similar. HIence BD: DG:: II: HB; also in the similar triangles ADG, AHK, AD: DG:: AH: HK; therefore (23.6) A) X BD: DG2:: AHll: IB:: AH2: All x IIB. If therefore we take between AD and BD, and between AH and HB, the mean proportionals M and N respectively, the foregoing analogy will become M2: DG2:: AH2: N2; hence (22.6) M: DG:: AH: N; consequently the rectangle M x N is equal to the rectangle AH x DG * aerefore ABC=ABG+ BCG+ ACG=AH x DG=-M x N=-V/(A) Y P',' / (AH x HB)= V (AD x BD x ib x AH.) CHAP. WI.] CONTENT OF LAND. 131 49.00 50.25 25.69 Sum 124.94 Half sum 62.47 log. 1.79567 13.47 1.12937 Remainders 12.22 1 08707 36.78 1.56561 2)5.57772 615 chains 2.78886 61.5 Acres- 61 A. 2 R. 2. What is the area of a triangular field whose sides measure 10.64, 12.28, and 9.00 chains? Ans. A. 2R. 26 P. 3. What quantity of land is contained in a triangle, the sides of which are 20, 30 and 40 chains? Ans. 29 A. OR. 7P PROBLEM VL Tofind the area of a trapezium, whenl one of the diagonals and the two perpendiculars, let fall on it from the opposite angles, are given. RULE. Multiply the sum of the perpendiculars by the diagonal, and half the product will be the area.* * DEMONSTRATION. The area of the triangle ABC (Fig. 70)-ACXBF; and the area of the triangle ADC= X; therefore the sum of these 132 CONTENT OF LAND. [CHAP. 1n. Note. —When all the sides and one of the diagonals are given, the trapezium will be divided into two triangles, the area of each of which may be found by the last problem. The sum of these areas will be the area of the trapezium. EXAMPLES. Fig. 70. 1. In a field ABCD, in the form of a trapezium, the diagoA nal AC measures 20.64 chains, F..-, C the perpendicular BF 6.96 chains, and DE 5.92 chains; required the area. Fig. 70. Ch. 6.96 5.92 12.88 20.64 5152 7728 2576 2)265.8432 132.9216 Ch. —13 A. 1 R. 6 P. 2. Required the area of a trapezium whose diagonal measures 16.10 ch. and the perpendiculars 6.80 ch. and 3.40 ch. Ans. 8 A. O R. 33a P. 3. The diagonal of a trapezium is 24 ch. and the perpendiculars are 8.27 ch., and 12.43 ch.; what is the area? Ans. 24 A. 3 R. 14 P. ACxBF ACxDE BF+DE areas, or the area of the trapezium ABCD=XAC. ~~~~~ +~ -- CHAP. III.] CONTENT OF LAND. 133 PROBLEM VII. To find lite area of a trapezium, whten all tite anjles and two opposite sides are given. Note. —When three of the angles are given, the fourth may be found, by subtracting their sum from 3600. RULE. Consider one of the given sides and its adjacent angles, or their supplements when their sum exceeds 1800, as the side and adjacent angles of a triangle, and find its double area by prob. 4. Proceed in the same manner with the other given side and its adjacent angles: Half the difference of the areas thus found will be the area of the trapezium.* EXAMPLES. 1. In a four-sided field ABCD, rig. 71. there are given the following bearings and distances, viz. AB, N. 240 E. dist. 6.90 ch.; BC,..... \ N. 640 40' E.; CD, S. 350 20' A E. dist. 11.50 ch.; and DA, S. 88~ W.: required the area. Fig. 71. From the given bearings, the angles may be found as follows: AD, N. 880 E. CB, S. 64~ 40' W. AB, N. 24 E. CD, S. 35 20 E. BAD= 640 BCD - 1000 00 * DEMONSTRATION. Let AB, CD, Fig. 71, be the given sides of the tray pezium ABCD. Produce DA, CB, to meet in E; then 2ABCD=2EDC — 2 EDC-2 EAB 2EAB or ABCD = 2. Hence the truth of the, rule is evident. 134 CONTENT OF LAND. [CuAP. HI. BC, N. 64~ 40' E. DC, N. 35~ 20' W. BA, S. 24 00 W. DA, S. 88 00 W. 40 40 123 20 180 00 180 00 ABC==139 20 ABC=56 40 Construction. Make AB=6.90, and draw DA, CB, making the angle DAB=640, and ABC =1390 20'; produce DA and make the angle EAF= 56~ 40'=-the given angle ADC; lay off AF=11.50=the given side BD, and parallel to AD draw FC, meeting BC in C; lastly draw CD parallel to AF, meeting AD in D; then will ABCD be the trapezium.* Calculation. The angle E=1800 —the sum of the angles BCD, ADC=230 20'. A r rad. Ar. Co. 0.00000 As rad.X sin. E, jsin. E. 230 20' Ar. Co. 0.40222 sin. FEABX sin. EBA, sin. EAB 1160 00' 9.95366 sin. EBA 40~ 40Q 9.81402 AB' ABAB 6.90 0.83885 AB2, j AB 6.90 0.83885 2 EAB 70.405 1.84760 * DE3MONSTRATION. By construction FC is parallel to AD and CD to AF; therefore (34.1) CD=AF and (29.1) the angle ADC-EAF; hence it is evident that the sides AB, CD, and the angles of the trapezium ABCD are respectively equal to the given sides and angles. CIAP. III.] CONTENT OF LAND. 135 Ai. E rad. Ar. Co. 0.00002 A.s rad.X sin. E, sin. E. 23~ 20' Ar. Co. 0.40222 sin. ECD 1000 00' 9.99335 sin. ECD X sin. EDC, in. ED C sin. EDC 56 40 9.92194 * * C1TYt, i CD 11.50 1.06070 CD 11.50 1.06070 2 EDC 274.731 2.43891 2 EAB 70.405 2ABCD 204.326 XBCD —102.163 Ch. —10 A. OR. 34.6P. 2. In a trapezium ABCD, the angles are, A=65~, B ^2-81~, C= 1200, and consequently D —94~; also the side AB -20 ch. and CD= 11 ch.: required the area. Ans. 22 A. 2 R. 27 P. 3. Required the area of a four-sided piece of land, bounded as follows: 1. N, 120 30' E. 2. N. 81 00 E. dist. 23.20 ch. 3. S. 36 00W. 4. N. 89 00 W. dist. 12.90 ch. Ans. 27A. 2R. 24P. PROBLEM VIII. To find the area of a trapezium when three sides and the two included angles are given. RUTLE. As radius, Is to the sine of one of the given angles; So is the rectangle of tlhe sides including this angle. To a certain quantity. 136 CONTENT OF LAND. [CHAP. III. As radius, Is to the sine of the other given angle; So is the rectangle of the sides including this other angle, To a second quantity. Take the difference between the sum of the given angles and 1800; Then, As radius, Is to the sine of this difference; So is the rectangle of the opposite given sides, To a third quantity. If the sum of the angles be less than 180~, subtract the third quantity from the sum of the other two, and half the difference will be the area of the trapezium. But if the sum of the given angles exceed 180~, add all the three quantities together, and half the suni will be the area.* Fig. 72. * DEMONSTRATION. Let ABCD (Fig.:P F 72 or 73) be the trapezium, having the ______ ___________/___/__ given sides, AD, AB, BC, and given " -\ /0 angles DAB, ABC. Complete the parallelograms ABCE, ABFD, and join ED, CF; then because EC, DF, are each parallel and equal to AB, they are (30.1) A B parallel and equal to each other, and (33.1) ECFD is a parallelogram; therefare ABFD=ABItG- G 3IIFD= (35.1) ABCE+ECFD=(34.1) ABCE+ 2ECD; to the first and last of these equals add ABCE, then ABFD+ ABCE = 2-ABCE-2ECD = 2ABCDE. But Fig,. 72, when the sum of the given angles DAB, ABC, is less than 180~, 2ABCDE -2ABCD+2EAD; therefore in this case ABFD+ABCE = 2ABCD + 2EAD; or ABFD + ABCE-2EAD = Fig. 73. 2ABCD. And, Fig. 73, when the suim o~f the given angles DAB, ABC, exceeds 1800, 2ABCDE 2ABCD-. 2EAD; therefore ABFD+ABCE= 2ABCD61c, cG 2EAD; or, ABFD+ABCE+2EAD=-2ABCD. But by prob. 3, one of the first two proportions gives 2BAD (= ABFD,)and the other gives 2 ABC (=ABCE;) also because the angle EAD is the A B CHAP. ImI.] CONTENT OF LAND. 137 EXAMPLES. I In a trapezium ABCD, there are given AD-23.32 ch., AB-= 25.70 ch., and BC= 15.84 ch., the angle DA B-= 64~, and ABC=82~: required the area. As rad. - Ar. Co. 0.00000: sin. DAB, 64 - - - - 9.95366 ADxAB AD 23.32 - 1.36773 AB 25.70 - - 1.40993 first quantity 538.66 - - - 2.73132 As rad. -Ar. Co. 0.00000: sin. ABC, 82~ 9.99575 AB 25.70 1.40993 ABxBC, BC 15.84 - 1.19975: secona quantity 403.12 - 2.60543 DAB 640 ABC 82 146 180 Difference 340 As rad. - Ar. Co. 0.00000: sin. difference 34 - - - - 9.74756 A 5AD 23.32 - - 1.36 73 BC 15.84 - - 1.19975 third quantity 206.55 - - - 2.31504 difference between the sum of the given angles and 1800, and the side EA=BC, the third proportion gives 2 EAD: hence the truth of the rule is manifest. 12* 8 138 CONTENT OF LAND. [CHAP.. III. 1st quantity 538.66 2d " 403.12 941.78 3d " 206.55 2)735.23 367.615 ch.=36 A. 3 R. 2 P. 2. What is the area of a four-sided lot of ground, three sides of which, taken in order, measure 6.15, 8.46, and 7.00 chains, respectively; the angle contained by the first and second sides 560, and that contained by the second and third sides 980 30'? Ans. 4 A. OR. 25 P. 3. One side of a quadrilateral piece of land bears S. 7] E. dist. 17.53 ch., the second, N. 87 E. dist. 10.80 ch. and the third, N. 251 E. dist. 12.92 ch.: what is the area? Ans. 21 A. 3 R. 2 P. PROBLEM IX. To find the area of a traprzoid. RULE. Multiply the sum of the parallel sides by their perpendicular distance, and half the product will be the area.* EXAMPLES. 1. Required the area of a trapezoid ABCD, of which the parallel sides AD, BC measure 6.14 and 9.48 chains, Fig. 74. * DEMONSTRATION. The trapezoid ABCD, Fig. c 74,=the triangle ABD+BDC=-(by prob. 2,) ADXBF BCXDE,,ADXBF D 2 2'=(because BF-DE,) A BCXBF-AD+BCXBF A B CHAP. III.] CONTENT OF LAND. 139 respectively, and their perpendicular distance BF or DE, 7.80 chains. Ch. 6.14 9.48 15.62 7.80 124960 10934 2)121.8360 60.9180 Ch.-= 6 A. 0 R. 15 P. 2. The parallel sides of a trapezoid are 12.41 and 8.22 chains, and their perpendicular distance 5.15 chains: required the area. Ans. 5 A. 1 R. 10 P. 3. Required the area of a trapezoid whose parallel sides are 11.34 and 18.46 chains, and their perpendicular distance 13.25 chains. Ans. 19 A. 2 R. 39 P. PROBLEM X. To find the area of a circle, or of an ellipsis.* RULE. Multiply the square of the circle's diameter, or the * If two pins be set upright in a plane, and a thread, the length of which is greater than twice the distance between the pins, having the ends tied together, be put about the pins; and if the point of a pin or pencil applied to the thread, and held so as to keep it uniformly tense, be moved round, till it return to the place from which the motion began; then the point of the pin or pencil will have described on the plane, a curved line called an Ellipsis. 140 CONTENT OF LAND. [CHAP. HII product of the two diameters of the ellipsis, by.7854, for the area.* Note 1. —If the diameter of a circle be multiplied by 3.1416, the product will be the circumference; also if the circumference be divided by 3.1416, the quotient will be the diameter. 2. If the area of a circle be divided by.7854, the square root of the quotient will be'the diameter. EXAMPLES. 1. How many acres are in a circle a mile in diameter? l mile -80 ch. 80 6400.7854 3141600 47124 5026.5600 Sq. Ch.=502 A. 2 R. 25 P. nearly. Or by Logarithms. Square of 80 80 log. 1.90309 { 80 1.90309.7854 -1.89509 5026.56 Sq. Ch. 3.70127 2. Required the area of an ellipsis, the longer diameter of which measures 5.36 ch. and the shorter 3.28 ch. * The demonstration of this rule is too abstruse to admit of a place in this work. The student who wishes to see a demonstration is referred to a treatise on Mensuration or Fluxions. ChAP. II.1 CONTENT OF LAND. 14] Ch. 5.36 3.28 4288 1072 1608 17.5808.7854 703232 879040 1406464 1230656 13.80796032 Sq. Ch.= 1 A. 1 R. 20.9 P. PROBLEM XI. Tite bearings and distances of the sides of a tract of land being given, to calculate the area. RULE. 1. Rule a table and head it as in the annexed example; observing that the letters E. D. D. and W. D. D., stand for East Double Departure and West Double Departure. 2. Find by prob. 12. chap. 1., the corrected differences of latitude and the departures, corresponding to the several sides, placing them in their proper places in the table. 3. When the departures corresponding to the first and last sides are of tlhe same name, add them togethler and place the sum opposite the first side, in the column of double departures, which is of that name; but when they are of deferent names, take their dffeence, and place it 142 CONTENT OF LAND. [CHAP. Ill. in the colunln of double departures, which is of the same name with the greater departure. Proceed in the same. manner with the departures corresponding to the first and second sides, placing the result opposite the second side; with those corresponding to the second and third sides, placing the result opposite the third side; and so on to the last. 4. Commencing with any side of the survey at pleaslre, assume any number whatever for a multiplier corresponding to that side, and place it in the column of multipliers, opposite to the side, marking it with the letter E, for east. If this multiplier and the double departure, corresponding to the next side, are of the same name, take their sum for the next multiplier, marking it with that name; but if they are of different names, take their difference, marking it with the name of the greater. Proceed in the same way with this multiplier and the next double departure; and so on till multipliers have been found corresponding to all the sides. 5. Multiply each of the corrected differences of latitude by its corresponding multiplier; and when the multiplier is east, place the product in that column of areas, which is of the same name with the difference of latitude; but when it is west, place the product in the column of areas, which is of a different name from that of the difference of latitude. 6. Add up the numbers in the columns of areas, and taking the difference of their sums, divide it by 2; the result will be the area of the survey.* ~ DEMONSTRATION. Let ABCDEFG, Fig. 79, be a plot of a survey; and let the east and west line AL, represent the assumed multiplier. From the points B and L draw BM parallel, and LM perpendicular to AL, meeting in CHAP. 111.] CONTENT OF LAND. 143 Note 1. —If the double departures have been correctly found, the sums of the numbers in the two columns, will be equal. Also, if the multipliers have been correctly obtwined, the sum or difference of the multiplier last found Fig. 79. N II b', h B L -~~r ~2. ~- --- a — --- --------- - -- - ---- --- M; and bisect BM by the meridian NS. Draw the other east and Wes lines, Cc, md, ne, rf, and ug,. and also the meridians.k, Bp Cn, Ds, ur Ew, and Gx. Then it is evident that the diffarences of latitude and the departures corresponding to the several sides will be as in the followingtable. Also according to the rule, Bh~ Av Bx, is the double departure corresponding to the first side; Bh+ Cl= Ck, is that for the second side; CI —Dm=Dp, is that for the third; and so on to the last.!Dist. N. S. E. W;. E. D. D. W.D.D Multipliers. N__. Areas. S. Areas. (B Ah Bh Bx Aa Bb, E 2Aab B - BC BI BCl Ck Bb+ Cc, E 2 BbcC CD Cm Dm Dp Cc+Dd,E 2 cdD DE Dq Eq En Dd+- Ee, E 2 DdeE EF Er Fr F s Ee~Ff, E 2 EefF FG Ft Gt Gu Ff+ Gg, E2 ofg G GA Gvl Av A i Gg+ Aa, E. 2 Gga A 144 CONTENT OF LAND. [CHAP. III. and the next double departure, according as they are of the same or different names, will be equal to the assumed multiplier. 2. It is best in general, to assume 0, for the first multiplier; as by so doing there is one multiplication less to be performed, and the other multipliers are mostly smaller numbers, than they would otherwise be. By construction, the assumed multiplier AL==Aa+aL =Aa+bM=Aa +Bb. By proceeding with this multiplier and the double departures, as directed in the rule, we shall evidently have the other multipliers as represented in the table. It is also plain that the products of the differences of latitude by these multipliers, will be as represented in the columns of north and south areas. The sum of the north areas is 2 BAGFfbB; and the sum of the south areas is 2BCDEFfbB. The difference of these is 2 ABCDEFGA; the half of which is the area of the survey. The preceding demonstration may easily be extended to the case in which the assumed multiplier is so small as to make the meridian NS pass through the survey. Thus, suppose Aw+HB to be the assumed multiplier, the meridian in this case coinciding with FtI. Then the multipliers will be equal to the differences between the above multipliers and the quantity, wa+hbII or its equal 2bII. We may therefore represent them, and the products, as in the following table; in which the multipliers are marked, and the products placed, in conformity with the rule. Dist. N. S. | Multipliers. N. Areas. S. Areas. AB Ah (Aa+ Bb) —2 bH1, E. 2 AafBb —2wabt1 -. _!_ i v —-.,y —.z',C BI l l(Bv+ c)- E. 2 )c C-2.bc K FG Ft 2 _H-(y+,) W. 2 Ffgt —2!GA (F 2 b- -(6 (4-A (a) W. 2 tffta —2 Giffc A if x Av|2!, of ( (r+ A ) \v.( CIAP. III.] CONTENT OF LAND. 145 3. Instead of assuming the first multiplier east, it might with equal propriety be assumed west. Also instead of finding the multipliers from the departures, they might be found in a similar manner, from the differences of latitude; using, in that case, the departures for multiplicands. 4. When one or two bearings or distances are omitted, they may be found by the problems in the last chapter; and in these cases the differences of latitude and departures are to be used as first obtained from the tables, there being no means of correcting them. EXAMPLES. 1. Given the bearings and distances of the sides of a tract of land as follows: 1st. 402~ E. 31.80 ch.; 2d. N. 540 E. 2.08 ch.; 3d. N. 29*~ E. 2.21 ch.; 4th. N. 284~ E. 35.35 ch.; 5th. N. 57~ W. 21.10 ch.; and 6th S. 470 W. 31.30 ch.; to the place of beginning. Required the area of the tract. The sum of the north areas is 2 AabB-2 Hbaw: and the sum or the south areas is 2 BCDEFfbB+2 wafF. —2 H1f F —2 AGFfa=2 BCDEFfbB+ 2 wafF-2 waf F —2 IIbaw - 2 AGFfa=2 BCDEFfbB-.2Hbaw-2AGFfa, =2BCDEFfbB- 2 AGFfa-2 Hbaw. If now the sum of the north areas be subtracted from that of the south areas, the remainder will be 2 BCDEIfb}d -2 AGFfa-2 AabB=2 BCDEFfbB —2 BAGFfb=2 ABCDEFGA. 13 T - 3Cor.Cor. _ ~ cD 1 Sta.| Bearings. |Dst. N. W. S. E.. iED.D.D.W.D.D. Multie N. Areas. S. Areas, I. _ 1S E 324.18 20.65.03 -05 24.21 20.70 2.15 -0.0 E. O U _.,- _.. -, _o ~, cD 2 |N. 54 E. -2.08.23 1.68 |.- 1,.~ -.~ |.. —|-LI_38|_||22 38 —|2 5 274| 7' rF 13 N. 291 E. 2.21 1.92 | 1.08.00.00 192 1.08 2.76 25.14E 48.2688 - C; 4 N.284 E. 35.35 31.003 17.00.04 1.051 30.96 17.05 1 18.13 43.27 E. 1339.6392 O 0 h N. 57 W. 21.10 11.49 17.69.02.03 11.47 17.66 0.61 42.66 E. 489.3102 O S.47. 31.30 21.34 22.89.03.04 21.3:7 22.85 40.51 2.15 E. 45-9455:s C? O | I!23.81 45.64 45.52 40.41 40.58.1'2.17 45.58 45.58 40.5151 43,27 43.27 1904.74561 45.9455 *T 6a 4< t45.52 40.41 45.9455.12 Er. $S:.17 Er. E. 2)1858.8001 o c * 10)929.40005 o D C) Area 92 A. 3 R. 30 P. z c. ~, 92.940005 o 4 _. E3':76 0 03.760020 40.F~~ ~~~~~ ~;n~~~~~~~~~ ~m~~ ~ $0.400800 o O ~C _O lcor. cor. Multi A Sta. Bearing. Dist. N. S. E. W. N.. E. N.W. E.D.D. W.D.D. pliers. N. Areas. |. re. _i — -I -l- --- -- J. - - N S P N. 750 E. 13.70 3.54 13.24.02.02 3.56 13.26 9.99. E-0 C 0.~ 2 N.20E. 10.309. 3.61.01.01 9.66 3.62 16.88 16.88 E 163.0608 O M - 3 East. 16.20 _ 16.20|.02.02|.02 16.22 19.84 3.22 36.72 E.7344 _ _ _ 4 S.331W. 35.30 29.44 119.49.05.05 29.39 19.44 34.94 33.50 E 984.5650 5 S. 76 W. 16.00 3.87 15.52.02.02 3.85 15.50 15.49 1.44 W 5.5440 G 6 |North. |9.00 9.00 I.01.01 9.01.01 11.51 16.93 W 152.5393 C D 7 S.84 W. 11.60' 1.21 11.541.02.02 1.19 11.52 20.79 28.44 W 33.8436 __Z_ _ -" —-' I_ -' 1, —------ 9 I 8O" * *65 0 8 N.531 W. 11.60 6.94 9.29.02.02 6.96 9.271 49.23 W 342.6408 D l N. 36 E.'19.3615.51 11.59.03.0/1559 11.61 2.34 46.89 W 728.6706 o 10 N. 221 E. 14.00 12.93 5.36.02.02,12.95 5.38 16.99 29.90 W 387L2050 cO 711 S4. 76 E.W 12.6001 2.175 11.68 02.02 2.73 11.70 17.08 12.828W 34.5486 4 ------ - *-.. 1 12 S. 15 W.l 10.8.5| 10.48 2.81.02!.01 10.46 2.80| 8.90 3.92 W 41.0032 "8 8 V. 10.62W 11.10 -8.020 10.08 3.27 6.079.99 W 100.6992 1 190.53|5757.6578576168E 161.93.281.26!57)s70-57.70 61.80 61.80l 92.021 92.02 379.8838 2595.)2073 6 57.57 61.68 379.8838.2s Er. N..25Er. E. 2'2215.7369 ~ * AreI 10 A. 3 R. 65P Sq. Ch. 1107.8684 4. ]~ ~' O o3ewCs t I-.... ~~ ~~ ~ 148 CONTENT OF LAND. LCHAP. IMl. 3. Given the boundaries of a tract of land as follow, tiz. 1st. S. 35~0~ WV. 11.20 ch.l; 2d. N. 45~ W. 24.36 ch.; 3d. d. 152~ E. 10.80 ch.; 4th. S. 77~ E. 16. ch.; 5th. N. 87 ~0 E. 21.50 ch.; 6th. S. 600 E. 14.80 ch.; South 10.91 ch.; 8th. N. 850 W. 29.28 ch.; to the place of beginning: required the area. Ans. 85 A. 3 R. 17 P. 4. Given the boundaries of a tract of land as follow, viz. 1st. N. 19~ E. 27 ch.; 2d. S. 77~ E. 22.75 ch.; 3d. S. 27~ E. 28.75 ch.; 4th. S. 520~ W. 14.50 ch.; 5th. S. 15i E. 19 ch.; 6th. West, 17.72 ch.; 7th. N. 36~ WV. 11.75 clh.; Sth. North, 16.07 ch.; 9th. N. 62~ WV. 14.88 ch.; to the place of beginning: required the area. Ans. 152 A. 2R. 6P. 5. Required tile area of a tract of land bounded as follows: 1st. S. 62' W. 7.57 ch.; 2d. N. 43~ W. 5.89 ch.; 3d North, 5.82 ch.; 4th. N. 933- W. 8.83 ch.; 5th. N. 48" E. 4.81 ch.; 6th. N. 12~ E. 4.66 ch.; 7th. N. 6240 E. 5.27 ch.; 8th. S. 6~0 E. 5.60 ch.; 9th. S. 40~0E. 5.87 ch.; 10th. East, 6,54 ch.; 11th. North, 5.52 ch.; 12th. N. 68~~ E. 3.10 ch.; 13th. S. 300 E. 7.90 ch.; 14th. S. 230~ Nr. 8.80 ch,.; 15th. S. 311- E. 6.42 ch.; 16th. S. 500~ W. 8.40 ch.; 17tth. N. 440~ W. 6.85 ch. to the place of beginning. Ans. 44 A. 2 R. 22 P. 6. Given the following field-notes to find the area of the survey; also the bearings and distance of the 3d side, wlhich were omitted to be taken on account of obstacles in the way. Cl. 1. S. 85~0 E. 23.30 2. S. 19 E. 31.12 3.. 4.N.64 WV 29.72 CHAP. M.] CONTENT OF LAND. 149 Ch. 5. N. 15~0 W. 22.46 6. N. 58 E. 25.94 7. S. 271 E. 6.60 Ains. Area 182 A. 0 R. 21.7 P. and the bearing and distance of the 3d side, S. 66~ 23' W. 28.06 -ch. 7. Being furnished with the field-notes of a tract of land, and requested to calculate the area, I found on examining them, that the figures expressing the angles of bearing of the 4th and 5th sides were so defaced as to be illegible: but as the remaining data are sufficient, the area is required. The field-notes are as follows: Ch. 1. S. 601`0 W. 10.34 2. N. 274 W. 17.88 3. N. 51 E. 15.85 4. N.- E. 9.61 5. S.- E. 19.18 6. S. 164 E. 22.21 7. S. 711 W. 16.66 8.N. 714 W. 5.76 Ans. 81 A. 2 R. 23 P. 8. In a survey, represented Fig. 81. Fin. 81, the corner at A was inaccessible, occasioned by the al overflowing of water; but being a tree, it can be seen from the adjacent corners B and L.. I therefore set my instrument at B and took the bearing to A, which I reversed, and set in my field-book as the first bearing. I then proceeded to take the bearings and distances of c the several sides to L; and at L, I took the bearing of the side LA. The field-notes being as follows, the length of the sides AB and LA, and the area are required. 13 * 150 CONTENT OF LAND. [CHAP. III. Nig. 80. AB, N. 514~ W. Ch. BC, S. 451 W. 15.16 CD, N. 50 W. 22.10 -. DE, North, 18.83 lAr" A m EF, N. 48 E. 22.60 FG,N.25k W. 20.17 GH, East, 26.57 III, S. 30i E. 22.86 IK, S. 44 W. 15.04 "4 KL, S. 47 E. 28.55,LA, S. 201 W. Ans. AB, 26.47 ch.; LA, 23.81 ch.; and the area 244 A. 3 R. 13 P. 9. In taking a survey of a tract of land bounded by six straight sides, Fig. 80, I was prevented going directly from the 3d to the 4th corner by a pond of water. I therefore set up two stakes near the edge of the pond, and took the bearing and distance from the 3d corner to the first stake, from the first stake to the second, and from the second to the 4th corner, and noted them in my field-book as all belonging to the 3d station bf the survey. The field-notes being as follows, the bearing and distance of the 3d side, and the area of the survey are required. 1. North, 7.81 Ch. 2. S. 76~0 E. 18.15 S. 52 W. 10.70) 3S. 71 W. 13.92 S. 334 E. 9.00) 4. N. 841 W. 27.12 5. N. 41 W. 22.00 6. East, 16.58 Ans. 3d side, S. 100 47' W. 28.42 ch.; and area 80A. OR. 25P. CHAP. III.] CONTENT OF LAND. 151 PROBLEM XII. To find the area, when of-sets are taken. RULE. 1. Find by the last problem, the area enclosed by the stationary lines and straight sides of the survey. 2. Subtract the stationary distance of each off-set, from that of the one immediately following; the remainilers will be the distances, intercepted on the stationary line, between each two adjacent off-sets. Place these under one another in a column as in the annexed examples. Also take the sums of each two adjacent off-sets, and place them in the next column, so as to correspond with the intercepted distances. 3. Multiply the sum of each two adjacent off-sets by their intercepted distance on the stationary line; then, half the sum of the products will be the area of the offsets on that line. 4. If there are off-sets on more than one statiorrary line, proceed in the same manner with the others. 5. When the stationary lines are within the boundary of the survey, add the areas of the off-sets to the area enclosed by the stationary lines and straight sides; but when the stationary lines are without the boundary, subtract the areas of the off-sets.* * DEMONSTRATION. Considering the boundary as straight between the ends of each two adjacent off-sets, it is plain that the area contained between the stationary line and boundary will be divided by the off-sets into trapezoides and triangles. Hence the truth of the rule is evident. 152 CONTENT OF LAND. [CHAP. II1. EXAMPLE 1, Fig. 82. Required the area of a piece of meadow, bounded on one side by a brook; the field-notes being as follows: Left-hand off-sets on the stat line. Stat. Dist. Off-sets. 1. N. 16~0E. 14.35 Ch. No. 1. 0.00 Ch. 0.30Ch. 2. East, 7.82 2. 0.95 0.84 3. S. 3H W. 14.45 Stat. line. 3. 2.03 0.86 4. N. 861 W. 11.07 4. 3.28 0.50 sN Fig. 82. 5. 5.20 1.80 6. 7.43 2.35 B i a~ 7. 8.98 1.45 2 /8. 10.46 1.08 t /i~~ 3.j 9. 11.71 1.85 ~ 10. 14.45 0.35 The area of the part / 6 _ ABCD will be found, g / 2 by the last problem, to j/ be 13A. 1 R. 11 P. ______ 5 _ AP.] CONTENT OF LAND. 153 Tofind the area of the off-sets. Sta. Dist.i Off-sets. Intercep. Sums of Nto. Ch. Ch. Dist. Ofsf-sets. Products. 1 0.00 0.30 2 0.95 0.84 0.95 1.14 1.0830 3 2.03 0.86 1.08 1.70 1.8360 4 3.28 0.50 1.25 1.36 1.7000 5 5.20 1.80 1.92 2.30 4.4160 6 7.43 2.35 2.23 4.15 9.2545 7 8.98 1.45 1.55 3.80 5.8900 8 10.46 1.08 1.48 2.53 3.7444 9 11.71 1.85 1.25 2.93 3.6625 10 14.45 0.35 2.74 2.20 6.0280_ 2)37.6144 18.8072 Ch. =1 A. 3R. 21P. A. R. P. Area of ABCD 13 1 11 Do. of off-sets 1 3 21 Whole area 15 0 32 EXAMPLE 2. Fig. 83. Required the area of a survey from the following field notes. U 154 CON'rENT OF LAND. [CHAP. 1II. Left hand off-sets. 1st. Stationary Line. 3d Stat Line. Sta. Dist. Off-sets. Sta. Dist. Off-sets. Ch. No. Ch. Ch. No. Ch. Ch. 1. N. 363O W. 30.00 1. 0.00 0.50 1. 0.00 0..55 2. N. 56~ E. 21.60 stat. line. 2. 6.10 3.40 2. 4.20 2.50 3. N. 269 E. 13.44 Do. 3. 10.15 3.10 3. 8.05 3.20 4. S. 71~ E. 18.96 Do. 4. 14.08 3.96 4. 15.15 2.45 5. S. 26~ E. 13.46 Do. 5. 19.29 2.70 5. 18.96 0.50 8. S. 45 W. 42.41 6. 21.60 0.55 4th Stat.. Line. 2d. Stat. Line. 1. 0.00 0.50 1. 0.00 0.55 2. 5.12 2.75 2. 13.44 0.55 3. 10.00 1.90 4. 13.46 0.70 The area within the stationary lines and straight sides, found by the last problem, is 1152.5381 square chains. To find the area of the of-sets. 1st. Stationary Line. Sta. Dist. Off-sets. Intercep. ISums ofl No. Ch. Ch. Dist. Off-sets. Products. 1 0.00 0.50 2 6.10 3.40 6.10 3.90 23.7900 3 10.15 3.10 4.05 6.50 26.3250 4 14.08 3.96 3.93 7.06 27.74581 5 19.20 2.70 5.12 6.66 34.0992 6 21.60 0.55 2.40 3.25 7.8000 2d. Stationary Line. Sta. Dist Off-sets. Intercep. Sums of{ No. Ch. Ch. Dist. Offsets. Products. 1 0.00 0.55 2 13.441 0.55 113.44 1.10 14.7840 CHAP. II1.] CONTENT OF LAND. 155 3d. Stationary Line. Sto. Dist. Off-sets. Intercep. Sums of No. Ch. Ch. Dist. Off-sets. Products 1 0.00 0.55 2 4.20 2.50 4.20 3.05 12.8100 3 8.05 3.20 3.85 5.70 21.9450 4 15.15 2.45 7.10 5.65 40.1150 5 18.96 0.50 3.81 2.95 11.2395 4th. Stationary line. I 0.00 0.50 2 5.12 2.75 5.12 3.25 16.6400 3 10.00 1.90 4.88 4.65 22.6920 4 13.46 0.70 3.46 2.60 8.9960 2)268.9815 Area of the off-sets 134.49075 Ch. Area within the stationary lines 1152.5381 1287.02885 Ch. 128.702885 Acr 4 2.811540 40 32.46160 Xrea of the survey, 128 A. 2 R. 32 P. 156 CONTENT OF LAND. [CHAP. III. EXAMPLE 3. Required the area of a meadow from the following field-notes. Left-hand off-sets on the stat. line. Sta. Dist. Off-sets. 1. N. 410~ E. 14.35 Ch. No. 1. 0.00 Ch1. 0.38 Ch, 2. S. 424 E. 14.71 Sta. line. 2. 2.65 2.35 3. S. 54 W. 16.32 3. 3.80 1.70 4. N. 321 W. 11.50 4. 6.00 2.75 5. 7.50 1.40 6. 9.60 3.20 7. 12.38 2.72 8. 14.71 0.42 Ans. Area 22 A. 3 R. 27 P. EXAMPLE 4. The following field notes are given, to find the area of the survey. Left-hand off-sets. On the 1st stat. line. On the 2d stat. line Sta. Dist. Off-sets. Sta. Dist. Off-sets Ch. No. Ch. Ch. No. Ch. Ch. 1. S. 69~0 E. 16.14 sta. line 1. 0.00 0.44 1. 0.00 0.31 2. S. 28 E. 9.38 Do. 2. 3.80 2.00 2. 2.67 2.94 3. S. 324 W. 21.20 3. 7.04 3.79 3. 6.20 2.62 4.N.48 W.22.47 4. 9.87 2.34 4. 9.38 0.39 5. N.261 E. 19.00 5. 13.24 3.00 6. 16.14 0.31 Ans. 56A. 2R. 18P. PROBLEM XIII. Given the hearing and distance of two stations from each other and the bearings of all the corners of a tract of land from these stations, to find the area of the tract. The method of doing this will be best explained by an example. CHAP. III.] CONTENT OF LAND. 157 EXAMPLE 1. Let ABCDEFGA, Fig. 84, represent a field, all the angles of which can be seen from two stations, 11 and I, without it. The bearing and distance of the stations, and the bearings of all the angles of the field, from each station, being as follow, it is required to find the area. Fig 84. D The station 1I bears from the station I, North, dist, 28. Ch. Bearings. - Bearings. HA S. 8110~E. IA N. 28~ E. HB S.8.51 E. IB N.424 E. TIC S.68 E. IC N. 51 E. HID S.58i E. ID N.71 E. IIE S. 35 E. IE S. 821 E. HF S.28 E. IF N. 73 E. HG S.40 E. J IG N. 60 E. Comstruction. Draw HI according to the given bearing and distance; and from the points 11 and I, draw IhA, 1IB, HC, &c., 14 158 CONTENT OF LAND. [CHAP il. and IA, IB, IC, &c. according to the given bearings; then will the intersections A, B, C, &c. of the corresI)onding bearings HA and IA, HB and IB, HC and IC, &c. be the angular points of the field. Calculation. In each of the triangles IHA, IHB, IHC, &c. we have the side IIH; and from the bearings of the sides, we have all the angles, to find the sides IA, IB, IC, &c. Then in each of the triangles, IAB, IBC, ICD, &c. we have two sides, and the included angle; whence the areas may be found by prob. III. From the sum of the areas of the triangles IAB, IBC, 1CD, and IDE, which is equal to the area IABCDEI, subtract the sum of the areas of the triangles IAG, IGF and IFE, which is equal to the area IAGFEI; the remainder will be the area of the field ABCDEFGA. Note.-In working the proportions for finding the sides IAf, IB, &c. it will be unnecessary, when the area only is required, to take out the natural numbers corresponding to the logarithms of those sides; because in the proportions for finding the areas it will be sufficient to know the logarithms of the sides, without knowing their real lengths. To find the log. of IA. As sin. HAI, 70~ 00' - - - - - 9.97299: sin. AII, 81 30 9.99520: II, 28 - - - - - - - - 1.44716 11.44236 - IA., - - - - log. 1.46937 CHAP. Ill.] CONTENT OF.LAND. 159 Tofind the log. of IB. As sin. HIBI 52~ 00' - - - - - 9.89653 sin. BHII 85 45 9.99880 IH 28 - 1.44716 11.44596: IB - - - - log. 1.54943 Tofind the log. of IC. As sin. HC1 600 30' - 9.93970 sin. CHI 68 00 - - - - - 9.96717 111: IH 28 - - - - - - 1.44716 11.41433 IC log. 1.47463 ToJind the log. of ID. As sin. HDI 50~ 45' 9.88896 sin. DHI 58 15 9.92960: IH 28 - 1.44716 11.37676 I: D - log. 1.48780 To find the log. of IE. As sin. HEI 470 00' 9.86413 sin. EHI 35 30 9.76395:.III 28 - - 1.44716 11.21111 IE - - - - log. 1.34698 160 CONTENT OF LAND. [CH1A. IU Tofind the log,. of IF. As sin. HFI 78~ 00' 9.99040 sin. FI1I 28 30 - - 9.67866: III 28 - 1.44716 11.12582 IF - - - - log. 1.13542 To find the log. of 1G. As sin. HGI 800 00' - 9.99335 sin. GiII 40 00 - 9.80807 ~:iI 28 - - 1.44716 11.25523 s IG - ------- log. 1.26188 Tdfind the double area of the triangle IAB. As rad. - - - - - 10.00000 sin. AIB 13~ 45' - 9.37600 * *IA x IB 5~ IA - log. 1.46937' I B - 1.54943:2IAB 248.2 2.39480 To find the dlouble area of the triangle IBC. As rad. - - - - - 10.00000 sin. BIC 9 15' -9.20613.:IB IC, r IB - ~ log. 1.549143 XIC - - 1.47463 2IBC 169.9 - - - - 2.20319 MAP. II.] CONTENT OF laND. 161 To find the double area of the triangle ICD. As rad. 10.00000 sin. CID 19~ 30' - 9.52350 * ICxID IC - - - log. 1.47463 ID.....-1.48780 2ICD, 306.15 - 2.48593 To find the double area of the triangle IDE. As rad. -10.00000 sin. DIE 26~ 30' - - - - - 9.64953 *IDx* ID E- - - - log. 1.48780 * *IE - - - - m- 1.34698 2IDE 305.007 - - - - - - 2.48431 To find the double area of the triangle IEF. As rad. - - - 10.00000 sin. EIF 240 00' - - - 9.60931 **IEXIF IE..log. 1.34698 *' IF ---- -— 1.13542 2IEF 123.511 - -.- - 2.09171 Tofind the double area of the triangle IFG. As rad 10.00000 sin. FIG 13~ 30' - 9.36818 IF - - - - log. 1.13549 IF X IG, IG = w..6188 * X XG --- - 1.26188 2IFG 58.274. - - 1.76548 14* X 162 CONTENT OF LAND. [CHAP. II. To find thie double area of the triangle IAG. As rad. - 10.00000 sin. AIG 310 30' - 9.71809 JAXIG, I - - log. 1.46937 IG - -.-1.26188 2IAG 281.412 - - - - - 2.44934 Ch. Ch. 2IAB - - - - 248.2 2IEF 123.511 2IBC - - - - 169.9 2IFG 58.274 2ICD - - - - 306.15 2IAG 281.412 2IDE - - - - 305.007 2 IAGFEI 463.197 2IABCDEI 1029.257 2IAGFEI 463.197 2ABCDEFGA 566.060 ABCDEFGA 283.03 Ch.-21 A. I R. 8 P. The bearings and distances of the sides, if required, might rd-dily be obtained. For, having found the distances IA, IB, we have in the triangle IAB, two sides, and an included angle; whence the angle IAB and side AB may be found. The angle IAB applied to the bearing of IA, will give the bearing of AB. In the same manier the bearings and distances of the other sides may be found. EXAMIPLE 2. Being required to calculate the area of a field, the owner of which refuses permission to go on it, I choose two stations, F and G, in the adjacent land, from whence CHAP. III.] CONTENT OF LA.ND. 163 all the angles of the field are visible. The bearing and distance of the stations, and the bearings of the angles, from each station, are as follow. What is the area of the field? The station G' bears from the station F, N. 43~ W. 20 ch. Bearings. i Bearings. FA N 251o E. GA S. 66~ E. FB N. 19 W. GB N. 23 E. FC N. 5 W. GC N. 38 E. FD N. 16 E. GD I. 60 E. FE N. 60 E.1 GE S. 84 E. Ans. 33 A. i R. 7P. PROBL'EM XIV. To find the area of a survey by protracting it, and dividing the plot into triangles and trapeziums. The method of doing this will be easily understood from the following example. EXAMPLE 1. Given the bearings and distances of the sides of a tract of land as follow: 1st. N. 500 E. 9.60 ch.; 2d. S. 32~ E. 16.38 ch.; 3d. S. 41~ W. 6.30 ch.; 4th. West, 8.43 ch.; 5th. N. 790~ V. 10.92 ch.; 6th. N. 50 E. 11.25 (ch.; 7th. S. 830 E. 6.48 ch.; to the place of. beginning Required the area. Fig. 75, is a plot of this survey: and by drawing the lines as in the plot, it is divided into two trapeziums AGFE, AEDF, and a triangle BDC. Measure the 164 CONTENT OF LAND. [CHAP. I1J. several bases and perpendiculars, on the same scale that was used in the protraction, and find the double areas of the triangle and trapeziums by probs. 2 and 6; the sum of these will be the double area of the survey. Bases. Perpens. EG 16.68 F 7 =203.662=2 AGFE Ab 4.71 — 203.66 28 — A G FE EB 19.17X dAc 5.85 = } 267.4215 2 AEDB EB 19.17 X Dd 8.10 BD 19.23 x Ce 5.16 - 99.2268=2 BDC 2)570.3111 ch.=2 ABCDEFG 285.15555 ch.- 28A. 2R. 2P. = the area required. EXAMPLE 2. The following field-notes are given to protract the survey and find the area. Ch. 1. N. 150 00' E. 20 2. N. 370 30' E. 10 3. East 7.50 4. S. 11 00' E. 12.50 5. South 13.50 6. West 10. 7. S. 360 30' W. 10. 8. N. 380~ 15' W. 8.50 Ans. 46 A. 2R. 9P CHAPTER IV. LAYING OUT AND DIVIDING LAND. PROBLEM I. To lay out a given quantity of land in a square form. RULE. Reduce the given quantity to chains or perches, and extract the square root, which will be the length of a side, of the same denomination to which the given quan - tity is reduced. EXAMPLES. 1. Required the side of a square that shall contain 9A. 3R. 28P. 40)28 Per. 4)3.7R. 9.925 A.=99.25 ch. Ch. 99.25(9.96 ch. the length of a side. 81 - 189)1825 1701 1986)12400 11916 484 166 166 IAYING OUT AND DIVIDING LAND. [CHAP. IV 2. Required the side of a square tract of land that shall contain 325 acres. Ans. 57 chains. PROBLEM II. To lay out a given quantity of land in a rectangularform, having one side giien. RULE. Divide the given content by the length of the given side, the quotient will be the length of the required side. EXAMPLES. 1. It is required to lay out 120 acres in a rectangular form, the length of one side being given, equal 100 perches. Acres. 120 4 480 40 1,00)192,00 192 Per. the length of the other side, 2. The length of a rectangular piece of land is 8 chains; what must be its breadth, that the content may be 5 acres? Ans. 6.25 chains CHAP. IV.1 LAYING OUT AND DIVIDING LAND. 167 PROBLEMI III. To hly out a given quantity of land in a rectangular form, having the length to the breadlh in a given ratio. RULE. As the less number of the given ratio, Is to the greater; So is the given area, To a fourth term.* The square root of this fourth term will be the length required. Having the length, the breadth may be found by the preceding problem. Or it may be found in the same manner as the length. Thus, As the greater number of the given ratio, Is to the less; So is the given area, To a fourth term. The square root of this fourth term will be the breadth required. EXNAMPLES. 1. It is required to lay out 864 acres in a rectangular corm, having the length to the breadth in the ratio of 5 to 3. * DEMONSTRATION. Let ABCD, Fig. 85, be a rect- Fig. 85. angle, and ABFE and AHGD be squares on the E greater and less sides respectively: then (1.6) AD: AE (AB):: the rectangle AC: square AF. Also D C AB: Al (AD):: the rectangle AC: square AG. Hence the truth of the rule is evident. A 13 168 LAYING OUT AND DIVIDING LAND. [CHAP. IV 864 A. 138240 P. Sq. P. Sq. P. As 3: 5: 138240: 230400 /230400 =480 Perches, the length required. Sq. P. Sq. P. As 5: 3 138240: 82944 1/82944 - 288 Perches, the breadth required. 2. It is required to lay out 27 A. 3 R. 20 P. in a rectangular form, having the length to the breadth in the ratio of 9 to 7. Ans. Length 75.725 P. Breadth 58.897 P. PROBLEM IV. To lay out a given quantity of land in a rectangular form, Ihaving the length to exceed t]he breadth by a given difference. RULE. To the given area, add the square of half the given difference of the sides, and extract the square root of the sum; to this root, add half the given difference for the greater side, and subtract it therefrom for the less.* Fig. 86. * DEMONSTRATION. Let ABCD, Fig. 86, be a G ~~ — G rectangle; in DC let DE be taken equal DA 6r BC, and let EC be bisected in F; then (6.2) DFP i,/, s =-DCXDE+FC22 - DCxAD+FC2 =the rectangle AC+the square of half the difference of the sides.' E} F I)DCO, DA; also DF+FC=DC, the greater side. C and DF-FC =DE or DA, the less side. This problem may be neatly constructed thus: Al (B take EC equal the given difference of the sides and bisect it in F; make EG perpendicular to EC and equal to the square root of the given area, and with the centre F and radius FG describe the arc DG meeting CE produced in D: make DA perpendicular to DC and equal to DE, and complete the rectangle ABCD, which will be the one required. Since (47.1) FG'2= EG2 +EF2-=the given area + tihe square of half the given difference of the sides, the truth of tih construction is plain. from the preceding demonstration. CHAP. IV.] LAYING OUT AND DIVIDING LAND. 169 EXAMPLES. 1. It is required to lay out 47 A. 2 R. 16 P. in a recta ngle, of which the length is to exceed the breadth by 80 perches. 2)80 P. 47 A. 2 R. 16 P.= 7616 Per. 1600 40 40 v 9216=96 half diff. add and subtract 40 1600) length 136 breadth 56 2. It is required to lay out 114 A. 2 R. 33.4 P. in a rectangular form, having the length to exceed the breadth by 15.10 ch. Ans. Length 42.25 ch. Breadth 27.15 ch. PROBLEM V. To lay out a given quantity of land in the form of a triangle or parallelogram, one side and an adjacent angle being given. RULE. For a triangle. As the rectangle of the given side and sine of the given angle, Is to twice the given area; So is radius, To the other side, adjacent to the given angle. Then having two sides and the included angle given, the other angles and side, if required, may be found by trig. case 3. 15 170 LAYING OUT AND DIVIDING LAND. [CHAP. IV. For a parallelogram. As the rectangle of the given side and sine of the given angle, Is to the given area; So is radius, To the other side, adjacent to the given angle.* EXAMPLES. Fig. 87. 1. Let AB, BC, Fig. 87, be two C sides of a tract of land; the bear// ing of AB is S. 872~ W. dist. 16.25 ch. and the bearing of BC, N. 27f~ E.; it is required to lay off 10 acres 7 / \ / by a straight line AD, running from A the point A, to the side BC. Bearing of BA, N. 87~o E. BC, N. 27i E. Angle B, 600 As ABXsin. B jAB 16.25 ch. - Ar. Co. 8.78915 sin. B, 60 0.0624..: twice the given area 200 sq. ch. - - - 2.30103::rad. - 10.00000: BD 14.21 ch. - 1.15265 * DEMONSTRATION. It is demonstrated, prob. 3, chap. 3, Content of Land, that rad.: sin. B:: ABXBD: 2ABD (see Fig. 87); therefore (1.6 cor.) rad.XAB: sin. BxAB:: ABXBD: 2ABD, or (16.5) sin. BxAB: 2ABD:: rad.XAB: ABXBD:: rad.: BD. Since ABDF is equal to 2ABD, the truth of the rule for the parallelogram is evident. This problem may be constructed as follows; take AB equal the given side and draw BC making the angle B equal to the given angle; make BE perpendicular to AB, and equal twice the given area of the triangle divided by the given side, or equal the given area of the parallelogram divided by the given side; and parallel to AB, draw EF cutting BC in D, and join DA; then will ABD be the triangle required; or complete the parallelogram ABDF, for the one required. The reason of the construction is plain. CMAP. IV.] LAYING OU1' AND DIVIDING LAND, 171 2. Given the side AB, Fig. 15, of a parallelogram, equal 20 ch. and the angle A 63~ 30'; required the side AC, that the content may be 21J acres. As AB x sin. A AB 20 ch. Ar. Co. 8.69897 X sin. A sin. A 63~ 30' 0.04821: the given area 215 sq. ch. - - - - 2.33244:: rad. - 10.00000: AC 12.01 ch. -1.07962 3. Given one side of a triangle, equal 30 perches, an angle adjacent to this side 71~ 15', and the area 2 acres; required the other side adjacent to the given angle. Ans. 22.53 perches. 4. Given one side of a parallelogram, equal to 32.26 ch., an angle adjacent to this side 83~ 30', and the area 74 acres; required the other side adjacent to the given angle. Ans. 23.09 ch. PROBLEM VI. Th7e area and base of a triangle being given, to cut of a given part of the area by a line running from the angle opposite the base. RIULE. As the given area of the triangle, Is to the area of the part to be cut off; So is the given base, To the base corresponding to that area.* * The truth of this rule is manifest from 1.O. 172 LAYING OUT AND DIVIDING LAND. [CHAP. IV. EXAMPLES. Fig. 88. 1. Given the area of the triangle ABC, Fig. 88, equal 650 square perches, and the length of the base AB, 40 perches; it is required to cut off 290 perches to/A,, B wards the angle A, by a line running from the angle C to the base. ABC. ADC. AB. AD. As 650: 290:: 40: 17.85 per. 2. In a triangle ABC, there are given the area 27 A. 1 R. 16 P. and the base AB 35.20 ch., to cut off 10 acres towards the angle B, by a line CD running from the angle C to the base: the part BD of the base is required. Ans. 12.87 ch. PROBLEM VII. The area and two sides of a triangle being given, to cut off a triangle containing a given area, by a line running from a given point in one of the given sides, and falling on the other. RULE. As the given area of the triangle, Is to the area of the part to be cut off; So is the rectangle of the given sides, To a fourth term. Divide this fourth term by the distance of the given point from the angular point of the two given sides; the quotient will be the distance of the required point from the same angle.* Fig. 89. * DEMONSTRATION. From the demonstration to prob. P C 3, chap. 3, we have, Fig. 89, rad.: sin. A:: ABX AC: 2ABC, and rad.: sin. A: APxAG: 2APG; therefore (11 & 16.5) 2AB: 2APG:: ABXAC: APxAG, or (15.5) ABC: APG:: ABxAC: APx o- B AG; hence the truth of the rule is manifest. CHAP. IV.] LAYING OUT AND DIVIDING LAND. 173 EXAMPLES. 1. Given the area of the triangle ABC, Fig. 89, 5 acres; the side AB 50 perches, the side AC 40 perches, and the distance of a point P from the angle A, 36 perches; it is required to find a point G to which, if a line be drawn from the point P, it shall cut off a triangle.APG containing 3 A. 0 R. 20 P. As the triangle ABC 800 sq. p. Ar. Co. 7.09691 the triangle APG 500 - - - - - - 2.69897 A;x/(AB 50 -.. 1.69897 AC 40 —---- 1.60206 APxAG - - 3.09691 AP 36 - - - - - - - log. 1.55630 AG 34.72 per. - 1.54061 2. Given the area of a triangle ABC, 12 A. 1 R. 23 P. the side AB 20 ch., the side AC 16.25 ch., and the distance of a point P in the side AB, from the angle A 8.50 ch.; it is required to find the distance AG of a point G in the line AC, so that a line drawn from P to G may cut off a triangle APG containing 3 acres. -Ans. 9.25 clh. PROBLEM VIII.'he area and base of a triangle being given, to cut off a triangle containing a given area, by a line running parallel to one of the sides. RUI,E. As the given area of the triangle, Is to the area of the triangle to be cut off; So is the square of the given base, To the square of the required base. 174 LAYING OUT AND DIVIDING LAND. [CHAP. Iv. The square root of the result will be the base of the required triangle.* EXAMPLES. Fig. 90. 1. Given the area of the triangle ABC, Fig. 90, 500 square perches, and G -\ the base AB 40 perches; it is required B * I) B to cut off 120 sq. per. towards the angle. A, by a line DG running parallel to the E''- -— *3 side BC. As the triangle ABC 500 - - - Ar. Co. 7.30103: the triangle ADG 120 - - - - - - 2.07918 *. AB2 JAB 40 - - - - - - 1.60206 AB 40 - - - - - - 1.60206 AD - - 2)2.58433 AD 19.6 per. - - - - - - - - - 1.29216 2. Given the area of a triangle ABC, 10 acres, and the base AB 25 ch., to find BD a part of the base, so that a line DG running from the point D, parallel to the side AC, may cut off a triangle BDG containing 4 1 acres. Ans. BD =16.77 ch.t * The truth of this rule is manifest from 19.6. This problem may be neatly constructed as follows: Let ABC, Fig. 90, be the given triangle, and AB the given base; on AB describe the semicircle AEB, and take AF to AB in the ratio of the part to be cut off, to the whole triangle; draw FE perpendicular to AB, meeting the semicircle in E, join AE, and make AD equal to AE; from D draw DG parallel to BC, and the thing is done. For, join EB, and we have, by similar triangles, AB: AE:: AE: AF; therefore (20.6 cor. 2) AB: AF:: AB2: AE2 (AD2):: [19.6] ABC: ADG. t If it be required to produce two sides of a given triangle so far that the triangle formed by these sides produced, and a line drawn between them parallel to the third side, may contain a given area, it may be done by the above rule. Thus, Fig. 90. ADG: ABC:: AD2: AB2. CHAP. IV.] LAYING OUT AND DIVIDING LAND. 175 PROBLEM IX. The bearings of two adjacent sides Fig. 91. AD, AE, Fig. 91, of a tract of land being given, to cut off a triangle C ABC containing a given area by a line BC running a given course. RULE. A B I From the given bearings of the lines, find the angles A, B, and C; then, As the rectangle of the sines of the angles A and B, Is to the rectangle of radius and sine of the angle C; So is twice the given area, To the square of the side AB.* In like manner the other sides may be found; or having found one side, the others may be found by trig. case 1. EXAMPLES. 1. Let the bearing of AD, Fig. 91, be N. 870 30' E. and of AE, N. 27~ 30' E.; it is required to cut off 10 acres by a line BC running N. 380 W. * The truth of this rule is evident from Fig. 92. the demonstration to prob. 4, chap. 3. Construction. Draw AD, AE, (Fig. 92,) according to the given bearings, and in c AD take AF equal the square root of the I given area, and on it describe the square AFGHII; make IE =AI, and draw ED, according to the reverse bearing of the division line BC, meeting AD in D; on AF D AD describe a semicircle, and produce. G-F to meet it in K, join AK and make,,, / AB equal to it; draw BC parallel to DE, and ABC will be the triangle required.'' For join IF, EF and KD; then (31.3, and K cor. 8.6) AD: AK (AB):: AK (AB): AF; or (cor. 19.6) AD: AF:: ADE: ABC; but (1.6) AD: AF:: ADE: AFE; therefore (11.5) ADE: ABC:: ADE: AFE, and consequently (9.5) ABC=AFE; but because AI=IE, AFE=2AFI=(41.1) AFGII; therefore ABC=AFGH=the given area of the triangle. 1 76 LAYING OUT AND DIVIDING LAND. [CHAP. IV. AD, N. 870 30' E. BA, S. 87~ 30' W. I CA, S. 27~ 30' W. AE, N. 27 30 E. BC, N. 38 00 W. ICB, S. 38 00 E. Angle A, 60 00 125 30 AngleC, 65 30 180 00 Angle B, 54 30 A 600 00' Ar. Co. 0.06247 As sin. A Xsin. B, B 54 30 0.08931 B 54 30 - 0.08931 C 65 30 - - 9.95902 rad. X sin. C, rad. 10.00000:: twice the given area, 200 sq. ch. - - 2.30103 AB2 - 2)2.41183 AB 16.07 -1.20591 2. Given the bearing of one side of a tract of land, S. 530 15' E., and the bearing of an adjacent side taken at the same angle, N. 550 00' E., to cut off 4 acres by a line running N. 40 00' W.; required the distance on the first side. Ans. 9.76 ch. Fig. 93. F PROBLEM X. The lbearings of three adjacewt sides, EA, AB, BF, Fig. 93 or 94, of a tract of land, and the length of the middle side |~ | L,'" $ AB, being given, to cut off a, / 8.-,',, t:: trapezoid ABCD, containing A. - 1. a given area, by a line DC, KPz,fg-.. E parallel to AB. RULE. From the given bearings find the angles A and B; add these together, and take the difference between their sum and 180~, and call it P. Then, As the product of' the sines of A and B, Ts to the product of radius and sine of P; CHAP. iv-.] LAYING OUT AND DIVIDING LAND. 1 77 So is twice the area to be cut off; To a fourth term. When the sum of the angles A and B is greater than 180~, add this fourth term to AB2; but when the sum of these angles is less than 180~, take the difference between this fourth term and AB2. The square root of the result will be DC. Then, As the sine of P, Is to the sine of B, So is the difference between DC and A13, To AD.* * DEMONSTRATION. Produce EA and FB, Fig. 95, to meet in P. Then (19.6) PDC: PAB: CD2: AB2, or (17.5) ABC): PAB:: CD2 —AB': AB2, or (A.5) PAB: ABCD:: AB2: CD2-AB2, or (15.16.5) 2PAB AB2:: 2ABCD: CD2 —AB2. But by the demonstration to prob. 4, chap. 3 LPAB: AB2:: sin. Ax sin. B: rad. x sin. P. Consequently, (11.5) sin. A x sin. B: rad, x sin. P:: 2ABCD: CD2-AB2. Fig. 94. B Fig. 95. F C -N,,,:.... Now it is plain that.- ".AB2, added to this 4th term, gives CD2. A p ___...... _similar demonstration a - applies when the sum of the angles A and B is less than 180~, as in Fig. 94_ The latter part of the rule does not require demonstration. Construiction. Draw AI, Fig. 93, perpendicular to AB, and mako it equ:al to the quotient of twice the given area divided by AB. From I, draw I11 parallel to AB meeting AE and BF, in G and II, and on GII, describe tlhe semicircle GMII. From A. draw AL parallel to BF; and make LEI perpeidicular to GII. Wllithl the distance IIM and centre IT, describe thle arcl MNi: and from N, draw ND parallel to AL. Lastly, draw DC parallel to AB, anid it will be the division line required. For join B1D, BG, and CG, Fig. 95. Then by similar triangles PG: PD:: GIL: DC:: (: G:II I:: 11:IlL:: DC: AB:: PC: PB. IIence (3.6) CG is parallel to BDL; anrd consequently the tritangle BDC is equal to B3' I). To each of these, add ABI), Then we lhuvec AB3CD=ABC. But it is llain firOn the construction that ABG is equal to the given area. linlce IIjCD is equal tc tthe given area. When\i the fium of the anales A and IB is less than Iv0.;as in Fii~. 94. 178 LAYING OUT AND DIVIDING LAND. [CIHAP. IT. EXAMPLES. 1. Given the bearing of EA, Fig. 93, West, AB, N. 10~ E. dist. 15 ch.; and BF, N. 58~ 30' E. to cut off 10 acres by a line CD, running parallel to AB. Required the length of the division line and the distance AD. AE, N. 900 E. BF, N. 580 30' E. A, 800 0' AB, N. 10'E. BA, S. 10 OW. B, 131 30 A -800 48~ 30 211 30 180 0 180 0 B=131 30 P-310 30' 4'ABsi. Axsn. i~ A, 800 00' Ar. Co. 0.00665 is sin. A sin. B B, 131 30 0.12554:twice the given area 200 sq. ch. - - - 2.30103: fourth term 141.68 - - - - - 2.15131 AB2= 225. As sin. P, 31430' - - Ar. Co. 0.28191: sin. B, 131 30 - 9.87446 DC — B, 4.15 - - 0.61805:AD 5.95 - - 0.77442 the semicircle must be described on AB; the point L must be deter mined by drawing GL parallel to FB: and the are MN must be described with the radius BM and centre B. The other parts of the construction are the same as before. CHAP. IV.] LAYING OUT AND DIVIDING LAND. 179 2. Given the bearings of three adjacent sides of a tract of land and the length of the middle one as follows: 1st. N. 200~ W.; 2d. N. 60~ 30' E. dist. 6 ch.; 3d. S. 61~ 30' E.; to cut off a lot containing 21 acres, by a line parallel to the 2d side. Required the length of the division line and the distance on the 1st side. Ans. Division line 8.70 ch.; distance on 1st side 3.45 ch. 3. Given as follows: 1st side N. 310 15' WT.; 2d N. 58~ 45' E. dist. 13.50 ch.; 3d S. 140 45' E.; to cut off 8 acres by a line parallel to the 2d side. The length of the division line and the distance on the 1st side are required.- Ans. Division line 11.61 ch.; distance on the 1st side 6.38 ch. PROBLEM XI. TLhe bearings of three adjacent sides Fig. 96. F EA, AB, BF, Fig. 96, of a tract ~of land, and the length of tlhe middle side AB being given, to -, " cut of a trapezium, ABCD, con-. taminng a given area, by a line \'.CD, running a given course. A.. RULE. Draw AS parallel to BF, meeting CD or CD produced, in S. From the given bearings, find the interior angles A, B, C, and D; add A and B together, and take the difference between their sum and 180~, and call it P. Then, As the product of the sines of C and D, Is to the product of radius and sine of P; So is twice the area to be cut of, To a fourth term. 180 LAYING OUT AND DIVIDING LAND. [CHAP. IV. Also, as the product of the shies of C and D, Is to the product of the sines of A and B; So is AB2, To a fourth term. When the sum of the angles A and B, is greater than 180~, add these two fourth terms together; but when the sum of these angles is less than 1800, take the difference of the fourth terms. The square root of the result will be CD. Then, < As sin. C:'sin. B:: AB: CS. The difference between CD and CS, gives DS. Then, As sin. P sin. C:: DS: AD.* Fig. 97. F * DEMONSTRATION. Produce EA and FB, Fig. 97, to meet in P; draw AR and BU, each parallel to CD, and let VW, also, a.... "s parallel to CD, make the trian-,gle PVW equal to PAB. Then (15.6) PA: PV:: PW: PB; \,..........'. / but (4.6) PA: PV:: AR: \,;~.:' ".: V/, VW, and PW: PB: YW: BU. Therefore (11.5) AR:'- a_._ VWV: YWV: BU, and hence (17.6) ARXBU VYW". But by trigonometry, As sin. ARB (sin. C): sin. B:: AB: AR, sin. AUB (sin. D): sin. A:: AB: BU. Hence (23.6) As sin. CXsin. D: sin. AXsin. B:: AB2: ARXBU; Or, sin. CXsin. D: sin. AXsin. B:: AB2: VW2. And by the demonstration to the rule in the last problem, we have sin. CX sin. D: rad. x sin. P:: 2VWCD (2ABCD): CD"~-VW2. The sum of these fourth terms gives CD2. The demonstration is similar, when the sum of the angles A and B is less than 180~. Construction. From B, Fig. 96, draw BU according to the reverse bearing of the division line DC, meeting EA or EA produced in U. Make AI perpendicular to AB, and equal to the quotient of twice the given area, divided by AB. Draw IG parallel to AB; GII parallel to BU; and UL parallel to BF On GIl, describe the semicircle GMII, and make LM perpendicular to GI; with the radius IIM and centre II, describe the arc MN; and from N, draw ND parallel to UL. From D, draw the division line DC parallel to BU. IDING LAND. 18] EXA RMPLES. 1. Let the bearings of EA, Fig. 96, be N. 800 30' W.; AB, North, dist. 12 ch.: and BF, N. 58~ E.; it is required to cut off 10 acres by a line DC, running N. 140 30' W. AE, S. 800 30' E. BF, N. 580 F. CB, S. 580 00' W. AB,N. 0 00 E. BA, S. 0 E. CD, S. 14 30 E. 80 30 58 C-720 3f0 180 0 180 A.- 990 30' B —122 DA, N. 80~ 30' W. A = 990 30f DC, N. 14 30 W. B= 122 00 D_-660 00' 221 30 180 00 P=41 30 s sin. Cysin. D C, 720 3 - - Ar. Co. 0.02058 D, 66 00 - 0.03927 d in. P, rad. —. 10.00000 Prad. X d P, 41 30 - 9.82126:: twice the given area 200 sq. ch. - 2.30103 fourth term 152.1 -2.18214 When the sum of the angles A and B is less than 1800, the semicir.e; must be described cn BU; the point L must be determined by drawing GL parallel to BF, and the arc MN must be described with the radius BM and centre B. The demonstration is exactly the same as that for the construction of the last problem. 16 182 LAYING OU AND DIVIDiNG LAND. [CHAP. IV. (x C 720 30' Ar. Co. 0.02058 As si. Cx sin. D, 1) 66 00 - 0.03927 A 99 30 - - - - - 9.9940X3 sin. Ax sin. B B 122 00 - - - 9.92842 AB 12 - - - - - 1.07918::AB2 AB 12 -.. - -1.07918: fourth term 138.24 - - - 2.14063 fourth term 152.10 DC= V 290.34= 17.04 As sin. C 72~ 30' Ar. Co. 0.02058: sin. B 122 00 9.92842:: AB 12 1.07298 CS 10.67 - - - - 1.02818 DC 17.04 DS 6.37 As sin. P 41~ 30' ---- - Ar. Co. 0.17874 sin. C 72 30 - - - 9.97942::DS 6.37.- _ 0.80414: AD 9.17- 0.96230 2, Given the bearings of three adjacent sides of a tract of' land and the length of the middle one as follow: 1st. N. 310 15' WV.; 2d. N. 58~ 45' E. dist. 13.50 ch.; 3d. S. 14~ 45' E.; to cut off 8 acres by a line from the 1st. side to the 3d. running S. 8'o~ 30' E.; requirea the length of the division line and the distance on the 1st side. Ans. Division line 12.76 ch.; distance 1st. side 2.69 ch. 3. Given as follow; 1st side, N. 74~ 45' W.; 2d. N. 37' E. (list. 17.24 ch.; 3d. N. 840 E.; to cut off a field containing 20 acres, bv a line from the 1st. side to the 3A. CnAP. IV.] LAYING OUT AND DIVIDING LAND. 183 running N. 20~ E. The length of the division line and the distance on the 1st side are required. Ans. Division line 19.68 ch.; distance on 1st side 14.01 oh. PROBLEM XII. The bearings of several adjacent sides, EA, AV, VW, WX, XB, BF, Fig. 98, of a tract of land, and the distance of tzch, except the first and last, being given, to cut off a given area, by a line DC, running a given course. RULE. Join AB and calculate the area of AVWXBA, and the bearing and distance of AB. Subtract the area of AVWXBA from the area to be cut off, the remainder will be the area ABCD. Then with the bearings of EA, AB, BF, DC, the distance AB, and the area of ABCD, proceed as in the last problem. EXAMIPLES. Px Fig. 98. 1. Let the bearing / of EA, Fig. 98, beN. 480 30' W.; AV, S. 78c W. dist. 8 ch.; VW, N. 260 30' W. dist. 11.08 ch.; WX, - - -.- - a N. 380 30' E. dist. Wx; -- 12.82 ch.; XB, S. 640 E. dist. 10.86 E ch.; and BF, S. 860 E. It is required to cut off 30 acres by a line DC, running N. 320 15' E. Stat. Bearing. IDiSt. N. S. E. i- - DD W.D.D. - Pliers.- IN. Ar. S. Areas.l AV S 80W 800 1 6; 783 I 1279 0.00! VW N. 262Wt 11.08 9.91 4.95 12.78 12.78 W 126.64>_{ WX N 381 F12.82 10.03 7-981 3.()3 9.75 W 97.7925 XB S. 61i E. 10.86 476.716 1 17.74 7.99 E 38.03241 BA (13.52) (4.96)5 4.80 12.79 E 172.9208 19.94 19.94 17.. I 17.74.57 25.57 435.3955. Sq. ch. 217.69775 184 LAALING OUT AND DIVIDING LAND. [CHIAP. IV. Sq. cf. Area to be cut off 300. Area of AVWXBA 217.69775 Area of ABCD 82.30225 As diff. of lat. of BA, 13.52 S. Ar. Co. 8.86902: dep. do. - - 4.96V. - - - - 0.69548::rad. - 10.00000 tang of bearing of BA, S. 20~ 9' WV. - - - 9.56450 As rad. - Ar. Co. 0.00000: sec. of bearing 200 9' - - - - 10.02743: diff. of lat. 13.52 - 1.13098 BA 14.40 - - - - - - - - - - - 1.15841 The angles, found from the bearings, are A= 111 21', B= 106 9', C-610 45', D-800 45' and P-370 30'. As sin. C x in. D C, 6145'- - - Ar. Co. 0.05508 D, 80 45 - - 0.00568: i rad. - 10.00000 rad. xsin. P, 37 30 9.78445:: twice the area ABCD 164.6 sq. ch. - - 2.21643: fourth term 115.25- - 2.06164 * C, 610 45' Ar. Co. 0.05508 As sin. Cxsin. D, 80 45 - 0.00568 A, 111 21 - - - - 9.96912 B B 106 9 - - - - 9.98251 ABAB, 14.40 - - - 1.15836i AB, 14.40 - - - - 1.15836 fourth term 213.36 - 2.32911 do. 165.25 DC= V/328.61 —18.13 CHAP. IV.] LAYING OUT AND DIVIDING LAND. 185 As sin. C, 610 45' Ar. Co. 0.05508 sin. B, 106 9 -. 9.98251:: AB, 14.40 - - - - - - - -1.15836: CS, 15.70 - - - - 1.19595 As sin. P, 37~ 30' Ar. Co. 0.21555 sin. C. 61 45 - 9.94492:: DS, 2.43 - - - - - - - - -0.38561: AD, 3.52 - - - - - - - - -0.54608 2. Given as follow; 1st side, N. 620 15' W.; 2d N. 190 E. dist. 18 ch.; 3d S. 77~ E. dist. 15.25 ch.; 4th S. 270 E.; to cut off 35 acres by a line, from the first side to the last, running N. 820 30' E. Required the length of the division line, and the distance on the first side. Ans. Division line 22.98 ch.; distance on 1st side 5.14 ch. PROBLEM XIII. The bearings of seve- N Fig. 99. ral adjacent sides, AB, BC, CD, DE, Fig. 99, of a tract of land, and the distance of each, ex- / / cept the last, being given, to cut off a given area by a line / All running from B the angle A, and.falling on the side DE. RULE. By Prob. 9, chap. t, change the bearings of all the given sides, so as to make the side, on which the division line is to fall, a meridian. 16* 2A 186 LAYING OUT AND DIVIDING LAND. [CHAP. IV With the given distances and chlange(l bearings, find the corresponding differences of latitude and departures; add together the numbers in each departure column, and take the difference of their sums, which will be the departure of the division line, and must be placed in the proper column, opposite said line. Then having all the departures, find the double departures, as in Prob. 11, of the last chapter. Find also the multipliers, beginning with the one to correspond with the division line, and fissuming it 0; multiply the known differences of latitude }1y their corresponding multipliers, and place the prolucts in the proper columns of north and south areas. Add together the products in each of the columns of areas, and subtract the less sum from the greater; take the difference between the remainder and double the area to be cut off, and divide it by the multiplier corresponding to the side on which the division line is to fall; the quotient will be the difference of latitude of this side, which place against it, in the column of north or south latitude, according as its changed bearing is north or south. Then add together the numbers in each latitude column, and take the difference of their sums, which will be the difference of latitude of' the division line, of the ssamle name with the less sum. With the difference of latitude and the departure of the division line, find, by Prob. 10, chap. 1, its changed bearing and its distance. Then find the true bearing by note to the rule in Prob. 9, chap. 1.* EXAMPLES. 1.'Let the bearing of AB be N. 62 1 W. 14.75 ch.; BC, N. 190 E. 27 ch.; CD, S. 770 E. 22.75 ch.; and DE, S. 27~ E.; it is required to cut off 70 acres by a line AH, running from the angle A and falling on the side DE. * The reason of this rule is sufficiently obvious without a demonstration. -!TAP. IV.1 LAYING OUT AND DIVIDING LAND. 187 1 1 1 co f1'Z 1 I I:- 1.1...(0 t o to - 0 0'~ C~o 00to- C As diff. lat. of I-A, 5.22 S. - - Ar. Co. 9.28233:dep. do. 28.33 W. - - 1.45225: rad. - --— 10.00000: tang. changed bear. of HA, S. 79~ 31' W 10.73458 Subtract 27 00 True bearing of HA, S. 52 34 W. 188 LAYING OUT AND DIVIDING LAND. [CHAP. Iv'. As rad. -Ar. Co. 0.00000 sec. changed bearing of HA, 79~ 34' - - 10.74210 diff. lat. 5.22 0.71767 dist. All, 28.83 - - - - - 1.45977 Hence AH, bears N. 520 34' E. dist. 28.83 ch. 2. Given as follow: 1st side S. 780 W. 8 ch.; 2d N. 261~ W. 11.08 ch.; 3d N. 38-i E. 12.82 ch.; 4th S. 640 E. 10.86 ch.; 5th S. 23i~ E.; to cut off 25 acres by a line running from the place of beginning, and falling on the 5th side; required its bearing and distance. Ans. N. 450 1' E. dist. 10.67 ch. PROBLEM XIV. Fig. 100. The side AB, C, CA of a triangular piece of ground being given, to divide it into two parts having a given ratio, by a line FE, / rrunningparallel to one of the sides x x i: as BC. RULE. As the sum of the numbers expressing the ratio of the parts, Is to that number of the ratio which corresponds to the part to be adjacent to A; So is the square BC, To the square of FE. Then, As BC: AB:: FE: AF.* * DEMONSTRATION. Let m to n be the ratio of the part AFE to the part FECB; then (18.5) m+n: m:: ABC: ADE: (19.6) BC2: FE2. Construction. On AB describe the semicircle AMB, and by Prob. 17, Page 36, divide AB in K, so that AK may be to KB in the given ratio of the nart AFE to the part FECB; draw KM perpendicular to AB, meeting the CHAP. IV.] LAYING OUT AND DIVIDING LAND. 189 EXAMPLES. 1. Let AB be 21.26 ch.; BC, 12.76 ch.; and AC, 19.30 ch.; it is required to divide the triangle by the line FE, parallel to BC, so that the part AFE may be to the part FECB as 2 to 3. As 5: 2:: 12.762: FE= 65.12704. FE - / 65.12704 =8.07. As 12.76: 21.26:: 8.07: AF= 13.45. 2. The three sides of a triangular piece of land, taken in order, measure 15, 10, and 13 chains respectively; it is required to divide it into two equal parts by a line parallel to the second side. What will be the length of the division line and its distance from the place of beginning, measured on the first side? Ans. Division line 7.07 ch.; dist. on 1st side 10.61 ch PROBLEM XV. The bearings and distances of the Fig. 101. sides AB, BC, CA, Fig. 101, of a triangular piece of ground being given, to divide it into two parts A having a given ratio, by a line FE,.. running a given course. RULE. As the product of the sines of F and E, Is to the product of the sines of B and C; semicircle in AM, and with the radius AM and centre A, describe the are MF. From F, draw the division line FE parallel to BC. Since, (35.3, and cor. 8.6) AB: A.M (AF):: AM (AF): AK, we have (20.6 cor. 2) AB: AK AB: AF2B: ABC: AFE. IIence the truth of the construction is evident 190 LAYING OUT AND DIVIDING LAND. [CHAP. IV, So is the square of BC, To a fourth term. Multiply this fourth term by that number of the ratio which corresponds to tile part to be adjacent to the angle A, and divide the product by the sum of the numbers expressing the ratio. The square root of the result will be FE. Then, As sin. A: sin. E:: FE: AF.* EXAMPLES. 1. Let the bearing of AB, be S. 824~ E. dist. 14.17 ch.; BC, N. 184~ W. 8.51 ch.; and CA, S. 611~ W. dist. 12.87 ch.; it is required to divide the triangle by the line FE, running N. 141~ E. so that the part AFE may be to the part FECB in the ratio of 2: 3. Fig. 102. n * DEMONSTRATION. Draw CG and BR, Fig. w 102, parallel to EF; and let VW, also parallel c