UDENTS CULTURE SOR ^ MATHEMATICS FOR STUDENTS OF AGRICULTURE A SERIES OF MATHEMATICAL TEXTS EDITED BY EARLE RAYMOND HEDRICK THE CALCULUS By Ellbry Williams Davis and William Charles Brenke. ANALYTIC GEOMETRY AND ALGEBRA By Alexander Ziwet and Louis Allen Hopkins. ELEMENTS OF ANALYTIC GEOMETRY By Alexander Ziwet and Louis Allen Hopkins. PLANE AND SPHERICAL TRIGONOMETRY By Alfred Monroe Kenyon and Louis Ingold. ELEMENTS OF PLANE TRIGONOMETRY By Alfred Monroe Kenyon and Louis Ingold. ELEMENTARY MATHEMATICAL ANALYSIS By John Wesley Young and Frank Millett Morgan. PLANE TRIGONOMETRY By John Wesley Young and Frank Millett Morgan. COLLEGE ALGEBRA By Ernest Brown Skinner. MATHEMATICS FOR STUDENTS OF AGRICULTURE AND GENERAL SCIENCE By Alfred Monroe Kenyon and William Vernon LOVITT. MATHEMATICS FOR STUDENTS OF AGRICULTURE By Samuel Eugene Rasor. THE MACMILLAN TABLES Prepared under the direction of Earle Raymond Hedrick. PLANE AND SOLID GEOMETRY By Walter Burton Ford and Charles Ammerman. CONSTRUCTIVE GEOMETRY Prepared under the direction of Earle Raymond Hedrick. JUNIOR HIGH SCHOOL MATHEMATICS By William Ledley Vosburgh and Frederick William Gentleman. MATHEMATICS FOR STUDENTS OF AGRICULTURE BY SAMUEL EUGENE RASOR PROFESSOR OF MATHEMATICS IN THE OHIO STATE UNIVERSITY NrtD gotft THE MACMILLAN COMPANY 1931 POINTED IN THE UNITED STATES OF AMERICA Copyright, 1921, Bt the macmillan company. All rights reserved no part of this book may be reproduced in any form without permission in writing from the publisher. Set up and electrotyped. Published September, 1921. NorbjDoti Press J. S. Gushing Co. Berwick & Smith Co. Norwood, Mass., U.S.A. PREFACE This book presents a year's work in mathematics for students taking agricultural courses in secondary, vocational, and technical schools and in colleges and universities. By omitting the more ele- mentary or the advanced topics, as may seem desirable, or by prop- erly selecting the exercises to be solved, the book will serve quite naturally for a half-year's work. The modem agriculturalist must devote a larger and larger part of his time to mental activity along scientific lines. To do this rationally he must have better methods for, and more practice in, estimating, computing, and comparing the materials and processes with which he works. The object of the present text is to furnish a better basis for effecting these ends. The text has been used in the form of mimeographed notes for several years in the classroom, and has been modified from time to time in the Hght of experience and of new conditions. It was soon found that it was not safe to assimie too much preparation on the part of students taking the course. Consequently, material is pre- sented in the text for study and practice in basic elementary pro- cesses in arithmetic, algebra, and geometry. Some study of this material will suffice for those students who have not had the proper preparation for the course as well as for those students who wish to review elementary facts and principles. The unifying element in the book is the attempt to make the prin- ciples of arithmetic, algebra, geometry, trigonometry, and graphic representation fimction with the student's interest and point of view. It is hoped that in this way he may find a natural tendency to use this most basic science in the solution at least of those problems which arise in his own sphere of activity. We may, therefore, expect the student to acquire in this way not less of mathematical value and content but more of it. The following are features of various chapters: Chapter I presents suggestions in drawing as a means of review- ing facts from geometry and connecting them naturally with such observations as the student can make; for example, estimating and computing the grade of a roof or a hill or a ditch or a railroad, the 910132 VI PREFACE projections of the sides of a farm contour on a base line, scale draw- ings of farms and farm buildings, and the composition and resolu- tion of forces graphically. In Chapter II a basis is presented for rational calculation and for estimation, including the question of accuracy. The topics consid- ered are percentage, profit and loss, bank discount, elementary men- suration of common plane figures and solids, including rules for esti- mating silos. In Chapter III a treatment of indirect measurement and propor- tion is given. This includes also the study of the sine, cosine, and tangent ratios and tables for them, with applications to simple prob- lems in farm surveying. A review of the main facts of elementary algebra may be obtained from Chapter IV. Graphic representation of quantities, including price curves, weight and age curves, leveling, etc., is presented in Chapter V. A rather fundamental treatment of compound interest, annuities, depreciation, and farm loans is given in Chapter IX. Chapter X treats averages and mixtures. It includes a discussion of dairy problems, arithmetic and geometric means, weighted aver- ages, the median, the mode, fertilizer mixtures, and cement mixtures. The chapter on Land Surveying (Chapter XIII) may be of inter- est from the standpoint of the applications and the new material. Chapter XIV is entitled Simple Machines. It presents a treat- ment of levers, the wheel and axle, the inclined plane, the screw, pulleys, safe loads for beams, work, horse-power, working day for a horse, draft, etc. The tables printed in the Appendix at the end of the book are adapted to the requirements of computing and estimating the exer- cises and examples. Throughout the text the exercises and illustra- tions are drawn largely from experience and the point of view of the agriculturahst. The author desires to acknowledge his indebtedness to Dean Al- fred Vivian and members of the faculty of the College of Agriculture of The Ohio State University, to R. D. Bohannan, Professor of Mathematics in The Ohio State University, and to E. R. Hedrick, Professor of Mathematics in the University of Missouri, for the in- spiration of their interest and suggestions. S. E. Rasor. The Ohio State University. CONTENTS PAGE Chapter I. Drawing Graphic Solutions 1 Chapter II, Computation Measurement 10 Chapter III. Indirect Measurement Trigonometry Survey- ing 44 Chapter IV. Review of Algebra 63 Chapter V. Graphic Representation of Quantities ... 88 Chapter VI. Graphs in Algebra 104 Chapter VII. Computation by Logarithms 124 Chapter VIII. The Progressions 139 Chapter IX. Compound Interest Annuities Depreciation . 149 Chapter X. Averages and Mixtures 167 Chapter XI. Geometry Mensuration . . . . . 184 Chapter XII. Oblique Triangles 200 Chapter XIII. Land Surveying 218 Chapter XIV. Simple Machines 235 Chapter XV. Composition and Resolution of Forces . . . 261 vu A LIST OF SIGNS AND SYMBOLS + , redid plus. , read minus. X , or , read times. -T- , read divided by. =, read equals, or is equal to. z^, read is not equal to. <, read is less than. >, read is greater than. .'., read therefore. ', redid feet, e.g. 2', means 2 feet. ", read inches, e.g. 2", means 2 inches. ^ ^' ' These are signs of aggregation. The expressions which they inclose are to be treated together as one quantity. [ ], Brackets. { }, Braces. a : b, read a is to b. Va, read square root of a. Va, read n'* root of a. a^, read a to the n^* power, or a exponent n. log oW, read logarithm of n to the base a. 'r, read perpendicular, or is perpendicular to Z, read angle, A, read angles. A, read triangle, A, read triangles. Kt. A, read right triangle. O, redid parallelogram. MATHEMATICS FOR STUDENTS OF AGRICULTURE CHAPTER I INTRODUCTION DRAWING GRAPHIC SOLUTIONS 1. Instruments and Materials. A large number of the exercises and problems in this book are to be solved graph- ically, i.e. by drawing appropriate diagrams. To do this work neatly and accurately, a few simple drawing instru- ments are essential. The materials and instruments needed for the work are described briefly below.* (a) Pencils. One medium hard pencil sharpened to a round point may be used for lettering. A hard pencil sharpened to a round point may be used for sketch- ing and marking points on paper. Another hard pencil, sharpened to a chisel-point (Fig. 1), may be used to draw straight lines. A piece of sandpaper may be used to whet the points. (6) A drawing hoard, about 12 or 15 by 18 inches. It should be made of soft wood and have at least . , , Fig. 1 one straight edge. (c) A T-square, 15 to 18 inches long. As shown in Fig. 2, a T-square consists simply of a wooden strip, Ay Those students who are taking courses in drawing are already in posses- sion of many of these instruments and materials needed for graphical work. 1 MATHEMATICS [I, 1 .A 1 called: th>bhead'f to. wfeich' is fastened a thin, straight-edged blade, B. ' ' ' . (d) Two trari'spa'irent Mangles. Each of these should have one right angle. The acute angles of one triangle should be 45, while the acute angles of the r other should be 30 and 60 (Fig. 3). ^ {e) A scale with the unit of length divided into tenths is to be preferred to one with the unit of length divided into eighths, twelfths, or sixteenths. (/) A "protractor, Fig. 3, 4 to 6 inches in diameter is recommended. It consists simply of angular graduations upon a semicircular piece of metal or some trans- -parent material. It is used for laying off and measuring angles upon a drawing. ig) Two pairs of compasses. One should be a pair of 6-inch pencil compasses, the other should be inking compasses of the same size. Qi) A pair of 6-inch hairspring di- viders. These are to be used in calculating lengths, distances, etc. on drawings made to scale. T-SQUARE Fig. 2 30 Triangle 45 Triangle Protractor Fig. 3 I, 2] INTRODUCTION 3 2. General Directions. All work must be done with neatness and accuracy. The acquisition of these two virtues in graphical work may require an unusual amount of time and effort. However, when they are once estab- lished as habits, much time will be saved in avoiding and in detecting errors, as well as in checking results. ABCDErGH/JKLMNOF^OnSTUVWXYZ obc defgh ijk/mnop qrstuv wxyz /234567a90&. Fig. 4. A Set of Letters and Figures All drawings should be done in pencil. After some train- ing in the use of the ruling pen, the drawings, if it is so desired, may be " inked in " with drawing ink. Figure 4 shows a set of letters and figures. They may be used in lettering descriptive work on a drawing instead of using ordinary writing. EXERCISES 1. Draw a straight line. For this purpose two things are required : (a) A plane or flat surface on which to draw the line. (6) A straightedge to be used as a guide. 2. State some facts about a straight line by testing a straightedge. To do this, select two points through which the straightedge is to pass, draw a line through them, reverse the straightedge, and redraw the line. How many points determine a straight line? How many points may two distinct straight lines have in common ? When does a straight line lie wholly in a plane? 3. On what surfaces, other than the plane, is it possible to lay a straight line so that each of its points coincides with the surface ? In this connection, examine a cylinder ; a cone ; a riding saddle ; a capstan. 4. The part of a straight line between two of its points is called a line segment or simply a segment, and is designated by its end points, 4 MATHEMATICS [I, 2 or sometimes by a small letter near the middle of the segment. Thus we speak of the ''segment AB'' or the "segment a." See Fig. 5. Fig. 5 On a given straight line construct a segment equal to a given seg- ment. Do this in two ways. (a) By the use of a pair of dividers. (b) By the use of a scale or a straightedge, perhaps an inch scale, or a centimeter scale. Which of these methods is the more accurate? How could the accuracy of the second method be improved? 5. Locate any four points A, B, C, D in the plane and draw the straight lines which any pair of them determines. In general, how many of these lines are there? Under what conditions are there less than this number of lines ? 6. Draw four straight lines at random. In general how many points of intersection are there? When are there less than this num- ber of intersections? What are the names of some of the possible figures obtained in this way? 7. Given two unequal segments a and b. Draw the segment equal to a-\-b. Equal to a b when a>b. Draw the segment equal to ab when a<^\/^ \a^ of 350' from an observer, subtends ^^ an angle of 22. ^^- ^ 5. How can a carpenter divide a 4f in. board into 3 equal strips without the use of fractions? See Fig. 9. 6. Divide a given line segment into seven equal parts. 4. Projection of a Line Segment. To project a given segment, as AB, Fig. 10, upon a line PQ drop perpendiculars to PQ from the end points of the segment. Then CD is the projection "of AB upon PQ and is called the horizontal * For suggestions on the architecture of farm buildings, see L. F. Allen, Rural Architecture, published by L. F. Allen, New York; Ekblaw, Farm Structures, Macmillan, 1916 ; G. G. Hill, Practical Suggestions for Farm Buildings, Wash., Gov. Ptg. Office, 1901 ; I. P. Roberts, The Farmstead, Macmillan, 1900. I, 4] INTRODUCTION projection in case PQ is horizontal. Projection on a vertical line is called vertical projection as BE, Fig. 10, in which AE is parallel to PQ, and BD perpendicular to it. c D ^ FiQ. 10 Similarly, PQ may be projected upon AB (produced if necessary) . In the case of a hill, or a grade, or a rafter, the horizontal projection is called the run of the rafter, etc., and the vertical projection is called the rise. The rise divided by run is called the slope; it is a measure of the steepness of the hill, or grade, or rafter. For example, if a rafter on a building Ridge Rise Slope =^^ Run 24' wide rises 6' at the comb of the roof, we say it rises 6' in 12' or 1 in 2, i.e. the slope is ^. If a hill rises 30' in a run of 100', it has a slope of 30 in 100, i.e. the grade is 30/100 or 30%. A 1% grade, as for a steam railroad, rises 1' in 100'. With reference to a roof, the word pitch is widely used ; it denotes rise of the roof divided by the span or whole width of the building (Fig. 11). Notice that the pitch of a roof is ha,lf gf the slope. 8 MATHEMATICS [I, 4 EXERCISES Draw each of the following figures on an enlarged scale and project each segment upon the other. Specify the projection in each case. Find the horizontal and the vertical projections, and find, by measuring, the per cent grade for each segment. c 3 Fig. 12 6. The altitude of an equilateral triangle whose side is 12' is pro- jected upon one of the adjacent sides. Find the projection. 7. Find, by drawing to scale, the side of a square whose diagonal is 10 rd. 8. Show how to bisect a given angle by using two carpenter's squares. [Hint. Use the geometric theorem that any point on the bisector of an angle is equidistant from the sides.] 9. Draw any oblique triangle. Draw the projections of two sides on the third side. 5. Graphic Resolution of Forces. A force may be repre- sented graphically by a straight line whose length represents the magnitude of the force, and whose direction is that of the force. See 213. In Fig. 13, F represents a force acting on W in the direction shown and at an angle BAC with the horizontal. Project F on a horizontal line through the point where the force is applied to W. In this way the horizontal projection Fig. 13 I, 5] INTRODUCTION 9 AB is said to be the horizontal component of the force F and the perpendicular BC is called the vertical component of the force. F is thus resolved into vertical and horizontal components. EXERCISES 1. A force of 10 pounds acts on a weight W at an angle of 30 with the horizontal. Draw the figure, making F 10 units in length. Project F on the horizontal line through the point of application of the force. By the scale, determine the magnitude of each component in pounds. What is thus the amount and the direction of the horizontal and the vertical pull on TF ? 2. In drawing a sled a horse pulls 200 pounds on his traces, which make an angle of 10 with the horizontal. Plot to scale the figure representing these conditions and determine in this way what hori- zontal force is employed in moving the sled along the ground and also what force tends to lift the sled vertically. 3. In moving a weight on a sled as in the previous exercise, it was found that the vertical pull was 40 pounds. Determine the pull on the traces by drawing to scale. What is the horizontal component which moves the sled? What is the effect of a "long hitch" in a case of this kind? 4. A weight of 40 lb. slides without friction down an incline which makes an angle of 30 with the horizontal. What force acting directly up the plane will keep the weight at rest ? [Hint. Draw as in the accompanying figure the segment AB 40 units in length and directly downward to represent the force due to the weight. Project this segment on the in- cline and call this projection AC. This is the component of the force down the plane. Now ZP=ZB = SO since S^ABC, PQR are similar. But ABC is half of an equilateral triangle and the side AC is one half oi AB. Thus AC rep- resents a force of 20 lb. Therefore a force of Pj^ ^^ 20 lb. acting directly up the plane will keep the weight at rest. The pressure on the incline is represented by CB perpendicular to it. Find it.] 5. What force is necessary to roll a barrel of cider weighing 200 lb. up an incline 12 ft. long into a wagon bed 3 ft. above the ground? CHAPTER II COMPUTATION MEASUREMENT 6. Measurements Are Approximations. Numbers met with in actual life are in general approximations. For example, if we should measure the width of a room and find the measurement to be 13' 6f ", obviously we could not be certain that this is the exact width. A more exact measure- ment could be obtained by using a ruler graduated to smaller divisions, for example to sixty-fourths of an inch. The width of the room is thus not measurable to an exact fraction of an inch. How nearly the measured width approximates the true width depends upon the wishes of the operator, upon his skill, upon the kind of instrument used, etc. Even the price of butter at 67^ per pound, while seemingly exact, is only approximate, if the weighing process be considered. Again, common fractions whose denominators contain factors other than 2 or 5 cannot be expressed exactly as terminating decimals. For example, one sixth of a dollar is 16.66 * ff = 17f^. The value may be expressed decimally to any degree of accuracy desired. Roots of numbers, while exact in themselves, are in general only approximations when expressed in decimal form. Thus, 1/6 = 0.17, or = 0.167, or 0.1667, etc., and V2 = 1.4, or =1.414, or =1.4142, depending upon the degree of accu- racy required. In all cases the number should express only the degree of accuracy demanded by the nature of the prob- lem. The cost of food to $0,001, or the volume of a barn to 1 cu. ft., are meaningless. 10 II, 7] COMPUTATION MEASUREMENT 11 7. Accuracy of Processes Involving Approximations. The accuracy or precision of a measurement is indicated by the number of significant * figures used to express the result and not by the number of decimal places. For example, 52.3 inches = 4.83 feet =1.35 meters = .293 rod according to the unit used. Notice that these four members represent the same length and are about equally accurate. They each contain three significant figures but in one case one decimal place is used, in another three decimal places. These numbers, and others like them, are said to be of three-figure accuracy. Such numbers may be obtained in measurements made with ordinary instruments and skill. Four-figure accuracy is sufficient for a great many engineer- ing operations. Five-figure accuracy results only from good instruments and great care. Thus a lot on Broadway, New York City, is measured to thousandths of a foot. Only experts can obtain six-figure accuracy, while eight-figure accuracy is as yet obtainable only in rare instances. In working with approximate numbers, the accuracy of calculated results can seldom equal that of the original numbers used. If a measured quantity is expressed in three figures, the result of operating with it should be expressed in general in not more than three figures by merely dropping unnecessary decimals. For example, 2.36 feet = 2.36X12 inches = 28.3 inches, not 28.32 inches. Here 2.36 feet as a measurement means that the true result is somewhere between 2.355 and 2.365. This is written (2.36 .005) feet. * All figures, other than zero, are significant in a number. A zero is significant unless used merely to locate the decimal point. In 206, 2.06, 0.206 the central zero is significant and these are called three-figured num- bers. In 0.0003 the zeros are not significant and this is called a one-figured number. A final zero may be significant ; thus 2.60 ft. means that the length measured is nearer 2.00 ft. than to 2.59 ft. or to 2.61 ft. If the result were written 2.6 ft., it would mean that the measurement had been made only to the nearest tenth of a foot. 12 MATHEMATICS [II, 7 Now (2.36 0.005) 12 inches = (28.32 0.06) inches, which shows that the final figure is in doubt. It should therefore be written 28.3 inches if no accompanying statement is made as to the error involved. However, it is usual to determine the doubtful place in a result from accompanying statements of probable error in the numbers used, and then drop unnecessary figures beyond the first doubtful one. 8. Percentage. The word per cent is a contraction of the Latin word per centum and means by the hundred, i.e. hundredths. It is written %. It furnishes a convenient way of making comparisons between quantities, rates, etc., by using a hundred as the basis. Thus, 6% means .06 or ^^ 135% means 1.35 or f^ Example. An analysis of a pound of alfalfa seed showed weed seed, 2^^ oz. ; dirt, Jf oz. ; good seed, 12^ oz. The relative amounts of weed seed, dirt, and good seed may be more easily recognized by stating them in hundredths of a pound, i.e. in per cents. The analysis thus shows weed seed 2^^ oz. = .16 lb. = 16% dirtif oz. = .06 1b.=6% good seed 12| oz. = .78 lb. =78% Computations involving per cent contain the following three elements. The Rate per cent, which is the number of hundredths taken. The Base, which is the number on which the hundredths are computed. The Percentage, which is the result of taking a certain per cent of a number. Rules. Base times Rate = Percentage. Percentage divided hy Base = Rate. Percentage divided hy Rate = Base. II, 8] COMPUTATION MEASUREMENT 13 Example 1. Find 8% of $150. 8% of $150 = .08 times $150 = $12.00 Example 2. What per cent of 750 is 25? 25 is ^%% of 750, but y^^j, =^^0 = -031 =3^%. Example 3. $1.05 is 15% of what amount? First Solution : By the rule, the amount required or the base = $1.05-^.15 = $7.00 Second Solution: If 15% of some number = $1.05, then 1% of that number = $.07, and 100%, or the number, =$7.00 Third Solution : Let x be the amount required. Then .15x = $1.05, whence a; = ^^^^ = $7.00 .15 The following equivalents may be found useful in esti- mating and in computing : 6% =.06 = ^. m% = i. 62i% = f. 6i% = .06i = A. 16|% = i. 66f % = f . 6f% = iV. 25% =i. 83i% = f 7|% = A. 33i% = f 87i% = |. 8i% = iV. 37i% = |. EXERCISES 1. Find 6i% of 300. Of 75. Of 18. Find 23% of 1684. |% of $24. 31% of 180 lb. 2. What per cent of 15 is 8? What per cent of 20 is 9? What per cent of 12| is 3 ? What per cent of 3| is 2| ? What per cent of 8 is 10 ? 3. What per cent of 371 feet is 5 feet? What per cent of 7 yards is 7 feet? What per cent of 2 miles is 528 feet? What per cent pure is an 18-karat gold ring? 4. 6 is 8% of what number? 4 is 10% of what? 18 is 2^% of what? 24 is 12% of what? 216 is 13% of what? 6 feet 2.8 inches is 22% of what length? 9.9 inches is 12^% of how many yards? 5. In a certain fertilizer for corn land, 47^% is cotton-seed meal, 43|% is phosphoric acid (phosphorus), and the rest kainit. How many pounds of each are there in a ton of the fertilizer? How much acid phosphate containing 16% phosphoric acid is re- quired to supply the phosphoric acid? 14 MATHEMATICS [II, 8 How much potash is there in the kainit required if kainit contains 13.54% potash? 6. Cooked eggs contain 12% protein and cooked roast beef contains 22.4% protein. If one egg weighs 2 oz., how many eggs will furnish as much protein as 1 pound of roast beef ? Which is the more economi- cal food with eggs at Q5^ per dozen and beef at 25^ per pound? 7. Ear corn shrinks in weight about 3% per month in drying. On Oct. 1 a man has 20 tons of corn. What should it weigh the following March 15? Which is the better proposition, to sell it "green" at $30 per ton or hold it till March 15 and sell it for $35 per ton, allowing 6% interest on the value of the corn? What "present" selling price (October 1) at 6% interest would be the equivalent of $35 per ton after drying (March 15) ? 8. Workmen strike for a 10% raise on their present scale of $2.25 per day. If they "stay out" 50 days, how long will it take them under the new scale to regain their loss, if they work 285 days to the year, not considering interest on the money involved ? 9. The champion Jersey cow (1914) (Sophie 19th of Hood Farm) produced 999.14 pounds of butter fat in 365 days. Her milk tested 5.69% butter fat. The champion Guernsey cow of the world, Murne Cowan (for a time champion of all breeds, 1915), produced 24,008 pounds of milk in 365 days, testing 4.57% butter fat. When Finderne Pride Johanna Rue (Holstein) completed her record in June, 1915, she became the world's champion dairy cow. She pro- duced, in 365 days, 28,403.7 pounds of milk containing 1176.47 pounds of butter fat. Compare these three breeds of dairy cows on the basis of the amount of milk produced ; on the basis of the amount of butter fat produced ; on the basis of the percentage of butter fat in the milk. 9. Trade Discounts. Profit and Loss. Merchants, jobbers, manufacturers, publishers, etc. usually have a fixed price, called the list price, from which they allow discounts to the trade, according to the credit of the customer, the amount purchased, the time of payment, etc. If there is more than one discount, it is understood that the first is the II, 9] COMPUTATION MEASUREMENT 15 discount from the list price, the second is the discount from the remainder, etc. The per cent of profit or loss in a transaction is usually reckoned on the cost. For example, if an article costing $6 is sold for $7.20, the gain is $1.20, i.e. it is ^ or 20% of the cost. EXERCISES 1. What is the cash cost of 12 bales of barb wire at $3.75, subject to a 10% discount, and a further discount of 2% if paid in 10 days? 2. Supphes for fences amounted to $58.75. If a discount of 12|% is allowed with an additional 2% off for cash, what is the cash cost of the supplies? 3. A dealer is offered 40% and 15% off on a bill of $126.83. What is the net price ? 4. Which is the better proposition, to pay a bill according to a straight discount of 25%, or according to successive discounts of 15%, 5%, 5%? How much better is it? 5. A dealer buys apples at $6.75 per barrel (3 bu.) and retails them at 75^ per peck. What is his gain or loss on 30 barrels, allowing 10% loss by decay ? 6. A renter raises 60 bu. per acre on 40 acres of corn where land is valued at $180 per acre. He sells the corn at $1.25 per bushel, pays a rental of 8% of the farm value, and estimates the cost of crop produc- tion at $45.75 per acre. How much and what per cent profit does he make on the cost of production? 7. A quantity of wheat was sold in succession by three dealers, each of whom made a profit of 4%. The last dealer received $1384, What did it cost the first dealer? 8. What per cent profit does a grocer make on sugar bought at 14^^ per pound, depreciated 10% from drying and downweight, and sold at 16)!f per pound? 9. A man buys 2000 bu. of potatoes at $1.45, sells f of them at 20% gain and the remainder at a loss of 1H%. Find his net rate of gain. 10. Goods are bought at 10% below list price and sold at 20% above list price. What is the per cent profit? 11. If a dealer gets discounts of 40% and 15%, what discount can he give on the list price to make 30% on his investment? 16 MATHEMATICS [11, 10 10. Simple Interest. Interest is money paid for the use of mone^^ The principal is the amount borrowed. The rate is the part of the principal to be paid for its use, usually for one year, even though it may be collected semiannually or quarterly if so specified. Thus the statement that a note or a bond is issued at 6% interest payable semiannually means that interest is payable semiannually at the rate of 6% per year. In computing simple interest, thirty days are considered an interest month and 360 days an interest year. The United States Government and some banks and business houses use the exact interest method. This method takes account of the exact number of days in the time interval, and proceeds upon the basis of 365 days to the year. Banks usually compute interest for the exact number of days. 11. Partial Pa3mients. Sometimes it is convenient to pay part of a note before it matures. Such payments should be indorsed on the back of the note, with the date and the amount of the payment. They are known as partial pay- ments. The Supreme Court of the United States, and nearly all of the states, use the following rule for partial payments, called the United States Rule, 1. Find the amount of the principal to the time of the first payment; if the payment equals or exceeds the interest, subtract the payment from the amount and treat the remainder as a new principal. 2. If the payment is less than the interest, find the amount of the same principal to the time when the sum of the payments shall equal or exceed the interest due, and subtract the sum of the payments from the amount. 3. Proceed in the same manner with the remaining payments until the time of settlement. II, 12] COMPUTATION MEASUREMENT 17 EXERCISES 1. $304.75 Chicago, 111., March 10, 1912. One year after date, for value received, I promise to pay to G. Botha, or order, Three Hundred Four and ^^q Dollars, with interest at 6%. B. Thoba. Indorsements : July 12, 1912, $100 ; Nov. 30, 1912, $100. What was due June 25, 1914? 2. $429.30 Columbus, O., April 13, 1913. On demand, I promise to pay Spencer and Arps, or order, Four Hun- dred Twenty-nine and y^/j Dollars, value received, with interest at 6%. R. Macklin. Indorsements: Oct. 2, 1913, $10; Dec. 14, 1914, $120; July 1, 1915, $25. What was due Jan. 1, 1916? 3. A note for $650, at 5%, is dated June 1, 1907, and has the follow- ing partial payments indorsed on it : Aug. 2, 1907, by labor, $5.00 ; Aug. 8, 1907, by labor, $2.00; Oct. 1, 1907, $95; June 1, 1908, $275; Dec. 1, 1908, $140. How much is due Jan. 1, 1909? 12. Present Worth. Bank Discoimt. The present worth of k dollars, payable at a given future time, is the sum that will amount to k dollars at the end of the time, at a specified rate of interest. The interest on the present worth of a debt due at a future time is called true discount. It is the difference between the debt and its present worth. Example. What is the present worth of $318 due in 1 year if money is worth 6% ? Solution : If x is the present worth, then the amount of x for 1 30*. at 6% is equal to $318, i.e. x(1.06)=$318, In any case, we have the following rule : Rule for Present Worth. To find the present worthy divide the given sum by the amount 0/ $1.00 for the given time at the given rate. 18 MATHEMATICS [II, 13 13. Bank Discount. When a loan is made at a bank, the interest is paid in advance, and the interest thus paid is called bank discount. The holder of a note which is due at a certain time may- want the value of the note before it is due. In this case he usually takes it to a bank where the note is discounted, i.e. the interest on the full value of the note at maturity, from the time of discount to the time of maturity, is subtracted from the amount of the note, and the remainder, called the proceeds, is paid to the holder of the note. The note then becomes the property of the bank. Example. A note of $600, dated March 1, 1908, interest 6%, to run 4 months, was discounted at 6%, March 20, 1908. Find the pro- ceeds. Amount of $600 for 4 mo. @ 6% $612.00 Interest on $612 @ 6% for 105 days, March 20 to July 3, (called the bank discount) 10.71 Proceeds, after deducting the bank discount, $601.29 EXERCISES 1. What is the present worth of $460.50 due in 3 yr. 9 mo. 18 days, if money is worth 6% ? Ans. $375. 2. Find the present worth of $375 due in 3 yr. 3 mo. at 6%. Ans. $313.81 3. Which is worth most, $320 in 12 mo., $310 in 6 mo., or $300 cash, if money is worth 6% ? Ans. The first. 4. A note of $700 dated Aug. 9, 1915, is due Feb. 12, 1916, and bears 6% interest. What is the present worth of this note on Oct. 8, 1915? What are the proceeds on that date if discounted in a bank at 7%? 5. A man buys a farm on March 25, 1911, agreeing to pay $2700, Oct. 1, 1911, and $2500, Jan. 1, 1912. If a discount of 7% is allowed for cash, how much will he gain by borrowing the money on his own note discounted at 6%? How much must he borrow? II, 14] COMPUTATION MEASUREMENT 19 6. A note of $560 to run 3 months, dated Aug. 4, 1914, bearing 5% interest, was discounted at 6% on Sept. 19. Find the proceeds of this note. 7. A man sold his farm for $6350, taking in payment a note at 5% interest due in 6 mo. He at once sold the note to a bank which discounted it at 6%. What were the proceeds, i.e. what did he get in cash for his farm ? 8. A note of $350 due in one year at 6% was discounted 54 days before it was due at 6%. What were the proceeds of the note? 14. Elementary Mensuration. Angles. In the following pages we shall treat angles, the lengths of Hnes and their ratios, the areas of surfaces, and the volumes of solids. If two straight lines proceed from the same point they form a plane angle. This point is the vertex of the angle. If there is only one angle at a vertex, the angle may be denoted by a capital letter at the ver- tex, as ZO, Fig. 15. Any angle may be de- noted by three letters, as ZAOPj Fig. 15, the middle letter being the vertex. An angle may be thought of as generated by a line turning about its end-point as a pivot. Thus, if the ray OP, Fig. 16, revolves about 0, the amount of the turn from OX to OP measures the angle XOP indicated by the arrow. , If the ray continues to rotate counter- clockwise until it is exactly opposite its original position it generates a straight ' angle, as XOX' , Fig. 16. If the ray continues till it reaches its original position, it generates a complete rotation, or one revolution. A unit of measure for angles is the degree. A complete rotation is 360 degrees, written 360. Let the student state the definitions of right angle, acute angle, obtuse angle. 20 MATHEMATICS [II, 15 15. Triangles. Let the student give definitions of triangle, equilateral triangle, isosceles triangle, right triangle (Fig. 17). The sum of the three angles of any triangle is two right angles, or 180. Oblique Triangle Right Triangle Fig. 17 It is customary, as shown in Fig. 17, to designate the sides of a triangle by the three small letters a, 6, c and the angles opposite these sides by the corresponding capital letters. A, B, C, respectively. In a right triangle the square of the hypotenuse equals the sum of the squares of the other two sides, i.e. from Fig. 18, a'-\-b\ 'c a _ b 1 I C di = a^ + l^ Fig. 18 This fact is called the Pythagoras theorem.* When the base and the altitude of a triangle are given the area is j A = ^base X altitude. When the three sides are given the area is A = \/s {s a){s b) {s c). * Tables of squares and square roots are found in the Appendix. They may prove of assistance in the computation. II, 15] COMPUTATION MEASUREMENT 21 where a, b, and c denote the three sides, respectively, and s = i (a+b+c). State in words the formula for the area of a triangle when the three sides are given. Example. What is the area of a triangle whose sides are 15.3, 16.4, and 17.7 rods. Solution : Let a = 15.3, h = 16.4, c = 17.7 ; then s=i(a+64-c)=24.7, s-a = 9.4, s-6 = 8.3, s-c = 7.0 Therefore A = V(24.7)(9.4)(8.3)(7.0) = 116.2 square rods. EXERCISES 1. A ladder 16 ft. long and standing on level ground just reaches a window 12 ft. from the ground. Assuming the wall to be perpen- dicular, how far from the base of the wall is the foot of the ladder ? 2. Find the length of the hypotenuse of a right triangle whose other two sides are equal and whose area is two acres. How many degrees are there in each of the acute angles of this triangle ? 3. Find the area of a triangle whose sides are 13, 18, and 21 in. 4. Find the altitude and the area of an isosceles triangle whose equal sides are 8 ft. long, and whose base is 6 ft. long. 5. Find the length of the longest umbrella that can be put di- agonally in a trunk which measures 20 X 20 X 34 inches. 6. The dimensions of a barn are 20' X 44' X 48'. What is the length of a wire which connects the upper corner at the eaves to the opposite lower corner? 7. A house is located 10 rods from a certain river, and a barn is 15 rods from this house and 19 rods from the same river. Find the point on the river equidistant from the house and the barn and find this distance. 8. A farmer wishes to run a pipe from some point on the river of Ex. 7 to the house and another pipe from the same point on the river to the barn. Where should it be laid to require the least length of pipe? Find X\m length. 22 MATHEMATICS [II, 16 16. Quadrilaterals. Let the student define quadrilateral, square, rectangle, parallelogram, trapezoid. (See Fig. 19.) Square Rectangle Trapezoid Parallelogram Fig. 19 The areas of such quadrilaterals may be found by the following rules. 1 . Square : Area = s^, where s = the length of one side. 2. Rectangle or Parallelogram: Area = base X altitude. 3. Trapezoid : Area = altitudeXhalf the sum of the parallel sides, i.e. _ (6+c) -" 2 ' where a is the altitude, and h and c are the parallel sides. EXERCISES 1. State in words the rule for the area of a trapezoid. 2. Find the area of a trapezoidal field whose parallel sides are 10 rods and 16 rods long and the distance between them is 8 rods. 3. Figure 20 as dimensioned represents the gable end of a building. Compute the length of the rafter OR and the area of ABDOC. What is the rise of the rafter per foot of its run? What is the pitch of the roof? (See 4.) 4. Show how a carpenter ''lays off" a t?t 90 rafter by use of a steel square, i.e. how he cuts the desired bevel on the rafter, making use of the ratio of rise to run. II, 17] COMPUTATION MEASUREMENT 23 5. Compute the pitch and the length of each rafter and the area of the entire gable end of the building as shown in Fig. 21. 6. ABC is a triangle, two of whose sides are AC = 17' and BC = 10'. The projection of BC on the base AB i^Q'. Draw the several cases possi- ble here and find the area of ABC for each case. Fig. 21 17. Polygons. Any plane figure bounded by straight lines is called a polygon. The sum of its sides is called the perimeter of the polygon. A polygon of five sides is called a pentagon] one of six sides, a hexagon ; one of eight sides, an octagon ; one of ten sides, a decagon, etc. A regular polygon is one whose sides are all equal, and whose angles, between adjacent sides, are all equal. The center of a regular polygon is the common center of its inscribed and circumscribed circles ( 19). A regular polygon can be divided into as many equal triangles as the polygon has sides. Each such tri- angle has one vertex at the center of the polygon, and its altitude, called the apothem of the polygon, is the distance from the center of the polygon to the center of a side, as OP, Fig. 22. An interior angle of a regular polygon, as B, Fig. 22, is equal to (n 2)180 -j-n where n is the number of sides of the polygon. Thus, for the regular pentagon, n = 5, and hence B=(5-2)180-^5 = 108. The area of a regular polygon is EPA Regulab Poltoons Fig. 22 apothemXthe perimeter. 24 MATHEMATICS [II, 17 EXERCISES 1. Find the area of a regular hexagon whose side is 1". 2. The side of a regular decagon is 1" and its apothem is 1.54" Find its area. 3. Prove the statement in the text that the interior angle of a regular polygon is equal to (n 2)180 -^n, by drawing the equal tri- angles AOE, EOD, etc. as in Fig. 22, and taking the sum of their in- terior angles less the angles at the center. 4. Find the number of degrees in an interior angle of a regular hexagon. A regular octagon. A regular decagon. A regular 48-gon. A regular 500-gon. 18. Irregular Areas. I. Straight Boundaries. If a tract is bounded by straight lines, i.e. if its perimeter is a polygon (Figs. 24-26), the area may be computed by projecting the sides of the polygon upon some convenient base line. The resulting trapezoids may be combined to obtain the required area. The area can also be computed by dividing the tract into triangles and measuring a sufficient number of dimensions of each triangle. II. Curved Boundaries. A curved boundary occurs when a tract touches a stream or a curved road. Stakes are set at points along the curved part, so that it will approximate closely to straight lines between the stakes. The perimeter then takes the form of an ^ irregular polygon and may f\ ^^^>v ^ be treated accordingly. ! j | ^v^d^/ | \ For example, in Fig. 23, | j I ' I ! stakes are set Sit A, B, C, p Q R s t u etc., along the boundary. ^^^' ^^ The sides AB, BC, etc. are now projected on a convenient base line, and the offsets from the stakes to the base line are meas- ured. The areas of the resulting trapezoids are then computed. II, 18] COMPUTATION MEASUREMENT 25 When the curved boundary is not too abrupt in any place, stakes may be set at regular intervals on the base Hne, and offsets to the curved boundary measured. The computation leads to the following general rule for offsets at regular intervals on the base line. Rule. Add all the offsets and subtract from this sum half the sum of the extreme offsets; multiply the remainder by the common distance between offsets. These methods form the basis for calculating areas in practical surveying. The base lines in surveying are usually the magnetic north-south lines. EXERCISES 1. Find the area of a triangular field ABC in which AE = 5 rods; EF = 9 rods; CE = 10 rods; F = 7rods. Fig. 24. [Hint. ^^ is the base line and AE and EF are the projections of AC and CB respectively upon it. The area of ABC is equal to the right triangle AEC plus the trapezoid CEFB minus the right triangle AFB.] 2. Find the area of A BCD in the diagram, cr taking PQ = 20 rods PD = 40 rods QR = 10 rods QC = 50 rods RS = 25 rods RA^ 15 rods ^B = 45 rods [Hint. The area of ABCD = PQCD+CQSB - DPRA-ARSB, Fig. 25. The area may also be found as follows : parallel to PS and extend PD and BS to meet it. Take the area of the outside rec- tangle and subtract from it the areas of the other outside figures.] 3. Find the area of the five-sided field ABCDE shown in Fig. 26, if xi/ = 5rods Ax = 15 rods Ey = 18 rods zu = 15 rods uv = 12 rods Du = 22 rods ?- p Q R S Fig. 25 Draw a line through C D 5- ^^\ [ j -^A^\ X V Z U V Fig. 26 yz = 10 rods ^2 = 4 rods Cy = llrods 26 MATHEMATICS [II, 18 4. In surveying a field between the road AG and the river (Fig. 27), the surveyor measured along AG and drove a stake every 100 feet, at B, C, etc. At right angles to the road, the distances to the river were measured as follows : From B, 100 ft. ; from C, 220 ft.; from D, 260 ft.; from E, 250 ft.; from F, 230 ft. Compute the approxi- mate area of the field and its value at $190 per acre. 6. Determine the area between a base line and an irregular boundary from the following table of offsets all on the same side of the base line. The distances are measured in feet between successive stakes on the base line. Give the result in square feet, and in acres. C D E F G Fig. 27 Distances. . . 202 156 467 159 239 Offsets. . . . 20.7 31.5 42.6 53.2 36.1 40.7 Ans. 49,698 sq. ft. 6. The following offsets were taken on the same side of a line at distances of 100 ft. apart. Find the areas between extreme boundaries and the base line. Distances . 100 200 300 400 500 600 700 -Offsets . . 55.1 76.1 83.3 79.9 69.7 59.5 83.3 70.5 19. Circles. The circle occurs very frequently in applied mathematics, perhaps oftener than any other geometric form. Let the student state the definitions of circle, center, radius, diameter, chord, arc. The area bounded by a chord and its arc is called a seg- ment, as AsB, Fig. 28. The area bounded by two radii and their intercepted arc is a sector, as OAsBO, Fig. 28. A straight line which touches a circle at only one point is a tangent, as TPS, Fig. 28. A straight line which intersects a circle in two points is a secant, as CF, Fig. 28. II, 20] COMPUTATION MEASUREMENT 27 A radius is perpendicular to the tangent at the point of contact, A polygon is inscribed in a circle when its sides are chords of the circle, and the circle is then circumscribed about the Fig. 28 polygon. A circle is inscribed in a polygon when each of the sides of the polygon is tangent to the circle. 20. Angles of a Circle. An angle formed by two radii is a central angle, as AOB, Fig. 28. An angle whose vertex lies on the circumference of a circle is an in- scribed angle, as QPR, Fig. 28. I. A central angle is said to be measured by the arc which it intercepts. This means that a given central angle contains a number of unit angles equal to the number of unit arcs in the inter- cepted arc. II. An inscribed angle is measured by one half of its inter- cepted arc. Thus, ZQPR = hixU the central angle whose arc is QR, Fig. 28. III. An angle between a tangent and a chord is measured by one half of the intercepted arc. Thus, ZQPT = hsi\i arc PQ, Fig. 28. A M B FiQ. 29 28 MATHEMATICS [II, 21 21. Relations between Radius, Arc, Circumference of a Circle. In any circle, the ratio of the circumference to the diameter is 3.1416, approximately. This ratio is represented by the Greek letter tt (pronounced pi), i.e. 7r = 3.1416 = 3i approximately. Then Circumference = 27rr, where r is the radius of the circle. [Note. The exact value of tt cannot be computed. Its value to the first ten decimal places is TT = 3.1415926535. Its value correct to over 700 decimal places has been computed. As a curiosity, we quote the value 77 = 3.1415926535897932384626433832795028841971693993751.] The length of any arc s of h degrees is (/i/360) 27rr, for we have, in Fig. 28, Arc g ^ h Circumference 360 ^ hence <"' ^ = 360 2'"'- Example 1. Find the circumference of a circle whose radius is 5'. By the formula, the circumference = 27r- 5' = 31.4'. Example 2. Find the length of the arc whose central angle is 30 in a circle whose radius is 5 ft. By definition, 30 of arc means /^o of the whole circumference. Then, we have arc of 30 =J^ X 27r X 4 ft. =5^ =2.62 ft. 360 6 22. Area of a Circle. Area of a Sector. Area of a Seg- ment. The area of any circle is equal to half the radius times the circumference, i.e. area of circle = - rX27rr = 7rr^. II, 22] COMPUTATION MEASUREMENT 29 The area of a sector of a circle whose arc is h degrees (Fig. 28) is /i/360 times the area of the circle, i.e. area of sector = ^^ irr'^. Example. Find the area of a sector whose arc is 30 in a circle whose radius is 5 ft. Area of sector =^-T;7r 25= j-^ = 6.55 sq. ft. The area of a segment of a circle, as ABsA, Fig. 28, is the area of the sector OAsB minus the area of the triangle OAB. EXERCISES 1. What is the diameter of a wagon wheel whose circumference is 150 in.? 2. How many revolutions per second is a 30-inch automobile wheel making when the speedometer registers 25 mi. per hour? 3. What is the area of the cross section of a 6-inch stove pipe? A 3-inch one ? 4. By using a carpenter's square find the diameter of a circular pipe having the same area of cross section as the sum of the areas of a 6-inch and an 8-inch circular pipe. See the ac- companying figure. Do the same for two pipes whose diameters are 5| in. and 8y\ in., respectively. [Hint. Use the theorem that similar surfaces are to each other as the squares of their like dimensions.] Fig. 30 5. Find the total pressure on the piston of a steam engine 8 inches in diameter, when the steam gauge registers 120 pounds per square inch. 6. A circular pond has a diameter of 75 yards. What is the cost at $3 per square yard of constructing a walk 40" wide around it ? 7. WTiat must be the length of a binding rod for a cylindrical silo 12 feet in diameter allowing 1 ft. for overlapping? 8. Find the length of the arc and the area of the corresponding sector whose central angle is 40 in a circle of radius 10 inches. 30 MATHEMATICS [II, 22 9. Find the area of the segment of a circle whose arc is 60, the radius being 8 inches. [Hint. The side opposite the 30 angle in a right triangle whose acute angles are 30, 60 is one half of the hypotenuse.] 10. The central angle whose arc is equal to the radius is often used as the unit of measure of angles. It is called a radian. Find the number of degrees in a radian. 11. Inscribe an equilateral triangle in a circle. Show the relation between the angles of the triangle and the arcs which they subtend. 12. Has a degree of arc the same length in two unequal circles? 13. The sizes of men's hats are indicated by the diameter of a circle whose perimeter is the distance around the head where the hat rests. What size of hat does a man need if the distance around his head meas- ures 22 inches ? 14. If you wear a 7\ iiat, what is the distance around your head? Verify. 23. Polyhedrons. A polyhedron is a solid bounded by planes. A polyhedron of four faces is called a tetrahedron; ICOSAHEDRON DODECAHEDRON OcTAHEDRON CUBB Tetrahedron Fig- 31. The Five Regular Sc one of six faces, a hexahedron; one of eight faces an octa- hedron; etc. (Figs. 31, 32). II, 24] COMPUTATION MEASUREMENT 31 24. Prisms. A prism is a polyhedron, two of whose faces, called its bases, are equal polygons in parallel planes, and whose other faces, called lateral faces, are parallelograms Right Prisms Oblique Prisms Fig. 32 whose vertices all lie in the bases. The altitude of a prism is the perpendicular distance between its bases. A prism whose bases are perpendic- ular to its lateral edges is called a right prism (Fig. 32). Otherwise, the prism is oblique. Prisms are called triangular, quad- rangular, etc., according as their bases are triangles, quadrilaterals, etc. A right section of a prism is a section perpendicular to the lateral edges of the prism, as ABODE, Fig. 33. A right prism is regular if its bases are regular polygons. Right Section Fig. 33 32 MATHEMATICS [II, 25 25. Parallelepipeds. A parallelepiped is a prism whose bases are parallelograms. Rectangular Parallelepiped Cube Oblique Parallelepiped Fig. 34 A rectangular parallelepiped is one whose six faces are all rectangles. For example, most boxes are rectangular paral- lelepipeds. A cube is a parallelepiped whose six faces are all squares. 26. Unit of Volume. The unit of volume is a cube whose edges are equal to the linear unit used in the measurement. 8 2Z ^ .7^0 Unit of Volume Fig. 35 The volume of any solid is the number of units of volume which it contains. II, 27] COMPUTATION MEASUREMENT 33 27. P5rramids. A pyramid is a polyhedron of which one face, called the base, is a polygon and the other faces are triangles having a common vertex, called the vertex of the pyramid. The altitude of a pyramid is the length of the perpendicular let fall from the vertex to the plane of the base. B C Regular Pyramid Fig. 36 A pyramid is regular if its base is a regular polygon whose center coincides with the foot of the perpendicular let fall from the vertex to the base. The slant height of a regular pyramid is the altitude of any one of the lateral faces, as VH, Fig. 36. Frustum of a Pyramid Fig. 37 A frustum of a pyramid is the portion of a pyramid in- cluded between the base and a section parallel to the base. The altitude of a frustum is the length of the perpendic- ular between the planes of its bases. 34 MATHEMATICS [II, 28 28. Cylinders. A cylindrical surface is a curved surface generated by a straight line which moves parallel to a fixed straight line and constantly touches a fixed curve not in the plane of the straight line. For convenience in drawing, as Cylindrical Surface Right Cylinder Fig. 38 Oblique Cylinder in the case of straight lines, it is usual to show only a limited portion of a cylindrical surface, but it really extends in- definitely in both directions. A cylinder is a solid bounded by a cylindrical surface and two parallel plane surfaces called the bases. The altitude of a cylinder is the perpendicular distance be- tween the planes of its bases. A right cylinder is one whose straight line generators are perpendicular to the bases. 29. Cones. A conical surface is the surface generated by a moving straight line which constantly touches a fixed curve and passes through a fixed point not in the plane of the curve. A cone is a solid bounded by a conical surface and a plane which cuts all positions of the generating line. This plane surface is called the base of the cone. II, 29] COMPUTATION MEASUREMENT 35 The altitude of a cone is the perpendicular distance from the vertex to the plane of the base. A right circular cone is one whose base is a circle and whose altitude meets the base at the center of the circle. Right Circular Cone Fig. 39 The slant height of a right circular cone is the distance along a generator from the vertex to the circle. A frustum of a cone is the portion of a cone included between the base and a plane parallel to the base. The frustum thus has two bases. The altitude of the frustum is the perpendicular distance between these bases. Frustum op a Cone Fig. 40 The lateral surface of a frustum of a cone is the portion of its lateral surface included between the planes of its bases. The slant height of a frustum of a right cone is the dis- tance between the bases measured along a generator. 36 MATHEMATICS [11, 30 30. The sphere. A sphere is a solid bounded by a surface all points of which are equally distant from a point within w- P ^.1 i D \ %b p' Sphere Fig. 41 called the center. A sphere may be generated by the revo- lution of a semicircle about its diameter as an axis. 31. Surface and Volume of Solids. 1. Prism: Volume= area of baseX altitude, = area of right sectionXlateral edge. Lateral Area of a Right Prism = perimeter of baseX altitude. 2. Cube: Volume = s^ where s = the length of a side. Entire Surface = Qs^. 3. Cylinder: Volume= area of baseX altitude = area of right sectionX slant height. Lateral Area of a Right Cylinder = perimeter of its baseX altitude. 4. Pyramid: Volume = one third of area of baseX altitude. Lateral Area of a Regular Pyramid = half perimeter of baseX slant height. II, 31] COMPUTATION MEASUREMENT 37 5. Cone: Volume = one third of area of baseXalti- tude. Lateral Area of Right Circular Cone = half perimeter of baseXslant height. 6. Frustum of a Pyramid or of a Cone: Volume = ~(b-]-B-^y/bB) o where a = altitude of frustum, b = area of upper base, B = area of lower base. Lateral Area of a Frustum of a Regular Pyramid or of a Right Circular Cone = slant heightXl/2 the sum of the perimeters of the bases = slant heightXthe perimeter of the mid-section. 7. The Sphere: Volume = 4:/Sirr^ where r = the radius. Surface = ATrr^, = the area of four great circles. 8. A Barrel : It may be considered as a double frustum of a cone and its volume and surface approximated accordingly. But for cir- cular staves a closer approximation is the following : Volume = -^ 7r/i(2 D^+d^), in which i) = bung diameter, d = head diameter, and /i = height. Fig. 42. Barrel 38 MATHEMATICS [II, 32 32. Similar Figures. In general, two figures are similar if they have the same shape. Two polygons are similar if they are mutually equiangular and if their pairs of corresponding sides are proportional* Similar polyhedrons are those that have the same number of faces, respectively similar and similarly placed and have their corresponding polyhedral angles f equal. Any two similar figures in a plane or in space can be placed in " perspective,'' i.e. so that straight lines joining correspond- ing points of the two figures will pass through a common Fig. 43. Perspective point, Fig. 43. Thus of two similar figures one is merely an enlargement of the other. A scale drawing, 3, and the surface it represents are thus similar. If each length in one figure is k times the corresponding length in the other figure, then each area in the first figure is fc2 times the corresponding area in the second. Also each volume in the first figure is k^ times the corresponding volume in the second. This may also be expressed as follows : Similar surfaces are to each other as the squares of their like dimensions. * To say that corresponding sides are proportional means that each length in one figure is a constant, k, times the corresponding length in the other figure. t The opening of three or more planes which meet at a common point is called a polyhedral angle. II, 32] COMPUTATION MEASUREMENT 39 Similar solids are to each other as the cubes of their like dimensions. Example. Two oranges have diameters of 2" and 3". Compare their surfaces and their volumes, respectively. Their volumes are to each other as 2^ is to 3^, i.e. as 8 to 27. Their surfaces as 2^ to 3^, i.e. as 4 to 9. EXERCISES 1. Find the capacity in gallons of a bucket 12" deep whose bottom and top diameters are 8" and 10". Find the whole surface of this bucket. 2. Find the diagonal of a barn 16' X 30' X 40'. 3. What is the side of a cube whose diagonal is 2"? 4. Find the volume in cubic feet of a conical haystack whose cir- cumference at the base is 55 ft. and whose slant height is 20 ft. 5. Find the number of gallons of vinegar in a barrel whose head diameter is 18 in., bung circumference 70 in., and height 35 in. 6. Find the volume of a regular pyramid with a square base 4 in. on each side, if the altitude of the pyramid is 12 in. Find also the lateral surface of this pyramid. 7. How much concrete is there in a circular silo whose walls are 8 in. thick, 12 ft. outside diameter and 30 ft. high? 8. How much water can be put in a round tank 8 ft. deep, if the inside diameters are 4 ft. at the bottom, and 5 ft. at the top? 9. The circumference of a sphere is 1 ft. 2 in. Find its area and its volume. 10. A farmer has two strings of drain tile 3 in. and 4 in. in diameter, respectively. He wishes to combine them into one equivalent tile. What should be its diameter ? Does a tile run full of water ? 11. What would be the size of a waterspout to be equivalent to three spouts 2 in., 4 in., and 5 in., in diameter? 12. How many 2-in. flues of an engine will equal in sectional area a smokestack 22 in. in diameter? 13. A farmer has a triangular field whose base is 20 rods and whose sides are 16 rods and 12 rods. Where, along the side 16 rods long, 40 MATHEMATICS [II, 32 should he run a fence parallel to the base in order to divide the field into two equal parts ? If the fence is run parallel to the base and starts at the middle of the side 16 rods long, how does it divide the field? 14. In a trapezoid the parallel sides 8 ft. and 12 ft. long are 16 ft. apart. Find the altitude of the triangle formed by extending the non-parallel sides of this trapezoid. What is the area of the trapezoid? 15. In the previous exercise, where should a line be run parallel to the parallel sides to divide the trapezoidal field into two equal parts? Ans. 7.2 rd. from the lower base. 16. Grindstones of Huron sandstone will safely stand a surface speed of 3600 ft. per minute. How many revolutions per minute (R. P. M.) may a 3Ht. stone be run? A 2-ft. stone? A 6-ft. stone? 17. A hollow, spherical steel shell is 1 in. thick and its outside diameter is 10 in. The specific gravity of steel is 7.8, and a cubic foot of water weighs 1000 oz. Find the weight of the shell. 18. The side of an equilateral triangle is 12 ft. Find its altitude and the area. If the area of an equilateral triangle is 400 sq. ft., find the side. Make a formula for area in terms of the side, x, say. 19. How much tin is there in a fimnel whose diameters are 1^ in. and 8 in. and whose height is 12 in.? What is the volume of this funnel? 20. Find the volume of a regular hexagonal pyramid 20 ft. high if each side of the hexagon is 4 ft. long. Find also the lateral surface of this pyramid. 21. How many cubic yards must be excavated in digging a ditch 200 rods long, 12 in. wide at the bottom, 6 ft. wide at the top, and a yard and a haK deep ? How much water would be discharged by such a ditch in 2 hr. time if it flows half-full at the rate of 1 ft. per second? 22. A chimney is to be constructed in the form of a frustum of a circular pyramid. The height is to be 160 ft., with outside diameters 7 ft. and 10 ft. The flue is to be 6 ft. in diameter throughout the entire height. How many 2x4x8 in. bricks will its construction require, allowing 10% for mortar? 23. Three oil cans are of the same height. Two of them are each 9 in. in diameter, and the other one is 13 in. in diameter. Compare the capacity of the large one with that of the other two. II, 32] COMPUTATION MEASUREMENT 41 24. A new grindstone is 30 in. in diameter and has a 3-in. face. What has it lost in weight when it is worn down to 26 in. in diameter, if sandstone weighs 2.42 times as much as an equal volume of water, and water weighs 62.5 lb. per cubic foot? 25. The frustum of a pyramid is 8 feet high, and its bases are equi- lateral triangles whose sides are each 3 feet and 4 feet, respectively. Find its voliune. 26. In the previous exercise, find the volume of the entire pyramid. 27. The height of a certain frustum of a right cone is the height of the entire cone. Compare the volumes of the frustum and the cone. 28. A lateral edge of a pyramid is 6 feet. At what distance from the vertex should this edge be cut by a plane parallel to the base to divide the pyramid into two parts which are to each other as 1 to 1? As 1 to 2? As 3 to 4? 29. Find the volume of a regular tetrahedron whose edge measures 3 inches. [Hint. The tetrahedron is a triangular pyramid.] 30. A wheat bin measures 4 ft. by 6 ft. by 10 ft. What are the dimensions of a similar bin that holds twice as much? 31. Find the area of the solid generated by an equilateral triangle that revolves about one of its sides if the length of the side is 6 inches. 32. If 0.203 of a gram of gold (specific gravity, 19.32) is plated over a spherical dome whose height is 3 cm., and the radius of whose base is 6 cm., what is the thickness of the gold? [Hint. 1 cc. of gold weighs 19.32 grams. Therefore .203^19.32 is the volume of the gold. This divided by the superficial area is the thickness.] 33. A steel sphere whose radius is .95 cm. weighs 28.25 grams. What is its density? Ans. 7.866 34. Funnels are best made with an angle of exactly 60. If such a funnel measures 8 cm. across the top, what size filter paper will fit it flush with the edge? 35. A rectangular block of wood 7.5 by 7.46 by 3.8 cm. weighs 152.7 grams. What is its specific gravity if 1 cc. of water weighs 1 gram? If this block is floated in water, to what depth will it sink? 42 MATHEMATICS [II, 33 33. Silos. A silo is a receptacle for preserving green feed, and silage is the material preserved. A silo is usually a cylindrical tank much higher than wide. Its size depends upon the number of animals, usually cattle, to be fed. The diameter of the silo must be of such size as to insure that the proper depth will be removed daily. Silage of all kinds de- teriorates unless it be fed regu- larly, evenly, and at a rate of not less than two inches depth daily. Removing five or six inches daily insures but little waste of feed. Experience has shown that the most satisfactory results are ob- tained by providing a horizontal feeding surface of about 5 square feet for each cow. See Ex. 9 below. To insure proper settling of the silage, excluding in this way the bacteria that cause decay, the height of the silo should seldom he less than 30 feet. On an average one ton of silage occupies 50 cubic feet. Thus, the diameter of a silo is controlled largely by the num- ber of cattle to be fed, while the height is gauged by the quantity to be fed. One cubic foot of silage per head is a widely used daily ration. FiQ. 44. A Cement Silo EXERCISES 1. What should be the height of a round silo for a herd of 25 cows, if each one is to be fed 40 lb. daily for 180 days? Solution : For 25 cows a horizontal feeding surface of 5 times 25 = 125 square feet is required. This is the area of a circle 12.6 feet II, 33] COMPUTATION MEASUREMENT 43 in diameter. But to feed 25 cows, each 40 lb. daily for 180 days, requires 25 X 40 X 180 lb. = 18,000 lb. = 90 tons. Since 1 ton occupies 50 cu. ft., 90 tons will require 4500 cu. ft. of space. Therefore the height is the capacity, 4500 cu. ft., divided by the cross-sectional area, 125 sq. ft., i.e. 36 ft. 2. What should be the size of a round silo to feed 30 cows 40 lb. each daily for 150 days, 39 lb. each daily for 100 days, and 15 lb. each daily for 75 days ? 3. What should be the size of a round silo for a herd of 20 cows, the ration being 40 lb. each daily for 100 days and 30 lb. each daily for 90 days? 4. Fattening steers are fed about 25 lb. each daily and calves 15 lb. each daily. How large a silo should a farmer construct to accommodate in this way 40 steers for 150 days and 20 calves for 100 days, estimating 40 lb. as the daily ration for an animal that requires 5 sq. ft. of hori- zontal feeding surface. 5. What is the capacity of a round silo 30 feet high and 15 feet in diameter? How many cows will it maintain? How many pounds daily can each be fed from this silo for 180 days? 6. What depth is fed daily in feeding 25 cows each 40 pounds per day ? If a smaller daily ration is contemplated, how should the above rule for size of a silo be changed, in order to insure proper feeding depth daily? 7. For a herd of 15 cows, each one fed 30 pounds daily for 175 days, what should be the minimum size of the silo? 8. A silo is 16 feet in diameter and 36 feet high. What is the least number of cows that must be kept to prevent the silage from spoiling, if each cow is fed 40 pounds daily ? 9. Explain the provision of 5 sq. ft. of horizontal feeding surface for each cow on the basis of a moderate daily feed of 333 lb,, an average of 50 cu. ft. of silage per ton, and a minimum of 2" depth removed daily. CHAPTER III INDIRECT MEASUREMENT TRIGONOMETRY SURVEYING 34. Proportion and Indirect Measurement. A propor- tion is an equality of ratios,* as 2 = 1., M = 12 Ib. ^ 4hr. ^ 20 mi. 3 12' $4 16 lb.' 7 hr. 35 mi. The proportion - = - may also be written in the form 6 a a:h = c: d. There are four terms in a proportion. If any three of them are known the fourth may be found. Thus, if we have a c multiplying each side by a, xv. oh X = c In stating a proportion where one of the terms is un- known it is convenient to put the unknown term first as just shown. To estimate the height of a ham, a farmer has an assistant * A ratio is the relation of one number to another of the same kind ex- pressed by their quotient. The ratio of a to 6 is a/b. Every fraction expresses the ratio of its numerator to its denominator and every integer expresses the ratio of itself to unity. 44 Ill, 34] INDIRECT MEASUREMENT 45 hold a staff NB vertically, Fig. 45. As he sights along AE to the top of the barn, the assistant marks the point B on M N Fig. 45. Indirect Measurement the staff. When a = BC, b = AC, and AD are measured, the distance x = DE may be computed from The height of the barn is then x plus the height of the ob- server. The principle used here forms the basis of nearly all practical methods of indirect measurement. EXERCISES 1. Find the height of a barn, when, as in Fig. 45, AD = 75', and the observer is 5' tall. 2', 6 = 5', 2. A post 12' high casts a shadow 30' long. How high is a tree whose shadow at the same time is 160' ? State the proposition from elementary geometry which is used in the solution of this problem. 3. At a distance of 62' from a building is a vertical post 10' high. By standing back of the post 6' and sighting over the end of a 4' vertical stick, the top of the post and the top of the building are in line. How high is the building? 4. A carpenter's square is 16" X 24", the blade is 1|" wide, and the arm is 2" wide. A man places the blade vertically on top of a post 5' high and 40' from a building. How high is the building, if the line of sight over the arm and blade just intersects the top of the building? 46 MATHEMATICS [in, 34 Fig. 46 5. The line of sight between two objects A and B, Fig. 46, is ob- structed by a building. The distance AB may now be estimated by measuring the distances AO and BO from some convenient point 0, laying off OC = OA and OD = OB, and finding DC. Show why. 6. The distance AB, Fig. 47, across a swamp or river may be estimated by use of a vertical staff to which is attached an arm CD at right angles to the staff and movable up and down along it. By sighting along EB and knowing AE, DE, DC, the distance AJ5 is found. If AE = 7', DC = S', and ED^^", find the distance across the swamp. Moreover, an accessible point B' may be located so that AB' = AB by simply re- volving the arm DC around the staff and sighting to B' in the line EB'. This device was used in the early days, and formed the basis of the surveying instruments used in those times. 7. A boy holds a pencil 6" long 2' from his eye so that it covers a chimney 200' distant. What is the height of the chimney? Draw a figure, and explain the process. 8. Draw a line segment. Divide it into three equal parts. Into five equal parts. 9. The following figure represents a diagonal scale used by drafts- men. The distance from to 10 is 1". .^' Fig. 47 10 987654321 Fig. 48. Diagonal Scale Explain the principle involved in the use of such a scale. Show how to set the dividers to the following distances : .27' 1.89", 1.04". .73' Ill, 36] INDIRECT MEASUREMENT 47 35. Angles and the Sides of a Right Triangle. A right triangle is formed by dropping a perpendicular line from any point B in the terminal of an angle, upon the initial line at C, Fig. 49. Thus a = BC is the side opposite the angle A, Fig. 49 h = AC is the side adjacent to the angle A, and AB is the hypotenuse of the right triangle. Now as the angle A changes, the sides of this right triangle will change. The amount of turn, or the size of the angle, thus depends upon, or is related to, the sides of the right triangle, as shown in the next few paragraphs. 36. Use of Instruments. Ratios. Angles. In the study of indirect measurement, 34, we saw that the ratios of sides and hypotenuse of right triangles play an important part in solving problems involving triangles. This is particularly true in finding heights and distances and in surveying. Where more exact methods are desired, a surveyor's transit, or a sextant, or some similar instrument, is used to find the ratio of the sides, instead of using a staff and sighting as was done in 34. For example, sup- pose we wish to know the height of a tree CB, as in Fig. 50. The angle A could ^''' ^^ be measured directly by means of one of these instruments. The ratio desired, a/h in this case, is called the tangent; it 48 MATHEMATICS [III, 36 can be found from tables which are specially prepared for this purpose. Suppose the angle at A is found to be 25 20'. Then from such a table (in the column for tangents, Table IX, Appendix), we find opposite 25 20' the value .4734. This means that a/b = .4734. If h is measured and is found to be 100', then a = (.4734) 100' = 47.34', the height required. Moreover, if we know the sides and therefore their ratio, such a table gives us the angle which corresponds to that ratio. For example, if the ratio a/b = .5467, we find from the table of tangents that this is the ratio for 28 40' ; hence A = 28 40'. 37. Trigonometric Ratios. For convenience these ratios in the right triangle have been given special names. Draw a triangle ABC right-angled at C, Fig. 51, naming, as is usual, the angles with the capitals and the sides opposite the angles by the corresponding small letters. FiQ. 51 Definitions. Tangent A is the ratio of the side opposite angle A to the side adjacent to A, or a/b in Fig. 51. Sine A is the ratio of the opposite side to the hypotenuse, or a/c in Fig. 51. Cosine A is the ratio of the adjacent side to the hypotenuse^ or h/c in Fig. 51. Abbreviations for these expressions are : tanA=-, sinA=-, cosA=' b c c These are called trigonometric ratios. In this same triangle write the similar ratios for tan B, sin B, cos B. Ill, 37] INDIRECT MEASUREMENT 49 EXERCISES Find the value of the unknown term in each of the following exercises, expressing this value decimally also. 1. 5 4. 23 = 2. 9 3. 27 = ,7 5. .4759 = 35.6 6. If tan 32 = -^, find a. 77. o Ans. a = (77.3) tan 32. 18 7. Solve, as in Ex. 6, (a) sin A = , for c. c (b) tan 25 =^,for6. / X cos 20 1 Q r (c) = 13, for a. 8. For a right triangle ABC, with sides a, b, c, as in Fig. 51, fill in the omitted entries in the following table from the geometric properties of these special triangles. Express the ratios decimally. Ex. a 6 c sin A cos A tan A angle A (a) 3 6 30 (b) 27 27 45 (c) 15 .5 60 id) 17 30 9. If o = 7, 6 = 24, find sin A, cos A, tan A. tan B. Also find sin B, cos B, 10. A rafter rises 5' in a horizontal run of 12'. What is the length of the rafter? What are the sine and the cosine and the tangent of the angle which this rafter makes with the horizontal? See Fig. 11. 11. The side of an equilateral triangle is 10.4'. Find its altitude and its area. If the side is s, find its altitude and its area in terms of s. 12. What is the length of the side of an equilateral triangle which contains 1 acre? 13. Is the sine a proper or an improper fraction? Why? Is the cosine greater than unity or less than unity? How large may the tangent become? 50 MATHEMATICS [III, 38 38. Tables. The values of the trigonometric ratios for angles differing by 10' have been computed to four decimal places and recorded in Table IX in the Appendix of this book. They are arranged with the angles from to 45 reading down the left side of the page and the name of the desired functions of these angles at the top of the page. Angles from 45 to 90 are read up the right side of the page and the name of the functions at the bottom of the page. By means of computed corrections these tables may be read to the nearest minute. These corrections are made according to the rule that relatively small changes in the angle are proportional to the corresponding changes in the value. The use of the tables is shown by the following examples.* Example 1. Find sin 37 20'. On the proper page of the table find 37 20' in the left-hand column marked Degrees at the top. Directly opposite 37 20' in the column marked Sine-Value at the top, we find .6065 Thus sin 37 20' = .6065 Similarly, sin 37 40' = .6111 Example 2. Find sin 37 24'. Find, in this case, sin 37 20' = .6065 sin 37 30' = .6088 Thus 10' change in the angle = a change of 23 in the value of the sine and therefore 4' change in the angle = a change of ^ of 23 = 9 in the sine, i.e. sin 37 20' = .6065 correction for 4' = 9 Hence sin 37 24' = .6074 Example 3. Find tan 37 47'. Here tan 37 40' = .7720 tan 37 50' = .7766 Therefore, for 10', the correction is 46. and for 7', the correction is j^ of 46 =32. Thus tan 37 40' = .7720 correction for 7' = 32 Therefore tan 37 47' = .7752 * For more extensive tables, giving values for every minute of arc to five decimal places, see, e.g., The Macmillan Tables. Ill, 38] INDIRECT MEASUREMENT 51 Example 4. Find cos 19 12'. We have cos 19 10' = .9446 cos 19 20' = .9436 and correction for 10' = 10 correction for 2' = 2. This correction is subtracted in the case of the cosine, since the valiie of the cosine becomes smaller as the angle increases. Thus cos 19 12' = .9444 Example 5. If sin A = .6870, find A. Not finding precisely this number in the table, take the one next nearest below it. Thus .6862 = sin4320' 6870 = sin A 8 = correction to be found for the angle. But .6884 = sin4330'. Therefore 22 = correction for 10' and 8 = correction for -^ of 10' = 4'. Hence .6870 = sin 43 20'-f4' = sin 43 24'. Example 6. If tan A = 1.6795, find A. 1.6753= tan 59 10', 1.6795 = tan A, 1.6864 = tan5920'. Therefore Ill = correction for 10' and 42 = correction for ^^ of 10' =4', whence 1.6795 = tan 59 24', i.e. A = 59 24'. EXERCISES With the table just explained, read and tabulate the sine, cosine, and tangent of each angle in the following table : Ex. Angle sin cos tan Ex. Angle sin cos tan 1 32 34' 7 71 9' 2 27 53' 8 27 7' 3 41 22' 9 81 29' 4 11 19' 10 68 41' 5 56 48' 11 84 16' 6 61 17' 12 64 54' 52 MATHEMATICS [HI, 39 39. Construction of Angles. An angle may be laid off directly by means of a protractor. However, it can be constructed more accurately by means of the value of its tangent obtained from a table. When squared paper * is used neither protractor nor compasses are necessary. Example 1. Given, tan A = .75, to construct A. Since 75 = -^^'jj = ^, lay off 4 squares horizontally, Fig, 52, and 3 squares vertically and draw the B y ^/ y^ a y h A ^ Fig. 52 hypotenuse AB. Then angle CAB is the required angle. Example 2. Construct K if sin ii[ = 5/13. Here the hypotenuse is 13 and the side opposite X is 5. The third side of the right triangle is therefore VlS^ 5^ = 12 and tan K = 5/12. The construction then follows as just shown. EXERCISES Draw a right triangle to which each of the following functions belongs and give the numerical value of each of the other two functions. 1. tanvl=j\. 3. sin 5=^. 6. sin/^ = .3 2. cosA: = t\. 4. tanfl' = 2i. 6. cosA=f. Construct the following angles from the value of their tangents : 7. 37. 9. 54 30'. 11. 84. 8. 20. 10. 5. 12. 60. 40. Solution of Right Triangles. To solve a right triangle is to compute the unknown parts. Two fundamentally different cases exist. I. When two sides are given. II. When a side and an acute angle are given. In the solution of problems, here as elsewhere, the given data should be drawn carefully to a convenient scale. This in itself is an approximate solution. * See 78. This is merely paper ruled into small squares or small rectangles. It can be purchased at most students' supply stores. III. 40] INDIRECT MEASUREMENT 53 EXERCISES Solve each of the following right triangles. It is assumed that the triangle is lettered as in Fig. 51 unless special lettering is given. 1. Given c = 12.5, A = 17"^ 24'. Find a, b, B. Solution : B = 90 - 17 24' = 72 36', hence -^=sinl724' = .2990; a = (12.5) (.2990) =3.74 12.5 = cosl724' = .9542, whence b = (12.5) (.9542) = 11.9 2. Given b = 140, B = 24 12'. Find a, c, A. Solution : A =90 -24 12' = 65 48' -^ =tan 65 48' = 2.2251 140 a = (140)(2.2251)=311.5 140 c 140 whence sin 24 12' = .4099, = 341.5 .4099 3. Given c = 340, 5 = 29. Ans. 4. Given b = 270, A = 54. Ans. 5. Given 6 = 42.4, B = 57 46'. Ans. 6. Given a = 3, 6 = 4. Fig. 54 A=61, 6 = 165, a = 298. B = 36, a = 372, c = 459. A =32 14', a = 26.7, c = 50.1 Solution whence Check: tan A=l = .75 A =36 52'. ^=sin3652' = .6000, c a2+b2 = c2, i.e. 32+42 = 52. 7. Given a = 37, 6 = 121. 8. Given c = 5.4, a = 2.6. [Hint. Use sin A =a/c.] 9. Given c= 63, 6 = 47. Ans. A = 17, 5 = 73, c = 127. Ans. A=2847', 5 = 61 13', 6=4.7 Ans. A = 41 45', B = 48 15', a = 42. 54 MATHEMATICS [III, 40 10. Two sides of a triangle are 37.4 and 29.8, and their included angle is 31 47', as shown in the figure. Find the third side, a, and the angles B and C of this triangle and the altitude, h, on AC. [Hint. Find p = AD and h in the right triangle ADB from p = (29.8) cos 31 47', etc. Then g = 37.4 -p.] 11. Ifc = 13.37,A = 1313', finda, fe,i5. 12. Given a = 12.78, c = 13.42 in a right triangle. Find the angles and the third side. 13. A tile drain has a fall of 4" to the rod. What angle does it make with the horizontal in the same vertical plane with the drain ? 14. What is the length of the rafters for a house 28' 4" wide if they have a rise of 1 in 2 of run, and project 16" beyond the eaves? What angle does each make with the horizontal in the same vertical plane with the rafter? 15. Each of two sides of an isosceles triangle is 27 rods long and the angle at the vertex is 47 38'. Find the base, altitude, and area. [Hint. Bisect the given angle and solve one right triangle.] 16. A corner post is anchored to a stone buried in the ground by a stay wire which makes an angle of 25 with the horizontal. If the tension in the wire is 200 pounds, find the vertical ^ ^ and the horizontal pull on the stone. 17. Two sides of a parallelogram are 12' and 15', and the included angle is ^ ib b a 30 35'. Find the two diagonals. ^^^' ^^ [Hint. Find the horizontal and the vertical projections of EC. Then AC follows from the right triangle AEC] 18. Two forces of 40 pounds and 50 pounds, respectively, act on a body at an angle of 40 52'. Find the resultant of these forces and the angle it makes with either force. [Hint. Find, as in the previous exercise, the diagonal of the par- allelogram whose sides represent to scale these forces.] 19. The draft of a certain plow is rated at 296 lb. What power must be supplied to operate it if the traces of the horses pulling it make an angle of 11 with the horizontal? Ill, 41] INDIRECT MEASUREMENT 55 41. Angles of Elevation, of Depression. The angle which a line from the eye to an object makes with a horizontal line in the same vertical plane is called an angle of eleva- tion if the object is above the eye of the observer and an angle of depression if the object is below the eye of the observer, Fig. 58. Fig. 58 Example. The angle of elevation of the top of a barn is 42 40' at a point 102.4' measured horizontally from the base. Find the height of the barn. From the right triangle, we have = tan 42M0' = . 9217 h'? 102.4' Hence ;i = (102.4) (.9217) = 94.38', the height required. EXERCISES 1. What is the angle of elevation of the sun when a 10' pole casts a shadow 8' long ? 2. What is the height of a tree if the angle of elevation of its top is 37 35' measured at a point 147.8' horizontally from its foot? 3. From the top of a cliff 167' high the angle of depression of a man on the plain is 27 28'. How far is the man from the base of the cliff? 4. A triangular field ABC contains one acre. If Z CAB = 52 8' and Z B = 90, find AB and BC. [Hint. Let AB = x and find BC in terms of x from tan CAB. The area in terms of x may now be found, etc.] 5. What is the rise in one third of a mile in a road which has a uniform grade of 2 30'? 56 MATHEMATICS [III, 41 6. A tower and a building stand on a level plane. At a window 23' from the ground, the angle of depression of the base of the tower is 13 30' and the angle of elevation of the top is 39 47'. Find the height of the tower, 7. Two men are lifting a stone by means of ropes. The ropes are in the same vertical plane. What is the weight of the stone if one man pulls 105 lb. in a direction 48 with the horizontal and the other one 85 lb. in a direction 67 with the horizontal? Ans. 156 lb. 8. A hillside farm contains 52 acres, actual area. It is located on a slope of 16. How much more is this than the horizontal surface underneath it and between the same vertical bounds? Ans. 2 acres. 9. At a point A the angle of elevation of the top of a tree was 32**. At B, 160' from A, the elevation of the tree was found to be 21. What is the height of the tree if A, B, and the tree are in a straight line and in the same level plane? [Hint. Drop a perpendicular AD on BE, Fig. 54. Z AED is equal to 32 -21 ^ ^^' ^ = 11. Find AD from the right triangle ABD. When AD is known, compute AE from the triangle AED. Then find EC] 10, Find the height of a precipice if its angles of elevation at two stations in a horizontal line with its base are 41 37' and 34 13' and the distance between stations is 145'. 42. Farm Surveys. Farms are surveyed by measuring distances and angles along the boundary of the farm and fixing the corners by permanent landmarks such as boulders, iron pins, concrete posts, etc. The data obtained in this way are then plotted to scale. The area of the survey is usually computed by drawing two axes north-south and east-west lines so that the scale diagram of the survey lies entirely to the right and above these lines respectively. Perpendiculars are drawn from each vertex to this north-south line, forming as many trapezoids {or triangles) as the survey has sides. Ill, 43] INDIRECT MEASUREMENT 57 43. Bearing of a Line or Course. Northing (Southing), Easting (Westing) of a Course. The bearing of a course means its direction by compass. N. 40 E., Fig. 61, means the direction reached after sighting north and then swinging the line of sight 40 toward the east, i.e. angle NOD = 4:0. When going around a plot or parcel of land to determine the lengths and bearings of all the sides, the surveyor bears north or south, east or west on each course, distances respectively equal to the projections of the course on these north-south, east-west lines. For example, the course OD, Fig. 61, bears ^ north a distance OH, called the north- ing (southing), and bears east a dis- tance HD, called its easting (west- ing). For a closed survey, the computation should show northings = southings, and eastings = westings. For the course OD, Fig. 61, northing 0H = OD cos 40, and easting HD = OD sin Jfi. In general, for any course we have, northing (southing) = length of course times cosine of its hearing, and easting (westing) = length of course times sine of its hearing. Northing (or southing) of a course is the difference of latitudes of its end points, easting (westing) is the difference of longitudes of its end points. 58 MATHEMATICS [in, 43 Example 1. Plot the following courses and compute their latitudes and longitudes, i.e. their northings or southings, j^ eastings or westings. b' A line is run from a point A, N. 35 E., 12 chains to a point B ; from B, S. 35 E., 10 chains to a point C; c' from C, S. 82 36' W., 12.7 chains to the place ^ of beginning. ^ Fig. 62 Then, AB', the northing on AB = 12 cos 35 = 9.83. B'B, the easting on AB = 12 sin 35 = 6.88 CC, the westing on CA = (12.7) sin 82 36' = 12.59 This may be arranged as follows : Course Bearing Distance Latitudes Longitudes Northing Southing Easting Westing AB BC CA N. 35 E. S. 35 E. S. 82 36' W. 12 ch. 10 ch. 12.7 ch. 9.83 8.19 1.63 6.88 5.74 12.59 Example 2. Plot the following survey to scale and find the area of the inclosed tract by projecting the courses on the north-south line through the vertex farthest west. Beginning at a post A, thence N. 3 E., 5.60 ch. to a stone 12" long and 5" square, thence S. 84 E., 7.50 ch. to a post, thence S. 10 30' E., c' 6.42 ch. to an iron pin, thence N. 80 30' W., 9.00 ch. to the place of beginning. The scale diagram is shown in Fig. 63. The d' -TrtrsJ^) computation may be arranged as follows : Fig. 63 For the first course AB, N. 3 E., 5.60 chains, we have latitude north, AB' = (5.60) cos 3 = (5.60) (.9986) = 5.592, longitude east, B'B = (5.60) sin 3 =(5.60) (.0523)= .287 Ill, 43] INDIRECT MEASUREMENT 59 These and similar results for the other courses give the following tabulation. Bearing Distance COMPUTED BALANCED Course Latitudes Longitudes Latitudes Longitudes North- ing South- ing East- West- ing N S E W AB BC CD DA N3E S84E S1030'E NSO'SO'W 5.60 ch. 7.50 ch. 6.42 ch. 9.00 ch. 5.592 1.485 .784 6.313 .293 7.443 1.170 8.877 5.596 1.491 .779 6.308 .287 7.435 1.164 8.886 28.52 7.077 7.0j7 7.077 .02 8.906 8.877 .029 8.877 7.087 7.087 8.886 8.886 Thus the error in latitudes is .02 ch. ; in longitudes the error is .029 ch. The survey is balanced by distributing these errors over the total perimeter, 28.52 ch. Therefore, error in latitudes for 1 chain = .02 error in longitudes for 1 chain 28.52 .029 28.52 .0007; .001; and error in AB, 5.60 ch. = (5.60) (.0007) = .004 for latitudes = (5.60) (.001) = .006 for longitudes. Similar results may be calculated and tabulated for the remaining courses BC, CD, DA. Since the northings are too small, add the corrections; similarly, subtract them from the southings. Likewise, balance the eastings and westings. This gives the balanced computation in the diagram above. The Area ABCD = C'B'BC+CDD'C' -ADD' -ABB' . C'B'BC = hC'B'(B'B-\-C'C) = h (.779) (.287+7.722) = 3.120 CDD'C = h C'D'iC'C+D'D) = h (6.308) (7.722+8.886) = 52.382 55.502 ADD' = hAD'' D'D = \ (1.491) (8.886) =6.625 ABB' = i A^' B'B = h (5.596) (.287) = .813 7.438 Hence, Area = 48.064 sq. ch. =4.8064 A. 60 MATHEMATICS [III, 43 EXERCISES 1. Find the number of acres in a triangular field whose sides are 14.6 ch., 17.3 ch., and 19.7 ch. 2. Find the area of a triangular field that has two sides 16.6 ch. and 31.2 ch., and the included angle 63 25'. 3. Find the area of a triangular field of which two angles are 67 33' and 48 28' and the included side is q 20.4 ch. [Hint. Find the third angle. Then find the altitude /i = (20.4) sin 48 28'; also find AK. Since angle B is found, KB is equal to ^-f-tan B. The area then follows from the base AB and the altitude /i.j 4. A man has a triangular shaped piece of land ABC as shown in the accompanying figure. The side AB bears N. 59 20' E., 27.30 rd. and BC bears N. 46 10' W., 26.80 rd. Find the length and the direction of AC and the area of the field assuming that the given measurements are correct. [Hint. Compute the longitudes EB and BK. Their difference is CD. Similarly compute DA. But CD and DA determine the direction of AC. The area ABC = BCDE+EBA-ACD.] Ans. CA bears S. 7 8' W. Plot the following surveys to scale, balance the computation for distribution of error of closure, and find the area. 5. Beginning at a stone in the center of the pike; thence south 81 i east, 114 perches to a post ; thence south 2 west, 62 perches to a stone ; thence south 88| west, 113f perches to an iron stake in the road; thence north 2 east, 80 perches to the place of beginning, containing 50 acres and 56 perches. 6. Beginning at a corner stone in the middle of the township road ; thence along said road north 48 west, 13 rods and 9 links, and north 20 west, 32 rods, and north 6 east, 54 rods to a corner in said road ; thence north 71 1 east, 138 rods; thence south 22 east, 16 rods to corner; thence south 45 west, 171 rods to the place of beginning, con- taining 50 acres and 38 rods of land, be the same more or less. Ill, 44] INDIRECT MEASUREMENT 61 7. Beginning at the southwest corner of fraction 24 ; thence N. 1 J E., 33.23 ch. to a stake in the west Hne of said fraction; thence S. 89 E., 8.154 ch. to a stone; thence N. 1^ E., 7.4 ch. to an iron pin in the north boundary line of said fraction; thence S. 89 E., 15.436 ch. to a stake ; thence S. 32^ W., 13.50 ch. ; thence S. 35p E., 5.54 ch. ; thence S. 48 E., 8.27 ch. to a stake ; thence S. 7^ W., 19.20 ch. to a stake in the south boundary Une of the lot; thence N. 89 W., 24.10 ch. to the place of beginning, containing 85.87 acres, be the same more or less. 44. Relations between the Sine, Cosine, Tangent. In any right triangle whose sides are a, b, c, Fig. 66, or, dividing by c^, \2 (!M But, by 37, this is (1) sia-A+cos^A = l. Since, by definition, sinA=-, cosA=-, tanA = r, it c c b follows, by dividing, that sinil (2) C0Si4 tani4. EXERCISES 1. Construct a right triangle whose sides are 3, 4, 5. Find the value of the sine, cosine, and tangent of each acute angle. Illustrate formulas (1) and (2) , 44, by use of this triangle. 2. Proceed as in Ex. 1 for a right triangle whose sides are 8, 15, 17. For one whose sides are 7, 24, 25. 3. Proceed as in Ex. 1 for a right triangle that has its hypotenuse c = 10, and an acute angle = 53 8'. 4. State sin A in terms of cos A. Also cos A in terms of sin A. 5. Given sin A = .6. Find cos A and tan A. [Hint. Construct a right triangle with one leg 6 and the hypotenuse 10. Find the other leg.] 62 MATHEMATICS [HI, 44 6. Given tan A = 1. Find sin A and cos A. 7. Given tan A = 1.732 Use a table to find sin A, cos A. Show that the following relations are true by changing one or both sides of the relation, making use of (1), (2), 44. 8. (sin A+cos A)2 = l+2 sin A cos A. [Hint. Square the parenthesis and reduce by (1), 44.] 9. f-^+tanA'\ f^-r-tan a) =1. \cos A / \cos A J 10. cos A tan A = l. sin A 11. sin2A(l+tan^A)=tan2A. 12. sin'* A cos^ A = sin^ A cos^ A . 13. (sinA+cos A)2+(sin A-cos A)2 = 2. ^ . 1 sin A _ cos A cos A 1+sinA [Hint. Clear the equation of fractions.] 45. Sine, Cosine, Tangent of Complementary Angles. In any right triangle, Fig. 67, A + 5 = 90. By definition. sin A = - = cos B, c A^^ ^>-^ sinB=-=cos A, c tanA-^ , .. b Fig. 67 h tan B Thus, the sine of an acute angle equals the cosine of its complement, etc. EXERCISES 1. Find sin 50, cos 50, tan 50 in terms of functions of 40. 2. Express each of the following as a function of the complementary angle. (a) sin 35. (c) cos 47. (e) tan 60. (6) tan 20. (d) cos 75. (/) sin 57 24'. CHAPTER IV REVIEW OF ALGEBRA 46. Use of Letters. Historically, the first use of algebra was to abbreviate explanations of arithmetic problems. Its modern use, however, is much broader. It includes the free use of symbols to state rules and to discuss problems and numbers which are stated in words. For example, the area of a triangle is the base multiplied by one half the altitude. This may be stated briefly as A = i ab, where A = area, a = altitude, b = base. 47. Equations. An equation is a statement that one expression is equal to another. This statement is written with the sign of equality, = , between the expressions. Thus the equation 3 a;+4= 15 is read, 3 x plus 4 equals, or is equal to, 15. If a value of an unknown quantity is found which makes the two sides of a given equation equal, it is said to satisfy the equation. This value is called a solution or a root of the equation. For example, if a:+4 = 5, the value x=l satisfies this equation, since, when x=l, both sides of the equation are equal. To determine finally whether a given value is a solution, the value should be substituted in the equxition to see if it does satisfy it. 63 64 MATHEMATICS [IV, 47 Example 1. Solve the equation 3x+4 = 10/ Subtracting 4 from each side of this equation gives 3x = 10-4, or 3x = 6. Dividing each side by 3 gives x = 2. This is the solution of the given equation. Substituting 2 for x in the given equation, we have 6+4 = 10. This means that the equation is satisfied, and hence 2 is a solution. Example 2. Solve the equation 5x_ x_29 6 4 12* Multiplying both sides of this equation by 12, the least common multiple of the denominators, gives 10 x -36 = 3 a: -29. Subtracting 3 x from each side and then adding 36 to each side, we find 10x-3x = 36-29, or 7 a: = 7, whence x = \. Substituting a: = 1 in the given equation gives oi -V = -f|=-V. Therefore x = 1 is a root of the equation. 48. Operations on Equations. Without disturbing the equality we may (1) Add or subtract the same number* on each side of the equation. This is usually performed by actually moving it to the other side of the equation and changing its sign. (2) Multiply or divide both sides of the equation by the same number * except that division by zero is excluded. * If any expression other than a simple number is used, it must at least represent a number. IV, 48] REVIEW OF ALGEBRA 65 EXERCISES Solve each of the following equations to find the value of the unknown letter or term. Check each solution. 1. 2xH-3 = 7. 4. -2a;+3=-7. 7. 12sinA+l=7. 2. 2a;-3 = 7. 5. 7x-15 = 3x-7. 8. 2 tan A +3 = 1+4 tan A. 3. 2a;+3=-7. 6. 6-5a = 3-a. 9. ^^x-\ = 2x-^. 10. The number of cows and sheep in a certain farmyard is 45, and there are four times as many sheep as cows. Find the number of each. 11. One angle is twice the size of another angle. If their sum is the complement of three times the larger one, find each of the two angles. Solve the similar problem obtained by substituting the word supplement for the word complement. 12. A does f of a piece of work in 10^ days. With the help of B they then finish the work in 4 days. How long would it take B alone to do the work? 13. Find the angle between the hands of a clock at quarter past four o'clock. At what time will they be exactly together? 14. The driver of a milk wagon is charged $28.56 for the entire load. He carries certified milk at 22^ per quart and 5 times as much milk in cans at 16 ff per quart. How much of each does he carry? 15. Find four consecutive integers whose sum is 94. 16. What are the dimensions of a right triangle whose area is 1.2 acres if the height is 3.7 times the base? 17. In a triangle one exterior angle is 108. If the ratio of the other two angles is 2.6, find the three angles of the triangle. 18. A miller has wheat worth $2.20 per bushel and another lot worth $3.00 per bushel. He wishes to mix these to make 20 bushels of wheat which shall be worth $2.40 per bushel. How much of each shall he take? 19. A farmer has a cow whose milk contains 3^% of butter fat (called a 3|% milk) and another one which gives a 5% milk. How shall he mix them to obtain 50 lb. of a 4% milk? 20. A man can seed a wheat field in 8 days. How long would it take him to do it with the aid of a helper who can do only | as much as he can? 66 MATHEMATICS [IV, 49 49. Addition. In business and in ordinary life it is often necessary to add positive and negative numbers or expres- sions. To add 8 x and 5 x is merely to subtract 5 x from 8 X. Longer expressions are added by grouping similar* terms. Example. Add &x %y-\-bz and 3a:+4 2/ 6z the sum is 9x 4?/ z EXERCISES Find the sum of the following : 1. 2a+36-4c, 5c-4a-6, and 6+5o-2c. 2. 4 x-5 y-2 z, 2 a;+3 y 7 z, and 9 xy-\-^ z. 3. 3A-5+2C, 5^-2B+C, and -^+3B-4C. 4. 5.6 x-2.7 y, 3.5 ar+4.2 y, and -2.2 a;-6.5 y. 5. 5x2 -2.5 m, -.5 a;2+4 m, and 1.45 x2+3.2 w. 50. Subtraction. To subtract one quantity from another, we proceed as in addition after changing the sign of the quantity to he subtracted. Example. From 4a:+ y l0z-\-2 take -x-6?/ + ll2-3 the difference is 5 x-\-7 y21 z+5 EXERCISES 1. From 6 a -7 6 -10 c take 12 a+5 6 -6 c. 2. From 15X-6-2?/ take 5 2/4-9- a;. 3. From2-7A2+A take3A-5A2+8. 4. From 1.5 J5-^2+2.8 take .9^-3^2 _i, 5. From 20 r- 12.5 s-|-4.2 t take 6 ^-2.4 r+9 s. 6. a2+24a-10-(16-5a+a2) = ? 7. x^-5x+7-{-2x^-5x+2) = ? 8 . Simplify 162 Vx + 5 Vx 24 7 Vx by combining similar terms. * Similar terms are those that have a common factor. For example 2x, 3x, ax, mnx are all similar with respect to x. IV, 52] REVIEW OF ALGEBRA 67 51. Removal of Signs of Grouping. When a negative sign precedes a parenthesis or other grouping -sign, change all positive or negative signs within, including the sign of the first term, when the sign of grouping is removed. But when a parenthesis is preceded by a positive sign, no change is made in the terms when the grouping sign is dropped. Example. 3x (5x2y)=3x5x-\-2y. -(P+Q-i2)= -P-Q+R. EXERCISES Remove the grouping signs and add similar terms in each of the following expressions. 1. 8x-(7-2x). 2. ilQx-7 y)-ix-4:y). 3. (2p-5g)-h(4p+g)+6Q. 4. A-(3B-2.5)-(B-1.5A). 5. 2a-(6-(2a-6)). 6. Solve 6 ?/ - (4 -2 t/) =20 for y. 7. Solve 12.5x-(10-2x) = 19-(x-2) for x. 8. Solve 7 x-[x-{5 a:-7)]+3 = 24 a;-9(H-x) for x. 52. Multiplication. The product of two numbers having like signs is positive, of two having unlike signs, negative. For example, three times 5 may be written (3)(5) = 5+5+5=15. In the same way (3)(-5) = (-5) + (-5) + (-5)= -15. The multiplier +3 is here a positive integer and the multi- plicand is added three times. If the multiplier is negative, the multiplicand should be subtracted. Thus (-3)(5)=-(+5)-(+5)-(+5)=-15, and (-3)(-5)=-(-5)-(-5)-(-5)=-(-15) = +15. Thus, generally, {a)(b) = ab and (-a)(-6) = a&, (a)(-6)=-a6 and i-a)(b)=-ab. 68 MATHEMATICS [IV, 53 53. Exponents. When some of the factors of a number are alike, it is often convenient to indicate the fact as follows : 2X4X4X4 = 2X43=128, 3 a ay y y = S a^y^. The small figure written above and to the right of an ex- pression denotes the number of equal factors and is called an exponent. EXERCISES Multiply 1. 4 0:2/ by 5 xy. 6. arri^cd^ by a?md^. 2. 2 a6c by fee. 7. -4 aSxz/ by -3 axV- 3. x^ by x^. 8. X* by xP. 4. 3 fe^c by 4 ahc^. 9. a^ by av. 6. 7 r7 by 2 s^ by 3 rV. 10. 32 by Z^. Carry out each of the following indicated multipUcations. 11. (a) 2(3+5). (fe) 2{x^-y). (c) 4a(a+6+ac). 12. a:(x2+l). 14. a2(a2_a+i). 13. x3(2x4-3a:3+4x). 15. x"(xi-+x3+"). 16. Multiply a:2+x2/+?/2 by x-?/. 17. What are the factors of a^ h^ as shown by Ex. 16? 18. Multiply a2-afe+fe2 by a+fe. 19. Multiply x'^+xy+y'^ by x^ xi/+i/2. 20. Multiply 3a6-2a2-2aby 2afe-2a-3fe-6. 54. Standard Type-forms. Multiplications may fre- quently be performed mentally. Some standard type-forms for this purpose are the following : (a+fe)(a-6) = a2-62. (a+6)(a+6) = a2+2a6+62. (a-6)(a-b) = a2-2a6+fe2 (a+6)3 = a3+3 0^6+3 ab''+h\ (a-6)3 = a3-3 a^fe+S a62-6^ (a-f 6+c)2 = a2+52_pc2+2 a?>+2 ac+2 6c. (a;+a)(a;+6) = x2+(a+6)a;-|-a6. IV, 56] REVIEW OF ALGEBRA 69 EXERCISES 1. State in words the type-forms given above. Carry out orally each of the following indicated multiplications. 2. {2x-3y)(2x-\-Zy). 11. {a+b+2y. 3. (x-3)(x+3). 12. {a-h-c)\ 4. ix^y-Sb){x^y-\-Sb). 13. (2x-a+3)2. 5. (a+3 6)(a-3 6). 14. (x+7)(x+2). 6. {x + [a+b])ix-[a+b]). 15. (a-3)(a+5). 7. ix-2y)\ 16. (a:+2)(x-3). 8. {Sx+2yy. 17. (3a6-x)(3a6+7x). 9. (x+2 2/)3. 18. (p'q- 10) (p'q+S). 10. (l-a)3. 19. (x2-3)(x2-4). 55. Division. The rules for signs hold in division as in multiplication, as shown in 52. EXERCISES 1. Divide y^ + 15 i/+36 by 2/ +4. 2. Divide x^+x^-f? x^-G x+8 by x2+2 x-f 8. 3. Divide A3- 1 by A-1. 4. Divide a^-\-b^c^-\-3 abc by a-\-bc. 6. Divide 1+x by 1 x. How many terms in the quotient? 6. Divide 1 by 1 +a: to four terms. 56. Factoring. The word factor means '' maker.*^ Fac- tors are thus merely numbers or expressions which make other numbers or expressions when multiplied together. The factors of 6, for example, are 2 and 3, since 2 times 3 is equal to 6. Factoring is the process of finding the factors of an ex- pression. Some expressions occur quite often in the ap- plications of algebra. The different cases may be grouped under so-called type-forms, such as taking out a common factor, grouping similar terms, the difference of two squares, etc. 70 MATHEMATICS [IV, 56 The following type-forms will aid in finding the factors of many expressions. ah-\-ac = a(b-\-c). a2-62=(a+6)(a-6). a'+2ah+={a+by. a'-2ah-\-={a-h)\ a^-h^= (a-h){a'-j-ah-\-). a3+63= (a-\-h){a^-ah+). x^-{-(a-\-h)x-\-ah= (x-\-a){x-\-h). Examples. A^-9 x^ = (A+3x){A-3x). 9 x2 - 30 xi/ +25 2/2 = (3 a: - 5 i/)2. EXERCISES Factor each of the following expressions by inspection. 1. a%^-xK 7. a:2+7a: + 12. 2. (x+yy-zK 8. a^+^ab-2l, 3. a2-(6+c)2. 9. x3 + l. 4. a^-(h-cy. 10. x^-y\ 5. 9x2+6x+l. 11. 8x3+27. 6. x2+4x2/+4?/2. 12. a3-27 6. Factor each of the following expressions by grouping. 13. axay-j-hx by = a{x y)+b{xy) = {a+b){xy). 14. 2/3+3 1/2+3 1/ + 1. 15. a^-ab^-a*b+bK 16. In x2+6 x+k, what is the value of k to make the expression a perfect square? 17. How much must be added to x2 +8 x to make it a perfect square ? How much to x2 +5 X to make it a perfect square? Resolve each of the following expressions into factors. 18. a^+Qa-\-S. 21. 1 -12 a6+35 a262. 19. x3-4x2+3x-12. 22. 48r2-10r-25. 20. 3a4-8a2-35. 23. 15 p^-jyr-2r\ 24. 4x2-13x+10. IV, 57] REVIEW OF ALGEBRA 71 57. The Binomial Theorem. From the following equal- ities we may observe the law for raising a binomial to positive, integral powers by inspection. (a+6)2 = a2-|-2a6+62. (a+6)3 = a3+3 a%-\-S ab'-{-b\ (a+6)4 = a4+4 a%+Q a^-\-4: a-^. {a-\-by = a'-\-5 a^fe+lO a%^-\-10 a%^-{-5 a+. We may now make the following observations. (1) The number of terms on the right of the equality sign is always one more than the exponent on the left. (2) The exponent of the first letter in the first term of the expansion is the same as the power on the left. This ex- ponent decreases by 1 in each succeeding term. The ex- ponent of the second letter, b, in the second term is 1, and increases by 1 in each succeeding term. (3) The first coefficient is 1 ; the second coefficient is exactly the power to which we are raising the binomial; and in general, the coefficient of any term may be found from the preceding coefficient by multiplying it by the exponent of a in that term and dividing this by one more than the exponent of b in that term. For example, (a+6)io = oio+10a96 + ^^^ -0862+ ... 1 * ^ = aio + 10 0*6+45 ab'-\-^^a''b^+ etc. o (a +6)" = a^ +na-% + ^ ' ^.^ ~ ^^ a^-%^+ etc. These facts make up the binomial theorem. It is true for all values of a and b and for n equal to any positive integer.* * The binomial theorem may be extended so that, under certain restric- tions, it will hold for other values of n, e.g. fractional and negative values of n. 72 MATHEMATICS [IV, 57 EXERCISES Expand each of the following binomials as indicated. 1. {a+by. 4. (a-2)3. 7. (x-V^)'- 2. (a+6)". 5. {a-2y. 8. ^| + |)'- 3. (a+fe)i6. 6. (x+Va)2. 10. Given two line-segments a and b. Con- struct a square on the segment a +6. Show by means of this square that {a+by = a^-\-b^+2 ab. Fig. 68 58. Fractions. A fraction is an indi- cated division of one number or expression called the numerator by another number or expression called the denominator. As in elementary arithmetic, the form of a fraction may be changed without thereby changing the value of the fraction. The value of a fraction is unchanged if both numerator and denominator are multiplied by the same expression, or divided by the same expression, excepting zero. Fractions may be reduced to lowest terms by dividing both numerator and denominator by their common factors. EXERCISES Reduce to lower terms each of the following expressions. 1 _6 2 3 ^!^.2-^. 4 * 32' ' 725' ' 2^3 ab^ ' 5. 13. 35 ax^y^ 28 a^x^y a%-ab\ a^b^ 15sin^a;tan3; 20 sin x tan^ x xy 6s2-s-2 . 9s2-4 10. 6. -mp-Q)(r-sy(t-iy 2S{p-q){s-r){l-t) ^ ax+ay-x-y ^^ t. X^-y^ X^-S x2-fx-6 12. x^-4xy-{-3y^ sin^ X cos^ X sin x-f-cos X 14. 12a^-aa;-6a^ 8a2-2ax-3a2' IV, 60] REVIEW OF ALGEBRA 73 59. Addition and Subtraction of Fractions. In order to odd or subtract fractions they must he reduced, as in elementary arithmetic, to the same denominator. This denominator should be the least one common to all of the denominators, rru 2 ,5_14, 15 29 ^^^^ 3+7-21 + 21 =21- Similarly, a^x^^^^^g^, h y by by by EXERCISES Perform the operations as indicated in each of the following expres- sions. 1. 2^3^4 9. X X x-l x+l 2. 3^15 V35 9; 10. 1+2^ 3 . a a^ a + 1 3. x-2 , 2x+3 3 ' 2 11. (a -6)2 4. 1+1. 6. 1- ah a 1 h 12. 3 1 5a 1 5 2 a-2 (a-2)2 6. * + 1 a+6 ah 13. 5 7,1. x^-2 x-\ x^^x-1 7. 1 1 a-\-h ah 14. x-2 x-\ , 3 x-l x-2 x2-3a:H-2 8. X ^ X ^ x-l x-\-\ 15. 5(a+6) 1 (a +6)2 a-h (a-6)2 60. Multiplication and Division of Fractions. As in arithmetic, to multiply one fraction by another multiply their numerators together for the numerator of the product and their denominators together for the denominator of the product. Any factors common to numerator and denominator may be cancelled before multiplying. To divide one fraction by another, invert the divisor and then multiply. 74 MATHEMATICS [IV, 60 When the division of one fraction by another is indicated in the form of a fraction, this fraction is called a complex fraction. a 2 ^, b a , c ad ad ^ 3 2^5 5 ^^^' -c = bd = -b''-c = Vc' 6=3><6=9' d 5 EXERCISES Perform each of the following indicated operations. ' (i--:)x-L- \ 9^/ xy 3^4^7^9 ^ 35 ^-39a ^ V^x^-^^r^^^i 15 g^ . -117 ., 18 6 27 62 3. "^^x ^ . 9 15^'^25_5r. 2 x2-2x-3 ' 32r ' 36 x'^-^y^ 10 4x2 . -12 ax x -^ . ^ > 10. x 2y x^y^ ay"^ 9y 5 r2^6t;+9 ^ t;-4 ^ ^^ 2 <2 . < 16-8y+y2 9-y2' ' a'+i^ *. a_^^' Simplify the following expressions. J , a -. _ 2jc _j_ 2 3 6 4 14. Z. 16. i- 18. ^ 111 5 1+- 1- x2 a2 x+y x y xy X x+y y 61. Simple Linear Equations of One Unknown Quantity. A linear equation is one in which there is no term of higher degree than the first. We shall now study some equations of the first degree which are more complicated than those given earlier in this IV, 61] REVIEW OF ALGEBRA 75 chapter but which may be solved by the same methods used there. Equations may be cleared of fractions by multiplying both sides of the equation by the least common multiple of the denominators. Example. Solve the following equation for x : 6a;-3 ^ 3x-2 2x4-1 X The equation is cleared of fractions by multiplying both sides by the least common denominator, (2 x-\-l){x). This gives x(6a:-3) = (3x-2)(2j:-fl), or 6x2-3x = 6x2-x-2. Collecting the like terms, we find -2x=-2 and thus x = l. The solution may be checked by substituting 1 for x in the original equation. EXERCISES Solve each of the following equations. 1.4(2.-1)-6(|-|)=.2 2. 4(x-|)2+6(x-f) = (2x-5)2+302. 3. One side of a right triangle is 5 inches and the other side is 2 inches shorter than the hypotenuse. How long is the hypotenuse? 4. When the radius of a circular plot of ground is increased 2 feet, the area is increased 88 sq. ft. Find the size of the original plot, using 7r = 22/7. 5. At what time between 3 and 4 o'clock are the hour and minute hands of a clock together? 6. Calculate when you were, or will he, half as old as your father. 76 MATHEMATICS [IV, 61 Solve each of the following equations, the last letters of the alphabet X, y, z, w, etc. being the unknowns. 7. ^+3 = 4. 3+?^^ 4 3 4 ^^+2 ^ 11. ax-j-bx = c-\-d. 9. (a+6)(x+2)=c. 12. y = ^%hll. 13. (a+2)(fc-0)-(a-z)(c+z)=2. 14. Solve the equation l+i for i. Also solve it for I. 15. Solve the following equation for F : C = i(F-32). This equation represents the relation between the readings on Centigrade and Fahrenheit thermometers. What is the Centigrade reading for 68 Fahrenheit? Change 25 C. to Fahrenheit reading. 16. A farmer made 100 lb. of fertilizer, using 95 parts acid phos- phate, 100 parts cottonseed meal, and 5 parts muriate of potash. He then decided to add enough acid phosphate to each 100 lb. of the original mixture to make the acid phosphate content 50% of the total. How much should he add ? Ans. 5 1b. [Hint. Let x = the number of pounds added. Then 50% of (100 + x) =47^ lb. -j-x.]. 17. In Ex. 16, how many pounds less of acid phosphate should be used to every 15 pounds of muriate of potash, in order to reduce the acid phosphate content of the mixture to 40%? 18. In the fertilizer of Ex. 16, how many pounds less of acid phos- phate shall be used to every 60 pounds of cottonseed meal in order to reduce the acid phosphate content of the mixture to 42.5%? 19. A fertilizer used in cotton fields consists of 62^% acid phos- phate, 30% dried blood, and 7^% muriate of potash. It is desired to raise the acid phosphate content to 70%. How many pounds of acid phosphate must be added to one ton to do this? IV, 62] REVIEW OF ALGEBRA 77 62. Quadratic Equations. A quadratic equation is one in which there is a term of the second degree, but none of higher degree. There may be terms of lower degree. For example, is a quadratic as it contains terms of the second and lower degree. Quadratics may be solved (a) by factoring, or (6) by completing the square. Example. Solve the equation 3 x^+G = 11 a: by factoring. Transpose all terms to the left side of the equation. Thus 3x2-llx+6 = 0. Factoring the left-hand side, we obtain (3x-2)(x-3)=0. But this means that each factor may be put equal to zero, since a product is zero if any of its factors are zero. Thus either 3a:-2 = 0, i.e. x = f, or else X 3 = 0, i.e. x = S. Check : Substituting f for x in the original equation gives 3.f+6 = ll.f, or V+6=- Substituting 3 for x in the same equation gives 3- 32+6 = 33. EXERCISES Solve each of the following quadratics by factoring. 1. 5x2+6x = 32. 2. 3x2-44x=-121. 3. 4x2+5x = 0. 4. x2-2 6x = 2ax-4a6. 5. 3x2 = 12. Notice, in this example, the double sign obtained on extracting the square root. 6. 11x2-10 = 5x2+44. 7. (2x+3)2 = 5x+39. 78 MATHEMATICS [IV, 63 63. Solution of the Quadratic by Completing the Square. The method of solution in this case may be observed in the following examples. Example 1. Solve the equation ^2+27 = 12 x for x by completing the square. Transpose the terms containing x and x^ to the left side of the equa- tion, those not containing an x to the right side of the equation. Thus, x^-\2x=-27. To make the left side a perfect square we must take half oj 12, square it, and add it to both sides of the equation. This gives a:2- 12 a: +36 =-27+36 = 9, or (x-6)2 = 9, whence Therefore a: = 6==3, i.e. x = 9 or x = S. Example 2. Solve the equation 7 x = 3x^6. Transposing the unknown terms to one side and the constant to the other side, we find 3a:2-7x = 6, or, dividing by 3, x^-^x = 2. Half the coefficient of x is ^. If we square this and add it to each side, we obtain or (x-|r=W- Thus a;-|==b-V, whence x = i-\-^-, or x=i^, i.e. x = 3 or x= |. EXERCISES Solve each of the following equations by completing the square and check the answers. 1. 2x2+4 = 9x. 4. 8p2_2p-15 = 0. 2. 3x2+5a:-12 = 0. 5. 5 tan^ A -8 tan A =21. 3. 4 22-20 2+25 = 0. 6. 12 sin^ A +6 = 17 sin A. IV, 64] REVIEW OF ALGEBRA 79 64. Simultaneous Equations. Two (or more) distinct equations which contain the same unknown quantities are called simultaneous equations, if they are considered at the same time. Solution of simultaneous equations is possible, in general, only when the number of given equations is the same as the number of unknown quantities. Example 1. The cost of a bushel of potatoes and a bushel of apples was $4.00, but the apples cost 20ff more than the potatoes. Find the price of each. Let a; = the price of a bushel of apples, and Then and Adding gives y=the price of a bushel of potatoes. a;+y = 400, x-y = 20. 2x = 420, x = $2.10 the price of apples, whence 2/ = $1.90 the price of potatoes. Either of the above equations taken singly would not be sufficient to determine x and y. Example 2. Solve the simultaneous equations \2x- 5?/ = 60, \7x-\2y = M2. Multiply the first equation by 7, the second one by 2. Thus 14 a: -351/ = 420, 14x-24i/ = 684. Now subtract and we have -111/ =-264, or 2/ = 24. For y in the first equation put 24, giving 2 X- 120 = 60, i.e. re = 90. Thus the set of values x = 90, i/ = 24 satisfies the second equation and is said to be the solution of the set of equations. 80 MATHEMATICS [IV, 64 EXERCISES Solve each of the following sets of simultaneous equations. 3 A +2 5 = 32, 1. 2. |5o+6 = 21, [2a+3& = 24. 3. {Sx+7y= \l2x-5y = 100, 88. 20A-3B = 1. 3x 19 2x +5 2/ = 13, -33 = ^^-^. 5. The standard daily ration for an adult laboring man should be about 16 oz. starch, 4 oz. fat, and 4 oz. albumen. These materials as found in bread, butter, and beef are as follows. Food Starch Fat Albumen Bread Butter Beef 54% 0% 0% 1% 83% 15% 9% 1% 15% Find the quantities of these foods required to make a daily ration. 6. If a dairy cow requires 1 lb. of protein to 6 lb. of carbohydrates in her food, what should be the proportions of dry peas and hay for her daily ration if the dry peas contain 10 lb. of protein and 32 lb. of carbo- hydrates per bushel of 60 lb., and the hay 88 lb, of protein and 880 lb. of carbohydrates per ton ? 65. Fractional Exponents. Thus far we have used only exponents that are positive integers. We defined a^ to mean aa, a" =aaa to n factors, etc. It is often convenient to use fractional exponents, such as x^/^. But X taken as a factor f times has no meaning. Since it is important that the symbols used should always follow the same rule, we shall, in the case of exponents, extend the definition for combining them so as to include fractional and negative values. That is, we will make the relation always true, whatever p and q may be. IV, 67] REVIEW OF ALGEBRA 81 Thus o}''^ shall mean a number such that Therefore, a}'"^ is the square root of a and a^''^ = Va. Likewise a^^* shall mean that fl3/4 . ^3/4 . fl3/4 . a^/^ = (^, i.e. a^^* IS ..>, 1. Parcel post rates in effect Jan. 1, 1914, are given in the following^ table. Exhibit this in graphic form. Weight of Parcel Not over 4 oz Over 4 oz. up to 1 lb. . . Over 1 lb. to 2 lb. . . . For each pound over 2 lb. up to 20 lb Rates 1st and 2d Zones 3d Zone 4th Zone 1 cent per ounce or fraction thereof 5ff li additional Si 2i additional 7^ 4:^ additional 2. Make a diagram of segments showing the periods of incubation of poultry as given in the following table. Name op Fowl Days Name op Fowl Days Name of Fowl Days Common Hen . Pheasant . . Duck, common 21 25 28 Pea Fowl Guinea . Goose . . 28 25 30 Partridge . . Duck, Barbary Turkey . . . 24 30 28 3. Show in graphic form the production of wheat in 1918 in the United States and in some other leading wheat-producing countries as given in the following table. Region Bushels Acreage United States Australia Canada France Germany Argentina 917,000,000 152,400,000 210,300,000 147,150,000 90,140,000 187,210,000 59,100,000 11,500,000 17,350,000 10,260,000 3,590,000 17,175,000 102 MATHEMATICS [V, 84 84. Area Charts. In example 2, 83, the table of areas given there 'was 'shdwn on a bar diagram. That table of areas will now be shown in various ways as follows. A. By a series of squares. The sides of the different squares are to each other as the square roots of the respective areas. Thus the side of the Russia Texas Germany France California Gt.Britain Ohio u Fig. 75. Comparing Areas by Squares square which represents the area of Russia is about three times that for Texas, etc. The squares are then constructed to a convenient scale. Is the comparison as effective in this form as in the bar diagram ? Why ? B. By sectors of circles. Inspection shows the area of Russia to be about twice that of all the others in this table combined. It should therefore Fig. 76. Comparing Areas by Sectors of Circles V, 84] GRAPHIC REPRESENTATION 103 occupy f of the circle. Texas is about -J the remainder, and is so represented on the circle. C By use of dot diagrams. Russia Texas Germany France Call- Great Ohio fomia Britain Sxplanation'. ^^H = 600,000 Sq.Mi. FiQ. 76 a. Dot Diagram D. By differently shaded areas. This is best illustrated by reference to ordinary maps of these countries, made to the same circle. EXERCISES 1. Exhibit in graphic form the following table of fertilizing materials and the per cent of each element carried. Material % NlTKOQEN % I*H08PH0Ric Acid % Potash Ground bone .... Dissolved bone . . . Cottonseed meal . . . Tankage Fish scrap Wood ashes 3.0 3.0 7.0 6.0 8.0 22.0 15.0 2.5 11.0 7.0 1.5 1.5 5.0 2. Make a dot diagram of the world's wheat production for 1914 given by the following data. The World 3,362,299,000 bushels. United States . Russia ... France ... Hungary (proper) India (British) . 891,017,000 bushels. 776,960,000 bushels. 319,667,000 bushels. 105,237,000 bushels. 312,032,000 bushels. CHAPTER VI GRAPHS IN ALGEBRA 85. The Equation and its Graph. We have seen so far that various relations between two quantities, as these quantities change, may be represented by graphs. In some cases we knew the law of this change as represented by an equation. For example, for lard at S5^ per pound, the law was : Cost = 35 times the number of pounds. The graphs were constructed by using two axes at right angles to each other. On these axes we used the number scales for various quantities such as cost, pounds, time, distance, etc. In general then we may use y and x to repre- sent these quantities and thus describe the relations between them in the language of algebra. This leads very often to an equation in y and x. Thus for lard at 30^ per pound, the cost, y, of any number of pounds, x, is y = S5x. 86. Rectangular Coordinates. It is thus evident that when the relation between two quantities is expressible by an equation, this relation may be exhibited by a corresponding graph. The following considerations are commonly accepted as the basis for such representation. Let us call the hori- zontal axis or scale, the x-axis, and the vertical axis, the y-axis. There being no restriction on these scales as to length, we give x and y each positive and negative values. 104 VI, 86] GRAPHS IN ALGEBRA 105 Let the two axes be drawn at right angles to each other. Their point of intersection at is called the origin, Fig. 77. They divide the plane into four parts, called quadrants, named as shown in the figure. From the origin to the right and from the origin up is regarded as positive in direction, but to the left, also down, is negative. We can now locate definitely any point in the plane (of course with reference to these two fixed lines). Thus the notation (2, 3) means the point located by going 2 units to the right on the x-axis and then from this place Second Quadrant |--1(2,3) (-4,-2)^- Third Quadrant First Quadrant C2,-3) Fourth Quadrant Fig. 77 3 units up parallel to the y-axis, as shown in Fig. 77. Like- wise, (2, 3) means to go 2 units to the right on the a:-axis and then 3 units down parallel to the 2/-axis; (4, 2) means 4 units to the left and 2 units down. The distance from the vertical axis is commonly called the abscissa of the point, and the distance from the a:-axis is known as the ordinate of the point. These two together are called the coordinates of the point. The notation for the coordinates of a point is thus very precise, viz. written in parenthesis with the x-coordinate or abscissa first and separated from the y-coor- dinate by a comma. 106 MATHEMATICS [VI, 86 EXERCISES 1. Locate the points (1, 5), (3, 7), (5, 2), (9, -3), and (12, -6). Draw lines connecting these points in the order given. 2. Draw the graph determined by connecting by a smooth curve the points (-3, -2), (-1, 0), (0, 1), (2, h), (5, 7), and (8, -11). 3. Let (0, 0) and (4, 0) be two vertices of an equilateral triangle. Find the coordinates of the third vertex in the first quadrant. What is the area of this triangle? What acute angles are involved in this triangle ? 4. Find the area of the triangle whose vertices are the points (2, 3), (5, 6), (4, 7), by dropping perpendiculars from each vertex on the y-axia. 87. Linear Equations and Their Graphs. Every equation of the first degree represents a straight line. We saw above that many price curves are straight lines. This is due to the fact that the cost is equal to the price of one article times the number of articles, i.e. (1) y=px, where y is the cost of x articles at p units each. But this equation is equivalent to the ordinary proportion, (2) P 1 Now in Fig. 78, AE represents the cost, p, of one article, since from (1) y = p when x = l. BD represents the cost of two articles, LC that of x or any number of articles. O .1 2 h Fig. 78. Linear Graph VI, 88] GRAPHS IN ALGEBRA 107 The triangles OAE, OBD, OCL are similar since CL (or y) _ PC (or x) AE (or p) OA (or 1) But this similarity of triangles can hold only when OL is a straight line, and the proportion (2) holds only when the equation (1) is of the first degree. Therefore an equation of the first degree represents a straight line. State the converse of this proposition, and show that it is true from Fig. 78. 88. Graph of an Equation. Let us construct the graph for the equation y = 2x+4. We first give a value to a:, as :c = 0, and find the corresponding value of y. In this case y = 4:. For x=l, y k Q, etc., as in the following table. X y 4 1 6 2 8 3 10 -1 2 -2 -3 - 2 -8 -12 Fig. 79 Now locate, as in Fig. 79, the points whose coordinates are these pairs of x and y values. Thus x = Q, 2/ = 4, i.e. (0,4) locates the point A on the diagram; ( 3, 2) locates B. It appears that the graph is a straight line. Why is it a straight line? 108 MATHEMATICS [VI, 88 EXERCISES Find several points on the graph of each of the following equations, and draw the graph. 1. y = x, y-\-x = 0. g ^4-^ = 1 * 2 "^ 2. y+x-2 = 0. 7. -^+^=1. 3. y = x-\-^. 4 6 4. 2x-Zy = Q. 8. |-?/ = l- 5. a:+?/ + l=0. 9. ?/-l=3(x-2). 10. What is the degree in x and y of each of the equations just plotted? What kind of graph does each represent? What is the reason for calling these equations linear equations? 11. Compare the straight lines y = x and y = x-\-^. If a: be given the same value in each equation, how are the corresponding ^/-values related? Try several values, say x = l, x = 2, a: = 4. Thus the values x = l, y = l, fix a point on the first line; the values x = \, ?/ = 4, fix a point on the second line. These ?/- values differ by 3. Try other values of x. 12. What happens to the equation when a straight line, say y = x-\-l, is raised three units parallel to itself? Raised five units? Lowered four units ? 13. Plot y = 2x. This is the equation of a straight line through the origin, since for x = 0,yisO. What is the equation of a straight line parallel to y = 2 x and 3 units above it ? How much larger is every y-vahie ? 14. Suppose that the cost of setting the type in a job of printing is $1.50, and the cost of paper and printing is ^^ per copy. Find the cost, y, of any number of copies, x (i.e. y = total cost and x = any number of copies). Draw the price curve from the cost equation. [Hint. In plotting the equation, let one vertical small space equal lOff and one horizontal small space equal 10 copies.] 15. The cost of setting the type for an order of printing is $1.00. The cost of material and printing is ^^ for each copy. What is the cost for any number of copies ? Draw the graph. Estimate from the graph the price of 150 impressions. Of 500 copies. VI, 89] GRAPHS IN ALGEBRA 109 16. For the job of printing in Ex. 15, another firm offers to disre- gard the initial cost of typesetting, and to print the copies for ^^ each. Find the cost for any number of copies. Draw the graph on the same diagram as for the previous example. 17. Of the two firms in the Exs. 15-16, which one will most cheaply furnish 100 copies? Which firm will give the greater number of copies for $1.25? For $2.00? How many copies will be printed just as cheaply by one firm as by the other ? What will they cost ? 18. The initiation fee in a certain society is $5.00 and the annual dues are $3.00. What is the cost of membership for any number of years? Draw the graph. Estimate from this graph the cost of mem- bership for 5 years. For 12 years. 19. The cost of mimeographing is quoted by a firm as follows : 25 copies of 1 page, 50ff; 50 copies of 1 page, 57^; 100 copies, 70^; 200 copies, 95 ff; 300 copies, 120^; 400 copies, 145^; 500 copies, 170^ Plot this data, and draw the price curve. What is the graph? 20. Draw a graph showing the relation between the Fahrenheit and Centigrade thermometers. Plot the Fahrenheit scale as the vertical axis and the Centigrade scale as the horizontal axis. The Fahrenheit scale is marked 32 at the freezing point, while is the freezing point on the Centigrade scale. Hence 32 must be subtracted from Fahrenheit readings before com- paring with the Centigrade reading. Find the difference between the readings for freezing and for boiling on each scale. What is the ratio between these two differences? Establish the algebraic relation, and plot the graph from it. Ans. f' = 9/5 C +32. 21. Calculate algebraically the Centigrade reading for 41 F., F., -20 F., +98 F., +68 F. Do your answers correspond to the graphical results? In the same way calculate, graphically and algebraically, the Fahren- heit readings for the following Centigrade readings : 14 C, 25 C, 20 C, -10 C. 89. Two Points Determine the Graph of a Straight Line. We have seen that every equation of the first degree in two variables represents a straight line. Since a straight line is fixed by any two points on it, to find the graph for an 110 MATHEMATICS [VI, 89 equation of the first degree, it is only necessary to find two points on the graph and draw the straight fine through them. Example. Plot the graph of the equation 3 a: 4i/ + 12 = 0. For x = 2,y is 4^, and for x = 1, ?/ is 2^. Locate the points (2, 4^), ( 1, 2^), and draw the straight Hne as in Fig. 80. 90. Intercepts. Since x = for every point on the y-axis, to find where any curve crosses the y-Sixis, we put x = 0, and find y. Similarly to find the x-inter- cept, put y = and find x. In the example just given, viz. 3 a: 4 1/ + 12 = 0, if x = 0, y is 3, and when 2/ = 0, X is 4. It is seen in Fig. 80 that the straight line passes through the points (4, 0) and (0, 3) on the axes. Such distances cut off on the axes by a curve or by a straight line are called the intercepts. 91. Simultaneous Solution of Equations. In solving simultaneous equations, for example, the pair 2x-y=\, 3x+2 2/=12, we find the pairs of values of x and y which satisfy both equations. To solve the given pair, multiply the first one by 2 and add the result to the second equation. This gives 7x=14, or x = 2. In either equation, if x = 2, y = 3. Thus (2, 3) is the set of values that satisfies both equations. Let us now see what this means with reference to the graphs of these equations, finding their intercepts, as in Fig. 81. Fig. 81 Plot them by Each line passes VI, 91] GRAPHS IN ALGEBRA 111 through the point (2, 3). Hence the coordinates for the point of intersection of two graphs is the set of values ob- tained by solving them simultaneously. EXERCISES Draw the graphs for each of the following pairs of equations. Find their solution from this diagram, and determine whether this agrees with their algebraic solution. 1. 2x-3y = 25, x+y = 5. 3. \ 5x+2y = S, ,2x-Sy=- 2x-3y = &, 12. 5. 5x+6 2/ = 7, 2x-4!/=-4. 4. J+I=^- 6. 3-4a: = y, _8x-\-2y = Q. 7. The equations of the sides of a triangle are x-y + l=0, 2x+2/ = 13, x+5y = n. Plot these equations and find each of the vertices of the triangle. 8. Find the length of each of the sides of the triangle of Ex. 7. [Hint. Use each side of the triangle as the hypotenuse of certain right triangles formed by drawing lines through the vertices parallel to the axes.] 9. A rectangular field is 32 rods longer than it is wide. The length of the fence around it is 308 rods. Find the dimensions of the field. 10. A steamer makes 6 mi. an hour against the current and 19^ miles an hour with the current. What is the rate of the current and what is the steamer's rate in still water? 11. The difference between two sides of a rectangular wheatfield is 30 rods. A farmer cuts a strip 5 rods wide around the field and finds the area of this strip to be 7^ acres. Find the dimensions of the field. 12. Given two boys A and B, a 30-lb. weight, and a teeter board. They find they will balance when B is 6 ft., and A 5 ft., from the ful- crum ; but if B places the 30-lb. weight on the board beside him, they will balance when B is 4 ft., and A 5 ft., from the fulcrum. Find the weight of each boy, using the law of levers ; that the weights balanced are inversely proportional to their respective distances from the fulcrum. 112 MATHEMATICS [VI, 92 92. Sine, Cosine, Tangent of an Obtuse Angle. Let XOB, Fig. 82, be an obtuse angle. From a point on OB drop a perpendicular, EC \js on XO produced. OC is then negative, |\^ 76, and OB is regarded as positive. The ! X~^ sine, cosine, tangent of XOB are now ^ ^ . _ - , , T, . . Fig. 82. Obtuse denned by the following ratios: Angle sin XOB= ^^, and is +, OB' ' cos XOB= , and is , CB tan XOB =-^^, and is . OC' It follows that the sine of the obtuse angle XOB is identically equal to the sine of its supplementary acute angle COB ; and that the cosine and the tangent of the obtuse angle XOB are numerically equal, respectively, to the cosine and the tangent of its supplementary acute angle COB, but opposite in sign. Hence, if A is any obtuse angle and 180 A its supple- ment, we have (1) sin {im-A) = sinA. (2) cos {1^0- A) =- cos A. (3) tan (180- A) = -fan A. Examples. sin 150 = sin (180 - 150) = sin 30 = + .5 cos 150= -cos (180 -150) = -cos 30= -.8660 tan 150 = -tan (180 - 150) = -tan 30 = - .5774 sin 110= sin (180 -110) =sin 70 = .9397 cos 110= -cos (180 -110) = -cos 70= -.3420 tan 110= -tan (180 - 110) = -tan 70= -2.747 VI, 93] GRAPHS IN ALGEBRA 113 93. Reading of Tables for Obtuse Angles. To find from the table the value of the sine, cosine, or tangent of an obtuse angle, find the value of the same function of the supplement of the angle, and for the cosine and the tangent prefix the negative sign. Examples, sin 1 12 10' = +sm 67 50' = + .9261 cos 112 10'= -cos 67 50'= -.3773 tan 112 10'= -tan 67 50'= -2.4545 Conversely, if cos A or tan A is negative, find from the table the angle whose cosine or tangent is respectively this same positive value. Then subtract this angle from 180. The result is A. Examples. Find A, if cos A = .8192 Look for +.8192 in the column of cosines. This corresponds to 35. Then .1 = 180 -35 = 145. Again, if tan A = .5774, find A. Look for +.5774 in the column of tangents. This corresponds to 30. Then A = 180 -30 = 150. If sin A has a given positive value, there are two angles which have this sine, one an acute angle and the other its supplement. EXERCISES 1. Given sin A = 3/5, where A is an obtuse angle. Find cos A , tan A . 2. Given tan A = 5/12, where A is an obtuse angle. Find sin A, cos A. 3. A flagstaff 80 feet high, on a horizontal plane, casts a shadow 110 feet long. Find the angle made by the sun's rays with the horizontal. Find the value of the following expressions from a table of natural functions. 4. sin 131 10' 6. tan 120 8. tan 95 15' 5. cos 150 20' 7. cos 95 15' 9. sin 95 15' Find A in each of the following expressions. 10. sin ^ = .4746 12. cos A = -.4746 14. tan A = -.4125 11. cos A = -.8090 13. tan A = -1.7321 15. cos A = -.5819 114 MATHEMATICS IVI, 94 94. Grade, Slope. In an earlier section, 4, we spoke of the pitch of a roof. It was defined as the rise of the roof divided by the width of the building. Applied to a hill, we saw that the rise of the hill divided by its run is its slope or its grade. Mention was made also of the fact that many grades, particularly in railroad construction, are spoken of in terms of per cent ; for example, a 5% grade is one that rises 5' in 100'. A little later, 36, this notion was made more precise in studying the right triangle. The rise of the hypotenuse of a right triangle divided by its run, i.e. the altitude divided by the base, was used as a measure of the base angle, and was called the tangent of this angle. The same notion of slope will be used in what follows for any straight line. 95. Slope of a Straight Line. Let a straight line pass through two points P={xi, yi) and Q = {x2, 2/2), Fig. 83. Drop perpendiculars parallel to the axes as shown. Then slope of PQ yi PA X2 Xt Thus the slope of a line between two points is equal to the difference of the y-coordinates of the points divided by the difference of their x-coordinates, subtracted in the same order. In Fig. 83 a, the slope of PQ is positive, since it is the rise divided by the run, in moving along the line to the right. Conversely, a rise divided by a run means a positive slope. Y Positive Slope Fig. 83 6. Negative Slope VI, 95] GRAPHS IN ALGEBRA 115 This corresponds to the tangent of the acute angle XKP which this Hne makes with the x-axis on the right side of the hne. But, in Fig. 83 6, the slope of PQ is negative, since it is the fall divided by the run, in moving to the right along the line. Conversely, fall divided by run means a negative slone. This corresponds to the tangent of the obtuse angle XKP which this line makes with the x-axis on the right side of the line, 92. Example 1. Find the slope of the line through the two points P = (3, 4) and Q = (8, 6). Find also the angle which this line makes with the X-axis, as angle XKP, Fig. 83 a. Solution : Taking the two points in the order P, Q, we find SlopeofPQ = |^ = ^ = +.4, or, if the points are taken in the order Q, P, we have 8-3 ^ ' so that the answer is the same in either case. Hence tan XKP = A, whence Z XKP = 21 48'. Example 2. Find the slope of the Une through the two points P = (3, 7) and Q = (10, 2). Find also the angle which this line makes with the X-axis on the right side of the line. Solution : Slope of PQ = ^^ = - I = - -7143, or tan XKP = - .7143, and Z XKP = 180 -31 12' = 148 48'. EXERCISES 1. Construct a line through (0, 0) whose slope is 3/4. [Hint. A slope +3/4 means a rise of 3 and a run of 4. Therefore begin at (0, 0), rise 3 units and run 4 units to the right. Connect the final point with (0, 0). The resulting line will have the slope 3/4.] 116 MATHEMATICS [VI, 95 2. Construct a line through (0, 0) whose slope is 3/4. [Hint. A negative slope means a fall and a run. Therefore, fall 3 units from (0, 0) and run 4 units to the right.] 3. Construct a slope of 5/7 from the point (1, 2). Also one of 5/7 from the same point. 4. Compare the slopes 3/4 and 4/3 from the same point (2, 2). Do you recognize the angle between these slopes? 5. Compare, as in exercise 4, the slopes 2/3 and +3/2, from the same point (3, 4). For each of the following pairs of points : (a) Plot the points. (fe) Draw the straight line through them. (c) Find the slope of the straight line. {d) Find the angle the line makes with the x-axis. 6. (1, 2) and (3, 3). 9. (-3, 2) and (2, -4). 7. (1, 2) and (3, -1). 10. (-3, -4) and (-1, -1). 8. (3, 2) and (-3, -4). 11. (-3, -4) and (-5, 1). 96. To Find the Equation of a Straight Line. Heretofore we have had certain equations given and our problem was to find the graphs of these equations. Our problem now is to find the equation when the graph is given. The problem here is to connect in some algebraic way the relation between the x-distance and the ^/-distance of a point which will hold for all points on the line. For example, if a point is located anywhere on the x-axis the one thing we can say of this point is that its i/-coordinate is always zero. The concise algebraic statement of this fact is the equation y = 0. Hence this is the equation of the x-axis, since it is the one statement that is true for any and all points on the a;-axis, and for no other point. What is the equation of the 2/-axis ? VI, 97] GRAPHS IN ALGEBRA 117 f(x,y) Again, let us find the equation of the Hne which bisects the first and third quadrants and passes through the origin. Let (x, y), Fig. 84, be any point on this Hne. The problem is to find some relation be- tween the ^/-distance and the x-distance which will be true for all points on this line. Since POA = 45, it follows at once that PA or 2/ is equal to OA or x. This fact is stated algebraically by the equation y=x, or x-y = 0. ^^ O A y = X Fig. 84 97. Line through Two Points. Line through One Point in a Given Direction. A straight line is fixed definitely in position if it passes through two fixed points, or if it passes through one fixed point in a given direction. The equation of the line in either case may be found by using the slope of the line. In one case the slope between the two given points may be found, in the other case it is given. The method will be shown by some particular examples. Example 1 . Find the equation of the straight line which passes through the point (3, 2) and has the slope 2. Solution: Plot the point (3, 2), Fig. 85. Draw a line through this point with the slope 2. Let P{x, y) be any point on the line. The slope from A to P is BP Fig. 85 But BP=PD-AC=y It follows that -2, and AB = 0D-0C = x-3. = 2. Hence this is the equation of the line. Clearing of fractions, we find y = 2x-4:. 118 MATHEMATICS [VI, 97 Attention is called to the form of this last equation, in which the slope of the line appears as the coefficient of x. Example 2. Find the equation of the straight line through (1, 2) and (5, 4). Solution: Plot these points and draw the line through them. Now take any point R on the line with coordinates {x, y), Fig. 86. Then slope P to Q = slope Q to R, R{x,y) or or 4-2_i/-4 5-1 a:-5' l_y-4 2 x-b' from which y x+|. What is the slope of the straight line? Does the equation of the straight line show this? Note the form of this equation. EXERCISES 1. Find the equations of the straight lines passing through the following points and having the given slopes. (a) Through (3, 4) with the slope 2/3. {h) Through (3, 4) with the slope -2/3. (c) Through (-3, 4) with the slope 6/5. id) Through (2, -3) with the slope -3. 2. What is the equation of a line parallel to the x-axis and four units above it ? Four units below it ? What is the equation of a line parallel to the ?/-axis and twenty units to the right of it ? Parallel to the ^/-axis and three units to the left ? 3. What is the slope of a straight line bisecting the first quadrant? Does this same line (extended) bisect the third quadrant? Why? Find the equation of the straight line bisecting the second and fourth quadrants. What is its slope? 4. Find the equation of a line through each of the following pairs of points. Reduce each equation to the form showing its slope. Find the x-intercept and the ^/-intercept in each case. (a) (3, 4) and (-2, 2). {d) (2, 4) and (1, -1). (6) (3, 2) and (5, 6). (e) (-15, -3) and (8/3, -7/9). (c) (-6, 1) and (-1, -5). (/) (-1, 0) and (4, -2). VI, 98] GRAPHS IN ALGEBRA 119 5. Write the equation of each of the straight Unes described below. (a) A line whose x-intercept is 4 and whose y-intercept is 7, i.e. a line through (4, 0) and (0, 7). (b) A line whose x-intercept is 3 and whose slope is 4/5. (c) A line whose y-intercept is 8/3 and whose slope is 2. 6. Find the equation of a Une through the origin (0, 0) and the point (5, 3). Of course there are no intercepts in this case. How does this fact appear in the equation? 7. Find the equation of a straight line whose x-intercept is twice as great as its 2/-intercept and which also passes through the point (-2, 3). [Hint. What is the slope of the line ?] 8. Find the equation of the straight hne which makes equal inter- cepts on the axes and which passes through the point (2, 4). 9. Find the equation of the straight line which makes an angle of 60 with the x-axis and which passes through the point (2, 3). [Hint. What function of the angle is the slope ?] 10. A ditch rises 1 ft. in 50 ft. Draw a diagram of it and find its equation. 11. Find all the points for which x = 2, i.e. draw the Une whose equation is a; = 2. Do the same for x= 2, 0, 1, 3, 3. 12. Draw the hues y = 0, 1, 1, 2, 3, 4, respectively. 98. Parallel Lines. Draw two non-vertical parallel lines as shown in Fig. 87. Select any two points on each line. The slopes are, respectively, ^ and ^. The triangles having these sides are similar. Hence T' ai _ fl2 61 62' ^^- ^^- Parallel and the two slopes are equal. There- fore, if two non-vertical lines are parallel, their slopes are equal. State and prove the converse. 120 MATHEMATICS [VI, 99 99. Perpendicular Lines. Draw two non-vertical, per- pendicular lines, Fig. 88. The slope of the one line is mi = -' (Why negative ?) Pi The slope of the other line is 02 But, as the triangles are similar. a2 hi 1 FiQ. 88 Perpendicu- hi ai ai LAR Lines __ J_ mi i.e. Hence, if two non-vertical lines are perpendicular to each other, the slope of one is the negative reciprocal of the slope of the other. EXERCISES 1. A line passes through the points (2, 5) and (6, 1). What is its slope ? What is the slope of a line perpendicular to it ? Ans. -3/2; 2/3. 2. What is the slope of the line whose equation is Is the point (4, 5) on this line ? What is the slope of a line perpen- dicular to it ? 3. Discuss the slopes of the following equations. (a) 2/ = 2a;+3, (c) ^=-ia;+5, (6) 2/ = 2a:+4, {d) 2/=-|x+6. 4. Find the equation of a line through the point (1, 2) parallel t0 2/ = 2x+3. [Hint. From the given equation, what is the slope ? Having found the slope, you can find the equation as required.] VI, 100] GRAPHS IN ALGEBRA 121 MUes V 3 / 2 / 1 / Hour* i 1 2 3 X 1 5. Find the slope and the intercepts on the axes for each of the following lines, and draw the lines. {a) 3x-4y = 12. {c)x+y = 5. ^ + ^=1 (6) 5x-{-2y^l0. (d) Qx+Sy = 2. ^^^2 3 6. Give the slopes of lines perpendicular to the lines whose equations are given in Ex. 5. 100. Motion Problems Involving Time, Rate, Distance. Many problems involving the element of time may be solved graphically. If a man walks 6 miles in 2 hours at a steady rate, he walks at the rate of 3 mi. per hour. This may be represented on a diagram. Fig. 89, by plotting time (in hours) on the horizontal axis and distance (in miles) on the vertical axis according to the relation distance = 3 times the number of hours traveled, or, what is the same thing, 2/ = 3x, if y means distance traveled and x means hours traveled. This gives the straight line OP, which represents geomet- rically the uniform motion of the man. Example. An automobile starts from Columbus, going 30 miles per hour. Two hours later a motorcyclist starts in pursuit at the rate of 54 miles per hour. If they continue at these rates, when and how far from the starting point will the motorcycle overtake the automobile ? In the diagram. Fig. 90, let each large horizontal space represent 1 hr., and each large vertical space 30 mi. Then in 1 hr. the auto goes 30 mi., in 2 hr. 60 mi., etc. The points representing this, i.e. (1 hr. 30 mi.), (2 hr. 60 mi.), etc., lie on a straight line OP. The motorcyclist starts at zero miles after 2 hours. This is represented by the point M. He goes 54 mi. in 1 hr. Hence his motion is repre- FiG. 89. Time- Dis- tance Graph Y MUes ,/ 150 / 120 r 90 60 30 A / ---y{/ /\ Vm Hours / 1/2 3 4 5 X Fig. 90 122 MATHEMATICS [VI, 100 sented by the graph MP. The point where these two graphs intersect shows the place and time of meeting. From the diagram the time is 4| hr. and the distance is 135 mi. Solve the problem algebraically and thus check the accuracy of the graphical solution. Of course, the graphical solution is not always exact. In most cases it is only approximate, but in any case it serves as a check and a guide to the algebraic solution. However, all measurements are only approximations, accurate only to a certain degree. In the preceding prob- lem, the rate of 30 mi. per hour is only an approximation- Thus our graphical, approximate solutions are frequently as accurate and rehable as the data from which these solu- tions are made. For example, in transferring degrees Fahrenheit to degrees Centigrade by the graphical process the approximate answer is often desirable. In fact very few thermometers except those made especially for scientific purposes are really accurate. Few thermometers give the correct temperature to within half a degree. In no case should an answer be expressed more accurately than the circumstances warrant. EXERCISES [Solve the following problems graphically.] 1. A starts for a town 12 mi. distant, walking at the rate of 3 mi. per hour. An hour and a half later B starts for the same place, driving at the rate of 7^ mi. per hour. When does B overtake A? Where is A when B reaches the town ? 2. In a mile race, A runs 6 yd. per second and B 5 yd. per second. If B starts 250 yd. ahead of A, who will win the race and how far ahead is the winner at the finish ? [Hint. Plot one small vertical space equal to 20 yd. and one small horizontal space equal to 5 sec] 3. At 6 A.M. a freight train leaves a station going north at the rate of 30 mi. per hour. At 9 a.m. an express train is due at this same VI, 100] GRAPHS IN ALGEBRA 123 station coming from the opposite direction at 60 mi. per hour. When and where shall these trains be ordered to pass ? 4. How long will it take a steamboat, going at the rate of 8 mi. per hour, to overtake a motor boat 10 mi. ofif, going directly away at the rate of 5 mi. per hour? 5. A train traveling 30 mi. per hour is followed 3 hr. after its start by a train making 50 mi. per hour. When and where will the fast train be ordered to pass the first one ? 6. Kansas City and Chicago are 498 mi. apart. Where do two trains pass, one starting from Chicago at 40 mi. per hour at the same time another leaves Kansas City at 50 mi. j>er hour? 7. A man rows downstream at the rate of 6 mi. per hour and rows back at the rate of 3 mi. per hour. How far can he go and return in 9 hours? [Hint. Construct a line representing the downward journey at 6 mi. per hour. Then, beginning at the point which represents 9 hr., construct the points which represent his position at each preceding hour, i.e. coimt back 1 hr. and up 3 mi. The intersection of these lines is the result required.] 8. A man goes from Chicago to Milwaukee on a train making 42^ mi. per hour and returns at once on a steamer making 17 mi. per hour. Find the distance between the cities if the trip requires 7 hr. 9. A freight train leaves a station at 12 m. At 1 : 04 p.m. a fast train making 40 mi. per hour leaves the same station in the same direc- tion and is ordered to pass the freight at 2 : 40 p.m. What is the rate of the freight train, and how far is the point of passing from the station ? 10. A man rides a wheel into the country at the rate of 8 mi. per hour. After riding a while an accident occurs and he walks back at the rate of 3 mi. per hour, reaching home 11 hr. after the start. How far did he go? CHAPTER VII COMPUTATION BY LOGARITHMS 101. Introduction. Computation is ordinarily carried on by the processes of multiplication, division, raising to powers, and extraction of roots. These arithmetic processes are often laborious by the ordinary methods. They may be greatly abridged by the use of tables of logarithms of num- bers. Logarithms of the trigonometric ratios are useful in trigonometric problems. The following table illustrates the principles involved. 102. Table of Exponents. The following table of expo- nents gives the first eight successive powers of 10. The table enables us to reduce multiplication and division of these numbers to addition and subtraction, respectively, due to the first law of exponents : Table of Exponents of 10 100=1. 103=1000. 10^=1,000,000. 101 = 10. 10^ = 10,000. 107 = 10,000,000. 102= 100. 105= 100,000. 108= 100,000,000. 103. Extension of the Table of Exponents. It occurs at once that a more complete table of exponents would be quite useful in performing multiplications and divisions. The table of 102 may be extended to contain numbers less than 1 by simply dividing 1 by 10 repeatedly and sub- 124 VII, 105] LOGARITHMS 125 trading 1 from the exponent each time. This conforms precisely to the way this table of exponents was made. Thus we find 102=100, 101=10, 100=1, 10-1 =^, 10-2 = 3^, etc. The gaps in such a table may be filled in by taking arith- metic means of successive exponents and geometric means, 144, of the corresponding numbers. For example, the arithmetic mean between 2 and 3 is 2.5. The number corresponding to this in the second column of the table of exponents must then be V(100)(1000), i.e. Vl00,000= 100 Vl6 = 316.2, which means that (lO)^'^ =316.2 In practice, however, these tables are computed by a more rapid process, which is too advanced to be discussed here. 104. Definition of a Logarithm. The exponents of 10 in the first column of the table of 102 are called the log- arithms of the numbers in the second column, and 10 is called the base of the system. In general, the logarithm of a number is the exponent by which the base is affected to produce the given number. Thus, if n is the number, x the exponent, and b the base, then b^=n and x = logi, n are equivalent expressions, where logj, n is read, the logarithm of n to the base b. For example, 3^ = 9 is the equivalent of logs 9 = 2; 4^ = 64 is the equivalent of log4 64 = 3. If 2=^ = 32, then x = log2 32, i.e. x = 5. 105. Logarithms to the Base 10. Any positive number except 1 might be used as a base for a system of logarithms. The system with 10 as base is most generally used; it is called the common system of logarithms. 126 MATHEMATICS [VII, 105 In this book, when no base is given, 10 will be understood to be the base ; and log N will mean logio -^ In general, logarithms are not integers but are composed of an integer and a fraction. The fractional part is usually written as a decimal fraction. Thus, since 102=100, i.e. log 100 =2, and 103 = 1000, i.e. log 1000 = 3, the logarithm of any number between 100 and 1000 is 2 plus a certain decimal. What is wanted is a means of find- ing this decimal part in each instance. These decimal parts have been determined and tabulated. Such a tabulation is called a table of logarithms. (See 108.) 106. Characteristic. Mantissa. The integral part of a logarithm is called the characteristic, and the decimal part is known as the mantissa. Mantissas are found from a table, characteristics are de- termined by inspection. Considerations of the tables of 102-103 lead to the following rule. Rule for Characteristic. The characteristic of loga- rithms of numbers greater than unity is one less than the num- ber of digits to the left of the decimal point. For example, log 324 = 2.5105, log 32.4=1.5105, log 3.24 = 0.5105 For positive numbers less than unity, the characteristic of the logarithm is negative but numerically one greater than the number of ciphers immediately following the decimal point. For example, log .00324= -3 and +.5105, written 7.5105-10, since 7-10=-3, log .0324= -2 and +.5105, written 8.5105-10. log .324= -1 and +.5105, written 9.5105-10. VII, 107] LOGARITHMS 127 Observe that while some characteristics are positive and some are negative, the mantissas are always positive. Dividing a number by 10 moves the decimal point one place to the left, and subtracts unity from the logarithm of the number. This does not affect the mantissa, but the characteristic is decreased by unity. Similarly, multiplying a number by 10 moves the decimal point one place to the right, and adds unity to the logarithm of the number. The mantissa remains unchanged, but the characteristic is increased by unity. For example, if (from a table) log 376 = 2.5752, then log 37.6 = 1.5752, log 3.76 =0.5752, log .376 = 9.5752-10, log .0376 = 8.5752-10, etc. 107. Properties of Logarithms. Since a logarithm is, by definition, an exponent, the rules for exponents apply to logarithms. Thus, since 10'^ 10*= 10^, 10-^10*= io-, (10^^)^=10^ (10)i/p=10/P, we have the following rules. I. The logarithm of the product of two numbers is equal to the sum of the logarithms of the factors of this product, i.e. log MN=log M-\-log N. For example, log 6 = log 2+log 3 since 6 = 2 times 3. II. The logarithm of the quotient of two numbers is equal to the logarithm of the numerator minus the logarithm of the denominator, i.e. log = log M log N. N For example, log (4/7) = log 4 -log 7. 128 MATHEMATICS [VII, 107 III. The logarithm of the pth power of a number is equal to p times the logarithm of the number, i.e. log {NP)=plog N. For example, log (2^) =3 log 2. IV. The logarithm of the rth root of a number is found by dividing the logarithm of the number by r, i.e. log ViV= - log N. r For example, log (7V2) = i log 7. EXERCISES Given log 2 = .3010, log 3 = .4771, log 5 = .6990, find the value of each of the following expressions. 1. log 4. 5. log 10. 9, log Vs. 2. log 6. 6. log 12. 10- log 8/9^ 4. log 9. 8. log Vs. ' ^ V34 . 6 y 108. The Table of Logarithms. It is supposed here that a table of mantissas to four decimal places is used. This meets the requirements of most of the calculating to be done in agriculture, and much of the calculating to be done in engineering. From a table of logarithms we can find : (a) the logarithm of a given number, (b) the number corresponding to a given logarithm. The general principles governing the use of tables may be explained by the following examples. The tables are found in the Appendix of this book, Table VIII. Example 1. Find the logarithm of 374. The characteristic is 2. Now look in the table in the left-hand column, headed A^ at the top, for 37. Then follow horizontally across to the mantissa in the column headed 4 at the top, where 5729 is found. This is the required mantissa. Thus log 374 = 2.5729 VII, 108] LOGARITHMS 129 Example 2. Find log 37. For log 37, we look for 37 in the column headed N. Opposite this, in the column headed is the mantissa .5682 Thus log 37 = 1.5682 Example 3. Find log 32.73 The characteristic is 1. This number cannot be found in our table, but we can obtain the mantissa for its logarithm by a process called interpolation. This assumes that to a small change in the number there corresponds a proportional change in the mantissa. The number 3273 lies 3/10 of the way from 3270 to 3280. Hence the mantissa for log 32.73 is similarly placed between log 3270 and log 3280. To obtain log 32.73, we find from the table Mantissa of log 3270 = .5145 Mantissa of log 3280 = .5159 Difference = 14 Oiu* desired mantissa is therefore .5145+3/10 of 14 = .5149 Hence we have log 32.73 = 1.5149 This may be arranged schematically as follows : Number Mantissa I 3270 5145 \ \ 3273 il Difference = 10 \ A|3273 iA / 14 = difference, 3280 5159 from which x: 14 = 3: 10, i.e. x = 3/10 of 14 = 4. Example 4. Fmd N if log iV = 0.8485 This mantissa cannot be found in our table, but we find the mantissas next below and above it, viz. 8482 and 8488. These mantissas corre- spond to 7050 and 7060. Thus since 8485 is 3/6 of the way from 8482 to 8488 we conclude that its corresponding number is 3/6 of the way from 7050 to 7060, i.e. 7055. Since the characteristic is O the number N is 7.055 Schematically this may be arranged as follows : Number Mantissa 1 1 7050 8482 \ \ A 3] Difference = 10 1 \ 8485 1/- /6 = difference^ \l 7060 8488 l/ m which x: 10 = 3:6, i.e. x = = 5. This added to 7050 gives N. 130 MATHEMATICS [VII, 108 EXERCISES 1. Find from the table the logarithms of each of the following numbers. (a) 47.3 (c) (.063)(417) (e) (.0349)1/3 (6) .0473 (d) (.0349)1/2 (/) (.00349)3/4 [Hint. To find log (.0349)1/2, i.e. ^ log (.0349), take ^ (8.5428 - 10), i.e. ^ (18.5428-20). This is equal to 9.2714 To divide by 3, write it 28.5428-30.] 2. Find the logarithms of each of the following numbers. (a) 243.6 [Hint. Find first log 243.0 The tabular difference or mantissa difference is 18. The correction for the fourth figure 6 is then ^jj of 18 = 11. This added to log 243.0 is the desired logarithm. Notice also that 11 is found in the 6th column at the right in the same row as 24.] (6) 3.874 (c) .8358 {d) 68.72 (e) 4657. 3. Find the numbers corresponding to each of the following log- arithms. (a) 1.4335 Ans. 27.13 (d) 3.4775 (6) 1.4341 Ans. 27.17 (e) 9.5687-10 (c) 8.4341-10 Ans. .02717 (J) 8.7741 109. Computation by Logarithms. The principal use of logarithms is in shortening computation. The outline of the work in the following examples suggests similar arrange- ments for other exercises. It is desirable to make first the outline of the work to be done, then look up all logarithms and enter them in their proper places. After this has been done, the necessary additions, subtractions, etc. are per- formed. This procedure saves time, and reduces very much the chances of error. Example 1. Find 25.66 X 18.43 (1) log 25.66 = 1.4092 (2) log 18.43 = 1.2655 Sum of (1) and (2) =2.6747 = log 472.8 Hence, the product =472.8 II, 109] Example 2. Find (1) (2) (3) Add (1) and (2) (4) (5) (6) Add (4) and (5) LOGARITHMS 131 devalue of 124^X^3843 .7687 X 49.56 log 124.8 = 2.0962 log .03843= 8.5847-10 = 10.6809-10 log .7687 = 9.8858-10 log 49.56 = 1.6951 = 11.5809-10 = 1.5809 (7) Subtract (6) from (3) =9.1000-10 = log (.1259) Hence, the value =.1259 Example 3. Find the value of x=(^748)i/3(29.36)^ log x = i log .3748+2 log 29.36-.i log 431.5 (1) log .3748 = 9.5738 - 10 = 29.5738 -30 (in order to divide by 3) (2) i log .3748= 9.8579-10 (3) log 29.36= 1.4677 (4) 2 log 29.36= 2.9354 (5) Add (2) and (4) =12.7933-10 = 2.7933 (6) log 431.5= 2.6350 (7) flog 431.5= 1.3175 (8) Subtract (7) from (5) =11.4758-10 = 1.4758 = log 29.91 Hence, x = 29.91 MISCELLANEOUS EXERCISES Find the value of each of the following expressions by means of logarithms. J . . (2348) (.4537)3 (c) ^2:83 ^ ^ (83.46) /^ri- "^ (d) ^/ "3547- (6) V13.84 A/TT^i^ (1.045)' 2, (^) V3979^-6854. ^^^ f 4^^^ ^Q^LS ^73.96; 3. The amount of a principal at compound interest for a certain time is given by the formula ^ ^ p, ^ _^^w in which A =the amount, P = the principal, r = the rate, n = the number of years or periods. 132 MATHEMATICS [VII, 109 Find by use of this formula, the amount of $598.40 for 15 years at 5% interest compounded annually. Compounded semiannually. Compounded quarterly. [Hint. For compounding semiannually, use n = twice the number of years and r = half the rate. For compounding quarterly, use n = four times the number of years and r = one fourth the rate.] 4. A field whose successive vertices are A, B, C, D, E is measured as follows: AB = 21 rods, BC = S9 rods, CD = 23 rods, DE = 30 rods, EA = 18 rods, BE = 35 rods, BD = 38 rods. Draw the figure to scale, divide it into triangles, and compute the area of each triangle in acres from the formula for the area in acres of a triangle, A_ Vs{sa){s b){sc) 160 where a, h, c are, respectively, the sides of the triangle, and s = \{a -\-h-\-c) . 5. What is the capacity in barrels (40 gallons) of a round cistern 12 feet deep and 5 feet in diameter? 6. The time of one vibration (one complete swing) of a pendulum is given by the formula T=WT/i, where I = length of the pendulum, g = attraction of gravitation, = 32.16 ft. =980.2 centimeters. Find the time of vibration of a pendulum 21.56" long. Of one 62.8 centimeters long. Find the length of a seconds pendulum, i.e. one which makes one vibration in one second. 7. How long is a pendulum that vibrates 3 times in 1 second? How long is it if it vibrates 4 times in 3 seconds? What is the time of vibration of a pendulum 40 ft. long? 8. A clock having a seconds pendulum gains 2 minutes per day. How much too short is the pendulum ? 9. The pendulum of a certain clock is of brass and is 36" long at 70 F. A brass rod will expand or contract approximately .00001052 of its length for each Fahrenheit degree change in temperature. What is the length of this pendulum at 40 F. ? How much will the clock gain in 1 week due to this change of temperature ? VII, 110] LOGARITHMS 133 10. A certain concrete sidewalk 578 feet long is laid in a region where the temperature varies from 20 F. to 105 F, What allow- ance for this change of temperature must be made in the length of the walk if concrete changes .00000668 of its length for each Fahrenheit degree change of temperature, if the concrete is laid when the tempera- ture is 50 F.? 11. Due to change in temperature a substance made of steel will change .0000065 of its length for each change of 1 F. What space should be left between the ends of steel rails 60 feet long when laid in a region where the temperature varies from 4 F. to 104 F., if the rails are laid when the temperature is 20 F. ? 110. The Slide Rule. Introduction. The slide rule is an apparatus for determining mechanically products, quotients, powers, roots. It consists of two principal pieces so made that one may slide upon the other. It is simply a table of logarithms arranged so that they may be added or subtracted conveniently, and even more rapidly than with the ordinary printed table. Of course, the logarithms are not printed on the slide rule, but each number simply stands in the position indicated by its logarithm. This gives a n \ \ \ \ I INI'' 1 2 S 4 5 6 7 8 9 10 Fig. 91. A Logarithmic Scale scale which is not uniform and which is called a logarithmic scale. Such a scale is shown in Fig. 91. This scale was graduated as follows : The distance PQ was divided into 1000 equal parts. Since log 2 = .301, 2 was placed at the 301st division, and as log 3 = .477, 3 was placed at the 477th division, etc. for the succeeding integers. To multiply numbers by the use of logarithms, we add the logarithms of the several factors and find the number cor- responding to the sum of the logarithms. Hence, to mul- tiply two numbers by use of the slide rule, it is only necessary 134 MATHEMATICS [VII, 110 to place the logarithm of one of the numbers on one piece of the rule, end to end with the logarithm of the other number on the other piece of the rule, and then read off as the product the graduation mark at the end of these combined lengths. For example, on the scale of Fig, 91 above, log 2, which is represented by the distance from 1 to 2, added to log 3, the distance from 1 to 3, gives log 6, the distance from 1 to 6. Thus, 2X3 = 6 on the scale. Show in the same way that 2X5 = 10. Also that (2.5) X (3.5) = ? Determine it first on the scale and compare with the result obtained by actual multiplication. Similarly (2.5)X(3.2)= ? 111. Description and History of the Slide Rule. One part of the rule is called the slide. It slides along a groove in the center of the other part called the stock. Some rules carry a runner for convenience in continued multiplication and division and squares and square roots. This is merely a sliding frame with a wire or line on it perpendicular to the length of the rule. The device of having one piece slide on the other was invented by William Ought red between 1620 and 1630, soon after the invention of logarithms by Napier in 1614. Fig. 92. The Mannheim Arrangement Practically all slide rules in present use are Mannheim rules. They are so named, not from any manufacturer, but from Lieutenant Mannheim of the French Army, who devised the present arrangement of the scales about 1850. Some higher priced rules have modifications of the Mannheim VII, 113] LOGARITHMS 135 arrangement, but the fundamental principles involved are the same as in the regular Mannheim rule. Fig. 92 shows the regular Mannheim arrangement.* 112. The Scales. Two scales are engraved along the upper edge of the groove. One, labeled A, is on the stock, the other, labeled B, is on the slide. These scales are identical. The slide simply makes it possible to add graphi- cally the segments on the scales. Along the lower edge of the groove are also two scales. One, C, is on the slide, the other, D, is on the stock. These two scales are likewise identical. Scales A and B range from 1 to 100 while C and D range only from 1 to 10. The range on one set being smaller, the number of divisions can be increased. Thus C and D give greater accuracy in reading them since there are more divisions. Notice also that every number on the A and B scales is the square of every number below it on the C and D scales, due to the fact that the unit on one is twice as great as on the other. This gives us at once a table of squares and square roots. The runner enables one to find corresponding numbers on the upper and lower scales, and also to hold a given position during a repeated operation. 113. Operations with the Slide Rule. To use a slide rule successfully, one must acquire the virtues of speed and accuracy. This requires practice, since the accuracy of a result often requires the estimation of a number falling be- tween the lines of division. * Fig, 92 is reproduced on a larger scale on the first fly leaf at the back of the book. By cutting out this leaf and carefully cutting up the figure, a slide rule can be made by the student. It will not be very accurate, but it will suffice to illustrate the principles. Slide rules, varying from the inexpensive pasteboard kind to the higher priced wood and celluloid ones used by engineers, may be had at practically all stores which carry draftsmen's supplies. 136 MATHEMATICS [VII, 113 A skillful operator, with a 10-inch slide rule, can always secure results accurate to three significant figures. This is accurate enough for most of the purposes of applied science. The beginner should use small numbers until he becomes familiar with the operations. 114. Multiplication. Multiply 2 by 3. Move the shde so as to put 1 of the B scale on 2 of the A scale ; then above 3 on the B scale read the product 6 on the A scale. This is simply adding logarithms. Notice that the 1 at either end of the B scale may he used. Try both ways. Use that 1 which brings the second factor under scale A . For example, multiply 65 by 3 using CD scales. If we set the left end 1 of C on 65 D, then 3 C is off the scale D. In this case put the right end 1 of C on 65 D which brings 3 C under 195. The decimal point is placed by inspection, i.e. by making an approximate mental computation to determine the num- ber of integers or the number of decimal places. For example, multiply 18.5 by 2.8. Set 1 C on 185 D and under 28 C read 518 D. But the product is roughly 18X3 = 54 ; therefore we know that there are only two places before the decimal point in the product of 18.5 and 2.8 Thus 518 read on D means 51.8 115. Division. Divide 6 by 2. Set 2 C on 6 D, under 1 C read the quotient 3 on D. This means merely sub- tracting logarithms. Also, divide 7.2 by 2.5. Set 25 C on 72 D, under 1 C read the quotient 2.9 on D. Therefore, To divide one number by another, set the divisor on scale C on the dividend on scale D, and under 1 C read the quotient on D. VII, 116] LOGARITHMS 137 116. Combined Multiplications and Divisions. Find the ^^^^^ ^^ (25.2) (9.6) 8.4 Set 84 C on 252 D, and under 9.6 C read the quotient 28.8 on D. Notice that 84 (7 on 252 D gives the quotient under 1 C but as this is to be multipHed by 96, 1 C is already on this quotient and hence we have only to read the product on scale D under 9.6 C. An important application of this is finding the fourth ter)n of a proportion. For example, from the proportion 8.4 : 25.2 = 9.6 :x, we find ^_ (25.2) (9.6) "^ SA The rule then to find the fourth term is obvious, viz. : To find the fourth term of a proportion, set the first term on the second, and under the third term read the fourth. In continued multiplications and divisions, the runner with its perpendicular indicator is quite convenient. Example 1. Find the value of 2 X 3 X 4. Set 1 C on 2 D, set runner on 3 C (marking this product). Now set 1 C at right end on runner and under 4 C read 24 on D. Thus 2X3X4 = 24. 56 Example 2. Find the value of - - 7X4 Set 7 C on 56 D, set runner on 1 C (marking the quotient), set 4 C on the runner and under 1 C read the result 2 on D. Example 3. Find the value of ?|^- 17X12 Set 17 C on 24 D, set runner on 51 C, set 12 C on rimner, under 1 C read result 6 on D. 138 MATHEMATICS [VII, 117 117. Squares and Square Roots. Scales C and D were constructed on a unit twice as large as that for scales A and B. Accordingly, every number on the A and B scales is the square of the number vertically below it on the C and D scales. Thus above 2 C we find 4 A, above 3 is 9, above 25 is 625, etc. Thus, To square any number, find the number on scale C and read its square vertically above it on scale B. To extract the square root of a number, find the number on scale B and read its square root vertically below it on scale C. Example 1. Find the value of 16 V3. Set 1 C on 3 ^, under 16 C read the result 27.7 on D. Example 2. Find the value of I6/V5. i^ = 16V5. V5 5 Set 5 C on 5 A, under 16 C read result 7.16 on D. Example 3. Find the area of a circle whose radius is 3 ft. Set 1 C on 3 D, above tt on B read the area 28.3 sq. ft., on A. EXERCISES Find the value of each of the following expressions. 1. 2. 3. (4.67) (8.32). (.341)(3.86). 31.5 . 16.2 , (97) (15) 23 5. 7. 8. 849 g (18.6) (3.4) 31.6 3.1416 27V19 37 (47.2)(18.1)(.47). on 9. Compute W, white oak beams, the safe load in tons, from the formula when uniformly distributed W = 2hd^ "3 I ' ' if b (the breadth in inches) is 3, d (the depth in inches) is 8, and I (the distance in inches between supports) is 192. 10. Find the area of a circle whose radius is 75 miUimeters. CHAPTER VIII THE PROGRESSIONS 118. Introduction. Progressions are among the oldest topics of all mathematics. Egyptian scholars had studied them thousands of years before the Christian era. By 2000 B.C. they had attained to a considerable knowledge of them, as evidenced in the papyrus of Ahmes. Progressions are important, not only in the solution of numerical problems, but also in theoretical developments. For example, out of the association of arithmetic and geo- metric progressions grew the subject of logarithms. 119. Arithmetic Progression. An arithmetic progression (written A. P.) is a succession of terms so related that any one, after the first, is obtained from the preceding one by adding a constant number. This constant is known as the common difference. Thus 2, 5, 8, 11, , with a common difference 3, 5, 1, 3, 7, , with a common difference 4, a, a-\-x, aH-2 x, , with a common difference x, are arithmetic progressions. Is 2, 3, 5, 4, 7, 11, , an arithmetic progression ? 120. Terms of an Arithmetic Progression. In such a succession of terms as just described, we are interested in the first term, known as a, the common difference, known as dj the last or nth term, known as ly the number of terms, known as n, and the sum of the terms, known as s. 139 140 MATHEMATICS [VITI, 120 Therefore, we may write the arithmetic progression in general as follows. 1st term 2d term 3d term 17th term nth or last term a, a-\-d, a-\-2d, ', a-\-lQ d, , a-{-(nl)d. It thus appears that any term is equal to the first term, plus the common difference multiplied by the number of the term less one, i.e. l=a-\-{n-l)d. For example, the 17th term = a+(17-l)d = a+16 d. EXERCISES 1. Given the A. P. 3, 7, 11, . What is the common difference? What is the 10th term? Begin with the 10th or last term and write the series backwards. 2. Do as in Ex. 1 for the A. P. 1, 4, 7, to 10 terms. 3. Do as in Ex, 1 for the A. P. a, a-\-x, a-\-2 x, to 10 terms. 4. If I is the last term and d the common difference, write the A. P. backwards. [Hint. If I is the last term, then ld is the next to the last, etc.] 5. Write the A. P. which exhibits the strokes of a clock striking the hours. 121. Siun of the Terms of an Arithmetic Progression. The simi of an A. P., i.e. the sum of all the terms taken, is found as follows. s=^a+{a-\-d)-\-{a-\-2 d)-\- ' to ntermsj or, written backwards, s = l^{l-d) + {l-2d)-\- to n terms. Addition of corresponding terms gives 2 s= (a+Z) + (a+0+ " taken n times = n(a+0. Thus, s=^ia+l). VIII, 121] THE PROGRESSIONS 141 Example 1. Find the total number of strokes made by a clock in striking 12 hours. Here a = l, 1 = 12, n = 12; therefore s = V(l+12) =78. Example 2. What is the total price paid for a lot bought on the installment plan of 25 f^ the first week, 50^ the second week, 75 jf the third week, etc. ; to continue for one year of 52 weeks? Here a = 25, d = 25, n = 52; hence the last term is Z = 25 + (51)25 = 1300 ff, 8 = ^ (a +0 = 26(25 + 1300) = 34,450ff = $344.50. EXERCISES Find the last term and the sum of all of the terms in each of the following A. P.'s. 1. 5, 9, 13, to 20 terms. Ans. Z = 81, s = 860. 2. 1, 3, 5, 7, to 25 terms. 3. 2i, 3|, 5, to 15 terms. 4. I, -J, -h ' to 16 terms. 5. Find the sum of the first 50 even integers beginning with 0. 6. A row of twenty trees is set at intervals of fifteen feet, beginning at a point forty feet from a house. How far is the last one from the house ? If a man carries a bucket of water from a well by the house to each tree, how far does he walk ? 7. Write an A. P. of nine terms for which the first term is 4 and the last term 20. [Hint. Use l = a + {n l)d where a = 4, Z = 20, etc.] 8. How far apart must 9 plants be set to be equally spaced in a row 16 ft. long? Ans. 2 ft. 9. If there are sixteen plants in a row 40 feet long, how far apart are they? 10. If the sum of an A.. P. is 18f, the first term |, the number of terms 10, find I and d. 11. A man deposits $75 in a savings bank. How much must this and later deposits be increased each year so that in 10 years his total deposit shall amount to $1875, not counting interest? 142 MATHEMATICS [VIII, 121 12. A farmer has $100 set aside for drilling a well. How deep could he have it drilled at 25^ for the first foot and for each additional foot 1 ^ more than the preceding one ? 13. A freely falling body descends 16 ft. the first second and ap- proximately 32 ft. more each second thereafter than in the preceding second. How deep is a well to the bottom of which a stone falls in 3 sec. ? In 6 sec. ? How far does it fall in the sixth second ? 14. A bomb is released from an airplane 1 mile high. How long does it take the bomb to reach the ground ? How far did it go the last second ? If the airplane is going 75 miles per hour, how far horizontally in advance of its starting point does the bomb hit the ground ? 15. According to the law of falling bodies as given in Ex. 13, what is the relation between distance, s, and time, t, i.e. between the sum of the A. P. and the number of terms? 122. Geometric Progression. A geometric progression (written G. P.) is a succession of terms so related that any term, after the first, is obtained by multiplying the preceding term by a constant number called the ratio. Thus 2, 8, 32, , with ratio 4, and 2, 6, 18, 54, , with ratio 3, are geometric progressions. What is the ratio for each of the following geometric progressions : 9,3,1,-1,...? 4, -12, 36, -108,-? 123. Terms of a Geometric Progression. In a geometric progression, we are interested in the first term, known as a, the ratio, known as r, the last or nth term, known as I, the number of terms, known as n, the sum of the terms, known as s. VIII, 124] THE PROGRESSIONS 143 Therefore, we may write, generally, the series of terms of a G. P. as follows. 1st term 2d term 3d term 17th term nth or last term a, ar, ar^, ar^^, ar^^. The exponent of r in any term is one less than the number of that term. The nth or last term is For example, the 17th term is ar^^. EXERCISES Which of the following sequences of numbers represent geometric progressions ? Find the ratio and the last term in each such progression. ! ij i 1> to 9 terms. 2. 2, 3, 5, 4, 7, to 6 terms. 3. 2, i ,V to 6 terms. 4. 7, .7, .77, to 5 terms. 5. 72, .72, .0072, to 5 terms. 6. (a+6), (a+6)2, (a+ft)', to 6 terms. 7. (a+fe), (a2-62), (^a^-ab^-a^+b^), to 4 terms. 8. V2, 2, Vs, to 7 terms. 9. If a and ar'^ are the end terms, respectively, of a G. P. of three terms, what is the middle term? What is the middle term if 4 and 9 are the end terms of a G. P. of three terms? 10. If 4 is the first term of a G. P. and ^^ is the fourth term, find the second and third terms. [Hint. Let o = 4, and ar^ = y ; find v.] 124. Sum of the Terms of a Geometric Progression. By definition the sum of n terms of a G. P. is s = a-\-ar-\-ar^-\- +ar"-i. Therefore rs = ar-]-ar'^-{-ar^-\- -\-ar^. Subtracting these gives s rs = a ar"^ = a rar^~^ = arlo rrtu a ar" a rl Thus s = = 1 r 1 r 144 MATHEMATICS [VIII, 124 Example. In the G. P. 6, 2, f, find the 6th term and the sum of 6 terms. Here a = 6, r = |, n = 6, l = aj^-^ = 6{^)^ = 6^lj = j\. Hence s is given either by the formula _ a-ar^ _ 6-6a)6 _728 1-r 1-^ 81 ' or by the formula o_ a-rl ^ Q-^' A -728. 1-r 1-i 81* EXERCISES 1. If a = 3, r = ^, n = 8, find i and s. 2. If a = 6, n = 6, Z= ff, find r and s. 3. How many terms of the G. P. 1^+2^+4^+ .- are necessary so that the sum shall be at least $40? Ans. 12 terms. 4. A blacksmith agrees to shoe a horse as follows : ^|if for the first nail driven, ^^ for the next, 1^ for the next, etc., for the 8 nails of one shoe, and a like amount for each of the other three shoes. What does he receive for the work? Ans. $2.55 5. If the first term of a G. P. of 5 terms is 2 and the last term 1250, find the ratio and the sum of 5 terms. 6. Find the sum of 7 terms of 1+2-1+2-2+. 7. If a = 2, n = 5, Z = 1250, find rand s. 8. If a = 17, Z = 459, n = 4, find r and s and write the series. 9. The common housefly will lay, incubate, and mature a litter of eggs about every 3^ weeks. What will be the number of descendants of one female fly in 14 weeks if there are 150 to a Utter, evenly distributed as to sex? 10. Find the amount of $1 at 4% compound interest for 1 year. For 2 years. For 3 years. For 4 years. [Hint. The rate is 4% = .04 and the amount is $1.04 for one year. This is the principal for the second year. The amount, therefore, for the second year is (1.04) times (1.04) = (1.04)2.] Do these results form a geometric progression? VIII, 124] THE PROGRESSIONS 145 11. Find the amount of p dollars at rate r compound interest for 1 year. For 2 years. For 3 years. For n years. Ans. The amount for n years = p(l +r)'*. [Hint. The amount of $1 for 1 year at rate r is $1(1+0 and the amount of $p is therefore p times this, i.e. $p(l+r). The amount for the second year is therefore (1+r) times the new principal $p(l+r), i.e. it is p(l+r)2.] 12. Find the amount of $50 for 8 years at 5% compound interest. [Hint. The computation is shortened by use of logarithms.] Ans. $50(1.05)'. 13. A man deposits $50 each year in a building and loan association at 5% compound interest. What is the total amount of interest and deposits at the end of 6 yr. ? [Hint. The first deposit is compounded at 5% for 6 yr. and thus amounts to 50(1.05)**. The second deposit amounts to 50(1.05)^ being on interest for 5 yr. The next deposit amounts to 50(1.05)*, etc. The total amount for the six years is thus 50(1.05) +50(1.05)2-1- ... +50(1.05)'. But this is a G. P. of six terms. Show that its sum is 50(1.05) [il^5|^]. Compute this amount by use of logarithms.] 14. Find the total amount of a yearly deposit of p dollars com- pounded at rate r for n years. rn _i_.^n_n Ans. p(l+r)[ii^^^ ij. 15. On a certain life insurance endowment policy for $1000, the insured agrees to pay $47.68 each year for 20 years, and he is to receive $1000 at the end of the period. What is the total amount of these deposits compounded annually at 3^%? What does the company obtain at the end of the period for carrying the risk, if it earns just 3|% on money in its possession? Ans. Amt. =$1395.44 16. A man buys a machine for $150. What amount of money should be deposited each year to draw 4% compound interest so that the amount of these deposits would purchase a similar new machine at the end of 12 years? Ans. $9.60 [Hint. Use the formula of Ex. 14, solving for p.] 146 MATHEMATICS [VIII, 124 17. What equal annual deposit at 4% compound interest will amount to $300 in 15 years? 18. What amount of money could profitably be expended for a harvesting machine which will last 15 years, if it saves annual labor bills of $40, money being worth 5%. [Hint. The total saving, with interest from the end of each year, is given by the sum of the geometric progression , s = 40(1.05)14+40(1.05)13+..^ +40 = 40[(1.05)l5-l]-^(.05). But this is the saving at the end oi 15 years. The principal p invested now would produce in 15 years the amount p(1.05)i^. It follows that p{1.05y^ = s. Solve for p, using logarithms.] Ans. p = $415.20 125. Infinite Geometric Progressions. Thus far we have considered only those progressions which have a limited number of terms. There are many problems and phenomena whose discussion calls for a series of an unlimited number of terms. For example, suppose an elastic ball falls 4 feet and rebounds half of that distance, then falls 2 feet and re- bounds half of that distance, etc., rebounding each time half of the distance fallen. The entire distance fallen would be the sum of the series 4+2+l+i+i+ to infinity. This is a G. P. in which the ratio is less than unity and the number of terms is unlimited. If, in a G. P., ri_J (l+r)--l 1 Hence the annual payment, p, for n years, which $1 will purchase is ,. . (\^rY ^"(l+r)"-l' and the annual payment, P, which %A will purchase is ^^ (l+r) 1 Example 1. What annual payment for 10 years will $500 purchase, if money is worth 5% ? SoTTTTioN- p- 500 (.05) (1.05)^" _ (500) (.05) (1.62890) SOLUTION, r- (io5)io_i (1.62890) -1 ' from a compound interest table. Therefore P = $64.75 EXERCISES 1. What is the present value of $600 due in 8 years if money is worth 4% compounded annually? Ans. $438.41 2. Solve the previous example if the interest is computed at 4% compounded semiannually. Ans. $437.07 3. Find the present value of an annuity of $53 receivable at the end of each year for 20 years, at 5%, (o) by logarithms and (6), 131, (6) by a compound interest table. Ans. $660.50 4. A man buys a farm and agrees to pay for it $1000 in cash and $500 at the end of each year for 10 years. If money is worth 6%, what would be the equivalent cash price of the farm ? Ans. $4680.44 5. What annual payments continued for 10 years are equal to $1000 cash, if money is worth 5%? Ans. $129.50 [Hint. Let p = the annual payment. Then from (6), since -=- ooo=p[a^g-.^]. from which p may be computed.] 156 MATHEMATICS [IX, 132 6. What annual payments for 20 years can be purchased for $500, money being valued at 5% ? Ans. $40.12 7. What semiannual payments continued for 20 years can be purchasedfor $500, at 5%? Ans. $19.92 8. A man buys a farm for $10,000 and wishes to pay for it by a series of equal annual payments of such amount as to pay both principal and interest in 30 years at 5%. Find the amount of the annual pay- ment. Ans. $650.50 [Hint. Find what annual payment $10,000 will purchase.] 9. A man secures $2400 for the purchase of farm equipment by a loan from the government. He agrees to pay $20 per month, including 6% interest on all sums remaining unpaid. How long will it take to pay off the loan and what will be the amount of the last payment ? Ans. 184 months ; last payment, $14.26 10. A farmer wishes to provide an income for old age. He esti- mates that at age 30 years he will have 35 years of productive activity ahead of him, and that he can save $250 per year during that time. This accumulation at 5% compound interest at age 65, together per- haps with $2500 cash available from the sale of his farm and tools, will purchase what annual payments for 15 years, if money is worth 5% ? Ans. $2416.27 11. Solve the previous exercise, estimating that the farmer can siave $100 per year for 30 years at 5% and that he will use this accumulation at age 60 years, plus $2500, to purchase an income for 20 years at 5%. What annual payments can he thus purchase? Ans. $733.73 133. Problem D. Sinking Fund. To find the annual payment that will amount to $A in n years. Such annual payments made to accumulate a given amount at a given time constitute a sinking fund. Let p denote the annual payment, and A the amount at the end of n years. Then, by (3), 130, the amount is (8) pi^^ = A, r whence IX, 134] COMPOUND INTEREST 157 EXERCISES 1. A farmer wishes to provide $1500 for the replacement of ma- chinery ten years hence. How much must he set aside each year at 6% compound interest to meet this contingency? Ans. $119.26 2. How much would the farmer of Ex. 1 need to set aside every six months to provide for the same amount, if the interest is compounded semiannually? Ans. $58.72 3. A township issues bonds for $100,000, payable at the end of 25 years, for road improvements. What sum must be set aside each six months to meet the payment of these bonds when due, if the sums set aside bear 4% interest? Ans. $1182.30 4. A man owes a note of $640 due at the end of 5 years with simple interest at 6%. He proposes to meet the payment of both principal and interest at the end of the time by depositing a certain amount every six months in a savings association, at 5% compound interest. Compute the amount to be thus deposited. Ans. $74.26 134. Perpetuities. Capitalization. When the payments of an annuity are continued indefinitely, it is called a per- petuity. Perpetuities are important in practical business affairs. An annual income from the rent of property, from an endowment fund, or from a fixed salary, has a definite cash equivalent. Likewise, an annual outgo continued indefinitely has a cash equivalent. The process of finding such cash equivalents is called capitalization. Problem E. To find the present value of a perpetuity whose payments are made annually. Let p = the annual payment due 1 yr. hence, and thereafter annually indefinitely. The present value of the first payment due in 1 year is p/a+r); The present value of the second payment due in 2 years isp/(l+r)2; The present value of the third payment due in 3 years isp/(l+r)3; 158 MATHEMATICS [IX, 134 and so forth indefinitely. The present value of the per- petuity is the sum of these present values of the separate payments. If s^ denote this sum, we have This is an infinite geometric series whose first term is p/(l+r), and whose ratio is 1/(1 +r). Hence, by 125 its sum is J. (11) 1 + r _p 1 r l+r EXERCISES 1. If money is worth 5%, what is the cash equivalent of a yearly income of $1500? Ans. $30,000. 2. What is the value of a farm which returns an income of $750 per year, continued indefinitely, money being worth 6%? Ans. $12,500. 3. What is the capitalized value of a fixed salary of $3000 a year, with money at 6% ? Ans. $50,000. 4. A railroad company pays a flagman $1200 per year. What sum could the company afford to pay to install automatic machinery to do away with the services of the flagman, with money at 4%? Ans. $30,000. 5. Is farm land which rents for $9.00 per acre selling too high at $250 per acre? What would be a reasonable rate of interest to allow in this case ? 135. Depreciation. Depreciation is the decrease in value of any perishable article which results from use and from advancing age. It may result also from progress in the arts and in methods of manufacture which renders an article obsolete, or out of date. Articles may become inadequate, and may have to be replaced, in the course of time, by others better adapted to the requirements. Thus, broadly speak- IX, 137] COMPOUND INTEREST 159 ing, depreciation includes wear and tear, obsolescence, and inadequacy. Annual depreciation is the annual theoretical decrease in value, expressed in money. 136. Methods of Determining Depreciation Charges. Various methods have been advocated to determine how much shall be charged to depreciation for a given article, or plant, or machine. It is convenient to measure deprecia- tion in terms of time, notwithstanding the fact that it does not necessarily depend wholly upon the passage of time. Periods of time are merely a convenient way of comparing the rapidity of depreciation with the rapidity of other occurrences. In 137-139 we shall consider three methods of computation. 137. The Straight Line Method. This method proceeds upon the theory that the annual depreciation of an article is uniform throughout its probable period of life. For example, if a harvester costs $120, and has a probable life of 10 years with no salvage value at the end of that time, the annual depreciation is merely one tenth of $120. If the salvage value is $15, then the annual charge for depreciation would be ($120- $15)/10= $10.50. Let X = the annual charge for depreciation on an article, y=its value or cost, y = its salvage value after a period of n years. Then, by the straight line method, we have (12) x=i:^. n EXERCISES 1. Show that all considerations of interest are disregarded in the straight line formula for estimating the depreciation charge. 160 MATHEMATICS [IX, 137 2. Draw a graph to show the value of the harvester of 137 at various times after its purchase. Why is (12) called the straight Une formula? 3. A house is worth $8000 and its estimated life is 40 years. Allow- ing $400 salvage value, what should be allowed each year for deprecia- tion, using the formula (12) ? Draw a graph. 4. What is the life of a wagon which depreciates about 5% of its original value each year ? Draw a graph. Ans. 20 years. 5. A Minnesota farm bulletin shows the following average depre- ciation of machinery, in per cents of the original cost. Machine % Depreciation Machine % Depreciation Threshing Outfit . 12.0 Gang Plows . . 7.40 Hay Loaders . . 11.78 Gasoline Engines 7.35 Manure Spreaders 11.67 Corn Cultivators 7.23 Corn Binders . . 10.03 Heavy Horses 6.17 Harrows . . . 8.72 Wagons .... 4.89 Grain Binders . . 7.91 Fanning Mill . . 4.58 Determine the life of each machine by the straight line formula, estimating the salvage value at 5% of the cost in each case. Estimate the value at the end of 6 years of a corn harvester costing originally $125. 6. What is the annual charge for depreciation and what is the probable life of a hay loader that costs $110, and that may be sold for $30 after 7 years' usage ? 7. If the cost of a machine is $100, its probable life 20 years, and its salvage value zero, what is the annual charge for depreciation, using formula (12) ? Construct a graph for this information, using time in years on the horizontal axis, and depreciation in dollars on the vertical axis. Construct a second graph, using time as before, but actual value of the machine on the vertical axis. 138. The Reducing Balance Method. This plan deducts for depreciation, at the end of each year, a certain constant percentage of the value of the article at the beginning of that year. This value itself decreases each year. Thus if an article costs $100 and if the depreciation each year is 10% IX, 139] COMPOUND INTEREST 161 of the value at the beginning of that year, its value at the end of the fourth year is found as follows : 10%of $100=$10; hence first balance = $100 -$10 =$90; 10% of $90 =$9; hence second balance = $90 $9 = $81 ; 10% of $81 = $8. 10; hence third balance = $81 -$8.10 =$72.90; 10% of $72.90 = $7.29; hence fourth balance = $72.90 -$7.29 = $64.61 This process may be continued till the balance reaches a given salvage value. It will be noticed that by this method the charge for depreciation is high in the first years of life of an article and comparatively low in the later years of its life. 139. The Reducing Balance Formula. Let x denote the constant fractional rate of depreciation which is to be de- ducted each year from the balance ; F, the original invest- ment ; y , the salvage value ; and n, the number of years before the salvage value is reached. Then, deducting x of V from V\ we have the value at the end of 1 yr. = F Fa; = F(l a:) ; thevalueattheendof2yr.= F(l-a:)-F(l-x)a: = F(l-x)2; the value at the end of 3 yr. = F(l xY ; the value at the end of n yr. = F(l x)". But this nth balance is the salvage value, F'. Thus (13) V{l-xY=r, i.e. (14) x=l-S/f. and (15) ,^ lo,V'-logV , log{l-x) 162 MATHEMATICS [IX, 139 Example 1. What is the value at the end of the fifth year of an automobile costing $500, if depreciation each year is 15% of the value at the beginning of the year? By (13), V' = Va-x)- = $500(1 -.15)5 = $500(.4436) =$221.80 Example 2. In what time will an automobile costing $500 de- preciate to $100, allowing 15% reduction each year on the value at the beginning of the year? By (15), = logl00-log500 log (1-.15) ^ 2.0000-2.6990 _ -(.6990) 9.9294-10 -(.0706) = 9.9 years = 10 years, approximately. 140. The Sinking Fund Method. By the sinking fund method, a fixed sum of money is set aside each year, and is allo'wed to accumulate at compound interest. It derives its name from the fact that it employs the principles ordi- narily used in estabHshing sinking funds to Hquidate in- debtedness and for purposes of replacements. Let us assume that a machine costs $1000, that its probable life is 25 years, and that its salvage value is zero. The problem then is to find what sum of money must be set aside each year in order that at the end of the 25 years the fund will have accumulated to $1000 at 5% compound interest. The annual fixed sum thus set aside in any year plus the interest accumulated in that year, represents the depreciation charge for that year. This problem has al- ready been discussed and solved in 133. In the preceding example the annual payment x is to be found that will accumulate to $1000 in 25 years at 5% compound interest. Sy (9)> 133, the annual payment is Vr ^ {l-\-r)--l' IX, 140] COMPOUND INTEREST 163 Hence, for the example just cited, (1000) (.05) ^ 120.952 = $20.95 per year. (1.05)25-1 ^ ^ EXERCISES 1. A horse 3 years old costs $250 and at age 28 has depreciated to $10 in value. Find the annual per cent of depreciation by the re- ducing balance formula. 2. Find the value of the horse of Ex. 1, by the reducing balance method, at age 5, at age 8, at age 13, at age 18, at age 25, using the per cent of depreciation found in Ex. 1. Draw a graph to represent these values. Does this computation of depreciation agree with the facts for the usual horse, i.e. is depreciation of a horse uniform over a period of years? Is it large at any time ? When is it largest ? 3. Use the straight line method to discuss Exs. 1 and 2. 4. What is the Ufe of a grain harvester costing $240, allowing $10 salvage value, and annual depreciation of 10%, using the reducing balance method? ^ns. 30 years. 5. Find by the reducing balance method the value of the harvester of Ex. 4 at the end of 2 years, 5 years, 10 years, 15 years, 20 years, 25 years. Plot the graph. Discuss these values, and decide where the depreciation is largest, where smallest, where it should be largest, etc. When is this binder worth $125, including salvage? 6. Use the straight line method to solve and discuss Exs. 4 and 5. 7. A farmer buys a manure spreader for $120. After using it for 3 years, he sells it for $75. Discuss the annual depreciation charge (a) by the straight Une method, (6) by the reducing balance method. 8. What annual payment should be set aside by the sinking fund method to replace a kitchen stove in 15 years, if it is valued at $75, and if money is worth 5%? 9. What amount should be set aside annually to replace an auto- mobile that costs $1500, at the end of five years, if money is worth 5%? 10. A farmer purchases an automobile for $2000, a harvester for $500, and a thresher for $1000, He assumes that the automobile will last for 6 years, the harvester 10 years, and the thresher 12 years. How much should be set aside annually to repurchase these machines when worn out, not counting salvage, if money is worth 5%? 164 MATHEMATICS [IX, 141 141. Summary of Interest Functions and Other Formulas. In the following formulas, n denotes the number of years or periods of time, and r denotes the rate of interest. I. i4 = />(!+ r)", the accumulation of %p compounded n years. II- P^ f. , x > ^^6 present worth of $A due in n years. (1+r)" (l+r)" 1 III. 5^= ^ ^ , the accumulation of $1 paid at end of each year for n years. ^ ^ , accumulation of $1 paid at the beginning of each year for n years. For the accumulation of $p paid annually, multiply III and IV by p. V. a^= ^ ^ , the present worth of an annual payment of $1 continued for n years. For an annual payment of $p, multiply by p. VI. J^= (l+r)"r ^^ annual income which $1 will fl^ (l+r)-l' purchase. 1 r VII. =- , the annual payment which will Sn\ (l+r)-l accumulate to $1. VIII. 5^=-, the present value of an annual payment of %p continued indefinitely. IX. x= , straight line depredation from V to V n in n years. IX, 141] COMPOUND INTEREST 165 X. a; = 1 ^J , reducing balance depreciation from V to F' in n years. Vr XI. x= ; , sinking fund allowance for replacement (1+r)" 1 of the original value, V, in n years. MISCELLANEOUS EXERCISES 1. Find the amount of $1000 at 4% compound interest for 10 years. 2. Compute the amount of $1000 at 4% compound interest for 10 years, compounded semiannually. 3. Compute the amount of $1000 at 4% compound interest for 10 years and 6 months. [Hint. (1.04)ioi may be computed readily by logarithms, but the commercial practice is to compute the compound interest to the end of the last even year, and then find the simple interest on this amount for the remaining fraction of a year.] 4. What sum at compound interest for 6 years at 4% will accumu- late to $1000? State this exercise in terms of present worth. 5. What rate at compound interest is a man receiving when $1000 will amount to $1935.28 in 15 years? 6. In what time will a sum of money at 5% compound interest double itself ? Ans. 14.2 years. 7. Find the amount of $1 at 5% compound interest for 1 year, 2 years, 3 years, 4 years, etc., to 10 years. Construct a graph showing this information, 8. Construct a graph showing the amount of $1 for 10 years as the rate varies. Use 2%, 2|%, 3%, etc., up to 6%. 9. How much will a man have saved at the end of 10 years if he deposits $60 semiannually in a savings bank, and interest is com- pounded at 5% semiannually? 10. What would be the cash value of a farm for which a man agreed to pay $1000 a year for 10 years, if money is worth 6% ? How much if money is worth 4|%? Ans. $7360.09; $7912.72 166 MATHEMATICS [IX, 141 11. A tractor that costs $1500 must be replaced in 10 years' time. If money can be accumulated at 5% interest, what sum should be set aside each year to meet the cost of a new one at the end of the time ? Ans. $119.26 12. What annual income for 30 years will $10,000 purchase, if money is worth 4|%? Ans. $613.92 13. What annual income for 15 years will $10,000 purchase, if money is worth 4^%? What, if money is worth 4%? Ans. $931.14; $899.41 14. If money is worth 5%, compare the following offers for a farm : (a) A rental of $600 per year for 20 years, or a cash price of $10,000. (b) A rental of $600 per year for 10 years plus $700 a year for the following 10 years, or a cash price of $12,500. CHAPTER X AVERAGES AND MIXTURES 142. Mixtures. In preparing many compounds, foods, fertilizers, concrete, etc., many different mixtures of sub- stances or of two varieties of the same substance, are in common use. The simplest case is that of mixing two varieties of the same substance that differ in percentage of strength of some important ingredient. As an illustration, we shall first show how to determine the proper amounts of different grades of milk to form milk of a desired standard of quality. 143. Dairy Problems. Milk is standardized according to its butter fat content. Such a standard milk may be obtained by adding milk of a higher per cent of butter fat when the per cent is too low or by the addition of skim milk when the per cent is too high. Cream is also used in the standardizing process. Water cannot be used, since it contains none of the other elements necessary in milk, such as sugar, casein, etc. The amount of water in milk is quickly determined by the specific gravity test.* Ordinary legal milk is about a 3% milk, i.e. 3% of its weight is butter fat. Skim milk is theoretically a 0% milk, but it usually contains a small trace of butter fat, from .1% to .5%. * See gallon in the Table of Equivalents, Table V, Appendix, where the weight of milk is shown. 167 168 MATHEMATICS \K, 143 Example. It is desired to add skim milk to 800 lb. of milk con- taining 4.5% of butter fat to form a standardized milk containing 4% of butter fat. How many pounds of skim milk must be added? Solution. Let x be the number of pounds of milk after adding skim milk. Then the amount of butter fat is .04 x ; hence .04 X = (.045) (800) lb. = 36 lb., ^^ a:=^=900 1b. .04 It follows that 100 lb. of skim milk should be added. Still another method is given in 148. EXERCISES 1. How many pounds of skim milk must be added to 200 pounds of 6% milk to standardize it to 4% milk? 2. How many pounds of skim milk must be added to 500 pounds of 5.8% milk to standardize it to 3% milk? 3. How much skim milk must be removed from 600 pounds of 3.2% milk to standardize it to 4% milk? 4. Given 200 lb. of 4% milk. Show how to standardize it to 5% milk by removing skim milk. 5. How much 20% cream can you make from 50 lb. of 6% milk? How much is it worth if butter sells at 56 ff per pound one pound of butter fat being sufficient for 116f % of its weight in butter? 6. In standardizing milk containing 3.5% butter fat to milk con- taining 4% butter fat, what per cent of the whole amount of milk is the skim milk removed? 7. How much each of 6% milk and skim milk should be mixed to obtam 120 pounds of 4% milk? 8. What are the proportions, and what are the amounts, of skim milk and of 5% milk required to obtain 50 pounds of 3% milk? 9. How many pounds of 5.2% milk must be added to 200 lb. of 3% milk to make 4% milk? Solution : Let a; = the number of pounds of 5.2% milk to be added. rp-L .04(200+0;) = .03 X 200+.052 x. Solve this equation for x. 10. How many pounds of 3.1% milk must be added to 54 lb. of 6.3% milk to make 4% milk? X, 144] AVERAGES AND MIXTURES 169 11. Find the proportions in which to mix 5% milk with 3.5% milk to make 4% milk. How many pounds of each should be taken to make 90 pounds of 4% milk? 12. How much 5.2% milk and skim milk containing .1% butter fat should be used to make 50 lb. of 3.9% milk? 13. How many pounds of .1% skim milk should be added to 5 gallons of 5.3% milk to produce 3.5% milk? 14. How much .1% skim milk must be removed from 780 lb. of 3% milk to raise the butter fat content to 4%? 16. I wish to standardize 26% cream to 20% cream by adding 4% milk. What are the proportions of cream and milk to be used? 16. How much 30% cream should be added to 500 lb. of 3% milk to make 5% milk? 17. What is the price per pound of butter fat when 20% cream sells for 150^ per gallon? 144. Arithmetic Mean, Geometric Mean, Average. The arithmetic mean, or simple average, of several numbers is their sum divided by their number. Thus the arithmetic mean or average of 4 and 9 is (4+9)/2=6|, of a and h is (aH-6)/2, of a, b, c is {a-\-b-\-c)/3. The arithmetic mean of the n numbers ai, a2, a^, , On is (I) j_ gi+Q2+fl3+ +gn . n For example, the average length of nine ears of com whose measure-- ments are 3", 4", 4", 6", 7", 7", 8", 9'', 10" is ^ ^ 3+4+4+6+7+7+8+9+10 ^q^ The geometric mean of two numbers is the square root of their product ; of three numbers, is the cube root of their product. The geometric mean of n numbers ai, 02, 03, -", dn is the nth root of their product, i.e. (2) G= Vfli 02-03 On. For example, the geometric mean of 4 and 9 is V4x9=c; the geometric mean of 4, 9, 12 is v/4x9x 12=7.56 170 MATHEMATICS \X, 144 The geometric mean of two numbers is very often called jhe mean proportional between the two numbers. An average, or a mean, is any representative type which is used to describe a whole collection of values or objects. Thus the collection of nine ears of corn noted above is described by saying that the average length of ear is 6|-". EXERCISES 1. A man walks a mile at the rate of 5 mi. per hour and a second mile at the rate of 3 mi. per hour. To what average rate for the two miles is this equivalent ? [Hint. Average time per mile = | {\+l) hr. =-s\ hr. This is 3| mi. per hour.] 2. What is the average rate for two miles the first of which is traveled at the rate of a miles per hour and the second at the rate of b miles per hour? ^^^ 2ab_^ a-\-b 3. What is the average rate for three successive miles traveled at the rates of 15, 20, 25 mi. per hour? Ans. 19.+ mi. per hour. 4. What is the average rate for four successive miles, the rates for which are 15, 20, 25, 30 mi., respectively? Ans. 21.+ mi. per hour. 5. A farmer makes 8% on his investment the first year. Adding this to his capital, he makes 10% on the amount the second year, and he makes 15% the third year. What is the average rate of increase for the three years? [Hint. Let r be the average rate and let p be the principal he starts with. Then the amount of p dollars at an average rate r for 3 years is p(l+r)3. This must equal the actual amount he has at the end of 3 years. This equahty may then be solved for r.] 6. What is the average rate of increase r of a certain capital if the annual rates are n, r^, Vz, respectively ? 145. Meaning and Use of an Average or Mean. We saw in 144 that any representative type or individual is an average or mean in so far as it represents, in some way, the whole collection to which it belongs. When we are X, 146] AVERAGES AND MIXTURES 171 concerned with groups of individuals we can seldom make a definite quantitative statement that surely applies to any given individual. But we can make statements that apply to the average of the group with a certainty that increases as the number of individuals increases. For example, the average healthy person 30 years old will live to be 67, though any given one of them may die sooner or later than that. The whole business of life insurance is based upon expecta- tions of this kind, which are based upon the experience of life insurance companies with hundreds of thousands of people. Again, we could not be certain of the true char- acteristics of Jersey cows from a single cow or from a few of them, but after a study of measurable characters of a large group of them, very definite statements can be made. To describe in a few words or symbols a whole group whose individuals we have measured, we put aside the measurements of any one individual and construct a sin- gle, intermediate number or characteristic. This gives some kind of average, which will be descriptive of the whole group. 146. Different Kinds of Averages. There are five dif- ferent kinds of averages in common use for different pur- poses, viz. : (1) the arithmetic mean ( 144), (2) the weighted arithmetic mean ( 147), (3) the geometric mean ( 144), (4) the mode ( 150), (5) the median ( 149). All of these are in general practical use. Such phrases as average rainfall, average rate of interest, average wage, average college student, all refer to different kinds of averages. 172 MATHEMATICS [X, 147 147. The Weighted Arithmetic Mean. A slight modifi- cation of the simple arithmetic mean is the weighted arith- metic mean. For example, the arithmetic mean of the lengths of a hundred ears of corn is merely the sum of these lengths divided by one hundred. However, these hundred measurements may be arranged in half-inch groups as follows. Inches 3.0 3.5 4.0 4.5 6.0 5.5 6.0 6.6 7.0 8 7.5 10 8.0 10 8.. 14 9.0 13 9.6 10 10.0 10.5 2 n.o 1 Frequencies or number of ears 2 2 2 3 4 6 7 The weighted arithmetic mean of these lengths is found by taking each group length times the number in the group. The sum of these products divided by the total number of ears is the average length, i.e. Av. length _ 1X3+2X3.5+2X4+ +13X9+10X9.5+ +2X10.5+1X11.0 100 = 7.7" In general, if Wi, m2, rris, , m^ represent the class marks {length of ear, temperature, per cent butter fat, etc.) and if Wi, W2, Ws, , Wr the frequencies {number of ears, gallons, pounds, etc.), then Wrmr _ Wr (3) Weighted Arithmetic Mean= ^^^^+^^^^+ ' ^^1 + ^2 + ^3 + Thus, the weighted arithmetic mean is obtained by multiply- ing each mark by the corresponding frequency, and dividing the sum of these products by the total number in the collection. The principal advantage of the weighted arithmetic mean over the simple average is that it is more easily computed. The weighted arithmetic mean is widely used in statistics, in mixing various compounds of different purities or of different temperatures, in determining results of civil serv- X, 148] AVERAGES AND MIXTURES 173 ice examinations where each subject is weighted, in mixing concrete from substances of known voids as sand and gravel, in locating centers of gravity of masses Wi, w^, Ws, , Wf lying along a straight line at distances Wi, W2, , w,., re- spectively, from a fixed point in that line (or lying in a plane at these distances from a fixed line), etc., etc. Example. Where, on a uniform bar whose own weight is neglected, is the center of gravity (balancing point) of weights of 1, 2, 3, 2, 2, 1 pounds, if they are placed at distances of 4", 6", 11", 12'', 20", 30", respectively, from one end of the bar? The required distance from that end is the weighted average A = 1X4+2X6+3 X11+2X12+2X20 + 1X30 _^3/, 1+2+3+2+2+1 148. Weighted Average of Grades of Milk. It is fre- quently necessary to know quickly the results of mixing grades of milk, or other substances of different purities or of different temperatures. Such problems may be stated in terms of weighted averages, as illustrated by the following example. Example. What kind of milk is a mixture of 3 gal. of 5.3% milk and 5 gal. of 3.1% milk? This is a problem in simple averages ( 144) if we allow the repetition of the 5.3% grade 3 times and of the 3.1% grade 5 times, i.e. Average = (5.3+5.3+5.3) + (3.1 +3.1+3.1+3.1 +3.1) ^3(5,^+5(M) =.3.9% milk. 3+5 Thus, generally, Wi pounds of an mi milk and W2 pounds of an mi milk will give an average milk Wi-\-W2 Clearing (4) of fractions and collecting, we have Wi(A mi)=W2(m2 A), or (5) "!! _ 1^2 A W2~ A mi 174 MATHEMATICS p:, 148 Equation (5) gives the ratio of the amounts of milk in terms of the kinds of milk. This leads at once to the so-called parallelogram method of standardizing mixtures. This is merely a graphic form of interpreting equation (5) as follows. Place the average in the center of the parallelogram, Fig. 93, and the marks of the substances concerned {per cent, strength, temperature, etc.), at the left-hand corners. Subtract diagonally, placing the remainders {taken positively) at the opposite right-hand corners respectively. These remainders are the ratios of the substances mixed which produce the given or the desired average. 'A Fig. 93. Dairymen's Parallelogram Example 1 should be added to produce a 3.5% milk? Let x= the number of pounds of 5% milk to be added, from the diagram How much 5% milk Therefore. 6^ .5^- -*-xlb. j|J!^ = j|, from which a: = 50 1b. Example 2. A farmer receives an order for 50 lb. of 3.6% milk. How many pounds each of 3.1% milk and 4.7% milk shall he take to fill the order? Let x = the number of pounds of 3.1% milk required and 50 a: = the number of pounds of 4.7% milk required. Therefore, from the diagram 8.1 p^^ ^-\ 1.1 > X lb. we have 4.7 ->50-a?. = -^ , from which 50-a: .5 a; = 34.4 lb X, 148] AVERAGES AND MIXTURES 175 EXERCISES 1. A farmer mixes 8 gal. of 3% milk, 10 gal. of 2.9% milk, 5 gal. of 3.5% milk, 4 gal. of 5.3% milk, 6 gal. of 3.7% milk. What per cent butter fat is the mixture? 2. What kind of milk must be added to the mixture of Ex. 1 to make 36gal. of 3.5%milk? 3. How much water must be added to a mixture of 3 pt. of 96% alcohol and 8 pt. of 78% alcohol to make the final mixture 60% alcohol? [Hint. Let x = the amount of water to be added.] 4. Masses of 4, 7, 9, 13 pounds are placed on a straight hne at distances of +2', 3', 4', +5', respectively, from a given point in that lijie. Where is the center of gravity with respect to this point? Ans. +3!' from the point. 6. Weights of 1, 2, 3, 4 pounds, respectively, are placed in order at the corners of a square 1' on each side. Find the center of gravity of this square, neglecting the weight of the square itself. [Hint. Let two sides of the square intersecting at the one pound weight be the x- and y-axes as in the accompanying figure. Find the x-distance of the center of gravity as in the pre- vious exercise. Similarly for the i/-distance.] 6. Find the center of gravity of an equi- lateral triangle on which weights of 1, 2, 3 pounds, respectively, are placed at each corner. 7. A candidate receives the following marks in a civil service examination. In penmanship, weighted 1, 90% ; in geography, weighted 2, 75% ; in arithmetic, weighted 5, 70% ; in U. S. History, weighted 3, 60% ; in English, weighted 2, 80%. Compare his standing with the candidate who has, for the same subjects and weights, marks of 85% in penmanship, 77% in geography, 60% in arithmetic, 70% in U. S. History, and 70% in English. [Hint. A weight of 1 in penmanship and of 2 in geography mean merely that geography is rated as twice as important as penmanship.] 8. Find the weighted arithmetic mean for the following table of measurements of 54 seeds. 2 lb Y 3 lb. 1 lb. 4 lb. Fig. 94 Length in mm. . . . 1.56 1.58 1.59 1.60 1.61 1.62 1.63 1.64 1.65 1.66 1.67 1.68 Number of Seeds . . . 1 ' 1 2 2 6 11 13 7 4 5 2 9. Solve Exs. 7-16, 143, by the parallelogram method. 176 MATHEMATICS [X, 149 149. The Median. The median of a collection is a value or type such that half of the other values exceed it and half are below it. If the numbers are arranged in numerical order in a column, the number that is halfway down the column is the median. It may be found also by pairing off the largest and the smallest and repeating the process until only one or two is left. If two are left, the median may then be taken halfway between them. Example 1. The median of 1, 3, 3, 5, 6, 8, 10 is 5. Example 2. The median wage of 825 workmen is that of the middlemost of the men when they are arranged in numerical order with respect to wages, i.e. if 412 receive more than $2.77 and 412 re- ceive less, then $2.77 is the median wage. The advantage of the median as a representative type is the fact that it can be determined easily. Very large or very small values do not affect it. However, it gives no special importance to extreme values and may be after all far removed from the type. 150. The Mode. The mode is the most frequent value. In the set of values 3, 4, 4, 5, 6, the mode is 4 since it occurs twice and the others only once. In a popular way, mode means fashion. Thus that type of citizen met oftener than any other is an average citizen. For a community that has three millionaires but all the other citizens in poverty, the arithmetic average for the financial condition of the community would give the impres- sion that all the citizens were well-to-do, while in reality the average citizen is in poverty. The mode is useful in describing such a situation. EXERCISES 1. Find the median length for the 100 ears of corn in the table of 147. 2. Find the median length of seed for the seeds tabulated in Ex. 8, 148. X, 151] AVERAGES AND MIXTURES 177 3. Which mean is meant in the following cases? Average-sized apple; average coloration of Rome Beauty apples; average price of butter; average wheat production per acre in the United States; average rate of depreciation of farm machinery; mean daily tempera- ture ; normal rainfall ; average student. 4. What per cent increase per year will double the attendance at a University in 10 years' time? [Hint. Use v^2. Why? Compute this value by logarithms.] 5. What is the average annual rate of increase of wheat production, if it has increased from 525 millions of bushels to 775 millions in 10 years' time? 6. What kind of averaged age is it, if out of a total of 100,000 people alive at age 10, about one half, viz. 51,230, are living at age 64? 7. What is the mode for the table of measurements in Ex. 8, 148? 151. Mixing Fertilizers. The use of commercial ferti- lizers is one method of returning to the soil plant foods needed by growing plants. These fertilizers contain nitrogen, phosphoric acid, and potash in various proportions for different crops and soils. It is quite often more satisfactory and much cheaper to buy the fertilizing materials and mix them on the farm. A fertilizer labeled 2-8-4 contains 2% nitrogen, 8% phos- phoric acid, and 4% potash. Nitrogen is obtained chiefly from nitrate of soda, dried blood, dried fish scrap, and cottonseed meal. Phosphoric acid (phosphorus) is obtained from ground bone, basic slag, ground phosphate rock, and acid phosphate. Potash is obtained from muriate of potash, sulphate of potash, and kainite. The usual compo- sition of fertilizing materials is about as follows. Chilean nitrate of soda contains 15% nitrogen ; acid phosphate contains about 16% phosphoric acid ; muriate of potash contains about 50% potash. More complete tables of fer- tilizing materials will be found in the Appendix, Tables III and IV. 178 MATHEMATICS \X, 151 Filler, usually in the form of sand or dry dirt, is added to make the proper per cent and to aid in distributing the fertilizer evenly. EXERCISES 1. A farmer wishes to make 1 ton of 2-8-4 fertilizer from Chilean nitrate, acid phosphate, and muriate of potash. (See 151.) How much of each ingredient must he use and how much filler must he add ? [Hint. The nitrogen content is 2% of 2000 pounds = 40 pounds. This must be obtained from Chilean nitrate which is only 15% nitro- gen. Therefore if x is the number of pounds of nitrate used, .15 x = 40 ; hence x = 267 pounds.] Ans. Filler = 573 pounds. 2. Show how to make one ton of 3-10-3 fertilizer, which is used for raising corn or wheat. Calculate its cost at 3f^ per pound for Chilean nitrate, 1^ per pound for acid phosphate, and 3^ per pound for muriate of potash. 3. Show how to make a 4-9-4 fertilizer, used for raising potatoes, from the materials of Ex. 2. Calculate the cost per ton. 4. A farmer has a ton of fertilizer containing 30% dried blood, 62 f% acid phosphate, and 7|% muriate of potash. How many pounds of acid phosphate must he add to this ton to raise the acid phosphate content of the mixture to 70%? Ans. 500 1b. 5. Compare a 2-8-2 fertilizer at $25.00 per ton and a 2-8-6 fertilizer at $30.00 per ton on the basis of the average cost per pound of plant food which they contain. What kind of average is this? 6. How many pounds of acid phosphate must be added to 400 lb. of a 2-6-4 fertilizer to convert it into a 2-8-4 fertilizer, if the acid phos- phate contains 16% phosphoric acid? Ans. 100 pounds. 7. How many pounds of acid phosphate and how many of muriate of potash must be added to a ton of a 2-6-4 fertilizer to convert it into a2-8-6fertiUzer? 8. A farmer mixes 500 pounds of a 2-9-5 fertiHzer and 300 pounds of a 2-6-2 fertilizer. What ratios do the ingredients of the mixture have? 9. Find the ratios of the ingredients for 200 lb. of a 2-6-4 fertilizer mixed with 300 lb. of a 1-2-1 fertilizer. X, 151] AVERAGES AND MIXTURES 179 In solving the following problems, use Tables III and IV, Appendix, which give the required amounts of fertilizing materials per acre to grow various crops, and which give the composition of these materials. 10. What quantities of dried blood (13% nitrogen) and bone meal (3% nitrogen, 24% phosphate) are needed for an acre of wheat ground which is strong in potash but which requires 12 lb. nitrogen and 20 lb. phosphate per acre ? Solution : Let x = \h. dried blood required. 2/ = lb. bone meal required. In X lb. dried blood there are .13 x lb. nitrogen, and in y lb. bone there are .03 y lb. nitrogen and .24 y lb. phosphate. Hence the total nitrogen is .13 X+.03 y = l2 lb., and the total phosphate is 24 2/ = 20 lb. Solving these two equations for x and y, we find x = 73 lb., y = 83 lb. 11. In Ex. 10, substitute nitrate of soda and phosphate of lime for dried blood and bone meal. 12. In Ex. 10, substitute com for wheat. 13. A gardener has a supply of wood ashes and a limited amount of hen manure mixed with an equal weight of leaves. How much dissolved bone should he buy to make a complete fertilizer for an acre of beets? Solution : Analysis of hen manure 2 2 1 % Analysis of leaves .7 .15 .3 % Average for equal amounts 1.35 1.08 .65% Analysis of dissolved bone 2 20 % Analysis of wood ashes 1 5 % The requirement for an acre of beets is 20 lb. nitrogen 25 lb. phosphate 35 lb. potash. Let x = no. of lb. of mixed hen manure and leaves required per acre, 2/ = no. of lb. of wood ashes per acre, 2 = no. of lb. of dissolved bone per acre. Then the total nitrogen is .0135 a; + .02 z = 20 lb., the total phosphate is .0108 x + .Ol 2/ + .20 2 = 25 lb., and the total potash is .0065 x + . 05 2/ =35 lb. Solving these equations, we find approximately, x = 1460 1b., 2/ = 510 1b., 2 = 21 lb. 180 MATHEMATICS pC, 151 14. A man wishes to raise an acre of potatoes. He has at hand 500 lb. farmyard manm^e and 100 lb. hen manure. These are to be mixed in this proportion when used with wood ashes to make a minimum complete fertilizer for the particular soil used. How many pounds of each are required? 15. Find the proportions of wood ashes and hen manure for an acre of celery on soil that requires no additional nitrogen. 16. A man has applied 4000 lb. of farmyard manure to an acre of tomato land. How much wood ashes, dried ferns, and hen manure, mixed in the proportions 4, 4, and 1, respectively, should he use to complete the f ertihzation ? 17. Find a complete fertiUzer made of acid bone and sulphate of ammonia for an acre of cabbage on clay land that requires no potash. 18. What quantities of floats, farmyard manure, and sulphate of potash are needed for raising an acre of onions ? 152. The Mixing of Concrete. Concrete is usually a mixture of cement, sand, and gravel (c, s, g), in various pro- portions, depending upon the use to which it is to be put. Since there is a certain per cent of air space {voids) in sand and gravel, it is necessary to calculate the various amounts of c, s, g to make a given volume, say 1 cu. yd., of solid con- crete. A common working rule is to allow 33^% voids in the sand and 45% voids in the gravel or crushed stone. Example. How many cubic yards each of cement, sand, and gravel, mixed in the ratio 1-2-4 (i.e. 1 part cement to 2 parts sand to 4 parts gravel) are required to make one cubic yard of solid concrete, allowing 33^% voids in the sand and 45% voids in the gravel? Let X = the cubic yards cement, y the cubic yards sand, = the cubic yards gravel required. Since the voids in the gravel are filled up in the finished concrete, the z cu. yd. of gravel count as only .55 z cu. yd. of solid concrete, and the y cu. yd. of sand count as only | y cu. yd. of solid, finished concrete. There being no voids in the cement, we have the equations : (1) x+f ?/ + .55 z = 1 (cu. yd. of concrete) ; X, 153] AVERAGES AND MIXTURES 181 (2) y = 2x and z=^4x from the ratios of the mixture. Substituting the values of y and z from (2) in (1), we have a: + ^-f2.2x = l. Solving this equation we obtain X = .22 cu. yd. cement, from which we find y = .4A cu. yd. sand, and z = .88 cu. yd. gravel. 153. Unit of Volume of Cement. Cement varies greatly in weight in different stages of compactness. Results show a variation from 86 to 123 pounds in the average weights of the same cement according as it was weighed sifted or packed in a barrel. The quantity of cement, in the pro- portion for mixing concrete, should therefore invariably be regulated by its weight. In case the proportions are by volume, then a definite weight or a definite number of packages of cement must be assumed as the unit volumes. As to the unit which should be selected for the volume of a cement barrel, the opinion of some fifty authorities (mem- bers of the American Society of Civil Engineers) differed greatly. An approximate average of all the figures sug- gested by them was that a barrel of cement should be 3.8(3.76) cubic feet, which corresponds to 100 pounds to the cubic foot. There are four sacks of cement in 1 barrel. EXERCISES Complete the following table giving the ingredients for 1 cu. yd. of solid concrete, allowing 33^ % voids in the sand, 40 % in the gravel. NintfBER Mixture Cement Sand Gravel Cu. Yd. Sacks 1 1-2-3 7.2 2 1-21-5 3 1-3-5 4 1-3-6 5 1-2-2 182 MATHEMATICS [X, 153 6. Give the ingredients for a cubic yard of 1-2-3 concrete allowing 20% voids in the sand and 40% in the crushed stone. 7. Solve as in Ex. 6 for a 1-3-5 concrete. 154. Proportions of Cement, Sand, Gravel for Various Purposes. The following are the most commonly accepted proportions of cement, sand, and gravel, respectively, for various purposes. (a) Rich : l-l|-3 for columns and structural parts under very heavy stress. (6) Standard : 1-2-4 for floors, beams, and columns under considerable stress. (c) Medium : l-2|-5 for walls, piers, sidewalks, etc., and not subject to great stress. (d) Lean : 1-3-6 for heavy mass work under compression only. 155. Fuller's Rule. For determining more quickly the amount of the ingredients in a cubic yard of concrete, the following formulas, known as Fuller's Rule, give rather good working results : (5) C= = the barrels of cement in 1 cu. yd. of concrete, (6) S= ^^^^^^-^^ = the cubic yards of sand in 1 cu. yd. of concrete, (7) G'=g(g)=: ^fa)(3-8) ^the cubic yards of gravel in 1 cu. vd. of concrete, where c = the number of parts of cement = 1 always, s = the number of parts of sand, Sr = the number of parts of gravel. To reduce barrels of cement to sacks, multiply the number of barrels by 4. X, 155] AVERAGES AND MIXTURES 183 EXERCISES Fill out the accompanying diagram according to Fuller's Rule and thus obtain some working rules. Parts in the FoRMtrLA 1. l-U-3 2. 1-2-4 3. 1-21-5 4. 1-3-6 Sacks Cement Cu. Yd. Cement Cu. Yd. Gravel 5. Compare the amounts of material necessary for 1 cu. yd. of a 1-2^-5 concrete mixed according to Fuller's Rule and according to the example of 152. CHAPTER XI GEOMETRY MENSURATION 156. Right Triangles. Some elementary mensuration rules were stated in Chapter II. The purpose of the present chapter is to simplify what has been given, and to give geometric rules and theorems of more advanced character. In a right triangle ABC, shown in Fig. 95, let the sides be denoted by a, h, c, in the usual manner ( 15), and let the perpendicular h, let fall from C on the side c, divide c into the two parts m and n (Fig. c 95). Then the following relations hold: (1) a'^ + b' = c\ (2) /i2 = mn, (3) mc, b^ Fig. 95 nc. Each of these theorems may be stated in words by the student : (1) is the Pythagoras Theorem ( 15) ; (2) and (3) may be obtained easily from the fact that the triangles ABC, ADC, and BDC are all similar ; (1) is obtained by adding the two parts of (3). 157. Oblique Triangles. In any triangle ABC (Fig. 96), the following theorems hold. An exterior angle made by any side and an adjacent side produced is equal to the sum of the two opposite interior angles. 184 XI, 158] GEOMETRY MENSURATION 185 The altitudes meet in a point 0, called the orthocenter. The medians, joining each vertex with the midpoint of the opposite side, meet in the center of gravity G. This point G trisects each median. The perpendiculars erected at the midpoints of the sides meet in a point Q which is the center of the circumscribed circle. In any triangle 0, G, and Q lie in a straight line, and G is two thirds of the way from to Q. The bisectors of the angles meet in a point P which is the center of the inscribed circle (Fig. 97). o c FiQ. 98 The bisector of an angle divides the opposite side into seg- ments which are proportional to the adjacent sides (Fig. 98). 158. Extension of the P5rthagoras Theorem for Any Triangle. In any triangle ABC (Figs. 99, 100), if h is the altitude on b, and x is the projection of c on b, it is shown r ^ ^. Fig. 99. a2 = 62 + c2-26x b C Fig. 100. o2 = 62 + c2 + 26x in plane geometry, by using the Pythagoras theorem on the two right triangles ABD and DCB, that 186 MATHEMATICS \XI, 158 (4) fl2 = 52_|_c2 _ 2 bx, if A is acute, or (5) a2= &2+C2+2 bx, if A is obtuse. Solving these equations for x, we may write also 2^2 _|_ ^2 _ q2 (6) X = , if A is acute, 2 o or (7) X = g'-^'-g' ^ if A is obtuse. 2 u 159. Computation of Altitudes. Area of a Triangle. If the three sides of a triangle are given, we may compute (a) The projections of any side upon any other side ( 158). (6) The altitudes. From Fig. 99 or 100, h = Vc' - x"^ where X is known from (6) or (7), 158. (c) The area. When the altitude has been found, the area follows at once. The formula for area of a triangle from 15, viz. : A = y/s{s a){s b) (sc), may be derived from 158. For, from Figs. 99, 100, V 26 J ,2 ^2_^2 ( b'+c''-ay _ 4c^-{+c''-ay , 4 62 ' or, factoring the numerator, ;,2_ (a+6+c)(6+c-a)(q-6+c)(a+6-c ) ^" 462 Let a+6+c = 2 s. Then we have a+6-c = 2s-2c = 2 (s-c), a-6+c = 2s-26 = 2 (s-6), -a+6+c=2 s-2 a = 2 (s-a). XI, 160] GEOMETRY MENSURATION 187 Therefore i,2_ 2 g 2 (s-a) 2 (s-b) 2 js-c) ^ 4b^ ""' and (8) h = Ws{s-a)(s-h){s-c). Hence the area, which is equal to hX-,is also equal to (9) Area = \/s{s a){s b) (s c). EXERCISES 1. The sides of a triangle are 9, 12, 15. Find the segments of the longest side made by bisecting the angle opposite it. 2. The sides of a triangle are 12, 14, 16. Draw it to scale using 14 as the base and find (a) The segments of the side 14 made by the bisector of the opposite angle. (h) The projections of the sides 12 and 16, respectively, on the side 14. (c) The altitude on the side 14. (d) The area of the triangle. 3. Find the area of a triangular pasture whose sides measure 19> 32, 25 rods, respectively. 160. The Circle. To the elementary facts about circles given in 19, 20, we may add the following theorems, which are usually proved in elementary geometry. a^ I. An angle between two secants is measured by one half of the difference d\ of the intercepted arcs. Thus, we have for example Z A PD ^^^ ^^^ = i (arc AD - arc BC), Fig. 101. II. An angle between two chords is measured by one half of the sum of the intercepted arcs. Thus, ZEOF = i (arc EF + a.rc GH), Fig. 101. 188 MATHEMATICS [XI, 160 III. If a tangent PT and a secant PAB he drown from any exterior point P to a circle (Fig. 102), the square of the tangent PT is equal to the product of the tivo segments of the secant, PA and PB ; i.e. (10) PT'=PA ' PB=PC'PD = a constant. This fact is of interest with refer- ence to visibility above the surface of the earth. Let PB, Fig. 103, be a diameter of the earth, PA the distance above the earth's surface, and PT the distance to the horizon. Then p PA PB = PT\ LetFA-fc;thenPB = /c+2r. AlsoletPr=d Therefore d'' = k{k-\-2r). Practically, h is small as compared with the diameter of the earth. Thus, k-\-2r may be written as 2 r. Hence ^^Q- i^^ d^=2 kr, or k= - (k and d in miles). 8000 But k is usually measured in feet. Therefore let k , from which 5280 and 7 ^ h ^ d^ 5280 8000^ h, the height fn/eef = f|f d'^=l d"^, approximately.* Therefore, the height in feet equals two thirds the square of the distance in miles. Example. A lighthouse is 240 feet high. How far is its hght visible at sea? SoiiUTioN : We have h = 240 ; hence 240 = I c/^ or (i = 19 miles. * See Exercise 9, 79. XI, 162] GEOMETRY MENSURATION 189 Fig. 105 161. Compound Curves. Arcs AB and AC of two circles, which are tangent to each other at A, Figs. 104-105, form a compound curve. The point of tangency A is the /^-=? transition point from one circle ^ ' to the other. Compound curves are used in the construction of rail- roads and highways where the track must be made to con- form to the physical features of the land. 162. The Area of a Segment of a Circle. In 22, mention was made of the area of a segment of a circle. In Fig. 106, let h = height of a segment, r = radius of the circle. Then OS = r h, and the angle SOP is given by (11) cos SOP = g|. The area of the sector OPQR and of the triangle OPR may now be found. The required area of the segment is their difference. EXERCISES 1. Find the diameter of a broken wheel a chord of which measures 16" and the height of the corresponding arc is 4". 2. The chord of an arc is 24" and the chord of half this arc is 13". Find the diameter of the circle. 3. Show that the following construction, Fig. 107, will divide a circle into three parts whose areas are equal. Trisect the radius at B and C. Draw a semi- A\ circle on OA as diameter. Erect perpendiculars BR, CP meeting the semicircle at R, P. With centers at O, draw the required circles with radii OP and OR. Fig. 107 190 MATHEMATICS [XI, 162 4. A line segment approximately equal to the circumference of a circle may be constructed as follows. Draw a diameter AD, Fig. 108, and COT = 30. Draw CT perpen- dicular to AD. From D draw DR perpendicular to AD and equal in length to three times the diameter. Then CR is approximately equal to the circumference. Show that the error in the cir- cumference, constructed in this way, is about 1 in 20,000. ^^^f. 108 Use the formula of 160 to solve the following exercises. 5. How high must an aeronaut rise above the earth in order to see a city 50 miles away? How high in order to see 100 mi. ? 6. How far can a man 6 ft. tall see when standing on a level plain? 7. The lookout in the rigging of a vessel is 100 feet above the water. How far can he see a beacon light known to be 150 feet high? 8. The Woolworth building in New York City is 750 feet high. How far can it be seen by the lookout on a vessel, if the lookout is 75 feet above the water ? 9. The curve ABC of a railroad track is com- posed of two circular arcs AB and BC, as in the accompanying figure. The center of BC is on the radius PB of AB. Prove that ABC is a compound curve, i.e. that ABf BC are arcs of tangent circles. 10. The track of a switch AOBC from a straight railroad track AD, as in the accompanying figure, is laid out as follows. Arc AO is constructed with center M, where AM is per- j^ pendicular to AD. MO is produced to N so that MO=ON and arc OB with center N is drawn. Show that AOB is a compound curve. Fig. 110 11. By sawing off 4" of the diameter of a log 12" in diameter what part of the area is sawed ? 12. The height of a segment of a circle is 3" and its base is 10". Find the radius and the area of the corresponding circle, and the area M of the segment. Ans. = 51". XI, 163] GEOMETRY MENSURATION 191 / F'/ \ ^ FiQ. 111. Ellipse with Axis Ai) Foci 163. The Ellipse. An ellipse is formed when a point moves so that the sum of its distances from two fixed points ls constant. The two fixed points are called foci, F and F' in Fig. 111. That diameter A A' on which the foci are located is called the major axis and is taken equal p to 2 a, i.e. OA = a. Similarly the other axis, called the minor axis, is perpendicular to A A' at and is taken equal to 2 6, i.e. OB = h (Fig. 111). An ellipse may be constructed by attaching the ends of a string to two pins fastened in a board or a piece of stiff cardboard and using the string as a loop, keep- ing it stretched by means of a pencil and allowing the pencil to move. Of course the length of the string is constant the ends being fastened. When the pencil is at B, Fig. 112, and pins at F and F\ the string takes the position FBF\ When the pencil is at A', the string reaches from F to A' and back to F\ With the pencil at A, we deduce AF = A'F'. In either case, then, the length of the string is equal to the length of the major axis, 2 a. Thus also FB = F'B = a. The perimeter of an ellipse is given approximately by the formula (12) P=tr(a+b).* The area of an ellipse is given exactly by the formula (13) A=Trab. * A more exact formula for the perimeter of an ellipse is L 4:\a + bl ^Q4.\a + bl ^ J The formula x(a -\- b) is very nearly correct if o 6 is small ; i.e. if the ellipse is nearly round. 192 MATHEMATICS [XI, 163 Example, Construct an ellipse by attaching the ends of a string 4' long to two pins 3^" apart fastened in a cardboard. By measurement, the major axis, 2 a = 4", FB = OA = 2", and OF = l\". Thus OB = V22 - (11)2 = .97 = 1" approximately. Hence a = 2", h = I", and we have Area = wab = (^)(2)(1) = 6.3 sq. in. Perimeter = 7r(a + b) = (^)3 = 9.4". Many things are elliptic in shape or in cross-section. For example, sections of cupola ventilators placed on the roof of a barn are ellipses. Many photograph mountings and in- terior decorations are elliptic. Many arches in buildings, bridges, etc., are elliptic. Heat- conducting pipes for furnaces have many sections that are elhptic. Fig. 112 The paths or orbits of the earth and the other planets about the sun are rather round ellipses with the sun at one focus. Hence the earth is nearer the sun during a part of the year and it moves faster then because of the increased attraction. When is the earth nearest the sun? How much nearer is it at that time ? EXERCISES 1. Find the area and the perimeter of an elliptic photograph 6" long and 4" wide. 2. A carpenter wishes to construct a frame for supporting the brick- work while laying up an arch over a grate, the arch to be half of an ellipse 46" wide with a rise of 11". Draw the ellipse to scale and show how to locate the foci. Find their distance from the center of the ellipse. 3. Compute the average velocity of the earth per second in its orbit about the sun, having given that its nearest and farthest distances from the sun are 90 millions and 93 millions of miles respectively. XI, 164] GEOMETRY MENSURATION 193 4. Elliptic gears will mesh with each other when revolving about their foci. At what distance from the end of the major axis should the hole for the center of revolution be drilled in an elliptic gear whose axes are 8" and 10"? Give an account of the elliptic gears used on a haypress to operate the plunger. 5. How large an opening in area and perimeter is necessary to install a circular ventilating cupola 36" in diameter on a roof which has a pitch of 5 in 24 ? How many 7" circular intakes will be necessary to equal approxi- mately the area of the opening in the roof? 164. Truncated Prism. A truncated prism is the part of a prism included between the base and a section made by a plane oblique to the base. The volume of a right truncated triangular prism, Fig. 113, is equal to the product of one third the sum of its lateral edges by the area of the base to which those edges are perpendic- ular, i.e. (14) Volume=\{a-{-b+c)Xbase. The same formula holds for any triangular truncated prism (Fig. 114) if the base be replaced by the area of a section perpendicular to the edges. Example. Corn is piled into the corner and along the side of a square building. It measures 4' high and 6' X 21' on the floor and 15' along the top crest. Find the volume of the pile. The pile takes the shape of a right truncated triangular prism. Accordingly its volume is 1(15+21+21) . i (4x6) =228 cu. ft. ^7 N M Fig. 114 Fig. 115 194 MATHEMATICS \X1, 164 EXERCISES 1. In excavating on the side of a hill for a building the following measurements were taken. From the surface to the bottom level of the cut 12', 9', 8', 6', respectively, at the four corners of the rectangular bottom which is to be 30' X 40'. Find the amount of dirt to be removed. [Hint. The shape of the required exca- vation will be that of two truncated right triangular prisms.] 2. What is the volume of the part of a rectangular barn under the roof and above the eaves if the barn measures 30' X 60' and the hip roof has a pitch of 1 in 3 on sides and ends ? Find also the angle made by the roof with the horizontal. Fig. 116 Fig. 117. Vertical Sections of a Ditch 3. Find the number of cubic yards of gravel in a pile 8' wide and 10' long if the sides and ends have a slope of 2 in 5. 4. A cut for a ditch has the dimensions shown in Fig. 117, which shows the longitudinal vertical section and the vertical cross-sections at each end. Find the volume of the earth to be removed. 165. The Sphere. Definitions. A section of the sphere made by a plane through the center of the sphere is called a great circle; if the plane does not pass through the center of the sphere the section so made is a small circle. Parallels of latitude on the surface of the earth are small circles. The meridians, which pass through both north and south poles, are great circles. XI, 166] GEOMETRY MENSURATION 195 A segment of the sphere is a portion of the volume of a sphere included between two parallel planes. It thus has two circular bases. In case one plane is tangent to the sphere, one base reduces to a point. The segment is then called a segment of one base. In Fig. 118, the part of the sphere above the plane DEF is a segment of one base; that be- tween DEF and ABC is a segment of two bases. A zone is a portion of the surface of a sphere included between two parallel planes (Figs. 118, 119). The altitude of the segment or of the zone is the perpen- dicular distance between the planes. 166. Volume of a Spherical Segment. Area of a Zone. (15) Volume = l'Trh{Z ri^-fS rzH/i'), where h denotes the altitude of the segment, and Tiy Ti denote the radii of the bases, respectively. (16) Lateral Area of Zone = 2Trrhf where r denotes the radius of the sphere. FiQ. 119 Example. A sphere 8" in radius is cut by two parallel planes, one passing 2" from the center, the ^ other 6" from the center. Find the volume of the ^ segment and the area of the zone between the two planes. The lateral area of the zone is 2 7rr/i = 2 7r(8)(4) =64 7r= 201 sq. in. Fig. 120 To find the volume, the radii n, r2 must be found. From the ac- companying figure r2=EF = V^pI^^ = V^^^ = V28. Volume = i nhiS n^+S ri'+h^) =| 7r4(180+84 + 16) =586 cu. in. 196 MATHEMATICS [XI, 166 EXERCISES 1. The diameter of a sphere is 6 feet. Find the surface of a segment whose height is 2 feet. 2. The radius of a sphere is 3 feet. Find the height of a zone whose area is equal to that of a great circle. 3. Find the area of a zone of one base whose radius is 2 feet and the height of the zone is 1 foot. 4. Find the volume of an apple whose circumference is 12 in. 5. Find the volume and the surface of a spherical segment, if the radii of its bases are 6' and 8', respectively, and its height is 3'. 6. How many pints of water can be put in a wash-basin in the shape of a segment of a sphere, if the distance across the top is 12 inches and the greatest depth is 4 inches? 7. What is the volume and the entire surface of a dishpan in the form of a segment of a sphere, if the distances across the top and bottom are, respectively, 20" and 16", and the depth is 5^"? 8. Compute the capacity of the milk-can shown in the diagram if it has the following dimensions, expressed in inches, D = 13", d = r', a = \", 6 = 4", h = l^". 9. How far from a sphere of radius 6" must a light be placed so that one sixth of the surface of the sphere will be illuminated ? Fig. 121 jS 167. Ellipsoid. The ellipsoid may be defined as the solid whose section by any plane is an ellipse, Fig. 122. (17) Volume = 4/3 irabc, where a, 6, c are the semi-axes. Fig. 122. Ellipsoid 168. Paraboloid of Revolution. This is a solid generated by revolving a portion of a parabola about its axis. Fig. 123. (18) Volume = T:fh/2, where h is the height of the paraboloid and r is the radius of the base. The surface of such a paraboloid is the ^^^ ^23 par- usual reflecting surface for headlights, etc. aboloid XI, 170] GEOMETRY MENSURATION 197 169. Anchor Ring or Torus. A torus or ordinary anchor ring is generated by a circle of radius r revolving about an axis in its plane (Fig. 124). If c is the distance of the center of the circle from the axis of rotation, and r is the radius of the circle, the vol- ume and the surface of the anchor ring are given by the formulas (19) FoZume=2ir2cr2, (20) Sur/ace = 4Tr2cr. Fia. 124. Anchor Ring B Fig. 125. a Prismoid 170. The Prismoid Formula. A solid is called a prismoid if it has for bases any two polygons in parallel planes and for faces quadrilaterals or triangles. Fig. 125. The altitude of a prismoid is the perpendicular distance between the planes of its bases. The mid-section of a prismoid is that section made by a plane parallel to and midway between its bases. The mid-section bisects the altitude and all of the lateral edges. The volume of any prismoid is given by the following simple formula /2i) Volume= {b-\- B-\-^ M)- , where h denotes the altitude of the prismoid, b and B denote the area-s of the two bases, and M denotes the area of the mid-section. If the area of one base, h, is zero, that base consists of a single point or a line, and the prismoid resembles a pyramid, a cone, or a wedge. For such a solid, M=J5/4, and the formula becomes i. If the bases, h and B, are equal, the prismoid resembles a prism or a cylinder, and the formula becomes V=hXB. 198 MATHEMATICS \X.l, 170 A frustum of a pyramid or of a cone is a prismoid for which 2 i.e. 4:M = h-\-B-\-2VbB since the mid-section of a frustum is similar to and midway between the bases * ; and the formula becomes h The prismoid formula apphes to the volume of a sphere, where h= B = 0, and M is the area of an equatorial section of the sphere. EXERCISES 1. Compute the volume of a football whose shape is an ellipsoid of semi-axes 3^ X 3^ X 5^. 2. Compute the solid contents in cubic inches of an egg whose larger end is a semi-ellipsoid whose semi-axes are |", ^|", and |f", and whose smaller end is a semi-eUipsoid whose semi-axes are V", ^|", and {^". 3. Find the volume of a reflecting surface whose form is a paraboloid, if its height is 5" and the radius of the base is 3^". 4. The cross-section of a sohd wrought-iron ring is a circle of 3" radius. The inner radius of the ring is 18". Find the surface and the volume of the ring. 5. Find the surface and volume of a life preserver whose shape is an anchor ring, formed by revolving a circle of a radius 3" about an axis 13" from the center of the circle. What weight of water would be displaced by such a preserver if it is submerged ? * If s and s' are corresponding sides of b and B, then (s + s')/2 is the cor- responding side of M. Therefore, ^ -^ and ^' ^ (8+sO/2 v^ (s + s')/2 VM Adding these two equations and reducing, we find 2 _ V^+VrB ^ 1 Vm XI, 170] GEOMETRY MENSURATION 199 6. A barrel made of circular staves has an end diameter of 18" and a bung diameter of 22". It is 34" high. What is its capacity in bushels, and in gallons? [Hint. Use (8), 31.] 10'6" Fig. 126. 10'6" Sections of a Railroad Cut 10'6" 7. If the barrel of Ex. 6 is standing on end, and if it is partly filled with a Hquid to a depth of 24", find the amount of the liquid in gallons. [Hint. Calculate the amount of the contents above the half-barrel by the prismoid formula.] 8. A railroad cut has the dimensions shown in Fig. 126, which shows the vertical section and three cross-sections. Find the number of cubic yards of earth to be re- moved in this cut. 9. Find the number of cubic yards of concrete in a pier having the dimensions as shown in Fig. 127, the bases being rec- tangles with semicircles. A Pier CHAPTER XII SOLUTION OF OBLIQUE TRIANGLES 171. Construction of Oblique Triangles with Ruler and Compass. A triangle consists of six parts three sides a, by c, and three angles A, B, C, respectively, opposite these sides. These parts are related as follows. I. A-fBH-C=180. II. The sum of any two sides is always greater than the third side. III. The larger side is opposite the larger angle, i.e. if a>6, then A>B, etc. If any three parts of a triangle are given, including always at least one side, the triangle may be constructed. The following cases arise. Case I. Given One Side and Two Angles. Let c he the given side, and let A and B he the given angles (Fig. 128). Take EF = c, Fig. 128, and construct Z A at Fig. 128 E and Z B at F, on the same side of c. Produce the sides of Z E Siud Z F till they meet at C. Then EEC is the required triangle. The third angle may be found, before constructing the triangle, from theorem I. If this angle is measured with a 200 XII, 171] OBLIQUE TRIANGLES 201 protractor after the construction is finished, the accuracy of the construction may be judged. Case II. Given Two Sides and the Included Angle. Let b and c be the two given sides and Z E their included angle (Fig. 129). FiQ. 129 Lay off AB = c. Fig. 129. At A, construct Z BAD= Z E, On AD take AC = b and draw CB. Then A ABC is the required triangle. Case III. Given the Three Sides. Let the three sides be a, b, c (Fig. 130). Take BC equal to a, Fig. 130. From B as a center, with a radius equal to c, describe an arc. From C as a center, with a radius equal to 6, describe an arc intersecting the n4 6 bZ / \ a Fig. 130 other arc at A. Draw AC and AB. Then BAC is the required triangle. Case IV. Ambiguous Case. Given Two Sides and an Angle Opposite One of Them. Let a and b be the given sides and A the angle opposite the side a (Fig. 131). Different constructions are possible according as a<6, a = b, a = the altitude on c, or a_ a^-\-c^-b^ cos >= 2ac = .3266 B = 7056'. C = 180-(A+5) = 61 57'. Check : a sin C =c sin A. log o = 1 .6243 log c = 1 .7050 log sin C = 9.9457 log sin A = 9.8649 1.5700 1.5699 The check agrees in the third place of mantissas. This is sufficient, since the values of the lengths of the sides a, 6, and c were given only to three significant figures. Finally, from (d), the area of the given triangle may be calculated as follows. log 6 = 1.7348 log c = 1.7050 log sin A = 9.8649 3.3047 log 2 = 0.3010 log Area = 3.0037 Area = 1009. EXERCISES In each of the following exercises determine the remaining parts, check the results, and find the area. 1. a = 14, 6 = 16, c = 10. Ans. A =60^ B = SV 47', C=38 13'. 2. a = 122, 6 = 146, c = 112. 3. a = 32, 6 = 47, c = 62. [Hint. From a figure, it is seen that C is obtuse.] 4. a = 132.4, 6 = 101.7, c = 84.50 5. A farmer has a four-sided, irregular field. If the comers are denoted by A, B, C, D taken in order around the field and if AB = 17, BC = 12, CD = 13, DA = 10, and BD = 1Q, find the area and the angle at A. 212 MATHEMATICS [XII, 180 Fig. 139 180. The Law of Segments. In 179, the triangle with three sides given was solved by use of the law of cosines. A second method for this case, better adapted to the use of logarithms, will now be given. Drop a perpendicular from A on a, dividing a into two seg- ments X and y, as shown in Fig. 139. From the resulting right triangles, we have (8) h^x^ = c^y'^, or x'^ y^^b^^c^, or (9) (x+2/)(x-2/) = (6+c)(&-c). From the figure, we see that (10) x-\-y = a. Dividing the sides of equation (9) by the sides of (10), in order, we obtain the relation (11) x-y (b+c)(b-c) Solving (10) and (11) by addition and subtraction, we find (12) a {b-\-c){b-c) 2 2a (13) y= (b+c)(b-c) 2a But we know that (14) COS C = b' and cos B=^ ; c hence B and C can be found. Then A = lSO-{A-\-B). The solution may be checked by the law of sines. XII, 180] OBLIQUE TRIANGLES 213 Example. Given a = 27, 6 = 24, c = 19, to find A, B, C. Formulas : (a) x+y = a. (6) ^-y^if>+C)(.b-C), a (Find X and y from (a) and (6) by addition and subtraction.) (d) cosB = y., cos C = Solution: x-\-y =27 ^_=(4^= 7.963 cos 5 = 9.52 19 log 9.52 = 0.9786 log 19 = 1 .2788 log cos B = 9.6998 3 = 59 5Q' A =76 48'. Check : b sin C = c sin B. log 6 = 1.3802 log sin C = 9.8360 1.2162 17.48 9.52 cosC= 17.48 24 log 17.48 = 1.2424 log 24 = 1.3802 log cos C = 9.8622 C = 43 16'. log c = 1.2788 log sin B = 9.9372 1.2160 EXERCISES Use the law of segments to solve each of the following triangles. Check your answers. 1. a = 42.6, 6 = 38.4, c = 27.2 2. = 39.42,6 = 47.82, c = 61.37 3. Use the law of segments to check Exs. 3, 4, 177. 4. Under what angle is a ship 850 feet long seen by a man who is standing on a wharf which is 600 feet from the bow of the ship and 950 feet from her tern? 214 MATHEMATICS \X.U, 181 181. Case IV. Given Two Sides and an Angle Opposite One of Them. Let the given parts be a, h, A. Construction of the given data to scale will usually indi- cate the possibilities of this case. See Case IV, 171, c c o Solution One Solution Two Solutions ^, Fig. 141. Possible Constructions for the Ambiguous Case // athe altitude on c, i.e. if a>h sin A. one solution if a = the altitude on c, i.e. if a = h sin A. no solution if ah, there is one solution. If Z A is obtuse, there is never more than one solution. A second angle, Z B here, is always ob- tained from the relation a sin B = h sin A. As found from the table, Z B is acute. Then B', Fig. 141, is the supplement of B since A B'CB is isosceles. This is often called the ambiguous case. Example. Given a = 20, c = 25, A = 52 40', to find h, B, C. The construction shows that two triangles ABC, ABC fulfill the conditions. Formulas : Fig. 142 (a) (ft) sin C = h = c sin A a asm B sin A (c) (d) h' = 180 -C. a sin B' sin A XII, 181] OBLIQUE TRIANGLES 215 Solution: From (a), log c = 1.3979 log sin A = 9.9004 sum = 1.2983 log a = 1 .3010 log sin C = 9.9973 C = 8340'. B = 4340'. From (6), log = 1.3010 log sin B = 9.8391 sum = 1.1401 log sin A = 9.9004 log 6 = 1.2397 6 = 17.4 Check : 6 sin C =c sin B. log 17.4 = 1.2405 log sin C = 9.9973 1.2378 Check : 6' sin C =c sin B'. log 13.0 = 1.1139 log sin r = 9.9973 1.1112 C = 9620'. (A+C)=31. C' = 180 B' = 1S0 From id), log a = 1.3010 log sin B' = 9.7118 sum = 1.0128 log sin A = 9.9004 log 6' = 1.1124 6' = 13.0 log 25 = 1.3979 log sin B = 9.8391 1.2370 log 25 = 1.3979 log sin B^= 9.71 18 1.1097 EXERCISES In each of the following exercises find the remaining parts and check the results. Determine first in each case whether there is one solution, or two solutions, or none. Find the area. 1. a = 50.8, 6 = 65.9, A =41 00'. Ans. B = 58 21', C=80 40', c = 76.4, B' = 12r 40', C' = 17 20', c'=23.1. Areas 1652, 499. 2. 6 = 249.8, c = 409.1,B = 37 38'. 3. 6 = 232, = 107,5=26 36'. 4. o = 518.1, c = 694.7, C = 115 42'. 5. 6 = 385.7, c = 491.2, B = 46 15'. 6. A man walking along a road due east sees a fort 10.2 miles away and bearing N. 57 E. If the guns of the fort have a range of 8 miles, how far may he go (o) before he comes within range of the guns, and (6) before he is again out of range? 216 MATHEMATICS \XU, 181 7. A 50-ft. chord of a circle subtends an angle of 100 at the center. In the larger segment, a triangle with one side 57.8 ft. is to be inscribed. Find the length of the third side and the area of the triangle. How many solutions are possible? 182. Summary. Case Given To Find Formulas Check I. One Side and Two C = C = 180-(A-\-B). Angles a = c sin A 177 as, c = 26.34, ^=80 15', fi=44 17'. 6 = sin C , c sin 5 6= sin C a sin B = b sin A. II. Two Sides and the a = IsT Method: for simple Included Angle as, 6 = 41.02, B = C = data. 178 a=V62-|_cJ-2 6ccosA. c =45.49, r> 6 sin A sin B= A =62 10'. C=180-(A+B). Cain B = bamC. 2d Method : y\ x=c cos A. e/.'V h=c sin A. y = bx. / X \y\ tan C = h/y. B = 180-{A-\-O. A b C Fig. 143 a- ^ . sinC a sin B = b sin A. III. Three Sides A = IsT Method: for simple as, a = 42.1. B = data. 179 6 = 54.3, c = 50.7 C = cos^-^+^^-"'. 2 6c cos5=etc. ; cos C= etc. 2d Method: Law of Seg- ments. 180 A 6/V x-\-y = a. ^..y-(b+cHb-c) a x = (add) y = (subtract) /. lA C a B Fig. 144 cosB=^; cosC=?- c 6 ^ = 180-(B+C). IV. Two Sides and an C = . ^ c sin A sm C= Angle Opposite B = 181 One of Them as, a = 20, c = 25, ^=52 40' 6 = Thus C = an acute angle, and C' = 180-C. fl = 180-(A+O. , a sin B and known as the 6 sin C=c sin B. Ambiguous Case sin A XII, 182] OBLIQUE TRIANGLES 217 MISCELLANEOUS EXERCISES 1. P and Q are separated by a hill. To find QP a straight Hne AB is run through Q such that AQ =756 ft., QB = 562 ft., Z QAP = 47 29', and Z QBP = 57 45'. Find QP. 2. A man observes the angle of elevation of a hill to be 29 18'. After walking one mile directly toward the hill on a level plane, he finds that the angle of elevation is 36 31'. Find the height of the hill. 3. The parallel sides of a trapezoid are 36 and 52 and the non- parallel sides are 23 and 27. Find the angles and the diagonals. 4. A surveyor has the following data. AB = 487 feet, BC = 179 feet, Z CAB = 2S. Show that something is wrong with the notes. 5. If a billiard table is a feet long and b feet wide, at what angle, /, must a ball be made to strike the long cushion in order that after striking each cushion in turn it will pass through its original position? Ans. tan/=-- a 6. A trough is 10 ft. long and its cross-section is a semi-circle whose radius is 1 ft. Find the number of gallons of water it contains when the water measures 4 in. deep. When 6 in. deep. When 9 in. deep. 7. A road OA is 9| miles long and makes an angle of 31 16' with a straight beach OX. From A two straight roads AB and AB' each 6 miles long run to the beach. Find the distance along the beach from O to the nearer of the points B and B'. (Harvard.) 8. What angle does y = 2x-\-6 make with the x-axis? What angle does 3 y 4 X = 12 make with the x-axis? 9. Plot the lines 2 x +3 i/ = 12 and 4 x -3 y = 12. Find their point of intersection and the angle each makes with the x-axis. 10. In the accompanying figure, O is the center of revolution of the crank pin of a steam engine, I is the connecting rod, P is the end of the sUde connected with the piston head, and A is the angle of revo- "^ ^Fia. 145 lution POR. What is OP, when ie = 6", i = 24", and A=30? AlsoifA=45? If A = 135? Ans. 29"; 27.9"; 19.4" CHAPTER XIII LAND SURVEYING 183. Introduction. To survey land means to measure distances and angles, to calculate from these measurements other distances and angles, to locate and mark the corners, to record these measurements and locations so that a plat or map can be made, and to find the area of the tract or parcel of land. Sometimes it is desirable to resurvey an original tract. The surveyor then seeks to find the original corners, to retrace the boundaries, and to replace any landmarks, such as posts, monuments, etc., that have decayed or have been destroyed. In addition to this he is often called upon to subdivide tracts of land into smaller parcels, as in city sub- divisions. 184. Instruments. For measuring distances surveyors usually use a steel tape 100 feet long, or a chain which is 66 feet = 4 rods long and which contains 100 links. For these and other units of measurement, see 185. The surveyor also uses pins to mark the successive applications of the tape or chain ; range poles for sighting and ranging out directions ; and a graduated rod or pole called a leveling- rod provided with a sliding target by which he can measure elevations and depressions, as in establishing the grade for a ditch. Almost all of the old land surveys of this country were made with the compass as the angle-measuring instrument. 218 XIII, 184] LAND SURVEYING 219 However, it has now been supplanted to a great extent by the transit. The transit (Fig. 146) is the instrument now most used by surveyors and engineers for measuring horizontal angles, and, with certain attachments, for measuring vertical angles Fig. 146 and distance, and for leveling. It is mounted on a tripod and consists essentially of the following three parts. (a) the telescope to determine direction. To the telescope is attached a level and a vertical graduated circle to determine angles of elevation and depression. 220 MATHEMATICS \XIU, 184 (6) the alidade, a horizontal circular piece provided with vertical supports to carry the telescope, with a graduated circle, and with precise levels so that this circle can be brought exactly into the horizontal plane. Angles in this plane may be read on this circle as the angular turn of the telescope. (c) the magnetic compass. 186. Surveyor's Measures. City lots are usually sur- veyed in terms of feet and square feet, with decimal parts of these units. However, in measuring larger pieces of land the following measures are used. 1 rod = 16i feet. = 1 perch= 1 pole (older plan). 1 chain (ch.) =4 rods = 66 feet. = 100 links ill). 1 link =7.92 inches. 1 Acre (A.) =160 square rods (sq. rd.). = 10 square chains (sq. ch.). 1 section = 640 acres. In some states Spanish-American units are used as follows : 1 Vara (va'-ra) = 33^ inches in Texas, = 33 inches in California, = 32.9927 inches in Mexico. 1 labor (la-bor') =a square whose side is 1000 varas, = 1,000,000 square varas, = 173 acres (approximately), = a quarter section (roughly). 1 league = a square whose side is 5000 varas, = 25 labors, = 6.8 square miles (approximately), 1 (Spanish) -4 ere = 5645 square varas (in Texas), XIII, 186] LAND SURVEYING 221 Since 100 links =1 chain, the older plan of writing both chains and hnks in a survey is discarded and distances are written in terms of decimals of a chain. For example, 11 chains 24 links is written 11.24 ch. Similarly, the area is written in decimals of an acre, as 2 acres 4 square chains = 2.4 A., 2 A. 47 sq. rd. = 2.294 A. 186. General Procedure as to Comers, Offsets, Obstacles, Measuring Slopes, etc. When a corner has been determined in a survey, and agreed upon by all parties concerned, it should be marked with stones, iron or concrete posts, etc., in such a way as to perpetuate its location as long as possible. In measuring distances two men, a leader and a follow er^ carry the tape or chain. The leader carries ten pins, and at each measurement he " sticks " a pin to mark it. On moving forward the follower picks up this pin. The process is repeated and tallied, both leader and follower keeping their own tallies. Since areas are computed as if the land were flat, it is necessary in measuring on slopes to keep the chain level (Fig. 147). It sometimes happens, on account of fences, brush, ponds, or other obstructions, that a measurement cannot be made c Ci \D Fig. 147. Measuring on a Sixjpe B D Fig. 149 on the desired line. The surveyor then measures a con- venient portion of a parallel to the required line at a certain perpendicular distance from it, as CD = BE^ Fig. 148. 222 MATHEMATICS fKIII, 186 An obstacle sometimes may be passed and the general direction continued beyond it by measuring an equilateral triangle about the obstacle as in Fig. 149. 187. Computing the Area. In 18 and 42, methods for finding the area of irregular shaped tracts have been discussed. These methods will now be discussed more fully, making use of the latitudes and longitudes of the corners of the sur- vey. These may be looked upon as the coordinates of the vertices of the survey and follow from the lengths and bear- ings of the courses. 188. Definitions. If the y-axis be a meridian and the X-axis an east-west line, then the ordinate of a point is called its latitude, its abscissa is its longitude. Let ABhe any course. Fig. 150. Draw a meridian through A and perpendiculars BB^, ^ A' to it and the y-Sixis. Then PA = OA' is the latitude of A, OB' is the latitude of B, and thus A'B'=the latitude- ci-^-- difference of the course AB. Similarly A A', BB' are the longitudes of A and B, respec- tively; and KB = the longitude-difference of the course AB. Latitude-difference north, i.e. a northing, is taken to be positive, a southing is taken to be negative. Similarly, longitude-difference east, i.e. an easting, is positive, a westing is negative. The double longitude of a course is the sum of the longi- tudes of its ends. Thus, in Fig. 150, the double longitude of AB is AA'+BB'. 189. Area when Lengths and Bearings of Courses are Given. By projecting the courses of a tract as ABODE A, Fig. 151, on a meridian, we see that to every course there XIII, 190] LAND SURVEYING 223 Fig. 151 corresponds a trapezoid. Then to find the area of ABCDEA we add the trapezoids CC'DD' and DD'EE' corresponding to courses which hear south and from this sum subtract the sum of the trapezoids correspond- ing to courses which hear north. The areas of the one set are called south areas, the others north areas. Again, the douhle area of any of these trap- ezoids as AA'BB' is numerically equal to the product of the double longitude of the corresponding course times its latitude-difference. Thus the double area of AA'BB'={AA'+BB')A'B\ We have, therefore, the following rule. Rule to Find the Double Area of a Tract. Multiply the douhle longitude of each course hy its latitude-difference, plac- ing the product in a column of north areas when the hearing is north, and in a column of south areas when the hearing is south. Add up the two columns and take the difference of the results. 190. Latitude-differences, Longitude-differences, and Double Longitudes. From Fig. 152, KA = AB cos KAB; KB = AB sin KAB, i.e. for any course, we have Latitude-difference = length of the course times cosine of its hearing, Longitude-difference = length of the course times sine of its hearing. From Fig. 152, let AB and BC be any two successive courses. Then the double longitude of BC= BB'+CC=(BB'+AA') -^KB+MC, That is, the double longitude of any course equals the douhle longitude of the preceding course, plus the longitude- difference of that course, plus the longitude-difference of the given course, eastings counted positive and westings negative. Fig. 152 224 MATHEMATICS fXIII, 190 If the reference meridian is taken one chain west of the most westerly corner and the survey begins at that corner, then for the first course the double longitude of AB= AA'+BB', =AA'^B'K+KB, = 1 + 1 + longitude-difference oi AB. Thus, double longitude of the first course equals 2 plus its longitude-difference. // the reference meridian posses through the most westerly corner, then the double longitude of the first course is merely its longitude-difference. 191. Balancing the Survey. For a closed survey we should have northings = southings, eastings = westings. If these are not equal, the difference is called the error of closure. This error, if not too large, may be distributed among the sides in proportion to their respective lengths so that the survey will close. This is called balancing the survey. This and other details of computation and its arrangement will be illustrated in the example, 192. 192. Illustrative Example. The field notes for a tract of land are given in the following table. Plot this data to scale, balance the survey for error of closure, and find the area. Course AB BC CD DE EA Beabing N. 40 E. N. 60 E. S. 20 E. S. 60 W. N. 38i W. Distance 7.18 ch. 9.20 ch. 15.22 ch. 11.42 ch. 12.72 ch. Plot the data to scale and take the reference meridian one chain west of the most westerly corner, i.e. AA'=1 ch. (Fig. 153). XIII, 192] LAND SURVEYING 225 Latitude-differences and lon- gitude-differences must next be computed and balanced. For the first course AB, we have, lat.-diff. = (7.18)cos40 = (7. 18) (.7660) = 5.50 long.-diff. =(7.18) sin 40 = (7.18)(.6428)=4.61 These and the results for the other courses may be tabulated as follows. FiQ. 153 Ck)UR8E Beabimg Distance (Chains) Lat.- DiFF. Long. -DlFP. N. S. E. W. AB N. 40 E. 7.18 5.50 4.61 BC N. 60 E. 9.20 4.60 7.97 CD S. 20 E. 15.22 14.30 5.21 DE S. 60 W. 11.42 5.71 9.89 EA N. 38| W. 12.72 9.99 7.87 55.74 20.09 20.01 17.79 17.76 20.01 17.76 Error .08 .03 The error in latitude-differences is .08, in longitude-differ- ences .03. The survey will be balanced or closed by distribut- ing these errors among the several courses in proportion to their lengths. The perimeter of the tract is 55.74 ch. ; hence .08 error in latitudes for 1 chain = error in longitudes for 1 chain = 55.74 .03 55.74 .0014, .0005 Therefore, for AB = 7.1S chains, the error in the latitude is assumed to be (7.18) (.0014) = .01, and the error in the lon- gitude is assumed to be (7.18) (.0005) = .00 226 MATHEMATICS [XIII, 192 Similar results may be computed and tabulated for the other courses BC, CD, DE, and EA. The northings are too large and each must be diminished by the corresponding corrections; the southings are too small and each must be increased ; similarly for the eastings and westings. On making these corrections the work will appear as in the table on page 227. To find the area, double longitudes are used. These are computed as follows from the rule in 190. long.-diff. of 1st course = 4.61 add 2. double longitude of 1st course = 6.61 long.-diff. of 1st course = 4.61 long.-diff. of 2d course = 7.97 double longitude of 2d course = 19.19 long.-diff. of 2d course = 7.97 long.-diff. of 3d course = 5.20 double longitude of 3d course= 32.36 long.-diff. of 3d course = 5.20 long.-diff. of 4th course = 9.90 double longitude of 4th course = 27.66 long.-diff. of 4th course = 9.90 long.-diff. of 5th course = 7.88 double longitude of 5th course = 9.88 Check: From a figure it is easily seen that the double longitude of the last course is numerically equal to its longi- tude-difference plus 2. This furnishes a check on the ac- curacy of the work. If the reference meridian is made to pass through the most westerly corner, then the double longitude of the last course is equal numerically to its longitude-difference. These values of the double longitudes may now be in- serted in the table and the double areas computed by the rule of 189. The work then takes the following final form. XIII, 192] LAND SURVEYING 227 ^ f (N 00 CO '"f 88 S5 1-H O 1 g < & CO 00 il CO S2 1 ( 1 05 i-H CO 00 (M II .9 ^28 g c3 s |i ,-( OS CO CO 00 CO rH CO CO GO 1 *2?3S5* ^' S8 S 2 < pq OS l^ I> W S OS (M '1< t>^ o OQ (N CO CO t^ 15 ^ ^ lO ^ OS 1 ^ OS l^ 00 00 g OS l> r-l C T-H t^ ,-H CO OS (N S '^ t> o 1-1 a 5 i' CO t^ 1-H 1-H o t ^ lO CO OS j; I?"" TjH OS 8 g^ 2S?3^S ^ 5^ i> OS u:) i-H (N s i tZ; ^OQOJ ^ 1 228 MATHEMATICS PCIII, 192 EXERCISES Plot the following described plats of land to scale, balance the sur- vey for error of closure, and find the area in acres. 1. Course AB BC CD DE EA Bearing N. 21 E. N. 65 E. S. 33 E. S. 65i W. N. 48 W. Distance 57.16 ch. 7.68 ch. 52.20 ch. 50.96 ch. 13.40 ch. Ans. 187.13 A. Course AB BC CD DE EA Bearing S. 23 E. S. 80 E. N. 38 E. N. 42 W. S. 60i W. Distance 48 ch. 52 ch. 46 ch. 62 ch. 64 ch. Ans. 666.36 A. Course AB BC CD DE EA Bearing N. 41 E. S. 70 E. S. 29 E. West N. 49i W. Distance 60 ch. 40 ch. 72 ch. 75 ch. 48 ch. Ans. 32.89 A. 4. Course AB BC CD DE EA Bearing N. 47 E. East S. 32 E. South N. 77 53' W. Distance 964 ch. 822 ch. 644 ch. 523 ch. 1961 ch. Ans. 235.79 A. 5. In the example of 192, find the length and bearing of AC and of AD. 6. Assuming that the measurements in the following description were all correctly made, find the length of the last course and compute the area. Beginning at an iron pin in the center of the township road and at the northeast corner of said tract, thence N. 89 W., 140.8 rods to a stone set in the center of the road, thence S. 29 10' W., 65.7 rods to a large boulder, thence S. 21 18' E., 42.3 rods to an 8" X 8" concrete post set in the ground on the south bank of the creek, thence S. 65 XIII, 194] LAND SURVEYING 229 36' E., 38.6 rods to an iron pin in the center of the Salem-Brookville turnpike, thence N. 59 14' E., 150.6 rods to an iron pin at the cross- roads, thence N. 2 W. to the place of beginning. 7. Balance the survey and compute its area from the following field notes, N. 27 E., 366.0 feet; N. 19 W., 354.0 feet; S. 69 W., 655.5 feet; S. 19 E., 275.3 feet ; S. 71 E., 486.0 feet. Ans. 6.732 A. 193. Locating by Metes and Bounds. The system of describing and locating lands by metes and bounds is used extensively over the world to-day and, quite naturally, was introduced into this country by the early settlers from Europe. To locate a desired area or region by landmarks, some permanent, or well-defined, point of beginning is first established. The boundary lines are then described in direc- tion and distance with reference to this point of beginning and with reference to more or less permanent natural objects such as trees, streams, well-established highways, as well as concrete monuments, stones, iron posts, etc., placed for the purpose. The directions are by magnetic compass and the distances are in terms of the surveyor's units. However, the fact that landmarks decay and change and, that the magnetic direction of the same line changes from year to yeary gives a transitory character to the platting of land in this way. 194. General Plan of Government Surveys. In 1785 practically all of the territory north and west of the Ohio River had been ceded to the United States Government by the withdrawal of state claims. A law passed by Congress May 20, 1785, provided that " The surveyors . . . shall proceed to divide the said territory into townships six miles square, by lines running due north and south and others crossing these at right angles, as near as may be." 230 MATHEMATICS [XIII, 194 Owing to the convergence of the meridians thiy, of course, was a mathematical impossibiUty. The phrase " as near as may be," however, has been broadly interpreted. Ac- cording to the provisions of this act and the acts of May 18, 1796, May 10, 1800, and Feb. 11, 1805, and to the rules of the commissioners of the general land office, a complete system exists, whose main features are as follows. First. Land is to be surveyed or marked out in divisions in the form of squares or rectangles. This has been done in the western states, including, roughly, all of the states of the Union except the original thirteen Colonies. Second. . To mark out the largest squares, north and south lines, called meridians, are first laid off 24 miles apart and marked with cornerstones, trees, or other per- manent objects. East and west lines, called base lines, are then run at right angles to these meridians at intervals of 24 miles. This .would divide the land up into 24-mile squares were it not for the convergence of the meridians towards the poles of the earth. This may be noticed on maps in geographies and in Fig. 154. Each 24-mile tract is then divided into sixteen nearly equal squares called townships by running north-south and east-west lines through the quarter points of the sides of the large tract. Certain meridians called principal meridians are located with great care. They govern the surveys of lands lying along them for considerable distances both toward the east and toward the west. The vertical rows of townships parallel to the principal meridian are called ranges. The first row on the east is called range No. 1 and is written RIE ; the second row east is No. 2, and is written R2E, etc. Similarly, standard base lines are located. The tiers of townships parallel to this base line are numbered north and south of it. A township in the first tier north of the base XIII, 194] LAND SURVEYING 231 line is township No. 1 and is written Tl N ; one in the second tier south is No. 2 and is written T2Sj etc. A township is thus identified by its number and its range from some standard base line and principal meridian. Fig. 154. Base Line and Principal Meridian TIN, Point out the following townships in Fig. 154. R2W; TSN, MW ; T2S, R7W ; T3S, RSE. Third. Townships are required to be divided into smaller squares called sections one mile each way and containing 640 acres, " as near as may be.'* These sections are num- bered as in Fig. 155, except the " first seven ranges " laid 6 5 4 3 2 1 36 30 24 18 12 6 31 32 33 34 35 36 7 8 9 10 11 12 35 29 23 17 11 5 30 29 28 27 26 25 18 17 16 15 14 13 34 28 22 16 10 4 19 20 21 22 23 24 19 20 21 22 23 24 33 27 21 15 9 3 18 17 16 15 14 13 30 29 28 27 26 25 32 26 20 14 8 2 7 8 9 10 11 12 31 32 33 34 35 36 31 25 19 13 7 1 6 5 4 3 2 1 Fig. 155 Fig. 156 Fig. 157 off by Hutchins west of the Pennsylvania line. (See 196.) These are numbered as in Fig. 156. In western Canada the sections are numbered as in Fig. 157. 232 MATHEMATICS [XIII, 195 195. Standard Parallels or Correction Lines. The eastern and western boundaries of townships are, as nearly as may- be, true meridians. Hence, when they are extended north- ward through several tiers of townships their convergence becomes considerable. At latitude 40 the convergence is about 6.7 feet per mile or over 40 feet to each township. To prevent this diminution in the size of townships to the north of the base line, standard parallels are run, along which six-mile measurements are made for a new set of townships. These lines are called correction lines for obvious reasons. The standard parallel or correction lines have been run at varying distances, the present distance being 24 miles. Public roads are usually built on the section lines, and wherever a north and south road crosses a correction line there is a " jog " in the road as shown on Fig. 154. 196. Principal Meridians and Base Lines of the United States. The first public land survey in the United States was started in 1786 by Thomas Hut chins, Geographer of the United States, who, with thirteen assistants, laid off a line from the southwest corner of Pennsylvania due north to a point on the north bank of the Ohio River. The exten- sion of this line north of the Ohio is known as Ellicott's Line. From this point on the north bank of the Ohio River, Hut chins started a line westward as a Base Line. Accord- ing to Congressional direction, he was to lay off meridians at intervals of six miles along this east- west " Geographer's line " and also parallels to this east-west line, every six miles. Each of these six-mile squares was to be divided into thirty-six square miles and each of these into ** quar- ters." The large squares were called " townships " after the New England word ' ' town. ' ' Hutchins and his assistants laid off only forty-two miles of the " Geographer's line," making XIII, 196] LAND SURVEYING 233 seven ranges of townships west of the Pennsylvania state boundary line. They were frightened away by the Indians. The Principal MeridianSj 194, are run from some initial point selected with great care and are located in latitude and longitude by astronomical means. More than thirty- two of these principal meridians have been surveyed at irregular intervals and of varying lengths. Some of them are numbered, while others are named. The first principal meridian (written. First P.M.) is the boundary between Indiana and Ohio ; the second is west of the center of Indiana, extending the entire length of the state ; the third extends the entire length of the state of Illinois and through the center ; the Tallahassee principal meridian, only twenty- three miles long, runs directly through that city. Other principal meridians are the Black Hills, Indian, Louisiana, Mount Diablo, San Bernardino, etc. A Base Line is run at right angles to the principal meridian at some point selected to begin the survey. Base lines are true geographical parallels, and as with the principal merid- ian, they are laid off with great care. The fourth P.M. in western Illinois and Wisconsin has two base lines, one at its southern extremity and the other on the boundary between Illinois and Wisconsin. The fifth P.M. and its base line carries the largest area embraced in any one system. This meridian extends from the mouth of the Arkansas River northward through Ar- kansas, Missouri, and Iowa. Its base line, known some- times as the Little Rock base line, passes a few miles south of that city in Arkansas. From this meridian and its base line all of Arkansas, Missouri, Iowa, North Dakota, and the major portions of Minnesota and South Dakota have been surveyed. This constitutes an area considerably larger than that of France and Great Britain and Ireland com- 234 MATHEMATICS fXIII, 196 bined. The most northern tier of townships from this base line is near the forty-ninth parallel, the boundary between the United States and Canada, and is numbered 163. The most northern township lies more than a thousand miles north of the base from which it was surveyed. There are nineteen tiers south of this Little Rock base line, so that the extreme length of this area is about 1122 miles. The most eastern range from this fifth P.M. is number 17 and its most western 104, making an east and west extent of 726 miles. CHAPTER XIV SIMPLE MACHINES 197. Introduction. Due to wider and more intense com- petition, both with man and nature, the modern farmer cannot expend his strength in developing mere mechanical power which may be created by a horse or a windmill or an engine. He must devote a larger and larger share of his time and energy to mental activity. The result of this tendency is to increase the number and the complexity of his machines. Thus a knowledge of mechanical principles has become a fundamental necessity in order that he may properly handle and care for machinery. 198. Motion. Force. Moment of a Force. Motion is change of position. That which produces motion or changes it (or tends to do so) is called force. Force and motion are transmitted and modified primarily by the use of machines. Simple examples of machines are a crowbar used to raise a heavy weight, a pair of pliers in use, an ordinary lift pump, a pulley, etc. If a force P (Fig. 158) is applied to the end of a lever work- ing over the fulcrum F, it tends to lift the weight W, thereby producing a tendency to rotate about T _._4^ ^i ^ F as an axis. This tendency to rotate tl^^^~" [ is called the moment of the force. ^^^^^--.^^ The perpendicular distance from the Fia. 158 axis .of rotation to the line of action of the force is called the moment arm. The moment of the force is measured by the force times the moment arm : 235 236 MATHEMATICS PCIV, 198 Moment of a Force = ForceXMoment Arm. From Fig. 158, Moment of P about the axis at F= PXa, Moment of W about the axis at F= TFXa' ; and for equihbrium (1) PXa=WXa\ 199. Levers. In every simple machine two forces are involved, viz. the resistance, or the force to be overcome, and the effort, or the force necessary to overcome the resist- ance. The relation between these two forces depends upon the nature and the dimensions of the machine. With levers, resistance and effort are inversely proportional to their distances from the fulcrum, i.e. W\W' = V \l, or Wl= W'V, i.e. Resistance (W) times its moment arm (I) = effort (W) times its moment arm (V). F I A w -l ' > F A B a\ I Z\ W w' ^ w w Fig. 159. Levers, showing Fulcrum and Moment Arm The beam balance, the common steelyard, scissors, pincers, the crowbar, etc. are familiar examples of levers. 200. Wheel and Axle. With the wheel and axle, resistance and effort are inversely proportional to the radii of the cylinder and wheel respectively, i.e. W:W'=r'\r, or Wr= W'r'. The windlass and the capstan are familiar examples of the wheel and axle. Wheels on separate axles, connected XIV, 203] SIMPLE MACHINES 237 by belts or by teeth in the circumference of the wheels, are modifications of the wheel and axle. 201. A General Principle. Work. If, in any simple machine, a body is made to move a distance d directly against a resist- ing force W, by an effort W that moves a distance d^ in the line of action of W\ the product W - d is equal to the product TF' d\ product is called the work done. w w Fig. 160. Wheel AND Axle This 202. The Inclined Plane. With the inclined plane, weight and effort are directly proportional to the length and the height of the incline, i.e. (2) W'l W' = l:h, or Wh= W'l w \A f h w \ FiQ. 161. Inclined Plane 203. The Screw. The screw is a modified form of the inclined plane. The effort is applied at the end of a lever and the length of the inclined plane is the distance traversed by the effort in one revolution, i.e. 2 irl, where I is the lever arm. The height of the incline is the distance d between two successive threads of the screw. This distance, d, is called the pitch of the screw. Here, the weight and effort are directly proportional to the distance traversed by the effort in one revolution, and the pitch of the screw, i.e. (3) W: W' = 2TTl:d, or Wd = 2irW'L 238 MATHEMATICS PCIV, 203 EXERCISES 1. What effort is required to lift a weight of 800 lb. by a lever whose weight arm is 6 in. and whose effort arm is 2 ft. 6 in. ? 2. A pump handle is 3 ft. 8 in. long and works on a pivot 4 in. from the end attached to the pump rod. What force is applied to the pump rod when the end of the handle is pushed down with a force of 10 lb. ? 3. A man wishes to lift a stone weighing 1000 lb. by means of a crowbar 6 ft. long. If he places the fulcrum 10 in. from the end of the bar, what force must be exerted at the end of the bar to raise the stone? In what direction must this force be applied? If the force is applied at a different angle, does it take more or less force? For example, if the bar stands at an angle of 45 when the force is applied, in what direction should the force be applied ? 4. What force is required to lift a load of 275 lb. on a wheelbarrow if the load is placed 18 in. back of the axle of the wheel and the ends of the handles are 5 ft. from the axle ? 5. The crank to the windlass of a well is 18 in. long and the cylinder upon which the rope is wound is 6 in. in diameter. How much force is necessary to lift a bucket of water weighing 48 lb., neglecting friction? How much if the friction is 20% of the force apphed? 6. Two men working at a capstan walk in a circle 8 ft. in diameter and each exerts a force of 60 lb. With each complete turn, 2 ft. of rope is pulled in. What is the pull along the rope, neglecting friction ? 7. A horse walking in a circle 15 ft. in diameter moves a house by means of a capstan 18 in. in diameter. The horse exerts a pull of 1200 lb. What is the resistance of the house if only f of the force exerted by the horse is available ? 8. A stone weighing 756 lb. is lifted into place in a wall by means of a cable wound about an axle of 5 in. radius. To this axle is rigidly attached a toothed wheel of 15 in. radius, and this toothed wheel is driven by another toothed wheel of 4 in. radius. This last wheel is turned by a crank 18 in. long. How much force must be applied to the crank, neglecting friction? How much, allowing 20% friction? 9. If a barrel of cider weighs 200 lb., what force is required to roll it up an incline 8 ft. long into a wagon 3 ft. high? How long would the incline have to be if a man could push only 40 lb. ? XIV, 203] SIMPLE MACHINES 239 10. A stone weighs 850 lb. What force is required to pull it on rollers up an incline 12 ft. long and place it upon a wagon 30 in. high? 11. What force is required to pull a loaded wagon weighing 3800 lb. up a slope 30 ft. high and 105 ft. long, neglecting friction ? 12. Ice is pulled up an incline 50 ft. long 20 ft. high into an ice house. What force, neglecting friction, is necessary to pull up a 200-lb. block? 13. A house that is being raised by jackscrews exerts a pressure of three tons on one of them. If the pitch of the screw is | in. and it is operated by a lever 2 ft. long, what force must be appUed at the end of the lever? 14. If the pitch of a jackscrew is f in. and the lever arm 30 in., what weight may be hfted by it with a force of 100 lb. ? 15. How far will a force of 180 lb. which moves 4 ft. lift a weight of 960 lb., if one third of the applied force is consumed in overcoming friction? Ans. 6 in. 16. What force, moving 6 ft. 8 in., will lift 512 lb. a distance of 1 ft. 3 in., if one third of the applied force is consumed in overcoming friction? Ans. 144 lb. 17. CD is a crowbar 6J ft. long, and is supported at F, 6 in. from C. How many pounds of force must be exerted by a man pressing down at D to raise a stone weighing 1800 lb. at A? To get the maximum of the force exerted at D, in what direction should it be applied ? 18. Suppose, in Ex. 17, that the crowbar stands at an angle of 30 with the horizontal. A man weighing 150 lb. sits at D on the end of the bar. What is the force communicated to the bar at D? How much will it Uft at C? 19. In pulling a nail from a board, as shown in the accompanying figure, find the resistance of the nail in starting if P=401b., D = 10", d = ir. 20. If a gate weighs 50 pounds and is 10' long, what is the pull on the upper hinge if it is 3' above the lower hinge ? [Hint. The center of mass is in the center of the gate. The system of forces then acts as a simple teeter board with the fulcrum on the lower hinge.] Ans. 83^ lb. 240 MATHEMATICS \X1Y, 204 204. The Pulley. The pulley consists of a grooved wheel capable of revolving about an axis, fixed into a framework called the block. When the axis of the pulley is fixed, the pulley is called a fixed pulley. It has no mechanical advantage,* its use being merely to change the direction of the force. If the pulley can ascend or descend, it is called a movable pulley and a mechanical advantage may be gained. In Fig. 163, the weight W is supported by the force P ap- plied to the cord ABDP fixed at A and passing under the pulley. The tension in the cord being everywhere the same, we have (4) P X lever arm CB=WX lever arm OB, But CB = 2 OB. Therefore D w Fig. 163. Sim- ple Movable Pulley (5) 2' or W=2P; and the mechanical advantage is 2, since W/P = 2. 205. System of Movable Pulleys. In Fig. 164 is shown a system of three movable pulleys each hanging from a fixed beam by a separate string. The tension at A in the cord under the lowest pulley is =W/2. The ten- sion at B in the cord under the second pulley is one half of the first tension, i.e. at A. Thus the tension at B is W 2 W 22 Fig. 164. System OF Pulleys * Mechanical advantage is the ratio of the resisting force W to the effort force P, i.e. W _ distance P moves P distance W moves mechanical advantage XIV, 206] SIMPLE MACHINES 241 The tension in the cord at C is one half that of the previous one, etc. to the nth pulley for which the tension in the cord supporting it is 17/2". Thus, for n pulleys arranged as in Fig. 164, we have P = TF/2". 206. Block and Tackle Cord around All Pulleys. In this system there are two blocks, A and B, Fig. 165, the upper of which is fixed and the lower movable, and each containing a number of pulleys or sheaves, each pulley mov- able around the axis of the block in which it lies. A single cord is attached to one of the blocks and passes alternately round the pulleys in the two blocks, the portions of the cord between successive pulleys being parallel. In the tackle shown in Fig. 165, there are three pul- leys in each block. The mechanical advantage of the block and tackle varies directly as the number of cords supporting the weight. For, as the cord passes round all the pulleys, its tension is the same throughout and equal to P. Then, if n be the number of cords at the lower block, nP will be the resultant total upward tension of the cords at the lower block, which must equal W. Therefore (6) nP = W, or w w Fig. 165. Block AND Tackle p=E. In the system of Fig. 165, there are six cords supporting the weight at the lower block. The mechanical advantage is therefore six. If the lower block is raised one foot, the force at P will move through six feet, and a given force at P, say 100 lb., will therefore raise a weight at W six times as great, or 600 lb. 242 MATHEMATICS [XIV, 206 EXERCISES 1. What force is necessary to raise a weight of 450 lb. by an arrange- ment of six pulleys in a block and tackle ? Arrange a block and tackle to lift 500 lb. if only 50 lb. is available as the power. 2. Find the force which will support a weight of 720 lb. with three movable pulleys, arranged as in Fig. 164. If 50 lb. is the available force, how many movable pulleys arranged as in Fig. 164 are necessary to lift 750 lb. ? 1550 lb. ? 3. A man weighing 145 lb. raises a weight of 448 lb. by a system of four movable pulleys arranged as in Fig. 164. What is his pressure on the ground as he raises the weight ? Ans. 11 7 lb . 4. How many movable pulleys are required in a block and tackle if 42 lb. is available with which to raise 336 lb. ? 207. Strength of Materials. In all farm buildings and farm implements the materials used are subjected to certain stresses and strains. The materials may be classified ac- cording to their physical structure under three headings : (1) wood ; (2) metal ; (3) stone, brick, cement, mortar, concrete. The kind of material used, and its design, must meet the demands made upon it without breaking. Two prime con- siderations in the use of materials for machines and buildings are the dze and shape of the material, as when used as a joist or a reaper tongue ; and the manner in which it is loaded. The strength of a beam is measured by the load it can safely withstand. The safe load which a beam will carry varies as each of the following factors. (1) Directly as the working fiber strength of its material. Each kind of material thus is given a rating or a constant. This constant provides also for a reasonable factor of safety. In the table presented below, 208, this factor of safety is six. This means merely that the safe load is one sixth of the breaking load. KIV, 208] SIMPLE MACHINES 243 (2) Directly as the breadth times the square of the depth for beams with rectangular cross-section. Wooden beams are usually of this form. (3) Inversely as the length. Thus we may write (7) safe load = constant (breadth) (depth)^ ^ length 208. Safe Loads for Beams under Bending Stress. There are at least four common ways of loading beams. We present the formulas for safe loads in these four cases, with the following notation. *S = safe load in pounds, d= depth in inches, 6 = breadth in inches, L = span in feet, A = constant depending on the material. The value of A is given in the following table for some com- mon materials. Material Const. A Material Const. A Yellow Pine . . . White Pine . . . Hemlock .... Chestnut .... 100 60 50 55 White Oak . . . Castiron .... Steel Limestone .... 75 167 889 8 Case I. Beam supported at one end and loaded at the other (Fig. 166). (8) bd'A 4 L Case II. Beam supported at one end M and uniformly loaded (Fig. 167). bd^A (9) S = 2L FiQ. 167 244 MATHEMATICS CSIV, 208 Case III. Beam supported at both ends and loaded in the middle (Fig. 168). L (10) Case IV. Beam supported at both ends and uniformly loaded (Fig. 169). (11) s=2Mi. Fig. 168 oooooomonnon Fig. 109 EXERCISES 1. A white oak timber 2 in. in width is to be used to support a hay- fork and carrier at the end of a barn. How deep must the timber be to carry safely a load of 1200 lb. applied 3 feet from the single support? Ans. (i = 9.8" = 10" piece to be used. 2. Compare a 2" XlO" and a 3" X8" timber as to cost and carrying capacity when used as a beam, edgewise. Do the same when used flatwise. 3. A steel beam 1|" wide is to be used to support a hayfork and carrier at the end of a barn. How deep must it be to carry safely a load of 1200 lb. applied 3 ft. from the single support? 4. What should be the maximum distance between supports for a 2" XlO" hemlock beam to safely sustain a middle load of 1000 lb. ? 5. What length of beam could be used in Ex. 4, if the load of 1000 lb. is uniformly distributed? 6. Find the ratio of the safe loads of two beams of wood, one being 10' long, 3"Xl2" in cross-section and having its load in the middle, and the other 8' long, 2" X8" in cross-section with its load uniformly distributed. 7. What should be the breadth and depth of a yellow pine rectangular beam 20' between the two end supports to carry safely a load of 1500 lb. at its center? [Hint. The selection must be governed by commercial sizes, as the dimensions of wooden beams are not fractional.] XIV, 209] SIMPLE MACHINES 245 209. Work. Work, in its scientific sense, is done only by a force in overcoming resistance. The mere push or pull of the force is not work. In order that work may be done, the point of application of the force must move. The work done is equal to the force applied multiplied hy the distance through which it acts. (See 201.) That is, (12) Work= ForceX Distance. To compute the work done multiply the force in pounds by the distance in feet. The product is expressed in foot-pounds. A foot-pound of work is said to be done when a force of one pound moves its point of application through a distance of one foot in the direction of the force. Example. How much work is done when a force of 20 pounds moves its point of application 40 feet in the direction of the force ? A force of 20 pounds acting through 40 feet is the equivalent of 800 pounds acting through 1 foot, i.e. 800 foot-pounds of work is said to be done. EXERCISES 1. How much work is done by a man weighing 150 pounds in climb- ing to the top of a barn 40 feet high? How much is done by a man weighing 120 pounds in walking up a hill 40 feet high? How much by a horse weighing 1200 pounds in walking up a hill 100 feet high? 2. Twenty pounds will pull a certain cart on a level road. How much work is done by a boy in pulling it a mile on this road ? 3. The draft of a certain plow {i.e. the force required to pull it) is 500 lb. How much work is done in plowing a furrow 100 rods long? How much work is done by this same plow in plowing a piece of land 60 rods long and 6 rods wide if the furrow slice is 12 inches wide? 4. What is the draft of a wagon, if 400,000 foot-pounds of work is done by a horse in drawing it a mile? 5. A baseball weighing 5x\ oz. was dropped from one of the windows at the top of Washington's monument and caught 4 ft. from the ground by a professional catcher. Compute the work done by the ball, assum- ing the window to be 545 feet from the ground. 246 MATHEMATICS fXIV, 209 6. A certain gas engine has a 4|-inch bore, a 5|-inch stroke, and makes 378 strokes per minute. How much work is done per minute if the average pressure against the piston head is 15 lb. per square inch? 210. Horse-power. If a man cuts 100 shocks of corn in a day of 10 hours, he does it at the rate of 10 shocks per hour. A horse which covers 21 miles in 3 hours has made the distance at the rate of 7 miles per hour. A belt which moves 50 feet per second has a rate of 3000 feet per minute. A horse which trots a mile in 2 minutes is covering the distance at the rate of 44 feet per second. If a force of 20 pounds acts through a distance of 300 feet for 5 minutes, it does 6000 foot-pounds of work but does it at the rate of 1200 foot-pounds per minute. These illustrations will make clear the following definition. A horse-power is the unit for the rate at which work is done. This unit is 33,000 foot-pounds per minute, i.e. one horse-power means that a body is moved by a force of 33,000 pounds in the direction of the force one foot in one minute. The value of this unit was adopted by James Watt as the result of experiments with strong dray horses in England. He was aware, however, that it is in excess of what can be done by an average horse over a full working day. 211. Working Day for a Horse. Tests show that a steady pull equal to one tenth of the weight of the horse, for a ten-hour daily service, at the walking rate of 2.5 miles per hour, is an average of effective service. One horse-power is developed by a horse pulling with a force of 150 pounds 2| miles per hour, since this means 5280 X2i 60 220 feet per minute. and 150 pounds 220 feet per minute is 33,000 foot-pounds per minute, or one horse-power. XIV, 211] SIMPLE MACHINES 247 However, this power will not be developed by the average horse at the present time, since to draw 150 pounds 2 J miles per hour and 10 hours per day, a horse should weigh from 1400 to 1800 pounds. This is larger than the ordinary horse used in America. The average horse will pull about 120 pounds during the average working day. Hence, the average horse will develop 5280X2^X120 ^ 25,400 foot-pounds per minute 60 = ^ horse-power. EXERCISES 1. A horse draws a load having a draft of 150 pounds at the rate of 3 miles per hour. What horse-power is he developing? At this rate should this horse be required to work full time ? 2. How much power is a horse developing when walking 4 miles per hour and pulling 75 pounds on his traces? 120 pounds? 3. A horse walks 2.5 miles per hour and pulls 120 pounds on his traces. How fast should he walk to develop the same power when the draft is increased to 160 pounds? 4. How much power does a 1600-pound horse develop when walk- ing 2.5 miles per hour if he pulls one tenth of his weight ? A 1200-pound horse? A 1000-pound horse? 5. If a horse weighs 1300 pounds and is pulling one tenth of its weight, how fast should it walk to develop one horse-power? If this horse walks 3 miles per hour, what horse-power is developed ? 6. What should be the weight of a horse which develops 1.3 horse- power in walking 3 miles per hour and pulhng about one tenth of its weight? 7. What power should be provided, allowing 25% loss of power due to friction, to unload hay at the rate of one half ton per minute if raised to a height of 30 feet? 8. How much power is necessary, if f of it is available, to raise corn in an elevator to a height of 50 feet at the rate of 3000 bushels per hour? (1 bushel of corn = 70 pounds.) 248 MATHEMATICS [XIV, 211 9. In filling a water tank 60 feet high, the water is raised at the rate of 200 gallons per minute. What horse-power engine should be provided, granting that only | of the power developed is available, and allowing in addition a 10% margin of safety for extraordinary pulls? 10. The maximum load on a hayfork is 800 pounds. What horse- power is required to raise this load 40 feet in 1 minute? How many horses each weighing 1000 pounds would be required to furnish this power if they walk 3 miles per hour and if only | of the power applied is available? 11. An elevator has a capacity of 1500 pounds. What power is necessary to operate this elevator through 200 feet in 2 minutes ? 12. Tests have shown that for the old-fashioned type of walking plow a furrow slice 8 Xl3 inches requires a draft of 673.3 pounds. Find (1) The horse-power required to draw this plow 2 miles per hour. (2) The weight of each of three horses to pull this plow 2 miles per hour and 10 hours per day. 13. For a modern walking plow, a furrow slice 8 Xl3 inches requires about 400 pounds draft. Find (1) The rate at which the three horses of the previous example may walk in drawing this plow 10 hours per day. (2) The horse-power required to draw this modern plow 2| miles an hour. 212. Draft. Draft is the amount of force required to move an object. The fundamental principle involved in overcoming resistance or in doing work is as follows. The force doing the work when multiplied by the distance through which it moves is equal to the resistance multiplied by the distance through which it moves. (See 201.) That is, in the case of raising a weight against the force of gravity, ForceXForce Distance = WeightX Weight Distance. The amount of increase or decrease of draft up or down a grade does not depend upon the draft on the level but de- pends only upon the weight of the load and upon the slope of the grade. For grades less than 15% the horizontal XIV, 212] SIMPLE MACHINES 249 distance is approximately equal to the actual length along the grade, i.e. the hypotenuse is approximately equal to the base of the right triangle. Example. What force is required to pull a load of 60 pounds along a 10% grade, neglecting ^ friction? R^ A 10% grade means 10 feet rise in 100 feet. The force P moves horizontally 100 feet and the weight rises 10 feet since /2Q = 60 pounds, BC = \0 ^'''' ^'^^ feet, and AB = 100 feet = AC approximately for grades of low per cent. Therefore PX100 = 60X10, or P = 6 pounds. Also, from similar triangles QP^BC . gP^ll QR AB' ' ' 60 100' and therefore QP, parallel to the incline, is equal to 6 pounds. This is the force necessary to keep the object from sliding down the inchne, neglecting friction. But this amounts finally to merely multiplying the load by the grade per cent to obtain the increased or decreased draft up or down a grade. Hence we have the following rule. For draft up a grade, add to the draft on the level a quantity obtained by multiplying the load by the grade per cent. For draft down a grade, subtract from the draft on the level the grade per cent of the total load. Example. The draft of 4000 pounds on a certain level road is 250 poimds. What is the draft on a 3% grade? 3% of 4000 lb. = 120 lb. Therefore 250 lb. + 120 lb. =370 lb., the draft up a 3% grade, and 250 lb. - 120 lb. = 130 lb., the draft dovm a 3% grade. 250 MATHEMATICS \X1Y, 212 EXERCISES 1. A total load of 3700 pounds has a draft of 108 pounds per ton on a certain level road. What is the draft up a 3% grade? Down a 5% grade? Up a 13% grade? Down a 15% grade? 2. A total load of 4500 pounds has a draft of 560 pounds up a 3% grade. What is the draft per ton on the level? What is the draft down a 5% grade for this same load? Up a 12% grade? 3. A load of 2600 pounds has a draft of 155 lb. on a level road. What is the per cent grade down which the draft is zero ? What would be the draft up this grade? How many degrees in the angle of eleva- tion for this grade ? 4. At what rate may a team of 1350-pound horses draw a total load of two tons on a level road, if the total draft is 165 lb. ? 5. What should be the size of a two-horse team to draw a load of 2 tons up a 4% grade 3 mi. per hour, if the draft on the level is 225 lb. ? CHAPTER XV COMPOSITION AND RESOLUTION OF FORCES 213. Graphic Representation of Forces. See 5. A force to be completely described must have a point of ap- plication, a direction, and a magnitude. Since a segment of a line may have all of these, it may be used to represent a force. In Fig. 171, let P be a force applied at and in the direction OP. Lay off on OP sl number of units equal to the magnitude of the force. When y the arrowhead is affixed to indicate the sense in the line of motion, the force is completely determined. For _ example, if a force of 10 lb. is repre- sented on a drawing by 1", then 30 ^^^'- ^'^^ lb. on the same drawing should be represented by 3". 214. Composition of Motions and of Forces. Parallelo- gram of Forces. The resultant of two or more forces is that force which singly will produce the same effect as the forces themselves when acting together. The components of the resultant are the individual forces which would together produce the same effect as the resultant. The composition of motions, or of forces, or of velocities, is merely the process of finding their resultant. The follow- ing cases arise. 251 252 MATHEMATICS [XV, 214 First. // two or more forces act in the same straight line, the resultant force is equal to their algebraic sum, and acts in the same line. If the forces are not all in the same sense, either sense may be selected as positive and the other as negative. Second. // the forces do not act in the same straight line, the parallelogram of forces is used to find their resultant, as follows. When a body is acted upon by two forces, we draw two line-segments, AD and AB, Fig. 172, to represent the forces, as in 213. If we com- plete the parallelogram DABC and draw the diagonal AC, that diagonal represents the resultant of the two given forces both in direction and in magnitude. Velocities may be combined to find a resultant velocity in precisely the same manner. 215. Computation of the Resultant of Two Given Forces. Let the two forces be Fi, F2 and be represented hy AB and AD, respectively (Fig. 173). Complete the par- allelogram DABC, as in 214, and let i2 = AC be the resultant. Let de- note the angle between the forces, and let a de- note the angle between Fi and the resultant, on AB. Then we have (1) BE =F2 cos e, (2) ^C = i^2sin^, I (3) AE=Fi-\-F2C0se, A B Fig. 172. Parallelogram OF Forces Project BC XV, 215] COMPOSITION OF FORCES 253 and (4) R= ylAE'+EC\ (5) tana=^, and R= ^^ AE sin a From these formulas, the resultant R and its direction a may be computed. EXERCISES 1. Three forces of 5 lb., 3 lb., and 2 lb., respectively, act upon a point in the same straight line in one sense, and two other forces of 8 lb. and 9 lb. act in the same line in the opposite sense. With what force and in which direction will the point move? Draw the conditions to the scale 1 lb. =^". 2. Two forces whose magnitudes are as 3 to 4 acting on a point at right angles to each other produce a resultant of 30 lb. What are the component forces ? What is the direction of the resultant ? Ans. 18 lb. and 24 lb. 3. Let ABC be a triangle, and D the middle point of the side BC. If the three forces, represented in magnitude and direction by AB, AC, and AD, act upon the point A, find the direction and magnitude of the resultant. Ans. 3 AD in the direction AD. 4. What single force would be sufficient to counteract a vertical force of 5 pounds and a horizontal force of 12 pounds acting on a body at the same instant? What would be the direction of this resultant? Ans. 13 pounds, 22 37' with the horizontal. 5. Find the magnitude and the direction of the resultant of two forces of 30 pounds and 40 pounds acting on a body at an angle of 60 with each other. ^ Dj -^c [Hint. In the accompanying figure, find ' / ^y"^/^'* BE and EC. Then CA may be found from the y o ^^^^ ^/ ! right triangle AEC] ly^^Kr M ' ^ OQ -^j jn 6. A river is flowing at the rate of 3 mi. per hour. How should a man, rowing a boat 5 miles per hour, point the boat up the river in order that his course may be directly across the river? What would be his rate directly across? What is the tiirm across if the river is 2 mi. wide? 254 MATHEMATICS [XV, 215 7. A man rows a boat across a river at the rate of 3.5 miles per hour. The river flows at the rate of 4.8 miles per hour. Find the speed of the boat. Where will it land along the opposite bank if the river is 2 miles wide? 8. Given Fi = 37.6, ^2 = 58.7, ^ = 58 20' (Fig. 173, 215), find R and a. 9. Given i^i = 37.6, ^2 = 58.7, ^ = 71 40', find R and a. 10. Given Fi = 62.7, ^2 = 46.8, ^ = 90, find R and a. 11. From an auto going 25 mi. per hour a man throws a stone at the rate of 60 ft. per second and at an angle of 45 with the line of motion of the auto. Find the speed and direction of the stone. 12. Two forces of 4 lb. and 3^2 lb. act at an angle of 45, and a third force of V42 lb. acts at right angles to their plane at the same point. Find their resultant. Ans. 10 lb. 216. Resolution of Forces. By the resolution of forces is meant the process of finding the components of given forces in two given directions. We saw in 215 that two forces, Fi and F2, acting at a point are equivalent to a single force, R, the resultant. It is evident then that the single force, R, acting along the diagonal AC of the parallelogram. Fig. 173, can be replaced by the two forces, Fi and F2, represented in magnitude and direction by two adjacent sides of the par- allelogram. The given total force AC may be used as the diagonal for an infinite number of parallelograms. For we can draw through A two lines AD and AB in any given directions. It follows that a single force may be resolved into com- ponents in any two given directions. 217. Rectangular Components of a Force. The most convenient components into which a force can be resolved are those whose directions are at right angles to each other. Thus let OX and OF be any two lines at right angles to XV, 217] COMPOSITION OF FORCES 255 each other and P, Fig. 175, any force acting at in the plane XOY. Completing the rectangle OMPN, we find the components along the two axes : (6) a:-component = OM = P cos a, (7) i/-component = ON = P sin a, where a is the angle which the di- rection of P makes with the x-axis. _ The X- and the ^/-components of a force are called its rectangular ^^- 175. Rectangular Components of a Force components. They are usually the horizontal and the vertical components. See also 5. Unless otherwise expressed, the components of a force will mean its rectangular components on a horizontal line and a vertical line. Velocities may be resolved into component velocities in precisely the same manner. EXERCISES 1. The tension in each of four traces by which two horses are drawing a plow is 50 lb. What are the horizontal and vertical components of the total force moving the plow if the traces make an angle of 14 with the horizontal ? 2. A ball rolls along the diagonal of a 6' Xl8' wagon bed at the rate of 10 feet per second. What is the speed of the ball when the wagon is moving forward at the rate of 12 feet per second? 3. Sleet is falling with a vertical speed of 16 feet per second. What is the velocity of the wind when the sleet falls at an angle of 60 with the horizontal ? What is the wind's velocity when the sleet falls at an angle of 45 with the horizontal ? Ans. 6.3 mi. per hour; 11 mi. per hour, approximately. 4. What must be the tension in a rope to support on a smooth ring a pig weighing 100 lb. between two posts 12 feet apart if the rope is allowed to sag 3 feet at the center? What is the tension when the sag is 2 feet? Ans. 112 lb., 158 lb. 256 MATHEMATICS [XV, 217 5. A weight W lies on a smooth inclined plane which makes an angle cc with the horizontal. Show that the components of W along and perpendicular to the plane are P = TF sin a, R = W cos a, where R is the reaction against the plane and P is the force acting parallel to the plane to keep the weight from slipping. 6. A weight of 10 lb. is held in position on a smooth inclined plane (inclination a) by a force of 2 lb. acting up the plane and by a force acting horizontally on the weight. Determine this horizontal force if sina = f. [Hint. Consider the components of all of the forces in the direction of the plane.] Ans. 5 lb. 7. A farmer sets three posts in the ground so as to form an equi- lateral triangle. A rope is passed around them the tension in which is 6 lb. Find the pressure on the posts. Ans. 6 V3 lb. 8. A barn is equipped with a hayfork, pulleys, etc. The rope lead- ing from the roof pulley to the floor pulley makes an angle of 70 with the horizontal. In drawing up hay the tension in the rope is 800 lb. and the rope beyond the floor pulley makes an angle of 10 with the horizontal. Find the vertical force necessary to hold the floor pulley to its fastening. Use scale diagrams as an approximate check. Consider the entire rope at all times in the same vertical plane. Study the following cases : (a) When the draft end of the rope is to the left of the floor pulley and 10 below the horizontal. (b) When it is to the left of the floor pulley but 10 above the hori- zontal. Fig. 177 XV, 217] COMPOSITION OF FORCES 257 (c) When it is to the right of the floor pulley and 10** above the horizontal. 9. A wind is blowing nearly from the west, as shown in the accom- panying figure, against the sail, AB, of a vessel going due north. What part of this wind force, WC, tends to drive the vessel north ? We may regard the wind force, WC, as the resultant of two forces, DC, perpendicular to the sail and tending to drive the vessel nearly northeast and WD parallel to it. But DC may be resolved into DE and EC, one of which tends to push the vessel sidewise, the other propels it northward. The side thrust is largely counter- acted by the water. 10. Show how the rigging of the vessel in the previous exercise may be shifted so that the same wind will propel the vessel south. Will the component propelling the vessel south be equal to the one propeUing it north? 11. Two equal rafters support a weight W at their upper end. Find the compression in each. Let the length of each ^ rafter be a and the horizontal distance between their lower ends be 2 h, Fig. 179. Let BW represent in direction and magnitude the weight W. From W draw WH parallel to AB and WK parallel to EB. Then BK and BH rep- resent the components of the force BW along the rafters BA and BE ( 216). Draw CE parallel to BL and extend AB to C. From the similar triangles BWK and CEB, BK a .' . r.T^ aW W 2Va2-62 i.e. BK = 2Va2-62 Find also the tension in the tie-rod along b, by taking the component of BK on the line AE. 12. Find the thrust in the equal rafters and the tension in the tie- rod for rafters 10 ft. long in a barn 16 ft. wide due to a total weight of 1001b. What are the thrust and the tension for rafters 9 ft. long in a barn 16 ft. wide? What is the efifect of short rafters? 258 MATHEMATICS [XV, 218 218. Triangle of Forces. The resultant of two forces F] and F2, Fig. 180, may be obtained more easily by drawing AB to represent the magnitude and direc- tion of one force and BC, similarly, for the other force. Then AC represents the re- sultant. The force CA (in direction from Cto A) is opposite in direction but equal in magnitude to the resultant of Fi and F2. // three forces he represented in magnitude and direction by the sides of a triangle, taken in order, they will he in equilibrium. 219. The Simple Jib Crane. For moving heavy weights, a machine called a crane is much used. It consists of a crane post AB, Si tie-rod BC, a jib CA. When loaded with a weight W attached at C, the point C is in equilibrium under the action of the three forces (a) the weight W, (6) the pull in the tie-rod, (c) the thrust in the jib. These form a triangle of forces in which any one of the forces is the resultant of the other two, and they may be repre- sented on a convenient scale by the three sides of a triangle, taken in order. In Fig. 181, CW represents the weight to scale. Draw WD parallel to AC and extend BC to D. Then on the same scale CD n^ represents the pull in the tie- ^ ^ rod and WD the thrust in the jib. The sense of the forces is obtained by following around the triangle in order, from C to IF to i) to C. Fig. 181. The Jib Crank XV, 219] COMPOSITION OF FORCES 259 Example. A weight of 10 tons is raised by a crane whose jib is 30' long and stands at an angle of 45 with the crane post. Find the thrust in the jib if the tie-rod is attached 20' from the bottom of the crane post. Since WCD is similar to ABC, C_ .,^^D 5 WD 10 30 20 ' ^10 Tona Fig. 182 Thus WD = 15 tons, the thrust in the jib. To compute CD we must know BC. This is found from the triangle ABC as follows : Draw BK perpendicular to AC. Then BK = 20 sin 45 = 14.1, AK = 20 cos 45 = 14.1, KC = 30 -14.1 = 15.9. Therefore BC = V(14.1)2 + (15.9)2 = 21.3 Since CD _BC ^21.3 10 20 20 ' it follows that CD = 10.6 tons, the pull in the tie-rod. When the weight is supported by a rope passed over pulleys at C and B and on to a drum on the crane post, the tension in the tie-rod is made less by an amount equal to the tension in the rope which supports the weight. This ten- sion is the weight itself. Graphically, if CW represents the weight and therefore the tension in its rope, CD is the whole tension in both tie-rod and rope. Mark off CH on CD and equal to CW. Then HD must be the tension in the tie-rod in this case. EXERCISES 1. A weight of 12 tons is suspended rigidly at the end of a jib 20 feet long by a tie-rod 15 feet long. Find the thrust in the jib and the tension in the tie-rod if the tie-rod is attached to the crane post. (a) 10 feet from its lower end. Ans. 24 tons ; 18 tons. (b) 13 feet from its lower end. (c) 15 feet from its lower end. (d) 20 feet from its lower end. 260 MATHEMATICS [XV, 219 2. In Ex. 1, find, for each of the different cases, the tension in the tie-rod when the weight is supported by a rope which passes over a pulley at C and is led parallel to the tie-rod over a pulley at B and on to a drum on the crane post. What happens in cases (c) and {d) ? 3. If AB = 12', AC = 8', and TF = 5 tons, supported rigidly at C, Fig. 181, find the thrust in the jib and the tension in the tie-rod when BC = (a) 8 feet; (6) 10 feet; (c) 12 feet; (d) 14 feet. 4. Discuss the previous exercise when the weight is supported by a rope led over a pulley at C, parallel to the tie-rod, and over a pulley atB. 5. The load suspended rigidly from the end of a jib is 5 tons. The compression in the jib is 12 tons and the inclination of the jib to the horizontal is 60. Find the tension in the tie-rod. (-,, 6. Discuss Ex. 5 when the jib is inclined to the horizontal at an angle of (a) 45; (b) 30. 7. A rod AB, 30" long, hinged at A and carry- ing a weight of 150 lb. at B, is held horizontally by a rope BC making an angle of 45 with AB. Find the tension in the rope. Let the segment BD = 150 lb. in the triangle BDE similar to ABC, in the accompanying figure. Then BE on the same scale is the tension in BC. What is DE ? Ans. Tension = 150 V2 lb. 8. Solve Ex. 7 if BC makes an angle of 135 with AB. Also if it makes an angle of 120 with it. [Hint. Let CB represent 150 lb. To find x -50.] 9. A weight of 24 lb. is suspended by two flexible strings one of which is horizontal and the other is inclined at an angle of 30 with the vertical. What is the tension in each string ? lb. \ Fig. 184 -^E Fig. 185 Ans. 8V3 1b.; I6V3 lb. 10. Hay is pulled up at the end of a bam into a mow by fork, pulleys, etc. The top pulley is attached to a beam AB fastened at A to the end of the barn. The beam is held horizontal by a rope attached at B and making an angle of 45 with AB. What is the tension in the rope if the top pulley is fastened to the beam 4' from A and if AB is 6' long, considering a load of 300 lb. on the pulley? XV, 220] COMPOSITION OF FORCES 261 Find also the thrust in the beam against the end of the barn. [Hint. Determine first what weight at B will produce the same moment as 300 lb. at the pulley.] 11. A rod 10" long, hinged at one end, supports a weight of 4 lb. S" from the attached end. The other end is free to move but is held horizontal by a string attached to this end and making an angle of 120 with the rod. Find the tension in the string. Also the pull in the rod. [Hint. Find first the equivalent weight at the end of the rod.] Solve also when the weight of 4 lb. is attached to the rod 6" from the hinged end. 12. A man weighing 200 lb. stands at the apex of a roof whose equal rafters are 10 ft. long on a building 16 ft. wide. Find the weight tend- ing to spread the walls and the thrust in the rafters. What would be the effect of shortening the rafters for the same width of the building ? 13. When a roof, as the one in Ex. 12, is covered uniformly with snow, what part of the weight of the snow tends to spread the walls ? 220. Resultant of Several Forces in a Plane Acting at a Point. First Method. The resultant may be found by repeated application of the triangle of forces. Let T^i, F2, F3, F4 be forces acting upon a particle at 0. The resultant of Fi and F2 is OB, Fig. 18G ; of OB and F3 is OC ; of OC and F^isOD. This process may be continued for any number of forces acting in the plane at a point. Thus OD is in magnitude and direction the resultant of the given forces. But this amounts to omitting the dotted lines and draw- ing AB, in sense and direction, equal to F2, BC similarly 262 MATHEMATICS CXV, 220 equal to F3, and CD similarly equal to F4, and connecting D to 0. This gives rise to a closed polygon called the polygon of forces. The polygon of forces perhaps reminds one of the dia- gram to scale of a closed survey in which the last course which closes the survey would be the resultant. This idea gives rise to a second method of finding the resultant. Second Method. Let forces Fi, F2, F3, F4, etc., in the same plane act on a particle at 0. Refer the forces to a pair of coordinate axes OX and OY. In surveying, these axes are the north-south' line and the east- west line. Resolve each force into its rectangular components with reference to the axes as shown in Fig. 187. For example, OA = Fi cos !, OB = Fi sin ai, etc., for each of the other forces. ^^^- ^^^ // a component acts upward or to the right it is positive; if downward or to the left it is negative. For example, OD and OG are negative. Let us denote by X the algebraic sum of the a:-components, and by Y the sum of the ^/-components. The system of forces is now reduced to a system of two forces at right angles to each other, whose resultant has a magnitude (8) i? = \/r+r. The direction of the resultant, a, is given by (9) Y tan a= X where a is the angle between the positive direction of R and the positive direction of the a;-axis. If X and Y are XV, 220] COMPOSITION OF FORCES 263 plotted properly, the quadrant in which a lies will be definitely determined. Example. Forces of 6, 8, 10 lb. act on a particle at angles of 30, 120, 225, respectively, with the horizontal. Find the magnitude and direction of the resultant. The details for taking the rectangular com- ponents may be arranged as follows. oTTa*'^ -7.1 Fig. 188 x-comp. j/-comp. Forces + - + - 6 5.2 3 8 4 6.9 10 7.1 7.1 Sums 5.2 11.1 9.9 7.1 Hence X=-5.9, 7 =+2.8 Lay off on the diagram X = OL = - 5.9 and F = L72 = 2.8. Then and OR tana = V2:8'+5J9' = 6.5, 2.8 -5.9 = -.4746 Look up in the table tan a' = .4746 and subtract from 180. Thus, a' = 25 23' with the negative end of a:-axis or a = 154 37', the direction of R with the positive x-axis. EXERCISES 1. Find graphically the resultant of the forces in the example of 220 by using the polygon of forces. 2. Forces of 2 lb., 3 lb., 4 lb. act along the sides of an equilateral triangle taken in order. Find the magnitude wid the direction and the line of action of the resultant. Find the resultants of the following systems of forces. Check graphically. 3. 100 lb. acting at an angle of with the horizontal, i.e. (100 lb., 0), and similarly (50 lb., 60), (200 lb., 180). Ans. (86.6 lb., 150). 264 MATHEMATICS PCV, 220 4. (4 lb., 60), (5 lb., 135), (6 lb., 210), (8 lb., 330). 5. (9 lb., 0), (7 lb., 50), (5 lb., 150), (3 lb., 225). 6. (250 lb., 0), (270 lb., 33), (160 lb., 114). Ans. (502 lb., 55). 7. (270 lb., 0), (132 lb., 54), (215 lb., 82), (138 lb., 330). Ans. (558 lb., 27). 8. (35 lb., 0), (24 lb., 132), (54 lb., 239). Ans. (297 lb., 253). 9. (60 lb., 0), (58 lb., 90), (86 lb., 292), (110 lb., 189). Ans. (42.3 lb., 247). 221. Resultant of Two Parallel Forces. In Fig. 189, let APy BQ represent two parallel forces P and Q acting at A and B respectively upon a body in the same direction. The resultant of P and Q is desired. At A and B introduce A N = BN', any two equal but op- posite forces. Since their re- sultant is zero, the resultant of P and Q is in no way disturbed. At A draw AP' the resultant oi AP and A N and similarly at B. Produce P'A and Q^B to meet at C. At C construct the parallelograms of forces exactly equal to those at A and B respectively. Now CN" = CN'" and thus neutralize each other while the components CP'", and CQ"\ equal respectively to P and Q, act in the line CK parallel to the lines of action at A and B. Thus, the re- sultant of P and Q is P+ Q. ^11 c N'" Fig. 189 Moreover, from similar triangles, AP AN CK AK' and BQ _ CK BN' BK' XV, 221] COMPOSITION OF FORCES 265 Dividing these two equations, member for member, and remembering that AN=BN', we have AP^BK^b BQ AK a Therefore, since AP represents the force P, and BQ repre- sents the force Q, it follows that (10) ^ = -, i.e. Pa^Qh. Q a A similar proof may be given for the case of unequal parallel forces acting in opposite directions. See also 198, 214. Both results may be combined as follows. The resultant of two parallel forces, acting in the same or opposite directions at the extremities of a rigid straight line, is parallel to the given forces, equal to their algebraic sum, and divides the given line into segments which are inversely pro- portional to the given forces. EXERCISES 1. A lever 10 ft. long is used to raise a weight of 300 lb. Where should the fulcrum be placed if a man's weight of 150 lb. is available at the other end of the lever? 2. The draft of a plow is 365 lb. It is drawn by two horses hitched to a doubletree of which the end clevis holes are 15 in. and 19 in., re- spectively, from the middle hole. How much does each horse pull? 3. Solve Ex. 2 for a draft of p lb. and end clevis holes a in. and h in. from the middle hole. 4. A horse weighs 1356 lb. of which 795 lb. are on the front feet and 561 lb. on the hind feet. If the distance between the front and hind feet, as measured at their points of contact with the ground, is 45 in., where does the line of gravitation fall between the feet, i.e. where is the line of action of the resultant of these two parallel forces? 5. A horse weighing 1250 lb. sustains 700 lb. on his front feet. The distance between fore and hind feet is 48 in. Where is the line of gravitation ? 266 MATHEMATICS [XV, 221 6. If the line of gravitation for the horse in the previous example falls 25 in. to the rear of the front feet, what is the distance between front and rear feet? 7. Let six parallel forces proportional to the numbers 1, 2, 3, 4, 5, 6 act at the points (-2, 0), (-1, 0), (0, 0), (1, 0), (4, 0), (-3, 0), re- spectively. Find the resultant and its point of application. 8. On a straight rod AP there are suspended 5 weights of 5, 15, 7, 6, 9 lb. at the points A, B, C, D, P, respectively, for which AB = S ft., BC = 6 ft., CD = 5 ft., DP = 4 ft. Find the resultant and its point of application. 9. A scantling 4" x4" Xl6' is supported 1 ft. from its end. What weight will it lift at the short end due to its own weight when horizontal if it weighs 4 lb. per linear foot? Solve also when the scantling stands at an angle of 30 with the horizontal. [Hint. The weight of the scantling acts at its center.] 10. A bar of uniform thickness, 10 ft. long and weighing 150 lb., is supported at its extremities in a horizontal position, and carries a weight of 400 lb. at a point distant 3 ft. from one ^ j^ extremity. Find the pressures on the points of "f support, i.e. at A and at D. ^ ^ 2 P [Hint. Let x = pressure at A, i.e. x lb. would 150 i6. just raise the lever AD. Taking moments about 4oo^z6. D, we have Fio. 190 x(10) ft. = (400)7 ft. + (150)5 ft., i.e. x = 355 lb. pressure at A.] 11. A bridge is 30 ft. long and weighs 15 tons. Find the pressure on the abutments when a wagon weighing 2 tons is one third of the way across. Ans. 8^ tons ; 8f tons. 12. The wheels of a wagon oarry a load of 800 lb. What extra horizontal pull is necessary to draw the load over an obstruction 2" high when the wheels are (a) 48" in diameter, (6) 42" in diameter, (c) 36" in diameter, {d) 30" in diameter? [Hint. The weight acts in the direction OB. Take moments about C, i.e. P XDC = Weight XAC] Ans. (a) 349 lb. XV, 221] COMPOSITION OF FORCES 267 13. A carriage wheel whose weight is W and radius r rests upon a level road. Show that the pull P necessary to draw the wheel over an obstacle, of height h, is 14. What height of obstacle may be cleared by the wheels of a wagon carrying a load of 800 lb. if the wheels are 36" high and an extra pull of 150 lb. is available? Ans. .31" 15. Two horses weighing, respectively, 1100 lb. and 1500 lb. are hitched to a doubletree 38" long. Where should the clevis hole be located so that each horse shall pull in proportion to his weight? 16. Figure 192 represents a doubletree with end holes A and B placed 36 inches apart and the center hole, C, 3 inches in front of the mid-point of AB. If the total draft of the load is 400 lb., what is each horse pulling when the doubletree is at an angle of 60" with the line of draft? [Hint. Compute A //, //K, and X^. Then Lx times AH must equal L2 times HE. Why ?] 17. Solve Ex. 16 when the angle made by the doubletree with the line of draft is 64 33'. 18. Solve Ex. 16, if the middle hole C is 4" in front of the center of AB. 19. Where should C be located so that the forces Li and L2 are equal for any angle the evener makes with the line of draft? Where should it be so that the horse ahead pulls the greater load? APPENDIX TABLE I. Drainage. A fall of 6" to T for each 100' is considered a good grade for farm drainage. Let A =the number of acres to be drained. d = depth in inches of water to be drained off in 24 hr. < = diameter of tile in inches. D = number of cubic feet of water discharged per second for a 1 % grade. X = the fall in inches per 100'. The following table gives D for tiles of different sizes. Tile diameter, t . . 4 6 8 10 12 15 20 Discharge, D . . 0.16 0.49 1.11 2.05 3.40 6.29 13.85 Then A=^DVi2x. a TABLE n. Weights of Produce. (The figures represent minimum weights per bushel.) Barley 48 lb. (in Ore., 46 lb. ; Ala., Ga., Ky., Pa., 47 lb. ; Cal., 50 lb. ; La., 32 lb.) Beans (white) . . . . 60 lb. Bran 201b. Buckwheat 48 lb. (in Cal., 40 lb. ; Ky., 56 lb. ; Ida., N. D., Okl., Ore., S. D., Texas, Wash., 42 lb. ; Kan., Minn., N. C, N. J., O., Tenn., 50 lb.) Clover Seed 60 lb. (in N. J., 64 lb.) Corn (ear) 70 lb., except in Miss., 72 lb. ; in Ohio, 68 lb. ; in Ind. after Dec. 1 and in Ky. after May 1, following the time of husk- ing, it is 68 lb. 270 MATHEMATICS Corn (shelled) .... 56 lb. (in Cal., 52 lb.) Corn (Meal) .... 50 lb. (in Ala., Ark., Ga., lU., Miss., N. C, Tenn., 48 lb.) Flaxseed 56 1b. Grass Seed (blue) ... 44 lb. Grass Seed (Hungarian) 50 lb. Hemp Seed 44 lb. Millet Seed 50 1b. Oats 32 lb. (in Ida., Ore., 36 lb. ; in Md., 26 lb;. in N. J., Va., 30 lb.) Onions 57 lb. Peas 601b. Potatoes (white) ... 60 lb. (in Md., Pa., Va., 56 lb.) Potatoes (sweet) . . . 55 lb. Rye 56 lb. (in Cal., 54 lb. ; in La., 32 lb.) Timothy Seed .... 45 lb. (in Ark., 60 lb. ; in N. D., 42 lb.) Turnips 55 lb. Wheat 60 1b. TABLE in. Pounds of Nitrogen; Phosphoric Acid; and Potash required per acre, as a minimum, to raise the following crops. Double the figures for maximum. Cereal Crops. Barley: 12; 20; 25. Buckwheat: 15; 30; 35. Corn and sorghum : 10 ; 35 ; 30. Oats and rye : 12 ; 20 ; 30. Wheat : 12; 20; 12. Garden Crops. Asparagus : 20 ; 30 ; 35. Cabbage and cauliflower : 40; 70; 90. Celery: 40; 50; 65. Cucumbers, muskmelon, pump- kin, squash, watermelon : 30 ; 50 ; 65. Egg plant : 40 ; 50 ; 90. Lettuce: 40; 50; 75. Onions: 45; 55; 80. Radishes: 15; 35; 45. Spinach: 15; 55; 40. Tomatoes: 25; 35; 40. Grasses. Lawns: 20; 25; 30. Meadow, millet: 15; 30; 35. Pasture: 15; 30; 40. Legumes. Alfalfa, clover : 5 ; 30 ; 40. Beans, peas : 5 ; 30 ; 35. Orchard Crops, Small Fruits. Apples, pears, quinces : 8 ; 30 ; 50. Blackberries : 15 ; 30 ; 40. Cherries, plums : 10 ; 35 ; 45. Currants, gooseberries : 10 ; 25 ; 40. Grapes : 8 ; 30 ; 45. Nursery stock : 10; 25; 30. Peaches: 15; 40; 55. Strawberries: 25; 55; 70. Roots, Tubers. Beets, turnips : 20; 25; 35. Carrots: 15; 35; 40. Horse radish : 15 ; 25 ; 35. Parsnips : 20 ; 55 ; 50. Potatoes : 30 ; 40 ; 65. APPENDIX 271 TABLE rV. Percentage of Nitrogen : Phosphoric Acid : Potash in Common Fertilizing Materials. (Analyses vary considerably. The following figures represent rather high grade material.) Nitrogenous Materials. Dried blood, 13 : : 0. Sodium nitrate, 15 : : 0. Ammonium sulphate, 20 : : 0. In analysis and guarantees, ammonia = about .82 nitrogen. Phosphatic Materials. Acid phosphate (superphosphate, dissolved stone, acid rock), : 16 : 0. Basic slag (Thomas slag, iron phosphate), 0:15:0. Floats (calcium phosphate, phosphate of lime, raw phos- phate, raw rock), : 12 : 0. Acid bone black, : 15 : 0. Combined Nitrogenous and Phosphatic Materials. Acid bone, 2 : 20 : 0. Bone meal, 3 : 24 : 0. Fish scrap (fish guano, fish tankage), 8:7:0. Ground tankage (meat guano, slaughterhouse refuse), 7:9:0. Steamed bone, 2 : 26 : 0. Manure cake, 4:4:0. Potash Materials. Muriate of potash (potassium chloride), : : 50. Kainit, : : 12. Sulphate of potash, : : 45. Combined Phosphatic and Potash Materials. Wood ashes, 0:1:5. Dry ferns, : .37 : 1.8. Combined Nitrogenous, Phosphatic, and Potash Materials. Corn stalks, .5:. 33: 1.67. Guano (Peruvian), 5:18:3. Cottonseed meal (cottonseed cake), 7:2.5:1.5. Farmyard manure, .5:. 5:. 5. Hen manure, dry, 2:2:1. Leaves, .7 : .15 : .3. Oat straw, .7 : .2 : 1.1. Pea straw, 1 : .3 : 1. Sheep manure, 2 : 1.5 : 1.5. TABLE V. Units and Equivalents in Weights and Measures. 1 acre 43,560 sq. ft. = 160 sq. rd. = 10 sq. ch. =40.47 ares (metric) 1 are (metric) 100 sq. meters = .01 hectare (ha.) = 119.6 sq. yd. = .02471 (1/40) acre 1 bag cement 94 cu. ft. =94 lb. 1 barrel 31^ gal. =4| cu. ft. 1 barrel cement 4 bags = 376 lb. =3.76 cu. ft. 1 barrel flour 196 lb. 1 barrel refined oil 42 gal. 1 barrel salt 200 lb. 1 brick 2" X4''' X8" 1 bushel (stroked) 2150.4 cu. in. = li cu. ft. = approximately 35.24 hters 272 MATHEMATICS 1 bushel (heaped) 2688 cu. in. to 2747 cu. in. 1 bushel of coal 80 lb, 1 centimeter (cm.) 3937 in. = 10 miUimeters 1 chain 66 ft. = 100 links (li.) 1 cubit 18 inches 1 cubic centimeter water . . 1 gram (at 4 C.) 1 cu. in 16.387 c.c. leu. ft =1/27 cu. yd. = .037 cu. yd. = .0283 cubic meter = 4/5 bu. =7^ gal. 1 cu. ft. brickwork 18 to 22 bricks 1 cu. ft cement 100 lb. 1 cu. ft. water 62,43 lb. = 1000 oz. approximately degrees Fahrenheit (F) . . . F = fC+32 degrees Centigrade (C) . . C = (F-32)f 1 dram (avoirdupois) . . . . 1/16 oz. 1 fathom 6 feet 1 foot 12 inches = 30.48 cm. 1 foot (board measure) ... a board 1 ft. square and 1 in. (or less) thick 1 furlong 40 rods 1 gallon 4 quarts = 3.7854 liters = 231 cu. in. 1 gallon 20% cream "... 8.4 lb. 1 gallon 22% cream .... 8.339 lb. = 1 gallon water 1 gallon milk (3^%) .... 8.6 lb. 1 gallon skim milk 8.65 lb. 1 gallon water 8,339 lb. 1 grain (avoirdupois) 065 gram 1 gram 15.432 grains = ,03527 oz. 1 gross ........ 12 dozen 1 hand 4 inches 1 hectare (ha.) 100 ares = 2.471 acres 1 hogshead 2bbl.=63gal. 1 inch 2,54 cm, =25.4 mm. 1 kilogram, or kilo (kg.) . . 1000 g. =2.20462 lb. 1 kilometer (km.) 1000 meters = .62 mi. 1 knot 1 nautical mile per hour = 1.15 mi. per hour 1 labor (la-bor') 1,000,000 sq. varas = 173 acres APPENDIX 273 1 league 3 nautical mi. =3.46 mi. 1 league (Spanish unit of area) 25 labors = 6.8 sq. mi. llink 7.92 in. = .66 ft. 1 liter 1000 cc.=244 cu. in. = 1.057 qt. (liquid) = .908 qt. (dry) 1 meter 39.37 in. =3.3 ft. .91 meter 1 yard 1 mil 1/6400 of a circumference = .001 radian, approximately 1 mile 5280 ft. =8 furlongs = 320 rd. = 1.6 kilometers 1 nautical mile 1.153 mile = l minute on equator 1 millimeter 039in. = .lcm. 1 oz. (avoirdupois) . . . . 28.35 grams 1 oz. (Troy) 31.1 grams 1 pace 3 feet 1 penny (English) $.0203 (before 1914) 1 perch = 1 pole 1 rod = 16^ feet 1 perch of stonework .... 240 cu. ft. TT (pi) 3.1416 (See p. 28.) 1 pole (see "perch") 1 pound (avoir.) 7000 grains = 16 oz. = . 4536 kilogram 1 pound English money ... 20 shimngs = $4,866 (before 1914) 1 pound (Troy) 5760 grains = 12 oz. 1 quart (hquid) 95 liter (.9464) 1 quart (dry) 1.1 liter 1 radian 180 -h7r = 57.3 = 1000 mils 1 rod =1 perch = 1 pole = 16^ feet 1 shiUing (EngHsh) .... $.243 (before 1914) 1 square foot 144 square inches 1 square inch 6.5 square centimeters 1 ton 2000 lb. 1 ton of hay 500 cu. ft. (varies from 450 cu. ft. to 550 cu. ft. depending on quality) 1 ton of coal 25 bu. heaped (2688 cu. in.) = 38f cu. ft. 1 vara (va'ra) 33^ in. in Texas 33 in. in Cal. and Mexico 1 yard 3 ft. = .91 meter 274 MATHEMATICS TABLE VI. American Experience MortaKty Table. At Number At Number At Number At Number Age. Surviving. Deaths. Age. Surviving. Deaths. Age. Surviving. Deaths. Age. Surviving. Deaths. 10 100,000 749 35 81,822 732 60 57,917 1,546 85 5,485 1,292 11 99,251 746 36 81,090 737 61 56,371 1,628 86 4,193 1,114 12 98,505 743 37 80,353 742 62 54,743 1,713 87 3,079 933 13 97,762 740 38 79,611 749 63 53,030 1,800 88 2,146 744 14 97,022 737 39 78,862 756 64 51,230 1,889 89 1,402 555 15 96,285 735 40 78,106 765 65 49,341 1,980 90 847 385 16 95,550 732 41 77,341 774 66 47,361 2,070 91 462 246 17 94,818 729 42 76,567 785 67 45,291 2,158 92 216 137 18 94,089 727 43 75,782 797 68 43,133 2,243 93 79 58 19 93,362 725 44 74,985 812 69 40,890 2,321 94 21 18 20 92,637 723 45 74,173 828 70 38,569 2,391 95 3 3 21 91,914 722 46 73,345 848 71 36,178 2,448 22 91,192 721 47 72,497 870 72 33,730 2,487 23 90,471 720 48 71,627 896 73 31,243 2.505 24 89,751 719 49 70,731 927 74 28,738 2,501 25 89,032 718 50 69,804 962 75 26,237 2,476 26 88,314 718 51 68,842 1,001 76 23,761 2,431 27 87,596 718 52 67,841 1,044 77 21,330 2,369 28 86,878 718 53 66,797 1,091 78 18,961 2,291 29 86,160 719 54 65,706 1,143 79 16,670 2,196 30 85,441 720 55 64,563 1,199 80 14,474 2,091 31 84,721 721 56 63,364 1,260 81 12,383 1,964 32 84,000 723 57 62,104 1,325 82 10,419 1,816 33 83,277 726 58 60,779 1,394 83 8,603 1,648 34 82.551 729 59 59,385 1,468 84 6,955 1,470 Certain Convenient Values for n = 1 to n=10 n 1/n Vn Vn ) n! l/n\ LoGio n 1 1.000000 1.00000 1.00000 1 1.0000000 0.000000000 2 0.500000 1.41421 1.25992 2 0.5000000 0.3010299f)6 3 0.333333 1.73205 1.44225 6 0.1G66667 0.477121255 4 0.250000 2.00000 1.58740 24 0.0416667 0.602059991 5 0.200000 2.23607 1.70998 120 0.0083333 0.698970004 6 0.166667 2.44949 1.81712 720 0.0013889 0.778151250 7 0.142857 2.64575 1.91293 5040 0.0001984 0.845098040 8 0.125000 2.82843 2.00000 40320 0.0000248 0.903089987 9 0.111111 3.00000 2.08008 362880 0.0000028 0.954242509 10 0.100000 3.16228 2.15443 3628800 0.0000003 1.000000000 The tables on the following pages are four-place tables of logarithms, trigonometric functions, powers and roots, and compound interest. More extensive tables on all these may be found in The Macmillan Tables, from which these are taken. TABLES 275 TABLE VII. Squares and Cubes, Square Roots, and Cube Roots. No. Squabe Cubs Square KoOT Cube Root No. Square Cube Square Root Cube Root 1 1 1 1.000 1.000 61 2,601 132,651 7.141 3.708 2 4 8 1.414 1.260 52 2,704 140,608 7.211 3.733 3 9 27 1.732 1.442 53 2,809 148,877 7.280 3.750 4 16 64 2.000 1.587 54 2,916 157,464 7.348 3.780 5 25 125 2.236 1.710 55 3,025 1(KJ,375 7.416 3.803 6 36 216 2.449 1.817 56 3,136 175,616 7.483 3.826 7 49 343 2.646 1.913 57 3,249 185,193 7.550 3.849 8 64 512 2.828 2.000 58 3,364 195,112 7.616 3.871 9 81 729 3.000 2.080 59 3,481 205,379 7.681 3.893 10 100 1,000 3.162 2.154 60 3,000 216,000 7.746 3.915 11 121 1,331 3.317 2.224 61 3,721 226,981 7.810 3.936 12 144 1,728 3.464 2.289 62 3,844 238,328 7.874 3.958 13 169 2,197 3.606 2.351 63 3,969 250,047 7.937 3.979 14 196 2,744 3.742 2.410 64 4,096 262,144 8.000 4.000 15 225 3,375 3.873 2.466 65 4,225 274,625 8.062 4.021 16 256 4,096 4.000 2.520 66 4,356 287,496 8.124 4.041 17 289 4,913 4.123 2.571 67 4,489 300,763 8.185 4.062 18 324 6,832 4.243 2.621 68 4,624 314,432 8.246 4.082 19 361 6,859 4.359 2.668 69 4,701 328,509 8.307 4.102 20 400 8,000 4.472 2.714 70 4,tX)0 343,000 8.367 4.121 21 441 9,261 4.583 2.759 71 5,041 357,911 8.426 4.141 22 484 10,648 4.090 2.802 72 5,184 373,248 8.485 4.160 23 529 12,167 4.796 2.844 73 5,329 389,017 8.544 4.179 24 576 13,8^4 4.899 2.884 74 5,476 405,224 8.602 4.198 25 625 15,625 5.000 2.924 75 5,625 421,875 8.660 4.217 26 676 17,576 5.099 2.962 76 6,776 438,976 8.718 4.236 27 729 19,683 5.1% 3.000 77 5,929 456,533 8.775 4.254 28 784 21,952 5.292 3.037 78 6,084 474,552 8.832 4.273 29 841 24,389 5.385 3.072 79 6,241 493,039 8.888 4.291 30 900 27,000 5.477 3.107 80 6,400 512,000 8.944 4.309 31 961 29,791 5.568 3.141 81 6,561 531,441 9.000 4.327 32 1,024 32,768 5.657 3.175 82 6,724 551,368 9.055 4.344 33 1,089 35,937 5.745 3.208 83 6,889 571,787 9.110 4.362 34 1,156 39,304 6.831 3.240 84 7,056 592,704 9.165 4.380 35 1,225 42,875 5.916 3.271 85 7,225 614,125 9.220 4.397 36 1,296 46,656 6.000 3.302 86 7,396 636,056 9.274 4414 37 1,369 50,653 6.083 3.332 87 7,569 658,503 9.327 4.431 38 1,444 54,872 6.164 3.362 88 7,744 681,472 9.381 4.448 39 1,521 59,319 6.245 3.391 89 7,921 704,969 9.434 4.465 40 1,600 64,000 6.325 3.420 90 8,100 729,000 9.487 4.481 41 1,681 68,921 6.403 3.448 91 8,281 753,571 9.539 4.498 42 1,764 74,088 6.481 3.476 92 8,464 778,688 9.592 4.514 43 1,849 79,507 6.557 3.503 93 8,649 804,357 9.644 4.531 44 1,936 85,184 6.633 3.530 94 8,836 830,584 9.695 4.547 45 2,025 91,125 6.708 3.557 95 9,025 857,375 9.747 4.563 46 2,116 97,336 6.782 3.583 96 9,216 884,736 9.798 4.579 47 2,209 103,823 6.856 3.609 97 9,409 912,673 9.849 4.595 48 2,304 110,592 6.928 3.634 98 9,604 941,192 9.899 4.610 49 2,401 117,649 7.000 3.659 99 9,801 970,299 9.950 4.626 50 2,500 125,000 7.071 3.684 100 10,000 1,000,000 10.000 4.642 276 Table VIII Four Place Logarithms N 1 2 3 4|$ 6 7 8 9 12 3 4 5 6 7 8 9 10 0000 0043 0086 0128 0170' 0212 0253 02f>4 0334 0374 4 8 12 17 21 25 29 33 37 11 12 13 14 15 16 17 18 19 0414 0792 1139 1461 1761 2041 2304 2553 2788 0453 0828 1173 1492 1790 2068 2330 2577 2810 0492 0864 1206 1523 1818 2095 2355 2(501 2833 0531 0899 1239 1553 1847 2122 2380 2625 2856 05(59 0934 1271 1584 1875 2148 2405 2648 2878 0607 0969 1303 1614 1903 2175 2430 2672 2900 0645 1004 1335 1644 1931 2201 2455 2695 2923 0682 1038 1367 1673 1959 2227 2480 2718 2945 0719 1072 1399 1703 1987 2253 2504 2742 29(57 0755 1106 1430 1732 2014 2279 2529 2765 2989 4 811 3 7 10 3 6 10 3 6 9 3 6 8 3 5 8 2 5 7 2 5 7 2 4 7 15 19 23 14 17 21 13 16 19 12 15 18 11 14 17 11 13 16 10 12 15 9 12 14 9 11 13 26 30 34 24 28 31 23 26 29 21 24 27 20 22 25 18 21 24 17 20 22 16 19 21 16 18 20 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 2 4 6 8 11 13 1 15 17 19 1 21 1 23 j 24 1 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 3222 3424 3617 3802 3979 4150 4314 4472 4624 4771 3243 3444 3636 3820 3997 4166 4330 4487 4(>3i) 4786 3263 3464 3655 3838 4014 4183 4346 4502 4(;54 4800 3284 3483 3674 3a56 4031 4200 4362 4518 4Wi9 3304 3502 3692 3874 4048 4216 4378 4533 4683 3324 3522 3711 3892 4065 4232 4393 4548 4698 3345 3.541 3729 3909 4082 4249 4409 45(54 4713 3.365 35(50 3747 3927 4099 4265 4425 4579 4728 3385 3.579 3766 3945 4116 4281 4440 4594 4742 3404 3598 3784 3962 4133 4298 4456 4609 4757 2 4 6 2 4 6 2 4 6 2 4 5 2 4 5 2 3 6 2 3 5 2 3 5 13 4 8 1012 8 10 12 7 9 11 7 9 11 7 9 10 7 8 10 6 8 9 6 8 9 6 7 9 14 16 18 14 16 17 13 15 17 12 14 16 12 14 16 11 13 15 11 12 14 11 12 14 10 12 13 4814 4829 4843 4857 4871 4886 4<)00 13 4 6 7 9 1 10 11 13 1 4914 5051 5185 5315 5441 5563 5682 5798 5911 4928 50(>5 5198 5328 5453 5575 5694 58m) 5922 4942 5079 5211 5340 5465 5587 5705 5821 5933 4955 5092 5224 5353 5478 5599 5717 5832 5944 4969 5105 5237 5366 5490 5611 5729 5843 5955 4983 5119 5250 5378 5502 5623 5740 5855 59f56 4997 5132 5263 5391 5514 5635 5752 5866 5977 5011 5145 5276 5403 5527 5647 5763 5877 5988 5024 5159 5289 5416 5539 5658 5775 5888 599<) 5038 5172 5302 5428 5.551 5670 5786 5899 6010 1 3 4 1 3 4 13 4 12 4 12 4 12 4 12 4 1 2 3 1 2 3 5 7 8 5 7 8 5 7 8 5 6 8 5 6 7 5 6 7 5 6 7 5 6 7 4 5 7 10 11 12 91112 9 1112 91011 910 11 81011 8 911 8 9 10 8 910 40 6021 0031 6042 6053 6064 6075 6085 6096 6107 6117 12 3 4 5 6 8 9 10 41 42 43 44 45 46 47 48 49 6128 6232 63a^ 6435 6532 6628 (J721 6812 6902 6138 6243 6345 6444 6542 6637 6730 ()821 ()911 6149 6253 6355 6454 (;551 6646 6739 6830 6920 7007 6160 6263 6365 6464 65()1 6656 6749 683i) 6928 7016 6170 6274 6375 6474 6571 6665 6758 (5848 6937 6180 6284 6385 6484 a580 6675 6767 6857 694(5 6191 6294 6395 6493 6590 (J084 6776 68(56 6955 6201 6,304 6405 6503 6599 6693 67a5 6875 69(W 6212 6314 6415 6513 6609 6702 6794 6884 6972 6222 6325 6425 6522 6618 6712 6803 6893 6981 12 3 12 3 12 3 12 3 12 3 12 3 12 3 12 3 12 3 4 5 6 4 5 6 4 6 6 4 5 6 4 5 6 4 6 6 4 5 6 4 5 6 4 4 5 7 8 9 7 8 9 7 8 9 7 8 9 7 8 9 7 7 8 7 7 8 7 7 8 6 7 8 50 6990 6998 7024 7033 7042 7050 7059 7067 1 2 3 3 4 6 1 6 7 8 51 52 53 54 7076 7160 7243 7324 7084 7168 7251 7332 7093 7177 7259 7340 7101 7185 7267 7348 7110 7193 7275 7.356 7118 7202 7284 7364 7126 7210 7292 7372 7135 7218 7300 7380 7143 7226 7308 7388 71.52 72.35 7316 73^)6 12 3 12 3 12 2 1 2 2 3 4 6 3 4 5 3 4 6 3 4 5 6 7 8 6 7 7 6 6 7 6 6 7 N 1 2 3 4 5 6 7 8 9 12 2 4 5 6 7 8 9 The proportional parts are stated in full for every tenth at the right-hand side. The logarithm of any number of four significant figures can be read directly by add- Table Till Four Place Logarithms 277 N 1 2 3 4 5 6 7 8 9 12 3 4 5 6 7 8 9 55 56 57 58 59 7404 7482 7559 7634 7709 7412 7490 7566 7642 7716 7419 7497 7574 7649 7723 7427 7505 7582 7657 7731 7435 7513 7589 7738 7443 7520 7597 7(572 7745 7451 7528 7(504 7679 7752 7459 7536 7612 7(586 7760 7466 7543 7619 7(5i)4 7767 7474 7551 7627 7701 7774 1 2 2 12 2 1 1 2 112 112 3 4 5 3 4 5 3 4 5 3 4 4 3 4 4 5 6 7 5 6 7 5 6 7 5 6 7 5 6 7 60 61 62 63 64 65 66 67 68 69 7782 7789 7860 7931 8000 8069 8136 8202 8267 8331 8395 7796 7803 7810 7882 7952 8021 8089 8222 8287 8;J51 7818 7825 7832 7839 7846 1 1 2 3 4 4 5 6 6 7853 7924 7993 8062 8129 8195 8261 8325 8388 7868 7938 8007 8075 8142 8209 8274 83;38 8401 7875 7945 8014 8082 8149 8215 8280 8:U4 8407 7889 7959 8028 8096 8162 8228 8293 8357 8420 7896 7;K56 8035 8102 8169 8235 8299 8363 8426 7903 7973 8041 8109 8176 8241 8306 8370 8432 7910 7980 8048 8116 8182 8248 8312 8376 8439 7917 7987 8055 8122 8189 8254 8319 8:^2 8445 112 112 112 1 1 2 112 112 112 112 112 3 3 4 3 3 4 3 3 4 3 3 4 3 3 4 3 3 4 3 3 4 3 3 4 3 3 4 5 6 6 5 5 6 5 6 6 5 5 6 5 5 6 5 5 6 5 5 6 4 5 6 4 5 6 70 W51 8457 M(\3 S470 8476 8482 8488 8494 8500 8.50(5 112 3 3 4 4 5 6 71 72 73 74 75 76 77 78 79 8513 8573 8633 8692 8751 8808 8865 8921 8976 8519 8579 8639 8698 8756 8814 8871 8927 8982 8525 8585 8645 8701 87(52 8820 8876 8932 8987 8531 8591 8651 8710 8768 8825 8882 8938 85)93 8537 8597 86.57 8716 8774 8831 8887 8943 8<)98 a543 8(503 8663 8722 8779 8837 8893 81H!) 9004 a549 8(509 8(569 8727 8785 8842 88W 8954 9009 a555 8(515 8675 8733 8791 8848 8904 89(50 <)015 8561 8(521 8681 8739 8797 8854 8910 896 40 1.5301 .0436 30 .04^6 .6397 .0437 .6401 22.iX)4 .3599 .99f0 .9996 30 1.5272 .0465 40 .04(55 .6677 .0466 .6(582 21.470 .3318 .9989 .9995 20 1.5243 .0495 50 .0494 .6940 .0495 .6945 20.206 .3055 .9988 .9995 10 1.5213 .0524 3 00' .0523 .7188 .0524 .7194 19.081 .2806 .9986 .9994 87 00' 1.5184 .0553 10 .0552 .7423 .0553 ,7429 18.075 .2571 .9985 .i]993 50 1.5155 .0582 20 .0581 .7645 .0582 .7652 17.169 .2348 .9983 .9993 40 1.5126 .0611 30 .0610 .7857 .0612 .78(55 16..350 .2135 .9981 .9992 30 1.5097 .0640 40 .0640 .8059 .0(541 .80(57 15.605 .1933 .9980 .9991 20 1.5068 .0669 50 .0669 .8251 .0070 .8261 14.924 .1739 .9978 .9990 10 1.5039 .0698 4 00' .0(598 .843(3 .0699 .8446 14.301 .1554 .9976 .9989 86 00' 1.5010 .0727 10 .0727 .8613 .0729 .8(524 13.727 .1376 .9974 .9989 50 1.4981 .0756 20 .0756 .8783 .0758 .8795 13.197 .1205 .9971 .9988 40 1.4952 .0785 30 .0785 .8946 .0787 .89(50 12.706 .1040 .9969 .9987 30 1.4923 .0814 40 .0814 .9104 .0816 .9118 12.251 .0882 .9967 .9986 20 1.4893 .0844 50 .0843 .9256 .0846 .9272 11.826 .0728 .9964. .9985 10 1.4864 .0873 5 00' .0872 .9403 .0875 .9420 11.430 .0580 .9962 .9f)83 85 00' 1.4835 .0902 10 .0901 .9545 .0^)04 .9563 11.059 .0437 .9959 .9982 50 1.4806 .0931 20 .0929 .%82 .0934 .9701 10.712 .0299 .9957 .9981 40 1.4777 .09()0 30 .0958 .9816 .0%3 .9836 10.385 .0164 .9954 .9980 30 1.4748 .0989 40 .0987 .9945 .0^)92 .)66 10.078 .0034 .9951 .9979 20 1.4719 .1018 50 .1016 .0070 .1022 .0093 9.7882 .9907 .9948 .9977 10 1.4690 .1047 6 00' .1045 .0192 .1051 .0216 9.5144 .9784 .9945 .9976 84 00' 1.4661 .1076 10 .1074 .0311 .1080 .0336 9.2553 .9(564 .9942 .9975 50 1.4632 .1105 20 .1103 .0426 .1110 .0453 9.0(^98 .9547 .9939 .9973 40 1.4603 .1134 30 .1132 .0539 .1139 .0567 8.7769 .9433 .9936 .9972 30 1.4573 .1164 40 .1161 .0(>48 .11(59 .0(578 8.5555 .9322 .t)932 .9ii71 20 1 .4544 .1193 50 .1190 .0755 .1198 .0786 8.3450 .9214 .9929 .9969 10 1.4515 .1222 7 00' .1219 .0859 .1228 .0891 8.1443 .9109 .9925 .9968 83 00' 1.4486 .1251 10 .1248 .0i>61 .1257 .0995 7.9530 .m)5 .9922 .99(5(5 50 1.4457 .1280 20 .1276 .10(50 .1287 .1096 7.7704 .8904 .9918 .9964 40 1.4428 .1309 30 .1305 .1157 .1317 .1194 7.5958 .8806 .9i)14 .9963 30 1.4399 .1338 40 .1334 .1252 .1346 .1291 7.4287 .8709 .9911 .9961 20 1.4370 .1367 50 .1363 .1345 .1376 .1385 7.2687 .8615 .9907 .9959 10 1.4341 .1396 8 00' .1392 .1436 .1405 .1478 7.1154 .8522 .9903 .9958 82 00' 1.4312 .1425 10 .1421 .1525 .1435 .1569 6.9682 .8431 .9899 .9^)56 50 1.4283 .1454 20 .1449 .1612 .1465 .1658 6.8269 .8342 .9894 .9954 40 1.4254 .1484 30 .1478 .1697 .1495 .1745 6.6912 .8255 .9890 .9952 30 1.4224 .1513 40 .1507 .1781 .1524 .1831 6.5(506 .8169 .9886 .9950 20 1.4195 .1542 50 .1536 .1863 .1554 .1915 6.4348 .8085 .9881 .9948 10 1.4166 .1571 9 00' .1564 .1943 .1584 .1997 6.3138 .8003 .9877 .9946 81 00' 1.4137 Value Logio Value Lo^io Value Logio Value Logio Degrees Radians Cosine Cotangent Tangent Sine Four Place Trigonometric Functions 279 [C liaracterist ics of Logarith ins omitted determine by the usual rule from the val ue] Radians Dkgrees Sine Tangent Cotangent Cosine V^alue Logio Value Lwgio Value L<>?io Value Lopio .1571 9 00' .1564 .m3 .1584 .1997 6..3138 .8003 .9877 .9^)46 81 00' 1.4137 .1600 10 .1593 .2022 .1614 .2078 6.1970 .7922 .9872 .9944 50 1.4108 .1(>29 20 .1622 .2100 .1644 .2158 6.0844 .7842 .98(58 .9942 40 1.4075 .1058 30 .1650 .2176 .1673 .2236 5.9758 .7764 .98(53 .f)940 30 1.4050 .1087 40 .1679 .2251 .1703 .2313 5.8708 .7(587 .9858 .9938 20 1.4021 .1716 50 .1708 .2324 .1733 .2389 5.7694 .7611 .9853 .9936 10 1.3992 .1745 10 00' .1736 .2397 .1763 .2463 5.6713 .75.37 .9848 .9934 80 00' 1.3963 .1774 10 .1765 .2468 .1793 .25.*5(5 5.57(54 .7464 .9843 .9931 50 1.3934 .1804 20 .171^ .2538 .1823 .2609 5.4815 .7391 .9838 .9929 40 1.35)04 .1833 30 .1822 .2606 .1853 .2(580 5.3955 .7320 .9833 .9927 30 1.3875 .1862 40 .1851 .2674 .1883 .27.50 5.3093 .7250 .9827 .9924 20 1.3846 .1891 .50 .1880 .2740 .1914 .2819 5.2257 .7181 .9822 .9922 10 1.3817 .1920 1100' .lf)08 .280(5 .1944 .2887 5.1446 .7113 .9816 .9919 79 00' 1.3788 .1949 10 .1937 .2870 .1974 .2953 5.0(558 .7047 .9811 .9917 .50 1.3759 .1978 20 .UK)5 .29;^ .2004 .3020 4.985)4 .6980 .9805 .9<)14 40 1.3730 .2007 30 .1994 .2997 .2035 .3085 4.9152 .6915 .9799 .9912 .30 1..3701 .2036 40 .2022 .3058 .2065 .3149 4.8430 .6851 .9793 .9909 20 1.3672 .20(>5 50 .2051 .3119 .2095 .3212 4.7729 .6788 .9787 .9907 10 1.3643 .2094 12 00' .2079 .3179 .2126 .3275 4.7046 .6725 .9781 .9904 78 00' 1.3614 .2123 10 .2108 .3238 .2156 .333(5 4.6382 .6664 .9775 .9901 50 1.3584 .2153 20 .2136 .329() .218() .3397 4.5736 .6(503 .9769 .9899 40 1.3555 .2182 30 .21(>4 .3353 .2217 .M.58 4.5107 .6542 .9763 .9896 30 1.3526 .2211 40 .2193 .3410 .2247 .3517 4.4494 .6483 .9757 .9893 20 1..3497 .2240 50 .2221 .34()6 .2278 .3576 4.3897 .6424 .9750 .9890 10 1.3468 .2269 13 00 .2250 .3521 .2309 .3634 4..3315 .6366 .9744 .9887 77 00' 1.3439 .2298 10 .2278 .3575 .2339 .3(591 4.2747 .6309 .9737 .9884 50 1.3410 .2327 20 .2'M) .3629 .2370 .3748 4.2193 .(5252 .9730 .9881 40 1.3381 .23.56 30 .2334 .3()82 .2401 .3804 4.1(553 .619(5 .9724 .9878 30 1.33.52 .2385 40 ,2.363 .37.">4 .2432 .3859 4.1126 .6141 .9717 .9875 20 1.3323 .2414 50 .2391 .3786 .2462 .3914 4.0(511 .(3086 .9710 .9872 10 1.325)4 .2443 14 00' .2419 .3837 .2493 .3^)08 4.0108 .6032 .9703 .9869 76 00' 1..3265 .2473 10 .2447 ..'{887 .2524 .4021 3.9(517 .5979 .9696 .9866 50 1.3235 .2502 20 .2476 .3937 .2555 .4074 3.913(5 .,5926 .%89 .9863 40 1.3206 .2531 30 .2504 .3986 .2586 .4127 3.8(567 .5S73 .9(581 .9859 30 1.3177 .25(K) 40 .2.532 .40:x5 .2617 .4178 3.8208 .5822 .9674 .985(5 20 1.3148 .2589 50 .2560 .4083 .2648 .4230 3.7760 .5770 .9(567 .9853 10 1.3119 .2618 1500' .2588 .41.30 .2(579 .4281 3.7321 .5719 .9659 .9&49 7600' 1.3090 .2647 10 .2()16 .4177 .2711 .4331 3.6891 .5(5(5!) .9(552 .9846 50 1.3061 .2676 20 .2(i41 .4223 .2742 .4381 3.(5470 .5619 .9644 .9843 40 1.3032 .2705 30 .2672 .4269 .2773 .44:30 3 6059 .5570 .9():36 .9839 30 1.3003 .2734 40 .2700 .4314 .2805 .4479 3.565(5 .5521 .9628 .9836 20 1.2974 .2763 50 .2728 ,4359 .2836 .4527 3.52(51 .5473 .9621 .9832 10 1.2945 .2793 16 00' .2756 .4403 .2867 .4575 3.4874 .5425 .9(513 .9828 74 00' 1.2915 .2822 10 .2784 .4447 .289*) .4(522 3.4495 ..5378 .9605 .9825 50 1.2886 .2851 20 .2812 .4491 .2931 .4(5(59 3.4124 ..5331 .959(5 .9821 40 1.28.57 .2880 30 .2.S40 .4533 .2962 .4716 3.3759 .5284 .9588 .9817 30 1.2828 .2909 40 .28(38 .457(> .25)94 .4762 3.3402 .52.38 .9580 .9814 20 1.2799 .2938 50 .28i)6 .4618 .3026 .4808 3.3052 .5192 .9572 .9810 10 1.2770 .2967 17 00' .2924 .4659 .;3057 .48.-13 3.2709 .5147 .9563 .9806 73 00' 1.2741 .2996 10 .2952 .4700 .3(J89 .4898 3.2371 .5102 .9555 .9802 50 1.2712 .3025 20 .2979 .4741 ..3121 .4943 3.2041 .5057 .9546 .9798 40 1.2683 .3054 30 .3007 4781 .3153 .4987 3.1716 .5013 .9537 .9794 30 1.2(354 ..3083 40 .3035 .4821 .3185 .5031 3.1397 .4i)(59 .9528 .975)0 20 1.2(325 .3113 50 .3062 .4861 .3217 .5075 3.1084 .4925 .9520 .978(5 10 1.2595 .3142 18 00' ..3090 .4900 .3249 .5118 3.0777 .4882 .9511 .9782 72 00' 1.2566 Value Logio Value LocTio Value Logio Value Logio Degrees Radians Cos NB Cotangent 'I'ang ENT Sine 280 Four Place Trigonometric Functions [Ctiaractoristics of Logarith ms omitted ietermine by the usual rule from the value] Radians Degkees Sine Tangent Cotangent Cosine Value Logio Value Logio Value Logio Value Logio .8142 18 00' .3090 .4900 .3249 .5118 3.0777 .4882 .9511 .9782 72 00' 1.2566 .8171 10 .3118 .4939 .3281 .5161 3.0475 .4839 .9502 .9778 50 1.2537 .8JU0 20 .3145 .4977 .3314 .5203 3.0178 .4797 .9492 .5)774 40 1.2508 .3229 30 .3173 .5015 .3346 .5245 2.9887 .4755 .9483 .9770 30 1.2479 .3258 40 .3201 .5052 .3378 .5287 2.9600 .4713 .9i74 .9765 20 1.2450 .3287 50 .3228 .5090 .3411 .5329 2.9319 .4671 .9465 .9761 10 1.2421 .3316 19 00' .3256 .5126 .3443 .5370 2.9042 .4630 .9455 .9757 71 00' 1.2392 .3.345 10 .3283 .5163 .3476 .5411 2.8770 .4589 .9446 .9752 50 1.2363 .3374 20 .3311 .5199 .3508 .5451 2.8502 .4549 .9436 .9748 40 1.2334 .3403 30 .3338 .52.35 .3541 .5491 2.8239 .4509 .9426 .9743 30 1.2305 .3432 40 .3365 .5270 .3574 .5531 2.7980 .4469 .9417 .9739 20 1.2275 .3462 50 .3393 .5306 .3607 .5571 2.7725 .4429 .9407 .9734 10 1.2246 .3491 20 00' .3420 .5341 ..3640 .5611 2.7475 .4389 .9397 .9730 70 00' 1.2217 .3520 10 .3448 .5375 .3673 .5650 2.7228 .4350 .9387 .9725 50 1.2188 .3.549 20 .;^75 .5409 .370(5 .5689 2.6985 .4311 .9377 .9721 40 1.2159 .3578 30 .3502 .5443 .3739 .5727 2.6746 .4273 .9367 .9716 30 1.2130 .3607 40 .3529 .5477 .3772 .5766 2.6511 .42M .9356 .9711 20 1.2101 .3636 50 .3557 .5510 .3805 .5804 2.6279 .4196 .9346 .9706 10 1.2072 .3665 21 00' .3584 .5543 .3839 .5842 2.6051 .4158 .9336 .9702 69 00' 1.2043 .3694 10 .3611 .5576 .3872 .5879 2.5826 .4121 .9325 .9697 50 1.2014 .3723 20 .3fi38 .5609 .3906 .5917 2.5605 .4083 .9315 .9692 40 1.1985 .3752 30 .3(565 .5(541 .3939 .5954 2.5386 .4046 .9304 .9687 30 1.1956 .3782 40 .3692 .5()73 .3973 .5991 2.5172 .4009 .9293 .9682 20 1.1926 .3811 50 .3719 .57(J4 .4006 .6028 2.4960 .3972 .9283 .9(577 10 1.1897 .3840 22 00' .3746 .5736 .4040 .6064 2.4751 .3936 .9272 .%72 68 00' 1.1868 .3869 10 .3773 .57(57 .4074 .6100 2.4545 .3900 .9261 .i)667 50 1.1839 .3898 20 .3800 .5798 .4108 .6136 2.4342 .3864 .9250 smi 40 1.1810 .3927 30 .3827 .5828 .4142 .6172 2.4142 .3828 .9239 .9(556 30 1.1781 .3956 40 .3854 .5859 .4176 .6208 2.3945 .3792 .9228 .9651 20 1.1752 .3985 50 .3881 .5889 .4210 .6243 2.3750 .3757 .9216 .9646 10 1.1723 .4014 23 00' .3907 .5919 .4245 .6279 2.3559 .3721 .9205 .9640 67 00' 1.1694 .4043 10 .39,-^ .5948 .4279 .6314 2.3369 .3686 .9194 .9635 50 1.1(565 .4072 20 .35)61 .5978 .4314 .6348 2.3183 .3652 .9182 .9629 40 1.1(536 .4102 30 .3987 .(5007 .4;M8 .6383 2.2998 .3617 .9171 .9624 30 1.160(5 .4131 40 .4014 .6036 .4:383 .(5417 2.2817 .3583 .9159 .9618 20 1.1577 .4160 50 .4041 .6065 .4417 .6452 2.2637 .3548 .9147 .9613 10 1.1548 .4189 24 00' .4067 .6093 .4452 .6486 2.2460 .3514 .9135 .9607 66 00' 1.1519 .4218 10 .4094 .6121 .4487 .6520 2 2286 .3480 .9124 .9602 50 1.1490 .4247 20 .4120 .6149 .4522 .6553 2.2113 .3447 .9112 .9596 40 1.14(51 .4276 30 .4147 .6177 .4557 .6587 2.1943 .3413 .9100 .9590 30 1.1432 .4305 40 .4173 .6205 .4592 .6620 2.1775 .3380 .9088 .9584 20 1.1403 .4334 50 .4200 .6232 .4628 .6654 2.1609 .3346 .9075 .9579 10 1.1374 .4363 25 00' .4226 .6259 .4663 .6687 2.1445 .3313 .9063 .9573 65 00' 1.1345 .4392 10 .4253 .6286 .4699 .6720 2.1283 .3280 .9051 .9567 50 1.1316 .4422 20 .4279 .6313 .4734 .6752 2.1123 .3248 .9038 .9561 40 1.1286 .4451 30 .4305 .6;mo .4770 .6785 2.0965 .3215 .9026 .9555 30 1.1257 .4480 40 .4331 .636(5 .4806 .6817 2.0809 .3183 .9013 .9549 20 1.1228 .4509 50 .4358 .6392 .4841 .6^50 2.0655 .3150 .9001 .9543 10 1.1195) .4538 26 00' .4384 .6418 .4877 .6882 2.0503 .3118 .8988 .9537 64 00' 1.1170 .4567 10 .4410 .6444 .4913 .6914 2.0353 .3086 .8975 .9530 50 1.1141 .4596 20 .4436 .6470 .4950 .(5946 2.0204 .3054 .8962 .9524 40 1.1112 .4625 30 .4462 .6495 .4986 .(5977 2.0057 .3023 .8949 .9518 30 1.1083 .4654 40 .4488 .6521 .5022 .7009 1.9912 .2991 .893(5 .9512 20 1.1054 .4(583 50 .4514 .6546 .5059 .7040 1.9768 .2960 .8923 .9505 10 1.1025 .4712 27 00' .4540 .6570 .5095 .7072 1.9626 .2928 .8910 .&499 63 00' 1.0996 Value Logio Value Lf>Pio Value Logio Value Logio Degrees Radians Cosine Cotangent Tangent Sine Four Place Trigonometric Functions 281 [Characteristics of Logarithms omi tted determine by the usual rule from the val le] Radians Deobees Sine Tangent Cotangent Cosine Value Logio Value Logio Value Logio Value Logio .4712 27 00' .4540 .6570 .5095 .7072 1.9(326 .2928 .8910 .9499 63 00' 1.0996 .4741 10 .45(36 .6595 .5132 .7103 1.9486 .2897 .8897 .9492 50 1.09(56 .4771 20 .4592 .6()20 .51(59 .7134 1.9347 .28(36 .8884 .948(5 40 1.0937 .4800 30 .4617 .(3(344 .5206 .7165 1.9210 .2835 .8870 .9479 30 1.0908 .4829 40 .4643 .m>s .5243 .7196 1.9074 .2804 .8857 .9473 20 1.0879 .4858 50 .4669 .6692 .5280 .7226 1.8940 .2774 .8843 .9466 10 1.0850 .4887 28 00' .4695 .6716 .5317 .72.57 1.8807 .2743 .8829 .9459 62 00' 1.0821 .4916 10 .4720 .()740 .5354 .7287 1.8(576 .2713 .8816 .9453 50 1.0792 .4945 20 .4746 .(3763 .5392 .7317 1.8546 .2683 .8802 .944(5 40 1.07(53 .4974 30 .4772 .0787 ..5430 .7348 1.8418 .2652 .8788 .9439 .'30 1.07.34 .5003 40 .4797 .(3810 .54(37 .7378 1.8291 .2(522 .8774 .9432 20 1.0705 .5032 50 .4823 .(5833 .5505 .7408 1.8165 .2592 .8760 .9425 10 1.0676 .5061 29 00' .4848 .(3856 .5543 .74:38 1.8040 .2.5(52 .8746 .9418 61 00' 1.0647 .5091 10 .4874 .6878 ..5581 .74(57 1.7917 .253:3 .87.32 .9411 50 1.0(517 .5120 20 .48)) .(5901 .5(319 .7497 1.775)6 .2503 .8718 .9404 40 1.0588 .5149 30 .4924 .()923 ..5(558 .752(3 1.7(575 .2474 .8704 .9397 :30 1.0.559 .5178 40 .4f)50 .&m .5(19(3 .75rAi 1.7.5.56 .2444 .8(589 .9390 20 1.05:30 .5207 50 .4975 .69(38 .5735 .7585 1.7437 .2415 .8675 .9383 10 1.0501 .623(3 30 00' liOOO .(5990 .5774 .7614 1.7:321 .2.38(5 .8(560 .9.375 60 00' 1.0472 .52(>5 10 .5025 .7012 .5812 .7(544 1.7205 .2:3,5(5 .8(546 .9:368 50 1.0443 .5294 20 .5050 .70.'3;3 .5851 .7(573 1.7090 .2:527 .8(331 .9:3(31 40 1.0414 .5323 30 .5075 .70.55 .5890 .7701 1.(5977 .2299 .8(316 .9353 30 1.0385 .5352 40 .5100 .707(5 .59;J0 .77:30 1.(58(54 .2270 .8(301 .9.34(5 20 1.0356 .5381 50 .5125 .7097 .5969 .7759 1.6753 .2241 .8587 .9338 10 1.0327 .Mil 31 00' MrrO .7118 .600<) .7788 1.(5643 .2212 .8572 .9331 59 00' 1.0297 .5440 10 .5175 .7139 .(3048 .781(5 1 .(55:34 .2184 .8557 .9323 50 1.02(58 .54()9 20 .5200 .71(50 .(5088 .7845 1.(5426 .2155 .8.542 .9315 40 1.02.39 .5498 30 .5225 .7181 .(5128 .787:5 1.6319 .2127 .8.526 .9:308 30 1.0210 .5527 40 .52.50 .7201 .61(58 .75)02 1.(3212 .2098 .8511 .9:300 20 1.0181 .5556 50 .5275 .7222 .6208 .7930 1.6107 .2070 .8496 .9292 10 1.0152 .5585 32 00' .5299 .7242 .6249 .7958 1.6003 .2042 .8480 .9284 58 00' 1.0123 .5<)14 10 .5324 .72(52 .(5289 .798(5 1.5900 .2014 .84(35 .927(3 50 1.00514 .5643 20 .5:548 .7282 .6.330 .8014 1.5798 .198(5 .8450 .9268 40 1.00(55 .5(i72 30 .5373 .7302 .6371 .8042 1.5(397 .1958 .8434 .92(30 30 1.0036 .5701 40 .5398 .7322 m\2 .8070 1.5,597 .19:30 .8418 .9252 20 1.0007 .5730 50 .5422 .7342 .6453 .8097 1.5497 .1903 .8403 .9244 10 .9977 .5760 33 00' .5446 .73(51 .6494 .8125 1.5399 .1875 .8387 .92.36 57 00' .9948 .5789 10 .5471 .7380 .(3536 .8153 1.5301 .1847 .8371 .9228 50 .95119 .5818 20 .5495 .7400 .(5577 .8180 1.5204 .1820 .8355 .9219 40 .9890 .5847 30 .5519 .7419 .(5(519 .8208 1.5108 .1792 .8339 .9211 30 .9861 .5876 40 .5544 .7438 .(5(5(51 .82:35 1.5013 .1765 .8323 .920:3 20 .9832 .5905 50 .5568 .7457 .(5703 .8263 1.4919 .1737 .8307 .9194 10 .9803 .5934 34 00' .5592 .7476 .6745 .82^)0 1.4826 .1710 .8290 .9186 56 00' .9774 .5963 10 .5616 .7494 .(5787 .8317 1.47:33 .1683 .8274 .9177 50 .9745 .5992 20 .5(m .7513 .68:50 .8344 1.4641 .1(556 .8258 .91(59 40 .9716 .6021 30 .5(3(34 .7.531 .6873 .8:371 1.4550 .1(329 .8241 .91(30 30 .9687 .6050 40 .5(388 .75;-30 .6916 .8398 1.44(50 .1602 .8225 .9151 20 .9(357 .(3080 50 .5712 .7568 .6959 .8425 1.4370 .1575 .8208 .9142 10 .9628 .6109 35 00' .5736 .7586 .7002 .8452 1.4281 .1548 .8192 .9134 55 00' .9599 .6138 10 .57(50 .7(304 .7046 .8479 1.4193 .1521 .8175 .9125 .50 .9570 .6167 20 ..5783 .7(322 .7089 .8506 1.4106 .1494 .8158 .9116 40 .9541 .6196 30 .5807 .7(540 .71:33 .8533 1.4019 .1467 .8141 .9107 30 .9512 .6225 40 .5831 .7(357 .7177 .8559 1.39:34 .1441 .8124 .9098 20 .9483 .6254 50 .5854 .7675 .7221 .8586 1.3848 .1414 .8107 .9089 10 .9454 .6283 36 00' .5878 .7692 .7265 .8613 1.3764 .1387 .8090 .9080 54 00' .9425 Value Log,o Value Lopio Value Logio Value Logio Degbees Uadians Cosine Cotangent Tangent Sine 282 Four Place Trigonometric Functions [Characteristics of Logarithms omitted determine by the usual rule from the val uel Radians Degbeee Sine Tangent Cotangent Cosine Value Logio Value Logio Value Logio Value Logio .6283 36 00' .5878 .7692 .7265 .8613 1.3764 .1387 .8090 .9080 54 00' .9425 .6312 10 .5901 .7710 .7310 .8639 1.3680 .13(51 .8073 .9070 50 .9396 .6341 20 .5925 .7727 .7355 .86(56 1.3597 .13(54 .8056 .90(51 40 .9367 .6370 30 .5948 .7744 .7400 .8692 1.3514 .1308 .8039 .9052 30 .9338 .6400 40 .5972 .7761 .7445 .8718 1.34.32 .1282 .8021 .9042 20 .9308 .6429 50 .5995 .7778 .7490 .8745 1.3351 .1255 .8004 .9033 10 .9279 .6458 37 00' .6018 .7795 .7536 .8771 1.3270 .1229 .7986 .9023 63 00' .9250 .6487 10 .6041 .7811 .7581 .8797 1.3190 .1203 .7i)(59 .9014 50 .9221 .6516 20 .(5065 .7828 .7627 .8824 1.3111 .1176 .7951 .9004 40 .9192 .6545 30 .6088 .7844 .7(573 .8850 1.3032 .1150 .7934 .89<)5 30 .9163 .6574 40 .6111 .7861 .7720 .8876 1.2954 .1124 .7916 .8985 20 .9134 .6(503 50 .6134 .7877 .7766 .8902 1.2876 .1098 .7898 .8975 10 .9105 .6632 38 00' .6157 .7893 .7813 .8928 1.2799 .1072 .7880 .8965 52 00' .9076 .66(51 10 .6180 .7910 .78(50 .8954 1.2723 .1046 .7862 .8955 50 .IKMT .6690 20 .6202 .7926 .7907 .8980 1.2647 .1020 .7844 .8945 40 .9018 .6720 30 .6225 .7941 .79.54 .9006 1.2572 .0994 .7826 .8935 30 .8988 .6749 40 .6248 .7957 .8002 .90,32 1.2497 .0968 .7808 .8925 20 .8959 .6778 50 .6271 .7973 .8050 .9058 1.2423 .0942 .7790 .8915 10 .8930 .6807 39 00' .6293 .7989 .8098 .9084 1.2349 .0916 .7771 .8905 5100' .8901 .68.36 10 .631(5 .8004 .8146 .9110 1.2276 .0890 .77.53 .8895 50 .8872 .68(55 20 .6338 .8020 .8195 .9i;55 1.2203 .08tJ5 .7735 .8884 40 .8843 .6894 30 .(5.3(51 .80;55 .8243 .91()1 1.2131 .0839 .7716 .8874 30 .8814 .6923 4(J .6383 .80.50 .8292 .9187 1.20.59 .0813 .7698 .8864 20 .8785 .6952 50 .6406 .8066 .8342 .9212 1.1983 .0788 .7679 .8853 10 .8756 .6981 40 00' .6428 .8081 .8391 .9238 1.1918 .0762 .7660 .8843 50 00' .8727 .7010 10 .6450 .som .8441 .9264 1.1847 .07:5(5 .7(542 .8832 50 .8698 .7039 20 .6472 .8111 .8491 .9289 1.1778 .0711 .7623 .8821 40 .8668 .7069 30 .(M94 .8125 .8541 .9315 1.1708 .0685 .7604 .8810 30 .8639 .7098 40 .6517 .8140 .8591 .9:341 1.1640 .0659 .7585 .8800 20 .8610 .7127 50 .6539 .8155 .8642 .9366 1.1571 .0634 .7566 .8789 10 .8581 .7156 4100' .6561 .8169 .8(593 .9392 1.1504 .0608 .7547 .8778 49 00' .8552 .7185 10 .6583 .8184 .8744 .0 .8745 30 .8465 .7272 40 .6648 .8227 .88^)9 .94M 1.1237 .0506 .7470 .8733 20 .8436 .7301 50 .6670 .8241 .8952 .9519 1.1171 .0481 .7451 .8722 10 .8407 .7330 42 00' .66<)1 .8255 .9004 .9544 1.1106 .0456 .7431 .8711 48 00' .8378 .7359 10 .6713 .8269 .9057 .9570 1.1041 .0430 .7412 .86{)9 50 .8348 .7389 20 .6734 .8283 .9110 .9595 1.0977 .0405 .7392 .8688 40 .8319 .7418 30 .6756 .8297 .9163 .9(521 1.0913 .0379 .7373 .8676 30 .8290 .7447 40 .6777 .8311 .9217 .9646 1.0850 .0354 .7353 .86(55 20 .8261 .7476 50 .(5799 .8324 .9271 .9671 1.0786 .0329 .7333 .8653 10 .8232 .7505 43 00' .6820 .8338 .9325 .9697 1.0724 .0303 .7314 .8641 47 00' .8203 .7534 10 .(5841 .8351 .9380 .9722 1.0661 .0278 .7294 .8629 50 .8174 .7563 20 .6862 .8365 .9435 .9747 1.059<) .0253 .7274 .8(518 40 .8145 .7592 30 .6884 .8378 .9490 .9772 1.0538 .0228 .7254 .8606 30 .8116 .7621 40 .6^)05 .8391 .9545 .9798 1.0477 .0202 .7234 .8594 20 .8087 .7650 50 .6926 .8405 .9601 .9823 1.0416 .0177 .7214 .8582 10 .8058 .7679 44 00' .6947 .8418 .9657 .9848 1.0.355 .0152 .7193 .8569 46 00' .8029 .7709 10 .6%7 .8431 .9713 .9874 1.0295 .0126 .7173 .8557 50 .79i)9 .7738 20 .6988 .8444 .9770 .98^)9 1.0235 .0101 .7153 .8545 40 .7970 .7767 30 .7009 .8457 .9827 .9924 1.0176 .0076 .7133 .85:32 30 .7941 .7796 40 .7030 .84(i9 .9884 .9949 1.0117 .0051 .7112 .8520 20 .7912 .7825 50 .7050 .8482 .9942 .9975 1.0058 .0025 .7092 .8507 10 .7883 .7854 45 00' .7071 .8495 1.0000 .0000 1.0000 .0000 .7071 .8495 45 00' .7854 Value Logio Value Logio Value Logio Value Logio Degrees Radians Cosine Cotangent Tangent Sine Table X Compound Interest: (l+r) 283 Amount of One Dollar Principal at Compound Interest After n Years n 21 1.9898 2.1911 2.4117 2.6533 3.2071 3.8697 1.5157 1.54(K) 1.5769 1.6084 l.(;40<) 1.6734 1.7069 1.7410 1.7758 1.6796 1.721() 1.7646 1.8087 1.85.39 1.9003 1.9478 1 .9)65 2.04('4 1.8603 1.9161 1.9736 2.0328 2.mm 2.1566 2.2213 2.2879 2..35<;() 2.0594 2.1.315 2.2061 2.2833 2.;3()32 2.44(X) 2.5316 2.6202 2.7119 2.2788 2.3699 2.4(J47 2.5633 2.6()58 2.7725 2.88.34 2.9987 3.1187 2.5202 2.6.337 2.7522 2.8760 3.0054 3.1407 3.2820 3.4297 3.5840 2.7860 2.92.53 3.0715 3.2251 3.3864 3.5557 3.7335 3.9201 4.1161 3.3996 3.6035 3.8197 4.0489 4.2919 4.5494 4.8223 5.1117 5.4184 4.1406 4.4304 4.7405 5.0724 5.4274 5.8074 6.21.39 6.(5488 7.1143 30 1.8114 2.0976 2.4273 2.8068 3.24:u 3.7453 4.3219 5.74^5 7.6123 31 32 33 34 35 36 37 38 39 1 .8476 1.S845 1.9222 1.9607 1.9999 2.0399 2.0807 2.1223 2.1647 2.1500 2.2038 2.2589 2.3153 2.3732 2.4325 2.4933 2.5557 2.6196 2.5001 2.5751 2.6523 2.7319 2.81.39 2.8983 2.9852 3.0748 3.1670 2.90.50 3.0067 3.1119 3.2209 3.3336 3.4503 3.5710 3.6960 3.8254 3.3731 3.5081 3.6484 3.7943 3.9461 4.1039 4.2681 4.4388 4.6164 3.9139 4.0900 4.2740 4.46(4 4.6673 4.8774 5.0969 5.3262 5.5659 5.8164 4.5380 4.7649 5.0032 5.2533 5.51(50 5.7918 6.0814 6.3855 6.7048 6.0881 6.4.5.34 6.8406 7.2510 7.(5861 8.1473 8.6361 9.1543 9.7035 8.1451 8.7153 9.3253 9.9781 10.6766 11.4239 12.22,36 13!0793 13.9948 40 2.2080 2.6851 3.2620 3.9.593 4.8010 7.0400 10.28.57 10.9029 11.5.570 12.2505 12.9855 13.7(546 14.5905 15.4659 16.3939 17.3775 14.9745 41 42 43 44 45 46 47 48 49 60 2.2522 2.2972 2.3432 2.3901 2.4379 2.4866 2.5.363 2.5871 2.6388 2.7522 2.8210 2.8915 2.9638 3.0379 3.1139 3.1917 3.2715 3.353:i 3.3599 3.4607 3.5645 3.6715 3.7816 3.8950 4.0119 4.1323 4.25(>2 4.0978 4.2413 4.3897 4.5433 4.7024 4.8669 5.0,373 5.2136 5.3961 5.5849 4.9931 5.1928 5.4005 5.6165 5.8412 6.0748 6..3178 6.5705 6.8333 7.1067 6.0781 6.3516 6.6374 6.9361 7.2482 7.5744 7.9153 8.2715 8.i437 9.0326 7.3920 7.7616 8.1497 8.5572 8.98.50 9.4343 9.1XXK) 10.4013 10.9213 11.4(574 16.0227 17.1443 18.3444 19.6285 21.0025 22.4726 24.0457 25.7280 27.5299 29.4576 2.6916 3.4371 4.3839 18.4202 284 Table XI Compound Discount: 1/(1 + r)^ Present Value of One Dollar Due at the End of n Years n 2^/0 ^h'/o 3% 31^0 4% 4^ CO - E = 00 0>- = E -a rrt = tr- t- EE =: - CD EE - i 1 io lO :: = ^ CO - -o 1 1 iG- = = -o CO 1 1 = E -- = = = ^ ^: -rp _ _ eo- - - tH = zz. ZI - E E -CO 00 = E <0 - 1 t- - - = E <0 1 E ^ Is 1 = i E E E 5 E i -^ ^ - - - - ^ < CQ O Q J THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO 50 CENTS ON THE FOURTH DAY AND TO $1.00 ON THE SEVENTH DAY OVERDUE. y'^f^M. ^T^ LD 21-100m-7.'39(402s) MM VM li 1,1,1 'nihi,!: ,|i 'lIlilH :1 v>l'ilii iHlj! lihi.'i.lii''-'* iMil 'i I. ;ti'!i|!!i||i!n "i 1 'ffl .11 lipi! '!iiii|li!i|! n I !P P P ii j|t|i||Hm ,n.i 'i iiliriiijij t 1 i-s iM|il!!!lir i m ! i ill lliiliilliliiilil!! ! ipliiiiii! mm li pffli mwm ft mmu 1 pllj mm iliililip Wmm 111 li 'Villllipll ^mmi 1 _ Mil iiii'iiiiiii iiili tl 'I wmm II