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Cleland, M.D., F.R.S., Professor of Anatomy and Physiology, Galway. 15. ZOOLOGY. By E. Ray Lankester, M.A. (Oxon.), London. 16 VEGETABLE ANATOMY AND PHYSIOLOGY. By J. H. Balfour, M.D., Edinburgh University. 17. SYSTEMATIC AND ECONOMIC BOTANY. By J.H. Balfour, M.D., Edinburgh University. 19. METALLURGY. By W. H. Greenwood, A. R.S.M. 2 vols. 20 NAVIGATION. By Henry Evers, LL.D., Plymouth. 21. NAUTICAL ASTRONOMY. By Henry Evers, LL.D. -2^ STE\M AND THE STEAM ENGINE— Land, Marine, and Locomotive. By H. Evers, LL.D., Plymouth. 23. PHYSICAL GEOGRAPHY. By John Young, M.D., Professor of Natural History, Glasgow University. ^ London, Edinburgh, and Harriot HUl Works, Glasgow. 2 ss X (-H ^ 3 ^ • U s 03 6 ^ X <-- ^ 'Id ° s u cC 2 < ^ s >> 2 B ^ w S '2 tS < u >3 f5 tu / CoIliitB' Jlbbiincei) /Smn^ Sm PURE IkTm AKITHMETIC. ALGEBKA, GEOMETRY, PLANE TEISONOMETEY. EDWARD ATKINS, B.Sc. (Lond.), HEAD-MASTER OK ST. MARTIN'S SCIBJNCB SCUOOL, LBICHSTER. LONDOxN' AND GLASGOW: WILLIAM COLLINS, SONS, & COMPANY. 1874. lAil rigkU reserved,] TO THE REVEREND WILLIAM FEY, M.A., HONORARY CANON OF PETERBOROUaH, AND SECRETARY OF THE LEICESTER ARCHIDIACONAL BOARD OF EDUCATION, ^hi0 Moxk t0 (BUtxtlti, AS A TRIBUTE OF RESPECT FROM ONE OF HIS OLD PUPILS. PREFACE. The special object of the present work is to meet the requirements of the Science and Art Department's Examina- tions in the first three stages of Pure Mathematics as set down in the Syllabus of the Science Directory. This will account for the arrangement of the subject matter. I hope, how- ever, that it will be found not unsuitable as a general class- book in Elementary Mathematics. In the Arithmetical Section my object has been to deduce the rules from first principles, avoiding as much as possible algebraical considerations. The Geometry consists simply of the first three books of Euclid's Elements with exercises, the only point calling for remark being the marginal notes. I have found it useful in class teaching to put down upon a blackboard the chief steps of the proposition — just those points, in fact, which it is necessary to retain in the memory ; and to encourage the pupil to depend upon himself for supplying the connecting links. This skeleton, as it were, of the proposition is placed in the margin, with the hope that it will be specially appreciated by many students of the industrial classes to whom the language of Euclid is ordinarily an insuperable barrier. Particular reference is here made to such of those classes as have grown up to almost manhood without any mathematical training. In the Algebraical Section of Stage I., the more difficult examples are set particularly for the exercise of thosa 6 PREFACE. students who take up the work of Stage II. without having previously used this book for Stage I. The amount of Plane Trigonometry included in this volume is small, extending only to the solution of triangles and the simpler cases of heights and distances. There is, however, sufficient to fully cover the requirements of Stage II. of the Government Syllabus. The treatment of the higher parts of Algebra and Plane Trigonometry, as well as that of Spherical Trigonometry, is reserved for a second volume. E. A. Leicester, November^ 1S73* CONTENTS. STAGE I. sectio:n' I.— akithmetic. Chap. I. — The Fundamental Principles and Rules Ap- plied TO Whole Numbers and Decimal Frac- tions, 9 „ II. — The Treatment of Fractions considered as Ratios, 16 ,j III. — Application to Concrete Quantities, . . 42 „ IV. — The Metric System, 49 „ V. — Proportion, c 61 „ VI. — Application to Ordinary Questions of Com- merce AND Trade, QS Miscellaneous Examples^ • • 85 SECTION II.— GEOMETRY. Euclid's Elements, Book I., 89 SECTION III.— ALGEBRA, Chap. I. — Elementary Principles, 141 ,, II. — Addition, Subtraction, Multiplication, and Division, 155 ,, III. — Involution and Evolution, .... 178 „ IV. — Greatest Common Measure, AND Least Common Multiple, 195 „ v.— Fractions 204 ,- VI.— Simple Equations and Problems Producing them, 209 CONTENTS. STAGE II. SECTION I.— GEOMETRY. Euclid's Elements, Book II., 23 1 Euclid's Elements, Book III., 254 SECTION II.— ALGEBRA. Chap. I. — Quadratic Equations, 296 „ IT. — Problems Producing Quadratic Equations, . 312 „ III. — Indices, , 316 „ IV.— Surds, 319 ,, V. — Ratio and Proportion, 327 SECTION III.— PLANE TRIGONOMETRY. Chap. I. — Modes of Measuring Angles by Degrees and Grades, 333 „ II. — The Goniometric Functions, .... 336 „ HI. — Contrariety of Signs — Changes of Magnitude AND Sign of the Trigonometrical Ratios through the Four Quadrants, ... . 342 „ IV. — Trigonometrical Ratios — continued. — Arith- metical Values of the Trigonometrical Ratios of 30°, 45°, 60°, &c., . . . 346 „ V. — Logarithms, 354 „ VI.— The Use of Tables, 358 ,, VII. — Properties of Triangles, .... 366 ,,VIII. — Solution of Right-Angled Triangles, . 374 „ IX. — Solution of Oblique- Angled Triangles, . 377 ■ „ X. — Heights and Distances, .... 381 Answers, 387 MATHEMATICS. FIRST STAGE. SECTION I. A.EITHMETIC. CHAPTER I. THE FUNDAMENTAL PRINCIPLES AND RULES APPLIED TO WHOLE NUMBERS AND DECIMAL FRACTIONS. Notation and Numeration. 1. We learn from elementary books on Aritlimetic, tliat figures have a local as well as an intrinsic value, and that the local value of a figure increases tenfold, or diminishes tenfold, according as its position is changed from right to left, or from left to right. Thus, commencing with the right hand figure of an ordinary number, the respective figures of the number stand for units, tens, hundreds, thousands, &c. ; or, beginning with the left hand figure, which, we will suppose, stands for thousands, the respective figures repre- sent thousands, hundreds, tens, units. Let us carry this principle a little further. Take the figures 68754, and suppose that 7 represents 7 units; the question then arises as to the number represented by 68754. Now, as 7 is the units' figure, we have evidently, by the above principle, 6 hundreds, 8 tens, 7 units; and further, remembering that the 10 ARITHMETIC. local value of a figure decreases tenfold for every remove to the right, the 5, on our supposition, must represent 5 tenths, and the 4 must represent 4 hundredths. Let us, as is usual in numbers thus represented, mark the units' figure by- placing a dot to the right of it. Thus, 35 7 "2 605 will then represent 3 hundreds, 5 tens, 7 units, 2 tenths, 6 hundredths, 5 ten-thousandths; and to take one other example, '3065, where the units' figure, though not expressed, is actually 0, will represent 3 tenths, 6 thousandths, 5 ten-thousandths. The dot is called the decimal point, and the digits to the right are called decimals, because they represent portions of the unit obtained by cutting it up into a number of equal parts, which is always some power of 10. It may be remarked, that 10 is called the first power of 10; 100, or 10 X 10, the second power, sometimes written 10^; 1000, or 10 X 10 X 10, the third power, written 10^, and so on. To make the subject clear, let us see what the decimals, •237, -2370, -0237 respectively represent. Now, the digits 2, 3, 7, in the first two decimals, are in exactly the same position with regard to the decimal point, and the respective digits in each have the same absolute value; moreover, the cipher affixed to the right of the decimal -2370, has no intrinsic value, and hence the two decimals, -237, and -2370, have the same absolute value. And since the reasoning is the same, no matter how many ciphers are affixed to the right, we get the following important principle : — The value of a decimal is not altered by affixing ciphers to the right. We will now compare the first and third of our examples, namely, the decimals -237 and -0237. The cipher which is here prefixed to the left, has again no intrinsic value ; but it has removed the digits 2, 3, 7, one stage to the right, and has, therefore, diminished their local value tenfold. The effect of prefixing the cipher, is therefore to diminish the absolute value of the decimal tenfold, and as every additional cipher so prefixed has a similar effect, we get another funda- mental principle, as follows : — The value of a decimal is diminished tenfold for every cipher prefixed. ADDITION AND SUBTRACTION. 11 After the above it is easy to see that we have only to remove the decimal point one place to the right or left, in order to increase or diminish respectively the value of a decimal tenfold. And if we allow the term decimal to in- clude numbers which are greater than unity, as 35*721, we may extend the principle thus: — A decimal may be divided by 10, by removing the dot one place to the left; by 100 or 10^ by removing the dot two places to the left; by 1000 or 10^ by removing the dot three places to the left, and so on ; and further, a decimal may be multiplied by 10, 100, 1000, &c., that is, by 10, 10^, 10^ &c., by removing the dot 1, 2, 3, &c., places respectively to the right. Thus, 6872-3476 divided by 10, 100, 1000 respec- tively, becomes 687*23476, 68*723476, 6*8723476; and mul- tiplied by the same becomes 68723*476, 687234*76, 6872347*6 respectively. Addition and Subtraction. 2. If the student has understood the preceding article, he will at once perceive that, provided we keep the units* figure under the units' figure in every case, there is no difference between the addition and subtraction of decimals, and the addition and subtraction of ordinary integers. All he has to take care of is that the decimal points are kept under each other. Ex. 1.— Add together 325*02, -647, 5*6073, -00214, 290, and 4*7001. Proceeding as in ordinary addition : 325*02 •647 5*6073 •00214 290^ _47001_ 625*97654 Ex. 2.— Take 6*291 from 18*3064. Proceeding as in ordinary subtraction: 18*3064 6*291 120154 12 ARITHMETIC. Ex. 3.— Find the difference between 15-02 and -6732. 15-02 •673 2 14-3468 Note. — In an example of this kind, where the number of decimal figures in the lower line exceeds the number in the upper, it is ad- visable to mentally supply ciphers to make up the deficiency in the upper line. This may be done, as we have seen, without altering the value of the upper line. Multiplication. 3. Suppose we have to multiply 2-935 by 6*34, and let us suppose the dot in eacb case removed to the extreme right. Then (Art. 1), we have multiplied the number 2*935 by 1000. and the number 6-34 by 100, and we have obtained the numbers 2935* and 634* respectively. As these numbers are integers, we may omit the dot, and write them 2935 and 634. Now 2935 x 634 = 1860790, but as we increased our original numbers one thousand and one hundredfold respec- tively, it is evident that our product is increased 1000 x 100, or one hundred thousandfold. Dividing, therefore, the above result, 1860790 by 100000, or what is the same thing (Art. 1), writing it 1860790* and removing the dot 5 places to the left, we get for our product of the numbers 2 '935 and 6-34 the result, 18-60790. We may remark that the number of decimal figures in the product, namely, 5, is the sum of the numbers of decimal figures in the two given numbers. We have, therefore, the following rule for multiplication : — Multiply the given numbers exactly as integers, regardless of the decimal points, and after the operation is finished, point off as many decimal figures in the product as there are together in the multiplier and multiplicand. Ex. 1.— Multiply 6-35 by -1703. •1703 6-35 8"5T5 5109 10218 1081405 DIVISION. 13 Now, the number of decimal figures in the multiplier and multiplicand together, is (4 + 2), or 6, and therefore we mark off 6 decimal figures in our product. This gives us 1-081405. Ex. 2.— Multiply -0063, by -017. •0063 •017 441 1071 And pointing off (4 + 3), or 7 decimal figures, we obtain for our product -0001071. Division. 4. Suppose we have to divide -76875 by 6-25. We will proceed as in the case of multiplication, by imagining the decimal points in each number removed to the extreme right. The numbers will then be 76875-, and 625*, or, omitting the dot, as they are now integers, they will be 76875, and 625. Proceed now as in ordinary division (which operation it is unnecessary to explain), and we get for our quotient 123. Now we must remember that we have increased our divi- dend 100,000 or 10^-fold, and that, consequently, our quotient will require to be diminished 10^-fold. This is done (Art. 1) by removing the dot of the number 123* five places to the left. But, before doing that, we know that the divisor has been increased 100 or 10^-fold, and on this account our quotient must be increased 10^-fold. This is done (Art. 1) by removing the dot two places to the right. Hence, to get the true quotient of -76875 by 6-25, we must remove the dot from the extreme right of the number 123*, five minus two, or three places to the left. Now three is the excess of the number of decimal figures in the dividend over the number in the divisor. We hence arrive at the following rule : — Proceed as in ordinary division, and when all the figures of the dividend have been brought down, and the remainder, if any, obtained^c cut off as many decimal figures in the 14 ARITHMETIC. quotient, as the number of decimal figures in the dividend exceeds the number in the divisor. Note. — ^When the number of decimal figures in the dividend is less than the number in the divisor, affix a sufficient number of ciphers to make the number of decimals in the dividend equal to the number in the divisor. After finishing the operation of ordinary division, there will be no decimal figures to cut off in the quotient. If there be a remainder, and the division carried on further, by affixing ciphers to the successive remainders, all the quotient figures thus obtained will be decimals. Ex. 1.— Divide 117-85088 by 6-272. 6-272)117-85088(1879 6272 55130 50176 49548 43904 56448 56448 We see that there are five decimal figures in the dividend, and three in the divisor, and so we cut ofiT (5 - 3) or 2 in the quotient. The answer is therefore 18-79. Ex. 2.— Divide 527-2 by -0008. Here it will be necessary to affix three ciphers to the dividend, and the operation will stand thus — '0008) 527-2000 659000 As there is no remainder, and the number of decimal figures in the dividend is equal to that in the divisor, we have none to cut oflf. The answer is therefore 659000. Ex. 3.— Divide 463-7 by 2-769 to four places of decimals. Here we must aflix two ciphers to the dividend, and the operation, as far as the ordinary remainder of long division, stands thus :— DIVISION. ] 5 2-769)463-700(167 2769 18680 16614 •20660 19383_ 1277 The quotient up to this point is integral ; but, as we have a remainder, we must continue the operation of division, first placing a dot at the right of the figures in the quotient, and affixing a cipher to the present and each successive re- mainder, until we have the requisite number of decimals in the quotient. Bj thus proceeding, it is easy to see we arrive at an answer — 167*4611. Ex. I. 1. Increase the numbers 4*523, 29, -02367, '07 respectively 10, 100, 1,000, 10,000-fold. 2. Divide by inspection the numbers 0*05, 1111, 4*0020, 45 respectively by 100, 10,000, 1,000, 10. 3. Express in words -3467, 34*67, -0003467, 3-467 ; and compare the values of the last three with the first. 4. Add together — (1.) 6*732, 14-9, -0064, 14*27006. (2.) -00291, -291, 29, 29100*9. (3.) -821, 29*60, 29*6, -0029. 5. By how much does 5 exceed 4-2763, and 16*021 exceed 12*70009] 6. Find the value of — (1.) 74*25 + -0067 - 3*0298 + 1-032 - 2-73. (2.) 3-276 - 8*2409 + 100326 - -00091. (3.) 2-5 - -00029 - 7-364 + 5-2791. 7. "What number added to four thousandths will give three n'lndredths, and what number subtracted from 8,000 units will give 291 units 29 hundredths? 16 ARITHMETIC. 8. Find value of — (1.) -3 X -3. (2.) 9-001 X 27-06. (3.) 0-403 x -009. (4.) -17 X -017 X 100. (5.) -3 X -005 x 6-4. (6.) (-4)2 X (-032)1 9. Find the quotient of — (1.) 79-4 by -397. (2.) 5-928 by 4742-4. (3.) 28 by -007. (4.) -6426 by 2-8. (5.) (-24)2 by 9-6. (6.) 1-806 by (1-9)1 10. Given the quotient -00073, the dividend 124-1, find the divisor when there is no remainder. 11. What is the value of — (1.) j2 - -815 j -r ) -201 + -039 - -002;] (2.) ) (-693)2 - (-307)2} ^ j-693 - -307 }1 12. If I add -061 to a certain number, and then divide the result by 290, I get "0009 for a quotient ; what is the number 1 CHAPTER II. THE TREATMENT OF FRACTIONS CONSIDERED AS RATIOS. 5. A Fraction is a part or parts of a whole. It is gener- ally expressed by two numbers, the one placed above the other and separated by a line. The lower number expresses the number of equal parts into which the whole quantity has been divided, and the upper number, how many of those parts are taken. Thus | is a fraction, and tells us that unity has been divided into 5 equal parts, and that we have taken 3 of those parts. The fraction | is read three fifths ; each of the equal parts into which unity has been divided being called a fifth. The denominator of a fraction is the lower number, and therefore shows the number of equal parts into which we have divided the unit. The numerator is the upper number, and tells us how many of these equal parts are taken. When the numerator is less than the denominator, the quantity expressed is actually less than a whole. The quantity is therefore a real or proper fraction. Again, when TREAT31EXT OF FRACTIONS CONSIDERED AS RATIOS. 17 tlie numerator and denominator are both integral numbers, the fraction is termed a simple fraction. Thus f , ly 1^, are both proper and simple fractions. It is however usual to include in the term fraction every expression which contains one or more simple fractions, with or without integral numbers. Thus I, I, V, 6i, 1, p, 4^ of A of^\, are all included in tlie term fraction. They are, moreover, called vulgar fractions to distinguish them from decimals, which, as will be shown further on, may be looked upon as fractions, according to the above definition, whose denominators are powers of 10, and not expressed but understood. It is convenient to classify fractions as follows : — (1.) A proper fraction is one whose numerator is less than 2l its denominator, as f, |, |f, — . (2.) An improper fraction is one whose numerator is not 5i less than its denominator, as f , f , -Jf , -5, (3.) A simple fraction is one whose numerator and denom- inator are both integral numbers, as |, f, 1%. (4.) A mixed number is a fraction expressed by an integer and a simple fraction, as 2^, 4|, 3f . (5.) A complex fraction has its numerator, or denominator, or both, in a fractional form, as ' , '^'■\ ~1 , f)}- cy 11 (G.) A compound fraction is a fraction of a quantity w^hicli is itself fractional, as a of 2i, 2^^ of G ^V o^ ^- 6. In the preceding article we have spoken of fractions in the ordinary way. We will now approach them from a different point of view. By the term ratio we understand the result of the com- parison of two quantities with regard to magnitude. There are two kinds of I'atios — ratio by difference or subtraction, 18 ARITHMETIC, and ratio by quotient or division. Tlius we may consider how much one quantity exceeds another, or we may consider how many times one quantity contains another. The former kind is called the arithmetical ratio, and the latter the geometrical ratio. We shall speak only of the latter. Definition. — ^The ratio between two quantities is that multiple, part, or parts which the former is of the latter. It is evident that a ratio can exist only between quantities of the same kind ; thus, we may compare 12 horses and 6 horses, but not 9 men and 4 miles. And if the quantities are reduced to the same denomination, we may treat the quantities as abstract, just as we find the quotient of one concrete quantity by another, by reducing them both to the same denomination, and dividing as if they were abstract quantities. Now, according to what has been stated above, the ratio of 12 to 7, or, as it is usually written, 12:7, is obtained by dividing 12 by 7 ; and this is the same thing as dividing unity or 1 into 7 equal parts, and computing how much 1 2 of such parts amount to. It hence follows that the fraction ^^^ is properly expressed by the ratio 12:7. The first term of a ratio is called the antecedent, and the second is called the consequent ; and hence we may consider a fraction as a ratio, the numerator being the antecedent of the ratio, and the denominator the consequent. When the antecedent is equal to the consequent, the ratio is said to be a ratio of equality ; and it is said to be a ratio of less or greater inequality according as the antecedent is less or greater than the consequent. Thus, 6 : 6 is a ratio of equality. 3 : 4 is a ratio of less inequality. 1 1 : 9 is a ratio of greater inequality. The student will therefore have no difficulty in assenting to* the following definitions : — (I.) A proper fraction, is a ratio of less inequality. (2.) An improper fraction is a ratio of equality or of greater inequality. (3.) A simple fraction is a ratio whose terms are integers. Thus, |- = 3 : 5 is a simple fraction. TREATMENT OF FRACTIONS CONSIDERED AS RATIOS. 10 (4.) A mixed number is a ratio of greater inequality, whose antecedent has been actually divided by its consequent, and the result expressed as an integer and simple fraction. Thus, 2f = V = 17 : 7. (5.) A complex fraction is a ratio, whose antecedent, or consequent, or both, are not integers. Tlius, ^-1, .:;, ^« = respectively to U : If, 2 : 7J, 9^ : 5 are coftiplex fractions. (6.) A compound fraction is an expression containing two or more ratios to be compounded together. Thus f of I- contains the ratios 3 : 4 and 7 : 9 to be compounded: -^ of -v of -J- contains the ratios 2^r : 7, ^7 61-9 ■" 3 : G|, 7^- : 9 to be compounded. 7. A fraction ivhose numerator and denominator are mtd- tiplied or divided hy the same quantity is not altered in value. Suppose, for example, we multiply the numerator and de- nominator of the fraction ? each by 4, we get f = i?-. Now the ratio of 3 I 7 is, from the definition of a ratio, four times as small as the ratio (3 x 4) ! 7 or 12 ; 7; and the ratio 12:7 is, for the same reason, four times as great as the ratio 12 : (7 X 4) or 12 : 28. It therefore follows that the ratio 3 ; 7 is exactly equal to the ratio 12 \ 28, and consequently 3. - 12 7 2 }J * Again, suppose we divided each term of the fraction ^^ ^7 ^» we get If = iV Now the ratio of 15 ; 27 is, from the definition of a ratio, three times as great as the ratio (15 -f 3) ; 27 or 5:27; and, again, the ratio 5 I 27 is three times as small as the ratio 5 ! (27 -r 3) or 5 ; 9. It therefore follows tliatthe vatio 15 : 27 = the ratio 5 ; 9, and consequently if = f. Cor. — An integer may be expressed as a fraction with any given denominator. For we may consider an integer as a ratio whose conser quent is 1, and we may multiply each term of this ratio by any given number without altering its value. 20 ARITHMETIC. Thus—G = ^ = e_>ijz. T 1 X 7 Or G = 4 = -«-^ 9 = A4. ' ^1X9 9 8. To multiply a fraction by a whole number, we may either multiply the numerator by the number, or divide the denominator by it. Ex. §- X 3 = ^^-^ - 2 4. QY we may proceed thus — a X 3 = -«- - -s. ^ 9 — 3 3 ^s lis evidently divisible by 3. Hence -^^lili. ~ .1 pajils - 6_fi_ . ' 2G ARITHMETIC. and each is now evidently divisible by 11. Hence Remark. — It may sometimes happen that, although we are able to tell by inspection some of the factors of numerator and denominator, none of them are common to both numerator and denominator. We cannot then strike them out, but we may use them to determine what would be left supposing they are struck out, and We may thus often come upon the G. C. M. of both ntfenerator and denominator. Ex. — Reduce aWj ^^ ^^^ lowest terms. Now 474 is even, and therefore divisible by 2 ; thus 474 -f 2 = 237. Again, 2133 is divisible by 9, for the sum of its digits, viz., (2 + 1 + 3 + 3) or 9 is so divisible ; thus, 2133 - 9 - 237. We have thus learned that, although 2 and 9 are not com- mon factors, 237 is a common factor, and, in fact, the G.C.M. Dividing the numerator and denominator of the mven fraction by 237, we get -^jz ^^ = ^ "^ -* -^ a 3 7 ^ 2. o •^ *^ 2 133 2133 — 237 9 We proceed now to give the general method of determin- ing the G.C.M. 15. To find the G.C.M. of two numbers. Rule. — Divide the greater number by the less, and if there be no remainder, the less is the G.C.M.; but if there be a remainder, make a divisor of this remainder, and a dividend of the first divisor ; if there be a remainder again, make a divisor of it, and a dividend of the preceding divisor, and so on until there be no remainder. The last divisor will be the G.C.M. Ex.— Find the G.C.M. of 282 and 799. The operation will stand thus — 282)799(2 or thus— 5_64 235)282(1 235 282 1 235 47)235(5 235 The G.C.M, is therefore 47. 282 799 5G4 235 282 235 47 235 235 I'ttfiATMENT Of* IPRACTIONS CONSIDERED A^ RATIOS. 27 The reason of this rule is easy to see. 47 divides 235, and it therefore divides 235 + 47 or 282. Hence it is a common divisor of 282 and 235, and it is therefore a common divisor of 282 and 282 x 2 + 235 or of 282 and 799. It is, moreover, the highest common divisor ; for every number which divides 799 and 282 must also divide 799 - 282 X 2 or 235 ; and hence every number which is a measure of 799 and 282 is also a measure of 235 and 282. Similarly, every number which divides 235 and 282 will also divide 282 - 235 or 47, and hence every measure of 799 and 282 is a measure of 47. But no higher number than 47 can divide 47, and therefore 47 is the G.C.M of 799 and 282. 16. It is now easy to see how any fraction may be reduced to its lowest terms. Ex. — Reduce 14 5 to its lowest terms. 148 703 592 111 148 111 37 111 111 Hence 37 is the G.C.M, and dividing numerator and denominator by it, we get 17. To find the G.C.M. of more than two numbers. The following rule needs no explanation : — Rule. — Find the G.C.M. of any two of the numbers, then the G.C.M. of this result and the third number, and so on. The last result will be the G.C.M. required. Ex.— Find the G.C.M. of 282, 987, 658, 1128. The operation will stand thus — 24 282 987 846 141 2 *94 658 564 141 94 4 1 2 47 1128 94 141 282 282 188 188 47 94 94 Hence the G.C.M. is 47. 128 ARITHMETIC. Ex. III. 1. Eesolve into prime factors 44, G4, 150, 252, 12G9, 4G2. 2. Find the prime factor of 1386, 1720, 4G08, 21175, 15972, 1425G. 3. A number is said to be a perfect number when it is equal to the sum of its aliquot parts. Show that the follow- ing are each perfect numbers — G, 28, 49G, 8128. 4. Show that 284 and 220 are a pair of amicable numbers; that is, such a ])air that each is equal to the sum of the divisors of the other, unity being here counted as a divisor. 5. Keduce, by inspection, to their lowest terms, the follow- ing fractions : a\ a in 3 5 12 1 _fi 3 2_ J^_^3^ ') T.5"J TFT' "¥o' y y 0? 3 9 505 123 3* /9 \ <) o 7 5 1 (i .-, 1 .-, :? 1 H 5 1_R 3 fi V -" • / T B () > 1 O H > 1 7 ( i ? 2 .V 2 ? .V .V "o J 2 o 2 * ('X\ 1 () « A 4 7_2 r^ ^04 inOLH _l"l7.3_ W-/ 2"1 6J UOOJ 133 1' 13 2? 3 o O 3 5 ? 7 137 13* ' G. Find the G.C.M. of (1.) 304, 323. (2.) 413, 448. (3.) 377, 533. (4.) 1866, 2832. (5.) 1189, 1517. (6.) 4374, 5103. (7.) 168, 378, 602. (8.) 539, 616, 792. (9.) 780, 1092, 2145. 7. Reduce to their lowest terms the following fractions — (^ \ 07 5 052 J7 02 206B 1302. 15 5 5 \^') T«^25> T22^T> 2oT4> 6Tirir 197 2> 2 7 1^9* /•2 \ 12 2 6 47 4 10 5j} 1_0_4 . 13 02. lA 9 V-"/ T8 3y> 6 5 3? T21d> T746> T344? 2106* 8. Show, without applying the rule for the G.C.M., that lAU = \h aud that lUn^ = Ih 9. It rains in a certain district 634 days out of every 2,219 ; express this fact in the simplest way possible. 10. Out of 1,659 men engaged in a battle, only 1,185 answered the roll-call in the evening ; express by a i-atio in its simplest terms the number missing in -relation to the whole. 11. There arc 40 numbers less than 100, and prime to it; what arc they? 12. A, B, C can do a piece of woik respectively in 318, 477, 795 hours; express the relative value of A, B, C as workmen by the simplest integral numbers possible. THE LEAST C03IM0N MULTIPLE, 2\) The Least Common Multiple. 13. It is often necessary to express fractions as equivalent fractions, having a common denominator; and it is, more- over, convenient to have this denominator as small as possible. Now, there are always an infinite number of numbers which will contain each of the given denominators as a factor, and our problem is therefore to obtain the least of such luimbers. Def. — The least common multiple (L.C.M.) of two or more numbers is the least number which contains each of the given numbers exactly. Rule. — Arrange the numbers in a line, ])utting one of them as a divisor. Strike out the greatest factor common to this divisor, and each of the numbers separately, and ])lace the several quotients in the line below ; at the same time bring down every number prime to the divisor. Kepeat the operation upon the second line, and so on until we have a line of numbers prime to each other. Multiply the several divisors and the numbers in the lowest line together, and their continued product will be the least common multiple. Ex. 1.— Find the L.C.M. of 12, IG, 3G, 45, 60. 36 12, 16, 36, 45, 60 5 1, 4, T, 5, 5^ I 4 1 1 Hence tlie L.C.M. is 36 x 5 x 4 = 720. Px. 2.^Find the L.C.M. of 15, 21, 60, 84, 140. 60 I 15, 21, 60,J4,_140 ^ r"ir7,"'""i, 7, 7' I 1, 1, 1 Hence the L.C.M. is 60 x 7 = 420. 19. Reduction of fractions to equivalent fractions having the least common denominator. It is evident that the least common denominator cannot be a less number than the L.C.M., and therefore the following rule needs no ex})lanation : — Rule. — Divide the L.C.M. of the given denominators by each denominator in turn, and multiply the corresponding uumerator by the quotient. TJie product thus obtained i^ 30 AIIITIIMETIC. the new numerator, and the L.C.M. is the least common denominator. Ex. 1. — Reduce I, -yV, ^, | /y to equivalent fractions, having the least common denominator. 12 3, 12, 8, 10 1, 1,2, 5 .-. The LCM. is 12 X 2 X 5 =: 120. Dividing 120 by the respective denominators, we get as quotients 40, 10, 15, 12; and hence the given fractions become Visa ^^ij> .■jJLLj^ ll^i_2. • 180 ' 1 ?0 * 1 20 > I JO ' Op J^-Sl. _/»jn_ lo..^ 1.32 ^* J T'i 0, 12 0, i 2 O, T 2l)' Ex. 2* Eeduce to their least common denominator the fol- lowing i—ij, J-i|, H, 13 S- * 45;30, 18,- 45, 63 Hence the L.C.M. is 45 x 2 x 7 4"2li^T~7 = 630. Dividing G30 by the respective denominators, we get for quotients 21, 35, 14, 10. Hence the required fractions are 11 X 2 1 , 13 X 3 5 5 2 X 14 go X t p . 231 4 .5_5 7 2 R^ 2 9 0, (J30 ' C30 ' 030 ' tJ30 ^'630^63 00 30 03 O Note 1. — The operation of dividing the L.C.M. of the denominators is often simpUfied by using the L.C.M. in its factorial form. Thus, in our present example, the L.C.M. is 45 x 2 x 7. Now it is easy to see that the quotient of this by 30 or 3 x 2 x 5 is 3 x 7 or 21, and that the quotient by 18 or 9 x 2 is 5 x 7 or 35, and so on. We thus avoid the process of long division. Note 2. — It is sometimes necessary, and generally advisable, espe- cially for beginners, to reduce the given fractions to their lowest tv-irms before applying the rule for the least common denominator. Thus, the least common denominator of the fractions f, -^^tj, II, taken as they are, is 60 ; whereas, if we reduce the second fraction to its lowest terms by striking out the factor 3, common to both numerator and denominator, the fractions become f , |, ^|, and the least common denominator is 20. If, however, the denominator of any such frac- tion not in its lowest terms is contained in the L.C.M. of the deno- minators, when the fractions are all in their lowest terms, it is unne- cessary to reduce the fraction to its lowest terms. We strongly recommend the beginner, however, to always com- mence by reducing the given fractions to their lowest terms. ADDITION OF FRACTIONS. 31 Ex. TV. 1. Find the L.C.M. of— (1.) 2, G, 8, 12. (2.) 4, 9, 10, 14. (3.) 15, 21, 40, 45. (4.) 12, 20, 35, 126. (5.) 39, 65, 52, 140. (6.) 37, 60, 222, 225. 2. Reduce to their least common denominalor — (1.) h h A, Ur (2.) h A, ih H. (3.) h h \h U' (-i.) I, •>'.., h (5.) h t\. -fW. H5. V^'/ T2 6> To 3J ^«y 3. A can run round a ring in three minutes, B in four minutes, and C in six minutes, and they start together. In how many minutes will they all be again at the starting point % 4. Arrange the fractions -j^^, .V/j, J 5, ^f, in order of mag- nitude. 5. Multiply the greatest of the fractions M, -f 2 J> TTi V 339. 6. Divide the least of the fractions /g, t\, ^g, by 6. 7. Keduce to a simple fraction the complex fraction having the greater of the fractions -?, J in the numerator, and the less in the denominator. 8. AYhich of the fractions J , -^^ is nearer to \ % 9. Show that — is less than ^. 9i 5i 10. Arrange in order of magnitude the following : — 4fof^ liofA 11 G-i l^of2f 13|ofi 3^ 11. Show that nine times the less of the fractions ^^) , ^* ^ is eight times the greater. 12. Show that the ratio 18 : 7 is a ratio of greater in- equality than the ratio 41 : 16. Addition of Fractions. 20. We have shown (Art. 6) that the numerator and deno- minator respectively represent the antecedent and consequent 32 ARITHMETIC. of a ratio ; and it is evident from the definition of a ratio (Art. 6) tlufct the sum of two ratios having tlie same conse- quent is equal to a ratio wliose antecedent is tlie sum of the given antecedents, and whose consequent is unaltered. Hence we have the following i-ule for addition of fractions — Rule. — Bring the given fractions to their least common denominator, add together the numerators thus obtained, and place under the sum the least common denominator. Ex. 1.— Add together |, i^, ^9_, |. The least common denominator is easily found to be 42. Dividing this by each of the given denominators, we get as quotients, 14, 2, 3, 7. — 2R 4_ 20 I 27 I S'i — 2 n4-2 + 2 7 + 35 — 42 ^ 42 ^ 4-2" + 42" " ^.y = ^-"^ = 2^4- = 2J-^. Ex. 2.~Add together' 31, 2|,'liJ, 4^.V The sum of the integral parts of the given fractions = 3 + 2 + 1 + 4 = 10. Hence, 3j- + 2^ + lij + 4^V - 10 + 4. J. -7 I J.L 4- 7_ The least common denominator is easily found to be 180. Dividing this by each of the given denominators we get as quotients 36, 20, 6, 15. Hence the required sum = 10 + lASjS + 7X20 ^ 1 I Xfi ^ 7X1 5 180 180 18 180 — 1 4- ^3 6. I 140 4. 6 I 106 = 10 + 3 6 + 14 + 6A±l_0_5 _ 10 + ^il 180 ^ "" = 10 + lit-j = lligj. Subtraction of Fractions. 21. After the preceding article there will be no difficulty in comprehending the following rule — E/ULE. — Bring the given fractions to their least common denominator, subtract the numerators thus obtained, and under the difference place the least common denominator. Ex. 1. — Subtract -J from ^. i — a— 7X5 _ 3Xn — 35 _ ?4 _ 3 5 — 24 — 1 1 « & - 4o' '4(r "" 4^' '•*" "" 4 -*«• 40 SUBTRACTION OF FRACTIONS. 53 Ex. 2.— Take C J from 9 J. 9J - Cj = 31- - J = 3 + 1^ - i^ = 3 + tV - i* = 3 + i^. Here the number to be subtracted is the greater. "We shall, however, take the less from the greater, and put the 7iegative sign to the remainder, meaning by this that the remainder has yet to be subtracted. Hence, 9j - 6f = 3 -- i^ = 2 + (1 - ij); now ij taken from unity or ^| evidently leaves ■^^. :, required result = 2 + -j^^ = 2^"'^. We will now give an example involving both addition and subtraction. Ex. 3.— Find the value of ^ - 7f + 4^^ - 1 t's- Whenever we have an expression involving both + and - signs, the simplest method is to add together all the quantities affected with the j)^"^^ sign, and likewise those affected with the minus sign; then taking the difference between these two sums, we place the sign of the greater sum before the result. Thus, taking first the integers, we have 6-7+4-1 = 10 — 8 = 2 (it must be remembered that the sign + is understood before a number which appears without a sign when it stands alone or at the head of an expression). Hence the given expression = 2 + ^ - f + t\ ~ tV = 2 + ^ ^^ '^ — 2^4 ^ fiX3 _ 5X2 — 2 + 1 2—8 + 1 5 — 1 36 36 36 36 36 = 22JLz^i2 = 2a = 2J. 36 "^^ * We shall give one more example in order to show how brackets are to be treated. Ex. 4.— Find the value of 7^ - (2| - 3i) + (4^ - \\) - (2A + 3tV). The general rule, which the student will better understand when we come to Algebra, is this — When a minus sign stands before a bracket, it changes all the signs within on removing the bracket ; but when a plus sign stands before the bracket, the latter may be removed without changing any of the signs within. 5 c 34 ARITHMETIC. Thus, taking tlie expression - (2^ - 3i) : — We must remember that the sign of 2^ when within the bracket is, according to a remark made in Ex. 3, understood to be plus. In fact, though it is unusual, we might write the expression thus : — ( + 24- - 3i^). Kow, remove the bracket, and it becomes - 2i + 3^. Again + (4J - li) = + 4j - l\. And - (2,^,- + 3tV) = - 2^0 ~ 3tV Hence our given expression — - 7i - 21 + 3i Hh 4i - H - 2^\ - 3tV = 14-8 + i -i + i+i - 1 -^3^ - tV = 6 + iXl 5 — 1^24 ^ 1X6 ^ 1X2 _ 1 X4 _ 3_Xj_2 _ iXl Q 120 120 120 120 120 120 120 = G + 15 —24 + 60 + 20 — 40 — 36 "10 — 6 + 9 5—110 120 Ex. Y. 1. Add together (1.) ^,1,1; (2.) |, ^V, ii ; (3.) h fV. tV 2. Find the sum of 3|, IT^i, 23^0, ly'V, §, H- 3. Add together 2j, ^, ^, 2| of H, 51 of 1^^. 4. Find the difference between (1.) | and f ; (2.) ^ and |; (3.) f and |. 5. Subtract (1.) 61 from S-^V; (2.) 3^ from 4^^; (3.) 1 " ^« fnnrv. ^T^ff 6. Take , '^» from If 2 ~ H 7. Find the value of 1/^^ - 2f + ^ - ^j. 8. Simplify the expression (2^ - li) - (3^ - 7-|). .9. By how much does 3^% - 1|- exceed 2^{J- - ^^^ ? 10. Take the difference of 6yV and 1 JJj from their sum. 11. Add the difference of the same two fractions to their sum. 31 of 1^ / 6 3^\ 12. Find the value of the expression J^- 2. - 1 -M MULTIPLICATION AND DIVISION OF FRACTIONS. 35 Multiplication of Fractions. 22. Rule. — Multiply together the numerators of the frac- tions for a new numerator, and the denominators for a new denominator. The reason of this rule is easily seen. Let it be required to find the product of J and J, or the value of f x J. Now, what is the meaning of multiplying the ratio 3 : 5 by tlio ratio 7 : 8 *? It means evidently that tlie ratio 3 : 5 is to be multiplied by 7, and the result divided by 8. Now (Art. 8) the ratio 3 : 5, when multiplied by 7, be- comes 3 X 7:5; and (Aii;. 9) the ratio 3x7:5, when divided by 8, becomes 3x7:5x8; and we have Jl X ^ =: -•''^" * » 5X0- And so on for any number of fractions. Hence the above rule. Ex 1. — Multiply together the fractions J, f, if. A X -^- X 1^ = 5 ^3X1 2 t> "^ '-^ ^ 6X8X26 Before actually performing the operation of multiplication, it is advisable to strike out any factor common to both numerator and denominator. We see that 5 is common to 5 and 25, 3 common to 3 and G, 4 common to 12 and 8, and we then have 5 X 4 X i| =1x1x3 = J 6 8 ii6 2X0X5 ^" The whole operation is sometimes written thus — 3 A X ^ X J-2 = ^ ^ '^ ^ '^ ^ = -v^-. 6 ^ 8 ^ 26 5XSXSS ''^ »' 2 2 5 Ex. 2.— Multiply together 2|, 3if, l^V, fl- 2J X 3^3- X 1^ X 51 = V X Vy^ X ti X U • s « 3 3 _ 1 8 X 1 Q N X s 1 X s_^ _ 9 _ 9 . 5XivjXj{t)Xs»i A Division of Fractions. 23. EuLE. — Invert the divisor, and proceed as in multi- plication. To explain this rule, let us endeavour to divide ^ by f . We ma^ evidently consider the required quotient as nothing 3G AniTHJIETia else tlian the ratio ^ : J, and tliis, by the reasoning of Art. 11, is equivalent to the ratio 7x8:9x5, and we hence get y » 9X6 Now, J is the divisor f when inverted, and hence the above rule. Ex. 1.— Divide t% by f . S 4 TO • 8 - T(T ^ 3 - f7^ - 5 ^5- Ex. 2.— Divide 1 J of 7i by 3^ of Sj. 1| of 71 -f 3L of 38- = I X V -^ ( V x V) = 5 X V X i\ X /p = ill = I^Vt. We have introduced a bracket on the right side of the first equality, for otherwise the sign -f affects only the first fraction ^3°. On the other side a bracket is unnecessary, for the sign -r standing before a compound fraction (not two fractions) aflfects the whole. Ex. 3. — Simplify the expression lA - 6| X 4i -f2f of24-rV The given expression = ij -^ V x V "^ (I x \Y) Ex. VI. 1. Find the sum, difierence, and product of 2^ and 1|. 2. Multiply the sum of the fractions 3|, 2-rV ^y their diflference. 3. Simplify the expression i (3^%)"- (Ijf /l -r- 1 3^^ - Ut [ • 4. Reduce to a simple fraction each of the following ex- pressions — (I.)ItV- 7i X 8| ^ 2f of20i. (2.)1tV - 7iof8| - 2f - 20i. 5. What is the diflference between (81 - 3j) and (5 J - 4tJj)1 r -n- • 1 4 , Y of 1| 6. Divide ^ by ^— ^^ 3 + :;^1 ' ■ 7i REDUCTION OF FRACTIONS TO DECIMALS. 37 1 7. Ileduce to a sim})le fraction 3 + 7 + t' F 8. Simi)lify the expressions (1.) 5^ ^"^ ^ ^^ 1-- ^^i of 14 of 21 ^ ' ^ ' »^)- D. Find the quotient of 103^2 by 30^ of ^gj. 10. The cost of 7f articles is ^£65^, what is the cost of eacli article ] 1 1. Find the cost of 89^ articles, when one cost .£4^^^. 12. The sum of two quantities is Zij^, and their dijQferonco is 6 }^ j required the greater. Reduction of Fractions to Decimals. 24. If we place a decimal point to the right of an integer, and add as many ciphers as we j^lease, it is clear, from Art. 1, that we do not alter its value. And hence a given ratio, as 3 : 8, is not altered in value by writing it 3-000 : 8 ; and fui-ther, dividing each of its terms by 8, according to the rule for division of decimals, it becomes '375 : 1. It therefore follows, putting each of these ratios in a fractional form, that .3. - 3:000 ^ ._^^ ^ .^^^ " H 1 We get, therefore, the following rule : — Ki'LE. — Place a decimal point to the right of the numera- tor, and add as many ciphers as may b© thought necessary. Divide the new numerator by the given denominator, accord- ing to the rule for division of decimals, and, if nocessary, add ciphers to the successive remainder until the division ter- minates, or until we have obtained as many decimal figures as required. 38 ARITHMETIC. Ex. 1. — Reduce ^V to a decimal. 32)5-0(-15625 32 ■ 180 160 200 192 80 64 160 . 160 Hence ^% =u -15625. Ex. 2. — Reduce ^rf ., to a decimal. 206)5-00(-01689i 296 2040 1776 2640 2368 2720 2664 560 296 264 It will be seen that we havef arrived at it remainder, 264^ exactly the same as the second remainder j and that^ there- fore, the quotient ^figures ^91 will continually repeat, and that the division will never terminate. We call 891 the recurring period of the decimal, and it is usual to indicate the fact of its recurrence by placing dots over its first aiid last figures, as above. We have, therefore, as a result, ^Jy =^ *01689i; NoTR. — It is easy to see that no fraction, reduced to its lowest fermSf ■whose denominator contains any prime factor, other than 2 or 5, can be expressed as a terminating decimal. For every terminating deci- mal is an exact number of tenths, hundredths, &c., and may, there* fore, bo transformed into a fraction, having some power of 10 as its REDUCTION OF DECIMALS TO FRACTIONS. 39 denominator. Now, if we wish to bring a fraction already in its lowest terms to an equivalent fraction having some power of 10 for its denominator, it can only be done by onultlplying its numerator and denominator by some integer ; and it is impossible to obtain any power of 10 by multiplication only, from a number which contains any prime factor, other than 2 or 5. Reduction of Terminating Decimals to Fractions. 25. Remembering (Art. 1) that any given terminating decimal may be considered as derived from an integer by diminishing it 10, or 10^, 10^, *fec. fold, we have •345 = 345 -f- 101 Hence, *345 is the value of the ratio 345 : 10^; and we , 1 Q,^ 345 345 have, also, -345 = — == j^. The following rule is, therefore, clear : — Rule. — Make a numerator of the integer, formed by tak- ing away the decimal point ; and for a denominator jjut 1, fol- lowed by as many ciphers as there are given decimal figures. Ex. 1.— -625 - T-VA = ^-^-i^ - f. 1000 8X125 ^ 100 20X6 ^O •0025 = T-^-^—^ =: 26 = -rl_ Reduction of Recurring or Circulating Decimals to Fractions. 26. There are two kinds, and it is convenient to treat them separately. (1.) Pure circulating deciinals, where the whole of the decimal figures recur. Rule. — Take away the decimal point and the dots, make a numerator of the integer thus obtained, and place under this as denominator as many nines as there are recurring figures. The following example will make this ride clear. Ex. — Reduce •207 to a fraction. The value of the decimals is evidently tiot altered by writing it -207207. Let us remove the decimal point three 40 AniTHMETIC. places to tlie riglit, 01%'wliat is the same thing (Art. 1), let us multiply the given decimal by 1000. We then get 207-207 as tlie value of 1000 times the given decimal. Now the number 207-207 includes the integer 207, and the given decimal -207; and it therefore follows that the integral part 207 is (1000 - 1), or 999 times the value of the given decimal. Hence, dividing it by 999, we get •207 = 207 ~ 999 - fgj. (2.) Mixed circulating deGimals. — Where part only of the figures recurs. E-ULE. — Take away the decimal point and the dots, sub- tract from the integer thus obtained the integer formed by the figures which do not recur, and make a numerator of the result. Then, for a denominator, place underneath as many nines as there are recurring figures, followed bv as many ciphers as there are figures which do not recur. We shall make this rule clear by the following example : — Ex.— Eeduce -24573 to a fraction. Let us remove the decimal point two places to the right,, thus, by Art. 1, multiplying the given decimal by 100 ; we, then get 100 times the value of -24573 = 24-573. Now, by case (1) above, 24-573 = 24f ^{ ; or reducing to an improper fraction, and noticing that 24 x 999 =• 24000 - 24, we have 24*573 = 24000 -- 24 + 5 73 — 2 4 5 7 3 " 2 4^ 9 9 9 9 9 9 * Hence, dividing this result by 100, we get *24573 = ^4r.n:i- 24. ^ IQO = 24573-24 / A ,u. 04 \ 999 99900 \-^^ i^« '^^' f Ex. 1.— -428571 = flff-U = ^-^-i^^^^-? = f. 999999 7X1 42867 ' ■Ry '9 700 — 2 7 9 — ,2 7 — 2fiR2 — 149X0X9 J^49 27. In arithmetical operations involving circulating deci- mals, and, indeed, any decimals having a large number of decimal figures, it is generally sufticient to obtain a result ijorrect to a given number of decimals. APPROXIMATE RESULTS. 41 1. In addition and sid)tr action we obtain this result most easily by using in our operation one or tioo more figures of the given decimals than are required in the result. Ex. 1. — Add together (correct to five places) the follow- ing:— •3026, 6-7294, -016, -4163729. •3026026 6-7294444 •0166666 •4163729 ^4650865. Ans. 7*46508. Note. — If our object is to obtain a decimal of five places which shall give the approximate sum of the given decimals we must write 7 '46509 as the approximate sum ; for 7 '46509 is nearer to the true value of 7-465086 &c. than 7 '46508. The general rule is to increase by 1 the decimal figure at which we stop, ivlien the next figure is 5 or above 5. Ex. 2. — Find the difference (correct to six places) of 3-0745 and 4-263, and express the approximate difiference by a deci- mal of five places. 4-26326326 3-0745 4545 l-1887T7¥l Hence the difference correct to six places is 1*188717, and the required approximate difference 1-18872. 2. In midtijilication and division of circulating decimals it is generally preferable to reduce the given decimals to frac- tions, bring out the result in a fractional form, and afterwards reduce this to a decimal. Ex. YII. 1. Express as a decimal the sum of the following fractions: — 3 1 .«» 7 3 .•> 7 1 2. Reduce to fractions the following decimals : — •35, ^026, •ie, -142857, -16, -4285714. 3. Find the value (correct to six places) of -237 + 3*816 — 6-0235 + 4-29 - -002 + 1-374. 42 ARITHMETIC, 4 Add together -62, -037, 2-476i, -8106, -7, -375. 5. Find the product and quotient of 3-5 4 by 4-3. 6. Simplify the expression — (4-6 X •428571 - 2-2 x -36) (1 - -16). _ In the next six examples the dots are signs of multiplica- tion, and you are required to give the values of the expres- sions correct to six places. ^ 1 1 1 1 1 , R 1 1 1 1 . "* '3 ~ 3^ + 5^ - 7^ + '^• C L J_ . 1 1 A ' ■I-2 *" 1-2-3 * h¥3^ ~ 1-2-3-4-5 "^ ^' 1n 1 1 1 1 r 10. 1 + g3- + — + -g-. + &c. } 239 ^ . + &c. 3-239* 12. 4 + -^^ + + -r-F + 'k—n + 1-2 2-3 34 4-5 5'6 6-7 CHAPTEE III. APPLICATION OF THE PRECEDING ARTICLES TO CONCRETE QUANTITIES. To find the Value of a Fraction of a Concrete Quantity. 28. Rule. — Multiply the concrete quantity by the nu- merator of the fraction, and divide the product by the deno- minator. VALUE OF A FRACTION OF A CONCRETE QUANTITY* 43 Ex. 1.— Find the value of | of 2 tons, 3cwt. 21 lbs. 3 f C) f^T%a O rt f 91 IVko (2 tons, 3 cwt. 21 lbs.) x 3 6 tons, 9 cwt. 2 gra. 7 lbs. = 18 cwt. 2 qrs. 1 lb. Ans, Ex. 2.— Find the value of 3 J- of £12. 6s. 2jd. X. s. (1. 3 times £12. 6s. 2Jd. = (£12. 6s. 2|d.) x 3 = 36 18 6f . ^^^- = 2 14 81 Hence, adding^ the value required =£39 13 3 J Ex. 3.— Find the value of | of 4 miles + I- of 3 fur. + /^ of 8 poles. Mile. Fur. Poles. Yards. 4 of4mnes = <-^'^^ = '-^^ = 1 5 28 3i- ^^ of 3 fur. = ^^-^L^L ^ ilJH. ^ 2 13 If /x of 8 poles = ^l^i^^ ^ » ^ Q 1 4H Hence, adding, the value required =2 3 4j| KoTE. — The addition of the yards is thus effected : — (3| + H + 4U) yds. = (8 + ^H^) yds. = (8 + \?A) yds: := 9f^ yds. = 1 polei(3i + |$-) yds; = 1 pole; (3 + ^— ^) yds. ^ 1 pold,4H ydsi Ex; VIII. find the values bf — 1. I of £1 ; ? of Is. ; ? of 12s. ; 5 of £3. 2. 5- of £5 ; ? of a guinea ; -^\ of 2s. 6d. ; % of a crown* 3: ^ of £1. 12s.; 2^ of £3. 10s.; 7^^ of £3. 4s. 51. 4. f of 1 ton ; I of 1 qr. ; J of 1 stone ; -^^ of 9 lbs. 5. 3f mile ; 4. of 3 fur. ; -? of 1 5 poles ; ^ of 2j of 1 2 yds. 8 2 " 51 G. ^Q- leap year; -,— lunar month; jTyT of 10 h. 15' 12^'. 7. 12 lbs. 3 oz. 7 dwt. 5 grs. x 3|; 10 oz. 4 gr. v 16j-. 44 ' AHITHMETIC. 8. 8f of 4 ac. 3 po. - -^\ of 1 sq. mile + If of 3 r. 20 sq. yds. 9. (5f of /^ - l^\) of 35^ 36^ 2^; (2^ - H of If) of 30^ 10. ^% of 1 lb. Troy + ^ of 1 lb. Troy - /^ of 1 lb. Avoir- dupois ( = 7000 grains). « 5^ of 3 soiineas ^^' ^ ""^ fifrfsriOd.^ ' ""^ 1 ton + f of 3 qrs. - ^V of 7 cwt. 12. 2t\ of 15 L 10' 131'' - ^V of 1 day, 12 b. ir 12j/. To Reduce a given Quantity to the Fraction of any other given Quantity of the same kind. 29. KuLE. — Reduce both the given quantities to the same denomination, and the fraction required will have the number of units in the first quantity as numerator, and that of the second quantity as denominator. Ex. 1. — Eeduce 3s. 8d. to the fraction of £1. 3s. 8d. = 44d, and£l = 240d. Hence, the fraction required = ^Vo = ih Or, better, thus : 3s. 8d. = 11 fourpences, and £1 = 60 fourpences. .'. Fraction required = J J. Note. — It is always best to keep the denominations to which the given quantities are reduced as high as possible. Ex. 2. — Reduce i of a moidore to the fraction of 2| guineas. 1 of a moidore = (| x 27)s., and 2^ guineas = (2 J x 21)s. 1 X 27 4- X 27 Hence, the fraction required = —- — -- = ^— -; , = l^ 11 ^ I ' ^ 2^x21 fx21 ^^-^1^' _ 1X9X2 _ T B ~ 6X7X7 — 245 Ex. 3.-— Reduce 3 cwt. 2 J- qrs. to the fraction of 4 cwt. 2 qrs. 4 lbs» 3 cwt. 2^^ qrs. = 14 J qrs., and 4 cwt. 2 qrs. 4 lbs. = 4 cwt. 2i qrs. = 18| qrs. 14 L J 2 7. .*. Fraction required = j-g^ = 7^ = ^VyVJ = h VALUE OP A DECIMAL OF A CONCRETE QUANTITY. 45 Ex. IX. Reduce — 1. Is. 8d. to the fraction of .£1 ; 7 id. to the fraction of 10s, 2. 2s. 4d. to the fraction of lOs. 8d.; Is. 7 id. to the frac- tion of 3s. 4 Id. 3. 3 qrs. 15 lbs. to the fraction of 1 ton; 2 stones 10 oz. to the fraction of 3 cwt. 4. 3 lbs. avoirdupois to troy weight; 10 lbs. 3 oz. 4 dwt. troy to avoirdupois. 5. 3 quii'es, 10 sheets to the fraction of 2 reams, 3 quires; 3 ft. 8i in. to the fraction of 3 yards. G. 30^ 3' 12'^ to the fraction of a right angle (-90^); 57^16' 21^^'' to the fraction of two right angles. What fraction is — 7. f ac. of 3i ac; 2i days of 17 weeks? •125 + 1-875 3-lGx 1-4 8. —rvrvTr^oK ~ acres of 19 \ poles : 7 yds. of 3i m ] •0140625 ^^ 2*375 x-i 9. ( JY rood + 7^ poles + ■ '^ [^^ ^^ yd. j of 3 acres 1 1^- oVg^l-of-^" pipes? 11. What fraction of his original income has a person left after paying a tax of 4d. in the £ 1 1 2. A garden roller is 2 ft. 6 in. wide, and it is rolled at the rate of 1 mile in 20 minutes : find in what fraction of a day a man will roll J of an acre. To Find the Value of a Decimal of a Concrete Quantity. 30. Rule. — Multiply the given decimal by the number of units in the concrete quantity when expressed in terms of one denomination, and the integral part of the result will be the number of units of this denomination. Tlien multiply the decimal part of this denomination by the number of units connecting it with the next lower and the integi^al part will be units of this latter denomination, and so on^ 46 ARITHMETIC. Ex. 1.— Find the value of -325 of £3. 10s, £3. 10s. = 70s,; proceeding then according to i^ule, we have : •325 70 22-750S. 12 9-OOd. Ans.: 22s. 9d. or £1. 2s. 9d. Ex. 2.— Find the value of -546875 of 3 tons. •546875 3 1-640625 tons. 20 12-812500 cwt. 4 3-2500 qrs. 28 200 50 7-00 lbs. Ans.: 1 ton, 12 cwt. 3 qrs. 7 lbs, Ex. 3.— Find the value of 6-66875 acres. 6-66875 acres. 2-67500 roods. 40 27-000 poles. Ans.: 6 acres, 2 roods, 27 poles. Ex. 4.— Find the value of -316 of £1, First Method. Second Method. 900 9 00 1 5_X 1 9 20 ■ 16 X^'o (3 o' 6-3333340s. Then by rule for reduction •3166667 nearly, 20 ^340s. 12 of fractions, 4-000008T. Ans. : 6s. 4d. £JS = 6s. 4d. Ans. The latter method is preferable when perfect accuracj^ is required. VALUE OF A DECIMAL OF A CONCRETE QUANTITY. 47 Ex. X. Find the value of — 1. X-375; £-98125; £-815625. 2. 416 of £3; -428571 of 6s. 5d. ; 8-571428 of 3s. Ofd. 3. -625 of 1 ton; -046875 of 3 tons; 4*39 of 1 cwt. 53 lbs. 4. 1-6671875 acres; -3475 rood; -076923 of 5 acres, 6 poles. 5. -2083 of 1 ream; -4583 quire; -383 of 3 reams,12 sheets. G. -3078125 pipe; -490625 tun; -37125 bushel. 7. Express in grains -142857 of 4 lbs. avoirdupois, and express the result in troy weight. 8. Express 10 oz. 3 dwt. 14 grs. as the decimal of 1 lb. avoirdupois. 9. What is the sum of -6 of 1 guinea, -083 of 1 crown, and -037 of £1. Os. 3d. 10. Find the value of (-20416 X 7-5\ 4 -^ -;2187r~") ^^^' "*" (*^^ ton- 769230 qrs.) x 26. 11. What is the value of (^ hr-) of 1 cwt. 35 lbs. 1 1 -I 12. Simplify / L-tA^ + LzJ^\ of 1 oz. 15 dwts. ' -^ \l .- -16 1 + -16/ To Reduce from one Denomination to the Decimal of another Denomination of the same kind. 31. E/ULE. — Bring the given quantity to the fraction of the proposed denomination, and reduce this fraction to a decimal, 48 ARITHMETIC. Ex. 1. — Reduce 3s. 3d. to the decimal of 8s. Ud. 3s. 3d. =26 three-halfpence, 8s. IJd. = 65 three-halfpence, and the fraction |f = 2->^-i-3 = | = -4. •^^ 6X13 6 Hence '4 is the decimal required. Ex, 2, — Keduce 3 qrs. 21 lbs. to the decimal of 2 cwt. 3 qrs. 3 qrs. 21 lbs. = 105 lbs. 2 cwt. 3 qrs. = 308 lbs. and the faction igf -^ i^-' = U = ^^^^ = -3409. 308 44X7 ** 11. Hence -3409 is the decimal required. Ex. 3. — Reduce 18s. 8Jd. to the decimal of £1. The rule may be applied most concisely as follows : — 4 12 20 1- •25 18-6875 •934375 .-. -934375 is the decimal required. The farthing is first reduced to the decimal of a penny, and the 8d. prefixed; then 8-25d. are reduced to the decimal of a shilling, and the 18s. prefixed; lastly, 18 •6875s. are reduced to the decimal of £1. Ex. 4. — Reduce 3 qrs. 21 lbs. to the decimal of 1 ton. 28 (1 21 3- 4 3-75 20 •9375 •046875 Hence '046875 is the decimal required. Ex. XI. Reduce — 1. 12s. 6d., 10s. 7id., lis. Ijd., 18s. 6^ d., each to the decimal of <£1. 2. 13s. Ojd., 10s. 8id., 9s. 6d., each to the decimal of 15s. 5jd. 3. 7s. 8fd. to the decimal of a guinea^ ancl 3s, 2^d. to the decimal of a moidore, • THE METRIC SYSTEM. 49 4. 1 qr. 7 lbs. to the decimal of 1 ton ; 3 cwt. 3 qrs. 20 lbs. to the decimal of 3 tons. 5. 3.V lbs. to the decimal of lu cwt.; 14 oz. to the decimal of 3 cwt. 2 qrs. G. 20 grs. to the decimal of 1 lb. Troy; 3 dwts. IG grs. to the decimal of 4 oz. 11 dwts. 7. 1 rood, 10 poles to the decimal of 1 acre; 3 roods, 15 l square yards to the decimal of 5 acres. 8. f horn's to the decimal of 10 weeks; 7 h. 18' to the decimal of 1 year (365 days). 9. Bring the sum of -r\ of 9 hours, | of 12i^ days, | of 7^ minutes, to the decimal of a week. 10. Express a pound troy as the decimal of a pound avoir- dupois. 11. Reduce the sum of G lbs. 6f oz. avoirdupois, and 8 oz. 6 dwts. 1 6 grs. to the decimal of 1 ton. 12. Express 2*36 of 4s. ~ -518 of 9s. 2d. + 1'4583 of Gd. as the decimal of £5. CHAPTEE IV. THE METRIC SYSTEM. 32. — The fundamental unit of the metric system is the metre, A metre is the ten-millionth, or ^7^7 part of 90° of the earth's meridian, and measures 39*3708 English inches. In order to express multiples and sub-multiples of this unit, and, indeed, of any unit in the metric system, we make use of one or more of the following prefixes : — Deka, . . 10 times. Deci, . . 10th. Hecto, . 100 „ Centi, . . 100th. Kilo, . . 1,000 „ Milli, . . 1,000th. Myria, . 10,000 „ We will arrange these prefixes and the word nnit in order according to their signification, thus — Myria, KilOy HectOy Deka, Unity Deciy Centiy Milli Now, as we j-e^d ihU line from left to right, it is evident ARITHMETIC. 4 7 8 Metre. Decim. Centim. Millim. 4 7 8 ^hat the words liave a signification decreasing tenfold in value; and as we read it from riglit to left, they have a signification increasing tenfold in value. It therefore follows that figures placed under the above words have a local as well as an intrinsic value ; and further, if when a figiire is wanting to complete the series, its place be filled up by a cipher, it will be seen that the local value corresponds exactly with the ordinary decimal notation. Moreover, we have only to place a mark (in fact, a decimal point) at the right of the figure standing under any of the words of the above memorial line, and the given quantity is at once expressed in the denomination corres- ponding to that figure. Thus, taking the metre as our unit : Myriani. Kilom. Dekam. Metres. Decira. Millim. 3 2 5 Myriam. Kilom. Ilectom. Dekam. - 3 2 5 = 3 '2054708 myriametres. '^-- 32-054708 kilometres. '^ 320-54708 hectometres. ^ 3205-4708 dekametres. =^ 32054-708 metres. := 320547-08 decimetres. = 3205470-8 centimetres. = 32054708- or 32054708 mniimetres. The following rule for expressing any quantity in terms of any one multiple or sub-multiple of the unit, or of the unit itself, is therefore evident : — Rule. — Put ciphers in the place of any multiple, unit, or sub-multiple absent in the series, and write the figures in close order, as in the ordinary decimal notation. Then 2:)lace a decimal point at the o'ight of the figure corresponding to the denomination in "which we wish to express the given quantity. Ex. 1. — Express 5 myriam. 3 hectom. 6 decim. as metres, Filling up with ciphers the vacant spaces, we have — Myriam. Kilom. Hectom. Dekam. Metres. Decim. 5 3 6 = 50300-6 metres. THE METRIC SYSTEM. 51 Ex. 2. — Express 3 dekam, 4 deciiii. as myriametres. Eilliiig lip with ciphers, we have — Myriam. Kilom. Hectom. Dekam. Metres. Pecim, Centira. Millim. 00030400 = 0*0030400 myriametres = 0*00304 myriametres. (The student will see that it was unnecessary here to extend the series beyond decimetres.) Ex. 3. — Expx-ess 13 metres 502 millimetres as kilometres. We may write the given quantity thus — Iviloni. Hectom. Deckam. Metres. Decim. Centim. Millim. 13 5 = 0*013502 kilometres. 2 "We have hitherto spoken only of the fundamental unit, and its multiples and sub-multiples. We shall hereafter (Aii}. 35) explain the priiacipal derived units, viz., the Gram, the Are, the Stere, the Litre, and the Franc ; but as the multiples and sub-multiples of these derived units bear the same relation respectively to the corresponding derived unit, as in the case of the fundamental unit, all the preceding remarks relative to the multiples and sub-multiples of the fundamental unit apply equally to those of the Gram, the Are, the Stere, the Litre, and the Franc. With regard to the units, multiples, and sub-multiples of square and cubic measure, prdperly so called, it is necessary to make a few remarks. 33. Square Measure. — The unit of square measure is the square metre ; and since the series myriam., hilom., hectom., dekam., metre., &c., decrease in value tenfold when read from left to right, and increases similarly when read from right to left, it follows that the series square myriam., square kilom., square hectom., square dekam., square metre, kc, will decrease or increase 10'^ or 100-fold. Hence we see that in square measure the multiples and sub-multiples increase or decrease successively 100-fold, and, therefore, when quan- tities in square measure are expressed by the ordinary decimal notation, each multiple or sub-multiple must occupy the place of two figures, a cipher being supplied when we have less than ten of any multiple or sub-multiple, and two ciphers when there is any blank in the series. 52 ARITHMETIC. Sq. kilom. Sq. hcctom. Sq. dekam. Sq. metre. Sq. centim. Ex. 10 3 15 3 5 Sq. kilom. Sq. hectom. Sq. dekam. Sq. metre. Sq. decim. Sq. centim. = 10 03 15 03 00 05 ^ 10-0315030005 square kilometres. = 1003-15030005 „ hectometres. = 100315-030005 „ dekametres. = 10031503-0005 „ metres. = 1003150300-05 „ decimetres. - 100315030005 „ centimetres. 34. Cubic Measure. — The unit of cubic measure is the cubic metre, and hence after the remarks in the last article, since 10* = 1000, when quantities in cubic measure are ex- pressed by the ordinary decimal notation, the units, mul- tiples, and sub-multiples must respectively occupy the place of three figures, ciphers being supplied to fill up blank spaces when necessary. Ex. 1. 325 cubic metres 51 cubic decimetres. = 325 „ „ 051 „ = 325-051 cubic metres. = 325051 cubic decimetres. Ex. 2. — 25 cubic metres 3 cubic decim. 40 cubic centim. = 25 cubic metres 003 cubic decim. 040 cubic centim, = 25-003040 cubic metres, = 25003-040 „ decimetres. = 25003040 „ centimetres. 35. Derived Units. — The principal derived units of the metric system are — 1 . The Gram, for measures of weight. The gra7}i is the weight of a cubic centimetre of distilled water at the temj)erature of 4° C 1 gram = 15*4323 grains, 1 gi-aiu = '0648 gram. 2. The Are, for land measure. The are is a square who^e side measure^ 10 metres ; it is therefore ec[ual tQ fv s^jiiarQ ilek^irietre, gr J 00 square metres, THE METRIC SYSTEM. 63 1 are = 119-6033 square yards, 1 hectare = 2*471 acres, 1 acre = *405 hectare. 3. The Stare, for fire-wood. The stere is equivalent to a cubic metre. It is therefore the solidity of a cube whose edge measures 1 metre. 4. The Litre, for measures of capacity. The litre is a capacity equal to the volume of a cube whose edge measures a decimetre or 10 centimetres. It is therefore equal to a cubic decimetre or 1,000 cubic centimetres, and 1,000 litres are equivalent to a cubic metre. 1 litre = -2201 gallon, 1 gallon = 1*543 litres, 11 gallons = 50 litres nearly. 5. The Franc, for money. The franc is a coin weighing 5 grams, and composed of an alloy, nine-tenths of which are silver and one-tenth copper. The following table exhibits at a glance the fundamental unit and the above derived units, together with the multiples and sub-multiples at present in use : — Table of the Metric System of Weights and Measures. Multiples. Units. Sub-Multiples. 10,000 1,000 100 10 METRE, Long Measure. 10th. lOOth. 1,000th. Myria. Kilo. Hecto. Deka. Deci. Centi. Milli. Myria. Kilo. Hecto. Deka. GRAM, AVeight. Deci. Centi. MiUi. Hecto. ARE, Land Measure. Centi. Deka. STERE, Solid Measure. Deci. Hecto. Deka. LITRE, Capacity. Deci. Centi. Milli. FRANC, Money. Decime. Centime. A quintal = 100 kilog. = 2 cwt. nearly ; a millier, or tonneau de mer, = 10 quintals = 20 cwt. nearly. 54 ' ARITHMETIC. Ex. XII. Examples upon the Multiples and Sub-Multiples of the Units. 1. Express each of the following as metres — 15 myriam.; 20 kilom.; 1 hectom.; 27 dekam.; 25 decim.; 100 centim.; 345 centim. ; 5294 millim. 2. How many centimetres in the following — 46 myriam.; 30 kilom.; 295 hectom.; 1*5 dekam.; 3'95 metres, 295 millim.? 3. Express according to the metric table — 20 kilog. 29 dekag.; 18 kilog. 85 decig.; 123 hectog. 13 centig.; 12 dekag. 296 millig.; 153 centig. 3 millig; 3427 millig. 4. How many decigrams in the following — 16 kilog. 12 centig; 25 hectog. 10 grams; 39,645 millig.; 20 kilog. 35 dekag. 5 grams'? 5. Express in ares — lOhectar. ; 296 centiar. ; 29 hectar. 3 centiar.; 3 hectar. 12 centiar.; 376,543 centiar. 6. How many square decimetres are there in the following — 100 sq. kilom.; 10 sq. hectom.; 5 sq. dekam.; 3498 sq. met. ; 46 sq. met. 1 7. How many steres in the following quantities — 15 dekas. ; 394 decis. ; 9 dekas. 2 decis. ; 186 dekas. 3 decis.; 3764 decis. ; 4 decis. ? 8. Express as cubic metres — 10,000 cubic decim. ; 1,234,567 cubic centim. ; 372,456,126 cubic millim.; 1,000,000 cubic centim.; 639 cubic centim.; 293 cubic decim. 9. Express as litres — 3kilol. 2 hectol. 3 litres; 4 decil.; 2 kilol. 3 millil.; 76,384 millil.; 2934centil.; 830 dekal. ; 34,576 decil. 10. Express as litres, as dekalitres, and as centilitres — 18 kilol. 3 hectol. 4 decil. 5 centil. 3 millil. 11. How many francs in the following sums of money — 100 cent.; 736 dec; 24,645 cent.; 5 cent. 25 dec; 1695 cent.? teE METRld SYSTEM. 55 1 2. How many centimes and how many decimes are expressed by the following — 13 francs; 7 fr. 13 c; 12 fr. 3 dec. 5 c; 29 c; 3 fr. 2 dec; 18 fr. 4 c.] Addition, Subtraction, Multiplication, and Division in the Metric System. 36. Since (Art. 33) all quantities in the metrical system may be expressed as one denomination by figures whose local as well as intrinsic values follow the decimal system of nota- tion, it is evident that when they are so expressed we may add, subtract, multiply, or divide them exactly as ordinary integers and decimals. Ex. 1. — Add together 49 metres 36 centim. ; 3 kilom. 2 dekam. 3 decim. ; 2 hectom. 3 metres 25 centim.; and 13 dekam. 327 millim. Metres. Millimetres. 49-36 49360 3020-3 3020300 203-25 or thus, by integers— 203250 130-327 130327 3403-237 3403237 Ans. : 3 kilom. 4 hectom. 3 metres, 2 decim. 3 centim. 7 millim. Ex. 2.— Subtract 3 kilog. 2 dekag. 37 millig. from 10 myriag. 25 grams, 369 millig. Kilograms. Milligrams. 100*025369 100025369 3-020037 or thus, by integers— 3020037 97-005332 97005332 Ans. : 9 myriag. 7 kilog. 5 grams, 3 decig. 3 centig. 2 millig. Ex. 3. — Multiply 12 dekasteres, 3 steres, 5 decisteres by 23. Steres. Decisteres. 123-5 1235 23 23 3705 or thus, by integers — 3705' 2470 2470 2840-5 steres* 28405 decisteres. Ans, : 284 dekasteres, 5 decisteres* 56 ARITHMETIC. Ex. 4. — Multiply 455,602 cubic centimetres by 36. Cubic Metres. Cubic Centimetres. 0-455602 455602 36 36 2733612 or tbus, by integers— 2733612 1366806 1366806 16-401672 16401672 Ans. : 16 cub. met. 401 cub. decim. 672 cub. centim. Ex. 5.— Divide 1369 kilol. 35 lit. 36 centil. by 72. Kilolitres. Centilitres. 136903536 «{! 1^369-0353^ ^ (9 T52Tr5(J4 or thus, by integers—'' ^ 1 8 19-01438 15211504 1901438 Ans.: 19 kilol. 1 dekal. 4 lit. 3 decil. 8 centil, or 19 kilol. 12 lit. 38 centil. Ex. 6. — How many times is 12 sq. dekam. 3 sq. met. 15 sq. decim. contained in 216 sq. dekam. 56 sq. met. 70 sq. decim. 1 Sq. Dekam. Sq. Dekam. Sq. Decim. Sq. Decim. 12-0315)216-5670(18 120315)2165670(18 120315 . 120315 "932520 or thus, by integers- -"5^2520 962520 962520 Ans.: 18. Ex. XIII. 1. Add together — (1.) 3 metres, 2 decim. 4 centim.; 18 metres, 219 millim. ; 4 kilom. 2 hectom. 3 dekam. 14 centim.; 12 kilom. 36 metres. (2.) 74006 hectom., 3216 kilom.; 12 myriam. 2167 metres. (3.) 4 sq. met. 42 sq. decim.; 12 sq. dekam. 18 sq. decim.; 82 sq. met. 3250 sq. decim.; 3*271 sq. met. (4.) 18 cub. met. 186 cub. decim.; 39-207365 cub. met. 30761 cub. centim.; 12 cub. met. 124-27 cub. decim. THE METRIC SYSTEM. 57 (5.) 25 kilog. 235 grams; 3072 centig.; 13 kilog. 51 grams, 63 millig. 8132-07 decig. (6.) 319 hectar. 4 ares, 51 centiar.; 93-712 hectar. 2375G-27 ares ; 6 hectar. 4 centiar. (7.) 3 steres, 5 decis.; 209 steres, 4 decis. ; 25-76 steres, 13-027 dekas. (8.) 51 kilol. 126 lit. 32 centil.; 123 lit. 3 centil. 15-02703 kilol.; 12 kilol. 3-27602 hectol. (9.) 161 fr. 35 c; 32 fr. 4 c; 8276 c; 10-26 fr.; 16 decimes, 5 c. 2. Subtract— (1.) 4 metres, 372 millim. from 16 hectom. 5-06 metres. (2.) 30765 centim. from 12 kilom. 4 metres, 9 millim. (3.) 3 sq. met. 89 sq. decim. from 1 sq. dekam. 7 sq. decim. (4.) 12-0324 cub. met. from 18 cub. met. 29 cub. millim. (5.) 39 grams, 65 millig. from 6 kilog. 12 gi-ams. (6.) 8 hectar. 19*08 ar. from 32 hectar. 70 ar. 2 centiar. (7.) 9 dekas. 6 decis. from 50 dekas. 2 decis. (8.) 6 kilol. 6 millil. from 700 kilol. 3 lit. 3 centil. (9.) 65 c. from 3 fr.; and 2 fr. 4 c. from 100 fr. 60 c. 3. Multiply— (1.) 10 metres, 35 millim. by 7, 11, 13. (2.) 18 kilom. 3-07 metres by 27, 48, 64. (3.) 3-0625 sq. met. by 16, 18, 35. (4.) 4 cub. met. 10 cub. decim. 5 cub. millim. by 19, 23, 26. (5.) 7364 hectog. 9-31 decig. by 15, 25, 20. (6.) 12 hectar. 3 centiar. by 30, 50, 40. (7.) 416 steres, 2-9 decis. by 100, 150, 60. (8.) 612305-06 litres by 12, 14, 16. (9.) 39 fr. 10 c. by 75, 105, 135. 4. A merchant OAved 1500 fr., and he gave in payment 69 metres of cloth at 3 fr. 4 c. per metre, 48 metres of silk at 8 fr. 65 c, 13*5 metres of calico at 75 c. How m;iich does he still owe ? 58 AUITHMETia 5. Make out the following bill — fr. c. 44 hectol. of oil, . . at 75 the litre. G6 kilog. 125 gi\ of sugar, „ 1 25 „ kilog. 375 gr. of pepper, . . „ 3 5 „ „ 128*75 hectog. of soap, . „ 1 75 „ „ 562 gr, 5 decig, of coffee, . „ 30 „ hectog. 6. Divide — (1.) 17 metres, 16 centim. by 11, 12, 13. (2.) 41 kilom. 82 dekam. by 15, 16, 17. (3.) 29 sq. met., 2740 sq. centim. by 14, 21, 35. (4.) 376-38 cub. met. by 9, 27, 45. (5.) 4 kilog. 14 dekag. 18 decig. by 22, 33, 55. (6.) 8 hectar. 58 ares by 65, 60, 55. (7.) 12 dekas. 1-2 decis. by 12, 13, 91. (8.) 36 myiial. 4 kilol. 16 lit. 7 decil. by 9, 18, 27. (9.) 7339 fr. 50 c. by 25, 30, 75. '''. Find the price of — (1.) A metre, when 2 met. 80 centim. cost 70 fr. (2.) A square decim., when 30 sq. met. cost 450 fr. 30 c. (3.) A cubic metre, when 15 cub. decim. cost 361 fr. 80 c. (4.) A hectometre, when 3 kilom. 125 metres cost 10 fr. 25 c. (5.) A kilog. of coffee, when 7 hectog. 50 grams cost 1 fr. 35 c. (6.) A hectare^ when 4265 fr. 2*50 c. is the price of 149 ares, 25 centiar* (7.) A stere, when 125 dekas. 4 decis. cost 20631 fr. 60 c. (8.) A decilitre, when 47 dekal. 5 litres cost 570 francs. (9.) A cub« centim., when 1 cubic metre cost 10,000 fr. 8. How many times is — (1.) 1 kilom. 470 met. 38 centim. contained in 36759*50 metres'? (2.) 12 sq. decim. 75 sq. centim. contained in lO sq* met* 20 sq. decim. ? (3.) 13 sq. met. 25 sq. decim. contained in 318 Sq. dekam.? THE METRIC SYSTEM. 69 (4.) 31 cub. met. 725 cub. decim. contained in 45684 cub. met. ? (5.) 345 millig. contained in 165 kilog. 6 hectog.l (6.) 275 centiar. contained in 396 hectar. ? (7.) 7 stores, 2*5 decis. contained in 29 dekas.? (8.) 4 kilog. 5 gi'ams, contained in 38 myriag. 4 kilog. 480 grams ] (9.) 8 centimes contained in 10 francs'? 9. A merchant bought 95 litres of wine for 118 fr. 75 c, and sold it at a loss of 10 c. per litre. What was the selling price per kilolitre ] 10. To make 12 suits of clothes, it required 40 metres of stuff 90 centim. wide. How much stuff will it take if the width is 80 centim. ] 11. How many cubic decimetres of iron are there in a bar- weighing 280 kilog. 368 grams, when one cubic centim. weighs 7 grams 788 millig. 1 12. An iron wire, 126 metres long, is cut into pieces 3 centim. 2*5 millim. long. How many pieces are there? Relation between the Metric Units and the Eng^lish System of Weights and Measures. 37. We shall work a few examples to show how quantities expressed in the metric system may be expressed in the English system, send vice versa. Ex. 1. — Reduce 10 kilom. 321 metres to English measure. 10 kilom. 321 metres = 10321 metres. = (10321 X 1-094) yards. - 10321 X 1.094 -1 - 17 60 mnes. = 6 miles 731-174 yards. Ex. 2. — Express 2 miles, 309 yards in the metric system. 2 miles, 309 yards =(2 x 1760 + 309) yards. = 3829 yards = f;^^ metres. = 3500 metres. = 3 kilom. 500 metres. 60 ARITltMETId. Ex. 3. — Eeduce 1 ton to kilograms, having given 1 gram = 15*4323 grains. 1 ton = 20 X 112 X 7000 grains. - 16-43 2 3 grams. = 1016050-7507 grams nearly. = 1016 kilog. 50 grams, 750*7 millig. nearly. Ex. 4. — Express £13. 17s. 4|d. in the pound and mil system. (£1 = 10 florins, 1 florin = 10 cents, 1 cent = 10 mils.) Keducing the given sum to the decimal of a pound, we have — £13. 17s. 4Jd. = £13*86875. = £13. 8 fl. 6 cent. 8f mil. Ex. XIY. 1. Express a mile in the metric system, having given that a metre = 39*3708 inches. 2. An are contains 1076*43 square feet. Reduce 53 ares 25 centiares to English measure. 3. The area of a room is 22 sq. met. 26 square decim. Express this in English measure (1 metre = 39*3708 inches). 4. A block of marble measures 3 feet, 3 inches in length, 2 feet, 6 inches in depth, and 3 feet, 9 inches in width. What is the solid content expressed in cuh. centim. 1 5. In 1235 litres how many gallons, when 50 litres =11 gallons nearly 1 6. Supposing a franc to be equivalent to 9|d., reduce £44. 13s. to francs. 7. Taking £1 sterling as equal to 25*22 francs, reduce £2. 13s. 7ld. to francs. 8. In 1852 France reaped about 47850000 hcctol. of wheat. Express this in gallons, assuming 1 gallon = 4*543 litres. 9. The ceiling of a room contains 83 sq. met. 53*96 sq. decim. What will be the expense of painting it at lOd. a square yard (1 metre = 1*094 yard)? 10. Find the cost of 2000 kilog. of sugar at fr. 50 c. per lb. PROPOKTION. 61 11. In England the unit of work is the foot-pound, and in the metric system it is the kilogram-metre. Reduce 62 metric units of work to English units, taking 1 gram = 15-4323 gi-ains, and 1 metre = 39*3708 inches. 12. The pressure of the atmosphere is 14| lbs. upon the square inch. Find the pressure in kilograms upon the square centimetre. CHAPTER V. PROPORTION. 38. Proportion is the equality of ratios. Thus, since the ratio 6:8 = f = J = i^, we have ratio 6:8 = ratio 15 : 20; and we say that the numbers 6, 8, 15, 20 form a proportion. We generally express the fact thus — 6 : 8 : : 15 : 20. It is easy to find hf trial that the product of the extreme terms is equal to the jjroduct of the means. Thus, we have 6 x 20 = 8. x 15. We may prove this property of the terms of a proportion to hold generally as follows : — Suppose we have given the proportion 12 : 21 : : 20 : 35. It follows, from our definition above, that \'\ — |2> ^^^^ multiplying each of these fractions by the product of their denominators, viz., by 21 x 35, we have ^f X 21 X 35 = fj X 35 X 21. Now (Art. 8), ^f X 21 = V = 12, and |§ x 35 = \o =, oq, and we hence have 12x35 = 20x21. Now, we have not in our reasoning taken into account the actual value of the terms of the given proportion ; and it is therefore evident that a similar result will follow from every proportion, and we may hence conclude generally : — In every proportion the product of the extremes i$ f^nal (q the product of the means. 62 ARITHMETIC. 39. Having given any three terms of a proportion, to find tlie remaining one. Since the product of the extremes is equal to the product of the means, the following rule is evident : — KuLE. — If the required term be a mean, divide the pro- duct of the extremes by the other mean ; but if the required term be an extreme, divide the product of the means by the other extreme. Ex. 1. — 28, 24, 30 are respectively the 1st, 3rd, and 4th terms of a proportion, required the 2nd term. "We have — 28 : required term : : 24 : 30 .-. required term - -^^-^ - ^-p - 35. Ex. 2. — 10, 45, 16 are respectively the 1st, 2nd, and 3rd terms, required the 4th term. We have — 10 : 45 : : 16 : required term .-. required term = ^^r^ = — i" = '^^' Ex. 3. — 2 hours, 45 minutes, 8 men are respectively the 1st, 2nd, and 3rd terms, required the 4th term. We must express (Art. 6) the 1st and 2nd terms in the same denomination, and the pToportion will stand thus — Min. Min. Men. 120 : 45 : : 8 : required term. Now, the ratio of the first two terms is the same as the ratio of the abstract numbers 120 and 45 ; and the 4th term must be of the same denomination as the 3rd term, otherwise the 3rd and 4th terms could not form a ratio. AVe have therefore — - Eequired term = --{~~' men = ^^ men = 3 men. Simple Proportion. 40. In Arithmetic we divide Proportion into Simple and Compound. Simple Proportion is the equality of two simple ratios, and therefore contains four simple terms ; and the usual problem is to find the fourth term, having given the Jirst three terms. SIMPLE PROPORTION. Go When we know the exact order of the given terms, the fourth term is, of course (Art. 38), found thus — Rule. — Multiply the 2nd and 3rd terms together, and divide by the 1st, The formal arrangement of the three given terms in their proper order is called the statement; and the only difficulty, therefore, in working a sum in Simple Proportion, or Single Rule of Three, as it is called, consists in stating it. We shall work a few examples to illustrate the mode of doing this. Ex. 1. — If 12 men earn £18, what will 15 men earn under the same circumstances ] We have hea^e two kinds of terms, men and earningSy and whatever ratio any given number of men bears to any second given number of men, it is evident that it must be equal to the ratio of the earnings of the first lot of men to the earnings of the second lot, and we may therefore write — Men. Men. 12 : 15 =- £18 : 2nd earnings. Men. Men. or 12 : 15 :: £18 : 2nd earnings. As the first two terms are of the same denomination, their ratio is not altered by treating them as abstract quantities, and the denomination of the 4th term must be the same as that of the 3rd. Hence we have — Ans. : = £^-^^ = £^-^ = £22. 10s. Ex. 2. — If 18 men do a piece of work in 25 days, in what time will 20 men do it % The two kinds of terms we have here to consider are men and time. In doing work we know that the tlTne will dimin- ish exactly as the number of men increases, and hence the ratio of the second lot of men to i}iQ first lot will be equal to the ratio of the given time to the time required. We therefore have — Men. Men. Days. 20 : 18 :: 25 : required time. .-. Ans. : = ^-^TT^ "^^Y^ = .^fi ^^J^ = 22-5 days. G4 ARITHMETIC. We Lave reasoned out the above examples thus to show that the working of problems in Rule of Three depends upon the ])rinciple of the equality of ratios. Practically, however, we proceed as follows : — Ex. 1. — If 12 men earn £18, what will 15 men earn under the same circumstances? We are required to find earnings^ and we therefore put down for the 3rd term the given earnings, thus — £ : :: 18 The question is with regard to 15 men instead of 12 men, and we know their earnings must be greatet*. We therefore place the greater of these terms in the 2nd place and the other in the 1st, and the statement becomes — Men. Men. £. 12 : 15 : 18 : required earnings. .*. as before — Ans. = £^-^77^ = £^^^~ - ^22. 10s. Ex. 2. — If 18 men do a piece of work in 25 days, in what time will 20 men do it ? We are required to find time, and we place therefore the given time, viz., 25 days, in the 3rd place. Again, the question is with regard to 20 men instead of 18 men. Kow, we know that 20 men require less time than 18 men to do a piece of work, and we hence place the less of these terms in the 2nd place. The statement then becomes — Men. Men. Days. 20 : 18 : : 25 : required time. ,'. as before — Ans. : = '-^~- days = '^-'- days = 22-5 days. Ex. XY. 1. If 12 articles cost £15, what will 624 cost? 2. What is the price of 35 loaves, when 29 loaves cost 15s. ShdA 3. If I get 140 metres of cloth for 541 ft*. 7^ c., what must I pay for 89 metres, 3 decim. ? COMPOUND PROPORTION. 65 4. If 4 cubic metres of water run into a cistern in 18 minutes, in what time will it be full, supposing it to be 4 metres long, 6 metres, 25 centim. deep, and 35 decim. wide ] 5. If the carriage of a parcel for the first 50 miles be Is. 3d., and if the rate be reduced by one-third for distances beyond, how far can the parcel be carried for Is. Td.l G. If a half-kilogram of sugar cost 1 fr. 10 c, what will be the cost of 3 kilog. 625 gleams. 1 7. There are two pieces of the same kind of cloth, measur- ing 43 yai-ds and 57 yards respectively, and the second costs £1. 9s. 2d. more than the first. What is the cost of the first? 8. A garrison of 720 men have provisions for 35 days, and after 7 days 120 more men arrive. How long will the i:)rovi- sions last ] 9. After paying 4d. in the pound income-tax a person has £299. 18s. 4d. left. What was the amount of his original income ] 10. Two clocks, one of which gains 3 minutes and the other loses 5 minutes per day, are put right at noon on Mon- day. What is the time by the second clock when the fii-st indicates 4 p.m. on the following Thursday'? 11. When will the hands of a clock be exactly 30 minute divisions apart between 2 and 3 o'clock ? 12. If I lend a friend £120 for 9 months, how long ought he to lend me £270? Compound Proportion. 41. Compound Proportion is an equality between ratios, one of which at least is a ratio compounded of two or more simple ratios. Arithmetical questions depending on Compound Proportion are generally said to belong to the Double Rule of Three ; and the proportion consists of an equality between a ratio, on the one hand, compounded of two or more simple ratios ; and, on the other hand, a simple ratio, whose consequent is required. The following examples will illustrate the method of T^or}^- ing questions in this rule : — QQ ARITHMETIC. Ex. 1.— If 12 horses eat 20 bushels of corn in 8 days, in what time will 24 horses eat 16 bushels'^ 24 horses : 12 horses ) o i . i ,. 20 bushels : 16 bushels / = = ^ ^^^^^ '• ''^l'^^'^'^ ^'"'^- Explanation. — We are required to find time ; and so, as in simple proportion, we put in the 3rd place the given time, viz., 8 days. Leaving, for the present, the quantity eaten out of considera- tio7i, we know that 24 horses requii-e less time to consume a given quantity of food than 12 horses do; we therefore place the less of these two terms in the 2nd place, and the other in the 1st place. (The statement up to this point is 24 horses : 12 horses : : 8 days : required time, and we might obtain 4 days as an answer, irrespective of the quantity eaten. We might now place this answer in the 3rd term of another simple proportion, and take the quantity eaten into consideration, irrespective of the number of horses, thus getting an answer depending both upon the number of horses and the quantity eaten. It is more convenient, however, to proceed thus :) Again, taking into consideration the quantity eaten, and leaving out of consideration the other given pair of terms, we see that less time is required to eat 16 bushels than to eat 20 bushels. We, therefore, jnit the less term in the 2nd place, and the other in the 1st. Now, treating the terms of the ratios which occupy the 1st and 2nd places as abstract quantities, and compound- ing them, we have : 24x20:12xl6::8 days : required time. .-. required time = ^ri^T^— days = 3-2 days. Ex. 2. — How much bread can I get for 9d. when wheat is at 18s. a bushel, if the foui-penny loaf weigh 3 lbs. when wheat is at 20s. a bushel i Proceeding as in Example 1, we have 18s * 20s I •• ^ ^^^* * weight required. Or, 4xl8:9x20::3 lbs. : weight required. ,-. weight requii^ed = --^i^r- ^^s. = 7-5 lbs. COMPOUND TROPORTION 67 Ex. XVI. 1. If 15 men can build a wall 81 feet long in 18 days, liow many men can build 135 feet of the same kind of wall in 30 days ] 2. In 4 days, 18 workmen can dig a ditch 162 yards long, 7 feet wide, and 12 feet deep. What must be the depth of a ditch which 45 workmen can dig in 7 days, supposing it to be 387 yards long and 5 feet wide 1 3. A traveller, going 15 hours a day, walks 1500 kilo- metres in 20 days. How far will he go in 30 days, walking 12 hours a day with the same velocity? Express your answer in English miles. 4. Two men are partners ; one puts in a capital of .£800, and receives as 6 months* profit £120. What is the capital of the other, who receives £3375 as 9 months' profit 1 5. Two tourists having spent £1. 16s. 8d. in 2^ days, meet three others with whom they continue their tour, and they spend while together £21. Is. 8d., at the same rate per day. Required how long they were in company. 6. If 16 men and 10 boys do a piece of work in 10 days, in how many days would 8 men and 18 boys do a piece 7 times as great, supposing the work of 5 boys equal that of 2 men? 7. Supposing the rate of carriage to be diminished one-thii-d after the first 50 miles, find the cost of carrying 16 cwt. for 40 miles, when 1 2 cwt. can be carried 1 00 miles for 4s. 2d. 8. A cistern is 8 metres, 4 decim. long, 1 metre, 8 centim. wide, and 275 centim. deep. Find the depth of another cis- tern of equal capacity whose length is 7 metres, 2 decim., and width 11 decim. 9. Persons whose incomes are less than £300 per annum are taxed upon £80 less than theii* income. Supposing 3 persons, having equal incomes, to pay £7 in the aggregate, at 4d. in the pound, find the total tax upon 14 persons, each havmg incomes 3 times as great. 10. If 12 horses eat 10 acres of gi'ass in 16 weeks, and 18 horses eat 10 acres in 8 weeks, how many hoi'ses would eat 40 acres in 6 weeks, the grass being supposed to grow uni- formly 1 68 ARITHMETIC. 11. A boat, propelled by 8 oars, whicli take 28 strokes per minute, goes at the rate of 9 J miles per hour. Find the rate of a boat propelled by 6 oars, which take 36 strokes per minute, the work done by each stroke of the latter being one-sixth less than that by each stroke of the former. 12. If 4 men and 1 women can do a piece of work in 8 days, which 1 2 women and 20 children can do in 4 days, in what time will 6 men, 18 women and 5 children do a work three times as great ] CHAPTEK VI APPLICATION TO ORDINARY QUESTIONS OP COMMERCE AND TRADE. Interest. 42. Interest is the money paid for the use of money. The Principal is the money lent, and the Amount is the sum of the interest and j)rincipal. The Rate of interest is the money paid for a given sum for a given time. £100 is in practice the given sum, and one year the given time. Thus, if <£4 be paid for the use of £100 for one year, the rate of interest is £4 per cent. ])ev annum, or as we generally say, 4 per cent. Simple Interest is interest calculated on the original principal only. Compound Interest is the interest which arises from adding the interest for each year to the principal of that year, and calculating interest for the next year upon the amount so obtained. , Simple Interest. 43. BuLE. — Multiply the prii:icipal by the rate per cent, and by the number of years ; then divide the product by 100, and the quotient will be the simple interest, SniPLE INTEREST. 69 Ex. 1. — Find the simple interest of £420 for 3 years at 5 per cent. £ 420 5 2 IOC) 3 £63-00 Ans.: £63. Reason for this process — £100 gains £5 in 1 year, and the question is to find the gain upon £420 in 3 years. Hence, proceeding as in Double Rule of Three, we have — 100 X 1 : 420 X 3 :: £5 : interest required. .*. interest required = £ 420 x _j_x_3^ which is exactly as stated in the rule. Ex. 2.— Find the simple interest of £352. Is. 8d. from March 16 to August 21, 1873, at 4 per cent. The following process will be easily underetood : — £ s. d, 352 1 8 4 £14-08 6 8 20 1-G(3S. 12_^ 8"00d. .*. £14. Is. 8d. is the interest for 1 year^ Now, from March 16 to August 21 are 158 days'; hencd we have — 365 days : 158 days :: £14. Is. 8d. : Interest requii-ed. .*. interest required = £6. Is. ll-j^jd. Ex. XYll. ) f Find the simple interest of— . \ 1. £350 for 4 years at 5 per cent, ! 2. £295. 2s. Id. for 3^ years at 4 per cent. 70 ARITHMETIC. 3. £375. 8s. 4d. for 2 J years at 4J^ per cent. 4. £160 from Feb. 1 to June 12, 1872, at 7|- percent, 5. £48 for 7 months at 1^ per cent, per month. . 6. £219. 4s. 2d. for 6 years at If per cent. In the six following examples, understand simple interest. 7. At what rate per cent, will £129. 8s. 4d. gain £6. 3s. 5ld. in 2|- years 1 8. A certain sum amounts in 3 years at 7|- per cent, to £289. 16s. 3|d. ; find the ori^ial sum. 9. In what time will £175. 6s. 3d. amount to £192. 7s. 10|-d. at 2| percent? 10. What sum will amount in 2 years 9 months at 4 per cent, to £427. 7s? 11. If £320 gain £9 in 13 months, in what time will £480 gain £6 at the same rate? 12. Find the interest of £29. 7s. 5d. for 6 months at 5^% per cent. Compound Interest. 44. Utile 1. — Find the interest for one year as in Simple Interest, and add it to the principal ; then find the interest for one year upon this amount reckoned as pfincijyal for the second year, and add it to the second year's principal, and so on. Subtract the original principal from the amount so obtained for the given number of years, and the result will be the compound interest required. Rule. 2. — Divide the given rate per cent, by 100, putting the result in a decimal form, and place U7iity before the decimal point. Kaise the number thus obtained to a power corresponding to the given number of years, multiply the principal by the result, and we get the amount for the given number of years. Thus, supposing 5 to be the rate per cent., we have, dividing by 100 and placing unity before the result, the number 1-05. Then, if the given number of years be 4, and £162 the principal, we have, according to rule, amount = £162 X (1-05)^ The first rule requii^es no explanation; the second mle may be explained thus : — COMPOUND INTEREST. 71 Ex. — Find the compound interest of £360 for 3 years at 4 per cent. Now, interest for .£100 for 1 year = £4, £1 „ =£-04; lience, amount of £1 „ = £1 + £-04 = £1-04. We thus see that the amount of £1 for 1 year at 4 per cent, is 1*04 times the original sum. It therefore follows that — Amount of the £1-04 for 1 year = 1*04 times £1*04, .-. amount of £1 „ 2 years =£(1-04 x 1*04) - £(1-04)2; and so, amount of £1 „ 3 years = 1*04 times £(1 -04)^ :^£(^04)^ hence, amount of £360 ,, „ = 360 times £(1-04)' = £360 X (1-04)^ The compound interest is then found by subtracting from this the original principal. Ex. 1. — Find the compound interest of £570 for 3 years at 5 per centi (We shall work this by Rule 1, and for convenience shall keep our quantities in a decimal form.) £ 570 5r £28*50 = interest for fii-st year. 570 £598*5 = principal for second year. 5 £29-925 - interest 598*5 £628*425 = principal for third year: 5 £31 '42125 = interest 628-425 £659*84625 = amount at end of third year, ) therefore 570 ~ original principal ; J subtracting, £8 9-84625 = compound interest for 3 years. 72 ARITHlilETia Ex. 2. — Find the compound interest o( £i'21. l'2s. Gd. for ■i years at 3 per cent. Now, £327. 12s. 6d. ^ £327-625. Hence, by Rule 2 — Amount -£327-625 x (1-03)^ :=: £368. 14s. lOfd. nearly. Ex. 3. — What sum of money, if i)ut out for 2 years at 4 per cent., will amount to £324. 9s. 7jd., compound interest being reckoned ? By Rule 2, the princijial may be found by dividing the amount by (1 -04)-. Now, given amount = £324. 9s. 74d. == £324*48. Hence principal required = £324-48 -f (1-04)- = £300.. Ex. XVIII. Find the compound interest of 1. £284 for 2 years at 4 per cent. 2. £312. 12s. 7|d. for 3 years at 5 per cent. 3. £283. 10s. for 2 years at 31 per cent. 4. £605. 12s. 6d. for 4 years at 4 per cent. 5. What is the difference between the simple and com- pound interest of £150 for 2 yeai^ at 6 per cent.? 6. Find the amount of £381. 1 florin 3 cents 5 mils for 3 years at 5 2)er cent. (£1 = 10 florins, 1 florin = 10 cents, 1 cent = 10 mils,) 7. Find the amount of £250 for 2 years at 4 j)er cent, per annum, interest being payable half-yearly. 8. What sum will amount in 3 years at i^ ^ler cent, com- l)ound interest to £200 ] 9. A town has 200,000 inhabitants, and it increases at the rate of 5 per cent, per annum ; And the number of inhabi- tants at the end of 3 years. 10. Find the difference in amount of £350 for 3 years at 4 per cent, simple interest, and £420 for 2 years at 5 per cent, compound interest. 11. HoAV much would a person who lays by £50 a year at 5 i)er cent, compound interest, draw out at the end of 4 years] 12. A person expects to receive £450 in 3 years ; what present sum is equivalent to this, reckoning compound inter- est at 4 per cent. ] DISCOUKT. 73 Discount. 45. When money is paid before it is due, the payee may, of course, put out the money at interest for the rest of the term, and thereby increase it. It therefore follows that the amount which ought to be paid for the discharge of an account before its proper time should be such a sum that, if put out at interest for the remainder of the term, will just amount to the original sum in question. Thus .£102. 10s. (interest being reckoned at 5 per cent, per annum) payable 6 months hence, would be fully dis- charged by paying £100 at once. For £100 in 6 months at 5 per cent, per annum would amount to £102. 10s. Hence, the payee ought to remit £2. 10s. from the full account. The amount remitted is called discount. It will be seen, therefore, that the discount on £102. 10s. due 6 months hence at 5 per cent, is £2. 10s. Bankers, however, are in the habit of charging interest instead of discount The banker's discount, therefore, on £100 due 6 months hence at 5 per cent, is £2. 10s. Hence, the (7^(6 discount on £102. 10s. due 6 months at 5 per cent, is the same as the hanker s discount on £100 under the same circumstances ; and bankers' discount on any given sum is in excess of the true discount. Tradesmen's bills are legally due three days after the term for which they are drawn is completed. This extension of time is called three days of grace. When a bill falls due on a Sunday, it is usual in England to meet it on the previous Saturday. Ex. 1. — Find the difference between the true discount and the banker's discount on£306 due 4 months hence at 6 per cent. Now, £100 would in 4 months gain \ of £6, or £2. Hence, the time discount on £102 due four months hence at G per cent, is £2, and therefore we have — £102 : £30G :: £2 : true discount required. Hence, true discount = ^^^VP "= '^^• Again, proceeding according to the rule for simple interest — -r^ , , ,. £306 xGxil- ^ , «^^,., Banker's discount = j^^j r= £f JS.= £6. 2s. 4jd. 74 Arithmetic. Hence, the excess of the banker's discount over the true discount is 2s. 4|d. Ex. 2. — A bill of .£350 drawn on March 15, at 6 months, is cashed on May 20, 1872 ; what is the banker's discount at 6 per cent. ? The bill is legally due on Sept 18, and from May 20 to Sept. 18 are 121 days. Now, the interest on <£350 for x year at 6 per cent, is easily found to be £21. Hence, 365 days : 121 days :: £21 . banker's discount requii^ed ; and /. banker's discount = £^,^!-^ = £6. 19s. UH. Percentages. 46. There are many questions which relate to ordinary commercial transactions which may be worked exactly as if we had to find the simple interest for one year — e.g., ques- tions in commission, brokerage, insurance^ &c. Commission is a sum of money charged, by ^n agent for buying or selling goods, at a certain rate per cent; upon the value of the goods; Brokeragei is similar to commission, but it is charged upoii money transactions instead of upon the sale of goods. Insurance is a sum charged per cent; Upon the value of property, the said value being paid to the insured in case of loss from causes as per agreement; Ex. 1.— Find the brokerage on £625; 5s; at 4 J per cent; £ s. d. 625 5 2501 312 12 d 6 28-13 12 2-72s. 12 6 8-70 Ans.: £28. 2s. Bx'^d. PERCENTAGES. 75 Ex. 2. — The rate of insurance is 4 J per cent., and the value of some pi-operty insured is worth £766. What will be the annual payment, so that, in case of fire, the owner may receive back his premium, as well as the value of his property 1 If, instead of paying £4. 5s. as insurance on every £100, he pays £4. 5s. upon every (£100 - £4. 5s.) or npon every £95. 15s. ; then, in case of fire, he will receive £100 for a damage of £95. 15s., and thus have the value of his property and the lamount of his premium. The problem is, therefore — If £4. 5s. is the premium on £95. 15s., what is the pre* mium on £766 ? Hence — £95. 15s. 1 £766 : : £4. 5s. : premitun required; And) therefore^ premium = £34. Ex. XIX. 1. Find the banker's discount on £4 12, dud months hence, at 6 per cent. 2. By how much ddes thes banker's discbunt bn £100^ due 3 months hence, at 5 per cent., exceed the true 1 3. What discount would be charged upon a bill drawii for £320, on April 15, at 4 months, and presented for payment on June 3 (discount at 7 per cent.) 1 4:i Find the discount on a bill drawn bn Aug. 3 for £200, at 6 months, and cashed on Sept. 10, discount being reckoned at 6 per cent. 5. A man buys gpbds for £250, being allowed G months credit, and he immediately feells them for the same amount, allowing 3 months credit. What does he gain by the trans- action, interest being reckoned at 5 per ceht. ? 6. Find the brokerage on £352. 17s. 6d. at 3 per ceni. 7. What is the brokerage on £4500 at -J per cent, i 8. What is the commission on the sale of goods to the amount of £850 at 5 per cent.? 9. A person insures for £1050 at 3f per cent. What is his annual premium 1 ARITHMETIC. 10. What "Will be the annual premium on pi*Operty "WoHh £965, so that the insured may obtain his premium back again with the vakie of his property, in case of loss — insurance being at 3 J per cent. I 11. The brokerage on a certain sum at J per cent, is £5. 10s. 7id. Find the sum. 12. Together with a commission of 4 per cent., goods cost a person £339. 8s. 5d. Find the cost price to the agent. Stocks and Shares. 47. "When a large amoimt of capital is to be raised, a company is generally formed, which raises the money by the issue of shares. We will suppose a person to hold a £100 share ; he will then be entitled to such a part of the profits of the company as £100 is of the whole capital. If there be a great demand for these shares, persons holding them may dispose of them for more than the nominal value, say for £106 ; whereas, if they are very little in demand, the seller may be glad to sell at, perhaps, £70. In the first instance, we should say that the shares were at 106, or at 6 preiniuni; and in the second instance, that the shares were at 70, or at 30 discount. If the selling price of the shares is £100, they are said to be at ^j>ar. So, when we read that the Three per Cent. Consols are quoted at 96|-, it means that an acknowledgment of indebted- ness on the part of the Government to the amount of £100, bearing interest at 3 per cent, per annum, may be bought for £96f. The buying and selling of stocks and shares is carried on by brokers, who charge a percentage from ^ to -|, sometimes upon the nominal value of the stock, but mostly upon the actual cash vahie. When brokerage is to be taken into ac- count in any example it will be mentioned, and it will be estimated by the first method, unless specified. Ex. 1.— What is the value of £1000 stock at 89tV per cent. ] No. of cents, stock = ^7 = 10 .-. value required - £89-r\ x 10 - £893. 2s. 6d. STOCKS AND SHARES. 77 Ex. 2.— What would be the cost of £1180 stock at 153^ per cent., including brokerage at i per cent.? No. of cents, stock = YcfcT = TVy and total cost of each cent. = £(153^ + i) = £153^. Hence, required cost - £153i x -^^f = £1811. 6s. Ex. 3. — What is the annual income arising from investing £6510 in tlie Four j^er Cents, at 93 ] Here, price of £100 stock - £93, and .-. No. of cents, stock for £6510 = ^^- = 70. Hence, annual income = £4 x 70 = £280. Ex. XX. 1. Find the cost of £750 stock at 92 J per cent. 2. I sell out £325 Three per Cent. Consols at 94. What do I get after allowing the broker i per cent, upon the cash he receives for the sale 1 3. Invest £5065. 9s. 9d. in the Three per Cent. Consols at 91f. 4. What would be the cost of £413. Is. 9d. reduced Three per Cents, at 92|, including a brokerage of \ per cent, upon the cost to the broker ] 5. What would be the proceeds of the sale of £6228 India Five per Cent, stock at 111^, deducting J jier cent, upon the selling price for brokerage 1 6. If I sell £8160 Spanish Three per Cent. Bonds at 31 per cent, and invest the proceeds, less J- per cent, brokerage, in Indian Railway Five per Cent. Stock at 107, what will be the dilfereiice in my annual income ] 7. What is the value sterling of $4000 American Bonds at 93f per cent. ($ = 4s. 6d.)? 8. Wliat is the value of |37,000 Unitea States Bonds at 93-j-v per cent. "J 78 ARITHMETIC. 9. Invest £954. Os. 7cl. in India Five per Cent. Stock at 11 Of, allowing i per cent, brokerage. 10. "What is the sterling value of 5000 francs Italian Bonds at 67 per cent. (Exchange 25 fr.)? 11. Find the sterling value of 6000 guilders Dutch 2 k per Cent. Bonds at 56^ (Exchange 12 guilders). 12. Invest £1025 in French Eentes at 51^, allowing i per cent, brokerage (Exchange 25 fr.). ^ Annuities. 48. Annuities are annual payments, the first payment being due at the end of a year. When an annuity is left untouched for a number of years, its amount is properly ob- tained by allowing compound interest. 49. To find the amount of an annuity of a given sum for a given time, at a given rate per cent. Let us suppose the annuity to be £1, the time 4 years, and 5 the rate per cent. The first payment will not be due till the end of a year ; so that at the end of 4 years it will have been accumulating for 3 years at compound interest; so the next payment, not being due till the end of the second year, will have been accumulating for 2 years at compound interest ; and so on, the last payment being made when due. Hence, the amount of an annuity of £1, for 4 years at 5 per cent., will be (reversing the above order) as follows : — £1 + amount of £1 for 1 year + amount of £1 for 2 years + amount of £1 for 3 years (compound interest being reckoned). And it is evident that for any other annuity we may follow the same method, and at the end multiply the sum by the number of £ in the annuity. Ex. — Find the amount of an annuity of £300 for 4 years at 5 per cent. We shall find the respective amoimts by Rule 2, Art, 44, Thus— ANNUITIES. £1*05 = amount for 1 year of £1 at 5 per cent. 1-05 525 105 iiM025 = „ „ 2 years 1-05 55125 11025_ 1-157625 = „ „ 3 years „ „ Hence, amount of annuity of £1 for 4 years — = £1 + £1-05 + £M025 + £M57625 = £4-310125. .-. amount of given annuity = £4-310125 x 300 = £1293. Os. 9d. 50. To find the present value of an annuity to continue for a given time. By the present value of an annuity to continue a given number of years is meant such a lump sum which, paid down at once, would, by accumulating at compound interest for the same time, amount to just the same sum as the annuity itself if it were allowed to accumulate. In the absence of alge- braical symbols, we shall best illustrate the method of finding such a lump sum by an example. Ex. — Find the present value of an annuity of £50 for 3 years at 4 per cent. As in Art. 49, we find the amount of the annuity — = {£1 + £1-04 + £(l-04)2[ X 50 = £156-08. Now, whatever be the present value required, we know that its amount in 3 years at 4 per cent, compound interest is found (Aii;. 44) by multiplying it by (1*04)^ or M24864. It, therefore, follows that if we know this amount before- liand we can find the present value by dividing it by 1-124864. Hence, present value = £156-08 4- 1-124864 = £138-755 nearly. 51. To find the present value of an annuity to continue for ever. 80 ARITHMETIC. It is evident that the sum we require is ene which, put out at interest, will annually produce a sum equal to that of the annuity itself. The problem then is simply this — Plaving given the interest for 1 year of a cei-tain sum, and the rate per cent., to find the principal. We have therefore the following rule : — KuLE. — Divide the given annuity by the rate per cen^., and multiply by 100, and the result is the present value. Ex. — How much must a gentleman invest at 5 per cent, in order to endow a charity with £60 a year. Pi*esent value of the annuity of £60 to continue for ever — There are many other questions connected with annuities which are, however, best left till the student has a knowledge of Logarithms. Ex. XXI. Find the amount of an annuity of — - 1. £120 for 3 years at 4 per cent. 2. £250 for 4 years at 4^ per cent. 3. £321 for 5 years at 5 per cent. 4. What is the present value of an annuity of £80, to continvie for six years, at 6 per cent. ] 5. A person who, according to the tables of mortality, is likely to live 10 years, wishes to insure an annual payment of £40 during life. What sum must he pay down, reckon- ing interest at 5 per cent. 1 (Give the result to four places of decimals.) G. A house produces a clear rental of £30. How many years' purchase is it worth, interest being reckoned at 5 per cent. 1 7. A gentleman invested a sum of money in the Three per Cent. Consols, in order that an annual payment of 7s. Gd. a year might be made in bread for ever. What sum did he invest 1 8. Find the present value of a pension of £120 a-year, payable half -y earl tj for 5 years, interest being at the rate pf 5 per cent, per annum. PROFIT AND LOSS. 81 9. A house, whicli ordinarily lets for £80 a-year, is leased for a term of four years, at a rent of £20, a certain sum being jmid in addition at the time of letting. Find this latter amount. 10. What is the present value of a freehold which pro- duces a clear rental of £50, but which cannot be entered upon for two years, reckoning interest at 5 per cent. 1 11. Find the annuity which in four years, at 4 per cent., will amount to £100. 12. A corporation borrows a sum of £3000 at 4 per cent. What annual payment will clear off the debt in ten years 1 (Give the result correct to four places of decimals.) Profit and Loss. 52. All questions involving the loss or gain per cent, by any transaction belong to this rule, and may be generally worked by Proportion. Ex. 1. — A man buys goods at 5s. and sells them at 5s. 8d. Find his gain per cent. The actual gain upon 5s. is 8d., and we are required to find the gain upon £100. Now 5s. : £100 : : 8d. : gain upon £100, ... gain upon £100 = M'ffff - £13^, or, required gain per cent = 13^. Ex. 2. By selling goods at 6s. 3d. there is a gain of 25' per cent. What will be the selling price to gain 10 per cent.] Now, selling price of goods which cost £100, so as to gain 25 per cent, is £125, and that to gain 10 per cent, is £110. Hence £125 : £110 : : 6s. 3d. : selling price required ; from which, selling price required = 5s. 6d. Ex. 3. — Find the cost price when articles sold at Is. 9d. entail a loss of 12^ per cent. Now, articles which cost £100 when sold at a loss of 12 J- per cent, must sell for £87^. Hence £87i- : Is. 9d. :: £100 : cost price required j from which, cost price = 2s, 5 F 82 ARITHMETIC, Ex. XXII. 1. Find tlie cost price of goods which are sold at a loss of 10 per cent, for 4s. lOid. 2. Goods which are sold for 7s. lid. entail a loss of 5 per cent. What should be the price to gain 30 per cent. 1 3. A tradesman reduces his goods 7 J- per cent. "What was the original price of an article which now fetches ^1. 7s. MJ 4. In what proportion must tea at 4s. 2d. be mixed with tea at 6s. a pound, so that a grocer may sell the mixture at 5s. 6d. and gain by the sale 10 per cent.? 5. A quantity of silk, after paying a duty of 12^ per cent., cost £54. Find the original cost price. 6. An innkeeper buys 37^ gallons of brandy at 14s. a gallon, and adds to it sufficient water to enable him to sell it at the same price and gain 12 per cent. How much water does he add ? 7. By selling goods at 8s. 2d. a tradesman gains 16| per cent. What will be the gain or loss per cent, by selling at 6s. lid. 8. A company has a capital of £750,000, and the working expenses for the year have been £42,123. 12s. 6d. What must have been the gross receipts in order that the share- holders may receive a dividend of 4 per cent. ? 9. If stock which is bought at 91^ is immediately sold at 9 If, what is the gain per cent."? 10. A person buys goods at 6 months' credit and sells them for cash at the nominal cost price immediately. What is his gain per cent.? (Interest 5 per cent.) 11. Goods are marked at a ready-money price and a credit price allowing 12 months. The credit price is £4. 9s. 3d., what is the ready-money price ? 12. Goods are now being sold at 10 per cent. loss. ! How' much per cent, must be put upon the selling price in order that they may be sold at 20 per cent, gain ? ESTIMATES. 83 Square Root and Cube Boot. 53, To avoid unnecessary repetition, the student is referred to the articles on Involution, Algebra, stage I., where the arithmetical principles and methods are explained. Estimates, 54. The following specimens will give the student an idea of what he may expect to meet with under the head of Esti- mates, It is usual, in ordinary transactions, to use certain abbx^eviations ; as cuh. for cubic measure, sup, for superficial measure, run. for running or lineal measure. Builders, too, are in the habit of calling twelfths of a foot — whether it be cubical, superficial, or lineal measure — by the name of inches. The names yards, feet, inches, are often written thus : yds., ', ", Ex. 1. — Digger, Bricklayer, and Mason. Yds. 25 232 5 IG G 14 9 33 12 5 19 Cub. Sup. ft Run. No. Run No. Diggings in trenches, filling-, wheeling, and carting away, 9" reduced common brickwork in mortar, pointed on both sides, ^y trimmer arches to hearths, Best blue brick on edge, paving in cement, 6" blue and red quarries do., 12" round blue coping bricks in cement, . . . Best red brick flat steps do.. Extra only to splayed brick angles to doors and windows, Do. to plinth in cement, Do. to sailing course to chimney, Do. to cornice to eaves of best red bricks, three courses, 2 core chimney flues, 1 set sink, •i do. stoves, 1 do. copper, 12 do. ornamental air gratings, Gable coping of hard stone 13" X 4i", twice splayed, 2 J rubbed hearth and back hearth, Solid York step 14" x 6", tooled, 2 knee stones to gables 18" x 14" x 10", ea. 1 apex stone do., 16" x 14" x 12",. 1 York stone sink 4' 0" X 1' 9", with rounded corners and hole cut for waste-pipe, 2 stone chimney-pieces with 7" X IJ", cham- fered jambs, mantel, and shelf, ea. Total, 1/ 4/2 2/3 4/6 3/10 2/4 1/2 0/2i 0/3 0/2 0/5^ 1/6 4/ 5/3 5/ 0/6 1/9 1/2 2/6 7/6 8/ 10/ 24/ 84 ARITHMETIC, Ex. 2. — Carpenter and Joiner. (A square = 100 sq. ft.) Sq. 21 Cub. Sup. Run. Sup. Run. No. Sup. Run. Sup. Run. Sup. Run, No. Fir framed-in roof timbers, Do. do., floors, Centreing to trimmer hearths, Do. to 3 openings 5 0" wide, with segment heads in 1^ brick wall ea. Do, to 4 do. 3' 0" wide do., ea. Labour in planing roof and floor-timbers, . . Do. in stop chamfering edges of do., 7" X ly ridge board, 3" X 1^" chamfered fillet to eaves, Inch clean red deal batten for boarded floors, wrought, 7" X 1" torus skirting, plugged, 2 labour to mitred margins to hearths, ... 16 mitres to skirting, ea. Inch deal treads and risers, ploughed, tongued, and screwed on 3 — 7" x 2^-" carriages 1^" wall-string housed, for treads and risers, 1^" close-string do., and sunk and beaded,.. 3'' X 2^" rounded oak hand-rail, French- polished, 9" X 1" beaded fascia, 2" — 6 panel doors, bead, flush, and square, 1^" do. do., 4|" X 3" rebated and beaded frame, 4|" X 2" do. do., 3" X 1^" moulding, mitred, 2" ovalo-moulded sashes, double hung, in deal-cased frames, with oak sunk and weathered sills, 1|" do. do., ^" mitred bead, 5^" X f" bead lining, 11 inch-beaded centre boards, Total, 2/9 2/8 0/4 2/ 1/6 0/Oi 0/Oi 0/3 0/3 0/9J 0/2 0/9 0/8 0/9 2/ 0/4 1/3 1/1 0/6 0/4^ 0/3 MISCELLANEOUS EXAMPLES. 86 MISCELLANEOUS EXAMPLES. {Selected from University and other Examination Papers.) 1. Show by an easy example tliat the division of one whole number by another is equivalent to a series of subtractions. Divide 1-02 by ^ J of -144. 2. If the Three per Cents, are at 91i, what interest does this give on £100 ? (Omit brokerage and fractions of a penny.) 3. How many lbs. in -321875 of a ton weight? Convert it into kilograms (omitting fractions), assuming that a cubic decimetre of distilled water weighs 15432*35 grains. 4. Eeduce to theii- simplest forms — r^^^z:: r^— and V3-5 - x/2-1 V^5'12 4- v^-03375 ^ V^80- y^ 5. Convert ^is ^^^ ^ decimal fraction, and find the vulgar fraction corresponding to the recurring decimal '22297. 6. Show, by proper attention to the value of the figures, in multiplying one number by another, that the order in which the figures of the multiplier are taken is of no importance. Multiply 6M43 by 47-982 correctly to three places of decimals, beginning with the left hand figure of the multiplier, and use as few figures as possible. 7. Extract the square root of 1095*61, and find to three places of decimals the value of V^ - r 8. Find the compound interest of £55 for one year, pay- able quarterly, at 5 per cent, per annum. A person bought into the Three per Cents, at 98, and after receiving three years' interest he sold at 90. How much per cent, on the sum invested did he gain or lose ? 86 ARITHMETIC. 9. Three gardeners working all day can plant a field in 10 days ; but one of them having other employment can only work half time. How long will it take them to complete the work ? 10. What fraction of a crown is f of 6s. 8d. 1 What is the value of f of a guinea ? Reduce llfd. to a decimal of a pound, correct to five places of decimals. 11. Reduce the expressions — Multiply 49if by SO^V, and add ^|^ to the result. Divide (2tV)' - 1 by {2^\y + S^V- 12. A bankrupt's estate amounts to £9 10. 3s. l|d., and his debts to £1875. What can he pay in the pound 1 and what will a creditor lose on a debt of £57 ? 13. A person having invested a sum of money in the Three per Cent. Consols received annually therefrom £233, after deducting the income-tax of 7d. in the pound. What is the sum of money? What can the stock be sold for when Consols are at 94-J-. 14. Find the value of -jm^^SiSioA and of 'UUi^- ■ 006 -vu i L z o 15. Prove the rule for finding the value of a circulating decimal, and divide 4*367 by the circulating decimal •052. 3 + /5 Reduce to its simplest form the quantity ^_^- _ 3-^5 8 - ^20 "" 8 + ^20 16. Three persons, A, B, C, hold a i)asture in common, for which they are to pay £30 per annum. A put in 7 oxen for 3 months ; B, 9 oxen for 5 months ; and C, 4 oxen for 12 months. How much rent ought each to jiay? 17. Calculate to four places of decimals the value of the . I of -31416 expression 7^~ V *93 MISCELLANEOUS EXAMPLES. 87 18. Find the least common multiple of 16, 24, and 30, and explain the method. 19. Wliat should be the price of English standard silver, 37-40ths fine, in order that the par of exchange between England and France should be 25 fr. 22 c. — 200 francs being coined from 1 kilogram of silver, 9-lOths fine? (1 kilog. = 15*434: grains). 20. A person buys 100 shares in a company for £3,500; after receiving four half-yearly dividends of 15s. 4d., 20s; lOd., 30s. 4d., and 38s. 9d. per share, he sells at a profit of 43 per cent. ; reckoning the simple interest of money at 4 per cent., how much above that interest has he gained? 21. The price of Three per Cent. Consols is 90| j what sum must be invested in order to purchase £24 per anmmi ; and what is the rate of interest on the money invested 1 22. Three partners in trade contribute respectively the sums of £438, £292, £730, with the agreement that each was to receive 5 per cent, on their respective investments, and that the remainder of the gains of the firm, if any, was to be divided between them in the proportion of the sums originally advanced. The whole gain of the firm was £200. What was each man's share ? 23. If 25 tons of goods are purchased for £37. 10s. and sold at 35s. a ton, what is the gain per ton ? At what rate per ton should the goods have been sold in order to obtain a profit of £9. 7s. 6d. ? 24. Find the value of /t of £3. 12s. ll|d. ; and find the fraction that 3 miles, 2 fur. 100 yards is of 12 leagues, 2 fur. 20 yards. 25. The sum of £9040. IGs. is placed in the Three and a Half per Cents, at 94 ; find the income obtained, allowing on the stock purchased -Jth per cent, to the broker, and ^^ per cent, for other expenses. 26. Express as a fraction '200123, and express as a recurring decimal '012 t '00i32. ARITHMETIC. 27. By the reduction of the income-tax from 7cl. in the pound to 5d. a person saves £28. 2s. 6d. a year ; what is his income 1 28. If 81 bushels of wheat are consumed by 56 men in 5 days, how long will 16 men take to consume 28 bushels 1 29. Find the square root of 4, and prove that v -694 = 83. 30. The periods of three planets which move uniformly in circular orbits round the sun are respectively 200, 250, and 300 days. Supposing that their positions relative to each other and to the sun to be given at any moment, determine how many days must elapse before they again have exactly the same relative positions. SECTION II. GEOMETRY. EUCLID'S ELEMENTS, BOOK L Definitions. 1 . A point is that which has position, but not magnitude. 2. A line is length without breadth. 3. The extremities of a line are points. 4. A straight line is that which lies evenly between its extreme points. 5. A superficies (or surface) is that which has only length and breadth. 6. The extremities of a superficies are lines. 7. A plane superficies is that in which any two points being taken, the straight line between them lies wholly in that superficies. 8. A plane angle is the inclination of two lines to one another in a plane, which meet together, but are not in the same direction. 9. A plane rectilineal angle is the inclination of two straight lines to one another, which meet together, but are not in the same straight line. Note. — When several angles are at one point B, any one of them is expressed by three letters, of which the middle letter is B, and the tirst letter is on one of the straight lines which contain the angle, and the last letter on the other line. DO GEOMETRY. Thus, the angle contained by the straight lines AB and BC is ex- pressed either by ABC or CBA, and the angle contained by AB and BB is expressed either by ABD or DBA. When there is only one angle at any given point, it may be expressed^^by the letter at that point, as the angle E. 10. When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle; and the straight line which stands on the other is called a perpendicular to it. 1 1 . An obtuse angle is that which is greater than a right angle. 12. An acute angle is that which is less than a right angle. 13. A term or boundary is the extremity of anything. 14. A figure is that which is enclosed by one or more boundaries. 15. A circle is a plane figure contained by one line, which is called the circumference, and is such, that all straight lines drawn from a certain point within the figure to the circumference are equal to one another. 1 6. And this point is called the centre of the circle, [and any straight line drawn from the centre to the circumference is called a radius of the circle]. DEFINITIONS. 91 1 7. A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference. 18. A semicircle is the figure contained by a diameter and the part of the circumference cut off by the diameter. 19. A segment of a circle is the figure contained by a straight line and the part of the circumference which it cuts off. 20. Rectilineal figures are those which are contained by straight lines. 21. Trilateral figures, or triangles, by three straight lines. 22. Quadrilateral figures, by four straight lines. 23. Multilateral figures, or polygons, by more than four straic;ht lines. 2 4. Of three-sided figures an equilateral triangle is that which has three equal sides. ^ 25. An isosceles triangle is that which has only two sides equal. 26. A scalene triangle is that which has three unequal sides. /\ 27. A right-angled triangle is that which has a riffht angle. 28. An obtuse-angled triangle is that which has an obtuse angle. 29. An acute-angled triangle is that which has three acute andes. 92 GEOMETRY. 30. Of four-sided figures, a square is tliat wliich has all its sides equal, and all its angles right angles. 31. An oblong is that which has all its angles right angles, but not all its sides equal. 32. A rhombus is that which has all its sides equal, but its angles are not right angles. 33. A rhomboid is that which has its opposite sides equal to one another, but all its sides are not equal, nor its angles right angles. 34. Parallel straight lines are such as are ^. in the same plane, and which being produced ever so far both ways do not meet. 35. A parallelogram is a four-sided figure of which the opposite sides are parallel ; and the diagonal is the straight line joining two of its opposite angles. All other four-sided figures are called trapeziums. Postulates. 1. Let it be granted that a straight line may be drawn from any one point to any other point. 2. That a terminated straight line may be produced to any length in a straight line. 3. And that a circle may be described from any centre, at any distance from that centre. Axioms. 1. Things which are equal to the same thing arc equal to one another. 2. If equals be added to equals the wholes are equal. EXPLANATION OP TERMS AND ABBREVIATIONS. 93 3. If equals be taken from equals the remainders are equal. 4. If equals be added to unequals the wholes are unequal. 5. If equals be taken from unequals the remainders are unequal. G. Things which are double of the same are equal to one another. 7. Things which are halves of the same are equal to one another. 8. Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another. 9. The whole is greater than its part. 10. Two straight lines cannot inclose a space. 11. All right angles are equal to one another. 12. If a straight line meet two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles, these straight lines being continually produced shall at length meet on that side on which are the angles which are less than two right angles. Explanation of Terms and Abbreviations. An Axiom is a truth admitted without demonstration. A Theorem is a truth which is capable of being de- monstrated from previously demonstrated or admitted truths. A Postulate states a geometrical process, the power of efiecting which is required to be admitted. A Problem proposes to efiect something by means of admitted processes, or by means of processes or constnic- tions, the power of effecting which has been previously demonstrated. A Corollary to a proposition is an inference which may be easily deduced from that proposition. The sign = is used to express equality. 4 means angle^ ^^nd A signifies triangle, . 94 GEOMETRY. The sign r> signifies '4s greater than," and <: '' is less than." + expresses addition ; thus AB + BC is the line whose length is the sum of the lengths of AB and BO. — expresses subtraction ; thus AB - BC is the excess of the length of the line AB above that of BC. AB- means the square described upon the straight line AB. AC=AB. BC=AB. ACandBC each=AB. .•.AB = =CA. :BC Proposition 1.— Problem. To describe an equilateral triangle on a given finite straight line. Let AB be the given straight line. It is required to describe an equilateral triangle on AB. Construction. — From the centre A, at the distance AB, describe the circle BCD (Post. 3). From the centre B, at the dis- tance BA, describe the circle ACE (Post. 3). From the point C, in which the circles cut one another, draw the straight lines CA, CB to the points A and B (Post. 1). Then ABC shall he an equilateral triangle. Proof. — Because the point A is the centre of the circle BCD, AC is equal to AB (Def 15). Because the point B is the centre of the circle ACE, BC is equal to BA (Def. 15). Therefore AC and BC are each of them equal to AB. But things which are equal to the same thing are equal to one another. Therefore AC is equal to BC (Ax. 1). Therefore AB, BC, and CA are equal to one another. Therefore the triangle ABC is equilateral, and it is de- scribed on the given straight line AB. Which was to he done. PROPOSITIONS. 95' Proposition 2.— Problem. From a given point to draw a straight line equal to a given straight line. Let A be the given point, and BC the given straight line. It is required to draw from the point A a straight line equal to BC. Construction. — From the point A to B draw the straight Draw ab. line AB (Post. 1). Upon AB describe the equilateral triangle DAB (Book I., A dabc- Prop 1) quilateral. Produce the straight lines DA, DB, to E and F (Post. 2). From the centre B, at the dis- tance BC, describe the circle CGH, meeting DF in G (Post. 3). From the centre D, at the dis- tance DG, describe the circle GKL, meeting DE in L (Post. 3). Then AL shall be equal to BC. Proof. — Because the point B is ^^^"^-J^^^^^ bc=bg. the centre of the circle CGH, BC is equal to BG (Def. 15). Because the point D is the centre of the circle GKL, DL i>l=dg. is equal to DG (Def. 15). But DA, DB, parts of them, are equal (Construction). da=db. Therefore the remainder AL is equal to the remainder BG iv/-^ = (Ax. 3). But it has been shown that BC is equal to BG. Therefore AL and BC are each of them equal to BG. ^ ^^ and But things which are equal to the same thing are equal to bg. ~ one another, therefore AL is equal to BC (Ax. 1). Therefore from the given point A a straight line AL has /. Ai^ - been dra^vn equal to the given straight line BC. Which was ^^' to he done. Proposition 3. — Problem. From the greater of two given straight lines to cut off a part equal to the less. Let AB and C be the two given straight lines, of which AB is the greater. 96 GEOMETRY. AE=AD. AD=C. AE and C each=AD. .'. AE=C. It is required to cut off from AB, the greater, a part equal to C, the less. Construction. — From the point A draw the straight line AD equal to C (I. 2). From the centre A, at the distance AD, describe the circle DEF, cutting ABinE (Post. 3). Then AE slmll he equal to C. Proof. — Because the point A is the centre of the circle DEF, AE is equal to AD (Def. 15). But C is also equal to AD (Construction). Therefore AE and C are each of them equal to AD. Therefore AE is equal to C (Ax. 1). Therefore, from AB, the greater of two given straight lines, a part AE has been cut off, equal to C, the less. Q,. E. F.^ AB=DE. AC=DF. ZBAC: /C EDF. Proposition 4.— Theorem. If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal to one another : they shall have their bases, or third sides, equal ; and the two triangles shall be equal, and their other angles shall be equal, each to each, viz., those to which the equal sides are opposite. Or, If two sides and the contained angle of one triangle be re- spectively equal to those of another, the triangles are equal in every respect. Let ABC, DEF be two triangles which have The two sides AB, AC, equal to the two sides DE, DF, each to each, viz., AB equal to DE, and AC equal to DF. And the angle BAC equal to the angle EDF : — then — The base BC shall be equal to the base EF ; The triangle ABC shall be equal to the triangle DEF ; ^ Q. E. F. is an abbreviation for quod erat faciendum, that ig " which ivas to he done," PROPOSITIONS. 97 And tliG other angles to wliich the equal sides are opposite, shall be equal, each to each, viz., the angle ABC to the angle DEF, and the angle ACB to the angle DFE. Proof. — For if the triangle ABC be applied to {or placed Suppose upon) the triangle DEF, ^ ^^ ' . iJiit upon So that the point A may be on the point D, and the ^ def. straight line AB on the straight line DE, The point B shall coincide with the point E, because AB is equal to DE (Hypothesis). Ajid AB coinciding with DE, AC shall coincide with DF, because the angle BAC is equal to the angle EDF (Hyp.). Therefore also the point C shall coincide with the point F, because the straight line AC is equal to DF (Hyp.). But the point B was proved to coincide with the point E. , Therefore the base BG shall coincide with the base EF. Because the point B coinciding with E, and C with F, if the base BC do not coincide with the base EF, two straight lines would enclose a space, which is impossible (Ax. 10). Therefore the base BC coincides with the base EF, and is bc=ef. therefore equal to it (Ax. 8). Therefore the whole triangle ABC coincides with the whole .*. A abo triangle DEF, and is equal to it (Ax. 8). = A def. And the other angles of the one coincide with the remain- z arc = ing angles of the other, and are equal to them, viz., the angle ^ acb*= ABC to DEF, and the angle ACB to DFE. ^ dfe. Therefore, if two triangles have, &c. (see Enunciation). Which was to he shown. Proposition 5.— Theorem. Tlie angles at the hose of an isosceles triangle are equal to one ano titer ; and if the equal sides he produced , the angles upon the other side of the hose shall also he equal. Let ABC be an isosceles triangle, of which the side AB is ab = ac. equal to the side AC. Let the straight lines AB, AC (the equal sides of the tri- angle), be produced to D and E. The angle ABC shall be equal to the angle ACB (angles at the hase)y 5 G 98 GEOMETRY. And the angle CBD shall be equal to the angle BCE {angles upon the other side of the base). Construction. — In BD take any point F. AG = AF. / \ From AE, the greater, cut off AG, equal to AF, the less (I. 3). Join FC, GB. Proof.— Because AF is equal to AG (Construction), and AB is equal to AC (Hyp.), Therefore the two sides FA, AC are equal to the two sides GA, AB, each to each ; And they contain the angle FAG, common to the two triangles AFC, AGB. .•.FC=GB Therefore the base FC is equal to the base GB (I. 4); = A AGB^ And the triangle AFC to the triangle AGB (I. 4) ; And the remaining angles of the one are equal to the z ABG~ remaining angles of the other, each to each, to which the z AFC= equal sides are opposite, viz., the angle ACF to the angle /. AGB. ^-g^^ ^^^^ ^^^ ^^gj^ j^^^ ^^ ^^^ ^^^^^ ^^^ ^ ^y And because the whole AF is equal to the whole AG, of which the parts AB, AC, are equal (Hyp.), BF=zCG. The remainder BF is equal to the remainder CG (Ax. 3), And FC was proved to be equal to GB ; Therefore the two sides BF, FC are equal to the two sides CG, GB, each to each. And the angle BFC was proved equal to the angle CGB; Therefore the triangles BFC, CGB are equal; and their other angles are equal, each to each, to which the equal sides are opposite (I. 4). i z GCB Therefore the angle FBC is equal to the angle GCB, and z BCF =' the angle BCF to the angle CBG. z CBG. ^j^(j since it has been demonstrated that the whole angle ABG is equal to the whole angle ACF, and that the parts of these, the angles CBG, BCF, are also equal, ii ^ ^n Therefore the remaining angle ABC is equal to the remain- mg augle ACB (Ax. 3), Which are the angles at the base of the triaugle ABC. PROPOSITIONS. 99 And it has been proved that the angle FBC is equal to the angle GCB (Dem. 11), Which are the angles upon the other side of the base, Therefore the angles at the base, &c. (see Enunciation). Which was to be shown. CoROLi^ARY. — Hence every equilateral triangle is also equiangular. Proposition 6. — Theorem. 7/ two angles of a triangle he equal to one anotlier, the ' sides also which subtend, or are op2)Osite to, the equal angles, shall be equal to one another. Let ABC be a triangle having the angle ABC equal to the angle ACB. The side AB shall be equal to the side AC. For if AB be not equal to AC, one of thenx is greater Suppose than the other. Let AB be the greater. ^^ ^^^'• Construction. — From AB, the greater, cut off a part DB, Make equal to AC, the less (I. 3). DB = Aa Join DC. Proof. — Because in the triangles DBC, ACB, DB is equal to AC, and BC is common to both, Therefore the two sides DB, BC are equal to the two sides AC, CB, each to each ; And the angle DBC is equal to the angle ACB (Hyp.) Therefore the base DC is equal to the base AB (L 4). And the triangle DBC is equal to the tri- angle ACB (I. 4), the less to the greater, ^ ^ A acb" which is absurd. Therefore AB is not unequal to AC, that is, it is equal to it. Wherefore, if two angles, &c. Q. E, D. * Corollary. — Hence every equiangular triangle is also * equilateral. ■" Q. E. D. is an abbreviation for quod erat demonstrandum, that is, **ivhich teas to be shown or proved" 100 GEOMETRY. Proposition 7.— Theorem. Upon the same base, and on the same side of it, there cannot he two triangles tlmt have their sides, lohich are terminated in one extremity of the base, equal to one another, and likewise those lohich are terminated in the other extremity. Let the triangles ACB, ADB, upon the same base AB, and on the same side of it, have, if possible, Their sides CA, DA, terminated in the extremity A of the base, equal to one another ; And their sides CB, DB, terminated in the extremity B of the base, likewise equal to one another. Case I. — Let the vertex of each triangle A J3 be without the other triangle. Construction. — Join CD. Proof. — Because AC is equal to AD (Hyp.), The triangle ADC is an isosceles triangle, and the angle ACD is therefore equal to the angle ADC (I. 5). But the angle ACD is greater than the angle BCD (Ax. 9). Therefore the angle ADC is also greater than BCD. Much more then is the angle BDC gi-eater than BCD. Again, because BC is equal to BD (Hyp.), The triangle BCD is an isosceles triangle, and the angle BDC is equal to the angle BCD (L 5). But the angle BDC has been shown to be gi'eater than the angle BCD (Dem. 5). Therefore the angle BDC is both equal to, and greater than the same angle BCD, which is impossible. Case II. — Let the vertex of one of the triangles fall within the other. Construction. — Produce AC, AD to E and F, and join CD. Proof. — Because AC is equal to AD (Hyp.), The triangle ADC is an isosceles triangle, and the angles ECD, FDC, upon the other side of its base CD, are equal to one another (I. 5). PROPOSITIONS. 101 But the angle ECD is greater tlian the angle BCD (Ax. 9). Therefore the angle FDC is likewise greater than BCD. Much more then is the angle BDC greater than BCD. / bcd. Again, because BC is equal to BD (Hyp.), The triangle BDC is an isosceles triangle, and the angle zbdc = BDC is equal to the angle BCD (I. 5). ^ ^^^• But the angle BDC has been shown to bo greater than the angle BCD. Therefore the angle BDC is both equal to, and greater than .-. zbdc the same angle BCD, which is impossible. ^ Tbcd. Therefore, upon the same base, . Proof. — ^Because AC is equal to CB -^ ^ ^ (Const.), and CD common to the two triangles ACD, BCD; The two sides AC, CD are equal to the two sides BC, CD, each to each ; And the angle ACD is equal to the angle BCD (Const.) ; Therefore the base AD is equal to the base DB (I. 4). Therefore the straight line AB is divided into two equal parts in the point D. Q, E. F. Proposition 11. — Problem* To dfaw a straight line at right angles to a given straight line from a given 'point hi the same. Let AB be the given straight line, and C a given point in it. It is required to draw a straight line from the point C at right angles to AB. Construction. — ^Take any point D in AC. Make CE equal to CD (I. 3). Upon DE describe the equilateral triangle DFE (I. 1). Join EC. y Then EC shall he at right angles to AB. Proof. — Because DC is equal to CE (Const.), and EC common to the two triangles DCE, ECE j ^ The two sides DC, CE, are equal a J) to the two sides EC, CE, each to each ; And the base DE is equal to the base EE (Const.) ; Therefore the angle DCE is equal to the angle ECE (I. 8) ; /^ I And they are adjacent angles. But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle (Def. 10) ; ^DCFr 104 GEOMETRY. ;. L DCF, Z ECF are right angles. Make ZABEa right L /CBE = ZEBA. and ZDBE== ZEBA. .-. Z DBE = Z CBE. Therefore eacli of tlie angles DCF, ECF is a right angle. Therefore from the given point C in the given straight line AB, a straight line FC has .been drawn at right angles to AB. Q.E.F. Corollary. — By help of this problem, it may be demon- strated that Two straight lines cannot have a common segment. If it be possible, let the two straight lines ABC, ABD, have the segment AB common to both of them. Construction. — From the point B, draw BE at right angles to AB (1. 11). Proof. — Because ABC is a CBE is equal to the angle EBA C the angle CD as ra- dius. Bisect FG inH. straight line, (Def. 10). Also, because ABD is a straight line, the angle DBE is equal to the angle EBA (Def. 10). Therefore the angle DBE is equal to the angle CBE. The less to the greater; which is impossible. Therefore two straight lines cannot have a common seg- ment. Proposition 12. — Problem. To draw a straight line perpendicular to a given straight line of unlimited lengthy from, a given point without it. Let AB be the given straight line, which may be produced to any length both ways, and let C be a point without it. It is requii-ed to draw from the point C, a straight line perpendicular to AB. Construction. — Take any point D upon the other side of AB. From the centre C, at the distance CD, describe the circle EGF, meet- ing AB in F and G (Post. 3). Bisect FG in H (I. 10). Join CF, CH, CG. Then CH shall he jyerpetulicular to AB. Proof* — Because FH is equal to HG (Const.), and HC fcommon to the two triangles FHC, GHC ; PROPOSITIONS. 105 The two sides FH, HG are equal to the two sides GH, HG, each to each ; And the base GF is equal to the base GG (Def. 15) ; Therefore the angle GHF is equal to the angle GHG (I. 8), /. adjacent and they are adjacent angles. ch'f.^cho But when a straight line, standing on another straight are equal, line, makes the adjacent angles equal to one another, each of the angles is called a right angle, and the straight line which stands on the other is called a perpendicular to it (Def. 10). Therefore, from the given point G, a perpendicular has been drawn to the given straight line AB. Q, E, F, Proposition 13. — Theorem. TJie angles wliich one straight line makes with another upon one side of it, are either two right angles, or are together equal to two right angles. Let the straight line AB make with GD, upon one side of it, the angles GBA, ABD. These angles shall either be two right angles, or shall to- gether be equal to two right angles. Proof. — If the angle GBA be equal to the angle ABD, each of them is a right angle (Def. 10). But if the angle GBA be not equal to the angle ABD, from the point B, draw BE at right angles to GD (I. 11). Therefore the angles GBE, EBD, are two right angles. Now the angle GBE is equal to the two angles GBA, ABE; to each of these equals add the angle EBD. D Make z CBE = z EBD = a right 2 . B C i5 B C Therefore the angles GBE, EBD, are equal to the three angles GBA, ABE, EBD (Ax. 2). Again, the angle DBA is equal to the two angles DBE, EBA ; to each of these equals add the angle ABG. .•.ZCBE+ z EBD = Z CBA-f Z ABE + z EBD, al- so z DBA + Z ABC Therefore the angles DBA, ABG, are equal to the three •= ^ ^^J angles DBE, EBA, ABG (Ax. 2). + z abc. lOG GEOMETRY, But the angles CBE, EBB have been shown to be equal to the same three angles ; And things which are equal to the same thing are equal to one another * .-. z CBE Therefore the angles CBE, EBD, are equal to the angles ± i III DBA, ABO (Ax. 1). + z ABC. But the angles CBE, EBD are two right angles. Therefore the angles DBA, ABC, are together equal to two right angles (Ax* 1). Therefore, the angles which one straight line, &c. Q. E. D, Proposition 14.— Theorem. If, at a point in a straight lin£, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall he in one and the same straight line. At the point B in the straight line AB, let the two straight lines BC, BD, upon the opposite sides of AB, make the ad- jacent angles ABC, ABD together equal to twO right angles. BD shall be in the same straight line with BC. For if BD be not in the same straight line with BC, let BE be in the same straight line with it. Proof. — Because CBE is a straight line, and AB meets it in B. Therefore the adjacent angles ABC, ABE are together equal to two right angles (I. 13). But the angles ABC, ABD, are also together equal to two right angles (Hyj^.) ; Therefore the angles ABC, ABE, are equal to the angles ABC, ABD (Ax. 1). Take away the common angle ABO* The remaining angle ABE is equal to the remaining angle = z ABD. ABD (Ax. 3), the less to the gi-eater, which is impossible ; Therefore BE is not in the same straisjht line with BC. Given z ABC + z ABD= two right angles. If possible^ let CBE be a straight line. z ABE PROPOSITIONS. 107 And, ill like manner, it may be demonstrated that no other can be in the same straight line with it but BD. Therefore BD is in the same straight line with BC. Therefore, if at a point, &c. Q. E. D, Proposition 15.— Theorem. If two straight lines cut one another, the vertical, 01* o])2)Osite angles shall he equal. Let the two straight lines AB, CD cut one another in the l)oint E. The angle AEC shall be equal to ^""\^^ angle DEB, and the angle CEB to ^^^v^ the angle AED. A E\~ B" Proof. — Because the straight line ^^ ^ cea 4- AE makes with CD, the angles CEA, ^''n^ht^ " AED, these angles are together equal to two right angles angles. (I. 13). Again, because the straight line DE makes with AB the ^ ^^^ , angles AED, DEB, these also are together equal to two right z deb == angles (I. 13). l^^l But the angles CEA, AED have been shown to be together equal to two tight angles. Therefore the angles CEA, AED are equal to the angles AED, DEB (Ax. 1). Take away the common angle AED. The remaining angle CEA is fequal to the remaining angle • ^ cE\ DEB (Ax. 3). = ^ DEB. In the same manner it can be shown that the angles CEB, AED are fequal. Therefore, if two straight lines^ &c. Q, E. D, Corollary 1. — From this it is manifest that if two straight lines cut one another, the angles which they make at thei point where they cut^ are together equal to four right angles. Corollary 2. — And, conse^iiently, that all the • angles made by any number of lihes ineeting in one point are together equal to four right angles, provided that no one of the angles be included in any other angle. 108 GEOMETRY. Proposition 16. — Theorem. // one side of a triangle he producedy the exterior angle shall be greater than either of the interior opposite angles. Let ABC be a triangle, and let its side BC be produced to D. The exterior angle ACD shall be greater than either of the interior opposite angles CBA, BAG. Mako /\ y/ Construction. — Bisect AC in E , .. , (I- 10). EF = BE. / \4 / Join BE, and produce it to F, making EF equal to BE (I. 3), and join FC. Proof. — Because AE is equal to EC, and BE equal to EF (Const.), AE, EB are equal to CE, EF, Cr\ each to each ; And the angle AEB is equal to the angle CEF, because they are opposite vertical angles (I. 15). Therefore the base AB is equal to the base CF (I. 4) ; And the triangle AEB to the triangle CEF (I. 4) ; And the remaining angles to the remaining angles, each to each, to which the equal sides are opposite. zBAE Therefore the angle BAE is equal to the angle ECF (I. 4). But tlie angle ECD is greater tlian the angle ECF (Ax. 9) ; ' _ Therefore the angle ACD is greater than the angle BAE. . z ACD In the same manner, if BC be bisected, and the side AC be produced to G, it may be proved that the angle BCG (or its equal ACD), is greater than the angle ABC. Therefore, if one side, &c. Q. E, D, Proposition 17.— Theorem. Any two angles of a triangle arc together less than two right angles. Let ABC be any triangle. Any two of its angles together shall be less than two right angles. -. L ECF. -^BAE. PROPOSITIONS. 1 09 Construction. -^Produce BC to D. Proof.— Because ACD is the ex- terior angle of the triangle ABC, it is greater than the inteiior and opposite angle ABC (I. 16). To each of these add the ancjle ACB. ^ ^ »' Therefore the angles ACD, ACB are greater than the ^ abc + angles ABC, ACB (Ax. 4). /.A^f ^ But the angles ACD, ACB arc together equal to two a"gies. right angles (I. 13); Therefore the angles ABC, ACB are together less than two right angles. In like manner, it may be proved that the angles BAC, ACB, as also the angles CAB, ABC are together less than two right angles. Therefore, any two angles, &c. Q, E. i>. Proposition 18.— Theorem. The greater side of every triangle is oj^j^osite the greater angle. Let ABC be a triangle, of which the side AC is greater ^^ > ^2^ than the side AB. The angle ABC shall be greater than the angle BCA. Construction. — Because AC is greater than AB, make AD equal to AB (I. 3), and join BD. B C Proof. — Because ADB is the exterior angle of the triangle ^ ^y)V, > BDC, it is greater than the interior and opposite angle BCD ^ i^cd, and z ADB = z ADD A (I. 16), But the angle ADB is equal to the angle ABD ; the triangle BAD being isosceles (I. 5), and Therefore the angle ABD is greater than the angle BCD >/bcd (or ACB). Much more then is the angle ABC greater t^an the angle ACB. Therefore, the greater side, tkc. Q. E. D, * no GEOMETRY, Proposition 19.— Theorem, The greater cmgle of every triangle is subtended hy the greater side, or has the greater side opposite to it, z' ABC >• -Let ABC be a triangle, of which the angle ABC is greater - ^^^- than the angle BCA; Tlie side AC shall be greater than the side AB. Proof. — If AC be not greater than AB, it must either be equal to or less than AB. It is not equal, for then the angle __^ ABC would be equal to the angle BCA ^ ^ (I. 5); but it is not (Hyp.) ; AC not = Therefore AC is not equal to AB. Neither is AC less than AB, for then the angle ABC would be less than the angle BCA (I. 18); but it is not (Hyp.); AC not < Therefore AC is not less than AB. And it has been proved that AC is not equal to AB ; Therefore AC is greater than AB. Therefore, the greater angle, &c. Q, E. D, Proposition 30. — Theorem. Any two sides of a triangle are together greater than the third side. Let ABC be a triangle ; Any two sides of it are together greater than the third side. Maiie Construction. — Produce BA to the point D, making AD •^^ =^^' equal to AC (I. 3), and join DC. Proof. — Because DA is equal to AC, the angle ADC is equal to the angle ACD (I. 5). But the angle BCD is greater than the angle ACD (Ax. 9); ^ 2SJ? ^ Therefore the angle BCD is greater than the angle ADC dKBDC). And because the angle BCD of the ^ I triangle DCB is greater than its angle BDC, and that the greater angle is subtended by the gi^eater side ; ■.DB>BC. ■** G Therefore the side DB is greater than the side BC (I. 19). PROPOSITIONS. Ill But BD is equal to BA and AC ; Therefore BA, AC are greater than BC. ac'^bc. In the same manner it may be proved that AB, BC are greater than AC ; and BC, CA greater than AB. Therefore any two sides, &c. Q, E, J), Proposition 21. — Theorem. If from the ends of the side of a triangle there he draimi iivo straight lines to a 2>oint within the triangle, these shall he less than the other two sides of the triangle, hut shall contain a greater angle. Let ABC be a triangle, and from the points B, C, the ends of the side BC, let the two straight lines BD, CD be drawn to the point D within the triangle; BD, DC shall be less than the sides BA, AC ; But BD, DC shall contain an angle BDC greater than the angle BAC. Construction. — Produce BD to E. Proof. — 1. Because two sides of a triangle are gi^eater than the third side (I. 20), the two sides BA, AE, of the triangle BAE are greater than BE. To each of these add EC. Therefore the sides BA, AC, are B greater than BE, EC (Ax. 4). Again, because the two sides OE, ED, of the triangle ^^' CED are greater than CD (I. 20), To each of these add DB. Therefore CE, EB are greater than CD, DB (Ax. 4). But it has been shown that BA, AC are greater than ^b^cd BE, EC ; + DB. Much more then are BA. AC greater than BD, DC. Proof. — 2. Again, because the exterior angle of a tiiangle is greater than the interior and opposite angle (I. 16), therefore BDC, the exterior angle of the triangle ^f^l^^ CDE, is greater than CED or CEB. ^^^^ ceb. For the same reason, CEB, the exterior angle of the tri- ^J^ ceb > angle ABE, is greater than the angle BAE or BAC, ^ bae. BA + AC > BE + 112. GEOMETRY. And it has been shown that the angle BDC is greater than CEB ; • I BDC Much more then is the angle BDC gi^eater than the angle >zBAC. BAG. Therefore, if from the ends, &c. Q, E. D, Proposition 22.—- Problem. To 7)iake a triangle of which the sides shall he equal to three given straight lines, hut any two whatever of these lines 7nust he greater than the third (I. 20). Let A, B, C be the three given straight lines, of which any two whatever are greater than the thii'd — namely, A and B greater than C, A and C greater than B, and B and C greater than A ; Jt is required to make a triangle of which the sides shall be equal to A, B, and C, each to each. Construction. — Take a straight line DE terminated at the point D, but unlimited to- wards E. Make DF equal to A, EG equal to B, and GH equal to C (I. 3). Erom the centre E, at the distance ED, describe the circle C DKL (Post. 3). andOH From the centre G, at the distance GH, describe the ^^■^'"^- circle HLK (Post. 3). Join KE, KG. Then the triangle KEG sliall Imve its sides equal to the three straight lines A, B, C. Proof. — Because the point E is the centre of the circle DKL, ED is equal to EK (Def. 15). But ED is equal to A (Const.) ; nc = A. Therefore EK is equal to A (Ax. 1). Again, because the point G is the centre of the circle HLK, GH is equal to GK (Def. 15). But GH is equal to C (Const.) ; GK = c. Therefore GK is equal to C (Ax. 1), PROPOSITIONS. 113 And FG is equal to B (Const.) ; > fg = B Therefore the three straight lines KF, FG, GK are equal to the three A, B, C, each to each. Therefore the triangle KFG has its three sides KF, FG, GK equal to the thi'ee given straight lines A, B, C. Q, E. i\ Proposition 23.— Problem. At a given 2)oint in a given straight liiie, to make a recti- lineal angle equal to a given rectilineal angle. Let AB be the given straight line, and A the given point in it, and DCE the given rectilineal angle. It is requii-ed to take an angle at the point A, in the straight line AB, equal to the rectilineal angle DCE. Construction. — In CD, CE, take any points D, E, and join DE. On AB constiTict a triangle AFG, the sides of which shall be equal to the three straight lines CD, DE, EC— namely, AF equal rl to CD, FG to DE, and AG to EC (1.22); Then the angle FAG shall he equal to the angle DCE. Proof. — Because DC, CE are equal to FA, AG, each to each, and the baso DE equal to the base FG (Const.), The angle DCE is equal to the angle FAG (I. 8). Therefore, at the given point A, in the given straight line AB, the angle FAG has been made equal to the given recti- lineal angle DCE. Q, E, F. Tlien z DCE = z FAG Proposition 24.— Theorem. If two triangles have two sides of the one equal to two sides of the other, each to each, hut the angle contained by the two sides of one of them greater than the angle contained by the two sides equal to them of the other, the base of that ivhich has th^ greater angle shall be greater tJian the base of the other. 5 U 114 GEOMETilY. Suppose DF > DE. Make Z EDG = z BAG. .•.BC=EG. and ^ EFG:: - EGF. .'.EG>EF. Let ABC, DEF, be two triangles which have The two sides AB, AC equal to the two DE, DF, each to each — namely, AB to DE,and AC to DF, But the angle BAC greater than the angle EDF; The base BC shall be greater than the base EF. Construction. — Let the side DF of the triangle DEF be , greater than its side DE. A. ^ Then at the point D, in the straight line ED, make the angle EDG equal to the angle BAC (I. 23). Make DG equal to AC or DF (I. 3). Join EG, GF. Proof. — Because AB is equal to DE (Hyp.), and AC to DG (Const.), the two sides BA, AC are equal to the two ED, DG, each to each ; And the angle BAC is equal to the angle EDG (Const.) ; Therefore the base BC is equal to the base EG (I. 4). And because DG is equal to DF (Const.), the angle DFG is equal to the angle DGF (L 5). But the angle DGF is greater than the angle EGF (Ax. 9); Therefore the angle DFG is greater than the angle EGF ; Much more then is the angle EFG greater than the angle EGF. And because the angle EFG of the triangle EFG is greater than its angle EGF, and that the greater angle is subtended by the greater side. Therefore the side EG is greater than the side EF (I. 19). But EG was proved equal to BC ; Therefore BC is greater than EF. Therefore, if two triangles, &c. Q. E. D. Proposition 25.— Theorem. If two triangles have two sides of the one equal to two sides of the other y each to each, hut the base of the one greater than the base of the other, the angle contained by the sides of that which has the greater base shall be greater tlmn the angle contained by the sides equal to tliem of the other. PKOPOSITIONS. 115 Let ABC, DEF, be two triangles, which have The two sides AB, AC equal to the two sides DE, DF, each to each — namely, AB to DE, and AC to DF, But the base BC greater than the base EF ; The angle BAC shall be greater than the angle EDF. Proof. — For if the angle BAC be not greater than the A. p angle EDF, it must either be equal to it or less. But the angle BAC is not equal to the angle EDF, for then the base BC would be equal to the base EF (I. 4), but it is not (Hyp.); Therefore the angle BAC is not equal to the angle EDF; ^.^^edf! Neither is the angle • BAC less than the angle EDF, for then the base BC would be less than the base EF (I. 24), but it is not (Hyp.), Therefore the angle BAC is not less than the angle EDF. <^/edf! And it has been proved that the angle BAC is not equal to the angle EDF; Therefore the angle BAC is greater than the angle EDF, Therefore, if two triangles, &c. Q, E. D. Proposition 26. — Theorem. If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side — namely, either the side adjacent to the equal angles in each, or the side opposite to them ; then shall the other sides he equal, eaxih to each ; and also the third angle of the one equal to the third angle of the other. Or, If two angles and a side in one triangle he respectively equal to two angles and a corresponding Me in another triangle, the tria7igles shall he equal in every respect. Let ABC, DEF be two triangles, which have • ^ The angles ABC, BCA equal to the angles DEF, EFD, each to each—namely, ABC to DEF, and BCA to EFD; Also one side equal to one side. Case 1. — First, let the sides adjacent to the ecjual angles ^c!^ef. in each be equal— namely, BC to EF; 116 GEOMETRY. Then shall the side AB be equal to DE, the side AC to DF, and the angle BAG to the angle EDF. For if AB be not equal to DE, one of them must be ft- ,j) greater than the othei*. Let AB be the greater of the two. Construction. — Make BG- equal to DE (I. 3), and join GC. Proof. — Because BG is equal to DE (Const.), and BC V is equal to EF (Hyp.), the two sides GB, BC are equal to the two sides DE, EF, each to each. And the angle GBC is equal to the angle DEF (Hyp.) ; Therefore the base GC is equal to the base DF (I. 4), And the triangle GBC to the triangle DEF (I. 4), And the other angles to the other angles, each to each, to which the equal sides are opposite; Therefore the angle GCB is equal to the angle DFE (I. 4). But the angle DFE is equal to the angle BCA (Hyp.) ; Therefore the angle GCB is equal to the angle BCA (Ax. 1), the less to the greater, which is impossible; Therefore AB is not unequal to DE, that is, it is equal to it; and BC is equal to EF (Hyp.) ; Therefore the two sides AB, BC are equal to the two sides DE, EF, each to each. And the angle ABC is equal to the angle DEF (Hyp.) ; Therefore the base AC is equal to the- base DF (I. 4), And the third angle BAC to the third angle EDF (I. 4). Case 2. — Next, let the sides which are opposite to the equal angles in each triangle be equal to one another — namely, AB equal to DE. Likewise in this case the other sides shall be equal, AC to A D DF, and BC to EF ; and also the angle BAC to the angle EDF. For if BC be not equal to EF, one of them must be greater than the other. Let BC be the greater of the two. H c B F Construction, — Make BH equal to EF (I. 3), and join AH, PROPOSITIONS. 117 Proof. — Because BH is equal to EF (Const.), and AB is bh = ef, equal to DE (Hyp.), the two sides AB, BH are equal to the ^^ ~ ^^' two sides DE, EF, each to each. And the angle ABH is equal to the angle DEF (Hyp.); ^ abh = Therefore the base AH is equal to the base DF (I. 4), And the triangle ABH to the triangle DEF (I. 4), And the other angles to the other angles, each to each, to which the sides are opposite ; Therefore the angle BHA is equal to the angle EFD (I. 4). . ^ ^j^^ But the angle EFD is equal to the angle BCA (Hyp.); =z efd Therefore the angle BHA is also equal to the angle BCA ~ ^ ^^^ (Ax. 1) ; _ That is, the exterior angle BHA of the triangle AHC, is equal to its interior and opposite angle BCA, which is impossible (I. 16) ; Therefore BC is not unequal to EF — that is, it is equal to it ; BC not and AB is equal to DE (Hyp.) ; to ef.* Therefore the two sides AB, BC are equal to the two sides DE, EF, each to each. And the angle ABC is equal to the angle DEF (Hyp.) ; Therefore the base AC is equal to the base DF (I. 4), And the third angle BAC is equal to the third angle EDF (I. 4). Therefore, if two triangles, &c. Q. E. D. i . Proposition 27.— Theorem. ' If a straight line faUimj upon two other straight lines make the altetviate angles equal to one another^ these two straight lines shall be 2:>aralleL Let the straight line EF, which falls upon the two straight lines AB, CD, make the alternate angles ^ 5/ 5^^ Given AEF, EFD, equal to one another. / ^\r^ z efd ' AB shall be parallel to CD. / ^^ For if AB and CD be not parallel, ^ A i> they will meet if produced, either / towards B, D, or towards A, C. Let them be produced, and meet towards B, D, in the point G. 118 GEOMETRY. L AEF> Proof. — Then GEF is a triangle, and its exterior angle ^ EFG, AEF is greater than the interior and opposite angle EFG and also (J. 16). = z EFG. But the angle AEF is also equal to EFG (Hyp.), which is impossible ; Therefore AB and CD, being produced, do not meet towards B, D. In like manner it may be shown that they do not meet towards A, C. But those straight lines in the same plane which being produced ever so far both ways do not meet are parallel (Def. 34) ; Therefore AB is parallel to CB. Therefore, if a straight line, &c. Q. E. D, Proposition 28.— Theorem. If a straight line falling upon two other straight lines ruahe the exterior angle equal to the interior and opposite upon the sarne side of the line, or make the interior angles upon the same side together equal to two right angles, the two straight lines shall he parallel to one another. Let the straight line EF, which falls upon the two straight lines AB, CD, make — The exterior angle EGB equal to the interior and opposite PA angle GHD, upon the same side ; Or make the interior angles on the -B same side, the angles BGH, GHD, together equal to two right angles ; C ^v^^ ^D ^B shall be parallel to CD. Proof 1 .—Because the angle EGB "^■P is equal to the angle GHD (Hyp.), And the angle EGB is equal to the angle AGH (I. 15) ; Therefore the angle AGH is equal to the angle GHD (Ax. 1), and these angles are alternate; Therefore AB is parallel to CD (I. 27). Proof 2.— Again, because the angles BGH, GHD are equal to two right angles (Hyp.), And the angles BGH, AGH are also equal to two riffht angles (I. 13). .*. z. AGH = z GHD PROPOSITIONS. 119 Therefore tlie angles BGH, AGH are equal to the angles .*. / bgh BGH, GHD(Ax. 1). =zbg2 Take away the common angle BGH. + ^ ^hd. Therefore the remaining angle AGH is equal to the remain- ing angle GHD (Ax. 3), and they are alternate angles. Therefore AB is parallel to CD (I. 27). Therefore, if a straight line, &c. Q, E, D, Proposition 29.— Theorem. If a straight line/all upon two parallel straightlineSyit makes the alternate angles equal to one another, and the exterior angle equal to the interior and opposite upon the same side ; and also the two interior angles upon the same side together equal to two right angles. Let the straight line EF fall upon the parallel straight lines AB, CD; The alternate angles AGH, GHD shall be equal to one another. The exterior angle EGB shall be p\ equal to GHD, the interior and op- \ posite angle upon the same side ; A A^ — B And the two interior angles on the \ same side BGH, GHD shall be to- ^ \, - ^ gether equal to two right angles. \ For if AGH be not equal to ^^ z agh > GHD, one of them must be greater ghd. than the other. Let AGH be the greater. (*>uppose.) Proof. — Then the angle AGH is greater than the angle GHD; to each of them add the angle BGH. Therefore the angles BGH, AGH are greater than the angles BGH, GHD (Ax. 4). But the angles BGH, AGH are together equal to two i-ight angles (1. 3). .^ ^^jj Therefore the andes BGH, GHD are less than two right + z ghd ^ ^ . Propositiou 38, — Theorem. Triarigles upon equal bases ^ and between the same parallels , are equal to one another. Let the triangles ABC, DEF, be on equal bases BC, EF, and between the same parallels BF, AD. The triangle ABC shall be equal to the triangle DEF. Construction. — Produce AD both ways to the points G, H. Through B draw BG par- f. A B allel to CA, and through F draw FH parallel to ED (L 31) Fisfurcs CiiiCA figures GBCA. DEFH, is a ^ c E F »efh Proof. — Then each of the \/ \ / \/ ^^i^^A and DEFH are 128 GEOMETRY. parallelogram (Def. 35), and they are equal to one another, be- cause they are on equal bases BO, EF, and between the same parallels BF, GH (I. 36); and the And the triangle ABC is half of the parallelogram Ir'J'half of C^BCA, because the diagonal AB bisects it (I. 34) ; these re- And the triangle DEF is half of the parallelogram «pectiveiy. DEFH, because the diagonal DF bisects it (I. 34). But the halves of equal things are equal (Ax. 7); Therefore the triangle ABC is equal to the triangle DEF. Therefore, triangles, &c. Q, E, D, Proposition 39.— Theorem. Equal triangles upon the same base, and on the same side o/itf are between the same parallels. Let the equal triangles ABC, DBC be upon the same base BC, and on the same side of it ; They shall be between the same parallels. Construction. — Join AD ; AD shall be parallel to BC. For if it is not, through A draw AE parallel to BC (I. 31), and join EC. Proof. — The triangle ABC is equal to the triangle EBC, because they are upon the same base BC, and between the same B c parallels BC, AE (I. 37). But the triangle ABC is equal to the triangle DBC (Hyp.) ; Therefore the triangle DBC is equal to the triangle EBC (Ax. 1), the greater equal to the less, which is impo^ible ; Therefore AE is not parallel to BC. In the same manner, it can be demonstrated that no line passing through A can be parallel to BC, except AD ; Therefore AD is parallel to BC. Therefore, equal triangles, &c. Q. E, D, Proposition 40. — Theorem. Equal triangles nxpon the same side of equal bases, tliat are in the same straight line, are between the same parallels. Let the equal triangles ABC, DEF, be upon the same side of equal bases BC, EF, in the same straight line BF, AE parallel to BC sup- pose. Then A DBC = A EBC, an ubsurdity. PROPOSITIONS. 129 Tlie triangles ABC, DEF shall be between the same parallels. Construction.— Join AD ; AD shall be parallel to BF. For if it is not, through A draw AG parallel to BF(I. 31), j^f to^^: and join GF. suppose. Proof.— The triangle ABC is equal to the triangle GEF, because they are upon equal bases BC, EF, and are between the same parallels BF, AG (I. 38). But the triangle ABC is equal to the triangle DEF ; Therefore the triangle DEF is equal to the triangle GEF (Ax. 1), the gi^eater equal to the less, which is impossible ; Therefore AG is not parallel to BF. In the same manner, it can be demonstrated that no line, passing through A, can be parallel to BF, except AD ; Therefore AD is parallel to BF. Therefore, equal triangles, &c. Proposition 41.— Theorem. If a 'parallelogram and a triangle he upon the same base, and between the same parallels^ the piarallelogram shall he double of the triangle. Let the parallelogram ABCD, and the triangle EBC be upon the same base BC, and between the same parallels BC, AE ; The parallelogi-ani ABCD shall be double of the triangle EBC. Construction. — Join AC. Proof. — The triangle ABC is equal to the triangle EBC, because they are upon the same base BC, and between the same parallels BC, AE (I. 37). But the parallelogram ABCD is double of the triangle And parai- ABC, because the diagonal AC bisects the parallelogram (I. 2^^^^b^ 130 GEOMETRY. Therefore the parallelogram ABCD is also double of the triangle EBC (Ax. 1). Therefore, if a parallelogram, &c. Q. E, J), Proposition 42.--Problem. To describe a parallelogram that shall he equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let ABC be the given triangle, and D the given recti- lineal angle ; It is required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D. Make BE A Y G CONSTRUCTION. BisCCt BC in E (I. = *^^ /ITT 7 10), and join AE. / Y / / At the point E, in the straight line z"cEF=D / i\ / /d ^-^j make the angle CEF equal to D / / \ / (^- ^^)- / / \/ Through A draw AEG parallel to B E c EC (I. 31). Through C draw CG parallel to EE (I. 31). Then EECG is the parallelogram required. Proof. — Because BE is equal to EC (Const.), the triangle ABE is equal to the triangle AEC, since they are upon equal bases and between the same parallels (I. 38) ; I^A^AEC Therefore the triangle ABC iS double of the triangle and also ' AEC. FECG = ^^t the parallelogram EECG is also double of the triangle 2 A AEC. AEC, because they are upon the same base, and between the same parallels (I. 41) ; Therefore the parallelogram EECG is equal to the triangle ABC (Ax. 6), And it has one of its angles CEF equal to the given angle D (Const.). Therefore a parallelogram FECG has been described equal to the given triangle ABC, and having one of its angles CEF equal to the given angle D. Q. E. F, A ABC: 3PJ10P0SITI0NS. 131 Proposition 43.— Theorem. The complements of the parallelograms which are about the dmgonal of any parallelogram are equal to one another. Let ABCD be a parallelogi-am, of whicli the diagonal is AC ; and EH, GF parallelograms about AC, that is, through which AC passes ; and BK, KD the other parallelograms, which make up the whole figure ABCD, and are therefore called the complements. The complement BK shall be equal to the complement KD. Proof. — Because ABCD is a parallelogram, and AC its diagonal, the triangle ABC is equal to the triangle ADC a adc.~ (I. 34). Again, because AEKH is a paral- lelogram, and AK its diagonal, the triangle AEK is equal to the triangle AHK (I. 34). For the like reason the triangle KGC is equal to the triangle KFC. Therefore, because the triangle AEK is equal to the triangle AHK, and the triangle KGC to KFC, The triangles AEK, KGC are equal to the triangles AHK, KFC (Ax. 2). But the whole triangle ABC was proved equal to the whole triangle ADC ; Therefore the remaining complement BK is equal to the .:.^bk = remaining complement KD (Ax. 3). Therefore, the complements, &c. Q. E. D, KD. Proposition 44.--Problem. To a given straight line to apply a parallelogram, lohich shall he equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let AB be the given straight line, C the given triangle, and D the given angle. 132 GEOMETRY. Make parallelo- jrram BEFG = A C, and z at B = L D, and EBAa straight line. HBandFE meet. It is required to apply to tlie straight line AB a parallelo- gram equal to tlie triangle C, and having an angle equal toD. Construction 1. — Make the parallelogram BEFG equal to the triangle C, and having the angle EBG equal to the angle D (I. 42) ; And let the parallelogram BEFG be made so that BE may be in the same straight line with AB. Produce EG to H. Through A draw AH parallel to BG or EF (I. 31). Join HB. Proof 1. — Because the straight line HE falls on the parallels AH, EF, the angles AHF, HFE are together equal to two right angles (I. 29). Therefore the angles BHF, HFE are together less than two right angles (Ax. 9). But straight lines which with another straight line make the interior angles on the same side together less than two right angles, will meet on that side, if produced far enough (Ax. 12) ; Therefore HB and FE shall meet if produced. Construction 2. — Produce HB and FE towards BE, and let them meet in K. Figures L13 = BF. Through K draw KL parallel to EA or FH (I. 31). Produce HA, GB to the points L, M. Then LB shall he the ioarallelograi)i required. Proof 2. — Because HLKF is a parallelogram, of which the diagonal is HK ; and AG, ME are the parallelograms about HK ; and LB, BF are the complements ; Therefore the complement LB is equal to the complement BF (I. 43). z D. PROPOSITIONS. 133 But BF is equal to the triangle C (Const.) ; But_ ^ Therefore LB is equal to the triangle C (Ax. 1). .-. lb = ' And because the angle GBE is equal to the angle ABM ^^* (I. 15), and likewise to the angle D (Const.) ; Therefore the angle ABM is equal to the angle D (Ax. 1). ^^^^^^^ ^ Therefore, the parallelogi-am LB is applied to the straight z gbe = line AB, and is equal to the triangle C, and has the angle ABM equal to the angle D. Q, E. F, Proposition 45. — Problem. To describe a parallelogram equal to a given rectilineal figure, and having an angle eqvxd to a given rectilineal angle. Let ABCD be the given rectilineal figure, and E the given rectilineal angle. It is required to describe a parallelogram equal to ABCD, and having an angle equal to E. Construction. — Join DB. A .. T) F Cx l i Describe the parallelo- gram FH equal to the triangle ADB, and having the angle FKH equal to the angle E (I. 42). To the straight line GH apply the parallelogram GM equal to the triangle DBC, and having the angle GHM equal to the angle E (I. 44). Then the figure FKML shall he the parallelogram required. Proof. — Because the angle E is equal to each of the angles FKH, GHM (Const.), Therefore the angle FKH is equal to the angle GHM (Ax. 1). Add to each of these equals the angle KHG; Tlierefore the angles FKH, KHG are equal to the angles KHG, GHM (Ax. 2). But FKH, KHG are equal to two right angles (I. 29); Therefore also KHG, GHM are equal to two right angles (Ax. 1). 134 GEOMETRY. And because at the point H, in the straight line GH, the two straight lines KH, HM, on the opposite sides of it, make the adjacent angles together equal to two right angles, Therefore KH is in the same straight line with HM (1. 14). And because the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal (I. 29). Add to each of these equals the angle HGL; Therefore the angles MHG, HGL are equal to the angles HGF, HGL (Ax. 2). But the angles MHG, HGL are equal to two right angles (L 29); Therefore also the angles HGF, HGL are equal to two right angles, And therefore FG is in the same straight line with GL (L 14). And because KF is parallel to HG, and HG parallel to ML (Const.); Therefore KF is parallel to ML (I. 30). And KM, FL are parallels (Const) ; Therefore KFLM is a parallelogram (Def. 35). And because the triangle ABD is equal to the jDarallelo- gram HF, and the triangle DBC equal to the parallelogram GM (Const.), Therefore the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM (Ax. 2). Therefore, the parallelogram KFLM has been described equal to the given rectilineal figure ABCD, and having the angle FKM equal to the given angle E. Q. E. F, Corollary. — From this it is manifest how to apply to a given straight line a parallelogram, which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure — namely, by applying to the given straight line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle ; and so on (I. 44). Proposition 46.— Problem. To describe a square upon a given straight line. Let AB be the given straight line; It is required to describe a square upon AB. PROPOSITIONS. 135 Construction. — From the point A draw AC at right angles to AB (I. 11), And make AD equal to AB (I. 3). Through the point D draw DE parallel to AB (I. 31). Through the point B di'aw BE parallel to AD (I. 31). Then ADEB shall he the square re- quired. Proof. — Because DE is parallel to AB, and BE parallel to AD (Const.), therefore ADEB is a parallelogram; Therefore AB is equal to DE, and AD to BE (I. 34). But AB is equal to AD (Const.); Therefore the four straight lines BA., AD, DE, EB are equal to one another (Ax. 1), And the parallelogram ADEB is therefore equilateral Likewise all its angles are right angles. For since the straight line AD meets the parallels AB, DE, the angles BAD, ADE are together equal to two right angles (I. 29). But BAD is a right angle (Const.) ; Therefore also ADE is a right angle (Ax. 3). But the opposite angles of parallelograms are equal (I. 34) ; Therefore each of the opposite angles ABE, BED is a right angle (Ax. 1) ; Therefore the figure ADEB is rectangular ; and it has it is^^rec been proved to be equilateral; therefore it is a square (Def. 30). Therefore, the figure ADEB is a square, and it is described .*.a square. upon the given straight line AB. Q. E. F. Corollary. — Hence every parallelogram that has onQ right angle has all its angles right angles. ADEB a B parallelo- gram. It is equi- lateral. Proposition 47.— Theorem. In any right-angled triangle, the square ivhich is described up(yn the side opposite to the right angle is equal to the squares described upon the sides which contain the right angle* 13G GEOMETRY. CAO is a straight line. BAH is a straight line. A ABD = A FBC. Hence })aralIelo- gram BL = square GB, and parallelo- gram CL = square HC. Let ABC be a liglit-angled triangle, Laving tlie right angle BAG ; The square described npon the side BC shall be equal to the squares described upon BA, AC. Construction. — On BC de- scribe the square BDEC (I. 46). On BA, AC describe the squares GB, HC (I. 46). Through A draw AL parallel to BD or CE (I. 31). Join AD, EC. Proof. — Because the angle BAC is a right angle (Hyp.), and that the angle BAG is also a right angle (Def. 30), The two straight lines AC, AG, upon opposite sides of AB, make with it at the point A the adjacent angles equal to two right angles ; Therefore CA is in the same straight line with AG (I. 14). Eor the same reason, AB and AH are in the same straight line. Now the angle DBC is equal to the angle EBA, for each of them is a right angle (Ax. 11); add to each the angle ABC. Therefore the whole angle DBA is equal to the whole angle EBC (Ax. 2). And because the two sides AB, BD are equal to the two sides FB, BC, each to each (Def. 30), and the angle DBA equal to the angle EBC ; Therefore the base AD is equal to the base EC, and the triangle ABD to the triangle FBC (I. 4). Now the parallelogram BL is double of the triangle ABD, because they are on the same base BD, and between the same parallels BD, AL (I. 41). And the square GB is double of the triangle EBC, because they are on the same base FB, and between the same parallels FB, GC (L 41). But the doubles of equals are equal (Ax. 6), therefore the parallelogram BL is equal to the square GB. In the same manner, by joining AE, BK, it can be shown that the parallelogram CL is equal to the square HC. 3?ROPO^iTlONS. 137 Therefore the whole square BDEC is equal to the two squares GB, HC (Ax. 2); And the square BDEC is described on the straight line is'A^+Aca BC, and the squares GB, HC upon BA, AC. Therefore the square described upon the side BC is equal to the squares described upon the sides BA, AC. Therefore, in any right-angled triangle, &c. Q. E. D. Proposition 48. — Theorem. If the square described iqwn one of the sides of a triangle he equal to the squares described upon the other two sides of it, the angle contained by these two sides is a right angle. Let the square described upon BC, one of the sides of the triangle ABC, be equal to the squares described upon the other sides BA, AC; The anijle BAC shall be a riffht an-gjle. Draw An f Construction. — From the point A draw AD at right right aiiglesto AC(I. 11). ^Jf^^^*^ Make AD equal to BA (I. 3), and join DC. (Do not Proof. — Because DA is equal to AB, the square on DA ^ba.Y^ is equal to the square on BA. To each of these add the square on AC. Therefore the squares on DA, AC are equal to the squares on BA, AC (Ax. 2). But because the angle DAC is a right angle (Const.), the square on DC is equal to the squares on DA, AC (I. 47), And the square on BC is equal to the squares on BA, AC (Hyp.) ; r^hew Therefore the square on DC is equal to the square on BC dc2 = (Ax. 1) ; and* And therefore the side DC is equal to the side BC. » DC = BC. And because the side DA is equal to AB (Const), and AC common to the two triangles DAC, BAC, the two sides DA, AC are equal to the two sides BA, AC, each to each. And the base DC has been proved equal to the ba.se BC ; 138 GEOMETRY. Hence _ Therefore the angle DAC is equal to BAG (I. 8), zBAC." But DAC is a right angle (Const.) ; Therefore also BAC is a right angle (Ax. 1). Therefore, if the square, &c. Q. E, D, EXERCISES. Prop. 1—15. 1. From the greater of two given straight lines to cut off a portion which is three times as long as the less. 2. The line bisecting the vertical angle of an isosceles triangle also bisects the base. 3. Prove Euc. I. 5, by the method of super-position. 4. In the figure to Euc. I. 5, show that the line joining A with the point of intersection of BG and FC, makes equal angles with AB and AC. 5. ABC is an isosceles triangle, whose base is BC, and AD is perpendicular to BC ; every point in AD is equally distant from B and C. 6. Show that the sum of the sum and difference of two given straight lines is twice the greater, and that the difference of the sum and difference is twice the less. 7. Prove the same property with regard to angles. 8. Make an angle which shall be three-fourths of a right angle. 9. If, with the extremities of a given line as centres, circles be dra-WTi intersecting in two points, the line joining the points of in- tersection will be perpendicular to the given line, and will also bisect it. 10. Find a point which is at a given distance from a given point and from a given line. 11. Show that the sum of the angles round a given point are together equal to four right angles. 12. If the exterior angle of a triangle and its adjacent interior angle be bisected, the bisecting lines will be at right angles. 13. If three points, A, B, C, be taken not in the same straight line, and AB and AC be joined and bisected by perpendiculars which meet in D, show that DA, DB, DO are equal to each father. Prop. 16—32. ^ 14. The perpendiculars from the angular points upon the opposite sides of a triangle meet in a point. 15. To construct an isosceles triangle on a given base, the sides being each of them double the given base. j EXERCISES ON THE PROPOSITIONS. 139 16» Describe an isosceles triangle having a given base, and whose vertical angle is half a right angle. 17. AB is a straight line, C and D are points on the same side of it ; find a i^oint E in AB such that the sum of CE and ED shall be a minimum. 18. Having given two sides of a triangle and an angle, construct the triangle. Examine the cases when there will be (1.) one solution; (2.) two solutions ; (3.) none. 19. Given an angle of a triangle and the sum and difference of the two sides including the angle, to construct the triangle. 20. Show that each of the angles of an equilateral triangle is two- thirds of a right angle, and hence show how to trisect a right angle. 21. If two angles of a triangle be bisected by lines drawn from the angular points to a given point within, then the line bisecting the third angle will pass through the same point. 22. The difference of any two sides of a triangle is less than the third side. 23. If the angles at the base of a right-angled isosceles triangle be bisected, the bisecting line includes an angle which is three halves of a right angle. 24. The sum of the lines drawn from any point within a polygon to the angular points is greater than half the sum of the sides of the polygon. Prop. 33—48. 25. Show that the diagonals of a square bisect each other at right angles, and that the square described upon a semi-diagonal is half the given square. 26. Divide a given line into any number of equal parts, and hence show how to divide a line similarly to a given line* 27. If D and E be respectively the middle points of the sides BC and AC of the triangle ABC, and AD and BE be joined, and intersect in G, show that GD and GE are respectively one-third of AD and BE. 28. The lines drawn to the bisections of the sides of a triangle from the opposite angles meet in a point. 29. Describe a square which is live times a given square. 30. Show that a square, hexagon, and dodecagon will fill up the space round a point. 31. Divide a square into three equal areas, by lines drawn parallel to one of the diagonals. 32. Upon a given straight line construct a regular octagon. 33. Divide a given triangle into equal triangles by lines drawn from one of the angles. 34. If any two angles of a quadrilateral are together equal to two right angles, show that the sum of the other two is two right angles. 35. The area of a trapezium having two parallel sides is equal to half the rectangle contained by the perpendicular distance between the parallel sides of the trapezium, and the sum of the parallel sides. 140 GEOMETRY. 36. The area of any trapezium is half the rectangle contained by one of the diagonals of the trapezium, and the sum of the perpendicu- lars let fall upon it from the opposite angles. 37. If the middle points of the sides of a triangle be joined, the lines form a triangle whose area is one-fourth that of the given triangle. 38. If the sides of a triangle be such that they are respectively the sum of two given lines, the difference of the same two lines, and twice the side of a square equal to the rectangle contained by these Imes, the triangle shall be right-angled, having the right angle opposite to the hrst-named side. 39. If a point be taken within a triangle such that the lengths of the perpendiculars upon the sides are equal, show that the area of the rectangle contained by one of the perpendiculars and the perimeter of the triangle is double the area of the triangle. 40. In the last problem, if be the given point, and OD, OE, OF the respective perpendiculars upon the sides BC, AC, and AB, show that the sum of the squares upon AD, OB, and DC exceeds the sum of the squares upon AF, BD, and CD by three times the square upon either of the perpendiculars. 41. Having given the lengths of the segments AF, BD, CE, in Problem 40, construct the triangle. 42. Draw a line, the square upon which shall be seven times the square upon a given line. 43. Draw a line, the square upon which shall be equal to the sum or difference of two given squares. 44. Reduce a given polygon to an equivalent triangle. 45. Divide a triangle into equal areas by drawing a line from a given point in a side. 46. Do the same with a given parallelogram. 47. If in the fig., Euc. I. 47, the square on the hypothenuse be on the other side, show how the other two squares may be made to cover exactly the square on the hypothenuse. 48. The area of a quadrilateral whose diagonals are at right angles is half the rectangle contained by the diagonals. SECTION HI. ALGEBRA. CHAPTER I. ELEMENTARY PRINCIPLES. 1. Algebra treats of numbers, the numbers being repre- sented by letters (symbols of quantity), affected with certain symbols of quality, and connected by symbols of operation. It is easy to see that these symbols of Quantity may be dealt with very much as we deal with concrete quantities in arithmetic. Thus, allowing the letter a to stand for the number of units in any quantity, and allowing also 2 a, 3 a, 4 a, tkc, to stand respectively for twice, thrice, four times, &c., as large a quantity as the letter a, it at once follows that we may perform the operations of addition, subtraction, multiplication, and division upon these symbols exactly as we do in ordinary arithmetic upon concrete quantities. For instance, 4 a and 6 a make 10 a, 9 a exceeds 5 a by 4 a, 15 a is 5 times 3 a, and 7 a is contained 8 times in 56 a. Neither is it necessary in these operations to state, or even to know the exact number of units for which any symbol of quantity stands, nor indeed the nature of these units ; it is simply sufficient that it is a symbol of quantity. Thus, in the science of chemistry, we use a weight called a crith ; and a person unacquainted with chemistry might not know wli ether a crith were a measure of length, weight, or capacity, or indeed whether it were a measure at all, yet he would at once allow that 6 criths and 5 critha JU"© H criths, that twicQ 4 crith^ c^rg 8 criths, 4q, 142 ALGEBRA. The Signs + and - as Symbols of Operation. 2. In purely aritlimetical operations, tlie signs + and - are respectively the signs of addition and subtraction. In this sense, too, they are used in algebra. * Thus, a + b means that b is to be added to a, and a — 1) means that b is to be subtracted from a. Hence, as long as a and b represent ordinary arithmetical numbers, a -v b admits of easy interpretation, as also does a -^ b, when b is not greater than a. But when b is greater than a, the expression a — b has no arithmetical meaning. By an extension, however, of the use of the signs + and — , we are able to give such expressions an intelligible signi- fication, whatever may be the quantities represented by a and 5. Positive and Negative Quantities.— The Signs + and - as Symbols of Affection or Quality, 8. Def, — A positive quantity is one which is affected with a + sign, and a negative quantity is one which is affected with a — sign. Let BA be a straight line, and O a point in the line; and suppose a person, starting , from O, to walk a miles in the "^ " -^ direction OA. Suppose also another person, starting from the same or any other point in BA to walk a miles in the direction OB. These persons will thus walk a miles each in exactly opposite directions. Now, we call one of these direc- tions positive (it matters not which) and the other negative. Let us take the direction OA as positive. We then have the first person walking a miles in a positive direction, and the second walking a miles in a negative direction. We represent these distances algebraically hj + a and — a respectively. It will therefore be seen that the signs + and — have no effect upon the magnitudes of quantities, but that they express the quality or affection of the quantities before which they stand. THE SUM OP ALGEBRAICAL QUANTITIES. 143 Again, suppose a person in business to get a profit of £6, while another suffers a loss of £6. We may express these facts algebraically in two ways. We may consider gain as positive, and loss as negative gain, and say that the former has gained + 6 pounds, while the latter has gained — 6 pounds. Or we may consider loss as positive, and gain as negative loss, and say that the former has lost — 6 pounds, while the latter has lost + 6 pounds. We hence see that the gain of + 6 is equal to a loss of - 6. and that a gain of — 6 is equal to a loss of + 6. The Sum of Algebraical Quantities. 4. Let a distance AB be measured to the right along the line AX. And let a further — *- -« ., distance BC be measured from B in the same direction. By the sum of these lines we mean the resulting distance of the point C from the original point A, that is to say, the distance AC. (It may be remarked that we add the line BC to the line AB by measuring BC in its own proper direction from the extremity B of AB. It is hardly necessary to remind the student that both lines are in the same straight line AX.) Let us represent the distances AB and BC by + a and + h respectively ; then the algebraical sum of the lines will be represented writing these quantities side by side, each with its own proper sign of affection. Thus the sum of the distances AB and BC is expressed by + a + h, or, as it is usual to omit the + sign of a positive quantity when the quantity stands alone or at the head of an algebraical expression, the sum of AB and BC is expressed by a + 5. Hence, the interpretation of a + 6 is that it represents the distance AC. Again, taking as above + a to represent the distance AB along the straight line AX, and ,^ <; j^ measured to the right, let a dis- ,^ ^, ^ tance BC be measured from B in ' * ^ the same straight line AX, but — c " ' ^ this time to the left. 144 ALGEBRA. Let the latter distance be represented by - 5. Then, on the principle above, AC is the sum of these dis- tances, and this sum is represented algebraically by + a — 6 or a - 6. (It "will be seen that the distance BC is again measured from B in its own proper direction^ and that the resultant distance AC of the point C is again the sum of the line AB and BC.) There will evidently be three cases, viz. : — 1. When the distance BC is less in magnitude than AB, in which case the point C is on the right of A, and the dis- tance AC is positive. 2. When the distance BC is equal in magnitude to AB, in which case the point C coincides with A, and the distance AC is zero. 3. When the distance BC is greater in magnitude than AB, the point C being then on the left of A, and the distance is negative. Now, a — h in all these cases represents the distance AC. It therefore admits of intelligible interpretation whether h be less than, equal to, or greater than a. And, since the distance AC is obtained in the first two cases by subtracting the distance BC from that of AB, and in the second case by subtracting as far as AB will allow of subtraction, and measuring the remainder to be subtracted in an opposite direction, it follows that — The sign - , which, standing before a letter, is a symbol of quality, becomes at once a symbol of subtraction in all cases when the quantity in question is placed immediately after any other given quantity with its proper sign of affection. Hence also we may conclude that the addition of a Tiega- tive quantity is equivalent to the subtraction of the corres- ponding positive quantity. 5. We may prove in a similar way that — The subtraction of a negative quantity is equivalent to the addition of the corresponding joosi^we quantity. Let, as before, + a represent the distance AB, measured from the point A to the rig] it ^ , , . , . and let it be required to sub- '^ a (* ^vact from f a the distance represented by - b. THE SUM OP ALGEBRAICAL QUANTITIES. 145 Now, in the last article, we added a distance to a given distance AB, by measuring the second distance in its own direction from the extremity B. We shall therefore be con- sistent if we subtract a. given distance — 6 by measuring this distance in a direction exactly opposite to its own direction, from the same extremity B. Now, the direction of — 6 is to the left. If, therefore, we measure a distance BC to the right, equal in magnitude to the distance to be subtracted, we obtain a distance AC which is correctly represented by a - ( - h). But AC is also correctly represented by a + b, and hence it follows that a-(-b)=a-\-b. We may apply the above principle to all magnitudes which admit of continuous and indefinite extension; as, for instance, to forces which pull and push, attract and repel ; to time past and time to come, to temperatures above zero and below zero, to money due and money owed, to distance up and dis- tance down, &c., in all which cases, having represented one by a quantity affected with a + sign, we may represent the other by a quantity affected with a - sign. 6. In expressing the sum of a number of quantities^ the order of the terms is immaterial. We will take, as our illustration, a body subject to various alterations of temperature, and we will suppose the tempera- ture of the body, before the changes in question, to be zero or 0°. Let. the temperature now undergo the following changes — viz., a rise of a°, a fall of 6°, a fall of c°, and, lastly, a rise of c^°. Let us consider a rise as positive, and therefore a fall as negative. We may then represent these changes respectively by + a, — 6, — c, + c?. And it is further evident that the resulting temperature will be represented by the sum of these quantities, which, as previously written, will be a - 6 — c + c?. But again, it is plain that the resulting temperature of the body will not be affected if these changes of temperature take place in the reverse order, or in any other order. Thus, suppose the temperature first falls c°, then rises a°, then rises d°, and, lastly, falls 6°, it is evident that the final temperature will be the same as before. And the sum of the quantities »- c, + a, -t- c^, —h, represents this final temperature, 14G ALGEBRA. Now, expressing the sum of these quantities by writing them (Art. 4) side by side with their proper signs of affection, and in the order in which they stand, we have — c + a + d — b for the sum. It therefore follows, since we might have chosen any other order of these terms with a similar result, that the sum of any number of quantities, + a, — h, — c, + cZ, may be expressed by writing the terms side by side with their proper signs of affection in any order whatever, Nevertheless, for convenience, and for other reasons, we write the terms generally in alphabetical order, or we arrange them according to the power (see Art. 16} of some particular letter. 7. We may sum up the results and remarks of the last four articles as follows : — 1. Positive and negative are used in exactly opposite senses. 2. The sign + before an algebraical quantity affirms the quality of the quantity as represented without the sign m question. Thus, + ( + a) = + a, and + ( — 5) = — 6. 3. The sign — before an algebraical quantity reverses the quality of the. quantity as represented without the sign in question. Thus, — ( + a) = — o, and — ( - 5) = +6. 4. The algebraical subtraction of a quantity is the same as the addition of the quantity with the sign of affection 5. The algebraical .addition of quantities is expressed by writing the quantities down side by side with their proper signs of affection. And they may be written down in any order, though we generally write them in alphabetical order, or arranged according to the power (Art. 16) of some letter. Brackets. r 8. Brackets — ( ), } }, [ ] — are used, for the most part, whenever we wish to consider an algebraical expression con- taining more than one term as a z(;Me. BRACKETS. 14:7 Thus, if WO wish to express that the quantity 3 « + 7 6 is to be added as a whole to 4 a, we write — 4a + (3 a + 7 6), and, while inclosed within brackets, we think and speak of S a + 7 5 as one quantity. Again, if we wish to express that 6 — c is to be subtracted from a, we write — ej — (5 - c). Let us consider what is the result of subtracting (b — c) from a. We may evidently, if we please, subtract b first, then afterwards - c from the quantity so obtained, without affecting this result. Now, we know by Art. 7 (4.) that this is equivalent fo adding the quantities - b and + c successively. Now, the sum of a, — 5, + c, is a - b + c. We have therefore a — (6 - c) = a - 6 + c. We observe that the sign of b within the brackets is -F, and that of c is - , whereas, in our final result, these sigiisa are both reversed. And we hence arrive at the following important principle : — When a minus sign stands before a bracket, its effect on removing the brackets is to reverse the sign of affection of every term within, Aiid it is evident that we may, by a similar course of reasoning, arrive at a principle equally important, viz. : — When a plus sign stands before a bracket, its effect on removing the bracket is to affirm the sign of affection of every term within. We shall show, in Art. 9, the use of brackets in expressing the product or quotient of quantities. Though, as stated above, brackets are, for the most part, used to group together as a whole a number of quantities, they are sometimes used to inclose single terms. Thus, in Art. 5, we have the expression a — ( - b). Now, the brackets are used here to express that the negative quantitij is to be subtracted as a negative quantity. And, in the same way, the expression a + (—6) indicates that the negative quantity ( - 6) is to be added to the quantity a. When one pair only is required we generally use the bi*ackets ( ) ; if, H8 ALGEBRA. however, a quantity already in brackets is to be inclosed in a second pair, we use { (, as in the expression — 3 a - I 6 5 + (4 c - c?) I . If a third pair be required we use the brackets [ ], and finally, we sometimes find it convenient to group a number of terms by means of a vinculum^ thus — 4:X -[^x - {by + {Zz " 7 a; - 2/)} + 2y\ It must be remembered that the vinculum has in an expression exactly the same force as brackets. 9. We shall, in this Article, show how to find the value of a few algebraical expressions, as illustrations of the foregoing principles : — Ex. l.—If a = 1, 6 = 2, c = 4, find the value of 3a + 55 + 7c. We have only to substitute the value of the letters in the given expression, putting a sign of multiplication to avoid ambiguity. Thus we have — 3a + 56 + 7c = 3xl + 5x2 + 7x4. = 3 + 10 + 28 = 41. Ex. 2.— If a; = 5, 2/ = 2, « = 6, find the value cf ^x + Zy - ^z. Here, 4a; + 32/-9;s = 4x5 + 3x2-9x6 '=20 + 6-54 = 26 - 54. The negative quantity is here the larger, and exceeds the positive quantity by 28, and hence (Art. 4 (3),) the result will be negative. We therefore have — 4a; + 32/-92;=-28. Ex. 3.--If aj = 1, 2/ = 3, ^ = 0, find the value of— 3aj-|22/ + (6« - 5 a; - 2/) I • The given expression — = 3xl-|2x3 + (GxO- 5x1-3) | PRODUCT OF TWO OR MORE QUANTITIES. 140 - 3 - j 6 + (0 - 2) j . = 3-(6-2)-3-4=-l. It may be advisable to simplify expressions of this kind before substituting the given value of the letters. The method of doing this, however, is deferred till we come to Chapter II. Ex. I. If a = 1, 5 = 2, c = 3, f^ = 0, e = 4, find the value of the following : — 1. 4c a + 2by3b + 7cyQa + 4cdy 4:c - 7e. 2. a + h + Cya — b + Cyb + c-a, a + b — c. 3. 3a + 7b-4.Cy2a + 7d + 3c,7a-lOb + 2c, 4. 6a- (2 6 - 3c), 7 b + (2 a - 4cd). 5. 3 c + (6 a - 7 6 + c), 26-|7c + (4cZ-Z;)|. 6. I 7 6 - (2 cZ - c) I - (4 6 - c + 6). If X = 2, 9/ = 3, z = 4:y find the sum and difierence of the following expressions : — 7. 3 03 — 4 y and 3 7/ - 4 a;. 8. 7 {x - y) and 4 (?/ - c). 9. 3 ic - (7 2/ + 4 ;^) and 8 y + 5 ;:; - 3 iT. 10. \2y - Qy + ^z and \2 y + Qt y + A:Z. 11. X - y - z and x - {y - z). 12. X - ( - 3 7/) and y - | 3 a; - ( - 3z)\. Product of Two or more Quantities. 10. The product of two or more quantities may be expressed in several ways. Tims, the product of a, 6, c may be Avritten as follows : — 1. abc, by placing the letters side by side without any sif/u between them. 2. a X b X c, by ])lacing bet«vcen them the sign x . 3. a , b , c, hy j)h\cing a dot between them. 150 ALGEBRA. 4. (a) (b) (c), by inclosing the quantities in brackets, writing them without any sign between the brackets. When, however, the quantities are negative, or either of them consists of more than one term, it is best to inclose such quantities in brackets, and, in most cases, it is necessary to do so. Thus, the product of a, - b, c would not be correctly expressed hy a - be (for this means, that the product of h and c is to be subtracted from a), but must be written a {- b) c. Again, the product of 2 a + 3 5 and a + 5 b cannot be written 2 a + 3ba + 5 6, as this expression means that three times the product of 6 x «^ is to be added to 2 a, and then 5 5 to the result. The product is correctly ex- pressed thus — {2a + 3b) (a + 5 b). (The student cannot be too strongly cautioned against leaving out brackets in cases of this kind. ) The Order of the Letters. 11. It is evident that a times b^b times a ; for, if we aii'ange ft rows of b things so as to have a horizontal rows and b vertical columns, we may either consider the number of rows or the number of columns. In the former case we have a times b things, and, in the latter, b times a things. We may therefore write the letters whose product we wish to express in any order. Thus, the product of a, b, c may be written in either of the following ways : — abCf acby bac, cba, bca, cab. For convenience, however, and for reasons which the student will see as he proceeds with the subject, we write them in the order of the alphabet, unless there happen to be special reasons to the contrary. Rule of Signs in Multiplication. 12. It was shown (Ai-t. 3) that a minus sign does not affect the magnitude of a quantity, but simjily its affection or QUOTIENT OF TWO QUANTITIES. I5l quality. It therefore follows that the product of + ^ and - h will be the same in magnitude as that of + a and + 6, or of ^ X h — that is, will be equal in magnitude to ah. But it is evident that the quality of the product will be the same as that of - h, for it is a times a negative quantity. We therefore have + a ( - 6) = - a6. So also, the product of - ^ny times the letter ai)pear^ as a factor. In this case, ^^^ therefore, we write cvK Wo call quantities of this form powers of the letter in question; and so, remembering that a may be considered as a^ on this principle, we have : — a, the first power of a ; a'^, the second power oi a; a^, the third power of a, &c. The figure written at the light hand at the top of the letter is called the index or exponent, and it is usual to call a", a^ respectively tlio square and cube of a. 154 ALGEBRA. from tlie fact tliey express respectively tKe area of a square whose side is a, and the volume of cube whose edge is a. 17. The square root of a quantity is that quantity which, w^hen raised to the second 2>ower or squared, will give the original quantity. _ It is generally written ^. Thus, x/l6 = 4, ^144 = 12. The cube root of a quantity is that quantity which, when raised to the third power or cubed, will give the original quantity. _ I t is generally written ^. Thus, 4^8 = 2, ^1 = 3, ;^1728 = 12. And so the fourth, fifth, &c., roots are indicated by the symbols tj , ^ , &c., respectively. 18. The dimensions of an algebraical quantity are the sum of the indices or exponents of the literal factors. Thus, the dimensions of 3 a'5V = 2 + 3 + 4 = 9, and the dimensions of ahar =1 + 1+2 = 4. 19. A homogeneous expression is one in which the dimen- sions of every term are the same. Thus, a^ + 3 a^6 + 3 aH^ + h^ is homogeneous, Whereas, 5 a + 3 a6 + 7 a^hc is not homogeneous, Ex. II Ifa = 2, 5 = 3,'r= 0, d = 1, Find the values of — 1. 6a^ + 3 5^ - 5 c"; a6 + ac + he, he + hd + cc7. 2. a^ + 3 a'^b + 3 a6- + ^^^ ; a^ + ^^^ + c^ - 3 abc. 3. (3a + 7 5) (4a - 9 6) ; {a" + 5") {a + b) {a - b). 4. 6 {2 a^ - 4 (2 h' - ^If^^^)] ; a'b + ab'' + a'c + ac^ + b~c + be". If a = 1, 5 = - 2, c = -- 3, cZ = 0, e = 4, Find the Values of— 5. (a + 5 + c + cZ + e)^ ; (cr + 2 aZ> + i" - c") + (a + 6 + c). a"" - d' c^ + 3 c\Z + 3 cd^ + cZ^ 6.. a^ + a'-'cZ + ad"^ + d^^ 6^-3 Z)'-^c + 3 bc^ 7. y/cr + e^ - /v/5 g^ + 6- "1 .1 ^ 'C^c^* + 3 c*^ + 3 c + r ah + ae ■\- bo ^ ADDtttOK, 155 8. 3 (4 J + 5 cY + 4 (c + e)- ; abcde. Ifoj = 3, y - 4, ;^ = 0, Find the value of — ^' (3a; - sl^fVff {2x + ^ar + f + 2). 10. {5 a;^ + 2(7/ 4- 2)2} {5 ar- 2(2/ + ;:;)-}. 11. a;^ + 2/* + z^. 12. (.^3 . 2/^) ^ ^ |3^^ + 3 (3^2 ^ 3^^ ^ 2/^)^1, CHAPTER II. ADDITION SUBTRACTION, MULTIPLICATION, AND DIVISION. Addition. 20. HuLE. — Arrange the terms of the given quantities so that like quantiti-es may be under each other ; add separately the positive and negative coefficients of each column ; take the difference and prefix the sign of the greater, and annex the common letter. (When the coefficients are all positive or all negative, we, of course, simply add them together and prefix the common sign for the coeffi- cient of the sum. ) Ex. 1. Ex. 2. 3a + 56-3c -3a + 76+ c - Ad 2c6 - 75 + 4c 2« - 26 + r)c + 3(^ 5a+ h-2c -7a-36 + 2c-Gf^ 4a-3 6 + 8c 2(j+56-8c+Gt^ Ans. 14 a- 46 + 7c Ans. - 6 « + 7 6 -^ d Ex. 3. Add together b x^ - 3 ?/^ + 3 y, G ?/" + 7 xij - 4 x, 4 icy + 6 flj - 5, - 2 a:^ - 3 x-?/ + 2. Arranging like quantities in each expression under vr.cli other, we have : — 5 ctr^ - 3 ?/- +3 7/ 7 ajy + G v/- - 4 .'x; 4 xj/ + (jx - 5 '-2 a r - 3xy + 2 Ans. 3ay^+8xi/+3y + 2x +3v/ -3 156 ALGEBRA. Ex. IIL Add together- 1. 3a - 2b, i a + 7 h, 2 a + 3b, a - 5 k 2. 9rr + 7 b% - 3a' + 4:b'\ a' + P, icr - 12b\ 3. a + b + c, 3a + 2b + 3c, -4:a+7b-c, 2b + 5c. L X - y - z, y - X - z, z - X ■- y, X ■\- y -V z. ^ 5. 3a^ - 4 a6 + 6 ¥, 7 ab - a- - W, 2 rr - 3 ab - 4 6-, 4 a^ + aZ> - 51 6. 2x^ - 7x- + 3, - ix^ + (jx' -2 X + 7, a;' - 2 x"^ - 4 a;, Gar*- 9a; - 12. 7. 2«2 + 7a6 + 35"^ - G a - 5 5 - 2, a- + 3f^ - 2 t + 9, 9a6 -2^-36 + 4, -3a2_i2a5 _ 35-+5^+ 105- 15. 8. cc^ - .oj^y - a;^;^ + xy"^ - xyz + ic^:;^, x^y - xy^ - xyz + if — 7/^;;; + yz^, cirz - cc?/;s - xz^ + t/",^; - 7/;s" + ::^. 9. cc^ + xh/ + a;^^, - £c^y — x^y'^ - xif, y^ + xy^ + a;"^^. 10. a^ + ab'^ + ac^ + 2 a-5 - 2 crc - 2 «5c, a-b + 5"^ + 5c- -t- 2 a5'- - 2 a5c — 2 5'^c, a-c + 5-c + c^ + 2 a5c - 2 ar - 2 5c-. 11. x^ - x]f -v x:^ - 3 o(?y + 3 a:^;*;, 3 a;-y/- + 3 ar;r + 3 xy'^z - 3 xyz' - 6 x'yz, y* - a^y - y:^ + 3 xy' - 3 x^yz, - 3 xif - 3 X7j^ — 3 ifz + 3 2/V + 6 a;?/-;^, z^ + ar^;^ -ifz-3 xyz + 3 ar^^;^, 3 a;?/^;^ + 3 a:5r* + 3 yV - 3 yz^ - 6 a;?/^-. 12. a^ - a% + 3 aV -I- ab'c - 3 rt5c2 - 5'^c, a^'b - a\ + 3 a5c^ - 3 ac^ - 5-c- + 5V, (^c - ^^c/ + 3 «c^ - 3 ac-c? + 5-c^ - 5^cc/, - a'* + d^d - 3 a^c- + 3 a(rd - ab'c + b'cd. Subtraction. 21. We have seen, Art. 7 (4.), that the subtraction of a quan- tity is equivalent to the addition of the same quantity with its sign of affection reversed. We therefore have the follow- ing rule : — Rule. — Change the sign of each term of the subtrahend, and proceed as in addition. Ex. 1. Ex. 2. 5 aj + 6 7/ - 3 c^ G (6^ - 3 a5 + 4 5^ 2 a ; - 2 7/ + 2 ;: ; --J a ' - 3 ab - 2 5- Ans. 3x + Sy -* 5 z, Ans. S cr + 6 b\ BRACKETS. 157 Ex. 3. a^ -{^ 3x ^1/ + Zxif + 2 if x^ — 3 ncF]/ ■{• 3 xy^ + y^. Ex. lY. 1. From 6a + 76 + 3c take 2a + 5&-2c. 2. From 2ic-3?/-8;^ take 6a;-52/-2;^. 3. Take 5a2 + 3a6 + 46- + 3a + 76 + 8 from 6a- + 3 J- - 2 a. 4. Take 6a^ + 8aV + ic^from 8a* + 6aV + 2x\ 5. Subtract the sum of the quantities a** + 2 arlr + 6^, a* - 2 a^6^ + ¥ from 6 a* + 8 ^^6^ + 6 6^ G. From x^ -\- y^ + z^ — 3 xyz take 4 x^ + ?/^ + 4 ;s^ + 3 a;-;:; + 3x^ - 3 xyz. 7. From 3 a;* + 3 aii.'^ - 9 aV + a^o; - a* take 2 x'* + 4 ax + 4 (x^x + a*. 8. Take a' - 5 a^^ + 7 ah'' - 2h^ from the sum of 2 a^ -- 9 a-6 + 11 a62 - 3 h^ and 6^ - 4 aft^ + 4 a^^ - dK 9. Subtract a + 6 + c + cZ from e +/+ ^ + A. 10. Take oj"* - 4 rc^y + 6 c^tf - 4 £c?/^ + y^ from a;'* + 4 a;^y 4- 6 ar?/^ + 4 xif + 2/*, and subtract the result from their sum. 11. Add together the given quantities in the last example, and subtract the result from 3 a;'* + \Oarif + 3y^, 12. Take a- + 6' + c^ + 2 a6 + 2 ac + 2 6c from 2 a- + 2 6^ + 4 rt6 - c\ Brackets — continued, 22. It was shown, in Ai-t. 8, that, when a quantity inclosed in brackets is to be added, we may remove the sign ( + ) of addition and the brackets without changing the sign of the terms within the brackets. On the contrary, when the quantity in brackets is to be subtracted, or has the sign minus before it, we must change the sign of every term within the brackets on removing the brackets and the sign of subtraction. We shall now see how to simplify expressions involving brackets connected by the signs of addition and subtractign ;— 158 ALGEBRA.* Ex. h Simplify (3a+5 6)-(G6-2c) + (-2a + & - 3 c). The given expression = 3fl^ + 56-664-2c-2a+6- 3 c, or adding together the like quantities, = a - c. Ex, 2, Reduce to its simplest form — a -^ {b - c) + \b + (a'-c)l - \{a - h) - cl, (When a pair of brackets is inclosed withia another pair, it is con- venient to remove the inner one first.) Hence the given expression — = a -^ h + c + \b -h a — cl - \a - b - ci _, „ ^. ,.« , . 3 aj X - 7 G .T - 9 Ex. 3. Simplify the expression -j — — - — + — - — -, ' The line separating the numerator and denominator of a fraction is a species of vinculum, since it serves to show that the whole numerator is to be divided by the whole denominator. Hence, on breaking up the two latter fractions into fractions having one term only in the numerator, we have — 3x x-r 6x-d 3 1 7 6 9 * 19 T— 2--'-8-^l^-2^''2^8^~8 = ^^¥. 23. As it is often necessary to inclose quantities within brackets, we shall now show how this is done. The following rule needs no explanation : — E-ULE. — When a number of terms is inclosed within brackets, if the sign placed before the brackets be + , the terms must be written down with their signs of affection ninchanged ; but, if the sign placed before the brackets be - , the sign of affection of every term placed within the brackets must be changed. Thus we may express rt + Z>-c~c?in any of the follow- ing ways : — a ■{■ b - G - d = a + {b - c - d) =^ {a + b) - (c + d) -a'-{-b + c + d) = (a + b - d) - c, kc, (When the word slgni^ used in future, the student is to under- etaad i?(//; oj afection, unless otherwise expressed. ) MULTIPLICATION, 150 Ex. V. Simplify the expressions — 1, 3x - {2x - y) ■}■ {6x + 3y-2z) - (Tx'-iy --3z), 2, (2 a\ + 2a'b + 2 air) - (2 a^ + o?h + aly" - ¥) + (a^ 3, 1 - (1 - 2 cc) - I 3 - (4 - 3 a;) I + | 5 - (4 cc - 4, 6 ar^ -^ (2 ^ 3 a; + ar^) + | - 7 + (5 a; - 8 aj - 2) | - (3 - 3 ~ 3 aj). Group together the terms of the four following expressions, so that — (i.) The first two and last two are inclosed in brackets. (ii.) The last three are inclosed in brackets. (iii.) The first three are so inclosed, and an inner paii* of brackets used to inclose the second and third. 5, a - 6 + c - c?. 6, -6a+76-3c + 5c?. 7, - 4 a^ + 12^2^ - \2xy' -^-if, 8, a' -- Z)^ - (^ + 3 abc. Add together — • 9. aa? + hxy + df, - doc^ - axy + ey^, hx^ + 3 cxy + /v/-, - 2aa;2 ^ <2>hy\ 10. ax - cy - ez, - bx -i- dy + /z, ex - ey - gz, - dx + fy + hz. Subtract — 11. (2 a + W aj - (3 6 + c) y + (4 c + fO « from (3 a - d) zz + {4: b - a) y + {5 c - h) z. 12. (y - z) a^ + (z - x) ab + (x - y) b^ from (y - a.*) a^ - {y - z) ab - (z - x)b\ Multiplication. 24. Bemembering the definition (Art. 14) of a coeflicient, it follows that the product of two terms having coefl[icients is found by multiplying the product of the coefficients by tlie product of the remaining factoid. Thus, the product of 4 a and 3 &, or 4 a x 3 6 = 12 a&. 160 ALGEBRA. Again, tlie terms to be nuiltiplied may contain like letters. Now, d^ = aaa, a^ = aaaaa^ and hence it follows that a^ X a^ = aaa x aaaaa = aaaaaaaa ^ a^. And, generally, we have a^ X a^ = aaa.., io m factors x aaa... to n factors, — aaa... to (m + n) factors = a"* + ". It therefore follows that the product of the powers of like letters is found by adding together the exponents of the like letters. Thus, a^ X a'^ ~ a^, and a^ x a = a^, &c. And, further, as regards the sign of the product, we have seen, in Art. 1 2, that like signs give + , and unlike signs give -. There are three things, then, which must be attended to in the multiplication of algebraical terms, viz. : — 1. The signs. — Like signs give +, and unlike signs give -. 2. The coefficients. — These are to be multiplied like ordin- ary numbers. 3. The letters. — The exponents of like letters are to be added together, and the powers so obtained written side by side with the unlike letters. Ex. 1. Ex. 2. Ex. 3. ^ a^h - 3 xyz - 5 ahd 2 ah x^v^ - 2 bcd^ Ans. 12 a^bc Ans. - Zo^t/z^ Ans. X^ahhd? 25. Whenever the multiplicand or multiplier, or both, contains more than one term, it is evident the product is found by multiplying each term of the multiplicand separately by each term of the multiplier, and adding together the separate products. Ex. 4. Ex. 5. 3 ^2 _ 7 a6 + 5 6- a^ - 3 ic^y + 3 a^ - ?/' ^ ah — 2xy Ms, 18 a^6 -^42 a'6' \ 30 ok^ r- 2 ^'y 1 6 ^f- ^#f\% xif MULTIPLICATION. 161 Ex. 6. a? + xy + 'i/ x^ — xy ■¥ y^ cc* + a^y + iJt^y^ x'f + xf + y ^ Ans, x^ + ic^i/^ + 2/*- Ex. 7. ar^ + (a + 5) cc + a6 X + c jc^ + (a + ft) ar^ + a6a; ca^ + (ac + 5c) a; + a5c a;* + (a + 6 + c) af^ + (a6 + ac + 6c) a; + abc. It will be observed that the terms in the above examples are all arranged according to the power of some letter. Thus, in Ex. 4, they are arranged according to the descend- ing powers of a; and in Exs. 5, 6, 7, they are arranged according to the descending powers of x. It matters not whether they are arranged according to ascending or descend- ing powers, and the result would be the same if they are not so arranged. It is then, however, much easier to collect like terms, as they generally fall under each other. When we come to division we shall find it necessary to arrange the terms according to the power of some letter. Ex. VI. 1. Multiply 3 a + 2b hy 4: a - 3 b, and 6 x + 7 y hj 3x - 5y. 2. Multiply oc^ + 2 a;^/ - 2 y^ hj x - 2 y, and 15 a^ + 17 ooy - 4 2/^ by 2 a; + y. 3. Multiply a^ + 2 ab + b^ hj a^ -- 2 ab + P, and a' + b^ by a" - b\ 4. Multiply a^ + ary + xif + y^ hy x - y, and the pro- duct by X* + y\ 5. Multiply a^ + b"^ + c^ — ab - ac - be hy a + b + c. G. Find the contiuued product of x'*' + y^j x^ + y^, x + y, X - y. 5 L 163 ALGEBRA. 7. Develop tlie expressions {a + by, and {x - yy, 8. Multiply5a=^+ 15^2+ 45« + 135 by a^- 2« - 3. 9. Multiply 1 + 3 a; - 7 ar by a? - 2, and a - xhy x^ + ax + a^. 10. Multiply a* - 2a^b + ^ orh^ - 2ah^ + 5nya^ + 2a^6 + Sa^i^ + 2a53 + h\ 11. Multiply a + 6aJ + ca:i^ + (^a;^ by ca; + / 12. Eind the product oi Q(? + px •¥ q and x - a, 13. Find the continued product oi x -v a, x + h, x ■{■ c. 28. We have explained in the last Article the general method to be pursued in Multiplication. There are, however, many algebraical expressions which may be mul- tiplied by inspection. I. Expressions of the form {a + hy. It is easily found by long multiplication that {a ■{■ hy = o? + lab + W, and this, expressed in ordinary language, may be read thus : — The square of the sum of two quantities is the square of the first ^ plus twice their product, plus the square of the second, (This rale is evidently true whether the quantities are j^osUlve or negative. ) Ex. 1. (3^ + 7by - (3ay + 2 (3«) (75) + (7 by = 9a- + 42a6 + 49 51 Ex. 2. {6x - 5)2 = (Qxy + 2 (6a;) ( - 5) + ( - 5)^ = 36a? - 60a; + 25. Ex.3.(a + b + cy= ^(a + b) + cV = (a + by + 2{a + b)c + (^ = (a" + 2ab + b'') + 2 {ac + be) + r = a^ + b^ + c" + 2ab + 2ac + 2bc, Remark. — It is a very common mistake with beginners to write down a^ + h^ as the square of (a 4- b), thereby leaving out twice the product of the quantities a and b. They should impress this for- mula, viz. : — {a + 6)2 = a^ + 2 a6 + 63 thoroughly upon the mind. MULTIPLICATION. 163 II. The form {a -h h) {a - b). It may be easily found by long multiplication that {a -h b) {a - b) = a^ - b^, which, in ordinary language, may be thus expressed : — llie product of the sum and difference of two quantities is the difference of the squares of the quantities. This formula may be applied in all cases where the terms of the quantities to be multiplied are of the same magni- tude, but when some of them differ in sign. Ex. 1. {a ■{■ b + c ■{■ d) {a ■\- b - c - d) = I (a + 6) + (c + cZ) I I (« + b) - (c + cZ) I = (a + by - (c + df = (a2+ 2«& + b') - (c- + 2c(Z + cP) = a^ + 6^ - c^ — cP + 2a6 - 2cd. Ex. 2. {a^- Sa^'b-h Sa¥- b^) (a^-h 3a'b + 3aP+ ¥) = j (a»+ 3a6-) - (3 a'b + P) j j {a^ + 3ab') + (3a'b + b') j = (a^ + 3a¥f - {3a'b + bj = (a« + Qa'U' + 9d'b') - {9a'b^ + Ga'^b^ + b') = a^-Sa'b^ + 3a^b'-b\ The principle to be adopted in all such cases is to find what terms in the given quantities are exactly alike, and put them first in brackets, when the remaining terms will fall into the proper form. Thus — (3a + 7b + 5c-d){3a-7b-h5c + d)= |(3ct + 5c) + (7 6 _ cZ) I I (3 ci + 5 c) - (7 6 - cZ) I . And(3a-75- 6c;+cZ)(3a + 76 + 50 + ^/)= |(3« + (Z) - (^^ + ^c)j j(3« + ^) + (76 + 5c) j. (III.) The form {x + a) {x + b). By multiplication we find that — {x + a) {x + b) = x^ + (a + b) X + ah, which in ordinary language may be thus expressed : — The product of two binomials containing a common term and an uncommon term, is the square of the common term, plus the sum of the uncommon terms multiplied into the common term, plus the product of the uncommon term. 164 ALGEBRA. Thus— Ix + Q)(x - 1) = aP+{6-l)x + 6 (- l) = x^ + 5x - 6. {x-5){x -7) = x^-{6 ■h7)x + {-5){- 7) = x' - 12 aj + 35. {x + 3a) (x - 5a) = x^ + {3 a - 6a) x + {3 a) ( - 5 a) = x^ " 2 ax — 16 a\ And so, (x + a + b) (x + c + d) = {x + a + b) (x + c + d) = x^ + (a + b + c + d) X + (a + b) (c + d). We may extend this formula to any number of factors. Thus, by multiplication, we find that — (x + a) (x + b) (x + c) = a^ + (a + b + c) x^ + (ab + ac + be) X + abc, {x + a) {x + b) (a; + c) (a? + (Z) - oj* + (« + & + c + d) a^ + (ab + ac + ad + be + bd + cd) a^ + {abc + abd + acd + bed) X + abed. Law of Formation of the Terms, 1. It will be seen that the coefficient of the first term is in each case unity ; that of the second term, the sum of the uncommon letters taken singly ; that of the third term, the sum of the uncommon letters taken two together ; that of the fourth term, the sum of the uncommon letters taken three together^ &c. 2. The power of the common letter is in the first term that of the number of the binomials, and it sinks one every term. Ex. (x + l){x + 2){x + 3) = a^ + (1 + 2 + 3)x'+(l + 2 + l + 3 + 2 + 3)a;+l + 2 + 3=af* + 6a;-+lla;+6, (IV.) The form (a + b + c + d + &c.)l By ordinary multiplication or otherwise we have (a + b + c + df = a^ ^b'' + c^ + d' + 2ab -v 2ac + 2ad + 2be + 2bd + 2cd ', and a similar result follows if we take a larger number of terms. We may hence deduce the follomng rule : — The square of the sum of any number of quantities is the sum of the squares of the quantities together with twice the sum of the products formed by multiplying the first quantity MULTIPLICATION. 1G5 into all that follow separately, then the second into all that follow, the third into all that follow, &c. Ex. 1. (1 + 2a; + Zx^f =\ + 4ar* + 9a;* - 1 + 4a; + lOar^ + 12a;»+ 9a;^ Ex. 2. (a'^ + 3a=6 + 3a62 + 6y = a« +9 a^^h' + 9 a^h^ = a« + 6a'^6 + 15a*62 + 20a«63+15fr6* + 6«6^ + 6^ Ex. VII. 1. Find by inspection the squares of a; - y, 3 « - 5 6, 4 c- + (^2, 3 a;2 _ 2 2/^. 2. Find the continued product of a ■\- h, or + Ir, a* + h*, and a - b. 3. Multiply wia; + ?i?/ by mx - ni/, and 5 a^ - 3 6- by 10 a^ + 6 61 4. Find the value of (a + b •{• c + d) (a -h + c - d), and of {a + b - c - d) {a + b + c + d), 5. Show that (ar + 2 a;y + ^t) (ar' - 2 a;?/ -{- y-) = a;* - 2ary + 2/*. G. Multiply X + 5 separately by a; + 1, a; + 2, a; - 3, a; - 5. 7. What is the continued product of a; - 2, a; + 3, a; - 6, X + 5"^ 8. If 2 s = « + 6 + c, find the value of 5 (5 - a) (s - b) («-c). 9. If 2 8 = a + b + c + d, what is the value of — (2 « - 2 a)'-^ + (2 5 -- 2 by + {2 8 - 2cy + {2s -2 dfl Prove that — 10. {ax + byY + {ex + di/Y + {ay - bxy + {cy - dxy = (a^ + 6* + c^ + (f ) (ar^ + f). 11. I {ac - bd)x + {ad + bc)yl^ + \ (ac - bd) y - {ad + 6c) a; I ^- = {a" + b') (r + c?^) (ar' + f). 12. {ax + by + czy = {a + b +c) {oar + bif + cx^) - ab (x - yy - ac {x - zy - be {y - zy. 166 ALGEBRA* 13. {x + a) (oj + h) (x ■{- c) = (x — a) (x - h) (x - c) -\- 2 {{a + b + c)x'^ + abcj. 14. {m + n + p + qY =r. i^m + ^if + {m + j)f + {m + qf + {n + ^)^ + {n + g)^ + (^ + <^)2 _ 2 (7/r + ti^ + 2^- + (7^). 15. {a? + 5^ + c^) (m^ + 'iV + p- + g^) =^ {am + 6?* + cp)- + {an - 6wi + cqY + (op - hq - cnif + (a^/ + 5p - c?i)l 16. {a - h) {b - c) + (5 - c) (c - a) + (c - a) (^ - ^>) = 3 {ab + ac + be) - {a + b + c)-. 17. (.'z; + 2/ + ^ + f*)^ - {x-y-z +aY = 4:{x + a) {'i/ + z). 18. {(« - bY + 4:ab} {{a + bY - i ab} {a' - b* -i- 19. (^2 + 6-) (C^ + C^2) :. (^ + ^,^)2 ^ (^^ ~ 5^)2^ 20. 4 {a^ + ¥) (c^ + dr) = {a + bY {c + ^)2 + {a - Z^)^ (c - cZ)2 + (a + 5)2 (c - c^)- + (a - 5)2 (c + cZ)l 21. 8 (ci^ + 5^) = (a + 5)^ + 6 (a^ ^ bf + (« - 5)^ 22. {x ~ 2/)' + {y^zY + {z-xY + 2{x^ y) {y - z) ^ 2{x - y){z- x) + 2{y ^ z){z'-x) = 0. 23. (o^ + 5 + c) { J + c - a) (a + c - 5) (a + 5 - c) = 4 a-6-, when a- + b'^ = c^. 24. {2 {ax^^ + 57/»')2 + (ay^ -. bx^} =. (a= + 5') {(«:*» + 2/'f + (^"^ - 2/*")'}- Division. S7. As in multiplication, we have especially three things to attend to, viz; : — 1. The signs. ^ We have learnt {Mt 13) that like signs give + , and unlike signs give - . 2. The coefficients » Uilderstanding here the numerical coefficients, it is plain that they may be divided as ordinary arithmetical quantities. 3. The letters. As the product of the quantities a and b is expressed by aby it follows that the quotient of ab by either of the factors a or 6 will give the remaining factor b or a respectively. DIVISION. 167 Thus, ah -■ a or — = h, and ah -r b or -— = a. a And so, xyz -r xz = y, and pqrs i- qs = pr. We may then conclude that when the divisor is contained as a factor in the dividend, the quotient is found by oimiting from the dividend those of its factors which constitute the divisor. If the divisor be not contained as an exact factor in the dividend, we may then express the quotient symbolically. Thus, xy -f ah = — */. ' ^ ah When, however, the dividend and divisor have a common factor, it is plain that we inay, as in arithmetic, strike out of the numerator and denominator of the symbolical quotient this common factor. Thus, 5abc^hd=^^^^. od d And 16 xyz -r 10 axz = ttt— ^ = f-^« 10 axz 5 a 28. A power of a quantity is divided by any other power of the same quantity by subtracting the index of the (Hvisor from that of the dividend, the quotient being that power of the quantity whose index is the remainder so obtained. 1. Let the power of the quantity in the dividend be the higlier. We have a^ = aaaaa, and a^ = aaa, .*. cr -r a^ = - aa =z a"- or . oKia -. Or, generally, m being greater than w, since a"* = aaa... to m factors, and a** - aaa...to n factors, we have— .m . u aaa...tom factors ,' / ' \r.^+^^r. ^m — » a -r a = = aaai..(m — ?i) factors = a • aaa...to7i factors 2. Let the power of the quantity in the dividend be the lower. 168 ALGEBRA. Suppose we have to divide a^ by al. Wehavea^ -^ dJ =^ ^^^^ := UZ \, daaaaaa aaa a? We may, however, so express the result that it shall agi'ee exactly with the proposition at the head of this article. For we may conceive of o? as representing the product of %inity and the quantity a^. We shall therefore be perfectly consistent if we allow o? to represent the quotient of unity by the quantity a^. We shall then have a" ^ = -^, and hence"we get from the __ a^ ^,. above result — Or, generally, m being less than n, we have — a"* aaa \to m factors 1 a" ^ aaa to n factors aaa (n - 7)i) factors = ^_^ ; or, using the notation just explained, a 3. Let the powers of the quantities in the dividend and divisor be equal. It is evident that their quotient is unity: Thus, ?.' = 22^ = 1 = 1. a* aaaa 1 And SO, — = 1. a"* If, however, we assume the principle proved in the two cases above to hold here, we have — a*" It follows therefore that a^ ~ \. Cor. — From the above interpretation of negative indices it follows that the same rules for multiplication and division of quantities involving them may be applied as in the case of positive indices. Thus,a^ X a-2 = aS x \ -- -' = a^-\ a^ o? 'And so, a^ - a~^ = a"*"^'^^ = al, "^ "" DIVISION. 169 Ex. 1. ?^ = ^xy. Ex. 2. ::^p:^ = - 3aV. JbiX. 3. ~ a^ - Sab + 2b, - la ^^ , ^x^y:^ - Qx'fz^ + 42a?2/V ^^' ^ 3^7^^ 4:0$ 2 U - + — , or 3 j^z^ yz xz = |i^2/"'^-' - 2y-h-'' + Ux-''z-\ o Ex. 5. Divide 3^2 + 13a6 - lOK^hja + 5 5. When the divisor, as in this example, contains more than one term, it is generally convenient to follow the method of arithmetical lon<; division. Thus — Ex. a + 56) 3 a' + 13 a6 + 15 a& - 2a6 - 2ah - I0¥{c - 10 6' - 10 6' !a - 26 6. a- Divide < + ab + %* + 6=), a,* + a-b" a* + ai'b by a' + + 6^ (a' + aW ab + - ab 6'. + 6= - a?b - a'b + b* - a?6' - aW a''6-' + a'6' + ab' ab^ + 6* + 6' It will be seen that in the last two examples care has been taken to keep the terms of the divisor, dividend, and successive remainders arranged according to the ascending or descending powers of some letter. In these cases we have arranged the terms according to the descending powers of a, and, as there- fore follows, according to the ascending powers of b. Want of care in this respect will often render the operation of find- ing the true quotient tiresome, if not impossible. The next two examples will illustrate this point. They may be 170 ALGEBRA. attempted first by the student, keeping tlie terms in the order, as given. Ex. 7. Divide 2 - 7 a; ~ 15a;2 by 5aj - 1. 3 a; - 2 Here we have the powers of x in the dividend descending, while in the divisor they are ascending. Arranging them in the divisor as in the dividend,, the operation is easy. Thus — y^^ + x^'^)o^ + 2/^ (^y - x^if + xif Q^ + a^y — a?y — xy^ - 1 + 5a;) 3,5a; - 1)- Divide o? + 2 -. 7a; 2 - 10 a; - 15a;^(- Or thuj 3a; 3a; - 15 ar^ + - 15 ar* - 15 a;^ 7a; + 2( 3a; — lOo; + 2 lOo; + 2 Ex. 8. 2/5 by o;- ' + 2/^^ xf + 2/^ xy^ + it We shall work the next example in two ways to illustrate, firstly, the above point again ; and, secondly, to show how the operation may be sometimes abbreviated by the use of brackets. Ex. 9. Divide a;' + 2/^ + ^ - 3 xyz by a; + 2/ + ^• a; + 2/ + z):i^ - 3 xyz + y^ + ^{aP -xy -xz + y^-yz + z^ aP + ary + orz -xhj. - xh/ ■ -aPz- - 3 xyz -xyz x:^ r^ - aPz + xy^ - -aPz 2 0;^^; 'Xyz- ■xyz^r x^ * -xyz-^r -xyz x£^ ^fz -fz- -yz' xz^ + xz'^ + yz^^ 2/^ + ^ ^ 'i' DIVISION. 171 Or thus, inclosing the last two terms of the divisor in brackets — X + {1/ + z))x^ - 3 xyz \' 7^ + z^{o(P - {i/ -^ z) x -i- (if -yz-\- z^) a^ + (y + z) x^ - (y + 2;) ar^ - 3 xyz -{y + z)x^-{y^+2yz-\-z^)x {f-yz-Viif)x + i^ + ^ if -yz+ z^)x + y^ + 2^ It will be seen in both the above operations that we have brought down the terms of the dividend only when the sub- trahends indicated they were required. This often prevents much useless repetition* Ex. tUL Find the quotient of — 1. 28 arb - 7 a6^ + U W hy 7 b; 3 x'lf - 12 xf by 3 xy, 2.-6 a'b + 15 a*62- 20 a^^' by - 3 ab; 4 xY + 6 xhf + 4 xy^ by 2 xy\ 3. aixr"*+ 6x"*y*+ cy- *» by re*" + ** ; «^'!!1~^" + Jx"*~*'?/** + c?/-" by aj"*~". 4. 30 x^ +2 xy - 12 7/- by 5 ic - 3 7/ ; and by 6 a: + 4 ^. 5. l + 2a;+3ar^ + 2jc^+a;*byl+a;+a:l 6. 12 - 19 a; - 21 aj^ by 7 rr - 3 ; and by 3 a; + 4. 7. a* - 4 ary'^ +12 xy^ - y*hy x - 3 y; ami by x-{^ y. 8. x^ - ^hj X - y; and x^ -h y^ hj x + y: 9. acx'^'^'* + adx'^y^ + kx^'g/^ + bdy"^.;^*^ by txT-^-di/^. 10. a« + a^62 + a^i* + 6^^ by a' - a'b -aW + Z^l 11. 6^ + a6 + Jc + ac by a + 6; and a^ + a6 + 5c + ac by a + c. 12. a + (rt + 6) a; + (^ + 5 + c) a;- + (a + 5 + c) a;^ + (6 + c) a;* + ca;^ by 1 + a; + ar + jc*. 13. a^ - jpa'* + qa? - f^a- + ^?a — 1 by a - 1. 172 ALGEBRA. 14. a^ + 6^ + c1 + (^^ - 2 {aW + arc" + ord? + 6V + t^cZ^ + c^c^^) - 8 a^cc? bya-6 + c + cZ; and that of this quotient by a - 6 - c - c?. 15. Show that the remainder, after the division of x^ - p^? + qx^ - rx + shj X — a^ is a* - jt?a^ + r/a' - ra + s. 16. Divide re* - 2/* by o;""^ — 2/~^ and x^ + cc"^ + 2 by X + a;~^ 17. Show that the quotient of 1 by 1 + a?, is 1 - cc + jc^ — cc^ + &c. ad infinitum, 18. Show that the quotient of 1 by 1 — 3 a; + 3 x' — a;' is 1 + 3 a; + 6 a;^ + 10 ic^ + 15 a;'* + &c. ocZ infinitum, 19. Divide {x + yY — 2 {x ■{- yY ^ ■\' z^hy x + y - z. 20. Divide (a - c)^ - 3 (a - c)^ (6 - c?) + 3 (a - c) (6 - (Z)^ - (6 - cZ)^ by a - 6 - c + cZ. Factors. 29. The ordinary method of finding the quotient of two algebraical quantities having been explained in the last article, we shall now proceed to show how, in certain cases, this method may be avoided, and the quotient written down at sight. It may be remarked at the outset, that the resolu- tion of algebraical expressions into their elementary factors is a subject of very great importance, and one which the student will do well to thoroughly master. (I.) The form x^ + 2 ax + a-. We have seen (Art. 26) that x"^ + 2 ax + a^ = (x + aY- Hence the suin of the squares of two quantities, together with twice the product of the quantities, is equal to the square of the sum of the quantities. And hence, any algebraical expression, which can be thrown into the form (ar + 2 aa; + a^), is of necessity a perfect square. Thus— ar' + 6 a; + 9 = a;- + 2 (3) aj + 32 = (aj + 3)1 a' - 10 a6 + 25 52 = a- + 2 a ( - 5 6) + ( - 5 5)= = {a - 5 5)1 16 aV - 56 ahxy + 49 hhf = (4 axY + 2(4: ax) {--7 by) + {-^ 7hyY = {iax- 7 hy)\ FACTORS. 173 3 nl/x" - 6 ahcxy + 3 achf -= ^ a (iV - 2 hcxy + cY) - 3 a{{hxy - 2 (6a^) (c^/) + (cyf] = 3 a {bx - cyf. (II.) The form o? - ^^l We have seen (Ai-t. 26, II.) that a'-h^ = {a + h) {a - h). Hence the difference of the squares of two quantities is equal to the product of the sum and difference of the quantities. Thus — aV - b^ = (axy - (byY = (ax + by) {ax — by). x'^y'= {aPy- iff = {x^ + f) (x^ - y") = (^ + 2/') (^ + y) (^ " y)- a" + b'' - c" - (P -h 2ab -- 2cd = (a^ + 2 ab + b') - {c' + 2cd + c/^) = {a + bf - (c + df = {{a + 6) + (c + d)} {{a + b) - {c + d)} = {a + b + c + d) {a + b — c - d). x^ + x'y^ + y* = {x^ + 2 aPy^ + y*) - (x^y^ = {^ + y'f - ^f ^ {{^ + y^) + ^y] {(a^" + ir) - ^y] = (a^ + xy + y^) {a^ - xy + y^). (III.) The form a;^ + jtx» + (7. This form evidently includes both the preceding, for the first form — viz., x^ + 2 ax + a^ is included, since (7 may be the square of half p ; and the second is included — viz., a^ - a^y since we may have p = 0, and q a negative square quantity. Now, the resolution into elementary rational factors of the quantity x^ ■{■ px •¥ q i^ not always possible ; but, since (Alii. 26, III), (x? + (a + b) x + ab = {x + a) (x + b), we have the following rule, when the quantity admits of resolution. Rule. — If the tliii-d term q of the quantity x^ + px + q can be broken up into two factors, a and by such that the sum of these will give the coefficient of Xy then the elementary factors of x^ + px -h q are x + a and x + b. Thus, a;^ + 7 a; + 12 = (a; + 3) (a; + 4); for the product of 3 and 4 is 12, an4 their ^um is 7, the coefficient of x. 174 ALGEBRA. xl - flj - 30 = (rw - 6) {x + 5); for the product of - 6 and 5 is — 30, and their sum is - 1. And so, x^ - ISx -h 32 = {x - 2) (x - 16), And a^ + 3 a6 - 108 6^ = a^ + (12 b - 9 h) a + {12 h) (-- db) = (a + 12b) {a - 9 6). (IV.) The form ax^ + bx + c. This is the general form of a trinomial. The following remarks, though equally applying to each of the three pre- ceding forms, are especially intended to be 'practically applied to trinomials not included by them. The above form will include such expressions as the fol- lowing:— 20 :«2 + lli^ _ 42, 6ar^ - 37aj + 55. It is evident that the product of the first terms of the factors will be the first term of the given trinomial, and that the product of the last terms of the factors will be the third term of the given trinomial. And, further, when the third term is negative, the last term of one factor must have the sign + , and the last term of the other the sign - ; but, when the third term is , positive, the last terms of the factors must have the same sign as the middle term. Thus, 12 ar^ - 31 oj - 30 - (4 aj + 3) (3 a; - 10). Here the factors of 12 cc^ are either 3 x and 4 a;, 6 aj and 2 5;, 12 a; and x, and the factors of 30 either 5 and 6, 3 and 10, 2 and 15, 1 and 30 ; and we must give a + sign to one of each of these latter pairs, and a - sign to the other. It is easily found on trial that, in order to obtain - 31 a; as the middle term, the factors of the trinomial must be 4 a? + 3 and 3 a; - 10. SowehavelOa^ - 41 a6 + 216- = (5a - 36) (2a -7 6), and aca^ + (ocZ + 6c) xy + bdif = (ax + by) (ca: + dy), (Y.) The forms x"" + y" and aj" - y^. We shall show in the next article that a rational in- tegral algebraical expression, involving x, contains a; - a as a factor when it vanishes on substituting a for x. Hence, a;"* + y^ and a;'* - y^ must each vanish on putting y for X, if they contain a; - 3/ as a factor, n being an integer. The former becomes y^ + y^ or 2 ?/", and the latter y" - y^ or 0. We therefore conclude that — xJ^ + 7/" does not contain x - y as a factor, and that — FACTORS. 175 X** - 2/** ^o®s contain re - 7/ as a factor, whether n be even or oc?c^. Again, on the same principle, they must each vanish if they contain x + ?/ ^^ ^ factor, on putting - y for x. The former becomes ( - yY + 2/", which vanishes when w isoddj and the latter becomes ( - yY - 2/", which vanishes when wis eve/i. Hence we conclude that — af* + 2/" contains x + yas a factor when n is odd, and x^ - 2/" contains cc + 2/ as a factor when n is even. Now, the quotient of either of these quantities by o^ + y or a; - 2/ ^^^ i^ ^^y particular case be found by long division. We thus find that — x^ + y^ =z (x + y) (x'^ - x^y + x^y"^ - xy^ + y^), x"^ - y^ = {x - y) (aj^ + a?y + xy'^ + 7/^), o? - if =- \x - y) {x^ -\- xy ■\- t/'). The law of formation of the co-factor in each case is easy to see ; and if we may assume this apparent law as generally true, we may conclude that, when an algebraical quantity is of the form cc" + y^ or ic" - y'\ and it contains x ■\- y or x - y as a factor, the law of formation of the co-factor is as follows: — Law of Formation of Co-Factor. 1. The terms are homogeneous, and of dimensions one degree lower than the given expression, the power of x in the first term being n - \, and diminishing each successive term by unity; and the power of y increasing each successive term by unity y and first appearing in the second term. 2. The coefficient of every term is unity. 3. The signs are alternately 4- and - , when a; + 2/ is the corresponding elementary factor ; and are all -f , when x - y is the corresponding elementary factor. Ex. 1. a^ -I- 32 = a^ + 2« = (a + 2) (a* + a' ' 2 -{- a^ • 2- -1- a • 2^ + 2^) = (a + 2) (a* + 2 a^ -t- 4 a^ -f- 8 a + 16). Ex. 2. a« - 6« = {aj - {h^ = (a^ 4- J/) (a« - h') = {a + h) (a^ -ab + b^)'{a - b) {a" + ab + b^ = (a + b) (a -- b) (V ^ ab + b'') {a^ + a6 + 5'). 30. The remainder of the division of a rational integral function ofxby:x. - a Qnay be found by putting a for x in the given function. 176 ALGEBRA. Def. — A function of x is an algebraical expression in- volving X ; and a rational integral function of x is an expres- sion of the form ax^ + bx''~^ + &c. + sx + t, where all the powers of x are integral and positive. Let/ (x)^ be a rational integral function of Xj and suppose Q to be the quotient, and H the remainder on dividing the function hj x - a. Then, evidently — Q (x - a) + B = f{x) identically. And this identity must hold for all values of x, and therefore holds when x = a. In this case we have Q {a - a) + R = f (a) + 7? =/(a) ori? =f{a). Now/ (a) is the result of putting a for x in the given func- tion, and is, as we have just shown, the remainder on dividing the given function hj x - a. Cor. 1. When there is no remainder, we must, of course, have/ (a) = 0. Hence, a given rational integral function of X vanishes when a is put for x, if it be divisible hy x-a, Ex. 1. The remainder, after the division of 2 ic^ — 5 x^ + 6 a; + 7 by re — 2 is 15. For, putting x = 2, we have — 2a;3- 5a;2 + 6aj + 7 - 2- 2^ - 5 • 2^ + 6 • 2 + 7 = 15, Ex. 2. The function— ic' - 2 oj^ + 5 a; — 52 is divisible by ic — 4. For, putting a; = 4, we have — cc»-2ar^ + 5aj-52 = 43-2-42 + 5-4-52 = 0. CoR. 2. Any rational integral function of x is divisible by X — \f when the sum of the coefficients of the terms is zero. For, putting aj = 1 in the given function, it is evident that it is reduced to the sum of its coefficients, which sum must be zero if the function be divisible by a; — 1. Ex. Each of the following functions is divisible by aj - 1, viz.: — 3 a;* -f. 7 a;^ - a;- + 12 a; - 21, 5 aj^ - 2 aj - 3, {a-h)x^ + (h — c)x + (c — a), (a + hf a;- — 4 ahx — (a- Hf. * The expression/ {x) must not be considered to mean the product; of/ and ic, but as a symbol used for convenience, FACTORS. 1 i i Cor. 3. Any rational integral function of x is divisible by X -h I, when the sum of the coefficients of the even powers of X is equal to the sum of the coefficients of the odd powers. (The term independent of x ia always to be considered as the co- efficient of an even power). Let ax'* + ax'' ~ * + &c. + rar + sa; + ^ be a rational in- tegral function of x. Put x =^ - I, then we have, if the function be divisible by 03 + 1 — a(«l)" +6(_l)n-i + &c. + r(- l)- + 5(-l) +^-0. Suppose 71 to be even, then evidently ( — 1)'*= (-1) ( — 1) { — l)...to an even number of factors = + 1. And so ( - 1)"-^ = ( - 1) ( - 1) ( - l)...to an odd number of factors = — 1 ; and so on. Hence we get a-6+&c. +r — s + t=Oy and this must evidently require the condition that the sum of the positive quantities is equal to the sum of the negative, and, therefore, that the sum of the coefficients of the even powers of X is equal to the sum of the coefficients of the odd j)owers. And a similar result will follow if we suppose n to be odd. Ex. Each of the following functions is divisible by jc + 1, viz.: — if3+5ar* + 7a;4-3, 5ar^-4a;* + 8ar-2a;-l, a^a;* - (a + 1) (a + 2) ar^ + 2 aj + 3 a + 4, 2^^ +{q +r)o(^ + {q + r) X + p, Ex. IX. Eesolve into elementary factors — 1. a;2 - 9 a^ 16 y* - 25 z\ 24 a^ - 54 6^, 8 »» - 27 y", 2. a;* — xy^, a^ - h^, xi/ + x^^y^ 2 x^y^z - 8 xy^^. 3. a* - 4 h\ cc* + o^y- + y\ a* - 2 aW + h\ a" + Ir - c' -^ 2aJx 4. a- + 62_c2^cZ2 ^ 2 a6- 2 cc7, a- ~ 6^ - c- + (/- + 2 he ■\- 2ad,a' - {b - c)\ 5. {x + If - (x + 2)^ {x + 5)2 - (x + 2)-, (2a + hf - (a - c)-. 6. {x^ - 2/')' + 4 (a;* + ar^r + y') (xPy^, {x" + y^f -^ ? (ar + ff sc'f-\'4: x*y\ 5 M 178 ALGEBRA. 7. x^ - ox " 70, ft3" + 11 oj 4- 10, fr - 15 ah + 56 h\ x^ - 4:x- 192. 8. rtV + aSa;?/ - 42 %% 3 «a;-— 2i: ax - GO «, 24 ac - 5 «c- + acl 9. Goj- - llo? -35, 8a;- + Gx - 135, 18a;- - 21a; - 72, 20a;= - 11a; - 42. 10: 3a^]/ + lOa^y + 3xf, 20 a;^ + 12 ax- + 25 hx^ ^ 1 5 abxj ni^^ + (piq + wzp) a; + ^. Write down the quotient of — 11. a;* - 16 by a; - 2, 3 a;*' + 96 by x + 2, a;^ - 27 by :^ - 3. 12. (a + Vf - (c + rZ)3 by rt + 6 + c + cZ, a^ - Z.^ + c^ - cZ- + 2 ac + 2 6(i by a + & + c - f/. Find the remainder after the division of — 13. x^ + f:^ + qx- + ra; 4- s by a; - rt, a;'* + a^ by ;^ - a\ 14. a;^ - 5a;^ + 7a; - 9 by a; + 3, x^ — 3a; + 7 by a; - 2. Show that — 15. 5a;^ - 3 a;^ + 7 a;- - 8 a; - 1 is divisible by a; — 1. 16. 2 a;* - 3a;^ + a;^ - 7 a; - 13 is divisible by a; + 1. CHAPTER III. Involution and Evolution. Involution. 31. Involution is the operation by which we obtain the "powers of quantities. This can of course be done by multi- plication, but the results obtained by the ^^ctual multiplication of simple forms enable us to develop without multiplica- tion more complex forms. As the subject requires the aid of the Binomial Theorem, we shall here show how to develop a few only of the more simple expressionB. SSt The power of a single tenn ia obtained by raising the JX^'Ol^UTION. 179 coefficient of the term to the power in question, and multijjhj' ing the exponents of the letters of the term by the exponent of the power in question. Thus, {a^f = a^ X a^ = aJ^ + ^ = a^ ^ ^, {ory' = or X a'"* x «"* to?i factors ^ a^^ + t^-*- w + ...to«tcrm8 And so, (iaWcy = 43a'-^W^V><' = Ua'bV\ 33, Development of the ^7iW,/) :=-- ft'^ + 6a'b + 10r*^^>5 ^ lOa^^' + 5 a5^ + 6». The following law of the formation of the terms is evident : — Law of Formation of Terms, 1. The first term contains a raised to the given power, and the power of a decreases by unity in each successive term, while the power of b (which first appears in the second term) increases by unity in each successive term, till it reaches the power of the given quantity. 2. The first coefficient is unity, and the coefficient of any term is found by multiplying the previous coefficient by the exponent of a in the previous term, and dividing the product by the number of terms hitherto developed. Ex. 1. (2aj+ Zyf = {2xy -^ 3(2aj)2(32/) + 3(2a;)(3 2^)2 + (3 7/)' = 8a;' + SGor^^/ + ^^^f + 272/'. Ex. 2. (a + 5 + c)' = {a + b + c)' = (« + bf + 3 (a + bfc + 3 (a + 5) c2 + c' = (a' + 3 a^b + Zab'' + b') + 3 (a? + 2ab + b'') c + 3(a + b) (^ -{- c^ = a^ + ¥ ^ c" ^ 3a^b + 3 a^c + 3ab^ + 363c + 3 «c' + ^bc^ + 6abc. (In the following example^ the above la"vy may be assumed a^ peneralli^ ime,} Ex. X. 1. Find tho values of {a'b% ( - 8 ab% (2 a^boY, { - »»y«»)^ 180 ALGEBRA. Expand — 2. {a + 35)*, (2a + h)\ {a - h)\ (3a - ih)\ 3. (2m + l)^ (5 a; + 2)^ (3 a - 4c)^ (- a - 6)1 4. h? + x + l)^ (3 ci - 6 + 4 c - cZ)^ (a + 2 6 - c)2 (3a+36 + 3c)l 5. (1 + a; - or^)', (ax + 6?/ + c;2;)^ | (a + 6)a;-(c+ 0^)2/1^ 6. (1 + a;)^ (1 + a; + a^)^ (a + 6a; + car)^ 7. (aaj - 6^)^ (3 a; + y)^ (3a; - yf, (ar^ + a;7/ + 2/=)» (a; - y)\ 8. (aJ* + a;^2/» + 2/T(^+ 2/)M^ - ^)', j(a + 6)»-4a6J'^'j(a - 6)^ + 4 a6 | . Simplify — 9. (a + 6 + c)^ - 3 (a + 6 + c)' c + 3 (a + 6 + c)r - c\ 10. (a - 6)3 + 3 (a - 6)^ (6 ^ c) + 3 (a - 6) (6 - cf + (6 - cf. 11. (1 + a; + Sar' + 30;^)' + (1 - a; +3ar^ - 3a;')l 12. j (a; + yf - (a;^ + y') \ ' - 27a)y (a;^ + f). Evolution. 34. Evolution is the operation by which we obtain the roots of quantities. Since the square or second power of a^ is (a^y or a^, we call (Art. 17) a^ the squa/re root or the second root of a^. And so, since the cube or ^/wVc^ power of a^ is (a^)^ or a^, we call a* the cube root or the ^/wVcZ root of a^. So, generally, since the nth power of a^ = (a*")"* = a*"", we call a"* the 72th root of a"*". Thus, we have s/aP = a^, ^^ = a\ ^'cT' = a"». Hence, in the case of quantities consisting of a single letter with a given exponent, when the given exponent contains as a factor the number indicating the root, we must divide the given exponent by this number, the quotient being the exponent of the root. (We shall see farther on that this rule holds when the given exponent ifi not so divisible, tJiQ root in this case bein§ palled a sur^)^ SQUARE ROOT. 181 35. Since the product of an eve^i number of negative /actors must give a positive result, and the product of an odd number of negative factors a negative result, it follows that — I. When the root is indicated by an even number — 1. The root of a positive quantity may be written either with a + or - sign. Thus, v^9^ = ±3 a, ^IQb' = ±2b, 2. The root of a negative quantity is impossible. Thus, V - tt^ V - a'*6^^, &c., are impossible quantities. II. When the root is indicated by an odd number, the root has always the sign of the given quantity. Thus, ^ - 27 b' = - 3 5^ ^32x'y = 2 x'y, (It may be remarked that the theory of impossible quantities forms an important branch of Algebra, which the student cannot yet enter upon. According to that theory, all quantities have as many roots as the number indicating the root. ) Square Root. 36. We shall now develop the method of finding the square root of a given quantity. Ex. 1. Find the square root of a- + 2ab + 5^. We know that a^ -\- 2ah ■\- h^ = {a + b)\ Hence, a + b is the square root of a^ + 2ab + 5^ or a^ -{■ {2 a + b)b. Now, it is evident that the first term a of the root is the square root of the first term a^ of the given quantity; and if this term be subtracted, there remains 2 ab + b^, from which to determine b the second term of the root. Now, b is contained in 2 ab + P or {2 a + b)b exactly {2 a + b) times. Hence it follows that the second term of the root is found by dividing the remainder by twice the first term of the root, and, if we wish to arrange our work in a way similar to long division, it is evident that we first take for our divisor 2 a + b, that is, twice the first term of the root added to the second tei-m, which mul- tiplied by b the second term, and subtracted, leaves no remainder. 182 ALGEBRA. Thus the whole process may be arranged as follows : — CI? 2a + b 2ab + b^ 2ab + W Ex. 2. Find the square root of a* + 2 a6 + 6^ + 2 ac + 2 6c + c". Now we know (Art. 26) that — o? ■\' 2ab + b'' + 2ac + 2bG + c'ovia + hY + 2{a + b)c + c^ = {a + b + cf. And if we compare the form (a + by + 2 {a + b)c -\- ^ with the form «^ + 2 a6 + 6^ it is evident that, having ob- tained as in the last example the first two terms a + 6, we shall by continuing the process obtain the third term. Thus — a- + 2 a6 + 6^ + 2 ac + 2 6c + c'(a + 6 + c d? + 6 26 + 2ab 2ab + + b^ + c 2ac + 2aG + 2 6c 2 6c + + c" c" We may deduce from the above examples the following general rule : — EULE. 1 . Arrange the terms of the given quantity according to the ascending or descending powers of some letter, and take the square root of the first term for the first terra of the quotient. 2. Subtract the square of the quotient, and bring down the next two terms of the given quantity. 3. Double the quotient, and jDlace the result as a trial divisor; then, f dividing the first of the terms brought down by this trial divisor to obtain the second term of the root, add the quotient so obtained to the first term of the root, and also to the trial divisor, to obtain a complete divisor. 4. Multiply the complete divisor by the second term of the root, and subtract the product, as in long division, from the terms brought down. 5. If there be any remainder or more terms to bring down, double the whole quotient for a trial divisor, and divide the SQUARE ROOT OF NUMERICAL QUANTITIES. 183 remainder by the new trial divisor, to obtain the third term of the root; and so on. Ex. 3. Find the square root of 1 - 4:X + lOa;^ - 12x'^+ Ox\ The terms are here arranged according to the ascending powers of x. Then proceeding according to rule, we have — 1 - 4 cc + 10 or - 12x3 +9x^(1 -2x + 3s(r 1 2-2aj -4aj+10a;2 - 4 a? + 4 ar 2 - 4a; + dxF Qa^ - Ux" + 9 x^ (jaP - I2a^ + 9x * The student will observe that twice the quotient is most easily obtained by bringing down the previous com2)lete divisor with its last term doubled. Square Boot of Numerical Quantities. 37. It is easy to apply the above method to numerical quantities. Since P = 1, 10^- = 100, 100^ = 10,000, 1,000^= 1,000,000, «fcc., it is evident that the square roots of numbers having less than three figures must contain one figure only ; That those having not less than three and less than five must contain two figures and two only ; Those having not less than five and less than seven must contain three figures and three only ; and so on. Hence it follows that, if a dot be placed over the units' figure, and over every alternate figure to the left, the num- ber of dots will give the number of figures in the square root. Thus, the square roots of the numbers 141376 and 1522756 have three and four figures respectively. In the number 141376 we call 14, 13, 76 respectively the first, second, and third periods. So in the number 152275G, the first, second, third, and fourth periods are respectively 1, 52, 27, 56. It is evident that the number of periods correspond to the number of figures in the square root, and it will be seen that the figures of each period are used in the operation for the coiTesponding figure of the root. 184 ALGEBRA. Ex. Find the square root of 565504. Pointing off the number, we find the first period to be 56. Now, the gi-eatest square in 56 is 49, and the square next greater is 64. Hence the number lies between the square of 700 and 800 ; and, following the algebraical method, 700 will be the first term of the root. The operation will stand thus — „ ^, c 565504 (700 + 50 + 2 = 753 490000 = a" 2a + b = 1400 + 50 =1450 75504 72500 = 2ab + P 2 a+2& + c = 1400 + 100+2 = 1502 3004" 3004 = 2ac + 2bc + c^ Or, omitting the useless ciphers, and bringing down one period of figures at a time, the operation will stand thus — 565504(752 49 145 755 725 1502 3004 3004 Ex. 1. Find the square root of 6091024. 6091024(2468 44 486 4928 39424 39421 Ex. 2. Find the square root of 83521. 8352i(289 4_ 48 435 384 569 5121 5121 SQUARE ROOT 01' A DECIMAL. l85 It will be observed that the second remainder, 51, is greater than the previous complete divisor, 48, and it might be supposed, therefore, that the second figure in the root should be 9 instead of 8. Now, the square of (« + 1) exceeds tho square of a by 2a + 1. Thus,Xa + If - o? ^ (p? + 2 a + I) - a" ^ 2 a ^- \. Hence it follows that, so long as any remainder is less than twice the corresponding number in the root + 1 , we may bo certain that we have taken the figure of the root sufficiently large. Thus, since the remainder is less than 28 x 2 + 1 or 57, we may be certain that 8 is the correct figure and not 9. Square Root of a Decimal. 38. It is evident that the square of any number containing one, two, three, &c., decimal figures, will contain two, four, six, : ■♦^ -*^ feX) g (M 55^ CD CO O CD 2 I I. I CO ^ CD CO I + I CO II II II I I + + CO CO CO ;<^ e^ H5co r5 f^ o It o + § O 02 « > cc K- J C3 O ^ ^^^ ^• *=^ J ^ <4H J^^ '^^ ^ P ^ fe (M ^-^ s ^ § -s c, ;^ ^ o B ^'^ ^ "^ ^^^^ s_^^ ^^ +f 1 ^s •i w cS C^ ■§ o o a -g^ o :g ^-^ ^ O 03 ^3 g S ^ ^ e.^ 8 O r^ _. ^ CO r^ ^ S3 • jii -^ ^ ^ ^ -+^ ^ ^ . (D S: ^ © Q o 3 ^ pi 189 2 o r-i - ^ , ?-( O TO O fH d i 4^ = 48 38503 33336 3x4x6 :== 72 ^ 6^ - 36 (^ 5556 ( 36 J 3 X 46^ 6348 3 X 46 X 8 = 1104 8^- 64 645904 5167232 5167232 Explanation. — The first two figures of the root are ob- tained as in Ex. 1 . We then treat the number they form, viz. , 46, as corresponding to a in the algebraical model, omitting useless ciphers. Obtaining then 3 a'*^ or 3 x 46^ ~ 6348, we find 8 to be the next figure of the root. Then writing under this, 3 a6 or 3 X 46 x 8 - 1104, and afterwards l? or 8- = 64, taking care as to the positions of the right-hand figures, and adding, we get 645904 as the complete divisor. Then as before. Remark. — It is unnecessary to be at the trouble to find the value of 3 x 46'"^ by ordinary multiplication. For re- ferring to the algebraical model, and writing here the succes- sive terms of the complete divisor, and adding, we have— ^ ( If we now again write down Jr Sun\ = 3 a'- + 3a6 + H^ T under this sum, and then add up }? ) the last four lines, we get — 3 a^ + 6 a6 + 3 b\ or 3 (a^ + 2ab + h') = 3 {a + by. This is three times the square of the first two terms of the root. It therefore follows that, if, as in the above example, after completing the operation for finding the first two figures of the cube root, we write under the complete divisor just ob- tained the value of the square of the second figure, and then ^d together the last four lines thus obtained, we get three 192 ALGEBRA. times the square of the quotient for a partial divisor by which to determine the next figure of the root. The four lines to be added are in the above example bracketed. This method will be found to materially shorten the work, for it may be similarly applied to find the trial divisor when the cu.be root consists of any number of figures Cube Root of a Decimal. 42. We know that the cube of any number containing one, two, three, &c., decimal figures will contain three, six, nine, &c., decimal figures respectively, and hence, conversely, every decimal considered as a cube must contain a number of decimal figures which is a multiple of threes and the number of decimal figures in the cube root must be one-third of the number contained in the given cube. It will then be necessary to add ciphers when the given number of decimal figures is not a multiple of 3. And by continuing the reasoning of Art. 37, if a dot be placed over the units' figure and over every third figui'e to the left, it will be sufiicient to bring down the decimal figures three at a time, putting a decimal point in the quotient when the first three are brought down. And further, if a^i integer be given which is not a pei-fect cube, we may proceed in the ordinary way till we arrive at a remainder, and then, putting a decimal point in the quotient, by affixing three ciphers to this and each successive re- mainder, approximate to the cube root as nearly as we please. Ex. 3. Find the cube root of 395-446904. 395-446904(7-34 . 343 3 X 7' = 147 "52446 3x7x3= 63 32 = S ^ 46017 15339 f 9J 3 X 73^ = 15987 73 X 4 rr 876 42 = 16 1607476 6429904 6429904 CUBE ROOT OP A DECIilAL. 193 43. We shall show farther on that when n + 2 figures of a cube root have been obtained by the ordinary method, n figures more may be obtained by dividing the remainder by the next trial divisor, provided that the whole number of figures in the root is 2 n + 2. We may apply this principle with advantage when we re- quire the cube root of number to a given number of decimals. Ex. Find the cube root of 87 to fiYQ places of decimals. The required cube root will evidently contain 6 figures, and since 6 here corresponds to 2 t* + 2 above, it is evident that n = 2, Hence, we shall find 4 (that is, n + 2) figures by the ordinary method, and then 2 more by division. The operation will stand thus — 87(4-43104 64 3x4^ = 48 23000 3x4x4 = 48 ■) 4? = 16 1 5296 ( 21184 16; 1816000 3 X 44^ = 5808 3 X 44 X 3 = 396 ^ 3« = H 584769 f 1754307 9; 61693000 3 X 443^ : 588747 3 X 443 X 1 1329 \ P = H 58887991 C 58887991 1) 280500900 58901283 235605132 44895768 Ans. 4-43104. Ex. XI. Find the square roots of — 1. 4a;VV, 16ay, ic* + 2aV + a' 2. 4 a;* - "^^ ^'^' • 00^4^.2 qa^s. ISar^y 29 A' - 30 aV + 25 «r/. N 194 ALGEBRA. 3. 25 a* - 30 a'b + 19 ^-6^ - 6 aP + b', 4. 1 -^ ix -\- lOa;^ - 20i^ + 25a:* - 24a;» + lGa;l 5. «2 + 2abx -h {2ac + ¥)x' + 2{ad + bc)x^ + {2bcl 6. «^Jc-« - 6aV«-i + 17a"aj-"-2 - 2iax'^''^ + IGct-^-^ 7. a;^ + 2 + x-\ orx^^^ - 2 + «-V. 8. 9a;-"* - 3a''x^ + 25 a^ - 30 ax^ +^ + 5a\ 4 Find the square roots of — • 9. 1296, 6241, 42849, 83521. 10. 10650-24, -000576, -1, ^W Give the values correct to four places of decimals of — ^ of -31416 V5 + V2 ^ VlO - 2 „ ,,,„ „ l-i. Tp?;; > — ; n: + "J > •> l*ll> Of V93 V5 - V2 VlO + 2 / 4 , V 32X6 Find the cube roots of — 13. SaWif, I25a^y,a^ + 6 a'b + 12 ab'^ + 8R 14. aj^2 + 9x'' + 6a;« - 99a;« - 42aj* + 441 a^ - 343. 15. a^ + 3xy + 3x7/^ + y^ - ^cxy - 3 car* - 3 nf + 3 c^a; + 3 chj - c^ 16. a^ + aj-3 + 3 (aj + x~% x^f/-^ + 3 xy-^ + 3 xy-"" + 1. Find the cube roots of — 17. 5849513501832, 1371-330631. 18. 20-346417; -037, tVtV Give the value of the following correct to four places of decimals : — T Q v'5'a2 + 4/-03375 1 1 y. — o,-— ■ m - v'-oT '4/4 + 4/2 + 1. .. tfm + ^-04 75 + 2 20. -::5i--;:^i — — r=^ of s V^ + v/-04 7 21. ( V7 + 2) ( V7 - 1), (5 + V3) (4 + V12). GREATEST COMMON MEASURE. 195 22. jirT6"j2] y '6'+ 15 ys: 03 ^3 6 + 2 75 J5'+ 1 " * 4 16 ' 4 ' _ 24. a^ (6 - c) - b^ (a- c) + c^ {a - h), where a = Vl % h^- - V^aixdc =~ 4/^027; CHAPTER lY. GREATEST COMMON MEASURE AND LEAST COMMON MULTIPLE, Greatest Common Measure. 44. In Arithmetic (page 24) we defined the G.C.M. of two or more numbers as their highest common factor. In Algebra the same definition will suffice, provided we understand by the term highest common /actor, the /actor 0/ highest dimen- sions (Art. 18). This, it need hardly be remarked, does not necessarily correspond to the factor of highest numerical value. 45. To^find the G.C.M. of two quantities. Rule. — Let A and B be the quantities, of which A is not of lower dimensions than B. Divide A by B, until a re- mainder is obtained of lower dimensions than B. Take this remainder as a new divisor, and the preceding divisor A as a new dividend, and divide till a remainder is again obtained of lower dimensions than the divisor ; and so on. The last divisor is the G.C.M. Before giving the general theory of the G.C.M. we shall work out a few examples. Ex. 1. Find the G.C.M. of or - 6aj - 27 and 2a;2_ ii^_ 03, According to the above rule, the operation is as follows ;— cc^ - 6a; - 27)2ar^- 11a;- 63(2 2ar'- 12a;- 54 ic - 9)a;2 - 6a; - 27(a; + 3 3x- 27 \ 3x - 27 .-. The G.C.M. is a; - 9. 196 ALGEBRA. Ex. 2. Find the G.O.M. of lOi^ + 31 a;^ « 63aj and Ucc» -f 51 a^ - 54 a;. We may tell by inspection that a; is a common factor, which we therefore strike out of both, only taking care to reserve it. The quantities then become — lOa:^ + 31a; - 63, andUa;^ + 51a; - 54. We may now proceed according to rule, taking the former as divisor. We see, however, that the coefficient of the first term of the dividend is not exactly divisible by the coefficient of the first term of the divisor. Multiply therefore (to avoid fractions) the dividend by such a number as will make it so divisible, viz., by 5. This will not afiect the G.C.M., as 5 is not a factor of the first expression, viz., 10 a;^ + 31 a; - 63. It may as well be here mentioned that the G.C.M. of two quantities cannot be affected by the multipli- <5ation or division of one of the quantities by any quantity which is not a measure of the other. We shall, for a similar reason, reject certain factors or introduce them into any of the remainders or dividends during the operation. (See Art. 47). Ux" •{- 61x - 64: ^ 10 a;^ + 31a; - 63)70 a;^ + 255 a; - 270(7 " 70a;^ + 217a; - 441 38a; + 171 Rejecting the factor 19 of this remainder, we have — 2a; + 9)10a;^ + 31a; - 63(5a; - 7 - 14 a; - 63 - 14 a; - 63 Hence, 2 a; + 9 is the last divisor, and multipljdng this by Xf the common measure struck out at the commencement, we find the aCM to be a; (2 a; + 9) or 2a;' + 9 a;. GREATESr COMMON MEASURE. 19T I I + + 1—1 I + »— I I CO <^, I I + I J-H 1 + CO lO I + CO r-4 tH (M 00 I I CO CD I I T-H CD + + ^ O cq o i—i I + 00 ^ cq -^ I i -^ r + 1 •Ij' T-i "5^ a 00 1:^ CO CO CO cq 1 I 00 o + I '^'^ O (M ^ 1—i iO CD ■H T—l + + 00 00 o o (?q ^ g CO rO ^.2 S tX3 'I^ s + r* ''i ^ >H T— « s J ^•k to 60 -* •3 2 ■I h^ r^ r-J 1 &< P^ 1 2 *-l3 ■* ^ CD 3 ^ I p CD -§ + 'S CD c3 '-' © 1—1 to ^ IS CD + + CO GO 00 H I C^^ ( •^ ( Oi "^ "^ ^ 1-1 r-H g cq o I— • o I. ' o ^ ■^: O CO (?q CD + + cq oi a 00 ^ o (M cq rH + + 00 ca 00 lO t^ r-i t^ CI CD 00 I i 00 00 CO I o o -« I r— I CO CD CO I a •J to •3 198 ALGEBRA. Dividing this remainder by 14371, and taking the quotient for a new divisor, we have — jc2-7a;-3)-72a^ + 679a;2-1009a;- 525(-72a: + 175 ^ -72a;^ + 504a;^+ 216 a; 175 ic^- 1225 a;- 525 175 ar^- 1225 a;- 525 :, x" - 7x - Sis the G.C.M. It will be seen that we have introduced and rejected factors during the operation in order to avoid fractional coefficients* This, as will be seen from the general theory, will not affect the result, provided that no factor thus introduced or rejected is a measure of the corresponding divisor or dividend, as the case may be. Theory of the Greatest Common Measure. 46. Let A and B be the two algebraical quantities, and the operation as indicated by the rule (Art. 45) be performed* Thus, let A be divided by B, with B)A(p quotient p and remainder C, Then j(?i? let B be divided by (7, with quotient ^, 0)B(q and remainder JD. Lastly, let G be qQ divided by Z>, with quotient r, and D)G(r remainder zero. ^^ Then we are required to show that ~7r D is the G.C.M. of A and B. (1.) i> is a common measure of A and B, Now, we have G = rl), B = qG + JD, A = pB + G. Hence, i> is a measure of (7, and therefore of qG, It is therefore a measure of qG + D or B. Hence, also, i> is a measure ofpB, and since it is also a measure of G, it must be a measure of pB + G or A, But we have shown it to be a measure of B, Hence, i) is a common measure of A and B. (2.) D is the G.C.M. of ^ and B, For every measure of A and B will divide A - pB or G; and hence every measure of ^ and B will divide B - qG ov JD, Now, JD cannot be divided by any quantity higher than JDf and, therefore, there cannot exist a measure of A and B h^her than J). Hence, D is the G.C.M. of A and B. 1:fifiORV OPTHE GftfiATESt COMMOK MEASURE. 199 47. A factor which does not contain any factor convnion to both A and B may he rejected at any stage of the 2)rocess, Let the operation stand thus : — B = 7nB ' suppose, B^A{p C)B\q pB qG;_ C = nC suppose, D)C\r r_D_ "0 where neither m nor n contains any fiictor common to A and B, It will be iari exercise for the student to show that B is the G.C.M. of^and^. 48. A factor^ which has no factor that the divisor has^ 7iiay he introduced into the dividend at any stage of the 2)rocess. The operation may stand thus — B)mA(2), where m has no factor that B has ; pB C)nB[q, where n has no factoi' that C has ; qc D)C{r rP As in Arts. 46, 47, it may be easily shown that D is the G.C.M. Both the above principles are made use of in working but Ex. 3 Art. 45. 49. When a bommon factor can be found by inspection, it is advisable to strike it out of the given expressions. Then, having found by the ordinary process the G.C.M. of the re- sulting quantities, we must multiply the G.C.M. so found by the rejected factbri • Thus, 4 cc is common to the qiiantities 4 a;^ — 20 a;- + 24 aij tind4a;^ 4- 16 ar^ - 84cc. Rejecting it, we get o? - 6 x + 6, and or + 4 a; - 21^ whose G.C.M. is easily found to be a; - 2. Multiplying by 4 a:, we find the required G.C.M. to be 4ar^ - 8a;. 200 ALGEBRA* 5O4 By a little ingenuity on the paft of the student in breaking up the given expressions into factors, the ordinary and often tedious process of finding the G.C.M. may be avoided. The limits of our space will allow us only one example. Ex. Find the G.C.M. of 3x^ + 4:x' - 10 x + 3, and 15 a;' + 47a;2 + 13 a; - 12. The first expression contains a; - 1 as a factor (Art. 30), for the sum of its coefficients is zero. The other factor may be obtained thus — 3a;^ + 4a;^-10a; + 3 = 3ar^- 3 x^ + 7a;2- 7 a;- 3a;+ 3 = 3a;2(a; - 1) + 7a;(a;- l)-3(a;- 1) - (3 a;2 + 7 a; - 3) (a; - 1). Now, 3 a;- + 7 a; •— 3 is not further resolvable, and a; - 1 is evidently (Art. 30) not a factor ofl5a;^ + 3ix^ + 13a;- 1 2. It is, therefore, very probable that 3 a;^ + 7 a; - 3 is the G.C.M. required. * We may test it thus — 15a;3 + 47a;2+ 13a;- 12 = 15a;3+ 35a;-- 15a;+ 12a;^ + 28a;- 12* = 5 a; ( 3 a;^ + 7 a; - 3 ) + 4 ( 3 a;2 + 7 a; - 3 ) = (5a; + 4) {Sx" + 7x- 3). Hence, 3a;^+ 7a;- 3 is the G.C.M. required. G.C.M. of Three or More Quantities. 51. The G.C.M. of three or more quantities may be found thus — Rule. — Find the G.C.M. of any two of the quantities, then the G.C.M. of the G.C.M. so found and a third quantity, and so on. The last found G.C.M. will be the G.C.M. required. Ex. XII. Find the G.C.M. of the following— 1. ar^ - 5 a; + 6 and a;^ + 3 a; — 18. 2. a;^ + 6 a;' +lla;+6 and a;' + 5 a;- + 7 a; + 3. 3. 2af»+ 10a;2- 18;B-90and 3 a;^ + 16a;2-26a; - 141. 4c, aP + (a + b €C + ab and x^ + {a + c) x + ac, 5. LEAST COMMON MULTIPLE. 201 6. a;^ - 4 J!c + 3 and aj' + 4 x- - 5. 7. 4ic'- 32a:« + 85 a; - 75 and 3 af» - 15ar^ + 15a; + 9, 8. 9 a;' - 3 a:y + 2 2/ - 4 and 6 x* - 4 a;^ - 9 a;?/^ + 6 ?/l 9. 48 a;* + 8 a:^ + 31 x" + 15 a; and 24 aj^ + 22 a;3 + 17 x" + 5 a;. 10. 15 o? + a'5 - 3 ah'' + 2h^ and 54 a-^^ - 24 b\ 11. 3a;'-(3c + c£+l)ar^-(2«4-6-3c-cZ+2)a; + 2 a + 6 and 2a:'-(a+6 + 2)a; + a + 6. 12. 6a:5 _ 4^1 - 11 ajS „ 3 a;^ - 3 a; - 1 and 4a;^ + 2a;' - 18 a;^ + 3 a; - 5. 13. ah + 2 0^ -- 3 b'' - 4:hc - ac " c^ and 9 ac + 2 a^ - 5a6 + 4c^ + 8 5c - 12 6^. 14. e'x' + e'' + x' + I and e'^'a;* - e-* + a;* - 1. 15. oa;* + {h + c) aP - ax '- h -^ c and ear^ - (f - g) x^ + (/ - e) a; - ^. 16. 4 a;* + 2 a;' + 4 a;^ + 39 a; - 9, 8 a;* + 20 a;^ + 51a; + 9, and 2 a;* + a;^ + 3 a;2 + 18 a;. 17. aa;^ - (c + 1) ar^ + (c + 1) a; - a, 5a;^ - (5 + cZ) a;' + (c + d)x^ — {c + e) X + €y and {c + I) oc^ + {d + 2) a;* ~ (c^ + 1) a;^ - (c + 2) x\ 18. a^ - 5' + c» + 3 ahc and a'' - h^ + c^ + 2 ac. Least Common Multiple. 52. When two or more algebraical expressions are arranged according to the powers of some letter, the expression of lowest dimensions which is divisible by each of the given expressions is called the L.C.M. 53. The L.C.M. of monomials and of expressions whose factors are apparent may be found by inspection. Ex. 1. Find the L.C.M. of a6, ac, acZ, Sc, hdy cd. If we form an expression, whose elementary factors con- tain each of the elementary factors of the given quantities, we shall evidently have a common multiple ; and if no ele- mentary factor of this expression is of a higher power than the highest power of the same factor in the given quantities, we shall get the L.C.M. Hence, the required L.C.M. = ahcd. 202 ALdl^BRA* Ex. 2. Find the L.C.M. of— {a - Z>) (5 - c), {a - h) {c -- a), (6 ^ c) (c - a)* Alls, {a - h) {b - c) {c - a). Ex. 3. Find the L.C.M. of a (x + 1), b (x" - 1), c (x^ + 2 X — 3),d(x^ + 4c X " 3)i We may write the given expressions thus — - a{x + 1), b{x + I) {oi - i), c{x -^ 1) (cd + 3), d (x + 1) (x -i- 3). Hence, the required L.C.Mi = abed (x - l) (x + 1) {x + 3). Ex. 4. Find the L.C.M. of a^ - aaj + a^^ a^ + ax + x", a' + xl, a^ - x\ Kow (Art. 29) a^ ■{- x^ = (a + x) {a^ - ax + ^), and a^ - a* = (ot - a) \a? -{• ax ■\- x'). Hence the required L.C.M. — ^ {a ■{- x) (a - x) (a^ -{• ax + x^) (cu^ - aa; + x) = a^ - x^. 64. !the L,d.M, qf two quantities is found by dividing their product by their G,C.M. Let a and b be the two quantities, and d the G.C.M. ; And suppose a - pd and b - qd. It is evident that p and ^contain no common factor. ^QWCQ pq is the L.C.M. of p and q ; and, therefore, no expres- sion of lower dimensions than pqd can possibly be divisible! by pd and qd. Hence pqd is the L.C.M. ot pd and qdy or of a and b, "No^pqd = pd X qd -■ d = a X b -r- d, and hence the rule: 55. To find the L.C.M. of three or more quantities. EuLB. — Find the L.C.M. bf two of the quantities, theri th0 L.C.M. of thd expressioii thus obtained and a third quantity, and so on. The last expression so found is the L.C.M* required. We shall prove this rule in the case of three t[uantities« Let tt, by c be the quantities, and ni be the L.C.M. of a iand. b. Then the L.C.M. of m and c is the L.C.M. required. For every common multiple of m and c is a common multiple of a^ 6, c. And every common multiple of a and 6 LEAST^ COMMON MULTIPLE. 203 taust contain the m, their least common multiple. Hence, every common multiple of a, 6, c must be a common multiple of 7?i and c, and the conA^erse is also true; Hence, the L.C.M. o£m and c is the L.C.M. of a, b, c. Ex. XIII. 4 Find the L.C.M. of— 1. axy^f 3 aVi/, 4 a^if, 6 ary^, 2. 5a^^^^6aV, 45V. 3. {a -- b){b - c), (6 - a) (a - c), (c - a) (c - 5). 4. ax {x + a), a^(aj - a), fc^ - a^* b, aP -h 3x + 2,x' + 4:x + dyX^ + 5x + e. 6» 0? - a; - 30, aj2 - llaj + 30, ar» - 25. r. 6ar^ + 37a; + 66, Sa;^ + 38a; + 35^ 12a;^ + 47a; + 40* a 5 (or^ - a; + 1), 6 (ar' + 1), 7 (o;^ + 1). 10. a;^ + (« + Z>) a; + a5, o;^ + (a + c) a; + ac, o;^ + {^ -^^ c)x -v be. 11. 1 - oj, 1 + a;, 1 + a;^ 1 + a;^, 1 + x^, 12. a;3 + 6a:' + 11a; + 6, a;^ - ^x" - 25a; + 150. 13* o? - 3a6 {a ^ b) ■- b\ a? - 6^ a^ + a^6 + a6l 14. X* - 1, 6a;5 + 5a;^ + 8a;' + 4ar^ + 2a; - 1. 15. a^ - 2o?b'^ + b\ a^ + 4a36 + 6^-6^ + 4a6' + 6S 16. 3a;3 _ 4^ .^ 1^ ^x^ _ 7aj + 5^ 4cc4 4. c a;^ + 10 a;. 17. Zx^ -v Qx- 24, or' - 12a; + 16^ 5a;^ - 122a; - 36. 18. a^ - ab\ P - a'b, aP - b\ o?b - a\ 19. 3a;* - 48, ^o? - 20, 30:^ - 16aj + 20. 20. a? '- 'if,x^ - 2 x'lf + 2/^ a;^ + or'y + a;?/* + 2/^. 2U ^ •\- ax^ -V aV + aV + a^x + a^ and a;'' ~ ax\ -f aV - o^x^ + a^a; - a\ 22. a2 ^ 52 - c^ - c^3 + 2a6 - 2cd and a' - 6- - c- + ci^ + 2ac^ - 2 6c. 204 ALGEfiRA. 23. ft' + i^ + c' - 3 a5c and {a + hf ^ 2 (a + b)c + c\ 24. (a + by - (c + c^)S (^ + cy - (b - d)\ {a + df - (6 - c)-, (c + dy - (ft - 5)', (6 + dy - (ft - c)^ and (6 + cy - (ft - (^)l CHAPTER V. Fractions, 56. It is unnecessary to repeat here the propositions relat- ing to fractions which were proved in Arithmetic, Chap. II. of this work. The student will see that, by substituting general symbols for the particular figures there used, the reasoning will equally hold. We shall work out a few examples to show the method of dealing with them in algebra. Ex. 1. Simplify the fraction g k — ^-je — • By inspection (Art. 30) we see that x - 3 is a factor of numerator and denominator. We have then — a:^ - 2a^ + a; -12 ^ a^ (a; ~ 3) + a; (a; - 3) + 4 (a? - 3 ) af^ + 2a; - 15 (a; - 3) (a; + 5) _ {x^ + X + 4) (x — 3) _ x^ + X + 4: (x - 3) (a; + 5) ST~5 ^^• 1 1 2ft Ex. 2. Find the value of a + b a - b d^ + b^' 1 1 2ft ^ (ft - 6) + (ft + b) __ 2ft ft + 6 "** ft - 6 a" + b^" (ft + 6) (ft - 6) a' + b' 2ft 2ft / 1 1 \ 2a (gg + h') , (g^ , 5^) 25^ 4ft5^ (ft^ - ^>^) (ft2 + 5^) ^ " ^' c?"::!^ "a* - 6^- FRACTIONS. 205 Ex. 3. Find the value of / ttt r + 77 ryr -v {a - b)(ci — c) (6 - a) (6 - c) a6 + (c _ a) (c - 6)- The second denominator has a factor, (5 - a), which differs from a factor, (a - 6), of the first denominator in sign only. "We shall therefore change the sign of the second fraction, and also of its first factor. This will not alter its value. And, similarly, we find that by changing the signs of each of the factors of the third denominator we shall have them in a form corresponding to factors of the first and second denomi- nators. The sign of the third fraction will not be changed, as the sign of the denominator will, on the whole, be unchanged. The given expression then will stand thus — he ac ah " (a -h){a- c) " {a - b) (b -- c) '^ loT^c) {b - c) he (h - c) - ac (a - c) + ah (a - h) " {a - b) {a - c) (6 - c) he (6 — c) - a-e + oc^ + a^h - ab^ = {a - 6) (a - c) (b-e) ' ^"' re-arranging, a^ (6 - c) - a (b^ - c2) + he (b - c) ,, ,. . ,. = — ^^ — , IK / r-71 \^ -y ^'^^^f dividing nume- (a - h) (a - c) (b - c) ° rator and denominator by 6 ~ c, ^ a^ - a{h + c) + he _ {a - h) (a " c) " (a - bfiaT^^^ ~ {a - b) (a - c) " _ ' Ex. 4. Simplify— f J _ 4 a^'x - 3 aa^ + a^ ]. ^ f ^2 . 4a^a; -{-^ax^ + x " ) I a ■¥ X ) \ a " X ) ' The given expression — a\a + x) - ia^x-^ 3aa;--ar^ a^a-x) + i a^x + 3 aar^ + x^ a + X a — X ■ 4 a'^x + 3 aar* - a;' X a + X (* - a? 206 ALGEBRA. — ■ — X , a + X _ {a - xf (a + x)^ _ {a - xf {a + x) X a + X a - X Ex. 5. Divide — 1— ' ^ '-r-^ - (^^ - ^Y^ (^ + 6) 1-1-4- --^Jby (a - 5) ] -% + ,-^o} Now — or, reducing — _ (« + 5) (aj - a) (aj + 5) Ex. XIY. Simplify the following expressions : — 1 a^ - 5a; + 4 a;^ - 3a? + 2 aj^ + 2 a; - 24' a."^ + 4a;^- 5* ^-^ 6a;2 + 29a; + 3.5 2a;3 + 7a; - 9 3. 14a;'^ + 39a; + 10' Sa;^ _ 3^ _ 4^. + 2 a^ - 2^262 + 5* 24a5 - 28a25 + 6a&^ - 75^ «^ - 4^26 + 4a62 - 6^' 6 a^ + 11 a6 - 21 6- 5.-K-^ 1 1 (* -1- 6 a - 1) a - h a + b' FRACTIONS. 207 , , b ah ah G. 7. 8. 9. a + h a - h ah - W cv" + ah 1 03-1 a; - 1 ij^l)^ 4 (ic- + 1) " 2 (x' + 1)-' _5 2__ _ _18 __ nx + 1 :c + 1 (.T + 2)- 5 (oj + 2) 5 (aj'^ + 1)' 11 11 14 cc+l aj + 3 (ic+ 1)-' ,Q 8 _ 4 2_ 16a! -f 14 _ lGj»-8 • (aj - 1)^ (a; - If Jx^f "^ 3 (ar - 1) 3(i-~-^TI) * 11. . X . -^ -. i_- ^ . ^ (« - 6) (d^ - c) (6 - a) (6 - c) (c - a) (c - h) t + ^ + 1 . (a - h) (a. - c) (6 - a) (6 - c) (c - a) {o - b) t + I + £!__. (a - b) (a — c) (6 - a) (b - c) (c - a) (c - b) ^ ^ 6^ ^ c^ (a - b) (a - c) {b - a) {b - c) (c - «) (c - 6)' "■ (F->)V')'(EM^)'a^o(F^)' 16. ■ ^■ 12. 13. 14. (a + ly + {b - cf + (a + cy - _ 2 _ _J_ ^ (a + 6) (6 - c) (a + c) « + c 6 - c « + 6' ( a + a; « - a; a" + ar J I a + x a-x a-" + a;- J ( a- - 6^ a + 6 j * I *" a? - ir a + h i ' 10.'{l-2_^._l^lx{ 2a^_2a=_j) I a- a-(a^ + ar) J I ar(a- - ar) ar J «^f-+l)+6=(£+l) + c^(?+l) ' c a 1) ah + be + cc^ 208 ALGEBRA. =Me-^p't>}-{(^:)'*_'}-{(f^:)'*'} a;- {y — z) + 'if- {z - x) + z- (x - ^z) 22. |Li^+ _I:i^l ~ / 1 + ^ -^-^'1 23, (a + 6) (a + c) (x -a) {a+ b) (6 - c) {x + b) /.3 (a + c) (6 — c) {x + c) 24. |?^'-^ + J6-^l - |?-^-i-il. ^ { a' a x^ ) \ cir a x ) \a - bf \a + b/ '' 26. 27. ^' + -i^-, - J±- - _12_ +2. (a + by (a - by a + b a - b 28. + y' - 2a;y ^ f_ {x - y) {a + aj) (x - y) {a + y) (a + 7/)- y X y X "^^ '^'^ 29 V - V '^- - y /"^ - — Y 30, 1 y + - SIMPLE EQUATIONS. 209 CHAPTER YI. Simple Equations. 57. "When two algebraical expressions are connected by the sign ( = ), they are said to form an equation. When the equality is such that it is true for all values of the letters in the given expressions, it is called an identity. Thus, {x + a){x-\-h)=x'+{a + h)x-vah\ ....,•, and {a + hf - (a - hf = 4.ah ] ^""^ i^^^^i^ies. 58. When the condition of equality is such that some one or other of the letters must have pai'ticular values or a limited number of values, the statement of equality is termed an equation of condition^ or, more briefly, an equation. Thus, it may be found on trial that the equality 4rx; + 2 = 3aj + 5 is true only when a; = 3. Such an expression is therefore an equation. '59. The letters of an equation to which particular or a limited number of values must be given are termed unknoivn quantities. Equations may contain one, two, three, or more unknown quantities. The determination of the particular value or values of the unknown quantities is called the solution of the equation, and each of the values which satisfies the equation is said to be a root of the equation. 60. The expressions on the left and right sides of the sign ( = ) are termed the first and second sides respectively. It follows, therefore, that — 1. If both sides of an equation he multiplied hy the same quantity, the equation still subsists. 2. If both sides be divided by the same quantity, the equa- iio7i still holds. 3. Any term may be transposed from one side to the other if the sign of the term be changed. Thus, if 3 aj + a = b, we must have also 3cc = 6 - a, 210 ALGEBRA. for this results from subtracting a from each side of the equation. 4. The equation holds if every term on both sides has its sign changed. Thus, if ax + b = ex - dj we may reason as follows : — The quantity (ax + b) looked upon as a whole is given equal to the quantity (ex — d) looked upon as a whole. If we change the qualities of these quantities, they will evidently be still equal. Hence, - (ax + b) ~ — (ex - d) or, - ax - b =^ —' ex + d. Now, this is the result of changing the sign of every term on both sides of the given equation. 5. The sides of an equation may be reversed without destroy- ing the equality. Thus, if mx + n = px + q/\\> must also follow that px + q = mx + n. 6. The sides of an equation may be raised to the same POWER, or we may extract the same root of both sides, and the equation still subsists. 61. Simple equations are those in which the unknown quantities are not higher than the first degTee, when the equations are reduced to a rational integral form. The following is the general method adopted in solving a simple equation involving only one unknown quantity — 1. Clear of fractions if necessary. 2. Transpose all the terms involving the unknovm quantity to the first side of the equation, and all the remaining terms to the second side. 3. Simplify both sides if necessary, and divide both sides by the coefficient of the unknown quantity. Ex. 1. Solve the equation 5fl3+6 = 3i:c+12. Transposing the terms, we have — 5a;-3a;=12-6. Now, simplifying, we get — » 2x = G; and dividing each side by the coefficient of the unknown quantity, viz., by 2, we have — ^ X = Q -r 2 = 3. SIMPLE EQUATIONS. 211 Verification. — Putting the value 3 for x in each side of the given equation, the first side becomes 5 x 3 + 6 or 21 ; and the second side becomes 3 x 3 + 12 or 21. The value of a: found therefore satisfies the given equation. -r^«^. X - 2 X ^^ £c-6^, Ex. 2. Given —j— H- j = 20 ^, find x. Clearing of fractions, by multiplying everi/ term on each side by the L.C.M. of the denominators, viz., by 6, we get — 3 (a; - 2) + 2 cc = 20 X 6 - 3 (a; - 6). (Beginners often neglect to multiply integral terms such as 20 by the L. CM.) or3aj - 6 + 2x = 120 - 3 aj + 18, or, transposing, 3aj + 2iw + 3 a; = 120 + 18 + 6, or, simplifying, Sx= 144, or dividing each side by 8, the coefficient of x, we have — a; = 18, the value required. (It will be good practice for the student to verify this result as in the last example). _ .4a;-. 21 ^^ 7 a; - 28 ^, d-7x Ex. 3. ^ — + 7| + 3 =x+ 3f g — + ^V It is sometimes convenient to first partially clear off frac- tions. Thus, multiplying each side by 72, v/e have — 72 ^4 a; — 21^ -^ ^ + 47 X 12 + 24 (7 a; ~ 28) - 72a; + 15 X 18 - 9 (9 - 7a:) + 6; 288 a* or —^ 72 X 3 + 564 + 168 a; - 672, = 72 a; + 270 - 81 + 63 a; + 6; or, transposing, 4H a; + 168 a; - 72 a; - 63 a; = 270 - 81 + 6 + 216 - 564 + 672; or, simplifying, 74^ a; = 519 ; or, multiplying eacli side l)y 7, • 519a; = 519 x 7; .-. a; = 7. 212 ALGEBRA. Ex. XIY. 1. 5a; + 2 = 2aj + 11. 2-4 + 6^8 -.^^• 3. 2x + a=3x-'b. 4. 3(a;-7) + 4a; = 2 (2a; - 4) + 2. ^a;-l X + 3 1x - 1 8a; 5. — TT— + 7— = ^ + 2^4" 6 ^12 8 (5 a; + 2) 2 a; - 1 _ 17a; - 2 5| + 80a ; ^' 3 "" 8 - 4 **" 7 • 9. ax + be = bx + ac, ^ X a X b abba ^^ X - a X - b _ ex - (^ 11. — ,— + = 2 + r — . 6 « ab 12. a&a; + 5^ = i'-a; +a^ 13. - + T +" = «& + «c + 5(?. a 6 c , , a; + 5 a + a; 14. = — 7 — . a b ^^ ax -v bx -v ex _ 15. 7 = a + 6 + c. ^/ o? - Zbx _, 5a; 6 Z>a; - 5 a' Z»a; + 4 a 16. X T. — ' - 6' = — + ^r— a - — 7 . a^ ^ a 2 a^ 4 a 17. •15 a; + -025 = •075 a; + -175. 18. i^Lj^ . -1^ . ?^4^ - .083. ABBREVIATED METHODS FOR PARTICULAR CASES. 213 20. (a; + a) {x + b) = (x + c) {x + d). 21. {x^a) {x-h) = (a; - ^r+~h)\ ^^ X " a x - b x - c'^ ^ A 1 1\ 23. —J— + + — ,- = 2 (- + T + -). be ac ab \a b c/ ^^ \ - ax I - bx 1 - coj /2 2 2\ 24. — T — + + r— = (- + r + -p- be ac ab \a o c^ Abbreviated Methods for Particular Cases. 62. When the unknown quantity is involved in both iiumerator and denominator, it is often convenient to reduce such fractions to mixed numbers, T. c, , X. 6a;-7 12a: + 18 ^ Ex. Solve the equation ■ o " 3 x - 6 ~ T>,. .. 6a;-7 ^"^19 By division we get ^ ^ ^ = 6 - ^-:^'^ 12 a; + 18 , 38 and — o K - = 4 + 3a;-5 ""^Saj-S* Hence the given equation becomes — / 19 \ / 38 X or, transposing, - ^^-^ = 3^ _ g + 2-6 + 4; 19 _ 38 . or, dividing each side by — 19, we have — _1 2 aj+2~~3a;-5' 214 ALGEBRA. Hence, multiplying each side by the L.C.M. of the de- nominators — 3x - 5 = -2(aj+2)= -2a;- 4, or3aj + 2a;= — 4 + 5, or 5 a; = 1, .•.^ = i. 63. When each side of an equation consists solely of a single fraction, the oiumerator of either fraction niay change places with the denominator of the other, a p Let ^ - " be the equation. Multiply each side by 6, then, by Art. 60 (1.)— a p pb •% X - -' X h, or a ~ - . h q ' q Divide each side by^, then, by Art. 60 (2.) — a ph ah - =: — + t?, or - = - p q ^' p q Here the denominator h of the first side of the given equation has changed places with the numerator p of the second side. And similarly we may show that 7 = - , where the other 6 a numerator and denominator have changed places. Cor. The two sides of an equation of the form . = - may he inverted, Q p For interchanging^; and h in the last result, viz., _- = — , a we get - = -, and therefore also, by Art. 60 (5.), we have a p (The student is cautioned against inverting the separate terms of the two sides of an equation when there are more than one term on each side. ) 64. When each side of an equation consists solely of a single fraction, we may perform the following operations : — ABBREVIATED METHODS FOR PARTICULAR CASES. 215 1. We may add or subtract tJie numerator and denominator of EACH fraction for a new numerator or denominator^ and retain either the original numerator or denominator for the other term of the fraction^ both sides being always similarly treated. Thus, if T = -, we have — (1.) -^— - -y-, (11-) -y- - -y-' (iii.) ^Jl_^ . 1L±J[^ (i,.) «_iLi . 2LZA or, (v.) we may have equations formed by inverting each of these. These results are easily obtained — For, since t = -, we have, adding unity to each side — a ^ p ^ a -h b p + a - + 1 = ^- + 1 or — j-~ = -. q q And so, by subtracting unity from each side, we get — a - b p - q , — f — r= , and so on. b q 2. We may take the sums of the numerator and denominator of each for new oiumerators or denominators, and the dif- ferences /or the other terms of the fraction ; and vice versa, both sides being always similarly treated, .a c a + b c + d Thus, if 7 = -7, we have also 7 = ,, ' b d^ - a -^ b c - d , a - b c - d and — -— T = — ; — >) a + c + a for we have lUst shown that — -. — = > ^ a - b V - ^ and — f — = -" b q 216 ALGEBRA. Hence, dividing equals by equals, we get— a •{■ h ^ a - h ^ p + q ^ p - q h ~ h ~ q r~ q ^ or T = , which is the first result. a - p - q' And inverting each side, we have, by Art. 63 (Cor.) — a - b P - q a + h ~ p + q ^ ^ ^ , - ,. mx ■\- a + h iiuc + a + c Ex. 1. Solve the equation -j - r -j- ^ 7hx - c - a nx — - a , mx •\- a -v h nx - c - d By Art. 63, we have = j ;. •^ ^ mx + a + c Qix - - a Then, by Art. 64 (1.), retaining the numerators and taking the differences for new denominators, we have — 771X + a -^ b oix — c — d b - c ~ b - c ■' or, multiplying each side by (b - c) — mx + a -{• b =^ 7ix - c " d; or, transposing — / X /7 7\ a + b + c + d {m " n)x = - {a + b + c + d); .\ x = _ Ex. 2. Solve -^^ ija + cc - is/a - X We may consider the quantity a as a fraction whose deno- minator is unity, or as - . Then, Art. 64 (2.), taking the swmand difference, v^^Q have— ^iJa + x_a+\ ^ 2Vr~^ ~ oT^^v'"'' tja + 03 ' a + \ — . — = 1 J or, squarmg — sja - X a - I (i + a;_a^ + 2a+l AfiMEVlATEt) METHODS FOR PARTICULAU CASES. 217 Again, taking the difference and suriiy we have — 2a; 4a 2a 2a^ + 2'^^ X _ 2a , 'a a^ + 1 2 a' a- + 1 65. We now give an example to show that sometimes the easy solution depends on an advantageous arrangement of the terms on the two sides of the equation. Ex. Solve V5~r4 + sjT^^ = 7. Transposing, we have — Va; + 4 = 7 - Va; - 3; squaring, then — a; + 4 = 49 - 14 Va; - 3 + (a; - 3); subtracting x from each side and transposing, then — 14 Va; - 3 = 49 - 3 - 4 = 42. .'. six - 3 = 3 ; or squaring, a; - 3 = 9; .-. a; = 3 + 9 = 12. Should the student commence by squaring at once, he will render the equation more complicated. Ex. XYI. 3a; + 7 3a; - 13 2. 3. 4. 6. 6. a; + 4 a; - 4 ' (a; - a) {x - h) = {x - c) (x • -4 3a; +13 3a; + 10 x 15 5 a; - 50 5' 1 - 25a; 3 - 2\x 28 - 5a; 10a; - 11 x 15 14 (a;- 1) 3 30 3 aj - 4 3a; - 13 _ 1 6a; + 5 18a; - 6 3' 3 5 - 2a; _ T 4a;= - 2 1 - 2a; 7 - 2a; 7 - 16a; + iy^ 2i8 ALGEBRA. .^- 13-2. -^ ^^ ^ —8— = ^^i 8 • g 6a; + 5 58i- + 14a ; _ ^ ' "3a; + 1 9 + 2a; ~ * 4a; - 9 6a; - 21 aj - 2^ 10 ^^ - '^ 2 - 14a ; 3^^ + a; ^ 10 - 3f a; _ 19 * 2 a; - 9 "^ 7 " 14 2 21' 11 1^ - 7a; - 6| _ 9a;^ - 12a; - 19 ^ 17 2a;-3 3a;-5 6aj-7' ■J ^ aa; + m + 1 ax + oi ax + m ax + m - I ax + n - 2 ax + m - 2 ax + n + 1 ax + n - V sjx -- a + b — ijx + a - b a — b. 13. Jx - a + b + Jx + a - b a + b 1 _ Vl - Vl - aJ , 1 + VI - n/1 - a; 15. J2x ■{• 10 + /N/2aj - 2 =• 6. „ 3 16. V8 - a; - "Ti-Zl, - x/l - a.. 17. ^l + Va; + ^1 - ^a; = 2. ,^ aa; + 1 + \/aV - 1 , aa; + 1 — v« ^ - 1 ci^ f« & a - 19. ^a; - V5 V^J - V« Vic 20. -7 7- = -7 7- + Va. ^Jx + s/a Va; - \/<^ /VaJ + » - / Jx - a _ , ^ 21. / —7 / ^-7 V. - a-. V Vc:j— o \J s/x + a PROBLEMS INVOLVING ONE UNKNOWN QUANTITY. 21D 1 2 + far' - la' 5 " ^- ^ -. 9. - 6 - 5a; + a^ " ^^ " a; - 3* ^^ ■ ^ a; + 5 ^ 2aj + 5 a;^ - 10 -^^* " ■ ^ **" ^T~4 = a; + 2 " a; + 3 + ^• 24 a; - - 2 ■ 4a; + 5 X + 1 ax + h ex -^ e ■V d ax + f Problems producing Simple Equations involving One Unknown Quantity. 66. To solve an algebraical problem we represent the re- quired or unknown quantity by a letter, as x, and then ex- j^ress the given conditions in algebraical language. Thus wo form an equation, the solution of which gives the required value of the unknown quantity. Ex. 1. My purse and money are together worth 24 shillings, and the money is worth seven times the purse. Find the value of each. Let X = the value in shillings of the purse, Then 1 x = „ „ money. Now, by problem the value of both together is 24 shillings. Hence we have — aj + 7 a; = 24 or 8a; = 24 .'. a; = 3, the value in shillings of the purse, and .*. also 7 a; = 7 x 3 = 21, „ money. Ex. 2. What number is that to which, if 36 be added, the sum shall be equal to 3 times the number It Let X — the number ; .*. a; + 36 = the sum when 36 is added, and 3 a; = 3 times the number. H6nce, by problem — a; + 36 = 3 a;, or a; - 3 a; = - 36, or - 2 a; = - 36; /.a; = — — - = 18, the number required. 220 ALGEBRA. Ex. 3. The distance between two towns is such that a train, whose speed is 30 miles an hour, takes 1 hour more in going 10 miles over 5 times the distance than a train whose speed is 20 miles an hour takes in going within 4 miles of 3 times the distance. Find the distance between the towns. Let X = the distance required in miles. Then 5 a; + 10 = 5 times the distance together with 10 miles, 5 a; + 10 and oa - *™^ ^ hours to travel this distance at 30 miles an hour. And so, — — - — = time in hours to travel 4 miles less Ji\) than 3 times the required distance, at 20 miles an hour. But by the problem the former of these times exceed the latter by 1 hour. Hence ^-^±-1^ - i^— _ A = 1. 30 20 From this we easily find a; = 28. Hence 28 miles is the distance required. Ex. 4. Find the price of an article, when as many can be bought for Is. 4d. as can be bought for 2s. after the price has been raised Id. Let X = the price required in pence; 1 /» Then — = number of articles bought for Is. 4d. X And a; + 1 = the raised price in pence. 24 = number of articles bought for 2s. at the raised X + 1 price. But, by the problem, the number of articles in each case is the same. Hence — = ~ — -, from which a; == 2. X X + I Hence 2d. is the price required. Ex. 5. A man sells geese at as many shillings each as the number he has, and having returned 5s., finds that if he had PROBLEMS INVOLVINa ONE UNKNOWN QUANTITY. 221 had 2 more to sell on the same condition, and had returned 3s., he would have had 38s. more. How many had he] Let X = the number required ; Then ar^ - 5 = number of shillings received. Also, on the second supposition, (a; + 2)' - 3 = number of shillings he would have received. Now, by the problem, this latter number is 38 more than the former. Hence (a; + 2)^ - 3 = a;^ - 5 + 38, from which we find X = 8. Ex. 6. A waterman finds that he can row 5 miles in f hour with the tide, and that it takes him 1^ hours to row the same distance against the tide when it is but half as strong. What is the velocity of the tide ] Let X = the velocity of the tide in miles per hour. Now the velocity of the boat when going with the tide ,= 5-1 = V. .'. Y - a; = velocity of the boat when there is no tide. Again, velocity of boat against the tide when it is half as strong = 5 -f H = V- 10 X .'. -^ + - :^ velocity of the boat, when there is no tide. Hence we have — 10 . a; 20 - , . , -_ + -_ = — a; ; from which O It o X = 2f. .*. The velocity required is 2 J miles per hour. Ex. XVII. 1. If I add 25 to 3 times a certain number, I obtain the same result as if I subtract 25 from 8 times the number. Find the number. 2. Divide 70 into 2 such parts that the one shall be as much above half the number as the other is above 15. 3. Divide £720 among A, B, and C, so that B may have twice (ig much as C, and A as much as B and C together. 222 ALGEBRA, 4. There are two trains, one of whicli goes 5 miles an hour faster than the other, and the former performs a journey of 100 miles, while the latter goes 75 miles. Find their respective I'ates, 5. A horse when let out for hire brings in a clear gain of 10s. per day, but costs Is. 6d. daily for food. At the end of 30 days his master had gained £11. lis. Required the number of days for which he was hired. 6. A and B have 4 guineas between them, and play at hazard. B loses | of his money, and afterwards gains ^V of what he then had. It is then seen that B has as much money as A had at the end of the first game. How much had each at first] 7. A and B have respectively an equal number of florins and crowns. B pays a debt of 4s. to A, and then A's money is just half B's. Find what each had. 8. A workman, instead of adopting the 9 hours' system, worked 10 hours daily, and had a corresponding rise of wages. By this means his wages were increased 4s. weekly. Find his original wages. 9. A person who has regular wages of 26s. weekly, think- ing to better himself, takes a job at higher wages. He is, however, put on half-time during 20 weeks of the year, and finds himself at the end of the year £4, 12s. worse off. Ee- quired his increased wages, 10. A company of men, arranged in a hollow square 4 deep, numbered 144. What was the number in a side of the square 1 11. In an examination paper there were two series of questions, and the questions of the second series carried each 3 marks more than those of the first series. A candidate who attempted 3 of the first series, obtaining half marks for them, and 5 of the second series, obtaining for these full marks, got altogether 80 marks. Find the number of marks attached to each question of the first series. 12. A grocer has tea at 3s. 4d. and at 4s. He sells altogether 64 pounds, thereby realizing £12. How much did he sell of each 1 PROBLEMS INVOLVING ONE UNKNOWN QUANTITY. 223 13. If 11 bo subtracted from 5 times a certain number, and tlie remainder divided by 6, the quotient will exceed by 2 the quotient obtained by subtracting 3 from 4 times the num- ber and dividing the remainder by 7. Find the number. 14. A garrison of 1,250 men were provisioned for 64 days; but after 22 days a certain number were called away, and it was found that the remaining provisions lasted the number left for 70 days. Find the number told off. 15. At a railway station £15 was taken for single fares, and £33. 15s. for returns. The number of return tickets exceeded the single tickets by 10, and the price of a return ticket was half as much again as a single ticket. Find the fare for a single journey, 16. In a tour lately made round the world, the distance travelled by water was 20,000 miles, and by land 8,000 miles; and the whole time taken was 220 days. Supposing the rate by water to be two-thirds of that by land, find the number of days travelled by land. 17. The distance between A and B is 32 miles. A person starting from A, at the rate of 4 miles an hour, meets another who started from B half an hour later, at a rate of 3i- miles an hour. At what point will they meef? 1 8. There are two clocks, one of which gains twice as much per day as ihe second loses, and they are set right at noon on Monday. When it is noon on Thursday by the first clock, it is 11-50 A.M. by the second. What is the gain per day of the first clock] 19. A draper raises his goods a certain rate per cent., and afterwards reduces them to the original price by lowering them 13^^ J per cent. Find the original rise per cent. 20. Bequired the distance between two towns such that a person can perfomi the journey one hour sooner when he walks 4 miles an hour than when he walks 3 J miles an hour] 21. The sum of £12. 15s. is paid away with an equal number of sovereigns, crowns, and sixpences, Bequired the number of each. 22. A walks along an inclined plane at a certain rate, and B walks along the base of the plane at a rate of one-third of 224 ALGEBRA. a mile per liour less than A. The inclination of the j)lane is such that A is always vertically over B, and that at the end of half an hour they are exactly five-sixths of a mile apart. Find the respective rates of A and B. 23. There is a direct road over a hill between two stations at the foot of each side. The distance on the one side from the foot to the top is 5 miles, and the road down the other side forms a right angle with the road up. It is also known to be 1 mile less down the hill than the direct distance by tunnel between the two stations. Find the distance down the hill. 24. Two trains, whose respective lengths are 122 and 98 yards, and the former of which is going at the rate of 35 miles an hour, pass each other in 30 seconds. Find the rate and relative direction of the second train. 25. A man bought a number of sheep for £1 32, and having lost 10, and sold 20 that were diseased at 6s. per head below cost price, disposed of the remainder for £116, thereby realizing his outlay. How many did he buy? 26. A boy spends 10s. in oranges and apples. The oranges were bought at 5 for 6d., and the apples at 3 for 2d.; and their number together amounted to 132. What did he spend on each? 27. If B is allowed 2 hours more time than A takes to do a piece of work, he will do 4 times as much, and if C is allowed 1 hour more than A, he also can do 4 times as much. More- over, D requires 4 hours more than A to do the piece of work. Also, the work done by A and B together is the same as that done by C and D together in the same time. Bequired the respective times for A, B, C, D to do a piece of work. 28. A person sells out £1,200 Three and a Half per Cent. stock, and invests the money in Two and a Half per Cents., whose price is 1 4 lower than the first-named stock. The loss in annual income is £7. Find the price of the first-named stock. 29. The banker's discount on a certain sum of money at 5 per cent, per annum is equal to the true discount on a sum JuSO larger, Find the sum, SIMULTANEOUS EQUATIONS. 225 30. An express train, which ought to porform its journey in 2|- hours, after having gone uniformly 80 miles, finds itself G minutes behind. However, by increasing the speed to as many miles per hour as there were miles in half the journey, it just arrived at its destination in time. Find the original speed of the train, and the length of the journey. 31. A vessel contains a quantity of spirit (sp. gr. -9) and water, and a cylinder of wood (sp. gr. -92), whose length is 10 inches, floats upright, so as. to be just covered by the spirit. Find how much of the cylinder floats in the water. 32. A mixture of hydrogen and oxygen is found to condense when fired, to 1 6 vols, of steam. Now, every 3 vols, of such a mixture is known to condense to 2 vols., when the original gases are in the proportion of 2 : 1. Find the quantity of each. 33. A mixture of 100 grams of sodic and potassic sulphates yielded a gram of baric sulphate. Now, each gram of sodic sulphate yields h gi^ams of baric sulphate, and each gram of potassic sulphate yields c grams of baric sulphate. Find the amount of sodic and potassic sulphates in the mixture. 34. If a oxen consume h acres of grass in c weeks, and a ' oxen consume h ' acres of grass in c ■ weeks, the grass growing uniformly, find the week's growth of an acre. 35. The freezing and boiling points of a common ther- mometer are marked 32° and 212° respectively; those on tlie centigrade thermometer are marked 0° and lOO''. At what temperature do the graduations agree ] 36. A person going at the rate of a miles per hour finds himself h hours too late when he has c miles farther to go. How much must he increase his speed to reach home in time ] Simultaneous Equations of the First Degree with two Unknown Quantities. 67. Suppose we have given the equation 3 a; - 4?/ = 5, then by ascribing to ?/ a series of values we get a correspond- ing series of values for x. Thus we may have ^ ~ i > , ^ ~ < 'r ? ,, '" rc y ^* g ^ y =^^r y = ^r y =^ n 22G ALGEBRA. Again, if niiotlier equation, as 4 x -v y ==-- 32, be given, \v'e may in the same way obtain a series of pairs of values whioii satisfy it. And, further, if the two equations are distinct and compatible, there is always a j^:)<:aV of values common to the two equations. This pair of values then satisfies both equa- tions, and the equations are called simultaneous equations. The methods of solving simultaneous equations will now be explained. First Method. — Equalize the coefficients of one of the un- known quantities in both equations, and add or subtract the equations so obtained, so as to obtain an equation with one unknown quantity, Ex. A:x + 37/= 17 (1)) 9ic -- 5?/ - 3 (2)/- From (1), multiplying each side by 9, we get — 36ic + 21y = 153 (3). And from (2), multiplying each side by 4, we get — 36a; - 20 y = 12 (4). (3) - (4), then 47 2/ = 141 Hence, substituting in (1), we have — 4a; + 3x3 = 17, from which we get — ■ X =. 2, Hence, the solution required is a: = 2, ?/ = 3. Second Method. — Express one of the imknown quantities in terms of the other by means of either equation, and sub- stitute its value in the other. Taking the same example, we have — 4a; + 37/ = 17 (1)1 9a; - 52/ - 3 (2) / * From (1) we have 4a;=: 17 - 2>y, ot x = — ~--^ — ^^...(3). Substituting this value of x in (2) we have — or, 153 - 27 y - 20 y = 12, from wliicli— 2/= 3. SIMULTANEOUS EQUATIONS, 227 Tlien from (3) we get x -^ ILZ-L^^ =. 2. Third Method. — Express mie of the unknown quantities in terms of the other by means of each equation, and equate the expression. We will this time express in each case y in terms of x, Wehave4a; + 3y = 17 (1) ) 9a; - 5y - 3 (2) J ' From (1) 3y := 17 - 4 07, or y - IL:iA^...(3). o And from (2) 5 y = 9aj ^ 3, ory = ^^ " ^ (4). Equating (3) and (4), then ^1-—^ = ^ ZL?, from which o 5 X = 2, Then from (3) by substitution, y = ^LzAjiJ = 3, o Simultaneous Equations of the First Degree of Three or more Unknown Quantities. 68. In the case of three unknown quantities we may obtain, from the three given equations, two equations with two unknown quantities, and then, by a similar method, from the two obtain an equation with one unknown quantity ; and a like method may be pursued for more than three unknown quantities. Ex. 3a; + y + 4cj = 25 (1) ix + Zy - ^z ^ - 3 (2) Qx -h 7y - Sz = 1 (3) From (1) 12a; + 4 y + 16;^ = 100 (4)) Andfrom(2) 12a; + 9y - 15;:; = - 9 (5) j (5)-(4),then 5y - ?^\z =-109 (6) Again, from (1) 6a; + 2^/ + 8^ = 50 (7) (3) - (7), then 5y - \^z ^ - 49 (8) (8) - (6), then 15;:; = GO .'. z ^ 4 ) 228 ALGEBRA. Hence from (8), by subcLltution — 5 2/ - 16 X 4 =-49, from which y - 3. And hence from (1), by substitution of the known values of y and ;: — 3 a; 4- 3 + 4 X 4 - 25, from which ic = 2. Hence the respective values of x^ y, ^, are 2, 3, 4. Ex. XYIII. 1. 6 X + 7/ = 22, 5 a; + 3 2/ - 27. 2. 4 a; - 3 7/ = 14, G a; + 5 2/ = 40. 3. 3 a; + 5 2/ = 44, 2/ - a; =^ 4. A _ . *^- _ 9 — . r ^ 9 9_ ^- 4 + 3 - -uj 5 + 10 - "TO- ^•6 + i = ^'4 + 6 = '- 7. 3 aj + 4 2/ + ;^ = 11, 2 a; + 2/ + 5 ;:; -= 19, 5aj+22/ + 3;2 = 18. 8. 7a; + 22/ + 3;:; = 20, 3a;-4y + 2;:; = l, 52/-2a;+7;^ = 29. 9. 2 a; - 5 2/ = 3, 3 y - 2 ;:; = - 1, 4 a; + 2 ;:;=:= 20. 10. aa; + ^2/ = ^> ^v^ + ^i2/ ~ ^i* 11. X •{• y = a, y + X =•■ h, X ■\- z = c. 12. ax + by = d, by + cz = e, ax* + cz = J*. 3x-2y 5x+2y , aj + 2?/ 3a;-22/_ , 13. —J— + — g = 5^, — ^ — ^^ 1. a ? + 2/ , ^-2/ _ ^ ^tJl 4. ?_r J( == -I ^^' 10 15 30' 15 10 30* a; 2/ ^ V SIMULTANEOUS EQUATIONS. 229 16. ^ + - = 5i — - = 0. 17. 4 (a; + 3 2/) - 3 (x + y) = i (a; + 2/) (^ + 3 y), ^Q + _1_ ^ 3. • X + y X + 3y 18. 73y - 5aj = (a; - 5 2/) (a: + 3 y), 2 5 7 X - by X -\- 3y 33 .n 1 1 11^11 19. - + - = a, - + - = -5, - + — = c. X y y z X z 20. 3a;-22/ = 0, 4y- 3;:; = 0, 55-6^^=14, 7 w - 2 ic = 3. 21. 3a;~27/=.5;:;-6y = 7a:-4;:; = l. 22. x + y + z=^ayy-{'Z-\-u=^hiZ-\-u-\-x = Cy u + X + y ~ d, ^^ oc y y z J X z 23. - + - = a, - + - = ^, — + - = c. 7/1 7i n p 1)1 p 24r. X + ay + hz = a^ y + az + hx = ^y z + ax + by = y, 25. ic + ay + a-;:; + a** = 0. i« + % + 6";s + P = 0, X + cy + crz + c' = 0. 2G. ic + ay + a-;s + a'w + a* = 0. X -r by + b^z + b^u + 5* = 0. X + cy + c^z + c^u + c* = 0. X + dy -h d^z + d^it + d* = 0. 27. a; + a?/ = «. 28. a b c , y + bz = p, X y z - + c« = y. a' 5' ^' _ V < + c/cc = ^. a""** y *** i ~" ?t + ex = e. a" 5" c" _ ,,/ X y z 29. a^i + 2^2 + 30^3 + &c. + nx^ = ai- ajj + 2a;3 + 3^4 + &c. + tz^i = aj. «« + 2ari + 3 a; + i.> "1 :• III- II Ml"! Nvuiiiciii j,..re re^pev:tivelv 54d. and o8d., or 4s. 6d. and ob. I'd. Ex. 2. Eind a fraction such that^ if we diminish its numerator by 1, it becomes equal to 4 ; and if we increase its denominator by 1 , it becomes equal to J < Let -be the required fraction. y Then, by problem, '^IT — ^ „ ) , • 1 ^^ , . •^ -^ y V ( , wnicli equations, when J X ^ 1 ( solved, give- ic - G, 2/ - 35. /. _1 is the fraction required. 35 Ex. 3. A's money, together with twice B's and thrice C's, amounts to £38 ; B's money, together with twice C's and thrice A's, to £35 ; and C's money, together with twice A's and thrice B's, also to £35. Eind the money of each. rrwOBLEMS TRODUOING SIMULTANEOUS EQUATIONS. 231 Let X, y, z be respectively the number of pounds each ias. Then we have — X + 2y + 2>z ^ Z^\ 2/ + 2 ;2; + 3 £C - 35 I- ; from which we get — ;:; + 2 a; + 3 7/ =- 35 ) a; ^ 5, 2/ = 6, ;:; = 7. Hence, £5, £6, <£7 are the respective moneys of A, B, and C. Ex. XIX. 1. A and B engage in play. A puts down lialf-a-crown to B's florin. They play twenty games, and then it is found that A has won 2 s. How many games did each win ? 2. There is a number, the sum of whose two digits is 1 0, and, if 36 be added to the number, the digits change places. 'V\^^c^ tlip mijnbf'r. '. ' ih.. ;.ii.l --H-:u- :,f I >. ;. sl..|i,\ It" !..■ l.:i.! j:n>.-l ;!.. (■••;, l<» r \'l}, iMT r"M\.. -M..; ^..|.' lir. •ills vv ...il.l \\'.i\ '■ iH-.;i, 1... .;.!. ..r-:.. I,. 4. iiiiglib Liiiie;:> ihc nuniL-raLur ui a< oc-rialii IracLiuii cAlccctis three times the denominator l)y 3, and five times the numer- ator added to twice the d<'ii(miln;ii<>i" ]ii;i,l<(^ 29. Find the fraction. 5. If the number of cows in a field were doubled, there would bo G5 cows and horses tog(ither ; but, if the number of horses be doubled, and that of the cows halved, there would be 4G. How many ai^e there of each ? 6. Thirty shillings are spent in brandy, and 42s. in gin ; 19 bottles being purchased in all. Had the 42s. been spent in brandy, and the 30s. in gin, 17 bottles only would have been bought. Find the cost per bottle of each. 7. A fishmonger receives 240 mackerel. He sells a certiiin number at 4 for a shilling, but the rest being seized as bad fish, and he being fined 10s., finds himself a loser by 9s. Had he sold them at 3 for a shilling, he would have been a gainer by 5s., if 13 more fish had been seized. How many did he sell, and what did he pay for the lot] :;. .\ 'jv !• l.:.s t. t 3.-. lb s. •lis i-J vv .mmI, n <• ll \,>\ 1VV |>;* Mi<' • liiMit iri< •s . ,.t' (■'. ..•!.. . Lis iiior-'. KilMJ t 1,..,U: ml ii \ >• 232 ALGEBRA. 8. Three persons invest tlieir money at 3, 4, 5 per cent, interest respectively. The total amount of interest is £38, and the interest of the first and third together is 2^ that of the second ; while the total interest would have been £34 had the rates been 5, 4, 3 per cent, respectively. Find the capital of each. 9. A toll-gate keeper receives 8s. 8d. for the toll of a number of horses, oxen, and sheep, the tolls for each being respectively 1^-d. Id., |-d. Had there been twice as many sheep and the number of horses diminished accordingly, he would have received 7s. 2d. Had the oxen passed through free, and the tolls for a horse and sheep respectively been 2d. and |-d., he would have received 9s. IJd. Find the number of each. 10. A, B, C start from the same place. B after a quarter of an hour doubles his rate, while C, who falls, after ten minutes diminishes his rate Jth. At the end of half an hour A is ^ mile before B, and i mile before C, and it is observed that the total distance which would have been walked by the three, had they each continued to walk imiformly from the first, is 6i miles. Find the original rate of each. 11. A, B, C, working 3, 4, 5 hours respectively, can do 2^^^:^ pieces of work; if they each work an hour more, they can finish an extra if of a piece ; and, if C does not work, the other two, working for 1 and 6 hours respectively, can together finish 1 piece. Find the time required for A, B, C to finish separately a piece of work. 12. There are three numbers such that, if the first be in- creased by 6, and the second diminished by 5, the product of the results is the product of the first two numbers ; if the second be increased by 2 and the third diminished by 3, the product of the results is the product of the second and third ; and, if the first be increased by 3 and the third diminished by 6, the product is that of the first and third. Find the numbers. 13. A person performed a journey of 22^ miles, partly by carriage at 10 miles an hour, partly by train at 36 miles an hour, and the remainder by walking at 4 miles an hour. He did the whole in 1 hour 50 minutes. Had he walked the PROBLEMS PRODUCING SIMULTANEOUS EQUATIONS. 233 first portion, and performed the last by carriage, it would have taken him 2 hours 30|- minutes. Find the respective distances by carriage, train, and walking. 14. A and B start from two places C and D, distant 28 miles, and it is found that A reaches T> 3 hours after they meet. Had the distance between C and D been 35 miles, A would have reached a point 28 miles from C 2 hours after he met B. Find the respective rates of A and B. 15. Three trains — a luggage, ordinary, and express — move along three parallel pairs of rails, the distance between the stations being 120 miles. The firet two start from the same station, and the express from the opposite. The luggage train, starting 2 hours first, is overtaken by the ordinary in 2 hours; and the express train, starting 1 hour after the ordinary, meets the luggage in 1 hour 7^- minutes. Had all three started from the same station, the ordinary would have been overtaken in 2 hours. Find the respective rates of the trains. 16. If (rtj, hiy Cj), (aoy h.2, Co), (^3, 63, C3) be the respective compositions by weight of three mixtures of three substances, and di, ch, d^ be the respective prices of the mixtures, find the price per unit of weight of each substance. 17. By alloying two ingots of gold in two given propor- tions, we form two new ingots of which the fineness of each is known. What is the fineness of each of the given ingots ] 18. A gi^oup of n persons play as follows: — The first 1 ' plays with the second and loses - of what he has, the second tlicn plays with the thii-d and loses - of what he has, the third then with the fourth, losin;^ - of what he has, &c., the 71 til with the first losin^? - of what he has. At the end they each have h. What had each at fii*st % 234 MATHEMATICS. SECOND STAGE. SECTION I. , n "E n ":\r t: T "R y. ECOLiD"^ lIl^Ji^ME^vT;!^. iiuuK ii. Definitions. i. A rectangle, or right-angled parallelogram, is said to be contained by any two of the straight lines which contain one of the right angles. 2. In any parallelogram, the figure which is composed of either of the parallelograms about a diameter, together with the two complements, is called a (jiioiiLO)i. Thus the i)arallelogram HG, to- gether with the complements AF, EC, is a gnomon, which is briefly ex- pressed by the letters AGK, or EHC, which are at the oj)posite angles of the parallelograms which make the gnomon. The rectangle under, or contained hy two lines, as AB and BC, is concisely cx^jressed thus : — AB, BC. PROPORTIONS. 235 Proposition 1.— Theorem. If there he two straight lines, one of loldch is divided into any number of 2^<^rtSj the rectangle contained hy the two straight lines is equal to the rectangles contained hy the un- divided livje, and the several parts of the divided line. Let A and BC be two straight lines; and let BC be divided into any parts in the points D, E ; The rectangle contained by the straight lines A and BC A • BC = shall be equal to the rectangle contained by A and BD, A • bd - together with that contained by A and DE, and that con- ^ • de - tained by A and EC. Construction. — From the point B draw BF at right angles toBC(I. 11), „ „ And make BG equal to A (I. 3). i j r I Through G dr«,w aH pnrall^l +o I III BCd. :^n. \ And tlii-«n!L;lt tli- |M»ii»r..> I »,E.< '.«lt':. ^^ Pkook. 'I'lirii tliH jvr«:iiigi'" !'• M i> ^,, equal U» tJM' i-tx-taiioU-s BK. \)\.. KH. ' ^ But BH is contained by A and i>C, I'or it is contained ^^^• by GB and BC, and GB is equal to A (Const.) ; And BK is contained by A and BD, for it is contained by GB and BD, and GB is equal to A ; And DL is contained by A and DE, because DK is equal to BG, which is equal to A (I. 34) * And in like manner EH is contained by A and EC ; Therefore the rectangle contained by A and BC is equal to the several rectaiigles contained by A and BD, by A and DE, and by A and EC. Therefore, if there be two straight lines, kd Q, E. D» Proposition 2.— Theorem. If a Btraighi line he divided into any two 2)arts, the rect- aiigles contained hy the lohole line and each of its jxirts are together equal to the square on the tohole line. Let the straight Ime AB be divided into an\- tw o parts in the point C ; 236 GEOMETRY. AB-BC -^-AB • AC = AB:\ The rectangle contained by AB and BC, together with tl.3 rectangle contained by AB and AC, shall be equal to the square on AB. Construction. — Upon AB describe the square ADEB (I. 46). Through C draw CF parallel to AD or BE (I. 31). Proof. — Then AE is equal to the rect- angles AF and CE. But AE is the square on AB ; Therefore the square on AB is equal to the rectangles AF and CE. And AF is the rectangle contained by BA and AC, for it is contained by DA and AC, of which DA is equal to BA : •And CE is contained by AB and BC, for BE is equal to AB. Therefore the rectangle AB, AC, together with the rect- angle AB, BC, is equal to the square on AB. Therefore, if a straight line, &c. Q. E. D, Proposition 3. — Theorem. If a straight line he divided into any two j^arts, the rect- angle contained hy the whole and one of the parts is equal to the square on that part, together ivith the rectangle contained by the two 2^arts. Let the straight line AB be divided into any two parts in the point C; The rectangle AB * BC shall be equal to the square on BC, together with the rectangle AC • CB. Construction. — Upon BC describe the square CDEB (I. 46). Produce ED to F ; and through A draw AF parallel to CD or BE (I. 31). Proof. — Then the rectangle AE is equal to the rectangles AD and CE. But AE is tlie rectangle contained by AB and BC, for it is contained by AB and BE, of which BE is equal to BC ; And AD is contained by AC and CB, for CD is equal to CB; PROPOSITIONS. 237 And CE is tlie square on BC. Therefore the rectangle AB, BC is equal to the square on BC, together with the rectangle AC, CB. Therefore, if a straight line, &c. Q. E. D. Proposition 4.— Theorem. If a straight line he divided into any two 2^ci'i'ts, the square on the tvhole line is equal to the sum of the squares on the tivo 2)artSy together with twice the rectangle contained by the j^cirts. Let the straight line AB be divided into any two parts inC; The square on AB shall be equal to the squares on AC and ab2 = CB, together with twice the rectangle contained by AC ^^A't.^Sp and CB. •'-^''*''^- Construction. — Upon AB describe the square ADEB (I. 4G), andjoinBD. ^ ^ ^ Through C draw CGF parallel to AD or BE (I. 31). Through G draw HCK parallel to AB ^- or DE (I. 31). Proof. — Because CF is parallel to AD, and BD falls upon them, Therefore the exterior angle BGC is "" * jj- equal to the interior and opposite angle ADB (I. 29). Because AB is equal to AD, being sides of a square, the angle ADB is equal to the angle ABD (I. 5); Therefore the angle CGB is equal to the angle CBG show first . . , V ^ ° ^ ^ tliat OK is ^Ax. Ijj a sqimVo Therefore the side BC is equal to the side CG (I. G). = ^^-^ But CB is also equal to GK, and CG to BK (I. 34); Therefore the figure CGKB is equilateral. It is likewise rectangidar. For since CG is parallel to BK, and CB meets them, the angles KBC and GCB are together equal to two right angles (I. 29). But KBC is a right angle (Const.), therefore GCB is a right angle (Ax. 3). Therefore also the angles CGK, GKB, opposite to these, are right angles (I. 34). Q / / 238 GEOMETRY. Therefore CGKB is rectangular ; and it has been proved ' eqnUateral; therefore it is a square; and it is upon the side CB. Co also ^or the same reason HF is also a square, and it is on £iF=AC2 the side HG, which is equal to AC (T. 34). Therefore HF and CK are the squares on AC and CB. And because the complement AG is equal to the com- ^„^ plement GE (I. 43), AG+GE And that AG is the rectangle cont^ned by AC and CG, =2AC-CB. ^j^^^ .^^ ^y ^Q ^^^^ Q;g^ Therefore GE is also equal to the rectangle AC, CB ; Therefore AG, GE are together equal to twice the rect- angle AC, CB; And HF, CK are the squares on AC and CB. Therefore the four figures HF, CK, AG, GE are equal to the squares on AC and CB, together with twice the rectangle AC, CB. But HF, CK, AG, GE, make up the whole figure ADEB, which is the square on AB ; .\ whole Therefore the square on AB is equal to the squares on ^jgye^or ^Q ^^^ ^jg ^^^ twice the rectangle AC • CB. +2 ac^cb: Therefore, if a straight line, &c. Q, E. D. Corollary. — From this demonstration it follows that the parallelograms about the diameter of a square are likewise squares. Proposition 5. — Theorem. If a straight line he divided into two equal parts, and also into two unequal parts, the rectangle contained hy the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line. Let the straight line AB be bisected in C, and divided unequally in D ; ad-db The rectangle AD, BD, together with the square on CP, icB2. ^^^^^ ^^ equal to the square on CB. Construction. — Upon CB describe the square CEFB (I. 46), and join BE. PROPOSITIONS. 239 Tlirougli D draw DHG parallel to CE or BF (I. 31). Tlijoagh H draw' KLM parallel to CB or EY, And through A draw AK par- allel to CL or BM. A, C -D^B Proof. — Then the complement CH is equal to the complement HF ^ (I. 43). To each of these add DM ; there- fore the whole CM is equal to the k g f whole DF (Ax. 2). ^^^ But CM is equal to AL (I. 3G), because AC is equal to CB al=cm (Hyp.); Therefore also AL is equal to DF (Ax. 1). To each of these add CH; therefore the whole AH is .•.ah= equal to DF and CH (Ax. 2). ^^+^^- But AH is contained by AD and BD, since DH is equal to DB (II. 4, cor.), And DF, together with CH, is the gnomon CMC ; Therefore the gnomon CMC is equal to the rectangle AD, DB. . • . cmg. To each of these equals add LG, which is equal to the Add^o^ * square on CD (II. 4, cor., and I • 34); ^^^^^°^ Therefore the gnomon CMG, together with LC, is equal to the rectangle AD, DB, together with the square on CD. But the gnomon CMC and LC make up the whole figure CEFB, which is the square on CB; Therefore the rectangle AD, DB, together with the square .-. CBs on CD, is equal to the square on CB. +CD2.^^ Therefore, if a straight line, &c. Q. E. D, Corollary. — From this proposition it is manifest that the difierence of the squares on two unequal lines AC, CD is equal to the j-ectangle contained by their sum and difference. Proposition 6. — Theorem. If a straight line he bisected, and pi^oduced to any pointy the rectangle contained by the whole line thus produced and the part of it produced, together with the square on half the line bisected, is equal to the square on the straight line which is made up of the half and the part produced. Let the straight line AB be bisected in C, and produced to D ; A ^' B n L H / K / 240 GEOMETRY. AD-DB The rectangle AD, DB, togetlicr with the square on CB, - cj)l. shall be equal to the square on CD. Construction. — Upon CD describe the square CEFD (I. 46), and join DE. Through B draw BHG parallel to CE or DF (I. 31). Through H draw KLM parallel k~ to AD or EF. And through A draw AK par- For allel to CL or DM. AL^CH Proof. — Because AC is equal ^ u jj- "" ' to CB (Hyp.), the rectangle AL is equal to CH (I. 36). But CH is equal to HF (I. 43), therefore AL is equal to HF (Ax. 14). .-.AM or To each of these add CM; therefore the whole AM is t^civia eq^^al to *lie gnomon CMG (Ax. 2). But AM is the rectangle contained by AD and DB, since DM is equal to DB (II. 4, cor.) ; Therefore the gnomon CMG is equal to the rectangle AD, DB (Ax. 1). Add to Add to each of these LG, which is equal to the square on ITcm^ CB (II. 4, cor., and I. 34) ; Therefore the rectangle AD, DB, together with the square on CB, is equal to the gnomon CMG and the figure LG. But the gnomon CMG and LG make up the whole figure CEFD, which is the square on CD ; .♦.AD-DB Therefore the rectangle AD, DB, together with the square 1 C£>2^ on CB, is equal to the square on CD. Therefore, if a straight line, &c. Q. E. B, Proposition 7.— Theorem. If a straight line he divided into any tivo 2yarts, the squares on the whole line and on one of the parts are equal to tivice the o^ectangle contained hy the whole and that party together with the square on the other part. Let the straight line AB be divided into any two parts in the point C; =^2^AB-BC -^^^ squares on AB and BC shall be equal to twice the + AC2. rectangle AB, BC, together with the square on AC, PROPOSITIONS. 241 Construction^. — Upon AB dencribo tho square ADEB (I. 4G), and join BD. Through C draw CGF parallel to AD or BE (I. 31). Through G draw HGK parallel to AB or DE (I. 31), Proof. — Then AG is equal to GE (I. 43). To each of these add CK ; therefore the whole AK is equal to the whole CE; Therefore AK and CE are double of AK. But AK and CE are the gnomon AKF, togetlicr with the square CK ; Therefore the gnomon AKF, together with the square CK, is double of AK. But twice the rectangle AB, BC is also double of AK, for BK is equal to BC (II. 4, cor.); Therefore the gnomon AKF, together with the square CK, is equal to twice the rectangle AB, BC. To each of these equals add HF, which is equal to the Add hf or square on AC (II. 4, cor., and I. 34); ^q^Jar^'^' Therefore the gnomon AKF, together with the squares CK and HF, is equal to twice the rectangle AB,BC, together with the square on AC. But the gnomon AKF, together with the squares CK and .-. ab2 HF, make up the whole figure ADEB and CK, which are ioAB b^ the squares on AB and BC; +AC-2. Therefore the squares on AB and BC are equal to twice the rectangle AB, BC, together with the square on AC. Therefore, if a straight line, &c. Q, E. D, For AK=CE. .AKF + CK - 2AB-BC. Proposition 8.— Theorem. If a straight line he divided into any two jmrts, four times the rectangle contained hy the whole line and one of the parts, together with the square on the other 2^ctft, is equal to the square on the straight line which is VlClde up af the whole line fLQid the first mentioned part. Let the straight line AB ^ divided intq J^njr t^yq part^ iv^ the point C; GEOMETRY. 4 AB • BC +AC8 = (AB+BC)2. For BN=CK =GR=RN . • . the four together = 4CK, And tlie rect- angles AG, MP, PL, RF, are equal to each other, and are toge- ther = 4 AG. Four times the rectangle AB, BC, together with the square on AC, sliall be equal to the square ou the straight line made up of AB and BC together. Construction. — Produce AB to D, so that BD may be equal to CB (Post. 2, and I. 3). Upon AD describe the square AEFD And construct two figures such as in ^ the preceding propositions. Proof. — -Because CB is equal to BD (Const.), CB to GK, and BD to KN (Ax. 1), For the same reason PR is equal toRO. And because CB is equal to BD, and GK to KIST, Therefore the rectangle CK is equal to BN, and GB to EN (I. 36). But CK is equal to BN, because they are the complements of the parallelogram CO (I. 43) ; Therefore also BN is equal to GR- (Ax. 1). Therefore the four rectangles BN, CK, GR, R^N" are equal to one another, and so the four are quadruple of one of them, CK. Again, because CB is equal to BD (Const.); And that BD is equal to BK, that is CG (II. 4, Cor., and I. 34); And that CB is equal to GK, that is GP (I. 34, and II. 4, cor.); Therefore CG is equal to GP (Ax. 1). And because CG is equal to GP, and PR to RO, The rectangle AG is equal to MP, and PL to RF (I. 36). But MP is equal to PL, because they are complements of the parallelogram ML (I. 43), and AG is equal to RF (Ax. 1) ; Therefore the four rectangles AG, MP, PL, RF are equal to one another, and so the four are quadruple of one of them, AG. And it was demonstrated that the four CK, BN, GR, and RN are quadruple of CK ; Therefore the eight rectangles which make up the gnomon AOH are quadruple of AK. PROPOSITIONS. 243 And because AK is the rectangle contained by AB and .-.Gnomon BC, for BK is equal to BC ; t^K+AG) Therefore four times tlie rectangle AB, BC is quadruple of ^^^^\j5(^ AK. But the gnomon AOH was demonstrated to be quadrnplo of AK; Therefore four times the rectangle AB, BC is equal to the gnomon AOH (Ax. 1). To each of these add XH, which is equal to the square Hence on AC (II. 4, cor., and I. 34); XHor'AC?, Tlierefore four times the rectangle AB, BC, together with the square on AC, is equal to the gnomon AOH and the square XH. But the gnomon AOH and the square XH make up the figure AEFD, which is the square on AD ; Therefore four times the rectangle AB, BC, together with ^ ab • bc the square on AC, is equal to the square on AD, that is, on = af = the line made up of AB and BC together. (ab+bc)-\ Therefore, if a straight line, &c. Q, E, J), Proposition 9.— Theorem. If a straight line be divided into two equal, a7ul also into two unequal parts, the squares on the two unequal ^Kirts are together double of the square on half the line, aiid of the square on the line bettveen the points of section. Let the straight line AB be divided into two equal parts in the point C^ and into two unequal parts in the point D ; The squares on AD and DB shall be together double of ads+dbq the squares on AC and CD. CD2)^^"^ Construction. — From the point C draw CE at right angles to AB (I. 11), and make it equal to AC or CB (I. 3), and join EA, EB. Through D draw DF parallel to CE (I. 31). Through F draw FG parallel to BA (I. 31), and join AF. Proof. — Because AC is equal to CE (Const.), the angle EAC is equal to the angle AEC (I. 5). 244 GEOMETRY. And because the angle ACE is a right angle (Const.), the angles AEC and EAC together make one right angle (I. 32), For and they are equal to one another ; AEC=j a Therefore each of the angles AEC and EAC is half a right right z - ^ '^ = CEB. angle. For the same reason, each of the angles CEB and EBC is half a right angle; .-. AEB is Therefore the whole anejle AEB is a ri^ht anoxic. And because the angle GEF is half a right angle, and the angle EGF a right angle, for it is equal to the interior and opposite angle ECB (I. 29), Therefore the remaining angle EFG is half a right angle ; Therefore the angle GEF is equal to the angle EFG, and EG=GF. the side EG is equal to the side GF (I. 6). Again, because the angle at B is half a right angle, and the angle FDB a right angle, for it is equal to the interior and opposite angle ECB (I. 29), Therefore the remaining angle BFD is half a right angle ; ^^^ Therefore the angle at B is equal to the angle BFD, and DF=DB. the side DF is equal to the side DB (I. 6). And because AC is equal to CE (Const.), the square on AC is equal to the square on CE ; Therefore the squares on AC and CE are double of the B„t square on AC, AE2= But the square on AE is equal to the squares on AC and CE, because the angle ACE is a right angle (I. 47); Therefore the square on AE is double of the square on AC. Again, because EG is equal to GF (Const.), the square on EG is equal to the square on GF ; Therefore the squares on EG and GF are double of the square on GF. So also But the square on EF is equal to the squares on EG and ^2 CDS. GF, because the angle EGF is a right angle (I. 47) ; Therefore the square on EF is double of the square on GF. And GF is equal to CD (I. 34); Therefore the square on EF is double of the square on CD^ But it has been demonstrated that the square on AE is al§o double of the square on AC ; EF2 = Therefore t^he squares on AE and EF are double of tli§ f^\ squares} pu AQ §.pd QJ), PROPOSITIONS. 2ib But the square on AF is equal to the squares on AE and But EF, because the angle AEF is a right angle (I. 47); Iaf^^^' Therefore the square on AF is double of the squares on AC =ad2+ and CD. =ad2+ But the squares on AD and DF are equal to the square ^^'^• on AF, because the angle ADF is a right angle (I. 47); Therefore the squares on AD and DF are double of the squares on AC and CD. And DF is equal to DB; therefore the squares on AD and .-. AD2+ DB are double of the squares on AC and CD. =^(AC2-fc Therefore, if a straight line, &c. Q.E.D. CD'-o Propositioii 10. — Theorem. If a straight line he bisected and produced to any 2>oint, tlie square on the whole line thus produced^ and the square on the jmrt of it produced, are together double of the square on half the line bisected, and of the square on the line made up of the half and the part produced. Let the straight line AB be bisected in C, and produced to D ; ;^5'+5,^^ The squares on AD and DB shall be together double of the CB-']. squares on AC and CD. Construction. — From the point C draw CE at right angles to AB, and make it equal to AC or CB (I. 11, I. 3), and join AE, EB. Through E draw EF parallel to AB, and through D draw DF parallel to CE (I. 31). Then because the straight line EF meets the parallels EC, FD, the angles CEF, EFD are equal to two right angles (I. 29); therefore the angles BEF, EFD are less than two right angles; therefore EB, FD will meet, if produced towards B and D (Ax. Let them meet in G, and join ^C AG. Proof. — Because AC is equal to CE (Const.), the angle CEA is equal to the angle EAC (I. 5). And the angle A CE is a right angle ; therefore each of the angles CEA and EAC is half a right angle (I. 32). 24G GEOMETRY. As in For tlie same reason each of the angles CEB and EBC is ^^^^' ' half a right angle ; ^i^'bt' z^ Therefore the whole angle AEB is a right angle. And because the angle EBC is half a right angle, the angle DBGr, which is vertically opposite, is also half a right angle (1.15); But the angle BDG is a right angle, because it is equal to the alternate angle DCE (I. 29); Therefore the remaining angle DGB is half a right angle, and is therefore equal to the angle DBG; BD=DG Therefore also the side BD is equal to the side DG (I. 6). ~" ' Again, because the angle EGF is half a right angle, and the angle at F a right angle, for it is equal to the opposite angle ECD (I. 34); Therefore the remaining angle FEG is half a right angle (I. 32), and therefore equal to the angle EGF; Also Therefore also the side GF is equal to the side FE (I. 6). "" * And because EC is equal to CA, the square on EC is equal to the square on CA; Therefore the squares on EC and CA are double of the square on CA. Ac?ainas But the square on AE is equal to the squares on EC and in Prop. 9. ^^ ^j^^^^. AE-^= Therefore the square on AE is double of the square '^^^■''- on AC. Again, because GF is equal to FE, the square on GF is equal to the sqviare on FE; Therefore the squares on GF and FE are double of the square on FE. But the square on EG is equal to the squares on GF and 'FE (I. 47); Therefore the square on EG is double of the square on FE. And FE is equal to CD (I. 34), And Therefore the square on EG is double of the square on CD. 2 cir ^^^^ i^ ^^^ been demonstrated that the square on AE is .•.AE'-^4- double of the square on AC; 2(A(S+ Therefore the squares on AE and EG are double of the GD^-)- squares on AC and CD. AE2H-EG3 ^^^^ *^i^ square on AG is equal to the squares on AE and =AG2; EG (I. 47); PROPOSITIONS. 247 2(AC-'+ Therefore the square on AG is double of the squares on /.ag2 AC and CD. ^b^'^ But the squares on AD and DG are equal to the square on .•.ad-+ AG(L47); SJ Therefore the squares on AD and DG are double of the cd- squares on AC and CD. And DG is equal to DB; therefore the squares on AD and DB are double of the squares on AC and CD. Therefore, if a straight line, CB, BD. Case I.— First, let AD fall within the triangle ABC. Proof. — Because the straight line CB is divided into two parts in the point D, The squares on CB and BD are equal to twice the rectangle contained by CB, BD, and the square on DC (II. 7). To each of these equals add the square on DA. Therefore the squares on CB, BD, DA are e(pial to twice •••CP"+ the rectangle iCB, BD, and the squares on AD and DC. ix\2)^ But the square on AB is equal to the squaies on BD and ^^J^^ DA, because the angle BDA is a right angle (I. 47) ; ncr^), And the square on AC is ecpial to the squares on AD and ^bs+ba^ DC (I. 47) ; =2CBDB Therefore tlie squares on CB and BA are equal to the "^ "* 250 GEOMETRY. square on AC, and twice the rectangle CB, BD; that is, the square on AC alone is less than the squares on CB and BA by twice the rectangle CB, BD. Case II. — Secondly, let AD fall without the triangle ABC. Proof. — Because the angle at D is a right angle (Const.), A the angle ACB is greater than a right angle (I. 16); Therefore the square on AB is equal to the squares on AC and CB, and twice the rectangle BC, CD (II. 12). To each of these equals add the square ■^ on BC. Therefore the squares on AB and BC are equal to the square on AC, and twice the square on BC, and twice the rectangle BC, CD (Ax. 2). But because BD is divided into two parts at C, The rectangle DB, BC is equal to the rectangle BC^ CD and the square on BC (II. 3) ; And the doubles of these are equal, that is, twice the rectangle DB, BC is equal to twice the rectangle BC, CD and twice the square on BC ; Therefore the squares on AB and BC are equal to the square on AC, and twice the rectangle DB, BC ; that is, the jl square on AC alone is less than the squares on AB and BC by twice the rectangle DB, BC. Case III. — Lastly, let the side AC be perpen- dicular to BC. Proof. — Then BC is the straight line between the perpendicular and the acute angle at B ; and it is manifest that the squares on AB and BC are equal to the square on AC, and twice the square on BC (I. 47, and Ax. 2). Therefore, in every tiiangle, &c. Q.E.D. Proposition 14 Problem. To describe a square that shall he equal to a given recti^ lineal figure. Let A be the given rectilineal figure* It is required to describe a square that bhall be equal to A. PROPOSITIONS. 251 Construction. — Describe the rectangular parallelogram BCDE equal to the rectilineal figure A (I. 45). If then the sides of it, BE, ED, are equal to one another . it is a square, and what was required is now done. But if they are not equal, produce one of them, BE, to F, and make EF equal to ED (I. 3). Bisect BF in G (1. 10), and from the centre G, at the dis- tance GB, or GF, describe the semicii-cle BHF ; Produce DE to H, and join GH ; Then the square described upon Ell shall he equal to the rectilineal figure A. Proof. — Because the straight line BF is divided into two equal parts in the point G, and into two unequal parts in the point E, The rectangle BE,EF, together with the square on GE, is equal to the square on GF (II. 5). But GF is equal to GH ; Therefore the i-ectangle BE,EF, together with the square on GE, is equal to the square on GH. But the square on GH is equal to the squares on GE and EH (I. 47) ; Therefore the rectangle BE,EF, together with the square on GE, is equal to the squares on GE and EH. Take away the square on GE, which is common to both ; Therefore the rectangle BE,EF is equal to the square on EH (Ax. 3). But the rectangle contained by BE and EF is the paral- lelogram BD, because EF is equal to ED (Const.); Therefore BD is equal to the square on EH. But BD is equal to the rectilineal figure A (Const.); Therefore the square on EH is equal to the rectilineal figure A. Therefore, a square has been made equal to the given rectilineal figiu-e A, viz., the square described on EH. Q.E.F, BE-EF +GE-I =GH-J =GE-i+ EH2. .-.BE-EF orBD =EH2. Hence EH2=A. 252 GEoMEi^itY. EXERCISES ON BOOK II. Prop. 1—11. 1 . Divide a given straight line into two sucli parts that the rect- angle contamed by them may be the greatest possible. 2. The sum of the squares of two straight lines is never less than twice the rectangle contained by the straight lines. 3. Divide a given straight line into two parts such that the squares of the whole line and one of the parts shall be equal to twice the square of the other part. 4. Given the sum of two straight lines and the difference of their squares, to find the lines. 5. In any triangle the difference of the squares of the sides is equal to the rectangle contained by the sum and difference of the parts into which the base is divided by a perpendicular from the vertical angle. C. Divide a given straight line into such parts that the sum of their squares may be equal to a given square. 7. If ABCD be any rectangle, A and C being opposite angles, and any point either within or without the rectangle — OA- + 00^* = 0B2 + 0D2.. 8. Let the straight line AB be divided into any two parts in the point C. Bisect CB in D, and take a point E in AC such that EC := CD. Then shall AD^ = AE^ + AC • CB. 9. If a point C be taken in AB, and AB be produced to D so that BD and AC are equal, show that the squares described upon AD and AC together exceed the square upon AB by twice the rectangle con- tained by AE and AC. 10. From the hypothenuse of a right-angled triangle portions are cut off equal to the adjacent sides. Show that the square on the middle segment is equal to twice the rectangle under the extreme segments. 11. If a straight line be divided into any number of parts, the square of the whole line is equal to the sum of the squares of the parts, together with twice the rectangles of the parts taken two and two together. 12. If ABC be an isosceles triangle, and DE be drawn parallel to the base BC, cuttuig in D and E either the side or sides produced, and EB be joined; prove that BE^ - BC • DE + CE^. Prop. 12—14. 13. In any triangle show that the sum of the squares on the sides is equal to twice the square on half the base, and twice the square on the line drawn from the vertex to the middle of the base. EXERCISES OX THE PROPOSITIONS. 253 14. If squares are described on the sides of any triangle, find tlio difference between the sum of two of the squares and the third square, and show from your result what this becomes when the angle opposite the third square is a right angle. 15. Show also what the difference becomes when the vertex of the triangle is depressed until it coincide with the base, 16. The square on any straight line drawn from the vertex of an isosceles triangle, together with the rectangle contained by the segments of the base, is equal to the square upon a side of the triangle. 17. If a side of a triangle be bisected, and a perpendicular drawn from the middle point of the base to meet the side, then the square of the altitude of the triangle exceeds the square upon half the base by twice the rectangle contained by the side and the straight line between the points of section of the side. 18. In any triangle ABC, if perpendiculars be drawn from each of the angles upon the opposite sides, or opposite sides produced, meet- ing them respectively in D, E, F, show that — BA^ + AC2 + CB2 =2AE-AC + 2CD-CB + 2BF-BA; all lines being measured in the same direction round the triangle. 19. Construct a square equal to the sum of the areas of two given rectilinear figures. 20. The base of a triangle is 63 ft., and the sides 25 ft. and 52 ft. respectively. Show that the segments of the base, made by a per- pendicular from the vertex, are 15 ft. and 48 ft. respectively, and that the area of the triangle is 630 sq. ft. 21. In the same triangle, show that the length of the line joining the vertex with the middle of the base is 22*9 ft. 22. A ladder, 45 ft. long, reaches to a certain height against a wall, but, when turned over without moving the foot, must be short- ened 6 ft. in order to reach the same height on the opposite side. •Supposing the width of the street to be 42 ft., show that the height to which the ladder reaches is 36 ft. 23. The base and altitude of a triangle are 8 in. and 9 in. respec- tively; show that its area is equal to a square whoso side is 6 in. Prove your result by construction. 24. On the supposition that lines can be always expressed exacthj h\ terms of some unit of length, what geometrical propositions may be deduced from the following algebraical identities ? — (1.) (a + 5)2 = a2 + 2rt6 + 5- (2.) (« + h) (a - fc) + 62 :=a2 (3.) (2a 4- 6) 6 + a^ = (« + by (4.) (a + 6)2 + &2 = 2 (rt + i) 6 + a^ (5.) 4(a + 6) 6 + />2 z= (a + 2 6^- (6.) (fi + 6)2 + (a - 6)2 = 2a2 + 2 62 (7.) (2a + 6)2 + 62 =2a2 H-2(a + 6)2 25i GEOMETRY, EUCLID'S ELEMENTS, BOOK IIL Pefinitions, 1. Equal circles are those of wliicli the diameters are equal, or from the centres of which the straight lines to the circum- ferences are equal, 2. A straight line is said to touch a circle, or to be a tangent to it, when it meets the circle, and being produced does not cut it. 3. Circles are said to touch one another, which meet, but do not cut one another, 4. Straight lines are said to be equally distant from the centre of a circle, when the perpendiculars drawn to them from the centre are equal. 5. And the straight line on which the greater perpendicular falls, is said to be farther from the centre. 6. A segment of a circle is the figure contained by a straight line and the cixxum- ference which it cuts off. 7. The angle of a segment is that which is contained by the straight line and the circvimference. 8. An angle in a segment is the angle contained by two straight lines drawn from any point in the circumference of the segment to the extremities of the straight line, which is the base of the segment. 9. An angle is said to insist or stand upon the circumference inter- cepted between the straight lines that contain the angle. PROPOSITIONS. 255 10. A sector of a circle is the figure con- tained by two straight lines drawn from the centre and the circumference between them. 11. Similar segments of circles are those which contain equal anojles. [Any portion of the circumference is called an rnr, and the chord of an arc is the straight line joining its extremities.] Proposition 1. — Problem. To find the centre of a given circle. Let ABC be tlie given circle. It is required to find its centre. Construction. — Draw within the circle any chord AB, and bisect it in D (I. 10). From the point D •draw DC at right angles to AB (I. 11). Produce CD to meet the circumference in E, and bisect CE in F (I. 10). Then the point F shall he the centre of the circle ABC. Proof. — For if F be not the centre, if possible let G be the centre ; and join GA, GD, GB. Then, because DA is equal to DB (Const.), and Du- com- mon to the two triangles ADG, BDG; The two sides AD, DG are equal to the two sides BD, DG, each to each ; And the base GA is equal to the base GB, being radii of the same circle; Therefore the angle ADG is equal to the angle BDG (I. 8), But when a straight line, standing on anothoi line, makes the adjacent angles equal to one aiiotln r, the angles is called a right angle (I. Def. ID). Suppose Gthe centre. straight ich of .ZADG : z BDG. 25 () GEOMETRY, Tlierefore the angle GDB is a right angle. But the angle FDB is also a right angle (Const.) ; -zFDiJ Therefore the angle GDB is equal to the angle FDB (Ax. 11), the less to the greater; which is impossible. Therefore G is not the centre of the circle ABO. In the same manner it may be shown that no point which is not in CE can be the centre. And since the centre is in CE, it must be in F, its point of bisection. Therefore F is the centre of the circle ABC : which w\as to be found. Corollary. — From this it is manifest that, if in a circle a straight line bisect another at right angles, the centre of thQ circle is in the line which bisects the other, Proposition 2. — Theorem, If any two 2^oints he taken in the circumference of a ch^cle^ the straight line which joins them shall fall tvithin the circle. Let ABC be a circle, and A and B any two points in the circumference. The straight line drawn from A to B shall fall within the circle. Construction. — Find D the centre of the circle ABC (III. 1), and join DA, DB. In AB take any point E ; join DE, and j)roduce it to the circumference in F. Proof. — Because DA is equal to DB, the angle DAB is equal to the angle DBA (I. 5). And because AE, a side of the triangle DAE, is produced to B, the exterior angle DEB is greater than the interior and opposite angle DAE (I. 16). But the angle DAE was proved to be equal to the angle DBE; Therefore the angle DEB is also greater than DBE. But the greater angle is subtended by ^he greater ^id^ (1.19); WH^-m, Therefore D? is greater th^ PB, PROPOSITIONS. 257 But DB is equal to DF ; therefore DF is greater than DE, aii^ tlic point E is therefore within the circle. In the same manner it may be proved that every point in AB lies within the circle. Therefore the straight line AB lies within the cii'cle. Therefore, if any two points, &:c, Q.E.D, Proposition 3.— Theorem. // a straight line drawn through the centre of a circle bisect a straight line in it which does not 2)ass through the centre^ it shall cut it at right angles ; and conversely ^ if it cut it at right angles, it shall bisect it. Let ABC be a circle, and let CD, a straight line drawn through the centre, bisect any straight line AB, which does not pass through the centre. CD shall cut AB at right angles. CoxsTiiucTioN. — Take E, the centre of the circle (III. 1), and join EA, EB. Proof. — Because AF is equal to FB (Hyp.), and FE common to the two triangles AFE, BFE, and the base EA equal to the base EB (I. Def. 15), Therefore the angle AFE is equal to the angle BFE (I. 8) ; Therefore each of the angles AFE, BFE is a right angle (I. def. 10); Therefore the straight line CD, drawn through the centre, ^^^p^^^- bLsecting another, AB, that does not pass through the centre, also cuts it at right angles. Conversely, let CD cut AB at right angles. CD shall bisect AB ; that is, AF shall bo equal to FB — the same construction being made. Proof. — Because the radii EA, EB are equal, the angle EAF is equal to the angle EBF (I. 5), And the angle AFE is equal to the angle BFE (Hyp.), Therefore, in the two triangles EAF, EBF, there are two angles in the one equal to two angles in the other, each to 258 GEOMETRY. each, and the side EF, which is opposite to one of the equal angles in each, is common to both ; Therefore their other sides are equal (I. 26); Therefore AF is equal to FB. Therefore, if a straight line, &c. Q,E.D, If they bisect each other, EF bi5?ectg AC cat rio-ht anj^les. And EF bisects BD at right unglcv •.ZFEA ;=ZFEB, Proposition 4. — Theorem. If in a circle two straight lines cut one another, which do not both pass through the centre, they do not bisect each other. Let ABCD be a circle, and AC, BD two straight lines in it, which cut one another at the point E, and do not both pass through the centre. AC, BD shall not bisect one another. If one of the lines pass through the centre, it is plain that it cannot be bisected by the other, which does not pass through the centre. But if neither of them pass through the centre, if possible, let AE be equal to EC, and BE to ED. Construction. — Take F, the centre of the circle (III. 1), and join EF. Because FE, a straight line drawn through the centre, bisects another line AC, which does not pass through the aL f ^^-^ centra (Hyp.), therefore it cuts it at riglit angles (III. 3) ; Therefore the angle FEA is a right angle. ^ Again, because the straight line FE bisects the straight line BD, which does not pass through the centre (Hyp.), therefore it cuts it at right angles (III. 3); Therefore the angle FEB is a right angle. But the angle FEA was shown to be a right angle ; Therefore the angle FEA is equal to the angle FEB, the less to the greater, which is impossible; Therefore AC, BD do not bisect each other. Therefore, if in a circle, FB Of all the straight lines FB, FC, FG-, &c., that can be ^^^^P^ drawn from F to the circumference, FA, which passes the'ieast. through the centre, shall be the greatest ; FD, the other part of the diameter AD, shall be the least; And of the others, FB, the nearer to FA, shall be greater than FC, the more remote ; and FC greater than FG. Construction. — Join BE, CE, GE. Tor . , BE+EF Proof. — Because any two sides of a triangle are greater crAF>BP ^ ^^ ^1^^^ ^jjQ ^^:^.^ g^j^^ j3j^^ j^j. ^^^ greater than BF (I. 20). But AE is equal to BE ; therefore AE, EF, that is, AF is greater than BF. Again, because BE is equal to CE, and EF common to the two triangles BEF, CEF, the two sides BE, EF are equal to the two CE, EF, each to each. But the angle BEF is greater than the angle CEF ; And Therefore the base FB is greater than the base FC ba.ser>F /T OIX ^baaeFC \ * "^ )' In the same manner it may be shown that FC is greater than FG. Ap^in Again, because GF, FE are greater than EG (I. 20), ^e5^^ and that EG is equal to ED ; and.-.>ED. Therefore GF, FE are greater than ED. .•.GF>FD Take away the common part FE, and the remainder GF is greater than the remainder FD (Ax. 5). PROPOSITIOKS. ^61 Therefore FA is the greatest, and FD the least, of all the straight lines from F to the circumference; and FB is greater than FC, and FC than FG. Also, there cannot be drawn more than two equal straight lines from the point F to the circumference, one on each side of the shortest line. Construction. — At the point E, in the straight line EF, make the angle FEH equal to the angle FEG (I. 23), and join FH. Proof. — Because EG is equal to EH, and EF common to ^^^"sles the two triangles GEF, HEF, the two sides EG, EF are and hep equal to tha two sides EH, EF, each to each ; fn^very And the angle GEF is equal to the angle HEF (Const.); respect. Therefore the base FG is equal to the base FH (I. 4). But, besides FH, no other straight line can be drawn from F to the circumference equal to FG. For, if it be possible, let FK be equal to FG ; Then, because FK is equal to FG, and FH is also equal to ^'J^^^^^ FG, therefore FH is equal to FK ; Jso = Wi, That is, a line nearer to that which passes through the i^poVibie. centre is equal to a line which is more remote; which is im- possible by what has been already shown. Therefore, if any point, &c. Q.E.D. Proposition 8.— Theorem. If any 2^oint he taken without a circle, and straight lines he draion from it to the circumference^ one of lohich passes through the centre ; of those which fall on the concave circum- ferencCy the greatest is tJuit which passes through the centrcy and of the resty that which is nearer to the one2^assing through the centre is always greater than one more remote ; hut of those which fall on the convex circumference, the least is that he- tween the j^oint without the circle and the diameter ; and of the rest, that which is nearer to the least is always less than one more remote ; and from the same point there can he drawn to the circumference two straight lines, and only two, which are equal to one another, one on each side of the least line. Let ABC be a circle, and D any pouit without it, and from D let the straight lines DA, DE, DF, DC be drawu 262 GEOMETRY. r>A>DE and <:DL For EM+MD or AD> ED. Also base ED >- base FD, Again. MK+KD >MD, or DG. .•.KD>- DG, and/.DG DC Therefore DG is the greatest straight line from D to the >DA. circumference, and DC is gi-eater than DB, and DB greater than DA (III. 7); But they But thcso lines are likewise equal, by hypothesis; which is --r impossible. Therefore E is not the centre of the circle ABC. In like manner it may be demonstrated that any other point than D is not the centre ; Therefore D is the centre of the circle ABC. Therefore, if a point, &c. Q,E,D, Proposition 10. — Theorem. One circumference of a circle cannot cut another in more than two points. Construction. — If it be possible, let the circumference If possible, _jf^ ABC cut the ciitjumference DEF in more than two points, viz., in the points B,G,F. Take the centre K of the circle ABC ' (III. 1), and join KB, KG, KF. Proof. — Because K is the centre of the circle ABC, the radii KB, KG, KF are all equal. the two ^nd because within the circle DEF there is taken the iiave tho point K, from which to the circumference DEF fall more than two equal straight lines KB, KG, KF, therefore K is the centre of the circle DEF (III. 9). But K is also the centre of the circle ABC (Const.) ; Therefore the same point is the centre of two circles which cut one another; which is impossible (III. 5). Therefore, one circumference, &c. Q.E.D. Proposition 11. — Theorem. If one circle touch another internally in any point, the straight line which joins their centres, being prodxiced, slmll pass through that ptoint Let the circle ADE touch the circle ABC internally in the same centre. PROPOSITIOICS. 265 point A; and let F bo the centre of the circle ABC, and G the centre of the circle ADE. The straight line which joins their centres, being produced, shall pass through the point of contact A. Construction. — For, if not, let it pass otherwise, if if not, possible, as FGDH. Join AF and AG. Proof. — Because AG, GF are greater than AF (I. 20), and AF is equal to HF (I. def. 15); ' Therefore AG, GF are greater than HF. Take away tlie common part GF, and the remainder AG is greater than the remainder HG. But AG is equal to DG (I. Def. 15); Therefore DG is greater than HG, the less than tlie greater; which is impossible. Therefore the straight line which joins the centres, being produced, cannot fall otherwise than upon the point A, that is, it must pass through it. Therefore, if one circle, &c. Q.E.D, AG>HG. But AG =DG. .*.DG=- HG. Proposition 12. — Theorem. If two circles touch each other externally in any jyoint, the straight line which joins their centres shall pass through that point. Let the two circles ABC, ADE touch each other externally in the point A; and let F be the centre of the circle ABC, and G the centre of the circle ADE; The straight line which joins their centres shall pass througli the point of contact A. Construction. — For, if not, let it pass otherwise, if possible, as FCDG. Join FA and AG. Proof. — Because F is the centre of the circle ABC, FA is equal to FC (I. Def. 15). And because G is the centre of the circle ADE, GA is equal to GD; ir not. 266 GEOMETRY. FGfs> FA+AG, but it is also leas. Therefore FA, AG are equal to FC, DG (Ax. 2). Therefore the whole FG is greater than FA, AG. But FG is also less than FA, AG (I. 20) ; wliich is im- possible. Therefore the straight line which joins the centres of the circles shall not pass otherwise than through the point A, that is, it must pass through it. Therefore, if two circles, &c. Q.E.D, Proposition 13.~Theorem. One circle cannot touch another in more points than one, whether it touch it internally or externally » I. First, let the circle EBF touch the circle ABC internally in the point B. Then EBF cannot touch ABC in any other point. If possible, Construction. — If it be possible, let EBF touch ABC in letittomai another point D; join BD, and draw GH bisecting BD at '"''"' right angles (I. 10, 11). tbcn GH passes throu<,'h the i)oint ofcoutact, which it does uot. Proof. — Because the two points B, D are in the circum- ference of each of the circles, the straight line BD falls withiii each of them (III. 2); Therefore the centre of each circle is in the straight line GH, which bisects BD at right angles (III. 1 cor.) Therefore GH passes through the point of contact (III. 11). But GH does not pass through the point of contact, because the points B, D are out of the line of GH; which is absurd. Therefore one circle cannot touch another internally in more points than one. PROPOSITIONS. 267 II. Next, let the circle ACK touch the cii'cle ABO externally in the point A. Then ACK cannot touch ABC in any other point. Construction. — If it be possible, let ACK touc?li ABC in if possible another point C. Join AC. ^„^ InCabof Proof. — Because the points A, C are in the circumference of the circle ACK, the straight line AC must fall within the cii'cle ACK (III. 2). But the circle ACK is without the circle ABC (Hyp.); ..... Therefore the straight line AC is without I J then the circle ABC. V / tSholIt But because the two points A, C are in "^s,^^^ ^^^ V^^p^'^^'^ the circumference of the circle ABC, the whiciiis straight line AC falls within the circle ABC (III. 2); which absurd, is absurd. Therefore one circle cannot touch another externally in more points than one. And it has been shown that one circle cannot touch an- other internally in more points than one. Therefore, one circle, &c. Q,E.D, Proposition 14.— Theorem. Equal straight lines in a circle are equally distant from the centre; and, conversely, those which are equally distant from the centre tire equal to one another. Let the straight lines AB, CD, in the circle ABDC, be equal to one another. Then they shall be equally distant from the centre. Construction. — Take E, the centre cif the cii'cle ABDC (III. 1). From E draw EF, EG, perpendiculars to AB, CD (I. 12). Join EA, EC. Proof. — Because the straight lin^ EF, passing through the centre, cuts the straight line AB, which does not pass through the centre, at right angles, it also bisects it (III. 3). Therefore AF is equal to FB, and AB is double of AF. For the like reason CD is double of CG. 268 GEOMETRY AF = CG, But AB is equal to CD (Hyp.); therefore AF is equal to CG (Ax. 7). And because AE is equal to CE, the square on AE is equal to the square on CE. But the squares on AF, FE are equal to the square on AE, because the angle AFE is a right angle (I. 47). For the like reason the squares on CG, GE are equal to the square on CE ; Therefore the squares on AF, FE are equal to the squares on CG, GE (Ax. 1). _ But the square on AF is equal to the square on CG, because AF is equal to CG; Therefore the remaining square on FE is equal to the remaining square on GE (Ax. 3) ; .•.EF=EG. And therefore the straight line EF is equal to the straight line EG. But straight lines in a circle are said to be equally distant from the centre, when the perpendiculars drawn to them from the centre are equal (III. Def. 4) ; Therefore AB, CD are equally distant from the centre. Conversely, let the straight lines AB, CD be equally dis- tant from the centre, that is, let EF be equal to EG; TTcre Then AB shall be equal to CD. and^ ' Proof. — The same construction being made, it may be -^^'^+.EF2 demonstrated, as before, that AB is double AF, and CD EG-'. " double of CG, and that the squares on EF, FA are equal to the squares on EG, GC. But the square on EF is equal to the square on EG, because EF is equal to EG (Hyp.); Therefore the remaining square on FA is equal to the remaining square on GC (Ax. 3), . •. AF = And therefore the straight line AF is equal to the straight CG, &c. Ym^ CG. But AB was shown to be double of AF, and CD double of CG; Therefore AB is equal to CD (Ax. 6); Therefore, equal straight lines, fk. And therefore BC is greater than FG. Conversely, let BC be gi-eater than FG. Then BC shall be nearer to the centre than FG, that is, the same construction being made, EH shall be less than EK. Proof. — Because BC is gi-eater than FG, BH is greater than FK. But the squares on BH, HE are equal to the squares on FK, KE; And the square on BH is gi-eater than the square on FK, because BH is gi-eater than FK; Therefore the square on HE is less than the square on KE, and the straight Jine EH less thau EK; 270 GE03IETRY. And therefore BC is nearer to the centre than FG (III, def. 5). Therefore, the diameter, &c. Q.E.D, Take any i)oint F in AE, then DF > DA and . •. > DC. Draw DH at right angles to HG, then DH < DA, and .'. < DK. Proposition 16.— Theorem. The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls ivithout the circle; and a straight line, mahing an acute angle ivith the diameter at its- extremity, cuts the circle. Let ABC be a circle, of which D is the centre, and AB a diameter, and AE a line drawn from A perpendicular to AB. The straight line AE shall fall without the circle. Construction. — In AE take any point F; join DF, and let DF meet the circle in C. Proof. — Because DAF is a right angle, it is greater than the angle AFl3 (I. 17); Therefore DF is greater than DA (I. 19). But DA is equal to DC; therefore DF is greater than DC. Therefore the point F is without the circle. In the same manner it may be shown that any other point in AE, except the point A, is without the circle. Therefore AE falls without the circle. Again, let AG make with the diameter the angle DAG less than a right angle. The line AG shall fall within the circle, and hence cut it. Construction. — From D draw DH at right angles to AG, and meeting the cir- cumference in K (I. 12). Proof. — Because DHA is a right angle, and DAH less than a right angle; Therefore the side DH is less than the side DA (I. 19). But DK is equal to DA; therefore DH is less than DK. Therefore the point H is within the circle. PROPOSITIONS. 271 Therefore tlie straight line AG cuts the circle. Therefore, the straight line, rG. is greater than FG; the part greater than the whole, which is impossible. Therefore FG is not per2)endicular to DE. In the same manner it may be shown that no other straight line from F is perpendicular to DE, but FC; therefore FC is perpendicular to DE. Therefore, if a straight line, ike. Q.EM, PROPOSITIONS. 273 take F the centre, out of the line. Proposition 19. — Theorem. If a straight line touch a circle, and from the 2yoint of con-- tact a straight line he drawn at right angles to the touching line, the centre of the circle shall he in that line. Let the straight line DE toucli the circle ABC in C, and from C let CA be drawn at right angles to DE. The centre of the circle shall be in CA. Construction. — For, if not, if possible, let F be tlie centre, if not. and join CF. Proof. — Because DE touches the circle ABC, and FC is drawn from the assumed centre to the point of contact. Therefore FC is perpendicular to DE (III. 18); Therefore FCE is a right angle. P -q- But the angle ACE is also a right angle (Const.); Therefore the angle FCE is equal to the angle ACE; the Then ^ less to the gi^eater, which is impossible. z fce ~ Therefore F is not the centre of the circle ABC. aif-bs"^'^^^ In the same manner it may be shown that no other point which is not in CA is the centre; therefore the centre is in CA. Therefore, if a stmight line, &c. Q.E.D, Proposition 20.— Theorem. The angle at the centre of a circle is douhle of the angle at the circumference, upon the same hase, that is, upo7i the same arc. Let ABC be a circle, and BEC an angle at the centre, and BAC an angle at the cir- cumference, which have the same arc BC for theii' base. The angle BEC shall be double of the angle BAC. Case I. — First, let the centre E of tlie circle be within the angle BAC. Construction.— Join AE, and produce it to the ference ii^ F, 5 9 cu'cuxa- 274 GEOMETRY. 1 BEF = /. EAB + Z EBA = 2zEAB. r-nd so z FEC = 2 z EAC. .'. ZBEC 5^:2 z BAG. .*. taking the differ- ence z BEC = 2 Z BAG, z BFD = 2 z BAD, Proof. — Because EA is equal to EB, the angle EAB is equal to the angle EBA (I. 5) ; Therefore the angles EAB, EBA are double of the angle EAB. But the angle BEE is equal to the angles EAB, EBA (I. 32); Therefore the angle BEE is double of the angle EAB. For the same reason the angle EEC is double of the angle EAC. Therefore the whole angle BEC is double of the whole angle BAG. Case II. — Next, let the centre E of the circle be without the angle BAC. Construction. — Join AE, and produce it to meet the circumference in F. Proof. — It may be demonstrated, as in the first case, that the angle EEC is double of the angle EAC, and that FEB, a part of EEC, is double of FAB, a part of FAC; Therefore the remaining angle BEC is double of the re- maining angle BAC, Thei'efore, the angle at the centre, &c. Q,E.D, Eroposition 21.— Theorem. /■ The angles in the same segment of a circle are equal to one another. Let ABCD be a circle, and BAD, BED angles in the same segment BAED. The angles BAD, BED shall be equal to one another. Case I. — First, let the segment BAED be greater than a semicircle. Construction. — Take F, the centre of the circle ABCD (III. 1), and join BE, DF. Proof. — Because the angle BED is at the Yh centre, and the angle BAD at the circum- ference, and that they have the same arc for their base, namely, BCD; Therefore the angle BED is double of the angle BAD j(III. 20). PROPOSITIONS. 275 For the same reason, the angle BFD is double of the angle and BED; Therefore the angle BAD is equal to the angle- BED (Ax. 7). B/^ Case II. — Next, let the segment BAED , be not greater than a semicircle. Construction. — Draw AF to the centre, and produce it to C, and join CE. Proof. — Then the segment BADC is greater than a semicircle, and therefore the angles BEC in it are equal by the first case. For the same reason, because the segment CBED is greater and than a semicircle, the angles CAD, CED are equal. j ced.~ Therefore the whole angle BAD is equal to the whole .-. z bad •angle BED (Ax. 2). Therefore, the angles in the same segment, Prop.* 23.^ ment AEB be applied to the segment CFD, so that the point A may be on the point C, and the straight line AB on the straight line CD, Then the point B shall coincide with the point D, because AB is equal to CD. And the straight line AB coinciding with CD, the seg- ment AEB must coincide with the segment CFD (III. 23); and is therefore equal to it. Therefore, similar segments, &c. Q.E.D, Proposition 25.— Problem. A segment of a circle being given, to describe the circle of which it is the segment. Let ABC be the given segment of a circle. 1 It is required to describe the cii'cle of which ABC is a segment. Construction. — Bisect AC in D (I. 10). From the point D draw DB at right angles to AC (I. 11), and join AB. Case I. — First, let the angles ABD, BAD be equal to one another. Then D shall be the centre of the circle required. 278 GEOMETRY. When zABD = /BAD, DA = DB and .-. Dthc centre. Make z BAE: A ABD. /. EA = EB and EA = EC. .-. EA = EE = EC, and there- fore E is the centre. Proof. — Because the angle ABD is equal to the angle BAD (Hyp.); Therefore DB is equal to DA (I. 6). But DA is equal to DC (Const.); Therefore DB is equal lo DC (Ax. 1). Therefore the three straight lines DA, DB, DC are all equal ; And therefore D is the centre of the circle (III. 9). Hence, if from the centre D, at the distance of any of the three lines, DA, DB, DC a circle be described, it will pass through the other two points, and be the circle required. Case II. — Next, let the angles ABD, BAD be not equal to one another. Construction. — At the point A, in the straight line AB, make the angle BAE equal to the angle ABD (I. 23); Produce BD, if necessary, to E, and join EC. Then E shall he the centre of the circle required. Proof. — Because the angle BAE is equal to the angle ABE (Const.), EA is equal to EB (I. 6). A-nd because AD is equal to CD (Const.), and DE is com- mon to the two triangles ADE, CDE, The two sides AD, DE are equal to the two sides CD, DE, each to each; And the angle ADE is equal to the angle CDE, for each of them is a right angle (Const.); Therefore the base EA is equal to the base EC (I. 4). But EA was shown to be equal to EB; Therefore EB is equal to EC (Ax. 1). Therefore the three straight lines EA, EB, EC are all equal; And therefore E is the centre of the circle (III. 9). Hence, if from the centre E, at the distance of any of the three lines EA, EB, EC, a circle be described, it will pass through the other two points, and be the circle required. In the^r^^ case, it is evident that, because the centre D is in AC, the segment ABC is a semicircle. In the second case, if the angle ABD be greater than BAD, the centre E falls without the segment ABC, which is therefore less than a semicircle; But if the angle ABD be less than the angle BAD, the PROPOSITIONS. 279 centre E falls within the segment ABC, which is therefore gi-eater than a semicircle. Therefore, a segment of a circle being given, the circle has been described of which it is a segment. Q.E.F, Proposition 26.— Theorem. In equal circles, equal angles stand upon equal arcs, ivhether they he at the centres or at the circumferences. Let ABC, DEF be equal circles, having the equal angles BGC, EHF at their centres, and BAC, EDF at theii' cir- cumferences. The arc BKC shall be equal to the arc ELF. Construction. — Join BC, EF. Proof. — Because the circles ABC, DEF are equal (Hyp.), the straight lines from their centres are equal (III. def. 1); Therefore the two sides BG, GC are equal to the two sides Trianglec EH, HF, each to each; l^^'^^l^ And the angle at G is equal to the angle at H (Hyp.); equal in Therefore the base BC is equal to the base EF (I. 4). spect. ^^' And because the angle at A is equal to the angle at D (Hyp.), A ^ ^ .4) The segment BAC is similar to the segment EDF (III. .-. ses- ^^^^' ^^/, . . and EDF And they are on equal straight lines BC, EF. are similar But similar segments of circles on equal straight lines are equal equal to one another (III. 24); fm^^^^ Therefore the segment BAC is equal to the segment EDF. .-. aVo But the whole circle ABC is equal to the whole circle ®^"*'* DEF (Hyp.); ♦ Therefore the remaining segment BKC is equal to the remaining segment ELF (Ax« 3)» 280 GEOMETRY. .*. arc BKC = arc ELF. Therefore the arc BKC is equal to the arc ELF. Therefore, in equal circles, &c. Q.E,D, Proposition 27. — Theorem. In equal circles, the angles which stand upon equal arcs are equal to one another, whether they be at the centres or at the circumferences. Let ABC, DEF be equal circles, and let the angles BGC, EHF, at their centres, and the angles BAC, EDF, at their circumferences, stand on equal arcs BC, EF. The angle BGC shall be equal to the angle EHF, and the angle BAC equal to the angle EDF. If one z irreater than the other, the corre- sponding arc is greater. Construction. — If the angle BGC be equal to the angle EHF, it is manifest that the angle BAC is also equal to the angle EDF (III. 20, ax. 7). But, if not, one of them must be the greater. Let BGC is be the greater, and at the point G, in the straight line BG, make the angle BGK equal to the angle EHF (I. 23). Proof. — Because the angle BGK is equal to the angle EHF, and that in equal circles equal angles stand on equal arcs, when they are at the centres (III. 26); Therefore the arc BK is equal to the arc EF. But the arc EF is equal to the arc BC (Hyp.) ; Therefore the arc BK is equal to the arc BC (Ax. 1); the less to the greater, which is impossible. Therefore the angle BGC is not unequal to the angle EHF; that is, it is equal to it. And the angle at A is half of the angle BGC, and the angle at D is half of the angle EHF (III, 20); PROPOSITIONS. 281 Therefore the angle at A is equal to the angle at D (Ax. 7). Therefore, in equal circles, &c. Q.E.D, Proposition 28. — Theorem. In eqiuil circles, equal chords cut off equal arcs, t/te greater equal to the greater, and the less equal to the less. Let ABC, DEF be equal circles, and BC, EF equal chords in them, which cut off the two greater arcs BAG, EDF, and the two less arcs BGC, EHF. The greater arc BAG shall be equal to the grcate:* arc EDF, and the less arc BGG equal to the less arc EHF. GoNSTRUCTiON. — Take K, L, the centres of the circljs TakeK (III. 1), and join BK, KG, EL, LF. centJis.^" Proof. — Because the circles ABG, DEF are equal, their radii are equal (III. def. 1). Therefore the two sides BK, KG are equal to the two sides EL, LF, each to each; And the base BG is equal to the base EF (Hyp.) ; Therefore the angle BKG is equal to the angle Trianpics ELF (I. 8). '^::^ But in equal circles equal angles stand on equal arcs, equal m GV6rv rC" when they are at the centres (III. 26) ; spect. Therefore the arc BGG is equal to the arc EHF. But the whole circle ABG is equal to the whole circle DEF (Hyp.); Therefore the remaining arc BAG is equal to the remain- ing arc EDF (Ax. 3). Therefore, in equal circles, (fcc. Q.KD* 282 GEOMETRY. Proposition 29. — Theorem. In equal circles equal arcs are subtended by equal chords. Let ABC, DEF be equal circles, and let BGC, EHF be equal arcs in them, and join BC, EF. The chord BC shall be equal to the chord EF. TakeK CONSTRUCTION. — Take K, L, the centres of the circles cluteef' (III. l),andjoinBK, KC,EL,LF. Then pROOF. — Because the arc BGC is equal to the arc zelf" EHF (Hyp.), the angle BKC is equal to the angle ELF ' (in. 27). And because the circles ABC, BEF are equal (Hyp.), their radii are equal (III. def. 1). Therefore the two sides BK, KC are equal to the two sides EL, LF, each to each; and they contain equal angles; and so base Therefore the base BC is equal to the base EF (I. 4). "" - ^-1 Therefore, in equal circles, &c. Q.E.D, BC EF. Proposition 30.— Problem. To bisect a given arc^ that is, to divide it into two equal parts. Let ABB be the given arc. It is required to bisect it. Construction. — Join AB, and bisect it in C (I. 10). From the point C draw CD at right angles to AB (I. 11), and join AD and DB. Then the arc ABD shall be bisected in the /;;X^ XX Proof. — Because AC is equal to CB »^6..o.vxvG ^— ^ -^ (Const.), and CD is common to the two tri- andCDB, A c « angles ACD, BCD; In the tri- anglesADG PROPOSITIONS. !^83 The two sides AC, CD are equal to the two sides BC, CD, each to each ; And the angle ACD is equal to the angle BCD, because each of them is a right angle (Const.); Therefore the base AD is equal to the base BD (I. 4). lasem" But equal chords cut off equal arcs, the greater equal to the greater, and the less equal to the less (III. 28) ; And each of the arcs AD, DB is less than a semicircle, because DC, if produced, is a diameter (III. 1, cor.); Therefore the arc AD is equal to the arc DB. Therefore, the given arc is bisected in D. Q.E.F, Proposition 31. — Theorem. In a circle^ the angle in a semicircle is a right angle; hut the angle in a seginent greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle* Let ABC be a circle, of which BC is a diameter, and E the centre ; and draw C A, dividing the circle into the seg- ments ABC, ADC, and join BA, AD, DC. The angle in the semicircle BAC shall be a right angle; The angle in the segment ABC, which is greater than a cemicii'cle, shall be less than a right angle; . The angle in the segment ADC, which is less than a semi- circle, shall be greater than a right angle. Construction. — Join AE, and produce BA to F. Proof. — Because EA is equal to EB (I. Def 15), The angle EAB is equal to the angle EBA(I. 5); And, because EA is equal to EC, The angle EAC is equal to the angle EC A; Therefore the whole angle BAC is equal 1/ \ N\\ ^ bae + to the two angles ABC, ACB (Ax. 2). \ e y ol- ' But F AC, the exterior angle of the tri- V / z abc T angle ABC, is equal to the two angles x^ ^ i acb = ABC, ACB (I. 32). al./'f. Therefore the angle BAC is equal to the angle FAC "ehta'is'o- (Ax. 1), -<. a right aujrle. 284 GEOMETRY. And therefore each of them is a right angle (I. Def. 10) ; Therefore the angle in a semicircle BAG is a right angle. And because the two angles ABC, BAG, of the triangle ABG, are together less than two right angles (I. 17), and that BAG has been shown to be a right angle; z ABC Therefore the angle ABG is less than a right angle. Therefore the angle in a segment ABG, greater than a semicircle, is less than a right angle. And, because ABGD is a quadrilateral figure in a circle, any two of its opposite angles are together equal to two right angles (III. 22); Therefore the angles ABG, ADG are together equal to two right angles. But the angle ABG has been shown to be less than a right angle ; Hence Therefore the angle ADG is greater than a right angle; z ADC > Therefore the ansjle in a segment ADG, less than a semi- R right . , . , ® -IT angle, by Circle, IS greater than a right angle. Prop. 32. Therefore, the angle, &c. Q.E.B. GoROLLARY. — From this demonstration it is manifest that, if one angle of a triangle be equal to the other two, it is a right angle. JFor the angle adjacent to it is equal to the same two angles (I. 32). And, when the adjacent angles are equal, they are right angles (I. def. 10). Proposition 32. — Theorem. The angles contained hy a tangent to a circle and a chord drawn from the point of contact are eqvxil to the angles in the alternate segments of the circle. Let EF be a tangent to the circle ABGD, and BD a chord drawn from the point of contact B, cutting the circle. The angles which BD makes with the tangent EF shall be equal to the angles in the alternate segments of the circle; That is, the angle DBF shall be equal to the angle in the Begment BAD, and the angle DBE shall be equal to the angle in the segment BGD. PROPOSITIONS. 285 Construction. — From the point B draw BA at riglit angles to EF (I. 11). Take any point C in the circumference BD, and join AD, DC, CB. Proof. — Because the straight line EF touches the circle ABCD at the point B (Hyp.), and BA is drawn at right angles *bo the tangent from the point of contact B (Const.), The centre of the circle is in BA (III. 19). Therefore the angle ADB, being in a semicircle, is a right Angle (III. 31). Therefore the other two angles BAD, ABD are equal to a right angle (I. 32). But ABF is also a right angle (Const.); Therefore the angle ABF is equal to the angles BAD, ABD. From each of these equals take away the common angle ABD ; Therefore the remaining angle DBF is equal to the re- maining angle BAD, which is in the alternate segment of the circle (Ax. 3). And because ABCD is a quadrilateral figure in a cii^cle, the opposite angles BAD, BCD are together equal to two right angles (III. 22). But the angles DBF, DBE are together equal to two right angles (I. 13); Therefore the angles DBF, DBE are together equal to the angles BAD, BCD. And the angle DBF has been shown equal to the angle BAD; Therefore the remaining angle DBE is equal to the angle BCD, which is in the alternate segment of the circle (Ax. 3). Therefore, the angles, &c. Q.E.D. The centre is in BA. . •. ADD is a rijjljt angle, and z BAD-f Z ABD = a rijfht angle = ABr. .-. z BAD = z Dcr. Also, Z BeD + Z BAD^ 2 right angles = z DBF + z DBE. .-. Z DBE = z BCD. Proposition 33.— Problem. Upon a given straight line to describe a segment of a circlcy containing an angle equal to a given rectilineal angle. Let AB be the given straight Jine, and C the given recti- Juieal angle, 2SG GEOMETRY, It is required to describe, on tlio given straiglit line AB, a segment of a circle, containing an angle equal to tlieangleC. Case I. — Let the angle C be a right angle. Construction. — Bisect AB in F (I. 10). From the centre F, at the distance FB, describe the semicircle AHB. Then AHB shall he the segment required. Proof. — Because AHB is a semicircle, the angle AHB in it is a right angle, and therefore equal to the angle C (HI. 31). Case II. — Let C be not a right angle. Construction. — At the point A, in the straight line AB, BAD =c; make the angle BAD equal to the angle C (I. 23). H Angle in a semicircle is a right angle. At point A make z and draw AE at right angles to AD. From F, middle of AB, draw perpendi- cular, meeting AK iu G. Then G is the centre of a circle pass- ing throngh A and B, From the point A draw AE at right angles to AD (I. 11). Bisect AB in F (L 10). From the point F draw FG at right angles to AB (I. 11), and join GB. Because AF is equal to BF (Const.), and FG is common to the two triangles AFG, BFG; The two sides AF, FG are equal to the two sides BF, FG, each to each ; And the angle AFG is equal to the angle BFG (Const.); Therefore the base AG is eqvial to the base BG (I. 4). And the circle described from the centre G, at the dis- tance GA, will therefore pass through the point B. Let this circle be described; and let it be AHB. The segment AHB slwdl contain an angle equal to the given Q'ectUineal angle C, ruoposiTioNs. 287 Proof. — Because from the point A, the extremity of the And ad iiameter AE, AD is drawn at right angles to AE (Const.); the drcie, Therefore AD touches the circle (III. 16, cor.) Because AB is drawn from the point of contact A, the angle DAB is equal to the angle in the alternate segment AHB (III. 32). But the angle DAB is equal to the angle C (Const.) ; and /. z Therefore the angle in the segment AHB is equal to the ^^ ^^^ = angle C (Ax. 1), c. Therefore, on the given straight line AB, the segment AHB of a circle has been described, containing an angle equal to the given angle C. Q.E,F, Proposition 34. — Problem. Froini a given circle to cut off a segment which shall contain an angle equal to a given rectilineal angle. Let ABC be the given circle, and D the given rectilineal angle. It is required to cut off from the circle ABC a segment that shall contain an angle equal to the angle D. CoNSTRUCTiox. — Draw the straight line EF touching the Draw tan- circle ABC in the point B (III. ^ ^ ^'^"^ ^^^^' 17); And at the point B, in the straight line BF, make the angle FBC equal to the angle D (I. 23). Then the segment BAC slmll contain an angle equal to the given angle D. Proof. — Because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B (Const.); Therefore the angle FBC is equal to the angle in the alter- nate segment BAC of the circle (III. 32). But the angle FBC is equal to the angle D (Const.); Therefore the angle in the segment BAC is equal to the •*. ^ b^ angle D (Ax. 1). = ^ ^^^ Therefore, from the given circle ABC, the segment BAC has been cut off, containing an angle equal to the given angle D. Q.E.F, 288 GEOMETRY, AE = EC. BE ED + EF2 r= AE-i + .'. BEED = AE2 = AEEC. Proposition 35. — Theorem. If two straight lines within a circle cut one another^ the rectangle contained by the segments of one of them shall be equal to the rectangle contained by the segments of the other. Let the two straight lines AC, BD cut one another in the point E, within the circle ABCD. The rectangle contained by AE and EC shall be equal to the rectangle contained by BE and ED. A<^ ^ Case I. — Let AC, BD pass each of them through the centre. Proof. — Because E is the centre, EA, EB, ' EC, ED are all equal (L def. 15); Therefore the rectangle AE, EC is equal to the rectangle BE, ED. Case II. — Let one of them, BD, pass through the centre, and cut the other, AC, which does not pass through the centre, at right angles, in the point E. Construction. — Bisect BD in E, then F is the centre of the circle; join AF. Proof. — Because BD, which passes through the centre, cuts AC, which does not pass through the centre, at right angles in E (Hyp.); Therefore AE is equal to EC (III. 3). And because BD is cut into two equal parts in the point F, and into two un- equal parts in the point E, The rectangle BE, ED, together with the square on EF, is equal to the square on FB (11. 5); that is, the square on AF. But the square on AF is equal to the squares on AE, EF (I. 47); Therefore the rectangle BE, ED, together with the square on EF, is equal to the squares on AE, EF (Ax. 1). Take away the common square on EF; Then the remaining rectangle BE, ED is equal to the remaining square on AE; that is, to the rectangle AE, EQ, ^iucQ Ai; is e(|ual to EC, PROPOSITIONS. 289 Case III. — Let BD, wliiclx passes tlirough the centre, cut the other AC, which does not pass through the centre, in the point E, but not at right angles. Construction. — Bisect BD in F, then F is the centre of the circle. [ ^^^^>k Join AF, and from F draw FG perpen- /j^ ^-^^^'^^^ I X e dicular to AC (I. 12). \ >2 Proof. — Then AG is equal to GC (III. 3). Therefore the rectangle AE, EC, together with the square on EG, is equal to the square on AG (II. 5). To each of these equals add the square on GF; Then the rectangle AE, EC, together witli the squares on EG, GF, is equal to the squares on AG, GF (Ax. 2). But the squares on EG, GF are equal to the square on EF; And the squares on AG, GF are equal to the square on AF (T. 47). Therefore the rectangle AE, EC, together with the square on EF, is equal to the square on AF; that is, the square on FB. But the square on FB is equal to the rectangle BE, ED, together with the square on EF (II. 5) ; Therefore the rectangle AE, EC, together with the square on EF, is equal to the rectangle BE, ED, together with the square on EF. Take away the common square on EF; And the remaining rectangle AE, EC is equal to the remaining rectangle BE, ED (Ax. 3). Case TV. — Let neither of the straight lines AC, BD pass through the centre. Construction. — Take the centre F (III. 1), and through E, the intei'section of the lines AC, BD, draw the diameter a\ GEFH. Proof. — Because the rectangle GE, EH is equal, as has been shown, to the rectangle AE, EC, and also to the rectangle BE, ED; Therefore the rectangle AE, EC is equal to the rectangle BE, ED (Ax. 1). Therefore, if two straight lines, ■ (a; + 2) ^ 15(2a; + 7) - 5 (6 a; + 16) , ^^'. (a; + 2) (a; + 4) (6 a; + 16) (2 a; + 7) ' '~~ . ■■ 2 25 or, simplifying, ^, ^ ^^ ^ g = 12^^74,^112 ' ADFECTED QUADRATICS. 299 or, clearing of fractions — 24 ar^ + 148 a; + 224 = 25x' + 150 aj + 200; or, transposing and reducing, -a;--2a;= — 24; or, changing the sign of each side, then — x" + 2x = 24; or, completing the square, ar^ + 2x + P = 24 + 1 = 25. Taking the square root of each side, we have — X + 1 = ±5 4aj=±5-l==4or-6. Ex. 3. Solve the equation ar*+6a;+25 = 0. We have o^ + 6 x = - 25; or, completing the square — af^ + 6a; + 32 = - 25 + 9 - - IG; ' or, extracting the square root of each side — a.^ + 3 = ± V^n[6 :^x= - 3 ± j~:ri6. As the quantity \/ - 16 has no exact or approximate value, the given equation has no real roots. The roots are therefore said to be imaginary. 4, Solution by breaking into factors. "We have seen (Stage I., Alg., Art* 30) that it is often easy to find by inspection the factors of quadratic expres- sions. We may make use of this knowledge to solve quad- i-atic equations. Ex. 1. Solve the equation x^ + 5 a; = 66. Transposing all to the firet Side, we have — ar + 5 a; - 66 = Oi And, resolving the first side into its element^i'y fiictors, we get— {x - 6) {x + 11) = 0. Now, if either of these factors is put equal to 0, the equation is satisfied. Hence^ we have, a; - 6 = 0, and x + 11 = 0; or, a; = 6, and a; = - IL :. 6 &,nd - li are the roots required. 300 ALGEBRA* Ex. 2. Given a;" - {a + h) x ■{■ ah - 0, find x. We have {x - o) {x - h) - 0. Now this equation is satisfied by making either of the factors = 0. . Hence, x - a - 0, ov x = a-A and, ic - 6 = 0, or a; = 6 ; J .*. ct, h are the roots required. Ex. I. 2. 5^2 + 6 = 86. 3. :^^^^l-,;i-^ = 34. 4. foj^- +1 = 6. 5. 3aj - :^ = 0. 6. 2 (a^ + 6^) - ic^= (ex - h)\ ' X x/24 + x" = 4 + ar^. 8. * \/aJ - c& = V ic - ^6^ + x^. X ' ^ x' 10. 1 1 aa; - Ja^x^ - 1 aa3 + Vav - 1 "'^• 11. ^a^ + a^ + ija? - a^ ^a^ + ar^ - ^a' - x' h. 12. X - sjx" - I ^' 13. a;^ - 5 a; + 6 = 0. 14. x^ - X = 72. 15. 3 a;- - a; = 2. 16. 4a;2 - 9a; = 28. 17. 6a;2 + a; - 35 = 0. 18. a;2 + 65 a; + 8 = 0. 19. a3(? + 5a; + c = 0. 20. "■-^^''''-'■ X 22. ^">-r«^. = " • * See Remark, page 304. ADFECTED QUADRATICS. 301 23. {a + hx) {ex + c?) = (a + 6) (c + dx). 24. [a - h) (x^ - h') = (a + b) {x - bf, 25. {a + b) {x — a) {x - b) = abx, o/> « , aJ b X a' - y^ JO. - + --=- + _ + — . X X a ab 27. (io(r - (c + d)x = bx- - -^,. a — b 28. a^^- - (« - 5)2 oj + a^ -- 6=^ar^ + {a? + a6 + b^fx + 51 29. -^-- - 31. 30 ^ + ^ + ^- = 2 a ; + 5 ' 2 aj + 3 13 ~ i^~^' 31. ?J^±i H- ,_^ . 5i. a; 2 a; + 9 ^ 32. ^1-1 + ^' " ^^ = 0. * a; + 3 a; + 8 33. 34. 35. 36. 37. 38. 3 a; - 7 5 a; + 3 ^ 5 4a;-2'^7a; + 4~9* ^x +^ 4a;-l_7aj+l 2'a;~-nr X - 2" ~ a; - 3 * 2a; + l_ a;+2 _ 7a; +8 a; + 2 4 a; + 4 4 a; + 13* 3a; _ 2 a; + 9 ^ 2 a; + 2 4 a;-2 a;+2"2a;-.r 3ar'-2a;+7 a? + 1 x - '^ 6 a;^ - 4 a; + 11 2 a;'^ + 14 a; - 9* 3 a^ + 10 _ ar^ + 2 a; + 3 ^ 2ar' 4- 2a; + 10 X X + 2 a; + 1 39. '^' J5 a; + G + Va; + 10 = 4. 40. >/3a; - 4 - J2x - 4 = x/^ 41. * Va^ + 6^ +^ + Jo" + 6- - a; = ^/iL^. N 6 ■ * See Bemark, page 3H 302 ALGEBRA. 42. * sj'2 a + X ■{- isji a - X — 2 ijx - a. 43. '" sjax + 6 + sibx + a ~ s/(a - b)x + a + b + 2 Jab. «■•J^"^-^/~^(^/^^/f)v.^. 45. C~^ - l\ . — ~-^ = /^ + iV ^ + ^ + c __ 2^ * \Z) /' b \6- /' ic 4G. (ft - h) (x + b - c) (x + c - a) + a-(x + b — c) + b'\x + c - a) + c\x + a - b =^ 0), 47, Show tliat, if x be real, the vahie oi x + - cannot lie x between 2 and - 2. X" 48. If, in the equation x - ^ = ct, the quantity x be real, show that a cannot be greater than 5. Equations which may be Solved like Quadratics. 5. Certain equations of a degree higher than the second may be solved like quadratics. It will be seen that, although it is impossible to lay down definite rules for the treatment of every such equation, the object to be attained is either (1.) To throw the equation into the form — X' + pX = q, when X is an expression containing the unknown quantity, and solving when possible this equation for X; or (2.), To strike out from each side a factor containing the unknown quantity, thus reducing the dimensions of the equation, and obtaining a value or values of the unknown quantity by equating this factor to zero. Ex. 1. Solve the equation a;'* + 144 = 25 x\ We have, transposing, £c'*-25a3^= -144; or, {xy - 25(0^) = - 144; * 3^9 Kemark, page 304, EQUATIONS WHICH MAY DE SOLVED LIKE QUADRATICS. 303 or, completing the square — • ^ 2 5 — +7.^ .-. x" = %^ ± I = 16 or 9. Hence, x = ± 4 or ± 3. The given equation has therefore four roots, being as many roots as the degree of the equation. Ex. 2. Solve a;« + 3 (K^ - 88. We have, (a^y + 3 (oc^) = 88 ; or, transposing, (a^Y + 3(aj3) - 88 = 0; or, (a^ - S){a^ + 11) = 0. Hence, the given equation is satisfied by — cbS _ 8 = 0, and also by ic^ + 1 1 = 0. We have then a;^ - 2» = 0, and ic^ + ( ^H)' - ; or, breaking into factors, we have — (aj ~ 2) (ar^ + a; • 2 + 2-) - 0, and (^ + .5^)|a;= - x'^lT+ (^ll)-l =0. From the^rs^ of these equations, we get — a; - 2 = or aj = 2, and a;^ + 2 a; + 4 = 0, from which two other roots may be found. And from the second equation, we get — x+ ^11 = 0, ora; = - ^11, and ar'-a;v^ll+ a3^121=0, which gives two other roots. We have therefore shown how to obtain the six roots of the given equation. Ex. 3. Solve— 23 ar^ - 75 a; - 6 a: V4ar»-9a;+ 9 + 40 = 0. Changing the sign of each side, and transposing, we have — 6a;>s/4'^"^9iT9 = 23^2 - 75 a; + 40; adding to each side 13ar^ — 9a; + 9, then — 13ar' _ 9a; + 9 + (jxsjia? - 9a; + 9 = 36a;2 - 84a; +49; or, (4a;2 - 9 a; + 9) + 2(3a;) n/4 ar^ - 9 a; + 9 + (3 a:)' =^[Qxf - 2(6a;)-7 + 7-; 304 ALGEBRA. or, ( x/4a;'^ - 9x + 9 + Sxf .=^ {6x - 7)1 /. J4:x' -dx + d + 3x = ± {(jx-7), (1.); or, taking the upper sign — Jlx' - dx + 9 = {6x - 7) - 3aj - 3 - 7. Hence, squaring each side — 4 a;2 - 9 a; + 9 = 9 cc^ - 42 ic + 49 ; or, 5 a;' - 33 aj + 40 = ; or, {x ^ 6) {6x - S) = 0. :. X = 6 or J. Again, taking the lower sign of ( 1), we have — V4'i--9a; + 9== - (6x - 7) - 3aj = - 9aj + 7 (2.); or, squaring — 4:x'^ - 9x + 9 = Slay" " l2Qx + 49; from which 77a]2 _ 117 aj + 40 = 0; or, (a; - 1) (77 aj - 40) = 0. :. x = 1 or ff . Hence, the roots of the given equation are 5, f , 1, ff. Remark. — If we proceed to verify these values of x, we shall find that the last two values — viz., 1 and 4f — will not satisfy the given equation unless we obtain the value of Ji a;"^ -- 9 a; + 9 by means of the equation from which these last roots were found. Thus from ( 2 ) we find, on putting 1 and f -?- successively for X — V4x*2 - 9aj + 9 = - 9 (1) + 7 = - 2, and >/4 a;'^ - 9 a; + 9 = - 9 (f -?) + 7 = 2^; and on svibstituting either of these values of JZo^^^^9x + 9 along with the corresponding value of x, the given equation is satisfied. Ex. 4. Solve— {x + b + c){x + h — c){b + c - x) = (a + b + c) {a + b - c) {b + c - a). By inspection we see that a is one of the roots. We shall therefore so arrange the equation as to be able to strike out ^ a; — a as a factor of each side (Ai't. 5). | EQUATIONS WHICH MAY BE SOLVED LIKE QUADRATICS. 305 We have — {x + b + c) (x + b - c) _ b + c (a + b -h c)(a + b - c) b + c - x' (x + by ~ c ^ _b + c - a ^' \a + by - d" ~ b + c ^ x' or, taking the difference of numerator and denominator — (x + by - (a + by _ X - a {a + by - c^ b + c — x' (x - a) (x -h 2b — a) x - a or, ^ i—^ 1 = , {a + by ■- c" b + c - X Dividing each side by aj - a, we have — X + 2b - a _ 1 (airSy—c'-b + c-x' and a; - a = 0, or a: =. a; Hence also (x + 2b - a) {b + c — x) = (a + by — c-; an ordinary quadratic from which two other roots may be determined. Ex. 5. * Solve the equation — ^ytya '^ iabx'^M ^ (a ' byx"". Dividing each side by as^, we have — aWx ^ - 4a^6^a;2^'^ = (a - by-, or, [abx^'^) - 4 a^b^ {abx'^) = (a - by ; or, completing the square — \abx^) - 4 a^b^ (abx^^) + [2 a^by = {a -by + 4: ab = (a + by. * We shall assume in this example that the laws of multiplication and division proved in Algebra, Stage I., Arts. 25, 27, hold for frac- tional indices, 5 V ^^^ ALGEBRA, Hence, extracting the square root — ' ahx^ -^^a'^lfi = ± (a + 6), /. abx^ = ± (a + 5) + 2a^6^ = (a^ + 5^)'or-(aJ-6^)2; Then, raising each side to the {2pq)\h power, we have— r + —A or (^ ^ -) Hence, taking the (p - 5')th root, we get — /I 1 \±n^ /I 1 \J_^ ^ == iTT + -T- b-^ or l-r - -T b-« /I 1 \i^2^ /I 1 \J_^ /I \\±i We have also, since the factor cc^ has been struck out — - x^ = 0, and :. x = as another solution, Ex. II. 1. 4 a;* - 11 05^ = 225. 2. 5 iB« - 17 x^ = 184. 3. cc^ + 5 aj-2 ^ 251. 4. (a:^ + 3 aj + 3)2 + 2 a^ = 189 - 6 a;, 5. 3a:-*^- 17 aj-^ = 1450. 6. a;" + ~ = 5. a;" 7. Vi~+~r2 + v'a? + 12 = 6, 8. x/5M^^ + 20 = aj2 + 9. 9. 1 ajV + - = 6 + b a:. EQUATIONS WHICH MAY BE SOLVED LIKE QUADRATICS, 307 10. (3 o; + 4) + ^3 o; + 4 - 12 = a U, x" + 2x = — 1-2 - 1. X + 4: s/a — X \/a " X __ sjx 12. 13. 9 CB + 24 Jx ^ -7= (-r + ^^i = 65. sIxX^Jx / 14. X - \!a ' Jh - X " Jh ' sjx ■¥ a, 1 1 a^ + 2ar» 15. {a - ^/a^ - ar^) (a + \/a* - ar^)" X* 1G.._-2J_ -^ =T-1- P + 9 1 • /»2 7j2 ITEJ"- . Ht l8, p + a? _ I a? ^ J ■ n/ x^ Na' + or 19. (a^" + 1) {x^- - If = 2 (a; + 1). a? _ 975 ^^ 7^0 21.-J2 {._3 2) i^^, i«+lliC+l X) XT 22. 6ar - 5a;- 8 VSa:^ + 5aj _ 4 = 12. 23. 2 .7;- + 5 x - 2 X -N/2af^ - 7 a; + 1 = 35. 24. 20ar^ - 9aaj-8a;A/5ar^-3aa; + 2a' = 1 a\ 25. ar'(a; - 2f + 6ar(a; - 2) = 24aj + 36 - 5a:l 26. 2ar + 20aj - ^3 ar^ - a; + 5 = 105. 27. 4a;-^ + 12a; - 2xsJ\9^ - 2a; + 19 = 30, 28. a;* + a;= + 1 = a(ar^ + a; 4- 1). 29. a;« - 1 = 0. f 308 ALGEBRA. 30. (x" - af + {ar - hf + {x^ - cf = 3 (a;2 - a) {x^ - h) {x" - c) + d ax^ - a{a + b + c)\ 31. {x - h) {x" - (?) = (a ■- h) (a" - c-). 32. {x - a) {x — h) (x ^ c) = (m — a) (m — h) {m - c). 33. {x - a) {x " h) (x - c) = (a + 5) (c^ + c) (6 + c). 34. (n - l)£c (ic^ + ax + c?) = a^ - ajl 35. (ax - 6)« + (coj - d^ = (a ^ cf^ - (5 + cZ)^ 36. cc^%a;2--i^ 1 = a((«2 + cc + 1). Simultaneous Quadratic Equations. 6. The following worked examples are given as speci- mens of tlie methods to be employed, but it must be under- stood that practice alone will give the student complete mastery over equations of this class. Ex. 1. Solve a;2 + 2/' = 20 (1.) ) X ^y = 6 (2.)/ As we have given the suTti of the unknown quantities, we shall work for the difference. From (2), multiplying each side by 2, we have — 2 a;' + 2 2/* = 40 and from ( 1 ), squaring, a.*' + 2 ccy + 2/^=36 Then, subtracting, ar^ - 2 ajy + y^ - 4 ; and, taking the square root, we have aj - 2/ = i 2 (3). (2) + (3), then 2i« = 8 or 4, and .'. cc = 4 or 2. (2) - (3), then 22/ = 4 or 8, and r. y ^ 2 or 4. Note. — Having found that a; = 4, j/ = 2, we might have told by inspection that the values a; = 2, y = 4, would also satisfy the given equations, for x and y are similarly involved in both equations. Ex. 2. Solve ar - 2/' = ^ • (1.) I ^y = ^ (2.)/ As we have given here the difference of the squares of the unknown quantities, it will be convenient to work for th^ stm of the squares. SIMULTANEOUS QUADRATIC EQUATIONS. 309 From (1.), squaring, x* - ^or\f + y- = 25, and fiom (2.), squaring, &c., 4ar^2/'^ := 144 Then, adding, a;* + 2ar^?/ + 2/* = 169 and taking the square root, a;^ + ?/2 _. + _ 13....... (3). (3.) + (1.), then, 2 ar = 18 or - 8, or ar = 9 or - 4, and .-. a; = ± 3 or ± 2 nT^I. ' (3.) - (1.), then, 2 7/2 = g or - 18, ory^ == 4 or -> 9, and .-. y = ± 2 or ± 3 V - 1. Note. — The student will see that the pairs of values which s atisfy the given equations are, x = S, y = 2; x = - 3, y — - 2; a; = 2 -v/ - 1, y - 3 V -T; X = - 2\/^n^, y = - S^/^^. Ex. 3. Solvea:2 + y = llaj (1.)) . f + x= lly (2.)/ Subtracting, then, oc^-y^ — x + y= llaj- 11 y; or, a^ - y^ = 12 (re - y). Now (aj — 2/) is a factor of each side, and hence, striking it out, we have — x + y = 12 (3.), and also x - y = (4.). Equations (3.) and (4.) may not be used as simultaneous equations, but each of tliem may be used in turn with either of the given equations. Thus, taking equations (3.) and (1.), we have — (1.) - (3.), a;^ - a;= llaj - 12, from which a; = 6 ± 2 sfS , and hence from (3.), by substitution, we easily get- 2/ = 6 + 2 ^/6; • Again, taking equations (4.) and (1.) — we have, from (4.) a; = y, and .*. from (1.), or ■{■ x ^ 11 aj or a;' = 10 a^, from which x = 10 or ; and so, from (4.), y = 10 or 0, 310 ALGEBRA* Hence, the pairs of values satisfying the given equations are — a; = 10, 2/ = 10 ; a; = 0, 2/ = ; x = Q + 2 sl\y =6-2 Sj, a; = 6 - 2 V6, 2/ = 6 + 2 V6. Note. — It is worth while remarking that when each of the terms of the given equations contain at least one of the unknown quantities, the values ic = 0, 2/ = will always satisfy. Ex. L Solve 3ar - 2xij = 55 (1.) ) ;x? - ^xy + ^f = 7 (2.) J Multiplying the equations together crosswise, we get — 55 cc^ - 275 ccy + 440 t/' = 21 of^ - 14 a;?/; or, transposing, 34 ic^ - 261 ajy + 440 ^/^ = ; or, (2 a; - 5 7/) (17 a; - 882/) = 0, from which 2 a; = 5 ?/, and 17 a; = 88 y. Each of these equations taken in twni with either (1.) or (2) will easily give the required values of x and y, Ex. 5. a;* + 2/* = 337 (1.), «J +2/ - 7 (2.) Erom ( 2 ), raising each side to i\iQ fourth power , we have — a;* + 4.a?y + 6 a;^^^ ^ 4^^ + 2/' = 2401; (3.) (3) - (l),then4a;^2/ + 6ar^2/' + ^^V^ = 2064; or, 2a?y + Z^y" + 2 xif =^ 1032; or, arranging, 2xy(x + yY - x'^y^ = 1032 ; but from (2), {x + yY = 49, and hence, 2 a;?y(49) - a^V = 1^32 ; or, ar2/' - 98 a;^ + 1032 = 0, from which xy = 1 2 or 86, (4.) From (2) and (4), x - y may now be easily obtained, and hence also the required values of x and y. Ex. III. 1. x + y =^ 5, xy = 6, 2. X - y — 2, xy = 15. 3. a;- + 2/' = 25, xy - 12. 4. ar + 2/" = 20, a; + 2/ = ^* SIMULTANEOUS QUADRATIC EQUATIONS. 31 1 5. or + y- = 29, a; - 2/ = 3. 6. ar» - 2/^ = 13, (ic - yY = 1. 1, a? --f = 21,xy = 18. 8. ar^ - t/" = 12, a; + 2/ = 6. 9. ar^ + 2/^ := 53, ar^ - 2/2 ^ - 45. 10. ar^ + a;?/ = 28, 2/* + ajy = 21. . ' 11. ar' + a;?/ + 2/^ = 19, a;2/ - af^ = - 3. 12. a; + 2/ = 13, Jx + Ijy = 5. 13. ar^ + a;?/ + 2/^ = 84, a; + J^j -{- y =^ IL 14. a;3 + 2/3 - 35, apy + X7f = 30. 15. -_ + - = a, + ^ = 5. aj 2/ af' 2/" 16. a; + 2/ = «{a; - y), o;^ + y^ = ^^. 17. a;* - 2/* = «j ar^ - 2/" = ^• 18. aj + 2/ = 5,a;3 + 2/^ = 35. 19. a; + 2/ = 5, ar* + 2/5 = 275. 20. a^ + 2/^= V(«^ + V)y ^y = G. 21. a; - 2/ = 2, a:' - 2/3 = 98. 22. xy{x Jt y) = a, x^y^{p(? + y^) = b, 23. a;7/(a; + y) = 30, ar2/^(a:« + y^) = 9900. 24. 4ar* - ^xy = 18, 52/^ - 2xy = 8. r; 1 1 30 25. a;^' + 2/^ = 35, aj^ + y^ =— r-i- a;^y^ 26. a^ + 2/^ = 3 a;, a;^ + y^* = a;. o^r ^ 216 . . 77 27. a;?/ + 6 = , a; + ?/ + 4 ^ ^. a;?/ X -^ y 28. (a; + t^)^ + 2(a; - 2/)' = 3 (a; + 2/) (^ - 2/)) a;' + 2/" = 10. 29. x" + 10a;y + 2/' = V {^^ - 2r)> a.-" + ^2/' - a; + 132/. 30. a;* - 2a;V + 2/* = 1 + ixy, a?{x + 1)'+ 2/'(2/ + 1) = xy, 9ar - 2/^ - 117 _ Ba;^ -2r + 1 31. 3a; + 2/ - 9 = 3^ — 7/ — 6 4a;+2/ + l 312 ALGEBRA. „„ a; + 2 ^ y + 2 _ ., X + 2 y - 2 _ ,, 33 _ V^.(.-y )^3^ 1_1^^^ ^r^ ^ f ^ {x - y) ' y X xy U, y ^2 x/5 a,-^ + 2/ + 3 = 32 - 5aj^ \y £c / ^y aj / 35. ? - 2^ -3= - f.^?^,3a; + 2/ = 7. y X ar - 2/^ 36. (a; + y) xy = c (hx + ay), xy {bx + ay) = a^y^ + abc (x ■{■ y — c). 37. 2/* = XT (ay - hx), x^ = ax - by, 38. ^/tIl±l.= ^EIpJ,x{y+iy = 3Q(f+r,^). y sjx 39. a; = » ^ - 7/ + ;2; a; + ;2; i» + 2/ 40. a; + 2/ + 5? = 6, a;^/ + a;;^; + 2/^ = 11, xyz - 6. 41. y? - yz - ^,y^ - xz = 0, z^ — xy ~ 0, 42. xyz = a^{x + y) = b'^{y + ^) = 0^(0? + J^). •»2 7i2 ^2 43. x'^ + y^ + z'^ = ^^ = ^^ = % x^ 2/ ^ 44. aj + 2/ + ^ + ^ = 4a + 45, a;?/ + a;;^ + 032* + 2/^ + 2/^^ + ;2?2^ = 6a^ + 12 a5 +6^^, xyz + aJ2/26 + a;;^;!* + yzu = ia^ + 12a^6 + 12a6^ + ib^, xyzu = a* + 4 a^6 + 6 a^b^ + 4 a6^ + 5^. 45. aj^2/^ + a;?/^;^; + xyz = a, y^^ + xy^z + a;2/«^ = 6, y?^ + a;^2/^ + a;2/^^ = c. 4G. (a; + y^ + ;s^ = 1125, X + y + z = 15, xy - 24. 47. If aa;^ = by^ = cz\ and - + - + -=«, show tliat •^ ' a; 2/ ^ aa;^ + 6?/^ + c;s^ = (a^ + b^ + c^ f a\ 1>R0BLEMS tnODUCINCJ QUADRATIC EQUATIONS. 313 48. Given i?=l+r, P=^(l- R-'^ M = PR^ show that R - A(p - Tr) "^^^ CHAPTER 11. Problems Producing Quadratic Equations. 7. We shall now discuss one or two problems whose solu- tions depend upon quadratic equations. Ex. 1. A person raised his goods a certain rate per cent., and found that to bring them back to the original price he must lower them 3 J less per cent, than he had raised them. Find the original rise per cent. Let X = the original rise per cent., then -rTTr: . 100 = the fall per cent to bring them to the original price. Hence, by problem — X - TTTT^ = 3j, which solved, gives aj = 20 or - IGf. The value a? = 20 is alone applicable to the problem. Re- membering, however, the algebraical meaning of the negative sign, it is easy to see that the second value, x = — 16|, gives us the solution of the following problem : — A person lowered his goods a certain rate per cent., and found that to bring them back to the original price he must raise them 3 J more per cent, than he had lowered them. Find the original fall per cent. The above solution tells us that the fall requii^ed is IGj per cent. Had we worked the latter problem first, we should have obtained ic = 16§ or - 20, the valuer = - 20 indicating the solution of the former problem. 314 ALGEBEA. Ex. 2. Find a number siicli that when multiplied by its deficiency from 100 the product is 196. Let X — the number, then 100 - a; = its deficiency from 100. Hence, by problem — X (100 - x) = 196, or x"" - 100 i» + 196 = 0; from which, x = 2 or 98. Both these values will be found to be consistent with the conditions of the problem. Ex. 3. The number of men required to build a house is such that, when four times the number is subtracted from three times the square of the number, the result is 160. Find the number of men. Let X = the number of men, then, by problem — Za? - 4 a; = 160, from which a; = 8 or - 6f . The value aj = 8 is alone applicable to the problem as it stands. If, however, we may conceive of 2^ fractional number of men — and this we may easily do here by supposing a hoy's work to be equal to f of a man's — we find that the second result gives us the solution of the following problem : — The number of men required to build a house is such that when four times the number is added to three times the square of the number, the result is 160. Find the number. The answer, as above indicated, is 6 men and 1 boy, where a boy is worth f of a man. The student will find, however, that in some cases there is no obvious interpretation to the second result, owing occasion- ally to the fact that certain terms are used in the problem to which the results will not apply, and indeed that the algebraical expression of the conditions of the problem is more general than the language of the problem itself. Ex. IV. 1. Find a number whose square is equal to the product of two other numbers, one of which is less by 6 than the required number, and the other greater by 9 than twice that number. 1>R0BLEMS Producing quadratic equations. 315 2. When the numerator and denominator of a certain fraction are each increased by unity the fraction is increased by T Jo, and when they are each diminished by unity the fraction is diminished by -y\. Find the fraction. 3. The mean proportional between the excess of a certain number above 21, and its defect from 37, is 28. Find the number. 4. A number of articles, which were bought for £4, cost each 3 shillings more than half as many shillings as there were articles. Find the number of articles. 5. There is a square court-y^rd, such that if its lengtli be increased by 10 feet, and its breadth diminished by 20 feet, its area would be 5,104 feet. Required the side of the square. 6. If the number of shillings given for an article be added to the number of articles which can be bought at the same price for 18 shillings, the result is 11. Find the price. 7. Two travellers set out to meet each other from two places 180 miles distant; the first goes 3 miles an hour, and the second goes 1 mile more per hour than one-fourth of the number of hours before they meet. When will they meet? 8. A farmer bought a number of calves, sheep, and pigs, the number of calves being equal to that of the sheep and pigs together. For the calves he gave 64s. a head, and for the sheep twice as many shillings as there were sheep. He paid £153. 12s. for the calves and sheep together, and £36. 1 2s. for the pigs — a pig costing as much as a sheep and calf together. Find the cost of the sheep per head. 9. There are two squares, and an oblong whose sides are equal to those of the squares, and it is noted that three times the area of the first square exceeds foiu' times the area of the oblong by 3 square feet, while twice the area of the square, together with three times the area of the rectangle, is 36 square feet. Required the sides of the squares. 10. The sum of two quantities is equal to 6 times the square of their product, and the sum of their cubes is equal to 36 times the product of their fifth powers. Find the quantities. 11. The solid content of a rectangular parallelepiped is 60 cubic feet, and the total area of the side is 98 square feet, while the sum of the edges is 48 feet. Requii^ed the dimensions. 316 ALGEBRA. 12. The products of the number of units of length in the sides of a polygon of n sides, when taken n - \, together are respectively a^, a^, ag, &c., a„. Required the lengths of the 13. A, B, and C can together do a piece of work in a day, and C's rates of work is the product of the rates of A and B. Moreover, C is one-fifth as good a workman as A and B together. Find the respective times required for A, B, C to do a piece of work. 14. The compound interest of a certain sum of money for 3 years is a, and the third year's interest is h. Find the principal and the rate per cent. 15. A owes B £a due m months hence, and also £h due n months hence. Find the equated time, reckoning interest at 5 per cent, per annum. 16. Find three quantities such that the sum of any two is equal to the reciprocal of the third. 1 7. Find three magnitudes, when the quotients arising from dividing the products of every two by the other are respec- tively a, 5, c. 18. The sum of three quantities is 9, the sum of theu' pro- ducts, taken two and two together, is 23, and their continued product is 15. Show that the three quantities are the roots of the equation a^ - 9 a^ + 23 a - 15 = 0. CHAPTER III. Indices. 8. "We shall reserve the discussion of the complete theory of Indices for Vol. II. , confining ourselves here to a few simple cases, and giving a few examples involving fractional and negative exponents. 9. Fractional exponents. Def. — The numerator of a fractional exponent indicates the powder to which the quantity must be raised, and the INDICES. 317 denominator the root which must be taken of the power so obtained. Thus, al = 4/a^ a^ = >«, all = >« =a=; m and generally a" = >/«"'. The above definition is that which follows at once if wr assume the law proved in Art. 24, page 159, viz., a"* x a"* = a"* + " to be true, whatever be the value of 7n and n. Thus we have — («t\u m in m a") = a^ X a'* X a" to 9i factors Hence, taking the Tith root, Let X = (a^)'' = 1^ i^^y, by Def., Art. 9. ... of = {a^y = (^^}, or, raising each side to the qih power, we have — pr Hence, taking the {qs)i\i root, we have x = a «« . 11. To show tluit a» x 5~« = (a6)«. Now,a^ X b^ ='V»^ ^'n/^ = ;/^^^= V{«^ = (a6)n, by Def., Aii;. 9. And so, a;r -^ 6'» = (r)"* Ex. 1. Multiply together aH^c^ by a'^icT. Adding together the indices of like letters, we have — • ^ + i = -^, i + i = A. i + i = ih Hence, the recjuired product is aH^'^^c^, 318 ALGEBRA, Ex, 2. Multiply x + x-y'-^ + y by a; — x^y'-^ + y. X + x^y'^ + 2/ aj _ x^y^ + y u? + x%f^ + rrv/ Ex. 3. Divide x- yhy oja - y^. x^ - y^)x - 2/{i^^ + x^y^i + 2/^ a; - tc%^ ' 2 i cc^2/^ 2 1. 12 __ aj«2/^ - 2/ ^ 2 Ex. V. Eind the value of — 1. {a'f,a%{a-'')-\{aWK 2. {a + x)^^{a^ + 2 «aj + aj^)^, {p^ - aj^)4 (a^ + x^)^. Multiply together — 3. a^ - a^x-i + jc^ and a^ + a*a;a + xs, 4. aj^ - 2/^anda;~^ 4- y~^, 5. aj + a;^2/^ + y and aj"" - a;~^ 2/"^ + ^'' • Divide — 6. aJ^-hhy a — 5^, ai - Z>^ by a" - ifi, 7. a;^ - xy^ + aj% - 2/^ by a;^ + x^^V' 8. a + 6+c — 3 a^6^c'J by a^ + 6^ + c^* 9. a^ - a^J-y- - a^6 + 6^ by a^6^'- - 6"^ SURDS. 319 Show that— 10. (or'*" + ar^")"*" = x"^ + " ^ x^-'' + x"""^ [ ^. XI, =^ - 2aj2i>? .< a?2j' - x^t } T^x^P9, aJ^ - ft2 m 1 m a- + 6^ 12. (x^^ - a*^) ^ « - a^") = (aj'^^ a-) (a:'^+ a^) (aj'^'+ af) {^ ^ oTY Find the square roots of — 13. a + b + c + 2 {a^h^ + ah^ + ifici), 14. 4:xy^ — 12aj% + 17ar^2/5 - 12a;^2/^ + 4a^. 15. a^S-^ - 4:ah-i - 8a-^6^ + ia^'b + 8. Find the cube roots of — 16. a;* + 9a;V + 6aj^ - 99ar^ - 42 aj^ + 441 a;* - 343. 17. a'y-^ + 3aJ*y"-^ + SaP^j-^ + 1. 18. ab(l + 3a-h^ + 3a-J6* + a-^6) (aJ-^ - 3a^6-' + 3a^6-4 + 1). CHAPTER lY. « Surds, 12. A surd quantity is one in which the root indicated cannot be denoted without the use of a fractional index. Thus, the following quantities are surds : — 320 ALGEBRA. ai, («= + «^)i, (a^ + b^ + c^)», (^^y, j^ Since, from what has been explained in the last chapter, these quantities may be written thus- it follows that surds may be dealt with exactly as we deal with their equivalent expressions with fractional indices. It is evident that rational quantities may be put in the form of surds, and conversely, expressions which have the /orm of surds may sometimes be rational quantities. Thus, a^ = 4/(^ = ^^; and ^a^ + Sa'b + 3a&~V¥ = ^{a + bf= a + k 13. A mixed quantity may he expressed as a surd. Thus, 3 4^5"= v^3"^ ^5"- .5^3^ x 5 = .^1357 and so, x spy = slx"^ . '^y = llx^'y, 14. Conversely, a surd may he expressed as a mixed quantity, when the root of any factor can be obtained. Thus, ViS"^ = sfda^b^ x 2a = s/da^b\ >J¥a^ 3abj2a. And ^(a' + hjx'if = ^{d' + h'fi^f x xy' - ^{a^ + byxy*^^^' = (a" + hjxy^^, 15. Fractional surd expressions may he so expressed that the surd portion may he integral. The process is called rationalizing the denominator, and is worth special notice. Ex 1 /^ = \^~^^ x/21 V2r It is much easier to find approximately the value of \l2\, and divide the result by 7, than to find the values of ^3 and ^7, and divide the former by the latter. Ex. 2. Eeduce to its simplest form /J j-—^ — / xy _ Ixyjb — c) _ *Jxy (b - c) su?vDS. 321 4 Ex. 3. Find the arithmetical value of p. 2-^3 The denominator is the difference of two quantities, one of hich is a quadratic surd. Now, we know that (2 - S) (2 + ^3) = 2^ - ( ^3)^ = 4 - 3 =^ 1, and hence we see that by multiplying numerator and denominator by the sum of the quantities in the denominator we can obtain the denominator in a rational form. Thus — ^- = ^(2 + J 3_^ _ 4(2+ JZ) ' 2 - V3 (2 - n/3)'(2 + V3) 2= - ( Jzf 4 (2. 73) ^4(2 W3),,(,, ^3) = 4(2 + 1-73205) = 4(3-73205.) = 14-92820. 4 ^ 4( 4x/2"'- ZslWf ' 4 n/2' + 3 x/3"' (4 x/2"+ 3 JZ) {i J'2 - 3 Vs) = ^(W2'-3x/3) ^ 4(4V2~^3v/3) ^ w^ /^ , ^ j^. (4 x/2)2 - (3 x/3)2 32-27 ' ^ " ^' We shall now give an example when the surds are not quadratic. a Ex 4. Rationalize the denominator of — i ^i* x^ — y^' Since {x^f - {y^f'' is (Art. 29, page 175) divisible by ^ - y^f it follows that the rationalizing factor is their quotient, wliich is easily found. 16. Surds may be reduced to a common index. Ex. 1. Express ^a and ]^b as surds having a common index. 1 1 Since ^a = a% and ^b = b», it follows that, by reducing the fractional indices to a common denominator, the n m given surds become respectively a"»«, 5'"«, or ^^/a^, *v^6'^ 5 X na so. 322 ALGEBRA. Ex. 2. Reduce ij'^b and ^x^if" to a common index. The least common denominator of the fractional indices of the given surds is 4 x 3 or 12. Hence we proceed as follows : — When the student has had a little practice, the first two steps of each of the operations may be omitted. 17. Addition and subtraction of similar surds, Def. Similar surds are those which have the same irra- tional factors. Ex. 1. Eind the sum of V^ 5 slW, - 2 Jib. We have — Vl2 + 5 n/27 - 2 x/tS = n/22 X 3 + 5 x/32 X 3 - 2 ^5^ x 3 = 2V5"+5x3/s/3"2x 5^3 - 2 V3 + 15 V3 - 10 V3 = (2 + 15 - 10) V3 - 7 vs: Ex. 2. Simplify— ja'b + 'lab'' + h' __ jd'b - V a^ - 2a6 + 6'-^ N' "^~r 2a6*^ + 6« 2a6 + 6^ The given expression — _ a + 5 ,~ a - b ,~ /a + b a -- b\ .-. ~ a-^b ^^ - -^rvb ^^ = v«-:r5 - -^rrv'^^ - (^ + 6)^ - (g - Z>) ^ 4a& (a - 5) (a + ^>) ' "^ ^ - a^ - 62 V^ 18. Multiplication and division of surds. The following examples will best illustrate these opera- tions ; — SURDS, 323 Ex. 1, Multiply a slx^yz by h sjxifu. We hq,ve, a ijm^yz x h ijx'i^u — ah ijx^yz x xy^n = ah sJxSjSiz = ahx^y^ sjuz, Ex. 2. Multiply a aJF ■¥ cjdhj a — JM, Arranging as in the case of rational quantities, we have — a ^b + c \ld a - Jhd a' Jh + ac Jd^ — ah Jd — cd sjh [d^ - cd) Jh + a{c - b) Jd Ex. 3. Divide a Jhhjh J a, Wehave,^-^.^4^ = '^=T V^^^ When the divisor is a compound quantity it will generally be the best to express the surds as quantities with fractional indices, and proceed as in ordinary division. 19,T/ie square root of a rational quantity cannot be 2^(iTtly rational and partly irrational. If possible, let ^Ja ~ ni + sfb; then, squaring, a = n}? ■\- 2,m Jh ■{- h; or, 2 7n ^b = a - (m^ + b); or, V6 = ^ - f -^ % that is, an irrational quantity is equal to a rational quantity, which is absurd, 20. To Jind the square root of a binomial ^ one of xoliose terms is a quadratic surd. Let a + tjb be the binomial. Assume ^J a + jjb = »Jx + ^y, (1); then, squaring, a + \fb = x + y + 2sJ xy, (2). Equating the rational and irrational parts (Art, 19), we have — 324 ALGEBRA. ,« + 2/ ^ «• (3.), ^Tidi2i\/xy = hjh ov ixy -^ h (4.) From (3) and (4) we easily find a? = \{a -v sj dr" - h), and y ^ \ (ci - s] d^ ^ b). Hence, from ( 1 ), the square root required is — ViT^'+'V^^-^ + sj\ (a - ^ (V' - 6). Note. — It is evident that, unless (a^ - 5) is a perfect square, our result is more complicated than the original expression, and therefore the above method fails in that case. Ex. 1. Find the square root of 14 + ^'J^, Let Vl4 + 6V5 = six + ^y (1.) Squaring, then, 14 + 6\/5 = ^ + 2/ + SVxy. Hence, equating the rational and irrational parts— ^ a; + 2/ = 14 (2.), 2Vr^ - 6\^5or4a;?/ = 180 (3). From (2) and (3) we easily find x = 9, y ~ 5. Hence the square root required is V^ + >v/5 or 3 + VB. Ex. 2. Find the square root of 39 + V 14967 Let >\/39~T~7l4% = six + Vy. Squaring, &c., we have, aj + 3/ = 39 ; and 4cxy = 1496. From these equations we easily find x - 22, y = IT. Hence, the square root required isv22+ Vl7. 21. The square roots of quantities of this hind may often he found by insjnction, Ex. 1. Find the square root of 19 + 8 V3. We shall throw this expression into the form a- + 2 ah + 6^, which we know is a perfect square. Dividing the irrational term by 2, we have 4^3. Now all we have to do is to break this up into two such factors that the sum of their squares shall be 19. The factors are evidently 4 and V3. SURDS. 326 Thus, we have 19 + 8^3^= (4)^ + 2 (4) Vs + (sfsy - (4 +_ V3)^ TJie square root is therefore 4+^3. Ex. 2. Find the square root of 29 + 12^57 We have 29 + 12 ^5"= (3)^ + 2 (3)2 J 5' + (2 ^/5)^ - (3 + 2 sk)\ The square root is therefore 3 + 2 fjd, Ex. VI. Express with fractional indices — 1. j^, ^'^b\ il^y ^^\ xy sf^d y^^ ijab Jxy \/a'* Keduce to entire surds — 3. 3 x/3, 4 N/f, I n/IS, 3 ^T, 4. 4.2^9.3"^, 4. 2~T |(f)"*- 5. 3V«6,«7^^(« + ..)J^-^^, Keduce to a common index — 6. .^/2,v'3. 7. 4/2, ^37 8. 2 V2, 3 v^5. 9. V^ ^g: 10. {a + aj)^? \/« - a:. 11. ap ^^ hv 'i- Simplify — 12. 712, ^^48, 3 ^^, \ ^GTR 13. x/4a^ + \a% l^cW + h\ J%r^^. N 64 a - , /I 27o : 7;? 3"^^ 7 ., n/ ~9 ^"' "*■ 3 a; ' 'S (x - «f (re + a) " ^26 ALGEBRA. Find the value of — 16. Vi2 + ^48 - 2 Va; x/56 + ^/fgsC 17. J^- ^W^ + ^ 'H 18. ^27 «''* + « 6^ - v'8a^- + « + 3 ^^GT^. Multiply — 19. a + Jab + bhy Ja - Jb, a^ + b^hy sja - Jb, 20. (aj + y)h by (oj + y)i, a + b s/dhj cir - ab Jd + b'd. 22. a^- + b^ 4- c^ + d} by a^ - 6 J" + c^ - di Divide— 23. a? + xy + y'^hj X + x^yi + y. 24. a^ - 7/^ by cc^ + y^i Kationalize the denominators of— • n/3' V7' n'5- 3 ' 4 1 . 26. 27. 28. 2 + n/3 3V2 - 2>/3* ^5 3 2 V"3 ^ V3 + J2 1 + J2 + Vs' x/2 + V 3 + Jd' Js - V2* a 1 ^ & " a;^ - 2/^ /y2 + \/3' X + a;%^ + y* Find the square roots of — 29. 11 + 4 V7, 8 + 2 Jl5, 30 - 10 s/d'^ 30. 8 + 2 s/T2, 9 - 6 >/2, 20 - 10 s/3. RATIO. , 327 CHAPTER V. RATIO AND PROPORTION. Ratio. 22. The student is referred to Chapter II. of the Arith- metic Section of this work for definitions and observations which need not be repeated here. 23. A ratio of greater inequaliti/ is diminished, and a ratio of less inequality is increased, hy increasing tlie terms of the ratio hy the same quantity. Let a : 5 or ^ be the ratio, and let each of its terms be a ■\' m m a + m . increased by m. It will then become -i Now, 5 -, as (a + m)b% (h -{- m) a, + m b^ ^ ' ^ ' or, as ab + bm %.ab ->r am; or, as bm 5 a^ii, or as 6 5 a. Hence the ratio -7-, is increased when b ^ a, that is, when b it is a ratio of less inequality; and is diminished "When b -^i a, that is, when it is a ratio of greater inequality. CoR. It may be shown in the same way that — A ratio of greater* inequality is increased, and a ratio of less inequality is diminished, by diminishing the terms of the ratio by the same quantity, 24. When the difference between the antecedent and con- sequent is small compared with either, the ratio of the higher powers of the terms is found by doubling, trebling, ^^^^ r X - = ^ X - or - = -,, b d b c d c c d .\ a: c :: b : d (alternando). (2.) 1 V T- = 1 -f -y or - = , ^ 6 d a c .*. b : a :: d : c (invertendo). Also, by Art. 64, page 214, we have — (3.) a + b : b :: c -h d : d (componendo), (i.) a - b : b : : c - d : d (dividendo), (5.) a - b : a :: c - d : c (convertendo). (6.) a ■{■ b : a - b :: c + d : c - d {componendo and divklendo), 27. If a : 6 : : c: d and e : / : : g : h, we may compound the proportions. Thuswehave, = r (l),and>=^ (2). (l)x(2),then,«^=|. ov ae : b/:: eg : dh. PROPORTION. 329 And ill the same way we may show that, if the correspond- ing terms of any number of proportions be multiplied together, the products will be proportional. 28. If three quantities are in continued 2yroportion, the first has to the third the duplicate ratio of what it has to the second. Let a, h, c be the given quantities in continued propor- tion; then — a _ b 6 ~ c -._ a a b a a a^ ^ Hence, - or - x - = 7 X y = fr,, ' c c b b V ,\ a : c :: a^ \ W. And, similarly, if a, b, c, d are four quantities in continued projjortion, a.: d :: a^ : b\ that is — The first has to the foui-th the triplicate ratio of what it has to the second; and so on, for any number of quantities. 29. We shall now give one or two examples of problems in Proportion. Ex. 1. If a: b::c: d, prove that ^3 ^ ^, = (^-^p-^) • Let r =^ -, = x; .\ a = bxy and c = dx. Hence, .^Ll^ = Ml^^jf = Jf^. t^^en, a^..- (a + cy (bx + dxy (6 + df J a^ + c^ /a + cV Ex. 2. It a : b :: c : d, prove that j j--=^ — .. . ^^ (f' - f> Jac ^ \/bd Let -J- = ~j = x; ,'. a - bxf and c = dx, ct Hence ^^ + ^_^^ + ^_^+l _ V^^.a; + J ^bd 'a - b '^ bx - b "^ a; - 1 ~ Jb(f,x - Jbl _ fjbx . dx + Jbd __ is/ac + Jbd tjbx.dx - fjbd \fac - ^bd' 330 ALGEBRA. Or, it may be worked tliiis — Since -- = --, we have /t- — / - , Hence, by Art. 26— CO + h _ ijac + \Jhcl Ex. 3. If a t & :: c : c? :: e :/, show that L a _ rmoT + QIC'' + pe'^Y r J. ct C e ,^. ^''i- = d=7='' W- a' _ C _ e' _ , ■■■!/- d^ -/'-"=• Hence, a^ = h^oG^, .*. wa*" = mh^'x^ \ c** = dVf :. n(f ~ nd^'x'' V, and .'.by addition, e^ = fV, :. pe'' = pf^'x'' j ma^ + TIC** + pe^ = (mJ** + Qid'' + p/'')x''. .'. Equating (1) and (2), we have — h \mJf + nd"^ + pf'/ Ex. YII; 1. Compare the ratios a ■{■ h : a - hj and a^ + Ir : ctr - h\ 2. Which is the greater of the ratios a + h : 2 a, and ^h:a + hi 3. What quantity must be subtracted from the consequent of the ratio a : 6 in order to make it equal to the ratio c-.d'l PROPORTION. 331 4. Compound the mtios 1 - x^ :l + y, cc - a;?/- : 1 + ar, and 1 :x - aP, 5. There are two numbers in the ratio of 6 : 7, but if 10 be added to each they are in the ratio of 8 : 9. Find the numbers. /» G. In what cases is a; + — >- or < 5 ? X 7. If = = , show that a + h + c ^ 0, X — y y — ^ z " X 8. Find the value of x when the ratio x + 2 a:x + 2 5 is the duplicate ratio of2a; + a + c:2a; + 6 + c. 9. Find x when the ratio x - h : x + 2 a — 5 is the triplicate ratio oi x - a : x -v a - h. ■.^-r/»«^ + V 2/ + ^ X •{■ Z . - ,/•,/. 10. If , = 7— — = i show that each of the frac- a -\- + c c + a' 4 i X + y -h z ., X y z tions IS equal to V , and that - = » = -• •i -i Ti< ^ c e , , . . , let + mc + 7^c IL Kt = -7 = 7i , thdii each IS equivalent to ,,-" — 'ff~7~~f\ hellce, show that — a h c 2z + 2x - y ~ 2x + 2y ^ z~ 2y + 2z^x X 'H '^ ^^®^ 2a""r'2T^7 " 2 6 + 2 c ^1^ " 2 c + 2 c* - 6* 12. If w :6 :: c : c?, then a ■\- h \ c + cZ : : x/a'"^ + a6 + 6" : si 6^ + cc/ + Or, 13. Find a fourth proportional to the quantities — a; + 1 ^ ■\' X ■\- 1 (c* + 1 tc'^n' a;^ - a; + l' S^"^^' 14. Find c in terms of a and & when — (1.) a \ a \\ a - b : b - c. (2.) a :b ::a - b :b - c. (3.) a : c : : rt - b : b - c. 33!J ALGEBRA. 15. If ay b, c are in continued proportion, show that 7 , hf he are also in continued proportion. 16. If a : h : : c : dy then — 17. From a vessel containing a cubic inches of hydrogen gas, b cubic inches are withdrawn, the vessel being filled up with oxygen at the same pressure. Show that if this opera- tion be repeated oi times successively, the quantitv of hydro- , , ^ , (^ _ 5)n gen remaining in the vessel is — „_i ■ cubic inches, when re- cti duced to the original pressure. 18. If, in Ex. 34, page 225, (^j, a.,, %), (Sj, bi, 5y), and (cj, C2, Cg) are corresponding terms respectively, show that SECTION III. PLANE TRIGONOMETEY. CHAPTER I. MODES OF MEASURING ANGLES BY DEGREES AND GRADES. 1. We are able to determine geometrically a right angle, and it might therefore be taken as the unit of angular mea- surement. Practically, however, it is too large, and so we take a determinate part of a right angle as a standard. In England we divide a right angle into 90 equal parts, called degrees, and we further subdivide a degree into 60 equal parts, called minutes, and again a minute into 60 equal parts, called seconds. This is the English or sexagesimal method. In France the right angle is divided into 100 equal parts, called grades, a grade into a hundred equal pai-ts, called minutes, and a minute into 100 equal pajts, called seconds. This is the French or centesimal method, and its advantages are those of the metric system generally. The symbols °, ', ", are used to express English degrees, minutes, seconds respectively, and the symbols ", \ '\ to express French grades, minutes, seconds respectively. Conversion of English and French Units. 2. Let D = the number of degrees in an angle, and G = the number of gi-ades in the same angle ; then --expresses the angle in terms of a right angle; and so also does -. >^^• 334 PLANE TRIGONOMETRY. Hence, D )0 G D - 100 "'■ 9 - G 10- D = 10^ = G -,l« .(1). andG == ^D = D + |d (2). Hence the following rules : — 1. To convert grades into degrees. From the number of grades SUBTRACT -yV) and the remain- der is the number of degrees. 2. To convert degrees into grades. To the number of degrees add \j and the sum is the num- ber of grades. Ex. 1. Convert 13^ 18' 75^' into English measure. No. of grades = 13-1875 Subtract -j\ of this = 1 -31875 .-. No of degrees - 11-86875 60 52/-12500 ' * 60 t 7^^-500 Ans. ir 52' 7"-5. Ex. 2. Convert 18° 7' 30'' into French measure, No. of degrees = 18*125 Add I of this == 2-0138 .-. No. of grades = 20-1388 Ans. 20^ 13' 88''-8. 3. An angle may be conceived to be generated by the revolution of a line about a fixed point. Thus — Let OA be an initial line, and let a line, OP, starting from OA, revolve with O as centre, and take up succes- sively the positions OPi, OPg, OP3, OP4. MODES OF MEASURING ANGLES. 335 Now the magnitude of an angle may be measm^ed by the amount of turning 7'equired to generate it. When, there- fore, the revolving line reaches the position OB, we may con- ceive an angle to have been generated whose magnitude is two right angles. And, further, when the revolving line assumes the positions OP3, OP4, the angles AOP3, AOP4 (the letters being read in the direction of revolu- tion) are angles whose mag- nitudes are each greater than two right angles. Indeed, when the revolving line again reaches the position OP, we may conceive an angle to have been generated whose magni- tude is four right angles. Lastly, if the revolution of the line OP be continued, we may conceive of angles being generated to whose magnitude there is no limit. Ex. I. 1. Express 39' 22' 30" in French measure, and 13M5^ 75^' in English measure. 2. One of the angles at the base of an isosceles triangle is 50"*. Express the vertical angle in grades. 3. Divide an angle of n degrees into two such parts that the number of degrees in one part may be twice the number of grades in the other. 4. Two angles of . a triangle are respectively a°, h^, express the other angle in degrees and grades. 5. If f of a right angle be the unit of measurement, ex- press an angle which contains 22*5 degrees. 6. Show how to reduce English seconds to French seconds. 7. If the unit of measurement be 8°, what is the value of 10^. 8. If two of the angles of a triangle be expressed in grades, and the third in degrees, they are respectively as the num- bers 5, 15, 18. Find the angles. 9. What is the value in degrees and grades of an angle 336 PLANE TRIGONOMETRY. which is the result of the revohition of a lino 3i times round. 10. In what quadrants are the following angles found : — 145°, 96^, 327°, 272^, 272°. 11. If a** be taken as the unit of angular measurement, express an angle containing h^, 12. What is the unit of measurement when a expresses - of a right angle 1 CHAPTER II. THE GONIOMETRIC FUNCTIONS. 4. It was formerly usual in works on Trigonometry to give the following definitions : — Let a circle be described from centre A, / 1^ with radius AB supposed to be unity, then — (1.) The sine of an arc BC is the perpen- dicular from one extremity, C, of the arc upon the diameter passing through the other extremity B. Thus CS is the sine of the arc BC. (2. ) The cosine of an arc is the sine of the complement of the arc. Thus, since DC is the complement of BC, S'C is the COSINE of the arc BC. (3.) The tangent of an arc BC is a line drawn from one extremity, B, of the arc touching the circle, and terminated in the diameter which passes through the other extremity, C, of the arc. Thus, BT is the tangent of the arc BC. (4.) The cotangent of an arc is the tangent of the complement of the arc. Thus, DT/ is the cotangent of the arc BC. (5. ) The secant of an arc BC is a line drawn from the centre through one extremity, C, of the arc, and terminated in the tangent at the other extremity. Thus, AT is the secant of the arc BC. (6.) The cosecant of an arc is the secant of the complement of the arc. Thus, AT ' is the cosecant of the arc BC, TRIGONOMETRICAL RATIOS. 337 (7.) The versed sine is that portion of the radius upon which the sine falls, which la included between the sine and the extremity of the arc. Thus, SB is the versed sine of the arc BC. (8.) The coversed sine is the versed sine of the complement of the arc. Thus, S'D w the coversed sine of the arc BC, (9. ) The suversed sine is the versed sine of the supplement of the arc. Thus, B ^S 25 the suversed sine of the arc BC. Representing the arc BC by A, it is usual to write the above func- tions thus : — Sin A, cos A, tan A, cot A, sec A, cosec A, vers A, covers A, suvers A. By mere inspection, the student will see that the following rela- tions hold : — (1.) Sin A = CS = ?^ r= ^^- = :^ = L_. 1 AS AT' cosec A (2.) CosA = S'C = -^ = A^. : ^ 1 AC (3.)TanA=BT = -^- = |^=--^^ = j^. (4.) Sin^ A + cos2 A = CS^ + S/C* = CS« + AS^ = AC^ = 1. (5.) Sec'^ A = AT2 = AB2 + BT» = 1 + taa'^ A. (6.) Cosec^ A = AT/2 = AD^ + DT/^ = i + cot^ A. / .^^rv A * -orp BT CS CS sin A (7.) Tan A = BT--r5=-ro= "oTTt = r* ^ ' AB AS S'C cos A (8.) Vers A = SB = AB - AS = AB - S'C = 1 - cos A. (9.) Covers A = S'D = AD - AS' = AD - CS = 1 - sin A. (10.) Suvers A = B'S = B'A + AS = 1 + cos A. It is more convenient, however, to define the sine, cosine^ &c., as in the next article, according to which definitions they are commonly called Trigonometrical Ratios. The student will see that if the above definitions be so far modified that, instead of the lines them- selves, the goniometric functions be taken as the ratios which the lines respectively bear to the radius, they are included in the defini- tions of the next article. Trigonometrical Ratios. 5. Let BAG be any angle, which we may denote by A, and P any point iu the line AC, Praw PM perpndicul£«r to AB. 5 Y 338 PLANE TRIGONOMETRY, Then (1.) Sin A == perpendicular ^ PM (2.) Cos A - (3,) Tan A = base Kyp- perpendicular AM AP* base a ) Cot A = ^a.se ^ '' perpendicular B PM AM* AM PM* (5.) Sec A = liyp- AP AM* ^ *^ perpendicular PM (7.) The versed sine is the remainder after subtract- ing the cosine from unity, or — vers A = 1 - cos A. (8.) The cover sed sine is the remainder after subtract- ing the sine from unity, or — covers A = 1 — sin A. (9.) The suversed sine is the sum of the cosine and unity, or — suvers A = 1 + cos A. The last three are not much used in practice. 6. Comparing (1.) and (6.), (2.) and (5.), (3.) and (4.), of the last article, we see at once that the sine and cosecant^ the cosine and secant, and the tangent and cotangent, are respectively each the reciprocal of the other. TRIGONOMETRICAL RATIOS. 339 We therefore have — 1 1 (1.) Sin A = .icosec A cosecA sin A (2.) Cos A = -i-., sec A = -i-r. ^ ' sec A cos A (3.) Tan A = - -~-r-, cot A = r^ ^ ^ cot A tan A Further — (4.) Sin^ A -f cos^ A = >^j. + ^^ = 5^, _ AF - ^ps - A- Hence also, transposing and taking the square root — (5.) Sin A = Jl - cos^ A. (6.) Cos A = VI - sin^ a: And again — (7.) Sec^A =-^-^,= AM- ^^ ^ AM- ^ "^ *^^'^- .ov ^ .A AP- PM2 + AM- ^ AM- (8.) Cosec- A = pjj5 = pjj, = ^ -^ P5P = 1 + cot- A. /nv m A I'M PM AM . . . (9.) Tan A = -^ = ;^ -^ -^^ = sin A ^ cos A ^ Bin A ~ cos A' /..xr. A AM AM PM (10.) Cot A = pjj = aF "^ AP " ^^'"^ "^ ^ ^^^ ^ _ cos A "" sin A* The student must make himself thoroughly master o£ the results in this article. 340 PLANE TRIGONOMETRY. 7. To express the trigonometrical ratios in terms of the sine, (1.) Cos A - Jl - sin^ A, by Art. 6 (6.) (2.) Tan A = J^, by Art. 6 (9.), sin A Jl - sin^ A (3.) Cot A = ^?^^, by Art. 6 (10.), ^ ^ sm A a/I - sin''* A sin A 1 cos A' (4.)SecA== -f., by Art. 6 (2.), Vl -sin^A (5.) Cosec A = -J--r, by Art. 6 (1.) ' ^ ^ sm A ^ ' Ex. If sin A = I, find the othet trigonometrical ratios. We have, cos A = n/1 - (|)^ = \/l - A = h . sin A , . ^ , tan A = -^ T = I V I = |. cos A ^ * 8. To express the trigonometrical ratios in terms of the cotangent. (1.) Sin A = — i- = - -J- , by Art. 6 (8.) cosec A vl + cor A (2.) Cos A = -— - = \ , by Art. 6 (7.) sec A VI + tan^ A 1 cot A J 1 VcoFaTi "^ cot'-^ A (^•)T'-A.,-^- TRIGONOMETRICAL RATIOS. 341 (4.) Sec A = -^- = 1 V--p^2L£=, by (2.) above, ^ ^ cos A J^^^^AT+^I ^ ^ ^ _ n/coj" a + 1 ~ cot A (5.) Cosec A = VI + cot^ A, by Art. 6 (8.) And in the same way the trigonometrical ratios may be expressed in terms of any one of them. Ex. II. 1. Given sin A = ^f, find the other trigonometrical ratios. 2. Given tan A = ||, find the remaining trigonometrical ratios. 3. If cot A = a, show that sin A = 4. If vers A = 6, then tan A = n/1 + a^ x/2 6 - 6^ 1-6 5. Construct by scale and compass an angle (1.) whose cosine is f ; (2.) whose tangent is |; (3.) whose secant is J2; (4.) whose cotangent is 2 + Jsi Prove the following identities : — 6. (Sin A + cos A)^ + (sin A - cos A)^ = 2. 7. Sec^ A + cos^ B . cosec^ B = cosec- B + sin^ A . sec^ A. 8. Sec^ A . cosec^ A = sec^ A + cosec^ A. 9. Sec A . cosec A = tan A + cot A. 10. Sin^A - cos^A = (sin^ A - cos^A) (sin* A + cos* A). , ^ Sec A + tan A _ cosec B - cot B Cosec B + cot B sec A — tan A sin* A + sin^Acos'A + cos* A 12, 1 + sin A cos A = = : — r r • 1 - sin A cos A 13. (.rcose + 7/ sine) (a: sine + 2/ cose) - (a; cose - ysine) (x sin — 2/ cos e) = 2 xij. 342 PLANE TRIGONOMETRY. 14. (a sin cos «/> + r cos 6 cos (/>) (5 sin 6 sin + r sin cos (p) — (b sin (jos<^ - r sin sin ) (a sin0 sin <^ + r cos sin <^) = rsin0(rcos0 + c&sin0). 15. If ic = Q* sin cos <^, 2/ = r sin sin y z - r cos 0, show that x^ + y^ + z^ ^ 7^. IG. If (t = b cos C + c cos B, & = tt cos C c = rt cos B + 6 cos A, show that a^ + b^ - c^ ^ 17. Given sin^A + 3 sin A = ^, find sin A. 18. Given cos^A — sin A = ^^^, find cos A. + c cos A, : 2a6cosC. 19. Solve sin A — cos A = V2' for sin A. 20. Find tan A, when tan A + 1 = sf^, sec A. 21. Given a cos A = 5 sin A + (x, find cot A. 22. Given tan^ A - 7 tan A + 6 = 0, find tan A. 23. Show that \l\ + 2 sin A cos A + ij\ — 2 sin A cos A = 2 cos A or 2 sin A, according as A is between 0° and 45°, or between 45° and 90°. 24. Given ??i sin^ A + Tzsin^B = a cos- A, m cos^ A + 7Z cos^ B = 5 sin^ A, find sin A and sin B. CHAPTEB III. CONTRARIETY OF SIGNS. — CHANGES OF MAGNITUDE AND SIGN OF THE TRIGONOMETRICAL RATIOS THROUGH THE FOUR QUADRANTS. 9. We have explained at some length the meaning and use of the signs + and - in algebra. They have a similar interpretation in trigonometry. 1. Lines. — Draw the horizontal ^ line A A, and draw BB ' at right angles, meeting it in O. Then considering O as ongin, (1.) All lines drawn to the right parallel to A A are called ^^osiVive, CONTRARIETY OP SIGNS. 343 and all lines drawn to the left parallel to A'A are called negative, (2.) All lines drawn upwards parallel to B'B are called positive^ and all lines drawn dovmwards parallel to B 'B are called negative. (3.) Lines drawn in every other dii*ection are considered positive, as is therefore the revolving line by which angles may be conceived to be gen- erated. 2. Angles. — A similar con- vention is made for angles. Let OA be an initial line, and let a revolving line about the centre O take up the positions OP and OP/. Then— (1.) That direction of revolution is considered positive which is contrary to that of the hands of a watch, and the angle generated is a ptositive angle. (The positive direction is then upvjards,) Thus, AOP is a positive angle. (2.) The negative direction of revolution is the same as that of the hands of a watch, and the angle thus generated is a negative angle. (The negative direction is then doumwards.) Thus, AOF is a ne- gative angle. p^ Hence, if the angles AOP and AOP^ be of the same magni- tude, and We have — ZAOP = thenZAOP/- - 10. We will examine the metrical ratios for angles greater than a right angle, and for negative angles. 344 PLANE TRIGONOMETRY. Let OP^, OP2, OPg, OP4 represent the position of tlie re-, volving line at any period of revolution in the several quadrants respectively, And let PiISTj, PgNg, Ps^g, P^IST^ be the respective per- pendiculars from the end of the revolving line upon the initial line. Then PjN^^, T^'^^y ^3^3? ^4^4 ^^'® respectively the per- pendiculars corresponding to the angles generated. Also, ONj, ONg, ON3, ON4 are respectively the bases of the right-angled triangles with respect to the angles in question. We have then in the second quadrant — Sin AOP, = M2, cos AOP2 = ^^, tan AOPo, = ?2^2 &c. It IS therefore evident that the relations between the trigo- nometrical ratios, which were proved to exist in Art. 7, also hold for angles in the second quadrant — that is, angles between 90° and 180°. And in the same way we may show that they hold for angles in the third, fourth, or any quadrant. And again, if we suppose the line to revolve in a oiegative direction, and take the position OP', we shall have P/N' the perpendicular corresponding to the negative angle AOP ', and ON ' the base. Hence, sin AOP' = ^, cos AOP/ = '±^^, tanAOF = ^,&c. And the relations proved in Art. 7 may be also similarly proved to exist here. Hence the relations proved in Art. 7 hold for any angles whatever. Changes of Magnitude and Sign of the Trigonometrical Ratios. 11. Let OP^, OP2, OP3, OP^ be positions of the revolving line in the several quadrants respectively; PiNp P2N2, CHANGES OF MAGNITUDE AND SIGN. 345 P3N3, "P^^^y the respective perpendiculars; and ON^, ONg, ON3, ON4, the bases of the corresponding right-angled triangles. Then— (1.) In the first quadrant — P N ON* SinAOP,= ^^^^^^^ tanAOPi = "o?^'^^^- At the commencement of the motion of the revolving line, the angle AOP^ = 0° ; Also, the perpendicular PjN^ = 0, And the base ON = OP^. Hence, we have — *SinO» = ^.0,cosO» = ^; = l, tan 0» = ^, = 0. As the revolving line moves from OA towards OB, P^N^ increases and ON^ diminishes ; and when it amves at OB, wehavePjN^ = OPj,andON"j = 0. But the angle generated is now a right angle. Hence we have — Sin90»=^?i = l,cos90° = ~=h OP ^ tan 90° = -^ - 00. Hence, as the angle increases from 0° to 90° — The sine changes in magnitude from to 1 and is + . Tlie cosine changes in magnitude from 1 to and is + . The tangent changes in magnitude from to 00 and is + . (2.) In the second quadrant — Here the perpe7idicular PgN^ is + , and the base ONo is - . * The student ought properly to look upon the values 0, 1, here obtained as the limiting values of the sinc^ cosincy and tangent respec- tively, when the angle is indefinitely diminished. 346 PLANE TRIGONOMETRY. Hence the sme during the second quadrant is + , the cosine is - , and the tangent is - . Again, as the revolving line moves from OB to OA', the perpendicular "^^^ diminishes until it becomes zero. Also, the base ONg increases in magnitude, until it finally coincides with OA', and .*. = — OPg. But the angle now described is 180°. Hence we have — Sin 180° = ^^ = 0> cos 180° = - ^-? = _ 1, tan 180° = - -^ = 0, &c. Hence in the second quadrant — The sine changes in magnitude from 1 to 0, and is positive. The cosine changes in magnitude from to 1, and is negative. The tangent changes in magnitude from oo to 0, and is negative. And in the same way may we trace the changes of magni- tude and sign in the third and fourth quadrants. Thus we shall find — (3.) In the third quadrant — The sine changes in magnitude from to 1, and is negative. The cosine changes in magnitude from 1 to 0, and is negative. The tangent changes in magnitude from to oo, and i^ positive. (4.) In the fourth quadrant — The sine changes in magnitude from 1 to 0, and is negative. The cosine changes in magnitude from to 1, and is positive. The tangent changes in magnitude from oo to 0, and is negative. Moreover, as the cosecant, secant, and cotangent are respectively the reciprocals of the sine, cosine, and tangent, it follows that their signs will follow respectively the latter, and that their magnitudes will be their reciprocals. CHAPTER TV. TRIGONOMETRICAL RATIOS CONTINUED. ARITHMETICAL VALUES OF THE TRIGONOMETRICAL RATIOS OF 30°, 45°, GO'*, &C. 12. To prove tkit sin A = cos (90' — A ), and that COS A = sin (90' - A), TRIGONOMETRICAL RATIOS. 347 Using the same figure as in Art. 5, we have — PM Sin A = — -— = cos APM. AP But Z APM = 90° - A, .-. SinA = cos(90« - A), and similarly — cos A =: sin (QO"* - A), tan A = cot (90« - A), cot A = tan (90^ - A), sec A = cosec (90" — A), cosec A = sec (90** - A). 13. Ratios of 45^ In the last figure, suppose Z PAM = 45% then also Z APM = 90^ - 45^^ = 45°. And hence Z PAM = Z APM, and .-.PM = AM (Euc. L, 6). Hence, also — AP or VaM- + P5F = x/2AM-or JTm:\ .-. AP = AM sj2ov PM n/27 Hence we have — PM 1 ... . = cos 4o , by Sin45» - ^^^^^ = AP' = PM^2 - 7f Art. 8. Tan 45°= tan PAM = Sec 45° = sec PAM = PM AM AP PM PM AMsJY AM = 1 = cot 45^ by Art. 8 = sf2r= cosec 45° AM by Art. 8. 14. Ratios of ^(f QXidi^O\ In the same diagram, suppose Z PAM = 30", then Z APM = 90'* - 30^ - 60°. Hence, if we conceive another triangle equal in every respect to APM to be described on the other side of AM, the whole would form an equilateral triangle whose side is AP. Hence, PM = ^^AP. 348 PLANE TRIGONOMETRY. Now AM = av/aF - PM^, .-. AM = N/AF^-^ilJ)^ J'S AP. We hence have — PTVr 1 AP Sin 30' ^ sinPAM = ^f = ^ = l= cos 60; by Art. 8. Cos 30° = cos PAM = AM AP ^AP , 2 ^^ _ x/3 Art. 8. AP = sin 60", by- Tan 30' = PM iAP AM - J3 AP JS ^ = 1 V3 = cot 60°, by Art. 8. 15. To show that sin (180" - A) = sin A, cos (180' - A) = - cos A, Let ZAOPi = A, And let the revol- ving line describe an angle AOP2 = 180^^ "TJT ¥7^ - A; Then ZA'0P2 = 180° - (180° - A) = A; Hence, Z AOP^ = A^OPg. Hence, also (Euc. I., 26), if P^N^, P2N0 be drawn perpen- dicular to AA^ P^N^ - PgNg, ON2 - - ON^, We have therefore — Sin (180° - A) = sin AOPg = ?,& = -?i^i = sin A. ^ ^ ^ OP2 OP^ Cos (180'' - A) - COSAOP2 ON, ON, . -opr-T)p; = -^^^^- And similarly — Tan (180° - A) = - tan A, cot (180° - A) = - cot A. Sec (180° -. A) = - sec A, cosec (180^^ - A) = cosec A. TRIGONOMETRICAL RATIOS. 349 16. To show that sin (180° + A) = - sin A, cos (180^ + A) = - cos A, Let AOPj = A, And let the revolving line take a position such that P1P3 is a straight line. Then, evidently, Z AOP3 = 180° + A. Then, as in last Article — P N - - P N 0N3 Hence — ONp ON. ON"i = - cos A. Sin (180^ + A) = sinAOPg = Cos (180^ + A) = cos AOP3 - ^p-- -^p- And similarly — Tan (180° + A) = tan A, cot (180° + A) = cot A. Sec (180° + A) = - sec A, cosec (180° + A) = - cosec A. 17. To show tlmt sin ( - A) = — sin A, and cos ( - A) = cos A, Let Z AOP = A, And let the revolving line describe an angle AOP' = -A. Then evidently, if PNP ' be drawn perpendicular to OA, we have (Euc. L, 26) P^N = - PK Hence — Sin (- A) = sin AOP' = P_^ OP' PN "OP = - sin A , Cos (- A) = cosAOP' = Qp7= Qp = cos A. And similarly — Tan ( - A) = - tan A, cot ( - A) = — cot A, Seg ( = A) = sec A, cosec (— A) = - cosec Ai 350 PLANE TRIGONOMETRY. Althougli the results of Arts. 12, 15, 16, 17 have been proved from diagrams where A is less than a right angle, the student will have no diffi- '1^ culty, if he has understood the proofs, in deducing the same results for angles of any magnitude whatever. 18. To show that tan - _ 1 - cos A sin A ' LetZAOC = A; Bisect it by the straight line OB, so that Z AOB = ^^ ^ and draw CD perpendicular to OA, meeting OB in E. (1). OD ED Then tan ^ = tan EOD - ?~. Now(Euc.YI.,2),^ - ?5:, and .• 00 EC ED ^ CD OD 00 + OD OC - OD 00 - 00 cos A CD' ^ CD(OC -- OD) 00* - OD'^ OC + OD = ^D(QC) - OD) CD- or CD OC sin A 1 - cos A sin A Q.KD, CoR 1. Hence, squaring — m 2 j^ _ (1 - cos A)' _ (1 — cos A)' _ 1 - cos A 2 sin^^S ~ 1 - cos^ A 1 + cos A' 1 - tan^ -o" A 2 cos A .-. Art. 64, page 215, 1 + tan^ -^ 1 •. Cos A = A 1 + tan* 2 .(1). TRIGONOMETRICAL RATIOS. 351 I ^, ») A. o A. • o -A. cos-^^ COS--— - sm---. Z 2i 2i sin^ -^ COS- ^ + sm^ -^ COS- •— = cos--- -sm-- (2.) = 2cos=~ - 1 (3.) = 2 (l - sin= ^) - 1 = 1 - 2sm= |-(4.) 19, To find the irlgonomeirical ratios of 15°, 75°, 120°, 135°, 150°. (1.) Batios 0/120". _ Sin 120° = sin (180° - 120°) = sin 60° = ^-|, Cos 120° --= - cos (180° - 120°) = - cos 60° = - I, Tan 120" = - tan(180° - 120°) = - tan60° = - V3,&c. (2.) Ratios of \^^\ Sin 150° = sin (180° - 150°) = sin 30° =i, Cos 150° rr. - cos (180° - 150°) =: - cos 30° = - -^, Tan 150° = - ten (180° - 150°) = - ton 30 = - -L V3' &C. Sin 15° = AzJ cos 15° = ^^^-^, &c. 352 PLANE TRIGONOMETRY. (3.) Ratios of 15°. By last Art., tan -~ = — ~ — ; put A = 30°, or -- = 15°, '^ 2 sm A __ 2 1 - yr XI, x.« iRO 1 - COS 30° 2 o /Q then tan 15 = -, — — -._ ^ = 2 - J3. sm 30 1 From this result we easily get, Art. 8, - 1 -— - cos (4.) Ratios of 76\ __ We have, sin 75° = sin (90° - 15°) = cos 15° = -^^^^-^, cos 75° = sin (90° - W) = sin 15° = ^^ 7 ^ 2^2 tan 75° ^ tan (90'' - 15°) = cot 15^ - ^—r, 2 ■— V 3 = 2 + n/3^ &c. (5.) i?a^ios 0/135°. We have, sinl35'= sin(180''- 135°) = sin45°= -^y v2 cos 135° = -008(180° - 135°) = - cos 45" JL ~ " a/2' tan 135° = - tan"(180° - 135°) = - tan 45' = — 1, &c. Ex. III. 1. Define a negative angle, and show that tan (—A) = - tan A, when A lies between - 90° and - 180°. 2. Trace the changes of sign of sin A . cos A through the four quadrants. 3. Trace the changes of sign of cos A + sin A, and of • PQS A - gi» A, as A gfenges from - 45** to 315°, TRIGONOMETRICAL RATIOS. 353 4. Assuming generally that cos 2 A = cos^ A - sin' A, trace the changes of sign of cos 2 A as A changes from - 45^ to 315^ 5. Write down the sines of 210^ 165^ 240°, - 120^ 6. Show that sin (90° + A) = cos A, and cos (90° + A) = - sin A, for any value of A from 0"" to 180°. A — 7. Assuming generally that 2 cos' -^c- = 1 + cos A, and r^- ..... ^ . , ^ A. "A - 2Bin^ — = 1 - cos A, show that V^cos o = - \/l + cos A and \f2'sm -^ = — Vl - cos A, when A lies between 360" and 540^ A 8. Given cos A = 1 - 2 sin' -,, show that sin A ^ . A A = 2 sin -^ cos-^ • j^ 0, Hence show that 2 cos i^ = - Ay 1 + sin A - Vl - sin A, when ^ lies between 135° and 225° Solve the following equations : — 10. Cos' A + f cos A = T«^. 11. Tan + 5cot0 = 6. 2 12. Sin A + sec A = —j^ + f 13. 2cos2A = 3 sin A. 14. Sin (A + B) = cos (A - B) = ^/i!- 15. Tan' A = 2 sin' A. 16. Sin (3 A + 75°) = cos (2 A - 15°). 5 17. Sec + cos = ;5~7r.- tan o. 18. Tan + cot = 4. 5 z 354 PLANE TKIGONOMETBY. CHAPTEE Y. LOGARITHMS. 20. Dep. — The logarithm of a number to a given Lase is the index of the power to which the base must be raised to obtain the number. Thus, we may obtain the numbers 1, 10, 100, 1,000, 10,000, the fifth figm-e. We then take the first three figures of the column headed 0, and affix the four figures of the column headed 6 in the horizontal line of the fii-st four figures of the number. Thus, log 79936 - 4-9027424 log -079927 = 2-9026935. It will be seen from the portion of the logarithmic table above extracted, that when the first three figures of the logarithm — viz., 902 — have been once printed, they are not 560 . PLANE TRIGONOMETRY. repeated, but must be understood to belong to every four, figures in each column, until they are superseded by higher figures, as 903. When, however, this change is intended to be made at any place not at the commencement of a horizontal row, the first of the four figures corresponding to the change is usually printed either in different type, or, as above, with a bar over it. Thus we have above 0031, indicating that from this point we must prefix 903 instead of 902. Thus, log 79-986 = 1-9030140, log -0079987 - 3-9030194. 26. To find the logarithm of a number not contained in the tables. Ex. Find the logarithm of 799-1635. Since'* the mantissa of the number 79916-35 is the same as the mantissa of the given number, and that the first five figures are contained in the tables, we may proceed as follows — (1.) Take out from the tables the mantissa corresponding to the number 79916. This is -9026337. (2.) Take out the mantissa of the next higher number in the tables— viz., 79917. This is -9026392. (3.) Find the difference between these mantissas. This is called the tabular difference, being the difference of the mantissse for a difference of unity in the numbers. We find tab. diff: -= -0000055, which we call D. (4.) Then assuming that small differences in numbers are proportional to the differences of the corresponding logarithms, we find the difference for -35 = -35 x -0000055 = -0000019, retaining only 7 figures. This is often called d. (5.) Now adding this value of d to the mantissa for the number 79916, we get the mantissa corresponding to the number 79916-35. (6.) Lastly, prefix to this mantissa the proper charac- teristic. The whole operation may stand thus — • M. of log 79916 = -9026337.' (1). M. of log 79917 = -9026392 Tabular difference or D" = -0000055 V -^ Thus log 79916*35 ^ log (100 x 799 '1635) = 2 -flog 799-1635. THE USE OF TABLES. 361 Hence, difference for '35 or cZ = -35 X -0000055 =-0000019 (2). Hence, adding (1.) and (2.) — M. of log 79916-35 = -9026356. .-. log 799-1635 = 2-9026356. or better thus, omitting the useless ciphers — M. of log 79916 = -9026337 M. of log 79917 = -9026392 .-. D = 55 Hence, cZ = -35 x 55 = 19 .-. M. of log 79916-35 = -9026356, as before. In the next article we shall show how the required dif- ference may be obtained by inspection from the tables. 27. Proportional parts. We saw in the example just worked that the tab. diff. (omitting the useless ciphers) is 55, and if we examine the table in Art. 25, we shall find the difference between the mantissse of any two consecutive numbers there to be 54 or 55 — generally 54. The number 54 is therefore placed in a separate column at the right of the table, and headed D. The student will understand that the tab. diflf. changes from time to time, and is not always 54 or 55. Now assuming as in (4.) of the last article, we have — Diff. for •1 = 54 X •1=5 Diff for -6 = 54 X •6 = 32 •2 = 54 X •2 = 11 „ -7 = 54 X •7 = 38 •3 = 54 X •3 = 16 „ -8 = 54 X •8 = 43 •4 = 54 X •4 = 22 . „ '9 = 54 X •9 = 49 •5 = 54 X •5 = 27 We find therefore the numbers 5, 11, 16, 22, 27, 32, 38, 43, 49 placed in a horizontal row at the bottom marked P, in the columns respectively headed 1, 2, 3, 4, &c. Hence, if we require the difference for (say) -7, we take out the number 38 from the horizontal row marked P, in- stead of being at the trouble to find it by actual computation. The following example will illustrate how we proceed when we require the diflerenco for a decimal containing more than one decimal figure. No explanation is needed. 362 PLANE TRIGONOMETRY. Ex. Find log 7994 -3726- M. of log 79943 - -9027804 Diff. for 7 = 38! „ 2 = 11 6 = ^ .-. M. of log 79943726 = -9027843 ^ Hence, log 7994-3726 = 3-9027843. 28. Having given the logarithm of a number to find the number. After the explanations of Art. 26, the method of working the following examples will be easily understood : — Ex. 1. Eind the number whose logarithm is 1-9030173. Taking from the tables the mantissoe next above and below, we have — •9030194 = M. of log 79987 •9030140 = M. of log 79986 (1). 54 = D. Again, -90301 73 = M. of log N (2). Hence, subtracting ( 1 ) from ( ^ ) — 33 - (/, the difference between the logarithms of the re- quired number and the next lower. 33 Now -— - — '61, the difference between the next lower 54 number and the required number. Hence^-9030173 - M. of log 79986-61 ; .-. 1-9030173 = log -7998661 ; .'. '7998661 is the number required. Ex. 2.^^ Find the value of lii^^f^J^. (1-32756)* We have — log N = 31og 1-023 + i-log -00123 - 4 log 1-32756. jS^ow, 3 log 1-023 = 3 X -0098756 = -0276268 1 log -00123 = \ (T3-0899051) = \ (4 + 1-0899051) - ^ T-2724763 * The logarithms used in this example are taken from the tables. TRIGONOMETRICAL TABLES. 363 /. adding, 3 log. 1-023 + i log -00123 =^3001031 Again, M. of log 13275 - -1230345 and difi: for -6 =- 196 .-. M. of log 13275-6 = -1230541 .-. 4 log 1-32756 = 4 X -1230541 = -4922164 Then, subtracting, log N" = 2-5078867 Hence we have, -5078867 = M. of log N, and -5078828 - M. of log 32202 ; 39 = d, also 135 = D, 39 and ^ = -29. .-. -5078867 = M. of log 32202-29; .-. 2-5078867 = log -03220229. Hence -03220229 is the number requii-ed. Trigonometrical Tables. 29. We use trigonometrical tables much in the same way as we do tables of ordinary logarithms of numbers. Tables have been formed of natural sines, cosines, &c., and also of logarithmic sines, cosines, &c. It is with the latter only we shall now deal, though many of our remarks apply equally to the former. As the values of the natural sines and cosines of all angles between 0° and 90° are (Art. 11) less than unity, it follows (Art. 21) that their logarithms are negative. To avoid, however, printing them in a negative form, and for other reasons, it is usual to add 10 to theii- real value, and hence in using them we must allow for this. The same thing is also done in the case of logaiithmic tangents, co- tangents, secants, and cosecants. We generally express the true logarithmic sine by log sin, and the tabular logarithmic sine by L sin. Hence, we have, log sin A - L sin A - 10, log cos A = L cos A - 10, kc. It muse be remembered in using the tables that, although (Art. 11) the sine, secant, and tangent of an angle increase as 364 Plane trigonometry. the angle increases from 0° to 90°, yet the cosine, cosecant, and cotangent dimi7iish as the angle increases. Hence, when any angle is not exactly contained in the tables, we must add the difference in the case of a sine, secant, or tangent; but subtract it in the case of a cosine, cosecant, or cotangent. And, conversely, when the given logarithm is not con- tained exactly in the tables, we must in the case of the sine, secant, or tangent take out the next lower tabular logarithm as corresponding to the angle next lower; but in the case of a cosine, cosecant, or cotangent, we must take out the next higher tabular logarithm as corresponding to the angle next lower in the tables. We shall assume that small differences in the angles are proportional to the corresponding differences of the logarithmic trigonometrical ratios Ex. 1. Find L sin 56° 28' 2r. Referring to tables, we have — Lsin56°28' Tab. diff. for 60'' or D = 836 24 60 ' Lsin56'28'24'' diff for 24'' or c^ = 836 = 9-9209393 334 Ex. 2. Find L cos 29" 31' 28': Now L cos 29° 31' Tab. diff for 60'' or D = - 716 28 9-9209727 = 9-9396253 diff for 28" . - ^^x716 = - 334 = 9-9395919 .\ Lcos29'31'28'' Ex. 3. Find the angle A, when L tan A = 9-8658585 We have 9-8658585 ^ L tan A. JSText lower, 9-8657702 = L tan 36° 17', 883 — difference or d, Also, 2648 = tab. diff. for 60" = D, And .-^ X 60" = -^ X 60" = 20". D 2648 Hence, 9-8658585 = Ltan 36' 17' 20". TRIGONOMETRICAL TABLES. 3G5 Ex. 4. Find the angle A, when L cot A = 10-0397936. We have, 10-0397936 = Lcot A, Next higher, 10-0399770 = L cot 42^ 22', 1834 = (liiFerence or d, Also, 2537 = tab. ditf. for 60'^ or D, And 4 X CO'' = M|| X 60'^ = 43". Hence, 10-0397936 = L cot 42° 22/ 43'^ Ex. V. : 1. Given log 47582 = 4-6774427, and log 47583 = 4-6774518, find log 47*58275. 2. Given log 5*2404 = -7193644, and log 524-05 = 2-7193727, find log -5240463. 3. Given log -5614 5 = T-7493 111, and log 56-146 = 1-7493188, find log Ay-05614581. 4. Given log 61683 = 4-7901655, and log_ 616-84 = 2-7901725, find the number whose logarithm is 2*7901693. 5. Find the value of (1*05)^'^, having given log 1-05 = •0211893, log 20789 = 4-3178336, and log 20790 = 4-3178545. 6. Find the compound interest of £120 for 10 years at 4 per cent, per annum, having given log 1-04 — -0170333, log 14802 = 4-1703204, and log 14803 = 4-1703497. 7. A corporation borrows £8,630 at 4 J per cent, compound interest, what annual payment will clear off the debt in 20 years? Log 1-045 = -0191163, log 4-1464 = -6176712, and log 4-1465 = -6176817. i 8. Find the value of-^ ^^^ ^ ^ , having given Log 1-032 = -0136797, log 34722 = 4-5406047. Log 3762 = 3-5754188, log 26202 = -4183344. Log 34721 = 4-5405922, log 26203 = -4183510. 36G PLANE TRIGONOMETRY. 9. Find Lsin 32"^ 28^ 31", having given Lsin 32° 28' = 9-7298197, and Lsin 32° 29' = 9-730bl82. 10. Find Lcosec 43° 48' 16'', having given L sin 43° 48' = 9-8401959, and L sin 43° 49' = 9-8403276. 11. Required the angle whose logarithmic cotangent is 10-1322449, having given L cot 36° 25,' - 10*1321127, L cot 36° 26^ - 10-1318483. 12. Construct a table of proportional parts, having given 163 as the tabular difference. 13. In what time will a sum of money double itself at 5 per cent, per annum, compound interests 14. Find x when 1*03* = 1*2143, having given that log 1-03 = -0128372, and log 12143 = 4*0843260. 15. Solve the equation 2^"^""^ — 40 = 9*2*, having given log 2 - -3010300. 16. Given L cos 32° 45' - 9*9341986, D = 752, find Lcos 32° 45' 12^; and Lsec 32° 45' 20,'.'. 17. Given Ltan 28° 38' = 10*2628291, D = 3003, find Ltan 28° 37.' 15'", and Lcot 28° 38/ 42'^ 18. Find the angle whose logarithmic cosine is 9*9590635, having given — Lcos 24° 29' = 9*9590805, Lcos 24° 31/ = 9*9589653. CHAPTER YIL PROPERTIES OF TRIANGLES. 30. The sines of the angles of a triangle are proportional to the opposite sides. We shall designate the sides opposite to the angles A, B, C, by the small letters a, 5, c, respectively. PROPERTIES OF TRIANGLES. 367 Dmw AD perpendicular to BC, or to BC produced. Then sin B = And sin C = AD _ AD AB ~ c AD AD AC " 6 •• .(1). .(2). IT /I \ /o\ smB AD AD h Hence, (1 ) ^ (2),-^—- = ^ =, _, ^ ' ^ sinC c c or sin B sin C It follows, therefore, from the symmetrical nature of this sin A sin B sin C n r n equation, that — : — = —i — = — • v--^-^- a b 31. In any triangle, cos C a^ + h^ -^ c" 2ab Taking the figures, of the last article, we have — (1.) Whe7i C ia an acute angle—-' By Euc. IL, 13, AB'^ = BC^ + AC^ - 2 BCCD, CD Now vjj = cos C, or CD = AC cos C. Hence we have, AB« = BC^ + AC^ - 2 BC.AC cos C, or r ^ a^ + h^ - 2ab cos C. •. cos C = 2a6 ' 3G8 PLANE TRIGONOMETRY. (2.) When C is an obtuse angle y as in the second figure — By Euc. IL, 12, AB^ = BC^ + AC^ + 2BC.CD; and CD = ACcosACD = ACcos(180°'-- C) = - ACcosG- Hence, AB^ - BC^ + AC^ + 2 BC ( - AC cos C) ; or, c^ = a^ + 6^ - 2 ah cos C. ri a^ + W - C^ , « /. cos U = — , as before. • c^ 2a6 From iheform of this result we have also — .; 5^ + c2 - a2 Cos A - CosB = 2 5c a" + c^ - y 2ac 32. To express the sine of any angle of a triangle in terms of the sides. We have — Sin^A = 1 - cos^ A = 1 - (^±^^y ^ ( 2 hcf - (b^ -r c" - ay {2bcy ^ {(b + cf - a'}{a^ - (5 - cf} " (2bcf ' (a + h + c){b + c — a) {a + c -b) (a + b -c), " (2bc)^ Hence, taking the square root, and taking the positive sign, because (Art. 11) the sin A is always positive when A is the angle of a triangle, we have — Sin A = -— ^(a + b -{■ c) {b + c - a) {a + c - b) {a + b - c). Let a + b + c = 2 Sf then b + c - a ~ 2 (s - a), a + c - b = 2 {s ^ b)y a + b - c = 2 (s - c). Hence, sin A = -^j- J2s.2{s - a),2{s -b).2 {s - c) 2 , . = -^ v^ (« - ^') (« - ^) {^ - c)- PROPERTIES OP TRIANGLES. 36ll From tlie form of this result we have also — Sin B = ^^ V5(*-a)(s ^ 6) (5 - c), sill C = -^ 4;\s - a) (« - 6) (« -"^. 33. To find the area of a triangle. Using the figures of Art. 30, we have — A ABC = i BC . AD, or, since AD - AC sin C, == iBC.ACsinC = |a6sinC (1). From Reform of this result we have also — A ABC = JacsinB (2). andAABC = ^^csinA (3). The results in (1), (2), (3) express the area of a triangle in terms of two sides, and the included angle. We will now express the area in terms of the three sides. We have — A ABC = J a6 sin C, or, by last Art., 2 , . = Ictb,-^ ^8(s - a) (s - h){8 - c) = Ji{8 - a){a " b) (^~c) (4). 34. To express the sine, cosine, and tangent of half an angle of a triangle in terms of the sides. We have, Art 18, 2 A . b- + c- -- a^ l-2sm ■:jr = cosA = 2"a ' ^ . A , 52 + c--a^ a^ - Ib-cY ...sin^-A 2 6c " 26c (a + c - b ) {a -{■ h - c) _ 2(s - 6 ).2(g - c). 2 6c - 2 6c" . , A {s -b){s- c) or,sm-2-= ^ • 5 A 1(8 -b) {s^ r) 2~= ^ 6c ^- ^- 370 PLANE TRIGONOMETRY. The form of tMs result gives us also — Sin3^_ / (^-«)(^-cT ^C _. I{s-a){s-b) Again, Art. 18, we have — „ .A P + c'-a" 2C0S-Y - 1 = cos A = ij-T ; or, 2cos- 2 = 1 + 2j^— = ^-^^^ (a + b + c) (b + c - a) 2s.2(s — a) ^ 2 bo " WFg ' or,cos^2 - bo A /7 2 ^ V" A fsis - a) cos TT - 5c The/orm of this result gives us also — .(2), B cos-rt Isis - b) C js{s - c) or Now, (1) -r (2), we have — Sin — t — f 2_ ^ / (g - h) (s - c) s{s - a) Cos^^^ ^' ^ ^ ^^ ' tan^ = ,/EZ553 (3). Theybrm of this result gives us also — tan ^-= /(7^ g) (. - c) ^^^C ^ /gZ"^)FZ^). 2 Sl s{s - b) ' 2 V s(5 - c) 35. In any triangle ABC, a = 6 cos C + c cos B. Using the figures of Art. 30, we have — (1.) When C is acute — BD =r AB cos B, and DC = AC cos C. ,-. BD + DC = AC cos C + AB cos B, or c^ = 5 cos C + c cos B. PROPERTIES OF TRIANGLES. 371 (2.) When C is obtuse— BD = AB cos B, and DC = AC cos ACD = AC cos (180^ - C) = - AC cos C. .-. BD - DC = AC cos C + AB cos B, ov a = b cos C + c cos B, as before. The form of this result gives us also — 6 = a cos C + c cos A, c = a cos B + 6 cos A. CoR. 1. Hence sin (B + C) = sin B cos C + cos B sin C. b c For we have, 1 = — cos C + — cos B, or, by Art. 30, > a a sin B ri sin C -o = - — - cos C -f — — - cos B, sm A sin A or, sin A = sin B cos C + sin C cos B. But sin A = sin (180° - A) = sin (B + C). .-. Sin (B + C) = sin B cos C + cos B sin C. CoR 2. Hence also — sin (B - C) = sin B cos C - cos B sin C. For, since this result has been proved for any angles of a triangle, it will be also true for a triangle which contains an angle supplementa/ry to B; that is, it will be true, if we put 180' - B for B. We then have- Sin (180° - B + C) = sin (180' - B) cos C + cos (180° - B) sin C. Butsin(180°-.B + C) = sin(180° - B"^C) = sin(B-C), sin (180° - B) = sin B, cos (180° - B) = - cos B. Hence, sin (B - C) = sin B cos C - cos B cos C. Note. — The results of Cor. 1 and Cor. 2 have been proved only for angles less than two right angles. We shall see iu Vol. II. they are generally true. 86. In any triangle ABC — Tan J (B - C) = |-^^cot i A We have, Art. 30, Sin B b sin B - sin C b - c "SmC" " ? ^^ sin~B + sin C = TTc ^^'' S72 PLANE TRIGONOMETRY, /B + B - C) Now, sin B = sill j - — k — + — s — j > ^^> ^7 ^^r. 1, Art. 35, B + C B-C B + C.B-C ,^, = sin — 2 — ^^^ — 2 — '^ ^^^ — 2 — ^^ — ^ — ( /• / B + C B - C ) and sin C = sin i — « — o — f > or, by Cor. 2, Art. 35, .B + C B-C B + C.B-C ,ox = sin _^_ cos— ^ cos _^_sin_^— (3). B + O B — O (2) - (3), then sin B - sin C = 2 cos — ^ — ^in — ^ — > B 4- B — O and (2) + (3), then sin B + sin C = 2 sin — ^ cos — « — ' ^ B + C . B-C • -D • /^ 2 cos — K — sin — K — smB - sinC 2 2 •'• slnB + sinC " ^ . B + C B"^^ 2 sin — ^ — cos — 2 — B-C B + C B-C /^, A\ = cot —^ — ' tan 2 — = ^^^ I ^^ "~ "9) tan - A B - C = tan 2" tan — ^ — W* ... . A B - C 6 -c (4) = (1), then tan -^ tan — ^ = ^-^' B-C 5-c ,, ^^^ .-.tan ^g— =6-T^c^*i^- ^•^•^• Ex. VI. 1. If a = 5, C = 30^ sin A = -i, find c. 2. Find a, having given 6 = 12, c=15, A = 60'. 3. Find tan A when a = 6, 6 = 7, c = 8. j4. What is the area of a triangle whose sides are 48, 52, 20? 5. Given two sides of a triangle to be 18 and 24, and the included angle 45°, find the area. PROPERTIES OF TRIANGLES. 373 6. Two of the sides of a triangle are as 2 : 1, and the iu- cluded angle is 60°, find the other angles. 7. In any triangle show that — 6 cos C - c cos B == si a?' - 4cbc cos B ^os C. sin B sin C 8. Show that the area of a triangle =. ^a\- sin A 9. An object is observed from two stations 100 yards apart, and the angles subtended by the distance between the object and either station are 45° and 60° respectively. Find the distance of the object from each station. 10. An observation is made from a point known to be distant 120 and 230 yards respectively from two trees, and the angle which the trees subtend is found to be 120°. Find the distance between the trees. 11. If a sin^ + 6 cos^ e = mA ' ^ , , , b sin^ + a cos^ ^ = n, > then _ + _-=_ + _. a tan = 6 tan 0, j ^^ ^ ''* ^ 1 2. If a/, 5/, c' be the sides of the triangle formed by joining the feefc of the perpendiculars from the angles A, B, C of the triangle ABC upon the opposite sides, then — ai , b' cl a" b^ c' 2abc 13. A perpendicular AD is drawn from the angle A of a triangle, meeting the opposite side BC and D ; and from D a perpendicular is dmwn to AC, meeting it in E. Show that DE = 6 sin C cos C. 14. Show that the length of AD in the last example — 6c sin A + ac sin B + a6sinC 3 a 15. Show that Js (s — a) (s - 6) (« — c) = J a6, when the triangle is right-angled at C. 16. Show that (a + b + c)- sin A sin B = (sin A + sin B + sin C)^ ab, 17. Show that in any triangle — cos^A + cos-B + cos-C + 2 cos A cos B cosC = 1. 374 PLANE TEIGONOMETRY. 18. The sides of a triangle ABC are in arithmetical progteg- 6^/3 Bion ; show that its area = — ^ J {2 a - b)(3b - 2 a). CHAPTER VIII. SOLUTION OP RIGHT-ANGLED TRIANGLES. 37. A triangle can always be determined when any three elements, with the exception of the three angles, are given. In the latter case we have only the same data as when two angles are given, for the third can always be found by sub- tracting the sum of the other two from two right angles (Euc. I., 32). Hence a right-angled triangle can always be determined when any two elements, other than the two acute angles, are given besides the right angle. And when one of the acute angles is given, the other may be obtained by subtracting it from a right angle. We have the following cases : — > Case 1, Wli^n the two sides containing the right angle are given, ^A. We shall take C as the right angle in every case. Now tan A = y or, L tan A — 10 = log a - logb; or, L tan A = 10 -f log a - log6...(l). This determines A, and we then have B = 90'' - A (2). Also, - = siii A, or log a loofc = Lsiil A - 10. .-. log c = 10 + log a - Lsin A , Hence the three elements, A^ B, c, are detei-mined* .(3). SOLUTION OP RIGHT-ANGLED TRIANGLES. 375 Case 2. Wlien the hypothenuae and a side are given. Let a be the given side. We have sin A = -, or L sm A - 10 = log a — log c. c »\ LsinA = 10 + log a - logc (1), and B = 90^ - A (2), also b^ = c^ - a^ = {c + a) {c - a) ; .-. log 5 = ^ {log(<^ + a) + log(c - a)} (3). Case 3. Wlien an acute angle and a side are given. Let A, a be the given angle and side. ThenB = 90^ - A (1), also - = tan B, or log 6 = L tan B - 10 + log a (2), and — - sin A, or log a - log c 5* L sin A - 10, or logc = 10 + log a - LsinAi Case 4. WIten the hypotJienuse and an acute angle are given* Let A be the given acute angle; WehaveB = 90° - A .;.. (1), Also- = sin A, or log a = logc + LsinA - 10... (2). c And- = cos A, or log 5 = log c + Lcos A - 10... .(3). c It is evident, from Art. 30, that when the angles only of a triangle are known, we can determine the ratio only of the three sides of the triangle to each other. Ex. 1. Given A = 23' 41', a = 35, solve the triangle; This is ail fexample of C^tse 3. We have B = 90' - 23^ 41' = G6' 19'. Again, log b - Ltan B - 10 - log a - Ltan 66' 19'' - 10 - log 35 = 10-3579092 - 10 + 1-5440680 = 1-9019772 = log 79-795 .-. b = 79-795. 376 PLANE TRIGONOMETRY. Also log c = 10 + log a - LsiiiA = 10 + log 35 - Lsin23°41' = 10 + 1-5440680 - 9-6038817 = 1-9401863 = log 87-134 .-. c = 87-134. Ex. 2. Given a - 214, & = 317, solve the triangle. This falls under Case 1. We have L tan A = 10 + log a - log h = 10 + log 214 ~ log 317 = 10 + 2-3304138 - log 2-5010593 = 9-8293545, Next lower in tables is 9-8292599 = Ltan 34' V\ .-. d = 946. Also, by tables, B = 2724, And^ X 60'/ ^ ^^44:^' = 21'^ nearly. D • 2724 ^ .-.Ltan A - Ltan 34'!' 21", or A = 34°1'21". Hence B = 90 - 34° 1/ 21" = 45° 58/ 39''. And similarly may c be determined. Ex. VII. 1. Given a = 32, A = 63° 45 ^ find h. Log 32 = 1-5051500, Loot 63° 45' = 9-6929750, Log 15780 = 4-1981070, log 15781 = 4-1901345. 2. Given c = 151, A = 37° 42/, find a. Log 151 = 2-1789769, Lsin 37° 42/ = 9-7864157, Log 92340 = 4-9653899, tab. difi: = 47. 3. Given c& = 60, c = 65, find 6, A. Log 2 - -3010300, log 3 - -4771213, Log 65 = 1-8129134, Lsin 67° 22/ = 9-9651953, Lsin 67° 23/ = 9-9652480. 4. Given a = 73, 6 = 84, find A, c. Log 73 = 1-8633229, Ltan 40° 59/ = 9-9389079, Log 84 = 1-9242793, Ltan 41° = 9-9391631, Lsin 40° 59/ = 9-8167975, Lsin 4P = 9*8169429, Log 111-288 = 2-0464479. SOLUTION OF OBLlQtJU-ANGLED TtllANGLES. 377 5. Given B = 71° 41' 10", c = 24, find h. Log 24 = 1-3802112, Lcos 18^ 18' = 9-9774609, Lcos 18** 19' = 9-9774191, log 2278-4 = 3-3576300, Log 2278-5 = 3-3576490. 6. Given a = 293, c = 751, find h. Log 1044 = 3-0187005, log 458 = 2*6608655, Log 691-49 = 2-8397830. 7. Given a = 12, A = 30^ find 6, c. 8. Given c = 10, B = 75°, find a, 6. 9. Given a = 17, c = 34, find A, h. 10. Given a = 5, 6 = 5 ^3, find A. 11. Given a = 28, B = 15°, find the length of the perpen- dicular from C on AB. 12. ^CD is the perpendicular from C on AB, and DE the perpendicular from D on BC. Given B = 60°, a = 20, find DE. CHAPTER IX. SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 38. Given the three aides of a triangle, to find tlie remaining parts. J2 ^ ^2 _ ^2 We have, Art. 31, cos A = — , from which A 2 he may be determined ; and from similar formulae we may find B and C. These formulae are not however adapted to loga- rithmic computation. We shall therefore find it generally advisable to use the formulae of Art. 34. We have, sin ^ = fj- U^ £, A _ 18(8 - a) 2 V be ' " A cos A f {s -b){8- c) S78 PLANEl TRiaONOilETRV. From either of these formulse we can determine A, and from similar formulse determine the other angles. 39. Given one side arid two angles, to Jlnd the remaining parts. Of course the third angle is at once known, Let a be the given side. We have, Ai-t. 30, b - "^"^^ ' ' sin A a sin C sm A hoth of which formula) are adapted to logarithmic com- putation. 40. Given two sides and the included angle, to find the remaining parts. Let h, c be the given sides, and A the included angle. I - c We have, Art. 36, tan |- (B - C) = t cot J A. This formula is adapted to logarithmic computation, and determines | (B - C). We know also |- (B + C), for it is the complement of J A. Hence, B and C are easily determined. . Then, Art. 30, a - —. — =p-> which determines a* sm jj Note. — When the two given sides are equal, the solution may be effected more easily hy drawing a perpendicular from the given angle upon the opposite side, and so bisecting it. By drawing a figure, it is easily seen that, in this case, B = C == 90* ~ i- A, and a = 2 6 cos B. 41. Given two sides and an angle opposite to one of them, to find the remaining parts. Let a, h, B be the given elements. Then we have. Art. 30, sin A - y sin B. (1.) Let the value of , sin B be unity. We then have sin A = 1 = sin 90°, and .'. A == 9i)^ Hence the triangle is right-angled at A. SOLLTIOX OF OBLIQUE-ANGLED TRIANGLES. 379 (2.) Let tlie value of , sin B be >• 1. We then have sin A >- 1, which is impossible. Hence, in this case, it is impossible to form a triangle with the given elements. (3.) Let the value of t sin B be -< 1. # Then, since, Art. 15, the sine- of an angle is the same as the sine of its supplement, there are two values of A which satisfy the equality, sin A = t sin B, and these values are supplementary. Let A, A' be the two values, then the relation between them is A + A' = 180°. If a is not greater than h, then A is not greater than B, and there is no doubt as to which value of A is to be taken. If, however, a is greater than h — that is, if the given angle is ojDposite to the less of the given sides, we must have A greater than B, and both values of A may satisfy this con- dition. This particular case, when the given angle is opposite to the less of the given sides, is called the ambiguous case. We will illustrate this geometrically, 42. Tlie amhigiums case. Let a, by B be given to construct the triangle. Draw the line BC equal to the given side a, and draw BA making an angle B with BC. Then with centre C and radius CA equal to b describe an arcs meeting BA or BA produced iit A and A', and join CA and CA'. Each of the triangles ABC, A'BC satisfies the given conditions. B^ For, in the triangle ABC, we have BC = a, AC = b, and Z ABC = B ; and in the triangle A'BCj we have BC = a, AC = 6, and Z A'BC = B. Again, the sides BA and BA' correspond to the two valuer 380 t>LAN^fi trftlGONOMETRY. of c which are obtained from the two values of A in the equality sin A = r sin B (see last Art.). And the angles ACB and A'CB correspond to the two values of C which would also be found. Cor. If a perpendicular CD be drawn from C upon AA' and if c' and c be the lengths of BA' and BA respectively, it may be easily shown that c' + c = 2 a cos B, and c' /^ c = 2 6 cos A. Ex. VIII. Solve the following triangles, having given — 1. 6 = 12, c = 6, A = 60°. 2. a = 18, 5 = 18 x/2;A = 30^ 3. a = 5 ^,b = 5 J2;c = | ( ^6 + J2). L a=12,B = 60°, G - 15°. 5. a= 3 J2 + J6,b = 6yC = 45°. 6. a = 10 ^/3; b = 15 x/27A = 45°. Given — 7. 6 = 251, c = 372, A = 40° 32', find B and C. Log 121 = 2-0827854, L cot 20° 16' = 10-4326795, Log 623 = 2-7944880, L tan 27° 44' = 9-7207827, Ltan27°45' = 9-7210893. S. a= 237, b = 341, B = 28° 24', find A. Log 237 = 2-3747483, L sin 28° 24' = 9-6772640, Log 341 = 2-5327544, L sin 19° 18' = 9-5191904, L sin 19° 19' = 9-5195510. 9. C = 26° 32', and a : 5 : : 3 : 5, find A, B. Log 2 = -3010300, L cot 13° 16' = 10-6275008, L tan 46° 40' = 10-0252805, L tan 46° 41 ' = 10-0255336. 10. ti = 14, 6 = 16, c = 18, find A, B. log 2 = -3010300, log 3 - -4771213, L tan 24° 5' = 9-6502809, Ltan 24° 6' - 9-6506199, L tan 29° 12' = 9-7473194, Ltan 29° 13' = 9-7476160. HEIGHTS AND DISTANCES. 381 11. a = 3, 5 = 2 A = 60°, find B, C, and c. Log 2 = -3010300, log 3 = -4771213, L sin 35° 15' = 9-7612851, L sin 35° 16' = 9-7614638, Log 1-3797 = -1397847, log 1-3798 = -1398161. 12. a = 5, 6 = 6, c = 7, find A. Log 2 = -3010300, Ltan 22° 12' = 9-6107586, Log 3 = -4771213, L tan 22° 13' = 96111196. 13. If c, c' be the two values of the third side in the ambiguous case when a, by A are given, show that — (c - c'Y + (c + c'Y tan- A = 4 a^. 14. If a, b, A are given, show from the equation — b^ + c^ - 2bc cos A = a^; that if c and c' be the two values of the third side — cc' = b^ - a^, and c + c' = 2 6 cos A. 15. Show also from the same equation that there is no ambiguous case when a = 6 sin A, and that c is impossible when a < 6 sin A. 16. Having a - 6, A, B, solve the triangle. 17. Given the ratios of the sides, and the angle A, solve the triangle. d> A 18. If in a triangle tan J (B — C) = tan- ^ cot ^, show that b cos (p = c. CHAPTER X. HEIGHTS AND DISTANCES. 43. "We shall now show how the principles of the previous chaptei-s are praictically applied in determining heights and distances. We have not space here to describe the instruments by which angles are pi-actically measured, but we shall assume that they can be measured to almost any degree of accuracy. 382 PLANE TRIGONOMETRY. AC tan ABC or tan 0. a 44. To find the height of an accessible object Let AC be the object, and let /iA any distance BC from its foot be measured. At B let the a^igle of elevation ABC be observed. Suppose BC = a, Z ABC = 0. Then, we have — AC BC /. AC = a tan e, the height required. lEx. Let a = 200, and = 30°. Then AC - 200 tan 30° = 200. ~ = -^^ ' v3 3 45. Tofmd the height of an inaccessible object ^A. At any point B in the horizontal plane of the base let the angle of elevation ABC be observed. Measure a convenient dis- tance BD in the straight line CB produced, and observe the angle of elevation ADO, Let BD = «, ZABC = 0, Z ADB - c/>. Then, Euc. I. 32, Z BAD = - <^. AB sinADB_AB sin sin(0 - 0)' Now, BD sin ADB AB or- — - sin BAD a AB = a. AC sm sin(0 — respectively. Show that the height of the second tower — , tan = A , - tane - tan respectively, and the angle they subtend is a. Show that if h be the height of the tower, the distance between the objects — = Ih vcosec*^ + cosec^ cos a. ANSWERS. I. — Page 15, 1. 45-23, 290, -2367, -7, &c. 2. '0005, IMl, -040020, -45, n> tU, l, I- 10. Uf 11. 3. 13. 28. 388 ANSWERS, III,— Page 28. 1. 4 X 11, 2«, 2 X 3 X 5^ 2- X 32 X 7, 3^ x 47, 2 X 3 X 7 X 11. 2. 2 X 7 X 32 X 11, 2^ X 5 X 43, 2^ x 3^ 5^ x 7 x IP, 22 X 3 X IP, 2^ X 3* X 11. 52 3 7 li _4_ 4. • IE. 2 A 1 ft 3 1 XT ' n 3 6., • 61 T> 8> yo> ITOJ y^ 32> 36> T6> TJ 3> Ttf^ y> 6> TT> 17. JLO -rll-T 36> 35> TyyT* 6. 19, 7, 13, 6, 41, 729, 14, 11, 39. 7. I, h t\, t\, if, Is t, f, if, ^, 5^, li. 9. 2 days out of every 7. 10. 2:7. 12. 15, 10, 6. IV.— Page 31. 1. 24, 1260, 2520, 1260, 5460, 33300. 24 f> «A _8 33. 24 36 35: 44. 3 36 5 6_ 231 234. • "6 0> "B'OJ 6^> "6 0^ 8 4> ¥T> ir4> ST ^ ^6T> ^ST? ^ 6 2^> IF 5 2 ^ t¥3. t'A. tVV; T^Vff, ^¥A, AVif, tWj; 4§1, /a, ?Va- 3. 12. 4. H. t\> ii. ih 5. 1065f. 6. Ttir- 7. H. 8. tV- 10. if, i, ,'5. v.— Page 34. 1; f, 2^L, 1^. 2. 29H. 3. UfJJ. 4. i, tV, tV 5. 2i, il, 314. 6. 2||. 7. /*• 8. 5^1^. 9. -ria- 10. 2jVt- 11. 12if 12. l,"^. VI.— Page 36. 1. H, H, Hh 2- 5AV 3. 5. 4. tJit, Trf^w- 5. 3,v- 6. 4. C'^sa R9662 I'lii '• Too* O. --TT2J> -^^^T^J' 9. 77. 10. £8. 12tVs, 11. ^352, 6s. 8d. 12, 20J, ARITHMETIC. 389 VIL— Page 41. 1. 2-203125. a 7^^, ^Vo, Uy h h f 3» 3703059239. 4. 5-098809263: 5: 15^, t\^ 6. 1. 7. 2-718281. 8. -321750. 9. '367879. 10. 1-015873. 11. 3141592. 12. -857142. VIII.— Page 43. 1. 13s. 4d., 7id., 2s. 8(1., £1. 5s. Sfd. 2. M. 7s. 6d., 6s., Is. l^^d., 3s. 9d. 3. X5. 17s. 4d., £8. 15s., £25. 3s. lly^d. 4. 17 cwt. 16 lbs., 16j lbs., 6? lbs., 2^ lbs. 5. 3 m. 2 f. 62f yds., 293^ yds., 4po. 1} yd., 82 J J yds. 6. 144 days, 32 days, 41irs. 54' 47". 7. 38 lbs, 7oz. 2dwt. 15^grs., 12dwt. 6;j«ggrs. 8. - 245 ac. 1 r. 27 po. 12^Vff Y^^- 9. 40° 3' 281", 35\ 10. 6153 grains. 11. 7t'j; 4 cwt. Iqr. 14 J lbs. 12. Iday3hrs. 8'24fH5j". IX.— Page 45. 4. 3 lbs. 7oz. 15dwts., S^Vf ll>s. 5. ^sVd, Hi- 6. IHJ, i^V 7. A, A. 8. 1185/^, ,HJTy. 9. WhW6\- 10. ^iJ^. 11. iS. 12. Tl4o^ayof24hrs. Wff. X.— Page 47. 1. 7s. 6d., 19s. 7Jd., 168. 3Jd. 2. XI. 58., 2s. 9d., XL 6s. 3d. 300 ANSWERS. 3. 12 cwt. 2 qrs., 2 cwt. 3 qrs. 7 lbs., 6 cwt. 1 qr. 25 lbs. 4. 1 ac. 2 r. 26f po., 13/o po., 1 r. 22 po. 5. 4 quires 4 sheets, 11 sheets, 23 quires 4| sheets. 6. 38 galls. 3 qts. 0\^ pts., 1 hhd. 60 galls. 2 qts. 1| pt. nearly, 1 pk. gall. 3 qts. 1| pt. nearly. 7. 4000 grains, 8 oz. 6 dwts. 16 grs. 8. '698 lbs. 9. 15s. 2d. 10. 24 tons 9 cwt. 2 qrs. 8 lbs. 11. 4 cwt. 1 qr. 10 lbs. 12. 3 oz. 14 dwts. XL— Page 48. 'l. -625, -53125, *55625, -928125. 2. -846153, -692307, -615384. 3. -36803, -11805. 4. -015625, -065476190. 5. -003125, -002232142857. 6. -003472, -040293. 7. '3125, -150625. 8. -0002232i42857j -00083. 9. 1-13085317460. 10. -82285714. 11. 12* -0544575 XII.— Page 54. 1 150000^ 20000^ 100, 270, 2-5, 1^ 3-45, 5-294. 2. 46000000^ 3000000j 2950000, 1500, 395, 29-5. 3. 2 myriag. kilog. 2 hectog. 9 dekag., 1 niyriag. 8 kilog. hectog. dekag. S grm. 5 decig., 1 myriag. 2 kilog. 3 hectog. dekag. grm. 1 decig; 3 fcentig.^ 1 hectog. 2 dekag. grm. 2 decig. 9 centig. 6 millig., 1 grm. 5 decig. 3 centig. 3 millig., 3 grm. 4 decig. 2 centig. 7 millig. 4. 160001-2, 25100, 396-45, 203550. 5. 1000, 2-96, 2900-03, 300-12, 3765-43. 6. 10000000000, 10000000, 50000, 349800, 4600. 7. 150, 39-4, 90*2, 1860*3, 3*764, -4. 8. 10, 1-234567, -372456126, 1, -000639, -293. 9. 3203, -4, 2000-003, 76-384, 29*34, 8300, 3457*6. 10. 18300*453, 1830*0453, 1830045*3. ARITHMETIC. 391 11. 1, 73-6, 246-45, 2-55, 16-95. 12. 1300, 130; 713, 71*3; 1235, 123*5; 2*9, 320, 32, 1804, 180-4. XIII.— Page 56. 1. 16287\)99 m., 10738767 m., 1322*371 sq. m., 69-548396 cub. m., 39129-99 gmi., 65632-02 ares, 368-93 st., 78603-982 Ut., 288-06 fr. 2. 1600-688 m., 11696*359 in., 96*18 sq. m., 5-967600029 cub. m., 5972-935 grm., 2450-94 ares, 409*6 st., 694003*024 lit., 2*35 fr., 98-56 fr. 3. (1) 70*245 m., 110*385 m., 130*455 m.; (2) 486082*89 m., &c.; (3) 49 sq. m., &c.; (4) 76*190000095 cub. m., &c.; (5) 11046013-965 gi%, (fee; (6) 36000*9 ar., &c. ; (7) 41629 St., ifec; (8) 7347660*72 lit., &c.; (9) 2932*50 fr., &c. 4. 864 fr. 91-5 c. 5. 3408 fr. If c. 6. (1) 1*56 m., 1-43 in., 1*32 m.; (2) 27788 m., 2613*75 m.y 2460 m.; (3) 2*0910 sq. m., Ac; (4) 41*82 cub. m.,&c.; (5) 188*263 grm., ifec; (6) 13*2 ar., &c.; (7) 10-01 st., &c.; (8) 40446*3 lit., &c.; (9) 293*58 fr., «fec. 7. 25 fr., -15010 fr., 24120 fr., 328 fr., 1*80 fr., 2857 fr. 64 c. nearly, 16*5 fr., 12 c, *01 fr. 8. 25, 80, 2400, 1440, 480000, 14400, 4, 96, 125. 9. 1150 fr. 10. 45. 11. 36. 12. 3877 nearly. XIV.— Page 60. 1. 1609*314 m. 2. 57319*8975 ft. 3. 239*613 sq. ft. 4. 862784. 5. 271*7. 6. 1128 fr. 7. 67 fr. 62 c. nearly. . 8. 1053268765 galls. a £4. 3s. 4cL 10. 2204 fr. 61 c. 11. 447-39. 12. 1*0392. XV.— Page 64. 1. £480. 2. 8s. Hid. 3. 344 fr. 48i c. 4. 39' 22i''. 5. 70. 6. 7 fr. 9ti c. 392 ANSWERS. 7. £4. 9s. 7d. 8. 31 days. 9. £305. 10. 3 li. 34if 11' P.M. 11. 32/ 43xV" past 2. 12. 4 months. XVI—Page 67. 1. 15. 2. 30|f ft. 3. 1125 miles (take 8 kilom. = 5 miles). 4. £15,000. 5. IHdays. 6. 92xV days 7. 2s. 8d. 8. 52-5 m. 9. £154. 10. 88 horses. 11. 7^ miles. 12. 12 days. XVII.— Page 69. 1. 4. 7. 0. £70. £i. 6s. 9f §d. £385. 2. £41. 6s. 2id. 6. £4. 4s. 8. £236. lis. 8d. 11. 5j months. 3. £39. 8s. 4id. 6. £21. 18s. 5d. 9. 3m- 12. 15s. 5id. * XVIII.— Page 72. 1. 3. 5. 7. 9. £23. 3s. 5-856d. 2. £49. 5s. 6-84d. £20. 3s. 10a49d. 4. £102. 17s. 4-941d. 10s. 9-6d. 6. £441. 2 fl. 1 c. 1-40 m, £270. 12s. l-929d. 8. £200 -f (1-045)'. £231525. 10. £71. Is. 11. £50{(l-05)3+ (1-05)2 + (1-05)} = £165. 10s. Hd. 12. £450 -f (1-04)1 XIX.— Page 75. 1. £12. 7s. 2-4d. 2. 3ifd. 3. £4. 13s. 3-38d. 4. £4. 17s. ll'66d. 5. £3. Os. 2T-Vo\d. 6. £10. lis. 8-7d. 7. £5. 12s. 6d. 8. £42. 10s. 9. £39. 7s. 6d. 10. £35. 11. £2212. 10s. 12. £326. 7s. S^Jd. ARITHMETIC. 393 XX.— Page 77. 1. £690. 18s. 9d. 2. £304. 14s. S^\d. 3. £5528. 10s. nearly. " 4. £382. 10s. lOd. 5. £6911. 3s. e^Vod. 6. £126. 14s. lOd. nearly. 7. £842. 12s. 6d. 8. £7768. 5s. 3jd. 9. £863. 7s. 6d. nearly. 10. £134. 11. £281. 5s. 12. 50,000 francs. XXI.—Page 80. 1. £374. lis. lOj'sd. 2. £1069. 10s. lljd. nearly. 3. £1773. 14s. 6id. 4, £393. 7s. 8id. nearly. 5. £308-8693. 6. 20. 7. 12s. lOd. 8. £522-0411. 9. £217-7937. 10. £1000 - (I -05)=. 11. £23. lis. nearly. 12. £369*8728. XXTI— Page 82. 1. 5s. 5d. 2. 10s. lOd.^ 3. £1. 10s. 4. 6 to 5. 5. £48. 6. 4^ gallons. 7. He loses 12J- per cent. 8. £72,123. 12s. 6d. 9. £tV3- 10. 2i. 11. £4. 5s. 12. 33j. Miscellaneous Examples. — Page 85. 1 10. 2. £3. 5s. lOd. \ 3. 721 lbs., 321 kilog. * 4. nAs + V^ ^ 5. -0078125, T^^V & 2933-73. 7. 33-1, 3-236. 8. £2. 16s. Did., he lost 12j per cent. 9. 12. 10. J, 6s., -04895. 11. 0, 4, 2500, l^V- 12. £29. 6s. 7jd., £0. 9s. Bid. 13. £8000 stock, £7530. 14. £-002, 34-3168. 15. 83-8967, 1-9387. , 16. £5ig, £1U3-, £12|3. 17. -2036. 18. 240. 19. 4s. 7-4d. per ounce. 20. £1763. Is. 394 ANSWERS. 21. £723, £3. 6s. 4d. 22. MO, £40, £100. 23. 5s., £1. 17s. 6d. 24. £1. Is. 5^ ll^Vx- 25. £336, ^\Vt. 26. /.VoV^, 9-69. 27. £3375. 28. 6/,-. 29. -3. . 30. 3000 days. ALGEBRA— STAGE I. I.— Page 149. 1. 8, 27, 6, - 16. 2. 6, 2, 4, 0. ^^ 3. 5, 11, - 7. 4. 11, 16. 5. ~ 2, - 15. 6. 8. 7. - 5, - 7. 8. - 11, 8. 9. 7, - 75. 10. 72, - 68* 11. -2,-3. 12. - 10, 32. II.— Page 154. 1. 51, 6, 3. 2. 35. 3. - 513, - 65. 4. - 1224, 30. 6. 0, 2. ' 6. 1, - 27. 7. 1, - 2. 8. 1591, 0. 9. 144. 10. 1521. 37 11. 145. 12. - >^yi029" III.— Page 156. 1. 10 ^ + 3 5. 2. 11 a". 3. 12 6 4- 8 c. 4. 0, 6. 8a2 + ah, 6. 3x' ^ a? - l^x - 2. 7. 4a6 - 4. 8. ix? ■{■ 'if + ^ — 3 xyz, 9. aj* + i!?]!^ + 2/*. 10. a^ + 6^ + c^ + 3 a^S + 3 aW' - a^c - ac^ - Wc - 6c' - 2 a6c. 11. cc'* + 7/^ + ;$;* - 4 33^2/ + 4 a3^;2J - 4a;?/^ - 4 2/^;^; + 4a;;2;^ - 4 2/;s^ + 6a;-2/2 -h 6 o;;??^ + 6 3/ V - 12 ar^^/^ + 12a;^^« - \2xyz\ 12. 0. IY.-^Page 157, 1. 4ct + 26 + 5c. 2. -4a;+2^-6«. 3. a' - 3 a6 - 6^ - 5 a - 7 6 - 8. ALGEBRA. 395 4. 2 a* - 2 arx^ + x\ 5. 4 a' + 8 a'b^ + 4 b\ 6. - 3(^5- 3««- Sarz^ Sxz". 7. cc* - o^c^ - 9 aV - 3 a^x - 2 a^ 8. 0. 9. -a-b-c—d + e+f+g+h. 10. 2 a:* - 8 a;3y + 12 or/- --Sx^ + 2y\ 11. a;* + 2 ary^ + y\ 12. or + b^ -2<^ + 2ab - 2ac - 2bc. v.— Page 159. 1. - a: + 82/ + ^. 2. a* - 6^ 3. 13 - 6a. 4. 5 ar - 3 a; - 7. 6. (a + 6) + (c - cZ), a - (6 — c + c?), |a _ (6 -c)}- cZ. 6. - (6a - 75)-.{3c - 5rf), -6a + (7 6 - 3(j + 6c0, -{6a- (76-3c)} + 5d. 7. - (4af^ - l2x'y) - (l2aj/ - aj/), - {4ai» - (I3af^2/ - 12a:^)}+ 4 2/^. 8. (a^ - &*) - (c^ - 3a5c), a^ - (6« + c» - 3a6c), {a» - (6' + c^)} + 3a5c. 9. - (rt - 6 + rf)a:* - (rt - i - 3 c) a:y ^ - (26 - d ^ e -/)2/'. 10. (a - 6 + c - cZ) a; - (c - c£ + e - /) y - (« - / + i7 - ^)^- 11. (a-b - d)x- {a - 7b - c)y- {b - c + d)z. 12. (2 - a;) a' + (a; ^ y)a6 + (y - z) b\ VI.— Page 161. 1. 12a« - a5 - 6 6^ 18ar - 9a;y - 35yl 2. jB» - 2 ajy" + 4 2/^, 30 x* + 49 a;*y + 9 ay» - 2/». 3. a* + 2 a=6^ + 6*, a* - 6*. 4. a?* - 3/*, «• - y". 6. a* + 6» + c* - 3 a6c. 6. a;» - 2^. 7. a» + 3a26 + 3a6= + b\ 8. 5a» + 5a* - 405a - 405, 896 ANSWERS. 9. - To;^ + 17aj- ^ 6x - 2, a^ - x\ 10. a« + 2 ci«^>2 + 3a^6^ + 2 a-6« + 6«. 11. af + (ac + b/)x + {h + f) ex- -\' (c^ + df) a^ ■{- cdxK 12. ^ + {p - a) x^ -{■ {q - ap) x — aq. 13. af^ + (ci + 6 + c) aj^ + (a6 ■{■ ac + ^c) a; + a6c. VII.— Page 165. 1. ar^ _ 2 a;^ + t/^ 9 a^ - 30 a6 + 25 h\ 16 c* + 8 c^cZ^ + cf^ 9 a;* - 12a;y + 4y*. 2. ^8 - h\ 3. mV - ^zV, 5 a* - 18 h\ 4. (a + cf - (6 + c^)^ (ci + hf - (c + (^)^. 6. a;2 + 4a; - 5, a;^ + 7 a; + 10, a;^ + 2a; - 15, ar^- 25. 7. a;* - 37 a;^ - 24a; + 180. 8. ^\-{4a262 - {a" - h" -c-)-}. 9. 4 (a^ + &2 ^ c^ + cZ^). VIII.--PAGE 171. I 4 f6= - a5 + 2 h\ xy - ^ y\ 2. 2 ft* - 5 d?h + Y ^^2^2^ 2 ar^ + 3 aji/ + 22/^. 8. oa;*"-" + ^a;-**?/^ + ca; -<"» + "> 2/^ ", 4. 6 a; + 4 7/, 5 a; - 3 2/. 5. 1 + a; + a;^. 6. - 3a; - 4, - 7a; + 3. 7. x^ - 3a;2?/ + 5a;i/2 + 27?/' + --^^, a; - 3?/ o^ - a;-?/ - 3a;v^ + 15 ?/' —, 8. a;* + x^y + ar?/^ + xy'^ + 2/*> ^* "" ^V + ^V "" ^V^ + 2/*' 9. oa;*" + hy\ 10. a' + a'b + ai'-^ + 61 11. 6 + c, a + 6. 12. « + 6a; + car'. 13. a* - (/? - 1) a' - (p - g - 1) a- - (;? - 1) a + 1. 14. The given expression is — (a + 6 + c - cZ) (« + 6 — c + cZ) (o& - 6 + c + c?) (a — 6 - c - c^); ALGEBRA. 397 16. - oi^'if - o^y, a? + a;-\ 17. {x + yf + (a? + yfz - (a; + y)z^ - is;^. 20. \a - c)2 - 2 (a - c) (6 - ^) + (6 - df. IX.— Page 177. 1. {X + 3a) {x - 3a), (42/^ + 5;r^)'(42/2 - 5;^), 6 (2a + 36) (2a- 3 6), {2a;-32/) {4a;2 + 6iC2/ + 92/'). 2. a; (a; - ?/) (ar^ + ajy + 2/'), (a - 6) (a + h) {a? + 6^) (a* + h% xy{x + y) {a^ - xy + f), 2 xfz {x + 2z){x - 2 z). 3. (a^ - 2 6^) (a^ + 2 6^), (^ + 0^2/ + 2/') (ar^ - a;^/ + y% (a + ft)'^ (a - 6)^ (a + 6 + c) (a + 6 - c). 4. (a + 6 + c + fZ) (a + 6 — c - c?), (a + 6 — c + i) (a - 6 + c + cZ), (a + 6 - c) (a - 6 + c). 5. 5 (2a; + 9), 3 (2 a; + 7), (3a + 5 - c) (a + 6 - c). 6. {x' + xy + if) [{x - y) {x^ - f) + 4ar^2/2 {x^ - a;?/ - 2/') [, {7? ■{■ xy •\- f) {^ - xy + y-) (x + yf (x - 2/)'. 7. {x - 10) (aj + 7), (a; + 1) {x + 10), (a - 7 6) (a - 8 b), (x - 16) (x + 12). 8. (aa; + 7 62/) (aaj - 6 by), 3 a (aj - 10) (a; + 2), ac {c - 8) (c + 3). 9. (3 a; H- 5) (2 a; - 7), (2 a; + 9) (4 a; - 15), 3 (2 a; + 3) (3 a; - 8), (4 a; - 7) (5 a; + 6). 10. xy{Zx-\-y){x+ 3 >/), a; (5 a; + 3 a) (4 a; + 5 6), (ma; + p) (nx + ^). 11. a;3 + 2ar^ + 4a; + 8, 3a;^ - 6 a;^ + 12a;- - 24a; + 48, a;* + 3 ar' + 9. 12. (a + 6)^ - (a + 6) (c + cZ) + (c + cZ)^ a - 6 + c + d, 13. a* + pa^ + q((/ + ra + s, 2a\ 14. - 102, 17. X.— Page 179. 1. a% - 27 aW, 16 a«6V, - x'yY. 398 ANSWERS. 2. a* + 12 a'^ + 54a-62 + loSaS^ + 81 h\ 16 a* + 32 a^^ + 24 aW + 8 aS^ + 6*^ a« - 5 a^6 + 10 a^h^ - 10 oW + 5 a6^ - h\ 27 a» - 108 «-6 + 144 ^6^ - 64 61 3. 16 m* + 32m3 + 24m2 + 8m + 1, 125 0^3 + \bOx\ + 60ic + 8, 81a* - 432 a^c + 216 aV - 1152 ac^ + 256 c*, - a3 „ 3 a^S - 3 aly" - h\ 4. cc* + 2a;3 + 3a;2 + 2aj + 1, 9 a^ + 52 ^ 16 c^ + d? - 6 a?:> + 24 6^c - 6 acZ - 8 6c + ^hd - ^cd, a^ + 4 6^ + c^ + 4 a6 - 2 ac - 4 6c, 9 (a^ + 6" + c^ + 2 6^6 + 2 ac + 2 6c). 6. 1 + 3 aj - 5 33^ + 3 03^ - £c^, (aa; + hyY + ?t(ax + hy) cz + &c. 6. 1 + 7 a; + 21 a;2 + 35aj3 + 35 a;* + 21 «;« + 7 a;« + aj^ (1 + i:c)* + 4(l + a;)V + 6 (1 + aj)V + 4 (1 + a;)a;« + c««, \a + 6a;)* + 4 (a + 6aj)^ cs^ + &c. 7. a«a;8 -• 8 alhx^y + 28 a^WxY - ^6 a«63aj«2/' + 70 a*6*ajV - 56 a^b'a^^ + 28 aWaPy' - 8a6%2 ^ js^s^ 729 aj« - 243 xY + 27 aj^^/* - 2/'> aj»- 3a;y + 3 a^2/« - yK 8. a^^ - 2 x'^y^ + y^ ' a^2 ^ 6 ^1^6^ + 15 Jb^ - 20 a«6« + 15 a*6« - 6 aW<> + 6^1 9. a^ + 3^26 + 3ab^ + 6^. 10. (a - c)l 11 2 (1 + 3 aPy. 12. 81 a^y (aj n- ^). XL— Page 193. 1. 2 a;i/V, 4 aV, ar^ + a% 2. 2 a;^ - 3 x'y + 5 a^y. 3. 5ft2 ^ 3a6 + 61 4. 1 - 2a; + 3a;- - 4a;3. 5. a + 6a; + cx'^ + da^. 6. aV - 3 aa;"-^ + 4:x''-\ 7. a; + a;-'^ aa;-^ - a-^x 8. 3 a;^ 5 a. 9. 36, 79, 207, 289. 10. 103-2, -024, -3, j. 11. 4-123, 1-224, -618, 3-732. 12, -0203, 4-6, 3-5036. 13. 2a6y, 5a;y a + 2 6. ALGEBRA. 399 14. x'^ + 3 oj- « 7. 15. X -{- y - c. 16. X + x-\ xy'^ + 1. 17. 18018, IMl. 18. 2-73, -3, -J. ■ 19. -5, -2599. j 20. 4/5^= 1-709. 21. 7-6457, 50-2487. 22. 4-4142, 3-7320. 23. -25. 24. 0. XII.— Page 200. 1. aj - 3. 2. ar + 4 a; + 3. 3. a; + 3. 4. aj + a. 5. or -^ ah + h\ 6. aj - 1. 7. a; - 3. 8. 3aj - 2. 9. 12 a;- + 5 a;. 10. 3 a + 2 5. 11. a; - 1. 12. 2 a:^ - 4 a.-^ + aj - 1. 13. 2a + 3b + c. 14. a;2 + 1. 15. x - 1. 16. 2aj2 - 3aj + 9. 17. 1. 18. a + b +c. XIII.— Page 203. 1. 12 a^a^f. 2. 60a-5V. 3. (a ^ b){b - c) {c - a), 4. a^x {x" - or). 5. (a; + 1) (aj + 2) {x + 3). 6. (a;- 6) {x -\-^){x'' 5). 7. (2 a; + 7) (3 aj + 8) (4 a; + 5). 8. 210 {d? + 1) (a^ + 1). 9. a" (x* + aV + a'). 10^ (x + a) (X + h) (x + c). 11. 1 - a^«. 12. (x + 1) (aj + 2) (x + 3) (aj - 5) (a; - 6) (aj + 5). 13. a (a - bf (a2 - a5 + b'). 14. (a; - 1) (x + 1) (x" + 1) (6 a^ + 5 a;2 + 2a; - 1). 15. {a' - by. 16. 2a;(a;2^i)(3^^3^_l^(2a;2^2a;-5)(2ar^-2aj + 5). 17. 3 (aj - 2)2 (aj + 4) (5ar^ + lOa;^ + 20a; + 18). 18. a'b''{a + b){a - b). 19. 15 (3 aj - 10) (a;* - 16). 20. (aP - 2^2)2 (^2 ^ ^2) (^4 ^ y)^ 21^ ^6 _ ^6^ 22. {a + b + c + d) (a -{- b -- c -^ d) (a - b - c + d). 23. (a + 6 + c) (a* + £3 ^ c^ - 3 abc). 24:. (a -{■ b + c -i- d) {b + c + d - a) (a + c + d- b) (a + b + d " c){a + b + c - d) (a + b - c-d). 400 ANSWERS. XIV.—Page 206. X ^ 6 x^ -2x + 2 3aj + 7 2ar^ + 9 Ix + 2' 5a;2 + 2a;-2' 4. ^(y - ^) - y' 1 g^ 2 + Jbf. 20. - a. 21. a' + 2 a. 22. f 23. - 2i. 24. «/ + b" A«7. l:, -J- J. 30. 5, - U- 33. 2, - iU- 36. 3, - le. 39. 15, - 1. 31. 3, - 5. 34. 5, i. 37. 2, 2i. 40. 2, 0. 32. 2, - 3i. 35. 8, - fi. 38. 4, - a. 41. 0, 2a6. 404 ANSWERS. 42. a, 3| a. 43. 0, 44. 1, (6 - a)h 45. a + 6 + c, - a — 6' II.— Page 306. 1. ±3. a 2, - ^V. 3. ±5, ± ^ 4. 2, - 5. 5. ± i. 6. ^2, ^ 7. 4, 69. 8. ± 4, ± Jl, 9. 3, - 4, 3 ± Jn. 10. I, 4. 11. - 2, - 2 ± J3. 12. I 13. 4, 5-, --^-. 13 ± x^l45). 14. 0, 6 - a. 15. ± i« x/3r 16. 2 - ^/6-2a^/6+ ^' 19. /i*l*JY. 20. ± 5, ± J-li. V" + ly 21. 2, - i, J(- 11 ± n/97). 22. 4, |. 23. 5, 3j, i(- 7 ± s/53). 24. 3a. 25. 3, - 2. 26. 4, - 14^, t (- 19 ± VTiT). 27. 3, - 2^, i|. 28. i (1 ± n/4^^^). 29. + 1, andiC* + ar= + 1 = 0. 30. ± /?L±^JL?. 31. e», i { s/r?~:nr^^ + 2 a6 + 6^ - (a - i-)}- 82. X = m is one solution. ALGEBRA. 405 33. X = a -^ h + c is one solution. 34. X = ^, 2ind a^ + ax -h x^ = 0. m 35. X = is one solution. a + c 36. a:^ - a; + 1 = gives two solutions. III.— Page 310. 1. X = 2, 3, 2/ = 3, 2. 2. a; = 5, - 3, 2/ = 3, - 5. 3. a; = 3, 4, 2/ = 4, 3: 4. a; = 2, 4, t/ = 4, 3. 5. a; = 5, - 2, 2/ = 2, - 5. 6. ± 7, ± 6. 7. ± 6, ± 3, 8. 4, 2. 9. ± 2, ± 7. 10. ± 4, ±3. 11. a; = ± 3,~y = ± 2, - -^. 12. a; = 9, 4, 2/ = 4, 9. 13. x = 2, 8, y = 8, 2. 14. a; = 3, 2, 2/ = 2, 3. ..;- _ 2 ^ ^ a ± J2¥'^^^~^ ^ ~ a^: J2U' - a^ (a + 1)6 _^ (a -1)6 16. ± n/2 {a^ + 1)' " 2 x/a^ + 1 ^' "- V^T^' - V 2 6 18. a: = 2, 3, y = 3, 2. 19. a; = 2, 3, 2/ = 3, 2. 20. a; = 2, 3, y = 3, 2 21. aj = 5, - 3, 2/ = 3, - 5, 22. X = ""-l ^^ ^ " ^ ^ 23. a: = 2, 3,y = 3, 2, v^4 cr» - 4 6 S4. a: = ± 8, + V n/I?^ 2^ = =t 2, ± x/|| 25. a; = 4, 9,y = 9, 4. 26. a? = 4, 1, y = 8, 0. 27. a;= 4, 3, 2/ = 3, 4. 28. ± 3, + 1. 29. a: = 0, 3, - il^l, 2/ = 0, 2, llSf , 406 ANSWERS. 30. 0, - 1. 31. x = 4, VaV, y = 3, - V5V. 32. a; = 5, 6, 2/ =t: 4. 33. aj = 3, 1, 2/ = 1, 0. 34. aj = 3, - V/, 2/ = 1, - If. 35. a: -2, 7,2/ = 4, -14. 36. aj + y = c, and xy = bx + ay, 2n. X = 0, a — h, y = 0, a — h, 38. V ± 9 ViJ, ^ ± ^H. 39. a; = , / (« + c - 6) (a + 6 - c)_ ^ 2 (6 + c - «) 40. 1, 2, 3. 41. ± 1, 0, 0. 42..=2, = « = 0,alsol=.(^.i,.l^. ^ A ^ 43. ^^. X = y = z = u = a + h. 45. 7' r, &c (a + 6 + c) * (a + 6 + c)^ 46. a; = 6, 4, 2/ = 4, 6, « = 5. IV.— Page 315. 1. 9, — 6. 2. Numerator 2f |, denominator 3. 3. 35 or 23. 4. 10, - 16. 5. 78, - 68 ft. 6. 2s., 9s. 7. 20 hours. 8. 48 shillings. V.—Page 318. 1. a^j a^, a% a. 2. a + x^ (a - xy. 3. a + a^x^ + a;l 4. x^^"^ - x'^y^. ALGEBRA. 407 5. 7. 9. 14. 16. 1. 2. 3. 5. 7. 10. 12. 13. 14. xy-'^ +x-hj + 1. 6. a' + ah^ + hi, at + i-t. X - xy, 8. a' + hi + c* - ahi - a^c'^ - bic^- a-b-i - Ji 13. a* + ji + «*• 2a;*2/^ - Sxyi + 2x^- 16. ai~ * - 2 + 2a~''bK a;* + 3a;?-7. 17. ar'y-J + 1. 18. (a* + fiJf. VI.— Page 325. x^, a^bi, x^yi, a^b^- ocya~ib~i, c^Sx~iy~i, a~mxn' V27, V24, x^% -^9. 4. ^/32, ^277 ^32, Vy. n/9^6, V^, ^a" - ar^. 6. ^4, ^7. ij^, iysT 8. V8, ^135". 9. "v^a^ ^^»- 4^(ti + xY, il{a - x)\ 11. a^,h 2 ^/l,2 >^J'6, 6 V7, 3 ^3T - - a, ^ — a'a;i P + 7 ^-~P , P2 X^ X + I 15. "3^ -JZx, X - -J{^-aJ. 16. 4V3, 5^7. 17. ^?^ - a6 + c sJaVa?. 18. (Sa'b -2a m + » - 12) ^. 19. d ~ 6. + b 20. (x -f •2/)*. h. - hHK 21. ; + d \/f ('^ + J)- 408 ANSWERS. 22. a - b^ + c^ - dt + 2 a^ci - 2 6'ci 23. X - x^-if^ + y: 24. o^ — 3(?y^ + x^y^ — xy + x^y^ - y^, 25. \ Vis, ± n/7, i 4/I75. 26. 3(2 - n/3), 1(3 V2 - 2sl%l{Jb^ ^3). ' 27. 1(2 + x/2 - x/6), -J- (2 s/3 + 3 x/2 - ^30), 5 + 2 ^6. 28. a {x^ - xhj^ + x^y^ — xy + x^y^ - y^), ^\ (3^ - 3\ 2^ + 3t. 2^ - 3 . 2 + 3^. 2t - 2*), ^ ^xy ^y'^ (oj - x^y^ + y), 29. 2 + Vt, V5 + x/3, 5 - V5. 30. V6 + a/2, V6 - V3, VT5 - V5". YIL— Page 330. 1. o^ 4- }? \o? - 5^ is the greater, if a >- 5, 2. The former. 3. 6 - ^. c 4. (1 - 2/) (1 + a:) : 1 + ic'. 5. 30, 35. 6. Less than 5 when x lies between 2 and 3 ; greater than 5 for all values beyond these limits. 8. X satisfies the equation — 4 ic^ -» (a + 5 _ 6 c)ic + 2 (c' - a6) = 0. 9.|. 10.1. 14. 25^ a, i;, ^^-. 2 a^ 2 a — PLANE TRIGONOMETRY. 409 PLANE TRIGONOMETRY. I.— Page 335. 1. 43-750^ ir 50' 3-03". 2. 88^ 88^88-8^ 3. i^n the irnmber of grades. 4. 180 - (a + -3^5) degrees, 200 - (i^o^ + 6) grades. 5. i 7. |. 8. 25^ 75^ 90°. 9. 576°, 648^. 10. 2nd, 1st, 4tli, 3rd, 4th. 11. « , ^. 12. -^-^. II— Page 341. 1. Cos A = -^%, tan A = y &c. 24 25 2. Sin A = — r= — , cos A = — >- , &c. n/T201' ^/l20l' 17. h 18. 3^, (2 v/89 ^ 5). 19. -^^^^• 20. 2 + ^3. 21. 00 , - ^,-^1 22. 1. 24 ^ ^~~~ 1 Sin A = FZK£^. III— Page 352. 2. From 0^ to 90°, + ; from 90° to 180°, - ; from 180° to 270°, + ; from 270° to 360°, -. 3. Cos A + sin A is + , + , - , - , as A lies between - 45° and 45°, 45° and 135°, 135^^ and 225°, 225° and 315° respectively. The corresponding signs of cos A - sin A are + , - , - , + . 4. Cos 2 A is +,-,+, - respectively, as in Question 3. ^' 2 J2 ' 2 ' *" 2 • 410 10. A = 60° ANSWERS. 11. A = 45°, and tan A = - 6. 12. A = 30°. 13. A = 30°. 14. A = 45°, B = 15°. 15. A = 0°, 45°, 135^ 16. A = 6°. 17. A = eO'^. 18. e = 2 ± J37 lY.— Page 358. I. 2, 3, 1, _ 1, - 2, 1-5. 2. I, 6, 1-5. 3. 1-0791812, 1-5563025, 1-6532125, 1-8750613, 2-6020600, -5740313, l-8239087,'2-8696761. 4. 1-3172901, 3-3172901, 2-3172901, 3-3172901. 5. 1, 1, 3,~3. 6. -020912, 2091200, -20912, 20912. 7. 1-3162760. 8. 3493-768. 9. -2427189. 10- 3-1303338. II. 6, 5. 12. 1-67. 1. 1-6774495. 4. -0616835. 7. £663-449. 10. 9-8401608. Y.— Page 365. 2. 6-7193696. 6. 2-07892. 8. 2620-248. 11. 36^26^30^^ YI.— Page 372. 3. 1-8213310. 6. ^177-63 nearly. 9. 9-7299222. 1. c = 12-5. 2. ^189. 4. 480. 5. 108 s/2: 9. C = 50 j6{JS^l),b : 10. a = 10 ^949. 5 3. Tan:| 6. B = 90^ C = 30°. 100 (v/S"" 1). PLANE TRIGONOMETRY. 411 VII.— Page 376. I. J = 15-78065. 2. a = 92-34057. 3. ^> = 25, A = 67'22'49^ 4. A = 40°59'32^c = 111-288. 5. h = 22-78438. 6. b = 691*49. 7. 5 = 12 ^3, c = 24. 8.a = 1(^6- V2), 6 = f(V6 + v/2). 9. A = 30°, b = 17 J3, 10. A = 30°. 11. CD = 7{J6 - ^2). 12. DE = 5^3. VIIL— Page 380. 1. a = 6 V3, B = 90°, G = 30°. 2. B = 45° or 135°, C = 105° or 15°, c = 9 ( ^6 ± ^2). 3. A = 60°, B = 45°, C = 75°. 4. A = 105°,^ = 6(3^2 - J6\c = 12{2- J3), 6. A = 75°, B = 60°, c - 2 ^6. 6. B = 60° or 120°, C = 75° or 15°, c = 5 (3± ^3). 7. B = 41° 59' 23'; C = 97° 28' 37^ 8. A = 19° 18' IK 9. A = 30° 3' 26", B = 123° 24' 34". 10. A = 48° 11' 22", B = 58° 24' 42". II. B = 35° 15' 52", C = 84° 44' 8", c = 137*9796. 12. A = 44° 24' 56". IX,— Page 384. 1. 200 (2 - -V3). 2. 60 73, 3. W3. 4. 6 V2. 412 ANSWEKS. 5. 6(73 - 1), 3 76(^3 - 1). 6.30(^3+1). 7. 1|5 73. 8. 1073. iriLLTAM COLLINS AND COMPANY, PRINTERS, GLASGOTV. William Collins, Sons, & Co.'s Educational "VTO^orks, - -»• COLLINS' SERIES OF FIRST-CLASS SCHOOL ATLASES, Carefully Constructed and Engraved from the best and latest AuthoritieSy and Beautifully Printed in Colours^ on Superfine Cream Wo-ve Paper, MODERN GEOGRAPHY— Grown Series. t. d. 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THE PRIMARY ATLAS OF PHYSICAL GEOGRAPHYj, 16 Maps, Demy 4to, 9 by 11 inches, Stiff Cover, ... ... ... 1 THE POCKET ATLAS OF PHYSICAL GEOGRAPHY, 16 Maps, on Guards, Demy 8vo, cloth, ... ... ... ... ... 2 London, Edinburgh, and Herriot Hill "Works, Glasgow. 'William Collins, Sons, & Go's Educational "Works. COLLINS^ SERIES OF SCHOOL ATLASES— Continued. HISTORICAL GEOGRAPHY. THE POCKET ATLAS OF HISTORICAL GEOGRAPHY, i6 Maps, 6^ by II inches, mounted on Guards, Imperial i6m<>, cloth, THE CROWN ATLAS OF HISTORICAL GEOGRAPHY, i6 Maps, with Letterpress Description by Wm. F. Collier, LL.D., Imperial i6mo, cloth, THE STUDENT'S ATLAS OF HISTORICAL GEOGRAPHY, 16 Maps, with Letterpress Description by Wm. F. Collier, LL.D., 8vo, cloth, ... 1 Roman Empire, Eastern and Western, 4th Century. 2 Europe, 6th Century, shewing Settle- ments of the Barbarian Tribes. 3 Europe, 9th Century, shewing Empire of Charlemagne. 4 Europe, 10th Century, at the Rise of the German Empire. 5 Europe, 12th Century, at the Time of the Crusaders. 6 Europe, 16th Century, at the Eve of the Reformation 7 Germany, 16th Century, Reformation and Thirty Years' War. d. 6 8 Europe, 17th and ISth Centuries. 9 Europe at the Peace of 1815. 10 Europe in 1870. 11 India, illustrating the Rise of the British Empire. 12 World, on M creator's Projection, shewing Voyages of Discovery. 13 Britain under the Romans. 14 Britain under the Saxons. 15 Britain after Accession of William the Conqueror. 16 France and Belgium, illustrating British History. 15 CLASSICAL GEOGRAPHY. THE POCKET ATLAS OF CLASSICAL GEOGRAPHY, Maps, Imperial i6mo, 64 by 11 inches, cloth lettered, THE CROWN ATLAS OF CLASSICAL GEOGRAPHY, 15 Maps, with Descriptive Letterpress, by Leonhard Schmitz, LL.D., Imperial i6mo, cloth lettered, THE STUDENT'S ATLAS OF CLASSICAL GEOGRAPHY, 15 Maps, Imperial 8vo, with Descriptive Lettei press, by Leonhard Schmitz, LL.D., cloth lettered, 9 Armenia, Mesopotamia, &c. 10 Asia Minor. 11 Palestine, (Temp. ChristL) 12 Gallix 13 Hispania. * 14 Germania, &c. 15 Britannia. 1 Orbis Veteribus Notus. 2 ^gyptus. 3 Regnum Alexandri Magni. 4 Macedonia, Thracia, &c. 5 Imperiura Romanum. 6 Grsecia. 7 Italia, (Septentrionalis.) 8 Italia, (Meridioualis.) Historical and Classical Atlas. THE STUDENT'S ATLAS OF HISTORICAL AND CLASSI- CAL GEOGRAPHY, consisting of 30 Maps as above, with Intro- ductions on Historical Geography by W. F. Collier, LL.D., and on Classical Geography by Leonhard Schmitz, LL.D., with a Copijus Index, Imperial 8vo, cloth. London, Edinburgh, and Herriot Hill V^orks, Glasgow "William Collins, Sons, & Go's Educational "Works. COLLINS' SERIES OF SCHOOL ATLASES— Continued. SCRIPTURE GEOGRAPHY. THE ATLAS OF SCRIPTURE GEOGRAPHY, i6 Maps, with Questions on each Map, Stiff Cover, THE POCKET ATLAS OF SCRIPTURE GEOGRAPHY, i6 Maps, 7^ by 9 inches, mounted on Guards, Imp. i6mo, cloth, 1 Ancient World, shewing probable Set- 9 Modern Rilestine. tlements of Descendants of Noah. 10 Physical Map of Palestine. 2 Countries mentioned in the Scriptures. 11 Journeys of the Apostle Paul. 3 Canaan in the time of the Patriarchs. "~ " 4 Journeyings of the Israelites. 5 Canaan as Divided, among the Twelve Tribes. 6 The Dominions of David and Solomon. 7 Babylonia. Assyria, Media, and Susiana. 8 Palestine in the Time of Christ. 12 Map shewing the prevailing Religions of the World. 13 The Tabernacle in the Wilderness. 14 Plans of Solomon's and Herod's Tem- ples. 15 Plan of Ancient Jerusalem. 16 Plan of Modern Jerusalem. BLANK PROJECTIONS AND OUTLINES. THE CROWN ATLAS OF BLANK PROJECTIONS, consisting of 16 Maps, Demy 4to, on Stout Drawing Paper, Stiff Wrapper, ... o 6 THE CROWN OUTLINE ATLAS, 16 Maps, Demy 4to, Stout Drawing Paper, Stiff Wrapper, ... ... ... ... ... o 6 THE IMPERIAL ATLAS OF BLANK PROJECTIONS, consisting of 16 Maps, Imperial 4.to, on Stout Drawing Paper, Stiff Wrapper, i 6 THE IMPERIAL OUTLINE ATLAS, 16 Maps, Imperial 410, Stout Drawing Paper, StifJ" Cover, ... ... ... ... ... i 6 A Specimen Map of any of the foregoing Atlases free on receipt oftivo Penny Stamps. SCHOOL-ROOM ^WALL MAPS. Printed in Colours, and Mounted on Cloth and Rollers^ Varnished. CHART OF THE WORLD, 5 ft. 2 in. by 4 ft. 6 in. , 20 o CENTRAL AND SOUTHERN EUROPE, 5 ft. 2 in. by 4 ft. 6 in., 20 o EUROPE, ASIA, AFRICA, NORTH AMERICA, SOUTH AMERICA, ENGLAND, SCOTLAND, IRELAND, PALES- TINE, INDIA, each 3 ft. by 2 ft. 5 in., 6 6 UNITED STATES OF AMERICA, 3 ft. 11 inches by 2 ft. 4 in., 8 6 COUNTY ^WALL MAPS. Printed in Colours^ and Mounted on Cloth and Rollers, Varnished. MIDDLESEX, LANCASHIRE, YORKSHIRE, WARWICK, DURHAM, CUMBERLAND, DERBYSHIRE, DORSET, GLOUCESTER, HAMPSHIRE, SOMERSET, STAFFORD, AND WILTS i each 54 in. by 48 in., 9 o CHART OF METRIC SYSTEM. CHART OF THE METRIC SYSTEM OF WEIGHTS AND MEASURES. Size 45 in. by 42 in., price, on Rollers, ... 9 "London, Edinburgh, and Herriot Hill "Works, Glasgow. ;j QAi-5"Y ^ "William Collins, Sons, & Co.'s Educational "Works. COLLINS' SERIES OF FIRST-CLASS SCHOOL ATLASES, Carefully Constructed and Engraved from the best and latest AuthoritleSy and Beautifully Printed in ColourSy on Superfine Cream Wave Paper, MODERN GEOGRAPHY— Crown Series. MY FIRST ATLAS, consisting of 12 Maps, 9 inches by 7^ inches, folded Svq^in Neat Wrapper, 6 1 The Hemispheres. 2 Europe. 3 Asia. 4 Africa. 5 North America. 6 South America. THE PRIMARY ATLAS, consisting of 16 Maps, 9 inches by 7^ inches, 4to, Stiff Wrapper, ... 7 England and Wales. 8 Scotland. 9 Ireland. 10 Central Europe. 11 Australia. 12 Palestine. 1 The Hemispheres. 2 Europe. 3 Asia. 4 Africa, 5 North America. 6 South America. 7 England and Wales. 8 Scotland. 9 Ireland. 10 Central Europe. 11 India. 12 Canada. 13 United States. " 14 Australia. 15 "New Zealand. 16 Palestine. THE POCKET ATLAS, consisting of 16 Maps, folded in 8vo, and mounted on Guards, cloth lettered, THE JUNIOR, OR YOUNG CHILD'S ATLAS, consisting of 1 6 Maps, 4to, with 16 pp. of (Questions on the Maps, in Neat Wrapper, THE SCHOOL BOARD ATLAS, consisting of 24 Maps, Crown 4to, cloth limp, THE PROGRESSIVE ATLAS, consisting of 32 Maps, 9 inches by 75 inches, 4to, cloth lettered, \., 1 The Hemispheres. 2 The World, (Mercator's Projection.) 3 Europe. 4 Asia. 5 Africa. 6 North America. 7 South America, 8 England and Wales. 9 Scotland. 10 Ireland. 11 France. 12 Holland and Belgium. 13 Switzerland. 14 Spain and Portugal. 15 Italy. 17 German Empire. 18 Austria. 19 Russia in Europe. 20 Turkey in.Europe, and Greece. 21 India. 22 Persia, Afghanistan^ and Beloochis- 23 Turkey in Asia. [tan. 24 Chinese Empire and Japan. 25 Arabia, Egypt, and Nubia. 26 Palestine. 27 Dominion of Canada. 28 United States. 29 West Indies. 30 Australia. 31 Ne\r South Wales, Victoria, and 32 New Zealand. [South Australia. 16 Sweden, Norway, and Denmark. THE CROWN ATLAS, consisting of 32 Maps, on Guards, with Index, 8vo, cloth lettered, ... ... ... ... ... ... 2 THE NATIONAL ATLAS, consisting of 32 Maps, 4to, with a Copious Index, cloth lettered, ... ... ... ... ... 2 London, Edinburgh, and Herriot Hill "Works, Glasgow. "William CoUins, Sons, & Co.'s Educational "Works. COLLINS' SERIES OF SCHOOL ATLASES-CONTINUED. MODERN GEOGRAPHY— Imperial Series. THE SELECTED ATLAS, consisting of i6 Maps, Imperial 4to, ii by 13 inches, Stiff Cover, ,,, 1 Tlie Hemispheres. 2 Europe. "■ 3 Asia. 4 Africa. 6 North America. 6 South Aineri(;a. 7 England and Wales. 8 Scotland. s. d, I d 9 Ireland. 10 Southern and Central Europe. 11 India. 12 Canada. 13 United States. 14 Australia. 15 New Zealand. 16 Palestine. THE PORTABLE ATLAS, consisting of 16 Maps, folded Imperial 8vo, cloth lettered, ... THE ADVANCED ATLAS, consisting of/32 Maps, Imperial 4to, cloth lettered, 1 Eastern and Western Hemispheres. 2 The World, (Mercator's Projection.) 8 Europe. 4 Asia. 6 Africa. 6 North America. 7 South America. 8 England and Wales. 9 Scotland. 10 Ireland. 11 France. 12 Holland and Belgium. 18 Switzerland. 14 Spain and Portugal 15 Italy. [the Baltic. 17 German Empire. 18 Austria. 19 Russia. 20 Turkey in Europe, and Greece. ^ India. 22 Persia, Afghanistan, and Beloochis- 23 Turkey in Asia. [tan. 24 Chinese Empire, and Japan. 25 Arabia, Egj'pt, Nubia, and Abys- 26 Palestine. [sinia. 27 Dominion of Canada. 28 United States. 29 West Indies and Central America. 30 Australia. 31 Victoria, New South Wales, and 16 Sweden and Norway, Denmark and |* 32 New Zealand. [South Australia. THE ACADEMIC ATLAS, consisting of 32 Maps, Imperial 410, with a Copious Index, cloth lettered, ... ... ... ... 5 o THE STUDENT*S ATLAS, consisting of 32 Maps, and 6 Ancient Maps, with a Copious Index, Imperial 8vo, cloth lettered, ... 6 o 93 Ancient Greece. 87 Historical Map of the British Is- 34 Ancient Roman Empire. lands from a.d. 1066. 35 Britain under the Romans. 88 France and Belgium, illustrating 36 Britain under the Saxons. British History. THE COLLEGIATE ATLAS, consisting of 32 Modern, 16 ?2istorIcal, and 2 Ancient Maps, mounted on Guards, with a Copious Index, Imperial 8vo, cloth lettered, 76 THE INTERNATIONAL ATLAS, consisting of 32 Modern, 16 Historical, and 14 Maps of Classical Geography, with Descriptive Letterpress on Historical Geography by W. F. Collier, LL.D. ; and on Classical Geography by L. Schmitz, LL.D., with Copious Indices, Imperial 8to, cloth mounted on Guards, 10 ^ London, Edinburgh, and Herriot Hill "Works, Glasgow.