IN MEMORIAM FLOR1AN CAJORI I MANUAL OF LOGARITHMS MANUAL OF LOGARITHMS TREATED IN CONNECTION WITH ARITHMETIC, ALGEBRA, PLANE TRIGONOMETRY, AND MENSURATION, FOR THE USE OF STUDENTS PREPARING FOR ARMY AND OTHER EXAMINATIONS BY G. F. MATTHEWS, B.A. LATE SCHOLAR OF ST. PETER'S COLLEGE, CAMBRIDGE MACMILLAN AND CO, AND NEW YORK 1890 [All rights reserved] PREFACE. THIS Manual is intended to supply a want that has daily become more apparent during many years' experience of preparing pupils for examina- tion. In the elementary text-books on Algebra and Trigonometry the subject is treated too shortly for practical purposes, and there is a great scarcity of examples. These failings I have endeavoured to remedy; and, to give the student accuracy and facility in his work, a very large num- ber of examples, over 1300 in all, have been introduced, among which will be found the more important of those that have been set during the last ten years in the examinations for entrance to Sandhurst, Wool- wich, and the Staff College. A few typical examples are worked out at full length in the course of the bookwork to assist the student and spare the tutor. The subject has been treated in connection with Arithmetic, Algebra, Plane Trigonometry, and Mensuration. Notwithstanding the care with which the examples have been worked out, there must neces- sarily be many errors in a work of this nature. I shall therefore esteem it a great favour if notification of these be made either to the publishers or myself. It is with many thanks that I acknowledge valuable suggestions from my friend and former college tutor Mr. J. D. H. Dickson, who so kindly consented to read through proof-sheets and to assist in making the book more useful to the student and the class-room. G. F. MATTHEWS. 98 SINCLAIR ROAD, W., September ; 1890. CONTENTS. CHAPTER I. SECTIONS. PAGE DEFINITIONS OF LOGARITHM, CHARACTERISTIC, MANTISSA, - 1-4 - I ARITHMETICAL OPERATIONS WITH NEGATIVE LOGARITHMS, - 5 - 4 ABBREVIATED PROCESSES OF MULTIPLICATION AND DIVISION, 6 - 6 CHAPTER II. FUNDAMENTAL PROPERTIES AND USES OF LOGARITHMS, 7-8 - 9 EASY PROBLEMS, 9 - ii (i) To FIND THE LOGARITHMS OF PRODUCTS, - 9 - ii (2) QUOTIENTS, - 9 - 12 (3) ,, ,, ,, POWERS AND ROOTS, - 9 - - 13 (4) ., ,, FACTORS, - 9 - - H (5) To FIND THE VALUES OF LOGARITHMIC EXPRESSIONS WITH- OUT THE AID OF GIVEN LOGARITHMS, - 9 - 16 (6) To DETERMINE THE CHARACTERISTICS OF LOGARITHMS, - 9 - 17 (7) To SOLVE EXPONENTIAL EQUATIONS, - ... 9 - - 18 CHAPTER III. THE SELECTION OF A BASE, 10 - 21 THE NAPIERIAN SYSTEM, ii - 21 THK COMMON SVSTRIW. T 2- I 7 ?.A CHAPTER IV. THE USE OF TABLES OF LOGARITHMS. PROPORTIONAL PARTS, - 18-21 - 28 CHAPTER V. RELATION BETWEEN THE LOGARITHMS OF NUMBERS TO DIFFER- ENT BASES, 22-25 - - 35 viii LOGARITHMS. CHAPTER VI. SECTIONS. INTEREST AND ANNUITIES, - 26-35 SIMPLE INTEREST, - - 27 COMPOUND INTEREST, 28 FORBORNE ANNUITIES, - 31 TERMINABLE ANNUITIES, 32 PERPETUAL ANNUITIES, - 33 RENEWAL OF LEASES, - 34 CHAPTER VII. APPLICATION OF LOGARITHMS TO PLANE TRIGONOMETRY. LOG- ARITHMIC RATIOS, - - - - 36-43 CHAPTER VIII. THE USE OF TABLES OF LOGARITHMIC RATIOS, - - 44-48 CHAPTER IX. REDUCTIONAL FORMULAE, 49 - SUBSIDIARY ANGLES, - 50 CHAPTER X. SOLUTION OF TRIANGLES, - 51-59 CHAPTER XI. HEIGHTS AND DISTANCES, - 60-65 - CHAPTER XII. APPLICATION OF LOGARITHMS TO MENSURATION, - 66 MISCELLANEOUS QUESTIONS. ( To be worked with Tables. ) EXAMPLES A, ARITHMETICAL, EXAMPLES B, TRIGONOMETRICAL, - ANSWERS, LOGARITHMS. CHAPTER I. Definitions of Logarithm, Characteristic, Mantissa, Arithmetical Operations. i. If a be a positive real quantity greater than unity the function a* can be shown to be a continuous function of x, susceptible of all positive values between o and +00. By this is meant that whatever real value, positive or negative, integral or fractional, be given to x, then (provided positive real roots only be admitted when x is fractional) a* will always equal some positive real quantity and will change in value continuously with x ; in other words, when any indefinitely small change is made in the value of x, there will always be a corresponding indefinitely small change in the value of a* ; and conversely, for every indefinitely small change in the value of a* there will always be a corresponding indefinitely small change in the value of x. Putting a*=y, these results may be conveniently stated thus : "When a is a constant real positive quantity greater than unity, x varies continuously with y, and conversely." Since y is here expressed as a function of x, it follows that x must also be some function of >-, and that, as the value ofy is determinate when x is known, so also, when y has this value given to it, among the corresponding values of x will be found that which determines the said value of y. x is in fact the index of that power of a which is equal to y, and we require some symbol to express this new function of y. The constant quantity being called the base, the new function is called the logarithm of y to the base a and is written log a y. Thus x = \og a y, and it is evident that cfy \ and x = \og a y ( express one and the same functional relation between the variables x andjy. Further, since log a y or x is the index of the power of a that is equal toy, it follows that alogaV =y is an algebmical identity< We therefore have the following Definition of a logarithm : " The logarithm of a number to a given base is the index of that power of the base which is equal to the number." LOGARITHMS. For example, the square of 3 is 9 (3 2 = 9), and therefore 2 is the logarithm of 9 to the base 3 ; or again, the logarithm of 10 to the base 100 is J, since the square root of 100 or iooi is equal to 10. 2. Now a is greater than unity and positive, so that a* increases continu- ously with x and is positive whatever x may be. But when x is negative and infinitely large numerically, cf or y is positive but indefinitely small, i.e. log a o= - oo . Also, so long as x is negative a* is always less than unity, and when x o it equals unity, for a=i. Therefore, since x and y vary continuously, as x increases from - oo to o, a x passes through all positive proper fractional values from o to i, and equals i when x = o y i.e. log ffl i=o. Again, as x increases from o to + oo , a x increases constantly from unity without limit, passing through all positive real values greater than unity till we have Iog ( + oo ) = + oo . Hence, though x, or the logarithm of the number y to the base a, may have any value between - oo and +00 i.e., may be either positive or negative, a x , or the number y, must be positive ; in other words, though we can always find the number corresponding with a given logarithm positive or negative, we can only find the logarithms of positive numbers. It is also clear that, whatever be the base, the logarithms of all numbers greater than unity are positive, less than unity negative. At the same time we see that, when the number is equal to the base, the logarithm is i, for a 1 = a or log tt # = i ; and when the number equals the reciprocal of the base, i.e. -, the logarithm is i, for a~ l = r- ] or log a ( - ) = - i. a \aj \aj The above relations between the values of numbers and their respective logarithms may be conveniently viewed in the following table Numbers ( + ve.), y- a x =o Logarithms ( + ve. or -ve.), x = \og a y= oo where F+F' '= I, so that F' is also a proper fraction, It is thus seen that any negative logarithm - (n + F), consisting of a negative fraction - F and of - n, a negative integer or zero, can be made to consist of the positive fraction F' (where F+F' ' = i) and the negative integer -( + x). The characteristic of the logarithm is - (n + i ), the mantissa F'. Example. Given \ogx= .32146872, arrange the logarithm in proper form. log* = - 3.2146872 = -3 -.2146872 = -3~(i -.7853128) Characteristic = 4. = -4 + . 7853128 Mantissa =.7853128. = 4.7853128 [Observe that the minus sign of the characteristic is written above and not before it, to avoid the confusion that would otherwise arise, since the decimal fraction that follows it is positive and not negative.] It is evident that when a logarithm is given as a decimal wholly negative, the characteristic of the logarithm is negative and arithmetically greater by unity than the portion of the negative logarithm to the left of the decimal point, while the mantissa is positive and equal to that decimal fraction which makes up unity together with the portion of the negative logarithm to the right of the decimal point. And the converse rule is readily deduced when a negative logarithm, properly expressed, is to be reduced to the wholly negative form. The decimal fraction that together with any other decimal fraction makes up unity may be written down at once from left to right by making up nines with every figure except the last which is not zero, when the figure is written down which makes up 10. Thus a logarithm expressed as a decimal wholly negative may be written down at once in proper logarithmic form, and vice-versa. E.g. .2468719= 1.7531281 - . 1 104030 = 2. 8895970 3.2415602= -2.7584398 I.3OIO3OO= - .6989700 5. Since negative logarithms are expressed as partly negative and partly positive, it is well to be able to perform the simple operations of addition, subtraction, multiplication, and division upon them while still retaining them in this form. The simple rule which guides all these operations is to " treat the mantissas arithmetically, the characteristics algebraically" DEFINITIONS. 5 (i) Addition of negative logarithms. Rule : Place the logarithms one under the other and proceed to add in the ordinary way. On arriving at the decimal point, the tens are carried on and added in algebraically with the several positive and negative characteristics, giving altogether a positive, negative, or zero result as the case may be. Examples. (i) 2.7864007 (ii) 4.8491642 3-24I95I5 3-6523150 _. 1 1 52404 .9324719 1.8413432 i. 50053 1 3 7.9849358 1.9344824 (ii) Subtraction of negative logarithms. Rule : Place the logarithms one under the other and proceed to subtract the mantissae arithmetically, borrowing as usual in the ordinary way, when necessary, on reaching the deci- mal point. The characteristics are then subtracted algebraically the one from the other, having previously paid back any borrowing that has taken place, by an algebraical addition of unity to the characteristic subtracted. Examples. (i) 3. 2801 562 ~\ 2.78632781 (paying back -2 becomes - I, and subtracting -I from -3 2.4938284) ^e result is -2). i) 2.8763405^1 4.44528621 (4 subtracted from 2 gives -2). 2.4310543^ (iii) Multiplication of negative logarithms by positive integers. Multiply the mantissa, and then the negative characteristic, adding in algebraic- ally with this latter product the tens, if any, that are carried on from the multiplication of the mantissa. Examples, (i) 2.7864258x24 (ii) 3.5117062x341 _ 4 .5117062 ). 5.1457032 (-3X34D+ 174 = 2849, 308742192 2046824? 15351186 174.49181 OBS. If the multiplier be negative, multiply by the corresponding positive number and change the sign of the result : E.g. 3.5117062 x -341 =-(3.5117062x341) = -(849.49181) = 849 -.49181 =848.50819. (iv) Division of negative logarithms by positive integers. Rule : If the negative characteristic be exactly divisible by the divisor, divide out at once in the ordinary way, the integral portion of the result being negative and the rest positive. 6 LOGARITHMS. If the negative characteristic be not exactly divisible, split it up into two portions, one negative and the other positive, the negative portion being the next integer arithmetically greater than the characteristic that is exactly divisible by the divisor. The quotient obtained by dividing this negative portion is then the negative characteristic of the result, while the compensatory positive portion is taken with the positive mantissa of the dividend to give on division the positive mantissa of the result. Examples. (1)6.2513248 + 3. ^1)5.0741213 + 4. (iii) 23.1175056+ 17. 3)6. 25 1 3248 4) -8 + 3.0741213 23.1175056 = -34+11.117056 2-0837749 2.7685303 J7 2 + >65397Q9 J 7 2. 6539709 . OBS. If the divisor be negative, divide by the corresponding positive number, and change the sign of the result after division, or of the dividend before division. 6. Multiplication and division of approximate decimals. Except in the few cases in which logarithms are wholly integral, they are given in the tables to a certain degree of approximation, and generally to 7 places of decimals. Now when approximate decimals are submitted to arithmetical operations, the results so obtained can never be correct to a greater degree of accuracy than are all those decimals that enter into the calculation. By the degree of accuracy is here meant the number of correct figures given, taking all the figures into consideration and not merely those that follow the decimal point. Hence, suppose a number correct to 5 figures is to be multiplied by another given correct to 8, the resulting product can be correct to 5 figures only, and the last three figures of the number given correct to 8 are useless and may be neglected in the multiplication. Again, if these two decimals, composed of 5 figures each, be multiplied together in the ordinary way, it will be found that certain columns are deficient, that is, that figures are absent from these columns, figures that would have appeared, had more figures been given in both the approximate decimals used. Hence the results obtained by adding up these deficient columns are useless. It naturally occurs to us that the process of multiplication might have been shortened, and that we might have adopted some method giving all the complete columns and excluding all the superfluous figures occurring in incom- plete columns. And this is the case. So, too, in the process of the division of approximate decimals, we are able to leave out superfluous figures and still obtain a result correct to the same degree of accuracy as are both divisor and dividend, and to the greatest degree of accuracy obtainable under the circum- stances. We will illustrate these processes by applying them to the multiplication and division of approximate logarithms. As regards addition and subtraction the end is attained by simply leaving out those figures that would appear in deficient columns, and the process needs no explanation. Logarithms, and other decimals, expressed exactly, and not approximately, may of course be treated as accurate to any number of figures, the correct figures not written being all of them zeros. DEFINITIONS. 7 (i) Multiplication of two positive logarithms. Example. Multiply 1.8836614 by 2.6180481. (i) Short Method. (2) Long Method. Rule: Multiply by the figures of the I.$8.3*$XH 1.8836614... multiplier in order, beginning on the extreme 2.6180481 2.6180481... left instead of the right, and cut off the ^767.228 i 8816614 ' fi S ures of the multiplicand from the right, 1:1301968 150 692912 one by one after each multiplication. In 188366 753 46456- 150692 150692 912.... 753 188366 150 113019684? i 37673228 4.9315158 4.93I5I58 -re- down the several products allowance must be made for the figures cut off, to the extent of carrying on the tens that would have been carried on had no figures been cut off, and the first figure written down in each line must be placed always in the same column. As regards the fixing of the decimal point, its position can be calculated in any line in the usual way by adding together the number of decimal figures in the multiplier and multiplicand that produce that line (of course neglecting the figures cut off), but perhaps this can be most conveniently done when multi- plying by the units figure of the multiplier, when one exists, for then we simply mark off as many decimal figures as are contained by that portion of the multiplicand multiplied. The last figure of the result obtained by this short method of multiplication may differ by a few units from the true figure in consequence of the omission of the next column, but this is the extent of the error. (ii) Division of two positive logarithms. Example. Divide 4-9315158 by 2.6180481. 2.6X^4^)4.9315158(1.8836614 2 6180481 2 3134677 2 0944384 2190293 2094438 orgcc Rule : Start the division in the ordinary Hg r^ j way, and after the first step in the division cut off the figures of the dividend from the right, I ^3 I 4 one by one a t e a c h successive step. In writing down the products allowance must be made as J 6o6 beforey^r the figures cut off. 1570 36 26 10 IO (iii) Multiplication and division of negative logarithms. When either or both of the logarithms are negative they must be converted into the wholly negative form previously to multiplication or division. The product or quotient will then be positive or negative according as both, or only one, is negative. 8 LOGARITHMS. EXAMPLES. II. 1. Convert the following logarithms into the wholly negative form : (i) 1.0072815, (ii) 3.1241793, (iii) -4.2840617, (iv) 2.5351600. 2. Express the following negative decimals in logarithmic form : (i) -.3124765, (ii) -2.9190618, (iii) -3.5, (iv) -2j. 3. Find the value of (i) 1.3876420 + 2.8561247 + 3.7201504 + 5.3876004 + 3, (ii) 4.2861720 + 3.1174628 + 6.5407106 + 2.5160208 + 3, (iii) .1876789+1.4027512 + 2.6171840 + 3.8405816, (iv) 4.6378315 + 2.8516720 + 3.4116712 + 3.9192117, (v) 5.7168497 + 1.3840795 + 2.1197197 + 4.2084006, (vi) 3.1196117+4.8533162 + 3.4024814 + 2.8461620, (vii) 4.6281479-6.2861504, (viii) 3.1750462-2.1686128, (ix) 2.0041060-3.1171628, (x) 3.4281025-4.5651526, (xi) .3176212-2.8607127, (xii) 4.1271616-3.2870122, (xiii) 2.8176404 -3.4688182 -2.6415287 + 1.41 1485* (xiv) 3.9641867-2.8451521-1.0067167-3.8976719, (xv) 4-3.4684254-2.6104602 + 3, (xvi) 3.2876406 + 5.3158452 + 2.1876717-4.7606156, (xvii) 2.4165314x7, (xviii) 3.5090067x2!, (xix) 1.9877617 xii, (xx) 4.2076842 x 72, (xxi) 2.8681184x113, '(xxii) 3. 2424860 x -99, (xxiii) 6.4282007 + 3, (xxiv) 2.5176861 + 7, (xxv) 18^.2150267+ -5, (xxvi) 1.6899129 X3J, (xxvii) 7. 24087 1 7 -7-2 j, (xxviii) 4.4465142 + 3.3, (xxix) 2.5176502x2.4045416, (xxx) 3.2164112 + 3.7176407, (xxxi) 1.1171115 + 2.8406712, (xxxii) 3.2082207x2.1176891, (xxxiii) 2.2461172x3.8406002, (xxxiv) 4.2895165 + 3.4671008. CHAPTER II. Fundamental Properties of the Logarithm. 7. Having defined the logarithm, we proceed to establish certain properties of the function, properties that render it invaluable as a means of facilitating arithmetical processes. We shall take b to represent the base, so that where the base is omitted from the logarithmic function it will be understood to be b. Prop. I. " The logarithm of a product of numbers equals the sum of the logarithms of the several factors." We have identically x - fi * x y ............ , and so on. By multiplication x.y.z...- #<* *+ lo s +!**+.. . log (x.y. r,...}=\ogx + \ogy + \ogz+... Q.E.D. Prop. II. ' ' The logarithm of a quotient of numbers equals the logarithm of the dividend diminished by that of the divisor." log a: "i y logy [ . By division - = &<* x ~ lo v ; Q.E.D. Prop. III. " The logarithm of any power or root of a number equals the logarithm of the number multiplied by the index of that power or root." We have x - b lo *, and x m = b m lo * * log x m = m log x. Q. E . D. Prop. IV. " The logarithm of the base itself is always unity." We have b l = b, .'. logj=i. Q.E.D. Prop. V. " The logarithm of unity is zero to any base." We have =i, .'. log&i=o. Q.E.D. 8. In Props. I, II, III are involved the important properties of the logarithmic function, and the argument for its great utility in facilitating the operations of multiplication and division and in finding the powers and roots of numbers. To render this part of the subject clearer we will connect these properties 9 I0 LOGARITHMS. of logarithms more directly with the corresponding theorems in indices, with which they are in reality identical, for logarithms, being simply and purely indices, admit as such of all the algebraical simplifications to which indices are subject. By the laws of indices, b x b n x b? x . . . = b m + n +P+- Reading these indices as the logarithms of numbers to the base b \ since b m is the number whose logarithm to the base b is equal to m, we deduce the following results. (i) The product of those numbers whose logarithms to the base b are m, n, /, etc., equals that number whose logarithm to the same base is m + n+fl+... (Prop. I.) (ii) The quotient of the numbers whose logarithms are m, n, equals that number whose logarithm is ;;/ - n. (Prop. II.) (iii) The power or root (index ;z) of the number whose logarithm is ;;/ equals that number whose logarithm is m multiplied by the index of the power or root. (Prop. III.) These results explain how logarithms may be utilized to effect products and quotients, and to find powers and roots of numbers. By (i) To find the product of certain numbers Add the logarithms of these numbers to- gether, and the required product will be the number whose logarithm is this sum of logarithms. By (ii) To find a quotient of numbers Subtract the logarithm of the divisor from that of the dividend, and the required quotient will be the number whose logarithm is this difference of logarithms. By (iii) To find a power or root of a number Multiply the logarithm of the number by the index of the power or root, and the required power or root will be the number whose logarithm is this product. In making practical use of these logarithmic properties we must of course be supplied with tables in which the logarithms of numbers are given to some degree of approximation and to a constant base. We then proceed as in the following examples (in which use is made of a table of logarithms calculated to the base 10). Examples, (i) Find the product 32 x 16 x 35. log 32= 1.505150x3 log 16 = 1.2041200 log 35 =1.5440680 /. by addition, log (32 x 16 x 35) =4.2533380 But log 1 7920 = 4. 25 33380, :. 32x16x35=17920. (ii) Find the value of 18069-7-57. log 1 8069 = 4. 2569341 log 57 =1.7558749 /. by subtraction, log (18069 -7-57) = 2.5010592 But log 317 = 2. 5010592, .'. 18069-7-57 = 317. FUNDAMENTAL PROPERTIES. (iii) Find the 4th power of 17. log 17 = 1.2304489 _ 4 /. multiplying by 4 (the index of 4th power), log I7 4 = But log 83521 =4.9217956, .'. 174 = 83521. (iv) Find the loth root of 59049. ^59049=^4.7712125 (X T \) /. multiplying by T V (^ Q index of the loth root), log 1 \/59O49 = .4771213 But log 3 = .4771213, .-. ^59049 = 3- In the extraction of roots the logarithmic function is of particular value, for the usual arithmetical processes extend only to square and cube roots and roots that may be extracted by a succession of these operations, and the Binomial Theorem is only practicable over a small range of numbers, whereas any roots whatever may be obtained approximately, and with great readiness, by the application of logarithms. Also, we shall find that the function enables us to solve approximately a certain class of equations called Exponential Equations, and thereby to effect the solution of a variety of questions in percentage and interest. 9. We will conclude this chapter with a few easy problems that depend for their solution directly upon the fundamental properties of the logarithm above explained, all logarithms requisite for the purposes of the questions being given. (i) To find the logarithms of products. Example. Find Iog 10 8o64, given Iog 10 2 = . 30 10300^ = -4771 213 \ = - 8450980 J Iog 10 8o64 = log (7 x 3 2 x 2 7 ) = Io gio7 + 2 Iogi 3 + 7 Iog 10 2 (Props. I, III.) = .8450980 + .9542426 + 2. 1072IOO = 3.9065506. EXAMPLES. III. i. Given Iog 10 2 = . 30 10300, Iog 10 3 = . 477 1213, Iog 10 7 = . 8450980, Iog 10 ii = 1.0413927, find (i) Iog 10 i728, (v) Iog 10 588, (ix) Iog 10 i875, (xiii) Iog 10 630, (xvii) Iog ]0 i9965, (ii) logioQS, (vi) Iog 10 285768o, (x) Iog 10 25200, (xiv) Iog 10 io78, (xviii) Iog l0 44000, (iii) Iog 10 675, (vii) Iog 10 5625, (xi) Iog 10 50o, (xv) Iog 10 66825, (xix) Iog 10 i4553, (viii) Iog 10 392, (xii) Iog 10 98784, (xvi) Iog 10 6o48, (XX) Iog 10 29282. [When the base is 10, logiO=i by Prop. IV, and log 2 + log 5 = log 10= I, so that fog 5 = r- log 2.] I2 LOGARITHMS. 2. Given a? + P=i, log 2 = .3010300) log (i +a) = .i 928998 k show that log (l +a + t>) = .3780762. log (i +) = .2622226 J 3. If log I025 = a, log 2 = /3, show that log 4100 = a + I2/S. (2) To find the logarithms of quotients. Examples, (i) Find log^ic^, given Iog 10 2 = . 3010300} = -4771213 = Iog 10 7 + 2 Iog 10 2 + 2 Iog 10 3 -2log 10 5 (Props. I, II, III.) = .8450980 + .6020600 + .9542426 - 1.3979400 = 1.0034606. (ii) Find Iog 10 .oi5, given Iog 10 2 = .3010300) Iog 10 3 = .4771213; = 2.1760913. [In such examples as (ii), "when the base is 10, it will subsequently be found sufficient to treat the number as a whole number, neglecting the decimal point, and then merely changing the characteristic of the result so obtained. Vide Props. VII, VIII, IX.] EXAMPLES. IV. i. Given Iog 10 2= .3010300, Iog 10 3= -477I2I3, Iog l0 7 = -8450980,) find = 1.0413927, Iog 10 is = 1.1139434, Iog 10 i7 = i. 2304489,,! (i) Iog 10 2.i, (ii) Iog 10 34.3, (iii) Iog 10 .i25, (iv) Iog 10 .ooo8, (v) Iog 10 .oo5, (vi) logu)- 4 -, (vii) Iog 10 3, (viii) (ix) Iog 10 3.^, (x) Iog 10 2.oi6, (xi) Iog 10 28.954, (xii) (xiii) Iog 10 .426, (xiv) Iog 10 22f, (xv) Iog 10 ,-_, (xvi) 30-851 (xvii) Iog 10 8.72, (xviii) Iog 10 '?' ' ., (xix) Iog 10 l7f|, (xx) 1.923076 FUNDAMENTAL PROPERTIES. 2. Given Iog e 2 = .693i47i8, log>3 = 1.09861229, find (i) Ioge4i> (ii) log e . 5, (iii) loge 3 ^, (iv) 3. Given Iog 10 2, Iog 10 3, Iog 10 7, find (0 2l ogio^+4log 10 ^+5log 10 ^, (ii) 2 Iog 10 6| + 5 Iog 10 14 + log * + 6 Iog 10 15, 9 I2i 14 ~ 3 Iog 10 y4 + 4 lgi<>2& - 3 Iog 10 7. 5. 4. If A, B, C be in H.P., then log (A + C), log (A - C), and log (A + C-2B) are in A. P. (3) To find the logarithms of powers and roots. Examples, (i) Find log^x/. 084, given Iog 10 2, Iog 10 3, Iog 10 7. = -log 10 .o84 (Prop. III.) = j 1 ( lo Sio7 + Iogio3 + 2 Iog 10 2 - 3 Iog 10 io) = -^(2.9242793) = 1.9282853. (ii) Find Iog 10 i 8^/7^1 2^/5, given Iog 10 2, Iog 10 3, Iog 10 7. -log]o(i8- 7- 12. 5) = Iog 10 i8 + i Iog 10 7 + J Iog 10 i2 + T V log 10 S = 1.7750829. EXAMPLES. V. Given the same logarithms as in question i, Ex. IV, find the following logarithms to the base 10 : (i) log (.0147)2, (ii) logN/126, (iii) log^s.l, (iv) log (|4^ } (v) logVTi, (vi) logAv/5, (vii) Iog2 ", (viii) Iog5~% (ix) log^/l, (x) log^/i, (xiii) log (2. i) 5 , (xvi) log (i. 75 A (xi) log(3irt (xii) log ^.0343, (xiv) logVii.j, (xvii) Iog(i47) 2 , (xv) log >/. 000207 36, (xviii) logf- 4 -^.)^, LOGARITHMS. (xix) log , (xx) log U. 0000024, (xxi) log V. 000735. \/.0245 (xxii) log \/.oi, (xxiii) log-J^/|^J", (xxiv) log (363 \/ 2)^, (xxv) log 7 V6.V/5, (xxvi) log- __, (xxvii) Vi-925 (xxviii) logV252\/4, (xxix) log -N/3v/4V5, (xxx) (xxxi) log-V^r' (xxxii) lo gX /^^-, (xxxiii) wfiw 3v7 ' (xxxiv) ]ogV(i5)ix(5i)f, (xxxv) logVi3/s/5-r/v/7, (xxxvi) log-[Y!i^x/i8 x \/2.il 1X25; J (xxxvii) log-[(3)*X3Ay?W2.el, (xxxviii) L 2. Given Iog e 2 = .69314718, Iog e 3= 1.09861229, find the logarithms to the base e of 3. Given Iog a 2 = . 1704321, find the logarithms to the base a of ! V^" 4 , 5> 4. Given log^r= 1.7140628, lo&?s 1. 4255632, find (1) 108 T5" (ii) Ios ^ 5. Prove that log e [log e {logX}]= I. (4) To find the logarithms of factors. Example. Given Iog 10 ^| =1.5650765^ logjo\^F= .661 1096 j-i find Iog 10 2, Iog 10 3, Iog 10 7. log .112= I.0492l8oJ These equations give Iog 10 2 + 2 Iog 10 3 - 2 Iog 10 7 = -. 4349235 } i Io gio3 + i Jogio? = . 66 1 1 096 V 4 Iog 10 2 + Iog 10 7 = 2. 0492 1 80 J Put Iog 10 2 = x, Iog l0 3 =y, log ]0 7 = s ; then .* + 2y - 30 = -. 4349 2 35l whence x or Iog 10 2 = . 3010300^ = 1.3222192 V j or Iog 10 3 = .4771213 ^ J s or Iog 10 7 = . 84 50980 J FUNDAMENTAL PROPERTIES. ! EXAMPLES. VI. 1. Given lo ]0 648 = 2.8115750! c ji 101/0864= 2. 9365137)' ^ I0gl 3 "^ 10gl 5 ' 2. Given Iog 10 18=1.25527251 Iog 10 i25 = 2.0959100 V, find the logarithms of the numbers from 2 to 9 inclusive. 21 = 1.3222 193 J 3. Given Iog 10 .oi2 = 2.0791812^ c ,, 14^018 = 2.2552725)' find lo & 2 and 10&03- 4. Given Iog 10 .oi3 = 2.1 1394341 fi , , log 10 .637=T.8Q4i394/' h ' Iogl 7 - nd ^o5o and Io glo35 o. 7. Given Iog 10 i25 =2.0969100, find Iog 10 i6 and log 10 (.ooO2) 5 . 8. Given Iog 10 4.2 = .6232493! Iog 10 .012 = 2.0791813 !-, find the logarithms of the nine digits to the base 10. Iog 10 .044 1 =2. 6444386 J 9. Given Iog 10 i = 1.6989700) log lo * = 1.5228787!' _ i _ . find the logarithms to the base 10 of \/6, f v 14.4, and T 7 ^ V27O x ^ v625- 10. Given Iog 10 i.4 = .1461280) Iogi i.5 = . 1 76o9i3/' find Iog 10 . 000315 and the value of 2 log 10 ^-+3 Iog 10 2 - 5 log ]0 i+ 5 Iog 10 i.6. 11. Givenlog 10 2 = .3010300! Iog 10 i3 = 1.1139434 V, find Iog 10 .ooo2o8 and Iog 10 2.oi6. = I.605664I J 12. Given Iog 10 | = 1.6989700! ' ' V, find Iog 10 i.5, Iog 10 2.s, and Iog 10 3.5. - Given - find 14. Express Iog ]0 .oo3i25 in terms of Iog 10 2. 15. Given jgi<>i8 = J' 25527 ^|, find the logarithms to the base 10 of 5, 6, 3, 450* '75 and l6 1 6. Given Iog 10 5. 76 = .7604226) log ]0 2 = .3010300 Iog 10 .0105 = 2.0211893 10 . . log ]0 2 = .3010300 V, find the logarithms of the digits above 2 to the base 10. ] e .9 = -x) e .96 = -y k e l. OI25 = S J 17. If log e log e .96 = -y k find Iog e 2, Iog e 3, and Iog e 5. log e 1 6 LOGARITHMS. 19. Given log ]n i.76 = .24 "55127"! c , , lo&9i =.9498777J' find logl 150 and 20. Given Iog 1ft i56 = 2. 21. Given logx/^ 2.0962321 j find logVa^ss T. 1255217.1 B 6 22. Given log e 54v/^ = 4- 4889840] loge^= .35981391' tje ) find the logarithms to the base e of tj2e and -[f.3 j^- 2 V g \. 23. Given log 10 2O= 1.3010300, find Iog 10 .oooi25 and the logarithm to the base 10 of .2 x .4 x .8 x ...to 10 factors . 5 x . 2. 5 x 1 2. 5 x ... to 6 factors ' (5) To find the values of logarithms and logarithmic expressions, no logs being given. Prop. VI. To show that log n a m - . og an (a w ) w =- logos' 1 , by Prop. Ill, = Examples, (i) Find the value of log 8 2 x /2. = - by Prop. IV. fi) Find the value of 6 log 10 -~4 log 10 i =log 10 i=a [When no logs are given, the only logs we are allowed to assume are log I which is always zero, and the logarithm of the base itself which is unity.] FUNDAMENTAL PROPERTIES. EXAMPLES. VII. 1. Find the values of (i) Iog 7 343 (ii) Iogvs4> (iii) Iog 2 79> (v) Iog 10 .oi, (vi) Iog 2 . 25 3.375> (vii) log^S, (ix) log 8X /2, (x) Iog 2 (|), (xi) logs / 25, (xiii) log 27 9v/3, (xiv) Iog 4 \/i6) (xv) log 4 V-5, (xvii) Iog 5 .04, (xviii) Iog 3 .i, (xix) Iog 2 .^i.6, 2. Find the values of (i) iolog 10 - + 7log 10 ^ + 4log 10 ^-, 2 18 25 log 10 >/27 + Iog 10 8 - log 10 N/iooo Iog 10 i.2 3lo glo i728 \ ( v ) logWS4 - 27 3. Show that lo glo ^ - 2 Iog 10 | + iogio^ = Iog 10 2 (6) 72; determine the characteristic of a logarithm. The mantissae of logarithms are always positive ; therefore, when a logarithm is not entirely integral, its characteristic is always the algebraically smaller of the two successive integers (whether positive or negative) between which the log- arithm lies : but these integers are by Prop. Ill respectively the logarithms of the same powers of the base ; hence, the characteristic of the logarithm of any number is the algebraically smaller of the indices of those successive powers of the base between which the number lies. 1 8 LOGARITHMS. Examples, (i) Find the characteristic of Iog 6 2o62. i 6 _6 36- 2062 lies between 6 4 and 6 5 , therefore 4 is the characteristic of Iog 6 2o62. [This method applies to the logarithms of all numbers greater than unity. ] 1296 = 6* 6 7776 = 6 5 (ii) Find the characteristic of Iog 12 .ooo23. 12)1. I2).o8333= I2' 1 .00023 lies between I2~ 3 and I2~ 4 , therefore 1 2). 00694= 12 ~ 2 4 is the characteristic of Iog 12 . 00023. 1 2). 00057 = 12 j- This me thod applies to the logarithms of .00004 = 12 4 numbers less than unity.] (iii) Find the characteristic of log 3 V.ooo7. Since 1 og 3 V. 0007 = ilog 3 . 0007, we find the characteristic of Iog 3 .ooo7 and then divide it by 5. Now, as in example (ii), the characteristic of Iog 3 .ooo7 is 7; therefore, since 7 = - 10 + 3, dividing - IO by 5 we find that the characteristic of log 3 Vooo7 is 2. EXAMPLES. VIII. i. Find the characteristics of (i) Iog 7 5473, (ii) Iog 5 .oi7, (iii) Iog 10 2i47, (iv) Iog 3 .o84, (v) Iog 2 2i.84, (vi) Iog 3 .i2, (vii) Iog 2 . 5 .oo6, (viii) Iog 25 .ooo2, (ix) Iog 12 >/35o, (x) Iog 7 \/.oi7, (xi) Iog 3 32i4, (xii) Iog 7 .oooi5, (xiii) Iog 2 \A.oi4, (xiv) Iogy246, (xv) Iog 1 . 5 (i3.2) ? , (xvi) Iog v i7 - , \/2 (xvii) Iog 2 . 5 v/.oooi7, (xviii) Iog 3 2. How many positive integers are there whose logs, to the base 3 have 6 for a characteristic? (7) To solve exponential equations. Exponential equations, soluble by means of logarithms, are of two classes ; (i) those in which we may proceed at once by taking logarithms, (2) those FUNDAMENTAL PROPERTIES. which must be reduced before taking logarithms. In the first class the signs + and - occur, if at all, only among the exponents : in the second class these signs occur between terms of the equation. Examples, (i) Solve the equation 3* . 2* = 4*+ 1 , given Iog 10 2 = . 30 10300! Equating the logarithm of the left hand side to that of the right, we have x Io gio3 + * logic 2 =(-*+!) Iog 10 4. x ( Io gio3 + logio 2 ~ Iogio4) = logi<>4 r - lQglo4 _ 2lQg 10 2 Iogio3 - = . 6020600 .1760913 (ii) Solve 4* +2* = 12, given Iog 10 2 and Iog 10 3- We have a 2 * + 2* - 12 = o, .'. (2* + 4)(2*-3) = oand 2* =3 or -4. Now 2* must be positive and cannot equal - 4, 2* = 3, i.e. x Iog 10 2 = Iog 10 3 and x = = 1.58496. iogio 2 EXAMPLES. IX. . Given Iog 10 2 = . 30 10300, log 7 = .84150980,1 logos = 4771213: iii = 1.0413927,} solve the e( i uations (vii) 8*. 125*-* = 2**+*. 5* (viii) 3 s *. 5 3*-4 =7 *-i a j,a- (ix) (x) (|)*+ 4 = 25^+2, ( xi ) l oglo 2 a: + 3 = 1.2221818, (xii) _ _j (xv) 3*-6. 3~*=5, (xvi) iS}' x -^ x = Si\ 2. Given log ]0 -= 1.69897, find x from the equation 20*= 100. 3- Given Iog 10 2, Iog lo3 , Iog 10 7, and that 10^5277 = 3.722387, solve the equations 4. Solve the equation ^36 = 1.3678, given Iog 10 2, Iog 10 3, and that log 13734. 546=4. 1378144. 5- Solve the equations (i) ^ = 8, x = 3y ; (ii) 4096* =8(64-*), (iii) 5* = 6^(2*). 6. Find the value of x from the equation i8 8 ' 4j: = (54^/2)^-2, using the base 7. Solve the equations a 8. Find x from the equation Iog w 2 +n ^= i -log n2+M (^+ i). 9. Solve a x (a x - i) = i. 20 LOGARITHMS. 10. Solve the equations y<* a 1< = a s I ! . Given (a + ) 2: V - 2 4o \ = ^ / 13. What is the smallest integral value of x for which (i^)* is greater than a million? given Iog 10 io. i = 1.0043214. 14. How many factors, each equal to , must be multiplied together that the product may be less than .000001 ? given Iog 10 2 = .3010300. 15. How many factors 3 1 . 3 2 . 3 3 ... must be taken that the product may just exceed ioo,oco? givenlog 10 3 = .477 I2I 3- 1 6. Find very nearly a 4th proportional to the 6th root of 9, the 4th root of 7, and the 5th root of 5; given Iog 10 2 =.30103, log w 7 = . 84510, \ Iogio3 = -47712, log 10 i55-6 = 2. 19201. / 17. The ist and I3th terms of a G.P. are 3 and 65 respectively; find the common ratio; given Iog 10 65= 1.8129134, ^1292.1592 = 3.1113160, logic 3= .4771213. 18. Given a 1 , a 3 , a 5 ... =/, find the number of factors a\ a 3 , a 5 , etc. 19. Given a 1 , a 2 . a?...a n =#, find the value of n. CHAPTER III. The Selection of a Base. 10. The selection of the base in compiling a system of logarithms might be quite arbitrary, but certain considerations tend to give prominence to two systems, called severally the Napierian and Common Systems. (I) The Napierian System. 11. This system, which derives its name from Napier, the inventor of logarithms, is calculated to the base I . ] [For calculating log e (or+ i) from logger. x< I.] [Logarithms to any other base may be substituted for the Napierian logarithms in the above, provided the series be multiplied by /*.] 2 4 LOGARITHMS. EXAMPLES. X. 1. Compute to 7 places of decimals the I5th and 25th roots of e. 2. Find to 5 places of decimals the numbers whose Napierian logarithms are .05, . 125, 1.998. 3. Calculate to 7 places of decimals the Napierian logarithms of (i) 1.007, (ii) 1.03, (iii) i.ooi, (iv) .998, (v) .999, (vi) .983. 4. Given that /* = .43429 for base 10 ; calculate Iog 10 999 and Iog 10 iooi. 5. Find the Napierian logarithm of nrVir correct to 16 places of decimals. 6. Find log e (f^J-) to 7 places of decimals, and deduce logeQ^nj-)- 7. Show that log e ioi - Iog e 99 = ^j very nearly. 8. Given log e io = 2.3O2585i j calculate Iog e i2 to the same number of figures. 9. Find Iog 10 7, Iog 10 n, and Iog 10 i3; given Iog 10 2 = .30103000! /* = . 43429448 /' 10. Given log e io = 2. 3025851, find Iog 10 i.i to 6 places of decimals. 11. Find Iog 10 3.ooi by a series, given Iog 10 3 = .4771213^ 12. Solve the equation 10*= 101 to 5 places of decimals ; given log e io = (II) The Common System. 12. In this system, also called the Briggsian System, the base is to, this number being the radix of the common scale of notation. There are great advantages in adopting for the base of a practical system of logarithms the number upon which our system of numerical notation is based. The advantages are these : (i) The mantissa of the logarithm is independent of the position of the decimal point in the number, and is the same for all numbers com- posed of the same significant figures in the same order. (ii) The characteristic of the logarithm is determinable at once by inspec- tion of the position of the decimal point in the number. Hence (iii) Our table of logarithms, complete in every way for practical use, need only give the mantissae corresponding with certain collections of sig- nificant figures in the numbers, without regard to the position of the decimal point ; and characteristics need not be tabulated. A table so formed will give not only the logarithms of integral numbers, but will, at one and the same time, supply the logarithms of all numbers partly or wholly fractional when expressed in the decimal form, numbers THE SELECTION OF A BASE. 25 whose logarithms to any other base either would require separate tabulation or must be obtained by a subtraction of the logarithms of the numerator and denominator of the corresponding vulgar fraction. Now it will presently appear that, when the logarithms of all the natural numbers from i to 100,000 have been calculated to 7 places of decimals, by the application of a certain principle called the theory of proportional parts the logarithms of numbers composed of any number of figures may be readily calculated to the same degree of approxima- tion. Hence, by the simple tabulation of 100,000 logarithms, we are supplied with the logarithms, correct to the yth figure after the decimal point, of all numbers (integral or fractional) composed of any num- ber of figures whatever. 13. We will proceed to prove the above important properties of common logarithms. Prop. VII. "The mantissae are the same for the common logarithms of all numbers which differ only in the position of the decimal point." Let C be the characteristic (integral), and M the mantissa (fractional), of the logarithm of any number x, so that Iog 10 #= C+M; then any number which differs from x only in the position of the decimal point may be represented by x x io n , where n is some positive or negative integer. Now logio(* x io w ) = Iog 10 * + Iog 10 io n = C+ M+ n = (C+n) + M- C' + M, C' being integral and M fractional. Hence the characteristic of log 10 C#x io w ) is C' (- C+n), and its mantissa M the same as that of Iog 10 jr. Q.E.D. Prop. VIII. "The characteristic of the common logarithm of a decimal number, partly or wholly integral, is zero or positive, and is one less than the number of digits in the integral portion." Let x be a decimal number having n digits in its integral portion, so that it is not less than lo"' 1 nor as great as io n , where n is some positive integer ; then not less than Icg^io' 1 nor as great as Iog 10 io w , .e. Iog 10 * n-i n, Iog 10 ^= (n i) + F (where /MS zero or some positive proper fraction). Hence the characteristic of Iog 10 x is n - i, i.e. is zero or positive, and is one less than , the number of digits in the integral portion of x. Q.E.D. Prop. IX. " The characteristic of the common logarithm of a decimal number, -wholly fractional, is negative and numerically one more than the number of ciphers preceding the first significant figure." Let or be a decimal number, wholly fractional, having n ciphers preceding its first sig- nificant figure, so that it is not less than 1-j nor as great as , where n is zero or some T|-v?i-rl icW positive integer ; then Iog 10 # is not less than Icg^io"^ 1 ' nor as great as Iog 10 io~ w , -(n+i) -, -(n+i) + F (where F is zero or some positive proper fraction). Hence the characteristic of Iog 10 .r is - (n + i), i.e. is negative and numerically one more than n, the number of ciphers preceding the first significant figure in the decimal value of x. Q.E.D. 2 6 LOGARITHMS, 14. To sum up the results given by the above propositions we have (i) The following rule for determining by inspection the characteristic of the common logarithm of any decimal number : Rule : When the decimal point does not come first in the number, the characteristic is positive and one less than the number of figures preceding the decimal point; when the deci- mal point does come first, it is negative and numerically one more than the number of ciphers immediately following the decimal point. (ii) The means of writing down the common logarithm of any decimal number when that of a number is given which differs from the former only in the position of the decimal point. The mantissa is, by Prop. VII., the same as that of the given logarithm, and the proper charac- teristic is prefixed in accordance with the above rule for characteristics. 15. We now see why it is sufficient, in seeking the common logarithm of any decimal number, to find the logarithm of the integral number composed of its significant figures. Example. Given Iog 10 2= .3010300"! ,- ,, Iogw3= .477I2I3/.' find I a- (X 48 and Iog lo4 8oo. Iog 10 48 = 4 logio 2 + log 10 3= 1.6812413, /. Iog 10 .ooo48 = 4.6812413, I 1 6. We can also determine by inspection the characteristic of the common logarithm of any root of a decimal number. Example. Find the characteristic of Iog 10 ^.000427. Iog 10 V. 000427 = i Iog 10 . 000427. Now the characteristic of Iog 10 . 000427 is 4, and putting -4 = - 6 + 2, for purposes of division by 3, we see that the characteristic of Iog 10 V. 000427 is 2. 17. By means of Props. VIII. and IX. we are also able to solve certain questions as to the position of the decimal point in the value of any numerical expression consisting of products and quotients. Example i. Given Iog 10 3 = .4771213, find the number of digits in the integral portion of 3(2-7) 5 . Let * = 3(2.7) 50 , then Iog 10 x = log 10 3 + 50 Iog 10 2. 7 = 22.045316. Hence, since the characteristic of Iog 10 # is 22, by Prop. VIII. the number of digits in the integral portion of x must be 23. Example ii. Given Iog 10 2 = .3010300, find the position of the first significant figure in the decimal value of \/(.ooi6) 20 . Let x = \/(.ooi6) 20 , then Iog 10 * = ^ Iog 10 (. ooi 6) = 19. 360800. Hence, since the characteristic of Iog 10 j; is Tg, by Prop. IX. the first significant figure must be the igth after the decimal point (there being 18 ciphers). THE SELECTION OF A BASE. EXAMPLES. XI. Find, by inspection, the characteristics of the following common logarithms : (i) log 31. 7, (ii) log 2467000, (iii) ^52115.32, (iv) log .0024, (v) log 8. 925, (vi) log 85000.9, (vii) log 2008, (viii) log. 00007, (ix) log .0067, (x) log. 75, (xi) logV.'oT, (xii) logN/.oooosJ, I' <"> 'o^T=l 27 (xvii) log V. i 100 , (xviii) logJ^-' ~ 2. Given Iog 10 867 50 = 4.9475 19; write down Iog 10 867.5, Iogio8.675, and Iog 10 .o8675. 3. Given Iog 10 8i2. 13=2.9096256; writedownlog 10 8i.2i3, Iog 10 8i2i3ooo, andlog 10 .ooo8i2i3. 4. Given Iog 10 2 = . 3010300, Iog 10 3 = -477I2I3, Iog 10 7 = .8450980 ; find (i) lo gio3-75> (") lo gio-5 6 25, (iii) Iog 10 .o625, (iv) Iog 10 i4.4, (v) Iog 10 2.45, (vi) Iog 10 22.4, (vii) lo glo . 000002 1, (viii) Iog 10 6.75. 5. Given Iog 10 8. 1617 = .9117806; find the numbers whose common logarithms are 1.9117806, 3.9117806, 2.0882194, 4.0882194. 6. How many figures are there in the integral portions of the numbers whose common logarithms are 3.00271, .28467, 6.98015, l^H? ? 7. What is the position of the first significant figure in the numbers whose common log- arithms are 1.34816, 4, - ( ) ? \ 4.2 , 8. How many digits are there in (i) 2^, given Iog 10 2 = .3010300 ; and in the integral portions of (ii) V(2.25) 60 , given Iog 10 150 = 2. 176091 3 ; (iii) \/(2.5) 32 , given Iog 10 2 = .3010300? 9. What is the position of the first significant figure in (i) (.I2) 2 *, given Iog 10 5 = .6989700, Iog 10 i. 5 = .176091 3; (ii) (.oo7)* r , given log ]0 7 = . 8450980; (in) ( 002 4) ^ given loc: 10 i. 2 = .0791812, Iog 10 i. 6 = .2041200? \/. 0000003 10. Given Iog 10 2, Iog 10 3 ; find the integral values between which x must lie that the integral part of (1.08)* may contain 4 digits. 11. The integral part of (3.98 1 P - 000 contains 6o,coo digits. Find Iog 10 398i correct to five decimal places. 12. Show that (ini) 100 is greater than 100, given Iog 10 2, Iog 10 3, Iog 10 7. CHAPTER IV. Tables. Their Application. 1 8. We have seen in the last chapter how a table of logarithms is compiled. The logarithms of numbers are first calculated to the base e, and then those to any other base are obtained by multiplying the former by a certain constant modulus. For common logarithms this modulus is .43429448. It was also there stated that, when the logarithms of the natural numbers up to a certain point have been calculated, by the application of a certain principle, called the Theory of Proportional Parts, those of all other numbers can be deduced to a degree of approximation that will depend upon the magnitude of the numbers to which, and the range over which, the principle is applied. We will first prove the theory as applied to the logarithms of numbers, and then discuss its accuracy and mode of application in the case of common logarithms. Prop. X. To show that, when the differences are small compared with the number, the change in the logarithm is approximately proportional to the change in the number. We have log z ( + d)- \og x n = \og x ^A = log/ 1 + -} n \ n] From this it is evident that, when d is so small when compared with n that all the powers of - n after the first may be neglected in the series, log x (n + d} - \og x n = -. d, n i. e. , log^ n + d)- log x n oc d, in other words, the increase in the logarithm is proportional to the increase in the number. Q.E.D. 19. Applying the proposition of the preceding article to common logarithms, we will suppose that our table contains the logarithms of numbers from i to 100,000 (so that n contains 5 figures), and that d is not greater than unity ; then, since /z is less than \ for common logarithms, ^. is less than J ( ) , and a fortiori less than -000000003 ; 2H" * \IO,000/ Q is less than - th of this, and so on ; 3 d 10,000 28 TABLES. THEIR APPLICATION. 29 each term of the series being less than * th of the preceding term. 10,000 Hence, at least as far as seven places of decimals, [For a table of logarithms from i to 1000 the theory of proportional parts will give results true at least to 3 places of decimals, while logarithms from 1000 to 10,000 will give results true to 5 places of decimals.] Taking d to be any decimal fraction, so that n + d is a mixed deci- mal number, we are able, by applying the principle of proportional parts, to obtain to seven places of decimals the logarithms of numbers in which the decimal point comes after the fifth figure, and thence, by merely altering the characteristics of the results, the logarithms of numbers with the decimal point holding any position in those num- bers. Or since, as should clearly be the case, the theory of propor- tional parts is not vitiated when d and n are both multiplied by any one and the same power of 10, the logarithms of integers containing seven or even eight figures can be calculated from the logarithms of numbers having the same first five figures and ciphers affixed to make the total number of figures the same in the two numbers, though d in this case is greater than unity : the characteristics can then be altered to suit the positions of the decimal point when the numbers are not integral. The theory of proportional parts is utilised not only for finding the logar- ithms of given numbers, but also for finding the numbers corresponding with given logarithms. In this latter case, when the logarithms are given to seven places of decimals, we can always get y-figure results, and when the differences are large between the successively tabulated logarithms we may get 8 figures, but after this point additional figures in the number do not affect the logarithms to seven places of decimals, so that, with y-figure logarithms, numbers having given logarithms can never be found correct to more than 8 figures. [With 5-figure logarithms we can get 5-figure results always, and never more than 6 figures.] 20. To show how this principle is applied in practice (i) To find the logarithm of a number of not more than 8 significant figures, (ii) To find the number corresponding with any logarithm not given exactly in the tables, we will take an example of each case, making use of tables that give the logarithms of numbers from i to 100,000 to 7 places of decimals. 345.66,69. We have 3 additional figures in the number whose logarithm is required, and therefore affix 3 ciphers to each of the given numbers, which will not affect their mantissae. 3 \ LOGARITHMS. The mantissae of the logarithms of 34566000 and 34567000 are .5386491 and .5386617 respectively: therefore the mantissa of the logarithm ot 34566269, which lies between 34566000 and 34567000, will have some value between .5386491 and .5386617 ; its first four figures will be 5386 and the remaining three will compose some number between 491 and 617 ; call it 491+^. Now arrange the numbers and their corresponding logarithms in the parallel columns, in ascending or descending order, and couple the quantities in the same way^ in each column^ one coupling on each side. The four differences placed outside the couplings are then four numbers in proportion. It does not really matter how the couplings are arranged, provided only they are made in the same way and in the same order in both columns, but it is advantageous to couple the two extremes together, and the mean with that extreme which will give a difference d outside the coupling, i.e. with the smallest of the three occurring in the d column. No. Log. . 7538$$ 1 7 looo jvjtf6269\ 126 . Thus, 1000:269:: 126: or IOOO \IO IOO IOOO Hence 491+^=525, log 34566269 = 7. 5386525, and log 345. 66269 = 2. 5386525. In an example of this kind it is not usual to work as fully as in the above illustration. The figures to the left of the lines indicated may be omitted, being the same in every line, and our calculation is much facilitated by the use of the table of differences generally given in a column in the logarithmic tables by the side of those logarithms to which the differences severally apply. On referring to the tables, we find 126 in the table of differences, and under this number its tenth parts worked out ready for use. The hundredths and thousandths are obtained by cutting off figures successively from the given tenths. 126 13 2C [Here we have the table of tenths for difference 126. It will be observed that 38 in tabulating the tenths, when the first figure left out is greater than 4, the figure at 3** CO which we stop is increased by unity, but there is no such increase when the first J 63 figure omitted is less than 5. The reason for this is that by this process the value 76 that is nearer to the true value is always taken, and that though the values taken 88 are sometimes too large, sometimes too small, in the end the deficiencies and 101 excesses tend to compensate one another.] "3 From the above table we see that T 2 (jths of the difference = 25 Toooths ) Total correction = 34 The working is generally written down thus : log 34566000 = 7. 538649 1 2 25 6 8 9 _J /. log 34566269 = 7. 5386525 TABLES. THEIR APPLICATION. The mantissa of the logarithm whose number is required lies between the two given mantissae : therefore the number lies between 86553 and 86554, and consists of 86553 with additional figures. Now the difference between the two given mantissae is 50* and we can only get two additional figures, since thousandths of the difference 50 will not affect the 7th figure of the mantissa. Call the number composed of these additional figures d, and arrange in parallel columns as before, affixing two ciphers. No. Log. 400 , /. 50^2600,1.6.^-52, , 50 47\ ) \2i) and the required number is .08655352, the decimal point being so placed that the characteristic may be 2. Here also if we are supplied with tables of differences our working is facilitated. 5. 10 [The difference between the logarithm whose number is required and the smaller 4|2O f the two given logarithms is 26. Now the nearest number to this in the table of difference 50 is 25, and this is seen to be ^ihs of the difference ; therefore 5 is the 6-0 first additional figure required. Also 26 - 25 = I, and this is ^ths of the difference 7i35 (rV tns w i tn one fig 111 " 6 cut off) 5 therefore 2 is the next figure required. Hence ^=52.] 8| 4 o 9l45 21. We will conclude the chapter by applying the processes of the last article to one or two examples. Example (i). Given log 2 = .3010300, log 7 = .8450980, ^90762 = 4.9579041, log 90763 = 4.9579088 ; find the value of \J ( 5 J to 6 places of decimals. v V 42 x 32 / Let x = A/f^A^J^Sy ; then log x = (log 2 9 4 + log 125 - log 42 - log 32) V V 42X32 = .9579053- ^300 /88 47^= 1 200, /so , 47(53\ / Mi/ \200 / Mi/ ' * = 9. 076226. Example (ii). Find a 3rd proportional to .0024 and 27 ; given log 2 = .3010300, log 30375 = 4.4825163, log 3 = -477 1 21 3, log 30376 = 4.4825306. Let x be the required proportional ; then .0024 : 27 : : 27 : x, x = jf- , .0024 and log x - 2 log 27 - log .0024 = 2.8627278-3.3802113 = 5.4825165. dooo 7306 i43=n6 log 3 1 50 =3.4983106 log 3- 1 598= .4996596 log 3- 1599= .4996733 log log 2984 = 3.4747988 log 907 14 = 4.9576743 log 90715 = 4. 957679 1 log 3 =.4771213 log i. 38 =.1398791 log 6. 2403 = .795205 5, , .... 1089 x. 01881 x. 405 lop 4 ^g 60 log i.i log 19 = 69 .6020600 -1.7781513 = .0413927 =.2787536 log .012644 = 2.1018845, (xiv) //6300X.OOII7X4T9 (2197)* log 2= .3OIO3OO log 13=1.1139434 log 90 =1.9542425 log 217.47 = 2.3373994, log 7=^.8450980 log .011 = 2.0413927 34 LOGARITHMS. 3. Calculate the product of the loth root of 5 by the 5th root of 10. log 2 = .3010300 log 18616 = 4.2698864 log 18617 = 4.2699097 4. Find a 4th proportional to the 5th power of n, the 4th power of 7, and the 5th power of S and calculate to 4 places of decimals the value of -?;* ^^L. ^22x70 log 2= .3010300 log 17814 = 4.2507614 log 3= -4771213 log 17815 = 4.2507858 log 7= .8450980 log 46588 = 4. 6082741 log ii = 1.0413927 log 46589 = 4.6682834 5. If fog* = .8567, what is the value of x ? log 26 log 2 = .3010300 log .13=1.1139434 log 16300 = 4.2121876, Z> = 266 6. If 33 = 7175.37, find .*. log 3 = -4771213 log 7 1. 753=1- 8558401, > = 6o 7. If 7* = 823542. 4, find x. log 7^.8450980 log 8.2354 = .9156847, > = 53 8. Calculate the value of I +e+e*+ ... + = 43 10. A solid cube of lead weighs 126.44 Ibs. 998 ozs. of water occupy one cubic foot, and a cubic foot of lead is 11.352 times as heavy as a cubic foot of water. Find the length of a side of the cube of lead correctly to 6 places of decimals of a foot. log .012644 = 2.1018845 log 56.311 = 1.7505932, Z> = 78 log i.i = .0413927 log 129 = 2.1105897 log 49-9 = 1.6981005 CHAPTER V. Variable Base. 22. We have hitherto confined ourselves to questions involving a constant base ; we will now investigate the relation that exists between the logarithms of numbers to different bases. Prop. XI. To prove Iog 6 a x log c = Iog c 0. We have b l ^ = a, ......... (in Hence ^ Io s<* = fl 10 *c& by (i) (iii) = logca by (ii) .'. Iog 6 a . log c = logcfl. Q.E.D. This result gives us \og b a = log c ax (,-^-A from which it is evident that when the log- arithms of numbers have been calculated to any base c, those to any other base b are got by multiplying the former by the constant quantity * . 23. In the same way as in Prop. XI it can be shown that and so on, for any number of logarithmic factors in which the number for each successive logarithm in the product is the base for the immediately preceding one. This result is easily remembered and applied, in consequence of the analogy it bears to the result obtained by compounding any number of ratios in which each consequent becomes the next antecedent. Thus a -x b * c * x x x y - a X X X ..* X X --- 9 bed y z z and reading the numerators as numbers, and the corresponding denominators as bases of logarithms, we have log ^ % lQgJf . i ogdf ...\ ogyX . \ ogz y = \ ogt a. Again, by Prop. XI, Iog 6 = and this corresponds with the identity c 2 4- Prop. XII. To prove Iog 6 a . log a = i. This of course follows from Prop. XI by putting c - a, since log a = I ; but it can be proved independently thus : 35 3 6 LOGARITHMS. Hence t> loe > a ' log 6 = a**"* by (i) (ii) I = b by (ii) \og b a.\og a b=i. Q.E.D. 2 5 . Example Ci). Given log^. 30 10300^ find ~ logl l8 ~ 2 logl 3 + logl 2 - *' 2552725 ~ - 8979445. Iog 10 25 2 log 10 5 1.3979400 Example (ii). Given Iog 8 9 = 1.056642 "j Iog 8 5 = .7739760 V, calculate the common logs, of the nine digits. Iog 5 7 =1.209062 J It will be sufficient to calculate Iog 10 2, Iog 10 3, and Iog 10 7, for the logarithms of the nine digits are easily expressed in terms of these. ' '- 56642 ' " ' (i) ) From (ii) lo glo2 = .30,0300 _&= .7739760, (ii)L substituting in (i) Iog 10 3 = . 477 1213 3 Io gio 2 Jogio7_ = , 2ogo 62 (iii) substituting in (iii) Iog 10 7^- 8450980 i-log 10 2 * whence Iog 10 4= 2log 10 2 =.6020600, Io gio5 = i - lo gio 2 = 6989700, Iog 10 6 = Iog 10 2 + Iog 10 3 = .7781513, logio 8 ^ 3logio 2 =.9030900, Igio9 = 2 Iog 10 3 = . 9542426. Example (iii). By what must logarithms to the base 2 be multiplied to find them to the baseS? Since Iog 8 ^ = Iog 2 # x Iog 8 2, the required multiplier is Iog 8 2 or log^, i.e. \. EXAMPLES. XIII. I. Prove that (i) x= Sy \Ja, (ii) \og a rX = \ag5 20, (vii) Iog 12 .7, (viii) Iog 5 i8, __ ( k ) 10 V3' 6 ' W log B .002I, (Xi) logj.4^, (Xii) Iog 4. Given Iog 10 5 = .6989700, find log 25 4O and Iog 40 25. 5- Given Iog 10 35 = 1.5440680, find Iog 10()0 (^). 6. Given Iog 10 5 = . 6989700, find Iog 2 . 5 io. VARIABLE BASE. 7. Given Iog 10 5.6= .7481880"! lgio73-5 - 1.8662873 V, find Iog 5 i2 and Iog 6 .o7. Iog 10 io. 8= 1.0334238 J Given [g89-\ find the logarithms to the base 10 of the first 4 digits. 9. Given logs ^3 = J f find the CO m mo n logarithm of 45- 10. Given Iog 10 ^= 3.6102407, log 10 j)/ = 2.2481883 ; find 37 12. The logarithm of a number to the base 4 is .35184, find its logarithm to the base 16. 13. By what must logarithms to the base ^2 be multiplied to find them to the base ^,'3? given Iog e 2 = .6931472, Iog e 3 = 1.0986123. 14. If x be the logarithm of a to the base b, what is the logarithm of a m to the base b n ? 15. Show that the logarithm of any number to the base a n is a mean proportional between its logarithms to the bases a and a n 1 6. Solve the equation ^ \og x & = 3 Iog 8 x. 17. If a, b, c be in G.P., prove that log^, logftA', log c -A^will be in H.P. 18. If a, b, c be in G.P., and logctf, log&c, log a & in A. P., then the common difference of the latter is i|. 19. If a, b, c be respectively the two sides and the hypotenuse of a right-angled triangle, then logfi+cfl + log c _ 6 a = 2 \og b+c a . log c -&0. 20. From the formula loge( ^ ) = *+ + +..., calculate Iog 10 5 to 5 places of decimals. \ I OC / 2 ^ Given log 1( /? = . 43429. 21. If Jr 3 = log^a, ^4 = 1 g^3 *=log a . n _ 2 * n _i, x^\Qg Xn _^x M then ^ . ^ 2 ...^ n = I. CHAPTER VI. Interest. Annuities Certain. I. Interest. 26. When a sum of money is lent for a time, the borrower pays to the lender a certain sum for the use of it. The sum lent is spoken of as the Capital or Principal: the sum paid for the use of it is called the Interest. The Amount is the Principal plus the total Interest at the end of the time for which the money was lent 27. (i) Simple Interest. The interest paid for the use of money is said to be Simple when it consists of a certain fixed sum paid at regular periods. It is generally reckoned at so much per cent, per annum. Thus 5 per cent. (5/ ) P er annum means $ paid annually on every .100 of Capital. If there be p periods of payment in the course of a year, 5 per cent. per annum would signify ^- paid every period on each ;ioo of Capital. Let P be the Principal, A the amount at the end of n years, and /the Interest accruing in the same time at 100 r per cent, per annum, so that r is the interest on i in i year ; then, for Simple Interest, A = P(i+nr), (i) I=nrP (2) 28. (ii) Compound Interest. Sometimes the borrower pays at the end of the whole time for which the money was lent a single sum to cover both Principal and Interest. In this case since the lender loses the use of the Interest as it accrues, it is clear that the borrower should pay interest on this also ; in other words, that this Interest should be added to the Principal as it becomes due, and that the borrower should proceed to pay Interest on the Principal so increased. In such a case the Principal or money lent is said to bear Compound Interest. It is reckoned in the same way as Simple Interest. When the Interest becomes due it is said to be convertible (into Principal), and the period between two successive times at which Interest becomes convertible is called the conversion-period. Let P be the Principal, A the Amount at the end of n periods, and 7 the Interest accruing in the same time ; while r is the interest on i for one conversion-period, and R=i+r-, then, for Compound Interest, 38 INTEREST. ANNUITIES CERTAIN. 39 - (I) (*) [The formulae given above for Simple and Compound Interest are appli- cable to questions of Present Worth and Discount, P being the Present Worth of a debt A> due in time , and / being the Discount] 29. Example (i). Find the amount of ^1000 invested for 21 years at 3/ per annum, Compound Interest, convertible half-yearly. r (the half-yearly interest on^i) = .oi5 R= 1.015, an d ^ = 42. Hence A = PR" = iooo(i.oi5) 42 and taking logs. log A = log 1000 + 42 log 1.015, log i. 015 = .0064660 42 129320 258640^ 42 log 1.015=7271572" log 1000 = 3, log ,4 =3. 27 1 572 Now log 1 868. 8 = 3. 27 1 563 9? /. ^=^1868.84. Example (ii). How long will it take for a sum of money to double itself at 6/ per annum, Compound Interest, convertible annually ? Here P= i, A = 2, R= 1.06. Let n years be the required time, then 2 = (i.o6) n , and taking logs. log 2 = n log 1.06, =I2 years . log 1. 06 .0253059 Example (iii). If the number of births and deaths be 3.5 and 1.2 per cent, respectively of the population at the beginning of each year, after how many years will the population be trebled ? The annual increase is 2.3 per cent, or .023 of the population at the commencement of the year. Let n years be the required time, then (i. 023)" = 3, and taking logs. n log 1.023 = log 3, n - - g 3 = 48. 3 years. EXAMPLES. XIV. \ [Compound Interest is understood unless the contrary is stated.] 1. Find the amount of^iooo in 10 years, allowing 5 per cent, per annum interest. log 2 = . 3010300, log 7 = .8450980, log 3 =.4771213, log i. 627 = .211893. Sandhurst. 2. In what time will a sum of money treble itself at 5 per cent, per annum ? Given log 2, log 3, and log 7. Sandhurst. 4 o LOGARITHMS. 3. Find, correct to a farthing, the present value of 10000 due 8 years hence at 5 per cent. per annum. Given log 2, log 3, log 7, log 67683 = 4. 8304796, log 67684 = 4. 8304860. Woolwich. 4. Find the amount of 5500 in 15 years at 5 per cent, per annum, giving the result in and the decimal of a . Given log 2, log 3, log 7, log 1 1 = 1.0413927, log 1. 1434- .0581982, log 1. 1435= .0582362. Woolwich. 5. Find by logarithms what the annual income will be if ^2700 stock be sold at 90 and re-invested in the 3 per cents, at 125. log 27 =1.4313638, log 25 =1.3979400, log 90= 1.9542425, log 972 = 2.9876663. Woolwich. 6. Find the total interest, payable half-yearly, on^ioo for 12 years at 5 per cent, per annum. log 2 = .3010300, log 18087 = 4.257367, log 41 = 1.6127839, log 18088 = 4.257391. 7. Find the present value of 1000 due 10 years hence, reckoning interest at 4 per cent. per annum. log 2 = .3010300, log 67557 = 4. 829670, log 130 = 2. 1 139434, log 67556 = 4.829664. 8. At what rate per cent, must money be lent that it may be doubled in 10 years? log 2 = .3010300, log 10717 = 4.0300732, D = 406. 9. How long will it take for 1000 to amount to 2500 at 5 per cent, per annum, con- vertible half-yearly ? log 2 = .30 1 0300, log 41 = 1.6127839. 10. A sum is laid out at 10 per cent, per annum, convertible annually, and another sum of double the amount at 5 per cent, per annum, convertible half-yearly. In what time will the two amounts be equal ? log II = 1.0413927, log 1025 = 3.0107239. 11. In how many years will the Interest on a given sum amount to double the Principal at 3^ per cent, per annum ? log 3 = -477i2i3, log 1 15 = 2. 0606978. 12. Find the amount of 1000 at the end of 10 years, allowing io/ per annum interest, convertible half-yearly. log 10225 = 4.0096633, log 15605 = 4. 193264. 13. A country trebles its population in a century. What is the increase in one year per million ? Given log 2, log 3, log 67 = 1.8260748. 14. What is the amount of one farthing for 500 years at 3 per cent, per annum ? log 103 = 2. 01 28372, log 26218 = 4.4185996. Z>=i65. 15. At what rate per cent, per annum will a given sum increase n-fold in a century? log 11 = 1.0413927, log 10242 = 4.0103848, Z> = 424. 16. A sum of money when put out to interest, payable half-yearly, amounts to 2316. IOS. in 5 years, and 2708. 5s. in 9 years. What is the rate of interest ? log 23165 = 4. 3648323, log 27082 = 4. 4326807, Z?=i6i, log 10197 = 4.0084724, D = 426. 17. On the birth of an infant icoo is invested in the Funds (2|/ payable half-yearly). Calculate what it will be worth when the child is 21 years old to the nearest shilling, log 10137 = 4.0059094, D = 429, log 1 7 745 = 4- 249076, Z? = 25. INTEREST. ANNUITIES CERTAIN. 41 18. A person borrowed ,11000 for two months at 5 per cent, per annum. At the end of the time the interest was added on, and the debt renewed for another two months. This was continually repeated till at the end of 2 years the debt and interest were paid. How much did this amount to ? log 2 = .3010300, log .011=2.0413927, log 90 =1.9542425, log 121. 51 =2.0846120, -> = 357. Woolwich. 19. Show that money will increase more than 5o-fold in a century at 4 per cent, per annum interest. log 2 = .30 1 0300. log 13 =1.1 13943. 20. The number of births in a town is 25 in every thousand of the population annually, and the deaths 20 in every thousand. In how many years will the population double itself? Given log 2, log 3, log 67 = 1.8260748. 21. A man borrows ^"1500 for 6 months and accepts a bill for ,1650 from a money lender. The bill is not met but is renewed every half-year at an increase of 20 per cent. After what time will the bill have amounted to at least ^7500 ? Given log 2, log 3, log 1 1. 22. A person puts out^iooo at 5 per cent, per annum interest, payable half-yearly, and each time that interest is paid adds one half of the same to his capital. Find to the nearest shilling the amount at the end of 20 years. Given log 2, log 3, log 16436 = 4.215796. . = 27. 23. A cottage at the beginning of a year was worth 2 50, but it was found that, by dilapid- ations, at the end of each year it had lost 10 per cent, of its value at the beginning of the year. After what number of years would the value of the cottage be reduced below ^25. log 3 = .4771213. Woolwich. 24. A young man on coming into his fortune at the age of 21 invests it in a bank which allows 5 per cent, per annum interest. At the end of each year he withdraws for his expenses a sum equal to fths of his 1st year's interest. At what age will he be penniless ? Given log 2, log 3, log 7. 25. A quantity o! water contained in a cubical cistern is found to lose by evaporation .04 of its volume in a day. The depth of the cistern is 6 ft., and a cubic foot of water weighs 1000 oz. Assuming the loss to take place by evaporation only, find to one decimal place what weight of water will be left in the cistern at the end of 10 days. Given log 2, log 3, log 14360 = 4.1571544, log 1 436 1 = 4. 1 57 1 847. Sandhurst. 26. If from a barrel full of spirit 5 per cent, be drawn and the deficiency made up with water, and the operation be repeated again and again, how soon will there be more water than spirit, and what will be the proportion of water to spirit then ? log 2= .3010300, log 48762 = 4. 68808. log 19 = 1.27875, 27. A person with a capital of ,10000, for which he receives interest at the rate of 5 per cent. per annum, spends/" 900 yearly. In how many years will he be ruined ? log 7 = .8450980, log 15 =1.1760913. 28. A man commences saving with the intention of putting by, every year, half as much again as he did the year before, and investing the same at 3 per cent, per annum. If he save 10 the first year, find how much he will have accumulated in to years, and the amount of his savings the last year. log 2=^3010300, log 23305 =4. 367449, D=ig, log 3= -4771213, ^97669 = 4.9897567, > = 45, log 47=1.6720979, ^11983 = 4.0785656, Z> = 362, log 103 = 2.0128372, ^38443 = 4.5848173, >=ri3- 4 2 LOGARITHMS. 29. On January I, 1880, I started saving a shilling every day, investing these savings at the end of every year at 2.\ per cent, per annum. What amount will have accumulated by the end of 21 years? log 1025 = 3.0107239, log 16795 = 4.225180, -> = 26, log 18087 = 4.257367, Z> = 24, log 1 1038 = 4.042890, D - 40. II. Annuities Certain. 30. An annuity is an annual payment of a given sum of money. An Annuity Certain may continue for a fixed number of years, when it is said to be terminable ; or may be vested in an individual and his heirs for ever, when it is said to be perpetual. Payments may, of course, be made periodically, so that the year consists of any number of equal periods. When this is the case it must be taken into account in the same way as the conversion-period is in questions of compound interest The Accumulated Value of a forborne annuity is the amount of the several instalments plus the compound interest on each for the period during which it has been forborne. The Purchase Price of an annuity is the sum of the present values of the several instalments. The Number of Years' Purchase of an annuity is the ratio of the purchase price to the annual instalment of the annuity. An annuity is Deferred or Not Deferred according as the first instalment is not or is due after the expiration of one period. (I) Forborne Annuities. 31. PROBLEM. To find the accumulated value of a forborne annuity. Suppose the annuity to be A for n successive periods of time, and r to be the interest i for one period, and suppose the last instalment to have been due x periods ago. Now the ?zth instalment + interest for x periods = AR X , the(-i)th (*+i) = AR x +\ the 1st (x + n-i) = AR x+n ~ l . Hence, by addition, F(the accumulated value) = AR X + AR X+I + ... + AR x+n ~ 1 = AR x (i+R+...+R n - 1 ) _AR x (R n -i) R- i INTEREST. ANNUITIES CERTAIN. 43 (COROLLARY. If the last instalment be only just due, putting x = o, we get V= ^~ r \ [ L It- 1 J Example. What is the accumulated value of an annuity of ^120 during 10 years, that lapsed 7 years ago, reckoning 4 per cent, per annum interest and half-yearly payments ? In this case A =60, x=i$, n = 2O, J?=i.O2, R-i To find R - i : log R = . 0086002 _ 20 logJ? 20 = .172004 Now log i. 4859- .171990 Z> = 29 )4o(i .'. JP 20 =1.48591, A9 and ^-1= .48591. Hence we have F= Mi-c .02 log V- log 60 + 14 log 1. 02 + log .48591 - log .02. log i. 02= .0086002 _ [4 344008 86002 I4logi.o2= .110403 log 60= 1.778151 log .48591 = 1.686556 1-575110 log. O2 = 2. 30 IO3O log V- 3.274080 Now log 1879.6 = 3. 274065 II 1 ... r= 1*79.67. (II) Terminable Annuities. 32. PROBLEM. To find Fthe purchase value, and Pthe number of years' purchase. (A) Deferred : Suppose the annuity to consist of n periodical payments otA, and the first payment to be made x periods hence ; and, as before, let R - 1 + r, where r is the interest on^i for one period. Then Present value of 1st instalment due in x periods = , 2nd ,, ,, (x+i) ,, =^+r 4?fh /.- . _\ A 99 99 /lll . *. "' fix+n-1- Hence, by addition, V (the purchase value) = 44 LOGARITHMS. If there be / payments in the course of one year, P (the no. of years' purchase) = -^ = ^ P:g+ra _~ \ r , (2) nn _ -i If the payments be annual, p-\ t and P +n _ 1{ , - (3) [ V is, of course, the amount of money which will realize the same as the annuity pay- ments provided the investments are made at the rate of interest reckoned in the sale of the annuity ; in other words, the rate reckoned in calculating V is the rate of interest the pur- chaser will make of his money, while replacing his capital, provided he can reinvest at the same rate.] (B) Not deferred : Putting x = i in (A), we have - * "> i R~ n and, for annual payments, P-, (3) K I (III) Perpetual Annuities. 33. PROBLEM. To find Fthe purchase value, and ./'the number of years' purchase. (A) Deferred : Putting = oo in (II), rr A / I \ A / T \ (2) , for annual payments ) P - -p^ry/ P _ ~\ (B) Not deferred : Putting n = oo in (II) and, for annual payments ; P-- .................................................................. (3) [Since rate per cent. = loor, it is clear from (3) that the number of years' purchase of a perpetual annuity with annual payments, to begin running at once, is - - - ; and, rate per cent. conversely, rate per cent. = - ^5 -- Also it is evident that the present number of years purchase value of a deferred perpetuity is the amount of money which, laid out at compound interest at the same rate, will purchase the perpetuity when possession is to be obtained.] HI. Renewal of Leases. 34. If, when/ years of a lease have to run, the tenant wishes to renew for a term p + n years, the sum he must pay is called the " fine for renewing n years of the lease." INTEREST. ANNUITIES CERTAIN. 45 Supposing A to be the net annual value of the estate, the fine clearly equals the present value of an annuity A, to vest after/ years, and to continue for n years. Hence the fine - If the object be merely to renew the original lease years, fine- '" A Rq ~ l 35- Example (i). Find the price that should be paid for an annuity of ^250 to commence in 3 years, and to continue for 10 years, allowing interest at 6 per cent. By (ID, (A>. ^=-44^1). Now A=2$o, x = 4, n=io, R= 1. 06, To find (I.o6)' 3 : log i. 06= .0253059 To find (i.o6)- 13 : log i. 06= .0253059 Now or = 1.9240823 log. 83961 = 1.9240776 Now 759177 253059 Iog(i.o6)- 13 = ^. 328977 or =1.671023 log .46883 = 1.671015 D= 9 )8o(9 8i_ (i.o6)- I3 = . 468839. (i.o6)- 3 = . 839619. .*. Hence (i.o6)- 8 -(i.o6)- 13 = .839619 -.468839 = .37078, r= ?g(. 37078), log r= log 250 + log . 37078 - log .06 = 3.1889050 log 1 544. 9 = 3- 1 889004 Z> = 28i )46o(2 ^=1544.92. 562 Example (ii). A person borrows /"looo ; what will be the amount of each instalment that both debt and interest may be repaid by 12 equal monthly instalments, allowing interest at the rate of 10 per cent, per annum ? By (II), (B), y= A ( l -R~ n \ wne re A = monthly instalment required. Now F=,iooo, = 12, R= i +r=%% (since r = interest on lzfiQ ie A- I0 ~~ "- in i month), To find i -IS Now " 12 Iog6i = 1.7853298 log 60= I.7 log fa= .0071785 - 12 log (|^)- 12 ---. 086142 or = 1.913858 log. 82008= 1.91386 Z?= ~~ and I - - 12 = . 820084, - 12 = . I799I6. 46 LOGARITHMS. Hence we have ^= l \ , log ^4 = log ioo-log6o-log. 179916= 1.9667789. But ^92.636=1.9667798 .'. ^=^"92.636. Example (iii). How long may I expect to live, if the reversionary interest in the fee simple of an estate that I hold for life producing ^200 a year be sold for ; 1500, allowing 5 per cent, interest ? By (III), (A), V- -, where my expectation of life is n years. Now ^=1500, ^ = 200, R =1.05, , L, (1.05)- (.05X1.05) Taking logs. wlog(i.o5) = log2-log 15 log. 05, , - ^g 2 - log 15 ~ ^g .05 - .4259687 _ 2Q r ars log 1. 05 .0211893 EXAMPLES. XV. 1. What is the accumulated value of a forborne annuity of ^150, that lapsed 2 years ago and should have been paid in half-yearly payments during 8 years ? log 1025 = 3.0107239, log 16386 = 4.214473, D = 26, log 1 1038 = 4.042890, D - 39. 2. What perpetuity will ^"2000 purchase so that possession may be had in 10 years, allowing interest at 4^ per cent. ? log 9= .954243, log 1045 = 3.0191163, log 13976 = 4. 145383, > = 3I- 3. In how many years will a debt of ^753. IDS. be discharged by annual payments of^ioo ; interest at 8 per cent. ? log 108 = 2.033424, log 397 12 = 4. 598922. 4. Find the present value of an annuity of ^"75 to vest in 10 years and then to continue for 15 ; interest at 4^ per cent. log 1045 = 3-0191 163, log 33 2 73 = 4- 5 220 92, log 64392 = 4. 808832, Z> = 7. 5. Find the present value at 4 per cent, per annum of a Fellowship of .300 a year for 6 years, payable half-yearly, the first payment being due in 6 months' time, log 102 = 2.0086002, log 78849 = 4. 896796, D - 6. 6. Find the present worth and the number of years' purchase of the Reversion to a Free- hold Estate of ; 1 200 a year after 30 years, reckoning interest at 6 per cent, log 2 = . 30 1 0300, log 3482 2 = 4. 54 1 8 54, log 12=1.0791812, log 2901 8 = 4.462668. log 106 = 2.0253059, 7. If I pay 13 \ years' purchase for a life-annuity, after how many years shall I be re- imbursed, allowing interest at 5 per cent? log 105 = 2.021 1893, log 325 = 2. 5 1 18834. 8. If 4! per cent, be the rate of interest reckoned, what sum must be paid now to receive a Freehold Estate of ^300 a year 12 years hence? log 2 = .3010300, log 1045 = 3.0191163, log 3 = .4771213, log 41080 = 4.613630. 9. How much must be paid annually that a debt of ^650 may be discharged in 20 years, allowing interest at 4 per cent. ? log 2= .3010300, log 47828 = 4. 679682, ^ = 9, log 13 =1.1 139434, log 45638 = 4. 659327, Z> = 9, log 5436 1 =4. 735287, D = S. INTEREST. ANNUITIES CERTAIN. 47 10. Find the number of years' purchase and the present value of the Fee Simple of a Free- hold Estate producing ^1315 per annum net, reckoning 4^ per cent, interest? log 2 = .3010300, log 29222 = 4.4657099, Z?=I49, log 9= .9542426, log 22222 = 4. 346783 1, .0=196. log 131 5 = 3- i 189258, 11. After how many years may I expect to acquire the Reversion to a Freehold Estate if I pay 5 years' purchase for it now, allowing 4 per cent. ? log 2 = .3010300. log 104 = 2.0170333. 12. A man 48 years old can buy an annuity of .150 for ^1812 i6s. Determine what is considered the expectation of life at 48, interest allowed at 5 per cent, log 2 = .30 1 0300, log 7= .8450980, log 3 = -477 1 21 3, log 1 1 872 = 4. 0745239. 13. Supposing a perpetuity to be worth 27 years' purchase, what must be paid for an annuity of ^500 to continue for 10 years ? log 27 = 1.43 1 3638, log 6951 1 =4. 842054, > = 6. log 28 =1.447 1 580, 14. An annuity of ^300 vests in 10 years' time : find the equivalent annuity vesting immediately and continuing for the same period, interest at 5 per cent, log 3= -4771213, log 18417 = 4.265219, D = 2$. log 155 = 2.021 1893, 15. The reversion of an estate in fee simple producing ^"60 a year is made over for the discharge of a debt of ^577 45. 5d. How soon ought the creditor to take possession, if he be allowed 5 per cent, per annum interest for his debt ? log 2 = .3010300, log 105 = 2.0211893, log 3 = -477i2i3, log 1 3853 = 4. 141 5438, Z> = 3i4 16. What is the value of the reversionary interest of an annuity of ,150 for 12 years after the next 8, 5^ per cent, interest being allowed ? log 1055 = 3. 0233525, log 65039 = 4. 813174, Z> = 7, log 341 15 = 4. 532945, Z>=I3. 17. If two joint proprietors have an equal interest in a freehold estate worth ^"2500 per annum, what annuity must the one allow the other during a term of 12 years that he may -buy him out and thus purchase to himself the whole freehold, allowing interest at 5 per cent, per annum ? log 105 = 2. 02 1 1 893, log 55683 = 4. 745723, D= 8, log 125 = 2.0969100, log 443 16 = 4. 646561, D= 6, log 28206 = 4. 450342, D = 1 5. 1 8. What will be the amount of an annuity of ^720 left unpaid for 26 years, allowing interest at 4 per cent, per annum, an instalment being just due ? log 104 = 2.0170333, log 27724 = 4. 442856, D- 16, log 180 = 2. 2552725, log 17724 = 4.2485617, Z> = 245, log 3 1 904 = 4. 5038451, Z?=i36. 19. How much must be paid annually that a debt of ^"1000 may be discharged in 20 years, interest at 5 per cent. ? log 105 = 2.021 1893, log 37689 = 4- 576215, log 5= -6989700, log 6231 1 =4. 794565, log 80243 = 4. 904407. 20. What difference does it make in the year whether a person receive his salary of ^600 quarterly or monthly, interest at 4.8 per cent. ? log 1012 = 3.0051805, log 10488 = 4. 020693 \ D _ log 1004 = 3.0017337, log 10490 = 4. 020776) 48 LOGARITHMS. 21. A loan of ^1000 is to be paid off in two years by equal quarterly payments. What is the amount of each payment, allowing interest at io/ ? log 82074 = 4.914206, D= 5, log 5= .6989700, log 17925 =4. 253459, Z? = 24, log 1025 = 3.0107239, log 1 3946 = 4. 1 44450, Z? = 3i. 22. The lease of an estate is granted for 7 years at a pepper-corn rent, with the condition that the tenant at the expiration of the lease may renew the same on paying a fine of ;ioo. What is the value of the landlord's interest in the estate immediately after any such renewal, allowing interest at the rate of 5 per cent, per annum ? log 105 = 2.0211893, log 4071=3.60970, log 14071 = 4. 148325, log 24564 = 4. 39030. 23. How many years' renewal will ^1009. 45. purchase of a 40 years' lease of an estate worth ^35 a Y ear at the expiration of 10 years, allowing 5/ interest? log 105 = 2.0211893. log 23 1 37 = 4. 364307, D=\g. log 87206 =4.940546, 24. If a perpetual annuity be worth 22| years' purchase, what annuity to continue for 8 years will ,2000 purchase ? log 2 = .3010300, log 70618 = 4.848915, D= 6, log 3 = -4771213, log 2938 1 =4.468067, Z>=i5, log 47 = 1.6720979, log 30253 = 4.480768, D = 14. 25. If I have to pay ,2150 when I am 21 years of age for an annuity of ^100 during my life, how long may I expect to live, 3 per cent, being the rate of interest reckoned ? log 2 = .30103, log 71 = 1.85126, log 103 = 2.01284. 26. Find to the nearest how much should be paid now for an annuity of $oo, the first instalment of which is paid to the annuitant five years hence, and the last instalment fifteen years hence, interest at 5/ . log 1.05 = .0211893, log 8.22702 = .915243, log 4.81027 = .682160. 27. In the case of a 30 years' lease of an estate whose annual rental is ,720, what fine must be paid in order to renew the lease after the expiration of 8 years, allowing interest at 6% ? log 109 = 2.0253059, log 277504 = 5.443270, log 17411=4.240823. 28. I buy the remainder of a lease, with 15 years to run, at 8 years' purchase. If I am only able to invest at 4 per cent., what interest shall I realise on the purchase money ? log 32=1.505150, log 18009 = 4.255490, Z> = 24, log 104 = 2.0170333, log 10630 = 4.026533, Z? = 4i. log 80094 = 4.903600, 29. How much money must be invested at Compound Interest that in 21 years it may purchase the Fee Simple of a freehold of ^200 net annual income, reckoning 4 per cent, in each case? log 5= .6989700, log 21941 =4. 341256, D = 20. log 104 = 2.0170333, 30. An estate whose clear annual value is 1800 is let on a 21 years' lease, renewable every seven years on payment of a fine ; what is the amount of the fine, allowing interest at 5 per cent. ? log 105 = 2.021 1893, log 50506 = 4. 703343, Z>= 9, log 35894 = 4- 555022, D=I2. CHAPTER VII. Application of Logarithms to Plane Trigonometry. 36. The trigonometrical ratios of angles are abstract numbers and continuous functions of the angle, that is, change continuously in value as the angle changes through any interval however small; hence logarithms can be applied to trigonometrical functions, so that we can treat of the logarithms of the trigonometrical ratios of angles ; and, since the logarithms of numbers vary continuously with the numbers when the base is positive and greater than unity (the numbers being then also positive), the logarithms of the positive trigonometrical ratios of angles are continuous functions of the angle and change continuously as the angle changes. These logarithms are called logarithmic ratios, e.g. the logarithm of the sine of A is called the logarithmic sine of A and written log sin A. The base adopted is 10, the base for common logarithms. 37. Now the trigonometrical ratios of angles are not always positive. For positive angles less than 90, i.e. for angles in the first quadrant, they are all positive, and we can therefore speak of the logarithms of all the trigonometri- cal ratios of angles in the first quadrant. But, corresponding with an angle of any magnitude, positive or negative, a positive angle less than 360 always exists whose trigonometrical ratios have all severally the very same values as those of the given angle, the bounding lines of the two angles being in the very same position. Again, corresponding with this positive angle less than 360, an angle can al- ways be found in the first quadrant whose trigonometrical ratios have all of them severally the same arithmetical values, though some of them will be of different sign. Hence whatever be the angles involved in a trigonometrical expression, the expression can always be reduced at once to one having the same form and involving only angles in the first quadrant, so that there will be no diffi- culty in applying logarithms to a trigonometrical expression involving any angles whatever, provided only the expression be adapted to logarithmic computation and be on the whole positive in value when angles lying in the first quadrant have been substituted for those occupying other positions. If the expression, after the reduction here spoken of, assumes a negative sign upon the whole, its logarithm cannot be taken ; but, if the problem be to find the value of the given expression, that of the corresponding positive expression can be found by means of logarithms, and, changing the sign of the result, we D 49 5 o LOGARITHMS. have the required value. Should the negative expression be one side of an equation to be solved by logarithms^ the other side must necessarily be also negative, and signs are changed on both sides before taking logarithms. Hence, finally, whatever angles be involved in a trigonometrical expression, ic can always be reduced to another, equally adapted to logarithms, in which the angles are all of them positive and less than 90, so that for all requisite pur- poses a table of logarithmic ratios need only give those of positive angles less than 90. To effect the necessary reduction for any angle that is not positive and less than 90, calculate the smallest positive angle having the same position and consequently all the same ratios. Call it A, an angle got by adding or subtracting 360 again and again ; then (i) if A lies between o and 90, it will be found in the tables ; (ii) if A lies between 90 and 180, iSo - A will lie between o and 90, and we have sin A = sin (i 80 - A), cosA = -cos (180-^), (iii) if A lies between 180 and 270, A- 180 will lie between o and 90, and we have sin A = - sin (A - 180), cos A = -cos (A- 1 80), tan A = tan (A - 180). (iv) if A lies between 270 and 360, 360-^ will lie between o and 90, and we have sin A = - sin (360 - A), cos A = cos (360 - A), tan A = - tan (360 - A). 38. The logarithmic ratios are positive or negative according as the trigonometrical ratios are greater or less than unity. Hence, among loga- rithmic ratios there will be as many negative as positive values, for, while numerically the sine and cosine cannot be greater than unity, the secant and cosecant cannot be less than unity, and of the tangent and cotangent, when one is greater the other is less than unity. Now, when the logarithmic ratios are negative, in accordance with the usual method of expressing negative logarithms they will have negative char- acteristics with positive mantissae. But, in the case of the logarithms of the trigonometrical ratios, we are unable to determine the characteristics by inspection as we could for the logarithms of decimal numbers ; hence th( characteristics must be tabulated and it must also be stated whether they an positive or negative. To avoid the recurrence of negative characteristics, th( logarithmic ratios are all increased by 10 before being tabulated, becoming thus tabular logarithmic ratios, which are positive over almost the whol< range o to 90, and at least between 10' and 89 50'. The tabular logarithm is written Z, e.g. the tabular logarithmic tangent of 20 would be written L tan 20. 39. It has been seen that the logarithmic ratios for the range o to 90 are LOG A RITHMIC RA 770 S. sufficient for all purposes ; it will now appear that really the tabulation of the range o to 45 gives all that is required, for any ratio of an angle A between 45 and 90 is equal to some ratio of the complementary angle 90 ^, which will lie between o and 45. We have L sin A = L cos (90 - A ), L cos A = L sin (90 - A), L tan A = L cot (90 - A), L cot^ = L tan (90 - A), L sec A = L cosec (90 - A), L cosec^ = L sec (90 - A). The subjoined extract from the tables will show how the above formulae are utilized in abbreviating our tabulation ; for instance, L tan 63 3' is given as 10.2937716, and is the same as L cot 26 57'; L cos 63 i' and L sin 26 59' are both equal to 9.6567987. 26 DEC. ' Sine. Diff. Cosec. Tang. Diff. Cotang. Secant. Diff. Cosine. ' I 2 3 9.6418420 9.6421009 9.6423596 9.6426182 2589 2587 2586 2583 10.3581580 10.3578991 10.3576404 10.3573818 9.6881818 9.6885023 9.6888227 9.6891430 3205 3204 3203 3201 10.3118182 10.3114977 10.3111773 10.3108570 10.0463398 10.0464015 10.0464631 10.0465249 617 616 618 617 9.9536602 9-9535985 9-9535369 9-9534751 60 59 58 57 57 ! 9-6563021 2484 10.3436979 9. 7062284 3126 10.2937716 10.0499262 643 9.9500738 1 58 9-6565505 2482 !0. 3434495 9.7065410 3 I2 5 10.2934590 10.0499905 643 9.9500095 2 59 9.6567987 2481 10.3432013 9.7068535 3124 10.2931465 10.0500548 643 9-9499452 I bo 9.6570468 2478 10.3429532 9.7071659 3122 10.2928341 10.0501191 644 9.9498809 ' Cosine. Diff. Secant. Cotang. Diff. Tang. Cosec. Diff. Sine. ' 63 DEC. It will be observed that in the tables the degrees of an angle lying within the range o to 45 appear in the left hand corner at the top of the page, while those of the complementary angle within the range 45 to 90 appear at the bottom in the right hand corner. Again, for angles from o to 45, the names of the ratios are at the top and the minutes are read downwards, while from 45 to 90 the ratios are at the bottom and the minutes are read upwards. 40. As the angle increases so also do its sine, secant, and tangent ; but the cosine, cosecant, and cotangent (the co-ratios) decrease as the angle increases. Hence also log sin A, log sec A, log tan A L sin A, L sec A, L tan A but log cos A, log cosec A, L cosec A, } Crease as the ang.e increase, 41. Certain trigonometrical ratios are reciprocal, i.e. have a product unity. In these cases the tabular logarithms will be together equal to 20. 5 2 LOGARITHMS. For sin A x cosec A = i, /. taking logs, log sin A + log cosec A = o, cos Ax sec A= i, log cos A+ log sec ^4 = o, tan^x cot ^4 = i, log tan A+ log cot ^4 = o, or, expressing in tabular logs., Z,siuA + L cosec ^ = 20, + Z sec A = 20, + L cot A = 20. Since the tabular logarithms of two reciprocal ratios have always a constant sum, viz. 20, their differences over any range must be the same, only in one case the difference will be an increase, while in the other it will be an equal decrease. It is for this reason that in the tables there is only one column of differences for each pair of reciprocal ratios. 42. When logarithms are applied to trigonometrical expressions, tabular logarithms are written down at once in every case of a logarithmic ratio, but compensation must be made with jo's if the sum of the coefficients of the tabular logarithms affected with the plus sign be not equal to that of the tabular logarithms having the minus sign. In the case of an equation the sum of the coefficients must be the same on both sides of the equation, other wise compensation will be necessary on that side on which there is a deficiency. Thus, log( j = log2 4- 2Z sin.* -log 3 -Lcosx- 10, and, taking logs, throughout the equation, we have 2 Z sin x - log 2 = J(log 2 -log i3) + Ztan 16 + 10. 43. From the foregoing articles it is clear that, when a tabular logarithmic ratio of an angle is known, we also know or can find at once (i) the same tabular logarithmic ratio of all angles for which the ratio has the same value as that of the given angle ; (ii) the tabular logarithms of the reciprocal ratios of all the same angles; (iii) the tabular logarithms of the complementary ratios of angles complemen- tary to those in (i) and (ii). For example, given Z sin 22 18' = 9. 5791616, find Z sin i5742 r (i) Z cosec 22i8'\ .... Z cosec 1 57 427 " Z cos 6742'\ ..... L sec 67 427 (m) We have Zsin K7 42M r _._ -. _ g , = 9 . 579l6l6 , Z sec 67 42' - 20 -Z cos 67 42' = 20 - Z sin 22 1 8' = 10.4208384. LOGARITHMIC RA TIOS. 5 3 It is seen that when there is a change to the reciprocal ratio, the angle re- maining unaltered, the required tabular logarithm is obtained by subtracting that given from 20 ; but when there is a change both to the complementary ratio and complementary angle, the tabular logar't im does not alter. EXAMPLES. XVI. 1. Given log 2 = .3010300, log 3 = .4771213 ; find the tabulated logarithms of (i) sin 30, (ii) sin 45, (iii) sec 30, (iv) sec 60, (v) tan 30, (vi) cot 45, (vii) sec 45, (viii) sin 90. 2. Given Z sin 60 = 9. 9375306 \ fi , , Z cos 60 = 9. 6989700! ' hr 10 S 2 3. Given that 4 sin 18 . sin 54 = i ~\ L sin 1 8 = 9.4899824 V, find Z sin 54. log 2= .3oio3ooJ 4. If Z sin 1 5 = 9.4129962 and log 2 = .3010300, find Z cos 15. 5. Write down the values of Iog 3 cot 60, Z 2 sin 30'. 6. (i) Given Z tan 35 22' = 9. 85 1 1 285, find Z tan 54 38'^ Z cot 54 38' V. Zcot3522'J (ii) Given Z cosec 117 46' = 10.0531293, find Z cosec 62 14'^ Z sin 62 14' I Z sec 27 46' T Z cos 27 46' J (iii) Given Z sin 44 53' = 9.8485989, find Z sin 135 7"j Z cosec 44 53' I Z sec 45 7'f Z cos 45 7'J 7. Transform the following equations into others involving tabular logarithms : (i) N/2 sec A - -\/ C S " , (") sine's/2 tan A = 3, (v) tan ^ . CHAPTER VIII. Tables of Logarithmic Ratios. 44. From the brief description of the tables in the last chapter it is seen how the angle corresponding with any given logarithmic ratio, or the logarith- mic ratio of any given angle, is found at once whenever the given quantity is contained exactly in the tables. When the given logarithmic ratio or angle does not appear exactly in the tables, but lies between two successive tabula- tions, then, as in the case of the logarithms of numbers, the corresponding angle or logarithmic ratio respectively can be found by the application of the principle of proportional parts, provided that over a small range the changes in the tabular logarithmic ratios are approximately proportional to the change in the angle. This, we will proceed to show, is the case, if only the angle is not very nearly o or 90. 45. It can be shown by Trigonometry that, if d be the circular measure of a very small angle (e.g. an angle not greater than i') so that sin d= tan d=d and cos d= i very approximately, and if be an angle not very near o or 90, we have sin(#+ d) - sin = sin dcosO- sin #(i - cos d) = dcos 0, approximately ; cos( + d) cos 9 = - sin d'sin - cos 6( i - cos d) = - d sin 0, approximately ; tan(0 + d) - tan = ^dsz 1467 ( 6828V 6 io88( 4i99 + = 327, Z cos 7529' = 9- 3990878, Z> = 4882, log 2= .3010300. 1 6. Find the values of (i) 2. 1078 cos 3 3/f, when A = 27io' ; log 21078 = 4.3238294 log 68066 = 4. 8329302, Z> = 64 Z cos 8i3o' = 9. 1697021 (ii) .02845 cos 3 -, when A = 35 i5'; lo 2 45 = 3-4540823 2 log 24628 = 4. 3914291, Z>=i77 Zcos 1 737' = 9.9791397, Z> = 40i .0076829 sin 2 log 76829 = 4.8855252 (iii) i, when A = 35I7'; lo g 86444 - 4-9367382 D = 50 cos A L cos 3517' = 9.91 18528 Zsin I738' = 9.4813342 (iv) 'tan- 13 */ 2 - ; 3 10 ' givenlog2, Iog3, (v) -cosec' 1 !? given Iog2, Iog3, Zsin 522'4o" = 8.97i8424 5 3 ' diff. ] " (vi) 2cos- 1 f^V; S iven log 75 =1.8750613 \4/ Z cos 2i28' = 9.9687773, VH, ^{(ijm TABL ES OF LO GA R1THMIC RA T10S. 17. Find the smallest positive values of the angles satisfying the equations (i) sin s 0=Vf ; given log 2, log;, Z sin 55I4' = 9. 9092371, Z> = 9io. (ii) 8 tan .* = 3 cos .* ; given log 3, Zsin I928' = 9.5227811, Z> = 3572. (iii) tan 3 =& ; given log 2, log 3, L tan 3645' = 9.8731668, D - 2634. (iv) 3 tan 6 = 8 cot ; given log 2, log 3, L cot s832 r = 9.7867520, # = 2837. (v) 3 sin 2 20 = 2^/2 ; given log 2, log 3, L sin 76V = 9. 987 1860, Z> = 3ii. (vi) 2 cos 4 = J sec ; given log 2, log 3, L cos 454i' = 9.8442432, Z> = 1293. (vii) 3sin 2 + 2sin0=i, given log 3, L sin I928' = 9. 5227811, Z> = 3572. (viii) sin0cos< = J\. given log 2, log 3, L sin 32 13' = 9. 7268269, Z> = 2OO4, sin0cos0 = $J' Zsin 739 / = 9- 1242477, Z> = (ix) sin x - 2 sin y\ t given log 2, Zsin 2633' = 9- 6502868, Z) = 2527, /J' Zsin6326' = 9.95i5389, Z>= 631. 5 9 18. Given Z tan 54i5'2o"= 10.1428185, and that the tabular difference for 10" = .0000444, find x from the equation 10 tan x = (tan 54i5'29") 8 . 19. Show that the smallest positive value of 6 which satisfies the equation 7 tan 2 + 8^/3 tan 0= i is 359'i6-2", having given log 2 = .3010300 L sin 3359' = 9- 7473743 Zsin 34 =9.7475617 CHAPTER IX. Reductional Formulae. 49. In order that expressions may be adapted to logarithmic computation they must be expressed as consisting of products and quotients. Hence, when logarithms are to be applied to trigonometrical expressions, these latter will frequently have to undergo reduction into a suitable form before any computation can take place. It will therefore be well to give a few of the simpler reductional formulae, all of which can be easily verified by the student, and which will assist him in working the more complicated examples. (A) i - sin *A = cos 2 A or I - cos 2 /? = sin 2 A, i + tan 2 ^4 = sec* A or sec 2 /? - i = tan 2 /?, i + cot 2 /? = cose'c 2 /? or cosec 2 ^ - i = cot 2 /?. (B) cos A cos j5+sin A sin B = cos (A + B), l+tan/? tan B cot A cot Z?+ 1 _ co\.Bco\.A From these we get (i) tan A + tan B - ^^ =^ T !> cos A cos .5 , sin A cos B , cos A cos .Z? sin ^4 sin B (ii) Putting A or B equal to 45, cos /? + sin A = x/2 sin (^4 + 45), tjz cos (45 - A ) , or ^/2 cos (^ - 45) ; cos A - sin A = ^/2 cos (^f + 45), or ^2 sin (45 -A); sin -<4 - cos A - v '2 sin (/? - 45). 60 REDUCTIONAL FORMULAE. 61 cos ^+ sin ^4 (iii) Putting /4 or B equal to 30 or 60, cos A + x/3 sin A = 2 sin (A + 30) or 2 cos (60 -^4), cos ^4 - V3 sin ,4 = 2 sin (30 - ,4) or 2 cos (^ + 60), x /3 sin ^ - cos A = 2 sin (A - 30), sin ,4 + x/3 cos ,4 = 2 cos(^ - 30), 2 cos (30 - A), or 2 sin (A + 60), sin A - ^/3 cos ^ = 2 sin (v2 - 60), ^3 cos ,4 - sin A - 2 cosM + 30) or 2 sin (60 - A]. _ ( 0+ = cot = cot (A 60). (iv) Putting = A, cos 2 A -sin 2 ^ = f i - 2 sinM = cos 2^ or I - cos 2.A = whence - i = cos 2^ or I + cos 2 A = _ ! = 2 COt ^4 . COt 2^4. 2 COt A (v) Putting .5 = 24, 3 sin ^4 - 4 sin 3 ^4 = sin 3^4 or 3-4 sin 2 .4 = sm ^ , sin -n( or 4 cos 2 ^4 - 3 = cos * ,. 3 cot'M - i 3 cot 2 ^4 - i sin A - sin B = 2 cos^-^ sin COS A + COS B - 2 COS COS , 2 2 cos B -cos A = 2 sin - sin - 2 2 62 LOGARITHMS. From these we get or 2 ^ cos 2 /? sin 2 ^ sin 2 sin A- sin Z? (D) cot ^4 + tan A = 2 cosec 2.4, cot A - tan ,4 = 2 cot 2^, cot A + tan /4 cot A -ten A = cos 2A. = . i + cos^ sin^4 2 Qt A_ 2"' From these we get I + cos A = sin A cot , i - cos A - sin A tan -, 2 I + sin A = cos A tan^45 + ^), I - sin A = cos ^ cot (45 + d\ _ i + cos^ 2' ~ \~___ ^^ A I A. A __j ^^ sec AtanA= tan (45 V cosec ^4 - cot A = tan , cosec A + cot A = cot , Subsidiary Angles. 50. Expressions may sometimes be adapted to logarithmic computation by the introduction of subsidiary angles. (i) To adapt \ / = 77 log 7 1354 = 4- 8534183 Z> = 6i log 90849 = 4. 9583202 D = 47 L tan 38i4' = 9.8964517 L sec 38i4 ; = 10. 1048555 L tan 38! 5' = 9. 8967116 Zsec38i5' = 10.1049550 log 562. 33 =2.7499913 log 713. 54 =2.8534183 7 54 i 6 /. log 3 or log 562. 337 = 2. 7499967 /. loga or ^713.541=2.8534189 Hence L tan = log b - log a + 10 = 9.8965778. To find Z sec : L tan L sec .'. 2599^= 1261 x 995 /7ii6 /955 ^=483 2599 5778\ I26l 995 8555 + ^ , 8555 + ^=9038 M5I7/ ^8555 f Hence L sec 6 = 10. 1049038. Therefore , log \/a 2 + 2 = log + Z sec 6- 10 = 2.8534189+10.1049038- 10 = 2.9583227 Now log 908. 49 = 2.9583202 Z> = 47 )25o(5 235 /. V 2 + ^ = 908. 495. (2) To adapt \/l> 2 + c 2 -2dccosA to logarithmic computation. Let *JP + c*- 2bc cosA=x; then (i) x 2 = P + c* - 2bc cos A = F + P - 2^(2 cos 2 ^ - i = (6 + c? - tfc cos 2 ! = (b + r) 2 { i - -^L_ cos 2 ! }' Now, t ^ c ^ being necessarily a proper fraction, we can put ^^cos'-ssm^orcosty in which case we have x - (b + c) cos B or (b + c) sin 6, respectively Zsin 6 or Zcos d can be found from (i), and the corresponding ectively is then substituted in (ii) to determine x. Or thus (ii), x> = 3 2 + c 2 - 2bc( I - 2 sin 2 -) = (b - cf + tfc sin^ Zsin 6 or Zcos d can be found from (i), and the corresponding value of Zcos0 or Zsin0 respectively is then substituted in (ii) to determine x. And putting -^ 2 sin 2 = tan 2 0, we have .# = (- c) sec 0, and the value of x can be determined from (i) and (ii) as before. 64 LOGARITHMS, Example. Compute the value of \/6 2 + c z -2t>ccosA when -8214, ^=3732, and A=6i$3'. Given log 4= .6020600 log 373 2 = 3- 57 194i6 L sin 5239' = 9-9003367 L sin 524o' = 9- 9OO433 l log 8214 = 3.9145547 log 7246 = 3-8600983 log 1 1946 = 4.0772225 L cos 5239' = 9. 7829614 L cos 5 2 4o' = 9- 7827958 L cos 3o56' = 9. 9333688 Since logarithmic tangents are not given we adopt the first mode. Put _4_ cos 2 - = sin 2 0, so that x = (b + c) cos 0. (b + c)- 2 log 4= .6020600 Iog82i4= 3.9145547 \ogc= log 3732= 3.5719416 2Z cos = 2Z cos 3056'3o"= 19. 8666618 2 26.9552181 2 log ( + <:)= 2 log 1 1 946= 8.1544450 2.L sin 6 = 18.8007731 Zsin0= 9.4003866 To find L cos 6 L cos /7958 1656(7958 + V 9 6i4 Hence Zsin 964(3866 \3367 Lcosd= 9.7828757 log ( + <:) = 4.0772225 - 10. log* = Zcos3o56' =9.9333688 diff. for 30" = - .0000379 Z cos 3056'3o"= 9.9333309 19.8666618 ' 964^=1656x465 ^=799 7958+^=8757 3.8600982, and # = 7246. (3) To solve the equation a sin x + b cos x-c. Put a = r cos 0\ so that tan = -, and the equation then becomes rsin (* + 0) = and # + can be calculated from (i) and (ii) respectively, and x is then the difference between these computed values. EXAMPLES. XVIII. I. Express in forms adapted to logarithmic computation, (iii) cos 5^ -cos 7^^ sin SA - sin 2 A ' , x cos 2/4 -cos4^_cos^ -cos 3/4 sin ^A -sin 2/4 sin 3/4 -sin /4 s sin 2a + sin 2/3 (ix) cos ^4+ cos 2A+ cos 3/4, (xi) cos 3* cos 2* + sin 4* sin*, (ii) tan 2 0-sin 2 0, ^ sin 7 A - sin A ^ sin 8 A - sin 2 A 1 / ., sin 3$ + 2 sin 5$ + sin 76 sin 50 + 2 sin 70 + sin 9$' sin _ ^ (x) cos^rcos(>/ + (xii) sin (a -/3) + sin (j8 -7) + sin (7 -a), REDUCTIONAL FORMULAE. (xiii) cosa + cosp + cosy + cos(a + B + y), (xiv) r 2 sin A - shi 2.A 2 sin A+ sin 2^4' / . tan 50 + tan 30 ^4 '""'tansa-tansV (xv.) tan ^ - tan -, (xvii) tan $A - tan 2A - tan A, (xviii) tan ^4 + tan (60 + ^4) + tan (120 (xix) cot sec 2 - cos cosec 0, (xx) (xxi) sin 3^4 cos 3 ^4 + cos 3^4 sin 3 ^4 , (xxii) <="> (xxv) 2 2 2. Compute the values of when ^ = o 2Q .. given log 2= .3010300 log 50165 = 4. 7004008 D - 87 (ii) .0085627 r ~ tan , when 2A = 374S'; S iven lo S 8 5627 = 4.9326107 i + tair^ log 67. 704 = 4. 8306 143 D = 64 Z sin 52 1 5' = 9. 8980060 (iii) . 000237 1 5 sm ^ ~ cos ^, when^ = 73io / 20 // ; Iog237i5 = 4.375 2 3i log 12701 =4. 1038379 Z>=342 Z tan 28io' = 9. 7287161 Z>=3O35 -, when A = I35I5'45"; g iven Io g 1 7 = 1.2304489 log 66770 = 4.8245814 D = 65 Z cos 4444' = 9-8514969 Z> = 1252 (v) 2 50000.9 I + cot ^, when 2,4 = 77; g iven Io g 2 5ooo= 4.3979400 Z>= 174 i+tan-^4 log 395 12= 4.5967290 Z>= 1 10 Z cot 383o' = 10.0993948 (vi) 1.00076(1 +tai&4), when A - 125; given log 10007 = 4.0003039 Z>= 434 ^30419 = 4.4831449 Z>=i 4 3 ^cos 55 = 9. 7585913 fvin A 3 /sin^(i-cos^) ,,^_ ^ p. given log 353 19 = 4. 5480084 Z>= 123 J V .0070639 ' when ^- 2I 5; log 52861 = 4 . 7231354 ^=82 Z sin 35 =9.7585913 Zsin723o' = 9.9794i95 sin2 ^ given log 175 = 2.2430380 3.5' Z sin 5243' = 9. 9007219 Z>=962 (ix) 32.574(tan^ + tan^), when A = 7i32'l given ^32574 = 4.5128711 Z^ = 25i8'/ log 1 1293 =4.0528093 Z>=385 Z sin 83 i o' = 9.9969040 Z cos 7 1 32' = 9. 5007206 Z cos 25i8' = 9. 9562081 (x) v/cosM - sin 4 ^4, when A - 34l6' ; given ^60494 = 4.7817123 Z?=72 Z cos 6832' = 9. 5634335 (xi) I1 -3i5sn^ when ^ = II5 45 '. given log 56575 = 4.7526246 i + cos^ log 1 6230 = 4. 2 103 1 85 Z>=268 Z sin 64! 5' = 9. 9545793 Z cos 57^2' = 9.7258229 D = 2012 66 LOGARITHMS. ii) tanM-tan 2 Z? h _ g ,, given log 37752 = 4- 5/69400 Z> = 1 15 (XU) sinM-sin^' WhCn A B I gJL } Z cos 74 8' = 9.4367980 Z cos 5328' = 9. 7747288 (xiii) A/3 sin ^- sin 3^ wheriA= ^ given log 38643 - 4.5870708 Z>= 113 > 3 cos A + cos 3^ Z tan 85 = 1 1.0580482 (xiv) sni 7Q + sin85 given log 50417= 4.7025770 Z>= 86 cos 50 + cos 105 ; L tan 77so' = 10.6542448 Zcos 73o'= 9.9962686 Zcos273o' = 9.9479289 -- when ^ = 6 o , Vcosec A + cot A 1 Vsec A + tan A ) given log 21439 = 4. 3312045 -#=203 Z tan 3i37' = 9. 7893023 ^=2829 L tan I322' = 9. 3758810 Z> = 5613 (xvi) (sin^+sinM/cos^ + cosM when A = ^^^ \ sin A - sin Bl \ cos A-cosJ g- 4 g28' / given log 14337 = 4- 1 564583 D = 303 L tan 3952 r = 9.9217602 .. sin 35 + sin 55 + sin 75 . given log 90383 = 4. 9560868 D = 48 1 ' sin 45 + sin 65 + sin 85 ' ^ sin 55 = 9.9133645 Z sin 65 = 9. 9572757 (xviii) J(i - tan A)\ when ^4 = i68'32" ; given log 63732 = 4.8043575 D - 69 L cos 1 68' = 9. 9825506 D = 366 Z cos 6i8' = 9. 6837430 .#=2293 3. Find the values of the positive angles less than 180 that satisfy the following equations : (i) sin # + cos #=1.2, given log 2, log 3, L sin 58 3' = 9.9286571, ^ = 787. (ii) sin # - cos .* = .3, given log 2, log 3, Zsin 12 14' = 9. 326 1174, ^=5823. (iii) cos .* - sin .* = .2118, given log 2, log 21 18 = 3.3259260, L cos 8i23'2o" = 9. 1753004, diff. 10" = 1390. (iv) sin .* - /y/3 cos .r = J, given log 2, log 3, L sin 935' = 9. 2213671, Z> = 7476. (v) ,y/3 sin x - i^ cos x, given log 625 = 2. 7958800, L cos 5i2o' = 9. 7957330, ^=1579. ( vi) 3 sin x + 4 cos x = 4. 3, given log 2, log 3, log 86 = 1 . 9344985, Zsin 59 1 9'= 9.9344988, Z tan 53 7' = 10. 1247266, D = 2632. (vii) 4 sin .r - 5 cos # = I, given Iog2, Iog4i = 1.6127839, Zsin 859'= 9-I93534I, diff. io Z tan 5i2o' = 10.0968034, D - 2590. (viii) tan0 + cot0 = 3^, given log 2, log 7, Z sin 345i' = 9.7569630, Z>=i8i5. (ix) tan x - cot x = 2|, given log 2, log 3, Zcot4i3-8'= 10.0511557, Z> = 2544. (x) i - tan 2 * = 7 tan x, given log 2, log 7, Z tan I556' = 9-4555 8 57 5 D = 4784. (xi) i + tan 2 * = 5 tan x, given log 2, Z sin 2334' = 9.6018600, D = 2895. (xii) 1-2 sin /4 + cos A = o, given log 2, Z tan 2633' = 9. 6986847, Z> = 3 1 59. REDUCTIONAL FORMULAE. 67 (xiii) i - sin A - cos A, given log 2, log 3, Z tan 33V = 9-8237981, D - 2738. (xiv) sec(* + 0) + sec(* - 0) = 2 sec 0, when = 140 ; given log 2, Z cos 40 = 9. 8842540, Z cos 57I2' = 9. 7337654, Z> = 1961. (xv) sin x + sin y= 1.24^ given log 62= 1.7923917 log 65= 1.8129134 log 1 24= 2.0934217 " j' = 9.8450181 Z Z tan 622o' = i o. 280445 1 Z Zsin622o' = 9.9472689 Z 3 / . 2 o _ 2 r 4. Find the acute angle whose tangent = \/ - m ~ sm , V sin 2 35-sm-25 given L sin 60 = 9. 9375306 L sin 80 = 9.9933515 Ztan46i3' = 10.0184499 D- 2529 5. If 2a = 36, find the acute angles satisfying the equation a cos 6 + b sin = ; given log 2, log 3, Z tan 33V = 9-8237981, # = 2738. 6. If sin = ;;z sin 01 find the principal trigonometrical ratios of and in forms adapted tan = n tan 0/' to logarithmic computation. 7. Given sin(0 + a) = #2sin0, find in terms adapted to logarithmic computation. . Find, by means of subsidiary angles, the values of , when a- 30. 4025, = 21.7856. Given log 21 785 = 4. 3381 576 Z>=i99 log 30402 = 4. 4800645 Z> = 1 43 log 37241 = 4.5710213 Z>=ii7 L tan 3548' = 9. 8580694 Z sec 3548' = 10.0909450 L tan 3549' = 9-858335? L sec 35^9' = 10.0910361 (i.i) \/ 2 + 2 , when a = 87.079, 3 = 129.384. Given log 87079 = 4. 9399 1 34 log 12938 = 4. 1 1 18671 Z> = 336 log 1 5595 = 4- 1 929854 Z> = 278 L cot 563' = 9.8281696 L sec 56 3' = 10.2530008 L cot 56V = 9.8278969 Zsec 56V = 10.2531885 3 , when a = .35991, = .24376. Given log 2437 6= 4. 3869624 Ztan346' = 9.83o62i3 D- 2721 Io g 35991 =4- 5561939 Z cos 346' = 9. 9180620 Z> = 856 log 43468 = 4. 638 1 697 Z> = 99 (iv) V 2 + c 1 - 2bc cos A t when b = 17. 14, c - 32. 36, ^4 = 4822'. Given log 2= .3010300 log 1714 = 3.2340108 log 495 = 2. 6946052 log 3236 = 3.5100085 Z cos 24!!' = 9.9601088 log 2457 5 = 4. 390493 5 Z>=i77 Z sin 60 1 3' = 9.9384747 Z cos 60 1 3' = 9.6961 130 Z sin 6oi4' = 9-938547 Z cos 6oi4' = 9.6958922 68 LOGARITHMS. when = 2139, log 4= .6020600 log 7956 = 3. 9006948 L cos 5744' = 9. 7274278 L cos 6i44' = 9-6753896 7.cos6i45' = 9.6751546 (v) (vi) 9. By introducing subsidiary angles adapt to logarithms the expressions o> \/^ w when = 104.28, log 4= .6020600 log 1 1326= 4.0540766 Zsin40i5'= 9.8103159 L tan 5946' = 10. 2344857 :58i7, -4 = ii528'. Given log 2139 = 3.3302108 log 5817 = 3.7646991 log 70080 = 4. 84 5 594 1 L sin 6i44' = 9.9448541 L sin6i45' = 9.9449220 = 217.54, A - 8o3o'. Given log 10428 = 4.0172010 log 21754 = 4-3375391 log 22499 = 4. D = 2904 >=62 CHAPTER X. Solution of Triangles. 51. Logarithms are applied to the Solution of Triangles, that is, are used for finding the remaining sides or angles when certain of them are given. To solve a triangle completely, of the six parts (3 sides and 3 angles), three must be known, but these must have values that are independent of one another. Now it is known that the three angles of a triangle are not independent in value, for they are always together equal to two right angles ; therefore it will not be sufficient to have only the three angles given, but a complete solution will be possible when (i) the three sides, (ii) two angles and a side, and (iii) two sides and an angle are given. The angles of a triangle are generally called A, B, C, and the sides respectively opposite to them a, , c. 52. We will first discuss the case of right-angled triangles. Let C be the right angle, so that C = 90; then (I) (2) (3) cos B cosA_\ or b = V~c sin A \_c cos B p sin B [_ccosA i.e. each side equals the hypot- enuse multiplied by the sine of the angle opposite to, or the cosine of the angle adjacent to the former. (4) ia.r\A~\_a cot B\~l) (5) _b cot B or b = C i.e. each side equals the other side multiplied by the tangent of the angle opposite to, or the cotangent of the angle adjacent to the former. In the right-angled triangle, one angle C ( = 90) is known; hence, in addition, of the remaining sides and angles two only need be given for the complete solution of the triangle, but these must not be the two angles. Case (i). Given the frvo sides a, b. Either A or B is found from (4) or (5) and the other angle is then known since 50. The hypotenuse c is then found from (2) or (3). 69 yo LOGARITHMS. Case (ii). Given one side and the hypotenuse, e.g. a, c. Either A or B is found from (2), then A + B = 90 gives the other angle. The other side b can then be found from (3), (4), or (5) ; or independently of the angles from (i). Case (iii). Given one angle and side, e.g. a, A. ^ = 90 -A. The other side b is found from (4) or (5), and the hypotenuse c from (2) or (3). Case (iv). Given one angle and the hypotenuse, e.g. c, A. B-fyf -A ; then a is found from (2), and b from (i), (3), (4), or (5). 53. When the triangle is not right-angled we have the following cases : (i) Given the three sides. (ii) Given two sides and the included angle. (iii) Given two sides and an angle not included. (iv) Given two angles and a side. In the solution of these triangles the following formulae are employed : A_^ (s-b)(s-c) sin^./!^ 2V be Putting tan 6 = - this formula becomes tan (2? - C) = tan(0 - 45)cot L C 2 Putting cos = c - it becomes tan \(B - C)= tan 2 ^ cot-.~l O 2 2 J , sin A _ sin B _ sin C \ j / --- 7 ~ - a b c (4) at^lP + c^-zfocosA, this formula being adapted to logarithms by introducing a sub- sidiary angle as in Art. 50. [In the above formulae the letters a, b, c are of course interchangeable, provided A, B, C be also interchanged in like manner.] Case I. 54- A. Given the three sides a, b, c. B. Solution. One of the formulae (i) is used to determine each of two of the angles A, B, C ; and the third angle is then known since A + B + C= 180. [OBS. When all the angles are required the tangent-formula is the most convenient to use since fewer logarithms are then required (4 instead of 6 on the right hand side) ; but, if SOLUTION OF TRIANGLES. 71 only one angle be wanted, there is no such advantage. Of course, when logs, are given for the purposes of any question, our selection of the formula must be guided by the data.] C. Example. If a - 217, b - 192, c - 89 ; find all the angles. Given log 32 = 1.5051500 L tan 465 5'= 10.0290779 ^ = 2532 log 57 = i-7558749 log 1 60 = 2. 204 1 200 Ztan3o59' = 9.7784875 # = 2862 log 249 = 2.3961993 W 92:? _^ J ~ a) JW 57- 160 2)498 249.32 249 = * Z tan = | (log 57 + log 160 - log 249 - log 32) + 10 = 10.0293228 But Ztan4655' = 10.0290779 2449 60 # = 2532)146940(58.0 202 5 6 and 840 L tan B = ^{log 32 + log 160 - log 249 - log 57) + 10 = 9.7785979 But L tan 3o59' = 9.7784875 To find C: and Ji.SC [OBS. In finding the values of A, B, C to the nearest second, , , - must be calcu- lated to the nearest tenth of a second, since the multiplication of these tenths by 2 may affect the seconds' units in the values of A, B> C. It is also evident that we need only know the ratios of the sides a, l>, c to one another, and that the numbers expressing the ratios a\b\c may be taken as the values of the sides themselves.] 72 LOGARITHMS. Case II. 55' A. Given two sides and the included angle, e.g. b, c, A. B. Solution. Formula (2) of Art. 53 gives %(B - C) ; then Adding and subtracting these, Z? and C are found. The third side a can then be found by means of formula (3) using the value just obtained of B or C, or independently by means of formula (4). C. Example. If = 23.46, c = 7.85, A =73I4'; find the remaining angles and side. Given log 1561 = 3.1934029 Zsin73i4' = 9.9811331 log 2346 = 3.3703280 Zcot3637' =10.1289428 log 3J3 1=3-495683 1 Ztan335i' = 9.8265323 ^ = 2730 log 22488 = 4. 35 1 9508 Zsin87i4'2o" = 9-9994955 Z>io"=io Z>=i 9 3 TofindZ?and C: ^^_ C) = ^cot| = '^o^?', .'. Ztan|(Z?-C) = log 15.61 -log 31.31 +Z cot 3637' = 9.8266626 But Z tan 335 1^ = 9.8265323 ~ Z> = 2730)78180(28.6 5460 Hence %(B- C) = 335i'28,6" and 17400 ' adding Z? = 87I4'28.6" _l63o an( j subtracting C- I93i'3i.4" To find a : sin A _ sinZ? /Taking the b, a portion of the formula, sinceX \ log b is given and not log c. ) = log 23.46 + Z sin 73I4' - Z sin 87I4'28.6" = 1.3519647 But log 22.488= 1.3519508 Z>= 193)1390(7 '35 1 Hence a- 22.4887. [OBS. In finding the angles B and C, only the ratio b : c need be given, in which case the numbers expressing the ratio can be used for the sides themselves ; but in finding the third side , the actual values of the sides must be given. This same method of solution is applicable when two sides , c are given and the difference B - C between their opposite angles. Formula (2) determines A, and thence B+C which = 180 A. B and C are then obtained by addition and subtraction of B+ C and B - C.] Case III. 56. A. Given two sides and an angle not included, e.g. a, b, A. B. Solution. Since the sides given are a and b, we take the a, b portion of formula (3) and thence determine the angled. Then C= 180- (A + J5) ; and the third side c is obtained by using the c portion of formula (3). SOLUTION OF TRIANGLES. 73 This is the case in which the solution may be ambiguous, that is, in which there may be two triangles with the given parts. [The Ambiguous Case.] First consider the angle A acute. A A Draw CA=b, CAX-A, and with centre C and radius = a describe a circle; there is ambiguity only when this circle cuts AX in two distinct points lying on the same side of A, neither of which coincides with A. The two triangles in the ambiguous case are ABC, AB'C ..A Now the perpendicular CD =b sin A. (i) ab. One solution AB"C. Not ambiguous. One solution AB'"C. A Not ambiguous. (In this case AB iv C has no angle A ; CAB iv = 180 - A. ) It will be seen from the above that (i) when a = b, there is no ambiguity ; (ii) when a does not equal b y the solution is ambiguous when the given angle (A) is opposite to the smaller side tinless the. triangle is right-angled. [When the triangle is right-angled the angle first found in the process of solution (i.e. B) comes out 90, so that LsinB = 10.] When the given angle is right or obtuse, taking the angle to be ADC or AB'C respec- tively^ it is clear from the figure that there can be no ambiguity, and in these cases the angle given is opposite to the greater side. In the ambiguous case there will be double values for each of the required parts. The A A acute value found for B (ABC] is taken from 180 to obtain its second value (AB'C). In A A each case A + B is taken from 180 to determine the third angle C (ACB or ACB'). To find the third side c (AB or AB'), either formula (3) of Art. 53 is used, or AD and DB can be calculated from the equations AD = b cos A, DB=a cos B, the two values of c being then the sum and difference of AD and DB. 74 LOGARITHMS. C. Example. If a = 47, =53, A = 2b42 f ; find the remaining angles and side. Given log 47 = 1.6720979 Zsin3642' =9.7764289 lo g 53=1-7242759 Zsin4222' =9.8285778 Z>=i385 log 77218 = 4. 8877 19 Zsin 79 4' =9.9920445 ^ = 244 log 77704 = 4. 89044 Zsin 54o'io" = 8.9947o89 Z>io" = 2i2i This is presumably an ambiguous case since the angle given is opposite to the smaller side. To find B\ sin A _ sin B sin B = b sin A (^ may k e necessar y here to reduce - beforeX \ taking logarithms to suit data. / .'. Z sin B = log b - log a + L sin A = Io g 53 - log 47 + L sin 36^2' = 9.8286069 But Z sin 4222' = 9. 8285778 -0=1385)17460(12.6 Hence j5 = 4222'i2.6"or 1385 3610 [The solution is ambiguous since B has 2770 not come out 90, and therefore we get a ~~g75 second value of B by taking the first value from 1 80.] To find C: A-\-B- 79 4' 12. 6" or ^ig^^", C- ioo55'47.4" or 54o'i2.6". To find c : sin A sin C (i) For the larger value of c we must take the larger value of C, /. log c- log 47 + Zsin ioo55'47.4"- Zsin 36^2' = log 47+ Zsin 79 4' 12. 6" -Zsin 36^2' = 1.8877186. Hence ^ = 77.218. (ii) The smaller value of c can be shown in like manner to be equal to 7.7704. [OBS. In finding the angles B t C only the ratio a : b need be given ; but for the third side c the actual values must be known.] Case IV. 57- A. Given two angles and a side, e.g., A, B, a. B. Solution. C= 180 -(A + B). The remaining sides b t c are determined by using formula (3) in the same way as in the example in the last article. ^ 58. The following table gives a list of the formulae used in the solution of triangles, and in finding their areas and the radii of their circumscribed, inscribed, and escribed circles. SOLUTION OF TRIANGLES. 75 A = Area of triangle, R = Radius of circumscribed circle, r=- Radius of inscribed circle, r a Radius of escribed circle, opposite to angle A. I. GIVEN 3 SIDES. II. GIVEN 2 SIDES AND INCLUDED ANGLE. III. GIVEN (i) 2 ANGLES AND A SIDE ; (n) 2 SIDES AND ANGLE NOT INCLUDED. be 2 V be ^=v 2 v sis -a) A /(, = 7 r V L -a)(s r- ,= - or \/^ M_) s-a V j-d- \ be sin A sin A _ sin Z? _ sin C" = 2, C=I20 Z tan 6ii7'= 10.261329 log 3 = .477121 Z tan 6ii8 / = 10.261629 7 8 LOGARITHMS. 23. If 3=I52 log 4. 132 =.6161603 40. The angles A, B, C of a triangle ABC are 40, 60, 80 respectively, and CD is drawn from C to the base bisecting the angle A CB ; find CD. AB = 100 inches L sin 40 = 9.8080675 log 2 = . 3010300 L sin 50 = 9. 8842540 log 5-73979 = -7588951 L sin 60 = 9-93753Q6 41. If b be to c as 1 1 to 10 and A - 3525', use the formula tan&(2? - C) = tan 2 ^ cot to find and C. log 1.1= .041393 Ztan I2i8'36" = 9.338891 L cos 2437'i2" = 9.958607 L cot I742'3o" = 10.495800 Ztan 828'56.5"= 9.173582 42. a - 3, b - 7, c - 8. Find C. log 75 = 1.8750613 L cot 496'22" = 9.9375306 43. The sides of a triangle are 7, n, 14 ; find the smallest angle. log 2 = .3010300 L tan I446' = 9.4209275 log 3 = '477I2I3 L tan I445' = 9.4204196 44. a- 12, b= 17, c = 23. Find A. ^364 = 2.5611014 L cos 15 14' = 9. 9844660 log 39 1 = 2. 592 1 768 diff. i ' = 344 45. ^ = 7, 3=io, c = 5. Find A. Iog2 = . 3010300 logii= 1.0413927 log 3 = -4771213 L cot 2oi6' = 10.4326795 Z> = 3886 46. The sides of a triangle are 32, 40, 66 ft. respectively ; find the greatest angle. log 207 = 2.3159703 L cot 66 1 8' = 9. 6424342 log 1073 = 3.0305997 diff. i' = 3433 47. The sides of a triangle are 25, 26, 27 ; find the largest angle. log 2 = .3010300 Ztan3i57' o" = 9. 7949455 log 3 = .4771213 L tan 3i56'5o" = 9.7948986 log 7 = .8450980 48. b - 5, c - 3, A - 120; find the other angles. log 4.8 = . 6812412 Ztan 8 1 2' = 9. 1586706 diff. 60" = .0008940 49. Two sides of a triangle are respectively 200 ft. and 115.462 ft. and the included angle is 30; find the other angles. log 4. 2269 =.6260220 log i. 57731 = -i979695 Ztan 1 5 = 9. 4280525 50. a- 55, 6 = 40, C= 120; find the other angles. log 3= .4771213 L cot 8447'2o" = 8. 9600075 log 19 =1.2787536 diff. io" = 2328 51. rt = 7, = 5, C = 4424'36"; find A and B. log 2 = . 3010300 L tan 22I2' = 9.6107586 tog 3 = -4771213 L tan 22I3' = 9.61 1 1 196 So LOGARITHMS. 52. a= 17, =13, C* = 407 ; 2o"; find ^ and B. log 2 = .3010300 log 3 = -477 1213 Z tan 203' = 9- 5622439 # = 3921 53. = 25, < = 7, ^ = 7344'; find B and C. log 75 = 1.8750613 L tan 3652' = 9.8750102 Z tan 3653' = 9.8752734 54. b = 19, a = 35, C = 577'3o"; find ^ and .#. log 15= 1.1760913 L cot 2833' = 10.2643323 dift. i' = 30o8 55. a- 14, = II, C = !34i'8"; find ^f and ^. log 120 = 2. 0791812 L cot 65o'4o" = 10.9207117 diff. io" = 1780 56. In a triangle ABC the angle A is 8644'. and the sides containing it are II ft. and 21 ft. Find the side opposite to A. log 2 = .3010300 L sin 4O42' = 9.8143131 log 231 = 2.3636120 L cos 4O42' = 9. 8797462 log 24255 = 4.3848013 L sin 4043' = 9.8144600 log 24256 = 4.3848192 L cos 4043' = 9-8796375 L cos 4322' = 9.8367447 57. If A = 30, AB = 5, BC= 3, find the remaining angles. log 12 = 1.0791812 58. Find the length of the side a of the triangle ABC, having given A = 653o', B- 7o4o', ^=123. log 123 =2.0902581 Z sin 653o' = 9. 9590229 log i. 6 1 74= .2088174 Z sin 435o' = 9. 8404593 log i. 6175= .2088443 59. Use the formulae where cos 6 = c to find the angles of the tri- angle whose sides a, b t c are respectively 10, 8, 4. log 2= .3010300 Z cos 2922'= 9. 9402670 &=7 11 log 15= 1.1760913 Z cos 785o' = 9. 2870480 60. If the vertical angle of a triangle be 120, the length of the line joining the vertex to the middle point of the base 10^/7 feet, and that of the line bisecting the vertical angle 24 feet ; find the sides and remaining angles. log 3= -477I2I Z sin 2324' = 9. 598952 log 19 =1.278754 Z sin 2325' = 9. 599244 CHAPTER XL Heights dnd Distances. 60. Problems in heights and distances are simply practical illustrations of the solution of triangles, and therefore the formulae used in solving them are those set forth in the last chapter. If we consider the positions of two points A and B, one (say A) at a higher level than the other (B\ the angles between the straight line AB and the horizontal lines through A and B in the same vertical plane are called respectively the angle of depression of B, and the angle of elevation of A. These two angles are of course equal. The angle of elevation of A (as viewed from B) is the angle through which the arm must be elevated from a horizontal position in order to point to A. The angle of depression of B (as viewed from A) is the angle through which the arm must be depressed from a horizontal position in order to point to B. The angle subtended at a point by any straight line is the angle con- tained by the two straight lines drawn from the point to the extremities of the straight line subtending it. Thus ACB is the angle subtended at C by the straight line AB. Two points are accessible to one another when no obstacle prevents the measurement of the direct distance between them. 6 1. Problem A. To find the height above the hori- zontal plane of an object standing upon the plane and accessible at its base. Data. Let AB be the object, its base being at A ; P the point of observation in the horizontal plane through A, Observations. Measure PA, BPA. Solution. BA (the required height )=PA tan BPA. and log BA = log PA + L tan BPA - 10, whence BA can be computed. P F 81 82 LOGARITHMS. 62. Problem B. To find the height above the horizontal plane of an inaccessible object. Case (i). By measurements in the same vertical plane with the object. Data. Let A be the object, P and Q two points of observa- tion in the same vertical plane with A and mutually accessible. Construction of figure. Draw AB perpendicular to PQ pro- duced, taking P and Q on the same side of B. Q Observations. Measure PQ, APB, AQB. PQ=PB-QB = AB cot APB - AB cot AQB Solution. TM. Aril . ,, . , ^ PQ sin APB sin AQB Thus AB (required height) = ^ , do(AQff-AP) and log .4^ = log /> + L sin ^^ + Ls\nAQB-L sin(A QB-APB) - 10, whence AB can be computed. (If />and Q had been on opposite sides of B, the only difference would have been a plus instead of a minus sign in the value of AB.) Case (ii) By measurements not in the same vertical plane with the object. Data. Let A be the object, P and Q two points of observation not in the same vertical plane with A but mutually accessible. Construction of figure. Draw AB perpendicular to the horizontal plane through P. Observations. Measure PQ, APB,APQ,AQP. Solution. PA = PQ} Q and therefore PAQ=iSo-(APQ + A QP), AB = PA sin APB A PQ sin A QPsin APB yiZ> sin PA Q and log AB = log PQ + L sin A QP+ L sin APB -LsinPAQ- 10, whence AB can be computed. 63. Problem C. To find the distance of an inaccessible object. A Data. Let A be the object, P and Q two points of observation mutually accessible. Observations. Measure P<2, APQ, AQP. Solution. PA = PQ* in PO - A , sin PA Q :. log PA = log PQ + L sin PQA - L sin PA Q, whence the distance PA can be computed. HEIGHTS AND DISTANCES. 83 64. Problem D. To find the distance between two accessible objects. Data. Let A and B be the two objects, both accessible from the point of observation P. Observations. Measure PA, PB, Solution. The distance AB can be computed as in Art. 55, two sides and the included angle being known in the triangle APB. 65. Problem E. To find the distance between two inaccessible objects. Case (i). By measurements in the same plane -with the objects. Data. Let A and B be the two objects, P and Q two points of observation mutually accessible and in the same plane with A and B. Observations. Measure PQ, APQ t BPQ, AQP, BQP. Solution. AP and BP can be computed from the triangles APQ and BPQ respectively by Art 57 ; and then, since APB = APQ-B$Q t the distance AB is obtained, as in Art. 55, from two sides and the included angle in the triangle APB. Case (ii). By measurements not in the same plane with the objects. In this case, in addition to the measurements of Case (i), we shall require the angle APB: the solution is then the same as in Case (i). EXAMPLES. XX. 1. A river PQ is 300 yards broad, and runs at the foot of a vertical cliff QR which subtends at the edge of the opposite bank an angle QPR of 25io'; find the height of the cliff. log 3= -4771213 Z tan 645o'= 10.3280372 log 1.4095 = .1490651 Z> = 3o8. Militia. 2. A lighthouse appears to a man in a boat 300 yards from its foot to subtend an angle of 62o'24. 7". Find in feet the height of the lighthouse. Iog3= .4771213 L tan 62o' = 9. 0452836 diff. I ' = 1 1 507 Sandhurst. 3. The shadow of a tower is observed to be half the known height of the tower, and some time after to be equal to the full height ; how much will the sun have gone down in the interval ? log 2 = . 3010300 L tan 6326' = 10.3009994 diff. i' = 3159 Sandhurst. 8 4 LOGARITHMS. A person wanting to calculate the height of a cliff, takes its angular altitude I23o', and then measures 950 yards in a direct line towards the base, when he is stopped by a river.; he then takes a second altitude and finds it 693o'. Find the height of the cliff. log 5= .6989700 Zsin I23o' = 9. 3353368 log 19=1.2787536 L cos 33 o' = 9.9235914 ^2296 = 3.3610566 Zcos2o3o' = 9.97i5876 Sandhurst. 5. From each of two ships, a mile apart, the angle is observed which is subtended by the other ship and a beacon on shore ; these angles are found to be 5225'i5" and 759'3" respectively. Find the distance of the beacon from each of the ships. log i. 2 1 97 = .0862530 L sin 75 9'3o" = 9. 9852635 log 1.2198 = .0862886 L sin 5225'i5" = 9.8990055 Woolwich. 6. AB is a horizontal line whose length is 400 yds.; from a point in the line between A and B a balloon ascends vertically, and afte? a certain time its altitude is taken simul- taneously from A and B-, at A it is observed to be 64 15', at B 482o'; find the height of the balloon when the observations are taken. log 2 =.3010300 L sin 64! 5' = 9-9545793 log 2. 29 1 49 = .46462 1 3 Z sin 482o' = 9. 8733352 L sin 6725' = 9. 9653532 Woolwich. 7. A man who is walking on a level plain towards a tower observes at a certain point that the elevation of the top of the tower is 10, and, after going 50 yds. nearer to the tower, that the elevation is 15. Find the height of the tower in yards to four places of decimals. L sin 15 = 9.4129962 log 25. 783= 1.4113334 Zcos 5 = 9. 9983442 ^25.784=1.4113503 .Woolwich. 8. A ship, sailing due north, observes two lighthouses bearing respectively N.E. and N.N.E. After sailing 20 miles the lighthouses are seen to be in a line due east ; find the dis- tance in miles between the lighthouses correct to four places of decimals. log 2= .3010300 log 11.715= 1.0687423 L tan 223o' = 9. 6 1 72243 log 1 1.716 = 1. 0687794 Woolwich. 9. The elevation of an object on the top of a tower 150 ft. high is found to be 5738' at a point 120 ft. from the base of the tower. Find the height of the object. log 12= 1.0791812 log 18933 = 4. 2772194 Z> = 230 L tan 5738' = 10. 1980454 10. The centre of the base of a tower which leans to the west is O, and P is an object at the top. From two points A (due east of O) and B (due west of O) P is observed to have the same altitude, viz. 5826'. The observer then walks from O due south to a point A" through a distance of 150 ft., and there finds that OA and OB subtend respec- tively at K the angles 3253' and 392i'. Find the height of /'above the ground, and its distance from the vertical line through O. log 2= .3010300 log 96977 = 4. 9866687 Z> = 44 log 3= -477i 2I 3 log 12299 = 4. 0898698 Z> = 353 7,tan3253'= 9.8105796 log 17901 =4.2528773 ^ = 243 Ztan392i' = 9.9137868 log 10998 = 4.0413137 Z> = 395 L tan 5826' = 10. 2 1 1 547 1 11. In order to calculate the height of a cliff, an observation is taken from a fixed position, and the angular altitude is found to be I23o' ; a second observation is taken from a point 950 yds. nearer to the cliff, and the angular altitude is found to be 693o'; find the height of the cliff, and the distance of the first station from its base. log 2= .3010300 log 19=1.2787536 Zsin I23o' = 9.3353368 log 8586 = 3-93379 L sin 203o' = 9. 5443253 log 2296 = 3. 36097 Z cos 2o3o' = 9. 97 1 5876 log 2297 = 3. 36116 Zsin 57 o' = 9. 9235914 HEIGHTS AND DISTANCES. 85 12. Two straight roads intersect at an angle of 30: from the point of junction two pedestrians A and B start at the same time, A walking along one of the roads at the rate of 5 miles an hour, B walking uniformly along the other road. At the end of 3 hours A and B are 9 miles apart. Show that there are two rates at which B may walk to fulfil the conditions, and determine the slower rate of the two. log 2 =.3010300 Zsin5627' =9-9208555 log 3 =.4771213 Zsin5626' =9.9207717 log 8. o 1 54 = . 9039248 L sin 2626'33" = 9. 6486522 13. A person in a balloon, which ascended vertically from the land at the sea level, finds the angle of depression of a ship at anchor to be 30; after descending again vertically for 600 ft. he finds the angle of depression to be 15; find the horizontal distance of the ship from the point of ascent. log 3 =.4771213 log i -9392 = -2876294 L cot 1 5= 10.5719475 14. In ascending a tower 150 ft. high a person observes from a window the depression of a point in the horizontal plane upon which the tower stands to be 48! 8'. When he reaches the top of the tower the depression of the same point is observed to be 562O'. Find the height of the window above the ground. log 2= .3010300 log 3= .4771213 tan 33*40'= 9.8235244 log 1 1214 = 4.04976 L tan 48i8' = 10.0501381 15. After climbing 1600 yards up a mountain side towards the summit in a direction making an angle of 38 1 2' with the horizontal plane, the summit is seen at an elevation of 6638'. Calculate the height of the mountain, its elevation at the foot being observed to be 532o'. log 2= .3010300 ^26562 = 4.4242608 D= 163 L sin I3i8' = 9.3618217 L sin 2826' = 9.6777309 L sin 532o' = 9.904241 1 16. The elevation of a tower at each of two points distant TOO yards from one another is 2622', and at a point midway between them 3O4o'. Find the height of the tower. log 2= .3010300 log 45156 = 4. 6547 1 55 Z>=96 L sin 2622' = 9.6474945 Zsin 4 1 8' = 8. 8749381 L sin 304o' = 9. 7076064 L sin 57 2' = 9. 9237554 17. Wishing to find the breadth of a river and being unable to walk any distance along the bank either way, I notice an object directly opposite to me^on the other bank and walk a distance of 400 ft. in a direction making an angle of 28I7' with the bank. The object is then seen in a direction making an angle of 78I2' with the bank. Determine the breadth of the river. Z sin 4955' = 9. 8837232 log 2= .3010300 log 14965 =4.1750767 = 290 18. The angle of elevation of a tower is 28i8' at a point A. After walking 270 ft. in a hori- zontal direction from A and at right angles to the line joining A to the base of the tower the elevation is seen to be i634'. Find the height of the tower. log 27 = 1.43 1 3638 log 96361 =4,983901 3 Zsin ii44' = 9.3o82590 L sin 28*18' = 9. 6758592 Zsin i634' = 9.455044i L sin 4452' = 9. 8484720 19. The car ol a balloon, C, is observed at A to have an elevation of 6648'. At a point B, 600 yards from A, the angle CBA is observed to be 5327'. CAB being 82I4', find the height of the balloon. log 600=2.7781513 Z sin 5327' = 9. 9048980 Iog634i4 = 4.8021851 Z sin 44I9' = 9.8442432 Z sin 6648' = 9. 9633795 86 LOGARITHMS. 20. From two points A and B on the bank of a river I observe two objects C and D at some distance from the bank upon the other side. The distance between A and B is 1000 yds. At A the angles CAB, DAB are observed to be respectively 7236' and 28io'; at B the angles CBA, DBA are found to be 4325' and 124 '42' respectively. Find the distance between C and D. L cos 22 13' = 9.9664987 log 7648 = 3.8835479 L sin 27 8' = 9. 6590246 log 1 8027 = 4. 25 59235 Zsin 43 2 5' = 9-837i456 log 2567 5 = 4. 4095 1 05 Zsin 55i8' = 9.9149479 log 13659 = 4- I354I89 L sin 6359' = 9-95359^5 log 2= .3010300 Zsin S75i' = 9-9277079 > = 794 L cos 575i' = 9. 7260240 Z 21. Walking in a horizontal direction from a point ^4 at which the elevation of an object is observed to be less than 30, I find on reaching B that the elevation is just doubled, and that at C it is trebled. A, B, C being in the same vertical plane with the object observed, AB 156 yards, and BC 109 yards, calculate the vertical height of the object. log 78=1.8920946 log 109 = 2.0374265 log 265 = 2.4232459 log 171 = 2.2329961 log 15233 = 4. 1827854 22. Standing directly in front of the centre one of three pillars of a building which are in the same vertical plane, and known to be 36 ft. apart, I observe the elevations of the pillars to be 3826' and 44I4'. What is my distance from the nearest pillar? log 36= 1.5563025 log 50645 = 4. 7045 37 Zsin 548' = 9. 0045634 Zsin3826' = 9.7935i35 Z sin 824o' = 9.9964330 Z cos44i4' = 9.8552192 23. A tower standing on a horizontal plane leans over towards the south. At equal distances due north and south of it, the elevations of its summit are 30 and 32 respectively. Calculate the inclination of the tower to the vertical. Z sin 2 = 8. 5428 1 92 Z tan 346' = 8. 8 1 84608 Zsin 32 = 9. 7242097 diff. i'= 19230 24. Three objects A, B, C are visible from a station D in the same plane, at which the sides of the triangle ABC subtend equal angles. Find AD ; given AB - 12 chains, AC- 6 chains, CAB = 46 34'. log 2= .30103 Zcot53i7' =9.87264 log 3= -47712 Ztani357'3o" = 9.39552 log 536 = 2. 72916 Z sin 5o4o'3o" = 9.88849 CHAPTER XII. Application to Mensuration. 66. Most of the formulae used in solving questions on the mensuration of plane and solid figures consist entirely of products and quotients, and are thus adapted to logarithmic computation. The following is a list of the more important formulae used in this branch of mathematics. (I) Any triangle. I. Triangles. Area= (i) \bh (i.e. \ base x height) (ii) (iii) - a)(s - 6)(s - + c). (2) Right-angled triangle. (C=go.) a (i.e. i product of the sides containing the right angle). to (3) Equilateral triangle. (Side, a.) Height = ^?, area = : r (the radius of the inscribed circle) = - tan 30 = -, 2 2^/3 R (the radius of the circumscribed circle) = - sec 30 = , (4) Isosceles triangle. (Each of equal sides = a. ) (i) Vertical angle 0, area = a 2 sin 6, (ii) Vertical angle 30 or 150, area = , (iii) Vertical angle 60 or 120, area = 87 88 LOGARITHMS. II. Quadrilaterals. (1) Square. (Side, a). Area = a 2 , (2) Rectangle. (Sides, a, b.) Area = ab, diagonal = a*j2. (3) Parallelogram. diagonal = a sec 6, where tan 6 = -. Vide Art. 50. Area= (i) bh (i.e. base x height) (ii) ab sin 0, diagonals = Va 2 + P 2at> cos 0. Vide Art. 50. (3) b (5) (4) Rhombus. Area = ^/' (i.e. ^ product of diagonals), or as for parallelogram, putting b-a ; . diagonals = 2a sin -, 2a cos - 2 2 [In the rhombus the diagonals bisect one another a/ right angles.'} (5) Trapezium or Trapezoid. Area. = %(a + &)& (i.e. mean length x height). (6) The area of any quadrilateral whose diagonals intersect at right angles equals half the product of the diagonals. III. Regular Polygons. (i) Hexagon. (Side, a.) Area = 6f^^3) (i.e. 6 equilateral triangles). (2) Polygon of n (Side, a. ) Area = cot 4 IV. Circles. Circumference of circle (radius, r) = 2irr, area of circle = ?rr 2 . Area of plane circular ring (radii, R, r) = ir(R + r}(R - r}. Arc of circular sector = a (2-rrr), area of circular sector = (i) - (ir;- 2 ), 360 (iii) \r-Q (d = circular measure of ct), (iv 'v- APPLICATION TO MENSURATION. 89 V. Polygons and Circles. (l) Regular polygon inscribed in circle. Area of polygon = n times OAB = ; cot or sin ^ , 4 n 2 n perimeter of polygon = n times AB = na or 2r sin , n = number of sides of polygon, C a - side of polygon, = radius of circle. O /N rA*J B \X-^* I -^ A (2) Circle inscribed in regular polygon. Area of polygon = n times perimeter of polygon = n times AB = na or znr tan , Area of polygon = times OAB = *^- cot 1^- or r 2 tan A ?Z VI. Rectangular Parallelepipeds. Volume of rectangular parallelepiped (edges, a, b, c) = ah, volume of cube (edge, a) = a 3 , diagonal of cube = V VII. Spheres. Surface of sphere (radius, r) = 47rr 2 , volume of sphere = ^Trr 3 . Curved Surface of spherical zone = zirrh. VIII. Prisms. Volume of prism = Bh (i.e. base x height). Volume of prismatic frustum "\ _ , //,/', p\ A or wedge / m (i.e. mean length x area of right section). 9 LOGARITHMS. IX. Cylinders. Volume of cylinder = irr^h, curved surface of cylinder = 2irrh, total surface of cylinder = 2trr(r + h). Volume of cylindrical shell = Trh(R-\-r}(R - r). Volume of cylindrical frustum = $(/+/')ir? a (i.e. mean length x area of right section), Curved surface of cylindrical frustum = (l+l')irr. X. Pyramids. Volume of pyramid = \Bh (i.e. & base x height). Volume of tetrahedron (edge, a}= , O/y/2 surface of tetrahedron = 2 v/3- Volume of octahedron (edge, a) - ^ , surface of octahedron = 2a 2 v /3- ,-()! ^-{- , where a, A are the areas of the top | -(.) and bottom of the frustum, and h is the height of the frustum. Volume of pyramidal frustum = A I XI. Cones. Volume of cone = ^Trr 2 . A, curved surface of cone = TT;-/, total surface of cone = 7rr(r+/). Volume of conical frustum = Curved surface of conical frustum = irl(R + r) APPLICATION TO MENSURATION. 91 100 links \ I00 ' 00 s * link f 1 22 yards} = '<*"" 4,8 4 o sq. yard^ j= I acre. 640 acres = I sq. mile. log TT = .4971499. EXAMPLES. XXL [Tables to be used.] A. MENSURATION OF PLANE FIGURES. 1. Two sides of a triangular field containing an obtuse angle are 127 yds. and 232 yds. respectively. Find to the nearest yard the length of the third side that the field may contain exactly an acre. 2. In a quadrangular field ABCD, ^' = 38.54 chains, BC = 24.16 chains, CD = 52 chains, DA = 35.08 chains, and the angle ACB is a right angle. Find its area in acres. 3. An equilateral triangle is inscribed in a square with one of its angular points coinciding with an angular point of the square. Find the ratio of the area of the triangle to the area of the square to three places of decimals. Staff College. 4. Find to three places of decimals the side of the equilateral triangle whose area equals that of the scalene triangle whose sides are 105, 116, and 143. 5. What is the height in inches of the isosceles triangle whose area is a square foot and ver- tical angle the unit of circular measure ? 6. On opposite sides of a base 120 yards long, two isosceles triangles are described whose vertical angles are respectively 38 15' and 8342'. Find the total area. 7. Find to the nearest sq. foot the area of a square whose side is 317.2857 feet. 8. What are the lengths of the diagonals of the rhombus whose acute angles are 6428', and whose area is 27 sq. inches. 9. A rhombus whose acute angles are 383o', and whose side is 12 inches long, has inscribed in it an isosceles triangle whose vertex coincides with one of these acute angles and whose base bisects the opposite sides. Find the area of this triangle. 10. Four equal rods, each 6 inches long, are hinged together so as to form a square. The rods are now turned about the hinges till opposite corners are 10 inches apart. Find the angles and area of the figure formed by the rods in this position. 11. The two parallel sides of a trapezium are 117 yds. 2 ft. and 172 yds. I ft. respectively, and the other sides are both 34 yds. long ; find the area to the nearest square foot. 12. Find the area of the trapezium whose parallel sides are respectively 112 ft. and 154 ft., and whose other sides make angles of 52 12' and 3748' with the greater of the given sides. 13. The two parallel sides of a trapezium are 89 feet apart, and the other sides make angles of 52I2' and 3736' with the greater of the two parallel sides, whose length is 254 ft. Find the area to the nearest square foot. 14. What would be the perimeter and area of a regular figure of 100 sides inscribed in a circle of 100 yards radius ? By how much does the area differ from that of the circle ? 15. Find to five places of decimals the ratio of the areas of the regular hexagon and octagon inscribed in any circle. 92 LOGARITHMS. 1 6. What is the number of sides in the regular polygon, the ratio of above inscribed and cir- cumscribed circles is most nearly equal to T ^ ? 17. Find the area and perimeter of the regular dodecagon inscribed in a circle of 6 ft. radius. 1 8. Find the area of the regular quindecagon inscribed in a circle of radius 5 ft. What ratio does it bear to that of the circumscribing quindecagon ? 19. What would be the difference between the areas enclosed by 500 yards of rope when held taut by 100 and 120 posts respectively, placed at equal distances along the circum- ference of a circle ? 20. Find in yards the radius of the circle whose area is half an acre. 21. Compute to the nearest square inch the area of a circle in which a chord 4 ft. in length subtends at the centre the ang'e iS36'. Staff College. 22. What is the length of the chord, in a circle of 10 ft. radius, which subtends an angle of 1 1 2 1 5' at the centre? 23. Express to the nearest second the angle which is subtended at the centre of a circle of 3 square inches area by an arc of I inch. 24. Compute to the nearest yard the length of that part of a circular railway curve which subtends an angle of 25 36' to a radius of a mile. 25. Find the distance in miles between two places on the equator which differ in longitude by 6i8', assuming the earth's equatorial diameter to be 7925.6 miles. 26. Find to the nearest square foot the area of the complete circle, whose sector of angle 5 contains an arc of 10 yards. 27. A circle is inscribed in a right-angled isosceles triangle. Find the ratio of the areas of the circle and triangle. Staff College. 28. In a circle of 10 ft. diameter a straight line 4 ft. long is placed. Compute to the nearest inch the lengths of the segments into which the circumference of the circle is thus divided. Staff College. 29. In what latitude will a correction of one second in time have to be reckoned for every furlong travelled east or west, taking the earth to be a sphere whose radius is 3957 miles? 30. Find the side of the equilateral triangle that can be inscribed in a circle whose area is 14 square inches. 31. Compute to the nearest square inch the area of the smaller segment into which a circle of 100 feet radius is divided by a chord of 37.25 feet. 32. Find to the nearest inch the length of the arc subtending an angle of 35 at the centre of a circle whose area is 1000 square yards. 33. After walking 200 yards round a circular pond, I notice that the point from which I started and an object in the centre of the pond lie in directions inclined at 32! 5' to each other. Compute the diameter and area of the pond. 34. Calculate the radius and area of the circle inscribed in the triangle whose sides are 131.16 ft- 373-75 ft., and 407.23 ft. respectively. 35. A circular plot of grass is surrounded by a walk 40 links wide, whose inner circumference is 2408 links ; find the number of acres contained in the walk. 36. Find to the nearest square inch the area of the equilateral triangle inscribed in the circle whose radius is 13.26 feet. APPLICATION TO MENSURATION, 93 37. Find the radius of the circle whose area is equal to what is left after cutting a sector of angle 4426' from a circle of 31.68 feet radius. 38. What is the area contained between the arc of a circular sector and the tangents at its extremities, the arc being 18^ inches long and the perimeter of the sector 35 inches ? 39. Find the length in inches of the circumference of the circle whose area is the one-mil- lionth of an acre. 40. What is the area of the segment of a circle of 8J inches radius which subtends an angle of i824' at the centre? 41. A railway curve is an arc of a circle of J mile radius. What is the shortest distance be- tween two stations whose distance apart along the line is looo yards? 42. Taking the latitude of St. Paul's to be 5i3o', what is its velocity in feet per sec. due to the earth's rotation ? (Diameter of earth = 7925.6 miles.) 43. Find in square inches the area of the segment of a circle, the arc being the tenth part of the whole circumference and the radius being 6 feet. 44. Compute in links the radius of the circle whose area is an acre. 45. Find the area of the segment which contains an angle of 38 12' on a base 8 feet long. 46. If, in a circle of 4 ft. radius, an arc of 10 ft. subtends a chord of 7.592 feet, find the value of TT to three places of decimals. 47. Two chords are drawn in a circle of 12 inches radius, cutting one another at right angles and subtending angles of 156 and 125 at the centre respectively ; find the area of the quadrilateral formed by joining their extremities. 48. Calculate the area and perimeter of the circle inscribed in a square the side of which is 359- 5678 feet. 49. Determine the diameter of the earth in geographical miles [60 to a degree of latitude], each degree subtending i at the centre of ihe earth. TT = 3. 1 4 1 59265 . . . Woolwich. 50. It is proposed to add to a square lawn, measuring 58 feet on a side, two circular ends, the centre of each circle being the point of intersection of the diagonals of the square. How much turf will be required for the purpose ? Woolwich. 51. What are the areas and perimeters of the two segments into which a circle of 13 ft. radius is cut by a chord of 2O ft. ? 52. An isosceles triangle whose vertical angle is 48I2' is inscribed in a circle of 18 ft. radius ; find the area between the triangle and circumference of the circle. 53. The arc of a semicircle is divided into two parts so that the chord of one is 5 times that of the other ; find the ratio of these parts. 54. A triangle whose sides are 17, 23, and 30 inches respectively has a circle inscribed in it, and in this circle a similar triangle is inscribed. Find the angles and area of this latter triangle. 55. Find the expense of paving a circular court 80 feet in diameter at 35. 40!. a square foot, leaving in the centre a space for a fountain in the shape of a hexagon, each side of which is a yard. 56. The chord of an arc of a circle is i8 inches, and the height of the arc is 6 inches ; find the length of the arc. 94 LOGARITHMS. 57. The perimeters of a circle, a square, and an equilateral triangle are each of them i foot. Find the area of each of these figures to the nearest hundredth of a square inch. 58. The side of an equilateral triangle is 200 ft. Find the radius of the circle circumscribing the triangle, and the area of the triangle to the nearest square inch. 59. The length of the arc of a sector is 13 feet 7 inches, and the angle of the sector is 56io'; find the area of the sector to the nearest square inch. 60. Two circles, whose diameters are 18.34 feet and 26.12 feet respectively, cut one another at an angle of 40; find the length of the common chord. B. MENSURATION OF SOLIDS. 61. The three conterminous edges of a rectangular parallelepiped are 3, 2.52, and 1.523 feet in length. Find the number of cubic inches of volume. Find also the cubical space inside a box of the same external dimensions, constructed of material ^th of a foot thick. 62. Compute the edge and diagonal of the cube whose volume is 100 cubic yards. 63. Find the length of the edge of a cubical block of stone containing 146 cub. yds. 716 cub. ins., and the number of sq. inches of surface. 64. What is the length of the side, to the nearest tenth of an inch, of a cubical cistern holding, when full, 2000 cub. ft. of water ? 65. The corners of a cube whose weight is found to be 7.38 Ibs. are ground down evenly and equally till the weight is reduced to 6.64 Ibs. Find the surface of the solid so formed, if looo cub. ins. of the material weigh 12.5 Ibs. 66. Find the radius of the sphere whose volume is 750 cub. ft. 67. Compute the radius of the sphere whose volume equals that of a cube of 12 inches radius. 68. How many square miles of the earth's surface lie in the tropics, i.e. between 22^ north and south of the equator, taking the diameter of the earth to be 7926 miles ? 69. Find the edge of the cubical block of lead which, when melted down, will make a million shot .125 inches in diameter. 70. Find the amount of material required to make a spherical balloon containing 10,000 cub. ft. of gas. 71. Find the radius of the sphere (l) whose volume = I cub. ft., (2) whose surface = I sq. ft. 72. How many cub. ft. of gas will be contained in a spherical balloon formed out of 180 sq. yds. of silk? 73. What would be the diameter of a spherical balloon made out of 112 yds. of canvas, 4^ ft. wide ? 74. A right triangular prism, whose edges are all equal, and a sphere are of equal volume. Compare their external surfaces. 75. How many cub. yds. of earth have been removed in boring a tunnel I mile 170 yards long, whose section is a semi-circle of 14 ft. radius ? 76. A right prism on a triangular base, each of whose sides is 21 inches, is such that a sphere, described within it, touches its five faces. Find the volume of the sphere, and of the space between it and the surface of the prism. APPLICATION TO MENSURATION. 95 77. Find the volume of a right triangular prism, the edges of whose base are 38.7, 49.2, and 40.3 ft. respectively, and whose height is 20 ft. 78. The vertical ends of a horizontal trough are parallel equilateral triangles, with 12 inches in each side, a side of each triangle being horizontal. If the distance between the ends be 6 ft., find (i) the number of cubic feet of water the trough will contain, (2) the number of gallons it will contain, it being given that a gallon of water weighs 10 Ibs. and a cubic foot of water 62.5 Ibs. 79. Determine the diameter of a cylindrical gas holder to contain 10 million cubic feet of gas, supposing the height to be made equal to the diameter ; and determine in tons the weight of iron plate, weighing 2| Ibs. per sq. ft., required in the construction of the gas holder, supposing it open at the bottom, and closed by a flat top. Woolwich. 80. A hollow pontoon has a cylindrical body 20 ft. long, and hemispherical ends, and is made of metal Jth of an inch thick. The outside diameter is 3 ft. 4 in. Find its weight, having given that a cubic inch of the metal weighs 4. 5 oz. Woolwich. 81. A right cylinder open at the top, with a diameter of 24 inches, weighs 167.5 Ibs. When filled with water it weighs 2131 Ibs. Find the height of the cylinder, it being given that a cubic foot of water weighs 62. 5 Ibs. 82. What is the weight of a cylinder formed of sheet iron \ inch thick, with an outer circum- ference of 10 ft. 7f ins. and a length of 3 ft. 6 ins. ? 240 cub. ins. of iron weigh 1000 oz. 83. A well 5 feet in diameter and 30 feet deep is to have a lining of bricks, fitting close together without mortar, 9 inches thick. Required approximately in Ibs. the weight of the bricks, supposing a brick 9 x 4^ x 3 ins. to weigh 5 Ibs. 84. A cylindrical pipe 14 feet long contains 396 cubic feet. Find its diameter, and the cost of gilding its surface at 9! d. per sq. ft. 85. A right circular cylinder is cut by two planes inclined to one another at an angle of 32 18', so that the areas of the two ends are each of them equal to 12 sq. feet, and the distance between their centres is 7 ft. : find the volume intercepted by the planes. 86. In a rectangular building with a wedge-shaped roof, whose ridge is parallel with the length of the building, there are cylindrical columns in a plane, at equal distances from one another and from the side walls of the building, and reaching from the ground to the roof. There are 6 of these columns, I2^f ft. in circumference : the height of the building is 80 ft. and of the walls 58 ft., while the width of the building is 122 ft. Find the total volume, and exposed surface, of the six columns. 87. Determine the number of cubic yards in a bank of earth on a horizontal rectangular base 60 ft. long and 20 ft. broad, the four sides of the bank sloping up to a ridge at an angle of 40 to the horizon. Woolwich. 88. How many cubic feet of earth must be dug out to form a trench 120 yards long, whose right section is a trapezium 7 feet deep, the inclination of the sides to the vertical being I220' and the breadth of the trench at the top 18 feet? 89. The Great Pyramid of Egypt was 481 feet high when complete, and its base was a square whose side was 764 feet long : find the volume to the nearest number of cubic yards. 90. Compute the solid content of the pyramid whose height is 6.99, each side of the triangular base being 4.18. 91. A pyramid on a square base has all its edges equal. If the exterior surface be 117.38 sq. inches, find its volume. 96 LOGARITHMS. 92. Find the total surface and volume of a pyramid on a hexagonal base, each side of which is 10 inches, the perpendicular height of the pyramid being 1 8 inches. Also find to the nearest second the angle of inclination to the base of the triangular faces. 93. A pyramid on a triangular base, whose sides are 12.7, 8.5, and 15.8 inches respectively, is cut by a plane parallel to the base and at a distance of 6 inches from it. If the height of the pyramid was 14 inches, find the volumes of the two portions into which the pyramid is divided. 94. A right pyramid, upon a square base whose side equals I foot, has its triangular faces inclined at 78 1 6' to the base. Find the inclination of the edges to the base, and the surface and volume of the pyramid. 95. A pyramid upon a regular hexagonal base, and with its triangular surfaces inclined at angles of 523o' with the base, stands upon an area of 15 sq. feet. What is its volume? 96. A conical hole is bored in a sphere, whose vertex is at the centre of the sphere and whose edge is circular. If the angle made by a straight line drawn from any point on the edge to the vertex with the plane of the circular edge be 6448', and the circumference of the sphere be 8 ft., find the volume removed to the nearest cubic inch. 97. What are the diameter and surface of the sphere of equal volume with the pyramid whose vertex is in one of the faces of a cube, and whose base is the opposite face : each edge of the cube being 13.7 ins.? 98. Compute the volume of the largest tetrahedron that can be formed out of a wooden sphere by planing down its surface, the circumference of the sphere being 217.64 inches. 99. Find the edge of the tetrahedron (i) whose volume, (2) whose surface equals that of a cube whose edge is 21.178 inches. 100. Compare the edges of the tetrahedron and octahedron that they may contain (i) equal volumes, (2) equal surfaces. 101. Find the edges and surfaces of the tetrahedron and octahedron that could be obtained by melting down a leaden spherical ball whose weight is 28.16 Ibs., supposing a cub. in. to weigh 6.6 oz. 102. What is the volume of the octahedron whose surface is 100 sq. inches? 103. Find the volume of the cone whose vertical angle is 7825', and diameter of base 8 inches. 104. The inclination of the slant height of a cone to its base is I425', and its height is 4 inches. What is the area of its curved surface? 105. What is the vertical angle of a right cone that its curved surface may be double that of the cylinder of the same base and height ? 106. If the vertical angle of a cone be 4327', and the diameter of its base 8 inches, find its volume and total external surface. 107. Compute to the nearest second the vertical angle of the cone in which the area of the curved surface is 3 times that of the plane surface. 108. What is the total surface of the right-angled cone whose volume is 394 cubic inches? 109. Find how many sq. yds. of canvas will be required to make a conical tent standing on an area of 100 sq. yds., and having its semi-vertical angle 383o'. no. Find the volume and the inclination to the vertical of the slant height of the conical tent that can be made out of 100 sq. yds. of canvas standing upon 50 sq. yds. of area. APPLICATION TO MENSURATION. 97 in. The vertical angle of a right cone is I2436', and its height is 17^ inches. Find its curved surface and volume. 112. The curved surface of a cone is 24 sq. ft., and its base is 18 sq. ft. Find the volume of the cone to the nearest cubic inch. 113. Find the height of the cone whose volume shall be 1000 cubic inches, if it stand upon a circular base whose radius is 10 inches. 1 14. Find the radius of the hemispherical bowl, which contains as much as a conical vessel whose vertical angle is 4248' and diameter of rim 8 inches. 115. The greatest cone that can be inserted in any given sphere has its vertical angle 60. Find the volume of the greatest cone for the sphere whose surface is 2148 sq. inches. 116. Find the volume of the largest cone that can be cut out of a sphere of 12 inches radius, the vertical angle of the cone being 72i8'. 117. Find the volume and surface of the solid generated by the revolution of an equilateral triangle about one of its sides, each side being 7.9 inches. 118. An isosceles triangle whose vertical angle is I564O', and whose equal sides are 15 inches long, revolves about its base. What are the volume and surface of the solid generated? 119. A regular hexagon, whose side is a foot, revolves about the straight line joining two opposite angular points. Find the volume of the solid generated in cubic inches. 1 20. A solid is made up of a right circular cylinder surmounted by a cone, on an equal base and of the same altitude. If the area of the common base be 10 square feet, and the vertical angle of the cone 683o', find the volume of the solid to the nearest cubic inch. 121. If S be the surface of a regular tetrahedron and / be the length of an edge, prove the formula log S = 2 log / + . 23856. Militia. 122. If Fis the volume of a sphere and A the area of its surface, prove that 3 log A - 2 log 6 + log TT + 2 log V. Calculate the value of A, if V= 796.325 cub. in. Militia. MISCELLANEOUS EXAMPLES. [Tables to be used.} A. Woolwich (1-5). 1. Find the value of (1)52.4574x3.78472, ^ 87.327 x 784.55 x .020868 (iii) (5.7432) 1 ' 24ti . .61659x58.844 2. Find (i) a 4th proportional to 1.3046, .01042, and 2.375, (ii) a mean proportional between 33.549 and 44.642. 3. How many terms of the series .04, .08, .16, .32, ..., will amount to 41943 ? 4. What is the amount of^iooo in 100 years at 5 per cent, per annum compound interest? 5. If the number of persons born in any year equals ^th of the whole population at the beginning of the year, and the number who die equals ^ih of it, find in how many years the population will be doubled. Staff College (6-10). 6. Find to three places of decimals the mean proportional between .0374 and 32310. 7. Find the cube root of .043758. 8. Compute to 5 places of decimals the value of N/* 8 + 3*, where x = .84729. 9. Employ logarithms to divide 39.8765 by V. 0000843, an d to compute 3 when a- .03857, the result in each case being given to the first place of decimals. 10. Calculate to the nearest penny the amount of ^"126. 8s. 6d. placed at 6 per cent, per annum, compound interest, for 20 years, convertible half-yearly. 1 1. Find the common logarithms of the following numbers : (i) 217.6328, (v) .5161205, (ix) 8400.827, (xiii) .002195976, (xvii) 1779.023, (ii) 16500.876, (vi) 8761.3577, (x) 113.1113, (xiv) 18030.15, (xviii) .11737017, (iii) 3.459125, (vii) 24.60908, (xi) .3510689, (xv) 2.5768643, (xix) 620.3151, (iv) .000784032, (viii) 59769'44 (xii) 2.852038, (xvi) 410428.4, (xx) .000007813248. 98 MISCELLANEOUS EXAMPLES. 99 12. Find the numbers whose common logarithms are (i) 3.2147067, (v) 2.7116210, (ix) 1.1071238, (xiii) 6.2361566, (xvii) -2.4625383, (ii) 1.8501042, (vi) 2.8517532, (x) 4.4236500, (xiv) 3.9987280, (xviii) -4.1047934, (iii) .9143314, (vii) 3.2400276, (xi) 1.3021811, (xv) 2.1685205, (xix) -.5682002, (iv) 4.2580703, (viii) 2.0003145, (xii) .5117097, (xvi) 2.0073841, (xx) -1.8394216. 13. Compute the following roots to 6 significant figures : (i) the 7th and I5th roots of . I, (iii) the cube root of .0000083825, (v) the 5th root of J\/2, (ii) the 5th root of 1000, (iv) the 7th root of (.00 18423 ) 10 . (vi) the nth root of 14. Find approximately the following proportionals : the mean proportional to (i) 35.76 and .004235, (ii) .003 and 3000000 ; the 3rd proportional to (iii) 31.13 and .02437, (iv) .082 and 7.4131 ; the 4th proportional to (v) .0081724, 3.17245, and .0001, (vi) .0076842, 32000, and .5. 15. Find the values of (i) Iog 10 (2o 7 .8 9 67) 2 <>, (ii) log 10 (M98x5620o\ 3 "V 7 ix. 000007 / v.0076 (vi) log l0 V(.iiinn) 13 , (xi) logu.g. 00719. 1 6. Compute the values of (i) (20.009) 5 , (vii) (.2692)-*, (*) (-5-90 4 , (xiii) (6327)*** (xvi) (1.418)^, 2-41 (Xix) (31.17) 1-39, 15 (xii) (ii) (151. 102) 3 , (v) (.oi86) 3 , (viii) (.07I7)- 2 , (xi) (- 2.089) ~ 3 , (xiv) (4.898)**, (xvii) (.00821)^", _5-4 (xx) (.00202) 4-s, (xxiii) \/9\/3\/2, x/32.25 J (iii) (7. 3001) - 4 , (vi) (.II24) 6 , (ix) (-n.04) 3 , (xii) (-20. 21) - 4 , 0031 (xviii) (.072) 39 f (xxi) IOO LOGARITHMS. (XXV) (xxvii) V- 000024 (xxx) 32.889^ .000246397 ' 9.864 x. 01234 .005678 x .0000876529' 1178 /' (xxviii) 30.284^0007 .0000842065 (xxxi) ' 7 9 J^J-iJ, (xxix) (xxxii) 14. .052^.028 (xxxiii) 49x(2i)Bxy7SQ (xxxiy) .003768(2.00?)* (xxxy) of ^.000317^ ''"* B 3o.oo76sV96.74 (xxxvi) (xxxvii) ^^4-^.026715, V3-789 ^13119.7 (xxxix) (xl) (27384 17. Find the value of (i) * 2 -*-56, when x = 310.427 (iii) 3^+14^-5, when.r= 72.823"; (v) ^ 3 -jc + 2^ 2 -2, when x= 21.513; (vii) (x-2f(x + $f, whenA-= 5.3212 (ix) (2.x- 2 - 5* - 12) 3 , when x = 7.2538 ; (xi) x 6 - lO-r 6 , when*= 1.7744; (xiii) %/x* - 5-* 2 + 4, when x = . 5768268 ; (xv) x/3189.718^ 8 , when x = 4. 10072 (xvii) V. 76^ + 63. 309-r 5 , whenjr= 7-39I34; - i), when x - 1.00008 ; when*= .418574; when ^ = 23.1525 ; ,when.r= 2.00765; when ^ = 35. 4848; x= .3625; when*= .021846; (xviii) ^{22.87 - \/8i5.0328^ 3 }, when * = 1 1.6038 ; ^14-753)^^(13^14^, when * = 2.17484; (2068.974)* (xx) 3* 6 - Jx 5 + 4x* - 5JC 3 - 2x z - 6x - 12, when x = 4. 107634. (iii) 317.68* = 74100, (vi) 3*. 2* +1 = V 2 f 1 8. Solve the following equations : (i) 2.03*= 10.2, (ii) i8i.2 z = .02, (iv) .171* = .051, (v) .001* = 221, (vii) = 1882384, (ix) \/. 0000286788 = . 123456, (xi) (si.S) 3 *-^ ^(.0076)2* (xiii) xyz- 317.24^ (xiv) (viii) (\/.ooo7i2) 2 *- 1 = 13.0156, (x) (.00761)* = .!, (xii) = 317 = 8276.5 = 12347 -24! .5 j- = .246879! (xv) = .453284^ (*+y}(y+z)= ._ = . 867091 j (y + z)(z+x) = (4.2) 16 19. Find the number which, multiplied by 604327, will give 2465816904306. 20. Extract the cube root of 949862087000. MISCELLANEOUS EXAMPLES. IOJ 21. How many digits are there in 2 100 and 3 64 ? 22. Find the number of digits in the integral portion of (4506. 23) 50 , and the position of the first significant figure in the decimal value of (iTT2"y)*' 23. What power of 2 is equal to 131072? 24. Compute the mean proportional between the side and diagonal of a square whose area is an acre. 25. Find Ioge4| when perann. Years ' per ann. (i) 20 at 4! (convertible annually), (ii) 15 at 5 (convertible annually), (iii) 50 3 (iv) 36 6 (v) 27 4 (vi) 18 ,, 3 (vii) 70 ,, 5 ,, (viii) 17 8 (ix) 100 ,, 4^ (x) 10 4 ., (xi) 23 10 (xii) 13 ,, 3^ (xiii) 28 3 ,, (xiv) 39 5 (xv) 69 ,, 7 (xvi) 81 4^ ,, (xvii) 56 ,, 4^- (convertible half-yearly), (xviii) 475- ,, 6 (convertible half-yearly), (xix) 42 ,, 5 ,, ,, (xx) 30 4 ,, 28. What sum will amount to ^"1000 at Compound Interest in v Per cent. v Per cent. Years ' perann. Years ' perann. (i) 17 at 4 (convertible annually), (ii) 12 at 3^ (convertible annually), (iii) 50 ,, 5 ,, ,, (iv) 10 4^ ,, (v) 100 ,, 5 (vi) 6 ,, 3 ,, ,, (vii) 20 ,, 6 ,, (viii) 15 4 (ix) 8 ,, 4 ,, (x) 21 4 (xi) 87 ,, 7 (xii) 72 ,, 3 (xiii) 35 3* ( xiv ) 26 9 (xv) 61 5 (xvi) 13 8 ,, (xvii) 10 ,, 4^ (convertible half-yearly), (xviii) 12 ,, 6 (convertible half-yearly), (xix) 7^, , 5 (xx) 21 4 29. At what rate per cent, per ann. will the following sums amount to^iooo, viz., Years. Years. (i) ,530 in 7 (convertible annually), (ii) ^100 in 10 (convertible annually), (iii) ^425,, 21 (iv);7i5 H (v) 200 ,, 32 (vi) ^350 13 (vii) 632 ,, 5^ (convertible half-yearly), (viii) 418 ,, 12 (convertible half-yearly), (ix) ^820 8^ (x) ^500 14? 102 LOGARITHMS. 30. In what time will the following sums amount to ^1000, viz., Per cent. per ann. (i) ^*3 2 5 at 5 (convertible annually), 8 (vii) ,815 3^ (convertible half-yearly), (ix)^20o,, 7 Per cent. per ann. (ii) ^450 at 4 (convertible annually), (iv)^ioo,, 4- ,, (vi) 270 ,, 10 ,, (viii) ,630 ,, 2\ (convertible half-yearly), (x) ^500 ,, 10 ,, 31. Find the time in which a sum of money will double itself at 2^, 4, 5, 8J, and 10 per cent. per ann., compound interest, respectively. 32. What sum of money will amount at compound interest to ;iooo in 6 years, and ^1250 in 8 years, and what rate of interest will be reckoned ? 33. At what rate per cent. , compound interest, will a sum of money quadruple itself once in a century ? 34. Find the accumulated values of forborne annuities of ^"100 in the following cases, pay- able annually : (i) for 12 years at 4^ per cent, per ann., (iii) ,,50 ,, 4 (v) 17 4* (vii) 18 ,, 6 (ix) 73 ., 4 (xi) 27 8 35. Find the present value of an annuity of^ioo (i) for 10 years at 3 per cent, per ann., (iii) 35 ,, 4 ,, (v) ,, 72 ,, 4^ (vii) ,, 89 3 (ix) ,, 51 ,, 4 (xi) 85 3 36. Find the annuity purchaseable with ^"looo (i) for 100 years at 3^ per cent, per ann., (iii) ,, 15 4i (v) 80 3 (vii) 64 4 (ix) 50 3 (xi) 81 5 (ii) for 21 years at 5 per cent, per ann., (iv) (vi) (viii) (x) (xii) 100 35 49 29 54 3 31 (ii) for 17 years at 3^per cent, per ann., (iv) (vi) (viii) (x) (xii) 100 26 44 60 96 (ii) for 21 years at 5 per cent, per ann., (iv) 25 4 (vi) 37 4* (viii) 99 3| (x) 18 5 (xii) 76 34- 37. What is the difference in value between a freehold and a 99 years' lease of a property worth ;loo per annum, taking the interest of money at 5 per cent, per annum? 38. If a debt of ^looo is to be paid off in 10 years by equal annual instalments, 5 per cent. being charged each year on the outstanding debt, find the amount of each instalment to the nearest penny. 39. An annuity of ^100 has remained unpaid for the last 21 years. What perpetuity is equivalent to the accumulated value, allowing 5 per cent, interest in each case? 40. For how many years has a certain annuity been unpaid if the accumulations at 5 per cent, be 21.58 times the value of the annuity. MISCELLANEOUS EXAMPLES. 103 B. i. Find (a) the Tabular Logarithmic Sines of (i) ii 4 36'54.6", (ii) 35 (iv) i842'24-o", (v) i63 (jS) the Tabular Logarithmic Cosines of (i) 5 ii9'2o.8", (ii) 4738'36.4", (iv) 3443'27-o", (v) 356i6'56.7", (7) the Tabular Logarithmic Tangents of (i) 3326'24.o", (ii) 2i635'52.6", (iv) 23420'49. i", (v) 78i8'30.o", (5) the Tabular Logarithmic Cotangents of (i) 53io' 4 o.6", (ii) io58'25.5", (iv) 25433'5i.3", (v) 36i7'24.o", (iii) (vi) 8257'i 4 .2"; (Hi) 272 2 7 ' 4 8.5", (vi) I72 9 'io.2"; (iii) I747'38.7", (vi) 87 o'43.3"; (iii) (vi) (iii) (vi) 2. Find the angles (a) whose Tabular Logarithmic Sines are (i) 9.6872304, (ii) 8.8645120, (iv) 9.8847125, (v) 9.9381029, (/3) whose Tabular Logarithmic Cosines are (i) 9.9692136, (ii) 9.3152164, (iv) 9.5242812, (v) 9.7098000, (7) whose Tabular Logarithmic Tangents are (i) 9.4361278, (ii) 10.2271613, (iv) 9-9972367, (v) 10.0178034, (8) whose Tabular Logarithmic Cotangents are (i) 10.5863078, (ii) 9.8119826, (iii) (iv) 9.58 2 35I5, ( v ) 8.9798217, (vi) when A = i6i8'4o"; A= 3 3, Find the values of (i) - - , .0342 9.7928147, 9.8545278 ; 9.8933790, 9-9405135 ; 10.1151415, 8.8794162 ; 10.2207100, 10.9408238. Staff College. A (v) \/i 2.118 tan 2 , (iv) \ r 3. 826 + .3942 cos 2 ^, ,, A= 5ii6'; A= 50; 2 , A= 32i2'2 4 ''; (vii) .00284/^7 cos 2 -, sin2A = ^', ~A . 91 56 + .4897 1 tan 3 , ^4 = 1522120; 3 (ix) cos*A - sin 4 A, ,, A - i>22$ 20 ; -tan 2 ^, A = 127! 5' and B = 45 2O/ ; I04 LOGARITHMS. (xv) _-, ^ = 250; Gooo6 (xvi) sin 3^4 sin 3 ^ + cos $A cos*A, when A - I33i7'- 4. Find the value of sin A + sin B + sin C, when A, J5, C are the angles of the triangle described in Euc. IV, 10. 5. If L tan A - 10. 5240134, find L sin A and Z cos y2. 6. Calculate, by introducing subsidiary angles, the value of V# 2 + 2 when (i) 0=i3i-573> = 34.21917; (ii) a- 16.0408, =18.1535; (iii) a- .717242, = 2.49801. 7. Solve the equations (i) \J sin 3 - = .26814, (ii) >/tan0= i|, (iii) tan 6 = \/7 cot 3 0, (iv) an 3 \ i 2 J' (v) 2 sin 2 + 3 cos 2 = 2, (vi) ^3 sin + cos = |, (vii) 12 sin ^r + 5 cos ^=13, (viii) tan(2^+ ^) = 3 8. Calculate to the nearest second the smallest positive angle (i) whose tangent equals 3, (ii) whose cosine equals - J. 9. Given cos $A = f I, find tan $A. 10. Find to the nearest second the angles of the isosceles triangle whose equal sides are double the base. 11. What acute angle has its sine to its cosecant in the ratio of 12 to 17 ? 12. If tan 2^4 = 2 ~" , find the value of A when 3 = . 13. Solve the equations - V by means of the tables of logarithmic ratios. i+xy~2^} 14. A computer, in referring to the tables, reads the logarithmic tangent by mistake for the cotangent and uses a value too great by .5316768. What is the angle? 15. Find all the positive and negative values of 6 less than 180 which satisfy the equation 4 sin 3 - sin 30 = . 1 6. If y = tan" 1 -^" 1 * ~^-- t find the value of y to the nearest second when x - .5127. MISCELLANEOUS EXAMPLES. 105 Staff College (17-44)- 17. The sides of a triangle being 87, 93, and 100 ft. in length, compute to the nearest inch the length oi the perpendicular drawn to the longest side from the opposite angle. 1 8. On the same base, 20 yds. in length, and on opposite sides of it, are an equilateral triangle and an isosceles triangle with the vertical angle 30. Compute to the nearest foot the length of the straight line joining the vertices of these triangles. 19. The radii of two intersecting circles being I and 2 feet, and their centres being 2 feet apart, find to the nearest inch the length of the straight line joining the points of intersection. 20. Each side of a parallelogram is 8 feet long, and its area is 46 sq. feet. Compute to the nearest minute the angles of the parallelogram. 21. Compute to the nearest second the angles of the two triangles which have two sides 17 and 12 feet long, and an angle 43i2'i2" opposite to the shorter of these sides. 22. Compute the remaining angles of a triangle wherein one angle is IO544'49", the side opposite to it 427 feet, and a side adjacent 250 feet. 23. P and Q are two points. An observer at A, where AP is perpendicular to PQ, measures the angle PAQ-'yf. He moves 100 yds. parallel to PQ to B and measures the angle ABP- 53. Compute the distance between P and Q to the nearest foot. 24. The summit of a wall 20 ft. high has, to an observer in the horizontal plane through its base, the angular elevation i836'. What is the distance of the point of observation from the tower ? If the observer is liable to an error of 30' of excess or defect in the measured elevation, within what limits can he be sure that his computed distance is correct ? 25. The angular elevation of an object above the horizon is taken at different points in a straight horizontal road. Its greatest elevation is 29! 7' and its elevation at a point in the road 200 yds. away from the former point of observation is i852'. Find the height of the object above the horizontal plane to the nearest foot. 26. The sides of a triangle being 580 and 483 feet long, and the angle opposite to the latter being 48 1 7 '23", find to the nearest second the two values of the angle opposite to the former side. 27. Compute to the nearest square foot the area of a triangle wherein sides 134 and 137 feet long include an angle Ii8i7'. 28. The lengths of the sides of a triangle being 34 ft., 46 ft., and 65 ft., compute to the nearest second the largest angle of the triangle. 29. A and B are points in the same horizontal line 1000 yds. apart, Pa. visible point. At A the angle is observed PAB = 2fi$', and at B the angle is observed PBA=2^^'. Find to the nearest foot the perpendicular distance of P from the line AB. 30. In a regular pentagon, whose sides are each 10 ft. long, compute to the nearest inch the length of a straight line drawn from an angular point to one of the more distant angular points. 31. A diagonal of a rectangle is 100 ft. long, and the angle which it makes with one of the sides of the rectangle is 34i8'22". Find to the nearest sq. ft. the area of the rectangle. 32. Two angles of a triangle being 22 1 8' 17" and 47i6'i8", and the shortest side being 222 ft. long, what is the length of the longest side ? I0 6 LOGARITHMS. 33. Sides of a triangle, 46 and 112 ft. long, include the angle I4329'. Compute to the nearest second the smallest angle of the triangle. 34. ABC being a triangle wherein the angle A is a right angle, a straight line AD is drawn bisecting. the right angle and meeting the opposite side BC in D. Find the length of AD to the nearest foot when AB is 34 ft. and AC is 56 ft. in length. 35. The shortest side of a right-angled triangle is 284 ft., and its smallest angle is i837'29". Find to the nearest foot the length of the hypotenuse. 36. In a plane triangle sides 320 and 562 feet in length include an angle I284'. Find the other angles, each to the nearest minute. 37. Compute to the nearest foot the radius of a circle inscribed in a triangle whose sides are 32, 56, and 80 feet in length. 38. A and B are two points in a horizontal plane. At A the elevation of a point C above the plane is I9i7' and at B it is i65', A and B being in the same vertical plane with C and on the same side of C. The height of C above the horizontal plane is 100 feet. Find to the nearest foot the distance AB. 39. AB is a vertical object, 50 ft. high, standing on ground of uniform slope. Measure BC, 200 ft., from the foot of the object up the slope, and let the elevation of A above the horizontal plane be observed at C to be I2I5'. Find the inclination of the ground to the horizon to the nearest minute. 40. Compute to the nearest second the acute angle A when tan A - 3 sin 38. 41. A side of a right-angled triangle being 214 yds. long, and the angle opposite to it 34! '21", find the length of the other side of the triangle to the nearest foot. 42. Compute to the nearest yard the length of the base of an isosceles triangle wherein the equal sides are each 190 yards in length, and each angle at the base is 31! 5'. 43. In the triangle ABC the side AC is 341 yards, BC is 237 yards, and the angle CAB is i8i7'i5". Find the length of the side AB to the nearest foot. 44. The lengths of the sides of a triangle being 37, 45, and 52 chains, find its area in acres, roods and perches to the nearest perch. Woolwich (45-59). 45. Prove that, to turn circular measure into seconds, we must multiply by 206265 ; and, to turn seconds into circular measure, we must multiply by .000004848, approximately. = 3.14159265...] 46. The value of the divisions on the outer rim of a graduated circle is 5', and the distance between two successive divisions is . I of an inch ; find the radius of the circle. A church spire whose height is known to be 45 feet subtends an angle of 9' at the eye ; find its distance approximately. 47- ^ = 3795 Y ds -j -B = 73 1 5' 1 5"> and C = 42i8'3o", find the other sides of the triangle. 48. b - 130, c - 63, and A - 42! 5'3o", find the other angles and the third side of the triangle. 49. Given, in feet, a - 10, = 24, ^ = 26, determine the angles and the area of the triangle in square feet. 50. Given a = 5 inches, b- 7 inches, ^4 = 3ii5', find the area of the larger triangle with these data. M ISC ELL A NE O US EX A MPLES. 107 51. The base of a triangle being 7 feet, and the base angles I2923' and 3836', find the length of the shortest side. 52. Two sides of a triangle are 2.7402 ft. and .7401 ft. respectively, and contain an angle 5927'5". Find the base and altitude of the triangle. 53. Given the difference between the angles at the base of a triangle I74S' and the sides sub- tending these angles 105.25 ft. and 76.75 ft.; find the angle included by the given sides. 54. In a circle which has a radius of 10 feet two chords Af>, CD are drawn at right angles to each other, and intersecting in 0. AO and CO are three and four feet respectively ; find the sides and angles of the quadrilateral A CBD formed by joining the extremities of the chords. 55. From a boat the angles of elevation of the highest and lowest points of a flagstaff, 30 ft. high, on the edge of a cliff are observed to be 46I2' and 44I3'; determine the height of the cliff and its distance. 56. The angular altitude of a lighthouse seen from a point on the shore is I23i'46", and from a point 500 ft. nearer to it is 2633'55". Required its height above the shore. 57. An observer in a balloon, when it is one mile high, observes the angle of depression of a conspicuous object on the horizontal ground to be 352o', then after ascending verti- cally and uniformly for 20 mins. he observes the angle of depression of the same object to be 554o'; find the rate of ascent of the balloon in miles per hour. 58. A tower which stands on a horizontal plane is 200 ft. high, and there is a small loophole in the tower at a certain height above the ground ; an observer is at a horizontal dis- tance from the tower of 300 ft., but stands on a mound so that his eye is 12 ft. above the ground on which the tower stands, and in that position the angles subtended at his eye by the portions of the tower above and below the loophole are equal ; find the height of the loophole from the ground. 59. An observer finds that from the doorstep of his house the angular elevation of the top of a church spire is 501, and that from the roof above the doorstep it is 40. The height of the roof above the doorstep being h, prove that the height of the top of the spire above the doorstep is equal to h cosec a . cos 40 . sin 50, and that the horizontal dis- tance of the top of the spire from the house is equal to h cosec a . cos 40 . cos $a. If h is 39 ft. and if a is equal to 7I7'39", calculate the height and the distance. 60. Solve completely the following right-angled triangles, C being the right angle : 0)^=127.38, = 250; >(ii) a- 10.7, ^- = 27.63; (iii) b= 8. 116, A = 34i8'24"; (iv) a =1000, A = 7235'; (v) c= 33-57, ^ 61. Given in a plane triangle A = 74i4 / 3o", B = 5i42'2o", c- 786.02, calculate the side a. 62. Given in a triangle # = 472.6, = 309.4, C T =65I4 / , find the area and the radius of the inscribed circle. 63. In a triangle ABC, ^C= 166.5 ft., ^C= 162.5 ft., the angle A = %?\$. Solve the triangle. 64. Iftan0 = _?., find 6; given = 7, = 3, C=II535'. a b 65. Given that in any triangle sin A + sin B + sin C = -, calculate the sum of the sines of the abc angles of the triangle whose sides are 31.7, 23.5, and 19.4 ft. respectively. io8 LOGARITHMS. 66. It is known that in any triangle _^ = ' a + b cos%(A-B) Use these formulae to solve coni pietely the triangle that can a-b sin \(A - B)) be inscribed in a circle of 10 ins. radius on a base 12 ins. long so that (i) the perimeter may be 30 ins., (ii) the difference of the two sides may be 4 ins. 67. Find the least angle of the triangle whose three sides are 200, 250, and 300 ft. respectively. 68. The base of an isosceles triangle is loo ft. and the vertical angle is 125; solve the triangle. 69. The sides of a triangle are 525 ft., 650 ft., and 777 ft. respectively. Determine its three angles. 70. Given two sides of a triangle, 102 ft. and 70 ft. long respectively, which contain an angle 9922', calculate the length of the base. 71. A tower 150 ft. high throws a shadow 35 ft. long on a horizontal plane. Find the sun's altitude to the nearest minute. 72. If the sides of a triangle are 51, 52, 53, find the area and the sine of the smallest angle. 73. Find to the nearest square foot the area of the rectilineal field ABCD, whose side AB is measured and found to be 250 yds., the following angles being observed : >A = 9O, CAB = 402o', DBA = 35i5', and CBA = 72. 74. If the longer diagonal of a parallelogram be 72 feet, one of its sides 58 feet, and the angle between the other side and the longer diagonal 42 12', find its greatest area. 75. The sides of a triangle are 15, 20, and 25 inches respectively. Find its area and smallest angle. 76. ABC is a triangle having the perpendiculars from A and B on opposite sides equal to 18 and 20 inches respectively. If AB = 30 inches, find the other sides and angles of the triangle. 77. The diagonals of a parallelogram are 186 and 78 yards, and include an angle of 34io'25". Find the area and perimeter of the parallelogram. 78. Given that one side of a triangle is double a second and that the two include an angle of I2724', find the other angles. 79. If the sides of a triangle be 7.152, 8.263, and 9.375 ft. respectively; find in inches the area of the triangle and the radius of its inscribed circle. 80. One of the sides of a rectangle is 1500 ft. long and subtends an angle of 35 17 '48" at either of the opposite angles ; find the length of the diagonal and the area of the rectangle. 81. A triangular plot has one side 106 ft. long, the adjacent angles being io5i6' and 3724'. Find the other sides. 82. Two sides of a triangle are 9 and 7 and the included angle is 3856'32"; find the base and remaining angles. 83. The vertical angle of a triangle is 120 and the difference of the sides is fths of the base ; find the other angles. 84. The sides of a quadrilateral are 135, 180, 150, and 125 yds., and the angle contained by the first two is a right angle. Determine the area of the figure to the nearest sq. ft. MISCELLANEOUS EXAMPLES. 109 85. The diagonals of a rhombus are 120 ft. and 195 ft. respectively. Find its angles. 86. If the altitude of an isosceles triangle be 3 times the base, find its angles. 87. One side of a triangular lawn is 172 feet long, its inclinations to the other sides being 703o' and 78 1 8'. Determine the other sides and area. 88. Two parallel chords of a circle whose radius is 50 yds., lying on the same side of the centre, subtend respectively 72 and 125 at the centre. Find to the nearest inch the distance between them. 89. Find the area to the nearest square foot of the largest triangle which has two sides equal to 175 and 1 60 feet respectively, and the angle opposite to the latter equal to 60. 90. Find the angles of the rhombus equal in area to one-third of a square described on an equal base. 91. Find the other angles of the triangle, two of whose sides, containing an angle of I23I2'24", are in the ratio of 13 to 17. 92. A triangular field has its sides 50, 60, and 70 yds. long respectively ; find its area to the nearest square foot. 93. Find the greatest angle of the triangle in which the perpendiculars drawn to the sides from opposite angular points are 3, 4, and 5 feet respectively. 94. The sides of a triangle are 31, 24, and II feet long respectively ; find the greatest angle and smallest altitude. 95. Find the angles of the right-angled triangle in which the straight line bisecting the right angle passes through a point of trisection of the hypotenuse. 96. The sides of a triangle are 1 12, 86, and 72 feet in length respectively. Find the greatest altitude of the triangle. 97. Two straight lines 200 and 300 ft. long include an angle of 5o26'; find the length of the straight line joining their extremities. 98. At a point in the side of a rectangular field, 20 yds. from the corner, the opposite side and the nearer of the two adjacent sides subtend angles of 353o' and 742o' respec- tively. Calculate the area of the field to the nearest square foot. 99. Find the angles of the isosceles triangle whose equal sides are 15 ft. long and whose area is 50 square feet. 100. A rectangle is 3 times as long as it is broad ; compute the angles between its diagonals to the nearest second. 101. A triangle has its base 175 feet long, and the adjacent angles 52i8' and 4O2i'. Find its area and shortest side. 102. The adjacent sides of a parallelogram are 75 and 115 ft., and the perpendicular from the point at which they meet to the diagonal is 45 ft. Calculate to the nearest second the angles of the parallelogram, and find its area. 103. A triangle has sides 98 and 172 feet long, and the angle opposite to the former is 20 12'; find the third side. no LOGARITHMS. 104. Find the lengths of the trisecting lines of the angle of an equilateral triangle whose side is 155 feet long, and the areas of the three triangles into which the whole triangle is divided. 105. Calculate to the nearest second the smallest angle of the triangle whose sides are 20.3, 13.5, and 25. 7 feet. 106. The sides of a triangle are 586 ft., 1212 ft., and 1600 ft.; find its area. 107. Compare the areas of a regular pentagon and hexagon described on equal bases. 108. A side of a triangle 118 feet long has an adjacent angle 329', and the opposite angle 546'; find the longest side of the triangle. 109. Two sides of a triangle, 3071 and 2846 feet respectively, contain an angle of 52 17'. Find its other angles and area. 1 10. Find the angles of the right-angled triangle, the sum and difference of whose sides are in the ratio 2:1. 111. The radius of a railway curve is 4 furlongs 2 chains, while the angle between the tan- gents at the two ends is I382o'. Calculate the lengths of the tangents, and the distance of the middle point of the curve from the intersection of the tangents in feet. 112. The angular altitude of a lighthouse seen from a point on the shore is I23i'46", and from a point 500 ft. nearer to it is 2633'55". Required its height above the shore. 113. AB is a vertical object on the horizontal plane CBD, and at the points C, D on oppo- site sides of theobject the elevations of A are observed, AC=l2i8' and ADB-i^\^. The distance CD is 40x3 yards : find the height of the object. 114. AB is a measured base 500 yards long, C is a visible object whose angular elevation above the horizontal plane at A is I2i6'. The angles are observed, CAB = ^\^> and CBA = u%i2'. Find the height of C above the horizontal plane at A to the nearest foot. 115. The angles subtended by a chimney shaft 150 ft. high, standing at one corner of a tri- angular yard, at the opposite corners are 252o' and 38 15' respectively, while the distance between these corners is 100 yards ; find the area of the yard. 116. From the top of a tower, whose height is 100 ft., the angles of depression of two small objects on the plain below, and in the same vertical plane with the tower, are observed and found to be 4325' and 12 12' respectively. Find the distance between them. 117. If a tower stands at the foot of a hill whose inclination to the horizon is io5o', and if from a point 100 ft. up the hill the tower subtends an angle of 55, find its height. 118. From the top of a hill I observe two cottages lying before me in the same direction, their angles of depression being 232o' and i8io' respectively. They are known to be mile apart, find the height of the hill. 119. Knowing that telegraph poles are placed at intervals of 20 yds. along the bank of a river, from a point on the opposite bank I observe that two of them, next but one to one an- other, lie in directions making angles 75i5' and 72 25' with the bank, one being to the right and the other to the left of the point of observation. Find the breadth of the river. 1 2O. From a ship sailing north two lighthouses are observed to lie due east. After an hour's time they are S.E. and S.S.E. respectively. The distance between the lighthouses being known to be 8 miles, find the speed of the ship. MISCELLANEOUS EXAMPLES. ! 1 1 121. A river is 300 yards broad and runs at the foot of a vertical cliff which subtends at the edge of the opposite bank an angle of 25io'. Find the height of the cliff. 122. At a point in a straight road I notice that two distant church spires are in a line making an angle of 7S45' with the road. A mile further on, the line joining them subtends an angle of I23O' while the more distant spire lies in a direction at right angles to the road ; find the distance apart. 123. At noon a column in the direction E.S.E. from an observer cast a shadow, the extrem- ity of which lay in the direction N. E. from him. The elevation of the column was found to be 45 and the length' of the shadow 80 feet ; determine the height of the column, and the altitude of the sun. 124. Wanting to know the breadth of a river I measure along the bank a base AB 250 feet long. At A the bearings of B and of a tree situated on the opposite bank are I244' and 6o33' E. of N. respectively ; and at B the tree bears 28' W. of N. Compute the breadth of the river to the nearest foot. 125. What is the distance from one another of the summits of two mountains, 3 miles and 2 miles high respectively, just visible the one from the other, taking the earth to be a sphere whose radius is 3957 miles ? 126. The height of the Peak of Teneriffe being 12170 ft., calculate the dip of the horizon to the nearest minute (neglecting refraction), and the distance of the visible horizon in miles. 127. Taking the earth to be a sphere of 7912 miles diameter, what will be the dip of the sea horizon to the nearest minute as seen from a mountain 3 miles high, making no allow- ance for terrestrial refraction. 128. From two points in the same straight line with the base of a tower, and in the same horizontal plane, the angles of elevation are observed to be 58I2' and 3i46'; find the height of the tower, the distance between the points of observation being 185 feef. 129. Two straight railroads are inclined to one another at an angle of 2Oi6'. At the same instant from their point of junction two engines start, one along each line. If one travel at the rate of 20 miles an hour, at what rate must the other travel so that after 3 hours the engines may be at a distance from each other of 30 miles ? 130. A, B, C are three points in a straight line on a level piece of ground. A vertical pole is erected at C ; the angle of elevation of its top as observed from A is 53o', and as observed from B io45'. The distance from A to B being 100 yds., find the distance BC and the height of the pole. 131. In order to ascertain the distance of an inaccessible object C, a person measures a length AB = 200 yds. in a convenient direction ; at A he observes the angle PAB = 60, and at B the angle PBA - ib92o'; find approximately the distance BP. What is the ex- tent of the error to which the result is liable, supposing there may be an error of i' in each angular measurement ? 132. ABC is a triangle on a horizontal plane on which stands a column CD, whose elevation at A is 503'2". AB is 100.62 ft, and BC, AC make with AB respectively angles of 4O35'i7" and 959'5o". Find the height of CD. 133. The angular elevation of a steeple at a place due south of it is 45, and at another place 650 ft. west of the former station it is 14! 7'. Find the height of the steeple. 134. Two cross roads meet a canal at angles of 373o' and 552o' respectively, and at points distant 3000 yards from one another. What would be the length of a road cut direct to the canal from their junction, and lying between the cross roads ? 112 135- LOGARITHMS. B starts to walk in a north-east direction from a station 400 yds. to the north of A at the rate of 90 yds. a minute ; how far and in what direction must A walk, starting simultaneously with B, in order to overtake him, walking at the rate of 120 yards a minute ? 136. A man places a ladder against a house so that it just reaches to the top. He observes that the ladder makes an angle of 7625'3o" with the ground in this position, and that on removing the foot of the ladder a distance of 10 ft., while the ladder itself rests against the wall in the same vertical plane as before, the angle is diminished by nio'2o". Find the height of the house. 137. From each of three points in the same horizontal plane, distant 65, 83, and 106 ft. apart, the elevation of a tower is observed to be 45. Find its height. 138. A hill, the sine of whose inclination is ^-, faces south; find the inclination of a road which travels up the hill in a north-easterly direction. ANSWERS. EXAMPLES. I. I. (i) x = \og 2 y, (iv) log a = ^, (ii) 2 = 3 \ogpq or 3 = 2 log g /, (iii) Iog 10 2 = . 30103, (v) Iog 10 7 = 845098, (vi) Iog 2 . 5 = - I . 2. (i) I0 1 ' 3 ^=2 (iv) r 16 =/, 5, (ii)x?= } (v) 3= i ' 4 , ("i) (vi) K 4. 24, .5, 4.642, i. ,778, 1.468, 31.623, 2. 154, 21.544. 6. The square root of the original base. 7. (i) 5, 15625, .2, (H) 8, N /2, #.5. 8. (i) 10, (ii) .25, (iii) 5- EXAMPLES. II. i. (i) -.9927185 (ii) -2.8758207, (iii) -3.7159383, (iv) -1.4648400. 2. (i) 1.6875235, (ii) 3.0809382, (iii) 4-5, (iv) 3.875. 3. (i) 2.3515175, (v) 3.4290495, (ix) 4-8869432, (xiii) 6.1187785, (xvii) 12.9157198 (xxi) 128.09738, (xxv) 3.5569947, (xxix) 4.435629, (xxxiii) 3.787334, (ii) 1.4603662, (vi) .2215713, (x) .8629499, (xiv) 4.2146460, , (xviii) 9.6497684, (xxii) 272.993886, (xxvi) 2.966376, (xxx) 2.590752, (xxxiv) 2.929802. (iii) 2.0481957, (vii) 2.3419975, (xi) 3.4569085, (xv) 4.9211144, (xix) 1.8653787, (xxiii) 2.1427336, (xxvii) 4.9276689, (xxxi) .761551, (iv) 2.8203864, (viii) 5.0064334, (xii) 8.8401494, (xvi) 8.0305419, (xx) 274.95326, (xxiv) 1.7882409, (xxviii) 2.923186, (xxxii) 6.794014, EXAMPLES. III. i- (i) 3-2375439, (v) 2.7693773, (ix) 3.2730013, (xiii) 2.7993406, (xvii) 4.3002694, (ii) 1.9912260, (vi) 6.4560138, (x) 4.4014006, (xiv) 3.0326187, (xviii) 4.6434527, H (iii) 2.8293039, (vii) 3.7501226, (xi) 2.6989700, (xv) 4.8249392, (xix) 4.1629526, (iv) 3.1373540, (viii) 2.5932860, (xii) 4.9946866, (xvi) 3.7816119, (xx) 4.4666008. "3 LOGARITHMS. I. (i) .3222193, (ii) i. 5352940, (iii) 1.0969100, (iv) 4.9030900, (v) 3.6989700, (vi) 2. 0667660, (vii) 5051500, (viii) . 4I3734I, (ix) .5772363, (x) . 3046341, (xi) I. 46I7I67, (xii) 2. 7433892, (xiii) 1.6300887, (xiv) I. 3569814, (xv) 2. 5IO7I9O, (xvi) I. 1653048, (xvii) .9408786, (xviii) I. 3947700, (xix) I.249II68, (xx) 3- 1892232. 2. (i) 1.50407740, (ii) f. 30685282, (iii) OI9I7075, (iv) 4- 20824053. 3- (i) .9085841, (ii) 1-3944794, (iii) I. 4087304. i. (i) 4- 3346346, (") .4200741, (v) .1250087, (vi) .3494850, (ix) I. 849485, (x) 1.8409596, (xiii) 1.6110965, (xiv) .2108715, (xvii) 4.3346346, (xviii) 1.7852474, (xxi) I. 5523267, (xxii) 1-3333333, (xxv) 1.0706253, (xxvi) 1-9593372, (a (xxix) I .8578305, (xxx) 6573238, (: (xxxiii) .6948141, (xxxiv) .4610087, (> (xxxvii) .1041808, (xxxviii) 1.5387740, (x 2. (i) . 43IS23H, (ii) 4.86028767, 3. (i) 3-4734774, (ii) .4507201, 4- (i) 4- 1484835, (") 3.5034842. EXAMPLES. EXAMPLES. V. (iii) .0214812, (vii) 1.969897, (xi) 1.8316161, (xv) 2.7722417, (xix) . 5369447, (xxiii) f. 3 1 36666, (xxvii) .6751920, (xxxi) 1.6410904, (xxxv) .5204397, (xxxix) 2.7780766, (iii) 1.82124095. (iii) .2776455. (iv) .218427, (viii) 1.860206, (xii) 1.5117647, (xvi) .1215190, (xx) 2.1267371, (xxiv) .5320500, (xxviii) .4336811, (xxxii) 1.9451523, (xxxvi) .1905862, (xl) 2.4802752. I. .4771213, .6989700. 2. 3- 5. 7. 9. II. log 2 = .30 1 0300, Iog6 = . 7781513, log 3 = .4771213, log 7 = .8450980, log 5 = .6989700, log 9 = .9542426. .3010300, .4771213. 2.5563025. 1.2041200, 19.5051500. .1556302, .2100295, 1.1810658. 4.3180634, .3046341. 13- -0457574, .8573326, 1.2697953. VI. log 4 = .6020600, log 8 = .9030900, 4. .8450980. 6. 1.6989700, 2.5440680. 8. Same as for question 2. 4.4983106, 5.6811595. .1760913, .3979400, .5440680. -(i + 5 log 2). 10. 12. 14. 15. 1 6. 17. 18. 20. 22. .6989700, .778i5 T 3, -4771213, 2.6532126, 2.8750613, 1.2041200. Same as for question 2. 1.0413927. 1.3802112, 1.4327022. .8465736, 3.2036149. 19. 2.1760913, 2.5976952. 21. 1.0292896, 2.6108769. 23. 4.09691, 3.87827. ANSWERS. II5 EXAMPLES. VII. I- (i) 3 (") 4, (i") &, (iv) .16, (v) 2, (vi) 1.5, (vii) 1.5, (viii) 1.75, (ix) .16, (x) 2, (xi) 6, (xii) i, (xiii) .83, (xiv) .6, (xv) 1.875, (xvi) I, (xvii) 2, (xviii) 2, (xix) 2, (xx) .3. 2. (i) I, (ii) o, (iii) 1.5, (iv) .5, (v) O, (vi) 2. EXAMPLES. VIII. 1. (i) 4, () 3, (i") 3, (iv) 3, (v) 4, (vi) 2, (vii) 6, (viii) 3, (ix) o, (x) 2, (xi) 7, (xii) 5, (xiii) 2, (xiv) II, (xv) 2, (xvi) 5, (xvii) 5, (xviii) i, (xix) 2, (xx) 2. 2. 1458. EXAMPLES IX. i. (i) # = .524252, (ii) # = .983974, (iii) # = -1.058746, (iv) # = -.762531, (v) # = 2.311457, (vi) # = 2.078224, (vii) #=1.062585, (viii) #=1.242073, '- M -889907, , 630930, 2. #=1.537244. 3. # = -3.31381, ^ = .< 4. #=13.734546. 5. (i) * = !, j>=!; (ii)# = &; (iii) # = . _ 2 o ,. _ log 2^3 . log 2^2 log /y/6 7 - * '^g^' J ~ "n^gT"' ~~ 7 ^- # = ;;. 9. X ~ -- - ' '^^ log a -, with similar values for y and z. 13. 1389. 14. 20. 16. 1.556. 17. 1.2921592. 19. - EXAMPLES. X. I. I.068939I, I.0408I08. 2. I.05I27, 1.13315, .99800. 3. (i). 0069756, (ii) .0295588, (iii) .0010096, (iv) 1.9988883, (v) 1.9989995, (vi) 1.9831929. 4- 2.99957, 3.00043. 5. .0020000006666670. 6. (i) .0049875, (ii) 1.9949874. 8. 2.4849067. 9. .8450980, 1.0413927, 1.1139434. 10. .041393. ii. .4772660. 12. # = 2.00432. EXAMPLES. XI. (i) i, (vii) 3, (xiii) 2, (ii) 7, (viii) 5, (xiv) i, (iii) 4, (ix) 3, (xv) 5, (iv) 3, (x) i, (xvi) 3, n6 LOGARITHMS. (v) o, (vi) 4, (xi) ^, (xii) 3, (xvii) 6, (xviii) 3. 2. 2.947519, .9475I9. 2.947519. 3. 1.9096256, 7.9096256, 4.9096256. 4- (i) .57403I3. (ii) 1.7501225, (iii) 2.7958800, (iv) 1.1583625, (v) .3891661, (vi) 1.3502480, (vii) 6.3222193, (viii) .8293038. 5. .81617, 8161.7, .012252, 12252. 6. 4, i, 7, 8. 7. ist, 4 th, 6th. 8. (i) 20, (ii) 4, (i") 5- 9. (i) 23rd, (ii) 25th, (iii) 3 1st. 10. Between 89 and 120. i. 3-59999- EXAMPLES. XII. . (i) 3.7520543, 5-650041 J (ii) 4-9173604, .00826723 ; (iii) 117 I 12 | 23 | 35 | 47 I 59 I 7Q I 82 | 94 | 105 J, 1.5703839, .0371853; i|2J 3 |4l5|6|7|8| 9 | (iv) 1.6186314, 4155.645; (v) 2.3194460, .4792313; (vi) 4.7286449, .05353531; (vii) 1^.9096825, 812.24; (viii) 7.1187647, 1314-56; (ix) 3.5648380, .03671821; (x) 217 I 22 | 43 I 65 | 87 | 109 | 130 | 152 | 174 | 195 .3010491; i ! 2 I 3 I 4 I 5 [ 6 | 7 I 8 | 9 (xi) 2.9005386, .0795315; (xii) .0133256; (xiii) 139 | 14 | 28 | 42 | 56 | 70 I 83 | 97 | in | 125 , 4.4941999; i " 2 | 3 I 4 I 5 I 6 | 7 I 8 | 9 (xiv) 5.0001042; (xv) 1.9967099, 1.0075944; (xvi) 2.2798950, 5.2801131, 1905.21. 2. (i) .8562428, (ii) .25, (iii) .9824394, (iv) .5875158, (v) 1.62681, (vi) .9932896, (vii) 1.930698, (viii) 1.565065, (ix) .374408, (x) 3.159818, (xi) .907144, (xii) 6.240325, (xiii) .001264450, (xiv) 2.174774. 3. 1.861646. 4. 46.58847, 178.1415. 5. 16.30076. 6. 8.081476. 7- 7- 8. 12818.3. 9. 10.0794. 10. .563119. EXAMPLES. XIII. 2. (i) 1.892789, (ii) 1.654175, (iii) 4.730587, (iv) .3529152, (v) 1.086033, (vi) 1.186087, (vii) 1.8564634, (viii) 1.795889, (ix) 1.6514556, (x) 4.168962, (xi) 1.3506808, (xii) .1223851, (xiii) 1.350681, (xiv) 9.20105. 3. .3010300, .4771213. 4. 1.146015, .872589. 5. .5440680. 6. .9116518. 7. 1.54396, 2.515839. ANSWERS. 117 o. lut; i u, lui: ^ -, .u~>g s - , n^g 8429'6". 10. 9.427172, 7429'i2". 12. L sin =9.6991158 Zcosec= 10.3008842 L cos = 9. 9374820 Zsec =10.0625180 L tan = 9. 76 1 6338 Zcot =10.2383662 14. Zsin0 = 9.8664913 Zcosecfl = 10.1335087 Z tan 6 = 10.0354585 L cot 6= 9.9645415 Z sec 0=io. 1689672 (i) 0=55i4'56", (v) e = 384'55"> (ii) .0246287, (vi) 4256'29", (ii) -r=i928'i6.4" (vi) = 454o'i-3"> (viii) 6 = i956'47-7", 0= (iii) .000864447, (iv) (vii) 6352'55.9". (in) = 3645'2i-9" (iv) = 583i'4-3"> (vii) & = I928'i6.4", (ix) * = 6326'5.8", j/ = 2633'54. i". 1 8. (i) sec 2 ^4 cosec 2 ^, (iv) ' cos (vii) 2cos(a + /3), EXAMPLES. XVIII. (ii) tan 2 ^ sin 2 0, (v) A "\A (viii) 4 sin A cos- cos , (ix) (iii) tan 6A, (vi) ^, sm7^ cos 2. A sin 2_ (x) (xi) cos jr cos 2^, (xiii) 4cos^costocosl^ (xiv) 222 (XXH) (xii) (xv) 4 cos 20 cos 4^, a , (xxv) cos 46. ANSWERS. I19 2. (i) 5.01654, (ii) .00677044, (iii) .000127011, (iv) .000667708, (v) 395122, (vi) 3.041917.- (vii) -5.28614, (viii) -.7045742, (ix) 112.9396, (x) .604946, (xi) 16.23075, (xii) 37.75256, (xiii) 38.64313, (xiv) 5.041787, (xv) .02143932, (xvi) -1.43377, (xvii) .903834, (xviii) .0637321. 3. (i) *=I33'7" (ii) *=57i4'5o.4", (iii) * = 3623'i2", (iv) x = 6935'38.6", (v) x = 84o'55.8", (vi) x = 6i i'i2", (vii) * = 6oi9'30.3", (viii) *= i725'3o", (ix) x = t (x) x 758'2i.7" or zero, (xi) x=- n47'2O.7", (xii) A (xiii) ^/ = 2237'ii.6", (xiv) x- H424'i6", (xv) x- io-j^'2g.g", jy=i745'54.7". 4. 46i3'37". 5- 2222'48.4"and45. 6. sing=\/ ( * + OT ? ( *"** ) m cos0 =X / V T V cos0 = (+!)(-!) m v (+iXff-l) tan0= % /^ + ;; y~^ ) tan=- I X / + OT ^- OT ). 7 - 21 m- i 2 8. (i) 37.24144, (ii) I55.9583, (i) -4346886, (iv) 24.57586, (v) 7008.071, (vi) 224.9947. 9. (i) If = a tan 0, the expression = - 2 cos * /cos 20 (ii) If = a cos (since a > b) and tan 3 - = tan 2 *, the expression = ! ft ft J2 sin - tan V 2 2 EXAMPLES. XIX. 2. 9335'. 3- 4. 5546'i6". 5. ^ = 79 6 / 2 3 . 4 ", 6. 78i7'39.6". 7. 72i2;59^, 8. 7o33'3o", 9- Io8 58' 6^ 4747 I " I926'3o". 6 154* 10. 90, 5i3'27.2", 4x/2. n. j9 = 7i44 / 29", 12. J? =108 12' 26", C = 48i5'3i". C= 4927'34", log a = .00001. f i826' 5.8", 13.-^ io826' 5.8", 14. 8i 45'2.5". 15. J 5=5ii8'2i"or i 53 7' 4 8. 4 ", 7.58947. C=884i'39"or 16. 17.1, 3.68. 17. 322i / 54' / J 8. 3856'32.9", 9Q, -7525750- >27', or I2622'. 20. ^^=1263.58, 21. 7548'54", AC- 767.72. I4n' 6". 22. 6ii7'22". 23. S= 455 / io.6", 24. 45i4'23". 25. 4928'32". 26. I3234'32". 27. 355'48.6". I2O L, ULrA Kl 1 fim o. 28. 7827'52.8". 29. i=* 2 2 $ 3 - 745o;38;;, 3-. 2633'55"> 6 3 26' 5". 32. B=Tg 6'23'J"! 74i3'5o", 35 1 6' i o". 34- A = io538 / 57", B = i538'57". 35- 2.529823. 36. ?=,$$$: 37- ^=6559', C= 4 i56'i2' ". 38. 60.38936. 39, , 28.47717 sq. ins. 40. 57-3979. 41. c=?#*is'. 42 ' 9 8 I2'44". 43- 293i'34". 44. 3028'i2.8". 45. 40 3 2'9". 4 6. I323 4 '32.2". 47- 6 3 53'46". 48. ?-2I I2 'l2^' 49- 30, 120. 50. r=f?$& 5I - ^ = 90 o'i9", 52. A = 8959'49* 53- J3=90 o O'23,f, 54. ^32 : 5 5 2'42 : : 55- A = I289'26", B= 389'26". 5 6. 24.25546. 57. B = 9333'26. 3" ft or 26 26' 33. f. 5 8. 161.7416. 59- A - io8i2'4o", 60. 40, 60, 20^/19, EXAMPLES. XX. i. 140.956 yds. 2. 100 ft. 3. i826'6". 4- 229.6 yds. 5- I mile, 1.219714 miles. 6. 229. 149 yds. 7- 25.7834yds. 8. 117157 miles. 9- 39- 333 ft- 10. 179.011 ft., 13.008 ft. ii. 229.65yds., 85.86 yds. 12. 2.6718 miles per hour. 13. 1939.2 ft. 14. 112.14 ft- 15. 2656.26yds. 16. 45 1 569 yds. 17- 1496.57 ft. 1 8. 96.361 ft. 19. 634. 14 yds. 20. 1365.9 yds. 21. 152.33 yds. 22. 50.645 ft. 23- 346'4.6". 24. 5. 36 chains. EXAMPLES. XXI. I. 354.415 yds. 2. 86.964 acres. 3. .464- 4- 117.772. 5- 16.235 ins. 6. 14400.84 sq. yds. 7- 100670 sq. ft. 8. 9.2542 ins., 5.8352 ins. 9. 33.6158 sq. ins. 10 -{n253' 5 7''; 3 3.i663s q .in. 11 - ^6388 sq. ft. 13. ,439. sq.ft. 12. 2705.253 sq. ft. H.{^ 9 2 5 :|/ y t', 20 . 68 , yd .i5. -91856. 121 i6. 31. 19. 2 sq. yds. 22. 16.6051 ft. 25. 435-7335 miles. 28. 49 ins., 328 ins. 31. 6268 sq. ins. 17. io8sq. ft., 37.27 ft. 20. 27.7545 yds. 23- 5837'56". 26. 371277 sq. ft. 29. 6 4 i5'i3". 32. 392 inches. 34- SS-S^Sft.", 8998.35s.ft. 35. 1.013465 acres. 37. 29.6606 ft. 40. .1831 sq. ins. 43. 105.06 sq. ins. 46. 3. 142. 49- 6875.5. 52. 55- 615.351 sq. ft. 833. 17s. 2*d. 38. 64. 737 sq. ins. 41. 947.056 yds. 44. 178.412 links. 47. 249.877 sq. ins. 50. 960.08 sq. ft. 53- 6.9576: i. 56. 23.697 inches. 58. 1 385. 64 in., 2494 1 535. in. 59. 13552 sq. ins. A* /I9896 cub. ins. 114851 cub. ins. fi (4.6416 yds. \ 8.0395 yds. 64. 151.2 inches. 67.. 7.4442 ins. 70. 2244.66 sq. ft. 73. 21.9382 ft. 76. 933. 2 c. ins. , 224. 4 c. ins. 77 79- 233,509ft. 82. 6 cwt. o qrs. 15! Ibs. 85. 80.6851 cub. ft. 88. 41503 cub. ft. 91. 66.3776 cub. ins. 65. 358.6682 sq. ins. 68. 75526230 sq. miles. 1 8. 76.2631 sq. ft., .956773. 21. 69290 sq. ins. 24. 786 yds. 27. .53901. 30. 3.65637 ins. 33. 1 98. 427 yds., 30923. 67 sq. yds. 36. 32891 sq. ins. 39. 8.8783 ins. 42. 947 ft. per sec. 45. 123.876 sq. ft. 48. 101543.4 s.ft., 1 129. 6i6ft. f 65.6 sq.ft., 465. 33 sq.ft., l ' 142.8185 ft., 78.8629ft. (-96 i' 9", 34i8'4", *\4940'47" 26.37 sq. ins. 57. 11.46, 9, 6.93. 60. 7- 3554 ft- 6 / 189.57 inches. *\ 215620 sq. inches. 66. 5.6363 ft. 69. 10.075 i ns > / 46. 8678 cub. ins. y 1285.878 sq. ins. 122. 415.474 cub. ins. 113- 9- 5493 inches. 116. 2141.6 cub. ins. 119. 5428.67 cub. ins. i. 3. 7. II. 12. 14. 15- 16. (i) 198.5366, 20. .3523864. (i) 2.3377244, (v) 1.7127511, (ix) 3.9243220, (xiii) 3.3416276, (xvii) 3.2501815, (i) 1639.482, (v) 5*4- 7792, (ix) .1279746, (xiii) 1722489, (xvii) .003447162, MISCELLANEOUS EXAMPLES. A. (ii) 39.405, (i) 8.82888. 4. 131500- 8. 1.51751- (ii) 4.2175071, (vi) 3.9425714, (x) 2.0535060, (xiv) 4-2559993, (xviii) 1.0695578, (ii) .7081156, (vi) .07108095, (x) 26524.67, (xiv) .009970755, (xviii) .00007856093, 2. (i) .0189694, (ii) 38.7. 5- 125-iyrs. 6. 34.762. 9- 4343-1, 672.2, , 10. 412. 8s. id. (iii) .5389662, (iv) 4-8943338, (vii) 1.3910954, (viii) 4.7764792, (xi) i. 5453923, (xii) .455 I 553 > (xv) .4110916, (xvi) 5.6132374, (xix) 2.7926124, (xx) 0.8928316, (iii) 8.209777, (iv) 18116.33, (vii) .001737911 , (viii) 100.0724, (xi) 20.05308, (xii) 3.248701, (xv) 147.4078, (xvi) .01017148, (xix) .2702712, (xx) .01447366. 13. (i) .719686, .857696, "(ii) 3-98107, (v) 1.04729, (i) .3891576, (v) .0038819, (i) 46.356952, (v) 8.5020252, (ix) l"8.533882, (vi) .979467. (ii) 94.8683, (vi) 2082195. (ii) 12.5419025, (vi) 2.2278352, (x) 1.9584873, (iii) .0203138, (iii) .000019078, (iii) i. 1 909054, (vii) 5.10348, (xi) 3-979376, (iv) .000123986, (iv) 670.171, (iv) 1.2497119, (viii) 12.3023115, (xii) 1.2860980. (i) 3207207, (ii) 3449933, (v) .00000643485, (vi) .0000020165, (ix) -1345-57, W 1219-97, (xiii) 11464820, (xiv) 5.40935, (xvii) .0197413, (xviii) -999791. (xxi) .919641, (xxii) .634713, (xxv) 62.5368, (xxvi) 33-3457, (xxix) .643237, (xxx) 4I475-24, (xxxiii) 3.32903, (xxxiv) .0117641, (iii) .000352115, (vii) 190.414, (xi) -.109694, (xv) .000664211, (xix) .00257122, (xxiii) 1.626835, (xxvii) 37-50885, (xxxi) 47-39I6, (iv) .000591684, (viii) 194-519, (xii) .00000599425, (xvi) 4.64204, (xx) 2420.9, (xxiv) .00000000000945096, (xxviii) 31933.18, (xxxii) 3.215162, (xxxv) .000154046, (xxxvi) 2.174773, (xxxvii) 125765.8, (xxxviii) .0000587254, (xxxix) .121104, (xl) .129557. ANSWERS. 123 17. (i) 95998-5' (") 154-63625, (iii) 16924.24, (v) 10858.52, (vi) .000320026, (vii) 2536.63, (ix) 184864, (x) 5815380, (xi) -294.5213, (xiii) 1.34756, (xiv) 4737-075, (xv) 4516032, (xvii) .00069865, (xviii) -35.4586, (xix) .02827142, (iv) 2310.43, (viii) -11697.97, (xii) .463638, (xvi) .0173972, (xx) 6946.36. 1 8. (1)^:^3.28004, (ii) * = -. 752371, ("i) *= i. 94638, (iv) * =1.68504, (v) # = -.7814641, (vi) * = -. 331588, (vii) *=i. 29977, (viii) # = -.0311154, (ix) # = 5, (x) # = 2.1186153, (xi) x = .012487, (xii) # = -1.212, (x =23025.4, ("#=.265667, (# = -529528, (xui)-[.y=. 00230828, (xiv)-U = -. 014845, (xv)-U = 529586, [z -5.96886, I* = .475183, U = 538307. 19. 3768478. 20. 9830. 21. 31, 31- 22. 183, the 8oth figure after the decimal. 23- The 1 7th power. 24. 82.7333 yds. 25. 1.504079. 26. 8, 1679616. 27. (i) 241.171, (v) 288.337, (ix) 8158.85, (xiii) 228.793, (xvii) 1208.6, (ii) 207.893, (vi) 170.243, (x) 148.024, (xiv) 670.475, (xviii) 1657.81, (iii) (vii) (xi) (XV) (xix) 558.493, 3042.64, 895.43, 10653.2, 795-808, (iv) 814.725, (viii) 370, (xii) 156.396, (xvi) 3535-25, (xx) 328.105. 28. (i) 513.373, (v) 7-6045, (ix) 703-185, (xiii) 299.977, (xvii) 491.934, (ii) 661.783, (vi) 837-484, (x) 438-834, (xiv) 106.393, (xviii) 690.466, (iii) (vii) (xi) (xv) (xix) 87-204, 311.805, 2.7772, 50.986, 690.466, (iv) 643.928, (viii) 555-265, (xii) 119.047, (xvi) 367-698, (xx) 435-304. 29. (i) 9-4937, (vi) 8.4106, (ii) 2.3293, (iii) (vii) 8.5195, (viii) 4.1587, 7.4026, (iv) 2.4252, (ix) 2.3484, (v) 5-I58, (x) 4.8379. 30. (i) 20.99 yrs., (vi) 1 3. 74 yrs., (ii) 23. 59 yrs., (iii) (vii) 5. 9 yrs., (viii) 8. 73 yrs., 1 8. 6 yrs., (iv) 52.31 yrs., (v) 7.89 yrs., (ix) 23. 4 yrs., (x) 7.1 yrs. 31- 28.07171-3., 17. 673 yrs., 14.207 yrs., 8. 497 yrs., 7 .273 yrs. 32. 512, 1 1. 8 per cent. 33- 34- 1.4 per cent. (i) 1546.4, (v) 2474. 1 7, (ix) 41289.9, (") 357L93> (vi) 6667.4, (x) 4521, (iii) (vii) (xi) 15266.7, 3090.57, 8735.1, (iv) 261003, (viii) 10854.1, (xii) i5453-8. 35- (i) 853.02, (v) 2128.81, (ix) 2161.75, (ii) 1265. 13, (vi) 1689.04, (x) 1892.93, (iii) (vii) (xi) 1866.46, 3093- 2 5> 3063.12, (iv) 1984-79, (viii) 1901.84, (xii) 2752.03. 36. (i) 36. 1 59, (v) 33. 1 12, (ix) 38.866, (ii) 77.996, (vi) 55.984, (x) 85. 546, (iii) 93. 1 14, (vii) 43. 538, (xi) 50.98, (iv) 64.012, (viii) 36.201, (xii) 37.765. 37- 15. 195. 5 \ ) 8. (i) 7i33'54", (ii) iO928'i6". (ii) 24.22513, (") (v) (viii) 5. 9-9813803, 9.4573669. (iii) 2.59894. 5 1 '25. 4", (iii) 6 = '-3", (vi) = 10. 12. ", 753i'2i", 2857'i8". =1.74398, '= -I45H9- 15- 45, 135, -20 4 2'i7", -I59i7'43"- 17. 896 inches. 1 8. 164 feet. 7553'29" or 104 6'3i", 2I 23. 322 feet. , / 6 3 4i'39 f/ , 2b 'ii6i8'2i". / 59.4288 feet, ^' \within 1.76 feet. 27. 8083 feet. 9. 1.360022. ii. 579'28.4". 14- 6i32'. 16. 737'i2". 19. 23 inches. 22 f 34 i7'55", t39 57'i6". 25. 525 feet. 28. I07 44 'i7". 29. 32. 35- 38. 728 feet. 548. 1 6 feet. 889 feet. 6 1 feet 951 feet. ANSWERS. 30. 194 inches. 33. io24'42". 36. 3335'> 39. i53'- 42. 325 yards. 125 31. 4656 sq. feet. 34. 30 feet. 37. 8 feet 40. 6i34' 4 ". 43- 337ft or 1 606 ft 44. 8 1 acres I rd. 31 po 46. 68.755 inches, 17189 feet. 47- 4028.5yds., 2831.7 yds* 4^- \ ^^ r /"\ ** f2237'ii.5", 49-i 6 7 22' 4 8.5", -93-5 9- [90, I2osq. ft 50. 17.1064 sq. inches. 51. 20.97616 ft. 12.44845 feet, I -713322 feet. 53- 90. f 5ft, i6ft, 5^1 5 ft., 12 ft, 54.-^ I2839'i3", 5i2o'47", [ii223'37", 6736'23". 55. 41 8. 4 ft, 430 ft 56. 200 feet. 57. 3.2 miles per hour. 58. 84 feet. 59. 159.422 ft, 215.676 ft 60. (i) A = 2659'59", B - 63o' i ", c - 280. 58 1 ; (ii) ^ = 2247'3", = 67i2'$7", ^ = 25.474; (iii) ^=554i'36", a = 5. 53774, ^- = 9.82528; (iv) =I72$', 6 = 3I3.7. c= 1048.05; (v) ^ = 6i47'2o", ^=15.8693, a = 29. 5823. {B i 54io'56" or I2549' 4", 61. 934-433- C-\ c = 733o' 4" or i5i'56", 196.8775 ft. or 6.68ft 64. 6242'53-3"- 65. 2.34506. 66. (i) |^=i33i4'54'; 5 3 = 9 5* 54", = 36 52' 12", a = 20 inches, b - 16 inches. 67. 70. 73- 79- 82. 8 5 . 88. 91. 132.771 feet. 342954 sq. feet. ,-g / base angles 273o', '\ sides 56.369 feet. 7i. 7652', 74. 4127.98 sq. feet 69. 72. 1170, .849056. 75. I5osq. inches, 3539'i8", / 2596 feet, ' ( 3178182 sq. feet QT / 168.62 feet, 51 '\ 106.1 6 feet. 20. 3968 ins., 1 8. 357 1 ins. ("4100.7 sq. inches, ^27.57 inches. 5.6568, 90, 5i3' 2 8". 83. 4 2 5o'22.5", i79'37.5". 84. 188900 sq. feet. f 63i2'54", \ii647' 6". 625 inches. 89. 10518 sq. feet 24i6'25.5", 323i'io.5". 92. 13227 sq. feet. Q fi f8o32'i5.6", 8o32'i5.6", 8 / 325. 130 feet, 312.984 6 '\i855'28.8". * 7 '{ 26357.4 sq. feet. feet, 90. I928'i6", i6o3i'44". 93. I 126 94- 120, 7-3752 feet. 97. 231.437 feet. 100. 7'48". IOT I 83 ' 468 feet or 3< 1239- 374 feet. 106. 301944 sq. feet. IOQ 3 25'4i", 68i7'i9", > 13456895 sq. feet. 112. 200 feet. 115. 27802 sq. feet. 118. 603.37 yds. 121. 140.956 yds. 124. 211 feet. 127. 2i4'. 1 102.91 yds., J \i9.5379yds. 133. 171. 12 feet. 136. 52.859 feet. LOGARITHMS. 95. 6326'6", 2633'54". 96. 86 feet. 98. 29344 sq. feet. 99- {f 23 ' 16 " ioi / 7852.7 sq. feet, OI * 1113.426 feet. 104. ( "8- 717 ft., * U740- 74 sq.ft., 107. .66221. ( i826' 6", IIO '{7i33'54". 113. 48.4345yds. 116. 356. 833 feet. 119. 68.9445yds. 122. 3332.45 yds. 125. 280 miles. 128. 1 86 feet. , /935.763yds., 6 'l3-o6 yds. 134- 1503-91 yds. 137. 53 feet. 102 I 5954'2o", i2o5'4o", ^ \7462. 35 sq. feet. 2410. 17 sq.ft., 7i 2q'47" 2740. 74 sq.ft. 1 08. 145.3596 feet. ill. 1054.83 ft., I93'9i ft. 1 14. 629 feet. 117. 114.197 feet. 1 20. 13.66 miles per hour. 123. 61.23 feet, 3725'45" 126. i58', 135 miles. 129. 25.97 or 1 1. 55 miles per hr. 132. 101.166 feet. / 1260 yds., I35 'l32 i' 4 o"E. ofN. PRINTED BY ROBERT MACLEHOSE, UNIVERSITY PRESS, GLASGOW. MESSRS. 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This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. .-.. J "-" J K- JUI On ,. >u C.BRARV USB MAR 5 W l nf ^ ^ W ^ 2 w**^ 30ct'fi2MU ^ia4^^^^^fta (]i&) X NOV3 1962 = REC'D L.J NGV 5 1962 LD 21A-50w-4,'6( (A9562slO)476B General Library Univettiry of California Berkeley YC 22429