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PURE IkTm
AKITHMETIC. ALGEBKA, GEOMETRY,
PLANE TEISONOMETEY.
EDWARD ATKINS, B.Sc. (Lond.),
HEAD-MASTER OK ST. MARTIN'S SCIBJNCB SCUOOL, LBICHSTER.
LONDOxN' AND GLASGOW:
WILLIAM COLLINS, SONS, & COMPANY.
1874.
lAil rigkU reserved,]
TO THE REVEREND
WILLIAM FEY, M.A.,
HONORARY CANON OF PETERBOROUaH,
AND SECRETARY OF THE
LEICESTER ARCHIDIACONAL BOARD OF EDUCATION,
^hi0 Moxk t0 (BUtxtlti,
AS A TRIBUTE OF RESPECT FROM ONE OF HIS
OLD PUPILS.
PREFACE.
The special object of the present work is to meet the
requirements of the Science and Art Department's Examina-
tions in the first three stages of Pure Mathematics as set down
in the Syllabus of the Science Directory. This will account
for the arrangement of the subject matter. I hope, how-
ever, that it will be found not unsuitable as a general class-
book in Elementary Mathematics.
In the Arithmetical Section my object has been to deduce
the rules from first principles, avoiding as much as possible
algebraical considerations.
The Geometry consists simply of the first three books
of Euclid's Elements with exercises, the only point calling
for remark being the marginal notes. I have found it
useful in class teaching to put down upon a blackboard
the chief steps of the proposition — just those points, in fact,
which it is necessary to retain in the memory ; and to
encourage the pupil to depend upon himself for supplying
the connecting links. This skeleton, as it were, of the
proposition is placed in the margin, with the hope that
it will be specially appreciated by many students of the
industrial classes to whom the language of Euclid is
ordinarily an insuperable barrier. Particular reference is
here made to such of those classes as have grown up to
almost manhood without any mathematical training.
In the Algebraical Section of Stage I., the more difficult
examples are set particularly for the exercise of thosa
6 PREFACE.
students who take up the work of Stage II. without having
previously used this book for Stage I.
The amount of Plane Trigonometry included in this
volume is small, extending only to the solution of triangles
and the simpler cases of heights and distances. There
is, however, sufficient to fully cover the requirements of
Stage II. of the Government Syllabus. The treatment of the
higher parts of Algebra and Plane Trigonometry, as well
as that of Spherical Trigonometry, is reserved for a second
volume.
E. A.
Leicester, November^ 1S73*
CONTENTS.
STAGE I.
sectio:n' I.— akithmetic.
Chap. I. — The Fundamental Principles and Rules Ap-
plied TO Whole Numbers and Decimal Frac-
tions, 9
„ II. — The Treatment of Fractions considered as
Ratios, 16
,j III. — Application to Concrete Quantities, . . 42
„ IV. — The Metric System, 49
„ V. — Proportion, c 61
„ VI. — Application to Ordinary Questions of Com-
merce AND Trade, QS
Miscellaneous Examples^ • • 85
SECTION II.— GEOMETRY.
Euclid's Elements, Book I., 89
SECTION III.— ALGEBRA,
Chap. I. — Elementary Principles, 141
,, II. — Addition, Subtraction, Multiplication, and
Division, 155
,, III. — Involution and Evolution, .... 178
„ IV. — Greatest Common Measure, AND Least Common
Multiple, 195
„ v.— Fractions 204
,- VI.— Simple Equations and Problems Producing
them, 209
CONTENTS.
STAGE II.
SECTION I.— GEOMETRY.
Euclid's Elements, Book II., 23 1
Euclid's Elements, Book III., 254
SECTION II.— ALGEBRA.
Chap. I. — Quadratic Equations, 296
„ IT. — Problems Producing Quadratic Equations, . 312
„ III. — Indices, , 316
„ IV.— Surds, 319
,, V. — Ratio and Proportion, 327
SECTION III.— PLANE TRIGONOMETRY.
Chap. I. — Modes of Measuring Angles by Degrees and
Grades, 333
„ II. — The Goniometric Functions, .... 336
„ HI. — Contrariety of Signs — Changes of Magnitude
AND Sign of the Trigonometrical Ratios
through the Four Quadrants, ... . 342
„ IV. — Trigonometrical Ratios — continued. — Arith-
metical Values of the Trigonometrical
Ratios of 30°, 45°, 60°, &c., . . . 346
„ V. — Logarithms, 354
„ VI.— The Use of Tables, 358
,, VII. — Properties of Triangles, .... 366
,,VIII. — Solution of Right-Angled Triangles, . 374
„ IX. — Solution of Oblique- Angled Triangles, . 377
■ „ X. — Heights and Distances, .... 381
Answers, 387
MATHEMATICS.
FIRST STAGE.
SECTION I.
A.EITHMETIC.
CHAPTER I.
THE FUNDAMENTAL PRINCIPLES AND RULES APPLIED
TO WHOLE NUMBERS AND DECIMAL FRACTIONS.
Notation and Numeration.
1. We learn from elementary books on Aritlimetic, tliat
figures have a local as well as an intrinsic value, and that
the local value of a figure increases tenfold, or diminishes
tenfold, according as its position is changed from right to
left, or from left to right. Thus, commencing with the right
hand figure of an ordinary number, the respective figures of
the number stand for units, tens, hundreds, thousands, &c. ;
or, beginning with the left hand figure, which, we will
suppose, stands for thousands, the respective figures repre-
sent thousands, hundreds, tens, units. Let us carry this
principle a little further. Take the figures 68754, and
suppose that 7 represents 7 units; the question then arises
as to the number represented by 68754. Now, as 7 is the
units' figure, we have evidently, by the above principle, 6
hundreds, 8 tens, 7 units; and further, remembering that the
10 ARITHMETIC.
local value of a figure decreases tenfold for every remove to
the right, the 5, on our supposition, must represent 5 tenths,
and the 4 must represent 4 hundredths. Let us, as is
usual in numbers thus represented, mark the units' figure by-
placing a dot to the right of it. Thus, 35 7 "2 605 will then
represent 3 hundreds, 5 tens, 7 units, 2 tenths, 6 hundredths,
5 ten-thousandths; and to take one other example, '3065,
where the units' figure, though not expressed, is actually 0,
will represent 3 tenths, 6 thousandths, 5 ten-thousandths.
The dot is called the decimal point, and the digits to the
right are called decimals, because they represent portions of
the unit obtained by cutting it up into a number of equal
parts, which is always some power of 10. It may be
remarked, that 10 is called the first power of 10; 100,
or 10 X 10, the second power, sometimes written 10^;
1000, or 10 X 10 X 10, the third power, written 10^, and
so on.
To make the subject clear, let us see what the decimals,
•237, -2370, -0237 respectively represent. Now, the digits
2, 3, 7, in the first two decimals, are in exactly the same
position with regard to the decimal point, and the respective
digits in each have the same absolute value; moreover, the
cipher affixed to the right of the decimal -2370, has no
intrinsic value, and hence the two decimals, -237, and -2370,
have the same absolute value. And since the reasoning is
the same, no matter how many ciphers are affixed to the
right, we get the following important principle : —
The value of a decimal is not altered by affixing ciphers
to the right.
We will now compare the first and third of our examples,
namely, the decimals -237 and -0237. The cipher which is
here prefixed to the left, has again no intrinsic value ; but it
has removed the digits 2, 3, 7, one stage to the right, and
has, therefore, diminished their local value tenfold. The
effect of prefixing the cipher, is therefore to diminish the
absolute value of the decimal tenfold, and as every additional
cipher so prefixed has a similar effect, we get another funda-
mental principle, as follows : —
The value of a decimal is diminished tenfold for every
cipher prefixed.
ADDITION AND SUBTRACTION. 11
After the above it is easy to see that we have only to
remove the decimal point one place to the right or left, in
order to increase or diminish respectively the value of a
decimal tenfold. And if we allow the term decimal to in-
clude numbers which are greater than unity, as 35*721, we
may extend the principle thus: —
A decimal may be divided by 10, by removing the dot one
place to the left; by 100 or 10^ by removing the dot two
places to the left; by 1000 or 10^ by removing the dot three
places to the left, and so on ; and further, a decimal may be
multiplied by 10, 100, 1000, &c., that is, by 10, 10^, 10^ &c.,
by removing the dot 1, 2, 3, &c., places respectively to the
right. Thus, 6872-3476 divided by 10, 100, 1000 respec-
tively, becomes 687*23476, 68*723476, 6*8723476; and mul-
tiplied by the same becomes 68723*476, 687234*76, 6872347*6
respectively.
Addition and Subtraction.
2. If the student has understood the preceding article, he
will at once perceive that, provided we keep the units* figure
under the units' figure in every case, there is no difference
between the addition and subtraction of decimals, and the
addition and subtraction of ordinary integers. All he has to
take care of is that the decimal points are kept under each
other.
Ex. 1.— Add together 325*02, -647, 5*6073, -00214, 290,
and 4*7001. Proceeding as in ordinary addition :
325*02
•647
5*6073
•00214
290^
_47001_
625*97654
Ex. 2.— Take 6*291 from 18*3064. Proceeding as in
ordinary subtraction:
18*3064
6*291
120154
12 ARITHMETIC.
Ex. 3.— Find the difference between 15-02 and -6732.
15-02
•673 2
14-3468
Note. — In an example of this kind, where the number of decimal
figures in the lower line exceeds the number in the upper, it is ad-
visable to mentally supply ciphers to make up the deficiency in the
upper line. This may be done, as we have seen, without altering
the value of the upper line.
Multiplication.
3. Suppose we have to multiply 2-935 by 6*34, and let us
suppose the dot in eacb case removed to the extreme right.
Then (Art. 1), we have multiplied the number 2*935 by 1000.
and the number 6-34 by 100, and we have obtained the
numbers 2935* and 634* respectively. As these numbers are
integers, we may omit the dot, and write them 2935 and
634. Now 2935 x 634 = 1860790, but as we increased our
original numbers one thousand and one hundredfold respec-
tively, it is evident that our product is increased 1000 x 100,
or one hundred thousandfold. Dividing, therefore, the
above result, 1860790 by 100000, or what is the same thing
(Art. 1), writing it 1860790* and removing the dot 5 places
to the left, we get for our product of the numbers 2 '935
and 6-34 the result, 18-60790. We may remark that the
number of decimal figures in the product, namely, 5, is the
sum of the numbers of decimal figures in the two given numbers.
We have, therefore, the following rule for multiplication : —
Multiply the given numbers exactly as integers, regardless
of the decimal points, and after the operation is finished,
point off as many decimal figures in the product as there
are together in the multiplier and multiplicand.
Ex. 1.— Multiply 6-35 by -1703.
•1703
6-35
8"5T5
5109
10218
1081405
DIVISION. 13
Now, the number of decimal figures in the multiplier and
multiplicand together, is (4 + 2), or 6, and therefore we
mark off 6 decimal figures in our product. This gives us
1-081405.
Ex. 2.— Multiply -0063, by -017.
•0063
•017
441
1071
And pointing off (4 + 3), or 7 decimal figures, we obtain for
our product -0001071.
Division.
4. Suppose we have to divide -76875 by 6-25. We will
proceed as in the case of multiplication, by imagining the
decimal points in each number removed to the extreme right.
The numbers will then be 76875-, and 625*, or, omitting the
dot, as they are now integers, they will be 76875, and 625.
Proceed now as in ordinary division (which operation it is
unnecessary to explain), and we get for our quotient 123.
Now we must remember that we have increased our divi-
dend 100,000 or 10^-fold, and that, consequently, our quotient
will require to be diminished 10^-fold. This is done (Art. 1)
by removing the dot of the number 123* five places to the
left. But, before doing that, we know that the divisor has
been increased 100 or 10^-fold, and on this account our
quotient must be increased 10^-fold. This is done (Art. 1)
by removing the dot two places to the right. Hence, to get
the true quotient of -76875 by 6-25, we must remove the dot
from the extreme right of the number 123*, five minus two,
or three places to the left. Now three is the excess of the
number of decimal figures in the dividend over the number
in the divisor. We hence arrive at the following rule : —
Proceed as in ordinary division, and when all the figures
of the dividend have been brought down, and the remainder,
if any, obtained^c cut off as many decimal figures in the
14 ARITHMETIC.
quotient, as the number of decimal figures in the dividend
exceeds the number in the divisor.
Note. — ^When the number of decimal figures in the dividend is less
than the number in the divisor, affix a sufficient number of ciphers
to make the number of decimals in the dividend equal to the number
in the divisor. After finishing the operation of ordinary division,
there will be no decimal figures to cut off in the quotient. If there
be a remainder, and the division carried on further, by affixing
ciphers to the successive remainders, all the quotient figures thus
obtained will be decimals.
Ex. 1.— Divide 117-85088 by 6-272.
6-272)117-85088(1879
6272
55130
50176
49548
43904
56448
56448
We see that there are five decimal figures in the dividend,
and three in the divisor, and so we cut ofiT (5 - 3) or 2 in
the quotient. The answer is therefore 18-79.
Ex. 2.— Divide 527-2 by -0008.
Here it will be necessary to affix three ciphers to the
dividend, and the operation will stand thus —
'0008) 527-2000
659000
As there is no remainder, and the number of decimal
figures in the dividend is equal to that in the divisor, we
have none to cut oflf. The answer is therefore 659000.
Ex. 3.— Divide 463-7 by 2-769 to four places of decimals.
Here we must aflix two ciphers to the dividend, and the
operation, as far as the ordinary remainder of long division,
stands thus :—
DIVISION. ] 5
2-769)463-700(167
2769
18680
16614
•20660
19383_
1277
The quotient up to this point is integral ; but, as we have
a remainder, we must continue the operation of division,
first placing a dot at the right of the figures in the quotient,
and affixing a cipher to the present and each successive re-
mainder, until we have the requisite number of decimals in
the quotient. Bj thus proceeding, it is easy to see we
arrive at an answer — 167*4611.
Ex. I.
1. Increase the numbers 4*523, 29, -02367, '07 respectively
10, 100, 1,000, 10,000-fold.
2. Divide by inspection the numbers 0*05, 1111, 4*0020,
45 respectively by 100, 10,000, 1,000, 10.
3. Express in words -3467, 34*67, -0003467, 3-467 ; and
compare the values of the last three with the first.
4. Add together —
(1.) 6*732, 14-9, -0064, 14*27006.
(2.) -00291, -291, 29, 29100*9.
(3.) -821, 29*60, 29*6, -0029.
5. By how much does 5 exceed 4-2763, and 16*021 exceed
12*70009]
6. Find the value of —
(1.) 74*25 + -0067 - 3*0298 + 1-032 - 2-73.
(2.) 3-276 - 8*2409 + 100326 - -00091.
(3.) 2-5 - -00029 - 7-364 + 5-2791.
7. "What number added to four thousandths will give three
n'lndredths, and what number subtracted from 8,000 units
will give 291 units 29 hundredths?
16 ARITHMETIC.
8. Find value of —
(1.) -3 X -3. (2.) 9-001 X 27-06. (3.) 0-403 x -009.
(4.) -17 X -017 X 100. (5.) -3 X -005 x 6-4.
(6.) (-4)2 X (-032)1
9. Find the quotient of —
(1.) 79-4 by -397. (2.) 5-928 by 4742-4. (3.) 28
by -007. (4.) -6426 by 2-8. (5.) (-24)2 by 9-6.
(6.) 1-806 by (1-9)1
10. Given the quotient -00073, the dividend 124-1, find
the divisor when there is no remainder.
11. What is the value of —
(1.) j2 - -815 j -r ) -201 + -039 - -002;]
(2.) ) (-693)2 - (-307)2} ^ j-693 - -307 }1
12. If I add -061 to a certain number, and then divide
the result by 290, I get "0009 for a quotient ; what is the
number 1
CHAPTER II.
THE TREATMENT OF FRACTIONS CONSIDERED AS RATIOS.
5. A Fraction is a part or parts of a whole. It is gener-
ally expressed by two numbers, the one placed above the
other and separated by a line. The lower number expresses
the number of equal parts into which the whole quantity has
been divided, and the upper number, how many of those
parts are taken. Thus | is a fraction, and tells us that
unity has been divided into 5 equal parts, and that we have
taken 3 of those parts. The fraction | is read three fifths ;
each of the equal parts into which unity has been divided
being called a fifth.
The denominator of a fraction is the lower number, and
therefore shows the number of equal parts into which we
have divided the unit.
The numerator is the upper number, and tells us how
many of these equal parts are taken.
When the numerator is less than the denominator, the
quantity expressed is actually less than a whole. The
quantity is therefore a real or proper fraction. Again, when
TREAT31EXT OF FRACTIONS CONSIDERED AS RATIOS. 17
tlie numerator and denominator are both integral numbers,
the fraction is termed a simple fraction.
Thus f , ly 1^, are both proper and simple fractions.
It is however usual to include in the term fraction every
expression which contains one or more simple fractions, with
or without integral numbers.
Thus I, I, V, 6i, 1, p, 4^ of A of^\, are all included
in tlie term fraction.
They are, moreover, called vulgar fractions to distinguish
them from decimals, which, as will be shown further on,
may be looked upon as fractions, according to the above
definition, whose denominators are powers of 10, and not
expressed but understood.
It is convenient to classify fractions as follows : —
(1.) A proper fraction is one whose numerator is less than
2l
its denominator, as f, |, |f, — .
(2.) An improper fraction is one whose numerator is not
5i
less than its denominator, as f , f , -Jf , -5,
(3.) A simple fraction is one whose numerator and denom-
inator are both integral numbers, as |, f, 1%.
(4.) A mixed number is a fraction expressed by an integer
and a simple fraction, as 2^, 4|, 3f .
(5.) A complex fraction has its numerator, or denominator,
or both, in a fractional form, as ' , '^'■\ ~1 ,
f)}- cy 11
(G.) A compound fraction is a fraction of a quantity w^hicli
is itself fractional, as a of 2i, 2^^ of G ^V o^ ^-
6. In the preceding article we have spoken of fractions in
the ordinary way. We will now approach them from a
different point of view.
By the term ratio we understand the result of the com-
parison of two quantities with regard to magnitude. There
are two kinds of I'atios — ratio by difference or subtraction,
18 ARITHMETIC,
and ratio by quotient or division. Tlius we may consider
how much one quantity exceeds another, or we may consider
how many times one quantity contains another. The former
kind is called the arithmetical ratio, and the latter the
geometrical ratio. We shall speak only of the latter.
Definition. — ^The ratio between two quantities is that
multiple, part, or parts which the former is of the latter.
It is evident that a ratio can exist only between quantities
of the same kind ; thus, we may compare 12 horses and 6
horses, but not 9 men and 4 miles. And if the quantities
are reduced to the same denomination, we may treat the
quantities as abstract, just as we find the quotient of one
concrete quantity by another, by reducing them both to the
same denomination, and dividing as if they were abstract
quantities.
Now, according to what has been stated above, the ratio of
12 to 7, or, as it is usually written, 12:7, is obtained by
dividing 12 by 7 ; and this is the same thing as dividing unity
or 1 into 7 equal parts, and computing how much 1 2 of such
parts amount to. It hence follows that the fraction ^^^ is
properly expressed by the ratio 12:7.
The first term of a ratio is called the antecedent, and the
second is called the consequent ; and hence we may consider
a fraction as a ratio, the numerator being the antecedent of
the ratio, and the denominator the consequent.
When the antecedent is equal to the consequent, the ratio
is said to be a ratio of equality ; and it is said to be a ratio
of less or greater inequality according as the antecedent is less
or greater than the consequent.
Thus, 6 : 6 is a ratio of equality.
3 : 4 is a ratio of less inequality.
1 1 : 9 is a ratio of greater inequality.
The student will therefore have no difficulty in assenting
to* the following definitions : —
(I.) A proper fraction, is a ratio of less inequality.
(2.) An improper fraction is a ratio of equality or of
greater inequality.
(3.) A simple fraction is a ratio whose terms are integers.
Thus, |- = 3 : 5 is a simple fraction.
TREATMENT OF FRACTIONS CONSIDERED AS RATIOS. 10
(4.) A mixed number is a ratio of greater inequality, whose
antecedent has been actually divided by its consequent, and
the result expressed as an integer and simple fraction.
Thus, 2f = V = 17 : 7.
(5.) A complex fraction is a ratio, whose antecedent, or
consequent, or both, are not integers.
Tlius, ^-1, .:;, ^« = respectively to U : If, 2 : 7J, 9^ : 5
are coftiplex fractions.
(6.) A compound fraction is an expression containing two
or more ratios to be compounded together.
Thus f of I- contains the ratios 3 : 4 and 7 : 9 to be
compounded: -^ of -v of -J- contains the ratios 2^r : 7,
^7 61-9 ■"
3 : G|, 7^- : 9 to be compounded.
7. A fraction ivhose numerator and denominator are mtd-
tiplied or divided hy the same quantity is not altered in value.
Suppose, for example, we multiply the numerator and de-
nominator of the fraction ? each by 4, we get f = i?-. Now
the ratio of 3 I 7 is, from the definition of a ratio, four times
as small as the ratio (3 x 4) ! 7 or 12 ; 7; and the ratio 12:7
is, for the same reason, four times as great as the ratio
12 : (7 X 4) or 12 : 28. It therefore follows that the ratio
3 ; 7 is exactly equal to the ratio 12 \ 28, and consequently
3. - 12
7 2 }J *
Again, suppose we divided each term of the fraction ^^ ^7 ^»
we get If = iV Now the ratio of 15 ; 27 is, from the definition
of a ratio, three times as great as the ratio (15 -f 3) ; 27 or
5:27; and, again, the ratio 5 I 27 is three times as small as
the ratio 5 ! (27 -r 3) or 5 ; 9. It therefore follows tliatthe
vatio 15 : 27 = the ratio 5 ; 9, and consequently if = f.
Cor. — An integer may be expressed as a fraction with any
given denominator.
For we may consider an integer as a ratio whose conser
quent is 1, and we may multiply each term of this ratio by
any given number without altering its value.
20 ARITHMETIC.
Thus—G = ^ = e_>ijz.
T
1 X 7
Or G = 4 = -«-^ 9 = A4.
' ^1X9 9
8. To multiply a fraction by a whole number, we may
either multiply the numerator by the number, or divide the
denominator by it.
Ex. §- X 3 = ^^-^ - 2 4. QY we may proceed thus —
a X 3 = -«- - -s.
^ 9 — 3 3
^s <o the first method —
The ratio 8 '. 9 will be evidently increased three times if
we multiply its antecedent by 3 ; this follows from the defi-
nition of a ratio. We thus get the ratio (8 x 3) ; 9 or 24 ; 9.
It therefore follows that f x 3 = -j^.
As to the seco7id method —
The ratio 8 : 9 will be evidently increased three times if
we make the consequent three times as small ; and we thus
get the ratio 8 ; (9 -h 3) or 8 ! 3 ; and hence it follows that
It may be remarked that the two results, \f and §, are of
exactly the same value (Art. 7), since the latter may be ob-
tained from the former by dividing each of its terms by 3.
In actual practice we sometimes pursue the first method,
and sometimes the second. If the denominator of the given
fi-action contains the multiplier as a factor, it is more con-
venient to use the second method, thus : —
.1.1 X 3 = -_u_ = UL.
1 a 12-^-3 4
On the other hand, when the denominator does not contain
the multiplier as a factor, we use the first method, thus : —
(1.) "^ X 5 = "^-^^ = 3_5,
^''la 13 13
(2.) « X 10 =
b 5
Here we see that the numerator and denominator have a
common factor 5, and therefore, by Art. 7, if we divide them
both by it, we have : —
vlO— eX2_in
TREATMENT OF FRACTIONS CONSIDERED AS RATIOS. 21
9. To divide a fraction by a whole number, we may either
multiply the denominator by the number, or we may divide
the numerator by it.
Ex. — Divide |f by 4.
We may proceed thus, (1.) if 4- 4 = -^^ = -fV-
K-^'/ 17 • ^ 17X1 ^ »
' As to the first method —
The ratio of 12 : 17 will evidently be diminished 4 times
if we divide its antecedent 12 by 4. We thus get the
ratio (12 -f- 4) : 17 or 3:17; and it therefore follows that
if - 4 = ^V
As to the second method —
The ratio of 12 : 17 can also be diminished 4 times by
increasing its consequent or divisor 4 times, so that we thus
get the ratio 12 : (17 x 4) or 12 : 68. It therefore follows
that if -f 4 = if. .
It may be remarked, as in Art. 8, that the two results, ^^
and ir|, have exactly the same value, for tlie latter can be
obtained from the former by multiplying each of its terms by
4 (see Ai-t. 7).
And again, in actual practice, we usually take the first
method when the numerator contains the divisor as a factor,
but not otherwise. Thus —
(1.) i-S ^ 6 = -La±B- = ^%.
V*"-/ ^3 • "^ 13X5 «^*
(.3.) if -f 8 = _iv.
^ / 1 ^ 1 7XU
Here it is convenient to divide the numerator and denomi-
nator by the common factor 4 (Art. 7).
We then have if -r 8 = — ^-v ^-r = — ^ = 3^.
^^ i7x(H-T-4) nxa -»*
10. To i-educe a mixed number to an improper fraction.
Looking at our deiinition of a mixed number (Art. G), the
following rule is evident :
Multiply the integral part by the denominator of the
fractional part, and add in the numerator ; this gives tlie
required numerator, and the denominator of the fractional
part is the required denominator.
22 ARITHMETIC.
Ex. — Eeduce 5^, 7| to improper fractions.
11. To reduce a complex fraction to its equivalent simj:)l0
fraction. Before stating a rule, let us take an example.
3-'-
Suppose we liave to reduce ^-^^ to an equivalent simple
fraction.
3?- 3X5 + 1 1^6
Kow, by. the last Art., tt, — — ^— = 77*
9
Again, the ratio M : V '^i^l ^ot be altered in value if we
multiply both its terms by the same quantity. Let us multiply
them by 9 and it becomes l6 x 9 : V ^ ^' Now, by Art.
8, M X 9 = T-fi2i9 and V x 9 = ^^17. ^ 47 ::= 47. The ratio
then becomes i^Ao : 47. We will again multiply the terms
of this ratio by the same quantity, viz. by 5, and we get the
ratio i^""^ X 5 : 47 X 5. Now, by Ai-t. 8, i^^ x 5 =
2A><JL = ^j5X9 =. 16 X 9. Hence the ratio M : V jg equi-
valent to the ratio 16 x 9 :47 x 5, and hence the
fi-action Jl = ^|-
47 47x5
sr
Now 16 and 9 are called the extreme terms of the complex
fraction -i-, and 5 and 47 are called its onean terms.
4 7'
y
AVe arrive then at the following rule : —
KuLE. — Bring the numerator and denominator to the form
of simple fractions, then multiply together the extreme terms
for a new numerator, and the mean terms for a new denomi-
nator.
7?- 7X5 + 1 rifi
mi^„„ — i _ fi _ s _ 36X9 _ 324 _q3
lllUS, 2x - 2 X p + 1 - TTT " r^Tg' - y^ - ':5y
13 P.
12. To reduce a compound fraction to its equivalent
simple fraction.
Let it bo required to find the simple fraction equivalent
to the compound fraction f of 4.
iTREAfMENT OF FRACTIONS CONSIDERED AS RATIOS. 23
Now J of f is the ratio 3 : 4, where the unit of this ratio
is ?. It is therefore, from the definition of ratio, equal to 3
times this unit divided by 4.
Now 3 times f = f x 3 = ^ (Art. 8),
And /. 3 times 4 -f 4 - ^-^^ V 4 = ^_x 3 = sj<x
^ n 7X44X7
Hence we arrive at the result —
And in the same way we might show that
J of J of J of 11 = -^ ^ AX 3 xji^
** -^ ^ 8X9x5x1
Hence the rule :—
Rule. — Multiply together the several numerators for a
new numerator, and the several denominators for a new
denominator.
Ex. 3J- of 21^ of 6 = AAiL+i of 2JL2_±_i of ^ =:
2^^ of t of f = ^^2<^J<Ji,
Ex. II.
1. Heduce the following to improper fractions —
2. Eeduce the integer 19 to sixths, tenths, thirteenths,
eighteenths, nineteenths, and twentieths.
3. Bring the following fractions to integers, and reduce
them respectively to fourths, sixths, eighths, tenths, twelfths,
and fourteenths —
4. Multiply the following fractions each by 10, 11, 12 —
5. By how much does 8 times the fraction ^^ exceed the
quotient of %V ^7 ^ ^
6. Divide the following fractions each by G, 7, 18 —
3_7 J 4 l5 3 7_ I03i 11^%-
7. Diminish the following ratios respectively 6, 7, 8-fold —
12:5,3^:4, 9-3. :2i.
24 ARITHMETIC.
8. Simplify the expressions —
(1.) i of f of I of U. (2.) -J of 12 of ii of A-
(3.) y\ of 3^ of J of 11. (4.) U of 2 of V of 1,V
off off. (5.) 2iof-JofHof4T-V
9. Reduce to simple fractions —
U n 5i_of 61 _2±_ 2^^ ., .31 . ,
^i'^SA' ' 3A ' 7i of 3if 9 J of 3i' ' ii '^'
10. What is tlie difference between 2f of 3^^ of 2iJ
iof 21 - 24 of 1-5.
Q6
and -*1J
U
11. Find the intepral value of
3jl of I- + A of 4
12. Add the sum of the fractions -^\ - -^ to their
difference.
13. To reduce a fraction to its lowest terms.
Def. a fraction is in its lowest terms when its numerator
and denominator have no common factor, or are prime to
each other. (Among such common factors, we include either
the numerator or denominator itself, when one of them
happens to be a divisor of the other.)
When the numerator and denominator have a common
factor, we may divide them both by it (Art. 7) without
altering the value of the fraction. Now, the highest common
factor of two or more numbers is called their greatest common
measure, usually written G.C.M.
Hence we have the following rule : —
Rule. — To reduce a fraction to its lowest terms, divide
the numerator and denominator by their G.C.M.
Ex. — Reduce ^| to its lowest terms.
We can easily see that 12 is the G.C.M. of 24 and 84 ;
hence, dividing each by 12, we get —
2j4 = 2_4_±_L2 =:: .2 thc fractlou required.
14. It is not, however, always easy to tell by inspection
the G.C.M. ; but, before giving a general method for deter-
mining it, it will be useful to make a few remarks as to tho
divisibility of numbers in certain cases.
^REATMEiC'T OF FRACTIONS C05^SIDERED AS RATIOS. 25
A number is divisible as follows : —
By 2, when it is even.
By 3, when the sum of its digits is divisible by 3.
By 4, w^hen the number formed by the last two iigures is
divisible by 4.
By 5, when it ends in 5 or 0.
By 6, when it is even, and is also divisible by 3.
By 8, when the number formed by the last three figures is
divisible by 8.
By 9, when the sum of its digits is divisible by 9.
By 10, when it ends in 0.
By 11, when the sum of the digits in the odd places (thai
is, the sum of the 1st, 3rd, 5th, ttc.) is equal to the sum of
the digits in the even places, or the one exceeds the other by
a multiple of 11.
By 12, when the number formed by its last two figures is
divisible by 4, and the sum of its digits is a multiple of 3.
We may add also : —
(1.) A number is divisible by 37, when it is composed of
digits Avhicli are repeated three times, or any multiple of
three times, as 111, 333, 444444, &c.
(2.) A number which has three figures repeated in the
same order is divisible by 7, 11, 13.
Thus 271271, 165165, 23023 are divisible by 7, 11, and
13 ; for the last may be written 023023.
(3.) A number which has four figures repeated in the same
order is divisible by 73 and 137.
Thus 53245324, 2760276 are both divisible by 73 and 137;
for the last may be written 02760276.
Hence a fraction may often then be reduced to its lowest
terms by gradually striking out factors determined by in-
spection.
Ex. — Reduce ^VVi "to its lowest terms.
Now 792 and 2244 arc each divisible by 4, for the ]iuin-
bcrs 92 and 44, which are formed by the last two figures of
each, are evidently so. Hence, dividing numerator and de-
nominator by 4, we have jlilil =_- . 7 n i -^ t ^ jjl" . each
" 2 2 44 2 2 44-^4 6(il>
lis evidently divisible by 3. Hence -^^lili. ~ .1 pajils - 6_fi_ . '
2G
ARITHMETIC.
and each is now evidently divisible by 11. Hence
Remark. — It may sometimes happen that, although we are able to
tell by inspection some of the factors of numerator and denominator,
none of them are common to both numerator and denominator. We
cannot then strike them out, but we may use them to determine
what would be left supposing they are struck out, and We may thus
often come upon the G. C. M. of both ntfenerator and denominator.
Ex. — Reduce aWj ^^ ^^^ lowest terms.
Now 474 is even, and therefore divisible by 2 ; thus
474 -f 2 = 237. Again, 2133 is divisible by 9, for the sum
of its digits, viz., (2 + 1 + 3 + 3) or 9 is so divisible ; thus,
2133 - 9 - 237.
We have thus learned that, although 2 and 9 are not com-
mon factors, 237 is a common factor, and, in fact, the
G.C.M. Dividing the numerator and denominator of the
mven fraction by 237, we get -^jz ^^ = ^ "^ -* -^ a 3 7 ^ 2.
o •^ *^ 2 133 2133 — 237 9
We proceed now to give the general method of determin-
ing the G.C.M.
15. To find the G.C.M. of two numbers.
Rule. — Divide the greater number by the less, and if there
be no remainder, the less is the G.C.M.; but if there be a
remainder, make a divisor of this remainder, and a dividend
of the first divisor ; if there be a remainder again, make a
divisor of it, and a dividend of the preceding divisor, and so
on until there be no remainder. The last divisor will be the
G.C.M.
Ex.— Find the G.C.M. of 282 and 799.
The operation will stand thus —
282)799(2 or thus—
5_64
235)282(1 235 282 1
235
47)235(5
235
The G.C.M, is therefore 47.
282
799
5G4
235
282
235
47
235
235
I'ttfiATMENT Of* IPRACTIONS CONSIDERED A^ RATIOS. 27
The reason of this rule is easy to see. 47 divides 235, and
it therefore divides 235 + 47 or 282. Hence it is a common
divisor of 282 and 235, and it is therefore a common divisor
of 282 and 282 x 2 + 235 or of 282 and 799.
It is, moreover, the highest common divisor ; for every
number which divides 799 and 282 must also divide 799 -
282 X 2 or 235 ; and hence every number which is a measure
of 799 and 282 is also a measure of 235 and 282. Similarly,
every number which divides 235 and 282 will also divide
282 - 235 or 47, and hence every measure of 799 and 282 is
a measure of 47. But no higher number than 47 can divide
47, and therefore 47 is the G.C.M of 799 and 282.
16. It is now easy to see how any fraction may be reduced
to its lowest terms.
Ex. — Reduce
14 5 to its lowest terms.
148
703
592
111
148
111
37
111
111
Hence 37 is the G.C.M, and dividing
numerator and denominator by it, we get
17. To find the G.C.M. of more than two numbers.
The following rule needs no explanation : —
Rule. — Find the G.C.M. of any two of the numbers, then
the G.C.M. of this result and the third number, and so on.
The last result will be the G.C.M. required.
Ex.— Find the G.C.M. of 282, 987, 658, 1128.
The operation will stand thus —
24
282
987
846
141
2 *94
658
564
141
94
4
1
2
47
1128
94
141
282
282
188
188
47
94
94
Hence the G.C.M. is 47.
128 ARITHMETIC.
Ex. III.
1. Eesolve into prime factors 44, G4, 150, 252, 12G9, 4G2.
2. Find the prime factor of 1386, 1720, 4G08, 21175,
15972, 1425G.
3. A number is said to be a perfect number when it is
equal to the sum of its aliquot parts. Show that the follow-
ing are each perfect numbers — G, 28, 49G, 8128.
4. Show that 284 and 220 are a pair of amicable numbers;
that is, such a ])air that each is equal to the sum of the
divisors of the other, unity being here counted as a divisor.
5. Keduce, by inspection, to their lowest terms, the follow-
ing fractions :
a\ a in 3 5 12 1 _fi 3 2_ J^_^3^
') T.5"J TFT' "¥o' y y 0? 3 9 505 123 3*
/9 \ <) o 7 5 1 (i .-, 1 .-, :? 1 H 5 1_R 3 fi
V -" • / T B () > 1 O H > 1 7 ( i ? 2 .V 2 ? .V .V "o J 2 o 2 *
('X\ 1 () « A 4 7_2 r^ ^04 inOLH _l"l7.3_
W-/ 2"1 6J UOOJ 133 1' 13 2? 3 o O 3 5 ? 7 137 13*
' G. Find the G.C.M. of
(1.) 304, 323. (2.) 413, 448. (3.) 377, 533. (4.) 1866,
2832. (5.) 1189, 1517. (6.) 4374, 5103. (7.) 168,
378, 602. (8.) 539, 616, 792. (9.) 780, 1092, 2145.
7. Reduce to their lowest terms the following fractions —
(^ \ 07 5 052 J7 02 206B 1302. 15 5 5
\^') T«^25> T22^T> 2oT4> 6Tirir 197 2> 2 7 1^9*
/•2 \ 12 2 6 47 4 10 5j} 1_0_4 . 13 02. lA 9
V-"/ T8 3y> 6 5 3? T21d> T746> T344? 2106*
8. Show, without applying the rule for the G.C.M., that
lAU = \h aud that lUn^ = Ih
9. It rains in a certain district 634 days out of every
2,219 ; express this fact in the simplest way possible.
10. Out of 1,659 men engaged in a battle, only 1,185
answered the roll-call in the evening ; express by a i-atio in
its simplest terms the number missing in -relation to the
whole.
11. There arc 40 numbers less than 100, and prime to it;
what arc they?
12. A, B, C can do a piece of woik respectively in 318,
477, 795 hours; express the relative value of A, B, C as
workmen by the simplest integral numbers possible.
THE LEAST C03IM0N MULTIPLE, 2\)
The Least Common Multiple.
13. It is often necessary to express fractions as equivalent
fractions, having a common denominator; and it is, more-
over, convenient to have this denominator as small as possible.
Now, there are always an infinite number of numbers which
will contain each of the given denominators as a factor, and
our problem is therefore to obtain the least of such luimbers.
Def. — The least common multiple (L.C.M.) of two or
more numbers is the least number which contains each of
the given numbers exactly.
Rule. — Arrange the numbers in a line, ])utting one of
them as a divisor. Strike out the greatest factor common to
this divisor, and each of the numbers separately, and ])lace
the several quotients in the line below ; at the same time
bring down every number prime to the divisor. Kepeat the
operation upon the second line, and so on until we have a
line of numbers prime to each other. Multiply the several
divisors and the numbers in the lowest line together, and
their continued product will be the least common multiple.
Ex. 1.— Find the L.C.M. of 12, IG, 3G, 45, 60.
36 12, 16, 36, 45, 60
5 1, 4, T, 5, 5^
I 4 1 1
Hence tlie L.C.M. is 36 x 5 x 4 = 720.
Px. 2.^Find the L.C.M. of 15, 21, 60, 84, 140.
60 I 15, 21, 60,J4,_140
^ r"ir7,"'""i, 7, 7'
I 1, 1, 1
Hence the L.C.M. is 60 x 7 = 420.
19. Reduction of fractions to equivalent fractions having
the least common denominator.
It is evident that the least common denominator cannot be
a less number than the L.C.M., and therefore the following
rule needs no ex})lanation : —
Rule. — Divide the L.C.M. of the given denominators by
each denominator in turn, and multiply the corresponding
uumerator by the quotient. TJie product thus obtained i^
30 AIIITIIMETIC.
the new numerator, and the L.C.M. is the least common
denominator.
Ex. 1. — Reduce I, -yV, ^, | /y to equivalent fractions, having
the least common denominator.
12
3, 12, 8, 10
1, 1,2, 5
.-. The LCM. is 12 X 2 X 5 =: 120.
Dividing 120 by the respective denominators, we get as
quotients 40, 10, 15, 12; and hence the given fractions
become
Visa ^^ij> .■jJLLj^ ll^i_2. •
180 ' 1 ?0 * 1 20 > I JO '
Op J^-Sl. _/»jn_ lo..^ 1.32
^* J T'i 0, 12 0, i 2 O, T 2l)'
Ex. 2* Eeduce to their least common denominator the fol-
lowing i—ij, J-i|, H, 13 S- *
45;30, 18,- 45, 63 Hence the L.C.M. is 45 x 2 x 7
4"2li^T~7 = 630.
Dividing G30 by the respective denominators, we get for
quotients 21, 35, 14, 10. Hence the required fractions are
11 X 2 1 , 13 X 3 5 5 2 X 14 go X t p . 231 4 .5_5 7 2 R^ 2 9 0,
(J30 ' C30 ' 030 ' tJ30 ^'630^63 00 30 03 O
Note 1. — The operation of dividing the L.C.M. of the denominators
is often simpUfied by using the L.C.M. in its factorial form. Thus,
in our present example, the L.C.M. is 45 x 2 x 7. Now it is easy
to see that the quotient of this by 30 or 3 x 2 x 5 is 3 x 7 or 21, and
that the quotient by 18 or 9 x 2 is 5 x 7 or 35, and so on. We thus
avoid the process of long division.
Note 2. — It is sometimes necessary, and generally advisable, espe-
cially for beginners, to reduce the given fractions to their lowest
tv-irms before applying the rule for the least common denominator.
Thus, the least common denominator of the fractions f, -^^tj, II, taken
as they are, is 60 ; whereas, if we reduce the second fraction to its
lowest terms by striking out the factor 3, common to both numerator
and denominator, the fractions become f , |, ^|, and the least common
denominator is 20. If, however, the denominator of any such frac-
tion not in its lowest terms is contained in the L.C.M. of the deno-
minators, when the fractions are all in their lowest terms, it is unne-
cessary to reduce the fraction to its lowest terms.
We strongly recommend the beginner, however, to always com-
mence by reducing the given fractions to their lowest terms.
ADDITION OF FRACTIONS. 31
Ex. TV.
1. Find the L.C.M. of—
(1.) 2, G, 8, 12. (2.) 4, 9, 10, 14. (3.) 15, 21, 40,
45. (4.) 12, 20, 35, 126. (5.) 39, 65, 52, 140.
(6.) 37, 60, 222, 225.
2. Reduce to their least common denominalor —
(1.) h h A, Ur (2.) h A, ih H. (3.) h h \h
U' (-i.) I, •>'.., h (5.) h t\. -fW. H5.
V^'/ T2 6> To 3J ^«y
3. A can run round a ring in three minutes, B in four
minutes, and C in six minutes, and they start together. In
how many minutes will they all be again at the starting
point %
4. Arrange the fractions -j^^, .V/j, J 5, ^f, in order of mag-
nitude.
5. Multiply the greatest of the fractions M, -f 2 J> TTi V
339.
6. Divide the least of the fractions /g, t\, ^g, by 6.
7. Keduce to a simple fraction the complex fraction having
the greater of the fractions -?, J in the numerator, and the
less in the denominator.
8. AYhich of the fractions J , -^^ is nearer to \ %
9. Show that — is less than ^.
9i 5i
10. Arrange in order of magnitude the following : —
4fof^ liofA 11
G-i l^of2f 13|ofi
3^
11. Show that nine times the less of the fractions ^^) ,
^* ^
is eight times the greater.
12. Show that the ratio 18 : 7 is a ratio of greater in-
equality than the ratio 41 : 16.
Addition of Fractions.
20. We have shown (Art. 6) that the numerator and deno-
minator respectively represent the antecedent and consequent
32 ARITHMETIC.
of a ratio ; and it is evident from the definition of a ratio
(Art. 6) tlufct the sum of two ratios having tlie same conse-
quent is equal to a ratio wliose antecedent is tlie sum of the
given antecedents, and whose consequent is unaltered. Hence
we have the following i-ule for addition of fractions —
Rule. — Bring the given fractions to their least common
denominator, add together the numerators thus obtained, and
place under the sum the least common denominator.
Ex. 1.— Add together |, i^, ^9_, |.
The least common denominator is easily found to be 42.
Dividing this by each of the given denominators, we get
as quotients, 14, 2, 3, 7.
— 2R 4_ 20 I 27 I S'i — 2 n4-2 + 2 7 + 35
— 42 ^ 42 ^ 4-2" + 42" " ^.y
= ^-"^ = 2^4- = 2J-^.
Ex. 2.~Add together' 31, 2|,'liJ, 4^.V
The sum of the integral parts of the given fractions =
3 + 2 + 1 + 4 = 10. Hence, 3j- + 2^ + lij + 4^V - 10 +
4. J. -7 I J.L 4- 7_
The least common denominator is easily found to be 180.
Dividing this by each of the given denominators we get as
quotients 36, 20, 6, 15.
Hence the required sum
= 10 + lASjS + 7X20 ^ 1 I Xfi ^ 7X1 5
180 180 18 180
— 1 4- ^3 6. I 140 4. 6 I 106
= 10 + 3 6 + 14 + 6A±l_0_5 _ 10 + ^il
180 ^ ""
= 10 + lit-j = lligj.
Subtraction of Fractions.
21. After the preceding article there will be no difficulty
in comprehending the following rule —
E/ULE. — Bring the given fractions to their least common
denominator, subtract the numerators thus obtained, and
under the difference place the least common denominator.
Ex. 1. — Subtract -J from ^.
i — a— 7X5 _ 3Xn — 35 _ ?4 _ 3 5 — 24 — 1 1
« & - 4o' '4(r "" 4^' '•*" "" 4 -*«•
40
SUBTRACTION OF FRACTIONS. 53
Ex. 2.— Take C J from 9 J.
9J - Cj = 31- - J = 3 + 1^ - i^
= 3 + tV - i* = 3 + i^.
Here the number to be subtracted is the greater. "We
shall, however, take the less from the greater, and put the
7iegative sign to the remainder, meaning by this that the
remainder has yet to be subtracted.
Hence, 9j - 6f = 3 -- i^ = 2 + (1 - ij); now ij
taken from unity or ^| evidently leaves ■^^.
:, required result = 2 + -j^^ = 2^"'^.
We will now give an example involving both addition and
subtraction.
Ex. 3.— Find the value of ^ - 7f + 4^^ - 1 t's-
Whenever we have an expression involving both + and
- signs, the simplest method is to add together all the
quantities affected with the j)^"^^ sign, and likewise those
affected with the minus sign; then taking the difference
between these two sums, we place the sign of the greater sum
before the result.
Thus, taking first the integers, we have 6-7+4-1 =
10 — 8 = 2 (it must be remembered that the sign + is
understood before a number which appears without a sign
when it stands alone or at the head of an expression).
Hence the given expression = 2 + ^ - f + t\ ~ tV
= 2 + ^ ^^ '^ — 2^4 ^ fiX3 _ 5X2 — 2 + 1 2—8 + 1 5 — 1
36 36 36 36 36
= 22JLz^i2 = 2a = 2J.
36 "^^ *
We shall give one more example in order to show how
brackets are to be treated.
Ex. 4.— Find the value of 7^ - (2| - 3i) + (4^ - \\)
- (2A + 3tV).
The general rule, which the student will better understand
when we come to Algebra, is this —
When a minus sign stands before a bracket, it changes all
the signs within on removing the bracket ; but when a plus sign
stands before the bracket, the latter may be removed without
changing any of the signs within.
5 c
34 ARITHMETIC.
Thus, taking tlie expression - (2^ - 3i) : —
We must remember that the sign of 2^ when within the
bracket is, according to a remark made in Ex. 3, understood
to be plus. In fact, though it is unusual, we might write the
expression thus : — ( + 24- - 3i^).
Kow, remove the bracket, and it becomes - 2i + 3^.
Again + (4J - li) = + 4j - l\.
And - (2,^,- + 3tV) = - 2^0 ~ 3tV
Hence our given expression —
- 7i - 21 + 3i Hh 4i - H - 2^\ - 3tV
= 14-8 + i -i + i+i - 1 -^3^ - tV
= 6 + iXl 5 — 1^24 ^ 1X6 ^ 1X2 _ 1 X4 _ 3_Xj_2 _ iXl Q
120 120 120 120 120 120 120
= G + 15 —24 + 60 + 20 — 40 — 36 "10 — 6 + 9 5—110
120
Ex. Y.
1. Add together (1.) ^,1,1; (2.) |, ^V, ii ; (3.) h fV. tV
2. Find the sum of 3|, IT^i, 23^0, ly'V, §, H-
3. Add together 2j, ^, ^, 2| of H, 51 of 1^^.
4. Find the difference between (1.) | and f ; (2.) ^ and |;
(3.) f and |.
5. Subtract (1.) 61 from S-^V; (2.) 3^ from 4^^; (3.)
1 " ^« fnnrv. ^T^ff
6. Take , '^» from
If 2 ~ H
7. Find the value of 1/^^ - 2f + ^ - ^j.
8. Simplify the expression (2^ - li) - (3^ - 7-|).
.9. By how much does 3^% - 1|- exceed 2^{J- - ^^^ ?
10. Take the difference of 6yV and 1 JJj from their sum.
11. Add the difference of the same two fractions to their
sum.
31 of 1^ / 6 3^\
12. Find the value of the expression J^- 2. - 1 -M
MULTIPLICATION AND DIVISION OF FRACTIONS. 35
Multiplication of Fractions.
22. Rule. — Multiply together the numerators of the frac-
tions for a new numerator, and the denominators for a new
denominator.
The reason of this rule is easily seen. Let it be required
to find the product of J and J, or the value of f x J.
Now, what is the meaning of multiplying the ratio 3 : 5 by
tlio ratio 7 : 8 *? It means evidently that tlie ratio 3 : 5 is to
be multiplied by 7, and the result divided by 8.
Now (Art. 8) the ratio 3 : 5, when multiplied by 7, be-
comes 3 X 7:5; and (Aii;. 9) the ratio 3x7:5, when
divided by 8, becomes 3x7:5x8; and we have
Jl X ^ =: -•''^"
* » 5X0-
And so on for any number of fractions. Hence the above
rule.
Ex 1. — Multiply together the fractions J, f, if.
A X -^- X 1^ = 5 ^3X1 2
t> "^ '-^ ^ 6X8X26
Before actually performing the operation of multiplication,
it is advisable to strike out any factor common to both
numerator and denominator. We see that 5 is common to 5
and 25, 3 common to 3 and G, 4 common to 12 and 8, and we
then have
5 X 4 X i| =1x1x3 = J
6 8 ii6 2X0X5 ^"
The whole operation is sometimes written thus —
3
A X ^ X J-2 = ^ ^ '^ ^ '^ ^ = -v^-.
6 ^ 8 ^ 26 5XSXSS ''^ »'
2 2 5
Ex. 2.— Multiply together 2|, 3if, l^V, fl-
2J X 3^3- X 1^ X 51 = V X Vy^ X ti X U
• s « 3 3
_ 1 8 X 1 Q N X s 1 X s_^ _ 9 _ 9
. 5XivjXj{t)Xs»i A
Division of Fractions.
23. EuLE. — Invert the divisor, and proceed as in multi-
plication.
To explain this rule, let us endeavour to divide ^ by f .
We ma^ evidently consider the required quotient as nothing
3G AniTHJIETia
else tlian the ratio ^ : J, and tliis, by the reasoning of Art. 11,
is equivalent to the ratio 7x8:9x5, and we hence get
y » 9X6
Now, J is the divisor f when inverted, and hence the
above rule.
Ex. 1.— Divide t% by f .
S 4
TO • 8 - T(T ^ 3 - f7^ - 5 ^5-
Ex. 2.— Divide 1 J of 7i by 3^ of Sj.
1| of 71 -f 3L of 38- = I X V -^ ( V x V)
= 5 X V X i\ X /p = ill = I^Vt.
We have introduced a bracket on the right side of the
first equality, for otherwise the sign -f affects only the first
fraction ^3°. On the other side a bracket is unnecessary, for
the sign -r standing before a compound fraction (not two
fractions) aflfects the whole.
Ex. 3. — Simplify the expression
lA - 6| X 4i -f2f of24-rV
The given expression = ij -^ V x V "^ (I x \Y)
Ex. VI.
1. Find the sum, difierence, and product of 2^ and 1|.
2. Multiply the sum of the fractions 3|, 2-rV ^y their
diflference.
3. Simplify the expression i (3^%)"- (Ijf /l -r- 1 3^^ - Ut [ •
4. Reduce to a simple fraction each of the following ex-
pressions —
(I.)ItV- 7i X 8| ^ 2f of20i.
(2.)1tV - 7iof8| - 2f - 20i.
5. What is the diflference between (81 - 3j) and (5 J - 4tJj)1
r -n- • 1 4 , Y of 1|
6. Divide ^ by ^— ^^
3 + :;^1 '
■ 7i
REDUCTION OF FRACTIONS TO DECIMALS. 37
1
7. Ileduce to a sim})le fraction 3 +
7 + t'
F
8. Simi)lify the expressions (1.) 5^ ^"^ ^ ^^ 1--
^^i of 14 of 21 ^ ' ^ ' »^)-
D. Find the quotient of 103^2 by 30^ of ^gj.
10. The cost of 7f articles is ^£65^, what is the cost of
eacli article ]
1 1. Find the cost of 89^ articles, when one cost .£4^^^.
12. The sum of two quantities is Zij^, and their dijQferonco
is 6 }^ j required the greater.
Reduction of Fractions to Decimals.
24. If we place a decimal point to the right of an integer,
and add as many ciphers as we j^lease, it is clear, from Art. 1,
that we do not alter its value. And hence a given ratio, as
3 : 8, is not altered in value by writing it 3-000 : 8 ; and
fui-ther, dividing each of its terms by 8, according to the rule
for division of decimals, it becomes '375 : 1. It therefore
follows, putting each of these ratios in a fractional form, that
.3. - 3:000 ^ ._^^ ^ .^^^
" H 1
We get, therefore, the following rule : —
Ki'LE. — Place a decimal point to the right of the numera-
tor, and add as many ciphers as may b© thought necessary.
Divide the new numerator by the given denominator, accord-
ing to the rule for division of decimals, and, if nocessary, add
ciphers to the successive remainder until the division ter-
minates, or until we have obtained as many decimal figures
as required.
38 ARITHMETIC.
Ex. 1. — Reduce ^V to a decimal.
32)5-0(-15625
32 ■
180
160
200
192
80
64
160 .
160 Hence ^% =u -15625.
Ex. 2. — Reduce ^rf ., to a decimal.
206)5-00(-01689i
296
2040
1776
2640
2368
2720
2664
560
296
264
It will be seen that we havef arrived at it remainder, 264^
exactly the same as the second remainder j and that^ there-
fore, the quotient ^figures ^91 will continually repeat, and
that the division will never terminate. We call 891 the
recurring period of the decimal, and it is usual to indicate
the fact of its recurrence by placing dots over its first aiid
last figures, as above.
We have, therefore, as a result, ^Jy =^ *01689i;
NoTR. — It is easy to see that no fraction, reduced to its lowest fermSf
■whose denominator contains any prime factor, other than 2 or 5, can
be expressed as a terminating decimal. For every terminating deci-
mal is an exact number of tenths, hundredths, &c., and may, there*
fore, bo transformed into a fraction, having some power of 10 as its
REDUCTION OF DECIMALS TO FRACTIONS. 39
denominator. Now, if we wish to bring a fraction already in its
lowest terms to an equivalent fraction having some power of 10 for
its denominator, it can only be done by onultlplying its numerator and
denominator by some integer ; and it is impossible to obtain any
power of 10 by multiplication only, from a number which contains
any prime factor, other than 2 or 5.
Reduction of Terminating Decimals to Fractions.
25. Remembering (Art. 1) that any given terminating
decimal may be considered as derived from an integer by
diminishing it 10, or 10^, 10^, *fec. fold, we have
•345 = 345 -f- 101
Hence, *345 is the value of the ratio 345 : 10^; and we
, 1 Q,^ 345 345
have, also, -345 = — == j^.
The following rule is, therefore, clear : —
Rule. — Make a numerator of the integer, formed by tak-
ing away the decimal point ; and for a denominator jjut 1, fol-
lowed by as many ciphers as there are given decimal figures.
Ex. 1.— -625 - T-VA = ^-^-i^ - f.
1000 8X125 ^
100 20X6 ^O
•0025 = T-^-^—^ =: 26 = -rl_
Reduction of Recurring or Circulating Decimals to
Fractions.
26. There are two kinds, and it is convenient to treat
them separately.
(1.) Pure circulating deciinals, where the whole of the
decimal figures recur.
Rule. — Take away the decimal point and the dots, make
a numerator of the integer thus obtained, and place under
this as denominator as many nines as there are recurring
figures.
The following example will make this ride clear.
Ex. — Reduce •207 to a fraction.
The value of the decimals is evidently tiot altered by
writing it -207207. Let us remove the decimal point three
40 AniTHMETIC.
places to tlie riglit, 01%'wliat is the same thing (Art. 1), let
us multiply the given decimal by 1000.
We then get 207-207 as tlie value of 1000 times the given
decimal. Now the number 207-207 includes the integer 207,
and the given decimal -207; and it therefore follows that the
integral part 207 is (1000 - 1), or 999 times the value of
the given decimal.
Hence, dividing it by 999, we get
•207 = 207 ~ 999 - fgj.
(2.) Mixed circulating deGimals. — Where part only of the
figures recurs.
E-ULE. — Take away the decimal point and the dots, sub-
tract from the integer thus obtained the integer formed
by the figures which do not recur, and make a numerator of
the result. Then, for a denominator, place underneath as
many nines as there are recurring figures, followed bv as
many ciphers as there are figures which do not recur.
We shall make this rule clear by the following example : —
Ex.— Eeduce -24573 to a fraction.
Let us remove the decimal point two places to the right,,
thus, by Art. 1, multiplying the given decimal by 100 ; we,
then get
100 times the value of -24573 = 24-573.
Now, by case (1) above, 24-573 = 24f ^{ ; or reducing
to an improper fraction, and noticing that 24 x 999 =•
24000 - 24, we have
24*573 = 24000 -- 24 + 5 73 — 2 4 5 7 3 " 2 4^
9 9 9 9 9 9 *
Hence, dividing this result by 100, we get
*24573 = ^4r.n:i- 24. ^ IQO = 24573-24 / A ,u. 04 \
999 99900 \-^^ i^« '^^' f
Ex. 1.— -428571 = flff-U = ^-^-i^^^^-? = f.
999999 7X1 42867 '
■Ry '9 700 — 2 7 9 — ,2 7 — 2fiR2 — 149X0X9 J^49
27. In arithmetical operations involving circulating deci-
mals, and, indeed, any decimals having a large number of
decimal figures, it is generally sufticient to obtain a result
ijorrect to a given number of decimals.
APPROXIMATE RESULTS. 41
1. In addition and sid)tr action we obtain this result most
easily by using in our operation one or tioo more figures of
the given decimals than are required in the result.
Ex. 1. — Add together (correct to five places) the follow-
ing:—
•3026, 6-7294, -016, -4163729.
•3026026
6-7294444
•0166666
•4163729
^4650865. Ans. 7*46508.
Note. — If our object is to obtain a decimal of five places which
shall give the approximate sum of the given decimals we must write
7 '46509 as the approximate sum ; for 7 '46509 is nearer to the true
value of 7-465086 &c. than 7 '46508. The general rule is to increase
by 1 the decimal figure at which we stop, ivlien the next figure is 5 or
above 5.
Ex. 2. — Find the difference (correct to six places) of 3-0745
and 4-263, and express the approximate difiference by a deci-
mal of five places.
4-26326326
3-0745 4545
l-1887T7¥l
Hence the difference correct to six places is 1*188717, and
the required approximate difference 1-18872.
2. In midtijilication and division of circulating decimals it
is generally preferable to reduce the given decimals to frac-
tions, bring out the result in a fractional form, and afterwards
reduce this to a decimal.
Ex. YII.
1. Express as a decimal the sum of the following fractions: —
3 1 .«» 7 3 .•> 7 1
2. Reduce to fractions the following decimals : —
•35, ^026, •ie, -142857, -16, -4285714.
3. Find the value (correct to six places) of -237 + 3*816 —
6-0235 + 4-29 - -002 + 1-374.
42
ARITHMETIC,
4 Add together -62, -037, 2-476i, -8106, -7, -375.
5. Find the product and quotient of 3-5 4 by 4-3.
6. Simplify the expression —
(4-6 X •428571 - 2-2 x -36) (1 - -16).
_ In the next six examples the dots are signs of multiplica-
tion, and you are required to give the values of the expres-
sions correct to six places.
^ 1 1 1 1 1 ,
R 1 1 1 1 .
"* '3 ~ 3^ + 5^ - 7^ + '^•
C L J_ . 1 1 A
' ■I-2 *" 1-2-3 * h¥3^ ~ 1-2-3-4-5 "^ ^'
1n 1 1 1 1 r
10. 1 + g3- + — + -g-. + &c.
}
239
^ . + &c.
3-239*
12. 4 + -^^ +
+ -r-F + 'k—n +
1-2 2-3 34 4-5 5'6 6-7
CHAPTEE III.
APPLICATION OF THE PRECEDING ARTICLES TO CONCRETE
QUANTITIES.
To find the Value of a Fraction of a Concrete Quantity.
28. Rule. — Multiply the concrete quantity by the nu-
merator of the fraction, and divide the product by the deno-
minator.
VALUE OF A FRACTION OF A CONCRETE QUANTITY* 43
Ex. 1.— Find the value of | of 2 tons, 3cwt. 21 lbs.
3 f C) f^T%a O rt f 91 IVko (2 tons, 3 cwt. 21 lbs.) x 3 6 tons, 9 cwt. 2 gra. 7 lbs.
= 18 cwt. 2 qrs. 1 lb. Ans,
Ex. 2.— Find the value of 3 J- of £12. 6s. 2jd.
X. s. (1.
3 times £12. 6s. 2Jd. = (£12. 6s. 2|d.) x 3 = 36 18 6f
. ^^^- = 2 14 81
Hence, adding^ the value required =£39 13 3 J
Ex. 3.— Find the value of | of 4 miles + I- of 3 fur. + /^
of 8 poles.
Mile. Fur. Poles. Yards.
4 of4mnes = <-^'^^ = '-^^ = 1 5 28 3i-
^^ of 3 fur. = ^^-^L^L ^ ilJH. ^ 2 13 If
/x of 8 poles = ^l^i^^ ^ » ^ Q 1 4H
Hence, adding, the value required =2 3 4j|
KoTE. — The addition of the yards is thus effected : —
(3| + H + 4U) yds. = (8 + ^H^) yds. = (8 + \?A) yds:
:= 9f^ yds. = 1 polei(3i + |$-) yds;
= 1 pole; (3 + ^— ^) yds. ^ 1 pold,4H ydsi
Ex; VIII.
find the values bf —
1. I of £1 ; ? of Is. ; ? of 12s. ; 5 of £3.
2. 5- of £5 ; ? of a guinea ; -^\ of 2s. 6d. ; % of a crown*
3: ^ of £1. 12s.; 2^ of £3. 10s.; 7^^ of £3. 4s. 51.
4. f of 1 ton ; I of 1 qr. ; J of 1 stone ; -^^ of 9 lbs.
5. 3f mile ; 4. of 3 fur. ; -? of 1 5 poles ; ^ of 2j of 1 2 yds.
8 2 " 51
G. ^Q- leap year; -,— lunar month; jTyT of 10 h. 15' 12^'.
7. 12 lbs. 3 oz. 7 dwt. 5 grs. x 3|; 10 oz. 4 gr. v 16j-.
44 ' AHITHMETIC.
8. 8f of 4 ac. 3 po. - -^\ of 1 sq. mile + If of 3 r. 20 sq. yds.
9. (5f of /^ - l^\) of 35^ 36^ 2^; (2^ - H of If) of 30^
10. ^% of 1 lb. Troy + ^ of 1 lb. Troy - /^ of 1 lb. Avoir-
dupois ( = 7000 grains).
« 5^ of 3 soiineas
^^' ^ ""^ fifrfsriOd.^ ' ""^ 1 ton + f of 3 qrs. - ^V of 7 cwt.
12. 2t\ of 15 L 10' 131'' - ^V of 1 day, 12 b. ir 12j/.
To Reduce a given Quantity to the Fraction of any
other given Quantity of the same kind.
29. KuLE. — Reduce both the given quantities to the same
denomination, and the fraction required will have the number
of units in the first quantity as numerator, and that of the
second quantity as denominator.
Ex. 1. — Eeduce 3s. 8d. to the fraction of £1.
3s. 8d. = 44d,
and£l = 240d.
Hence, the fraction required = ^Vo = ih
Or, better, thus :
3s. 8d. = 11 fourpences,
and £1 = 60 fourpences.
.'. Fraction required = J J.
Note. — It is always best to keep the denominations to which the
given quantities are reduced as high as possible.
Ex. 2. — Reduce i of a moidore to the fraction of 2| guineas.
1 of a moidore = (| x 27)s., and 2^ guineas = (2 J x 21)s.
1 X 27 4- X 27
Hence, the fraction required = —- — -- = ^— -; , = l^ 11 ^ I
' ^ 2^x21 fx21 ^^-^1^'
_ 1X9X2 _ T B
~ 6X7X7 — 245
Ex. 3.-— Reduce 3 cwt. 2 J- qrs. to the fraction of 4 cwt.
2 qrs. 4 lbs»
3 cwt. 2^^ qrs. = 14 J qrs., and 4 cwt. 2 qrs. 4 lbs. = 4 cwt.
2i qrs. = 18| qrs.
14 L J 2 7.
.*. Fraction required = j-g^ = 7^ = ^VyVJ = h
VALUE OP A DECIMAL OF A CONCRETE QUANTITY. 45
Ex. IX.
Reduce —
1. Is. 8d. to the fraction of .£1 ; 7 id. to the fraction of 10s,
2. 2s. 4d. to the fraction of lOs. 8d.; Is. 7 id. to the frac-
tion of 3s. 4 Id.
3. 3 qrs. 15 lbs. to the fraction of 1 ton; 2 stones 10 oz.
to the fraction of 3 cwt.
4. 3 lbs. avoirdupois to troy weight; 10 lbs. 3 oz. 4 dwt.
troy to avoirdupois.
5. 3 quii'es, 10 sheets to the fraction of 2 reams, 3 quires;
3 ft. 8i in. to the fraction of 3 yards.
G. 30^ 3' 12'^ to the fraction of a right angle (-90^);
57^16' 21^^'' to the fraction of two right angles.
What fraction is —
7. f ac. of 3i ac; 2i days of 17 weeks?
•125 + 1-875 3-lGx 1-4
8. —rvrvTr^oK ~ acres of 19 \ poles : 7 yds. of 3i m ]
•0140625 ^^ 2*375 x-i
9. ( JY rood + 7^ poles + ■ '^ [^^ ^^ yd. j of 3 acres 1
1^- oVg^l-of-^" pipes?
11. What fraction of his original income has a person left
after paying a tax of 4d. in the £ 1
1 2. A garden roller is 2 ft. 6 in. wide, and it is rolled at
the rate of 1 mile in 20 minutes : find in what fraction of a
day a man will roll J of an acre.
To Find the Value of a Decimal of a Concrete Quantity.
30. Rule. — Multiply the given decimal by the number of
units in the concrete quantity when expressed in terms of one
denomination, and the integral part of the result will be the
number of units of this denomination. Tlien multiply the
decimal part of this denomination by the number of units
connecting it with the next lower and the integi^al part
will be units of this latter denomination, and so on^
46 ARITHMETIC.
Ex. 1.— Find the value of -325 of £3. 10s,
£3. 10s. = 70s,; proceeding then according to i^ule, we have :
•325
70
22-750S.
12
9-OOd. Ans.: 22s. 9d. or £1. 2s. 9d.
Ex. 2.— Find the value of -546875 of 3 tons.
•546875
3
1-640625 tons.
20
12-812500 cwt.
4
3-2500 qrs.
28
200
50
7-00 lbs. Ans.: 1 ton, 12 cwt. 3 qrs. 7 lbs,
Ex. 3.— Find the value of 6-66875 acres.
6-66875 acres.
2-67500 roods.
40
27-000 poles. Ans.: 6 acres, 2 roods, 27 poles.
Ex. 4.— Find the value of -316 of £1,
First Method. Second Method.
900 9 00
1 5_X 1 9
20 ■ 16 X^'o (3 o'
6-3333340s. Then by rule for reduction
•3166667 nearly,
20
^340s.
12 of fractions,
4-000008T. Ans. : 6s. 4d. £JS = 6s. 4d. Ans.
The latter method is preferable when perfect accuracj^ is
required.
VALUE OF A DECIMAL OF A CONCRETE QUANTITY. 47
Ex. X.
Find the value of —
1. X-375; £-98125; £-815625.
2. 416 of £3; -428571 of 6s. 5d. ; 8-571428 of 3s. Ofd.
3. -625 of 1 ton; -046875 of 3 tons; 4*39 of 1 cwt. 53 lbs.
4. 1-6671875 acres; -3475 rood; -076923 of 5 acres, 6 poles.
5. -2083 of 1 ream; -4583 quire; -383 of 3 reams,12 sheets.
G. -3078125 pipe; -490625 tun; -37125 bushel.
7. Express in grains -142857 of 4 lbs. avoirdupois, and
express the result in troy weight.
8. Express 10 oz. 3 dwt. 14 grs. as the decimal of 1 lb.
avoirdupois.
9. What is the sum of -6 of 1 guinea, -083 of 1 crown,
and -037 of £1. Os. 3d.
10. Find the value of
(-20416 X 7-5\
4 -^ -;2187r~") ^^^' "*" (*^^ ton- 769230 qrs.) x 26.
11. What is the value of
(^ hr-)
of 1 cwt. 35 lbs. 1
1
-I
12. Simplify / L-tA^ + LzJ^\ of 1 oz. 15 dwts.
' -^ \l .- -16 1 + -16/
To Reduce from one Denomination to the Decimal of
another Denomination of the same kind.
31. E/ULE. — Bring the given quantity to the fraction of
the proposed denomination, and reduce this fraction to a
decimal,
48 ARITHMETIC.
Ex. 1. — Reduce 3s. 3d. to the decimal of 8s. Ud.
3s. 3d. =26 three-halfpence,
8s. IJd. = 65 three-halfpence,
and the fraction |f = 2->^-i-3 = | = -4.
•^^ 6X13 6
Hence '4 is the decimal required.
Ex, 2, — Keduce 3 qrs. 21 lbs. to the decimal of 2 cwt. 3 qrs.
3 qrs. 21 lbs. = 105 lbs.
2 cwt. 3 qrs. = 308 lbs.
and the faction igf -^ i^-' = U = ^^^^ = -3409.
308 44X7 ** 11.
Hence -3409 is the decimal required.
Ex. 3. — Reduce 18s. 8Jd. to the decimal of £1.
The rule may be applied most concisely as follows : —
4
12
20
1-
•25
18-6875
•934375 .-. -934375 is the decimal required.
The farthing is first reduced to the decimal of a penny, and
the 8d. prefixed; then 8-25d. are reduced to the decimal of a
shilling, and the 18s. prefixed; lastly, 18 •6875s. are reduced
to the decimal of £1.
Ex. 4. — Reduce 3 qrs. 21 lbs. to the decimal of 1 ton.
28
(1
21
3-
4
3-75
20
•9375
•046875
Hence '046875 is the decimal required.
Ex. XI.
Reduce —
1. 12s. 6d., 10s. 7id., lis. Ijd., 18s. 6^ d., each to the
decimal of <£1.
2. 13s. Ojd., 10s. 8id., 9s. 6d., each to the decimal of
15s. 5jd.
3. 7s. 8fd. to the decimal of a guinea^ ancl 3s, 2^d. to the
decimal of a moidore, •
THE METRIC SYSTEM. 49
4. 1 qr. 7 lbs. to the decimal of 1 ton ; 3 cwt. 3 qrs. 20 lbs.
to the decimal of 3 tons.
5. 3.V lbs. to the decimal of lu cwt.; 14 oz. to the decimal
of 3 cwt. 2 qrs.
G. 20 grs. to the decimal of 1 lb. Troy; 3 dwts. IG grs.
to the decimal of 4 oz. 11 dwts.
7. 1 rood, 10 poles to the decimal of 1 acre; 3 roods, 15 l
square yards to the decimal of 5 acres.
8. f horn's to the decimal of 10 weeks; 7 h. 18' to the
decimal of 1 year (365 days).
9. Bring the sum of -r\ of 9 hours, | of 12i^ days, | of 7^
minutes, to the decimal of a week.
10. Express a pound troy as the decimal of a pound avoir-
dupois.
11. Reduce the sum of G lbs. 6f oz. avoirdupois, and
8 oz. 6 dwts. 1 6 grs. to the decimal of 1 ton.
12. Express 2*36 of 4s. ~ -518 of 9s. 2d. + 1'4583 of Gd. as
the decimal of £5.
CHAPTEE IV.
THE METRIC SYSTEM.
32. — The fundamental unit of the metric system is the
metre, A metre is the ten-millionth, or ^7^7 part of 90° of
the earth's meridian, and measures 39*3708 English inches.
In order to express multiples and sub-multiples of this unit,
and, indeed, of any unit in the metric system, we make
use of one or more of the following prefixes : —
Deka, . . 10 times. Deci, . . 10th.
Hecto, . 100 „ Centi, . . 100th.
Kilo, . . 1,000 „ Milli, . . 1,000th.
Myria, . 10,000 „
We will arrange these prefixes and the word nnit in order
according to their signification, thus —
Myria, KilOy HectOy Deka, Unity Deciy Centiy Milli
Now, as we j-e^d ihU line from left to right, it is evident
ARITHMETIC.
4
7
8
Metre.
Decim.
Centim.
Millim.
4
7
8
^hat the words liave a signification decreasing tenfold in value;
and as we read it from riglit to left, they have a signification
increasing tenfold in value.
It therefore follows that figures placed under the above
words have a local as well as an intrinsic value ; and further,
if when a figiire is wanting to complete the series, its place
be filled up by a cipher, it will be seen that the local value
corresponds exactly with the ordinary decimal notation.
Moreover, we have only to place a mark (in fact, a
decimal point) at the right of the figure standing under any
of the words of the above memorial line, and the given
quantity is at once expressed in the denomination corres-
ponding to that figure.
Thus, taking the metre as our unit :
Myriani. Kilom. Dekam. Metres. Decira. Millim.
3 2 5
Myriam. Kilom. Ilectom. Dekam.
- 3 2 5
= 3 '2054708 myriametres.
'^-- 32-054708 kilometres.
'^ 320-54708 hectometres.
^ 3205-4708 dekametres.
=^ 32054-708 metres.
:= 320547-08 decimetres.
= 3205470-8 centimetres.
= 32054708- or 32054708 mniimetres.
The following rule for expressing any quantity in terms of
any one multiple or sub-multiple of the unit, or of the unit
itself, is therefore evident : —
Rule. — Put ciphers in the place of any multiple, unit, or
sub-multiple absent in the series, and write the figures in close
order, as in the ordinary decimal notation. Then 2:)lace a
decimal point at the o'ight of the figure corresponding to the
denomination in "which we wish to express the given quantity.
Ex. 1. — Express 5 myriam. 3 hectom. 6 decim. as metres,
Filling up with ciphers the vacant spaces, we have —
Myriam. Kilom. Hectom. Dekam. Metres. Decim.
5 3 6
= 50300-6 metres.
THE METRIC SYSTEM. 51
Ex. 2. — Express 3 dekam, 4 deciiii. as myriametres.
Eilliiig lip with ciphers, we have —
Myriam. Kilom. Hectom. Dekam. Metres. Pecim, Centira. Millim.
00030400
= 0*0030400 myriametres = 0*00304 myriametres.
(The student will see that it was unnecessary here to
extend the series beyond decimetres.)
Ex. 3. — Expx-ess 13 metres 502 millimetres as kilometres.
We may write the given quantity thus —
Iviloni.
Hectom. Deckam. Metres. Decim. Centim.
Millim.
13 5
= 0*013502 kilometres.
2
"We have hitherto spoken only of the fundamental unit, and
its multiples and sub-multiples. We shall hereafter (Aii}. 35)
explain the priiacipal derived units, viz., the Gram, the Are,
the Stere, the Litre, and the Franc ; but as the multiples and
sub-multiples of these derived units bear the same relation
respectively to the corresponding derived unit, as in the case
of the fundamental unit, all the preceding remarks relative
to the multiples and sub-multiples of the fundamental unit
apply equally to those of the Gram, the Are, the Stere, the
Litre, and the Franc.
With regard to the units, multiples, and sub-multiples of
square and cubic measure, prdperly so called, it is necessary
to make a few remarks.
33. Square Measure. — The unit of square measure is the
square metre ; and since the series myriam., hilom., hectom.,
dekam., metre., &c., decrease in value tenfold when read from
left to right, and increases similarly when read from right
to left, it follows that the series square myriam., square
kilom., square hectom., square dekam., square metre, kc,
will decrease or increase 10'^ or 100-fold. Hence we see that
in square measure the multiples and sub-multiples increase
or decrease successively 100-fold, and, therefore, when quan-
tities in square measure are expressed by the ordinary
decimal notation, each multiple or sub-multiple must occupy
the place of two figures, a cipher being supplied when we
have less than ten of any multiple or sub-multiple, and two
ciphers when there is any blank in the series.
52 ARITHMETIC.
Sq. kilom. Sq. hcctom. Sq. dekam. Sq. metre. Sq. centim.
Ex. 10 3 15 3 5
Sq. kilom. Sq. hectom. Sq. dekam. Sq. metre. Sq. decim. Sq. centim.
= 10 03 15 03 00 05
^ 10-0315030005 square kilometres.
= 1003-15030005 „ hectometres.
= 100315-030005 „ dekametres.
= 10031503-0005 „ metres.
= 1003150300-05 „ decimetres.
- 100315030005 „ centimetres.
34. Cubic Measure. — The unit of cubic measure is the
cubic metre, and hence after the remarks in the last article,
since 10* = 1000, when quantities in cubic measure are ex-
pressed by the ordinary decimal notation, the units, mul-
tiples, and sub-multiples must respectively occupy the place
of three figures, ciphers being supplied to fill up blank
spaces when necessary.
Ex. 1. 325 cubic metres 51 cubic decimetres.
= 325 „ „ 051 „
= 325-051 cubic metres.
= 325051 cubic decimetres.
Ex. 2. — 25 cubic metres 3 cubic decim. 40 cubic centim.
= 25 cubic metres 003 cubic decim. 040 cubic centim,
= 25-003040 cubic metres,
= 25003-040 „ decimetres.
= 25003040 „ centimetres.
35. Derived Units. — The principal derived units of the
metric system are —
1 . The Gram, for measures of weight.
The gra7}i is the weight of a cubic centimetre of distilled
water at the temj)erature of 4° C
1 gram = 15*4323 grains, 1 gi-aiu = '0648 gram.
2. The Are, for land measure.
The are is a square who^e side measure^ 10 metres ; it is
therefore ec[ual tQ fv s^jiiarQ ilek^irietre, gr J 00 square metres,
THE METRIC SYSTEM.
63
1 are = 119-6033 square yards, 1 hectare = 2*471 acres,
1 acre = *405 hectare.
3. The Stare, for fire-wood.
The stere is equivalent to a cubic metre. It is therefore
the solidity of a cube whose edge measures 1 metre.
4. The Litre, for measures of capacity.
The litre is a capacity equal to the volume of a cube
whose edge measures a decimetre or 10 centimetres. It is
therefore equal to a cubic decimetre or 1,000 cubic centimetres,
and 1,000 litres are equivalent to a cubic metre.
1 litre = -2201 gallon, 1 gallon = 1*543 litres, 11 gallons
= 50 litres nearly.
5. The Franc, for money.
The franc is a coin weighing 5 grams, and composed of an
alloy, nine-tenths of which are silver and one-tenth copper.
The following table exhibits at a glance the fundamental
unit and the above derived units, together with the multiples
and sub-multiples at present in use : —
Table of the Metric System of Weights and Measures.
Multiples.
Units.
Sub-Multiples.
10,000
1,000
100
10
METRE,
Long Measure.
10th.
lOOth.
1,000th.
Myria.
Kilo.
Hecto.
Deka.
Deci.
Centi.
Milli.
Myria.
Kilo.
Hecto.
Deka.
GRAM,
AVeight.
Deci.
Centi.
MiUi.
Hecto.
ARE,
Land Measure.
Centi.
Deka.
STERE,
Solid Measure.
Deci.
Hecto.
Deka.
LITRE,
Capacity.
Deci.
Centi.
Milli.
FRANC,
Money.
Decime.
Centime.
A quintal = 100 kilog. = 2 cwt. nearly ; a millier, or tonneau de mer, = 10 quintals
= 20 cwt. nearly.
54 ' ARITHMETIC.
Ex. XII.
Examples upon the Multiples and Sub-Multiples of the
Units.
1. Express each of the following as metres —
15 myriam.; 20 kilom.; 1 hectom.; 27 dekam.; 25 decim.;
100 centim.; 345 centim. ; 5294 millim.
2. How many centimetres in the following —
46 myriam.; 30 kilom.; 295 hectom.; 1*5 dekam.; 3'95
metres, 295 millim.?
3. Express according to the metric table —
20 kilog. 29 dekag.; 18 kilog. 85 decig.; 123 hectog.
13 centig.; 12 dekag. 296 millig.; 153 centig. 3 millig;
3427 millig.
4. How many decigrams in the following —
16 kilog. 12 centig; 25 hectog. 10 grams; 39,645 millig.;
20 kilog. 35 dekag. 5 grams'?
5. Express in ares —
lOhectar. ; 296 centiar. ; 29 hectar. 3 centiar.; 3 hectar.
12 centiar.; 376,543 centiar.
6. How many square decimetres are there in the following —
100 sq. kilom.; 10 sq. hectom.; 5 sq. dekam.; 3498 sq.
met. ; 46 sq. met. 1
7. How many steres in the following quantities —
15 dekas. ; 394 decis. ; 9 dekas. 2 decis. ; 186 dekas. 3 decis.;
3764 decis. ; 4 decis. ?
8. Express as cubic metres —
10,000 cubic decim. ; 1,234,567 cubic centim. ; 372,456,126
cubic millim.; 1,000,000 cubic centim.; 639 cubic centim.;
293 cubic decim.
9. Express as litres —
3kilol. 2 hectol. 3 litres; 4 decil.; 2 kilol. 3 millil.;
76,384 millil.; 2934centil.; 830 dekal. ; 34,576 decil.
10. Express as litres, as dekalitres, and as centilitres —
18 kilol. 3 hectol. 4 decil. 5 centil. 3 millil.
11. How many francs in the following sums of money —
100 cent.; 736 dec; 24,645 cent.; 5 cent. 25 dec;
1695 cent.?
teE METRld SYSTEM. 55
1 2. How many centimes and how many decimes are expressed
by the following —
13 francs; 7 fr. 13 c; 12 fr. 3 dec. 5 c; 29 c; 3 fr. 2
dec; 18 fr. 4 c.]
Addition, Subtraction, Multiplication, and Division
in the Metric System.
36. Since (Art. 33) all quantities in the metrical system
may be expressed as one denomination by figures whose local
as well as intrinsic values follow the decimal system of nota-
tion, it is evident that when they are so expressed we may
add, subtract, multiply, or divide them exactly as ordinary
integers and decimals.
Ex. 1. — Add together 49 metres 36 centim. ; 3 kilom. 2
dekam. 3 decim. ; 2 hectom. 3 metres 25 centim.; and 13
dekam. 327 millim.
Metres. Millimetres.
49-36 49360
3020-3 3020300
203-25 or thus, by integers— 203250
130-327 130327
3403-237 3403237
Ans. : 3 kilom. 4 hectom. 3 metres, 2 decim. 3 centim. 7
millim.
Ex. 2.— Subtract 3 kilog. 2 dekag. 37 millig. from 10
myriag. 25 grams, 369 millig.
Kilograms. Milligrams.
100*025369 100025369
3-020037 or thus, by integers— 3020037
97-005332 97005332
Ans. : 9 myriag. 7 kilog. 5 grams, 3 decig. 3 centig. 2 millig.
Ex. 3. — Multiply 12 dekasteres, 3 steres, 5 decisteres by 23.
Steres. Decisteres.
123-5 1235
23 23
3705 or thus, by integers — 3705'
2470 2470
2840-5 steres* 28405 decisteres.
Ans, : 284 dekasteres, 5 decisteres*
56 ARITHMETIC.
Ex. 4. — Multiply 455,602 cubic centimetres by 36.
Cubic Metres. Cubic Centimetres.
0-455602 455602
36 36
2733612 or tbus, by integers— 2733612
1366806 1366806
16-401672 16401672
Ans. : 16 cub. met. 401 cub. decim. 672 cub. centim.
Ex. 5.— Divide 1369 kilol. 35 lit. 36 centil. by 72.
Kilolitres. Centilitres.
136903536
«{!
1^369-0353^ ^ (9
T52Tr5(J4 or thus, by integers—'' ^ 1 8
19-01438
15211504
1901438
Ans.: 19 kilol. 1 dekal. 4 lit. 3 decil. 8 centil, or 19 kilol.
12 lit. 38 centil.
Ex. 6. — How many times is 12 sq. dekam. 3 sq. met. 15
sq. decim. contained in 216 sq. dekam. 56 sq. met. 70 sq.
decim. 1
Sq. Dekam. Sq. Dekam. Sq. Decim. Sq. Decim.
12-0315)216-5670(18 120315)2165670(18
120315 . 120315
"932520 or thus, by integers- -"5^2520
962520 962520
Ans.: 18.
Ex. XIII.
1. Add together —
(1.) 3 metres, 2 decim. 4 centim.; 18 metres, 219 millim. ;
4 kilom. 2 hectom. 3 dekam. 14 centim.; 12 kilom. 36
metres.
(2.) 74006 hectom., 3216 kilom.; 12 myriam. 2167
metres.
(3.) 4 sq. met. 42 sq. decim.; 12 sq. dekam. 18 sq. decim.;
82 sq. met. 3250 sq. decim.; 3*271 sq. met.
(4.) 18 cub. met. 186 cub. decim.; 39-207365 cub. met.
30761 cub. centim.; 12 cub. met. 124-27 cub. decim.
THE METRIC SYSTEM. 57
(5.) 25 kilog. 235 grams; 3072 centig.; 13 kilog. 51
grams, 63 millig. 8132-07 decig.
(6.) 319 hectar. 4 ares, 51 centiar.; 93-712 hectar. 2375G-27
ares ; 6 hectar. 4 centiar.
(7.) 3 steres, 5 decis.; 209 steres, 4 decis. ; 25-76 steres,
13-027 dekas.
(8.) 51 kilol. 126 lit. 32 centil.; 123 lit. 3 centil. 15-02703
kilol.; 12 kilol. 3-27602 hectol.
(9.) 161 fr. 35 c; 32 fr. 4 c; 8276 c; 10-26 fr.; 16
decimes, 5 c.
2. Subtract—
(1.) 4 metres, 372 millim. from 16 hectom. 5-06 metres.
(2.) 30765 centim. from 12 kilom. 4 metres, 9 millim.
(3.) 3 sq. met. 89 sq. decim. from 1 sq. dekam. 7 sq. decim.
(4.) 12-0324 cub. met. from 18 cub. met. 29 cub. millim.
(5.) 39 grams, 65 millig. from 6 kilog. 12 gi-ams.
(6.) 8 hectar. 19*08 ar. from 32 hectar. 70 ar. 2 centiar.
(7.) 9 dekas. 6 decis. from 50 dekas. 2 decis.
(8.) 6 kilol. 6 millil. from 700 kilol. 3 lit. 3 centil.
(9.) 65 c. from 3 fr.; and 2 fr. 4 c. from 100 fr. 60 c.
3. Multiply—
(1.) 10 metres, 35 millim. by 7, 11, 13.
(2.) 18 kilom. 3-07 metres by 27, 48, 64.
(3.) 3-0625 sq. met. by 16, 18, 35.
(4.) 4 cub. met. 10 cub. decim. 5 cub. millim. by 19, 23, 26.
(5.) 7364 hectog. 9-31 decig. by 15, 25, 20.
(6.) 12 hectar. 3 centiar. by 30, 50, 40.
(7.) 416 steres, 2-9 decis. by 100, 150, 60.
(8.) 612305-06 litres by 12, 14, 16.
(9.) 39 fr. 10 c. by 75, 105, 135.
4. A merchant OAved 1500 fr., and he gave in payment 69
metres of cloth at 3 fr. 4 c. per metre, 48 metres of silk
at 8 fr. 65 c, 13*5 metres of calico at 75 c. How
m;iich does he still owe ?
58 AUITHMETia
5. Make out the following bill —
fr. c.
44 hectol. of oil, . . at 75 the litre.
G6 kilog. 125 gi\ of sugar, „ 1 25 „ kilog.
375 gr. of pepper, . . „ 3 5 „ „
128*75 hectog. of soap, . „ 1 75 „ „
562 gr, 5 decig, of coffee, . „ 30 „ hectog.
6. Divide —
(1.) 17 metres, 16 centim. by 11, 12, 13.
(2.) 41 kilom. 82 dekam. by 15, 16, 17.
(3.) 29 sq. met., 2740 sq. centim. by 14, 21, 35.
(4.) 376-38 cub. met. by 9, 27, 45.
(5.) 4 kilog. 14 dekag. 18 decig. by 22, 33, 55.
(6.) 8 hectar. 58 ares by 65, 60, 55.
(7.) 12 dekas. 1-2 decis. by 12, 13, 91.
(8.) 36 myiial. 4 kilol. 16 lit. 7 decil. by 9, 18, 27.
(9.) 7339 fr. 50 c. by 25, 30, 75.
'''. Find the price of —
(1.) A metre, when 2 met. 80 centim. cost 70 fr.
(2.) A square decim., when 30 sq. met. cost 450 fr. 30 c.
(3.) A cubic metre, when 15 cub. decim. cost 361 fr.
80 c.
(4.) A hectometre, when 3 kilom. 125 metres cost 10 fr.
25 c.
(5.) A kilog. of coffee, when 7 hectog. 50 grams cost
1 fr. 35 c.
(6.) A hectare^ when 4265 fr. 2*50 c. is the price of 149
ares, 25 centiar*
(7.) A stere, when 125 dekas. 4 decis. cost 20631 fr. 60 c.
(8.) A decilitre, when 47 dekal. 5 litres cost 570 francs.
(9.) A cub« centim., when 1 cubic metre cost 10,000 fr.
8. How many times is —
(1.) 1 kilom. 470 met. 38 centim. contained in 36759*50
metres'?
(2.) 12 sq. decim. 75 sq. centim. contained in lO sq* met*
20 sq. decim. ?
(3.) 13 sq. met. 25 sq. decim. contained in 318 Sq. dekam.?
THE METRIC SYSTEM. 69
(4.) 31 cub. met. 725 cub. decim. contained in 45684
cub. met. ?
(5.) 345 millig. contained in 165 kilog. 6 hectog.l
(6.) 275 centiar. contained in 396 hectar. ?
(7.) 7 stores, 2*5 decis. contained in 29 dekas.?
(8.) 4 kilog. 5 gi'ams, contained in 38 myriag. 4 kilog.
480 grams ]
(9.) 8 centimes contained in 10 francs'?
9. A merchant bought 95 litres of wine for 118 fr. 75 c, and
sold it at a loss of 10 c. per litre. What was the selling
price per kilolitre ]
10. To make 12 suits of clothes, it required 40 metres of
stuff 90 centim. wide. How much stuff will it take if
the width is 80 centim. ]
11. How many cubic decimetres of iron are there in a bar-
weighing 280 kilog. 368 grams, when one cubic centim.
weighs 7 grams 788 millig. 1
12. An iron wire, 126 metres long, is cut into pieces 3 centim.
2*5 millim. long. How many pieces are there?
Relation between the Metric Units and the Eng^lish
System of Weights and Measures.
37. We shall work a few examples to show how quantities
expressed in the metric system may be expressed in the
English system, send vice versa.
Ex. 1. — Reduce 10 kilom. 321 metres to English measure.
10 kilom. 321 metres = 10321 metres.
= (10321 X 1-094) yards.
- 10321 X 1.094 -1
- 17 60 mnes.
= 6 miles 731-174 yards.
Ex. 2. — Express 2 miles, 309 yards in the metric system.
2 miles, 309 yards =(2 x 1760 + 309) yards.
= 3829 yards = f;^^ metres.
= 3500 metres.
= 3 kilom. 500 metres.
60 ARITltMETId.
Ex. 3. — Eeduce 1 ton to kilograms, having given 1 gram
= 15*4323 grains.
1 ton = 20 X 112 X 7000 grains.
- 16-43 2 3 grams.
= 1016050-7507 grams nearly.
= 1016 kilog. 50 grams, 750*7 millig. nearly.
Ex. 4. — Express £13. 17s. 4|d. in the pound and mil
system.
(£1 = 10 florins, 1 florin = 10 cents, 1 cent = 10 mils.)
Keducing the given sum to the decimal of a pound, we
have —
£13. 17s. 4Jd. = £13*86875.
= £13. 8 fl. 6 cent. 8f mil.
Ex. XIY.
1. Express a mile in the metric system, having given that
a metre = 39*3708 inches.
2. An are contains 1076*43 square feet. Reduce 53 ares
25 centiares to English measure.
3. The area of a room is 22 sq. met. 26 square decim.
Express this in English measure (1 metre = 39*3708 inches).
4. A block of marble measures 3 feet, 3 inches in length,
2 feet, 6 inches in depth, and 3 feet, 9 inches in width. What
is the solid content expressed in cuh. centim. 1
5. In 1235 litres how many gallons, when 50 litres =11
gallons nearly 1
6. Supposing a franc to be equivalent to 9|d., reduce
£44. 13s. to francs.
7. Taking £1 sterling as equal to 25*22 francs, reduce
£2. 13s. 7ld. to francs.
8. In 1852 France reaped about 47850000 hcctol. of
wheat. Express this in gallons, assuming 1 gallon = 4*543
litres.
9. The ceiling of a room contains 83 sq. met. 53*96 sq.
decim. What will be the expense of painting it at lOd. a
square yard (1 metre = 1*094 yard)?
10. Find the cost of 2000 kilog. of sugar at fr. 50 c. per lb.
PROPOKTION. 61
11. In England the unit of work is the foot-pound, and in
the metric system it is the kilogram-metre. Reduce 62
metric units of work to English units, taking 1 gram =
15-4323 gi-ains, and 1 metre = 39*3708 inches.
12. The pressure of the atmosphere is 14| lbs. upon the
square inch. Find the pressure in kilograms upon the square
centimetre.
CHAPTER V.
PROPORTION.
38. Proportion is the equality of ratios.
Thus, since the ratio 6:8 = f = J = i^,
we have ratio 6:8 = ratio 15 : 20;
and we say that the numbers 6, 8, 15, 20 form a proportion.
We generally express the fact thus —
6 : 8 : : 15 : 20.
It is easy to find hf trial that the product of the extreme
terms is equal to the jjroduct of the means.
Thus, we have 6 x 20 = 8. x 15.
We may prove this property of the terms of a proportion
to hold generally as follows : —
Suppose we have given the proportion 12 : 21 : : 20 : 35.
It follows, from our definition above, that \'\ — |2> ^^^^
multiplying each of these fractions by the product of their
denominators, viz., by 21 x 35, we have
^f X 21 X 35 = fj X 35 X 21.
Now (Art. 8), ^f X 21 = V = 12, and |§ x 35 = \o =, oq,
and we hence have 12x35 = 20x21.
Now, we have not in our reasoning taken into account the
actual value of the terms of the given proportion ; and it is
therefore evident that a similar result will follow from every
proportion, and we may hence conclude generally : —
In every proportion the product of the extremes i$ f^nal (q
the product of the means.
62 ARITHMETIC.
39. Having given any three terms of a proportion, to find
tlie remaining one.
Since the product of the extremes is equal to the product
of the means, the following rule is evident : —
KuLE. — If the required term be a mean, divide the pro-
duct of the extremes by the other mean ; but if the required
term be an extreme, divide the product of the means by the
other extreme.
Ex. 1. — 28, 24, 30 are respectively the 1st, 3rd, and 4th
terms of a proportion, required the 2nd term.
"We have — 28 : required term : : 24 : 30
.-. required term - -^^-^ - ^-p - 35.
Ex. 2. — 10, 45, 16 are respectively the 1st, 2nd, and 3rd
terms, required the 4th term.
We have — 10 : 45 : : 16 : required term
.-. required term = ^^r^ = — i" = '^^'
Ex. 3. — 2 hours, 45 minutes, 8 men are respectively the
1st, 2nd, and 3rd terms, required the 4th term.
We must express (Art. 6) the 1st and 2nd terms in the
same denomination, and the pToportion will stand thus —
Min. Min. Men.
120 : 45 : : 8 : required term.
Now, the ratio of the first two terms is the same as the ratio
of the abstract numbers 120 and 45 ; and the 4th term must
be of the same denomination as the 3rd term, otherwise the
3rd and 4th terms could not form a ratio.
AVe have therefore — -
Eequired term = --{~~' men = ^^ men = 3 men.
Simple Proportion.
40. In Arithmetic we divide Proportion into Simple and
Compound. Simple Proportion is the equality of two simple
ratios, and therefore contains four simple terms ; and the usual
problem is to find the fourth term, having given the Jirst
three terms.
SIMPLE PROPORTION. Go
When we know the exact order of the given terms, the
fourth term is, of course (Art. 38), found thus —
Rule. — Multiply the 2nd and 3rd terms together, and
divide by the 1st,
The formal arrangement of the three given terms in their
proper order is called the statement; and the only difficulty,
therefore, in working a sum in Simple Proportion, or Single
Rule of Three, as it is called, consists in stating it.
We shall work a few examples to illustrate the mode of
doing this.
Ex. 1. — If 12 men earn £18, what will 15 men earn
under the same circumstances ]
We have hea^e two kinds of terms, men and earningSy and
whatever ratio any given number of men bears to any
second given number of men, it is evident that it must be
equal to the ratio of the earnings of the first lot of men to
the earnings of the second lot, and we may therefore write —
Men. Men.
12 : 15 =- £18 : 2nd earnings.
Men. Men.
or 12 : 15 :: £18 : 2nd earnings.
As the first two terms are of the same denomination, their
ratio is not altered by treating them as abstract quantities,
and the denomination of the 4th term must be the same as
that of the 3rd.
Hence we have —
Ans. : = £^-^^ = £^-^ = £22. 10s.
Ex. 2. — If 18 men do a piece of work in 25 days, in what
time will 20 men do it %
The two kinds of terms we have here to consider are men
and time. In doing work we know that the tlTne will dimin-
ish exactly as the number of men increases, and hence the ratio
of the second lot of men to i}iQ first lot will be equal to the
ratio of the given time to the time required. We therefore
have —
Men. Men. Days.
20 : 18 :: 25 : required time.
.-. Ans. : = ^-^TT^ "^^Y^ = .^fi ^^J^ = 22-5 days.
G4 ARITHMETIC.
We Lave reasoned out the above examples thus to show that
the working of problems in Rule of Three depends upon the
])rinciple of the equality of ratios. Practically, however,
we proceed as follows : —
Ex. 1. — If 12 men earn £18, what will 15 men earn under
the same circumstances?
We are required to find earnings^ and we therefore put
down for the 3rd term the given earnings, thus —
£
: :: 18
The question is with regard to 15 men instead of 12 men,
and we know their earnings must be greatet*. We therefore
place the greater of these terms in the 2nd place and the
other in the 1st, and the statement becomes —
Men. Men. £.
12 : 15 : 18 : required earnings.
.*. as before —
Ans. = £^-^77^ = £^^^~ - ^22. 10s.
Ex. 2. — If 18 men do a piece of work in 25 days, in what
time will 20 men do it ?
We are required to find time, and we place therefore the
given time, viz., 25 days, in the 3rd place.
Again, the question is with regard to 20 men instead of
18 men. Kow, we know that 20 men require less time than
18 men to do a piece of work, and we hence place the less of
these terms in the 2nd place. The statement then becomes —
Men. Men. Days.
20 : 18 : : 25 : required time.
,'. as before —
Ans. : = '-^~- days = '^-'- days = 22-5 days.
Ex. XY.
1. If 12 articles cost £15, what will 624 cost?
2. What is the price of 35 loaves, when 29 loaves cost
15s. ShdA
3. If I get 140 metres of cloth for 541 ft*. 7^ c., what must
I pay for 89 metres, 3 decim. ?
COMPOUND PROPORTION. 65
4. If 4 cubic metres of water run into a cistern in 18
minutes, in what time will it be full, supposing it to be 4
metres long, 6 metres, 25 centim. deep, and 35 decim. wide ]
5. If the carriage of a parcel for the first 50 miles be
Is. 3d., and if the rate be reduced by one-third for distances
beyond, how far can the parcel be carried for Is. Td.l
G. If a half-kilogram of sugar cost 1 fr. 10 c, what will be
the cost of 3 kilog. 625 gleams. 1
7. There are two pieces of the same kind of cloth, measur-
ing 43 yai-ds and 57 yards respectively, and the second costs
£1. 9s. 2d. more than the first. What is the cost of the
first?
8. A garrison of 720 men have provisions for 35 days, and
after 7 days 120 more men arrive. How long will the i:)rovi-
sions last ]
9. After paying 4d. in the pound income-tax a person has
£299. 18s. 4d. left. What was the amount of his original
income ]
10. Two clocks, one of which gains 3 minutes and the
other loses 5 minutes per day, are put right at noon on Mon-
day. What is the time by the second clock when the fii-st
indicates 4 p.m. on the following Thursday'?
11. When will the hands of a clock be exactly 30 minute
divisions apart between 2 and 3 o'clock ?
12. If I lend a friend £120 for 9 months, how long ought
he to lend me £270?
Compound Proportion.
41. Compound Proportion is an equality between ratios,
one of which at least is a ratio compounded of two or more
simple ratios.
Arithmetical questions depending on Compound Proportion
are generally said to belong to the Double Rule of Three ;
and the proportion consists of an equality between a ratio,
on the one hand, compounded of two or more simple ratios ;
and, on the other hand, a simple ratio, whose consequent is
required.
The following examples will illustrate the method of T^or}^-
ing questions in this rule : —
QQ ARITHMETIC.
Ex. 1.— If 12 horses eat 20 bushels of corn in 8 days, in
what time will 24 horses eat 16 bushels'^
24 horses : 12 horses ) o i . i ,.
20 bushels : 16 bushels / = = ^ ^^^^^ '• ''^l'^^'^'^ ^'"'^-
Explanation. — We are required to find time ; and so, as
in simple proportion, we put in the 3rd place the given
time, viz., 8 days.
Leaving, for the present, the quantity eaten out of considera-
tio7i, we know that 24 horses requii-e less time to consume a
given quantity of food than 12 horses do; we therefore place
the less of these two terms in the 2nd place, and the other
in the 1st place.
(The statement up to this point is 24 horses : 12 horses : : 8 days
: required time, and we might obtain 4 days as an answer, irrespective
of the quantity eaten. We might now place this answer in the 3rd
term of another simple proportion, and take the quantity eaten into
consideration, irrespective of the number of horses, thus getting an
answer depending both upon the number of horses and the quantity
eaten. It is more convenient, however, to proceed thus :)
Again, taking into consideration the quantity eaten, and
leaving out of consideration the other given pair of terms, we
see that less time is required to eat 16 bushels than to eat 20
bushels. We, therefore, jnit the less term in the 2nd place,
and the other in the 1st.
Now, treating the terms of the ratios which occupy the
1st and 2nd places as abstract quantities, and compound-
ing them, we have :
24x20:12xl6::8 days : required time.
.-. required time = ^ri^T^— days = 3-2 days.
Ex. 2. — How much bread can I get for 9d. when wheat
is at 18s. a bushel, if the foui-penny loaf weigh 3 lbs. when
wheat is at 20s. a bushel i
Proceeding as in Example 1, we have
18s * 20s I •• ^ ^^^* * weight required.
Or, 4xl8:9x20::3 lbs. : weight required.
,-. weight requii^ed = --^i^r- ^^s. = 7-5 lbs.
COMPOUND TROPORTION 67
Ex. XVI.
1. If 15 men can build a wall 81 feet long in 18 days, liow
many men can build 135 feet of the same kind of wall in 30
days ]
2. In 4 days, 18 workmen can dig a ditch 162 yards long,
7 feet wide, and 12 feet deep. What must be the depth of a
ditch which 45 workmen can dig in 7 days, supposing it to
be 387 yards long and 5 feet wide 1
3. A traveller, going 15 hours a day, walks 1500 kilo-
metres in 20 days. How far will he go in 30 days, walking
12 hours a day with the same velocity? Express your
answer in English miles.
4. Two men are partners ; one puts in a capital of .£800,
and receives as 6 months* profit £120. What is the capital
of the other, who receives £3375 as 9 months' profit 1
5. Two tourists having spent £1. 16s. 8d. in 2^ days, meet
three others with whom they continue their tour, and they
spend while together £21. Is. 8d., at the same rate per day.
Required how long they were in company.
6. If 16 men and 10 boys do a piece of work in 10 days,
in how many days would 8 men and 18 boys do a piece 7
times as great, supposing the work of 5 boys equal that of
2 men?
7. Supposing the rate of carriage to be diminished one-thii-d
after the first 50 miles, find the cost of carrying 16 cwt. for
40 miles, when 1 2 cwt. can be carried 1 00 miles for 4s. 2d.
8. A cistern is 8 metres, 4 decim. long, 1 metre, 8 centim.
wide, and 275 centim. deep. Find the depth of another cis-
tern of equal capacity whose length is 7 metres, 2 decim., and
width 11 decim.
9. Persons whose incomes are less than £300 per annum
are taxed upon £80 less than theii* income. Supposing 3
persons, having equal incomes, to pay £7 in the aggregate, at
4d. in the pound, find the total tax upon 14 persons, each
havmg incomes 3 times as great.
10. If 12 horses eat 10 acres of gi'ass in 16 weeks, and 18
horses eat 10 acres in 8 weeks, how many hoi'ses would eat
40 acres in 6 weeks, the grass being supposed to grow uni-
formly 1
68 ARITHMETIC.
11. A boat, propelled by 8 oars, whicli take 28 strokes per
minute, goes at the rate of 9 J miles per hour. Find the rate
of a boat propelled by 6 oars, which take 36 strokes per
minute, the work done by each stroke of the latter being
one-sixth less than that by each stroke of the former.
12. If 4 men and 1 women can do a piece of work in 8
days, which 1 2 women and 20 children can do in 4 days, in
what time will 6 men, 18 women and 5 children do a work
three times as great ]
CHAPTEK VI
APPLICATION TO ORDINARY QUESTIONS OP COMMERCE
AND TRADE.
Interest.
42. Interest is the money paid for the use of money.
The Principal is the money lent, and the Amount is the
sum of the interest and j)rincipal.
The Rate of interest is the money paid for a given sum for
a given time. £100 is in practice the given sum, and one
year the given time. Thus, if <£4 be paid for the use of £100
for one year, the rate of interest is £4 per cent. ])ev annum,
or as we generally say, 4 per cent.
Simple Interest is interest calculated on the original
principal only.
Compound Interest is the interest which arises from
adding the interest for each year to the principal of that
year, and calculating interest for the next year upon the
amount so obtained. ,
Simple Interest.
43. BuLE. — Multiply the prii:icipal by the rate per cent,
and by the number of years ; then divide the product by
100, and the quotient will be the simple interest,
SniPLE INTEREST. 69
Ex. 1. — Find the simple interest of £420 for 3 years at
5 per cent.
£
420
5
2 IOC)
3
£63-00 Ans.: £63.
Reason for this process —
£100 gains £5 in 1 year, and the question is to find the
gain upon £420 in 3 years.
Hence, proceeding as in Double Rule of Three, we have —
100 X 1 : 420 X 3 :: £5 : interest required.
.*. interest required = £ 420 x _j_x_3^ which is exactly as
stated in the rule.
Ex. 2.— Find the simple interest of £352. Is. 8d. from
March 16 to August 21, 1873, at 4 per cent.
The following process will be easily underetood : —
£ s. d,
352 1 8
4
£14-08 6 8
20
1-G(3S.
12_^
8"00d.
.*. £14. Is. 8d. is the interest for 1 year^
Now, from March 16 to August 21 are 158 days'; hencd
we have —
365 days : 158 days :: £14. Is. 8d. : Interest requii-ed.
.*. interest required = £6. Is. ll-j^jd.
Ex. XYll. )
f
Find the simple interest of— . \
1. £350 for 4 years at 5 per cent, !
2. £295. 2s. Id. for 3^ years at 4 per cent.
70 ARITHMETIC.
3. £375. 8s. 4d. for 2 J years at 4J^ per cent.
4. £160 from Feb. 1 to June 12, 1872, at 7|- percent,
5. £48 for 7 months at 1^ per cent, per month. .
6. £219. 4s. 2d. for 6 years at If per cent.
In the six following examples, understand simple interest.
7. At what rate per cent, will £129. 8s. 4d. gain £6. 3s.
5ld. in 2|- years 1
8. A certain sum amounts in 3 years at 7|- per cent, to
£289. 16s. 3|d. ; find the ori^ial sum.
9. In what time will £175. 6s. 3d. amount to £192. 7s.
10|-d. at 2| percent?
10. What sum will amount in 2 years 9 months at 4 per
cent, to £427. 7s?
11. If £320 gain £9 in 13 months, in what time will
£480 gain £6 at the same rate?
12. Find the interest of £29. 7s. 5d. for 6 months at 5^%
per cent.
Compound Interest.
44. Utile 1. — Find the interest for one year as in Simple
Interest, and add it to the principal ; then find the interest
for one year upon this amount reckoned as pfincijyal for the
second year, and add it to the second year's principal, and so
on. Subtract the original principal from the amount so
obtained for the given number of years, and the result will
be the compound interest required.
Rule. 2. — Divide the given rate per cent, by 100, putting
the result in a decimal form, and place U7iity before the
decimal point. Kaise the number thus obtained to a power
corresponding to the given number of years, multiply the
principal by the result, and we get the amount for the given
number of years.
Thus, supposing 5 to be the rate per cent., we have,
dividing by 100 and placing unity before the result, the
number 1-05. Then, if the given number of years be 4,
and £162 the principal, we have, according to rule, amount
= £162 X (1-05)^
The first rule requii^es no explanation; the second mle
may be explained thus : —
COMPOUND INTEREST. 71
Ex. — Find the compound interest of £360 for 3 years at
4 per cent.
Now, interest for .£100 for 1 year = £4,
£1 „ =£-04;
lience, amount of £1 „ = £1 + £-04 = £1-04.
We thus see that the amount of £1 for 1 year at 4 per cent,
is 1*04 times the original sum. It therefore follows that —
Amount of the £1-04 for 1 year = 1*04 times £1*04,
.-. amount of £1 „ 2 years =£(1-04 x 1*04)
- £(1-04)2;
and so, amount of £1 „ 3 years = 1*04 times £(1 -04)^
:^£(^04)^
hence, amount of £360 ,, „ = 360 times £(1-04)'
= £360 X (1-04)^
The compound interest is then found by subtracting from
this the original principal.
Ex. 1. — Find the compound interest of £570 for 3 years
at 5 per centi
(We shall work this by Rule 1, and for convenience shall
keep our quantities in a decimal form.)
£
570
5r
£28*50 = interest for fii-st year.
570
£598*5 = principal for second year.
5
£29-925 - interest
598*5
£628*425 = principal for third year:
5
£31 '42125 = interest
628-425
£659*84625 = amount at end of third year, ) therefore
570 ~ original principal ; J subtracting,
£8 9-84625 = compound interest for 3 years.
72 ARITHlilETia
Ex. 2. — Find the compound interest o( £i'21. l'2s. Gd. for
■i years at 3 per cent.
Now, £327. 12s. 6d. ^ £327-625.
Hence, by Rule 2 —
Amount -£327-625 x (1-03)^ :=: £368. 14s. lOfd. nearly.
Ex. 3. — What sum of money, if i)ut out for 2 years at 4
per cent., will amount to £324. 9s. 7jd., compound interest
being reckoned ?
By Rule 2, the princijial may be found by dividing the
amount by (1 -04)-.
Now, given amount = £324. 9s. 74d. == £324*48.
Hence principal required = £324-48 -f (1-04)- = £300..
Ex. XVIII.
Find the compound interest of
1. £284 for 2 years at 4 per cent.
2. £312. 12s. 7|d. for 3 years at 5 per cent.
3. £283. 10s. for 2 years at 31 per cent.
4. £605. 12s. 6d. for 4 years at 4 per cent.
5. What is the difference between the simple and com-
pound interest of £150 for 2 yeai^ at 6 per cent.?
6. Find the amount of £381. 1 florin 3 cents 5 mils for
3 years at 5 2)er cent.
(£1 = 10 florins, 1 florin = 10 cents, 1 cent = 10 mils,)
7. Find the amount of £250 for 2 years at 4 j)er cent, per
annum, interest being payable half-yearly.
8. What sum will amount in 3 years at i^ ^ler cent, com-
l)ound interest to £200 ]
9. A town has 200,000 inhabitants, and it increases at the
rate of 5 per cent, per annum ; And the number of inhabi-
tants at the end of 3 years.
10. Find the difference in amount of £350 for 3 years at
4 per cent, simple interest, and £420 for 2 years at 5 per
cent, compound interest.
11. HoAV much would a person who lays by £50 a year at
5 i)er cent, compound interest, draw out at the end of 4 years]
12. A person expects to receive £450 in 3 years ; what
present sum is equivalent to this, reckoning compound inter-
est at 4 per cent. ]
DISCOUKT. 73
Discount.
45. When money is paid before it is due, the payee may,
of course, put out the money at interest for the rest of the
term, and thereby increase it. It therefore follows that the
amount which ought to be paid for the discharge of an
account before its proper time should be such a sum that, if
put out at interest for the remainder of the term, will just
amount to the original sum in question.
Thus .£102. 10s. (interest being reckoned at 5 per cent,
per annum) payable 6 months hence, would be fully dis-
charged by paying £100 at once. For £100 in 6 months at
5 per cent, per annum would amount to £102. 10s. Hence,
the payee ought to remit £2. 10s. from the full account. The
amount remitted is called discount.
It will be seen, therefore, that the discount on £102. 10s.
due 6 months hence at 5 per cent, is £2. 10s.
Bankers, however, are in the habit of charging interest
instead of discount The banker's discount, therefore, on
£100 due 6 months hence at 5 per cent, is £2. 10s.
Hence, the (7^(6 discount on £102. 10s. due 6 months at 5
per cent, is the same as the hanker s discount on £100 under
the same circumstances ; and bankers' discount on any given
sum is in excess of the true discount.
Tradesmen's bills are legally due three days after the term
for which they are drawn is completed. This extension of
time is called three days of grace. When a bill falls due on
a Sunday, it is usual in England to meet it on the previous
Saturday.
Ex. 1. — Find the difference between the true discount and
the banker's discount on£306 due 4 months hence at 6 per cent.
Now, £100 would in 4 months gain \ of £6, or £2.
Hence, the time discount on £102 due four months hence at
G per cent, is £2, and therefore we have —
£102 : £30G :: £2 : true discount required.
Hence, true discount = ^^^VP "= '^^•
Again, proceeding according to the rule for simple
interest —
-r^ , , ,. £306 xGxil- ^ , «^^,.,
Banker's discount = j^^j r= £f JS.= £6. 2s. 4jd.
74 Arithmetic.
Hence, the excess of the banker's discount over the true
discount is 2s. 4|d.
Ex. 2. — A bill of .£350 drawn on March 15, at 6 months, is
cashed on May 20, 1872 ; what is the banker's discount at 6
per cent. ?
The bill is legally due on Sept 18, and from May 20 to
Sept. 18 are 121 days.
Now, the interest on <£350 for x year at 6 per cent, is easily
found to be £21. Hence,
365 days : 121 days :: £21 . banker's discount requii^ed ;
and /. banker's discount = £^,^!-^ = £6. 19s. UH.
Percentages.
46. There are many questions which relate to ordinary
commercial transactions which may be worked exactly as
if we had to find the simple interest for one year — e.g., ques-
tions in commission, brokerage, insurance^ &c.
Commission is a sum of money charged, by ^n agent for
buying or selling goods, at a certain rate per cent; upon the
value of the goods;
Brokeragei is similar to commission, but it is charged upoii
money transactions instead of upon the sale of goods.
Insurance is a sum charged per cent; Upon the value of
property, the said value being paid to the insured in case of
loss from causes as per agreement;
Ex. 1.— Find the brokerage on £625; 5s; at 4 J per cent;
£ s. d.
625 5
2501
312 12
d
6
28-13 12
2-72s.
12
6
8-70 Ans.: £28. 2s. Bx'^d.
PERCENTAGES. 75
Ex. 2. — The rate of insurance is 4 J per cent., and the
value of some pi-operty insured is worth £766. What will
be the annual payment, so that, in case of fire, the owner
may receive back his premium, as well as the value of his
property 1
If, instead of paying £4. 5s. as insurance on every £100,
he pays £4. 5s. upon every (£100 - £4. 5s.) or npon every
£95. 15s. ; then, in case of fire, he will receive £100 for a
damage of £95. 15s., and thus have the value of his property
and the lamount of his premium.
The problem is, therefore —
If £4. 5s. is the premium on £95. 15s., what is the pre*
mium on £766 ?
Hence —
£95. 15s. 1 £766 : : £4. 5s. : premitun required;
And) therefore^ premium = £34.
Ex. XIX.
1. Find the banker's discount on £4 12, dud months
hence, at 6 per cent.
2. By how much ddes thes banker's discbunt bn £100^ due
3 months hence, at 5 per cent., exceed the true 1
3. What discount would be charged upon a bill drawii for
£320, on April 15, at 4 months, and presented for payment
on June 3 (discount at 7 per cent.) 1
4:i Find the discount on a bill drawn bn Aug. 3 for £200,
at 6 months, and cashed on Sept. 10, discount being reckoned
at 6 per cent.
5. A man buys gpbds for £250, being allowed G months
credit, and he immediately feells them for the same amount,
allowing 3 months credit. What does he gain by the trans-
action, interest being reckoned at 5 per ceht. ?
6. Find the brokerage on £352. 17s. 6d. at 3 per ceni.
7. What is the brokerage on £4500 at -J per cent, i
8. What is the commission on the sale of goods to the
amount of £850 at 5 per cent.?
9. A person insures for £1050 at 3f per cent. What is
his annual premium 1
ARITHMETIC.
10. What "Will be the annual premium on pi*Operty "WoHh
£965, so that the insured may obtain his premium back
again with the vakie of his property, in case of loss — insurance
being at 3 J per cent. I
11. The brokerage on a certain sum at J per cent, is
£5. 10s. 7id. Find the sum.
12. Together with a commission of 4 per cent., goods cost
a person £339. 8s. 5d. Find the cost price to the agent.
Stocks and Shares.
47. "When a large amoimt of capital is to be raised, a
company is generally formed, which raises the money by the
issue of shares. We will suppose a person to hold a £100
share ; he will then be entitled to such a part of the profits
of the company as £100 is of the whole capital. If there be
a great demand for these shares, persons holding them may
dispose of them for more than the nominal value, say for
£106 ; whereas, if they are very little in demand, the seller
may be glad to sell at, perhaps, £70. In the first instance,
we should say that the shares were at 106, or at 6 preiniuni;
and in the second instance, that the shares were at 70, or at
30 discount. If the selling price of the shares is £100, they
are said to be at ^j>ar.
So, when we read that the Three per Cent. Consols are
quoted at 96|-, it means that an acknowledgment of indebted-
ness on the part of the Government to the amount of £100,
bearing interest at 3 per cent, per annum, may be bought for
£96f.
The buying and selling of stocks and shares is carried on
by brokers, who charge a percentage from ^ to -|, sometimes
upon the nominal value of the stock, but mostly upon the
actual cash vahie. When brokerage is to be taken into ac-
count in any example it will be mentioned, and it will
be estimated by the first method, unless specified.
Ex. 1.— What is the value of £1000 stock at 89tV per
cent. ]
No. of cents, stock = ^7 = 10
.-. value required - £89-r\ x 10 - £893. 2s. 6d.
STOCKS AND SHARES. 77
Ex. 2.— What would be the cost of £1180 stock at 153^
per cent., including brokerage at i per cent.?
No. of cents, stock = YcfcT = TVy
and total cost of each cent. = £(153^ + i) = £153^.
Hence, required cost - £153i x -^^f = £1811. 6s.
Ex. 3. — What is the annual income arising from investing
£6510 in tlie Four j^er Cents, at 93 ]
Here, price of £100 stock - £93,
and .-. No. of cents, stock for £6510 = ^^- = 70.
Hence, annual income = £4 x 70 = £280.
Ex. XX.
1. Find the cost of £750 stock at 92 J per cent.
2. I sell out £325 Three per Cent. Consols at 94. What
do I get after allowing the broker i per cent, upon the cash
he receives for the sale 1
3. Invest £5065. 9s. 9d. in the Three per Cent. Consols
at 91f.
4. What would be the cost of £413. Is. 9d. reduced Three
per Cents, at 92|, including a brokerage of \ per cent, upon
the cost to the broker ]
5. What would be the proceeds of the sale of £6228 India
Five per Cent, stock at 111^, deducting J jier cent, upon the
selling price for brokerage 1
6. If I sell £8160 Spanish Three per Cent. Bonds at 31
per cent, and invest the proceeds, less J- per cent, brokerage,
in Indian Railway Five per Cent. Stock at 107, what will
be the dilfereiice in my annual income ]
7. What is the value sterling of $4000 American Bonds
at 93f per cent. ($ = 4s. 6d.)?
8. Wliat is the value of |37,000 Unitea States Bonds at
93-j-v per cent. "J
78 ARITHMETIC.
9. Invest £954. Os. 7cl. in India Five per Cent. Stock at
11 Of, allowing i per cent, brokerage.
10. "What is the sterling value of 5000 francs Italian
Bonds at 67 per cent. (Exchange 25 fr.)?
11. Find the sterling value of 6000 guilders Dutch 2 k per
Cent. Bonds at 56^ (Exchange 12 guilders).
12. Invest £1025 in French Eentes at 51^, allowing i per
cent, brokerage (Exchange 25 fr.). ^
Annuities.
48. Annuities are annual payments, the first payment
being due at the end of a year. When an annuity is left
untouched for a number of years, its amount is properly ob-
tained by allowing compound interest.
49. To find the amount of an annuity of a given sum for
a given time, at a given rate per cent.
Let us suppose the annuity to be £1, the time 4 years,
and 5 the rate per cent. The first payment will not be due
till the end of a year ; so that at the end of 4 years it will
have been accumulating for 3 years at compound interest;
so the next payment, not being due till the end of the second
year, will have been accumulating for 2 years at compound
interest ; and so on, the last payment being made when due.
Hence, the amount of an annuity of £1, for 4 years at 5
per cent., will be (reversing the above order) as follows : —
£1 + amount of £1 for 1 year + amount of £1 for 2
years + amount of £1 for 3 years (compound interest being
reckoned).
And it is evident that for any other annuity we may follow
the same method, and at the end multiply the sum by the
number of £ in the annuity.
Ex. — Find the amount of an annuity of £300 for 4 years
at 5 per cent.
We shall find the respective amoimts by Rule 2, Art, 44,
Thus—
ANNUITIES.
£1*05 = amount for 1 year of £1 at 5 per cent.
1-05
525
105
iiM025 = „ „ 2 years
1-05
55125
11025_
1-157625 = „ „ 3 years „ „
Hence, amount of annuity of £1 for 4 years —
= £1 + £1-05 + £M025 + £M57625 = £4-310125.
.-. amount of given annuity = £4-310125 x 300 = £1293.
Os. 9d.
50. To find the present value of an annuity to continue
for a given time.
By the present value of an annuity to continue a given
number of years is meant such a lump sum which, paid down
at once, would, by accumulating at compound interest for the
same time, amount to just the same sum as the annuity itself
if it were allowed to accumulate. In the absence of alge-
braical symbols, we shall best illustrate the method of finding
such a lump sum by an example.
Ex. — Find the present value of an annuity of £50 for 3
years at 4 per cent.
As in Art. 49, we find the amount of the annuity —
= {£1 + £1-04 + £(l-04)2[ X 50 = £156-08.
Now, whatever be the present value required, we know
that its amount in 3 years at 4 per cent, compound
interest is found (Aii;. 44) by multiplying it by (1*04)^ or
M24864.
It, therefore, follows that if we know this amount before-
liand we can find the present value by dividing it by
1-124864.
Hence, present value = £156-08 4- 1-124864 = £138-755
nearly.
51. To find the present value of an annuity to continue
for ever.
80 ARITHMETIC.
It is evident that the sum we require is ene which, put
out at interest, will annually produce a sum equal to that of
the annuity itself. The problem then is simply this —
Plaving given the interest for 1 year of a cei-tain sum,
and the rate per cent., to find the principal.
We have therefore the following rule : —
KuLE. — Divide the given annuity by the rate per cen^.,
and multiply by 100, and the result is the present value.
Ex. — How much must a gentleman invest at 5 per cent, in
order to endow a charity with £60 a year.
Pi*esent value of the annuity of £60 to continue for ever —
There are many other questions connected with annuities
which are, however, best left till the student has a knowledge
of Logarithms.
Ex. XXI.
Find the amount of an annuity of — -
1. £120 for 3 years at 4 per cent.
2. £250 for 4 years at 4^ per cent.
3. £321 for 5 years at 5 per cent.
4. What is the present value of an annuity of £80, to
continvie for six years, at 6 per cent. ]
5. A person who, according to the tables of mortality, is
likely to live 10 years, wishes to insure an annual payment
of £40 during life. What sum must he pay down, reckon-
ing interest at 5 per cent. 1 (Give the result to four places
of decimals.)
G. A house produces a clear rental of £30. How many
years' purchase is it worth, interest being reckoned at 5 per
cent. 1
7. A gentleman invested a sum of money in the Three
per Cent. Consols, in order that an annual payment of 7s. Gd.
a year might be made in bread for ever. What sum did he
invest 1
8. Find the present value of a pension of £120 a-year,
payable half -y earl tj for 5 years, interest being at the rate
pf 5 per cent, per annum.
PROFIT AND LOSS. 81
9. A house, whicli ordinarily lets for £80 a-year, is leased
for a term of four years, at a rent of £20, a certain sum
being jmid in addition at the time of letting. Find this
latter amount.
10. What is the present value of a freehold which pro-
duces a clear rental of £50, but which cannot be entered upon
for two years, reckoning interest at 5 per cent. 1
11. Find the annuity which in four years, at 4 per cent.,
will amount to £100.
12. A corporation borrows a sum of £3000 at 4 per cent.
What annual payment will clear off the debt in ten years 1
(Give the result correct to four places of decimals.)
Profit and Loss.
52. All questions involving the loss or gain per cent, by
any transaction belong to this rule, and may be generally
worked by Proportion.
Ex. 1. — A man buys goods at 5s. and sells them at 5s. 8d.
Find his gain per cent.
The actual gain upon 5s. is 8d., and we are required to
find the gain upon £100.
Now 5s. : £100 : : 8d. : gain upon £100,
... gain upon £100 = M'ffff - £13^,
or, required gain per cent = 13^.
Ex. 2. By selling goods at 6s. 3d. there is a gain of 25'
per cent. What will be the selling price to gain 10 per
cent.]
Now, selling price of goods which cost £100, so as to gain
25 per cent, is £125, and that to gain 10 per cent, is £110.
Hence £125 : £110 : : 6s. 3d. : selling price required ;
from which, selling price required = 5s. 6d.
Ex. 3. — Find the cost price when articles sold at Is. 9d.
entail a loss of 12^ per cent.
Now, articles which cost £100 when sold at a loss of 12 J-
per cent, must sell for £87^.
Hence £87i- : Is. 9d. :: £100 : cost price required j
from which, cost price = 2s,
5 F
82 ARITHMETIC,
Ex. XXII.
1. Find tlie cost price of goods which are sold at a loss of
10 per cent, for 4s. lOid.
2. Goods which are sold for 7s. lid. entail a loss of 5 per
cent. What should be the price to gain 30 per cent. 1
3. A tradesman reduces his goods 7 J- per cent. "What
was the original price of an article which now fetches
^1. 7s. MJ
4. In what proportion must tea at 4s. 2d. be mixed with
tea at 6s. a pound, so that a grocer may sell the mixture at
5s. 6d. and gain by the sale 10 per cent.?
5. A quantity of silk, after paying a duty of 12^ per cent.,
cost £54. Find the original cost price.
6. An innkeeper buys 37^ gallons of brandy at 14s. a
gallon, and adds to it sufficient water to enable him to sell it
at the same price and gain 12 per cent. How much water
does he add ?
7. By selling goods at 8s. 2d. a tradesman gains 16| per
cent. What will be the gain or loss per cent, by selling at
6s. lid.
8. A company has a capital of £750,000, and the working
expenses for the year have been £42,123. 12s. 6d. What
must have been the gross receipts in order that the share-
holders may receive a dividend of 4 per cent. ?
9. If stock which is bought at 91^ is immediately sold at
9 If, what is the gain per cent."?
10. A person buys goods at 6 months' credit and sells
them for cash at the nominal cost price immediately. What
is his gain per cent.? (Interest 5 per cent.)
11. Goods are marked at a ready-money price and a credit
price allowing 12 months. The credit price is £4. 9s. 3d.,
what is the ready-money price ?
12. Goods are now being sold at 10 per cent. loss. ! How'
much per cent, must be put upon the selling price in order
that they may be sold at 20 per cent, gain ?
ESTIMATES.
83
Square Root and Cube Boot.
53, To avoid unnecessary repetition, the student is referred
to the articles on Involution, Algebra, stage I., where the
arithmetical principles and methods are explained.
Estimates,
54. The following specimens will give the student an idea
of what he may expect to meet with under the head of Esti-
mates, It is usual, in ordinary transactions, to use certain
abbx^eviations ; as cuh. for cubic measure, sup, for superficial
measure, run. for running or lineal measure. Builders, too,
are in the habit of calling twelfths of a foot — whether it be
cubical, superficial, or lineal measure — by the name of inches.
The names yards, feet, inches, are often written thus :
yds., ', ",
Ex. 1. — Digger, Bricklayer, and Mason.
Yds.
25
232
5
IG
G
14
9
33
12
5
19
Cub.
Sup.
ft
Run.
No.
Run
No.
Diggings in trenches, filling-, wheeling, and
carting away,
9" reduced common brickwork in mortar,
pointed on both sides,
^y trimmer arches to hearths,
Best blue brick on edge, paving in cement,
6" blue and red quarries do.,
12" round blue coping bricks in cement, . . .
Best red brick flat steps do..
Extra only to splayed brick angles to doors
and windows,
Do. to plinth in cement,
Do. to sailing course to chimney,
Do. to cornice to eaves of best red bricks,
three courses,
2 core chimney flues,
1 set sink,
•i do. stoves,
1 do. copper,
12 do. ornamental air gratings,
Gable coping of hard stone 13" X 4i", twice
splayed,
2 J rubbed hearth and back hearth,
Solid York step 14" x 6", tooled,
2 knee stones to gables 18" x 14" x 10", ea.
1 apex stone do., 16" x 14" x 12",.
1 York stone sink 4' 0" X 1' 9", with rounded
corners and hole cut for waste-pipe,
2 stone chimney-pieces with 7" X IJ", cham-
fered jambs, mantel, and shelf, ea.
Total,
1/
4/2
2/3
4/6
3/10
2/4
1/2
0/2i
0/3
0/2
0/5^
1/6
4/
5/3
5/
0/6
1/9
1/2
2/6
7/6
8/
10/
24/
84
ARITHMETIC,
Ex. 2. — Carpenter and Joiner. (A square = 100 sq. ft.)
Sq.
21
Cub.
Sup.
Run.
Sup.
Run.
No.
Sup.
Run.
Sup.
Run.
Sup.
Run,
No.
Fir framed-in roof timbers,
Do. do., floors,
Centreing to trimmer hearths,
Do. to 3 openings 5 0" wide, with segment
heads in 1^ brick wall ea.
Do, to 4 do. 3' 0" wide do., ea.
Labour in planing roof and floor-timbers, . .
Do. in stop chamfering edges of do.,
7" X ly ridge board,
3" X 1^" chamfered fillet to eaves,
Inch clean red deal batten for boarded floors,
wrought,
7" X 1" torus skirting, plugged,
2 labour to mitred margins to hearths, ...
16 mitres to skirting, ea.
Inch deal treads and risers, ploughed,
tongued, and screwed on 3 — 7" x 2^-"
carriages
1^" wall-string housed, for treads and
risers,
1^" close-string do., and sunk and beaded,..
3'' X 2^" rounded oak hand-rail, French-
polished,
9" X 1" beaded fascia,
2" — 6 panel doors, bead, flush, and square,
1^" do. do.,
4|" X 3" rebated and beaded frame,
4|" X 2" do. do.,
3" X 1^" moulding, mitred,
2" ovalo-moulded sashes, double hung, in
deal-cased frames, with oak sunk and
weathered sills,
1|" do. do.,
^" mitred bead,
5^" X f" bead lining,
11 inch-beaded centre boards,
Total,
2/9
2/8
0/4
2/
1/6
0/Oi
0/Oi
0/3
0/3
0/9J
0/2
0/9
0/8
0/9
2/
0/4
1/3
1/1
0/6
0/4^
0/3
MISCELLANEOUS EXAMPLES. 86
MISCELLANEOUS EXAMPLES.
{Selected from University and other Examination Papers.)
1. Show by an easy example tliat the division of one
whole number by another is equivalent to a series of
subtractions.
Divide 1-02 by ^ J of -144.
2. If the Three per Cents, are at 91i, what interest does
this give on £100 ? (Omit brokerage and fractions of a penny.)
3. How many lbs. in -321875 of a ton weight? Convert
it into kilograms (omitting fractions), assuming that a cubic
decimetre of distilled water weighs 15432*35 grains.
4. Eeduce to theii- simplest forms — r^^^z:: r^— and
V3-5 - x/2-1
V^5'12 4- v^-03375 ^
V^80- y^
5. Convert ^is ^^^ ^ decimal fraction, and find the vulgar
fraction corresponding to the recurring decimal '22297.
6. Show, by proper attention to the value of the figures, in
multiplying one number by another, that the order in which
the figures of the multiplier are taken is of no importance.
Multiply 6M43 by 47-982 correctly to three places of
decimals, beginning with the left hand figure of the multiplier,
and use as few figures as possible.
7. Extract the square root of 1095*61, and find to three
places of decimals the value of
V^ - r
8. Find the compound interest of £55 for one year, pay-
able quarterly, at 5 per cent, per annum.
A person bought into the Three per Cents, at 98, and
after receiving three years' interest he sold at 90. How
much per cent, on the sum invested did he gain or lose ?
86 ARITHMETIC.
9. Three gardeners working all day can plant a field in 10
days ; but one of them having other employment can only
work half time. How long will it take them to complete
the work ?
10. What fraction of a crown is f of 6s. 8d. 1 What is
the value of f of a guinea ? Reduce llfd. to a decimal of a
pound, correct to five places of decimals.
11. Reduce the expressions —
Multiply 49if by SO^V, and add ^|^ to the result.
Divide (2tV)' - 1 by {2^\y + S^V-
12. A bankrupt's estate amounts to £9 10. 3s. l|d., and his
debts to £1875. What can he pay in the pound 1 and what
will a creditor lose on a debt of £57 ?
13. A person having invested a sum of money in the Three
per Cent. Consols received annually therefrom £233, after
deducting the income-tax of 7d. in the pound. What is the
sum of money? What can the stock be sold for when Consols
are at 94-J-.
14. Find the value of -jm^^SiSioA and of 'UUi^-
■ 006 -vu i L z o
15. Prove the rule for finding the value of a circulating
decimal, and divide 4*367 by the circulating decimal •052.
3 + /5
Reduce to its simplest form the quantity ^_^-
_ 3-^5
8 - ^20
"" 8 + ^20
16. Three persons, A, B, C, hold a i)asture in common,
for which they are to pay £30 per annum. A put in 7
oxen for 3 months ; B, 9 oxen for 5 months ; and C, 4 oxen
for 12 months. How much rent ought each to jiay?
17. Calculate to four places of decimals the value of the
. I of -31416
expression 7^~
V *93
MISCELLANEOUS EXAMPLES. 87
18. Find the least common multiple of 16, 24, and 30, and
explain the method.
19. Wliat should be the price of English standard silver,
37-40ths fine, in order that the par of exchange between
England and France should be 25 fr. 22 c. — 200 francs being
coined from 1 kilogram of silver, 9-lOths fine? (1 kilog.
= 15*434: grains).
20. A person buys 100 shares in a company for £3,500;
after receiving four half-yearly dividends of 15s. 4d., 20s;
lOd., 30s. 4d., and 38s. 9d. per share, he sells at a profit of
43 per cent. ; reckoning the simple interest of money at 4 per
cent., how much above that interest has he gained?
21. The price of Three per Cent. Consols is 90| j what sum
must be invested in order to purchase £24 per anmmi ; and
what is the rate of interest on the money invested 1
22. Three partners in trade contribute respectively the
sums of £438, £292, £730, with the agreement that each
was to receive 5 per cent, on their respective investments,
and that the remainder of the gains of the firm, if any, was
to be divided between them in the proportion of the sums
originally advanced. The whole gain of the firm was £200.
What was each man's share ?
23. If 25 tons of goods are purchased for £37. 10s. and
sold at 35s. a ton, what is the gain per ton ?
At what rate per ton should the goods have been sold
in order to obtain a profit of £9. 7s. 6d. ?
24. Find the value of /t of £3. 12s. ll|d. ; and find the
fraction that 3 miles, 2 fur. 100 yards is of 12 leagues, 2
fur. 20 yards.
25. The sum of £9040. IGs. is placed in the Three and a
Half per Cents, at 94 ; find the income obtained, allowing
on the stock purchased -Jth per cent, to the broker, and ^^
per cent, for other expenses.
26. Express as a fraction '200123, and express as a
recurring decimal '012 t '00i32.
ARITHMETIC.
27. By the reduction of the income-tax from 7cl. in the
pound to 5d. a person saves £28. 2s. 6d. a year ; what is his
income 1
28. If 81 bushels of wheat are consumed by 56 men in 5
days, how long will 16 men take to consume 28 bushels 1
29. Find the square root of 4, and prove that v -694 = 83.
30. The periods of three planets which move uniformly in
circular orbits round the sun are respectively 200, 250, and
300 days. Supposing that their positions relative to each
other and to the sun to be given at any moment, determine
how many days must elapse before they again have exactly
the same relative positions.
SECTION II.
GEOMETRY.
EUCLID'S ELEMENTS, BOOK L
Definitions.
1 . A point is that which has position, but not magnitude.
2. A line is length without breadth.
3. The extremities of a line are points.
4. A straight line is that which lies evenly between its
extreme points.
5. A superficies (or surface) is that which has only length
and breadth.
6. The extremities of a superficies are lines.
7. A plane superficies is that in which any two points
being taken, the straight line between them lies wholly in
that superficies.
8. A plane angle is the inclination of two lines to one
another in a plane, which meet together, but are not in the
same direction.
9. A plane rectilineal angle is the inclination
of two straight lines to one another, which meet
together, but are not in the same straight line.
Note. — When several angles are at one point B, any one of them is
expressed by three letters, of which the middle letter is B, and the
tirst letter is on one of the straight lines which contain the angle,
and the last letter on the other line.
DO
GEOMETRY.
Thus, the angle contained by the straight lines AB and BC is ex-
pressed either by ABC or CBA, and the angle contained by AB and
BB is expressed either by ABD or DBA. When there is only one
angle at any given point, it may be expressed^^by the letter at that
point, as the angle E.
10. When a straight line standing on another
straight line makes the adjacent angles equal
to one another, each of the angles is called a
right angle; and the straight line which stands
on the other is called a perpendicular to it.
1 1 . An obtuse angle is that which is greater
than a right angle.
12. An acute angle is that which is less
than a right angle.
13. A term or boundary is the extremity of anything.
14. A figure is that which is enclosed by one or more
boundaries.
15. A circle is a plane figure contained by
one line, which is called the circumference,
and is such, that all straight lines drawn
from a certain point within the figure to the
circumference are equal to one another.
1 6. And this point is called the centre of the circle, [and
any straight line drawn from the centre to the circumference
is called a radius of the circle].
DEFINITIONS.
91
1 7. A diameter of a circle is a straight line drawn through
the centre, and terminated both ways by the circumference.
18. A semicircle is the figure contained by a diameter and
the part of the circumference cut off by the diameter.
19. A segment of a circle is the figure contained by a
straight line and the part of the circumference which it
cuts off.
20. Rectilineal figures are those which are contained by
straight lines.
21. Trilateral figures, or triangles, by three straight lines.
22. Quadrilateral figures, by four straight lines.
23. Multilateral figures, or polygons, by more than four
straic;ht lines.
2 4. Of three-sided figures an equilateral triangle
is that which has three equal sides. ^
25. An isosceles triangle is that which has only
two sides equal.
26. A scalene triangle is that which has three
unequal sides.
/\
27. A right-angled triangle is that which has
a riffht angle.
28. An obtuse-angled triangle is that which
has an obtuse angle.
29. An acute-angled triangle is that which
has three acute andes.
92 GEOMETRY.
30. Of four-sided figures, a square is tliat wliich
has all its sides equal, and all its angles right angles.
31. An oblong is that which has all its angles
right angles, but not all its sides equal.
32. A rhombus is that which has all its sides
equal, but its angles are not right angles.
33. A rhomboid is that which has its
opposite sides equal to one another, but all its
sides are not equal, nor its angles right angles.
34. Parallel straight lines are such as are
^. in the same plane, and which being produced
ever so far both ways do not meet.
35. A parallelogram is a four-sided figure of which the
opposite sides are parallel ; and the diagonal is the straight
line joining two of its opposite angles. All other four-sided
figures are called trapeziums.
Postulates.
1. Let it be granted that a straight line may be drawn from
any one point to any other point.
2. That a terminated straight line may be produced to any
length in a straight line.
3. And that a circle may be described from any centre, at
any distance from that centre.
Axioms.
1. Things which are equal to the same thing arc equal to
one another.
2. If equals be added to equals the wholes are equal.
EXPLANATION OP TERMS AND ABBREVIATIONS. 93
3. If equals be taken from equals the remainders are equal.
4. If equals be added to unequals the wholes are unequal.
5. If equals be taken from unequals the remainders are
unequal.
G. Things which are double of the same are equal to one
another.
7. Things which are halves of the same are equal to one
another.
8. Magnitudes which coincide with one another, that is,
which exactly fill the same space, are equal to one another.
9. The whole is greater than its part.
10. Two straight lines cannot inclose a space.
11. All right angles are equal to one another.
12. If a straight line meet two straight lines, so as to
make the two interior angles on the same side of it taken
together less than two right angles, these straight lines being
continually produced shall at length meet on that side
on which are the angles which are less than two right
angles.
Explanation of Terms and Abbreviations.
An Axiom is a truth admitted without demonstration.
A Theorem is a truth which is capable of being de-
monstrated from previously demonstrated or admitted
truths.
A Postulate states a geometrical process, the power of
efiecting which is required to be admitted.
A Problem proposes to efiect something by means of
admitted processes, or by means of processes or constnic-
tions, the power of effecting which has been previously
demonstrated.
A Corollary to a proposition is an inference which may be
easily deduced from that proposition.
The sign = is used to express equality.
4 means angle^ ^^nd A signifies triangle, .
94
GEOMETRY.
The sign r> signifies '4s greater than," and <: '' is less
than."
+ expresses addition ; thus AB + BC is the line
whose length is the sum of the lengths of
AB and BO.
— expresses subtraction ; thus AB - BC is
the excess of the length of the line AB above
that of BC.
AB- means the square described upon the
straight line AB.
AC=AB.
BC=AB.
ACandBC
each=AB.
.•.AB =
=CA.
:BC
Proposition 1.— Problem.
To describe an equilateral triangle on a given finite straight
line.
Let AB be the given straight line.
It is required to describe an equilateral triangle on AB.
Construction. — From the centre
A, at the distance AB, describe the
circle BCD (Post. 3).
From the centre B, at the dis-
tance BA, describe the circle ACE
(Post. 3).
From the point C, in which the
circles cut one another, draw the
straight lines CA, CB to the points A and B (Post. 1).
Then ABC shall he an equilateral triangle.
Proof. — Because the point A is the centre of the circle
BCD, AC is equal to AB (Def 15).
Because the point B is the centre of the circle ACE, BC
is equal to BA (Def. 15).
Therefore AC and BC are each of them equal to AB.
But things which are equal to the same thing are equal to
one another. Therefore AC is equal to BC (Ax. 1).
Therefore AB, BC, and CA are equal to one another.
Therefore the triangle ABC is equilateral, and it is de-
scribed on the given straight line AB. Which was to he done.
PROPOSITIONS. 95'
Proposition 2.— Problem.
From a given point to draw a straight line equal to a given
straight line.
Let A be the given point, and BC the given straight line.
It is required to draw from the point A a straight line
equal to BC.
Construction. — From the point A to B draw the straight Draw ab.
line AB (Post. 1).
Upon AB describe the equilateral triangle DAB (Book I., A dabc-
Prop 1) quilateral.
Produce the straight lines DA, DB, to E and F (Post. 2).
From the centre B, at the dis-
tance BC, describe the circle CGH,
meeting DF in G (Post. 3).
From the centre D, at the dis-
tance DG, describe the circle GKL,
meeting DE in L (Post. 3).
Then AL shall be equal to BC.
Proof. — Because the point B is ^^^"^-J^^^^^ bc=bg.
the centre of the circle CGH, BC is
equal to BG (Def. 15).
Because the point D is the centre of the circle GKL, DL i>l=dg.
is equal to DG (Def. 15).
But DA, DB, parts of them, are equal (Construction). da=db.
Therefore the remainder AL is equal to the remainder BG iv/-^ =
(Ax. 3).
But it has been shown that BC is equal to BG.
Therefore AL and BC are each of them equal to BG. ^ ^^ and
But things which are equal to the same thing are equal to bg. ~
one another, therefore AL is equal to BC (Ax. 1).
Therefore from the given point A a straight line AL has /. Ai^ -
been dra^vn equal to the given straight line BC. Which was ^^'
to he done.
Proposition 3. — Problem.
From the greater of two given straight lines to cut off a part
equal to the less.
Let AB and C be the two given straight lines, of which
AB is the greater.
96
GEOMETRY.
AE=AD.
AD=C.
AE and C
each=AD.
.'. AE=C.
It is required to cut off from AB, the greater, a part equal
to C, the less.
Construction. — From the point A
draw the straight line AD equal to C
(I. 2).
From the centre A, at the distance
AD, describe the circle DEF, cutting
ABinE (Post. 3).
Then AE slmll he equal to C.
Proof. — Because the point A is the centre of the circle
DEF, AE is equal to AD (Def. 15).
But C is also equal to AD (Construction).
Therefore AE and C are each of them equal to AD.
Therefore AE is equal to C (Ax. 1).
Therefore, from AB, the greater of two given straight lines,
a part AE has been cut off, equal to C, the less. Q,. E. F.^
AB=DE.
AC=DF.
ZBAC:
/C EDF.
Proposition 4.— Theorem.
If two triangles have two sides of the one equal to two sides
of the other, each to each, and have also the angles contained
by those sides equal to one another : they shall have their bases,
or third sides, equal ; and the two triangles shall be equal, and
their other angles shall be equal, each to each, viz., those to
which the equal sides are opposite. Or,
If two sides and the contained angle of one triangle be re-
spectively equal to those of another, the triangles are equal in
every respect.
Let ABC, DEF be two triangles which have
The two sides AB, AC, equal to the two sides DE, DF,
each to each, viz., AB equal to
DE, and AC equal to DF.
And the angle BAC equal to
the angle EDF : — then —
The base BC shall be equal
to the base EF ;
The triangle ABC shall be
equal to the triangle DEF ;
^ Q. E. F. is an abbreviation for quod erat faciendum, that ig " which
ivas to he done,"
PROPOSITIONS. 97
And tliG other angles to wliich the equal sides are opposite,
shall be equal, each to each, viz., the angle ABC to the angle
DEF, and the angle ACB to the angle DFE.
Proof. — For if the triangle ABC be applied to {or placed Suppose
upon) the triangle DEF, ^ ^^
' . iJiit upon
So that the point A may be on the point D, and the ^ def.
straight line AB on the straight line DE,
The point B shall coincide with the point E, because AB
is equal to DE (Hypothesis).
Ajid AB coinciding with DE, AC shall coincide with DF,
because the angle BAC is equal to the angle EDF (Hyp.).
Therefore also the point C shall coincide with the point F,
because the straight line AC is equal to DF (Hyp.).
But the point B was proved to coincide with the point E. ,
Therefore the base BG shall coincide with the base EF.
Because the point B coinciding with E, and C with F, if
the base BC do not coincide with the base EF, two straight
lines would enclose a space, which is impossible (Ax. 10).
Therefore the base BC coincides with the base EF, and is bc=ef.
therefore equal to it (Ax. 8).
Therefore the whole triangle ABC coincides with the whole .*. A abo
triangle DEF, and is equal to it (Ax. 8). = A def.
And the other angles of the one coincide with the remain- z arc =
ing angles of the other, and are equal to them, viz., the angle ^ acb*=
ABC to DEF, and the angle ACB to DFE. ^ dfe.
Therefore, if two triangles have, &c. (see Enunciation).
Which was to he shown.
Proposition 5.— Theorem.
Tlie angles at the hose of an isosceles triangle are equal to
one ano titer ; and if the equal sides he produced , the angles upon
the other side of the hose shall also he equal.
Let ABC be an isosceles triangle, of which the side AB is ab = ac.
equal to the side AC.
Let the straight lines AB, AC (the equal sides of the tri-
angle), be produced to D and E.
The angle ABC shall be equal to the angle ACB (angles
at the hase)y
5 G
98 GEOMETRY.
And the angle CBD shall be equal to the angle BCE
{angles upon the other side of the base).
Construction. — In BD take any point
F.
AG = AF. / \ From AE, the greater, cut off AG,
equal to AF, the less (I. 3).
Join FC, GB.
Proof.— Because AF is equal to AG
(Construction), and AB is equal to AC
(Hyp.),
Therefore the two sides FA, AC are
equal to the two sides GA, AB, each to
each ;
And they contain the angle FAG, common to the two
triangles AFC, AGB.
.•.FC=GB Therefore the base FC is equal to the base GB (I. 4);
= A AGB^ And the triangle AFC to the triangle AGB (I. 4) ;
And the remaining angles of the one are equal to the
z ABG~ remaining angles of the other, each to each, to which the
z AFC= equal sides are opposite, viz., the angle ACF to the angle
/. AGB. ^-g^^ ^^^^ ^^^ ^^gj^ j^^^ ^^ ^^^ ^^^^^ ^^^ ^ ^y
And because the whole AF is equal to the whole AG, of
which the parts AB, AC, are equal (Hyp.),
BF=zCG. The remainder BF is equal to the remainder CG (Ax. 3),
And FC was proved to be equal to GB ;
Therefore the two sides BF, FC are equal to the two sides
CG, GB, each to each.
And the angle BFC was proved equal to the angle
CGB;
Therefore the triangles BFC, CGB are equal; and their
other angles are equal, each to each, to which the equal sides
are opposite (I. 4).
i z GCB Therefore the angle FBC is equal to the angle GCB, and
z BCF =' the angle BCF to the angle CBG.
z CBG. ^j^(j since it has been demonstrated that the whole angle
ABG is equal to the whole angle ACF, and that the parts of
these, the angles CBG, BCF, are also equal,
ii ^ ^n Therefore the remaining angle ABC is equal to the remain-
mg augle ACB (Ax. 3),
Which are the angles at the base of the triaugle ABC.
PROPOSITIONS. 99
And it has been proved that the angle FBC is equal
to the angle GCB (Dem. 11),
Which are the angles upon the other side of the base,
Therefore the angles at the base, &c. (see Enunciation).
Which was to be shown.
CoROLi^ARY. — Hence every equilateral triangle is also
equiangular.
Proposition 6. — Theorem.
7/ two angles of a triangle he equal to one anotlier, the '
sides also which subtend, or are op2)Osite to, the equal angles,
shall be equal to one another.
Let ABC be a triangle having the angle ABC equal to
the angle ACB.
The side AB shall be equal to the side AC.
For if AB be not equal to AC, one of thenx is greater Suppose
than the other. Let AB be the greater. ^^ ^^^'•
Construction. — From AB, the greater, cut off a part DB, Make
equal to AC, the less (I. 3). DB = Aa
Join DC.
Proof. — Because in the triangles DBC, ACB, DB is
equal to AC, and BC is common to both,
Therefore the two sides DB, BC are equal
to the two sides AC, CB, each to each ;
And the angle DBC is equal to the angle
ACB (Hyp.)
Therefore the base DC is equal to the base
AB (L 4).
And the triangle DBC is equal to the tri-
angle ACB (I. 4), the less to the greater, ^ ^ A acb"
which is absurd.
Therefore AB is not unequal to AC, that is, it is equal to it.
Wherefore, if two angles, &c. Q. E, D. *
Corollary. — Hence every equiangular triangle is also *
equilateral.
■" Q. E. D. is an abbreviation for quod erat demonstrandum, that is,
**ivhich teas to be shown or proved"
100 GEOMETRY.
Proposition 7.— Theorem.
Upon the same base, and on the same side of it, there cannot
he two triangles tlmt have their sides, lohich are terminated in
one extremity of the base, equal to one another, and likewise
those lohich are terminated in the other extremity.
Let the triangles ACB, ADB, upon the same base AB,
and on the same side of it, have, if possible,
Their sides CA, DA, terminated in the
extremity A of the base, equal to one
another ;
And their sides CB, DB, terminated in
the extremity B of the base, likewise
equal to one another.
Case I. — Let the vertex of each triangle
A J3 be without the other triangle.
Construction. — Join CD.
Proof. — Because AC is equal to AD (Hyp.),
The triangle ADC is an isosceles triangle, and the angle
ACD is therefore equal to the angle ADC (I. 5).
But the angle ACD is greater than the angle BCD (Ax. 9).
Therefore the angle ADC is also greater than BCD.
Much more then is the angle BDC gi-eater than BCD.
Again, because BC is equal to BD (Hyp.),
The triangle BCD is an isosceles triangle, and the angle
BDC is equal to the angle BCD (L 5).
But the angle BDC has been shown to be gi'eater than the
angle BCD (Dem. 5).
Therefore the angle BDC is both equal to, and greater than
the same angle BCD, which is impossible.
Case II. — Let the vertex of one of the
triangles fall within the other.
Construction. — Produce AC, AD to
E and F, and join CD.
Proof. — Because AC is equal to AD
(Hyp.),
The triangle ADC is an isosceles triangle,
and the angles ECD, FDC, upon the other
side of its base CD, are equal to one another (I. 5).
PROPOSITIONS. 101
But the angle ECD is greater tlian the angle BCD (Ax. 9).
Therefore the angle FDC is likewise greater than BCD.
Much more then is the angle BDC greater than BCD. / bcd.
Again, because BC is equal to BD (Hyp.),
The triangle BDC is an isosceles triangle, and the angle zbdc =
BDC is equal to the angle BCD (I. 5). ^ ^^^•
But the angle BDC has been shown to bo greater than the
angle BCD.
Therefore the angle BDC is both equal to, and greater than .-. zbdc
the same angle BCD, which is impossible. ^ Tbcd.
Therefore, upon the same base, <kc. Q. E. D.
Proposition 8. — Theorem.
If two triangles have two sides of the one eqnal to two sides
of the other, each to each, and liave likewise their bases equal,
the angle which is contained by the two sides of the one shall
be equal to the angle contained by the two sides, equal to them,
of the other. Or,
If two triangles have three sides of the one respectively equal
to the three sides of the other, they are equal in every resjyect,
those angles being equal which are ojyj^osite to the eqioal sides.
Let ABC, DEF be two triangles which have
The two sides AB, AC equal to the two sides DE, DF, ^^'«» ^^
each to each, viz., AB to DE, and AC to DF, ac = df.
And the base BC equal to the base EF. ^^^ ef.
The angle BAC shall be equal to the angle EDF.
Proof. — For if the triangle ABC be applied to the triangle
DEF,
So that the point B may be on E, and the straight line BC
on EF,
The point C shall coin- j^
cide with the point F,
because BC is equal to EF
(Hyp.).
Therefore, BC coincid-
ing with EF, BA and AC
shall coincide with ED
and DF.
For if the base BC coincides with the base EF,
102
GEOMETRY.
But the sides BA, AC, do not coincide with the sides
ED, DF, but have a different situation, as EG, GF,
Then upon the same base, and on the same side of it, there
mil be two triangles, which have their sides terminated in
one extremity of the base equal to one another, and likewise
their sides, which are terminated in the other extremity.
But this is impossible (I. 7).
.*. BA, AC Therefore, if the base BC coincides with the base EF, the
i^cmncide sidcs BA, AC must coincide with the sides ED, DF.
ed'^df Therefore the angle BAC coincides with the angle EDF,
and is equal to it (Ax. 8).
Also the triangle ABC coincides with the triangle DEI^*
and is therefore equal to it in every respect (Ax. 8)i
Therefore, if two tiiangles, &c. Q, E. B,
Make
AE = AD.
ADEFe-
quilateral;
/. L DAF
= I EAF.
Proposition 9. — Problem.
To bisect a given rectilineal angle, that is, to divide it into two
equal parts.
Let BAC be the given rectilineal angle.
It is required to bisect it;
Construction. — Take any point D in AB.
From AC cut off AE equal to AD (I. 3).
Join DE.
Upon DE, on the side remote from A, de-
scribe an equilateral triangle DEF (I. 1).
O JoinAF.
27ien the straight line AF shall bisect the angle BAC.
Proof.— Because AD is fequal to AE (Const.), and AF is
common to the two triangles DAF, EAF;
The two sides DA, AF are equal to the two sides EA,
AF, each to each ;
And the base DF is equal to the base EF (Const.);
Therefore the angle DAF is equal to the angle EAF (I. S):
Therefore the given rectilineal angle BAC is bisected by
the straight line AF. Q. E. F.
Proposition 10.— Problem.
To bisect a given finite straight line, that is, to divide it inta
two equal parts.
PROPOSITIONS.
103
c
Make
AABCe-
\
quilateral
■\
and
\
/ACD =
ZBCD.
;. AD s
DB.
Let AB be the given straight line.
It is required to divide it into two equal parts.
Construction. — Upon AB describe the
equilateral triangle ABC (I. 1).
Bisect the angle ACB by the straight line
CD (I. 9).
Then AB sliall be cut into two equal 2JC('rts
in the point T>.
Proof. — ^Because AC is equal to CB -^ ^ ^
(Const.), and CD common to the two triangles ACD, BCD;
The two sides AC, CD are equal to the two sides BC,
CD, each to each ;
And the angle ACD is equal to the angle BCD (Const.) ;
Therefore the base AD is equal to the base DB (I. 4).
Therefore the straight line AB is divided into two equal
parts in the point D. Q, E. F.
Proposition 11. — Problem*
To dfaw a straight line at right angles to a given straight
line from a given 'point hi the same.
Let AB be the given straight line, and C a given point in it.
It is required to draw a straight line from the point C at
right angles to AB.
Construction. — ^Take any point D in AC.
Make CE equal to CD (I. 3).
Upon DE describe the equilateral triangle DFE (I. 1).
Join EC. y
Then EC shall he at right angles
to AB.
Proof. — Because DC is equal to
CE (Const.), and EC common to
the two triangles DCE, ECE j ^
The two sides DC, CE, are equal a J)
to the two sides EC, CE, each to each ;
And the base DE is equal to the base EE (Const.) ;
Therefore the angle DCE is equal to the angle ECE (I. 8) ; /^ I
And they are adjacent angles.
But when a straight line, standing on another straight
line, makes the adjacent angles equal to one another, each of
the angles is called a right angle (Def. 10) ;
^DCFr
104
GEOMETRY.
;. L DCF,
Z ECF are
right
angles.
Make
ZABEa
right L
/CBE =
ZEBA.
and
ZDBE==
ZEBA.
.-. Z DBE
= Z CBE.
Therefore eacli of tlie angles DCF, ECF is a right angle.
Therefore from the given point C in the given straight line
AB, a straight line FC has .been drawn at right angles to
AB. Q.E.F.
Corollary. — By help of this problem, it may be demon-
strated that
Two straight lines cannot have a common segment.
If it be possible, let the two straight lines ABC, ABD,
have the segment AB common to
both of them.
Construction. — From the point
B, draw BE at right angles to AB
(1. 11).
Proof. — Because ABC is a
CBE is equal to the angle EBA
C
the angle
CD as ra-
dius.
Bisect FG
inH.
straight line,
(Def. 10).
Also, because ABD is a straight line, the angle DBE is
equal to the angle EBA (Def. 10).
Therefore the angle DBE is equal to the angle CBE. The
less to the greater; which is impossible.
Therefore two straight lines cannot have a common seg-
ment.
Proposition 12. — Problem.
To draw a straight line perpendicular to a given straight
line of unlimited lengthy from, a given point without it.
Let AB be the given straight line, which may be produced
to any length both ways, and let C be a point without it.
It is requii-ed to draw from the point C, a straight line
perpendicular to AB.
Construction. — Take any point
D upon the other side of AB.
From the centre C, at the distance
CD, describe the circle EGF, meet-
ing AB in F and G (Post. 3).
Bisect FG in H (I. 10).
Join CF, CH, CG.
Then CH shall he jyerpetulicular to AB.
Proof* — Because FH is equal to HG (Const.), and HC
fcommon to the two triangles FHC, GHC ;
PROPOSITIONS.
105
The two sides FH, HG are equal to the two sides GH,
HG, each to each ;
And the base GF is equal to the base GG (Def. 15) ;
Therefore the angle GHF is equal to the angle GHG (I. 8), /. adjacent
and they are adjacent angles. ch'f.^cho
But when a straight line, standing on another straight are equal,
line, makes the adjacent angles equal to one another, each of
the angles is called a right angle, and the straight line which
stands on the other is called a perpendicular to it (Def. 10).
Therefore, from the given point G, a perpendicular has
been drawn to the given straight line AB. Q, E, F,
Proposition 13. — Theorem.
TJie angles wliich one straight line makes with another upon
one side of it, are either two right angles, or are together equal
to two right angles.
Let the straight line AB make with GD, upon one side of
it, the angles GBA, ABD.
These angles shall either be two right angles, or shall to-
gether be equal to two right angles.
Proof. — If the angle GBA be equal to the angle ABD,
each of them is a right angle (Def. 10).
But if the angle GBA be not equal to the angle ABD,
from the point B, draw BE at right angles to GD (I. 11).
Therefore the angles GBE, EBD, are two right angles.
Now the angle GBE is equal to the two angles GBA, ABE;
to each of these equals add the angle EBD.
D
Make
z CBE =
z EBD =
a right 2 .
B C i5 B C
Therefore the angles GBE, EBD, are equal to the three
angles GBA, ABE, EBD (Ax. 2).
Again, the angle DBA is equal to the two angles DBE,
EBA ; to each of these equals add the angle ABG.
.•.ZCBE+
z EBD =
Z CBA-f
Z ABE +
z EBD, al-
so z DBA
+ Z ABC
Therefore the angles DBA, ABG, are equal to the three •= ^ ^^J
angles DBE, EBA, ABG (Ax. 2). + z abc.
lOG
GEOMETRY,
But the angles CBE, EBB have been shown to be equal
to the same three angles ;
And things which are equal to the same thing are equal
to one another *
.-. z CBE Therefore the angles CBE, EBD, are equal to the angles
± i III DBA, ABO (Ax. 1).
+ z ABC. But the angles CBE, EBD are two right angles.
Therefore the angles DBA, ABC, are together equal to
two right angles (Ax* 1).
Therefore, the angles which one straight line, &c. Q. E. D,
Proposition 14.— Theorem.
If, at a point in a straight lin£, two other straight lines,
upon the opposite sides of it, make the adjacent angles together
equal to two right angles, these two straight lines shall he in
one and the same straight line.
At the point B in the straight line AB, let the two
straight lines BC, BD, upon the
opposite sides of AB, make the ad-
jacent angles ABC, ABD together
equal to twO right angles.
BD shall be in the same straight
line with BC.
For if BD be not in the same
straight line with BC, let BE be in
the same straight line with it.
Proof. — Because CBE is a straight line, and AB meets
it in B.
Therefore the adjacent angles ABC, ABE are together
equal to two right angles (I. 13).
But the angles ABC, ABD, are also together equal to two
right angles (Hyj^.) ;
Therefore the angles ABC, ABE, are equal to the angles
ABC, ABD (Ax. 1).
Take away the common angle ABO*
The remaining angle ABE is equal to the remaining angle
= z ABD. ABD (Ax. 3), the less to the gi-eater, which is impossible ;
Therefore BE is not in the same straisjht line with BC.
Given
z ABC +
z ABD=
two right
angles.
If possible^
let CBE be
a straight
line.
z ABE
PROPOSITIONS. 107
And, ill like manner, it may be demonstrated that no
other can be in the same straight line with it but BD.
Therefore BD is in the same straight line with BC.
Therefore, if at a point, &c. Q. E. D,
Proposition 15.— Theorem.
If two straight lines cut one another, the vertical, 01* o])2)Osite
angles shall he equal.
Let the two straight lines AB, CD cut one another in the
l)oint E.
The angle AEC shall be equal to ^""\^^
angle DEB, and the angle CEB to ^^^v^
the angle AED. A E\~ B"
Proof. — Because the straight line ^^ ^ cea 4-
AE makes with CD, the angles CEA, ^''n^ht^ "
AED, these angles are together equal to two right angles angles.
(I. 13).
Again, because the straight line DE makes with AB the ^ ^^^ ,
angles AED, DEB, these also are together equal to two right z deb ==
angles (I. 13). l^^l
But the angles CEA, AED have been shown to be
together equal to two tight angles.
Therefore the angles CEA, AED are equal to the angles
AED, DEB (Ax. 1).
Take away the common angle AED.
The remaining angle CEA is fequal to the remaining angle • ^ cE\
DEB (Ax. 3). = ^ DEB.
In the same manner it can be shown that the angles CEB,
AED are fequal.
Therefore, if two straight lines^ &c. Q, E. D,
Corollary 1. — From this it is manifest that if two
straight lines cut one another, the angles which they make at
thei point where they cut^ are together equal to four right
angles.
Corollary 2. — And, conse^iiently, that all the • angles
made by any number of lihes ineeting in one point are
together equal to four right angles, provided that no one of
the angles be included in any other angle.
108 GEOMETRY.
Proposition 16. — Theorem.
// one side of a triangle he producedy the exterior angle
shall be greater than either of the interior opposite angles.
Let ABC be a triangle, and let its side BC be produced to D.
The exterior angle ACD shall be greater than either of the
interior opposite angles CBA, BAG.
Mako /\ y/ Construction. — Bisect AC in E
, .. , (I- 10).
EF = BE. / \4 / Join BE, and produce it to F,
making EF equal to BE (I. 3), and
join FC.
Proof. — Because AE is equal to
EC, and BE equal to EF (Const.),
AE, EB are equal to CE, EF,
Cr\ each to each ;
And the angle AEB is equal to the angle CEF, because
they are opposite vertical angles (I. 15).
Therefore the base AB is equal to the base CF (I. 4) ;
And the triangle AEB to the triangle CEF (I. 4) ;
And the remaining angles to the remaining angles, each to
each, to which the equal sides are opposite.
zBAE Therefore the angle BAE is equal to the angle ECF
(I. 4).
But tlie angle ECD is greater tlian the angle ECF
(Ax. 9) ; ' _
Therefore the angle ACD is greater than the angle BAE.
. z ACD In the same manner, if BC be bisected, and the side AC be
produced to G, it may be proved that the angle BCG (or its
equal ACD), is greater than the angle ABC.
Therefore, if one side, &c. Q. E, D,
Proposition 17.— Theorem.
Any two angles of a triangle arc together less than two right
angles.
Let ABC be any triangle.
Any two of its angles together shall be less than two right
angles.
-. L ECF.
-^BAE.
PROPOSITIONS. 1 09
Construction. -^Produce BC to D.
Proof.— Because ACD is the ex-
terior angle of the triangle ABC, it is
greater than the inteiior and opposite
angle ABC (I. 16).
To each of these add the ancjle
ACB. ^ ^ »'
Therefore the angles ACD, ACB are greater than the ^ abc +
angles ABC, ACB (Ax. 4). /.A^f ^
But the angles ACD, ACB arc together equal to two a"gies.
right angles (I. 13);
Therefore the angles ABC, ACB are together less than
two right angles.
In like manner, it may be proved that the angles BAC,
ACB, as also the angles CAB, ABC are together less than
two right angles.
Therefore, any two angles, &c. Q, E. i>.
Proposition 18.— Theorem.
The greater side of every triangle is oj^j^osite the greater angle.
Let ABC be a triangle, of which the side AC is greater ^^ > ^2^
than the side AB.
The angle ABC shall be greater than
the angle BCA.
Construction. — Because AC is
greater than AB, make AD equal to
AB (I. 3), and join BD. B C
Proof. — Because ADB is the exterior angle of the triangle ^ ^y)V, >
BDC, it is greater than the interior and opposite angle BCD ^ i^cd,
and
z ADB =
z ADD
A
(I. 16),
But the angle ADB is equal to the angle ABD ; the
triangle BAD being isosceles (I. 5), and
Therefore the angle ABD is greater than the angle BCD >/bcd
(or ACB).
Much more then is the angle ABC greater t^an the angle
ACB.
Therefore, the greater side, tkc. Q. E. D,
* no GEOMETRY,
Proposition 19.— Theorem,
The greater cmgle of every triangle is subtended hy the
greater side, or has the greater side opposite to it,
z' ABC >• -Let ABC be a triangle, of which the angle ABC is greater
- ^^^- than the angle BCA;
Tlie side AC shall be greater than the side AB.
Proof. — If AC be not greater than
AB, it must either be equal to or less
than AB.
It is not equal, for then the angle
__^ ABC would be equal to the angle BCA
^ ^ (I. 5); but it is not (Hyp.) ;
AC not = Therefore AC is not equal to AB.
Neither is AC less than AB, for then the angle ABC
would be less than the angle BCA (I. 18); but it is not
(Hyp.);
AC not < Therefore AC is not less than AB.
And it has been proved that AC is not equal to AB ;
Therefore AC is greater than AB.
Therefore, the greater angle, &c. Q, E. D,
Proposition 30. — Theorem.
Any two sides of a triangle are together greater than the
third side.
Let ABC be a triangle ;
Any two sides of it are together greater than the third side.
Maiie Construction. — Produce BA to the point D, making AD
•^^ =^^' equal to AC (I. 3), and join DC.
Proof. — Because DA is equal to AC, the angle ADC is
equal to the angle ACD (I. 5).
But the angle BCD is greater than the angle ACD (Ax. 9);
^ 2SJ? ^ Therefore the angle BCD is greater than the angle ADC
dKBDC).
And because the angle BCD of the
^ I triangle DCB is greater than its angle
BDC, and that the greater angle is
subtended by the gi^eater side ;
■.DB>BC. ■** G Therefore the side DB is greater
than the side BC (I. 19).
PROPOSITIONS. Ill
But BD is equal to BA and AC ;
Therefore BA, AC are greater than BC. ac'^bc.
In the same manner it may be proved that AB, BC are
greater than AC ; and BC, CA greater than AB.
Therefore any two sides, &c. Q, E, J),
Proposition 21. — Theorem.
If from the ends of the side of a triangle there he draimi
iivo straight lines to a 2>oint within the triangle, these shall he
less than the other two sides of the triangle, hut shall contain
a greater angle.
Let ABC be a triangle, and from the points B, C, the ends
of the side BC, let the two straight lines BD, CD be drawn
to the point D within the triangle;
BD, DC shall be less than the sides BA, AC ;
But BD, DC shall contain an angle BDC greater than
the angle BAC.
Construction. — Produce BD to E.
Proof. — 1. Because two sides of a
triangle are gi^eater than the third side
(I. 20), the two sides BA, AE, of the
triangle BAE are greater than BE.
To each of these add EC.
Therefore the sides BA, AC, are B
greater than BE, EC (Ax. 4).
Again, because the two sides OE, ED, of the triangle ^^'
CED are greater than CD (I. 20),
To each of these add DB.
Therefore CE, EB are greater than CD, DB (Ax. 4).
But it has been shown that BA, AC are greater than ^b^cd
BE, EC ; + DB.
Much more then are BA. AC greater than BD, DC.
Proof. — 2. Again, because the exterior angle of a
tiiangle is greater than the interior and opposite angle
(I. 16), therefore BDC, the exterior angle of the triangle ^f^l^^
CDE, is greater than CED or CEB. ^^^^ ceb.
For the same reason, CEB, the exterior angle of the tri- ^J^ ceb >
angle ABE, is greater than the angle BAE or BAC, ^ bae.
BA + AC
> BE +
112. GEOMETRY.
And it has been shown that the angle BDC is greater
than CEB ;
• I BDC Much more then is the angle BDC gi^eater than the angle
>zBAC. BAG.
Therefore, if from the ends, &c. Q, E. D,
Proposition 22.—- Problem.
To 7)iake a triangle of which the sides shall he equal to three
given straight lines, hut any two whatever of these lines 7nust
he greater than the third (I. 20).
Let A, B, C be the three given straight lines, of which
any two whatever are greater than the thii'd — namely, A
and B greater than C, A and C greater than B, and B and
C greater than A ;
Jt is required to make a triangle of which the sides shall
be equal to A, B, and C, each to each.
Construction. — Take a straight line DE terminated at
the point D, but unlimited to-
wards E.
Make DF equal to A, EG
equal to B, and GH equal to
C (I. 3).
Erom the centre E, at the
distance ED, describe the circle
C DKL (Post. 3).
andOH From the centre G, at the distance GH, describe the
^^■^'"^- circle HLK (Post. 3).
Join KE, KG.
Then the triangle KEG sliall Imve its sides equal to the three
straight lines A, B, C.
Proof. — Because the point E is the centre of the circle
DKL, ED is equal to EK (Def. 15).
But ED is equal to A (Const.) ;
nc = A. Therefore EK is equal to A (Ax. 1).
Again, because the point G is the centre of the circle
HLK, GH is equal to GK (Def. 15).
But GH is equal to C (Const.) ;
GK = c. Therefore GK is equal to C (Ax. 1),
PROPOSITIONS.
113
And FG is equal to B (Const.) ; > fg = B
Therefore the three straight lines KF, FG, GK are equal
to the three A, B, C, each to each.
Therefore the triangle KFG has its three sides KF, FG,
GK equal to the thi'ee given straight lines A, B, C.
Q, E. i\
Proposition 23.— Problem.
At a given 2)oint in a given straight liiie, to make a recti-
lineal angle equal to a given rectilineal angle.
Let AB be the given straight line, and A the given point
in it, and DCE the given rectilineal angle.
It is requii-ed to take an angle at the point A, in the
straight line AB, equal to the rectilineal angle DCE.
Construction. — In CD, CE,
take any points D, E, and join
DE.
On AB constiTict a triangle
AFG, the sides of which shall be
equal to the three straight lines
CD, DE, EC— namely, AF equal rl
to CD, FG to DE, and AG to EC
(1.22);
Then the angle FAG shall he equal to the angle DCE.
Proof. — Because DC, CE are equal to FA, AG, each to
each, and the baso DE equal to the base FG (Const.),
The angle DCE is equal to the angle FAG (I. 8).
Therefore, at the given point A, in the given straight line
AB, the angle FAG has been made equal to the given recti-
lineal angle DCE. Q, E, F.
Tlien
z DCE =
z FAG
Proposition 24.— Theorem.
If two triangles have two sides of the one equal to two sides
of the other, each to each, hut the angle contained by the two
sides of one of them greater than the angle contained by the two
sides equal to them of the other, the base of that ivhich has th^
greater angle shall be greater tJian the base of the other.
5 U
114
GEOMETilY.
Suppose
DF > DE.
Make Z
EDG =
z BAG.
.•.BC=EG.
and
^ EFG::
- EGF.
.'.EG>EF.
Let ABC, DEF, be two triangles which have
The two sides AB, AC equal to the two DE, DF, each to
each — namely, AB to DE,and AC to DF,
But the angle BAC greater than the angle EDF;
The base BC shall be greater than the base EF.
Construction. — Let the side DF of the triangle DEF be
, greater than its side DE.
A. ^ Then at the point D, in
the straight line ED, make
the angle EDG equal to the
angle BAC (I. 23).
Make DG equal to AC
or DF (I. 3).
Join EG, GF.
Proof. — Because AB is
equal to DE (Hyp.), and AC to DG (Const.), the two sides
BA, AC are equal to the two ED, DG, each to each ;
And the angle BAC is equal to the angle EDG (Const.) ;
Therefore the base BC is equal to the base EG (I. 4).
And because DG is equal to DF (Const.), the angle DFG
is equal to the angle DGF (L 5).
But the angle DGF is greater than the angle EGF (Ax. 9);
Therefore the angle DFG is greater than the angle EGF ;
Much more then is the angle EFG greater than the angle
EGF.
And because the angle EFG of the triangle EFG is greater
than its angle EGF, and that the greater angle is subtended
by the greater side.
Therefore the side EG is greater than the side EF (I. 19).
But EG was proved equal to BC ;
Therefore BC is greater than EF.
Therefore, if two triangles, &c. Q. E. D.
Proposition 25.— Theorem.
If two triangles have two sides of the one equal to two sides
of the other y each to each, hut the base of the one greater than
the base of the other, the angle contained by the sides of that
which has the greater base shall be greater tlmn the angle
contained by the sides equal to tliem of the other.
PKOPOSITIONS. 115
Let ABC, DEF, be two triangles, which have
The two sides AB, AC equal to the two sides DE, DF,
each to each — namely, AB to DE, and AC to DF,
But the base BC greater than the base EF ;
The angle BAC shall be greater than the angle EDF.
Proof. — For if the angle
BAC be not greater than the A. p
angle EDF, it must either be
equal to it or less.
But the angle BAC is not
equal to the angle EDF, for
then the base BC would be
equal to the base EF (I. 4), but
it is not (Hyp.);
Therefore the angle BAC is not equal to the angle EDF; ^.^^edf!
Neither is the angle • BAC less than the angle EDF, for
then the base BC would be less than the base EF (I. 24), but
it is not (Hyp.),
Therefore the angle BAC is not less than the angle EDF. <^/edf!
And it has been proved that the angle BAC is not equal
to the angle EDF;
Therefore the angle BAC is greater than the angle EDF,
Therefore, if two triangles, &c. Q, E. D.
Proposition 26. — Theorem.
If two triangles have two angles of the one equal to two
angles of the other, each to each, and one side equal to one
side — namely, either the side adjacent to the equal angles in
each, or the side opposite to them ; then shall the other sides he
equal, eaxih to each ; and also the third angle of the one equal
to the third angle of the other. Or,
If two angles and a side in one triangle he respectively equal
to two angles and a corresponding Me in another triangle, the
tria7igles shall he equal in every respect.
Let ABC, DEF be two triangles, which have • ^
The angles ABC, BCA equal to the angles DEF, EFD,
each to each—namely, ABC to DEF, and BCA to EFD;
Also one side equal to one side.
Case 1. — First, let the sides adjacent to the ecjual angles ^c!^ef.
in each be equal— namely, BC to EF;
116
GEOMETRY.
Then shall the side AB be equal to DE, the side AC to
DF, and the angle BAG to the angle EDF.
For if AB be not equal to DE, one of them must be
ft- ,j) greater than the othei*. Let AB
be the greater of the two.
Construction. — Make BG-
equal to DE (I. 3), and join GC.
Proof. — Because BG is
equal to DE (Const.), and BC
V is equal to EF (Hyp.), the two
sides GB, BC are equal to the two sides DE, EF, each to
each.
And the angle GBC is equal to the angle DEF (Hyp.) ;
Therefore the base GC is equal to the base DF (I. 4),
And the triangle GBC to the triangle DEF (I. 4),
And the other angles to the other angles, each to each, to
which the equal sides are opposite;
Therefore the angle GCB is equal to the angle DFE (I. 4).
But the angle DFE is equal to the angle BCA (Hyp.) ;
Therefore the angle GCB is equal to the angle BCA (Ax.
1), the less to the greater, which is impossible;
Therefore AB is not unequal to DE, that is, it is equal to
it; and BC is equal to EF (Hyp.) ;
Therefore the two sides AB, BC are equal to the two sides
DE, EF, each to each.
And the angle ABC is equal to the angle DEF (Hyp.) ;
Therefore the base AC is equal to the- base DF (I. 4),
And the third angle BAC to the third angle EDF (I. 4).
Case 2. — Next, let the sides which are opposite to the
equal angles in each triangle be equal to one another — namely,
AB equal to DE.
Likewise in this case the other sides shall be equal, AC to
A D DF, and BC to EF ; and also the
angle BAC to the angle EDF.
For if BC be not equal to
EF, one of them must be greater
than the other. Let BC be the
greater of the two.
H c B F Construction, — Make BH
equal to EF (I. 3), and join AH,
PROPOSITIONS. 117
Proof. — Because BH is equal to EF (Const.), and AB is bh = ef,
equal to DE (Hyp.), the two sides AB, BH are equal to the ^^ ~ ^^'
two sides DE, EF, each to each.
And the angle ABH is equal to the angle DEF (Hyp.); ^ abh =
Therefore the base AH is equal to the base DF (I. 4),
And the triangle ABH to the triangle DEF (I. 4),
And the other angles to the other angles, each to each, to
which the sides are opposite ;
Therefore the angle BHA is equal to the angle EFD (I. 4). . ^ ^j^^
But the angle EFD is equal to the angle BCA (Hyp.); =z efd
Therefore the angle BHA is also equal to the angle BCA ~ ^ ^^^
(Ax. 1) ; _
That is, the exterior angle BHA of the triangle AHC, is
equal to its interior and opposite angle BCA, which is
impossible (I. 16) ;
Therefore BC is not unequal to EF — that is, it is equal to it ; BC not
and AB is equal to DE (Hyp.) ; to ef.*
Therefore the two sides AB, BC are equal to the two sides
DE, EF, each to each.
And the angle ABC is equal to the angle DEF (Hyp.) ;
Therefore the base AC is equal to the base DF (I. 4),
And the third angle BAC is equal to the third angle
EDF (I. 4).
Therefore, if two triangles, &c. Q. E. D.
i .
Proposition 27.— Theorem. '
If a straight line faUimj upon two other straight lines make
the altetviate angles equal to one another^ these two straight
lines shall be 2:>aralleL
Let the straight line EF, which
falls upon the two straight lines
AB, CD, make the alternate angles ^ 5/ 5^^ Given
AEF, EFD, equal to one another. / ^\r^ z efd '
AB shall be parallel to CD. / ^^
For if AB and CD be not parallel, ^ A i>
they will meet if produced, either /
towards B, D, or towards A, C.
Let them be produced, and meet towards B, D, in the
point G.
118 GEOMETRY.
L AEF> Proof. — Then GEF is a triangle, and its exterior angle
^ EFG, AEF is greater than the interior and opposite angle EFG
and also (J. 16).
= z EFG. But the angle AEF is also equal to EFG (Hyp.), which is
impossible ;
Therefore AB and CD, being produced, do not meet
towards B, D.
In like manner it may be shown that they do not meet
towards A, C.
But those straight lines in the same plane which being
produced ever so far both ways do not meet are parallel
(Def. 34) ;
Therefore AB is parallel to CB.
Therefore, if a straight line, &c. Q. E. D,
Proposition 28.— Theorem.
If a straight line falling upon two other straight lines ruahe
the exterior angle equal to the interior and opposite upon the
sarne side of the line, or make the interior angles upon the same
side together equal to two right angles, the two straight lines
shall he parallel to one another.
Let the straight line EF, which falls upon the two straight
lines AB, CD, make —
The exterior angle EGB equal to the interior and opposite
PA angle GHD, upon the same side ;
Or make the interior angles on the
-B same side, the angles BGH, GHD,
together equal to two right angles ;
C ^v^^ ^D ^B shall be parallel to CD.
Proof 1 .—Because the angle EGB
"^■P is equal to the angle GHD (Hyp.),
And the angle EGB is equal to the angle AGH (I. 15) ;
Therefore the angle AGH is equal to the angle GHD
(Ax. 1), and these angles are alternate;
Therefore AB is parallel to CD (I. 27).
Proof 2.— Again, because the angles BGH, GHD are
equal to two right angles (Hyp.),
And the angles BGH, AGH are also equal to two riffht
angles (I. 13).
.*. z. AGH
= z GHD
PROPOSITIONS. 119
Therefore tlie angles BGH, AGH are equal to the angles .*. / bgh
BGH, GHD(Ax. 1). =zbg2
Take away the common angle BGH. + ^ ^hd.
Therefore the remaining angle AGH is equal to the remain-
ing angle GHD (Ax. 3), and they are alternate angles.
Therefore AB is parallel to CD (I. 27).
Therefore, if a straight line, &c. Q, E, D,
Proposition 29.— Theorem.
If a straight line/all upon two parallel straightlineSyit makes
the alternate angles equal to one another, and the exterior angle
equal to the interior and opposite upon the same side ; and
also the two interior angles upon the same side together equal to
two right angles.
Let the straight line EF fall upon the parallel straight
lines AB, CD;
The alternate angles AGH, GHD shall be equal to one
another.
The exterior angle EGB shall be p\
equal to GHD, the interior and op- \
posite angle upon the same side ; A A^ — B
And the two interior angles on the \
same side BGH, GHD shall be to- ^ \, - ^
gether equal to two right angles. \
For if AGH be not equal to ^^ z agh >
GHD, one of them must be greater ghd.
than the other. Let AGH be the greater. (*>uppose.)
Proof. — Then the angle AGH is greater than the angle
GHD; to each of them add the angle BGH.
Therefore the angles BGH, AGH are greater than the
angles BGH, GHD (Ax. 4).
But the angles BGH, AGH are together equal to two i-ight
angles (1. 3). .^ ^^jj
Therefore the andes BGH, GHD are less than two right + z ghd
^ ^ <two right
angles. ^ angles.
But if a straight line meet two straight lines, so as to
make the two interior angles on the same side of it taken
together less than two right angles, these straight lines being
120
GEOMETRY.
Hence
AB and
CD meet,
and are
parallel.
.-. zAGHj
not une-
qual to
L GHD.
and
zEGB =
zGHD,
also
ZBGH +
ZGHD =
two right
angles.
continually produced, shall at length meet on that side on
which are the angles which are less than two right angles
(Ax. 12);
Therefore the straight lines AB, CD will meet if produced
far enough.
But they cannot meet, because they are parallel straight
lines (Hyp.) ;
Therefore the angle AGH is not unequal to the angle
GHD — that is, it is equal to it.
But the angle AGH is equal to the angle EGB (I. 15) ;
Therefore the angle EGB is equal to the angle GHD
(Ax. 1).
Add to each of these the angle BGH.
Therefore the angles EGB, BGH, are equal to the angles
BGH, GHD (Ax. 2).
But the angles EGB, BGH, are equal to two riglit angles
(I. 13).
Therefore also BGH, GHD, are equal to two right angles
(Ax. 1).
Therefoije, if a straight line, <fec. Q, E. D,
I AGH or
L AGK =
L GHF,
.*. z AGK
= z GKD.
Proposition 30* — Theorem.
Straight lines which are parallel to the same straight lines
are parallel to one another.
Let AB, CD be each of them parallel to EF ;
AB shall be parallel to CD.
Construction. — Let the straight line GHK cut AB,
EF, CD.
Proof. — Because GHK cuts the par-
allel straight lines AB, EF, the angle
AGH isequal to the angle GHF (1. 29).
Again, because GK cuts the parallel
straight lines EF, CD, the angle GHF
is equal to the angle GKD (I. 29).
And it was shown that the angle
AGK is equal to the angle GHF ;
Therefore the angle AGK is equal to the angle GKD
(Ax. 1), and they are alternate angles ;
E
A
F
B
D
C
PROPOSITIOJCS. 121
• Therefore AB is parallel to CD (I. 27).
Therefore, straight lines, &c. Q, K D,
Proposition 31. — Problem.
To draw a straight line through a given point, 2^ctrallel to a
given straight litie.
Let A be the given point, and BC the given straight line.
It is required to draw a straight line through the point A,
parallel to BC.
Construction. — Tn BC take any
point D, and join AD.
At the point A, in the straight line
AD, make the angle DAE equal to the B J) 5 Make
angle ADC (I. 23). z adc.^
Produce the straight line EA to F.
Then EF sliall he parallel to BC.
Proof. — Because the straight line AD, which meets the Tij^y are
two straight lines BC, EF, makes the alternate angles EAD, angles.
ADC equal to one another ;
Therefore EF is parallel to BC (I. 27).
Therefore, the straight line EAF is drawn through the
given point A, parallel to the given straight line BC. Q.E. F.
Proposition 32. — Theorem.
1/ a side of any triangle he jyroduced, the exterior angle is
equal to the two interior and ojyj^osite angles ; and the three
interior angles of every triangle are equal to two right amjles.
Let ABC be a triangle, and let one of its sides BC be pro-
duced to D;
The exterior angle ACD shall be equal to the two interior
and opposite angles CAB, ABC ;
And the three interior angles
of the triangle— namely, ABC,
BCA, CAB, shall be equal to
two right angles. ^^_ ^ ■ 5
Construction. — Through the ^ Make
point C, draw CE parallel to AB (I. 31). toAfi!^ ^
122 GEOMETRY.
Tiien Proof. — Because AB is parallel to CE, and AC meets
z ACE,^ "^^^G^^j *^^ alternate angles BAG, ACE are equal (I. 29).
and ' Again, because AB is parallel to CE, and BD falls upon
z ECD = "fcliem, the exterior angle ECD is equal to the interior and
z ABC. opposite angle ABC (I. 29).
But the angle ACE was shown to be equal to the angle
BAC;
,\ z ACD Therefore the whole exterior angle ACD is equal to the
+ z ABC. two interior and opposite angles BAC, ABC (Ax. 2).
Add To each of these equals add the angle ACB.
Therefore the angles ACD, ACB are equal to the three
angles CBA, BAC, ACB (Ax. 2).
But the angles ACD, ACB are equal to two right angles
(1.13);
Therefore also the angles CBA, BAC, ACB are equal to
two right angles (Ax. 1).
Therefore, if a side of any triangle, &c. ' Q. E, D,
Corollary 1. — All the interior angles of any rectilineal
figure^ together with four right angles, are equal to twice as
many right angles as the figure has sides.
For any rectilineal figure ABCDE can, by drawing
straight lines from a point F within the figure to each angle,
be divided into as many triangles as the
figure has sides.
And, by the preceding proposition, the
angles of each triangle are equal to two
right angles.
Therefore all the angles of the triangles
are equal to twice as many right angles
as there are triangles; that is, as there are sides of the
figure*
But the same angles are equal to the angles of the figure,
together with the angles at the point F ;
And the angles at the point F, which is the common
vertex of all the triangles, are equal to four right angles (I.
15, Cor. 2) ;
Therefore all the angles of the figure, together with four
right angles, are equal to twice as many right angles as the
figure has sides.
I PROPOSITIOI^S. 123
Corollary 2. — All the exterior angles of any rectilineal
figure are together equal to four right angles.
The interior angle ABC, with its adjacent exterior angle
ABD, is equal to two right angles (I. 1 3) ;
Therefore all the interior, together with all the exterior
angles of the figure, are equal to twice as many right angles
as the figure has sides.
But all the interior angles, together
with four right angles, are equal to
twice as many right angles as the
figure has sides (I. 32, Cor. 1) ;
Therefore all the interior angles,
together with all the exterior angles,
are equal to all the interior angles
and four right angles (Ax. 1).
Take away the interior angles which are common ;
Therefore all the exterior angles are equal to four right
angles (Ax^ 3).
Proposition 33.— Theorem.
Tlie straight lines which join the extremities of two equal
and pofrallel straight lines towards the same parts are also
them^sehes equal and parallel.
Let AB and CD be equal and parallel straight lines joined
towards the same parts by the straight lines AC and BD )
AC and BD shall be equal and parallel.
Construction. — Join BC.
Proof. — Because AB is parallel to CD, and BC meets
them, the alternate angles ABC, BCD are equal (I. 29).
Because AB is equal to CD, and BC common to the two
triangles ABC, DCB, the two sides A b
AB, BC are equal to the two sides
DC, CB, each to each ;
And the angle ABC was proved
to be equal to the angle BCD ;
Therefore the base AC is equal to the base BD (I. 4), ^J^^
And the triangle ABC is equal to the triangle BCD (I. 4),
And the other angles are equal to the other angles, each to
each, to which the equal sides are opposite ;
Z ABC:
z BCD.
124 GEOMETRY.
*" AOB - Therefore the angle ACB is equal to the angle CBD.
t: CBD."" And because the straight line BC meets the two straight
lines AC, BD, and makes the alternate angles ACB, CBD
equal to one another ;
Therefore AC is parallel to BD (I. 27) ; and it was shown
to be equal to it.
Therefore, the straight lines, &c. Q, E. D.
Proposition 34. — Theorem.
The opposite sides and angles of a parallelogram are equal
to one another, and the diagonal bisects the parallelogram — that
is, divides it into two equal parts.
Let ACDB be a parallelogram, of which BC is a diagonal ;
The opposite sides and angles of the figure shall be equal
to one another,
And the diagonal BC shall bisect it.
Proof. — Because AB is parallel to CD, and BC meets
them, the alternate angles ABC, BCD
are equal to one another (I. 29) ;
Because AC is parallel to BD, and
BC meets them, the alternate angles
_ ACB, CBD are equal to one another
° (I. 29) ;
Therefore the two triangles ABC, BCD have two angles,
ABC, BCA in the one, equal to two angles, BCD, CBD in
the other, each to each ; and the side BC, adjacent to the
equal angles in each, is common to both triangles.
Therefore the other sides are equal, each to each, and the
third angle of the one to the third angle of the other —
.'. AB =^ namely, AB equal to CD, AC to BD, and the angle BAC to
bd', zbac the angle CDB (I. 26).
= z CDB, j^^^ because the angle ABC is equal to the angle BCD,
and the angle CBD to the angle ACB,
and Therefore the whole angle ABD is equal to the whole
' ' '^'^ angleACD(Ax. 2).
And the angle BAC has been shown to be equal to the
angle BDC; therefore the opposite sides and angles of a
parallelogram are equal to one another.
Also the diagonal bisects it.
ABD
= Z ACD,
PROPOSITIONS.
125
For AB being equal to CD, and BC common,
The two sides AB, BC are equal to the two sides CD and
CB, each to each.
And the angle ABC has been shown to be equal to the
angle BCD ;
Therefore the triangle ABC is equal to the triangle BCD ^^abc =3
(I. 4),
And the diagonal BC divides the parallelogram ABCD
into two equal parts.
Therefore, the opposite sides, &c. Q, E, B,
also
A -
A BCD,
Proposition 35. — Theorem,
Parallelograr)is wpon the same hose, and between the same
joarallels, are equal to one another. ,- ^
Let the parallelograms A BCD, EBCF be on the same base
BC, and between the same parallels AF, BC;
The parallelogram ABCD shall be equal to the parallelo-
gi\am EBCF.
Case 1.— If the sides AD, DF of the ^
parallelograms ABCD, DBCF, opposite
to the base BC, be terminated in the
same point D, it is plain that each of the
parallelograms is double of the triangle
DBC (I. 34), and that they are therefore equal to one
another (Ax. 6).
Case 2. — But if the sides AD, EF, opposite to the base
BC, of the parallelograms ABCD, EBCF, be not terminated
iu the same point, then —
Proof. — Because
ABCD is a parallelo-
gi'am, AD is equal to
BC (I. 34). B c
For the same reason EF is equal to BC;
Therefore AD is equal to EF (Ax. 1), and DE is common;
Therefore the whole, or the remainder, AE, is equal to the
whole, or the remainder, DF (Ax. 2, or 3),
And AB is equal to DC (I. 34).
Therefore the two EA, AB are ^(jual to W^ two FD, DC,
each to each;
AD = BC.
EF = BC.
•.AE=Dr
Hence
126 GEOMETRY,
And the exterior angle FDC is equal to the interior EAB
(I. 29);
Therefore the base EB is equal to the base EC (I. 4),
a^'e'^ = And the triangle EAB equal to the triangle EDO (I. 4).
^ ^^^' Take the triangle EDO from the trapezium ABCE, and
from the same trapezium ABCE, take the triangle EAB, and
the remainders are equal (Ax. 3),
That is, the parallelogram ABCD is equal to the parallelo-
gram EBCE.
Therefore, parallelograms, &c. Q, E, D.
Proposition 36.— -Theorem,
Parallelograms upon equal bases, and between the same
parallels, are equal to one another.
Let ABCD, EEGH be parallelograms on equal bases BC,
EG, and between the same parallels AH, BG;
The parallelogram ABCD shall be equal to the parallelo-
gram EEGH,
Construction. — Join BE, CH.
Proof. — Because BC is equal
to EG (Hyp.), and EG to EH
BC = EH, i/___iX_J ^l Therefore BC is equal to EH
^ c i^ Q (^x. 1) ; and they are parallels,
and joined towards the same parts by the straight lines
BE, CH.
But straight lines which join the extremities of equal and
parallel straight lines towards the same parts, are themselves
equal and parallel (I. 33) ;
Therefore BE, CH are both equal and parallel;
EBcnr" Therefore EBCH is a parallelogram (Def 35),
paraiieio- And it is equal to the parallelogram ABCD, because they
cquai'each ^re on the same base BC, and between the same parallels BC,
AH (I. 35).
Eor the like reason, the parallelogram EEGH is equal to
the same parallelogram EBCH;
Therefore the parallelogram ABCD is equal to the parallelo-
gram EEGH (Ax. 1).
Therefore, parallelograms, &c. Q, E, D,
and
UE = CH.
of the
PROPOSITIONS. 127
Proposition 37.— Theorem.
Triangles upon the smne base, ami between the same parallels y
are equal to one another.
Let the triangles ABC, DBC be on the same base BC, and
between the same parallels AD, BC ;
The triangle ABC shall be equal to the triangle DBC.
Construction. — Produce AD both ways, to the points E, F.
Through B draw BE parallel to CA, and through C draw
CF parallel to BD (I. 31).
Proof. — Then each of the
figures EBCA, DBCF, is a paral^
lelogram (Def. 35), and they are
equal to one another, because the}
are on the same base BC, and
between the same parallels BC,EF
(I. 35.);
And the triangle ABC is half of the parallelogram EBCA, and the
because the diagonal AB bisects it (I. 34) ; are^resjic-
And the triangle DBC is half of the parallelogram DBCF, ]^l^^^^^^^^
because the diagonal DC bisects it (I. 34).
But the halves of equal things are equal (Ax. 7) ;
Therefore the triangle ABC is equal to the tiiangle DBC,
Therefore, triangles, &o, Q, E, i>.
Propositiou 38, — Theorem.
Triarigles upon equal bases ^ and between the same parallels ,
are equal to one another.
Let the triangles ABC, DEF, be on equal bases BC, EF,
and between the same parallels BF, AD.
The triangle ABC shall be equal to the triangle DEF.
Construction. — Produce AD both ways to the points
G, H.
Through B draw BG par- f. A B
allel to CA, and through F
draw FH parallel to ED
(L 31)
Fisfurcs
CiiiCA
figures GBCA. DEFH, is a ^ c E F »efh
Proof. — Then each of the \/ \ / \/ ^^i^^A and
DEFH are
128
GEOMETRY.
parallelogram (Def. 35), and they are equal to one another, be-
cause they are on equal bases BO, EF, and between the
same parallels BF, GH (I. 36);
and the And the triangle ABC is half of the parallelogram
Ir'J'half of C^BCA, because the diagonal AB bisects it (I. 34) ;
these re- And the triangle DEF is half of the parallelogram
«pectiveiy. DEFH, because the diagonal DF bisects it (I. 34).
But the halves of equal things are equal (Ax. 7);
Therefore the triangle ABC is equal to the triangle DEF.
Therefore, triangles, &c. Q, E, D,
Proposition 39.— Theorem.
Equal triangles upon the same base, and on the same side
o/itf are between the same parallels.
Let the equal triangles ABC, DBC be upon the same base
BC, and on the same side of it ;
They shall be between the same parallels.
Construction. — Join AD ; AD shall be parallel to BC.
For if it is not, through A draw AE
parallel to BC (I. 31), and join EC.
Proof. — The triangle ABC is equal to
the triangle EBC, because they are upon
the same base BC, and between the same
B c parallels BC, AE (I. 37).
But the triangle ABC is equal to the triangle DBC (Hyp.) ;
Therefore the triangle DBC is equal to the triangle EBC
(Ax. 1), the greater equal to the less, which is impo^ible ;
Therefore AE is not parallel to BC.
In the same manner, it can be demonstrated that no line
passing through A can be parallel to BC, except AD ;
Therefore AD is parallel to BC.
Therefore, equal triangles, &c. Q. E, D,
Proposition 40. — Theorem.
Equal triangles nxpon the same side of equal bases, tliat
are in the same straight line, are between the same parallels.
Let the equal triangles ABC, DEF, be upon the same side
of equal bases BC, EF, in the same straight line BF,
AE parallel
to BC sup-
pose.
Then
A DBC =
A EBC, an
ubsurdity.
PROPOSITIONS.
129
Tlie triangles ABC, DEF shall be between the same
parallels.
Construction.— Join AD ; AD shall be parallel to BF.
For if it is not, through A draw AG parallel to BF(I. 31), j^f to^^:
and join GF. suppose.
Proof.— The triangle ABC is equal to the triangle GEF,
because they are upon equal bases BC, EF, and are between
the same parallels BF, AG
(I. 38).
But the triangle ABC is
equal to the triangle DEF ;
Therefore the triangle DEF
is equal to the triangle GEF
(Ax. 1), the gi^eater equal to
the less, which is impossible ;
Therefore AG is not parallel to BF.
In the same manner, it can be demonstrated that no line,
passing through A, can be parallel to BF, except AD ;
Therefore AD is parallel to BF.
Therefore, equal triangles, &c.
Proposition 41.— Theorem.
If a 'parallelogram and a triangle he upon the same base,
and between the same parallels^ the piarallelogram shall he
double of the triangle.
Let the parallelogram ABCD, and the triangle EBC be
upon the same base BC, and between the same parallels
BC, AE ;
The parallelogi-ani ABCD shall be double
of the triangle EBC.
Construction. — Join AC.
Proof. — The triangle ABC is equal to
the triangle EBC, because they are upon
the same base BC, and between the same
parallels BC, AE (I. 37).
But the parallelogram ABCD is double of the triangle And parai-
ABC, because the diagonal AC bisects the parallelogram (I. 2^^^^b^
130 GEOMETRY.
Therefore the parallelogram ABCD is also double of the
triangle EBC (Ax. 1).
Therefore, if a parallelogram, &c. Q. E, J),
Proposition 42.--Problem.
To describe a parallelogram that shall he equal to a given
triangle, and have one of its angles equal to a given rectilineal
angle.
Let ABC be the given triangle, and D the given recti-
lineal angle ;
It is required to describe a parallelogram that shall be
equal to the given triangle ABC, and have one of its angles
equal to D.
Make BE A Y G CONSTRUCTION. BisCCt BC in E (I.
= *^^ /ITT 7 10), and join AE.
/ Y / / At the point E, in the straight line
z"cEF=D / i\ / /d ^-^j make the angle CEF equal to D
/ / \ / (^- ^^)-
/ / \/ Through A draw AEG parallel to
B E c EC (I. 31).
Through C draw CG parallel to EE (I. 31).
Then EECG is the parallelogram required.
Proof. — Because BE is equal to EC (Const.), the triangle
ABE is equal to the triangle AEC, since they are upon equal
bases and between the same parallels (I. 38) ;
I^A^AEC Therefore the triangle ABC iS double of the triangle
and also ' AEC.
FECG = ^^t the parallelogram EECG is also double of the triangle
2 A AEC. AEC, because they are upon the same base, and between the
same parallels (I. 41) ;
Therefore the parallelogram EECG is equal to the triangle
ABC (Ax. 6),
And it has one of its angles CEF equal to the given angle
D (Const.).
Therefore a parallelogram FECG has been described equal
to the given triangle ABC, and having one of its angles CEF
equal to the given angle D. Q. E. F,
A ABC:
3PJ10P0SITI0NS. 131
Proposition 43.— Theorem.
The complements of the parallelograms which are about the
dmgonal of any parallelogram are equal to one another.
Let ABCD be a parallelogi-am, of whicli the diagonal is
AC ; and EH, GF parallelograms about AC, that is, through
which AC passes ; and BK, KD the other parallelograms,
which make up the whole figure ABCD, and are therefore
called the complements.
The complement BK shall be equal to the complement
KD.
Proof. — Because ABCD is a parallelogram, and AC its
diagonal, the triangle ABC is equal to the triangle ADC a adc.~
(I. 34).
Again, because AEKH is a paral-
lelogram, and AK its diagonal, the
triangle AEK is equal to the triangle
AHK (I. 34).
For the like reason the triangle
KGC is equal to the triangle KFC.
Therefore, because the triangle
AEK is equal to the triangle AHK, and the triangle KGC
to KFC,
The triangles AEK, KGC are equal to the triangles
AHK, KFC (Ax. 2).
But the whole triangle ABC was proved equal to the
whole triangle ADC ;
Therefore the remaining complement BK is equal to the .:.^bk =
remaining complement KD (Ax. 3).
Therefore, the complements, &c. Q. E. D,
KD.
Proposition 44.--Problem.
To a given straight line to apply a parallelogram, lohich
shall he equal to a given triangle, and have one of its angles
equal to a given rectilineal angle.
Let AB be the given straight line, C the given triangle,
and D the given angle.
132
GEOMETRY.
Make
parallelo-
jrram
BEFG =
A C, and
z at B =
L D, and
EBAa
straight
line.
HBandFE
meet.
It is required to apply to tlie straight line AB a parallelo-
gram equal to tlie triangle C, and having an angle equal
toD.
Construction 1. — Make the parallelogram BEFG equal
to the triangle C, and having the angle EBG equal to the
angle D (I. 42) ;
And let the parallelogram BEFG be made so that BE may
be in the same straight line with AB.
Produce EG to H.
Through A draw AH parallel to BG or EF (I. 31).
Join HB.
Proof 1. — Because the straight line HE falls on the
parallels AH, EF, the angles AHF, HFE are together equal
to two right angles (I. 29).
Therefore the angles BHF, HFE are together less than
two right angles (Ax. 9). But straight lines which with
another straight line make the interior angles on the same
side together less than two right angles, will meet on that
side, if produced far enough (Ax. 12) ;
Therefore HB and FE shall meet if produced.
Construction 2. — Produce HB and FE towards BE,
and let them meet in K.
Figures
L13 = BF.
Through K draw KL parallel to EA or FH (I. 31).
Produce HA, GB to the points L, M.
Then LB shall he the ioarallelograi)i required.
Proof 2. — Because HLKF is a parallelogram, of which
the diagonal is HK ; and AG, ME are the parallelograms
about HK ; and LB, BF are the complements ;
Therefore the complement LB is equal to the complement
BF (I. 43).
z D.
PROPOSITIONS. 133
But BF is equal to the triangle C (Const.) ; But_ ^
Therefore LB is equal to the triangle C (Ax. 1). .-. lb = '
And because the angle GBE is equal to the angle ABM ^^*
(I. 15), and likewise to the angle D (Const.) ;
Therefore the angle ABM is equal to the angle D (Ax. 1). ^^^^^^^ ^
Therefore, the parallelogi-am LB is applied to the straight z gbe =
line AB, and is equal to the triangle C, and has the angle
ABM equal to the angle D. Q, E. F,
Proposition 45. — Problem.
To describe a parallelogram equal to a given rectilineal
figure, and having an angle eqvxd to a given rectilineal angle.
Let ABCD be the given rectilineal figure, and E the
given rectilineal angle.
It is required to describe a parallelogram equal to ABCD,
and having an angle equal to E.
Construction. — Join
DB. A .. T) F Cx l i
Describe the parallelo-
gram FH equal to the
triangle ADB, and
having the angle FKH
equal to the angle E
(I. 42).
To the straight line
GH apply the parallelogram GM equal to the triangle DBC,
and having the angle GHM equal to the angle E (I. 44).
Then the figure FKML shall he the parallelogram required.
Proof. — Because the angle E is equal to each of the angles
FKH, GHM (Const.),
Therefore the angle FKH is equal to the angle GHM
(Ax. 1).
Add to each of these equals the angle KHG;
Tlierefore the angles FKH, KHG are equal to the angles
KHG, GHM (Ax. 2).
But FKH, KHG are equal to two right angles (I. 29);
Therefore also KHG, GHM are equal to two right angles
(Ax. 1).
134 GEOMETRY.
And because at the point H, in the straight line GH, the
two straight lines KH, HM, on the opposite sides of it, make
the adjacent angles together equal to two right angles,
Therefore KH is in the same straight line with HM (1. 14).
And because the straight line HG meets the parallels
KM, FG, the alternate angles MHG, HGF are equal (I. 29).
Add to each of these equals the angle HGL;
Therefore the angles MHG, HGL are equal to the angles
HGF, HGL (Ax. 2).
But the angles MHG, HGL are equal to two right angles
(L 29);
Therefore also the angles HGF, HGL are equal to two
right angles,
And therefore FG is in the same straight line with GL
(L 14).
And because KF is parallel to HG, and HG parallel to
ML (Const.);
Therefore KF is parallel to ML (I. 30).
And KM, FL are parallels (Const) ;
Therefore KFLM is a parallelogram (Def. 35).
And because the triangle ABD is equal to the jDarallelo-
gram HF, and the triangle DBC equal to the parallelogram
GM (Const.),
Therefore the whole rectilineal figure ABCD is equal to
the whole parallelogram KFLM (Ax. 2).
Therefore, the parallelogram KFLM has been described
equal to the given rectilineal figure ABCD, and having the
angle FKM equal to the given angle E. Q. E. F,
Corollary. — From this it is manifest how to apply to a
given straight line a parallelogram, which shall have an angle
equal to a given rectilineal angle, and shall be equal to a given
rectilineal figure — namely, by applying to the given straight
line a parallelogram equal to the first triangle ABD, and
having an angle equal to the given angle ; and so on (I. 44).
Proposition 46.— Problem.
To describe a square upon a given straight line.
Let AB be the given straight line;
It is required to describe a square upon AB.
PROPOSITIONS.
135
Construction. — From the point A draw AC at right
angles to AB (I. 11),
And make AD equal to AB (I. 3).
Through the point D draw DE parallel
to AB (I. 31).
Through the point B di'aw BE parallel
to AD (I. 31).
Then ADEB shall he the square re-
quired.
Proof. — Because DE is parallel to
AB, and BE parallel to AD (Const.),
therefore ADEB is a parallelogram;
Therefore AB is equal to DE, and AD to BE (I. 34).
But AB is equal to AD (Const.);
Therefore the four straight lines BA., AD, DE, EB are
equal to one another (Ax. 1),
And the parallelogram ADEB is therefore equilateral
Likewise all its angles are right angles.
For since the straight line AD meets the parallels AB,
DE, the angles BAD, ADE are together equal to two right
angles (I. 29).
But BAD is a right angle (Const.) ;
Therefore also ADE is a right angle (Ax. 3).
But the opposite angles of parallelograms are equal (I. 34) ;
Therefore each of the opposite angles ABE, BED is a right
angle (Ax. 1) ;
Therefore the figure ADEB is rectangular ; and it has it is^^rec
been proved to be equilateral; therefore it is a square
(Def. 30).
Therefore, the figure ADEB is a square, and it is described .*.a square.
upon the given straight line AB. Q. E. F.
Corollary. — Hence every parallelogram that has onQ
right angle has all its angles right angles.
ADEB a
B parallelo-
gram.
It is equi-
lateral.
Proposition 47.— Theorem.
In any right-angled triangle, the square ivhich is described
up(yn the side opposite to the right angle is equal to the squares
described upon the sides which contain the right angle*
13G
GEOMETRY.
CAO is a
straight
line.
BAH is a
straight
line.
A ABD =
A FBC.
Hence
})aralIelo-
gram BL
= square
GB, and
parallelo-
gram CL
= square
HC.
Let ABC be a liglit-angled triangle, Laving tlie right
angle BAG ;
The square described npon the
side BC shall be equal to the
squares described upon BA, AC.
Construction. — On BC de-
scribe the square BDEC (I. 46).
On BA, AC describe the squares
GB, HC (I. 46).
Through A draw AL parallel
to BD or CE (I. 31).
Join AD, EC.
Proof. — Because the angle BAC
is a right angle (Hyp.), and that the angle BAG is also a
right angle (Def. 30),
The two straight lines AC, AG, upon opposite sides of
AB, make with it at the point A the adjacent angles equal
to two right angles ;
Therefore CA is in the same straight line with AG (I. 14).
Eor the same reason, AB and AH are in the same straight
line.
Now the angle DBC is equal to the angle EBA, for each
of them is a right angle (Ax. 11); add to each the angle ABC.
Therefore the whole angle DBA is equal to the whole angle
EBC (Ax. 2).
And because the two sides AB, BD are equal to the two
sides FB, BC, each to each (Def. 30), and the angle DBA
equal to the angle EBC ;
Therefore the base AD is equal to the base EC, and the
triangle ABD to the triangle FBC (I. 4).
Now the parallelogram BL is double of the triangle ABD,
because they are on the same base BD, and between the
same parallels BD, AL (I. 41).
And the square GB is double of the triangle EBC, because
they are on the same base FB, and between the same parallels
FB, GC (L 41).
But the doubles of equals are equal (Ax. 6), therefore the
parallelogram BL is equal to the square GB.
In the same manner, by joining AE, BK, it can be shown
that the parallelogram CL is equal to the square HC.
3?ROPO^iTlONS. 137
Therefore the whole square BDEC is equal to the two
squares GB, HC (Ax. 2);
And the square BDEC is described on the straight line is'A^+Aca
BC, and the squares GB, HC upon BA, AC.
Therefore the square described upon the side BC is equal
to the squares described upon the sides BA, AC.
Therefore, in any right-angled triangle, &c. Q. E. D.
Proposition 48. — Theorem.
If the square described iqwn one of the sides of a triangle
he equal to the squares described upon the other two sides of
it, the angle contained by these two sides is a right angle.
Let the square described upon BC, one of the sides of the
triangle ABC, be equal to the squares described upon the
other sides BA, AC;
The anijle BAC shall be a riffht an-gjle. Draw
An f
Construction. — From the point A draw AD at right right
aiiglesto AC(I. 11). ^Jf^^^*^
Make AD equal to BA (I. 3), and join DC. (Do not
Proof. — Because DA is equal to AB, the square on DA ^ba.Y^
is equal to the square on BA.
To each of these add the square on AC.
Therefore the squares on DA, AC are
equal to the squares on BA, AC (Ax. 2).
But because the angle DAC is a right
angle (Const.), the square on DC is equal to
the squares on DA, AC (I. 47),
And the square on BC is equal to the
squares on BA, AC (Hyp.) ; r^hew
Therefore the square on DC is equal to the square on BC dc2 =
(Ax. 1) ; and*
And therefore the side DC is equal to the side BC. » DC = BC.
And because the side DA is equal to AB (Const), and AC
common to the two triangles DAC, BAC, the two sides
DA, AC are equal to the two sides BA, AC, each to
each.
And the base DC has been proved equal to the ba.se BC ;
138 GEOMETRY.
Hence _ Therefore the angle DAC is equal to BAG (I. 8),
zBAC." But DAC is a right angle (Const.) ;
Therefore also BAC is a right angle (Ax. 1).
Therefore, if the square, &c. Q. E, D,
EXERCISES.
Prop. 1—15.
1. From the greater of two given straight lines to cut off a portion
which is three times as long as the less.
2. The line bisecting the vertical angle of an isosceles triangle also
bisects the base.
3. Prove Euc. I. 5, by the method of super-position.
4. In the figure to Euc. I. 5, show that the line joining A with
the point of intersection of BG and FC, makes equal angles with
AB and AC.
5. ABC is an isosceles triangle, whose base is BC, and AD is
perpendicular to BC ; every point in AD is equally distant from B
and C.
6. Show that the sum of the sum and difference of two given
straight lines is twice the greater, and that the difference of the sum
and difference is twice the less.
7. Prove the same property with regard to angles.
8. Make an angle which shall be three-fourths of a right angle.
9. If, with the extremities of a given line as centres, circles be
dra-WTi intersecting in two points, the line joining the points of in-
tersection will be perpendicular to the given line, and will also
bisect it.
10. Find a point which is at a given distance from a given point
and from a given line.
11. Show that the sum of the angles round a given point are
together equal to four right angles.
12. If the exterior angle of a triangle and its adjacent interior
angle be bisected, the bisecting lines will be at right angles.
13. If three points, A, B, C, be taken not in the same straight line,
and AB and AC be joined and bisected by perpendiculars which meet
in D, show that DA, DB, DO are equal to each father.
Prop. 16—32.
^ 14. The perpendiculars from the angular points upon the opposite
sides of a triangle meet in a point.
15. To construct an isosceles triangle on a given base, the sides
being each of them double the given base.
j EXERCISES ON THE PROPOSITIONS. 139
16» Describe an isosceles triangle having a given base, and whose
vertical angle is half a right angle.
17. AB is a straight line, C and D are points on the same side of
it ; find a i^oint E in AB such that the sum of CE and ED shall be a
minimum.
18. Having given two sides of a triangle and an angle, construct
the triangle. Examine the cases when there will be (1.) one solution;
(2.) two solutions ; (3.) none.
19. Given an angle of a triangle and the sum and difference of the
two sides including the angle, to construct the triangle.
20. Show that each of the angles of an equilateral triangle is two-
thirds of a right angle, and hence show how to trisect a right angle.
21. If two angles of a triangle be bisected by lines drawn from the
angular points to a given point within, then the line bisecting the
third angle will pass through the same point.
22. The difference of any two sides of a triangle is less than the
third side.
23. If the angles at the base of a right-angled isosceles triangle be
bisected, the bisecting line includes an angle which is three halves
of a right angle.
24. The sum of the lines drawn from any point within a polygon
to the angular points is greater than half the sum of the sides of the
polygon.
Prop. 33—48.
25. Show that the diagonals of a square bisect each other at right
angles, and that the square described upon a semi-diagonal is half
the given square.
26. Divide a given line into any number of equal parts, and hence
show how to divide a line similarly to a given line*
27. If D and E be respectively the middle points of the sides BC
and AC of the triangle ABC, and AD and BE be joined, and intersect
in G, show that GD and GE are respectively one-third of AD and BE.
28. The lines drawn to the bisections of the sides of a triangle
from the opposite angles meet in a point.
29. Describe a square which is live times a given square.
30. Show that a square, hexagon, and dodecagon will fill up the
space round a point.
31. Divide a square into three equal areas, by lines drawn parallel
to one of the diagonals.
32. Upon a given straight line construct a regular octagon.
33. Divide a given triangle into equal triangles by lines drawn from
one of the angles.
34. If any two angles of a quadrilateral are together equal to two
right angles, show that the sum of the other two is two right angles.
35. The area of a trapezium having two parallel sides is equal to
half the rectangle contained by the perpendicular distance between
the parallel sides of the trapezium, and the sum of the parallel sides.
140 GEOMETRY.
36. The area of any trapezium is half the rectangle contained by
one of the diagonals of the trapezium, and the sum of the perpendicu-
lars let fall upon it from the opposite angles.
37. If the middle points of the sides of a triangle be joined, the
lines form a triangle whose area is one-fourth that of the given
triangle.
38. If the sides of a triangle be such that they are respectively the
sum of two given lines, the difference of the same two lines, and
twice the side of a square equal to the rectangle contained by these
Imes, the triangle shall be right-angled, having the right angle
opposite to the hrst-named side.
39. If a point be taken within a triangle such that the lengths of
the perpendiculars upon the sides are equal, show that the area of the
rectangle contained by one of the perpendiculars and the perimeter of
the triangle is double the area of the triangle.
40. In the last problem, if be the given point, and OD, OE, OF
the respective perpendiculars upon the sides BC, AC, and AB,
show that the sum of the squares upon AD, OB, and DC exceeds the
sum of the squares upon AF, BD, and CD by three times the square
upon either of the perpendiculars.
41. Having given the lengths of the segments AF, BD, CE, in
Problem 40, construct the triangle.
42. Draw a line, the square upon which shall be seven times the
square upon a given line.
43. Draw a line, the square upon which shall be equal to the sum
or difference of two given squares.
44. Reduce a given polygon to an equivalent triangle.
45. Divide a triangle into equal areas by drawing a line from a
given point in a side.
46. Do the same with a given parallelogram.
47. If in the fig., Euc. I. 47, the square on the hypothenuse be on
the other side, show how the other two squares may be made to cover
exactly the square on the hypothenuse.
48. The area of a quadrilateral whose diagonals are at right angles
is half the rectangle contained by the diagonals.
SECTION HI.
ALGEBRA.
CHAPTER I.
ELEMENTARY PRINCIPLES.
1. Algebra treats of numbers, the numbers being repre-
sented by letters (symbols of quantity), affected with certain
symbols of quality, and connected by symbols of operation.
It is easy to see that these symbols of Quantity may be
dealt with very much as we deal with concrete quantities in
arithmetic. Thus, allowing the letter a to stand for the
number of units in any quantity, and allowing also 2 a, 3 a,
4 a, tkc, to stand respectively for twice, thrice, four times,
&c., as large a quantity as the letter a, it at once follows
that we may perform the operations of addition, subtraction,
multiplication, and division upon these symbols exactly as we
do in ordinary arithmetic upon concrete quantities. For
instance, 4 a and 6 a make 10 a, 9 a exceeds 5 a by 4 a, 15 a
is 5 times 3 a, and 7 a is contained 8 times in 56 a.
Neither is it necessary in these operations to state, or even
to know the exact number of units for which any symbol of
quantity stands, nor indeed the nature of these units ; it is
simply sufficient that it is a symbol of quantity. Thus, in
the science of chemistry, we use a weight called a crith ;
and a person unacquainted with chemistry might not know
wli ether a crith were a measure of length, weight, or
capacity, or indeed whether it were a measure at all, yet he
would at once allow that 6 criths and 5 critha JU"© H criths,
that twicQ 4 crith^ c^rg 8 criths, 4q,
142 ALGEBRA.
The Signs + and - as Symbols of Operation.
2. In purely aritlimetical operations, tlie signs + and -
are respectively the signs of addition and subtraction. In
this sense, too, they are used in algebra. *
Thus, a + b means that b is to be added to a, and a — 1)
means that b is to be subtracted from a.
Hence, as long as a and b represent ordinary arithmetical
numbers, a -v b admits of easy interpretation, as also does
a -^ b, when b is not greater than a. But when b is greater
than a, the expression a — b has no arithmetical meaning.
By an extension, however, of the use of the signs + and
— , we are able to give such expressions an intelligible signi-
fication, whatever may be the quantities represented by
a and 5.
Positive and Negative Quantities.— The Signs + and -
as Symbols of Affection or Quality,
8. Def, — A positive quantity is one which is affected
with a + sign, and a negative quantity is one which is
affected with a — sign.
Let BA be a straight line, and O a point in the line;
and suppose a person, starting ,
from O, to walk a miles in the "^ " -^
direction OA. Suppose also another person, starting from
the same or any other point in BA to walk a miles in the
direction OB. These persons will thus walk a miles each in
exactly opposite directions. Now, we call one of these direc-
tions positive (it matters not which) and the other negative.
Let us take the direction OA as positive. We then have
the first person walking a miles in a positive direction, and
the second walking a miles in a negative direction. We
represent these distances algebraically hj + a and — a
respectively.
It will therefore be seen that the signs + and — have no
effect upon the magnitudes of quantities, but that they express
the quality or affection of the quantities before which they
stand.
THE SUM OP ALGEBRAICAL QUANTITIES. 143
Again, suppose a person in business to get a profit of £6,
while another suffers a loss of £6. We may express these
facts algebraically in two ways. We may consider gain as
positive, and loss as negative gain, and say that the former
has gained + 6 pounds, while the latter has gained — 6
pounds. Or we may consider loss as positive, and gain as
negative loss, and say that the former has lost — 6 pounds,
while the latter has lost + 6 pounds. We hence see that
the gain of + 6 is equal to a loss of - 6. and that a gain of
— 6 is equal to a loss of + 6.
The Sum of Algebraical Quantities.
4. Let a distance AB be measured to the right along the
line AX. And let a further — *- -« .,
distance BC be measured from
B in the same direction. By the sum of these lines we mean
the resulting distance of the point C from the original point
A, that is to say, the distance AC.
(It may be remarked that we add the line BC to the line AB by
measuring BC in its own proper direction from the extremity B of
AB. It is hardly necessary to remind the student that both lines
are in the same straight line AX.)
Let us represent the distances AB and BC by + a and
+ h respectively ; then the algebraical sum of the lines will
be represented writing these quantities side by side, each
with its own proper sign of affection.
Thus the sum of the distances AB and BC is expressed by
+ a + h, or, as it is usual to omit the + sign of a positive
quantity when the quantity stands alone or at the head of
an algebraical expression, the sum of AB and BC is expressed
by a + 5.
Hence, the interpretation of a + 6 is that it represents the
distance AC.
Again, taking as above + a to represent the distance AB
along the straight line AX, and ,^ <; j^
measured to the right, let a dis- ,^ ^, ^
tance BC be measured from B in ' * ^
the same straight line AX, but — c " ' ^
this time to the left.
144 ALGEBRA.
Let the latter distance be represented by - 5.
Then, on the principle above, AC is the sum of these dis-
tances, and this sum is represented algebraically by + a — 6
or a - 6.
(It "will be seen that the distance BC is again measured from B in
its own proper direction^ and that the resultant distance AC of the
point C is again the sum of the line AB and BC.)
There will evidently be three cases, viz. : —
1. When the distance BC is less in magnitude than AB,
in which case the point C is on the right of A, and the dis-
tance AC is positive.
2. When the distance BC is equal in magnitude to AB,
in which case the point C coincides with A, and the distance
AC is zero.
3. When the distance BC is greater in magnitude than
AB, the point C being then on the left of A, and the distance
is negative.
Now, a — h in all these cases represents the distance AC.
It therefore admits of intelligible interpretation whether h
be less than, equal to, or greater than a.
And, since the distance AC is obtained in the first two
cases by subtracting the distance BC from that of AB, and
in the second case by subtracting as far as AB will allow of
subtraction, and measuring the remainder to be subtracted in
an opposite direction, it follows that —
The sign - , which, standing before a letter, is a symbol
of quality, becomes at once a symbol of subtraction in all
cases when the quantity in question is placed immediately
after any other given quantity with its proper sign of
affection.
Hence also we may conclude that the addition of a Tiega-
tive quantity is equivalent to the subtraction of the corres-
ponding positive quantity.
5. We may prove in a similar way that —
The subtraction of a negative quantity is equivalent to the
addition of the corresponding joosi^we quantity.
Let, as before, + a represent the distance AB, measured
from the point A to the rig] it ^ , , . , .
and let it be required to sub- '^ a (*
^vact from f a the distance represented by - b.
THE SUM OP ALGEBRAICAL QUANTITIES. 145
Now, in the last article, we added a distance to a given
distance AB, by measuring the second distance in its own
direction from the extremity B. We shall therefore be con-
sistent if we subtract a. given distance — 6 by measuring this
distance in a direction exactly opposite to its own direction,
from the same extremity B.
Now, the direction of — 6 is to the left. If, therefore, we
measure a distance BC to the right, equal in magnitude to
the distance to be subtracted, we obtain a distance AC which
is correctly represented by a - ( - h). But AC is also
correctly represented by a + b, and hence it follows that
a-(-b)=a-\-b.
We may apply the above principle to all magnitudes which
admit of continuous and indefinite extension; as, for instance,
to forces which pull and push, attract and repel ; to time
past and time to come, to temperatures above zero and below
zero, to money due and money owed, to distance up and dis-
tance down, &c., in all which cases, having represented one
by a quantity affected with a + sign, we may represent the
other by a quantity affected with a - sign.
6. In expressing the sum of a number of quantities^ the
order of the terms is immaterial.
We will take, as our illustration, a body subject to various
alterations of temperature, and we will suppose the tempera-
ture of the body, before the changes in question, to be zero
or 0°. Let. the temperature now undergo the following
changes — viz., a rise of a°, a fall of 6°, a fall of c°, and, lastly,
a rise of c^°. Let us consider a rise as positive, and therefore
a fall as negative. We may then represent these changes
respectively by + a, — 6, — c, + c?.
And it is further evident that the resulting temperature
will be represented by the sum of these quantities, which, as
previously written, will be a - 6 — c + c?.
But again, it is plain that the resulting temperature of the
body will not be affected if these changes of temperature take
place in the reverse order, or in any other order. Thus,
suppose the temperature first falls c°, then rises a°, then rises
d°, and, lastly, falls 6°, it is evident that the final temperature
will be the same as before. And the sum of the quantities
»- c, + a, -t- c^, —h, represents this final temperature,
14G ALGEBRA.
Now, expressing the sum of these quantities by writing them
(Art. 4) side by side with their proper signs of affection,
and in the order in which they stand, we have — c + a + d
— b for the sum.
It therefore follows, since we might have chosen any other
order of these terms with a similar result, that the sum of
any number of quantities, + a, — h, — c, + cZ, may be
expressed by writing the terms side by side with their proper
signs of affection in any order whatever,
Nevertheless, for convenience, and for other reasons, we
write the terms generally in alphabetical order, or we arrange
them according to the power (see Art. 16} of some particular
letter.
7. We may sum up the results and remarks of the last
four articles as follows : —
1. Positive and negative are used in exactly opposite
senses.
2. The sign + before an algebraical quantity affirms the
quality of the quantity as represented without the sign m
question.
Thus, + ( + a) = + a, and + ( — 5) = — 6.
3. The sign — before an algebraical quantity reverses the
quality of the. quantity as represented without the sign in
question.
Thus, — ( + a) = — o, and — ( - 5) = +6.
4. The algebraical subtraction of a quantity is the same as
the addition of the quantity with the sign of affection
5. The algebraical .addition of quantities is expressed by
writing the quantities down side by side with their proper
signs of affection. And they may be written down in any
order, though we generally write them in alphabetical order,
or arranged according to the power (Art. 16) of some
letter.
Brackets.
r
8. Brackets — ( ), } }, [ ] — are used, for the most part,
whenever we wish to consider an algebraical expression con-
taining more than one term as a z(;Me.
BRACKETS. 14:7
Thus, if WO wish to express that the quantity 3 « + 7 6 is
to be added as a whole to 4 a, we write —
4a + (3 a + 7 6),
and, while inclosed within brackets, we think and speak of
S a + 7 5 as one quantity.
Again, if we wish to express that 6 — c is to be subtracted
from a, we write —
ej — (5 - c).
Let us consider what is the result of subtracting (b — c)
from a. We may evidently, if we please, subtract b first,
then afterwards - c from the quantity so obtained, without
affecting this result.
Now, we know by Art. 7 (4.) that this is equivalent fo
adding the quantities - b and + c successively.
Now, the sum of a, — 5, + c, is a - b + c.
We have therefore a — (6 - c) = a - 6 + c.
We observe that the sign of b within the brackets is -F,
and that of c is - , whereas, in our final result, these sigiisa
are both reversed. And we hence arrive at the following
important principle : —
When a minus sign stands before a bracket, its effect on
removing the brackets is to reverse the sign of affection of
every term within,
Aiid it is evident that we may, by a similar course
of reasoning, arrive at a principle equally important,
viz. : —
When a plus sign stands before a bracket, its effect on
removing the bracket is to affirm the sign of affection of
every term within.
We shall show, in Art. 9, the use of brackets in expressing
the product or quotient of quantities.
Though, as stated above, brackets are, for the most part,
used to group together as a whole a number of quantities,
they are sometimes used to inclose single terms. Thus, in
Art. 5, we have the expression a — ( - b). Now, the
brackets are used here to express that the negative quantitij
is to be subtracted as a negative quantity. And, in the same
way, the expression a + (—6) indicates that the negative
quantity ( - 6) is to be added to the quantity a. When one
pair only is required we generally use the bi*ackets ( ) ; if,
H8 ALGEBRA.
however, a quantity already in brackets is to be inclosed in
a second pair, we use { (, as in the expression —
3 a - I 6 5 + (4 c - c?) I .
If a third pair be required we use the brackets [ ], and
finally, we sometimes find it convenient to group a number
of terms by means of a vinculum^ thus —
4:X -[^x - {by + {Zz " 7 a; - 2/)} + 2y\
It must be remembered that the vinculum has in an
expression exactly the same force as brackets.
9. We shall, in this Article, show how to find the value of
a few algebraical expressions, as illustrations of the foregoing
principles : —
Ex. l.—If a = 1, 6 = 2, c = 4, find the value of
3a + 55 + 7c.
We have only to substitute the value of the letters in the
given expression, putting a sign of multiplication to avoid
ambiguity.
Thus we have —
3a + 56 + 7c = 3xl + 5x2 + 7x4.
= 3 + 10 + 28 = 41.
Ex. 2.— If a; = 5, 2/ = 2, « = 6, find the value cf
^x + Zy - ^z.
Here, 4a; + 32/-9;s = 4x5 + 3x2-9x6
'=20 + 6-54
= 26 - 54.
The negative quantity is here the larger, and exceeds the
positive quantity by 28, and hence (Art. 4 (3),) the result will
be negative.
We therefore have —
4a; + 32/-92;=-28.
Ex. 3.--If aj = 1, 2/ = 3, ^ = 0, find the value of—
3aj-|22/ + (6« - 5 a; - 2/) I •
The given expression —
= 3xl-|2x3 + (GxO- 5x1-3) |
PRODUCT OF TWO OR MORE QUANTITIES. 140
- 3 - j 6 + (0 - 2) j .
= 3-(6-2)-3-4=-l.
It may be advisable to simplify expressions of this kind
before substituting the given value of the letters. The
method of doing this, however, is deferred till we come to
Chapter II.
Ex. I.
If a = 1, 5 = 2, c = 3, f^ = 0, e = 4, find the value of the
following : —
1. 4c a + 2by3b + 7cyQa + 4cdy 4:c - 7e.
2. a + h + Cya — b + Cyb + c-a, a + b — c.
3. 3a + 7b-4.Cy2a + 7d + 3c,7a-lOb + 2c,
4. 6a- (2 6 - 3c), 7 b + (2 a - 4cd).
5. 3 c + (6 a - 7 6 + c), 26-|7c + (4cZ-Z;)|.
6. I 7 6 - (2 cZ - c) I - (4 6 - c + 6).
If X = 2, 9/ = 3, z = 4:y find the sum and difierence of the
following expressions : —
7. 3 03 — 4 y and 3 7/ - 4 a;.
8. 7 {x - y) and 4 (?/ - c).
9. 3 ic - (7 2/ + 4 ;^) and 8 y + 5 ;:; - 3 iT.
10. \2y - Qy + ^z and \2 y + Qt y + A:Z.
11. X - y - z and x - {y - z).
12. X - ( - 3 7/) and y - | 3 a; - ( - 3z)\.
Product of Two or more Quantities.
10. The product of two or more quantities may be
expressed in several ways.
Tims, the product of a, 6, c may be Avritten as follows : —
1. abc, by placing the letters side by side without any
sif/u between them.
2. a X b X c, by ])lacing bet«vcen them the sign x .
3. a , b , c, hy j)h\cing a dot between them.
150 ALGEBRA.
4. (a) (b) (c), by inclosing the quantities in brackets,
writing them without any sign between the brackets.
When, however, the quantities are negative, or either of
them consists of more than one term, it is best to inclose such
quantities in brackets, and, in most cases, it is necessary to
do so.
Thus, the product of a, - b, c would not be correctly
expressed hy a - be (for this means, that the product of
h and c is to be subtracted from a), but must be written
a {- b) c. Again, the product of 2 a + 3 5 and a + 5 b
cannot be written 2 a + 3ba + 5 6, as this expression means
that three times the product of 6 x «^ is to be added to 2 a,
and then 5 5 to the result. The product is correctly ex-
pressed thus —
{2a + 3b) (a + 5 b).
(The student cannot be too strongly cautioned against leaving out
brackets in cases of this kind. )
The Order of the Letters.
11. It is evident that a times b^b times a ; for, if we aii'ange
ft rows of b things so as to have a horizontal rows and b
vertical columns, we may either consider the number of rows
or the number of columns. In the former case we have
a times b things, and, in the latter, b times a things.
We may therefore write the letters whose product we
wish to express in any order.
Thus, the product of a, b, c may be written in either
of the following ways : —
abCf acby bac, cba, bca, cab.
For convenience, however, and for reasons which the
student will see as he proceeds with the subject, we write
them in the order of the alphabet, unless there happen to be
special reasons to the contrary.
Rule of Signs in Multiplication.
12. It was shown (Ai-t. 3) that a minus sign does not affect
the magnitude of a quantity, but simjily its affection or
QUOTIENT OF TWO QUANTITIES. I5l
quality. It therefore follows that the product of + ^ and
- h will be the same in magnitude as that of + a and + 6,
or of ^ X h — that is, will be equal in magnitude to ah.
But it is evident that the quality of the product will be the
same as that of - h, for it is a times a negative quantity.
We therefore have + a ( - 6) = - a6.
So also, the product of - <x and + h, will have the same
magnitude as that of + a and + h — that is, its magnitude
will be ah. But since, to take a geometrical illustration, a
line of length, h, drawn negatively once, negatively twice, <kc.,
must give a negative result, the quality of the product in
question must be negative.
"We therefore have (- a) {+ h) = - ah.
Again, the product of - a and - b will be equal in
magnitude to ah, and its quality will be evidently the
reverse of the quality of the product of - a and + b, and
the quality of the latter product is above shown to be
negative. The product of -ax - 6 will be therefore
positive.
Hence we have ( - a) ( - 5) = + ah.
Collecting these results, and remembering that ( + a)
( + 5) = + ah, we have —
{+ a) {+ h) = + ah.
{+ a) {- h) = - ah.
(-«)(+&)= -ah,
( - a) ( - 6) = + ah.
We have then the following rule of signs in multiplica-
tion : —
Rule. — Like signs give + , and unlike signs give - .
Quotient of Two Quantities.
13. The quotient of two quantities is expressed in either
of the two following ways : —
1. By placing the divisor under the dividend, sepai'ated by
a line.
Thus, the quotient of a by 6 = ^.
1S2. ALGEBRA;
2. By placing the divisor after the dividend with a sign
-f between them ; thus, a ^ h.
When either of the quantities is negative, it is better, if the
second be used, to inclose the negative quantity, if the
divisor, in brackets.
Thus, the quotient of a by - 5 may be written a -r — h,
but it is better written a -■ {- h).
And when either of the quantities contains more than one
term, it must'always be inclosed in brackets when expressed
by the second method.
Thus, the quotient of « + 2 6 by 2 c = (a + 2 5) -f 2 c,
not a + 2 5 -r 2 c, for this would mean that the quotient of
2 6 by 2 c is to be added to a.
Rule of Signs in Division.
It IS evident from the last article that —
(+ a6) -f (+ a) = + h.
{- ab) ^ {+ a) =^ - b,
(- ab) -T {- a) = + b,
{+ ab) - {- a) = - b.
We have therefore the following rule : —
KuLE. — In division, as in multiplication, like signs give
+ , and unlike signs give - .
Coefficients.
14. When a quantity can be broken up into two factors,
each of those factors is called a coefficient of the other.
Thus, taking the quantity 4 abc, we see that 4 is the
coefficient of abc, 4 b that of ac, 4 c that of ab, 4 be that of
a, &c. We call 4 a numerical coefficient ; but when the
coefficient contains a letter or letters we call it a literal
coefficient. We often speak of the numerical coefficient of
a quantity as the coefficient of the quantity. It is, moreover,
unusual to write down unity as a numerical coefficient, and
so, when an algebraical quantity has no expressed numerical
coefficient, we may, conversely, consider unity as such.
POWERS. 153
15. Like quantities are generally defined as those which
differ only in their numerical coefficients. Thus, Si x, 3 y,
Aixyz, 12 ah are respectively like to the quantities lOxc, 5 y,
6 xyz, 4 ah, while 3 x^, 6 xy, 7 5^ are called unlike.
It is sometimes convenient, however, to consider as like
quantities those which may contain perhaps one common
letter only, though containing other letters which are not
common. Thus, for the purposes of addition and subtrac-
tion, we may consider 3 ax and 4 hx as like quantities,
having 3 a and 4 h respectively as their coefficients. Now,
their sum, since they are respectively equivalent to 3 a and
4 h times x, will be equivalent to (3 a + 4 6) times x, and
may then be written (3 a + ^h) x.
Hence we learn that addition and subtraction of quanti-
ties having any common factor may, considering the un-
common factor as coefficient, be expressed according to the
following rule : —
Add together the coefficients of the common factor, and take
the sum as a new coefficient of the common factor.
And a similar rule will apply to the subtraction of such
quantities.
Thus, we have the sum of 4 axy, - 2 hey, 3 dey,
= (iax - 2hc + 3 de) y*
Powers.
16. We have seen (Art* 10) that ahc means that a is to be
multiplied by h, and the product by c. And so, if we wish
to exj^ress that a is to bo multiplied by a, and this product
again by a, we might write the expression thus : aaa.
Instead of this, however, wo usually place a small figure at
the^p and on the right hand af the letter a, to indicate how
^^^,^>^ny times the letter ai)pear^ as a factor. In this case,
^^^ therefore, we write cvK Wo call quantities of this form
powers of the letter in question; and so, remembering that a
may be considered as a^ on this principle, we have : —
a, the first power of a ; a'^, the second power oi a; a^, the
third power of a, &c. The figure written at the light hand
at the top of the letter is called the index or exponent, and it
is usual to call a", a^ respectively tlio square and cube of a.
154 ALGEBRA.
from tlie fact tliey express respectively tKe area of a square
whose side is a, and the volume of cube whose edge is a.
17. The square root of a quantity is that quantity which,
w^hen raised to the second 2>ower or squared, will give the
original quantity. _
It is generally written ^. Thus, x/l6 = 4, ^144 = 12.
The cube root of a quantity is that quantity which,
when raised to the third power or cubed, will give the
original quantity. _
I t is generally written ^. Thus, 4^8 = 2, ^1 = 3,
;^1728 = 12. And so the fourth, fifth, &c., roots are
indicated by the symbols tj , ^ , &c., respectively.
18. The dimensions of an algebraical quantity are the
sum of the indices or exponents of the literal factors.
Thus, the dimensions of 3 a'5V = 2 + 3 + 4 = 9, and
the dimensions of ahar =1 + 1+2 = 4.
19. A homogeneous expression is one in which the dimen-
sions of every term are the same.
Thus, a^ + 3 a^6 + 3 aH^ + h^ is homogeneous,
Whereas, 5 a + 3 a6 + 7 a^hc is not homogeneous,
Ex. II
Ifa = 2, 5 = 3,'r= 0, d = 1,
Find the values of —
1. 6a^ + 3 5^ - 5 c"; a6 + ac + he, he + hd + cc7.
2. a^ + 3 a'^b + 3 a6- + ^^^ ; a^ + ^^^ + c^ - 3 abc.
3. (3a + 7 5) (4a - 9 6) ; {a" + 5") {a + b) {a - b).
4. 6 {2 a^ - 4 (2 h' - ^If^^^)] ; a'b + ab'' + a'c +
ac^ + b~c + be".
If a = 1, 5 = - 2, c = -- 3, cZ = 0, e = 4,
Find the Values of—
5. (a + 5 + c + cZ + e)^ ; (cr + 2 aZ> + i" - c") + (a +
6 + c).
a"" - d' c^ + 3 c\Z + 3 cd^ + cZ^
6..
a^ + a'-'cZ + ad"^ + d^^ 6^-3 Z)'-^c + 3 bc^
7. y/cr + e^ - /v/5 g^ + 6-
"1 .1 ^ 'C^c^* + 3 c*^ + 3 c + r
ah + ae ■\- bo ^
ADDtttOK, 155
8. 3 (4 J + 5 cY + 4 (c + e)- ; abcde.
Ifoj = 3, y - 4, ;^ = 0,
Find the value of —
^' (3a; - sl^fVff {2x + ^ar + f + 2).
10. {5 a;^ + 2(7/ 4- 2)2} {5 ar- 2(2/ + ;:;)-}.
11. a;^ + 2/* + z^.
12. (.^3 . 2/^) ^ ^ |3^^ + 3 (3^2 ^ 3^^ ^ 2/^)^1,
CHAPTER II.
ADDITION SUBTRACTION, MULTIPLICATION, AND DIVISION.
Addition.
20. HuLE. — Arrange the terms of the given quantities so
that like quantiti-es may be under each other ; add separately
the positive and negative coefficients of each column ; take
the difference and prefix the sign of the greater, and annex
the common letter.
(When the coefficients are all positive or all negative, we, of course,
simply add them together and prefix the common sign for the coeffi-
cient of the sum. )
Ex. 1. Ex. 2.
3a + 56-3c -3a + 76+ c - Ad
2c6 - 75 + 4c 2« - 26 + r)c + 3(^
5a+ h-2c -7a-36 + 2c-Gf^
4a-3 6 + 8c 2(j+56-8c+Gt^
Ans. 14 a- 46 + 7c Ans. - 6 « + 7 6 -^ d
Ex. 3. Add together b x^ - 3 ?/^ + 3 y, G ?/" + 7 xij - 4 x,
4 icy + 6 flj - 5, - 2 a:^ - 3 x-?/ + 2.
Arranging like quantities in each expression under vr.cli
other, we have : —
5 ctr^ - 3 ?/- +3 7/
7 ajy + G v/- - 4 .'x;
4 xj/ + (jx - 5
'-2 a r - 3xy + 2
Ans. 3ay^+8xi/+3y + 2x +3v/ -3
156 ALGEBRA.
Ex. IIL
Add together-
1. 3a - 2b, i a + 7 h, 2 a + 3b, a - 5 k
2. 9rr + 7 b% - 3a' + 4:b'\ a' + P, icr - 12b\
3. a + b + c, 3a + 2b + 3c, -4:a+7b-c, 2b + 5c.
L X - y - z, y - X - z, z - X ■- y, X ■\- y -V z. ^
5. 3a^ - 4 a6 + 6 ¥, 7 ab - a- - W, 2 rr - 3 ab - 4 6-,
4 a^ + aZ> - 51
6. 2x^ - 7x- + 3, - ix^ + (jx' -2 X + 7, a;' - 2 x"^ - 4 a;,
Gar*- 9a; - 12.
7. 2«2 + 7a6 + 35"^ - G a - 5 5 - 2, a- + 3f^ - 2 t + 9,
9a6 -2^-36 + 4, -3a2_i2a5 _ 35-+5^+ 105- 15.
8. cc^ - .oj^y - a;^;^ + xy"^ - xyz + ic^:;^, x^y - xy^ - xyz +
if — 7/^;;; + yz^, cirz - cc?/;s - xz^ + t/",^; - 7/;s" + ::^.
9. cc^ + xh/ + a;^^, - £c^y — x^y'^ - xif, y^ + xy^ + a;"^^.
10. a^ + ab'^ + ac^ + 2 a-5 - 2 crc - 2 «5c, a-b + 5"^ + 5c- -t-
2 a5'- - 2 a5c — 2 5'^c, a-c + 5-c + c^ + 2 a5c - 2 ar - 2 5c-.
11. x^ - x]f -v x:^ - 3 o(?y + 3 a:^;*;, 3 a;-y/- + 3 ar;r + 3 xy'^z
- 3 xyz' - 6 x'yz, y* - a^y - y:^ + 3 xy' - 3 x^yz, - 3 xif
- 3 X7j^ — 3 ifz + 3 2/V + 6 a;?/-;^, z^ + ar^;^ -ifz-3 xyz +
3 ar^^;^, 3 a;?/^;^ + 3 a:5r* + 3 yV - 3 yz^ - 6 a;?/^-.
12. a^ - a% + 3 aV -I- ab'c - 3 rt5c2 - 5'^c, a^'b - a\ +
3 a5c^ - 3 ac^ - 5-c- + 5V, (^c - ^^c/ + 3 «c^ - 3 ac-c? + 5-c^
- 5^cc/, - a'* + d^d - 3 a^c- + 3 a(rd - ab'c + b'cd.
Subtraction.
21. We have seen, Art. 7 (4.), that the subtraction of a quan-
tity is equivalent to the addition of the same quantity with
its sign of affection reversed. We therefore have the follow-
ing rule : —
Rule. — Change the sign of each term of the subtrahend,
and proceed as in addition.
Ex. 1. Ex. 2.
5 aj + 6 7/ - 3 c^ G (6^ - 3 a5 + 4 5^
2 a ; - 2 7/ + 2 ;: ; --J a ' - 3 ab - 2 5-
Ans. 3x + Sy -* 5 z, Ans. S cr + 6 b\
BRACKETS. 157
Ex. 3.
a^ -{^ 3x ^1/ + Zxif + 2 if
x^ — 3 ncF]/ ■{• 3 xy^ + y^.
Ex. lY.
1. From 6a + 76 + 3c take 2a + 5&-2c.
2. From 2ic-3?/-8;^ take 6a;-52/-2;^.
3. Take 5a2 + 3a6 + 46- + 3a + 76 + 8 from 6a- + 3 J- - 2 a.
4. Take 6a^ + 8aV + ic^from 8a* + 6aV + 2x\
5. Subtract the sum of the quantities a** + 2 arlr + 6^,
a* - 2 a^6^ + ¥ from 6 a* + 8 ^^6^ + 6 6^
G. From x^ -\- y^ + z^ — 3 xyz take 4 x^ + ?/^ + 4 ;s^ + 3 a;-;:;
+ 3x^ - 3 xyz.
7. From 3 a;* + 3 aii.'^ - 9 aV + a^o; - a* take 2 x'* + 4 ax
+ 4 (x^x + a*.
8. Take a' - 5 a^^ + 7 ah'' - 2h^ from the sum of 2 a^ --
9 a-6 + 11 a62 - 3 h^ and 6^ - 4 aft^ + 4 a^^ - dK
9. Subtract a + 6 + c + cZ from e +/+ ^ + A.
10. Take oj"* - 4 rc^y + 6 c^tf - 4 £c?/^ + y^ from a;'* + 4 a;^y
4- 6 ar?/^ + 4 xif + 2/*, and subtract the result from their sum.
11. Add together the given quantities in the last example,
and subtract the result from 3 a;'* + \Oarif + 3y^,
12. Take a- + 6' + c^ + 2 a6 + 2 ac + 2 6c from 2 a- + 2 6^
+ 4 rt6 - c\
Brackets — continued,
22. It was shown, in Ai-t. 8, that, when a quantity inclosed
in brackets is to be added, we may remove the sign ( + ) of
addition and the brackets without changing the sign of the
terms within the brackets. On the contrary, when the
quantity in brackets is to be subtracted, or has the sign
minus before it, we must change the sign of every term
within the brackets on removing the brackets and the sign
of subtraction.
We shall now see how to simplify expressions involving
brackets connected by the signs of addition and subtractign ;—
158 ALGEBRA.*
Ex. h Simplify (3a+5 6)-(G6-2c) + (-2a + &
- 3 c).
The given expression = 3fl^ + 56-664-2c-2a+6-
3 c, or adding together the like quantities,
= a - c.
Ex, 2, Reduce to its simplest form —
a -^ {b - c) + \b + (a'-c)l - \{a - h) - cl,
(When a pair of brackets is inclosed withia another pair, it is con-
venient to remove the inner one first.)
Hence the given expression —
= a -^ h + c + \b -h a — cl - \a - b - ci
_, „ ^. ,.« , . 3 aj X - 7 G .T - 9
Ex. 3. Simplify the expression -j — — - — + — - — -,
' The line separating the numerator and denominator of a
fraction is a species of vinculum, since it serves to show that
the whole numerator is to be divided by the whole denominator.
Hence, on breaking up the two latter fractions into fractions
having one term only in the numerator, we have —
3x x-r 6x-d 3 1 7 6 9 * 19
T— 2--'-8-^l^-2^''2^8^~8 = ^^¥.
23. As it is often necessary to inclose quantities within
brackets, we shall now show how this is done.
The following rule needs no explanation : —
E-ULE. — When a number of terms is inclosed within
brackets, if the sign placed before the brackets be + , the
terms must be written down with their signs of affection
ninchanged ; but, if the sign placed before the brackets be - ,
the sign of affection of every term placed within the brackets
must be changed.
Thus we may express rt + Z>-c~c?in any of the follow-
ing ways : —
a ■{■ b - G - d = a + {b - c - d) =^ {a + b) - (c + d)
-a'-{-b + c + d) = (a + b - d) - c, kc,
(When the word slgni^ used in future, the student is to under-
etaad i?(//; oj afection, unless otherwise expressed. )
MULTIPLICATION, 150
Ex. V.
Simplify the expressions —
1, 3x - {2x - y) ■}■ {6x + 3y-2z) - (Tx'-iy --3z),
2, (2 a\ + 2a'b + 2 air) - (2 a^ + o?h + aly" - ¥) + (a^
3, 1 - (1 - 2 cc) - I 3 - (4 - 3 a;) I + | 5 - (4 cc -
4, 6 ar^ -^ (2 ^ 3 a; + ar^) + | - 7 + (5 a; - 8 aj - 2) |
- (3 - 3 ~ 3 aj).
Group together the terms of the four following expressions,
so that —
(i.) The first two and last two are inclosed in brackets.
(ii.) The last three are inclosed in brackets.
(iii.) The first three are so inclosed, and an inner paii* of
brackets used to inclose the second and third.
5, a - 6 + c - c?.
6, -6a+76-3c + 5c?.
7, - 4 a^ + 12^2^ - \2xy' -^-if,
8, a' -- Z)^ - (^ + 3 abc.
Add together —
• 9. aa? + hxy + df, - doc^ - axy + ey^, hx^ + 3 cxy + /v/-,
- 2aa;2 ^ <2>hy\
10. ax - cy - ez, - bx -i- dy + /z, ex - ey - gz, - dx +
fy + hz.
Subtract —
11. (2 a + W aj - (3 6 + c) y + (4 c + fO « from (3 a - d) zz
+ {4: b - a) y + {5 c - h) z.
12. (y - z) a^ + (z - x) ab + (x - y) b^ from (y - a.*) a^
- {y - z) ab - (z - x)b\
Multiplication.
24. Bemembering the definition (Art. 14) of a coeflicient,
it follows that the product of two terms having coefl[icients
is found by multiplying the product of the coefficients by tlie
product of the remaining factoid.
Thus, the product of 4 a and 3 &, or 4 a x 3 6 = 12 a&.
160 ALGEBRA.
Again, tlie terms to be nuiltiplied may contain like
letters.
Now, d^ = aaa, a^ = aaaaa^ and hence it follows that
a^ X a^ = aaa x aaaaa = aaaaaaaa ^ a^.
And, generally, we have
a^ X a^ = aaa.., io m factors x aaa... to n factors,
— aaa... to (m + n) factors = a"* + ".
It therefore follows that the product of the powers of like
letters is found by adding together the exponents of the like
letters.
Thus, a^ X a'^ ~ a^, and a^ x a = a^, &c.
And, further, as regards the sign of the product, we have
seen, in Art. 1 2, that like signs give + , and unlike signs
give -.
There are three things, then, which must be attended to
in the multiplication of algebraical terms, viz. : —
1. The signs. — Like signs give +, and unlike signs
give -.
2. The coefficients. — These are to be multiplied like ordin-
ary numbers.
3. The letters. — The exponents of like letters are to be
added together, and the powers so obtained written side by
side with the unlike letters.
Ex. 1. Ex. 2. Ex. 3.
^ a^h - 3 xyz - 5 ahd
2 ah x^v^ - 2 bcd^
Ans. 12 a^bc Ans. - Zo^t/z^ Ans. X^ahhd?
25. Whenever the multiplicand or multiplier, or both,
contains more than one term, it is evident the product is
found by multiplying each term of the multiplicand
separately by each term of the multiplier, and adding
together the separate products.
Ex. 4. Ex. 5.
3 ^2 _ 7 a6 + 5 6- a^ - 3 ic^y + 3 a^ - ?/'
^ ah — 2xy
Ms, 18 a^6 -^42 a'6' \ 30 ok^ r- 2 ^'y 1 6 ^f- ^#f\% xif
MULTIPLICATION. 161
Ex. 6.
a? + xy + 'i/
x^ — xy ■¥ y^
cc* + a^y + iJt^y^
x'f + xf + y ^
Ans, x^ + ic^i/^ + 2/*-
Ex. 7.
ar^ + (a + 5) cc + a6
X + c
jc^ + (a + ft) ar^ + a6a;
ca^ + (ac + 5c) a; + a5c
a;* + (a + 6 + c) af^ + (a6 + ac + 6c) a; + abc.
It will be observed that the terms in the above examples
are all arranged according to the power of some letter.
Thus, in Ex. 4, they are arranged according to the descend-
ing powers of a; and in Exs. 5, 6, 7, they are arranged
according to the descending powers of x. It matters not
whether they are arranged according to ascending or descend-
ing powers, and the result would be the same if they are not
so arranged. It is then, however, much easier to collect like
terms, as they generally fall under each other. When we
come to division we shall find it necessary to arrange the
terms according to the power of some letter.
Ex. VI.
1. Multiply 3 a + 2b hy 4: a - 3 b, and 6 x + 7 y hj
3x - 5y.
2. Multiply oc^ + 2 a;^/ - 2 y^ hj x - 2 y, and 15 a^ +
17 ooy - 4 2/^ by 2 a; + y.
3. Multiply a^ + 2 ab + b^ hj a^ -- 2 ab + P, and a'
+ b^ by a" - b\
4. Multiply a^ + ary + xif + y^ hy x - y, and the pro-
duct by X* + y\
5. Multiply a^ + b"^ + c^ — ab - ac - be hy a + b + c.
G. Find the contiuued product of x'*' + y^j x^ + y^, x + y,
X - y.
5 L
163 ALGEBRA.
7. Develop tlie expressions {a + by, and {x - yy,
8. Multiply5a=^+ 15^2+ 45« + 135 by a^- 2« - 3.
9. Multiply 1 + 3 a; - 7 ar by a? - 2, and a - xhy x^ +
ax + a^.
10. Multiply a* - 2a^b + ^ orh^ - 2ah^ + 5nya^ + 2a^6
+ Sa^i^ + 2a53 + h\
11. Multiply a + 6aJ + ca:i^ + (^a;^ by ca; + /
12. Eind the product oi Q(? + px •¥ q and x - a,
13. Find the continued product oi x -v a, x + h, x ■{■ c.
28. We have explained in the last Article the general
method to be pursued in Multiplication. There are,
however, many algebraical expressions which may be mul-
tiplied by inspection.
I. Expressions of the form {a + hy.
It is easily found by long multiplication that
{a ■{■ hy = o? + lab + W,
and this, expressed in ordinary language, may be read thus : —
The square of the sum of two quantities is the square of
the first ^ plus twice their product, plus the square of the
second,
(This rale is evidently true whether the quantities are j^osUlve or
negative. )
Ex. 1. (3^ + 7by - (3ay + 2 (3«) (75) + (7 by = 9a-
+ 42a6 + 49 51
Ex. 2. {6x - 5)2 = (Qxy + 2 (6a;) ( - 5) + ( - 5)^ =
36a? - 60a; + 25.
Ex.3.(a + b + cy= ^(a + b) + cV
= (a + by + 2{a + b)c + (^
= (a" + 2ab + b'') + 2 {ac + be) + r
= a^ + b^ + c" + 2ab + 2ac + 2bc,
Remark. — It is a very common mistake with beginners to write
down a^ + h^ as the square of (a 4- b), thereby leaving out twice
the product of the quantities a and b. They should impress this for-
mula, viz. : —
{a + 6)2 = a^ + 2 a6 + 63
thoroughly upon the mind.
MULTIPLICATION. 163
II. The form {a -h h) {a - b).
It may be easily found by long multiplication that
{a -h b) {a - b) = a^ - b^,
which, in ordinary language, may be thus expressed : —
llie product of the sum and difference of two quantities is
the difference of the squares of the quantities.
This formula may be applied in all cases where the terms
of the quantities to be multiplied are of the same magni-
tude, but when some of them differ in sign.
Ex. 1. {a ■{■ b + c ■{■ d) {a ■\- b - c - d)
= I (a + 6) + (c + cZ) I I (« + b) - (c + cZ) I
= (a + by - (c + df = (a2+ 2«& + b') - (c- + 2c(Z + cP)
= a^ + 6^ - c^ — cP + 2a6 - 2cd.
Ex. 2. {a^- Sa^'b-h Sa¥- b^) (a^-h 3a'b + 3aP+ ¥)
= j (a»+ 3a6-) - (3 a'b + P) j j {a^ + 3ab') + (3a'b + b') j
= (a^ + 3a¥f - {3a'b + bj
= (a« + Qa'U' + 9d'b') - {9a'b^ + Ga'^b^ + b')
= a^-Sa'b^ + 3a^b'-b\
The principle to be adopted in all such cases is to find
what terms in the given quantities are exactly alike, and put
them first in brackets, when the remaining terms will fall
into the proper form. Thus —
(3a + 7b + 5c-d){3a-7b-h5c + d)= |(3ct + 5c)
+ (7 6 _ cZ) I I (3 ci + 5 c) - (7 6 - cZ) I .
And(3a-75- 6c;+cZ)(3a + 76 + 50 + ^/)= |(3« + (Z)
- (^^ + ^c)j j(3« + ^) + (76 + 5c) j.
(III.) The form {x + a) {x + b).
By multiplication we find that —
{x + a) {x + b) = x^ + (a + b) X + ah,
which in ordinary language may be thus expressed : —
The product of two binomials containing a common term and
an uncommon term, is the square of the common term, plus
the sum of the uncommon terms multiplied into the common
term, plus the product of the uncommon term.
164 ALGEBRA.
Thus—
Ix + Q)(x - 1) = aP+{6-l)x + 6 (- l) = x^ + 5x - 6.
{x-5){x -7) = x^-{6 ■h7)x + {-5){- 7) = x' - 12 aj
+ 35.
{x + 3a) (x - 5a) = x^ + {3 a - 6a) x + {3 a) ( - 5 a) = x^
" 2 ax — 16 a\
And so,
(x + a + b) (x + c + d) = {x + a + b) (x + c + d)
= x^ + (a + b + c + d) X + (a + b) (c + d).
We may extend this formula to any number of factors.
Thus, by multiplication, we find that —
(x + a) (x + b) (x + c) = a^ + (a + b + c) x^ + (ab + ac
+ be) X + abc,
{x + a) {x + b) (a; + c) (a? + (Z) - oj* + (« + & + c + d)
a^ + (ab + ac + ad + be + bd + cd) a^ + {abc + abd + acd
+ bed) X + abed.
Law of Formation of the Terms,
1. It will be seen that the coefficient of the first term is
in each case unity ; that of the second term, the sum of the
uncommon letters taken singly ; that of the third term, the
sum of the uncommon letters taken two together ; that of the
fourth term, the sum of the uncommon letters taken three
together^ &c.
2. The power of the common letter is in the first term that
of the number of the binomials, and it sinks one every term.
Ex. (x + l){x + 2){x + 3) = a^ + (1 + 2 + 3)x'+(l +
2 + l + 3 + 2 + 3)a;+l + 2 + 3=af* + 6a;-+lla;+6,
(IV.) The form (a + b + c + d + &c.)l
By ordinary multiplication or otherwise we have (a + b + c
+ df = a^ ^b'' + c^ + d' + 2ab -v 2ac + 2ad + 2be +
2bd + 2cd ', and a similar result follows if we take a larger
number of terms.
We may hence deduce the follomng rule : —
The square of the sum of any number of quantities is the
sum of the squares of the quantities together with twice the
sum of the products formed by multiplying the first quantity
MULTIPLICATION. 1G5
into all that follow separately, then the second into all that
follow, the third into all that follow, &c.
Ex. 1. (1 + 2a; + Zx^f =\ + 4ar* + 9a;*
- 1 + 4a; + lOar^ + 12a;»+ 9a;^
Ex. 2. (a'^ + 3a=6 + 3a62 + 6y
= a« +9 a^^h' + 9 a^h^
= a« + 6a'^6 + 15a*62 + 20a«63+15fr6* + 6«6^ + 6^
Ex. VII.
1. Find by inspection the squares of a; - y, 3 « - 5 6, 4 c-
+ (^2, 3 a;2 _ 2 2/^.
2. Find the continued product of a ■\- h, or + Ir, a* + h*,
and a - b.
3. Multiply wia; + ?i?/ by mx - ni/, and 5 a^ - 3 6- by 10 a^
+ 6 61
4. Find the value of (a + b •{• c + d) (a -h + c - d), and
of {a + b - c - d) {a + b + c + d),
5. Show that (ar + 2 a;y + ^t) (ar' - 2 a;?/ -{- y-) = a;* -
2ary + 2/*.
G. Multiply X + 5 separately by a; + 1, a; + 2, a; - 3, a; - 5.
7. What is the continued product of a; - 2, a; + 3, a; - 6,
X + 5"^
8. If 2 s = « + 6 + c, find the value of 5 (5 - a) (s - b)
(«-c).
9. If 2 8 = a + b + c + d, what is the value of —
(2 « - 2 a)'-^ + (2 5 -- 2 by + {2 8 - 2cy + {2s -2 dfl
Prove that —
10. {ax + byY + {ex + di/Y + {ay - bxy + {cy - dxy =
(a^ + 6* + c^ + (f ) (ar^ + f).
11. I {ac - bd)x + {ad + bc)yl^ + \ (ac - bd) y - {ad
+ 6c) a; I ^- = {a" + b') (r + c?^) (ar' + f).
12. {ax + by + czy = {a + b +c) {oar + bif + cx^) -
ab (x - yy - ac {x - zy - be {y - zy.
166 ALGEBRA*
13. {x + a) (oj + h) (x ■{- c) = (x — a) (x - h) (x - c) -\-
2 {{a + b + c)x'^ + abcj.
14. {m + n + p + qY =r. i^m + ^if + {m + j)f + {m + qf
+ {n + ^)^ + {n + g)^ + (^ + <^)2 _ 2 (7/r + ti^ + 2^- + (7^).
15. {a? + 5^ + c^) (m^ + 'iV + p- + g^) =^ {am + 6?* + cp)-
+ {an - 6wi + cqY + (op - hq - cnif + (a^/ + 5p - c?i)l
16. {a - h) {b - c) + (5 - c) (c - a) + (c - a) (^ - ^>) =
3 {ab + ac + be) - {a + b + c)-.
17. (.'z; + 2/ + ^ + f*)^ - {x-y-z +aY = 4:{x + a) {'i/ + z).
18. {(« - bY + 4:ab} {{a + bY - i ab} {a' - b* -i-
19. (^2 + 6-) (C^ + C^2) :. (^ + ^,^)2 ^ (^^ ~ 5^)2^
20. 4 {a^ + ¥) (c^ + dr) = {a + bY {c + ^)2 + {a - Z^)^
(c - cZ)2 + (a + 5)2 (c - c^)- + (a - 5)2 (c + cZ)l
21. 8 (ci^ + 5^) = (a + 5)^ + 6 (a^ ^ bf + (« - 5)^
22. {x ~ 2/)' + {y^zY + {z-xY + 2{x^ y) {y - z) ^
2{x - y){z- x) + 2{y ^ z){z'-x) = 0.
23. (o^ + 5 + c) { J + c - a) (a + c - 5) (a + 5 - c) =
4 a-6-, when a- + b'^ = c^.
24. {2 {ax^^ + 57/»')2 + (ay^ -. bx^} =. (a= + 5') {(«:*» + 2/'f
+ (^"^ - 2/*")'}-
Division.
S7. As in multiplication, we have especially three things
to attend to, viz; : —
1. The signs.
^ We have learnt {Mt 13) that like signs give + , and unlike
signs give - .
2. The coefficients »
Uilderstanding here the numerical coefficients, it is plain
that they may be divided as ordinary arithmetical quantities.
3. The letters.
As the product of the quantities a and b is expressed by
aby it follows that the quotient of ab by either of the factors
a or 6 will give the remaining factor b or a respectively.
DIVISION. 167
Thus, ah -■ a or — = h, and ah -r b or -— = a.
a
And so, xyz -r xz = y, and pqrs i- qs = pr.
We may then conclude that when the divisor is contained
as a factor in the dividend, the quotient is found by oimiting
from the dividend those of its factors which constitute the
divisor.
If the divisor be not contained as an exact factor in the
dividend, we may then express the quotient symbolically.
Thus, xy -f ah = — */.
' ^ ah
When, however, the dividend and divisor have a common
factor, it is plain that we inay, as in arithmetic, strike out of
the numerator and denominator of the symbolical quotient
this common factor.
Thus, 5abc^hd=^^^^.
od d
And 16 xyz -r 10 axz = ttt— ^ = f-^«
10 axz 5 a
28. A power of a quantity is divided by any other power
of the same quantity by subtracting the index of the (Hvisor
from that of the dividend, the quotient being that power of
the quantity whose index is the remainder so obtained.
1. Let the power of the quantity in the dividend be the
higlier.
We have a^ = aaaaa, and a^ = aaa,
.*. cr -r a^ = - aa =z a"- or .
oKia -.
Or, generally, m being greater than w, since
a"* = aaa... to m factors, and a** - aaa...to n factors,
we have—
.m . u aaa...tom factors ,' / ' \r.^+^^r. ^m — »
a -r a = = aaai..(m — ?i) factors = a •
aaa...to7i factors
2. Let the power of the quantity in the dividend be the
lower.
168 ALGEBRA.
Suppose we have to divide a^ by al.
Wehavea^ -^ dJ =^ ^^^^ := UZ \,
daaaaaa aaa a?
We may, however, so express the result that it shall agi'ee
exactly with the proposition at the head of this article.
For we may conceive of o? as representing the product of
%inity and the quantity a^. We shall therefore be perfectly
consistent if we allow o? to represent the quotient of unity
by the quantity a^.
We shall then have a" ^ = -^, and hence"we get from the
__ a^ ^,.
above result —
Or, generally, m being less than n, we have —
a"* aaa \to m factors 1
a" ^ aaa to n factors aaa (n - 7)i) factors
= ^_^ ; or, using the notation just explained,
a
3. Let the powers of the quantities in the dividend and
divisor be equal.
It is evident that their quotient is unity:
Thus, ?.' = 22^ = 1 = 1.
a* aaaa 1
And SO, — = 1.
a"*
If, however, we assume the principle proved in the two
cases above to hold here, we have —
a*"
It follows therefore that a^ ~ \.
Cor. — From the above interpretation of negative indices
it follows that the same rules for multiplication and division
of quantities involving them may be applied as in the case of
positive indices.
Thus,a^ X a-2 = aS x \ -- -' = a^-\
a^ o?
'And so, a^ - a~^ = a"*"^'^^ = al, "^ ""
DIVISION. 169
Ex. 1. ?^ = ^xy.
Ex. 2. ::^p:^ = - 3aV.
JbiX. 3. ~ a^ - Sab + 2b,
- la
^^ , ^x^y:^ - Qx'fz^ + 42a?2/V
^^' ^ 3^7^^
4:0$ 2 U
- + — , or
3 j^z^ yz xz
= |i^2/"'^-' - 2y-h-'' + Ux-''z-\
o
Ex. 5. Divide 3^2 + 13a6 - lOK^hja + 5 5.
When the divisor, as in this example, contains more than
one term, it is generally convenient to follow the method of
arithmetical lon<; division. Thus —
Ex.
a + 56)
3 a'
+ 13 a6
+ 15 a&
- 2a6
- 2ah
- I0¥{c
- 10 6'
- 10 6'
!a -
26
6.
a-
Divide <
+ ab +
%* +
6=),
a,* + a-b"
a* + ai'b
by a' +
+ 6^ (a'
+ aW
ab +
- ab
6'.
+ 6=
- a?b
- a'b
+ b*
- a?6' -
aW
a''6-' +
a'6' +
ab'
ab^
+ 6*
+ 6'
It will be seen that in the last two examples care has been
taken to keep the terms of the divisor, dividend, and successive
remainders arranged according to the ascending or descending
powers of some letter. In these cases we have arranged the
terms according to the descending powers of a, and, as there-
fore follows, according to the ascending powers of b. Want
of care in this respect will often render the operation of find-
ing the true quotient tiresome, if not impossible. The next
two examples will illustrate this point. They may be
170
ALGEBRA.
attempted first by the student, keeping tlie terms in the order,
as given.
Ex. 7. Divide 2 - 7 a; ~ 15a;2 by 5aj - 1.
3 a; - 2
Here we have the powers of x in the dividend descending,
while in the divisor they are ascending. Arranging them in
the divisor as in the dividend,, the operation is easy. Thus —
y^^ + x^'^)o^ + 2/^ (^y - x^if + xif
Q^ + a^y
— a?y — xy^
- 1 + 5a;)
3,5a; - 1)-
Divide o? +
2 -. 7a;
2 - 10 a;
- 15a;^(-
Or thuj
3a;
3a;
- 15 ar^ +
- 15 ar*
- 15 a;^
7a; + 2(
3a;
—
lOo; + 2
lOo; + 2
Ex. 8.
2/5 by o;-
' + 2/^^
xf + 2/^
xy^ + it
We shall work the next example in two ways to illustrate,
firstly, the above point again ; and, secondly, to show how the
operation may be sometimes abbreviated by the use of brackets.
Ex. 9. Divide a;' + 2/^ + ^ - 3 xyz by a; + 2/ + ^•
a; + 2/ + z):i^ - 3 xyz + y^ + ^{aP -xy -xz + y^-yz + z^
aP + ary + orz
-xhj.
- xh/ ■
-aPz-
- 3 xyz
-xyz
x:^
r^
- aPz + xy^ -
-aPz
2 0;^^;
'Xyz-
■xyz^r
x^
*
-xyz-^r
-xyz
x£^
^fz
-fz-
-yz'
xz^ +
xz'^ +
yz^^
2/^ +
^
^
'i'
DIVISION. 171
Or thus, inclosing the last two terms of the divisor in
brackets —
X + {1/ + z))x^ - 3 xyz \' 7^ + z^{o(P - {i/ -^ z) x -i- (if -yz-\- z^)
a^ + (y + z) x^
- (y + 2;) ar^ - 3 xyz
-{y + z)x^-{y^+2yz-\-z^)x
{f-yz-Viif)x + i^ + ^
if -yz+ z^)x + y^ + 2^
It will be seen in both the above operations that we have
brought down the terms of the dividend only when the sub-
trahends indicated they were required. This often prevents
much useless repetition*
Ex. tUL
Find the quotient of —
1. 28 arb - 7 a6^ + U W hy 7 b; 3 x'lf - 12 xf by
3 xy,
2.-6 a'b + 15 a*62- 20 a^^' by - 3 ab; 4 xY + 6 xhf
+ 4 xy^ by 2 xy\
3. aixr"*+ 6x"*y*+ cy- *» by re*" + ** ; «^'!!1~^" + Jx"*~*'?/** +
c?/-" by aj"*~".
4. 30 x^ +2 xy - 12 7/- by 5 ic - 3 7/ ; and by 6 a: + 4 ^.
5. l + 2a;+3ar^ + 2jc^+a;*byl+a;+a:l
6. 12 - 19 a; - 21 aj^ by 7 rr - 3 ; and by 3 a; + 4.
7. a* - 4 ary'^ +12 xy^ - y*hy x - 3 y; ami by x-{^ y.
8. x^ - ^hj X - y; and x^ -h y^ hj x + y:
9. acx'^'^'* + adx'^y^ + kx^'g/^ + bdy"^.;^*^ by txT-^-di/^.
10. a« + a^62 + a^i* + 6^^ by a' - a'b -aW + Z^l
11. 6^ + a6 + Jc + ac by a + 6; and a^ + a6 + 5c +
ac by a + c.
12. a + (rt + 6) a; + (^ + 5 + c) a;- + (a + 5 + c) a;^ +
(6 + c) a;* + ca;^ by 1 + a; + ar + jc*.
13. a^ - jpa'* + qa? - f^a- + ^?a — 1 by a - 1.
172 ALGEBRA.
14. a^ + 6^ + c1 + (^^ - 2 {aW + arc" + ord? + 6V +
t^cZ^ + c^c^^) - 8 a^cc? bya-6 + c + cZ; and that of this
quotient by a - 6 - c - c?.
15. Show that the remainder, after the division of x^ - p^?
+ qx^ - rx + shj X — a^ is a* - jt?a^ + r/a' - ra + s.
16. Divide re* - 2/* by o;""^ — 2/~^ and x^ + cc"^ + 2 by
X + a;~^
17. Show that the quotient of 1 by 1 + a?, is 1 - cc + jc^
— cc^ + &c. ad infinitum,
18. Show that the quotient of 1 by 1 — 3 a; + 3 x' — a;'
is 1 + 3 a; + 6 a;^ + 10 ic^ + 15 a;'* + &c. ocZ infinitum,
19. Divide {x + yY — 2 {x ■{- yY ^ ■\' z^hy x + y - z.
20. Divide (a - c)^ - 3 (a - c)^ (6 - c?) + 3 (a - c) (6 - (Z)^
- (6 - cZ)^ by a - 6 - c + cZ.
Factors.
29. The ordinary method of finding the quotient of two
algebraical quantities having been explained in the last
article, we shall now proceed to show how, in certain cases,
this method may be avoided, and the quotient written down
at sight. It may be remarked at the outset, that the resolu-
tion of algebraical expressions into their elementary factors
is a subject of very great importance, and one which the
student will do well to thoroughly master.
(I.) The form x^ + 2 ax + a-.
We have seen (Art. 26) that x"^ + 2 ax + a^ = (x + aY-
Hence the suin of the squares of two quantities, together
with twice the product of the quantities, is equal to the
square of the sum of the quantities.
And hence, any algebraical expression, which can be thrown
into the form (ar + 2 aa; + a^), is of necessity a perfect square.
Thus—
ar' + 6 a; + 9 = a;- + 2 (3) aj + 32 = (aj + 3)1
a' - 10 a6 + 25 52 = a- + 2 a ( - 5 6) + ( - 5 5)= =
{a - 5 5)1
16 aV - 56 ahxy + 49 hhf = (4 axY + 2(4: ax) {--7 by)
+ {-^ 7hyY = {iax- 7 hy)\
FACTORS. 173
3 nl/x" - 6 ahcxy + 3 achf -= ^ a (iV - 2 hcxy + cY) -
3 a{{hxy - 2 (6a^) (c^/) + (cyf] = 3 a {bx - cyf.
(II.) The form o? - ^^l
We have seen (Ai-t. 26, II.) that a'-h^ = {a + h) {a - h).
Hence the difference of the squares of two quantities is
equal to the product of the sum and difference of the
quantities. Thus —
aV - b^ = (axy - (byY = (ax + by) {ax — by).
x'^y'= {aPy- iff = {x^ + f) (x^ - y")
= (^ + 2/') (^ + y) (^ " y)-
a" + b'' - c" - (P -h 2ab -- 2cd
= (a^ + 2 ab + b') - {c' + 2cd + c/^)
= {a + bf - (c + df
= {{a + 6) + (c + d)} {{a + b) - {c + d)}
= {a + b + c + d) {a + b — c - d).
x^ + x'y^ + y* = {x^ + 2 aPy^ + y*) - (x^y^
= {^ + y'f - ^f
^ {{^ + y^) + ^y] {(a^" + ir) - ^y]
= (a^ + xy + y^) {a^ - xy + y^).
(III.) The form a;^ + jtx» + (7.
This form evidently includes both the preceding, for the
first form — viz., x^ + 2 ax + a^ is included, since (7 may be
the square of half p ; and the second is included — viz.,
a^ - a^y since we may have p = 0, and q a negative square
quantity.
Now, the resolution into elementary rational factors of the
quantity x^ ■{■ px •¥ q i^ not always possible ; but, since
(Alii. 26, III),
(x? + (a + b) x + ab = {x + a) (x + b),
we have the following rule, when the quantity admits of
resolution.
Rule. — If the tliii-d term q of the quantity x^ + px + q
can be broken up into two factors, a and by such that the
sum of these will give the coefficient of Xy then the
elementary factors of x^ + px -h q are x + a and x + b.
Thus, a;^ + 7 a; + 12 = (a; + 3) (a; + 4); for the product
of 3 and 4 is 12, an4 their ^um is 7, the coefficient of x.
174 ALGEBRA.
xl - flj - 30 = (rw - 6) {x + 5); for the product of - 6
and 5 is — 30, and their sum is - 1.
And so, x^ - ISx -h 32 = {x - 2) (x - 16),
And a^ + 3 a6 - 108 6^ = a^ + (12 b - 9 h) a +
{12 h) (-- db) = (a + 12b) {a - 9 6).
(IV.) The form ax^ + bx + c.
This is the general form of a trinomial. The following
remarks, though equally applying to each of the three pre-
ceding forms, are especially intended to be 'practically
applied to trinomials not included by them.
The above form will include such expressions as the fol-
lowing:— 20 :«2 + lli^ _ 42, 6ar^ - 37aj + 55.
It is evident that the product of the first terms of the
factors will be the first term of the given trinomial, and that
the product of the last terms of the factors will be the third
term of the given trinomial.
And, further, when the third term is negative, the last
term of one factor must have the sign + , and the last term
of the other the sign - ; but, when the third term is ,
positive, the last terms of the factors must have the same
sign as the middle term.
Thus, 12 ar^ - 31 oj - 30 - (4 aj + 3) (3 a; - 10).
Here the factors of 12 cc^ are either 3 x and 4 a;, 6 aj and 2 5;,
12 a; and x, and the factors of 30 either 5 and 6, 3 and 10,
2 and 15, 1 and 30 ; and we must give a + sign to one of each
of these latter pairs, and a - sign to the other. It is easily
found on trial that, in order to obtain - 31 a; as the middle term,
the factors of the trinomial must be 4 a? + 3 and 3 a; - 10.
SowehavelOa^ - 41 a6 + 216- = (5a - 36) (2a -7 6),
and aca^ + (ocZ + 6c) xy + bdif = (ax + by) (ca: + dy),
(Y.) The forms x"" + y" and aj" - y^.
We shall show in the next article that a rational in-
tegral algebraical expression, involving x, contains a; - a as
a factor when it vanishes on substituting a for x.
Hence, a;"* + y^ and a;'* - y^ must each vanish on putting
y for X, if they contain a; - 3/ as a factor, n being an integer.
The former becomes y^ + y^ or 2 ?/", and the latter y" -
y^ or 0. We therefore conclude that —
xJ^ + 7/" does not contain x - y as a factor, and that —
FACTORS. 175
X** - 2/** ^o®s contain re - 7/ as a factor, whether n be
even or oc?c^.
Again, on the same principle, they must each vanish if
they contain x + ?/ ^^ ^ factor, on putting - y for x.
The former becomes ( - yY + 2/", which vanishes when w isoddj
and the latter becomes ( - yY - 2/", which vanishes when wis eve/i.
Hence we conclude that —
af* + 2/" contains x + yas a factor when n is odd, and
x^ - 2/" contains cc + 2/ as a factor when n is even.
Now, the quotient of either of these quantities by o^ + y or
a; - 2/ ^^^ i^ ^^y particular case be found by long division.
We thus find that —
x^ + y^ =z (x + y) (x'^ - x^y + x^y"^ - xy^ + y^),
x"^ - y^ = {x - y) (aj^ + a?y + xy'^ + 7/^),
o? - if =- \x - y) {x^ -\- xy ■\- t/').
The law of formation of the co-factor in each case is easy
to see ; and if we may assume this apparent law as generally
true, we may conclude that, when an algebraical quantity is of
the form cc" + y^ or ic" - y'\ and it contains x ■\- y or x - y
as a factor, the law of formation of the co-factor is as follows: —
Law of Formation of Co-Factor.
1. The terms are homogeneous, and of dimensions one
degree lower than the given expression, the power of x in the
first term being n - \, and diminishing each successive term
by unity; and the power of y increasing each successive term
by unity y and first appearing in the second term.
2. The coefficient of every term is unity.
3. The signs are alternately 4- and - , when a; + 2/ is the
corresponding elementary factor ; and are all -f , when x - y
is the corresponding elementary factor.
Ex. 1. a^ -I- 32 = a^ + 2« = (a + 2) (a* + a' ' 2 -{- a^ • 2-
-1- a • 2^ + 2^) = (a + 2) (a* + 2 a^ -t- 4 a^ -f- 8 a + 16).
Ex. 2. a« - 6« = {aj - {h^ = (a^ 4- J/) (a« - h') =
{a + h) (a^ -ab + b^)'{a - b) {a" + ab + b^ = (a + b) (a -- b)
(V ^ ab + b'') {a^ + a6 + 5').
30. The remainder of the division of a rational integral
function ofxby:x. - a Qnay be found by putting a for x in
the given function.
176 ALGEBRA.
Def. — A function of x is an algebraical expression in-
volving X ; and a rational integral function of x is an expres-
sion of the form ax^ + bx''~^ + &c. + sx + t, where all
the powers of x are integral and positive.
Let/ (x)^ be a rational integral function of Xj and suppose
Q to be the quotient, and H the remainder on dividing the
function hj x - a.
Then, evidently —
Q (x - a) + B = f{x) identically.
And this identity must hold for all values of x, and
therefore holds when x = a.
In this case we have Q {a - a) + R = f (a)
+ 7? =/(a)
ori? =f{a).
Now/ (a) is the result of putting a for x in the given func-
tion, and is, as we have just shown, the remainder on
dividing the given function hj x - a.
Cor. 1. When there is no remainder, we must, of course,
have/ (a) = 0. Hence, a given rational integral function
of X vanishes when a is put for x, if it be divisible hy x-a,
Ex. 1. The remainder, after the division of 2 ic^ — 5 x^ +
6 a; + 7 by re — 2 is 15.
For, putting x = 2, we have —
2a;3- 5a;2 + 6aj + 7 - 2- 2^ - 5 • 2^ + 6 • 2 + 7 = 15,
Ex. 2. The function—
ic' - 2 oj^ + 5 a; — 52 is divisible by ic — 4.
For, putting a; = 4, we have —
cc»-2ar^ + 5aj-52 = 43-2-42 + 5-4-52 = 0.
CoR. 2. Any rational integral function of x is divisible by
X — \f when the sum of the coefficients of the terms is zero.
For, putting aj = 1 in the given function, it is evident
that it is reduced to the sum of its coefficients, which sum
must be zero if the function be divisible by a; — 1.
Ex. Each of the following functions is divisible by
aj - 1, viz.: —
3 a;* -f. 7 a;^ - a;- + 12 a; - 21, 5 aj^ - 2 aj - 3,
{a-h)x^ + (h — c)x + (c — a), (a + hf a;- — 4 ahx — (a- Hf.
* The expression/ {x) must not be considered to mean the product;
of/ and ic, but as a symbol used for convenience,
FACTORS. 1 i i
Cor. 3. Any rational integral function of x is divisible by
X -h I, when the sum of the coefficients of the even powers of
X is equal to the sum of the coefficients of the odd powers.
(The term independent of x ia always to be considered as the co-
efficient of an even power).
Let ax'* + ax'' ~ * + &c. + rar + sa; + ^ be a rational in-
tegral function of x.
Put x =^ - I, then we have, if the function be divisible
by 03 + 1 —
a(«l)" +6(_l)n-i + &c. + r(- l)- + 5(-l) +^-0.
Suppose 71 to be even, then evidently ( — 1)'*= (-1)
( — 1) { — l)...to an even number of factors = + 1.
And so ( - 1)"-^ = ( - 1) ( - 1) ( - l)...to an odd
number of factors = — 1 ; and so on.
Hence we get a-6+&c. +r — s + t=Oy and this
must evidently require the condition that the sum of the
positive quantities is equal to the sum of the negative, and,
therefore, that the sum of the coefficients of the even powers
of X is equal to the sum of the coefficients of the odd j)owers.
And a similar result will follow if we suppose n to be odd.
Ex. Each of the following functions is divisible by jc + 1,
viz.: —
if3+5ar* + 7a;4-3, 5ar^-4a;* + 8ar-2a;-l,
a^a;* - (a + 1) (a + 2) ar^ + 2 aj + 3 a + 4, 2^^ +{q +r)o(^
+ {q + r) X + p,
Ex. IX.
Eesolve into elementary factors —
1. a;2 - 9 a^ 16 y* - 25 z\ 24 a^ - 54 6^, 8 »» - 27 y",
2. a;* — xy^, a^ - h^, xi/ + x^^y^ 2 x^y^z - 8 xy^^.
3. a* - 4 h\ cc* + o^y- + y\ a* - 2 aW + h\ a" + Ir - c' -^
2aJx
4. a- + 62_c2^cZ2 ^ 2 a6- 2 cc7, a- ~ 6^ - c- + (/- + 2 he
■\- 2ad,a' - {b - c)\
5. {x + If - (x + 2)^ {x + 5)2 - (x + 2)-, (2a + hf
- (a - c)-.
6. {x^ - 2/')' + 4 (a;* + ar^r + y') (xPy^, {x" + y^f -^
? (ar + ff sc'f-\'4: x*y\
5 M
178 ALGEBRA.
7. x^ - ox " 70, ft3" + 11 oj 4- 10, fr - 15 ah + 56 h\ x^
- 4:x- 192.
8. rtV + aSa;?/ - 42 %% 3 «a;-— 2i: ax - GO «, 24 ac -
5 «c- + acl
9. Goj- - llo? -35, 8a;- + Gx - 135, 18a;- - 21a; - 72,
20a;= - 11a; - 42.
10: 3a^]/ + lOa^y + 3xf, 20 a;^ + 12 ax- + 25 hx^ ^
1 5 abxj ni^^ + (piq + wzp) a; + ^.
Write down the quotient of —
11. a;* - 16 by a; - 2, 3 a;*' + 96 by x + 2, a;^ - 27 by
:^ - 3.
12. (a + Vf - (c + rZ)3 by rt + 6 + c + cZ, a^ - Z.^ + c^
- cZ- + 2 ac + 2 6(i by a + & + c - f/.
Find the remainder after the division of —
13. x^ + f:^ + qx- + ra; 4- s by a; - rt, a;'* + a^ by
;^ - a\
14. a;^ - 5a;^ + 7a; - 9 by a; + 3, x^ — 3a; + 7 by a; - 2.
Show that —
15. 5a;^ - 3 a;^ + 7 a;- - 8 a; - 1 is divisible by a; — 1.
16. 2 a;* - 3a;^ + a;^ - 7 a; - 13 is divisible by a; + 1.
CHAPTER III.
Involution and Evolution.
Involution.
31. Involution is the operation by which we obtain the
"powers of quantities. This can of course be done by multi-
plication, but the results obtained by the ^^ctual multiplication
of simple forms enable us to develop without multiplica-
tion more complex forms. As the subject requires the aid of
the Binomial Theorem, we shall here show how to develop a
few only of the more simple expressionB.
SSt The power of a single tenn ia obtained by raising the
JX^'Ol^UTION. 179
coefficient of the term to the power in question, and multijjhj'
ing the exponents of the letters of the term by the exponent
of the power in question.
Thus, {a^f = a^ X a^ = aJ^ + ^ = a^ ^ ^,
{ory' = or X a'"* x «"* to?i factors ^ a^^ + t^-*- w + ...to«tcrm8
And so, (iaWcy = 43a'-^W^V><' = Ua'bV\
33, Development of the ^7iW,/<mr^A,and^j5A powers of a + 6.
We know (Art, 26, 1.) that (a + bf = a" -^ 2ab + b\
.-. (a + by = (a? + 2 a6 + b^) {a-\-b)=.a^ + Z a^b + 3ab^ + b];
also, (a + by = {a^ + 3 ab^ + aP + b^) (a + ^) = a* +
and (a + bf = (a^ + 4 a^^ + 6 ^-^^ + 4 aS^ + Z^^) (a + Z>)
:=-- ft'^ + 6a'b + 10r*^^>5 ^ lOa^^' + 5 a5^ + 6».
The following law of the formation of the terms is evident : —
Law of Formation of Terms,
1. The first term contains a raised to the given power, and
the power of a decreases by unity in each successive term,
while the power of b (which first appears in the second term)
increases by unity in each successive term, till it reaches the
power of the given quantity.
2. The first coefficient is unity, and the coefficient of any
term is found by multiplying the previous coefficient by the
exponent of a in the previous term, and dividing the product
by the number of terms hitherto developed.
Ex. 1. (2aj+ Zyf = {2xy -^ 3(2aj)2(32/) + 3(2a;)(3 2^)2
+ (3 7/)' = 8a;' + SGor^^/ + ^^^f + 272/'.
Ex. 2. (a + 5 + c)' = {a + b + c)' = (« + bf +
3 (a + bfc + 3 (a + 5) c2 + c' = (a' + 3 a^b + Zab'' + b')
+ 3 (a? + 2ab + b'') c + 3(a + b) (^ -{- c^ = a^ + ¥ ^ c"
^ 3a^b + 3 a^c + 3ab^ + 363c + 3 «c' + ^bc^ + 6abc.
(In the following example^ the above la"vy may be assumed a^
peneralli^ ime,}
Ex. X.
1. Find tho values of {a'b% ( - 8 ab% (2 a^boY, { - »»y«»)^
180 ALGEBRA.
Expand —
2. {a + 35)*, (2a + h)\ {a - h)\ (3a - ih)\
3. (2m + l)^ (5 a; + 2)^ (3 a - 4c)^ (- a - 6)1
4. h? + x + l)^ (3 ci - 6 + 4 c - cZ)^ (a + 2 6 - c)2
(3a+36 + 3c)l
5. (1 + a; - or^)', (ax + 6?/ + c;2;)^ | (a + 6)a;-(c+ 0^)2/1^
6. (1 + a;)^ (1 + a; + a^)^ (a + 6a; + car)^
7. (aaj - 6^)^ (3 a; + y)^ (3a; - yf, (ar^ + a;7/ + 2/=)» (a; - y)\
8. (aJ* + a;^2/» + 2/T(^+ 2/)M^ - ^)',
j(a + 6)»-4a6J'^'j(a - 6)^ + 4 a6 | .
Simplify —
9. (a + 6 + c)^ - 3 (a + 6 + c)' c + 3 (a + 6 + c)r - c\
10. (a - 6)3 + 3 (a - 6)^ (6 ^ c) + 3 (a - 6) (6 - cf +
(6 - cf.
11. (1 + a; + Sar' + 30;^)' + (1 - a; +3ar^ - 3a;')l
12. j (a; + yf - (a;^ + y') \ ' - 27a)y (a;^ + f).
Evolution.
34. Evolution is the operation by which we obtain the
roots of quantities.
Since the square or second power of a^ is (a^y or a^, we call
(Art. 17) a^ the squa/re root or the second root of a^.
And so, since the cube or ^/wVc^ power of a^ is (a^)^ or a^, we
call a* the cube root or the ^/wVcZ root of a^.
So, generally, since the nth power of a^ = (a*")"* = a*"", we
call a"* the 72th root of a"*".
Thus, we have s/aP = a^, ^^ = a\ ^'cT' = a"».
Hence, in the case of quantities consisting of a single letter
with a given exponent, when the given exponent contains as
a factor the number indicating the root, we must divide the
given exponent by this number, the quotient being the
exponent of the root.
(We shall see farther on that this rule holds when the given exponent
ifi not so divisible, tJiQ root in this case bein§ palled a sur^)^
SQUARE ROOT. 181
35. Since the product of an eve^i number of negative
/actors must give a positive result, and the product of an odd
number of negative factors a negative result, it follows that —
I. When the root is indicated by an even number —
1. The root of a positive quantity may be written either
with a + or - sign.
Thus, v^9^ = ±3 a, ^IQb' = ±2b,
2. The root of a negative quantity is impossible.
Thus, V - tt^ V - a'*6^^, &c., are impossible quantities.
II. When the root is indicated by an odd number, the root
has always the sign of the given quantity.
Thus, ^ - 27 b' = - 3 5^ ^32x'y = 2 x'y,
(It may be remarked that the theory of impossible quantities forms an
important branch of Algebra, which the student cannot yet enter
upon. According to that theory, all quantities have as many roots as
the number indicating the root. )
Square Root.
36. We shall now develop the method of finding the
square root of a given quantity.
Ex. 1. Find the square root of a- + 2ab + 5^.
We know that a^ -\- 2ah ■\- h^ = {a + b)\
Hence, a + b is the square root of a^ + 2ab + 5^ or a^
-{■ {2 a + b)b.
Now, it is evident that the first term a of the root is the
square root of the first term a^ of the given quantity;
and if this term be subtracted, there remains 2 ab + b^,
from which to determine b the second term of the root.
Now, b is contained in 2 ab + P or {2 a + b)b exactly
{2 a + b) times. Hence it follows that the second term of
the root is found by dividing the remainder by twice the
first term of the root, and, if we wish to arrange our work
in a way similar to long division, it is evident that we
first take for our divisor 2 a + b, that is, twice the first
term of the root added to the second tei-m, which mul-
tiplied by b the second term, and subtracted, leaves no
remainder.
182 ALGEBRA.
Thus the whole process may be arranged as follows : —
CI?
2a + b 2ab + b^
2ab + W
Ex. 2. Find the square root of a* + 2 a6 + 6^ + 2 ac
+ 2 6c + c".
Now we know (Art. 26) that —
o? ■\' 2ab + b'' + 2ac + 2bG + c'ovia + hY +
2{a + b)c + c^ = {a + b + cf.
And if we compare the form (a + by + 2 {a + b)c -\- ^
with the form «^ + 2 a6 + 6^ it is evident that, having ob-
tained as in the last example the first two terms a + 6, we shall
by continuing the process obtain the third term. Thus —
a- + 2 a6 + 6^ + 2 ac + 2 6c + c'(a + 6 + c
d?
+
6
26
+
2ab
2ab
+
+
b^
+
c
2ac +
2aG +
2 6c
2 6c
+
+
c"
c"
We may deduce from the above examples the following
general rule : —
EULE.
1 . Arrange the terms of the given quantity according to the
ascending or descending powers of some letter, and take the
square root of the first term for the first terra of the quotient.
2. Subtract the square of the quotient, and bring down
the next two terms of the given quantity.
3. Double the quotient, and jDlace the result as a trial
divisor; then, f dividing the first of the terms brought down
by this trial divisor to obtain the second term of the root,
add the quotient so obtained to the first term of the root,
and also to the trial divisor, to obtain a complete divisor.
4. Multiply the complete divisor by the second term of
the root, and subtract the product, as in long division, from
the terms brought down.
5. If there be any remainder or more terms to bring down,
double the whole quotient for a trial divisor, and divide the
SQUARE ROOT OF NUMERICAL QUANTITIES. 183
remainder by the new trial divisor, to obtain the third term
of the root; and so on.
Ex. 3. Find the square root of 1 - 4:X + lOa;^ - 12x'^+ Ox\
The terms are here arranged according to the ascending
powers of x. Then proceeding according to rule, we have —
1 - 4 cc + 10 or - 12x3 +9x^(1 -2x + 3s(r
1
2-2aj -4aj+10a;2
- 4 a? + 4 ar
2 - 4a; + dxF Qa^ - Ux" + 9 x^
(jaP - I2a^ + 9x *
The student will observe that twice the quotient is most
easily obtained by bringing down the previous com2)lete
divisor with its last term doubled.
Square Boot of Numerical Quantities.
37. It is easy to apply the above method to numerical
quantities.
Since P = 1, 10^- = 100, 100^ = 10,000, 1,000^= 1,000,000,
«fcc., it is evident that the square roots of numbers having less
than three figures must contain one figure only ;
That those having not less than three and less than five
must contain two figures and two only ;
Those having not less than five and less than seven must
contain three figures and three only ; and so on.
Hence it follows that, if a dot be placed over the units'
figure, and over every alternate figure to the left, the num-
ber of dots will give the number of figures in the square root.
Thus, the square roots of the numbers 141376 and
1522756 have three and four figures respectively.
In the number 141376 we call 14, 13, 76 respectively the
first, second, and third periods. So in the number 152275G,
the first, second, third, and fourth periods are respectively
1, 52, 27, 56.
It is evident that the number of periods correspond to the
number of figures in the square root, and it will be seen that
the figures of each period are used in the operation for the
coiTesponding figure of the root.
184 ALGEBRA.
Ex. Find the square root of 565504.
Pointing off the number, we find the first period to be 56.
Now, the gi-eatest square in 56 is 49, and the square next
greater is 64. Hence the number lies between the square
of 700 and 800 ; and, following the algebraical method, 700
will be the first term of the root.
The operation will stand thus — „ ^, c
565504 (700 + 50 + 2 = 753
490000 = a"
2a + b = 1400 + 50 =1450 75504
72500 = 2ab + P
2 a+2& + c = 1400 + 100+2 = 1502 3004"
3004 = 2ac + 2bc + c^
Or, omitting the useless ciphers, and bringing down one
period of figures at a time, the operation will stand thus —
565504(752
49
145 755
725
1502 3004
3004
Ex. 1. Find the square root of 6091024.
6091024(2468
44
486
4928 39424
39421
Ex. 2. Find the square root of 83521.
8352i(289
4_
48 435
384
569 5121
5121
SQUARE ROOT 01' A DECIMAL. l85
It will be observed that the second remainder, 51, is
greater than the previous complete divisor, 48, and it might
be supposed, therefore, that the second figure in the root
should be 9 instead of 8.
Now, the square of (« + 1) exceeds tho square of a by
2a + 1.
Thus,Xa + If - o? ^ (p? + 2 a + I) - a" ^ 2 a ^- \.
Hence it follows that, so long as any remainder is less than
twice the corresponding number in the root + 1 , we may bo
certain that we have taken the figure of the root sufficiently
large.
Thus, since the remainder is less than 28 x 2 + 1
or 57, we may be certain that 8 is the correct figure and
not 9.
Square Root of a Decimal.
38. It is evident that the square of any number containing
one, two, three, &c., decimal figures, will contain two, four,
six, <fec., decimal figures respectively ; and, hence, conversely,
every decimal considered as a square must contain an even
number of decimal figures, and its square root must contain
half this even number of figures. It will then be necessary
to add a cipher when the given number of decimal figures
is odd.
Further, since decimals and integers follow the same system
of notation, it is evident that if a dot be placed over the
units^ figure of the given number, the pointing off may bo
performed with regard to the integi-al part exactly as in
integers, there being no necessity to point ofi* the decimals,
only taking care to bring them down in pairs, and putting a
decimal point in the quotient when the first pair is brought
down.
And again, if an integer be given which is not a i)ei*fect
square, we may, by afiixing to the right of it a decimal point
and an even number of ciphers, gradually approximate to the
square root as nearly as we please.
186 Algebra.
Ex. 3. Find the square root of 1-8225.
l-8225(l-35
1
23
82
69
265
1825
1325
Ex. 4. Eind the square root of 247 to four places of
decimals.
Instead of adding a decimal point and eight ciphers to
the right of the given number, we will proceed in the
ordinary way till we arrive at a remainder. Then putting a
decimal point in the quotient, we shall add two ciphers to
this and each successive remainder.
247(15-7162
1
25
147
125
307
2200
2149
3141
5100
3141
31426
195900
188556
314322
734400
628644
105756
39. It will be shown hereafter that wheri n + 1 figures of
a square root have been obtained by the ordhmry method, n
figures more may be obtained by dividing the remainder by
the number formed by taking twice the quotient already
obtained, 2^'^ovided that the whole number ofi figures in the
root is 2 n + 1 .
Ex. Find the square root of 29 to six places of deci-
mals.
Cube root.
'187
The square root required will evidently contain seven
figures. We shall therefore find the first ybitr figures by the
ordinary method, and the other three by the above method.
Thus—
29(5-3851()4
25
103
400
309
1068
9100
8544
10765
55606
53825
10770
17750 then by division,
10770
69800
64620
51800
43080
Ans. 5-385164.
8720
Cube Root.
40. "We will next develop the method of finding the cube
root of a quantity.
£x. 1. Find the cube root of a^ + 3 a'b + 3 a6' + h\
We know (Art.. 33) that {a + hf =^ it? + ^ cch +
3 a}? + 5^. Hence a + 6 is the cube root of a^ + 3 oirh +
3 a}? + h\
We see then that, the quantity being arranged according
to the powers of a, the first term a of the cube root
is the cube root of the first term of the given quantity;
and if this term d^ be subtracted, there remains Zd^b +
3a6- + 61
We see again that if this remainder be divided by 3 a^, its
first term gives h the second term of the root, and, further, if
it be divided by 6, we get 3 d^ + 3 a6 + 6^ as a quotient.
If we wish therefore to arrange the whole process in a %vay
l88* algebha.
similar to ordinary division, it is evident that we must
write as a divisor 3 a^ + 3 a& + Wy in order that after
multiplication by the quotient figure h we may obtain a
quantity which when subtracted shall leave no remainder.
The operation will then stand thus —
a^ + 3 a^^ + 3 ah^ + h\a + h
3 «2 + 3 a6 + 6^ 3 ^^6 + 3 ah^ + 6*
^a?h + 3ah^ + h^
We call 3 or the trial divisor, because by means of it we
search for the second term of the cube root. Having ob-
tained this second term, we then form the complete divisor
3c62 + 3«6 + 61
Ex. 2. Find the cube root of 8 a;» - 36 a?y + Uxif - 27 if,
^a^ -SQx'i/ + 54:xif - 27 f{2x -3y
Sx'
12£c2_ iSxij + dif -36ar^i/ + 540^2/" -272/^
-36ar^y + 54V-2 7y^
Explanation. — We find the cube root of 8 cc^ to be 2 x.
This is then the first term of the quotient, and corresponds to
a in the previous example. We now require 3 a* for a trial
divisor. This, of course, = 3(2 a?)^ = 12a;l Subtracting the
first term of the given quantity and dividing the first term
of the remainder by this trial divisor, we obtain - 3 y for
the quotient. This forms the second term of the root, and
corresponds to 5 in the last example. We now easily obtain
3 ab and b^.
Thus, 3a6 - 3 {2x){- 32/) = - ISxy, and b^ = (-3^)-
Hence, the complete trial divisor, corresponding to 3 a* +
oab + 6^ in the last example, = 12 ocF - 18 a:?/ + 9y\
Multiplying now by - 3 y the . quotient, we obtain
- 36 x^y +54 xy^ - 27 y^ which subtracted leaves no re-
mainder.
Hence, 2 aj.-.Sy^is the.cube.root.
CUBE ROOT OF NUMERICAL QUANTITIES.
I
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190 ' ALGEBRA.
Ex. 1, Find the cube root of 262144.
262144(60 ^^ 4 - 64
216000
3 a- - 10800 46144
3ab = 720
b' = 16
3^2 + 3a5 + 6^ = 11536 46144
Explanation. — Pointing off the given number, we find
the first period to be 262, and that the cube root consists of
two figures. Now, the greatest perfect cube in 262 is 216,
which is the cube of 6. Hence, the given number lies between
the cubes of 60 and 70 ; and following the algebraical method,
60 will be the first term of the cube. This, we see, corre-
sponds to a in the algebraical method.
We first then subtract the cube of a — viz., 216000, which
leaves as a remainder 46144.
We now wiite down 3 a^ or 3 (60)3 ^ 10800, which is the
trial divisor for determining b. Dividing then by this value of
3 aPy we find 6 = 4, which is the second term of the cube root.
We next obtain 3 a6 = 3 (60) (4) = 720, and b^ = A^ = IQ,
and so by addition we get 3a^ + 3 ab + b^ ~ 11536, which
is the complete divisor. Multiplying this then by the
quotient figure, we subtract the product, and, there being no
remainder, we find the cube root to be 64.
We may omit the useless ciphers in the above operation,
if, remembering the local value of figures when numbers are
expressed in ordinary notation, we take care to place the
right-hand figure of the value of 3 ab one place to the right
of the corresponding figure of the value of 3 a^ ; and also to
place the right hand figure of b^ one figure fiu*ther still to the
right.
The operation will tbeu stand thus —
262144(64
2iq
3 X 62 =; 108 46144
3x6x4= 72
4^ ^ 16
1153G 46144
CUBE ROOT OF NUMERICAL QUANTITIES. 101
Ex. 2. Find the cube root of 102503232.
102503232(468
_64_
3 >i 4^ = 48 38503
33336
3x4x6 :==
72 ^
6^ -
36 (^
5556 (
36 J
3 X 46^
6348
3 X 46 X 8 =
1104
8^-
64
645904
5167232
5167232
Explanation. — The first two figures of the root are ob-
tained as in Ex. 1 . We then treat the number they form, viz. ,
46, as corresponding to a in the algebraical model, omitting
useless ciphers. Obtaining then 3 a'*^ or 3 x 46^ ~ 6348, we
find 8 to be the next figure of the root. Then writing under
this, 3 a6 or 3 X 46 x 8 - 1104, and afterwards l? or 8- = 64,
taking care as to the positions of the right-hand figures, and
adding, we get 645904 as the complete divisor. Then as before.
Remark. — It is unnecessary to be at the trouble to find
the value of 3 x 46'"^ by ordinary multiplication. For re-
ferring to the algebraical model, and writing here the succes-
sive terms of the complete divisor, and adding, we have—
^ ( If we now again write down Jr
Sun\ = 3 a'- + 3a6 + H^ T under this sum, and then add up
}? ) the last four lines, we get —
3 a^ + 6 a6 + 3 b\ or 3 (a^ + 2ab + h') = 3 {a + by.
This is three times the square of the first two terms of
the root.
It therefore follows that, if, as in the above example, after
completing the operation for finding the first two figures of
the cube root, we write under the complete divisor just ob-
tained the value of the square of the second figure, and then
^d together the last four lines thus obtained, we get three
192 ALGEBRA.
times the square of the quotient for a partial divisor by which
to determine the next figure of the root.
The four lines to be added are in the above example
bracketed. This method will be found to materially shorten
the work, for it may be similarly applied to find the trial
divisor when the cu.be root consists of any number of figures
Cube Root of a Decimal.
42. We know that the cube of any number containing one,
two, three, &c., decimal figures will contain three, six, nine,
&c., decimal figures respectively, and hence, conversely, every
decimal considered as a cube must contain a number of
decimal figures which is a multiple of threes and the number
of decimal figures in the cube root must be one-third of the
number contained in the given cube. It will then be necessary
to add ciphers when the given number of decimal figures is
not a multiple of 3.
And by continuing the reasoning of Art. 37, if a dot be
placed over the units' figure and over every third figui'e to
the left, it will be sufiicient to bring down the decimal figures
three at a time, putting a decimal point in the quotient when
the first three are brought down.
And further, if a^i integer be given which is not a pei-fect
cube, we may proceed in the ordinary way till we arrive at
a remainder, and then, putting a decimal point in the quotient,
by affixing three ciphers to this and each successive re-
mainder, approximate to the cube root as nearly as we please.
Ex. 3. Find the cube root of 395-446904.
395-446904(7-34 .
343
3 X 7' = 147 "52446
3x7x3= 63
32 = S ^
46017
15339 f
9J
3
X
73^
=
15987
73
X
4
rr
876
42
=
16
1607476
6429904
6429904
CUBE ROOT OP A DECIilAL.
193
43. We shall show farther on that when n + 2 figures of
a cube root have been obtained by the ordinary method, n
figures more may be obtained by dividing the remainder by
the next trial divisor, provided that the whole number of
figures in the root is 2 n + 2.
We may apply this principle with advantage when we re-
quire the cube root of number to a given number of decimals.
Ex. Find the cube root of 87 to fiYQ places of decimals.
The required cube root will evidently contain 6 figures,
and since 6 here corresponds to 2 t* + 2 above, it is evident
that n = 2, Hence, we shall find 4 (that is, n + 2) figures
by the ordinary method, and then 2 more by division.
The operation will stand thus —
87(4-43104
64
3x4^ =
48
23000
3x4x4 =
48 ■)
4? =
16 1
5296 (
21184
16;
1816000
3 X 44^ =
5808
3 X 44 X 3 =
396 ^
3« =
H
584769 f
1754307
9;
61693000
3 X 443^
: 588747
3 X 443 X 1
1329 \
P =
H
58887991 C
58887991
1)
280500900
58901283
235605132
44895768
Ans. 4-43104.
Ex. XI.
Find the square roots of —
1. 4a;VV, 16ay, ic* + 2aV + a'
2. 4 a;* - "^^ ^'^' • 00^4^.2 qa^s.
ISar^y
29 A' - 30 aV + 25 «r/.
N
194 ALGEBRA.
3. 25 a* - 30 a'b + 19 ^-6^ - 6 aP + b',
4. 1 -^ ix -\- lOa;^ - 20i^ + 25a:* - 24a;» + lGa;l
5. «2 + 2abx -h {2ac + ¥)x' + 2{ad + bc)x^ + {2bcl
6. «^Jc-« - 6aV«-i + 17a"aj-"-2 - 2iax'^''^ + IGct-^-^
7. a;^ + 2 + x-\ orx^^^ - 2 + «-V.
8. 9a;-"* - 3a''x^ + 25 a^ - 30 ax^ +^ + 5a\
4
Find the square roots of — •
9. 1296, 6241, 42849, 83521.
10. 10650-24, -000576, -1, ^W
Give the values correct to four places of decimals of —
^ of -31416 V5 + V2 ^ VlO - 2 „ ,,,„ „
l-i. Tp?;; > — ; n: + "J > •> l*ll> Of
V93 V5 - V2 VlO + 2
/ 4 ,
V 32X6
Find the cube roots of —
13. SaWif, I25a^y,a^ + 6 a'b + 12 ab'^ + 8R
14. aj^2 + 9x'' + 6a;« - 99a;« - 42aj* + 441 a^ - 343.
15. a^ + 3xy + 3x7/^ + y^ - ^cxy - 3 car* - 3 nf
+ 3 c^a; + 3 chj - c^
16. a^ + aj-3 + 3 (aj + x~% x^f/-^ + 3 xy-^ + 3 xy-"" + 1.
Find the cube roots of —
17. 5849513501832, 1371-330631.
18. 20-346417; -037, tVtV
Give the value of the following correct to four places of
decimals : —
T Q v'5'a2 + 4/-03375 1
1 y. — o,-— ■
m - v'-oT '4/4 + 4/2 + 1.
.. tfm + ^-04 75 + 2
20. -::5i--;:^i — — r=^ of s
V^ + v/-04 7
21. ( V7 + 2) ( V7 - 1), (5 + V3) (4 + V12).
GREATEST COMMON MEASURE. 195
22. jirT6"j2] y '6'+ 15 ys:
03 ^3 6 + 2 75 J5'+ 1
" * 4 16 ' 4 ' _
24. a^ (6 - c) - b^ (a- c) + c^ {a - h), where a = Vl %
h^- - V^aixdc =~ 4/^027;
CHAPTER lY.
GREATEST COMMON MEASURE AND LEAST COMMON MULTIPLE,
Greatest Common Measure.
44. In Arithmetic (page 24) we defined the G.C.M. of two
or more numbers as their highest common factor. In Algebra
the same definition will suffice, provided we understand by
the term highest common /actor, the /actor 0/ highest dimen-
sions (Art. 18). This, it need hardly be remarked, does not
necessarily correspond to the factor of highest numerical value.
45. To^find the G.C.M. of two quantities.
Rule. — Let A and B be the quantities, of which A is not
of lower dimensions than B. Divide A by B, until a re-
mainder is obtained of lower dimensions than B. Take this
remainder as a new divisor, and the preceding divisor A as a
new dividend, and divide till a remainder is again obtained
of lower dimensions than the divisor ; and so on. The last
divisor is the G.C.M.
Before giving the general theory of the G.C.M. we shall
work out a few examples.
Ex. 1. Find the G.C.M. of or - 6aj - 27 and 2a;2_ ii^_ 03,
According to the above rule, the operation is as follows ;—
cc^ - 6a; - 27)2ar^- 11a;- 63(2
2ar'- 12a;- 54
ic - 9)a;2 - 6a; - 27(a; + 3
3x- 27 \
3x - 27
.-. The G.C.M. is a; - 9.
196 ALGEBRA.
Ex. 2. Find the G.O.M. of lOi^ + 31 a;^ « 63aj and Ucc»
-f 51 a^ - 54 a;.
We may tell by inspection that a; is a common factor,
which we therefore strike out of both, only taking care to
reserve it. The quantities then become —
lOa:^ + 31a; - 63, andUa;^ + 51a; - 54.
We may now proceed according to rule, taking the
former as divisor. We see, however, that the coefficient
of the first term of the dividend is not exactly divisible by
the coefficient of the first term of the divisor. Multiply
therefore (to avoid fractions) the dividend by such a number
as will make it so divisible, viz., by 5. This will not afiect
the G.C.M., as 5 is not a factor of the first expression,
viz., 10 a;^ + 31 a; - 63.
It may as well be here mentioned that the G.C.M.
of two quantities cannot be affected by the multipli-
<5ation or division of one of the quantities by any
quantity which is not a measure of the other. We
shall, for a similar reason, reject certain factors or introduce
them into any of the remainders or dividends during the
operation. (See Art. 47).
Ux" •{- 61x - 64:
^
10 a;^ + 31a; - 63)70 a;^ + 255 a; - 270(7
" 70a;^ + 217a; - 441
38a; + 171
Rejecting the factor 19 of this remainder, we have —
2a; + 9)10a;^ + 31a; - 63(5a; - 7
- 14 a; - 63
- 14 a; - 63
Hence, 2 a; + 9 is the last divisor, and multipljdng this by
Xf the common measure struck out at the commencement, we
find the aCM to be a; (2 a; + 9) or 2a;' + 9 a;.
GREATESr COMMON MEASURE.
19T
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198 ALGEBRA.
Dividing this remainder by 14371, and taking the quotient
for a new divisor, we have —
jc2-7a;-3)-72a^ + 679a;2-1009a;- 525(-72a: + 175
^ -72a;^ + 504a;^+ 216 a;
175 ic^- 1225 a;- 525
175 ar^- 1225 a;- 525
:, x" - 7x - Sis the G.C.M.
It will be seen that we have introduced and rejected factors
during the operation in order to avoid fractional coefficients*
This, as will be seen from the general theory, will not affect
the result, provided that no factor thus introduced or rejected
is a measure of the corresponding divisor or dividend, as the
case may be.
Theory of the Greatest Common Measure.
46. Let A and B be the two algebraical quantities, and the
operation as indicated by the rule (Art. 45) be performed*
Thus, let A be divided by B, with B)A(p
quotient p and remainder C, Then j(?i?
let B be divided by (7, with quotient ^, 0)B(q
and remainder JD. Lastly, let G be qQ
divided by Z>, with quotient r, and D)G(r
remainder zero. ^^
Then we are required to show that ~7r
D is the G.C.M. of A and B.
(1.) i> is a common measure of A and B,
Now, we have G = rl), B = qG + JD, A = pB + G.
Hence, i> is a measure of (7, and therefore of qG, It is
therefore a measure of qG + D or B. Hence, also, i> is a
measure ofpB, and since it is also a measure of G, it must be
a measure of pB + G or A, But we have shown it to be a
measure of B, Hence, i) is a common measure of A and B.
(2.) D is the G.C.M. of ^ and B,
For every measure of A and B will divide A - pB or G;
and hence every measure of ^ and B will divide B - qG ov
JD, Now, JD cannot be divided by any quantity higher than
JDf and, therefore, there cannot exist a measure of A and B
h^her than J). Hence, D is the G.C.M. of A and B.
1:fifiORV OPTHE GftfiATESt COMMOK MEASURE. 199
47. A factor which does not contain any factor convnion to
both A and B may he rejected at any stage of the 2)rocess,
Let the operation stand thus : —
B = 7nB ' suppose,
B^A{p C)B\q
pB qG;_
C = nC suppose, D)C\r
r_D_
"0
where neither m nor n contains any fiictor common to A
and B,
It will be iari exercise for the student to show that B is the
G.C.M. of^and^.
48. A factor^ which has no factor that the divisor has^ 7iiay
he introduced into the dividend at any stage of the 2)rocess.
The operation may stand thus —
B)mA(2), where m has no factor that B has ;
pB
C)nB[q, where n has no factoi' that C has ;
qc
D)C{r
rP
As in Arts. 46, 47, it may be easily shown that D is the
G.C.M.
Both the above principles are made use of in working but
Ex. 3 Art. 45.
49. When a bommon factor can be found by inspection, it
is advisable to strike it out of the given expressions. Then,
having found by the ordinary process the G.C.M. of the re-
sulting quantities, we must multiply the G.C.M. so found by
the rejected factbri
• Thus, 4 cc is common to the qiiantities 4 a;^ — 20 a;- + 24 aij
tind4a;^ 4- 16 ar^ - 84cc.
Rejecting it, we get o? - 6 x + 6, and or + 4 a; - 21^
whose G.C.M. is easily found to be a; - 2.
Multiplying by 4 a:, we find the required G.C.M. to be
4ar^ - 8a;.
200 ALGEBRA*
5O4 By a little ingenuity on the paft of the student in
breaking up the given expressions into factors, the ordinary
and often tedious process of finding the G.C.M. may be
avoided. The limits of our space will allow us only one
example.
Ex. Find the G.C.M. of 3x^ + 4:x' - 10 x + 3, and
15 a;' + 47a;2 + 13 a; - 12.
The first expression contains a; - 1 as a factor (Art. 30),
for the sum of its coefficients is zero. The other factor may
be obtained thus —
3a;^ + 4a;^-10a; + 3 = 3ar^- 3 x^ + 7a;2- 7 a;- 3a;+ 3
= 3a;2(a; - 1) + 7a;(a;- l)-3(a;- 1)
- (3 a;2 + 7 a; - 3) (a; - 1).
Now, 3 a;- + 7 a; •— 3 is not further resolvable, and a; - 1
is evidently (Art. 30) not a factor ofl5a;^ + 3ix^ + 13a;-
1 2. It is, therefore, very probable that 3 a;^ + 7 a; - 3 is the
G.C.M. required. *
We may test it thus —
15a;3 + 47a;2+ 13a;- 12 = 15a;3+ 35a;-- 15a;+ 12a;^ + 28a;- 12*
= 5 a; ( 3 a;^ + 7 a; - 3 ) + 4 ( 3 a;2 + 7 a; - 3 )
= (5a; + 4) {Sx" + 7x- 3).
Hence, 3a;^+ 7a;- 3 is the G.C.M. required.
G.C.M. of Three or More Quantities.
51. The G.C.M. of three or more quantities may be found
thus —
Rule. — Find the G.C.M. of any two of the quantities, then
the G.C.M. of the G.C.M. so found and a third quantity, and
so on. The last found G.C.M. will be the G.C.M. required.
Ex. XII.
Find the G.C.M. of the following—
1. ar^ - 5 a; + 6 and a;^ + 3 a; — 18.
2. a;^ + 6 a;' +lla;+6 and a;' + 5 a;- + 7 a; + 3.
3. 2af»+ 10a;2- 18;B-90and 3 a;^ + 16a;2-26a; - 141.
4c, aP + (a + b €C + ab and x^ + {a + c) x + ac,
5.
LEAST COMMON MULTIPLE. 201
6. a;^ - 4 J!c + 3 and aj' + 4 x- - 5.
7. 4ic'- 32a:« + 85 a; - 75 and 3 af» - 15ar^ + 15a; + 9,
8. 9 a;' - 3 a:y + 2 2/ - 4 and 6 x* - 4 a;^ - 9 a;?/^ + 6 ?/l
9. 48 a;* + 8 a:^ + 31 x" + 15 a; and 24 aj^ + 22 a;3 + 17 x"
+ 5 a;.
10. 15 o? + a'5 - 3 ah'' + 2h^ and 54 a-^^ - 24 b\
11. 3a;'-(3c + c£+l)ar^-(2«4-6-3c-cZ+2)a;
+ 2 a + 6 and 2a:'-(a+6 + 2)a; + a + 6.
12. 6a:5 _ 4^1 - 11 ajS „ 3 a;^ - 3 a; - 1 and 4a;^ + 2a;'
- 18 a;^ + 3 a; - 5.
13. ah + 2 0^ -- 3 b'' - 4:hc - ac " c^ and 9 ac + 2 a^ -
5a6 + 4c^ + 8 5c - 12 6^.
14. e'x' + e'' + x' + I and e'^'a;* - e-* + a;* - 1.
15. oa;* + {h + c) aP - ax '- h -^ c and ear^ - (f - g) x^ +
(/ - e) a; - ^.
16. 4 a;* + 2 a;' + 4 a;^ + 39 a; - 9, 8 a;* + 20 a;^ + 51a;
+ 9, and 2 a;* + a;^ + 3 a;2 + 18 a;.
17. aa;^ - (c + 1) ar^ + (c + 1) a; - a, 5a;^ - (5 + cZ) a;' +
(c + d)x^ — {c + e) X + €y and {c + I) oc^ + {d + 2) a;* ~
(c^ + 1) a;^ - (c + 2) x\
18. a^ - 5' + c» + 3 ahc and a'' - h^ + c^ + 2 ac.
Least Common Multiple.
52. When two or more algebraical expressions are arranged
according to the powers of some letter, the expression of
lowest dimensions which is divisible by each of the given
expressions is called the L.C.M.
53. The L.C.M. of monomials and of expressions whose
factors are apparent may be found by inspection.
Ex. 1. Find the L.C.M. of a6, ac, acZ, Sc, hdy cd.
If we form an expression, whose elementary factors con-
tain each of the elementary factors of the given quantities,
we shall evidently have a common multiple ; and if no ele-
mentary factor of this expression is of a higher power than
the highest power of the same factor in the given quantities,
we shall get the L.C.M.
Hence, the required L.C.M. = ahcd.
202 ALdl^BRA*
Ex. 2. Find the L.C.M. of—
{a - Z>) (5 - c), {a - h) {c -- a), (6 ^ c) (c - a)*
Alls, {a - h) {b - c) {c - a).
Ex. 3. Find the L.C.M. of a (x + 1), b (x" - 1),
c (x^ + 2 X — 3),d(x^ + 4c X " 3)i We may write the given
expressions thus — -
a{x + 1), b{x + I) {oi - i),
c{x -^ 1) (cd + 3), d (x + 1) (x -i- 3).
Hence, the required L.C.Mi = abed (x - l) (x + 1)
{x + 3).
Ex. 4. Find the L.C.M. of a^ - aaj + a^^ a^ + ax + x",
a' + xl, a^ - x\
Kow (Art. 29) a^ ■{- x^ = (a + x) {a^ - ax + ^),
and a^ - a* = (ot - a) \a? -{• ax ■\- x').
Hence the required L.C.M. —
^ {a ■{- x) (a - x) (a^ -{• ax + x^) (cu^ - aa; + x) = a^ - x^.
64. !the L,d.M, qf two quantities is found by dividing
their product by their G,C.M.
Let a and b be the two quantities, and d the G.C.M. ;
And suppose a - pd and b - qd.
It is evident that p and ^contain no common factor.
^QWCQ pq is the L.C.M. of p and q ; and, therefore, no expres-
sion of lower dimensions than pqd can possibly be divisible!
by pd and qd.
Hence pqd is the L.C.M. ot pd and qdy or of a and b,
"No^pqd = pd X qd -■ d = a X b -r- d, and hence the rule:
55. To find the L.C.M. of three or more quantities.
EuLB. — Find the L.C.M. bf two of the quantities, theri
th0 L.C.M. of thd expressioii thus obtained and a third
quantity, and so on. The last expression so found is the
L.C.M* required.
We shall prove this rule in the case of three t[uantities«
Let tt, by c be the quantities, and ni be the L.C.M. of a
iand. b.
Then the L.C.M. of m and c is the L.C.M. required.
For every common multiple of m and c is a common
multiple of a^ 6, c. And every common multiple of a and 6
LEAST^ COMMON MULTIPLE. 203
taust contain the m, their least common multiple. Hence,
every common multiple of a, 6, c must be a common multiple
of 7?i and c, and the conA^erse is also true; Hence, the
L.C.M. o£m and c is the L.C.M. of a, b, c.
Ex. XIII. 4
Find the L.C.M. of—
1. axy^f 3 aVi/, 4 a^if, 6 ary^,
2. 5a^^^^6aV, 45V.
3. {a -- b){b - c), (6 - a) (a - c), (c - a) (c - 5).
4. ax {x + a), a^(aj - a), fc^ - a^*
b, aP -h 3x + 2,x' + 4:x + dyX^ + 5x + e.
6» 0? - a; - 30, aj2 - llaj + 30, ar» - 25.
r. 6ar^ + 37a; + 66, Sa;^ + 38a; + 35^ 12a;^ + 47a;
+ 40*
a 5 (or^ - a; + 1), 6 (ar' + 1), 7 (o;^ + 1).
10. a;^ + (« + Z>) a; + a5, o;^ + (a + c) a; + ac, o;^ +
{^ -^^ c)x -v be.
11. 1 - oj, 1 + a;, 1 + a;^ 1 + a;^, 1 + x^,
12. a;3 + 6a:' + 11a; + 6, a;^ - ^x" - 25a; + 150.
13* o? - 3a6 {a ^ b) ■- b\ a? - 6^ a^ + a^6 + a6l
14. X* - 1, 6a;5 + 5a;^ + 8a;' + 4ar^ + 2a; - 1.
15. a^ - 2o?b'^ + b\ a^ + 4a36 + 6^-6^ + 4a6' + 6S
16. 3a;3 _ 4^ .^ 1^ ^x^ _ 7aj + 5^ 4cc4 4. c a;^ +
10 a;.
17. Zx^ -v Qx- 24, or' - 12a; + 16^ 5a;^ - 122a; - 36.
18. a^ - ab\ P - a'b, aP - b\ o?b - a\
19. 3a;* - 48, ^o? - 20, 30:^ - 16aj + 20.
20. a? '- 'if,x^ - 2 x'lf + 2/^ a;^ + or'y + a;?/* + 2/^.
2U ^ •\- ax^ -V aV + aV + a^x + a^ and a;'' ~ ax\ -f
aV - o^x^ + a^a; - a\
22. a2 ^ 52 - c^ - c^3 + 2a6 - 2cd and a' - 6- - c-
+ ci^ + 2ac^ - 2 6c.
204 ALGEfiRA.
23. ft' + i^ + c' - 3 a5c and {a + hf ^ 2 (a + b)c
+ c\
24. (a + by - (c + c^)S (^ + cy - (b - d)\ {a + df
- (6 - c)-, (c + dy - (ft - 5)', (6 + dy - (ft - c)^ and
(6 + cy - (ft - (^)l
CHAPTER V.
Fractions,
56. It is unnecessary to repeat here the propositions relat-
ing to fractions which were proved in Arithmetic, Chap. II.
of this work. The student will see that, by substituting
general symbols for the particular figures there used, the
reasoning will equally hold. We shall work out a few
examples to show the method of dealing with them in algebra.
Ex. 1. Simplify the fraction g k — ^-je — •
By inspection (Art. 30) we see that x - 3 is a factor of
numerator and denominator. We have then —
a:^ - 2a^ + a; -12 ^ a^ (a; ~ 3) + a; (a; - 3) + 4 (a? - 3 )
af^ + 2a; - 15 (a; - 3) (a; + 5)
_ {x^ + X + 4) (x — 3) _ x^ + X + 4:
(x - 3) (a; + 5) ST~5 ^^•
1 1 2ft
Ex. 2. Find the value of
a + b a - b d^ + b^'
1 1 2ft ^ (ft - 6) + (ft + b) __ 2ft
ft + 6 "** ft - 6 a" + b^" (ft + 6) (ft - 6) a' + b'
2ft 2ft / 1 1 \
2a
(gg + h') , (g^ , 5^) 25^ 4ft5^
(ft^ - ^>^) (ft2 + 5^) ^ " ^' c?"::!^ "a* - 6^-
FRACTIONS. 205
Ex. 3. Find the value of / ttt r + 77 ryr -v
{a - b)(ci — c) (6 - a) (6 - c)
a6
+ (c _ a) (c - 6)-
The second denominator has a factor, (5 - a), which differs
from a factor, (a - 6), of the first denominator in sign only.
"We shall therefore change the sign of the second fraction,
and also of its first factor. This will not alter its value.
And, similarly, we find that by changing the signs of each
of the factors of the third denominator we shall have them in
a form corresponding to factors of the first and second denomi-
nators. The sign of the third fraction will not be changed, as
the sign of the denominator will, on the whole, be unchanged.
The given expression then will stand thus —
he ac ah
" (a -h){a- c) " {a - b) (b -- c) '^ loT^c) {b - c)
he (h - c) - ac (a - c) + ah (a - h)
" {a - b) {a - c) (6 - c)
he (6 — c) - a-e + oc^ + a^h - ab^
= {a - 6) (a - c) (b-e) ' ^"' re-arranging,
a^ (6 - c) - a (b^ - c2) + he (b - c) ,, ,. . ,.
= — ^^ — , IK / r-71 \^ -y ^'^^^f dividing nume-
(a - h) (a - c) (b - c) °
rator and denominator by 6 ~ c,
^ a^ - a{h + c) + he _ {a - h) (a " c)
" (a - bfiaT^^^ ~ {a - b) (a - c) " _ '
Ex. 4. Simplify—
f J _ 4 a^'x - 3 aa^ + a^ ]. ^ f ^2 . 4a^a; -{-^ax^ + x " )
I a ■¥ X ) \ a " X ) '
The given expression —
a\a + x) - ia^x-^ 3aa;--ar^ a^a-x) + i a^x + 3 aar^ + x^
a + X a — X
■ 4 a'^x + 3 aar* - a;'
X
a + X (* - a?
206 ALGEBRA.
— ■ — X ,
a + X
_ {a - xf (a + x)^ _ {a - xf {a + x)
X
a + X a - X
Ex. 5. Divide —
1— ' ^ '-r-^ - (^^ - ^Y^
(^ + 6) 1-1-4- --^Jby (a - 5) ] -% + ,-^o}
Now —
or, reducing —
_ (« + 5) (aj - a) (aj + 5)
Ex. XIY.
Simplify the following expressions : —
1 a^ - 5a; + 4 a;^ - 3a? + 2
aj^ + 2 a; - 24' a."^ + 4a;^- 5*
^-^ 6a;2 + 29a; + 3.5 2a;3 + 7a; - 9
3.
14a;'^ + 39a; + 10' Sa;^ _ 3^ _ 4^. + 2
a^ - 2^262 + 5* 24a5 - 28a25 + 6a&^ - 75^
«^ - 4^26 + 4a62 - 6^' 6 a^ + 11 a6 - 21 6-
5.-K-^ 1 1
(* -1- 6 a - 1) a - h a + b'
FRACTIONS. 207
, , b ah ah
G.
7.
8.
9.
a + h a - h ah - W cv" + ah
1 03-1 a; - 1
ij^l)^ 4 (ic- + 1) " 2 (x' + 1)-'
_5 2__ _ _18 __ nx + 1
:c + 1 (.T + 2)- 5 (oj + 2) 5 (aj'^ + 1)'
11 11 14
cc+l aj + 3 (ic+ 1)-'
,Q 8 _ 4 2_ 16a! -f 14 _ lGj»-8
• (aj - 1)^ (a; - If Jx^f "^ 3 (ar - 1) 3(i-~-^TI) *
11. . X . -^ -. i_- ^ . ^
(« - 6) (d^ - c) (6 - a) (6 - c) (c - a) (c - h)
t + ^ + 1 .
(a - h) (a. - c) (6 - a) (6 - c) (c - a) {o - b)
t + I + £!__.
(a - b) (a — c) (6 - a) (b - c) (c - a) (c - b)
^ ^ 6^ ^ c^
(a - b) (a - c) {b - a) {b - c) (c - «) (c - 6)'
"■ (F->)V')'(EM^)'a^o(F^)'
16. ■ ^■
12.
13.
14.
(a + ly + {b - cf + (a + cy - _ 2 _ _J_ ^
(a + 6) (6 - c) (a + c) « + c 6 - c « + 6'
( a + a; « - a; a" + ar J I a + x a-x a-" + a;- J
( a- - 6^ a + 6 j * I *" a? - ir a + h i '
10.'{l-2_^._l^lx{ 2a^_2a=_j)
I a- a-(a^ + ar) J I ar(a- - ar) ar J
«^f-+l)+6=(£+l) + c^(?+l)
' c a 1) ah + be + cc^
208 ALGEBRA.
=Me-^p't>}-{(^:)'*_'}-{(f^:)'*'}
a;- {y — z) + 'if- {z - x) + z- (x - ^z)
22. |Li^+ _I:i^l ~ / 1 + ^ -^-^'1
23,
(a + 6) (a + c) (x -a) {a+ b) (6 - c) {x + b)
/.3
(a + c) (6 — c) {x + c)
24. |?^'-^ + J6-^l - |?-^-i-il.
^ { a' a x^ ) \ cir a x )
\a - bf \a + b/ ''
26.
27. ^' + -i^-, - J±- - _12_ +2.
(a + by (a - by a + b a - b
28.
+ y' - 2a;y ^ f_
{x - y) {a + aj) (x - y) {a + y) (a + 7/)-
y X y X "^^ '^'^
29 V - V '^- - y /"^ - — Y
30,
1
y + -
SIMPLE EQUATIONS. 209
CHAPTER YI.
Simple Equations.
57. "When two algebraical expressions are connected by the
sign ( = ), they are said to form an equation.
When the equality is such that it is true for all values of
the letters in the given expressions, it is called an identity.
Thus, {x + a){x-\-h)=x'+{a + h)x-vah\ ....,•,
and {a + hf - (a - hf = 4.ah ] ^""^ i^^^^i^ies.
58. When the condition of equality is such that some one
or other of the letters must have pai'ticular values or a
limited number of values, the statement of equality is termed
an equation of condition^ or, more briefly, an equation.
Thus, it may be found on trial that the equality
4rx; + 2 = 3aj + 5
is true only when a; = 3. Such an expression is therefore
an equation.
'59. The letters of an equation to which particular or a
limited number of values must be given are termed unknoivn
quantities.
Equations may contain one, two, three, or more unknown
quantities.
The determination of the particular value or values of the
unknown quantities is called the solution of the equation,
and each of the values which satisfies the equation is said to
be a root of the equation.
60. The expressions on the left and right sides of the sign
( = ) are termed the first and second sides respectively. It
follows, therefore, that —
1. If both sides of an equation he multiplied hy the same
quantity, the equation still subsists.
2. If both sides be divided by the same quantity, the equa-
iio7i still holds.
3. Any term may be transposed from one side to the other
if the sign of the term be changed.
Thus, if 3 aj + a = b, we must have also
3cc = 6 - a,
210 ALGEBRA.
for this results from subtracting a from each side of the
equation.
4. The equation holds if every term on both sides has its
sign changed.
Thus, if ax + b = ex - dj we may reason as follows : —
The quantity (ax + b) looked upon as a whole is given
equal to the quantity (ex — d) looked upon as a whole. If
we change the qualities of these quantities, they will evidently
be still equal.
Hence, - (ax + b) ~ — (ex - d)
or, - ax - b =^ —' ex + d.
Now, this is the result of changing the sign of every term on
both sides of the given equation.
5. The sides of an equation may be reversed without destroy-
ing the equality.
Thus, if mx + n = px + q/\\> must also follow that
px + q = mx + n.
6. The sides of an equation may be raised to the same
POWER, or we may extract the same root of both sides, and
the equation still subsists.
61. Simple equations are those in which the unknown
quantities are not higher than the first degTee, when the
equations are reduced to a rational integral form.
The following is the general method adopted in solving a
simple equation involving only one unknown quantity —
1. Clear of fractions if necessary.
2. Transpose all the terms involving the unknovm quantity
to the first side of the equation, and all the remaining terms
to the second side.
3. Simplify both sides if necessary, and divide both sides
by the coefficient of the unknown quantity.
Ex. 1. Solve the equation 5fl3+6 = 3i:c+12.
Transposing the terms, we have —
5a;-3a;=12-6.
Now, simplifying, we get — »
2x = G;
and dividing each side by the coefficient of the unknown
quantity, viz., by 2, we have — ^
X = Q -r 2 = 3.
SIMPLE EQUATIONS. 211
Verification. — Putting the value 3 for x in each side of the
given equation, the first side becomes 5 x 3 + 6 or 21 ; and
the second side becomes 3 x 3 + 12 or 21. The value of a:
found therefore satisfies the given equation.
-r^«^. X - 2 X ^^ £c-6^,
Ex. 2. Given —j— H- j = 20 ^, find x.
Clearing of fractions, by multiplying everi/ term on each
side by the L.C.M. of the denominators, viz., by 6, we get —
3 (a; - 2) + 2 cc = 20 X 6 - 3 (a; - 6).
(Beginners often neglect to multiply integral terms such as 20 by
the L. CM.)
or3aj - 6 + 2x = 120 - 3 aj + 18, or, transposing,
3aj + 2iw + 3 a; = 120 + 18 + 6, or, simplifying,
Sx= 144,
or dividing each side by 8, the coefficient of x, we have —
a; = 18, the value required.
(It will be good practice for the student to verify this result as in
the last example).
_ .4a;-. 21 ^^ 7 a; - 28 ^, d-7x
Ex. 3. ^ — + 7| + 3 =x+ 3f g — + ^V
It is sometimes convenient to first partially clear off frac-
tions. Thus, multiplying each side by 72, v/e have —
72 ^4 a; — 21^
-^ ^ + 47 X 12 + 24 (7 a; ~ 28)
- 72a; + 15 X 18 - 9 (9 - 7a:) + 6;
288 a*
or —^ 72 X 3 + 564 + 168 a; - 672,
= 72 a; + 270 - 81 + 63 a; + 6;
or, transposing,
4H a; + 168 a; - 72 a; - 63 a;
= 270 - 81 + 6 + 216 - 564 + 672;
or, simplifying,
74^ a; = 519 ; or, multiplying eacli side l)y 7,
• 519a; = 519 x 7;
.-. a; = 7.
212 ALGEBRA.
Ex. XIY.
1. 5a; + 2 = 2aj + 11.
2-4 + 6^8 -.^^•
3. 2x + a=3x-'b.
4. 3(a;-7) + 4a; = 2 (2a; - 4) + 2.
^a;-l X + 3 1x - 1 8a;
5. — TT— + 7— = ^ +
2^4" 6 ^12
8 (5 a; + 2) 2 a; - 1 _ 17a; - 2 5| + 80a ;
^' 3 "" 8 - 4 **" 7 •
9. ax + be = bx + ac,
^ X a X b
abba
^^ X - a X - b _ ex - (^
11. — ,— + = 2 + r — .
6 « ab
12. a&a; + 5^ = i'-a; +a^
13. - + T +" = «& + «c + 5(?.
a 6 c
, , a; + 5 a + a;
14. = — 7 — .
a b
^^ ax -v bx -v ex _
15. 7 = a + 6 + c.
^/ o? - Zbx _, 5a; 6 Z>a; - 5 a' Z»a; + 4 a
16. X T. — ' - 6' = — + ^r— a - — 7 .
a^ ^ a 2 a^ 4 a
17. •15 a; + -025 = •075 a; + -175.
18. i^Lj^ . -1^ . ?^4^ - .083.
ABBREVIATED METHODS FOR PARTICULAR CASES. 213
20. (a; + a) {x + b) = (x + c) {x + d).
21. {x^a) {x-h) = (a; - ^r+~h)\
^^ X " a x - b x - c'^ ^ A 1 1\
23. —J— + + — ,- = 2 (- + T + -).
be ac ab \a b c/
^^ \ - ax I - bx 1 - coj /2 2 2\
24. — T — + + r— = (- + r + -p-
be ac ab \a o c^
Abbreviated Methods for Particular Cases.
62. When the unknown quantity is involved in both
iiumerator and denominator, it is often convenient to reduce
such fractions to mixed numbers,
T. c, , X. 6a;-7 12a: + 18 ^
Ex. Solve the equation ■ o " 3 x - 6 ~
T>,. .. 6a;-7 ^"^19
By division we get ^ ^ ^ = 6 - ^-:^'^
12 a; + 18 , 38
and — o K - = 4 +
3a;-5 ""^Saj-S*
Hence the given equation becomes —
/ 19 \ / 38 X
or, transposing, - ^^-^ = 3^ _ g + 2-6 + 4;
19 _ 38 .
or, dividing each side by — 19, we have —
_1 2
aj+2~~3a;-5'
214 ALGEBRA.
Hence, multiplying each side by the L.C.M. of the de-
nominators —
3x - 5 = -2(aj+2)= -2a;- 4,
or3aj + 2a;= — 4 + 5,
or 5 a; = 1,
.•.^ = i.
63. When each side of an equation consists solely of a single
fraction, the oiumerator of either fraction niay change places
with the denominator of the other,
a p
Let ^ - " be the equation.
Multiply each side by 6, then, by Art. 60 (1.)—
a p pb
•% X - -' X h, or a ~ - .
h q ' q
Divide each side by^, then, by Art. 60 (2.) —
a ph ah
- =: — + t?, or - = -
p q ^' p q
Here the denominator h of the first side of the given equation
has changed places with the numerator p of the second side.
And similarly we may show that 7 = - , where the other
6 a
numerator and denominator have changed places.
Cor. The two sides of an equation of the form . = - may
he inverted,
Q p
For interchanging^; and h in the last result, viz., _- = — ,
a
we get - = -, and therefore also, by Art. 60 (5.), we have
a p
(The student is cautioned against inverting the separate terms of
the two sides of an equation when there are more than one term on
each side. )
64. When each side of an equation consists solely of a
single fraction, we may perform the following operations : —
ABBREVIATED METHODS FOR PARTICULAR CASES. 215
1. We may add or subtract tJie numerator and denominator
of EACH fraction for a new numerator or denominator^ and
retain either the original numerator or denominator for the
other term of the fraction^ both sides being always similarly
treated.
Thus, if T = -, we have —
(1.) -^— - -y-, (11-) -y- - -y-'
(iii.) ^Jl_^ . 1L±J[^ (i,.) «_iLi . 2LZA
or, (v.) we may have equations formed by inverting each of
these.
These results are easily obtained —
For, since t = -, we have, adding unity to each side —
a ^ p ^ a -h b p + a
- + 1 = ^- + 1 or — j-~ = -.
q q
And so, by subtracting unity from each side, we get —
a - b p - q ,
— f — r= , and so on.
b q
2. We may take the sums of the numerator and denominator
of each for new oiumerators or denominators, and the dif-
ferences /or the other terms of the fraction ; and vice versa,
both sides being always similarly treated,
.a c a + b c + d
Thus, if 7 = -7, we have also 7 = ,,
' b d^ - a -^ b c - d
, a - b c - d
and — -— T = — ; — >)
a + c + a
for we have lUst shown that — -. — = >
^ a - b V - ^
and — f — = -"
b q
216 ALGEBRA.
Hence, dividing equals by equals, we get—
a •{■ h ^ a - h ^ p + q ^ p - q
h ~ h ~ q r~ q ^
or T = , which is the first result.
a - p - q'
And inverting each side, we have, by Art. 63 (Cor.) —
a - b P - q
a + h ~ p + q
^ ^ ^ , - ,. mx ■\- a + h iiuc + a + c
Ex. 1. Solve the equation -j - r -j-
^ 7hx - c - a nx — - a
, mx •\- a -v h nx - c - d
By Art. 63, we have = j ;.
•^ ^ mx + a + c Qix - - a
Then, by Art. 64 (1.), retaining the numerators and taking
the differences for new denominators, we have —
771X + a -^ b oix — c — d
b - c ~ b - c ■'
or, multiplying each side by (b - c) —
mx + a -{• b =^ 7ix - c " d; or, transposing —
/ X /7 7\ a + b + c + d
{m " n)x = - {a + b + c + d); .\ x = _
Ex. 2. Solve -^^
ija + cc - is/a - X
We may consider the quantity a as a fraction whose deno-
minator is unity, or as - .
Then, Art. 64 (2.), taking the swmand difference, v^^Q have—
^iJa + x_a+\ ^
2Vr~^ ~ oT^^v'"''
tja + 03 ' a + \
— . — = 1 J or, squarmg —
sja - X a - I
(i + a;_a^ + 2a+l
AfiMEVlATEt) METHODS FOR PARTICULAU CASES. 217
Again, taking the difference and suriiy we have —
2a; 4a
2a 2a^ + 2'^^
X _ 2a ,
'a a^ + 1
2 a'
a- + 1
65. We now give an example to show that sometimes the
easy solution depends on an advantageous arrangement of
the terms on the two sides of the equation.
Ex. Solve V5~r4 + sjT^^ = 7.
Transposing, we have —
Va; + 4 = 7 - Va; - 3; squaring, then —
a; + 4 = 49 - 14 Va; - 3 + (a; - 3);
subtracting x from each side and transposing, then —
14 Va; - 3 = 49 - 3 - 4 = 42.
.'. six - 3 = 3 ; or squaring,
a; - 3 = 9; .-. a; = 3 + 9 = 12.
Should the student commence by squaring at once, he will
render the equation more complicated.
Ex. XYI.
3a; + 7 3a; - 13
2.
3.
4.
6.
6.
a; + 4 a; - 4 '
(a; - a) {x - h) = {x - c) (x •
-4
3a; +13 3a; + 10 x
15 5 a; - 50 5'
1 - 25a; 3 - 2\x 28 - 5a; 10a; - 11 x
15 14 (a;- 1) 3 30 3
aj - 4 3a; - 13 _ 1
6a; + 5 18a; - 6 3'
3 5 - 2a; _ T
4a;= - 2
1 - 2a; 7 - 2a; 7 - 16a; + iy^
2i8 ALGEBRA.
.^- 13-2. -^ ^^ ^ —8— = ^^i 8 •
g 6a; + 5 58i- + 14a ; _ ^
' "3a; + 1 9 + 2a; ~ *
4a; - 9 6a; - 21 aj - 2^
10 ^^ - '^ 2 - 14a ; 3^^ + a; ^ 10 - 3f a; _ 19
* 2 a; - 9 "^ 7 " 14 2 21'
11 1^ - 7a; - 6| _ 9a;^ - 12a; - 19 ^ 17
2a;-3 3a;-5 6aj-7'
■J ^ aa; + m + 1 ax + oi ax + m
ax + m - I ax + n - 2 ax + m - 2
ax + n + 1
ax + n - V
sjx -- a + b — ijx + a - b a — b.
13.
Jx - a + b + Jx + a - b a + b
1 _ Vl - Vl - aJ ,
1 + VI - n/1 - a;
15. J2x ■{• 10 + /N/2aj - 2 =• 6.
„ 3
16. V8 - a; - "Ti-Zl, - x/l - a..
17. ^l + Va; + ^1 - ^a; = 2.
,^ aa; + 1 + \/aV - 1 ,
aa; + 1 — v« ^ - 1
ci^ f« & a -
19.
^a; - V5 V^J - V« Vic
20. -7 7- = -7 7- + Va.
^Jx + s/a Va; - \/<^
/VaJ + » - / Jx - a _ , ^
21. / —7 / ^-7 V. - a-.
V Vc:j— o \J s/x + a
PROBLEMS INVOLVING ONE UNKNOWN QUANTITY. 21D
1 2 + far' - la' 5 "
^- ^ -. 9. - 6 - 5a; + a^ " ^^ " a; - 3*
^^ ■ ^ a; + 5 ^ 2aj + 5 a;^ - 10
-^^* " ■ ^ **" ^T~4 = a; + 2 " a; + 3 + ^•
24
a; -
-
2 ■
4a;
+
5
X
+
1
ax
+
h
ex -^ e
■V d ax + f
Problems producing Simple Equations involving One
Unknown Quantity.
66. To solve an algebraical problem we represent the re-
quired or unknown quantity by a letter, as x, and then ex-
j^ress the given conditions in algebraical language. Thus wo
form an equation, the solution of which gives the required
value of the unknown quantity.
Ex. 1. My purse and money are together worth 24 shillings,
and the money is worth seven times the purse. Find the
value of each.
Let X = the value in shillings of the purse,
Then 1 x = „ „ money.
Now, by problem the value of both together is 24 shillings.
Hence we have —
aj + 7 a; = 24
or 8a; = 24
.'. a; = 3, the value in shillings of the purse,
and .*. also 7 a; = 7 x 3 = 21, „ money.
Ex. 2. What number is that to which, if 36 be added, the
sum shall be equal to 3 times the number It
Let X — the number ;
.*. a; + 36 = the sum when 36 is added,
and 3 a; = 3 times the number.
H6nce, by problem —
a; + 36 = 3 a;,
or a; - 3 a; = - 36,
or - 2 a; = - 36;
/.a; = — — - = 18, the number required.
220 ALGEBRA.
Ex. 3. The distance between two towns is such that a train,
whose speed is 30 miles an hour, takes 1 hour more in going
10 miles over 5 times the distance than a train whose speed
is 20 miles an hour takes in going within 4 miles of 3 times
the distance. Find the distance between the towns.
Let X = the distance required in miles.
Then 5 a; + 10 = 5 times the distance together with 10 miles,
5 a; + 10
and oa - *™^ ^ hours to travel this distance at 30
miles an hour.
And so, — — - — = time in hours to travel 4 miles less
Ji\)
than 3 times the required distance, at 20 miles an hour.
But by the problem the former of these times exceed the
latter by 1 hour.
Hence ^-^±-1^ - i^— _ A = 1.
30 20
From this we easily find a; = 28.
Hence 28 miles is the distance required.
Ex. 4. Find the price of an article, when as many can be
bought for Is. 4d. as can be bought for 2s. after the price
has been raised Id.
Let X = the price required in pence;
1 /»
Then — = number of articles bought for Is. 4d.
X
And a; + 1 = the raised price in pence.
24
= number of articles bought for 2s. at the raised
X + 1
price.
But, by the problem, the number of articles in each case is
the same.
Hence — = ~ — -, from which a; == 2.
X X + I
Hence 2d. is the price required.
Ex. 5. A man sells geese at as many shillings each as the
number he has, and having returned 5s., finds that if he had
PROBLEMS INVOLVINa ONE UNKNOWN QUANTITY. 221
had 2 more to sell on the same condition, and had returned
3s., he would have had 38s. more. How many had he]
Let X = the number required ;
Then ar^ - 5 = number of shillings received.
Also, on the second supposition,
(a; + 2)' - 3 = number of shillings he would have received.
Now, by the problem, this latter number is 38 more than
the former.
Hence (a; + 2)^ - 3 = a;^ - 5 + 38, from which we find
X = 8.
Ex. 6. A waterman finds that he can row 5 miles in f
hour with the tide, and that it takes him 1^ hours to row
the same distance against the tide when it is but half as
strong. What is the velocity of the tide ]
Let X = the velocity of the tide in miles per hour.
Now the velocity of the boat when going with the tide
,= 5-1 = V.
.'. Y - a; = velocity of the boat when there is no tide.
Again, velocity of boat against the tide when it is half as
strong = 5 -f H = V-
10 X
.'. -^ + - :^ velocity of the boat, when there is no tide.
Hence we have —
10 . a; 20 - , . ,
-_ + -_ = — a; ; from which
O It o
X = 2f.
.*. The velocity required is 2 J miles per hour.
Ex. XVII.
1. If I add 25 to 3 times a certain number, I obtain the
same result as if I subtract 25 from 8 times the number.
Find the number.
2. Divide 70 into 2 such parts that the one shall be as
much above half the number as the other is above 15.
3. Divide £720 among A, B, and C, so that B may have
twice (ig much as C, and A as much as B and C together.
222 ALGEBRA,
4. There are two trains, one of whicli goes 5 miles an
hour faster than the other, and the former performs a journey
of 100 miles, while the latter goes 75 miles. Find their
respective I'ates,
5. A horse when let out for hire brings in a clear gain of
10s. per day, but costs Is. 6d. daily for food. At the end of
30 days his master had gained £11. lis. Required the
number of days for which he was hired.
6. A and B have 4 guineas between them, and play at
hazard. B loses | of his money, and afterwards gains ^V of
what he then had. It is then seen that B has as much
money as A had at the end of the first game. How much
had each at first]
7. A and B have respectively an equal number of florins
and crowns. B pays a debt of 4s. to A, and then A's money
is just half B's. Find what each had.
8. A workman, instead of adopting the 9 hours' system,
worked 10 hours daily, and had a corresponding rise of
wages. By this means his wages were increased 4s. weekly.
Find his original wages.
9. A person who has regular wages of 26s. weekly, think-
ing to better himself, takes a job at higher wages. He is,
however, put on half-time during 20 weeks of the year, and
finds himself at the end of the year £4, 12s. worse off. Ee-
quired his increased wages,
10. A company of men, arranged in a hollow square 4 deep,
numbered 144. What was the number in a side of the
square 1
11. In an examination paper there were two series of
questions, and the questions of the second series carried each
3 marks more than those of the first series. A candidate who
attempted 3 of the first series, obtaining half marks for them,
and 5 of the second series, obtaining for these full marks, got
altogether 80 marks. Find the number of marks attached to
each question of the first series.
12. A grocer has tea at 3s. 4d. and at 4s. He sells
altogether 64 pounds, thereby realizing £12. How much did
he sell of each 1
PROBLEMS INVOLVING ONE UNKNOWN QUANTITY. 223
13. If 11 bo subtracted from 5 times a certain number,
and tlie remainder divided by 6, the quotient will exceed by 2
the quotient obtained by subtracting 3 from 4 times the num-
ber and dividing the remainder by 7. Find the number.
14. A garrison of 1,250 men were provisioned for 64 days;
but after 22 days a certain number were called away, and it
was found that the remaining provisions lasted the number
left for 70 days. Find the number told off.
15. At a railway station £15 was taken for single fares,
and £33. 15s. for returns. The number of return tickets
exceeded the single tickets by 10, and the price of a return
ticket was half as much again as a single ticket. Find the
fare for a single journey,
16. In a tour lately made round the world, the distance
travelled by water was 20,000 miles, and by land 8,000 miles;
and the whole time taken was 220 days. Supposing the rate
by water to be two-thirds of that by land, find the number
of days travelled by land.
17. The distance between A and B is 32 miles. A person
starting from A, at the rate of 4 miles an hour, meets another
who started from B half an hour later, at a rate of 3i- miles
an hour. At what point will they meef?
1 8. There are two clocks, one of which gains twice as much
per day as ihe second loses, and they are set right at noon
on Monday. When it is noon on Thursday by the first clock,
it is 11-50 A.M. by the second. What is the gain per day of
the first clock]
19. A draper raises his goods a certain rate per cent., and
afterwards reduces them to the original price by lowering them
13^^ J per cent. Find the original rise per cent.
20. Bequired the distance between two towns such that a
person can perfomi the journey one hour sooner when he
walks 4 miles an hour than when he walks 3 J miles an hour]
21. The sum of £12. 15s. is paid away with an equal
number of sovereigns, crowns, and sixpences, Bequired the
number of each.
22. A walks along an inclined plane at a certain rate, and
B walks along the base of the plane at a rate of one-third of
224 ALGEBRA.
a mile per liour less than A. The inclination of the j)lane is
such that A is always vertically over B, and that at the end
of half an hour they are exactly five-sixths of a mile apart.
Find the respective rates of A and B.
23. There is a direct road over a hill between two stations
at the foot of each side. The distance on the one side from
the foot to the top is 5 miles, and the road down the other
side forms a right angle with the road up. It is also known
to be 1 mile less down the hill than the direct distance by
tunnel between the two stations. Find the distance down
the hill.
24. Two trains, whose respective lengths are 122 and 98
yards, and the former of which is going at the rate of 35 miles
an hour, pass each other in 30 seconds. Find the rate and
relative direction of the second train.
25. A man bought a number of sheep for £1 32, and having
lost 10, and sold 20 that were diseased at 6s. per head below
cost price, disposed of the remainder for £116, thereby
realizing his outlay. How many did he buy?
26. A boy spends 10s. in oranges and apples. The oranges
were bought at 5 for 6d., and the apples at 3 for 2d.; and
their number together amounted to 132. What did he spend
on each?
27. If B is allowed 2 hours more time than A takes to do
a piece of work, he will do 4 times as much, and if C is allowed
1 hour more than A, he also can do 4 times as much. More-
over, D requires 4 hours more than A to do the piece of work.
Also, the work done by A and B together is the same as that
done by C and D together in the same time. Bequired
the respective times for A, B, C, D to do a piece of work.
28. A person sells out £1,200 Three and a Half per Cent.
stock, and invests the money in Two and a Half per Cents.,
whose price is 1 4 lower than the first-named stock. The loss
in annual income is £7. Find the price of the first-named
stock.
29. The banker's discount on a certain sum of money at
5 per cent, per annum is equal to the true discount on a sum
JuSO larger, Find the sum,
SIMULTANEOUS EQUATIONS. 225
30. An express train, which ought to porform its journey
in 2|- hours, after having gone uniformly 80 miles, finds itself
G minutes behind. However, by increasing the speed to as
many miles per hour as there were miles in half the journey,
it just arrived at its destination in time. Find the original
speed of the train, and the length of the journey.
31. A vessel contains a quantity of spirit (sp. gr. -9) and
water, and a cylinder of wood (sp. gr. -92), whose length
is 10 inches, floats upright, so as. to be just covered by the
spirit. Find how much of the cylinder floats in the water.
32. A mixture of hydrogen and oxygen is found to condense
when fired, to 1 6 vols, of steam. Now, every 3 vols, of such
a mixture is known to condense to 2 vols., when the original
gases are in the proportion of 2 : 1. Find the quantity of
each.
33. A mixture of 100 grams of sodic and potassic sulphates
yielded a gram of baric sulphate. Now, each gram of sodic
sulphate yields h gi^ams of baric sulphate, and each gram of
potassic sulphate yields c grams of baric sulphate. Find the
amount of sodic and potassic sulphates in the mixture.
34. If a oxen consume h acres of grass in c weeks, and a '
oxen consume h ' acres of grass in c ■ weeks, the grass growing
uniformly, find the week's growth of an acre.
35. The freezing and boiling points of a common ther-
mometer are marked 32° and 212° respectively; those on tlie
centigrade thermometer are marked 0° and lOO''. At what
temperature do the graduations agree ]
36. A person going at the rate of a miles per hour finds
himself h hours too late when he has c miles farther to go.
How much must he increase his speed to reach home in time ]
Simultaneous Equations of the First Degree with two
Unknown Quantities.
67. Suppose we have given the equation 3 a; - 4?/ = 5,
then by ascribing to ?/ a series of values we get a correspond-
ing series of values for x.
Thus we may have ^ ~ i > , ^ ~ < 'r ? ,, '" rc y ^*
g ^ y =^^r y = ^r y =^ n
22G ALGEBRA.
Again, if niiotlier equation, as 4 x -v y ==-- 32, be given, \v'e
may in the same way obtain a series of pairs of values whioii
satisfy it. And, further, if the two equations are distinct and
compatible, there is always a j^:)<:aV of values common to the
two equations. This pair of values then satisfies both equa-
tions, and the equations are called simultaneous equations.
The methods of solving simultaneous equations will now
be explained.
First Method. — Equalize the coefficients of one of the un-
known quantities in both equations, and add or subtract the
equations so obtained, so as to obtain an equation with one
unknown quantity,
Ex. A:x + 37/= 17 (1))
9ic -- 5?/ - 3 (2)/-
From (1), multiplying each side by 9, we get —
36ic + 21y = 153 (3).
And from (2), multiplying each side by 4, we get —
36a; - 20 y = 12 (4).
(3) - (4), then 47 2/ = 141
Hence, substituting in (1), we have —
4a; + 3x3 = 17, from which we get — ■
X =. 2,
Hence, the solution required is a: = 2, ?/ = 3.
Second Method. — Express one of the imknown quantities
in terms of the other by means of either equation, and sub-
stitute its value in the other.
Taking the same example, we have —
4a; + 37/ = 17 (1)1
9a; - 52/ - 3 (2) / *
From (1) we have 4a;=: 17 - 2>y, ot x = — ~--^ — ^^...(3).
Substituting this value of x in (2) we have —
or, 153 - 27 y - 20 y = 12, from wliicli—
2/= 3.
SIMULTANEOUS EQUATIONS, 227
Tlien from (3) we get x -^ ILZ-L^^ =. 2.
Third Method. — Express mie of the unknown quantities
in terms of the other by means of each equation, and equate
the expression.
We will this time express in each case y in terms of x,
Wehave4a; + 3y = 17 (1) )
9a; - 5y - 3 (2) J '
From (1) 3y := 17 - 4 07, or y - IL:iA^...(3).
o
And from (2) 5 y = 9aj ^ 3, ory = ^^ " ^ (4).
Equating (3) and (4), then ^1-—^ = ^ ZL?, from which
o 5
X = 2,
Then from (3) by substitution, y = ^LzAjiJ = 3,
o
Simultaneous Equations of the First Degree of Three
or more Unknown Quantities.
68. In the case of three unknown quantities we may
obtain, from the three given equations, two equations
with two unknown quantities, and then, by a similar method,
from the two obtain an equation with one unknown
quantity ; and a like method may be pursued for more than
three unknown quantities.
Ex. 3a; + y + 4cj = 25 (1)
ix + Zy - ^z ^ - 3 (2)
Qx -h 7y - Sz = 1 (3)
From (1) 12a; + 4 y + 16;^ = 100 (4))
Andfrom(2) 12a; + 9y - 15;:; = - 9 (5) j
(5)-(4),then 5y - ?^\z =-109 (6)
Again, from (1) 6a; + 2^/ + 8^ = 50 (7)
(3) - (7), then 5y - \^z ^ - 49 (8)
(8) - (6), then 15;:; = GO
.'. z ^ 4
)
228 ALGEBRA.
Hence from (8), by subcLltution —
5 2/ - 16 X 4 =-49,
from which y - 3.
And hence from (1), by substitution of the known values
of y and ;: —
3 a; 4- 3 + 4 X 4 - 25,
from which ic = 2.
Hence the respective values of x^ y, ^, are 2, 3, 4.
Ex. XYIII.
1. 6 X + 7/ = 22, 5 a; + 3 2/ - 27.
2. 4 a; - 3 7/ = 14, G a; + 5 2/ = 40.
3. 3 a; + 5 2/ = 44, 2/ - a; =^ 4.
A _ . *^- _ 9 — . r ^ 9 9_
^- 4 + 3 - -uj 5 + 10 - "TO-
^•6 + i = ^'4 + 6 = '-
7. 3 aj + 4 2/ + ;^ = 11, 2 a; + 2/ + 5 ;:; -= 19,
5aj+22/ + 3;2 = 18.
8. 7a; + 22/ + 3;:; = 20, 3a;-4y + 2;:; = l,
52/-2a;+7;^ = 29.
9. 2 a; - 5 2/ = 3, 3 y - 2 ;:; = - 1, 4 a; + 2 ;:;=:= 20.
10. aa; + ^2/ = ^> ^v^ + ^i2/ ~ ^i*
11. X •{• y = a, y + X =•■ h, X ■\- z = c.
12. ax + by = d, by + cz = e, ax* + cz = J*.
3x-2y 5x+2y , aj + 2?/ 3a;-22/_ ,
13. —J— + — g = 5^, — ^ — ^^ 1.
a ? + 2/ , ^-2/ _ ^ ^tJl 4. ?_r J( == -I
^^' 10 15 30' 15 10 30*
a; 2/ ^ V
SIMULTANEOUS EQUATIONS. 229
16. ^ + - = 5i — - = 0.
17. 4 (a; + 3 2/) - 3 (x + y) = i (a; + 2/) (^ + 3 y),
^Q + _1_ ^ 3. •
X + y X + 3y
18. 73y - 5aj = (a; - 5 2/) (a: + 3 y),
2 5 7
X - by X -\- 3y 33
.n 1 1 11^11
19. - + - = a, - + - = -5, - + — = c.
X y y z X z
20. 3a;-22/ = 0, 4y- 3;:; = 0, 55-6^^=14,
7 w - 2 ic = 3.
21. 3a;~27/=.5;:;-6y = 7a:-4;:; = l.
22. x + y + z=^ayy-{'Z-\-u=^hiZ-\-u-\-x = Cy
u + X + y ~ d,
^^ oc y y z J X z
23. - + - = a, - + - = ^, — + - = c.
7/1 7i n p 1)1 p
24r. X + ay + hz = a^ y + az + hx = ^y z + ax + by = y,
25. ic + ay + a-;:; + a** = 0.
i« + % + 6";s + P = 0,
X + cy + crz + c' = 0.
2G. ic + ay + a-;s + a'w + a* = 0.
X -r by + b^z + b^u + 5* = 0.
X + cy + c^z + c^u + c* = 0.
X + dy -h d^z + d^it + d* = 0.
27. a; + a?/ = «. 28. a b c ,
y + bz = p, X y z
- + c« = y. a' 5' ^' _ V
< + c/cc = ^. a""** y *** i ~"
?t + ex = e. a" 5" c" _ ,,/
X y z
29. a^i + 2^2 + 30^3 + &c. + nx^ = ai-
ajj + 2a;3 + 3^4 + &c. + tz^i = aj.
«« + 2ari + 3 a; + <fcc. + wa^n-i = ^n-
230 ALGEBRA*
X "fl t X 1/ Z X y Z \ \ 1
30. - + -/ + - =- ^ + - + - = - + -+-,-=-+ y- + -.
a b c c a G a a b c
Problems producing Simultaneous Equations of the
First Degree.
69. Tliere are certain problems which may be solved with
much greater facility by the introduction of more than one
unknown quantity.
Ex. 1. Nine men and seven women receive together
£3. 2s. 8d., and three men receive lOd. more than four women.
Find the receipts of each.
Let X and y respectively represent in pence the receij^ts of
a man and woman.
Then, since £3. 2s. 8d. :^^ 75 2d., expressing the Conditions
of the problem in •vlorebrpipal iMno-nasfp
'.» .'• V // - 7:»"j I S..|v ii,r^ these e«jiiHtji.iis.
;; .-• - ! // ii» . A . hii.U •
litjiiof. lii'" r(,'Ccl[>i.> "1 :• III- II Ml"! Nvuiiiciii j,..re re^pev:tivelv
54d. and o8d., or 4s. 6d. and ob. I'd.
Ex. 2. Eind a fraction such that^ if we diminish its
numerator by 1, it becomes equal to 4 ; and if we increase its
denominator by 1 , it becomes equal to J <
Let -be the required fraction.
y
Then, by problem, '^IT — ^ „ ) , • 1 ^^ , .
•^ -^ y V ( , wnicli equations, when
J X ^ 1 ( solved, give-
ic - G, 2/ - 35.
/. _1 is the fraction required.
35
Ex. 3. A's money, together with twice B's and thrice C's,
amounts to £38 ; B's money, together with twice C's and
thrice A's, to £35 ; and C's money, together with twice A's
and thrice B's, also to £35. Eind the money of each.
rrwOBLEMS TRODUOING SIMULTANEOUS EQUATIONS. 231
Let X, y, z be respectively the number of pounds each
ias. Then we have —
X + 2y + 2>z ^ Z^\
2/ + 2 ;2; + 3 £C - 35 I- ; from which we get —
;:; + 2 a; + 3 7/ =- 35 )
a; ^ 5, 2/ = 6, ;:; = 7.
Hence, £5, £6, <£7 are the respective moneys of A, B,
and C.
Ex. XIX.
1. A and B engage in play. A puts down lialf-a-crown
to B's florin. They play twenty games, and then it is found
that A has won 2 s. How many games did each win ?
2. There is a number, the sum of whose two digits is 1 0,
and, if 36 be added to the number, the digits change places.
'V\^^c^ tlip mijnbf'r.
'. ' ih.. ;.ii.l --H-:u- :,f I >. ;. sl..|i,\
It" !..■ l.:i.! j:n>.-l ;!.. (■••;, l<»
r \'l}, iMT r"M\.. -M..; ^..|.' lir.
•ills vv ...il.l \\'.i\ '■ iH-.;i, 1... .;.!.
..r-:.. I,.
4. iiiiglib Liiiie;:> ihc nuniL-raLur ui a< oc-rialii IracLiuii cAlccctis
three times the denominator l)y 3, and five times the numer-
ator added to twice the d<'ii(miln;ii<>i" ]ii;i,l<(^ 29. Find the
fraction.
5. If the number of cows in a field were doubled, there
would bo G5 cows and horses tog(ither ; but, if the number of
horses be doubled, and that of the cows halved, there would
be 4G. How many ai^e there of each ?
6. Thirty shillings are spent in brandy, and 42s. in gin ;
19 bottles being purchased in all. Had the 42s. been spent
in brandy, and the 30s. in gin, 17 bottles only would have
been bought. Find the cost per bottle of each.
7. A fishmonger receives 240 mackerel. He sells a
certiiin number at 4 for a shilling, but the rest being seized
as bad fish, and he being fined 10s., finds himself a loser by
9s. Had he sold them at 3 for a shilling, he would have
been a gainer by 5s., if 13 more fish had been seized. How
many did he sell, and what did he pay for the lot]
:;.
.\ 'jv !•
l.:.s t.
t 3.-.
lb s.
•lis i-J vv
.mmI, n
<• ll
\,>\ 1VV
|><M- «v
"III,., riinl
l,,vv,.,.-.
.,1 1
1m- S.
>;* Mi<'
• liiMit iri<
•s . ,.t' (■'.
..•!..
. Lis
iiior-'.
KilMJ t
1,..,U:
ml
ii \ >•
232 ALGEBRA.
8. Three persons invest tlieir money at 3, 4, 5 per cent,
interest respectively. The total amount of interest is £38,
and the interest of the first and third together is 2^ that of
the second ; while the total interest would have been £34 had
the rates been 5, 4, 3 per cent, respectively. Find the
capital of each.
9. A toll-gate keeper receives 8s. 8d. for the toll of a
number of horses, oxen, and sheep, the tolls for each being
respectively 1^-d. Id., |-d. Had there been twice as many
sheep and the number of horses diminished accordingly, he
would have received 7s. 2d. Had the oxen passed through
free, and the tolls for a horse and sheep respectively been 2d.
and |-d., he would have received 9s. IJd. Find the number
of each.
10. A, B, C start from the same place. B after a quarter
of an hour doubles his rate, while C, who falls, after ten
minutes diminishes his rate Jth. At the end of half an hour
A is ^ mile before B, and i mile before C, and it is observed
that the total distance which would have been walked by the
three, had they each continued to walk imiformly from the
first, is 6i miles. Find the original rate of each.
11. A, B, C, working 3, 4, 5 hours respectively, can do
2^^^:^ pieces of work; if they each work an hour more, they
can finish an extra if of a piece ; and, if C does not work,
the other two, working for 1 and 6 hours respectively, can
together finish 1 piece. Find the time required for A, B,
C to finish separately a piece of work.
12. There are three numbers such that, if the first be in-
creased by 6, and the second diminished by 5, the product of
the results is the product of the first two numbers ; if the
second be increased by 2 and the third diminished by 3, the
product of the results is the product of the second and third ;
and, if the first be increased by 3 and the third diminished
by 6, the product is that of the first and third. Find the
numbers.
13. A person performed a journey of 22^ miles, partly by
carriage at 10 miles an hour, partly by train at 36 miles an
hour, and the remainder by walking at 4 miles an hour. He
did the whole in 1 hour 50 minutes. Had he walked the
PROBLEMS PRODUCING SIMULTANEOUS EQUATIONS. 233
first portion, and performed the last by carriage, it would
have taken him 2 hours 30|- minutes. Find the respective
distances by carriage, train, and walking.
14. A and B start from two places C and D, distant 28
miles, and it is found that A reaches T> 3 hours after they
meet. Had the distance between C and D been 35 miles, A
would have reached a point 28 miles from C 2 hours after
he met B. Find the respective rates of A and B.
15. Three trains — a luggage, ordinary, and express — move
along three parallel pairs of rails, the distance between the
stations being 120 miles. The firet two start from the same
station, and the express from the opposite. The luggage
train, starting 2 hours first, is overtaken by the ordinary in
2 hours; and the express train, starting 1 hour after the
ordinary, meets the luggage in 1 hour 7^- minutes. Had all
three started from the same station, the ordinary would have
been overtaken in 2 hours. Find the respective rates of the
trains.
16. If (rtj, hiy Cj), (aoy h.2, Co), (^3, 63, C3) be the respective
compositions by weight of three mixtures of three substances,
and di, ch, d^ be the respective prices of the mixtures, find the
price per unit of weight of each substance.
17. By alloying two ingots of gold in two given propor-
tions, we form two new ingots of which the fineness of each
is known. What is the fineness of each of the given
ingots ]
18. A gi^oup of n persons play as follows: — The first
1 '
plays with the second and loses - of what he has, the second
tlicn plays with the thii-d and loses - of what he has, the
third then with the fourth, losin;^ - of what he has, &c., the
71 til with the first losin^? - of what he has. At the end
they each have h. What had each at fii*st %
234
MATHEMATICS.
SECOND STAGE.
SECTION I.
, n "E n ":\r t: T "R y.
ECOLiD"^ lIl^Ji^ME^vT;!^. iiuuK ii.
Definitions.
i. A rectangle, or right-angled parallelogram, is said to
be contained by any two of the straight lines which contain
one of the right angles.
2. In any parallelogram, the figure which is composed of
either of the parallelograms about a diameter, together with
the two complements, is called a (jiioiiLO)i.
Thus the i)arallelogram HG, to-
gether with the complements AF,
EC, is a gnomon, which is briefly ex-
pressed by the letters AGK, or EHC,
which are at the oj)posite angles of
the parallelograms which make the
gnomon.
The rectangle under, or contained hy two lines, as AB and BC, is
concisely cx^jressed thus : — AB, BC.
PROPORTIONS. 235
Proposition 1.— Theorem.
If there he two straight lines, one of loldch is divided into
any number of 2^<^rtSj the rectangle contained hy the two
straight lines is equal to the rectangles contained hy the un-
divided livje, and the several parts of the divided line.
Let A and BC be two straight lines; and let BC be
divided into any parts in the points D, E ;
The rectangle contained by the straight lines A and BC A • BC =
shall be equal to the rectangle contained by A and BD, A • bd -
together with that contained by A and DE, and that con- ^ • de -
tained by A and EC.
Construction. — From the point B draw BF at right angles
toBC(I. 11), „ „
And make BG equal to A (I. 3). i j r I
Through G dr«,w aH pnrall^l +o I III
BCd. :^n. \
And tlii-«n!L;lt tli- |M»ii»r..> I »,E.< '.«lt':. ^^
Pkook. 'I'lirii tliH jvr«:iiigi'" !'• M i> ^,,
equal U» tJM' i-tx-taiioU-s BK. \)\.. KH. ' ^
But BH is contained by A and i>C, I'or it is contained ^^^•
by GB and BC, and GB is equal to A (Const.) ;
And BK is contained by A and BD, for it is contained
by GB and BD, and GB is equal to A ;
And DL is contained by A and DE, because DK is equal
to BG, which is equal to A (I. 34) *
And in like manner EH is contained by A and EC ;
Therefore the rectangle contained by A and BC is equal
to the several rectaiigles contained by A and BD, by A and
DE, and by A and EC.
Therefore, if there be two straight lines, kd Q, E. D»
Proposition 2.— Theorem.
If a Btraighi line he divided into any two 2)arts, the rect-
aiigles contained hy the lohole line and each of its jxirts are
together equal to the square on the tohole line.
Let the straight Ime AB be divided into an\- tw o parts
in the point C ;
236
GEOMETRY.
AB-BC
-^-AB • AC
= AB:\
The rectangle contained by AB and BC, together with tl.3
rectangle contained by AB and AC, shall
be equal to the square on AB.
Construction. — Upon AB describe the
square ADEB (I. 46).
Through C draw CF parallel to AD or
BE (I. 31).
Proof. — Then AE is equal to the rect-
angles AF and CE.
But AE is the square on AB ;
Therefore the square on AB is equal to the rectangles AF
and CE.
And AF is the rectangle contained by BA and AC, for it
is contained by DA and AC, of which DA is equal to BA :
•And CE is contained by AB and BC, for BE is equal
to AB.
Therefore the rectangle AB, AC, together with the rect-
angle AB, BC, is equal to the square on AB.
Therefore, if a straight line, &c. Q. E. D,
Proposition 3. — Theorem.
If a straight line he divided into any two j^arts, the rect-
angle contained hy the whole and one of the parts is equal to
the square on that part, together ivith the rectangle contained
by the two 2^arts.
Let the straight line AB be divided into any two parts in
the point C;
The rectangle AB * BC shall be equal to the square on BC,
together with the rectangle AC • CB.
Construction. — Upon BC describe the square CDEB
(I. 46).
Produce ED to F ; and through A
draw AF parallel to CD or BE (I. 31).
Proof. — Then the rectangle AE is
equal to the rectangles AD and CE.
But AE is tlie rectangle contained by
AB and BC, for it is contained by AB
and BE, of which BE is equal to BC ;
And AD is contained by AC and CB, for CD is equal to CB;
PROPOSITIONS. 237
And CE is tlie square on BC.
Therefore the rectangle AB, BC is equal to the square on
BC, together with the rectangle AC, CB.
Therefore, if a straight line, &c. Q. E. D.
Proposition 4.— Theorem.
If a straight line he divided into any two 2^ci'i'ts, the square
on the tvhole line is equal to the sum of the squares on the tivo
2)artSy together with twice the rectangle contained by the j^cirts.
Let the straight line AB be divided into any two parts
inC;
The square on AB shall be equal to the squares on AC and ab2 =
CB, together with twice the rectangle contained by AC ^^A't.^Sp
and CB. •'-^''*''^-
Construction. — Upon AB describe the square ADEB
(I. 4G), andjoinBD. ^ ^ ^
Through C draw CGF parallel to AD or
BE (I. 31).
Through G draw HCK parallel to AB ^-
or DE (I. 31).
Proof. — Because CF is parallel to AD,
and BD falls upon them,
Therefore the exterior angle BGC is "" * jj-
equal to the interior and opposite angle ADB (I. 29).
Because AB is equal to AD, being sides of a square, the
angle ADB is equal to the angle ABD (I. 5);
Therefore the angle CGB is equal to the angle CBG show first
. . , V ^ ° ^ ^ tliat OK is
^Ax. Ijj a sqimVo
Therefore the side BC is equal to the side CG (I. G). = ^^-^
But CB is also equal to GK, and CG to BK (I. 34);
Therefore the figure CGKB is equilateral.
It is likewise rectangidar.
For since CG is parallel to BK, and CB meets them, the
angles KBC and GCB are together equal to two right
angles (I. 29).
But KBC is a right angle (Const.), therefore GCB is a
right angle (Ax. 3).
Therefore also the angles CGK, GKB, opposite to these,
are right angles (I. 34).
Q
/
/
238 GEOMETRY.
Therefore CGKB is rectangular ; and it has been proved
' eqnUateral; therefore it is a square; and it is upon the
side CB.
Co also ^or the same reason HF is also a square, and it is on
£iF=AC2 the side HG, which is equal to AC (T. 34).
Therefore HF and CK are the squares on AC and CB.
And because the complement AG is equal to the com-
^„^ plement GE (I. 43),
AG+GE And that AG is the rectangle cont^ned by AC and CG,
=2AC-CB. ^j^^^ .^^ ^y ^Q ^^^^ Q;g^
Therefore GE is also equal to the rectangle AC, CB ;
Therefore AG, GE are together equal to twice the rect-
angle AC, CB;
And HF, CK are the squares on AC and CB.
Therefore the four figures HF, CK, AG, GE are equal
to the squares on AC and CB, together with twice the
rectangle AC, CB.
But HF, CK, AG, GE, make up the whole figure ADEB,
which is the square on AB ;
.\ whole Therefore the square on AB is equal to the squares on
^jgye^or ^Q ^^^ ^jg ^^^ twice the rectangle AC • CB.
+2 ac^cb: Therefore, if a straight line, &c. Q, E. D.
Corollary. — From this demonstration it follows that the
parallelograms about the diameter of a square are likewise
squares.
Proposition 5. — Theorem.
If a straight line he divided into two equal parts, and also
into two unequal parts, the rectangle contained hy the unequal
parts, together with the square on the line between the points of
section, is equal to the square on half the line.
Let the straight line AB be bisected in C, and divided
unequally in D ;
ad-db The rectangle AD, BD, together with the square on CP,
icB2. ^^^^^ ^^ equal to the square on CB.
Construction. — Upon CB describe the square CEFB
(I. 46), and join BE.
PROPOSITIONS. 239
Tlirougli D draw DHG parallel to CE or BF (I. 31).
Tlijoagh H draw' KLM parallel to CB or EY,
And through A draw AK par-
allel to CL or BM. A, C -D^B
Proof. — Then the complement
CH is equal to the complement HF ^
(I. 43).
To each of these add DM ; there-
fore the whole CM is equal to the k g f
whole DF (Ax. 2). ^^^
But CM is equal to AL (I. 3G), because AC is equal to CB al=cm
(Hyp.);
Therefore also AL is equal to DF (Ax. 1).
To each of these add CH; therefore the whole AH is .•.ah=
equal to DF and CH (Ax. 2). ^^+^^-
But AH is contained by AD and BD, since DH is equal
to DB (II. 4, cor.),
And DF, together with CH, is the gnomon CMC ;
Therefore the gnomon CMC is equal to the rectangle AD, DB. . • . cmg.
To each of these equals add LG, which is equal to the Add^o^ *
square on CD (II. 4, cor., and I • 34); ^^^^^°^
Therefore the gnomon CMG, together with LC, is equal
to the rectangle AD, DB, together with the square on CD.
But the gnomon CMC and LC make up the whole figure
CEFB, which is the square on CB;
Therefore the rectangle AD, DB, together with the square .-. CBs
on CD, is equal to the square on CB. +CD2.^^
Therefore, if a straight line, &c. Q. E. D,
Corollary. — From this proposition it is manifest that the
difierence of the squares on two unequal lines AC, CD is
equal to the j-ectangle contained by their sum and difference.
Proposition 6. — Theorem.
If a straight line he bisected, and pi^oduced to any pointy the
rectangle contained by the whole line thus produced and the
part of it produced, together with the square on half the line
bisected, is equal to the square on the straight line which is
made up of the half and the part produced.
Let the straight line AB be bisected in C, and produced to D ;
A
^' B n
L
H
/
K
/
240 GEOMETRY.
AD-DB The rectangle AD, DB, togetlicr with the square on CB,
- cj)l. shall be equal to the square on CD.
Construction. — Upon CD describe the square CEFD
(I. 46), and join DE.
Through B draw BHG parallel
to CE or DF (I. 31).
Through H draw KLM parallel k~
to AD or EF.
And through A draw AK par-
For allel to CL or DM.
AL^CH Proof. — Because AC is equal ^ u jj-
"" ' to CB (Hyp.), the rectangle AL is equal to CH (I. 36).
But CH is equal to HF (I. 43), therefore AL is equal to
HF (Ax. 14).
.-.AM or To each of these add CM; therefore the whole AM is
t^civia eq^^al to *lie gnomon CMG (Ax. 2).
But AM is the rectangle contained by AD and DB, since
DM is equal to DB (II. 4, cor.) ;
Therefore the gnomon CMG is equal to the rectangle
AD, DB (Ax. 1).
Add to Add to each of these LG, which is equal to the square on
ITcm^ CB (II. 4, cor., and I. 34) ;
Therefore the rectangle AD, DB, together with the square
on CB, is equal to the gnomon CMG and the figure LG.
But the gnomon CMG and LG make up the whole figure
CEFD, which is the square on CD ;
.♦.AD-DB Therefore the rectangle AD, DB, together with the square
1 C£>2^ on CB, is equal to the square on CD.
Therefore, if a straight line, &c. Q. E. B,
Proposition 7.— Theorem.
If a straight line he divided into any tivo 2yarts, the squares
on the whole line and on one of the parts are equal to tivice
the o^ectangle contained hy the whole and that party together
with the square on the other part.
Let the straight line AB be divided into any two parts in
the point C;
=^2^AB-BC -^^^ squares on AB and BC shall be equal to twice the
+ AC2. rectangle AB, BC, together with the square on AC,
PROPOSITIONS.
241
Construction^. — Upon AB dencribo tho square ADEB
(I. 4G), and join BD.
Through C draw CGF parallel to AD or
BE (I. 31).
Through G draw HGK parallel to AB
or DE (I. 31),
Proof. — Then AG is equal to GE
(I. 43).
To each of these add CK ; therefore the
whole AK is equal to the whole CE;
Therefore AK and CE are double of AK.
But AK and CE are the gnomon AKF, togetlicr with the
square CK ;
Therefore the gnomon AKF, together with the square CK,
is double of AK.
But twice the rectangle AB, BC is also double of AK, for
BK is equal to BC (II. 4, cor.);
Therefore the gnomon AKF, together with the square CK,
is equal to twice the rectangle AB, BC.
To each of these equals add HF, which is equal to the Add hf or
square on AC (II. 4, cor., and I. 34); ^q^Jar^'^'
Therefore the gnomon AKF, together with the squares
CK and HF, is equal to twice the rectangle AB,BC, together
with the square on AC.
But the gnomon AKF, together with the squares CK and .-. ab2
HF, make up the whole figure ADEB and CK, which are ioAB b^
the squares on AB and BC; +AC-2.
Therefore the squares on AB and BC are equal to twice
the rectangle AB, BC, together with the square on AC.
Therefore, if a straight line, &c. Q, E. D,
For
AK=CE.
.AKF
+ CK -
2AB-BC.
Proposition 8.— Theorem.
If a straight line he divided into any two jmrts, four times
the rectangle contained hy the whole line and one of the parts,
together with the square on the other 2^ctft, is equal to the
square on the straight line which is VlClde up af the whole line
fLQid the first mentioned part.
Let the straight line AB ^ divided intq J^njr t^yq part^ iv^
the point C;
GEOMETRY.
4 AB • BC
+AC8 =
(AB+BC)2.
For
BN=CK
=GR=RN
. • . the four
together
= 4CK,
And
tlie rect-
angles
AG, MP,
PL, RF,
are equal
to each
other, and
are toge-
ther =
4 AG.
Four times the rectangle AB, BC, together with the square
on AC, sliall be equal to the square ou the straight line
made up of AB and BC together.
Construction. — Produce AB to D, so that BD may be
equal to CB (Post. 2, and I. 3).
Upon AD describe the square AEFD
And construct two figures such as in ^
the preceding propositions.
Proof. — -Because CB is equal to BD
(Const.), CB to GK, and BD to KN
(Ax. 1),
For the same reason PR is equal
toRO.
And because CB is equal to BD, and GK to KIST,
Therefore the rectangle CK is equal to BN, and GB to
EN (I. 36).
But CK is equal to BN, because they are the complements
of the parallelogram CO (I. 43) ;
Therefore also BN is equal to GR- (Ax. 1).
Therefore the four rectangles BN, CK, GR, R^N" are equal
to one another, and so the four are quadruple of one of them,
CK.
Again, because CB is equal to BD (Const.);
And that BD is equal to BK, that is CG (II. 4, Cor., and
I. 34);
And that CB is equal to GK, that is GP (I. 34, and II. 4,
cor.);
Therefore CG is equal to GP (Ax. 1).
And because CG is equal to GP, and PR to RO,
The rectangle AG is equal to MP, and PL to RF (I. 36).
But MP is equal to PL, because they are complements of
the parallelogram ML (I. 43), and AG is equal to RF (Ax. 1) ;
Therefore the four rectangles AG, MP, PL, RF are equal
to one another, and so the four are quadruple of one of them,
AG.
And it was demonstrated that the four CK, BN, GR, and
RN are quadruple of CK ;
Therefore the eight rectangles which make up the gnomon
AOH are quadruple of AK.
PROPOSITIONS. 243
And because AK is the rectangle contained by AB and .-.Gnomon
BC, for BK is equal to BC ; t^K+AG)
Therefore four times tlie rectangle AB, BC is quadruple of ^^^^\j5(^
AK.
But the gnomon AOH was demonstrated to be quadrnplo
of AK;
Therefore four times the rectangle AB, BC is equal to the
gnomon AOH (Ax. 1).
To each of these add XH, which is equal to the square Hence
on AC (II. 4, cor., and I. 34); XHor'AC?,
Tlierefore four times the rectangle AB, BC, together with
the square on AC, is equal to the gnomon AOH and the
square XH.
But the gnomon AOH and the square XH make up the
figure AEFD, which is the square on AD ;
Therefore four times the rectangle AB, BC, together with ^ ab • bc
the square on AC, is equal to the square on AD, that is, on = af =
the line made up of AB and BC together. (ab+bc)-\
Therefore, if a straight line, &c. Q, E, J),
Proposition 9.— Theorem.
If a straight line be divided into two equal, a7ul also into
two unequal parts, the squares on the two unequal ^Kirts are
together double of the square on half the line, aiid of the square
on the line bettveen the points of section.
Let the straight line AB be divided into two equal parts
in the point C^ and into two unequal parts in the point D ;
The squares on AD and DB shall be together double of ads+dbq
the squares on AC and CD. CD2)^^"^
Construction. — From the point C draw CE at right
angles to AB (I. 11), and make it equal
to AC or CB (I. 3), and join EA, EB.
Through D draw DF parallel to CE
(I. 31).
Through F draw FG parallel to BA
(I. 31), and join AF.
Proof. — Because AC is equal to CE
(Const.), the angle EAC is equal to the angle AEC (I. 5).
244 GEOMETRY.
And because the angle ACE is a right angle (Const.), the
angles AEC and EAC together make one right angle (I. 32),
For and they are equal to one another ;
AEC=j a Therefore each of the angles AEC and EAC is half a right
right z - ^ '^
= CEB. angle.
For the same reason, each of the angles CEB and EBC is
half a right angle;
.-. AEB is Therefore the whole anejle AEB is a ri^ht anoxic.
And because the angle GEF is half a right angle, and the
angle EGF a right angle, for it is equal to the interior and
opposite angle ECB (I. 29),
Therefore the remaining angle EFG is half a right angle ;
Therefore the angle GEF is equal to the angle EFG, and
EG=GF. the side EG is equal to the side GF (I. 6).
Again, because the angle at B is half a right angle, and
the angle FDB a right angle, for it is equal to the interior
and opposite angle ECB (I. 29),
Therefore the remaining angle BFD is half a right angle ;
^^^ Therefore the angle at B is equal to the angle BFD, and
DF=DB. the side DF is equal to the side DB (I. 6).
And because AC is equal to CE (Const.), the square on
AC is equal to the square on CE ;
Therefore the squares on AC and CE are double of the
B„t square on AC,
AE2= But the square on AE is equal to the squares on AC and
CE, because the angle ACE is a right angle (I. 47);
Therefore the square on AE is double of the square on AC.
Again, because EG is equal to GF (Const.), the square on
EG is equal to the square on GF ;
Therefore the squares on EG and GF are double of the
square on GF.
So also But the square on EF is equal to the squares on EG and
^2 CDS. GF, because the angle EGF is a right angle (I. 47) ;
Therefore the square on EF is double of the square on GF.
And GF is equal to CD (I. 34);
Therefore the square on EF is double of the square on CD^
But it has been demonstrated that the square on AE is
al§o double of the square on AC ;
EF2 = Therefore t^he squares on AE and EF are double of tli§
f^\ squares} pu AQ §.pd QJ),
PROPOSITIONS.
2ib
But the square on AF is equal to the squares on AE and But
EF, because the angle AEF is a right angle (I. 47); Iaf^^^'
Therefore the square on AF is double of the squares on AC =ad2+
and CD. =ad2+
But the squares on AD and DF are equal to the square ^^'^•
on AF, because the angle ADF is a right angle (I. 47);
Therefore the squares on AD and DF are double of the
squares on AC and CD.
And DF is equal to DB; therefore the squares on AD and .-. AD2+
DB are double of the squares on AC and CD. =^(AC2-fc
Therefore, if a straight line, &c. Q.E.D. CD'-o
Propositioii 10. — Theorem.
If a straight line he bisected and produced to any 2>oint, tlie
square on the whole line thus produced^ and the square on the
jmrt of it produced, are together double of the square on half
the line bisected, and of the square on the line made up of the
half and the part produced.
Let the straight line AB be bisected in C, and produced to D ; ;^5'+5,^^
The squares on AD and DB shall be together double of the CB-'].
squares on AC and CD.
Construction. — From the point C draw CE at right angles
to AB, and make it equal to AC or CB (I. 11, I. 3), and join
AE, EB.
Through E draw EF parallel to AB, and through D draw
DF parallel to CE (I. 31). Then because the straight line
EF meets the parallels EC, FD, the angles CEF, EFD are
equal to two right angles (I. 29); therefore the angles BEF,
EFD are less than two right angles;
therefore EB, FD will meet, if
produced towards B and D (Ax.
Let them meet in G, and join ^C
AG.
Proof. — Because AC is equal
to CE (Const.), the angle CEA is equal to the angle EAC
(I. 5).
And the angle A CE is a right angle ; therefore each of the
angles CEA and EAC is half a right angle (I. 32).
24G GEOMETRY.
As in For tlie same reason each of the angles CEB and EBC is
^^^^' ' half a right angle ;
^i^'bt' z^ Therefore the whole angle AEB is a right angle.
And because the angle EBC is half a right angle, the angle
DBGr, which is vertically opposite, is also half a right angle
(1.15);
But the angle BDG is a right angle, because it is equal to
the alternate angle DCE (I. 29);
Therefore the remaining angle DGB is half a right angle,
and is therefore equal to the angle DBG;
BD=DG Therefore also the side BD is equal to the side DG (I. 6).
~" ' Again, because the angle EGF is half a right angle, and
the angle at F a right angle, for it is equal to the opposite
angle ECD (I. 34);
Therefore the remaining angle FEG is half a right angle
(I. 32), and therefore equal to the angle EGF;
Also Therefore also the side GF is equal to the side FE (I. 6).
"" * And because EC is equal to CA, the square on EC is equal
to the square on CA;
Therefore the squares on EC and CA are double of the
square on CA.
Ac?ainas But the square on AE is equal to the squares on EC and
in Prop. 9. ^^ ^j^^^^.
AE-^= Therefore the square on AE is double of the square
'^^^■''- on AC.
Again, because GF is equal to FE, the square on GF is
equal to the sqviare on FE;
Therefore the squares on GF and FE are double of the
square on FE.
But the square on EG is equal to the squares on GF and
'FE (I. 47);
Therefore the square on EG is double of the square on FE.
And FE is equal to CD (I. 34),
And Therefore the square on EG is double of the square on CD.
2 cir ^^^^ i^ ^^^ been demonstrated that the square on AE is
.•.AE'-^4- double of the square on AC;
2(A(S+ Therefore the squares on AE and EG are double of the
GD^-)- squares on AC and CD.
AE2H-EG3 ^^^^ *^i^ square on AG is equal to the squares on AE and
=AG2; EG (I. 47);
PROPOSITIONS.
247
2(AC-'+
Therefore the square on AG is double of the squares on /.ag2
AC and CD. ^b^'^
But the squares on AD and DG are equal to the square on .•.ad-+
AG(L47); SJ
Therefore the squares on AD and DG are double of the cd-
squares on AC and CD.
And DG is equal to DB; therefore the squares
on AD and DB are double of the squares on AC
and CD.
Therefore, if a straight line, <kc. Q.E.D.
Proposition 11.— Problem.
To divide a given straight line into two parts, so that the
rectangle contained by the whole and one of the parts shall be
equal to the square on the other part.
Let AB be the given straight line.
It is required to divide AB into two parts, so that the
rectangle contained by the whole and one of the parts shall
be equal to the square on the other pai-t.
Construction. — Upon AB describe
the square ABDC (I. 46).
Bisect AC in E (I. 10), and join BE.
Produce CA to F, and make EF equal
to EB (I. 3).
Upon AF describe the square AFGH
(I. 46).
Produce GH to K.
Then A B shall be divided in II, so
that the rectangle AB, BHis equal to the
square on AH,
Proof. — Because the straight line AC is bisected in E,
and produced to F,
The rectangle CF, FA, together with the square on AE, cF'FA +
is equal to the square on EF (II. 6). tSV^
But EF is equal to EB (Const.); =eb-
Therefore the rectangle CF, FA, together with the square
on AE, is equal to the square on EB.
But the square on EB is equal to the squares on AE and =a1}3+
AB, because tho angle EAB is a right angle (I. 47); ^^^'
248
GEOMETIIT?.
Therefore the rectangle CF, FA, together with the square
on AE, is equal to the squares on AE and AB.
Take away the square on AE, which is common to both ;
Therefore the remaining rectangle CF, FA is equal to the
square on AB (Ax. 3).
But the figure FK is the rectangle contained by CF and
FA, for FA is equal to FG;
And AD is the square on AB;
.FK=AB Therefore the figure FK is equal to AD.
Take away the common part AK, and the remainder FH
is equal to the remainder HD (Ax. 3).
But HD is the rectangle contained by AB and BH, for
A.B is equal to BD;
And FH is the square on AH ;
Therefore the rectangle AB,BH is equal to the square on
AH.
Therefore the straight line AB is divided in H, so that the
rectangle AB, BH is equal to the square on AH. Q.E.F,
CFFA.
=AB2.
Take away
AK, then
FH=HD
or ABBH
=AH^
BD-'=BC2
+CD2+
2BCCD.
Proposition 12.— Theorem.
In ohtuse-angted triangles if a perpendicular he drawn from
either of the acute angles to the opjmsite side produced, the
square on the side subtending the obtuse angle is greater than
the squares on the sides containing the obtuse angle, by twice
the rectangle contained by the side on which, when piroduced,
the perpendicular falls, and the straight line intercepted with-
out the triangle, between tJie perpendicular and the obtuse angle.
Let ABC be an obtuse-angled triangle, having the obtuse
angle ACB; and from the point A let AD
be drawn perpendicular to BC produced.
The square on AB shall be greater
than the squares on AC and CB by twice
the rectangle BC, CD.
Proof. — Because the straight line BD
is divided into two parts in the point C,
The square on BD is equal to the
squares on BC and CD, and twice the rectangle BC, CD
(II. 4).
To each of these equals add the square on DA;
I^ROPOSITIOKB. 249
Therefore the squares on BD and DA are equal to the .-.bdsh da^
squares on BC, CD, DA, and twice the rectangle BC, CD. =bc2+
But the square on BA is equal to the squares on BD and (^K!+
DA, because tjie angle at D is a right angle (I. 47); +2BCCD
And the square on CA is equal to the squares on CD and ^5^"^
DA (I. 47) ; +2BCCD.
Therefore the square on BA is equal to the squares on BC
and C A, and twice the rectangle BC, CD ; that is, the square
on BA is greater than the squares on BC and CA by twice
the rectangle BC, CD.
Therefore, in obtuse-angled triangles, &c. Q.E.D.
Proposition 13.— Theorem.
In everi/ triangle, the square on the side subtending an acute
angle is less tlian the squares on the sides containing that angle,
by twice the rectangle contained by either of these sides, and
the straight line intercepted between the perpendicular let fall
on it from the ppp)osite angle and the acute angle.
Let ABC be any triangle, and the angle at B an acute
angle ; and on BC, one of the sides containing it, let fall the
perpendicular AD from the opposite angle (I. 12).
The square on AC, opposite to the angle B, shall be less AC2=CB2
than the squares on CB and BA, by twice the rectangle ^b^b1>
CB, BD.
Case I.— First, let AD fall within the
triangle ABC.
Proof. — Because the straight line CB is
divided into two parts in the point D,
The squares on CB and BD are equal
to twice the rectangle contained by CB,
BD, and the square on DC (II. 7).
To each of these equals add the square on DA.
Therefore the squares on CB, BD, DA are e(pial to twice •••CP"+
the rectangle iCB, BD, and the squares on AD and DC. ix\2)^
But the square on AB is equal to the squaies on BD and ^^J^^
DA, because the angle BDA is a right angle (I. 47) ; ncr^),
And the square on AC is ecpial to the squares on AD and ^bs+ba^
DC (I. 47) ; =2CBDB
Therefore tlie squares on CB and BA are equal to the "^ "*
250
GEOMETRY.
square on AC, and twice the rectangle CB, BD; that is, the
square on AC alone is less than the squares on CB and BA
by twice the rectangle CB, BD.
Case II. — Secondly, let AD fall without the triangle ABC.
Proof. — Because the angle at D is a right angle (Const.),
A the angle ACB is greater than a right
angle (I. 16);
Therefore the square on AB is equal
to the squares on AC and CB, and twice
the rectangle BC, CD (II. 12).
To each of these equals add the square
■^ on BC.
Therefore the squares on AB and BC are equal to the
square on AC, and twice the square on BC, and twice the
rectangle BC, CD (Ax. 2).
But because BD is divided into two parts at C,
The rectangle DB, BC is equal to the rectangle BC^ CD
and the square on BC (II. 3) ;
And the doubles of these are equal, that is, twice the
rectangle DB, BC is equal to twice the rectangle BC, CD
and twice the square on BC ;
Therefore the squares on AB and BC are equal to the
square on AC, and twice the rectangle DB, BC ; that is, the
jl square on AC alone is less than the squares on AB
and BC by twice the rectangle DB, BC.
Case III. — Lastly, let the side AC be perpen-
dicular to BC.
Proof. — Then BC is the straight line between the
perpendicular and the acute angle at B ; and it is
manifest that the squares on AB and BC are equal
to the square on AC, and twice the square on BC (I. 47,
and Ax. 2).
Therefore, in every tiiangle, &c. Q.E.D.
Proposition 14 Problem.
To describe a square that shall he equal to a given recti^
lineal figure.
Let A be the given rectilineal figure*
It is required to describe a square that bhall be equal to A.
PROPOSITIONS.
251
Construction. — Describe the rectangular parallelogram
BCDE equal to the rectilineal figure A (I. 45).
If then the sides of it, BE, ED, are equal to one another .
it is a square, and what
was required is now
done.
But if they are not
equal, produce one of
them, BE, to F, and
make EF equal to ED
(I. 3).
Bisect BF in G (1. 10), and from the centre G, at the dis-
tance GB, or GF, describe the semicii-cle BHF ;
Produce DE to H, and join GH ;
Then the square described upon Ell shall he equal to the
rectilineal figure A.
Proof. — Because the straight line BF is divided into two
equal parts in the point G, and into two unequal parts in
the point E,
The rectangle BE,EF, together with the square on GE,
is equal to the square on GF (II. 5).
But GF is equal to GH ;
Therefore the i-ectangle BE,EF, together with the square
on GE, is equal to the square on GH.
But the square on GH is equal to the squares on GE and
EH (I. 47) ;
Therefore the rectangle BE,EF, together with the square
on GE, is equal to the squares on GE and EH.
Take away the square on GE, which is common to both ;
Therefore the rectangle BE,EF is equal to the square on
EH (Ax. 3).
But the rectangle contained by BE and EF is the paral-
lelogram BD, because EF is equal to ED (Const.);
Therefore BD is equal to the square on EH.
But BD is equal to the rectilineal figure A (Const.);
Therefore the square on EH is equal to the rectilineal
figure A.
Therefore, a square has been made equal to the given
rectilineal figiu-e A, viz., the square described on EH.
Q.E.F,
BE-EF
+GE-I
=GH-J
=GE-i+
EH2.
.-.BE-EF
orBD
=EH2.
Hence
EH2=A.
252 GEoMEi^itY.
EXERCISES ON BOOK II.
Prop. 1—11.
1 . Divide a given straight line into two sucli parts that the rect-
angle contamed by them may be the greatest possible.
2. The sum of the squares of two straight lines is never less than
twice the rectangle contained by the straight lines.
3. Divide a given straight line into two parts such that the squares
of the whole line and one of the parts shall be equal to twice the
square of the other part.
4. Given the sum of two straight lines and the difference of their
squares, to find the lines.
5. In any triangle the difference of the squares of the sides is equal
to the rectangle contained by the sum and difference of the parts
into which the base is divided by a perpendicular from the vertical
angle.
C. Divide a given straight line into such parts that the sum of their
squares may be equal to a given square.
7. If ABCD be any rectangle, A and C being opposite angles, and
any point either within or without the rectangle — OA- + 00^*
= 0B2 + 0D2..
8. Let the straight line AB be divided into any two parts in the
point C. Bisect CB in D, and take a point E in AC such that EC
:= CD. Then shall AD^ = AE^ + AC • CB.
9. If a point C be taken in AB, and AB be produced to D so that
BD and AC are equal, show that the squares described upon AD and
AC together exceed the square upon AB by twice the rectangle con-
tained by AE and AC.
10. From the hypothenuse of a right-angled triangle portions are
cut off equal to the adjacent sides. Show that the square on the
middle segment is equal to twice the rectangle under the extreme
segments.
11. If a straight line be divided into any number of parts, the
square of the whole line is equal to the sum of the squares of the
parts, together with twice the rectangles of the parts taken two and
two together.
12. If ABC be an isosceles triangle, and DE be drawn parallel to
the base BC, cuttuig in D and E either the side or sides produced,
and EB be joined; prove that BE^ - BC • DE + CE^.
Prop. 12—14.
13. In any triangle show that the sum of the squares on the sides
is equal to twice the square on half the base, and twice the square on
the line drawn from the vertex to the middle of the base.
EXERCISES OX THE PROPOSITIONS. 253
14. If squares are described on the sides of any triangle, find tlio
difference between the sum of two of the squares and the third
square, and show from your result what this becomes when the
angle opposite the third square is a right angle.
15. Show also what the difference becomes when the vertex of the
triangle is depressed until it coincide with the base,
16. The square on any straight line drawn from the vertex of an
isosceles triangle, together with the rectangle contained by the
segments of the base, is equal to the square upon a side of the
triangle.
17. If a side of a triangle be bisected, and a perpendicular drawn
from the middle point of the base to meet the side, then the square
of the altitude of the triangle exceeds the square upon half the base
by twice the rectangle contained by the side and the straight line
between the points of section of the side.
18. In any triangle ABC, if perpendiculars be drawn from each of
the angles upon the opposite sides, or opposite sides produced, meet-
ing them respectively in D, E, F, show that —
BA^ + AC2 + CB2 =2AE-AC + 2CD-CB + 2BF-BA;
all lines being measured in the same direction round the triangle.
19. Construct a square equal to the sum of the areas of two given
rectilinear figures.
20. The base of a triangle is 63 ft., and the sides 25 ft. and 52 ft.
respectively. Show that the segments of the base, made by a per-
pendicular from the vertex, are 15 ft. and 48 ft. respectively, and
that the area of the triangle is 630 sq. ft.
21. In the same triangle, show that the length of the line joining
the vertex with the middle of the base is 22*9 ft.
22. A ladder, 45 ft. long, reaches to a certain height against a
wall, but, when turned over without moving the foot, must be short-
ened 6 ft. in order to reach the same height on the opposite side.
•Supposing the width of the street to be 42 ft., show that the height
to which the ladder reaches is 36 ft.
23. The base and altitude of a triangle are 8 in. and 9 in. respec-
tively; show that its area is equal to a square whoso side is 6 in.
Prove your result by construction.
24. On the supposition that lines can be always expressed exacthj
h\ terms of some unit of length, what geometrical propositions may
be deduced from the following algebraical identities ? —
(1.) (a + 5)2 = a2 + 2rt6 + 5-
(2.) (« + h) (a - fc) + 62 :=a2
(3.) (2a 4- 6) 6 + a^ = (« + by
(4.) (a + 6)2 + &2 = 2 (rt + i) 6 + a^
(5.) 4(a + 6) 6 + />2 z= (a + 2 6^-
(6.) (fi + 6)2 + (a - 6)2 = 2a2 + 2 62
(7.) (2a + 6)2 + 62 =2a2 H-2(a + 6)2
25i GEOMETRY,
EUCLID'S ELEMENTS, BOOK IIL
Pefinitions,
1. Equal circles are those of wliicli the diameters are equal,
or from the centres of which the straight lines to the circum-
ferences are equal,
2. A straight line is said to touch
a circle, or to be a tangent to it,
when it meets the circle, and being
produced does not cut it.
3. Circles are said to touch one
another, which meet, but do not cut
one another,
4. Straight lines are said to be equally
distant from the centre of a circle, when
the perpendiculars drawn to them from the
centre are equal.
5. And the straight line on which the
greater perpendicular falls, is said to be
farther from the centre.
6. A segment of a circle is the figure
contained by a straight line and the cixxum-
ference which it cuts off.
7. The angle of a segment is that which is contained by
the straight line and the circvimference.
8. An angle in a segment is the angle
contained by two straight lines drawn
from any point in the circumference of
the segment to the extremities of the
straight line, which is the base of the
segment.
9. An angle is said to insist or
stand upon the circumference inter-
cepted between the straight lines that contain the
angle.
PROPOSITIONS.
255
10. A sector of a circle is the figure con-
tained by two straight lines drawn from
the centre and the circumference between
them.
11. Similar segments of circles
are those which contain equal
anojles.
[Any portion of the circumference is called an rnr, and the chord
of an arc is the straight line joining its extremities.]
Proposition 1. — Problem.
To find the centre of a given circle.
Let ABC be tlie given circle.
It is required to find its centre.
Construction. — Draw within the circle any chord AB,
and bisect it in D (I. 10).
From the point D •draw DC at right
angles to AB (I. 11).
Produce CD to meet the circumference
in E, and bisect CE in F (I. 10).
Then the point F shall he the centre of the
circle ABC.
Proof. — For if F be not the centre, if
possible let G be the centre ; and join GA,
GD, GB.
Then, because DA is equal to DB (Const.), and Du- com-
mon to the two triangles ADG, BDG;
The two sides AD, DG are equal to the two sides BD,
DG, each to each ;
And the base GA is equal to the base GB, being radii of
the same circle;
Therefore the angle ADG is equal to the angle BDG (I. 8),
But when a straight line, standing on anothoi
line, makes the adjacent angles equal to one aiiotln r,
the angles is called a right angle (I. Def. ID).
Suppose
Gthe
centre.
straight
ich of
.ZADG
: z BDG.
25 () GEOMETRY,
Tlierefore the angle GDB is a right angle.
But the angle FDB is also a right angle (Const.) ;
-zFDiJ Therefore the angle GDB is equal to the angle FDB
(Ax. 11), the less to the greater; which is impossible.
Therefore G is not the centre of the circle ABO.
In the same manner it may be shown that no point which
is not in CE can be the centre.
And since the centre is in CE, it must be in F, its point
of bisection.
Therefore F is the centre of the circle ABC : which w\as
to be found.
Corollary. — From this it is manifest that, if in a circle a
straight line bisect another at right angles, the centre of thQ
circle is in the line which bisects the other,
Proposition 2. — Theorem,
If any two 2^oints he taken in the circumference of a ch^cle^
the straight line which joins them shall fall tvithin the circle.
Let ABC be a circle, and A and B any two points in the
circumference.
The straight line drawn from A to B shall fall within the
circle.
Construction. — Find D the centre of the circle ABC
(III. 1), and join DA, DB.
In AB take any point E ; join DE, and
j)roduce it to the circumference in F.
Proof. — Because DA is equal to DB,
the angle DAB is equal to the angle
DBA (I. 5).
And because AE, a side of the triangle
DAE, is produced to B, the exterior angle
DEB is greater than the interior and
opposite angle DAE (I. 16).
But the angle DAE was proved to be equal to the angle
DBE;
Therefore the angle DEB is also greater than DBE.
But the greater angle is subtended by ^he greater ^id^
(1.19);
WH^-m, Therefore D? is greater th^ PB,
PROPOSITIONS. 257
But DB is equal to DF ; therefore DF is greater
than DE, aii^ tlic point E is therefore within the
circle.
In the same manner it may be proved that every point in
AB lies within the circle.
Therefore the straight line AB lies within the cii'cle.
Therefore, if any two points, &:c, Q.E.D,
Proposition 3.— Theorem.
// a straight line drawn through the centre of a circle bisect
a straight line in it which does not 2)ass through the centre^ it
shall cut it at right angles ; and conversely ^ if it cut it at right
angles, it shall bisect it.
Let ABC be a circle, and let CD, a straight line drawn
through the centre, bisect any straight line AB, which does
not pass through the centre.
CD shall cut AB at right angles.
CoxsTiiucTioN. — Take E, the centre of the circle (III. 1),
and join EA, EB.
Proof. — Because AF is equal to FB
(Hyp.), and FE common to the two triangles
AFE, BFE, and the base EA equal to the
base EB (I. Def. 15),
Therefore the angle AFE is equal to the
angle BFE (I. 8) ;
Therefore each of the angles AFE, BFE
is a right angle (I. def. 10);
Therefore the straight line CD, drawn through the centre, ^^^p^^^-
bLsecting another, AB, that does not pass through the centre,
also cuts it at right angles.
Conversely, let CD cut AB at right angles.
CD shall bisect AB ; that is, AF shall bo equal to FB
— the same construction being made.
Proof. — Because the radii EA, EB are equal, the angle
EAF is equal to the angle EBF (I. 5),
And the angle AFE is equal to the angle BFE
(Hyp.),
Therefore, in the two triangles EAF, EBF, there are two
angles in the one equal to two angles in the other, each to
258
GEOMETRY.
each, and the side EF, which is opposite to one of the equal
angles in each, is common to both ;
Therefore their other sides are equal (I. 26);
Therefore AF is equal to FB.
Therefore, if a straight line, &c. Q,E.D,
If they
bisect each
other,
EF bi5?ectg
AC cat
rio-ht
anj^les.
And EF
bisects BD
at right
unglcv
•.ZFEA
;=ZFEB,
Proposition 4. — Theorem.
If in a circle two straight lines cut one another, which
do not both pass through the centre, they do not bisect each
other.
Let ABCD be a circle, and AC, BD two straight lines in
it, which cut one another at the point E, and do not both
pass through the centre.
AC, BD shall not bisect one another.
If one of the lines pass through the centre, it is plain that
it cannot be bisected by the other, which does not pass through
the centre.
But if neither of them pass through the centre, if possible,
let AE be equal to EC, and BE to ED.
Construction. — Take F, the centre of the circle (III. 1),
and join EF.
Because FE, a straight line drawn
through the centre, bisects another line
AC, which does not pass through the
aL f ^^-^ centra (Hyp.), therefore it cuts it at riglit
angles (III. 3) ;
Therefore the angle FEA is a right
angle.
^ Again, because the straight line FE bisects the straight
line BD, which does not pass through the centre (Hyp.),
therefore it cuts it at right angles (III. 3);
Therefore the angle FEB is a right angle.
But the angle FEA was shown to be a right angle ;
Therefore the angle FEA is equal to the angle FEB, the
less to the greater, which is impossible;
Therefore AC, BD do not bisect each other.
Therefore, if in a circle, <fcc. Q.R D,
pPvOrosiTiONS, 259
Proposition 5. — Theorem,
If two circles cut one another, they shall not have the same
centre.
Let the two circles ABC, CDG cut one another in the
points B, C.
They shall not have the same centre.
For, if it be possible, let E be their centre ; join EC, and
draw any straight line EFG, meeting the ^
circles in F and G.
Proof. — Because E is the centre of the A/
circle ABC, EC is equal to EF (I. Def.
15).
And because E is the centre of the
circle CDG, EC is equal to EG.- \:s^^ ^^ and =eg.
But EC was shown to be equal to EF;
Therefore EF is equal to EG (Ax. 1), the less to the greater, .-.EFrrEa
which is impossible ;
Therefore E is not the centre of the circles ABC, CDG.
Therefore, if two circles, &c, Q.E.D,
Proposition 6.— Theorem.
If 07ie circle touch another internally, they slkoll not have
the same centre.
Let the circle CDE touch the circle ABC internally in the
point C.
They shall not have the same centre.
Construction. — For, if it be possible, let F be their centre; if they have
join FC, and draw any straight line FEB,
meeting the circles in E and B.
Proof. — Because F is the centre of
the circle ABC, FC is equal to FB (I Def.
15). A|
And because F is the centre of the
circle CDE, FC is equal to FE.
But FC was shown to be equal to FB ;
Therefore FE is equal to FB (Ax. 1), the less to the greater, .-.FErrFB
which is impossible ;
Therefore F is not the centre of the circles ABC, CDE,
Therefore, if one circle, &c, Q,E,D,
2 GO GEOMETRY.
Proposition 7. — Theorem.
If any 'point he taken in the diameter of a circle, lohich is
not the centre of all the straight lines which can he drawn
from this point to the circumference, the greatest is that in
which the centre is, and the other part of that diameter is the
least ; and, of any others, that which is nearer to the straight
line which passes through tJie centre is always greater than
one more remote ; and from the sam^e point there can he drawn
to the circumference two straight lines, and only two, which
are equal to one another, one 07i each side of the shortest line.
Let A BCD be a circle, AD its diameter, and E its
centre, in which let any point F be taken which is not the
centre.
FA>FB Of all the straight lines FB, FC, FG-, &c., that can be
^^^^P^ drawn from F to the circumference, FA, which passes
the'ieast. through the centre, shall be the greatest ;
FD, the other part of the diameter AD, shall be the least;
And of the others, FB, the nearer to FA, shall be greater
than FC, the more remote ; and FC greater than FG.
Construction. — Join BE, CE, GE.
Tor . ,
BE+EF Proof. — Because any two sides of a triangle are greater
crAF>BP ^ ^^ ^1^^^ ^jjQ ^^:^.^ g^j^^ j3j^^ j^j. ^^^ greater
than BF (I. 20).
But AE is equal to BE ; therefore
AE, EF, that is, AF is greater than BF.
Again, because BE is equal to CE,
and EF common to the two triangles BEF,
CEF, the two sides BE, EF are equal to
the two CE, EF, each to each.
But the angle BEF is greater than the angle CEF ;
And Therefore the base FB is greater than the base FC
ba.ser>F /T OIX
^baaeFC \ * "^ )'
In the same manner it may be shown that FC is greater
than FG.
Ap^in Again, because GF, FE are greater than EG (I. 20),
^e5^^ and that EG is equal to ED ;
and.-.>ED. Therefore GF, FE are greater than ED.
.•.GF>FD Take away the common part FE, and the remainder GF
is greater than the remainder FD (Ax. 5).
PROPOSITIOKS. ^61
Therefore FA is the greatest, and FD the least, of all the
straight lines from F to the circumference; and FB is greater
than FC, and FC than FG.
Also, there cannot be drawn more than two equal straight
lines from the point F to the circumference, one on each side
of the shortest line.
Construction. — At the point E, in the straight line EF,
make the angle FEH equal to the angle FEG (I. 23), and
join FH.
Proof. — Because EG is equal to EH, and EF common to ^^^"sles
the two triangles GEF, HEF, the two sides EG, EF are and hep
equal to tha two sides EH, EF, each to each ; fn^very
And the angle GEF is equal to the angle HEF (Const.); respect.
Therefore the base FG is equal to the base FH (I. 4).
But, besides FH, no other straight line can be drawn from
F to the circumference equal to FG.
For, if it be possible, let FK be equal to FG ;
Then, because FK is equal to FG, and FH is also equal to ^'J^^^^^
FG, therefore FH is equal to FK ; Jso = Wi,
That is, a line nearer to that which passes through the i^poVibie.
centre is equal to a line which is more remote; which is im-
possible by what has been already shown.
Therefore, if any point, &c. Q.E.D.
Proposition 8.— Theorem.
If any 2^oint he taken without a circle, and straight lines he
draion from it to the circumference^ one of lohich passes
through the centre ; of those which fall on the concave circum-
ferencCy the greatest is tJuit which passes through the centrcy
and of the resty that which is nearer to the one2^assing through
the centre is always greater than one more remote ; hut of those
which fall on the convex circumference, the least is that he-
tween the j^oint without the circle and the diameter ; and of
the rest, that which is nearer to the least is always less than
one more remote ; and from the same point there can he drawn
to the circumference two straight lines, and only two, which
are equal to one another, one on each side of the least line.
Let ABC be a circle, and D any pouit without it, and
from D let the straight lines DA, DE, DF, DC be drawu
262
GEOMETRY.
r>A>DE
and
<:DL
For
EM+MD
or AD>
ED.
Also
base ED
>- base FD,
Again.
MK+KD
>MD, or
DG.
.•.KD>-
DG,
and/.DG
<KD, &c.
to tlie circumference, of which DA passes through the
centre.
Of those which fall on the concave part of the circum-
ference AEFC, the greatest shall be DA, which passes
through the centre, and the nearer to it shall be greater
than the more i^emote, viz., DE greater than DF, and DF
greater than DC.
But of those which fall on the convex circumference GKLH,
the least shall be DG between the point D and the diameter
AG-, and the nearer to it shall be less than the more remote,
viz., DK less than DL, and DL less than DH.
Construction. — Take M, the centre of the circle ABC
(III. 1), and join ME, MF, MC, MH, ML, MK.
Proof. — Because any two sides of a
triangle are greater than the third side,
EM, MD are greater than ED (I. 20).
But EM is equal to AM ; therefore
AM, MD are greater than ED— that
is, AD is greater than ED.
Again, because EM is equal to FM,
and MD common to the two triangles
EMD, FMD ; the two sides EM, MD
are equal to the two sides FM, MD,
each to each ;
But the angle EMD is greater than
the angle FMD ;
Therefore the base ED is greater than
the base FD (I. 24).
In like manner it may be shown that
FD is greater than CD;
Therefore DA is the greatest, and DE greater than DF
and DF greater than DC.
Again, because MK, KD are greater than MD (I. 20),
and MK is equal to MG;
The remainder KD is greater than the remainder GD —
that is, GD is less than KD.
And because MK, DK are drawn to the point K within
the triangle MLD from M and D, the extremities of its side
MD;
Therefore MK,DK are less than ML, LD (I. 21).
PROPOSITIONS. 263
But MK is equal to ML ; therefore the remainder KD is
less than the remainder LD.
In like manner it may be shown that LD is less than HD.
Therefore DG is the least, and KD less than DL, and DL
less than DH.
Also, there can be drawn only two equal straight lines from
the point D to the circumference, one on each side of the
least line.
Construction. — At the point M, in the straight line MD,
make the angle DMB equal to the angle DMK (I. 23), and
join DB;
Proof. — Because MK is equal to MB, and MD common Triangles
to the two triangles KMD, BMD; the two sides KM, MD andBMD
are equal to the two sides BM, MD, each to each; ?irev2r^
And the angle DMK is equal to the angle DMB (Const) ; respects
Therefore the base DK is equal to the base DB (I. 4).
But, besides DB, no other straight line can be drawn from
D to the circumference equal to DK.
For, if it be possible, let DN be equal to DK.
Then, because DN is equal to DK, and DB is also equal I)N=dk
to DK, therefore DB is equal to DN (Ax. 1); =d^*
That is, a line nearer to the least is equal to one which is T^^c^ is
more remote ; which is impossible by what has been already
shown*
Therefore, if any point, &c. Q.EiD,
Proposition 9.~-Theorem.
If a point he taken within a circle^ from which there can he
drawn more tlian two equal straight lines to the circumference^
that point is the centre of the circle, ^
Let the point D be taken within the circle ABC, from
which to the circumference there can be
drawn more than two equal straight lines,
viz., DA, DB, DC.
The point D shall bo the centre of the
circle.
Construction. — For if not, let E be
the centre ; join DE, and produce it to
the circumference in F and C
264 GEO:.IETRY.
Proof.— ThenFG is adiameterof the circle ABC (I.Def. 17).
If D be not And because in FG, the diameter of the circle ABC, there
the centre. -^ ^^-^^^ ^j^^ p^-^^^ jy^ ^^^ ^-^^ ^^^^^^ .
DG->DC Therefore DG is the greatest straight line from D to the
>DA. circumference, and DC is gi-eater than DB, and DB greater
than DA (III. 7);
But they But thcso lines are likewise equal, by hypothesis; which is
--r impossible.
Therefore E is not the centre of the circle ABC.
In like manner it may be demonstrated that any other
point than D is not the centre ;
Therefore D is the centre of the circle ABC.
Therefore, if a point, &c. Q,E,D,
Proposition 10. — Theorem.
One circumference of a circle cannot cut another in more
than two points.
Construction. — If it be possible, let the circumference
If possible, _jf^ ABC cut the ciitjumference DEF in
more than two points, viz., in the points
B,G,F.
Take the centre K of the circle ABC
' (III. 1), and join KB, KG, KF.
Proof. — Because K is the centre of
the circle ABC, the radii KB, KG, KF
are all equal.
the two ^nd because within the circle DEF there is taken the
iiave tho point K, from which to the circumference DEF fall more
than two equal straight lines KB, KG, KF, therefore K is
the centre of the circle DEF (III. 9).
But K is also the centre of the circle ABC (Const.) ;
Therefore the same point is the centre of two circles which
cut one another; which is impossible (III. 5).
Therefore, one circumference, &c. Q.E.D.
Proposition 11. — Theorem.
If one circle touch another internally in any point, the
straight line which joins their centres, being prodxiced, slmll
pass through that ptoint
Let the circle ADE touch the circle ABC internally in the
same
centre.
PROPOSITIOICS.
265
point A; and let F bo the centre of the circle ABC, and G
the centre of the circle ADE.
The straight line which joins their centres, being produced,
shall pass through the point of contact A.
Construction. — For, if not, let it pass otherwise, if if not,
possible, as FGDH. Join AF and AG.
Proof. — Because AG, GF are greater
than AF (I. 20), and AF is equal to HF
(I. def. 15); '
Therefore AG, GF are greater than HF.
Take away tlie common part GF, and the
remainder AG is greater than the remainder
HG.
But AG is equal to DG (I. Def. 15);
Therefore DG is greater than HG, the less than tlie
greater; which is impossible.
Therefore the straight line which joins the centres, being
produced, cannot fall otherwise than upon the point A, that
is, it must pass through it.
Therefore, if one circle, &c. Q.E.D,
AG>HG.
But AG
=DG.
.*.DG=-
HG.
Proposition 12. — Theorem.
If two circles touch each other externally in any jyoint, the
straight line which joins their centres shall pass through that
point.
Let the two circles ABC, ADE touch each other externally
in the point A; and let F be the centre of the circle ABC,
and G the centre of the circle ADE;
The straight line which joins
their centres shall pass througli
the point of contact A.
Construction. — For, if not,
let it pass otherwise, if possible,
as FCDG. Join FA and AG.
Proof. — Because F is the
centre of the circle ABC, FA
is equal to FC (I. Def. 15).
And because G is the centre of the circle ADE, GA is
equal to GD;
ir not.
266
GEOMETRY.
FGfs>
FA+AG,
but it is
also leas.
Therefore FA, AG are equal to FC, DG (Ax. 2).
Therefore the whole FG is greater than FA, AG.
But FG is also less than FA, AG (I. 20) ; wliich is im-
possible.
Therefore the straight line which joins the centres of the
circles shall not pass otherwise than through the point A,
that is, it must pass through it.
Therefore, if two circles, &c. Q.E.D,
Proposition 13.~Theorem.
One circle cannot touch another in more points than one,
whether it touch it internally or externally »
I. First, let the circle EBF touch the circle ABC internally
in the point B.
Then EBF cannot touch ABC in any other point.
If possible, Construction. — If it be possible, let EBF touch ABC in
letittomai another point D; join BD, and draw GH bisecting BD at
'"''"' right angles (I. 10, 11).
tbcn
GH passes
throu<,'h
the i)oint
ofcoutact,
which it
does uot.
Proof. — Because the two points B, D are in the circum-
ference of each of the circles, the straight line BD falls withiii
each of them (III. 2);
Therefore the centre of each circle is in the straight line
GH, which bisects BD at right angles (III. 1 cor.)
Therefore GH passes through the point of contact (III. 11).
But GH does not pass through the point of contact,
because the points B, D are out of the line of GH; which
is absurd.
Therefore one circle cannot touch another internally in
more points than one.
PROPOSITIONS. 267
II. Next, let the circle ACK touch the cii'cle ABO
externally in the point A.
Then ACK cannot touch ABC in any other point.
Construction. — If it be possible, let ACK touc?li ABC in if possible
another point C. Join AC. ^„^ InCabof
Proof. — Because the points A, C are
in the circumference of the circle ACK, the
straight line AC must fall within the cii'cle
ACK (III. 2).
But the circle ACK is without the circle
ABC (Hyp.); .....
Therefore the straight line AC is without I J then
the circle ABC. V / tSholIt
But because the two points A, C are in "^s,^^^ ^^^ V^^p^'^^'^
the circumference of the circle ABC, the whiciiis
straight line AC falls within the circle ABC (III. 2); which absurd,
is absurd.
Therefore one circle cannot touch another externally in
more points than one.
And it has been shown that one circle cannot touch an-
other internally in more points than one.
Therefore, one circle, &c. Q,E.D,
Proposition 14.— Theorem.
Equal straight lines in a circle are equally distant from the
centre; and, conversely, those which are equally distant from
the centre tire equal to one another.
Let the straight lines AB, CD, in the circle ABDC, be
equal to one another.
Then they shall be equally distant from the centre.
Construction. — Take E, the centre cif the cii'cle ABDC
(III. 1).
From E draw EF, EG, perpendiculars to AB, CD (I. 12).
Join EA, EC.
Proof. — Because the straight lin^ EF, passing through
the centre, cuts the straight line AB, which does not pass
through the centre, at right angles, it also bisects it (III. 3).
Therefore AF is equal to FB, and AB is double of AF.
For the like reason CD is double of CG.
268 GEOMETRY
AF = CG, But AB is equal to CD (Hyp.); therefore AF is equal to
CG (Ax. 7).
And because AE is equal to CE, the square on AE is
equal to the square on CE.
But the squares on AF, FE are equal to
the square on AE, because the angle AFE
is a right angle (I. 47).
For the like reason the squares on CG,
GE are equal to the square on CE ;
Therefore the squares on AF, FE
are equal to the squares on CG, GE
(Ax. 1). _
But the square on AF is equal to the square on CG,
because AF is equal to CG;
Therefore the remaining square on FE is equal to the
remaining square on GE (Ax. 3) ;
.•.EF=EG. And therefore the straight line EF is equal to the straight
line EG.
But straight lines in a circle are said to be equally distant
from the centre, when the perpendiculars drawn to them
from the centre are equal (III. Def. 4) ;
Therefore AB, CD are equally distant from the centre.
Conversely, let the straight lines AB, CD be equally dis-
tant from the centre, that is, let EF be equal to EG;
TTcre Then AB shall be equal to CD.
and^ ' Proof. — The same construction being made, it may be
-^^'^+.EF2 demonstrated, as before, that AB is double AF, and CD
EG-'. " double of CG, and that the squares on EF, FA are equal to
the squares on EG, GC.
But the square on EF is equal to the square on EG,
because EF is equal to EG (Hyp.);
Therefore the remaining square on FA is equal to the
remaining square on GC (Ax. 3),
. •. AF = And therefore the straight line AF is equal to the straight
CG, &c. Ym^ CG.
But AB was shown to be double of AF, and CD double
of CG;
Therefore AB is equal to CD (Ax. 6);
Therefore, equal straight lines, <fec. Q.E.D.
PROPOSITIONS.
269
Proposition 15.— Theorem.
The diameter is the greatest straight line in a circle; and,
of all others, that which is nearer to the centre is always greater
than one more remote; and the greater is nearer to the centre
than the less.
Let ABCD be a circle, of which AD is a diameter, and E
the centre ; and let BC be nearer to the centre than FG.
Then AD shall be greater than any straight line CB,
which is not a diameter ; and BC shall be greater than FG.
Construction. — From the centre E draw EH FK per-
pendiculars to BC, FG (I. 12), and join
EB, EC, EF.
Proof. — Because AE is equal to BE,
and ED to EC,
Therefore AD is equal to BE, EC.
But BE, EC are greater than BC
(1.20);
Therefore also AD is greater than BC.
And because BC is nearer to the centre
(Hyp.), EH is less than EK (III. Def. 5).
But, as was demonstrated in the preceding proposition, ^ .
BC is double of BH, and FG double of FK, and the squares eh--j+ HTi2
on EH, HB are equal to the squares on EK, KF. kf^^^ ^
But the square on EH is less than the square on EK,
because EH is less than EK ;
Therefore the square on HB is greater than the square on
KF, and the straight line BH greater than FK; hb>fk.
And therefore BC is greater than FG.
Conversely, let BC be gi-eater than FG.
Then BC shall be nearer to the centre than FG, that is,
the same construction being made, EH shall be less than EK.
Proof. — Because BC is gi-eater than FG, BH is greater
than FK.
But the squares on BH, HE are equal to the squares on
FK, KE;
And the square on BH is gi-eater than the square on FK,
because BH is gi-eater than FK;
Therefore the square on HE is less than the square on
KE, and the straight Jine EH less thau EK;
270
GE03IETRY.
And therefore BC is nearer to the centre than FG (III,
def. 5).
Therefore, the diameter, &c. Q.E.D,
Take any
i)oint F in
AE,
then
DF > DA
and . •. >
DC.
Draw DH
at right
angles to
HG, then
DH < DA,
and .'. <
DK.
Proposition 16.— Theorem.
The straight line drawn at right angles to the diameter of a
circle, from the extremity of it, falls ivithout the circle; and a
straight line, mahing an acute angle ivith the diameter at its-
extremity, cuts the circle.
Let ABC be a circle, of which D is the centre, and AB a
diameter, and AE a line drawn from A perpendicular to
AB.
The straight line AE shall fall without the circle.
Construction. — In AE take any point F; join DF, and
let DF meet the circle in C.
Proof. — Because DAF is a right angle,
it is greater than the angle AFl3 (I. 17);
Therefore DF is greater than DA
(I. 19).
But DA is equal to DC; therefore DF
is greater than DC.
Therefore the point F is without the
circle.
In the same manner it may be shown that any other point
in AE, except the point A, is without the circle.
Therefore AE falls without the circle.
Again, let AG make with the diameter the angle DAG
less than a right angle.
The line AG shall fall within the circle, and hence cut it.
Construction. — From D draw DH at
right angles to AG, and meeting the cir-
cumference in K (I. 12).
Proof. — Because DHA is a right angle,
and DAH less than a right angle;
Therefore the side DH is less than the
side DA (I. 19).
But DK is equal to DA; therefore DH
is less than DK.
Therefore the point H is within the circle.
PROPOSITIONS. 271
Therefore tlie straight line AG cuts the circle.
Therefore, the straight line, <fec. Q.E.D.
Corollary. — From this it is manifest that the straight
line which is drawn at right angles to the diameter of a
circle, from the extremity of it, touches the circle (III. Def.
2) ; and that it touches it only in one point, because if it did
meet the circle in two points it would fall within it (III. 2).
Also it is evident that there can be but one stmight line
which touches the circle in the same point.
Proposition 17.— Problem.
To draw a straight line from a given point, either loithoiit
or in the circumference, which shall touch a given circle.
First, let the given point A be without the given circle
BCD.
It is required to draw from A a straight line which sliall
touch the given circle.
Construction. — Find the centre E of the circle (III. 1),
and join AE.
From the centre E, at the distance
EA, describe the circle AFC
From the point D draw DF at right
angles to EA (I. 11), and join EBF
and AB.
Then AB shall touch tJie circle BCD.
Proof. — Because E is the centre of
the circles AFG, BCD, EA is equal to ^ -^ EA ed re-
EF, and ED to EB; ^ "i^^F^'^liB
Therefore the two sides AE, EB are equal to the two sides ^id ^' e
FE, ED, each to each; common;
And the angle at E is common to the two triangles AEB,
FED;
Therefore the base AB is equal to the base FD, and the
triangle AEB to the triangle FED, and the other angles to
the other angles, each to each, to which the equal sides are
opposite (I. 4); .^ ^ y^i3P^
Therefore the angle ABE is equal to the angle FDE. %ith^'^^^
But the angle FDE is a right angle (Const.); angL*
272
GEOMETRY.
.'. AB
touches
the circle.
Therefore the angle ABE is a right angle (Ax. 1).
And EB is drawn from the centre (Const.)
But the straight line drawn at right angles to a diametei*
of a circle, from the extremity of it, touches the circle (III.
16, cor.);
Therefore AB touches the circle, and it is drawn from the
given point A.
Next, let the given point be in the circumference of tlie
circle, at the point D.
Draw DE to the centre E, and DF at right angles to DE ;
Then DF touches the circle (III. 16, cor.)
Therefore, from the given points A and D, straight lines,
AB and DF, have been drawn, touching the given circle
BCD. Q.E.F,
If not, sup-
pose FG
l)erpendi-
cular.
Proposition 18.—Theorem.
If a straight line touch a circle, the straight line drawn
from the centre to the point of co7itact shall be ^:)er/)e?icZicw^r
to the line touching the circle.
Let the straight line DE touch the circle ABC in the
point C; take the centre F (III. 1), and draw the straight
line FC.
FC shall be perpendicular to DE.
Construction. — For, if not, let FG be d^rawn from the
A point F perpendicular to DE, meeting
the circumference in B.
Proof. — Because FGC is a right angle
(Hyp.), FCG is an acute angle (I. 17),
and to the greater angle the greater side
is opposite (I. 19);
Therefore FC is greater than FG.
Then must " ' ~^^"^ But FC is equal to FB; therefore FB
iB>rG. is greater than FG; the part greater than the whole, which
is impossible.
Therefore FG is not per2)endicular to DE.
In the same manner it may be shown that no other straight
line from F is perpendicular to DE, but FC; therefore FC
is perpendicular to DE.
Therefore, if a straight line, ike. Q.EM,
PROPOSITIONS.
273
take F the
centre, out
of the line.
Proposition 19. — Theorem.
If a straight line touch a circle, and from the 2yoint of con--
tact a straight line he drawn at right angles to the touching
line, the centre of the circle shall he in that line.
Let the straight line DE toucli the circle ABC in C, and
from C let CA be drawn at right angles to DE.
The centre of the circle shall be in CA.
Construction. — For, if not, if possible, let F be tlie centre, if not.
and join CF.
Proof. — Because DE touches the circle
ABC, and FC is drawn from the assumed
centre to the point of contact.
Therefore FC is perpendicular to DE
(III. 18);
Therefore FCE is a right angle. P -q-
But the angle ACE is also a right angle (Const.);
Therefore the angle FCE is equal to the angle ACE; the Then ^
less to the gi^eater, which is impossible. z fce ~
Therefore F is not the centre of the circle ABC. aif-bs"^'^^^
In the same manner it may be shown that no other point
which is not in CA is the centre; therefore the centre is in
CA.
Therefore, if a stmight line, &c. Q.E.D,
Proposition 20.— Theorem.
The angle at the centre of a circle is douhle of the angle at
the circumference, upon the same hase, that is, upo7i the same
arc.
Let ABC be a circle, and BEC an angle
at the centre, and BAC an angle at the cir-
cumference, which have the same arc BC
for theii' base.
The angle BEC shall be double of the
angle BAC.
Case I. — First, let the centre E of tlie
circle be within the angle BAC.
Construction.— Join AE, and produce it to the
ference ii^ F,
5 9
cu'cuxa-
274
GEOMETRY.
1 BEF =
/. EAB +
Z EBA
= 2zEAB.
r-nd so z
FEC =
2 z EAC.
.'. ZBEC
5^:2 z BAG.
.*. taking
the differ-
ence
z BEC =
2 Z BAG,
z BFD =
2 z BAD,
Proof. — Because EA is equal to EB, the angle EAB is
equal to the angle EBA (I. 5) ;
Therefore the angles EAB, EBA are double of the angle
EAB.
But the angle BEE is equal to the angles EAB, EBA
(I. 32);
Therefore the angle BEE is double of the angle EAB.
For the same reason the angle EEC is double of the angle
EAC.
Therefore the whole angle BEC is double of the whole
angle BAG.
Case II. — Next, let the centre E of the
circle be without the angle BAC.
Construction. — Join AE, and produce it
to meet the circumference in F.
Proof. — It may be demonstrated, as in
the first case, that the angle EEC is double
of the angle EAC, and that FEB, a part of
EEC, is double of FAB, a part of FAC;
Therefore the remaining angle BEC is double of the re-
maining angle BAC,
Thei'efore, the angle at the centre, &c. Q,E.D,
Eroposition 21.— Theorem.
/■
The angles in the same segment of a circle are equal to one
another.
Let ABCD be a circle, and BAD, BED angles in the same
segment BAED.
The angles BAD, BED shall be equal to one another.
Case I. — First, let the segment BAED
be greater than a semicircle.
Construction. — Take F, the centre of the
circle ABCD (III. 1), and join BE, DF.
Proof. — Because the angle BED is at the
Yh centre, and the angle BAD at the circum-
ference, and that they have the same arc for
their base, namely, BCD;
Therefore the angle BED is double of the angle BAD
j(III. 20).
PROPOSITIONS.
275
For the same reason, the angle BFD is double of the angle and
BED;
Therefore the angle BAD is equal to the
angle- BED (Ax. 7). B/^
Case II. — Next, let the segment BAED ,
be not greater than a semicircle.
Construction. — Draw AF to the centre,
and produce it to C, and join CE.
Proof. — Then the segment BADC is
greater than a semicircle, and therefore the angles
BEC in it are equal by the first case.
For the same reason, because the segment CBED is greater and
than a semicircle, the angles CAD, CED are equal. j ced.~
Therefore the whole angle BAD is equal to the whole .-. z bad
•angle BED (Ax. 2).
Therefore, the angles in the same segment, <fec. Q.E.D.
= z BED.
Proposition 22.— Theorem.
TJie ojyposite angles of any quadrilateral figure inscribed in
a circle are together eqvAxl to two right angles.
Let ABCD be a quadrilateral figure inscribed in the circle
ABCD.
Any two of its opposite angles shall be together equal to
two right angles.
Construction. — Join AC, BD.
Proof. — Because the three angles of
every triangle are together equal to two
right angles (I. 32),
The three angles of the triangle CAB,
namely, CAB, ACB, ABC, are together jij^
equal to two right angles.
But the angle CAB is equal to the angle
CDB, because they are in the same seg-
ment CDAB (III. 21);
And the angle ACB is equal to the angle ADB, because
they are in the same segment ADCB ;
Therefore the two angles CAB, ACB are together equal to
the whole angle ADC (Ax. 2).
Z CAB +
z ACB -f-
z ABC =
2 right
angles.
The first
two toge-
ther =
z CDB +
z ADB =
Z ADC.
276
GEOMETRY.
.-. L ADC
+ /. ABC
= 2 right
angles.
To each of these equals add the angle ABC ;
Therefore the three angles CAB, ACB, ABC are equal to
the two angles ABC, ADC.
But the angles CAB, ACB, ABC are together equal to
two right angles (I. 32);
Therefore also the angles ABC, ADC are together equal
to two right angles.
In like manner it may be shown that the angles BAD,
BCD are together equal to two right angles.
Therefore, the opposite angles, &c. Q,E,D,
Proposition 23.— Theorem.
If possible,
Upon the same straight line, and on the same side of ity there
cannot he two similar segments of circles not coinciding with
one another.
If it be possible, on the same straight line AB, and on
the same side of it, let there be two similar segments of
circles ACB, ADB not coinciding with
one another.
Construction. — Then, because the circle
ACB cuts the circle ADB in the two points
A, B, they cannot cut one another in any
other point (III. 10);
Therefore one of the segments must fall
within the other.
Let ACB fall within ADB; draw the straight line BCD,
and join AC, AD.
Proof. — Because the segment ACB is similar to the seg-
ment ADB (Hyp.), and that similar segments of circles
contain equal angles (III. Def 11);
Therefore the angle ACB is equal to the angle
ADB; that is, the exterior angle of the triangle ACD,
and oppo- equal to the interior and opposite angle; which is impos-
'/.ADC. sible (I. 16).
Therefore, there cannot be two similar segments of circles
on the same straight line, and on the same sjde of it, whiqh
flo not coincide. Q.JS,J)^
exterior
z ACB =
interior
PROPOSITIONS.
277
Propositioii 24. — Theorem. i
Similar segments of circles upon equal straight lines are I
equal to one another. '
Let AEB, CFD be similar segments of circles upon the
equal straight lines AB, CD.
The segment AEB shall P They arc
be equal to the segment ^ ^ ^ "'^ because
CFD. X N. / _\ they must
Proof. — For if the seg-A B c i> Prop.* 23.^
ment AEB be applied to the
segment CFD, so that the point A may be on the point C,
and the straight line AB on the straight line CD,
Then the point B shall coincide with the point D, because
AB is equal to CD.
And the straight line AB coinciding with CD, the seg-
ment AEB must coincide with the segment CFD (III. 23);
and is therefore equal to it.
Therefore, similar segments, &c. Q.E.D,
Proposition 25.— Problem.
A segment of a circle being given, to describe the circle of
which it is the segment.
Let ABC be the given segment of a circle. 1
It is required to describe the cii'cle of which ABC is a
segment.
Construction. — Bisect AC in D (I. 10).
From the point D draw DB at right angles to AC (I. 11),
and join AB.
Case I. — First, let the angles ABD, BAD be equal to
one another.
Then D shall be the centre of the circle required.
278
GEOMETRY.
When
zABD =
/BAD,
DA = DB
and
.-. Dthc
centre.
Make
z BAE:
A ABD.
/. EA =
EB
and EA =
EC.
.-. EA =
EE = EC,
and there-
fore E is
the centre.
Proof. — Because the angle ABD is equal to the angle
BAD (Hyp.);
Therefore DB is equal to DA (I. 6).
But DA is equal to DC (Const.);
Therefore DB is equal lo DC (Ax. 1).
Therefore the three straight lines DA, DB, DC are all
equal ;
And therefore D is the centre of the circle (III. 9).
Hence, if from the centre D, at the distance of any of the
three lines, DA, DB, DC a circle be described, it will pass
through the other two points, and be the circle required.
Case II. — Next, let the angles ABD, BAD be not equal
to one another.
Construction. — At the point A, in the straight line AB,
make the angle BAE equal to the angle ABD (I. 23);
Produce BD, if necessary, to E, and join EC.
Then E shall he the centre of the circle required.
Proof. — Because the angle BAE is equal to the angle
ABE (Const.), EA is equal to EB (I. 6).
A-nd because AD is equal to CD (Const.), and DE is com-
mon to the two triangles ADE, CDE,
The two sides AD, DE are equal to the two sides CD,
DE, each to each;
And the angle ADE is equal to the angle CDE, for each
of them is a right angle (Const.);
Therefore the base EA is equal to the base EC (I. 4).
But EA was shown to be equal to EB;
Therefore EB is equal to EC (Ax. 1).
Therefore the three straight lines EA, EB, EC are all
equal;
And therefore E is the centre of the circle (III. 9).
Hence, if from the centre E, at the distance of any of the
three lines EA, EB, EC, a circle be described, it will pass
through the other two points, and be the circle required.
In the^r^^ case, it is evident that, because the centre D
is in AC, the segment ABC is a semicircle.
In the second case, if the angle ABD be greater than
BAD, the centre E falls without the segment ABC, which is
therefore less than a semicircle;
But if the angle ABD be less than the angle BAD, the
PROPOSITIONS. 279
centre E falls within the segment ABC, which is therefore
gi-eater than a semicircle.
Therefore, a segment of a circle being given, the circle has
been described of which it is a segment. Q.E.F,
Proposition 26.— Theorem.
In equal circles, equal angles stand upon equal arcs, ivhether
they he at the centres or at the circumferences.
Let ABC, DEF be equal circles, having the equal angles
BGC, EHF at their centres, and BAC, EDF at theii' cir-
cumferences.
The arc BKC shall be equal to the arc ELF.
Construction. — Join BC, EF.
Proof. — Because the circles ABC, DEF are equal (Hyp.),
the straight lines from their centres are equal (III. def. 1);
Therefore the two sides BG, GC are equal to the two sides Trianglec
EH, HF, each to each; l^^'^^l^
And the angle at G is equal to the angle at H (Hyp.); equal in
Therefore the base BC is equal to the base EF (I. 4). spect. ^^'
And because the angle at A is equal to the angle at D
(Hyp.),
A ^ ^ .4)
The segment BAC is similar to the segment EDF (III. .-. ses-
^^^^' ^^/, . . and EDF
And they are on equal straight lines BC, EF. are similar
But similar segments of circles on equal straight lines are equal
equal to one another (III. 24); fm^^^^
Therefore the segment BAC is equal to the segment EDF. .-. aVo
But the whole circle ABC is equal to the whole circle ®^"*'*
DEF (Hyp.);
♦ Therefore the remaining segment BKC is equal to the
remaining segment ELF (Ax« 3)»
280
GEOMETRY.
.*. arc
BKC =
arc ELF.
Therefore the arc BKC is equal to the arc ELF.
Therefore, in equal circles, &c. Q.E,D,
Proposition 27. — Theorem.
In equal circles, the angles which stand upon equal arcs are
equal to one another, whether they be at the centres or at the
circumferences.
Let ABC, DEF be equal circles, and let the angles BGC,
EHF, at their centres, and the angles BAC, EDF, at their
circumferences, stand on equal arcs BC, EF.
The angle BGC shall be equal to the angle EHF, and the
angle BAC equal to the angle EDF.
If one z
irreater
than the
other, the
corre-
sponding
arc is
greater.
Construction. — If the angle BGC be equal to the angle
EHF, it is manifest that the angle BAC is also equal to the
angle EDF (III. 20, ax. 7).
But, if not, one of them must be the greater. Let BGC
is be the greater, and at the point G, in the straight line
BG, make the angle BGK equal to the angle EHF
(I. 23).
Proof. — Because the angle BGK is equal to the angle
EHF, and that in equal circles equal angles stand on equal
arcs, when they are at the centres (III. 26);
Therefore the arc BK is equal to the arc EF.
But the arc EF is equal to the arc BC (Hyp.) ;
Therefore the arc BK is equal to the arc BC (Ax. 1); the
less to the greater, which is impossible.
Therefore the angle BGC is not unequal to the angle
EHF; that is, it is equal to it.
And the angle at A is half of the angle BGC, and the
angle at D is half of the angle EHF (III, 20);
PROPOSITIONS. 281
Therefore the angle at A is equal to the angle at D
(Ax. 7).
Therefore, in equal circles, &c. Q.E.D,
Proposition 28. — Theorem.
In eqiuil circles, equal chords cut off equal arcs, t/te greater
equal to the greater, and the less equal to the less.
Let ABC, DEF be equal circles, and BC, EF equal chords
in them, which cut off the two greater arcs BAG, EDF, and
the two less arcs BGC, EHF.
The greater arc BAG shall be equal to the grcate:*
arc EDF, and the less arc BGG equal to the less arc
EHF.
GoNSTRUCTiON. — Take K, L, the centres of the circljs TakeK
(III. 1), and join BK, KG, EL, LF. centJis.^"
Proof. — Because the circles ABG, DEF are equal, their
radii are equal (III. def. 1).
Therefore the two sides BK, KG are equal to the two
sides EL, LF, each to each;
And the base BG is equal to the base EF (Hyp.) ;
Therefore the angle BKG is equal to the angle Trianpics
ELF (I. 8). '^::^
But in equal circles equal angles stand on equal arcs, equal m
GV6rv rC"
when they are at the centres (III. 26) ; spect.
Therefore the arc BGG is equal to the arc EHF.
But the whole circle ABG is equal to the whole circle
DEF (Hyp.);
Therefore the remaining arc BAG is equal to the remain-
ing arc EDF (Ax. 3).
Therefore, in equal circles, (fcc. Q.KD*
282 GEOMETRY.
Proposition 29. — Theorem.
In equal circles equal arcs are subtended by equal chords.
Let ABC, DEF be equal circles, and let BGC, EHF be
equal arcs in them, and join BC, EF.
The chord BC shall be equal to the chord EF.
TakeK CONSTRUCTION. — Take K, L, the centres of the circles
cluteef' (III. l),andjoinBK, KC,EL,LF.
Then pROOF. — Because the arc BGC is equal to the arc
zelf" EHF (Hyp.), the angle BKC is equal to the angle ELF
' (in. 27).
And because the circles ABC, BEF are equal (Hyp.),
their radii are equal (III. def. 1).
Therefore the two sides BK, KC are equal to the two
sides EL, LF, each to each; and they contain equal angles;
and so base Therefore the base BC is equal to the base EF (I. 4).
"" - ^-1 Therefore, in equal circles, &c. Q.E.D,
BC
EF.
Proposition 30.— Problem.
To bisect a given arc^ that is, to divide it into two equal
parts.
Let ABB be the given arc.
It is required to bisect it.
Construction. — Join AB, and bisect it in C (I. 10).
From the point C draw CD at right angles to AB (I. 11),
and join AD and DB.
Then the arc ABD shall be bisected in the
/;;X^ XX Proof. — Because AC is equal to CB
»^6..o.vxvG ^— ^ -^ (Const.), and CD is common to the two tri-
andCDB, A c « angles ACD, BCD;
In the tri-
anglesADG
PROPOSITIONS. !^83
The two sides AC, CD are equal to the two sides BC, CD,
each to each ;
And the angle ACD is equal to the angle BCD, because
each of them is a right angle (Const.);
Therefore the base AD is equal to the base BD (I. 4). lasem"
But equal chords cut off equal arcs, the greater equal to
the greater, and the less equal to the less (III. 28) ;
And each of the arcs AD, DB is less than a semicircle,
because DC, if produced, is a diameter (III. 1, cor.);
Therefore the arc AD is equal to the arc DB.
Therefore, the given arc is bisected in D. Q.E.F,
Proposition 31. — Theorem.
In a circle^ the angle in a semicircle is a right angle; hut
the angle in a seginent greater than a semicircle is less than a
right angle; and the angle in a segment less than a semicircle
is greater than a right angle*
Let ABC be a circle, of which BC is a diameter, and E
the centre ; and draw C A, dividing the circle into the seg-
ments ABC, ADC, and join BA, AD, DC.
The angle in the semicircle BAC shall be a right angle;
The angle in the segment ABC, which is greater than a
cemicii'cle, shall be less than a right angle; .
The angle in the segment ADC, which is less than a semi-
circle, shall be greater than a right angle.
Construction. — Join AE, and produce BA to F.
Proof. — Because EA is equal to EB (I. Def 15),
The angle EAB is equal to the angle
EBA(I. 5);
And, because EA is equal to EC,
The angle EAC is equal to the angle
EC A;
Therefore the whole angle BAC is equal 1/ \ N\\ ^ bae +
to the two angles ABC, ACB (Ax. 2). \ e y ol- '
But F AC, the exterior angle of the tri- V / z abc T
angle ABC, is equal to the two angles x^ ^ i acb =
ABC, ACB (I. 32). al./'f.
Therefore the angle BAC is equal to the angle FAC "ehta'is'o-
(Ax. 1),
-<. a right
aujrle.
284 GEOMETRY.
And therefore each of them is a right angle (I. Def. 10) ;
Therefore the angle in a semicircle BAG is a right angle.
And because the two angles ABC, BAG, of the triangle
ABG, are together less than two right angles (I. 17), and
that BAG has been shown to be a right angle;
z ABC Therefore the angle ABG is less than a right angle.
Therefore the angle in a segment ABG, greater than a
semicircle, is less than a right angle.
And, because ABGD is a quadrilateral figure in a circle,
any two of its opposite angles are together equal to two right
angles (III. 22);
Therefore the angles ABG, ADG are together equal to two
right angles.
But the angle ABG has been shown to be less than a right
angle ;
Hence Therefore the angle ADG is greater than a right angle;
z ADC > Therefore the ansjle in a segment ADG, less than a semi-
R right . , . , ® -IT
angle, by Circle, IS greater than a right angle.
Prop. 32. Therefore, the angle, &c. Q.E.B.
GoROLLARY. — From this demonstration it is manifest that,
if one angle of a triangle be equal to the other two, it is a
right angle.
JFor the angle adjacent to it is equal to the same two
angles (I. 32).
And, when the adjacent angles are equal, they are right
angles (I. def. 10).
Proposition 32. — Theorem.
The angles contained hy a tangent to a circle and a chord
drawn from the point of contact are eqvxil to the angles in the
alternate segments of the circle.
Let EF be a tangent to the circle ABGD, and BD a chord
drawn from the point of contact B, cutting the circle.
The angles which BD makes with the tangent EF shall
be equal to the angles in the alternate segments of the circle;
That is, the angle DBF shall be equal to the angle in the
Begment BAD, and the angle DBE shall be equal to the
angle in the segment BGD.
PROPOSITIONS.
285
Construction. — From the point B draw BA at riglit
angles to EF (I. 11).
Take any point C in the circumference
BD, and join AD, DC, CB.
Proof. — Because the straight line EF
touches the circle ABCD at the point B
(Hyp.), and BA is drawn at right angles
*bo the tangent from the point of contact
B (Const.),
The centre of the circle is in BA (III.
19).
Therefore the angle ADB, being in a semicircle, is a right
Angle (III. 31).
Therefore the other two angles BAD, ABD are equal to a
right angle (I. 32).
But ABF is also a right angle (Const.);
Therefore the angle ABF is equal to the angles BAD, ABD.
From each of these equals take away the common angle ABD ;
Therefore the remaining angle DBF is equal to the re-
maining angle BAD, which is in the alternate segment of
the circle (Ax. 3).
And because ABCD is a quadrilateral figure in a cii^cle,
the opposite angles BAD, BCD are together equal to two
right angles (III. 22).
But the angles DBF, DBE are together equal to two
right angles (I. 13);
Therefore the angles DBF, DBE are together equal to
the angles BAD, BCD.
And the angle DBF has been shown equal to the angle
BAD;
Therefore the remaining angle DBE is equal to the angle
BCD, which is in the alternate segment of the circle (Ax. 3).
Therefore, the angles, &c. Q.E.D.
The centre
is in BA.
. •. ADD is
a rijjljt
angle,
and
z BAD-f
Z ABD =
a rijfht
angle
= ABr.
.-. z BAD
= z Dcr.
Also,
Z BeD +
Z BAD^
2 right
angles
= z DBF
+ z DBE.
.-. Z DBE
= z BCD.
Proposition 33.— Problem.
Upon a given straight line to describe a segment of a circlcy
containing an angle equal to a given rectilineal angle.
Let AB be the given straight Jine, and C the given recti-
Juieal angle,
2SG
GEOMETRY,
It is required to describe, on tlio given straiglit line AB, a
segment of a circle, containing an angle equal to tlieangleC.
Case I. — Let the angle C be a right
angle.
Construction. — Bisect AB in F
(I. 10).
From the centre F, at the distance
FB, describe the semicircle AHB.
Then AHB shall he the segment required.
Proof. — Because AHB is a semicircle, the angle AHB
in it is a right angle, and therefore equal to the angle C
(HI. 31).
Case II. — Let C be not a right angle.
Construction. — At the point A, in the straight line AB,
BAD =c; make the angle BAD equal to the angle C (I. 23).
H
Angle in a
semicircle
is a right
angle.
At point A
make z
and draw
AE at right
angles to
AD.
From F,
middle of
AB, draw
perpendi-
cular,
meeting
AK iu G.
Then
G is the
centre of a
circle pass-
ing
throngh A
and B,
From the point A draw AE at right angles to AD (I. 11).
Bisect AB in F (L 10).
From the point F draw FG at right angles to AB (I. 11),
and join GB.
Because AF is equal to BF (Const.), and FG is common
to the two triangles AFG, BFG;
The two sides AF, FG are equal to the two sides BF, FG,
each to each ;
And the angle AFG is equal to the angle BFG (Const.);
Therefore the base AG is eqvial to the base BG (I. 4).
And the circle described from the centre G, at the dis-
tance GA, will therefore pass through the point B.
Let this circle be described; and let it be AHB.
The segment AHB slwdl contain an angle equal to the given
Q'ectUineal angle C,
ruoposiTioNs. 287
Proof. — Because from the point A, the extremity of the And ad
iiameter AE, AD is drawn at right angles to AE (Const.); the drcie,
Therefore AD touches the circle (III. 16, cor.)
Because AB is drawn from the point of contact A, the
angle DAB is equal to the angle in the alternate segment
AHB (III. 32).
But the angle DAB is equal to the angle C (Const.) ; and /. z
Therefore the angle in the segment AHB is equal to the ^^ ^^^ =
angle C (Ax. 1), c.
Therefore, on the given straight line AB, the segment
AHB of a circle has been described, containing an angle
equal to the given angle C. Q.E,F,
Proposition 34. — Problem.
Froini a given circle to cut off a segment which shall contain
an angle equal to a given rectilineal angle.
Let ABC be the given circle, and D the given rectilineal
angle.
It is required to cut off from the circle ABC a segment
that shall contain an angle equal to the angle D.
CoNSTRUCTiox. — Draw the straight line EF touching the Draw tan-
circle ABC in the point B (III. ^ ^ ^'^"^ ^^^^'
17);
And at the point B, in the
straight line BF, make the angle
FBC equal to the angle D (I. 23).
Then the segment BAC slmll
contain an angle equal to the
given angle D.
Proof. — Because the straight line EF touches the circle
ABC, and BC is drawn from the point of contact B (Const.);
Therefore the angle FBC is equal to the angle in the alter-
nate segment BAC of the circle (III. 32).
But the angle FBC is equal to the angle D (Const.);
Therefore the angle in the segment BAC is equal to the •*. ^ b^
angle D (Ax. 1). = ^ ^^^
Therefore, from the given circle ABC, the segment BAC
has been cut off, containing an angle equal to the given
angle D. Q.E.F,
288
GEOMETRY,
AE = EC.
BE ED +
EF2 r=
AE-i +
.'. BEED
= AE2 =
AEEC.
Proposition 35. — Theorem.
If two straight lines within a circle cut one another^ the
rectangle contained by the segments of one of them shall be
equal to the rectangle contained by the segments of the other.
Let the two straight lines AC, BD cut one another in the
point E, within the circle ABCD.
The rectangle contained by AE and EC shall be equal to
the rectangle contained by BE and ED.
A<^ ^ Case I. — Let AC, BD pass each of them
through the centre.
Proof. — Because E is the centre, EA, EB,
' EC, ED are all equal (L def. 15);
Therefore the rectangle AE, EC is equal
to the rectangle BE, ED.
Case II. — Let one of them, BD, pass through the centre,
and cut the other, AC, which does not pass through the
centre, at right angles, in the point E.
Construction. — Bisect BD in E, then
F is the centre of the circle; join AF.
Proof. — Because BD, which passes
through the centre, cuts AC, which does
not pass through the centre, at right angles
in E (Hyp.);
Therefore AE is equal to EC (III. 3).
And because BD is cut into two equal
parts in the point F, and into two un-
equal parts in the point E,
The rectangle BE, ED, together with the square on EF,
is equal to the square on FB (11. 5); that is, the square on
AF.
But the square on AF is equal to the squares on AE, EF
(I. 47);
Therefore the rectangle BE, ED, together with the square
on EF, is equal to the squares on AE, EF (Ax. 1).
Take away the common square on EF;
Then the remaining rectangle BE, ED is equal to the
remaining square on AE; that is, to the rectangle AE, EQ,
^iucQ Ai; is e(|ual to EC,
PROPOSITIONS.
289
Case III. — Let BD, wliiclx passes tlirough the centre, cut
the other AC, which does not pass through
the centre, in the point E, but not at
right angles.
Construction. — Bisect BD in F, then
F is the centre of the circle. [ ^^^^>k
Join AF, and from F draw FG perpen- /j^ ^-^^^'^^^ I X e
dicular to AC (I. 12). \ >2
Proof. — Then AG is equal to GC
(III. 3).
Therefore the rectangle AE, EC, together with the square
on EG, is equal to the square on AG (II. 5).
To each of these equals add the square on GF;
Then the rectangle AE, EC, together witli the squares on
EG, GF, is equal to the squares on AG, GF (Ax. 2).
But the squares on EG, GF are equal to the square on EF;
And the squares on AG, GF are equal to the square on
AF (T. 47).
Therefore the rectangle AE, EC, together with the square
on EF, is equal to the square on AF; that is, the square on FB.
But the square on FB is equal to the rectangle BE, ED,
together with the square on EF (II. 5) ;
Therefore the rectangle AE, EC, together with the square
on EF, is equal to the rectangle BE, ED, together with the
square on EF.
Take away the common square on EF;
And the remaining rectangle AE, EC is equal to the
remaining rectangle BE, ED (Ax. 3).
Case TV. — Let neither of the straight
lines AC, BD pass through the centre.
Construction. — Take the centre F
(III. 1), and through E, the intei'section
of the lines AC, BD, draw the diameter a\
GEFH.
Proof. — Because the rectangle GE,
EH is equal, as has been shown, to the
rectangle AE, EC, and also to the rectangle BE, ED;
Therefore the rectangle AE, EC is equal to the rectangle
BE, ED (Ax. 1).
Therefore, if two straight lines, <fec. Q,E.D,
5 T
Bisect BD
in F the
centre.
Draw FG
at right
angles to
AC.
.'. AG =
GC.
Now,
AEEC +
EG-J
= AG2.
290
GEOMETKY.
.-. AD DC
+ EB2 =
BD2+EB2.
.-.AD DO
= BD2.
Draw EF
perpendi-
cular to
AC.
Proposition 36.— Theorem.
If from a point without a circle two straight Ihus he drawn,
one of which cuts the circle, and the other touches it; the rect-
angle contained by the whole line which cuts the circle, and the
part of it without the circle, shall he equal to the square on the
line which touches it.
Let D be any point without the circle ABC, and let DC A,
DB be two straight lines drawn from it, of which DC A cuts
the circle, and DB touches it.
The rectangle AD, DC shall be equal to the square on DB.
Case I. — Let DC A pass through the
centre E, and join EB.
Proof. — Then EBD is a right angle
(IIL 18),
And because the straight line AC is
bisected in E, and produced to D, the rect-
angle AD, DC, together with the square on
EC, is equal to the square on ED (II. 6).
But EC is equal to EB;
Therefore the rectangle AD, DC, together
with the square on EB, is equal to the square
on ED.
But the square on ED is equal to the squares on EB, BD,
because EBD is a right angle (L 47);
Therefore the rectangle AD, DC, together
with the square on EB, is equal to the
squares on EB, BD.
Take away the common square on EB ;
Then the remaining rectangle AD, DC
is equal to the square on DB (Ax. 3).
Case II.— Let DCA not pass through
the centre of the circle ABC.
Construction. — Take the centre E
(III. 1), and draw EF perpendicular to
AC (L 12), and join EB, EC, ED.
Proof. — Because the straight line EF,
which passes through the centre, cuts the straight line AC,
which does not pass through the centre, at right angles, it
also bisects it (III. 3) ;
PROPOSITIONS.
291
Therefore AF is equal to FC.
And because the straight line AC is bisected in F and
produced to D, the rectangle AD, DC, together with the
square on FC, is equal to the square on FD (II. 6).
To each of these equals add the square on FE ;
Therefore the rectangle AD, DC, together with the squares
on CF, FE, is equal to the squares on DF, FE (Ax. 2).
But the squares on CF, FE are equal to the square on CE,
because CFE is a right angle (I. 47) ;
And the squares on DF, FE are equal to the square on
DE;
Therefore the rectangle AD, DC, together with the square
on CE, is equal to the square on DE.
But CE is equal to BE;
Therefore the rectangle AD, DC, together with the square
on BE, is equal to the square on DE.
But the square on DE is equal to the squares on DB, BE,
because EBD is a nght angle (I. 47) ;
Therefore the rectangle AD, DC, together with the square
on BE, is equal to the squares on DB, BE.
Take away the common square on BE ;
Then the remaining rectangle AD, DC is equal to the
square on DB (Ax. 3).
Therefore, if from any point, &c. Q. E, D, ^
Corollary. — If from any point without
a circle there be drawn two straight lines
cutting it, as AB, AC, the rectangles con-
tained by the whole lines and the parts of
them without the circle, are equal to one
another; namely, the rectangle BA, AE
is equal to the rectangle CA, AF; for each
of them is equal to the square on the
straight line AD, which touches the circle.
.-. AF =
FC.
.♦. AD-DC
-f FC2 =
FD2.
.-. ADDC
+ EC2 =
DE2.
.-. AD-Db
+ BE2 =
DB2-fBE2
.-. ADDC
= DB2.
Proposition 37. — Theorem.
If from a 'point without a circle there he draion two straight
lines, one of which cuts the circle, and the other meets it, and if
the rectangle contained by tlie whole line which cuts the circle,
and the part of it without the circle, he equal to the sqiiare on
292
GEOMETRY.
Draw DE
touching
the circle.
Then
DE = DB.
And tri-
angles
DBF and
DEF are
equal in
everj' re-
spect.
/. DBF is
aright
angle ;
and there-
fore DB
touches
the cir ?le.
the line which meets the circle^ the line which meets the circle
shall touch it.
Let any point D be taken without the circle ABC, and
from it let two straight lines, DCA, DB, be drawn, of which
DC A cuts the circle, and DB meets it; and let the rectangle
AD, DC be equal to the square on DB.
Then DB shall touch the circle.
Construction. — Draw the straight line DE, touching the
circle ABC (III. 17);
Find F the centre (III. P and join FB,
FD, FE.
Proof. — Then FED is a right angle
(III. 18).
And because DE touches the circle ABC,
and DCA cuts it, the rectangle AD, DC
is equal to the square on DE (III. 36).
But the rectangle AD, DC is equal to
the square on DB (Hyp.);
Therefore the square on DE is equal to
the square on DB (Ax. 1);
Therefore the straight line DE is equal to the straight line
DB.
And EF is equal to BF (I. De£ 15);
Therefore the two sides DE, EF are equal to the two
sides DB, BF, each to each ;
And the base DF is common to the two triangles DEF,
DBF;
Therefore the angle DEF is equal to the angle DBF (I. 8).
But DEF is a right angle (Const.) ;
Therefore also DBF is a right angle (Ax. 1).
And BF, if produced, is a diameter; and the straight line
which is drawn at right angles to a diameter, from the ex-
tremity of it, touches the circle (III. 16, Cor.);
Therefore DB touches the circle ABC.
Tlierefore, if from a point, (fee. Q.E.B, *
EXERCISES ON THK PROPOSITIONS. 293
EXERCISES ON BOOK III.
Prop. 1—15.
1 . Two straight lines intersect. Describe a circle passing tlirdugli
the point of intersection and two other points, one in each straight
line.
2. If two circles cut each other, any two parallel straight lines
drawn through the points of section to cut the circumferences are
equal.
3. Show that the centre of a circle may be found by drawing per-
pendiculars from the middle points of any two chords.
4. Through a given point, which is not the centre, draw the least
line to meet the circumference of a given circle, whether the given
point be within or without the circle.
5. The sum of the squares of any two chords in a circle, together
with four times the sum of the squares of the perpendiculars on them
from the centre, is equal to twice the square of the diameter.
6. With a given radius, describe a circle passing through the
centre of a given circle and a point in its circumference.
7. If two chords of a circle are given in magnitude and position,
describe the circle.
8. Describe a circle which shall touch a given circle in a given
point, and shall also touch another given circle.
9. If, from any point in the diameter of a circle, straight lines be
drawn to the extremities of a parallel chord, the squares of these
lines are together equal to the squares of the segments into which
the diameter is divided.
10. If two circles touch each other externally, and parallel dia-
meters be drawn, the straight line joining extremities of these dia-
meters will pass through the point of contact.
11. Draw three circles of given radii touching each other.
12. If a circle of constant radius touch a given circle, it will
always touch the same concentric circle.
13. If a chord of constant length be inscribed in a circle, it will
always touch the same concentric circle.
14. The locus of the middle points of chords parallel to a given
straight line is a line drawn through the centre perpendicular to the
parallel chords.
Prop. 1G— 30.
15. Show that the two tangents from an external point are equal
in length.
IG. Draw a'tangent to a given circle, making a given angle with a
given straight line.
17. If ^ polygon having an even numoer of sides bo inscribed in a
circle, the sums of the alternate angles are equal.
294 GEOMETRY.
18. If such a polygon be described about a circle, the sums of the
alternate sides are each equal to half the perimeter of the polygon.
19. If a polygon be inscribed in a circle, the sum of the angles in
the segments exterior to the polygon, together with two right
angles, is equal to twice as many right angles as the polygon has
sides.
20. Draw the common tangents to two given circles.
21. From a given point draw a straight line cutting a given circle,
so that the intercepted segment of the line may have a given length.
22. The straight line which joins the extremities of equal arcs
towards the same parts are parallel.
23. Any parallelogram described about a circle is equilateral, and
any parallelogram inscribed in a circle is rectangular.
24. Two opposite sides of a quadrilateral circumscribing a circle
touch the circle at extremities of a diameter. Show that the area of
the quadrilateral is equal to one-half the rectangle contained by the
diameter, and the sum of the other sides.
Prop. 31—37.
25. A tangent is drawn to a circle of 21 inches diameter from a
point 17 '5 inches from the centre. Find the length of the tangent.
26. Show that a man 6 feet high, standing at the sea level, has a
view of 3 miles (approximately) in every direction, along a horizontal
plane passing through his eye.
27. The angle between a tangent to a circle and the chord through
the point of contact is equal to half the angle which the chord sub-
tends at the centre.
28. From a given point P, within or without a circle, draw a
straight line cutting the circle in A and B such that PA shall be
three-fourths of PB.
Ex. Let the circle be of 1*5 inches radius, and point P 3*5 inches
from its centre. Prove your construction by scale.
29. The greatest rectangle which can be inscribed in a circle is a
square whose area is equal to half that of the square described upon
the diameter as side.
30. If the base and vertical angle of a triangle remain constant in
magnitude while the sides vary, show that the locus of the middle
point of the base is a circle.
31. Given the vertical angle, the difference of the two sides con-
taining it, and the difference of the segments of the base made by a
perpendicular from the vertex, to construct the triangle.
32. Show that the locus of the middle point of a straight line,
which moves with its extremities upon two straight lines at right
angles to each other, is a circle.
33. Show how to produce a straight line, that the rectangle con-
tained by the given line, and the whole line thus produced, may be
equal to the square of the part produced.
Ex, If the length of the given line be 2 inches, show geometrically
that the length of the part in'oduced is (V^ + 1) inches.
EXERCISES ON THE PROPOSITIONS. 295
34. Given the height and chord of a segment of a circle to find the
I'adius of the circle.
Ex. If the chord be 24 inches, and the height of the segment be 4
inches, show that the radius of the circle is 20 inches.
35. Show that the locus of the middle points of chords which pass
through a fixed point is the circle described as diameter upon the
line joining the fixed point and the centre of the given circle.
36. Let ACDB be a semicircle whose diameter is AB, and AB, BC
any two chords intersecting in P; prove that
AB2 = DAAP + CBBP.
MATHEMATICS.
SECOND STAGE.
SECTION II.
ALGEBRA.
CHAPTEE I.
QUADRATIC EQUATIONS.
1. Equations of this class, when reduced to a rational in-
tegral form, contain the square of the unknown quantity, but
no higher powers.
When the equation contains the square only of the unknown
quantity, and not the first power, it is called a pure quadratic.
Thus, aj2 _ 25 - 0, 4ic2 ^ 10 = 19, 5aj2 := 180, are ;mre
quadratics. When the equation contains the square of the
unknown quantity, as well as the first j^ower, it is called an
adjected quadratic.
Thus, ^ - 5aj = 6,cc2-aj--30 = 0, 2a;2 + aj+3 = 6,
are adfected quadratics.
Pure Quadratics.
2. To solve these, we treat them exactly as we do simple
equations, until we obtain the value of the square of the un-
known quantity ; then, taking the square root of each side,
we obtain the value of the unknown quantity. It will be
ADFECTED QUADRATICS. 297
seen (Stage I., Alg. ; Art. 35) that the unknown quantity in a
pure quadratic has always two vahies, equal in magnitude,
but of op2)osit6 sign.
Ex. 1. Given 3 a;- + 12 = 687, find x.
We have 3a;^ = 687 - 12 == 675, or a;- = 225.
.-. X = ±15.
2a; + 7 2a; - 7 56 ^ ,
Ky 2 Given ~— • - — » find x,
1.x. ^. uiven 2.^2 __ 7^. 2ar^ + 7a; ~ 6a;- - 1)9'
Brinffino: the fractions on the first side to a common de-
nominator, we have —
(2 a; + 7)^ - (2 g ; - 7 )^ _ 56
a; (4 a;- - 49) ~ Qx' - 99'
56 a; 56
.^^''a;(4a;2 - 49) ~ 6 ar^ - 99'
^^' 4ar^ - 49 " 6 ar^ - 99 ^ clearing of fractions;
then, G or - 99 = 4 a;^ - 49, from which
a;2 = 25
a; = ± 5.
Adfected Quadratics.
3. Solution by completing tite square.
Suppose we have given the equation ar^ + 2 aa; = 3 a-, to
find a;.
It will be remembered that (a;- + 2 ax + rr) is a perfect
square, viz. , the square of {x + a). It is evident, then, that the
first side of our equation will become a perfect square by the
addition of cfc^ as a third term.
Adding, then, «- to each side of the given equation, we
have —
a;- + 2 aa; + a^ = 4 a-, or
{x + ay = i a\
Taking now the squai'o root of each side, we have—
X + a = ± 2 a.
.-.a; = ± 2 a — a.
= a OT - 3 a.
298 ALGEBRA.
"We may remark that the quantity c?^ added to the expres-
sion a? •\- 2 ax in order to make it a ijerfect square, is the
square of half the coefficient of x. The operation itself is
called comiiilethig the square.
An adfected quadratic may therefore be solved as
follows : —
1. Reduce and arrange it until all the terms involving x
are on the first side, and the coefiflcient of-s? is unity,
2. Complete the square by adding to each side the
SQUARE OF HALF the coefficient of x.
3. Take the square root of each side, put a double sign to
the second side, and transpose the term of the first side not
involving x.
Ex. 1. Solve the equation 3 aj^ + 18 a; + 4- = 52.
We have Z x^ + 18a5 = 52 - 4 =48; or, dividing each
side by 3, x^ + Q x ~ 16.
Here, the coefficient of x is 6, the half of which is 3.
Adding then the square of 3 to each side, to complete tht
square, we have —
a;2 + 6 a; + 32 = 16 + 9 = 25.
Taking the square root of each side, we have-^-
a; + 3 = ±5.
.*. a; = ± 5 - 3 = 2 or - 8*
Ex. 2. Given ?-±-^ . ^-±J; . i^-±4 _ ^A^-^.
a; + 4 a; + 2 2a; + 7 6 a; + 16
find X,
We have (l L.\ ^ A _ __i_ \ ^ A - ^-^^)
V a; + 4/ V^ a; + 2/ V 2a; + 7/
- (2 - -i"— V
V 6x + 16/^
. ,.« .11 15 5
or, smipliiyinff, — = —■ - -,:
V ^ -^ ^'a; + 2 a; + 4 6a; + 16 2 a; + 7'
(a; + 4) >■ (a; + 2) ^ 15(2a; + 7) - 5 (6 a; + 16) ,
^^'. (a; + 2) (a; + 4) (6 a; + 16) (2 a; + 7) '
'~~ . ■■ 2 25
or, simplifying, ^, ^ ^^ ^ g = 12^^74,^112 '
ADFECTED QUADRATICS. 299
or, clearing of fractions —
24 ar^ + 148 a; + 224 = 25x' + 150 aj + 200;
or, transposing and reducing, -a;--2a;= — 24;
or, changing the sign of each side, then —
x" + 2x = 24;
or, completing the square, ar^ + 2x + P = 24 + 1 = 25.
Taking the square root of each side, we have —
X + 1 = ±5
4aj=±5-l==4or-6.
Ex. 3. Solve the equation ar*+6a;+25 = 0.
We have o^ + 6 x = - 25;
or, completing the square —
af^ + 6a; + 32 = - 25 + 9 - - IG; '
or, extracting the square root of each side —
a.^ + 3 = ± V^n[6
:^x= - 3 ± j~:ri6.
As the quantity \/ - 16 has no exact or approximate
value, the given equation has no real roots. The roots are
therefore said to be imaginary.
4, Solution by breaking into factors.
"We have seen (Stage I., Alg., Art* 30) that it is often
easy to find by inspection the factors of quadratic expres-
sions. We may make use of this knowledge to solve quad-
i-atic equations.
Ex. 1. Solve the equation x^ + 5 a; = 66.
Transposing all to the firet Side, we have —
ar + 5 a; - 66 = Oi
And, resolving the first side into its element^i'y fiictors, we
get—
{x - 6) {x + 11) = 0.
Now, if either of these factors is put equal to 0, the
equation is satisfied.
Hence^ we have, a; - 6 = 0, and x + 11 = 0;
or, a; = 6, and a; = - IL
:. 6 &,nd - li are the roots required.
300 ALGEBRA*
Ex. 2. Given a;" - {a + h) x ■{■ ah - 0, find x.
We have {x - o) {x - h) - 0.
Now this equation is satisfied by making either of the
factors = 0. .
Hence, x - a - 0, ov x = a-A
and, ic - 6 = 0, or a; = 6 ; J
.*. ct, h are the roots required.
Ex. I.
2. 5^2 + 6 = 86.
3. :^^^^l-,;i-^ = 34. 4. foj^- +1 = 6.
5. 3aj - :^ = 0. 6. 2 (a^ + 6^) - ic^= (ex - h)\
' X x/24 + x" = 4 + ar^.
8. * \/aJ - c& = V ic - ^6^ + x^.
X ' ^ x'
10.
1
1
aa; - Ja^x^ - 1 aa3 +
Vav - 1 "'^•
11.
^a^ + a^ + ija? - a^
^a^ + ar^ - ^a' - x'
h.
12.
X - sjx" - I ^'
13. a;^ - 5 a; + 6 = 0.
14.
x^ - X = 72.
15. 3 a;- - a; = 2.
16.
4a;2 - 9a; = 28.
17. 6a;2 + a; - 35 = 0.
18.
a;2 + 65 a; + 8 = 0.
19. a3(? + 5a; + c = 0.
20.
"■-^^''''-'■
X
22.
^">-r«^. = "
•
* See Remark, page 304.
ADFECTED QUADRATICS. 301
23. {a + hx) {ex + c?) = (a + 6) (c + dx).
24. [a - h) (x^ - h') = (a + b) {x - bf,
25. {a + b) {x — a) {x - b) = abx,
o/> « , aJ b X a' - y^
JO. - + --=- + _ + — .
X X a ab
27. (io(r - (c + d)x = bx- - -^,.
a — b
28. a^^- - (« - 5)2 oj + a^ -- 6=^ar^ + {a? + a6 + b^fx + 51
29. -^-- - 31.
30 ^ + ^ + ^- = 2 a ; + 5
' 2 aj + 3 13 ~ i^~^'
31. ?J^±i H- ,_^ . 5i.
a; 2 a; + 9 ^
32. ^1-1 + ^' " ^^ = 0.
* a; + 3 a; + 8
33.
34.
35.
36.
37.
38.
3 a; - 7 5 a; + 3 ^ 5
4a;-2'^7a; + 4~9*
^x +^ 4a;-l_7aj+l
2'a;~-nr X - 2" ~ a; - 3 *
2a; + l_ a;+2 _ 7a; +8
a; + 2 4 a; + 4 4 a; + 13*
3a; _ 2 a; + 9 ^ 2 a; + 2 4
a;-2 a;+2"2a;-.r
3ar'-2a;+7 a? + 1 x - '^
6 a;^ - 4 a; + 11 2 a;'^ + 14 a; - 9*
3 a^ + 10 _ ar^ + 2 a; + 3 ^ 2ar' 4- 2a; + 10
X X + 2 a; + 1
39. '^' J5 a; + G + Va; + 10 = 4.
40. >/3a; - 4 - J2x - 4 = x/^
41. * Va^ + 6^ +^ + Jo" + 6- - a; = ^/iL^.
N 6 ■
* See Bemark, page 3H
302 ALGEBRA.
42. * sj'2 a + X ■{- isji a - X — 2 ijx - a.
43. '" sjax + 6 + sibx + a ~ s/(a - b)x + a + b + 2 Jab.
«■•J^"^-^/~^(^/^^/f)v.^.
45. C~^ - l\ . — ~-^ = /^ + iV ^ + ^ + c __ 2^
* \Z) /' b \6- /' ic
4G. (ft - h) (x + b - c) (x + c - a) + a-(x + b — c) +
b'\x + c - a) + c\x + a - b =^ 0),
47, Show tliat, if x be real, the vahie oi x + - cannot lie
x
between 2 and - 2.
X"
48. If, in the equation x - ^ = ct, the quantity x be real,
show that a cannot be greater than 5.
Equations which may be Solved like Quadratics.
5. Certain equations of a degree higher than the second
may be solved like quadratics. It will be seen that, although
it is impossible to lay down definite rules for the treatment
of every such equation, the object to be attained is either
(1.) To throw the equation into the form —
X' + pX = q,
when X is an expression containing the unknown quantity,
and solving when possible this equation for X; or (2.), To
strike out from each side a factor containing the unknown
quantity, thus reducing the dimensions of the equation,
and obtaining a value or values of the unknown quantity
by equating this factor to zero.
Ex. 1. Solve the equation a;'* + 144 = 25 x\
We have, transposing, £c'*-25a3^= -144;
or, {xy - 25(0^) = - 144;
* 3^9 Kemark, page 304,
EQUATIONS WHICH MAY DE SOLVED LIKE QUADRATICS. 303
or, completing the square —
• ^ 2 5 — +7.^
.-. x" = %^ ± I = 16 or 9.
Hence, x = ± 4 or ± 3.
The given equation has therefore four roots, being as many
roots as the degree of the equation.
Ex. 2. Solve a;« + 3 (K^ - 88.
We have, (a^y + 3 (oc^) = 88 ; or, transposing,
(a^Y + 3(aj3) - 88 = 0;
or, (a^ - S){a^ + 11) = 0.
Hence, the given equation is satisfied by —
cbS _ 8 = 0, and also by ic^ + 1 1 = 0.
We have then a;^ - 2» = 0, and ic^ + ( ^H)' - ;
or, breaking into factors, we have —
(aj ~ 2) (ar^ + a; • 2 + 2-) - 0, and
(^ + .5^)|a;= - x'^lT+ (^ll)-l =0.
From the^rs^ of these equations, we get —
a; - 2 = or aj = 2,
and a;^ + 2 a; + 4 = 0, from which two other roots may be
found.
And from the second equation, we get —
x+ ^11 = 0, ora; = - ^11,
and ar'-a;v^ll+ a3^121=0, which gives two other roots.
We have therefore shown how to obtain the six roots of
the given equation.
Ex. 3. Solve—
23 ar^ - 75 a; - 6 a: V4ar»-9a;+ 9 + 40 = 0.
Changing the sign of each side, and transposing, we have —
6a;>s/4'^"^9iT9 = 23^2 - 75 a; + 40;
adding to each side 13ar^ — 9a; + 9, then —
13ar' _ 9a; + 9 + (jxsjia? - 9a; + 9 = 36a;2 - 84a; +49;
or, (4a;2 - 9 a; + 9) + 2(3a;) n/4 ar^ - 9 a; + 9 + (3 a:)'
=^[Qxf - 2(6a;)-7 + 7-;
304 ALGEBRA.
or, ( x/4a;'^ - 9x + 9 + Sxf .=^ {6x - 7)1
/. J4:x' -dx + d + 3x = ± {(jx-7), (1.);
or, taking the upper sign —
Jlx' - dx + 9 = {6x - 7) - 3aj - 3 - 7.
Hence, squaring each side —
4 a;2 - 9 a; + 9 = 9 cc^ - 42 ic + 49 ;
or, 5 a;' - 33 aj + 40 = ;
or, {x ^ 6) {6x - S) = 0.
:. X = 6 or J.
Again, taking the lower sign of ( 1), we have —
V4'i--9a; + 9== - (6x - 7) - 3aj = - 9aj + 7 (2.);
or, squaring —
4:x'^ - 9x + 9 = Slay" " l2Qx + 49;
from which 77a]2 _ 117 aj + 40 = 0;
or, (a; - 1) (77 aj - 40) = 0.
:. x = 1 or ff .
Hence, the roots of the given equation are 5, f , 1, ff.
Remark. — If we proceed to verify these values of x, we
shall find that the last two values — viz., 1 and 4f — will not
satisfy the given equation unless we obtain the value of
Ji a;"^ -- 9 a; + 9 by means of the equation from which these
last roots were found.
Thus from ( 2 ) we find, on putting 1 and f -?- successively
for X —
V4x*2 - 9aj + 9 = - 9 (1) + 7 = - 2,
and >/4 a;'^ - 9 a; + 9 = - 9 (f -?) + 7 = 2^;
and on svibstituting either of these values of JZo^^^^9x + 9
along with the corresponding value of x, the given equation
is satisfied.
Ex. 4. Solve—
{x + b + c){x + h — c){b + c - x)
= (a + b + c) {a + b - c) {b + c - a).
By inspection we see that a is one of the roots. We shall
therefore so arrange the equation as to be able to strike out ^
a; — a as a factor of each side (Ai't. 5). |
EQUATIONS WHICH MAY BE SOLVED LIKE QUADRATICS. 305
We have —
{x + b + c) (x + b - c) _ b + c
(a + b -h c)(a + b - c) b + c - x'
(x + by ~ c ^ _b + c - a
^' \a + by - d" ~ b + c ^ x'
or, taking the difference of numerator and denominator —
(x + by - (a + by _ X - a
{a + by - c^ b + c — x'
(x - a) (x -h 2b — a) x - a
or, ^ i—^ 1 = ,
{a + by ■- c" b + c - X
Dividing each side by aj - a, we have —
X + 2b - a _ 1
(airSy—c'-b + c-x' and a; - a = 0, or a: =. a;
Hence also (x + 2b - a) {b + c — x) = (a + by — c-; an
ordinary quadratic from which two other roots may be
determined.
Ex. 5. * Solve the equation —
^ytya '^ iabx'^M ^ (a ' byx"".
Dividing each side by as^, we have —
aWx ^ - 4a^6^a;2^'^ = (a - by-,
or, [abx^'^) - 4 a^b^ {abx'^) = (a - by ;
or, completing the square —
\abx^) - 4 a^b^ (abx^^) + [2 a^by
= {a -by + 4: ab = (a + by.
* We shall assume in this example that the laws of multiplication
and division proved in Algebra, Stage I., Arts. 25, 27, hold for frac-
tional indices,
5 V
^^^ ALGEBRA,
Hence, extracting the square root — '
ahx^ -^^a'^lfi = ± (a + 6),
/. abx^ = ± (a + 5) + 2a^6^ = (a^ + 5^)'or-(aJ-6^)2;
Then, raising each side to the {2pq)\h power, we have—
r + —A or (^ ^ -)
Hence, taking the (p - 5')th root, we get —
/I 1 \±n^ /I 1 \J_^
^ == iTT + -T- b-^ or l-r - -T b-«
/I 1 \i^2^ /I 1 \J_^
/I \\±i
We have also, since the factor cc^ has been struck out — -
x^ = 0, and :. x = as another solution,
Ex. II.
1. 4 a;* - 11 05^ = 225.
2. 5 iB« - 17 x^ = 184.
3. cc^ + 5 aj-2 ^ 251.
4. (a:^ + 3 aj + 3)2 + 2 a^ = 189 - 6 a;,
5. 3a:-*^- 17 aj-^ = 1450.
6. a;" + ~ = 5.
a;"
7. Vi~+~r2 + v'a? + 12 = 6,
8. x/5M^^ + 20 = aj2 + 9.
9. 1 ajV + - = 6 + b a:.
EQUATIONS WHICH MAY BE SOLVED LIKE QUADRATICS, 307
10. (3 o; + 4) + ^3 o; + 4 - 12 = a
U, x" + 2x = — 1-2 - 1.
X + 4:
s/a — X \/a " X __ sjx
12.
13. 9 CB + 24 Jx ^ -7= (-r + ^^i = 65.
sIxX^Jx /
14. X - \!a ' Jh - X " Jh ' sjx ■¥ a,
1 1 a^ + 2ar»
15.
{a - ^/a^ - ar^) (a + \/a* - ar^)"
X*
1G.._-2J_ -^ =T-1-
P + 9 1 • /»2 7j2 ITEJ"- . Ht
l8,
p + a? _ I a? ^ J
■ n/ x^ Na' + or
19. (a^" + 1) {x^- - If = 2 (a; + 1).
a? _ 975 ^^ 7^0
21.-J2 {._3 2) i^^,
i«+lliC+l X) XT
22. 6ar - 5a;- 8 VSa:^ + 5aj _ 4 = 12.
23. 2 .7;- + 5 x - 2 X -N/2af^ - 7 a; + 1 = 35.
24. 20ar^ - 9aaj-8a;A/5ar^-3aa; + 2a' = 1 a\
25. ar'(a; - 2f + 6ar(a; - 2) = 24aj + 36 - 5a:l
26. 2ar + 20aj - ^3 ar^ - a; + 5 = 105.
27. 4a;-^ + 12a; - 2xsJ\9^ - 2a; + 19 = 30,
28. a;* + a;= + 1 = a(ar^ + a; 4- 1).
29. a;« - 1 = 0.
f
308 ALGEBRA.
30. (x" - af + {ar - hf + {x^ - cf
= 3 (a;2 - a) {x^ - h) {x" - c) + d ax^ - a{a + b + c)\
31. {x - h) {x" - (?) = (a ■- h) (a" - c-).
32. {x - a) {x — h) (x ^ c) = (m — a) (m — h) {m - c).
33. {x - a) {x " h) (x - c) = (a + 5) (c^ + c) (6 + c).
34. (n - l)£c (ic^ + ax + c?) = a^ - ajl
35. (ax - 6)« + (coj - d^ = (a ^ cf^ - (5 + cZ)^
36. cc^%a;2--i^ 1 = a((«2 + cc + 1).
Simultaneous Quadratic Equations.
6. The following worked examples are given as speci-
mens of tlie methods to be employed, but it must be under-
stood that practice alone will give the student complete
mastery over equations of this class.
Ex. 1. Solve a;2 + 2/' = 20 (1.) )
X ^y = 6 (2.)/
As we have given the suTti of the unknown quantities, we
shall work for the difference.
From (2), multiplying each side by 2, we have —
2 a;' + 2 2/* = 40
and from ( 1 ), squaring, a.*' + 2 ccy + 2/^=36
Then, subtracting, ar^ - 2 ajy + y^ - 4 ;
and, taking the square root, we have aj - 2/ = i 2 (3).
(2) + (3), then 2i« = 8 or 4, and .'. cc = 4 or 2.
(2) - (3), then 22/ = 4 or 8, and r. y ^ 2 or 4.
Note. — Having found that a; = 4, j/ = 2, we might have told by
inspection that the values a; = 2, y = 4, would also satisfy the given
equations, for x and y are similarly involved in both equations.
Ex. 2. Solve ar - 2/' = ^ • (1.) I
^y = ^ (2.)/
As we have given here the difference of the squares of the
unknown quantities, it will be convenient to work for th^
stm of the squares.
SIMULTANEOUS QUADRATIC EQUATIONS. 309
From (1.), squaring, x* - ^or\f + y- = 25,
and fiom (2.), squaring, &c., 4ar^2/'^ := 144
Then, adding, a;* + 2ar^?/ + 2/* = 169
and taking the square root, a;^ + ?/2 _. + _ 13....... (3).
(3.) + (1.), then, 2 ar = 18 or - 8, or ar = 9 or - 4,
and .-. a; = ± 3 or ± 2 nT^I.
' (3.) - (1.), then, 2 7/2 = g or - 18, ory^ == 4 or -> 9,
and .-. y = ± 2 or ± 3 V - 1.
Note. — The student will see that the pairs of values which s atisfy
the given equations are, x = S, y = 2; x = - 3, y — - 2; a; = 2 -v/ - 1,
y - 3 V -T; X = - 2\/^n^, y = - S^/^^.
Ex. 3. Solvea:2 + y = llaj (1.)) .
f + x= lly (2.)/
Subtracting, then, oc^-y^ — x + y= llaj- 11 y;
or, a^ - y^ = 12 (re - y).
Now (aj — 2/) is a factor of each side, and hence, striking it
out, we have —
x + y = 12 (3.),
and also x - y = (4.).
Equations (3.) and (4.) may not be used as simultaneous
equations, but each of tliem may be used in turn with either
of the given equations.
Thus, taking equations (3.) and (1.), we have —
(1.) - (3.), a;^ - a;= llaj - 12,
from which a; = 6 ± 2 sfS ,
and hence from (3.), by substitution, we easily get-
2/ = 6 + 2 ^/6;
• Again, taking equations (4.) and (1.) —
we have, from (4.) a; = y,
and .*. from (1.), or ■{■ x ^ 11 aj or a;' = 10 a^,
from which x = 10 or ;
and so, from (4.), y = 10 or 0,
310 ALGEBRA*
Hence, the pairs of values satisfying the given equations
are —
a; = 10, 2/ = 10 ; a; = 0, 2/ = ;
x = Q + 2 sl\y =6-2 Sj,
a; = 6 - 2 V6, 2/ = 6 + 2 V6.
Note. — It is worth while remarking that when each of the terms
of the given equations contain at least one of the unknown quantities,
the values ic = 0, 2/ = will always satisfy.
Ex. L Solve 3ar - 2xij = 55 (1.) )
;x? - ^xy + ^f = 7 (2.) J
Multiplying the equations together crosswise, we get —
55 cc^ - 275 ccy + 440 t/' = 21 of^ - 14 a;?/;
or, transposing, 34 ic^ - 261 ajy + 440 ^/^ = ;
or, (2 a; - 5 7/) (17 a; - 882/) = 0,
from which 2 a; = 5 ?/, and 17 a; = 88 y.
Each of these equations taken in twni with either (1.) or
(2) will easily give the required values of x and y,
Ex. 5. a;* + 2/* = 337 (1.),
«J +2/ - 7 (2.)
Erom ( 2 ), raising each side to i\iQ fourth power , we have —
a;* + 4.a?y + 6 a;^^^ ^ 4^^ + 2/' = 2401; (3.)
(3) - (l),then4a;^2/ + 6ar^2/' + ^^V^ = 2064;
or, 2a?y + Z^y" + 2 xif =^ 1032;
or, arranging, 2xy(x + yY - x'^y^ = 1032 ;
but from (2), {x + yY = 49,
and hence, 2 a;?y(49) - a^V = 1^32 ;
or, ar2/' - 98 a;^ + 1032 = 0,
from which xy = 1 2 or 86, (4.)
From (2) and (4), x - y may now be easily obtained, and
hence also the required values of x and y.
Ex. III.
1. x + y =^ 5, xy = 6,
2. X - y — 2, xy = 15.
3. a;- + 2/' = 25, xy - 12.
4. ar + 2/" = 20, a; + 2/ = ^*
SIMULTANEOUS QUADRATIC EQUATIONS. 31 1
5. or + y- = 29, a; - 2/ = 3.
6. ar» - 2/^ = 13, (ic - yY = 1.
1, a? --f = 21,xy = 18.
8. ar^ - t/" = 12, a; + 2/ = 6.
9. ar^ + 2/^ := 53, ar^ - 2/2 ^ - 45.
10. ar^ + a;?/ = 28, 2/* + ajy = 21. . '
11. ar' + a;?/ + 2/^ = 19, a;2/ - af^ = - 3.
12. a; + 2/ = 13, Jx + Ijy = 5.
13. ar^ + a;?/ + 2/^ = 84, a; + J^j -{- y =^ IL
14. a;3 + 2/3 - 35, apy + X7f = 30.
15. -_ + - = a, + ^ = 5.
aj 2/ af' 2/"
16. a; + 2/ = «{a; - y), o;^ + y^ = ^^.
17. a;* - 2/* = «j ar^ - 2/" = ^•
18. aj + 2/ = 5,a;3 + 2/^ = 35.
19. a; + 2/ = 5, ar* + 2/5 = 275.
20. a^ + 2/^= V(«^ + V)y ^y = G.
21. a; - 2/ = 2, a:' - 2/3 = 98.
22. xy{x Jt y) = a, x^y^{p(? + y^) = b,
23. a;7/(a; + y) = 30, ar2/^(a:« + y^) = 9900.
24. 4ar* - ^xy = 18, 52/^ - 2xy = 8.
r; 1 1 30
25. a;^' + 2/^ = 35, aj^ + y^ =— r-i-
a;^y^
26. a^ + 2/^ = 3 a;, a;^ + y^* = a;.
o^r ^ 216 . . 77
27. a;?/ + 6 = , a; + ?/ + 4 ^ ^.
a;?/ X -^ y
28. (a; + t^)^ + 2(a; - 2/)' = 3 (a; + 2/) (^ - 2/)) a;' + 2/" = 10.
29. x" + 10a;y + 2/' = V {^^ - 2r)> a.-" + ^2/' - a; + 132/.
30. a;* - 2a;V + 2/* = 1 + ixy, a?{x + 1)'+ 2/'(2/ + 1)
= xy,
9ar - 2/^ - 117 _ Ba;^ -2r + 1
31. 3a; + 2/ - 9 =
3^ — 7/ — 6 4a;+2/ + l
312 ALGEBRA.
„„ a; + 2 ^ y + 2 _ ., X + 2 y - 2 _ ,,
33 _ V^.(.-y )^3^ 1_1^^^
^r^ ^ f ^ {x - y) ' y X xy
U, y ^2 x/5 a,-^ + 2/ + 3 = 32 - 5aj^
\y £c / ^y aj /
35. ? - 2^ -3= - f.^?^,3a; + 2/ = 7.
y X ar - 2/^
36. (a; + y) xy = c (hx + ay),
xy {bx + ay) = a^y^ + abc (x ■{■ y — c).
37. 2/* = XT (ay - hx), x^ = ax - by,
38. ^/tIl±l.= ^EIpJ,x{y+iy = 3Q(f+r,^).
y sjx
39. a; = » ^ -
7/ + ;2; a; + ;2; i» + 2/
40. a; + 2/ + 5? = 6, a;^/ + a;;^; + 2/^ = 11, xyz - 6.
41. y? - yz - ^,y^ - xz = 0, z^ — xy ~ 0,
42. xyz = a^{x + y) = b'^{y + ^) = 0^(0? + J^).
•»2 7i2 ^2
43. x'^ + y^ + z'^ = ^^ = ^^ = %
x^ 2/ ^
44. aj + 2/ + ^ + ^ = 4a + 45,
a;?/ + a;;^ + 032* + 2/^ + 2/^^ + ;2?2^ = 6a^ + 12 a5 +6^^,
xyz + aJ2/26 + a;;^;!* + yzu = ia^ + 12a^6 + 12a6^ + ib^,
xyzu = a* + 4 a^6 + 6 a^b^ + 4 a6^ + 5^.
45. aj^2/^ + a;?/^;^; + xyz = a,
y^^ + xy^z + a;2/«^ = 6,
y?^ + a;^2/^ + a;2/^^ = c.
4G. (a; + y^ + ;s^ = 1125,
X + y + z = 15,
xy - 24.
47. If aa;^ = by^ = cz\ and - + - + -=«, show tliat
•^ ' a; 2/ ^
aa;^ + 6?/^ + c;s^ = (a^ + b^ + c^ f a\
1>R0BLEMS tnODUCINCJ QUADRATIC EQUATIONS. 313
48. Given i?=l+r, P=^(l- R-'^ M = PR^
show that R - A(p - Tr) "^^^
CHAPTER 11.
Problems Producing Quadratic Equations.
7. We shall now discuss one or two problems whose solu-
tions depend upon quadratic equations.
Ex. 1. A person raised his goods a certain rate per cent.,
and found that to bring them back to the original price he
must lower them 3 J less per cent, than he had raised them.
Find the original rise per cent.
Let X = the original rise per cent.,
then -rTTr: . 100 = the fall per cent to bring them to
the original price.
Hence, by problem —
X - TTTT^ = 3j, which solved, gives
aj = 20 or - IGf.
The value a? = 20 is alone applicable to the problem. Re-
membering, however, the algebraical meaning of the negative
sign, it is easy to see that the second value, x = — 16|,
gives us the solution of the following problem : —
A person lowered his goods a certain rate per cent., and
found that to bring them back to the original price he must
raise them 3 J more per cent, than he had lowered them.
Find the original fall per cent.
The above solution tells us that the fall requii^ed is IGj
per cent.
Had we worked the latter problem first, we should have
obtained ic = 16§ or - 20, the valuer = - 20 indicating
the solution of the former problem.
314 ALGEBEA.
Ex. 2. Find a number siicli that when multiplied by its
deficiency from 100 the product is 196.
Let X — the number,
then 100 - a; = its deficiency from 100.
Hence, by problem —
X (100 - x) = 196, or x"" - 100 i» + 196 = 0;
from which, x = 2 or 98.
Both these values will be found to be consistent with the
conditions of the problem.
Ex. 3. The number of men required to build a house is such
that, when four times the number is subtracted from three
times the square of the number, the result is 160. Find the
number of men.
Let X = the number of men,
then, by problem —
Za? - 4 a; = 160, from which
a; = 8 or - 6f .
The value aj = 8 is alone applicable to the problem as it
stands. If, however, we may conceive of 2^ fractional number
of men — and this we may easily do here by supposing a hoy's
work to be equal to f of a man's — we find that the second
result gives us the solution of the following problem : —
The number of men required to build a house is such that
when four times the number is added to three times the
square of the number, the result is 160. Find the number.
The answer, as above indicated, is 6 men and 1 boy,
where a boy is worth f of a man.
The student will find, however, that in some cases there is
no obvious interpretation to the second result, owing occasion-
ally to the fact that certain terms are used in the problem
to which the results will not apply, and indeed that the
algebraical expression of the conditions of the problem is
more general than the language of the problem itself.
Ex. IV.
1. Find a number whose square is equal to the product of
two other numbers, one of which is less by 6 than the required
number, and the other greater by 9 than twice that number.
1>R0BLEMS Producing quadratic equations. 315
2. When the numerator and denominator of a certain
fraction are each increased by unity the fraction is increased
by T Jo, and when they are each diminished by unity the
fraction is diminished by -y\. Find the fraction.
3. The mean proportional between the excess of a certain
number above 21, and its defect from 37, is 28. Find the
number.
4. A number of articles, which were bought for £4, cost
each 3 shillings more than half as many shillings as there
were articles. Find the number of articles.
5. There is a square court-y^rd, such that if its lengtli be
increased by 10 feet, and its breadth diminished by 20 feet,
its area would be 5,104 feet. Required the side of the square.
6. If the number of shillings given for an article be added
to the number of articles which can be bought at the same
price for 18 shillings, the result is 11. Find the price.
7. Two travellers set out to meet each other from two
places 180 miles distant; the first goes 3 miles an hour, and
the second goes 1 mile more per hour than one-fourth of the
number of hours before they meet. When will they meet?
8. A farmer bought a number of calves, sheep, and pigs,
the number of calves being equal to that of the sheep and
pigs together. For the calves he gave 64s. a head, and for
the sheep twice as many shillings as there were sheep. He
paid £153. 12s. for the calves and sheep together, and £36.
1 2s. for the pigs — a pig costing as much as a sheep and calf
together. Find the cost of the sheep per head.
9. There are two squares, and an oblong whose sides are
equal to those of the squares, and it is noted that three times
the area of the first square exceeds foiu' times the area of the
oblong by 3 square feet, while twice the area of the square,
together with three times the area of the rectangle, is 36
square feet. Required the sides of the squares.
10. The sum of two quantities is equal to 6 times the
square of their product, and the sum of their cubes is equal to 36
times the product of their fifth powers. Find the quantities.
11. The solid content of a rectangular parallelepiped is 60
cubic feet, and the total area of the side is 98 square feet,
while the sum of the edges is 48 feet. Requii^ed the
dimensions.
316 ALGEBRA.
12. The products of the number of units of length in the
sides of a polygon of n sides, when taken n - \, together are
respectively a^, a^, ag, &c., a„. Required the lengths of the
13. A, B, and C can together do a piece of work in a day,
and C's rates of work is the product of the rates of A and B.
Moreover, C is one-fifth as good a workman as A and B
together. Find the respective times required for A, B, C to
do a piece of work.
14. The compound interest of a certain sum of money for
3 years is a, and the third year's interest is h. Find the
principal and the rate per cent.
15. A owes B £a due m months hence, and also £h due
n months hence. Find the equated time, reckoning interest
at 5 per cent, per annum.
16. Find three quantities such that the sum of any two is
equal to the reciprocal of the third.
1 7. Find three magnitudes, when the quotients arising from
dividing the products of every two by the other are respec-
tively a, 5, c.
18. The sum of three quantities is 9, the sum of theu' pro-
ducts, taken two and two together, is 23, and their continued
product is 15. Show that the three quantities are the roots
of the equation a^ - 9 a^ + 23 a - 15 = 0.
CHAPTER III.
Indices.
8. "We shall reserve the discussion of the complete theory
of Indices for Vol. II. , confining ourselves here to a few simple
cases, and giving a few examples involving fractional and
negative exponents.
9. Fractional exponents.
Def. — The numerator of a fractional exponent indicates
the powder to which the quantity must be raised, and the
INDICES. 317
denominator the root which must be taken of the power so
obtained.
Thus, al = 4/a^ a^ = >«, all = >« =a=;
m
and generally a" = >/«"'.
The above definition is that which follows at once if wr
assume the law proved in Art. 24, page 159, viz., a"* x a"*
= a"* + " to be true, whatever be the value of 7n and n.
Thus we have —
(«t\u m in m
a") = a^ X a'* X a" to 9i factors
Hence, taking the Tith root,
Let X = (a^)'' = 1^ i^^y, by Def., Art. 9.
... of = {a^y = (^^},
or, raising each side to the qih power, we have —
pr
Hence, taking the {qs)i\i root, we have x = a «« .
11. To show tluit a» x 5~« = (a6)«.
Now,a^ X b^ ='V»^ ^'n/^ = ;/^^^= V{«^
= (a6)n, by Def., Aii;. 9.
And so, a;r -^ 6'» = (r)"*
Ex. 1. Multiply together aH^c^ by a'^icT.
Adding together the indices of like letters, we have — •
^ + i = -^, i + i = A. i + i = ih
Hence, the recjuired product is aH^'^^c^,
318 ALGEBRA,
Ex, 2. Multiply x + x-y'-^ + y by a; — x^y'-^ + y.
X + x^y'^ + 2/
aj _ x^y^ + y
u? + x%f^ + rrv/
Ex. 3. Divide x- yhy oja - y^.
x^ - y^)x - 2/{i^^ + x^y^i + 2/^
a; - tc%^
' 2 i
cc^2/^
2 1. 12
__
aj«2/^ - 2/
^ 2
Ex. V.
Eind the value of —
1. {a'f,a%{a-'')-\{aWK
2. {a + x)^^{a^ + 2 «aj + aj^)^, {p^ - aj^)4 (a^ + x^)^.
Multiply together —
3. a^ - a^x-i + jc^ and a^ + a*a;a + xs,
4. aj^ - 2/^anda;~^ 4- y~^,
5. aj + a;^2/^ + y and aj"" - a;~^ 2/"^ + ^'' •
Divide —
6. aJ^-hhy a — 5^, ai - Z>^ by a" - ifi,
7. a;^ - xy^ + aj% - 2/^ by a;^ + x^^V'
8. a + 6+c — 3 a^6^c'J by a^ + 6^ + c^*
9. a^ - a^J-y- - a^6 + 6^ by a^6^'- - 6"^
SURDS. 319
Show that—
10. (or'*" + ar^")"*" = x"^ + " ^ x^-'' + x"""^ [ ^.
XI, =^<a;p + oj? > - 2aj2i>?
.< a?2j' - x^t } T^x^P9,
aJ^ - ft2
m 1 m
a- + 6^
12. (x^^ - a*^) ^ « - a^")
= (aj'^^ a-) (a:'^+ a^) (aj'^'+ af) {^ ^ oTY
Find the square roots of —
13. a + b + c + 2 {a^h^ + ah^ + ifici),
14. 4:xy^ — 12aj% + 17ar^2/5 - 12a;^2/^ + 4a^.
15. a^S-^ - 4:ah-i - 8a-^6^ + ia^'b + 8.
Find the cube roots of —
16. a;* + 9a;V + 6aj^ - 99ar^ - 42 aj^ + 441 a;* - 343.
17. a'y-^ + 3aJ*y"-^ + SaP^j-^ + 1.
18. ab(l + 3a-h^ + 3a-J6* + a-^6) (aJ-^ - 3a^6-'
+ 3a^6-4 + 1).
CHAPTER lY.
«
Surds,
12. A surd quantity is one in which the root indicated
cannot be denoted without the use of a fractional index.
Thus, the following quantities are surds : —
320 ALGEBRA.
ai, («= + «^)i, (a^ + b^ + c^)», (^^y, j^
Since, from what has been explained in the last chapter,
these quantities may be written thus-
it follows that surds may be dealt with exactly as we deal
with their equivalent expressions with fractional indices.
It is evident that rational quantities may be put in the
form of surds, and conversely, expressions which have the
/orm of surds may sometimes be rational quantities.
Thus, a^ = 4/(^ = ^^;
and ^a^ + Sa'b + 3a&~V¥ = ^{a + bf= a + k
13. A mixed quantity may he expressed as a surd.
Thus, 3 4^5"= v^3"^ ^5"- .5^3^ x 5 = .^1357
and so, x spy = slx"^ . '^y = llx^'y,
14. Conversely, a surd may he expressed as a mixed
quantity, when the root of any factor can be obtained.
Thus, ViS"^ = sfda^b^ x 2a
= s/da^b\ >J¥a^ 3abj2a.
And ^(a' + hjx'if = ^{d' + h'fi^f x xy'
- ^{a^ + byxy*^^^' = (a" + hjxy^^,
15. Fractional surd expressions may he so expressed that
the surd portion may he integral.
The process is called rationalizing the denominator, and is
worth special notice.
Ex 1 /^ = \^~^^ x/21 V2r
It is much easier to find approximately the value of \l2\,
and divide the result by 7, than to find the values of ^3
and ^7, and divide the former by the latter.
Ex. 2. Eeduce to its simplest form /J j-—^ —
/ xy _ Ixyjb — c) _ *Jxy (b - c)
su?vDS. 321
4
Ex. 3. Find the arithmetical value of p.
2-^3
The denominator is the difference of two quantities, one of
hich is a quadratic surd.
Now, we know that (2 - S) (2 + ^3) = 2^ - ( ^3)^
= 4 - 3 =^ 1, and hence we see that by multiplying
numerator and denominator by the sum of the quantities
in the denominator we can obtain the denominator in a
rational form.
Thus — ^- = ^(2 + J 3_^ _ 4(2+ JZ)
' 2 - V3 (2 - n/3)'(2 + V3) 2= - ( Jzf
4 (2. 73) ^4(2 W3),,(,, ^3)
= 4(2 + 1-73205) = 4(3-73205.)
= 14-92820.
4 ^ 4( 4x/2"'- ZslWf
' 4 n/2' + 3 x/3"' (4 x/2"+ 3 JZ) {i J'2 - 3 Vs)
= ^(W2'-3x/3) ^ 4(4V2~^3v/3) ^ w^ /^ , ^ j^.
(4 x/2)2 - (3 x/3)2 32-27 ' ^ " ^'
We shall now give an example when the surds are not
quadratic.
a
Ex 4. Rationalize the denominator of — i ^i*
x^ — y^'
Since {x^f - {y^f'' is (Art. 29, page 175) divisible by
^ - y^f it follows that the rationalizing factor is their
quotient, wliich is easily found.
16. Surds may be reduced to a common index.
Ex. 1. Express ^a and ]^b as surds having a common
index.
1 1
Since ^a = a% and ^b = b», it follows that, by
reducing the fractional indices to a common denominator, the
n m
given surds become respectively a"»«, 5'"«, or ^^/a^, *v^6'^
5 X
na so.
322 ALGEBRA.
Ex. 2. Reduce ij'^b and ^x^if" to a common index.
The least common denominator of the fractional indices of
the given surds is 4 x 3 or 12. Hence we proceed as
follows : —
When the student has had a little practice, the first two
steps of each of the operations may be omitted.
17. Addition and subtraction of similar surds,
Def. Similar surds are those which have the same irra-
tional factors.
Ex. 1. Eind the sum of V^ 5 slW, - 2 Jib.
We have —
Vl2 + 5 n/27 - 2 x/tS
= n/22 X 3 + 5 x/32 X 3 - 2 ^5^ x 3
= 2V5"+5x3/s/3"2x 5^3
- 2 V3 + 15 V3 - 10 V3
= (2 + 15 - 10) V3 - 7 vs:
Ex. 2. Simplify—
ja'b + 'lab'' + h' __ jd'b -
V a^ - 2a6 + 6'-^ N' "^~r
2a6*^ + 6«
2a6 + 6^
The given expression —
_ a + 5 ,~ a - b ,~ /a + b a -- b\ .-.
~ a-^b ^^ - -^rvb ^^ = v«-:r5 - -^rrv'^^
- (^ + 6)^ - (g - Z>) ^ 4a&
(a - 5) (a + ^>) ' "^ ^ - a^ - 62 V^
18. Multiplication and division of surds.
The following examples will best illustrate these opera-
tions ; —
SURDS, 323
Ex. 1, Multiply a slx^yz by h sjxifu.
We hq,ve, a ijm^yz x h ijx'i^u — ah ijx^yz x xy^n
= ah sJxSjSiz = ahx^y^ sjuz,
Ex. 2. Multiply a aJF ■¥ cjdhj a — JM,
Arranging as in the case of rational quantities, we have —
a ^b + c \ld
a - Jhd
a' Jh + ac Jd^
— ah Jd — cd sjh
[d^ - cd) Jh + a{c - b) Jd
Ex. 3. Divide a Jhhjh J a,
Wehave,^-^.^4^ = '^=T V^^^
When the divisor is a compound quantity it will generally
be the best to express the surds as quantities with fractional
indices, and proceed as in ordinary division.
19,T/ie square root of a rational quantity cannot be 2^(iTtly
rational and partly irrational.
If possible, let ^Ja ~ ni + sfb;
then, squaring, a = n}? ■\- 2,m Jh ■{- h;
or,
2 7n ^b = a - (m^ + b);
or, V6 = ^ - f -^ %
that is, an irrational quantity is equal to a rational quantity,
which is absurd,
20. To Jind the square root of a binomial ^ one of xoliose
terms is a quadratic surd.
Let a + tjb be the binomial.
Assume ^J a + jjb = »Jx + ^y, (1);
then, squaring, a + \fb = x + y + 2sJ xy, (2).
Equating the rational and irrational parts (Art, 19), we
have —
324 ALGEBRA.
,« + 2/ ^ «• (3.),
^Tidi2i\/xy = hjh ov ixy -^ h (4.)
From (3) and (4) we easily find a? = \{a -v sj dr" - h),
and y ^ \ (ci - s] d^ ^ b).
Hence, from ( 1 ), the square root required is —
ViT^'+'V^^-^ + sj\ (a - ^ (V' - 6).
Note. — It is evident that, unless (a^ - 5) is a perfect square, our
result is more complicated than the original expression, and therefore
the above method fails in that case.
Ex. 1. Find the square root of 14 + ^'J^,
Let Vl4 + 6V5 = six + ^y (1.)
Squaring, then, 14 + 6\/5 = ^ + 2/ + SVxy.
Hence, equating the rational and irrational parts—
^ a; + 2/ = 14 (2.),
2Vr^ - 6\^5or4a;?/ = 180 (3).
From (2) and (3) we easily find x = 9, y ~ 5.
Hence the square root required is V^ + >v/5 or 3 + VB.
Ex. 2. Find the square root of 39 + V 14967
Let >\/39~T~7l4% = six + Vy.
Squaring, &c., we have, aj + 3/ = 39 ;
and 4cxy = 1496.
From these equations we easily find x - 22, y = IT.
Hence, the square root required isv22+ Vl7.
21. The square roots of quantities of this hind may often
he found by insjnction,
Ex. 1. Find the square root of 19 + 8 V3.
We shall throw this expression into the form a- + 2 ah
+ 6^, which we know is a perfect square.
Dividing the irrational term by 2, we have 4^3. Now
all we have to do is to break this up into two such factors
that the sum of their squares shall be 19. The factors are
evidently 4 and V3.
SURDS. 326
Thus, we have 19 + 8^3^= (4)^ + 2 (4) Vs + (sfsy
- (4 +_ V3)^
TJie square root is therefore 4+^3.
Ex. 2. Find the square root of 29 + 12^57
We have 29 + 12 ^5"= (3)^ + 2 (3)2 J 5' + (2 ^/5)^
- (3 + 2 sk)\
The square root is therefore 3 + 2 fjd,
Ex. VI.
Express with fractional indices —
1. j^, ^'^b\ il^y ^^\
xy sf^d y^^
ijab Jxy \/a'*
Keduce to entire surds —
3. 3 x/3, 4 N/f, I n/IS, 3 ^T,
4. 4.2^9.3"^, 4. 2~T |(f)"*-
5. 3V«6,«7^^(« + ..)J^-^^,
Keduce to a common index —
6. .^/2,v'3. 7. 4/2, ^37
8. 2 V2, 3 v^5. 9. V^ ^g:
10. {a + aj)^? \/« - a:. 11. ap ^^ hv 'i-
Simplify —
12. 712, ^^48, 3 ^^, \ ^GTR
13. x/4a^ + \a% l^cW + h\ J%r^^.
N 64 a
- , /I 27o : 7;? 3"^^ 7 .,
n/ ~9 ^"' "*■ 3 a; ' 'S (x - «f (re + a) "
^26
ALGEBRA.
Find the value of —
16. Vi2 + ^48 - 2 Va; x/56 + ^/fgsC
17. J^- ^W^ + ^ 'H
18. ^27 «''* + « 6^ - v'8a^- + « + 3 ^^GT^.
Multiply —
19. a + Jab + bhy Ja - Jb, a^ + b^hy sja - Jb,
20. (aj + y)h by (oj + y)i, a + b s/dhj cir - ab Jd + b'd.
22. a^- + b^ 4- c^ + d} by a^ - 6 J" + c^ - di
Divide—
23. a? + xy + y'^hj X + x^yi + y.
24. a^ - 7/^ by cc^ + y^i
Kationalize the denominators of—
• n/3' V7' n'5-
3 ' 4 1 .
26.
27.
28.
2 + n/3 3V2 - 2>/3* ^5
3 2
V"3
^ V3 + J2
1 + J2 + Vs' x/2 + V 3 + Jd' Js - V2*
a 1 ^ & "
a;^ - 2/^ /y2 + \/3' X + a;%^ + y*
Find the square roots of —
29. 11 + 4 V7, 8 + 2 Jl5, 30 - 10 s/d'^
30. 8 + 2 s/T2, 9 - 6 >/2, 20 - 10 s/3.
RATIO. , 327
CHAPTER V.
RATIO AND PROPORTION.
Ratio.
22. The student is referred to Chapter II. of the Arith-
metic Section of this work for definitions and observations
which need not be repeated here.
23. A ratio of greater inequaliti/ is diminished, and a ratio
of less inequality is increased, hy increasing tlie terms of the
ratio hy the same quantity.
Let a : 5 or ^ be the ratio, and let each of its terms be
a ■\' m
m
a + m .
increased by m. It will then become -i
Now, 5 -, as (a + m)b% (h -{- m) a,
+ m b^ ^ ' ^ '
or, as ab + bm %.ab ->r am; or, as bm 5 a^ii, or as 6 5 a.
Hence the ratio -7-, is increased when b ^ a, that is, when
b
it is a ratio of less inequality; and is diminished "When
b -^i a, that is, when it is a ratio of greater inequality.
CoR. It may be shown in the same way that —
A ratio of greater* inequality is increased, and a ratio of less
inequality is diminished, by diminishing the terms of the ratio
by the same quantity,
24. When the difference between the antecedent and con-
sequent is small compared with either, the ratio of the higher
powers of the terms is found by doubling, trebling, <fec., their
difference.
Let a + X 1 a ov be the ratio, where x is small
a
compared with a,
rr^ (a-^xf a^ -^ 2 ax + or - 2 x ,
Then ^ ^-^ = 5 = 1 + — nearly =
Ot 0i Oi
a -h 2 X ,
nearly.
328 ' ALGEBRA.
(a + xY a^ + 3 a^x + 3 aaP + a? ^ Sx ,
^^ 7-^ = 3 = 1 + — nearly =
a + 3a; , ,
nearly; and so on.
Ex. (1002)2 : (1000)2 = 1004 : 1000 nearly.
(1002)3 : (1000)3 = 1006 : 1000 nearly.
Proportion.
25. Proportion, as has been already said, is the relation
of equality expressed between ratios.
Thus, the expression a :h = c: d,
ov a: b :: c:d,
a c
is called a proportion.
26. The following results are easily obtained : —
,^.ci. a c,, a b c b a b
(1.) bince r = T}> ^^^^ r X - = ^ X - or - = -,,
b d b c d c c d
.\ a: c :: b : d (alternando).
(2.) 1 V T- = 1 -f -y or - = ,
^ 6 d a c
.*. b : a :: d : c (invertendo).
Also, by Art. 64, page 214, we have —
(3.) a + b : b :: c -h d : d (componendo),
(i.) a - b : b : : c - d : d (dividendo),
(5.) a - b : a :: c - d : c (convertendo).
(6.) a ■{■ b : a - b :: c + d : c - d {componendo and
divklendo),
27. If a : 6 : : c: d and e : / : : g : h, we may compound
the proportions.
Thuswehave, = r (l),and>=^ (2).
(l)x(2),then,«^=|.
ov ae : b/:: eg : dh.
PROPORTION. 329
And ill the same way we may show that, if the correspond-
ing terms of any number of proportions be multiplied together,
the products will be proportional.
28. If three quantities are in continued 2yroportion, the first
has to the third the duplicate ratio of what it has to the
second.
Let a, h, c be the given quantities in continued propor-
tion; then —
a _ b
6 ~ c
-._ a a b a a a^ ^
Hence, - or - x - = 7 X y = fr,,
' c c b b V
,\ a : c :: a^ \ W.
And, similarly, if a, b, c, d are four quantities in continued
projjortion, a.: d :: a^ : b\ that is —
The first has to the foui-th the triplicate ratio of what it
has to the second; and so on, for any number of quantities.
29. We shall now give one or two examples of problems
in Proportion.
Ex. 1. If a: b::c: d, prove that ^3 ^ ^, = (^-^p-^) •
Let r =^ -, = x; .\ a = bxy and c = dx.
Hence, .^Ll^ = Ml^^jf = Jf^. t^^en, a^..-
(a + cy (bx + dxy (6 + df
J a^ + c^ /a + cV
Ex. 2. It a : b :: c : d, prove that j j--=^ — ..
. ^^ (f' - f> Jac ^ \/bd
Let -J- = ~j = x; ,'. a - bxf and c = dx,
ct
Hence ^^ + ^_^^ + ^_^+l _ V^^.a; + J ^bd
'a - b '^ bx - b "^ a; - 1 ~ Jb(f,x - Jbl
_ fjbx . dx + Jbd __ is/ac + Jbd
tjbx.dx - fjbd \fac - ^bd'
330 ALGEBRA.
Or, it may be worked tliiis —
Since -- = --, we have /t- — / - ,
Hence, by Art. 26—
CO + h _ ijac + \Jhcl
Ex. 3. If a t & :: c : c? :: e :/, show that
L
a _ rmoT + QIC'' + pe'^Y
r J. ct C e ,^.
^''i- = d=7='' W-
a' _ C _ e' _ ,
■■■!/- d^ -/'-"=•
Hence, a^ = h^oG^, .*. wa*" = mh^'x^ \
c** = dVf :. n(f ~ nd^'x'' V, and .'.by addition,
e^ = fV, :. pe'' = pf^'x'' j
ma^ + TIC** + pe^ = (mJ** + Qid'' + p/'')x''.
.'. Equating (1) and (2), we have —
h \mJf + nd"^ + pf'/
Ex. YII;
1. Compare the ratios a ■{■ h : a - hj and a^ + Ir : ctr - h\
2. Which is the greater of the ratios a + h : 2 a, and
^h:a + hi
3. What quantity must be subtracted from the consequent
of the ratio a : 6 in order to make it equal to the ratio c-.d'l
PROPORTION. 331
4. Compound the mtios 1 - x^ :l + y, cc - a;?/- : 1 + ar,
and 1 :x - aP,
5. There are two numbers in the ratio of 6 : 7, but if 10 be
added to each they are in the ratio of 8 : 9. Find the
numbers.
/»
G. In what cases is a; + — >- or < 5 ?
X
7. If = = , show that a + h + c ^ 0,
X — y y — ^ z " X
8. Find the value of x when the ratio x + 2 a:x + 2 5 is
the duplicate ratio of2a; + a + c:2a; + 6 + c.
9. Find x when the ratio x - h : x + 2 a — 5 is the
triplicate ratio oi x - a : x -v a - h.
■.^-r/»«^ + V 2/ + ^ X •{■ Z . - ,/•,/.
10. If , = 7— — = i show that each of the frac-
a -\- + c c + a'
4 i X + y -h z ., X y z
tions IS equal to V , and that - = » = -•
•i -i Ti< ^ c e , , . . , let + mc + 7^c
IL Kt = -7 = 7i , thdii each IS equivalent to ,,-" — 'ff~7~~f\
hellce, show that —
a h c
2z + 2x - y ~ 2x + 2y ^ z~ 2y + 2z^x
X 'H '^
^^®^ 2a""r'2T^7 " 2 6 + 2 c ^1^ " 2 c + 2 c* - 6*
12. If w :6 :: c : c?, then
a ■\- h \ c + cZ : : x/a'"^ + a6 + 6" : si 6^ + cc/ + Or,
13. Find a fourth proportional to the quantities —
a; + 1 ^ ■\' X ■\- 1 (c* + 1
tc'^n' a;^ - a; + l' S^"^^'
14. Find c in terms of a and & when —
(1.) a \ a \\ a - b : b - c.
(2.) a :b ::a - b :b - c.
(3.) a : c : : rt - b : b - c.
33!J ALGEBRA.
15. If ay b, c are in continued proportion, show that
7 , hf he are also in continued proportion.
16. If a : h : : c : dy then —
17. From a vessel containing a cubic inches of hydrogen
gas, b cubic inches are withdrawn, the vessel being filled up
with oxygen at the same pressure. Show that if this opera-
tion be repeated oi times successively, the quantitv of hydro-
, , ^ , (^ _ 5)n
gen remaining in the vessel is — „_i ■ cubic inches, when re-
cti
duced to the original pressure.
18. If, in Ex. 34, page 225, (^j, a.,, %), (Sj, bi, 5y), and
(cj, C2, Cg) are corresponding terms respectively, show that
SECTION III.
PLANE TRIGONOMETEY.
CHAPTER I.
MODES OF MEASURING ANGLES BY DEGREES AND GRADES.
1. We are able to determine geometrically a right angle,
and it might therefore be taken as the unit of angular mea-
surement. Practically, however, it is too large, and so we
take a determinate part of a right angle as a standard.
In England we divide a right angle into 90 equal parts,
called degrees, and we further subdivide a degree into 60
equal parts, called minutes, and again a minute into 60 equal
parts, called seconds. This is the English or sexagesimal
method.
In France the right angle is divided into 100 equal parts,
called grades, a grade into a hundred equal pai-ts, called
minutes, and a minute into 100 equal pajts, called seconds.
This is the French or centesimal method, and its advantages
are those of the metric system generally.
The symbols °, ', ", are used to express English degrees,
minutes, seconds respectively, and the symbols ", \ '\ to express
French grades, minutes, seconds respectively.
Conversion of English and French Units.
2. Let D = the number of degrees in an angle,
and G = the number of gi-ades in the same angle ;
then --expresses the angle in terms of a right angle;
and so also does -. >^^•
334 PLANE TRIGONOMETRY.
Hence,
D
)0
G D
- 100 "'■ 9 -
G
10-
D
= 10^
= G
-,l«
.(1).
andG == ^D = D + |d (2).
Hence the following rules : —
1. To convert grades into degrees.
From the number of grades SUBTRACT -yV) and the remain-
der is the number of degrees.
2. To convert degrees into grades.
To the number of degrees add \j and the sum is the num-
ber of grades.
Ex. 1. Convert 13^ 18' 75^' into English measure.
No. of grades = 13-1875
Subtract -j\ of this = 1 -31875
.-. No of degrees - 11-86875
60
52/-12500
' * 60
t 7^^-500
Ans. ir 52' 7"-5.
Ex. 2. Convert 18° 7' 30'' into French measure,
No. of degrees = 18*125
Add I of this == 2-0138
.-. No. of grades = 20-1388
Ans. 20^ 13' 88''-8.
3. An angle may be conceived to be generated by the
revolution of a line about a fixed point. Thus —
Let OA be an initial line, and let a line, OP, starting
from OA, revolve with O as centre, and take up succes-
sively the positions OPi, OPg, OP3, OP4.
MODES OF MEASURING ANGLES. 335
Now the magnitude of an angle may be measm^ed by
the amount of turning 7'equired to generate it. When, there-
fore, the revolving line reaches the position OB, we may con-
ceive an angle to have been
generated whose magnitude
is two right angles. And,
further, when the revolving
line assumes the positions
OP3, OP4, the angles AOP3,
AOP4 (the letters being read
in the direction of revolu-
tion) are angles whose mag-
nitudes are each greater than two right angles. Indeed,
when the revolving line again reaches the position OP, we
may conceive an angle to have been generated whose magni-
tude is four right angles. Lastly, if the revolution of the
line OP be continued, we may conceive of angles being
generated to whose magnitude there is no limit.
Ex. I.
1. Express 39' 22' 30" in French measure, and 13M5^
75^' in English measure.
2. One of the angles at the base of an isosceles triangle is
50"*. Express the vertical angle in grades.
3. Divide an angle of n degrees into two such parts that the
number of degrees in one part may be twice the number of
grades in the other.
4. Two angles of . a triangle are respectively a°, h^, express
the other angle in degrees and grades.
5. If f of a right angle be the unit of measurement, ex-
press an angle which contains 22*5 degrees.
6. Show how to reduce English seconds to French seconds.
7. If the unit of measurement be 8°, what is the value
of 10^.
8. If two of the angles of a triangle be expressed in grades,
and the third in degrees, they are respectively as the num-
bers 5, 15, 18. Find the angles.
9. What is the value in degrees and grades of an angle
336
PLANE TRIGONOMETRY.
which is the result of the revohition of a lino 3i times
round.
10. In what quadrants are the following angles found : —
145°, 96^, 327°, 272^, 272°.
11. If a** be taken as the unit of angular measurement,
express an angle containing h^,
12. What is the unit of measurement when a expresses -
of a right angle 1
CHAPTER II.
THE GONIOMETRIC FUNCTIONS.
4. It was formerly usual in works on Trigonometry to give the
following definitions : —
Let a circle be described from centre A,
/ 1^ with radius AB supposed to be unity, then —
(1.) The sine of an arc BC is the perpen-
dicular from one extremity, C, of the arc
upon the diameter passing through the other
extremity B.
Thus CS is the sine of the arc BC.
(2. ) The cosine of an arc is the sine of the
complement of the arc.
Thus, since DC is the complement of BC,
S'C is the COSINE of the arc BC.
(3.) The tangent of an arc BC is a line
drawn from one extremity, B, of the arc
touching the circle, and terminated in the diameter which passes
through the other extremity, C, of the arc.
Thus, BT is the tangent of the arc BC.
(4.) The cotangent of an arc is the tangent of the complement of
the arc.
Thus, DT/ is the cotangent of the arc BC.
(5. ) The secant of an arc BC is a line drawn from the centre through
one extremity, C, of the arc, and terminated in the tangent at the
other extremity.
Thus, AT is the secant of the arc BC.
(6.) The cosecant of an arc is the secant of the complement of the arc.
Thus, AT ' is the cosecant of the arc BC,
TRIGONOMETRICAL RATIOS. 337
(7.) The versed sine is that portion of the radius upon which the
sine falls, which la included between the sine and the extremity of
the arc.
Thus, SB is the versed sine of the arc BC.
(8.) The coversed sine is the versed sine of the complement of the
arc.
Thus, S'D w the coversed sine of the arc BC,
(9. ) The suversed sine is the versed sine of the supplement of the arc.
Thus, B ^S 25 the suversed sine of the arc BC.
Representing the arc BC by A, it is usual to write the above func-
tions thus : — Sin A, cos A, tan A, cot A, sec A, cosec A, vers A,
covers A, suvers A.
By mere inspection, the student will see that the following rela-
tions hold : —
(1.) Sin A = CS = ?^ r= ^^- = :^ = L_.
1 AS AT' cosec A
(2.) CosA = S'C = -^ = A^. :
^ 1 AC
(3.)TanA=BT = -^- = |^=--^^ = j^.
(4.) Sin^ A + cos2 A = CS^ + S/C* = CS« + AS^ = AC^ = 1.
(5.) Sec'^ A = AT2 = AB2 + BT» = 1 + taa'^ A.
(6.) Cosec^ A = AT/2 = AD^ + DT/^ = i + cot^ A. /
.^^rv A * -orp BT CS CS sin A
(7.) Tan A = BT--r5=-ro= "oTTt = r*
^ ' AB AS S'C cos A
(8.) Vers A = SB = AB - AS = AB - S'C = 1 - cos A.
(9.) Covers A = S'D = AD - AS' = AD - CS = 1 - sin A.
(10.) Suvers A = B'S = B'A + AS = 1 + cos A.
It is more convenient, however, to define the sine, cosine^ &c., as
in the next article, according to which definitions they are commonly
called Trigonometrical Ratios. The student will see that if the
above definitions be so far modified that, instead of the lines them-
selves, the goniometric functions be taken as the ratios which the
lines respectively bear to the radius, they are included in the defini-
tions of the next article.
Trigonometrical Ratios.
5. Let BAG be any angle, which we may denote by A,
and P any point iu the line AC, Praw PM perpndicul£«r
to AB.
5 Y
338
PLANE TRIGONOMETRY,
Then (1.) Sin A ==
perpendicular ^ PM
(2.) Cos A -
(3,) Tan A =
base
Kyp-
perpendicular
AM
AP*
base
a ) Cot A = ^a.se
^ '' perpendicular
B
PM
AM*
AM
PM*
(5.) Sec A =
liyp-
AP
AM*
^ *^ perpendicular PM
(7.) The versed sine is the remainder after subtract-
ing the cosine from unity, or —
vers A = 1 - cos A.
(8.) The cover sed sine is the remainder after subtract-
ing the sine from unity, or —
covers A = 1 — sin A.
(9.) The suversed sine is the sum of the cosine and
unity, or —
suvers A = 1 + cos A.
The last three are not much used in practice.
6. Comparing (1.) and (6.), (2.) and (5.), (3.) and (4.), of
the last article, we see at once that the sine and cosecant^
the cosine and secant, and the tangent and cotangent, are
respectively each the reciprocal of the other.
TRIGONOMETRICAL RATIOS. 339
We therefore have —
1 1
(1.) Sin A = .icosec A
cosecA sin A
(2.) Cos A = -i-., sec A = -i-r.
^ ' sec A cos A
(3.) Tan A = - -~-r-, cot A = r^
^ ^ cot A tan A
Further —
(4.) Sin^ A -f cos^ A = >^j. + ^^ = 5^,
_ AF
- ^ps - A-
Hence also, transposing and taking the square root —
(5.) Sin A = Jl - cos^ A.
(6.) Cos A = VI - sin^ a:
And again —
(7.) Sec^A =-^-^,= AM- ^^ ^ AM- ^ "^ *^^'^-
.ov ^ .A AP- PM2 + AM- ^ AM-
(8.) Cosec- A = pjj5 = pjj, = ^ -^ P5P
= 1 + cot- A.
/nv m A I'M PM AM . . .
(9.) Tan A = -^ = ;^ -^ -^^ = sin A ^ cos A
^ Bin A
~ cos A'
/..xr. A AM AM PM
(10.) Cot A = pjj = aF "^ AP " ^^'"^ "^ ^ ^^^ ^
_ cos A
"" sin A*
The student must make himself thoroughly master o£ the
results in this article.
340 PLANE TRIGONOMETRY.
7. To express the trigonometrical ratios in terms of the
sine,
(1.) Cos A - Jl - sin^ A, by Art. 6 (6.)
(2.) Tan A = J^, by Art. 6 (9.),
sin A
Jl - sin^ A
(3.) Cot A = ^?^^, by Art. 6 (10.),
^ ^ sm A
a/I - sin''* A
sin A
1
cos A'
(4.)SecA== -f., by Art. 6 (2.),
Vl -sin^A
(5.) Cosec A = -J--r, by Art. 6 (1.) '
^ ^ sm A ^ '
Ex. If sin A = I, find the othet trigonometrical ratios.
We have, cos A = n/1 - (|)^ = \/l - A = h
. sin A , . ^ ,
tan A = -^ T = I V I = |.
cos A ^ *
8. To express the trigonometrical ratios in terms of the
cotangent.
(1.) Sin A = — i- = - -J- , by Art. 6 (8.)
cosec A vl + cor A
(2.) Cos A = -— - = \ , by Art. 6 (7.)
sec A VI + tan^ A
1 cot A
J
1 VcoFaTi
"^ cot'-^ A
(^•)T'-A.,-^-
TRIGONOMETRICAL RATIOS. 341
(4.) Sec A = -^- = 1 V--p^2L£=, by (2.) above,
^ ^ cos A J^^^^AT+^I ^ ^ ^
_ n/coj" a + 1
~ cot A
(5.) Cosec A = VI + cot^ A, by Art. 6 (8.)
And in the same way the trigonometrical ratios may be
expressed in terms of any one of them.
Ex. II.
1. Given sin A = ^f, find the other trigonometrical
ratios.
2. Given tan A = ||, find the remaining trigonometrical
ratios.
3. If cot A = a, show that sin A =
4. If vers A = 6, then tan A =
n/1 + a^
x/2 6 - 6^
1-6
5. Construct by scale and compass an angle (1.) whose
cosine is f ; (2.) whose tangent is |; (3.) whose secant is J2;
(4.) whose cotangent is 2 + Jsi
Prove the following identities : —
6. (Sin A + cos A)^ + (sin A - cos A)^ = 2.
7. Sec^ A + cos^ B . cosec^ B = cosec- B + sin^ A . sec^ A.
8. Sec^ A . cosec^ A = sec^ A + cosec^ A.
9. Sec A . cosec A = tan A + cot A.
10. Sin^A - cos^A = (sin^ A - cos^A) (sin* A + cos* A).
, ^ Sec A + tan A _ cosec B - cot B
Cosec B + cot B sec A — tan A
sin* A + sin^Acos'A + cos* A
12, 1 + sin A cos A = = : — r r •
1 - sin A cos A
13. (.rcose + 7/ sine) (a: sine + 2/ cose) - (a; cose - ysine)
(x sin — 2/ cos e) = 2 xij.
342
PLANE TRIGONOMETRY.
14. (a sin cos «/> + r cos 6 cos (/>) (5 sin 6 sin </> + r sin cos (p)
— (b sin (jos<^ - r sin sin </>) (a sin0 sin <^ + r cos sin <^)
= rsin0(rcos0 + c&sin0).
15. If ic = Q* sin cos <^, 2/ = r sin sin <f>y z - r cos 0, show
that x^ + y^ + z^ ^ 7^.
IG. If (t = b cos C + c cos B, & = tt cos C
c = rt cos B + 6 cos A, show that a^ + b^ - c^ ^
17. Given sin^A + 3 sin A = ^, find sin A.
18. Given cos^A — sin A = ^^^, find cos A.
+ c cos A,
: 2a6cosC.
19. Solve sin A — cos A =
V2'
for sin A.
20. Find tan A, when tan A + 1 = sf^, sec A.
21. Given a cos A = 5 sin A + (x, find cot A.
22. Given tan^ A - 7 tan A + 6 = 0, find tan A.
23. Show that \l\ + 2 sin A cos A + ij\ — 2 sin A cos A
= 2 cos A or 2 sin A, according as A is between 0° and 45°, or
between 45° and 90°.
24. Given ??i sin^ A + Tzsin^B = a cos- A,
m cos^ A + 7Z cos^ B = 5 sin^ A, find sin A and sin B.
CHAPTEB III.
CONTRARIETY OF SIGNS. — CHANGES OF MAGNITUDE AND SIGN OF
THE TRIGONOMETRICAL RATIOS THROUGH THE FOUR QUADRANTS.
9. We have explained at some length the meaning and
use of the signs + and - in
algebra. They have a similar
interpretation in trigonometry.
1. Lines. — Draw the horizontal
^ line A A, and draw BB ' at right
angles, meeting it in O. Then
considering O as ongin,
(1.) All lines drawn to the right
parallel to A A are called ^^osiVive,
CONTRARIETY OP SIGNS.
343
and all lines drawn to the left parallel to A'A are called
negative,
(2.) All lines drawn upwards parallel to B'B are called
positive^ and all lines drawn dovmwards parallel to B 'B are
called negative.
(3.) Lines drawn in every other dii*ection are considered
positive, as is therefore the
revolving line by which angles
may be conceived to be gen-
erated.
2. Angles. — A similar con-
vention is made for angles. Let
OA be an initial line, and
let a revolving line about the
centre O take up the positions
OP and OP/. Then—
(1.) That direction of revolution is considered positive
which is contrary to that of the hands of a watch, and the
angle generated is a ptositive angle.
(The positive direction is then upvjards,)
Thus, AOP is a positive angle.
(2.) The negative direction of revolution is the same as that
of the hands of a watch, and the angle thus generated is a
negative angle.
(The negative direction is then doumwards.)
Thus, AOF is a ne-
gative angle. p^
Hence, if the angles
AOP and AOP^ be
of the same magni-
tude, and
We have —
ZAOP =
thenZAOP/- -
10. We will
examine the
metrical ratios for
angles greater than a
right angle, and for negative angles.
344 PLANE TRIGONOMETRY.
Let OP^, OP2, OPg, OP4 represent the position of tlie re-,
volving line at any period of revolution in the several
quadrants respectively,
And let PiISTj, PgNg, Ps^g, P^IST^ be the respective per-
pendiculars from the end of the revolving line upon the
initial line.
Then PjN^^, T^'^^y ^3^3? ^4^4 ^^'® respectively the per-
pendiculars corresponding to the angles generated.
Also, ONj, ONg, ON3, ON4 are respectively the bases of
the right-angled triangles with respect to the angles in
question.
We have then in the second quadrant —
Sin AOP, = M2, cos AOP2 = ^^,
tan AOPo, = ?2^2 &c.
It IS therefore evident that the relations between the trigo-
nometrical ratios, which were proved to exist in Art. 7,
also hold for angles in the second quadrant — that is, angles
between 90° and 180°.
And in the same way we may show that they hold for
angles in the third, fourth, or any quadrant.
And again, if we suppose the line to revolve in a oiegative
direction, and take the position OP', we shall have P/N' the
perpendicular corresponding to the negative angle AOP ', and
ON ' the base.
Hence, sin AOP' = ^, cos AOP/ = '±^^,
tanAOF = ^,&c.
And the relations proved in Art. 7 may be also similarly
proved to exist here.
Hence the relations proved in Art. 7 hold for any angles
whatever.
Changes of Magnitude and Sign of the Trigonometrical
Ratios.
11. Let OP^, OP2, OP3, OP^ be positions of the revolving
line in the several quadrants respectively; PiNp P2N2,
CHANGES OF MAGNITUDE AND SIGN. 345
P3N3, "P^^^y the respective perpendiculars; and ON^, ONg,
ON3, ON4, the bases of the corresponding right-angled
triangles.
Then—
(1.) In the first quadrant —
P N ON*
SinAOP,= ^^^^^^^
tanAOPi = "o?^'^^^-
At the commencement of the motion of the revolving line,
the angle AOP^ = 0° ;
Also, the perpendicular PjN^ = 0,
And the base ON = OP^.
Hence, we have —
*SinO» = ^.0,cosO» = ^; = l,
tan 0» = ^, = 0.
As the revolving line moves from OA towards OB, P^N^
increases and ON^ diminishes ; and when it amves at OB,
wehavePjN^ = OPj,andON"j = 0. But the angle generated
is now a right angle. Hence we have —
Sin90»=^?i = l,cos90° = ~=h
OP ^
tan 90° = -^ - 00.
Hence, as the angle increases from 0° to 90° —
The sine changes in magnitude from to 1 and is + .
Tlie cosine changes in magnitude from 1 to and is + .
The tangent changes in magnitude from to 00 and is + .
(2.) In the second quadrant —
Here the perpe7idicular PgN^ is + ,
and the base ONo is - .
* The student ought properly to look upon the values 0, 1, here
obtained as the limiting values of the sinc^ cosincy and tangent respec-
tively, when the angle is indefinitely diminished.
346 PLANE TRIGONOMETRY.
Hence the sme during the second quadrant is + , the
cosine is - , and the tangent is - .
Again, as the revolving line moves from OB to OA', the
perpendicular "^^^ diminishes until it becomes zero. Also,
the base ONg increases in magnitude, until it finally coincides
with OA', and .*. = — OPg. But the angle now described
is 180°.
Hence we have —
Sin 180° = ^^ = 0> cos 180° = - ^-? = _ 1,
tan 180° = - -^ = 0, &c.
Hence in the second quadrant —
The sine changes in magnitude from 1 to 0, and is positive.
The cosine changes in magnitude from to 1, and is negative.
The tangent changes in magnitude from oo to 0, and is negative.
And in the same way may we trace the changes of magni-
tude and sign in the third and fourth quadrants.
Thus we shall find —
(3.) In the third quadrant —
The sine changes in magnitude from to 1, and is negative.
The cosine changes in magnitude from 1 to 0, and is negative.
The tangent changes in magnitude from to oo, and i^ positive.
(4.) In the fourth quadrant —
The sine changes in magnitude from 1 to 0, and is negative.
The cosine changes in magnitude from to 1, and is positive.
The tangent changes in magnitude from oo to 0, and is negative.
Moreover, as the cosecant, secant, and cotangent are
respectively the reciprocals of the sine, cosine, and tangent, it
follows that their signs will follow respectively the latter,
and that their magnitudes will be their reciprocals.
CHAPTER TV.
TRIGONOMETRICAL RATIOS CONTINUED. ARITHMETICAL VALUES
OF THE TRIGONOMETRICAL RATIOS OF 30°, 45°, GO'*, &C.
12. To prove tkit sin A = cos (90' — A ), and that
COS A = sin (90' - A),
TRIGONOMETRICAL RATIOS.
347
Using the same figure as in Art. 5, we have —
PM
Sin A = — -— = cos APM.
AP
But Z APM = 90° - A,
.-. SinA = cos(90« - A),
and similarly —
cos A =: sin (QO"* - A),
tan A = cot (90« - A),
cot A = tan (90^ - A),
sec A = cosec (90" — A),
cosec A = sec (90** - A).
13. Ratios of 45^
In the last figure, suppose Z PAM = 45% then also
Z APM = 90^ - 45^^ = 45°. And hence Z PAM = Z APM,
and .-.PM = AM (Euc. L, 6).
Hence, also —
AP or VaM- + P5F = x/2AM-or JTm:\
.-. AP = AM sj2ov PM n/27
Hence we have —
PM 1 ... .
= cos 4o , by
Sin45» - ^^^^^ = AP' = PM^2 - 7f
Art. 8.
Tan 45°= tan PAM =
Sec 45° = sec PAM =
PM
AM
AP
PM
PM
AMsJY
AM
= 1 = cot 45^ by Art. 8
= sf2r= cosec 45°
AM
by Art. 8.
14. Ratios of ^(f QXidi^O\
In the same diagram, suppose Z PAM = 30", then
Z APM = 90'* - 30^ - 60°.
Hence, if we conceive another triangle equal in every
respect to APM to be described on the other side of AM, the
whole would form an equilateral triangle whose side is AP.
Hence, PM = ^^AP.
348
PLANE TRIGONOMETRY.
Now AM = av/aF - PM^, .-. AM = N/AF^-^ilJ)^
J'S
AP.
We hence have —
PTVr 1 AP
Sin 30' ^ sinPAM = ^f = ^ = l= cos 60; by Art. 8.
Cos 30° = cos PAM =
AM
AP
^AP ,
2 ^^ _ x/3
Art. 8.
AP
= sin 60", by-
Tan 30' =
PM iAP
AM - J3
AP
JS
^ = 1 V3 = cot 60°, by
Art. 8.
15. To show that sin (180" - A) = sin A,
cos (180' - A) = - cos A,
Let ZAOPi = A,
And let the revol-
ving line describe an
angle AOP2 = 180^^
"TJT
¥7^ - A;
Then ZA'0P2 =
180° - (180° - A) = A;
Hence, Z AOP^ = A^OPg.
Hence, also (Euc. I., 26), if P^N^, P2N0 be drawn perpen-
dicular to AA^ P^N^ - PgNg, ON2 - - ON^,
We have therefore —
Sin (180° - A) = sin AOPg = ?,& = -?i^i = sin A.
^ ^ ^ OP2 OP^
Cos (180'' - A) - COSAOP2
ON, ON, .
-opr-T)p; = -^^^^-
And similarly —
Tan (180° - A) = - tan A, cot (180° - A) = - cot A.
Sec (180° -. A) = - sec A, cosec (180^^ - A) = cosec A.
TRIGONOMETRICAL RATIOS.
349
16. To show that sin (180° + A) = - sin A,
cos (180^ + A) = - cos A,
Let AOPj = A,
And let the revolving
line take a position such
that P1P3 is a straight
line.
Then, evidently, Z AOP3
= 180° + A.
Then, as in last Article —
P N - - P N
0N3
Hence —
ONp
ON.
ON"i
= - cos A.
Sin (180^ + A) = sinAOPg =
Cos (180^ + A) = cos AOP3 - ^p-- -^p-
And similarly —
Tan (180° + A) = tan A, cot (180° + A) = cot A.
Sec (180° + A) = - sec A, cosec (180° + A) = - cosec A.
17. To show tlmt sin ( - A) = — sin A, and
cos ( - A) = cos A,
Let Z AOP = A,
And let the revolving line
describe an angle AOP'
= -A.
Then evidently, if PNP '
be drawn perpendicular to
OA, we have (Euc. L, 26)
P^N = - PK
Hence —
Sin (- A) = sin AOP' =
P_^
OP'
PN
"OP
= - sin A ,
Cos (- A) = cosAOP' = Qp7= Qp = cos A.
And similarly —
Tan ( - A) = - tan A, cot ( - A) = — cot A,
Seg ( = A) = sec A, cosec (— A) = - cosec Ai
350
PLANE TRIGONOMETRY.
Althougli the results of Arts. 12, 15, 16, 17 have been
proved from diagrams where A is less than a right angle, the
student will have no diffi-
'1^ culty, if he has understood
the proofs, in deducing the
same results for angles of
any magnitude whatever.
18. To show that tan -
_ 1 - cos A
sin A '
LetZAOC = A;
Bisect it by the straight line OB, so that Z AOB = ^^ ^
and draw CD perpendicular to OA, meeting OB in E.
(1).
OD ED
Then tan ^ = tan EOD - ?~.
Now(Euc.YI.,2),^ - ?5:, and .•
00 EC
ED ^ CD
OD 00 + OD
OC - OD 00 - 00 cos A
CD'
^ CD(OC -- OD)
00* - OD'^
OC + OD
= ^D(QC) - OD)
CD-
or
CD
OC sin A
1 - cos A
sin A
Q.KD,
CoR 1. Hence, squaring —
m 2 j^ _ (1 - cos A)' _ (1 — cos A)' _ 1 - cos A
2 sin^^S ~ 1 - cos^ A 1 + cos A'
1 - tan^ -o" A
2 cos A
.-. Art. 64, page 215,
1 + tan^ -^
1
•. Cos A =
A
1 + tan* 2
.(1).
TRIGONOMETRICAL RATIOS. 351
I ^,
») A. o A. • o -A.
cos-^^ COS--— - sm---.
Z 2i 2i
sin^ -^ COS- ^ + sm^ -^
COS- •—
= cos--- -sm-- (2.)
= 2cos=~ - 1 (3.)
= 2 (l - sin= ^) - 1 = 1 - 2sm= |-(4.)
19, To find the irlgonomeirical ratios of 15°, 75°, 120°,
135°, 150°.
(1.) Batios 0/120". _
Sin 120° = sin (180° - 120°) = sin 60° = ^-|,
Cos 120° --= - cos (180° - 120°) = - cos 60° = - I,
Tan 120" = - tan(180° - 120°) = - tan60° = - V3,&c.
(2.) Ratios of \^^\
Sin 150° = sin (180° - 150°) = sin 30° =i,
Cos 150° rr. - cos (180° - 150°) =: - cos 30° = - -^,
Tan 150° = - ten (180° - 150°) = - ton 30 = - -L
V3'
&C.
Sin 15° = AzJ cos 15° = ^^^-^, &c.
352 PLANE TRIGONOMETRY.
(3.) Ratios of 15°.
By last Art., tan -~ = — ~ — ; put A = 30°, or -- = 15°,
'^ 2 sm A __ 2
1 - yr
XI, x.« iRO 1 - COS 30° 2 o /Q
then tan 15 = -, — — -._ ^ = 2 - J3.
sm 30 1
From this result we easily get, Art. 8,
- 1
-— - cos
(4.) Ratios of 76\ __
We have, sin 75° = sin (90° - 15°) = cos 15° = -^^^^-^,
cos 75° = sin (90° - W) = sin 15° = ^^ 7 ^
2^2
tan 75° ^ tan (90'' - 15°) = cot 15^ - ^—r,
2 ■— V 3
= 2 + n/3^ &c.
(5.) i?a^ios 0/135°.
We have, sinl35'= sin(180''- 135°) = sin45°= -^y
v2
cos 135° = -008(180° - 135°) = - cos 45"
JL
~ " a/2'
tan 135° = - tan"(180° - 135°) = - tan 45'
= — 1, &c.
Ex. III.
1. Define a negative angle, and show that tan (—A)
= - tan A, when A lies between - 90° and - 180°.
2. Trace the changes of sign of sin A . cos A through
the four quadrants.
3. Trace the changes of sign of cos A + sin A, and of
• PQS A - gi» A, as A gfenges from - 45** to 315°,
TRIGONOMETRICAL RATIOS. 353
4. Assuming generally that cos 2 A = cos^ A - sin' A,
trace the changes of sign of cos 2 A as A changes from - 45^
to 315^
5. Write down the sines of 210^ 165^ 240°, - 120^
6. Show that sin (90° + A) = cos A, and cos (90° + A)
= - sin A, for any value of A from 0"" to 180°.
A —
7. Assuming generally that 2 cos' -^c- = 1 + cos A, and
r^- ..... ^ .
, ^ A. "A -
2Bin^ — = 1 - cos A, show that V^cos o = - \/l + cos A
and \f2'sm -^ = — Vl - cos A, when A lies between 360"
and 540^
A
8. Given cos A = 1 - 2 sin' -,, show that sin A
^ . A A
= 2 sin -^ cos-^ •
j^
0, Hence show that 2 cos i^ = - Ay 1 + sin A
- Vl - sin A, when ^ lies between 135° and 225°
Solve the following equations : —
10. Cos' A + f cos A = T«^.
11. Tan + 5cot0 = 6.
2
12. Sin A + sec A = —j^ + f
13. 2cos2A = 3 sin A.
14. Sin (A + B) = cos (A - B) = ^/i!-
15. Tan' A = 2 sin' A.
16. Sin (3 A + 75°) = cos (2 A - 15°).
5
17. Sec + cos = ;5~7r.- tan o.
18. Tan + cot = 4.
5 z
354 PLANE TKIGONOMETBY.
CHAPTEE Y.
LOGARITHMS.
20. Dep. — The logarithm of a number to a given Lase
is the index of the power to which the base must be raised to
obtain the number.
Thus, we may obtain the numbers 1, 10, 100, 1,000, 10,000,
<fec., by raising the base 10 to the powers 0, 1, 2, 3, 4, &c.,
respectively ; and hence, by the above definition, we have —
Log 1 = 0, log 10 = 1, log 100 = 2, log 1,000 = 3, (tc,
10 10 10 10
the suffix 1 being added to the word log to indicate that the
base is 10. It is usual, however, in common logarithms to
omit this suffix ; and hence, when there is no base expressed,
the student will understand 10.
Again, the numbers 1, 2, 4, 8, 16, &c., may be obtained
by finding the values of 2^, 2\ 2'"^, 2^, 2^, &c., respectively,
and hence we have by definition —
Log 1 = 0, log 2 == 1, log 4 = 2, log 8 =: 3, &c.
2 2 2 2
So also we find log 16 = 2, log 125 = 3, log 81 = 4, (kc.
4 5 3
Ex, Find log 256, log 216, and the logarithm of 9 to base
4 36 •
s/3.
Log 256 = log 4* = 4, by definition.
4 4
Log 216 = log 68 .= log {Q^Y = log 36^ = #, by definition.
36 36 36 36
Log 9 = log 32 =--• log (\/3)^ = 4, by definition.
Characteristics of Ordinary Logarithms.
21. Def. — The characteristic of a logarithm is the integral
part of the logarithm, and the fractional part (generally ex-
pressed as a decimal) is called the mantissa.
CHARACTERISTIC'S OF ORDINARY LOGARITHMS. 355
1, lumbers containing integer digits.
Every number containing n digits in its integral part must
lie between 10""*^ and 10",
Thus, 6 lies between ID® and 10^ 29 lies between 10^ and 10^,
839 lies between 10^ and 10% &c.
Hence the ordinary logarithms of all numbers having n
integer digits lies between (ii - 1) and 7i.
The integral portion or characteristic of the logarithm of a
number having n integer digits is therefore (n - 1).
Hence we have the following rule : —
Rule 1. — The characteristic of the logarithm of a number
having integer digits is one less than the number of integer
digits.
Thus, the characteristics of the logarithms of 32, 713*54,
8-7168, 56452, 73607*9 are respectively 1, 2, 0, 4, 4.
2. Numbers less tJian unity expressed as decimals.
All such numbers having n zeros immediately after the
decimal point lie between T^and - ^ ^ , or between 10"**
Thus, '3 Ilea between I ^d •!, or between 1 and r^, or 10** and
•027 lies between '1 and '01, or between — and -— , or 10~
10 lO'*
10
and 10- 9;
•000354 lies between '001 and '0001, or between -L and -— -,
10* 10*
or 10 ~ ^ and 10 "~ *, and so on.
Hence, by Def., Art. 20, the logarithm of any number hav-
ing n zeros immediately after the decimal point lies between
- n and - {n + 1). Hence, the logarithm is negative, and
the integral part of this negative quantity is n. It is how-
ever usual to write all the man tissue of logarithms as positive
quantities, and the negative integral part of the logarithm
will be the next higher negative integer, viz., - (n -\- 1).
We have therefore the foHowing rule : —
Rule 2. — Tlie cliaracteristic of the logarithm of a number
less than unity, and expressed as a decimal, is the negativs
356 PLANE TRIGONOMETRY.
integer next greater than the number of zeros immediately
after the decimal point.
Thus, the characteristics of the logarithms of -3, -0076,
•02535, -7687, are respectively -1,-3,-2,-1.
22. The logarithm of the product of two numbers is the
SUM of the logarithms of the numbers.
Let m and n be the numbers, and let a be the base.
Since m and n must be each some power of a, integral or
fractional, positive or negative, assume —
Z. y( Then, by definition of a logarithm,
X = log^ m, and y = log n.
Nowwe have mn = a' . a^ = a*+^, and hence, by definition,
log 7nn = X + y; vre therefore have —
( log (mn) = log m + log n. Q.E.D.
Cor. This proposition may be extended to any number of
factors.
Thus, log^ {mnpq) = log^ m + log^ n + log^jp + log^ q.
23. The logarithm of the quotient of two numbers is found
by SUBTRACTING the logarithm of the denominator from the
logarithm of the numerator.
Assuming, as in the last Art,, we have —
ic = log^ m, y = log^ n.
7/1/ Ui
Also, — = -^ = a""" ^, and hence, by definition,
^ n o? ■ ' *' '
m
loff — = aj - 7/: we therefore have—
^a n
24. The logarithm of the power of a number is found by
i^LTiPLYiNG the logarithm of the number by the index of the
power.
CHARACTERISTICS OP ORDINARY LOGARITHMS. 357
Let it be required to find log N^,
Assume N = a*, and therefore x ~ log iT.
We have N^ = (a')^ = a'", and hence, by definition,
log^ 2P = px = p log^ iV^. Q.KD.
Ex. 1.— Given log 2 = -3010300, and log 3 = 4771213,
find the logarithms of 18, 15, '125, 6*75.
Log 18 = log (2 X 3-) = log2 + 2 log 3 = -3010300 +
2(-4771213) = 1-2552726.
Log 15 = log (3 X V) .^ log 3 + log 10 - log 2 =
•4771213 + 1 - -3010300 = M760913.
Log -125 = log(^^3) = log 1 - 3 log 2 = - 3 X -3010300
= - -9030900 = - 1 +J1 - -9030900) = - 1 + -0969100,
or, as usually written, = 1-0969100.
Log6«75 = logi^ = Iog|^ = 31og3-21og2 = 3(-4771213)
- 2(-3010300) =-• -8293039.
Ex. 2. Find the logarithm of (^•^) ^ ('375)* ^^^.
(2-43)5 X (-032)H
given log 2 and log 3.
We have —
LogN = i log 2-4 + 4 log -375 - 5 log 2-43
-(-^) log -032.
1 , 2» X 3 ■ , 3 ^ , 3» 1 , 2'
" 2 ^"^-ro- ■*■ ^ ^"^2^ - ^ ^^^TO^ ^ 3 ^'^ 103
^ J (3 log 2 + log 3 - log 10) + 4 (log 3 - 3 log 2)
•- 5 (5 log 3 - 2 log 10) + ^ (5 log 2 - 3 log 10)
= (I - 12 + log 2 + (I + 4 - 25) log 3
+ (_ 1 + 10 - 1) log 10
= - V X -3010300 - V X -4771213 + V x 1
= - 2-6590983 - 9-7809867 + 8-5 = - 3-9400850
= 4 + (4 - 3-9400850) = 4" -0599150.
358 PLANE TRIGONOMETRY.
Ex. IV.
1. Find the logaritlim to base 4 of the following numbers:
16, 64, 2, -25, -0625, 8.
2. Find the value of log 32, log 25, log -729.
8 ^^ -81
3. Given log 2 - -3010300, and log 3 - 4771213, find
the logarithms of 12, 36, 45, 75, -04, 3-75, -6, -074.
4. Given log 20763 = 4-3172901, what is the logarithm
of 2-0763, 2076-3, -020763, -0020763?
5. Write down the characteristics of the common logarithms
of 29-6, -25402, -0034, 6176-003.
6. Given log 20-912 ^ 1-3203956, what numbers corre-
spond to the following logarithms:— 2-3203956, 6-3203956,
T-3203956, 4-3203956?
7. Given log 20-713 = 1-3162430, and log 20714 -
3-3162640, find log -2071457.
8. Given log 3-4937 = -5432856, and log 3-4938 =
•5432980, find the number whose logarithm is 3-5432930.
9. Given log 1-05 = -0211893, log 2-7 = 1-4313638, log
135 = 2-1303338, find the value oflog^^ll^^JI^.
(1'05)
10. Given log 18 == 1*2552725, and log 2-4 = -3802112,
find the value of log -00135.
11. What are the characteristics of log 1167, and log 1965?
12. Having log 2 = -3010300, and log 3 = -4771213, find
X when 18* = 125.
CHAPTEE YI.
THE USE OF TABLES.
25, Tables have been formed of the logarithms of all num-
bers from 1 to 100,000, and we shall now show how they are
THE USE OF TABLES.
iftO
practically used. "We shall not enter here upon the method
of forming the tables themselves.
The following is a specimen of the way in which the
logarithms of numbers are usually tabulated : — '■
No.
1
2
3
4
6
6
7
8
9
D.
7990
9025468
6522
6577
6631
5685
5740
6794
6848
6903
6957
91
6011
6066
6120
6174
6229
6283
6337
6392
6446
6500
92
6555
6609
6663
6718
6772
6826
6881
6935
6989
7044
93
7098
7152
7207
7261
7315
7370
7424
7478
7533
7587
94
7641
7696
7750
7804
7859
7913
7967
8022
8076
8130
54
95
8185
8239
8293
8348
8402
8456
8511
8565
8619
8674
96
8728
8782
8836
8891
8945
8999
9054
9108
9162
9217
97
9271
9325
9380
9434
9488
9542
9597
9651
9f05
9760
98
9814
9868
9923
9977
0031
0085
0140
0194
0248
0303
99
9030357
0411
0466
0520
0574
0628
0683
0737
0791
0846
8000
0900
0954
1008
1063
1117
1171
1226
1280
1334
1388
«•{
54 a;{
5
11
16
22
27
32
38
43
49
Thus, if the number consist of four figures only, we have
simply to copy out the figures in the column headed 0, prefix
a decimal point, and the proper characteristic.
Ex. Log 7991 = 3-9026011, log 7*995 = -9028185.
When we speak of a number consisting of four figures only,
we include such numbers as '003654, -07682, &c., the num-
ber of zeros immediately following the decimal points not
being counted.
Thus, log -07997 = J-9029271
log -007992 = 3-9026555.
When the number contains five figures, as, for instance,
79936, we look along the line containing the first four figures
— ^viz., 7993 — of the number until the eye rests upon the
column headed 6> the fifth figm-e. We then take the first
three figures of the column headed 0, and affix the four
figures of the column headed 6 in the horizontal line of the
fii-st four figures of the number.
Thus, log 79936 - 4-9027424
log -079927 = 2-9026935.
It will be seen from the portion of the logarithmic table
above extracted, that when the first three figures of the
logarithm — viz., 902 — have been once printed, they are not
560 . PLANE TRIGONOMETRY.
repeated, but must be understood to belong to every four,
figures in each column, until they are superseded by higher
figures, as 903. When, however, this change is intended to
be made at any place not at the commencement of a
horizontal row, the first of the four figures corresponding to
the change is usually printed either in different type, or, as
above, with a bar over it. Thus we have above 0031,
indicating that from this point we must prefix 903 instead
of 902.
Thus, log 79-986 = 1-9030140,
log -0079987 - 3-9030194.
26. To find the logarithm of a number not contained in the
tables.
Ex. Find the logarithm of 799-1635.
Since'* the mantissa of the number 79916-35 is the same as
the mantissa of the given number, and that the first five figures
are contained in the tables, we may proceed as follows —
(1.) Take out from the tables the mantissa corresponding
to the number 79916. This is -9026337.
(2.) Take out the mantissa of the next higher number in the
tables— viz., 79917. This is -9026392.
(3.) Find the difference between these mantissas. This is
called the tabular difference, being the difference of the
mantissse for a difference of unity in the numbers. We find
tab. diff: -= -0000055, which we call D.
(4.) Then assuming that small differences in numbers are
proportional to the differences of the corresponding logarithms,
we find the difference for -35 = -35 x -0000055 = -0000019,
retaining only 7 figures. This is often called d.
(5.) Now adding this value of d to the mantissa for the
number 79916, we get the mantissa corresponding to the
number 79916-35.
(6.) Lastly, prefix to this mantissa the proper charac-
teristic.
The whole operation may stand thus — •
M. of log 79916 = -9026337.' (1).
M. of log 79917 = -9026392
Tabular difference or D" = -0000055
V -^ Thus log 79916*35 ^ log (100 x 799 '1635) = 2 -flog 799-1635.
THE USE OF TABLES. 361
Hence, difference for '35 or cZ
= -35 X -0000055 =-0000019 (2).
Hence, adding (1.) and (2.) —
M. of log 79916-35 = -9026356.
.-. log 799-1635 = 2-9026356.
or better thus, omitting the useless ciphers —
M. of log 79916 = -9026337
M. of log 79917 = -9026392
.-. D = 55
Hence, cZ = -35 x 55 = 19
.-. M. of log 79916-35 = -9026356, as before.
In the next article we shall show how the required dif-
ference may be obtained by inspection from the tables.
27. Proportional parts.
We saw in the example just worked that the tab. diff.
(omitting the useless ciphers) is 55, and if we examine the
table in Art. 25, we shall find the difference between the
mantissse of any two consecutive numbers there to be 54 or
55 — generally 54. The number 54 is therefore placed in a
separate column at the right of the table, and headed D.
The student will understand that the tab. diflf. changes from time
to time, and is not always 54 or 55.
Now assuming as in (4.) of the last article, we have —
Diff. for
•1 = 54 X
•1=5
Diff for -6 = 54 X
•6 = 32
•2 = 54 X
•2 = 11
„ -7 = 54 X
•7 = 38
•3 = 54 X
•3 = 16
„ -8 = 54 X
•8 = 43
•4 = 54 X
•4 = 22
. „ '9 = 54 X
•9 = 49
•5 = 54 X
•5 = 27
We find therefore the numbers 5, 11, 16, 22, 27, 32, 38, 43,
49 placed in a horizontal row at the bottom marked P, in
the columns respectively headed 1, 2, 3, 4, &c.
Hence, if we require the difference for (say) -7, we take
out the number 38 from the horizontal row marked P, in-
stead of being at the trouble to find it by actual computation.
The following example will illustrate how we proceed
when we require the diflerenco for a decimal containing more
than one decimal figure. No explanation is needed.
362 PLANE TRIGONOMETRY.
Ex. Find log 7994 -3726-
M. of log 79943 - -9027804
Diff. for 7 = 38!
„ 2 = 11
6 = ^
.-. M. of log 79943726 = -9027843
^ Hence, log 7994-3726 = 3-9027843.
28. Having given the logarithm of a number to find the
number.
After the explanations of Art. 26, the method of working
the following examples will be easily understood : —
Ex. 1. Eind the number whose logarithm is 1-9030173.
Taking from the tables the mantissoe next above and below,
we have —
•9030194 = M. of log 79987
•9030140 = M. of log 79986 (1).
54 = D.
Again, -90301 73 = M. of log N (2).
Hence, subtracting ( 1 ) from ( ^ ) —
33 - (/, the difference between the logarithms of the re-
quired number and the next lower.
33
Now -— - — '61, the difference between the next lower
54
number and the required number.
Hence^-9030173 - M. of log 79986-61 ;
.-. 1-9030173 = log -7998661 ;
.'. '7998661 is the number required.
Ex. 2.^^ Find the value of lii^^f^J^.
(1-32756)*
We have —
log N = 31og 1-023 + i-log -00123 - 4 log 1-32756.
jS^ow, 3 log 1-023 = 3 X -0098756 = -0276268
1 log -00123 = \ (T3-0899051)
= \ (4 + 1-0899051) - ^ T-2724763
* The logarithms used in this example are taken from the tables.
TRIGONOMETRICAL TABLES. 363
/. adding, 3 log. 1-023 + i log -00123 =^3001031
Again, M. of log 13275 - -1230345
and difi: for -6 =- 196
.-. M. of log 13275-6 = -1230541
.-. 4 log 1-32756 = 4 X -1230541 = -4922164
Then, subtracting, log N" = 2-5078867
Hence we have, -5078867 = M. of log N,
and -5078828 - M. of log 32202 ;
39 = d,
also 135 = D,
39
and ^ = -29.
.-. -5078867 = M. of log 32202-29;
.-. 2-5078867 = log -03220229.
Hence -03220229 is the number requii-ed.
Trigonometrical Tables.
29. We use trigonometrical tables much in the same way
as we do tables of ordinary logarithms of numbers.
Tables have been formed of natural sines, cosines, &c., and
also of logarithmic sines, cosines, &c. It is with the latter
only we shall now deal, though many of our remarks apply
equally to the former.
As the values of the natural sines and cosines of all
angles between 0° and 90° are (Art. 11) less than unity, it
follows (Art. 21) that their logarithms are negative. To
avoid, however, printing them in a negative form, and for
other reasons, it is usual to add 10 to theii- real value, and
hence in using them we must allow for this. The same
thing is also done in the case of logaiithmic tangents, co-
tangents, secants, and cosecants.
We generally express the true logarithmic sine by log sin,
and the tabular logarithmic sine by L sin.
Hence, we have, log sin A - L sin A - 10,
log cos A = L cos A - 10, kc.
It muse be remembered in using the tables that, although
(Art. 11) the sine, secant, and tangent of an angle increase as
364
Plane trigonometry.
the angle increases from 0° to 90°, yet the cosine, cosecant,
and cotangent dimi7iish as the angle increases.
Hence, when any angle is not exactly contained in the
tables, we must add the difference in the case of a sine,
secant, or tangent; but subtract it in the case of a cosine,
cosecant, or cotangent.
And, conversely, when the given logarithm is not con-
tained exactly in the tables, we must in the case of the sine,
secant, or tangent take out the next lower tabular logarithm
as corresponding to the angle next lower; but in the case of a
cosine, cosecant, or cotangent, we must take out the next
higher tabular logarithm as corresponding to the angle next
lower in the tables.
We shall assume that small differences in the angles are
proportional to the corresponding differences of the logarithmic
trigonometrical ratios
Ex. 1. Find L sin 56° 28' 2r.
Referring to tables, we have —
Lsin56°28'
Tab. diff. for 60'' or D = 836
24
60 '
Lsin56'28'24''
diff for 24'' or c^ =
836
= 9-9209393
334
Ex. 2. Find L cos 29" 31' 28':
Now L cos 29° 31'
Tab. diff for 60'' or D = - 716
28
9-9209727
= 9-9396253
diff for 28"
. - ^^x716 = - 334
= 9-9395919
.\ Lcos29'31'28''
Ex. 3. Find the angle A, when L tan A = 9-8658585
We have 9-8658585 ^ L tan A.
JSText lower, 9-8657702 = L tan 36° 17',
883 — difference or d,
Also, 2648 = tab. diff. for 60" = D,
And .-^ X 60" = -^ X 60" = 20".
D 2648
Hence, 9-8658585 = Ltan 36' 17' 20".
TRIGONOMETRICAL TABLES. 3G5
Ex. 4. Find the angle A, when L cot A = 10-0397936.
We have, 10-0397936 = Lcot A,
Next higher, 10-0399770 = L cot 42^ 22',
1834 = (liiFerence or d,
Also, 2537 = tab. ditf. for 60'^ or D,
And 4 X CO'' = M|| X 60'^ = 43".
Hence, 10-0397936 = L cot 42° 22/ 43'^
Ex. V.
: 1. Given log 47582 = 4-6774427, and log 47583 =
4-6774518, find log 47*58275.
2. Given log 5*2404 = -7193644, and log 524-05 =
2-7193727, find log -5240463.
3. Given log -5614 5 = T-7493 111, and log 56-146 =
1-7493188, find log Ay-05614581.
4. Given log 61683 = 4-7901655, and log_ 616-84 =
2-7901725, find the number whose logarithm is 2*7901693.
5. Find the value of (1*05)^'^, having given log 1-05 =
•0211893, log 20789 = 4-3178336, and log 20790 = 4-3178545.
6. Find the compound interest of £120 for 10 years at 4
per cent, per annum, having given log 1-04 — -0170333, log
14802 = 4-1703204, and log 14803 = 4-1703497.
7. A corporation borrows £8,630 at 4 J per cent, compound
interest, what annual payment will clear off the debt in 20
years?
Log 1-045 = -0191163, log 4-1464 = -6176712, and
log 4-1465 = -6176817. i
8. Find the value of-^ ^^^ ^ ^ , having given
Log 1-032 = -0136797, log 34722 = 4-5406047.
Log 3762 = 3-5754188, log 26202 = -4183344.
Log 34721 = 4-5405922, log 26203 = -4183510.
36G PLANE TRIGONOMETRY.
9. Find Lsin 32"^ 28^ 31", having given Lsin 32° 28' =
9-7298197, and Lsin 32° 29' = 9-730bl82.
10. Find Lcosec 43° 48' 16'', having given L sin 43° 48' =
9-8401959, and L sin 43° 49' = 9-8403276.
11. Required the angle whose logarithmic cotangent is
10-1322449, having given L cot 36° 25,' - 10*1321127, L cot
36° 26^ - 10-1318483.
12. Construct a table of proportional parts, having given
163 as the tabular difference.
13. In what time will a sum of money double itself at 5 per
cent, per annum, compound interests
14. Find x when 1*03* = 1*2143, having given that log
1-03 = -0128372, and log 12143 = 4*0843260.
15. Solve the equation 2^"^""^ — 40 = 9*2*, having given
log 2 - -3010300.
16. Given L cos 32° 45' - 9*9341986, D = 752, find Lcos
32° 45' 12^; and Lsec 32° 45' 20,'.'.
17. Given Ltan 28° 38' = 10*2628291, D = 3003, find
Ltan 28° 37.' 15'", and Lcot 28° 38/ 42'^
18. Find the angle whose logarithmic cosine is 9*9590635,
having given —
Lcos 24° 29' = 9*9590805,
Lcos 24° 31/ = 9*9589653.
CHAPTER YIL
PROPERTIES OF TRIANGLES.
30. The sines of the angles of a triangle are proportional
to the opposite sides.
We shall designate the sides opposite to the angles A, B,
C, by the small letters a, 5, c, respectively.
PROPERTIES OF TRIANGLES. 367
Dmw AD perpendicular to BC, or to BC produced.
Then sin B =
And sin C =
AD _ AD
AB ~ c
AD AD
AC " 6 ••
.(1).
.(2).
IT /I \ /o\ smB AD AD h
Hence, (1 ) ^ (2),-^—- = ^ =, _,
^ ' ^ sinC c c
or
sin B sin C
It follows, therefore, from the symmetrical nature of this
sin A sin B sin C n r n
equation, that — : — = —i — = — • v--^-^-
a b
31. In any triangle, cos C
a^ + h^ -^ c"
2ab
Taking the figures, of the last article, we have —
(1.) Whe7i C ia an acute angle—-'
By Euc. IL, 13, AB'^ = BC^ + AC^ - 2 BCCD,
CD
Now vjj = cos C, or CD = AC cos C.
Hence we have, AB« = BC^ + AC^ - 2 BC.AC cos C,
or r ^ a^ + h^ - 2ab cos C.
•. cos C =
2a6 '
3G8 PLANE TRIGONOMETRY.
(2.) When C is an obtuse angle y as in the second figure —
By Euc. IL, 12, AB^ = BC^ + AC^ + 2BC.CD;
and CD = ACcosACD = ACcos(180°'-- C) = - ACcosG-
Hence, AB^ - BC^ + AC^ + 2 BC ( - AC cos C) ;
or, c^ = a^ + 6^ - 2 ah cos C.
ri a^ + W - C^ , «
/. cos U = — , as before. •
c^ 2a6
From iheform of this result we have also — .;
5^ + c2 - a2
Cos A -
CosB =
2 5c
a" + c^ - y
2ac
32. To express the sine of any angle of a triangle in terms
of the sides.
We have —
Sin^A = 1 - cos^ A = 1 - (^±^^y
^ ( 2 hcf - (b^ -r c" - ay
{2bcy
^ {(b + cf - a'}{a^ - (5 - cf}
" (2bcf '
(a + h + c){b + c — a) {a + c -b) (a + b -c),
" (2bc)^
Hence, taking the square root, and taking the positive sign,
because (Art. 11) the sin A is always positive when A is the
angle of a triangle, we have —
Sin A = -— ^(a + b -{■ c) {b + c - a) {a + c - b) {a + b - c).
Let a + b + c = 2 Sf then b + c - a ~ 2 (s - a),
a + c - b = 2 {s ^ b)y a + b - c = 2 (s - c).
Hence, sin A = -^j- J2s.2{s - a),2{s -b).2 {s - c)
2 , .
= -^ v^ (« - ^') (« - ^) {^ - c)-
PROPERTIES OP TRIANGLES. 36ll
From tlie form of this result we have also —
Sin B = ^^ V5(*-a)(s ^ 6) (5 - c),
sill C = -^ 4;\s - a) (« - 6) (« -"^.
33. To find the area of a triangle.
Using the figures of Art. 30, we have —
A ABC = i BC . AD, or, since AD - AC sin C,
== iBC.ACsinC
= |a6sinC (1).
From Reform of this result we have also —
A ABC = JacsinB (2).
andAABC = ^^csinA (3).
The results in (1), (2), (3) express the area of a triangle in
terms of two sides, and the included angle.
We will now express the area in terms of the three sides.
We have —
A ABC = J a6 sin C, or, by last Art.,
2 , .
= Ictb,-^ ^8(s - a) (s - h){8 - c)
= Ji{8 - a){a " b) (^~c) (4).
34. To express the sine, cosine, and tangent of half an
angle of a triangle in terms of the sides.
We have, Art 18,
2 A . b- + c- -- a^
l-2sm ■:jr = cosA = 2"a '
^ . A , 52 + c--a^ a^ - Ib-cY
...sin^-A 2 6c " 26c
(a + c - b ) {a -{■ h - c) _ 2(s - 6 ).2(g - c).
2 6c - 2 6c"
. , A {s -b){s- c)
or,sm-2-= ^ •
5
A 1(8 -b) {s^ r)
2~= ^ 6c ^- ^-
370 PLANE TRIGONOMETRY.
The form of tMs result gives us also —
Sin3^_ / (^-«)(^-cT ^C _. I{s-a){s-b)
Again, Art. 18, we have —
„ .A P + c'-a"
2C0S-Y - 1 = cos A = ij-T ;
or, 2cos- 2 = 1 + 2j^— = ^-^^^
(a + b + c) (b + c - a) 2s.2(s — a) ^
2 bo " WFg '
or,cos^2 - bo
A /7
2 ^ V"
A fsis - a)
cos TT -
5c
The/orm of this result gives us also —
.(2),
B
cos-rt
Isis - b) C js{s - c)
or
Now, (1) -r (2), we have —
Sin — t — f
2_ ^ / (g - h) (s - c) s{s - a)
Cos^^^ ^' ^ ^ ^^ '
tan^ = ,/EZ553 (3).
Theybrm of this result gives us also —
tan ^-= /(7^ g) (. - c) ^^^C ^ /gZ"^)FZ^).
2 Sl s{s - b) ' 2 V s(5 - c)
35. In any triangle ABC, a = 6 cos C + c cos B.
Using the figures of Art. 30, we have —
(1.) When C is acute —
BD =r AB cos B, and DC = AC cos C.
,-. BD + DC = AC cos C + AB cos B, or
c^ = 5 cos C + c cos B.
PROPERTIES OF TRIANGLES. 371
(2.) When C is obtuse—
BD = AB cos B,
and DC = AC cos ACD = AC cos (180^ - C) = - AC cos C.
.-. BD - DC = AC cos C + AB cos B,
ov a = b cos C + c cos B, as before.
The form of this result gives us also —
6 = a cos C + c cos A, c = a cos B + 6 cos A.
CoR. 1. Hence sin (B + C) = sin B cos C + cos B sin C.
b c
For we have, 1 = — cos C + — cos B, or, by Art. 30, >
a a
sin B ri sin C -o
= - — - cos C -f — — - cos B,
sm A sin A
or, sin A = sin B cos C + sin C cos B.
But sin A = sin (180° - A) = sin (B + C).
.-. Sin (B + C) = sin B cos C + cos B sin C.
CoR 2. Hence also —
sin (B - C) = sin B cos C - cos B sin C.
For, since this result has been proved for any angles of a
triangle, it will be also true for a triangle which contains an
angle supplementa/ry to B; that is, it will be true, if we put
180' - B for B. We then have-
Sin (180° - B + C) = sin (180' - B) cos C
+ cos (180° - B) sin C.
Butsin(180°-.B + C) = sin(180° - B"^C) = sin(B-C),
sin (180° - B) = sin B, cos (180° - B) = - cos B.
Hence, sin (B - C) = sin B cos C - cos B cos C.
Note. — The results of Cor. 1 and Cor. 2 have been proved only
for angles less than two right angles. We shall see iu Vol. II. they
are generally true.
86. In any triangle ABC —
Tan J (B - C) = |-^^cot i A
We have, Art. 30,
Sin B b sin B - sin C b - c
"SmC" " ? ^^ sin~B + sin C = TTc ^^''
S72 PLANE TRIGONOMETRY,
/B + B - C)
Now, sin B = sill j - — k — + — s — j > ^^> ^7 ^^r. 1, Art. 35,
B + C B-C B + C.B-C ,^,
= sin — 2 — ^^^ — 2 — '^ ^^^ — 2 — ^^ — ^ — ( /•
/ B + C B - C )
and sin C = sin i — « — o — f > or, by Cor. 2, Art. 35,
.B + C B-C B + C.B-C ,ox
= sin _^_ cos— ^ cos _^_sin_^— (3).
B + O B — O
(2) - (3), then sin B - sin C = 2 cos — ^ — ^in — ^ — >
B 4- B — O
and (2) + (3), then sin B + sin C = 2 sin — ^ cos — « — '
^ B + C . B-C
• -D • /^ 2 cos — K — sin — K —
smB - sinC 2 2
•'• slnB + sinC " ^ . B + C B"^^
2 sin — ^ — cos — 2 —
B-C
B + C B-C /^, A\
= cot —^ — ' tan 2 — = ^^^ I ^^ "~ "9)
tan -
A B - C
= tan 2" tan — ^ — W*
... . A B - C 6 -c
(4) = (1), then tan -^ tan — ^ = ^-^'
B-C 5-c ,, ^^^
.-.tan ^g— =6-T^c^*i^- ^•^•^•
Ex. VI.
1. If a = 5, C = 30^ sin A = -i, find c.
2. Find a, having given 6 = 12, c=15, A = 60'.
3. Find tan A when a = 6, 6 = 7, c = 8.
j4. What is the area of a triangle whose sides are 48, 52, 20?
5. Given two sides of a triangle to be 18 and 24, and the
included angle 45°, find the area.
PROPERTIES OF TRIANGLES. 373
6. Two of the sides of a triangle are as 2 : 1, and the iu-
cluded angle is 60°, find the other angles.
7. In any triangle show that —
6 cos C - c cos B == si a?' - 4cbc cos B ^os C.
sin B sin C
8. Show that the area of a triangle =. ^a\-
sin A
9. An object is observed from two stations 100 yards
apart, and the angles subtended by the distance between the
object and either station are 45° and 60° respectively. Find
the distance of the object from each station.
10. An observation is made from a point known to be
distant 120 and 230 yards respectively from two trees, and
the angle which the trees subtend is found to be 120°. Find
the distance between the trees.
11. If a sin^ + 6 cos^ e = mA ' ^ , , ,
b sin^ + a cos^ ^ = n, > then _ + _-=_ + _.
a tan = 6 tan 0, j ^^ ^ ''* ^
1 2. If a/, 5/, c' be the sides of the triangle formed by joining
the feefc of the perpendiculars from the angles A, B, C of the
triangle ABC upon the opposite sides, then —
ai , b' cl
a" b^ c' 2abc
13. A perpendicular AD is drawn from the angle A of a
triangle, meeting the opposite side BC and D ; and from D a
perpendicular is dmwn to AC, meeting it in E. Show that
DE = 6 sin C cos C.
14. Show that the length of AD in the last example —
6c sin A + ac sin B + a6sinC
3 a
15. Show that Js (s — a) (s - 6) (« — c) = J a6, when the
triangle is right-angled at C.
16. Show that (a + b + c)- sin A sin B
= (sin A + sin B + sin C)^ ab,
17. Show that in any triangle —
cos^A + cos-B + cos-C + 2 cos A cos B cosC = 1.
374
PLANE TEIGONOMETRY.
18. The sides of a triangle ABC are in arithmetical progteg-
6^/3
Bion ; show that its area = — ^ J {2 a - b)(3b - 2 a).
CHAPTER VIII.
SOLUTION OP RIGHT-ANGLED TRIANGLES.
37. A triangle can always be determined when any three
elements, with the exception of the three angles, are given.
In the latter case we have only the same data as when two
angles are given, for the third can always be found by sub-
tracting the sum of the other two from two right angles
(Euc. I., 32).
Hence a right-angled triangle can always be determined
when any two elements, other than the two acute angles, are
given besides the right angle. And when one of the acute
angles is given, the other may be obtained by subtracting it
from a right angle.
We have the following cases : — >
Case 1, Wli^n the two sides containing the right angle are
given,
^A. We shall take C as the right angle
in every case.
Now tan A =
y
or, L tan A — 10 = log a - logb;
or, L tan A = 10 -f log a - log6...(l).
This determines A, and we then have
B = 90'' - A (2).
Also, - = siii A, or log a
loofc = Lsiil A - 10.
.-. log c = 10 + log a - Lsin A ,
Hence the three elements, A^ B, c, are detei-mined*
.(3).
SOLUTION OP RIGHT-ANGLED TRIANGLES. 375
Case 2. Wlien the hypothenuae and a side are given.
Let a be the given side.
We have sin A = -, or L sm A - 10 = log a — log c.
c
»\ LsinA = 10 + log a - logc (1),
and B = 90^ - A (2),
also b^ = c^ - a^ = {c + a) {c - a) ;
.-. log 5 = ^ {log(<^ + a) + log(c - a)} (3).
Case 3. Wlien an acute angle and a side are given.
Let A, a be the given angle and side.
ThenB = 90^ - A (1),
also - = tan B, or log 6 = L tan B - 10 + log a (2),
and — - sin A, or log a - log c 5* L sin A - 10,
or logc = 10 + log a - LsinAi
Case 4. WIten the hypotJienuse and an acute angle are given*
Let A be the given acute angle;
WehaveB = 90° - A .;.. (1),
Also- = sin A, or log a = logc + LsinA - 10... (2).
c
And- = cos A, or log 5 = log c + Lcos A - 10... .(3).
c
It is evident, from Art. 30, that when the angles only of a
triangle are known, we can determine the ratio only of the
three sides of the triangle to each other.
Ex. 1. Given A = 23' 41', a = 35, solve the triangle;
This is ail fexample of C^tse 3.
We have B = 90' - 23^ 41' = G6' 19'.
Again, log b - Ltan B - 10 - log a
- Ltan 66' 19'' - 10 - log 35
= 10-3579092 - 10 + 1-5440680
= 1-9019772 = log 79-795
.-. b = 79-795.
376 PLANE TRIGONOMETRY.
Also log c = 10 + log a - LsiiiA
= 10 + log 35 - Lsin23°41'
= 10 + 1-5440680 - 9-6038817
= 1-9401863 = log 87-134
.-. c = 87-134.
Ex. 2. Given a - 214, & = 317, solve the triangle.
This falls under Case 1.
We have L tan A = 10 + log a - log h
= 10 + log 214 ~ log 317
= 10 + 2-3304138 - log 2-5010593
= 9-8293545,
Next lower in tables is 9-8292599 = Ltan 34' V\
.-. d = 946.
Also, by tables, B = 2724,
And^ X 60'/ ^ ^^44:^' = 21'^ nearly.
D • 2724 ^
.-.Ltan A - Ltan 34'!' 21",
or A = 34°1'21".
Hence B = 90 - 34° 1/ 21" = 45° 58/ 39''.
And similarly may c be determined.
Ex. VII.
1. Given a = 32, A = 63° 45 ^ find h.
Log 32 = 1-5051500, Loot 63° 45' = 9-6929750,
Log 15780 = 4-1981070, log 15781 = 4-1901345.
2. Given c = 151, A = 37° 42/, find a.
Log 151 = 2-1789769, Lsin 37° 42/ = 9-7864157,
Log 92340 = 4-9653899, tab. difi: = 47.
3. Given c& = 60, c = 65, find 6, A.
Log 2 - -3010300, log 3 - -4771213,
Log 65 = 1-8129134, Lsin 67° 22/ = 9-9651953,
Lsin 67° 23/ = 9-9652480.
4. Given a = 73, 6 = 84, find A, c.
Log 73 = 1-8633229, Ltan 40° 59/ = 9-9389079,
Log 84 = 1-9242793, Ltan 41° = 9-9391631,
Lsin 40° 59/ = 9-8167975, Lsin 4P = 9*8169429,
Log 111-288 = 2-0464479.
SOLUTION OF OBLlQtJU-ANGLED TtllANGLES. 377
5. Given B = 71° 41' 10", c = 24, find h.
Log 24 = 1-3802112, Lcos 18^ 18' = 9-9774609,
Lcos 18** 19' = 9-9774191, log 2278-4 = 3-3576300,
Log 2278-5 = 3-3576490.
6. Given a = 293, c = 751, find h.
Log 1044 = 3-0187005, log 458 = 2*6608655,
Log 691-49 = 2-8397830.
7. Given a = 12, A = 30^ find 6, c.
8. Given c = 10, B = 75°, find a, 6.
9. Given a = 17, c = 34, find A, h.
10. Given a = 5, 6 = 5 ^3, find A.
11. Given a = 28, B = 15°, find the length of the perpen-
dicular from C on AB.
12. ^CD is the perpendicular from C on AB, and DE the
perpendicular from D on BC. Given B = 60°, a = 20,
find DE.
CHAPTER IX.
SOLUTION OF OBLIQUE-ANGLED TRIANGLES.
38. Given the three aides of a triangle, to find tlie remaining
parts.
J2 ^ ^2 _ ^2
We have, Art. 31, cos A = — , from which A
2 he
may be determined ; and from similar formulae we may find
B and C. These formulae are not however adapted to loga-
rithmic computation. We shall therefore find it generally
advisable to use the formulae of Art. 34.
We have, sin ^ = fj- U^ £,
A _ 18(8 - a)
2 V be '
" A
cos
A f {s -b){8- c)
S78 PLANEl TRiaONOilETRV.
From either of these formulse we can determine A, and from
similar formulse determine the other angles.
39. Given one side arid two angles, to Jlnd the remaining
parts.
Of course the third angle is at once known, Let a be the
given side.
We have, Ai-t. 30, b - "^"^^
' ' sin A
a sin C
sm A
hoth of which formula) are adapted to logarithmic com-
putation.
40. Given two sides and the included angle, to find the
remaining parts.
Let h, c be the given sides, and A the included angle.
I - c
We have, Art. 36, tan |- (B - C) = t cot J A.
This formula is adapted to logarithmic computation, and
determines | (B - C).
We know also |- (B + C), for it is the complement of J A.
Hence, B and C are easily determined. .
Then, Art. 30, a - —. — =p-> which determines a*
sm jj
Note. — When the two given sides are equal, the solution may be
effected more easily hy drawing a perpendicular from the given angle
upon the opposite side, and so bisecting it. By drawing a figure, it
is easily seen that, in this case, B = C == 90* ~ i- A, and a = 2 6 cos B.
41. Given two sides and an angle opposite to one of them, to
find the remaining parts.
Let a, h, B be the given elements.
Then we have. Art. 30, sin A - y sin B.
(1.) Let the value of , sin B be unity.
We then have sin A = 1 = sin 90°, and .'. A == 9i)^
Hence the triangle is right-angled at A.
SOLLTIOX OF OBLIQUE-ANGLED TRIANGLES. 379
(2.) Let tlie value of , sin B be >• 1.
We then have sin A >- 1, which is impossible.
Hence, in this case, it is impossible to form a triangle
with the given elements.
(3.) Let the value of t sin B be -< 1. #
Then, since, Art. 15, the sine- of an angle is the same as
the sine of its supplement, there are two values of A which
satisfy the equality, sin A = t sin B, and these values are
supplementary.
Let A, A' be the two values, then the relation between
them is A + A' = 180°.
If a is not greater than h, then A is not greater than B,
and there is no doubt as to which value of A is to be taken.
If, however, a is greater than h — that is, if the given angle
is ojDposite to the less of the given sides, we must have A
greater than B, and both values of A may satisfy this con-
dition. This particular case, when the given angle is opposite
to the less of the given sides, is called the ambiguous case.
We will illustrate this geometrically,
42. Tlie amhigiums case.
Let a, by B be given to construct the triangle.
Draw the line BC equal to the
given side a, and draw BA making
an angle B with BC.
Then with centre C and radius
CA equal to b describe an arcs
meeting BA or BA produced iit
A and A', and join CA and CA'.
Each of the triangles ABC,
A'BC satisfies the given conditions. B^
For, in the triangle ABC, we have BC = a, AC = b, and
Z ABC = B ; and in the triangle A'BCj we have BC = a,
AC = 6, and Z A'BC = B.
Again, the sides BA and BA' correspond to the two valuer
380 t>LAN^fi trftlGONOMETRY.
of c which are obtained from the two values of A in the
equality sin A = r sin B (see last Art.).
And the angles ACB and A'CB correspond to the two
values of C which would also be found.
Cor. If a perpendicular CD be drawn from C upon AA'
and if c' and c be the lengths of BA' and BA respectively, it
may be easily shown that c' + c = 2 a cos B, and c' /^ c =
2 6 cos A.
Ex. VIII.
Solve the following triangles, having given —
1. 6 = 12, c = 6, A = 60°.
2. a = 18, 5 = 18 x/2;A = 30^
3. a = 5 ^,b = 5 J2;c = | ( ^6 + J2).
L a=12,B = 60°, G - 15°.
5. a= 3 J2 + J6,b = 6yC = 45°.
6. a = 10 ^/3; b = 15 x/27A = 45°.
Given —
7. 6 = 251, c = 372, A = 40° 32', find B and C.
Log 121 = 2-0827854, L cot 20° 16' = 10-4326795,
Log 623 = 2-7944880, L tan 27° 44' = 9-7207827,
Ltan27°45' = 9-7210893.
S. a= 237, b = 341, B = 28° 24', find A.
Log 237 = 2-3747483, L sin 28° 24' = 9-6772640,
Log 341 = 2-5327544, L sin 19° 18' = 9-5191904,
L sin 19° 19' = 9-5195510.
9. C = 26° 32', and a : 5 : : 3 : 5, find A, B.
Log 2 = -3010300, L cot 13° 16' = 10-6275008,
L tan 46° 40' = 10-0252805, L tan 46° 41 ' = 10-0255336.
10. ti = 14, 6 = 16, c = 18, find A, B.
log 2 = -3010300, log 3 - -4771213,
L tan 24° 5' = 9-6502809, Ltan 24° 6' - 9-6506199,
L tan 29° 12' = 9-7473194, Ltan 29° 13' = 9-7476160.
HEIGHTS AND DISTANCES. 381
11. a = 3, 5 = 2 A = 60°, find B, C, and c.
Log 2 = -3010300, log 3 = -4771213,
L sin 35° 15' = 9-7612851, L sin 35° 16' = 9-7614638,
Log 1-3797 = -1397847, log 1-3798 = -1398161.
12. a = 5, 6 = 6, c = 7, find A.
Log 2 = -3010300, Ltan 22° 12' = 9-6107586,
Log 3 = -4771213, L tan 22° 13' = 96111196.
13. If c, c' be the two values of the third side in the
ambiguous case when a, by A are given, show that —
(c - c'Y + (c + c'Y tan- A = 4 a^.
14. If a, b, A are given, show from the equation —
b^ + c^ - 2bc cos A = a^;
that if c and c' be the two values of the third side —
cc' = b^ - a^, and c + c' = 2 6 cos A.
15. Show also from the same equation that there is no
ambiguous case when a = 6 sin A, and that c is impossible
when a < 6 sin A.
16. Having a - 6, A, B, solve the triangle.
17. Given the ratios of the sides, and the angle A, solve the
triangle.
d> A
18. If in a triangle tan J (B — C) = tan- ^ cot ^, show that
b cos (p = c.
CHAPTER X.
HEIGHTS AND DISTANCES.
43. "We shall now show how the principles of the previous
chaptei-s are praictically applied in determining heights and
distances.
We have not space here to describe the instruments by
which angles are pi-actically measured, but we shall assume
that they can be measured to almost any degree of accuracy.
382
PLANE TRIGONOMETRY.
AC
tan ABC or tan 0.
a
44. To find the height of an accessible object
Let AC be the object, and let
/iA any distance BC from its foot be
measured.
At B let the a^igle of elevation
ABC be observed.
Suppose BC = a, Z ABC = 0.
Then, we have —
AC
BC
/. AC = a tan e, the height required.
lEx. Let a = 200, and = 30°.
Then AC - 200 tan 30° = 200. ~ = -^^ '
v3 3
45. Tofmd the height of an inaccessible object
^A. At any point B in the
horizontal plane of the base
let the angle of elevation
ABC be observed.
Measure a convenient dis-
tance BD in the straight line
CB produced, and observe
the angle of elevation ADO,
Let BD = «, ZABC = 0, Z ADB - c/>.
Then, Euc. I. 32, Z BAD = - <^.
AB sinADB_AB sin</>
sin(0 - 0)'
Now,
BD
sin ADB AB
or- — -
sin BAD a
AB = a.
AC
sm <t>
sin(0 — </))*
(!)•
Again, -v-g = sin ABC = sin 0, .\ AC = ABsin0,
or, from (1) AC = a!!^L^^?L^.
sm (o - <p)
Ex. Let BD = 120, = UU", ^ = 45^
Then, AC = 120.^!^.^;^^,^^^
sm (60 - 45 )
HEIGHTS AND DISTANCES.
383
^ 120
120.
sin 60° sin 45
sin 15'
; — = 120.
n/3 - 1
2^/2'
V3 -^ 1
60 (3 + V3).
46. To find the height of an inaccessible object when it is not
convenient to measure any distance in a line with the base of
the object.
Let a distance BD be
measured in any direction in
the same horizontal plane as
BC, and let the angles ABC,
ABD, ADB be observed.
Let BD = a, Z ABC = a
ZABD = ^,ZADB = y.
Tlien, Euc. I., 32, Z BAD
- 108° - (^ + y).
A.B ^ sin ADB _
^^^^' BD sin BAD ~ 2
sm y
AB
a
AC
sin (^ + y)
L {180°
AB =
(/3 + 7)}
,or
Again, - . = sin ABC = sin a,
A-XJ
a.-^-- (1).
sm (^ + y) ^ '
*. AC = AB.sin a.
or from (1), AC
a ? - " -— 7
sin (/3 + y)'
47. To find the distance of an
object by observation from the top
of a lower ivhose height is
hiovyii.
Let B be the object in the same B
horizontal plane with c the foot
of the tower, and let the angle of deprcBsxon DAB be
observed.
;^t AC = A; Z DAB = e,
381 PLANP TRIGONOHETRY.
Then we have, ?^ = cot ABC = cot DAB, or?? = cot o,
AC h
,', BC = 7i cot 0, the distance required.
Cor. Suppose B to be not in the same level with C, Let
m be the height of B above C, then B is on the same level
with a point which is 7b - m from A. . *. BC = {h - 711) cot 0,
Ex. IX.
1. Find the height of a tower 200 yards distant when it
subtends an angle of 15°.
2. From the top of a tower, the angle of depression of a
point in the horizontal plane at the fbot of the tower was 30°,
Given the height to be 60 ft., find the distance of the point.
3. The angle of depression of two consecutive milestones
in a direct line with the summit of a hill were observed to be
60^ and 30°. Find the height of the hHl.
4. An object is observed from a ship to be due E. After
sailing due S. for six miles it is observed to be N.E. Find
the distance from the last position of the ship.
5. There is an object A, and two stations B and C are
taken in the same plane. Given that BC = 50, Z ABC =
G0^ Z ACB = 30°, find AB, AC.
6. The elevation of a tower is found to be 45°, and on ap-
proaching GO feet nearer the elevation is 75°. Find the height
of the tower.
7. Wishing to know the breadth of a river I observed an
object on the opposite bank, and, having walked along the side
of the river a distance of 100 yards, found the angle subtended
by the object and my first station to be 30°. Find the
breadth of the river.
8. A person, standing exactly opposite to the centre of an
oblong which measures 16 ft. by 12 ft., and such that the line
drawn from the centre to his eye is at right angles to the
oblong, observes that the diagonal subtends an angle of
60°. Find his distance.
9. The angles of elevation of the summit of one tower,
whose height is h^ are observed from the base and summit of
HEIGHTS OP DISTANCES. 385
another, and found to be e and </> respectively. Show that
the height of the second tower —
, tan
= A , -
tane - tan</)
10. From the top of a tower the angles of depression of two
objects in a direct line, and whose distance from each other
is a, are o, /3 respectively. Show that the height of the
tower —
cot /3 ~ cot a
11. A person, having walked a distance a from one comer
along a side of an oblong, observes that the side immediately
behind him subtends an angle a, and the side in front an
angle A Show that the dimensions of the oblong are —
a tan a, a (1 + tan a cot p).
12. Three points A, B, C form a triangle whose sides are a,
h, c respectively, and a person standing at a point S, such that
SA is at right angles to BC, observes that the side AC sub-
tends an angle 0. Show that the distance of S from B —
= -«— J{a^ + c^ - b'^f + (cr + b'^ — c^ cot" e.
13. A person walks a yards from A to E along AB the side
of a triangle ABC, and observes that the angle AEC = « ;
he also walks b yards from B to F along BA, and observes
Z CFB = p. Having given AB - c, find BC and AC.
14. A tower is observed from three stations A, B, C, in a
straight line not meeting the tower, to subtend angles «, p, y
respectively. Show that if AB = a, BC = b, the height of
the tower —
^a
ab(a + b)
a cot*"* y — (a + 6) cot'- /3 + 6 cot'- a
When are the conditions impossible?
15. From two stations whose distance apart is a, and which
are due W. and due S. respectively of one end of a wall, the
angles subtended by the wall are each a. Show that tho
length of the wall is a sin o.
38G PLANK TRIGONOMETRY.
16. The angles of elevation of the top of a tower, whose
height is h and standing on a hill, are «, /3, when observed
from two stations a miles distant, and in a direct line up
# the hill. Show that if be the slope of the hill —
a sin a sin /3
cos
h sin (j3 - a)
17. The elevation of a tower was observed to be «, but
on walking in the horizontal plane a distance a at right
angles to the line joining the first position and the foot of
the tower, the elevation was /?. Show that the height of the
tower was —
Vcot^ p - cot^ a
18. The angles of depression of two objects in the same
horizontal plane, as seen from the top of a tower, are and </>
respectively, and the angle they subtend is a. Show that if
h be the height of the tower, the distance between the
objects —
= Ih vcosec*^ + cosec^ </» ~ ^ cosec cosec </> cos a.
ANSWERS.
I. — Page 15,
1. 45-23, 290, -2367, -7, &c.
2. '0005, IMl, -040020, -45, <fec.
3. Three thousand four hundred and sixty-seven thou-
sandths; thirty-four, and sixty-seven hundredths; three
thousand four hundred and sixty-seven milliouths; three,
and four hundi-ed and sixty-seven thousandths.
4. 35-90846, 29130-19391, 60-0239.
5. -7237, 3-32091.
6. 69-5289, 5-06679, -41481.
7. -026, 7708-71.
8. -09, 24-356706, -003627, -289, -0096, -00016384.
9. 200, -00125, 4000, -2295, -006, -5002, «tc.
10. 170000. 11. 4-97 ike, 1. 12. -2.
IL— Page 23.
1 '-? 1^ X9P-1 A.**-! 10 1 3L?J?,
-*"• 7 ' 9 * 10 ' 17 ' 999* 64 *
2. ''V, VD^ w, w, w, w-
3. G, 7, 12, 11, 9, 14; V. V, V. Vo", W, W-
4. 7|, Ij, 7i, 16t, 71i, 31i, &c. 5. 52.
6. 15, A. H. «, 17A» 1«; \hi^
7. h z\, H; ih ^f, U; A. Vtf». «•
8. ,»„ 6, 1, 1, C. 9. i, H> n> tU, l, I-
10. Uf 11. 3. 13. 28.
388 ANSWERS,
III,— Page 28.
1. 4 X 11, 2«, 2 X 3 X 5^ 2- X 32 X 7, 3^ x 47,
2 X 3 X 7 X 11.
2. 2 X 7 X 32 X 11, 2^ X 5 X 43, 2^ x 3^ 5^ x 7 x IP,
22 X 3 X IP, 2^ X 3* X 11.
52 3 7 li _4_ 4. • IE. 2 A 1 ft 3 1 XT ' n 3 6.,
• 61 T> 8> yo> ITOJ y^ 32> 36> T6> TJ 3> Ttf^ y> 6> TT>
17. JLO -rll-T
36> 35> TyyT*
6. 19, 7, 13, 6, 41, 729, 14, 11, 39.
7. I, h t\, t\, if, Is t, f, if, ^, 5^, li.
9. 2 days out of every 7.
10. 2:7. 12. 15, 10, 6.
IV.— Page 31.
1. 24, 1260, 2520, 1260, 5460, 33300.
24 f> «A _8 33. 24 36 35: 44. 3 36 5 6_ 231 234.
• "6 0> "B'OJ 6^> "6 0^ 8 4> ¥T> ir4> ST ^ ^6T> ^ST? ^ 6 2^> IF 5 2 ^
t¥3. t'A. tVV; T^Vff, ^¥A, AVif, tWj; 4§1, /a, ?Va-
3. 12. 4. H. t\> ii. ih 5. 1065f.
6. Ttir- 7. H. 8. tV- 10. if, i, ,'5.
v.— Page 34.
1; f, 2^L, 1^. 2. 29H. 3. UfJJ.
4. i, tV, tV 5. 2i, il, 314. 6. 2||.
7. /*• 8. 5^1^. 9. -ria-
10. 2jVt- 11. 12if 12. l,"^.
VI.— Page 36.
1. H, H, Hh 2- 5AV 3. 5.
4. tJit, Trf^w- 5. 3,v- 6. 4.
C'^sa R9662 I'lii
'• Too* O. --TT2J> -^^^T^J'
9. 77. 10. £8. 12tVs,
11. ^352, 6s. 8d. 12, 20J,
ARITHMETIC.
389
VIL— Page 41.
1. 2-203125. a 7^^, ^Vo, Uy h h f
3» 3703059239. 4. 5-098809263: 5: 15^, t\^
6. 1. 7. 2-718281. 8. -321750.
9. '367879. 10. 1-015873.
11. 3141592. 12. -857142.
VIII.— Page 43.
1. 13s. 4d., 7id., 2s. 8(1., £1. 5s. Sfd.
2. M. 7s. 6d., 6s., Is. l^^d., 3s. 9d.
3. X5. 17s. 4d., £8. 15s., £25. 3s. lly^d.
4. 17 cwt. 16 lbs., 16j lbs., 6? lbs., 2^ lbs.
5. 3 m. 2 f. 62f yds., 293^ yds., 4po. 1} yd., 82 J J yds.
6. 144 days, 32 days, 41irs. 54' 47".
7. 38 lbs, 7oz. 2dwt. 15^grs., 12dwt. 6;j«ggrs.
8. - 245 ac. 1 r. 27 po. 12^Vff Y^^-
9. 40° 3' 281", 35\
10. 6153 grains.
11. 7t'j; 4 cwt. Iqr. 14 J lbs.
12. Iday3hrs. 8'24fH5j".
IX.— Page 45.
4. 3 lbs. 7oz. 15dwts., S^Vf ll>s.
5. ^sVd, Hi- 6. IHJ, i^V
7. A, A. 8. 1185/^, ,HJTy.
9. WhW6\- 10. ^iJ^.
11. iS. 12. Tl4o^ayof24hrs.
Wff.
X.— Page 47.
1. 7s. 6d., 19s. 7Jd., 168. 3Jd.
2. XI. 58., 2s. 9d., XL 6s. 3d.
300 ANSWERS.
3. 12 cwt. 2 qrs., 2 cwt. 3 qrs. 7 lbs., 6 cwt. 1 qr. 25 lbs.
4. 1 ac. 2 r. 26f po., 13/o po., 1 r. 22 po.
5. 4 quires 4 sheets, 11 sheets, 23 quires 4| sheets.
6. 38 galls. 3 qts. 0\^ pts., 1 hhd. 60 galls. 2 qts. 1| pt.
nearly, 1 pk. gall. 3 qts. 1| pt. nearly.
7. 4000 grains, 8 oz. 6 dwts. 16 grs. 8. '698 lbs.
9. 15s. 2d. 10. 24 tons 9 cwt. 2 qrs. 8 lbs.
11. 4 cwt. 1 qr. 10 lbs. 12. 3 oz. 14 dwts.
XL— Page 48.
'l. -625, -53125, *55625, -928125.
2. -846153, -692307, -615384. 3. -36803, -11805.
4. -015625, -065476190. 5. -003125, -002232142857.
6. -003472, -040293. 7. '3125, -150625.
8. -0002232i42857j -00083. 9. 1-13085317460.
10. -82285714. 11.
12* -0544575
XII.— Page 54.
1 150000^ 20000^ 100, 270, 2-5, 1^ 3-45, 5-294.
2. 46000000^ 3000000j 2950000, 1500, 395, 29-5.
3. 2 myriag. kilog. 2 hectog. 9 dekag., 1 niyriag. 8
kilog. hectog. dekag. S grm. 5 decig., 1 myriag. 2 kilog.
3 hectog. dekag. grm. 1 decig; 3 fcentig.^ 1 hectog. 2
dekag. grm. 2 decig. 9 centig. 6 millig., 1 grm. 5 decig.
3 centig. 3 millig., 3 grm. 4 decig. 2 centig. 7 millig.
4. 160001-2, 25100, 396-45, 203550.
5. 1000, 2-96, 2900-03, 300-12, 3765-43.
6. 10000000000, 10000000, 50000, 349800, 4600.
7. 150, 39-4, 90*2, 1860*3, 3*764, -4.
8. 10, 1-234567, -372456126, 1, -000639, -293.
9. 3203, -4, 2000-003, 76-384, 29*34, 8300, 3457*6.
10. 18300*453, 1830*0453, 1830045*3.
ARITHMETIC. 391
11. 1, 73-6, 246-45, 2-55, 16-95.
12. 1300, 130; 713, 71*3; 1235, 123*5; 2*9, 320, 32,
1804, 180-4.
XIII.— Page 56.
1. 16287\)99 m., 10738767 m., 1322*371 sq. m., 69-548396
cub. m., 39129-99 gmi., 65632-02 ares, 368-93 st., 78603-982
Ut., 288-06 fr.
2. 1600-688 m., 11696*359 in., 96*18 sq. m., 5-967600029
cub. m., 5972-935 grm., 2450-94 ares, 409*6 st., 694003*024
lit., 2*35 fr., 98-56 fr.
3. (1) 70*245 m., 110*385 m., 130*455 m.; (2) 486082*89
m., &c.; (3) 49 sq. m., &c.; (4) 76*190000095 cub. m., &c.;
(5) 11046013-965 gi%, (fee; (6) 36000*9 ar., &c. ; (7) 41629
St., ifec; (8) 7347660*72 lit., &c.; (9) 2932*50 fr., &c.
4. 864 fr. 91-5 c. 5. 3408 fr. If c.
6. (1) 1*56 m., 1-43 in., 1*32 m.; (2) 27788 m., 2613*75
m.y 2460 m.; (3) 2*0910 sq. m., Ac; (4) 41*82 cub. m.,&c.;
(5) 188*263 grm., ifec; (6) 13*2 ar., &c.; (7) 10-01 st., &c.;
(8) 40446*3 lit., &c.; (9) 293*58 fr., «fec.
7. 25 fr., -15010 fr., 24120 fr., 328 fr., 1*80 fr., 2857 fr.
64 c. nearly, 16*5 fr., 12 c, *01 fr.
8. 25, 80, 2400, 1440, 480000, 14400, 4, 96, 125.
9. 1150 fr. 10. 45. 11. 36. 12. 3877 nearly.
XIV.— Page 60.
1. 1609*314 m. 2. 57319*8975 ft. 3. 239*613 sq. ft.
4. 862784. 5. 271*7. 6. 1128 fr.
7. 67 fr. 62 c. nearly. . 8. 1053268765 galls.
a £4. 3s. 4cL 10. 2204 fr. 61 c.
11. 447-39. 12. 1*0392.
XV.— Page 64.
1. £480. 2. 8s. Hid. 3. 344 fr. 48i c.
4. 39' 22i''. 5. 70. 6. 7 fr. 9ti c.
392 ANSWERS.
7. £4. 9s. 7d. 8. 31 days. 9. £305.
10. 3 li. 34if 11' P.M. 11. 32/ 43xV" past 2.
12. 4 months.
XVI—Page 67.
1. 15. 2. 30|f ft.
3. 1125 miles (take 8 kilom. = 5 miles).
4. £15,000. 5. IHdays. 6. 92xV days
7. 2s. 8d. 8. 52-5 m. 9. £154.
10. 88 horses. 11. 7^ miles. 12. 12 days.
XVII.— Page 69.
1.
4.
7.
0.
£70.
£i. 6s. 9f §d.
£385.
2. £41. 6s. 2id.
6. £4. 4s.
8. £236. lis. 8d.
11. 5j months.
3. £39. 8s. 4id.
6. £21. 18s. 5d.
9. 3m-
12. 15s. 5id.
*
XVIII.— Page 72.
1.
3.
5.
7.
9.
£23. 3s. 5-856d. 2. £49. 5s. 6-84d.
£20. 3s. 10a49d. 4. £102. 17s. 4-941d.
10s. 9-6d. 6. £441. 2 fl. 1 c. 1-40 m,
£270. 12s. l-929d. 8. £200 -f (1-045)'.
£231525. 10. £71. Is.
11. £50{(l-05)3+ (1-05)2 + (1-05)} = £165. 10s. Hd.
12. £450 -f (1-04)1
XIX.— Page 75.
1. £12. 7s. 2-4d. 2. 3ifd.
3. £4. 13s. 3-38d. 4. £4. 17s. ll'66d.
5. £3. Os. 2T-Vo\d. 6. £10. lis. 8-7d.
7. £5. 12s. 6d. 8. £42. 10s.
9. £39. 7s. 6d. 10. £35.
11. £2212. 10s. 12. £326. 7s. S^Jd.
ARITHMETIC. 393
XX.— Page 77.
1. £690. 18s. 9d. 2. £304. 14s. S^\d.
3. £5528. 10s. nearly. " 4. £382. 10s. lOd.
5. £6911. 3s. e^Vod. 6. £126. 14s. lOd. nearly.
7. £842. 12s. 6d. 8. £7768. 5s. 3jd.
9. £863. 7s. 6d. nearly. 10. £134.
11. £281. 5s. 12. 50,000 francs.
XXI.—Page 80.
1. £374. lis. lOj'sd. 2. £1069. 10s. lljd. nearly.
3. £1773. 14s. 6id. 4, £393. 7s. 8id. nearly.
5. £308-8693. 6. 20.
7. 12s. lOd. 8. £522-0411.
9. £217-7937. 10. £1000 - (I -05)=.
11. £23. lis. nearly. 12. £369*8728.
XXTI— Page 82.
1. 5s. 5d. 2. 10s. lOd.^ 3. £1. 10s.
4. 6 to 5. 5. £48. 6. 4^ gallons.
7. He loses 12J- per cent. 8. £72,123. 12s. 6d.
9. £tV3- 10. 2i. 11. £4. 5s.
12. 33j.
Miscellaneous Examples. — Page 85.
1 10. 2. £3. 5s. lOd. \
3. 721 lbs., 321 kilog. * 4. nAs + V^ ^
5. -0078125, T^^V & 2933-73.
7. 33-1, 3-236. 8. £2. 16s. Did., he lost 12j per cent.
9. 12. 10. J, 6s., -04895. 11. 0, 4, 2500, l^V-
12. £29. 6s. 7jd., £0. 9s. Bid. 13. £8000 stock, £7530.
14. £-002, 34-3168. 15. 83-8967, 1-9387. ,
16. £5ig, £1U3-, £12|3. 17. -2036. 18. 240.
19. 4s. 7-4d. per ounce. 20. £1763. Is.
394 ANSWERS.
21. £723, £3. 6s. 4d. 22. MO, £40, £100.
23. 5s., £1. 17s. 6d. 24. £1. Is. 5^ ll^Vx-
25. £336, ^\Vt. 26. /.VoV^, 9-69.
27. £3375. 28. 6/,-. 29. -3. . 30. 3000 days.
ALGEBRA— STAGE I.
I.— Page 149.
1. 8, 27, 6, - 16. 2. 6, 2, 4, 0. ^^ 3. 5, 11, - 7.
4. 11, 16. 5. ~ 2, - 15. 6. 8. 7. - 5, - 7.
8. - 11, 8. 9. 7, - 75. 10. 72, - 68*
11. -2,-3. 12. - 10, 32.
II.— Page 154.
1. 51, 6, 3. 2. 35. 3. - 513, - 65.
4. - 1224, 30. 6. 0, 2. ' 6. 1, - 27. 7. 1, - 2.
8. 1591, 0. 9. 144. 10. 1521.
37
11. 145. 12. -
>^yi029"
III.— Page 156.
1. 10 ^ + 3 5. 2. 11 a". 3. 12 6 4- 8 c. 4. 0,
6. 8a2 + ah, 6. 3x' ^ a? - l^x - 2. 7. 4a6 - 4.
8. ix? ■{■ 'if + ^ — 3 xyz, 9. aj* + i!?]!^ + 2/*.
10. a^ + 6^ + c^ + 3 a^S + 3 aW' - a^c - ac^ - Wc - 6c'
- 2 a6c.
11. cc'* + 7/^ + ;$;* - 4 33^2/ + 4 a3^;2J - 4a;?/^ - 4 2/^;^; + 4a;;2;^
- 4 2/;s^ + 6a;-2/2 -h 6 o;;??^ + 6 3/ V - 12 ar^^/^ + 12a;^^«
- \2xyz\ 12. 0.
IY.-^Page 157,
1. 4ct + 26 + 5c. 2. -4a;+2^-6«.
3. a' - 3 a6 - 6^ - 5 a - 7 6 - 8.
ALGEBRA. 395
4. 2 a* - 2 arx^ + x\ 5. 4 a' + 8 a'b^ + 4 b\
6. - 3(^5- 3««- Sarz^ Sxz".
7. cc* - o^c^ - 9 aV - 3 a^x - 2 a^ 8. 0.
9. -a-b-c—d + e+f+g+h.
10. 2 a:* - 8 a;3y + 12 or/- --Sx^ + 2y\
11. a;* + 2 ary^ + y\
12. or + b^ -2<^ + 2ab - 2ac - 2bc.
v.— Page 159.
1. - a: + 82/ + ^. 2. a* - 6^ 3. 13 - 6a.
4. 5 ar - 3 a; - 7.
6. (a + 6) + (c - cZ), a - (6 — c + c?),
|a _ (6 -c)}- cZ.
6. - (6a - 75)-.{3c - 5rf), -6a + (7 6 - 3(j + 6c0,
-{6a- (76-3c)} + 5d.
7. - (4af^ - l2x'y) - (l2aj/ - aj/),
- {4ai» - (I3af^2/ - 12a:^)}+ 4 2/^.
8. (a^ - &*) - (c^ - 3a5c), a^ - (6« + c» - 3a6c),
{a» - (6' + c^)} + 3a5c.
9. - (rt - 6 + rf)a:* - (rt - i - 3 c) a:y
^ - (26 - d ^ e -/)2/'.
10. (a - 6 + c - cZ) a; - (c - c£ + e - /) y
- (« - / + i7 - ^)^-
11. (a-b - d)x- {a - 7b - c)y- {b - c + d)z.
12. (2 - a;) a' + (a; ^ y)a6 + (y - z) b\
VI.— Page 161.
1. 12a« - a5 - 6 6^ 18ar - 9a;y - 35yl
2. jB» - 2 ajy" + 4 2/^, 30 x* + 49 a;*y + 9 ay» - 2/».
3. a* + 2 a=6^ + 6*, a* - 6*. 4. a?* - 3/*, «• - y".
6. a* + 6» + c* - 3 a6c. 6. a;» - 2^.
7. a» + 3a26 + 3a6= + b\
8. 5a» + 5a* - 405a - 405,
896 ANSWERS.
9. - To;^ + 17aj- ^ 6x - 2, a^ - x\
10. a« + 2 ci«^>2 + 3a^6^ + 2 a-6« + 6«.
11. af + (ac + b/)x + {h + f) ex- -\' (c^ + df) a^ ■{- cdxK
12. ^ + {p - a) x^ -{■ {q - ap) x — aq.
13. af^ + (ci + 6 + c) aj^ + (a6 ■{■ ac + ^c) a; + a6c.
VII.— Page 165.
1. ar^ _ 2 a;^ + t/^ 9 a^ - 30 a6 + 25 h\ 16 c* + 8 c^cZ^ + cf^
9 a;* - 12a;y + 4y*.
2. ^8 - h\ 3. mV - ^zV, 5 a* - 18 h\
4. (a + cf - (6 + c^)^ (ci + hf - (c + (^)^.
6. a;2 + 4a; - 5, a;^ + 7 a; + 10, a;^ + 2a; - 15, ar^- 25.
7. a;* - 37 a;^ - 24a; + 180.
8. ^\-{4a262 - {a" - h" -c-)-}.
9. 4 (a^ + &2 ^ c^ + cZ^).
VIII.--PAGE 171.
I 4 f6= - a5 + 2 h\ xy - ^ y\
2. 2 ft* - 5 d?h + Y ^^2^2^ 2 ar^ + 3 aji/ + 22/^.
8. oa;*"-" + ^a;-**?/^ + ca; -<"» + "> 2/^ ",
4. 6 a; + 4 7/, 5 a; - 3 2/. 5. 1 + a; + a;^.
6. - 3a; - 4, - 7a; + 3.
7. x^ - 3a;2?/ + 5a;i/2 + 27?/' + --^^,
a; - 3?/
o^ - a;-?/ - 3a;v^ + 15 ?/' —,
8. a;* + x^y + ar?/^ + xy'^ + 2/*> ^* "" ^V + ^V "" ^V^ + 2/*'
9. oa;*" + hy\ 10. a' + a'b + ai'-^ + 61
11. 6 + c, a + 6. 12. « + 6a; + car'.
13. a* - (/? - 1) a' - (p - g - 1) a- - (;? - 1) a + 1.
14. The given expression is —
(a + 6 + c - cZ) (« + 6 — c + cZ) (o& - 6 + c + c?) (a — 6 - c - c^);
ALGEBRA. 397
16. - oi^'if - o^y, a? + a;-\
17. {x + yf + (a? + yfz - (a; + y)z^ - is;^.
20. \a - c)2 - 2 (a - c) (6 - ^) + (6 - df.
IX.— Page 177.
1. {X + 3a) {x - 3a), (42/^ + 5;r^)'(42/2 - 5;^),
6 (2a + 36) (2a- 3 6), {2a;-32/) {4a;2 + 6iC2/ + 92/').
2. a; (a; - ?/) (ar^ + ajy + 2/'),
(a - 6) (a + h) {a? + 6^) (a* + h%
xy{x + y) {a^ - xy + f),
2 xfz {x + 2z){x - 2 z).
3. (a^ - 2 6^) (a^ + 2 6^), (^ + 0^2/ + 2/') (ar^ - a;^/ + y%
(a + ft)'^ (a - 6)^ (a + 6 + c) (a + 6 - c).
4. (a + 6 + c + fZ) (a + 6 — c - c?),
(a + 6 — c + i) (a - 6 + c + cZ),
(a + 6 - c) (a - 6 + c).
5. 5 (2a; + 9), 3 (2 a; + 7), (3a + 5 - c) (a + 6 - c).
6. {x' + xy + if) [{x - y) {x^ - f) + 4ar^2/2 {x^ - a;?/ - 2/') [,
{7? ■{■ xy •\- f) {^ - xy + y-) (x + yf (x - 2/)'.
7. {x - 10) (aj + 7), (a; + 1) {x + 10),
(a - 7 6) (a - 8 b), (x - 16) (x + 12).
8. (aa; + 7 62/) (aaj - 6 by), 3 a (aj - 10) (a; + 2),
ac {c - 8) (c + 3).
9. (3 a; H- 5) (2 a; - 7), (2 a; + 9) (4 a; - 15),
3 (2 a; + 3) (3 a; - 8), (4 a; - 7) (5 a; + 6).
10. xy{Zx-\-y){x+ 3 >/), a; (5 a; + 3 a) (4 a; + 5 6),
(ma; + p) (nx + ^).
11. a;3 + 2ar^ + 4a; + 8, 3a;^ - 6 a;^ + 12a;- - 24a; + 48,
a;* + 3 ar' + 9.
12. (a + 6)^ - (a + 6) (c + cZ) + (c + cZ)^ a - 6 + c + d,
13. a* + pa^ + q((/ + ra + s, 2a\ 14. - 102, 17.
X.— Page 179.
1. a% - 27 aW, 16 a«6V, - x'yY.
398 ANSWERS.
2. a* + 12 a'^ + 54a-62 + loSaS^ + 81 h\
16 a* + 32 a^^ + 24 aW + 8 aS^ + 6*^
a« - 5 a^6 + 10 a^h^ - 10 oW + 5 a6^ - h\
27 a» - 108 «-6 + 144 ^6^ - 64 61
3. 16 m* + 32m3 + 24m2 + 8m + 1,
125 0^3 + \bOx\ + 60ic + 8,
81a* - 432 a^c + 216 aV - 1152 ac^ + 256 c*,
- a3 „ 3 a^S - 3 aly" - h\
4. cc* + 2a;3 + 3a;2 + 2aj + 1,
9 a^ + 52 ^ 16 c^ + d? - 6 a?:> + 24 6^c - 6 acZ - 8 6c
+ ^hd - ^cd,
a^ + 4 6^ + c^ + 4 a6 - 2 ac - 4 6c,
9 (a^ + 6" + c^ + 2 6^6 + 2 ac + 2 6c).
6. 1 + 3 aj - 5 33^ + 3 03^ - £c^, (aa; + hyY + ?t(ax + hy) cz + &c.
6. 1 + 7 a; + 21 a;2 + 35aj3 + 35 a;* + 21 «;« + 7 a;« + aj^
(1 + i:c)* + 4(l + a;)V + 6 (1 + aj)V + 4 (1 + a;)a;« + c««,
\a + 6a;)* + 4 (a + 6aj)^ cs^ + &c.
7. a«a;8 -• 8 alhx^y + 28 a^WxY - ^6 a«63aj«2/' + 70 a*6*ajV
- 56 a^b'a^^ + 28 aWaPy' - 8a6%2 ^ js^s^
729 aj« - 243 xY + 27 aj^^/* - 2/'>
aj»- 3a;y + 3 a^2/« - yK
8. a^^ - 2 x'^y^ + y^
' a^2 ^ 6 ^1^6^ + 15 Jb^ - 20 a«6« + 15 a*6« - 6 aW<> + 6^1
9. a^ + 3^26 + 3ab^ + 6^. 10. (a - c)l
11 2 (1 + 3 aPy. 12. 81 a^y (aj n- ^).
XL— Page 193.
1. 2 a;i/V, 4 aV, ar^ + a% 2. 2 a;^ - 3 x'y + 5 a^y.
3. 5ft2 ^ 3a6 + 61 4. 1 - 2a; + 3a;- - 4a;3.
5. a + 6a; + cx'^ + da^. 6. aV - 3 aa;"-^ + 4:x''-\
7. a; + a;-'^ aa;-^ - a-^x 8. 3 a;^ 5 a.
9. 36, 79, 207, 289. 10. 103-2, -024, -3, j.
11. 4-123, 1-224, -618, 3-732.
12, -0203, 4-6, 3-5036. 13. 2a6y, 5a;y a + 2 6.
ALGEBRA. 399
14. x'^ + 3 oj- « 7. 15. X -{- y - c.
16. X + x-\ xy'^ + 1. 17. 18018, IMl.
18. 2-73, -3, -J. ■ 19. -5, -2599. j
20. 4/5^= 1-709. 21. 7-6457, 50-2487.
22. 4-4142, 3-7320. 23. -25. 24. 0.
XII.— Page 200.
1. aj - 3. 2. ar + 4 a; + 3. 3. a; + 3.
4. aj + a. 5. or -^ ah + h\ 6. aj - 1.
7. a; - 3. 8. 3aj - 2. 9. 12 a;- + 5 a;.
10. 3 a + 2 5. 11. a; - 1. 12. 2 a:^ - 4 a.-^ + aj - 1.
13. 2a + 3b + c. 14. a;2 + 1. 15. x - 1.
16. 2aj2 - 3aj + 9. 17. 1. 18. a + b +c.
XIII.— Page 203.
1. 12 a^a^f. 2. 60a-5V.
3. (a ^ b){b - c) {c - a), 4. a^x {x" - or).
5. (a; + 1) (aj + 2) {x + 3). 6. (a;- 6) {x -\-^){x'' 5).
7. (2 a; + 7) (3 aj + 8) (4 a; + 5).
8. 210 {d? + 1) (a^ + 1). 9. a" (x* + aV + a').
10^ (x + a) (X + h) (x + c). 11. 1 - a^«.
12. (x + 1) (aj + 2) (x + 3) (aj - 5) (a; - 6) (aj + 5).
13. a (a - bf (a2 - a5 + b').
14. (a; - 1) (x + 1) (x" + 1) (6 a^ + 5 a;2 + 2a; - 1).
15. {a' - by.
16. 2a;(a;2^i)(3^^3^_l^(2a;2^2a;-5)(2ar^-2aj + 5).
17. 3 (aj - 2)2 (aj + 4) (5ar^ + lOa;^ + 20a; + 18).
18. a'b''{a + b){a - b). 19. 15 (3 aj - 10) (a;* - 16).
20. (aP - 2^2)2 (^2 ^ ^2) (^4 ^ y)^ 21^ ^6 _ ^6^
22. {a + b + c + d) (a -{- b -- c -^ d) (a - b - c + d).
23. (a + 6 + c) (a* + £3 ^ c^ - 3 abc).
24:. (a -{■ b + c -i- d) {b + c + d - a) (a + c + d- b)
(a + b + d " c){a + b + c - d) (a + b - c-d).
400 ANSWERS.
XIV.—Page 206.
X ^ 6 x^ -2x + 2
3aj + 7 2ar^ + 9
Ix + 2' 5a;2 + 2a;-2'
4. ^(y - ^) - y' 1 g^ 2<t 25
x{y + z) - y^^ 7? - y'^' * d^ - P^ d^ - i
8.
a + 6* " 2(a; - 1) (a;^ + If'
^ 10
(x + 1) (a; + 2)2 (a;2 + 1/
9 8a; - 20 ^^ 8 a;^ + 8
(x + 1)2 (aj + 3)* • (x - \f (a^ + 1/
11. 0. 12. 0. 13. 1. 14. a + 6 + c. 15. 1. 16. 0.
17.
-
16a'a;
(»2 - ar')-'
18. "; -
a5
a6
+ J
- I
;;. 19.1
■•
20. :
21.
F
8
- y)ky -
«)(a;
-«)■
22
1
23.
¥
1
- a) (a; +
6)(x
+ c)
24.
a; + 5 a.
25.
1.
26.
4 a
2J3
~6y
27.
t
f*y^
t
^)
28.
(i-
+ a;) (cs +
yf
29.
1.
30. 1^.
XV.
.—Page 212.
1.
3.
2. 4.
3. a +
h.
4. 5.
5.
6.
6. 3.
7. 5.
8.2.
ALGEBRA. 401
9. c. 10. a + h. II. a + b + c. 12. "'+f + ^'.
6
13. abc. 14. - (a + 6). 15. abc. 16. 1^(^_1).
3 6 - 4 a
17. 2. 18. 8. 19. \.
20. '^^ - «° ,
a + — c — d
21. «' + «* + ^\ 22. 1 .
a + b abc
23. a + 6 + c.
24. 1 .
a + b + c
XVL— Page 217.
L 6. 2. "* - '^ ^
& 40.
4. flM- 5- - t'sV-
6. -i.
7. H. a 3.
9. - H.
10. 8. 11. IJi.
12. S-'^-^
2 a
13. -«'-f. i4.2(;-jy-
(i^jy-
15. 3. 16. - 8. 17. 1.
'«■ ".:.'
19. ( > + Jbf. 20. - a.
21. a' + 2 a.
22. f 23. - 2i. 24. «/<f;^)-±J«/.^l^g.
XYIL— Page 221.
L 10. 2. 45, 25. 3. £360, £240, £120.
4. 20, 15 miles per hour. 5. 24.
6. 36s., 48s. 7. 12. 8. 36s. 9. 30s.
10. 13. 11. 10. 12. 24 lbs., 40 lbs.
5 2c
402 ANSWERS.
13. 13. 14. 750. 15. 15s. 16. 46^^.
17. 18 miles from A. 18. -i^^hw
19. 15 per cent. 20. 28 miles. 21. 10.
22. 4-J-, 4 miles per hour. 23. 12 miles.
24. 20 miles per hour in same direction.
25. 120. 26. 6s., 4s. 27. 2, 1, |, 6 hours.
28. 98 per cent. 29. £1,000. 30. 38 ^^ miles per houx
31. 2 inches. 32. 1 6 vols, of hydrogen, 8 of oxygen,
33 1 - ^QQg 1005- 1 OA ab'c - a'bc'
b - c cc'(ab - ab)
orb
35. - 40°. 36.
c - ah
XVIII.— Page 228.
1. 3, 4. 2. 5, 2. 3. 3, 7. 4. 4, 5.
5. 12, 8. 6. 2, 6. 7. 1, 2, 3. 8. 1, 2, 3.
9 4 12. 10. ^^Ji_^i ^^1 - <^i< ^
11. X = "^^-4--^, to. 12. X = d+f-e^
2 2 a
13. 4, 5. 14. 3, 8. 15. 4, 8. 16. 3, 4.
17. 3, 5. 18. 8, 1. 19. X = ? ^, &c.
20. 2, 3, 4, 1. 21. 3, 4, 5. 22. x = (^ + c + d-b ^ ^^^
23. X = ^(» + <'-^)^ &c. 30. a; = y = « = 1.
XIX.— Page 231.
1. 12, 8. 2. 37. 3. 6 lbs., 5 stones.
4. -f. 5. 17 horses, 24 cows.
ALGEBRA. 403
6. 6s., 3s. 7. 220 and £2. Us.
8. 200, 300, 400. 9. 48, 23, 18. 10. 5, 3, ^.
11. 4, 6, 8 hours. 12. 6, 10, 18. 13. 7i, 12, 3.
14. 4, 3 miles per hour. 15. 16, 32, 48 miles per hour.
16. Oiic + hii/ + CiZ = dj, a^ + 603/ + CgSJ = cZj,
d-^c + h^ -^ c^ = d^, from which x, y, z.
ALGEBRA— STAGE IL
I.— Page 300.
t ±3.
4. ±i.
2. ± 4.
5. ± 2.
3. ±5.
6. ± (a + 6).
7. ± 1.
8. ± Va' - b^.
9. ±>J2ab- b'
''■^s~^-^
"■ * " 4'
' + 1
+ 3 &■
12. ± '^ V.
13. 2, 3.
14. 9, - 8.
15. 1, - |.
18. - 5, - |.
16. 4, - J.
19. ^ (6 n/6^ -
17. - i, - f
4 ac).
20. 5, 3^.
23. 1, «^ - (« +
oc
21. 7, - 6.
22. 6, - |.
24 b, a.
25. « + J, «* ,,
+
26. a, 6.
a-b a — b
93 a'' + a6 + Zi''
a - 6
29 4 ^?..
■'°- a-b '
a" - cd> + b"
A«7. l:, -J- J.
30. 5, - U-
33. 2, - iU-
36. 3, - le.
39. 15, - 1.
31. 3, - 5.
34. 5, i.
37. 2, 2i.
40. 2, 0.
32. 2, - 3i.
35. 8, - fi.
38. 4, - a.
41. 0, 2a6.
404 ANSWERS.
42. a, 3| a. 43. 0,
44. 1,
(6 - a)h
45. a + 6 + c, -
a — 6'
II.— Page 306.
1. ±3. a 2, - ^V. 3. ±5, ± ^
4. 2, - 5. 5. ± i. 6. ^2, ^ 7. 4, 69.
8. ± 4, ± Jl, 9. 3, - 4, 3 ± Jn. 10. I, 4.
11. - 2, - 2 ± J3. 12. I
13. 4, 5-, --^-. 13 ± x^l45). 14. 0, 6 - a.
15. ± i« x/3r 16. 2 - ^/6-2a^/6+ ^'
19. /i*l*JY. 20. ± 5, ± J-li.
V" + ly
21. 2, - i, J(- 11 ± n/97). 22. 4, |.
23. 5, 3j, i(- 7 ± s/53). 24. 3a.
25. 3, - 2. 26. 4, - 14^, t (- 19 ± VTiT).
27. 3, - 2^, i|. 28. i (1 ± n/4^^^).
29. + 1, andiC* + ar= + 1 = 0. 30. ± /?L±^JL?.
31. e», i { s/r?~:nr^^ + 2 a6 + 6^ - (a - i-)}-
82. X = m is one solution.
ALGEBRA. 405
33. X = a -^ h + c is one solution.
34. X = ^, 2ind a^ + ax -h x^ = 0.
m
35. X = is one solution.
a + c
36. a:^ - a; + 1 = gives two solutions.
III.— Page 310.
1. X = 2, 3, 2/ = 3, 2. 2. a; = 5, - 3, 2/ = 3, - 5.
3. a; = 3, 4, 2/ = 4, 3: 4. a; = 2, 4, t/ = 4, 3.
5. a; = 5, - 2, 2/ = 2, - 5. 6. ± 7, ± 6.
7. ± 6, ± 3, 8. 4, 2. 9. ± 2, ± 7.
10. ± 4, ±3. 11. a; = ± 3,~y = ± 2, - -^.
12. a; = 9, 4, 2/ = 4, 9. 13. x = 2, 8, y = 8, 2.
14. a; = 3, 2, 2/ = 2, 3.
..;- _ 2 ^ ^
a ± J2¥'^^^~^ ^ ~ a^: J2U' - a^
(a + 1)6 _^ (a -1)6
16. ±
n/2 {a^ + 1)' " 2 x/a^ + 1
^' "- V^T^' - V 2 6
18. a: = 2, 3, y = 3, 2. 19. a; = 2, 3, 2/ = 3, 2.
20. a; = 2, 3, y = 3, 2 21. aj = 5, - 3, 2/ = 3, - 5,
22. X = ""-l ^^ ^ " ^ ^ 23. a: = 2, 3,y = 3, 2,
v^4 cr» - 4 6
S4. a: = ± 8, + V n/I?^ 2^ = =t 2, ± x/||
25. a; = 4, 9,y = 9, 4. 26. a? = 4, 1, y = 8, 0.
27. a;= 4, 3, 2/ = 3, 4. 28. ± 3, + 1.
29. a: = 0, 3, - il^l, 2/ = 0, 2, llSf ,
406 ANSWERS.
30. 0, - 1. 31. x = 4, VaV, y = 3, - V5V.
32. a; = 5, 6, 2/ =t: 4. 33. aj = 3, 1, 2/ = 1, 0.
34. aj = 3, - V/, 2/ = 1, - If.
35. a: -2, 7,2/ = 4, -14.
36. aj + y = c, and xy = bx + ay,
2n. X = 0, a — h, y = 0, a — h,
38. V ± 9 ViJ, ^ ± ^H.
39. a; = , / (« + c - 6) (a + 6 - c)_
^ 2 (6 + c - «)
40. 1, 2, 3. 41. ± 1, 0, 0.
42..=2, = « = 0,alsol=.(^.i,.l^.
^ A ^
43.
^^. X = y = z = u = a + h.
45. 7' r, &c
(a + 6 + c) * (a + 6 + c)^
46. a; = 6, 4, 2/ = 4, 6, « = 5.
IV.— Page 315.
1. 9, — 6. 2. Numerator 2f |, denominator 3.
3. 35 or 23. 4. 10, - 16. 5. 78, - 68 ft.
6. 2s., 9s. 7. 20 hours. 8. 48 shillings.
V.—Page 318.
1. a^j a^, a% a. 2. a + x^ (a - xy.
3. a + a^x^ + a;l 4. x^^"^ - x'^y^.
ALGEBRA.
407
5.
7.
9.
14.
16.
1.
2.
3.
5.
7.
10.
12.
13.
14.
xy-'^ +x-hj + 1. 6. a' + ah^ + hi, at + i-t.
X - xy, 8. a' + hi + c* - ahi - a^c'^ - bic^-
a-b-i - Ji 13. a* + ji + «*•
2a;*2/^ - Sxyi + 2x^- 16. ai~ * - 2 + 2a~''bK
a;* + 3a;?-7. 17. ar'y-J + 1. 18. (a* + fiJf.
VI.— Page 325.
x^, a^bi, x^yi, a^b^-
ocya~ib~i, c^Sx~iy~i, a~mxn'
V27, V24, x^% -^9. 4. ^/32, ^277 ^32, Vy.
n/9^6, V^, ^a" - ar^. 6. ^4, ^7.
ij^, iysT 8. V8, ^135". 9. "v^a^ ^^»-
4^(ti + xY, il{a - x)\ 11. a^,h
2 ^/l,2 >^J'6, 6 V7, 3 ^3T
- - a, ^ — a'a;i
P + 7 ^-~P ,
P2
X^
X + I
15.
"3^
-JZx,
X -
-J{^-aJ.
16.
4V3,
5^7.
17. ^?^
- a6
+ c
sJaVa?.
18.
(Sa'b
-2a
m + »
- 12) ^.
19.
d ~
6.
+ b
20. (x -f
•2/)*.
h.
- hHK
21.
; + d \/f ('^ +
J)-
408 ANSWERS.
22. a - b^ + c^ - dt + 2 a^ci - 2 6'ci
23. X - x^-if^ + y:
24. o^ — 3(?y^ + x^y^ — xy + x^y^ - y^,
25. \ Vis, ± n/7, i 4/I75.
26. 3(2 - n/3), 1(3 V2 - 2sl%l{Jb^ ^3). '
27. 1(2 + x/2 - x/6), -J- (2 s/3 + 3 x/2 - ^30), 5 + 2 ^6.
28. a {x^ - xhj^ + x^y^ — xy + x^y^ - y^),
^\ (3^ - 3\ 2^ + 3t. 2^ - 3 . 2 + 3^. 2t - 2*),
^ ^xy ^y'^ (oj - x^y^ + y),
29. 2 + Vt, V5 + x/3, 5 - V5.
30. V6 + a/2, V6 - V3, VT5 - V5".
YIL— Page 330.
1. o^ 4- }? \o? - 5^ is the greater, if a >- 5,
2. The former. 3. 6 - ^.
c
4. (1 - 2/) (1 + a:) : 1 + ic'. 5. 30, 35.
6. Less than 5 when x lies between 2 and 3 ; greater than
5 for all values beyond these limits.
8. X satisfies the equation —
4 ic^ -» (a + 5 _ 6 c)ic + 2 (c' - a6) = 0.
9.|. 10.1. 14. 25^ a, i;, ^^-.
2 a^ 2 a —
PLANE TRIGONOMETRY. 409
PLANE TRIGONOMETRY.
I.— Page 335.
1. 43-750^ ir 50' 3-03". 2. 88^ 88^88-8^
3. i^n the irnmber of grades.
4. 180 - (a + -3^5) degrees, 200 - (i^o^ + 6) grades.
5. i 7. |. 8. 25^ 75^ 90°. 9. 576°, 648^.
10. 2nd, 1st, 4tli, 3rd, 4th. 11. « , ^. 12. -^-^.
II— Page 341.
1. Cos A = -^%, tan A = y &c.
24 25
2. Sin A = — r= — , cos A = — >- , &c.
n/T201' ^/l20l'
17. h 18. 3^, (2 v/89 ^ 5). 19. -^^^^•
20. 2 + ^3. 21. 00 , - ^,-^1 22. 1.
24 ^ ^~~~
1 Sin A = FZK£^.
III— Page 352.
2. From 0^ to 90°, + ; from 90° to 180°, - ; from 180° to
270°, + ; from 270° to 360°, -.
3. Cos A + sin A is + , + , - , - , as A lies between - 45°
and 45°, 45° and 135°, 135^^ and 225°, 225° and 315°
respectively. The corresponding signs of cos A -
sin A are + , - , - , + .
4. Cos 2 A is +,-,+, - respectively, as in Question 3.
^' 2 J2 ' 2 ' *" 2 •
410
10. A = 60°
ANSWERS.
11. A = 45°, and tan A = - 6.
12. A = 30°. 13. A = 30°.
14. A = 45°, B = 15°. 15. A = 0°, 45°, 135^
16. A = 6°. 17. A = eO'^. 18. e = 2 ± J37
lY.— Page 358.
I. 2, 3, 1, _ 1, - 2, 1-5. 2. I, 6, 1-5.
3. 1-0791812, 1-5563025, 1-6532125, 1-8750613,
2-6020600, -5740313, l-8239087,'2-8696761.
4. 1-3172901, 3-3172901, 2-3172901, 3-3172901.
5. 1, 1, 3,~3.
6. -020912, 2091200, -20912, 20912. 7. 1-3162760.
8. 3493-768. 9. -2427189. 10- 3-1303338.
II. 6, 5. 12. 1-67.
1. 1-6774495.
4. -0616835.
7. £663-449.
10. 9-8401608.
Y.— Page 365.
2. 6-7193696.
6. 2-07892.
8. 2620-248.
11. 36^26^30^^
YI.— Page 372.
3. 1-8213310.
6. ^177-63 nearly.
9. 9-7299222.
1. c = 12-5. 2. ^189.
4. 480. 5. 108 s/2:
9. C = 50 j6{JS^l),b :
10. a = 10 ^949.
5
3. Tan:|
6. B = 90^ C = 30°.
100 (v/S"" 1).
PLANE TRIGONOMETRY. 411
VII.— Page 376.
I. J = 15-78065. 2. a = 92-34057.
3. ^> = 25, A = 67'22'49^
4. A = 40°59'32^c = 111-288.
5. h = 22-78438. 6. b = 691*49.
7. 5 = 12 ^3, c = 24.
8.a = 1(^6- V2), 6 = f(V6 + v/2).
9. A = 30°, b = 17 J3, 10. A = 30°.
11. CD = 7{J6 - ^2). 12. DE = 5^3.
VIIL— Page 380.
1. a = 6 V3, B = 90°, G = 30°.
2. B = 45° or 135°, C = 105° or 15°, c = 9 ( ^6 ± ^2).
3. A = 60°, B = 45°, C = 75°.
4. A = 105°,^ = 6(3^2 - J6\c = 12{2- J3),
6. A = 75°, B = 60°, c - 2 ^6.
6. B = 60° or 120°, C = 75° or 15°, c = 5 (3± ^3).
7. B = 41° 59' 23'; C = 97° 28' 37^
8. A = 19° 18' IK
9. A = 30° 3' 26", B = 123° 24' 34".
10. A = 48° 11' 22", B = 58° 24' 42".
II. B = 35° 15' 52", C = 84° 44' 8", c = 137*9796.
12. A = 44° 24' 56".
IX,— Page 384.
1. 200 (2 -
-V3).
2. 60 73,
3. W3.
4. 6 V2.
412 ANSWEKS.
5. 6(73 - 1), 3 76(^3 - 1).
6.30(^3+1). 7. 1|5 73.
8. 1073.
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THE POCKET ATLAS OF HISTORICAL GEOGRAPHY, i6
Maps, 6^ by II inches, mounted on Guards, Imperial i6m<>, cloth,
THE CROWN ATLAS OF HISTORICAL GEOGRAPHY, i6
Maps, with Letterpress Description by Wm. F. Collier, LL.D.,
Imperial i6mo, cloth,
THE STUDENT'S ATLAS OF HISTORICAL GEOGRAPHY,
16 Maps, with Letterpress Description by Wm. F. Collier, LL.D.,
8vo, cloth, ...
1 Roman Empire, Eastern and Western,
4th Century.
2 Europe, 6th Century, shewing Settle-
ments of the Barbarian Tribes.
3 Europe, 9th Century, shewing Empire
of Charlemagne.
4 Europe, 10th Century, at the Rise of
the German Empire.
5 Europe, 12th Century, at the Time of
the Crusaders.
6 Europe, 16th Century, at the Eve of
the Reformation
7 Germany, 16th Century, Reformation
and Thirty Years' War.
d.
6
8 Europe, 17th and ISth Centuries.
9 Europe at the Peace of 1815.
10 Europe in 1870.
11 India, illustrating the Rise of the
British Empire.
12 World, on M creator's Projection,
shewing Voyages of Discovery.
13 Britain under the Romans.
14 Britain under the Saxons.
15 Britain after Accession of William
the Conqueror.
16 France and Belgium, illustrating
British History.
15
CLASSICAL GEOGRAPHY.
THE POCKET ATLAS OF CLASSICAL GEOGRAPHY,
Maps, Imperial i6mo, 64 by 11 inches, cloth lettered,
THE CROWN ATLAS OF CLASSICAL GEOGRAPHY, 15 Maps,
with Descriptive Letterpress, by Leonhard Schmitz, LL.D., Imperial
i6mo, cloth lettered,
THE STUDENT'S ATLAS OF CLASSICAL GEOGRAPHY, 15
Maps, Imperial 8vo, with Descriptive Lettei press, by Leonhard
Schmitz, LL.D., cloth lettered,
9 Armenia, Mesopotamia, &c.
10 Asia Minor.
11 Palestine, (Temp. ChristL)
12 Gallix
13 Hispania. *
14 Germania, &c.
15 Britannia.
1 Orbis Veteribus Notus.
2 ^gyptus.
3 Regnum Alexandri Magni.
4 Macedonia, Thracia, &c.
5 Imperiura Romanum.
6 Grsecia.
7 Italia, (Septentrionalis.)
8 Italia, (Meridioualis.)
Historical and Classical Atlas.
THE STUDENT'S ATLAS OF HISTORICAL AND CLASSI-
CAL GEOGRAPHY, consisting of 30 Maps as above, with Intro-
ductions on Historical Geography by W. F. Collier, LL.D., and on
Classical Geography by Leonhard Schmitz, LL.D., with a Copijus
Index, Imperial 8vo, cloth.
London, Edinburgh, and Herriot Hill V^orks, Glasgow
"William Collins, Sons, & Go's Educational "Works.
COLLINS' SERIES OF SCHOOL ATLASES— Continued.
SCRIPTURE GEOGRAPHY.
THE ATLAS OF SCRIPTURE GEOGRAPHY, i6 Maps, with
Questions on each Map, Stiff Cover,
THE POCKET ATLAS OF SCRIPTURE GEOGRAPHY, i6
Maps, 7^ by 9 inches, mounted on Guards, Imp. i6mo, cloth,
1 Ancient World, shewing probable Set- 9 Modern Rilestine.
tlements of Descendants of Noah. 10 Physical Map of Palestine.
2 Countries mentioned in the Scriptures. 11 Journeys of the Apostle Paul.
3 Canaan in the time of the Patriarchs. "~ "
4 Journeyings of the Israelites.
5 Canaan as Divided, among the Twelve
Tribes.
6 The Dominions of David and Solomon.
7 Babylonia. Assyria, Media, and Susiana.
8 Palestine in the Time of Christ.
12 Map shewing the prevailing Religions
of the World.
13 The Tabernacle in the Wilderness.
14 Plans of Solomon's and Herod's Tem-
ples.
15 Plan of Ancient Jerusalem.
16 Plan of Modern Jerusalem.
BLANK PROJECTIONS AND OUTLINES.
THE CROWN ATLAS OF BLANK PROJECTIONS, consisting
of 16 Maps, Demy 4to, on Stout Drawing Paper, Stiff Wrapper, ... o 6
THE CROWN OUTLINE ATLAS, 16 Maps, Demy 4to, Stout
Drawing Paper, Stiff Wrapper, ... ... ... ... ... o 6
THE IMPERIAL ATLAS OF BLANK PROJECTIONS, consisting
of 16 Maps, Imperial 4.to, on Stout Drawing Paper, Stiff Wrapper, i 6
THE IMPERIAL OUTLINE ATLAS, 16 Maps, Imperial 410, Stout
Drawing Paper, StifJ" Cover, ... ... ... ... ... i 6
A Specimen Map of any of the foregoing Atlases free on receipt oftivo Penny Stamps.
SCHOOL-ROOM ^WALL MAPS.
Printed in Colours, and Mounted on Cloth and Rollers^ Varnished.
CHART OF THE WORLD, 5 ft. 2 in. by 4 ft. 6 in. , 20 o
CENTRAL AND SOUTHERN EUROPE, 5 ft. 2 in. by 4 ft. 6 in., 20 o
EUROPE, ASIA, AFRICA, NORTH AMERICA, SOUTH
AMERICA, ENGLAND, SCOTLAND, IRELAND, PALES-
TINE, INDIA, each 3 ft. by 2 ft. 5 in., 6 6
UNITED STATES OF AMERICA, 3 ft. 11 inches by 2 ft. 4 in., 8 6
COUNTY ^WALL MAPS.
Printed in Colours^ and Mounted on Cloth and Rollers, Varnished.
MIDDLESEX, LANCASHIRE, YORKSHIRE, WARWICK,
DURHAM, CUMBERLAND, DERBYSHIRE, DORSET,
GLOUCESTER, HAMPSHIRE, SOMERSET, STAFFORD,
AND WILTS i each 54 in. by 48 in., 9 o
CHART OF METRIC SYSTEM.
CHART OF THE METRIC SYSTEM OF WEIGHTS AND
MEASURES. Size 45 in. by 42 in., price, on Rollers, ... 9
"London, Edinburgh, and Herriot Hill "Works, Glasgow.
;j
QAi-5"Y
^
"William Collins, Sons, & Co.'s Educational "Works.
COLLINS' SERIES OF FIRST-CLASS SCHOOL ATLASES,
Carefully Constructed and Engraved from the best and latest AuthoritleSy and
Beautifully Printed in ColourSy on Superfine Cream Wave Paper,
MODERN GEOGRAPHY— Crown Series.
MY FIRST ATLAS, consisting of 12 Maps, 9 inches by 7^ inches,
folded Svq^in Neat Wrapper,
6
1 The Hemispheres.
2 Europe.
3 Asia.
4 Africa.
5 North America.
6 South America.
THE PRIMARY ATLAS, consisting of 16 Maps, 9 inches by 7^
inches, 4to, Stiff Wrapper, ...
7 England and Wales.
8 Scotland.
9 Ireland.
10 Central Europe.
11 Australia.
12 Palestine.
1 The Hemispheres.
2 Europe.
3 Asia.
4 Africa,
5 North America.
6 South America.
7 England and Wales.
8 Scotland.
9 Ireland.
10 Central Europe.
11 India.
12 Canada.
13 United States. "
14 Australia.
15 "New Zealand.
16 Palestine.
THE POCKET ATLAS, consisting of 16 Maps, folded in 8vo, and
mounted on Guards, cloth lettered,
THE JUNIOR, OR YOUNG CHILD'S ATLAS, consisting of 1 6
Maps, 4to, with 16 pp. of (Questions on the Maps, in Neat Wrapper,
THE SCHOOL BOARD ATLAS, consisting of 24 Maps, Crown
4to, cloth limp,
THE PROGRESSIVE ATLAS, consisting of 32 Maps, 9 inches by
75 inches, 4to, cloth lettered, \.,
1 The Hemispheres.
2 The World, (Mercator's Projection.)
3 Europe.
4 Asia.
5 Africa.
6 North America.
7 South America,
8 England and Wales.
9 Scotland.
10 Ireland.
11 France.
12 Holland and Belgium.
13 Switzerland.
14 Spain and Portugal.
15 Italy.
17 German Empire.
18 Austria.
19 Russia in Europe.
20 Turkey in.Europe, and Greece.
21 India.
22 Persia, Afghanistan^ and Beloochis-
23 Turkey in Asia. [tan.
24 Chinese Empire and Japan.
25 Arabia, Egypt, and Nubia.
26 Palestine.
27 Dominion of Canada.
28 United States.
29 West Indies.
30 Australia.
31 Ne\r South Wales, Victoria, and
32 New Zealand. [South Australia.
16 Sweden, Norway, and Denmark.
THE CROWN ATLAS, consisting of 32 Maps, on Guards, with
Index, 8vo, cloth lettered, ... ... ... ... ... ... 2
THE NATIONAL ATLAS, consisting of 32 Maps, 4to, with a
Copious Index, cloth lettered, ... ... ... ... ... 2
London, Edinburgh, and Herriot Hill "Works, Glasgow.
"William CoUins, Sons, & Co.'s Educational "Works.
COLLINS' SERIES OF SCHOOL ATLASES-CONTINUED.
MODERN GEOGRAPHY— Imperial Series.
THE SELECTED ATLAS, consisting of i6 Maps, Imperial 4to, ii
by 13 inches, Stiff Cover, ,,,
1 Tlie Hemispheres.
2 Europe.
"■ 3 Asia.
4 Africa.
6 North America.
6 South Aineri(;a.
7 England and Wales.
8 Scotland.
s. d,
I d
9 Ireland.
10 Southern and Central Europe.
11 India.
12 Canada.
13 United States.
14 Australia.
15 New Zealand.
16 Palestine.
THE PORTABLE ATLAS, consisting of 16 Maps, folded Imperial
8vo, cloth lettered, ...
THE ADVANCED ATLAS, consisting of/32 Maps, Imperial 4to,
cloth lettered,
1 Eastern and Western Hemispheres.
2 The World, (Mercator's Projection.)
8 Europe.
4 Asia.
6 Africa.
6 North America.
7 South America.
8 England and Wales.
9 Scotland.
10 Ireland.
11 France.
12 Holland and Belgium.
18 Switzerland.
14 Spain and Portugal
15 Italy. [the Baltic.
17 German Empire.
18 Austria.
19 Russia.
20 Turkey in Europe, and Greece.
^ India.
22 Persia, Afghanistan, and Beloochis-
23 Turkey in Asia. [tan.
24 Chinese Empire, and Japan.
25 Arabia, Egj'pt, Nubia, and Abys-
26 Palestine. [sinia.
27 Dominion of Canada.
28 United States.
29 West Indies and Central America.
30 Australia.
31 Victoria, New South Wales, and
16 Sweden and Norway, Denmark and |* 32 New Zealand. [South Australia.
THE ACADEMIC ATLAS, consisting of 32 Maps, Imperial 410,
with a Copious Index, cloth lettered, ... ... ... ... 5 o
THE STUDENT*S ATLAS, consisting of 32 Maps, and 6 Ancient
Maps, with a Copious Index, Imperial 8vo, cloth lettered, ... 6 o
93 Ancient Greece. 87 Historical Map of the British Is-
34 Ancient Roman Empire. lands from a.d. 1066.
35 Britain under the Romans. 88 France and Belgium, illustrating
36 Britain under the Saxons. British History.
THE COLLEGIATE ATLAS, consisting of 32 Modern, 16 ?2istorIcal,
and 2 Ancient Maps, mounted on Guards, with a Copious Index,
Imperial 8vo, cloth lettered, 76
THE INTERNATIONAL ATLAS, consisting of 32 Modern, 16
Historical, and 14 Maps of Classical Geography, with Descriptive
Letterpress on Historical Geography by W. F. Collier, LL.D. ; and
on Classical Geography by L. Schmitz, LL.D., with Copious Indices,
Imperial 8to, cloth mounted on Guards, 10 ^
London, Edinburgh, and Herriot Hill "Works, Glasgow.