Constructive Text-book of Practical Mathematics

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JBMRIES

III II 1 1
07903

)

CONSTRUCTIVE TEXT-BOOK

OF

PRACTICAL MATHEMATICS

HY

HORACE WILMER MARSH

Head of Department of Mathematics School of Science
and Technology, Pratt Institute

Volume IV

%

TECHNICAL TRIGONOMETRY

FIRST EDITION '. . .V -
FIRST THOUsiA-N'D

• t. " *. *

NEW YORK

JOHN WILEY & SONS, Ino.

London: CHAPMAN & HALL, Limited

1914

1 A . !.

7: york!
F'Ur.i,K. LIBHARYj

078651

T»LDE* F<3UN»AT»0N».

Copyright, 1914,

BY

HORACE WILMER MARSH

• •

• •

•• • •

• ••

• • • • • •

• - « -

• ••

• • • ••* - •

-•••• ••• •

THE SCIENTIFIC PRESS

ROBERT DRUMMOND AND COMPANY

BROOKLYN, N. Y.

PREFACE

h

i

f

The aim of this book is to give the student a clear,
intelligent, rational, and useable knowledge of the trigo-
nometry underlying his technical studies and the industries.

An explanation of the method employed and of the
purpose of the Mathematics Work-book is given in the
preface of Volume II of the Constructive Text-book to
which the reader is referred.

This text had its beginning in the Spring of 1897, in a
student's notes on a series of talks by the author, to a
class in trigonometry. During the years intervening, the
constantly changing and enl^rgtag mimeograph text has
been tested for one term of e*L*?h year, trie test for the past
eight years having been made with five classes). five times
a week. :

The book is not, therefore,; an expression of precon-
ceived notions of trigonometric. siibjecVsiaGter or method
of presentation, but is rather the resultant of persistent,
psychological, pedagogical, and experimental study of the
capacities, acquirements, natural inclinations, and neces-
sities of the student who would learn trigonometry.
1 Prolonged study of the subject only increased the

• author's conviction of the simplicity of the principles
underlying this branch of mathematics. Comparative
^ study of conventional texts and students' needs, only served
to emphasize the confusion and complexity of the usual
subject-matter and arrangement.

vi PREFACE

With a belief in the idea that that knowledge is of most
worth which moves the feelings most profoundly, an endeavor
was made to determine the subject-matter, method, and
arrangement to which the student most readily responds.
The result is a trigonometry which is somewhat intimately
related to the manufacturing, technical industries rather
than to the navigator's art, the height of a castle, or the
determination of the numerical value of the various func-
tions of certain angles without the use of a table; a trigo-
nometry in which a correct result in all problems is of
supreme importance and is therefore determined by the
slide-rule and by a table of not less than six places so that
it shall be possible to compute with accuracy the length or
other dimension of a piece of material, when such length
is expressed in one integral figure and the piece is to be
machined to the ten-thousandth of an inch. Such a result
is radically different from the determination of the " height
of a tree on the river bank " (by a method which no forester
would use), and gives both student and teacher a radically
different conception of trigononjetry.

Trigonometric jnvWepaat^f - " ruins " and distances of
" inacce&$&ld !' W&gs,\*w?ricK 'centuries ago challenged the
mathemattdan^ fail p& \rifcipst even the school boy of this
generation whp^lf.Jt&»plan§ % tp enter the industries, must
design, or build^^r jfeptf^i^V)r .superintend the operation of
high powered',* $\itbtiiaticy and complicated machinery in
which a comparatively trifling error in computation or
construction might wreck a machine costing thousands of
dollars.

Teachers accustomed to the conventional will miss the
familiar trigonometric juggling which for centuries has
separated the study of the right triangle from the oblique
and has given every student a vague notion that an oblique
triangle can be solved only after one has " worried through "
a hundred or more, puzzling identities.

For these omissions the author offers no excuse exceot

PREFACE Yii

the statements of the preceding paragraphs. Some of the
other unique features of the text have their explanation
in the following quotation from Woodberry's Heart of Man:

" In setting forth first principles, the elabora-
tion of a more highly organized knowledge may be
felt as an obscuration of truth, an impediment to cer-
tainty, a hindrance in the effort to touch and
handle the essential matter, and for this reason a
teacher dispenses with much in his exposition."

For the style, arrangement of subject-matter, reading
of proof, aid in preparation of index, and other assistance,
I gratefully acknowledge my many obligations to my wife,
Annie Griswold Fordyce Marsh.

Horace Wilmer Marsh.

Brooklyn, New York,
August 24, 1913.

CONTENTS

PAGE

The Work-book 3

CHAPTER I

Logarithms

Section 1, Logarithm of a Number Greater than Unity. Section
2, Logarithm of a Number Less than Unity. Section 3,
Naperian or Hyperbolic Logarithms. Section 4, Logarithm
of a Product. Section 5, Logarithm of a Quotient. Section
6, Logarithm of a Power. Section 7, Logarithm of a Root.
Section 8, Solution of an Exponential Equation. Section 9,
Model Solutions 11

CHAPTER II
The Right Triangle

Section 1, Functions of an Acute Angle. Section 2, Solution of a
Right Triangle. Section 3, The Isosceles Triangle. Section
4, Applied Problems 46

CHAPTER III

The Oblique Triangle

Section 1, Functions of any Angle. Section 2, The Four Laws of
Solution. Section 3, Solution of Triangles. Section 4,
Applied Problems 117

ix

x CONTENTS

CHAPTER IV
Relation of Functions

PAGE

Section 1, Functions of Compound Angles. Section 2, Functions
of Multiple Angles 187

CHAPTER V

Use of the Slide Rule

Section 1, Scales. Section 2, Sines and Tangents. Section 3,
Powers and Roots. Section 4, Multiplication. Section 5,
Division. Section 6, Reciprocals. Section 7, Trigonometric
Computations < 200

REFERENCE TABLES

I. Measures of Length 220

II. Measures of Area 221

III. Measures of Volume 222

IV. Measures of Weight 223

V. Decimal Equivalents 224

VI. U. S. and Metric Equivalents •. 225

VII, Wire-gage Sizes 226

TECHNICAL TRIGONOMETRY

TECHNICAL TRIGONOMETRY

THE WORK-BOOK

1. Description. The constructive form of this text
requires that each student shall prepare the work daily
in accordance with the suggestions, questions, and direc-
tions in numerous developing exercises, which force a student
not only to do his own thinking but to express his thought
in written form.

For this purpose the mathematics' work-book is used,
which both in size and form is the result of sixteen years
of experiment with over 2000 students. It consists of a
note-book cover, the daily record sheet described in a
subsequent paragraph, and 250 removable sheets of
16-pound, unruled, linen paper measuring 5JX8-| inches,
with fasteners for attaching the sheets to the back cover
and the student's written work to the front cover.

2. Instruments. In order/ to prepare the work in a
satisfactory manner and to secure the greatest educational
benefit each student will require the following equipment:

12-inch triangular scale with U. S. and metric gradu-
ations,
medium lead pencil,
ink and pencil erasers,
fountain pen,
ruling pen,
compasses,

3

4 TECHNICAL TRIGONOMETRY 3

protractor,

red, black, and India ink,

slide-rule.

3. Value of Careful Work. It is obvious that skill is
never acquired by careless, indifferent effort. Therefore,
as in manual training, the desideratum in every exercise,
example, and problem in this text, is perfection in the
finished work.

The instructions in the two subsequent paragraphs and
throughout the text have accordingly been written to
stimulate each student to a serviceable ideal of excellence
and efficiency and to give him an increasing ability and
enthusiasm for its realization.

4. Instructions for Work-book Entries. (1) What to
do First. On the inside of the front cover of the work-
book write your full name, home and rooming address,
and name of school and course.

Attach all record sheets except one, to the back cover
under the blank sheets.

(2) Use of Ink. With the exception of the drawing
and the first exercise in lettering, all work is to be done
directly with pen and ink whether in the classroom or
outside.

(3) Date, and Page Number. Enter date on which
work is prepared, in the upper right corner of the page
about one and one-half inches from the top.

Number each page in the lower right corner whea
finished.

(4) Headings. Enter all work under the same heading
as in the text or as otherwise specified.

Begin paragraph numbers and headings about one-half
inch from the left margin.

(5) Lettering. The greater legibility of lettered head-
ings and the practical value of the ability to letter neatly
and rapidly, justify the requirement that title pages,

4 THE WORK-BOOK 5

chapter, section, and paragraph headings, and problem
titles shall be lettered. The best style of letter for this
purpose, because the simplest and the most easily made, is
the Engineering News alphabet shown below and used
quite generally in drafting rooms throughout the country.
The only principle involved in learning to use this
alphabet is that each letter consists of straight lines, or
arcs of circles, or both, and that the width and the height
are the same.

ABCDEFGHI JKLMNOPQRSTUVWXYZ a
1234567890 1i 2|3f4jf5£6& 7.089

ABCDEFGHIJKLMNOPQRSTUVWXYZ 1234667690
f 1M "H £ « K ||Vi-§f-~,

abcdeFghijklmnopqrsfuvwxyz

Be sure to observe that W is not an inverted M , nor M
an inverted W. Observe also particularly how R and
G are made.

(6) Title Pages. Letter title pages in India ink
without punctuation, for the subject and for each chapter.
Insert these in the work-book preceded by a blank sheet.

(7) Spacing. Indicate a new topic both by heading
and by extra space. Keep all work in straight lines with
no irregular spacing between words or lines.

If straight lines are difficult without a guide, rule a
page of the work-book in India ink with lines from three-
sixteenths to one-fourth of an inch apart, and place it
under the page when writing.

Examples and problems are best separated by extra
space only, but if preferred the separation may be em-
phasized by a hair-line not over two inches long, lightly
drawn with straight-edge and ruling pen.

6 TECHNICAL TRIGONOMETRY 5

(8) Drawing. Drawings which are to be measured
should be drawn lightly and carefully with pencil and straight-
edge, and after measurement should be inked in with India
ink and a ruling pen. Other drawings may be done directly
in India ink.

As soon as possible learn to draw a light, smooth, drafts-
man's line.

(9) Symbols. Make parentheses, equality signs, and
other symbols carefully: parentheses with regular curves
of the same height as the quantities inclosed; the lines
of the equality sign exactly the same length and about
one-eighth of an inch long.

Learn to make comparatively small, neat figures and
to draw free-hand, smooth, light, straight lines when
performing the four fundamental operations.

Work slowly, seriously, and steadily, and thereby
become expert by avoiding careless mistakes.

5. Instructions for the Record Sheet. Six daily record
sheets, a half year's supply, are furnished with the work-
book. These have columns for the instructor's stamp
and for the daily entry by the student, of date, paragraph
and problem numbers, and number of hours spent in
outside preparation of studies.

Submit Work for Inspection as Follows: On the
first day prepare a record sheet by filling in the blanks as
indicated. At the top of the time columns letter the
names of the studies in which outside preparation is
required, as Math., Phys., C. L. (chemistry laboratory), etc.

Observe that the record sheet provides for a complete
record of the mathematics work and is a time sheet fop
all studies.

Whenever work is to be submitted attach it to the
record sheet on which fill in the entries denoted by the
column headings, making no entries in the remarks column.

When an additional record sheet is needed place it on
top of those already filled.

9 THE WORK-BOOK 7

6. Excuse for Non-performance and Absence. If

unable to do assigned work, present to the instructor at
the beginning of the period a written excuse with date,
assignment, exact reason for failure, and signature. Enter
the date on the record sheet and write the word " Excuse "
in the remarks column.

In case of absence enter the date of each day's absence
and write " Absent " in the remarks column.

7. Collection and Distribution of Work-books, (a)
Collection. On the stroke of the bell at the beginning
of the period each student will pass his work-book along
the row in reverse order from which the chairs or desks
are numbered, being sure to place it on top of the books
passed to him.

The student receiving the books of the last row will
collect each row's books and will place them in the file.

Work-books may be taken from the classroom only when
permission is noted on the record sheet by the instructor.

(6) Distribution. At the beginning of the mathe-
matics period the collector will place the books at the
end of the rows so that each student may remove his book
from the pile as it is passed.

Books of absentees will be reported directly to the
instructor's desk by the collector. In the collector's absence
the next student in the row will attend to the books.

8. Inspection. The remarks column on the record
sheet is for the instructor's stamp. When the dater is
used instead of the " accepted " stamp it signifies that
the work is incomplete, or unsatisfactory, or incorrect.
Changes in such work unless obvious or indicated in the
book, must be arranged with the instructor before the
close of the period.

9. Corrected Work. Incorrect or rejected work is due in
correct form at the beginning of the next mathematics period.

Make corrections in red ink on the same page with the
incorrect work. If numerous mistakes have been made

8 TECHNICAL TRIGONOMETRY 10

prepare a new page and insert it following the incorrect
one on which write in red ink " Corrected on next page,"
with date of correction.

10. Solutions of Problems. In paragraph 94, a solu-
\ Hon formula is defined and explained.

In deriving solution formulas and in all other work in
the form of a logical proof or demonstration, number each
equation consecutively at the left in a vertical column,
with Arabic numerals inclosed in a parenthesis.

Separate equations from specified operations by a
horizontal rippled line not shorter than half an inch. It
must follow each equation except those taken from the
text or formed from the conditions specified in the problem.

The following model solution shows how to number,
how to use the rippled line, and how to abbreviate.

Problem. — Formulate the area of a right triangle when
the hypotenuse and an adjacent angle are known.

nh 1

(1) A A = -<r- ~~~~ ~w AreaA = ^ base X Alt

(2) But a = c cos <j> — ~~ — Side Adj Acute Z

(3) And b = c sin <t> — ^ Side Opp Acute Z

<a\ a a c sin <l>Xc cos <t> Q , .

(4) .\ AA = ^-s Subs Ax

/r\ a a c 2 sin <t> cos <f> n .„

(5) .'. A A = 1 r Def Exp

11. Indication of Results. Indicate final results in all
problems by double underlining with parallel hair-lines
not over one-sixteenth of an inch apart. Do not write the
word " Answer."

On the same line with the result write a statement in
initial capitals specifying exactly what the result repre-
sents, whether diameter of pulley, radius of curve, helix
angle, instantaneous voltage, etc.

IS THE WORK-BOOK 9

12. Index. At the end of the school year arrange all
the work-books of the year in order, make an alphabetic
index for your combined book, cut the work-book cover
in two through the back, and bind all together with one
pair of fasteners.

On the front cover attach a label about 4"X5" with
a line border and lettered title enumerating the subjects
covered, as suggested in the facsimile label below:

MATHEMATICS WORK-BOOK

FIRST YEAR

ALGEBRA

GEOMETRY

TRIGONOMETRY

WRITTEN BY

PRATT INSTITUTE

1911-1912 S.M.D,

A student's work-book containing the first year's work.

TECHNICAL TRIGONOMETRY

MARSH'S MATHEMATICS

DAILY RECORD SHEET

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CHAPTER I

LOGARITHMS

Section 1, Logarithm of a Number Greater than Unity.
Section 2, Logarithm op a Number Less than Unity.
Section 3, Naperian or Hyperbolic Logarithms. Sec-
tion 4, Logarithm of a Product. Section 5, Logarithm
of a Quotient. Section 6, Logarithm of a Power.
Section 7, Logarithm of a Root. Section 8, Solu-
tion of an Exponential Equation.

13. Two Ways of Multiplying. 100 may be multi-
plied by 1000 as follows:

(1) 100X1000=100000,

(2) 10 2 X 10 3 = 10 6 = 100000.

Observe that in (2) the product is obtained by addition
of the exponents of the powers of 10 which equal 100 and
1000. It is therefore possible to multiply together numbers
which are integral powers of 10, by addition of the exponents
of these powers.

In like manner 472 may be multiplied by 67.5 by addi-
tion of the exponents of the powers of 10 which equal 472
and 67.5. But here are two difficulties:

What powers of 10 equal 472 and 67.5, and what does
10 equal when raised to the sum of these two powers?

The answer to this question and the removal of the
difficulties follow:

100 = 102
472 = 10 2 +
1000= 10 3
67.5 = 10 1 +
10=10!

11

^12 TECHNICAL TRIGONOMETRY 14

472 is greater than the 2d power of 10 and less than the
3rd; 67.5 is greater than the 1st power of 10 and less than
the 2d.

Therefore 472 = 10 2 +a decimal

an c = I rvl 4*a decimal

t

A table of logarithms is an arrangement of numbers in
sequence with the decimal parts of the powers of 10 which
equal the numbers.

Taking the decimals from the table we have:

472 = io 2 - 673942

67.5 = 1Q1-82M04

Therefore 472 X 67.5 = 10 4 503246

By the table io 4 -503246 = 3 1860#

Therefore 472 X 67.5 = 31860.

In practice the labor of this method of multiplication
is reduced by setting down only the exponents of 10 as
follows:

472 X 67.5 = 31860 2.673942

1.829304
4.503246

In like manner by the use of a table of logarithms, one
number may be divided by another and any power or root
of a number determined.

14. What a Logarithm is. Logarithms are used as a
means of shortening and simplifying the mathematical
processes of multiplication, division powers, and roots.

In every system of logarithms all numbers are regarded
as powers of another number which is called the base of
the system.

Therefore the definition of a logarithm:

The logarithm of a number is the exponent of the power
to which the base of the system must be raised to equal the
number.

15 LOGARITHMS 13

In the system of logarithms in common use, called the
common or Briggs' System, the base of the system is 10.
Hence in this system all numbers are regarded as powers of 10.

Consider any number, as 306. In the Briggs' system
the logarithm of 306 is the exponent of the power to which
the base 10 must be raised to equal 306.

Now 102 = 100 and 10** = 1000. But 306 is greater
than 100 and less than 1000. Therefore in order to obtain
306 from 10, 10 must be raised to a power between the second
and the third.

Therefore 306 = 10 2 +a decimal

As shown 2+a decimal is the exponent of the power
to which the base 10 must be raised to equal 306.

But by definition the exponent of the power to which
the base must be raised to equal a given number is the
logarithm of that number.

Therefore log 306 = 2+a decimal.

This decimal is given in the table of the Logarithms of
Numbers.

§ 1. THE LOGARITHM OF A NUMBER GREATER THAN

UNITY

15. A Number Having Three Figures. Direction I.
In the table of the Logarithms of Numbers find 306 in the
column headed N.

Place the index finger of the left hand directly under
306 and move the hand to the right in a horizontal line
until it is under the number in the column headed 0. This
number with the two figures prefixed (called leading figures)
which are immediately above the blank space to the left
of it in the same column, is the decimal part of the logarithm
306.

Therefore io 2 - 485721 = 306 ; log 306 = 2.485721.

14 TECHNICAL TRIGONOMETRY 16

The log of 306 therefore consists of two parts:

(1) Integral, called the characteristic,

(2) Decimal, called the mantissa.

The decimal part only, is given in the tables.

16. Accurate Use of the Tables. The above direction
for using the index finger of the left hand is given in order
to secure speed and accuracy in the use of the tables. It
makes possible the unobstructed use of the right hand for
writing the figures from the tables and the index finger can
be kept in its position on the page until the required number
has been written from the table and the written number compared
with the printed number.

Another excellent method is to move a straight-edge
or a blank sheet of paper up or down the page until the
required number can be read just above the upper edge.

17. Reason for the Characteristic. The integral part
of the logarithm of 306 is 2, but the number of figures in
306 is 3.

In this instance the integral part of the logarithm is
one less than the number of integral figures in 306, the
natural number.

This is also true regarding any other number, for example
4798.

This number is greater than 10 3 and less than 10 4 .
Therefore as in the case of 306 the integral part of the
logarithm is one less than the number of integral figures
in the natural number.

18. Characteristic of the Logarithm of a Number
Greater than Unity. Rule 1. The characteristic (integral
part) of the logarithm of any number greater than 1 is one
less in unit value than the number of integral figures in
that number.

By this rule the characteristic of the logarithm of any
number greater than unity may be determined without

20 LOGAEITHMS 15

the necessity of locating 'it with respect to integral powers
of 10. I

19. Position of Decimal Point. Since the division of
a number by 10 is made by moving the decimal point
one place to the left and division by 100 is made by moving
the decimal point two places to the left and so on, the position
of the decimal point affects the characteristic only.

For example, log 306 = 2.485721

log 30.6 = 1.485721
log 3.06 = .485721
log .306 =*I. 485721
log .0306= 2.485721

Each number being one-tenth the preceding represents
one less integral power of 10. Therefore, every shift of
decimal point to the left means one less integral power
of 10 and every shift to the right, one more integral power
of 10.

In both cases the mantissa is the same. In reading
mantissas, therefore, disregard decimal points in the natural
numbers.

20. A Negative Characteristic. The mantissas of the
logarithms of all numbers are positive but the character-
istics may be positive or negative. They are positive for
numbers greater than unity; they are negative for decimals.

It is therefore impossible to denote the logarithm of a
decimal by a minus sign written in the usual position since
that would indicate the entire logarithm as negative.
Accordingly negative characteristics are indicated by a minus
sign above them.

Thus log .401 = 1.603144.

* The minus sign above the characteristic is used to denote that
the characteristic is negative but that the mantissa is positive.

16

TECHNICAL TRIGONOMETRY

21

21. Examples. About one inch and a half from the
top of the page rule the following table with the columns
long enough for twelve entries and sufficiently wide for the
numbers in this paragraph and in paragraphs 23 and 28.

Table I
LOGARITHMS

3 Figures.

4 Figures.

>4 Figures.*

Number.

Logarithm.

Number.

Logarithm.

Number.

Logarithm.

In the first column enter the following numbers: 375,
189, 208, 784, 999, 118, 510, 619, 200, 907, 666, 103.
In the second column enter:

(1) The characteristics (determined by rule I), and

(2) The mantissas (determined by direction I).

22. A Number Having Four Figures. Direction II.
To read the logarithm of a number consisting of four figures,
as for example, 5848, find the first three figures, 584, in
the N column.

Place the index finger of the left hand immediately
under 584 and move the hand in a horizontal line to the
right until it is under the number in the column headed 8.
This number with the leading figures prefixed from the
zero column is the mantissa of the log 5848.

The mantissa of the logarithm of any number containing
four figures is always to be found in the same horizontal
line as the first three figures, in the column headed by the
fourth figure of the number.

* This symbol means " greater than."

24 LOGARITHMS 17

23. Examples. In Table I enter the following numbers
and their logarithms : 2084, 3009, 6000, 1289, 9104, 1059,
7855, 8899, 5123, 7248, 6167, 1070,

24. A Number Containing any Number of Figures
Consider the number 84678. This number is greater than
84670 and less than 84680.

log 84680=4.927781
and log 84670=4.927730

10 = 51

That is, a difference of 10 in these natural numbers
corresponds to a difference of 51 in their logarithms.

In other words a difference of 10 between 84670 and
84680 is expressed by a difference of 51 in their logarithms.

Since the difference of 10 in the natural numbers equals
a difference of 51 in their logarithms, a difference of 1 in

natural numbers equals a difference of ttt of 51, or 5.1 in

logarithms. .

Therefore a difference of 8 in natural numbers equals
a difference of 8 times 5.1, or 40.8 in logarithms.

Therefore the logarithm of 84678 is 40.8 greater than
the logarithm of 84670.

Therefore log 84678=4.927771.

The same result would have been obtained by multi-
plying 51 (the difference between log 84670 and log
84680) by 8 the last figure of the given number 84678,
with a decimal point before that figure.

A continuance of this investigation with numbers of
more than four figures will give the following rule:

Rule. The logarithm of a number having any number
of figures equals the logarithm of the first four figures as
given in the tables, plus the product obtained by multiply-

18 ' TECHNICAL TRIGONOMETRY 25

ing the difference between the log of a number 1 greater
than the first four figures of the given number, by all the
remaining figures of the given number with a decimal
point before them.

Therefore, Direction III: To find the logarithm of a
number consisting of more than four figures, for example,
642147, find the logarithm of the first four figures, as in
Direction II.

Then multiply the difference between the mantissa
of the next higher number of four figures, 6422, and the
mantissa of 6421 (the first four figures), by the remaining
figures of the given number with a decimal point before
them.

Add this product to the logarithm of the first four figures,
placing it for performing the operation of addition so that
the right integral figure of it is under the right figure of the
mantissa of the logarithm to which it is to be added.

Thus log 6422 = 807670

log 6421 = 807603
difference = 67

.47

469

268

31.49
. 807603

log 642147 = 5.807634

25. Tabular Difference. In some tables the average
difference between successive mantissas, called tabular
difference, is given in a column headed D. In using such
tables it is customary not to ascertain the exact difference
by subtraction, but to multiply the tabular difference in the
same horizontal line as the mantissa of the given number
(if there is no tabular difference in the same horizontal line,
use the tabular difference immediately above), by all the

27 LOGARITHMS 19

figures of the natural number except the first four, with a
decimal point before them.

26. Proportional Parts. In modern tables of log-
arithms the last column is used for the proportional parts
which must be added to the mantissa when the number
has more than four figures.

The entries in the column are as follows:

The numbers at the top of each group are differences,
those at the left are the figures of the natural number,
and those in the several groups are the proportional parts
required for a fifth figure of a natural number, provided the
mantissa table gives readings for the first four significant
figures.

If the reading is for a sixth figure, one figure must be
pointed off in the number read; if the seventh figure, two
places must be pointed off, and so on. This is due to the
fact that the value of a figure in the sixth place is only
one-tenth of its value in the fifth place, and its value in
the seventh is only one-hundredth of its value in the fifth
place.

The column of proportional parts saves time and labor
and does away with the necessity of multiplying the dif-
ferences as required in the preceding paragraph.

In the Author's six-place tables, proportional parts are
given at the bottom of the page, arranged with differ-
ences at the top and figures of the natural number at the
left.

27. How to Use the Table of Proportional Parts. In

this paragraph the logarithm of 7145863 will be determined

by

(1) Direction III,

(2) The table of proportional parts.

(1) The logarithm of the first four figures 7145, is
6.854002.

20 TECHNICAL TRIGONOMETRY 27

Multiplying the difference 61, from the D column by
the remaining figures with a decimal point before them
we have:

.863

61 6.854002

863 53

5178 .\ log 7145863 = 6 . 854055

52.643

(2) The readings from the table give:

6 . 854002
49
37
18

6.854055
In the reading:

49 was the reading for 8, the fifth figure of the given
number, and was therefore set down directly under the
mantissa;

37 was the reading for 6, the sixth figure of the given
number, and was therefore shifted one place to the right;

18 was the reading for 3, the seventh figure of the given
number, and was therefore shifted two places to the right
when set down for addition.

Observe that in the result only 6 places were retained
because the readings were taken from a 6-place table.

Always use exact difference. An inspection of any page
of the table of the logarithms of numbers will show that
the entries in the D column are not exact but are the average
differences between successive mantissas for one or more lines.

Therefore, whenever a reading is taken which requires
the use of the table of proportional parts, determine the dif-
ference between the last figure of the mantissa which is read
and the last figure of the mantissa in the next column, and
take the other figures of the difference from the D column*

29 LOGARITHMS 21

When reading in the 9 column use the last figure of the
following zero column to determine the exact difference.

It is frequently the case that the table of proportional
parts gives no difference corresponding to the actual dif-
ference between the successive mantissas. No rule can
be given as to what difference to use in such cases, as some-
times the next larger difference will give a more nearly
correct reading, and sometimes the next smaller, but this
could not be determined without a table which gives the
next figure of the mantissas. To secure uniformity of
results it is suggested that the next smaller difference be
invariably used under such circumstances.

In other words, when the exact difference is not in the
proportional parts table, read proportional part under the
next smaller difference.

Therefore when using the table of proportional parts

(1) Note the actual difference between the last figure of
the mantissa read and the last figure of the mantissa in the
next column, and take the other figures from the D column.

(2) If this difference is not given in the proportional
parts table, take P.P. readings under the next smaller
difference.

28. Examples. In Table I enter the following numbers
and their logarithms, taking all readings from the table.

Show all computations on a page following Table I.

70829, 238495.06, 1007.61, 500.0089, 8799920.678,
4900.3240, 30005, 800062, 4300091, 699.714, 5.30792,
19.9049.

29. Miscellaneous Examples. On the same page with
Table I rule Table II for the entry of the examples below
and those in paragraph 34. Make entries as in Table I and
show all computation.

11007, 30000, 54431, 472, 7980, 3466145, 427214.96,
1097, 20.0084, 7290.632, 3.008, 47.979.

22 TECHNICAL TRIGONOMETRY 30

§ 2. THE LOGARITHM OF A NUMBER LESS THAN

UNITY

30. Sign of the Characteristic. Suppose that we wish
to determine the characteristic of the logarithm of .1; in
other words suppose we wish to know what power of 10
will equal .1.

Now

10° = 1

and

10! = 10

but

1 = 10 times .1

and

10 = 100 times .1

When raised to the zero power, therefore, 10 becomes
10 times .1 and when raised to the first power it becomes
100 times .1.

As even a zero power gives a result ten times too large
it is evident that 10 can equal .1 only when raised to a
power less than 0, and therefore to some negative power.

This power may be determined as follows:

•l-^-io-s

01 =-— =— = 10- 2
100 10 2 '

001 = — — - = — = 10 - 3
1000 10 3 '

Therefore .1 can be obtained by raising 10 to a power
indicated by the exponent — 1, that is, .1 = 10" 1 .

But by definition, the exponent of the power to which
the base of the system must be raised to equal a given
number, is the logarithm of the number.

Therefore (a) log .1 = — 1

and (6) log .01 = -2

and (c) log .001 = -3.

32 LOGARITHMS 23

In (a) it is seen that .1 is obtained from 10 by raising
10 to a power indicated by the exponent — 1.

In (6) .01 is obtained by raising 10 to a power indicated
by the exponent —2.

In (c) .001 is obtained by raising 10 to a power indicated
by the exponent —3.

By continuing this analysis with any decimal it would
be found that the characteristic of the logarithm of a deci-
mal is always negative.

31. Unit Value of Characteristic. The unit value of
the characteristic may be determined from a consideration
of some decimal, for example .306.

It is evident that .306 is greater than .1 and less than 1.

But log .1= —1 and log 1=0.

Therefore .306 can be obtained from 10 by raising 10
to a power indicated by an exponent greater than — 1,
and less than 0, i.e., —1+ a positive decimal; in other
words the logarithm of .306 equals — 1+a decimal.

Therefore log .306 = 1.485721 .

The minus sign is placed immediately above the character-
istic to show that the characteristic only, is negative, while the
mantissa is positive.

It must never be placed in front of but always above
the characteristic.

In the preceding work it has been shown that the char-
acteristic of the log .306 is — 1. But the first figure in .306
is one place from the decimal point. In this instance
therefore, as in (a), (6), and (c),the characteristic of the
logarithm of the decimal is negative and in unit value
equals the number of places of the first significant figure
from the decimal point.

32. .Another Illustration. The decimal .042 is less than
.1 which equals 10" l and is greater than .01 which equals
10- 2 .

The power therefore to which 10 must be raised to give

24 TECHNICAL TRIGONOMETRY 33

.042 is less than the power indicated by the exponent— 1
and is greater than the power indicated by the exponent— 2.

But by definition, the exponent of the power to which 10
must be raised to equal a given number is the logarithm
of the number.

Therefore the logarithm of .042 is less than —1 and
greater than —2.

Therefore .042 = 10~ 2+adecimal .

(See also paragraph 234.)

33. Characteristic of Logarithm of a Number Less than
Unity. It is evident from the preceding illustrations that
the characteristic of the logarithm of a decimal is negative
and in unit value equals the number of places of the first
significant figure of the decimal from the decimal point.

The first significant figure of .042 is two places from the
decimal point; the characteristic is —2^ Therefore

Rule II. The characteristic of the logarithm of a number
less than 1, in sign is negative and in unit value is equal to
the number of places of the first significant figure of the given
number from the decimal point.

34. Examples. Under the heading Decimals enter the
following with their logarithms in Table II and on a sub-
sequent page show all computations:

.428, .0343, .00072, .0700486, .00009915671, .000008409,
.17, .0056007, .95064, .70147, .314087, .0127078.

*

Show also the readings for the logarithms of the follow-
ing numbers:

1. 1.009678. 2. 581.639.

3. .00049637. 4. .072385.

5. 3891.739. 6. .8276309.

7. 18.46708. 8. 4.72381.

9. .008372008. 10. 542.39X10-*.

36 LOGARITHMS 25

35. Summary. Following is a summary of the essen-
tial facts presented in the preceding pages of this
chapter.

1. Logarithms are exponents.

2. A logarithm consists of two parts:

(a) Integral, called the characteristic, ascertained by
inspection;

(6) Decimal, called the mantissa, given in the tables.

3. The characteristic of the logarithm of a number
greater than 1, in sign is always positive and in unit value
is one less than the number of integral figures in the
number.

4. The characteristic of the logarithm of a number
less than 1, in sign is always negative and in unit value is
equal to the number of places of the first significant figure
of the number from the decimal point.

5. The mantissa of a number is not affected by the
position of the decimal point in that number.

6. The mantissa of the logarithm of a number having
less than four figures is in the column in the same hori-
zontal line with the given number.

7. The mantissa of the logarithm of a number having
four figures is in the horizontal line with the first three
figures of the number, in the column headed by the fourth
figure of the given number.

8. The logarithm of a number having more than four
figures is obtained by adding to the logarithm of the first
four figures, either

(1) the product of the tabular difference by the remain-
ing figures preceded by a decimal point; or

(2) the readings from the table of proportional parts.

36. Antilogarithms. The natural number corresponding
to any given logarithm is called an antilogarithm. It can
be obtained by the inverse of the processes employed in
determining the logarithm of a natural number.

26 TECHNICAL TRIGONOMETRY 36

In the work-book enter the following:

Antilog 2.589703 =

Disregarding the characteristic 2, find the leading
figures 58 in the table of logarithms of numbers in the
column.

Place the index finger of the left hand under these figures
and move it down the column until it is under 9, the
third figure of the given mantissa .589703.

Move the finger to the right along the line or the line
above, until it is under the mantissa next smaller than
.589703.

Write this mantissa in the work-book under the given
mantissa.

After it write an equality sign followed by the natural
number to which it corresponds, the first three figures
of which are in the N column, in the same horizontal line,
and the fourth figure is at the top of the column in which
the mantissa was read.

(1) Subtract the second mantissa from the given man-
tissa.

Determine the figure of the natural number to which
the remainder is equal, by the P.P. table as follows:

Subtract the mantissa read, from the mantissa in the
next column.

Under this difference in the table of proportional parts,
find the number equal to or smaller than the given remain-
der. Enter this number under the remainder obtained
in (1).

The figure at the left end of the line in which the number
was read is the fifth figure of the natural number.

Write the five figures now found, as the antilog*
2.589703 as follows: log" 1 * 2.589703 = .

* Antilog is symbolized log -1 , the dash 1 signifying anti. Observe
that the symbol is not minus 1, but dash 1.

38 LOGARITHMS 27

37. Position of Decimal Point. The number obtained
in paragraph 36 may be pointed off by reference to the
characteristic 2.

Does its positive sign indicate that the natural number of
whose logarithm it is the characteristic, is integral or decimal?

What is indicated by its unit value?

Point off the required number of integral figures and under
the work now entered write reason for the operation as follows :

Sign of the characteristic 2 is (state whether + or — ).

Therefore the natural number is (state whether greater
or less than 1).

The unit value of the characteristic is (state how many).

Therefore the natural number has (state how many
integral figures).

38. Examples. The following arrangement is suggested
for the work of this paragraph:

1.806259

248 =*6401

11

7 =

= 1

40

41 E

= 6

.640116

What to do when the leading figures change in the line

and when a difference gives no reading is shown below:

3.700364

58 =5016

6 =0

60
60=7

.00501607

Determine and fill in the omitted entries in the table on
page 28, showing all readings.

♦This symbol is a combination of a dash (from log -1 meaning
antilog) and an equality sign. It means and should be read " whose
antilog equals."

28

TECHNICAL TRIGONOMETRY

89

Table III
ANTILOGARITHMS

No.

Logarithm.

Number.

No.

Logarithm.

Number.

1
2
3

4
5
6

2 . 706184
2.706184
1 . 580462
3.418051
3.280096
4.121873

7

8

9

10

11

12

2.900173
1 . 607585
3 . 526319
4.470008
1 . 342004
.087267

§ 3. NAPERIAN OR HYPERBOLIC LOGARITHMS

39. Systems of Logarithms. There are two kinds or
systems of logarithms:

1. Common or Briggs' logarithms.

2. Natural, hyperbolic, or Naperian logarithms.

In elementary, practical computation by logarithms,
common or Briggs' logarithms are used almost exclusively
and are the logarithms always meant when one speaks of
logarithms.

These two systems have the following bases:

Briggs' or common, base 10.

Naperian or hyperbolic, base e = 2.718284.

The logarithms of numbers in these systems are denoted
as follows:

Common, log 28.341.

Naperian, log« 28.341 or hyp* log 28.341.

In the computations of higher mathematics where it
is understood that the logarithms employed in formulas
are to base e, a common logarithm is denoted by subscript.

Thus logio a.

40. Naperian or Hyperbolic Logarithms. Many engi-
neering formulas have been derived by higher mathematics
and therefore involve logarithms to base e.

* " hyp " means hyperbolic.

41

LOGARITHMS

29

When such logarithms are required they may be deter-
mined in two ways:

(1) From a table of hyperbolic logarithms.

(2) By multiplying the common logarithms by 2.3026,
generally used as 2.3.

From (2) it will be noted that the logarithm of any
number to base e equals 2.3 times the logarithm of the number
to base 10.

Thus log« a = 2.3 log 10 a.

Solve the equation for log™ a.

Therefore to reduce Briggs' logarithms to Naperian
multiply by what?

To reduce Naperian to Briggs* divide by what?

41. Examples. Any number whose use as a multiplier
converts one quantity into another quantity is called a
coefficient, conversion factor, constant, or modulus.

The computation of the omitted entries in the following
table will give practice in the use of the conversion factor
2.3 in transforming common or Briggs' logarithms into
Naperian or hyperbolic logarithms.

Rule the table in the work-book and compute and fill
in the omitted entries.

Table IV

USE OF CONVERSION FACTOR

- ■
Ex.

Number.

Log

Log fl

1

296.4

2

31.065

3

5.834

4

1796.3

—

5

85478

6

60.007

7

138.009

8

4.6342

9

71324.6

10

5009.61

30

TECHNICAL TRIGONOMETRY

42

42. How the Modulus is obtained. Following is a
proof that the Naperian logarithm of a number equals
2.3 times its common logarithm.

Given n any positive number,

a = logett,

and & = log 10 n.

Prove a = 2.3026&,

or logen = 2.3026 log 10 n.

(1)

a = logett
and 5=log 10 n,

► —\/n^\/\/n^\xw

-Hyp

(2)

e a = n
and 10* = n

. ./N^V/N^N/N^V/N/-

~Def log

(3)

e a = 10 & ~~—

~=ity Ax

(4)

e = 10 T

-Root Ax

(5)

... log 10 6=-

-Def log

■

(6)

b

log 10 6

-Mul Ax and Div Ax

(7)

But 6 = 2.71828—

7

—Notation

(8)

b

Subs Ax

log 10 2.71828

(9)

But log 2.71828 = .434294

Table

(10)

b

7 = 2.30266 Subs Ax

4

.43429

(11)

.-. log«n= 2.3026

log 10 n~

— (1) and Subs Ax

The number in the denominator of (10) is known in
mathematics as the modulus of the common system of
logarithms because by its use Naperian logarithms are reduced
to common logarithms.

43. Hyperbolic Logarithm of a Decimal. . A simple
way of determining the logarithm of a decimal to base e

44 LOGARITHMS 31

by the use of the conversion factor 2.3026 or 2.3 is to
multiply the mantissa and characteristic separately as shown
below. Observe that the characteristic is written following
the mantissa.

Required the hyp log .09685.

.986100-2
2.3 2.3

2958300
1972200

2.2680300-4.6
4.6

3. 668030
hyp log .09685=5.6668030

44. Ways of Denoting Ciphers. When several ciphers
precede the first significant figure of a decimal they may be
denoted by:

(1) One cipher and a subscript whose unit value equals

the number of ciphers.

(2) Indicated multiplication of the significant figures

by 10 with a negative exponent having the same
number of units as the number of decimal places.

(3) Indicated multiplication of the significant figures

as a decimal, by 10 with a negative exponent hav-
ing the same number of units as the number of
ciphers immediately following the decimal point.

Thus .000498 and .0000007345 may be written

(1) .0 3 498 and .0 6 7345, or

(2) 498X10- 6 and 7345X10" 10 , or

(3) .498X10" 3 and .7345 X 10 ~ 6 .

If a number ends in several ciphers it may be denoted
by an indicated multiplication by 10 with an exponent
whose unit value equals the number of ciphers.

Thus 1832000000 is conveniently denoted by 1832 X10 6 . i

32

TECHNICAL TRIGONOMETRY

45

45. Examples. Compute the Naperian logarithms of
the following numbers, arranging the work as in paragraph
43.

1. .0864.

4. .0 3 731.

7. .0 4 80951.

10. .0 3 9812003.

13. .0 3 720041.

16. .00826054.

2. .48064.

5. .005614.

8. .0062149.
11. .817216.
14. .631142.
17. .049671.

3. .32196.

6. .073485.

9. .0«50009.
12. .0981467.
15. .0 6 480072.
18. .0029346.

§ 4. LOGARITHM OF A PRODUCT

46. How Logarithms are Used. Logarithms are expo-
nents. Their use is therefore governed by the laws of
exponents in the following operations:

Multiplication,

Division,

Involution,

Evolution,

Solution of an Exponential.

47. Multiplication by Logarithms. Since in the Briggs*
system all numbers are regarded as powers of 10, the
multiplication of two or more factors is only the multi-
plication of two or more 10s whose exponents are the
logarithms of the respective factors.

The exponents (logarithms) are therefore added.

Thus

Therefore

43.4X2.91 = 10* 637490 X 10 .463893
101.637490

10 .463893
1Q2.101383

43.4X2.91 = 10 2101383

47 LOGAEITHMS 33

What this product is, may be determined by reading
the antilogarithm of 2.101383 from the table.

Thus 2.101383

059 =; 1262

324

310 =9

140
138=4

The characteristic 2 denotes three integral figures.

Therefore 43.4X2.91 = 126.294, which is exactly the
same as would be obtained by arithmetical multiplication.

In practice 9 logarithms are never shown as powers of 10
but are set down directly.

Thus 43.4X2.91 = 126.294

1.637490

.463893

2.101383

059 b 1262

324

310 =9

140
138=4

One or More Factors Decimal.

If one or more of the factors of a product are decimals
and their logarithms therefore have negative characteristics,
there are three ways of doing the work:

(1) Set down the logarithms directly with their char-

acteristics as determined.

(2) Both add and denote the subtraction of 10, before

adding the logarithms.

(3) Set down the logarithms of the decimal, increased

by 10.

34 TECHNICAL TRIGONOMETRY 48

Thus

2454.1 X. 00568

(1)

18

3.389875

g. 754348

1 . 144241

(2)

18

3.389875

7.754348-10

11.144241-10

(3)

18

3.389875

7.754348

♦11.144241

3951 =1393

291

281 =9

100

94=3

13.939*

Whenever logarithms having negative characteristics are
to be added or subtracted it is suggested that they be set down
as shown in (3).

48. Examples in Multiplication., Solve the following
as indicated in the preceding paragraph:

1. 10.09X687. 2. 381.56X16.9217.

3. 7.298X1.654. 4. .5341X13.908.

6. 181.96X31.148. 6. .00715X1.0083.

7. 5.037X236.84. 8. 29.53 X. 42159.

9. .0*31287X12.64. 10. .0428961 X. 084507.

11. .81965 X .0 3 7964 X 15.823 X .073854.

12. .0 4 7864X 1384.19X .78113X 144.58X .0834.

* Two integral figures are pointed off in the result because the
actual characteristic is 1, characteristic 11 being 10 too large.

60

LOGARITHMS

35

§ 5. LOGARITHM OF A QUOTIENT

49. Division by Logarithms, or the Logarithm of a
Fraction. In the division of algebraic quantities the
exponents of the quantities in the divisor are subtracted
from the exponents of the same quantities in the dividend.

Therefore, in division by logarithms the logarithm of the
divisor is subtracted from the logarithm of the dividend.

A fraction is an indicated division of the numerator
by the denominator.

Therefore the value of a fraction may be determined by
subtracting the logarithm of the denominator from the logarithm
of the numerator and by reading the antilogarithm of the
remainder.

When the logarithm of the dividend is smaller than the
logarithm of the divisor it should be increased by one or
more 10s, as may be necessary.

Thus

(1)

3.728 -s- 145.73 = .02558
10.571476

(2)

2.163549

460
89

8.407927

0^=2558

26
17 = 1

90
84=5

60. Examples in Division.

by logarithms:

1. 12.803 -f- 1.728.
3. 3.7201-5-14.96.

.00568 = .0 5 231449
2454.1 '

7.754348 75
3.389893 18

4.364455 —
363 =2314
92

170
169=9

Solve the following examples

2. 39.1857-5-7.264.
4. .082305-5-4.7812.

36

TECHNICAL TRIGONOMETRY

60

5.

7.

9.

11.

13.

16.

17.

19.

21.

23.

26.

27.

29.

31.

33.

54.8607
50.7002*

.0»1208
.0 4 7862'

.970008
1.20705'

42.984
.90358*

.30876
.00043581*

.083451
.0041486*

11.6309
.72184 *

45.8671
318.0075'

.014592
.0056847*

21.8429
178.0096'

1396.75
23815.62'

.063581
.0043217*

398.41X10-'
.54189X10* '

15.7847 X. 35685
.048129X47.842*

5173.96 X 10 ~ 5
5300.48 X.0 6 6814'

.084056 X. 0054734

6.

8.

10.

12.

14.

16.

18.

20.

22.

24.

26.

28.

30.

32.

34.

1.9684
8.17201*

.00835
.74802*

.98305
14.0091'

1.72914
16.3429'

182.639
73.6425*

514.292
3084.43'

.3183
.074129*

.00096425
4 72638 #

.0 6 90723
.0 4 61895*

.44571
.0059412*

408.039
3423.086*

.0 3 465125
14962.9X10-*'

.076485 X 10 -»
1 7.632 X. 340084*

.017358X5.63419
496.431 X. 0054866'

.914184X30.639X84.69
.071945 X. 18726X44.069'

36. 17.28 X

,021968X1.70345

52 LOGARITHMS 37

§ 6. LOGARITHM OF A POWER

51. Involution by Logarithms. Involution is the
process of finding any power of a number.

The process and the reason for it should be evident
from the following:

(3.968) 3 = (10 698672 ) 3 = 62.476

.598572

3

1.795716

672 =6247

44

41 =6

30
28=4

In practice the power of 10 is not shown.

Thus ( . 09867) 4 =.0 4 94785

2.994185

4

5.976740

17=9478

23

23=5

The characteristic of the product is 5 because 4x2=8: and
8+3 which was carried, gives 5.

52. Multiply Characteristic and Mantissa, Separately.

When the exponent of the power of a decimal has more than
one figure the computation is best made by separate multi-
plication of mantissa and characteristic.

38 TECHNICAL TRIGONOMETRY 63

Thus .006348- 65 = .037300

.802637-3
.65 .65
4013185
4815822
.52171405-1.95
1.95

2.571714

09^ =3730
5 =0

50
j46=4

In the two preceding illustrations the logarithm of the
given number was multiplied by the exponent of the power
because to raise a number to a power is only to raise 10
with some exponent to the given power. In Chapter XIII of
the Technical Algebra is it shown that any power of a mo-
nomial is obtained by multiplying its exponent by the ex-
ponent of the power. Therefore the following

Rule for Logarithm of a Power. To obtain any
power of a number, multiply its logarithm by the exponent of the
power and read the antilogarithm.

53. Examples in Powers. Determine the following
powers by logarithms as shown in the illustrations of the
two preceding paragraphs. The first four examples, as
will be seen, are different powers of ir.

1. 3.1415926 2 .

2. 3.1415926 8

3. 3.1415926*.

4. 3.1415926*

6. .71834 4 .

6. .05738 13 .

'• e-)*

* C-) •

9. 713.604 28 .

10. .0 3 74209».

64

LOGARITHMS

39

11 f-l^tV

V-75631/ '

15.

/ .8621 1X42.19
\ 2.296

.2986 X

)•

/ 1.2986X.124 \<

7 * V .76201 / *

19. 1942.683*.

21. 37.634*X8.0097*.

23. 4317.92* X 72834*.

25. .63591 *-s- 2.7314*.

27. (.49518* X 6.701 58*) '

29.

.54631«X34.605»
596.43* X. 0009248*

/ 7.143<X.82176» y
* V .007824* / "

33. 8.0053942* -6.7084*.

/ .04318 y

\. 012685/ *

14. (29.608X.60312)*.

16.

r

18.

20.
22.
24.
26.

28.
30.
32.
34.

/.08433X10'V

3.406

)'•

/ 27045X10- 6 \*
\ 2.298 / '

15761.8*.

.08379* X. 92046*.

(2.7803»X.84507«) 2 .

(1.7046'X3.95167)*.

17.964*X.00952X1728

34.762*

(

xX9.345A *
4.0073* / '

21 . 436 4- .04573 *
208.92*-f-5.7814»'

1 9.783 X. 24685'- 1.9632*
5609.41*

§ 7. LOGARITHM OF A ROOT

64. Evolution by Logarithms. In Chapter XIII of the
Technical Algebra it is shown that a root is only a power
whose exponent is fractional.

Write a rule for the determination of any root of a
number by logarithms.

How to Extract the Root of a Decimal when
Root Index is Positive and Evenly Divisible. When the
root of a decimal is required and the root index is posi-
tive and evenly divisible in the negative characteristic, the

40 TECHNICAL TRIGONOMETRY 54

division may be performed with the negative characteristic
written in its usual place preceding the mantissa.

Thus ^.0072861 3)5.862504 498

1.287501 6

V.0 3 50187 2 )1.700591 531

2.350296 60

What to do when the Root Index is Negative, or
when not Evenly Divisible. When the root index is nega-
tive, or not evenly divisible in a negative characteristic,
there are several ways of determining the quotient as
shown in subsequent paragraphs.

The two best are the following:

(1) When the root index is positive and not evenly divi-
sible, both add to the logarithm of the given number, and
indicate the subtraction from it, of the smallest integral
multiple of the root index which will eliminate the negative
characteristic. Divide the result by the root index.

(2) When the root index is negative, even though it is
evenly divisible in the characteristic, find the excess of
the negative characteristic over the positive mantissa.
Divide the result by the root index.

(1) Positive Root Index, not evenly Divisible.

2 V.0007154 = .042896

1.854549
4.6

2.3 )1.454549 -4.6
.632413 -2
356=4289
57
j>l=6

In the preceding illustration, the smallest integral multiple
of 2.3 which will eliminate 3 is 2X2.3 =4.6.

55

LOGARITHMS

41

4.6 was therefore added and its subtraction denoted, before
the division was performed.

The characteristic of the quotient, as shown, is 2.

(2) Negative Root Index.

-3

V.0 7 7584= 236.247

-3)-

8.879898
7.120102 *

2.373367

280 =2362
87
74 =4

130
129=7

66. Examples in Roots. Compute the following roots
by logarithms as shown in the illustrations of paragraph 54:

1. V.002948.
3. V714629.
5. 1,8 V.004157.
7. V3.1415926.
9. "V.27806.

11. V.004248.

13. Vl294.63.

15. V.0 4 609218.

17. "V.012983.

IS. "V29.856.

21. V.017648.

23. "V.0063057.

2. V.052915.

4. V545.17.

6. 4 V.061208.

8. V.7821.

10. V2.38007.

12. V.867502.

14. V.096317.

16. V.12.00087.

18. "V.71405.

20. "V.0 5 78201.

22. Vl604.905.

24. V.091872.

* —7.120102 is the excess of minus over plus in 8.879898 and is
obtained by subtracting .879898 from 8.

42

TECHNICAL TRIGONOMETRY

56

25. N/341. 738.

27. "V.013968.

29. "V.0085621.

81. V.83176.

33. V2.85009 X 3 V.044216.

26. V29.6294.
28. Vl78.4219.
30. 3 ' 2 V4498.706.
32. V37.6421.
34. 175 - 2 V. 061287.

35. V. 0017298 xV21.0634xVl42.833.

§ 8. SOLUTION OF AN EXPONENTIAL EQUATION

56. Definition and Illustration. An exponential equation
is one in which the unknown quantity is the exponent.

Thus 24.5* = 12.298

Equations of this kind are solved as follows:
(1) Base Greater than Unity.

x=

• • X

log 24.5* =
■ log 24.5 =

X-
X-

=log 12.298
- log 12.298
log 12.298
log 24.5
1.089835
1.389166

552
283

.037028

.142702

318

31

119

187

19S

1

187

.037360

.142754

.142754

1.894606

593=

7845

13

lis

2

.78452

56 LOGARITHMS 43

(2) Base Decimal. Since negative numbers have no
logarithms, the following illustrates what to do when both
of the known quantities are decimal and when therefore the
algebraic sum of characteristic and mantissa is negative:

*»»

A- *

V o=

a

o=

.0543° =.2347

1.370513

2.734800

- .629487
- 1.265200

.629487
1.265200

(3) One Term of Fraction, Negative. When only
one number is decimal, the operations are as follows:

1.342 6 =. 08763
2.942653

b =

b=

6=-

.127753

-1.057347
.127753

1.057347

.127753

As shown b is negative. The value of the fraction is
therefore computed by logarithms and the antilog is written
with a minus sign before it.

In practice, work like that in the preceding illustrations
should be set down as follows:

(1) 24.5*= 12.298

1.089835 552

1.389166 283

and so on as shown in (1),

44 TECHNICAL TRIGONOMETRY 57

(2) ,0543° = .2347

1.370513
2.734800

.629487
1.265200

and so on as shown in (1).

57. When the Exponent is Negative. In case the
unknown exponent is negative, the operations are as follows:

1.342-*= .08763
2.942653

-6=

6 =

.127753

1.057347

.127753

Observe that the last equation is obtained from the
second equation by multiplying by —1.

58. Examples. Solve the following equations:

1. 19.87* =345.2. 2. .2467* = 12.675.

3. 3.7983 r =449.31. 4. .00434* - .097504.

6. .01432* = .18335. 6. .645001* =27.6404.

7. .5264"° =1.9234. 8. .01872-* =.80441.

9. .0734"* = 1.11054. 10. .7087 * =22.007.

11. .00651" 2 * =3.7841. 12. 56.908' =.15925.

13. .4084* =7.0382. 14. .07329 r = .23194.

15. 7.6824 c =1.70321. 16. .92215 r = 2.2973.

17. .04664* =3.7163. 18. .0584* =2.4556.

19. .2097 r =5.9216. 20. 3.8116* =.09675.

21. 4.2356' = .95064. 22. .007582* =1.35674.

23. .08251* = .75263. 24. .58124* = .06352.

69 LOGARITHMS 45

59. Summary of Laws with Formulas. There are five
laws for logarithmic computation, which may be formulated
as follows:

I. Logarithm of a Product, ab.

log (ab) =log a+log b.

a
II. Logarithm of a Fraction, -.

b

a
log- » log a - log b.

D

in. Logarithm of a Power, a b .

log a b =b log a.

IV. Logarithm of a Root, V a.

log Va=— , •
b

V. Logarithm of an Exponential, a 1 =b.

z log a =log b.

If either b or a is decimal,

* logb
log a*

If both b and a are decimal, or greater than 1,

logb

u-

log a'

Note. For a more comprehensive treatment of powers
and roots and logarithmic computation, see Marsh's Tech-
nical Algebra, Chapter XV, §§ 9 and 10.

* See (3) page 43. t See (2) page 43.

CHAPTER II
TIIE RIGHT TRIANGLE

Section 1, Functions of an Acute Angle. Section 2, Solu-
tion of a Right Triangle. Section 3, The Isosceles
Triangle. Section 4, Applied Problems.

§ 1. FUNCTIONS OF AN ACUTE ANGLE

60, Construction. Enter the headings in the work-book.
Place the straight-edge across the page and without

moving it, draw base lines of unequal length about half
an inch apart, for three right triangles, and sufficiently long
to fill nearly the width of the page. Draw these lines sc
that there shall be room below for Table V.

On each base line carefully draw a right triangle by
sliding your 30-60 triangle along the straight-edge, making
the acute angle A the same size in each.

Denote each right angle by C and the other acute angles
byB.

Denote the sides by small letters corresponding to the
capital letters at the opposite vertices, and number each
triangle as indicated in the table.

Rule the table in the work-book, with heavy marginal
lines.

61. Tabulation and Comparison of Results. With a
centimeter or 64th scale, measure each of the sides of the
triangles and enter the measurements and ratios in the
table as indicated.

46

61

THE RIGHT TRIANGLE

47

Table V

STUDY OF FUNCTIONS

Sides.

Fig. I.

Fig. II.

Fig. III.

a =

No.

6 =
c =

Ratios.

Fraction.

Decimal.

Fraction.

Decimal.

Fraction.

Decimal.

a

1

C

2

b

c

3

a
b~

4

c
a

5

c

6

a

Reduce each ratio to a decimal of four places, increasing
the fourth decimal figure by 1 when the fifth figure is 5
or greater.

When the entries in the table are completed write the
answers to the following questions:

Is there any difference in the decimal value of - or of

any of the other five ratios because of the difference in the
size of the three triangles?

It is proved in geometry that mutally equiangular
triangles are similar.

Are triangles 1, 2, and 3 similar? Why?

What therefore is the geometric reason why the ratios
of the same homologous sides are all equal to each other?

48

TECHNICAL TRIGONOMETKY

62

62. Ratios Dependent. Draw a right triangle, Fig. IV,
having angle A considerably larger or smaller (as convenient)
than angle A in the previous figures.

Measure the sides and enter the measurements in

Table V a , the ruling for which may be determined from

Table V.

»
Do the six ratios in Table V a have the same respective

decimal values as in Table V?

Is this because of the difference in the size of the triangles
or because the acute angles in Fig. IV are not the same as
in Figs. I, II, and III?

Therefore does the numerical value of the ratios of the
sides of a right triangle depend upon the size of the triangle,
or does it depend upon the size of the acute angles?

63. Sides with Reference to an Acute Angle. Under
paragraph number and heading, rule Table VI as here
shown and complete the entries.

Table VI

FUNCTIONS OF ACUTE ANGLES OF A RIGHT TRIANGLE

Nos. of Ratios.

Ratios.

Sides with Respect to Angle A.

1

2

a

opp

C

c

hypot
adj

hypot

3

4

5

6

65

THE EIGHT TEIANGLE

49

64. A Function. A function is a quantity whose value
depends upon the value of another. Write this definition.

The ratio of the sides of a right triangle are functions
of what parts of the triangle?

There are how many of these ratios?

Therefore in a right triangle there are how many func-
tions of what parts of the triangle?

65. Names and Definitions of the Six Functions. Fol-
lowing is a table containing the names, abbreviations, and
definitions of the six functions. These definitions are the
basis of the study of trigonometry and of trigonometric
computation, and should be thoroughly memorized.

Table VII
DEFINITIONS OF FUNCTIONS

No.

1
2
3
4
5
6

Name of
Function.

Sine

Cosine

Tangent

Cosecant

Secant

Cotangent

•Abbre-
viation.

sm= - =

COS= — =*

tan= t^

csc= — =

sec= t =

cot= — =

Ratio.

a

Definition.

Mnemonic

Opp Side

C~~

Hypot

Sinophy

b

Adj Side

c~~

Hypot

Cosadhy

a

Opp Side

b"

Adj Side

Tanopad

e

1

a

sin

e

1

b~

cos

b

1

a

tan

Formation of
Mnemonic.

Sin, Opp side,
Hypot

Cos, Adj side,
Hypot

Tan, Opp side,
Adj side

A good way to fix in mind the last three functions is to
memorize the fact that the cotangent and tangent are
reciprocals, and that no co-function is the reciprocal of any
other co-function.

50 TECHNICAL TRIGONOMETRY 66

66. Explanation of Table VII. The table shows that
the six functions of an angle are the

sine, cosine, tangent, cosecant, secant, cotangent.

Do not fail to notice that the abbreviation of each
function is written with a small initial letter and is not
followed by a period. They should never be written in any
other way.

By the definitions in the table, if 6 is one of the acute
angles of a right triangle,

a . side opposite the angle

bine u = ; :

hypotenuse

side adjacent to the angle

Cosine 6 =

hypotenuse
side opposite the angle

Tangent = ., ,. ,

side adjacent

Cosecant = -r

Secant

sine
1

Cotangent =

cosine
1

tangent*

Thus in the figure which shows a triangle with the right
angle C, we have the following:

1 2

A C

csc A=— ,
a

A C
SeC ^ = T";

cot A=— .
a

The functions in column 2 are in what relation respec-
tively to the functions in column 1?

68

THE RIGHT TRIANGLE

51

67. Functions of Both Acute Angles. Draw a right
triangle, Fig. V, making the hypotenuse the base.

Denote the right angle by and the acute angles by T
and 7.

Denote the sides by small letters corresponding to the
capitals at the opposite vertices of the triangle.

Rule Table VIII and make the entries indicated, in terms
of sides as lettered on the figure.

Table VIII
FUNCTIONS AND CO-FUNCTIONS

Function.

Ratio.

Function.

Ratio.

sin T =

•

COS 7 =

cos !T =

sin 7 =

tan!T =

cot 7 =

esc T =

sec 7 =

cot 7 =

tan 7 =

secT =

esc V —

68. Meaning of Co-Functions. In Fig. V, 7+!T=how
many degrees by geometry?

Therefore 7 and T are in what relation?

T is what of 7?
7 is what of 77

cos 7 equals what?
sin T equals what?

The cosine of an angle is therefore the sine of an angle
which is in what relation to the given angle?
The cosine is therefore what angle's sine?
" co " is therefore an abbreviation for what word?

Therefore the cosecant is what angle's secant?

The cotangent is what angle's tangent?

Therefore the co-functions are what angle's functions?

52

TECHNICAL TRIGONOMETKY

69

69. Three Functions, Three Reciprocals. The table
below separates the six functions into three functions and
their reciprocals.

Rule it in the work-book under paragraph heading and
number, and complete the entries by reference to Fig. I.

Table IX

FUNCTIONS AND RECIPROCALS

Function.

Ratio.

Reciprocal.

Ratio.

sin A =
cos A =
tan A =

CSC A =
sec A =
cot A =

70. Derivation of Other Functions from Sine and Cosine.

1. DERIVATION OF TANGENT

Enter headings and figure in the work-book.
In terms of sides as lettered,

(1) sin !T=what?*

(2) cos!T=what?

Divide the first equation by
T the second and simplify the
second member.

t

But -r = what function of 77
h

(5)

sin T
cos T

= what other function of T.

Solve (5) as answered in your book, for sin T.
for cos T.

Solve it

* Number all equations and specify authorities as shown in para-
graph 10.

72 THE BIGHT TRIANGLE 53

2. DERIVATION OF COSECANT

esc T = what? (See definition.)

In your answer, substitute from equation (1) and
simplify.

Therefore the cosecant of an acute angle equals the
ratio of what sides with respect to the angle?

To what is the secant equal in terms of sides?

To what is the cotangent equal in terms of sides?

71. Numerical Value of sin 2 +cos 2 . In Fig.2

sin H= what?
cos £T=what?

Perform the operations necessary to obtain

sin 2 £T+cos 2 #.*

In the second number apply the law for the square of
the hypotenuse and finish the demonstration.

72. How to Read the Functions of an Angle. In this
paragraph are directions and models for reading

Natural functions,
Logarithmic functions.

1. NATURAL FUNCTIONS

The natural functions of an angle are read from the
table of natural functions which is so arranged that sines
and cosines are on the left, and tangents and cotangents
are on the right, only.

Sines and Tangents. When reading sines and tangents,
read degrees at the top, minutes at the left.

Cosines and Cotangents. When reading cosines and
cotangents, read degrees at the bottom, minutes at the
right.

* Observe the position of the exponent to denote the square of
the function.

54 TECHNICAL TRIGONOMETRY 72

Proportional Parts. When the angle is expressed in
degrees, minutes and seconds, the reading for seconds equals
the number of seconds times proportional part for 1 second.

When reading sines and tangents add this product to the
degrees and minutes reading, because sines and tangents
increase as the angles increase.

When reading cosines and cotangents subtract this product
from the degrees and minutes reading, because cosines and
cotangents decrease as the angles increase.

Following are the natural functions of 28° 12' 24":

4.26
sin 28° 12' 24" 24

.472551 T704
102 852

.472653

cos 28° 12' 24"

.881303

55

.881248*

tan 28° 12' 24"

. 536195

151

. 536346

cot 28° 12' 24"

1.86499

51

102.24

2.31
24

924
462

55.44

6.28
24

2512
1256

150.72

2.13
24

852

1.86448* 426

51.12

The last column shows the proportional part for 1"
multiplied by the given number of seconds.

* Observe that the seconds' reading from the proportional parts table
is subtracted because a co-function is being read.

74

THE RIGHT TRIANGLE

55

2. LOGARITHMIC FUNCTIONS

A logarithmic function is the logarithm of the natural
function. In the tables, 10 has been added to all logarithmic
functions in order to avoid negative characteristics.

When reading logarithmic sines and tangents read
degrees and seconds at top, minutes at the left, and add
seconds' reading in proportional parts table.

When reading logarithmic cosines and cotangents, read
degrees and seconds at bottom, minutes at the right, and
subtract seconds' reading in proportional parts table.

Following are the logarithmic functions of 28° 12' 24":

sine

tangent

9.674527
16

9.674543

9 . 729424
20

9 . 729444

cosine

9.945103

5*

cotangent

9 . 945098

10.270576
20*

10.270556

73. Caution. The natural function table gives no
direct reading for co-functions of angles expressed in degrees
only, since there is no zero-minutes' line.

Therefore if the natural cosine or cotangent of 40° is
required, the reading must be taken for 39° 60'.

The logarithmic function table gives no direct reading
for the co-functions of an angle expressed in degrees and
minutes only, since there is no zero-seconds' column.

Therefore if the logarithmic cosine or cotangent of 40°
12' is required, the reading must be taken for 40° 11' 60".

74. Examples in Reading Functions. Fill in the omitted
entries in the following table, showing all readings and
computations:

* See tooizote, page 54.

56

TECHNICAL TRIGONOMETRY

75

Table X
READING OF NATURAL AND LOGARITHMIC FUNCTIONS

No.

Angle.

nat. sine

log. sine

nat. tan

log. tan.

1
2
3
4
5

30°

48° 26'

61° 18' 30"

45°

75° 43' 52"

•

No.

Angle.

nat. cos

log. cos

nat. cot

log. cot

1
2
3

4
5

30°

48° 26'

61° 18' 30"

45°

75° 43' 52"

§ 2. SOLUTION OF A RIGHT TRIANGLE

75. Application of Definitions. Under this heading,
near the left margin draw to scale a right triangle, Fig. VI,
with basal acute angle A,
base =3 inches, hypotenuse = 5 inches.

By definition sin A = what?

cos A = what?
tan A = what?

Number the equations consecutively, and specify
authorities.

How many different quantities in equations (1), (2),
and (3) respectively?

In (1) how many quantities are unknown?

Can (1) therefore be solved?

In (2) how many quantities are unknown?

Can (2) therefore be solved?

78 THE RIGHT TRIANGLE 57

In (3) how many quantities are unknown?

Can (3) therefore be solved?

Therefore which equation only, is solvable?

76. Solution for the Acute Angles. Write the solvable
equation, substitute the known values, and reduce the second
member to a decimal of six places, annexing or prefixing
ciphers if necessary.

This decimal equals what function of the basal acute
angle?
• Is it a natural function or a logarithmic function?

In what table may it therefore be read?

Head it in the table and write the value of the corre-
sponding angle.

By geometry write the value of the remaining acute
angle in Fig. VI, showing the computation.

77. Solution for the Perpendicular. If an angle is
known, are its functions known?

Where may they be read?

In (1) of paragraph 75 make the substitution now pos-
sible and solve for the unknown side.

Determine the unknown side by geometry also, and by
another equation of paragraph 75.
Show all readings and computations.

78. Tabulation and Comparison of Results. When
the unknown parts of a triangle are determined by the
solution of equations containing a function of an angle,
the triangle is said to be solved trigonometrically.

When the unknown parts are determined by a theorem
from geometry the solution is called geometric; when deter-
mined by measurement the solution is called graphic.

Measure the perpendicular and the acute angles in Fig.
VI and enter the measurements together with the values
obtained by computation, in the following table:

58

TECHNICAL TKIGONOMETEY

79

Table XI

SOLUTION OF FIG. VI

Part.

Graphic.

Geometric.

Trigonometric.

In this table do not enter the data.

79. Deductions from the Preceding Solution. The

unknown parts of a right triangle can be obtained

(1) Graphically, by scale and protractor.

(2) Geometrically, by square hypotenuse and sum angles

triangle.

(3) Trigonometrically, by solving equations obtained

by the definitions of the functions of either acute
angle.

Fig. 3. — Protractor.

80. Definition of Trigonometry. The word trigonometry
means the measurement of a trigon or three-angled figure,
and therefore of a triangle.

81 THE RIGHT TRIANGLE 59

Trigonometry is that branch of mathematics which
treats of the functions of an angle, and of the solution of
a triangle.

The sides and the angles of a triangle are called its parts.

Solution is possible only when not less than three parts
are known, one of which must be a side.

81. Instructions for Solution of Examples, (a) Leave a
margin of not less than one and one-half inches at the top
of each page and number each in the lower right corner
when finished.

(6) Draw triangles only when an aid to solution. They
should not be sketched, but drawn to scale and the scale
specified. As soon as possible solve examples without
drawing the figure.

(c) When triangles are drawn to illustrate any of the
examples in right triangles letter them as follows:

Right angle, C\ Acute angles, A and B.

In an isosceles triangle denote the equal angles by A
and B, and the other angle by C.

(d) Double underline all results with a straight-edge,
with lightly drawn, parallel lines, a sixteenth of an inch
apart.

(e) In all operations with logarithms retain no more
decimal places than are given in the tables, and increase by
1 the last figure retained when the following figure is 5 or
greater.

(/) Write all the formulas necessary to solution and
set down characteristics as far as possible before using
the tables. This saves time and gives an increasing ability
to detail the work.

(g) Study every example to determine the shortest
and simplest method of solution.

(h) If a factor or divisor has only one figure multiply
or divide arithmetically and do not use logarithms.

60 TECHNICAL TRIGONOMETRY 82

82. Solution of a Right Triangle. Enter in the work-
book the following example as here shown.

In solving, first find A by geometry; then compute
the unknown sides by the functions of B or A whose form-
ulation contains but one unknown quantity.

Following is a suggested form of entry for finding 6.
Solve for the other unknown part in a similar way,
making entry in the work-book in the same form.
Compute 6 and c also by natural numbers.

1. a=281 . D 6

g=35° 18' tan#=-

A= /. 6=atan5

6= log a =

c= + log tan B=

log b =
/. b =

83. Comparative Solution. Solve the following right
triangles graphically, geometrically, and trigonometrically.

In the trigonometrical solution compute each unknown
part both by natural numbers and by logarithms.

2. 6=18 3. fc = 34 4. A =30°

q=244 c=124 c=100

J3=* B= a=*

* A= A= B=

c = a= 6 =

Below is a suggested arrangement for the solution of
examples 2, 3, and 4:

Graphic. Geometric.

a=

6=

4.

A =30°
c=100

a=
6=

85

THE RIGHT TRIANGLE

61

Trigonometric.

Natural Numbers.

sin A = —

sin A =
c=
a =

c

c sin A

Logarithms.

log c =

log sin ii =

log a=

a=

84. Comparison of Results. Under paragraph heading,
rule the following table and make entries indicated.

Table XII
COMPARISON OF RESULTS

Ex.

Unknown
Part.

Graphic.

Geometric.

Trigonometric.

Nat'l Nos.

Logarithms.

2
3
4

85. Model Solution. The following arrangement giving
only the natural equations, readings and computation,
is suggested for all work involving logarithms.

6
a

14.97
20.84

5=35° 41' 27"
A=54°18'33'i
c=25.659

89° 59' 60"
35° 41' 27"

54° 18' 33"

62

TECHNICAL TRIGONOMETRY

86

tan Z?=—

a

*B=35°41/

11.175222
1 . 318898

9.856324

293=35° 41' 20"
31=7''

sin A=—

a
c

a

sin A
25.65

11.318898
9 . 909651

1.409247
087

160
153

70
68:

2565
;9

86. Slide-Rule Check. The star in the preceding
model denotes a slide-rule reading. In all subsequent
work it is urged that each example and problem be com-
pletely formulated and slide-rule readings taken before
using the tables for the logarithmic computation.

Following are the diagrammatic settings for the read-
ings on a Mannheim rule:

tan B = —
a

c—

a

sin A

c

20.84

1

1

D

14.97

R

T

35° 41'

tl

A

S

20.84
54° 18' 33"

25.65

tl

In the left diagram the ratio — is set up on C and D

a

and the runner is then moved to the C index but the reading

is not taken. The slide is then reversed, all indexes are

aligned, and the angle is read on the T scale under the

runner.

t Denotes the index, by which is meant the beginning and the end
of any logarithmic scale.

87

THE RIGHT TRIANGLE

63

In the right diagram the ratio

a

is set up on A and

sin A

*S with the slide reversed and the value of c is read on A
over the slide index. Observe that C and D are used with
a tangent formula; A and B scales or A and S alone with
a sine formula.

87. Examples in the Right Triangle. Solve the follow-
ing triangles in accordance with the instructions in para-
graph 81.

5. 6=318
c=600

8. a =300.08
c =375.19

11. 6 =.6145
A =17° 4"

14. a =.42916
c=.5

17. B =29° 0' 14"
a = 3.8419

20. & =29.307
c =40.659

23. a =4.799
& =3.8452

26. a =.79603
A =38° 14' 51

29. 6=234.64
a = 192.087

32. c = 123.478
6=100.0084

36. c =9.87007
a =7.2009

38. A =17° 49' 13"
c =791.53

41. c =.008154
B =45° 45'

//

6. A =51° 3' 13"
a =4.3971

9. 6=7009.24
c =9000.03

12. c =4.9075
6=4.0009

16. A =11° 11' 11"
c = lll.ll

18. B =80° 18' 8"
c =2.4913

21. A =54° 8.5'
c =6630.49

24. a =999.88
6=1291.05

27. 6 = .60521
A =41° 10' 1"

30. 5=67° 18' 19"
6=302.450

33. A =78°
a = 172.009

36. 6=7.41056
a =9.73196

39. c = 10000.91
6=429.35

42. a =.098714
c =1.173006

7. B =23° 22' 12"
o=3.14159

10. B =20° 58' 59"
c =.012584

13. A =71° 21' 11"
6=34.721

16. 6 = .0060157
c = .00723

19. c=3V7~
6=6.1

22. a =7 ^.40106
c=9V 72145

26. A =70° 5.3'
6 = .9 2 Vl03

28. c =2a

8.29 6 =63.17

31. a = .58967
B =34° 18' 49"

34. B =60° 54'
a = 100

37. J5 =42° 48' 52"
c = 1410.095

40. 5=34° 15' 8"
c =791.53

43. A =39° 18' 4"
a =.073196

64 TECHNICAL TRIGONOMETKY 88

88. Direct Method of Deriving the Formulas. In
Chapter II of the Technical Algebra, it was shown in the
study of forces that if

F=any force,
y=the vertical component,
»=the horizontal component,

0=the angle at which the force acts with the hori-
zontal,

then

(1) y=F sin

(2) and x=Fcos

Enter these equations in the work-book.

Draw a triangle showing F, y, x, and as an acute angle.

Is y opposite, or adjacent to 0?

Is x opposite, or adjacent to 0?

F is what side of the triangle?

I. Side Opposite. The side opposite either acute
angle of a right triangle, equals the hypotenuse times what
Junction of the angle?

opp = hypot X what?

II. Side Adjacent. The side adjacent to either acute
angle of a right triangle equals the hypotenuse times what
function of the angle?

adj = hypot X what?

III. Hypotenuse. Copy the two equations at the
beginning of this paragraph and solve as follows:

Solve (1) for F.

Solve (2) for F.

Therefore the hypotenuse of a right triangle equals
the side opposite an acute angle, divided by what function
of the angle?

THE RIGHT TRIANGLE 65

The hypotenuse of a right triangle equals the side adjacent
to an acute angle divided by what function of the angle?

IV. Opposite and Adjacent in terms op Tangent.
The tangent of either acute angle of a right-triangle equals
the ratio of what sides with respect to the angle?

tan = what?

Solve this equation for opp.

Solve it for adj.

Therefore the side opposite either acute angle of a right
triangle equals the side adjacent times what function of
the angle?

opp = adj X what?

The side adjacent to either acute angle of a right triangle
equals the side opposite divided by what function of the
angle?

89. Summary of Formulas for Solution of a Right
Triangle. In the preceding paragraph it is required that
two formulas De written for each of the sides of a right
triangle.

Rewrite these six formulas under the headings.

Opposite,
Adjacent,
Hypotenuse.

Include with them, three formulas for an acute angle
of a right triangle, under the headings

Sine,

Cosine,

Tangent.

66

TECHNICAL TRIGONOMETRY

90

90. Examples. Solve
the formulas of paragraph

44. c=Vl3496.58
B =42° 34' 29"

46. c = -# 25684.5
a = ^20000.9

48. 5=5° 28' 7"
c = .0*43092

50. 718 a =43.723
A =50° 16' 6"

62. b = .73425
c = .97008

64. £=50° 18'
c =.65967

66. 5=61° 19' 5"
b = 165.708

68. c =.87426
a =.60009

60. 6=12Vl56.78
c =210.43

62. £=27°0'25"
a = .04561" 2

64. 6 = 143.78
a=34.008 2

66. A =48° 17' 29"
c =7.6342

68. A =63° 14' 51"
a =.04368"*

70. £=60°
6=9000

72. a = .86715
c = 1.40601

74. c = .089126«
b = .06834

the following right triangles by
89:

46. B =81° 18' 3"
a = .540006

47. 6 = .0 3 76428
c = .0012075

49. 3.128"-' = .08738
£=21° 34'

51. a =.008349
c = .034251

53. A =70° 14' 12"
c= 1342.51

55. A =36° 48' 2"
a =.0590073

67. £=38 ° 17' 45"
c=Vl 140.09

69. 6=V l4.296
a = ^28.702

61. A =80° 51'
a =.42(549.6)*

63 c= 58.9622
a =41.705 2

65. 6 =V. 09005
c =.56074

67. A =8 0° 12' 1 3"
6=V96.84

69. 5=45
a = 1200

71. £=75 ° 35'
c = -#184.007

73. a = .004286
b = .007312

75. c=26
a =34.438

92 THE EIGHT TRIANGLE 67

76. a =36 77. c=5a

c =1.7208 6=146.09

78. B =39° 58JT 79. A =24° 43.4'

c =34.76^12.907 6 = 3498.3 XHT 3

80. B =70° 20.6' 81. 6 = .0973

805.7 a =2185.42 c

a=—
5

§ 3. THE ISOSCELES TRIANGLE

91. How Solved. An isosceles triangle has two equal
sides, and therefore two «qual angles as proved in geometry.

Draw an isosceles triangle.

From the vertex of the angle formed by the equal sides,
draw a perpendicular to the opposite side.

Does the perpendicular bisect the opposite side, or does
it divide it into unequal segments?

Letter the sides and angles of the triangle as specified
in paragraph 81.

The two triangles into which the triangle is divided by
the perpendicular, are what kind of triangles?

One acute angle of each of the two triangles, is what
part of C?

One side of each of the two triangles, is what part of c?

Therefore if three parts of an isosceles triangle are
known, one of them being a side, C can be determined by
solving one of the right triangles for what part of C?

c can be determined by solving one of the right triangles
for what part of c?

By reference to the figure, write a formula for C/2.

Write a formula for c/2 and solve it for c.

Write directions for the solution of an isosceles triangle.

92. Examples. Compute the unknown parts of the
following isosceles triangles. Draw figures only when it
seems difficult to solve without a figure.

68

TECHNICAL TRIGONOMETRY

92

1.

c=24
6=30

2.

C=52°16'1S"
a = 17.293

3.

C=29°17'21"
c =34.628

4.

B =30° 11' 15"
6 = 136.759

6.

A =67° 4"
6=751.682

6.

c =208.4
6=186.52

7.

A =46° 35.2'
a =.349607

8.

c =7.0009
6=5.3287

9.

B =51° 29.5'
o=341.007

10.

c = 1429.075
C =61° 58.1'

11.

a =1342.001
c =897.008

12.

14.

16.

18.
20.

•

22.
24.
26.

C =27° 28.7'
a = 1000.08

13.

a =39.8V94.006
c =425.36

£=39° 2'
a=V.01368

C=50°
6 = ^648.9

15.

c =^1728.5
a = 10.764

17.

6 = ^.00762
c=.9732

B =61° 42.3'
c =.843^729.4

c=^.4672
A =30° 19.1'

19.

6=32.8lV.8762
C =48° 5.8'

21.

a =.469^.8724
B =16° 29.5'

17.389-* = .9834
A =34.3°

23.

a=24V7.892
£=27° 19' 31"

6 =.09^6.318
c=.1475

c =218-^412.96
A =41° 18' 21"

25.

a = 40.82V 9607
C=78°49'

27.

a =346.79
A =32° 24' 40"

28.

6=41.2085
C=96°42'31"

29.

(7=63° 9' 8"
c = 1074.56

30.

A =40° 51' 39"
c = 1.08479

31.

a = 1009.805
c =2148.62

32.

£=18° 27' 43"
a=731.96Vl8.291

33.

A =31° 47' 51"
6=78.549

34.

C=59°5'17"
a = 1094.68

M

THE EIGHT TRIANGLE

69

§ 4. APPLIED PROBLEMS

V

93. Notation for Angles. Figures drawn to illustrate
problems are best lettered in significant symbols with angles
denoted by letters of the Greek alphabet as in the tech-
nical applications of mathematics. The Greek letters
commonly used are as follows:

Letter.

Greek Name.

English Equivalent.

a

alpha

a

e

theta

th

*

phi

ph

8

delta

d

94. Solution Formulas. In the solution of problems,
draw a figure to scale for each problem whenever possible,
and specify the scale.

Measure the required parts accurately and set down
the measurements under the heading, Graphic Solution.
When this has been done, write or derive a formula whose
first member is the required unknown part and whose second
member contains only those quantities or functions of
quantities, or both, for which numerical values are given in
the problem. In some of the problems the formula for the
unknown may be written directly; in others it will be
necessary to write two or more equations from which the
formula for the required unknown part may be derived by
substitution from one equation into the other.

When the conditions of a problem do not give a right
triangle, draw a perpendicular to form a right triangle
having one known side.

Sometimes solution is possible by the right triangle
method only when several perpendiculars are drawn.

70 TECHNICAL TRIGONOMETRY 95

Whatever the conditions, take no readings until a formula
has been derived whose first member is the required unknown
part and whose second member contains only quantities or
functions of quantities for which numerical values may be
substituted from the data.

In order to distinguish a formula of this kind from others,
it will be known as the solution formula. Therefore the fol-
lowing definition:

A solution formula is a formula whose first member is
the required unknown quantity or function of it, and whose
second member contains only those quantities or functions
of quantities for which numerical values may be substituted
from the data.

In all work make the following your habitual practice:

(1) Study each problem to determine the simplest
method of solution.

(2) Be sure that the solution formulas are correct, that
the readings have been correctly copied from the tables, and
that no mistake has been made in the computations.

95. Problems. Solve the following problems as speci-
fied in the preceding paragraph.

1. Dividers. The legs of a pair of dividers h% inches long,
are set at an angle of 30° 17'.

Determine the distance between the points.

2. Wedge Angle. A wedge measuring 16 inches along the
sides, is 4 inches thick.

Determine the angle at the point of the wedge.

3. Radius from Chord and Central Angle. Compute the
radius of a circle in which a chord 16 feet long subtends an angle
of 26° 18' 14" at the center.

4. Slope of Roof. The ridge-pole of a roof is 15f feet above
the center of the attic floor and the attic is 60 feet wide.

95 THE EIGHT TRIANGLE 71

What is the slope of the roof? (The slope is the tangent of
the angle of inclination to the horizontal.)

S. Perimeter of a Triangle. An equilateral triangle is cir-
cumscribed about a circle whose radius is 21 inches.

Compute the perimeter of the triangle.

6. Height of a Hill. From the top of a hill, the angles of de-
pression of two successive mile-stones on a straight level road
leading to the foot of the hill, are 15° and 35" respectively.

Compute the height of the hill.

7. Distance between Objects. From an elevation 146.9 feet
high, the angles of depression of two objects situated in the same
horizontal line with the base of the elevation and on the same
aide, are 41° 23' and 60° 56'.

Compute the distance between the two objects.

8. Length of Wire. From one edge of an excavation 33 feet
in width, the angle of elevation of the top of a wall on the opposite
edge of the excavation, is 59°.

Compute the length of the wire required to reach from the
point of observation to the top of the wall.

9. Height of Plane. From an aeroplane directly above a
town, the angle of depression of another town 18 miles away on
level ground, was 16".

How high was the aeroplane at the moment the observation
was taken?

10. Degree of Grade. A road of uniform grade along the
side of a mountain, rises 500 feet in a distance of 1 mile, meas-
ured along the grade.

Compute the angle of the grade.

72

TECHNICAL TRIGONOMETRY

95

11. Height of Monument. A building 107.8 feet high stands
on the same level with a monument. The angles of depression
of the top and the base of the monument as read from the top
of the building, are 5° 15 and 6° 41' respectively. <

Compute the height of the monument.

12. Distance. A straight level road leads to the foot of a
hill 165 feet high. From the top of the hill the angles of de-
pression of two objects in direct line on the road, are 15° and 12°
respectively.

Compute the distance between the two objects.

13. Pitch of a Roof. Fig. 5 shows the gable end of a building
with one side of a roof having both sides of equal pitch.

Fig. 5.

The pitch of a roof of this type equals its total rise h, divided by
W, the entire width of the building.

Formulate and compute the pitch and all the angles shown on
the figure, when h = 8' 4" and W = 24' 5".

14. Use of a Base-Line. In order to determine the distance of
an object to which a direct line could not be run, a base-line of 240
feet was measured, from each end of which the angular deviation
of the object from the base-line, was read.

What was the distance of the object from the nearer end of the
base-line, if one angle was 56° 14' and the other 90°?

95

THE RIGHT TRIANGLE

73

15. Height of a Balloon. A balloon 2? meters in horizontal
diameter, when sighted by an observer at a distance of 1800 meters
on level ground, subtends an angle of three minutes.

O YT-robtended angle

Fig. 6.

Compute the height of the balloon at. the moment of the obser-
vation.

16. Height on a Slope. In order to determine the height of a
tree on sloping ground, a base-line 50 feet long was run from the foot
of the tree, directly up the hill, the slope of which was .734. From
the end of this base-line, the angle of elevation of the top of the
tree was 38°.

Compute the height of the tree.

17. Roof Pitches. Compute and fill in the omitted entries in
Table XIII for each of the roofs shown in Fig. 7.

Table XIII

ROOF PITCHES

No.

Pitch.

Rise per Ft.
Run.*

»angle of pitch.

Exact.

Approz.

1
2
3
4
5
6
7

* Run is measured on the horizontal.

74

TECHNICAL TRIGONOMETRY

95

yi Pitch

X Pitch

# Pitch

X Pitch

H Pitch

X Pitch

l.Pitch

Fig. 7.

18. Angle Lay-Out with Compasses. In Fig. 8, AO is 2 inches

long. Point V is determined by des-

v cribing an arc with the compasses from

as a center. The angle A is to be laid

out by drawing VA.

> Write a general formula for VOm

Fig. 8.

terms of AO and A.

95

THE RIGHT TRIANGLE

75

Compute and tabulate the compass setting VO for the angles ,
in the table.

Table XIV

ANGLE LAYOUT

No.
1

AO.

A.

VO.

No.

AO.

A.

VO.

3

10°

11

3

48°30 /

2

3

15°

12

3

50°

3

3

20°

13

3

55°

4

3

25°

14

3

60°

5

3

30°

15

3

65°

6

3

35°

16

3

66°

7

3

40°

17

3

70°

8

3

45°

18

3

75°

3

32°

19

3

80°

10

3

21°

20

3

51° 45'

19. Angle Lay-Out with Steel Square. In work with wood an
angle is conveniently laid out with a carpenter's steel square as
shown in Fig. 9.

The long end of the square is the blade; the short end is the
tongue.

Fig. 9.

Formulae 0, using the notation

db — distance on blade,
(4= distance on tongue.

Compute 6 when the square is placed on a board as illustrated.

20. Compute and tabulate the omitted entries in the following
table.

76

TECHNICAL TRIGONOMETRY

96

Table XV
STEEL SQUARE LAYOUT

No.

«»■

*t

9.

No.

h-

d v

9.

1

6

12

11

18

6

2

8

12

12

15

5

3

9

12

13

15

7

4

11

12

14

8

6

5

13

12

15

12

4

6

15

12

16

13J

7

7

15

10

17

42

8

18

10

18

25

9

18

9

19

50

10

18

8

20

45

21. Rise and Run of a Rafter. The rise of a rafter is the vertical
distance one end is above the other. The rise per foot run is the
vertical distance the rafter rises in 1 foot of horizontal distance.

By reference to Table XV compute and tabulate the steel square
settings for each pitch.

Test the correctness of your computation by making a model
of one of the pitches.

*

96. The Mariner's Compass. The problems immedi-
ately following require a knowledge of the points of the
compass. In order to understand them and to be able
to draw the figures, either make a copy of a compass card
or draw the following card:

Draw six concentric circles with radii of the following lengths
and with circumferences as specified:

i inch light, 1 inch medium, lj inches heavy, 1A inches heavy,
2 inches medium, 2A inches light.

Divide each quadrant arc into eight equal parts, denoting the
divisions of the arcs by lines drawn radially to the circumference
of the 2A-inch circle, and discontinued between the 1- and 1 A-inch
circumferences.

96 THE EIGHT TRIANGLE 77

Between the circumferences of the 1ft- and 2-inch circles indi-
cate the four cardinal points by heavy lettered capitals N, E, S, W.

Beginning at N, between the circumferences of the I- and 1J-
inch circles letter the points of division of the arc of the first quad-

Fig. 10.— Mariner's Compass Card, Old Style.

rant, clockwise, in the following order in line with the respective
radial lines:
NbyE, NNE, ^NEbyN, NE, NEbyE, ENE, EbyN.
Letter the points of division of the arcs of the other quadrants
in a similar way, determining the letters by reference to the first
quadrant.

78 TECHNICAL TRIGONOMETRY M

The completed figure if correctly drawn and lettered,
represents the card of the Mariner's Compass. Each of
the thirty-two equal divisions is called a point or rhumb.
These may be subdivided into halves and quarters.

FlO. 11. — Mariner's Compass Card, New Style.

This cut is reproduced from a drawing made and photographed by

Alumnus Mr. Cullings, A. C. '11.

The compass was invented by the Chinese over four
thousand years ago. As might be expected, their name
for it has no reference to the northward trend of the needle,
but means " needle pointing south."

(See Encyclopedia Britannica.)

97

THE EIGHT TRIANGLE

79

97. Problems. Solve the following as directed in para-
graph 94.

1. From a lighthouse L, two ships S and & are sighted NW and
N 18° E respectively. K '

From Si at the same instant, S is observed bearing SW.

If LSi is 17.34 miles, compute the distance between the ships.

To draw the figure, mark the
point L.

Through L draw with pencil a
very light, short, north and south
line.

From L draw a line representing
the direction of S from L as speci-
fied in the problem.

Draw a line representing the di- s
rection of Si from L.

Draw a light north and south line
through S, and from S draw a line
representing Si's direction from S.
At its point of intersection with the
line representing Si's direction from
L, mark the point &.

Draw SSiL in India ink. The size of the figure will depend on
the distance laid off for SL.

2. Three cruisers A, B, and C are so located that from A, C is
S 20° E and B is S 44° W. From B, C is E 20° N.

BC equals 4.795 miles. , v

Compute BA and CA.

From A draw a light south line, and lines representing the direc-
tion of C and B from A. Mark B at a convenient distance from A
on the line representing its direction from A, and from B draw a
light east line.

3. At 2:45 p.m., from a ship sailing E 43° N, at the rate of 8 miles
per hour, a fort is observed E 51° S. At 5 p.m. the fort is seen S
26° W.

Compute the distance of the fort from the ship at the time of
each observation.

Fig. 12.

80 TECHNICAL TRIGONOMETRY 97

4. From a ship, a rock and a lighthouse were observed in the
same straight line bearing N 28° E. After the ship had sailed 9.6
miles NW, the rock was due E, the lighthouse E 35° N.

Compute the distance of the rock from the lighthouse.

6. From a ship sailing due north, two lighthouses are observed
due E. After sailing 8$ miles, the lighthouses are respectively SE
and ESE.

Compute the distances from the ship at the first observation.

6. Two vessels sail from the Canary Islands, N 42° W and S
48° W at rates of 19£ and 17 miles per hour respectively.

Compute the distances between them at the end of one hour and
a half, and the bearing of the second vessel from the first.

7. A motor boat sails due north at a uniform rate. At a certain
time a lighthouse is sighted 8.71 miles away to the NW. Twenty
minutes later, the lighthouse bears SW.

Compute the speed of the boat in knots.*

8. A vessel sailing NNW is observed at 3:30 p.m. from a light-
house 6 miles away, to be ENE. At 4:10 p.m. the vessel is
due N.

Compute the rate of sailing in miles per hour.

9. From a ship sailing due W a lighthouse is sighted at 10:30
a.m. bearing SW, 13 miles distant. At 11:30 a.m. the lighthouse
bears SE.

How many knots is the ship making?

10. From a ship sailing due W at the rate of 17.8 miles an hour,
a headland is observed at 11 :46 a.m. bearing N by W. At 12:23 p.m.
the headland bears E by N.

Determine the distance of the headland from the ship at the time
of each observation.

11. From a ship sailing due E at a uniform rate, a lighthouse
9.28 miles distant is sighted at 9:48 a.m., bearing due S.

At 10:38 a.m. the lighthouse bears S 28° 23' W.
* Compute the rate of sailing both in knots and miles per hour,
and the bearing of the lighthouse at 12 m.

* A knot=l nautical mile per hour.

97 THE RIGHT TRIANGLE 81

12. A cruiser is sighted from a fort at 8:55 a.m. bearing NNW,
9 miles away. At 10:50 a.m. the cruiser sights the fort bearing
E 7i° N.

If the cruiser was sailing at uniform speed on a course WSW,
what was the rate of sailing and the distance of the cruiser from
the fort?

13. Two lighthouses L\ and L 2 are 14.625 miles apart on a line
bearing W 18° N. A battleship steaming N 8° W with a speed
of 20 miles per hour, sights L\ at 2:42 a.m. bearing E 21° N and L%
bearing N 43i° W.

If the battleship changes neither course nor speed, at what time
will it be in direct line with the lighthouses and what will be its
distance from each?

14. From a ship sailing at uniform speed over a course bearing
ENE a lighthouse and a headland are sighted in a straight line SSE
at 12:10 p.m. At 12:15 p.m. the lighthouse is S 17.5° W and the
headland S 7.5 E.

How many knots is the ship making if the distance between the
lighthouse and the headland is 4£ miles?

15. From two telescopes mounted 275 feet apart on the deck
of a battleship, an object is sighted at sea. The angular deviation
of the object from the line joining the telescopes is 110° from one
telescope and 68° from the other.

Compute the distance of the object from the nearer telescope.

16. A spire whose shape is a regular hexagonal pyramid, is
129.3 feet high. A side of the base measures 5.7 feet.

Compute the slant height, the lateral area, and the angle formed
by the base and the face.

17. The diagonal of an inscribed square equals 9\/l8 feet.
Compute the perimeter, area, and the length of a median of an

equilateral triangle circumscribed about the same circle.

18. Sewer Section. The figure represents a cross-section of a
standard egg-shaped sewer.

The radii of the top, sides, and bottom are in the ratio of
1 : 3 : £ respectively.

82

TECHNICAL TRIGONOMETRY

97

Fig. 13.

This means that

AC=3EV
and LT=\EV

Compute the area of the cross-section when the radius of the
top is 20 inches.

State the advantages of a sewer of this shape, over a circular
sewer.

19. Radius of Curve. The inner rail of a trolley track runs
9| feet from the curbstone. At a corner where the street is deflected
through an angle of 70° the inner rail runs 4\ feet, radially, from the
corner.

Compute the radius of the curve on which the track passes the
corner.

20. Equilibrium on an Inclined Plane. Figure 14 shows
an inclined plane on which

L = length of plane;
h = height of plane;
a = angle of inclination to horizontal;
P=» force required to overcome friction and balance W;
/coefficient of friction;

THE RIGHT TRIANGLE

W-the weight;
F=the friction.

F = WfCQHa

Compute P when* =35 J°, (F=275, rm.u.

J=.l2.

21. Bevel Gears. Figs. 15 and 16 show bevel gears with their
shafts at right angles, the smaller gear being called the pinion.

r^

K

c.

y

i ^_ __

..a- A \ j ^A>a

•4.-4$

j f

-Da — — >i

a—ACB—the angle of increment;

ff=ACBi —the angle of decrement;

j=vlC/J —center angle of gear;

Si -FCE, =ccnter angle of pinion;

X — face angle of gear;
AF =diameter of gear;
D» =AF+2k -diameter of blank;

e — i— 0— angle of working depth.

84 TECHNICAL TRIGONOMETRY 9T

Formulate the following: tana, tan p, tan i, tan S,, X.
Formulate A in terms of AB and S, and substitute in the formula
for D*

Fig. 16.

22. Bevel Gear Computation. Two shafts at right angles
to each other are connected by a pair of 18-pitch bevel gears with
a velocity ratio of 2 to 3.

The larger gear has 72 teeth.

Compute the number of teeth of the pimion and all the other
dimensions of both gears as follows:

(1) Number of teeth varies as the velocity ratio.

number of teeth

(2) Pitch diameter =

(3) AB-
(i)AB=-

1
pitch
1

diametral pitch "

r+-

.157

pitch pitch

(5) J-?

(6) 8, =90-?

(7) iC-Vf+5

(8) ..

(9) t-

97 THE RIGHT TRIANGLE 85

(10) x=
(ID e=

(12) h =

(13) £> =

23. Spiral Gear Cutter. The number of teeth* in the cutter
required for cutting the teeth of a spiral gear, may be determined
from the following law:

No. teeth in gear

Number of cutter =

cube of cos spiral angle*

Formulate this law and compute the number of cutter required
for a 17-tooth spiral gear with a spiral angle of 45°.

24. Spiral Gear Formulas. Fig. 17 shows a spiral gear and
the different pitches which are used in its design and manufacture.

N =number of teeth in gear:
P n — normal circular pitch of gear

= pitch of hob or cutter;
a = angle of inclination of teeth to axis of gear
= spiral angle;
OR= axis of gear;
P' =real pitch of gear

= circular pitch;
P" — pitch in the direction of the axis of gear

= pitch parallel to axis;
D v = diameter of gear in the pitch line
= pitch diameter,
= diameter of pitch circle;
P = number of teeth per inch of pitch diameter.

= diametral pitch.
d= depth of teeth;
Dj = diameter of blank
= outside diameter;
s= addendum;

(1) Formulate P f in terms of P and a.

(2) Formulate P" in terms of P and a.

* Called number of cutter.

TECHNICAL TRIGONOMETRY

Formulate the following laws and number the formulas con-
secutively with the two preceding, with lettered heading for each.

(3) The diameter of the blank for a spiral gear equals the diam-
eter of the pitch circle plus twice the addendum.

(4) The diameter of the pitch circle equals the number of teeth
in the gear times the circular pitch divided by jr.

(5) The addendum equals the reciprocal of the pitch of the
cutter.

(6) The normal circular pitch equals ir divided by the diametral
pitch of the cutter.

(7) The pitch diameter equals the number of teeth divided by
the product of the diametral pitch and the cosine of the spiral angle.

(3) The diameter of the blank equals the reciprocal of the diam-
etral pitch times the following sum:

2 +

number of teeth
cosine of spiral angle"

(9) The depth of the teeth equals 2.15708 divided by the diam-
etral pitch.

35. Outside Diameter of a Spiral Gear. In the formulas
of Problem 24 make the following substitutions:

In (3) substitute from (4) and (5).

In the formula thus obtained substitute for circular pitch from
(1), and for normal circular pitch from (6).

97 THE RIGHT TRIANGLE 87

Use the formula resulting from these four substitutions for com-
puting the outside diameter of a spiral gear which is to have 24 teeth
at an angle of 50° with the axis, cut with a 16-pitch cutter.

26. Spiral Gear Computation. A spiral gear with 20 teeth at
an angle of 60° is to be hobbed with a cutter whose diametral pitch
is 3.

Formulate and compute:

(a) circular pitch;

(6) pitch in direction of gear axis;

(c) pitch diameter;

(d) outside diameter;

(e) depth of teeth.

27. Apothem. Formulate the apothem of any regular polygon
in terms of the side of the polygon and an angle of the polygon.

Solve the formula for the side.

28. Regular Polygon Formulas. Derive general formulas in
terms of radius and central angle, for the side, apothem, and area of
a regular polygon.

Solve each formula for R.

29. Regular Polygon Computations. Compute the side,
apothem, and area of all regular polygons from 5 to 12 sides inclu-
sive, when each has a radius of unity.

30. Pentagon Lay-Out. Regular polygons of a required radius
or side, are readily laid out with the compasses by the aid of a cir-
cumscribed circle.

For radius, use formulas of Problem 28.

Lay out a regular pentagon whose side is 5.57 inches.

31. Regular Polygon Lay-Out. Lay out the following regular
polygons:

(1) Hexagon with side = If inches.

(2) Octagon with radius =2A inches.

(3) 9-gon with side =lft inches.

(4) 10-gon with side = 1 A inches.

(5) Equilateral triangle with radius =2.24 inches.

88

TECHNICAL TRIGONOMETRY

97

32. Side of a Hexagon. Prove by a trigonometric method that
the side of a regular inscribed hexagon is equal to the radius.

Prove it also by a geometric method.

33. Area of a Right Triangle. Draw a right triangle and denote
the sides and angles in conventional notation. Derive formulas
for the area when

(1) one leg and adjacent angle are known,

(2) one leg and opposite angle are known, ,

(3) two legs are known.

Compute the areas of three of the triangles in the examples on
pages 63 or 66, to which these three formulas apply.

34. A 30-60 Triangle. By a trigonometric method prove that
when one leg of a right triangle is equal to half the hypotenuse,
the acute angles are 30° and 60°.

35. Moon's Radius. The moon at a distance of 238,840 miles
from the earth subtends an angle of 31.12'.

Derive a general formula for the radius of the moon and compute
the radius from the data.

Fig. 18.

36. Chord of a Circle. Derive a formula in terms of radius
and central angle for a chord of a circle.

Solve the formula for R.

37. Modulus of a Triangle. The ratio of any side of a plane
triangle to the sine of the opposite angle is called the^ modulus of
the triangle; that is

a b c

modulus =

sin A sin B sin C*

97

THE RIGHT TRIANGLE

89

One of the best checks for the solution of a triangle is to deter-
mine whether these three ratios have the same value.
This is called checking by modulus.

In the figure, is the center of the
circle circumscribed about the triangle
ABC.

Draw an enlarged figure in the work-
book.

Draw OA, forming an angle a with
a radius perpendicular to b.

Then sin a equals what? Fig. 19.

Prove a=B, substitute in the sine equation, and determine the

relation between the modulus - — -and the diameter D.

sin 2?

Therefore the modulus of any plane triangle is equal to what?

38. In a triangle, c =7.219 and C =58° 34' 42".
Compute the diameter of the circumscribed circle.

39. In a triangle, b = 12.495, C=40° 7' 8", and A =75° 29'.
What is the circumference and the area of the circumscribed

circle?

40. Length of an Arc. The lengths of arcs of 60° or less may
be determined with negligible error by the following approximate
formula:

TZ) =

8VS-VD

o
Fig. 20.

In the figure, OS is a 12-
inch radius perpendicular to
VD, and— FD=60°.

Compute the length of
— VD by the formula. (Num-
ber of degrees in V may be
determined by ^SD.)

* This symbol denotes an arc.

90 TECHNICAL TRIGONOMETRY 97

Determine the closeness of the approximation of the result by
use of the exact mathematical formula

n vnD

~ VD =m* D =m°>

in which n equals the number of degrees of arc.

Is the error by the first formula negligible as stated?

41. Arc Computation. By the approximate formula of the
preceding problem determine the length of an arc of 50° 40', having
a radius of 9| inches. Check by the second formula.

42. General Formula for Length of an Arc. Derive a general
approximate formula for the length of an arc of 60° or less which
has no unknowns in the second member; in other words formulate
VS and VD for any arc and radius and substitute in the first formula
of the Problem 40.

43. Arc Greater than 60°. When an arc is greater than 60°,
half its length may be computed by formula and the result doubled.

By the formula of Problem 42 determine the length of an arc
of 95° 12' and check by the exact formula, using the same radius
as in Problem 41.

44. Helix Angle. The thread of a screw is called a helix. The
angle of the helix is the angle of deviation of the helix from a line
perpendicular to the axis of the screw. This angle is determined
from the diameter of the screw and the advance of the screw in one
revolution.

Gfrco

t*«5^ — .Advance

Fig. 21.

advance of screw

(1) What function of angle of helix =— ; ; ?

circumference of screw

Express as a formula.

The pitch of a screw is the reciprocal of the number of threads
per inch.

(2) Advance =what in terms of pitch? Formulate.

(3) Circumference =what in terms of diameter? Formulate.
Substitute in formula (1) from (2) and (3).

OT

THE RIGHT TRIANGLE

91

45. Helix Angle Computation. Determine and enter the helix
angle for>each diameter and pitch in the following table in which
D is the diameter of the screw in inches, and the numbers at the
tops of the columns denote the number of threads to the inch.

Tabulate angles to the nearest minute only.

Table XVI

ANGLE OF HELIX

No.

D.

3.

4.

5.

6.

1

i

2

i

3

1

4

1»

5

11

6

2

46. Acme Thread. The drawing shows
a sectional view of an acme thread.

Formulate and compute angle 0.

Fig. 22.

47. B. & S. Worm Thread. The draw-
ing shows a sectional view of a Brown &
Sharpe worm thread.

Formulate and compute angle 0.

Fig. 23.

48. Thread Measurement by Caliper. In measuring the
width of a thread at the point, a caliper may be used as shown in
Fig. 24.

92

TECHNICAL TRIGONOMETRY

97

Fig. 24.

Derive a formula for W in terms of h, V, and angle A.
Compute W, when h = A, A =29°, and 7=4.

49. Thread Measurement by a Wire. The

drawing shows a sectional view of the wire
measurement of a Brown & Sharpe 29-degree
worm thread.

Derive a formula in terms of P and $ = 29°,
for R the radius of the wire when resting in the
thread groove and just flush with the tops of the
threads as shown.

Compute R when P =2.

Fig. 25.

60. Sprocket Wheel. The drawing shows a sprocket wheel
with pitch circle C.

Fig. 26-

97

THE RIGHT TRIANGLE

93

Derive a formula for D, the diameter of the pitch circle, in
terms of and P the length of the chord.

Compute D when the sprocket has 8 teeth, and P=3 inches.

51. Dovetail. The drawing shows a sectional view of a dove-
tail.

Fig. 27.

Derive a formula for w in terms of v, h, and 0.

Compute w from the given dimensions when A=f", 0=3.5".

62. Angle of Taper. A No. 13 Pratt & Whitney taper reamer
has the following dimensions:

diameter large end 1.2.59",
diameter small end 1.009",
length of reamer 12".

Fig. 28.

Derive a formula for the angle of the taper with the center line,
and compute the angle.

53. Two-Point Ball Bearing.

D=the diameter of the ball

circle,
d— diameter of the balls,
n = number of balls,
180°

d=Dsin

n

-.003.

Compute D
when n=21 and d=3125".

Fig. 29.

94

TECHNICAL TRIGONOMETRY

97

54. Chordal Distance. Forty-seven holes are to be drilled at
equal distances around a circular plate 21 inches in diameter, with
their centers at a distance of 1£ inches from the edge.

Derive a formula for the chordal distance between centers and
compute the distance from the data.

55. Diagonal of Bar. Derive a trigonometric formula for the
diagonal of a cross-section of a square bar.

Compute the diagonal when the bar is 3| inches square.
Check by some other formula.

56. Reamer Computation. Formulate and determine the square
which can be milled on the end of a reamer shank whose diameter
is 2A inches.

Check by square hypotenuse.

67. Diagonal. Derive a general trigonometric formula for the
distance across the corners of a bar measuring 3} by 2A inches in
cross-section. Compute and check.

58. Diameter of Pulley. To deter-
mine the diameter of a broken pulley,
the chord c, and a the height of the
arc were measured in order to com-
pute a.

Prove a = e and derive a formula for
D the diameter of the pulley.

Compute D when a = 1 A inches and
c=d inches.

59. Numerical Value of Sine 2 +
Cosine 2 . Draw any large right tri-
angle and determine the approximate
numerical value of sin 2 J?+cos 2 -B when

B is one of the a.cute angles.

(1) by measurement with the protractor, and computation from

the readings in the table;

(2) by measurement of the sides which form the sine and cosine

ratios, and computation.

60. Determine the exact numerical value of sin 2 +cos 2 by demon-
station as follows:

97

THE EIGHT TRIANGLE

93

sin B equals what by definition?

cos B equals what?

.*. sin*2?+cos , 2? equals what?

In these equations substitute no numerical values and take no
readings from the tables, but simplify the second member of the last
equation by applying the theorem on the square of the hypotenuse.

61. By the table determine the numerical value of

sin 2 a+COS 2 a

when « has the Mowing values: 30°, 45°, 12° 16', 25° 31' 24".

62. Tangent in Terms of Sine and Cosine. If 6 is an acute
angle of a right triangle, derive a formula for tan in terms of
sin e and cos 0, from the definitions of these functions.

63. By readings from the table, test the formula of Problem 62
when e has the following values:

41° 12' 36", 32° 17', 109° 42.8'.

64. Diagonal of Trapezoid. In the figure,

A =28°
BA =1.298"
h = .05"
h is perpendicular to BA.

Fig. 31.

Derive a formula for BC, and compute its numerical value.

65. Characteristic. Explain in what way the characteristic
of the logarithm of 64097 can be determined from 10* and 10 8 .

66. Height of Mountain. From the deck of a ship the angle
of elevation of the top of a mountain is 46°, and from the masthead
it is 44°.

The height of the mast is 120 feet.

Determine the height of the mountain above the deck.

96 TECHNICAL TRIGONOMETRY 97

67. Diameter of Hexagonal Nut. The
long diameter of a hexagonal nut is 1.75 inches.

Formulate and compute the short diame-
ter.

68. Diameter of Shaft. The diameter of
a shaft in inches is expressed by the formula

Fig. 32.

^

in which T.M. = torsional moment of resistance, and /= shearing
stress.

By logarithms calculate the diameter required when

T.M. =25000 and /=8000.

69. By formula sin 2 A =2 sin A cos A. Use the table to deter-
mine whether this is true when A =29° 13'.

70. Length of a Guy- Wire. A telegraph pole 60 feet high, stand-
ing vertically on level ground, is to be braced by a wire, one end
of which is to be wound around the pole at a distance of 6 inches
from the top.

The other end of the wire is to be fastened to a stake driven
into the ground at a distance of 75 feet from the foot of the pole
and inclined from the pole at an angle of about 30°.

If the stake is 2 feet long and the wire is to wind around it at
a distance of 4 inches from the top, derive a formula (general) for
the length of the wire and compute the length in round numbers
when 30 inches is allowed for winding.

71. Compass Lay-Out of a 29° Angle. Describe a circle with
any radius and draw its diameter.

Take one-fourth the diameter in the compasses and from one
extremity of the diameter strike off an arc of the circumference
on each side. From the points thus determined draw lines to the
other extremity of the diameter.

Determine whether these lines form a 29° angle:

(1) by the protractor,

(2) by trigonometry.

97

THE RIGHT TRIANGLE

97

72. Setting of a Cutter.

d= distance cutter should be off
center;

R = radius of work;

C= radial depth of cut;

a = radial angle of cutter.

Formula: d = (R— C)sina.

Draw an enlarged figure and derive the
formula.

The angle of a cutter on the side which
cuts the radial side of the groove is 28°.
How much off center must the cutter be
placed in order to cut a groove { inch deep,
measured radially, in a piece . of work 2J
inches in diameter?

Fig. 33.

73. Set-Over. In turning a taper the distance the tailstock
must be moved to one side is called the set-over.

^Set-over

Fig. 34.

When the taper is denoted on the drawing in degrees the set-over
must be read in a table of set-overs or must be computed by trig-
onometry.

Compute the set-over for the piece shown above when
a = 50.

74. Formula for Set-Over. Derive a general formula for set-
over in terms of length of piece and angle of taper.

75. Set-Over Computation. Compute and tabulate to four
decimal places the set-over per inch for tapers of from one to five
degrees inclusive, increasing by one-half degree.

98

TECHNICAL TRIGONOMETRY

97

76. Flange Angle. The drawing shows two pipes of the same
diameter joined by a flange.

Fig. 35.

Prove that the flange angle of N° equab a, the angle through
which the pipe is deflected.

Write a general formula for L in terms of T, Z>, and N.

78. Spacing a Bolt Circle. The drawing shows a metal plate
with a bolt circle. For what distance in thousandths must the

Fig. 36.

compasses be set to divide the circumference of the bolt circle into
13 equal parts when the radius is 3A inches?

« THE EIGHT TRIANGLE 99

79. Compass Setting for Bolt Circle. la terms of B and n
derive a general formula for the compass setting to divide a cir-
cumference of any radius R into any number of equal parte n,

80. Bolt-Circle Computation. By the formula of the preced-
ing problem compute and tabulate to thousandths tbe compass
settings for radii 1, 2, 3, and 4, when the number of bolts is 6, 8,
12, 15, 17, and 20.

81. Strength of Current in Tangent Galvanometer,
formula of a tangent galvanometer is
. rR tan S

The

;h i=the current strength;
r = the radius of the coil;

H = the horizontal component of the earth's magnetism;
fi=the angle of deflection of the needle;
n =the number of turns of the coil.

Fig. 37.
=25 cm., 7i =5, H = .1580 dyne, and 6 =38° 17'.

82, Compute 9 to the nearest minute when R =
L=.029, £ = .000052, by the formula

6,78651

100

TECHNICAL TTUGONOMETBY

97

83. Track Turnout The cut shows a turnout from the

straight track K in which the dotted line denotes the center of
the gage.

R = radius of turnout to gage center;
fl=frog angle;
F=0S =frog distance;
G =gage.

Write a trigonometric formula for F £
in terms of R, 6, and G.

Compute and tabulate F when G = C
4' 8J" and the other quanities have the
values here tabulated.

Table XVII
FROG DISTANCE

«.

R.

P.

6° 18'
4° 48'

1088
1421

84. Diameter of Ball Race. Fig. 39 shows a ball race of
diameter D.

r THE BIGHT TRIANGLE

Notation: 7) = diameter of race;
d = diameter of balls;
s =distanee between balls;
. n— number of balls.

Formulate « in terms of n.

86. Ball Race Computa-
tion, Compute the diameter
of the ball race for 24 balls
| inch in diameter.

86. Diameter of BaH Race
with Clearance. In Problem
84 if s denotes the clearance
allowed each ball,

BK=what part of d+

what part of s?

Substitute for RV in (1) of
Problem 84 and for CR in
terms of D c the diameter of
the ball race to the center of

Also formulate D.

Fio. 39.
the balls, and solve for D c .

87. Ball-Race Formulas. Solve the final formula of Problem
86 for the following:

(1) the clearance, s;

(2) the radius of the balls, r;

(3) the diameter of the balls, d;

102

TECHNICAL TRIGONOMETRY

97

(4) sin

180*

n

(5) the number of balls, n;

(6) the diameter of the ball race, D.

88. Compute the diameter of a ball race to the center of the
balls for 30 balls A of an inch in diameter, allowing a clearance
of .125 of an inch.

89. Compute and tabulate the omitted entries in the following
table:

Table XVIII
BALL-RACE COMPUTATION

No.

d.

n.

c.

*.

D c-

D.

1

A"

20

.or'

2

8

.15"

2.1"

3

i"

16

1.5

4

i"

. 185"

3.45"

5

12

. 125"

4"

90. Wire Measurement of Standard 60° Sharp Thread.

Fig. 41.

In the measurement of a V-thread a wire is used as shown
in the figure.

Formulate c in terms of the diameter of the wire and the thread
angle.

91. Pipe-Bend Computation. Fig. 42 shows the construction
necessary to determine V.

97

THE EIGHT TEIANGLE

103

Formulate and compute V to the nearest ten-thousandth of
an inch

when

A=lf";

0=40°;
27=2";

£=5A".

Fig. 42.

92. Metric Wire-Gage.* The cut shows a section of a metric
wire-gage. As illustrated the gage was worked out by the assump-
tion of a §° angle in a metal plate, one side of the angle being
parallel to the edge of the plate. A wire which, when placed in
the angle as far as it will go, has its point of contact with the
parallel side of the angle exactly 3 millimeters from the vertex, is
called number 3; one whose point of contact is 8 millimeters is
called number 8, etc.

As assigned, determine whether the diameters tabulated are
correct.

* Gage designed by Mr. Merritt W. Griswold of Highwood, N. J.

104

TECHNICAL TRIGONOMETRY

87

INTERNATIONAL STANDARD

_ w h* ga CH •* ►-* jo

m

OMKrWSS-fGiil"

ss

m « 8

« ft ^ £ A r

IS

iiiiMiirLn

ORDER WIRE BY THIS GAUGE
FOUNDED ON ANGLE OF 1-DEGREE

•*ta CO (► Ol 09 ~* oo «o °

ItU

rpTTf

i i rn'im

rnry

il Mill 1 1 1 1 1 1 1 I ! | f M| II

Fig. 43.

Table XIX
INTERNATIONAL STANDARD METRIC WORLD GAGE

Diameter.

Diameter.

No.

No.

No.

No.

mm*

Inches.

mm.

Inches.

1

0.01745

0.00068

1

19

3.31622

0.13056

19

2

0.05236

0.00206

2

20

3.66530

. 14430

20

3

0.10472

0.00412

3

21

4.03183

0.15873

21

4

0.17454

0.00687

4

22

4.41581

0.17385

22

5

0.26181

0.01031

5

23

4.81725

0.18965

23

6

0.36653

0.01443

6

24

5.23614

0.20614

24

7

0.48870

0.01924

7

25

5 . 67248

0.22332

25

8

0.62833

0.02473

8

26

6 . 12628

0.24119

26

9

0.78542

0.03092

9

27

6 . 59753

0.25974

27

10

0.95995

0.03779

10

28

7.08624

0.27898

28

11

1 . 15195

0.04535

11

29

7 . 59240

0.29891

29

12

1.36139

0.05359

12

30

8.11612

0.31952

30

13

1.58829

0.06253

13

31

8.65708

• . 34082

31

14

1.83263

0.07215

14

32

9.21560

0.36281

32

15

2.09445

0.08245

15

33

9.79158

0.38549

33

16

2.37371

0.09345

16

34

10.03851

0.40885

34

17

2.67043

0.10513

17

35

35

18

2.98460

0.11750

18

1

THE RIGHT TRIANGLE 105

A

106

TECHNICAL TRIGONOMETRY

97

93. Clearance in a Ball Race. By reference to Fig. 44 for-
mulate and compute c the total clearance, and the distance between
each ball when equally spaced.

94. Angles of Bevel-Gear. In Fig. 45 formulate 0, <f>, and a
in terms of the known dimensions as lettered.

95. Lay-Out of Angles by Two-foot Rule. Angles are some-

Fig. 46.

Fig. 47.

times laid out by a two-foot rule, opened to different widths foi
different angles.

Formulate, compute, and tabulate to hundredths of an inch,
the distances the ends must be opened for all angles from 5° to 90°
inclusive, increasing by 5°.

Check by the traverse table.

96. Lay-Out of Angles by the Compasses. By the traverse
table determine and tabulate the compass settings n order to lay
out the angles of Problem 95 with the large compasses in your
drawing set.

97

THE RIGHT TRIANGLE

107

97. Voltage from an A.C.* Generator. V*'hen a coil of wire
is rotated in a uniform electric field, an alternating current is
generated, the voltage of which fluctuates from zero to maximum,
t j zero, to maximum oppositely, to zero,

These changes are best exhibited in the curve shown in the
figure which is called the sine curve.

The complete succession of changes from 0° to 360° is called
a cycle.

e =the instantaneous voltage for any part of the cycle;
E m ^ =the maximum voltage produced by an A.C. gen-
erator;
6= the phase angle or number of degrees through which
the coil has rotated.

Fig. 48.

The formula for e in terms of E^^ and may be derived from
the following vector diagram, so called because it represents the
revolution of a radius vector f or vector E^^, counter-clockwise,
the vectorial angle being denoted by 6.

As shown, e is four things:

(1) the instantaneous e.mi.;

(2) the side opposite in the right

triangle;

(3) the perpendicular drawn to pro-

ject ^max on the horizontal;

(4) the vertical component of E^^^

Write the formula for e in terms of i^ma* and 0.
98. The mflMninn voltage of an A.C. generator is 25,000 volts.
Compute the voltage at f of the cycle.

* A.C. means alternating current.
t Moving radius.

Fia. 49.

108

TECHNICAL TRIGONOMETRY

97

99. Rate of Change of an A.C. Current. The rate of change
Og of an A.C. current in amperes at any point in the cyele, equals
2t times the frequency / times the maximum current /„„, times
the cosine of the angle of the cycle 8.

The maximum current /„ equals 1.414 times the effective
current 1&.

Formulate and substitute in the preceding formula.

100. Lead or Lag. An alternating current may lag behind
the voltage, or lead the voltage, the amount of lead or lag being
expressed in degrees and denoted by e or <f>.

The angle of lag may be determined from the following law:

(1) The tangent of the angle of lag equals the ratio of
reactance to resistance R.

(2) Reactance equals 2ir times the frequency f, times the
inductance L in henry s.

Formulate both laws and in the tangent formula substitute
for reactance.

101, Draw a right triangle showing 8 and sides marked react-
ance and resistance.

102. Compute the lag when

the reactance is 10 and the

Illustrate by right triangle
drawn to scale.

103. Compute the lag when
the resistance is 8, the induc-
tance .005 henry, and the fre-
quency 130.

Illustrate by right triangle
drawn to scale.

104. Power Factor. The

cosine of the angle of lead or
lag in an A.C. circuit is called
the power factor.
Compute the power factor in Problems 102 and 103.

Fio. 50.

97

THE EIGHT TRIANGLE

109

Fig. 51.

105. Compute the power factor when the resistance is 6, the
frequency 100, and the inductance .004 henry.

106. Formula for Power Factor. The figure shows a right
triangle with the following notation:

P =true power of an alternating cur-
rent circuit in watts;
/ = current in amperes;
E= voltage;
4* = the phase angle.

Formulate P in terms of the other
quantities and solve for the power factor.

Explain why cos <t> is called the power factor.

107. Phase Angle. Compute the phase angle in an alternating
current circuit of 12 amperes under a voltage of 120 volts, when
the watt-meter stands at 1350.

108. What is the power factor when the power delivered is
2000, and the ampere-meter and volt-meter show readings of 20
and 110 respectively?

Show diagram.

109. A Vector Diagram. A vector is a line of known or com-
putable length and direction.

A vector quantity is one whose graphic representation is a
vector.

Any graphic representation of vector quantities is called a
vector diagram, of which the figure here shown is an illustration.

110

TECHNICAL TRIGONOMETRY

97

OB=E BV=e, 0(7=7, CD=t;
4> = phase angle of voltage;
= difference between phase angle of voltage
and phase angle of current.

Formulate i in terras of 7, 4>, and 0.
Solve the formula for 7.
Formulate E and solve for e.

110. Compute i when 0=60°, 0=35°, and 7=45, the symbols
having the same significance as in Problem 109.

111. From the vector diagram of Problem 109, compute 7
when <t> =25° 32', =5.2°, and i =95.5.

112. Instantaneous Current and Voltage. The diagram shows

the current leading the .vol-
tage.

Project 7 and E on OV
and formulate the instan-
taneous current and the
instantaneous voltage.

113. Compute 7 when

and 0=21.3°.

Fig. 53.

f=23.5, = 15° 15',

114. A Roof Truss. The sketch shows a wooden roof truss
with a span of 150 feet, the rafters being at an angle of 30° 45'

with the horizontal. The tie-rods a and b, and the struts c and d,
divide the chord AR into three equal parts. The tie-rods a and
b are perpendicular io the rafters OR and OA.

Derive solution formulas and compute the following:

AY, DR,

a %

6, c, d, 07), AL t and LY.

97

THE RIGHT TRIANGLE

111

115. A surveyor measured an irregular field in the following
manner: from A he ran a line E 72° N, 40 chains; thence N 48°
W, 55 chains; thence W 8° S, 7 chains; thence S 13° W, 15

Fig. 55. — Surveyors' Compass.

chains; thence S 17° E, 29 chains; thence to A the point of
starting.

Draw the survey with great care to the scale iV' = l chain,
naming the successive stations (points from which the readings
were taken) by capital letters in their order from A. (See Fig.
56.)

Compute the bearing (direction) of the line which closed the
survey; also the area of the field.

112

TECHNICAL TRIGONOMETRY

97

The area may be determined by dividing the field into triangles,
o by drawing lines through the extreme northern and southern, and
eastern and western points, so as to form a rectangle whose area
exclusive of the field, may be determined by triangulation.

The traverse table may be used to lessen the labor of computation.

m C

Fig. 56.

The following problems are stated in the language of
the original deeds given in conveyance of the property.

Plot * each survey and compute the approximate
acreage.

116. Beginning at a point; thence south 81 §° east 114 perches
to a post and stones; thence south 2° west 62 perches to post
and stones; thence south 88 J ° west 113 J perches to knot in road;

* Plot means draw to scale.

97

THE EIGHT TRIANGLE

113

thence north 2° east SO perches to the place of beginning, con-
taining 50 acres and 56 perches.

117. Beginning at an original corner of stake and atones; thence
north 48° east 141j perches to a post corner; thence north 42°
west 14 perches to a corner in edge of Ball's pond; thence north
45° east 21 perches to a stake and stones comer in the edge of
Ball's pond; thence south 42° east 121 perches to a post in swamp;
thence south 45" west 162J perches to a corner; thence south

Fm. 57. — Surveyors' Chain and Tripod.

42° west 113rV perches to the place of beginning, containing
113 acres and 28 perches be the same more or less.

118. Beginning at a point in the center of the road; thence
along said road north 48° west 13 rods and 9 links, and north
20° west 32 rods, and north 6° east 54 rods to corner in said road;
thence north 71J°'east 138 rods; thence south 22° east 16 rods
to corner; thence south 45° west 171 rods to the place of beginning,
containing 50 acres and 38 rods of land, be the same more or

114

TECHNICAL TBIGONOMETKY

119. Beginning at a hemlock the southeast comer hereof;
thence north 4° east 100 perches to stone corner; thence north
86° west 116 perches to stone comer; thence south 5" west 19.7
perches to center of road ; thence along center of said road south

45° east 25 7 perches; thence south 70° east 15 perches, south
52° east 16 perches; south 33° east 14 perches; thenee south 60°
east 16 perches; south 45° east 18 perches; south 19J° east
10 perches; south 9° west 10 perches; thence south 86° east
29ft perches to the beginning, containing 46rV aeres more or

120. Beginning at a point in the old turnpike road; thence
along said road north 28° east 25 perches; thence north 43° east
25 perches; thence north 47° east 33 perches to a post; thence
south 91 perches to a post; thence west 9 perches; thence north
87° west 52 perches to the place of beginning, containing 17 acres
more or less.

121. Beginning in the center of the road; thence along said
road south 51J° west 15 rods; thence south 21J" west 17J rods;
thence south 31° east 32 rods and 4 links to a comer; thence north
85J" west 43§ rods to a corner; thence north 4J° west 68 rods to
a comer; thence north 491° east 34 rods to a corner; thence south

41J° east, 46 rods to the place of
beginning, containing 22 acres of
land, be the same more or less.

122, Commencing at an iron
corner near a rock; thence north
89f° west, 79ftftF rods to an iron
corner; thence south 21° 39' west
28ft rods to a post; thence by
the same south 893° e astj 90""°
rods, to the aforesaid east line of
lands; thence north 3£ B west
26ft rods to the place of ' begin-
ning, containing 14-^r acres of
land, be the same more or less.

123. Beginning at a stake in
the Carpenter road; thence along

i" east 35 rods to a corner; thence south 10j°

Fid. 58.— Steel Tape.
said road north

97

THE EIGHT TRIANGLE

115

east 16 rods and 13 links to a corner in the wall; thence south
66° west, 29 rods and 17 links to a stake; thence by the same
north 24° west 27 rods and 2 links to the place of beginning, con-
taining i{ acres of land, be the same more or has.

124. Beginning at a post and stones the northwest corner
hereof; thence south 196 perches along the dividing line to a post
and stones the southwest comer hereof; thence east 131 perches
to a post and stones the southeast corner hereof in the dividing
line; thence north 63A perches to a post and stones the south-
east corner of a piece of land; thence west 65tSf perches along the
dividing line to a post and stones corner;
thence north 132A perches along the west
line to a post and stones corner; the north-
east corner hereof; thence west 65A perches
along the dividing line between lots Kos.
95 and 112 to the place of beginning, con-
taining 106 acres more or less.

126. Beginning at a hemlock tree the
southeast corner hereof; thence north 4°
east 100 perches to a stone corner; thence
north 86° west 116 perches to a stono
corner; thence south 5" west 19A perches
to center of road leading from Montrose
to Silver Lake; thence along the center of
said road the following courses and dis-
tances, south 45" east 25-rV perches; thence
south 70° east 15 perches; thence south
45° east 18 perches; thence south 19J°
east 10 perches; thence south 9° west 10 ""•"^ *<**wm
perches; thence south 86° east 29A perchea
to the place of beginning, containing 46-nr acres of land.

126. Beginning at a point in the old turnpike road, the south-
east corner of lot; thence along said road north 28° east 25 perches;
thence north 43° east 25 perches; thence north 47° east 33 perches
to a post; thence by other lands south 91 perches to a post; thence
west along the lot 9 perches; thence along the same north 67°
west 52 perches to the place of beginning, containing 17 acres,
more or less.

116 TECHNICAL TRIGONOMETRY 97

127. Beginning in the center of the road; leading from Harford
to New Milford; thence along said road south 3f ° east
29 rods and 21 links; thence 27J° west 20 rods; thence
south lOf ° west 24 rods; thence north 71° west 22
rods and 20 links; thence south 4|° west 25 rods
and 5 links to stake and stones; thence along lands
north 85£° west 35 rods and 18 links to stake and
stones; thence north 4J° east 77 rods to stake and
stones; thence south 85%° east 43J rods to the place
of beginning, containing 22 acres, more or less,
being the southern half of lot known as the wood
lot.

128. Beginning at a point in the center of the
road, 20 feet from a shade tree near grange hall;
thence south 72£° east 35 A perches; thence by the

j5 same south 56£° east 13A perches to a corner in
£ the center of the said road; thence south 11J° east
% 121 perches to the center of the Page road; thence by
£f the center of the same north 88° west 6 rods; thence
p§ north 70° west 8 rods; thence north 66° west 9 rods;
I thence north 80° west 6 rods to a corner in the center
S of the road; thence north 16° east 6 rods to a corner;
g thence north 42f ° west 40 rods to a corner; thence

north 18i° east 101 rods to the place of beginning;

containing 40 acres and 134to perches be the same more

or less.

129. Beginning at the northwest corner of lot,
north 67£° east, 38 feet to a corner in the west line of
an alley 10 feet in width; thence along the western
bounds of said alley north 10^° west, 33A feet to a
corner; thence still on the west side of said alley north
23° west, 8f feet to the southeast corner of lot;
thence along the southern bounds of the lot, south
60° 10' west, 45J feet to a corner in the eastern
boundary of street; thence along the east bounds of
said street, south 22^° east, 37 feet to the place of
beginning.

f^

CHAPTER III

THE OBLIQUE TRIANGLE

«v<

Section 1, Functions of any Angle. Section 2, The Four
Laws of Solution. Section 3, Solution of Triangles.
Section 4, Applied Problems.

§ 1. FUNCTIONS OF ANY ANGLE

98. Introduction. An oblique triangle is a triangle
which is not right-angled. It therefore includes the isos-
celes triangle, except the special case in which the angles
opposite the equal sides are each 45°.

An oblique triangle may be solved by the laws of the
right triangle, if a perpendicular is drawn from the vertex
of one of its angles so as to form a right triangle, one side
of which is a known side of the given oblique triangle.
This method generally necessitates several transformations
and operations, and for this reason oblique triangles are
solved by the theorems which follow.

After a little practice it will be possible to determine
from the problem itself, which law to use in order to find
any required unknown part. Before this is possible, each
should be tried in succession until an equation of only
one unknown quantity is obtained.

99. Angles Named by Quadrants. In the discussion
of the right triangle in the preceding chapter, the functions
were defined with reference to the acute angles of a right
triangle.

It is now necessary to define them with reference to
any angle in any kind of triangle.

117

118 TECHNICAL TRIGONOMETEY 99

Under the chapter and paragraph headings draw a
unit * circle.

Draw a horizontal and a vertical diameter.

Beginning with the quadrant above the horizontal and
to the right of the vertical diameter, number the quadrants
in Roman notation in a counter-clockwise direction, placing
the numbers within the quadrants close to the horizontal
diameter.

Suppose an angle of less than 90° to be placed with its
vertex in coincidence with the center of the circle, one of its
sides in coincidence with the segment of the horizontal
diameter at the right of the center, and the other side in
the direction in which the quadrants were numbered.

The second side would then lie in what quadrant?

Therefore an angle of from 0° to 90° may be called an
angle of what quadrant?

Suppose the side of the angle, not in coincidence with
the horizontal diameter, to revolve in the plane of the circle
about the center in a counter-clockwise direction; when
the angle becomes greater than 90° and less than 180°,
in what quadrant will the moving side lie?

Therefore an angle of from 90° to 180° may be called
an angle of what quadrant?

An angle of from 180° to 270° may be called an angle of
what quadrant?

An angle of from 270° to 360° may be called an angle
of what quadrant?

An angle of from 360° to 450° may be called an angle
of what quadrant?

An acute angle is therefore an angle of what quadrant?
An obtuse angle is therefore an angle of what quadrant?
The angles of a triangle therefore are angles of either
of what auadrants?

* A circle whose radius is unity.

100

THE OBLIQUE TRIANGLE

119

100. Functions of any Angle. Under paragraph heading,
draw a horizontal line about one and one-half inches long.
Suppose this line to be pivoted at its left extremity and to
revolve in a counter-clockwise direction. By a light line
represent its position after revolution through an arc of
less than 90°.

From its extremity draw a construction perpendicular
to the line in its initial position.

Mark the sides of the right triangle thus formed, as
follows:

Radius vector (moving radius). Denote on figure by
R F, printed without the triangle, midway between vertices.

Perpendicular (from extremity moving radius to other
side angle). Denote on figure by symbol -L.

Projection (of radius vector on other side angle) . Denote
on figure by Proj.

Represent by 6 the angle formed by the radius vector
and its initial position.

Table XX
GENERAL DEFINITIONS, FUNCTIONS

Functions.

Special Definition

for Acute Angles

Right Triangle.

Genera] Definition.

sin =

C08 =

tan«
csc=»
sec =
cot =

By definition Sine, sin 6 equals the ratio of what sides
with reference to angle 0?

Therefore sin equals the ratio of what lines as marked
on the figure?

120 TECHNICAL TKIGONOMETEY 101

Therefore the sine of an angle may be denned as the
ratio of what lines?

In like manner define each of the other functions in terms
of sides as marked, entering definitions in Table XX
ruled in the work-book.

101. Convention of Signs of Functions. Mathemati-
cians have adopted the following conventions regarding the
signs of the functions of an angle in the different quadrants:

The radius vector in any position is positive.

The projection when to the right from the point of revolu-
tion is positive.

The projection when to the left from the point of revolu-
tion is negative.

The perpendicular when above the projection of the radius
vector is positive.

The perpendicular when below the projection of the
radius vector is negative.

Summary: Positive, Right, Above.

Negative, Left, Below.

102. Signs in the First Quadrant. Draw a first quad-
rant angle, 0i.

By the general definitions in Table XX, write the
equation of each of the six functions of the angle 0i, in
the first quadrant, naming sides as marked on your figure
in paragraph 100, and writing before each term of each
ratio the proper sign, + or — , as determined by the con-
vention of signs in paragraph 101.

Enter these signs on the figure.

By the Law of Signs, mark the proper sign after each
of the ratios in the six equations.

103. Signs in the Second Quadrant. Draw a figure
showing the initial line and radius vector after revolution
into the second quadrant.

104

THE OBLIQUE TRIANGLE

121

Fig. 61.

From the extremity of the radius vector draw a per-
pendicular to the initial line
produced.

Mark the sides of the tri-
angle as in paragraph 100,
and enter on the figure the
signs of the sides as deter-
mined by paragraph 101.

By the general definitions in Table XX, write all the
functions of the angle 02 formed by the initial line and the
radius vector, writing before each term of each ratio the
proper sign.

Following each ratio write its sign as determined by the
Law of Signs in division.

104. Signs in the Third and Fourth Quadrants. In

the same way represent the initial line, and the radius
vector in the 3rd, and in the 4th quadrant, projecting each
radius vector on the initial line produced if necessary.
Dimension 0s and 04 on the figures as shown in Fig. 61.
Write functions and signs as in the previous sections.

Summarize results for all quadrants in the following
table ruled in the work-book.

Table XXI
SIGNS OF FUNCTIONS

1

Functions.

Signs in Quadrants.

I.

II.

III.

IV.

Sine

Cosine

Tangent

Cosecant

Secant

Cotangent

122

TECHNICAL TRIGONOMETRY

105

105. Use of the General Definitions. In subsequent
mathematical work even when right triangles are involved,
functions of all angles necessary to solution are best defined
in terms of the general definitions.

In one of the preceding paragraphs, an acute angle was
found to be an angle of what quadrant?

Therefore all functions of an acute angle have what
sign?

An obtuse angle was found to be of what quadrant?

Therefore what functions of an obtuse angle are positive?

What are negative?

Therefore what sign must be prefixed to the cosine
and tangent of an obtuse angle when read from the table?

(Underline answer so that it will be more easily remem-
bered.)

106. Three Additional Functions. The following table
shows the nine functions of an angle.

Table XXII
NINE FUNCTIONS OF AN ANGLE

Function.

General
Definition.

Function.

Definition.

sin

COS

tan

vers
ext-sec

perpendicular

radius vector
projection

radius vector
perpendicular

projection
1— cosine
secant — 1

CSC
8CC

cot
covers

1

sine
1

cosine
1

tangent
1 —sine

107. Line Functions of an Angle. The names of the
functions of an angle become significant only when the
functions are defined as lines instead of ratios. By many
of the old mathematicians and by many teachers of mathe-

107

THE OBLIQUE TEIANGLE

123

matics of the preceding century, a function of an angle
was defined as a function of the arc which measures the angle
in a unit * circle.

The figure shows a unit circle with the line functions
of the angle 0, as determined
by the definitions which fol-
ic w.

All arcs were supposed to
be swept out or originated
from the right extremity of
the horizontal diameter, by a
radius rotating counter-clock-
wise. Therefore in Fig. 62, O
denotes the origin and T the
terminus of the arc OT which
measures the angle 0.

In reading through the
following definitions be sure to
study the figure until you see clearly that each function
meets every requirement of the definition.

Fig. 62.

Definitions op Line Functions.

The sine of an arc is a perpendicular from the terminus
of the arc to the diameter at the origin of the arc.

The tangent of an arc is a tangent to the arc at its origin,
terminating in the orig?n at one extremity, and at the other
in a radius or diameter produced from or through the
terminus of the arc.

The secant of an arc is the segment of a diameter pro-
duced through or from the terminus of the arc, and lying
between the tangent and the center of the circle.

The versine of an arc equals 1 minus the cosine of the
arc.

* A unit circle is a circle whose radius is unity.

124 TECHNICAL TRIGONOMETRY 108

Observe on the figure:

The cosine is the complement's sine.

The cotangent is the complement's tangent.

The cosecant is the complement's secant.

The coversine is the complement's versine.

In Fig. 62

C = center of unit circle;

0= origin;

T= terminus;

0= angle;
Of = arc measuring 0;

a = complementary angle,
LT = complementary arc;
TjF = sin ^0:T= sin 0;
AO = tan —0!F= tan 0;
AC = sec ^OT = sec 0;
FO= vers <~-OT= vers 0;
i4!T=ext-sec^^Or=ext-sec 0;
TH = cos ^OT = cos = sin ^TL = sin a;
BL = cot ^OT = cot = tan ~TL = tan a;
.BC = esc ^0r= esc 0=sec-^rL=seca;
HL == cvs ^Or = cvs = vers ^TL = vers a.

In the work-book draw a unit circle showing the line
functions, and write down the notation and the equations.

§ 2. THE FOUR LAWS OF SOLUTION

108. Law of Sines.

In any plane triangle the sines of the angles are pro-
portional to the opposite sides.
Case I. An Acute Triangle.
Case II. An Obtuse Triangle.

109 THE OBLIQUE TRIANGLE 125

In Case I draw an acute-angled triangle, and a con-
struction line from any vertex perpendicular to the opposite
side.

Write the conclusion with reference to the two angles
adjacent to the side to which the perpendicular was drawn.

In the conclusion what is the first quantity required?

What does this quantity equal, in terms of the general
definition of the sine in Table XX?

In the same form express the value of the second quantity
required in the conclusion, and finish the demonstration.

In Case II draw an obtuse triangle and a construction
line from the vertex of either acute angle perpendicular
to the opposite side produced.

By the general definition of sine, write the sine of each
given angle adjacent to the produced side, and finish the
demonstration.

109. Law of Segments.

In any plane triangle, if a perpendicular is drawn
to the longest side from the vertex of the opposite angle,
the difference of the segments of that side equals the
product of the sum and the difference of the other two sides,
divided by the longest side.

(In writing the " difference of the other two sides " in
the conclusion, be sure to make the greater of those sides
the first term.)

Letter the figure as follows:

h = perpendicular;
g = greater segment,
s = shorter segment.

Denote the sides of the given triangle by small letters
corresponding to the capitals at the opposite vertices.

Express the conclusion as an equation, naming parts
as lettered.

126 TECHNICAL TRIGONOMETRY HO

By square hypotenuse, g 2 equals what?
And s 2 equals what?

Subtract the second equation from the first, factor
both members and finish the demonstration.

When the demonstration is finished draw a triangle
in which a is the longest side, and formulate the law.

Draw a triangle in which b is the longest side and formu-
late the law.

Draw a triangle in which c is the longest side and formu-
late the law.

110. Law of Cosines.

In any plane triangle, the square of any side equals
the sum of the squares of the other two sides, minus twice
the product of the two into (times) the cosine of their
included angle.

Case I. The Square of a Side Opposite an Acute
Angle.

Case II. The Square of a Side Opposite an Obtuse
Angle.

(Do not neglect to write the hypothesis and the conclu-
sion for each case.)
By geometry the square of the given side in Case I
equals what?

(If necessary, see theorem, " Square side opposite acute
angle.")

By comparison of the resulting equation with the con-
clusion, determine what substitution is necessary and finish
the demonstration for Case I.

Demonstrate Case II in a similar manner.

111. Law of Tangents.

In any plane triangle, the sum of any two sides
divided by their difference, equals the tangent of one-half
the sum of the opposite angles divided by the tangent of
one-half their difference.

110 THE OBLIQUE TEIANGLE 127

Figure, Hypothesis, and Conclusion.

In order to demonstrate, two construction lines are neces-
sary, and it is therefore suggested that one page
be used for the figure, hypothesis, and conclu-
sion, and a second page for the demonstration.
Probably the best figure will be one about the
same shape as shown in Fig. 63, with c about
1 inch in length and b about 3 inches.

In formulating the conclusion * use the
fractional form for the ratios.

By reference to the conclusion it will be
seen that it is a proportion. But the quantities
constituting this proportion are lines.

It is therefore necessary to form two tri-
angles, two sides of which are the quantities yiq 63.
required in the first member of the conclusion.

But if four quantities as sides of triangles are to form
a proportion, the triangles must be proved to be in what
relation to each other?

Construction. The first quantity required in the con-
clusion is the sum of the two sides, but there is no line in the
figure which represents this sum.

Such a line must therefore be constructed.

Through A, produce c to V, making AV = b.

On BV, dimension b+c.

Draw VC.

From A, lay off on 6, AT=c.

On 6, dimension b—c.

From B, draw through T, a line terminating in VC at 0.

Determination of the Two Triangles. In the figure as
completed, the line which is the sum of the two given

* To avoid any tendency to confusion in signs, set down the greater
side and the greater angle as the first term of both the sum and the
difference.

128 TECHNICAL TRIGONOMETRY 110

sides is a side of what two triangles? (In answering name
triangles by letters at the vertices.)

The line which is the difference of the two given sides
is a side of what two triangles?

The two triangles which shall be used therefore, are one
of the first two and one of the second two which you have
named.

Demonstration. Keeping in mind the construction,
determine by demonstration the two triangles which are
in the relation in which their homologous sides are pro-
portional.

The necessary relation can be shown by proving that
two of the four triangles are mutually equiangular. The
proof will be simplified by denoting the equal angles by the
same number, one being primed.

The proportion when finally written, must have the seg-
ments of VC for the second ratio.

By reference to the conclusion it will be seen that for
the ratio of the two segments of VC, must be substituted

sum opp angles B+C

tan tan — - —

diff opp angles t B—C'
tan tan — - —

That this substitution may be possible,

(1) Prove that each segment is the side of a right
triangle.

(2) Write the value of the first segment in terms of BO
and the tangent of the angle to which it is opposite.

(3) Write the value of the second segment in terms of
BO and the tangent of the angle to which it is opposite.

In the equation whose second member is the ratio of
the segments, substitute these values and simplify.

If you will now express B in terms of VBO and CBO

113 THE OBLIQUE TRIANGLE 129

as numbered on your figure, it should be possible to finish
the demonstration without further suggestion.

112. Formulation of Law of Tangents. In the solution
of triangles it will soon be found:

(1) that this law can be used only when two sides and
the included angle are known,

(2) that the denominator of the second fraction is there-
fore always the unknown quantity, because when one angle
of a triangle is known the sum of the other two angles is
known, but their difference is unknown.

This being the case, the law is best written in the follow-
ing form:

sum opp angles

,.«. . , , (din kwn sds) tan

diff Zs opp kwn sds 2

2 sum known sides

Draw a triangle in which b and c are known, with b>c,
and formulate the law. Write greater side and greater
angle first in both sum and difference.

Draw a triangle in which a and b are known, with b>a,
and formulate the law.

Draw a triangle in which a and c are known, with c>a 9
and formulate the law.

§ 3. SOLUTION OF TRIANGLES.

113. Examples in Oblique Triangles.

1. a =42 inches,

6=32 inches,
A =80°.

Find B, C, and c.

Using scale and protractor draw the triangle with the greatest
possible accuracy.
Solve it graphically.

From what vertex must a perpendicular be drawn to the oppo-

130

TECHNICAL TRIGONOMETRY

113

site side in order to form two right triangles, one side of each of
which is a known side of the given triangle?

Draw this perpendicular and formulate * and compute the
unknown parts of the given triangle by the laws of the right triangle.

Disregarding the perpendicular, solve * the given triangle by
one of the laws of the oblique triangle.

Enter the final results in the work-book in the following form:

Table XXIII
COMPARISON OF RESULTS

Solution of Example 1.

Unknown Parts.

Graphic.

Laws Rt. Triangle.

Laws Obi. Triangle.

B =

C =

c =

2.

c = 16
6=24
5=68° 30'

Solve the triangle: (1) graphically;

(2) by laws of the right triangle;

(3) by laws of the oblique triangle.

3. c = 124

A =49° 21' 12"
B =61° 40"

Solve by a law of the oblique triangle.

4. 6 = 14

c = 17
a=19

Solve the triangle : ( 1 ) graphically ;

(2) by laws of the right triangle;

(3) by laws of the oblique triangle.

* See paragraph 94.

114 THE OBLIQUE TRIANGLE 131

To solve by laws of the right triangle, draw a perpendicular
to side a from the opposite vertex. Denote by s one of the seg-
ments of a, and denote the other segment in terms of a and s.

Write two values for the square of the perpendicular and finish
the solution.

In solving the triangle by a law of the oblique triangle, it will
be found that two different laws will apply.

Solve it by each of these laws.

State which necessitates the less work and which seems the
more convenient to use in this instance.

5. a =50

6=43
c=65

Compute B by two different laws of the oblique triangle.

6. c = 1.284

a = 1.5076
6 = 1.6399

Compute A by each of the two laws of the oblique triangle
which will apply.

In this example which of these laws necessitated the less work,
and which now seems preferable? Is it the same law which was
used in example 5?

7. 6=5179.2

C =57° 27' 31"
a =3890.5

Solve by one law of the oblique triangle.

114. Application of the Laws of the Oblique Triangle.

In Example 1 how many sides and angles included, oppo-
site, or adjacent were given? In solving the triangle for B
what law of the oblique triangle was used?

In Example 2 how many sides and angles included,
opposite, or adjacent were given? In solving for C what
law of the oblique triangle was used?

In Example 3 how many sides and angles included or

132

TECHNICAL TRIGONOMETRY

115

adjacent were given? In solving for a or b what law of the
oblique triangle was used?

Tabulate in the work-book as follows:

Table XXIV
SOLUTION OF THE OBLIQUE TRIANGLE

Ex.

No. of Parts Given.

Law Applied.

Other Law Obi.
As Applicable.

Sides.

Zs Adj Opp or
Incl.

1
2
3

4
5
6

7

■

-

115. How to Solve Oblique Triangles. The following
table is a logical deduction from the entries in Table XXIV,
and is a key to the solution of oblique triangles.

Table XXV
CASES AND LAWS

Case.

Given.

Use.

1

2
3

4

1 side and 2 angles: 1
Find 3rd angle, then /

2 sides and an opposite angle

2 sides and the included angle

Or, find sum other two angles, then

3 sides j

Law of Sines

Law of Sines

Law of Cosines
Law of Tangents
Law of Segments
Law of Cosines

116. The Ambiguous Case. When two sides and an
opposite angle are known and a second angle is therefore
determined by the law of sines, two solutions are some-
times possible. This is owing to the fact that the sine of
an angle and its supplement are both positive. It is there-

117

THE OBLIQUE TRIANGLE

133

fore impossible to determine from the sine itself whether
the angle whose sine has been determined by computation,
is acute or obtuse.

The following shows under what conditions the angle
is acute, or either acute or obtuse.

Suppose the given sides are a and 6, and A is given:

(1) If a>b, and A is obtuse, B is acute. Why?

(2) If a>b, and A is acute, B is acute. Why?

(3) If a<6, and A is acute, B is acute or obtuse,

and there are two solutions. Why?

It will be evident that (3) is the so-called ambiguous
case. Therefore when two sides and the angle opposite the
shorter side are given, the angle opposite the longer side
may be acute or obtuse, and two solutions must be shown.

117. Model Solutions. The following arrangement of
work is recommended for the examples in the next para-
graph.

9.

LAW OF SINES

B =31° 15' 29"

179° 59' 60"

C = 100°

131 15 29

a =1000.9

48° 44' 31"

A =

c =

6 =

asinC

C " • A

sin A

r asin#

sin A

391

3.000391

3.000000

9.715047

9.993351

31

12.993742

12.715469

9.876072

9.876072

3.117670

2.839397

03=1311
67

52=6908
45

66=2

44=7

(c+b)(c-b)
g-S= —

134 TECHNICAL TRIGONOMETKY 117

In the preceding model 9 denotes the number of the example.
Observe it is made larger than the other figures. The entries
above the single underline are the example; those between the single
and the double underline are the results obtained by computation.

Observe that the logarithms are set down in the same order
as the quantities in the formulas, and that —10 is not shown
following logarithms of decimals.

LAW OF SEGMENTS

The best law when these sides are given.

38. 6 = 127.51 c =

a =234.8 6 =

c = 195.14 c+6 =

2g =

2s-
s =

Cos£=|

CosC=4
b

LAW OF TANGENTS

The best law when two sides and included angle are given.

43. a=9.&39 179° 59' 60" a =

6=8.797 41 16 12 6 =

C=41°16'12" a-6 =

A = a +& =

C- A + B _

- 2

/ M+ A+B A ~ B

, A-B M)tm ~ -T- =

tanT r" ^Tb a*-

* To obtain A, add the half difference to the half sum.

t To obtain B, subtract the half difference from the half-sum.

118

THE OBLIQUE TRIANGLE

135

118. Examples. Solve the following triangles by the
laws of the oblique triangle. Time will be saved by writing
all the formulas required for the complete solution of any
problem before taking any readings from the tables.

8.

a =98.726

9.

5=31° 15' 29"

6 = 129.799

C = 100°

C=71°14'

a = 1000.9

10.

(7=28° 55'

11.

6=518.26

A =131° 17' 43"

c =694.50

6=7841

= 1000

12.

(7 = 14° 34' 2"

13.

c = .920007

A =100° 29' 7"

a = .80049

a =3.98

6=1.394

14.

£=67° 49"

15.

£=44° 45.9'

C =70° 7' 14"

C =70° 7' 14"

c =.09536

6 =.7627

16.

6 =.00945

17.

a =26.45

a = .00748

£ =39° 49' 49"

c =.01196

C = lll°25'

18.

a =725.18

. 19.

c =7.6321

c =943.06

6 = 10.842

6 = 1286.09

A =75° 13.6'

20.

6 = .0073

21.

a =4.28679

a =.0129

c =5.70082

C = 134°50'

6=6.2349

22.

C =112° 19"

23.

B =72° 18.9'

B =36° 41"

C=91°4'5"

a=5

6 =.3768

24.

A =45°

25.

a = .47923

£=45°

c = .82095

c=100

6 = 1.008

26.

c = .07534

27.

a = .7293

6 = .12569

6 = .685

A =116° 24.5'

C = 139° 6.7'

28.

a = .003377

29.

c =.22968

£=39° 17' 17"

a = .37002

6 =.005693

£ = 108° 10' 11"

136 TECHNICAL TRIGONOMETRY 118

30.

0=35° 35' 35"
a=70°18'2"
Side opp. a =1000.8

31.

c =.99989
6 = 1.756
A =42° 24.3'

32.

c =28.4167
a =42.5095
6=47.684

33.

6=67.143
c =41.112
A =42° 12' 3"

34.

a =2.1763
c =3.4286
6=4.6075

35.

0=3009.9'
A =81° 4.6'
6=5187.3

36.

a = .00049
6 = .001223
c = .000968

37.

a =29.738
c =32.482
£= 15° 25.7'

38.

6 = 127.51
a =234.8
c =195.14

39.

a = 100.478

c = 119.738

B =56° 49.4'

40.

B =40° 12' 13"
A =71° 18' 18"
c = 1.728 meters

41.

6=99.857
c =299.735
A =123° 9.4'

42.

A =11° 11' 18"
5=21° 13' 16"
a =.14468

43.

a =9.639
6=8.797
C= 41° 16.2'

44.

a =.017829
6 = .01034
c =.01283

45.

6 =.03805
c = .02094
A =15° 42' 8"

46.

a =5.684]
6=8.432]
c =4.212

47.

A =44° 30' 10"
6 =4.62
c=3.98

48.

a =6.892
6=3.648
A =44° 13' 40"

49.

A =33° 42' 30"
B =48° 36' 10"
6=6.542

50.

a =432.6

B =22° 54' 32"

C = 18°47'54"

51.

a =34.2
6=68.9
C=43°22'10"

52.

a =7.634
6 =8.621
c = 14.462

53.

B=34°20'
C = 15°59'10"
a =.7854

118

THE OBLIQUE TKIANGLE

137

64.

6 =473.08
A =45° 45' 46"
c= 621.42

66.

a =.2347
6 =.1759
c=.3614

66.

a =207.008
c =545.305
B =31° 20' 17"

67.

a =64.71
6=55.42
c= 94.55

68.

a =64.3001
B =43° 20' 5" ,
c =21° 8' 10"

69.

6=4.0003
c =1.0902
B =47° 32' 27"^

60.

a=8
6=5
C =32° 25' 40"

61.

a =4320.21

6=7432.4

C=73°58'21"_

62.

a =44.689
6=73.901
c =28.42

63.

a =35
6=32
c = 19

64.

A =127° 14' 28"
£=26° 39' 12"
c = 14.7263

66.

c = 1126.38
6=742.675
B =47° 36' 22"

66.

a =6422.8
6 = 11978
c = 14722

67.

a =84.76
6=72.339
c = 128.607

68.

a = 1433.06
6 = 1217.95
£=77° 18' 24"

69.

a = 1387.56
6 - 1592.49
£ = 113° 41' 30"

70.

A =15° 24' 38"
£=37°0'45"
c = .431986

71.

6 = .29817
c = .37605
C = 110°54'

72.

6 = .098726
c = .346507
A =121° 4' 13",

73.

a =.78629
c = 1.00307
C = 104°16'22"

74.

a =.6068
6 =.497145
c = .720008

75.

a = 1666.84
6 = 1781.75
c =2100.046

76.

a =47.819

B =24° 17' 32"

C=39°28'15"

77.

a =67.8
c =52.976
C=34°17'35"

138 TECHNICAL TRIGONOMETRY 118

78.

a =346.72

79.

6=73.476

c =278.914

c =41 .896

B =52° 33' 33"

A =26° 19' 24"

80.

a =425.6718

81.

6=9.426

6 =392.784

c =7.8941

c =378.56

a =3.75

82.

#=63° 46.9'

83.

C=75°49'16"

A =49° 39' 10"

6=4.168

c = 1.3165

c=6.25

84.

6=2.5

86.

6=2.51625

c=3

c =3.0126

A =33° 10' 1"

A =33° 10' 1"

86.

6=7.5346

87.

6=7.5

c = 10.4237

c = 10.4

a =5.8376

a =5.8

88.

A =55° 35' 15"

89.

6=160.12

5=25° 25'

c =84.73

a = 185.13

5=51° 30' 5"

90.

A =45° 30' 10"

91.

a =76.45

6=5.637

6=83.21

c = 10.89

c =40.94

92.

a =7.592

93.

a =451 .25

6=4.813

6=302.29

B =46° 23' 35"

c =254.131

94.

a =420.95

95.

a =322.28

£=75° 24' 10"

c =610.009

c =40° 15' 40" ,

B =41° 10' 5"

96.

6=75.3709

97.

a =311.78

c =60.42

6=645.9

A =55° 10' 14"

c =407.6

98.

a =73

99.

6=25

6=82

c=81

c=91

A =49° 51' 10"

100

. a =.17964

101

. a = 19834.7

5=34° 14' 6"

C=69°51'33"

C=52°17'29"

B =33° 33' 33"

118

THE OBLIQUE TRIANGLE

139

102.

o = 12
c = 16
£ = 111° 12' 13"

103.

A =59° 42'
c = 12136
6=76548

104.

a =22
6=25
c-30

105.

6=9541
c=6725
o=4372

106.

a = 1098
6=625
£=15° 10' 12"

107.

a = 1106

C=36°10'4"

£=49°11'1"

108.

a = 19
6 = 15
c=25

109.

= 11

6 = 12
C=67°10'15"

110.

o =746.891

6=809.46

C=47°19'16"

111.

o = 11.5062
6=7.756
c =8.10111

112.

c = 12.041
A =9.441
5=71° 10' 6"

113.

c = 12.041
6 = 14.714
B =75° 30' 16"

114.

a = 14056.7
6=19214.5
(7 = 17° 41' 10"

115.

A =24° 31' 10"
5=76° 4' 9"
c =214.0

116.

a =785.4

6=310.32

c=741

117.

0=21.420
6=52.065
c=39

118.

~c=5.25

A =40° 5'

B =35° 10' 12"

119.

c = 1.4328
6=5.2167
5=62° 18' 5"

120.

a = 15.97
6=6.342
c=70°32'4"

121.

o=3.2576
6=4.625
c =2.7542

122.

c =3.165
a =5.634
£ = 18° 7' 2"

123.

a =4.1645
6=8.5312
c =5.28905

124.

A =16° 52' 41"

C=89°50'9"

0=4.57631

125.

A =72° 20'
6 = 10.563
c =8.6747

140 TECHNICAL TRIGONOMETRY 118

126.

a « 15.454

127.

6 = .0258429

c =65.8509

c = .0857581

A =45° 10' 12"

5=27° 18' 53"

128.

a = .10568

129.

a = .02565

6 =.085754

6 =.06288

c = .0725681

c = .052575

130.

a =64.3001

131.

6 =4.0003

/?=43°20'5"

c = 1.0902

C=21°8'10"

B =47° 32' 27"

132.

a=8

133.

a =4320.21

6=5

6=7432.4

c =32° 25' 40"

C =73° 58' 21"

134.

a =42.689

135.

a =25

6=78.901

6=40

c =66.42

c=37

136.

a =16.84

137.

a =2.465

B =48° 39' 51"

c =3.326

C=37°16'27"

A =54° 28' 3"

138.

A =32° 49' 16"

139.

a=9

6 = 12

6 = 16

c=15

c=20

140.

c=39°53'43"

141.

a = .7856

a = 116.84

6 =.4694

6 = 177.38

c = .5967

142.

a = 14.2546

143.

a =2.42591

£=45° 15' 2"

C = 1.5628

C=27°13'5"

A =49° 50' 16"

144.

£ = 13° 14' 15"

145.

a =32

a =25

6=20

c = 16

c=40

146.

C=44°15'19"

147.

a = .02567

a = 13.145

6 =.03226

6=29.116

c =.04323

148.

B =43° 44' 53"

149.

a =49.6748

A =81° 3' 10"

6=36.4297

c =3.4789

C=42°9'54"

118

THE OBLIQUE TRIANGLE

141

160.

a=800
6=496.72
c =647.925

151.

C=42°3'16"
a =92.746
6 = 49.565 .

162.

o =.43476
6 = 1.00978
c =.74321

153.

A =49° 14' 14"
c = 14.3422 *
a =7.92476

164.

a =47.819
5=34° 14' 16"
C=28°17'25"

155.

a =89.11
(7=67° 11' 56"
c =75.4809

166.

a =416.8
c =346.91
B =49° 33' 17"

157.

6=75.819
c =66.411
A =42° 21' 11"

158.

a =501
6=734.6
c =894.81

159.

c =416.85
a =518.3
6=682.706

160.

a = 196.77
(7=81° 2.5'
£:=77°0' 12"

161.

a = .0046897
6 = .0011159
5=39° 14' 24"

162.

0=449.678
6=334.0053
c =650.545

163.

6=720.095
a = 1048.63
C = 44°32' 40"

164.

C=129°39'52"
6 = 18
a=14

165.

c = .0043756
a = .0070529
6 = .0093147

166.

A =80° 27' 21"
£=27° 34' 17"
c = 187.63

167.

a =489.712
6 = 1231.96
A =14° 27' 55"

168.

c =.47961
a =.28546
£=35° 24' 48"

169.

c =321.78
a =417.038
A =57° 51' 46"

170.

c =576.284
6=429.006
a =307.815

171.

a =843.78
6=486.48
C = 110°27'43"

172.

a =34.193
6=75
A =15°

173.

c =71.83
A =43° 18"
B =70° 21.5'

142

TECHNICAL TRIGONOMETRY

118

174.

a =28

175.

a =37.2393

c = 12

£=20°

£=39° 16' 8"

c =24.8056

176.

a =.17638

177.

a = 17

6 = .31424

6 = 13

c = .£6352

c=24

178.

a =532.089

179.

a =329.076

A =35° 22' 12"

6 =465.0853

B =102° 45'

A =35° 44' 35"

180.

6=16.8347

181.

c = .296008

a =24.2369

6 =.377651

C=29°28' 30"

A =56° 10'

182.

c =.321096

183.

a =25.7961

a =.462537

6 = 10.4384

6 =.21267

c = 18.3129

•

184.

a =67.25

185.

6=72.145

(7=76° 31' 15"

a =68.62

£-54° 29' 22"

B =67° 55' 21"

186.

a =52.651

187.

a = .3141

6=28.309

6 =.7850

c=49°0'27"

c=.5106

188.

6=461.21

189.

a = .21313

c =263.29

b =.42098

A =109° 27' 59"

c =.38615

190.

a = .8059

191.

a =99

6 = .5182

6=85.06

c=.67

A =46° 15.8'

192.

6 = 156

193.

a =52

C=44°58'19"

c=66

A =52° 35' 21"

B =99° 3'

191

a=16

195.

6=3526.85

c=8

a =2759.3

6 = 11

C = 110°21'5"

196.

c =723.4

197.

a = .002347

B =42° 38' 50"

6 = .003289

A =78° 0' 30"

£=62° 47.6'

119

THE OBLIQUE TRIANGLE

143

198.

a = .37284

199.

6=97

c=.9276

a =72

B =49° 42' 37"

C = 124° 25' 30"

200,

a = .05456

201.

a =.64295

6 =.07289

6 = .48763

c = .0825

c = .523079

202.

a =5.874

203.

a =34.73

C =38° 6' 40"

6 = 16.8

5=25° 13' 10"

B =56° 58' 12"

204.

a = .50794

205.

a =3.845

6 =.68357

6=6.832

C=28°34'20"

c =4.751

206.

a =6.785

207.

a = .059832

6=3.217

6 = .065118

C = 44°16'10"

c =.084795

208.

a =.2968

209.

4 =129° 0' 6"

b = .342009

8 = 14° 5' 4"

30.4" c = .3961

5.096 f = 40.9607 feet.

119. The Fundamental and Necessary Facts of
Trigonometry. One can make little claim to a knowledge
of trigonometry and can use it only in a haphazard and
uncertain manner, or by " rule-of-thumb," unless the
principles involved in the following problems are thoroughly
understood. Trigonometry, like other branches of human
knowledge, is the application of a few basal and essential
notions or principles to a special kind of work. The special
work of trigonometry is the solution of triangles, and in
this paragraph the fundamental principles which make
this work possible are, as it were, brought up to the surface
so that they may be isolated, related, and mastered. This
paragraph is therefore the most important of any in the
book and should be studied until everything called for
has been carefully and correctly worked out in the work-
book and has become perfectly familiar.

1. State the advantages of the use of logarithms.

2. Define an exponent.

144 TECHNICAL TRIGONOMETRY 119

3. Define a logarithm.

4. Show why the characteristic of the logarithm of a number
greater than 1, is 1 less in unit value than the number of integral
figures in the given number.

5. Show why the characteristic of the logarithm of a number
less than 1, in sign is negative, and in unit value is equal to the
number of places of the first significant figure from the decimal
point.

6. Prove a = 1.

7. Prove — =a~ n .

ar

8. Show why the logarithm 1 =0.

9. Show why the logarithm 10 = 1.

10. Show why the logarithm of a product = the sum of the
logarithms of the factors.

11. Show why the logarithm of a fraction = the logarithm of
the numerator— the logarithm of the denominator.

12. Explain why a decimal point in a number affects only
the characteristic of its logarithm, and does not affect the mantissa.

13. Show how to solve 8.6* =9000.

14. In two columns show the two systems of logarithms with
the various names for each.

15. Formulate the exact relation of a Naperian logarithm to
a common logarithm.

16. From the sine and the cosine as formulated, derive the
other four functions.

17. Prove that sin 2 0+cos 2 = 1.

18. Formulate sin in terms of cos 0. Formulate cos in
terms of sin 0.

19. Formulate tan in terms of sin and cos and prove
the formula.

20. Show that the cos is the complement's sin.

21. Enter in a table the definitions of the six functions.

22. Draw a right triangle, and enter in a table the definitions
of the six functions of each of the acute angles.

23. By a table, show three functions and three reciprocal
functions.

24. Show how to solve an isosceles triangle by the laws of
the right triangle.

119 THE OBLIQUE TRIANGLE 145

25. State and illustrate the necessities to the solution of any
triangle.

26. What is an oblique triangle?

27. Write the Laws of the Oblique Triangle.

28. Demonstrate the Law of Sines.

29. Demonstrate the Law of Segments.

30. Demonstrate the Law of Cosines.

31. Demonstrate the Law of Tangents.

32. Show by a table, when each of these laws is to be applied.

33. In a table, define each of the six functions, specially and
generally.

34. Show the signs of the functions in the first quadrant.

35. Show the signs of the functions in the second quadrant.

36. Show the signs of the functions in the third quadrant.

37. Show the signs of the functions in the fourth quadrant.

38. By a table, give the signs of the functions in all the
quadrants.

39. When will the law of sines apply? Illustrate.

40. When will the law of segments apply? Illustrate.

41. When will the law of tangents apply? Illustrate.

42. When will the law of cosines apply? Illustrate.

43. What is the greatest numerical value of the sine? Show why.

44. What is the greatest numerical value of the cosine? Show
why.

45. What is the greatest numerical value of the tangent? Show
why.

46. sin0°«=what? sin 90°= what?
cos 0° = what? cos 90° = what?
tan 0° = what? tan 90° = what?

47. The sine of an obtuse angle always has what sign?

48. The cosine of an obtuse angle always has what sign?

49. The tangent of an obtuse angle always has what sign?

60. The reciprocals of these functions always have what signs?
51. State exactly how you would read any function of an
obtuse angle.

62. Determine two angles each having the sine = .607483.

9

63. Construct an angle whose cotangent =— and determine

its sine, cosine, and tangent.

146 TECHNICAL TRIGONOMETRY 120

120. An Anti-function. In trigonometric formulation it
is sometimes convenient to use an inverse- or anti-function
symbol. This symbol is dash 1.

Thus sin" 1 denotes the anti- or inverse-sine.
tan" 1 denotes the anti- or inverse-tangent.

Although written in the position of an exponent, the
symbol is not an exponent and the line preceding the 1
is a dash and not a minus.

The symbol when used on a trigonometric function, may
always be interpreted by " the angle (or arc) whose is."
In reading, the name of the function is inserted between
the words whose and is.

Thus

sin" 1 is read the angle whose sine is,
tan" 1 is read the angle whose tangent is,
cos" 1 is read the angle whose cosine is.

The equation,

4> = tan 1 r-

is read

<f> equals the angle whose tangent is a over 6.
The expression

r sin ( tan" 1 -^— J,

is read

a

r times the sine of the angle whose tangent is — .

b

Therefore, if

sec0 = -^y,

then

-i2.25
= sec -3-7-.

122 THE OBLIQUE TRIANGLE 147

In the solution of trigonometric problems it is sometimes
possible to solve only by the use of a particular function of
an angle when the angle itself is not known but must first
be determined from some other function.

When this is the case the anti-function notation may
be used to advantage in the solution formula.

121. Problems. Express the following in the inverse-
function notation, formulating the angles in each:

1. cos a = .7896, 6. d sin <t> = .968,

2. sina=— , 7. 5cos0=-,

c c

3 96
3 * COt e = ITo^ 8# ® S ~~ r ^ tan e =5 * 89 >

r
4. tan <l>=7rj 9. h sec 6 = 1.44 — Id,

ov

6. esc d - .4284 X 7.66, 10. ^^— = <*& sin |.

§4. APPLIED PROBLEMS.

122. Suggestions for Solution. The solution of the
problems in the next paragraph will be facilitated by the
following procedure:

(1) Read the problem through carefully.

(2) Enter data near the right margin.

(3) Draw the figure to scale.

(4) Denote angles by the customary Greek letters and
opposite sides by the corresponding or other lower case
English letters. Sides opposite and <t> may be denoted by
t and p respectively.

(5) Analyze the problem; that is, determine what
parts are known, whether

One side and two angles,
Two sides and an opposite angle,
Two sides and included angle
Three sides.

148

TECHNICAL TRIGONOMETRY

123

(6) Observe whether the problem is best solved by-
laws of the right triangle or of the oblique.

(7) Derive the solution formulas necessary for complete
solution.

94.

If you do not know what a solution formula is, see paragraph

(8) Take readings from the table, and solve.

(9) Check by the slide-rule.

123. Problems. Solve the following problems.

1. Determination of Phase Constant The figure illustrates
a symmetrical armature winding distribution.

x = number of slots per pole,
c =LN = volts per slot,
E = VS = volts per pole,

A = — = phase constant.
xe

6= phase difference between
adjacent slots.

Prove K =

Fig. 64.

. xO

. 0'
xsin —

61

Draw a radius perpendicular to LN and determine e in terms
of r and 0.

Prove ON perpendicular to SV, and determine E in terms of
r and 0.

Substitute for E and c in the formula for K.

2. Velocity Diagram. In Fig. 65.

OPL is the path of a projectile,
RA is tangent to the curve at P,
PN is perpendicular to RA at P,
and PM is perpendicular to RL.

123

THE OBLIQUE TRIANGLE

149

Prove (1) tana =

NM
PM'

(2) PM a mean proportional to NM and RM . (Trigo-
nometric proof, only.)

Fig. 65.

3. Weight. A triangular brass plate .386 inch thick, measures
13.85"X19.04"X 11.367".

What is the weight of the plate?

4. Functions of an Angle. Write the numerical values with
proper signs, of all the functions of 151° 28' 34".

6. Power Factor. In an alternating current circuit the cosine
of the phase angle is the power factor.

Fig. 66.

With the dimensions shown in the cut and 0=34° 30', com-
pute the power factor cos #.

150

TECHNICAL TRIGONOMETRY

123

6. True Power. In the vector diagram, Fig. 67,

# = 110,
# 2 =57,
^=80.

7. Resistance Drop.
6, determine <t>

when

and

Compute the true power
whose formula is

W=IE cos 6
when 7=8.

With conditions the same as in Problem

J-8,

/-CO,
L = .01S,

Ez=2ttJLL

Also compute #4, the resistance drop.

8. Distance at Sea. Two telescopes are mounted 180 feet
apart on the deck of a battleship. A lighthouse being sighted
its bearing with the line joining the telescopes is read from both
telescopes at the same moment. One angle is 87° 14', the other
94° 24'.

Compute the distance of the lighthouse from the nearer tele-
scope.

9. Range of Guns. A battleship located 1000 yards E 17° 34'
N from a fort in the harbor, sights a belligerent cruiser N 11°. 15' W.

At the same moment word is received by wireless from the
fort that the cruiser bears N 4° 23' E.

For what distance must the guns of both fort and battleship
be ranged, in order to fire on the cruiser.

10. Arch Computation. A circular arch is to be erected to
carry a bridge across a stream 230 feet wide, the other data
being as follows:

Piers, 18 feet high, erected at the water's edge.

133

THE OBLIQUE TRIANGLE

151

Face of piers, eloping backward at an angle of 7° from the
perpendicular.

Distance of springers from face, 15 inches.

Height of crown of intrados above water, 57 feet.
Find: (1) the span,

(2) the rise,

(3) the radius of curvature.

Fio. 68.

11. Width of Stream. A person goes 89 yards up a slope
of 2' 3" in 7.5 feet, directly back from the bank of a river, and
reads the angle of depression, 2\°, of the water's edge on the
opposite side.

Find the width of the river.

12. A Vector. Two forces act on a body at the same point.
One of them equals 341 lbs. and makes an angle of 39.8° with
the resultant which equals 119.6 lbs.

At what angle does the other force act and what does it equal?

13. System of Two Heavy Bodies. In Fig. (
are two weights attached to a rope which i
as shown.

152

TECHNICAL TRIGONOMETRY

123

0=35°,

* = 75°,
Wi =357 lbs.,
TT 2 =4691bs.,

0=32.1.

Compute the tension in the rope by the formula

T =

g WiW^m fl+sin 4 )
Wi+W t

14. Trolley Route. A trolley line is to be built in open, com-
paratively level country, joimng two towns A and B, 40.5 miles
apart, the second town lying N 10° 30' W from the first. The
route is to pass through one of two other towns C and D, located
as follows from A :

C, N 21° E, llf miles,
Z), W 15° N, 19J miles.

Which is the shorter route from A to B and how much?

15. Distance between Two Inaccessible Objects. To deter-

Fig. 70.

mine the distance between A and B, both being across a river,
the following measurements were taken:

CD
ACB
BCD
BDA
CDA

225,
74° 15',
44° 27',
66° 54',
39° 38'.

123 THE OBLIQUE TRIANGLE 153

Solve (1) graphically,

(2) by laws oblique triangle

16. Area of a Parallelogram. The diagonals of a parallel-
ogram are 81 and 106 and cross each other at an angle of 29° 18'.

Find the area of the parallelogram.

17. Diagonal of a Parallelogram. Two of the sides of a parallel-
ogram measure 23.4 and 31.75 and the angle between them is
123° 17.2'.

Determine (1) the lengths of the diagonals; (2) the area.

18. Distance, Course, and Rate of a Ship. From a window
of the Wolf-rock lighthouse 116 J feet above sea-level, a ship
whose angle of depression is 9° 5.6', is sighted N 32° 7.3' W. An

Fig. 71.

hour and fifteen minutes later the ship is E 3° N, and its angle
of depression is 12° 8.2'.

Find: (1) the distance of the ship from the lighthouse at the
time of the first observation,

(2) its course,

(3) its rate of sailing.

19. Dimension Problem in Tool Work.* Given locating hole
L with £| JH at 17° 24' with the 1 . WF is || to JH and .183"
from it.

ST is also given, making an Z of 4° 54' with the horizontal.

Intersection of ST and BF is .209" above center of O.

* Figure, problem, and solution submitted by alumnus Mr. R. F.
Hetzel, S.M.D., '12.
t £. means center line.

154

TECHNICAL TRIGONOMETRY

123

Required to find distance x from intersection of ST and WV
to center line EF.

Draw PK 1 to JH, and AB 1 to EF through intersection A
of ST and WV.

Solve A DOC finding side DC.
Solve A PC# finding side PC.
Subtract DC from PC finding PD.
ZPAD =90°+ A PAZ- Z DAB = 90°+ 17° 24' -4° 54',

Z PAD = 102° 30'.

123

THE OBLIQUE TRIANGLE

155

Solve APAD finding side AD.
Solve AABD finding side AB.

A B=x.

Compute AB also, from a solution formula.

20. Diameter of a Wire. The figure shows a wire in a metric
gage.

Compute the diameter and the size of the wire by Brown and
Sharpe gage when ilf=5 millimeters,. (See Reference tables.)

Fig. 73.

21. Gun on Pedestal Mount. With the dimensions as given
on the figure compute AB, the length of the elevating screw,
when the gun is elevated 15° and when depressed 5°.

Fig. 74 .

156

TECHNICAL TRIGONOMETRY

113

22. Electric Sign, An electric light sign is to be made in the
shape of a regular pentagon. The aide of the pentagon is 12 feet.
How many 55-watt lamps will be required, placed 8" center
to center on the sides of the pentagon and on its diagonals?

How much current will be required
to operate the lamps, on a 110-volt
direct circuit? (See Marsh's " Indus-
trial Mathematics," Chapter XII.)

23. Sheet Metal Computation.
The figure shows a view of a cylin-
drical pipe of diameter D, intersecting
■ a roof.

Compute and fill in the omitted
mtries in the table.

Fig. 75.

Table XXVI
SHEET METAL CALCULATION

Ha.

Pitrh.*

«

D

a

■-■

s

7

e

B

10

1

6

h

E

2

1

4

u

E

3

1
3

h

E

4

2

k

B

5

2

3

h

n

(i

3

4

h

F.

7

1

h

E

J3 THE OBLIQUE TRIANGLE

24. Calculations of Ball-Bearings.* In the figure

D=^", G=.70446",
# = .1612".

•Problem, figures, solution, and Table XXVII submitted by
alumnus Mr. M. J, Hallam, S,M4>., 1909,

158

TECHNICAL TRIGONOMETRY

133

Find dimensions F and T.

(1)

(2)

(3)
(4)
(5)

(1)

(2)
(3)
(4)

-K

b = ( // -'- )cos 20° =.00495 cos 20° = 00465".

sin a = — = ' =38° 20' 30"
// .1612 '

<£=90°-(a+20°) =31° 39' 30",

c=// sin # = .08460,

F = (P+2b)-2c,
= (1.133+2X.00465)-2X.08460 = .9731" .

O .07812 naMiH „

«=90°-(a+20°) =63° 37' 59",
e = .70446 sin <£ = .63117,
7 T =2e+7 = 2X.63117+.1027 =1.36504".

Carefully check the formulas and computations.

25. Bail-Bearing Formulas. By reference to the figure in
Problem 24 derive solution formulas for F and T.

26. Checking B all-Bearing Calculations. Check the T and
F entries in the following table by the formulas of Problem 25.

Table XXVII
BALL BEARINGS

No.

V

D

H

G

O

H

T

P

1

.070

. 79583

.07812

.1612

1.50501

1.1121

2

.228

.81145

. 08593

.1773

1.68567

1.25472

3

.3533

. 93407

. 10156

.2096

2.02894

1.51908

4

.3927

1 . 10235

.11718

.2418

2.37251

1.78414

5

.3931

1.27037

. 13281

.2741

2.67725

2.01014

\

123

THE OBLIQUE TRIANGLE

159

27. Rectification of a Circumference.* To rectify, by deriva-
tion means to make straight. The rectification of a circumference
is therefore the problem of the determination of a straight line
whose length equals the circumference. Since the circumference
and the diameter are incommensurable, exact rectification is
impossible. The following
method is said to give a result
in which the error is not
greater than .00001.

Draw a semicircumference
and its diameter.

Bisect the semicircumfer-
ence at V.

From V draw a tangent
terminating in a tangent from
one extremity of the diameter.

From V as a center, with
a radius equal to the radius
of the semicircumference, de-
scribe an arc cutting the semi-
circumference on the side of
V toward the tangent. (Show
only a short portion of the
arc to represent the point of
intersection.)

Through the point of intersection, draw a line from V ter-
minating in the tangent at S.

From the other extremity of the diameter draw a tangent
to F, in an opposite direction from F, equal to the diameter.

Then SF equals one-half the given circumference.

Construct a full-page figure in the work-book, by the instruc-
tions given.

Measure the length of SF, double the measurement, and by
computation of 27rr, determine the degree of approximation to
the circumference.

F s-X— irr.

Fig. 77.

* Figures and proofs in problems 27, 28, and 29 submitted by
Mr. M. E. Meinecke, A.E., 1909, when a student.

160 TECHNICAL TRIGONOMETRY 123

Proof.

(1) y=r-m,

(2) m=r tan 30°,

(3) .-. y=r(l-tan30°),

(4) z = V±r*+y* = rV4+(l-tan30°)»,

(5) <£=90°+a,

a=t an--=tan-^— j,

(7) .-. ^gO^+tan-^ 1 ^^!),

(8) a; = V4r 2 +,? 2 -4rzcos <£,

(9) /. a =r V4+4+(l -tan 30) 2 -4\/4+(l -tan 30°) 2

(6)

cos

gOM-taa-- 1 -* 811300

)],

(10) a; =rV8+.178634+4V4.178634 cos 78° 4' 3"*,

(11) a-=rV9.906582,

(12) x=3.l47r

(13) /. x±*wr .

Study this proof through carefully, checking every statement,
reading, and computation.

Write it in the work-book with hypothesis, conclusion, and
authority for every statement including the complete transforma-
tions in (9) and (10).

Compute the value of x in (10) with a 6-place table (or greater),
compute 27jt, and determine how closely 2x approximates
2wr+. 00001.

It should be evident that any arc of known magnitude and
radius may be rectified by this method.
Explain how.

28. Rectification of an Arc by Rankine's Method. An arc

of unknown magnitude may be easily rectified by the following
method:

* = This symbol, as here used, denotes '* approximately equals."

123

THE OBLIQUE TRIANGLE

161

Draw any arc, using a large radius so as to make a good figure.

Bisect the chord of the arc.

Produce the chord to A, a distance equal to half its length.

At V, the extremity of the arc through which the chord is
produced, draw a tangent to the arc.

From A as a center with three times the half chord as a radius
describe an arc cutting the tangent in (7, in the direction of the
axe.

ri v R=Y C«=X.

Fig. 78.

Then VC is a trifle less than the length of the given arc, the

1
difference in the rectification of an arc of 60° being — of the

yuu

length of the arc.

For arcs of 60° or under, no distribution of this error need
be made.

Construct a full page figure as directed.

Proof.

9 a 2 a

(1) -a 2 =x 2 + T -2xX„ cos a,
4 4 2

(2) a=2rsin— ,

(3) « = 180°-|,

162

TECHNICAL TRIGONOMETRY

123

(4)

2x4r 2 sin 2 -=x 2 +rx

f2Xsin|xcos|j
=x 2 +x(rsin0),

(5)

r / d> r 2

x = — - sin 4>±<J8r* sin 2 tj+"j sin 2 <j>,

(6) x =r [ -- sin <£ =bsin |^8+cos 2 |J ,

(7) or x -r [sin |-J ^3+sin |) (3 -sin |) -| sin *J ,

Let 4> = 1 Radian =57° 17.747'.

(8) x = r[.47942G V3.479426X2.520574 - .420736],

(9) x =r(1.419795 -.420736),

(10) /. x = .999059r (always smaller).

(11) /. x=y.

Make the same study of this proof and write it, as in Prob-
lem 27.

Determine the degree of accuracy of the method for a 30° arc
of a 3 inch circle.

29. Determination of an Arc of a Circle Equal to a Given

c Straight Line. Draw a circle with
the given line tangent to it at the
point from which the arc is to be
determined, terminating in 0, the
point of tangency and lying on
the same side of as the arc.

From lay off on the tangent a
distance OL equal to one-fourth the
length of the tangent.

From L with a radius equal
to three-fourths the length of the
tangent, cut the circumference at R.

Then RO is the arc required,
within the limits of error specified
in the preceding problem.

Fig. 79.

Construct a full-page figure as directed.

123

THE OBLIQUE TRIANGLE

163

To prove OC = ^ OR, produce OR through to A, making

0A =

OR

From A as a center with A R as a radius describe an arc cutting
the tangent OC. Show that this arc cuts OC at (7, and refer to
the demonstration of the preceding problem.

30. Drafting Problem.*

A D

Fig. 80.

a =23°,

47=2.8",
CJ5=3.1".

In the figure A V and BC are each perpendicular to OB. VD
is perpendicular to OC. It is required to pass a circle through
the points V, D, and C, and to determine its diameter

(a) graphically,

(b) trigonometrically.

For the trigonometric solution draw DC and a perpendicular
form V to CB.

31. Radius of a Crushing Roll. Fig. 81 shows two rolls
for crushing coal or ore.

Draw an enlarged figure in the work-book and formulate R
in terms of T, W, and a

* Figure and problem submitted by alumnus Mr. Leroy Whitcomb,
S.M.D., 1903.

164

TECHNICAL TRIGONOMETRY

123

when R -radius of roll,

T= distance between contact points of material with the

rolls,
JT- distance between crushing faces of rolls,
a -angle with center line, of a radius to outer point of
material contact with face of roll.

If your formula is not in terms of versine, transform it into
a formula for R in terms of vers a.

32. Resultant Field Intensity.*

#i = first field intensity,

#2 = second field intensity,
di = distance first field from conductor,
efo = distance second field from conductor,
7 = current,

Formulate H in terms of di, cfc, and a\

* From Steinmetz' Electrical Engineering.

123

THE OBLIQUE TRIANGLE

165

33. Exact Length of Open Belt. Draw a figure the
full width of the page with the construction and notation
shown in the cut.

Fig. 83.

Observe that

Length of Belt = arc of contact on J+arc of contact on
II+2VT.

Therefore prove:

(1) VT = Vd 2 -(R-r) 2 .

(2) Arc contact on / = #( w+2 sin- 1 * —r-^).

(3) Arc contact on 7/ = r(7r— 2 sin- 1 — -i— J.

(4) Length L =

Prove (2) as follows:

Formulate sin a in terms of R, r, and d.
Prove a = ZDOT, and substitute.

From describe a unit circle with radius r', cutting
OD and OT in x and y.

*This symbol in connection with a trigonometric function is not

an exponent, but denotes the inverse or anti-sine and should be read

R—r
the arc whose sine is — r— .

166 TECHNICAL TRIGONOMETRY 123

Then Z DOT is measured by~^xy ?

.\ sin/^\r?/= sin DOT Def. sin^-sin Unit O

(a) .\ ^xy = sin - l what?

From C=2icR.
and c = 2vr\

s~*xy r'
P rove ^DT = R-

Solve this equation for ^DT, substituting the numerical
value of r'.

Yov^xy substitute from (a) and finish the demon-
stration.

Be sure to number all statements and specify all
authorities.

Work out the demonstration of (3) in the same form,
omitting nothing.

Under the lettered heading, Total Length Open Belt,
write the complete formula in its simplest form.

Formulate the length of an open belt when the pulleys
are the same size.

34. Approximate Length Open Belt. Owing to the
fact that a belt is an elastic connector and will therefore
stretch, the following approximate formula is frequently
used to determine the length of an open belt over pulleys
of different size:

L = H(R+r)+2d.

This formula may always be used when the pulleys

R-r
are in such ratio that — -r- does not exceed .16. Within

a

this limit, the length is slightly in excess, and beyond it

a trifle short.

By the formula compute the approximate length when

d = 18'4",

/* = 9i" f
r = 4f".

123

THE OBLIQUE TEIANGLE

167

35. Actual Length. Using the same data as in problem
34 compute the exact length of an open belt, making no
allowance for crowning.

36. Open Belt Computation. Formulate, compute, and
tabulate the omitted entries as assigned:

Table XXVII
LENGTH OPEN BELTS

XT —

j

T>

R-r

. R-r

Length

No.

a

K

r

d

d

Approx

Exact

1

12'

n

6

2

15'

Hi

5i

3

10' 6"

12

71

4

18'

12i

61

5

11' 6"

10

7

6

8' 3"

iof

51

7

9' 4"

16

9

37. Exact Length Crossed Belt. Draw a large figure
with the construction and notation shown in the cut.

Fig. 84.

Prove:

(1) VT

(2) Arc contact on J

r(t+2 sin- 1 ^).

168

TECHNICAL TRIGONOMETRY

123

(3) Arc contact on II = rlv+2 sin- 1 — t— J.

(4) L = ?

Write the complete demonstration in the same form
as in Problem 33, numbering all statements and specifying
all authorities.

38. Approximate Length Crossed Belt. The following
is a convenient formula for the approximate length of a

R+r
crossed belt on pulleys of unequal size when , is not

greater than .23.

L=3f(ft+r)+2d.

Why is the constant 3|?

Compute the length when the data are the same as in
problem 34.

39. Exact Length Crossed Belt. With data the same
as in problem 34 formulate and compute the exact length
of a crossed belt with an allowance of .065" for crowning.

40. Compute and tabulate the omitted entries as assigned :

Table XXVIII
LENGTH CROSSED BELTS

No.

d.

R.

r.

R +r

d '

. R+r

81D — -z — .

a

Length.

Approx.

Exact.

1
2
3
4
5
6
7
8
9
10

w

10}'
12'

Hi'

10' 8"

15'

16i'

12' 3"

18'

17i'

12"
12"
Hi"
13"
12|"
10.5"
12i"
14"
15"
18"

7"

7i"

6J"

8"

8i"

7f"

8i"
9|"

10"

12i"

i

THE OBLIQUE TRIANGLE

169

41. Die-casting Design.* The figure with the dimen-
sions given, shows a problem in the design of a die casting.
This problem was presented to the designers of a large
manufacturing company with the statement that it could
not be solved from the data. The computation of k and
/ cost the company $ 26.

Draw a large figure showing the three radii and the given
dimensions.

Draw d, e, and k, enter the angles as shown, and write
all the formulas necessary to the determination of h and /.

Compute h and/to the nearest hundred thousandth.

•Problem and drawing submitted by Mr. Edw. J. Utz, S.M.D., '13.

170

TECHNICAL TRIGONOMETRY

128

42. Position of Cross-head. In the figure,

c represents the crank of an engine,
0, the crank center,
r, the connecting rod,
S, a point on the cross-head,
4>, the angle of revolution.

Fig. 86.

Formulate OS in terms of c, r, and <£.

Compute OS
when r=6'4"

c = 15"
and <£=30°.

43. Concrete Standpipe. Fig. 87 represents the trian-
gular corner of a building with a concrete standpipe of

Fig. 87.

the same thickness as the wall of the building and having
an outside diameter of 19 feet.

123

THE OBLIQUE TRIANGLE

171

C, Ci, D, and D\ denote the center of steel rods, C and
Ci being at the outside circumference of the standpipe,
and D and D\ being in the center of the walls as shown.
The walls of both building and standpipe are 15 inches
thick and tf> = 36° 30".

Required the distance from C to the corner of the build-
ing, A.

44. Zone of a Circle. Fig. 88 shows a belt or zone of

Fig. 88.

a circle, formed by two parallel chords. Derive a solution
formula for the area of the belt when a and c are known.
Compute the area

— *i"

when c = 2f", and a = 5^

Problem submitted by alumnus Mr. Neal C. Pike, A.C., '11.

45. Sewer Construction. Figure 89 shows the center
line, £, of a sewer in two streets which deflect through an
angle of 84° 42' 4" at the curved corner. It is required
to join the sewers with the arc of a circle having the same
central angle as the curve joining the house lines.

Since the center line of the sewer in Aqueduct Avenue
is 28 feet from the house line, and in Tremont Avenue 40
feet from the house line, it will be evident that the required

172

TECHNICAL TRIGONOMETRY

123

radius R x , is shorter than the Tremont Avenue radius Rt,
and that, therefore, the new center is at some distance c,
radially, toward Tremont Avenue from the old center. The
perpendicular distance of the center of the common curve
required, from the Aqueduct Avenue radius R a , is denoted
on the figure by h.

Center Line of Sewer

Tremont Avenue ~&

House Line

a-84°4sV l

Old Center

Required :

(1) R x the radius of the common curve joining the
pipe centers,

(2) The distance h,

(3) The distance b,

(4) The length of the curved portion of the sewer.

To derive a formula for (1), formulate R x in terms of
Rt and c; also in terms of R a and b. Equate the values,
and for c and b substitute in terms of h and a.

Problem submitted by alumnus Mr. Emil Hohn, A.E., '01.

133

THE OBLIQUE TRIANGLE

173

46. Sewer and Concrete Section. In Fig. 90 is shown
a cross-section of the same sewer as in Problem 45. Com-
pute the sectional area both of sewer and concrete for a
sewer 29" by 40", taking the other dimensions for this size
from Table XXIX.

47. Sewer Computations. The following table is from
the book of Standard Details of Construction, issued by
the President of the Borough of the Bronx, New York City,
1913.

Table XXIX
PIPE AND CONCRETE SEWER SECTIONS

in

u

i!

IfKaj

O

5
1*

3
1*=

Offset.

u

A.

B.

2fl"X40"

ft"

4' ft"

1' 111"

2' 101"

7*'

5A"

2' 3("

12,82

32"x44"

ft"

ft I)"

2' 1"

3' OA"

7*"

ft*"

2' 51"

14.00

34"X46"

8"

.y »"

2' 2"

3' 2"

9"

8"

2' 6H"

14.78

38"X50"

8"

6' ll"

2' 7"

3' 2 A"

9"

fi*"

2' 8A"

19.08

40"X53"

H"

tv r

2' 8"

3' 4H"

9"

<i"

2' 10A"

20.33

42"X56"

X"

B'6"

2' 9"

3'»i"

12"

tii"

3' 0j"

21.43

By the formulas of Problem 40 determine whether the
tabulated areas are correct, and compute sectional pipe

174 TECHNICAL TRIGONOMETRY 1»

48. Roller Bearings. The inner diameter of the roller
bearing in Fig. 91- is 1$ inches. The clearance between
the rollers and the outside of the race is .003".

Compute

(1) diameter of rollers,

( 2) outside diameter of race.

123

THE OBLIQUE TRIANGLE

175

49. Offset Pipe-bend. Fig. 92 is a sketch of an offset
bend reproduced from a customer's print for transmission
to the shop. The bend is made from lap-welded steel
pipe with steel flanges welded on each end and the latter
machine-finished all over before bending. It is of course

i— m ■ m ■ m » — ■ ■■ t- -r -i -i — — .,

3E-

Fig. 92.

necessary to know the amount of pipe required for
the job, before the flanges are finished, and in order
to make the computation the center line is assumed
as constant during bending, which is very nearly the
fact.

Fig. 93 is a working diagram from which to calculate

176

TECHNICAL TRIGONOMETEY

123

the length. The pipe is finished to this dimension on a
lathe and then given to the bender, who heats such por-
tions of it as are necessary, straps it to a table and bends
it with the aid of an engine, rope, tackle, etc. This is no
job for a novice, and is always done by an experienced and

-f f-TT-

_A ±& :l

Fig. 93.

high-priced man. The pipe will approximate closely the
curve shown, and the over-all length between flanges
(16' 10J"), will usually be within ^" to J".

(1) Draw both figures to scale in the work-book, as
large as the page permits, and on Fig. 93 fill in the omitted
dimensions d, c, a, and b.

(2) Formulate and compute 4>, AE, 0, FH, 0, <~*QF,
and total length OV as sent to the bender.

50. Offset Bend. In the preceding problem the length of
an offset bend was determined from a given radius. This
problem presents a pipe-bend in which it is required to
join the two pipes by a reversed curve as illustrated in Fig.
94, having only the length L, the offset c, and the straight
ends a and b.

123

THE OBLIQUE TRIANGLE

177

Draw a large figure and a perpendicular bisector of the
chord FB, terminating in the perpendicular BD at 0.

Formulate and compute the radius OB; also the total
length of the bent piece on the center line SFT

when

and

a
b
c
L

2' 2",
1' 7",
3' 5",
11' 9".

Fig. 94.

51. Offset Bend. Fig. 95 illustrates another type of
pipe-bend in which the straight portion of the pipe at the
center is 12 inches long and is bent at an angle of 45° with
the horizontal. Draw an enlarged figure to scale, locating
first DE and FG, and the straight center portion ST.

At S and T draw perpendiculars terminating in DE
and FG as shown.

Describe arcs and draw the straight ends.

Formulate and compute R, the total length from flange
to flange, and the offset d. (Take a, b, and L from Prob. 50.)

52. Pipe-bend. Fig. 96 shows a pipe center which
enters a room at V with a straight length of 2' 3".

178

TECHNICAL TRIGONOMETRY

123

Fig. 95.

Fig. 96.

123

THE OBLIQUE TRIANGLE

179

Draw Fig. 96 to an enlarged scale and from the dimen-
sions given formulate and compute 8, 6, and the total length
from to V.

53. Pipe-bend. A pipe-bend in which three different
radii are used is shown in Fig. 97.

Fig. 97.

Formulate and compute 0, a, the three arcs, and the
total length from flange to flange.

180

TECHNICAL TRIGONOMETRY

123

64. Pipe-bend. Fig. 98 shows the center line of a
pipe-bend whose curved portion is three arcs of the same
radius.

TT»

Fig. 98.

Derive a solution formula for the length from flange to
flange. Compute the length from the dimensions given.

55. Pipe-bend. From the dimensions given in Fig. 99
formulate and determine the angles a, 8, and therefore <f>.

§:i-A'«__L

r<-

i

Fig. 99.

Also formulate and determine p and X, and therefore 0.
Compute the total length of the piece.

123

THE OBLIQUE TRIANGLE

181

56. Pipe-bend. In the piece shown in Fig. 100 the
straight ends are of equal length.

Determine a and b, and the length of the piece as
dimensioned.

Fig. 100.

57. Pipe-bend. The problem illustrated in Fig. 101
was first solved by a man who had a motto " Any figure
which can be laid out exact can be figured exact, but
if construction is approximate the calculation will be,
also."

Carefully make the construction, and work through the
solution here given, and determine whether it is correct.

If possible, simplify it by deriving solution formulas
and performing no unnecessary computations.

In working the problem use Greek letters for angles
as far as possible, number all equations, and specify all
authorities.

182

TECHNICAL TRIGONOMETRY

123

Construction.

Lay off LK=W 5 J".

KA=T\0\ ±LK.

AB=U" _L KA.

BE at a slope of 2\" in 12"(i.e. ED = 2*", BD = 12").

BH =2' 10" JL BE.

HC± BH at H.

r — 1 * : — **

_ Ai_ -J.^81<*P®*W ** !*"

T I I I jL^" — — -»

Fig. 101.

Describe arc KST with 2' 10"R.

From F with # = 5' 8" describe arc to cut CH, and from

intersection describe arc TJP, R = 2' 10".
Draw FB, FH, FC, etc.

(/. Figure can be laid out exact.)

123 THE OBLIQUE TRIANGLE 183

Solution.

Slope = 2^" in 12".

.-. tan Ztffl£=|j=. 20833.

.'. ZEBD = 11°46'.

Now Z EBH =90°
:. ZDBH=90°- ZEBD=Q0°- 11° 46'=78° 14'.

tan ZABF =4?=^? =4.303571.

AB 14

.-. ZABF =76° 55'.

ZABD=ABF+FBD
.'. ZFBD=90 o -76°55' = 13°5'.

ZFBH = ZFBD+ZDBH = 13° 5'+78° 14'=91° 19'.

BF 2 =AB 2 +AF 2 = 196+3630.0625.
.•. BF = V3826.0625 = 61 . 8551.

In A FBH

HF=Vb~H 2 +BF 2 -2BHxBF cos ZFBH

= V34 2 +61.8551 2 +2X34X61.8551X.022948

= V4982.0625+96.5311 = V5078.5936.

A HF= 71.264.

sin ZBHF = BF

sin ZHBF HF

. ,-,„_, 61.8551 X. 999736 Qcni .

.". sin ZBHF= _, noA =.86914.

71.264

.-. ZBHT = 60°22'.

ZCHF = 90°- ZBHF =90° -60° 22' = 29° 38'.

smZHCF HF

sin ZCHF FC

. „__, 71. 264 X. 494448 _ 1QO
/. sin HCF= ^s =.5182.

OO

.-. Z#CF= 148° 47'.
L CFH = 180° - ( Z F HC + Z HCF)

(178° 250

= 180°-(29° 38'+148° 47') = 1° 35'.

184 TECHNICAL TRIGONOMETRY 123

ZHFB = 18Q°-(ZFBH+ ZBHF)

(151° 410

= 180°- (91° 19'+60° 220=28° 19'.
ZAFB = 90 o -ZABF=90°-76 o 55' = 13°5'.
Z AFC = (sum of preceding) =42° 59'.
ZSFT = 90° -AFC = 90° -42° 59'=47° 1'.
ZTFK = 90°+ ZSFT =90° +4:7° l' = 137° 1'.

Now Z PCF= Z PCJ+ Z JCF
But Z PC J = Z EBD = 11° 46'
And Z JCF =99- Z AFC=47° 1'
/. Z PCF = 11° 46'+47° l' = 58°47'.

tvt jin FCXsmZHFC 68X.027631

Now #C = . svjj t, — = — ahaaao — = 3.7578.

sin Z FHC .494448

/. BP=HC= 3.7578.

Circumference O 2' 10" R = 213.628.

., arcPJr = 58.78333X213.628 = 34r

137.02X213.628
360

And arc TSK = ™' ^^ iJ '"° = 82|".

Center line = £P+arc PJT+arc TSK+KL.

= 3f"+34£"+82|"+3' 5f " = 13' 6£".

Problem, drawing, and solution submitted by alumnus Mr. A. A.
Fraser, A.E., '01, who furnished all pipe-bend problems with the
exception of Problems 50 and 51.

123

THE OBLIQUE TRIANGLE

185

68. Walk Computation. The figure shows a plot of
ground measuring 100' X 200', with a walk 25 feet wide
diagonally across it.

Required the distance x.

Ffg. 102.

Suggestions.

tan a = what?

(1) Solve for x.

cos 0=what?

(2) Solve for x.

Prove d=a and substitute in (2).
Eliminate x in (1) and (2) and simplify.
But tan a equals the ratio of what other functions
of x?

Substitute and solve for sin a.
Finish the problem.

59. Determination of a Very Small Angle. In trigo-
nometric computations it is sometimes necessary to deter-
mine a more exact value of a very small angle than is
possible by usual methods. The figure shows a right
triangle in which b and c are known, and A supposedly
very small is required.

186 TECHNICAL TRIGONOMETRY 128

Suggestions.

Bisect A, making segments t and s.

From the figure formulate the theorem that the bisector
of an angle of a triangle divides the opposite side into segments
proportional to the respectively adjacent sides.

Take the proportion by composition and by alternation.

Substitute for t+s in terms of c and b.

Introduce c+b under the radical and simplify.

Prove

tan

A_ /c^b
2 \c+b

Fig. 103.

60. An .Old Survey. Plot the following survey, com-
pute the exact acreage and the reduction in the tillage
area by a road 12 feet wide diagonally through it as illus-
trated in Fig. 102.

" Beginning at the northeasterly corner of lot No. 91,
thence south 30 chains and 48 links; thence east 32 chains
and 81 links to a hemlock stake marked and numbered
92, 93, 100, and 101; thence north 30 chains and 48 links
to a small beach marked and numbered 84, 85, 92, and
93; thence west 32 chains and 81 links to the place of
beginning containing 100 acres of land more or less, as per
John Burns' survey thereof made in the year 1776."

CHAPTER IV
RELATION OF FUNCTIONS

Section 1, Functions of Compound Angles. Section 2,

Functions op Multiple Angles.

§1. FUNCTIONS OF COMPOUND ANGLES

124. Definition and Formulas. Functions of compound
angles are functions of the algebraic sum of two or more
angles.

The functions discussed in this chapter are of value
both in the subsequent/ study of mathematics and in its
applications.

The functions of compound angles will be presented
in the following order, a and p being any two angles,

and 0=a—p.

1. sin (a+P) =sin a cos 0+cos a sin p.

2. cos (a+P) = cos a cos 0— sin a sin p.

o +or , / , „x tan «+tan p

3. tan {a+P) =z — 7 1 jr.

v ' 1— tana tan p

4. sin (a— p) =sin a cos p — cos a sin p.

5. cos (a— P) =cos a cos j8+sin a sin 0.

« 4. ( o\ tan a -tang

6. tan [pL—P) = r-T-7 1 x-.

1+tanatang

7. cot (a+j8)= — 7 — , , Q .

v cot a+cot /3

, . cota cot P+l

8. cot (a—P)= —5 7 — .

cot p — cot a

187

188 TECHNICAL TRIGONOMETRY 125

9. sin (a+0)+sin (a— p) =2 sin a cos 0.

10. sin (a+p)— sin (a— 0) =2 cos a sin 0.

11. cos (a+0)+cos (a— j8) =2 cos a cos 0.

12. cos (a— /3) — cos (a+/3) =2 sin a sin 0.

13. sin (a+j8) sin (a— j3) =sin 2 a— sin 2 0.

14. cos (a+jS) cos (a—jS) =cos 2 a— sin 2 0.

15. sin 0+sin 0=2 sin $(tf>+0) sin Ktf>-0)-

16. sin <£-sin = 2 cos \(<t>+B) sin £(tf>-0)-

17. cos <£+cos 0=2 cos §(#+0) cos \(<l>— 0).

18. cos 0— cos = -2 sin Ktf>+0) sin §(<£-0).

19. tana+tanj8 = -.

cos a cos p

OA 4. * * Sil1 (<* — P)

20. tan a — tan = -.

cos a cos j8

21. l-tau«tanfl= Cos(a+ fl

COS a COS /3

™ * , a x o cos (a— /3)

22. l+tanatan0 = ^-.

cos a cos p

23.

24.

25.

sin (g+ft) _ tan a+tan ft
sin (a—p) tan a— tan 0"

cos (a+j8) _ 1 — tana tan ft
cos (a— j8) 1+tan « tan ft '

tan|(<ft+0) _ sin <ft+sin
tan 5 (<£ — 0) sin <j> — sin 0"

125. sin (a+ft). Sine of the Sum of Two Angles.

The sine of the sum of two angles equals the sine of the first
into the cosine of the second, plus the cosine of the first into
the sine of the second.

Draw a large figure and write hypothesis and conclu-
sion.

125 RELATION OF FUNCTIONS 189

From some point N in OV draw NA perpendicular
to OQ.

From the same point N draw NM perpendicular to OS.

To which of the given angles, is the angle between the
perpendiculars equal? Denote on the figure by the same
symbol as used for that given angle.

Draw MB perpendicular to OQ, and MF perpendicular
to NA.

Fig. 104.

By definition, sin (a+/3) equals what ratio?
For NA substitute in terms of MB and NF and express
the second member as the sum of two fractions.
Group as follows:

. ,, , m MB .NF

sin (A+B)= 77T7+

ON ' ON'

By reference to the conclusion and the figure determine
by what quantity the first fraction must be both multi-
plied and divided, and by what quantity the second frac-
tion must be both multiplied and divided. Finish the
demonstration.

By the use of the tables make a test of the formula
when a = 28° 35'

and 0=47° 50' 32".

190 TECHNICAL TRIGONOMETRY 129

126. cos (a+0). Cosine of the Sum of Two Angles.

The cosine of the sum of two angles equals the cosine of the
first into the cosine of the second minus the sine of the first
into the sine of the second.

Demonstrate.

Make a test of the formula

when a = 17° 29' 23"

and = 54° 40' 38"

127. tan (a+p). Tangent of the Sum of Two Angles.

The tangent of the sum of two angles equals the fraction whose
numerator is the tangent of the first plus the tangent of the
second and whose denominator is unity minus the tangent
of the first into the tangent of the second.

Write the value of tan (a+P) in terms of sin (a+P)
and cos (a+P).

Substitute from the formulas obtained in paragraphs 125
and 126, and divide both numerator and denominator of
the second member by the first term of the denominator.

Test the formula, using the same values of the angles
as in paragraph 126.

128. sin (a— p). Sine of the Difference of Two Angles.

The sine of the difference of two angles equals the sine of the
first into the cosine of the second minus the cosine of the first
into the sine of the second.

From N, some point in OV, draw NT±OQ and
NM±OV, terminating in OS at M. (See Fig. 105.)

Through M draw a line || NT, terminating in OQ at L.

From N draw a perpendicular terminating in ML
at R.

On the figure denote by a the angle which is equal to a.

131 EELATION OF FUNCTIONS 191

Demonstrate.

Test the formula, using the same angles as in para-
graph 126.

Fig. 105.

120. cos (a— 0). Cosine of the Difference of Two
Angles* The cosine of the difference of two angles equals
the cosine of the first into the cosine of the second plus the
sine of the first into the sine of the second.

Demonstrate.

Test with the same angles as in the preceding paragraph.

130. tan (a— j3). Tangent of the Difference of Two
Angles. The tangent of the difference of two angles equals
the fraction whose numerator is the tangent of the first minus
the tangent of the second, and whose denominator is unity
plus the tangent of the first into the tangent of the second.

Demonstrate and test with any two angles.

131. cot (a+0). Cotangent of the Sum of Two
Angles. The cotangent of the sum of two angles equals the
fraction whose numerator is the cotangent of the first into the
cotangent of the second, minus unity, and whose denominator
is the cotangent of the second plus the cotangent of the first.

Demonstrate and test
when a = 41° 39' 15"

and 0=33° 42'.

192 TECHNICAL TRIGONOMETRY 132

132. cot (a—p). Cotangent of the Difference of Two
Angles. The cotangent of the difference of two angles equals
the fraction whose numerator is the cotangent of the first into
the cotangent of the second, plus unity, and whose denomina-
tor is the cotangent of the second minus the cotangent of the
first.

Demonstrate and test with the angles of paragraph 131.

133. sin (a+j8)+sin (a-0). Sine of Sum Plus Sine
of Difference. The sine of the sum of two angles plus the
sine of the difference equals twice the sine of the first into the
cosine of the second.

Demonstrate and test with any two angles.

134. sin (A+B) -sin (A-B). Sine of Sum Minus Sine
of Difference. The sine of the sum of two angles minus the
sine of the difference equals twice the cosine of the first into
the sine of the second.

Demonstrate and test.

136. cos (A+B)+cos (A-B). Cosine of Sum Plus
Cosine of Difference. The cosine of the sum of two angles
plus the cosine of the difference equals twice the cosine of
the first into the cosine of the second.

Demonstrate and test.

136. cos (A— B)— cos (A+B). Cosine of Difference
Minus Cosine of Sum. The cosine of the difference of two
angles minus the cosine of the sum equals twice the sine of the
first into the sine of the second.

Demonstrate and test.

137. sin (A+B) sin (A-B). Sine of Sum Times Sine
of Difference. The sine of the sum of two angles into the
sine of the difference equals the cosine square of the second
minus the cosine square of the first.

Demonstrate and test.

140 RELATION OF FUNCTIONS 193

138. cos (A+B) cos (A-B). Cosine of Sum Times
Cosine of Difference. The cosine of the sum of two angles
into the cosine of the difference equals the cosine square of the
first minus the sine square of the second.

Demonstrate and test.

139. sin 4>+sin 0. Sum of the Sines of Two Angles.

The sum of the sines of two angles equals twice the sine of
half the sum of the angles into the cosine of half the difference
of the angles.

In the formula of paragraph 133, substitute 4> for a+fi
and for a— p.

Solve a+P = <t>,

and a— =

for a and 0, and substitute.

140. sin <j>— sin 0. Difference of the Sines of Two
Angles. The difference of the sines of two angles equals twice
the cosine of half the sum of the angles into the sine of half
the difference of the angles.

Demonstrate by the same method as in paragraph 139.

Demonstrate also by the following method:

Draw any angle <£, for convenience about 60°, having
sides not less than three inches long. Denote vertex by 0.

With one of these sides as a radius describe a quadrant
arc.

Mark <j> on the intercepted arc AB, named in a counter-
clockwise direction.

From the side terminating in A, lay off in a counter-
clockwise direction an angle 0, for convenience about 15°,
intersecting the arc at C.

What angle in the figure equals <£— 0?

Bisect this angle, producing the bisector to the arc.

194 TECHNICAL TRIGONOMETRY HI

Mark —^— on the angle whose side passes through B.

Draw the chord BC cutting the bisector in D.
Prove BC perpendicular to the bisector.
In the two simultaneous equations

4>+6 = s*

and 4>—d=d*

eliminate and solve for — .

£t ■

A I a

Denote the result on the figure by — «— .

To the common side of angles 4> and 0, draw the per-
pendiculars BR, DS, and CT.

From D draw a parallel to the side whose extremity is A,
terminating in BR at V.

From C draw a parallel to DV, terminating in DS.

Formulate sin <j> and sin 6.

To what therefore is their difference equal?

Reduce the second member to the form

2^ F

OB'

Finish the demonstration.
Test, using any angles.

141. cos <t>+cos 6. Sum of the Cosines of Two Angles.

The sum of the cosines of two angles equals twice the cosine
of half the sum of the angles into the cosine of half the difference
of the angles.

Demonstrate and test.

* s denotes the sum of the two angles, d denotes their difference.

146 RELATION OF FUNCTIONS 195

142. cos <t>— cos 0. Difference of the Cosines of Two
Angles. The difference of the cosines of two angles equals
minus twice the sine of half the sum of the angles into the sine
of half the difference.

Demonstrate and test.

143. tan a+tan 0. Sum of the Tangents of Two
Angles. The sum of the tangents of two angles equals the
sine of the sum of the angles, over the cosine of the first into
the cosine of the second.

Use the formula of paragraph 125.

Test with any angles. v

144. tan a— tan 0. Difference of the Tangents of Two
Angles. The difference of the tangents of two angles equals
the sine of the difference of the angles, over the cosine of the
first into the cosine of the second.

Use the formula of paragraph 128.
Test with any two angles.

145. 1— tan a tan (3. One minus the Product of the
Tangents. One minus the product of the tangents of two
angles equals the cosine of their sum, over the cosine of the
first into the cosine of the second.

Demonstrate and test

when a = 47°10'28"

and = 15° 36'.

146. 1+tan a tan j3. One Plus the Product of the
Tangents. One plus the product of the tangents of two
angles equals the cosine of their difference, over the cosine
of the first into the cosine of the second.

Demonstrate and test
when a = 47° 10' 28"

and j3 = 68°22'45".

196 TECHNICAL TRIGONOMETRY 147

sin (a+p)

147. — — -> Sine Sum over Sine Difference. The

sin (a— f$)

sine of the sum of two angles over the sine of the difference
equals the suni of the tangents of the angles, over the difference
of the tangents of the angles.

Demonstrate and test
when a = 31° 18' 44"

and = 83° 36' 15".

COS (a+<8)

148. -. Cosine Sum over Cosine Difference.

COS (a — P)
The cosine of the sum of two angles over the cosine of the dif-
ference equals one minus the product of the tangents of the
angles , all over one plus the product of the tangents.

Demonstrate and test
when a = 56° 42' 10"

and = 63° 25'.

tan

2

149. -. Tangent Half Sum over Tangent Half

<f) — u

tan— —
2

Difference. The tangent of half the sum of two angles over

the tangent of half their difference equals the sum of their sines

all over the difference of their sines.

Demonstrate and test
when = 30°

and = 50° 40' 30".

151 RELATION OF FUNCTIONS 197

§2. FUNCTIONS OF MULTIPLE ANGLES

150. Definition and Formulas. A multiple angle is a

product obtained by multiplying an angle by any number.
Thus, if the angle is 6,

26, 30, \B*, etc.

are multiple angles of 0.

The functions of multiple angles will be presented in
the following order:

1. sin 2a = 2 sin a — cos a.

2. cos 2a = cos 2 a — sin 2 a.

= 2 cos 2 a — 1
= 1 — 2 sin 2 a.

2 tan a

3. tan2a =

1 — tan 2 a'

4. sin - = ^2""2 COSa *

5. cos 2 = \2 5 ° 0S ""

6. tan 1=^1+-^-^.

151. sin 2a. Sine of Twice an Angle. The sine of
twice an angle equals twice the sine of the angle minus the
cosine of the angle.

* Fractions of angles are sometimes called submultiple angles.

198 TECHNICAL TRIGONOMETRY 152

Demonstrate by the same figure and formula as in para-
graph 125, and let p=a.

Test the formula
when a = 73° 18' 29".

152. cos 2a. Cosine of Twice an Angle. The cosine
of twice an angle equals the square of the cosine of the angle
minus the square of the sine of the angle.

Formulate cos (a+j8), let j3=a, and substitute.

But sin 2 a+cos 2 a = what?

By reference to the conclusion as formulated in para-
graph 150, determine the substitutions required to obtain
the formulas given for cos 2a.

Test all three formulas

when a = 31° 43'.

153. tan 2a. Tangent of Twice an Angle. The tangent
of twice an angle equals twice the tangent of the angle divided
by one minus the square of the tangent of the angle.

Demonstrate and test when a has the same values
as in paragraphs 151 and 152.

154. sin -. Sine of Half an Angle. The sine of half

an angle equals the square root cf one-half the difference

between unity and the cosine of the angle.

8
In the second formula of paragraph 150 let a=-.

Substitute - for a, and for cos 2 in terms of sin 2 .

166. cos -. Cosine of Half an Angle. The cosine of

half an angle equals the square root of one-half the sum of
unity and the cosine of the angle.

156 EELATION OF FUNCTIONS 199

The demonstration may be determined from the pre-
ceding paragraph.

8
156. tan -. Tangent of Half an Angle. The tangent

of half an angle equals the square root of the sum of unity
and the reciprocal of the square of the tangent of the angle,
minus the reciprocal of the tangent of the angle.

In the third formula of paragraph 150, substitute ~

5
for a and solve for tan ~.

CHAPTER V

USE OF THE SLIDE-RULE*

Section 1, Scales. Section 2, Sines and Tangents. Section 3,
Powers and Roots. Section 4, Multiplication. Section
5, Division. Section 6, Reciprocals. Section 7, Trig-
onometric Computation.

§ 1. SCALES

157. Description. A slide-rule is so named from the
fact that one piece of the rule slides in the grooves of the
fixed outside pieces.

The lowest priced rules are known as plain Mannheim
rules, so called because the scale arrangement is the
same as was devised in 1850 by Lieutenant Mannheim of
the French Army. The higher priced rules have modifica-
tions of the plain Mannheim arrangement, all of which are
designed to simplify the use of the rule by lessening the
slide and runner movement.

Therefore while it is a fact that all computations which
are possible on any slide-rule can be made on a plain
Mannheim, it is also the fact that many of the ordinary
trigonometric computations can be made in less time,
with fewer settings, and more accurately, on rules having
one or more inverted scales and which instead of being
channeled are made of three pieces with scales on both

* For a more comprehensive treatment of the slide-rule with
photographic reproductions, diagrams, and explanations of all settings,
and numerous examples, see Marsh's Technical Algebra, Part I, pp.
279-349.

200

158

USE OF THE SLIDE-RULE

201

sides. Rules thus made are called duplex because of the
two faces.

Reference has just been made to an inverted scale.
This is a scale graduated from right to left, while a direct
scale is graduated from left to right.

158. Graduations. A study of the following table will
help to explain how the scales on a slide-rule are graduated.

No.

1

2

3

4

5

6

7

8

9

10

log.

.301

.477

.602

.699

.778

.845

C03

.954

1.0

The upper line shows the integers from 1 to 10 in
succession. The lower line contains their respective loga-
rithms. Observe that while the difference between any
two successive numbers in the upper line is unity, in the
lower line the difference continually decreases. Thus the
difference between the logarithm of 3 and the logarithm
of 2 is .176, but between the logarithm of 9 and the
logarithm of 8 the difference is only .051.

An equal parts scale is a scale whose graduations are
equally spaced.

A logarithmic scale is a scale whose graduations are
proportional to the logarithms of numbers on an equal
parts scale, and therefore exhibit a decreasing interval
along the scale.

A plain Mannheim rule is distinguished by having
four direct, logarithmic scales on its face, and no more, —
two on the stock or fixed part of the rule and two on the
slide or movable part. These four scales in order are called
A, B,C, and D. A and B are exactly alike and are mounted
twice in the length of the rule. C and D are exactly alike
and are mounted once in the length of the rule. The
points where a logarithmic scale begins and ends are called
indexes. A and B therefore have left, center, and right

202 TECHNICAL TRIGONOMETRY 159

indexes. C and D have left and right indexes only. As
the graduations are logarithmic, the interval between them
continually decreases along the rule, with the result that
within a short distance it is necessary to change the unit
of graduation to avoid crowding. Thus on A and B, there
are the following units:

From 1 to 2, .02;

2 to 5, .05;
5 to 10, .1.

On C and D there are the following units:

From 1 to 2, .01 ;

2 to 4, .02;
4 to 10, .05.

On S, the sine scale, there are the following units:

From 1° to 10°, 5';
10° to 20°, 10';
20° to 40°, 30';
40° to 70°, 1°;
70° to 80°, 2°.

On T, the tangent scale, there are the following units:

From 5° 42' 38" to 20°, 5';

20° to 45°, 10'.

169. Readings. Following are summarized instructions
for the various slide-rule readings in the computations of
trigonometry which involve

Sines, ' Cube Roots,

Tangents, Logarithms,

Squares, Multiplication,

Square Roots, Division,

Cubes, Reciprocals.

162 USE OF THE SLIDE-RULE 203

§ 2. SINES AND TANGENTS

160. Sines. On a Mannheim rule, a sine is read as
follows:

Unreversed Slide.

(a) Turn the rule over and move the slide so as
to place the given angle on the S scale, under the index
mark on the upper edge of the notch at the right end of
the rule, or under the index mark on the transparent plate
if the rule is so made.

(b) Reverse the rule and move the runner to the right
A index.

(c) Read the sine under the runner, on the B scale.

Reversed Slide.

(a) Reverse the slide.

(b) Align indexes.

(c) Move runner to the given angle on 8.

(d) Read sine on A under the runner.

161. Pointing Off a Sine. Place point cipher before
all sine readings on left B or left A; place point only, before
all sine readings on right B or right A except sin 90°, which
equals 1.

162. Tangents. (1) Angles Less Than 5° 42' 38".
As may be verified by a table of tangents, the tangents of
angles less than 5° 42' 38" have the same value to three
places as the sines. They are therefore read with the S
and B, or S and A scales.

(2) Angles between 5° 42' 38" and 45°. Tangents of
angles within the range of the tangent scale are read as
follows:

204 TECHNICAL TRIGONOMETRY 162

Unreversed Slide

(a) Turn the rule over and move the slide so as to
place the given angle on T under the index mark on the
transparent plate, or on the lower edge of the notch in the
left end of the rule.

(6) Turn the rule over and move the runner to the
left D index.

(c) Under the runner read the tangent on C

Reversed Slide

(a) Reverse the slide.
(6) Align indexes.

(c) Move runner to the given angle on T.

(d) Read tangent on D under the runner.

(3) Acute Angles Greater than 45°.

Reversed Slide

(a) Reverse the slide.

(6) Move runner to the D index.

(c) Move slide so as to place under the runner, 90°
minus the given angle on T.

(d) Move runner to slide index.

(e) Read tangent on D under the runner.

Inverted Slide

(a) Invert * the slide.

(6) Align indexes.

(c) Move runner to 90° minus the given angle on T.

(d) Read tangent given angle on D under the runner.

* Invert means reverse and insert with the ends interchanged.

166 USE OF THE SLIDE-RULE 205

(4) Obtuse Angles. The tangent of an obtuse angle
equals minus the tangent of the supplement. Therefore

(a) Subtract the given angle from 180°.

(6) Read tangent of the resulting acute angle as specified
in (2) or (3), and place minus before the reading.

Tangent on a Duplex

(a) Align indexes.

(6) If angle is acute and less than 5° 42' 38" move
runner to angle on S and read sine and therefore tangent
on A under the runner.

(c) If angle is acute and between 5° 42' 38" and 45° move
runner to angle on Tand read tangent on D under the runner.

(d) If angle is acute and between 45° and 80°, move
runner to 90° minus given angle on T and read tangent
on CI under the runner.

163. Pointing Off a Tangent. Place point directly
before all tangent readings of angles from 5° 42' 38" to
45° except tan 45° which equals 1; place point immedi-
ately following the first figure of all tangent readings of
angles from 45° to 80°.

164. Summary of Tangent Readings. Slide Unreversed.
Angles Less Than 5° 42' 38" : Read tangent as a sine, on

B under the right A index.
Angles from 5° 42' 38" to 45°: Read tangent on C, over

the left D index.
Angles from 45° to 80°: Set on T, 90° minus the angle;

read tangent on D, under the right C index.
Angles from 80° to 90° : Read in a table of tangents.
Obtuse Angles. Read the tangent of the supplement.
Place minus before the reading.

165. Integral Figures. In all slide-rule computation,
results are pointed off by inspection when possible. It is

i
■i

206 TECHNICAL TRIGONOMETRY 166

frequently impossible and since the number of figures which
can be determined is usually limited to the first three or
four, the arithmetical rules for pointing off from the number
of decimal places involved, cannot be employed.

Results are therefore pointed off from the number of
integral figures in the numbers used.

The integral figures of all numbers are determined as
follows:

(1) Number Greater Than 1. The number of integral
figures in a number greater than unity equals the number
of figures which precede the decimal point.

(2) Number Less Than 1. The number of integral figures
in a decimal equals minus the number of ciphers between the
decimal point and the first significant figure of the decimal.

Thus 12.72 has 2 integral figures;

.6 has zero integral figures;

.0296 has— 1 integral figures;
.00348 has— 2 integral figures.
.000092 has— 4 integral figures.

§ 3. POWERS AND ROOTS

166. Squares. The square of a number is read with
no movement of the slide, as follows :

(1) Move the runner to the number on the D scale.

(2) Read the square on the A scale under the runner.

The explanation of the reading of a square is simple.
Since the 1 to 10 graduation of the A scale is exactly half
the length of the 1 to 10 graduation of the D scale, any
length under the runner on A from the left index repre-
sents twice the logarithm of the aligned number on D.
The graduation reading at the end of the length on A is
therefore the square of the aligned graduation reading
on D.

169

USE OF THE SLIDE-RULE

207

167. Pointing Off a Square. (1) The number of integral
figures in a square on left A is one less than twice the number
of integral figures in the given number.

(2) The number of integral figures in a square on right
A is twice the number of integral figures in the given
number.

If n denotes the number of integral figures in the number
whose square is wanted, the rule may be tabulated as
follows:

SQUARES

Square.

Integral Figures.

On left A
On right A

2n-l
2n

168. Square Roots. The square root of a number may-
be read on the slide-rule as follows:

(1) Separate the given number or consider it to be
separated into periods of two figures each in both directions
from the decimal point.

(2) If the first period at the left has one significant
figure, set the runner to the number on left A; if the first
period has two significant figures, set the runner to the
number on right A.

(3) Read the square root under the runner on D.

169. Pointing Off. First Method.

Each period in the number gives one figure in the root.
A period of two ciphers gives one cipher in the root.

Second Method. If n denotes the number of integral
figures in the number whose square root is to be read,
the number of integral figures in a square root may be
tabulated as on page 208.

208

TECHNICAL TRIGONOMETRY

170

SQUARE ROOTS

Number.

Integral Figures.

On left A.
On right A

n+1

2

n

2

170. Cubes. A cube requires three times the logarithm
of a number, and is therefore read as follows:

On Regular Scales.

(1) Move the runner to the number on Z).

Twice the logarithm is then under the runner on A.

(2) Bring slide index to the runner.

(3) Move the runner to the number on B.

(4) Read the cube on A under the runner.

On K or Cube Scale. A polyphase rule is one which
has a cube scale. This is a regular logarithmic scale mounted
three times in the length of the rule. Therefore when
the runner is set at any given number on Z), three times
the logarithm of the number is under the runner on K.

171. Pointing Off a Cube.

ON D AND A SCALES

Cube read on.

Slide Projecting
to the

Intergal Figures
in Cube.

Left A

Right

3n*-2

Left

3n-l

Right A

Right

3n-l

Left

3n

* n denotes the number of integral figures in the given number.

172

USE OF THE SLIDE-RULE

209

The preceding table applies only when readings on left A are
the cubes of numbers on the left half of D, and when readings
on right A are cubes of numbers on the right half of D.

ON K SCALE

Cube Read on

Integral Figures
in Cube.

K,

3n-2
3n-l

172, Cube Roots. Reference to the method of reading
a cube on the regular scales will explain the following
instructions for reading a cube root:

On Regular Scales. (1) Consider the given number as
separated into periods of three figures each in both directions
from the decimal point.

(2) If the first period has one significant figure, set
the runner to the given number on left A.

If the first period has more than one significant figure,
set the runner to the given number on right A.

(3) Keeping the runner fixed, move the slide until the
number on B under the runner and on D under the slide
index are exactly the same.

This number is the cube root of the number to which
the runner was set on A. (See paragraph 170.)

On the Cube or K Scale. (1) Separate the given number
into periods of three figures each in both directions from
the decimal point.

(2) If the first period has one significant figure, set the
runner to the given number on Ki; if two, set the runner
to the number on K2; if three, set the runner to the number
on K3.

(3) Read cube root under the runner on D.

210 TECHNICAL TRIGONOMETRY 173

173. Pointing Off a Cube Root. Each period in the
given number makes one figure in the root.

174. Logarithms. Logarithms are read on the L scale
which, on all ten-inch rules, has 500 equally spaced grad-
uations. Therefore on a Mannheim rule,

(a) Set the slide index to the number on D.

(6) Read mantissa of the logarithm under the L notch
index.

By reference to the table on page 201 it will be evident
that when the runner is moved to any number on Z), the-
mantissa of the logarithm of the number is under the runner
on L; conversely, when the runner is moved to any number
on L, its antilogarithm is under the runner on D,ona duplex.

175. Pointing Off a Logarithm. Point off by the char-
acteristic, the same as when a table of logarithms is used.

§ 4. MULTIPLICATION

176. Scales. Numbers may be multiplied on A and
B, or on C and D. Since C and D have a greater number
of graduations, better readings are obtained from them,
and the following instructions are therefore for C and D.

The same instructions apply to A and B.

(1) Set the runner to the first factor on D.

(2) Set the left C index to the runner. Look along
the rule to see whether the second factor on C is off (pro-
jects beyond) the rule. If it is, move the right C index
to the runner.

(3) Set the runner to the second factor on C.

(4) Read the product on D under the runner.

177. Pointing Off a Product. The number of integral
figures in a product read with the C and D scales when
the slide projects to the left, equals the sum of the integral
figures in the factors; when the slide projects to the right,
the number of integral figures equals one less than the

179 USE OF THE SLIDE-RULE 211

sum of the integral figures in the factors. This may be
tabulated as follows:

INTEGRAL FIGURES IN C AND D PRODUCT

Projection of Slide.

Integ. Figs, in Prod.

Left
Right

Sum Integ. Figs. Factors
Sum Integ. Figs. Factors — 1

178. Two or More Numbers and a Trigonometric
Factor.

On a Mannheim.

One Factor a Tangent. (1) Take the product of the
arithmetical factors on C and D. Do not read the product
but keep it under the runner.

(2) Being sure not to disturb the runner, reverse the
slide and bring the T index to the runner.

(3) Move runner to given angle on T and read product
on D. Observe that in (1) the logs of the arithmetical
factors were brought end to end; in (2) and (3) the log tan
was added.

One Factor a Sine. Use A y B, and S scales and pro-
ceed exactly as when one factor is a tangent.

§ 6. DIVISION

179. How to Divide. Division is performed on the
slide-rule by setting runner and slide so that the logarithm
of the divisor may be subtracted from the logarithm of
the dividend, as follows:

(1) Set the runner to the dividend on D.

(2) Set the divisor on C under the runner.

(3) Set the runner to the C index.

(4) Read the quotient on D under the runner.

C and D settings may also be interchanged.

212

TECHNICAL TRIGONOMETRY

180

Observe and remember:

(a) The quotient is always 01% the same scale as the
dividend,

(6) The quotient is always read at the index of the scale
on which the divisor is taken.

180. Pointing Off a Quotient. The number of integral
figures in a quotient equals the following:

(1) The number of integral figures in the dividend minus
the number of integral figures in the divisor when the slide
projects to the left.

(2) One more than the difference between the number
of integral figures in dividend and divisor when the slide
projects to the right.

POINTING OFF RULES FOR DIVISION

Scales.

Direction of Slide.

Integral Figures in
Quotient.

CandD
A and B

Left

Difference

Right

Difference +1

CI and D
A and BI

Left

Difference +1

Right

Difference

§ 6. RECIPROCALS

181. Reciprocal of a Number. Since the reciprocal of
a number is one divided by the number, it may be read
as follows:

On Direct Scales:

(1) Move the runner to the given number on D.

(2) Move the C index to the runner.

(3) Read the reciprocal on C at the D index.

183 USE OF THE SLIDE-RULE 213

Observe that

(a) The reciprocal is read on C because the dividend
1 was taken on C;

(6) The reciprocal is read at the D index because the
divisor was taken on D.

If D and C settings are interchanged so that the runner
is set to the D index and the number on C aligned with it,
the quotient is on D at the C index because the dividend
1 is on D and the divisor, the given number, is on C

On an Inverted Scale:

(1) Align indexes.

(2) Move the runner to number on D.

(3) Read the reciprocal on CI under the runner.

182. Reciprocal of a Tangent. On Direct Scales:

(1) Align angle on T with the D index.

(2) Read reciprocal on D at the T index.

On an Inverted Scale:

(1) Align indexes.

(2) Move runner to angle on T.

(3) Read reciprocal on CI under the runner.

183. Reciprocal of a Sine.

On Direct Scales:

(1) Align angle on S with the A index.

(2) Read reciprocal on A at the S index.

On Inverted Slide:

(1) Insert the slide with ends reversed and trigono-

metric scales to the front.

(2) For reciprocal of sine move runner to given angle

on &

(3) Read reciprocal of sine on A under the runner.

(4) For reciprocal of tangent move runner to given

angle on T.

(5) Read reciprocal of tangent on D under the runner.

214

TECHNICAL TRIGONOMETRY

184

Observe that a reciprocal is read as a quotient by
dividing 1 by the given number on direct scales; or is
read with indexes aligned, on an inverted scale, from the
number on a direct scale.

If the rule is a plain Mannheim and therefore has no
inverted scales, they can be obtain :d by inverting the slide.

§ 7. TRIGONOMETRIC COMPUTATIONS

184. Functions. In the following paragraphs are dia-
grams and explanations of trigonometric computations of
common occurrence:

FUNCTIONS

Wanted.

Read.

sin
tan
cos

cote

sec 6

esc

vers
covrs

ext-sec

See Paragraph 160

See Paragraph 162

sin (9O°-0)

1

tan
1

sin (9O°-0)
1

sin

l-sin(9O°-0)

1 —sin

1 x

sin (9O°-0)

186. Solution of Triangles.

I.

, a 591
tanA= 6 = 784-

c

784

1

f

D

591

R

1

T

37° 1'

45°

185

USE OF THE SLIDE-RULE

215

An angle is read on T under the runner when all indexes
are aligned and the runner is set to the tangent on D. It
is therefore necessary to obtain the decimal value of the

.. 591 n
ratio =gr on D.

But a quotient is always on the same scale as the dividend.
Therefore the preceding diagram shows:

(1) Runner at 591 on D.

(2) 784 on C under the runner.

(3) Runner moved to C index.

(4) Slide reversed, indexes aligned, angle read on T

under the runner.

tt * p h 244

II. tanij = — = -tt-

a 18

Tangent ratio greater than 1; therefore beyond T
scale limit.

Therefore invert ratio, giving.

tan A —

244*

This ratio equals .0 and is therefore below the lower
limit of the tangent scale. It must therefore be read
on the sine scale.

Thus

A

18

R

1

B

244

1

S

4° 14'

90°

An angle is read on S under the runner when all indexes
are aligned and the runner is set to the sine on A.

It is therefore necessary to obtain the decimal value

of the ratio -— - on A.
244

216

TECHNICAL TRIGONOMETEY

186

But a quotient is always on the same scale as the dividend.
Therefore the runner must be set at 18 on A.

Now when 18 on A is aligned with 244 on B shall the
runner be moved to the index of B which gives the quotient
on right A, or to the B index which will give the quotient
on left At This is determined as follows:

All sines read on right A when indexes are aligned,
with slide reversed in a plain Mannheim, are immediately
preceded by a decimal point; all sines on left A under the
same conditions, are preceded by .0.

18
244

gives .0.

Therefore the preceding diagram shows:

(1) Runner at 18 on A.

(2) 244 on B under the runner.

(3) Runner moved to B index under left A.

(4) Slide reversed, indexes aligned, angle read on

S under the runner.

186. Division by a Sine.

a 4.3971

C = ~ T=-r

sin A sin 51° 3' 13"'

A
S

4.3971
51° 3' 13"

5.65
90°

187. Division by a Tangent.

59.85

d =

23 tan 14° 30'*

c

10.03

59.85

T

5° 42' 38"

14° 30'

D

1

23

•

189

USE OF THE SLIDE-RULE

217

The diagram shows:

(1) Runner set at 23 on D.

(2) Slide reversed, with left index to runner.

(3) Runner moved to 14° 30' on T.

(4) Slide reversed and 59.85 on C moved to the runner.

(5) Runner moved to D index and quotient read on

C under the runner.

Observe that the quotient was read on C because the
dividend 59.85 was taken on C.

188. Reading Involving a Square.

40.8X3960 2

W 2 =

3965.5 2
3960

W .no/ 3960 V

A

40.7

Quotient 8

B

40.8

C

3965.5

1

D

3960.

The diagram shows:

(1) Runner set at 3960 on Z>.

(2) 3965.5 on C moved to runner.

(3) Square of quotient on A over C index, not read

because not wanted.

(4) Runner moved to 40.8 on B and product read

on A under the runner.

189. Final Suggestions. Study the principles involved
every time a setting is made, so as to become independent
of all diagrams and instructions as soon as possible.

Apply no rules for pointing off except in an emergency.

If in doubt about a reading, set the rule again or set
it another way.

218

TECHNICAL TRIGONOMETRY

189

Perseverance and practice will give an enthusiasm for
the slide-rule and an increasing appreciation of its value
as an instant and infallible detector of an error in com-
putation by the tables and as a marvellous saver of time
and nervous energy.

GREEK ALPHABET

Greek

Greek

English

Greek

Greek

English

letter

name

equivalent

letter

name

equivalent

A a

Alpha

a

Nv

Nu

n

B/J

Beta

b

nt

Xi

X

ry

Gamma

g

O o

Omicron

6

A8

Delta

d

Uv

Pi

P

E<

Epsilon

8

PP

Rho

r

Z£

Zeta

z

S <r

Sigma

s

H 7]

Eta

e

T T

Tau

t

Theta

th

Y v

Upsilon

u

I i

Iota

■

i

<P <f>

Phi

ph

Kk

Kappa

k

x x

Chi

ch

A A

Lambda

1

* \p

Psi

ps

MfJL

Mu

m

O a)

Omega

In the applications of mathematics, angles are commonly denoted
by the small or lower case letters of the Greek alphabet. Those most
generally used are a, 6, <j>, and 5.

REFERENCE TABLES

220

REFERENCE TABLES

TABLE I.— LENGTH
United States and English Linear Measure

Quantity.

Unit.

Symbol.

12 inches
3 ft.

5} yd.

16} ft.

» l ;
40 rd.

8 fur. ]

320 rd.

5280 ft. J

Ifoot

1 yard
r Irod

1 pole
I 1 perch

1 furlong*

1 mile

ft.

yd.

rd.
P.

fur.

mi.

-

* Seldom used.

Surveyor's Long Measure

Quantity.

Unit.

Symbol.

7.92 inches
25 U.
4rd. \
66 ft./
80 ch.

llink
Irod

1 chain
1 mile

li.
rd.

ch.
mi.

Nautical Measure

Quantity.

6 feet
120 fathoms
7} cable-lengths
880 fathoms
1 . 153 statute miles

1 minute of circum-
ference of earth
3 geographic miles
20 leagues
60 geographic miles
69 . 16 statute miles
1 knot

}

Unit.

1 fathom

1 cable-length

1 statute mile

1 geographic or nautical mile

1 league

1 degree of earth's circumfer-
ence

1 nautical mile per hour.

REFERENCE TABLES

221

TABLE I.— Continued.
Metric Linear Measure

— —

1 meter =39.37 inches.

Quantity.

Unit.

Symbol.

10 mm.*

1 centimeter

cm.

10 cm.

1 decimeter

dm.

10 dm.

1 meter

m.

10 m.

1 dekameter

Dm.

10 Dm.

1 hectameter

Hm.

10 Hm.

1 kilometer

Km.

10 Km.

1 myriameter

Mm.

* mm. denotes millimeters.

TABLE II.— AREA
United States and English Square Measure

Quantity.

— ■

Unit.

144 square inches

9 sq. ft.

30} sq. yd. \

272} sq. ft. J

160 sq. rd. \

43560 sq. ft. J

1 square foot
1 square yard

1 square rod
1 acre

Surveyor's Square Measure

Quantity.

Unit.

625 sq. li.

16 sq. rd.

10 sq. ch.
640 A.

36 sq. mi.

1 sq. rd.
1 sq. ch.
1 acre
1 sq. mile
1 township
(IT. S. public lands)

Metric Square Measure

1

Quantity.

Unit.

100 sq. mm.
100 sq. cm.
100 sq. dc.
100 sq. m.
100 sq. Dk.
100 sq. Hk.
100 sq. Km.

1 sq. centimeter
1 sq. decimeter
1 sq. meter
1 sq. decameter
1 sq. hektometer
1 sq. kilometer
1 sq. myriameter

222

REFERENCE TABLES

TABLE III.— VOLUME

Cubic Measure

Quantity.

Unit.

1728 cubic inches
27 cu. ft.
24} cu. ft.*

1 cu. ft.

1 cu. yd.

1 perch* (masonry)

* In some states. Not a legal standard. Should always be specified.

Dry Measure

Quantity.

Unit.

2 pints

8qt.

4pk. \
2150.42 cu. in. /

2688 cu. in.
2218 . 19 cu. in.

1 quart
1 peck

tj « ( 1 bushel (struck)

1 1 bushel (heaped)
Eng. 1 bushel (struck)

Liquid Measure

Quantity.

Unit.

4 gills
2pt.
4qt.
231 cu. in.
271i cu. in.
3l| gal.
2 bbl.
63 gal.
42 gal.

}
}

1 pint
1 quart

1 gallon (U. S.)

1 Imperial gallon (Eng.)*
1 barrel t

1 hogshead

1 bbl. refined oil

* The exact imperial gallon has a capacity of 277.274 cubic inches,
t The size of a barrel or cask varies to such an extent that the capacity in gallons
is sometimes stamped on the outside.

REFERENCE TABLES

223

TABLE IV.— WEIGHT
Avoirdupois Weight

Quantity.

Unit.

Symbol.

16 drams

1 ounce

OZ.

16 oz.

1 pound

lb.

100 lbs.

1 hundredweight

cwt.

20cwt. \
2000 lbs. J

1 short ton (U. S.)

T.

112 lbs.

1 quarter (Eng.)

qr.

20 qrs. \
2240 lbs. J

1 long ton (Eng.)

T.

2204.6 lbs.

1 metric ton

T.

Troy Weight

Quantity.

Unit.

Symbol.

24 grains
20 pwt.
12 oz. 1
5760 grains J

1 pennyweight
1 ounce

1 pound

pwt.
oz.

lb.

Volume and Weight op Water

Unit.

Equivalent.

Approximate.

Exact.

1 gal. (U. S.)

1 cu.£t.

lhter
1 kilogram
1 kilo J

/ . 134 cu.ft.
1 8i lbs.
/ 7| gal.
1 62.4 lb.

2.21b.

. 13368 cu.ft.
8.3356* lb.
7.480517 gal.
62.35471* lb.

2. 204622 fib.

* At 62° Fahrenheit, barometer 30".

t By Act of Congress. Distilled (pure) water at maximum density, barom-
eter 30".

224

REFERENCE TABLES

TABLE V.— DECIMAL EQUIVALENTS OF PARTS OF AN

INCH

Fraction.

Decimal.

Fraction.

Decimal.

Fraction.

Decimal.

A

.01563

ft

.32813

ft

.70313

A

.03125

ft

.34375

a

.71875

A

.04688

«

. 35938

ft

.73438

1-16

.0625

3-8

.375

3-4

.75

A

.07813

ft

.39063

a

.76563

A

.09375

a

.40625

if

.78125

A

.10938

a

.42188

a

.79688

1-8

.125

7-16

.4375

13-16

.8125

A

. 14063

ft

.45313

ft

.82813

A

. 15625

a

.46875

if

.84375

ft

.17188

a

.48438

ft

.85938

3-16

.1875

1-2

.5

7-8

.875

ft

.20313

ft

.51563

«

.89063

A

.21875

a

.53125

»

.90625

ft

.23438

«

.54688

tt

.92188

1-4

.25

9-16

.5625

16-16

.9375

ft

.26563

ft

.57813

ft

.95313

A

.28125

a

.59375

ft

.96875

ft

. 29688

w

.60938

ft

.98438

6-16

.3125

6-8

ft

u

ft

11-16

.625

.64063
.65625
.67188
.6875

1

1.00000

REFERENCE TABLES

225

TABLE VI.— U. S. AND METRIC EQUIVALENTS

TT • a.

Equivalent.

Unit.

Approximate.

Exact.

1 acre

.40 hectares

.4047

1 bushel

35 i liters

35.24

1 cm.

.39 in.

.3937

1 cc.

.08 cu. in.

.0610

1 cu. ft.

.028 cum.

.0283

1 cu. in.

16.4 cc.

16.387

1 cum.

35.3 cu. ft.

35.31

1 cum.

1.3 cu. yd.

1.308

1 cu. yd.

.76 cum.

.7645

1ft.

30J cm.

30.48

1 gal. (U. S.)

3.8 liters

3.785

1 grain

.065 g.

.0648

1 gram
1 hectare

15| gr.

15.43

2.5 acres

2.471

1 inch

2 . 5 cm.

2.54

lkilo

2.2 lbs.

2.205

1 km.

.62 mile

.6214

1 liter

.91 qt. (dry)
1.1 qt. (liq.)

.9081

1 liter

1.057

1 meter

3.3 ft.

3.281

1 mile

1 . 6 km.

1.6093

1 mm.

.039 in.

.03937

1 oz. (avoir.)

28J g.

28.35

1 oz. (troy)

31 g.

31.10

1 peck

8.8 liter

8.809

1 pint (liq.)

.47 1.

.4732

1 pound

.45 kg.

.4536

1 qt. (dry)

1.1 1.

1.101

1 qt. (liq.)

.95 1.

.9464

1 scm.

.16 sq. in.

.1550

1 sq. ft.

.093 sm.

.0929

1 sq. in.

6 . 5 scm.

6.452

1 sq. mile

260 Ha.

259.

1 sm.

1.2 sq. yd.

1.196

1 sm.

11 sq. ft.

10.76

1 sq. rod

25 . 3 sm.

25.293

1 sq. yd.

.84 sm.

.8361

1 ton (U. S.)

.91 m. ton

.9072

1 ton (Eng.)

1 m. ton

1.017

1 ton (metric)

1.1 t. (U. S.)

1.102

1 ton (metric)

.98 t. (Eng.)

.9842

lyd.

.91 m.

.9144

See also, page 400.

226

REFERENCE TABLES

TABLE X.— WIRE GAGE SIZES
In Decimals op an Inch

Washburn

& Moen

British

Number

of
Gauge.

Birming-
ham or
Stubs Iron
Wire

American

or Brown &

Sharpe

Wire

Manufac-
turing Co.
andJohnA.
Roebling's

Trenton

Iron Co.

Wire

Gauge.

American

Screw Co.

Screw

Wire

Imperial or

English

Legal

Standard

Gauge.

Gauge.

Sons Co.

Gauge.

Wire.

Wire

Gauge.

Gauge.

0000000

.500

000000

.4600

.464

00000

.4300

.450

.432

0000

.454

.460000

.3938

.400

.400

000

.425

.409642

.3625

.360

.0315

.372

00

.380

.364796

.3310

.330

.0447

.348

.340

.324861

.3065

.305

.0578

.324

1

.300

.289297

.2830

.285

.0710

.300

2

.284

.257627

.2625

.265

.0842

.276

3

.259

.229423

.2437

.245

.0973

.252

4

.238

.204307

.2253

.225

.1105

.232

5

.220

. 181940

.2070

.205

.1236

.212

6

.203

. 162023

.1920

.190

.1368

.192

7

.180

.144285

.1770

.175

.1500

.176

8

.165

. 128490

.1620

.160

.1631

.160

Q

.148

.114423

.1483

.145

.1763

.144

10

.134

. 101897

.1350

.130

.1894

.128

11

.120

.090742

.1205

.1175

.2026

.116

12

.109

.080808

.1055

.105

.2158

.104

13

.095

.071962

.0915

.0925

.2289

.092

14

.083

.064084

.0800

.0806

.2421

.080

15

.072

.057068

.0720

.070

.2552

.072

16

.065

.050821

.0625

.061

.2684

.064

17

.058

.045257

.0540

.0525

.2816

.056

18

.049

.040303

.0475

.045

.2947

.048

19

.042

.035890

.0410

.040

.3079

.040

20

.035

.031961

.0348

.035

.3210

.036

21

.032

.028462

.03175

.031

.3342

.032

22

.028

.025346

.0286

.028

.3474

.028

23

.025

.022572

.0258

.025

.3605

.024

24

.022

.020101

.0230

.0225

.3737

.022

25

.020

.017900

.0204

.020

.3868

.020

26

.018

.015941

.0181

.018

.4000

.018

27

.016

.014195

.0173

.017

.4132

.0164

28

.014

.012641

.0162

.016

.4263

.0148

29

.013

.011257

.0150

.015

.4395

.0136

30

.012

.010025

.0140

.014

.4526

.0124

31

.010

.008928

.0132

.013

.4658

.0116

32

.009

.007950

.0128

.012

.4790

.0108

33

.008

.007080

.0118

.011

.4921

.0100

34

.007

.006305

.0104

.010

.5053

.0092

35

.005

.005615

.0095

.0095

.5184

.0084

36

.004

.005000

.0090

.009

.5316

.0076

37

.004453

.0085

.0085

.5448

.0068

38

.003965

.0080

.008

.5579

.0060

39

.003531

.0075

.0075

.5711

.0052

40

.003144

.0070

.007

.5842

.0048

From the Cambria Handbook.

INDEX

A.C. generator, 107

Acme thread, 91

Acute angles, functions of, 46-56

Ambiguous case, 132-133

Angle layout, 74^75

Angles, compound, 187-196

multiple, 197-199

notation* 69

of depression and elevation, 71
Anti-function, 146-147
Apothem, 87
Applied Problems, right triangle

69-116
Arc functions, 122-124
Arc, length of, 89-90
Arch computations, 150-151
Armature winding, 148

B

Ball bearing, 93

Ball race, MXH02, 105-106

B & S worm-thread, 91

Bevel gears, 83-85

Bolt circle, 98-99

t Caliper measurement of thread,
91-92
Checking by modulus, 88-89

Circular pitch, 85
Circumscribed circle, diameter, 89
Clearance in ball race, 101-102,

105-106
Co-functions of an angle, 51
Comparative solution, 60-62
Compass, Mariners, 76-78

surveyor's, 111 «
Compound angles, 187-196
Cosecant, derivation of, 53
Cutter spiral gear, 85
Cutter setting, 97
Cycle, 107

D

Diametral pitch, 85
Definitions, direct scale, 201

equal parts scale, 201

function, 48

functions, 120

inverted scale, 204
slide, 201

line functions, 123-124

logarithmic scale, 201

oblique triangle, 117

polyphase rule, 208

power factor, 108

six functions, 49

slide-rule, 200

solution formula, 70

trigonometry, 58

227

228

INDEX

E

Examples, instructions for solu-
tion, 59

inverse-functions, 147
isosceles triangle, 68
logarithms, 16-17, 21, 24, 28-29,

32, 34-36, 38-39, 41-42, 44
oblique triangle, 129-131, 135-

143
right triangle, 63, 66-68

Extradosed arch, 150-151

Facts of trigonometry, 143-145
Flange angle, 98

Formulas, functions, compound
angles, 187-188
multiple angles, 197

spiral gears, 85-87
Frog angle distance, 100
Function, definition, 48
Functions, compound angles, 187-
196

definition, 49

general definitions, 120

multiple angles, 197-199

of acute angles, 46-56

of an angle, 117-124

of an arc, 122-124

relation of, 187-199

G

Gears, bevel, 83-85
Greek alphabet, 218

H

Helix angles, 91
Hexagon, 88

Instructions, solution of examples,
59
solution of problems, 69-70

Inverse-function, 146-147
Isosceles triangle, examples, 68

how solved, 67

solution, 67-68

Law of, cosines? 126
segments, 125-126
tangents, 126-129
sines, 124-125
Layout, angles, 74-75
regular polygon, 87
Lead or lag, 108
Line functions, 122-124
Logarithmic functions, reading of,

55-56
Logarithms, 11-45
antilogarithms or log- 1 , 25-28
base in higher mathematics,

28
base of common and Napierian

systems, 13, 28
Briggs', 13, 28
common, 13, 28
conversion factor, 29-31
definition,
logarithm, 12
table of logarithms, 12
division, 35-36
examples, 16-17, 21, 24, 28-29,

32, 34-36, 38-39, 41-42, 44.
exponential equations, 42-44
formulas for operations, 45
how to reduce common to

Napierian, 28-31
how to take readings, 13-14
hyperbolic, 28-32
logarithm of a decimal, 22-24
mantissa, 25
meaning of, hyp, 28

log- 1 , 26-27
model solutions, exponential '
equation, 42-44

INDEX

229

Logarithms, model solutions, di-
vision, 35

multiplication, 33-34

powers, 37-38

roots, 40-41
modulus of common system,

29-31
multiplication, 32-34
Naperian, 28-32
Natural, 28

negative characteristic, 22-25
number greater than unity,

13-21
powers, 37-39
proportional parts, 19-21
roots, 39-42
summary of laws, 45
systems of, 28
table of proportional parts,

19-20

M

Mariner's compass, 76-78
Measures of, area, 221

length, 220

volume, 222

weight, 223
Metric wire-gage, 103-104
Model readings, logarithmic func-
tions, 55-56

natural functions, 53-54
Model solutions, 133-134

right triangle, 61
Modulus of a triangle, 88
Multiple angles, 197-199
Natural functions, reading of, 53-

54
Nine functions, 122

N

Notation, angles, 69
isosceles triangle, 59
right triangle, 59

O

Oblique triangle, ambiguous case,
132-133
cases and laws, 132
examples, 129-131,'%35-143
laws of solution, 124-129
model solutions, 133-134
problems, 148-186
suggestions for solution, 147-
148

Offing at sea, 150

Path of projectile, 148-149
Phase angle, 107
Pipe-bends, 175-184
Pitch diameter, 85
Pitch of a roof, 72, 73, 74
Points of compass, 77-78
Power factor, 108-109, 149
Problems, Acme thread, 91

angle layout, 106

angle 29°, layout, 96

Angle layout with compasses,
7^-75

angle of taper, 93

Apothem, 87

arc computation, 90

arch computation, 150-151

area hexagon, 88

area parallelogram, 153

area right triangle, 88

ball bearings, 93, 157-158

ball race, 100-102, 105-106

B & S worm-thread, 91

Base line, 72

belt, 165-168

bevel-gear angles, 106

bevel gears, 83-85

bolt circle, 98-99

caliper measurement, 91

chord of a circle, 88

230

INDEX

Problems, chordal distance, 94
circumference through thrto

points, 163
compass layout, 106
concrete standpipe, 170-171
crossed belt, 167-168
cross-head, 170
crushing roll, 163-164
current strength, 99
cutter for spiral gears, 85
degree of grade, 71
diagonal of a bar, 94
diagonal parallelogram, 153
diameter pulley, 94
diameter wire, 155
die-casting design, 169
dimension in tool-work, 153-155
distance, 72, 152
distance between objects, 71
dividers, 70
dovetail, 93
electric sign, 156
equilibrium on inclined plane,

82
field intensity, 164
flange angle, 98
force computation, 151
frog angle distance, 100
fundamental facts of trigonor

etry, 143-145
gun elevation, 155
guy-wire, 96
height, 73
height of, balloon, 73

hill, 71

monument, 72

mountain, 85
helix angle, 90-91
hexagonal nut, 96
instantaneous voltage, 110
John Burns' survey, 186
lag, 108
lead, 108

Problems, length of an arc, 89-90
length of belts, 165-168
length of a wire, 71
metric wire-gage, 103-104
model solution, 8
modulus of a triangle, 88
moon's radius, 88
oblique triangle, 148-186
offing at sea, 150
offset pipe-bend, 175-177
old survey, 186
open belt, 165-167
perimeter of triangle, 71
phase angle, 109
phase constant, 148
pipe-bend, 102-103
pipe-bends, 175-184
pitch of roofs, 72, 73, 74
position of cross-head, 170
power factor, 108-109, 149
radius from chord and rise, 70
radius of curve, 82
range of a gun, 150
Rankine's method, 160-161
rate of change in A.C. circuit,

108
reactance, 108
reamer computation, 94
rectification 159-162
regular polygons, 87
resistance drop, 150
right triangle, 69-116
rise and run of rafter, 76
roller bearings, 174
roof pitches, 72, 73, 74
roof truss, 110
sailing, 79-81
shaft diameter, 97
sin 2 +cos 2 , 94-95
surveys, 112-116
set-over, 97
setting of a cutter, 97
sewer construction, 171-173

INDEX

231

Problems, sewer section, 81-82

sheet metal computation, 156

side of a hexagon, 88

slope of roof, 70

spiral gear cutter, 85

spiral gears, 85-87

sprocket wheel, 92-93

system of heavy bodies, 151-152

taper reamer, 93

30-60 triangle, 88

track trunout, 100

trapezoid, 95

trolley route, 152

true power, 150

two-point ball bearing, 93

vector, 151

vector diagram, 109-110

velocity diagram, 148-149

very small angle, 185-186

voltage, 107

walk computation, 185

wedge angle, 70

weight of a plate, 149

width of a stream, 151

wire measurement of thread, 92

wire measurement sharp V-
thread, 102

Wolf-rock lighthouse, 153

zone of a circle, 171
Protractor, 58
Pulley diameter, 94

R

Radius vector, 119
Rafter, rise and run, 76
Range of a gun, 150
Ranging, pins, 115

pole, 116
Reactance, 108
Regular polygon problems, 87
Resistance drop, 150
Rhumb, 78

Right triangle, 46-116

area, 87, 88

direct derivation of formulas,
64-65

examples, 63, 66-68

model solution, 61

problems, 69-116

solution, 56-67
Roller-bearing, 174

S

Set-over, 97

Sewer calculation, 171-173
Signs, convention of 119, 121
Sin 2 +cos 2 , 53, 94-95
Slide-rule, 200-218
cube roots, 209-210
cubes, 208-209
cube scale, 208-209
division, 211-212
functions, 214
graduation, 201-202
integral figures, 205-206
logarithms, 210
Mannheim, 200
Multiplication, 210-211
pointing off, cube, 208-209

cube root, 208-209

division, 212

logarithm, 210

multiplication, 210-211

product, 210-211

quotient, 212

sine, 203

square, 207

square root, 207-208

tangent, 205
powers and roots, 206-210
quotient, 212
reciprocals, 212-214
scales, 200-202
sine divisor, 216
sine-factor, 211

232

INDEX

Slide-rule, sines, 203

solution of triangles, 214-217

square factor, 217

square roots, 207-208

squares, 206-207

tangent divisor, 216-217

tangent factor, 211

tangents, 203-206

trigonometric computations,
214-218
Spiral gear, cutter, 85

formulas, 85-87
Solution formula, definition, 70
Sprocket wheel ,92-93
Steel tape, 114

Suggestions for solution, 147-148
Surveyor's, chain, 113

compass, 111
Surveys, 112-116
Symbols, dash — 1, 146

Tables
area, 221

decimal equivalents, 224
length, 220

U. S. and metric equivalents,
225

Tables

volume, 222

weight, 223

wire gage, 226
Tangent, derivation of, 52

galvanometer, 99
Taper, angle of, 93
Thread, acme, 91

B. & S. worm, 91

wire-measurement of, 92
Triangle, isosceles, 67-68

modulus of , 88

oblique, 117-186

oblique, problems, 148-186

right, 46-116

right, problems, 69-116
Trigonometry, definition, 58
True power, 150

Vector diagram, 109-110
V-thread, 102

W

Wire-gage, 226

Wire measurement V-thread, 102

Work-book, 3-10

if

1

THE NEW YORK PUBLIC LIBRARY

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