Plane and spherical trigonometry, and Four-place tables of logarithms

■If. 1

ilif 1

TRIGC

1 liWuiRjuu'JunMflwwinBpQOM

ffTtfWWKWMWQQWmim

yWnTBBtmffllDCCTHHBlUOwHM

Q-/45-3

CORNELL

UNIVERSITY

LIBRARY

Given to the
COLLEGE OF ENGINEERING

by*

Mr. J. R. Geib.

Date Due

^gbgy^

-m

TH38fr

Cornell University Library
QA 531.G76

Plane and spherical trigonometry, and Fou

3 1924 004 031 708

Cornell University
Library

The original of this book is in
the Cornell University Library.

There are no known copyright restrictions in
the United States on the use of the text.

http://www.archive.org/details/cu31 924004031 708

MATHEMATICAL TEXTS

FOR SCHOOLS AND COLLEGES

EDITED BY

PERCEY F. SMITH, Ph.D.

PROFESSOK OF MATHEMATICS IN THE SHEFFIELD
SCIENTIFIC SCHOOL OF YALE UNIVERSITY

PLANE AND SPHERICAL
TRIGONOMETRY

AND FOUR-PLACE TABLES OF
LOGARITHMS

BY
WILLIAM ANTHONY GRANVILLE, Ph.D., LL.D.

PRESIDENT OF PENNSYLVANIA COLLEGE

GINN AND COMPANY

BOSTON • NEW YORK ■ CHICAGO • LONDON
ATLANTA • DALLAS • COLUMBUS ■ SAN FRANCISCO

Entered at Statioxers' Hall

COPYRIGIIT, 1009, BY
WILLIAM ANTHONY GRANVILLE

ALL EIGHTS RESERVED
815.10

GTNN AND COMPANY ■ PRO-
PRIETORS • BOSTON • U.S.A.

PREFACE

It has been the author's aim to treat the subject according to the
latest and most approved methods. The book is designed for the use
of colleges, technical schools, normal schools, secondary schools, and
for those who take up the subject without the aid of a teacher.
Special attention has been paid to the requirements of the College
Entrance Board. The book contains more material than is required
for some first courses in Trigonometry, but the matter has been so
arranged that the teacher can make such omissions as will suit his
particular needs.

The trigonometric functions are defined as ratios ; first for acute
angles in right triangles, and then these definitions are extended to
angles in general by means of coordinates. The student is first
taught to use the natural functions of acute angles in the solution
of simple problems involving right triangles. Attention is called to
the methods shown in §§ 23-29 for the reduction of functions of
angles outside of the first quadrant. In general, the first example!
given under each topic are worked out, making use of the natural
functions. A large number of carefully graded exercises are given,
and the processes involved are summarized into working rules
wherever practicable. Illustrative examples are worked out in
detail under each topic.

Logarithms are introduced as a separate topic, and attention is
called to the fact that they serve to minimize the labor of com-
putation. Granville's Four-Place Tables of Logarithms is used.
While no radical changes in the usual arrangement of logarithmic
tables have been made, several improvements have been effected
which greatly facilitate logarithmic computations. Particularly
important is the fact that the degree of accuracy which may be
expected in a result found by the aid of these tables is clearly
indicated. Under each case in the solution of triangles are given
two complete sets of examples, — ■ one in which the angles are ex-
pressed in degrees and minutes, and another in which the angles
are expressed in degrees and the decimal part of a degree. This
arrangement, which is characteristic of this book, should be of great

vi PREFACE

advantage to those secondary schools in which college preparation
involving both systems is necessary.

To facilitate the drawing of figures and the graphical checking
of results a combined ruler and protractor of celluloid is furnished
with each copy of the book, and will be found on the inside of the
back cover.

In Spherical Trigonometry some simplifications have been intro-
duced in the application of Napier's rule of circular parts to the
solution of right spherical triangles. The treatment of oblique
spherical triangles is unique. By making use of the Principle of
Duality nearly one half of the work usually required in deriving
the standard formulas is done away with, and the usual six cases
in the solution of oblique spherical triangles have been reduced to
three. An attempt has been made to treat the most important appli-
cations of Spherical Trigonometry to Geodesy, Astronomy, and Navi-
gation with more clearness and simplicity than has been the rule in
elementary treatises.

The author's acknowledgments are due to Professor John C. Tracy
for many valuable suggestions in the treatment of Spherical Trigo-
nometry, to Messrs. L. E. Armstrong and C. C. Perkins for verifying
the answers to the problems, and to Mr. S. J. Berard for drawing

the figures.

W. A. GRANVILLE

CONTENTS

PLANE TRIGONOMETRY

CHAPTER I

TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES. SOLUTION OF

EIGHT TRIANGLES

Section Page

1; Trigonometric functions of an acute angle defined 1

2. Functions of 45°, 30°, 60° ... . ... . 4

3. Solution of right triangles ........ 7

4. General directions for solving right triangles .... .... 7

5. Solution of isosceles triangles . . . . . ... 13

6. Solution of regular polygons . . ... 14

7. Interpolation 16

8. Terms occurring in trigonometric problems 19

CHAPTER II

TRIGONOMETRIC FUNCTIONS OF ANY ANGLE

9. Generation of angles ... . . 24

10. Positive and negative angles 24

11. Angles of any magnitude ... . .... 25

12. The four quadrants 25

13. Rectangular coordinates of a point in a plane . . .... 26

14. Distance of a point fro m the origin .... ... . . 27

15. Trigonometric functions of any angle defined .... ... 28

16. Algebraic signs of the trigonometric functions 29

17. Given the value of a function, to construct the angle 29

18. Eive of the trigonometric functions expressed in terms of the sixth . 34

19. Line definitions of the t rigonometric functions .36

20. Changes in the valuesoFTthe functions as the angle varies 38

21. Angular measure . 43

22. Circular measure . . . . . 43

23. Reduction of trigonometric functions to functions of acute angles . 47

24. Functions of complementary angles . . 47

25. Reduction of functions of angles in the second quadrant 47

26. Reduction of functions of angles in the third quadrant .... 51

27. Reduction of functions of angles in the fourth quadrant 53

28. Reduction of functions of negative angles 55

29. General rule for reducing the functions of any angle ...... 57

viii PLANE TKIGONOMETKY

CHAPTER III

RELATIONS BETWEEN THE TRIGONOMETKIC FUNCTIONS
Section Page

30. Fundamental relations between the functions 59

31. Any function expressed in terms of each of the other five functions . 60

CHAPTER IV
TRIGONOMETRIC ANALYSIS

32. Functions of the sum and of the difference of two angles 63

33. Sine and cosine of the sum of two angles 63

34. Sine and cosine of the difference of two angles 66

35. Tangent and cotangent of the sum and of the difference of two angles 68

36. Functions of twice an angle in terms of the functions of the angle . . 69

37. Functions of multiple angles . . . . 70

38. Functions of an angle in terms of functions of half the angle ... 72

39. Functions of half an angle in terms of the cosine of the angle ... 72

40. Sums and differences of functions .... 73

41. Trigonometric identities 75

CHAPTER V

GENERAL VALUES OF ANGLES. INVERSE TRIGONOMETRIC FUNCTIONS.
TRIGONOMETRIC EQUATIONS

42. General value of an angle . . ... ... . . 79

43. General value for all angles having the same sine or the same cosecant 79

44. General value for all angles having the same cosine or the same secant 81

45. General value for all angles having the same tangent or the same co-

tangent . . 82

46. Inverse trigonometric functions ... . . 84

47. Trigonometric equations . . . .89

48. General directions for solving a trigonometric equation 90

CHAPTER VI

GRAPHICAL REPRESENTATION OF TRIGONOMETRIC FUNCTIONS

49. Variables . . .... ...

50. Constants

51. Functions

52. Graphs of functions

53. Graphs of the trigonometric functions . .

54. Periodicity of the trigonometric functions .

93
93
93
93

95
97

65. Graphs of the trigonometric functions plotted by means of the unit circle 97

CONTENTS ii

CHAPTER VII

SOLUTION OF OBLIQUE TRIANGLES
Section Page

56. Relations between the sides and angles of a triangle 101

57. Law of sines . 102

58. The ambiguous case 104

59. Law of cosines 108

60. Law of tangents . ... ... Ill

61. Trigonometric functions of the half angles of a triangle . ... 113

62. Formulas for finding the area of an oblique triangle 117

CHAPTER VIII

THEORY AND USE OF LOGARITHMS

63. Need of logarithms in Trigonometry 119

64. Properties of logarithms 121

65. Common system of logarithms . . 124

66. Rules for determining the characteristic of a logarithm 125

67. Tables of logarithms 128

68. To find the logarithms of numbers from Table I 129

69. To find the number corresponding to a given logarithm .... 133

70. The use of logarithms in computations . . 135

71. Cologarithms 137

72. Change of base in logarithms . . ... . 138

73. Exponential equations ... 140

74. Use of the tables of logarithms of the trigonometric functions . . . 141

75. Use of Table II, angle in degrees and minutes . 142

76. To find the logarithm of a function of an angle ... ... 143

77. To find the acute angle corresponding to a given logarithm . . . 144

78. Use of Table III, angle in degrees and the decimal part of a degree 147

79. To find the logarithm of a function of an angle 148

80. To find the acute angle corresponding to a given logarithm .... 149

81. Use of logarithms in the. solution of right triangles 152

82. Use of logarithms in the solution of oblique triangles 158

Case I. When two angles and a side are given . . .... 158

Case II. When two sides and the angle opposite one of them are given

(ambiguous case) 161

Case III. When two sides and the included angle are given . . . 164

Case IV. When all three sides are given 167

83. Use of logarithms in finding the area of an oblique triangle .... 170

84. Measurement of land areas 172

85. Parallel sailing 173

86. Plane sailing 174

87. Middle latitude sailing 175

CONTENTS

CHAPTER IX

ACUTE ANGLES NEAR 0° OB 90°
Section Pare

88. Limits of and as x approaches the limit zero .... 178

x x

89. Functions of positive acute angles near 0° and 90° 179

90. Rule for finding the functions of acute angles near 0° 180

91. Rule for finding the functions of acute angles near 90° 181

92. Rules for rinding the logarithms of the functions of angles near 0°

and 90° 182

93. Consistent measurements and calculations 183

CHAPTER X

RECAPITULATION OF FORMULAS

List of formulas in Plane Trigonometry . . ... . . 189-191

SPHERICAL TRIGONOMETRY

CHAPTER I
EIGHT SPHERICAL TRIANGLES

1. Correspondence between the parts of a triedral angle and the parts of

a spherical triangle . . ... 193

2. Properties of spherical triangles .... 194

3. Formulas relating to right spherical triangles 196

4. Napier's rules of circular parts . . . . 199

5. Solution of right spherical triangles ... . 200

6. The ambiguous case. Two solutions " 203

7. Solution of isosceles and quadrantal triangles 204

CHAPTER II
OBLIQUE SPHERICAL TRIANGLES

8. Fundamental formulas ... . 206

9. Law of sines 206

10. Law of cosines 207

11. Principle of Duality 208

12. Trigonometric functions of half the supplements of the angles of a

spherical triangle in terms of its sides 210

13. Trigonometric functions of the half sides of a spherical triangle in

terms of the supplements of the angles 214

CONTENTS xi

Section Pagb

14. Napier's analogies . 215

15. Solution of oblique spherical triangles . 216

16. Case I. (a) Given the three sides ... . 217

17. Case I. (6) Given three angles ... 218

18. Case II. (a) Given two sides and their included angle . . 219

19. Case II. (b) Given two angles and their included side 222

20. Case III. (a) Given two sides and the angle opposite one of them

(ambiguous case) 224

21. Case III. (6) Given two angles andtHe side opposite one of them

(ambiguous case) 226

22. Length of an arc of a circle in linear units 228

23. Area of a spherical triangle ... . 229

CHAPTER III

APPLICATIONS OF SPHERICAL TRIGONOMETRY TO THE CELESTIAL AND
TERRESTRIAL SPHERES

24. Geographical terms . . 231

25. Distances between points on the surface of the earth ... . . 232

26. Astronomical problems ... . . 235

27. The celestial sphere 235

28. Spherical coordinates . 237

29. The horizon and meridian system . . 238

30. The equator and meridian system . 240

31. Practical applications 241

32. Relation between the observer's latitude and the altitude of the celes-

tial pole 242

33. To determine the latitude of a place on the surface of the earth . . 242

34. To determine the time of day 247

36. To find the time of sunrise or sunset 250

36. To determine the longitude of a place on the earth 250

37. The ecliptic and the equinoxes .... 253

38. The equator and hour circle of vernal equinox system 253

39. The system having for reference circles the ecliptic and the great circle

passing through the pole of the ecliptic and the vernal equinox . 256

40. The astronomical triangle 258

41. Errors arising in the measurement of physical quantities .... 259

CHAPTER IV
RECAPITULATION OP FORMULAS

42. Right spherical triangles 262

43. Relations between the sides and angles of oblique spherical triangles 262

44. General directions for the solution of oblique spherical triangles . . 264

45. Length of an arc of a circle in linear units 264

46. Area of a spherical triangle 264

PLANE TRIGONOMETRY

CHAPTER I

TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES
SOLUTION OF RIGHT TRIANGLES

1. Trigonometric functions of an acute angle defined. We shall assume
that the student is familiar with the notion of the angle between two
lines as presented in elementary Plane Geometry. For the present
we will confine ourselves to the consideration of
acute angles.

Let EAD be an angle less than 90°, that is, an
acute angle. From B, any point in one of the sides
of the angle, draw a perpendicular to
the other side, thus forming a right tri-
angle, as ABC. Let the capital letters
A, B, C denote the angles and the small
letters a, b, c the lengths of the corre-
sponding opposite sides in the right tri-
angle.* We know in a general way from Geometry that the sides
and angles of this triangle are mutually dependent. Trigonometry
begins by showing the exact nature of this dependence, and for this
purpose employs the ratios of the sides. These ratios are called
trigonometric functions. The six trigonometric functions of any
acute angle, as A, are denoted as follows :

sin A,

read

"sine of A";

COS, A,

read

"cosine of A";

tan A,

read

"tangent of A";

esc A,

read

"cosecant of A";

sec A,

read

" secant of A " ;

cot A,

read

"cotangent of A."

* Unless otherwise stated the hypotenuse of a right triangle will always be denoted by c
and the right angle by C.

1

PLANE TEIGONOMETKY

These trigonometric functions (ratios) are defined as follows (see
figure) :

hypotenuse / c

opposite side / a

(1) sin^=-p— =-

s ' hypotenuse \ c

adjacent side / b\

(2) cosA = — - (= - ;

v ' hypotenuse \ c j

(4) esc A =

opposite side V a

m ccci- h yp° tenuse / - c

^ ' adjacent side \ b

opposite side / a\ ,„ adjacent side / b

(3) tanA = -^ — — - = r ! (6) cot A = J — — = -

v ' adjacent side \ bj ' opposite side \ a

The essential fact that the numerical value of any one of these
functions depends upon the magnitude only of the angle A, that is,
is independent of the point B from which the perpendicular upon
< the other side is let fall, is easily established.*

These functions (ratios) are of fundamental importance in the
study of Trigonometry. In fact, no progress in the subject is pos-
sible without a thorough knowledge of the above six definitions.
They are easy to memorize if the student will notice that the three
in the first column are reciprocals respectively of those directly
opposite in the second column. For,

a 1 1

sin A = — = - = ;

c o esc A

a

A h 1 1 ■

cos A = - = - = ;

c c sec A

A ° 1 1 •

csc A = - = - = — — - ;
a a sin A

s6c A = - = -? = ;

b b cos A

tan^ = t = 7 = — TT
b b cot A

a

cot A = - = - =

a a tan^4

* For, let B' he any other point in AD, and B" any point in AE. Draw the perpendicu-
lars B'C and B"C" to AE and AD respectively. The three triangles ABC, AB'C, AB"C'\

are mutually equiangular since they are right-
angled and have a common angle at A. Therefore
they are similar, and we have

BC = B'C' = B"C"
AB AB' AB"

But each of these ratios defines the sine of A.
In the same manner we may prove this property
for each of the other functions. This shows that
the size of the right triangle we choose is imma-
terial; it is only the relative and not the actual
lengths of the sides of the triangle that are of
importance.

The student should also note that every one
of these six ratios will change in value when the
angle A changes in size.

FUNCTIONS OF ACUTE ANGLES

If we apply the definitions (1) to (6) inclusive to the acute angle
B, there results

sin B = - ;
c

cos B = — ;
c

tan B = - ;

a

CSC B

=

P

sec B

=

cots

=

a

Comparing these with the functions of the angle A, we see that

sin A = cos B ;
cos A = sin B ;
tan vl = cot B ;

esc ^4. = sec B ;
sec A = esc £ ;
cot A = tanB.

Since A + B = 90° (i.e. 4 and B are complementary) the above
results may be stated in compact form as follows :

Theorem. A function of an acute angle is equal to the co-function *
of its complementary acute angle.

Ex. 1. Calculate the functions of the angle A in the right triangle where
o = 3, 6 = 4.

Solution, c = Va 2 + 6 2 = VsT+16 = V25 = 6. '

Applying (1) to (6) inclusive (p. 2),

sin A = | ; esc A = % ;

cos A = | ; sec A = | -,

tan.4 = | ; cot A = J .

Also find all functions of the angle B, and com- A
pare results.

6=4

Ex. 2. Calculate the functions of the angle B in the right triangle where
a = 3, c = 4.

Solution, b = Vc*~^a? = Vl6 - 9 = V7.

sin B =

V7

4 '

6=V7

cos B = - ;
4

tan2? = ;

esc B = — — ;

V7

4
sec B = - ;
3'

cotJS =

V7

Also find all functions of the angle A, and compare results.

* Sine and cosine are called co-functions of each other. Similarly tangent and cotangent,
also secant and cosecant, are co-functions.

4 PLANE TRIGONOMETRY

Ex. 3. Calculate the functions of the angle A in the right triangle where

a = 2 mn, b = m 2 — re-
solution.

c = Va 2 + 6 2 = Vi m% 2 + m* - 2 m 2 n 2 + n 4
= Vm 4 + 2 to 2 k 2 + n* = m 2 + n 2 .

6=m 2 -n 2

sin .4. =
cos^l =
tan^l =

2 mn
m 2 + n 2 '
m 2 — n 2
m 2 + n 2 '

2mn

esc A =
seoA =
cot A =

?n 2

+

n 2

2

mn

m 2

+

n 2

m 2

—

ri>

m 2

-

n 2

2 ran

m 2 — n 2

Ex. 4. In a right triangle we have given sin .4. = £ and a = 80 ; find c.
Solution. From (1), p. 2, we have the formula

smA = --

Substituting the values of sin A and a that are given, there results

4_80
5~"c" ;
and solving, c = 100. Ans.

2. Functions of 45°, 30°, 60°. These angles occur very frequently
in problems that are usually solved by trigonometric methods. It is
therefore important to find the values of the
trigonometric functions of these angles and to
memorize the results.

(a) To find the functions of 45°.
Draw an isosceles right triangle, as ABC.
This makes

angle A = angle B = 45°.

6=i

Since the relative and not the actual lengths of the sides are of
importance, we may assign any lengths we please to the sides satis-
fying the condition that the right triangle shall be isosceles.

Let us choose the lengths of the short sides as unity, i.e. let a = 1
and 6 = 1.

Then c = Va 2 + 6 2 = V2, and we get

sin 45° = -^=; csc45° = V2;

V2

cos 45° = -4=; sec 45° = V2;

V2

tan 45° = 1 ; cot 45° = 1.

FUNCTIONS OF ACUTE ANGLES

(6) To find the functions of 30° and 60°.

Draw an equilateral triangle, as ABD. Drop the perpendicular BC
from B to AD, and consider the triangle ABC, where

angle A = 60° and angle ABC = 30°.
Again take the smallest side as unity, i.e. let 6 = 1. This makes
c = AB = AD = 2A C = 2 b = 2,
and a = Vc 2 — 6 2 = V4 - 1 = Vs. Therefore

sin 60° = ~; csc60° = -?=;
2 V3

cos 60° =

see 60° = 2 :

tan 60° = V3 ; cot 60° = -==•

V3

Similarly, from the same triangle,

sin 30° = - ;

esc 30° = 2 ;

V3
cos 30° = -^ ;

sec 30° = — =;
V3

tan 30° = ^=;

cot 30° = V3.

V3'

Writing the more important of these results in tabulated form,*
we have

Angle

30°

45°

60°

sin

±=.50
2

V2

^=.86 +
2

cos

^=.86 +
2

V2

± = .50
2

tan

^ = .57 +
V3

1

V3 = 1.73 +

The cosecant, secant, and cotangent are easily remembered as being
the reciprocals of the sine, cosine, and tangent respectively.

* To aid the memory we observe that the numbers in the first (or sine) row are respec-
tively VI, V2, Vi ; each divided by 2.

The second (or cosine) row is formed by reversing the order in the first row.

The last (or tangent) row is formed by dividing the numbers in the first row by the
respective numbers in the second row.

PLANE TRIGONOMETRY

The student should become very familiar with the 45° right tri-
angle and the 30°, 60° right triangle. Instead of memorizing the
above table we may then get the values of the functions directly
from a mental picture of these right triangles.

Ex. 5. Given a right triangle where A = 60°, a = 100 ; find c.
Solution. Since we know A (and therefore also any function of A), and the
sine of A involves a, which is known, and c, which is wanted, we can find c by
using the formula

sinA = *. by(l),p. 2

V3
Substituting o = 100, and sin A = sin 60° = — from the

above table, we have

V3 100

Clearing of fractions and solving for c, we get
200 200

V3 1-7 +

; 117.6 + . Ans.

What is the value of B ? Following the method illustrated above, show that
6 = 58.8 + .

EXAMPLES

Only right triangles are referred to in the following examples.

1. Calculate all the functions of the angle A, having given a = 8, b = 15.

Ans. &inA = ~fj, cosA = \fy, tanjl = T 8 j, etc.

2. Calculate the functions of the angle B, having given a = 5, c = 7.

. . . V5 _ 5 . „ V24

Ans. sm B = , cos B = - , tan B = , etc.

7 7 5

3. Calculate the functions of the angle A, having given 6 = 2, c =vTl.

a rr 2 V7

Ans. \—, — =, — -, etc.
\11 Vn 2

4. Calculate the functions of the angle B, having given a = 40, c = 41.

Ans - ? 9 t. It. i% etc.

5. Calculate the functions of the angle A, having given a — p, 6 = q.

P 9 P

Ans.

Vp 2 + 32 Vp 2 + q 2 1

, etc.

6. Calculate the functions of the angle A, having given a = Vm 2 + mn,

c = m + n. Vm 2 + mn Vmn + n 2 [m

Ans. , , -»/— , etc.

m + n m + n V n

7. Calculate the functions of the angle B, having given a = Vm^+li 2 "

c = m + n. V2mn Vm 2 + n 2 / 2«n

Ans. — , , -./_- etc.

m + n m + n \m 2 + n 2

FUNCTIONS OF ACUTE ANGLES 7

8. Given sin.4 = f, c = 200.5; calculate a. Ans. 120.3.

9. Given cosA = .44, c = 30.5; calculate ft. Ans. 13.42.

10. Given tan A = 1 j-, & = ff; calculate c. Ans. T 9 T Vl30.

11. Given A = 30°, a = 25; calculate c. Also find B and b.

Ans. c = 50, B = 60°, 6 = 25 V&

12. Given B = 30°, c = 48 ; calculate 6. Also find A and a.

Ans. 5 = 24, A = 60°, a = 24 V3.

13. Given £ = 45°, 6 = 20 ; calculate c. Also find A and a.

^ms. c = 20 V2, 4 = 45°, a = 20.

3. Solution of right triangles. A triangle is composed of six parts,
three sides and three angles. To solve a triangle is to find the parts
not given. A triangle can be solved if three parts, at least one of
which is a side, are given.* A right triangle has one angle, the right
angle, always given ; hence a right triangle can be solved if two sides,
or one side and an acute angle, are given. One of the most impor-
tant applications of Trigonometry f is the solution of triangles, and
we shall now take up the solution of right triangles.

The student may have noticed that Examples 11, 12, 13, of the
last section were really problems on solving right triangles.

When beginning the study of Trigonometry it is important that
the student should draw the figures connected with the problems as
accurately as possible. This not only leads to a better understanding
of the problems themselves, but also gives a clearer insight into the
'meaning of the trigonometric functions and makes it possible to test
roughly the accuracy of the results obtained. For this purpose the
only instruments necessary are a graduated ruler and a protractor.
A protractor is an instrument for measuring angles. On the
inside of the back cover of this book will be found a Granville's
Transparent Combined Ruler and Protractor, with directions for
use. The ruler is graduated to inches and centimeters and the
protractor to degrees. The student is advised to make free use of
this instrument.

4. General directions for solving right triangles.

First step. Draw a figure as accurately as possible representing
the triangle in question.

Second step. When one acute angle is known, subtract it from 90°
to get the other acute angle.

* It is assumed that the given conditions are consistent, that is, that it is possible to
construct the triangle from the given parts.

t The name Trigonometry is derived from two Greek "words which taken together mean
" I measure .a triangle."

8 PLANE TRIGONOMETRY

Third step. To find an unknown part, select from, (1 ) to (6), p. 2,
a formula involving the unknown pari and two known parts, and
then solve for the unknown part.

Fourth step. Check the values found by noting whether they satisfy
relations different from those already employed in the third step.
A. convenient numerical check is the relation,

a 2 = c 2 - b 2 = (c + b) (c - b).
Large errors may be detected by measurement.

Since the two perpendicular sides of a right triangle may be taken
as base and altitude, we have at once

Area of a right triangle = %r- •

In the last section the functions 30°, 45°, 60°, were found. In more
advanced treatises it is shown how to calculate the functions of angles
in general.

We will anticipate some of these results by making use of the
following table where the values* of the trigonometric functions
for each .degree from 0° to 90° inclusive are correctly given to
four or five significant figures.

In looking up the function of an angle between 0° and 45° inclu-
sive, we look for the angle in the extreme left-hand vertical column.
The required value of the function will be found on the same hori-
zontal line with the angle, and in the vertical column having that
function for a caption at the top. Thus,

sin 15° = .2588,

cot 41° = 1.1504, etc.

Similarly, when looking up the function of an angle between 45°
and 90° inclusive we look in the extreme right-hand vertical column.
The required value of the function will be found on the same hori-
zontal line with the angle as before, but in the vertical column hav-
ing that function for a caption at the bottom. Thus,

cos 64° = .4384,

sec 85° = 11.474, etc.

When we have given the numerical value of the function of an
angle, and wish to find the angle itself, we look for the given num-
ber in the columns having the given function as a caption at the top

* Also called the natural values of the trigonometric functions in contradistinction to
their logarithms (see Tables II and III of Granville's Four-Place Tables of Logarithms).

FUNCTIONS OF ACUTE ANGLES

Table A

NATURAL VALUES OF THE TRIGONOMETRIC FUNCTIONS

Angle

sin

COS

tan

cot

sec

CSC

0°

.0000

1.0000

.0000

CO

1.0000

CO

90°

1°

.0175

.9998

.0175

57.290

1.0002

57.299

89°

2°

.0349

.9994

.0349

28.636

1.0006

2S.654

88°

3°

.0523

.9986

.0524

19.081

1.0014

19.107

87°

4°

.0698

.9976

.0699

14.301

1.0024

14.336

86°

5°

.0872

.9962

.0875

11.430

1.0038

11.474

85°

6°

.1045

.9945

.1051

9.5144

1.0055

9.5668

84°

7°

.1219

.9925

.1228

8.1443

1.0075

8.2055

83°

S°

.1392

.9903

.1405

7.1154

1.0098

7.1853

82°

9°

.1564

.9877

.1584

6.3138

1.0125

6.3925

81°

10°

.1736

.9848

.1763

5.6713

1.0154

5.75SS

80°

11°

.1908

.9816

.1944

5.1446

1.0187

5.2408

79°

12°

.2079

.9781

.2126

4.7046

1.0223

4.8097

78°

13°

.2250

.9744

.2309

4.3315

1.0263

4.4454

77°

14°

.2419

.9703

.2493

4.0108

1.0306

4.1336

76°

15°

.2588

.9659

.2679

3.7321

1.0353

3.8637

75°

16°

.2756

.9613

.2S67

3.4874

1.0403

3.6280

74°

17°

.2924

.9563

.3057

3.2709

1.0457

3.4203

73°

18°

.3090

.9511

.3249

3.0777

1.0515

3.2361

72°

19°

.3256

.9455

.3443

2.9042

1.0576

3.0716

71°

20°

.3420

.9397

.3640

2.7475

1.0642

2.9238

70°

21°

.3584

.9336

.3839

2.6051

1.0711

2.7904

69°

22°

.3746

.9272

.4040

2.4751

1.0785

2.6695

68°

23°

.3907

.9205

.4245

2.3559

1.0864

2.5593

67°

24°

.4067

.9135

.4452

2.2460

1.0946

2.4586

66°

25°

.4226

.9063

.4663

2.1445

1.1034

2.3662

65°

26°

.4384

.8988

.4877

2.0503

1.1126

2.2812

64°

27°

.4540

.8910

.5095

1.9626

1.1223

2.2027

63°

28°

.4695

.8829

.5317

1.8807

1.1326

2.1301

62°

29°

.4848

.8746

.5543

1.8040

1.1434

2.0627

61°

30°

.5000

.8660

.5774

1.7321

1.1547

2.0000

60°

31°

.5150

.S572

.6009

1.6643

1.1666

1.9416

59°

32°

.5299

.8480

.6249

1.6003

1.1792

1.8871

58°

33°

.5446

.8387

.6494

1.5399

1.1924

1.8361

57°

34°

.5592

.8290

.6745

1.4826

1.2062

1.7883

56°

35°

.5736

.8192

.7002

1.4281

1.2208

1.7434

55°

36°

.5878

.8090

.7265

1.3764

1.2361

1.7013

54°

37°

.6018

.7986

.7536

1.3270

1.2521

1.6616

53°

38°

.6157

.7880

.7813

1.2799

1.2690

1.6243

52°

39°

.6293

.7771

.8098

1.2349

1.2868

1.5890

51°

40°

.6428

.7660

.8391

1.1918

1.3054

1.5557

50°

41°

.6561

.7547

.8693

1.1504

1.3250

1.5243

49°

42°

.6691

.7431

.9004

1.1106

1.3456

1.4945

48°

43°

.6820

.7314

.9325

1.0724

1.3673

1.4663

47°

44°

.6947

.7193

-.9657

1.0355

1.3902

1.4396

46°

45°

.7071

.707]

1.0000

1.0000

1.4142

1.4142

45°

cos

sin

cot

tan

CSC

see

Angle

10

PLANE TRIGONOMETRY

or bottom. If we find it in the column having the given function
as a top caption, the required angle will be found on the same
horizontal line and in the extreme left-hand column. If the given
function is a bottom caption, the required angle will be found in
the extreme right-hand column.

Thus, let us find the angle x, having given tan x = .7536.

In the column with tan as top caption we find .7536. On the
same horizontal line with it, and in the extreme left-hand column, we
find the angle x = 37°.

Again, let us find the angle x, having given

sin a; = .9816.

In the column with sin as bottom caption we find .9816. On the
same horizontal line with it, and in the extreme right-hand column,
we find the angle x = 79°.

The following examples will further illustrate the use of the table.

Ex. 1. Given A = 36°, c = 2W ; solve the right triangle. Also find its area.
Solution. First step. Draw a figure of the triangle indicating the known and

unknown parts.

&=?

Second step. B = 90°-A = 90°- 35°= 56°.

Third step. To find a use formula (1), p. 2,

namely, a

sin A = -■
c

Substituting the value of sin A = sin 35°

= .5736 (found from the table) and c = 267,

we have „

.6736 = —.

267

Solving for a, we get a = 153.1.*

» Multiplying, sin 36° = .5736

267
40152
34416
11472
a =153.1512
Since oar table gives not more than the first four significant figures of the sine of an
angle, it follows, in general, that all but the first four significant figures of the product are
doubtful. The last three figures of the above product should therefore be omitted, for the
result will not be more accurate if they are retained. To illustrate this in the above example,
suppose we take the sine of 35° from a five-place table, that is, a table which gives the first
live significant figures of the sine. Then

sin 35° = .57358

267

401506
344148
114716
0=153.14586

Comparing, we see that the two values of a agree in the first four significant figures only.
Hence we take a = 153.1.

FUNCTIONS OF ACUTE ANGLES
To find 6 use formula (2), p. 2, namely,

11

cos .4 =

Substituting as before, we have

.8192 = — ,
267

since from the table cos A = cos 35° = .8192. Hence

6 = 218.7.

Fourth step. By measurements we now check the results to see that there
are no large errors. As a numerical check we find that the values of a, 6, c sat-
isfy the condition c 2 = a 2 + &2.

To find the area of the triangle, we have

ab 153.1x218.7 ,.,,.,

Area = — = = 16,741.

2 2 '

Ex. 2. A ladder 30 ft. long leans against the side
of a building, its foot being 15 ft. from the building.
What angle does the ladder make with the grc jnd ?

Solution. Our figure shows a right triangle
with hypotenuse and side adjacent to the re-
quired angle (= x) given. Hence

cosx = ^ = £=.5 = .5000.

This number is found in the column having cos at the bottom and opposite
60°. Hence x = 60°- Ans.

We shall now derive three formulas by means of which the work
of solving right triangles may be simplified. From (1), (2), (3), p. 2,

a
sin A = -i or,
c

a = o sin A ;

. b
cos A = - > or,
o

b = e cos A ;

tan A = - 1 or,

a = bt&nA.

These results may be stated as follows :

(7) Side opposite an acute angle = hypotenuse x sine of the angle.

(8) Side adjacent an acute angle = hypotenuse x cosine of the angle.

(9) Side opposite an acute angle = adjacent side x tangent of the angle.

12

PLANE TEIGONOMETKY

EXAMPLES
Solve the following right triangles (C = 90°).

No.

Given Parts

Required Parts

Area

1

^ = 60°

6 = 4

5 = 30°

c = 8

a = 6.928

13.856

2

4 = 30°

a = 3

£ = 60°

c = 6

6 = 5.196

7.794

3

a = 6

c= 12

^1 = 30°

5 = 60°

6= 10.39

31.18

4

a = 4

6 = 4

^1 = 45°

£ = 45°

c = 5.657

8

5

a = 2

c = 2.8284

A = 45°

£ = 45°

6 = 2

2

6

a = 51.303

c= 150

4 = 20°

£ = 70°

6 = 140.95

3615.6

7

5 = 51°

c = 250

A = 39°

a = 157.3

5 = 194.3

15282

8

A = Z6°

c= 1

5 = 54°

a= .5878

6 = .809

.2378

9

c = 43

a = 38.313

A= 63°

£ = 27°

6 = 19.52

373.9

10

6 = 9.696

e = 20

^ = 61°

£ = 29°

a = 17.492

84.8

11

a = 137.664

c = 240

^4 = 35°

£ = 55°

6 = 196.6

13532

12

^1 = 75°

a = 80

5 = 15°

6 = 21.43

c = 82.82

857

13

;1 = 25°

a = 30

£ = 65°

6 = 64.336

c = 70.99

965

14

£ = 55°

6 = 10

4 = 35°

a = 7.002

c = 12.208

35

15

£ = 15°

6 = 20

A = 75°

a = 74.64

c = 77.28

746.5

16

a = 36.4

6= 100

^1 = 20°

£ = 70°

c = 106.4

1820

17

a = 23.315

6 = 50

^1 = 25°

£ = 65°

c= 55.17

583

18

a= 17.1

c = 50

^1 = 20°

£ = 70°

6 = 46.985

402

19

4= 10°

6 = 30

5=80°

a = 5.289

c = 30.46

79

20

.4 = 20°

c = 80

£ = 70°"

a = 27.36

6 = 75.176

1028

21

£ = 86°

6= .08

.4 = 4°

a = .00559

c = .0802

.0002

22

£ = 32°

c = 1760

A = 58°

6 = 932.62

a= 1492.5

696968

23

a = 30.21

c = 33.33

4 = 65°

£ = 25°

6 = 14.085

213

24

a = 13.395

6 = 50

A = 15°

£ = 75°

c = 51.77

335

25

6 = 93.97

c= 100

^1 = 20°

£ = 70°

a = 34.2

1607

26. A tree is broken by the wind so that its two parts form with the ground a
right-angled triangle. The upper part makes an angle of 35° with the ground,
and the distance on the ground from the trunk to the top of the tree is 50 ft.
Find the length of the tree. Ans. 96.05 ft.

27. In order to find the breadth of a river, a dis-
tance AB was measured along the bank, the point A
being directly opposite a tree C on the other side. If
the angle ABC was observed to be 55° and AB 100 ft.,
find the breadth of the river. Ans. 142.8 ft.

28. Two forts defending a harbor are 2 mi. apart.
From one a, hostile battleship is observed due south
and from the other 15° east of south. How far is the
battleship from the nearest fort ? Ans. 7.464 mi.

29. A vessel whose masts are known to reach 100 ft. above her water line
subtends in a vertical plane an angle of 5° to an observer in a rowboat. How
far is the boat from the vessel ? Ans. 1143 ft.

FUNCTIONS OF ACUTE ANGLES

13

30. The vertical central pole of a circular tent is 20 ft. high, and its top is
fastened by ropes 40 ft. long to stakes set in the ground. How far are the
stakes from the foot of the pole, and what is the inclination of the ropes to the
ground? Ans. 34.6ft; 30°.

31. A wedge measures 10 in. along the side and the angle at the vertex is 20°.
Find the width of the base. Ans. 3.47 in.

32. At two points A, B, 400 yd. apart on a straight horizontal road, the
summit of a hill is observed ; at A it is due north with an elevation of 40°, and at
B it is due west with an elevation of 27°. Find the height of the hill.

Ans. 522.6 ft.

5. Solution of isosceles triangles. An isosceles triangle is divided
by the perpendicular from the vertex to the base into two equal
right triangles ; hence the solution of an isosceles triangle can be
made to depend on the solution of one of these right triangles. The
following examples will illustrate the method.

Ex. 1. The equal sides of an isosceles triangle are each 40 in. long, and the
equal angles at the base are each 25°. Solve the triangle and find its area.

Solution. B = 180° - (A + C) = 180° - 50° = 130°. Drop the perpendicular
BD to AC. B

AD = AB cos A

= 40 cos 25° by (8), p. 11

w

= 40 x .9063
= 36.25.
Therefore

by (7), p. 11
by (9), p. 11

AC = 2 AD = 72.50 in.
To find the area we need in addition the altitude BD.

BD = AB smA = 40 sin 25°

= 40 x. 4226 = 16. 9.
Check : BD = AD tan 25° = 36.25 x .4663 = 16.9.
Also, Area = \AC x BD = 612.6 sq. in.

Ex. 2. A barn 60 ft. wide has a gable roof whose rafters are 30 V2 ft. long.

What is the pitch of the roof, and how far above the eaves is the ridgepole ?

Solution. Drop a perpendicular from

B to AD. Then

AC 30 1
cos X = = =

-^B 30 V2 V2

Hence x = 45° = pitch of the roof.
Also, BC = 45 sin x by (8), p. 11
1

= 30V2-

vS

Check : AB = V^ C 2 + JSG 2 = V(30)* + (30)* :

= 30 ft.

= height of the ridgepole
above the eaves.

V1800 = 30 V2.

14 . PLANE TKIGONOMETKY

EXAMPLES

1. The equal sides of an isosceles triangle are each 12 in. long, and the angle
at the vertex is 120°. Find the remaining parts and the area.

Ans. Base = 20.78 in.; base angles = 30°; area = 62.35 sq. in.

2. The equal angles of an isosceles triangle are each 35°, and the base
is 393.18 in. Find the remaining parts.

Ans. Vertex angle = 110° ; equal sides = 240 in.

3. Given the base 300 ft. and altitude 150 ft. of an isosceles triangle ; solve
the triangle.

Ans. Vertex angle = 90° ; equal angles = 45° ; equal sides = 212. 13 ft.

4. The base of an isosceles triangle is 24 in. long and the vertex angle is
48° ; find the remaining parts and the area.

Ans. Equal angles = 06° ; equal sides = 29.5 in. ; area = 323.4 sq. in.

5. Each of the equal sides of an isosceles triangle is 50 ft. and each of its
equal angles is 40°. Find the base, the altitude, and the area of the triangle.

Ans. Alt. = 32.14 ft. ; base = 76.6 ft. ; area = 1231 sq. ft.

6. The base of an isosceles triangle is 68.4 ft. and each of its equal sides is
100 ft. Find the angles, the height, and the area.

Ans. 40°, 70°; 93.97 ft.; 3213.8 sq.ft.

7. The base of an isosceles triangle is 100 ft. and its height is 35.01 ft. Find
its equal sides and the angles. Ans. 61.04 ft.; 35°, 110°.

8. The base of an isosceles triangle is 100 ft. and the equal angles are each
65°. Find the equal sides, the height, and the area.

Ans. 118.3 ft; 107.23 ft; 5361.5 sq.ft.

9. The ground plan of a barn measures 40 x 80 ft. and the pitch of the roof is
45° ; find the length of the rafters and the area of the v? hole roof, the horizontal
projection of the cornice being 1 ft. Ans. 29.7 ft. ; 4870.5 sq. ft.

6. Solution of regular polygons.
Lines drawn from the center of a
regular polygon of n sides to the
vertices are the radii of the circum-
scribed circle and divide the poly-
gon into n equal, isosceles triangles.
The perpendiculars from the center
to the sides of the polygon are the
radii of the inscribed circle and
divide these n equal isosceles tri-
angles into 2 n equal right triangles.

Hence the solution of a regular polygon depends on the solution

of one of these right triangles.

FUNCTIONS OF ACUTE ANGLES

15

From Geometry we know that the central angle ABC ■
in the right triangle ABD the

180°

360°
n

; hence

Also,

angle x =

AD = - = half the length of one side,

AB = R = radius of circumscribed circle,
BD = r = radius of inscribed circle,
p = no = perimeter of polygon,

pr

~ = area of polygon.

EXAMPLES

1. One side of a regular decagon is 10 in. ; find radii of inscribed and circum-
scribed circles and area of polygon.

Solution. Since n = 10, in this example we have

x = -

180° _ 180°
"IF

= 18°.

Then R = ■

and
Check
Also,

hence

r = -

sin 18°

5

.3090
5

= 16.18 in.,

= 15.39 in.

tan 18° .3249

: r = R cos 18° = 16.18 x .9511
= 15.39.
p = 10 x 10 = 100 in.
= perimeter of polygon ;
pr _ 100 x 15.39
2 ~ 2

= 769. 5 sq. in.
= area.

2. The side of a regular pentagon is 24 ft. ; find R, r, and area.

Ana. 20.42 ft.; 16.52 ft. ; 991.2 sq. ft.

3. Find the remaining parts of a regular polygon, having given

(a) n = 9, c = 12. Arts. R = 17.54 ; r = 16.48 ; area = 889.9.

(b) n = 18, R = 10. r = 9.848 ; c = 3.472 ; area = 307.7.

(c) n = 20, R = 20. r = 19.75 ; c = 6.256 ; area = 1236.
(d)n=12, r=8. iJ = 8.28; c = 4.29; area = 206.

4. The side of a regular hexagon is 24 ft. Find the radii of the inscribed and
circumscribed circles ; also find the difference between the areas of the hexagon
and the inscribed circle, and the difference between the areas of the hexagon and
the circumscribed circle. Ans. R = 24 ft.; r = 20.8 ft.; 138.4 sq. ft.; 312 sq. ft.

5. If c be the side of a regular polygon of n sides, show that

x, 1 180 ° A
R = -c esc and r ■■

2 n

1 x 180°
- c cot

2 71

16

PLANE TKIGONOMETKY

6. If r be the radius of a circle, show that the side of the regular inscribed
180°
polygon of n sides is 2 r sin , and that the side of the regular circumscribed

• „ ♦ 180 ° n

polygon is 2 r tan

7. Interpolation. In the examples given so far we have needed the
functions of such angles only as were explicitly given in our table ;
that is, the number of degrees in the angle involved was given by a
whole number. It is evident that such will not always be the case.
In general, our problems will involve angles expressed in degrees
and parts of a degree, as 28.4°, 5.63°, 10° 13', 72° 27.4', 42° 51' 16", etc.

In order to find from the table the numerical value of the function
of such an angle not given in the table, or to find the angle corre-
sponding to a given numerical value of some function not found in
the table, we use a process called interpolation. This is based on the
assumption that a change in the angle causes a proportional change
in the value of each function, and conversely, provided these changes
are small* To illustrate ; from the table we have
sin 38° = .6157
sin 37° = .6018

Subtracting, .01 39 = difference for one degree ;

that is, at 37° a change of one degree in the angle causes a change
in the value of the sine of .0139. If, then, x is any other small change
in the angle from 37°, and d the corresponding change in the value
of the sine, we must have, near 37°,

V:x\: .0139 : d,
.-. d = .0139x,
if x is expressed in the decimal parts of a degree.

For example, let us tabulate the values of the sines of all angles
from 37° to 38° at intervals of 0.1 of a degree.

. . sin 37.1° = .6018 + .0014 = .6032
.-. sin 37.2° = .6018 + .0028 = .6046

sin 37.3° = .6018 + .0042 = .6060
.-. sin 37.4° = .6018 + .0056 = .6074

sin 37.5° = .6018 + .0070 = .6088
.-. sin37.6° = .0018 + .0083 = .6101
,-. sin 37.7° = .6018 + .0097 = .6115

sin 37.8° = .6018 + .0111 = .6129
.-. sin 37.9° = .6018 + .0125 = .6143

* This condition is most important. The change in value of the cotangent for one degree
is very large when the angle is very small. In this ease the table -would therefore lead to
very inaccurate results if interpolation -was used for cotangents of small angles (see
Chapter IX, p. 178).

X

d

0.1°

.0014

0.2°

.0028

0.3°

.0042

0.4°

.0056

0.5°

.0070

0.6°

.0083

0.7°

.0097

0.8°

.0111

0.9°

.0125

FUNCTIONS OP ACUTE ANGLES 17

The following examples will further illustrate the process of
interpolating.

(a) To find the function of a given angle when the angle is not
found in the table.

Ex. 1. Find sin 32.8°.

Solution. The sine of 32.8° must lie between sin 32° and sin 33°. Prom the
table on p. 9,

sin 33° = .5446
sin 32° = .5299

.0147 = difference in the sine (called the
tabular difference) corresponding to a difference of 1° in the angle. Now in
order to find sin 32.8°, we must find the difference in the sine corresponding
to .8° and add it to sin 32°, for the sine will be increased by just so much when
the angle is increased from 32° to 32.8°. Denoting by d the difference corre-
sponding to .8°, we have

1° : .8° : : .0147 : d,
or, d = .0118.

Hence .sin 32° = . 5299

d - .0118 = difference for .8°

sin 32.8° = .5417. Ans.

Ex. 2. Eind tan 47° 25'.

Solution. The tangent of 47° 25' must lie between tan 47° and tan 48°. Erom
the table,

tan48° = l.llQ6
tan 47° = 1.0724

.0382 = tabular difference correspond-
ing to a difference of 60' (=1°) in the angle. Denoting by d the difference
corresponding to 25', we have

60' : 25' : : .0382 : d,
or, d = .0159.

Hence tan 47° = 1.0724

d = .0159 = difference for 25'.

.-. tan 47° 25' = 1.0883. Ans.

Ex. 3. Eind cos 68. 57°.

Solution. The cosine of 68. 57° must lie between cos 68° and cos 69°. From
the table,

cos 68° = .3746
cos 69° = .3584

.0162 = tabular difference correspond-
ing to a difference of 1° in the angle. Denoting by d the difference corresponding
to. 57°, we have

l°:.57°::.0162:d,
or, d = .0092.

18 PLANE TRIGONOMETRY

Since the cosine decreases as the angle increases, this difference must be
subtracted* from cos 68° in order to get cos 68. 57°

Hence cos 68° = .3746

d = .0092 = difference for. 57°.

.-. cos 68. 57° = .3654. Ans.

(S) To find an angle when the given numerical value of a function
of the angle is not found in the table.

Ex. 4. Find the angle whose tangent is .4320.

Solution. This problem may also be stated : Having given tan x = .4320, to
find the angle x. We first look up and down the columns with tan at top or
bottom, until we find two numbers between which .4320 lies. These are found
to be .4245 and .4452, the former being tan 23° and the latter tan 24°. We then
know that the required angle x must lie between 23° and 24°. To find how far
( = y) beyond 23° the angle x lies, we first find the difference between tan 23°
and tan x ; thus,

tan x = . 4320
tan 23° = .4245

.0075 = difference in the tangent corre-
sponding to the excess of the angle x over 23° ; denote this excess by y. Also,

tan 24° = .4452
tan 23° = .4245

.0207 = tabular difference correspond-
ing to a difference of 1° in the angle. Then, as before,

1° :y:\. 0207:. 0075,
or, y = .36°.

Hence x = 23° + "y = 23.36°. Ans.

In case we want the angle expressed in degrees and minutes, we can either
multiply .36° by 60, giving 21.6' so that the required angle is 23° 21.6', or else
we can find y in minutes at once by using instead the proportion

60' : y : : .0207 : .0075,
or, y = 21.6'.

Hence x = 23° + y = 23° 21.6'. Ans.

EXAMPLES
1. Verify the following :

(a) sin 51.6° = .7836. (f) esc 80.3° = 1.0145. (k) sec 25° 2.5' = 1.1038.

(b) tan 27.42° = .5188. (g) sin 43° 18' = .6858. (1) esc 72° 54' = 1.0463.

(c) cos79.9° = .1753. (h) cos 84° 42' = .0924. (m) sin 58° 36.2' = .8536.

(d) cot 65.62° = .4532. (i) tan 31° 7.8' = .6040.

(e) sec 12.37° = 1.0238. (j) cot 11° 43.4' =4.8263.

* In the case of the sine, tangent, and secant this difference is always added, because
these functions increase when the angle increases (the angle being acute). In the case of
the cosine, cotangent, and cosecant, however, this difference is always subtracted, because
these functions decrease when the angle increases. It is always the function of the smaller
of the two angles that this difference is added to or subtracted from.

TEEMS IN TRIGONOMETRIC PROBLEMS

19

2. Find the angle x, having given

(a) sin x = .5280.

(b) tana; = .6344.

(c) sec x = 1.2122.

(d) cos x = . 9850.

(e) cot x= 3.5249.

(f) esc x = 1.7500.

(g) sin x = . 9425.
(h) cos x = . 2118.

(i) tanx = 1.1652.

(j) cot x = .0803.
(k) sec x = 4.6325.

(1) esc x = 1.2420.

(m) sin x = .7100.

(n) cos x = .9999.

(o) tan x = . 9845.

(p) cot x = 8.6892.

Ans.

x = 31.87°.
x = 32.39°.
x = 34.41°.
x = 9.93°.
x = 15.85°.
x = 34.85°.
x = 70° 28.96'.
x = 77.77°.
x = 49° 21.4'.
x = 85.41°.
x = 77.51°.
x = 53.63°.
x = 45° 14.3'.
x = 0° 30'.
x = 44°33'.
x = 6°36.1'.

8. Terms occurring in trigonometric problems. The vertical line
at a point is the line which coincides with the plumb line through
that point.

A horizontal line at a point is a line which is perpendicular to the
vertical line through that point.

A vertical plane at a point is a plane which contains the vertical
line through that point.

The horizontal plane at a point is the plane which is perpendicu-
lar to the vertical line through that point.

A vertical angle is one lying in a vertical plane.
A horizontal angle is one lying in a horizontal plane.
The angle of elevation of an object above
the horizontal plane of the observer is the ver-
tical angle between the line
drawn from the observer's
eye to the object, and a hori-
zontal line through the eye.

The angle of depression of an object below the horizontal plane

of the observer is the vertical
angle between the line drawn
from the observer's eye to the
object, and a horizontal line
through the eye.

The horizontal distance between
two points is the distance from
one of the two points to the vertical line drawn through the other.

Horizontal Line

Horizontal Line

20

PLANE TKIGONOMETEY

The vertical distance between two points is the distance from one
of the two points to the horizontal plane through the other.

Thus, let BC be the vertical line
at B, and let the horizontal plane
at A cut this vertical line in C ;
then AC is called the horizontal
distance between A and B and BC
the vertical distance.
The Mariner's Compass is divided into 32 equal parts ; hence
each part = 360° -*- 32 = 11£°. The following figure shows how
the different divisions are designated. North, south, east, and west
are called the cardinal points, and on paper these directions are
usually taken as upward, downward, to the right, and to the left
respectively. The direction of an object from an observer at C may
be given in several ways. Thus, A in the figure is said to bear N.E.
by E. from C, or from C the bearing of A is N.E. by E. In the
same way the bearing of C from A is S.W. by W. The point A is
3 points north of east and 5 points east of north. Also, E. 33|° N.
means the same as N.E. by E.

In order to illustrate the application of the trigonometric functions
(ratios) to the solution of practical examples, we shall now give a
variety of problems on
finding heights, dis- ^ ^

tan ces , angles , areas , etc.
In solving these prob-
lems it is best to follow
some definite plan. In
general we may proceed
as follows :

(a) Construct a draw-
ing to some convenient
scale which will show
the relations between
the given and the re-
quired lines and angles.

(b) If necessary draw
any auxiliary lines that
will aid in the solution, and decide on the simplest steps that will
solve the problem.

(c) Write down the formulas needed, make the calculations, and
check the results.

EXAMPLES

21

EXAMPLES

Solve the following right triangles (C = 90°).

No.

Given Parts

Required Parts

1

a = 60

c = 100

A = 36° 52'

£ = 53° 8'

6 = 80

2

a = 16.98

c = 18.7

A = 65° 14'

£ = 24° 46'

6 = 7.833

3

a = 147

c= 184

A =53° 2'

£ = 36° 58'

6 = 110.67

4

A = 34° 15'

a = 843.2

£ = 55° 45'

c = 1498.5

6 = 1238.7

5

4 = 31° 14.2'

c = 2.934

£ = 58° 45.8'

a = 1.521

6 = 2.509

6

£ = 47.26°

c = 4.614

^ = 42.74°

a = 3.131

6 = 3.389

7

A = 23.5°

c = 627

£ = 66.5°

a = 250

6 = 575

8

A = 28° 5'

c =. 2280

£ = 61° 55'

a = 1073

6 = 2011

9

.5 = 43.8°

6 = 50.94

A = 46.2°

a = 53.13

c = 73.6

10

£ = 6° 12.3'

c = 3721

^1 = 83° 47. 7'

a = 3699

6 = 402.2

11

a = .624

c= .91

A = 43° 18'

£ = 46° 42'

6 = .6623

12

a= 5

6 = 2

^. = 68° 12'

£ = 21° 48'

c = 5.385

13

a = 101

6 = 116

A = 41° 3'

£ = 48° 57'

c = 153.8

14

^ = 43.5°

c = 11.2

£ = 46.5°

a = 7.71

6 = 8.124

15

£ = 68° 50'

a = 729.3

4 = 21° 10'

6 = 1884

c = 2020

16

4 = 58.65°

c = 35.73

£ = 31.35°

a = 30.51

5= 18.59

17

B = 10.85°

c = .7264

A = 79.15°

a = .7134

6 = .1367

18

a = 24.67

6 = 33.02

A = 36° 46'

£ = 53° 14'

c = 41.22

19

B = 21° 33' 51"

a = .821

A = 68° 26' 9"

6 = .3244

c = .8827

20

^. = 74°0'18"

c = 275.62

£ = 15° 59' 42"

a = 264.9

6 = 75.93

21

A = 64° 1.3'

6 = 200.05

£ = 25° 58.7'

a = 410.6

c = 456.7

22

6 = .02497

c = .04792

.4 = 58° 36'

£ = 31° 24'

a = .0409

23

6 = 1.4367

c = 3.4653

A = 65° 30'

£ = 24° 30'

a = 3.153

24. The length of a kite string is 250 yd., and the angle of elevation of the
kite is 40°. Pind the height of the kite, supposing the line of the kite string to
be straight. Ans. 160.7 yd.

25. At a point 200 ft. in a horizontal line from the foot of a tower the angle
of elevation of the top of the tower is observed to be 60°. Pind the height of
the tower. Ans. 346 ft.

26. A stick 10 ft. in length stands vertically on a horizontal plane, and the length
of its shadow is 8.391 ft. Pind the angle of elevation of the sun. Ans. 50°.

27. Prom the top of a rock
that rises vertically 80 ft. out of
the water the angle of depression
of a boat is found to be 30° ; find
the distance of the boat from the
foot of the rock. Ans. 138.57 ft.

28. Two ships leave the same
dock at the same time in direc-
tions S.W. by S. and S.E. by E.
at rates of 9 and 9.5 mi. per hour
respectively. Find their distance apart after 1 hr.

Ans. 13.1 mi.

22 PLANE TEIGONOMETEY

29. From the top of a tower 120 ft. high the angle of depression of an obj
on a, level with the base of the tower is 27° 43'. What is the distance of
object from the top and bottom of the tower 1 Ans. 258 ft., 228

30. A ship is sailing due east at the rate of 7.8 mi. an hour. A headlanc
observed to bear due north at 10.37 a.m. and 33° west of north at 12.43 p,
Find the distance of the headland from each point of observation.

Ans. 25.22 mi., 30.08 I

31. A ship is sailing due east at a uniform rate of speed. At 7 a.m. a lig]
house is observed bearing due north, 10.32 mi. distant, and at 7.30 a.m. it bei
18° 13' west of north. Find the rate of sailing of the ship and the bearing
the lighthouse at 10 a.m. Ans. 6.79 mi. per hour, 63° 8' W. of

32. From the top of a tower the angle of depression of the extremity of
horizontal base line 1000 ft. in length, measured from the foot of the tower,
observed to be 21° 16' 37"- Find the height of the tower. Ans. 389.5 :

33. The length of the side of a regular octagon is 12 in. Find the radii of t
inscribed and circumscribed circles. Ans. 14.49 in., 15.69 i

34. What is the angle of elevation of an inclined plane if it rises 1 ft. in
horizontal distance of 40 ft. ? Ans. 1°2I

35. A ship is sailing due N.E. at the rat^ of 10 mi. an hour. Find the rate
which she is moving due north. Ans. 7.07 mi. per lion

36. A ladder 40 ft. long may be so placed that it will reach a window 33 i
high on one side of the street, and by turning it over without moving its foi
it will reach a window 21 ft. high on the other side. Find the breadth of tl
street. Ans.' 56.64 f

37. At a point midway between two towers on a horizontal plane the angl
of elevation of their tops are 30° and 60° respectively. Show that one tower
three times as high as the other.

38. A man in a balloon observes that the bases of two towers, which are
mile apart on a horizontal plane, subtend an angle of 70°. If he is exactly abo'
the middle point between the towers, find the height of the balloon.

Ans. 3770 f

39. In an isosceles triangle each of the equal angles is 27° 8' and each of tl
equal sides 3.088. Solve the triangle. Ans. Base = 5.49

40. What is the angle of elevation of a mountain slope which rises 238 ft. i
a horizontal distance of one eighth of a mile ? Ans. 19° fit

41. If a chord of 41.36 ft. subtends an arc of 145° 37', what is the radius i
the circle ? Ans. 21.65 f

42. If the diameter of a circle is 3268 ft. , find the angle at the center sul
tended by an arc whose chord is 1027 ft. Ans. 36° 37.8

43. From each of two stations east and west of each other the angle >
elevation of a balloon is observed to be 45°, and its bearings N.W. and N.l
respectively. If the stations are 1 mi. apart, find the height of the balloon.

Ans. 3733 f

EXAMPLES

23

44. In approaching a fort situated on a plain, a reconnoitering party finds at
one place that the fort subtends an angle of 10°, and at a place 200 ft. nearer
the fort that it subtends an angle of 15°. How high is the fort and what is the
distance to it from the second place of observation ?

Bint. Denoting the height by y and the distance by x, we have

y = x tan 15", by (9), p. 11

also, y = (x + 200) tan 10°. . by (9), p. 11

Solve these two simultaneous equations for x and y, substituting the values of tan 15°
and tan 10° from the table on p. 9. Ang % _ ggg t t. r y = 103 ft.

45. A cord is stretched around two wheels with radii of 7 ft. and 1 ft. respec-
tively, and with their centers 12 ft. apart. Prove that the length of the cord is
12V3+10:rft.

46. A flagstaff 25 'ft. high stands on the top
of a house. From a point on the plain on which
the house stands, the angles of elevation of the
top and the bottom of the flagstaff are observed
to be 60° and 45° respectively. Find the height
of the house. Ans. 34.15 ft.

47. A man walking on a straight road ob-
serves at one milestone a house in a direction
making an angle of 30° with the road, and at
the next milestone the angle is 60°. How far
is the house from the road ? Ans. 1524 yd.

48. Find the number of square feet of pave-
ment required for the shaded portion of the
streets shown in the figure, all the streets being
50 ft. wide.

OQ7KA

Ans. =^ + 7500 = 24094.

V3

CHAPTEE II

TRIGONOMETRIC FUNCTIONS OF ANY ANGLE

9. Generation of angles. The notion of an angle, as usually pre-
sented in Elementary Geometry, is not general enough for the pur-
poses of Trigonometry. We shall have to deal with positive and
negative angles of any magnitude. Such a conception of angles may
be formed as follows :

An angle may be considered as generated by a line which first coin-
cides with one side of the angle, then revolves about the vertex, and
finally coincides with the other side.

initial side

This line is called the generating line of the angle. In its first
position it is said to coincide with the initial side of the angle, and
in its final position with the terminal side of the angle.

Thus, the angle A OB is generated by the line OP revolving about
in the direction indicated from the initial side OA to the terminal
side OB.

10. Positive and negative angles. In the above figures the angles
were generated by revolving the generating line counter-clockwise ;
mathematicians have agreed to call such angles positive. Below are

angles having the same initial and terminal sides as those above, but
the angles are different since they have been generated by revolving
the generating line clockwise ; such angles are said to be negative.*

* The arcs with arrowheads will be drawn full when indicating a positive angle, and
dotted when indicating a negative angle.

24

TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 25

11. Angles of any magnitude. Even if angles have the same initial
and terminal sides, and have been generated by rotation in the same
direction, they may be different. Thus, to generate one right angle,
the generating line rotates into the position OB as shown in Fig. a.
If, however, the generating line stops in the position OB after
making one complete revolution, as shown in Fig. b, then we have
generated an angle of magnitude five right angles ; or, if two com-
plete revolutions were first made, as shown in Fig. c, then we have

Cl

Fie. a

Fig. b

Fig. c

generated an angle of magnitude nine right angles ; and so on indefi-
nitely. This also shows that positive angles may have any magnitude
whatever. Similarly, by making complete revolutions clockwise, it
is seen that negative angles may have any magnitude.*

12. The four- quadrants. It is customary to divide the plane about
the vertex of an angle into four parts called quadrants, by passing
two mutually perpendicular lines through the vertex. Thus, if is
the vertex, the different quadrants are named as indicated in the
figure below, the initial side being horizontal and drawn to the right.

An angle is said to be (or lie) in
a certain quadrant when its ter-
minal side lies in that quadrant.

In the figures shown on the pre-
vious page, only the least positive
and negative angles having the
given initial and terminal sides
are indicated by the arcs. As a
matter of fact there are an infinite
number of positive and negative angles in each case which have
the same initial and terminal sides, all differing in magnitude by
multiples of 360°. The following examples will illustrate the
preceding discussion.

Second
Quadrant

Third
Quadrant

First
Quadrant

initial side

Fourth
Quadrant

> Thus, the minute hand of a clock generates - 4 rt. d every hour, i.e. - 96 rt. A every day.

26

PLANE TRIGONOMETRY

EXAMPLES

1. Show that 1000° lies in the fourth quadrant.

Solution. 1000° = 720° + 280° = 2 x 360° + 280°. Hence we make two com-
plete revolutions in the positive direction and 280° beyond, and the terminal-
side of 280° lies in the fourth quadrant.

2. Show that — 568° lies in the second quadrant.

Solution. — 568° = — 360° — 208°. Hence we make one complete revolution
in the negative direction and 208° beyond in the negative direction, and the
terminal side of — 208° lies in the second quadrant.

3. In what quadrants are the following angles ?

(a) 225°. (e) 651°. (i) 540°. (m) 1500°.

(b) 120°. (f) - 150°. (j) 420°. (n) 810°.

(c) - 315°. (g) - 75°. (k) - 910°. (o) - 540°.

(d) -240°. (h) -1200°. (1) -300°. (p) 537°.

13. Rectangular coordinates of a point in a plane. In order to define
the functions of angles not acute, it is convenient to introduce the

notion of coordinates. Let A"'A" be a
horizontal line and Y'Y a line per-
pendicular to it at the point O. Any
point in the plane of these lines
(as P) is determined by its distance
_l> j an d direction from each of the perpen-
diculars A*'A' and Y'Y. Its distance
from Y'Y (as NP = a) is called the
abscissa of the point, and its distance
from A" A (as MP = b) is called the ordinate of the point.
Abscissas measured to the right of Y'Y are positive.
Abscissas measured to the left of F'Fare negative.
Ordinates measured above X'X are positive.
Ordinates measured below X'X are negative.
The abscissa and ordinate taken to- r '

gether are called the coordinates of the II I

point and are denoted by the symbol (-»+) (+,+)

(a, b).

The lines X'X and Y'Y are called the
axes of coordinates, X'X being the axis of
abscissas or the axis ofX, and FT the axis
of ordinates or the axis of Y; and the point
O is called the origin of coordinates.

The axes of coordinates divides the plane into four parts called
quadrants (just as in the previous section), the figure indicating the
proper signs of the coordinates in the different quadrants.

x±±=i

X'

III

X

IV

TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 27

\

->X

To plot a point is to locate it from its coordinates. The most con-
venient way to do this is to first count off from along A" A' a num-
ber of divisions equal to the abscissa, to the right or left according
as the abscissa is positive or negative.
Then from the point so determined -|— -
count off a number of divisions equal ~
to the ordinate, upward or downward
according as the ordinate is positive x '
or negative. The work of plotting
points is much simplified by the use
of coordinate or plotting paper, con- (V3)

structed by ruling off the plane into

equal squares, the sides being parallel
to the axes. Thus, to plot the point (4, — 3), count off four divisions
from on the axis of X to the right, and then three divisions
downward from the point so determined on a line parallel to the
axis of Y. Similarly, the following figures show the plotted points
(- 2, 3), (- 3, - 4), (0, 3).

tr 1 —

Y'

1

> J

=t

JL-

— 1 1—

—

t\

1

Xf

(-3(-41

T..

■^II 1

-4-

(6,4).

*X

14. Distance of a point from the origin. Represent the abscissa of
a point P by a and the ordinate by b, and its distance from the
Yjl ' i i origin by h. Then

h = Va 2 + b%

since h is the hypotenuse of a right
triangle whose sides are a and b.
Although h may be either positive
or negative, it will be sufficient for
our purposes to treat it as being
always positive.

In order to become familiar with the notion of coordinates, the
student should plot a large number of points.

■*-x

28

PLANE TRIGONOMETRY

EXAMPLES

1. (a) Plot accurately the points (5, 4), (- 3, 4), (- 2, - 4), (6, - 1), (6, 0),
(- 5, 0), (0, 4), (0, - 3).

(b) What is the distance of each point from the origin ?

Ans. Vil, 5, 2 VI, etc.

2. Plot accurately the points (1, 1), (- 1, - 1), (-1, 1), (Vz, 1), (Vz, - 1),
( — V5, — 1), and find the distance of each one from the origin.

3. Plot accurately the points (V2, 0), (- 6, - 10), (3, - 2 V5), (10, 3), (0, 0),

(0, -V5), (3, -5), (-4, 5).

15. Trigonometric functions of any angle defined. So far the six

trigonometric functions have been defined only for acute angles
(§ 1, p. 2). Now, however, we shall give a new set of definitions
which will apply to any angle whatever, and which agree with the
definitions already given for acute angles.

&

Angle in first quadrant

Y \

i

>X

&■

Yi

<

B

B

1

3V

P

<

»

<

2

O

¥'■

X'

+x

Angle in second quadrant

jf-

n

n

l

<

\

,

(,

\

A

/

r*

s

A

3

1

^\

B

B

r

T

¥X

Angle in third quadrant

Angle in fourth quadrant

Take the origin of coordinates at the vertex of the angle and the
initial side as the axis of A'. Draw an angle XOB in each quadrant.

From any point P on the terminal side OB of the angle draw PQ
perpendicular to the initial side, or the initial side produced. In
every case OQ is the abscissa and QP the ordinate of the point P.

TBIGONOMETKIC FUNCTIONS OF ANY ANGLE 29

Denoting by XOB any one of these angles, their functions are
defined as the following ratios :

op ordinate op hypotenuse

(10) smXOB = ^- = r ; ) (13) cscXOg = — »= .. x — ;

v ' OP hypotenuse v ' QP ordinate

(11) cos XOB = ^ =

hypotenuse
_ qp _ ordinate _
^ ' OQ abscissa '

OP hypotenuse

(14) secXOB = ^=-^—. ;

v ' OQ abscissa

00 abscissa *

(15) cotJCOg = ^= .. „ •*
x ' QP ordinate

To the above six functions may be added the versed sine (written ver-
sin) and coversed sine (written coversin), which are defined as follows :

veisia XOB = 1 — cos XOB ; coversin XOB = 1 — sin. XOB.

16. Algebraic signs of the trigonometric functions. Bearing in mind
the rule for the algebraic signs of the abscissas and ordinates of points
given in § 13, p. 26, and remembering that the hypotenuse OP is
always taken as positive (§ 14, p. 27), we have at once, from the defi-
nitions of the trigonometric functions given in the last section, that :

In I Quadrant, all the functions are positive.

In II Quadrant, sin and esc are positive ; all the rest are negative.

In III Quadrant, tan and cot are positive; all the rest are negative.

In IV Quadrant, sec and cos are positive ; all the rest are negative.

These results are also exhibited in the following
Eulb fok Signs

II J

VI

sin +

CSC +

all +

tan +

cos +

iirs

cot +

sec +

A y

All functions not indicated in each quadrant are negative.
This rule for signs is easily memorized if the student remembers that
reciprocal functions of the same angle must necessarily have the same
sign, i.e. s in anc l esc have the same sign,

cos and sec have the same sign,
tan and cot have the same sign.

17. Having given the value of a trigonometric function, to construct
geometrically all the angles which satisfy the given value, and to find
the values of the other five functions. Here we will make use of the

* As in acute angles it is seen that the functions in one column are the reciprocals of
the functions in the other.

30

PLANE TRIGONOMETRY

notion of coordinates, assuming as before that each angle has its
vertex at the origin, and its initial side coinciding -with the axis of X.
It remains, then, only to fix the terminal side of each angle, or, what
amounts to the same thing, to determine one point (not the origin)
in the terminal side. When one function only is given, it will appear

that two terminal sides satisfying the
given condition may be constructed. Thus,
if we have given tanx = f,'we may write

tan x =

*x

2 — 2 _ ordinate
— 3 abscissa '

(12), p. 29

Fig. a

and hence, taking tan x = § , one ter-
minal side is determined by the origin
and (3, 2), giving the angle XOB (in
the first quadrant).

2

— - j is determined by the

The other terminal side, taking tan x

origin and (— 3, — 2), giving the angle XOB' (in the third quadrant)
Hence all the angles x * which satisfy the condition

tan x = §

have the initial side OX
and the terminal side OB,

or, have the initial side OX
and the terminal side OB'.

Fig. a
Fig.J

Let us now determine the values of all the
functions. From Fig. a,

OP = y/~OQ 2 + QP 2 = V9 + 4 = Vl3 (always positive).

Hence by § 15, p. 29, from Fig. a,
2

sin XOB =
cos XOB =
taxi XOB =

Vl3'
3

Vl3'

2

3 !

esc XOB =
sec XOB =

Vl3

2 !
Vl3

3 ''

cot XOB = - ■

2i

* It is evident that, corresponding to each figure, there are an infinite number of both
positive and negative angles differing by multiples of 360° which satisfy the given condition.

TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 31
Similarly, irom Fig. b,

sin XOB' ■■

2
Vis'

cos XOB' = -

Vl3

tan X OB' =

3'

esc XOB' = ■
sec XOB' = ■
cot XOB' = :

Vl3

2 !
Vl3

Or, denoting by a; any angle which satisfies the given condition,
we may write down these results in more compact form as follows :

sin x = ±

Vis'

cos x = ± . — ;
Vl3

tan x = — ;

o

csc x = ±

sec a; = ±

3

cot a; = — ■

Vl3
2 5
Vl3

The method is further illustrated in the following examples :

Ex. 1. Having given sin x = — \, construct the angle x. Also find the values
of the other five functions.

Solution. Here we may write,.by (10), p. 29,

1 _ — 1 _ ordinate
3 3 hypotenuse

sin a; :

(hypotenuse always positive).

Since abscissa = ± V(hypot.) 2 — (ord.) 2 _= ± V9 — 1 = ± 2 V2, one terminal
side is determined by the origin and ( — 2 V2, — 1) , ( y, ,

giving the angle XOB in the third quadrant. Here

*X

sin XOB = -
cos XOB
tan XO.B

The other terminal side is determined by the origin and (2 V2, — 1) giving
Y4-

V

csc XOB = - 3;

2V2

- ~3~ ;

sec XOB =

2 V2

1

2 Vi'

cot XOB = 2 V2.

the angle XOB' in fourth quadrant. Here
1

*X

sin XOR = -
cos XOB' =
tan XOB' =

3'
2j/2
~3~ ;
1

2V2'

cscXOB / = -3;

3

sec XOB' = ;

2V2

cotXaB' = -2\/2.

32 PLANE TRIGONOMETRY

Or, denoting by x any angle which satisfies the given condition, we have

sin x = ; esc x = — 6 ;

3

2V2 ,. 3

cos x = T —^— ; sec x = =F — -= ,

3 2V2

tanx = ± ; cotx = ±2\/2.

2V2

771

Ex. 2. Having given cot x = — , find all the other functions of x.
n

Solution. Here we may write, by (15), p. 29,

m — m abscissa

cotx = — = = — ,

n — n ordinate

and hypotenuse = Vm s + n 2 .

Hence one terminal side is determined by the origin and (ra, n), and the
other terminal side by the origin and (— m, — n). Therefore

n Vm ! + n 2

sin x = ± ■ ; esc x = ± ;

Vm" + n 2 n

m Vm 2 + n 2

cos x = + — ; secx = ± i

Vm 2 + n 2 ra

tanx = — ; cotx = — •
m »

EXAMPLES

In each of the following examples construct geometrically the angle x, and
compute the values of all the functions of x.

Given.

1. sin x = - •
5

1

2. cos x =

3

3. cotx = -3.t

5

4. sec x =

3

„ 13,

5. esc x = —
5 c

* When m and n have the same sign, x represents angles in the first and third quadrants.
"When m and n have opposite signs, x represents angles in the second and fourth quadrants.
3 3

t cot X = — 3 = =

1-1

6.

a

tan x = - ■

b

7.

sin x = c.

8.

a 2 -6 2

a 2 + 6 2

9.

esc x ■= — V3.

10.

m
cos x = — •

11.

tan x = — V 1

12.

2

sin x =

3

13.

tan x = 2. 5.

14.

sec x = p.

TKIGONOMETRIC FUNCTIONS OF ANY ANGLE 33

ANSWERS

Quad-
rant

sin

cos

tan

CSC

sec

cot

1.

I

3

5

4
5

3

4

5
3

5
4

4
3

II

3
5

4
5

3

4

5
3

5
4

4
~3

2.

II

2V^
3

1
3

-2V2

3

2V2

-3

1

2 V2

HI

2V2
3

1
3

2V2

3

2 V2

-3

1

2V2

3.

II

1

VTo

3

VTo

1
3

VTo

VTo

-3

IV

1
VTo

3
VTo

1
3

-VTo

VTo
3

-3

4

3

4

5

5

3

4.

II

5

5

~3

4

3

i

i

3

4

5

5

3

III

5

5

3

4

3

4

5

12

6

13

13

12

5.

I

13

13

12

5

12

~5~

5

12

5

13

13

12

II

13
a

_ 13

6
X

12
a
6

c

4. .. ..

5

12

If

6

Va 2 + 6 2

Va 2 +& 2

b

Va 2 + 6 2
c

Va 2 +6 2
± Vl-c 2

a
1
c

1 6
1
Vl-c 2

a

7

Vl-c 2

Vl-c 2

c

8.

I,

2 aft
a 2 +6 2

a 2 -6 2

2 aft
a 2 -ft 2

a 2 + 6 2
± lab

a 2 + 6 2

a 2 -& 2
± 2a&

IV

a 2 + 6 2

a 2 -ft 2

9.

III,
IV

1

Vs

*4

m
c

1

-Vs

c

-1-

c
m

±V2

10

y/&-Tr&

Vc 2 -m 2

1 m

c

m

Vt?-m*

Vc*-m 2

11.

rv

VTi
4

V2

-V7

4

VTi

4

1

V7

12.

in,

IV

2
3

V5

2

±Vl

3
2

3

¥ vi

^

13.

hi

5
V29

2
V29

5
2

V29

V29

±_ 2~

2
5

Vp 2 -1

1
p

+ *

P

1

-1

14.

±Vp 2 -l

Vp 2 -!

Vp 2 -i

34

PLANE TEIGONOMETEY

18. Five of the trigonometric functions expressed in terms of the
sixth. For this purpose it is again convenient to use the definitions
of the functions which depend on the notion of coordinates ( § 13,
p. 26). The following examples will illustrate the method.

Ex. 1. Express, in terms of sin a;, the other five functions of x.
sin x ordinate

Solution. Since sin a; = •

hypotenuse

by (10), p. 29

abscissa = ± V (hypotenuse) 2 — (ordinate) 2
— ± Vl — sin 2 x.

Hence, by definitions,*

±Vl-si» ! a:

sin x = sin x ;

sin a;

cos x = ± Vl — sin 2 a; j
tan x = ± -

sec x = ± ■

smx

vl — sin 2 x

cot x = ±

Vl — sin 2 x
Vl — sin 2 x

Ex. 2. Express, in terms of tan x, the other five functions of x.

tan x _ ordinate
1 abscissa

Solution. Since tan x :

hypotenuse = ± V (abscissa) 2 + (ordinate) 2

Hence

= ±Vl + tan 2 x.

Vl+ tan 2 x

esc x = ± ■

tanx

Vl + tan 2 x
tan x = tan x ;

secx = ± Vl + tan 2 x;

1

cotx =

tanx

Ex. 3. Having given secx = f , find the values of the other five functions.

„,..„. 5 hypotenuse , „ , x

Solution. Since secx = - = ^^ , by (14), p. 29

4 abscissa

ordinate = ± V(hypotenuse) 2 — (abscissa) 2
= ±V25-16
= ±3.

Hence

sin x = ± f ;
cos x = J ;
tan x = ± | ;

cscx =± -j

sec x = j ;
cot x = ± $ •

* It is convenient to draw a right triangle (as above) to serve as a check on the numer-
ical part (not the algebraic signs) of our work. We then refer to the definitions of the
functions of an acute angle (p. 2) where

adjacent side corresponds to the abscissa,
and opposite side corresponds to the ordinate.

TK1G0N0METK1C FUNCTIONS OP ANY ANGLE 35

EXAMPLES
1. Express, in terms of cos x, the other five functions of x.

Ans. sin x = ± Vl — cos 2 x ; cscx = ±
cosx=cosx; secx =

Vl — cos 2 x
1
cosx'

Vl— cos 2 x cosx

tanx=± ; cotx = ±-

Vl — cos 2 x

2. Express, in terms of cotx, the other fire functions of x.

Ans. sin x = ± — ; esc x = ± Vl + cot 2 x ;

Vl+cot^x

cotx Vl + cot 2 x

cos x = ± — ; sec x = ± ;

Vl + cot 2 x cotx

tan x = ; cotx = cotx.

cotx

3. Express, in terms of secx, the other five functions of x.

Vsec 2 x — 1 sec x

Ans. sin x = ± ; cscx = ± ■

secx Vsec 2 x-1

1

cos x = ; sec x = sec x :

secx

tanx = ± Vsec 2 x — 1 ; cotx = ± — •

Vsec 2 x — 1

4. Express, in terms of esc x, the other five functions of x.

Ans. sin x = ; esc x = esc x ;

cscx

Vcse 2 x — 1 cscx

cosx = ± ; secx = ±

oscx Vcse 2 x-1

tan x = 4- ; cotx = ± Vcsc 2 x — 1 .

Vcsc 2 x — 1

5. Having given sec x = — ^-, find the values of the other five functions of x.

Ans. sin x = ± |f ; cscx = ± |J;
cos x = — ^y ; sec x = — lg- ;
tanx = T "V"' cotx ~ ^ A-

6. Having given sin x = a, find the values of the other functions of x.

Ans. sin x = a ; esc x = - ;

a

cos

x = ± Vl - a? ; see x = ± — —

Vl- a 2

a Vl-a 2
tan x = ± , ; cot x = ± ■

VT^Hi? a

36

PLANE TRIGONOMETRY

7. Having given cot x = V2, find the values of the other functions of x.

Ans. sina; = ± — ; cscx = ±v3;

cosx = ±^-; secs = ±-y-;
tan x = ; cot x = V2.

19. Line definitions of the trigonometric functions. The definitions of
the trigonometric functions given in § 15, p. 29, are called the ratio
definitions. From these we shall now show how the functions of any

C a-

Angle in first quadrant

Angle in second quadrant

C -c

Angle in third quadrant

Angle in fourth quadrant

angle may be represented by the numerical measures of the lengths
of lines drawn as shown above in connection with a unit circle (i.e.
a circle with radius unity).

TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 37

Applying these ratio definitions, we get

QP

smAOP = 7Jp-^T ) = QP '

00

cos AOP = -P-— = 00 :

OP(=l) v '

txnAOP = ^ = AT * = AT;

sec .lOP =

OQ OA (= 1)

OP_ OT *

OQ ~ OA (= 1)

cot .40P = -^ = — — — — ' = BC:
QP OB (= 1)

.„„ OP OC t

esc .40P = = -— — ' = OC.

QP OB (= 1)

From these results the so-called line definitions of the trigono-
metric functions may be stated as follows :

The sin equals the length of the perpendicular drawn from the
extremity of the terminal radius to the horizontal diameter.

The cos equals the length of the line drawn from, the center to the
foot of this perpendicular.

The tan equals the length of a line drawn tangent to the circle
from, the right-hand extremity of the horizontal diameter and meet-
ing the terminal radius produced.

The sec equals the distance from the center to the point of inter-
section of this tangent with the terminal radius produced.

The cot equals the length of a line drawn tangent to the circle from,
the upper extremity of the vertical diameter and meeting the terminal
radius produced.

The esc equals the distance from the center to the point of intersec-
tion of this cotangent with the terminal radius produced.

Algebraic signs must, however, be attached to these lengths so as
to agree with the rule for the signs of the trigonometric functions
on p. 29. We observe that

sin and tan are positive if measured upward from the horizontal
diameter, and negative if measured downward ;

cos and cot are positive if measured to the right of the vertical diam-
eter, and negative if measured to the left ;

sec and esc are positive if measured in the same direction as the
terminal side of the angle, and negative if measured in the opposite
direction.

* Since triangles OQP and OAT are similar,
t Since triangles OQP and OBC are similar.

38

PLANE TRIGONOMETBY

20. Changes in the values of the functions as the angle varies.

(a) The sine. Let x denote the variable angle A OP.

As x decreases, the sine decreases through the values QiP lt Q2P1,

etc., and as x approaches zero as a limit, the sine approaches zero

as a limit. This is written .,, .

sin 0=0.

As x increases from 0° and approaches 90° as a limit, the sine is
positive, and increases from zero through the values Q S P S , QJPi, etc.,
and approaches OB (= 1) as a limit. This is written

sin 90°= 1.

As x increases from 90° and approaches 180° as a limit, the sine
is positive and decreases from OB (= 1) through Q 5 P 5 , etc., and ap-
proaches zero as a limit. This is written

sin 180° = 0.

As x increases from 180° and approaches 270° as a limit, the sine

is negative and increases in nu-
merical value from zero through
Q s P e , etc., and approaches the
limit OB' {=—!). This is written

sin 270° = - 1.

As x increases from 270°
and approaches 360° as a limit,
the sine is negative and de-
creases in numerical value from
OB' (= — 1) through Q 7 P 7 , etc.,
and approaches the limit zero.
This is written

sin 360° = 0.

(b) The cosine. Using the last figure, we see that as x decreases,
the cosine increases through the values OQ u OQ 2 , etc., and as x ap-
proaches zero as a limit, the cosine approaches the limit OA (= 1).

This is written „„ n o ,

cos 0=1.

As x increases from 0° and approaches 90° as a limit, the cosine
is positive and decreases from OA (= 1) through the values OQ 8 , OQ t)
etc., and approaches the limit zero. This is written

cos 90° = 0.

TEIGONOMETKIC FUNCTIONS OF ANY ANGLE 39

As x increases from 90° and approaches 180° as a limit, the cosine
is negative and increases in numerical value from zero through OQ 6 ,
etc., and approaches the limit OA'(=— 1). This is written

cos 180° = — 1.

As x increases from 180° and approaches 270° as a limit, the
cosine is negative and decreases in numerical value from OA' (= — 1)
through OQ 6 , etc., and approaches the limit zero. This is written

cos 270° = 0.

As x increases from 270° and approaches 360° as a limit, the cosine
is positive and increases from zero through OQ 1} etc., and approaches
the limit OA (= 1). This is written

cos 360° = 1.

(c) The tangent. Let x denote the variable angle AOT.

As x decreases, the tangent decreases through the values A T 1} A T 2 ,
etc., and as x approaches zero as a limit, the tangent approaches the
limit zero.' This is written
tan 0° = 0.

t

As x increases from 0° and approaches
90° as a limit, the tangent is positive and
increases from zero through the values
AT % , AT 4 , etc., without limit, i.e. beyond
any numerical value. This is written
tan 90° = + co.*

Now suppose the angle x to be equal
. to the angle A OP and let it approach
90° as a limit ; then the corresponding
tangent A T e is negative and increases
in numerical value without limit. This
is written

tan 90° = — ao.

We see, then, that the limit of the tangent will be + oo or — oo
according as x is increasing or decreasing as it approaches the limit
90°- As one statement these last two results are written

tan 90° = oo,
when, as in this book, no distinction is made for the manner in which
the angle approaches the limit 90°.

* + oois read plus infinity.
— oo is read minus infinity.
oo is read simply infinity.

40 PLANE TRIGONOMETRY

As x increases from 90° and approaches 180° as a limit, the tangent
is negative and decreases in numerical value from — oo through
AT 6 , AT h , etc., and approaches the limit zero. This is written

tan 180° = 0.

As x increases from 180° and approaches 270° as a limit, the
tangent is positive and increases from zero through A T„, A T 4 , etc.,
without limit. This is written

tan 270° = oo.

As x increases from 270° and approaches 360° as a limit, the

tangent is negative and decreases in numerical value from — oo

through AT 6 , AT B , etc., and approaches the limit zero. This is

written

tan 360° = 0.

(d) The secant. Using the last figure, we see that as x, decreases,
the secant decreases through the values OT x , OT it etc., and approaches
OA (= 1) as a limit. This is written

sec 0° = 1.

As x increases from 0° and approaches 90° as a limit, the secant
is positive and increases from OA (= 1) through OT it OT i} etc., with-
out limit. This is written

sec 90° = oo.

As x increases from 90° and approaches 180° as a limit, the secant
is negative and decreases in numerical value from — oo through 01\,
OT 5 , etc., and approaches minus OA (= — 1) as a limit. This is written

sec 180° = — 1.

As x increases from 180° and approaches 270° as a limit, the secant
is negative and increases in numerical value from minus OA (= — 1)
through OT s , OT i} etc., without limit. This is written

sec 270° = oo.

As x increases from 270° and approaches 360° as a limit, the secant
is positive and decreases from + oo through OT 6 , OT 5 , etc., and ap
proaches the limit OA (= 1). This is written

sec 360° = 1.

TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 41

(e) The cotangent. Let x denote the -variable angle AOC.

As x decreases, the cotangent increases through the values BC lt
BC 2 , etc., and as x ap-
proaches 0° as a limit, the
cotangent increases with-
out limit. This is written

cot0°

QO.

As x increases from 0°
and approaches 90° as a
limit, the cotangent is posi-
tive and decreases from
+ co through the values
BC S , BC it etc., and approaches the limit zero. This is written

cot 90° = 0.

As x increases from 90° and approaches 180° as a limit, the cotan-
gent is negative and increases in numerical value from zero through
BC S , BC S , etc., without limit. This is written

cot 180° = oo.

As x increases from 180° and approaches 270° as a limit, the cotan-
gent is positive and decreases from + co through BC Z , BC it etc., and
approaches the limit zero. This is written

cot 270° = 0.

As x increases from 270° and approaches 360° as a limit, the cotan-
gent is negative and increases in numerical value from zero through
BC h , BC e , etc., without limit. This is written

cot 360° = oo.

(/) The cosecant. Using the last figure, we see that as x decreases,
the cosecant increases through the values OC^, OC 2 , etc., and as x
approaches 0° as a limit, the cosecant increases without limit. This
is written

csc0°

: OO.

As x increases from 0° and approaches 90° as a limit, the cosecant
is positive and decreases from + co through 0C S) 0C 4 , etc., and ap-
proaches the limit OB(=l). This is written

esc 90° = 1.

42

PLANE TKIGONOMETRY

As x increases from 90° and approaches 180° as a limit, the cose-
cant is positive and increases from OB (= 1) through OC b) OC & , etc.,
without limit. This is written

esc 180° = oo.

As x increases from 180° and approaches 270° as a limit, the cose-
cant is negative and decreases in numerical value from — oo through
0C 8 , OCi, etc., and approaches the limit minus Ofi(=— 1). This is
written esc 2 70° — — 1 .

As x increases from 270° and approaches 360° as a limit, the
cosecant is negative and increases in numerical value from minus
OB(= — 1) through OC s , OC 6 , etc., without limit. This is written

esc 360° = oo.

These results may be written in tabulated form as follows : *

0°

90°

180°

270°

360°

sin

1

-1

cos

1

-1

1

tan

00

00

cot

oo

oo

oo

sec

2

00

-1

oo

1

CSC

oo

1

oo

-1

CO

It is of importance to note that as an angle varies its
sine and cosine can only take on values between — 1 and + 1 inclusive ;
tangent and cotangent can take on any values lohatever ;
secant and cosecant can take on any values whatever, except those
lying between — 1 and + 1.

Prove the following :

(a) sinO° + cos 90° = 0.

(b) sin 180° + cos 270° = 0.

(c) cosO° + tanO° = l.

(d) tanl80° + cot90° = 0.

(e) sin 270° - sin 90° = - 2.

EXAMPLES

(f) cos 0° + sin 90° = 2.

(g) cos 180° + sin 270° = - 2.
(h) sec0° + csc90° = 2.

(i) sec 180° - sec 0° = - 2.
(]') cos 90° -cos 270° = 0.

(k) sin 90° + cos 90° + esc 90° + cot 90° = 2.
(1) cos 180° + sec 180° + sin 180° + tan 180° = - 2.
(m) tan 360° - sin 270° - esc 270° + cos 360° = 3.

* The above table is easily memorized if the student will notice that the first four columns
are composed of squares of four blocks each, in which the numbers on the diagonals are the
same ; also the first two columns are identical with the next two if 1 be replaced by — 1 ;
also the first and last columns arc identical.

TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 43

2. Compute the values of the following expressions :

(a) a sin 0° + 6 cos 90° - c tan 180°. Ans. 0.

(b) a cos 90° -6 tan 180° + c cot 90°. 0.

(c) a sin 90° - b cos 360° + (a - b) cos 180°. 0.

(d) (a 2 - & 2 ) cos 360° - 4 ab sin 270°. a 2 + 4 ab - & 2 .

21. Angular measure. There are two systems in general use for the
measurement of angles. For elementary work in mathematics and
for engineering purposes the system most employed is

Degree measure, or the sexagesimal system. * The unit angle is one
degree, being the angle subtended at the center of a circle by an arc
whose length equals ^^ of the circumference of the circle. The degree
is subdivided into 60 minutes, and the minute into 60 seconds.
Degrees, minutes, and seconds are denoted by symbols. Thus 63
degrees 15 minutes 36 seconds is written 63° 15' 36"- Reducing the
seconds to the decimal part of a minute, the angle may be written
63° 15.6'. Reducing the minutes to the decimal part of a degree, the
angle may also be written 63.26°. f It has been assumed that the
student is already familiar with this system of measuring angles,
and the only reason for referring to it here is to compare it with
the following newer system.

22. Circular measure. The unit angle is one radian, being the angle
subtended at the center of a circle by an arc whose length equals the
length of the radius of the circle.

Thus, in the figure, if the length of the
arc AB equals the radius of the circle, then

angle A OB = 1 radian.

The circular measure of an angle is its
magnitude expressed in terms of radians.
This system was introduced early in the
last century. It is now used to a cer-
tain extent in practical work, and is
universally used in the higher branches of mathematics.

Both of the above systems will be used in what follows in this book. %

* Invented by the early Babylonians, -whose tables of weights and measures were based
on a scale of 60. This was probably due to the fact that they reckoned the year at 360 days.
This led to the division of the circumference of a circle into 360 degrees. A radius laid off
as a chord would then cut off 60 degrees.

t To. reduce seconds to the decimal part of a minute we divide the number of seconds by
60. Similarly, we reduce minutes to the decimal part of a degree. See Conversion Tables on
p. 17 of Granville's Four-Place Tables of Logarithms.

X A third system is the Centesimal or French System. The unit is one grade, being ,J 5 of
a right angle. Each grade is divided into 100 minutes and each minute into 100 seconds. This
system has not come into general use.

44 PLANE TRIGONOMETRY

Now let us find the relation between the old and new units. From
Geometry we know that the circumference of a circle equals 2 ttR ;
and this means that the radius may be measured off on the circumfer-
ence 2 7r times.* But by the above definition each radius measured
off on the circumference subtends an angle of one radian at the center,
and we also know that the angles about equal 360°. Therefore

2 it radians = 360°,

7r radians = 180°,

. ,. 180° 180°

1 radian = = , or,

7r 3.1416

(16) 1 radian = 57.2957° +.

It therefore follows at once that :

To reduce radians to degrees, multiply the number of radians by
57.2957 {=f}

To reduce degrees to radians, divide the number of degrees by
57.2957 (=^)-

Since 360 degrees = 2 tt radians,

it ,. 3.1416 ,.

1 degree = — - radian = radian, or,

loO lou

(17) 1 degree = .01745 radian.

Hence the above rules may also be stated as follows :

To reduce radians to degrees, divide the number of radians by

■ 01746 ( = iio)-

To reduce degrees to radians, multiply the number of degrees by

• 01745 ( = iio)-

The student should now become accustomed to expressing angles
in circular measure, thus :

TT

360° = 2 7r radians, 60° = — radians,

o

180° = ir radians, 30° = — radians,

6

' 90° = — radians, 45° = ^ radians,

3tt

IT

270° = — radians, 15° = — radians, etc.

* The student should carefully observe that we do not lay off these radii as chords.

TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 45

When writing the trigonometric functions of angles expressed in
circular measure it is customary to omit the word " radians," thus :

sin (it radians) is written simply sin tt and = sin 180°,

/7T \ 7T

tan / — radians I is written simply tan — and = tan 90°,
cot/-— radians j is written simply cot — and = cot 135°,

4

'•< I -r- radians J is written simply cos — and = cos 150°,

esc (1 radian) is written simply esc 1 and = esc 57.29°,
sec (J- radian) is written simply sec ^ and = sec 28.65°, etc.

Since the number of times that the radius of a circle can be meas-
ured off on an arc of the same circle determines the number of
radians in the angle subtended at the center by that arc, we have

,„„. ,_ , , ,. . , length of subtending arc

(18) Number of radians in angle = — = ; — rr- 2

v / length of radius

Hence, knowing any two of the three quantities involved, the third
may easily be found.

Ex. 1. What is the circular measure of the angle subtended by an arc of length

3. 7 in. if the radius of the circle is 2 in. ? Also express the angle in degrees.

Solution. Substituting in (18), we have

3 7
Number of radians = -^— — 1.85. Ans.

2

To reduce this angle to degrees, we have, from (16),
1.85 x 57.2957° = 105.997°. Ans.

Ex. 2. What is the radius of a circle in which an arc of length 64 in.

subtends an angle of 2.5 radians?

64
Solution. Substituting in (18), 2.5 = — ,

R

E = 25.6 in. Ans.

EXAMPLES

1. In what quadrant does an angle lie * if its sine and cosine are both nega-
tive ? if sine is positive and cosine negative ? if sine is negative and cosine posi-
tive ? if cosine and tangent are both negative ? if cosine is positive and tangent
negative ? if sine and cotangent are both negative ? if sine is negative and secant
positive ?

2. What signs must the functions of the acute angles of a right triangle
have? Why?

* That is, in what quadrant will its terminal side lie 1

46 PLANE TRIGONOMETRY

3. What functions of an angle of an oblique triangle may be negative ? Why f

4. In what quadrant do each of the following angles lie ?

6tt it lie Utt l\ir 15 tt it + 2 3tt + 2 1 _ _6

."l2 ; _ ; _ T ; ^" ; 4 ; 16 ; 6 ; 5- ' ' 1' '2'

5. Determine the signs of the six trigonometrical functions for each one of
the angles in the last example.

6. Express the following angles in degrees :

,,.1.2'. 25 . 8t.«- + 1. 3 . 28. 3T + 2

Ans. 74.4844°; 28.6478°; 120°; -143.239°; -67.5°;
39.549°; -171.887°; -160.4279°; 130.92°.

7. Express the following angles in circular measure: 22 J°; 60°; 135°;
-720°; 990°; -120°; -100.28°; 45.6°; 142° 43.2'; -243.87°; 125° 23' 19"
(1°=. 01 745333). Ans. 0.3926; 1.0470; 2.3558; -12.5640; 17.2755; -2.0940;

-1.7499; .7957; 2.4905; -4.2555; 2.1880.

8. Express in degrees and in radians :

(a) Seven tenths of four right angles.

(b) Five fourths of two right angles.

(c) Two thirds of one right angle.

Ans. (a) 252°, ?JL; (b)225°,^; (c) 60°, £.
5 4 3

9. Eind the number of radians in an angle at the center of a circle of radius
25 ft., which intercepts an arc of 37^ ft. Ans. 1.5.

10. Find the length of the arc subtending an angle of 4£ radians at the center
of a circle whose radius is 25 ft. Ans. 112^ ft.

11. Find the length of the radius of a circle at whose center an angle of 1.2
radians is subtended by an arc whose length is 9.6 ft. .4ns. 8 ft.

12. Find the length of an arc of 80° on a circle of 4 ft. radius. Ans. 5.6 ft.

13. Find the number of degrees in an angle at the center of a circle of radius
10 ft. which intercepts an arc of 5 it ft. Ans. 90°.

14. Find the number of radians in an angle at the center of a circle of radius
3-t 2 t inches, which intercepts an arc of 2 ft. Ans. 7.64.

15. How long does it take the minute hand of a clock to turn through — if

radians ? 50

Ans. — mm.

IT

16. What angle in circular measure does the hour hand of a clock describe
in39min.22isec? A - II rad.

64

17. A wheel makes 10 revolutions per second. How long does it take to turn
through 2 radians, taking tt = V ? Ans. ^J^ sec.

18. A railway train is traveling on a curve of half a mile radius at the rate
of 20 mi. per hour. Through what angle has it turned in 10 sec. ?

Ans. 6-j*r degrees.

19. The angle subtended by the sun at the eye of an observer is about half a
degree. Find approximately the diameter of the sun if its distance from the
observer be 90,000,000 mi. Ans. 785,400 mi.

TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 47

23. Reduction of trigonometric functions to functions of acute -angles.
The values of the functions of different angles are given in trig-
onometric tables, such, for instance, as the one on p. 9. These
tables, however, give the trigonometric functions of angles between
0° and 90° only, while in practice we sometimes have to deal with
positive angles greater than 90° ' and with negative angles. We
shall now show that the trigonometric functions of an angle of any
magnitude whatever, positive or negative, can be expressed in terms
of the trigonometric functions of a positive angle less than 90°,
that is, of an acute angle. In fact, we shall show, although this is
of less importance, that the functions of any angle can be found in
terms of the functions of a positive angle less than 45°.

In the next eighteen sections x and y denote acute angles.

24. Functions of complementary angles. To make our discussion
complete we repeat the following from p. 3.

Theorem. A function of an acute angle is equal to the co-function of
its complementary acute angle. ,

Ex. Express sin 72° as the function of a positive angle less than 45°.
Solution. Since 90° — 72° = 18°, 72° and 18° are complementary, and we get
sin 72° = cos 18°. Ans.

EXAMPLES

1. Express the following as functions of the complementary angle :

(a) cos 68°. (e) cot 9. 167°. (i) esc 62° 18'.

(b) tan 48.6°. (f) sin 72° 61' 43". .. oot .2_7r

(c) sec 81° 16'. tr U ' ' 6 '

(g) C °V (k) sin 1.2.

K) Sm 3 ' (h) sec 19° 29.8'. (1) tan 66° 22.3'.

2. Show that in a right triangle any function of one of the acute angles
equals the co-function of the other acute angle.

3. If A, B, C are the angles of any triangle, prove that

sin£A = cos %(B + C).

25. Reduction of functions of angles in the second quadrant.

First method. In the unit circle whose center is O (see figure on
next page), let A OP' be any angle in the second quadrant. The func-
tions of any such angle are the same as the corresponding functions
of the positive angle AOP' = 180° - P'OQ'. Let x be the measure of
the acute angle P'OQ', and construct AOP = P'OQ' = x.

48

PLANE TKIGONOMETKY

Now draw the lines representing all the functions of the supple-
mental angles x and 180° — x. From the figure

angle QOP = angle P'OQ',
OP = OP'.

by construction
equal radii

Therefore the right triangles OPQ and OP'Q' are equal, giving

OQ'=OQ.

But OQ'= cos (180° - x) and
OQ = cos x ; hence cos (180° — x)
equals cos x in numerical value.

Since they have opposite
signs, however, we get

cos (180° — x) = — cos x.

Also, from the same triangles,
Q'p' = QP.

But Q'P' = sin (180° — x) and QP = sin x, and since they have the

<

B

^\

P/

T

/r

A

1 qr

2

V'

same sign, we get

sin (180° — x) = sin x.

Similarly, the two right triangles OTA and OT'A may be proven
equal, giving

or,

AT' = AT and OV = OT,
tan (180° — a;) = — tan x and sec (180° — x) = — sec x.

In the same manner, by proving the right triangles OBC and OBC'
equal, we get

or,

BC = BC and OC = OC,
cot (180° — x) = — cot a; and esc (180° — x) = esc x.

esc (180° — x) = esc a; ;
sec (180° — a;) = — sec x ;
cot (180° — x) = — cot a;.

Collecting these results, we have
sin (180° — x) = sin x ;
cos (180° — x) = — cos a; ;
tan (180° — x) = — tan a; ;

Hence we have the

Theorem. The functions of an angle in the second quadrant equal
numerically the same-named functions of the acute angle between its
terminal side and the termdnal side of 180°. The algebraic signs,
however, are those for an angle in the second quadrant.

TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 49

Ex. 1. Express sin 123° as the function of an acute angle, and find its value.
Solution. Since 180° - 123° = 57°,

sin 123° = sin (180° - 57°) = sin 57° = .8387 (p. 9). Ans.

Ex. 2.' Find the value of sec

6

Solution, sec — = sec 150° = sec (180° — 30°) = - sec 30° = Ans.

6 K ' V3

Ex. 3. Find tan 516°.

Solution. 516° is an angle in the second quadrant, for 516° — 360° = 156°.
Hence tan 516° = tan 156°* = tan (180° - 24°) = - tan 24° = - .4452. Ans.
Second method. The angle A OP' may also be written 90° -f y, where y
measures the acute angle BOP'. Since the angles BOP' and P'OQ' are com-
plementary, we have, from theorem on p. 47,

sin x = cos y ; esc x = sec y ;

cos x = sin y ; secx=cscy;

tanx=coty; cotx = tany.

Since 180° — x = 90° + y, we get, combining the above results with the results
on the previous page,

sin (90° + y) = cos y ; esc (90° + y) = sec y ;

cos (90° + y) = — sin y ; sec (90° + y) = — esc y ;

tan (90° + V) = - cot y ; cot (90° + y) = — tan y.

Hence we have the

Theorem. The functions of an angle in the second quadrant equal numerically
the co-named functions of the acute angle between its terminal side and the termi-
nal side of 9<P. The algebraic signs, however, are those for an angle in the second
quadrant.

Ex. 4. Find the value of cos 109°.
Solution. Since 109° = 90° + 19°,

cos 109° = cos (90° + 19°) = - sin 19* = - .3256. Ans.

Ex. 5. Find the value of cos — — ■

4

Solution. — = 855° = 720°-+ 135°.

4
Therefore

cos — = cos 855° = cos 135° = cos (90° + 45°) = - sin45° = • Ans.

4 V2

The above two methods teach us how to do the same thing, namely,
how to find the functions of an angle in the second quadrant in terms
of the functions of an acute angle. The first method is generally to
be preferred, however, as the name of the function does not change,
and hence we are less likely to make a mistake.

* The above theorem was proven for an angle of any magnitude whatever whose termi-
nal side lies in the second quadrant. The generating line of the angle may have made one
or more complete revolutions before assuming the position of the terminal side. In that
case we should first (if the revolutions have been counter-clockwise, i.e. in the positive
direction) subtract such a multiple of 360" from the angle that the remainder will be a
positive angle less than 3C0°.

50

PLANE TRIGONOMETRY

EXAMPLES

1. Construct a table of sines, cosines, and tangents of all angles from 0° to
180° at intervals of 30°.

Ans.

0°

30°

60°

90°

120°

150°

180°

sin
cos
tan

1

1
2

V3
2
1

Vs

V3
2
1
2

V3

1

CO

V3
2
1
2

-V3

1

2

V3
2
1

"Vs

-1

2. Construct a table of sines, cosines, and tangents of all angles from 90° to
180° at intervals of 15°, using table on p. 9.

Ans.

90°

105°

120°

135°

150°

165°

180°

sin
cos
tan

1.0000
0.0000

CO

.9659

- .2588

- 3.7321

.8660
- .5000
-1.7321

.7071

-.7071

-1.0000

.5000
-.8660

-.5774

.2588
-.9659
-.2679

0.0000

-1.0000

0.0000

3. Construct a table of sines, cosines, and tangents of all angles from 90° to
135° at intervals of 5°.

4. Express the following as functions of an acute angle :

(a) sin 138°.

(b) tan 883°.

4 7T

(e) cot

v ' 5

(h)

. 13ir

sin

5

(c) cos 165° 20'.

(f) cot 170.48°.

«

cos 2.58.

(d) sec 102° 18'.

(g) esc 317°

U)

tan 1.5.

5. Find values of the

f ollowing :

(a) sin 128° = .788.

(b) cos 160°= -.9397.

. . . Sir
(g) sin—.

(m)
(n)

cot 95° 14'.
esc 126° 42.8'

(c) tanl35°=-l.

(d) sec— = -2.
v ' 3

/ \ . llir ..

(e) cot = — 1.

(h) tan 108° 15'.
(i) cos 173° 9.4'.

(j).tan

\ti 6

(k) cos 496.7°.

(o)
(P)

(q)

(r)

. 7tt

sm

9

cos 500°.

tan 870°.

sec 1.9°.

(f) esc 835°= 1.1034.

(1) sec 168.42°.

W

tanl.

6. Express the following as functions of an acute angle less than 45°

(a) sin 106° = cos 16°. llrr-

(b) cos 148.3° = -cos 31.7°. ( e ) csc^-.

(c) tan 862°. ^ 23?

(d) sec 794° 52'.

.„. -GO TV

(f ) COS

TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 51

sec x ;
cot (180° + a) = cot a;.

26. Reduction of functions of angles in the third quadrant.

First method. In the unit circle whose center is 0, let A OP' be any
angle in the third quadrant. The functions of any such angle are
the same as the corresponding functions of the positive angle
A OP' = 180° + Q'OP'. Let * be the measure of the acute angle
Q'OP', and construct AOP — Q'OP' = x.

Now drawing the lines representing all the functions of the angles
x and 180° + x, we get, just as in the previous case,

sin (180° + x) = — sin x ; esc (180° + x) — — esc x ;

cos (180° + x) — - cos x ; sec (180° + x) -.

tan (180° + x) = tan x ;

Hence we have the

Theorem. The functions of an
angle in the third quadrant equal
numerically the same-named func-
tions of the acute angle between its
terminal side and the terminal side
of 180°. The algebraic signs, how-
ever, are those for an angle in the
third quadrant.

Ex. 1. Express cos 217° as the func-
tion of an acute angle, and find its value. -B '
Solution. Since 217° - 180° = 37°,

cos 217° = cos (180° + 37°) = - cos 37° = - .7986. Ans.

Ex. 2. Find value of esc 225°.

Solution, esc 225° = esc (180° + 45°) = -esc 45° = -V2. Ans.

Ex. 3. Find value of sin 600°.

Solution. 600° is an angle in the third quadrant, for 600° - 360° = 240°.

V3
Hence sin 600° = sin 240° = sin (180° + 60°) = - sin 60° = - — . Ans.

Second method. The angle A OP' may also be. written 270° - y, where y meas-
ures the acute angle P'OB'. Since the angles P'OB' and Q/OP' (=AOP) are
complementary, we have, from theorem on p. 47, combined with the above
results, remembering that 180° + x = 270° - y,

sin (270° - y) = - cos y ; esc (270° - y) = - sec y ;

cos (270° - y) = - sin y ; sec (270° - y) = - esc y ;

tan (270° - y) = cot y ; cot (270° - y) = tan y.

Hence we have the

Theorem. The functions of an angle in the third quadrant equal numerically
the co-named functions of the acute angle between its terminal side and the termi-
nal side of 270°- The algebraic signs, however, are those of an angle in the third
quadrant.

^X

P/

T

1 Q '

1

rfx\

A

V/6

Q

/ V*

p

R=l .

52

PLANE TRIGONOMETRY

Ex. 4. Find sin 269°.

Solution. Since 270° - 11° = 259°,

sin 259° = sin (270° - 11°) = - cos 11° = - .9816. Ans.
As in the last case, the first method is generally to be preferred.

EXAMPLES

1. Construct a table of sines, cosines, and tangents of all angles from 0° to
270° at intervals of 45°.

Ans.

0°

45°

90°

135°

180°

225°

270°

sin

1

1

1
V2

1

-1

cos

1

l

1

"V2

-1

1
V5

tan

l

00

-1

l

oo

2. Construct a table of sines, cosines, and tangents of all angles from 180° to
270° at intervals of 15°, using table on p. 9.

Ans.

180°

195°

210°

225°

240°

255°

270°

sin
cos
tan

- 1.0000

- .-2588

- .9659
.2679

- .5000

- .8660

.5774

-.7071

-.7071

1.0000

- .8660

-.5000

1.7321

- .9659

- .2588
3.7321

- 1.0000

00

3. Construct a table of sines, cosines, and tangents of all angles from 136° to
270° at intervals of 5°.

4. Express the following as functions of an acute angle :

(a) tan 200°.

(b) sin 583°.

(c) cos 224° 26'.

(d) sec 260° 40'.

5. Find values of the following :

(a) tan 235° = 1.4281.

(b) cot 1300° = 1.1918.

(c) sin 212° 16'.

,,. 4ff 1

(d) cos — =

V ' 3 2

. . 7tt 2

(e)sec~=- — . (j) csc

(f) sin 609°.

(e) cot
o

(f) csc 4.3.

(g) sin 128°.

13tt

(g) cos ^--

(h) tan 4.

... 29 7T

(l) cot—-.

21 7T

(k) sin 228.4°.
(1) tan 255° 27.

(h) cos 998.7°.

... . 16tt

(i) sin—.

... 8tt

(3) cos -g--

(m) cot 185° 52'.

(n) cos 587°.

(o) esc

7

(p) sin 262° 10'.

(q) cos 204.86°.

(r) tan

TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 53

6. Express the following as functions of an acute angle less than 45° :

, x 17tt

(a) cos — .

(b) tan 236.5°.

(c) sin 594°.

,,, 5 JT

(d) sec

23 7T

(e) sin-

i

(f) cos 260° 53.4'.

27. Reduction of functions of angles in the fourth quadrant.

First method. As before, let A OP 1 be any angle in the fourth
quadrant. The functions of any-
such angle are the same as the C^
corresponding functions of the
positive angle A OP' = 360° —
P'OQ. Let x be the measure of
the acute angle P'OQ, and con-
struct A OP = P'OQ = x. Now,
drawing the lines representing
all the functions of the angles x
and 360° — x, we get, just as in
the previous eases,

sin (360° - as) = — sin x ;
cos (360° — x) = cos x ;
tan (360° - *) = — tan x ;

esc (360° — x) = — esc x ;
sec (360° — x) = see x ;
cot (360° — x) = — cot x ;

Hence we have the

Theorem. The functions of an angle in the fourth quadrant equal
numerically the same-named functions of the acute angle between its
terminal side and the terminal side of 360°. The algebraic signs,
however, are those for an angle in the fourth quadrant.

Ex. 1. Express sin 327° as the function of an acute angle, and find its value.
Solution. Since 360° - 327° = 33°,

sin 327° = sin (360° - 33°) = - sin 33° = - . 5446. Ans.

Ex. 2. Find value of cot — •

o

Solution, cot— = cot 300°= cot (360° -60°) = -cot 60° = - — • Ans.

Ex. 3. Find value of cos 1000°.

Solution. This is an angle in the fourth quadrant, for 1000° - 720° = 280°.

Hence cos 1000° = cos 280° = cos (360° - 80°) = cos 80° =. 1736. Ans.

54

PLANE TRIGONOMETRY

Second method. The angle A OP' may also be written 270° + y, where y
measures the acute angle B'OP'. Since the angles B'OP' and P'OQ are com-
plementary, we have, from theorem on p. 47, combined with the above results,
remembering that 360° - x = 270° + y,

sin (270° + y) =- cosy;
cos (270° + y) = siny;
tan (270° + y) =-coty;

esc (270° + y) = - sec y ;
sec(270° + ^) = oscy;
cot(270° + y) = -tanjr.

Hence we have the

Theorem. The functions of an angle in the fourth quadrant equal numerically
the co-named functions of the acute angle between its terminal side and the termi-
nal side of 270°. The algebraic signs, however, are those of an angle in the fourth
quadrant.

IItt

Ex. 4. Find value of cos ;

Solution, cos— = cos 330°= cos (270° +60°) = sin 60° = — • Ans.
6 ^

As before, the first method is generally to be preferred.

EXAMPLES

1. Construct a table of sines, cosines, and tangents of all angles from 180° to
360° at intervals of 30°.

Ans.

180°

210°

240°

270°

300°

330°

360°

1

V3

Vs

1

sin

~2

2

— 1

2

2

V3

1

1

V3

1

cos

— 1

2

2

2

2

tan

1
V3

V^

00

-V3

1

~V3

2. Construct a table of sines, cosines, and tangents of all angles from 270° to
360° at intervals of 15°, using table on p. 9.

Ans.

270°

285°

300°

315°

330°

345°

360°

sin
cos
tan

-1.0000

00

-.9659

.2588

-3.7321

-.8660

.5000

-1.7321

-.7071

.7071

-1.0000

-.5000

.8660

-.5774

-.2588

.9659

-.2679

1.0000

3. Construct a table of sines, cosines, and tangents of all angles from 270° to
860° at intervals of 5°.

, TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 55

4. Express the following as functions of an acute angle :

(a) sin 289°.

(b) cos 322.4°.

(c) tan 295° 43'.

(d) cot 356° 11'.
•

5. Find values of the following :

(a) sin275° = -.9962.

(b) cos 336° = .9135.

(c) tan 687° = -.6494.

(d) cot 1055°.

(e) sec 295° 52.6'.

... . 17 TT

(f) sin

v ; 9

(e) sm655°.

/« ' 9 ""

(f) esc — -

(g) sin 275. 5°.

(g) esc 5.2.

(h) cos

4

... 11 ir

(i) csc-— -

(h) cos

18 1

(j) fe&

5tt

(i) sec 246°.

0) tan — -

(k) sin 27 5° 22'.

(1) cot 348°.

(m) tan 660°.

/ x 13,r

(n) sec

v ' 8

. . . hir

(o) sin — .

28. Reduction of functions of negative angles. Simple relations exist
between the functions of the angle x and — x where x is any angle
whatever. It is evident that x and — x will lie, one in the first quad-
rant and the other in the fourth quadrant, as angles A OP and A OP'

B „, , B

in the right-hand figure ; or, one will lie in the second quadrant and
the other in the third quadrant, as the angles A OP and A OP' in the
left-hand figure. In either figure, remembering the rule for signs
(§ 16, p. 29), we get

QP=-QP',
OQ = OQ,
AT = -AT',
OT = OT',
BC=- BC,
OC =-OC,

. sin x = — sin (— x) ;
cos x = cos (— x) ;
, tan x = — tan (— x) ;
. sec x = sec (— x) ;
. cot x = — cot (— x) ;
, csc x = — csc (— x).

We may write these results in the form

sin (— x) = — sin x ; esc (— x) — — csc x ;

cos (— x) = cos x ; sec (— x) = sec x ;

tan (— x) = — tan x ; cot(— x) = — cot x.

56

PLANE TRIGONOMETRY

Hence we have the

Theorem. The functions of — x equal numerically the same-named
functions of x. The algebraic sign, however, will change for all
functions except the cosine and secant*

Ex. 1. Express tan ( — 29°) as the function of an acute angle, and find its value.
Solution, tan (-29°) = -tan 29° = -.5543. Ans.

Ex. 2. Find value of sec (- 135°).

Solution, sec ( - 135°) = sec 135° = sec (180° - 45°) = - sec 46° = - Vi. Ans.

Ex. 3. Find value of sin (- 540°).

Solution. sm(-540°) = -sin54Q° = -sin(360 o +180 o ) = -sinl80°=0. Ans.

EXAMPLES

1. Construct a table of sines, cosines, and tangents of all angles from 0° to
360° at intervals of 30°. Ans.

2. Find values of the following :

(a) tan (-33°) = -.6494.

V3

(b) sin(-60°) = -— •

V2

(c) cos (-135°) = -

(d) cot(- 259°).
(e)sec(-j).
(f) sin (-1231°).

to) «*(-!£).

(h) sin (-1000°).
(i) cos (-2.3).

M-t)-

(k) sin (-176.9°).
(1) cos (-88° 12.7')

(m) tan/--^V
(n) cot(- 842°).

* Another method for reducing the functions of a negative angle consists in adding such
a multiple of + 360° to the negative angle that the sum becomes a positive angle less than
360°. The functions of this positive angle -will be the same as the functions of the given
negative angle, since their terminal sides will coincide. To illustrate :

Ex. Find value of cos (- 240°).

Solution. Adding + 360° to - 240° gives + 120°.

Hence cos(- 240°)= cos 120° = cos (180° -60°)= -cos 60°=- J. Ans.

Angle

sin

cos

tan

0°

1

1

Vs

1

— 30°

2

2

"Vi

-60°

_Vs

2

1
2

-Vs

-90°

-1

00

-120°

VI

2

1
_ 2

V5

- 150°

1

2

Vs

2

1
V§

- 180°

-1

-210°

1
2

_V3
2

1

"vs

-240°

Vs
2

1
~2

-Vs

-270°

1

00

-300°

Vs

2

1
2

V3

-330°

1
2

VI
2

1

Vs

-360°

1

TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 57

29. General rule for reducing the functions of any angle to the functions
of an acute angle. The results of the last seven sections may be stated
in compact form as follows, x being an acute angle.*

General Rule.

I. Whenever the angle is 180° ±x or 360° ± x, the functions of the
angle are numerically equal to the same-named functions of x.

II. Whenever the angle is 90° ± x or 270° ± x, the functions of the
angle are numerically equal to the co-named functions of x.

III. In any case the sign of the result is the same as the sign of the
given function taken in the quadrant where the given angle lies.

The student is advised to use I wherever possible, since the liabil-
ity of making a mistake is less when the name of the function
remains unchanged throughout the operation. Work out examples
from pp. 50-56, applying the above general rule.

Angle

sin

cos

tan

cot

sec

CSC

90°

95°

100°

1.0000
.9962
.9848

0.0000
- .0872
-.1736

00

-11.430
-5.6713

0.0000
-.0875
-.1763

CO

-11.474
-5.7688

1.0000
1.0038
1.0154

EXAMPLES
1. Construct a table for every five degrees from 90° to 180°.

Ans.

2. Construct a table as in Ex. 1 for every 15° from 180° to 270°.

3. Construct a table as in Ex. 1 for every 10° from 0° to — 90°.

4. Reduce the following to functions of x :

(a) sin (x— 90°) = — cosx.f

(b) cos (x — 7r) = — cos x.

(c) tan I — x 1 = cot x.

(d) cot(x - 2tt) = cotx. (j) cot(- x - 8tt).

(e) sec (x — 180°) = — sec x.

(f ) esc ( — X ] = — sec x.

(g) sin (x - 270°).
(h) cos(— x — ir).

(i) tan^x--^j-

(k) sec(x-630°).
(1) esc (x - 720°).

* In case the given angle is greater than 360° we assume that it has first heen reduced to
a positive angle less than 360° hy the subtraction of some multiple of 360°. Or, if the given
angle is negative, "we assume that it has been reduced to a positive angle by the theorem on
p. 66.

t Since x is acute, % — 90° w a negative angle. Hence sin (x - 90°) = — sin (90° — x) = — cos a*

58 PLANE TRIGONOMETRY

5. Find values of the following :

(a) cos 420°. (i) cot 549° 39'. , p)seo !£5.

(b) sin7«8°. (j) esc 387° 68'. v/ 6

/ 4ir\ „, / 11 t\ (q) sin (-5.3).

(0)B ec(- T j. (k)sec(-— j.

(d) cot (-240°)

(r) cos ( — 1 •

n tan /_ Stt\ W • \ 12/

(e)csc^. ^ i ' <s)tan(-^V

(6) cso 3 (m) sin (-830°). W \ 12/

(f) tan 7. 5. 9ir (t) sec (- 123.8°).

(g) sin (-2.8). (n) cos — . (u) sin (- 256° 19.6')
(h) cos 952.8°. (o) cot 1020°. (v) cos(- 98° 3r).

6. Prove the following :

(a) sin 420° • cos 390° + cos ( - 300°) • sin ( - 330°) = 1.

(b) cos 570° • sin 510° - sin 330° • cos390° = 0.

(c) a cos (90° -z) + b cos (90° + x) = (o - 6) sin x.

(d) m cos I a;) ■ sinl x) = msinx cosx.

(e) (a - 6) tan (90° - x) + (a + 6) cot (90° + x) = (a - b) cotx - (a + 6) tan x.

(f) sinl — 1- x]sin(ir + x) + cosl — (- x)cos(7r — x) = 0.

(g) cos(7T+x)cosl y\ — sm(7r + x)sml y\ = co&xsmy — smxcosy.

(h) tan a; + tan(— y) — tan(-7r — y) = tanx.

(i) cos (90° + a) cos (270° - a) - sin (180° - a) sin (360° - a) = 2 sin^a.

sin (180° -y) 1

u; sin (270° -y) v "' sin2(270°-2/)

(k) 3 tan 210° + 2 tan 120° = - V3. (n) tan ^ (2 w + x) = tan ^ x.

(1) 5 sec" 135° - 6 cotf 300° = 8. (o) esc J (x - 2 n-j i = - sec \ x.

(m) cos £ (x - 270°) = sin \ x, (p) cos ^ (y - 810°) = - sin £ y.

CHAPTEE III

RELATIONS BETWEEN THE TRIGONOMETRIC FUNCTIONS

30. Fundamental relations between the functions. From the defi-
nitions (and footnote) on p. 29 we have at once the reciprocal
relations

sin*
1

(19)

sin x = ,

esc x

CSC* =

(20)

1

cos X = ,

^ sec*

sec* =

(21)

1
tan* = ,

rnt r

cot* =

COS*

1

tan*'

Making use of the unit circle, we shall now derive five more very-
important relations between the functions.
In the right triangle QPO

QP , , OQ

tan x = — — i and cot x = — ■

Substituting the functions
equivalent to QP and OQ, we get

, s sin* cos*

(22) tan-* = , cot* = -

v ' cos* sin*

Again, in the same triangle,

B

cot

/^

N(5^

T

/

44

a

f

f

t&x\

COS

A

I

Q

v_

R=\

(23)

In triangle OAT,

(24)

In triangle OCB,

(25)

QP +OQ =OP , or,
sin 2 * + cos 2 * = 1.
OA 2 + Zf 2 = Of 2 , or,
1 + tan 2 *=sec 2 *.

OB* + BC 2 = OC\ or
1 + cot 2 * = csc 2 *.

While in the above figure the angle a- has been taken in the first
quadrant, the results hold true for any angle whatever, for the above

59

60 PLANE TRIGONOMETRY

proofs apply to any one of the figures on p. 36 without the change
of a single letter.

While it is of the utmost importance to memorize formulas
(19) to (25), p. 59, as they stand, the student should also learn the
following formulas where each one of the functions is expressed
explicitly in terms of other functions.

(26) sin x =

v ' cscr

(19), p. 59

(27) sin x = ± Vl — cos 2 x* Solving (23), p.

59, for sin x

(28) cos x =

v ' sec x

(20), p. 59

(29) cos x — ± Vl — sin 2 Jr. Solving (23), p.

59, for cos x

(30) tan* = '• '

v ' cot X

(21), p. 59

(31) tan x = ± Vsec 2 x — 1. Solving (24), p.

59, for tan x

sin x sin x =fc Vl — cos 2 x
(3°) tan.r — — — '

v ' cos at iVl-sin 2 * cos*

[From (22), P- 59 ; also (29) and (27)-]

(33) csc x = —

v : ' sin x

(19), p. 59

(34>. csc x = ± Vl + cot 2 x. Solving (25), p.

59, for csc x

(35) sec x =

v J cos x

(20), p. 59

(36) sec x = ± Vl + tan 2 x. Solving (24), p.

59, for sec x

1

(37) cot x =

v ' tan x

(21), p. 59

(38) cot x = =fc Vcsc 2 x — 1. Solving (25), p.

59, for cot x

, „ cos x cos x ± Vl — sin 2 x

(39) cot X = = = ;

.

sinx ±Vl-cos 2 * Slnx

[From (22), p. BS ; also (27) and (29).]

31. Any function expressed in terms of each of the other five functions.

By means of the above formulas we may easily find any function
in terms of each one of the other five functions as follows :

* The double sign means that we get two values for some of the functions unless a con-
dition is given which determines whether to choose the plus or minus sign. The reason for
this is that there are two angles less than 360° for which a function has a given value.

TRIGONOMETRIC RELATIONS 61

Ex. 1. Find sin x in terms of each of the other five functions of x.
(a) SlnX = cic"£' from (26)

(b) sin x = ± Vl - cos 2 a, from (27)

(c) sini = substitute (34) in (a)

±V l + cot2a;

(d) sinx = ± -Jl L_ = ± sec 2 x-l substitute (28) in (b)

\ sec 2 a; sec a;

(e) sin a: = = ta " X ■ Substitute (37) in (c)

+ L 1 ±Vtan 2 x + l

\ tan 2 a;

Ex. 2. Find cos a; in terms of each of the other five functions.

(a) ° 0SX = iic^' from (28)

(b) cosx = ± Vl - sin 2 x, from (29)

1

(c) cosx = substitute (36) in (a)

±Vl + tan 2 x

(d) cos x = ± A /l- 1 = ±Vcsc2a ~ 1 , substitute (26) in (b)

\ CSC 2 X CSC X

, . 1 cotx

(e) cosx = _____ = • Substitute (30) m (c)

± I 1 ±Vcot 2 x + l

\ cot 2 x

Ex. 3. Find tan x in terms of each of the other five functions.

(a) tanx = — — , from (30)

cotx

tt>) tan x = ± Vsec 2 x - 1, from (31)

(c) tanx = - . from (32)

±Vl — sin 2 x

± Vl - cos 2

(d) tanx = , from (32)

COS X

(e) tanx = ■ Substituting (38) in (a)

± Vcsc 2 x — 1

Ex. 4. Prove that sec x — tan x • sin x = cos x.

Solution. Let us take the first member and reduce it by means of the formu-
las (26) to (39), p. 60, until it becomes identical with the second member. Thus

1 sinx .

sec x — tan x ■ sin x = sin x

cos x cos x

[Since sec x = and tan x = I
cos x cos x \

1 — sin 2 x cos 2 x

cos x cos x

= cosx. Ans.

(23), p. 50

* Usually it is best to change the given expression into one containing sines and cosines
only, and then change this into the required form. Any operation is admissible that does
not change the value of the expression. Use radidals only when unavoidable.

62

PLANE TRIGONOMETRY

Ex. 5. Prove that sin x (sec x + osc x) — cos x (sec x — esc x) = sec x esc x.
Solution, sin x (sec x + esc x) — cos x (sec x — esc x)

["

= sinx! 1 1 — cosxl ; )

\cosx sun/ \cosx sin a;/

sinz J

Since secx= and osc a:

cosx

sinx cosx

cosx sinx

sin x cos x
cos x sin x
sin 2 x + cos 2 x

cos x sin x

_1 1

cos x sin x

cos x sin x
= sec x esc x. 4ns,

(23), p. 59

EXAMPLES

1. Find sec x in terms of each of the other five functions of x.
Ans.

1

- 1 ± Vl + tan 2 x , -
oosx ±Vl-sin 2 x

± Vcof 2 X + 1

cscx

cotx ±Vcsc 2 x-l

2. Find cot x in terms of each of the other five functions of x.

1 / — : r ±Vl-sin 2 x

Ans. , ±vcsc 2 x — 1, : 1

cosx

tanx " sinx ±Vl — cos 2 x ±Vsec 2 x-

3. Find esc x in terms of each of the other five functions of x.
Ans.

- , ± Vl + cot 2 X , —

sinx ±Vl-

±Vtan 2 x + l

secx

cos 2 x

tan x ± Vsec 2 x — 1

4. Prove the following :

(a) cos x tan x = sin x.

(b) sin x sec x = tan x.

(c) sin y cot y = cos y.

(d) (1 + tan 2 y) cos 2 y = 1.

(e) sin 2 A + sin 2 4 tan 2 A = tan 2 4.

(f ) cot 2 4 - cos 2 A = cot 2 4 cos 2 4 .

(g) tan A + cot.4 = sec A esc A.

(h) cos 4. esc A = cot 4.

(i) cos 2 A — sin 2 A = 1-2 sin* A.

(j) cos 2 4 - sin* A = 2 cos 2 A - 1.

(k) (l + cot 2 B)sin 2 .B = l.

(1) (csc 2 ^ - 1) sin 2 A = cos 2 A.

(m) see 2 A + cso 2 ^. = sec 2 JL esc 2 A.

(n) cos 4 C - sin* C + 1 = 2 cos 2 C.

(o) (sinx + cosx) 2 + (sinx — cosx) 2 = 2.
(p) sin 8 x cos x -4- cos 3 x sin x = sin x cos x.
(q) sin 2 B + tan 2 B = sec 2 B — cos 2 #.
sin i/

(r) coty +

= esc y.

1 4- cos j/

(s) cos B tan B + sin B cot B = sin B + cos B.

(t) see x cso x (cos 2 x — sin 2 x) = cot x — tan x.

. . cosC sinC . _ _

(u) - — - — - + — = sin C + cosC.

1 — tan C 1 — cot C

sin z 1 + cos 2

(v) 1- -;—. = 2 cscz.

1 + cos z sin z

CHAPTEE IV

TRIGONOMETRIC ANALYSIS

32. Functions of the sum and of the difference of two angles. We now

proceed to express the trigonometric functions of the sum and dif-
ference of two angles in terms of the trigonometric functions of the
angles themselves.* The fundamental formulas to be derived are :

(40) sin (x + y) = sin x cos y -+- cos x sin y.

(41) sin (x — y) = sin x cos y — cos x sin y.

(42) cos (x + y) = cos x cos y — sin x sin y.

(43) cos (x — y) — cos x cos y + sin x sin y.

33. Sine and cosine of the sum of two angles. Proofs of formulas (40)
and (42). Let the angles x and y be each a positive angle less than
90°. In the unit circle whose center is 0, lay off the angle AOP = x
and the angle POQ = y. Then the angle A OQ = x + y.

O O E C O E

In the first figure the angle x + y is less than 90°, in the second
greater than 90°.

Draw QC perpendicular to OA. Then

(a) sin (x + y) = CQ, and

(b) cos (x + y) = OC.

Draw QD perpendicular to OP. Then

(c) sw.y=DQ, and

(d) cos y = OD.f

* Since x and y are angles, their sum x + y and their difference x - y are also angles.
Thus if x = 61° and y = 23°, then x + y =84° and x - y = 38°. The student should observe that
sin (x + y) is not the san.e as sin x + sin y, or cos(x - y) the same as cos x - cos y, etc.

t The student will see this at once if the book is turned until OP appears horizontal.

63

64 PLANE TRIGONOMETRY

Draw DE perpendicular and DF parallel to OA. Then angle
DQF = angle AOP (= x), having their sides perpendicular each to
each. From (a),

(e) sin (x + y) = CQ = CF + FQ = ED + FQ.

ED being one side of the right triangle OED, we have

ED = OD • sin x. from (7), p. 11

But from (d), OD = cos y. Therefore

(f ) ED = sin x cos y.

FQ being one side of the right triangle QFD, we have

FQ = DQ- cos x. from (8), p. 11

But from (c), DQ = sin y. Therefore

(g) FQ = cos x sin y.

Substituting (f) and (g) in (e), we get

(40) sin(x + y) = sin x cosy + cos x sin y.

To derive (42) we use the same figures. Erom (b),

(h) cos (x + y) = OC = OE -CE = OE - FD*

OE being one side of the right triangle OED, we have

OE = OD cos x. from (8), p. 11

But from (d), OD =-. cos y. Therefore

(i) OE = cos x cos y.

FD being a side of the right triangle QFD, we have

FD = DQ sin x. from (7), p. 11

But from (c), DQ = sin y. Therefore

(j) FD = sin x sin y.

Substituting (i) and (j) in (h), we get

(42) cos (* + y) = cos jr cosy — sinx siny.

In deriving formulas (40) and (42) we assumed that each of the
angles x and y were positive and less than 90°. It is a fact, however,
that these formulas hold true for values of x and y of any magnitude
whatever, positive or negative. The work which follows will illus-
trate how this may be shown for any particular case.

* When x + y is greater than 90°, OC is negative.

TRIGONOMETRIC ANALYSIS 65

Show that (42) is true when x is a positive angle in the second quadrant and y
a positive angle in the fourth quadrant.

Proof. Let x = 90° + x' and y = 270° + y'*; then x + y = 360° + (x' + y') and

(k) x' = x - 90°, y' = y- 270°, x'+y' = x + y- 360°.

cos (x + y) = cos [360° + (x' + y^] = cos (x' + y0 by § 29, p. 57

= cos x' cos y' — sin x' sin y' by (42)

= cos (x - 90°) cos (y - 270°) - sin (x - 90°) Sin (y - 270°) from (k)
= sinx(— siny) — (— cosxcosy) by §29, p. 57

= cos x cos y — sin x sin y. Q. e. d.

Show that (40) is true when x is a positive angle in the first quadrant and y
a negative angle in the second quadrant.

Proof. Let x = 90° - x' and y = - 180° - y'; then x + y = - 90° - (x' + y')
and

(1) x'=90°-x, y' = -180°-y, s' + y' = -90°-(x + y).

sin(x + y) = sin [- 90° - (x' + y^] = - cos(x' + y') by § 28, p. 56

= — [cos x' cos y' — sin x' sin y 7 ] by (40)

= -[cos(90°-x)cos(-180°-y)-sin(90°-x)sin(-180°-y)] from(l)
= — [sinx(— cosy) -- cosxsiny] by §29, p. 57

= sin x cos y + cos x sin y. q. e.d.

EXAMPLES

1. Find sin 75°, using (40) and the functions of 45° and 30°.
Solution. Since 75° = 45° + 30°, we get from (40)

sin 75° = sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30°

1 V5 1 1 . K

= from p. 5

V2 2 V2 2

V3 + 1 .
= Ans.

2V2

2. Find cos(x + y), having given sinx = f and einy = -fg,x and y being
positive acute angles.

Solution. By the method shown on p. 30 we get first

sin x = f , cos x = f , sin y = -^, cosy = |§.

Substituting these values in (42), we get

aoB{x + V ) = t-ii-i- T % = ft- Ans.

* The student should note that x" and y' are acute angles.

66 PLANE TRIGONOMETRY

3. Show that cos 75° = — , using the functions of 46° and 30°.

2V2

4. Prove that sin 90° = 1 and cos 90° = 0, using the functions of 60° and 30°.

5. If tan x = I and tan y = ^ ¥ , find sin (x + y) and cos (x + y) when x and y
are acute angles. Ans. sin (x + y) = £ , cos (x + y) = f .

6. By means of (40) and (42) express the sine and cosine of 90° + x, 180° + x,
270° + x, in terms of sin a; and cosx.

7. Verify the following :

(a) sin(45° + s) = C ° S ^, Sina; . (c) sin (y + 3 Q°) = ^ sin ^ + C0S ^ ■

(0)^(60°+^=°°^-^^. (d)co S( B + 45° ) = COS ^_ sinJ? .

8. Find sin (A + B) and cos (A + B), having given sin A = \ and sin B = %.

+ V5 + 2V3 -tVl5 — 2

Ans. sm(A + B) = ± ± , cos(A + B) = ±-^ — 1

34. Sine and cosine of the difference of two angles. Proofs of formulas

(41) and (43). It was shown in the last section that (40) and (42)
hold true for values of x and y of any magnitude whatever, positive
or negative. Hence (41) and (43) are merely special cases of (40)
and (42) respectively. Thus, from (40),

sin (x + y) = sin x cos y + cos x sin y.
Now replace y by — y. This gives

(a) sin (x — y) = sin x cos (— y) + cos x sin (— y).

But cos (— y) = cos y, and sin (— y) = — 6in y. from p. 55

Substituting back in (a), we get

(41) sin (x — y) = sin x cos y — cos x sin y.

Similarly, from (42),

cos (x + y) = cos x cos y — sin x sin y.

Now replace y by — y. This gives

(b) cos (x — y) = cos x cos (— y) — sin x sin (— y).

But cos (— y) = cos y, and sin (— y) = — sin y. from p. 55

Substituting back in (b), we get

(43) cos (x — y) = cos xcosy + sin * sin y.

TRIGONOMETRIC ANALYSIS 67

EXAMPLES

1. Find cos 15°, using (43) and the functions of 45° and 30°.
Solution. Since 15° = 45° — 30°, we get from (43)

cos 15° = cos (45° - 30°) = cos 45° cos 30° + sin 45° sin 30°

_j_ V£ j_ i

_ V2"' 2 V2'2

_ Vi + i

— 7=- - Ans.

2V2

Work out the above example, taking 15° = 60° — 45°.

2. Prove sin (60° + x) — sin (60° - x) = sin x.

Solution. sin (60° + x) = sin 60° cos x + cos 60° sin x. by (40)

sin (60° - x) = sin 60° cos x - cos 60° sin x. by (41 )

.-. sin (60° + x) — sin (60° — x) = 2 cos 60° sin x by subtraction

= 2 • J ■ sin x = sin x. .4ns.

V§ — 1

3. Show that sin 15° = -, using the functions of 45° and 30°.

2V2

■

4. Find sin(x — y) and cos(x — y), having given sinx = ^ and sin?/ = -^,

x and y being acute angles.

. . . , 2V2-V15 , , 2V30 + 1
Ans. sm(x-y) = — , cos(x-j/) = —

5. Find sin(x — y) and cos(x — y), having given tanx = ^ and tany = J, x
and y being acute angles. Ans. sin (x — y) = fci cos ( x — V) = If-

6. By means of (41) and (43) express the sine and cosine of 90° — x, 180° — x,
270° — x, 360° — x, in terms of sinx and cosx.

7. Verify the following :

(a) sin(45° - x) = ™*^* . (c) sin( V - 30°) = ^^~ C0Sy .

(b) CO s(60°-^)= C °^ 2+ 2 V5sin ^. W cos(i ? + 45° ) = £^^.

(e) sin (60° + x) - sin x = sin (60° - x).

(f ) cos (30° + y) - cos (30° -y) = - sin y.

(g) cos (45° + x) + cos (45° — x) = V2 cos x.
(h) sin (45° + P) - sin (45° - P) = V2 sinP.
(i) cos (Q + 45°) + sin (Q - 45°) = 0.

(j) sin (x + y) sin (x — y) — sin 2 x — sin 2 y.

(k) sin (x + 2/ + z) = sinx cos j/ cos z + cos x sin 2/ cos z

+ cos x cos y sin z — sin x sin y sin z.

Hint, sin (x + y + s) -= sin (a; + y) cos 2 + cos (x + y) sin z.

68 PLANE TRIGONOMETRY

35. Tangent and cotangent of the sum and of the difference of two
angles. From (22), p. 59, and (40) and (42), p. 63, we get

sin (x + y) sin x cos y + cos x sin y

\e\W ( 3T I iy) ~~ ■ — — — •

^ y ' cos (a; + y) cos x cos y — sin x sin y
Now divide both numerator and denominator by cos x cos y. This

sin x cos y cos x sin y
cos x cos y cos x cos y

S lves sin x cos y cos x sin y

tan (x + y) =

cos x cos y sin a; sin y
cos X cos ^ cos x cos y
sin x siny
cos x cos y
sin a; sin y
cos x cos y

sin a: , siny

Since = tan x and = tan y, we get

cos a; cosy

,..-. , tanx + tany

(44) tan (* + ?) = ■

1 — tan x tan y

In the same way, from (41) and (43) we get

tanjr — tany

(45) tan (* — y) = — —

v ' v ' 1 + tan x tan y

From (22), p. 59, and (40) and (42), p. 63, we get

cos (x + y) cos x cos y — sin x sin y

cot (a; + y) = . ) J ! = r-^-

v 7 sin (a; + y) sin a; cos y + cos x sin y

Now divide both numerator and denominator by sin x sin y. This

sdves

° cos a; cosy sin a: sin y

, , . sin x sin w sin x sin w

cot (x + y) = -, 2 __£

v ' sin a; cos y cos a; sin y

sin x sin y sin a; sin y
cos aj cos y

-. :— ^ — 1

sin x sin y

cos y cos x
sm y sin x

. cosx cosy

Since — — • = cot x, and — — - = cot y, we get
smx smy

, ABS ., , N cotjrcoty — 1

(46) cot (x + y) = —

v ' K ' cot y + cot x

TEIGONOMETKIC ANALYSIS 69

In the same way, from (41) and (43) we get

,._. L , . cot x cot y + 1

(47) cot (x — y) = 1-!— .

v ' v ' cot y — cot x

Formulas (40) to (47) may be written in a more compact form as

' "' "" ' ' sin (x ± y) = sin x cos y ± cos x sin y,

cos (x ±y) = cos x cos y zp sin x sin y,

, . . tan x ± tan y

tan(a; ± y) = ■ — =^-i

v ' 1 zp tan x tan y

. . . . cot x cot «il

cot (x ± y) = r - ^ —

^ ' cot jr ± cot x

The formulas derived in this chapter demonstrate the Addition
Theorem for trigonometric functions, namely, that any function of
the algebraic sum of two angles is expressible in terms of the func-
tions of those angles.

EXAMPLES

1. Find tan 16°, using (45) and the functions of 60° and 45°.
Solution. Since 15° = 60° - 45°, we get from (45)

tanl5° = tan(60°-45°) = ^an 60° - tan 45° = Vg-1 = g _ ^ ^
1 + tan 60° tan 45° i + V§
Work out above example, taking 15° = 45° — 30°.

2. Find tan (a; + y) and tan(x — y), having given tanx = \ and tany = \.

Ans. tan (x + y) = fy, tan (x — y) = f .

3. Find tan 75° from the functions of 45° and 30°. Ans. 2 + V&

4. Verify the following :

1 + tanx sin(x + y) tanx + tan?/

(a) tan (45° + x) = (f ) -^— =

1 — tan x sm (x — y) tan x — tan y

(h)'cot (y -45°) = i±^. <*) tanx + tan, = ^|±^.

(c) tan( 4 -60°) = ^^- W ^ " COt * = SfsS "

1 + V3tan^ cos(x + y)
V3cotB-l (i) 1- tanxtany = * '-.

(d) cot (P + 30°) = VdCOti * -i • oos * cos V

cotB + V3 ... . D ,„ n cos(P+Q)
(l) cot P cot — 1 = ! — •

(e) tan (x ±45°) + cot (xt 45°) = 0. w/ ^ sin P sin Q

36. Functions of twice an angle in terms of the functions of the angle.

Formulas (40) to (47) were shown to hold true for all possible values
of x and y ; hence they must hold true when x equals y.
To find sin 2 x we take (40),

sin (x + y) = sin x cos y + cos x sin y.

70 PLANE TRIGONOMETRY

Replace y by x. This gives

sin (x + x) = sin x cos x + cos x sin a;, or

(48) sin 2 Jf = 2 sin x cos jr.

To find cos 2 a; we take (42),

cos (x + y) = cos a; cos y — sin a; sin y.

Replace y by a;. This gives

cos (x + x) = cos a; cos a; — sin a; sin x, or

(49) cos 2 x = cos 2 x — sin 2 *.

Since cos 2 x = 1 — sin 2 x, (49) may be written

(49 a) cos 2 x — 1 — 2 sin 2 jr.

Or, since sin 2 x = 1 — cos 2 x, (49) may also be written

(49b) cos2* = 2 cos 2 * — 1.

To find tan 2 x we take (44),

_ tan x + tan y
*■ ^' 1 — tan a; tan y

Replace y by x. This gives

. tan x + tan a;

tan ( x + x) = ■ ; > or

^ ' 1 — tan x tan x

2 tan jc

( 5 °) Un2x =T-^Fx-

37. Functions of multiple angles. The method of the last section
may readily be extended to finding the functions of nx in terms of
the functions of x.

To find sin 3 x in terms of sin x we take (40),

sin (x + y) = sin x cos y + cos x sin y.

Replace y by 2 x. This gives
sin (x + 2 x) = sin x cos 2 x + cos x sin 2 x, or

sin 3 x = sin x (cos 2 x — sin 2 x) + cos x (2 sin x cos x) by (49), (48)
= 3 sinx cos 2 x — sin 8 x

= 3sinx(l— sin 2 x)— sih'x by (23), p. 5&

= 3 sin x — 4 sin 8 x. Ans.

To rind tan 4 x in terms of tan x, we take (44), (50),

,_ _ . 2 tan 2 x 4 tan x (1 — tan 2 x) .
tan4x=tan(2x + 2x) = 1 _ t?in22a! = 1 _ 6tan2a; + tan 4. Ans.

V

>

TBIGONOMETBIC ANALYSIS 71

EXAMPLES
2
1. Given sinx = — — , x lying in the second quadrant; find sin 2x, cos 2x,

tan 2 x. "V6

2
Solution. Since sin x = — and x lies in the second quadrant, we get, using

method on p. 30, "^

sinx=- — , cosx = , tanx = — 2.

VE V6

Substituting in (48), w6 get

4

2 / 1 \

sin2x = 2smxcosx = 2 ( — ) = —

VE\ VEJ

5

Similarly, we get cos2x =— § by substituting in (49), and tan 2x = £ by
substituting in (50).

2. Given tan x = 2, x lying in the third quadrant ; find sin 2 x, cos 2 x, tan 2 x.

-4ns. sin2x= f, cos2x = — §, tan2x = — $.

3. Given tan x = - ; find sin 2 x, cos 2 x, tan 2 x.

6 2a6 6 2 -a 2 2 aft

.4ns. ±

a 2 + 6 2 6 2 + a 2 6 2 - a 2

4. Show that cos3x = 4 cos 8 x — 3 cosx.

* ou , w . o .i 3 tan .4- tan 8 ^4

5. Show that tan 3 A =

1-3 tan 2 ^i

6. Show that sin 4 x = 4 sin x cos x — 8 sin 8 x cos x.

7. Show that cos4P = 1-8 sin 2 £ + 8 sin* P.

8. Show that sin 5x = 5 sinx — 20 sin 8 x + 16 sin 6 x.

9. Show that tan (45° + x) — tan (45° - x) = 2 tan 2 x.

10. Show that tan (45° + G) + tan (45° - C) = 2 sec 2 C.

11. Verify the following :

. „ 2tanx ._ „ csc 2 x

(a) sm2x = — • (f) sec2x = — -•

w 1 + tan 2 x csc 2 x - 2

1 — tan 2 x (g) 2 esc 2 a = sec s esc s.

(b) cos2x= l + tan2a .- (h 1 ) cot* -tan* = 2 cot 2*.

(c) tanP + cotP = 2 esc 2P. (i) cos 2x = 2_-sec 2 X -

(d) cos2x = cos 4 x - sin 4 x. sec 2 x

(e) (sinx + cos x) 2 = 1 + sin 2 x. (j) (sin x - cos x) 2 = 1 - sin 2 x.

12. In a right triangle, C being the right angle, prove the following -.

OB sin 2 .i-sin 2 P „. tan2 ,_ 2ab ,

(b) sin( J 4-P) + eos24 = 0. (i) sin 2 .4 = sin 2 P.

(c) sin(4-P) + sin(24 + C) = 0. (j)sin2X=^.

(d) (sin4 - sinP) 2 + (cos4 + cosP) 2 = 2. c 2

(e ) J«±I + A /3= 4===L- (k)cos24= 62 - a2

(f ) tan £ = cot A + cos C. n\ gm 3 4 =

(g) cos 2 4 + cos 2 P = 0.

c a
3 oW - a 8

72 PLANE TRIGONOMETRY

38. Functions of an angle in terms of functions of half the angle.

From (48), p. 70,

v ' sin 2 a; = 2 sm a; cos «.

Eeplaee 2xby x, or, what amounts to the same thing, replace *
by ^ • This gives

(51) sin * = 2 sin - cos - •

From (49), p. 70, cos 2 a; = cos 2 x — sin 2 x.

Replace 2 x by x, or, what amounts to the same thing, replace x
by ^ • This gives

X X

(52) cos x = cos 2 sin 3 - •

ft &

From (50), p. 70, tan 2 a; = 1 _ t ^ a . •

Replace 2 x by a;, or, what amounts to the same thing, replace x

X

— ■ This gives

2tan-
(53) tanx = .

1 _ tan 2 £
2

39. Functions of half an angle in terms of the cosine of the angle.

From (49 a) and (49 b), p. 70, we get

2 sin 2 x = 1 — cos 2 x,
and 2 cos 2 a: = 1 + cos 2 x.

Solving for sin x and cos x,

sin a; = ±

ll— cos 2 a:

N — 5

-±>j£

i j. + cos 2 a;
and cos x = ± -\l

Replace 2 x by x, or, what amounts to the same thing, replace x

by - • This gives -~ % ,

,_., . x 1 1 — cos x
(54) sil--=± A J - ,

(55) cos|.c±-y|-

+ cos*

\

TEIGONOMETRIC ANALYSIS
To get tan - we divide (54) by (55). This gives

73

. x

sm-

, x 2

tan- = -_

cos-

N

■ cos a;

-J-

+ eosa;

i) or,

(56)

tan - = db \\- —

2 \l +

COS X

+ COS X

Multiplyin g both nu merator and denominator of the right-hand
member by Vl + cos x, we get *

. V""

(

Verify t
sin 82° *

\

x sin x

tan- = ;

2 1 + cos x

or, muj.Jjjiying both numerator and denominator by Vl — cos x,
we get

(58)

X 1 — cos X

tan — = .

2 sinx

Since tangent and cotangent are reciprocal functions, we have at
once from (56), (57), and (58),

(59)
(60)
(61)

40. Sums and differences of functions. From p. 63,

(40) sin (x + y) = sin x cos y + cos x sin y.

(41) sin (x — y) = sin x cos y — cos x sin y.

(42) cos (x + y) = cos x cos y — sin x sin y.

(43) cos (x — y) = cos x cos y + sin a; sin y.

X

cot— —

± N

1 + cos X

2

1 — cos X

X

cot— =

2

1 + COS X

sinx

X

cot- —
2

sinx

1-c

osx

t -\ / l — COS3? VTT

cos x Vl— co&ne sin x

-cos a; 1 + cosa; I+coe-r
The positive sign vr> ly of the radical is taken since 1 + cos x can never be negative and
tan — and sin x alwajgo a-ve like signs.

74 PLANE TRIGONOMETRY

First add and then subtract (40) and (41). Similarly, (42) and (4$.'^
This gives

(a) sin (x + y) + sin (a; — y) = 2sina;cosy. Adding(40)and(41)

(b) sin (x + y) — sin (x — y)= 2 cos x sin y. Subtracting (4 l)from (40)

(c) cos(x + y) + cos(x — y)= 2 cos x cosy. Adding (42) and (43)

(d) cos(sc + y) — cos(a; — y)=— 2sina;siny. Subtracting(43)from(42)

Let x + y =A x + y = A

and x — y = B x — y =B

Adding, 2x=A+B Subtracting, 2 y = A — B

x = $(A+B). y = i(A-B).

Now replacing the values of x + y, x — y, x, y in tern/ ' f A and
B in (a) to (d) inclusive, we get I

(62) sin A + sin B = 2 sin | (4 + B) cos \(A — B)*

(63) sinX — sinB= 2 cos|(A + B) sin §(.4 — B).

(64) cos^ + cosB= 2cos|(A + B)v)s\{A — B).

(65) cos.4 — cos.B = — 2 &i.n.\(A + B) sin§(.4 — B).

Dividing (62) by (63), member for member, we obtain

sin^+sinB _ 2 sin £ (A +B) cos J {A —B)
siaA — sin.B — 2 cos £ (A + B) sin %(A—B)
_sinK£-KB) cos^(^--B)

But

(66)

cot^(A-B)

oos^(A+B) sin %(A—B)

= tan £ (4 + J3) cot i (A — B).

1

tan ±(A—B)

Hence

sin .4 + sinB _tanl(A + B)
amA — sinB ~~ tan|(4 — B)

EXAMPLES

1

1. Find sin 22 £°, having given cos 45° =

VI

* /r

Solution. From (64), sin- = ± -» / —

cosx

Let x = 45°, then - = 22 £°, and we ,

11--

sin 22£° =

V2 1

\

= !V2~
2

4ns.

TEIGONOMETEIG ANALYSIS 75

2. Reduce the sum sin 7 x + sin 3 a; to the form of a product.
Solution. From (62),

sin^. + sinB = 2sin^(^ + B)cos{-(A-B).

Let4 = 7x, B = 3x. Then A + B = lOx, and^.-.B = 4x.
Substituting back, we get

sin7a; + sin3x = 2sm5xcos2x. Ans.

3. Find cosine, tangent, and cosecant of 22j .

Xna. - V2+V2, Vis - 1,

2 Va'WI

4. Find sine, cosine, and tangent of 15°, having given cos 30° =

Ans. - V2-V3; - V2+V3"; -xl—
2 2 \| 2 +

V3

5. Verify the following: 2 2 >2+V3

(a) sin 82° + sin 28° = cos 2°. ,\(- x X V_ 1

(b) sin50° + sinl0° = 2sin30°cos20°. (S) \ Sm 2~ COS 2) - 1 ~ smx -

(c) cos 80° - cos 20°= - 2 sin 60° sin 30°. x sin £ x

(d) cos 5x + cos9x = 2 cos7xcos2x. ' ' an 4 — 1 + cos 1 x '

,. sin7x — sin5x „ sinl-r

( e ) — ; T- = tanx - (i\ cot - - * .

0087^ + 00850; (,C °4 1-cos^x
sin 33° + sin 3° _ sin4 + sin _g ,

(f) cos33° + cos3°- tan18 - 0) cos ^ +cosB = to 2^ + B) -

a; 1

6. Find sine, cosine, and tangent of -, if cosx = -and x lies in the first

quadrant. V3 Vfi V2

4m - IT IT 1"

7. Find sine, cosine, and tangent of - , if cos x = a.

Il — a k + a /l-o

^ s - ± V^- ,± V^-' ± Vi+^-

8. Express sine, cosine, and tangent of 3 x in terms of cos 6 x.

/l — cos 6 x Vl + cos 6 x /l — cos6x

Ans . ±> /— ± , iVl + cosex'

9. In a right triangle, being the right angle and c the hypotenuse, prove
the following :

„B c — a ... a-b , A—B

(a) sin"- = - — -■ (d) - — - = tan
w 2 2c a + b

(b) (cos I + sin I J= £±i • (e) tan | =

w 2 2c

41. Trigonometric identities. A trigonometric identity is an expres-
sion which states in the form of an equation a relation which holds
true for all values of the angles involved. Thus, formulas (26) to (39),
p. 60, are all trigonometric identities, since they hold true for all

76 PLANE TRIGONOMETRY

values of x; also formulas (62) to (66), p. 74, are identities, since they
hold true for all values of A and B. In fact, a large part of the work
of this chapter has consisted in learning how to prove identities by-
reducing one member to the form of the other, using any known
identities (as in Ex. 4, p. 61).

Another method for proving an identity is to reduce both members
simultaneously, step by step, using known identities, until both mem-
bers are identical in form. No general method can be given that will
be the best to follow in all cases, but the following general directions
will be found useful.

General directions for proving a trigonometric identity.*

First step. If multiple angles, fractional angles, or the sums or
differences of angles are involved, reduce all to functions of single
angles only f and simplify.

Second step. If the resulting members are not readily reducible to
the same form, change all the functions into sines and cosines.

Third step. Clear of fractions and radicals.

Fourth step. Change all the functions to a single function. In case
the second step has been taken, this means that we change to sines only
or to cosines only. The two members may now easily be reduced to the
same form.

Ex. 1. Prove the identity

1 + tan 2 x tan x = seo 2 x.

Solution. Since tan 2 x = and sec 2 x = = , the

equation becomes : l-tan»x cos2x sec*x-sm 2 x

First step. 1 +

1 — tan 2 x cos 2 x — sin 2 x
l + tan 2 x_ 1

1 — tan 2 x cos 2 x — sin 2 x
sin 2 a;

1 +

cos 2 a 1

sin 2 x cos 2 x — sin 2 x
cos 2 x
cos 2 x + sin 2 x 1

or, simplifying,

Second step.

or, simplifying,

cos 2 x — sin 2 x cos 2 x — sin 2 x

Third step. cos 2 x + sin 2 x = 1,

° r 1 = 1. from (23), p. 59

* In -working out examples under this head it will appear that it is not necessary to take
all of the steps in every case, nor will it always be round the best plan to take the steps in
the order indicated.

t For instance, replace sin 2 a: by 2 sin a: cos x, tan 2 a; by - — anx , cos (a; + y) by cos x
cosy — sinarsiny, etc. 1 — tana;

TRIGONOMETRIC ANALYSIS 77

Ex. 2. Prove ^ (x + y) _ tan a + tan y
sin (x — y) tan x — tan y

Solution. Since sin (x + y) = sin x cos y + cos x sin y, and sin (x — y) = sin x cos y
- cos x sin y, the equation becomes :

First step.

sin x cos y + cos x sin y
sin x cos 2/ — cos x sin 2/

tan x + tan y
tan x — tan y
sin x sin 2/

Second step.

sin x cos 2/ + cos x sin 2/

cos x cos 2/

sin x cos j/ — cos x sin j/

sin x sin j/

cos x cos y

Simplifying,

sin x cos y + cosx sin y

sin x cos y + cos x sin y

sin x cos ?/ — cos x sin 2/

sin x cos 2/ — cos x sin 2/

or,

1 =

:1.

EXAMPLES
Prove the following identities :

1. tan x sin x + cos x = sec x.

2. cot x — sec x esc x (1 — 2 sin 2 x) = tan x.

3. (tanx +. cot x) sin x cosx = 1.
sin 2/ 1 — cos 2/

4.

1 + cos y sin y

-£.

, . ■ smA . . .

5. -\/ = sec4 — tan.4.

. + sin^i

6. tan x sin x cos x + sin x cos x cot x = 1.

7. cot 2 x = cos 2 x + (cot x cosx) 2 .

8. (sec 2/ + osc y) (1 — cot 2/) = (sec y — esc y) (1 + cot 2/).

9. sin 2 z tan z + cos 2 z cot z + 2 sin z cos z = tan z + cot z.

10. sin 6 x + cos 6 x = sin 4 x + cos 4 x — sin 2 x cos 2 x.

11. sinBtan 2 .B + csc.Bsec 2 .B = 2tanJBsec£+ cscB — sinB.

12. Work out (a) to (v) under Ex. 4, p. 62, following the above general
directions.

13. cos(x + 2/)cos(x — y) = cos 2 x — sin 2 y.

14. sin (A + B) sin (A - B) = cos 2 B - cos 2 A.

15. sin (x + y) cos y — cos (x + 2/) sin y = sin x.
cos (x — y) _ 1 + tan x tan y

16.

cos (x + y) 1 — tan x tan y
sin (4 - .B)

17. tan^A — tanB =

18. cotx + cot 2/ =

cos J. cos I?
sin (x + 2/)

sin x sin 2/
19. sin x cos (2/ + z) — sin 2/ cos (x + z) = sin (x - y) cos z.

78 PLANE TRIGONOMETRY

20. *"■('- »> + *■"*»= t»n g. .
1 — tan (0 — (p) tan cj>

21. cos (x — y + z) = cos x cos y cos z + cos x sin y sin z — sin x cos y sin z

+ sin x sin ?/ cos z.

22. sin (x + y — z) + sin (x + z — y) + sin (y + z — x)

= sin (x + y + z) + 4 sin x sin y sin z.

23. cos x sin (y — z) + cos y sin (z — x) + cos z sin (x — y)-= 0.

24. Work out (a) to (k) under Ex. 7, p. 67, and (a) to (j), under Ex. 4, p. 69,
following the above general directions.

25 l + sin2x = Aanx + iy ^ ^ = l-sin2X

1 — sin 2 x \tan x — 1/ cos 2 .4

sin2x „. cos 8 x + sin 8 x 2 — sin 2 x

26.

tan x =

1 + cos 2 x

30.

27.

sin2x

31.

1 — cos 2 x

28.

cot.4 — 1 /l — sin 2A
cot4 + l~ Vl + sin2vi

32.

33.

sin 3 x = 4 sin x sin (60° + x) sin (60°

-■).

34.

sin 4 x

= 2 cos 2 x.

sin2x

35.

sin 4 .B = 4 cos 2 B sin B cos B.

cos x + sin x 2

sin 3 x — sin x

cos 3 x + cos x
sin 3 x — sin x
cos x — cos 3 x

= tan x.
cot 2 x.

36. Work out (a) to (j) under Ex. 11, p. 71, following the above general
directions.

37. sin 9x — sin7x = 2 cos8xsinx.

1-tan 2 ?

38. cos7x + cos5x= 2cos6x cosx. .„ 2

46. = cosx.

39. sin5x-sin2x =eot 7x 1+tan 2 *

= uui • • o

cos 2 x — cos 5x2 z

acx I ■ x , X V i , • 2 tan -

40. sin - + cos - = 1 + sin x. 2

\ 2 2/ 47. — = sinx.

., 1+seoy „ „y 1 + tan 2 -

41. — ^ = 2cos 2 ^. 2

sec y 2 x

48. 1 + tan x tan - = sec x.

._ sin ^.+ sin B 1, , _ 2

42. -Z - = -cot-(4-B).

cos^-cosB 2 X 49. .tan? + 2sin^cotx = sinx.

2 2

43.

1 + tan -

cos 9 2 60. 1 + cot? - = .

1 ~ Sine 1-tan*' 2 Slna:tan l

2 xx

44. cos3or — cos7ar=2sin5arsin2cr. tan 2 - + cot 2 -

t K . , x

51.

2 2 l"+cos 2 x

45. cot - + tan - = 2 esc x. tan 2 - - cot 2 - 2 cos x

2 2 2 2

52. Work out (a) to (f) under Ex. 5, p. 76, following the above general
directions.

CHAPTEE V

GENERAL VALUES OF ANGLES. INVERSE TRIGONOMETRIC FUNC-
TIONS. TRIGONOMETRIC EQUATIONS

42. General value of an angle. Since all angles having the same
initial and terminal sides have the same functions, it follows that
we can add 2 v to the angle or subtract 2 it from the angle as many
times as we please without changing the value of any function.
Hence all functions of the angle x equal the corresponding func-
tions of the angle

2 mr + x,

where n is zero or any positive or negative integer.

The general value of an angle having a given trigonometric func-
tion is the expression or formula that includes all angles having this
trigonometric function. Such general values will now be derived for
all the trigonometric functions.

43. General value for all angles having the same sine or the same co-
secant. Let x be the leas't positive angle whose sine has the given
value a, and consider first the case when a is positive.

Construct the angle x (= XOP), as on p. 31, and also the angle
it — x (= XOP 1 ), having the
same value a for its sine. Then
every angle whose terminal
side is either OP or OP' has
its sine equal to a, and it is
evident that all such angles
are found by adding even
multiples of 7r to, or subtract-
ing them from, x and it — x.

Let n denote zero, or any
positive or negative integer. When n is even, mr + x includes all
the angles, and only those, which have the same initial and terminal
sides as x (= XOP). Therefore, when n is even,

(.4) mr + x = mr + (- l)"x*

* The factor (- 1)» is positive for all even values of n and negative for all odd values of n.

79

80

PLANE TRIGONOMETRY-

Again, when n is odd, n — 1 is even, and (n — 1) 7r + (nr — x) in-
cludes all the angles, and only those, which have the same initial
and terminal sides as it — x (= XOP 1 ). But when n is odd,

(*)

(n — 1) it + (it — x) = Jwr — a; = mr + (— l) n x.

From (4) and (B) it follows that the expression nir +(— l) n a; for
all values of n includes all the angles, and only those, which have

the same initial and terminal
sides as x and 7r — x.

In case a is negative, ir — x
will be negative, as shown in
the figure, but the former line
of reasoning will still hold
true in every particular.

Since sine and cosecant
are reciprocal functions, it
follows that the expression
for all angles having the
same cosecant is also nir + (— V) n x. Hence

(67) mr + (- l)"x

is the general value of all the angles, and only those, which have the
same sine or cosecant as x*

This result may also be expressed as follows :

sin x — sin[n7T + (— l)"Jf]i
esc x = csc[n7T + (— I)"*]-

o _

Ex. 1. Find the general value of all angles having the same sine as ——

Solution. Let x = — in (67). This gives
4

mr + (— 1)"

4 '

Ans.

Ex. 2. Find the four least positive angles whose cosecant equals 2.
Solution. The least positive angle whose cosecant = 2 is

This gives

-• Let x = - in (67).
6 6

717T + (— 1)«-

* In deriving this rale we have assumed x to be the least positive angle having the given
sine. It follows immediately from the discussion, however, that the rule holds true if we
replace x Dy an angle of any magnitude whatever, positive or negative, which has the given
sine. The same observation applies to the rules derived in the next two sections

GENEEAL VALUES OF ANGLES
When n = 0, we get

81

*= 30°.

When n = 1, we get
When n = 2, we get
When n = 3, we get

2ir +

150°.

: 390°.

3tt--= 510°. Ana.

44. General value for all angles having the same cosine or the same
secant. Let x be the least positive angle whose cosine has the given
value a, and consider first the case when a is positive. Construct
the angle x (=XOP), and also
the angle — x (= XOP'), hav-
ing the same value a for its
cosine. Then every angle
whose terminal side is either
OP or OP' has its cosine equal
to a, and it is evident that all
such angles are found by adding
even multiples of ir to, or sub-
tracting them from, x and — x.
Let n denote zero, or any
positive or negative integer. Eor any value of n,

(A) 2 wrr + x *

includes all the angles, and only those, which have the same initial
and terminal sides as x (= XOP). Similarly,

(B) 2 nir — x
includes all the angles, and
only those, which have the
same initial and terminal sides
as -x(=XOP f ).

In case a is negative, the
same line of reasoning still
holds true.

Since cosine and secant are
reciprocal functions, it follows
that the same discussion holds for the secant. Hence, from (A)a,nd(B),

(68) 2 nv ± x

is the general value of all the angles, and only those, which have the
name cosine or secant as x.

* in is even and 2» - 1 is odd for all values of n.

82

PLANE TRIGONOMETRY

This result may also be expressed as follows :

cos x = cos (2 mr d= x),
sec x = sec (2 rm ± jc) .

Ex. 1. Given aosA = — ; find the general value of A. Also find the five

least positive values of A. ~v2

Solution. The least positive angle whose cosine = — is If we let

3*- V2 4

x = — in (68), we get

4 4 = 2nx±^.

When n = 0,
When ji = 1,
When 7i = 2,

A

4

=

135°.

A

= 2tt

±

3?r_

225°

or

495°.

A

= 4tt

±

3?r_
4

685°

or

855°.

Arts.

45. General value for all angles having the same tangent or the same
cotangent. Let x be the least positive angle whose tangent has the
given value a, and consider first the case when a is positive.

Construct the angle x (= XOP), and also the angle it + x (= XOP')
having the same value a for its tangent. Then every angle whose

terminal side is either OP or
OP' has its tangent equal to a,
and it is evident that all such
angles are found by adding
even multiples of ir to, or
subtracting them from, x and
7T + x.

Let n denote zero, or any
positive or negative integer.
When n is even,

w

nir + x

includes all the angles, and only those, which have the same initial
and terminal sides as x (= XOP).

Again, when n is odd, n — 1 is even, and

(B) (n — 1)7T +(ir + x)= mr + x

includes all the angles, and only those, which have the same initial
and terminal sides as ir + x (—XOP 1 ).

In case a is negative, the sfime line of reasoning still holds true.

GENEKAL VALUES OF ANGLES

83

Since tangent and cotangent are reciprocal functions, it follows
that the same discussion holds for the cotangent. Hence, from (A)
and (-B), for all values of n,

(69) rnr + x

is the general value of all the
angles, and only those, which
have the same tangent or
cotangent as x.

This result may also be
stated as follows:

tan* = tan (tot + x),
cot x = cot (mr + x).

EXAMPLES

1. Given sin A = £ ; find the general value of A. Also find the four least

positive values of A.

Ans. mr + (- !)»-;, 30°, 150°, 390°, 510°.

V§

2. Given cos^i = — ; find the general value of A. Also find all values of A

a n "■ it llr

numerically less than 2 it. Ans. i mr ± — ; ±—,± — —

3. Given tan .4 = 1 ; find the general value of A. Also find the values of A

numerically less than 4 tt.

t jt 3tt bir 7ir 9ir Utt 13tt lbir

44 44 44 44 4

. «. . ~ 1 i ., mr . -v 7r

4. Given sin 2 x = - ; show that s = — - + (— 1)» — .

2 2 12

5. Given cos 3 x ■■

1 . t . . 2mr , 2tt

- ; show that x = ±

2' 3 9

6. In each of the following examples find the general values of the angles,
having given

Ans. A = mr + (- 1)» ( ± ^j =

(a) sin A = ±1.

(b) cotx=±Vl$.

(c) cosy=±-.

(d) tan.B = ±l.

(e) cscG = ±-^/2.

•■ mr ±

2

x = nir ±

6

(f) sec^=± ■

,/3

# = mr ± -•

O

4

4

A = mr ± — •

84 PLANE TEIGONOMETEY

7. Given sin x = and tan x = ; find the general value of x.

2 V3

Solution. Since sin a; is — and tans is +, x must lie in the third quadrant.
The smallest positive angle in the third quadrant which satisfies the condition

sin x = — is — , and this angle also satisfies tan x =

2 6 ^

Hence x = 2 wr -\ ■ Ans.

6

„, _Ji. In each of the following examples find all the positive values of x less
than 2 ir which satisfy the given equations.

(a) cosa; = — — • Ans. —,

V2 4 4

.... 1 J 3r 6x 7ir

(») sinx = ±— • .

V2 4 4 4 4

ir 27r 4t 5tt

I'T' T'T'

ir 3tt 5-ir Itt

T T' T' T

67T 77T

T'T"

8'T'

w 57r 7w llrr

46. Inverse trigonometric functions. The value of a trigonometric
function of an angle depends on the value of the angle ; and con-
versely, the value of the angle depends on the value of the function.
If the angle is given, the sine of the angle can be found ; if the sine
is given, the angle can be expressed. It is often convenient to rep-
resent an angle by the value of one of its functions. Thus, instead
of saying that an angle is SO , we may say (what amounts to the same
thing) that it is the least positive angle whose sine is ^. We then con-
sider the angle as a function of its sine, and the angle is said to be
an inverse trigonometric function, or anti-trigonometric, or inverse
circular function, and is denoted by the symbol

sin -11 -, or, arc sin £*,

read "inverse sine of \," or, "arc (or angle) whose sine is $."
Similarly, cos -1 a; is read "inverse cosine of x," tan" 1 ?/ is read "in-
verse tangent of y," etc. If a is the value of the tangent of the
angle x, we are now in a position to express the relation between a

* Symbol generally used in Continental books.

(c)

tana

= ±V5.

(d)

COtX:

= ±1.

(e)

COSX :

__V3
2

(f)

secx :

= 2.

(g)

sin x :

1

= ± 2-

INVEKSE TRIGONOMETRIC FUNCTIONS 85

and x in two different ways. Thus, tanx = a, meaning the tangent
of the angle x is a; or, x = tan -1 a, meaning x is an angle ivhose
tangeni is a. The student should note carefully that in this connec-
tion — 1 is not an algebraic exponent, but is merely a part of the
mathematical symbol denoting an inverse trigonometric function.
tan -1 a does not denote

(tana.) -1 = >

J tan a

but does denote each and every angle whose tangent is a.

The trigonometric .functions (ratios) are pure numbers, while the
inverse trigonometric functions are measures of angles, expressed in
degrees or radians.

Consider the expressions

tan x = a, x — tan -1 a.

In the first we know that for a given value of the angle x, tan x
(or a) has a single definite value.

In the second we know from (69), p. 83, .that for a given value
of the tangent a, tan -1 a (or x) has an infinite number of values.

Similarly, for each of the other inverse trigonometric functions.
Hence :

The trigonometric functions are single valued, and the inverse trigo-
nometric functions are many valued.

The smallest value numerically of an inverse trigonometric func-
tion is called its principal value.* For example, if

tan x =1,

then the general value of x is, by (69), p. 83,

77"

x = tan -1 1 = mr + — >
4

where n denotes zero or any positive or negative integer, and

7 = 45°
4

is the principal value of x.

* Hence, if sin x, esc a?, tan x, or cot x is positive, the principal value of x lies between
and -^ j if sina:, cscx, tana:, or cota: is negative, the principal value of x lies between

and - 1 •

7T 7T

If cosx or sec a: is positive, the principal value of x lies between ^ and — -^^ prefer-
ence being given to the positive angle.

86 PLAJSTE TRIGONOMETRY

Similarly, if cos x = \, then by (68), p. 81,

x = cos J ;r = 2 nir ± — >
Z o

tr
where the principal value of x is — = 60°.

o

Since the sine and cosine of an angle cannot be less than — 1 nor
greater than + 1, it follows that the expressions

sin -1 a. and cos -1 a

have no meaning unless a lies between — 1 and + 1 inclusive. Simi-
larly, it is evident that the expressions .

sec -1 a and esc -1 a

have no meaning for values of a lying between — 1 and + 1.

Any relation that has been established between trigonometric
functions may be expressed by means of the inverse notation.

Thus, we know that

cos x = Vl — sin 2 a;. (29), p. 60

This may be written

(A) x = cos -1 Vl— sin 2 a;.

Placing sin a; = a, then x = sin -1 a, and (A) becomes

sin -1 a = cos -1 Vl — a 2 .
Similarly, since cos 2 x = 2 cos 2 a; — 1, (49 b), p. 70

we may write

(B) 2x = cos- 1 (2 cos 2 x - 1).

Placing cos a; = c, then x = cos _1 c, and (B) becomes
2cos- 1 c = cos- 1 (2c 2 -l).

Since we know that the co-functions of complementary angles are
equal, we get for the principal values of the angles that

sin -1 a + cos -1 a = —> < a S 1

tan -1 b + cot- 1 5 = ^> < b

sec -1 g + esc -1 c = — • 1 < c

INVERSE TRIGONOMETRIC FUNCTIONS 87

We shall now show how to prove identities involving inverse
trigonometric functions for the principal values of the angles.

Ex. 1. Prove the identity

m + n

(a) tan _1 m + tan- 1 ?! = tan -1 -

1 — mn
Proof. Let

(b) ji = tan -1 m and B = tan _1 m.

(c) Then tan 4 = m and tan .B = n.

Substituting from (b) in first member of (a), we get

A + B = tan- 1 — — — ,
1 — mn

or, what amounts to the same thing,

(d) tan(4 + £) = -^±^.

1— mn

But from (44), p. 68,

. . _. tanA + tanB

(e) tanM+B) =

w v 1- tan^tanJS

Substituting from (c) in second member of (e), we get

(f) tan(^ + B) = ^±^.

1 — mn

Since (d) and (f) are identical, we have proven (a) to be true.

Ex. 2. Prove that

(g) sin- 1 ! + cos- 1 !! = sin- 1 ^.
Proof. Let

(h) .4 = sin -1 ! an< ^ B=cos- 1 ^.

(i) Then sin .4 = f and cosB=\%.

(j) Also cos 4 = f and sinJB = T 8 T .*

Substituting from (h) in first member of (g), we get
A + B= sin-i^,
or, what amounts to the same thing,
(k) sm(A + B) = $l

But from (40), p. 63,
(1) sin (A + B) = sin A cos B+ cos A sin B.

Substituting from (i) and (j) in second member of (1), we get
(m) sin(4 + -B) = !-H + W T = i£.

Since (k) and (m) are identical, we have proven (g) to be true.

The following example illustrates how some equations involving
inverse trigonometric functions may be solved.

* Found by method explained on p. 30.

88 PLANE TRIGONOMETRY

Ex. 3. Solve the following equation for x :

tan~ 1 2a + tan _1 3x = —
4

Solution. Take the tangent of both sides of the equation. Thus *

tan (tan- 1 2 x + tan -1 3 x) = tan — ,

tan (tan- 1 2 x) + tan (tan- 1 3 x) , , ....

or, — = 1, from (44), p. 68

1 - tan (tan- 1 2 x) tan (tan- 1 3 x)

2x + 3x

or, = 1.

' l-2x-3x

Clearing of fractions and solving for x, we get

x = £ or — 1.

x = £ satisfies the equation for the principal values of tan- 1 2 x and tan -1 3x.
x = — 1 satisfies the equation for the values

tan-!(- 2) = 116.57°,
tan- 1 (-3) = -71. 67°.

EXAMPLES

1. Express in radians the general values of the following functions :

(a) sin- 1 — -■ Ans. nic + (— 1)" — • (e) tan- 1 — = . Ans. mr -\

V2 4 V3 6

(b)dn-i(-^). n»-(-l)-J. (i)Un-i(±Vs). wr±|.

(c) cos- 1 — — 2ri7r±^- (g) cot-i(± 1). mr±~.

2 o 4

(d)ooB-i(-i). !w±j- (hJcot-^-LY TC1r + J.

2. Prove the following :

(a) tan -1 a — tan- 1 6 = tan- 1 (h) 2 tan- 1 a = tan- 1

l + a& 1-a 2

2a

(b) 2tan- 1 a= sin- 1 -■ (i) sin-ia = cos-iVl- a 2 .

1 + a 2 '

(c) 2 sin- 1 a = cos- 1 (1 - 2 a 2 ). (j) sin- i a = tan- 1

Vl-a 2

(d) tan- 1 * = sin- 1 — — ==.. (k) tan- 1 * = cos- 1

Vl + a 2 Vl + a 2

(e) tan- 1 tan- 1 = - . (1) sin- 1 - + sin- 1 — = sin- 1 — .

n m + n 4 6 17 85

(f) cos- 1 - + tan- 1 - = tan- 1 -^ (m) cos- 1 - + cos- 1 — = cos- 1 — .

5 5 11- v ' 5 13 65

2 12 119

(g) 2 tan- 1 - = tan- 1 — . (n) tan- 1 - + tan- 1 — = tan- 1 - •

3 5 7 13 9

* The student should remeuiber that tan-' 2 a: and tan- 1 3 a are measures of angles.

TKIGONOMETRIC EQUATIONS 89

3. Solve the following equations :

(a) tan- 1 ! + tan-!(l - x) = tan-M-).

(b) tan-i* + 2 coWx = — .

o

... .1- 1 _x + 1 ir

(c) tan- 1 [■ tan- 1 = -.

x + 2 x+2 4

(d) cos- 1 \- tan- 1 =

x 2 + 1 x 2 - 1 3

(e) tan- 1 ^-i^ + tan- 1 ^li = tan- 1 (-7).

a: — 1 a;

(f) tan- 1 (x + 1) + tan- 1 (x - 1) = tan- 1 — .

ol

(g) sin -!x + sin- 1 2 x = — .

o

(h) sm- 1 - + sin- 1 — = - .
x x 2

4. Find the values of the following :

2a

Ans,

1

x = -•
2

x = V§.

X = ± \l

x=± V3.

x = 2.

x = -8, l

14

x = ± 13.

(a) sin (tan- 1 — ).
\ 12/

Ans.

5

±13"

(d) cos (2 cos -1 a).

(b) cot (2 sin- 1 -V

7
±24-

(e) tan (2 tan- 1 a).

(c) sin/ tan -1 - + tan~

i>

1

(f) cos(2tan- 1 a).

1-a 2
1 + a 2 '

47. Trigonometric equations. By these we mean equations involv-
ing one or more trigonometric functions of one or more angles. For
instance,

2cos 2 x + V3sina; +1 =

is a trigonometric equation involving the unknown angle x. We have
already worked out many problems in trigonometric equations. Thus,
Examples 1-8, pp. 83, 84, are in fact examples requiring the solution
of trigonometric equations.

To solve a trigonometric equation involving one unknown angle
is to find an expression (the student should look up the general
value of an angle, p. 85) for all values of the angle which satisfy the
given equation.

No general method can be given for solving trigonometric equa-
tions that would be the best to follow in all cases, but the following
general directions (which are similar to those given on p. 76 for
proving identities) will be found useful.

90 PLANE TKIGONOMETKY

48. General directions for solving a trigonometric equation.*

First step. If multiple angles, fractional angles, or the sums or
differences of angles are involved, reduce all to functions of a single
anglerf and simplify.

Second step. If the resulting expressions are not readily reducible
to the same function, change all the functions into sines and cosines.

Third step. Clear of fractions and radicals.

Fourth step. Change all the functions to a single function.

Fifth step. Solve algebraically (by factoring or otherwise) for the
one function now occurring in the equation, and express the general
value of the angle thus found by (67), (68), or (69). Only such values
of the angle which satisfy the given equation are solutions.

Ex. 1. Solve the equation

cos 2 x sec x + sec x + 1 = 0.

Solution. Since cos 2 x = cos 2 x — sin 2 x, we get

First step. (cos 2 x — sin 2 x) sec x + sec x + 1 = 0.

Second step. Since sec x = 1 this becomes

cosx

cos 2 x — sin 2 x 1 _

cosx cosx —

Third step. cos 2 x — sin 2 x + 1 + cosx = 0.

Fourth step. Since sin 2 x = 1 — cos 2 x, we have

cos 2 x — 1 + cos 2 x + 1 + cosx = 0,

or, 2 cos 2 x + cosx = 0.

Fifth step. cos x (2 cos x + 1) = 0.

Placing each factor equal to zero, we get

cos x = 0,

or, from (68), p. 81, x = cos- 1 *) = 2 rwr ± - •

Also, 2 cos x + 1 = 0,

cosx = — J,

./ 1\ „ , 2tt
or, x = cos- 1 1 — -I = 2rnr ± — ■

Hence the general values of the angles which satisfy the equation are
2 nir ± — and 2 nw ± — •

£t O

The positive angles less than 2 w which satisfy the equation are then
ir lir 4ir 37r

2' T' T' T*

* In working out examples under this head it will appear that it is not necessary to take
all of the steps in every case, nor will it always he found the best plan to take the steps in
the order indicated.

t For instance, replace cos 2 a; by cos 2 x — sin2 a:, sin(a; + T\ by —

v 4 ' vs

TEIGONOMETEIC EQUATIONS 91

Ex. 2. Solve the equation

2 sin 2 x + V3 cos x + 1 = 0.

Solution. Since sin 2 x = 1 — cos 2 x, we get

Fourth step. 2 — 2 cos 2 x + Vs cos x + 1 = 0,

or, 2 cos 2 x — Vs cos x — 3 = 0.

Fifth step. This is a quadratic in cos x. Solving, we get

R V3

cos x = V3 or

2

Since no cosine can be greater than 1, the first result, cosx =V3, cannot be
used. From the second result,

■i-f)=

5,r A

2 mr ± — - • Ana.
6

EXAMPLES
Solve each of the following equations :

1. sin 2 x = l. Ans. x=mr+( — 1)"( ±— l*=mr±

2. csc 2 x = 2. x = n7r + (-l)»(±-)=mr±

3. tan 2 x = 1.

4. cot 2 x = 3.

5. cos 2 x = -■

4

4

6. sec 2 x = - -

3

7. 2sin 2 x + 3cosx = 0.

8. cos 2 or — sin 2 a = - ■

2

9. 2 Vs cos 2 a = sin a.

10. sin 2 y — 2 cos y + - = 0.

4

11. sin.4 + cos.d = V2.

12. 4sec 2 y-7tan 2 # = 3.

13. tanB + cotB = 2.

14. tan 2 x - (1 4 V3) tan x + Vs = 0.

» Since the principal value of a: = sin- 1 1 = - and of i = sin- 1 (-!)=- — >

X = mr ± — •
4

7T

x = mr ± - ■
6

■k
x = nir ± — •
o

TT

x = mr ± — ■
6

2tt
x = 2 mr ± —— ■

ir
a = mr ± — ■
6

a = mr + (- 1)»

7T
I'

y = 2 w ± - •
o

A = 2 rwr + - •
4

7T

2/ = mr ± - ■

1? = mr H

4

7T IT

x = mr + — i mr + .—
4 o

92 PLANE TEIGONOMETEY

15. cot 2 x+( V5 H — )cotx + l=0. Ana. x = mr -\ — -, rnr -\ — -•

\ V3/ 6 3

16. tan 2 x + cot 2 x = 2. x =.mr ± - •

4

17. tan / x + -) = 1 + sin 2 x. x = mr, mr — ^ •

18. csex cotx = 2 VI. i = 2mr + - ■

6

19. sin- = csc a; — cotx. x = 2 mr.

2

20. csc y + cot y = Vs. y = 2mr + -■

o

21. 3 (sec 2 a + cot 2 a) = 13. a = nir ± -, nir ± - ■

o o

Find all the positive angles less than 360° which satisfy the following
equations :

22. cos 2x + cosx = - 1. Ans. x = 90°, 120°, 240°, 270°.

23. sin 2 x — cos 2 x — sin x + cos x = 0. x = 0°, 90°, 210°, 330°.

V3

24. sin (60° - x) - sin (60° + x) = x = 240°, 300°.

A

V3

25. sin (30° + x) - cos (60°+ x) = x = 210°, 330°.

26. tan (45° - x) + cot (45° - x) = 4. x = 30°, 150°, 210°, 330°.

27. cos 2 x = cos 2 x. x = 0°, 180°.

28. 2 sin y = sin 2y. y = 0°, 180°.

29. sin x + sin 2 x + sin 3 x = 0. x = 0°, 90°, 120°, 180°, 240°, 270°.

30. tan x + tan 2 x = tan 3 x. x = 0°, 60°, 120°, 180°, 240°, 300°

31. secx — cotx = cscx — tanx. x = 45°, 225°.

32. sin 4 x - cos 3 x = sin 2 x. x = 30°, 90°, 150°, 2 10°, 270°, 330°.

33. Vl + sin x — Vl — sin x = 2 cos x.

34. sin 4 x + cos 4 x = - ■

8

35. sec(x+120°)+sec(x-120°) = 2cosx.

36. sin (x + 120°) + sin (x + 60°) = - ■

37. sin y + sin 3 y = cos y — cos 3 y.

Find the general value of x that satisfies the following equations :

38. cosx = and tanx = 1. Ans. x = (2n + lW + -■

V2 v ' 4

39. cot x = — V3 and csc x = — 2. x = 2mr — -■

6

40. Find positive values of A and B which satisfy the equations

cos(A-S) = - and sin(^+B) = l. Ans. —and-.

2 '2 12 4

41. Find positive values of A and B which satisfy the equations

tan(4-B) = l and sec(^ + £) = -^- ^ns. ^1 and — .

V5 24 24

CHAPTER VI
GRAPHICAL REPRESENTATION OF TRIGONOMETRIC FUNCTIONS

49. Variables. A variable is a quantity to which an unlimited
number of values can be assigned. Variables are usually denoted
by the later letters of the alphabet, as x, y, z.

50. Constants. A quantity whose value remains unchanged is
called a constant. Numerical or absolute constants retain the same
values in all problems, as 2, 5, v7, ir, etc. Arbitrary constants are
constants whose values are fixed in any particular problem. These
are usually denoted by the earlier letters of the alphabet, as
a, b, c, etc.

51. Functions. A function of a variable is a magnitude whose
value depends on the value of the variable. Nearly all scientific
problems deal with quantities and relations of this sort, and in the
experiences of everyday life we are continually meeting conditions
illustrating the dependence of one quantity on another. Thus, the
weight a man is able to lift depends on his strength, other things
being equal. Hence we may consider the weight lifted as a function
of the strength of the man. Similarly, the distance a boy can run
may be considered as a function of the time. The area of a square
is a function of the length of a side, and the volume of a sphere is
a function of its diameter. Similarly, the trinomial

x* -7x -6

is a function of x because its value will depend on the value we
assume for x, and

sin.A, cos 2 A, tan —
Z

are functions of A.

52. Graphs of functions. The relation between the assumed values
of a variable, and the corresponding values of a function depending
on that variable, are very clearly shown by a geometrical representa-
tion where the assumed values of the variable are taken as the
abscissas, and the corresponding values of the function as the

93

94

PLANE TRIGONOMETRY

ordinates of points in a plane (see § 13, p. 26). A smooth curve
drawn through these points in order is called the graph of the
function. Following are

General directions for plotting the graph of a function.

First step. Place y equal to the function.

Second step. Assume different values for the variable (= x) and
calculate the corresponding values of the function (= y), writing
down the results in tabulated form.

Third step. Plot the points having the values of x as abscissas
and the corresponding values of y as ordinates.

Fourth step. A smooth curve draion through these points in order
is called the graph of the function.

Ex. 1. Plot the graph of 2 re- 6.

Solution. First step. Let y — 2 x — 6.

Second step. Assume different values for x and compute the corresponding
values of y. Thus, if

x = 0, y = -6;
x = l, y = -4;
x = 2, y = -2;

x = - 1, y = -8;
x = -2, y = -10;

etc.

Arranging these results in tabulated form, the first two columns give the
corresponding values of x and y when we assume positive values of x, and the

J I '

X ^n —

■/• *

~

-- .

— /-!■

_

T" " 1 "

1— yfl

. — — (

X

V

X

y

-6

- 6

1

-4

-1

- 8

2

-2

-2

-10

3

-3

-12

4

2

-4

-14

5

4

-5

-16

6

6

-6

-18

etc.

etc.

etc.

etc.

last two columns when we assume nega-
tive values of x. For the sake of sym-
metry x = is placed in both pairs
of columns.
Third step. Plot the points found.

Fourth step. Drawing a smooth curve through these points gives the graph
of the function, which in this case is a straight line.

GEAPHICAL EEPEESENTATION

95

Ex. 2. Plot the graph of x 2 - 2 x - 3.

Solution. Firststep. Letj/=x 2 — 2x— 3.

Second step. Computing y by assum-
ing values of x, we find the following
table of values.

X

V

X

y

-3

-3

1

-4

- 1

2

-3

-2

5

3

-3

12

4

5

-4

21

5

12

etc.

etc.

6

21

etc.

etc.

Third step. Plot the points found.

Fourth step. Drawing a smooth curve
-through these points gives the graph of
"the function.

53. Graphs of the trigonometric functions
trigonometric function
we assume values for
the angle ; the circular
measures of these angles
are taken as the ab-
scissas, and the corre-
sponding values of the
function found from
the table on p. 9 are
taken as the ordinates
■of points on the graph.

Ex. 1. Plot the graph
•of sinx.

Solution. First step. Let
y = sin x.

Second step. Assuming
values of x differing by
30°, we calculate the corre-
sponding values of y from
the table on p. 9. In tabu-
lating the results it will be
noticed that the angles are
expressed both in degree
measure and in circular

\YA i

i

--

-

1

1 1

\

f

/

-

/

/

v'

JL

i°

/!

/ j

1

Y'

To find the graph of a

X

y

X

y

0°

0°

30°

7T

6

.50

- 30°

IT

6

- .50

60°

7T

3

.86

-.- 60°

3

- .86

90°

7T
2

1.00

- 90°

IT

~ 2

-1.00

120°

2tt

T

.86

-120°

2tt
_ ~3~

- .86

150°

hit

IT

.50

- 150°

5tj-
~6

- .50

180°

7T

-180°

— 7T

210°

6

- .50

-210°

7tt
6

.50

240°

47T

T

- .86

-240°

4?r

.86

270°

3ir
2

-1.00

-270°

Sir

2

1.00

300°

hit

IT

- .86

-300°

5tt
~~3~

.86

330°

Htt
6

- .50

-330°

Htt
6

.50

360°

2tt

-360°

-2tt

96

PLANE TRIGONOMETRY

measure. It is most convenient to use the degree measure of an angle when
looking up its function, while in plotting it is necessary to use its circular
measure.

Third step. In plotting the points we must use the circular measure of the
angles for abscissas. The most convenient way of doing this is to lay off
distances ir = 3. 1416 to the right and left of the origin and then divide each
of these into six equal parts. Then when

Also when

X =

o,

V =

0;

X =

6'

y =

.50=AB;

X =

8*

y =

.86 =CD;

X =

7T

— )

2

y =

1.00 -EF; etc.

X = -

7T
~ — )

y = -

- .50= OB; etc

*-x

Fourth step. Drawing a smooth curve through these points, we get the graph
of sin x for values of x between — 2 ir and 2 it. It is called the sine curve or
sinusoid.

Discussion, (a) Since sin (x ± 2 ir) = sin x, it follows that

'y = sin x = sin (x ± 2 ir),

that is, the graph is unchanged if we replace x by x ± 2 w. This means, how-
ever, that every point' is moved a distance 2ir to the right or left. Hence the
arc PNMLO may be moved parallel to XX' until P falls at 0, N at F, M at I,
etc., that is, into the position OFIJK, and it will be a part of the curve in its
new position. In the case of the sine curve it is then only necessary to plot
points, say, from x — — ir to x = ir, giving the arc or double undulation MLOFI.
The sine curve consists of an indefinite number of such arcs extending to the
right and left.

(b) From the graph we see that the maximum value of sinx(=j/) is
1(=EF=QN, etc.) and the minimum value is — l(=SJ = IiL, etc.),
while x can take on any value whatever.

(c) Since the graph crosses the axis of x an infinite number of times, we see
that the equation

sin x =

has an infinite number of real roots, namely, x = 0, ± 2 ir, ± 4 tt, etc.

GRAPHICAL REPRESENTATION

97

54. Periodicity of the trigonometric functions. From the graph of
sin x in the above example we saw that as the angle increased from
to 2 7r radians, the sine first increased from to 1, then decreased
from 1 to — 1, and finally increased from — 1 to 0. As the angle
increased f rqin 2 it radians to 4 ir radians, the sine again went through
the same series of changes, and so on. Thus the sine goes through
all its changes while the angle changes 2 ir radians in value. This,
is expressed by saying that the period of the sine is 2 ir.

Similarly, the cosine, secant, or cosecant passes through all its.
changes while the angle changes 2 it radians.

The tangent or cotangent, however, passes through all its changes
while the angle changes by -it radians. Hence, the period of the sine,
cosine, secant, or cosecant is 2 it radians; while the period of the
tangent or cotangent is it radians.

As each trigonometric function again and again passes through
the same series of values as the angle increases or decreases uni-
formly, we call them periodic functions.

55. Graphs of trigonometric functions plotted by means of the unit
circle. The following example will illustrate how we may plot the
graph of a trigonometric function without using any table of
numerical values of the function for different angles such as given
on p. '9.

Ex. 1. Plot the graph of sin x.

Solution. Let y = sin x. Draw a unit circle.

Divide the circumference of the circle into any number of equal parts (12 in

1.

i

j

''

J

i

T>

1

i

T-'

/ j

'

■

L

-A

—

a

i

I

a

■ ■

£

-

*A

n

N 1

this case). At the several points of division drop perpendiculars to the horizon-
tal diameter. Then the sine of the angle A OB, or, what amounts to the same

thing,

sine of arc AB = QB,

sine of arc AE = NE,

sine of arc A J = OJ, etc.

It is evident that if we take the lengths of the arcs as the abscissas and the
corresponding lengths of the perpendiculars as the ordinates of points in a plane,
these points will lie on the graph of sin x. If we choose the same scale as in

98

PLANE TEIGONOMETEY

Ex. 1, p. 96, the two graphs could be made to coincide, but in this example
the unit of length chosen is larger. The main features of the two graphs of
sin x are the same, however, the discussion being the same for both.

In In

Circle Graph
"When x = arc zero = zero,

x = arc.A.B = OA,
x = axcAC = OC,
x = arc^.D = OE,
x = arc^4.E = OG,
x = a,rc AF = 01,
x = a,vc AG = OK,
x = arc-4.iT = OL,

In In

Circle Graph
y = zero = zero ;
y = QB =AB;
y = PC =CD;
y=OD = EF;
y = NE = GH;
y = MF=IJ;
y = zero = zero ;
y = MH=LM, etc.

EXAMPLES

1. Plot the graphs of the following functions :

(a) x + 2.

"■ ' x + 1

(o) x 2 -4x + 3.

(b) 3x-6.

(p) x 3 — 4 x.

(e) 2x + l.

(i) 2-.

(d) as*.

(e) x 8 .

<*>x-

(q) x 3 -2x+l.

(j) log 10 x.
(k) 2 x 2 - i.

(r) x 8 -7x + 6.

(1) 8-x 2 .

(s) x 8 - 5x- 12

< g) x-2-

(m) 6 + 5 x + x 2 .

(t) x 4 -l.

(n) x 2 - 3 x + 2.

(u) x 5 - 2.

2. Plot the graph of cos x.

Solution. Let y = cos x. The cosine curve is found to be as follows :

■jP - - tp: ::

ilil | I 1 1 | ) 1 j — L -[~-

> jid ]/-L i ^v

t il T^

V'.^.J.,..— __.,/...,!,, , ,. ,,\.

. JL- 3_^t5i.-»T

• J M£s st j_ii_ -Ji^fO- - -jJ?<

J L I. J\ - 4\

s SL'-t-- S * _ J? V

*^^ &^

~~[

I 1

"1 \

r ' r

■ ■ ■ ■ ! -£\\ 1

To plot the graph of cos x by means of the unit circle we may use the
circle on p. 97. Taking the abscissas as arcs zero, AB, AC, AD, etc., and
the corresponding ordinates as OA, OQ, OP, zero, etc., respectively, we will
get points lying on the cosine curve.

3. Plot the graph of tan x.

Solution. The tangent curve is shown on next page.

GRAPHICAL REPRESENTATION

99

i

1 i i

J 1 1 1

3 r A. . -

1

i

|] I '

-jM hr

tlr

_ '_ 'l '

1

' ''

j t

--

i 11

: - :

/- 1

I i

•

l

1 _/ i

i

/■" t"

■Is *

! i 1

gar

i i

r i

J

i i

jiQl

M

A S=5

Hit !

-MF-f-

/ j.jr|_.

' L£ L w

— *3«L-2,.

ir

P I

r

U :_l

/"- fr-

V ^_CD^

3 X

-' ! -r-i-

7h!-r

1

i i

_j_

1 li

1 : ! " J

* T

/-a

i
i

/

li

i

1 J

i /

fl

}

,-i/

J

III ' J

£ r .

,|

7i i

1 /

f 1 | 7

|

-

J UL

---

-

i 1

-L L i I

1 ! ltF"

-h4-

i 1

_J_ i it

.J —

II

jj

1 1

i ;|

1

4r

1 1

ji

I

L

1

. J_

"i

J f

J "T

)|

1

' ■ ■

r 1

l 1

To construct the tangent curve from the unit circle shown, we have

In In In Is

Circle Graph Circle Graph

When i = arc zero = zero,

x = a,TcAB = OA,

x = arc .AC = OC,

x = arc AD = OD,

x = arc AE = OE,

y = zero = zero ;
y = AM = AB ;
y = AN = CD ■
y = qo = oo ;
y = AQ =EF, etc.

4. Plot the graph of sec x.

Solution. The secant curve is given below.

— +H ±

rl

"+t

+m-

±rt£

+Hi^-

u

1 1

1 ■ i i

1 : i

1 | II

_it :

1 1

- ' - ' 4 -i -

ii 1 !

' / E

i r

f
1
I

I !

| 1 [

4. J

r

i / i

1 1 L

ft

v

1
l

I

/'

H 1 ' !
2 i 31

s

I 1 V^ Va

Y 4^

,

13 I" C

* E"

iSij -I 1j

!

.4 isi_a

v I"? ■*■

sT . :

j

r i

;

9 I'

3^

p —

T- -A-

!

■

IE. _ 1

V !_■-'- . -;J-J

-in - —

\ i

M

- ! — i L'J-

, \ \ %

1 t-

— —

J~L <

Ai,--

.._ .j

! \

1 ii

■■ljh-j-f- -f-

-h-i

fi-,+r

iliV

+-]]-

'

_U i ! -'JL

j 1 i . II

nf y*

! 1 ! !j

Using the unit circle, we have

In In

Circle Graph
When x — arc zero = zero,

x = a,rc AB = OB,
x = arc-AC = OD,
x = axe AD = OF,
x = arcAE = OG,

In In

Circle Graph
y = OA = OA ;
y= OM=BC;
y= ON -DE;
y = co = oo;
y = OQ = GS, etc.

100 PLANE TRIGONOMETRY

5. Plot the cotangent curve.

6. Plot the cosecant curve.

-

--|l-ll -'--}— -

^h--|iH-K^^^T

11 il 1 i !

L_. JiJ LL

-[--

1

! 1 1 ! !

1

-H"9 : -1

__IL \ .

1 IV i

! ' ij ji i ! ' ! i t

I /! ill

1

- ' '-/i W-M

i
i

1 ! / »l 1 I. i , 1 ■ \R

1 i _fcl . PZlJ

-r- -J/ 1!- U U

-H-ff

I ■ / H 1 • I ' ■ 1 ■ 5 B

^/ 1 i i n

J-I-U-*

* 1 1 »

/ [i .u : _ ,. :.,. !\;

:_,jll|Ijjl u..:

H D K

>/i [' ' ! I ! : I [^a

^-^3^\

r ia: i ,l S-~ ul'-' '- I !

'El J ?w

to- u-^ar : f, ; ;

IS"! ! 1 T

'V

1 i - -> 3 4. ^

II flT [

L -i

w - -H -

^^ ■ ' _(. -

X i 1

A

i i ! U * a 3

*■<*-* I zt

1 1 , r

! ^

]| J

!' i i^**^^fc-

T ^^^

I

# \

i > j-

[

ij 1/ i

! f I

\ !■-

i|_

i / \

t

Ii / \

r

ui-L * L

1 ii

• / r

^m-hT 4

'1 i

i ! L :

-!-!-!- -jt-h-i H-l-hvl-l-H

P-M-hlifH- 1 - 1

-

7. Draw graphs of (a) sin a; + cos s, (b) cos x — sin x, (c) sin 2 a;, (d) tan 2 a;,
(e) sin x cos x.

CHAPTER VII

SOLUTION OF OBLIQUE TRIANGLES

56. Relations between the sides and angles of a triangle. One of the
principal uses of Trigonometry lies in its application to the solution
of triangles. That is, having given three elements of a triangle (sio}es
and angles) at least one of which must be a side, to find the others.
In Plane Geometry the student has already been taught how to solve
triangles graphically. That is, it has been shown how to construct
a triangle, having given

Case I. Two angles and one side.
Case II. Two sides and an opposite angle.
Case III. Two sides and the included angle.
Case IV. Three sides

From such a construction of the required triangle the parts not
given may be found by actual measurement with a graduated ruler
and a protractor. On account of the limitations of the observer
and the imperfections of the instruments used, however, the results
from such measurements will, in general, be only more or less rough
approximations. After having constructed the triangle from the
given parts by geometric methods, it will be seen that Trigonometry
teaches us how to find the unknown parts of the triangle to any
degree of accuracy desired, and the two methods may then serve as
checks on each other.

The student should always bear in mind, when solving triangles,
the two following geometrical properties which are common to all
triangles :

(70) The sum of the three angles equals 180°.

(71) The greater side lies opposite the greater angle, and conversely.

The trigonometric solution of oblique triangles depends upon the
application of three laws, — the law of sines, the law of cosines,
and the law of tangents, to the derivation of which we now turn
our attention.

101

102

PLANE TRIGONOMETRY

57. Law of sines. The sides of a triangle are proportional to the
sines of the opposite angles.

Proof. Fig. 1 represents a triangle all of whose angles are acute,
while Fig. 2 represents a triangle, one angle of which is obtuse (as A).

Fig. 1

Fig. 2

Draw the perpendicular CD(= h) on AB or AB produced. From
either figure, using the right triangle ACD,

W

■ A k

sm.A = -•
b

Tin Fig. 2, 8in^=sin(180°-^)=sinC^l-D--.l

Also, using the right triangle BCD,

(-B)

Dividing (A) by (B) gives

or, by alternation in proportion
(C)

Similarly, by drawing ]
(D)
(E)

Writing (C), (D), (E) as a single statement, we get the law of sines.

sin .8 =

h

a

sin A
sinB

a

V

,

ion,
a

b

sini?

sin A

endiculi

irs from A and B w<

*

c

and

sin.B

sin C

c

a

respectively.

sinC

sin A '

(72)

sin A sin B sin C

SOLUTION OF OBLIQUE TEIANGLES

103

Each of these equal ratios has a simple geometrical meaning, as may be shown
if the Iujjo of sines is proved as follows :

Circumscribe a circle about the triangle ABC as shown in the figure, and
draw the radii OB, OC. Denote the radius of the circle by B. Draw OM per-
pendicular to BG.

Since the inscribed angle A is measured by one half of the arc BG and
the central angle BOC is measured by the whole arc BC, it follows that
the angle BOC = 2 A, or,

angle BOM = A.

Then BM = R sinBOM=R sin A, by (7), p. 11
and a = 2BM = 2RsinA,

a

or,

2R:

sin A
In like manner it may be shown that

2R = and 2K = — — .

sin B sin C

Hence, by equating the results, we get
a be

sinB

2R =

sin A

sinC

The ratio of any side of a triangle to the sine of the opposite angle is numer-
ically equal to the diameter of the circumscribed circle.

It is evident that a triangle may be solved by the aid of the law
of sines if two of the three known elements are a side and its opposite
angle. The case of two angles and the included side being given,
may also be brought under this head, since by (70), p. 101, we may
find the third angle which lies opposite the given side.

Ex. 1. Given A = 65°, B = 40°, a = 60 ft. ; solve the triangle.
Solution. Construct the triangle. Since two angles are given we get the
third angle at once from (70), p. 101. Thus,

C = 180° - (A + B) = 1S0° - 105° = 75°.

Since we know the side a and its opposite angle A we may use the law of
sines, but we must be careful to choose such ratios in (72) that only one un-
known quantity is involved. Thus, to find the side 6 use

a b

sin .4 sini?

Clearing of fractions and solving for the only
unknown quantity 6, we get

t a sini?

o= ■ ■ •
sin A

Substituting the numerical values of sin A and sin B from the table on p. 0,
and a = 50 ft, we get 5 x 0.6428

6 =

0.9063

= 35.46 ft.

104

PLANE TEIGONOMETEY

Similarly, to find the side c, use

a c

sin A sin G

Clearing of fractions and solving for c, we get

asinO 50 x 0.9659

sin .4

0.9063

= 53.29 ft.

By measurements on the figure we now check the results to see that there
are no large errors.

Since we now know all the sides and angles of the triangle, the triangle is
said to be solved.

58. The ambiguous case. When two sides and an angle opposite
one of them are given, the solution of the triangle depends on the
law of sines. We must first find the unknown angle which lies
opposite one of the given sides. But when an angle is determined
by its sine, it admits of two values which are supplements of each
other ; hence either value of that angle may be taken unless one is
excluded by the conditions of the problem.

Let a and b be the given sides and A
(opposite the side a) the given angle.

If a>b, then by Geometry A>B, and
B must be acute whatever be the value
of A, for a triangle can have only one
obtuse angle. Hence there is one, and
only one, triangle that will satisfy the given conditions.

If a = b, then by Geometry A=B, both A
and B must be acute, and the required triangle
is isosceles.

If a<b, then by Geometry A< B, and A
must be acute in order that the triangle shall
be possible ; and when A is acute it is evident
from the figure that the two triangles ACB and ACB' will satisfy

the given conditions provided a
is greater than the perpendicular
CP ; that is, provided
a >b smA.

The angles ABC and AB'C are
supplementary (since Z B'BC
= Z BB'C) ; they are, in fact, the

supplementary angles obtained (using the law of sines) from the

formula

sinB =

b sin A

. SOLUTION OF OBLIQUE TEIANGLES 105

That is, we get the corresponding acute value B from a table of-
sines, and the supplementary obtuse value as follows :

B' = 1W°-B.

If, however, a = b sin A = CP, then sinB = 1, B = 90°, and the tri-
angle required is a right triangle.

If <z< 5 sin 4 (that is, greater than CP), then sinS>l, and the
triangle is impossible.

These results may be stated in compact form as follows :

Two solutions : If A is acute and the value of a lies between b and
b sin A.

No solution : If A is acute and a <Lb sin A, or if A is obtuse and
a < b or a = b.

One solution : In all other cases.

The number of solutions can usually be determined by inspection
on constructing the triangle. In case of doubt find the value of b sin A
and test as above.

Ex. 1. Given a = 21, 6 = 32, A = 115° ; find the remaining parts.
Solution. In this case a <b and A > 90° ; hence the triangle is impossible
and there is no solution.

Ex. 2. Given a = 32, 6 = 86, A = 30° ; find the remaining parts.
Solution. Here 6 sin 4 = 86 x \ = 43; hence a<bsinA, and there is no
solution.

Ex. 3. Given a = 40, 6 = 30, A = 75° ; find the remaining parts.
Solution. Since a > 6 and A is acute there is one solution only.

r>y me law

a b

C

*/

sin. A sins'

or,

. „ 6sin^. 30 x .9659

sin B = = —

a 40

.-. sinB = .7244,

X

or,

B = 46.4°, the only admissible
value of B.

A\75°

??v

A C;

= i

Then

C = 180° - (A + B) = 180° - 121.4° = 58.6°.

To find C, we get, by the law of sines,

sin C sin A

_ a sin C __ 40 x .8535
C ~~ sin^i .9659

Check the results by measurements on the figure.

= 35.3.

106

PLANE TRIGONOMETRY

Ex. 4. Solve the triangle, having given b = 16, a = 12, A = 52°.

Solution. Here bsinA = 15 x .7880 = 11.82 ; hence, since A is acute and a
lies between 6 and 6 sin .4, there are two solutions. That is, there are two tri-
angles, ACBi and A CB 2 , which satisfy the given conditions. By the law of sines,

a b

sin A
sin Bi ■

sinBi
b sin A

16 x .7880

= .9850.

a 12

This gives Bi = 80.07°, and the supplementary angle B 2 = 180° — B x = !
Let us first solve completely the triangle AB^C.

d = 180° - (A + B{) = 47.93°.
a c\

Bythe law of sines,

or,

sin A

sinCi
a sin Ci

12 x .7423
.7880

sin A
Now, solving the triangle AB^C,

C 2 = 180° - {A + B 2 ) = 28.07°.
By the law of sines,

Slll^l Bill U 2

12 x .4706

:11.3.

sin A
or, c 2 =

The solutions then are :

sinC 2
a sin C 2

sin .4.

.7880

= 7.2.

Tor triangle ABiC
B,_= 80.07°,
Ci = 47.93°,
c 1 = 11.3.

Tor triangle AB^C
JB 2 = 99.93°,
C 2 = 28.07°,
c 2 = 7.2.

Check the results by measurements on the figure.

In the ambiguous case care should be taken to properly combine
the calculated sides and angles.

EXAMPLES

1. Find the number of solutions in the following triangles, having given :

(a) a = 80, b = 100, A = 30°. Ans. Two.

(b) o = 50, 6 = 100, A = 30°. One.

(c) a = 40, 6 = 100, A = 30°. None.

(d) a = 13, b = 11, A = 69°. One.

(e) a = 70, 6 = 75, A = 60°. Two.

(f) a = 134, 6 = 84, B = 52°. None.

(g) a = 200, b = 100, A = 30°. One.

2. Solve the triangle, having given a = 50, A = 65°, B = 40°.

4ns. C = 75°, 6 = 35.46, c = 53.29.

3. Solve the triangle, having given 6 = 7.07, A = 30°, C = 105°.

Ans. B = 45°, a = 5, c = 9.66.

SOLUTION OF OBLIQUE TRIANGLES

107

4. Solve the triangle, having given c = 9.56, 4 = 45°, B = 60°.

4ns. C = 75°, a = 7, 6 = 8.57.

5. Solve the triangle when c = 60, A = 50°, B = 75°.

Ans. = 55°, 6 = 70.7, a = 56.1.

6. Solve the triangle when a = 550, A = 10° 12', B = 46° 36'.

4ns. C = 123° 12', 6 = 2257.4, c = 2600.2.

7. Solve the triangle when a = 18, 6 = 20, A = 55.4°.

4ns. JBi = 66.2°, d = 58.4°, Cj = 18.6 ;
.B 2 = 113.8°, C 2 = 10.8°, c 2 = 4.1.

8. Solve the triangle when a = 3 V2, 6 = 2 V5, 4 = 60°.

Ans. C = 75°, B = 45°, c = 4.73.

9. Solve the triangle when 5 = 19, c = 18, C = 15° 49'.

Ans. B x = 16° 43', A 1 = 147° 28', a t = 35.5 ;
B 2 = 163° 17', 4 2 = 54', a 2 = 1.04.

10. Solve the triangle when a = 119, 6 = 97, A = 50°.

4ns. B = 38.6°, C = 91.4°, c = 155.3.

11. Solve the triangle when a = 120, 6 = 80, A = 60°.

4ns. B = 35.3°, C = 84.7°, c = 137.9.

12. It is required to find the horizontal distance from a
point 4 to an inaccessible point B on the opposite bank of a
river. We measure off any convenient horizontal distance as
AC, and then measure the angles CAB and ACB.

Let 4C=283feet, angle C45=38°, and angle 4 CB= 66. 3°.
Solve the triangle ABC for the side AB. Ans. 267 .4 ft.

13. A railroad embankment stands on a horizontal plane
and it is required to find the distance from a point 4 in the
plane to the top B of the embankment. Select a point C at

the foot of the embankment lying in the
same vertical plane as 4 and B, and meas-
ure the distances 4 C and CB, and the angle
BAG. Let 4C = 48.5 ft, BC = 84 ft., and
angle B4C = 21.5°. Solve the triangle for
the side AB.

Ans. 127.2 ft.

14. A tree 4 is observed from two points
B and C, 270 ft. apart, on a straight road. The angle BCA is 55° and the angle
CBA=6b°. Find the distance from the tree to the nearer point B. Ans. 255.4 ft.

15. To determine the distance of a hostile fort 4 from a place B, a line BC
and the angles ABC and BCA were measured and found to be 1006.62 yd., 44°,
and 70° respectively. Find the distance AB. Ans. 1035.5 yd.

16. A triangular lot has two sides of lengths 140.5 ft. and 170.6 ft., and the
angle opposite the former is 40°. Find the length of a fence around it.

4ns. 353.9 ft., or 529.6 ft.

17. Two buoys are 64.2 yd. apart, and a boat is 74.1 yd. from the nearer
buoy. The angle between the lines from the buoys to the boat is 27.3°. How
far is the boat from the further buoy ? 4ns. 120.3 yd.

108

PLANE TEIGONOMETE.Y

18. Prove the following for any triangle :
(a) a = b cos C + c eosB,
b = a cosC + c cos A,
c = a cosB + b cos A.

1 sin C + c 2 sin B

(b) V&esin.BsinC7 :

b + c

(c)
(d)

sin A + 2 sin B _ sin C

a + 2b c

sin? A — m sin 2 B sin 2 C

a 2 - m& 2 c 2

(e) a sin (B - C) + b sin (C - 4) + c sin (.4 - B) = 0.

19. If R is the radius of the circumscribed circle, prove the following for
any triangle [s = \ (a + b + c)] :

(a) B(smA + sinB + sin C) = s.

(b) be = 4 £ 2 (cos 4 + cos B f-os C).
1 1

(o)

■ + •

1

4R

s — a s — 6 s — c
20. Show that in any triangle

8 Vs (s — a) (s - b) (s — c)

a + b __ cos\(A-B)
c sin ^ C

59. Law of cosines. In any triangle the square of any side is equal
to the sum of the squares of the other two sides minus twice the product
of these two sides into the cosine of their included angle.

Proof. Suppose we want to find the side a in terms of the other
two sides b and c and their included angle A.

t-;s-

When the angle A is acute (as in Fig. 1) we have, from Geometry,
UB i = AC* + AB i -2ABxAD )

[The square of the side opposite an acute angle equals the sum"]
of the squares of the other two sides minus twice the product I
of one of those sides into the projection of the other upon it.J

or,

But
Hence

a 2 = b 2 +c*-2cAD.
AD= b cos A .
a 2 = b 2 + c 2 — 2 be cos A.

(8), p. 11

SOLUTION OF OBLIQUE TKIANGLES 109

When the angle .1 is obtuse (as in Fin. 2) we have, from Geometry,
_.-C _b" T _i* ^ Z ^5 v -
CB* =AC* +AB* + 2AB X AD,

I ine square of the side opposite an obtuse angle equals the siuu"|
I of the squares of the other two skies plus twice the product I
Lof one of those sides into the projection of the other upon it. J

or. a s =b a +i"+2cAD.

But A D = b cos DA C (8\ p. 11

= &eos^l$0 : — .1)
= — b cos A. Hence in any ease

(73) a 3 =& s + c 2 -2dccos.i.

Similarly, we may find

(74) 6 s = a* + c 3 — 2 ac cos B.
^75) c* = a s + b i -2abcosC*

Observe that if A = 90°. then o os -4 = 0. and (73 > becomes o 3 = b 2 4- r 2 ,
which is the known relation between the sides of a right triangle

where -1 is the right angle.

Solving (73), (J4 1 , (75 > for the cosines of the angles, we get

&* + £*_<,*
(76) cosA —

2 be

tf + c*-

&

2ac

a s +6 s -

-c*

(77) cosB =

(78) cosC = 2Qb

These formulas are useful in finding the angles of a triangle,
having given its sides.

Formulas ^73 >. (74), (75) may be used for finding the third side
of a triangle when two sides and the included angle are given. The
other angles may then be found either by the law of sines or by
formulas (76). (77 (78.

* Since a and A. 6 and 18, e and C stand for any - "1e of a triangle and the opposite angle,
from any formula expressing a general relation between those parts another formula may
be deduced by i-AtinytMjj lie letter « eyelical order. Thus, in 73 by changing a to ft, 6 to
c r e to t. and A to Bwv obtain 74 ; and in 74 by .'ti.mdn; 6 toco toa,a to t. and.fi to C

we get 1 75 . This is a great help in mem -■rijfr.^ s. ■me sots ot" formulas.

<r , '*

110 PLANE TRIGONOMETRY

Ex. 1. Having given A = 47°, 6 = 8, c = 10 ; solve the triangle.

Solution. To find the side o use (73).

a 2 = 6 2 + c 2 - 2 6ccos4

= 64 + 100 - 2 x 8 x 10 x .6820

= 64.88.

.-. a = V54.88 = 7.408.

.=io -3

To find the angles C and B use the law of sines.

1nS = »^ = 8x - 7814 = .7896. ,.5 = 52.2°.
a 7.408

sin C = i^i = 10 X - 7314 = .9873. ,. C = 80.8°.
a 7.408

To check our work we note the fact that A + B + C = 47°+ 52.2°+ 80.8°= 180°.

Ex. 2. Having given a = 7, 6 = 3, c = 5; solve the triangle.

Solution. Using formulas (76), (77), (78) in order to find the angles, we get

cos^ = &2 + c2 - a2 = 32 + 62 - 72 = -l = -.5000. ...1 = 120°.
2 6c 2-3-5 2

cosB = " 2 + c 2 -6 2 = 7 2 +5 2 -3 2 = 18 = 92g6 . B =

2ac 2-7-5 14

cosC= a2 + &2 - c2 = 72 + 82 - 62 = l 1 = .7857. ,. C = 38.2°.
2 ab 2-7-3 14

C/iecfc : A + B+C = 120° + 21.8° + 38.2° = 180°.

EXAMPLES

1. Having given a = 30, 6 = 54, C = 46° ; solve the triangle.

Ans. 4 = 33.1°, B= 100.9°, c = 39.56.

2. Having given 4 = 60°, 6 = 8, c = 5 ; find a and the cosines of the angles
B and C. Ans. 7, \, \\.

3. Having given a = 33, c = 30, B = 35.4° ; find 4 and O.

Ans. 4 = 80.7°, C=63.9°.

4. Having given a = 4, 6 = 7, c = 10 ; solve the triangle.

4ns. A = 18.2°, B = 33.1°, C = 128.7°.

5. Having given a = 21, 6 = 24, c = 27 ; solve the triangle.

Ans. A = 48.2°, B = 68.4°, O = 73.4°.

6. Having given a = 2, 6 = 3, c = 4; find the cosines of the angles A, B, C.

Ans - h ib ~ i-

7. Having given a = 77.99, 6 = 83.39, C = 72° 15' ; solve the triangle.

Ans. A = 51° 15 ', B = 66° 30', c = 95.24.

8. If two sides of a triangle are 10 and 11 and the included angle is 50°, find
the third side. Ans. 8.92.

SOLUTION OF OBLIQUE TKIAXGLES

111

9. The two diagonals of a parallelogram are 10 and 12 and they form an
angle of 49.3° ; find the sides. Ana. 10 and 4.68.

10. In order to find the distance between two
objects, A and B, separated by a pond, a station
C was chosen, and the distances CA = 426 yd.,
CB = 322. 4 yd. , together with the angle A CB = 68. 7°,
were measured. Find the distance from A to B.

Ans. 430.85 yd.

11. A ladder 52 ft. long
is set 20 ft. from the foot
of an inclined buttress, and
reaches 46 ft. up its face.
Find the inclination of the face of the buttress.

Ans. 95.9°.

12. Under what visual angle is an object 7 ft. long seen by an observer
whose eye is 5 ft. from one end of the object and 8 ft. from the other end ?

Ans. 60°.

13. Two stations, A and B, on opposite sides of a mountain, are both visible
from a third station C. The distance AC = 11.5 mi., BC = 9.4 mi., and angle
ACB = 59.5°. Find the distance between A and B. Ann. 10.5 mi.

14. Prove the following for any triangle :

(a) a(b 2 +c 2 )cosA + 6(c 2 + a 2 ) cosB+ c(a? + V>)cosC--

„ „ 6 + c cosB + cosC
(b)

3a6c.

a 1 — cos.4

(c) a + 6 + c = (6 + c) cos A + (c + a) cosB +(a + b) cos C.
cos A . cosB cosC a 2 + 6 2 + c 2

(d)

• + -

a b c 2abc

(e) a 2 + ft 2 + c 2 = 2 (aft cos C + bc cos A + ca cosB).

60. Law of tangents. The sum of any two sides of a triangle is to
their difference as the tangent of half the sum, of their opposite angles
is to the tangent of half their difference.

Proof. By the law of sines,

a b

sin A sinj}

and by division and composition in proportion,

a + b sin A + sin B

w

sin A — sinB

But from (66), p. 74,

sin A + sinB

(B)

sinA — sinB

tan \(A +B)
tan £(4 -B)

112 PLANE TEIGONOMETEY

Hence equating (4) and (B), we get

(79) a + b = tan K^+*) *
^ ' a — b tan §(4 — S) '

a- -t i a + c tani(4+C)

Similarly, we get = *-* 1»

■" 5 a-c tan£(4-C)

b + c ^ tan \ (B + C) t
6 -c - tan £(.B-C)'

When two sides and the included angle are given, as a, b, C, the
law of tangents may be employed in finding the two unknown
angles A and B.% Since a + b, a — b, A+B (=180°— C), and there-
fore also tan £(4 +B), are known, we clear (79) of fractions and
solve for the unknown quantity tan \ (A — B). This gives

(80) tan §(4-.B) = ll-? tan 1(4+*). '

We shall illustrate the process by means of an example.

Ex. 1. Having given a = 872. 5, b = 632.7, C = 80° ; solve the triangle.
Solution. a + b = 1605.2, a - 6 = 239.8, A + B = 180°- C= 100°, and
i {A + B) =50°.

From (79), since tan £ {A + B) = tan 50° = 1.1918,

tan h(A - B) = - — -tan 1(4 + B) = =^^- x 1.1918 = .1899.
5 a + b 5V ; 1505.2

.-. £ (A - B) = 10.6°.

Adding this result to £ (A + B) = 50° gives

A = 60.6°.
Subtracting the result from ^ (A + B) = 50° gives
B B = 39.4°.

To find the side c, use the law of sines. Thus,

asinC 872.5 x .9848 no „ „

c = — = = yob./.

sinJ. .8712

We will now derive formulas for solving triangles having three
sides given, which are more convenient than (76), (77), (78), p. 109.

* If b > a, then B>A, making a - & and A - B negative. The formula still holds true, but

to avoid negative quantities it is better to write the formula in form -, = -' — i\-= ~ *

° b-a tan %(B -A)

t These may also be found by changing the letters in cyclical order (see footnote, p. 109).

% When logarithms are used in solving triangles, having given two sides and the included
angle, the law of tangents, which involves products, is to be preferred to the law of cosines,
which involves sums.

SOLUTION OF OBLIQUE TRIANGLES 113

61. Trigonometric functions of the half angles of a triangle in terms

of its sides. Denote half the sum of the sides of a triangle (i.e. half
the perimeter) by s. Then

(4) 2 s = a + b + c.

Subtracting 2 c from both sides,

2s — 2c = a + b + c — 2c, or,
(-B) 2 (s - c) = a + b - c.

Similarly,

(C) 2(s- b)=a-b + c,

(D) 2(s — a) = —a + b + c.

In (49a), (49b), p. 70, replace 2x by A, and, what amounts to
the same thing, x by \A. This gives

(E) 2sin 2 J^=l-cos^,

(F) 2cos 2 £.4=l+eos,4.

b 2 + c 2 — a 2
But from (76), p. 109, cos^l = — ! hence (E) becomes'

(fl) 2ain .^ =1 _L±i!_JL

_ 2Sc-ft 2 -c 2 + a il
~~ 2 be

^ a 2 -(b 2 -2bc + c 2 )
~ 2bc

_ g' _ (b - c ) 2

~ 2 6c

_ (a + b — c)(a — b + c)
~ 2 be

[a2 — (5 — c) 2 being the product of the sum and difference of a and 6 — c]

2(s-c)2(s-i) , ._ ._

( 81) .-. 8 inM = ^ME±

114 PLANE TRIGONOMETRY

Similarly, (F) becomes

J2 + c 2 _ a 2

2 be + b 2 + e 2 - a 2
~ 2 be

_ (b + e) 2 - a 2

2 be
_ (b + e + a)(b + c - a)

2bc
_ 2s-2(s-a)
2 be

(82) .:cosU = ^EE^.

citi JL A

Since tan \A — t—-> we get, by substitution from (81) and (82),

COS ft- J*.

(83) tanM.xBS.
v ' N s(s— a)

Since any angle of a triangle must be less than 180°, \A must be
less than 90° and all the functions of \A must be positive. Hence
only the positive signs of the radicals in (81), (82), (83) have been
taken.

Similarly, we may get

sin i5 = ^3S3, sin^= > J

\s(s-b) f

tan \B =

(s — a)(s — c)

|( s - a )( s "

-b)

i ab

\s(s-e)

I ab '

(s — a)(s -

-b)

»"-Y s(s-b) ' tan ^ C= N ,( s _ )

There is then a choice of three different formulas for finding the
value of each angle. If half the angle is very near 0°, the formula
for the cosine will not give a very accurate result, because the cosines
of angles near 0° differ little in value ; and the same holds true of
the formula for the sine when half the angle is very near 90°. Hence
in the first case the formula for the sine, in the second that for the
cosine, should be used. In general, however, the formula for the tan-
gent is to be preferred.

* Also found by changing the letters in cyclical order.

SOLUTION OF OBLIQUE TEIANGLES 115

When two angles, as A and B, have been found, the third angle,
C, may be found by the relation A+B + C = 180°, but it is best to
compute all the angles from the formulas, so that we use the sum of
the angles as a test of the accuracy of the results.

It is customary to use a second form of (83), found as follows :

* \ s(s — a)

= J . / T - g )Q-6)0 Z "c)
' \ s(s-a) 2

[Multiplying both numerator and denomina-l
tor of the fraction under the radical by s~ a. J

_ l 1 -«)(s- &)0-0

s — a N s

Denoting the radical part of the expression by r,

(84) r = -\J-i ^ ii 1 , and we get

(85) tan|A= — - — Similarly,

(86) tan|B =

s—b

r

(87) tan | C =

By proving one of the last three formulas geometrically it may be shown that
r is the radius of the inscribed circle. C

Proof. Since angle NA = \A,

NO M A

(A) tan^: '

AN
If « denotes half the perimeter, we have
2s=AN+NB+BL+LC+CM+MA. A ^ ^J?

But NB = BL, CM = LC, MA = AN; therefore

2 s = 2 AN + 2BL + 2LC,
or, s = AN+(BL + LC) = AN + a.

This gives AN = s — a.

NO
Substituting in (A), tan J A =

Comparing this result with (85) and (84) shows that

N Q =r = J {S ~ a) {S ~ b) (8 ~~^

* When logarithms are used in solving triangles, having given the three sides, formulas
(84). (85). (86). (87), which involve products, are more convenient than the law of cosines,
which involves sums.

116 PLANE TRIGONOMETRY

Ex. 1. Solve the triangle whose sides are 13, 14, 16.
Solution. Let a = 13, 6 = 14, c = 16.

Then

2s = a + 6 + c = 42,

or,

s = 21.

Also,

s — a = 8, s — 6 = 7, s — c = 6.

From (84),

r _ /(s-a)(s-6)(s-c)_ te.7.6

From (85),

tani^= r = 4 = 1 =.5000.

2 o „ B <>

. . J 4 = 26.56°, or A = 53.12°.

From (86), tan 1 B = ~^— = - = .5714.

v " * s-6 7

.-. |S= 29.74°, or B = 59.48°.

From (87), tan 1 C = — ^— = - = - = .6667.

v " * s-c 6 3

. . £ C = 33.69°, or C = 67.38°.
Cftecfc : 4 + B + C = 53. 12° + 59.48° + 67.38° = 179.98°.*

EXAMPLES

1. Solve Examples 1, 3, 8, p. 110, using the law of tangents.

2. Solve Examples 4, 5, 6, p. 110, using formulas (84), (86), (86), (87), p. 116.

3. Prove the following for any triangle :

(a) (a + b) sin J. C = c cos \{A-B).

(b) tan£Btan^C = 6 + C ~ a .

i 3 6 + c + a

(c) 6 cos 2 £ + c cos 2 ^B = s.

(d) (6 + c - a) tan ^ A = (c + a - 6) tan ^ B.

(e) & = (a + o) 2 sin 2 J C + (a - 6) 2 cos 2 £ C.

(f) c (cos A + cosB) = 2 (a + 6) sin 2 ^ C.
cos 2 £ A _ a (s - a)

™ cos 2 £ B _ 6 (s - 6) '

(h) 6 sin 2 — he sin 2 — = s — a.

(i) a cos £ 2? cos J C esc ^ A = s.

(j) sin^d. = 2 sin — cos — = — Vs (s — a)(s — b) (s — c).

* The error .02° arises from the fact that we used a four-place table. If we had used a
table giving the tirst five significant figures of the tangent, the error would have been less ;
if a six-place table, still less, etc. For ordinary purposes, however, the results we get, using
a four-place table, are sufficiently accurate.

SOLUTION OF OBLIQUE TRIANGLES

11'

4. If B and r denote the radii of the circumscribed and inscribed circles
respectively, prove the following for any triangle :
o sin £ B sin £ C
cos^A

abe i ,f

(a) r =

(b) R.

(c) r + — +^ = r-^-
oc ca ao 2riJ

a&c

4V*(s- a)(s- 6)(s -c) 2 Ysin.4 sinBsinC

(e) after = 4 B (s — a) (s — 6) (s — c).

62. Formulas for finding the area of an oblique triangle.

Case I. When two sides and the included angle are known.
Let b, c, and A be known. Take c as the base. Denote the altitude
by h and the area by S. Then, by Geometry,

S = \ ch.
But h = b sin A (from (7), p. 11); hence

(88)
Similarly,

S = \ be sin A.

S = \ac sin B = \ ab sin C.

c a

The o,rea, of a triangle equals half the product of any two sides
■multiplied by the sine of the included angle.

Ex.1. Find the area of a triangle, having given 6 = 20 in. , c = 15 in. , A = 60°.
Solution. Substituting in (88),

1

6c sin A = - x 20 x 15 x — = 75 Vz sq. in. Ans.
2 2 2

Case II. When the three sides are known.
sin A = 2 sin J^4 cos \ A

](s-b)(s-c) ^ \s(s-a)

2 /

= ^ V *( s - a )0- *) («-")■

Substituting this value of sin ^4 in (88), we get

(89) s = ^s(s-a)(s- b)(s-c).

Ex. 2. Having given a = 13, 6 = 14, c = 15 ; find the area.
Solution. S = ^ (a + b + c) = 21, s — a = 8, s-b = 7, 8 — c = 6.
Substituting in (89),

8 = V*(*-a)(8-&)(8-c) = V21 x 8 x 7 x6 = 84. Ans.

(51), p. 72

i ( 81 )> ( 82 ).
\ pp. 113, 114

118 PLANE TRIGONOMETRY

Case III. Problems which do not fall under Cases I or II directly
may be solved by Case I, if we first find an additional side or angle
by the law of sines.

Ex. 3. Given a = 10 Vs, b = 10, A = 120° ; find the area of the triangle.
Solution. This does not come directly under either Case I or Case II, but,
by the law of sines,

bsinA_W x £V3 1

sin B = ■

10V3 2

Therefore B = 30° and C = 180° - (A + B) = 30°.

Since we now have the two sides a and 6 and the included angle C, the prob-
lem comes under Case I, and we get

S = ^a6sinC= l x 10V3X 10 x ^ = 25V3. Ans.

EXAMPLES

1. Find the areas of the following triangles, having given

Ans. 240.
17.32.
193.18.
30 V3.
600.
17*.

15,541.7.
6V6.
30,600.

11,981 or 2347.8.
45.75.
10.4.

2. Show that the area of a parallelogram equals the product of any two adja-
cent sides multiplied by the sine of the included angle.

3. Find a formula for the area of an isosceles trapezoid in terms of the par-
allel sides and an acute angle.

4. Show that the area of a quadrilateral equals one half the product of its
diagonals into the sine of their included angle.

5. The base of an isosceles triangle is 20, and its area is 100 -=- V3; find its
angles.

6. Prove the following for any triangle :

, . „ abc 2 abc I A B C\

(a) S = -— . (e) S = I cos — cos — cos — 1 •

v 4fl w a + 6 + c\ 2 2 2/

(b) S = rs. a s 62

(c) S = Rr( S inA + sin B+ sin C). <*> S = 1 sin 2 B + J sin 2A '

(d) S = £ a 2 sin B sin C csc 4.

(a) a = 40,

6 = 13,

c = 37.

(b) 6 = 8,

c = 5,

A = 60°.

(c) 6 = 10,

c = 40,

A = 75°.

(d) a = 10,

6 = 12,

C = 60°.

(e) a = 40,

c = 60,

B = 30°.

(f) a = 7,

c = 5V2,

B = 135°.

(g) 6 = 149,

A = 70° 42',

, B = 39° 18'

(h) a = 5,

6 = 6,

e = 7.

(i) a = 409,

6 = 169,

c = 510.

(j) a = 140.5,

6 = 170.6,

A = 40°.

(k) c = 8,

J? = 100.1°,

C = 31.1°.

(1) a = 7,

c = 3,

A = 60°.

CHAPTER VIII
THEORY AND USE OF LOGARITHMS

63. Need of logarithms* in Trigonometry. Many of the problems
arising in Trigonometry involve computations of considerable length.
Since the labor connected with extensive and complicated calcula-
tions may be greatly lessened by the use of logarithms, it is advan-
tageous for us to use them in much of our trigonometric work.
Especially is this true of the calculations connected with the solution
of triangles. We shall now give the fundamental principles of log-
arithms and explain the use of logarithmic tables.

Definition of a logarithm. The power to which a given number
called the base must be raised to equal a second number is called the
logarithm of the second number.

Thus, if

(A ) b x = N, (exponential form)

then x = the logarithm of N to the base b. This statement is written
in abbreviated form as follows :

(B) x = logi,iV. (logarithmic form)

(A) and (B) are then simply two different ways of expressing the
same relation between b, x, and N.

(A) is called the exponential form.

(B) is called the logarithmic form.

The fact that a logarithm is an exponent may be emphasized by
writing (^1) in the form

(base) Iog = number.
For example, the following relations in exponential form, namely,
3 2 =9, 2 5 = 32, a) 8 =i, *"=»,

are written respectively in the logarithmic form

2 = log 3 9, 5 = log 2 32, 3 = logjl, y=log x z;

* Logarithms were invented by John Napier (1550-1617), Baron of Merehiston in Soot-
land, and described by him in 1614.

119

120 PLANE TRIGONOMETRY

where 2, 5, 3, y are the logarithms (exponents),

3, 2, £, x are the bases, and
9, 32, \, z are the numbers respectively.

Similarly, the relations

25* = V25 = 5, io-s^ = _*_ = . 001,

8' = -V& = S/64 = 4, 6°=^ = 1

are written in logarithmic form as follows :

^=log 25 5, -3 = log 10 .001, | = log 8 4, = log b L

EXAMPLES

1. In the following name the logarithm (exponent), the base, and the number,
and write each in logarithmic form : 2 3 = 8, i 2 = 16, 5 2 = 25, 3 8 = 27, 3* = 81.

Solution. In the first one, 3 = logarithm, 2 = base, 8 = number ; hence
log 2 8 = 3. Ana.

2. Express the following equations in logarithmic form : (y) 2 = ^, "v / 125 = 5,
2-* = -^, 10-* = .01,p»=g.

3. Express the following equations in the exponential form : log 4 64 = 3,
log 7 49 = 2, log 6 216 = 3, log™ .0001 = - 4, log 4 2 = J, log„a = l, log o l = 0,
logi,a = c.

4. When the base is 2, what are the logarithms of the numbers 1, 2, |-, 4, ^,
8, 64, 128 ?

5. When the base is 5, what are the logarithms of the numbers 1, 5, 25, 125,

6. When the base is 10, what are the logarithms of the numbers 1, 10, 100,
1000, 10,000, .1, .01, .001, .0001?

7. When the base is 4 and the logarithms are 0, 1, 2, 3, — 1, — 2, £, what
are the numbers ?

8. What must be the bases when the following equations are true :

Jog64 = 2? logl21 = 2? log625 = 4? log3jL = _2?

9. When the base is 10, between what integers do the logarithms of the fol-
lowing numbers lie : 83, 251, 1793 ?

Solution. Since logi 10 = 1 and log 10 100 = 2, and 83 is a number lying be-
tween 10 and 100, it follows that logi 83 = a number lying between 1 and 2.

Similarly, logi 251 = a number lying between 2 and 3,

logiol793 = a number lying between 3 and 4.

THEORY AND USE OF LOGARITHMS 121

10. Verify the following :

(a) log 10 1000 + logiolOO + logiolO + log 10 l = 6.

(*>) loglOiV + lo glOTTTT - lo glO T tfV?F = °-

(c) logio.001 - log 10 .01 + log 10 .l = - 2.

(d) log 2 8 - 3 log 8 2 + log 2 1 = 2.

(e) 2 log a a + 2 log a - + log a 1 = 0.

a

(f) 2 log 4 2 + £ log 2 4 - log 2 2 = 1.

(g) logs 3 + log 8 £ - logs 81 = - 5.
(h) 3 log 27 3 - £ log 8 27 + log 9 3 = £.
(i) 4 logi 6 4 + 2 logt-jlj + £ log 2 16 = 0.
(j) 2 log 8 64 - log 7 49 + £ log 55 V = 1.
(k) log 8 64 + log 4 64 + log 2 64 = 11.

(1) log 5 25 - log 5 125 + 2 log 5 5 = 1.
(m) 2 log 36 6 - log 6 36 + logs ■& = - 3.

64. Properties of logarithms. Since a logarithm is simply a new
name for an exponent, it follows that the properties of logarithms
must be found from the laws in Algebra governing exponents.

Theorem I. The logarithm of the product of two factors equals the
sum of the logarithms of the two factors.

Proof. Let the two factors be M and N, and let x and y be their
logarithms to the common base b. Then

(A) log„M=x, and \og b N=y.

Writing these in the exponential form,

(B) b x = M, and b« = N.

Multiplying together the corresponding members of equations (B),

b x+ »=MN.
Writing this in the logarithmic form gives

log b MN = x + y = log b M + log^iV. from (A)

By successive applications this theorem may evidently be ex-
tended to the product of any number of factors as follows :

log b MNPQ = log b M- NPQ = log, M + log b NPQ . Th. I

= log 6 M+ log h N + log„PQ
= log b M +- log 6 2\r+ log„P + log b Q.

122 PLANE TRIGONOMETRY

Theorem II. The logarithm, of the quotient of two numbers is equal
to the logarithm of the dividend minus the logarithm of the divisor.
Proof. As in Theorem I, let

(A) \og b M=x, and log b N=y.
Writing these in the exponential form,

(B) b x = M, and b« = N.

Dividing the corresponding members of equations (j5), we get

M

N
Writing this in logarithmic form gives

M
lo g& ^ = x - V = l°g b ikf - tog b N. from (A)

Theorem III. The logarithm of the pth power of a number is equal
to p times the logarithm of the number.
Proof. Let log b N=x.

Then b x = N.

Raising both sides to the pth power,

ftp* _ ]SfP_

Writing this in logarithmic form gives

log;,iV p = px = p\og b N.

Theorem IV. The logarithm of the rth root of a number is equal
to the logarithm, of the number divided by r.

Proof. Let log 6 iV=x.

Then b x = N.

Extracting the rth root of both sides,

X 1

V = N r .
Writing this in logarithmic form gives

\ogJ = X - = l ^i^-log b N.
r r r

from the preceding four theorems it follows that if we use the
logarithms of numbers instead of the numbers themselves, then the
operations of multiplication, division, raising to powers, and extract-
ing roots are replaced by those of addition, subtraction, multiplicar
tion, and division respectively.

THEORY AND USE OF LOGARITHMS 123

Ex. 1. Find the value of logio V.Q01.

Solution, log™ V.001 = £ log l0 .001 Th. IV

b la s (c + d)^
Ex. 2. Write logi^ — — - — - in expanded form.

Solution. log 6 <fe±^ = Ilog 6 ^±^

= i \ loga a 3 + log 6 (c + d)» - log 6 c 2 j Th. I, II

= J{31ogio+Jlogi(c + (i)-21og»c^. Th. Ill, IV

"When no base is indicated we mean that the same base is to be used through-
out. Thus, the relation

Th. IV

log JJ^+rt = 1 j 3 loga + l log(c + d) _ 2 logc |

holds true for any number used as the base. For the sake of convenience we
shall call the left-hand member of an equation like the last one the contracted
form of the logarithmic expression, and the right-hand member the expanded form.

Ex. 3. Write 3 log (x + 1) + 3 log (x - 1) + \ log x - 2 log (x 2 + 1) in the con-
tracted form.

Solution. 3 log (x + 1) + 3 log (x — 1) + \ log x - 2 log (x 2 + 1)

= log(x + l) 3 + log(x - l) 3 + logx* - log(x 2 + l) 2

. (X + l) 3 (x-l) 3 X* . Vx(x 2 -1) 3 „

= log — = log '— ■ Ans.

(x 2 +l) 2 s (x 2 + l) 2

Another form of the answer is found as follows :

. ^fr 2 - 1 ) 3 _ ,„„ /*(s 2 -l) 6 \*_ 1 ,„„ x (*» - I) 6

(x 2 +l) 2 5 \(x 2 + l) 4 / 2 8

(X 2 + l)*

EXAMPLES

1. Verify the following :

(a) logm VlOOO + log 10 Vm = \. (e) log 2 V§ + log s (J) 2 = - £.

(b) log 10 (. I) 4 - log 10 VM\ = - 3. (f ) log 2 (. 5)» - log 4 -^16 = - J^.

(c) logio V^ + i ogl0 VlO = 0. (g) logs ^*X> + logn -2 / 121 = J„a.

(d) log 10 ^T00-log 10 (.01) 2 = ^. (h) log 8 (2)s + log, (^)1=1.

2. Write the following logarithmic expressions in expanded form : *

... . aft sin C ... . Vp 2 (1 - q)

(b) log (f) log

(c) logP(l + r)». (g) log

Vp(l + q)

(m + n) 8 2
Vm — n(l + s)

tVd *\ c Vo-6/

a 3 6 2 c*
* To verify your results, reduce them back to the original form.

124 PLANE TRIGONOMETRY

3. Write the following logarithmic expressions in contracted form :

(a) 21ogx + £log2/ - 31ogz.

(b) 31og(l - x) - 2 log (2 + x) + logc.

(c) glog(x - 1) - 1 logx - £ log (s + 2) + logc.

(d) log y - £ log (y« + 4) + log c.

(e) £{21og(s - 1) + 31og(x + 1) + £logx - f log(x2 + 1)|.

65. Common* system of logarithms. Any positive number except
unity may be taken as the base, and to every particular base chosen
there corresponds a set or system of logarithms. In the common
system the base is 10, being the one most convenient to use with
our decimal system of numbers. In what follows the base is usually
omitted when writing expressions in the logarithmic form, the base 10
being always understood. Thus log 10 100=2 is written log 100 =2, etc.

The logarithm of a given number in the common system is then
the answer to the question :

What power of 10 will equal the given number ?

The following table indicates what numbers have integers for
logarithms in the common system.

Exponential Form Logarithmic Form

Since

10 4 =10,000 we have

log 10,000

= 4

10 3 =1000

log 1000

= 3

10 2 =100

log 100

= 2

10 1 =10

log 10

= 1

10° =1

logl

=

10" 1 = .l

log.l

= -1

10- 2 = .01

log .01

= -2

io- 3 = .001

log .001

= -3

10-* = .0001

log .0001

= -4

etc., etc.

Assuming that as a number increases its logarithm also increases,
we see that a number between 100 and 1000 has a logarithm lying
between 2 and 3. Similarly, the logarithm of a number between .1
and .01 has a logarithm lying between — 1 and — 2. In fact the
logarithm of any number not an exact power of 10 consists, in gen-
eral, of a whole-number part and a decimal part.

* Also called the Briggs System, from Henry Briggs (1556-1631), professor at Gresham
College, London, and later at Oxford. He modified the new invention of logarithms so as
to make it convenient for practical use.

THEORY AND USE OF LOGARITHMS 125

Thus, since 4587 is a number lying between 10 8 and 10 4 , we have

log 4587 = 3 + a decimal.

Similarly, since .0067 is a number lying between 10 -3 and 10 -2 ,

log .0067 = - (2 + a decimal)
= — 2 — a decimal.

For practical reasons the logarithm of a number is always written
in such a form that the decimal part is positive. When the loga-
rithm as a whole is negative, the decimal part may be made positive
by adding plus unity to it. Then, so as not to change the value of
the logarithm, we add minus unity to the whole part. Thus in the

last example, , A/ .„_ , „ s , . ,,

^ ' log .0067 = (— 2) + (— a decimal)

= (— 1 — 2) + (1 — a decimal)

= — 3 + a new decimal.

To emphasize the fact that only the whole part of a logarithm is
negative, the minus sign is usually written over the whole part.

For example, log .004712 = - 2.3268

= - 2 - .3268

= (_l_2) + (l-.3268)

= 3.6732.

The whole-number part of a logarithm is called the characteristic
of the logarithm.

The decimal part of a logarithm is called the mantissa of the
logarithm.

Thus if log 357 = 2.5527 and log .004712 = 3.6732, 2 and - 3 are
the characteristics and .5527 and .6732 the mantissas.

From the previous explanations and by inspection of the table on
the opposite page we get the following :

66. Rules for determining the characteristic of a logarithm.

The characteristic of a number greater than unity is positive, and
one less than the number of digits in the number to the left of the
decimal point.

The characteristic of a number less than unity is negative, and is
one greater numerically than the number of zeros between the decimal
point and the first significant figure of the number.

Ex. Write down the characteristics of the logarithms of the numhers 27,683,
466.2, 9.67, 436,000, 26, .04, .0000612, .7963, .8, .0012.

Ans. 4, 2, 0, 6, 1, - 2, - 6, - 1, - 1, - 3.

126 PLANE TRIGONOMETRY

Theorem V. Numbers with the same significant part * (and which
therefore differ only in the position of the decimal point) have the
same mantissa.

Proof. Consider, for example, the numbers 54.37 and 5437.

Let 10* = 54.37.

If we multiply both members of this equation by 100 (= 10 2 ), we
have 102 . 1Qx = 1Qx+2 = 5437;

or, x + 2 = log 5437.

Henee the logarithm of one number differs from that of the other
merely in its whole part (characteristic).

Thus, if log 47,120 = 4.6732,

then log 47.12 = 1.6732,

and log .004712 = 3.6732.

Special care is necessary in dealing with logarithms because of
the fact that the mantissa is always positive, while the character-
istic may be either positive or negative. When the characteristic
is negative it is best for practical reasons to add 10 to it and
write — 10 after the logarithm, thus giving the logarithm a new
form without change of value. Thus, if

(A) log .0249 = 2.3962,

we add 10 to — 2, giving 8 in the place of the characteristic, and
counteract this by writing — 10 after the logarithm ; that is

(B) log .0249 = 8.3962 - 10.

In case we wish to divide a logarithm having a negative character-
istic by an integer (as is sometimes required in applying Theorem IV,
p. 122), it is convenient to add and subtract 10 times that integer.
Thus in case we wish to divide such a logarithm by 2, we add and
subtract 20 ; if by 3, we add and subtract 30 ; and so on. Suppose
we want to divide the logarithm of .0249, which is 2.3962, by 3. We
would then add and subtract 30, so that

(C) log .0249 = 28.3962 - 30,

a form more convenient than (A ) or (S) when we wish to divide the
logarithm by 3. Thus,

i log .0249 = $ (28.3962 - 30) = 9.4654 - 10.

* The significant part of a number consists of those figures which remain when we
ignore all initial and final zeros. Thus, the significant part of 24,000 is 24 ; of 6.050 is 605 ;
of .00907 is 907 ; of .00081070 is 8107.

THEORY AND USE OF LOGARITHMS 127

This result may be written in form (^4) by adding the 9 in front
to the — 10 at the end, giving — 1 = 1 as the characteristic. Hence

$ log .0249 = 1.4654.

Another method for dividing a logarithm which has a negative
characteristic will now be illustrated. Suppose we wish to divide
2.3962 (=- 2 + 0.3962) by 2. We get at once

21-2 + 0.3962

- 1 + 0.1981 = 1.1981.

in case we wish to divide by 3 (as in the above example), we first
add and subtract 1 in order to make the negative characteristic
exactly divisible by 3. Thus,

3 |-3 + 1.3962

- 1 + 0.4654 = 1.4654.

The following examples will illustrate the best methods for
performing the four fundamental operations of Arithmetic on
logarithms.

Case I. Addition of logarithms.

(a) To add two logarithms having positive characteristics, as 3.2659 and

1.9866.

3.2659

1.S

5.2525

This is in no way different from ordinary addition.

(b) To add two logarithms, one having a negative characteristic, as 4.2560
and 2.8711.

4.2560 or, 6.2560 - 10

2.8711 2.8711

1.1271 9.1271-10

i.e. 1.1271

Since the mantissas (decimal parts) are always positive, the carrying figure 1 from
the tenth's place is positive. Hence in adding the first way, 1— 4 + 2 = — 1=1
will be the characteristic of the sum.

(c) To add two logarithms having negative characteristics, as 2.4069 and
1.9842.

2.4069 or, 8.4069 - 10

1.9842 9.9842 - 10

2.3911 18.3911 - 20

i.e. 2.3911

128

PLANE TRIGONOMETRY

Case II. Subtraction of logarithms.

(a) To subtract logarithms having positive characteristics.

From 5.6233 From

2.4673

or,

12.4673-10

ss 3.8890 take

3.7851

3.7851

1.7343

2.6822

8.6822 - 10

i.e.

2.6822

In the first example we have ordinary

subtraction.

In the second we subtract

a greater logarithm from a smaller one and the result as a whole is negative,
(b) To subtract logarithms having negative characteristics.

From
take

2.1163
3.4492
4.6671

or,

12.1163-10
7.4492 - 10
4.6671

From
take

1.6899
1.9083

or,

9.6899 - 10
1.9083

3.7816

i.e.

7.7816 - 10
3.7816

From
take

2.1853
1.7442

or,

18.1853-20
9.7442-10

2.4411

i.e.

8.4411 - 10
2.4411

Case III.

Multiplication of logarithms

by numbers.

Multiply
by

0.6842
5

by

Multiply

2.7012
3

8.7012 -

10
3

3.4210

4.1036

i.e.

26.1036-30
4.1036

In the second example the carrying figure from tenth's place is + 2. Add-
ing this + 2 to — 2 x 3 gives 2 — 6 = — 4 = 4 for the characteristic. ■.

Case IV. Division of logarithms by numbers.

(a) Divide 3.8530 by 2.

2 1 3. 8530
1.9265
(b) Divide 2.2411 by 3.

Here we first add and then we subtract 30, writing the logarithm in the form
28.2411-30. 3128.2411-30

i.e.

9.4137
1.4137

10

67. Tables of logarithms. The common system (having the base 10)
of logarithms is the one used in practical computations. For the
convenience of the calculator the common logarithms of numbers up
to a certain number of significant figures have been computed and
arranged in tabulated forms called logarithmic tables. The common
system has two great advantages.

THEORY AND USE OF LOGARITHMS 129

(^4) The characteristic of the logarithm of a number may be written
down on mere inspection by following the rules on p. 125.

Hence, as a rule, only the mantissas of the logarithms of numbers
are printed in the tables.

(B) The logarithms of numbers having the same significant part
have the same mantissa (Th. V, p. 126).

Hence a change in the position of the decimal point in a number
affects the characteristic alone, and it is sufficient to tabulate the
mantissas * of integers only. Thus,

log 3104 = 3.4920, log 31.04 = 1.4920,

log .03104 = 2.4920, log 310,400 = 5.4920 ;

in fact, the mantissa of any number whatever having 3104 as its
significant part will have .4920 as the mantissa of its logarithm.

Table I, pp. 2, 3,f gives immediately the mantissas of the loga-
rithms of all numbers -whose first significant figure is 1 and whose
significant part consists of four or fewer digits ; and on pp. 4, 5 are
found the mantissas of the logarithms of all numbers whose first
significant figure is greater than 1 and whose significant part cpn-
sists of three or fewer digits.

68. To find the logarithms of numbers from Table I, pp. 2-5.

When the first significant figure of the number is 1, and there are
four or fewer digits in its significant part, follow

Rule I. First step. Determine the characteristic by inspection,
using the rule on p. 125.

Second step. Find in the vertical column N, Table I, pp. 2, S, the
first three significant figures of the number. The mantissa required
is in the same horizontal row with these figures and in the vertical
column having the fourth significant figure at the top (and bottom).

Ex. 1. Find log 1387.

Solution. First step. From the rule on p. 125 we see that the characteristic
will be + 3, that is, one less than the number of digits (four) to the left of the
decimal point.

Second step. On p. 2, Table I, we find 138 in column N. The required man-
tissa will be found in the same horizontal row with 138 and in the vertical col-
umn which has 7 at the top. This gives the mantissa .1421.

Therefore log 1387 = 3.1421. Ans.

* In order to save space the decimal point in front of each mantissa is usually omitted
In the tables.

t The tables referred to in this book are Granville's Fow-Place Tables of Logarithms
(Ginn & Company).

130 PLANE TRIGONOMETRY

If the significant part of the number consists of less than four
digits, annex zeros until you do have four digits.

Ex. 2. Find log 17.

Solution. First step. By the rule on p. 125 the characteristic is found to be 1.

Second step. To find the mantissa of 17 we look up the mantissa of 1700. On
p. 3, Table I, we locate 170 in column N. The required mantissa is found in
the same horizontal row with 170, and in the vertical column having at the
top. This gives the mantissa .2304.

Therefore log 17 = 1.2304. Ans.

Ex. 3. Find log. 00152.

Solution. First step. By the rule on p. 125 we find that the characteristic is
— 3, that is, negative, and one greater numerically than the number of zeros
(two) immediately after the decimal point.

Second step. Locate 152 in column N, Table I, p. 3. In the same horizon-
tal row with 152 and in the vertical column with at the top we find the required
mantissa .1818.

Therefore log.00152 = 3.1818 = 7.1818 - 10. Ans.

To find the logarithm of a number when the first significant figure
of the number is greater than 1 and there are three or fewer digits in
its significant part, follow

Rule II. First step. Determine the characteristic by rule on p. 125.

Second step. Find in the vertical column N, Table I, pp. 4, 5, the
first two significant figures of the number. The mantissa required
is in the horizontal row with these figures and in the vertical column
having the third significant figure at the top (and bottom).

Ex. 4. Find log 5.63.

Solution. First step. The characteristic here is zero.

Second step. On p. 4, Table I, we locate 56 in column N. In the horizon-
tal row with 56 and in the vertical column with 3 at the top we find the required
mantissa .7505.

Therefore log 5.63 = 0.7505. Ans.

If the significant part of the number consists of less than three
digits, annex zeros until you do have three digits.

Ex. 5. Find log 460,000.

Solution. First step. The characteristic is 5.

Second step. On p. 4, Table I, we locate 46 in column N. In the horizontal
row with 46 and in the vertical column with at the top we find the required
mantissa .6628.

Therefore log 460,000 = 5.6628. Ans.

THEORY AND USE OF LOGARITHMS 131

Ex. 6. Find log. 08.

Solution. First step. The characteristic is — 2.

Second step. Using 800, we find that the mantissa is .9031.

Therefore log .08 = 2.9031 = 8.9031 - 10. Ans.

Ex. 7. Find (a) log 1872, (b) log 5, (c) log .7, (d) log 20,000, (e) log 1.808,

(f) log. 000032, (g) log .01011, (h) log 9.95, (i) log 17.35, (j) log. 1289, (k) log 2500,
(1) log 1.002.

Ans. (a) 3.2723, (b) 0.6990, (c)_1.8461, (d) 4.3010, (e) 0.2572, (f) 5.5051,

(g) 2.0048, (h) 0.9978, (i) 1.2393, (j) 1.1103, (k) 3.3979, (1) 0.0009.

When the first significant figure of a number is 1 and the num-
ber of digits in its significant part is greater than 4, its mantissa
cannot be found in Table I ; nor can the mantissa of a number be
found when its first significant figure is greater than 1 and the num-
ber of digits in its significant part be greater than 3.

By intei-polation,* however, we may, in the first case, find the
mantissa of a number having a fifth significant figure ; and in the
second case, of a number having a fourth significant figure. In this
book no attempt is made to find the logarithms of numbers with
more significant figures, since our four-place tables are in general
accurate only to that extent.

We shall now illustrate the process of interpolation by means of
examples.

Ex. 8. Find log 2445.

Solution. By rule on p. 125 the characteristic is found to be 3. The required
mantissa is not found in our table. But by Rule II, p. 130,

log 2450 = 3.3892

and log 2440 = 3.3874

Difference in logarithms = .0018

Since 2445 lies between 2440 and 2450, it is clear that its logarithm must lie
between 3.3874 and 3.3892. Because 2445 is just halfway between 2440 and
2450 we assume that its logarithm is halfway between the two logarithms, t
We then take half (or .5) of their difference, .0018 (called the tabular difference),
and add this to log 2440 = 3.3874. This gives

log 2445 = 3.3874 + .5 x .0018 = 3.3883.

If we had to find log 2442, we should take not half the difference, but .2 of
the difference between the logarithms of 2440 and 2445, since 2442 is not half-
way between them but two tenths of the way.

* Illustrated by examples on pp. 16-19 In the ease of trigonometric functions.

t In this process of interpolation we have assumed and used the principle that the
increase of the logarithm is proportional to the increase of the number This principle is
not strictly true, though for numbers whose first significant figure is greater than 1 the
error is so small as not to appear in the fourth decimal place of the mantissa. For numbers
whose first significant figure is 1 this error would often appear, and for this reason Table I,
pp. 2, 3, gives th« mantissas of all such numbers exact to four decimal places.

132

PLANE TRIGONOMETRY

In order to save work in interpolating, when looking up the loga-
rithms of numbers whose mantissas are not found in the table, each
tabular difference occurring in the table has been multiplied by .1, .2,
.3, • • ■ , .9, and the results are printed in the large
right-hand column with " Prop. Parts " (propor-
tional parts) at the top. Thus, on p. 4, Table I, the
first section in the Prop. Parts column shows the
products obtained when multiplying the tabular
differences 22 and 21* by .1, .2, .3, . - ., .9. Thus,

.1 x 22 = 2.2

.1 x 21 = 2.1

.2 x 22 = 4.4

.2 x 21 = 4.2

.3 x 22 = 6.6

.3 x 21 = 6.3

.4 x 22 = 8.8

.4 x 21 = 8.4

.5 X 22 = 11.0

.5 x 21 = 10.5

etc.

etc.

Hence

Extra
Digit

Difference

22

21

1

2.2

2.1

2

4.4

4.2

3

6.6

6.3

4

8.8

8.4

5

11.0

10.5

6

13.2

12.6

7

15.4

14.7

8

17.6

16.8

9

19.8

18.9

To find the logarithm of a number whose mantissa is not found in the
table, t use

Rule III. First step. Find the logarithm of the number, using
only the first three (or four) digits of its significant part when look-
ing up the mantissa.%

Second step. Subtract the mantissa just found from the next
greater mantissa in the table to find the corresponding tabular
difference.

Third step, hi the Prop. Parts column locate the block correspond-
ing to the tabular difference found. Under this difference and oppo-
site the extra digit § of the number will be found the proportional
part of the tabular difference which should be added to the extreme
right of the logarithm found in the first step. The sum will be the
logarithm of the given number.

Ex. 9. Find log 28.64.

Solution. Since the mantissa of 2864 is not found in our table, this example
comes under Rule III, the extra digit being 4.

First step. log 28.60 = 1.4564 Rule II

Second step. log 28.70 = 1.4579 Rule II

Tabular difference = 15 ||

* These are really .0022 and .0021, it being customary to drop the decimal point.

t That is, a number whose logarithm cannot be found by Rule I or Rule II, because its
significant part contains too many digits.

J When the first significant figure is 1, use the first four digits, following Rule I ; when
the first significant figure is greater than 1, use the first three digits, following Rule II.

§ In finding log 4836, for instance, 6 is called the extra digit, or, in finding log 14,835 the
extra digit is 5.

II The tabular difference = .0015, but the decimal point is usually omitted in practice.

THEOKY AND USE OF LOGARITHMS 133

Third step. About halfway down the Prop. Parts column on p. 4 we find the
block giving the proportional parts corresponding to the tabular difference 15.
Under 16 and opposite the extra digit 4 of our number we find 6.0. Then

log 28.60 = 1.4564

6 Prop. Part

log 28.64 = 1.4570. Ans.

Ex. 10. Pind log. 12548.

Solution. Since the mantissa of 12,548 is not found in our table, this example
comes under Rule III, the extra digit being 8.

First step. log . 12540 = 1. 0983

Second step. log . 12550 = 1.0986

Tabular difference = 3

Third step. In the Prop. Parts column on p. 2 we find the block giving the
proportional parts corresponding to the tabular difference 3. Under 3 and oppo-
site the extra digit 8 we find 2.4(= 2). Then

log. 12540 = 1.0983

2 Prop. Part

log. 12548 = 1.0985. Ans.

Ex. 11. Verify the following :

(a) log 4583 =3.6612. (e) log 1000. 7 =3.0003.

(b) log 16.426 = 1.2155. (f) . log 724,200 =5.8598.

(c) log .09688 = 2.9862. (g) log 9.496 =0.9775.

(d) log. 10108 = 1.0047. (h) log .0004586 = 4.6614.

69. To find the number corresponding to a given logarithm, use

Rtti/e IV. On pp. 2-5, Table I, look for the mantissa of the
given logarithm. If the mantissa is found exactly in the table, the
first significant figures of the corresponding number are found in
the same row under the N column, while the last figure is at the top
of the column in which the mantissa was found. Noting what the
characteristic in the given logarithm is, place the decimal point so as
to agree with the rule on p. 125.

In case the mantissa of the given logarithm is not found exactly in the
table we must take instead the following steps :

First step. Locate the given mantissa between two mantissas in the
tables.

Second step. Write down the number corresponding to the lesser of
the two mantissas. This will give the first three (or four) significant
figures of the required number.

Third step. Find the tabular difference between the two mantissas
from the table, and also the difference between the lesser of the two
and the given mantissa.

134 PLANE TRIGONOMETRY

Fourth step. Under the Prop. Parts column find the block cor-
responding to the tabular difference found. Under this tabular
difference pick out the proportional part nearest the difference found
between the lesser mantissa and the given mantissa, and to the left
of it will be found the last {extra) figure of the number, which figure
we now annex.

Fifth step. Noting what the characteristic of the given logarithm
is, place the decimal point so as to agree with the rule on p. 125.

Ex. 12. Find the number whose logarithm is 2.1892.

Solution. The problem may also be stated thus : find x, having given

log x = 2. 1892.

On p. 3, Table I, we find this mantissa, . 1892 exactly, in the same horizon-
tal row with 154 in the N column and in the vertical column with 6 at the top.
Hence the first four significant figures of the required number are 1546. Since
the characteristic is 2, we place the decimal point so that there will be three
digits to the left of the decimal point, that is, we place it between 4 and 6. Hence

x = 154.6. Ans.

Ex. 13. Find the number whose logarithm is 4.8409.

Solution. That is, given log * = 4.8409, to find x. Since the mantissa .8409 is
not found exactly in our table, we follow the last part of Rule IV.

First step. The given mantissa, .8409, is found to lie between .8407 and .8414
on p. 4, Table I.

Second step. The number corresponding to the lesser one, that is, to .8407,
is 693.

Third step. The tabular difference between .8407 and .8414 is 7, and the dif-
ference between .8407 and the given mantissa .8409 is 2.

Fourth step. In the Prop. Parts column under the block corresponding to the
tabular difference 7, we find that the proportional part 2.1 is nearest to 2 in
value. Immediately to the left of 2.1 we find 3, the (extra) figure to be annexed
to the number 693 found in the second step. Hence the first four significant
figures of the required number are 6933.

Fifth step. Since the characteristic of the given logarithm is 4, we annex one
zero and place the decimal point after it in order to have five digits of the num-
ber to the left of the decimal point. Hence

x = 69,330. Ans.

Ex. 14. Find the numbers whose logarithms are (a) 1.8055, (b) 1.4487,
(c)0.2164, (d) 2.9487, (e)2.0529, (f) 5.2668, (g) 3.9774, (h)4.0010, (i) 8.4430 -10,*
(j) 9.4975 - 10.

Ans. (a) 63.9, (b) .281, (c) 1.646, (d) 888.6, (e) .011295, (f) 184,850,
(g) 9493, (h) .00010023, (i) .02773, (j) .3144.

* By (A), (B), p. 126, 8.4430-10= 2.4430.

THEORY AND USE OF LOGARITHMS 135

70. The use of logarithms in computations. The following examples
will illustrate how logarithms are used in actual calculations.

Ex. 1. Calculate 243 x 13.49, using logarithms.
Solution. Denoting the product by x, we may write

x = 243 x 13.49.

Taking the logarithms of both sides, we get

log a; = log 243 + log 13.49. Th. I, p. 121

Looking up the logarithms of the numbers,

log 243 = 2.3856 Rule II, p. 130

log 13.49 = 1.1300 Rule I, p. 129

Adding, logx = 3.5156

By Rule IV, p. 133, x = 3278. Ans.

Ex. 2. Calculate

Solution. Let x =

76,420
1375 x .06423

76,420
Then log x = log 1375 + log .06423 - log 76,420

Th. I, p. 121, and Th. II, p. 122
logl375= 3.1383 ' Rule I, p. 129

log. 06423 = 8.8077 - 10 Rule III, p. 132

Adding, 11.9460 - 10

log 76,420 = 4.8832 Rule III, p. 132

Subtracting, logx= 7.0628-10

or, logx= 3.0628.

By Rule IV, p. 133, x = .0011555. Ans.

Ex. 3. Calculate (5.664)».

Solution. Let x = (5.664) 3 .

Then log x = 3 log 5.664. Th. Ill, p. 122

log 5. 664 = 0. 7531 Rule III, p. 132

Multiplying by 3, §

logx = 2.2593
By Rule IV, p. 133, x = 181.67. Ans.

Ex. 4. Calculate -v^.7182.

Solution. Let * = "8^7182 = (.7182)*.

Then logx = J log. 7182. Th. IV, p. 122

log .7182 = 1.8662 Rule III, p. 132

= 29.8562-30. (b), Case IV, p. 128

Dividing by S, 3 129.8562-30

logx =9.9521 -10
= 1.9521.
By Rule IV, p. 133, x = .8956. Ans.

136

PLANE TRIGONOMETRY

Ex. 5. Calculate

Solution. Let
Then

Dividing by 2,

'I V7194 x~87
"N 98,080,000 '

i= N

98,080,000

|~ (7194)*x 87 ")*
" L 98,080,000 J '

Adding,

log a; = ^ [£ log7194 + log 87 - log 98,080,000],

log 7 194 = 3.8569

2 1 3. 8569

I log 7 194 = 1.9285

log 87 = 1.9395

3.8680

or,

Subtracting,
Dividing by 3,

13.8680 - 10
log 98,080,000 = 7.9916

5.8764 - 10

25.8764 - 30

3 |25.8764-30

logx = 8.6255- 10

= 2.6255.
.-. x = .04222. Ans.

(a), Case II, p. 128
(b), Case IV, p. 128

Ex. 6. Calculate
Solution. Let

8

x 62.'

'3 x .052

56 x

8.793

8
x = —

x 62.73 x

.052

Then

50 x 8.793
logx = [log 8 + log 62.73 + log. 052] - [log 56 + log 8.793].

log 8= 0.9031

log 62. 73= 1.7975

log .052 = 8.7160-10
log numerator = 11.4166 — 10

log denominator = 2.C924

logz= 8.7242-10

= 2.7242.

.-. x = .05299. Ans.*

log 56= 1.7482

log 8.793 = 0.9442

log denominator = 2.6924

* Instead of looking up the logarithms at once when we write down log 8, log 62.73, etc.,
it is better to write down an outline or skeleton of the computation before using the tables
at all. Thus, for above example,

log 8= 0. log 56=1.

log 02.73= 1. log 8.793=0.

log .062= 8. -10 log denominator =

log numerator =

log denominator =

logx =
.\ x =

Tt saves time to look up all the logarithms at once, and, besides, the student is not so apt
to forget to put down the characteristics.

THEORY AND USE OF LOGARITHMS 137

71. Cologarithms. The logarithm of the reciprocal of a number is
called its cologarithm (abbreviated colog). Hence if N is any positive
number,

colog N = log — = log 1 - log N Th. II, p. 122

= — log N = — log N,

That is, the cologarithm of a number equals minus the logarithm
of the number, the minus sign affecting the entire logarithm, both
characteristic and mantissa. In order to avoid a negative mantissa
in the cologarithm, it is customary to subtract the logarithm of the
number from 10 — 10. Thus, taking 25 as the number,

colog 25 = log ^ 3 = log 1 — log 25.

But log 1 = 0,

or, what amounts to the same thing,

log 1 = 10.0000 - 10.
Also, log 25 = 1. 3979

colog 25= 8.6021-10

Since dividing by a number is the same as multiplying by the
reciprocal of the number, it is evident that when we are calculating
by means of logarithms we may either subtract the logarithm of a
divisor or add its cologarithm. When a computation is to be made in
which several factors occur in the denominator of a fraction, it is
more convenient to add the cologarithms of the factors than to sub-
tract their logarithms. Hence

Rule V. Instead of subtracting the logarithm of a divisor, we
may add its cologarithm. The cologarithm of any number' is found
by subtracting its logarithm from 10.0000 — 10.

Ex. 1. Find colog 52. 63.

Solution. 10.0000 - 10

log 52.63= 1.7212
colog 52. 63= 8.2788-10. Ans. Eule V

Ex. 2. Find colog. 016548.

Solution. 10.0000 - 10

log .016548 = g.2187 - 10
colog. 016548= 1.7813. Ans. Rule V

Thus we see that the cologarithm may be obtained from the loga-
rithm by subtracting the last significant figure of the mantissa from
10 and each of the others from 9.

138 PLANE TRIGONOMETRY

In order to show how the use of cologarithms exhibits the written
work in more compact form, let us calculate the expression in Ex. 6,

namel y> 8 x 62.73 x -052

X 56 x 8.793

Solution. Using cologarithms,

logx = log 8 + log 62.73 + log .052 + colog 56 + colog 8.793.

log 8= 0.9031
log 62. 73= 1.7975
log. 052= 8.7160-10

colog 56= 8.2518-10 since log 56 = 1.7482

colog 8. 7 93 = 9.0558 - 10 since log 8. 793 = 0. 9442
logx = 28.7242 -30

= 2.7242.
.-. x = .05299. Ans.

Calculate the following expressions, using logarithms :

3. 9.238 x .9152. Ans. 8.454. n1 /.08726\* . „,„

4.336.8 + 7984. .04218. Ul Umj " 4 «--«> la

5. (.07396)5. .000002213. 12. (5 38.2 x .000 5969)*. .8678.

e _ 15.008x0843 ^ ^ u/TsTMyi

i7//31.63\ 3
\ll29-j-

.06376 x 4.248

7. V2. 1.414.

8. 1/ 5. 1.495. 14 - [w)

9. ^.02305. .2846. 15 . ^ x ^ x VM.

14. (» V.

.6443.
.7036.

10 . J^ 6 . .1606. 16. 1^1* -7.672.

962 52.37

17

18.

(-2563) x .03442 An. .2415.

714.8 x (-.511)

121.6 x(- 9.025) _ Q725

(- 48.3) x 3662 x (- .0856)

72. Change of base in logarithms. We have seen how the logarithm
of a number to the base 10 may be found in our tables. It is some-
times necessary to find the logarithm of a number to a base different
from 10. For the sake of generality let us assume that the loga-
rithms of numbers to the base a have been computed. We wish to
find the logarithm of a number, as N, to a new base b ; that is, we
seek to express \og b N in terms of logarithms to the base a.

Suppose log 6 iV = x,

that is, b x = N.

* From the definition of a logarithm, p. 119, it is evident that a negative number can have
no logarithm. If negative numbers do occur in a computation, they should be treated as if
they were positive, and the sign of the result determined by the rules for signs in Algebra,
irrespective of the logarithmic work. Thus, in Example 16 above, we calculate the value'
of 401.8 -f- 52.37 and write a minus sign before the result.

THEORY AND USE OF LOGARITHMS 139

Taking the logarithms of both sides of this equation to the base a,

log a fr T = log AT,
or , x log„6 = log„JV. Th. Ill, p. 122

Solving, x = l p^.

But log 6 N = x. By hypothesis.

(90) ■•• log 6 * =

10g a AT

log»&

Theorem VI. The logarithm of a number to the new base b equals
the logarithm, of the same number to the original base a, divided by
the logarithm of b to the base a.

This formula is also written in the form

log„2\r = 7l/.log„iV,
where M = is called the modulus of the new system with

respect to the original one*

This number M does not depend on the particular number N, but
only on the two bases a and I.

In actual computations a = 10, since the tables we use are com-
puted to the base 10,

Ex. Find logs 21.

Solution. Here N = 21, b = 3, a = 10. Substituting in (90),

log 3 21 = ^U« = 2.771. Ans.
logi 3 .4771

EXAMPLES

1. Verify the following :

(a) log 2 7 = 2.807. (e) log,, 8 = 0.0464. (i) logs 10 =2.096.

(b) logs 4 = 1.262. (f) log 8 5 = 0.7740. (j) log 5 100 = 2.86.
(o) log 4 9=1.585. (g) log, 14 = 1.366. (k) log„.l= -2.096.
(d) log 6 7 = 1.209. (h) logj 102 = 2.873. (1) log 5 .01= -2.86.

2. Find the logarithm of T 7 T in the system of which 0.5 is the base.

3. Find the base of the system in which the logarithm of 8 is %.

4. Prove logs a ■ log„6 = 1.
1

5. Prove logjylO =

logioiV

* If, then, we have given the logarithms of numbers to a certain base a, and we wish to
find the logarithms of the same numbers to a new base b, we multiply the given logarithms

by the constant multiplier (modulus) .V== T - Thus, having given the common loga-

rithms (base 10) of numbers, wc can reduce them to the logarithms of the same numbers to

the base e(= 2.718) by multiplying them by M= = 2.3020.

140 PLANE TKIGONOMETEY

73. Exponential equations. These are equations in which the un-
known quantities occur in the exponents. Such equations may often
be solved by the use of logarithms, as illustrated in the following
examples :

Ex. 1. Given 81* = 10 ; find the value of x.
Solution. Taking the logarithms of both members,

log 81* = log 10,

or, a; log 81 = log 10. Th. Ill, p. 122

, . log 10 1.0000 . _ ft . .

Solving, x = — — = = 0.524. Ans.

s ' log 81 1.9085

Ex. 2. Express the solution of

a 1x + S x = c

in terms of logarithms.

Solution. Taking the logarithms of both members,

l0g a 2a; + S _|_ log ft* = logc. Th. I, p. 121

(2 x + 3) log a + x log 6 = log a Th. Ill, p. 122

2 x log a + 3 log a + x log b = log c.

x (2 log a + log 6) = log c — 3 log a.

jE = log e -81oga j Ans
2 log a + log 6

Ex. 3. Solve the simultaneous equations

(A) 2* • 3 v = 100.

(B) x + y = 4.

Solution. Taking the logarithms of both members of (A), and multiplying (B)
through by log 2, we get

x log 2 + y log 3 = 2 Th. I, III, p. 122

x log 2 + y log 2 = 4 log 2
Subtracting, y (log 3 — log 2) = 2 — 4 log 2

2-41og2 2-1.2040

Solving, y =

log3-log2 .4771 -.3010

• 796 ° A KO
y = = 4.52.

.1761

Substituting back in (B), we get a; = — .52.

EXAMPLES
1. Solve the following equations :

(a) 5* = 12.

Ans.

1.54.

(g) (1.3)* =7.2.

Ans.

7.53.

(b) 7* = 25.

1.65.

i l

(c) (0.4)-* =7.

2.12.

(h) (0.9)* 2 = (4.7) s.

0.45.

(d) 10*- 1 = 4.

1.602.

(i) 7*+« = 5.

-2.1729.

(e) 4*-! = 5*+!.

- 13.43.

(j) 22*+s-6*-i = 0.

9.5414

(f) 4* = 40.

2.66.

THEORY AND USE OF LOGARITHMS^ 141

2. Solve the following simultaneous equations :

(a) 4* ■ Zv = 8, Ans. x = .9005, (o) 2* • 2v = 2 22 , 4ns. x = 13,
2*-8<' = 9. ?/ = .7565. x - y = i. y = Q.

(b) 3* • 4» = 15,552, x = 5, (d) 2* • 3» = 18, x = 1,
4* ■ 5* = 128,000. y = 3. 5* • 7? = 245. y = 2.

3. Indicate the solution of the following in terms of logarithms :

(a) A = P(r + 1)*. Ans. x = ]2iAsL^I.

log(r + l)

(b)a*+»« = fc x = _l ± &.

\ log a

(c) a 1 • by = m, x :

logc
log d log m — log 6 log n

log a log <J — log 6 log c
c*-di< = ji. ^ logalogn -logc log m

log a log d — log 6 log c
(d) a 2;c - 3 -a 8 i'- 2 = a 8 , x=5,

3x + 2y=17. j/ = l.

74. Use of the tables of logarithms of the trigonometric functions.
On p. 9 the values of the trigonometric functions of angles from
0° to 90° were given in tabulated form. When we are using loga-
rithms in calculating expressions involving these trigonometric func-
tions it saves much labor to have the logarithms of these functions
already looked up for us and arranged in tabulated form.* Two com-
plete sets of such logarithms of the trigonometric functions are
given. Table II, pp. 8-16, should be used when the given or re-
quired angle is expressed in degrees, minutes, and the decimal part
of a minute ; and Table III, pp. 20-37, when the given or required
angle is expressed in degrees, and the decimal part of a degree. - ) -
In both tables the following directions hold true :

Angles between 0° and 45° are in the extreme left-hand column
on each page,! and the logarithm of the function of any angle will
be found in the same horizontal row with it and in the vertical col-
umn with the name of the function at the top ; that is, sines in the
first column, tangents in the second, cotangents in the third, and
cosines in the fourth, counting from left to right.

* To distinguish between the two kinds of tables, that on p. 9 is called a Table of
Natural Functions, while the logarithms of these functions arranged in tabulated form is
called a Table of Logarithmic Functions.

t The division of the degree into decimal parts, instead of using minutes and seconds,
has much to recommend it theoretically, and is also regarded vith favor by many expert
computers. In fact, a movement towards the adoption of such a system of subdivision is
not only gaining headway in France and Germany, but is making itself felt in America.

J The angles increase as we read downwards.

10 = 1.7941.

p. 16

= 2.5363.

p. 8

= 0.5905.

p. 31

10 = 1.9994.

p. 25

142 PLANE TRIGONOMETRY

Angles between 45° and 90° are in the extreme right-hand column
on each page,* and the logarithm of the function of any angle will
be found in the same horizontal row with it and in the vertical col-
umn with the name of the function at the bottom ; that is, cosines
in the first column, cotangents in the second, tangents in the third,
and sines in the fourth, counting from left to right.

In order to avoid the printing of negative characteristics, the

number 10 has been added to every logarithm in the first, second,

and fourth columns (those having log sin, log tan, and log cos at the

top). Hence in writing down any logarithm taken from these three

columns — 10 should be written after it. Logarithms taken from the

third column, having "log cot" at the top, should be used as printed.

Thm -

' log sin 38° 30' = 9.7941

log cot 0° 10' = 2.5363

log tan 75.6° = 0.5905

log cos 2.94° = 9.9994

75. Use of Table II, pp. 8-i6, the given or required angle being
expressed in degrees and minutes, f This table gives the logarithms
of ' the sines, cosines, tangents, and cotangents of all angles from
0° to 5° and from 85° to 90° for each minute on pp. 8-12 ; and on
pp. 13-16, from 5° to 85° at intervals of 10 minutes.

The small columns headed "diff. V" immediately to the right of
the columns headed " log sin " and " log cos " contain the differences,
called tabular differences, in the logarithms of the sines and cosines
corresponding to a difference of 1' in the angle. Similarly, the small
column headed "com. diff. V" contains the tabular differences for
both tangent and cotangent corresponding to a difference of 1' in
the angle. It will be observed that any tabular difference is not in
the same horizontal row with a logarithm, but midway between the
two particular logarithms whose difference it is. Of course that
tabular difference should always be taken which corresponds to
the interval in which the angle in question lies. Thus, in finding
log cos 78° 16', the tabular difference corresponding to the interval
between 78° 10' and 78° 20' is 6.1.

* The angles increase as we read upwards.

t In case the given angle involves seconds, first reduce the seconds to the decimal part
of a minute by dividing by 60. Thus,

88° 18' 42" = 88" 18.7', since 42" <= $$ = .7' ;

2° 0' 16" = 2" 0.27', since 16" = \§ = .266'

If the angle is given in degrees and the decimal parts of a degree, and it is desired tc
use Table II, the angle may be quickly found in degrees and minutes by making use of the
Conversion Table on p. 17.

THEORY AND USE OF LOGARITHMS 143

76. To find the logarithm of a function of an angle when the angle is
expressed in degrees and minutes, use

RuiiE VI. When the given angle is found exactly in Table II, the
logarithm of the given function of the angle is immediately found in
the same horizontal row and in the vertical column having the given
function at the top when the angle is less than 45", or at the bottom
when the angle is greater than 45°.

In case the given angle is not found exactly in the table we should
take the following steps :

(a) Write down the logarithm of the same function of the next
less angle found in the table, and also the corresponding tabular
difference for 1'.

(b) To find the correction necessary, multiply- this tabular differ-
ence by the excess in minutes of the given angle over the angle whose
logarithm was written down.

(c) If sine or tangent, add } , , . , . ^

Z-. . . f- this correction.*

If cosine or cotangent, subtract J

This rule, as well as the next three, assumes that the differences
of the logarithms of functions are proportional to the differences of
their corresponding angles. Unless the angle is very near 0° or 90°,
this is in general sufficiently exact for most practical purposes.

Ex. 1. Find log tan 32° 30'.

Solution. On p. 15, Table II, we find the angle 32° 30' exactly ; hence, by
Eule VI, we get immediately from the table

log tan 32° 30' = 9.8042 - 10. Ans.

Ex. 2. Find log cot 88° 17'.

Solution. On p. 9, Table II, we find the angle 88° 17' exactly ; hence, by
Kule VI, we get at once

log cot 88° 17' =8.4767 -10. Ans.

Ex. 3. Find log sin 23° 26'.

Solution. The exact angle 23° 26' is not found in Table II ; but then, by
Rule VI, from p. 14,

log sin 23° 20' = 9. 5978 - 10 Tab . mtt . = 2 .9

corr. for 6' = 17 Excess = 6

log sin 23° 26' = 9. 5995 -10. Ans. Corr - =rM

* The sine and tangent increase as the angle increases, hence we add the correction ;
the cosine and cotangent, however, decrease as the angle increases, hence we Bubtract the
correction. Of course this is true only for acute angles.

14-1 PLANE TRIGONOMETRY

Ex. 4. Find log cos 54° 42' 18".

Solution. Since 18" is less than half a minute, we drop it, and from p. 16,
Table II, by Rule VI,

log cos 54° 40' = 9. 7622 - 10 Tab - diff - = L8

j. „, , Excess = 2

corr. for 2' = 4 Corr = £ g

logcos54°42'= 9.7618 -10. Ans. l.e.=4

Ex. 5. Find log cot 1° 34.42'.

Solution. From p. 9, Table II, by Rule VI,

log cot 1° 34' =1.6630 Tab.diff.= 46

corr. for .4' = 18 Excess = .4

log cot 1° 34.4' =1.5612. Ans. Corr - = 18 - 4

When the angles are given in the table at intervals of 10', it is only
necessary to take our angle to the nearest minute, while if the angles
are given for every minute, we take our angle to the nearest tenth of
a minute. Thus, in Ex. 4, we find cos 54° 42', dropping the seconds ;
and in Ex. 5 we find log cot 1° 34.4', dropping the final 2.

Ex. 6. Verify the following :

(a) log tan 35° 50' = 9.8586 - 10. (g) log cos 27° 28' = 9.9480 - 10.

(b) logsin61° 58' =9.9458 -10. (h) log cot 61° 49' = 9.8957 - 10.

(c) log tan 82° 3' 20"= .8550. (i) log sin 85° 57' = 9.9989 - 10.

(d) log cos 44° 32' 50"= 9.8528 -10. (j) log cot 45° 0' 13" = 0.0000.

(e) log tan 1° 53.2' = 8.5178 - 10. (k) log sin 120° 24.3' = 9.9358 - 10.
(i) log tan 87° 15.6' =1.3201. (1) log tan 243° 42' 16" = 0.3060.

77. To find the acute angle in degrees and minutes which corresponds
to a given logarithmic function, use

Rule VII. When the given logarithmic function is found exactly
in Table II, then the corresponding angle is immediately found in the
same horizontal row, to the left if the given function is written at
the top of the column, and to the right if at the bottom.

In case the given logarithmic function is not found exactly in the
table we should take the following steps :

(a) Write down the angle corresponding to the next less logarithm
of the same function found in the table, and also the corresponding
tabular difference for 1'.

(b) To find the necessary correction in minutes divide this tabular
difference into the excess of the given logarithmic function over the
one written down.

(c) If sine or tangent, add 1 , . . *
' . > this correction.*

If cosine or cotangent, subtract J

* See footnote, p. 143.

THEOEY AND USE OF LOGARITHMS 145

In searching the table for the given logarithm, attention must be
paid to the fact that the functions are found in different columns
according as the angle is less or greater than 45°. If, for example,
the logarithmic sine is found in the column with "log sin" at the
top, the degrees and minutes must be taken from the left-hand.
column, but if it is found in the column with " log sin " at the
bottom, the degrees and minutes must be taken from the right-hand
column. Similarly, for the other functions. Thus, if the logarithmic
cosine is given, we look for it in two columns on each page, the one
having " log cos " at the top and also the one having " log cos " at
the bottom.

Ex. 7. Find the angle whose log tan = 9.6946 — 10.

Solution. This problem may also be stated as follows : haying given log tan x
= 9.6946 — 10 ; to find the angle x. Looking up and down the columns having
" log tan " at top or bottom, we find 9.6946 exactly on p. 15, Table n, in the
column with " log tan " at top. The corresponding angle is then found in the
same horizontal row to the left and is x = 26° 20'.

Ex. 8. Find the angle whose log sin = 9.6652 — 10.

Solution. That is, having given log sinx = 9.6652 — 10 ; to find the angle x.
Looking up and .down the columns having " log sin " at top or bottom, we do
not find 9.6652 exactly ; but (Rule VII) the next less logarithm in such a column
is found on p. 15, Table II, to be 9.6644, which corresponds to the angle 27° 30 / ,
and the corresponding tabular difference for 1' is 2.4. Hence

log sins = 9.6652 — 10 Tab. dlff.l'l Excess |Corr.
log sin 27° 30" = 9.6644 - 10 —

72

excess =

° 8

Since the function involved is the sine, we add this correction, giving
x = 27° 30' +3' =27° 33'. Ant.

Ex. 9. Find the angle whose log cos = 9.3705 — 10.

Solution. That is, having given log cos x = 9.3705 — 10 ; to find the angle x.
Looking up and down the columns having " log cos " at top or bottom, we do
not find 9.3705 exactly ; but (Rule VII) the next less logarithm in such a col-
umn is found on p. 13, Table II, to be 9.3682, which corresponds to the angle
76° 3<y, and the corresponding tabular difference for V is 5.2. Hence

logcosx = 9.3705 - 10 Tab.difl.riExcesslCorr.
log cos 76° 30' = 9.3682 - 10 — ' ^? L-^—

excess = 23

■:i

Since the function involved is the cosine, we subtract this correction, giving
x = 76° 30' -4' = 76° 26'. Ans.

146 PLANE TRIGONOMETRY

Ex . 10. Given log tan x = 8. 7670 - 10 ; find x.

Solution. By Rule VII the next less logarithmic tangent is found on p. 11,
Table II.

logtanx = 8.7570 - 10 Tab. diff.l' [Excess! Corr.

22 5.0 .2

log tan 3° 16' = 8.7665-10
excess = 5

Hence x = 3° 16' + .2' = 3° 16.2'. Ans.

44
6

13.0

Ex. 11. Given cotx = (1.01) 6 ; find x.
Solution. Taking the logarithms of both sides,

log cot x= 5 log 1.01. Th. Ill, p. 122

But log 1.01 = 0.0043

and, multiplying by 5, 5

log cotx = 0.0215 ; to find x.

By Rule VII the next less logarithmic cotangent is found on p. 16, Table II.

log cotx = 0.0215 Tab.diff.l' [Excess! Corr.

log cot 43° 40' = 0.0202 2.6 \ 13.0 | 5

excess = 13

Hence x = 43° 40' - 5' = 43° 36'. Ans.

Ex. 12. Verify the following :

(a) H log sin x = 9.5443 - 10, then x = 20° 30'.

(b) If log cosx = 9.7531 - 10, then x = 55° 30'.

(c) If log tan x = 9.9570 - 10, then x = 42° 10'.

(d) If log cotx = 1.0034, then x = 5° 40'.

(e) If log sin x = 8.0436 - 10, then x = 0° 38'.

(f) If log cosx = 8.7918 - 10, then x = 86° 27'

(g) If log tan x = 9.5261 - 10, then x = 18° 34'.
(h) If log cotx = 0.6380, then x = 12° 58'.
(i) If log sin x = 9.9995 - 10, then x = 87° 16'.*
(j) If log cosx = 8.2881 - 10, then x = 88° 53.3'.
(k) If log tan x = 2. 1642, then x = 89° 36.4'. _
(1) If log tanx = 7.9732 - 10, then x = 0°32.3'.

(m) If log sin x = 9.8500 - 10, then x = 45° 4'.
(n) If log cos x = 9.9000 - 10, then x = 37° 25'.
(o) If log tan x = 0.0036, then x = 45° 14'.
(p) If log cotx = 1.0000, then x = 5° 43'.
(q) If log cotx = 3.9732, then x = 89° 27.7'.

* "When there are several angles corresponding to the given logarithmic function, we
choose the middle one.

THEORY AND USE OF LOGARITHMS 141

EXAMPLES

Use logarithms when making the calculations in the following examples :

1. Given 184sinS x = (12.03)2 cos 57° 20' ; find x.
Solution. First we solve for sin j;. giving

„„,._ » /(ia.08}» eoe57° ay

V 184

Taking the logarithms of both sides,

log sin x = ^ [2 log 12.03 + log cos 57° 20' + colog 184].

2 log 12.03= 2.1606 since log 12.03 = 1.0803

log cos 57° 20' = 9. 7322 - 10

colog 184 = 7.7352 - 10 since los* 184 = 2.2648

19.6280

-20

3 |29.0280

-30

log sin j

= 9.8700

-10

. I

= 4S 3 44'.

-las.

2.

Given

cosi = (.9854)*;

find x.

3

Calcnl

k 4.236 cos 52°
ate

1{K

Ans. 5^45'.

" 13.087 sin 48^ 5' " 26 ° 9-

4. Given 1.5 cot 82 D = x- sin 12° 15'; find x. .9968.

Bint. First solve for x, giving

' i j cot sa°

_ \ sin li" 15' '

5. Given 50 tan i = -^.2584 ; find x. 0° 49'.

„ _ , , . sin 24° 13' cot 58° 2"

6. Calculate 84-ir

cos 33° 17' tan 19° 58' ' " "

7. Calculate Vcos 10° 5' ta n 73° 11'. 1.805.

„„,,<. (sin33°18'V<Vcot71°20'

8. Calculate 0044°°

10.658 tan 63° 54' ■""-«—

9. Given 3 cot x = ¥7; ; find x. 72° 45'.

10. Given sinz = (.9361) 10 ; find jr. 31° 6'.

11. Given 2.3 tan x = (1.002) la5 ; find x. 29P 24'.

78. Use of Table HI, pp. 20-37, the given or required angle being
expressed in degrees and the decimal part of a degree.* This table gives,
on pp. 20-29, the logarithms of the sines, cosines, tangents, and
cotangents of all angles from 0° to 5°. and from 85° to 90° for every
hundredth part of a degree ; and on pp. 30-37 from 5° to 85° for
every tenth of a degree.

The tabular differences between the logarithms given in the table
are given in the same manner as were the tabular differences in
Table II, and the general arrangement is the same.

■ In case the angle is given in degrees, minutes, and seconds, and it is desired to use
Table TTX, we may quickly reduce the angle to degrees and the decimal part of a degree by
using the Conversion Table on p. 17.

148 PLANE TRIGONOMETRY

79. To find the logarithm of the function of an angle when the angle
is expressed in degrees and the decimal part of a degree, use

Rule VIII. When the given angle is found exactly in Table III,
the logarithm of the given function of the angle is immediately
found in the same horizontal row and in the vertical column having
the given function at the top when the angle is less than J/.5°, or at
the bottom when the angle is greater than 45°.

In case the given angle is not found exactly in the table we should

take the following steps :

(a) Write down the logarithm of the same function of the next
less angle * found in the table and note the tabular difference which
follows.

(b) In the Prop. Parts column locate the block corresponding to
this tabular difference. Under this difference and opposite the extra
digit of the given angle will be found the proportional part of the
tabular difference (that is, the correction).

(c) If sine or tangent, add 1 , . .

v ' z. > this correction.^

If cosine or cotangent, subtract )

Ex. 1. Find log sin 27. 4°.

Solution. On p. 34, Table III, we find the angle 27.4° exactly; hence, by
Rule VIII, we get at once

log sin 27.4° = 9.6629 - 10. Ana.

Ex. 2. Find log cot 3. 17°.

Solution. On p. 26, Table III, we find the angle 3.17° exactly; hence, by
Rule VIII, we get immediately from the table

log cot 3. 17° = 1.2566. Ans.

Ex.3. Find log tan 61.87°.

Solution. The exact angle 61.87° is not' found in our tables. But then, by
Rule VIII, the next less angle is 61.8°, the extra digit of the given angle being 7,
and we have, from p. 34, Table III,

log tan 61.8° = 10.2707 - 10.

The tabular difference between log tan 61.8° and log tan 61.9° is 18. In the
Prop. Parts column under 18 and opposite the extra digit 7 we find the pro-
portional part 12. 6 ( = 13). Then

log tan 61. 80° =0.2707

13 Prop. Part.

log tan 61.87° = 0.2720. Ana.

* This "next less angle" will not contain the last (extra) digit of the given angle.
■t See footnote, p. 143.

THEORY AND USE OF LOGARITHMS 149

Ex. 4. Find log cot2.158°.

Solution. The exact angle 2.158° is not found in our tables. But then, by-
Rule VIII, the next less angle is 2. 15°, the extra digit of the given angle being 8,
and we have, from p. 24, Table III,

log cot 2. 15° =1.4255.

The tabular difference between log cot 2. 15° and log cot 2. 16° is 20. In the
Prop. Parts column under 20 and opposite the extra digit 8 we find the propor-
tional part 16. Then

log cot 2. 150° =1.4255

16 Prop. Part,
log cot2.158° = 1.4239. Ann.

Ex. 5. Verify the following :

(a) log tan 37.6° = 9.8865 - 10. (g) log tan 88.564° = 1.6009.

(b) log sin 63.87° = 9.9532 - 10. (h) log cos20.03° = 9.9729 - 10.

(c) log cot 1.111° = 1.7123. (i) log sin 89.97° = 0.0000.

(d) log sin 0.335° = 7.7669 - 10. (j) log cot34.84° = 0.1574.

(e) log cos45.68° = 9.8443 - 10. (k) log sin 155.42° = 9.6191 - 10.

(f) log tan 3.867° = 8.8299 - 10. (1) log tan 196.85° = 9.4813 - 10.

80. To find the acute angle in degrees and decimal parts of a degree
which corresponds to a given logarithmic function, use

Rule IX. When the given logarithmic function is found exactly in
Table III, then the corresponding angle is immediately found in the
same horizontal row ; to the left, if the given function is written at
top of the column, and to the right if written at the bottom.

In case the given logarithmic function is not found exactly in the
table we should take the following steps :

(a) Locate the given logarithm between two of the logarithms of
the same function given in the tables.

(b) The lesser angle of the two angles corresponding to these
logarithms will be the required angle complete except for the last
digit. Write this angle down with the corresponding logarithmic
function.

(c) Find the difference between the logarithm just written down
and the given logarithm, also noting the corresponding tabular
difference in the table.

(d) In the Prop. Parts column, under this tabular difference, pick
out the proportional part nearest the difference found in (c), and to
the left of it will be found the last (extra) digit of the required angle,
which we now annex.

150 PLANE TRIGONOMETRY

Ex. 6. Having given log tan x = 9. 5364 — 10 ; to find the angle x.

Solution. Looking up and down the columns having "log tan" at top oi
bottom, we do not find 9.5364 exactly. But then, by liule IX, we locate n
between 9.5345 and 9.5370, on p. 32, Table III. Except for the last digit the
required angle will be the lesser of the two corresponding angles, that is, 18.9°.
Then

log tan 18. 9° =9. 6345 -10
log tan x = 9.5364-10

19 = difference.

The corresponding tabular difference being 25, we find in the Prop. Parts col-
umn that 20 is the proportional part under 25 which is nearest 19. To the left
of 20 is the last (extra) digit 8 of the required angle. Hence x = 18.98°. Ans.

Ex. 7. Having given log cosx = 8.6820 — 10 ; find x.

Solution. On p. 25, Table III, we locate 8.6820 between 8.6810 and 8.6826.
Except for the last digit, the required angle must be the lesser of the.two cor-
responding angles, that is, 87.24°. Then

log cos 87.24° = 8.6826 - 10
log cosx = 8.6820 - 10

6 = difference.

The corresponding tabular difference being 16, we find in the Prop. Parts col-
umn that 6.4 is the proportional part under 16 which is nearest 6. To the left
of 6.4 is the last (extra) digit 4 of the required angle. Hence x = 87.244°. Ans.

Ex. 8. Verify the following :

(a) If log sin x = 9.6371 - 10, then x = 25.7°-

(b) If log cosx = 9.9873 - 10, then x = 13.8°.

(c) If log tanx = 8.9186 - 10, then x = 4.74°.

(d) If log cot x = 1.1697, then x = 3.96°.

(e) If log sinx = 9.5062 - 10, then x = 18.67°

(f) If log cosx = 9.9629 - 10, then x = 23.35°.

(g) If log tanx = 9.8380 - 10, then x = 34.55°.
(h) If logcotx = 9.3361 - 10, then x = 77.77°.

(i) If log sinx = 8.6852 - 10, then x = 2.776°.
(j) If log cos x = 9.9995 - 10, then x = 2.74°.
(k) If log tanx = 7.2642 - 10, then x = 0.105°.
(1) If log cotx = 1.7900, then x = 0.929°.
(m) If log sinx = 9.5350 - 10, then x = 20.05°.
(n) If log cosx = 9.8000 - 10, then x = 50.88°.
(o) If log tanx = 0.0035, then x = 45.23°.
(p) If log cotx = 2.0000, then x = 0.573°.
(q) If log sinx = 0.0000, then x = 90°.
(r) If log tan x = 0.0000, then x = 45°.

THEORY AND USE OF LOGARITHMS

151

EXAMPLES

Use logarithms when making the calculations in the following examples:

1. Given tan x = (1.018)!2; find x.

Solution. Taking the logarithms of both sides,

log tan x = 12 log 1.018. Th. Ill, p. 122

But
and, multiplying by 12,

log 1.018 = 0.0077
12

log tanx = 0.0924

On p. 36 we locate 0.0924 between 0.0916 and 0.0932. Then

log tan 51.0° =0.0916
log tan x = 0.0924

8 = difference.

The tabular difference is 16. In the Prop. Parts column under 16 we find 8.0
exactly. To the left of 8.0 we find the last digit 5 of the required angle. Hence
x = 51.05°. Ans.

2. Given 56.4 tan 5 x = (18.65) 5 cos69.8°; finds.
Solution. First we solve for tan x, giving

5/(18.65)5 cos 69.8°

tan x = \ :

\ 56.4

Taking the logarithms of both sides,

log tan a; = \ [5 log 18.65 + log cos69.8°+ colog 56.4].

5 log 18.65= 6.3535

log cos 69. 8°= 9.5382 -10

colog 56.4 = 8.2487 - 10

24.1404-20

5 |54.1404-50

log tan x = 10.8281 - 10.

.-. x = 81.55°. Ans.

3. Given cosx =V.9681; find x.

, , 26.52 tan 33.86°
4 - Calculate i0086eot88g63O -

5. Given V§ sin 48.06° = x 8 cos 2. 143° ; find X.
Hint. First solve for x, giving

cos 2.143"

_ J 1 /VI sin 48.06°

6. Given 5 cot x = ^.4083 ; find x.
V83 cos 52.82°

7. Given sin x ■■

(13. 382) 2

- ; find X.

8. Calculate V361 tan 87. 5° sin 9. 53°

since log 18.65 = 1.2707
since log 56.4 = 1.7513

Ans. 10.25°.
9.745.

1.0885.

81.56°.
1.762°.
37.

152 PLANE TRIGONOMETRY

81. Use of logarithms in the solution of right triangles. Since the
solutions of right triangles involve the calculation of products and
quotients, time and labor may be saved by using logarithms in the
computations. From p. 7 we have the following :

General directions for solving right triangles.

First step. Draw a figure as accurately as possible representing
the triangle in question.

Second step. When one acute angle is known, subtract it from, 90°
to get the other acute angle.

Third step. To find an unknown part, select from (1) to (6), p. 2,
a formula involving the unknown part and two known parts, and
then solve for the unknown part*

Fourth step. Check the values found by seeing whether they
satisfy relations different from those already employed in the third
step. A convenient numerical check is the relation

a 2 = c 2 - b 2 = (c + b) (p - S).f
Large errors may be detected by measurement.

For reference purposes we give the following formulas from p. 8
and p. 11. , t

Area of a right triangle = — •

(7) Side opposite an acute angle = hypotenuse x sine of the angle.

(8) Side adjacent an acute angle = hypotenuse x cosine of the angle.

(9) Side opposite an acute angle = adjacent side x tangent of the angle.

It is best to compute the required parts of any triangle as far as
possible from the given parts, so that an error made in determining
one part will not affect the computation of the other parts.

* This also includes formulas (7), (8)» (9). on p. 11.

t When we want the hypotenuse, the other two sides being given, this formula is not
ffell adapted to logarithmic computation, since

c= Vaa + b%

and we have a summation under the radical that cannot he performed hy the use of our
logarithmic tables. If, however, we have the hypotenuse c and one side (as b) given to find
the other side a, then

a= Vc 2 -& 2 = V(c-6)(c + 6),

and we have a product under the radical. The factors c-b and c + b of this product are
easily calculated hy inspection, and then we can use logarithms advantageously. Thus

log a = £ [log (c-b) + log (c + ft)].

t In case a or 6 is not given, or both a and 6 are not given, we first find what we need
from the known parts, as when solving the triangle, so that we can use the above formula
for finding the area.

THEOEY AND USE OF LOGAEITHMS

153

In trigonometric computations it sometimes happens that the
unknown quantity may be determined in more than one way.
When choosing the method to be employed it is important to keep
in mind the following suggestions :

(a) An angle is best determined from a trigonometric function
which changes rapidly, that is, one having large tabular differences,
as the tangent or cotangent.

(b) When a number is to be found (as the side of a triangle) from
a relation involving a given angle, it is best to employ a trigonometric
function of the angle which changes slowly, as the sine or cosine.

As was pointed out on pp. 13, 14, the solution of isosceles
triangles and regular polygons depends on the solution of right
triangles.

The following examples will illustrate the best plan to follow in
solving right triangles by the aid of logarithms.

Ex. 1. Solve the right triangle if A = 48° 17', c = 324. Also find the area.
Solution. First step. Draw a figure of the triangle indicating the known and
unknown parts.

Second step. B=90°-A = 41° 43 > .

Third step. To find a use a = c sin .A.

Taking the logarithms of both sides,

log a = log c + log sin A.

Hence, from Tables I and II,*

logc= 2.5105
log sin A - 9.8730 - 10
loga = 12.3835 -10

= 2.3835.
.-. a = 241.8.

To find 6 use b = c cos^i .

Taking the logarithms of both sides,

log 6 = log c + log cos A.

Hence, from Tables I and II,

logc= 2.5105
log cos A = 9.8231 - 10
log 6 = 12.3336 -10
= 2.3336

.-. 6 = 215.6.

« If we wish to use Table III instead of Table II, we reduce 17' to the decimal of a degree.

Thus,

A= 48° 17' = 48.28°.

154

PLATTE TRIGONOMETRY

Fourth step. To,check these results numerically, let us see if a, b, c satisfy

the equation

a 2 = c 2 - 6 2 = (c + 6) (c - 6),

or, using logarithms, 2 log a =,log(c + 6) + log(c — 6),

that is, log a = £ [log (c + 6) + log (c — 6)].

Here c + b = 639.6 and c - 6 = 108.4.

log(c + 6) = 2.7321

log(c- 6) = 2.0350

2 log a = 4.7671

loga= 2.3835.

Since this value of log a is the same as that obtained above, the answers are
probably correct.

To find the area use formula

. ab

Area = —
2

log area = log a + log b - log 2.

loga = 2.3835

log 6 = 2.3336

4.7171

log 2 = 0.3010

log area = 4.4161

.-. area = 26,070.

Ex. 2. Solve the right triangle, having given 6 = 15.12, c = 30.81.
Solution. Here we first find an acute angle ; to find A use

ooaA = -. (2), p. 2

log cos .4 = log 6 — log c. .

log 6 = 11. 1796 -10
loge= 1.4887
logcos^= 9.6909-10

. . A = 60° 36'. from Table II, p. 15

B = 90° - A = 29° 24'.

Hence

15.12

To find a we may use

a = bt&nA.
log a = log b + log tan .4.

logo =1.1796.

log tan 4. = 0.2491

log a = 1.4287

by (9), p. 11

a = 26.84.

THEORY AND USE OF LOGARITHMS 155

To check the work numerically, take

o*=(e + 6)(c-6),
or > log a = £ [log (c + 6) + log (c - &)] .

Here c + b = 45.93 and c-b = 16.69.

log(c + 6) = 1.6621

log(c - 6) = 1.1956

2 log a = 2.8578

loga = 1.4288.

This we see agrees substantially with the above result.

Ex. 3. Solve the right triangle, having given B = 2.325°, a = 1875.3.
Solution. A = 90° - B = 87.676°

sin^ = -. by(l), p. 2

Solving for the unknown side c,

_ a
smA
logo = loga — log sin A,

Hence, from Tables I and III,*

B

log a = 13.2731-

- 10

ii

log sin A= 9.9996-

- 10

C3

logc= 3.2735
.-. c = 1877.

A ti c

tan.4 = -•
6

by (3), p. 2

Solving for the unknown side b,

b= a .

tan A
log b = log a — log tan A

loga= 13.2731 -10
log tan A = 11.3915-10
log 6= 1.8816

.-. 6=76.13.

To check the work we may use formulas

a?=(c + b) (c - 6),
or, b — csinB, by (7), p. 11

since neither one was used in the above calculations.

* If -we wish to use Table II instead of Table III, we reduce 2.325° to degrees and min-
utes. Thus, B= 2.325° = 2° 19.5'.

156

PLANE TEIGONOMETEY

EXAMPLES
Solve the following right triangles (C= 90°), using logarithmic Tables I and II. «

No.

Given

Parts

Req

jired Parts

1

A = 43° 30'

c = 11.2

B = 46° 30'

a= 7.709

6 = 8.124

2

B = 68° 50'

a = 729.3

A = 21° 10'

6= 1883.5

c = 2019.5

3

B = 62° 56'

6 = 47.7

A = 27° 4'

a = 24.37

c = 53.56

4

a = .624

c = .91

A = 43° 18'

B = 46° 42'

6 = .6623

5

A = 72° 7'

a = 83.4

B = 17° 53'

6 = 26.91

c = 87.64

6

6 = 2.887

e = 5.11

B = 34° 24'

A = 55° 36'

a = 4.216

7

A = 52° 41'

b = 4247

B = 37° 19'

a = 5571

c = 7007

8

a = 101

6 = 116

A = 41° 2'

5 = 48° 58'

c= 153.8

9

A = 43° 22'

a = 158.3

£ = 46° 38'

6=167.6

c = 230.5

10

a = 204.2

c = 275.3

A = 47° 53'

B = 42° 7'

6= 184.7

11

B = 10° 51'

c = .7264

A = 79° 9'

a = .7133

6 =.1367

12

a = 638.5

6 = 501.2

4 = 51° 53'

B = 38° r

c = 811.7

13

6 = .02497

c = .04792

A = 58° 36'

B = 31° 24'

a = .0409

14

B = 2° 19' 30"

a = 1875.3

-4 = 87° 40'

30"

6=76.13

c = 1877

15

£ = 21° 33' 51"

a = .8211

A = 68° 26'

9"

6 = .3245

c = .8829

16

^1 = 74° 0' 18"

c = 275.62

B = 15° .59'

4^'

a = 264.9

6 = 75.95

17

B = 34° 14' 37"

6 = 120.22

A = 55° 45'

23"

a =176.57

c = 213.6

18

a= 10.107

6 = 17.303

A = 30° 17.6'

B = 59° 42.4'

c = 20.04

19

a = 24.67

6 = 33.02

4 = 36° 46'

B = 53° 14'

c = 41.22

20

^1 = 78° 17'

= 203.8

B = 11° 43'

6 = 42.27

c = 208.15

21. Find areas of the first five of the above triangles.

Ans. (1) 31.32; (2) 686,900; (3) 681.3; (4) .2067; (5) 1122.5.

Solve the following isosceles triangles where A, B, C are the angles and
a, b, c the sides opposite respectively, o and 6 being the equal sides.

22. Given A = 68° 57', 6 = 35.09. Ans. C= 42° 6', c = 25.21.

23. Given B = 27° 8', c = 3.088. Ans. C =125° 44', a = 1.735.

24. Given C = 80° 47', 5 = 2103. Ans. A = 49° 36.5', c = 2725.4.

25. Given a = 79.24, c = 106.62. Ans. A = 47° 43', C = 84° 34'.

26. Given C = 151° 28', c = 95.47. Ans. A = 14° 16', a = 49.25.

27. One side of a regular octagon is 24 ft. ; find its area and the radii of the
inscribed and circumscribed circles. Ans. Area = 2782, r = 28.97, R = 31.36.

* For the sake of clearness and simplicity, one set of triangle examples is given which
are adapted to practice in using Table II, the given and required angles being expressed in
degrees and minutes ; and another set is given on p. 157 for practice in the use of Table III,
the given and required angles being expressed in degrees and the decimal part of a degree.
There is no reason why the student should not work out the examples in the first set, using
Table III, and those in the second set, using Table II, if he so desires, except that it may
involve a trifle more labor. This extra work of reducing minutes to the decimal part of a
degree, or the reverse, may be reduced to a minimum by making use of the Conversion
Tables on p. 17. It is possible, however, that an answer thus obtained may differ from the
one given here by one unit in the last decimal place. This practice of giving one set of
triangle examples for each of the Tables II and III will be followed throughout this book
when solving triangles.

THEORY AND USE OF LOGARITHMS 157

Solve the following right triangles (C= 90°), using logarithmic Tables I and III.

No.

Given

Parts

Required Parts

28

a = 5

6 = 2

4 = 68.2°

B = 21.8°

c = 5.385

29

£ = 32.17°

c = .02728

A = 57.83°

a = .02309

6 = .01452

30

A = 58.65°

c = 35.73

B = 31.35°

a = 30.51

6=18.59

31

A = 22.23°

6 = 13.242

£ = 67.77°

a = 6.413

c = 14.31

32

6 = .02497

c = .04792

A = 58.6°

£ = 31.4°

a = .0409

33

a = 273

6 = 418

4=33.15°

£ = 56.85°

c = 499.3

34

£ = 23.15°

6=75.48

A = 66.85°

a =176.5

c = 191.9

35

A= 31.75°

a = 48.04

B = 58.25°

6 = 77.64

c = 91.28

36

6 = 512

c = 900

A= 55.32°

£ = 34.68°

a = 740.2

37

a= 52

c = 60

A = 60.06°

£ = 29.94°

6 = 29.94

38

A = 2.49°

a = .83

£ = 87.51°

6 = 19.085

c = 19.107

39

4 = 88.426°

6 = 9

£= 1.574°

a = 327.5

c = 327.6

40

£ = 4.963°

6 =.07

4 = 85.037°

a = .8062

c = .8092

41

£ = 85.475°

c = 80

4 = 4.525°

a = 6.313

6 = 79.74

42

a = 100.87

6 = 2

4 = 88.864°

£= 1.136°

c = 100.9

43. Find the areas of the first five of the above triangles.

4ns. (28) 5; (29) .0001677; (30) 283.6; (31) 35.84; (32) .00051.

44. The perimeter of a regular polygon of 11 sides is 23.47 ft. Find the
radius of the circumscribed circle. Ans. 3.79 ft.

45. Two stations are 3 mi. apart on a plain. The angle of depression of one
from a balloon directly over the other is observed to be 8° 15'. How high is the
balloon? 4ns. .435 mi.

46. A rock on the bank of a river is 130 ft. above the water level. From a
point just opposite the rock on the other bank of the river the angle of elevation
of the rock is 14° 30' 21". Find the width of the river. 4ns. 502.5 ft.

47. A rope 38 ft. long tied to the top of a tree 29 ft. high just reaches the
level ground. Find the angle the rope makes with the tree. 4ns. 40° 15'.

48. A man 5 ft. 10 in. high stands at a distance of 4 ft. 7 in. from a lamp-post,
and casts a shadow 18 ft. long. Find the height of the lamp-post. 4ns. 7.32 ft.

49. The shadow of a vertical cliff 113 ft. high just reaches a boat on the sea
93 ft. from its base. Find the altitude of the sun. 47is. 50° 33'.

50. The top of a tree broken by the wind strikes the ground 15 ft. from the
foot of the tree and makes an angle of 42° 28' with the ground. Find the origi-
nal height of the tree. 4ns. 34.07 ft.

51. A building is 121 ft. high. From a point directly across the street its
angle of elevation is 65° 3'. Find the width of the street. 4ns. 56.3 ft.

52. Given that the sun's distance from the earth is 92,000,000 mi., and the
angle it subtends from the earth is 32'. Find diameter of the sun.

4ns. About 856,400 mi.

53. Given that the radius of the earth is 3963 mi. , and that it subtends an
angle of 57' at the moon. Find the distance of the moon from the earth.

4ns. About 239,017 mi.

158 PLANE TKIGONOMETEY

54. The radius of a circle is 12,732, and the length of a chord is 18,321. Find
the angle the chord subtends at the center. Ans. 02° 2'.

55. If the radius of a circle is 10 in., what is the length of a chord which
subtends an angle of 77° 17' 40" at the center ? Ana. 12.488 in.

56. The angle between the legs of a pair of dividers is 43°, and the legs are
7 in. long. Find the distance between the points. Ans. 5. 13 in.

82. Use of logarithms in the solution of oblique triangles. As has

already been pointed out, formulas involving principally products,
quotients, powers, and roots are well adapted to logarithmic compu-
tation; while in the case of formulas involving in the main sums
and differences, the labor-saving advantages of logarithmic compu-
tation are not so marked. Thus, in solving oblique triangles, the

law of sines

a b o

sin. A sin 12 sinC
and the law of tangents

tan I (A - B) = |^| tan i(A + B),

are well adapted to the use of logarithms, while this is not the case
with the law of cosines, namely,

a 2 = 6 2 + c 2 — 2 be cos A.

In solving oblique triangles by logarithmic computation, it is con-
venient to classify the problems as follows :

Case I. When two angles and a side are given.
Case II. When two sides and the angle opposite one of them are
given (ambiguous case).

Case III. When two sides and included angle are given.
Case IV. When all three sides are given.

Case I. When two angles and a side are given.

First step. To find the third angle, subtract the sum of the two

given angles from 180°.

Second step. To find an unknown side, choose a pair of ratios from

the law of sines

a b c

smA sin .8 sinC

which involve only one unknown part, and solve for that part.
Check : See if the sides found satisfy the law of tangents.

THEORY AND USE OE LOGARITHMS

159

Ex. 1. Having given 6 = 20, A = 104°, B = 19° ; solve the triangle.
Solution. Drawing a figure of the triangle on which we indicate the known
and unknown parts, we see that the problem comes under Case I.

First step. C = 180° - (A + B) = 180° - 123° = 57°.

Second step. Solving

sin A sin 2J

6 sin A

for a, we get

a :

or,

sinB
log a = log 6 + log sin A — log sin B.

log 6= 1.3010
log sin A = 9. 9869 - 10 *

11.2879 - 10
log sin B = 9.5126 - 10

loga= 1.7753

a = 59.61.

Solving

sin B sin C

for c, we get

c =

bsinC

or,

sin J?
log c = log 6 + log sin C — log sin B.

log 6= 1.3010
log sin C = 9.9236

10

Check :

Here,

11.2246-10
log sin B = 9.5126-10
logc= 1.7120

c = 51.52.

a + c = 111.13, a-c = 8.09;

A + C = 161°, A - C = 47° ;

£(4 + C) = 80°30', £(^L-C) = 23° 30'.

tan 1 (A -C) = ^ tan $(A + C),
or, log tan £ (A — C) = log (a — c) + log tan \ (A + C) — log (a + c)

log(a-c)= 0.9079
log tan $(A + C) = 10.7764-10
11.6843 - 10

log(a + c) = 2.0458
log tan \ (A - C) = 9. 6386 - 10

.-. £(4-C) = 23°31',

which substantially agrees with the above results.

« sin A = sin 104° = sin (180° - 104°) = sin 76° . Hence log sin 104° = log sin 76" = 9.9869 - W.

160

PLANE TBIGONOMETKY

EXAMPLES
Solve the following oblique triangles, using logarithmic Tables I and II.

No.

Given Parts

Required Parts

1

a=10

4 =38°

.5=77° 10'

C=64°60'

6=15.837

c=14.703

2

a =795

4 = 79° 59'

5=44°41'

C=55°20'

6=567.6

c=664

3

6 =.8037

5 =52° 20'

C=101°40'

4 = 26°

a =.445

c=.9942

4

c = .032

4 = 36° 8'

B=U°21'

C=99°25'

a = . 01913

6=. 02272

5

6=29.01

4 = 87°40'

C=33°15'

6 =59° 5'

a=33.78

c=18.54

6

a =804

4 = 99° 55'

5=45°1'

C=35°4'

6=577.3

c = 468.9

7

a=400

4 = 54° 28'

C=60°

.5=65° 32'

6=447.4

c=425.7

8

c=161

A = 35° 15'

C=123°39'

5=21°6'

a=111.6

6=69.62

9

a=5.42

5=42°17.3'

C=82°28.4'

A = 55° 14.3'

6=4.439

c = 6.542

10

6=2056

4 = 63°52.8'

5=70°

C=46°7.2'

a =1964.7

c= 1577.3

11

a=7.86

5=32°2'62"

C=43°25'26"

4 = 104°31'42"

6=4.309

c=5.583

12

5=8

4=80°

5=2°15'46"

C= 97° 44' 14"

a =199. 53

c=200.73

Solve the following oblique triangles, using logarithmic Tables I and III.

No.

Given Parts

Eeq

uired Parts

13

a = 500

A = 10.2°

6 = 46.6°

C = 123.2°

6 = 2051

c = 2363

14

a = 45

.4 = 36.8°

C = 62°

5 = 81.2°

6 = 74.25

c = 66.33

15

6 = .085

5 = 95.6°

C = 24.2°

A = 60.2°

a = .0741

c = .035

16

6 = 5685

5 = 48.63°

C = 83.26°

4 = 48.11°

a = 5640

c = 7523

17

c = 7

4 = 59.58°

C = 60°

5 = 60.42°

a = 6.971

6 = 7.03

18

c = .0059

5 = 75°

C = 36.87°

4 = 68.13°

a = .00913

6 = .0095

19

a = 76.08

5 = 126°

C = 12.44°

4 = 41.56°

6 = 92.8

c = 24.7

20

a = 22

4 = 3.486°

5 = 73°

C= 103.514°

6 = 346

c = 351.8

21

6 = 8000

4 = 24.5°

B = 86.495°

C = 69.005°

a = 3324

c = 7483

22

6 = 129.38

4 = 19.42°

C = 64°

5 = 96.58°

a = 43.29

c = 117.05

23

c = 95

4 = 2.086°

5 = 112°

C = 65.914°

a = 3.788

6 = 96.5

24

5 = 132.6

4 = 1°

C = 75°

5 = 104°

a = 2.385

c = 131.98

25. A ship S can be seen from each of two points 4 and 5 on the shore.
By measurement 45 = 800 ft., angle SAB = 67° 43', and angle S54 = 74° 21'.
Find the distance of the ship from 4. 4ns. 1253 ft.

26. Two observers 5 mi. apart on a plain, and facing each other, find that
the angles of elevation of a balloon in the same vertical plane with them-
selves are 65° and 58° respectively. Find the distances of the balloon from
the observers. 4ns. 4.607 mi.; 4.45 mi.

27. One diagonal of a parallelogram is 11.237, and it makes the angles 19° 1'
and 42° 54' with the sides. Find the sides. 4ns. 4.15 and 8.67.

28. To determine the distance of a hostile fort 4 from a place 5, a line
BC and the angles 45C and 5C4 were measured and found to be 322.6 yd.,
60° 34', 56° 10' respectively. Find the distance 45. 4ns. 300 yd.

THEORY AND USE OF LOGARITHMS 161

29. From points A and B at the bow and stern of a ship respectively, the
foremast, C, of another ship is observed. The points A and B are 300 ft.
apart, and the angles ABC and BAG are found to be 65.46° and 112.85° respec-
tively. What is the distance between the points A and G of the two ships ?

Ans. 9254 ft.

30. A lighthouse was observed from a ship to bear N. 34° E. ; after the ship
sailed due south 3 mi. it bore N. 23° E. Eind the distance from the lighthouse
to the ship in each position. Ans. 6.143 mi. and 8.792 mi.

31. In a trapezoid the parallel sides are 15 and 7, and the angles one of them
makes with the nonparallel sides are 70° and 40°. Eind the nonparallel sides.

Ans. 8 and 5.47.

Case II. When two sides and the angle opposite one of them are given,
as a, b, A (ambiguous case *) .

First step. Using the law of sines as in Case J, calculate log sinB.

If log sinB = 0, sinB = 1, B = 90° j it is a right triangle.

If log sinB > 0, sinB > 1 (impossible); there is no solution.

If log sinB< and b < a, only the acute value of B found from
the table can be used ; there is one solution.^

If log sinB < O and 6 > a, the acute value of B found from the table
and also its supplement, should be used ; and there are two solutions, t

Second step. Find C (one or two values according as we have one
or two values of B) from

C = 180°-(A+B).

Third step. Find c (one or two values), using law of sines.
Check : Use law of tangents.

Ex. 1. Having given a = 36, 6 = 80, A = 28° ; solve the triangle.
Solution. In attempting to draw a figure of the triangle, the construction
appears impossible. To verify this, let us find log sin B in order to apply our tests.

First step. Solving -. — r = ^-^= for sinB,
* Bini sin_B

. _ ftsin^l
sm B = ,

or, log sin B = log 6 + log sin A — log a.

log 6= 1.9031
log sin A = 9.6716 - 10
11.5747 - 10
logq= 1.5563
log sin B = 10.0184 -10
= 0.0184.
Since log sinB > 0, sinB > 1 (which is impossible), and there is no solution.

* In this connection the student should read over 5 58, pp. 104, 105.

t For if b<a, B must be less than A, and hence B must be acute.

t Since 6>o, A must be acute, and hence B may be either acute or obtuse.

162

PLANE TKlGOJSfOMETRY

Ex. 2. Having given o = 7.42, b = 3.39, A = 106°; solve the triangle.

Solution. Draw figure.

First step. From law of sines,

. _ osin.4

sm B = >

a

log sin B = log 6 + log sin A — log a.

log 6= 0.5302
log sin A = 9.9849 - 10 *
10.5151 - 10

logq= 0.8704
log sin B= 9.6447-10

.-. B = 26°11'.

Using Table II

Since log sin B < and 6 < a, there is only one solution.
Second step. O = 180° - (A + B) = 180° - 131° 11' = 48° 49'.
Third step. By law of sines,

a sin C

sini
or, log c = log a + log sin C — log sin .A.

loga= 0.8704
log sin C = 9.8766 - 10
10.7470 - 10
log sin 4 = 9.9849 - 10

logc= 0.7621
.-. c= 5.783.

Check : Use law of tangents.

tan£(C - B) = ^^tan £( C + B),
or, log tan £ (C - B) = log (c - 6) + log tan £ (C + B) - log (c + 6).

Substituting, we find that this equation is satisfied.

Ex. 3. Given a = 732, 6 = 1015, A = 40° ; solve the triangle.

Solution. It appears from the construction of the triangle that there are

two solutions.

First step. By law of sines,

. _ 6 sin A

BinB = ,

a

or, log sin B = log 6 + log sin A — logo.

log 6= 3.0065
log sin A = 9.8081-10
= 12.8146 - 10

logq= 2.8645
log sin B= 9.9501-10

* Sin A = sin 105° = sin ( 180° - 105°) = sin 75° . Hence log sin A = log sin 75° = 9.9849 - 10.

THEORY AND USE OF LOGARITHMS

163

Since log BinB < and 6 > a, we have two solutions, which test verifies our
construction. From Table II we find the first value of B to be

B 1 = 6Z°3'.

Hence the second value of B is

B 2 = 180° - B 1 = 116° 57'.
Second step. C x = 180° - {A + B t ) = 180° - 103° 3" = 76° 57' ;
C 2 = 180° - (A + B 2 ) = 180° - 156° 57' = 23° 3'.

Third step. From law of sines,

Cl =

o sin Ci

or,

sin A
log c x = log a + log sin Ci — log sin .4.

loga= 2.8645
log sin d = 9.9886 - 10
12.8531 - 10
log nmA = 9.8081 - 10
logc 1= 3.0450

In the same manner, from
we get

.-. Ci =

c 2 =

1109.3.
a sin C 2

sin A
C2 = 445.9.

Cheek: Use tan£(C - B) = -tan£(C + B) for both solutions.

EXAMPLES

Solve the following oblique triangles, using logarithmic Tables I and IL

No.

Given Pakts

Required Parts

1

a=50

c=66

.4 = 123° 11'

Impossible

2

a=5.08

6=3.59

A = 63° 50^

_B=39°21'

C=76°49 /

c=5.511

3

a=62.2

6=74.8

A =27° 18'

^ 1 =33°28 /
B 2 =146°32'

C 1 = 119°14'
C 2 =6°10 /

c 1= 118.32
02=14.567

4

6=.2337

c=.1982

5=109°

A = 17° 41'

C = 53°19'

a =.07508

5

a=107

c=171

C=31°53'

A = 19° 18'

B=128°49 /

6=252.2

6

6=3069

c=1223

C=55°52'

Impossible

7

6=6.161

c=6.84

B=44°3'

4 1 =68°47 /
4 2 =23°7 /

Ci = 67°10 /
C 2 =112°50'

ai=6.92
02=2.913

8

a=8.656

c=10

A = 59° 57'

JB=30°3'

C=90°

6=5.009

9

a=214.56

6=284.79 _B=104°20'

4=46° 53'

C=28°47'

c=141.5

10

a=32.16

c =27.08

C'=62°24'

4i = 70°12'
4 2 =109°48'

Bi=57°24'
.B 2 =17°48 /

6 1= 28.79
62 = 10.45

11

6=811.3

c= 606.4

£=126° 6' 20"

A = 16° 44' 40"

C=37°10'

a=289.2

164 PLANE TRIGONOMETRY

Solve the following oblique triangles, using logarithmic Tables I and III.

No.

Given Pakts

Required Pakts

12

a = 840

6 = 485

-4 = 21.6°

5 = 12.21°

C = 146.29°

c = 1272

13

a = 72.63

5 = 117.48

4 = 80°

Impossible

14

= 177

6 = 216

A = 35.6°

5 t = 45.27°
5 2 = 134.73°

C 1 = 99.13°
C 2 = 9.67°

ci = 300.3
c 2 = 51.09

15

6 = 9.399

c = 9.197

5 = 120.4°

A = 2.02°

C= 57.58°

a = .3841

16

6 = .048

c = .0621

5 = 57.62°

Impossible

17

5 = 19

c = 18

C=15.8°

A x = 147.5°
4 2 = 0.9°

5i = 16.7°
5 2 = 163.3°

a x = 35.52
a 2 = 1.0385

18

a = 55.55

c = 66.66

C = 77.7°

.4 = 54.5°

5 = 47.8°

6=50.54

19

o = 34

c = 22

= 30.35°

4i = 51.37°
4 2 = 128.63°

5 1 = 98.28°

5 2 = 21.02°

&i = 43.07
6 2 = 15.613

20

a = 528

6 = 252

4 = 124.6°

5 = 23.14°

C = 32.26°

c = 342.3

21

6 = 91.06

c = 77.04

5 = 51.12°

4 = 87.69°

= 41.19°

a = 116.88

22

a = 17,060 6 = 14,050

5 = 40°

A r = 51.32°

Ci = 88.68°

Ci = 21,850

4 2 = 128.68°

C 3 = 11.32°

c 2 = 4290

23. One side of a parallelogram is 35, a diagonal is 63, and the angle between
the diagonals is 21° 37'. Find the other diagonal. Ans. 124.62.

24. The distance from 5 to C is 145 ft. , from A to C is 178 ft., and the angle
ABC is 41° 10'. Find the distance from A to 5. Ans. 259.4 ft.

25. Two buoys are 2789 ft. apart, and a boat is 4325 ft. from the nearer buoy.
The angle between the lines from the buoys to the boat is 16° 13'. How far is
the boat from the further buoy ? Ans. 6667 ft.

Case III. When two sides and the included angle are given, as a, b, C*
First step. Calculate a+b, a—b; also \{A+B)from A+B=180°— c.
Second step. From law of tangents,

tani(4-5)=^|tani(,l+5) ;

we find ^(A—B~). Adding this result to \(A + B) gives A, and sub-
tracting it gives B.

Third step. To find side c use law of sines ; for instance,

a sin C
sin 4
Chech: Chech by law of sines, \ that is, see if

log a — log sinA= log b — log sinB = log c — log sin C.

* In case any other two sides and included angle are given, simply change the cyclic
order of the letters throughout. Thus, if b, c, A are given, use

tan - (B - O = ^^ tan - (B + C) , etc.
2 b + c 2 V

t From law of sines,

a b c

s4n A sin B sin C

THEORY AND USE OF LOGARITHMS

165

Ex. 1. Having given a = 540, 6 = 420, C = 52° 6' ; solve the triangle, using
logarithms from Tables I and II.

Solution. Drawing a figure of the triangle on which we indicate the known
and unknown parts, we see that the problem comes under
Case II, since two sides and the included angle are given.

First step.

a= 540

540

180°

6 = 420

420

C= 52° 6'

a + 6 = 960

a - 6 = 120

A + B = 127° 54'
. ^(A + B)= 63° 57'.

Second step.

tan -(A —
2 V

B) = a_6 tan 1 M + J

' + 6 2 V

or,

Adding,
Third step.

log tan %(A-B) = log (o - 6) + log tan %(A + B)- log (a + 6).

log (a -6)= 2.0792
log tan $(A + B)= 10.3108 - 10
12.3900 - 10

log(o + 6)= 2.9823
log tan i (A-B)= 9.4077 - 10

.-. %(A-B) = U°2V

\(A + B) = 63° 57'
i(A-B) = U°2\-

A = 78° 18'.

a sin C
sin .A

63° 57'

14° 2 1'

Subtracting, B=49°36'.

From

sin G sin A

log c = log a + log sin C — log sin A.

log

loga =
sinC =

2.7324
9.8971 - 10
12.6295-10

log

sin A =
logc =
.-. c =

9.9909-10
2.6386
435.1.

Check :

By law of sines,

log a:
log sin A ■■

= 12.7324-10
= 9.9909-10

log 6, = 12.6232 -
log sin B= 9.8817 -
2.7415

10
10

2.7415

Ex. 2. Having given a = 167, c
arithms from Tables I and III.

= 82, B = 98°

; s

Solution. First step.

a = 167

c= 82

o + c = 249

a

167

82

-c= 85

logc = 12.6386 -10
log sin C = 9.8971 -10
2.7415

solve the triangle, using log-

188°

A + C= 82°
.-. ^(A+C)= 41°. >

166

PLANE TRIGONOMETRY

Second step.

ta.n-(A-C)= a

— -tani(4 + C),
a + c 1

log tan £(4 - C) = log(a - c) + logtan£(4 + C) ■

log(a-c)= 1.9294
logtan£(4 + C) = 9.9392-10

log(a + c)

log(a + c):

11.8686 ■
2.3962

10

Adding,
Third step.

logtan£(4-C) = 9.4724-

-10

.-. £(4-C) = 16.53°

£(4 + C) = 41.00°

^(4-C) = 16.53°

4 = 57.53°.

41.00°

16.53°

Subtracting, C = 24.47°.

, osinJS

. b a

sin A

from

sin 2? sin A

b = ?

log 6 = log a + log sin B — log sin A.

logo = 2.2227
log sin B = 9.9958-10 *

12.2185 - 10
log sin A = 9.9262-10
log 6= 2.2923
.-. 6 = 196.

Check : By law of sines,
logos = 12.2227 -10

log sin A = 9.9262 -10
2.2965

which substantially agree.

log 6 = 12.2923 -10
log sin B = 9.9958 - 10
2.2965

logc = 11.9138 -10
log sin C = 9.6172 - 10
2.2966

EXAMPLES
Solve the following oblique triangles, using logarithmic Tables I and II.

No.

Given Parts

Required Parts

1

o=27

c = 15

5=46°

A = 100° 57'

C=33°3'

6=19.78

2

0=486

6=347

C=51°36'

4 = 83° 16'

B=45°9'

c=383.5

3

6=2.302

e = 3.567

4=62°

B=39°16'

C = 78°44'

0=3.211

4

0=77.99

6=83.39

C=72°16'

4 = 51° 14.5'

5=56° 29.5'

c = 95.24

5

o=0.917

6=0.312

0=33° 7.2'

4 = 132° 18.4

' B= 14° 33.4'

c=.6775

6

a=.3

6=.363

C=124°56'

A = 24° 41.8'

5=30° 22.2'

c=.5886

7

6=1192.1

c=356.3

A =26° 16'

U=143°29'

C=10°15'

a=886.6

8

a=7.4

c = 11.439

J3=82°26'

A =35° 2'

C=62°32'

6=12.777

9

a=53.27

6=41.61

C=78°33'

4 = 59° 16.5'

.6=42° 10.5'

c = 60.74

10

6 =.02668

c = . 05092

A = 115° 47'

B=21°1.2'

C=43°11.8'

o=. 06699

11

a = 51.38

c=67.94

B= 79° 12' 36"

4 = 40° 52.7'

C= 59° 64.7'

6=77.12

12

5=V3

c = V3

A =35° 53'

B= 93° 28.5'

C= 50° 38.5'

a=1.313

* sin B= sin 98°= sin (180° -98°)= sin 82°. .-. log sin 98°= log sin 82° =9.9958- 10.

THEORY AND USE OF LOGARITHMS 167

Solve the following oblique triangles, using logarithmic Tables I and III.

No.

Given Parts

Required Parts

13

a = 17

6= 12

C = 59.3°

A = 77.2°

£ = 43.5°

c = 14.99

14

a = 55.14

6 = 33.09

C = 30.4°

A = 117.4°

B = 32.2°

c = 31.43

15

6 = 101

c = 158

A = 37.38°

B = 38.26°

C = 104.36°

a = 99.04

16

a = 101

6 = 29

C = 32.18°

A = 136.4°

B = 11.42°

c= 78

17

c = 45

6 = 29

A = 42.8°

B = 39.72°

C = 97.48°

a = 30.84

18

a = .085

c = .0042

B = 56.5°

A = 121.07°

C = 2.43°

6 = .08276

19

6 = .9486

c = .8852

A = 84.6°

B = 49.88°

C = 45.52°

a = 1.235

20

6 = 6

c = 9

A = 88.9°

B = 34.03°

C= 57.07°

a= 10.72

21

a = 12

6 = 19

C = 5.24°

A = 8.84°

B = 165.92°

c = 7.132

22

a = 42,930

c = 73,480

.5 = 24.8°

X = 27.56°

C= 127.64°

6 = 38,920

23. In order to find the distance between two objects, A and B, separated
by a swamp, a station C was chosen, and the distances CA = 3825 yd., CB =
3476 yd., together with the angle -4CB = 62° 31', were measured. "What is
the distance AB ? Ans. 3800 yd.

24. Two trains start at the same time from the same station and move along
straight tracks that form an angle of 30°, one train at the rate of 30 mi. an
hour, the other at the rate of 40 mi. an hour. How far apart are the trains at
the end of half an hour ? Ans. 10.27 mi.

25. In a parallelogram the two diagonals are 5 and 6 and form an angle of
49° 18'. Find the sides. Ans. 5.004 and 2.339.

26. Two trees A and B are on opposite sides of a pond. The distance of A
from a point C is 297.6 ft., the distance of B from C is 864.4 ft., and the angle
ACB is 87.72°. Find the distance AB. Ans. 903 ft.

27. Two stations A and B on opposite sides of a mountain are both visible
from a third station C. The distances AC, BC, and the angle ACB were meas-
ured and found to be 11.5 mi., 9.4 mi., and 59° 31' respectively. Find the dis-
tance from A to B. Ans. 10.535 mi.

28. From a point 3 mi. from one end of an island and 7 mi. from the other
end the island subtends an angle of 33° 55.8'. Find the length of the island.

Ans. 4.814 mi.

29. The sides of a parallelogram are 172.43 and 101.31, and the angle included
by them is 61° 16'. Find the two diagonals. Ans. 152.33 and 238.3.

30. Two yachts start at the same time from the same point, and sail, one due
north at the rate of 10.44 mi. an hour, and the other due northeast at the rate
of 7.71 mi. an hour. How far apart are they at the end of 40 minutes ?

Ans. 4.927 mi.

Case IV. When all three sides a, b, c are given.

First step. Calculate s = ^ (a + b + e), s — a, s — b, s — c.

Second step. Find log r from

, = ^ (s-")(s-l>)(s-°) . (84 ) to (87), p. 115

168

PLANE TRIGONOMETRY

Third step. Find angles A, B, C from

r

tan \A

tan ^ B = ■

r r

— - > tan A C = —
— b 2

— o

s — a
Check : SeeifA + B + C= 180°.

Ex. 1. Having given a = 51, 6 = 65, c = 20 ; solve the triangle.

Solution. Drawing a figure of the triangle on which we indicate the known
and unknown parts, we see that since the three sides are given, the problem
comes under Case IV.

First step, a = 51

6= 65

c= 20

2 s = 136

Hence

Second step.

s = 68
a= 51
a = 17

s = 68

6 = 65

s-6= 3

s = 68

c = 20

s - c = 48

j(s - a) (s

-b)(s-e)

or,

logr = £[log(s - a) + log(s - 6) + log(s - c) - logs].

From the table of logarithms,

6 = 65

Third step. From the formula tan £ A

log(s- a) = 1.2304

log (s- 6) = 0.4771

log (s-c) = 1.6812

3.3887

logs =1.8325

2 1 1.5562

logr = 0.7781

s — a

log tan ^ A = log r — log (s — a).

logr = 10.7781 -10
log(s-q)= 1.2304

or,

From the formula tan £ B =

\A= 19.44°

and .4=38.88°

Check : A + B+C= 179.98".

log tan £4= 9.5477-10

using Table II*

£J. = 19°27',

A - 38° 54'.

s — 6

log tan £ _B = logr — log(s

-6).

logr = 10.7781 -10

log (s -6)= 0.4771

log tan £ B = 10.3010 - 10

£.8 = 63° 26',

using Table II

B = 126° 52'.

ad, we get

, $B= 63.43°,

5 C= 7.12°,

, J3 = 126.86°,

C= 14.24°.

THEORY AND USE OF LOGARITHMS

169

From the formula tan £ O = ■

Check :

log tan I G = log r — log (s — c).

logr = 10.7781 -10
log (s-c)= 1.6812
log tan £ C = 9.0969 - 10
^ C = 7° 8'.
C = 14° 16'.

A = 38° 54'

£ = 126° 52'

C = 14° 16'

A + B + C = 180° 2'

Using Table II

EXAMPLES

Solve the following oblique triangles, using logarithmic Tables I and II.

No.

Given Parts

Required Parts

1

a = 2

6 = 3

c = 4

.4 = 28° 58'

£ = 46° 34'

C = 104° 28'

2

a = 2.5

6 = 2.79

c = 2.33

A = 57° 38'

B = 70° 28'

O = 51° 54'

3

a = 5.6

6 = 4.3

c = 4.9

^4 = 74° 40'

B = 47° 46'

C = 57° 34'

4

a= 111

6 = 145

c = 40

A = 27° 20'

B = 143° 8'

C = 9° 32'

5

a = 79.3

6 = 94.2

c = 66.9

.4. = 65° 56'

B = 79° 44'

C = 44° 20'

6

a= 321

6 = 361

c = 402

A = 49° 24'

B = 58° 38'

C = 71° 58'

7

a= .641

b = .529

c = .702

A = 60° 52'

£ = 46° 6'

C = 73°2'

8

a = 3.019

6 = 6.731

c = 4.228

A = 18° 12'

£ = 135° 52'

C = 25° 56'

9

a= .8706

6 = .0916

c = .7902

A = 149° 50'

£ = 3° 2'

C = 27° 10'

10

a = 73

6 = 82

c = 91

4 = 49° 34'

£ = 58° 46'

C = 71° 38'

11

a = 1.9

6 = 3.4

c = 4.9

A = 16° 26'

£ = 30° 24'

C = 133° 10'

12

o= .21

6 = .26

c = .31

A = 42° 6'

£ = 56° 6'

C = 81°48'

13

a = 513.4

6 = 726.8

c = 931.3

A = 33° 16'

£ = 50° 56'

C = 95°48'

14

a=V5

6 = V6

c=V7

X = 51° 52'

£= 59° 32'

C = 68° 34'

Solve the following oblique triangles, using logarithmic Tables I and III.

No.

Given Parts

Required Parts

15

a = 4

6 = 7

c = 6

^. = 34.78°

£ = 86.42°

C= 58.82°

16

a = 43

6 = 50

c = 57

A = 46.82°

£ = 57.98°

C = 75.18°

17

a= .23

6= .26

c = .198

A = 58.44°

£ = 74.38°

C= 47.18°

18

a = 61.3

6 = 84.7

c = 47.6

A = 45.2°

£ = 101.38°

C = 33.44°

19

a = .0291

6= .0184

c = .0358

^.= 54.06°

£ = 30.8°

C = 95.16°

20

a= 705

6 = 562

c = 639

A = 71.56°

£ = 49.14°

C = 59.32°

21

a = 56

6 = 43

c = 49

^. = 74.68°

£ = 47.78°

C = 57.56°

22

a = 301.9

6 = 673.1

c = 422.8

A = 18.2°

£= 135.86°

C = 25.94°

23

a = 2.51

6 = 2.79

c = 2.33

A = 57.88°

£ = 70.3°

C= 51.84°

24

a = 80

6=90

c = 100

^ = 49.46°

£ = 58.76°

G = 71.78°

170 PLANE TRIGONOMETRY

25. The sides of a triangular field are 7 rd., 11 rd., and 9.6 rd. Find the
angle opposite the longest side. Ans. 81° 22'.

26. A pole 13 ft. long is placed 6 ft. from the base of an embankment, and
reaches 8 ft. up its face. Find the slope of the embankment. Ans. 44° 2'.

27. Under what visual angle is an object 7 ft. long seen when the eye of the
observer is 5 ft. from one end of the object and 8 ft. from the other end ?

Ans. 60°.

28. The distances between three cities, A, B, and C, are as follows ;
AB = 166 mi., AC = 72 mi., and BC = 186 mi. B is due east from A. In
what direction is from A ? Ans. N. 4° 24' W. or S. 4° 24' W.

29. Three towns, A, B, and C, are connected by straight roads. AB = 4 mi.,
BC = 5 mi. , A C = 1 mi. Find the angle made by the roads AB and BC.

Ans. 101.55°.

30. The distances of two islands from a buoy are 3 and 4 mi. respectively.
If the islands are 2 mi. apart, find the angle subtended by the islands at the buoy.

Ans. 28.96°.

31. A point P is 13,581 ft. from one end of a wall 12,342 ft. long, and
10,025 ft. from the other end. What angle does the wall subtend at the point P ?

Ans. 60.86°.

83. Use of logarithms in finding the area of an oblique triangle. From
§ 62, p. 117, we have the following three eases.

Case I. When two sides and the included angle are given, use one

of the formulas

/oox aft sinC be sin A ac sin B

(88) S= , S= , S = ,

2 2 2

where S = area of the triangle.

Ex. 1. Given a = 25.6, 5 = 38.2, C = 41° 56' ; find the area of the triangle.

„ , „ a&sinC

Solution. S =

2

logS = logo + logo + logsinC - k>g2.

loga= 1.4082

log 6= 1.5821

log sin C = 9.8249 - 10

12.8152 - 10

log2 = 0.3010

logS= 12.5142-10

= 2.5142.

.-. 8 = 326.8. Ans.

Case II. When the three sides are given, use formula

(89) S = Vs(s-a)(s-&)(s-c),

where S = area of the triangle,

and s = ] s (a + b + c).

THEORY AND USE OF LOGARITHMS 171

Ex. 2. Find the area of a triangle, having given a =12.53, 6 = 24.9, c =18.91.
Solution, a = 12.53 Hence

6 = 24.9

s = 28.17 s = 28.17

s = 28.17

c = 18.91

a = 12.53 6 = 24.9

c = 18.91

2 s = 56.34

s-a = 15.64 s - b = 3.27

s-c= 9.26

s = 28.17.

S = Vs (s - a) (s - 6) (s - c).

log S = £ [log s + log (s - a) + log (s

-6) + log(s-

logs = 1.4498

log(s

- a) = 1.1942

log (J

i-6) = 0.5145

logO

i - c) = 0.9666

2 1 4.1251

logS = 2.0626

.-. 8 = 115.5. Ans.

«)]

Case III. ^irea problems which do not fall directly under Cases I
or II may be solved by Case I if we first find an additional side or
angle by the law of sines.

Ex. 3. Given A = 34° 22', B = 66° 11', c = 78.35 ; find area of triangle.
Solution. This does not now come directly under either Case I or Case II. But
C = 180° - (A + B) = 180° - 100° 33' = 79° 27'.

And, by law of sines,

csin.4
smC
log a = log c + log sin A — log sin C.

logc= 1.8941
log sin 4 = 9.7517 -10
11.6458 - 10
log sin C = 9.9926 - 10

loga= 1.6532

Now it comes under Case I.

ac sin B

o = ■

2
logS = loga + logc + logsinJJ - log 2.

loga= 1.6532
logc= 1.8941
log sin B = 9.9614-10
13.5087 - 10
log 2 = 0.3010
logS = 13.2077 -10
= 3.2077

.-. S = 1613.3. Ans.

172

PLANE TRIGONOMETRY

EXAMPLES

Find the areas of the following oblique triangles, using Tables I and II for
the first ten and Tables I and III for the rest.

No.

Given Paets

Area

i

a = 38

c = 61.2

B = 67° 56'

1078

2

6 = 2.07

.4 = 70°

B = 36° 23'

3.257

3

6 = 116.1

c= 100

A = 118° 16'

5113

4

a = 3. 123

A = 53° 11'

B = 13° 57'

1.354

5

6 = .43,9

A = 76° 38'

C = 40° 35'

.0686

6

a = .3228

c= .9082

B = 60° 16'

.1273

7

c = 80.25

B = 100° 5'

C = 31° 44'

4494

8

a = .010168

6= .018225

C = 11° 18.4'

.000018155

9

a = 18.063

A = 96° 30'

B = 35°

70.55

10

6= 142.8

c = 89.0

a = 95

4174

11

a = 100

B = 60.25°

C = 54.5°

3891

12

a = 145

6 = 178

5 = 41.17°

12,383

13

a = 886

6 = 747

C = 71.9°

314,600

14

a =266

6 = 352

C = 73°

44,770

15

a = 960

6 = 720

= 25.67°

149,730

16

a = 79

6 = 94

c = 67

2604

17

a = 23.1

6 = 19.7

c = 25.2

215.9

18

a = 5.82

6 = 6

c = 4.26

11.733

19. The sides of a field ABCD are AB = 37 rd. , BC = 63 rd. , and DA = 20 rd.,
and the diagonals A C and BD are 75 rd. and 42 rd., respectively. Required the
area of the field. 1570 sq. rd.

20. In a field ABCD the sides AB, BC, CD, and DA are 155 rd., 236 rd.,
252 rd., and 105 rd., respectively, and the length from A to C is 311 rd. Find
the area of the field. • 29,800 sq. rd.

21. The area of a triangle is one acre ; two of its sides are 127 yd. and
150 yd. Find the angle between them. 30° 32'.

22. Given the area of a triangle =12. Find the radius of the inscribed
circle if a = 60 and B = 40° 35.2'.

84. Measurement of land areas. The following examples illustrate
the nature of the measurements made by surveyors in determining
land areas, and the usual method employed for calculating the area
from the data found. The Gunter's chain is 4 rd., or 66 ft., in length.
An acre equals 10 sq. chains, or 160 sq. rd.

EXAMPLES

1. A surveyor starting from a point A runs N. 27° B. 10 chains to B, thence
N.E. by E. 8 chains to C, thence S. 5° W. 24 chains to D, thence N. 40° 44'
W. 13.94 chains to A. Calculate the area of the field ABCD.

Solution. Draw an accurate figure of the field. Through the extreme west-
erly point of the field draw a north-and-south line. From the figure, area

THEORY AND USE OF LOGARITHMS

173

iM

ABCD = area trapezoid* GCDE — (area trapezoid QOBF + area triangle
.F25.4 + area triangle ABE) = 13.9 acres. Ans.

2. A surveyor measures S. 50°25'E. 6.04 chains, thence
S. 58°10'W. 4.15 chains, thence N. 28° 12' W. 5.1 chains,
thence to the starting point. Determine the direction and
distance of the starting point from the last station, and
find area of the field inclosed.

Ans. N. 39° 42' E. 2 chains ; 1.66 acres.

3. One side of a field runs N. 83° 30' W. 10.5 chains,
the second side S. 22° 15' W. 11.67 chains, the third side
N. 71° 45' E. 12.9 chains, the fourth side completes the
circuit of the field. Find the direction and length of the
fourth side, and calculate the area of the field.

Ans. N. 25° 1' E. 6.15 chains ; 8.78 acres.

4. From station No. 1 to station No. 2 is S. 7° 20' W.
4.57 chains, thence to station No. 3 S. 61° 55' W. 7.06
chains, thence to station No. 4 N. 3° 10' E. 5.06 chains,
thence to station No. 5 N. 33° 50' E. 3.25 chains, thence
to station No. 1. Find the direction and distance of
station No. 1 from No. 5, and calculate the area of the
field inclosed. Ans. E. 1° 15' N. 4.7 chains ; 3.55 acres.

85. Parallel sailing. When a vessel sails due east or due west, that
is, always travels on the same parallel of latitude, it is called parallel
sailing. The distance sailed is the departure,^ and it is expressed in

geographical % miles. Thus, in the figure, arc AB is the departure
between A and B. The latitudes of A and B are the same, i.e. arc
EA = angle EOA = arc QB = angle QOB. The differenqe in longitude

» From Geometry the area of a trapezoid equals one half the sum of the parallel sides
times the altitude. Thus, area GCDE = \(GC + ED) GE.

t The departure between two meridians is the arc of a parallel of latitude comprehended
between those meridians. It diminishes as the distance from the equator increases.

X A geographical mile or knot is the length of an arc of one minute on a great circle of
the earth.

174

PLANE TRIGONOMETRY

of A and B = arc EQ. The relation between latitude, departure, and
difference in longitude may be found as follows : By Geometry,

axoAB DA DA A A „ , ,_., ,

= = = cos 0^4 -D =t= cos A OE = cos latitude.

arcEQ OE OA

(90)

arc AB = arcEQ cos latitude, or,
departure

Diff. long. =

cos latitude

EXAMPLES

1. A ship whose position is lat. 25° 20' N., long,
knots. Find the longitude of the place reached.

Solution. Here departure = 140,

and latitude = 25° 20' N.

6° 10' W. sails due west 140

Substituting in above formula (90),
diff. long. =

140

cos 25° 20'
log 140 = 12.1461 -10
log cos 25° 20' = 9.9561 - 10
log diff . long. = 2.1900

diff. long. = 154.9' = 2° 34.9'.
Hence longitude of place reached = 36° 10' + 2° 34.9' = 38° 44.9' W. Ana.

2. A ship in lat. 42° 16' N., long. 72° 16' W., sails due east a distance of
149 geographical miles. "What is the position of the point reached ?

Ans. Long. 68° 55' W.

3. A vessel in lat. 44° 49' S., long. 119° 42' E., sails due west until it reaches
long. 117° 16' E. Eind the departure. Ans. 103.6 knots.

4. A ship in lat. 36° 48' N., long. 56° 15' W., sails due east 226 mi. Find the
longitude of the place reached. Ans. Long. 51° 33' W.

5. A vessel in lat. 48° 54' N., long. 10° 55' W., sails due west until it is in
long. 15° 12' W. Find the number of knots sailed. Ahs. 168.9 knots.

86. Plane sailing. When a ship sails in such a manner as to cross

successive meridians at the same angle,,
it is said to sail on a rhumb line. This ,
angle is called the course, and the distance
between two places is measured on p
rhumb line. Thus, in the figure, if* 5 ^
ship travels from A to B on a rhumb lint.

arc AB = distance,
angle CAB = course,
arc CB = departure,
arc A C = difference in latitude
between A and B.

(.North Pole)

p

gqOM

,aW

THEORY AND USE OF LOGAEITHMS 175

An approximate relation between the quantities involved is ob-
tained by regarding the surface of the earth as a, plane surface, that
is, regarding A CB as a plane right triangle, the angle A CB being the
right angle. This right triangle is called the triangle of plane sailing.

From this plane right triangle we get

CB = ABsinA, and ' -""'"■'

A C = AB cos A ; or,

(91) Departure = distance x sin course, and

(92) Diff. lat. = distance x cos course.

If AB is long, the error caused by neglecting
the curvature of the earth will be too great to
make these results of any value. In that case
AB may be divided into parts, such as AE, EG, GI, IB (figure on
p. 174), which are so small that the curvature of the earth may be
neglected.

EXAMPLES

1. A ship sails from lat. 8° 45' S., on a course N. 36° E. 345 geographical mi.
Find the latitude reached and the departure made.

Solution. Here distance = 345 and course = 36°.
.-. departure = 345 sin 36°. diff. lat. = 345 cos 36°.

log345 = 2.5378 log345 = 2.5378

log sin 36° = 9.7692-10 log cos 36° = 9.9080 -10

log departure = 2. 3070 log diff. lat. = 2.4458

.-. departure = 202.8 mi. Ans. diff. lat. = 279. 1' = 4° 39. V.

As the ship is sailing in a northerly direction she will have reached latitude
go 45 / _ 4 o 39 y _ 4 o 5 9 x s _ Ans

2. A ship sails from lat. 32° 18' N., on a course between N. and W., a dis-
tance of 344 mi., and a departure of 103 mi. Find the course and the latitude
reached. Ans. Course K 17° 25' W. , lat. 37° 46' N.

3. A ship sails from lat. 43° 45' S., on a course N. by E. 2345 mi. Find the
latitude reached and the departure made.

Ans. Lat. 5° 25' S., departure = 457.5 mi.

4. A ship sails on a course between S. and E. 244 mi., leaving lat. 2° 52' S.,
and reaching lat. 6° 8' S. Find the course and the departure.

Ans. Course S. 56° 8' E., departure = 202.6 mi.

87. Middle latitude sailing. Here we take the departure between
two places to be measured on that parallel of latitude which lies
halfway between the parallels of the two places. Thus, in the
figure on p. 174, the departure between A and B is LM, measured
on a parallel of latitude midway between the parallels of A and B-

176 PLANE TRIGONOMETRY

This will be sufficiently accurate for ordinary purposes if the run
is not of great length nor too far away from the equator. The mid-
dle latitude is then the mean of the latitudes of A and B. The
formula (90) on p. 174 will then become

departure

(93) Diff. long. =

cos mid. lat.

EXAMPLES

1. A ship in lat. 42° 30' N., long. 58° 51' W., sails S. 33° 45' E. 300 knots.
Find the latitude and longitude of the position reached.

Solution. We know the latitude of the starting point A. To get the latitude
of the final position B, we first find diff. in lat. from (92). This gives

diff. lat. = 300 cos 33° 45'.
log 300 = 2.4771
log cos 33° 45' = 9.9198 - 10
log diff. lat. = 2.3969

diff. lat. = 249.4' =4° 9.4'.

Since the ship sails in a southerly direction, she will
,, , 7 ... havereaohedlatitude=42 30'-4°9.4'=38°20.6'N. Ans.

To get the longitude of B we must first calculate the
departure and middle latitude for substitution in (93). From (91)

departure = 300 sin 33° 45'.

log 300= 2.4771
log sin 33° 45' = 9.7448 - 10
log departure = 2.2219
departure = 166.7'.
Middle latitude = J (42? 30' + 38° 20.6') = 40° 26.3'.

"1 ftfi 7

Substituting in (93), diff . long. = '■

8 v '' 5 cos 40° 25.3'

log 166.7= 12.2219-10
log cos 40° 25. 3' = 9.8815 - 10
log diff. long. = 2.3404

diff. long. = 219' = 3° 39'.

Since the ship sails in an easterly direction, she will have reached longitude
= 58° 51' - 3° 39' = 65° 12' W. Ana.

2. A vessel in lat. 26°15'N., long. 61°43'W., sails N.W. 253 knots. Find
the latitude and longitude of the position reached.

Ans. Lat. 29° 13.9' N. ; long. 65° 5.1' W.

3. A ship leaves lat. 31° 14' N., long. 42° 19' W., and sails E.N.E. 325 mi. Find
the position reached. Ans. Lat. 33° 18.4' N. ; long. 36° 24' W.

4. Leaving lat. 42° 30' N. , long. 58° 51' W., a battleship sails S. E. by S. 300 mi.
Find the place reached. Ans. Lat. 38° 21' N. ; long. 55° 12' W.

THEORY AND USE OF LOGARITHMS 177

6. A ship sails from a position lat. 49° 56' N., long. 15° 16' W., to another
lat. 47° 18' N., long. 20° 10' W. Find the course and distance.

Ans. Course, S. 50° 53' W. ; distance = 250.5 mi.

Hint. The difference in latitude and the difference in longitude are known, also the
middle latitude.

6. A torpedo boat in lat. 37° N. , long. 32° 16' W. , steams N. 36° 56' W. , and
reaches lat. 41° N. Find the distance steamed and the longitude of the position
reached. Ans. Distance = 300.3 mi.; long. 36°" 8' W.

7. A ship in lat. 42° 30' N., long. 58° 51' W., sails S.E. until her departure is
163 mi. and her latitude 38° 22' N. Eind her course and distance and the longi-
tude of the position reached.

Ans. Course, S. 33° 19' E. ; distance = 296.7 mi. ; long. 55° 17' W.

8. A cruiser in lat. 47°44'K, long. 32°44'W., steams 171 mi. N.E. until her
latitude is 50° 2' N. Find her course and the longitude of the position reached.

Ans. Course, N. 36° 11' E. ; long. 30° 10' W.

9. A vessel in lat. 47° 15' N. , long. 20° 48' W. , sails S. W. 208 mi. , the departure
being 162 mi. Find the course and the latitude and longitude of the position
reached. Ans. Course, S. 51° 9' W. ; lat. 45° 4.5' N. ; long. 24°42'W.

CHAPTEE IX

ACUTE ANGLES NEAR 0° OR 90°

88. When the angle x approaches the limit zero, each of the ratios

, , approaches unity as a limit, x being the circular measure

xx

of the angle.

Proof. Let be the center of a circle whose radius is unity. Let
arc AP = x, and let arc AP' = x in numerical value. Draw PP', and
let PT and P'T be the tangents drawn to the circle at P and P'.
From Geometry

But

PQP' < PAP' < PTP'.

PQP' = PQ -j- QP' = 2 sin x in numerical value,
PAP' = PA + AP' = 2 x in numerical value,
and PTP' = PT+ TP' = 2 tan x in numerical value.

Substituting in (A).

2 sin a; < 2 a; < 2 tan x.

Dividing through by 2, we have

(B) sin* < x < tan x,
which proves that

If x be the circular measure of
an acute angle, it will always lie be-
tween sin x and tan x, being greater
than sin x and less than tan x.

Dividing (B) through by sin x, we get

x 1

K

<

sin x cos x
11 we now let x approach the limit zero, it is seen that

x

limit.
x=0 sin a;

must lie between the constant 1 and " ml *' ,

x =° cos a;

178

which is also 1.

ACUTE ANGLES NEAR 0° OE 90° 179

Hence limit -?— = 1, or,

SIM

(C) limit f^f = 1

Similarly, if we divide (B) through by tan x, we get

x
cos x < - < 1.

tana;

As before, if x approaches zero as a limit,

limit tana:

a:= g;

must lie between the constant 1 and hmit cos x w hich is also 1.

x=0 ' .

Hence limit -^- = 1, or,

x =° tana; '

(D) limit ^1 = 1

The limits (C) and (D) are of great importance both in pure and
applied mathematics. These results may be stated as follows :

When x is the circular measure of a very small angle we may
replace sin x and tan x in our calculations by x.

89. Functions of positive acute angles near 0° and 90°. So far we

have assumed that the differences in the trigonometric functions are
proportional to the differences in the corresponding angles. While
.this is not strictly true, it is in general sufficiently exact for most
practical purposes unless the angles are very near 0° or 90°. In
using logarithms we have also assumed that the differences in the
logarithms of the trigonometric functions are proportional to the
differences in the corresponding angles. This will give sufficiently
accurate results for most purposes if we use Tables II or III in
the tables and confine ourselves to angles between .3° (= 18') and
89.7° (= 89° 42') inclusive. If, however, we have an angle between
0° and .3°(= 18') or one between 89.7° (= 89° 42') and 90°, and are
looking for exact results, it is evident that the ordinary method will
not do. For example, the tabular difference (Table II) between the
logarithmic sine, tangent, or cotangent of 8' and the logarithm of
the corresponding functions of 9' is 512, while between 9' and 10' it
is 457. If we interpolate here in the usual way it is evident that

180 PLANE TRIGONOMETRY

our results will be only approximately correct. In case it is desired
to obtain more accurate results we may use the principle established
in the last section, namely :

We may replace sin x and tan x in our calculations by x when x is
a very small angle and is expressed in circular measure.

Prom a table giving the natural functions of angles, we have

sin 2.2°= 0.03839 = 0.0384,
tan 2.2°= 0.03842 = 0.0384.
Also 2.2° = 0.0384 radians.

Hence it is seen that in any calculation we may replace the sine
or tangent of any angle between 0° and 2.2° by the circular measure
of the angle without changing the first four significant figures of
the result. Also since

cos 87.8°= sin (90°- 87.8°) = sin 2.2°= 0.0384,
cot 87.8° = tan (90° - 87.8°) = tan 2.2° = 0.0384,
and 2.2° = 90° - 87.8° = 0.0384 radians,

we may replace the cosine or cotangent of any angle between 87.8°
and 90° by the circular measure of the complement of that angle.
We may then state the following rules :

90. Rule for finding the functions of acute angles near 0°.

sin x = circular measure of x,*
tanx = circular measure of x,

cot x = — ; — !

circular measure oj x

cos x is found from the tables in the usual way.^

* The following equivalents may be used for reducing an angle to circular measure
(radians), and in other computations.

1° = Ton radians.

1° = 0.0174533 radians, log 0.0174533= 8.2419 - 10.

1' = 0.0002909 radians, log0.0002909= 6.4637 - 10.

1" = 0.0000048 radians, log0.0000048 = 4.6856- 10.
180°

=^-=57.29578° = 1 radian, log57.29578= 1.7581.

T =3.14159 log tt= 0.4971.

= %p approximately.

t esc a: and sec a: are simply the reciprocals of sinx and cos a; respectively.

ACUTE ANGLES NEAR. 0° OB, 90° 181

91. Rule for finding the functions of acute angles near 90°.

cos x = circular measure of the complement of x,*
cot x = circular measure of the complement of x,

1

circular measure of the complement of x

sin x is found from the tables in the usual way.f

Since any function of an angle of any magnitude whatever, posi-
tive or negative, equals some function of a positive acute angle, it is
evident that the above rules, together with those on p. 57, will suffice
for finding the functions of angles near ±90°, ±180°, ±270°, ±360°.

Ex. 1. Find sine, tangent, and cotangent of 42'.
Solution. Reducing the angle to radians,

42' = 42 x 0.0002909 radians = 0.01222 radians.

Therefore sin 42' = 0.01222,

tan 42' = 0.01222,

cot 42' = — - — = 81.833. Ans.
0.01222

Ex. 2. Find cosine, cotangent, and tangent of 89° 34.6'.
Solution. The complement of our angle is 90° - 89° 34.6' = 25.4'.
Reducing this remainder to radians,

25.4' = 25.4 x 0.0002909 radians = 0.00739 radians.

Therefore cos 89° 34. 6' = 0. 00739,

cot 89° 34. 6' =0.00739,

tan 89° 34. 6' = — - — = 135. 32. Ans.
0.00739

When the function of a positive acute angle near 0° or 90° is
given, to find the angle itself we reverse the process illustrated
above. For instance:

Ex. 3. Eind the angle subtended by a man 6 ft. tall at a distance of 1225 ft.

Solution. _ lI ■ | 6

From the figure tan x = T -$$-g. " 122 6

But, since the angle is very small, we may replace tan x by x, giving

x — j^-g radians = 0.0049 radians.
Or, reducing the angle to minutes of arc, we get

0.0049 „ -la a, a —

z — minutes of arc = 16.8 . Ans.

0.0002909

* If the angle is given in degrees, subtract it from 90° and reduce the remainder to cir.
cular measure (radians). If the angle is given in circular measure (radians), simply subtract

it from - (= 1.57079).
2
t esc x and sec a; are simply the reciprocals of sin x and cos x respectively.

182 PLANE TRIGONOMETRY

92. Rules for finding the logarithms of the functions of angles near 0°
and 90°.* For use in logarithmic computations trie rules of the last
two sections may be put in the following form :

If the angle is given in degrees, minutes, and seconds, it should first
be reduced to degrees and the decimal part of a degree (see Conver-
sion Table on p. 17 of Tables').

Rule I. To find the logarithms of the functions of angles near 0°.

log sin x° = 2.2419 + log re.f

log tan x" = 2.2419 + log x.

log cot x° = 1.7581 — log x.%

log cos x" is found from the tables in the usual way.

Rule II. To find the logarithms of the functions of an angle near 90°.

log cos x° = 2.2419 + log (90 — x).
log cot x" = 2.2419 + log (90 — x).
log tanx" = 1.7581 — log (90 — x).
log sin x° is found from the tables in the usual way.

Ex. 1. Find log tan 0.045°.

Solution. As is indicated in our logarithmic tables, ordinary interpolation
will not give accurate results in this case. But from the above rule,

log tan 0.045° = 2.2419 + log 0.045
= 2.2419 + 2.6532.
.-. log tan 0.045° =4.8951. Ans.

On consulting a much larger table of logarithms, this result is found to be
exact to four decimal places. Interpolating in the ordinary way, we get

log tan 0.045° = 4.8924,

which is correct to only two decimal places.

* These rules will give results accurate to four decimal places for all angles between 0°
and 1.1° and between 88.9° and 90°.

t Since 1 degree = 0.017453 radians, the circular measure of

x degrees = 0.017453 • x radians.
Hence, from p. 180, sin x° = 0.017453 ■ x,

and log sin x° = log 0.017453 + log x

= 2.2419+ log x.

% From p. 180, cot x° = ,

C 017453 • x
and log cot x" = - log 0.017453 — log x

= 1.7581 -log a;.

ACUTE ANGLES NEAR 0° OR 90° 183

Ex. 2. Find log tan 89.935°.
Solution. From the above rule,

log tan 89.935° = 1.7581 - log (90 - 89.935)
= 1.7581 -log 0.065
= 1.7581-2.8129.

.-. log tan 89.935° =2.9452. Ans.

If the tangent itself is desired, we look up the number in Table I correspond-
ing to this logarithm. This gives

tan 89.935° =881.4.

93. Consistent measurements and calculations. In the examples
given so far in this book it has generally been assumed that the
given data were exact. That is, if two sides and the included angle
of a triangle were given, as 135 ft., 217 ft., and 25.3° respectively, we
have taken for granted that these numbers were not subject to errors
made in measurement. This is in accordance with the plan followed
in the problems that the student has solved in Arithmetic, Algebra,
and Geometry. It should not be forgotten, however, that when we
apply the principles of Trigonometry to the solution of practical prob-
lems, — engineering problems, for instance, — it is usually necessary
to use data which have been found by actual measurement, and there-
fore are subject to error. In taking these measurements one should
carefully see that they are made with about the same degree of
accuracy. Thus, it would evidently be folly to measure one side of
a triangle with much greater care than another, for, in combining
these measurements in a calculation, the result would at best be no
more accurate than the worst measurement. Similarly, the angles of
a triangle should be measured with the same care as the sides.

The number of significant figures in a measurement is supposed
to indicate the care that was intended when the measurement was
made, and any two measurements showing the same number of sig-
nificant figures will, in general, show about the same relative care in
measurement. If the sides of a rectangle are about 936 ft. and 8 ft.,
the short side should be measured to at least two decimal places. A
neglected 4 in the tenths place will alter the area by 374 sq. ft.

The following directions will help us to make consistent measure-
ments and avoid unnecessary work in our calculations.

1. Let all measured lines and calculated lines show the same num-
ber of significant figures, as a rule.

2. When the lines show only one significant figure, let the angles
read to the nearest 5°.

184 PLANE TRIGONOMETRY

3. When the lines show two significant figures, let the angles read
to the nearest half degree.

4. When the lines show three significant figures, let the angles
read to the nearest 5'.

5. When the lines show four significant figures, let the angles
read to the nearest minute.

EXAMPLES

1. The inclination of a railway to the horizontal is 40'. How many feet does
it rise in a mile 1 Arts. 61.43.

2. Given that the moon's distance from the earth is 238,885 mi. and subtends
an angle of 31' 8" at the earth. Find the diameter of the moon in miles.

Ana. 2163.5.

3. Given that the sun's distance from the earth is 92,000,000 mi. and subtends
an angle of 32' 4" at the earth. Find the sun's diameter. Ans. 858,200 mi.

4. Given that the earth's radius is 3963 mi. and subtends an angle of 57' 2"
at the moon. Find the distance of the moon from the earth. Ans. 238,833 mi.

5. Given that the radius of the earth is 3963 mi. and subtends an angle of 9"
at the sun. Find the distance of the sun from the earth. Ans. 90,840,000 mi.

6. Assuming that the sun subtends an angle of 32' 4" at the earth, how far
from the eye must a dime be held so as to just hide the sun, the diameter of a
dime being | in. 1 Ans. 76.6 in.

7. Find the angle subtended by a circular target 5 ft. in diameter at the
distance of half a mile. Ans. 6' 30.6".

MISCELLANEOUS EXAMPLES

1. A balloon is at a height of 2500 ft. above a plain and its angle of elevation
at a point in the plain is 40° 35'. How far is this point from the balloon ?

Ans. 3843 ft.

2. A tower standing on a horizontal plain subtends an angle of 37° 19.5' at
a point in the plain distant 369.5 ft. from the foot of the tower. Find the height
of the tower. Ans. 281.8 ft.

3. The shadow of a steeple on a horizontal plain is observed to be 176.23 ft.
when the elevation of the sun is 33.2°. Find the height of the tower.

Ans. 115.3 ft.

4. From the top of a lighthouse 112.5 ft. high, the angles of depression of
two ships, when the line joining the ships passes through the foot of the light-
house, are 27.3° and 20.6° respectively. Find the distance between the ships.

Ans. 81 ft.

5. From the top of a cliff the angles of depression of the top and bottom of a
lighthouse 97.25 ft. high are observed to be 23° 17' and 24° 19' respectively.
How much higher is the cliff than the lighthouse ? Ans. 1947 ft.

6. The angle of elevation of a balloon from a station due south of it is
47° 18.5', and from another station due west of the former and 671.4 ft. from it
the elevation is 41° 14'. Find the height of the balloon. Ans. 1000 ft.

MISCELLANEOUS EXAMPLES 185

7. A ladder placed at an angle of 75° with the street just reaches the sill
of a window 27 ft. above the ground on one side of the street. On turning the
ladder over without moving its foot, it is found that when it rests against a wall
on the other side of the street it is at an angle of 15° with the street. Find
the breadth of the street. Ans. 3424 ft.

8. A man traveling due west along a straight road observes that when he
is due south of a certain windmill the straight line drawn to a distant church
tower makes an angle of 30° with the direction of the road. A mile farther on
the bearings of the windmill and church tower are N.E. and N.W. respectively.
Find the distances of the tower from the windmill and from the nearest point
on the road. Ans. 2.39 mi., 1.37 mi.

9. Standing at a certain point, I observe the elevation of a house to be 45°,
and the sill of one of its windows, known to be 20 ft. above the ground, sub-
tends an angle of 20° at the same point. Find the height of the house.

Ans. 54.94 ft.

10. A hill is inclined 36° to the horizon. An observer walks 100 yd. away
from the foot of the hill, and then finds that the elevation of a point halfway
up the hill is 18°. Find the height of the hill. Ans. 117.58 yd.

11. Two straight roads, inclined to one another at an angle of 60°, lead from
a town A to two villages B and C ; B on one road distant 30 mi. from A, and
C on the other road distant 15 mi. from A. Find the distance from B to C.

Ans. 25.98 mi.

12. Two ships leave harbor together, one sailing N.E. at the rate of 7^ mi.
an hour and the other sailing north at the rate of 10 mi. an hour. Prove that
the distance between the ships after an hour and a half is 10.6 mi.

13. A and B are two positions on opposite sides of a mountain; C is a point
visible from A and B. From A to C and from B to C are 10 mi. and 8 mi.
respectively, and the angle BCA is 60°. Prove that the distance between A and
Bis 9. 165 mi.

14. A and B are two consecutive milestones on a straight road and C is
a distant spire. The angles ABC and BAG are observed to be 120° and 45°
respectively. Show that the distance of the spire from A is 3.346 mi.

15. If the spire C in the last example stands on a hill, and its angle of ele-
vation at A is 15°, show that it is .866 mi. higher than A.

16. If in Example 14 there is another spire D such that the angles DBA
and DAB are 45° and 90° respectively and the angle DAC is 45°, prove that
the distance from C to D is very nearly 2f mi.

17. A and B are consecutive milestones on a straight road ; C is the top of a
distant mountain. At A the angle CAB is observed to be 38° 19'; at B the
angle CBA is observed to be 132° 42', and the angle of elevation of C at B is
10° 15'. Show that the top of the mountain is 1243.7 yd. higher than B.

18. A base line AB, 1000 ft. long, is measured along the straight bank of a
river; C is an object on the opposite bank; the angles BAC and CBA are
observed to be 65° 37' and 53° 4' respectively. Prove that the perpendicular
breadth of the river at C is 829.8 ft.

186 PLANE TRIGONOMETRY

19. The altitude of a certain rook is observed to be 47°, and after walking
1000 ft. towards the rook, up a slope inclined at an angle of 82° to the horizon,
the observer finds that the altitude is 77°. Prove that the vertical height of the
rock above the first point of observation is 1084 ft.

20. A privateer 10 mi. S.W. of a harbor sees a ship sail from it in a direc-
tion S. 80° E., at a rate of 9 mi. an hour. In what direction and at what
rate must the privateer sail in order to come up with the ship in 1^ hr. ?

Ans. N. 76° 66' E. 13.9 mi. per hour.

21. At the top of a chimney 150 ft. high, standing at one corner of a tri-
angular yard, the angle subtended by the adjacent sides of the yard are 80° and
46° respectively, while that subtended by the opposite side is 30°. Show that
the lengths of the sides are 150 ft., 86.6 ft., and 106.8 ft. respectively.

22. A person goes 70 yd. up a slope of 1 in 3J- from the edge of a river, and
observes the angle of depression of an object on the opposite bank to be 2J°.
Find the breadth of the river. Ans. 422.18 yd.

23. A flagstaff h ft. high stands on the top of a tower. From a point in
the plain on which the tower stands the angles of elevation of the top and bot-
tom of the flagstaff are observed to be a and /3 respectively. Prove that the

,.,..,, . fttanfl .. . ftsinfl-cosar.

height of the tower is - — ft., i.e. — : — - ft.

tan a — tan /3 sin (a — /3)

24. The length of a lake subtends at a certain point an angle of 46° 24', and
the distances from this point to the two extremities of the lake are 840 and
290 ft. Find the length of the lake. • Ans. 255.8 ft.

25. From the top of a cliff h ft. high the angles of depression of two ships at
sea in a line with the foot of the cliff are a and /3 respectively. Show that the
distance between the ships is h (cot/3 — cot or) ft.

26. Two ships are a mile apart. The angular distance of the first ship from
a fort on shore, as observed from the second ship, is 35° 14' 10" ; the angular
distance of the second ship from the fort, observed from the first ship, is 42° 11'
53" Find the distance in feet from each ship to the fort.

Ans. 3121 ft., 3684 ft.

27. The angular elevation of a tower at a place due south of it is a, and at

another place due west of the first and distant d from it, the elevation is /3.

Prove that the height of the tower is

d . dsina- sin/3

i.e. —

Vcot 2 |3- cot 2 a Vsin (a — /3) • sin (a + /3) ,

28. To find the distance of an inaccessible point C from either of two points
A and B, having no instruments to measure angles. Prolong. CA to o, and OB
to b, and join AB, Ab, and Ba. Measure AB, 500; aA, 100; aB, 560; bB,
100; and Ab, 550. Ans. 500 and 636.

29. A man stands on the top of the wall of height h and observes the angular
elevation a of the top of a telegraph post ; he then descends from the wall
and finds that the angular elevation is now /3 ; prove that the height of the post

exceeds the height of the man by h —

sin ((3 - a)

MISCELLANEOUS EXAMPLES 187

30. Two inaccessible points A and B are visible from D, but no other point
can be found whence both are visible. Take some point C, whence A and B
can be seen, and measure CD, 200 ft.; ADC, 89°; ACD, 50° 30'. Then take
some point E, whence D and B are visible, and measure DE, 200 ; BDE, 54° 30' ;
BED, 88° 30'. At D measure ADB, 72° 30'. Compute the distance AB.

Ans. 345.4 ft.

31. The angle of elevation of an inaccessible tower situated on a horizontal
plane is 63° 26' ; at a point 500 ft. farther from the base of the tower the ele-
vation of its top is 32° 14'. Find the height of the tower. Ans. 460.5 ft.

32. To compute the horizontal distance between two inaccessible points A
and B, when no point can be found whence both can be seen. Take two points
C and D, distant 200 yd., so that A can be seen from C, and B from D. From
C measure CF, 200 yd. to F, whence A can be seen ; and from D measure DE,
200 yd. to E, whence B can be seen. Measure AFC, 83°; ACD, 53° 30' ; ACF,
54° 31'; BDE, 54° 30'; BDC, 156° 25'; DEB, 88° 30'. Ans. 345.3 yd.

33. A tower is situated on the bank of a river. From the opposite bank the
angle of elevation of the tower is 60° 13', and from a point 40 ft. more distant
the elevation is 50° 19'. Find the breadth of the Tiver. Ans. 88.9 ft.

34. A ship sailing north sees two lighthouses 8 mi. apart, in a line due west ;
after an hour's sailing one lighthouse bears S.W. and the other S.S.W. Find
the ship's rate. Ans. 13.6 mi. per hour.

35. A column in the north temperate zone is east-southeast of an observer,
and at noon the extremity of its shadow is northeast of him. The shadow is
80 ft. in length, and the elevation of the column at the observer's station is 45°.
Find the height of the column. Ans. 61.23 ft.

36. At a distance of 40 ft. from the foot of a tower on an inclined plane the
tower subtends an angle of 41° 19' ; at a point 60 ft. farther away the angle sub-
tended by the tower is 23° 45'. Find the height of the tower. Ans. 56.5 ft.

37. A tower makes an angle of 113° 12' with the inclined plane on which it
stands ; and at a distance of 89 ft. from its base, measured down the plane, the
angle subtended by the tower is 23° 27'. Find the height of the tower.

Ans. 51.6 ft.

38. From the top of a hill the angles of depression of two objects situated in
the horizontal plane of the base of the hill are 45° and 30° ; and the horizontal
angle between the two objects is 30°. Show that the height of the hill is equal
to the distance between the objects.

39. I observe the angular elevation of the summits of two spires which appear
in a straight line to be a, and the angular depressions of their reflections in still
water to be /3 and y. If the height of my eye above the level of the water be c,
then the horizontal distance between the spires is

2 c cos'q sin (ft — 7)
sin (j3 - or) sin (7 - a)

40. The angular elevation of a tower due south at a place A is 30°, and at a
place B, due west of A and at a distance a from it, the elevation is 18°. Show

a
that the height of the tower is — - •

V2V5+2

.188 PLANE TRIGONOMETRY

41. A boy standing c ft. behind and opposite the middle of a football goal
sees that the angle of elevation of the nearer crossbar is A and the angle
of elevation of the farther one is B. Show that the length of the field is
c(tan.4 cot-B — 1).

42. A valley is crossed by a horizontal bridge whose length is I. The sides
of the valley make angles A and B with the horizon. Show that the height of

the bridge above the bottom of the valley is

cot^l + cotB

43. A tower is situated on a horizontal plane at a distance a from the base

of a hill whose inclination is a. A person on the hill, looking over the tower,

can just see a pond, the distance of which from the tower is 6. Show that, if

the distance of the observer from the foot of the hill be c, the height of the

be sin a

tower is

a + 6 + c cos or

44. From a point on a hillside of constant inclination the angle of inclination
of the top of an obelisk on its summit is observed to be a, and a ft. nearer to
the top of the hill to be /3 ; show that if h be the height of the obelisk, the incli-
nation of the hill to the horizon will be

cos-

x fa sin a: sin /3 ">
\h sin(/3-a)J

CHAPTEE X

RECAPITULATION OF FORMULAS

Plane Trigonometry
Right triangles, pp. 2-1 1.

(1) sin A =

(2) cos A =

(3) tan A = ^

(7) Side opposite an acute angle

= hypotenuse x sine of the angle.

(8) Side adjacent an acute angle

= hypotenuse x cosine of the angle.

(9) Side opposite an acute angle

= adjacent side x tangent of the angle.

Fundamental relations between the functions, p. 59.

(19) sin x =

(20) cos a; =

(21) tana; =

(22) tanx =

cos X

(23) sin 2 x + cos 2 x = 1.

(24) sec 2 a; = 1 + tan 2 a;.

csc a;

1
sec a;

1
cot X
sin x

csc x =

sec x =

cot x =

SIM
1

cos a;

1
tana;

cos x

cot x = —

sin x

(25) csc 2 x = 1 + cot 2 x.

Functions of the sum and of the difference of two angles, pp. 63-69.

(40) sin (x + y) = sin x cos y + cos x sin y.

(41) sin (x — y) = sin x cos y — cos x sin y.

(42) cos (x + y) = cos x cos y — sin x sin y.

(43) cos (a; — y) = cos a; cos y + sin x siny.

180

-y

(44)

tan (a; + y) =

(45)

tan (x — y) =

(46)

cot (x + y) =

(47)

cot (x — y) =

190 PLANE TRIGONOMETRY

tan x + tan y
1 — tan x tan y

tan x — tan y
1 + tan x tan y
cot * cot ?/ — 1

cot y + cot x
cot x cot y + 1

cot y — cot a;

Functions of twice an angle, p. 70.

(48) sin 2 x = 2 sin x cos a;.

(49) cos 2 a; = cos 2 aj — sin 2 x.

Functions of an angle in terms of functions of half the angle, p. 72.

X X

(51) sin x = 2 sin - cos - ■

Zl -J

(52)

cos a;

= cos 2 --

■ 2 •"

(53)

tan x

2tan-

Zi

1 - tan 2

a;

2

Fund

ions 01

. X

sin-

X

cos ^

i half an

angle, pp.

72-73-

(58)

(59)

(54)

— cos a;
2

tan- =

(55)

. Il + coscc

- ± N 2 •

cot ^ =

1 — COS :

sin a;

■¥-

,, m . x ) l — cos a: x 1+cosa;

(56) tan- = ±% Tw ^ (60) oot g =— j^-.

(57) tan2 = -5»«_. (61) cot^^^.
v y 2 1 +.cos x K ' 2 1 — cos x

Sums and differences of functions, p. 74.

(62) sin A + siaB = 2 sin $(A + B) cos £ (A — B).

(63) sin A—sinB = 2 cos £ (4 + B) sin £ (^ - B).

(64) cos4 +cos.B = 2cos£(J. + .B)cos£(,4 — B).

(65) cos;! — cos£ = — 2 sin £(4 + £)sin J (4 — 5).
.„„. sin^l +sin.B _ tan'^-(^ + B)

^ ' smA — sinB ~~ tan ^(4 — B) '

cos a;

cos x

RECAPITULATION OF FORMULAS 191

Law of sines, p. 102.

a

(72)

sin A sinB sin C

Law of cosines, p. 109.

(73) a 2 = i 2 +c 2 - 2 be cos A.

Law of tangents, p. 112.

*• ' a — b tan^(A-B)

Functions of the half angles of a triangle ia terms of the sides,
pp. 113-115.

s = I (a + b + e).

(81)
(82)
(83)
(84)

sin \ A -

-x ( *

-m-

0)

-\

be

cos \ A -

->I4

! — a)
be

tsm^A =

->l<*

-b)(s-
s(s — a)

«)

r=\l

(s — a)(s — b) (s — c)

s

(85) tan£ A =

s — a

r

(86) taniiJ = — -

r

(87) tan£C = s _

Area of a triangle, p. 117.

(88) S = ^bc sin A.

(89) S = Vs (« — a) (s — b) (s — c),

SPHERICAL TRIGONOMETRY

CHAPTEE I

RIGHT SPHERICAL TRIANGLES

1. Correspondence between the face angles and the diedral angles of a
triedral angle on the one hand, and the sides and angles of a spherical
triangle on the other. Take any triedral angle O—A'B'C and let a
sphere of any radius, as OA, be described about the vertex as
a center. The intersections of this sphere with the faces of the

triedral angle will be three arcs of great circles of the sphere, form-
ing a spherical triangle, as ABC. The sides (arcs) AB, BC, CA of
this triangle measure the face angles A' OB', B'OC", CO A' of the
triedral angle. The angles ABC, BCA, CAB, are measured by the
plane angles which also measure the diedral angles of the triedral
angle; for, by Geometry, each is measured by the angle between
two straight lines drawn, one in each face, perpendicular to the
edge at the same point.

Spherical Trigonometry treats of the trigonometric relations be-
tween the six elements (three sides and three angles) of a spherical
triangle ; or, what amounts to the same thing, between the face and
diedral angles of the triedral angle which intercepts it, as shown in
the figure. Hence we have the

Theorem. From any property of triedral angles an analogous prop-
erty of spherical triangles can be inferred, and vice versa.

193

194

SPHERICAL TRIGONOMETRY

It is evident that the face and diedral angles of the triedral angle
are not altered in magnitude by varying the radius of the sphere;
hence the relations between the sides and angles of a spherical tri-
angle are independent of the length of the radius.

The sides of a spherical triangle, being arcs, are usually expressed
in degrees.* The length of a side (arc) may be found in terms of
any linear unit from the proportion

circumference of great circle : length of arc : : 360° : degrees in arc.

A side or an angle of a spherical triangle may have any value

from 0° to 360°, but any spherical triangle can always be made to

depend on a spherical triangle having
each element less than 180°.

Thus, a triangle such as ADEBC
(unshaded portion of hemisphere in
figure), which has a side ADEB greater
than 180°, need not be considered, for
its parts can be immediately found
from the parts of the triangle ABC,
each of whose sides is less than 180°.
For arc ADEB = 360° — arc AB, angle
CAD = 180° - angle CAB, etc. Only

triangles whose elements are less than 180° are considered in this book.
2. Properties of spherical triangles. The proofs of the following

properties of spherical triangles may be found in any treatise on

Spherical Geometry:

(a) Either side of a spherical triangle is less than the sum of the
other two sides.

(b) If two sides of a spherical triangle
are unequal, the angles opposite them are
unequal, and the greater angle lies opposite
the greater side, and conversely.

(c) The sum of the sides of a spherical
triangle is less than 360°.f

(d) The sum of the angles of a spher-
ical triangle is greater than 180° and less
than 540°. ±

* One of the chief differences between Plane Trigonometry and Spherical Trigonometry
is that in the former the sides of triangles are expressed in linear units, while in the latter
alt the parts are usually expressed in units of arc, i.e. degrees, etc.

t In a plane triangle the sum of the sides may have any magnitude.

t In a plane triangle the sum of the angles is always equal to 180°,

EIGHT SPHERICAL TRIANGLES 195

(e) If A'B'C is the polar triangle* of ABC, then, conversely,
ABC is the polar triangle of A'B'C

(/) In two polar triangles each angle of one is the supplement
of the side lying opposite to it in the other. Applying this to the
last figure, we get

A = 180° - a',

B = 180° - V,

C = 180° - c',

A' = 180° -a,

B' = 180°-b,

C = 180° - c.

A spherical triangle which has one or more right angles is called
a right spherical triangle.

EXAMPLES

1. Find the sides of the polar triangles of the spherical triangles whose angles
are as follows. Draw the figure in each case.

(a) A = 70°, B = 80°, C = 100°. Arts, a' = 110°, V = 100°, c' = 80°.

(b) A = 56°, B = 97°, C = 112°.

(c) A = 68° 14', B = 52° 10', C = 98° 44'.

(d) A = 115.6°, B = 89.9°, C = 74.2°.

2. Find the angles of the polar triangles of the spherical triangles whose sides
are as follows:

(a) a = 94°, 6 = 52°, c = 100°. Ans. A' = 86°, B' = 128°, C = 80°.

(b) a = 74° 42', 6 = 95° 6', c = 66° 25'.

(c) a = 106.4°, 6 = 64.3°, c = 51.7°.

3. If a triangle has three right angles, show that the sides of the triangle are
quadrants.

4. Show that if a triangle has two right angles, the sides opposite these angles
are quadrants, and the third angle is measured by the opposite side.

5. Find the lengths of the sides of the triangles in Example 2 if the radius of
the sphere is 4 ft.

3. Formulas relating to right spherical triangles. Erom the above
Examples 3 and 4, it is evident that the only kind of right spherical
triangle that requires further investigation is that which contains
only one right angle.

In the figure shown on the next page let ABC be a right spherical
triangle having only one right angle, the center of the sphere being
at O. Let C be the right angle, and suppose first that each of the
other efemenAl is less than 90°, the radius of the sphere being unity. .

le |rt

*Th&poZ«r Irwhgle of any spherical triangle is constructed by describing arcs of great
circles aboi^tfthe jjrtices of the original triangle as poles.

196

SPHERICAL TRIGONOMETRY

Pass an auxiliary plane through. B perpendicular to OA, cutting
0.4 at E and OC at £>. Draw BE, BD, and BE. BE and DE are
each perpendicular to OA ;

[If a straight line is J_ to a plane, it is _L to every line in trie plane.] N

therefore angle BED = angle A. The plane BDE is perpendicular to

the plane AOC

' [" If a straight line is J_ to a plane, every plane"!
Lpassed through the line is X to the first plane. J

hence BD, which is the intersection of the planes BDE and BOC, is
perpendicular to the plane AOC,

["If two intersecting planes are each X to a third"] .
Lplane, their intersection is also X to that plane.J

and therefore perpendicular to OC and DE. (&V t

In triangle EOD, remembering that angle EOD = b, we have
07?

OD

= cos b,

,Si

<: !

or, clearing of fractions,

{A) OE = OD ■ cos b.

But OE = cos o (= cos EOB),

and OD = cos a (== cos DOB).

Substituting in (,1), we get

(1) cos c — cos a cos b.

In triangle BED, remembering that angle BE D = angle A, we have
BD .

= SHI yl ,

BE '

or, clearing of fractions,

(23) BD =BE- sin A.

But BD = sin a (= sin DOB),

and BE = sin c (= sin EOB).

RIGHT SPHERICAL TRIANGLES 197

Substituting in (B), we get

(2) sin a = sin c sin A.

Similarly, if we had passed the auxiliary plane through A perpen-
dicular to OB,

(3) sin b = sin c sin B.
Again, in the triangle BED,

(O cos ^=ff-

DE

But DE = OD sin b, from sin b =

OD
OD = cos a (= cos DOB),

and .B.E = sin c (= sin £0.8).

Substituting in (C),

,_,. ODsinS sinJ

(D) cos .4 = — ; = cos a ■ —.

sin c sin c

But from (3), — — = sin B. Therefore

v '' sine

(4) cos A — cos a sin B.

Similarly, if we had passed the auxiliary plane through A perpen-
dicular to OB,

(5) cos B — cos 6 sin A.

The above five formulas are fundamental ; that is, from them we
may derive all other relations expressing any one part of a right
spherical triangle in terms of two others. For example, to find a
relation between A, b, c, proceed thus :

Erom (4), cos A = cos a sin B

_ cos c sin b
cos b sin e

[Since cos a = ^^ from (1), and sin B= — — from (3).
cos b sm c J

sin b cos c

cos b sin c

(6)

.*. cos A

= tan b cot c.

Similarly, we may get

(7)

COSB:

= tan a cot c.

(8)

sin b ■.

= tan a cot A.

(9)

sin a :

= tan 6 cot B.

(10)

COSC :

= cot A cot B.

198

SPHERICAL TRIGONOMETRY

These ten formulas are sufficient for the solution of right spher-
ical triangles. In deriving these formulas we assumed all the
elements except the right angle to be less than 90°. But the formu-
las hold when this assumption is not made. ' For instance, let us
suppose that a is greater that 90°. In this case the auxiliary plane
BDE will cut CO and AO produced beyond the center 0, and we
have, in triangle EOD,

OE
(E) cos DOE (= cos b) =

But

and

OD

OE = cos EOB = — cos A OB = — cos c,
OD = cos DOB = — cos COB — — cos a.

Substituting in (E), we get

cos o ,

cos b = > or cos o = cos a cos o,

cos a

which is the same 8T (1).

Likewise, the other formulas will hold true in this case. Similarly,
they may be shown to hold true in all cases.

If the two sides including the right angle are either both less or

o 6

both greater than 90° (that is, cos a and cos b are either both positive
or both negative), then the product

(F)

cos a cos b

will always be positive, and therefore cose, from (1), will always
be positive, that is, c will always be less than 90°. If, however, one
of the sides including the right angle is less and the other is greater
than 90°, the product (F), and therefore also cos c, will be negative,
and c will be greater than 90°.

Hence we have

Theorem I. If the two sides including the right angle of a right
spherical triangle are both less or both greater than 90°, the hypotenuse

EIGHT SPHEEIGAL TRIANGLES 199

is less than 90° ; if one side is less and the other is greater than 90°,
the hypotenuse is greater than 90°.

-T7. ,,, t ,„, • „ cos A ., . , cos-B

From (4) and (5), sinB = > and smi = r- -

x ' ' cos a eoso

Since A and B are less than 180°, sin A and sinB mnst always be
positive. But then cos A and cos a must have the same sign, that is,
A and a are either both less than 90° or both greater than 90°. Simi-
larly, for B and b. Hence we have

Theorem II. In a right spherical triangle an oblique angle and the
side opposite are either both less or both greater than 90°.

4. Napier's rules of circular parts. The ten formulas derived in the
last section express the relations between the three sides and the two
oblique angles of a right spherical triangle. All these relations may
be shown to follow from two very useful rules discovered by Baron
Napier, the inventor of logarithms.

For this purpose the right angle (not entering the formulas) is
not taken into account, and we replace the hypotenuse and the two
B B,

b b

oblique angles by their respective complements ; so that the five
parts, called the circular parts, used in Napier's rules are a, b, A c ,
c c , B c . The subscript c indicates that the complement is to be
used. The first figure illustrates the ordinary method of represent-
ing a right spherical triangle. To emphasize the circular parts
employed in Napier's rules, the same triangle might be represented
as shown in the second figure. It is not necessary, however, to draw
the triangle at all when using Napier's rules ; in
fact, it is found to be more convenient to simply
write down the five parts in their proper order as on j^ s e

the circumference of a circle, as shown in the third
figure (hence the name circular parts). ,

Any one of these parts may be called a middle
part; then the two parts immediately adjacent to it are called adja-
cent parts, and the other two opposite parts. Thus, if a is taken as
a middle part, B c and b are the adjacent parts, while c c and A c are
the opposite parts.

200

SPHERICAL TRIGONOMETRY

Napier's rules of circular parts.

Rule I. The sine of any middle part is equal to the product of the
tangents of the adjacent parts.

Rule II. The sine of any middle part is equal to the product of the
cosines of the opposite parts.

These rules are easily remembered if we associate the first one
with the expression " tan-adj." and the second one with " cos-opp." *

Napier's rules may be easily verified by applying them in turn to
each one of the five circular parts taken as a middle part, and com-
paring the results with (1) to (10).

For example, let c c be taken as a middle part ; then A c and B c are
the adjacent parts, while a and b are the opposite parts.

Then, by Rule I, gin ^ = Un ^ tan ^

c ° or, cos c = cot A cot B ;

A c b c which agrees with (10), p. 197.

By Rule II, sin c,. = cos a cos b,

I, a or, cos c = cos a cos b ;

which agrees with (1), p. 196.

The student should verify Napier's rules in this manner by taking
each one of the other four circular parts as the middle part.

Writers on Trigonometry differ as to the practical value of Napier's
rules, but it is generally conceded that they are a great aid to the
memory in applying formulas (1) to (10) to the solution of right
spherical triangles, and we shall so employ them.

5. Solution of right spherical triangles. To solve a right spherical
triangle, two elements (parts) must be given in addition to the right
angle. For the sake of uniformity we shall continue to denote the
right angle in a spherical triangle ABC by the letter C.

General directions for solving right spherical triangles.

£

b

-y-

First step. Write down the five circular parts as in first figure.

Second step. Underline the two given parts and the required un-
known part. Thus, if A c and a are given to find b, ive underline all
three as is shown in the second figure.

* Or by noting that a is the first vowel in the words " tangent'* and " adjacent," while o
is the first vowel in the words " cosine " and " opposite."

EIGHT SPHERICAL TRIANGLES

201

Third step. Pick out the middle part (in this case b) and cross the
line under it as indicated in the third figure.

Fourth step. Use Rule I if the other tivo parts are adjacent to the
middle part (as m case illustrated), or Rule II if they are opposite,
and solve for the unknoivn part.

Check : Check with that rule which involves the three required parts*

Careful attention must be paid to the algebraic signs of the func-
tions when solving spherical triangles ; the cosines, tangents, and
cotangents of angles or arcs greater than 90° being negative. When
computing with logarithms we shall write (n) after the logarithms
when the functions are negative. If the number of negative factors
is even, the result will be positive ; if it is odd, the result will be
negative and (n) should be written after the resulting logarithm.
In order to be able to show our computations in compact form, we
shall write down all the logarithms of the trigonometric functions
just as they are given in our table ; that is, when a logarithm has a
negative characteristic we will not write down — 10 after it. f

Ex. 1. Solve the right spherical triangle, having given B -
Solution. Follow the above general directions.

33° 50', a = 108°.

To find A
c c

A&

Using Rule II

sin .4.,, = cos B c cos a
cos A = sin B cos a
log sin 5 = 9.7457
log cos a = 9.4900 (n)
log cosA = 9.2357 (n)
.-. 180°-A±=80°6'
and A = 99° 54'.

To find b

n, :

b_ a

Using Rule I

sin a = tan B c tan 6
tan b = sin a tan B
log sin a = 9.9782
log tan B= 9.8263
log tan 6 = 9.8045
.-. 6 = 32° 31'.

To find c

Cc

$

b
Using Rule I

sin B c = tan c c tan a
cote = COS.B cot a
log cos# = 9.9194
log cot a = 9.5118 (n)
log cot c = 9.4312 (n)
. . 180° - c = 74° 54'
and c = 105° 6'.

The value of log cos A found is the same as that found in our first computa-
tion. The student should observe that in checking our work in this example

* Thus, in above case, Ac and a are given ; therefore we underline the three required
parts and cross b as the middle part. Applying Rule II, cc and B e being opposite parts, we
get sin b = cos Cc cos B e , or, sin 6 = sin cein.fi.

t For example, as in the table, we will write log sin 24° = 9.6093. To be exact, this should
be written log sin 21° = 9.6093 -10, Or, logsin24°= f.6093.

+ Since cos A is negative, we get the supplement of A from the table.

202

SPHERICAL TRIGONOMETRY

c c

it was not necessary to look up any new logarithms. Hence the check in this
case is only on the correctness of the logarithmic work.*

Check: Using Rule I

sin .A,, = tan 5 tan c„
A . jj cos A = tan b cote

~^ C • log tan 6 = 9.8045

log cot c = 9.4312 (n)
b_ a log cos^l = 9.2357 (n)

In logarithmic computations the student should always write down
an outline or skeleton of the computation before using his logarithmic
table at all. In the last example this outline would be as follows :

log sin B =

log cos o =

log cos A =

.: 180° -A =

and A =

(n)
(n)

log sin a =

log tan B =

log tan 6 =

.-. 6 =

log cos B =

log cot a =

log cot c =

.. 180°- c =

and c =

(71)

(n)

It saves time to look up all the logarithms at once, and besides it
reduces the liability of error to thus separate the theoretical part of
the work from that which is purely mechanical. Students should be
drilled in writing down forms like that given above before attempt-
ing to solve examples.

Ex. 2. Solve the right spherical triangle, having given c = 70° 30', A = 100°.

Solution. Follow the general directions.

To find a To find b To find B

b a

-r-

Using Rule II

sin a = cos c c cosmic
sin a = sin c sin A
log sine = 9.9743
log sin A = 9.9934
log sin a = 9.9677
.-. 180°- at =68° 10'
and a = 111° 50'.

*

J3o

b_

Using

a

Rule I

sin.A =

tanb =

log cos A -

log tan c =

: tan 6 tan c c
= cos A tan c
= 9.2397 (»)
= 0.4509

log tan 6 = 9. 6906 (n)
.-. 180° -6 = 26° 8'
and 6 = 153° 52'.

6 o

Using Rule I

sin c c = tan A c tan B c
cot B = cos c tan A
log cose = 9.5235
log tan^L = 0.7537 (n)
log cot B = 0.2772 (n)
.-. 180°-B=27°51'
and B = 152° 9'.

The work of verifying the results is left to the student.

* In order to be sure that the angles and sides have been correctly taken from the tables,
in such an example as this, we should use them together with some of the given data in
relations not already employed.

t Since a is determined from its sine, it is evident that it may have the value 68° 10' found
from the table, or the supplementary value 111 50' . Since A > 90°, however, we know from
Th. II, p. 199, that a > 90° ; hence a= 111 50' is the only solution.

EIGHT SPHERICAL TRIANGLES

203

6. The ambiguous case. Two solutions. When the given parts of
a right spherical triangle are an oblique angle and its opposite side,
there are two triangles which
satisfy the given conditions.
For, in the triangle ABC, let
C = 90°, and let .4 and CB ,i<^ J J'

(= a) be the given parts. If

we extend AB and AC to A',

it is evident that the triangle A'BC also satisfies the given condi-
tions, since BCA' = 90°, A' = A, and BC = a. The remaining parts
in A'BC are supplementary to the respective remaining parts in
ABC. Thus

A'B = 180°- c, A'C = 180°- b, A'BC = 180° -ABC.

This ambiguity also appears in the solution of the triangle, as is
illustrated in the following example :

Ex. 3. Solve the bright spherical triangle, having given A = 105° 59', a =
128° 33'.

Solution. We proceed as in the previous examples.

To find b

sin 6 =
sin 6 =

a_

tan a tan^. c
tan a cot A

log tan a = 0.0986 (n)
iogcot^L= 9.4570 (n)
log sin 6 = 9.5556

.-. b = 21° 4', or,
180°- 6 =158° 56' = 6'.*

To find B

Cc

4-°

sin A c = cos a cos B c

. „ cos A

sin B =

cos a

log cos A = 9.4399 (n)

log cos a = 9.7946 (n)

log sin B = 9.6453

.-. B = 26° 14', or,

180° -B =153° 46'= B'.

To find c

A c

b a

sin a = cos^l,, cos c„

sin a

sm c = ■

smA

log sin a = 9.8932

log sin^l = 9.9828

log sine = 9. 9104

.-. c'=54°27', or,

180°-c'=125°33'=c.t

Hence the two solutions are :

1. 6 = 21° 4', c = 125° 33', B = 26° 14' (triangle ABC) ;

2. V = 158° 56', . c' = 54° 27', B' = 153° 46' (triangle A'BC).

It is not necessary to check both solutions. We leave this to the student.

* Since sin B is positive and B is not known, we cannot remove the ambiguity. Hence
both the acute angle taken from the table and its supplement must be retained.

t The two values of B must be retained, since b has two values which are supplementary-
t Since a > 90° and b has two values, one > and the other < 90°, it follows from Th. I,
p. 198, that c will have two values, the first one < 90° and the second > 90°.

204

SPHEKICAL TEIGONOMETEY

EXAMPLES

Solve the following right spherical triangles :

No.

Given

Parts

Keqdieed Parts

1

a = 132° 6'

6 = 77° 51'

A = 131° 27'

B = 80° 55'

c = 98° 7'

2

a = 159°

c = 137° 20'

A = 148° 5'

B = 65° 23'

6 = 37° 54'

3

A = 50° 20'

B = 122° 40'

a = 40° 42'

6 = 134° 31'

c = 122° V

4

a = 160°

6 = 38° 30'

A = 149° 41'

B = 66° 44'

c = 137° 20'

5

£ = 80°

6 = 67° 40'

4 = 27° 12'

a = 25° 25'

c = 69°54'; or,

A'= 152° 48'

a'= 154° 35'

c'= 110° 6'

6

B = 112°

c = 81° 50'

A = 109° 23'

a = 110° 58'

6 = 113° 22'

7

a = 61°

B = 123° 40'

A = 66° 12'

6 = 127° 17'

c = 107° 5'

8

a = 61° 40'

b = 144° 10'

4 = 72° 29'

B = 140° 38'

c = 112° 38'

9

A = 99° 50'

a = 112°

5 = 27° 7'

b = 25° 24'

c= 109° 46'; or,

B'= 152° 53'

&'= 154° 36'

c'= 70° 14'

10

6 = 15°

c = 152° 20'

A = 120° 44'

a = 156° 30'

B = 33° 53'

11

A = 62° 59'

B = 37° 4'

a = 41° 6'

6 = 26° 25'

c = 47° 32'

12

A = 73° 7'

c = 114° 32'

a = 60° 31'

B = 143° 50'

b = 147° 32'

13

B = 144° 54'

6 = 146° 32'

4 = 78° 47'

a = 70° 10'

c = 106°28'; or,

A'= 101°. 13'

a'= 109° 50'

c'= 73° 32'

14

B = 68° 18'

c = 47° 34'

A = 30° 32'

a = 22° 1'

6 = 43° 18'

15

A = 161° 52'

b = 131° 8'

a = 166° 9'

.B = 101° 49'

c = 50° 18'

16

a = 113° 25'

6 = 110° 47'

X = 112° 3'

B = 109° 12'

c = 81° 54'

17

a = 137° 9'

B = 74° 51'

A = 135° 3'

6 = 68° 17'

c = 105° 44'

18

.4 = 144° 54'

B = 101° 14'

a = 146° 33'

6 = 109° 48'

c = 73° 35'

19

a = 69° 18'

c = 84° 27'

^ = 70°

B = 75° 6'

6 =74° 7'

20. For more examples take any two parts in the above triangles and solve
for the other three.

7. Solution of isosceles and quadrantal triangles. Plane isosceles
triangles -were solved by dividing each one into two equal right tri-
angles and then solving one of the right triangles. Similarly, we
may solve an isosceles spherical triangle by dividing it into two sym-
metrical right spherical triangles by an arc drawn from the vertex
perpendicular to the base, and then solving one of the right spheri-
cal triangles.

A quadrantal triangle is a spherical triangle one side of which is
a quadrant (= 90°). By (/), p. 195, the polar triangle of a quad-
rantal triangle is a right triangle. Therefore, to solve a quadrantal
triangle we have only to solve its polar triangle and take the sup-
plements of the parts obtained by the calculation.

Ex. 1. Solve the triangle, having given c = 90°, a = 07° 38', b = 48° 50'.

Solution. This is a quadrantal triangle since one side c = 90°. We then find
the corresponding elements of its polar triangle by (/), p. 195. They are C'=90°,
A' = 112° 22', B' = 131° 10'. We solve this right triangle in the usual way.

EIGHT SPHERICAL TRIANGLES

205

Construct the polar (right) triangle.
Given A' = 112° 22', B' = 131° 10':

To find a'

Cc'

4

#

sin 4 ' c = cos a' cos .Be

. . cos A'

cos a =

sinB'

log cos 4'= 9.5804- (n)

log sin B' =08767

log cos a' =9.7037 (n)

180° - a' = 59° 38'.

a' = 120° 22'.

Similarly, we get

6' = 135° 23', c' = 68° 55'.

Hence in the given quadrantal tri-
angle we have

A = 180° - a' = 5C° 38',
B = 180° - V = 44° 37',
G = 180° - c' = 111° 5'.

EXAMPLES

Solve the following quadrantal triangles :

No.

Given Parts

Required Parts

1

A = 139° 6 = 143°

c = 90°

a = 117° 1' JS = 153° 42' C = 132° 34'

2

A = ib° 30' B = 139°20'

c = 90°

a = 57° 22' 6 = 129° 42' C = 57°53'

3

a = 30° 20' C = 42°40'

c = 90°

A = 20°V £ = 141° 30' 6 = 113° 17'

4

B=70°12' C=106°25'

c = 90°

A = 33° 28' a = 35° 4' 6 = 78° 47'

5

A = 105° 53' a = 104° 54'

c = 90°

B = 69°16' 6 = 70° C = 84°30'; or
B=110°44' 6 = 110° C = 95°30'

Solve the following isosceles spherical triangles :

No.

Given Parts

Required Parts

6

a = 54° 20' c = 72° 54' A

= B

6 = 54° 20' A = B = 57° 59' C = 93° 59'

7

a = 54°30' C = 71° A

= B

b = 54° 30' A = B = 67° 30' c = 56° 26'

8

a = 66°29' A = B = 50°1T

6 = 66° 29' c=lll°30' C = 128°42'

9

c = 156°40' C = 187°46' A

= B

a = b = 79° or 101°
A = B = 199° 34'

Prove the following relations between the elements of a right spherical
triangle (O = 90°):

10. eosMsin 2 e = sin(e + a)sin(c — a). 13. sin (6 + c) = 2 cos 2 ^ A cos 6 sin c.

11. tan a cos c = sin 6 cot B. 14. sin (c — 6) = 2 sin 2 £ 4 cos 6 sin c.

12. sin 2 X = cos 2 B + sin 2 a sin 2 B.

CHAPTEE II

OBLIQUE SPHERICAL TRIANGLES

8. Fundamental formulas. In this chapter some relations between
the sides and angles of any spherical triangle (whether right angled
or oblique) will be derived.

9. Law of sines. In a spherical triangle the sines of the sides are
proportional to the sines of the opposite angles.

Proof. Let ABC be any spherical triangle, and draw the arc CD
perpendicular to AB. There will be two cases according as CD falls

C

upon AB (first figure) or upon AB produced (second figure). For
the sake of brevity let CD =p, AD = n, BD = m, angle ACD = x,
angle BCD — y.

In the right triangle ADC (either figure)

(A) sis. p = sin b sin A.

In the right triangle BCD (first figure)
(B) sinp == sin a sin B.

This also holds true in the second figure, for

sin DBC = sin (180° -B) = sinB,

Equating the values of sinp from (A) and (B),
sin a sinB = sin b sin A,
or, dividing through by sin^i sinB,

sin a sin b

Eule II, p. 200
Eule II, p. 200

(C)

sin A

sinB
206

OBLIQUE SPHERICAL TRIANGLES 207

In like manner, by drawing perpendiculars from A and B, we get

, „ sin b sine

(D) - T — = - T —- , and

x / Gin R cin fl

sin B sin C
sin c _ sin a
sinC sin.4'

. . sine sin a ,. ,

( E) —. — - = — — 7 , respectively.

Writing (C), (D), (E) as a single statement, we get the law of sines,

. sin a sin b sin c *

^ sin.4 — sinl? — sinC

10. Law of cosines. In a spherical triangle the cosine of any side
is equal to the product of the cosines of the other two sides plus the
product of the sines of these two sides and the cosine of their included
angle.

Proof. Using the same figures as in the last section, we have in
the right triangle BDC,

cos a = cosp cos m Rule II, p. 200

= cos p cos (c — n)
= cosp { cos c cos n + sin c sin n j

(A) = cosp cos c cos n + cosp sin c sin n.

In the right triangle ADC,

(B) cosp cos n = cos b.

' cos b

Whence cos p = >

cosn

and, multiplying both sides by sin n,

cos b , .

(C) cospsmn = sin n = cos b tan n.

K ' ± cos n

But cos A = tan n cot b, or, Rule I, p. 200

( D) tan n = tan b cos A.

Substituting value of tan re from (D) in (C), we have
. (E) cosp sin n = cos b tan b cos A = sin boos A.

Substituting the value Of cosp cos re from (B) and the value of
cosp sin re from (E) in (A), we get the law of cosines,

(F) cos a = cos b cos c + sin 5 sin c cos ^1.

» Compare with the law of sines in Granville's Flame Trigonometry, p. 102.

208 SPHERICAL TRIGONOMETRY

Similarly, for the sides b and c we may obtain
(£) cos b = cos c cos a + sin c sin a cos B,

(fl) cos c = cos a cos b + sin a sin b cos C.

11. Principle of Duality. Given any relation involving one or more
of the sides a, b, c, and the angles A, B, C of any general spherical
triangle. Now the polar triangle (whose sides are denoted by a', b', c',
and angles by A', B', C") is also in this case a general spherical
triangle, and the given relation must hold true for it also ; that is,
the given relation applies to the polar triangle if accents are placed
upon the letters representing the sides and angles. Thus (F), (G),
(Zf) of the last section give ns the following law of cosines for the
polar triangle :

(A) cos a' = cos b' cos c' + sin V sin c' cos A '.

(E) cos V = cos c' cos a' + sin c' sin a' cos B'.

(C) cos c' = cos a' cos b' + sin a' sin b' cos C.

But by (/), p. 195,

a' = 180°-^, b' = 180°-B, c' = 180°-C,

A' = 180°-a, B' = 180°-b, C" = 180°-c.

Making these substitutions in (^4), (B), (C), which refer to the
polar triangle, we get

(D) cos (180° - A) = cos (180° - B) cos (180° - C)

+ sin (180° - B) sin (180° - C) cos (180° - a),

(E) cos (180° - B) = cos (180° - C) cos (180° - A )

+ sin (180° - C) sin (180° - A) cos (180° - b),

(F) cos (180° - C) = cos (180° - A) cos (180° - B)

+ sin (180° - A) sin (180° - B) cos (180° - c),

which involve the sides and angles of the original triangle.

The result of the preceding discussion may then be stated in the
following form :

Theorem. In any relation between the parts of a general spherical
triangle, each pari may be replaced by the supplement of the opposite
part, and the relation thus obtained will hold true.

OBLIQUE SPHERICAL TKIANGLES

209

The Principle of Duality follows when the above theorem is applied
to a relation involving one or more of the sides and the supplements
of the angles (instead of the angles themselves).

Let the supplements of the angles of the triangle be denoted by
a, /?, y * ; that is,

a = 180°- A, j8 = 180°-£, y = 180°-C,
or, A =180°- a, B = 180°-p, C = 180°-y.

When we apply the above theorem to a rela-
tion between the sides and supplements of the
angles of a triangle, we, in fact,

replace a by a (= 180° — A),

replace b by /3 (= 180° -B),

replace e by y (= 180° - C),

replace a (=180° -A) by 180° -(180°- a)=a,

replace £ (= 180° -B) by 180°- (180°- b) = b,

replace y (=180°-C) by 180° -(180°- c) = c,

or, what amounts to the same thing, interchange the Greek and
Soman letters. For instance, substitute

A =180° -a, B = 180°- ft C = 180°-y

in (F), (G), (H) of the last section. This gives the law of cosines
for the sides in the new form

(12)
(13)
(14)

cos a = cos b cos c — sin b sin c cos a,
cos b = cos c cos a — sin c sin a cos jG,
cos c = cos a cos b — sin a sin b cos y.

[Since cos A = cos (180° — <*)*= - cos a, etc.]

If we now apply the above theorem to these formulas, we get the
law of cosines for the angles, namely,

(15) cos a = cos /? cos y — sin /? sin y cos a,

(16) cos fl = cos y c° 8 a — sin y sin a cos b,

(17) cos y = cos a cos /? — sin a sin cos c,

* a, /3, y are then the exterior angles of the triangle, as shown in the figure.

210 SPHERICAL TRIGONOMETRY

that is, we have derived three new relations between the sides and
supplements of the angles of the triangle.* We may now state the

Principle of Duality. If the sides of a general spherical triangle are
denoted by the Roman letters a, b, c, and the supplements of the cor-
responding opposite angles by the Greek letters a, /?, y, then, from
any given formula involving any of these six parts, we may write
down a dual formula by simply interchanging the corresponding Greek
and Roman letters.

The immediate consequence of this principle is that formulas in
Spherical Trigonometry occur in pairs, either one of a pair being the
dual of the other.

Thus (12) and (15) are dual formulas ; also (13) and (16), or (14)
and (17).

If we substitute

A = 180°- a, B = 180°- ft C = 180°-y

in the law of sines (p. 207), we get

sin a _ sin b sin c
sin a sin /? sin y

[Since sin A = sin (180° — a) = sin a, etc.]

Applying the Principle of Duality to this relation, we get

sin a _ sin /3 _ sin y
sin a sin b sin c

which is essentially the same as the previous form.

The forms of the law of cosines that we have derived involve
algebraic sums. As these are not convenient for logarithmic calcu-
lations, we will reduce them to the form of products.

12. Trigonometric functions of half the supplements of the angles of
a spherical triangle in terms of its sides. Denote half the sum of the
sides of a triangle (i.e. half the perimeter) by s. Then

(A) 2s = a + b + c,

or, s = ^(a + b + c).

* If we had employed the interior angles of the triangle in our formulas (as has been the
almost universal practice of writers on Spherical Trigonometry), the two sets of cosine
formulas would not have been of the same form. That the method used here has many
advantages will become more and more apparent as the reading of the text is continued.
Not only are the resulting formulas much easier to memorize, but much labor is saved in
that, when'we have derived one set of formulas for the angles (or sides), the corresponding
set of formulas for the sides (or angles) may be written down at once by mere inspection
by applying this Principle of Duality. The great advantage of using this Principle of
Duality was first pointed out by MSbius (1790-1868).

OBLIQUE SPHERICAL TRIANGLES 211

Subtracting 2 c from both sides of (^4),

2s — 2c = a + b + c — 2 c, or,

(B) s-c = ) s (a + b-c).

Similarly,

(C) s-b = $(a-b + c), and

(-D) « -« = $(- a + 5 + c) = £(Z» + c - a).

From Plane Trigonometry,

(E) 2 sin 2 J a = 1 — cos a,

(F) 2 cos 2 £ a = 1 + cos a.

But from (12), p. 209, solving for cos a,

cos b cos c — cos a
cos a =

hence (E) becomes

2sin 2 i

sin b sin c

1-

cos 5 cos c — cos a

sin b sin c

sin

J sin c — cos 5 cos c

+

cos a

sin & sin c

cos a — (cos b cos c — sin

b sin c)

sin 5 sin c
_ cos a — cos (b + c)

sin 6 sin o
_ — 2 sin £ (a + 6 + e) sin £ (a — 5 — c) #

or,

sin 6 sin c

(G) 2sinH« = 2sin ^ + & + c)sill2 - (& + C - a) -

v sin 5 sin c

[Since sin § (a — b — c)= — sin \(—a + b + c)= — sin £(& + c — a).]

Substituting from (4) and (Z>) in ((?), we get

i 2 i

sin s sin (s — a)

sin" A- a = : — r^ » or,

J sin 6 sin c

,,„. . , I sins sin (s — a)

(18) sin|a = >J ^-r-\ '-

K ' \ sin b sine

*Let -4 = a .4=a

£=6+c B=b+c

A+B=a+b+c A—B=a—b—c

l(A + B)=t(a + b + c). i(A-B)=i(a-b-c).

Hence, substituting in (65), P. 74, Granville's Plane Trigonometry, namely,

cos .4 — cos B= — 2Bini(A + B)sin^{A-B),
we get cos a — cos (6 + c)= — 2 sin i (a + b + c) sin J(a - b - c).

212 SPHERICAL TRIGONOMETRY

Similarly, (F) becomes

. . , cos b cos o — cos a

2 cos 2 A « = H ^-r— :

2 sin b sin c ,

sin 6 sin c + co s & cos c — cos a
sin b sin c

_ cos, (b — o) — cos a
sin 6 sin c

— 2 sin 1 (a + b — o) sin A (& — c — a) *

= — : ^ *-^ > or,

sin 6 sin c

2 sin i- (a + b — c) sin A (a — b + c)
(H) 2cos 2 Aa = * K . / . — ^ z -

v ' J sin b sin o

[Since sin J (b - c - a) = - sin J (- b + c + a) = - sin J (a - b + c).]

Substituting from (B) and (C) in (H), we get

sin (s — c) sin (s — b)

cos 2 A« = * — : — '— -. — 1 or,

sin b sm c

, , , /sin (s — &) sin (s — c)

(19) cosia=-<J i — . I .. -•

v ' \ sin 6 sin c

sin -i- #
Since tan \a = -—; we get from this, by substitution from (18)

and (19),
(20)

T sin s sin (s — a) f

tania=-Aj-r-7 , N \ , } ■'

\ sin (s — b ) sin ( s — c)

* Let ^4=&-c A=b-c
B=a B= a

A+B=a+b— c A — B= b—c—a

^A + B)=^(a+b-c). i(^-B)=i(6-c-a).

Hence, substituting in formula (65), found on p. 74, Granville's Plane Trigonometry,
namely,

cos A - cos B= - 2 sin £(^1 + B) sin J(^4 - .B),

we get cos(fc-c) — cosa=-2sin£(ct + &-c)sin£(6-c — a).

t In memorizing these formulas it will be found an aid to the memory to note the fact
that under each radical

(a) only the sine function occurs.

(b) The denominators of the sine and cosine formulas involve those two sides of the tri-
angle which are not opposite to the angle sought.

(c) "When reading the numerator and denominator of the fraction in the tangent formula,
£ comes first and then the differences

s — a, s-6, s — c,

in cyclical order ; s and the first difference occurring also in the numerator of the cor-
responding sine formula, while the last two differences occur in the numerator of the
corresponding cosine formula.

OBLIQUE SPHERICAL TRIANGLES 213

In like manner, we may get
(21) sinl^ = - N j:

sin s sin (s — V)

sin c sin a

(22) cosf/?=^j!i

sin (s — c) sin (s — a)
sin c sin a

<*> -^H jr ,in(s " 6) :

Also

» sin (s — c) sin (s — a _)

.„.. . , I sin s sin (s — c)

( 24 ) sin ly = 'V : . ,. ■

v ' * r N sin a sin b

(25) cos|y = j!^iZLf)i^li),

' > sin a sin 6

.„„. , sin s sin (s — c)

v y 2r \sin(s — a)sin(s —

6)

In solving triangles it is sometimes more convenient to use other
forms of (20), (23), (26). Thus, in the right-hand member of (20),
multiply both the numerator and denominator of the fraction under
the radical by sin(s — a). This gives

r sm s sm 2 (s — a)

tania; = A

2 Ns

i sin (s — a) sin (*■ — V) sin (s — o)

. \_~^ sins

= sin (s—a) '<J -^—

v ' >sin(s— o

a) sin (s— b)ein (s—c)
# _ Isin (s — a) sin (s — b) sin (s — c)

Let tan A d

1 y sm s

sin (s — a)
then tan \a= ,_ , , •

2 tan \ a

Similarly, for tan \ /? and tan \ y. Hence

, . |sin(s — a)sin(s— 6) sin (s — c)
(27) tan|^>J-i l s l

(28) tan | a =

sin (s — a)
tan|d

» -w-tS^

(30) tan|y =

2 '

sin (s — c)
tan|d

■• It may be shown that d = diameter of the circle inscribed in the spherical triangle.

214 SPHEKICAL TRIGONOMETRY

13. Trigonometric functions of the half sides of a spherical triangle in
terms of the supplements of the angles. By making use of the Princi-
ple of Duality on p. 208, we get at once from formulas (18.) to (30),
by replacing the supplement of an angle by the opposite side and each
side by the supplement of the opposite angle, the following formulas :

(33)

(42)
(43)

, , . , Isina sin (a — a)

(31) sin§a = -0 —± L,

v ' \ sin/? sin y

(32) cos!a = -y

sin (<y — p) sin (<r — y)

sin /? sin y

sin o" sin (a — a)

tan I a = \H— -, „ s . ;

2 Nsin(o-— B)sinCo- —

(er-/?)sin(o--y)'

,„.-. . , . Isin a sin (a — B)

(34) sin | b = aJ : ^ ^ ,

v ' > sin y sin a

(35) cos | b = Jsin(o--y)sin(o--a)

s ' > sin v sin a

/„„n i , I sin a sin (a — B)

<*> tan ' &= N sin(o--y)sin(a-a) '

(37) sin| C=A J Sin<rsin ^E l),
K ' y sin a sin p

(38) cos|c=^

sin (a — a) sin (<r — /?)

sin a sin fj

(39\ ■ tan 1 c - i l ~~ sin °" Sin ( °" ~ y ~

(39) tan , c _ \ sin((r _ a)sin((r _ ^ •

(40) tanlg* => j!^- a)sin ( a -^ S in ^-y)

/.-v i sin(o-— a)

(41) tan|a= v J ,

K ' tan Jo

smcr

sin (a- — /?)
tan|6= /■ /' .
tan 1 5

taD2 ^ tan 16 '

where or == £ (a -f /3 + y)

• =£(180°-,4+180 o -.B+180 o -C)
= 270°- l(A+B + C).

What we have done amounts to interchanging the corresponding
Greek and Roman letters.

* it may be shown that S is the supplement of the diameter of the circumscribed circle.

OBLIQUE SPHERICAL TRIANGLES 215

14. Napier's analogies. Dividing (20) by (23), we get

tan \a _ I sin s sin (s — a) I sin s sin (s — b) ~

N sin (s — c) sin (s — a)

or,

'<

sin s sin (s —

a)

sin(s

— b) sin (s

-o) "

sin s sin (s —

a)

sin(s

— 6) sin (s — c)

sin s sin (s —

b )

sin J a
cos I- a

cos £ j3 \ sin (s — c) sin (s — a)

sin £ a cos J /? _ sin (s — a)
cos £ a sin £ fi sin (s — i)

By composition and division, in pro-portion,

sin \ a cos J /? + cos ^ « sin J fi _ sin (,s — «) + sin (s — &)
sin J a cos £ /3 — cos \ a sin J /3 sin (s — a) — sin (s — b)

Erom (40), (41), p. 63, and (66), p. 74, Granville's Plane Trigo-
nometry, the left-hand member equals

smQa+J^
sin(i«- J/3)'
and the right-hand member

sin (s — a) + sin(s — &) _ tan J [s — q. + (s — 5)] _ tan Jc #
sin(s — a) — sin (s — 6) tan J [s — a —(s — 5)] tan J (b — a)

Equating these restilts,we get, noting that tan ^(b— a)=— tan£(a— 5),

sin J (a + /J) _ tan ^c

sin J (a — /?) tan J (a — 6) '

<»> «-»<— >~=i&g-K

In the same manner we may get the two similar formulas for
tan J (b — c) and tan \ (c — a).
Multiplying (20) and (23), we get

I sin s sin (s — a) sin s sin (s — b)

tan 1 a tan 18= "V^-7 , x \ . — ^ AJ^ r-^— 7 — ^>

* ¥ r \ sin (s — J) sin (s — c) N sin (s — c) sin (s — a)

sin 1 a sin J # sin s

cos £ a cos J /? sin (s — c)

By composition and division, in proportion,

cos J « cos J 8 — sin J or sin J 8 sin (s — c) — sin s
cos £ a cos £ /3 + sin J a sin J S sin (s — c) + sin s

• Fiir»-a + »-6=2«-o-J=o + 6 + c — o-6 = f, and s-a-$ + b=b — a.

216 SPHERICAL TRIGONOMETRY

From (42), (43), p. 63, and (66), p. 74, Granville's Plane Trigo-
nometry, the left-hand member equals

cos (\ a + % /?) .
cos (i a - i /8) '
and the right-hand member

sin (.i — c) — sin s _ tan ^ (s — e — s) _ tan \ (— c) *
sin (s — c) + sin s tan ^(s — c + s) tan £ (a. + b)

Equating these results, we get, noting that tan £ (— c) = — tan ^ ?,

cos £ (a + j3) _ tan ^ c

cos i (« — y8) tan J (a + 6) '

(45) tani(a + ft) = - C ° Si(a ~ ^ tan j c.
K J K ' cos J (a + 0) 2

In the same manner we may get the two similar formulas for
tan \ (b + e) and tan £ (c + »)•

By making use of the Principle of Duality on p. 208,we get at once
from formulas (44) and (45),

(46) u»l(.-/0— £}g^t»l y ,

(47) tan|(a + ^) = -^^ta„| y .

By changing the letters in cyclic order we may at once write down
the corresponding formulas for tan ^ (fi — y), tan ^ (y — a), tan £ (B+y),
and tan i (y + «)■

The relations derived in this section are known as Napier's analogies.

Since cos £(«-&) and tan h 7 = tan i ( 180 ° - c ) = tan ( 90 ° ~h c )
= cot J C are always positive, it follows from (47) that cos \(a + b)
and tan \ (a + B) always have opposite signs ; or, since tan £(« + /3)
= tan ^(180° -4 + 180° -.8) = tan £[360° -(,4 + £)] = tan [180°-
£ {A + B)] = — tan£ (A + B), we may say that cos £ (a + b) and
tan |(^[ + B) always have the same sign. Hence we have the

Theorem. In a spherical triangle the sum of any two sides is less
than, greater than, or equal to 180°, according as the sum of the cor-
responding opposite angles is less than, greater than, or equal to 180°.

15. Solution of oblique spherical triangles. We shall now take up the
numerical solution of oblique spherical triangles. There are three
cases to consider with two subdivisions under each case.

* For s-c-s = -c,

and s — c + s = 2s-c = a + b + c-c=a + b.

OBLIQUE SPHERICAL TRIANGLES

217

Case I.
Case II.
Case III.

16. Case I.
namely,

(27)
(28)
(29)
(30)
to find a, /?,

(a) Given the three sides.

(b) Given the three angles.

(a) Given two sides and their included angle.

(b) Given two angles and their included side.

(a) Given two sides and the angle opposite one of them.

(b) Given two angles and the side opposite one of them.

(a) Given the three sides. Use formulas from p. 213,

tan^rf

jsin (s — a) sin (s — b) sin (s — c)
\ sins

tan I a =

tan ^ d

tan $ fi =
tan ^ y =

sin (s — a)
tan ^-<2

sin (s — b)
tan £ d

sin (s — c)

tan I d

y, and therefore A, B, C, and check by the law of sines,

sin a sin b sin c
sin A sinB sin C

Ex. 1. Given a = 60°, b = 137° 20', c = 116° ; find A, B, C.

Solution.

a = 60°

6 = 137° 20'

c = 116°

2S:

313° 20'

8 = 156° 40'.

s - a = 96° 40'.

s - 6 = 19° 20'.

s - c = 40° 40'.

To find A use (28)

logsin(s-a) = 9.9971

log tan £ d = 9^

To find log tan ^ d use (27)

log sin (s- a) = 9.9971

log sin (s- 6)= 9.5199

log sin (s-c) = 9.8140

29.3310

log sin s = 9.5978

2| 19.7332

log tan ^ d = 9.

log tan ^.a = 0.1305
i or = 53° 29'.
or = 106° 58'.
. ..4=180°-106°58'=73°2'.

Check: log sin a = 9.9375
log sin^. = 9.9807

To find Buse (29)

log sin (s- 6) = 9.5199

; tan \ d = 9J

log tan ^/3 = 9.6533
\ /3 = 24° 14'.
= 48° 28'.
. B=180°-48°28'=131°32'.

log sin 6 = 9.8311

log sin B = 9.8743

9.9568

To find C use (30)
log sin (s -c) = 9.8140
log tan £• d = 9.8666
log tan J y = 9.9474
£ Y = 41° 32'.
7 = 83° 4'.
•. C=180°-83°4'=96°56'.

log sin c = 9.9537
log sin C = 9J.

This checks up closer than is to be expected in general. There may be a
variation of at most two units in the last figure when the work is accurate.

218

SPHERICAL TRIGONOMETRY

EXAMPLES

Solve the following oblique spherical triangles :

No.

Given Parts

Required Parts

1

a = 38°

6=51°

c = 42°

^.= 51° 58'

B = 83° 54'

C = 58° 53'

2

a = 101°

6 = 49°

c = 60°

A=Ui° 32'

B = 27° 52'

C = 32° 28'

3

a = 61°

6 = 39°

c = 92°

^1=35° 32'

B = 24° 42'

C = 138° 24'

4

a = 62° 20'

b = 54° 10'

c = 97° 50'

^1=47° 22'

B = 42° 20'

C = 124° 38'

5

a = 58°

6 = 80°

c = 96°

^4 = 55° 58'

B=74°14'

C = 103° 36'

6

a = 46° 30'

b = 62° 40'

c = 83° 20'

yl=43°58'

5 = 58° 14'

C=108°6'

7

a = 71° 15'

6 = 39° 10'

c = 40° 35'

A =130° 36'

B = 30°26'

C = 31°26'

8

a = 47° 30'

6 = 55° 40'

c = 60° 10'

.4=56° 32'

B = 69° 7'

C = 78°58'

9

a = 43° 30'

6 = 72° 24'

c = 87° 50'

4 = 41° 27'

.8= 66° 26'

C = 106° 3'

10

a =110° 40'

b = 45° 10'

c = 73° 30'

.4=144° 27'

B = 26° 9'

C = 36°35'

17. Case I. (6) Given the three angles. Use formulas from p. 214,
namely,*

= l sin(,r-« ) sin(o--flsin ifL - x ) >
> sin <r

sin (a-. — a)

(40)

(41)
(42)
(43)

tan \a =

tan £8

sin (er — )8)

tan *b— — -^ — — r^ >
tan £8

tan^c

_ sin (o- — y)

tan £ 8

io ,/stmZ a, b, c ; and check by the law of sines,

sin a _ sin b _ sin c
sin A sin B sinC

Ex. 1. Given A = 70°, B = 131° 10', C = 94° 50'; find a, 6, c.
Solution. Here we use the supplements of the angles.

To find log tan \ S use (40)

a = 180°- A = 110°
/3 = 180°-JB = 48° 50'
7 = 180°- C = 85° 10'
2 <r = 244°
<r = 122°.
a- a= 12°.
<r - /3 = 73° 10'.
<r _ 7 = 36° 50'.

log sin (<r - a) - 9.3179

log sin (<r - (8) = 9.9810

log sin (<r - y) = 9.7778

29.0767

log sin <r = 9.9284

2 | 19 1483

log tan £ <5= 9.5742

* These formulas may be -written down at once from those used in Case I, (a), p. 217, by
simply interchanging the corresponding Greek and Roman letters.

OBLIQUE SPHERICAL TRIANGLES

219

To find a use (41)

log sin (a- - a) = 9.3179
log tan £ S = 9.5742
logtan|a = 9.7437
\ a = 29°.
.-. a = 58°.

To find 6 use (42)

log sin (<r-0) = 9.9810

log tan % 8 = 9.5742

log tan£& = 0.4068

%b = 68° 36'.

6 = 137° 12'.

Check :

log sin a = 9.9284

log sin 4 = 9.9730

9.9554

log sin 6 = 9.8321

log sin£ = 9.8767

9.9554

To find c use (43)

log sin (o-- 7) = 9.7778
log tan \ 5 = 9.5742
logtan£c = 0.2036
£c = 57°58'.
c = 115° 56'.

log sin c = 9.9539

log sin C = 9. 9985

9.9554

EXAMPLES

Solve the following oblique spherical triangles :

No.

Given Parts

Required Parts

1

4=75°

£ = 82°

C = 61°

a = 67° 52'

6 = 71° 44'

c = 57°

2

4=120°

£ = 130°

C = 80°

a = 144° 10'

.6 =148° 49'

c = 41° 44'

3

4 = 91° 10'

£ = 85° 40'

C = 72° 30'

a = 89° 51'

6 = 85° 49'

c = 72° 32'

4

4 = 138° 16'

£ = 31° 11'

C = 35°53'

a = 100° 5'

6 = 49° 59'

c = 60° 6'

5

A =78° 40'

£ = 63° 50'

C = 46°20'

a = 39° 30'

6 = 35° 36'

c = 27° 59'

6

4=121°

£ = 102°

C = 68°

a = 130° 50'

6 = 120° 18'

c = 54° 56'

7

4=130°

£ = 110°

C = 80°

a = 139° 21'

6 = 126° 58'

c = 56° 52'

8

4 = 28°

£ = 92°

C = 85°26'

a = 27° 56'

6 = 85° 40'

c = 84° 2'

9

4= 59° 18'

£ = 108°

C = 76°22'

a = 61° 44'

6= 103° 4'

c = 84° 32'

10

4=100°

£=100°

C = 50°

a =112° 14'

6 = 112° 14'

c = 46°4'

18. Case II. (a) Given two sides and their included angle, as a, b,

C. Use formulas on p. 216, namely,

(46)

(47)

, / „x sin 4- (a — b) ,

tani(a: — /?) = :— 7-7 rf tan Ay,

^ v ^ y sin£(a + &) 2 "

cos 1 (a — b) ^

tan £(« + £) = y) — -rftaniy,

^ v ^' cos £(« + £) '

to ^jwZ a and (i and therefore A and B ; and from p. 215 use (44)

solved for tan \ e, namely,

sin j(a + /3)tan j- (a — b)

(44)

tan i c =

sin£(a — /3)

to find c. Cheek by the law of sines

Ex. 1. Given a = 64° 24', 6 = 42° 30', C = 58° 40' ; find 4, £, c.

Solution. 7 = 180° - C = 121° 20'. .-. i 7 = 60° 40'.

a= 64° 24'
6 = 42° 30'

a + b = 106° 54'
.-. \ (a + 6) = 53° 27'.

a = 64° 24'
5 = 42° 30'
a - 6 = 21° 54'

.-. i (a -6) = 10° 57'.

220

SPHERICAL TRIGONOMETRY

To find i (a- p) use (46)

log sin b (a - b) = 9. 2786

log tan^7 = 0.2503

9.5289

log sin -^ (a + 6) = 9.9049

log tan-£(« - p) = 9.6240 (n)

.-. £(a-/3) = -22°49'.*

To find A and B

b(a + p)= 108° 49'
i(a-p)=- 22° 49'
Adding, a = 86°

Subtracting, p = 131° 38'.
.-. A = 180° - a = 94°.
5 = 180° - = 48° 22'.

ToJtnd^(a: + /3) use (47)

log cos £ (a- 6) = 9.9920

log tan 4- 7= 0.2503

10.2423

log cos£ (a + 6) = 9.7749

log tan b(a + p)= 0.4674 (n)

180° -•£(« + £) = 71° ll'.t

.-. | (a +0) = 108° 49'.

To find c use (44)

logsin£(a + jS)= 9.9761

log tan | (a - 6) = 9.2867

19.2628

log sin £ (a- - /3) = 9.5886 (n)

logtan£c = 9.6742 J

£c = 25°17'.

. . c = 50° 34'.

C%ec& : log sin a = 9.9551
log sin A = 9S

log sin 6 :
log sin B -.

9.9562

= 9.8297

: 9.8735

9.9562

log sin c = 9.8878

log sin C = 9.9315

9.9563

If o only is wanted, we may find it from the law of cosines, (14),
p. 209, without previously determining A and B. But this formula
is not well adapted to logarithmic calculations. Another method is
illustrated below, which depends on the solu-
tion of right spherical triangles, and hence
requires only those formulas which follow
from applying Napier's rules of circular
parts, p. 200.

Through B draw an arc of a great circle per-
pendicular to A C, intersecting A C at D. Let

BD=p, CD = m, AD = n.

Applying Rule I, p. 200, to the right spherical triangle BCD, we
have cos C = tan m cot a, or,

(.4) tan m = tan a cos C.

Applying Rule II, p. 200, to BCD,

cos a = cos to cos^, or,
(B) cosp = cos a sec to.

* Since tan £ (a -8) is negative, £(ar — 0) may be an angle in the second or fourth quad-
rant. But a > b, therefore A> B and a < 8, since a and 8 are the supplements of A and B.
Hence \{a — 8) must be a negative angle numerically less than 90°.

t Here £(a + 8) must be a positive angle less than 180°. Since tan $ (a +,8) is negative,
$(« + 8) must lie in the second quadrant, and we get its supplement from the table.

} tan J c is positive, since sin j (a- a) is negative and there is a minus sign before the fraction.

OBLIQUE SPHERICAL TRIANGLES
Applying the same rule to ABB,

cos c = cos n cosjj, or,
(C) cosp = cos o sec n.

Equating (B) and (C),

cos o sec n = cos a sec m, or,
cos c = cos a sec »i cos n.

But ?t = J — m ; therefore

(-D) cos c = cos a. sec m cos (i — m).

i

Now c may be computed from (4) and (Z>), namely,

(48) tan m = tan a cos C,

221

(49)

cos c

_ cos a cos (6 — m)

cos m

Ex. 2. Given a = 98°, 6 = 80°, C = 110° ; find c.
Solution. Apply the method just explained.

To find b — m use (48)

log tan a = 0.8522 (n)
log cos C = 9.5341 (n)
log tan m = 0.3863
m = 67° 40'.
.-. b-m = 12° 20'.

To find c use (49)

log cos a = 9.1436 (n)
;cos(6-ro) = 9.98 99
19.1335
log cos m = 9.5798
log cos c = 9. 5537 (to)
180° - c = 69° 2'.
c = 110° 58'.

EXAMPLES

Solve the following oblique spherical triangles :

No.

Given Parts

Required Parts

1

6 = 137° 20'

c = 116° A = 70°

£ = 131° 17' C = 94°48' a = 57° 57'

2

a = 72°

6 = 47° C = 33°

A = 121° 33' £ = 40° 57' c = 37°26'

3

a = 98°

c = 60° £ = 110°

^4=87° C=60°51' b = 111° 17'

4

6 = 120° 20'

c = 70°40' ^4 = 50°

B - 134° 57' C = 50° 41' a = 69° 9'

5

a = 125° 10'

b = 153° 50' C = 140° 20'

A = 147° 29' £ = 163° 9' c=76°8'

6

a = 93° 20'

6 = 56° 30' C = 74° 40'

.4 = 101° 24' £ = 54° 58' c=79°10'

7

6 = 76° 30'

c = 47° 20' A = 92° 30'

£ = 78° 21' C = 47°47' a = 82° 42'

8

c = 40°20'

a = 100° 30' £ = 46° 40'

A = 131° 29' C = 29° 33' 6 = 72° 40'

9

6 = 76° 36'

c = 110° 26' ^1 = 46° 50'

£ = 57° 38' C = 125° 32' a = 57° 8'

10

a = 84° 23'

b = 124° 48' C = 62°

^1=68° 27' £ = 129° 53' c= 70° 52'

222

SPHERICAL TRIGONOMETRY

19. Case II. (b) Given two angles and their included side, as A, B, c.

Use formulas * o% pp. 215, 216, namely,

(44) tan Ha - &) = - ^ ti" T ff tan £ c >

(45) tani(a + b) = - °° S \ j" ~ ff tan£ e ,

to ./iwrf a and b ; and from p. 216, use (46) solved for tan £ y, namely,

sin $(a + b) tan ■$■ (a — /?)

(46)

tan £ y ■■

sin £ (a — 6)
to ./md y and therefore C. Check by the law of sines.

Ex. 1. Given c = 116°, A = 70°, B = 131° 20' ; find a, 6, C.
Solution, a = 180° - ^ = 110°, and /3 = 180° - 5 = 48° 40'.

a = 110°
p = 48° 40'

a + /3 = 158°40'
.-. £(a + /3) = 79°20'.

To find i (a - 6) use (44)

logsin£(a-/3) = 9.7076

log tan ^-c = 0.2042

9.9118

logsin £(<* + P) = 9.9924

log tan \ (a - b) = 9.9194 (n)

.-. £(a-6) = -39°43'.t

a = 110°
ft = 48° 40'
a-p = 61°20'
£(<*-/3) = 30°40'.

c = 116°.
£c = 58°.

To find a and b

To find i (a + 6) use (45)

log cos i(a~ p)= 9.9346

log tan £ c = 0.2042

10.1388

log cos -J- {a + p) = _9 1 2674

log tan £ (a + 6) = 0.8714 (n)

180° -£(a + b) = 82° 21'.

.. |(a + &) = 97°39'.

To find C use (46)

log sin £ (a + 6) = 9.9961

log tan £(a - /3) = 9.7730

19.7691

log sin £ (o - 6) = 9.8055 (n)

log tan £7 = 9.9636

£7 = 42° 36'.

7 = 85° 12'.

.-. = 180° - 7 = 94° 48'.

Cftecfc : log sin a = 9.9281 log sin 6 = 9.8308 log sin c = 9.9537
logsin^. = 9.9730 log sin B = 9.8756 logsinC = 9.E
9.9651

% (a + 6) = 97° 39'
| (a -ft) = - 39° 43'

Adding,

Subtracting,

a = 67° 56'
6 = 137° 22'.

9.9562

9.9552

* Same as those used in Case II, (a), p. 219, with Greek and Roman letters interchanged,
t Since A < B it follows that a < b, and } (a - 6) is negative.

OBLIQUE SPHERICAL TRIANGLES

223

If C only is wanted, we can calculate it without previously
determining a and b, by dividing the given triangle into two right
spherical triangles, as was illustrated on
p. 220.

Through B draw an are of a great
circle perpendicular to AC, intersecting
AC at D. Let BD=p, angle ABB = x,
angle CBD = y. Applying Rule I of
Napier's rules, p. 200, to the right spher-
ical triangle ABD, we have

cos o = cot x cot A, or,

(A) cot x = tan A cos c.

Applying Rule II, p. 200, to ABD, we have
cos A = oosp sin x, or,

(B) cosp = cos A esc x.

Applying the same rule to CBD,

cos C = cosp sin y, or,

(C) cosp = cos C esc y.

Equating (B) and (C),

cos C esc y = cos A esc a;, or,
cos C = cos vl esc x sin ?/.

But y = B — x ; therefore

(D) cos C = cos A esc a; sin (B — x).

Now C may be computed from (A) and (D), namely,
(50) cot x = tan .4 cos c.

cos 4 sin (B — x)

(51)

cos C = ■

sin x

Ex. 2. Given ^. = 35° 46', B = 115° 9', c = 51° 2' ; find 0.
Solution. Apply the method just explained.

To find B- x use (50)

log tan .4 = 9.8575

log cose = 9. 7986

log cot a; = 9.6561

x = 65° 38'.

.-. B - x = 49° 31'.

To find C use (51)

log cos A = 9.9093

log sin (B - x) = 9.8811

19.7904

log sin x = 9. 9595

log cos C= 9.8309

C = 47°21'.

224

SPHERICAL TEIGONOMETEY

EXAMPLES

Solve the following oblique spherical triangles :

No.

Given Parts

Required Parts

1

.4=67° 30'

S = 46° 50'

c=74°20'

a = 63° 15' 6 = 43° 53' C = 95°l'

2

B = 98° 30'

(7 = 67° 20'

a = 60° 40'

6 = 86° 40' c = 68°40' ^=59° 44'

3

C=110°

^1=94°

6 = 44°

a = 1 14° 10' c = 120° 46' B = 49° 34'

4

C = 70°20'

B = 43° 50'

o = 50° 46'

6 = 32° 59' c = 47°45' ^. = 80° 14'

5

A =78°

B = 41°

c = 108°

a = 95°38' 6 = 41°52' C = 110°49'

6

£ = 135°

C = 50°

a = 70° 20'

6 = 120° 16' c = 69° 20' ^.= 50° 26'

7

A =31° 40'

C=122°20'

6 = 40° 40'

a = 34° 3' c=64°19' .8 = 37° 40'

8

A = 108° 12'

£ = 145° 46'

c =126° 32'

a = 69° 5' 6 =146° 25' C = 125°12'

9

A = 130° 36'

B = 30° 26'

c = 40° 35'

a =71° 15' 6 = 39° 10' C = 31°26'

10

.4=51° 58'

E = 83° 54'

c = 42°

a = 38° 6=51° C=58°53'

20. Case III. (a) Given two sides and the angle opposite one of them,
as a, b, B (ambiguous case *).

From the law of sines, p. 207, we get

sin a sin B

(11)

sin 4 :

sin b

which gives A^. To find C we use, from p. 216, formula (46), solved
for tan \ y, namely,

(46)

tan | y = —

sin|(a + ft)tan§(q-ff)

sin I (a — 6)
To find c, solve (44), p. 215, for tan £ c, namely,

(44)

Chech by the law of sines.

tnn i r _ sinj(a + fltan|( a -&)
sin§(a-£)

Ex. 1. Given a = 58°, 6 = 137° 20', B = 131° 20'; find A, C, c.
Solution.

To find A use (11)

log sin a = 9.9284

log sin B = 9.8756

19.8040

log sin 6 = 9.8311
log sin A= 9.9729

. . A 1 = 69° 58',
A 2 = 180° - Ai = 110° 2'.

a =58°
6 = 137° 20'

a =58°
6= 137° 20'

o + 6=195°20' o-6=-79°20'
\ (o+ 6) = 97° 40'. \ (a -6) =-39° 40'.

= 180° -.8 = 48° 40'.

Since a<b and both Ai and A 2
are<JB, it follows that we have two
solutions.

* Just as in the corresponding case in the solution of plane oblique triangles (Granville's
Plane Trigonometry, pp. 105 t 161), there may be two solutions, one solution, or no solution,
depending on the given data.

t Since the angle A is here determined from its sine, it is necessary to consider both of
the values found. If a>b then A>B; and if a<b then^<£. Hence [next page]

OBLIQUE SPHEEICAL TRIANGLES

225

First solution, a x = 180° - A-y = 110° 2'.

a x = 110° 2'
P= 48° 40'
a x + (8 = 158° 42'

£(ai + |8)=79°21'.

To^nd d use (46)

log sin £ (a + 6) = 9.9961

log tan \ {a x -p) = 9.7733

19.7694

log sin t(a-b)= 9.8050 (n)

log tan ^7i = 9.9644

£71 = 42° 39'.

71 = 85° 18'.

.-. d = 180° - 7l = 94° 42'.

Cftecfc : log sin a = 9. 9284 log sin 6
log sin A x — 9.9729 log sin B
9.9555

<*! = 110° 2'
j3= 48° 40'
ai-/3= 61° 22'
£(ai-/?)=30°41'.

To find ci use (44)

log sin £ (orj + p) = 9.9924
log tan I (a - 6) = 9.9187 (n)

19.9111

log sin £ (a! - |3) = 9.7078

log tan £c x = 10.2033

•Jci = 57°57'-
.-. Ci = 115° 54'.

= 9.8311

= 9.8756

9.9555

log sin ci = 9.9541

log sin Ci = 9.9985

9.9556

Second solution. a 2 = 180° — X 2 = 69 e 58'.

or 2 = 69° 58'

|8= 48° 40'

Q-2 + P = 118° 38'

£(a 2 + ,3) = 59°19'.

To find C 2 wse (46)

logsin£(a + &) = 9.9961

log tan \ (or 2 - p) = 9.2743

19.2704

log sin £ (a - 6) = 9.8050 (m)

log tan £72= 9.4654

■£-72 = 16° 17'.

7 2 = 32° 34'.

.-. C 2 = 180° - 72 = 147° 26'.

Check : log sin a = 9.9284 log sin 6
log sin _4 2 = 9.9729 log sin B
9.9555

a 2 = 69° 58'

p = 48° 40'

a 2 - p = 21° 18'

£(a 2 -/3) = 10°39'.

To find C\ use (44)

logsin£(a- 2 + (3) = 9.9345
log tan £ (a - 6) = 9.9187 (n)

19.8532

log sin £ (ar 2 - p) - 9.2667

log tan \ c 2 = 10.5865

£c 2 = 75°28'.

. . c 2 = 150° 56'.

: 9. 831 1 log sin c 2 = 9. 6865

: 9.8756 log sin C 2 = 9. 7310

9.9555 9.9555

If the side c or the angle C is wanted without first calculating the
value of A, we may resolve the given triangle into two right triangles
and then apply Napier's rules, as was illustrated under Cases II, (a),
and II, (b), pp. 220, 223.

Theorem. Only those values of A should be retained which, are greater or less than B
according as a is greater or less than b.

If log sin A = a positive number, there will be no solution.

226

SPHERICAL TRIGONOMETRY

EXAMPLES

Solve the following oblique spherical triangles :

No.

Given Parts

Required Pakts

1

a =43° 20'

6=48° 30' 4 = 58° 40'

B!=68°46' C 1= 70°46'
£ 2 =111°14' C 2 = 14°29'

ci=49°16'
c 2 = ll°36'

2

a =56° 40'

6=30° 50' 4 = 103° 40'

£=36° 35' C=52°

c=42°39'

3

a =30° 20'

6=46° 30' A =36° 40'

£i = 59°4' Ci = 97°39'
B2 = 120°56' C 2 =28°5'

C! = 56°57'
c 2 =23°28'

4

6=99° 40'

c=64°20' £=95° 40'

C=65°30' 4 = 97°20'

a=100°45'

5

a =40°

6=118° 20' 4 = 29° 40'

£ 1= 42°40' Ci = 159°54'
£ 2 = 137°20' C 2 =50°21'

Ci = 153°30'
c 2 =90°10'

6

a = 115°20'

c = 146°20' C=141°10'

Impossible

7

a=109°20'

c=82° A = 107° 40'

C=90° £=113°35'

6=114° 50'

8

6=108° 30'

c=40°50' C=39°50'

£i=68°18' 4i = 132°34'
£ 2 = 111°42'4 2 = 77°5'

ai=131°16'
o 2 =95° 50'

9

a =162° 20'

6 = 15° 40' £=125°

Impossible

10

a =65°

c = 138°10'^=42°30'

0=146° 38' £=55° 1'

6=96° 34'

21. Case III. (6) Given two angles and the side opposite one' of them,
as A, B, b (ambiguous case *).

From the law of sines, p. 207, we get

BVO.A sin b

(11)

sin a =

sinB

which gives a.| To find c we use, from p. 215, the formula % (44),
solved for tan £ o, namely,

tan 1 ^ Bin§(a + ff)tan§(q-&)
2 sm\{a-P)

(44)

To find 0, solve (46), p. 216, for tan\ y, namely,

, , , sin§(a + &)tan§(a — 8)

(46) tan|y = aV TV ^ ^.

v J * * sin|(a — b)

Check by the law of sines.

* Just as in Case II, (&), we may have two solutions, one solution, or no solution, depending
on the given data.

t Since the side is here determined from its sine, it is necessary to examine both of the
values found. If A > B then a > b ; and if A < B then a < b. Hence we have the

Theorem. Only those values of a should be retained which are greater or less than b
according as A is greater or less than B.

If log sin a = a positive number, there will be no solution.

% Same as those used in Case III, (a), p. 224, when the Greek and Roman letters are
interchanged.

OBLIQUE SPHERICAL TRIANGLES

227

Ex. 1. Given A = 110°, B = 131° 20', 6 = 137° 20' ; find a, c, C.
Solution, a = 180° -A = 70°, and /3 = 180°-B = 48° 40'.

To find a use (11)

log sin A= 9.9730
log sin 6 = 9.8311
19.8041
log sin B= 9.8756
log sin a = 9.9285

.-. ai = 58° 1',
a! = 121° 59'.

. = 180° ■

a = 70°
,8= 48° 40'
-/3= "

a = 10°
j3 = 48°40'

First solution.

(i!= 58° 1'
6 = 137° 20'

a! + 6 = 195° 21'
|(a 1 + 6) = 97°41'.

To find Ci use (44)

log sin £ (a + j3) = 9. 9346
log tan \ (en - 6) = 9.9187 (n)

19.8533

log sin \ (a - |3) = 9. 2674

log tan £ci = 10. 5859

£ci = 75°27'.
.-. ci = 150° 54'.

Cheek: log sin ai = 9.9285 log sin 6 = 9.8311 log sin ci = 9.6

log sin A = 9.9730 log sin B = 9.8756 log sin Ci = 9. 7314
9.9555

a + /3 = 118°40' <x-/3 = 21°20'
b (a + 0) = 59° 20'. \ {a - p) = 10° 40'.

Since A<B and both ai and a 2
are < 6, it follows that we have two
solutions.

ai= 58° V
6 = 137° 20'
ai - 6 = - 79° 19"
£(ai-6) = -39°40'.

Toured Ci use (46)

logsin£(ai + 6) = 9.9961

log tan % (a - j3) = 9.2750

19.2711

log sin £ (ai - 6) = 9.8050 (n)

log tan £ 7i= 9.4661

£71 = 16° 18'.
71 = 32° 36'.
.-. Ci = 180° -71 = 147° 24'.

9.9555

9.9555

Second solution. This gives c 2 = 64° 8', and C 2 = 85° 18'.

Remembering that a 2 = 121° 59', we may now check the second solution.

Check : log sin a 2 = 9. 9285 log sin 6 = 9. 8311 log sin c 2 = 9. 9542

log sin A = 9.9730 log sin B = 9.8756 log sin C 2 = 9.9985

9.9555 9.9555 9.9557

Hence the two solutions are

ai = 58° 1' ci = 150° 54' Ci = 147° 23',

and

a 2 = 121°59' c 2 = 64°8' C 2 = 85° 18'.

If the angle C or the side is wanted without first computing a,
we may resolve the given triangle into two right triangles and then
apply Napier's rules, as was illustrated under Cases II, (a), and
II, (b), pp. 220, 223.

228

SPHERICAL TRIGONOMETRY

EXAMPLES

Solve the following oblique spherical triangles :

No.

Given Parts

Required Parts

1

4 = 108° 40'

.8=134° 20'

a =145° 36'

6 = 154° 45' c=34°9'

C=70°18'

2

£=116°

C=80°

c=84°

6 = 114° 50' 4 = 79° 20'

a =82° 56'

3

4 = 132°

£=140°

6=127°

Oi=67° 24' d = 164°6'
a 2 =112°36'C 2 = 128°21'

Ci = 160°6'
c 2 =103°2'

4

4=62°

C=102°

a =64° 30'

c=90° £=63° 43'

6=66° 26'

5

4 = 133° 50'

£=66° 30'

a = 81° 10'

Impossible

6

5=22° 20'

C = 146°40'

e = 138°20'

6=27° 22' 4 = 47° 21'

a = 117°9'

7

4 = 61° 40'

C=140°20'

c=150°20'

a 1= 43°3' £ 1 = 89°24'
a 2 = 136°57'£ 2 =26°59'

6! = 129°8'
6 2 =20°36'

8

£=73°

C=81°20'

6= 122° 40'

Impossible

9

£=36° 20'

C=46°30'

6=42° 12'

4 1 = 164°44' <i 1 = 162°88'
4 2 =119°17' a 2 = 81°17'

C! = 124°41'
c 2 =55°19'

10

4 = 110° 10'

£=133° 18'

a=147°6'

6=155° 5' c=32°59'

C=70°16'

22. Length of an arc of a circle in linear units. From Geometry we
know that the length of an are of a circle is to the circumference of

the circle as the number of degrees in the
arc is to 360. That is

L:2itR::N: 360, or,
VRN

Is — ■ ' ' '-' *

180

L = length of arc,

N = number of degrees in arc,

R = length of radius.

In case the length of the arc is given
to find the number of degrees in it, we instead solve for N, giving

(53)

N =

180 L
TTR

Considering the earth as a sphere, an arc of one minute on a great
circle is called a geographical mile or a nautical mile.* Hence there
are 60 nautical miles in an arc of 1 degree, and 360 x 60 = 21,600
nautical miles in the circumference of a great circle of the earth. If
we assume the radius of the earth to be 3960 statute miles, the length

* In connection with a ship's rate of sailing a nautical mile is also called a knot.

OBLIQUE SPHERICAL TRIANGLES 229

of a nautical mile (= 1 min. = 5 \ of a degree) in statute miles will
be, from (52),

3.1416 x 3960 x A 11S .
L- m = 1.15 mi.

Ex. 1. Eind the length of an arc of 22° 30' in a circle of radius 4 in.
Solution. Here N = 22° .30' = 22. 5°, and B = 4 in.

Substituting in (52), L = 3 - 1416 x 4 x 22.5 = ^ .^ ^

180

Ex. 2. A ship has sailed on a great circle for 5J hr. at the rate of 12 statute
miles an hour. How many degrees are there in the arc passed over ?
Solution. Here L = 5£ x 12 = 66 mi. , and B = 3960 mi.

Substituting in (53), N = 18 ° X 66 = .955° = 57.3'. Ans.

23. Area of a spherical triangle. From Spherical Geometry we
know- that the area of a spherical triangle is to the area of the sur-
face of the sphere as the number of degrees in its spherical excess *
is to 120. That is,

Area of triangle : 4 ttR 2 :: E : 720, or,

TTIPE

(54) Area of spherical triangle :

180

In case the three angles of the triangle are not given, we should first
find them by solving the triangle. Or, if the three sides of the tri-
angle are given, we may find E directly by Lhuilier's formula, f namely,

(55) tan \E = Vtan | s tan § (s — c )tan§(s — 6)tan|(s— c),

where a, h, c denote the sides and s = \ (a + b + c).

The area of a spherical polygon will evidently be the sum of the
areas of the spherical triangles formed by drawing arcs of great cir-
cles as diagonals of the polygon.

Ex. 1. The angles of a, spherical triangle on a, sphere of 25-in. radius are
A = 74° 40', B = 67° 30', C = 49° 50'. Eind the area of the triangle.
Solution. Here E = (A + B+C)- 180° = 12°.

c w + *• • /ka\ a 3.1416 x (25)2 x 12 19no . .
Substituting in (54), Area = — = 130.9 sq. in. Ans.

* The spherical excess (usually denoted by E) of a spherical triangle is the excess of the
sum of the angles of the triangle over 180°. Thus, if A, B, and C are the angles of a spher-
ical triangle,

E=A + B+C-1W>.

t Derived in more advanced treatises.

230 SPHERICAL TRIGONOMETRY

EXAMPLES

1. Find the length of an arc of 5° 12' in a circle whose radius is 2 ft. 6 in.

Ans. 2.72 in.

2. Find the length of an arc of 76° 30' in a circle whose radius is 10 yd.

Ans. 13.17 yd.

3. How many degrees are there in a circular arc 15 in. long, if the radius
is 6 in.? Ans. 143° 18'.

4. A ship sailed over an arc of 4 degrees on a great circle of the earth each
day. At what rate was the ship sailing ? Ans. 11.515 mi. per hour.

5. Find the perimeter in inches of a spherical triangle of sides 48°, 126°, 80°,
on a sphere of radius 25 in. Ans. 110.78 in.

6. The course of the boats in a yacht race was in the form of a triangle
having sides of length 24 mi., 20 mi., 18 mi. If we assume that they sailed on
arcs of great circles, how many minutes of arc did they describe ?

Ans. 53.85 min.

7. The angles of a spherical triangle are A = 63°, B = 84° 21', C = 79°; the
radius of the sphere is 10 in. What is the area of the triangle ?

Ans. 80.88'sq. in.

8. The sides of a spherical triangle are a — 6.47 in. , 6 = 8.39 in. , c = 9.43 in. ;

the radius of the sphere is 25 in. What is the area of the triangle ?

Ans. 26.9 sq. in.
Hint. Find E by using formula (55)'

9. In a spherical triangle A = 75° 16', B = 39° 20', c = 26 ft. ; the radius of
the sphere is 14 ft. Find the area of the triangle. Ans. 158.45 sq. ft.

10. Two ships leave Boston at the same time. One sails east 441 mi. and the
other 287 mi. E. 38° 21' N. the first day. If we assume that each ship sailed
on an arc of a great circle, what is the area of the spherical triangle whose ver-
tices are at Boston and at the positions of the ships at the end of the day ?

Ans. 41,050 sq. mi.

11. A steamboat traveling at the rate of 15 knots per hour skirts the entire
shore line of an island having the approximate shape of an equilateral triangle
in 18 hr. What is the approximate area of the island ? Ans. 4651.1 sq. mi.

12. Find the areas of the following spherical triangles, having given

(a) a = 47° 30', 6 = 55° 40', c = 60° 10'; B = 10 ft. Ans. 42. 96 sq. ft.

(b) a = 43° 30', 6 = 72° 24', c = 87° 50'; R = 10 in. 59.21 sq. in.

(c) A = 74° 40', B = 67° 30', C = 49° 50'; B = 100 yd. 2094 sq. yd.

(d) A = 112° 30', B = 83° 40', C = 70° 10'; B = 25 cm. 941.4 sq. cm.

(e) a = 64° 20', b = 42° 30', C = 50° 40'; B = 12 ft. 46.74 sq. ft.

(f) C = 110°, A = 94°, 6 = 44° ; B = 40 rd. 2056.5 sq. rd

(g) a = 43° 20', b = 48° 30', A = 58° 40'; B = 100 rd. 19.76 acres,
(h) A = 108° 40', B = 134° 20', o = 145° 36'; B = 3960 mi. 36,466,667 sq. mi.

CHAPTEE III

APPLICATIONS OF SPHERICAL TRIGONOMETRY TO THE CELESTIAL
AND TERRESTRIAL SPHERES

24. Geographical terms. In what follows we shall assume the earth
to be a sphere of radius 3960 statute miles.

The meridian of a place on the earth is that great circle of the
earth which passes through the place and the north and south poles.

(North pole)

N

W
(West)

S
(South pole)

Thus, in the figure representing the earth, NGS is the meridian of
Greenwich, NBS is the meridian of Boston, and NCS is the merid-
ian of Cape Town.

The latitude of a place is the arc of the meridian of the place ex-
tending from the equator to the place. Latitude is measured north
or south of the equator from 0° to 90°. Thus, in the figure, the
arc LB measures the north latitude of Boston, and the arc TC
measures the south latitude of Cape Town.

The longitude of a place is the arc of the equator extending
from the zero meridian * to the meridian of the place. Longitude is

* As in this case, the zero meridian, or reference meridian, is usually the meridian pass-
ing through Greenwich, near London. The meridians of Washington and Paris are also
used as reference meridians.

231

232

SPHERICAL TRIGONOMETRY

measured east or west from the Greenwich meridian from 0° to 180°.
Thus, in the figure, the are MT measures the east longitude of Cape
Town, while the arc ML measures the west longitude of Boston.
Since the arcs MT and ML are the measures of the angles MNT and
MNL respectively, it is evident that we can also define the longitude
of a place as the angle between the reference meridian and the
meridian of the place. Thus, in the figure, the angle MNT is the
east longitude of Cape Town, while the angle MNL is the west
longitude of Boston.

The bearing of one place from a second place is the angle between
the arc of a great circle drawn from the second place to the first
place, and the meridian of the second place. Thus, in the figure, the
bearing of Cape Town from Boston is measured by the angle CBN
or the angle CBL, while the bearing of Boston from Cape Town is
measured by the angle NCB or the angle SCB.*

25. Distances between points on the surface of the earth. Since we
know from Geometry that the shortest distance on the surface of a

(_~North pole)
N

(South pole)

sphere between any two points on that surface is the arc, not greater
than a semicircumference, of the great circle that joins them, it is
evident that the shortest distance between two places on the earth
is measured in the same way. Thus, in the figure, the shortest

* The bearing or course of a ship at any point is the angle the path of the ship makes
with the meridian at that point.

APPLICATIONS OF SPHERICAL TRIGONOMETRY 233

distance between Boston and Cape Town is measured on the arc
BC of a great circle. We observe that this arc BC is one side of
a spherical triangle of which the two other sides are the arcs BN
and CN. Since

arc BN = 90°- arc LB = 90°- north latitude of Boston,
arc CN = 90° + arc TC = 90° + south latitude of Cape Town,

and angle BNC = angle MNL + angle MNT
= west longitude of Boston

+ east longitude of Cape Town
= difference in longitude of Boston and Cape Town,

it is evident that if we know the latitudes and longitudes of Boston
and Cape Town, we have all the data necessary for determining two
sides and the included angle of the triangle BNC. The third side
BC, which is the shortest distance between Boston and Cape Town,
may then be found as in Case II, (a), p. 219.

In what follows, north latitude will be given the sign + and south
latitude the sign — .

Rule for finding the shortest distance between two points on the earth
and the bearing of each from the other, the latitude and longitude of each
point being given.

First step. Subtract the latitude of each place from 90°.* The
results will be the two sides of a spherical triangle.

Second step. Find the difference of longitude of the two places by
subtracting the lesser longitude from the greater if both are E. or both
are W., but add the two if one is E. and the other is W. This gives
the included angle of the triangle.^

Third step. Solving the triangle by Case II, (a), p. 219, the third
side gives the shortest distance between the two points in degrees of
arc,% and the angles give the bearings.

* Note that this is algebraic subtraction. Thus, if the two latitudes were 25° N. and
42° S., we would get as the two sides of the triangle,

90" -25° =65° and 90° -(-42°) =90° + 42° =132°.

t If the difference of longitude found is greater than 180°, we should subtract it from
360° and use the remainder as the included angle.

t The number of minutes in this arc will be the distance between the two places in geo-
graphical (nautical) miles. The distance between the two places in statute miles is given
by the formula 3.1416 x 3960 x iv"

Mi= ' I

180
where N= the number of degrees in the arc.

234 SPHERICAL TRIGONOMETRY

Ex. 1. Find the shortest distance along the earth's surface between Boston
(lat. 42°21'N., long. 71° 4' W.) and Cape Town (lat. 33° 56' S., long. 18° 26' E.),
and the bearing of each city from the other.

Solution. Draw a spherical triangle in agreement with
the figure on p. 232.

First step.

c = 90° -42° 21' = 47° 39',

6 = 90° - (- 33° 56') = 123° 56'.

Second step.

N = 71° 4' + 18° 26' = 89° 30' = difference in long.

Third step. Solving the triangle by Case II, (a), p. 219, we get
n = 68° 14' = 68.23° = 4094 nautical miles,
= 52° 43',
and JB = 116°43'.

Hence a ship sailing from Boston to Cape Town on the arc of a great circle
sets out from Boston on a course S. 63° 17' E. and approaches Cape Town on a
course S. 52° 43' E.*

EXAMPLES

1. Find the shortest distance between Baltimore (lat. 39° 17' N., long. 76° 37'
W.) and Cape Town (lat. 33° 56' S., long. 18° 26' E.), and the bearing of each
from the other. Ans. Distance = 180° — 65° 48' = 6852 nautical miles,

S. 64° 58' E. = bearing of Cape Town from Baltimore,
N. 57° 42' W. = bearing of Baltimore from Cape Town.

2. What is the distance from New York (lat. 40°43'N., long. 74° W.) to
Liverpool (lat. 53° 24' N., long. 3° 4' W.)? Find the bearing of each place from the
other. In what latitude will a steamer sailing on a great circle from New York
to Liverpool cross the meridian of 50° W. , and what will be her course at that
point 1 Ans. Distance = 47° 50' = 2870 nautical miles,

N. 75° 7' W. = bearing of New York from Liverpool,

N. 49° 29' E. = bearing of Liverpool from New York.

Lat. 51° 13' N., with course N. 66° 54' E.

3. Find the shortest distance between the following places in geographical
miles :

(a) New York (lat. 40° 43' N. , long. 74° W. ) and San Francisco (lat. 37° 48' N. ,
long. 122° 28' W.). Ans. 2230.

(b) Sandy Hook (lat. 40° 28' N. , long. 74° 1' W. ) and Madeira (lat. 32° 28' N. ,
long. 16°55'W.). Ans. 2749.

' (c) San Francisco (lat. 37° 48' N., long. 122° 28' W.) and Batavia (lat. 6° 9' S.,

long. 106° 53' E.). Ans. 7516.

(d) San Francisco (lat. 37° 48' N., long. 122° 28' W.) and Valparaiso (lat.

33° 2' S., long. 71° 41' W.) Ans. 5109.

* A ship that sails on a great circle (except on the equator or a meridian) must be con-
tinually changing her course. If the ship in the above example keeps constantly on the
course S. 63° 17' E., she will never reach Cape Town.

APPLICATIONS OF SPHERICAL TRIGONOMETRY 235

4. Find the shortest distance in statute miles (taking diameter of earth as
7912 mi.) between Boston (lat. 42° 21' N., long. 71° 4' W.) and Greenwich (lat.
51° 29' N.), and the bearing of each place from the other.

Ans. Distance = 3275 mi. ,

K 53° 7' E. = bearing of Greenwich from Boston,
N. 71° 39' W. = bearing of Boston from Greenwich.

5. As in last example, find the shortest distance between and bearings for
Calcutta (lat. 22° 33' N, long. 88° 19' E.) and Valparaiso (lat. 33° 2' S., long.
71°42'"W.). Ans. Distance = 11,012.5 mi.,

S. 64° 20.5' E. = bearing of Calcutta from Valparaiso,
S. 54° 54. 5' W. = bearing of Valparaiso from Calcutta.

6. Find the shortest distance in statute miles from Oberlin (long. 82° 14' W.)
to New Haven (long. 72° 55' W.), the latitude of each place being 41° 17' N.

Ans. 483.3 mi.

7. From a point whose latitude is 17° N. and longitude 130° W. a ship sailed
an arc of a great circle over a distance of 4150 statute miles, starting S. 54°20' W.
Find its latitude and longitude, if the length of 1° is 69£ statute miles.

Ans. Lat. 19° 42' S., long. 178° 21' W.

26. Astronomical problems. One of the most important applications
of Spherical Trigonometry is to Astronomy. In fact, Trigonometry
was first developed by astronomers, and for centuries was studied
only in connection with Astronomy. We shall take up the study of
a few simple problems in Astronomy.

27. The celestial sphere. When there are no clouds to obstruct
the view, the sky appears like a great hemispherical vault, with the
observer at the center. The stars seem to glide upon the inner sur-
face of this sphere from east to west,* their paths being parallel cir-
cles whose planes are perpendicular to the polar axis of the earth,
and having their centers in that axis produced. Each star | makes
a complete revolution, called its diurnal (daily) motion, in 23 hr.
56 min., ordinary clock time. We cannot estimate the distance of
the surface of this sphere from us, further than to perceive that it
must be very far away indeed, because it lies beyond even the
remotest terrestrial objects. To an observer the stars all seem to be
at the same enormous distance from him, since his eyes can judge
their directions only and not their distances. It is therefore natural,
and it is extremely convenient from a mathematical point of view,
to regard this imaginary sphere on which all the heavenly bodies
seem to be projected, as having a radius of unlimited length. This

* This apparent turning of the sky from east to west is in reality due to the rotation of
the earth in the opposite direction, just as to a person on a swiftly moving train the objects
outside seem to be speeding by, while the train appears to be at rest. The sky is really mo-
tionless, while the earth is rotating from west to east.

t By stars we shall mean fixed stars and nebula? whose relative positions vary so slightly
that it takes centuries to make the change perceptible.

236

SPHERICAL TRIGONOMETRY

sphere, called the celestial sphere, is conceived of as having such
enormous proportions that the whole solar system (sun, earth, and
planets) lies at its center, like a few particles of dust at the center of
a great spherical balloon. The stars seem to retain the same relative
positions with respect to each other, being in this respect like places
on the earth's surface. As viewed from the earth, the sun, moon,
planets, and comets are also projected on the celestial sphere, but
they are changing their apparent positions with respect to the stars
and with respect to each other. Thus, the sun appears to move east-
ward with respect to the stars about one degree each day, while the
moon moves about thirteen times as far.

The following figure represents the celestial sphere, with the
earth at the center showing as a mere dot.

( North

celestial "i

•pole)

(North
point AT ; —
of 1VK
horizon)

( South
S point

of
horizon)

'p' (South
celestial
pole)

Z'

(Nadir)

The zenith of an observer is the point on the celestial sphere
directly overhead. A plumb line held by the observer and extended
upwards will pierce the celestial sphere at his zenith {Z in figure).

The nadir is the point on the celestial sphere which is diametric-
ally opposite to the zenith (Z' in the figure).

The horizon of an observer is the great circle on the celestial
sphere having the observer's zenith for a pole ; hence every point
on the horizon (SWNE in the figure) will be 90° from the zenith
and from the nadir. A plane tangent* to a surface of still water

* On account of the great distance, a plane passed tangent to the earth at the place of
the observer will cut the celestial sphere in a great circle which (as far as we are concerned)
coincides with the observer's horizon.

APPLICATIONS OF SPHERICAL TRIGONOMETRY 237

at the place of the observer will cut the celestial sphere in his
horizon.

All points on the earth's surface have different zeniths and horizons.

Every great circle passing through the zenith will be perpendicular
to the horizon ; such circles are called vertical circles (as ZMHZ' and
ZQSP'Z' in figure).

The celestial equator or equinoctial is the great circle in which the
plane of the earth's equator cuts the celestial sphere (EQWQ' in
the figure).

The poles of the celestial equator are the points (P and P' in the
figure) where the earth's axis, if produced, would pierce the celestial
sphere. The poles may also be defined as those two points on the
sky where a star would have no diurnal (daily) motion. The Pole
Star is near the north celestial pole, being about 1|° from it. Every
point on the celestial equator is 90° from each of the celestial poles.

All points on the earth's surface have the same celestial equator
and poles.

The geographical meridian of a place on the earth was defined
as that great circle of the earth which passes through the place and
the north and south poles. The celestial meridian of a point on the
earth's surface is the great circle in which the plane, of the point's geo-
graphical meridian cuts the celestial sphere (ZQSP'Z'Q'NP in the
figure). It is evidently that vertical circle of an observer which passes
through the north and south points of his horizon. All points on the
surface of the earth which do not lie on the same north-and-south
line have different celestial meridians.

The hour circle of a heavenly body is that great circle of the celes-
tial sphere which passes through the body * and through the north
and south celestial poles. In the figure PMDP' is the hour circle of
the star M. The hour circles of all the heavenly bodies are contin-
ually changing with respect to any observer.

The spherical triangle PZM, having the north pole, the zenith, and
a heavenly body at its three vertices, is a very important triangle in
Astronomy. It is called the astronomical triangle.

28. Spherical coordinates. When learning how to draw (or plot) the
graph of a function, the student has been taught how to locate a
point in a plane by measuring its distances from two fixed and mutu-
ally perpendicular lines called the axes of coordinates, the two dis-
tances being called the rectangular coordinates of the point.

« By this is meant that the hour circle passes through that point on the celestial sphere
where we see the heavenly body projected.

238 SPHERICAL TRIGONOMETRY

If we now consider the surface to be spherical instead of plane, a
similar system of locating points on it may be employed, two fixed
and mutually perpendicular great circles being chosen as reference
circles, and the angular distances of a point from these reference
circles being used as the spherical coordinates of the point. Since
the reference circles are perpendicular to each other, each one of
them passes through the poles of the other.

In his study of Geography the student has already employed such
a system for locating points on the earth's surface, for the latitude
and longitude of a point on the earth are really the spherical coordi-
nates of the point, the two reference circles being the equator and
the zero meridian (usually the meridian of Greenwich). Thus, in the
figure on p. 231, we may consider the spherical coordinates of Boston
to be the arcs ML (west longitude) and LB (north latitude) ; and of
Cape Town the spherical coordinates would be the arcs M T (east
longitude) and TC (south latitude). Similarly, we have systems of
spherical coordinates for determining the position of a point on the

celestial sphere, and

(Zenith) e

Z we shall now take up

^^ ' ^ := \~-^^ the study of the more

/A l \ ,-'" ]\ nnportant of these.

J J i / \ // \ 29. The horizon and

( North „ K/ I i / \ I \

pole) r /t:JI I /Z \ / \ meridian system. In

i /b "\ j /f V M \ this case the two fixed

£/ _. -"-^;rTi£^-----/4-"-' ! -- \ and mutll ally perpen-

J^l ^:<lK a r<!L.jy-..T ,~"Jo dicular great circles of

. \ / _L_^ — reference are the hori-

Horizon w Ml H zon of tne observer

( Sunset)

(SHWNE) and his
meridian (SM 2 ZPN), and the spherical coordinates of a heavenly
body are its altitude and azimuth.

The altitude of a heavenly body is its angular distance above the
■ horizon measured on a vertical circle from 0° to 90°.* Thus the
altitude of the sun M is the arc HM. The distance of a heavenly
body from the zenith is called its zenith distance {ZM in the figure),
and it is evidently the complement of its altitude. The altitude of
the zenith is 90°. The altitude of the sun at sunrise or sunset is zero.

The azimuth of a heavenly body is the angle between its vertical
circle and the meridian of the observer. This angle is usually

* At sea the altitude is usually measured by the sextant, while on laud a surveyor's
transit is used.

APPLICATIONS OF SPHERICAL TRIGONOMETRY 239

measured along the horizon from the south point westward to the
foot of the body's vertical circle.* Thus the azimuth of the sun M
' is the angle SZH, which is measured by the arc SH. The azimuth
of the sun at noon is zero and at midnight 180°. The azimuth of
a star directly west of an observer is 90°, of one north 180°, and of
one east 270°.

Knowing the azimuth and altitude (spherical coordinates) of a
heavenly, body, we can locate it on the celestial sphere as follows.
Erom the south point of the horizon, as S (which may be considered
the origin of coordinates, since it is an intersection of the reference
circles), lay off the azimuth, as SH. Then on the vertical circle
passing through H lay off the altitude, as HM. The body is then
located at M.

Ex. 1. In each of the following examples draw a figure of the celestial sphere
and locate the body from the given spherical coordinates.

Azimuth

A Ititude

Azimuth

Altitui

(a)

45°

45°

(J)

0°

0°

0»)

60°

30°

(k)

180°

0°

(o)

90°

60°

(1)

0°

90°

(d)

120°

75°

(m)

90°

0°

(e)

180°

55°

(n)

270°

0°

(f)

225°

0°

(o)

360°

0°

(g)

300°

60°

(P)

330°

45°

(a)

315°

15°

(q)

75°

75°

(i)

178°

82°

W

90°

90°

Since any two places on the earth have, in general, different merid-
ians and different horizons, it is evident that this system of spher-
ical coordinates is purely local. The sun rises at M x on the eastern
horizon (altitude zero), mounts higher and higher in the sky, on a
circle (M X M 2 M^) parallel to the celestial equator, until it reaches the
observer's meridian M 2 (at noon, when its altitude is a maximum),
then sinks downward to M z and sets on the western horizon.

Similarly, for any other heavenly body, so that all are continually
changing their altitudes and azimuths. To an observer having the
zenith shown in the figure, a star in the northern sky near the north
pole will not set at all, and to the same observer a star near the south
pole will not rise at all. If its path for one day were traced on the
celestial sphere, it would be a circle (as ABC) with its center in the
polar axis and lying in a plane parallel to the plane of the equator.

* That is, azimuth is measured from 0° to 360° clockwise.

240

SPHERICAL TRIGONOMETRY

• -, s

30. The equator and meridian system. In this case the two fixed
and mutually perpendicular great circles of reference are the celestial
equator (EQDWQ 1 ) and the meridian of the observer (NPZQSP'Z'Q 1 ) ;
and the spherical coordinates of a heavenly body are its declination
and hour angle.

The declination of a heavenly body is its angular distance north or
south of the celestial equator measured on the hour circle of the

body from 0° to 90°.*
Thus, in the figure, the
arc DM is a measure of
the north declination
of the star M. North
declination is always
considered positive and
south declination nega-
tive. Hence the decli-
nation of the north
pole is + 90°, while
that of the south pole
is - 90°.

The declinations of
the sun, moon, and
planets are continually
changing, but the dec-
lination of a fixed star changes by an exceedingly small amount
in the course of a year. The angular distance of a heavenly body
from the north celestial pole, measured on the hour circle of
the body, is called its north polar distance (PM in figure). The
north polar distance of a star is evidently the complement of its
declination.

The hour angle of a heavenly body is the angle between the merid-
ian of the observer and the hour circle of the star measured 'west-
ward, from the meridian from 0° to 360°. Thus, in the figure, the
hour angle of the star M is the angle QPD (measured by the arc
QD). This angle is commonly used as a measure of time, hence the
name hour angle. Thus the star M makes a complete circuit in
24 hours ; that is, the hour angle QPD continually increases at the
uniform rate of 360° in 24 hours, or 15° an hour. For this reason
the hour angle of a heavenly body is usually reckoned in hours from

* The declinations of the sun, moon, planets, and some of the fixed stars, for any time of
the year, are given in the Nautical Almanac or American Epkemeris, published by the United
States government.

APPLICATIONS OP SPHEEICAL TRIGONOMETRY 241

to 24, one hour being equal to 15°.* When the star is at M x (on
the observer's meridian) its hour angle is zero. Then the hour
angle increases until it becomes the angle M V PM (when the star is
at M). When the star sets on the western horizon its hour angle
becomes M^PM* Twelve hours after the star is at M 1 it will be at
M 3 , when its hour angle will be 180° (= 12 hours). Continuing on
its circuit, the star rises at M 4 and finally reaches M 1} when its hour
angle has become 360° (= 24 hours), or 0° again.

Knowing the hour angle and declination (spherical coordinates) of
a heavenly body, we can locate it on the celestial sphere as follows.
From the point, as Q, where the reference circles intersect, lay off the
hour angle (or arc), as QD. Then on the hour circle passing through
D lay off the declination, as DM. The body is then located at M.

Ex. 1. In each of the following examples draw a figure of the celestial sphere
and locate the body from the given spherical coordinates.

Hour angle Declination

Hour angle Declination

(a)

45°

N. 30°

U)

00°

S. 45°

(°)

60°

N. 60°

W

0°

0°

(e)

90°

S. 45°

(1)

180°

0°

(d)

120°

S. 30°

(m)

90°

N. 90°

(«)

180°

N. 50°

(n)

270°

0°

(f)

5hr.

N. 75°

(o)

12 hr.

S. 10°

(g)

15 hr.

-25°

(P)

3hr.

+ 80°

(h)

6hr.

+ 79°

(q)

9hr.

-45°

«

Ohr.

-90°

M

20 hr.

+ 60°

31. Practical applications. Among the practical applications of
Astronomy the most important are :

(a) To determine the position of an observer on the surface of the
earth {i.e. his latitude and longitude).

(b) To determine the meridian of a place on the surface of the
earth.

(c) To ascertain the exact tim.e of day at the place of the observer.

(d) To determine the position of a heavenly body.

The first of these, when applied to the determination of the place
of a ship at sea, is the problem to which Astronomy mainly owes its
economic importance. National astronomical observatories have been

* On account of the yearly revolution of the earth ahout the sun, It takes the sun about
4 minutes longer to make the circuit than is required by any particular fixed star. Hence
the solar day is about 4 minutes longer than the sidereal (star) day, but each is divided into
24 hours ; the first giving hours of ordinary clock time, while the second gives sidereal hours,
which are used extensively in astronomical work. When speaking of the sun's hour angle
it shall be understood that it is measured in hours of ordinary clock time, while the hour
angle of a fixed star is measured in sidereal hours. In either case 1 hour = 15°.

242

SPHERICAL TRIGONOMETRY

established, and yearly nautical almanacs are being published by the
principal nations controlling the commerce of the world, in order to
supply the mariner with the data necessary to determine his position
accurately and promptly.

32. Relation between the observer's latitude and the altitude of the
celestial pole. To an observer on the earth's equator (latitude zero)
the pole star is on the horizon ; that is, the altitude of the star is
zero. If the observer is traveling northward, the pole star will grad-
ually rise ; that is, the latitude of the observer and the altitude
of the star are both increasing. Finally, when the observer reaches
the north pole of the earth his latitude and the altitude of the star
have both increased to 90°. The place of the pole in the sky then

depends in some way on the observer's latitude, and we shall now
prove that the altitude of a celestial pole is equal to the latitude of
the observer.

Let be the place of observation, say some place in the northern
hemisphere ; then the angle QCO (or arc QO) measures its north lati-
tude. Produce the earth's axis CP until it pierces the celestial sphere
at the celestial north pole. A line drawn from in the direction (as
OP 2 ) of the celestial north pole will be parallel to CP 1} since the
celestial north pole is at an unlimited distance from the earth (see
§ 27, p. 235). The angle NOP?, measures the altitude of the north
pole. But CO is perpendicular to ON and CQ is perpendicular -to
OP 2 (since it is perpendicular to the parallel line CP-i) ; hence the
angles NOP 2 and QCO are equal, and we find that the altitude of
the pole as observed at is equal to the latitude of 0.

33. To determine the latitude of a place on the surface of the earth.
If we project that part of the celestial sphere which lies above the

APPLICATIONS OF SPHERICAL TRIGONOMETRY 243

Horizon O

( Observer)

horizon on the plane of the observer's celestial meridian, the horizon
will be projected into a line (as NS), and the upper half of the celes-
tial equator will also be projected into a line (as OQ). From the last
section we know that the latitude of the observer equals the altitude
of the elevated celestial pole (arc NP in figure), or, what amounts to
the same thing, equals the angular distance between the zenith and
the celestial equator (arc ZQ in figure). If then the elevated pole
could be seen as a definitely
marked point in the sky, the
observer's latitude would be
found by simply measuring
the angular distance of that
pole above the horizon. But
there are no fixed stars visible
at the exact points where the
polar axis pierces the celestial
sphere, the so-called polar star being about 1^° from the celestial
north pole. Following are some methods for determining the lati-
tude of a place on the surface of the earth.

First method. To determine latitude by observations on circumsolar
stars. The most obvious method is to observe with a suitable instru-
ment the altitude of some star near the pole (so near the pole that
it never .sets ; as, for instance, the star whose path in the sky is
shown as the circle ABC in figure, p. 238) at the moment when it
crosses the meridian above the pole, and again 12 hours later, when
it is once more on the meridian but below the pole. In the first
case its elevation will be the greatest possible ; in the second, the
least possible. The mean of the two observed altitudes is evidently
the latitude of the observer. Thus, in the figure on this page, if
NA is the maximum altitude and NB the minimum altitude of
the star, then

NA + NB

= NP = altitude of pole

= latitude of place of observation.

Ex. 1. The maximum altitude of a star near the pole star was observed to be
54° 16', and 12 hours later its minimum altitude was observed to be 40° 24'.
What is the latitude of the place of observation ?

Solution. 64° 16' + 40° 24' = 94° 40'.

Therefore

94° 40'

= 47° 20' = altitude of north pole

= north latitude of place of observation.

244

SPHERICAL TRIGONOMETRY

Horizon

( Observer )

Second method. To determine latitude from the meridian altitude

of a celestial body whose declination is known. The altitude of a star

M is measured when it is on
z

M(.star) the observer's meridian. If we

subtract this meridian altitude
(arc SM in figure) from 90°,
we get the star's zenith dis-
tance (ZM). In the Nautical
Almanac we now look up the
star's declination at the same
instant ; this gives us the arc
QM. Adding the declination of the star to its zenith distance, we get

QM + MZ = QZ = NP = altitude of pole = latitude of place.

Therefore, when the observer is on the northern hemisphere and
the star is on the meridian south of zenith,

North latitude = zenith distance + declination*

If the star is on the meridian between the zenith and the pole (as
at M" t), we will have

North latitude = NP = ZQ = QM" — ZM"

= declination — zenith distance.

M(Star)

P(South
pole)

If the observer is on the
southern hemisphere and the
star M is on his meridian
between the zenith and south
pole, we would have

South latitude

= SP' = SM— MP'
= SM- (90°- QM)
= altitude — co-declination,

if we consider only the numerical value of the declination.

In working out examples the student should depend on the figure
rather than try to memorize formulas to cover all possible cases.

Ex. 2. An observer in the northern hemisphere measured the altitude of a
star at the instant it crossed his celestial meridian south of zenith, and found
it to be 63° 40'. The declination of the star for the same instant was given by
the Nautical Almanac as 21° 15' N. What was the latitude of the observer ?

* If the star is south of the celestial equator (as at M' ), the same rale will hold, for then
the declination is negative (south), and the algebraic sum of the zenith distance and decli-
nation will still give the arc QZ.

t Maximum altitude, if a circumpolar star.

APPLICATIONS OF SPHEEICAL TRIGONOMETRY 245

63° 40',

M (Star)

Solution. Draw the semicircle NZSO. Lay off the arc SM -
which locates the star at M. Since the dec-
lination of the star is north, the celestial
equator may be located by laying off the
arc M Q = declination = 21° 15' towards the
south. The line QO will then be the pro-
jection of the celestial equator, and OP,
drawn perpendicular to QO, will locate the
north pole P.

Zenith distance = ZM = 90° - SM (alt.)
= 90° - 63° 40' = 26° 20'.
.-. North latitude of observer = NP = ZQ = ZM (zen. dist.) + MQ (dec.)
= 26° 20' + 21° 15' = 47° 35'.

Third method. To determine latitude when the altitude, declination,
and hour angle of a celestial body are known. Referring to the astro-
nomical (spherical) triangle PZM, we see that

side MZ

= 90°- HM (alt.)
= co-altitude,

the altitude of the star
being found by measure-
ment. Also

side PM

= 90° - DM (dec.)
= co-declination,

the declination of the star being found from the Nautical Almanac.
Angle ZPM = hour angle, which is given. This hour angle will
be the local time when the observation is made on the sun. We then
have two sides and the angle opposite one of them given in the
spherical triangle PZM. Solving this for the side PZ, by Case
III, (a), p. 224, we get

Latitude of observer = NP = 90° — PZ.

Ex. 3. The declination of a star is 69° 42' N. and its hour angle 60° 44'. "What is
the north latitude of the place if the altitude of the star is observed to be 49° 40' ?
Solution. Referring to the above figure, we have, in this example,
side MZ = co-alt. = 90° - 49° 40' = 40° 20',
side PM = co-dec. = 90° - 69° 42' = 20° 18',
angle ZPM = hour angle = 60° 44'.

Solving for the side PZ by Case III, (a), p. 224, we get side PZ =47° 9'=co-lat.
.-. 90° - 47° 9' = 42° 51' = north latitude of place.

The angle MZP is found to be 27° 53'; hence the azimuth of the star
(angle SZH) is 180° - 27° 53' = 152° 7'.

Horizon

246

SPHERICAL TRIGONOMETRY

EXAMPLES

1. The following observations for altitude have been made on some north
circumpolar star. What is the latitude of each place ?

Maximum altitude

Minimum altitude

North latitude

(a) New York

50° 46'

30° 40'

Ans.

40° 43'

(b) Boston

44° 22'

40° 20'

42° 21'

(c) New Haven

58° 24'

24° 10'

41° 17'

(d) Greenwich

64° 36'

38° 22'

51° 29'

(e) San Francisco

55° 6'

20° 30'

37° 48'

(f) Calcutta

24° 18'

20° 48'

22° 33'

2. In the following examples the altitude of some heavenly body has been
measured at the instant when it crossed the observer's celestial meridian. What
is the latitude of the observer in each case, the declination being found from
the Nautical Almanac ?

Hemisphere Meridian altitude

(a) Northern

(b) Northern

(c) Northern

(d) Northern

(e) Northern

(f) Northern

(g) Southern
(h) Southern
(i) Southern
(j) Southern

60°

75° 40'
43° 27'
38° 6'
50°

28° 46'
67°

45° 26'
72°
22° 18'

Declination
N. 20°
N. 32° 13'
S. 10° 52'
S. 44° 26'
N. 62°
N. 73° 16'
S. 59°
S. 81° 48'
S. 8° *
N. 46° 25'

Body is
S. of zenith
S. of zenith
S. of zenith
S. of zenith
N. of zenith
N. of zenith
S. of zenith
S. of zenith
N. of zenith
N. of zenith

Ans.

Latitude
50° N.
46° 33' N.
35°41'N.
7° 28' N. "
22° N.
12° 2' N.
36° S.
37° 14' S.
26° S.
21° 17' S.

3. In the following examples the altitude of" some heavenly body not on the
observer's celestial meridian has been measured. The hour angle and declination
are known for the same instant. Find the latitude of the observer in each case.

Hemisphere

Altitude

Declination

Hour angle

Latitude

(a) Northern

40°

N. 10°

50°

Ans. 27° 2' N.

(b) Northern

15°

S. 8°

65°

35°38'N.

(c) Northern

52°

N. 19°

2hr.

48° 16' N.

(d) Northern

64° 42'

N. 24° 20'

345°

3°34'N.
or 46° 36' N.

(e) Northern

0°

S. 5°

5hr.

71° 22' N.

(f) Northern

25°

0°

21 hr.

53°18'N.

(g) Northern

0°

N. 11° 14'

68° 64'

No solution

(h) Northern

9° 26'

0°

72° 22'

57°14'N.

(i) Southern

38°

S. 12°

52°

33° 56' S.
or 4° 8' S.

(j) Southern

19°

N. 7°

3hr.

52° 56' S.

(k) Southern

46° 18'

S. 15° 23'

326°

49° 14' S.

(1) Southern

0°

N. 14°

38°

72° 26' S.

(m) Southern

57° 36'

0°

2hr.

12° 50' S.

APPLICATIONS OP SPHEEICAL TEIGONOMETRY 247

34. To determine the time of day. A very simple relation exists
between the hour angla of the sun and the time of day at any place.
The sun appears to move from east to west at the uniform rate of
15° per hour, and when the sun is on the meridian of a place it is
apparent noon at that place. Comparing,

Hour angle of sun

0°

15°

30°

45°

90°
180°
195°
210°
270°
300°
360°

Time of day
Noon

1 P.M.

2 P.M.

3 P.M.
6 P.M.

Midnight

1 A.M.

2 A.M.
6 A.M.
8 A.M.

Noon

The hour angle of the sun M is the angle at P in the astronomical
(spherical) triangle PZM. We may find this hour angle (time of

(North p
pole) *2

(North) JJ

S( South)

day) by solving the astronomical triangle for the angle at P, provided
we know three other elements of the triangle.

248 SPHEEICAL TRIGONOMETRY

DM = declination of sun, and is found from the Nautical Almanac.
.'. Side PM = 90° — DM = co-declination of sun.

HM = altitude of sun, and is found by measuring the angular dis-
tance of the sun above the horizon with a sextant or transit.

.". side MZ = 90° — HM = co-altitude of sun.

NP = altitude of the celestial pole
= latitude of the observer (p. 243).

.'. Side PZ = 90°— NP = co-latitude of observer.

Hence we have

Rule for determining the time of day at a place whose latitude is
known, when the declination and altitude of the sun at that time and
place are known.

First step. Take for the three sides of a spherical triangle

the co-altitude of the sun,
the co-declination of the sun,
the co-latitude of the place.

Second step. Solve this spherical triangle for the angle opposite the
first-mentioned side. This will give the hour angle in degrees of the
sun, if the observation is made in the afternoon. If the observation is
made in the forenoon, the hour angle will be 360° — the angle found.

Third step. When the observation is made in the afternoon the time

of day will be

hour angle

IS

When the observation is made in the forenoon the time of day will be
( *our angle _ 12 y M

Ex. 1. In New York (lat. 40°43'N.) the sun's altitude is observed to be
30° 40'. Having given that the sun's decimation is 10° N. and that the observa-
tion is made in the afternoon, what is the time
of day?

Solution. First step. Draw the triangle.
Side a = co-alt. = 90° - 30° 40' = 59° 20'.

Side 6 = co-dec. = 90° - 10° = 80°.

Sun

PoU Side c = co-lat. = 90° - 40° 43' = 49° 17'.

Second step. As we have three sides given, the solution of this triangle comes
under Case I, (a), p. 217. But as we only want the angle A (hour angle), some

APPLICATIONS OF SPHEEICAL TKIGONOMETEY 249

labor may be saved by using one of the formulas (18), (19), (20), pp. 211, 212.
Let us use (18),

a= 59° 20'

6= 80°

c= 49° 17'

. . /sin s sin (s — a)

sin \a = -t/ — -A ' ,

\ sin o sin c

log sin £ a: = £ [log sin s + log sin (s - a) - {log sin 6 + log sin c}].

2s = 188° 37'

s = 94° 19'.

« - a = 34° 59'.

log sin s = 9. 9988 log sin 6=9. 9934

log sin (s - a) = 9.7584 log sin e= -9.8797

log numerator = 19.7572 log denominator = 19.8731

log denominator = 19.8731
9.8841
2 | 19.8841
log sin £ a = 9.9421
Ja=61°4'.
a = 122° 8'.
.-. A = 180° - a = 57° 52' = hour angle of sun.

Third step. Time of day = — p.m. = 3 hr. 51 min. p.m. Ans.

15

EXAMPLES

1. In Milan (lat. 45°30'N.) the sun's altitude at an afternoon observation
is 26° 30'. The sun's declination being 8° S., what is the time of day ?

Ans. 2 hr. 33 min. p.m.

2. In New York (lat. 40° 43' N.) a forenoon observation on the sun gives
30° 40' as the altitude. What is the time of day, the sun's declination being
10° S.? Ans. 9 hr. 46 min. a.m.

3. A mariner observes the altitude of the sun to be 60°, its declination at the
time of observation being 6° N. If the latitude of the vessel is 12° S., and the ob-
servation is made in the morning, find the time of day. Ans. 10 hr. 24 min. a.m.

4. A navigator observes the altitude of the sun to be 35° 23', its declination
being 10° 48' S. If the latitude of the ship is 26° 13' N., and the observation is
made in the afternoon, find the time of day. Ans. 2 hr. 45 min. p.m.

5. At a certain place in latitude 40° N. the altitude of the sun was found to
be 41° If its declination at the time of observation was 20° N., and the obser-
vation was made in the morning, how long did it take the sun to reach the
meridian ? Ans. 3 hr. 31 min.

6. In London (lat. 51° 31' N.) at an afternoon observation the sun's altitude
is 15° 40'. Find the time of day, given that the sun's declination is 12° S.

Ans. 2 hr. 59 min. p.m.

7. A government surveyor observes the sun's altitude to be 21°. If the latitude
of his station is 27° N. and the declination of the sun 16° N. , what is the time of
day if the observation was made in the afternoon 1 Ans. 4 hr. 57 min. p.m.

8. The captain of a steamship observes that the altitude of the sun is 26° 30'. If
he is in latitude 45° 30' K and the declination of the sun is 18° N., what is the time
of day if the observation was made in the afternoon ? Ans. 4 hr. 41 min. p.m.

250

SPHEEICAL TKIGONOMETKY

35. To find the time of sunrise or sunset. If the latitude of the place
and the declination of the sun is known, we have a special case of the
preceding problem ; for at sunrise or sunset the sun is on the hori-
zon and its altitude is zero. Hence the co-altitude, which is one side
of the astronomical triangle, will be 90°, and the triangle will be a
quadrantal triangle (p. 204). The triangle may then be solved by the
method of the last section or as a quadrantal triangle.

EXAMPLES

1. At what hour will the sun set in Montreal (lat. 45° 30' N.), if its declina-
tion at sunset is 18° N. ? Ans. 7 hr. 17 min. p.m.

2. At what hour will the sun rise in Panama (lat. 8° 57' N.), if its declination
at sunrise is 23° 2' S.? Ans. 6 hr. 15 min. a.m.

3. About the first of April of each year the declination of the sun is 4° 30' N.
I ind the time of sunrise on that date at tie following places :

(a) New York (lat. 40° 43' N.). Ans. 5 hr. 45 min. a.m.

(b) London (lat. 51° 31' N.). 5 hr. 37 min. a.m.

(c) St. Petersburg (lat. 60° N.). 5 hr. 29 min. a.m.

(d) New Orleans (lat. 29° 58' N.). 5 hr. 50 min. a.m.

(e) Sydney (lat. 33° 52' S.). 6 hr. 12 min. a.m.

36. To determine the longitude of a place on the earth. From the
definition of terrestrial longitude given on p. 231 it is evident that

the meridians on the
earth are projected into
hour circles on the
celestial sphere. Hence
the same angle (or arc)
which measures the
angle between the celes-
tial meridians, (hour
circles) of the place
of observation and of
Greenwich may be
taken as a measure of
the longitude of the
place. Thus, in the fig-
ure, if PQP' is the me-
ridian (hour circle) of
Greenwich and PDP' the meridian (hour circle) of the place of
observation, then the angle QPD (or arc QD) measures the west
longitude of the place. If PMP' is the hour circle of the sun, it is
evident that

APPLICATIONS OP SPHERICAL TRIGONOMETRY 251

angle QPM = hour angle of sun for Greenwich
= local time at Greenwich ;

angle DPM = hour angle of sun for observer

= local time at place of observation.

Also, angle QPM — angle DPM = angle QPD = longitude of place.

Hence the lo'ngitude of the place of observation equals the differ-
ence * of local times between the standard meridian and the place in
question. Or, in general, we have the following

Rule for finding longitude : The observer's longitude is the amount
by which noon at Greenwich is earlier or later than noon at the place
of observation. If Greenwich has the earlier time, the longitude of the
observer is east ; if it has the later time, then the longitude is west.

We have already shown (p. 248) how the observer may find his
own local time. It then remains to determine the Greenwich time
without going there. The two methods which follow are those in
general use.

First method. Find Greenwich time by telegraph (wire or wireless').
By far the best method, whenever it is available, is to make a direct
telegraphic comparison between the clock of the observer and that
of some station the longitude of which is known. The difference
between the two clocks will be the difference in longitude of the
two places.

Ex. 1. The navigator on a battleship has determined his local time to be
2 hr. 25 min. p.m. By wireless he finds the mean solar time at Greenwich to
be i hr. 30 min. p.m. What is the longitude of the ship ?
Solution. Greenwich having the later time,

4 hr. 30 min.

2 hr. 25 min.

Subtracting, 2 hr. 5 min. = west longitude of the ship.

Eeducing this to degrees and minutes of arc,

2 hr. 5 min.
15
Multiplying, 31° 15' = west longitude of ship.

Second method. Find Greenwich time from a Greemvich chronom-
eter. The chronometer is merely a very accurate watch. It has been
set to Greenwich time at some place whose longitude is known, and
thereafter keeps that time wherever carried.

* This difference in time is not taken greater than 12 hours. If a difference in time be-
tween the two places is calculated to he more than 12 hours, we subtract it from 24 hours
and use the remainder instead as the difference.

252 SPHERICAL TRIGONOMETRY

Ex. 2 An exploring party have calculated their local time to be 10 hr. a.m.
The Greenwich chronometer which they cany gives the time as 8 hr. 30 min. a.m.
What is their longitude ?

Solution. Greenwich has here the earlier time.

10 hr.

8 hr. 30 min.
Subtracting, 1 hr. 30 min. = 22° 30' = east longitude.

EXAMPLES

1. In the following examples we have given the local time of the observer
and the Greenwich time at the same instant. Find the longitude of the observer
in each case.

Observer's Corresponding Longitude

local time Greenwich time of observer

(a) Noon. 3 hr. 30 min. p.m. Ans. 52°30'W.

(b) Noon. 7 hr. 20 min. a.m. 70° E.

(c) Midnight. 10 hr. 15 min. p.m. 20° 15' E.

(d) 4 hr. 10 min. p.m. Noon. 62° 30' E.

(e) 8 hr. 25 min. a.m. Noon. 53°45'W.

(f) 9 hr. 40 mill. p.m. Midnight. 36° W.

(g) 2,hr. 15 min. p.m. 11 hr. 20 min. a.m. 43°45'E.
(h) 10 hr. 26 min. a.m. 5 hr. 16 min. a.m. 77°30'E.

(i) 1 hr. 30 min. p.m. 7 hr. 45 min. p.m. 93° 45' W.

(j) Noon. Midnight. 180° W. orE.

(k)0hr. p.m. 6hr. a.m. 180° E. orW.

(1) 5 hr. 45 min. a.m. 7 hr. 30 min. p.m. 153° 45' E.

(m) 10 hr. 55 min. p.m. 8 hr. 35 min. a.m. 145° W.

2. If the Greenwich time is 9 hr. 20 min. p.m., January 24, at the same instant
that the time is 3 hr. 40 min. a.m., January 25, at the place of observation, what
is the observer's longitude ? Ans. 96° E.

3. The local time is 4 hr. 40 min. a.m., March 4, and the corresponding Green-
wich time is 8 hr. p.m., March 3. What is the longitude of the place ?

Ans. 130° E.

4. In the following examples we have given the local time of the observer
and the local time at the same instant of some other place whose longitude is
known. Eind the longitude of the observer in each case.

Observer's Corresponding time and Longitude

local time longitude of the other place of observer

(a) 2hr. p.m. 5 hr. p.m. at Havana (long. 82° 23' W.) Ans. 127°23'W.

(b) 10 hr. a.m. 3 hr. p.m. at Yokohama (long. 139° 41' E.) 64°41'E.

(c) 5 hr. 20 min. p.m. 11 hr. 30 min. p.m. at Glasgow (long. 4° 16' W.) 96°46'W.

(d) 8hr.25min.A.M. 6hr.35min. a.m. at VeraOuz(long.96°9'W.) 68°39'W.

(e) 9hr. 45min. p.m. Midnight at Batavia (long. 106° 52' E.) 73°7'E.

(f) 7hr. 40 min. p.m. Noon at Gibraltar (long. 5° 21' W.) 109° 39' E.

(g) 4 hr. 50 min. p.m. Noon at Auckland (long. 174° 50' E.) 112°40'W.

APPLICATIONS OF SPHERICAL TRIGONOMETRY 253

6. What is the longitude of each place mentioned in the examples on p. 240,
the Greenwich time for the same instant being given below ?

Example, p. 249

(a) Ex. 3

(b) Ex; 4

(c) Ex. 5

(d) Ex. 7

(e) Ex. 8

Greenwich time
2 hr. 12 min. p.m.
i hr. 52 min. p.m.
6 hr. 9 min. a.m.
10 hr. 33 min. p.m.
6 hr. 25 min. p.m.

Longitude of place
Ans. 57° W. long, (vessel)
31° WW. long, (vessel)
50° E. long, (observer)
84° W. long, (surveyor)
26° W. long, (ship)

37. The ecliptic and the equinoxes. The earth makes a complete
circuit around the sun in one year. To us, however, it appeal's as if
the sun moved and the earth stood still, the (apparent) yearly path
of the sun among the stars being a great circle of the celestial sphere
which we call the ecliptic. Evidently the plane of the earth's orbit

Earth

cuts the celestial sphere in the ecliptic. The plane of the equator and

the plane of the ecliptic are inclined to each other at an angle of about

23^° (= e), called the obliquity of the ecliptic (angle LVQ in figure).

The points where the ecliptic intersects the celestial equator are
called the equinoxes. The point where the sun crosses the celestial
equator when moving northward (in the spring, about March 21) is
called the vernal equinox, and the point where it crosses the celestial
equator when moving southward (in the fall, about September 21)
is called the autumnal equinox.

If we project the points V and A in our figure on the celestial
sphere, the point V will be projected in the vernal equinox and the
point A in the autumnal equinox.

38. The equator and hour circle of vernal equinox system.* The two
fixed and mutually perpendicular great circles of reference are in

• Sometimes called the equator system.

254

SPHERICAL TRIGONOMETRY

(North celestial polej
P

this case the celestial equator (QVQ') and the hour circle of the ver-
nal equinox (PVP') f also called the equinoctial colure ; and the spher-
ical coordinates of a
heavenly body are its
declination and right
ascension.

The declination of a
heavenly body has al-
ready been defined on
p. 240 as its angular
distance north or south
of the celestial equator
measured on the hour
circle of the body from
0° to 90°, positive if
north and negative if
south. In the figure
DM is the north decli-
nation of the star M.
The right ascension of a heavenly body is the angle bet-ween the
hour circle of the body and the hour circle of the vernal equinox
measured eastward from the latter circle from 0° to 360°, or in hours
from to 24. In the figure, the angle VPD (or the arc VD) is the
right ascension of the star M. The right ascensions of the sun,
moon, and planets are continually changing.* The angle LVQ (= e)
is the obliquity of the ecliptic (= 23£°).

Ex. 1. In each of the following examples draw a figure of the celestial sphere
and locate the body from the given spherical coordinates.

Bight ascension

Declination

Bight ascension

Declination

(a)0°

0°

(j) 90°

0°

(b) 180°

0°

(k) 270°

0°

(c) 90°

N. 90°

(1) 90°

S. 90°

(d) 46°

N. 45°

(m) 46°

S.45°

(e) 60°

N. 60°

(n) 90°

S. 30°

(f) 120°

+ 30°

(o) 240°

+ 60°

(g) 300°

-60°

(p) 330°

-45°

(h) 12 hr.

+ 45°

(q) 6 hr.

+ 15°

(i) 20 hr.

0°

(r) 9hr.

-75°

* The right ascensions of the sun, moon, and planets may be found in the Nautical
Almanac for any time of the year.

APPLICATIONS OF SPHERICAL TRIGONOMETRY 255

Pole

Ex. 2. The right ascension of a planet is 10 hr. 40 min. and its declination
S. 6°. Find the angular distance from this planet to a fixed star whose right
ascension is 3 hr. 20 min. and decimation N. 48°.

Solution. Locate the planet and the star on the celestial sphere. Draw the
spherical triangle whose vertices are at the north
pole, the planet, and the fixed star. Then

Angle A = difference of right ascensions
= 10 hr. 40 min. - 3 hr. 20 min.
= 7 hr. 20 min. = 110°.
Side 5 = co-declination of star

= 90° _ 48° = 42°.
Side c = co-declination of planet

= 90° - (- 6°) = 96°. To find side a.

As we have two sides and the included angle
given, the solution of this triangle comes under Case II, (a), p. 219. Since a only
is required, the shortest method is that illustrated on p. 220, the solution depending
on the solution of right spherical triangles. On solving, we get a= 107° 48'. Ans.

39. The system having for reference circles the ecliptic and the great
circle KVK' passing through the pole of the ecliptic and the vernal

(Sorth celestial pole)
P

Planet

equinox.* The spherical coordinates of a heavenly body in this case
are its latitude and longitude.^

The latitude of a heavenly body is its angular distance north or
south of the ecliptic, measured on the great circle passing through

* Sometimes called the ecliptic system.

t Sometimes called celestial latitude and longitude in contradistinction to the latitude
and longitude of places on the earth's surface (terrestrial latitude and longitude), which were
defined on p. 231, and which have different meanings.

256 SPHEEICAL TRIGONOMETRY

the body and the pole of the ecliptic. Thus, in the figure, the arc TM
measures the north latitude of the star M.

The longitude of a heavenly body is the angle between the great
circle passing through the body and the pole of the ecliptic, and the
great circle passing through the vernal equinox and the pole of the
ecliptic, measured eastward from the latter circle from 0° to 360°.
In the figure, the angle VKT (or the arc VT) is the longitude of the
star M. The latitudes and longitudes of the sun, moon, and planets
are continually changing. The angle LVQ (= e) is the obliquity of
the ecliptic (= 23£° = arc KP).

Since the ecliptic is the apparent yearly path of the sun, the celes-
tial latitude of the sun is always zero. The declination of the sun,
however, varies from N. 23£° (= arc QV) on the longest day of the
year in the northern hemisphere (June 21), the sun being then the
highest in the sky (at L), to S. 23£° (arc Q'L') on the shortest day
of the year (December 22), the sun being then the lowest in the sky
(at L'). The declination -of the sun is zero at the equinoxes (March
21 and September 21).

Ex. 1. In each of the following examples draw a figure of the celestial sphere
and locate the body from the given spherical coordinates.

Celestial longitude

Celestial latitude

Celestial longitude

Celestial latitude

(a) 0°

0°

(j) 90°

0°

(b) 90°

N. 90°

(k) 180°

0°

(c) 180°

N. 45°

(1) 0°

S. 60°

(d) 270°

0°

(m) 60°

N. 30°

(e) 45°

S. 30°

(n) 120°

N. 45°

(f) 135°

+ 15°

(o) 270°

-75°

(g) 315°

+ 60°

(p) 30°

-60°

(h) 6hr.

-45°

(q) 9hr.

0°

(i) 15 hr.

+ 45°

(r) 18 hr.

+ 30°

Ex. 2. Given the right ascension of a star 2 hr. 40 min. and its declination
24° 20' N., find its celestial latitude and longitude.

Solution. Locate the star on the celestial sphere. Consider the spherical
triangle KPM on the next page.

Angle KPM = Z Q'PV + Z VPD
= 90° + right ascension
= 90° + 2 hr. 40 min.
= 90° + 40° = 130°.
Side PM = co-declension
= 90°-24°20'
= 65° 40'.

and

APPLICATIONS OP SPHEEICAL TRIGONOMETRY 257

Side KP = LQ = e = 23° 30'.
To find side KM = co-latitude of the star,
angle PKM = co-longitude of the star.

(North pole)
P

As we have two sides and the included angle given, the solution of this tri-
angle comes under Case II, (a), p. 219. Solving, we get

Side KM = 81° 52' and Z PKM = 44° 52'.
.-. 90° - KM = 90° - 81° 52' = 8° 8' = TM = latitude of star,
and 90° - Z PKM = 90° - 44° 52' = 45° 8' = VT = longitude of star.

EXAMPLES

1. Find the distance in degrees between the sun and the moon when their
right ascensions are respectively 12 hr. 39 min., 6 hr. 56 min., and their declina-
tions are 9° 23' S., 22° 50' N. Ans. 90°.

2. Find the distance between Regulus and Antares, the right ascensions
being 10 hr. and 16 hr. 20 min.. and the polar distances 77° 19' and 116° 6'.

Ans. 99° 56'.

3. Find the distance in degrees between the sun and the moon when their
right ascensions are respectively 15 hr. 12 min., 4 hr. 45 min., and their decli-
nations are 21° 30' S., 5° 30' N. Ans. 152° 23'.

4. The right ascension of Sirius is 6 hr. 39 min., and his declination is
16°31'S.; the right ascension of Aldebaran is 4hr. 27 min., and his declination
is 16° 12' N. Find the angular distance between the stars. Ans. 46° 8'.

5. Given the right ascension of a star 10 hr. 50 min., and its declination
12° 30' N., find its latitude and longitude. Take e = 23° 30'.

Ans. Latitude = 18° 24' N., longitude = 168° 53'.

6. If the moon's right ascension is 4 hr. 16 min. and its declination 6° 20' N.,
what is its latitude and longitude ?

Ans. Latitude = 14° 43' N., longitude = 62° 58'.

258

SPHERICAL, TRIGONOMETRY

7. The sun's longitude was 69° 40'. What was its right ascension and decli-
nation ? Take e = 23° 27'.

Ans. Right ascension = 3 hr. 50 min., declination = 20° 5' N.
Hint. The latitude of tlie sun is always zero, since it moves in the ecliptic. Hence in the
triangle KPM (figure, p. 257), A"-l/= 90°, and it is a quadrantal triangle. This triangle may
then be solved by the method explained on p. 204.

8. Given the sun's declination 16° 1' N. , find the sun's right ascension and
longitude. Take e = io 3 27'

Ans. Right ascension = 9 hr. 14 min., longitude = 136° 7'.

9. The sun's right ascension is 14 hr. 8 min. ; find its longitude and declina-
tion. Take e = 23° 27'. Ans. Longitude = 214° 16', declination = 12° 56' S.

10. Find the length of the longest day of the year in latitude 42° 17' N.

Ans. 15 hr. 6 min.

Hint. This will he the time from sunrise to sunset when the sun is the highest in the sky,
that is, when its declination is 23° 27' X.

11. Find the length of the shortest day in lat. 42° 17' N. Ans. 8 hr. 64 min.
Hint. The sun will then he the lowest in the sky, that is, its declination will he 23° 27' S.

12. Find the length of the longest day in New Haven (lat. 41° 19' N.). Take
e = 23° 27'. Ans. 15 hr.

13. Find the length of the shortest day in New Haven. Ans. 9 hr.

14. Find the length of the longest day in Stockholm (lat. 59° 21' N.). Take
e = 23° 27'. Ans. 18 hr. 16 min.

15. Find the length of the shortest day in Stockholm. Ans. 5 hr. 48 min.

40. The astronomical triangle. TVe have seen that many of our
most important astronomical problems depend on the solution of

(North
pole) t\

(North) jy

S( South)

the astronomical triangle PZM. In any such problem the first
thing to do is to ascertain which parts of the astronomical triangle

APPLICATIONS OF SPHERICAL TRIGONOMETRY 259

are given or can be obtained directly from the given data, and which
are required. The different magnitudes which may enter into such
problems are

HM = altitude of the heavenly body,

DM = declination of the heavenly body,
angle ZPM = hour angle of the heavenly body,
angle SZM = azimuth of the heavenly body,

NP = altitude of the celestial pole
= latitude of the observer.

*

As parts of the astronomical triangle PZM we then have
side MZ = 90° - HM = co-altitude,
side PM = 90° — DM = co-declination,
side PZ = 90° - NP = co-latitude,

angle ZPM = hour angle,

angle PZM = 180° — azimuth (angle SZM)*

The. student should be given practice in picking out the known
and unknown parts in examples involving the astronomical tri-
angle, and in indicating the case under which the solution of the
triangle comes.

For instance, let us take Ex. 15, p. 261.

r Latitude = 51° 32' N
.-. side PZ = 90°- 51° 32'= 38° 28'.
Altitude = 35° 15'.
.-. side ikfZ = 90°-35°15'=54°45'.
Declination = 21° 27' N.
.-. side MP = 90°- 21° 27'= 68° 33'.
Required : Local time = hour angle = angle ZPM.

Since we have three sides given to find an angle, the solution of
the triangle comes under Case I, (a), p. 217. This gives angle ZPM
= 59° 45'= 3 hr. 59 min. p.m.

41. Errors arising in the measurement of physical quantities. f Errors
of some sort will enter into all data obtained by measurement. For
instance, if the length of a line is measured by a steel tape, account
must be taken of the expansion due to heat as well as the sagging of
the tape under various tensions. Or, suppose the navigator of a ship

* When the heavenly body is situated as in the figure. If the body is east of the ob-
server's meridian, we would have angle PZM= azimuth — 180°. \

1 Iu this connection the student is advised to read § 93 in Granville's Plane Trigonometry.

Given parts

260 SPHERICAL TRIGONOMETRY

at sea is measuring the altitude of the sun by means of a sextant. The
observed altitude should be corrected for errors due to the following
causes :

1. Dip. Owing to the observer's elevation above the sea level
(on the deck or bridge of the ship), the observed altitude will be too
great on account of the dip (or lowering) of the horizon.

2. Index error of sextant. As no instrument is perfect in con-
struction, each one is subject to a certain constant error which is
determined by experiment.

S. Refraction of light. Celestial bodies appear higher than they
really are because of the refraction of light by the earth's atmos-
phere. This refraction will depend on the height of the celestial
body above the horizon, and also on the state of the barometer and
thermometer, since changes in the pressure and temperature of the
air affect its density.

Jf. Semidiameter of the sun. As the observer cannot be sure where
the center of the sun is, the altitude of (say) the lower edge' of the
sun is observed and to that is added the known semidiameter of the
sun for that day found from the Nautical Almanac.

5. Parallax. The parallax of a celestial body is the angle sub-
tended by the radius of the earth passing through the observer, as
seen from the body. As viewed from the earth's surface, a celestial
body appears lower than it would be if viewed from the center, and
this may be shown to depend on the parallax of the body.

We shall not enter into the detail connected with these correc-
tions, as that had better be left to works on Eield Astronomy ; our
purpose here is merely to call the attention of the student to the
necessity of eliminating as far as possible the errors that arise when
measuring physical quantities.

For the sake of simplicity we have assumed that the necessary cor-
rections have been applied to the data given in the examples found
in this book.

MISCELLANEOUS EXAMPLES

1. The continent of Asia has nearly the shape of an equilateral triangle.
Assuming each side to be 4800 geographical miles and the radius of the earth
to be 3440 geographical miles, find the area of Asia.

Ans. About 13,333,000 sq. mi.

2. The distance between Paris (lat. 48° 50' N.) and Berlin (lat. 52° 30' N.) is
472 geographical miles, measured on the arc of a great circle. What time is it
at Berlin when it is noon at Paris ? Ans. 44 rain, past noon.

3. The altitude of the north pole is 46°, and the azimuth of a star on the
horizon is 135°. Find the polar distance of the star. Ans. 60°.

APPLICATIONS OF SPHERICAL TRIGONOMETRY 261

4. What will be the altitude of the sun at 9 a.m. in Mexico City (lat. 19°26'N.),
if its declination at that time is 8° 23' N. ? Ans. 45° 5'.

5. Find the altitude of the sun at 6 hr. a.m. at Munich (lat. 48° 9' N.) on the
longest day of the year. Ans. Altitude = 17° 16'.

6. Find the time of day when the sun bears due east and due west on the
longest day of the year at St. Petersburg (lat. 59° 56' N.).

Ans. 6 hr. 58 min. a.m., 5 hr. 2 min. p.m.

7. What is the direction of a wall in lat. 52° 30' N. which casts no shadow at
6 a.m. on .the longest day of the year ?

Ans. 75° 1 1', reckoned from the north point of the horizon.

8. Find the latitude of the place at which the sun rises exactly in the north-
east on the longest day of the year. Ans. 65° 45' N.

9. Find the latitude of the place at which the sun sets at 10 hr. p.m. on the
longest day. Ans. 63° 23' N. or S.

10. Given the latitude of the place of observation 52° 30' N. , the declination
of a star 38°, its hour angle 28° 17'. Find the altitude of the star.

Ans. Altitude = 65° 33'.

11. Given the latitude of the place of observation 51° 19' N., the polar dis-
tance of a star 67° 59', its hour angle 15° 8'. Find the altitude and azimuth of
the star. Ans. Altitude = 58° 23', azimuth = 27° 30'.

12. Given the declination of a star 7° 54' N. , its altitude 22° 45', its azimuth
50° 14'. Find the hour angle of the star and the latitude of the observer.

Ans. Hour angle = 45° 41', latitude = 67° 59' N.

13. The latitude of a star is 51° N., and its longitude 315°. Find its declina-
tion. Take e = 23° 27'. Ans. Declination = 32° 23' N.

14. Given the latitude of the observer 44° 50' N., the azimuth of a star 41° 2',
its hour angle 20°. Find its declination. Ans. Declination = 20° 49' N.

15. Given the latitude of the place of observation 51° 32' N. , the altitude of
the sun west of the meridian 35° 15', its declination 21° 27' N. Find the local
time. Ans. 3 hr. 69 min. p.m.

CHAPTER IV

RECAPITULATION OF FORMULAS

Spherical Trigonometry

42. Right spherical triangles, pp. 196-197.

(1) cos c = cos a cos b,

(2) sin a = sin c sin A,

(3) sin b = sin sin B,

(4) cos A = cos a sin B,

(5) cos B = cos 6 sin A,

(6) cos ^4 = tan b cot c,

(7) cos B = tan a cot c,

(8) sin b = tan a cot A,

(9) sin a = tan b cot /J,
(10) cos c = cot ,1 cot B.

General directions for solving right spherical triangles by Napier's
rules of circular parts are given on p. 200.

Spherical isosceles and quadrantal triangles are discussed on p. 204.

43. Relations between the sides and angles of oblique spherical tri-
angles, pp. 206-216.

a = 180°-^, /J = 180°-j3, y = 180°-r;.

d = diameter of inscribed circle.

S = 180° — diameter of circumscribed circle.

or

Law of

sines

, p. 207.

(ii;

sin a.

sini
sin B

sine

sin^l

sinC

sin a

sin b

sine

sin a: sin/3 siny

Law of cosines for the sides, p. 209.

(12) cos a = cos b cos « — sin b sin c cos a,

2fi2

RECAPITULATION OF FORMULAS
Law of cosines for the angles, p. 209.
(15) cos a = cos B cos y — sin 8 sin y cos a.

Functions of ^ a, ^ B, £ y in terms of the sides, pp. 211-213
/iq\ - . /sin * sin (s — o)

(18) smj«= A/ r-rA '-

\ sin 6 sine

263

(19)

(20)

(27)
{•28)
(29)
(30)

cos

tan

i a . ^ J*^~( s - fe)sin(s -^7)
\ sin b sin <r

, a ^ r~sin^sin(g - a)~~
\sin (s — 6)sin(s — c)

tan J<2 = -v

sin (s — a)

sin(s — q)sin(s — 6)sin(s — e)

sins

tan ^ a
tanjjg

tan J y =

tan \d

sin(s — b)

tan | rf

sin (s — c)

tan Jd

Functions of the half sides in terms of a, B, y, p. 214.
(31)

sin

I sin a- sin (<r — a

£a = "V ^-^

s > sin B sin y

).

(32)
(33)

(40)
(41)
(42}
(43)

1 sin j8 sin -y

sin <r sin (a- — «)

cos J

tan * a = \iinT<r-i8)sin(«r-y)

sin (a- — g)sin(a- — ft)sin(tr — y)
sina-

tan^a =
tanji =

tan j =

sin (<r — a)
tanjS

sin (o- — B)
tan^S

sin (o- — y)
tan JS

264 SPHEEICAL TRIGONOMETRY

Napier's Analogies, p. 215.

(44) tan i (a-i)=-^Jfc|l taili< ..
v ' sin \ (a + /3) 2

(45) tan }(« + b) = - C ° S * ^ ~ g tan fr C .
v ' v y cos i (a + /3) a

(46) tan i(g - /?) =~ Sln t^ 7 ?? tan jy-
v ' * v ^ y sin J (a + *) 2 '

(47) tanK« + ffl=- C0S tl a I?i tai1 ^-
v ' a v ^ cos £ (a + b) 2 '

44. General directions for the solution of oblique spherical triangles,
pp. 216-227.

Case 1. (a) Given ths three sides, p. J 17.

(V) Given the three angles, p. 218.
Case II. (a) Given two sides (uid their included angle, p. 219.

(b) Given tiro angles and their included side, p. 222.

Case III. (a) Given tiro sides and the angle opposite one of them
p. 224.
(b) Given tiro angles and the side opposite one of them
p. 226.

45. Length of an arc of a circle in linear units, p. 228.

{5Z) L ~ 180

N = number of degrees in angle.

46. Area of a spherical triangle, p. 229.

(54) A!ea = l!r

E=A+B + C -180°.

(55) tan^ E = Vtan £ s tan ^ (s — a) tan ±(s — b) tan %(s — c)

FOUR-PLACE TABLES OF
LOGARITHMS

COMPILED BY

WILLIAM ANTHONY GEANVILLE, Ph.D., LLD.

PBESIDENT OF PENNSYLVANIA COLLEGE

GINN AND COMPANY

BOSTON • NEW YORK • CHICAGO • LONDON
ATLANTA ■ DALLAS • COLUMBUS ■ SAN FRANCISCO

Entered at Statioxebs' Haix

COPYRIGHT, 1908, BY

WILLIAM ANTHONY GKANVILLE

ALL RIGHTS RESERVED

715.12

gfct gtfcen«ii« >ctt«

GINN AND COMPANY • PRO-
PRIETORS • BOSTON ■ U.S.A.

CONTENTS

Pages

Table I. Logarithms of Numbers 1-5

EULES FOR FINDING THE LOGARITHMS OF THE TRIGONO-
METRIC Functions of Angles near 0° and 90° . . 6

Table II. Logarithms of the Trigonometric Func-
tions, the Angle being expressed in Degrees
and Minutes 7-16

Conversion Tables for Angles 17

Table III. Logarithms of the Trigonometric Func-
tions, the Angle being expressed in Degrees and
the Decimal Part of a Degree 19-37

Table of Natural Values of the Trigonometric

Functions for Every Degree 38

Table I

FOUR-PLACE LOGARITHMS OP NUMBERS

This table gives the mantissas of the common logarithms (base 10)
of the natural numbers (integers) from 1 to 2000, calculated to four
places of decimals.

A logarithm found from this table by interpolation may be in
error by one unit in the last decimal place.

TABLE I. LOGARITHMS OF NUMBERS

No.

100

101
102
103

104
105
106

107
108
109
110
111
112
113

114
115
116

117
118
119
120

121

122
123

124
125
126

127
128
129
130
131
132
133

134
135
136

137
138
139
140
141
142
143

144
145
146

147
148
149
150

No.

0000

0043
0086
0128

0170
0212
0253

0294
0334
0374

0414

0453
0492
0531

0569
0607
0645

0682
0719
0755

0792

0828
0864
0899

0934
0969
1004

1038
1072
1106

1139

1173
1206
1239

1271
1303
1335

1367
1399
1430
1461
1492
1523
1553

1584
1614
1644

1673
1703
1732
1761

0004

0048
0090
0133

0175
0216
0257

0298
0338
0378

0418

0457
0496
0535

0573
0611
0648

0686
0722
0759

0795

0831
0867
0903

0938
0973
1007

1041
1075
1109

1143

1176
1209
1242

1274
1307
1339

1370
1402
1433
1464
1495
1526
1556

1587
1617
1647

1676
1706
1735
1764

0009

0052
0095
0137

0179
0220
0261

0302
0342
0382

0422

0461
0500
0538

0577
0615
0652

0689
0726
0763

0799

0835
0871
0906

0941
0976
1011

1045
1079
1113

1146

1179
1212
1245

1278
1310
1342

1374
1405
1436
1467
1498
1529
1559

1590
1620
1649

1679
1708
1738
1767

0013

0056
0099
0141

0183
0224
0265

0306
0346
0386

0426

0465
0504
0542

0580
0618
0656

0693
0730
0766

0803

0839
0874
0910

0945
0980
1014

1048
1082
1116

1149

1183
1216
1248

1281
1313
1345

1377
1408
1440
1471
1501
1532
1562

1593
1623
1652

1682
1711
1741
1770

0017

0060
0103
0145

0187
0228
0269

0310
0350
0390

0430

0469
0508
0546

0584
0622
0660

0697
0734
0770

0806

0842
0878
0913

0948
0983
1017

1052
1086
1119

1153

1186
1219
1252

1284
1316
1348

1380
1411
1443
1474
1504
1535
1565

1596
1626
1655

1685
1714
1744
1772

0022

0065
0107
0149

0191
0233
0273

0314
0354
0394

0434

0473
0512
0550

0588
0626
0663

0700
0737
0774

0026 0030

0069
0111
0154

0195
0237
0278

0318
0358
0398

0438

0477
0515
0554

0592
0630
0667

0704
0741
0777

0810 0813

0846
0881
0917

0952
0986
1021

1055
1089
1123

1156

1189
1222
1255

1287
1319
1351

1383
1414
1446
1477
1508
1538
1569

1599
1629
1658

1688
1717
1746
1775

0849
0885
0920

0955
0990
1024

1059
1093
1126

1159

1193
1225
1258

1290
1323
1355

1386
1418
1449
1480
1511
1541
1572

1602
1632
1661

1691
1720
1749
1778

0073
0116
0158

0199
0241
0282

0322
0362
0402

0441

0481
0519
0558

0596
0633
0671

0708
0745
0781

0817

0853
0888
0924

0959
0993
1028

1062
1096
1129

1163

1196
1229
1261

1294
1326
1358

1389
1421
1452
1483
1514
1544
1575

1605
1635
1664

1694
1723
1752
1781

0035

0077
0120
0162

0204
0245
0286

0326
0366
0406

0445

0484
0523
0561

0599
0637
0674

0711
0748
0785

0821

0856
0892
0927

0962
0997
1031

1065
1099
1133

1166

1199
1232
1265

1297
1329
1361

1392
1424
1455
1486
1517
1547
1578

1608
1638
1667

1697
1726
1755
1784

0039

0082
0124
0166

0208
0249
0290

0330
0370
0410

0449

0488
0527
0565

0603
0641
0678

0715
0752
0788

0824

0860
0896
0931

0966
1000
1035

1069
1103
1136

1169

1202
1235
1268

1300
1332
1364

1396
1427
1458
1489
1520
1550
1581

1611
1641
1670

1700
1729
1758
1787

Prop. Farts

to

a

u

s

a

~v

o&

1.0
16

2.0
2.6
3.0
3.5

4.0
4.5

4

1

0.4

2

0.8

3

1.2

4

1.6

5

2.0

6

2.4

7

2.8

H

3.2

9

3.6

3

1

0.3

2

0.6

3

OS)

4

1.2

5

1.5

6

1.8

7

2.1

8

2.4

9

2.7

2

0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8

TABLE I. LOGARITHMS OF XUMBEES

Ho.

150
151

152
153

154
155
156

157
15S
159
160
161
162
163

164
165
166

167
16S
169
170
171
172
173

174
175
176

177
ITS
179
180

1S1
1S2

1S3

1S4

1S5
1S6

1S7
1SS
1S9
190
191
192
193

19+
195
196

197
198
199
200

Ho.

o ; i j 2

1761 1764

1790
ISIS
1847

1S75
1903
1931

1959
19S7
2014
2041

1767 1770 1772

1793 1796]

1S21 1S24
1S50J 1853

1S7S 1SS1

1906
1934

1962
1989

2017

2044

2068
2095
2122

214S
2175
2201

2227
2253
2279

1909'
1937

1965|
1992 j

2019

2047

2304

2071
209S
2125

2151
2177
2204

2230
2256
22S1

179S

1S27
1S55

1SS4I
1912
1940

1967
1995

2022

1801
1S30

1S5S

1SS6
1915
1942

1970
199S

2025

2049 2052

2307

2330

2355 j
23S0 |

2405
2430[

2455

24S0J
2504

2529!

2074
2101
2127J

2154
2 ISO
2206

2232|
2258
22S4 1

2076 2079
2103) 2106
2130 2133

2156
21S3|
2209J

2235
2261|
22S7

2159
21S5
2212

22.

2263
22S9
23l0i 2312) ~23l5

iwo
235S
23S3

240S
2433
245S

24S2

2507
2531

i553j_:

25
2601 1

2625,

26481
2672
2695|

271SJ
2742
27651

2335
2360
23S5

2410
2435
2460

24S5
2509

2338

2363
23SS

2340
2365
2390

2413 2415
243S 2440
2463 2465

24S7
2512
2536

2490
2514

253S

255S, 2560

2579 25S2
2603 260i
2627 2629

2651
2674
2697

2721
2744
2767

27SS

2810
2S33
2856

287S
2900
2923

2945
2967
29S9

3010

2790

2S13
2S35
2S58

2S80
2903
2925

2947
2969
2991

3012

2562

2653
2676
2700

2723
2746
2769'

25S4
260S

2632

2655
2679
2702

2725
2749

2792 2794

2815
283S
2860

2SS3
2905
2927

2949

2971
2993|

2817
2840
2862

25S6
2610
2634

265S
26S1
2704

272S
2751
2774

2797

2S19

2S42
2865

2885 2SS7
2907 2909
2929: 2931

2951] 2953
2973 i 29!
299512997

30151 3017; 3019

8 i 9

1775 177S| 17S1 17S4 1787

1S04 1S07 1S10! 1S13; 1816
1S33 1S56 ! 1S3S 1S41I 1844
1S61 1S64 1S67! 1S70 1S72

1SS9S 1S92 1S95 1S9S
1917 1920 1923' 1926
19451 194S 1951 1953

1973,
2000|
202S

1976
2003
2030]

197S
2006
2033

2055; 2057, 2060

20S2
2109
2135

2162

21SS
2214

20S4
211l|

213S,

2164 1

219l|

2217

19S1
2009

2036

1901
1928
1956

19S4
2011

203S

1

2240; 2243
2266 2269
2292! 22941

20S
2114

2140

2167
21931
2219

2245
2271
2297i

2063 2066

2090
2117

2143'

21 70 1

2196!
}>->■>

2IW
2274!
22991

231;

20

2343
236S|

2393

241S
2443
2467,

2492
2516
2541-

2345
2370
2395

2420
2445!

2470

2494 1
2519
2543 i

2348

2375'
239S!

2423'
244S

2472!

2325
2350
2375
2400

242.

2450

2475

I

2497! 2499
2521! 2524
2545 254S

2092
2119
2146

2172
2198
2225

2251

2276
2302
2327
2353
237S
2403

242S

245
2477

2502

2526
2550

2565: 2567 25 70 25 72

25S9
2613
2636

2660 1
26S3I
2707|

2730' 2732

2755
277S ;
2S01

2591
2615,
2639;

2662!
26S6I
2709

27761

2594 25961
2617 2620
2641 2643

2665| 2667
26SS 2690
2711 2714,

2735 2737!
275S 2760;
27Sl!27S3!

2574
259S
2622
2646

2669
2693
2716

2739
2762

27S5

2799'

2S04 2806

2522
2S44
2S67

2SS9
291l|

2934J

2956!
297S
2999

3021

2S24,
2S47
2869J

2S9ll
2914
2936|

295S'
2980!
3002!

2S26
2S49|
2S7l|

2S94 1
2916
293S

2S2S
2S51
2874

2S96
2918
2940

2960| 2962
29S2I29S4
3004 3006

3023 302S302S

2808

2831
2853
2876

2898
2920
2942

2964
2986
3008

3030

6

8 ! 9

Prop. Farts

H

3

0.3
0.6

OS

is

1.5
1.8

2J.
2.4
2.7

2

1

0.2

2

0.4

3

0.6

4

0.8

S

1.0

6

1.2

7

1.4

s

1.6

9

1.S

TABLE I. LOGARITHMS OF NUMBERS

No.

1

2

3

4

5

6

7

8

9

Prop. Farts

20

21
22
23

24
25
26

27
28
29
30
31
32
33

34
35
36

37
38
39
40

41
42
43

44
45
46

47
48
49
50

51
52
53

54
55
56

57
5S
59
60
61
62
63

64
65
66

67
68
69
70

3010

3032

3054

3075

3096

3118

3139

3160

3181

3201

bo
•it
•a
d

a

M

Difference

3222
3424
3617

3802
3979
4150

4314
4472
4624

3243
3444
3636

3820
3997
4166

4330
4487
4639

3263
3464
3655

3838
4014
4183

4346
4502
4654

3284
3483
3674

3856
4031
4200

4362
4518
4669

3304
3502
3692

3874
4048
4216

4378
4533
4683

3324
3522
3711

3892
4065
4232

4393
4548
4698

3345
3541
3729

3909
4082
4249

4409
4564
4713

3365
3560
3747

3927
4099
4265

4425
4579
4728

3385
3579
3766

3945
4116

4281

4440
4594
4742

3404
3598
3784

3962
4133
4298

4456
4609

4757

1
2
3
4
5
6
7
8
9

22

2.2
4.4
6.6
8.8
11.0
13.2
15.4
17.6
19.8

21

2.1
4.2
6.3
8.4
10.5
12.6
14.7
16.8
18.9

4771

4786

4800

4814

4829

4843

4857

4871

4886

4900

4914
5051
5185

5315
5441
5563

5682
5798
5911

4928
5065
5198

5328
5453
5575

5694
5809
5922

4942
5079
5211

5340
5465
5587

5705
5821
5933

4955
5092
5234

5353
5478
5599

5717
5832
5944

4969
5105
5237

5366
5490
5611

5729
5843
5955

4983
5119
5250

5378
5502
5623

5740
5855
5966

4997
5132
5263

5391
5514
5635

5752
5866
5977

5011
5145
5276

5403
5527
5647

5763

5877
5988

5024
5159
5289

5416
5539
5658

5775
5888
5999

5038
5172
5302

5428
5551
5670

5786
5900
6010

1
2
3
4
5
6
7
8
9

20

2.0
4.0
6.0
8.0
10.0
12.0
14.0
16.0
18.0

19

1.9
3.8
5.7
7.6
9Ji
11.4
13.3
15.2
17.1

6021

6031

6042

6053

6064

6075

6085

6096

6107

6117

1
2
3
4
5
6
7
8
9

18

1.8
3.6
5.4
7.2
9.0
10.8
12.6
14.4
16.2

17

1.7
3.4
5.1
6.8
8.5
10.2
11.9
13.6
15.3

6128
6232
6335

6435
6532
6628

6721
6812
6902

6138
6243
6345

6444
6542
6637

6730
6821
6911

6149
6253
6355

6454
6551
6646

6739
6830
6920

6160
6263
6365

6464
6561
6656

6749
6839
6928

6170
6274
6375

6474
6571
6665

6758
6848
6937

6180
6284
6385

6484
6580
6675

6767
6857
6946

6191
6294
6395

6493
6590
6684

6776
6866
6955

6201
6304
6405

6503
6599
6693

6785
6875
6964

6212
6314
6415

6513
6609
6702

6794
6884
6972

6222
6325
6425

6522
6618
6712

6803
6893
6981

1

2
3

4
5
6
7
8
9

16

1.6
3.2
4.8
6.4
8.0
9.6
11.2
12.8
14.4

15

1.5

3.0

4.5

6.0

7.5

9.0

10.5

12.0

133

6990

6998

7007

7016

7024

7033

7042

7050

7059

7067

7076
7160
7243

7324
7404
7482

7559
7634
7709

7084
7168
7251

7332
7412
7490

7566
7642
7716

7093
7177
7259

7340
7419
7497

7574
7649
7723

7101
7185
7267

7348
7427
7505

7582
7657
7731

7110
7193

7275

7356
7435
7513

7589
7664
7738

7118
7202
7284

7364
7443
7520

7597
7672

7745

7126
7210
7292

7372
7451
7528

7604
7679
7752

7135
7218
7300

7380
7459
7536

7612
7686
7760

7143
7226
7308

7388
7466
7543

7619
7694
7767

7152
7235
7316

7396
7474
7551

7627
7701

7774

1

2
3

4
5
6
7
8
9

14

1.4
2.8
4.2
5.6
7.0
8.4
9.8
11.2
12.6

13

1.3
2.6
3.9
5.2
6.5
7.8
9.1
10.4
11.7

7782

7789

7796

7803

7810

7818

7825

7832

7839

7846

7853
7924
7993

8062
8129
8195

8261
8325
8388

7860
7931
SOOO

8069
8136
8202

8267
833,1
8395

7868
7938
8007

8075
8142
8209

8274
8338
"8401

7875
7945
8014

8082
8149
8215

8280
8344
8407

7882
7952
8021

8089
8156
8222

8287
8351
8414

7889
7959
8028

8096
8162
8228

8293
8357
8420

7896
7966
8035

8102
8169
8235

8299
8363
8426

7903
7973
8041

8109
8176
8241

8306
8370
8432

7910
7980
8048

8116
8182
8248

8312
8376
8439

7917
7987
8055

8122
8189
8254

8319
8382
8445

1
2
3
4
5
6
7
8
9

12

1.2
2.4
3.6
4.8
6.0
7.2
8.4
9.6
10.8

11

1.1
2.2
3.3
4.4
5.5
6.6
7.7
8.8
9.9

8451

8457

8463

8470

8476

8482

8488

8494

8500

8506

No.

1

2

3

4

5

6

7

8

9

TABLE I. LOGARITHMS OF NUMBERS

No.

1

2

3

4

5

6

7

8

9

Prop. Parts

70

71

72
73

74
75
76

77
78
79
80
81
82
83

84
85
86

87
88
89
90

91
92
93

94
95
96

97
98
99
100

8451

8457

8463

8470

8476

8482

8488

8494

8500

8506

Ex.
dig.

Difference

8513
8573
8633

8692
8751
8808

8865
S921
8976

8519
8579
8639

8698
8756
8814

8871
8927
8982

8525
8585
8645

8704
8762
8820

8876
8932
8987

8531
8591
8651

8710
8768
8825

8882
8938
8993

8537
8597
8657

8716
8774
8831

8887
8943
8998

8543
8603
8663

8722
8779
8837

8893
8949
9004

8549
8609
8669

8727
8785
8842

8899
8954
9009

8555
8615
8675

8733
8791
8848

8904
8960
9015

8561
8621
8681

8739
8797
8854

8910
8965
9020

8567
8627
8686

8745
8802
8859

8915
8971
9025

1
2

3

4
5
6
7
8
9

10

1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0

9

0.9

1.8
2.7'
3.6
4.5
5.4
6.3
7.2
8.1

1
2
3
i
5
6
7
8
9

1
2
3
4
5
6
7
8
9

8

0.8
1.6
2.4
3.2
4.0
4.8
5.6
6.4
7.2_

6

0.6
1.2
1.8
2.4
3.0
3.6
4.2
4.8
5.4

7
0.7
1.4
2.1
2.8
3.5
4.2
4.9
5.6
6.3

9031

9036

9042

9047

9053

9058

9063

9069

9074

9079

9085
9138
9191

9243
9294
9345

9395
9445
9494

9090
9143
9196

9248
9299
9350

9400
9450
9499

9096
9149
9201

9253
9304
9355

9405
9455
9504

9101
9154
9206

9258
9309
9360

9410
9460
9509

9106
9159
9212

9263
9315
9365

9415
9465
9513

9112
9165
9217

9269
9320
9370

9420
9469
9518

9117
9170
9222

9274
9325
9375

9425
9474
9523

9122
9175
9227

9279
9330
9380

9430
9479
9528

9128
9180
9232

9284
9335
9385

9435
9484
9533

9133
9186
9238

9289
9340
9390

9440
9489
9538

5

0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5

9542

9547

9552

9557

9562

9566

9571

9576

9581

9586

9590
9638
9685

9731
9777
9823

9868
9912
9956

9595
9643
9689

9736
9782
9827

9872
9917
9961

9600
9647
9694

9741
9786
9832

9877
9921
9965

9605
9652
9699

9745
9791
9836

9881
9926
9969

9609
9657
9703

9750
9795
9841

9886
9930
9974

9614
9661
9708

9754
9800
9845

9890
9934
9978
0022

9619
9666
9713

9759
9805
9850

9894
9939
9983

9624
9671
9717

9763
9809
9854

9899
9943
9987

9628
9675
9722

9768
9814
9859

9903
9948
9991

9633
9680
9727

9773
9818
9863

9908
9952
9996

1
2
3

4
5
6
7
8
9

4

0.4
' 0.8
1.2
1.6
2.0
2.4
2.8
3.2
3.6

0000

0004

0009

0013

0017

0026

0030

0035

0039

No.

1

2

3

4

5

6

7

8

9

RULES FOR FINDING THE LOGARITHMS OF THE

TRIGONOMETRIC FUNCTIONS OF ANGLES

NEAR 0° AND 90°

The derivation of the following rules will be found on page 182,
Granville's Plane Trigonometry.

If the angle is given in degrees, minutes, and seconds, it should
first be reduced to degrees and the decimal part of a degree. For
this purpose use the conversion table on page 17.

Rule I. To find the Logarithms of the Functions of an Angle
near 0°.*

log sin x° = 2.2419 + logx.
log tan x° = 2.2419 + log x.
log cot x° = 1.7581 - logx.
log cos x° is found from the tables in the usual way.

Rule II. To find the Logarithms of the Functions of an Angle
near 90°. \

log cos x° = 2.2419 + log (90 — x).
log cotx = 2.2419 + log (90 - x).
log tan x° = 1.7581 - log (90 - x).
log sin x° is found from the tables in the usual way.

These rules ■will give results accurate to four decimal places for
all angles between 0° and 1.1° and between 88.9° and 90°.

* Example 1, page 182, Granville's Plane Trigonometry, illustrates the application
of this rule.

t Example 2, page 183, Granville's Plane Trigonometry, illustrates the application
of this rule.

Table II

FOUR-PLACE LOGARITHMS OF TRIGONOMETRIC

FUNCTIONS, THE ANGLE BEING EXPRESSED

IN DEGREES AND MINUTES

This table gives the common logarithms (base 10) of the sines,
cosines, tangents, and cotangents of all angles from 0° to 5° and
from 85° to 90° for each minute ; and from 5° to 85° at intervals
of 10 minutes, all calculated to four places of decimals. In order
to avoid the printing of negative characteristics, the number 10 has
been added to every logarithm in the first, second, and fourth
columns (those having log sin, log tan, and log cos at the top).
Hence in writing down any logarithm taken from these three
columns — 10 should be written after it. Logarithms taken from
the third column (having log cot at the top) should be used as
printed.

A logarithm found from this table by interpolation may be in
error by one unit in the last decimal place, except for angles
between 0° and 18' or between 89° 42' and 90°, when the error may
be larger. In the latter cases the table refers the student to the
formulas on page 6 for more accurate results.

TABLE II. LOGARITHMIC SINES

0°

Angle

log sin

difi. 1'

log tan

com.
difi.l'

log cot

log cos

0° 0'

0° 1'

10.0000

90° 00'

89° S9 7

6.4637

6.4637

3.5363

10.0000

0° 2'

6.7648

c

6.7648

S

3.2352

10.0000

89° 58'

0° 3'

6.9408

=

6.9408

"

3.0592

10.0000

89° 57'

0° 4'

7.0658

£

7.0658

£

2.9342

10.0000

89° 56'

0° 5'

7.1627

Ss°'

7.1627

'— —

2.8373

10.0000

89° 55'

0° 6'

7.2419

7.2419

2.7581

10.0000

89° 54'

0° 7'

7.3088

7.3088

s"o

2.6912

10.0000

89° 53'

0° 8'

7.3668

ts^ «

7.3668

si 3

2.6332

10.0000

89° 52'

0° 9'
0° 10'

0° 11'

7.4180

— ^ -
a -- -

u -_ i-

« 3 o

7.4180

O P c

a - =
*« ° b

-5' H ®

<D ED

2.5820

10.0000

89° 51'
89° 50'
89° 49'

7.4637

7.4637

2.5363

10.0000

7.5051

7.5051

2.4949

10.0000

0° 12'

7.5429

7.5429

2.4571

10.0000

89° 48'

0° 13'

7.5777

7.5777

Sir*

2.4223

10.0000

89° 47'

0° 14'

7.6099

■Elf

7.6099

■sss

2.3901

10.0000

89° 46'

0° 15'

7.6398

7.6398

ogfl

2.3602

10.0000

89° 45'

0° 16'

7.6678

bo

7.6678

2.3322

10.0000

89° 44'

0° 17'

7.6942

7.6942

2.3058

10.0000

89° 43'

0° IS'

7.7190

7.7190

2.2810

10.0000

89° 42'

0° 19'
0°20'

0° 21'

7.7425

235
223
211

7.7425

235
223
212

2.2575

10.0000

89° 41'
89° 40'
89° 39'

7.7648

7.764S

2.2352

10.0000

7.7859

7.7S60

2.2140

10.0000

0° 22'

7.8061

202

7.8062

202

2.1938

10.0000

89° 38'

0° 23'

7.8255

194

7.8255

193

2.1745

10.0000

89° 37'

0° 24'

7.8439

184

7.8439

184

2.1561

10.0000

89° 36'

0° 25'

7.8617

178

7.8617

1(8

2.1383

10.0000

89° 35'

0° 26'

7.87S7

170

7.S787

170

2.1213

10.0000

89° 34'

0° 27'

7.8951

164

7.8951

164

2.1049

10.0000

89° 33'

0° 28'

7.9109

158

7.9109

158

2.0891

10.0000

89° 32'

0° 29'
0°30'

0° 31'

7.9261

152
147
143

7.9261

152
148
142

2.0739

10.0000

89° 31'
89° 30'

89° 2<y

7.940S

7.9409

2.0591

10.0000

7.9551

7.9551

2.0449

10.0000

0° 32'

7.9689

138

7.9689

138

2.0311

10.0000

S9° 28'

0° 33'

7.9822

133

7.9823

134

2.0177

10.0000

89° 27'

0° 34'

7.9952

130

7.9952

129

2.0048

10.0000

89° 26'

0° 35'

8.0078

126

8.0078

126

1.9922

10.0000

89° 25'

0° 36'

8.0200

122

8.0200

122

1.9800

10.0000

89° 24'

0° 37'

8.0319

119

8.0319

119

1.9681

10.0000

89° 23'

0° 3S'

8.0435

116

8.0435

116

1.9565

10.0000
10.0000

89° 22'

0° 39'
0°40'

0° 41'

8.0548

113

110
107

8.0548

113
110
107

1.9452

89° 21'
89° 20'

89° W

8.0658

8.0658

1.9342

10.0000

8.0765

8.0765

1.9235

10.0000

0° 42'

8.0870

105

8.0870

105

1.9130

10.0000

89° 18'

0° 43'

8.0972

102

8.0972

102

1.9028

10.0000

89° 17'

0° 44'

8.1072

100

8.1072

100

1.8928

10.0000

89° 16'

0° 45'

8.1169

97

8.1170

98

1.8830

10.0000

89° 15'

0° 46'

8.1265

96

8.1265

95

1.8735

10.0000

89° 14'

0° 47'

8.1358

93

8.1359

94

1.8641

10.0000

89° 13'

0° 48'

8.1450

92

8.1450

91

1.8550

10.0000

89° 12'

0° 49'
0°50'

0° 51'

8.1539

89
88
86

8.1540

90
87
86

1.8460

10.0000

89° 11'
89° 10'

89° 9'

8.1627

8.1627

1.8373

10.0000

8.1713

8.1713

1.8287

10.0000

0° 52'

8.1797

84

8.1798

85

1.8202

10.0000

89° 8'

0° 53'

8.1880

83

8.1880

82

1.8120

9.9999

89° 7'

0° 54'

8.1961

81

8.1962

82

1.8038

9.9999

89° 6'

0° 55'

8.2041

80

8.2041

79

1.7959

9.9999

89° 5'

0° 56'

8.2119

78

8.2120

79

1.7880

9.9999

89° 4'

0° 57'

8.2196

77

8.2196

76

1.7804

9.9999

89° 3'

0° 58'

8.2271

75

8.2272

76

1.7728

9.9999

89° 2'

0° 59'
0° 60'

8.2346

75
73

8.2346

74
73

1.7654

9.9999

89° 1'
89° C

8.2419

8.2419

1.7581

9.9999

log cos

diff.l'

log cot

com.
difi.l'

log tan

log sin

Angle

89°

COSINES, TANGENTS, AND COTANGENTS

1°

Angle

log sin

diS.l'

log tan

com.
difi.l'

log cot

log cos

1° 0'

1° 1'

8.2419

71

8.2419

72

1.7581

9.9999

88° 60'

88° 59'

8.2490

8.2491

1.7509

9.9999

1° 2'

8.2561

8.2562

71

1.7438

9.9999

88° 58'

1° 3'

8.2630

69

8.2631

69

1.7369

9.9999

88° 57'

1° 4'

8.2699

69

8.2700

69

1.7300

9.9999

88° 56'

1° 5'

8.2766

67

8.2767

67

1.7233

9.9999

88° 55'

1° 6'

8.2832

66

8.2833

66

1.7167

9.9999

88° 54'

1° 7'

8.2898

66

8.2899

66

1.7101

9.9999

88° 53'

1° 8'

8.2962

64

8.2963

64

1.7037

9.9999

88° 52'

1° 9'
1° 10'

1° 11'

8.3025

63
63

62

8.3026

63
63

61

1.6974

9.9999

S8° 51'
88° 50'

88° 49'

8.3088

8.3089

1.6911

9.9999

8.3150

8.3150

1.6850

9.9999

1° 12'

8.3210

60

8.3211

61

1.6789

9.9999

88° 48'

1° 13'

8.3270

60

8.3271

60

1.6729

9.9999

88° 47'

1° 14'

8.3329

59

8.3330

59

1.6670

9.9999

88° 46'

1° 15'

8.3388

59

8.3389

59

1.6611

9.9999

88° 45'

1° 16'

8.3445

57

8.3446

57

1.6554

9.9999

88° 44'

1° 17'

8.3502

57

8.3503

56

1.6497

9.9999

88° 43'

1° 18'

8.355S

56

8.3559

56

1.6441

9.9999

88° 42'

1° 19'
1°20'

1° 21'

8.3613

55
55
54

8.3614

55
55

54

1.6386

9.9999

88° 41'
88° 40'

88° 39'

8.3668

8.3669

1.6331

9.9999

8.3722

8.3723

1.6277

9.9999

1° 22'

8.3775

53

8.3776

1.6224

9.9999

88° 38'

1° 23'

8.3828

53

8.3829

53

1.6171

9.9999

88° 37'

1° 24'

8.3880

52

8.3881

52

1.6119

9.9999

88° 36'

1° 25'

8.3931

51

8.3932

51

1.6068

9.9999

88° 35'

1° 26'

8.3982

51

8.3983

51

1.6017

9.9999

88° 34'

1° 27'

8.4032

50

8.4033

50

1.5967

9.9999

88° 33'

1° 28'

8.4082

50

8.4083

50

1.5917

9.9999

88° 32'

1° 29'
1°30'

1° 31'

8.4131

49
49

48

8.4132

49
49

48

1.5868

9.9999

88° 31'
88° 30'

88° 29'

8.4179
8.4227

8.4181

1.5819

9.9999

8.4229

1.5771

9.9998

1° 32'

8.4275

48

8.4276

47

1.5724

9.9998

88° 28'

1° 33'

8.4322

47

8.4323

47

1.5677

9.9998

88° 27'

1° 34'

8.4368

46

8.4370

47

1.5630

9.9998

88° 26'

1° 35'

8.4414

46

8.4416

46

1.5584

9.9998

88° 25'

1° 36'

8.4459

45

8.4461

45

1.5539

9.9998

88° 24'

1° 37'

S.4504

45

8.4506

45

1.5494

9.9998

88° 23'

1° 38'

8.4549

45

8.4551

45

1.5449

9.9998

88° 2'2'

1° 39'
1°40'
1° 41'

8.4593

44
44
43

8.4595

44
43
44

1.5405

9.9998

88° 21'
88° 20'

88° 19'

8.4637

8.4638

1.5362

9.9998

8.4680

8.4682

1.5318

9.9998

1° 42'

8.4723

43

8.4725

43

1.5275

9.9998

88° 18'

1° 43'

8.4765

42

8.4767

42

1.5233

9.9998

88° 17'

1° 44'

8.4807

42

8.4809

42

1.5191

9.9998

88° 16'

1° 45'

8.4848

41

8.4851

42

1.5149

9.9998

88° 15'

1° 46'

8.4890

42

8.4892

41

1.5108

9.9998

88° 14'

1° 47'

8.4930

40

8.4933

41

1.5067

9.9998

88° 13'

1° 48'

8.4971

41

8.4973

40

1.5027

9.9998

88° 12'

1° 49'
1°50'

1° 51'

8.5011

40
39
40

8.5013

40
40
39

1.4987

9.9998

88° 11'
88° 10'

88° 9'

8.5050

8.5053

1.4947

9.9998

8.5090

8.5092

1.4908

9.9998

1° 52'

8.5129

39

8.5131

39

1.4869

9.9998

88° 8'

1° 53'

8.5167

38

8.5170

39

1.4830

9.9998

88° 7'

1° 54'

8.5206

39

8.5208

38

1.4792

9.9998

88° 6'

1° 55'

8.5243

37

8.5246

38

1.4754

9.9998

88° 5'

1° 56'

8.5281

38

8.5283

37

1.4717

9.9998

88° 4'

1° 57'

8.5318

37

8.5321

38

1.4679

9.9997

88° 3'

1° 58'

8.5355

37

8.5358

37

1.4642

9.9997

88° 2'

1° 59'
1°60'

8.5392

37 '
36

8.5394

36
37

1.4606

9.9997

88° 1'
88° 0'

8.5428

8.5431

1.4569

9.9997

log cos

difi.l'

log cot

com.
difi.l'

log tan

log sin

Angle

88°

10

TABLE II. LOGARITHMIC SINES

2°

Angle log sin difi.l' log tan dig}/ log cot log cos

0'

1'

2'
3'

4'
5'
6'
7'
8'

•r

10'

ir

2° 12'
2° 13'
2° 14'
2° IS'
2° 16'
2° 17'
2° 18'
2° Vf
2° 20'
2° 21'
2° 22'
2° 23'
2° 24'
2° 25'
2° 26'
2° 27'
2° 28'
2° 29'
2° 30'
2° 31'
2° 32'
2° 33'
2° 34'
2° 35'
2° 36'
2° 37'
2° 38'
2° 39'
2° 40'
2° 41'
2° 42'
2° 43'
2° 44'
2° 45'
2° 46'
2° 47'
2° 48'
2° 49'
2° 50'
2° 51'
2° 52'
2° 53'
2° 54'
2° 55'
2° 56'
2° 57'
2° 58'
2° 59'
2° 60'

8.5428

8.5464
8.5500
8.5535
8.5571
8.5605
8.5640
8.5674
8.5708
8.5742

8.5776

8.5809
8.5842
8.5875
8.5907
8.5939
8.5972
8.6003
8.6035
8.6066

8.6097

8.6128
8.6159
8.6189
8.6220
8.6250
8.6279
8.6309
8.6339
8.6368

8.6397

8.6426
8.6454
8.6483
8.6511
8.6539
8.6567
8.6595
8.6622
8.6650

8.6677

8.6704
8:6731
8.6758
8.6784
8.6810
8.6837
8.6863
8.6889
8.6914

8.6940

8.6965
8.6991
8.7016
8.7041
8.7066
8.7090
8.7115
8.7140
8.7164

8.7188

36
.16
35
36
34
35
34
34
34
34
33
33
33
33
32
33
31
32
31
31
31
31
30
31
30
29
30
30
29
29
29
28
29
28

27
28
27
27
27
27

27
26
26
25
26
25
26
25
25
25
24
25
25
24
24

8.5431

8.5467
8.5503
8.5538
8.5573
8.5608
8.5643
8.5677
8.5711
8.5745

8.5779

8.5812
8.5845
8.5878
8.5911
8.5943
8.5975
8.6007
8.6038
8.6070

8.6101

8.6132
8.6163
8.6193
8.6223
8.6254
8.6283
8.6313
8.6343
8.6372

8.6401

8.6430
8.6459
8.6487
8.6515
8.6544
8.6571
8.6599
8.6627
8.6654

8.6682

8.6709
8.6736
8.6762
8.6789
8.6815
8.6842
8.6868
8.6894
8.6920

8.6945

8.6971
8.6996
8.7021
8.7046
8.7071
8.7096
8.7121
8.7145
8.7170

8.7194

36
36
35
35
35
35
34
34
34
34
33
33
33
33
32
32
32
31
32
31
31
31
30
30
31
29
30
30
29
29
29

28
29
27

27
28
27
27
26
27
26
27
26
36
26
25
26
25
25
25
25
25
25
24
25
24

1.4569

1.4533
1.4497
1.4462
1.4427
1.4392
1.4357
1.4323
1.4289
1.4255

1.4221

1.4188
1.4155
1.4122
1.4089
1.4057
1.4025
1.3993
1.3962
1.3930
1.3899

1.3868
1.3837
1.3807
1.3777
1.3746
1.3717
1.3687
1.3657
1.3628

1.3599

1.3570
1.3541
1.3513

1.3485
1.3456
1.3429
1.3401
1.3373
1.3346

1.3318

1.3291
1.3264
1.3238
1.3211
1.3185
1.3158
1.3132
1.3106
1.3080

1.3055

1.3029
1.3004
1.2979
1.2954
1.2929
1.2904
1.2879
1.2855
1.2830

1.2806

9.9997

9.9997
9.9997
9.9997
9.9997
9.9997
9.9997
9.9997
9.9997
9.9997

9.9997

9.9997
9.9997
9.9997
9.9997
9.9997
9.9997
9.9997
9.9997
9.9996

9.9996

9.9996
9.9996
9.9996
9.9996
9.9996
9.9996
9.9996
9.9996
9.9996

9.9996

9.9996
9.9996
9.9996
9.9996
9.9996
9.9996
9.9995
9.9995
9.9995

9.9995

9.9995
9.9995
9.9995
9.9995
9.9995
9.9995
9.9995
9.9995
9.9995

9.9995

9.9995
9.9995
9.9995
9.9994
9.9994
9.9994
9.9994
9.9994
9.9994

9.9994

log cos difi. 1' log cot fljg™ f' log tan log sin

8T

87° 60'

87° 59'
87° 58'
87° 57'
87° 56'
87° 55'
87° 54'
87° 53'
87° 52'
87° 51'
87° 50'
87° 49'
87° 48'
87° 47'
87° 46'
87° 45'
87° 44'
87° 43'
87° 42'
87° 41'
87° 40'
87° Z<¥
87° 38'
87° 37'
87° 36'
87° 35'
87° 34'
87° 33'
87° 32'
87° 31'
87° 30'
87° 29'
87° 28'
87° 27'
87° 26'
87° 25'
87° 24'
87° 23'
87° 22'
87° 21'
87° 20'
87° 19'
87° 18'
87° 17'
87° 16'
87° 15'
87° 14'
87° 13'
87° 12'
87° 11'
87° 10'
87° 9'
87°
87°
87°
87°
87°
87°
87°
87°
87°

Angle

COSINES, TANGENTS, AND COTANGENTS

11

a

o

Angle

log sin

difi.l'

log tan

com.
difi.l'

log cot

log cos

3° 0'

3° 1'

8.7188

24

8.7194

24

1.2806

9.9994

86° 60'

86° 59'

8.7212

8.7218

1.2782

9.9994

3° 2'

8.7236

24

8.7242

24

1.2758

9.9994

86° 58'

3° 3'

8.7260

24

8.7266

24

1.2734

9.9994

86° 57'

3° 4'

8.7283

23

8.7290

24

1.2710

9.9994.

86° 56'

3° 5'

8.7307

24

8.7313

23

1.2687

9.9994

86° 55'

3° 6'

8.7330

23

8.7337

24

1.2663

9.9994

86° 54'

3° 7'

8.7354

24

8.7360

23

1.2640

9.9994

86° 53'

3° 8'

8.7377

23

8.7383

23

1.2617

9.9994

86° 52'

3° 9'
3° 10'

3° 11'

8.7400

23
23
22

8.7406

23
23
23

1.2594

9.9993

86° 51'
86° 50'

86° 49'

8.7423

8.7429

1.2571

9.9993

8.7445

8.7452

1.2548

9.9993

3° 12'

8.7468

23

8.7475

23

1.2525

9.9993

86° 48'

3° 13'

8.7491

23

8.7497

22

1.2503

9.9993

86° 47'

3° 14'

8.7513

22

8.7520

23

1.2480

9.9993

86° 46'

3° 15'

8.7535

22

8.7542

22

1.2458

9.9993

86° 45'

3° 16'

8.7557

22

8.7565

23

1.2435

9.9993

86° 44'

3° 17'

S.7580

23

8.7587

22

1.2413

9.9993

86° 43'

3° 18'

8.7602

22

8.7609

22

1.2391

9.9993

86° 42'

3° 19'
3° 20'

3° 21'

8.7623

21
22
22

8.7631

22
21
22

1.2369

9.9993

86° 41'
86° 40'

86° 39'

8.7645

8.7652

1.2348

9.9993

8.7667

8.7674

1.2326

9.9993

3° 22'

8.7688

2L

8.7696

22

1.2304

9.9993

86° 38'

3° 23'

8.7710

22

8.7717

21

1.2283

9.9992

86° 37'

3° 24'

8.7731

21

8.7739

22

1.2261

9.9992

86° 36'

3° 25'

8.7752

21

8.7760

21

1.2240

9.9992

86° 35'

3° 26'

8.7773

21

8.7781

21

1.2219

9.9992

86° 34'

3° 27'

8.7794

21

8.7802

21

1.2198

9.9992

86° 33'

3° 28'

8.7815

21

8.7823

21

1.2177

9.9992

86° 32'

3° 29'
3° 30'

3° 31'

8.7836

21
21
20

8.7844

21

21
21

1.2156

9.9992

86° 31'
86° 30'

86° 29'

8.7857

8.7865

1.2135

9.9992

8.7877

8.7886

1.2114

9.9992

3° 32'

8.7898

21

8.7906

20

1.2094

9.9992

86° 28'

3° 33'

8.7918

20

8.7927

21

1.2073

9.9992

86° 27'

3° 34'

8.7939

21

8.7947

20

1.2053

9.9992

86° 26'

3° 35'

8.7959

20

8.7967

20

1.2033

9.9992

86° 25'

3° 36'

8.7979

20

8.7988

21

1.2012

9.9991

86° 24'

3° 37'

8.7999

20

8.8008

20

1.1992

9.9991

86° 23'

3° 38'

8.8019

20

8.8028

20

1.1972

9.9991

86° 22'

3° 39'
3° 40'

3° 41'

8.8039

20
20
19

S.S048
8.S067

20
20
20

1.1952

9.9991

86° 21'
86° 20'
86° 19'

8.8059

1.1933

9.9991

8.8078

8.80S7

1.1913

9.9991

3° 42'

8.8098

20

8.8107

20

1.1893

9.9991

86° 18'

3° 43'

8.8117

19

8.S126

19

1.1874

9.9991

86° 17'

3° 44'

8.8137

20

8.8146

20

1.1854

9.9991

86° 16'

3° 45'

8.8156

19

8.8165

19

1.1835

9.9991

86° 15'

3° 46'

8.8175

19

8.8185

20

1.1815

9.9991

86° 14'

3° 47'

8.8194

19.

8.8204

19

1.1796

9.9991

86° 13'

3° 48'

8.8213

19

8.8223

19

1.1777

9.9990

86° 12'

3° 49'
3° 50'

3° 51'

8.8232

19
19
19

8.8242

19
19
19

1.1758

9.9990

86° 11'
86° 10'

86° 9'

8.8251

8.8261

1.1739

9.9990

8.8270

8.8280

1.1720

9.9990

3° 52'

8.8289

19

8.8299

19

1.1701

9.9990

86° 8'

3° 53'

8.8307

18

8.8317

18

1.1683

9.9990

86° 7'

3° 54'

8.8326

19

8.8336

19

1.1664

9.9990

86° 6'

3° 55'

8.8345

19

8.8355

19

1.1645

9.9990

86° 5'

3° 56'

8.8363

18

8.8373

18

1.1627

9.9990

86° 4'

3° 57'

8.8381

18

8.8392

19

1.1608

9.9990

86° 3'

3° 58'

8.8400

19

8.8410

18

1.1590

9.9990

86° 2'

3° 59'
3° 60'

8.8418

18
18

8.8428

18
18

1.1572

9.9990

86° 1'
86° 0'

8.8436

8.8446

1.1554

9.9989

log cos

difi.l'

log cot

com.
diff.1'

log tan

log sin

Angle

86°

12

TABLE II. LOGARITHMIC SINES

Angle

0'

1'
2'
3'
4'
5'
6'
V
8'
9'
10'
11'
12'
4° 13'
4° 14'
4° 15'
4° 16'
4° 17'
4° 18'
4° 19'
4° 20'
4° 21'
4° '22'
4° 23'
4° 24'
4° 25'
4° 26'
4° 27'
4° 28'
4° 29'
4° 30'
4° 31'
4° 32'
4° 33'
4° 34'
4° 35'
4° 36'
4° 37'
4° 38'
4° 39'
4° 40'
4° 41'
4° 42'
4° 43'
4° 44'
4° 45'
4° 46'
4° 47'
4° 48'
4° 49'
4° 50'
4° 51'
4° 52'
4° 53'
4° 54'
4° 55'
4° 56'
4° 57'
4° 58'
4° 59'
4° 60'

log sin (lift. 1

8.8436

8.8454
8.8472
8.8490
8.8508
8.8525
8.8543
8.8560
8.8578
8.8595

8.8613

8.8630
8.8647
8.8665
8.8682
8.8699
8.8716
8.8733
8.8749
8.8766

8.8783

8.8799
8.8816
8.8833
8.8849
8.8865
8.8882
8.8898
8.8914
8.8930

8.8962
8.8978
8.8994
8.9010
8.9026
8.9042
8.9057
8.9073
8.9089

8.9104

8.9119
8.9135
8.9150
8.9166
8.9181
8.9196
8.9211
8.9226
8.9241

8.9256

8.9271
8.9286
8.9301
8.9315
8.9330
8.9345
8.9359
8.9374
8.9388

8.9403

]8
18
18
18
17
18
17
18
17
18

17
18
17
17
17
17
10
17
17
16
17
17
16
16
17
IG
10
10
16
16
16
10
16
10
10
15
10
10
15
15
16
15
10
15
15
15
15
15
15
15
15
15
14
15
15
14
15
15
15

log tan

8.8446

8.8465
8.8483
8.8501
8.8518
8.8536
8.8554
S.8572
8.8589
8.8607

8.8624

8.8642
8.8659
8.8676
8.8694
8.8711
8.8728
8.8745
8.8762
8.8778

8.8795

8.8812
8.8829
8.8845
8.8862
8.8878
8.8895
8.8911
8.8927
8.8944

8.8960

8.8976
8.8992
8.9008
8.9024
8.9040
8.9056
8.9071
8.9087
8.9103

8.9118

8.9134
8.9150
8.9165
8.9180
8.9196
8.9211
8.9226
8.9241
8.9256

8.9272

8.9287
8.9302
8.9316
8.9331
8.9346
8.9361
8.9376
8.9390
8.9405

8.9420

com.
cliff. 1'

19
18
18
17
18
18
18
17
18
17
18
17
17
18
17
17
17
17
16
17

17
10
17
16
17
16
16
17
16
16
16
16
16
16
16
15
10
16
15
16
16
15
15
16
15
15
15
15
16
15
15
14
15
15
15
15
14
15
15

log cos diff.l' log cot d "jft™i

85°

log cot

1.1554

1.1535
1.1517
1.1499
1.1482
1.1464
1.1446
1.1428
1.1411
1.1393

1.1376

1.1358
1.1341
1.1324
1.1306
1.1289
1.1272
1.1255
1.1238
1.1222

1.1205

1.1188
1.1171
1.1155
1.1138
1.1122
1.1105
1.1089
1.1073
1.1056

1.1040

1.1024
1.1008
1.0992
1.0976
1.0960
1.0944
1.0929
1.0913
1.0897

1.0882

1.0866
1.0850
1.0835
1.0820
1.0804
1.0789
1.0774
1.0759
1.0744

1.0728

1.0713
1.0698
1.0684
1.0669
1.0654
1.0639
1.0624
1.0610
1.0595

1.0580

log tan

log cos

9.9989

9.9989
9.9989
9.9989
9.9989
9.9989
9.9989
9.9989
9.9989
9.9989

9.9989

9.9988
9.9988
9.9988
9.9988
9.9988
9.9988
9.9988
9.9988
9.9988

9.9988

9.9987
9.9987
9.9987
9.9987
9.9987
9.9987
9.9987
9.9987
9.9987

9.9987

9.9986
9.9986
9.9986
9.9986
9.9986
9.9986
9.9986
9.9986
9.9986

9.9986

9.9985
9.9985
9.9985
9.99S5
9.9985
9.9985
9.9985
9.9985
9.9985

9.9985

9.9984
9.9984
9.9984
9.9984
9.9984
9.9984
9.9984
9.9984
9.9984

9.9983

85° 60'

85° 59'
85° 58'
85° 57'
85° 56'
85° 55'
85° 54'
85° 53'
85° 52'
85° 51'
85° 50'

85° 49'
85° 48'
85° 47'
85° 46'
85° 45'
85° 44'
85° 43'
85° 42'
85° 41'
85° 40'
85° 39'
85° 38'
85° 37'
85° 36'
85° 35'
85° 34'
85° 33'
85° 32'
85° 31'
85° 30'

85° 29'
85° 28'
85° 27'
85° 26'
85° 25'
85° 24'
85° 23'
85° 22'
85° 21'
85° 20'

85° 19'
85° 18'
85° 17'
85° 16'
85° 15'
85° 14'
85° 13'
85° 12'
85° 11'
85° 10'

85° 9'

85°

85°

85°

85°

85°

85°

85°

85°

85°

7'
6'
5'
4'
3'
2'
1'
0'

log sin Angle

J

COSINES, TANGENTS, AND COTANGENTS

13

5°-15°

Angle

log sin

diff.l'

log tan

com.
diff.l'

log cot

log cos

diff.l'

5° 0'

5° 10'

8.9403

14.2

8.9420

14.3

1.0580

9.9983

.1

85° 0'

84° 50'

8.9545

8.9563

1.0437

9.9982

5° 20'

8.9682

13.7

8.9701

13.8

1.0299

9.9981

.1

84° 40'

5° 30'

8.9816

13.4

8.9836

13.5

1.0164

9.9980

.1

84° 30'

5° 40'

8.9945

12.9

8.9966

13.0

1.0034

9.9979

.1

84° 20'

5° SO'
6° 0'
6° 10'

9.0070
9.0192

12.5
12.2
11.9

9.0093

12.7
12.3
12.0

0.9907

9.9977

.2
.1
.1

84° 10'
84° 0'
83° 50'

9.0216

U.97S4

9.9976

9.0311

9.0336

0.9664

9.997S

6° 20'

9.0426

11.5

9.0453

11.7

0.9547

9.9973

.2

83° 40'

6° 30'

9.0539

11.3

9.0567

11.4

0.9433

9.9972

.1

83° 30'

6° 40'

9.0648

10.9

9.067S

11.1

0.9322

9.9971

.1

83° 20'

6° SO'
7° 0'
7° 10'

9.0755

10.7
10.4
10.2

9.0786

10.8
10.5
10.4

0.9214

9.9969

.2
.1
.2

83° 10'
83° 0'

82° SC

9.0S59

9.0891

0.9109

9.9968

9.0961

9.0995

0.9005

9.9966

7° 20'

9.1060

9.9

9.1096

10.1

0.S904

9.9964

.2

82° 40'

7° 30'

9.1157

9.7

9.1194

9.8

0.8806

9.9963

.1

82° 30'

7° 40'

9.12S2

9.5

9.1291

9.7

0.8709

9.9961

.2

82° 20'

7° SO'
8° 0'

S° 10'

9.1345

9.3
9.1
8.9

9.138S

9.4
9.3
9.1

0.8615

9.9959

.2
.1
.2

82° 10'
82° 0'
81° 50'

9.1436

9.1478

0.8522

9.9958

9.1525

9.1569

0.8431

9.9956

8° 20'

9.1612

8.7

9.1658

8.9

0.8342

9.9954

.2

81° 40'

8° 30'

9.1697

8.5

9.1745

8.7

0.825S

9.9952

.2

81° 30'

8° 40'

9.1781

8.4

9.1831

8.6

0.8169

9.99S0

.2

81° 20'

8° SO'
9° 0'

9° 10'

9.1863

8.2
8.0
7.9

9.1915

8.4
8.2
8.1

0.S08S

9.9948

.2
.2
.2

81° 10'
81° 0'

80° 50'

9.1943

9.1997

0.8003

9.9946

9.2022

9.2078

0.7922

9.9944

9° 20'

9.2100

7.8

9.2158

8.0

0.7842

9.9942

.2

80° 40'

9° 30'

9.2176

7.6

9.2236

7.8

0.7764

9.9940

.2

80° 30'

9° 40'

9.22S1

7.5

9.2313

7.7

0.7687

9.9938

.2

80° 20'

9° 50'
10° 0'

10° 10'

9.2324

7.3
7.3
7.1

9.2389

7.6
7.4
7.3

0.7611

9.9936

.2
.2
.3

80° 10'
80° 0'

79° 50'

9.2397

9.2463

0.7537

9.9934

9.2468

9.2536

0.7464

9.9931

10° 20'

9.2538

7.0

9.2609

7.3

0.7391

9.9929

.2

79° 40'

10° 30'

9.2606

6.8

9.2680

7.1

0.7320

9.9927

.2

79° 30'

10° 40'

9.2674

6.8

9.2750

7.0

0.72SO

9.9924

79° 20'

10° 5C
11° 0'
11° 10'

9.2740

6.6
6.6
6.4

9.2S19

6.9
6.8
6.6

0.7181

9.9922

.2
.3
.2

79° 10'
79° 0'

78° SO'

9.2806

9.2887

0.7113

9.9919

9.2870

9.2953

0.7047

9.9917

11° 20'

9.2934

6.4

9.3020

6.7

0.6980

9.9914

.3

7S° 40'

11° 3<y

9.2997

6.3

9.3085

6.5

0.6915

9.9912

.2

78° 3C

11° 40'

9.3058

6.1

9.3149

6.4

0.68S1

9.9909

.3

78° 20'

11° SO'
12° 0'

12° 10'

9.3119

6.1
6.0
5.9

9.3212

6.3
6.3
6.1

0.6788

9.9907

.3
.3

78° IV
78° 0'

77° 50'

9.3179

9.3275

0.6725

9.9904

9.3238

9.3336

0.6664

9.9901

12° 20'

9.3296

5.8

9.3397

6.1

0.6603

9.9899

77° 40'

12° 30'

9.3353

5.7

9.3458

6.1

0.6542

9.9896

.3

77° 3C

12° 40'

9.3410

5.7

9.3S17

5.9

0.6483

9.9893

77° 20'

12° 50'
13° 0'
13° 1C

9.3466

5.6
5.5
5.4

9.3576

5.9
5.8
5.7

0.6424

9.9S90

.3
.3
.3

77° 10'
77° 0'
76° 50'

9.3521

9.3634

0.6366

9.9887

9.3S75

9.3691

0.6309

9.98S4

13° 20'

9.3629

5.4

9.3748

5.7

0.6252

9.9881

76° 40'

13° 30'

9.3682

5.3

9.3804

5.6

0.6196

9.9878

76° 30'

13° 4C

9.3734

5.2

9.3859

5.5

0.6141

9.9875

76° 20'

13° SO-
14° 0'
14° 10'

9.3786

5.2

5.1
5.0

9.3914

5.5
5.4
5.3

0.6086

9.9872

.3
.3
.3
.4
.3

76° 10'
76° 0'

75° SC

9.3837

9.3968

0.6032

9.9869

9.38S7

9.4021

0.5979

9.9866

14° 20'

9.3937

5.0

9.4074

5.3

0.5926

9.9863

75° 40'

14° 30'

9.3986

4.9

9.4127

5.3

0.5873

9.9859

75° 3(f

14° 40-

9.4035

4.9

9.4178

5.1

0.S822

9.9856

75° 20'

14° 50'
15° 0'

9.4083

4.8
4.7

9.4230

5.2
5.1

O.S770

9.98S3

.3

.4

75° 10'
75° 0'

9.4130

9.4281

0.5719

9.9849

log cos

diff.l'

log cot

com.
diff. 1'

log tan

log sin

diff.l'

Angle

75°-85°

u

TABLE II. LOGARITHMIC SINES

15°-25°

Angle

log sin

cliff. 1'

log tan

com.
difi.r

log cot

log cos

cliff. 1'

15° 0'

15° 10'

9.4130

4.7

9.4281
9.4331

5.0

0.5719

9.9849

.3

75° 0'

74° 50'

9.4177

0.5669

9.9846

15° 20'

9.4223

4.6

9.4381

5.0

0.5619

9.9843

.3

74° 4C

15° 3V

9.4269

4.6

9.4430

4.9
4.9

0.5570

9.9839

.4

74° 30'

15° 40'

9.4314

9.4479

0.5521

9.9836

.3

74° 2V

15° SO'
16° 0'
16° 10'

9.4359

4.5
4.4
4.4

9.4527

4.8
4.8
4.7

0.5473

9.9832

.4
.4
.3

74° 10'
74° 0'
73° 50'

9.4403

9.4575

0.5425

9.9828

9.4447

9.4622

0.5378

9.9825

16° 20'

9.4491

4.4

9.4669

4.7

0.5331

9.9821

.4

73° 40'

16° 30'

9.4533

4.2

9.4716

4.7

0.5284

9.9817

.4

73° 30'

16° 40'

9.4576

4.3

9.4762

4.6

0.5238

9.9814

.3

73° 20'

16° SO'
17° 0'

17° 10'

9.4618

4.2
4.1
4.1

9.4808

4.6
4.5
4.5

0.5192

9.9810

.4
.4
.4

73° 10'
73° 0'

72° 5C

9.4659

9.4853

0.5147

9.9806

9.4700

9.4898

0.5102

9.9802

17° 20'

9.4741

4.1

9.4943

4.5

0.5057

9.9798

.4

72° 40-

17° 30'

9.4781

4.0

9.4987

4.4

0.5013

9.9794

.4

72° 30*

17° 40'

9.4821

4.0

9.5031

4.4

0.4969

9.9790

.4

72° 20'

17° 50'
18° 0'
18° 1C

9.4861

4.0
3.9
3.9

9.5075

4.4
4.3
4.3

0.4925

9.9786

.4
.4
.4

72° 10'
72° 0'
71° 50'

9.4900

9.5118

0.4882

9.9782

9.4939

9.5161

0.4839

9.9778

18° 20'

9.4977

3.8

9.S203

4.2

0.4797

9.9774

.4

71° 40'

18° 30'

9.5015

3.8

9.5245

4.2

0.4755

9.9770

.4

71° 30'

18° 40'

9.5052

3.7

9.5287

4.2

0.4713

9.9765

.5

71° 2C

18° 50'
19° 0'

19° 10'

9.5090

3.8
3.6
3.7

9.5329

4.2

4.1

4.1

0.4671

9.9761

.4
.4
.5

71° 10'
71° 0'

70° 50'

9.5126

9.5370

0.4630

9.9757

9.5163

9.5411

0.4589

9.9752

19° 20'

9.5199

3.6

9.5451

4.0

0.4549

9.9748

.4

70° 40'

19° 30'

9.5235

9.5491

4.0

0.4509

9.9743

.5

70° 30'

19° 40'

9.5270

9.5531

4.0

0.4469

9.9739

.4

70° 20'

19° 50'
20° 0'

20° IV

9.5306

3.6
3.5
3.4

9.5571

4.0
4.0
3.9

0.4429

9.9734

.5
.4
.5

70° 10'
70° 0'

69° 50'

9.5341

9.5611

0.4389

9.9730

9.5375

9.5650

0.4350

9.9725

20° 20'

9.5409

3.4

9.5689

3.9

0.4311

9.9721

.4 !

69° 40'

20° 30'

9.5443

9.5727

3.8

0.4273

9.9716

.5 1

69° 30'

20° 40'

9.5477

3.3
3.3
3.3

9.5766

3.9

0.4234

9.9711

.5

69° 20'

20° SO'
21° 0'

21° 10'

9.5510

9.5804

3.8
3.8
3.7

0.4196

9.9706

.5
.4
.5

69° 10'
69° 0'

68° 50'

9.5543

9.5842

0.4158

9.9702

9.5576

9.5879

0.4121

9.9697

21° 20'

9.5609

9.5917

3.8

0.4083

9.9692

.5

68° 40'

21° 30'

9.5641

3.2

9.5954

3.7

0.4046

9.9687

.5

68° 30'

21° 40'

9.5673

3.2

9.5991

3.7

0.4009

9.9682

.5

68° 20'

21° 50'
22° 0'
22° 1C

9.5704

3.1
3.2
3.1

9.6028

3.7
3.6
3.6

0.3972

9.9677

.5
.5
.5

68° 10'
68° 0'

67° 50'

9.5736

9.6064

0.3936

9.9672

9.5767

9.6100

0.3900

9.9667

22° 20'

9.S798

3.1

9.6136

3.6

0.3864

9.9661

.6

67° 40'

22° 30'

9.S828

3.0

9.6172

3.6

0.3828

9.9656

.5

67° 30'

22° 40'

9.5859

.3.1

9.6208

3.6

0.3792

9.9651

.5

67° 20'

22° 50'
23° 0'
23° 10'

9.5889

3.0
3.0
2.9

9.6243

3.5
3.6
3.5

0.3757

9.9646

.5
.6
.5

67° 10'
67° 0'
66° 50'

9.5919

9.6279

0.3721

9.9640

9.5948

9.6314

0.3686

9.9635

23° 20'

9.5978

3.0

9.6348

3.4

0.3652

9.9629

.6 !

66° 40'

23° 30'

9.6007

2.9

9.6383

3.5

0.3617

9.9624

.5

66° 30'

23° W

9.6036

9.6417

3.4

0.3583

9.9618

.6

66° 20'

23° 50'
24° 0'

24° 10'

9.6065

2.9

2.8
2.8

9.6452

3.5
3.4
3.4

0.3548

9.9613

.5
.6
.5

66° 10-
66° 0'

65° 5C

9.6093

9.6486

0.3514

9.9607

9.6121

9.6520

0.3480

9.9602

24° 20"

9.6149

2.8

9.6553

3.3

0.3447

9.9596

.6

65° W

24° 30'

9.6177

2.8

9.6587

3.4

0.3413

9.9590

.6

65° 3C

24° 40'

9.6205

2.8

9.6620

3.3

0.3380

9.9584

.6

65° 2C

24° 50'
25° 0'

9.6232

2.7
2.7

9.6654

3.4
3.3

0.3346

9.9579

.5
.6

65° 10'
65° 0'

9.6259

9.6687

0.3313

9.9573

log cos

diff. 1'

log cot

com.
die. V

log tan

log sin

diff. 1'

Angle

6

>5°-75

COSINES, TANGENTS, AND COTANGENTS

15

25°-3S

1°

Angle

log sin

diff.1'

log tan

com.

diff.r

log cot

log cos dij

1.1'

25° 0'

25° 10'

9.6259

2.7

9.6687

3.3

0.3313

9.9573

6

65° 0'

64° 5C

9.6286

9.6720

0.3280

9.9567

25° 20 /

9.6313

2.7

9.6752

3.2 "

0.3248

9.9561

6

64° 40'

25° 30'

9.6340

2.7

9.6785

3.3

0.3215

9.9555

6

64° 30'

25° 40'

9.6366

2.6

9.6817'

3.2

0.3183

9.9549

6

64° 20'

25° 50"
26° 0'
26° 1C

9.6392

2.6
2.6
2.6

9.6850

3.3
3.2
3.2

0.3150

9.9543

6
6

7

64° 10'
64° 0'
63° 50'

9.6418

9.6882

0.3118

9.9537

9.6444

9.6914

0.3086

9.9530

26° 20'

9.6470

2.6

9.6946

3.2

0.3054

9.9524

6

63° 40"

26° 30'

9.6495

2.5

9.6977

3.1

0.3023

9.9518

6

63° 30'

26° 40'

9.6521

2.6

9.7009

3.2

0.2991

9.9512

6

63° 20-

26° SW
27° 0'

27° 10'

9.6546

2.5
2.4
2.5

9.7040

3.1
3.2
3.1

0.2960

9.9505

7
6

7

63° 10'
63° 0'

62° 50'

9.6570

9.7072

0.2928

9.9499

9.6595

9.7103

0.2897

9.9492

27° 20'

9.6620

2.5

9.7134

3.1

0.2866

9.9486

6

62° 40'

27° 3^

9.6644

2.4

9.7165

3.1

0.2835

9.9479

7

62° 30'

27° 4C

9.6668

2.4

9.7196

3.1

0.2804

9.9473

6

62° 20'

27° 5V
28° 0'

28° 10'

9.6692

2.4
2.4
2.4

9.7226

3.0
3.1

3.0

0.2774

9.9466

7
7
6

62° 10 7
62° 0'
61° 50'

9.6716

9.7257

0.2743

9.9459

9.6740

9.7287

0.2713

9.9453

28° 20'

9.6763

2.3

9.7317

3.0

0.2683

9.9446

7

61° W

28° 3CK

9.6787

2.4

9.7348

3.1

0.2652

9.9439

7

61° 3^

28° 4C

9.6810

2.3

9.7378

3.0

0.2622

9.9432

7

61° 20'

28° 5C
29° 0'

29° 1C

9.6833

2.3
2.3
2.2

9.7408

3.0
3.0
2.9

0.2592

9.9425

7

7
7

61° 1C
61° C
60° 50"

9.6856

9.7438

0.2562

9.9418

9.6878

9.7467

0.2533

9.9411

29° 20'

9.6901

2.3

9.7497

3.0

0.2503

9.9404

7

60° 40'

29° 30'

9.6923

2.2

9.7526

2.9

0.2474

9.9397

7

60° 30"

29° W

9.6946

2.3

9.7556

3.0

0.2444

9.9390

7

60° 20'

1<P 50"
30° 0'

30° 1C

9.6968

2.2
2.2
2.2

9.7585

2.9

2.9
2.9

0.2415

9.9383

7
8

7

60° 10'
60° C

59° 5C

9.6990

9.7614

0.2386

9.9375

9.7012

9.7644

0.2356

9.9368

30° 20'

9.7033

2.1

9.7673

2.9

0.2327

9.9361

7

59° 40-

30° 30'

9.7055

2.2

9.7701

2.8

0.2299

9.9353

8

59° 3C

30° 40'

9.7076

2.1

9.7730

2.9

0.2270

9.9346

7

59° 2C

30° 50'
31° 0'

31° 10'

9.7097

2.1
2.1
2.1

9.7759

2.9
2.9
2.8

0.2241

9.9338

8
.7
.8

59° 1C
59° W

58° 5C

9.7118

9.7788

0.2212

9.9331

9.7139

9.7816

0.2184

9.9323

3i° 2<y

9.7160

2.1

9.7845

2.9

0.2155

9.9315

8

58° 40'

31° 3C

9.7181

2.1

9.7873

2.8

0.2127

9.9308

7

58° 30'

31° 40 7

9.7201

2.0

9.7902

2.9

0.2098

9.9300

.8

58° 20'

31° 50'
32° 0'

32° 10'

9.7222
9.7242
9.7262

2.1
2.0
2.0

9.7930

2.8
2.8
2.8

0.2070

9.9292

8
8
8

58° 10-
58° 0'

57° 5C

9.7958

0.2042

9.9284

9.7986

0.2014

9.9276

32° 2(y

9.7282

2.0

9.8014

2.8

0.1986

9.9268

8

57° 40'

32° 3<y

9.7302

2.0

9.8042

2.8

0.1958

9.9260

8

57° 30'

32° 4C

9.7322

2.0

9.8070

2.8

0.1930

9.9252

8

57° 20"

32° 5C
33° 0'

33° 10-

9.7342

2.0
1.9
1.9

9.8097

2.7

2.8
2.8

0.1903

9.9244

8
8
8
9
8
8
9
8
9
8
9
9

57° 1C
57° C
56° 5C

9.7361

9.8125

0.1875

9.9236

9.7380

9.8153

0.1847

9.9228

33° 2W

9.7400

2.0

9.8180

2.7

0.1820

9.9219

56° 40'

33° 30'

9.7419

1.9

9.8208

2.8

0.1792

9.9211

56° 30"

33° 40'

9.7438

1.9

9.8235

2.7

0.1765

9.9203

56° 20'

33° sor
34° 0'
34° IV

9.7457

1.9
1.9
1.8

9.8263

2.8
2.7

2.7

0.1737

9.9194

56° 1C
56° 0'

55° sty

9.7476

9.8290

0.1710

9.9186

9.7494

9.8317

0.1683

9.9177

34° 20-

9.7513

1.9

9.8344

2-7

0.1656

9.9169

55° 4C

34° 3C

9.7531

1.8

9.8371

2 7

0.1629

9.9160

55° 3C

34 e W

9.7550

1.9

9.8398

2.7

0.1602

9.9151

55° 20'

34° 50-
35° 0'

9.7568

1.8
1.8

9.8425

2.7
2.7

0.1575

9.9142

9

8

55° 1C
55° 0'

9.7586

9.8452

0.1548

9.9134

log cos

diff.1'

log cot

com.
diff.1'

log tan

log sin dil

5.1'

Angle

55°-65

1°

16

TABLE II. LOGARITHMIC SINES

35°- 45°

Angle

log sin difi. 1

l°gtan ffii

log cot

log COS

difi. 1'

35° 0'

35° 10'
35° 2C
35° 30'
35° W
35° SO'
36° 0'
36° 10'
36° 20'
36° 30'
36° 40'
36° SO'
37° 0'
37° 10'
37° 20'
37° 30'
37° W
37° SO-
38° 0'
38° 10'
38° 20'
38° 30'
38° 40'
38° 50'
39° 0'

39° i<y

39° 20'
39° 30'
39° 40'
39° SO'
40° 0'
40° IV
40° 20'
40° 30'
40° 4C
40° SV
41° 0'
41° 10'
41° 20'
41° 30'
41° 40'
41° SO'
42° 0'
42° 10'
42° 20'
42° 30'
42° 40'
42° SO'
43° 0'
43° 10'
43° 20'
43° 30'
43° 40'
43° SO'
44° 0'
44° 10'
44° 20'
44° 30'
44° 40'
44° SO'
45° 0'

9.7S86

9.7604
9.7622
9.7640
9.76S7
9.7675

9.7692

9.7710
9.7727
9.7744
9.7761
9.7778

9. 7795

9.7811
9.7828
9.7844
9.7861
9.7877

9.7893

9.7910
9.7926
9.7941
9.7957
9.7973

9.7989

9.8004
9.8020
9.8035
9.8050
9.8066

J.S081

9.8096
9.8111
9.812S
9.8140
9.815S
9.8169

9.8184
9.8198
9.8213
9.8227
9.8241

9.S255

9.8269
9.8283
9.8297
9.8311
9.8324

9.8338

9.8351
9.8365
9.8378
9.8391
9.8405

9.8418

9.8431
9.8444
9.8457
9.8469
9.8482

9.8495

log cos difi. 1

1.8
1.8
1.8
1.7
1.8
1.7
1.8
1.7
1.7
1.7
1.7
1.7
1.6
1.7
1.6
1.7
1.6
1.6
1.7
1.6
1.5
1.6
1.6
1.6
1.5
1.6
1.5
1.5
1.6
1.5
1.5
1.5
1.4
1.5
1.5
1.4
1.5
1.4
1.5
1.4
1.4
1.4
1.4
1.4
1.4
1.4
1.3
1.4
1.3
1.4
1.3
1.3
1.4
1.3
1.3
1.3
1.3
1.2
1.3
1.3

9.8452

9.8479
9.8506
9.8533
9.8S59
9.8586

9.8613

9.8639
9.8666
9.8692
9.8718
9.874S

9.8771

9.8797
9.8824
9.8850
9.8876
9.8902

9.8928

9.8954
9.8980
9.9006
9.9032
9.905 S

9.9084

9.9110
9.9135
9.9161
9.9187
9.9212

9.9238

9.9264
9.9289
9.9315
9.9341
9.9366

9.9392

9.9417
9.9443
9.9468
9.9494
9.9519

9.9544

9.9570
9.9S95
9.9621
9.9646
9.9671

9.9697

9.9722
9.9747
9.9772
9.9798
9.9823
9.9848

9.9874
9.9899
9.9924
9.9949
9.9975

0.0000

2.7

2.7
2.7
2.6
2.7
2.7
2.6
2.7
2.6
2.6
2.7
2.6
2.6
2.7
2.6
2.6
2.6
2.6
2.6
2.6
2.6
2.6
2.6
2.6
2.6
2.5
2.6
2.6
2.5
2.6
2.6
2.5
2.6
2.6
2.5
2.6
2.5
2.6
2.5
2.6
2.5
2.5
2.6
2.5
2.6
2.5
2.5
2.6
2.5
2.5
2.5
2.6
2.5
2.5
2.6
2.5
2.5
2.5
2.6
2.5

0.1S48

9.9134

0.1S21
0.1494
0.1467
0.1441
0.1414

9.912S
9.9116
9.9107
9.9098
9.9089

0.1387

0.1361
0.1334
0.1308
0.1282
0.1255

0.1229

0.1203
0.1176
0.1150
0.1124
0.1098

0.1072

0.1046
0.1020
0.0994
0.0968
0.0942

0.0916

0.0890
0.0865
0.0839
0.0813
0.0788

0.0762

0.0736
0.0711
0.0685
0.0659
0.0634

0.0608

0.0583
0.0SS7
0.0532
0.0506
0.0481

0.04S6

0.0430
0.0405
0.0379
0.03S4
0.0329

0.0303

0.0278
0.02S3
0.0228
0.0202
0.0177

0.0152

0.0126
0.0101
0.0076
0.0051
0.0025

0.0000

log cot ^i, log tan

45°- 55°

9.9080

9.9070
9.9061
9.9052
9.9042
9.9033

9.9023

9.9014
9.9004
9.899S
9.898S
9.8975

9.8965

9.8955
9.8945
9.8935
9.8925
9.8915

9.8905

9.889S
9.8884
9.8874
9.8864
9.8853

9.8843

9.8832
9.8821
9.8810
9.8800
9.8789

9.8778

9.8767
9.8756
9.874S
9.8733
9.8722

9.8711

9.8699
9.8688
9.8676
9.8665
9.8653

9.8641

9.8629
9.8618
9.8606
9.8594
9.8582

9.8569

9.85S7
9.8S4S
9.8S32
9.8520
9.8S07

9.8495

.9
.9

.9

.9

.9

.9

1.0

.9

.9

1.0

.9

1.0

.9

1.0

.9

1.0

1.0

1.0

1.0

1.0

1.0

1.0

1.0

1.0

1.0

1.1

1.0
1.0

1.1

1.0

1.1
1.1

1.1

1.0

1.1
1.1
1.1
l.l

1.1

1.2
1.1
1.1
1.2
1.1
1.2
1.1
1.2
1.2
1.2
1.1
1.2
1.2
1.2
1.3
1.2
1.2
1.3
1.2
1.3
1.2

log sin

difi. V

55° 0'

54° 50'
54° 40'
54° 30'
54° 20'
54° 10'
54° 0'
53° SO'
53° 40'
53° 30'
53° 20'
53° 10'
53° 0'
52° SO'
52° 40'
52° 30'
52° 20'
52° 10'
52° 0'
51° 50'
51° 40'
51° 30'
51° 20'
51° 1C
51° 0'
50° SO'
50° 40'
50° 30'
50° 20'
50° 10'
50° 0'
49° 50'
49° 40'
49° 30'
49° 20'
49° 10'
49° 0'
48° SO'
48° 40'
48° 30'
48° 20'
48° 10'
48° 0'
47° 50'
47° 40'
47° 30'
47° 20'
47° 10'
47° 0'
46° 50'
46° 40'
46° 30'
46° 20'
46° 10'
46° 0'
45° 50'
45° 40'
45° 30'
45° 20'
45° 10'
45° 0'

Angle

CONVERSION TABLES FOR ANGLES

17

To change fkom Minutes and Seconds into the Decimal
Parts of a Degree or into Radians

From seconds

1"=

2"=

3"=

4"=

5" =

6"=

7"=

8"=

9"=

10"=

20"=

30"=

40"=

50"=

0.00028°=
0.00056° =
0.00083°=
0.00111°=
0.00139°=
0.00167°=
0.00194°=
0.00222°=
0.00250°=
0.00278° =
0.00556°=
0.00833°=
0.01111°=
0.01389°=

0.0000048 Rad.
0.0000097
0.0000145
: 0.0000194
: 0.0000242
; 0.0000291
0.0000339
: 0.0000388
0.0000436
0.0000485
0.0000970
0.0001454
0.0001939
0.0002424

From minutes

l'=0.017°=

2'= 0.033°=

3'=0.050°=

4' =0.067°=

5' =0.083°=

6'=0.100°=

7'=0.117°=

8'=0.133°=

9'=0.150°=

10'=0.167°=

20'=0.333°=

30'=0.500°=

40' =0.667°=

50'=0.833°=

0.00029 Rad.

:0.00058 "

0.00087 "

: 0.00116 "

: 0.00145 "

: 0.00175 "

0.00204 "

:0.00233 "

0.00262 "

0.00291 "

0.00582 "

0.00873 "

0.01164 "

0.01454 "

From degrees into
radians

1°=0
2°=0.
3°=0
4°=0
5°=0
6°=0
7°=0
8°=0
9°=0

10° =0.

20° =0.

30° =0.

40° =0.

50° =0.

.01745 Rad.

03491

.05236

.06981

.08727

.10472

.12217

13963

.15708

17453

34907

52360

69813

87266

to change from decimal parts of a degree into mlnutes

and Seconds

0.0000° =

0.000' =

0"

0.20° = 12.0' = 12'

0.60° = 36.0' = 36'

0.0001° =

0.006' =

0.36"

0.21° = 12.6' = 12' 36"

0.61° = 36.6' = 36' 36"

0.0002° =

0.012' =

0.72"

0.22° = 13.2' = 13' 12"

0.62° = 37.2' = 37' 12"

0.0003° =

0.018' =

1.08"

0.23° = 13.8' = 13' 48"

0.63° = 37.8' = 37' 48"

0.0004° =

0.024' =

1.44"

0.24° = 14.4' = 14' 24"

0.64° = 38.4' = 38' 24"

0.0005° =

0.030' =

1.80"

0.25° = 15.0' = 15'

0.65° = 39.0' = 39'

0.0006° =

0.036' =

2.16"

0.26° = 15.6' = 15' 36"

0.66° = 39.6' = 39' 36"

0.0007° =

0.042' =

2.52"

0.27° = 16.2' = 16' 12"

0.67° = 40.2' = 40' 12"

0.0008° =

0.048' =

2.88"

0.28° = 16.8' = 16' 48"

0.68° = 40.8' = 40' 48"

0.0009° =

0.054' =

3.24"

0.29° = 17.4' = 17' 24"

0.69° = 41.4' = 41' 24"

0.0010° =

0.06(r =

3.60"

0.30° = 18.0' = 18'

0.70° = 42.0' = 42'

0.001° =

0.06' =

3.6"

0.31° = 18.6' = 18' 36"

0.71° = 42.6' = 42' 36"

0.002° =

0.12' =

7.2"

0.32° = 19.2' = V¥ 12"

0.72° = 43.2' = 43' 12"

0-003° =

0.18' =

10.8"

0.33° = 19.8' = 19' 48"

0.73° = 43.8' = 43' 48"

0.004° =

0.24' =

14.4"

0.34° = 20.4' = 20' 24"

0.74° = 44.4' = 44' 24"

0.005° =

0.30' =

18.0"

0.35° = 21.0' = 21'

0.75° = 45.0' = 45'

0.006° =

0.36' =

21.6"

0.36° = 21.6' = 21' 36"

0.76° = 45.6' = 45' 36"

0.007° =

0.42' =

25.2"

0.37° = 22.2' = 22' 12"

0.77° = 46.2' = 46' 12"

0.008° =

0.48' =

28.8"

0.38° = 22.8' = 22' 48"

0.78° = 46.8' = 46' 48"

0.009° =

0.54' =

32.4"

0.39° = 23.4' = 23' 24"

0.79° = 47.4' = 47' 24"

0.010° =

0.60' =

36.0"

0.40° = 24.0' = 24'

0.80° = 48.0' = 48'

0.01° =

0.6' =

36"

0.41° = 24.6' = 24' 36"

0.81° = 48.6' = 48' 36"

0.02° =

1.2' =

1' 12"

0.42° = 25.2' = 25' 12"

O.S2° = 49.2' = 49' 12"

0.03° =

1.8' =

1' 48"

0.43° = 25.8' = 25' 48"

0.83° = 49.8' = 49' 48"

0.04° =

2.4' =

2' 24"

0.44° = 26.4' = 26' 24"

0.84° = 50.4' = 50' 24"

0.05° =

3.0' =

3'

0.45° = 27.0' = 27'

0.85° = 51.0' = 51'

0.06° =

3.6' =

3' 36"

0.46° = 27.6' = 27' 36"

0.86° = 51.6' = 51' 36"

0.07° ' =

4.2' =

4' 12"

0.47° = 28.2' = 28' 12"

0.87° = 52.2' = 52' 12"

0.08° =

4.8' =

4' 48"

0.48° = 28.8' = 28' 48"

0.88° = 52.8' = 52' 48"

0.09° =

5.4' =

5' 24"

0.49° = 29.4' = 29' 24"

0.89° = 53.4' = 53' 24"

0.10° =

6.0' =

6'

0.50° = 30.0' = 30'

0.90° = 54.0' = 54'

0.11° =

6.6' =

6' 36"

0.51° = 30.6' = 30' 36"

0.91° = 54.6' = 54' 36"

0.12° =

7.2' =

7' 12"

0.52° = 31.2' = 31' 12"

0.92° = 55.2' = 55' 12"

0.13° =

7.8' =

7' 48"

0.53° = 31.8' = 31' 48"

0.93° = 55.8' = 55' 48"

0.14° =

8.4' =

8' 24"

0.54° = 32.4' = 32' 24"

0.94° = 56.4' = 56' 24"

0.15° =

9.V =

9'

0.55° = 33.C = 33'

0.95° = 57.C = 57'

0.16° =

9.6' =

9'36"

0.56° = 33.6' = 33' 36"

0.96° = 57.6' = 57' 36"

0.17° =

10.2' =

10' 12"

0.57° = 34.2' = 34' 12"

0.97° = 58.2' = 58' 12"

0.18° =

10.8' =

10' 48"

0.58° = 34.8' = 34' 48"

0.98° = 58.8' = 58' 48"

0.19° =

11.4' =

11' 24"

0.59° = 35.4' = 35' 24"

0.99° = 59.4' = 59' 24"

0.20° =

12.CC =

12'

0.60° = 36.0' = 36'

1.00° = 60.0' = 60'

Table III

FOUR-PLACE LOGARITHMS OF TRIGONOMETRIC

FUNCTIONS, THE ANGLE BEING EXPRESSED

IN DEGREES AND THE DECIMAL

PART OF A DEGREE

This table gives the common logarithms (base 1(T) of the sines,
cosines, tangents, and cotangents of all angles from 0° to 5°, and
from 85° to 90° for every hundredth part of a degree, and from 5°
to 85° for every tenth of a degree, all calculated to four places of
decimals. In order to avoid the printing of negative characteristics,
the number 10 has been added to every logarithm in the first,
second, and fourth columns ^those having log sin, log tan, and
log cos at the top). Hence in writing down any logarithm taken
from these three columns — 10 should be written after it. Loga-
rithms taken from the third column (having log cot at the top)
should be used as printed.

A logarithm found from this table by interpolation may be in
error by one unit in the last decimal place, except for angles
between 0° and 0.3° or between 89.7° and 90°, when the error may
be larger. In the latter cases the table refers the student to the
formulas on page 6 for more accurate results.

19

20

TABLE III. LOGARITHMIC SINES

0°

Angle

log sin

diS.

log tan

com.
diK.

log cot

log cos

Prop. Parts

0.00°

0.01°

.

10.0000

90.00°

89.99°

an

6.2419

6.2419

3.7581

10.0000

0.02°

6.5429

6.5429

3.4571

10.0000

89.98°

ti

Difference

0.03°

6.7190

6.7190

3.2810

10.0000

89.97°

0.04°

6.8439

45

6.8439

to

3.1561

10.0000

89.96°

H

0.05°
0.06°

6.9408
7.0200

3

m

?

6.9408
7.0200

CO

u

3.0592
2.9800

10.0000
10.0000

89.95°
89.94°

—

79

78

77

0.07°

7.0870

7.0870

9i

2.9130

10.0000

89.93°

1

9

7.9
15 8

7.8
15.6

7.7
15 4

0.08°

7.1450

d

7.1450

d

2.8550

10.0000

89.92°

3

23.7

23.4

23.1

0.09°
0.10°

7.1961

7.1961

.9

2.8039

10.0000

89.91°
89.90°

4
5
6

31.6
39.5
47.4

31.2
39.0
46.8

30.8

38.5
46.2

7.2419

7.2419

2.7581

10.0000

0.11°

7.2833

>

7.2833

'Ed

2.7167

10.0000

89.89°

l

55.3
63.2

54.6
62.4

53.9
61.6

0.12°

7.3211

7.3211

I— 1

2.6789

10.0000

89.88°

9

71.1

70.2

69.3

0.13°

7.3558

u

a

7.3558

t-i

2.6442

10.0000

89.87°

76

75

74

0.14°

7.3880

bo

7.3880

ttD

2.6120

10.0000

89.86°

1

7.6

7.5

7.4

0.15°

7.4180

£

7.4180

a

2.5820

10.0000

89.85°

2

15.2

15.0

14.8

0.16°

7.4460

pj

7.4460

s

2.5540

10.0000

89.84°

3

22.8

22.5

22.2

4

30.4

30.0

29.6

0.17°

7.4723

7.4723

2.5277

10.0000

89.83°

5

38.0

37.5

37.0

0.18°

7.4971

■J —

7.4972

3) ft

2.5028

10.0000

89.82°

b

7

45.6
53 2

45.0
52.5

44.4
51.8

0.19°

7.5206

a o

7.5206

2.4794

10.0000

89.81°

8

60.8

60.0

69.2

0.20°

0.21°

7.5429

33
3. S

7.5429

.3 3
8. a

2.4571

10.0000

89.80°

89.79°

9

68.4

73

67.5

72

66.6

71

7.5641

7.5641

2.4359

10.0000

0.22°

7.5843

7.5843

SS

2.4157

10.0000

89.78°

1

7.3

7.2

7.1

0.23°

7.6036

7.6036

4J ' H

2.3964

10.0000

89.77°

2

14.6

14.4

14.2

•H co

it

21.9

21.6

21.3

0.24°

7.6221

7.6221

2.3779

10.0000

89.76°

4

29.2

28.8

28.4

0.25°

7.6398

.9 «

7.6398

2.3602

10.0000

89.75°

i>
6

36.5
43.8

36.0
43.2

35.5
42.6

0.26°

7.6568

"d w

7.6569

"t3 to

2.3431

10.0000

89.74°

7

51.1

50.4

49.7

O fl

8

58.4

57.6

56.8

0.27°
0.28°

7.6732
7.6890

7.6732
7.6890

2.3268
2.3110

10.0000
10.0000

89.73°
89.72°

9

65.7

64.8

03.9

0.29°
0.30°

0.31°

7.7043

142

7.7043
7.7190

142

2.2957

10.0000

89.71°
89.70°

89.69°

1
2
3

69

6.9
13.8
20.7

68

6.8
13.6

20.4

67

6.7
13.4
20.1

7.7190

2.2810

10.0000

7.7332

7.7332

2.2668

10.0000

0.32°

7.7470

138

7.7470

138

2.2530

10.0000

89.6S°

4

27.6
34.5

27.2
34.0

26.8
33.5

0.33°

7.7604

134

7.7604

134

2.2396

10.0000

S9.67°

6

41 4

40.8

40.2

130

130

7

483

47.6

46.9

0.34°

7.7734

7.7734

2.2266

10.0000

89.66°

8

55.2

54.4

53.6

0.35°

7.7859

125

7.7860

126

2.2140

10.0000

89.65°

9

62.1

61.2

60.3

0.36°

7.7982

123
119

7.7982

122
119

2.2018

10.0000

89.64°

66

65

64

0.37°

7.8101

7.8101

2.1899

10.0000

89.63°

1

6.6

6.5

6.4

0.38°

7.8217

116

7.8217

116

2.1783

10.0000

89.62°

2

13.2

13.0

12.8

0.39°
0.40°

7.8329

112
110

7.8329

112
110

2.1671

10.0000

89.61°
89.60°

4
5

19.8
26.4
33.0

19.5
26.0
32.5

19.2
25.6
32.0

7.8439

7.8439

2.1561

10.0000

108

108

6

39 6

39

38.4

0.41°

7.8547

7.8547

2.1453

10.0000

89.59°

7

46.2

45.5

44.8

0.42°

7.8651

104

7.8651

104

2.1349

10.0000

89.58°

8

52.8

52.0

51.2

0.43°

7.8753

102
100

7.8754

103
99

2.1246

10.0000

89.57°

9

59.4

58.5

57.6

0.44°

7.8853

7.8853

2.1147

10.0000

89.56°

63

62

61

0.45°

7.8951

98

7.8951

98

2.1049

10.0000

89.55°

1

6.3

6.2

6.1

0.46°

7.9046

95

7.9046

95

2.0954

10.0000

89.54°

9
it

12.6
18.9

12.4

186

18.3

0.47°

7.9140

94

7.9140

94

2.0860

10.0000

89.53°

4

5

25.2
31.5

24.8
31.0

24.4
30.5

0.48°

7.9231

91

7.9231

91

2.0769

10.0000

S9.52°

6

37.8

37.2

36.6

0.49°
0.50°

7.9321

90
87

7.9321

90

88

2.0678

10.0000

89.51°
89.50°

7
8
9

44.1
50.4
56.7

43.4
49.6
55.8

42.7
48.8
54.9

7.9408

7.9409

2.0591

10.0000

log cos

diS.

log cot

com.
diK.

log tan

log sin

Angle

89°

COSINES, TANGENTS, AND COTANGENTS

21

0°

Angle

log sin

dig.

log tan

com.
diH.

log cot

log cos

Prop. Parts

0.50°

0.51°

7.9408

86

7.9409

86

2.0591

10.0000

89.50°

89.49°

'So

7.9494

7.9495

2.0505

10.0000

0.52°

7.9579

85
82
82

7.9579

84
83
81

2.0421

10.0000

89.48°

t3

Difference

0.53°

7.9661

7.9662

2.0338

10.0000

89.47°

"5

0.54°

7.9743

7.9743

2.0257

10.0000

89.46°

w

0.55°
0.56°

7.9822
7.9901

79
79
76

7.9823
7.9901

80
78
77

2.0177
2.0099

10.0000
10.0000

89.45°
89.44°

60

59

58

0.57°

7.9977

76

7.9978

2.0022

10.0000

89.43°

i

9

6.0
12

5.9
11 8

5.8
11 6

0.58°

8.0053

8.0053

1.9947

10.0000

89.42°

3

18.0

17.7

17.4

0.59°

8.0127

74

8.0127

74

1.9873

10.0000

89.41°

4

24.0

23.6

23.2

73

73

5

30

29 5

29

0.60°

0.61°

8.0200

72
71

8.0200

72

1.9800

10.0000

89.40°

89.39°

6
7

8

36.0
42.0
4S.0

35.4
41.3

47.2

34.8
40.6
46 4

8.0272

8.0272

1.9728

10.0000

0.62°

8.0343

8.0343

1.9657

10.0000

89.38°

9

54.0

53.1

52.2

0.63°

8.0412

69
68

8.0412

69
69

1.9588

10.0000

89.37°

57

56

55

0.64°

8.0480

8.0481

1.9519

10.0000

89.36°

1

57

5.6

55

0.65°

8.0548

8.0548

67

1.9452

10.0000

89.35°

2

11.4

11.2

11.0

0.66°

S.0614

66

8.0614

66

1.9386

10.0000

89.34°

3

17.1

16.8

16.5

65

66

4

22.8

22.4

22.0

0.67°

8.0679

8.0680

1.9320

10.0000

89.33°

5

28.5

28.0

27.5

0.68°

8.0744

65

8.0744

64

1.9256

10.0000

89.32°

6

7

34.2
39.9

33.6
39.2

33.0
38.5

0.69°
0.70°

0.71°

8.0807

63
63
61

8.0807

63
63
62

1.9193

10.0000

89.31°
89.30°

89.29°

8
9

45.6
51.3

44.8
50.4

44.0
49.5

8.0870

8.0870

1.9130

10.0000

54

53

52

8.0931

8.0932

1.9068

10.0000

0.72°

8.0992

61

8.0992

60

1.9008

10.0000

89.28°

1

5.4

5.3

5.2

0.73°

8.1052

60
59

8.1052

60
59

1.S948

10.0000

89.27°

2
3

10.8
16.2

10.6
15.9

10.4
15.6

0.74°

8.1111

8.1111

1.8889

10.0000

89.26°

4

21.6

21.2

20.8

0.75°

8.1169

58

8.1170

59

1.8830

10.0000

89.25°

5
fi

27.0
32.4

26.5
31.8

26.0

31.2

0.76°

8.1227

8.1227

1.8773

10.0000

89.24°

7

37.8

37.1

36.4

57

57

8

43.2

42 4

41.6

0.77°

8.1284

8.1284

1.8716

10.0000

89.23°

9

48.6

47.7

46.8

0.78°

8.1340

56

8.1340

56

1.8660

10.0000

89.22°

0.79°
0.80°

0.81°

8.1395

55
55
53

8.1395

55
55

54

1.8605

10.0000

89.21°
89.20°

89.19°

1
2
3
4
R

51

5.1
10.2
15.3
20.4
25.5

50

5.0
10.0
15.0
20.0
25.0

49

4.9
9.8
14.7
19.6

24.5

8.1450

8.1450

1.8550

10.0000

8.1503

8.1504

1.S496

10.0000

0.82°

8.1557

54

8.1557

53

1.8443

10.0000

89.18°

0.83°

8.1609

52

8.1610

1.8390

10.0000

89.17°

6

30.6

30.0

29.4

52

52

7

35.7

35.0

34.3

0.84°

8.1661

8.1662

1.8338

10.0000

89.16°;

8

40.8

40.0

39.2

0.85°

8.1713

52

8.1713

51

1.8287

10.0000

89.15°

9

45.9

.45.0

44.1

0.86°

8.1764

51
50

8.1764

51
50

1.8236

10.0000

89.14°

48

47

46

0.87°

8.1814

8.1814

1.8186

9.9999

89.13°

1

4.8

4.7

4.6

0.88°

8.1863

49

8.1864

50

1.8136

9.9999

89.12°

2
3

4
5
6

7

9.6
14.4
19.2
24.0
28.8
33.6

9.4
14.1
18.8
23.5
28.2
32.9

9.2

13:8

18.4
23.0
27.6
32.2

0.89°
0.90°

0.91°

8.1912

49
49
48

8.1913

49
49
48

1.8087

9.9999

89.11°
89.10°

89.09°

8.1961

8.1962

1.8038

9.9999

8.2009

8.2010

1.7990

9.9999

0.92°

8.2056

47

8.2057

47

1.7943

9.9999

89.08°

8
9

38.4
43.2

37.6
42.3

36.8

0.93°

8.2103

47
47

8.2104

47
46

1.7896

9.9999

89.07°

0.94°

8.2150

8.2150

1.7850

9.9999

89.06°

45

44

43

0.95°

8.2196

46

8.2196

46

1.7804

9.9999

89.05°

1

2
3

4.5
9.0
13 5

4.4
8.8
13 2

4.3
8.6
12.9

0.96°

8.2241

45

8.2242

46

1.7758

9.9999

89.04°

0.97°

8.2286

8.2287

1.7713

9.9999

89.03°

4
5

18.0
W.,5

17.6
22,0

17.2
21.5

0.98°

8.2331

45

8.2331

44

1.7669

99999

89.02°

6

27.0

26.4

25.8

0.99°
1.00°

8.2375

44
44

8.2376

45
43

1.7624

9.9999

89.01°
89.00°

7
8
9

31.5
36.0
40.5

30.8
35.2
39.6

30.1
34.4
38.7

8.2419

8.2419

1.7581

9.9999

log cos

diH.

log cot

com.
diH.

log tan

log sin

Angle

89°

22

TABLE III. LOGAKITHMIC SINES

Angle

1.00°

1.01°
1.02°
1.03°

1.04°
1.05°
1.06°

1.07°
1.08°
1.09°
1.10°
1.11°
1.12°
1.13°

1.14°
1.15°
1.16°

1.17°
1.18°
1.19°
1.20°
1.21°
1.22°
1.23°

1.24°
1.25°
1.26°

1.27°
1.28°
1.29°
1.30°
1.31°
1.32°
1.33°

1.34°
1.35°
1.36°

1.37°
.1.38°
1.39°
1.40°

1.41°
1.42°
1.43°

1.44°
1.45°
1.46°

1.47°
1.48°
1.49°
1.50°

log sin difi

8.2419

8.2462
8.2505
8.2547

8.2589
8.2630
8.2672

8.2712
8.2753
8.2793

8.2832

i.2872
(.2911
!.2949

S.29S8
S.3025
(.3063

S.3100
(.3137
(.3174

8.3210

8.3246
8.3282
8.3317

8.3353
8.3388
8.3422

8.3456
8.3491
S.3524

8.3558

8.3591
8.3624
8.3657

8.3689
8.3722
8.3754

8.3786
8.3817
8.3S48

5.3880

S.3911
8.3941
8.3972

8.4002
8.4032
8.4062

8.4091
8.4121
8.4150

8.4179

log cos

43
43
42

42

41
42
40

41
40
39
40
39
38
39

37
38
37

37
37
36
36
36
35
36

35
34

34

35
33
34
33
33
33
32

33
32

32

31
31

32
'31
30
31
30

30
30
29

30
29
29

dig.

log tan

8.2419

8.2462
8.2505
8.2548

8.2590
8.2631
8.2672

8.2713
8.2754
8.2794

S.2833

8.2S73
8.2912
S.2950

8.2988
8.3026
8.3064

8.3101
8.3138
8.3175

S.3211

8.3247
8.32S3
8.3318

8.3354
8.3389
8.3423

8.3458
8.3492
8.3525

8.3559

8.3592
8.3625
8.3658

8.3691
8.3723
8.3755

8.3787
8.3818
8.3850

S.38S1

8.3912
8.3943
8.3973

8.4003
8.4033
8.4063

8.4093
8.4122
8.4152

8.4181

log cot

com.
difi.

43
43
43
42

41
41

41

41
40
39
40
39
38
38

38
38
37

37
37
36
36
36
35
36

35
34
35

34
33
34
33
33
33
33

32
32

32

31
32
31
31
31
30
30

30
30
30

29
30

log cot

1.7581

1.7538
1.7495
1.7452

1.7410
1.7369

1.7328

1.7287
1.7246
1.7206

1.7167

1.7127
1.7088
1.7050

1.7012
1.6974
1.6936

1.6899
1.6862
1.6825

1.6789

1.6753
1.6717
1.66S2

1.6646
1.6611
1.6577

1.6542
1.6508
1.6475

1.6441

1.6408
1.6375
1.6342

1.6309
1.6277
1.6245

1.6213
1.6182
1.6150

1.6119

1.6088
1.6057
1.6027

1.5997
1.5967
1.5937

1.5907
1.5878
1.5848

1.5819

com. ,

difi. l0 & tan
88°

log cos

9.9999

9.9999
9.9999
9.9999

9.9999
9.9999
9.9999

9.9999
9.9999
9.9999

9.9999

9.9999
9.9999
9.9999

9.9999
9.9999
9.9999

9.9999
9.9999
9.9999

9.9999

9.9999
9.9999
9.9999

9.9999
9.9999
9.9999

9.9999
9.9999
9.9999

9.9999

9.9999
9.9999
9.9999

9.9999
9.9999
9.9999

9.9999
9.9999
9.9999

9.9999

9.9999
9.9999
9.9999

9.9999
9 9999
9.9999

9 9999
9.9999
9.9999

9.9999

log sin

89.00°

88.99°
88.98°
88.97°

88.96°
88.95°
88.94°

88.93°
88.92°
88.91°
88.90°

88.89°

88.86°
88.85°
88.84°

88.83°
88.82°
88.81°
88.80°
88.79°
88.78°
88.77°

88.76°
88.75°
8S.74°

88.73°
88.72°
8S.71°
88.70°
88.69°
88.68°
88.67°

88.66°
88.65°
88.64°

88.63°
88.62°
88.61°
88.60°

88.59°
88.58°
88.57°

88.56°
88.55°
88.54°

88.53°
88.52°
88.51°
88.50°

Angle

Flop. Parts

Difference

43

42

4.3

4.2

8.6

8.4

12.9

12.6

17.2

16.8

21.5

21.0

25.8

25.2

30.1

29.4

34.4

33.6

38.7

37.8

41

40

4.1

4.0

8.2

8.0

12.3

12.0

16.4

16.0

20.5

20.0

24.6

24.0

28.7

28.0

32.8

32.0

36.9

36.0

39

38

3.9

3.8

7.8

7.6

11.7

11.4

15.6

15.2

19.5

19.0

23.4

22.8

27.3

26.6

31.2

30.4

35.1

34.2

37

36

3.7

3.6

7.4

7.2

11.1

10.8

14.8

14.4

18.5

iao

22.2

21.6

25.9

25.2

29.6

28.8

33.3

32.4

33

34

3.4

3.3

6.8

6.6

10.2

9.9

13.6

13.2

17.0

16.5

20.4

19.8

23.8

23.1

'27.2

26.4

30.6

29.7

31

30

3.1

30

6.2

6.0

9.3

9.0

12.4

12.0

15.5

15.0

18.6

18.0

21.7

21.0

24.8

24.0

27.9

27.0

35

3 5
7.0
10.5
14.0
17.5
21.0
24.5
28.0
31.5

32

3.2
6.4
9.6
12.8
16.0
19.2
22.4
25.6

29

2.9
5.8
8.7
11.6
14.5
17.4
20.3
23.2
26.1

COSINES, TANGENTS, AND COTANGENTS

23

1°

Angle

log sin

dig.

log tan

com.

difi.

log cot

log COB

Prop. Farts

1.50°

8.4179

29

8.4181

29

1.5819

9.9999

88.50°

•U

1.51°

8.4208

8.4210

1.5790

9.9998

88.49°

bo

1.52°

8.4237

29

8.4238

28

1.5762

9.9998

88.48°

13

DifierAnnn

1.53°

8.4265

28
28

8.4267

29

28

1.5733

9.9998

88.47°

1.54°

8.4293

8.4295

1.5705

9.9998

88.46°

W

1.55°
1.56°

8.4322
8.4349

29
27
28

8.4323
8.4351

28
28
28

1.5677
1.5649

9.9998
9.9998

88.45°
88.44°

1.57°

8.4377

8.4379

1.5621

9.9998

88.43°

1.58°

8.4405

26

8.4406

27

1.5594

9.9998

88.42°

1.59°
1.60°

1.61°

8.4432

27
27
27

8.4434

28
27
27

1.5566

9.9998

88.41°
88.40°

88.39°

1

2
3

29

2.9

5.8
8.7

28

2.8
5.6
8.4

8.4459

8.4461

1.5539

9.9998

8.4486

8.4488

1.5512

9.9998

1.62°

8.4513

27

8.4515

27

1.5485

9.9998

88.38°

4

11.6

11.2

1.63°

8.4540

27
27

8.4542

27
26

1.5458

9.9998

88.37°

5
6

14.5
17.4

14.0
16.8

1.64°

8.4567

8.4568

1.5432

9.9998

88.36°

7

20.3

19.6
22.4
25.2

1.65°

8.4593

26
26

8.4595

27
26

1.5405

9.9998
9.9998

88.35°
88.34°

8
9

26.1

1.66°

8.4619

26

8.4621

26

1.5379

1.67°

8.4645

8.4647

1.5353

9.9998

88.33°

27

26

1.68°

8.4671

26

8.4673

26

1.5327

9.9998

88.32°

1

2
3
4
5
6

2.7
5.4
8.1
10.8
13.5
16.2

2.6
5.2
7.8
10.4
13.0
15.6

1.69°
1.70°

1.71°

8.4697

26
26
25

8.4699

26
26
25

1.5301

9.9998

88.31°
88.30°

88.29°

8.4723

8.4725

1.5275

9.9998

8.4748

8.4750

1.5250

9.9998

1.72°

8.4773

25

8.4775

25

1.5225

9.9998

88.28°

7
8
9

18.9
21.6
24.3

18.2
20.8
23.4

1.73°
1.74°

8.4799
8.4824

26
25

8.4801
8.4826

26
25

1.5199
1.5174

9.9998
9.999S

88.27°
88.26°

1.75°

8.4848

24

8.4851

25

1.5149

9.9998

88.25°

25

24

1.76°

8.4873

25
25

8.4875

24
25

1.5125

9.9998

•

88.24°

1
2

2.5
5.0

2.4
' 4.8

1.77°

8.4898

8.4900

1.5100

9.9998

88.23°

3

7.5
10.0
12.5

7.2
9 6

1.78°

8.4922

24

8.4924

1.5076

9.9998

88.22°

5

12.0

1.79°

8.4947

25
24

8.4949

25

24

1.5051

9.9998

88.21°

6

7
8
9

15.0
17.6
20.0
22.5

14.4
16.8
19.2
21.6

1.80°

1.81°

8.4971

24

8.4973

24

1.5027

9.9998

88.20°

88.19°

8.4995

8.4997

1.5003

9.9998

1.82°

8.5019

24

8.5021

24

1.4979

9.9998

88.18°

1.83°

8.5043

24

8.5045

24

1.4955

9.9998

88.17°

23

22

23

23

1

2.3

2.2

1.84°

8.5066

8.5068

1.4932

9.9998

88.16°

2

4.6

4.4

1.85°

8.5090

24

8.5092

24

1.4908

9.9998

88.15°

3
4
5

6.9
9.2
11.5

6.6

1.86°

8.5113

23

8.5115

23

1.4885

9.9998

88.14°

11.0

1.87°

8.5136

23

8.5139

1.4861

9.9998

88.13°

6

7

13.8
16.1

13.2
15.4

1.88°

8.5160

24

8.5162

23

1.4838

9.9998

88.12°

8

18.4

17.6

1.89°
1.90°
1.91°

8.5183

23
23
22

8.5185
8.5208

23
23
23

1.4815

9.9998

88.11°
88.10°

88.09°

9

20.7

19.8

1

21

2.1

8.5206

1.4792

9.9998

8.5228

8.5231

1.4769

9.9998

1.92°

8.5251

23

8.5253

22

1.4747

9.9998

88.08°

2

4.2

1.93°

8.5274

23
22

8.5276

23
22

1.4724

9.9998

88.07°

3
4

6.3

8.4

1.94°

8.5296

8.5298

1.4702

9.9998

88.06°

5

Q

10.5
12 6

1.95°

8.5318

22

8.5321

23

1.4679

9.9997

88.05°

1

14.7

1.96°

8.5340

22
23

8.5343

22
22

1.4657

9.9997

88.04°

8
9

16.8
18.9

1.97°

8.5363

8.5365

1.4635

9.9997

88.03°

1.98°

8.5385

22

8.5387

22

1.4613

9.9997

88.02°

1.99°
2.00°

8.5406

21
22

8.5409

22
22

1.4591

9.9997

88.01°
88.00°

8.5428

8.5431

1.4569

9.9997

log cos

diS.

log cot

com.
difi.

log tan

log sin

Angle

88°

24

TABLE III. LOGARITHMIC SINES

2°

Angle

log sin

diff.

log tan

com.
diff.

log cot

log cos

Proj

. Farts

2.00°

2.01°

8.5428

22

8.5431

22

1.4569

9.9997

88.00°

87.99°

to

©

S.5450

8.5453

1.4547

9.9997

2.02°

S.5471

SI

8.5474

21

1.4526

9.9997

87.98°

Td

2.03°

8.5493

22
21

8.5496

22
21

1.4504

9.9997

87.97°

u

2.0V 5

8.5514

8.5517

1.4483

9.9997

87.96°

w

1=1

2.0.S°
2.06°

8.5535
8.5557

21
22
21

8.5538
8.5559

21
21
21

1.4462
1.4441

9.9997
9.9997

87.95°
87.94°

22

2.07°

8.5578

8.5580

1.4420

9.9997

87.93°

1

2

2.2

4 4 i

2.08°

8.5598

20

8.5601

21

1.4399

9.9997

87.92°

3

6.8 i

2.09°

8.5619

21

8.5622

21

1.4378

9.9997

87.91°

4

8.8

2.10°

2.11°

21
21

21
21
20
21

87.90°

87.89°

5
6
7
g

11.0

13.2 >
15.4
17 6 i

8.5640

8.5643

1.4357

9.9997

8.5661

8.5664

1.4336

9.9997

2.12°

8.5681

21

8.5684

1.4316

9.9997

87.88°

9

19.8

2.13°

8.5702

20

8.5705

20

1.4295

9.9997

87.87°

21

2 1 r

2.14°

8.5722

8.5725

1.4275

9.9997

87.86°

1

2.15°

8.5742

20

8.5745

20

1.4255

9.9997

87.85°

2

4.2

2.16°

8.5762

20
20

8.5765

20

1.4235

9.9997

87.84°

3
4
5

6.3
8.4
10.5

2.17°

8.5782

8.5785

1.4215

9.9997

87.83°

2.18°

8.5802

20

8.5805

20

1.4195

9.9997

87.82°

6

7

12.6
14 7

2.19°
2.20°

2.21°

8.5822

20
20
20

8.5825

20
20
20

1.4175

9.9997

87.81°
87.80°

87.79°

8
9

16.8
18.9

8.5842

8.5845
8.5865

1.4155

9.99-97

20

8.5862

1.4135

9.9997

2.22°

8.5881

19

8.5884

19

1.4116

9.9997

87.78°

1

2.0

2.23°

8.5901

20
19

8.5904

20
19

1.4096

9.9997

87.77°

2
3

4.0 (
6.0

2.24°

8.5920

8.5923

1.4077

9.9997

87.76°

4

8.0

2.25°

8.5939

19

8.5943

20

1.4057

9.9997

87.75°

5

Q

10.0

2.26°

8.5959

20
19

8.5962

19
• 19

1.4038

9.9997

87.74°

7
8

14.0
16.0

2.27°

8.5978

19

8.5981

19

1.4019

9.9997

87.73°

9

18.0

2.28°

8.5997

8.6000

1.4000

9.9997

87.72°

2.29°
2.30°

2.31°

8.6016

19
19
19

8.6019

19
19
19

1.3981

9.9997

87.71°
87.70°

87.69°

1
2
3

19

1.9
3.8
5.7

8.6035

8.6038

1.3962

9.9996

8.6054

8.6057

1.3943

9.9996

2.32°

8.6072

18

8.6076

19

1.3924

9.9996

87.68°

4
5

7.6

9.B !;
11.4
13.3
15.2

2.33°
2.34°

8.6091
8.6110

19
19 '

8.6095
8.6113

19
18

1.3905
1.3887

9.9996
9.9996

87.67°
87.66°

6
7
8

2.35°
2.36°

8.6128
8.6147

18
19
18

8.6132
8.6150

19

18
19

1.3868
1.3850

9.9996
9.9996

87.65°
87.64°

9

17.1

18

1.8

2.37°

8.6165

8.6169

1.3831

9.9996

87.63°

1

2.38°

8.6183

18

8.6187

18

1.3813

9.9996

87.62°

2

3.6

2.39°
2.40°

8.6201

18
19

8.6205

18
18

1.3795

9.9996

87.61°
87.60°

3

4
5

5.4
7.2
9.0

8.6220

8.6223

1.3777

9.9996

18

19

6

7

10.8
12.6

2.41°

8.6238

8.6242

1.3758

9.9996

87.59°

2.42°

8.6256

18

8.6260

1.3740

9.9996

87.58°

8

14.4

2.43°

8.6274

18

8.6277

17

1.3723

9.9996

87.57°

9

16.2

2.44°

8.6291

17

8.6295

18

1.3705

9.9996

87.56°

17

2.45°

8.6309

18

8.6313

18

1.3687

9.9996

87.55°

1

1.7

2.46°

8.6327

18
17

8.6331

18
17

1.3669

9.9996

87.54°

2
3

3.4

5.1

2.47°

8.6344

8.6348

1.3652

9.9996

87.53°

4
5
6

6.8
8.5
10.2

2.48°

8.6362

18

8.6366

18

1.3634

9.9996

87.52°

2.49°
2.50°

8.6379

17
18

8.6384

18
17

1.3616

9.9996

87.51°
87.50°

7
8
9

11.9
13.6
15.3 i

8.6397

8.6401

1.3599

9.9996

log cos

diff.

log cot

com.
diff.

log tan

log sin

Angle

87°

COSINES, TANGENTS, AND COTANGENTS

25

2°

Angle

log sin

cliff.

log tan

com.
diS.

log cot

log cos

Prop. Faits

2.50°

2.51°
2.52°
2.53°

2.54°
2.55°
2.56°

2.57°
2.58°
2.59°
2.60°
2.61°
2.62°
2.63°

2.64°
2.65°
2.66°

2.67°
2.68°
2.69°
2.70°

2.71°
2.72°
2.73°

2.74°
2.75°
2.76°

2.77°
2.78°
2.79°
2.80°
2.81°
2.82°
2.83°

2.84°

2.85°
2.86°

2.87°
2.88°
2.89°
2.90°

2.91°
2.92°
2.93°

2.94°
2.95°
2.96°

2.97°
2.98°
2.99°
3.00°

8.6397

17
17
18
17

17
17

17

17
16
17
17
16
17
16

17
16
16

17
16
16
16
16
16
16

15
16
16

16
15
16
15
16
15
15

15
16
15

15
15
15
15
15
15
14

15
15
14

15
15
14

8.6401

17
18
17
17

17
17
17

17
17
16
17
17
16
17

16
17
16

16
16
17
16
16
16
16

15
16
16

16
15
16
15
16
15
16

15
15
15

15
15
15
15
15
15
15

15
15
14

15
14
15

1.3599

9.9996

87.50°

87.49°
87.48°
87.47°

87.46°
87.45°
87.44°

87!43°
87.42°
87.41°
87.40°
87.39°
87.38°
87.37°

87.36°
87.35°
87.34°

87.33°
87.32°
87.31°
87.30°

87.29°
87.28°
87.27°

87.26°
87.25°
87.24°

87.23°
87.22°
87.21°
87.20°

87.19°
87.18°
87.17°

87.16°
87.15°
87.14°

87.13°
87.12°
87.11°
87.10°

87.09°
87.08°
87.07°

87.06°
87.05°
87.04°

87.03°
87.02°
87.01°
87.00°

£

60
•a

a
m

W

B
a
n

iH

•r-t

8.6414
8.6431
8.6449

8.6466
8.6483
8.6500

8.6517
8.6534
8.6550

8.6418
8.6436
8.6453

8.6470
8.6487
8.6504

8.6521
8.6538
8.6555

1.3582
1.3564
1.3547

1.3530
1.3513
1.3496

1.3479
1.3462

1.3445

9.9996
9.9996
9.9996

9.9996
9.9996
9.9996

9.9996
9.9996
9.9996

i

2
3
4
5
6
7
8
9

18

1.8

3.6

5.4

7.2

9.0
10.8
12.6
14.4
16.2 i

8.6567

8.6571

1.3429

9.9996

8.6584
8.6600
8.6617

8.6633
8.6650
8.6666

8.6682
8.6699
8.6715

8.6588
8.6605
8.6621

8.6638
8.6654
8.6671

8.6687
8.6703
8.6719

1.3412
1.3395
1.3379

1.3362
1.3346
1.3329

1.3313
1.3297
1.3281

9.9995
9.9995
9.9995

9.9995
9.9995
9.9995

9.9995
9.9995
9.9995

1
2
3
4
5
6
7
8
9

17

1.7
3.4
5.1
6.8
8.5
10.2
11.9
13.6
15.3

8.6731

8.6736

1.3264

9.9995

8.6747
8.6763
8.6779

8.6795
8.6810
8.6826

8.6842
8.6858
8.6873

8.6752
8.6768
8.6784

8.6800
8.6815
8.6831

8.6847
8.6863
8.6878

1.3248
1.3232
1.3216

1.3200
1.3185
1.3169

1.3153
1.3137
1.3122

9.9995
9.9995
9.9995

9.9995
9.9995
9.9995

9.9995
9.9995
9.9995

1
2
3
4
5
6
7
8
9

16

1.6
3.2
4.8
6.4
8.0
9.6
11.2
12.8
14.4

8.6889

8.6894

1.3106

9.9995

8.6904
8.6920
8.6935

8.6950
8.6965
8.6981

8.6996
8.7011
8.7026

8.6909
8.6925
8.6940

8.6956
8.6971
8.6986

8.7001
8.7016
8.7031

1.3091
1.3075
1.3060

1.3044
1.3029
1.3014

1.2999
1.2984
1.2969

9.9995
9.9995
9.9995

9.9995
9.9995
9.9995

9.9995
9.9995
9.9994

1
2
3
4
5
6
7
8
9

15

1.6
3.0
4.5
6.0
7.6
9.0
10.6
12.0
13.5

8.7041

8.7046

1.2954

9.9994

1
2
3
4
5
6
7
8
9

14

1.4
2.8
4.2
6.6
7.0
8.4
9.8
11.2
12.6

8.7056
8.7071
8.7086

8.7100
8.7115
8.7130

8.7144
8.7159
8.7174

8.7061
8.7076
8.7091

8.7106
8.7121
8.7136

8.7150
8.7165
8.7179

1.2939
1.2924
1.2909

1.2894
1.2879
1.2864

1.2850
1.2835
1.2821

9.9994
9.9994
9.9994

9.9994
9.9994
9.9994

9.9994
9.9994
9.9994

8.7188

8.7194

1.2806

9.9994

log cos

diS.

log cot

com.
diS.

log tan

log sin

Angle

87°

26

TABLE III. LOGARITHMIC SINES

3°

Angle

log sin

difi.

log tan

com.
difi.

log cot

log cos

Prop.

Farts

3.00°

3.01°
3.02°
3.03°

3.04°
3.05°
3.06°

8.7188

14
15

14
14

15
14

14

8.7194

14
15
14
15

14
14
14

1.2806

9.9994

87.00°

86.99°
86.98°
86.97°

86.96°
86.95°
86.94°

■ »H

bo

■a

03
u

m

o

a

at

s

m

P

8.7202
8.7217
8.7231

8.7245
8.7260
8.7274

8.7208
8.7223
8.7237

8.7252
8.7266

8.7280

1.2792
1.2777
1.2763

1.2748
1.2734
1.2720

9.9994
9.9994
9.9994

9.9994
9.9994
9.9994

3.07°
3.08°
3.09°
3.10°
3.11°
3.12°
3.13°

8.7288
8.7302
8.7316

14 '

14

14

14

14

14

14

8.7294
8.7308
8.7323

14

15
14
14
14
14
13

1.2706
1.2692
1.2677

9.9994
9.9994
9.9994

86.93°
86.92°
86.91°
86.90°

86.89°
86.88°
86.87°

1

2
3
4
5
6
7
8
9

1.5

3.0

4.5

6.0

7.5

9.0

10.5

12.0

13.5

8.7330

8.7337

1.2663

9.9994

8.7344
8.7358
8.7372

8.7351
8.7365
8.7379

1.2649
1.2635
1.2621

9.9994
9.9994
9.9994

3.14°
3.15°
3.16°

8.7386
8.7400
8.7413

14
13
14

8.7392
8.7406
8.7420

14
14
14

1.2608
1.2594
1.2580

9.9993
9.9993
9.9993

86.86°
S6.85°
86.84°

3.17°
3.18°
3.19°
3.20°

3.21"
3.22°
3.23°

8.7427
8.7441
8.7454

14
13
14
14
13
13
■ 14

8.7434
8.7448
8.7461

14
13
14
13
14
13
14

1.2566
1.2552
1.2539

9.9993
9.9993
9.9993

86.83°
86.82°
86.81°
86.80°

86.79°
86.78°
S6.77°

1
2
3

4
5
6
7
8
9

14

1.4
2.8
4.2
5.6
7.0
8.4
9.8
11.2
12.6

8.7468

8.7475

1.2525

9.9993

8.7482
8.7495
8.7508

8.7488
8.7502
8.7515

1.2512
1.2498
1.2485

9.9993
9.9993
9.9993

3.24°
3.25°
3.26°

8.7522
8.7535
8.7549

13
14
13

8.7529
8.7542
8.7556

13
14
13

1.2471
1.2458
1.2444

9.9993
9.9993
9.9993

86.76°
86.75°
86.74°

3.27°
3.28°
3.29°
3.30°

3.31°
3.32°
3.33°

3.34°
3.35°
3.36°

8.7562
8.7575
8.7588
8.7602
8.7615
8.7628
8.7641

8.7654
8.7667
8.7680

13
13
14
13
13
13
13

13
13
13

8.7569
8.7582
8.7596

13
14
13
13
13
13
13

13
13
13

1.2431
1.2418
1.2404

9.9993
9.9993
9.9993

86.73°
86.72°
86.71°
86.70°
86.69°
86.68°
86.67°

86.66°
86.65°
86.64°

1

2
3

4
5
6
7
8
9

~13"

1.3
2.6
3.9
5.2
6.5
7.8
9.1
10.4
11.7

8.7609

1.2391

9.9993

8.7622
8.7635
8.7648

8.7661
8.7674
8.7687

1.2378
1.2365
1.2352

1.2339
1.2326
1.2313

9.9993
9.9993
9.9993

9.9993
9.9993
9.9993

3.37°
3.38°
3.39°
3.40°

3.41°
3.42°
3.43°

8.7693
8.7705
8.7718

12
13
13
13
12
13
13

8.7700
8.7713
8.7726

13
13
13
12
13
13
13

1.2300
1.22S7
1.2274

9.9992
9.9992
9.9992

86.63°
86.62°
86.61°
86.60°

86.59°
86.58°
86.57°

1

2
3
4
5
6

~12~

1.2
2.4
3.6
4.8
6.0
7.2

8.7731

8.7739

1.2261

9.9992

8.7744
8.7756
8.7769

8.7751
8.7764
8.7777

1.2249
1.2236
1.2223

9.9992
9.9992
9.9992

3.44°
3.45°
3.46°

8.7782
8.7794
8.7807

12
13

12

8.7790
8.7802
8.7815

12
13
12

1.2210
1.2198
1.2185

9.9992
9.9992
9.9992

86.56°
86.55°
86.54°

7
8
9

8.4
9.6
10.8

3.47°
3.48°
3.49°
3.50°

8.7819
8.7832
8.7844

13
12
13

8.7827
8.7840
8.7852

13
12
13

1.2173
1.2160
1.2148

9.9992
9.9992
9.9992

86.53°
86.52°
86.51°
86.50°

8.7857

8.7865

1.2135

9.9992

log COB

difi.

log cot

com.
difi.

log tan

log sin

Angle

86°

COSINES, TANGENTS, AND COTANGENTS

Angle

3.50°

3.51°
3.52°
1.53°

3.54°

3.55°
3.56°

3.57°
5.58°
3.59°
3.60°
3.61°
5.62°
3.63°

3.64°
3.65°
3.66°

3.67°
3.6S°
3.69°
3.70°

o.71°
3.72°
3.73°

3.74°
3.75°
3.76°

3.77°
3.7S°
3.79°
3.80°
3.81°
3.S2°
3.S3°

3.S4°

3.85°
3.36°

3.S7°
3.SS°
3.S9°
3.90°
3.91°
3.92°
3.93°

3.94°
3.95°
3.96°

3.97°
3.98°
3.99°
4.00°

log sin difi.

S.7S57

S.7S69
S.7SS1
S.7S94

S.7906
S.791S
&7930

S.7943
S.7955
S.7967

S.7979 I

S.7991
S.S003
S.S015

S.S027
S.S039
S.S051

S.S062

S.S074
S.80S6

S.S09S
S.S109
S.S121
S.S133

S.S144
S.S156
S.S168

S.S179
S.S191
S.S202

S.S213

S.S22.

S.8236

S.S24S

S.S259
S.S270
S.S2S1

S.S293
aS304|

S.S315

S.8526 I

S.S34S
S.S359

S.S370
S.S3S1
S.S392

S.S403 I
S.S414 |

S.S425

S.S436

13
13
13
13

19
13
13

13
13
13
13
13
13
13

ia

13
11

13
13
19
11
13
19
11

13
13

11

19
11
11
13
11
13
11

11
11

19

11
11
11
11
11
11
11

11
11
11

11
11
11

log cos i difi.

log tan !

com.
difi.

i log cot

S.7S65

8.7S77
S.7S90
a?902

S.7914

S.7927 ,
S.7939 |

S.7951
S.7963

S.7975

S.798S

s.sooo|

S.S012 '
S.S024

S.S036 I
S.S04S I
S.S059

S.S071 S

S.S0S3

S.S095

S.S107

8.8119 |
S.S130 !

S.S142

S.S154
S.S165
S.S177

S.S1SS
S.S200
S.S212

S.S223 !

S.S234
S.S246
S.S257

S.S269
S.S2S0
S.S291

S.S302 |

S.S314

S.S325

S.S336

S.S347
S.S35S
S.S370 |

S.S3S1
S.S392
S.8403

S.S4H!
S.S425 '
S.8436

S.S446

19
13
13
13

13
13
19

13
13
13
19
19
19
13

13
11
13

13
13
13
19
11
13
19

II

19
II

13

19
11
II
19
11
13

11
11
11

13
11
11
11
11
13
11

11

11
11

II
11
10

1.2135

1.2123
1.2110
1.209S

1.2086
1.2073
1.2061

1.2049

1.2037
1.2025

1.2012

1.2000
1.19S8
1.1976

1.1964
1.1952
1.1941

1.1929
1.1917
1.1905

1.1S93

1.1SS1
1.1S70
1.1S5S

1.1S46
1.1S35
1.1 S23

1.1S12
1.1S00
1.1 7SS

1.17

1.1766

1.1754
1.1743

I 1.1731

1.1720

J 1.1709

: 1.169S

I 1.16S6

1.1675

1.1664

I 1.1653
i 1.1642
[ 1.1630

! 1.1619
1.160S
1.1597

| 1.15S6

i 1.1575

1.1564

1.1554

log cot

com.
difi.

log cos

9.9992

9.9992
9.9992
9.9992

9.9992
9.9992
9.9992

9.9992
9.9992
9.9991

9.9991

9.9991
9.9991
9.9991

9.9991
9.9991
9.9991

9.9991
9.9991
9.9991

9.9991

9.9991
9.9991
9.9991

9.9991
9.9991
9.9991

9.9991
9.9991
9.9990

9.9990

9.9990
9.9990
9.9990

9.9990
9.9990
9.9990

9.9990
9.9990
9.9990

9.9990

9.9990
9.9990
9.9990

9.9990
9.9990
9.9990

9.9990
9.9990
9.99S9

9.99S9

log tan log sin

86°

86.50°

S6.49°
S6.4S°
S6.47°

86.46°

86.45°
86.44°

S6.43°
S6.42
S6.41°
86.40°
S6.39°
S6.3S°
S6.37°

S6.36°
S6.35°
86.34°

S6.33°
S6.32°
S6.31°
86.30°

86.29°
S6.2S°
S6.27°

86.26°
86.25°
S6.24°

S6.23°
S6.22°
S6.21°
86.20°
86.19°
S6.1S°
86.17°

86.16°
S6.15°
S6.14°

S6.13°
86.12°
S6.11°
86.10°
S6.09
S6.0S
S6.07

S6.06°
S6.05°
S6.04°

S6.03°
S6.02°
S6.01°
86.00°

Angle

Prop. Farts

T* -

13

i-i

2.6
3.9
5.2
6.5
7.S
9.1
10.4
11.7

12

! 1.2
2.4
3.6

I 4.S
6.0

i 7.2

I °" 4
9.6

10.S

11

1.1

2.2

as

4.4
5.5
6.6

S.S
9.9

10

1

1.0

'">

2.0

o

3.0

4

4.0

5

5.0

6

6.0

l

7.0

S

8.0

9

9.0

28

TABLE III. LOGAKITHMIC SINES

Angle

log sin

difi.

log tan

com.
difi.

log cot

log COB

Prop. Parts

4.00°

4.01°
4.02°
4.03°

4.04°
4.05°
4.06°

4.07°
4.08°
4.09°
4.10°
4.11°
4.12°
4.13°

4.14°
4.15°
4.16°

4.17°
4.18°
4.19°
4.20°
4.21°
4.22°
4.23°

4.24°
4.25°
4.26°

4.27°
4.28°
4.29°
4.30°
4.31°
4.32°
4.33°

4.34°
4.35°
4.36°

4.37°
4.38°
4.39°
4.40°
4.41°
4.42°
4.43°

4.44°
4.45°
4.46°

4.47°
4.48°
4.49°
4.50°

8.8436

8.8447
8.8457
8.8468

8.8479
8.8490
8.8500

8.8511
8.8522
8.8532

8.8543

8.8553
8.8564
8.8575

8.8585
8.8595
8.8606

8.8616
8.8627
8.8637

8.8647

8.8658
8.8668
8.8678

8.8688
8.8699
8.8709

8.8719
8.8729
8.8739

8.8749

S.8759
5.8769
5.8780

S.S790
S.8799
S.8809

5.8819

5.8829
5.8839

8.8849

5.8859
5.8869
5.8878

8.8898
8.8908

8.8917
8.8927
8.8937

8.8946

n

10

11
ii

n

10

11
11

10

11

10

11
11

10
10

11

10

11

10
10

11

10
10
10

11

10
10

10
10
10
10
10

11

10

9
10
10

10
10
10
10
10
9
10

10
10

9

10
10
9

8.8446

8.8457
8.8468
8.8479

8.8490
8.8501
8.8511

8.8522
8.8533
8.8543

8.8554

5.8565
5.8575
5.8586

5.8596
5.8607
5.8617

5.8628
5.8638
5.8649

8.8659

8.8669
8.8680
8.8690

8.8700
8.8711
8.8721

8.8731
8.8741
8.8751

8.8762

8.8772
8.8782
8.8792

8.8802
8.8812
8.8822

8.8832
8.8842
8.8852

8.8862

8.8872
8.8882
8.8891

8.8901
8.8911
8.8921

8.8931
8.8940
8.8950

8.8960

n
n
n
n

n

10

n
n

10

n
n

10

n

10

11

10

11

10

11

10
10

11

10
10

11

10
10

10
10

11

10
10
10
10

10
10

10

10
10
10
10
10
9
10

10
10
10

9
10
10

1.1554

9.9989

1.1543
1.1532
1.1521

1.1510
1.1499
1.1489

1.1478
1.1467
1.1457

9.9989
9.9989
9.9989

9.9989
9.9989
9.9989

9.9989
9.9989
9.9989

1.1446

9.9989

1.1435
1.1425
1.1414

1.1404
1.1393
1.1383

1.1372
1.1362
1.1351

9.9989
9.9989
9.9989

9.9989
9.9989
9.9989

9.9988
9.9988
9.9988

1.1341

9.9988

1.1331
1.1320
1.1310

1.1300
1.1289
1.1279

1.1269
1.1259
1.1249

9.9988
9.9988
9.9988

9.9988
9.9988
9.9988

9.9988
9.9988
9.9988

1.1238

9.9988

1.122S
1.1218
1.1208

1.1198
1.1188
1.1178

1.1168
1.1158
1.1148

9.9988
9.9988
9.9988

9.9988
9.9987
9.9987

9.9987
9.9987
9.9987

1.1138

9.9987

1.1128
1.1118
1.1109

1.1099
1.1089
1.1079

1.1069
1.1060
1.1050

9.9987
9.9987
9.9987

9.9987
9.9987
9.9987

9.9987
9.9987
9.9987

1.1040

9.9987

86.00°

85.99°
85.98°
85.97°

85.96°
85.95°
85.94°

85.93°
85.92°
85.91°
85.90°

85.89°
85.88°
85.87°

85.86°
85.85°
85.84°

85.83°
85.82°
85.81°
85.80°

85.79°
85.78°
85.77°

85.76°
85.75°
85.74°

85.73°
85.72°
85.71°
85.70°

85.69°
85.68°
85.67°

85.66°
85.65°
85.64°

85.63°
85.62°
85.61°
85.60°

85.59°
85.58°
85.57°

85.56°
85.55°
85.54°

85.53°
85.52°
85.51°
85.50°

log cos difi. log cot Sj?' log tan log sin Angle

85 p

be

■■a
a

M

w

11

1.1

2.2
3.3
4.4
5.5
6.6
7.7
8.8
9.9

10

1

1.0

2

2.0

3

3.0

4

4.0

6

6.0

6

6.0

7

7.0

8

8.0

9

9.0

9

1

0.9

2

1.8

3

2.7

4

3.6

5

4.5

6

5.4

7

6.3

8

7.2

9

8.1

COSINES, TANGENTS, AND COTANGENTS

29

Angle

4.50°

4.51°
4.52°
4.53°

4.54°
4.55°
4.56°

4.57°
4.58°
4.59°
4.60°
4.61°
4.62°
4.63°

4.64°
4.65°
4.66°

4.67°
4.68°
4.69°
4.70°

4.71°
4.72°
4.73°

4.74°
4.75°
4.76°

4.77°
4.78°
4.79°
4.80°
4.81°
4.82°
4.83°

4.84°
4.85°
4.86°

4.87°
4.8S°
4.89°
4.90°
4.91°
4.92°
4.93°

4.94°
4.95°
4.96°

4.97°
4.98°
4.99°
5.00°

log sin

8.8946

3.8956
i.8966
5.8975

i.8985
S.8994
5.9004

S.9013
S.9023
S.9032

8.9042

diff.

8.9051
8.9060
8.9070

8.9079
8.9089
8.9098

8.9107
8.9116
8.9126

8.9135

8.9144
8.9153
8.9162

8.9172
8.9181
8.9190

8.9199
8.9208
8.9217

8.9226

8.9235
8.9244
8.9253

8.9262
8.9271
8.9280

8.9289
8.9298
8.9307

8.9315

8.9324
8.9333
8.9342

8.9351
8.9359
8.9368

8.9377
8.9386
8.9394

8.9403

10
10
9
10

9
10
9

10

9

10

10
9

10

10
9
9
9
9

10

log tan

8.8960

8.8970
8.8979
8.8989

8.8998
8.9008
8.9018

8.9027
8.9037
8.9046

8.9056

8.9065
8.9075
8.9084

8.9093
8.9103
8.9112

8.9122
8.9131
8.9140

8.9150

8.9159
8.9168
8.9177

8.9186
8.9196
8.9205

8.9214
8.9223
8.9232

8.9241

log cos diS. log cot

8.9250
8.9260
8.9269

8.9278
8.9287
8.9296

8.9305
8.9313
8.9322

8.9331

8.9340
8.9349
8.9358

8.9367
8.9376
8.9384

8.9393
8.9402
8.9411

8.9420

com.
difi.

10
9

10
9

10
10
9

10
9
10

10
9
10

9
9
10
9
9

10
9

9

9
9
9
9
10
9
9

dm.

log cot

1.1040

1.1030
1.1021
1.1011

log cos

9.9987

9.9987
9.9986
9.9986

1.1002 9.9986
1.0992 9.9986
1.0982 9.9986

1.0973
1.0963
1.0954

1.0944

1.0935
1.0925
1.0916

1.0907
1.0897
1

1.0878
1.0869
1.0860

1.0850

1.0841
1.0832
1.0823

1.0814
1.0804
1.0795

1.0786
1.0777
1.0768

1.0759

1.0750
1.0740
1.0731

1.0722
1.0713
1.0704

1.0695
1.0687
1.0678

9.9986
9.9986
9.9986

9.9986

9.9986
9.9986
9.9986

9.9986
9.9986
9.9986

9.9986
9.9985
9.9985

9.9985

9.9985
9.9985
9.99S5

9.9985
9.9985
9.9985

9.9985
9.9985
9.9985

9.9985

1.0669

1.0660
1.0651
1.0642

1.0633
1.0624
1.0616

1.0607
1.0598
1.0589

9.9985
9.9985
9.9985

9.9984
9.9984
9.9984

9.9984
9.9984
9.9984

9.9984

9.9984
9.9984
9.9984

9.9984
9.9984
9.9984

9.9984
9.9984
9.9984

1.0580

log tan

85°

9.9983

log sin

85.50°

85.49°
85.48°
85.47°

85.46°
85.45°
85.44°

85.43°
85.42°
85.41°
85.40°

85.39°
85.38°
85.37°

85.36°
85.35°
85.34°

85.33°
85.32°
85.31°
85.30°

85.29°
85.28°
85.27°

85.26°
85.25°
85.24°

85.23°
85.22°
85.21°
85.20°
85.19°
85.18°
85.17°

85.16°
85.15°
85.14°

85.13°
85.12°
85.11°
85.10°

85.09°
85.08°
85.07°

85.06°
85.05°
85.04°

85.03°
85.02°
85.01°
85.00°

Angle

Prop. Farts

bo

&

sa

10

1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0

9

1

0.9

2

1.8

3

2.7

4

3.6

5

4.5

6

5.4

7

6.3

8

7.2

9

8.1

8

1

0.8

2

1.6

3

2.4

4

3.2

5

4.0

6

4.8

7

6.6

8

6.4

9

7.2

30

TABLE III. LOGARITHMIC SINES

5°-10°

Angle

log sin

dig.

log tan

com.
difi.

log cot

log cos

difi.

Prop. Parts

5.0°

5.1°

S.9403

86

8.9420

86

1.0580

9.9983

85.0°

84.9°

£

■ i-t

8.9489

8.9506

1.0494

9.9983

5.2°

8.9573

84

8.9591

85

1.0409

9.9982

84.8°

T*

Difference

5.3°

8.9655

81

8.9674

83

82

1.0326

9.9981

84.7°

a

5.4°

8.9736

8.9756

1.0244

9.9981

84.6°

w

5 5°

8.9816

80

8.9836

80

1.0164

9.9980

84.5°

5.6°

8.9894

78
76

8.9915

79
77

1.0085

9.9979

84.4°

62

61

60

5.7°

8.9970

8.9992

1.0008

9.9978

84.3°

i

6.2
12.4

6.1
12.2

6.0
12.0

5.8°

9.0046

76

9.0068

76

0.9932

9.9978

84.2°

3

18.6

18.3

18.0

5.9°

9.0120

74

72

9.0143

75

73

0.9857

9.9977

84.1°

4

24.8
31.0

24.4
30.5

24.0
30.0

6.0°

9.0192

72

9.0216

73

0.9784

9.9976

84.0°

6

37.2

36.6

36.0

6.1°

9.0264

9.0289

0.9711

9.9975

83.9°

7
8

43.4
49.6

42.7
48.8

42.0
48.0

6.2°
6.3°

9.0334
9.0403

69

9.0360
9.0430

70

0.9640
0.9570

9.9975
9.9974

83.8°
83.7°

9

55.8

54.9

54.0

69

69

W

58

57

6.4°

9.0472

9.0499

0.9501

9.9973

83.6°

1

5.9

5.8

5.7

6.5°

9.0539

67

9.0567

0.9433

9.9972

83.5°

2

11.8

11.6

11.4

6.6°

9.0605

66
65

9.0633

66
66

0.9367

9.9971

83.4°

3

4

17.7
23.6

17.4
23.2

17.1
22.8

6.7°

9.0670

9.0699

0.9301

9.9970

83.3°

5

29.5

29.0

28.5

6.8°

9.0734

64

9.0764

65

0.9236

9.9969

83.2°

6

7

35.4
41.3

34.8
40.6

34.2
39.9

6.9°

9.0797

63
62

9.0828

63

0.9172

9.9968

83.1°

8

47.2

46.4

45.6

7.0°

7.1°

9.0859

61

9.0891

63

0.9109

9.9968

83.0°

82.9°

9

53.1

51.3

56

55

54

9.0920

9.0954

0.9046

9.9967

7.2°

9.0981

61

9.1015

61

0.8985

9.9966

82.8°

1

5.6

5.5

5.4

7.3°

9.1040

59
59

9.1076

61
59

0.8924

9.9965

82.7°

2
3

11.2
16.8

11.0
16.5

10.8
16.2

7.4°

9.1099

9.1135

0.8865

9.9964

82.6°

4

22.4

22.0

21.6

7.5°

9.1157

58

9.1194

59

0.8806

9.9963

82.5°

5
6

28.0
33.6

27.5
33.0

27.0
32.4

7.6°

9.1214

57
57

9.1252

58

0.8748

9.9962

82.4°

7
8

39.2
448

38.5
440

37.8
43.2

7.7°
7.8°

9.1271
9.1326

55

9.1310
9.1367

57

0.8690
0.8633

9.9961
9.9960

82.3°
82.2°

9

50.4

49.5

48.6

7.9°

9.1381

55
55

9.1423

56
55

0.8577

9.9959

82.1°

53

52

51

8.0°

8.1°

9.1436

53

9.1478

55

0.8522

9.9958

82.0°

81.9°

1
2
3

5.3
10.6
15.9

5.2
10.4
15.6

5.1
10.2
15.3

9.1489

9.1533

0.8467

9.9956

8.2°

9.1542

53

9.1587

54

0.8413

9.9955

81.8°

4
1

21.2
26.5

20.8
26.0

20.4
25.5

8.3°

9.1594

52
52

9.1640

53

0.8360

9.9954

81.7°

6

7

31.8
37,1

31.2
36.4

30.6
35.7

8.4°

9.1646

9.1693

0.8307

9.9953

81.6°

8

42.4

41.6

40.8

8.5°

9.1697

51

9.1745

52

0.8255

9.9952

81.5°

9

47.7

46.8

45.9

8.6°

9.1747

50
50

9.1797

52
51

0.8203

9.9951

81.4°

50

49

48

8.7°

9.1797

9.1848

0.8152

9.9950

81.3°

1

5.0

4.9

4.8

8.8°

9.1847

50

9.1898

50

0.8102

9.9949

81.2°

2

10.0

9.8

9.6

8.9°
9.0°

9.1°

9.1895

48
48
48

9.1948

50
49
49

0.8052

9.9947

2

81.1°
81.0°

80.9°

3
4
5
6

7

15.0
20.0
25.0
30.0
35.0

14.7
19.6
24.5
29.4
34.3

14.4
19.2
24.0

28.8
33.6

9.1943

9.1997

0.8003

9.9946

9.1991

9.2046

0.7954

9.9945

9.2°

9.2038

47

9.2094

48

0.7906

9.9944

80.8°

8

40.0

39.2

38.4

9 3°

9.2085
9.2131

47

9.2142
9.2189

48

0.7858
0.7811

9.9943
9.9941

80.7°
80.6°

9

45.0

44.1

43.2

9.4°

46

47

2

47

46

45

9.5°

9.2176

45

9.2236

47

0.7764

9.9940

80.5°

1

4.7

4.6

4.5

9.6°

9.2221

45
45

9.2282

46
46

0.7718

9.9939

2

80.4°

2
3

9.4
14.1

9.2
13.8

9.0
13.5

9.7°

9.2266

9.2328

0.7672

9.9937

80.3°

4
5

18.8
23.5

18.4
23.0

18.0
22.5

9.8°

9.2310

9.2374

46

0.7626

9.9936

80.2°

6

28.2

27.6

27.0

9.9°
10.0°

9.2353
9.2397

43
44

9.2419

45
44

0.7581

9.9935

S0.1°
80.0°

7
8
9

32.9
37.6
42.3

32.2
36.8
41.4

31.5
36.0
40.5

9.2463

0.7537

9.9934

log cos

difi.

log cot

com.
din".

log tan

log sin

difi.

Angle

80°- 85°

COSINES, TANGENTS, AND COTANGENTS

31

10°-15°

Angle

10.0°

10.1°
10.2°
10.3°

10.4°
10.5°
10.6°

10.7°
10.8°
10.9°
11.0°

11.1°
11.2°
11.3°

11.4°
11.5°
11.6°

11.7°
11.8°
11.9°
12.0°

12.1°
12.2°
12.3°

12.4°
12.5°
12.6°

12.7°
12.8°
12.9°
13.0°
13.1°
13.2°
13.3°

13.4°
13.5°
13.6°

13.7°
13.8°
13.9°
14.0°
14.1°
14.2°
14.3°

14.4°
14.5°
14.6°

14.7°
14.8°
14.9°
15.0°

log sin

9.2397

9.2439
9.2482
9.2524

9.2565
9.2606
9.2647

9.2687
9.2727
9.2767

9.2806

9.2845
9.2883
9.2921

9.2959
9.2997
9.3034

9.3070
9.3107
9.3143

9.3179

9.3214
9.3250
9.3284

9.3319
9.3353
9.3387

9.3421
9.3455
9.3488

9.3521

9.3554
9.3586
9.3618

9.3650
9.3682
9.3713

9.3745
9.3775
9.3806

9.3837

9.3867
9.3897
9.3927

9.3957
9.3986
9.4015

9.4044
9.4073
9.4102

9.4130

diH.

log tan

com.
diS.

9.2463

9.2507
9.2551
9.2594

9.2637
9.2680
2.2722

9.2764
9.2S05
9.2846

9.2887

9.2927
9.2967
9.3006

9.3046
9.3085
9.3123

9.3162
9.3200
9.3237

9.3275

9.3312
9.3349
9.3385

9.3422
9.3458
9.3493

9.3529
9.3564
9.3599

9.3634

9.3668
9.3702
9.3736

9.3770
9.3804
9.3837

9.3870
9.3903
9.3935

9.3968

9.4000
9.4032
9.4064

9.4095
9.4127
9.4158

9.4189
9.4220
9.4250

9.4281

log cot

0.7537

0.7493
0.7449
0.7406

0.7363
0.7320
0.7278

0.7236
0.7195
0.7154

0.7113

0.7073
0.7033
0.6994

0.6954
0.6915
0.6877

0.6838
0.6800
0.6763

0.6725

0.6688
0.6651
0.6615

0.6578
0.6542
0.6507

0.6471
0.6436
0.6401

0.6366

0.6332
0.6298
0.6264

0.6230
0.6196
0.6163

0.6130
0.6097
0.6065

0.6032

0.6000
0.5968
0.5936

0.5905
0.5873
0.5842

0.5811
0.5780
0.5750

0.5719

log COB

9.9934

9.9932
9.9931
9.9929

9.9928
9.9927
9.9925

9.9924
9.9922
9.9921

9.9919

9.9918
9.9916
9.9915

9.9913
9.9912
9.9910

9.9909
9.9907
9.9906

9.9904

9.9902
9.9901
9.9899

9.9897
9.9896
9.9894

9.9892
9.9S91
9.9S89

9.9887

9.9885
9.9884
9.9882

9.9880
9.9878
9.9876

9.9875
9.9873
9.9871

9.9869

9.9867
9.9865
9.9863

9.9861
9.9859
9.9857

9.9855
9.9853
9.9851

9.9849

diH

80.0°

79.9°
79.8°
79.7°

79.6°
79.5°
79.4°

79.3°
79.2°
79.1°
79.0°

78.9°
78.8°
78.7°

78.6°
78.5°
78.4°

78.3°
78.2°
78.1°
78.0°

77.9°
77.8°
77.7°

77.6°
77.5°
77.4°

77.3°
77.2°
77.1°
77.0°

76.9°
76.8°
76.7°

76.6°
76.5°
76.4°

76.3°
76.2°
76.1°
76.0°

75.9°
75.8°
75.7°

75.6°
75.5°
75.4°

75.3°
75.2°
75.1°
75.0°

Prop. Parts

Difference

44

43

4.4

4.3

8.8

8.6

13.2

12.9

17.6

17.2

22.0

21.5

26.4

25.8

30.8

30.1

35.2

34.5

39.6

38.8

41

40

4.1

4.0

8.2

8.0

12.3

12.0

16.4

16.0

20.5

20.0

24.6

24.0

28.7

28.0

32.8

32.0

36.9

36.0

38

37

3.8

3.7

7.6

7.4

11.4

11.1

15.2

14.8

19.0

18.5

22.8

22.2

26.6

25.9

30.4

29.6

34.2

33.3

35

34

3.5

3.4

7.0

6.8

10.5

10.2

14.0

13.6

17.5

17.0

21.0

20.4

24.5

23.8

28.0

27.2

31.5

30.6

32

31

3.2

3.1

6.4

6.2

9.0

9.3

12.8

12.4

16.0

15.5

19.2

18.6

22.4

21.7

25.6

24.8

28.8

27.9

29

28

2.9

2.8

5.8

5.6

8.7

8.4

11.6

11.2

14.5

14.0

17.4

16.8

20.3

19.6

23.2

22.4

26.1

25.2

42

4.2
8.4
12.6
16.S
21.0
25.2
29.4
33.6
37.8

39

3.9
7.8
11.7
15.6
19.5
23.4
27.3
31.2
35.1

36

3.6
7.2
10.8
14.4
18.0
21.6
25.2
28.8
32.4

33

3.3
6.6
9.9
13.2
16.5
19.8
23.1
26.4
29.7

30

3.0
6.0
9.0
12.0
15.0
18.0
21.0
24.0
27.0

2

0.2
0.4
0.6

0.8
1.0
1.2
1.4
1.6
1.*

log cos difi. log cot ,^ m- log tan log sin diH. Angle

75°-80°

32

TABLE III. LOGARITHMIC SINES

15°- 20"

Angle

log sin

diS.

log tan

com,

diff.

log cot

log cos

diff.

2
2
2
2

2
2
2

2
2
3
2
2
2
2

3
2

2

2
3
2
2
3
2
2

3

2
3

2
2
3
2
3
2
3

2
3
3

2
3
2
3
3
2
3

3
2
3

3
2
3

diff.

Prop. Farts

15.0°

15.1°
15.2°
15.3°

15.4°
15.5°
15.6°

15.7°
15.8°
15.9°
16.0°
16.1°
16.2°
16.3°

16.4°
16.5°
16.6°

16.7°
16.8°
16.9°
17.0°

17.1°
17.2°
17.3°

17.4°
17.5°
17.6°

17.7°
17.8°
17.9°
18.0°

18.1°
18.2°
18.3°

18.4°
18.5°
18.6°

18.7°
18.8°
18,9°
19.0°
19.1°
19.2°
19.3°

19.4°
19.5°
19.6°

19.7°
19.8°
19.9°
20.0°

9.4130

38
38
28
38

27
27
27

27
27
26
27
26
26
26

25

26
25

25
25
25
25
25
34
24

24

24
24

24
23
34
23
23
23
33

23
22
23

32
32
32
22
22
22
21

22
21
23

21
21
21

9.4281

30
30
30
29

30
29

29

29
29
29

28
29
28
28

28
28
27

28
27
27
27
27
27
27

26
27
26

26
26
26
25
26
26
25

25
25
25

25
25
25
24
25
34
24

24
25
23

34
24
34

0.5719

9.9849

75.0°

74.9°
74.8°
74.7°

74.6°
74.5°
74.4°

74.3°
74.2°
74.1°
74.0°
73.9°
73.8°
73.7°

73.6°

73.5°
73.4°

73.3°
73.2°
73.1°
73.0°

72.9°
72.8°
72.7°

72.6°

72.5°
72.4°

72.3°
72.2°
72.1°
72.0°

71.9°

71.8°
71.7°

71.6°
71.5°
71.4°

71.3°
71.2°
71.1°
71.0°
70.9°
70.8°
70.7°

70.6°
70.5°
70.4°

70.3°
70.2°
70.1°
70.0°

a

bD

3

09

a

M

w

Difference

9.4158
9.4186
9.4214

9.4242
9.4269
9.4296

9.4323
9.4350
9.4377

9.4311
9.4341
9.4371

9.4400
9.4430
9.4459

9.4488
9.4517
9.4546

0.5689
0.5659
0.5629

0.5600
0.5570
0.5541

0.5512
0.5483
0.5454

9.9847
9.9845
9.9843

9.9841
9.9839
9.9837

9.9835
9.9833
9.9831

1

2
3

4
5
6
7
8
9

30

3.0
6.0
9.0
12.0
15.0
18.0
21.0
24.0
27.0

29

2.9
5.8
8.7
11.6
14.5
17.4
20.3
23.2
26.1

9.4403

9.4575

0.5425

9.9828

9.4430
9.4456
9.4482

9.4508
9.4533
9.4559

9.4584
9.4609
9.4634

9.4603
9.4632
9.4660

9.4688
9.4716
9.4744

9.4771
9.4799
9.4826

0.5397
0.5368
0.5340

0.5312
0.5284
0.5256

0.5229
0.5201
0.5174

9.9826
9.9824
9.9822

9.9820
9.9817
9.9815

9.9813
9.9811
9.9808

1
2
3
4
5
6
7
8
9

28

2.8
5.6
8.4
11.2
14.0
16.8
19.6
22.4
25.2

27

2.7
6.4
8.1
10.8
13.5
16.2
18.9
21.6
24.3

9.4659

9.4853

0.5147

9.9806

9.4684
9.4709
9.4733

9.4757
9.4781
9.4805

9.4829
9.4853
9.4876

9.4880
9.4907
9.4934

9.4961
9.4987
9.5014

9.5040
9.5066
9.5092

0.5120
0.5093
0.5066

0.5039
0.5013
0.4986

0.4960
0.4934
0.4908

9.9804
9.9801
9.9799

9.9797
9.9794
9.9792

9.9789
9.9787
9.9785

1
2
3
4
5
6
7
8
9

26

2.6
5.2
7.8
10.4
13.0
15.6
18.2
20.8
23.4

25

2.5
5.0
7.6
10.0
12.5
15.0
17.5
20.0
22.5

9.4900

9.5118

0.4882

9.9782

9.4923
9.4946
9.4969

9.4992
9.5015
9.5037

9.5060
9.5082
9.5104

9.5143
9.5169'
9.5195

9.5220
9.5245
9.5270

9.5295
9.5320
9.5345

0.4857
0.4831
0.4805

0.4780
0.4755
0.4730

0.4705
0.4680
0.4655

9.9780
9.9777
9.9775

9.9772
9.9770
9.9767

9.9764
9.9762
9.9759

1
2
3
4
5
6
7
8
9

24

2.4
4.8
7.2
9.6
12.0
14.4
16.8
19.2
21.6

23

2.3
4.6
6.9
9.2
11.5
13.8
16.1
18.4
20.7

9.5126

9.5370

0.4630

9.9757

9.5148
9.5170
9.5192

9.5213
9.5235
9.5256

9.5278
9.5299
9.5320

9.5394
9.5419
9.5443

9.5467
9.5491
9.5516

9.5539
9.5563
9.5587

0.4606
0.4581
0.4557

0.4533
0.4509
0.4484

0.4461
0.4437
0.4413

9.9754
9.9751
9.9749

9.9746
9.9743
9.9741

9.9738
9.9735
9.9733

1

2
3
4
5
6
7
8
9

22

2.2
4.4
6.6
8.8
11.0
13.2
15.4
17.6
19.8

21

2.1
4.2
6.3
8.4
10.5
12.6
14.7
16.8
18.9

9.5341

9.5611

0.4389

9.9730

log cos

difi.

log cot

com.

diff.

log tan

log sin

Angle

70°-75°

COSINES, TANGENTS, AND COTANGENTS 33

20°-25°

Angle

20.0°

20.1°
20.2°
20.3°

20.4°
20.5°
20.6°

20.7°
20.8°
20.9°
21.0°
21.1°
21.2°
21.3°

21.4°
21.5°
21.6°

21.7°
21.8°
21.9°
22.0°

22.1°
22.2°
22.3°

22.4°
22.5°
22.6°

22.7°
22.8°
22.9°
23.0°
23.1°
23.2°
23.3°

23.4°
23.5°
23.6°

23.7°
23.8°
23.9°
24.0°

24.1°
24.2°
24.3°

24.4°
24.5°
24.6°

24.7°
24.8°
24.9°
25.0°

log sin

9.5341

9.5361
9.5382
9.5402

9.5423
9.5443
9.5463

9.5484
9.5504
9.5523

9.5543

9.5563
9.5583
9.5602

9.5621
9.5641
9.5660

9.5679
9.5698
9.5717

9.5736

9.5754
9.5773
9.5792

9.5810
9.5828
9.5847

9.5865
9.5883
9.5901

9.5919

9.5937
9.5954
9.5972

9.5990
9.6007
9.6024

9.6042
9.6059
9.6076

9.6093

9.6110
9.6127
9.6144

9.6161
9.6177
9.6194

9.6210
9.6227
9.6243

9.6259

diS.

log tan

9.5611

9.5634
9.5658
9.5681

9.5704
9.5727
9.5750

9.5773
9.5796
9.5819

9.5842

9.5864
9.5887
9.5909

9.5932
9.5954
9.5976

9.5998
9.6020
9.6042

9.6064

9.6086
9.6108
9.6129

9.6151
9.6172
9.6194

9.6215
9.6236

9.6257

9.6279

9.6300
9.6321
9.6341

9.6362
9.6383
9.6404

9.6424
9.6445
9.6465

9.6486

9.6506
9.6527
9.6547

9.6567
9.6587
9.6607

9.6627
9.6647
9.6667

com.
difi

9.6687

log cos difi. log cot °.°. m * log tan

22
22
21
22

2]
22
21

21
21
22
21
21
20
21

21
21
20

21
20
21

20
21
20
20

20
20
20

20

log cot

0.4389

0.4366
0.4342
0.4319

0.4296
0.4273
0.4250

0.4227
0.4204
0.4181

0.4158

0.4136
0.4113
0.4091

0.4068
0.4046
0.4024

0.4002
0.3980
0.3958

0.3936

0.3914
0.3892
0.3871

0.3849
0.3828
0.3806

0.3785
0.3764
0.3743

log cos

9.9730

9.9727
9.9724
9.9722

9.9719
9.9716
9.9713

9.9710
9.9707
9.9704

9.9702

9.9699
9.9696
9.9693

9.9690
9.9687
9.9684

9.9681
9.9678
9.9675

9.9672

0.3721

0.3700
0.3679
0.3659

0.3638
0.3617
0.3596

0.3576
0.3555
0.3535

0.3514

0.3494
0.3473
0.3453

0.3433
0.3413
0.3393

0.3373
0.3353
0.3333

0.3313

9.9669
9.9666
9.9662

9.9659
9.9656
9.9653

9.9650
9.9647
9.9643

9.9640

9.9637
9.9634
9.9631

9.9627
9.9624
9.9621

9.9617
9.9614
9.9611

9.9607

9.9604
9.9601
9.9597

9.9594
9.9590
9.9587

9.9583
9.9580
9.9576

9.9573

difi,

log sin difi. Angle

70.0°
69.9°
69.8°
69.7°

69.6°
69.5°
69.4°

69.3°
69.2°
69.1°
69.0°

68.9°
68.8°
68.7°

68.6°
68.5°
68.4°

68.3°
68.2°
68.1°
68.0°
67.9°
67.8°
67.7°

67.6°
67.5°
67.4°

67.3°
67.2°
67.1°
67.0°

66.9°
66-8°
66.7°

66.6°
66.5°
66.4°

66.3°
66.2°
66.1°
66.0°

65.9°

65.8°
65.7°

65.6°
65.5°
65.4°

65.3°
65.2°
65.1°
65.0°

Prop. Parts

Difference

23

2.3

4.6

6.9

9.2

11.6

13.8

16.1

18.4

20.7

21

2.1

4.2
6.3
8.4
10.6
12.6
14.7
16.8
18.9

19

1.9
3.8
5.7
7.6
9.5
11.4
13.3
15.2
17.1

17

1.7

3.4

5.1

6.8

8.5

10.2

11.9

13.6

15.3

2

0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8

3

0.3
0.6
0.9
1.2
1.5
1.8
2.1
2.4
2.7

22

2.2
4.4
6.6
8.8
11.0
13.2
15.4
17.6
19.8

20

2.0
4.0
6.0
8.0
10.0
12.0
14.0
16.0
18.0

18

1.8
3.6
5.4
7.2
9.0
10.8
12.6
14.4
16.2

16

1.6
3.2

4.8
6.4
8.0
9.6
11.2
12.8
14.4

4

0.4
0.8
1.2
1.6
2.0
2.4
2.8
3.2
3.6

65°-70°

34

TABLE III. LOGARITHMIC SINES

25°-30°

Angle

log sin

difi.

log tan

com.
difi.

log cot

log cos

difi.

Prop. Paits

25.0°

25.1°
25.2°
25.3°

9.6259

17
16
16
16

9.6687

19
20
20
19

0.3313

9.9573

4
3
4
4

65.0°

64.9°
64.8°
64.7°

+3

'So

■a
d

Difference

9.6276
9.6292
9.6308

9.6706
9.6726
9.6746

0.3294
0.3274
0.3254

9.9569
9.9566
9.9562

25.4°
25.5°
25.6°

9.6324
9.6340
9.6356

16
16
15

9.6765
9.6785
9.6804

20
19
20

0.3235
0.3215
0.3196

9.9558
9.9555
9.9551

3
4
3

64.6°
64.5°
64.4°

w

20

19

25.7°
25.8°
25.9°
26.0°
26.1°
26.2°
26.3°

9.6371
9.6387
9.6403

16
16
15
16
15
16
15

9.6824
9.6843
9.6863

19
20
19
19
19
19
19

0.3176
0.3157
0.3137

9.9548
9.9544
9.9540

4
4
3
4
4
4
3

64.3°
64.2°
64.1°
64.0°

63.9°
63.8°
63.7°

1

2
3
4
5
6
7
8
9

2.0
4.0
6.0
8.0
10.0
12.0
14.0
16.0
18.0

1.9
3.8
5.7
7.6
9.5
11.4
13.3
16.2
17.1

9.6418

9.6882

0.3118

9.9537

9.6434
9.6449
9.6465

9.6901
9.6920
9.6939

0.3099
0.3080
0.3061

9.9533
9.9529
9.9525

26.4°
26.5°
26.6°

9.6480
9.6495
9.6510

15

15
16

9.6958
9.6977
9.6996

19
19
19

0.3042
0.3023
0.3004

9.9522
9.9518
9.9514

4
4
4

63.6°
63.5°
63.4°

1

18

1.8

17

1.7

26.7°
26.8°
26.9°
27.0°

27.1°
27.2°
27.3°

9.6526
9.6541
9.6556
9.6570
9.6585
9.6600
9.6615

15
15
14
15
15
15
14

9.7015
9.7034
9.7053

19
19
19
18
19
19
18

0.2985
0.2966
0.2947

9.9510
9.9506
9.9503

4
3
4
4
4
4
4

63.3°
63.2°
63.1°
63.0°

62.9°
62.8°
62.7°

2
3

4
5
6

7
8
9

3.6
5.4
7.2
9.0
10.8
12.6
14.4
16.2

3.4
5.1
6.8
8.5
10.2
11.9
13.6
15.3

9.7072

0.2928

9.9499

9.7090
9.7109
9.7128

0.2910
0.2891
0.2872

9.9495
9.9491
9.9487

27.4°
27.5°
27.6°

9.6629
9.6644
9.6659

15
15
14

9.7146
9.7165
9.7183

19
18
19

0.2854
0.2835
0.2817

9.9483
9.9479
9.9475

4
4
4

62.6°
62.5°
62.4°

1

2

16

1.6
3.2

15
1.5
3.0

27.7°
27.8°
27.9°
28.0°
28.1°
28.2°
28.3°

9.6673
9.6687
9.6702

14
15
14
14
14
15
14

9.7202
9.7220
9.7238

18
18
19
18
18
18
19

0.2798
0.2780
0.2762

9.9471
9.9467
9.9463

4
4
4
4
4
4
4

62.3°
62.2°
62.1°
62.0°
61.9°
61.8°
61.7°

3
4

5
6

7
8
9

4.8
6.4
8.0
9.6
11.2
12.8
14.4

4.5
6.0
7.5
9.0
10.5
12.0
13.5

9.6716

9.7257

0.2743

9.9459

9.6730
9.6744
9.6759

9.7275
9.7293
9.7311

0.2725
0.2707
0.2689

9.9455
9.9451
9.9447

28.4°
28.5°
28.6°

9.6773
9.6787
9.6801

14
14
13

9.7330
9.7348
9.7366

18
18
18

0.2670
0.2652
0.2634

9.9443
9.9439
9.9435

4

4
4

61.6°
61.5°
61.4°

1
2
3

14

1.4
2.8
4.2

13

1.3
2.6
3.9

28.7°
28.8°
28.9°
29.0°
29.1°
29.2°
29.3°

9.6814
9.6828
9.6842

14
14
14
13
14
13
14

9.7384
9.7402
9.7420

18
18
18
17
18
18
18

0.2616
0.2598
0.2580

9.9431
9.9427
9.9422

4
5
4
4
4
4
5

61.3°
61.2°
61.1°
61.0°
60.9°
60.8°
60.7°

4
5
6
7
8
9

5.6
7.0
8.4
9.8
11.2
12.6

5.2
6.5
7.8
9.1
10.4
11.7

9.6856

9.7438

0.2562

9.9418

9.6869
9.6883
9.6896

9.7455
9.7473
9.7491

0.2545
0.2527
0.2509

9.9414
9.9410
9.9406

3

4

29.4°
29.5°
29.6°

29.7°
29.8°
29.9°
30.0°

9.6910
9.6923
9.6937

9.6950
9.6963
9.6977

13

14
13

13
14
13

9.7509
9.7526
9.7544

9.7562
9.7579
9.7597

17
18
18

17
18
17

com.
difi.

0.2491
0.2474
0.2456

0.2438
0.2421
0.2403

9.9401
9.9397
9.9393

9.9388
9.9384
9.9380

4
4
5

4
4
5

60.6°
60.5°
60.4°

60.3°
60.2°
60.1°
60.0°

1
2
3
4
5
6
7
8
9

0.3
0.6
0.9
1.2
1.5
1.8
2.1
2.4
2.7

0.4
0.8
1.2
1.6
2.0
2.4
2.8
3.2
3.6

9.6990

9.7614

0.2386

9.9375

log cos

difi.

log cot

log tan

log sin

difi.

Angle

60°-65°

COSINES, TANGENTS, AND COTANGENTS

35

30°- 35°

Angle

log sin

difi.

log tan

com.
diS.

18
17

18

17
18
17

17

18
17
17
17
17
17

17
17
17

17
17
17
17
17
16
17

17
16

17

16
17
16
17
16

17
16

17

16
17
16
16
17
16
16

16

16

16
16
16

log cot

log cos

difi.

Prop. Parts

30.0°

30.1°
30.2°
30.3°

30.4°
30.5°
30.6°

30.7°
30.8°
30.9°
31.0°

31.1°
31.2°
31.3°

31.4°
31.5°
31.6°

31.7°
31.S°
31.9°
32.0°
32.1°
32.2°
32.3°

32.4°
32.5°
32.6°

32.7°
32.S°
32.9°
33.0°

33.1°
33.2°
33.3°

33.4°
33.5°
33.6°

33.7°
33.8°
33.9-"
34.0°
34.1°
34.2°
34-3°

34.4°
34.5°
34.6°

34.7°
34.8°
34.9°
35.0°

9.6990

13
13
13
13

13
13
12

13
13
12
13
13
12
12

13
12
12

13
12
12
12
12
12
12

12

12
12

12

12
12
11
12

12

12
11

12
11
11
11
11

11
11

11

11

9.7614

0.2386

9.9375

4
4
5
4

5

4
5

4
5
4
5
4
5
5

4
5

5

4
5
5
5
4
5
5

5
5

4

5
5
5
5
5
5
5

5
5
5

5
5
5
5
6
5
5

5
5
6

5
5
5

60.0°

59.9°
59.8°
59.7°

59.6°
59.5°
59.4°

59.3°
59.2°
59.1°
59.0°

58.9°
58.8°
58.7°

58.6°
5S.5°
58.4°

58.3°
58.2°
58.1°
58.0°

57.9°
57.8°
57.7°

57.6°
57.5°
57.4°

57.3°
57.2°
57.1°
57.0°

56.9°
56.8°
56.7°

56.6°
56.5°
56.4°

56.3°
56.2°
56.1°
56.0°

55.9°
55.8°
55.7°

55.6°
55.5°
55.4°

55.3°
55.2°
55.1°
55.0°

I

■&

■a

03

w
1

2
3
4
5
6
7
8
9

Difference

9.7003
9.7016
9.7029

9.7042
9.7055
9.7068

9.7080
9.7093
9.7106

9.7632
9.7649
9.7667

9.7684
9.7701
9.7719

9.7736
9.7753
9.7771

0.2368
0.2351
0.2333

0.2316
0.2299
0.2281

0.2264
0.2247
0.2229

9.9371
9.9367
9.9362

9.9358
9.9353
9.9349

9.9344
9.9340
9.9335

18

1.8
3.6
5.4
7.2
9.0
10.8
12.6
14.4
16.2

17
1.7
3.4
5.1
6.8
8.5
10.2
11.9
13.6
15.3

9.7118

9.77S8

0.2212

9.9331

9.7131
9.7144
9.7156

9.7168
9.7181
9.7193

9.7205
9.7218
9.7230

9.7805
9.7822
9.7839

9.7856

9.7S73
9.7S90

9.7907
9.7924
9.7941

0.2195
0.217S
0.2161

0.2144
0.2127
0.2110

0.2093
0.2076
0.2059

9.9326
9.9322
9.9317

9.9312
9.9308
9.9303

9.9298
9.9294
9.9289

1
2
3
4
5
6
7
8
9

16

1.6
3.2
4.8
6.4
S.O
9.6
11.2
12.8
14.4

9.7242

9.795S

0.2042

9.9284

9.7254
9.7266
9.7278

9.7290
9.7302
9.7314

9.7326
9.7338
9-7349

9.7975
9.7992
9.S00S

9.8025
9.S042
9.S0S9

9.8075
9.8092
9.8109

0.2025
0.200S
0.1992

0.1975
0.1958
0.1941

0.1925
0.1908
0.1891

9.9279
9.9275
9.9270

9.9265
9.9260
9.9255

9.9251
9.9246
9.9241

1
2
3
4
5
6
7
8
9

13

1.3
2.6
3.9
5.2
6.5
7.8
9.1
10.4
11.7

12

1.2
2.4
3.6
4.8
6.0
7.2
8.4
9.6
10.8

9.7361

9.S125

0.1875

9.9236

9.7373
9.7384
9.7396

9.7407
9.7419
9.7430

9.7442
9.7453
9.7464

9.8142
9.8158
9.8175

9.8191
9.820S
9.8224

9.8241
9.8257
9.8274

0.1S58
0.1S42
0.1825

0.1809
0.1792
0.1776

0.1759
0.1743
0.1726

9.9231
9.9226
9.9221

9.9216
9.9211
9.9206

9.9201
9.9196
9.9191

1
2
3
4
5
6
7
8
9

11

1.1
2.2
3.3
4.4
5.5
6.6
7.7
8.8
9.9

9.7476

9.8290

0.1710

9.9186

9.7487
9.7498
9.7509

9.7520
9.7531
9.7542

9.7553
9.7564

9.7575

9.8306
9.S323
9.S339

9.8355
9.8371
9.83SS

9.8404
9.8420
9.8436

0.1694
0.1677
0.1661

0.1645
0.1629
0.1612

0.1596
0.1 5S0
0.1564

9.9181
9.9175
9.9170

9.9165
9.9160
9.9155

9.9149
9.9144
9.9139

1

2
3
4
5
6
7
8
9

5

0.5
1.0
1.6
2.0
2.5
3.0
3.5
4.0
4.5

6

0.6
1.2
1.8
2.4
3.0
3.6
4.2
4.8
5.4

9.7586

9.8452

0.1548

9.9134

log cos

difi.

log cot

com.
difi.

log tan

log sin

difi.

Angle

55°-60°

36

TABLE III. LOGARITHMIC SINES

35°-40°

Angle

log sin

difi.

log tan

com.
difi.

log cot

log COB

difi.

Prop. Farts

35.0°

35.1°
35.2°
35.3°

9.7586

11

10

n
11

9.8452

16
16
17
16

0.1548

9.9134

6
5
5
6

55.0°

54.9°
54.8°
54.7°

4J
•iH

bn

Difference

9.7597
9.7607
9.7618

9.8468
9.8484
9.8501

0.1532
0.1516
0.1499

9.9128
9.9123
9.9118

35.4°
35.5°
35.6°

9.7629
9.7640
9.7650

n

10
11

9.8517
9.8533
9.8549

16
16
16

0.1483
0.1467
0.1451

9.9112
9.9107
9.9101

5
6
5

54.6°
54.5°
54.4°

w

17

16

35.7°
35.8°
35.9°
36.0°
36.1°
36.2°
36.3°

9.7661
9.7671
9.7682

10

11

10

11

10
10
11

9.8565
9.8581
9.8597

16
16
16
16
15
16
16

0.1435
0.1419
0.1403

9.9096
9.9091
9.9085

5
6
5
6
5
6
6

54.3°
54.2°
54.1°
54.0°

53.9°
53.8°
53.7°

1

2
3
4
5
6
7
8
9

1.7
3.4
6.1
6.8
8.5
10.2
11.9
13.6
15.3

1.6

3.2

4.8

6.4

8.0

9.6

11.2

12.8

14.4

9.7692

9.8613

0.1387

9.9080

9.7703
9.7713
9.7723

9.8629
9.8644
9.8660

0.1371
0.1356
0.1340

9.9074
9.9069
9.9063

36.4°
36.5°
36.6°

9.7734
9.7744
9.7754

10
10
10

9.8676
9.8692
9.8708

16
16
16

0.1324
0.1308
0.1292

9.9057
9.9052
9.9046

5
6
5

53.6°
53.5°
53.4°

1

15

1.5

36.7°
36.8°
36.9°
37.0°

37.1°
37.2°
37.3°

9.7764
9.7774
9.7785

10

11

10
10
10
10
10

9.8724
9.8740
9.8755

16
IS
16
16
16
15
16

0.1276
0.1260
0.1245

9.9041
9.9035
9.9029

6
6
6
5
6
6
6

53.3°
53.2°
53.1°
53.0°

52.9°
52.8°
52.7°

2
3
4
5
6
7
8
9

3.0
4.5
6.0
7.5
9.0
10.5
12.0
13.5

9.7795

9.8771

0.1229

9.9023

9.7805
9.7815
9.7825

9.8787
9.8803
9.8818

0.1213
0.1197
0.1182

9.9018
9.9012
9.9006

37.4°
37.5°
37.6°

9.7835
9.7844
9.7854

9
10
10

9.8834
9.8850
9.8865

16
15
16

0.1166
0.1150
0.1135

9.9000
9.8995
9.8989

5
6
6

52.6°
52.5°
52.4°

1
2

11

1.1
2.2

10

1.0
2.0

37.7°
37.8°
37.9°
38.0°
38.1°
38.2°
38.3°

9.7864
9.7874
9.7884

10
10

9
10
10

9
10

9.8881
9.8897
9.8912

16
15
16
16
15
16
15

0.1119
0.1103
0.1088

9.8983
9.8977
9.8971

6
6
6
6
6
6
6

52.3°
52.2°
52.1°
52.0°
51.9°
51.8°
51.7°

3
4
5
6
7
8
9

3.3
4.4
5.5
6.0
7.7
8.8
9.9

3.0
4.0
5.0
6.0
7.0
8.0
9.0

9.7893

9.8928

0.1072

9.8965

9.7903
9.7913
9.7922

9.8944
9.8959
9.8975

0.1056
0.1041
0.1025

9.8959
9.8953
9.8947

38.4°
38.5°
38.6°

9.7932
9.7941
9.7951

9
10
9

9.8990
9.9006
9.9022

16
16
15

0.1010
0.0994
0.0978

9.8941
9.8935
9.8929

6
6
6

51.6°
51.5°
51.4°

1
2
3

9

0.9
1.8
2.7

38.7°
38.8°
38.9°
39.0°

39.1°
39.2°
39.3°

9.7960
9.7970
9.7979

10
9

10
9
9

10
9

9.9037
9.9053
9.9068

16
15
16
15
16
15
16

0.0963
0.0947
0.0932

9.8923
9.8917
9.8911

6
6
6
6
6
6
7

51.3°
51.2°
51.1°
51.0°
50.9°
50.8°
50.7°

4
5
6
7
8
9

3.6
4.5
5.4
6.3
7.2
8.1

9.7989

9.9084

0.0916

9.8905

9.7998
9.8007
9.8017

9.9099
9.9115
9.9130

0.0901
0.0885
0.0870

9.8899
9.8893
9.8887

5

0.5
1.0
1.5
2.0

6

0.6
1.2

1.8
2.4

39.4°
39.5°
39.6°

9.8026
9.8035
9.8044

9
9
9

9.9146
9.9161
9.9176

15
15
16

0.0854
0.0839
0.0824

9.8880
9.S874
9.8868

6
6
6

50.6°
50.5°
50.4°

1
2
3
4

39.7°
39.8°
39.9°
40.0°

9.8053
9.8063
9.8072

10
9
9

9.9192
9.9207
9.9223

15
16
15

0.0808
0.0793
0.0777

9.8862
9.8855
9.8849

7
6
6

50.3°
50.2°
50.1°
50.0°

5
G
7
8
9

2.5
3.0
3.5
4.0

4.5

3.0
3.6

4.2
4.8
5.4

9.8081

9.9238

0.0762

9.8843

log cos

dig.

log cot

com.
difi.

log tan

log sin

difi.

Angle

50°-55°

•

COSINES, TANGENTS, AND COTANGENTS

37

40°-45°

Angle

log sin

difi.

log tan

com.
difi.

log cot

log cos

difi.

Ifrop. Parts

40.0°

40.1°
40.2°
40.3°

40.4°
40.5°
40.6°

40.7°
40.8°
40.9°
41.0°

41.1°
41.2°
41.3°

41.4°
41.5°
41.6°

41.7°
41.8°
41.9°
42.0°

42.1°
42.2°
42.3°

42.4°
42.5°
42.6°

42.7°
42.8°
42.9°
43.0°
43.1°
43.2°
43.3°

43.4°
43.5°
43.6°

43.7°
43.8°
43.9°
44.0°
44.1°
44.2°
44.3°

44.4°
44.5°
44.6°

44.7°
44.8°
44.9°
45.0°

9.80S1

9.8090
9.8099
9.8108

9.8117
9.8125
9.8134

9.8143
9.8152
9.8161

9.8169

9.8178
9.8187
9.8195

9.S204
9.8213
9.8221

9.8230
9.8238
9.8247
"9^8255

9.8264
9.8272
9.8280

9.8289
9.8297
9.8305

9.8313
9.8322
9.8330

9.8338

9.8346
9.8354
9.8362

9.8370
9.8378
9.8386

9.8394
9.8402
9.8410

9.8418

9.8426
9.8433
9.8441

9.8449
9.S457
9.8464

9.8472
9.8480
9.8487

9.8495

9.9238

9.9254
9.9269
9.9284

9.9300
9.9315
9.9330

9.9346
9.9361
9.9376

9.9392

9.9407
9.9422
9.9438

9.9453
9.9468
9.9483

9.9499
9.9514
9.9529
9.9544

9.9560
9.9575
9.9590

9.9605
9.9621
9.9636

9.9651
9.9666
9.9681

9.9697

9.9712
9.9727
9.9742

9.9757
9.9772
9.9788

9.9803
9.9818
9.9833

9.9848

9.9864
9.9879
9.9894

9.9909
9.9924
9.9939

9.9955
9.9970
9.9985

10.0000

16
15
15
16

15
15
16

15
15
16
15

15
16
15

15
15
16

15
15

15
16
15
15
15

16
15
15

15
15

16
15
15

15
15

15
16
15

15
15
15
16
15
15
15

15
15
16

15
15
15

0.0762

9.8843

0.0746
0.0731
0.0716

0.0700
0.06S5
0.0670

0.0654
0.0639
0.0624

9.8836
9.8830
9.8823

9.8817
9.8810
9.8804

9.8797
9.8791
9.8784

0.0608

9.877S

0.0593
0.0578
0.0562

0.0547
0.0532
0.0517

0.0501
0.04S6
0.0471

9.8771
9.8765
9.8758

9.8751
9.8745
9.8738

9.8731
9.8724
9.8718

0.0456

9.8711

0.0440
0.0425
0.0410

0.0395
0.0379
0.0364

0.0349
0.0334
0.0319

0.0303

0.0288
0.0273
0.0258

0.0243
0.0228
0.0212

0.0197
0.01S2
0.0167

0.0152

0.0136
0.0121
0.0106

0.0091
0.0076
0.0061

0.0045
0.0030
0.0015

0.0000

9.8704
9.8697
9.8690

9.8683
9.8676
9.8669

9.8662
9.8655
9.8648

9.8641

9.8634
9.8627
9.8620

9.8613
9.8606
9.8598

9.8591
9.8584
9.8577

9.8569

9.8562
9.8555
9.8547

9.8540
9.8532
9.8525

9.8517
9.8510
9.8502

9.8495

50.0°

49.9°
49.8°
49.7°

49.6°
49.5°
49.4°

49.3°
49.2°
49.1°
49.0°

48.9°
48.8°
48.7°

48.6°
48.5°
48.4°

48.3°
48.2°
48.1°
48.0°
47.9°
47.8°
47.7°

47.6°
47.5°
47.4°

47.3°
47.2°
47.1°
47.0°
46.9°
46.8°
46.7°

46.6°
46.5°
46.4°

46.3°
46.2°
46.1°
46.0°

45.9°
45. S°
45.7°

45.6°
45.5°
45.4°

45.3°
45.2°
45.1°
45.0°

log cos difi. log cot ^' log tan log sin difi. Angle

45°-50°

H
n

16

1.6

3.2
4.8
6.4
8.0
9.6
11.2
12.8
14.4

15

1.5

3.0

4.5

6.0

7.5

9.0

10.5

12.0

13.5

9

0.9
1.8
2.7
3.6
4.5
5.4
6.3
7.2
8.1

8

0.8
1.6
2.4
33.
4.0
4.8
5.6
6.4
7.2

7
0.7
1.4
2.1
2.8
3.5
4.2
4.9
5.6
6.3

6

0.6
1.2
1.8
2.4
3.0
3.6
4.2
4.8
5.4

38

TABLE OF TRIGONOMETRIC FUNCTIONS

TABLE OF NATURAL VALUES OF THE TRIGO-
NOMETRIC FUNCTIONS

Angle

sin

COS

tan

cot

sec

CSC

0°

.0000

1.0000

.0000

00

1.0000

oo

90°

1°

.0175

.9998

.0175

57.290

1.0002

57.299

89°

2°

.0349

.9994

.0349

28.636

1.0006

28.654

88°

3°

.0523

.9986

.0524

19.081

1.0014

19.107

87°

4°

.0698

.9976

.0699

14.300

1.0024

14.336

86°

5°

.0872

.9962

.0875

11.430

1.0038

11.474

85°

6°

.1045

.9945

.1051

9.5144

1.0055

9.5668

84°

7°

.1219

.9925

.1228

8.1443

1.0075

8.2055

83°

8°

.1392

.9903

.1405

7.1154

1.0098

7.1853

82°

9°

.1564

.9877

.1584

6.3138

1.0125

6.3925

81°

10°

.1736

.9848

.1763

5.6713

1.0154

5.7588

80°

11°

.1908

.9816

.1944

5.1446

1.0187

5.2408

79°

12°

.2079

.9781

.2126

4.7046

1.0223

4.8097

78°

13°

.2250

.9744

.2309

4.3315

1.0263

4.4454

77°

14°

.2419

.9703

.2493

4.0108

1.0306

4.1336

76°

15°

.2588

.9659

.2679

3.7321

1.0353

3.8637

75°

16°

.2756

.9613

.2S67

3.4874

1.0403

3.6280

74°

17°

.2924

.9563

.3057

3^2709

1.0457

3.4203

73°

18°

.3090

.9511

.3249

^3^022$

1.0515

3.2361

72°

19°

.3256

. .9455

.3443

^9TS42

1.0576

3.0716

71°

20°

.3420

.9397

.3640

2.7475

1.0642

2.9238

70°

21°

.3584

.9336

.3839

2.6051

1.0711

2.7904

69°

22°

.3746

.9272

.4040

2.4751

1.0785

2.6695

68°

23°

.3907

.9205

.4245

2.3559

1.0864

2.5593

67°

24°

.4067

.9135

.4452

2.2460

1.0946

2.4586

66°

25?

.4226

.9063

.4663

2.1445

1.1034

2.3662

65°
64°

26°

.4384

.8988

.4877

2.0503 .

1.1126

2.2812

27°

.4540

.8910

.5095

1.9626

1.1223

2.2027

63°^

28°

.4695

.8829

.5317

1.8807

1.1326

2.1301

62°

29°

.4848

.8746

.5543

1.8040

1.1434

2.0627

61°

30°

.5000

.8660

.5774

1.7321

1.1547

2.0000

60°

31°

.5150

.8572

.6009

1.6643

1.1666

1.9416

59°

32°

.5299

.8480

.6249

1.6003

1.1792

1.8871

58°

33°

.5446

JS-587

.6494—

— 1.5399-

-1,1924

1.8361

57°

34°

.5592

.8290

.6745

1.4826

1.2062

1.7883

56°

35"

.5736

.S192

.7002

1.4281

1.2208

.1.7434 .

-55°

36°

.5878

.8090

J265

1.3764

1.2361

1.7013

54°

37°

.6018

.7986

<Cj5_3£

1.3270

1.2521

1.6616

53°

38°

.6157

.7S80

.7813

1.2799

1.2690

1.6243

52°

39°

.6293

.7771

.8098

1.2349

1.2868

1.5890

51°

40°

.6428

.7660

.8391

1.1918

1.3054

1.5557

50°

41°

.6561

.7547

.8693

1.1504

1.3250

1.5243

49°

42°

.6691

.7431

.9004

1.1106

1.3456

1.4945

48°

43°

.6S20

.7314

.9325

1.0724

1.3673

1.4663

47°

44°

.6947

.7193

.9657

1.0355

1.3902

1.4396

46°

45°

.7071

.7071

1.0000

1.0000

1.4142

1.4142

45°

cos

sin

cot

tan

CSC

sec

Angle

Date Due

JAN I t> 13"

1

I

" "W

i

i

•