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marked below.
PLANE AND SPHERICAI
TRIGONOMETRY
AND FOUR-PLACE TABLES OP
LOGARITHMS
BY
WILLIAM ANTHONY GEANVILLE, TH.D., LLD.
FOKMEULY PRESIDENT OF PENNSYLVANIA COLLEGE
GINN AND COMPANY
BOSTON NEW YORK CHICAGO LONDON
.TLANTA DALLAS COLUMBUS SAN FRANCISCO
ENTERED AT STATIONERS* HALL
COPYRIGHT, 1908, BY
WILLIAM ANTHONY GRANVILLE
ALL RIGHTS RESERVED
PRINTED IN THE UNITED STATES OF AMERICA
327.4
GINN 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 witho$ 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 denned 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 examples
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
front 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
bhree. 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.
5 W. A. GRANVILLE
CONTENTS
PLANE TRIGONOMETRY
CHAPTER I
TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES. SOLUTION OF
RIGHT 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
0. Generation of angles 24
% " X 10. Positive and negative angles 24
Angles of any magnitude 25
The four quadrants 25
ctangular coordinates of a point in a plane 26
ce of a point from 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. Five of the trigonometric functions expressed in terms of the sixth . 34
19. Line definitions of the trigonometric functions 36
20. Changes in the values of the functions as the angle varies 88
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 61
27. Reduction of functions of angles in the fourth quadrant 53
28. Reduction of functions of negative angles 56
29. General rule for reducing the functions of any angle 67
vii
viii CONTENTS
CHAPTER in
RELATIONS BETWEEN THE TRIGONOMETRIC FUNCTIONS
SBOTion 1 PAOX
80. Fundamental relations between the functions 59
81. Any function expressed in terms of each of the other five functions . 60
CHAPTER IV
TRIGONOMETRIC ANALYSIS
82. Functions of the sum and of the difference of two angles 63
88. Sine and cosine of the sum of two angles 63
84. Sine and cosine of the difference of two angles 66
$5. Tangent and cotangent of the sum and of the difference of two angles 68
86. Functions of twice an angle in terms of the functions of the angle . . 69
87. Functions of multiple angles 70
88. 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 OP ANGLES. INVERSE TRIGONOMETRIC FUNCTIONS.
TRIGONOMETRIC EQUATIONS
42. General value of an angle 79
48. 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 OP TRIGONOMETRIC FUNCTIONS
49. Variables 98
50. Constants 93
51. Functions . . . * 93
53. Graphs of functions 93
58. Graphs of the trigonometric functions 95
54. Periodicity of the trigonometric functions 97
55. Graphs of the trigonometric functions plotted by means of the unit circle 97
CONTENTS fe
CHAPTEK VII
SOLUTION OF OBLIQUE TBIANGLE8
SECTION PAGE
66. Relations between the sides and angles of a triangle 101
57. Law of sines 102
68. The ambiguous case . . . . , 104
69. Law of cosines 106
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
68. Need of logarithms in Trigonometry 119
64. Properties of logarithms 121
66. Common system of logarithms 124
66. Rules for determining the characteristic of a logarithm 126
67. Tables of logarithms 128
68. To find the logarithms of numbers from Table I 1^9
69. To find the number corresponding to a given logarithm 188
70. The use of logarithms in computations 186
71. Cologarithms 187
72. Change of base in logarithms 188
78. Exponential equations 140
74. Use of the tables of logarithms of the trigonometric functions ... 141
76. Use of Table II, angle in degrees and minutes 142
76. To find the logarithm of a function of an angle 148
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 162
82. Use of logarithms in the solution of oblique triangles 168
Case I. When two angles and a side are given 168
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
88. Use of logarithms in finding the area of an oblique triangle .... 170
84. Measurement of land areas ' 172
86. Parallel sailing 178
86. Plane sailing 174
87. Middle latitude sailing .176
CONTENTS
CHAPTER IX
ACUTE ANGLES NEAR OE 90
SECTION PAGE
88. Limits of and as x approaches the limit zero 178
x x
89. Functions of positive acute angles near and 90 179
90. Rule for finding the functions of acute angles near 180
91. Rule for finding the functions of acute angles near 90 181
92. Rules for finding the logarithms of the functions of angles near
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
RIGHT SPHERICAL TRIANGLES
1. Correspondence between the parts of a tried ral 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
6. Solution of right spherical triangles 200
6. The ambigiious 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
SECT-TOW PAGE
14. Napier's analogies 216
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 and the 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 OP SPHERICAL TRIGONOMETRY TO THE CELESTIAL AND
TERRESTRIAL SPHERES
24. Geographical terms 231
26. 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
81. 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
84. 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 260
37. The ecliptic and the equinoxes 263
38. The equator and hour circle of vernal equinox system 268
39. The system having for reference circles the ecliptic and the great circle
passing through the pole of the ecliptic and the vernal equinox . 266
40. The astronomical triangle 268
41. Errors arising in the measurement of physical quantities .... 269
CHAPTER IV
V 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
46. Length of an arc of a circle in linear units 264
46. Area of a spherical triangle 264
PLANE TRIGONOMETRY
CHAPTEK 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 HOfion 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 y By C denote the angles and the small
letters a, b y 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 y read " cosecant of A ";
sec A, read " secant of A " ;
cot -4, read "cotangent of A."
* Unless otherwise stated the hypotenuse of a right triangle will always be denoted bye
and the right angle by C.
1
PLANE TRIGONOMETBY
These trigonometric functions (ratios) are defined as follows (see
figure) :
,v
(1)
v /
opposite side / a\
-~- - = -);
cj
hypotenuse \
adjacent side / b\
(2) cos As:- - (=-)>
v ' hypotenuse \ c]
opposite side
hypotenuse/ c\
= J . - ( = -);
opposite side \ a/
/ c\
( = ih
\ b]
adjacent side
A
C0t ^ =
hypotenuse
adjacent side
adjacent side / b
opposite "
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,
sin A
1
esc A
A
esc A = - = - = - ;
a a sin A
A
cos A - - =
c c sec A
b
A
sec A = - = - =
b b cos A
4. A
tan A =- = T =
b b cot A ?
a
cot A
a a tan A
b
* For, let ' be any other point in AD, and B" any point in AE. Dr&w the perpendicu-
lars jB'C" and J3"C" to AE and ^/> respectively. The three triangles ABC, AB'C', AB"C",
&TQ 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 7
8. Given Bin A = f , c = 200.5 ; calculate a. Ana. 120.8.
9. Given cos A = .44, c = 30.5; calculated. Ana. 13.42.
10. Given tan A = ty, & = ff; calculates Ana.
11. Given A = 30, a = 25 ; calculate c. Also find 5 and 6.
-4n. c = 60, JS = 60, 6 = 26 V5.
12. Given B = 30, c = 48; calculate 6. Also find A and a.
. 6 = 24, A = 60, a = 24 \/3.
13. Given J5 = 45, 6 = 20 ; calculate c. Also find A and a.
s. c = 20 V2, -4 = 46, 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 | 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 arid 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 party 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 noting whether they satisfy
relations different from those already employed in the third step.
A convenient numerical check is the relation,
a* = c*-b*=*(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 = &
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 to 90 inclusive are correctly given to
four or five significant figures.
In looking up the function of an angle between 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 a* 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 Qranville's Four-Place Tables of Logarithm).
FUNCTIONS OF ACUTE ANGLES
TABLE A .
NATURAL VALUES OF THE TRIGONOMETRIC FUNCTIONS
Angle
sin
008
tan
cot
sec
080
.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
14.301
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
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
10306
4.1336
76
15
.2588
.9659
.2679
3.7321
1.0353
3.8637
75
16
.2756
.9613
.2867
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
.8572
.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
.7071
1.0000
1.0000
1.4142
1.4142
45
008
sin
cot
tan
C8C
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 t 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
sinz == .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 = 35, c = 267 ; 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 = W-A = 90 - 35 = 55.
Third step. To find a use formula (1), p. 2,
namely,
sin A = - -
c
6-?
Multiplying,
Substituting the value of sin A = sin 86
= .6736 (found from the table) and c = 267,
we have a
.5736 = -:!-.
267
Solving for a, we get a = 163.1.*
sin 35 = .5736
267
40152
34416
11472
(1=153.1512
Since our 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
a = 153.14586
Comparing, we see that the two values ot 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
Substituting as before, -we have
cos-4 = -.
c
.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 2 _ 2 , 2
To find the area of the triangle we have
ab 163.1x218.7
Area = - =
= 16,741.
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 ground ?
Solution. Our figure shows a right triangle
with hypotenuse and side adjacent to the re-
quired angle ( = x) given. Hence tf\X
cosx = $ = J = .6 = .6000.
This number is found in the column having cos at the bottom and opposite
60. Hence x = 60. Ana.
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 = -> or,
f> c
a = c sin A ;
cos A = - ) or,
c
C
b = c cos A ;
a
tan A = - > or,
b
a = b tan A.
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 TRIGONOMETRY
EXAMPLES
Solve the following right triangles (C = 90).
No.
GIVEN
PARTS
REQUIRED PARTS
AREA
1
^4 = 60
6 = 4
J3=30
c = 8
a = 6.928
13.866
2
-4 = 30
a = 3
# = 60
c = 6
6 = 6.196
7.794
3
a=6
c= 12
.4 = 30
5 = 60
6 = 10.39
31.18
4
a = 4
6 = 4
.4 = 46
J5 = 45
c = 6.667
8
5
a = 2
c = 2.8284
^4 = 45
J5 = 46
6 = 2
2
6
a = 61.303
c = 160
^4 = 20
= 70
6 = 140.96
3616.6
7
B = 61
c = 260
.4 = 39
a = 157.3
6 = 194.3
15282
8
J. = 36
c = l
#=64
a =.5878
6 = .809
.2378
9
c = 43
a = 38.313
^=63
= 27
6 = 19.62
373.9
*
10
6 = 9.696
c = 20
^4 = 61
5 = 29
a = 17.492
84.8
11
a = 137.664
c = 240
-4 = 35
J3 = 65
6 = 196.6
13532
12
^ = 76
a = 80
=15
6 = 21.43
c = 82.82
857
13
^4 = 25
a = 30
= 65
6 = 64.336
c = 70.99
965
14
B = 65
6 = 10
.4 = 35
a = 7.002
c = 12.208
35
15
J3=15
6 = 20
A = 76
a = 74.64
c = 77.28
746.5
16
a = 36.4
6= 100
.4 = 20
# = 70
c = 106.4
1820
17
a = 23.315
6 = 60
^4 = 26
E = 65
c= 65.17
583
18
a = 17.1
c = 50
^4 = 20
= 70
6 = 46.985
402
19
,4=10
6 = 30
J? = 80
a = 5.289
c = 30.46
79
20
.4 = 20
c = 80
5=70
a = 27. 36
6 = 76.176
1028
21
= 86
6 =.08
-4 = 4
a = .00559
c = .0802
.0002
22
B = 32
c = 1760
A = 68
6 = 932.62
a= 1492.6
696968
23
a = 30.21
c = 33.33
^4 = 65
# = 25
6 = 14.085
213
24
a = 13.395
6 = 60
A = 15
JB = 76
c= 61.77
335
25
6 = 93.97
c= 100
^. = 20
5=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 60 ft.
Find the length of the tree. -4ns. 96.06 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 66 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 16 east of south. How far is the
battleship from the nearest fort ? A ns. 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 6 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. 84. 6 ft.; 80.
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, J5, 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. 522.6ft.
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.
AD = ABcosA
= 40 cos 26 by (8), p. 11
= 40 x .9063
= 36.25. A*
Therefore , AC = 2 AD = 72.50 in.
To find the area we need in addition the altitude BD.
BD = AB sin A = 40 sin 26 by (7), p. 11
= 40 x .4226 = 16.9.
Check : BD - AD tan 25 = 36.25 x .4663 = 16.9. by (9), p. 11
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
AB 30 \/2 V2
Hence x = 45 = pitch of the roof.
Also, BC = AB sin x by (8), p. 11
-=:
V
= 30 ft.
= height of the ridgepole
above the eaves.
Check : AB =
2 = V(30)2 + (30)2 = Vl800 = 30 VS.
14
PLANE TKIGOKOMETRY
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 = 62.36 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 = 00; equal sides = 29.5 in. ; area = 32:1.4 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; 93.97ft.; 3213.8 sq.ft.
7. The base of an isosceles triangle is 100 ft. and its height is 85.01 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
66. Find the equal sides, the height, and the area.
Ans. 118.3ft; 107.23ft; 5361. 6 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 whole 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 A BC = ; hence
in the right triangle ABD the
, 180
angle x =
n
Also, AD = - = half the length of one side,
AB = R = radius of circumscribed circle,
BD = r == radius of inscribed circle,
p = nc = perimeter of polygon,
* = 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
180 = 180 = 13 o
n 10
Then R = ^ - = -JL_ = 16.18 in.,
sin 18 .3090
and r = - = 6 _ = 16.39 in.
tan 18 .3249
Check : r = R cos 18 = 16. 18 x .9511
= 15.39.
Also, p = 10 x 10 = 100 in.
= perimeter of polygon ;
, pr 100 x 15.39 _ n
hence = = 769.5 sq. in.
= area.
2. The side of a regular pentagon is 24 ft. ; find E, r, and area.
Ans. 20.42ft.; 16.52ft.; 991.2 sq. ft.
3. Find the remaining parts of a regular polygon, having given
(a) n = 9, c = 12. Ans. 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. R = 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.
6. If c be the side of a regular polygon of n sides, show that
_ 1 180 . 1 180
E = - c esc and r = - c cot
2 n 2 n
16 PLANE TBIGONOMETKY
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
o + 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,
1 : x : : .0139 : d
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.
x d
0.1 .0014 .-. sin 37.1 = .6018 + .0014 = .6032
0.2 .0028 ^ .-. sin 37.2 = .6018 + -0028 = .6046
0.3 .0042 .-. sin 37. 3 = .6018 + .0042 = .6060
0.4 .0066 .-. sin 37.4 = .6018 + .0066 = .6074
0.6 .0070 .-. sin 37.6 = .6018 + .0070 = .6088
0.6 .0083 .-. sin 37. 6 = .6018 -f .0083 = .6101
0.7 .0097 , .-. sin 37. 7 = .60 18 + .0097 = .6116
0.8 .0111 .-. sin 37.8 = .6018 -f .0111 = .6129
0.9 .0126 .-. 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 case the table would therefore lead to
very inaccurate results it interpolation was used for cotangents of email angles (see
Chapter JX, p. 178).
FUNCTIONS OF 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 82 and sin 33. From the
table on p. 9,
sin 33 = .5446
sin 32 = .5209
.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 82 to 32.8. Denoting by d the difference corre-
sponding to .8, we have
or, d = .0118.
Hence sin 32 -.5299
d = .Q118 = difference for .8
.-. sin 32.8 = .6417. Ans.
Ex. 2. Find tan 47 26'.
Solution. The tangent of 47 26' must lie between tan 47 and tan 48. From
the table,
tan 48 = 1.1 106
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 26", we have
60' : 25' : : .0382 : d,
or, d = .0159.
Hence tan 47 = 1.0724
__ _ d = .0159 = difference for 26'.
.-. tan 47 26' = 1.0883. Ans.
Ex. 8. Find cos 68. 57.
Solution. The cosine of 68.67 must lie between cos 68 and cos 69. From
the table,
cos 68 = .3746
cos 69 = .3584
.0162 = tabular difference cqrrespond-
ing to a difference of 1 in the angle. Denoting by d the difference corresponding
to. 67, we have
l:.67::.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. 67.
Hence cos 68 = .3746
d = .0092 = difference for. 67.
.-. cos 68. 67 = .3654. Ana.
(#) 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 tana; = .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 = .4320
tan 23 = J345
.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 -f 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':?/ :: .0207 : .0075,
or, y = 21.C'.
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 C 18' = .6858. (1) esc 72 54' = 1.0463.
(c) cos 79. 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.
TEEMiS IN TKiGONOMETKiC PROBLEMS
19
2. Find the angle x, having given
(a) sin x = .6280.
(b) tan x = . 6344.
(c) sec x = 1.2122.
(d) cos x = .9860.
(e) cot x= 3. 6249.
(f) esc x = 1.7600.
(g) sin x = .9426.
(h) cos x = . 2118.
(i) tanx = 1.1662.
(j) cot x = .0803.
(k) sec x = 4.6326.
(1) esc x = 1.2420.
(m) sin = .7100.
(n) cos x = . 9999.
(o) tan x = . 9845.
(p) cot x = 8.6892.
Ana. x = 31.87.
x = 32.39.
x = 34.41.
x = 9.93.
x = 16.85.
x = 34.85.
x = 70 28.96'.
x = 77.77.
x = 4921.4 / .
x = 85.41.
x = 77.61.
x = 53.63.
x = 45 14.3'.
x = 30'.
x = 44 33'.
B = 036.1'.
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- m
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 TKIGONOMETKY
The vertical distance between two points is the distance from one
of the two points to the horizontal plane through the other.
Thus, let EC 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 EC
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 KE. 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| K
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-
tances, 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 62'
B = 63 8'
6 = 80
2
a = 16.98
c = 18.7
A 66 14'
B = 24 46'
6 = 7.883
3
a = 147
c= 184
.4 = 63 2'
B = 86 68'
6 = 110.67
4
4 = 84 15'
a = 843.2
= 56 46'
c = 1498.6
6=1238.7
5
A = 81 14.2'
c = 2.934
B =68 45. 8'
a = 1.521
6 = 2.609
6
B = 47.26
c = 4.614
^1 = 42.74
a = 3. 131
6 = 3.389
7
.4 = 23.6
c = 627
J3 = 66.6
a = 260
6 = 675
8
A = 28 6'
c = 2280
B = 61 56'
a = 1073
6 = 2011
9
B = 43.8
6 = 60.94
A = 46.2
a= 68.18
c = 73.6
10
B = 6 12.3'
c = 3721
^1 = 83 47. 7'
a = 8699
6 = 402.2
11
a = .624
c=.91
A = 43 18'
5 = 46 42'
6 = .6623
12
a = 6
6 = 2
A = 68 12'
B = 21 48'
c = 6.385
13
a = 101
6=116
^1 = 41 3'
B = 48 67'
c = 163.8
14
A = 43.6
c = 11.2
J5 = 46.5
a =7.71
6 = 8.124
15
B = 68 60'
a =729. 3
A = 21 10'
6 = 1884
c = 2020
16
A = 68.66
c = 36.73
5 = 81.36
a = 30. 61
6= 18.59
17
B = 10.85
c = .7264
A = 79.15
a = .7134
6 =.1867
18
a = 24.67
6 = 38.02
A = 36 46'
B = 68 14'
c = 41.22
19
# = 21 33' 61'
' a = .821
J. = 68 26' 9"
b = .3244
c = .8827
20
^ = 740'18"
c = 276.62
J5=1569'42"
a = 264.9
6 = 76.93
21
u4 = 641.3 /
6 = 200.06
5 = 25 68. 7'
a = 410.6
c = 466.7
22
6 = .02497
c = .04792
.A = 68 36'
J5 = 31 24'
a = .0409
23
6 = 1.4367
c = 3.4653
A = 65 30'
B = 24 30'
a = 8.168
24. The length of a kite string is 260 yd. , and the angle of elevation of the
kite is 40. Find the height of the kite, supposing the line of the kite string to
be straight. -4ns. 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. Find 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 8.891 ft. Find the angle of elevation of the sun. Ans. 60.
27. From 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 9.6 mi. per hour
respectively. Find their distance apart after 1 hr.
22 PLANE TRIGONOMETRY
29. From the top of a tower 120 ft. high the angle of depression of aa object
on a level with the base of the tower is 27 48'. What is the distance of the
object from the top and bottom of the tower ? Ana. 268 ft., 228 ft.
30. A ship is sailing due east at the rate of 7.8 mi. an hour. A headland is
observed to bear due north at 10.37 A.M. and 33 west of north at 12.43 P.M.
Find the distance of the headland from each point of observation.
Ana. 26.22 mi., 30.08 mi.
31. A ship is sailing due east at a uniform rate of speed. At 7 A.M. a light-
house is observed bearing due north, 10.32 mi. distant, and at 7.30 A.M. it bears
18 18' west of north. Find the rate of sailing of the ship and the bearing of
the lighthouse at 10 A.M. Ans. 6.79 mi. per hour, 63 8' W. of N.
32. From the top of a tower the angle of depression of the extremity of a
horizontal base line 1000 ft. in length, measured from the foot of the tower, is
observed to be 21 16' 37". Find the height of the tower. Ans. 389.6 ft.
33. The length of the side of a regular octagon is 12 in. Find the radii of the
inscribed and circumscribed circles. Ans. 14.49 in., 16.69 in.
34. What is the angle of elevation of an inclined plane if it rises 1 ft. in a
horizontal distance of 40 ft. ? Ans. 1 26'.
35. A ship is sailing due N.E. at the rate of 10 mi. an hour. Find the rate at
which she is moving due north. Ans. 7.07 mi. per hour.
36. A ladder 40 ft. long may be so placed that it will reach a window 83 ft.
high on one side of the street, and by turning it over without moving its foot
it will reach a window 21 ft. high on the other side. Find the breadth of the
street. Ans. 66.64ft.
37. At a point midway between two towers on a horizontal plane the angles
of elevation of their tops are 30 and 60 respectively. Show that one tower is
three times as high as the other.
38. A man in a balloon observes that the bases of two towers, which are a
mile apart on a horizontal plane, subtend an angle of 70. If he is exactly above
the middle point between the towers, find the height of the balloon.
Ans. 3770ft.
39. In an isosceles triangle each of the equal angles is 27 8' and each of the
equal sides 3.088. Solve the triangle. Ans. Base = 6.496.
40. What is the angle of elevation of a mountain slope which rises 238 ft. in
a horizontal distance of one eighth of a mile ? Ans. 19 60'.
41. If a chord of 41.36 ft. subtends an arc of 146 37', what is the radius of
the circle? Ans. 21.66ft.
42. If the diameter of a circle is 8268 ft., find the angle at the center sub-
tended by an arc whose chord is 1027 ft. Ans. 86 37.8'.
43. From each of two stations east and west of each other the angle of
elevation of a balloon is observed to be 46, and its bearings N.W. and N.E.
respectively. If the stations are 1 mi. apart, find the height of the balloon.
Ans. 3783ft.
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 16. How high is the fort and what is the
distance to it from the second place of observation ?
Hint. Denoting the height by y and the distance by #, we have
y = x tan 15, by (9), p. 11
also, y = (x -f 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. An$ x _ 3g6 f ^ y _ 10 g f t
I* 200
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+ lOTrft.
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 ang^s 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. 34.15ft.
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 00. 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.
28750
Ans.
+ 7600 = 24094.
CHAPTEK 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.
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
O 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
TKIGONOMETRIC 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 live right angles ; or, if two com-
plete revolutions were first made, as shown in Fig. c, then we have
FIG. a
FIG. 6
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
First
Quadrant
Quadrant
Third
initial side
Fourth
Quadrant
Quadrant
Thus, the minute hand of a clock generates - 4 rt. A 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 608 lies in the second quadrant.
Solution. 668 = 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) 051. (i) 640. (in) 1500.
(b) 120. (f) - 150. (j) 420. (n) 810.
(c) - 316. (g) - 75. (k) - 910. (o) - 640.
(d) -240. (h) -1200. (1) -300. (p) 637.
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 0. Any
point in the plane of these lines
(as 7 J ) is determined by its distance
and direction from each of the perpen-
diculars X'X and Y'Y. Its distance
from y'F(as 7VP = ) is called the
abscissa of the point, and its distance
from X'X (as MP = /;) is called the ordinate of the point.
Abscissas measured to the right of Y'Y are positive.
Abscissas measured to the left of Y'Y are negative.
Ordinates measured above X'X are positive.
Ordinates^measured below X'X are negative.
The abscissa and ordinate taken to- Y
gether are called the coordinates of the II
point and are denoted by the symbol (->+)
i V
A7
a ' P(<*b)
1
b ;
'
3T'
IV
The lines X'X and Y'Y are called the p
axes of coordinates, X'X being the axis of _
abscissa s or th e a xis of A, an d Y'Y the axis j j j
ofordinates or the axis of Y; and the point
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,
TKIGONOMETK1C FUNCTIONS OF ANY ANGLE 27
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 X'X 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 I
count off a number of divisions equal
to the ordinate, upward or downward
according as the ordinate is positive %'
or negative. The work of plotting
points is much simplified by the use
of coordinate, or plotting paper, con-
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 F. Similarly, the following figures show the plotted points
(- 2, 3), (- 3, - 4), (0, 3).
(-2,3)
(-3,-^
(0,8)
14. Distance of a point from the origin. Eepresent the abscissa of
a point P by a and the ordinate by />, and its distance from the
Y* origin by It. Then
iw) * = V^TT 2 ,
since A is the hypotenuse of a right
triangle whose sides are a and b.
+X 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.
28
PLANE TRIGONOMETRY
EXAMPLES
1. (a) Plot accurately the points (6, 4), (- 3, 4), (- 2, - 4), (5, - 1), (6, 0),
<- 6, 0), (0, 4), (0, - 3).
(b) What is the distance of each point from the origin ?
Ans. Vil, 5, 2 \/6, etc.
2. Plot accurately the points (1, 1), (- 1, - 1), (- 1, 1), (V3, 1), (V3, - 1),
( V, 1), and find the distance of each one from the origin.
3. Plot accurately the points (V2, 0), (- 6, - 10), (3, - 2 V2), (10, 3), (0, 0),
(0, -V6), (3, -6), (-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, arid which agree with the
definitions already given for acute angles.
o
Angle in first quadrant
Angle in second quadrant
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 X. 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.
TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 29
Denoting by XOB any one of these angles, their functions are
defined as the following ratios :
OP ordinate
(11)
v '
(12)
v '
abscissa
OP
OQ
(14)
(15)
op __ kypotenuse
"""" QP ~~ ordinate '
__ OP _ hypotenuse t
~~ OQ "~ abscissa '
no abscissa *
QP "" ordinate
hypotenuse
ordinate
abscissa '
To the above six functions may be added the versed sine (written ver-
sin) and co versed sine (written coversin), which are defined as follows :
versin XOB = 1 cos XOB ; coversin X OB = 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
RULE Foii SIGNS
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
81 & n > 1<e ' sin and esc have the same sign,
cos and sec have the same sign,
tan and cot liave 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 tan x = , we may write
__ 2 _ 2 _ ordinate
tail X TT ~ : ; j
3 3 abscissa
(12), p. 29
] and hence, taking tan# = , one ter-
| minal side is determined by the origin
1 and (3, 2), giving the angle XOB (in
FIG. a the first quadrant).
2
The other terminal side, taking tan x = - ? is determined by the
o
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 j
and the terminal side OB, J
have the initial side OX '
and the terminal side OB 1 . *
Fig. a
Fig. b
Let us now determine the values of all the
functions. From Fig. a,
OP = 'V'OQ 2 + <2> a = V9 + 4 = Vl3 (always positive).
Hence by 15, p. 29, from Fig. a,
sin XOB =
cos XOB =
tanA r OB = ^;
Vl3'
3
CSC XOB =
sec XOB =
Vl3
^~ 5
Vl3
cot A" OB = --
* It Is evident that, corresponding to each figure, there are an infinite number of both
positive and negative angles diif ering by multiples of 360 which satisfy the given condition.
TKIGONOMETB1C FUNCTIONS OF ANY ANGLE 31
Similarly, from Fig. &,
csc XOB 1 =
smXOB 1 = ~
cos XOB' =
Vis'
3
ViV
3
9*
Or, denoting by x any angle which satisfies the given condition,,
we may write down these results in more compact form as follows :
o
Sill X = ;
COS X =
tan .r = - ;
Via'
3
csc x =
sec x
Vis
3 5
3
COt tf = -
The method is further illustrated in the following examples :
Ex. 1. Having given sin x = * , construct the angle x. Also find the values
)f tlie other five functions.
Solution. Here we may write, by (10), p. 29,
1 1 ordinate
sinx = - = =
(hypotenuse always positive).
3 3 hypotenuse
Since abscissa = -i V(hypot.) a (ord.) 2 _= A/9 1 = 2 A/2, one terminal
ide is determined by the origin and ( 2 V2, 1),
jiving the angle XOI? in the third quadrant. Here
sin XOB = - - ;
3
-^Z
cos XOB = -
1
cscA'O/J 3 ;
2 A 2 '
cotJTOJ?= 2 A/2.
2 V2
The other terminal side is determined by the origin and (2 A X 2, - 1) giving
yx the angle XOB' in fourth quadrant. Heie
1
! o
csc XO B' = - 3 ;
-**X
cos XOB' =
t&uXOB' = -
2 A/2
~3~ ;
1
sec XOB' =
2V2
2 A/2'
82 PLANE TRIGONOMETRY
Or, denoting toy x any angle which satisfies the given condition, we have
sin x = ;
3
cos x = =F :
tan x =
2\/2
csc x = 3 ;
3
=T -p]
2V2
= 2 \/2.
Ex. 2. Having given cotx = , find all the other functions of x.
n
Solution. Here we may write, by (15), p. 29,
cotx = = -
n n ordinate
and hypotenuse = Vra 2 -f n 2 .
Hence one terminal side is determined by the origin and (m, n), and the
other terminal side by the origin and ( m, n). Therefore
sin x =
tan x =
sec x =
Vm 2 + n 2
*
cotx = *
n
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 = -
a
6. tan x = -
11. tanx = -V7.
2.
cosx
7.
sin x
= c.
12. sin x =
3
3
84-
a 2 -6 2
1 ^ tan T* 9
c:
T
a 2 + 6 2
4.
secx
6
_.
9.
cscx
= -V3.
14. secx=p.
5.
cscx
= -.
10.
cosx
= .
* When m and n haye the same sign, x represents angles in the first and third quadrants.
When m and n hare opposite signs, x represents angles in the second and fourth quadrants*
-3.= =
1 **-l'
TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 33
ANSWERS
Quad-
rant
sin
cos
tan
080
860
cot
1.
n
I
II
8
6
3
6
2 V2
4
6
4
6
1
3
4
8
4
6
3
5
3
3
5
4
6
4
3
4
3
4
3
1
z.
3
2\/2
3
1
2 v 2
2V2
3
3
2 V2
1
III
~3~
1
3
3
2 V2
1
2 V2
v/Tn
Vio
2V2
TV
Vio
i
Vio
3
3
1
VTo
3
Vio
3
4.
5.
a
II
III
I
II
Vio
4
6
4
__
6
18
6
13
j_
Vio
3
6
3
6
12
13
12
"" 13
b
3
4
~3
4
3
6
12
6
~12
a
6
4
6
4
13
6
18
6
Va 2 + b 2
3
6
~8
6
3
13
12
13
12
Va 2 +6 2
8
4
3
4
12
T
12
6
6
7
"" Va 2 + 62
Va 2 +& 2
-v/1 /^2
6
c
a
1
6
1
a
vr^
I,
2a&
a 2 -6
""" Vl~c 2
2a6
c
a 2 4-& 2
Vl-c 2
a 2 4-6 2
c
a 2 -6 2
9.
in
IV
III,
IV
^ a 2 + 6 2
1
vi
Vc 2 w 2
a^ + bs
J5
T V
W
^a 2 -^ 2
^
Vc 2 -w 2
2ab
-V5
, c
a 2 -6 2
^
c
- 1 2a6
iV2
m
n,
c
Vl4
C
V5
x
m
VT"
Vc^-m 2
4
m
T 4
_L ______
Vc 2 m a
1
12.
1Q
rv
m,
IV
i,
1 4
2
3
5
T ~r
T^
3
2
, 2
VS
6
Vli
3
2
V29
v
^
v&>
V7
V6
2
2
14,
in
"*" V29
vy - 1
"*" V29
1
2
i -v/ria T
1 6
1 P
^ 2
5
1
P !
P
v p 1
"Vp'-l
^
Vp 2 -l
34
PLANE TRIGCXNOMETKY
18. Fire 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
1
Solution. Since sin x = -
hypotenuse
by (10), p. 29
abscissa = V(hypotenuse) 2 - (ordinate) 2
= Vl sin 2 x.
Hence, by definitions,*
sin x = sin x ;
esc x = -
1
tanx = -
Vl - sin 2 x
sinx
sec x =
cot x =
sinx
Ex. 2. Express, in terms of tan x, the other five functions of x.
tan x ordinate
Solution. Since tan x = -
abscissa
Hence
hypotenuse = V(abscissa) 2 -f (ordinate) 2
tan 2 x.
sin x =
tanx
tan 2 x
cos x = .
.
vl + tan 2 x
tari x = tan x ;
cotx =
tanx
Ex. 3. Having given secx = |, find the values of the other five functions.
p. 29
Solution. Since sec x = 5 = ^ypo^nuse ^ fey
4 abscissa
ordinate = V(hypotenuse) 2 (abscissa) 2
Hence
1 Q
= 5.
sin x = ^ ;
cos x = | ;
tanx = |;
3
cscx = f
sec x = | ;
cot x = j
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.
TK1GONOMETKIC FUNCTIONS OF ANY ANGLE 35
EXAMPLES
1. Express, in terms of cos x, the other five functions of x.
An*, sin x Vl cos 2 x ;
cos x = cos x ;
Vl cos 2 x
2. Express, in terms of cot x, the other fire functions of x.
Ans. sin x = - ; esc x = Vl -f cot a x ;
cotx
tan x =
cotx
cotx = cotx.
3. Express, in terms of secx, the other five functions of x.
cos x = ;
secx
tan x = db Vsec 2 x 1 ; cot x =
4. Express, in terms of esc x, the other five functions of x.
Ans. sin x = ; cscx = cscx ;
tan x =
5. Having given secx = -*/-, find the values of the other five functions of x.
Ans. sin x = \fy ; cscx = -J-J;
cos x = ^ ; secx = Y-;
tan x = =p ~V ; cotx = T A.
6. Having given sin x = a, find the values of the other functions of x.
cscx = - ;
a
Ans. sin x = a ;
cos x = Vl a 2 ; sec x =
tan x = .
86
PLANE TKiaONOMETEY
7. Having given cot x = Vi, find the values of the other functions of x.
Ans. sin x ; esc x = V5 ,
V3
tanx = - ; cot x = V2.
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
cot
Angle in first quadrant
Angle in second quadrant
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 (Le.
a circle with radius unity).
TEIGONOMETBIC FUNCTIONS OF ANY ANGLE 87
Applying these ratio definitions, we get
Q.P
^T) = QP '
00
OP OT *
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.
PLANE TKIGONOMETEY
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 Qi-Pi,
etc., and as x approaches zero as a limit, the sine approaches zero
as a limit. This is written . ^ _ Q
As x increases from and approaches 90 as a limit, the sine is
positive, and increases from zero through the values Q 8 P 8j QtP*, 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 6 P 6 , etc., and approaches the
limit OB\=-T). 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 (= 1) through Q 7 P 7 , etc.,
and approaches the limit zero.
This is written
sin 360 = 0.
(#) The cosine. Using the last figure, we see that as x decreases,
the cosine increases through the values OQi, OQ Z , etc., and as x ap-
proaches zero as a limit, the cosine approaches the limit OA (= 1).
This is written
cosO =
As x increases from and approaches 90 as a limit, the cosine
is positive and decreases from 6 A (= 1) through the values OQ 8 ,
etc., and approaches the limit zero. This is written
cos 90 = 0.
TRIGONOMETRIC 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 B ,
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 OQt, 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 ly 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 ly A 7 T 2 ,
etc., and as x approaches zero as a limit, the tangent approaches the
limit zero. This is written
tan = 0.
As x increases from and approaches
90 as a limit, the tangent is positive and
increases from zero through the values
AT 8) AT i, etc., without limit, i.e. beyond
any numerical value. This is written
tan 90 = + oo.*
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 AT% is negative and increases
in numerical value without limit. This
is written
tan 90 = oo.
We see, then, that the limit of the tangent will be -f 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 = 00,
when, as in this book, no distinction is made for the manner in which
the angle approaches the limit 90.
4- oo is read plus infinity.
oo is read minus infinity.
co 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 <x> through
ATi, AT 6 , 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 8 , A T 4 , etc.,
without limit. This is written
tan 270 = 00.
As x increases from 270 and approaches 360 as a limit, the
tangent is negative and decreases in numerical value from oo
through AT^ AT 6 , 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 O 7\, OT 2 , etc., and approaches
OA (= 1) as a limit. This is written
secO=l.
As x increases from and approaches 90 as a limit, the secant
is positive and increases from OA (= 1) through OT 8 , OT^ etc., with-
out limit. This is written
sec 90 = 00.
As x increases from 90 and approaches 180 as a limit, the secant
is negative and decreases in numerical value from oo through OT 6 ,
OT 5 , etc., and approaches minus OA (= 1) as a limit. This is written
sec 180 = - 1.
As x increases from 180 andjapproaches 270 as a limit, the secant
is negative and increases in numerical value from minus OA (= 1)
through 0r 8 , 0r 4 , 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 4- oo through OT^ OT 6 , 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 2) etc., and as x ap-
proaches as a limit, the
cotangent increases with-
out limit. This is written
cot = oo.
As x increases from
and approaches 90 as a
limit, the cotangent is posi-
tive and decreases from
4- oo through the values
BC 8) BC, 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
J3C 6 , BC Q) etc., without limit. This is written
cot 180 = 00.
As x increases from 180 and approaches 270 as a limit, the cotan-
gent is positive and decreases from -f- oo through BC Sy BC 4 , 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 6} BC 6) etc., without limit. This is written
cot 360 = QO.
(/) The cosecant. Using the last figure, we see that as x decreases,
the cosecant increases through the values OC ly OC 2 , etc., and as x
approaches as a limit, the cosecant increases without limit. This
is written ^
csc =00.
As x increases from and approaches 90 as a limit, the cosecant
is positive and decreases from -f- oo through OC 8 , OC 4 , etc., and ap-
proaches the limit 0J5(=1). This is written
esc 90 = 1.
42
PLANE TBIGONOMETKY
As x increases from 90 and approaches 180 as a limit, the cose-
cant is positive and increases from OB(= 1) through <9<7 B , <9C 6 , 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
OC 8 , 0C 4 , etc., and approaches the limit minus OB(= 1). This is
written csc270 = -l.
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 6 , OC 6 , etc., without limit. This is written
esc 360 = oo.
These results may be written in tabulated form as follows : *
90*
180^
270*
360
sin
1
'0
-1
cos
1
-1
1
tan
00
CO
cot
00
CO
00
sec
1
oo
-1
oo
1
CSC
00
1
oo
-1
oo
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 whatever ;
secant and cosecant can take on any values whatever, except those
lying between 1 and + !
1. Prove the following :
(a) sin -f cos 90 = 0.
(b) sin 180 + cos 270 = 0.
(c) cosO + tanO = l.
(d) tan 180 -f cot 90 = 0.
(e) sin 270 -sin 90 = -2.
EXAMPLES
(f) cos + sin 90 = 2.
(g) cos 180 + sin 270 = - 2.
(h) secO + esc 90 = 2.
(i) sec 180 - sec = - 2.
(j) cos 90 - cos 270 = 0.
(k) sin 90 + cos 90 -f esc 90 -f cot 90 = 2.
(1) cos 180 + sec 180 + sin 180 -f tan 180 = - 2.
(m) tan 360 - sin 270 - esc 270 4- 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 are identical.
TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 43
2. Compute the values of the following expressions :
(a) a sin + b cos 90 - c tan 180. Ans. 0.
(b) a cos 90 b tan 180 + c cot 90. 0.
(c) a sin 90 - b cos 360 + (a - b) cos 180. 0.
(d) (a 2 - #>) cos 360 ~ 4 ab sin 270. a 2 + 4 a& - 6 s .
21. Angular measure. There are two systems iii 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 ^\-Q 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 AOB = 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.
t A third system is the Centesimal or French System. The unit is one grade, being X & B 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 TrR ;
and this means that the radius may be measured off on the circumfer-
ence 2 TT 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 TT radians = 360,
TT radians = 180,
180 180
>
(16) 1 radian = 57.2957 + .
It therefore follows at once that :
To reduce radians to degrees, multiply the number of radians by
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,
(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
To reduce degrees to radians, multiply the number of degrees by
The student should now become accustomed to expressing angles
in circular measure, thus :
360 = 2 TT radians, 60 = ~ radians,
o
180 = TT radians, 30 = ~ radians,
b
7T 7T
90 = radians, 45 = -7 radians,
A 4
270 = - radians, 15 = ^ radians, etc-
.4 1.4
* The student should carefully observe that we do not lay off these radii as chords.
TEIGOKOMETEIC 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 (TT radians) is written simply sin TT and = sin 180,
(7T \ 7T
radians ) is written simply tan and = tan 90,
2, I A
cot( radians) is written simply cot--- and = cot 135,
cosf - 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
. s w , . length of subtending arc
(18) Number of radians in angle = - - -^
^ ' 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 = = 1.86. Ans.
2
To reduce this angle to" degrees, we have, from (16),
1.86 x 67.2957=: 105.907. 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), 2.6 = ,
R
R = 26.6 in. Ana.
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?
46 PLANE TBIGONOMETKY
3. What functions of an angle of an oblique triangle may be negative ? Why ?
4. In what quadrant do each of the following angles lie ?
iZ _ I _ 1Z 147r __ H 15?r . ^ + 2 . _ 37T-+ 2 .o.l. __ i . _ 6
12 ' ~~0' ~3~' T~ ; 4 ' 16 ' 6 ' 6 ' J 4'* J 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 :
13- * 27r - 25- 37r . 7r + 1 . 3 . Q 6 . 8ff + 2
L3 '2'T' ' ~T' ~6~' ' ' 6
Ana. 74.4844; 28.6478; 120; -143.239; -67.6;
39.649; -171.887; -160.4279; 130.92.
7. Express the following angles in circular measure : 22^ ; 60 ; 136 ;
-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; 17.2765; -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, ^; (b) 225, ^~; (c) 60, |.
9. Find the number of radians in an angle at the center of a circle of radius
26 ft., which intercepts an arc of 37 i ft. Ans. 1.6.
10. Find the length of the arc subtending an angle of 4 J radians at the center
of a circle whose radius is 26 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. Ans. 8 ft.
12. Find the length of an arc of 80 on a circle of 4 ft. radius. Ans. 6.6 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 6 TT ft. Ans. 90.
14. Find the number of radians in an angle at the center of a circle of radius
ST\ 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.
7T
16. What angle in circular measure does the hour hand of a clock describe
in89min. 22* sec.? Ans _ !Z ra d.
64
17. A wheel makes 10 revolutions per second. How long does it take to turn
through 2 radians, taking TT = V ? Ans. ^fa 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- 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. 786,400 mi
TRIGONOMETRIC FUNCTION 8 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
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 ?/ 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 46.
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". , .. , 2 TT
v ' ("rt cot
(c) ecSloiO'. E> W 6
(d)Bin* 6 (k) sin 1.2.
v ' 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, J5, C are the angles of any triangle, prove that
sin J^i = cos (J5 -f (7).
25. Reduction of functions of angles in the second quadrant.
First method. In the unit circle whose center is (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 A OP' = 180 P'OQ'. Let x be the measure of
the acute angle P'OQ', and construct AOP = P'OQ' = x.
48
PLANE TKIGONOMETBY
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' =3 QP.
But Q'P' = sin (180 x) and QP = sin x, and since they have the
same sign, we get . ,. OAO .
' sm (180 x) = sm x.
Similarly, the two right triangles OTA and 0TM may be proven
equal, giving
or,
' = AT and 07* = OT,
tan (180 #) = tan x and sec (180 x) = sec cc.
In the same manner, by proving the right triangles OBC and OBC 1
equal, we get , = , =
or
BC , = BC and oc , = OC}
cot (180 x) = cot x and esc (180 x) = esc #.
Collecting these results, we have
sin (180 -x)= since;
cos (180 x) = cos x ;
tan (180 - x) = - tan x ;
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 terminal side of 180. The algebraic signs,
however, are those for an angle in the second quadrant.
esc (180 x) = esc x ;
sec (180 x) = sec x ;
cot (180- a;) = ~ cot a.
TEIGONOMETEIC FUNCTIONS OP ANY ANGLE 49
Ex. 1. Express sin 128 as the function of an acute angle, and find its value.
Solution. Since 180 - 123 = 67,
sin 123 = sin (180 - 57) = sin 67 = .8887 (p. 9). Ans.
Ex. 2. Find the value of sec .
5- 6 2
Solution, sec - = sec 160 = sec (180 -. 30) = - sec 80 = --== Ans.
6 V
Ex. 3. Find tan 616.
Solution. 516 is an angle in the second quadrant, for 616 860 = 156.
Hence tan 616 = tan 156* = tan (180 - 24) = - tan 24 = - .4462. Ans.
Second method. The angle A OP' may also be written 90 -f y, where y
measures the acute angle J30P'. Since the angles BOP" and P / OQ f are com-
plementary, we have, from theorem on p. 47,
sin x = cos y ; esc x = sec y ;
cosx = sin2/; secx=cscy;
tanx=cot2/; cotz = tany.
Since 180 - x = 90 4- 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 -f y) =- coty ; cot(90 + y) =- tany.
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 90P. 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 = - .3266. Ans.
Ex. 6. Find the value of cos .
Solution. = 866 = 720 + 186.
Therefore
cos *i![ = cos 856 = cos 136 = cos (90 + 45) = - sin 46 = - -L . Ans.
4 -v/2
The aboye 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 300 from the angle that the remainder will be *
positive angle less than 360.
50
PLANE TBIGONOMETRY
EXAMPLES
1. Construct a table of sines, cosines, and tangents of all angles from to
180 at intervals of 30.
Ans.
30
60
90
120
150
180
1
V3
V8
1
sin
__ _
1
"2
2
2
2
V8
1
1
Va
cos
1
1
2
2
2
2
tan
1
V3
Vi
00
-V5
1
~Vs
2. Construct a table of sines, cosines, and tangents of all angles from 90 to
180 at intervals of 16, using table on p. 9.
Ans.
90
105
120
135
150
165
180
sin
1.0000
.9669
.8660
.7071
.5000
.2588
0.0000
cos
0.0000
- .2588
-.5000
-.7071
-.8660
-.9659
-1.0000
tan
00
- 3.7321
-1.7321
-1.0000
-.5774
-.2679
0.0000
3. Construct a table of sines, cosines, and tangents of all angles from 90 to
186 at intervals of 5.
4. Express the following as functions of an acute angle :
(a) sin 138.
(b) tan 883.
(c) cos 165 20'.
(d) sec 102 18'.
(e) cot .
(f) cot 170. 48.
(g) esc 817.
18 IT
(h) sin
t
(i) cos 2. 58.
(j) tan 1.6.
6. Find values of the following :
(a) sin 128 =.788.
(b) cos 160= -.9397.
(c) tan 185 =-1.
(e) cot = -1.
(f) esc 886= 1.1034.
6. Express the following as functions of an acute angle less than 46 :
(a) sin 106 = cos 16. 11 TT
(b) cos 148.3 = - cos 31. 7. (e) CSC "12" '
(c) tan 862 23 TT
(d) sec 794 62'. <*' cos ~g~'
. , . 87T
(g) sin 7 .
(m) cot 96 14 r .
(n) esc 126 42.8'.
(h) tan 108 15'.
( . . ITT
(i) "cos 173 9. 4'.
9
/j\ f"QT)
(p) cos 600.
U/ ta " rt
(q) tan 870.
(k) cos 496. 7.
(r) sec 1.9.
(1) sec 168.42.
(s) tanl.
TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 51
26. Reduction of functions of angles in the third quadrant.
First method. In the unit circle whose center is <9, 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
.4 OP' = 180+ Q'OP'. Let x be the measure of the acute angle
Q'OP 1 , and construct A OP = Q'OP' = x.
Now drawing the lines representing all the functions of the angles
x and 180 + x y we get, just as in the previous case,
sin (180 + ) = - sin x ; esc (180 -f- x) = - esc x ;
cos (180 -f x) = cos x ; sec (180 + rr) = sec x ;
tan (180 + x) = tan x ; cot (180 + a) = cot 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 tin*
third quadrant.
Ex. 1. Express cos 217 as the func-
tion of an acute angle, and find its value.
Solution. Since 217 - 180 = 37,
cos217 = cos(180 4- 37) = - cos37 = - .7986. Ana.
Ex. 2. Find value of esc 226.
Solution, esc 226 = esc (180 + 46) = - esc 46 = - V5. An*.
Ex. 3. Find value of sin 600.
Solution. 600 is an angle in the third quadrant, for 600 - SCO = 240.
V5
Hence sin 600 = sin 240 = sin (180 + 60) = - sin 60 = - . Ana.
Second method. The angle A OP" may also be written 270 - y, where y meas-
ures the acute angle P'05'. Since the angles P'0#' and Q'OP' (= A OP) are
complementary, we have, from theorem on p. 47, combined with the above
results, remembering that 180 + x = 270 - y,
sin (270 y) = cosy ; csc(270 y) = secy ;
cos(270 - y) = - siny ; sec(270 - y) = - cscy ;
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 27 OP. The algebraic signs, however, are those of an angle in the third
quadrant.
62
PLANE TRIGONOMETRY
Ex. 4. Find Bin 269.
Solution. Since 270 - 11 = 269,
sin 269 = sin (270 - 11) = - cos 11 = - .9816. Ana.
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 to
270 at intervals of 46.
-4ns.
45
90
135
180
225
270
1
1
1
sin
1
j
V2
V2
V2
1
1
1
cos
1
-1
V2
V2
v/2
tan
1
00
-1
1
CO
2. Construct a table of sines, cosines, and tangents of all angles from 180 to
270 at intervals of 16, using table on p. 9.
Ans.
180
195
210
225
240
255
270
sin
- .2588
-.5000
-.7071
-.8660
- .9659
- 1.0000
cos
-1.0000
- .9669
- .8660
- .7071
-.5000
- .2688
tan
.2679
.6774
1.0000
1.7821
3.7321
00
3. Construct a table of sines, cosines, and tangents of all angles from 136 to
270 at intervals of 6.
4. Express the following as functions of an acute angle :
(a) tan 200.
(b) sin 683.
(c) cos 224 26'.
(d) sec 260 40'.
tl^.
6
(f) esc 4. 3.
(g) sin 128.
5. Find values of the following :
(a) tan 235 = 1.4281.
(b) cot 1300 = 1.1918.
(c) sin 212 16'.
... 47T 1
(d)cos = --.
o &
,.. __ 7 IT 2
(f) sin 609.
(h) cos 998.7.
... . 167T
(.) sm_.
0) cos^.
/ \ 187r
(m) cot 185 62'.
(g) COS-.
(n) cos 687.
(h) tan 4.
29 7T
87T
(o) esc.
(i) cot .
(p) sin 262 10'.
... 21 7T
(q) cos 204.86.
(j) esc-.
, v . 97T
*
(r) tan .
(k) sin 228.4.
v/ 8
(1) tan 256 27. 8'.
TRIGONOMETBIC FUNCTIONS OF ANY ANGLE 53
6. Express the following as functions of an acute angle less than 46 :
(c) sin 594.
(d) sec^.
17 ir
(a) cos .
(b) tan 286. 6.
(.) sin.
(f) cos 260 63.4'.
27. Reduction of functions of angles in the fourth quadrant.
First method. As before, let A OP' be any angle in the fourth
quadrant. The functions of any
such angle are the same as the
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 cases,
sin (360 x) = sin x ;
cos (360 aj) = cos x ;
tan (360 - x) = tan x ;
esc (360 - x) = esc x ;
sec (360 x) = sec x ;
cot (360 - x) = - cot 2;
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 = - .6446. Ana.
Ex. 2. Find value of cot-
o
Solution, cot = cot 300 = cot (360 - 60) = - cot 60 = . ATM.
3 V8
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. Ana.
54 PLANE TRIGONOMETBY
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 -f V,
sin (270 + y) = - cosy;
cos (270 + y) = sin y ;
tan (270 + y) = - cot y j
esc (270 -f y) == sec y ;
sec (270 -f I/) = cscy;
cot (270 + y) = - tan y.
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.
Ex. 4. Find value of cos
llTT
6
Solution, cos = cos 830 = cos (270 -f 60) = sin 60 = . Ans.
6 2
As before, the first method is generally to fee preferred.
EXAMPLES
1. Construct a table of sines, cosines, and tangents of all angles from 180 to
860 at intervals of 30.
Ans.
180
210
240
270
300
330
360
1
V3
Va
1
sin
1
2
2
2
2
V3
1
1
V3
cos
1
2
2
2
2
tan
1
V5
00
-V8
1
V3
V3
2. Construct a table of sines, coshies, and tangents of all angles from 270 to
360 at intervals of 16, using table on p. 9.
Ans.
270
285
300
315
330
345
360
sin
-1.0000
-.9659
-.8660
-.7071
-.5000
-.2688
cos
.2588
.6000
.7071
.8660
.9659
1.0000
tan
00
-8.7321
-1.7321
-1.0000
-.5774
-.2679
3. Construct a table of sines, cosines, and tangents of all angles from 270 to
860 at intervals of 6.
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 296 43'.
(d) cot 866 11'.
5. Find values of the following :
(a) sin 276 = -.9962.
(b) cos 336 = .9136.
(c) tan 687 = -.6494.
(d) cot 1056.
(e) sec 296 62.6'.
... . 17 7T
(f) sm
v ' 9
(e) sin 655.
(f) CSC
(g) sin 275. 6.
(g) csc'6.2.
16 TT
(h) cos
(i) esc
(j) tan
llTT
,uv 18 IT
(h) cos ,
(i) sec 246.
(j) tan-
(k) sin 275 22'.
(1) cot 348.
(m) tan 660.
(n) seci^~
(o) sin
67T
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 1
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'J .-. s i n x = sin ( x) ;
/. cosx = cos( x);
.*. tan x = tan ( x) ;
.-. secx = sec( #);
OQ = OQ,
AT=z-AT',
OT = 07 7 ',
BC = - BC\
OC =-0C",
. cot x = cot ( x) ;
. esc x = esc ( x).
We may write these results in the form
sin ( x) = sin x ; esc ( x) = esc x ;
cos ( x) = cos a ; sec ( #) = sec x ;
tan( #) = tana;; cot (#) = cot x.
66
PLA1STE 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 = -.6643. Ana.
Ex. 2. Find value of sec (- 136).
Solution, sec (- 136) = sec 135 = sec (180 -46) = -sec 45 = -\/2. Ans.
Ex. 3. Find value of sin ( 540).
Solution, sin (-640) = -sin 640 = -sin (360+ 180) = -sin 180= 0. Ans.
EXAMPLES
1. Construct a table of sines, cosines, and tangents of all angles from to
- 860 at intervals of 30. Ans.
2. Find values of the following :
(a) tan (-83) = -.6494.
(c) cos (-136) = -
(d) cot(- 269).
(e)sec(-l).
(f) sin (-1231).
(h) sin (-1000).
(i) cos (-2.8).
(k) sin (-176.9).
(1) cos (-88 12. 70
/ v *
(m) t
(n) cot (-842).
Angle
sin
cos
tan
1
1
V3
1
30
~2
2
~v
-60
_Vs
1
-Vs
2
2
-90
-1
00
- 120
Vs
1
V3
2"
2
- 160
i
Vs
1
2
2
Vi
- 180
-1
-210
1
Vs
1
2
2
~vl
-240
V5
1
-Vs
2
2
-270
1
00
-300
Vs
1
v
2
2
-880
1
2
Vs
2
1
V8
-360
1
* Another method for reducing the functions of a negative angle consists in adding such
a multiple of + 300 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(~240o)=cosl20 cos(180 -eo>)-coseoo-- J. Ana.
TRIGONOMETBIC 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 i x or 360 i #, 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. 60-56, applying the above general rule.
EXAMPLES
1. Construct a table for every five degrees from 90 to 180.
Ans.
Angle
sin
cos
tan
cot
800
080
90
1.0000
0.0000
00
0.0000
00
1.0000
96
.9962
- .0872
-11.430
-.0876
-11.474
1.0038
100
.9848
-.1786
-6.6713
-.1763
-6.7688
1.0164
2. Construct a table as in Ex. 1 for every 16 from 180 to 270.
3. Construct a table as in Ex. 1 for every 10 from to 90.
4. Reduce the following to functions of x :
(a) sin (x 90) = cosx.f
(b) cos (x TT) = cos x.
(c) tan I x j = cot*.
(d) cot (x - 2 ir) = cot x.
(e) sec (x - 180) = - sees.
(f) c
(g) sin (x - 270).
(h) cos( x - ir).
(j) COt(-X-87T).
(k) sec (x- 630).
(1) esc (x - 720).
* In case the given angle is greater than 360 we assume that it has first been reduced to
a positive angle less than 360 by the subtraction of some multiple of 300. 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, x 90 is a negative angle. Hence sin (x 90) = sin (90 x ) - cos x.
58 PLANE TRIGONOMETRY
6. Find values of the following :
(a) cos 420. (i) cot 649 39'. ( } gec * 3 ?r
(b) sin 768. (j) esc 387 68'. w 6
(q) sin(- 6.3).
(
(d) cot (-240).
(f) tan 7.6. 9ir (t) sec (- 123.8).
(g) sin ( - 2.8). W cos T ' ( U ) sin (- 266 19.6
(h) cos 962. 8. (o) cot 1020. (v) cos (-98 31').
6. Prove the following :
(a) sin 420 - cos 390 + cos ( - 300) - sin ( - 330) = 1.
(b) cos 670 sin 610 - sin 330 cos 390 = 0.
(c) a cos (90 - x) + b cos (90 + x) = (a 6) sin x.
(d) m cost -- x) sinf -- x) = rrcsinx cosx.
(e) (a - 6) tan (90 - x) + (a + b) cot (90 + x) = (a - b) cot x - (a + 6) tan x.
(f) sin ( h x jsin (TT -f x) + cos ( h x j cos (IT x) = 0.
(g) cos (TT+X) cos ( y\ sin (TT + )sin ( y \ = cosx siny sinx cosy.
(h) tan x 4- tan ( y) tan (TT y} = tan x.
(i) cos (90 + a) cos (270 - a) - sin (180 - a) sin (360 - a) = 2 sin2 a .
sin (180 -y) 1 __
VJ; sin (270 - y) v ^' sin* (270 - y)
(k) 8 tan 210 -f 2 tan 120 = - Vs. (n) tan J (2 TT -H x) = tan x.
(1) 6sec2 135 - 6 cot 2 300 = 8. (o) cso J (x - 2 ?r) = - sec J x.
(m) cos (x - 270) = sin x, (p) cos %(y - 810) = - sin J y.
CHAPTER 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
(19)
sin*
_ 1
~"~ CSC * '
CSC*
= ~
1
i
(20)
cos*
sec*
secjc v
cos*
(21)
tan*
1
cot*
1
cot*'
tan*
cot
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 = , and cot x =
Substituting the functions
equivalent to QP and OQ, we get
, v sin*
(22) tan* = - 1
v ' cos*
Again, in the same triangle,
cos*
-
sin*
or,
(23)
In triangle OA T,
(24)
In triangle OCJB,
(25)
While in the above figure the angle x has been taken in the first
quadrant, the results hold true for any angle whatever, for the above
59
Z 2 + AT* = Or 2 , or,
l-t-tan a jr=sec 2 *.
OC 2 , or
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 = L . (19), p. 59
v ' CSCJT v " *
(27) sin x = =t Vl-cos 2 *.* Solving (23), p. 59, for sin x
(28) coijr = JL. (20), p. 59
OCv X
(29) cos x = Vl sin 2 jr. Solving (23), p. 59, for cos x
(30)tan* = -L. (21), p. 59
(31) tan x = db Vsec 2 x 1. Solving (24), p. 59, for tana;
,^. sin x sin x Vl cos 2 x
(32) tan* =
v ' i
COS * . ^/i __ g j n 2 x COS *
[Prom (22), p. 59 ; also (29) and (27).]
( 33 ) C8C * = 5HT (),P.B
(34) esc x = Vl + cot 2 x. Solving (25), p. 59, for esc x
(35) sec x = -L- . (20), p. 59
v ' cos x v /? r
(36) sec x = db VI + tan 2 x. Solving (24), p. 59, for sec x
(21), p. 59
(38) cot x = d= Vcsc 2 *--!. Solving (25), p. 59, for cot x
,^ ^ C08jr cos* db Vl sin 2 x
(39) cot x = - =
sin*
[From (22), P. 59 ; 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 lor
this is that there are two angles less than 3GO* for which a function has a given value.
TRiGONOMETKIC KELATIONS
Gl
Ex. 1. Find sin x in terms of each of the other five functions of x.
I
(a) sinx =
C8CX
(b) sinx = VI cos a x,
(c) sin x = -
from (26)
from (27)
substitute (84) in (a)
substitute (28) in (b)
Substitute (87) in (c)
tan 2 x
Ex. 2. Find cos x in terms of each of the other five functions.
(a) cos x =
secx
(b) cosx = vl sin 2 x,
(c) cos x =
Vcot 2 x + 1
from (28)
from (29)
substitute (36) in (a)
substitute (26) in (b)
Substitute (30) in (c)
cot 2 x
Ex. 8. Find tanx in terms of each of the other five functions.
(a) tan x =
cotx
(b) tan x = Vsec 2 x - 1,
sinx
from (80)
from (81)
from (32)
from (82)
Substituting (38) in (a)
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
sec x tan x sin x =
cosx cosx
I Since sec x and tan x
I cos a: cos a;
__ 1 sin 2 x ___ cos 2 X
"~ cosx "~ cosx
= cosx. Ans.
(28), p. 69
* 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 radicals only when unavoidable.
62
PLANE TRIGONOMETRY
Ex, 6. Prove that sin x (sec x -f esc x) cos x (sec x esc x) = sec x esc x.
Solution, sin x (sec x + csc x) cos x (sec x esc x)
I Since sec x =
cos a:
__ sinx
~~ cosx
sinx
\cosx
and CBCX--
1 \ /I 1 \
. )_ cosx : )
sin x/ \cos x sin x/
i I
sin a; J
| 1 l{
COBX
sinx
cos x
cos x sin x
sin 2 x -f cos 2 x
cos x sin x
_1 1
cos x sin x
cos x sin x
= sec x esc x. -4n.
(88), p. 69
PLES
1. Find sec x in terms of each of the other five functions of x.
Ans.
cosx
, Vl + tan 2 x ,
cotx
db Vcsc a x 1
2. Find cot x in terms of each of the other five functions of x.
Ana.
1
tanx ' sinx Vl-cos 2 x iVsec^x-l
3. Find esc x in terms of each of the other five functions of x.
t
Ana. , v 1 -f- cot 2 x , -
secx
Vl COS 2 X
4. Prove the following :
(a) cosx tanx = sinx. (h)
(b) sin x sec x = tan x. (i)
(c) sin y cot y = cos y. (j)
(d) (1 -f tan 2 y) cos 2 y = 1. (k)
(e) sin 2 A + sin 2 A tan 2 A = tan 2 A. (1)
(f) cot 2 A cos 2 J[ = cot 2 A cos 2 A. (m)
(g) tan A + cot A = sec A esc A. (n)
(o) (sin x -f cos x) 2 -f (sin x
(p) sin 8 x cos x -|- cos 8 x sin x =
(q) sin*.B -f tan 2 B = sec 2 B -
siny
tanx
cosA cacA = cot A.
cos 2 A - sin 2 A = 1-2 sin 2 A.
cos 2 A - sin 2 A = 2 cos 2 A - 1.
(1 + cot 2 !?) sin 2 5 = 1.
(esc 2 A - 1) sin 2 A = cos 2 A.
sec 2 A -f esc 2 A = sec 2 A esc 2 A
cos* C - sin* C + 1 = 2 cos 2 C.
cosx) a = 2.
= sinx cosx.
- cos 2 B.
, .
(r) cot y H
= esc y.
1 -h cosy
(s) cos B tan B -f- sin B cot B
(t) sec x esc x (cos 2 x sin 2 x)
cosC sin (7
= sin B + cos B.
= cot x tan x,
smz
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 -4- y) = sin x cos y -f- cos x sin y.
(41) sin (x y) = sin JT.COS y cos x sin y.
(42) cos (x + y) = cos x cos y sin x sin y.
(43) cos (x y) = cos JT 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 A OP = x
and the angle POQ^=y. Then the angle A OQ = x -j- y.
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 -f- y) = CQ, and
(b) cos (a -f y) = OC.
Draw QD perpendicular to OP. Then
(c) sin y = DQ, and
(d) cos y = OD.f
* Since a; and y are angles, their sum x + y and their difference x y are also angles.
Thus if a?= 61 and y = 23, then x + y =84 and a; - y = 38. The student should observe that
Bin (a? + y) is not the same as sin x + sin y, or cos (a: 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 A OP (= z), having their sides perpendicular each to
each. From (a),
(e) sin (x -f 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), J9Q = siny. Therefore
(g) FQ cos x sin y.
Substituting (f) and (g) in (e), we get
(40) sin ( x + y) = sin x cos y + cos jr sin #.
To derive (42) we use the same figures. From (b),
(h) cos (x 4- y) = OC = OF - CE = 6>E - FIX*
0I being one side of the right triangle 0JZ>, we have
OE = OD cos x. from (8), p. 11
But from (d), OD cos y. Therefore
(i) 0J = 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 ( x -j- y) = cos x cos y sin x sin y.
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 negatire.
TKIGONOMETEIC 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 -f x' and y = 270 -f y'*; then x + y = 860 -f (x' + y^ and
(k) x'=x-90, y'=y-270, x'+y'=34-y-860.
cos (x -f y) = cos [360 -f (x' + y')] = cos (x' + y f ) by 29, p. 67
= 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. 67
= cos x cos y sin x sin y. Q. s. 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 = - ISO -?/'; then x + y - - 90 - (x' + y 7 )
and
(1) x'=90-x, y'=-180-y, x'+y' = -90--(x + y).
sin (x + y) = sin [ - 90 - (x' + y')] = - cos (x' + y'} by 28, p. 66
= [cos x' cos y' sin x' sin y'] by (40)
==-[cos(90 -x)cos(-180 -y)-shi(90 -x)sin(-180-y)] from(l)
= [sin x ( cos y) cos x sin y] by 29, p. 67
= sin x cos y -f- cos a? sin y . Q. E. D.
EXAMPLES
1. Find sin 76, using (40) and the functions of 46 and 80.
Solution Since 76 = 45 + 30, we get from (40)
sin 76 = sin (45 -f 80) = sin 46 cos 80 + cos 46 sin 80
1 V5 1 1
= : --- [-'- from p. 6
V2 2 V2 2
V8 + 1
Ans.
2. Find cos (x + y), having given sin x = $ and sin y = ^, x and y being
positive acute angles.
Solution. By the method shown on p. 80 we get first
sinx = f, cosx = $, siny = -j^, cosy = }.
Substituting these values in (42), we get
cos(x + y) = I if - f A = H- -**
* The student should note that a^ and y' are acute angles.
66 PLANE TRIGONOMETRY
3. Show that cos 76 = - ^r- , using the functions of 46 and 80.
4. Prove that sin 90 = 1 and cos 00 = 0, using the functions of 60 and 80.
6. If tanx = f and tan y = ^ ? , find sin (x -f y) and cos(x -f y) when x and y
are acute angles. Ana. sin (x + y) = f , cos (x -f y) = .
6. By means of (40) and (42) express the sine and cosine of 90 + x, 180 + x,
270 -f x, in terms of sinx and cosx.
7. Verify the following :
(a)
8. Find sin (^1 -f B) and cos (A 4- J5), having given sin J. = ^ and sin B = .
B) = , eo-M + B) =
6 6
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 -f y) = sin x cos y -f- 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) = sin y. from p. 56
Substituting back in (a), we get
(41) sin (x y) = sin x cos y cos * 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(jr y) = cos* cosy 4- sin .r sin y.
TRIGONOMETRIC ANALYSIS 67
EXAMPLES
1. Find cos 16, using (43) and the functions of 46 and 30.
Solution. Since 15 = 46 - 80, we get from (43)
cos 16 = cos (46 - 80) = cos 46 cos 30 + sin 46 sin 80
\/3 1
i
Ans.
2\/2
Work out the above example, taking 16 = 60 - 46.
2. Prove sin (60 -f x) - sin (60 - x) = sin x.
Solution. sin (60 4- x) - sin 60 cos x -f cos 60 sin x. by (40)
sin (60 - x) = sin 60 cos x - cos 60 sin x. by (41 )
/. sin (60 -f x) sin (60 x) = 2 cos 60 sin x by subtraction
= 2 sin x = sin x. Ans.
Vs i
3. Show that sin 16 = - , using the functions of 46 and 30.
2V2
4. Find sin(x y) and cos(x y), having given sinx = J and siny = ^,
x and y being acute angles.
. . . 2\/2-VT6 , . 2V80 + 1
Ans. sin (x y) - , cos (x y) = - -
5. Find sin(x - y) and cos(x - y), having given tanx = j and tany = |, x
and y being acute angles. ,4ns. sin(x - y) = ^, cos(x - y) = %$,
6. By means of (41) and (43) express the sine and cosine of 90 x, 180 x,
270 x, 360 x, in terms of sin x and cos x.
7. Verify the following :
(e) sin (60 + x) - sin x = sin (60 - x).
(f ) cos (80 + y) - cos (30 - y) = - sin y.
(g) cos (46 + x) + cos (46 - x) = \/2 cos x.
(h) sin (46 + P) - sin (46 - P) = V2 sin P.
(i) cos (Q + 46) 4- sin (Q - 46) = 0.
(j) sin (x + y) sin (x - y) = sin 2 x - sin 2 !/.
(k) sin (x + y -f 2) = sinx cos y cos z -f coax sin y cosz
+ cos x cos y sin z sin x sin 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 -f y) __ sin x cos y 4- cos x sin y
\ T- y)
Now divide both numerator and denominator by cos x cos y. This
sin x cos y cos a; sin y
cos x cos y cos x cos y
S lves sin x cos y cos a; sin y
cos x cos y sm x sm y
cos a; cos y cos a; cos y
sin a; sin y
cos x cos y
sin a; smy
cos x cos y
. sin a; , siny
Since = tan x and = tan ?/, we get
cos a; cosy
(44) t.(* + ,)
v x v x
In the same way, from (41) and (43) we get
tan* tany
(45) tan(* y) = ~
v x v x 1 -f- tan JT tan y
From (22), p. 59, and (40) and (42), p. 63, we get
cos (x + y) cos x cos y sin a? sin y
^ ' ~~ sin (x -f- y) ~~ sin a; cos y + cos a; sin y
Now divide both numerator and denominator by sin x sin y. This
cos x cos y sin a; sin y
sin a; sin y sin a; sin y
sin a; cos y cos x sin y
sin a; sin y sin a; sin y
cos a; cos y
sin x sin y
cosy cos a;
sin y sin x
~. cos a? , , cosy
Since = cot a:, and -; - = cot y, we get
sin a? ' smy y ' &
/^\
(46)
TRIGONOMETRIC ANALYSIS 69
In the same way, from (41) and (43) we get
COtJCCOty + 1
,._. ^, N
(47) cot(x y) =
v ' v '
cot y--
Formulas (40) to (47) may be written in a more compact form as
loliows : _.
cos (a; y) = cos cc cos y qp sin cc sin y,
, , , N tan x tan y
tan(x y) == - - - - ^->
v ' 1 zp tan a; tan y
,. cot x cot y IE 1
cot (xy) = - . T
v JJ
The f ormulaA 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 16 = 60 45, we get from (45)
tanl6^tan(60- 4 5)= ^n 60 - tan 45 = VS-!^.^ ^
1 + tan 60 tan 46 i + V
Work out above example, taking 16 = 46 30.
2. Find tan (x -f y) and tan (x y), having given tan x = ^ and tan y = .
Arts, tan (x + y) = ^, tan (x y) = f .
3. Find tan 75 from the functions of 46 and 30. Ans. 2 -f- Vs.
4. Verify the following :
(a) tan(45 + x) = . (f)
. .
1 tan x sin (x y) tan x tan y
Va rot 7? - 1 (i) 1 - tanx tan y =
dCQt ^
-
(d) cot( + 80) = . cosx cosy
cotB + V3 cot P cot O i- cos ( p +Q)
(e) tan (x 46) + cot (T 46) = 0. u/w-^^ v ^""
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 2x we take (40),
sin(# 4- y) = sin a; eos y + cos x siny.
70 PLANE TBIGONOMETEY
Eeplace y by x. This gives
sin (x + x) = sin x cos x -f- cos x sin x 9 or
(48) sin 2 * = 2 sin x cos jr.
To find cos 2xwQ take (42),
cos (x + y) = cos # cos y sin x sin y.
Eeplace y by a. This gives
cos (x -f #) = cos a? cos a? sin x sin a, OP
(49) cos 2 * = cos 8 x sin 8 jr.
Since cos 2 a = 1 sin 2 x, (49) may be written
(49 a) cos 2 x = 1 2 sin 8 jr.
Or, since sin 2 a; = 1 cos 2 #, (49) may also be written
(49 b) cos 2 JT = 2 cos 8 JT 1.
To find tan 2 a; we take (44),
, , t . tan x 4- tan y
tan (x 4- y) = *-
v yj 1 tan x tan y
Replace y by x. This gives
_ tan x 4- tan x
^ ' 1 tan # tan a
(50)
1 - tan 8 *
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 a; in terms of sin x we take (40),
sin (x + 'y) sin x cos y -j- cos x sin y.
Eeplace y by 2x. This gives
sin(x + 2#) = sinxcos2a; -f cossin2a, or
sin 3 x =^in x (cos 2 x sin 2 #) -f- cos x (2 sin x cos x) by (49), (48)
sin a; cos 2 a? sin 8 a;
3 sin x (1 sin 2 x) sin'sc by (23), p. 59
3 sin x 4 sin 8 a;. Ans.
To find tan 4 x in terms of tana;, we take (44), (50),
2tan2x 4 tan a; (1 - tan 2 a;)
^ - ^-~ = q ^T~4 . 4. < '
1 tan 2 2a; 1 6 tan 2 a; -f- tan 4 a;
TRIGONOMETRIC ANALYSIS 71
EXAMPLES
2
1. Given sinx = - x lying in the second quadrant; find sin 2 as, cos 2x,
tan2x. ^
2
Solution. Since sin x = and x lies in the second quadrant, we get, using
method on p. 30, "^
2 1
sinxn -, cosx=- , tan = 2.
V6 V6
Substituting in (48), we get
sin 2 x = 2 sin x cos x = 2 ( ) =
V6\ vV 6
Similarly, we get cos2x = $ by substituting in (49), and tan 2x = J by
substituting in (50).
2. Given tanx = 2, x lying in the third quadrant ; find sin 2x, cos 2x, tan 2x.
Ans. sin 2 x = $, cos 2 x = f , tan 2 x = J.
3. Given tan x = - ; find sin 2 x, cos 2 x, tan 2 x.
2a& & 2 -a 2 2a&
ft 2 - a 2
4. Show that cosSx = 4 cos 8 x 8 cosx.
* eru 4.1 A * o A 3 tan .4 tan 8 -4
6. Show that tan 8 A =
6. Show that sin 4 x = 4 sin x cos x 8 sin 8 x cos x.
7. Show that cos 4 1? = 1 8 sin 2 J5 -f 8 sin 4 B.
8. Show that sin 6 x = 6 sinx - 20 sin s x + 16 sin 6 x.
9. Show that tan (46 + x) - tan (45 - x) = 2 tan 2 x.
10. Show that tan (46 + C) + tan (45 - C) = 2 sec 2 C.
11. Verify the following :
, x . _ 2 tanx ,.. _ csc 2 x
(a) sm 2 x = -- (f) sec 2 x = - - --
v 1 + tan 2 x v ' csc 2 x - 2
_ * ~~ tan2x fe) 2 esc 2 s = sec s esc s.
(b) cos 2 x - ___. ^ cou _ tan* = 2 cot 2 *.
(c) tan P + cot P = 2 esc 2 P. ( . } cog 2 x = ?_
(d) cos2x = cos*x sin*x. sec 2 x
(e) (sin x + cos x) 2 = 1 -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 :
2a&
OT, - ., .
(a) cos2 = -- (h)
v ; 2 2 v '
6 - a 2
(b) sin(^L-J5) + cos2^L = 0. (i) sin 2 ^1 = sin 2 B.
(c) sin(^ - B) + sin(2^ + C) = 0. . , __ 2a6
(d) (sin^l - sin5) 2 -f (cos^l + cos) 2 = 2. u; c 2 '
(k)
(f) tan B = cot^l -f cosC. (1) sin 8>4 =
(g)
72 PLANE TRIGONOMETRY
38. Functions of an angle in terms of functions of half the angle.
From (48), p. 70, . .
v " r ' sin 2 x = 2 sin x cos a.
Replace 2 x by #, or, what amounts to the same thing, replace x
by - This gives
^
(51) sin*= 2sin-cos-
From (49), p. 70, cos 2 a; = cos 2 x sin 2 a%
Replace 2 x by oj, or, what amounts to the same thing, replace x
by ^ . This gives
(52) cos x = cos 2 - sin 2 - -
From (50), p. 70, tan 2x =
Replace 2 x by a;, or, what amounts to the same thing, replace x
by|- This gives
2 2tan
(53)
l-tan 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
2sm 2 # = l cos 2 a;,
and 2 cos 2 x = 1 + cos 2 x.
Solving for sin x and cos x, _
. . , 1
since = AH
cos 2 x
^ 9
- * , |l-fcos 2x
and cos x = "V
Replace 2 aj by x, or, what amounts to the same thing, replace x
by ;r This gives
. or . |l COSJT
(54) sin-= =
, mw ^ X , l + COSJC
(55) cos- -
TBIGCXNOMETB1C ANALYSIS 78
x
To get tan- we divide (54) by (55). This gives
=.> or.
cos*
Multiplying both numerator and denominator of the right-hand
member by Vl -f- cos #, we get *
(57) tan- = .
v ' 2 1 + cos*
or, multiplying both numerator and denominator by Vl cos x,
we get
s**^ ^ x I
(58) tan - =
v '
2 sin*
Since tangent and cotangent are reciprocal functions, we have at
once from (56), (57), and (58),
/.^v x ll + cos*
(59) cot-==fc-\J T ^
v 2 N 1 cos *
/^N ' 1 + COS *
(60) ^0 =
.
2 smjr
//x ^
(61) cot- =
2 1 cos x
40. Sums and differences of functions. From p. 63,
(40) sin (# 4- y) = sin x cos y 4- cos x sin y.
(41) sin (a; y) = sin x cos y cos x sin y.
(42) cos (a; -f- y) = cos x cos y sin x sin y.
(43) cos (x y) = cos x cos y -f- sin x sin y.
cosx **l+cofla?*
The positive sign only of the radical is taken since I + COQX can never be negative and
tan and sin x always have like signs.
74 PLANE TRIGONOMETRY
First add and then subtract (40) and (41). Similarly, (42) and (43),
This gives
(a) sin (x 4- y) -f- sin (x y) = 2sin#cosy. Adding (40) and (41)
(b) 8in(x + y) sin(x i/) = 2 cos x sin y. Subtracting(41)from(40)
(c) cos (x -f y) -f- cos (x y) = 2 cos # cos y. Adding (42) and (43)
(d) cos(aj-hy) cos(# y)= 2 sin x sin y. Subtracting(43)f rom (42)
Let x 4- y ==A x -f y = A
x y = B x y = B
Adding, 2x =A+B Subtracting, 2 y = ^4 B
Now replacing the values of a; + y, a; y, x, y in terms of A and
J5 in (a) to (d) inclusive, we get
(62) sin4 + sinB = 2 sin|(A + B) cos|(4 - B).
(63) sin A sinB = 2 cos|(A + B) sin|(A B).
(64) cos A -f cos B = 2 cos | (A + B) cos |(A B).
(65) cos A - cosB = 2 sini(A + B) sinful - B).
Dividing (62) by (63), member for member, we obtain
sin./l-f-sin J B_2sm(v4 -f 5) cos %(A B)
sin A sin ~~ 2 cos (A -f B) sin (A B)
_ sin (A +B) cos % (A B)
"~ cos^(^4 +jB) 'sin J(U B)
= tan J (^1 + J3)cot %(A- B).
But cot ^ (^ - J5) = 7 - - Hence
2 v 7 tan $(A B)
sin .4 + sinB __ tan |(4 + B)
^ ' sinA - sinB ~ tan|(4 - B) '
EXAMPLES
1. Find sin 22^, having given cos 46 =
V2
Solution. From (54), sin- = ^j 1 "" COSa .
2 \ 2
Let x = 46, then - = 22^, and we get
2
sin22^=^| = lV2~V2. An*.
TKIGONOMETBIC ANALYSIS 75
2. Reduce the sum sin 7 x + sin 8 x to the form of a product.
Solution. From (62),
sin A + sin B = 2 sin J ( A -f B) cos (A B).
Let-4=7x, JB = 3x. Then ^ -f # = 10 x, and^L- = 4x.
Substituting back, we get
sin 7 x + sin 8 x = 2 sin 6x cos 2 x. Ans.
3. Find cosine, tangent, and cosecant of 22^.
Ans. - V2+V2, V5 - 1, 2
vs
4. Find sine, cosine, and tangent of 16, having given cos 30 = --
5. Verify the following :
(a) sin 82 -f sin 28 = cos 2. / . x _ x\ 2 _ _ .
(b) sin 60 + sin 10 = 2 sin 30 cos 20. (g ' \* m 2 C S 2/ ~" Sm X '
(c) cos 80- cos 20= - 2 sin 60 sin 30. x sin J x
(d) cos 6x -f cos9x = 2 cos7x cos2x. ' ' an 5 "~ i ^_ cos4-x'
8 in7x-sin6x = _
w W
Bin 38+ sin 3 _ .
0)
X 1
6. Find sine, cosine, and tangent of -, if cosx = - and x lies in the first
quadrant. V3 V5 \/2
x 3 ' 8 ' 2
7. Find sine, cosine, and tangent of - , if cos x = a.
2 ?
1-a /I -fa /I a
, V ^ViTV
8. Express sine, cosine, and tangent of 3 x in terms of cos 6 x.
A /I cos6x Vl + cos 6 x /I - cosGx
AM - \ 2 ' i Vl + cos6x'
9. In a right triangle, C being the right angle and c the hypotenuse, prove
the following :
. . . n B c a ... a b AB
(a) 81D *_ = _. (d) _ = tu, -5-
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 TEIGONOMETEY
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 loth 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 -f tan 2 x tan x = sec 2 x.
2 tan x 11
Solution. Since tan 2 x = and sec 2 x = = , the
equation becomes : 1 ~ tan ' x cos 2 x se 2 x ~
1 tan 2 x cos 2 x sin 2 x
l-ftan 2 x__ 1
1 tan 2 x cos 2 x sin 2 x
or, simplifying,
Second step. ^-1 = - , (22), p. 69
siB 2 x cos 2 x - sin 2 x
COS 2 X
... . cos 2 x -f sin 2 x 1
or, simplifying, = .
cos 2 x sin 2 x cos 2 x sin 2 x
Third step. cos 2 x + sin 2 x = 1,
or 1 = 1. from (83), p. 69
* 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 found the best plan to take the steps in
the order indicated.
t For instance, replace sin 2 a; by 2 sin a? cos ar, tan 2 x by ^ r^- cos (a? + y) by cos*
cos y- sin a: sin y, etc. i-tan x
TRIGONOMETKIC ANALYSIS 77
Ex . 2 . Prove h(* + = tans + 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 :
-,. . . sin x cos y + cos x sin y tan x -f tan y
jfiTst step. - = - .
sin x cos y cos x sin y tan x tan y
sin x sin y
~ _ sin x cos y + cos xsiny _ cos x cos y
sin x cos y cos x sin y sin x sin y
Simplifying,
cos x cos y
sin x cos y 4- cosx sin y sin x cos y + cos x sin y
sin x cos y cos x sin y sin x cos y cos x sin y
or, 1 = 1.
EXAMPLES
Prove the following identities :
1. tan x sin x -f cos x = sec x.
2. cot x sec x esc x (1 2 sin 2 x) = tan x.
3. (tan x -j- cot x) sin x cos x = 1.
sin y _ 1 cos y
siny
= sec^l tan^d.
6. tan x sin x cos x -f sin x cos x cot x = 1.
7. cot 2 x = cos 2 x 4- (cot x cos x) 2 .
8. (sec y 4- esc y) (1 cot y) = (sec y esc y) (1 4- cot y).
9. sin' 2 z tan z 4- cos 2 z cot z -I- 2 sin z cos z = tan z + cot z.
10. sin 6 x 4- cos 6 x = sin 4 x 4- cos*x sin 2 x cos 2 x.
11. sin B tan 2 B 4- esc B sec 2 B = 2 tan J5 sec U 4- esc B - sin B.
12. Work out (a) to (v) under Ex. 4, p. 62, following the above general
directions.
13. cos(x 4- y) cos(x y) = cos 2 x sin 2 y.
14. sin (^4 4- B) sin (A - B) = cos 2 B - cos 2 A.
15. sin (x -f y) cos y cos (x -|- y) sin y = sin x.
cos (x y) 14- tan x tan y
16. =
cos (x 4- y) 1 tan x tan y
17.
cos A cosB
18.
sm x sin y
19. sin x cos (y -f 2) sin y cos (x + z) = sin (x y) cos z.
T8 PLANE TRIGONOMETRY
20.
1 tan (0 0) tan
21. cos(x y -f 2) = cosxcosy cosz -f cos a; sin y sin * sin x cos ]/ sin 2
+ sin x sin y cos z.
22. sin(x + y - z) + sin(x + z - y) -f sin(y -f * - x)
. = sin(x + y + ) -f 4 sin x sin y sin 2.
23. cosxsin(y - z) -f cosy 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. i
following the above general directions.
26. 1 + dnax /Ugx + lV ^ = 1 - sin 2J
l-sin2x \tanx -I/ cos2^t
sin 2 x OA cos 8 x -f sin a x 2 - sin 2 x
26. tanx= -- . 30. - : - = - - --
1 -f cos 2 x cos x + sin x 2
^ sin2x sin3x-sinx
27. cotx = -- 31. - = tanx.
1 cos 2 x cos 3 x -f- cos x
28. - = l. 32. - = cot2*.
cot ^1 + 1 \l + sin2^ cosx cosSx
33. sin 3 x = 4 sin x siri (60 + x) sin (60 - x).
... sin4x n ft
34. -- 2 cos 2 x.
sin2x
35. sin 4 B = 4 cos 2 5 sin cos B.
36. Work out (a) to (j) under Ex. 11, p. 71, following the above general
directions.
37. sin 9x sin7x = 2 cos 8 x sin x. ., _x
1 tan 2 -
38. cos7x-f cos6x = 2cos6xcosx.
__ sin 6x sin2x 7x
39. - = cot --
cos 2 x cos 6 x 2
40. (sin ~ + cos ? ) = 1 -f sin x.
\ 2 2/
48. 1 + tan x tan = sec x.
2
49. tan 5 + 2 sin^ ? cotx = sinx.
22
sinxtan-
44. cos3a-cos7a=2sin6asin2o:. 2 1 -f
'
46. cot + tan = 2cscx. tan' - - cot^ ? " 2cosx
22 22
52. Work out (a) to (f) under Ex. 6, 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 TT to the angle or subtract 2 TT 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
8 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 least 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
TT 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 &, and it is
evident that all such angles
are found by adding even
multiples of TT to, or subtract-
ing them from, x and TT x.
Let n denote zero, or any
positive or negative integer. When n is even, mr -f x includes all
the angles, and only those, which have the same initial and terminal
sides as x (= XOP). Therefore, when n is even,
mr
x = mr + ( !)*.*
The factor (- 1) la positive for all even values of n and negative for all odd values of *.
79
80
PLANE TRIGONOMETRY
Again, when n is odd, n 1 is even, and (n 1) TT -f (TT x) in-
cludes all the angles, and only those, which have the same initial
and terminal sides as TT x (== XOP 1 ). But when n is odd,
(J5) (n 1) TT + (TT #) = TITT x = TITT -|- ( !)"#
From (.4) and (#) it follows that the expression HIT + ( l) n # for
all values of n includes all the angles, and only those, which have
the same initial and terminal
sides as x and TT x.
In case a is negative, TT 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 mr + ( l) n x. Hence
(67) 7HT + ( l) n JT
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 jr= sin [nxr + ( I)"*],
esc x= esc [nrr + ( I)"*].
Ex. 1. Find the general value of all angles having the same sine as
o _ 4
Solution. Let x = in (67). This gives
, / !x 37r A
mr + ( l) n Ans.
4
Ex. 2. Find the four least positive angles whose cosecant equals 2.
Solution. The least positive angle whose cosecant = 2 is Let x = in (67).
This gives 6 6
nir +(-!)"
6
* In deriving this rule 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 by 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.
GENERAL VALUES OF ANGLES
81
When n = 0, we get
When n = 1, we get
When n = 2, we get
When n = 3, we get
J-1W.
3*.- - = 610 An8.
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 TT to, or sub-
tracting them from, x and x.
Let n denote zero, or any
positive or negative integer. For any value of n y
includes all the angles, and only those, which have the same initial
and terminal sides as x (= XOP). Similarly,
(B) 2mr x
includes all the angles, and
only those, which have the
same initial and terminal sides
as -x(=XOP').
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 (^l)and(-B),
(68) 2 rnr =fc x
is the general value of all the angles, and only those, which have the
same cosine or secant as x.
* 2n i even and 2 n - 1 is odd for all values of n.
82
PLANE TRIGONOMETRY
This result may also be expressed as follows :
cos x = cos (2 mr =t jr),
sec* = sec(2n7rdb x).
Ex. 1. Given cos A = ; find the general value of A. Also find the five
least positive values of A. ^ 2
Solution. The least positive angle whose cosine = ~ is If we let
V2 *
x = in (68), we get
4
When n 0,
When n 1,
When n = 2,
~.
=^=135.
4
= 2 IT = 225 or 496.
4
= 686 or 855. Ans.
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 TT -f x (= XOP f )
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 TT to, or
subtracting them from, x and
7T + X.
Let n denote zero, or any
positive or negative integer.
When n is even,
(^4) HTT + 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 l)?r -J-(TT + #)= mr -f- #
includes all the angles, and only those, which have the same initial
and terminal sides as TT -f x (= XOP*).
In case a is negative, the same line of reasoning still holds true
GEKEKAL 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) TOT + x
is the general value of all the
angles^ and only those, whioh
have the same tangent or
cotangent as x. jp~
This result may also be
stated as follows :
tan x = tan (nir + x) 9
cot x = cot (mr + *).
EXAMPLES
1. Given sin A = i; find the general value of A. Also find the four least
positive values of A.
V3
Ana. nir + ( -!)-; 30, 160, 390, 610.
6
2. Given cos A = ; find the general value of A. Also find all values of A
numerically less than 2 IT. Ana. 2 nir - ; , r-
6 o o
3. Given tan .4 = 1 ; find the general value of A. Also find the values of A
numerically less than 4 v.
7T7T 37T67T 7w97T llTTlSTT 16 V
_; _,__,_,,_,-_,_,-_.
4. Given sin2a; = i; show that x = + (- 1) .
_ ^, rt ,
6. Given cos Sx = -- ; show that x = -
2 39
6. In each of the following examples find the general values of the angles,
having given
(a)
(b)
(c)
(d)
(e)
An*. A =
.
-.
:-
vS
84 PLANE TRIGONOMETRY
7. Given sin x = and tan x = ; find the general value of x.
2 V3
Solution. Since sin a; is and tanx is -f , 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 =
26 V8
Hence x = 2 mr + Ana.
8. In each of the following examples find all the positive values of x less
than 2 w which satisfy the given equations.
1 A v 77r
(a) cosx= Ana. , .
vS 4 4
v . , 1 TT STT STT 7*r
(b)smx = _,_,_,_.
7T 2 7T 4 7T 6 7T
_,_,_,_.
... , TT STT STT 7w
(d)ootx = l. _, v , __,__.
V STT 7ir
.(e)co 8 x = - - - fl -, T .
,*v TT STT
(f) secx = 2. -, - .
o o
/ v - , 1 TT STT 7?r ll?r
tt)BU.= s . _,_,_,_.
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 30, 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 funation, or anti-trigonometric, or inverse
circular function, and is denoted by the symbol
sin- 1 ^, or, arc sin J*,
read "inverse sine of ," or, "arc (or angle) whose sine is J."
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.
INVERSE TRIGONOMETRIC FUNCTIONS 85
and * in two different ways. Thus, tan x = a, meaning the tangent
of the angle x is a ; or, x = tan~ l a, meaning x is an angle ivhose
tangent 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 = >
v ' 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# =1,
then the general value of x is, by (69), p. 83,
x = tan" 1 1 = mr 4- -7 >
4
where n denotes zero or any positive or negative integer, and
is the principal value of x.
* Hence, if sin x, esc a?, tan ar, or cot x is positive, the principal value of x lies between
and ~ ; if sin x, esc x, tan ar, or cot x is negative, the principal value of x lies between
If cos a? or sec a: is positive, the principal value of x lies between ^ and -^i prefer-
tnee being given to the positive angle.
86 PLANE TEIGO^OMETKY
Similarly, if cos x = J, then by (68), p. 81,
x = cos" 1 ^ = 2 HIT ->
& tj
7T
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 -j- 1 inclusive. Simi-
larly, it is evident that the expressions
sec" 1 ^ and csc" 1 ^
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 a ic. (29), p. 60
This may be written
(A) x = cos"" 1 Vl sin 2 a?.
Placing sinx = a, then x = sin- 1 ^, 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 x = c, then x = cos^c, and (B) becomes
2 cos-^c = cos- 1 (2 c 2 1).
Since we know that the co-functions of complementary angles are
equal, we get for the principal values of the angles that
7T
sin-" 1 ^ 4- cos-^a = ~> < a ^ 1
&
tan- 1 b + cot- 1 ***^, 0< b
&
sec- J c -h csc- 1 *? = - 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 -f n
(a) tan- 1 ??* -f tan -1 n = tan- 1 -
1 inn
Proof. Let
(b) A = tan -1 m and B = tan- 1
(c) Then tan .4 = m and tan B = n.
Substituting from (b) in first member of (a), we get
1 mn
or, what amounts to the same thing,
(d) tan (A + B) =
1 mn
But from (44), p. 68,
/ v / A T>V tan A> + tan B
(e) tan (A + B) .
1- tan^ltanJ?
Substituting from (c) in second member of (e), we get
(f ) tan (A -f B) =
Since (d) and (f) are identical, we have proven (a) to be true.
Ex. 2. Prove that
Proof. Let
(h) A sin- 1 ! and J?= cos- 1 -^.
(i) Then sin -4 = $ and cosl?=^.
(j) Also cos .4 = f and smJ5 = ^y.*
Substituting from (h) in first member of (g), we get
or, what amounts to the same thing,
But from (40), p. 63,
(1) sin (A -f B) = sin A cos B + cos A sin B.
Substituting from (i) and (j) in second member of (1), we get
(m) sin (A -f B) = J J| -f J ^ = j j.
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. 90*
88 FLAKE TRIGONOMETRY
Ex. 8. Solve the following equation for x ;
IT
1 4*
Solution. Take the tangent of both sides of the equation. Thus*
tanftan-^x + tan-iSx) = tan-,
tan(tan- 1 2x)-f tan(tan-i3g) , , ....
or i ^ ^ ^^ = i from (44), p. 68
1 1- tan (tan- 1 2 x) tan (tan- 1 8 x) \ /*
2x-f8x
l-2x.3x = L
Clearing of fractions and solving for x, we get
x = J or 1.
x = satisfies the equation for the principal values of tan- 1 2 x and tan- 1 8 x.
x = 1 satisfies the equation for the values
tan- 1 (-2) = 116. 67,
tan- 1 (-8) = -71. 57.
EXAMPLES
1, Express in radians the general values of the following functions :
1 TT 1 TT
(a) sin- 1 = . Ana. nir + ( !)"- (e) tan- 1 -. Ans. rnr H
V2 4 V3 6
(f)tan-i(V5).
4 / O
\/8 7T
(c) cos- l - . 2 UTT i - (g) cot- 1 ( i 1).
2. Prove the following :
(a) tan- l a - tan- 1 b = tan- 1 ~ . . (h) 2 tan- 1 a = tan- 1 -
l-fa6 ^ ' 1-a 2
2 ^
(b) 2 tan- i a = sin- J (i) sin- 1 a = cos- l Vl - a 2 .
1-f a 2
Gt
(c) 28m- 1 a = cos-^l 2 a 2 }. (j) sin- 1 a= tan- 1
(d) tan-ia = sin- 1 -=== (k) tan- 1 a = cos- 1 -
/l -fa 2 V 1-h a 2
/\ i f- _. t 7w. HIT /tv 1 8 ..8 .,77
(e) tan- 1 tan- 1 = - . (1) sin- 1 - + sin- 1 = sin- 1 .
v/ n m-fn4 w 6 17 86
(f) cos- 1 - -f tan- 1 - = tan- 1 -- 1 (m) cos- 1 - -f cos- 1 - = cos- 1 --.
5 5 11 5 18 65
(g) 2tan-i? = tan- 1 . (n) tan- 1 ^ -f tan- 1 = tan- 1 -.
85 7 18 V
* Th student should remember that tan- 1 2 a; and tan- 1 3 a; are measures of angles.
TBIGONOMETBIC EQUATIONS
3. Solve the following equations :
(a) tan-ix + tan-i(l - x) = tan-i-V Ant. z = -.
(b) tan-ix + 2 cot-*x = . x = VS.
3
x+24
(e) tan-i- + tan-i- = tan-* (- 7). x - 2.
x 1 x
(f) tan-i(x + l)-J-tan-i(x-l)=tan-i~. s = ~
ol
(g) sin~ix + 8in-i2x = -. x =
3 14
(h) sin-i ^ + S i n -i 1? = E. x = 13.
x x 2
4. Find the values of the following :
(a) sin (tan- 1 ) . -4ns. (d) cos (2 cos- 1 a). ^1 ws. 2 a 2 - 1.
\ 12/ 13
(b) cot (2 sin-i ?) ~ (e) tan (2 tan-i a). J^L .
(c) sin(tan-i- 4- tan-i-V -i=. (f) cos(2 tan-^
\ 2 8/ y%
**.
1 -f a 2
47. Trigonometric equations. By these we mean equations involv-
ing one or more trigonometric functions of one or more angles. For
instance,
2 cos 2 a; + Vs sin x -f 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 solye a trigonometric equation involving one unknown angle
LS to find an expression (the student should look up the general
sralue 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 TRIGONOMETRY
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
angle j ,f o>nd 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 -f 1 = 0.
Solution. Since cos 2 x = cos 2 x siri 2 x, we get
First step. (cos 2 x sin 2 x) sec x-fsecx4-l = 0.
Second step. Since sec x = this becomes
cosx
cos 2 x sin 2 x 1
___________ _^. _|_ j ___ y >
COS X COS X
Third step. cos 2 x sin 2 x -f 1 + cosx = 0.
Fourth step. Since sin 2 x = 1 cos 2 x, we have
COS 2 X 1 -f COS 2 X + 1 + COSX = 0,
or, 2 cos 2 x 4- 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^O = 2 nir - -
2
Also, 2 cos x + 1 = 0,
cosx = - J,
or, x = cos-M - -) = 2n7r db -
Hence the general values of the angles which satisfy the equation are
2n7r~ and 2nir .
2 3
The positive angles less than 2 IT which satisfy the equation are then
w 2ir 4ir Sir
a' 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 be found the best plan to take the steps in
the order indicated.
t For instance, replace cos 2 a by cos** - sin* ar, sin (* + -) by x * cos x , etc.
\ *' V
TEIGONOMETRIC EQUATIONS 91
Ex. 2. Solve the equation
2 sin 2 x + V5 cosx + 1 = 0.
Solution. Since sin 2 a: = 1 cos 2 x, we get
Fourth step. 2-2 cos 2 x -f V cosx + 1 = 0,
or, 2 cos a x - V3 cos x - 3 = 0.
Fi/M step. This is a quadratic in cos x. Solving, we get
/5 "^
cos x = V3 or ---
Since no cosine can be greater than 1, the first result, cosx = V, cannot be
used. From the second result,
EXAMPLES
Solve each of the following equations :
1. sin 2 x = l. Ans. x=n?r+(- W-V=n7r -.
2. csc 2 x = 2. z=n7r + (~
3. tan 2 x = 1. x = IMF -
4
4. COt 2 X = 3. X = W7T -
6
c 1 ^
5. cos 2 x = - x = n-Tr
_ _ 4 7T
6. sec 2 x = - x = nTr .
3 6
27T
7. 2 sin 2 x -f 3 cosx = 0. X = 2 rnr -
3
1 7T
8. cos 2 or sin 2 cr = - a = nir - .
2 i
9. 2 V cos 2 a = sin a. a= mr + (- l) n -
o
1 fl-
lC. sin 2 y 2 cosy + - = 0. y = 2 n?r -
4 3
11. sin^l -|- cos^l = V^. A = 2mr + ^.
12. 4 sec 2 y 7 tan 2 y = 3. y = nx -
o
13. tanB-f cotB=2. # = n7r-f--
4
14. tan 2 x (1 -f \/3) tan x 4- V = 0. x = nir + - , nir -f ^
* Since the principal value of x sin- 1 1 and of a; = sin- 1 (- 1)
92 PLANE TEIGONOMETEY
/ r 1 \ A 6ir 2ir
v vs/ co x ~~ ' "* x ~~ n7r "*" "0"" ' ntr + ^*~
16. tan 2 x -f cot 2 x = 2. x = nir db -
17. tan / x -h ) = 1 4- sin 2 x. x = nir, nir .
18. cscxcotx = 2Vs. x = 2nir-.
x 6
19. sin - = esc x cot x, x = 2 nir.
20. esc y -f cot y V5. y = 2 nir 4- - .
3
21. 3 (sec 2 a + cot 2 a) = 18. a=nir~, nir-.
o 8
Find all the positive angles less than 860 which satisfy the following
equations :
22. cos 2x 4- cosx = - I. Ana. x = 90, 120, 240, 270.
23. sin 2 x - cos 2 x - sin x 4- cos x = 0. x = 0, 90, 210, 330.
24. sin (60 - x) - sin (60 + x) = . x = 240, 300.
2
25. sin (30 4- x) - cos(60+ x) = x = 210, 330.
2
26. tan (46 - x) 4- cot (46 - x) = 4. x = 30, 160, 210, 380.
27. cos 2 x = cos 2 x. x = 0, 180.
28. 2 sin y = sin 2 y. y = 0, 180.
29. sin x -1- sin 2 x 4- sin 3 x = 0. x = 0, 90, 120, 180, 240, 270.
30. tan x 4- tan 2 x = tan 3 x. x = 0, 60, 120, 180, 240, 300
31. secx cotx = cscx tanx. x = 45, 225.
32. sin 4x - cos Sx = sin 2x. x =80, 90, 160, 2 10, 270, 380.
33. Vl -f sin x Vl - sin x = 2 cosx.
34. sin*x 4- cos*x = -.
8
35. sec (x 4-120) + sec(x -120) = 2 cosx.
36. sin (x 4- 120) 4- sin (x 4- 60) = ? .
37. sin y -f sin 8 y = cos y cos 8 y.
Find the general value of x that satisfies the following equations :
38. cosx = and tan x = 1. Ana. x = (2n-fl)ir4--.
V2 4
39. cotx = Vs and esc x = 2. x = 2 nir
6
40. Find positive values of A and B which satisfy the equations
cos (4 - B) = - and sin(^l 4- B) = - . Ans. and -.
2 2 12 4
41. Find positive values of A and B which satisfy the equations
O OK -. I A *>
tan (^4 - B) = 1 and sec (A 4- B) = . Ans. ^^ and .
Vs 24 24
CHAPTEK 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, Vl, TT, 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 depen deuce of one quantity on another. Thus, the
we If/ Jit 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
is a function of x because its value will depend on the value we
assume for x y and
A
am A, cos 2A, tan
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 TKIGONOMETEY
ordinates of points in a plane (see 13, p. 20). 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 tJie function (=?/), writing
down the results -in tabulated form.
Third step. Plot the 2><>wts having the values of x as abscissas
and the corresponding values of y as ordhiates.
Fourth step. A smooth, curve drawn through, these points in order
is called the graj>h of the function,
Ex. 1. Plot the graph of 2x 0.
Solution. First step. Let?/ 2x 6.
Second step. Assume different values for x and compute the corresponding
values of y. Thus, if
x = 0, y = - ;
x= 1, y = -4;
x = 2, y=-2;
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
X
y
x
y
-(>
- 6
1
__4
- 1
- 8
2
- 2
2
-10
3
-3
- 12
4
2
-4
- 14
6
4
- 5
-16
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.
GRAPHICAL REPRESENTATION
95
Ex. 2. Plot the graph of x 2 - 2 x - 3.
Solutit n. First step . Le t y = x* 2x - 3.
Second step. Computing y by assum-
ing values of x, we find the following
table of values.
X
y
X
y
- 3
-3
1
- 4 ;
- 1
2
_3
2
5
3
o ;
-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 011 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
To find the graph of a
X
y
x
y
7T
30
__
.60
- 30
- .50
6
7T
7J-
60
.86
-< 60
- .86
3
;}
90
7T
1.00
- 90
7T
-1.00
2
2
120
2?r
.8(5
-120
27T
- .86
3
3
150
r>7r
.50
- 150
f>7T
- .50
6
()
180
TT
-180
TT
210
TTT
- .50
- 210
7?T
.60
6
6
240
47T
- .86
-240
47T
.86
3
3
270
37T
- 1.00
-270
37T
1.00
2
2
300
5?r
- .86
- 300
57T
.86
3
3
330
llTT
- .50
-830
llTT
.60
6
6
360
27T
-360
-27T
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 oft
distances TT - 3.1416 to the right and left of the origin and then divide each
of these int( six equal parts. Then when
Also when
x =
x= 0,
7T
tV
TT
s'
X = -,
0;
y =
.50 -AB;
^ =
.8<>=CZ>;
y = 1
[.00 =EF', etc.
y = ~
.50 = 6r/f ; etc.
Fourth step. Drawing a smooth curve through these points, we get the graph
of sin x for values of x between 2 *r and 2 TT. It is called the sine curve or
sinusoid.
Discussion, (a) Since sin (x 2 TT) sin x, it follows that
y = sin x = sin (x 2 TT),
that is, the graph is unchanged if we replace x by x 2 TT. This means, how-
ever, that every point is moved a distance 2 TT to the right or left. Hence the
arc PNMLO may be moved parallel to XX' until P falls at O, 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 = TT to x = Tr^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(=y) is
1 ( = EF QN, etc.) and the minimum value is 1(=SJ = RL, 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?r, ^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 TT 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 from 2?r radians to 4 TT radians, the sine again went through
the same series of changes, and so QH. Thus the sine goes through
all its changes while the angle changes 2 TT radians in value. This
is expressed by saying that the^wW of the sine is 2TT.
Similarly, the cosine, secant, or cosecant passes through all its
changes while the angle changes 2 TT radians.
The tangent or cotangent, however, passes through all its changes
while the angle changes by TT radians. Hence, the period of the sine,
cos hie, secant, or cosecant is 2 TT radians; while the period of the
tangent or cotangent is TT 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 as.
Solution. Let y = sin x. Draw a unit circle.
Divide the circumference of the circle into any number of equal parts (12 in
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 ,4 1? = QJ?,
sine of arc^U? = NE,
sine of arc AJ = 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 hi
98
PLANE TRIGO^STOMETEY
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 = archil? = OA,
x = arc^ C = OC,
x = -ATC AD = OE,
x = sacAE = OG,
x = &TcAF = O/,
x = 'ArcAG = OK,
x - we AH = OL,
IN IN
CIRCLE GRAPH
y = zero = zero ;
y = QB = AB ;
y = PC = CD-
y = OD = EF;
y = NE = Gil ;
y = MF=IJ-
y = zero = zero ;
y = MH ~ LM, etc.
EXAMPLES
1. Plot the graphs of the following functions :
(a) x + 2.
(b) Zx -0.
(c) 2x + l.
(d) *'.
(e) x*.
(K) -
x-2
x-2
x + 1 '
(i) 2-.
(j) logioX.
(k) 2x 2 -4.
(1) 8-x*.
(m) 4- f>x + ;
(n) x 2 -3x +
(o) x 2 -4x + 3.
(p) x 3 -4x.
(q) a*-2z+l.
(r) x 8 -7x + 6.
(s) x 8 - 5x- 12.
(t) x* - 1.
(u) x 6 - 2.
2. Plot the graph ot cos x.
Solution. Let y = cos x. The cosine curve is found to be as follows :
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
o page.
GRAPHICAL REPRESENTATION
74
99
To construct the tangent curve from the unit circle shown,
IN IN
CIRCLE GRAPH
IN IN
CIRCLE GRAPH
When
x arc zero = zero, y ~ zero = zero ;
x = arc AC = 0(7, ?/ = AN = CD ;
x = arc -AD OD, y = oo oo ;
x = arc .A E = OE, y = A Q = JF, etc.
4. Plot the graph of sec x.
Solution. The secant curve is given below.
FA
i r
Using the unit circle, we have
IN IK
CIRCLE GRAPH
When x = arc zero = zero,
x = arc A B = OB,
x arc AC OD,
x = arc -AD = OF,
x = arc^l^ = 0(7,
IK IN
CIRCLE GRAPH
y=ON=DE-,
y = oo =00;
y = OQ = GH, etc.
100 PLANE TRIGONOMETRY
6. Plot the cotangent curve.
6. Plot the cosecant curve.
7. Draw graphs of (a) sin x -f cos x, (b) cos x sin x, (c) sin 2 x, (d) tan 2 x,
(e) sin x cos x.
CHAPTER
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 (sides
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 TKIGONOMETBY
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).
Draw the perpendicular CD(= h) on AB or AB produced. From
either figure, using the right triangle A CD,
,.. . . h
(A) sin A = -
7 b
Fin Fig. 2, sin A = sin (180 -A)~ sin CAD** -!
Also, using the right triangle BCD,
h
a
Dividing (A) by (J5) gives
sin A __ a
b
or, by alternation in proportion,
a b
v ' sin A smB
Similarly, by drawing perpendiculars from A and B we get
be
<*)
sinC
c a
sin C sin A
and
respectively.
Writing (C), (D), (E) as a single statement, we get the law of sines,
(72)
b
sin A ~" eintf
SOLUTION OF OBLIQUE TKIANGLES
103
Each of these equal ratios has a simple geometrical meaning, as may be shown
if the law of sines is proved as follows :
Circumscribe a circle about the triangle ABC as shown in the figure, and
draw the radii 0#, OC. Denote the radius of the circle by R. Draw OM per-
pendicular to BC.
Since the inscribed angle A is measured by one half of the arc BC and
the central angle BOG is measured by 'the whole arc .BC, it follows that
the angle BOC = 2A, or,
angle BOM = A.
Then BM = E sinBOM = R sin4, by (7), p. 11
and a = 2BM =
or,
In like manner it may be shown that
2# = -A- and 2R = -^--.
sin B sin C
Hence, by equating the results, we get
a _ b _ c
sin B
212 =
sin .4
sin C
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 = 50 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) = 180 - 106 = 76.
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 b use
a = 6
sin A sin B
Clea/ing of fractions and solving for the only
unknown quantity 6, we get
asinB
~~ sin A
c=?
Substituting the numerical values of &mA and sinJB from the table on p. 9,
and a = 60 ft, we get
60 x 0.6428
0.0063
= 36.46 ft
104
PLANE TRIGONOMETRY
Similarly, to find the side c, use
a __ c
sin-4 "" sinC*
Clearing of fractions and solving for c, we get
asinC 60x0.9669
BinA
0.9063
= 68.29 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 A CB and A OB 1 will satisfy
the given conditions provided a
is greater than the perpendicular
CP ; that is, provided .
a>b sin A.
B
The angles ABC and AB'C are
supplementary (since Z.&BC
= Z BB'C) ; they are, in fact, the
supplementary angles obtained (using the law of sines) from the
formula
SOLUTION OF OBLIQUE TKIANGLES 105
That is, we get the corresponding acute value B from a table of
sines, and the supplementary obtuse value as follows :
If, however, a = b sin A = CP, then sin B = 1, B = 90, and the tri-
angle required is a right triangle.
If a<&sin^(that is, greater than CP), then sinJS>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 < b 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 = 116 ; 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 = 82, b = 86, A = 30 ; find the remaining parts.
Solution. Here 6 sin A = 86 x ^ = 43; hence a < 6 sin 4, and there is ne
solution.
Ex. 3. Given a = 40, b = 30, A 75 ; find the remaining parts.
Solution. Since a > b and A is acute there is one solution only.
By the law of sines,
sin A sin I?
bsinA 30 x. 9669
.-. sin B =.7244,
or, B = 46.4, the only admissible
value of B.
Then C = 180 - (A + B) = 180 - 121.4 = 68.6.
To find C, we get, by the law of sines,
c _ a
sinC~ sin-4 '
asinC 40 x .8635 or
or, c = ; = - - = 36.8.
' sin .4 .9669
Check the results by measurements on the figure.
106 PLANE TBIGONOMETBY
Ex. 4. Solve the triangle, having given 6 = 16, a = 12, A = 52.
Solution. Here 6 sin .4 = 16 x .7880 = 11.82; hence, since A is acute and a
lies between 6 and 6 sin A, there are two solutions. That is, there are two tri-
angles, A CBi and A CB Z , which satisfy the given conditions. By the law of sines,
a = b
sin A sin#i
. _ ftsin^L 16 x. 7880 _ DCrt
or. sin BI = = = .9860.
a 12
This gives BI = 80.07, and the supplementary angle B 2 = 180 - BI = 99.93.
Let us first solve completely the triangle AE\C.
Ci = 180 ~(A + B$ = 47.93.
a GI
By the law of sines,
SlIl^L SlIlUi
n oi C1- 1O vy 7>4O^
'sin^L sind
12
.7880
Now, solving the triangle AB$C,
<7 2 = 180 - (A + B 2 ) = 28.07.
By the law of sines, 2
sinC 2
a sin C 2 _ 12 x .4706 - ----- c r - H
~ .7880 - '
The solutions then are :
For triangle ABiC For triangle AB Z C
JB!=80.07, J5 2 =99.93,
Ci = 47.93, C 2 = 28.07,
Cirrll.3. C 2 = 7.2.
Check the results by measurements on the figure.
In the ambiguous ease 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. Ana. Two.
(b) a = 60, 6 = 100, A = 30. One.
(c) a = 40, 6 = 100, A = 30. None.
(d) a = 13, 6 = 11, A = 69. One.
(e) a = 70, b = 76, A = 60. Two.
(f ) a = 184, 6 = 84, B = 62. None.
(g) a = 200, b = 100, A = 30. One.
2. Solve the triangle, having given a = 60, A 66, B = 40.
Ana. C = 75, ft = 36.46, c = 68.29.
3. Solve the triangle, having^given 6 = 7.07, A = 30, C = 106.
Am. B = 46, a = 6, c = 9.0&
SOLUTION OF OBLIQUE TKIANGLES
107
4. Solve the triangle, having given c = 9.56, A = 45, B = 60.
Ans. C = 75, a = 7, 6 = 8.67.
5. Solve the triangle when c = 60, A = 60, B = 75.
Ans. C = 65, 6 = 70.7, a = 56.1.
6. Solve the triangle when a = 650, A = 10 12', B = 46 30'.
,4ns. C = 123 12', 6 = 2257.4, c = 2600.2.
7. Solve the triangle when a = 18, 6 = 20, A = 55.4.
.4ns. B l = 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 \/2, 6 = 2 \/S, A = 60.
.4ns. (7 = 75, B = 45, c = 4.73.
9. Solve the triangle when 6 = 19, c = 18, C = 15 49'.
^Lns. B l = 16 43', A^ = 147 28', a t = 35.6 ;
7>' 2 = 163 17', A 2 = 54', a 2 = 1.04.
10. Solve the triangle when a - 119, 6 = 97, A = 50.
Ans. B = 38.6, C = 91.4, c = 155.3.
11. Solve the triangle when a = 120, 6 = 80, A = 00.
.4ns. B = 35.3, C = 84.7, c = 137.9.
12. It is required to find the horizontal distance from a
point A to an inaccessible point B on the opposite bank of a
river. We measure off any convenient horizontal distance as
A C, and then measure the angles CA B and A CB.
Let .40=283 feet, angle CAB=3S, and angle ^4 C#= 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 A 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 A and B, and meas-
ure the distances A C and CB, and the angle
BAC. Let AC = 48.6 ft., 7^(7 = 84 ft., and
angle BAC = 21.b. Solve the triangle for
the side AB.
Ans. 127.2ft.
14. A tree A is observed from two points
B and C, 270 ft. apart, on a straight road. The angle EGA is 55 and the angle
CBA = 66. Find the distance from the tree to the nearer point B. A ns. 255.4 ft.
16. To determine the distance of a hostile fort A from a place #, 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. Ana. 1035.6 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.
Ans. 353.9 ft., or 629.6 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 ? Ans. 120.3 yd.
108
PLANE TRIGONOMETKY
18. Prove the following for any triangle :
(a) a = b cos C -f c cos B,
b = a cos C + c cos ^4,
c = a cos J5 -|- 6 cos A.
(b) V&c sin B sinC = ^ r l
1 ; 6 + c
sin J -f 2 sin # __ sin (7
sin 2 ./! - m sin 2 7? __ sin 2 C
a 2 - m& 2 ~~ c 2
(e) a sin (7? - C) + b sin (C A) -f c sin (A - B) = 0.
19. If R is the radius of the circumscribed circle, prove the following for
any triangle [s = (a 4- b -f c)] :
(a) Tt (sin ^4 + sin B -f- sin C) = .
(b) 6c = 4/i 2 (cos .4 -h cosB cosC).
20. Show that in any triangle
59. Law of cosines. Jn aw// 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 I and c and their included angle A.
When the angle A is acute (as in Fig. 1) we have, from Geometry,
PThe square of the side opposite an acute angle equals the sum"!
I of the squares of the other two sides minus twice the product I
l_of one of those sides into the projection of the other upon it.J
or,
But
Hence
AD =b cos A.
a a = 6 a -f c* 2 be cos A.
(8), p. H
SOLUTION OF OBLIQUE TRIANGLES 109
When the angle A is obtuse (as in Fig. 2) we have, from Geometry,
f 2ABXAD,
r~The square of the side opposite an obtuse angle equals the sum"!
I of the squares of the other two sides plus twice the product I
|_of one of those sides into the projection of the other upon it.J
or, a*=b 2 +c*+ 2cAD.
But A D = b cos DA C (8), p. 11
= b cos (180 -4)
= b cos A . Hence in any case
(73) fl 3 = & 2 + c 2 - 2 be cos A.
Similarly, we may find
(74) & a = a 2 + c 2 - 2 ac cosB.
(75) c 2 = a 2 + & 2 - 2 ab cos C.*
Observe that if A = 90, then cos A = 0, and (73) becomes a 2 = b* + c*,
which is the known relation between the sides of a right triangle
where A is the right angle.
Solving (73), (74), (75) for the cosines of the angles, we get
(76)
(77) COS 5=^-1-:
v ' 2ac
(78) cosC=^ "*" * "
2 aft
These formula,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, b and JB, c and C stand for any side of a triangle and the opposite angle^
from any formula expressing a general relation between these parts another formula may
be deduced by changing the letters in cyclical order. Thus, in (73) by changing a to 6, b to
c, c to a, and A to B we obtain (74) ; and in (74) by changing b to c, c to a, a to 6, and B to C
we get (75). This is a great help in memorizing some sets of formulas.
110 PLANE TRIGONOMETRY
Ex. 1. Having given A 47, 6 = 8, c = 10 ; solve the triangle.
Solution. To find the side a use (73).
a 2 = &2 + C 2_ 26ccos^4
= 64 + 100 - 2 x 8 x 10 x .6820
= 64.88.
.-. a = V54.88 = 7.408.
7?
c ~ 10 To find the angles C and B use the law of sines.
8iD B = = ' = . 7898. ,.=52.2.
7.408
10 x. 7314
o. .. C = oO.o .
a 7.408
To check our work we note the fact that A + B -f C = 47 -f 62. 2 -f 80.8= 180.
Ex. 2. Having given a = 7, 6 = 3, c = 6 ; solve the triangle.
Solution. Using formulas (76), (77), (78) in order to find the angles, we get
2ac 2-7-6 14
= = .9286. , J) =
a 2 4. 7,2 __ C 2 72 i 32 _ 52 11
cos C = 2LJ? - 1 = ^_^ - 1 = if = .7867. .-. C = 38.2.
2ab 2-7.3 14
C^ecA: : A 4- J5 + (7 = 120 + 21.8 + 38.2 = 180.
EXAMPLES
1. Having given a = 30, 6 = 54, C = 46 ; solve the triangle.
Ana. A = 33.1, B= 100.9, c = 39.66.
2. Having given A = 60, 6 = 8, c = 6 ; find a and the cosines of the angles
B and C. Ans. 7, }, j J.
3. Having given a = 33, c = 30, B = 36.4 ; find A and C.
-4ns. ^ = 80.7, C=63.9.
4. Having given a = 4, 6 = 7, o*= 10 ; solve the triangle.
Ans. A = 18.2, B = 33.1, C = 128.7.
5. Having given a = 21, b = 24, c = 27 ; solve the triangle.
Ana. A = 48.2, B = 68.4, C = 73.4.
6. Having given a = 2, 6 = 3, c = 4; find the cosines of the angles A, B y C.
7. Having given a = 77.99, 6 = 88.39, C = 72 16'; solve the triangle.
Ans. A = 61 16% B = 66 30', c = 96.24.
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 TRIANGLES
111
9. The two diagonals of a parallelogram are 10 and 12 and they form an
angle of 49.8 ; find the sides. Ans. 10 and 4.68.
10. In order to find the distance between two
objects, A and 2?, separated by a pond, a station
C was chosen, and the distances CA = 426 yd.,
CJB = 822.4 yd., together with the angle AC B = 68. 7,
were measured. Find the distance from A to B.
Ans. 430.85yd.
11. A ladder 62 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. 96.9.
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 J5, 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. Am. 10.6 mi.
14. Prove the following for any triangle :
(a) a (6 2 -f c 2 ) cos A + 6 (c 2 + a 2 ) cos B + c (a 2 + 6 2 ) cos G = 8 abc.
6 -f c _ cos B 4- cos C
a 1 cos.4
(c) a -f b + c = (6 + c) cos A + (c -f a) cos B -f (a -f 6) cos O.
cos A cosB cos C _ a 2 + 6 2 -f- c 2
abc 2 abc
(e) a 2 + 6 2 + c 2 = 2 (ab cos C -f be COB A -f 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 sinB
and by division and composition in proportion,
a + b __ sin A -f sin B
^ ' a b sin 4 sin J5
But from (66), p. 74,
sin .4 4- sin^B _ tan
^ ^ sin A sin5 tan
B)
112 PLANE TBIGONOMETBY
Hence equating (A) and (B), we get
a - b ~~ tan }(4 - S)
. ., , i.
Similarly, we get
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 + />, a-b,A+B (=180- C), and there-
fore also tanj(v4 +J3), are known, we clear (79) of fractions and
solve for the unknown quantity tan (A -B). This gives
(80) tan \(A -5) = J^| tan |(4 +*).
We shall illustrate the process by means of an example.
Ex, 1. Having given a = 872.6, b = 632.7, C = 80; solve the triangle.
Solution, a + o = 1505.2, a - 6 = 239.8, A + B = 180 - (7 = 100, and
From (79), since tan J (4 + JB) = tan 50 = 1.1918,
Adding this result to ( A + B) = 60 gives
XN - A = 60.6.
Subtracting the result from (4-f B) = 50 gives
B = 39.4.
To find the side c, use the law of sines. Thus,
?L?lB^- 872 - 6 x - 9848 - 986 2
- __ - _ ^^^
We will now derive formulas for solving triangles having three
sides given, which are more convenient than (76), (77), (78), p. 109.
If 6 > a, then J9 > A, making a - 6 and A - B negative. The formula still holds true, but
o + a tan t(B + A)
to avoid negative quantities it is better to write the formula in form -^a^ t&n^(B- A)'
t These may also be found by changing the letters in cyclical order (see footnote, p. 109).
t 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 TKIANGLES 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
(A) 2s = 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,
(Z>) 2 (s a) = - a + b 4- c.
In (49a), (49b), p. 70, replace 2x by A, and, what amounts to
the same thing, x by A . This gives
(E) 2 sin 2 .4 =1- cos ,4,
(F) 2cos 2 .4=l-f cos A
#2 -I- C 2 <y 2
But from (76), p. 109, cos ,4 = - - J hence (E) becomes
2i be
(O)
^2 - (^ - 2 be -f
_ (a + ^ ~ c) (<*- b + c)
~ 2 be
*^-^) 9 being the product of the sum and difference of a and b-c.']
114 PLANE TRIGONOMETRY
Similarly, (F) becomes
2 be
2 be
c ""
(82)
Since tan ^4 = ~-> we get, by substitution from (81) and (82),
COS A- ^i
(83)
;- b) (s - c)
Since any angle of a triangle must be less than 180, A must be
less than 90 and all the functions of J A must be positive. Hence
only the positive signs of the radicals in (81), (82), (83) have been
taken.
Similarly, we may get
. , r,
tan 5 =
tan 1C =
.
c)
There is t'hen 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 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 ordor.
SOLUTION OF OBLIQUE TRIANGLES 115
When two angles, as A and J5, have been found, the third angle,
C, may IDC 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)*
[Multiplying both numerator and denomina-1
tor of the fraction under the radical by s - a. J
1
s a N
(-a)( -&)(-<
Denoting the radical part of the expression by r,
(84) r = xT yv /v ' , and we get
i s
(85) tan \ A = Similarly,
S Q
(86) tan \B = ?,
S "
(87) tan|C = ^-.
s c
By proving one of the last three formulas geometrically it may be shown that
r is the radius ot the inscribed circle. C
Proof. Since angle NA = % A, ^r^J^
Ml
W
^O\
If 8 denotes half the perimeter, we have ' *""
A*
But NB = BL, CM = LC,MA = AN-, therefore
2s = 2AN+2BL + 2LC,
or, s = AN + (BL + LC) = AN + a.
This gives AN = 8 a.
Substituting in (A), tan \A __
Comparing this result with (85) and (84) shows that
* 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 TRIGONOMETEY
Ex. 1. Solve the triangle whose sides are 13, 14, 16.
Solution. Let a = 18, 6 = 14, c = 16.
Then 2s = a
or, a = 21.
Also, s-a = 8, s-6 = 7, s - c = 6.
Fro m(8 4), r
From (88), tan %A = -? -- = - = .5000.
s ct 8 2
.-. %A = 26.56, or A = 63.12.
From (86), tan JS = ^ = - = .6714.
8 7
.-. J B = 29.74, or J5 = 59.48.
From (87), tan \ C = -^- = - = - = .6667.
.-. C = 33.69, or C = 67.38.
Check : A -f J5 + (7 = 63. 12 + 69.48 -f 67.38 = 179.98 *
EXAMPLES
1. Solve Examples 1, 3, 8, p. 110, using the law of tangents.
2. Solve Examples 4, 6, 6, p. 110, using formulas (84), (85), (86), (87), p. 115.
3. Prove the following for any triangle :
(a) (a
(c) 6 cos 2 C + c cos 2 B = s.
(d) (6 + c - a) tan A = (c -f a - 6) tan B
(e) c 2 = (a + b) 2 sin 2 C -f (a - &) 2 cos 2 J C.
(f) c (cos^l -f cos B) = 2 (a + 6) sin 2 ^ 0.
cos 2 1 A a (s a)
(g)
cos 2 %B b(s -b)
C 1 7?
(h) 6sin 2 h csin 2 =. s - a.
2 2
(i) a cos B cos C esc .4 = .
^ ^ 2
(j) sin A = 2 sin cos = v(s a) (s b) (s c).
2 2 oc
* The error .02 arises from the fact that we used a four-place table. If we had used a
table giving the first live 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
117
4. If E and r denote the radii of the circumscribed and inscribed circles
respectively, prove the following for any triangle :
a sin ^Bsin^C . . 1 J_ 4. * *
~ cos kA be ca ab
4Vs(s - a)(s - b)(s -c) 2 \siiiyl smBsinC
(e) abcr = 4 R (s - a) (s - b) (s - c).
62. Formulas for finding the area of an oblique triangle.
CASE I. When two sides and the included angle are known.
Let by c, and A be known. Take c as the base. Denote the altitude
by h and the area by 8. Then, by Geometry,
S \ ch.
But h b sin A (from (7), p. 11); hence
(88) S = \bcain A.
Similarly, S = % ac sin B = | ab sin C.
C C
The area of a triangle equals half the product of ant/ tuco sides
multiplied by the sine of t/i e included angle.
Ex. 1. Find the area of a triangle, having given b = 20 in., c = 16 in., A = 60.
Solution. Substituting in (88) t
1 1 V3 r
S = -be sin A = - x 20 x 15 x = 75 V3 sq. in. Ans.
22 2 4
CASE II. When the three sides are known.
sin A = 2 sin %A cos $A (51), p. 72
pp. 113, 114
Substituting this value of sin A in (88), we get
(89) fc s
Ex. 2. Having given a = 13, 6 = 14, c = 15 ; find the area.
Solution. S = (a-|-b-fc) = 21, -a = 8, 8-6 = 7, -c = 6.
Substituting in (89),
= Vs(8-a)(-&)(8-c) = V21 x 8 x 7 x 6 = 84.
118 PLANE TRIGONOMETRY
CASE III. Problems which do not fall under Cases I or II directly
may be solved by Case /, if we first find an additional side or angle
by the law of sines.
Ex. 3. Given a = 10 \/3, 6 = 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,
. _, frsin^L 10 x
C1 Tl /? _ ___ _
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 0, the prob-
lem comes under Case I, and we get
5= a6sinC= xlOVxlOx = 258. Ans.
EXAMPLES
1. Find the areas of the following triangles, having given
Ans. 240.
17.32.
193. 18.
30 V3.
600.
17J.
15,541.7.
6\/6.
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 -f- \/3 j find its
(a) a = 40,
6 = 13,
c = 37.
(b) b = 8,
c = 5,
A = 60.
(c) b = 10,
c = 40,
A = 75.
(d) a = 10,
6 = 12,
C = 60.
(e) a = 40,
c = 60,
B = 30.
(f)a = 7,
c = 5 V2,
# = 135.
(g) b = 149,
J. = 70 42',
5 = 39 18'.
(h) o=5,
6 = 6,
c = 7.
(i) a = 409,
6 = 169,
c = 510.
(j) a = 140. 5,
6 = 170.0,
A = 40.
(k) c = 8,
5 = 100.1,
= 31.1.
(1) a = 7,
c = 3,
vl = 60.
6. Prove the following for any triangle :
, x abc . . _ 2 abc / A B C\
(a) S = (e) S = r - I cos cos cos ) .
1 4# v 7 a + 6 + c\ 2 2 2/
(b) S = rs. a 2 6 2
(c) S = Br( 8 in A + sin B+ sin C). <*> S = I sin 2 B + 4 sin * A '
(d) 5 = i a 2 sin B sin C cscA
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
(^4 ) 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 = Iog 6 TV". (logarithmic form)
(A) and (B) are then simply two different ways of expressing the
same relation between b, Xj 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 (A) in the form
(base) log = number.
For example, the following relations in exponential form, namely,
3* =9, 2' = 32, ft)'=i, *=,
are written respectively in the logarithmic form
2 = Iog 8 9, 5 = log 2 32, 3
* logarithms were invented by John Napier (1550-1617), Baron of Merchlgton in Soot-
land, and described by "him in 1614.
119
120 PLANE TEIGONOMETliY
where 2, 5, 3, y are the logarithms (exponents),
3, 2, , x are the bases, and
9, 32, J, z are the numbers respectively.
Similarly, the relations
are written in logarithmic form as follows :
i = log a6 5, -3 = log 10 .001, = log 8 4, = log 6 l.
EXAMPLES
1. In the following name the logarithm (exponent), the base, and the number,
and write each in logarithmic form : 2 3 = 8, 4* = 16, 5 2 = 26, 3 = 27, 3* = 81.
Solution. In the first one, 8 = logarithm, 2 = base, 8 = number ; hence
logs 8 = 3. Ana.
2. Express the following equations in logarithmic form : ( J) 2 = J , v^!26 = 6,
2-* = ^, 10-' =.01,^=0.
3. Express the following equations in the exponential form : Iog 4 64 = 8,
Iog 7 49 = 2, Iog 6 216 = 8, Iog 10 .0001 = - 4, log 4 2=J, log a a = l, log a l = 0,
lo&a = c.
4. When the base is 2, what are the logarithms of the numbers 1, 2, J, 4, J,
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, 8, - 1, - 2, , what
are the numbers ?
8. What must be the bases when the following equations are true :
Iog64 = 2? log 121 = 2? Iog625 = 4? log^ = -2?
9. When the base is 10, between what integers do the logarithms of the fol-
lowing numbers lie : 83, 261, 1708?
Solution. Since logiolO = 1 and logiolOO = 2, and 83 is a number lying be-
tween 10 and 100, it follows that Iog 10 88 = a number lying between 1 and 2.
Similarly, logic 261 = & number lying between 2 and 8,
Iogi 1798 = a number lying between 3 and 4.
THEORY AND USE OF LOGARITHMS 121
10, Verify the following:
(a) logiolOOO 4 logiolOO + log 10 4- logic 1 = 6.
(b) logio T V
(c) logio.OOl - logio.Ol + Iog 10 .l = - 2.
(d) Iog 8 8 - 3 Iog 8 2 + logil = 2.
(e) 2 Iog a + 2 log a - + log a 1 = 0.
a
(f) 2 Iog 4 2 + i Iog 2 4 - Iog 2 2 = 1.
(g) logs 8 + logs J - logs 81 = - 6.
(h) 3 Iog 27 3 - logs 27 -f logo 3 = .
(i) 4 Iog l6 4 -f 2 Iog 4 ^ + i Iog 2 16 = 0.
(j) 2 Iog 8 64 - Io g7 49 + i logg^y = 1.
(k) Iog 8 64 -f Iog 4 64 -f- Iog 2 64 = 11.
(1) logs 26 - logs 126 + 2 Iog 6 6 = 1.
(m) 2 log M 6 - Iog 6 86 -f log e ^ = - 8.
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 b M=x, and log fe JV = y.
Writing these in the exponential form,
(B) b* = M, and b v = N.
Multiplying together the corresponding members of equations (B),
Writing this in the logarithmic form gives
log & MN = x + y = log 6 M + Iog 6 #. from (.4)
By successive applications this theorem may evidently be ex-
tended to the product of any number of factors as follows:
logft MNPQ = Iog 6 M NPQ = Iog 6 M + Iog 6 NPQ Tk I
= log 6 M+ log, AT + log 6 PQ
log 6 M -h Iog 6 AT 4- log 6 P 4 log><2.
122 PLANE TRIGONOMETEY
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) log b M=x, and Iog b N = y.
Writing these in the exponential form,
(J5) b x = M, and b* = N.
Dividing the corresponding members of equations (j5), we get
M
Writing this in logarithmic form gives
f rom ( 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 ff = N.
Kaising both sides to the pth power,
b px = NP.
Writing this in logarithmic form gives
log b N p =px=p Iog 6 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 b N=x.
Then b x = N.
Extracting the rth root of both sides,
X 1
If a= N r .
Writing this in logarithmic form gives
From the preceding four theorems it follows that if we use the
^ogarithms 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, multiplica-
tion, aild division respectively.
THEOEY AND USE OF LOGAEITHMS 128
Ex. 1. Find the yalue of logio V.001.
Solution, logio V JOOI = J Iog 10 .001 Th. IV
Ex. 2. Write log b ^ ' C a - in expanded form.
Th. IV
= i j log* a 9 + log ft (c + d)* - Iog 6 c 2 1 Th. I, II
= J J3 Iog 6 a -f \ log b (c + d) - 2 Iog 6 c^. Th. Ill, IV
When no base is indicated we mean that the same base is to be used through-
out. Thus, the relation
1 jsioga + ^log(c + d) - 21ogc
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 -f 1) + 3 log (x - 1) + i log x - 2 log (x 2 -f 1) in the con-
tracted form.
Solution. 3 log (x + 1) -f 3 log (x - 1) 4- $ log x - 2 log (x 2 -f 1)
= log (x -f I) 8 + log( - I) 8 -f- logx* - log(x 2 + I) 2
_ O g _____ _ og^-j-y^-.
Another form of the answer is found as follows :
_1. *(x-l)
2 - (x 2 + l)*
EXAMPLES
1. Verify the following :
(a) Iog 10 VlOOO + log 1Q V.01 = J. (e) Iog 2 V -f logs ( J) 2 ^ - .
(c) logio v^ -f- logio VlO = 0. (g) logs ^125 + logn ^121 = y .
2. Write the following logarithmic expressions in expanded form : *
/ \ T I ^ \^ ~~ ) \ ~~ / f(*\ IACT
(a) log-^/-i - i e ) 10 S\
lr o ~ C*
.,. . absmC /f . ,
(b) log (f) 1<
(c) logP(H-r). (g) log^
8)
To verify your results, reduce them back to the original form.
124 PLANE TRIGONOMETRY
3. Write the following logarithmic expressions in contracted form :
(a) 2 logx + ^ logy - 3 logz.
(b) 31og(l - x) - 21og(2 + x) + logc.
(c) $ log(x - 1) - logx - log(x + 2) + loge.
(d) logy - J log (2/ 2 + 4) -]- logc.
(e) {21og(x - 1) + 81og(x + 1) -f Jlogx - log(x 2 + 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 Iog 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 8 =1000 log 1000 =3
10 2 =100 log 100 =2
10 1 =10 log 10 =1
10 =1 logl =0
lO' 1 ^.! log.l =-1
10~ 2 = .01 log .01 =-2
10- 8 = .001 log .001 =-3
10- 4 = . 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 lias 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 tlie 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 -f a decimal.
Similarly, since .0067 is a number lying between 10~ 8 and 10~ a ,
log .0067 = (2 -f- 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,
= ( _ 2) + ( _ a decimal)
= (- 1 - 2) -f (1 - a decimal)
3 -f- 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 numbers 27,688,
466.2, 9.67, 436,000, 26, .04, .0000612, .7963, .8, .0012.
An*. 4, 2, 0, 5, 1, - 2, - 6, - 1, - 1, - 8.
126 PLANE TBIGONOMETRY
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 1Q2 . 10 x == 10 x+a = 5 437)
or, x + 2 = log 5437.
Hence 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
(J5) 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 forai more convenient than (A) or (B) when we wish to divide the
logarithm by 3. Thus,
4 log .0249 = J (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 6.060 Is 605 ;
Of .00007 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
J 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
2-2 + 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 3.2669 and
1.9866.
3.2669
1.9866
6.2626
This is in no way different from ordinary addition.
(b) To add two logarithms, one having a negative characteristic, as 4.2660
and 2.8711.
4.2660 or, 6.2660 - 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
2.8911 18.3911-20
i.e. 5.3911
128 PLANE TRIGONOMETRY
CASE II. Subtraction of logarithms.
(a) To subtract logarithms having positive characteristics.
From 6.6283 From 2.4673 or, 12.4673 10
3.8890 take 3.7851 8.7861
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 2.1163 or, 12.1163-10
take 8.4492 7.4492 - 10
4.6671 4.6671
From 1.6899 or, 9.6899-10
take 1.9083 1.9083
8.7816 7.7816-10
i.e. 3.7816
From 2.1853 or, 18.1853-20
take 1.7442 9.7442-10
2.4411 8.4411 - 10
i.e. 2.4411
CASE III. Multiplication of logarithms by numbers.
Multiply 0.6842 Multiply 2.7012 or, 8.7012 - 10
by 6 by 3 3
3.4210 4.1036 26. 1036 - 80
i.e. 4.1086
In the second example the carrying figure from tenth's place is -f 2. Add-
ing this -f 2 to - 2 x 3 gives 2 - 6 = - 4 = 4 for the characteristic.
CASE IV. Division of logarithms by numbers.
(a) Divide 8.8530 by 2.
2 1 8. 8530
1.9266
(b) Divide 2.2411 by 8.
Here we first add ana then we subtract 30, writing the logarithm in the form
28.2411-80. 8|28.2411-80
9.4137 - 10
i.e. 1.4137
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.
THEOEY AND USE OF LOGARITHMS 129
(A) 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. 1'26).
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,t gives immediately the mantissas of the logar
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 con-
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 9 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, 3, 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 1887.
Solution. First step. From the rule on p. 125 we see that the characteristic
will be -f 3, that is, one less than the number of digits (four) to the left of the
decimal point.
SecoTid 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 = 8.1421. 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 Fo-wr-Place Tables of Logarithms
(Qinn & Company).
130 PLANE TEIGONOMETEY
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 = 1.2804. Ans.
Ex. 3. Find log. 00162.
Solution. First step. By the rule on p. 126 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 162 in column JV, Table I, p. 3. In the same horizon-
tal row with 162 and in the vertical column with at the top we find the required
mantissa .1818.
Therefore log .00162 = 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 5.68.
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 66 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. 6. 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= 6.6628. 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 .9081.
Therefore log .08 = 2.9031 = 8.9031 - 10. Ana.
Ex. 7. Find (a) log 1872, (b) log 6, (c) log .7, (d) log 20,000, (e) log 1.808.
(f) log.000032, (g) log. 01011, (h) Iog9.95, (i) log 17.35, (j) log. 1289, (k) log 2500,
(1) log 1.002.
Ans. (a) 8.2728, (b) 0.6990, (c)_1.8451, (d) 4.8010, (e) 0.2572, (f) 6.6051,
(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 interpolation,* 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. 126 the characteristic is found to be 8. 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 2446 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
2460 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 -h .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 2446, since 2442 is not half-
way between them but two tenths of the way.
* Illustrated by examples on pp. 16-19 in the case 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 numhers 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 the mantissas of all such numbers exact to four decimal places.
132
PLANE TRIGONOMETKY
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,
.1x22= 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
6
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. In 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 = 16 U
* 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.
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 .0016, but the decimal point is usually omitted in practice.
THEORY AND USE OF LOGAEITHMS 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 16.
Under 16 and opposite the extra digit 4 of our number we find 6.0. Then
log 28.60 = 1.4664
6 Prop. Part
log 28. 64 = 1.4570. Ans.
Ex. 10. Find log. 12648.
Solution. Since the mantissa of 12,648 is not found in our table, this example
comes under Rule III, the extra digit being 8.
First step. log . 12640 = 1. 0983
Second step. log . 12660 = 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. 12640 = 1.0983
2 Prop. Part
log. 12648 =1.0985. Ans.
Ex. 11. Verify the following :
(a) log 4583 =3.6612. (e) log 1000.7 =8.0003.
(b) log 16.426 = 1.2155. (f) log 724,200 =6.8698.
(c) log .09688 = 2.9862. (g) log 9.496 = 0.9775.
(d) log. 10108 = 1.0047. (h) log .0004686 = 4.6614.
69. To find the number corresponding to a given logarithm, use
RULE 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
v
log* = 2. 1892.
On p. 3, Table I, we find this mantissa, . 1892 exactly, in the same horizon-
tal row with 164 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 = 164.6. Ans.
Ex. 18. Find the number whose logarithm is 4.8409.
Solution. That is, given log x = 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 \s 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-af ter 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) 1.8065, (b) 1.4487,
(c) 0.2164, (d) 2.9487, (e) 2.0529, (f) 6.2668, (g) 8.9774, (h) 4. 0010, (i) 8. 4430 -10,*
(j) 9.4976 - 10.
Ans. (a) 63.9, (b) .281, (c) 1.646, (d) 888.6, (e) .011296, (f) 184,860,
(g) 9498, (h) .00010023, (i) .02773, (j) .3144.
By (A), (J?), 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 18.49, using logarithms.
Solution. Denoting the product by x, we may write
x = 243 x 18.49.
Taking the logarithms of both sides, we get
log x = log 248 -f log 18.49. Th. I, p. 121
Looking up the logarithms of the numbers,
% log 243 = 2.8866 Rule II, p. 130
log 13.49 = 1.1300 Rule I, p. 129
Adding, logx = 8.6166
By Rule IV, p. 133, x = 3278. Ana.
^ 0^1 i * 1376 x -06423
Ex. 2. Calculate
76,420
. .. T 1376 x .06423
Sol^on. Letx = ^
Then Iqg x = log 1375 + log .06423 - log 76,420
Th. I, p. 121, and Th. II, p. 122
log 1376 = 3.1383 Rule I, p. 129
log. 06423 = 8.8077 - 10 Rule III, p. 132
Adding, 11.9460 - 10
log 76,420 = 4.8882 Rule III, p. 132
Subtracting, logx = 7.0628-10
>r, logx= 3.0628.
By Rule IV, p. 133, x = .0011666. Ans.
Ex. 3. Calculate (6.664).
Solution. Let x = (6.664) 8 .
Then log x = 3 log 6.664. Th. Ill, p. 122
log 6.664 = 0.7631 Rule III, p. 132
Multiplying by 3, 3
logx = 2. 2693
By Rule IV, p. 133, x = 181.67. Ans.
Ex. 4. Calculate ^.7182.
Solution. Let *= ^?7182 = (.7182)*.
Then logx = J log. 7 182. Th. IV, p. 122
log .7182 = 1.8662 Rule III, p. 182
= 29.8662 - 80. (b), Case IV, p. 128
Dividing by 8, 8 [29.8662-80
logx = _9.9521 -10
= 1.9621.
By Rule IV, p. 188, x = .8966. Ana.
136 PLANE TRIGONOMETRY
Ex. 6. Calculate
fV7194 x 87
N "98,080,000 '
3IV7194 x 87 ["(7194)* x 87]*
Solution. Let x = \ 9g Q8Q OOQ = [ 98 080 000 J "
Then logx = [$ log 7 194 + log 87 - log 98,080,000] .
log 7 194 = 3.8569
Dividing by 2, 2 1 3. 8569
log7194 = 1.9285
log 87= 1.9395
Adding, 3.8680
or, 13.8680 - 10 "(a), Case II, p. 128
log 98,080,000 = 7.9916
Subtracting, 5.8764 - 10
or, 25.8764 - 30 ;b), Case IV, p. 128
Dividing by 8, 3 [25.8764-30
log x = 8.6255- 10
= 2.6265.
.-. x = .04222, Ana.
T, a />, , , 8 * 62 ' 73 X - 052
Ex.6. Calculate ^^ .
- , ,. T t 8 x 62.78 x .052
Solution. Let * = ____.
Then log x = [log 8 + log 62,73 + log .062] - [log 66 + log 8. 793]
log 8 = 0.9031 log 56 = 1.7482
log 62. 73= 1.7976 log 8. 793 = 0.9442
log . 062 = 8.7100-10 log denominator = 2. 6924
log numerator = 11.4166-10
log denominator = 2.6924
logx= 8.7242-10
= 2.7242.
.-. x = .05299. Ans.*
* 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. Iog56l.
log 62.73- 1. log 8.793 0.
log .052 8. 10 log denominator =
log numerator =
log denominator = _________
loga?
It 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.
THEOBY 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 fronPlO 10. Thus, taking 25 as the number,
colog 25 = log 5 = log 1 log 25.
But log 1 = 0,
or, what amounts to the same thing,
log 1 = 10.0000 - 10.
Also, log 25 = J..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
KULE 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 62. 68.
Solution. 10.0000-10
log 52.63= 1.7212
colog 62.63 = ~8.2788 - 10. Ans. Rule V
Ex. 2. Find colog. 016648,
Solution. 10.0000 - 10
log .016648 := 8.2187 - 10
colog. 016648= 1.7813. Ana. 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,
namely > __8x 62.73 x. 052
X 56 x 8.793
Solution. Using cologarithms,
logx = log 8 + log 62.73 + log .062 + colog 66 4- colog 8.793.
log 8= 0.9031
log 62. 73= 1.7976
log .062= 8.7160-10
colog 66 = 8.2618 - 10 since log 66 = 1.7482
colog 8. 793 = 9.0668 - 10 since log 8. 793 = 0.9442
logx = 28. 7242 -30
= 2.7242.
.-. x = .06299. Ans.
Calculate the following expressions, using logarithms :
3.
4.
6.
6.
9.238 x. 9152. Ans. 8.454.
336.8^-7984. .04218.
(.07396) 5 . .000002213. 12.
15.008 x. 0843
/ 872( Y Ans 5010
V.1321/' AHS ' - 6010 *
(538.2 x. 0006969)*. .8678.
ir/73l63y
".06376 x 4.248 18 '
\V429J' ' 63L
7.
8.
y&
1.495. 14t
'
9.
^.02305.
.2846. 15 .
v^ x v'S x Viol. .7036.
10.
3/J03296
\ 7.962 "
.1606. 16.
- 401 ' 8 .* 7672
62.87
17.
(-2563) x .03442
-4ns. .2415.
714.8 x (-.611)
18.
121.6 x (-9.025)
.0725.
(- 48.3) x 3662 x (- .0866)'
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 log & JV in terms of logarithms to the base a.
Suppose Iog 6 N=x,
that is, b* = 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+ 62.37 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,
we get 1 j x I jy
or, x log a & = log a ^V: Th. Ill, p. 122
Solving, x = 2J2 .
But log b N = 'x. By hypothesis.
(90) .-. lo gb N= l ^2.
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 ofb to the base a.
This formula is also written in the form
= M .
where M = : - - is called the modulus of the new system with
Iog a 6 J y
respect to the original one*
This number M does not depend on the particular number JV, but
only on the two bases a and b.
In actual computations a = 10, since the tables we use are com-
puted to the base 10.
Ex. Find Iog 8 21.
Solution. Here N = 21, 6 = 3, a = 10. Substituting in (90),
loga^^ 2.7J1. Ans.
B8 Iogio3 .4771
^ EXAMPLES
1. Verify the following :
(a) Iog 2 7 = 2.807. (e) Iog 9 8 = 0.9464. (i) logs 10 =2.096.
(b) Iog 8 4 = 1-262. (f) Iog 8 6 = 0.7740. (j) Iog 6 100 = 2.86.
(c) Iog 4 9 = 1.585. (g) Iog 7 14 = 1.356. (k) Iog 8 .l= - 2.096.
(d) logs 7 = 1.209. (h) logs 102 = 2.873. (1) Iog 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. Piove log& a Iog 6 = 1.
6 f Prove logjvlO =
* 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) M y Thus, having given the common loga-
rithms (base 10) of numbers, we can reduce them to the logarithms of the same numbers to
the base e(= 2.718) by multiplying them by M = -. - - = 2.3026.
10 gio e
140 PLANE TBIGONOMETBY
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, x log 81 = log 10. Th. Ill, p. 122
Ex. 2. Express the solution of
in terms of logarithms.
Solution. Taking the logarithms of both members,
log a ax+8 + log 6* = logc. Th. I, p. 121
(2x -f 3) logo + x logft = logc. Th. in, p. 122
2icloga-f Sloga + xlogb = logc.
x (2 log a + log 6) = log c 3 log a.
2 log a + log b
Ex. 8. 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 Iog2 + y Iog8 = 2 Th. I, m, p. 122
x Iog2 + y Iog2 = 4 Iog2
Subtracting, y (log 8 log 2) = 2 4 log 2
Solving, 2-41og2^ 2-1.2040
Iog8-log2 .4771 -.3010
.7960
Substituting back in (B), we get x = - .62.
EXAMPLES
1. Solve the following equations :
(a) & = 12. Ans. 1.54. (g) (1.8)* = 7.2. Ans. 7.68.
(b) 7* = 26. 1.65. I _i
(c) (0.4)~* =7. 2.12. (h) (0.9)* = (4. 7) . 0.46.
(d) 10*-i = 4. 1.002. fl) 7*+s=6. -2.1729.
(e) 4*-i ... 5*+^ _ 13 43 (j) 220:+ a _ 0*-i = o. 9.5414
(f) & = 40. 2.66.
THEORY AND USE OF LOGARITHMS 141
2. Solve the following simultaneous equations :
(a) 4* 3* = 8, Ans. x = .9006, (c) 2* 2f = 222, ATM. x = 13,
2*-8v = 9. y = .7666. x - y = 4. y = 9.
(b) 3* 4* = 16,662, x = 6, (d) 2* 3* = 18, x = 1,
4*. 6^ = 128,000. y = 3. 6* 7* = 245. y = 2.
3. Indicate the solution of the following in terms of logarithms:
(a) A = P(r + 1)*. Ans. x =
(b) a *^*=6. x
(c) a- 6* = m, x =
log a log d log 6 log c
c*.<^ = 7i. y = logalogn-logclogm
log a log d log b log c
(d) a2*-8 . a 8y-2 _ a 8 ? = 6^
3x-f2y=17. y = l.
74. Use of the tables of logarithms of the trigonometric functions.
On p. 9 the values of the trigonometric functions of angles from
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. f
In both tables the following directions hold true :
Angles between and 45 are in the extreme left-hand column
on each page,t 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 with 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.
$ The angles increase as we read downwards.
142 PLANE TEIGONOMETEY
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.
ThuSj log sin 38 30' = 9.7941 - 10 = 1.7941. p. 16
log cot 10' = 2.5363 = 2.5363. p. 8
log tan 75.6 = 0.5905 = 0.5905. p. 31
log cos 2.94 = 9.9994 - 10 = 1.9994. p. 25
75. Use of Table II, pp. 8-16, 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
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 . 1'" 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. 1'" 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" = f $ = .7' ;
2 0' 16" = 2 0.27', since 16" = | = -266 1
If the angle is given in degrees and the decimal parts of a degree, and it is desired to
use Table II, the angle may be quickly found in degrees and minutes by making use of the
Conversion Table on p. 17.
THEOEY AND USE OF LOGAEITHMS 148
76. To find the logarithm of a function of an angle when the angle is
expressed in degrees and minutes, use
RULE 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 4&y r a t 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 I 1 .
(#) 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 1 J7 *
T/. r I* 118 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 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 82 30' exactly; hence, by
Rule VI, we get immediately from the table
log tan 82 80' = 9.8042 - 10. Ana.
Ex. 2. Find log cot 88 17'.
Solution. On p. 9, Table II, we find the angle 88 17' exactly ; hence, by
Rule VI, we get at once
log cot 88 17' = 8.4767 - 10. Ana.
Ex. 8. Find log sin 28 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. 6978 - 10 Tab< ^ _ 2 . 9
corr. for 6' = 17 Excess 6
log sin 23 26' =9. 6995 -10. Ans. Corr - ~ 17 - 4
* 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 subtract the
correction. Of course this is true only for acute angles.
144 FLAKE TRIGONOMETRY
Ex. 4. Find log cos 54 42' IS''.
Solution. Since 18" is less than half a minute, we drop it, and from p. 16,
Table II, by Rule VI,
log cos 64 40' = 9.7622 - 10 Tab - diff -= ^
* A Excess = 2
corr. for 2= 4 Corr ^55
log cos 64 42' = 9.7618 - 10. Ana. i.e.-4
Ex. 6. 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 A' 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 36 60' = 9.8586 - 10. (g) log cos 27 28' = 9.9480 - 10.
(b) log sin 61 58' = 9. 9458 - 10. (h) log cot 51 49' = 9. 8957 - 10.
(c) log tan 82 3' 20"= .8550. (i) log sin 85 57' = 9.9989 - 10.
(d) log cos 44 82' 50"= 9.8628 - 10. (j) log cot 45 0' 13" = 0.0000.
(e) log tan 1 63.2' = 8.5178 - 10. (k) log sin 120 24.3' = 9.9358 - 10.
(f) log tan 87 15.6' = 1.3201. (1) log tan 243 42' 15" = 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 //, 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 I 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 *\ *
/ ;' * _ \ this correction.*
If cosine or cotangent, subtract J
* See footnote, p. 148.
THEORY 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 : having 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. 16, Table II, 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 sin a; = 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 9.6662 exactly ; ljut (Rule VII) the next less logarithm in such a column
is found on p. 16, Table II, to be 9.6644, which corresponds to the angle 27 30',
and the corresponding tabular difference for V is 2.4. Hence
log sin x = 9.6652 - 10
log sin 27 30'= 9.6644 - 10
excess =8
Since the function involved is the sine, we add this correction, giving
x = 27 30' + 3' = 27 33'. Ans.
Ex. 9. Find the angle whose log cos = 9.3705 10.
Solution. That is, having given log cosx = 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 9.8706 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 30', and the corresponding tabular difference for 1' is 6.2. Hence
log cos x - 9.3705 - 10
log cos 76 30' = 9.3682 - 10
excess =23
Since the function involved is the cosine, we subtract this correction, giving
z = 76 30' -4' = 76 26'. Ans.
146 PLANE TEiGONOMETKY
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.
log tan x = 8. 7670 - 10
log tan 3 16'= 8. 7666 -10
excess =6
Hence x = 8 16' + .2' = 3 16.2'. Ana.
Ex. 11. Given cot a; = (1.01) 5 ; find x.
Solution. Taking the logarithms of both sides,
log cot x = 6 log 1.01. Th. Ill, p. 122
But log 1.01 = 0.0043
and, multiplying by 6, 6
log cotx = 0.0216 ; to find x.
By Rule VII the next less logarithmic cotangent is found on p. 16, Table II.
log COt X = 0.0216 Tab.diff.l' I Excess I Corr.
log cot 43 40' = 0.0202 2.6 I 13.0 | 5
excess =18 iM
Hence x = 48 40' - 6' = 43 36'. Ans.
Ex. 12. Verify the following :
(a) If log sin x = 9.6443 - 10, then x = 20 30'.
(b) If log cosx = 9.7631 - 10, then x = 66 30'.
(c) If log tanx = 9.9570 10, then x = 42 10'.
(d) If log cotx = 1.0034, then x = 6 40'.
(e) If log sin x = 8.0486 - 10, then x = 38'.
(f) If log cosx = 8.7918 - 10, then x = 86 27'.
(g) If log tan x = 9.6261 - 10, then x = 18 34'.
(h) If log cotx = 0.6380, then x = 12 68'.
(i) If log sin x = 9.9996 - 10, then x = 87 16'.*
(j) If log cosx = 8.2881 - 10, then x = 88 63.3'.
(k) If log tan x = 2.1642, then x = 89 86.4'.
(1) If log tan x = 7.9732 - 10, then x = 32.3'.
(m) If log sin x = 9.8600 - 10, then x = 46 4'.
(n) If log cosx = 9.9000 - 10, then x = 87 26'.
(o) If log tanx = 0.0036, then x = 46 14'.
(p) If log cot x = 1.0000, then x = 6 43'.
(q) If log cotx = 3.9782, then x = 89 27.7'.
* When there are several angles corresponding to the given logarithmic function, we
choose the middle one.
THEOKY AND USE OF LOGARITHMS
147
EXAMPLES
Use logarithms when making the calculations in the following examples :
1. Given 184 sin 8 x = (12.03) 2 cos 67 20' ; find x.
Solution. First we solve for sin x, giving
" -
12. 03) 2 cos 57 20'
--
since log 12. 03 = 1.0803
since log 184 = 2.2648
Taking the logarithms of both sides,
log sin x = J [2 log 12.03 + log cos 67 20' -f colog 184].
2 log 12.03= 2.1606
log cos 57 20' = 9. 7322 - 10
colog 184 = 7.7352 - 10
19.0280 - 20
3 1 20. 6280 -30
log sin x= 9.8760-10
.-. x = 4844 / . Ans.
2. Given cosx = (.9854)* ; find x.
4. 236 cos 62 19'
3. Calculate
13. 087 sin 48 6'
4. Given 1.6 cot 82 = x 2 sin 12 15' ; find x.
Hint. First solve for x, giving
x-
5. Given 60 tanx = v.2684 ; find x.
_ _ . . ^ sin 24 13" cot 58 2'
6. Calculate
cos 33 17' tan 19 58'
7. Calculate VcoslO6 / tan 73 11'.
8. Calculate
(sin83 18 / )8Vcot71 20 /
10. 668 tan 63 64'
9. Given 3 cot x = VTi ; find x.
10. Given sinx = (.9361) 10 ; find x.
11. Given 2.3 tanx = (1.002) 126 ; find x.
.2061).
.0968.
49'.
.8426.
1.806.
.004422.
72 46'.
31 6'.
29 24'.
78. Use of Table III, 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 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 ease the angle is given in degrees, minutes, and seconds, and it is desired to use
Table 111, 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 4&y or a t
the bottom when the angle is greater than 4^>
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 , . .
w ; ' . \ ' > 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. Ans.
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 = 1.2666. Ans. Y- n '?' > ''
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 ( = 18). Then
log tan 81. 80 =0.2707
18 Prop. Part.
log tan 61.87 = 0.2720. Ans.
* 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 cot 2. 168.
Solution. The exact angle 2. 168 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 2.16* and log cot 2. 16 is 20. In the
Prop. Parts column under 20 and opposite the extra digit 8 we find the proper-
tional part 16. Then
log cot 2. 150 =1.4255
16 Prop. Part.
log cot 2. 158 =1.4239. Ans.
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 cos 20.03 = 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 cot 84. 84 = 0.1574.
(e) log cos 45. 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 = 9.4818 - 10.
80. To find the acute angle in degrees and decimal parts of a degree
which corresponds to a given logarithmic function, use
EULE 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. 6364 10 ; to find the angle x.
Solution. Looking up and down the columns having "log tan" at top or
bottom, we do not find 9.6364 exactly. But then, by Rule IX, we locate it
between 9.6346 and 9.6370, 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 = 9.6346 - 10
log tan x = 9.6364-10
19 = difference.
The corresponding tabular difference being 26, we find in the Prop. Parts col-
umn that 20 is the proportional part under 26 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. 26, 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 cos x = 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 sin x = 9.6062 - 10, then x = 18.67.
(f) If log cosx = 9.9629 - 10, then x = 23.36.
(g) If log tana; = 9.8380 - 10, then x = 34.66.
(h) If log cot x = 9.3361 - 10, then x = 77.77.
(i) If log sin x - 8.6862 - 10, then x = 2.776
(j) If log cosx = 9.9996 - 10, then x = 2.74.
(k) If log tanx = 7.2642 - 10, then x = 0.106.
(1) If log cotx = 1.7900, then x = 0.929.
(m) If log sinx = 9.5860 - 10, then x = 20.06.
(n) If log cosx = 9.8000 - 10, then x = 60.88.
(o) If log tanx = 0.0036, then x = 46.23.
(p) If log cotx = 2.0000, then x = 0.678.
(q) If log sin x = 0.0000, then x = 90.
(r) If log tan x = 0.0000, then x = 46.
THEORY AND USE OF LOGARITHMS 151
EXAMPLES
Use logarithms when making the calculations in the following examples .
1. Given tanx = (1.018)" ; find x.
Solution. Taking the logarithms of both sides,
log tan x = 12 log 1.018. Th. Ill, p. 122
But log 1.018 = 0.0077
and, multiplying by 12, 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 6 of the required angle. Hence
E = 61.06. Ans.
2. Given 66.4 tan 5 x = (18. 66)* cos 69. 8; find x.
Solution. First we solve for tanx, giving
. |X _8. 66)6 cos 69.8
tan x = - ' v
66.4
Taking the logarithms of both sides,
log tan x = [6 log 18.65 + log cos 69.8+ colog 66.4].
6 log 18.65 = 6.3636 since log 18.66 = 1.2707
log cos 69. 8 = 9. 6382 - 10
colog 66.4= 8.2487 -10 since log 66.4 = 1.7613
24.1404-20
6| 64. 1404- 60
log tanx = 10.8281 - 10.
/. x = 81.66. Ans.
3. Given cos x=V.9681; find x. Ans. 10.26.
. - . . . 26. 62 tan 33.86
4. Calculate ^rr^^- 9.746.
5. Given Vi sin 48.06 = x 8 cos 2. 148 ; find x. 1.0886.
Hint. First solve for a?, giving
'/Vi sin 48.00
X "Af oos2.143
6. Given 6 cotx = #.4083 ; find x. 81.66.
7. Given 8 inx = ; find. 1.762=.
8. Calculate V861tan87. 6 sin 9. 68. 87.
162 PLANE TEIGONOMETRY
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*=c*-b 2 = (c + b) (c - &).f
Large errors may be detected by measurement.
For reference purposes we give the following formulas from p. 8
and p. 11.
Area of a right triangle = 22 .
(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
well adapted to logarithmic computation, since
and we have a summation under the radical that cannot be performed by 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 8 -&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
asily calculated by inspection, and then we can use logarithms advantageously. Thus
log a = i [log (c - b) + log (c + b)].
* In case a or 6 is not given, or both a and b 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 LOGARITHMS
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=W-A = 41 48'.
Third step. To find a use - a = c sin A.
Taking the logarithms of both sides,
log a = log c + log sin .4.
Hence, from Tables I and II,*
logc= 2.5106
log sin A = 9.8730-10
loga= 12.3836-10
= 2.3835.
.-. a = 241.8.
To find 6 use 6 = c cos .4.
Taking the logarithms of both sides,
log 6 = log c + log COB A.
Hence, from Tables I and II,
logc= 2.6105
log cos^. = 9.8231-10
log b = 12. 3336 -10
= 2.8336
.-. 6 = 216.6.
* If ir e wish to use Table m instead of Table II, we reduce 17' to the decimal of a degree
T!ra8f <4~ 48 17' -
154
PLANE TRIGONOMETRY
Fourth step. To check these results numerically, let us see if a, &, c satisfy
the equation
or, using logarithms, 2 log a = log(c + b) 4- log(c 6),
that is, log a = J [log (c + 6) -f- log (c - &)].
Here c -f 6 = 689.6 and c - b = 108.4.
log (c + 6) = 2. 7321
log (c- 6) = 2.0360
2 log a = 4. 7671
log a = 2.3886.
Since this value of log a is the same as that obtained above, the answers are
probably correct.
To find the area use formula
A *
Area =
log area = log a -f- log 6 log 2.
log a = 2.3836
log 6 = 2. 3336
4.7171
log 2 = 0.8010
log area = 4.4161
.. area = 26,070.
Ex. 2. Solve the right triangle, having given 6 = 16.12, c = 30.81.
Solution. Here we first find an acute angle ; to find A use
6-15.12
cos^ = -. (2), p. 2
log cos .4 = log b logc.
log 6 = 11. 1796 -10
logc= 1.4887
log cos A= 9.6909-10
.-. A = 60 36'. from Table II, p. 16
Hence B = 90 - A = 29 24'.
To find a we may use
log a = log 6 + log tan-4.
log b =1.1796.
log tan ,4 = 0.2491
log a =1.4287
(9), p. 11
.-. a = 26.84.
THEORY AND USE OF LOGARITHMS 155
To check the work numerically, take
a* = (c + b)(c-b),
or, log a = [log (c + 6) + log (c - 6)].
Here c 4- & = 46.93 and c - 6 = 15.69.
log (c + 6) = 1.6621
log(c- b) ='1.1956
2 log a = 2.8678
log a = 1.4288.
This we see agrees substantially with the above result.
Ex. 3. Solve the right triangle, having given B = 2.326, a = 1875.8.
Solution. A = W-B = 87.676
sin^ = -.
Solving for the unknown side c,
_ a
sin A
log c = log a log sin A.
Hence, from Tables I and III,*
log a = 13.2731 -10
log sin A = 9.9996-10
logc= 3.2736
.-. c = 1877.
= by(8), p. 2
Solving for the unknown side ft,
tan A
log 6 = log a log tan A
log a = 13.2731 -10
log tan A = 11.3916-10
log 6= 1.8816
/. 6= 76.13.
To check the work we may use formulas
or, 6 = c sinJB, 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 i educe 2.325 to degrees and min-
utes. Thus, 5= 2.3250 = 2 19.5'.
156
PLANE TEIGONOMETEY
EXAMPLES
Solve the following right triangles (C= 90), using logarithmic Tables I and II.*
No.
GIVEN
PARTS
REQUIRED 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 = 1888.6
c = 2019.6
3
B = 62 66'
6 = 47.7
A - 27 4'
a = 24.37
c = 63.66
4
a = .624
c = .91
A = 43 18'
B = 46 42'
6 = .6623
5
A = 72 T
a = 83.4
= 17 63'
6 = 26.91
c = 87.64
6
b = 2.887
c = 5.11
B = 34 24'
A = 66 36'
a = 4.216
7
A = 62 41'
6 = 4247
B = 37 19'
a = 6671
c = 7007
8
a = 101
6= 116
A = 41 2'
B = 48 68'
c = 163.8
9
A = 48 22'
a = 168.3
B = 46 38'
6 = 167.6
c = 230.6
10
a = 204.2
c = 276.3
-4 = 47 63'
B - 42 7'
6= 184.7
11
B = 10 51'
c = .7264
A = 79 9'
a = .7133
6 =.1367
12
a = 688.6
6 = 601.2
A = 61 63'
B = 38 7'
c = 811.7
13
b = .02497
c = .04792
A = 68 36'
5 = 31 24'
a = .0409
14
B = 2 19' 80"
a = 1876.3
A = 87 40' 30"
6 = 76.13
c = 1877
15
5 = 21 83' 61
" a = .8211
A = 68 26' 9"
b = .8246
c = .8829
16
A = 74 0' 18"
c = 276.62
B = 16 69' 42"
a = 264.9
6 = 75.96
17
B = 34 14' 37
" b = 120.22
A - 66 46' 23"
a = 176. 67
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
J. = 36 46'
B = 63 14'
c = 41.22
20
A = 78 17'
a = 203. 8
B = 11 43'
6 = 42.27
c = 208.16
21. Find areas of the first five of the above triangles.
Ans. (I) 31.32; (2) 686,900; (3) 681.3; (4) .2067; (6) 1122.6.
Solve the following isosceles triangles where A, B, C are the angles and
a, 6, c the sides opposite respectively, a and 6 being the equal sides.
22. Given A = 68 67', 6 = 36.09. Ans. C = 42 6', c = 26.21.
23. Given B = 27 8', c = 3.088. Ans. C = 126 44', a = 1.786.
24. Given C = 80 47', 6 = 2103. Ans. A = 49 36.6', c = 2725.4.
25. Given a = 79.24, c = 106.62. Ans. A = 47 43', C = 84 84'.
26. Given C = 161 28', c = 96.47. Ans. A = 14 16', a = 49.26.
27. One side of a regular octagon is 24 ft. ; find its area and the radii of the
inscribed and circumscribed circled. Ans. Area = 2782, r = 28.97, B = 81.36.
* 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
A - 68.2
B = 21.8
c = 6.385
29
= 32.17
c = .02728
A - 67.83
a = .02809
6 = .01462
30
A = 68.66
c = 36.73
B = 31.85
a = 30. 61
6= 18.69
31
A = 22.28
6 = 13.242
= 67.77
a = 6.418
c = 14.81
32
6 = .02497
c = .04792
A - 68.6
= 81.4
a = .0409
33
a = 273
6 = 418
A= 33.16
= 66.85
c = 499.3
34
=23.16
6=76.48
A = 66.86
a = 176.6
c = 191.9
35
^=31.76
a = 48.04
B = 68.26
6 = 77.64
c = 91.28
36
6 = 612
c = 900
A= 66.82
= 34.68
a = 740.2
37
a =62
c= 60
A = 60.06 ,
=29.94
6 = 29.94
38
A = 2.49
a = .83
7> > = 87.61
6 = 19.085
c= 19.107
39
^4 = 88.426
6 = 9
= 1.674
a = 327.5
c = 827.6
40
= 4.963
6= .07
A = 85.037
a = .8062
c = .8092
41
B = 86.476
c = 80
^1 = 4.625
a = 6.313
6= 79.74
42
a = 100.87
6 = 2
A = 88.864
B= 1.186
c = 100.9
43. Find the areas of the first five of the above triangles.
Ans. (28) 6; (29) .0001677; (30) 288.6; (31) 36.84; (32) .00061.
44. The perimeter of a regular polygon of 11 sides is 28.47 ft. Find the
radius of the circumscribed circle. Ans. 3.79 ft.
45. Two stations are 8 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? Ans. .485 mi.
46. A rock on the bank of a river is 180 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 80' 21". Find the width of the river. Ans. 602.6 ft.
47. A rope 88 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. Ans. 40 16'.
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. Ans. 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. Ans. 60 33'.
50. The top of a tree broken by the wind strikes the ground 16 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. Ana. 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. Ans. 66.8 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.
Ans. About 866,400 mi.
53. Given that the radius of the earth is 8963 mi. , and that it subtends an
angle of 67" at the moon. Find the distance of the moon from the earth.
Ans. 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. Ana. 92 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? Ans. 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. 6.13 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 _ ^ _ c
sin A sin B sin C
and the law of tangents
tan -4 -B = '~ tan
are well adapted to the use of logarithms, while this is not the case
with the law of cosines, namely,
a 2 = 2 -f 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
sin A sin B sin C
which involve only one unknown party and solve for that part.
Check : See if the sides found satisfy the law of tangents.
THEORY AND USE OF 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 = 67.
Second step. Solving = - for a, we get
sin A sin B
6 sin .A
log a = log 6 -f log sin A log sin B.
log 6= 1.3010 B
log sin A 9.9869 - 10 *
11.2879 - 10
log sin B = 9.6126-10
loga= 1.7763
a = 69.61.
Solving
sin B sin C
for c, we get
b sin C
log c = log 6 -f log sin C log sin B.
log 6= 1.3010
log sin C = 9.9236 - 10
11.2246-10
log sin B = 9.6126 - 10
logc= 1.7120
c = 61.62.
Check :
A + C = 161,
C)= 80 80',
a c = 8.09 ;
A - C = 47 ;
Here,
or,
tan (A - C) = - - tan (A + C),
log tan (A -C) = log (a - c) + log tan (A -f C) - log(a + c).
log(a-c)= 0.9079
log tan ( A + C ) = 10.7764-10
11.6843 - 10
log (a + c) = 2.0468
logtan(A-C)= 9.6386-10
which substantially agrees with the above results.
im sin 10* sin (180 - 104)= sin 76. Hence log Bin 104- log Bin 76"= 9.9869 - 10.
160
PLANE TRIGONOMETRY
EXAMPLES
Solve the following oblique triangles, using logarithmic Tables I and II.
No.
GIVEN PARTS
REQUIRED PARTS
1
a=10
,4=88
J5=7710'
C=6460 X
6=15.837
c=14.708
2
a=795
,4 = 79 69'
#=4441'
(7=65 20'
6=567.6
c=664
3
6=.8037
J?=6220'
C=10140'
^4 = 26
a =.446
c=.9942
4
c = .032
J.=368'
=4427'
C=9925'
a = . 01913
6=. 02272
5
6=29.01
vi = 8740'
(7=3315'
B =59 6'
a =33. 78
c=18.64
6
a =804
^1 = 99 65'
7?=451'
C=354'
6=577.3
c = 468.9
7
a =400
.4 = 54 28'
0=60
J5=6532'
6=447.4
c = 426.7
8
c=161
-4 = 86 16'
C= 12339 /
J5=216'
a=111.6
6 = 69.62
9
a = 5. 42
I?=4217.3'
C = 8228.4'
J. = 6514.3'
6=4.439
c = 6.542
10
6 = 2056
^. = 63 62. 8'
=70
C=467.2'
a= 1964.7
c= 1677.3
11
a=7.86
.B=322'52"
O=4325'26"
A = 104 31' 42"
6 = 4.309
c=6.583
12
6=8
J. = 80
/*=215'46"
0=97 44' 14"
a = 199. 63
c= 200.73
Solve the following oblique triangles, using logarithmic Tables I and III.
No.
GIVEN PARTS
REQUIRED PARTS
13
a
= 500
.4 = 10.2
B
= 46.6
=
123.2
6 =
2051
c
= 2363
14
a
= 46
A = 36.8
C
= 62
B
81.2
6 =
74.25
c
= 66.33
15
6
= .085
= 95.6
C
= 24.2
A =
60.2
a =
.0741
c
= .035
16
6
= 5686
I? = 48. 63
C
= 83.26
A =
48.11
a =
6640
c
= 7523
17
c
= 7
.4 = 60.58
c
= 60
B =
60.42
a =
6.971
6
= 7.03
18
c
= .0059
5 = 75
c
= 36.87
A =
68.13
a =
.00913
6
= .0096
19
a
= 76.08
J* = 126
c
= 12.44
A =
41.56
6 =
92.8
c
= 24.7
20
a
= 22
,4 = 3.486
B
= 73
C =
103 514
6 =
346
c
= 351.8
21
6
= 8000
A = 24. 5
B
= 86.495
C
69.006
a =
3324
c
= 7483
22
6
= 129.38
A= 19.42
C
= 64
B =
96.58
a =
43.29
c
= 117.05
23
c
= 96
A = 2.086
B
= 112
SI
65.914
a =
3.788
6
= 96.5
24
6
= 132.6
^4 = 1
C
= 75
7* =
104
a =
2.385
c
= 131.98
25. A ship S can be seen from each of two points A and B on the shore.
By measurement AB = 800 ft., angle SAB = 67 43 X , and angle SBA = 74 21'.
Find the distance of the ship froni A. Ans. 1268 ft.
26. Two observers 6 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 66 and 68 respectively. Find the distances of the balloon from
the observers. Ans. 4.607 mi.; 4.45 mi.
27. One diagonal of a parallelogram is 11.237, and it makes the angles 19 V
and 42 64' with the sides. Find the sides. Ans. 4.16 and 8.67.
28. To determine the distance of a hostile fort A from a place B, a line
BC and the angles ABC and BCA were measured and found to be 822.6yd.,
60 84', 56 10' respectively. Find the distance A B. Ana. 800 yd.
THEOKY AND USE OF LOGARITHMS 161
29. From points A and B at the bow and stern of a ship respectively, the
foremast, 0, of another ship is observed. The points A and B are 300 ft.
apart, and the angles ABC and BAG are found to be 66.46 and 112.86 respec-
tively. What is the distance between the points A and C of the two ships ?
Ans. 9254ft.
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. -Find the distance from the lighthouse
to the ship in each position. Ans. G.143 mi. and 8.792 mi.
31. In a trapezoid the parallel sides are 16 and 7, and the angles one of them
makes with the nonparallel sides are 70 and 40. Find the nonparallel sides.
Ans. 8 and 6.47.
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 I, calculate log sinB.
If log sinB = 0, sinB = 1, B = 90 ; 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 < and b > a, the acute value of B found from the table
and also its supplement, should be used ; and there are two solutions.^.
Second step. Find C (one or two values according as we have one
or two values of B) from
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.
step. Solving __ = - f or sin B,
a
or, log sin B = log 6 -f log sin A log a.
log 6= 1.9031
log sin A = 9.6716 - 10
11.6747 - 10
loga= 1.6563
log sin B = 10.0 184- 10
= 0.0184.
Since log sin B > 0, sinB > 1 (which is impossible), and there is no solution.
In this connection the student should read over 58, pp. 104, 105.
t For if 6<a, B must be less than A, and hence B must be acute.
% Since b>a, A must be acute, and hence B may be either acute or obtuse.
162
PLANE TRIGONOMETRY
Ex. 2. Having given a = 7.42, b = 3.89, A = 106; solve the triangle.
Solution. Draw figure.
First step. From law of sines,
b sin A
or,
6=3.39
sin B
a
log sin B = log 6 + log sin A log a.
log 6= 0.6302
log sin A = 9.9849-10*
10.5161-10
log a = 0.8704
logsinJ5 = 9.6447-10
.-. B = 26ir. Using Table II
Since log sin B < and 6 < a, there is only one solution.
Second step. C - 180 - (A + B) = 180 - 131 ir = 48 49'.
Third step. By law of sines,
a sin C
sin A
or, log c = log a -f log sin C log sin^t.
loga= 0.8704
log sin C = 9.8766 - 10
10.7470 - 10
log sin A = 9.9849 - 10
logc= 0.7621
.-. c= 6.783.
Check : Use law of tangents.
tan J(C - B) = ^j-| tan ( C 4- B),
or, log tan (C - J5) = log (c - 6) + log tan J (C + J?) - log (c -f &).
Substituting, we find that this equation is satisfied.
Ex. 3. Given a = 732, 6 = 1016, 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
or,
Bl
log sin B = log 6 -f log sin .4 log a.
log 6= 3.0066
log sin A = 9.8081 - 10
= 12.8146 -10
loga= 2.8646
log sin B- 9.9601-10
j. .-! J* f
w -^o*?.
* Sin ^ = sin 105 = sin (180 - 105) - sin 75. Hence log sin A m log sin 75 - 9.9849 - 10.
THEORY AND USE OF LOGARITHMS
168
Since log sin B < and b > a, we have two solutions, which test verifies our
construction. From Table II we find the first value of B to be
B l = 63 3'.
Hence the second value of B is
Second step.
Third step. From law of sines,
2 = 180- J?! = 116 p7'.
Ci = 180 - ( A + BJ = 180 - 103 3' = 76 67' ;
C 2 = 180 - (A -f # 2 ) = 180 - 166 67' = 23 3'.
a sin Ci
or,
log Ci = log a 4- log sin Ci log sin -4.
loga = 2.8645
log sin C\ = 9.9886 10
12.8531 - 10
log sin A = 9.8081 -10
logci = 3". 0460
In the same manner, from
we get
= 1109.3.
_ a sin C 2
- -
sin A
c-b
Check : Use tan (C - #) = tan J (G + B) for both solutions.
EXAMPLES
Solve the following oblique triangles, using logarithmic Tables I and II.
No.
GIVEN PARTS
KBQUIRED PARTS
1
a=50
c = 66 A = 12&>1!'
Impossible
2
a = 5.08
6 = 3.59 A =63 50'
1> > =3921 /
C = 7649' c = 6.611
3
a = 62.2
6=74.8 ^1 = 27 18'
7? 1 = 3328 /
Cirrll^U 7 Ci=118.32
7? 2 = 14632 /
C 2 = 610' C 2 = 14.667
4
6=.2337
c = .1982 #=109
^ = 174r
C = 5319' a =.07608
5
a =107
c=171 C=3153 /
^1 = 19 18 /
B =128 49' 6 = 262.2
6
6=3069
c=1223 C=6552'
Impossible
7
6=6.161
c=6.84 J5=443 /
^! = 6847'
C! = 6710' ai=6.92
^ 2 = 237 /
C 2 =11260' 02=2.913
8
a=8.666
c=10 ^L = 6967 /
J5=303 /
C=90 6=6.009
9
a=214.66
6 =284. 79 B =104 20'
A = 46 63'
C=2847' c=141.6
10
a =32. 16
c=27.08 C=6224 /
^1 = 70 12'
J5 1 =6724' 6 1= 28.79
^2=109 48'
B 2 = 1748' 62 = 10.46
11
6=811.3
c= 606.4 5=1266 / 20 //
^1 = 16 44' 40"
0=37 10' a=289.2
164 PLANE TRIGONOMETRY
Solve the following oblique triangles, using logarithmic Tables I and III.
No.
GIVEN PARTS
REQUIRED PARTS
12
a = 840 6 = 486 ^=21.6
# = 12.21 C = 146.29 c = 1272
13
a = 72.63 6 = 117.48 ^1 = 80
Impossible
14
a = 177 6 = 216 .A = 86.6
B! = 46.27 Ci = 99.13 Cl = 300.3
# 2 = 134.73 <7 2 = 9.67 c 2 = 61.09
15
6 = 9.399 c = 9.197 #=120.4
^1 = 2.02 (7=67.68 a =.3841
16
6 = .048 c = .0621 5 = 57.02
Impossible
17
6 = 19 c = 18 (7=15.8
A l = 147.6 J?i = 16.7 ai = 35.62
A 2 = 0.9 # 2 = 163.3 a 2 = 1.038S
18
a = 55.65 c= 66.66 = 77.7
^1 = 64.6 # = 47.8 6 = 50.54
19
a = 34 c = 22 (7 = 30.35
4i = 51.87 #1 = 98.28 6 1 = 43.07
A 2 = 128.63 # 2 = 21.02 6 2 = 15.612
20
a = 628 6 = 262 .A = 124.6
# = 23.14 (7 = 32.26 c = 342.3
21
6 = 91.06 c = 77.04 B = 61. 12
^. = 87.69 (7 = 41.19 a = 116.88
22
a = 17,060 6 = 14,050 B = 40
AI = 51.32 <7i = 88.68 c t = 21,86C
4 8 = 128.08 02 = 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 # to (7 is 145 ft., from A to C is 178 ft., and the angle
ABC is 41 10'. Find the distance from A to #. Ans. 269.4 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 h
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)fromA+B=18Q--C l
Second step. From law of tangents,
we find ^(A #). Adding this result to ^(A + J5) gives A, and sub-
tracting it gives B.
Third step. To find side c use law of sines; for instance)
a sin C
c = :
sin A
Check : Check 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 6, c, A are given, use
* From law of sines.
tan-OB-C)- --la:
2 b + c *
a b c
sin A sin B sin C
THEORY AND USE OF LOGARITHMS
165
Ex. 1. Having given a = 640, 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 =640
6 = 420
a + & = 960
640
420
a-& = 120
180
C= 62 V
=: 127 64'
= 63 67'.
Second step.
420
or,
Adding,
Third step.
log tan (A - B) = log (a - b) + log tan J (A + B) - log (a + 6)
log (a -6)= 2.0792
log tan \ (A + B) = 10.3108 - 10
12.3900 - 10
log (a + 6)= 2.9823
log tan (A - B) = 9.4077 - 10
-5) = 14 21'
63 67'
14 2r
Subtracting, B= 49 36'.
(A - B) = 14 21'
A = 78 18'.
asinC
From
sin C
a
sin A
log c = log a + log sin C log sin A.
loga= 2.7324
log sin C = 9.8971 - 10
12.0296-10
log sin A = 9.9909-10
logc= 2.6386
/. c = 436.1.
Check : By law of sines,
log a = 12.7324 - 10 log 6 = 12.6232 - 10
log sin^L = 9.9909 - 10 log sinl? = 9.8817 -10
2.7416 2.7416
log c = 12. 6386 -10
log sin C = 9.8971-10
2.7416
Ex. 2. Having given a = 167, c = 82, B = 98 ; solve the triangle, using log
arithms from Tables I and III.
Solution. First step.
a = 167
167
ISO
c= 82
82
B= 98
a + c = 249
a-c= 86
A + C= 82
" it-
4 + C)= 41.
166
PLANE TRIGONOMETKY
Second step.
or,
- C) = log(a - c) + log tan %(A + C) - log(a -f c)
log(a-c) = 1.9294
log tan (A + C) = 9.9392 - 10
11.8686 - 10
log(a 4- c) = 2.3962
log tan $ (A - C) = 9.4724 - 10
) = 41.00
Adding,
T%ird ttcp.
6 =
A = 67.63.
a sin I?
sin A
41.00
16.63
Subtracting, 0=24.47.
6 a
from
sin B sin A
log 6 = log a -f log sin B log sin A.
loga= 2.2227
log sin B = 9.9958-10 *
12.2186 - 10
log sin A = 9.9262 10
log 6= 2.2923
.-. b = 196.
Check : By law of sines,
loga = 12.2227 - 10 log 6 = 12.2923 - 10
log sin^l = 9.9262-10 log sin B = 9.9958-10
2.2966 2.2966
which substantially agree.
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.
1
GIVEN PARTS
REQUIRED PARTS
a=27 c=15 =46
^1 = 100 67' 0=33 3' 6=rl9.78
2
a=486 6=347 C=6136'
^=83 15' # = 45 9' c = 383.6
3
6=2.302 c=3.567 .A = 62
J5=3916 / 0=78 44' a=3.211
4
a=77.99 6=83.39 0=72 16'
^1 = 51 14.5' jB=5629.5 / c=96.24
5
a=0.917 6=0.312 0=33 7.2 X
^1 = 132 18.4' #=14 33.4' c=.6775
6
a=.3 6=.363 C=12456 /
A = 24 41.8' JB=3022.2' c = .5886
7
6=1192.1 c=356.3 ^. = 2616 /
1^=143 29' (7=10 15' a= 886.6
8
a=7.4 c=11.439 #=8226'
^1 = 35 2' (7=62 32' 6=12.777
9
a=53.27 6=41.61 C=7833 /
A = 69 16.5' J3=4210.6' c=60.74
10
6=.02668 c=.06092 ^ = 11647'
B=211.2' 0=43 11.8' a=.06699
11
12
a= 61.88 c=67.94 J5=7912 / 36 //
6=V5 c=V ^l=8668 /
^4 = 40 52.7' C = 6954.7' 6=77.12
B=9328.5 / 0=60 38.6' a= 1.313
gin JB- tin 98 - sin (180 - 98)- sin 82. . . log sin 98 - log sin 82< = 9.9968 - 10.
THEOBY 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 =
69.3
A =
77.2
B =
43.6
c
= 14.99
14
a
= 65.14
b =
33.09 C =
30.4
A =
117.4
B =
82.2
c
= 31.48
15
b
= 101
c =
158 A =
37.38
B =
38.26
C =
104.36
a
= 99.04
16
a
= 101
b =
29 C =
32.18'
A =
136.4
B =
11.42
c
= 78
17
c
= 45
b =
29 A =
42.8
B =
39.72
C =
97.48
a
= 80.84
18
a
= .086
c =
.0042 B =
66.6
A =
121.07
C =
2.43
b
= .08276
19
b
= .9486
c =
.8862 A =
84.6
B =
49.88
C =
45.52
a
= 1.286
20
b
= 6
c =
9 A-
88.9
B =
34.03
c =
67.07
a
= 10.72
21
a
= 12
6 =
19 C =
6.24
A =
8.84
B =
165.92
c
= 7.132
22
a
= 42,930
c =
73,480 B =
24.8
A =
27.56
C =
127.64
b
= 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 = 3826 yd., CB =
8476 yd., together with the angle ACB = 62 31', were measured. What is
the distance A B ? 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 80 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 6 and 6 and form an angle of
49 18'. Find the sides. Ans. 6.004 and 2.839.
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
u4CBis87.72. Find the distance AB. Ans. 903ft.
27. Two stations A and B on opposite sides of a mountain are both visible
from a third station C. The distances A (7, BC, and the angle ACB were meas-
ured and found to be 11.6 mi., 9.4 mi., and 69 3r respectively. Find the dis-
tance from A to B. Ans. 10.636 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 66.8'. 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. 162.33 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, 6, c are given.
First step. Calculate s = (a + b -f c), s a, s b, s c.
Second step. Find log r from
r =
a) (s b) (s c)
(84) to (87), p. 115
168 PLANE TRIGONOMETRY
Third step. Find angles A, B, C from
T T T
tan kA = ? tan \B = -> tan 1C =
2 s a 3 s b 2 s c
Check : See if A + B + C = jf0 .
Ex. 1. Having given a = 61, 6 = 66, 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 = 61
Hence
b = 65
8 = 68
s = 68
8 = 68
c= 20
a =61
6 = 66
c = 20
2s = 136
8-a = 17
8-6= 3
s - c = 48
s = 68.
Second step.
i-a)(8-b)(8-c)
_ ,
logr = [log(s - a) + log(s - b) + log (a - c) - logs].
From the table of logarithms,
TTwrd step. From the formula tan A =
log (s -a) = 1.2304
log(s- 6) =0.4771
log (s -c) = 1.6812
3.3887
log s = 1.8325
2 | 1.6662
logr = 0.7781
r
s a
log tan J. = logr - log (a a).
logr = 10. 7781 -10
log(s - a) = 1.2304
log tan .4= 9.6477-10
using Table II*
or,
A = 88 54'.
From the formula tan J B =
log tan"^ JB = logr - log (a 6).
logr = 10. 7781 -10
log(s - 6) = 0.4771
log tan J- B = 10.3010 - 10
= 63 26',
J5 = 12662 / .
using Table II
* If we use Table III instead, we get
i A 1 19.44, $
and w438.88,
: ^ + B H- C= 179.98.
63.43,
126.86,
} C= 7.12,
C 14.24.
THEORY AND USE OF LOGARITHMS 169
From the formula tan ^ C =
log tan J C = log r log (s c).
logr= 10.7781 -10
log(s - c) = 1.6812
logtanC= 9.0969-10
= 14 16'. Using Table II
Check: A- 38 64'
B = 126 62'
C = 14 16'
A + B + C = 180 2'
EXAMPLES
Solve the following oblique triangles, using logarithmic Tables I and II.
No.
1
GIVEN PARTS
REQUIRED PARTS
a = 2 6 = 3 c = 4
A = 28 68' B = 46 34' C = 104 28'
2
a = 2.6 6 = 2.79 c = 2.33
A = 67 38' B = 70 28' C = 61 64'
3
a = 6.6 6 = 4.3 c = 4.9
A = 74 40' B = 47 46' C = 67 84'
4
a = 111 6 = 146 c = 40
A = 27 20' 5 = 143 8' C = 9 32'
5
a = 79.3 6 = 94.2 c = 66.9
A = 66 66' B = 79 44' C = 44 20'
6
a = 321 6 = 361 c = 402
J. = 49 24' B = 68 38' C = 71 68'
7
a =.641 6 =.629 c = .702
A = 60 62' B = 46 6' (7 = 73 2'
8
a = 3.019 6 = 6.731 c = 4.228
^4 = 18 12' B = 136 62' C = 26 66'
9
a =.8706 6 = .0916 c = .7902
A = 149 60' B = 3 2' C = 27 10'
10
a = 73 6 = 82 c = 91
A = 49 34' = 68 46' C = 7 1 38'
11
a = 1.9 6 = 3.4 c = 4.9
4 = 16 26' I? = 30 24' C = 133 10'
12
a =.21 6 = .26 c = .31
^L = 42 6' B =66 6' = 81 48'
13
14
a = 618.4 6 = 726.8 c = 931.3
a=V6 6=V6 c=V7
A = 33 16' B = 60 56' C = 96 48'
A = 61 62' B = 69 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
A
= 34.78
5 = 86.42
C =
68.82
16
a
= 43
6 =
60
c =
67
A
= 46.82
= 67.98
c-
76.18
17
a
= .23
6 =
.26
c =
.198
A
= 68.44
B = 74.38
c =
47.18
18
a
= 61.3
6 =
84.7
c =
47.6
A
= 46.2
B = 101.38
c =
38.44
19
a
= .0291
6 =
.0184
c =
.0368
A
= 64.06
B = 30.8
c =
96.16
20
a
= 706
6 =
662
c =
639
A
= 71.66
J5 = 49.14
c =
69.32
21
a
= 66
6 =
43
c =
49
A
= 74.68
5 = 47.78
n
67.66
22
a
= 301.9
6 =
673.1
c =
422.8
A
= 18.2
5=186.86
c =
26.94
23
a
= 2.61
6 =
2.79
c =
2.33
A
= 57.88
B = 70.3
c =
61.84
24
a
= 80
6 =
90
c =
100
A
= 49.46
B = 68.76
c =
71.78
1TO 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 18 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 6 ft. from one end of the object and 8 ft. from the other end ?
Ans. 60.
28. The distances between three cities, -4, JB, 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 C from A ? Ans. N. 4 24' W. or S. 4 24' W.
29. Three towns, A, JB, and C, are connected by straight roads. AB = 4 mi.,
BC = 6 mi., AC = 7 mi. Find the angle made by the roads AB and BC.
Ans. 101.56.
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 13,681 ft. from one end of a wall 12,342 ft. long, and
10,026 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 cases.
CASE I. When two sides and the included angle are given, use one
of the formulas
, ^ ab sin C be sin A
(88) S = __, S= __,
where S = area of the triangle.
Ex. 1. Given a = 25.6, 6 = 38.2, C = 41 66' ; find the area of the triangle.
, .. a a&sinC
Solution. 8 =
log S = log a -f log b + log sin C log 2.
loga= 1.4082
log 6= 1.5821
log sin C = 9.8249 - 10
. 12.8152 - 10
log 2= 0.3010
log 8= 12.5142 -10
= 2.5142.
.-. 8 = 326.8. Ans.
CASE II. When the three sides are given, use formula
(89) s=V s ( s -)(s-6)(5-c),
where S = area of the triangle,
and s=*a + b + c.
THEORY AND USE OF LOGARITHMS 171
Ex. 2. Find the area of a triangle, having given a = 12.63, b = 24.9, c = 18.91.
Solution, a = 12.63 Hence
6 = 24.9 5 = 28.17 8 = 28.17 = 28.17
c = 18.91 a= 12.63 ft = 24.9 c= 18.91
25 = 56.34 8- a = 15.64 8-6= 3.27 8-c= 9.26
* = 28.17.
q -
logS = | [logs + log(3 - a) + log(s - 6) + log(s - c)],
logs = 1.4498
log (s- a) = 1.1942
log (s- b) = 0.6145
log (-c) = 0.9666
2 1 4. 1261
log 5 = 2.0626
.-. 5 = 116.6. Ans.
CASE III. Area 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,
__ c sin A
sinC
log a = log c -f log sin A log sin (7.
logc= 1.8941
log sin A= 9.7617-10
11.6458-10
log sin C = 9.9926-10
loga= 1.6532
Now it comes under Case I.
__
_
log 5 = log a -f logc -f logsinJB Iog2.
loga= 1.6682
logc= 1.8941
logsinS = 9.9614 - 10
13.6087 - 10
Iog2= 0.3010
log 8 = 13.2077 -10
= 3.2077
/. 3= 1613.8. 4n,
172
PLANE TEIGONOMETEY
EXAMPLES
Find the areas of the following oblique triangles, using Tables I and II foi
the first ten and Tables I and III for the rest.
No.
GIVEN PARTS
AREA
1
a = 38
c = 61.2
B = 67 66'
1078
2
b = 2.07
.4 = 70
B = 36 23'
3.267
3
6= 116.1
c = 100
A = 118 16'
6113
4
a = 3. 123
A = 63 ir
J?=1367'
1.364
5
b = .439
A = 76 38'
C = 40 36'
.0686
6
a = .3228
c = .9082
5 = 60 16'
.1273
7
c = 80.26
B = 100 6'
C = 31 44'
4494
8
a = .010168
6 =.018226
O = 11 18.4'
.000018166
9
a = 18.063
J. = 96 30'
J5 = 36
70.66
10
b = 142.8
c = 89.6
a = 96
4174
11
a= 100
B = 60.26
C = 64.6
3891
12
a= 146
6=178
# = 41.17
12,383
13
a = 886
6=747
C= 71.9
314,600
14
a =266
6 = 862
C = 73
44,770
15
a = 960
6 = 720
C = 26.67
149,730
16
a = 79
6 = 94
c = 67
2604
17
a = 23.1
6 = 19.7
c = 26.2
216.9
18
a = 6.82
6 = 6
c = 4.26
11.733
19. The sides of a field ^#CDare J.#=37rd., J5C = 63rd., and 1^4 = 20 rd.,
and the diagonals A C and BD are 76 rd. and 42 rd., respectively. Required the
area of the field. 1670 sq. rd.
20. In a field ABCD the sides AB, BC, CD, and DA are 166 rd., 236 rd.,
262 rd., and I06rd., respectively, and the length from A to C is 311 rd. Find
ttie 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 E. 10 chains to B, thence
N.E. by E. 8 chains to C, thence S. 6 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
ABCD = area trapezoid* GCDE - (area trapezoid GCBF -f area triangle
FBA -f area triangle ADE) = 13.9 acres. Ans.
2. A surveyor measures S. 60 25' E. 6. 04 chains, thence
S. 68 10' W. 4.16 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 16' 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 V E. 6.15 chains ; 8.78 acres.
4. From station No. 1 to station No. 2 is S. 7 20' W.
4.67 chains, thence to station No. 3 S. 61 66' W. 7.06
chains, thence to station No. 4 N. 3 10' E. 6.06 chains,
thence to station No. 6 N. 33 W E. 3.25 chains, thence
to station No. 1. Find the direction and distance of
station No. 1 from No. 6, 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
(North pole)
P
1)
geographical t 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 difference in longitude
* From Geometry the area of a trapezoid equals one halt the sum of the parallel sides
times the altitude. Thus, area GCDE **(GC + ED) GE.
t The departure hetween two meridians is the arc of a parallel of latitude comprehended
between those meridians. It diminishes as the distance from the equator increases.
$ A geographical mile or knot is the length of an arc of one minute on a great circle oi
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,
&TCAB DA DA ,,.,-
777; = TT^T = TTT = cos OAD = coaAOE = cos latitude.
arc-EJQ OE OA
.-. arc AB = arcJEJQ cos latitude, or,
departure
(90) Diff. long. =
cos latitude
EXAMPLES
1. A ship whose position is lat. 26 20' N. , long. 36 10' W. sails due west 140
knots. Find the longitude of the place reached.
Solution. Here departure = 140,
and latitude = 25 20' N.
Substituting in above formula (90),
diff. long. = - 14
cos 25 20'
log 140 = 12. 1461 -10
log cos 25 20' = 9.9561 - 10
log difl long. = 2.1900
diff. long. = 154.9' = 2 34.9'.
Hence longitude of place reached = 36 10' + 2 34.9' = 38 44.9' W. Ans.
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 65' W.
3. A vessel in lat. 44 49' S., long. 119 42' E., sails due west until it reaches
long. 117 16' E. Find 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.
6. 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. Ans. 168.9 knots.
86. Plane sailing. When a ship sails in such a manner as to cross
(North pole) successive meridians at the same angle,
p j.t is said to sail on a rhumb line. This
angle is called the course, and the distance
between two places is measured on a
rhumb line. Thus, in the figure, if a
ship travels from A to B on a rhumb line,
arc.4jB= distance,
angle CAB = course,
arc CB = departure,
arc A C = difference in latitude
between A and B.
THEORY AND USE OF LOGARITHMS 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
Depart.
arid C
AC=ABcosA ; or,
(91) Departure = distance x sin course, and
(92) Diff. lat. = distance x cos course. S
If A B 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 46' S., on a course N. 36 E. 346 geographical mi.
Find the latitude reached and the departure made.
Solution. Here distance = 346 and course = 36.
.-. departure = 346 sin 36. diff. lat. = 346 cos 36.
log 346 = 2. 6378 log 346 = 2. 6378
log sin 36 = 9.7692-10 log cos 36 = 9.9080 - 10
log departure = 2.3070 log diff. lat. = 2.4468
.-. departure = 202.8 mi. Ans. diff. lat. = 279. 1' = 4 39. 1'.
As the ship is sailing in a northerly direction she will have reached latitude
8 46' -4 39.1' =4 6.9' S. Ans.
2. A ship sails from lat. 32 18' N., on a course between N. and W., a dis-
tance of 844 mi., and a departure of 103 mi. Find the course and the latitude
reached. ' Ans. Course N. 17 26' W., lat. 37 46' N.
3. A ship sails from lat. 43 46' S., on a course N. by E. 2346 mi. Find the
latitude reached and the departure made.
Ans. Lat. 6 26' S., departure = 467.6 mi.
4. A ship sails on a course between S. and E. 244 mi., leaving lat. 2 62' S.,
and reaching lat. 6 8' S. Find the course and the departure.
Ans. Course S. 66 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 61' 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 #, we first find diff. in lat. from (92). This gives
4 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
i, - D art -- -" have reached latitude = 4230'-4 9.4'=3820.6'N. Ans.
To get the longitude of 7>* 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 = (42 30' + 38 20. 6') = 40 26.3'.
Substituting in (93), diff. long.
log 166.7= 12.2219-10 l
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
= 68 51' - 3 39' = 55 12' W. A ns.
2. A vessel in lat. 2616'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. 800 mi.
Find the place reached. Ans. Lat. 38 21' N. ; long. 56 12' W.
THEORY AND USE OF LOGARITHMS 177
5. A ship sails from a position lat. 49 66' N M long. 16 16' W M to another
lat. 47 18' N., long. 20 10' W. Find the course and distance.
Ans. Course, S. 50 53' W. ; distance = 250.6 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. 3216'W., steams N. 3666 / W., and
reaches lat. 41 N. Find the distance steamed and the longitude of the position
reached. Ans. Distance = 300.3 mi.; long. 30 8' W.
7. A ship in lat. 42 30' N., long. 68 51' W., sails S.E. until her departure is
163 mi. and her latitude 38 22' N. Find her course and distance and the longi-
tude of the position reached.
Ans. Course, S. 33 W E. ; distance = 296.7 mi. ; long. 66 17' W.
8. A cruiser in lat. 4744'N., long. 32 44' W., steams 171 mi. N.E. until her
latitude is 60 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 OR 90
88. When the angle x approaches the limit zero, each of the ratios
sin* tan* . _ ,. .. . . ,. . ,
- , - , approaches unity as a limit, x being the circular measure
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 f , and
let PT and P'T be the tangents drawn to the circle at P and P'.
From Geometry
(A) PQP' < PAP 1 < PTP'.
But PQP' = PQ + QP f =: 2 sin x in numerical value,
PAP' = PA -f AP' = 2 x in numerical value,
and PTP 1 = PT 4- !TP f = 2 tan # in numerical value.
Substituting in (A).
2 sin x < 2 x < 2 tan aj.
Dividing through by 2, we have
(B) sinjc < JIT < tana:,
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
sm x cos x
If we now let x approach the limit zero, it is seen that
limit _JL_
Xtss sin a?
must lie between the constant 1 and
178
which is also 1.
ACUTE ANGLES NEAK OE 90 179
Hence ^j* -? = 1, or,
(C) limit 2i!i = L
v / x=0 x
Similarly, if we divide (2?) through by tan x, we get
cos x < < 1.
tanx
As before, if x approaches zero as a limit,
limit tana;
must lie between the constant 1 and lit cos ^ which is also 1.
Hence 'i-i't^hr 1 ' ''
(D) limit ^ = i.
\ / JT=0 X
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 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 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
and .3(= 18 f ) 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, thfe 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 TlilGONOMETBY
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.
From 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 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 #,*
tanx = circular measure ofx 9
cot x = . >
circular measure of x
cos x is found from the tables in the usual way.^f
The following equivalents may be used for reducing an angle to circular measure
(radians), and in other computations.
I-. ^ radians.
10-0.0174633 radians, log 0.0174533 - 8.2419 -10.
1' * 0.0002909 radians, log 0.0002909 6.4637 - 10.
I" = 0.0000048 radians, log 0.0000048 = 4.6856 - 10.
~p 57.29578 - 1 radian, Iog57.29578 1.7581.
T 3.14159 log ir = 0.4971.
=8 y approximately.
t esc x and sec x are simply the reciprocals of sin x and cos x respectively.
ACUTE ANGLES NEAK Oli 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."\
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'= - = 81.838. 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 = 0.00789 radians.
Therefore cos 89 34.6' = 0.00739,
cot 89 34. 6' =0.00739,
tan 89 84. 6' = - = 185. 32. Ana.
0.00739
When the function of a positive acute angle near or 90 is
given, to find the angle itself we reverse the process illustrated
above. For instance:
Ex. 3. Find the angle subtended by a man 6 ft. tall at a distance of 1225 ft.
Solution.
From the figure tanx = T
But, since the angle is very small, we may replace tan x by x, giving
x = T / 7 y radians = 0.0049 radians.
Or, reducing the angle to minutes of arc, we get
0.0049
0.0002909
minutes of arc = 16.8'. Ana.
* 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).
t ceo* and sec x are simply the reciprocals of sin x and cos x respectively.
182 PLANE TKIGOKOMETEY
92. Rules for finding; the logarithms of the functions of angles near
and 90.* For use in logarithmic computations the 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 -f log a.f
log tan x = 2.2419 -f log x.
log cot x = 1.7581 - log x.\
log cosx 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 -f log (90 x).
log tan x = 1.7581 log (90 x).
log sin x is found from the tables in the usual way.
Ex. 1. Find log tan 0.046.
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 -f 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
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 a?,
and , log sin x = log 0.017453 -f logo?
-2.2419 + log*.
, From p. 180, rt ^._^_,
and log cot x = - log 0.017453 - log x
-1.7581 -log x.
ACUTE ANGLES NEAK OR 90 183
Ex. 2. Find log tan 89. 93 6.
Solution. From the above rule,
log tan 89.936 = 1.7581 - log (90 - 89.936)
= 1.7681 -lpgO.066
= 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. 936 =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 TEIGONOMETEY
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 ? Ans. 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.
Ans. 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. 868,200 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. 238,838 mi.
5. Given that the radius of the earth is 8968 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 f in. ? Ans. 76.6 in.
7. Find the angle subtended by a circular target 5 ft. in diameter at the
distance of half a mile. Ana. 6' 80.6".
MISCELLANEOUS EXAMPLES
1. A balloon is at a height of 2600 ft. above a plain and its angle of elevation
at a point in the plain is 40 86'. How far is this point from the balloon ?
Ans. 3843ft.
2. A tower standing on a horizontal plain subtends an angle of 37 19.6' at
a point in the plain distant 369.6 ft. from the foot of the tower. Find the height
of the tower. Ana. 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 88.2. Find the height of the tower.
Ana. 115.8ft.
4. From the top of a lighthouse 112.6 ft. high, the angles of depression of
two ships, when the line joining the ships passes through the foot of the light-
house, are 27.8 and 20.6 respectively. Find the distance between the ships.
Ana. 81 ft.
5. From the top of a cliff the angles of depression of the top and bottom of a
lighthouse 97.26 ft. high are observed to be 23 17' and 24 19' respectively.
How much higher is the cliff than the lighthouse ? Ana. 1947 ft
6. The angle of elevation of a balloon from a station due south of it is
47 18.6', 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. 1000ft
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 16 with the street. Find
the breadth of the street. Ans. 84.24ft.
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 80 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
MI the road. Ans. 2.89 mi., 1.87 mi.
9. Standing at a certain point, I observe the elevation of a house to be 46,
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. 64.94ft.
10. A hill is inclined 86 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. 117.68yd.
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 80 mi. from A, and
C on the other road distant 16 mi. from A. Find the distance from B to C.
Ans. 26.98 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 JB. 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
B is 9. 166 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 46
respectively. Show that the distance of the spire from A is 8.346 mi.
15. If the spire C in the last example stands on a hill, and its angle of ele-
vation at A is 16, 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 46 and 90 respectively and the angle DAG is 46, 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 88 19'; at B the
angle CBA is observed to be 182 42', and the angle of elevation of C at B is
10 W. Show that the top of the mountain is 1248.7 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 BAG and CBA are
observed to be 66 87' and 68 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 rock is observed to be 47, and after walking
1000 ft. towards the rock, up a slope inclined at an angle of 32 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 30 and
46 respectively, while that subtended by the opposite side is 30. Show that
the lengths of the sides are 160 ft., 86.6 ft., and 106.8 ft. respectively.
22. A person goes 70 yd. up a slope of 1 in 3 J 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 ft respectively. Prove that the
, . T . , . h tan ft , . h sin 8 - cos a .
height of the tower is - ft., i.e. 1 ft.
tana tan ft sin (a ft)
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 346 and
290 ft. Find the length of the lake. Ans. 266.8 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 ft respectively. Show that the
distance between the ships is h (cot ft cot a) 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 36 14' 10" ; the angular
distance of the second ship from the fort, observed from the first ship, is 42 11'
63". Find the distance in feet from each ship to the fort.
Ans. 3121 ft., 3634 ft.
27. The angular elevation of a tower at a place due south of it is or, and at
another place due west of the first and distant d from it, the elevation is ft.
Prove that the height of the tower is
d . dBina-smft
zn:> 1.6.
Vcot 2 /3-cot 2 a Vsin(a-/3) .sin(
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 a, and CB
to 6, and join AB, Ab, and Ba. Measure AB, 600; a A, 100; aB, 560; 6J?,
100 ; and Ab, 650. Ana. 600 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 ft ; prove that the height of the post
, .. . . , . . A . , .sin 8> cos a
exceeds the height of the man by h
am (ft -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 D
can be seen, and measure CD, 200 ft.; ADC, 89; ACD, 60 30'. Then take
some point 18, whence D and B are visible, and measure DE, 200 ; BDE, 64 80' ;
BED, 88 30'. At D measure ADB, 72 30'. Compute the distance AB.
Ans. 345.4ft.
31. The angle of elevation of an inaccessible tower situated on a horizontal
plane is 0326 X ; at a point 600 ft. farther from the base of the tower the ele-
vation of its top is 82 14'. Find the height of the tower. Ans. 460.6 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; A CD, 63 80'; ACF,
64 31'; BDE, 64 30'; BDC, 166 26'; DEB, 88 30'. Ans. 346.3yd.
33. A tower is situated on the bank of a river. From the opposite bank the
angle of elevation of the tower is 00 13', and from a point 40 ft. more distant
the elevation is 50 19'. Find the breadth of the river. 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 46'. Find the height of the tower. Ans. 66.6 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. 61.6ft.
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 46 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 2 (* sin (/3 7)
sin ( a) sin (7 a)
40. The angular elevation of a tower du' 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
n
that the height of the tower is -
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
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 --
5 cot^l-fcot
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 b. 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 + b + c cos a
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 <*, and a ft. nearer to
the top of the hill to be ft ; show that if h be the height of the obelisk, the incli-
nation of the hill to the horizon will be
h sin(/3 a)J
CHAPTER X
RECAPITULATION OF FORMULAS
PLANE TRIGONOMETRY
Right triangles, pp. 2-11.
(1) sin A = - (4) csc J. = -
C CL
(2) cos A = - (5) sec A = \
v ' c v ' b
b
(3) tan A = - - (6) cot ,4 = -
O Cb _
(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 o the angle.
Fundamental relations between the functions, p. 59.
O
1
1
/OA\ ftf\a r _
cscx
1
sinaj
1
\+iv) COS X -
/O1 \ f nn o -
sec a;
1
cosaj
1
{"}-) tan a? -
/99^ f.n rr =
cot a? '
sin x
rrf: .^ -
tana?
cos x
Sill i
(23) sin 2 a; -f cos 2 a; = 1.
(24) sec 2 a; = 1 -f tan 2 ic. (25) csc 2 a; = 1 + cot 2 a;.
Functions of the sum and of the difference of two angles, pp. 63-69.
(40) sin (x + y) = sin a? cos y + cos x sin y.
(41) sin (x y) = sin x cos y cos x sin ^/.
(42) cos (a; 4- 1/) = cos a; cos y sin x sin y.
(43) cos (a; y) = cos x cos y 4- sin a; sin y.
180
190 PLANE TRIGONOMETRY
x ...
(44)
(45)
v '
, , . cot x cot
(46) cot (a? 4- y) = T~r
v ' v y/ coty 4- cotx
,,_. cot # cot y4-l
(47) cot (aj y) = - - - *-r --
v ' v *// cot y cot a?
Functions of twice an angle, p. 70.
(48) sin 2 x = 2 sin x cos #.
(49) cos 2x = cos 2 a? sin 2 a:.
,,~rvx , 2 tan x
(50) tan 2x = - - - -
v / 1 tan 2 x
Functions of an angle in terms of functions of half the angle, p. 7
X X
(51) sin x = 2 sin - cos -
Zt
(52) cos x = cos 2 ^ sin 2 -
2 tan
(53) tan x =
l-tan'f
Functions of half an angle, pp. 7273.
* ^ . ,. . x , I J. cos x x *T \ j ~~~ cos a?
(64) sin ^ =^ 2 ' ( 58 > tan 2 =
sn a?
XKCTX * _i_ 1-fCOSX X 1+COSa
(56) cos -\ 2 - (59) cot - = ^^^^
(66) tan| = Jf^^- (60) *f ^3LS25.
v J 2 M 1 4- cos x v 7 2 sin #
sinx ,_. x a: sin a;
,~^ /n +. ,
(57) tan ~ = r * (61) cot - = -
v J 214- cos x v / 2 1
cos a?
Sums and differences of functions, p. 74.
(62) sin A 4- sin B = 2 sin (4 -f- J5) cos
(63) sin .4 sin B = 2 cos (^ -f ^) sin %(A B).
(64) cos A 4- cos JB = 2 cos (.4 4- #)cos (4 5).
(65) cos.4 cos B = 2 sin
sin A 4- sniff _ tan %(A + B)
^ ' sin A sin B "" tan ^ ^4 ff
RECAPITULATION OF FORMULAS 191
Law of sines, p. 102.
(72) a = b = C
^ * sin A smB sinC
Law of cosines, p. 109.
(78) a 2 = b 2 + c 2 - 2 fo cos ^4.
Law of tangents, p. 112.
a -\-b tan
a b tan (^4 B)
Functions of the half angles of a triangle in terms of the sides,
pp. 113-115. s = l(a + b + c).
(81) sin i A =
(82) cosM = \(^F-^-
i be
(83) tani^ = >J^
*(* - a)
(84) r = > J('-)('j*Klll^.
/>*
(85) tan %A=-
s
(86) tan | ^ = -_
(87) tan^C^^
Area of a triangle, p. 117.
(88) S =
(89) S = Vs (s a) (s ft) (s - c).
SPHERICAL TRIGONOMETRY
CHAPTER 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 O 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", C'OA' 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. Prom any property of triedral angles an analogous prop-
erty of spherical triangles can be inferred, and vice versa.
193
194
SPHEEICAL 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 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 A DEB 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 A DEB = 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. t
* 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
all 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 q -ft', C = 180 - c',
A ' = 180 - #, B' = 180 - ft, 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. Ans. a' = 110, V = 100, c' = 80.
(b) A = 66, B = 97, C = 112.
(c) A = 68 14', B = 52 10', C = 98 44'.
(d) A = 115.6, = 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, K' = 128, C" = 80.
(b) a = 74 42', b = 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. From 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 0. Let C be the right angle, and suppose first that each of the
other elements i s less than 90, the radius of the sphere being unity.
* The polar triangle of any spherical triangle is constructed by describing arcs of great
circles about the vertices of the original triangle as poles.
196
SPHERICAL TRIGONOMETBY
Pass an auxiliary plane through B perpendicular to OA, cutting
OA at E and OC at D. Draw BE, BD, and DE. BE and DE are
each perpendicular to OA ;
[If a straight line is J_ to a plane, it is JL to every line in the plane.]
therefore angle BED = angle A. The plane BDE is perpendicular to
the plane A OC ; straight line is J. to a plane, every plane!
[passed through the line is _L to the first plane, j
hence BD, which is the intersection of the planes BDE and BOC, is
perpendicular to the plane AOC, ** ^
rlf two intersecting planes are each _L to a third"!
[plane, their intersection is also _L to that plane, j
and therefore perpendicular to OC and DE.
In triangle EOD, remembering that angle EOD = b, we have
OE
OD
= cos b,
But
and
or, clearing of fractions,
(A) OE = OD-cosb.
OE = cos c (= cos EOB),
OD = cos a (= cos DOB).
Substituting in (.4), we get
(1) cos c = cos a cos b.
In triangle BED, remembering that angle BED = angle A, we have
BD
5i
or, clearing of fractions,
(B) BD
But J3D = sin a (= sin DOB),
and BE = sin c (= sin
EIGHT 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 = sine sin B. .
Again, in the triangle BED,
JDE
(C) cos.4=-.
But DE = OD sin b, from sin b = -~~
OD = cos a (= cos DOB),
and BE = sin c (= sin EOB).
Substituting in (C),
, . OD sin 5 sin b
(D) cos ^4 = : = cos a
v sine sine
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 b 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 :
From (4), cos A = cos a sin B
__ cos c sin b
cos b sine
[Since cos a*= ^ from (1), and sin J5= from (3).l
L cost sine J
__ sin b cos c
cos b sin c
(6) .'. cos A = tan b cot c.
Similarly, we may get
(7) cos B = tan a cot c.
(8) sin b = tan a cot A.
(9) sin a = tan & cot B.
(10) cos c = cot ^1 cotB
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,
(E)
But
and
OF
cos DOE (= cos b) = -
OE = cos EOB = cosAOB = -~ cos c,
OD = cos DOB = cos COB = cos a.
Substituting in (E), we get
7 cos c
cos b = ? or cos c = cos a cos b,
cos a '
which is the same as (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
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 cos c, 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 oot/i greater than 90, the hypotenuse
EIGHT SPHEEICAL TE1ANGLES 199
is less than 90 ; if one side is less and the other is greater than 90,
the hypotenuse is (jr eater than 90.
^ ,,x i ,*v r, COS ,4 , . t CQSB
Jrom (4) and (5), smJ5 = ? and sin A = -
x 7 v ' cos a cos
Since A and B are less than 180, sin A and sin B must 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 splierical 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
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 c9 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 ^ fl 2^
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 e are
the opposite parts.
200
SPHERICAL TRIGONOMETRY
Kapler'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 Eule I, gin ^ = tan ^ tan ^
c * or, cos c = cot A cot B ;
A 9 B 9 which agrees with (10), p. 197.
By Rule II, sin c c = cos a cos b,
b 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.
So
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, we 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."
RIGHT 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 Mule I if the other two parts are adjacent to the
middle part (as in case illustrated}, or Rule II if they are opposite,
and solve for the unknown 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 = 33 60', a = 108.
Solution. Follow the above general directions.
To find A
Using Rule II
sin AC cos B c cos a
cos A = sin B cos a
log sin JB = 9. 7467
log cos a = 9.4900 (n)
log cos A = 9.2867 (n)
and
A = 09 64'.
To find b
Using Rule I
sin a = tan B c tan 6
tan 6= sin a tan JB
log sin a = 9. 9782
log tan B= 9.8263
log tan b = 9.8046
.-. b = 82 31'.
To find c
c c
b a
Using Rule I
sinJ5 c = tan c c tan a
cote = cos .Boot a
log cos B = 9. 9194
log cot a = 9. 5118 (n)
log cote = 9. 4312 (n)
/. 180 - c = 74 64'
and c = 106 6 X .
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 being opposite parts, we
get sin 6 = cos c cos JS, or, sin b = sin c sin B.
t For example, as in the table, we will write logjsin 24* 9.6003. To be exact, this should
be written log sin 24 0.6003 - 10, or, log sin 24 1.6003.
$ Since cos .4 is negative, we get the supplement of A from the table.
202
SPHERICAL TRIGONOMETRY
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
C?
sin AC = tan b tan c
^ jj o cos A = tan b cot c
"*~* log tan 6= 9.8045
log cot c = 9.4312 (n)
b_ a log cos A = 9.2367 (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 a =
(n)
log sin a =
log tan B =
log cos A =
.-. 180 -A =
and A =
(n)
log tan b =
.: b =
log cos B =
log cot a =
log cot c =
... 180- c =
and c =
(n)
(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
c c
Using Rule II
sin a = cos c c cos A c
sin a = sin c sin A
log sin c = 9.9743
log sin ^ = 9. 9934
log sin a = 9. 9677
.-. 180-at = 6810 /
and a = 111 50'.
To find b
c c
Using Rule I
sin^4 c =: tan 6 tan c c
tan b = cos A tan c
log "bos A = 9.2397 (n)
log tan c = 0.4609
log tan 6 = 9. 6906 (n)
.-. 180-& = 268'
and b = 163 62'.
To find B
c.
Using Rule I
sin c c = tan A c tan B c
cotB cose tan A
log cose = 9.6286
log tan ^1 = 0. 7637 (n)
log cot = 0.2772 (n)
.-. 180- =27 61'
and B = 162 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,Ve~^hould 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 60' . Since A > 90, however, we know from
Th. II, p. 199, that a > 90 ; hence a 111 60' is the only solution. *
RIGHT 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 A and CB
(= 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 1 = 90, A 1 = 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- ft, vl'JSC = 180 4JBC.
This ambiguity also appears in the solution of the triangle, as is
illustrated in the following example :
Ex. 3. Solve the right spherical triangle, having given A = 106 59', a =
128 33'.
Solution. We proceed as in the previous examples.
Tofindb To find B
^
a
sin 6 = tan a tan A c
sin b = tan a cot A
log tan a = 0.0986 (n)
log cot A = 9.4670 (n)
log sin 6 = 9.6666
.-. 6 = 214 / , or,
180 -6 =158 66' = 6'.*
sin AC = cos a cos B c
. _ cos^.
Sln B =
cos a
log cos A = 9.4399 (n)
log cos a = 9.7946 (n)
log sin B = 9. (5463
.-. B - 26 14', or,
To find c
sine =
sin a = cosmic cos c c
sin a
sin A
log sin a = 9.8932
= 9.9828
log sine = 9.9104
.-, c'= 64 27', or,
180-c / =126 33 / =c.$
Hence the two solutions are :
1. 6 = 21 4', c = 126 33',
B = 26 14' (triangle ABC) ;
B' = 163 46' (triangle A'BC).
2. 6' = 168 66', cf = 64 27',
It is not necessary to check both solutions. We leave this to the student.
* Since sinj? 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 must be retained, since b has two values which are supplementary-
{ 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
SFHEBICAL TKIGOSTOMETRY
EXAMPLES
Solve the following right spherical triangles :
No.
GIVEN PARTS
REQUIRED PARTS
1
a = 132 0' 6 = 77 51'
^. = 131 27' = 80 56' c = 987'
2
a = 159 c = 137 20'
A = 148 5' B = 66 23' b = 87 64'
3
A = 60 20' B = 122 40'
a = 40 42' 6 = 134 31' c = 122 7'
4
a = 160 b = 38 30'
A = 149 41' B = 60 44' c =* 137 20'
5
B = 80 b = 67 40'
A = 27 12' a = 25 26' c = 69 64' ; or,
J.'=15248' a'=15435' c'= 110 6'
6
B = 112 c = 81 50'
A = 109 23' a = 110 58' b = 113 22'
7
a = 61 B = 123 40'
^1 = 66 12' b = 127 17' c = 1076'
^8
a = 61 40' & = 144 10'
.4=72 29' =140 38' c = 11238'
9
J. = 99 60' a = 112
B = 27 7' & = 25 24' c = 109 46' j or,
'= 152 53' 6'= 154 36' c'= 70 14'
10
6 = 16 c = 15220'
A = 120 44' a = 156 30' = 33 63'
>11
A = 62 69' B = 37 4'
a = 41 6' 6 = 26 25' c = 47 32'
12
J. = 73 7' c = 114 32'
a = 60 31' B = 143 50' 6 = 147 32'
13
J5 = 144 64' b = 146 32'
A = 78 47' a = 70 10' c = 106 28'; or,
A'= 101 13' a'= 109 60' c'= 73 32'
14
J? = 68 18' c = 47 34'
A = 30 32' a = 22 1' b = 43 18'
15
A = 161 62' 6 = 131 8'
a = 166 9' J5 = 10149' c = 60 18'
16
a = 113 25' 6 = 110 47'
^1 = 112 3' 5 = 109 12' c = 8164'
>17
a = 137 9' B = 74 61'
^ = 135 3' 6 = 68 17V c = 105 44'
18
A = 144 64' B = 101 14'
a = 146 33' 6 = 109 48' c = 73 35'
19
a = 69 18' c = 84 27'
A = 70 B = 75 6' b= 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 = 67 88', 6 = 48 60 / .
Solution. This is a quadrantal triangle since one side c = 90. We then find
the corresponding elements of its polar triangle by (/), p. 196. They are C'=90,
A' = 112 22 X , J?' = 131 10'. We solve this right triangle in the usual way.
RIGHT SPHERICAL TRIANGLES
205
Construct the polar (right) triangle.
Given A' = 112 22', & = 181 KT:
To find a'
Co'
&'
B' c
= cosa'cosJSc
, cos A'
cos a' = - - .
sm'
log cos ^'-9. 6804 (n)
log sin '=9. 8767
log cos a' =9. 7037 (n)
180 -a' =69 38'.
a' = 120 22'.
Similarly, we get
&' = 135 23', c' = 68 65'.
Hence in the given quadrantal tri-
angle we have
A = 180 - a' = 69 38',
B = 180 - V - 44 37',
C = 180 - c' = 111 6'.
EXAMPLES
Salve the following quadrantal triangles :
No.
GIVEN PARTS
REQUIRED PARTS
1
2
3
4
5
A = 139 6 = 143 c = 90
A = 45 30' = 139 20' c = 90
a = 30 20' (7 = 42 40' c = 90
# = 70 12' C = 106 25' c = 90
A = 106 53' a = 104 64' c = 90
a = 117l' B = 153 42' C= 132 34'
a = 57 22' 6 = 129 42' C = 67 53'
< 4=201 / #=141 30' 6 = 113 17'
.4 = 33 28' a - 35 4' 6 = 78 47'
B = 69 16' 6 = 70 C = 84 30'; or
#=110 44' 6=110 C = 9530'
Solve the following isosceles spherical triangles :
No.
GIVEN
PARTS
REQUIRED PARTS
6
a = 64
20'
c =
72 64'
^
= B
b
= 64
20' A = B = 57
69' C
= 93 69'
7
a = 64
30'
c =
71
A
= B
b
= 54
30' A = B = 67
30' c
= 66 26'
8
,. _ fJAO OQ' A _ JD 60 1 7'
b
= 66
2g' Q ui 3Q'
C
= 12842 ;
9
c=16640 / 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 (C = 90):
10. cos 2 A sin 2 c = sin (c + a) sin (c a). 13. sin (6 + c) = 2 cos 2 % A cos 6 sin c.
11. tan a cos c = sin b cot B. 14, sin (c b) = 2 sin 2 J J. cos 6 yin c.
12. sin 2 4 =s 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 9
--c-
upon AB (first figure) or upon AB produced (second figure). For
the sake of brevity let CD = p, AD = n, BD = m, angle A CD = x,
angle BCD = y.
In the right triangle ADC (either figure)
(A) sinp = sin b sin A.
In the right triangle BCD (first figure)
(B) sinjt? = sin a sin B.
This also holds true in the second figure, for
sin DEC = sin (180 B) = sin B.
Equating the values of sinp from (A) and (J3),
sin a sinB = sin b sin .4,
or, dividing through by sin A sin 5,
sin a sin b
Eule II, p. 200
Rule II, p. 200
(C)
sin A
206
OBLIQUE SPHEKICAL TRIANGLES 207
In like manner, by drawing perpendiculars from A and B, we get
. . sin b sine .
(D) - - = - - ? and
v ' smC
sine sin a A . .
respectzyely.
Writing, (C), (D), (K) as a single statement, we get the law of sines,
^ sin a sin b sin c
' '
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
= cosp cos(c n)
= cosp \ cos c cos n + sin c sin n \
(^4) = cos p cos c cos n -|- cos^? sin c sin n.
In the right triangle A DC,
(B) cosp cos n = cos b.
TTTl. COS *
Whence cos p = - >
COS7&
and, multiplying both sides by sin n,
(C) cos p sin n = -- sin n = cos b tan n.
v ' cos 7&
But cos A = tan ^ cot /;, or, Rule I, p. 200
(D) tan n = tan b cos ^4.
Substituting value of tanw- from (D) in (C), we have
(2?) cos^? sin n = cos b tan # cos ^4 = sin b cos A.
Substituting the value of cos p cos n from (B) and the value of
cosjt) sin^lrom (E) in (^4), we get the law of cosines,
(.F) cos a = cos & cos c -f sin 5 sin c cos -4.
* Compare with the law of sines in Qranville's Plane Trigonometry, p. 102.
208 SPHEKICAL TKIGONOMETKY
Similarly, for the sides I and c we may obtain
(<7) cos b = cos c cos a -f sin c sin a cos B,
(H) cos c = cos a cos b -f- sin a sin & 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 1 ,
and angles by A\ J3', 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), (6r),
(H) of the last section give us the following law of cosines for the
polar triangle :
(^4) cos a' cos b 1 cos c' -f sin b' sin c 1 cos A '.
(B) cos b' = cos c' cos a f -f sin c' sin a' cos B'.
(C) cos c' = cos a' cos b 1 + si n tt ' si n ^' cos C* f
But by (/), p. 195,
a' = 180-^, ft' = 180 -5, c' = 180-C,
^' = 180 -a, 5' = 180 -ft, C f = 180-c.
Making these substitutions in (.4), (J5), (C), which refer to the
polar triangle, we get
(D) cos (180 -4) = cos (180 -) cos (180 - C)
+ sin (180 - B) sin (180 - C) cos (180 - a),
(JE) cos (180 -5)= cos (180 -C) cos (180-^)
+ sin (180 - C) sin (180 - A) cos (180 - ft),
(F) cos (180 - C) = cos (180 - 4) 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 part may be replaced by the supplement of the opposite
party and the relation thus obtained will hold true.
OBLIQUE SPHERICAL TRIANGLES
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, B 9 y * ; that is,
or ^=
a J3 = -
= 180 -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-^),
replace b by /3 (= 180 ),
replace c by y (= 180- C),
replace a (= 180 -4) by 180- (180- a) = a,
replace j8 (= 180 - B) by 180 - (180 ft) = ft,
replace y (= 180 -C) by 180 - (180- c) = c,
or, what amounts to the same thing, interchange the Greek and
Roman letters. For instance, substitute
A =180 -a,
= lSO-p,
in (F), (<7), (H) of the last section. This gives tne law of cosines
for the sides in the new form
(12)
(13)
(14)
cos a = cos b cos c sin 6 sin c cos a,
cos ft = cos c cos a sin c sin a cos /?,
cos c = cos a cos ft sin a sin ft cos y.
[Since cos A = cos (180 a)= 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 ft cos y sin ft sin y cos a,
(16) cos ft = cos y cos a sin y sin a cos ft,
(17) cos y = cos a cos ft sin a sin /? cos c,
* or, 0, 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, ft, 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
.4 = 180- or, J5 = 180-/3, C = 180-y
in the law of sines (p. 207), we get
sin a __ sinb _ sin r._
sin a sin /3 sin y
[Since sin A sin (180 - <*)= sin a, etc.]
Applying the Principle of Duality to this relation, we get
sin a __ sin/? __ siny
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
(-4) 2 s = 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 Mobius (1790-1868).
OBLIQUE SPHERICAL TRIANGLES . 211
Subtracting 2 c from both sides of (A),
2s 2c = a + b + c 2c,OT.
(B) s-c = $(a + b-c).
Similarly,
(C) s b = (a b + c), and
(2>) s - a = i ( a + & + c) = J(& + c a).
From Plane Trigonometry,
() 2 sin 2 J a = 1 - cos a,
(F) 2 cos 2 i a: = 1 + cos a.
But from (12), p. 209, solving for cos a,
cos b cos c cos a
COS tf = - : : - J
sin & sin <?
hence (E) becomes
n . H cos & cos c cos a
2 sm 2 1 # = 1 - : T -
* sm b sin c
___ sin 6 sin c cos b cos c -f cos a
sin & sin c
__ cos a (cos ft cos c sin b sin c)
~~~ sin b sin c
cos fl ~ CQS (^ ~f- c)
sin ^> sin c
___ 2 sin %(a-}-l-{-c) sin J (a b c) *
~ i . > or
sin b sin c
(G)
x 7 J sin ^ s
[Since sin J(- 6 <)= sin ^(-a-f 6 + c)= -sin J(6 + c-a).]
Substituting from (^) and (D) in (G), we get
. , sin s sin (s a)
= - -~ - -
/-N
(18)
*Lt
or,
sm b sm c
|8ins8in(s a)
= '
A-B=a-b-c
c). $(A - /?)= i(a - b - c).
Henct, substituting in (05), p. 74, Granville's Plane Trigonometry, namely,
cos A - cos= - 2 sin l(A + E)8in $(A - B),
we get cos a - cos (6 + c) = - 2 sin (a + b + c) sin j (a - & - c).
212 SPHEEICAL TRIGONOMETRY
Similarly, (F) becomes
, . , cos b cos c cos a
2 cos 2 4 a = H -- -
2 sin sin c
- cos cos c cos &
sin b sin c
_ cos (b c) cos a
sin 6 sin c
2 sin ^ (a -f ft 0) sin %(b c a) *
"~~ sin b sin c !
* sin
[Since sin (6 - c - a) = sin (- I + c + a) = - sin (a - b + c).]
Substituting from (B) and (C) in (//), we get
_ sin (5 c) sin (5 b)
cos 2 la = - * : L-. * - L > or,
* sin b sin c
Since tan = - > we get from this, by substitution from (18)
and (19), COS5a
A+B=a+bc A-S=b-c-a
Let
Hence, substituting in formula (05), found on p. 74, Granville's Plane Trigonometry,
namely,
cos A * cos B* - 2 sin $(A + ) sin i(^l - 5),
we get cos(& - c)- cos a= - 2 sin J(a + 6 - c) sin J(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.
(6) 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,
s comes first and then the differences
a -a, *-6, -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 TKIANGLES 218
In like manner, we may get
,~-v .1/1 lain s sin (s 6)
(21) sinifi = -0 - : - \ - ^,
v ' ar M sine sin a
/~~v i I sin (s c) sin (5 a)
(22) **/*= N sincsina '
(23) .-i ,l 8ins8in(s-d)
Also
(24) -.-i )6ins8in(5-c)
,~^ 1 8in ( 5 fl ) 8in ( S '
(25) . cosjy -vl -^ '- ^
,^ N , c)
(26)
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
( sins sin 2 (s a)
tan i a = ^-r- N . , , N / , r
* > sin (s a) sin (s o) sm (s c)
. , . BITLS
= Sin (5 a) A -: ; - r~ : - - rr -. ; - r
v y Msm(s a)sm(,5 b)ain(s c)
Let
sm s
,, , . sin (5 a)
then tan i a = ^ r-^
2 tan ct
Similarly, for tan /? and tan y. Hence
(27)
> It may be shown that d = diameter of the circle inscribed in the spherical triangle.
214 SPHEEICAL TKIGONOMETEY
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 :
(40)
. [sin <r sin (<r a)
(31) 6in * a =N
sin/?siny
, . sin (a B) sin (a y)
(32) cosJ0=>l * ^ ^ 1 - L
sin ft sin y
, . , sin cr sin (cr a)
(33) tan \ a = \\-r-, - m \ , t
v ' Msm<r ?sincr
, - m , <
(<r /?)sin(cr y)
it |sin<rsin((r-
(34) sinj&=-0
v x X
sin y sin a
/^x 1 1 l ein (" y) sin (" a )
(35) cosj&=^- A *) \ . >
sin y sin a
/rtw . . , I sin O" sin (<r y)
(37) sin|c=Aj : \ r; .
x x > sin a sin ^
/ON i sin (" a ) 8in (^
(38) cos | c = AJ
sin a sin ft
where r = %(a + ft + y)
= j (180 - ^ + 180 - -f 180 - C)
What we have done amounts to interchanging the corresponding
Greek and lioman letters.
* It may be shown that 8 is the supplement of the diameter of the circumscribed circle.
OBLIQUE SPHERICAL TRIANGLES 215
14. Napier's analogies. Dividing (20) by (23), we gt
tan a 1 sin s sin (s
) .
tan % ft Nsin (5 ft) sin (s
->
cos or
sin 5 sin (5
a)
sin (s ft) sin (*
'-)
sinfr/3
cos
sin %a c
N
os^
sin 5 sin (5
*)
sin (s c) sin (s
ft sin (5 a)
-a)
or,
__
Hence .
cos a sin ^ p sin (s ft)
By composition and division, in proportion,
sin ^ a cos ^ /3 + cos ^ or sin ^ ft __ sin (s a) + sin (s ft)
sin a cos /3 cos a sin /3 ~~ sin (5 a) sin (* ft)
From (40), (41), p. 63, and (66), p. 74, Granville's Plane Trigo-
nometry, the left-hand member equals
and the right-hand member
sin (g a) + sin(s ft) _ tan j- [g a -f (g ft)] _ tan \c *
sin(5 a) sin (s ft) tan ^[s a (s ft)] tan (ft a)
Equating these results, we get, noting that tan (ft a) = tan ( ft),
sin ^ (<r -f jg) _ tan ^c
sini(or /3) ~" tan^(a ~ ft)' r>
In the same manner we may get the two similar formulas for
tan ^ (ft c) and tan (c a).
Multiplying (20) and (23), we get
sin s sin (s a) sin s sin (s ft)
tan a tan 0= AJ-^T - ,. \ . L . r X^-r - r-V~; ^- T >
2 a ^ 1 sin (s ft) sm (5 c) N sin (5 c) sin (s a)
sn asm/3 sin s
cos ^ a cos /? sin (s c)
By composition and division, in proportion,
cos ^ a cos ^ /? sin ^ # sin ^ /? ___ sin (s c) sin s
cos a cos /? + sin a sin /? ~~ sin (s c) -f sin s
* For -a + *-&=B2s-a-& = a + 6 + c a-6 = c, and s-a-* + 6
216 SPHEEICAL TRIGONOMETRY
From (42), (43), p. 63, and (66), p. 74, Granville's Plane Trigo-
nometry, the left-hand member equals
and the right-hand member
sin (s c) sin s __ tan ^ (s c s) __ tan |( c) *
sin (s c) -f sin 5 tan J (5 c -f- s) tan J (a + b)
Equating these results, we get, noting that tan ( c) = tan
cos ^ (g -f /?) __ tan ^c
cos -(<* - 0) tan J(a -f b) ' r '
(45)
In the same manner we may get the two similar formulas for
tan (b + c) and tan (c -f- a).
By making use of the Principle of Duality on p. 208, we get at once
from formulas (44) and (45),
()
(47)
By changing the letters in cyclic order we may at once write down
the corresponding formulas for tan (/? y), tan J (y a), tan J (/3-{-y),
and tan i (y + a).
The relations derived in this section are known as Napier's analogies.
Since cos J (a - b) and tan J y = tan fc ( 180 - 6') = tan (90 - i C)
= cot J- C are always positive, it follows from (47) that cos \(a -f- U)
and tan (a 4- /?) always have opposite signs ; or, since tan \ (a -f ft)
= tan (180 --4 -f 180 -5) = tan J. [360- (A + J5)] = tan~[180-
i (^l -f #)] = tan J (^4 -k -B), we may say that cos J- (a + b) and
tan |(yl -h 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 c-*~c,
and 8~c + s** 2-c
OBLIQUE SPHERICAL TRIANGLES
217
CASE I. (a) Given the three sides.
(b) Given the three angles.
CASE II. (a) Given two sides and their included angle.
(b) Given two angles and their included side.
CASE III. (a) Given two sides and the angle opposite one of them.
(b) Given two angles' and the side opposite one of them.
16. Case I. (a) Given the three sides. Use formulas from p.
lamely,
/sin (s a) sin (s b) sin (s c)
tand = \l - * * '-
(27)
(28)
(29)
(30)
ofinda, /3, y,
tan a =
sins
sin (s a)
tan \d
. sin (s b)
tan J == v / >
2 tan d
sin (5 c)
tan J. y = v _ x ,
2 ' tan d
therefore A, B, C, and check by the law of sines,
sin a __ sin b __ sin c
sin A sinJ? sin (7
Ex. I. Given a = 60, 6 = 137 20', c = 116; find A, B, C.
Solution.
a = 60
6 = 137 20'
c = 116
2 s = 313 20'
s = 166 40'.
s - a = 96 40'.
s - b = 19 20'.
s - c = 40 4(X.
To find log tan % d use (27)
log sin (s -a) = 9.9971
log sin (s- b) = 9.5199
log sin (s c) = 9.8140
29.3310
log sin = 9.6978
2 1 19.7382
log tan Jd= 9.8666
To find A use (28)
log sin (s- a) = 9.9971
logtan^tf = 9.8666
To find B use (29)
log sin (s- 6) = 9.5199
log tan ^d = 9.8666
To find C use (80)
log sin (s-c) = 9.8140
log tan ^d = 9.8666
log tan | a = 0.1305
\ a = 53 29'.
a = 106 68'.
.^=180 -10658'=732'.
Check: log sin a = 9.93'
log tan J/3 = 9.6533
= 48 28'.
.-. =180 -4828'=13132'.
75 log sin 6 = 9.8311
37 log sin B = 9. 8743
log tan ^ 7 = 9.9474
7 = 83 4'.
> ..C=180-83 4'=9666'.
log sine = 9.9637
log sin = 9. 9969
9.9568 9.9568 9.9668
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
SPHEEICAL TRIGONOMETRY
EXAMPLES
Solve the following oblique spherical triangles :
No.
GIVEN PARTS
REQUIRED PARTS
'l
a = 38
6=61
c = 42
A =51 68' J5 = 8364' C = 5863'
'2
a = 101
6 = 49
c = 60
A=U2 32' B = 2752' C = 3228'
> 3
a = 61
6 = 39
c = 92
.4=35 32' = 24 42'' C =138 24'
'4
a = 62 20'
6 =54 10'
c = 97 50'
^1=47 22' J5 = 4220' C = 12438'
5
a = 58
6 = 80
c = 96
A =65 58' JB=7414' 0=103 36'
6
a = 46 30'
b = 62 40'
c = 83 20'
^1=43 68' B = 58 14' (7=108 6'
7
a = 71 15'
b = 39 10'
c = 4036'
J.= 13036' j5 = 3026' (7 = 31 26'
8
a = 47 30'
b = 55 40'
c = 60 10'
A =66 32' J5 = 697' (7 = 78 68'
9
a = 43 30'
6 = 72 24'
c = 87 50'
J.= 4127' J5 = 6626' (7 = 106 3'
10
a =110 40
' 6 = 45 10'
c = 73 30'
^1=144 27' J5 = 269' C = 3635 /
17. Case I. (&) Given the three angles. Use formulas from p.
namely,*
(40)
tanj8~Y"< ff ~ g > 8in( ! r ~fl 8in(<ir "y>,
sin (or #)
sin<r
(41)
(42)
(43) < *-- tan 8
to find a, b, c ; and check by the law of sines,
sin a ___ sin b __ sin c
sin A sin B sin (7
Ex. 1. Given ^L = 70, B = 131 10', C y = 94 60'; find a, 6, c.
Solution. Here we use the supplements of the angles.
= 180- -4 = 110
^ = 180-^= 48 50'
y = 180- C = 85 10'
2<r=244
<r = 122. *
<r - a = 12. "
*-p= 73 10'. -
<r-y= 36 60'.
To find log tan % 5 use (40)
log sin (<r- a) = 9.3179 vx
log sin (<r - j8) = 9.9810 w
logsin((T - y) = 9.7778 *
29.0767 ;
logsin<r= 9.9284-
2 1 19. 1483
log tan $8 1= 9.6742 V
* These formulas may be written down at once from those used in Case I, (a), p. 217, by
glmply interchanging the corresponding Greek and Roman letters.
OBLIQUE SPHERICAL TRIANGLES
219
To find a use (41)
log sin (a- - a) = 9.3179
log tan 8 = 9. 5742
To find b use (48)
log sin (<r /3) = 9.9810
logtan5 = 9.5742
To find c use (48)
log sin (o- 7) = 9.7778
log tan 5 = 9. 6742
logtana = 9.7437
a = 29.
.-. a = 68.
Check : log sin a = 9.92
log sin ^1 = 9.97
log tan|6 = 0.4068
%b = 68 36'.
6 = 137 12'.
84 log sin 6 = 9.8321
30 log sin B = 9. 8767
log tan c = 0.2036
c = 5758'.
c = 115 56'.
log sine = 9.9539
log sin (7 = 9. 9985
9.9564 9.9554 9.9654
EXAMPLES
Solve the following oblique spherical triangles :
No.
GIVEN
PARTS
REQUIRED PARTS
I
A =
75
B =
82
ri
61
a
= 67 52'
b = 71 44'
c
= 57
*
A =
120
B =
130
o =
80
a
= 144 10'
b = 148 49'
c
= 41 44'
A =
91 10'
B =
85 40'
c =
72 30'
a
= 89 61'
b = 85 49'
c
= 72 32'
*4
^1=138 16' B =
31 11'
c
35 53'
a
= 100 5'
b = 49 59'
c
= 60 6'
5
A
78 40'
T>
63 50'
c =
46 20'
a
= 39 30'
b = 35 36'
c
= 27 69'
6
A-
121
B =
102
(1 __
68
a
= 130 50'
b = 120 18'
c
= 54 66'
7
A =
130
B =
110
c =
80
a
= 139 21'
b = 126 58'
c
= 66 62'
8
A =
28
B =
92
=
85 26'
a
= 27 66'
b = 85 40'
c
= 84 2'
9
A-
59 18'
B =
108
s~i
76 22'
a
= 61 44'
b =103 4'
c
= 84 32'
10
A =
100
B =
100
/-Y
50
a
= 112 14'
b = 112 14'
c
= 46 4'
18. Case II. (a) Given two sides and their included angle, as a,
C. Use formulas on p. 216, namely,
,,6,
to find a and {$ and therefore A and B ; and from, p. 215 use (44)
wived for tan -J- c, namely,
(44)
tan * c = -
^
c. Check by the law of sines.
Ex. 1. Given a = 64 24', 6 = 42 30', C = 68 40' ; find A , B, c.
Solution, y = 180 - C = 121 20'. .-. 7 = 60 40',
a= 64 24' a = 6424 7 ^
6= 42 30' 6 = 42 30V
a + b = 106 54 r
/. -J- (a + 6) = 63 27'.
a - 6 = 21 54'.
/. \ (a -6) = 10 57'.
220
SPHERICAL TRIGONOMETRY
To find | (a - /8) use (46)
log sin (a -6) = 0.2786
log tan j- 7 = 0.2603
9.5289
log sin j- (a + 6) = 9. 9049
log tan (a - /3) = 9.6240 (n)
To find %(a + p) use (47)
log cos (a- 6)= 9.9920
= 0.2603
10.2423
log cos(a -f &) = 9.7749
log tan % (a + ) = 0.4674 (n)
180 - (a + /3) = 71ll'.t
= 108 49'.
To find A and B
(a + /3)= 108 49 /
|(q~/3)=- 22 49'
Adding, a = 86
Subtracting, /3 = 131 38'.
.-. J. = 180 - a = 94.
.8 = 180 - = 48 22'.
To find c use (44)
log sin -J- (a + /3) = 9. 9761
log tan ^(a - 6) = 9.2867
19.2628
log sin % (a - 0) = 9.6886 (n)
log tan c= 9.6742}
c = 2617'.
.-. c = 60 34 X .
Check : log sin a = 9. 9561
log sin A = 9.9989
9.9662
log sin b = 9.8297 log sin c = 9.8878
log sin B = 9.8736 log sin C = 9.9316
9.9662 9.9663
If c 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
c / / \ 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
Applying Rule I, p. 200, to the right spherical triangle BCD, we
nave cos C = tan m cot a, or,
(A) tan m = tan a cos C.
Applying Rule II, p. 200, to BCD,
cos a s=s cos m cosp, or,
(JB) cos^> = cos a sec m.
Since tan(<r-0) is negative, $( 13) may be an angle in the second or fourth quad'
rant. But a > 6, therefore A > B and <r </3, since a and /3 are the supplements of A and B.
Hence |(*- ft) must be a negative angle numerically less than 90.
t Here ${* + 0) must be a positive angle less than 180. Since tanJ(r + /3) is negative,
}( + ft) must lie in the second quadrant, and we get its supplement from the table.
} tan } c is positive, since sin i(*-/3) is negative and there is a minus sign before the fraction.
OBLIQUE SPHERICAL TRIANGLES
Applying the same rule to ABD,
cos c = cos n cosp, or,
(C) cosjp = cos c sec n.
Equating (B) and (C),
cos c sec n = cos a sec m, or,
cos c = cos a sec w cos n.
But n = 6 m ; therefore
(D) cos c = cos a sec m cos (b m).
Now c may be computed from (A) and (D), namely,
(48) tan m = tan a cos C,
(49)
221
cos a cos (6 m)
cos c = i '-
cos m
Ex. 2. Given a = 98, b = 80, C = 110; find c.
Solution. Apply the method just explained.
To find b -m use (48)
log tan a = 0,8622 (n)
log cos C = 9.6841 (n)
log tan m = 0.3863
m = 6740'.
.-. 6 - m = 12 20'.
c use (49)
log cos a = 9. 1436 (n)
log cos (b - m) = 9.9899
19.1335
log cos m = 9.6798
log cose = 9.6637 (n)
180 - c = 69 2'.
c = 11068'.
EXAMPLES
Solve the following oblique spherical triangles :
No.
GIVEN PARTS
REQUIRED PARTS
J
6 = 137 20' c = 116 ^i = 70
B = 131 17' C = 94 48' a = 67 57'
2
a = 72 6 = 47 C = 33
A = 121 33' B = 40 67' c = 37 26'
d*
a = 98 c = 60 5 = 110
vl=87 C =60 61' b = 111 17'
4
b = 120 20' c = 70 40' A = 60
T> 1 QJ.O K 1 ?' r~* /SO*"* 4- 1 ' (L ~~~ fiQ^ ft'
5
a = 126 10' 6 = 163 60' C = 140 20'
.4 = 147 29' 5 = 163 9' c=768'
6
a =93 20' 6 = 66 80' = 74 40'
J. = 101 24' B = 64 68' c=7910 /
7
b = 76 30' c = 47 2<X A = 92 30'
B=7821' C = 4747' a = 82 42'
8
c = 40 20' a = 100 30' B = 46 40'
A = 131 29' C = 29 33' 6 = 72 40 7
9
b = 76 36' c = 1 10 26' A = 46 50'
J5 = 5738' C = 126 32' a = 57 8'
10
a = 84 23' 6 = 124 48' C = 62
A = 68 27' B = 129 53' c = 70 62'
222
SPHERICAL TRIGONOMETRY
19. Case II. (b) Given two angles and their included side, as A, B, c.
Use formulas * on pp. 215, 216, namely,
to find a and b ; and from p. 216, use (46) solved for tan | y, namely,
(46) taniv = - Sin ^ (ffi + ^ )taili(Q: ~^,
^ sin ^ (a b)
to find y and therefore C. Check by the law of sines.
Ex. 1. Given c = 116, J. = 70, B = 131 20' ; find a, 6, C.
Solution, a = 180 - A = 110, and = 180 - = 48 40'.
a =110
a = 110
ff = 48 40'
a + j8 = 168 40'
.-. J (a + 0) = 70 20'.
To./M(a<-&) wse (44)
log sin | (a - /3) = 9.7076
log tan ^c = 0.2042
9.9118
log sin % (a + /3) = 9.9924
log tan | (a - 6) = 9.9194 (n)
ff = 48 40'
a - = 6120'
(a - jS) = 30 40'.
c = 116.
To find a and b
I (a + b} = 97 39'
Adding,
Subtracting,
a = 57 66^
b = 137 22'.
To find (a + 6) use (45)
log cos -J- (a - ft) = 9.9846 J
log tan e = 0.2042
10.1388
log cos (a -f /3) = 9.2674
log tan \(a + b)= 0.8714 (n)
180- | (a -f-6) = 82 21'.
.-. \ (a + 6) = 97 39'.
To find C use (46)
log sin (a + b) = 9.9961
log tan (a - ) = 9.7730
19.7691
log sin -J- (a - 6) = 0.8066 (n)
log tan 7= 9.9636
7 = 42 36'.
7 = 86 12'.
Cfcecfc: logsina = 9.9281 logsinfc *= 9.8308 logsinc = 9.9637
log sin A = 9. 9730 log sin B = 9.8766 log sin C = 9. 9986
9.9561 9.9662 9.9652
* Same as those used in Case II, (a), p. 219, with Greek and Boman letters interchanged
t Since A<Bit follows that a < fc, and } (a b) is negative.
OBLIQUE SPHERICAL TEIANGLES
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 arc of a great
circle perpendicular to AC, intersecting
AC at D. Let BD = p, angle ABD=x,
angle CBD = y. Applying Rule I of
Napier's rules, p. 200, to the right spher-
ical triangle ABD, we have
cos c cot x cot A, or,
(A ) , cot x = tan A cos c.
Applying Eule II, p. 200, to ABD, we have
cos A = cos p sin x, or,
(J5) cosjp = cos A esc x.
Applying the same rule to CBD,
cos C = cos p sin y, or,
(C) cos jt? = cos C esc y.
Equating (B) and (C),
cos (7 esc y = cos A esc a?, or,
cos C = cos A esc # sin y.
But y = 5 x ; therefore
(Z>) , cos C = cos J. esc x sin (B x).
Now C may be computed from (A) and (D), namely,
(50) cot x = tan -4 cos c.
cos-4 sin(B jc)
= - -i - L .
(51)
cos C =
smjc
Ex. 2. Given A = 36 46', B = 116 9', c = 61 2' ; find C.
Solution. Apply the method just explained.
To find B-xuse (50)
log tan ^. = 9. 8576
log cos c = 9.7986
log cot a; = 9. 6561
x = 65 38'.
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 0= 9.8309
C = 47 21'.
224
SPHERICAL TRIGONOMETRY
EXAMPLES
Solve the following oblique spherical triangles :
No.
GIVEN PARTS
REQUIRED PARTS
'l
^1 = 67 30' # = 46 60' c=7420'
a = 63 15' 6 = 43 68' C = 96l'
'2
B = 98 30' C = 67 20' a = 60 40'
6 = 86 40' c = 6840' .4=59 44'
'3
= 110 ^1=94 6 = 44
a = 114 10' c = 120 46' # = 49 34'
4
C = 70 20' B = 43 50' a = 50 46'
6 = 32 59' c = 474,r .4 = 80 14'
5
^4=78 J5 = 41 c = 108
a = 95 38' 6 = 41 52' C = 11049'
fc6
J5 = 136 (7=60 a =70 20'
6 = 120 16' c = 69 20' ^1=60 26'
7
A =31 40' C=12220' 6 = 40 40'
a = 34 3' c'=6419' J5 = 3740'
8
A = 108 12' B= 145 46' c=12082'
a = 69 6' 6 = 146 26' C = 125 12'
9
A = 130 36' J5 = 30 26' c = 40 35'
a =71 15' b = 39 10' C = 3126'
10
A = 61 68' B - 83 64' c = 42
a = 38 b = 51 C = 68 63'
20. Case III. (a) Given two sides and the angle opposite one of them,
as a, bj B (ambiguous case *) .
From the law of sines, p. 207, we get
/<<N . A sin a sin B
(11) sm A = : -
N ' smfc
which gives A~\. To find C we use, from p. 216, formula (46), solved
for tan - y, namely,
(46) tnnlv- *>( + )***( -fl
v y 2r sin|(a 6)
To find c, oZve (44), jt?. 215, for tan^c, namely,
Check by the law of sines.
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
a=58 a=68
6= 137 20' 6= 137 20'
log sin B= 9.8766
a-f-&=19520' a-&=-7920'
1 / i "h\ nTO /i ri' 3 in t\\ ooo /i A^
19.8040
log sin 6= 9.8311
(tt-t-O) 97 40 . yvfl 0) o9 40 .
log sin A= 9.9729
80 -Ai=: 110 2'.
Since a<6 and both AI and A%
are < JB, it follows that we have two
solutions. *$
or, ^1
* Just as in the corresponding case in the solution of plane oblique triangles (Granville's
Plane Trigonometry, pp. 105, 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>&', and if a < b then A<B. Hence [next page]
OBLIQUE SPHERICAL TBIANGLES
First solution, ai = 180 - AI = 110 2'.
226
0= 48 40'
+ = 158 42'
/9 = 48 40'
- /5 = 61 22'
-# = 80041'.
To find Ci use (46)
log sin % (a + 6) = 9.9961
log tan (ai /3) = 9. 7733
To find Ci use (44)
log sin % (&I 4- /3) = 9.9924
log tan ^- (a - 6) = 9.9187 (n)
19.7694
log sin (a b) = 9.8050 (?i)
19.9111
log sin % (<*i - ft = 9. 7078
log tan -^7! = 9.9644
71 = 42 39'.
7i = 85 18'.
.-. Ci = 180 - 71 = 94 42'.
Check : log sin a = 9. 9284 log sin I
log sin A i = 9. 9729 log sin B
log tan ci = 10.2033
|c 1 =5767 / .
.-. ci = 116 64'.
= 9.8311 log sin Ci = 9. 9541
= 9.8766 log sin C x = 9.9985
9.9565 9.9555 9.9666
/Second solution. a z = 180 - A 2 = 69 68'.
a 2 = 69 58'
p= 4840 /
or 2 + ft = 118 88'
a 2 = 69 58'
/3 = 48 40 X
- p = 21 18 7
To find C z use (46)
log sin ^ (a + 6) = 9.9961
log tan (# 2 - j8) = 9.2743
19.2704
log sin | (a - 6) = 9.8050 (n)
log tan 72= 9.4654
72 = 32 34 X .
180 - 72 = 147 26 7 .
To find ci use (44)
log sin (a 2 + 0) = 9. 9345
log tan-J-(a - 6) = 9.9187 (n)
19.8532
log sin -J- (a 2 - /3) = 9.2667
logtan-J-c a = 10.6865
Check : log sin a = 9.9284
log sin A% = 9.9729
9.9555
log sin 6 = 9.8311
log sin B = 9. 8756
9.9555
.-. c 2 = 160 66'.
log sin c 2 = 9.6866
log sin C a = 9. 7810
9.9555
If the side <? 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, (&), pp. 220, 223.
Theorem. Only those values of A should be retained which are greater or less than 3
according as a is greater or less than 6.
# log sin A** a positive number, there wiil be no solution.
226
SPHERICAL TRIGONOMETRY
EXAMPLES
Solve the following oblique spherical triangles :
No.
GIVEN PARTS
REQUIRED PARTS
1
a=4320'
6 =48 30'
^t = 5840'
B 1 =6846'
Ci=7046'
c 1 = 4916 /
#2 = 111 14'
C 2 = 1429'
c 2 = ll36'
2
a=6640'
b =30 50'
A = 103 40'
=36 35'
C=52
c=4239'
3
a =30 20'
6=46 30'
^1=36 40'
#i=69 4'
Ci = 9739 /
Ci = 5657'
2 =120 56'
C 2 = 286'
C 2 = 2328'
4
&=9940'
c=6420'
B=9540'
C=6530'
^. = 97 20'
a =100 45'
5
a=40
6=118 20'
4 = 29 40'
5i = 4240'
C! = 15954'
ci = 16330 /
5 2 = 13720'
C 2 = 6021'
c 2 =9010 /
6
a=11620'
c=14620'
C=14110'
Impossible
7
a=10920'
c=82
^1 = 107 40'
C=90
J?=11335 /
6=11450 y
8
b =108 30'
c=4050'
0=39 60'
J5i=6818' .
4i = 13234'
ai = 13116'
J5 2 = lll42' -
4 2 = 775'
a 2 = 9660'
9
a=16220'
6 = 15 40'
B=125
Impossible
10
a =55
c = 138 10' A =42 30'
C=14638'
5=550 r
6 =96 34'
21. Case III. (2>) Given two angles and the side opposite one of them,
as Aj B, b (ambiguous case *).
Prom the law of sines, p. 207, we get
sin 4 shift
sin a =
: -
sin 5
which gives a.f To find c we use, from p. 215, the formula \ (44),
solved for tan^c, namely,
(44)
tall 5 c
5
To find C, solve (46), p. 216, for tan^y, namely,
Check by the law of sines.
* Just as in Case II, (5), we may have two solutions, one solution, or no solution, depending
on the given data.
f Since the side is here determined from its sine, it is necessary to examine both of the
values found. If A > E then a > b ; and if A < E 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 sina- 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 p = 180-J5 = 48 40'.
To find a use (11)
log sin A = 9.9730
log sin 6 = 9.8311
19.8041
logsinJB = 9.8756
log sin a = 9.9285
.-. ai = 58 1',
a 2 = 180 - ai = 121 59'.
a- 70
ft = 48 4Q/
a = 70
= 48 40'
First solution.
ai= 58 r
5 = 137 20'
6 = 195 21'
i use (44)
log sin -J- (a + j3) = 9. 9346
log tan 4 (ai - 6) = 9.9187 (n)
19.8533
log sin -J- (a - /3) = 9.2674
log tan ci = 10. 5859
c 1 = 75 27'.
.-. ci = 150 54'.
a + /8 = 118 40' a - /3 = 21 20'
(<* + ) = 69 20 X . (a-/3) = 10 40 X .
Since ^1<B and both ai and 02
are < 6, it follows that we have two
solutions.
ai= 58 r
6 = 137 20'
01 - 6 = - 79 IV
-6) = -39 40'.
To find Ci use (46)
log sin -J- (a! + b) = 9.9961
logtan|-(a - ft) = 9.2750
19.2711
log sin (ai - 6) = 9.8050 (n)
log tan 71= 9.4661
71 = 16 18'.
71 = 32 36'.
... Ci = 180 -7i = 147 24'.
Check: log sin ai = 9.9285 log sin b = 9.8311 log sin ci = 9.6869
log sin A = 9.9730 log sin 5 = 9.8756 log sin Ci = 9.7314
9.9555 9.9555 9.9565
Second solution. This gives c 2 = 64 8', and <7 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 b ~ 9.8311 log sin c 2 = 9.9542
log sin A = 9.9730 log sin B = 9.8756 log sin C 2 = 9.9985
9.9655 9.9565 9.9567
Hence the two solutions are
ai = 58 1' Ci = 150 64' d = 147 28',
md a 2 = 121 59' c 2 = 64 8' C 2 = 85 18'.
If the angle C or the side c is wanted without first computing a,
Tve may resolve the given triangle into two right triangles and then
ipply Napier's rules, as was illustrated under Cases II, (a), and
a, (5), pp. 220, 223.
228
SPHEEICAL TRIGONOMETRY
EXAMPLES
Solve the following oblique spherical triangles :
No.
GIVEN PARTS
REQUIRED PARTS
1
A = 108 40' J5= 134 20' a= 146 36'
6=154 45' c=349' 0=70 18'
2
=116 (7=80 c=84
6=114 50' ^ = 7920' a=8256'
3
^ = 132 5=140 6=127
ai=6724' Ci = 1646' d=160 6'
a a = 112 36' <7 a = 128 21' c 2 = 103 2'
4
^4=62 0=102 a=6430'
c=90 5=63 43' 6=66 26'
5
^1 = 133 SO' =66 30' a=8110 /
Impossible
6
=2220' C=14640' e=13820'
6=2722' ^L = 4721' a=117 9'
7
.4 = 61 40' C= 140 20' c=15020'
ai=433' 7?i=8924' 6 1 = 1298 /
a 2 = 1 36 57' E 2 = 26 59' 6 2 = 20 36'
8
5=73 C=8120' 6= 122 40'
Impossible
9
5=3620' C=4630' 6=42 12'
^x=16444' ai = 16238' ci=12441'
^2=119 17' a 2 =8117' c 2 =6519'
10
A = 110 10' B= 133 18' a=1476'
6= 155 5' c=3269' 0=70 16'
22. Length of an arc of a circle in linear units. From Geometry we
know that the length of an arc of a circle is to the circumference of
the circle as the number of degrees in the
arc is to 860. That is
L :27rR::N:360, or,
7TRN
~~ 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 JV, giving
(53)
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. = ^ of a degree) in statute miles will
be, from (52),
3.1416 x 3960 x^.
loO
Ex. 1. Find the length of an arc of 22 80' in a circle of radius 4 in.
Solution. Here N = 22 30' = 22.6, and R = 4 in.
i- ... .- /=v T 3.1416 x 4 x 22.5
Substituting in (52), L = -- = 1.67 in. Ana.
180
Ex. 2. A ship has sailed on a great circle for f>J- hr. at the rate of 12 statute
miles an hour. How many degrees are there in the arc passed over ?
Solution. Here L = 6 x 12 = 66 mi. , and R = 3960 mi.
1 QA y &a
Substituting in (53), ^= _____ = .965== 57.3'. A*.
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 720. That is,
Area of triangle : 4 7r.fi 2 : : E : 720, or,
(54) Area of spherical triangle =
loO
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,! namely,
(55) tan \ E = Vtanf stan \(s a)tan J(s 6)tanj(s c),
sphere a, b, c denote the sides and s = % (a -\- 1 + c).
The area of a spherical polygon will evidently be the sum of the
ireas of the spherical triangles formed by drawing arcs of great cir-
3les as diagonals of the polygon.
Ex. 1. The angles of a spherical triangle on a sphere of 26-in. radius are
<4 = 74 4<X, B = 67 30', C = 49 50'. Find the area of the triangle.
Solution. Here E = (A + B + C) - 180 = 12.
Substituting in (54), Area = 8 - 1416x Wxl2 = lg() g gq ^ An$
loU
* The spherical excess (usually denoted by E) of a spherical triangle is the excess of the
turn of the angles of the triangle over 180. Thus, if A, B, and C are the anglaa of a spher-
cal triangle.
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. 13.17yd.
3. How many degrees are there in a circular arc 16 in. long, if the radius
is 6 in.? Ans. 143 IS'.
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. 11.616 mi. per hour.
5. Find the perimeter in inches of a spherical triangle of sides 48, 126, 80,
on a sphere of radius 26 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. 63.86 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. , b = 8.39 in. , c = 9.43 in. ;
the radius of the sphere is 26 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 = 76 16', B = 39 20', c = 26 ft. ; the radius of
the sphere is 14 ft. Find the area of the triangle. Ans. 168.46 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. 41,060 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. 4661.1 sq. mi.
12. Find the areas of the following spherical triangles, having given
(a) a = 47 30', 6 = 56 40', c = 60 10'; R = 10 ft. Ans. 42.96 sq. ft.
(b) a = 43 30', b = 72 24', c = 87 60'; E = 10 in. 69.21 sq. in.
(c) A = 74 40', B = 67 &)', C = 49 60'; R = 100 yd. 2094 sq. yd.
(d) A = 112 30', B = 83 40', C = 70 10'; R = 26 cm. 941.4 sq. cm.
(e) a = 64 20', b = 42 30', C = 50 40'; E = 12 ft. 46.74 sq. ft.
(f) C - 110, A = 94, b = 44 ; B = 40 rd. 2056.5 sq. rd
(g) a = 43 20', b = 48 30', A = 68 40'; R = 100 rd. 19.76 acres,
(h) A = 108 40', B = 134 20', a = 145 36'; E = 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
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 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-
ng through Greenwich, near London. The meridians of Washington and Paris are also
used as reference meridians.
232
SPHERICAL TRIGONOMETRY
measured east or west from the Greenwich meridian from to 180.
Thus, in the figure, the arc M T 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 1 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
S
(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
EC of a great circle. We observe that this arc EC 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 = north latitude of Boston,
arc CN = 90 4- arc TC = 90 + south latitude of Cape Town,
and angle BNC angle MNL -f angle MNT
= west longitude of Boston
-f 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
J5C, 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 -f and south
latitude the sign .
Rule for finding the shortest distance between two points on the earth
_nd 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. %19, 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 85 and 90-(-42)90 + 42 s 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.
$ 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 JT
L 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. 4221'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 89^
6 = 90 - (_ 33 66') = 123 66'.
Second step.
N=714'+ 18 26' = 89 30' = difference in long.
Third step. Solving the triangle by Case II, (a), p. 219, we get
n = 08 14' = 68.23 = 4094 nautical miles,
C= 6243 / ,
and B = 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
courses. 6243'E.*
EXAMPLES
1. Find the shortest distance between Baltimore (lat. 39 17' N., long. 76 37'
W.) and Cape Town (lat. 33 60' S., long. 1826 / E.), and the bearing of each
from the other. Ans. Distance = 180 66 48' = 6862 nautical miles,
S. 64 68' E. = bearing of Cape Town from Baltimore,
N. 5742 / W. = bearing of Baltimore from Cape Town.
2. What is the distance from New York (lat. 4043'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 ? 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. 61 13' N., with course N. 66 64' 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. 4028'N., long. 74 1' W.) and Madeira (lat. 3228'N.,
long. 1666 / W.). Ans. 2749.
(c) San Francisco (lat. 37 48' N., long. 122 28' W.) and Batavia (lat. 6 9' S.,
long. 106 63' E.). Ans. 7616.
(d) San Francisco (lat. 3748'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 3. 63 17' *, she will never reach Cape Tovn.
APPLICATIONS OF SPHEEICAL TEIGONOMETEY 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. ,
N. 53 V E. = bearing of Greenwich from Boston,
N. 7139 / 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. 6' 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 65' 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 J 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 f 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 onlij 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 nebulae 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.
(Zenith)
z
(North _
celestial P/
.pole)
(North
poin
horizon)
-\A (South
}S point
of
horizon)
* '(South
celestial
pole)
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 1 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 TEIGONOMETEY 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 1 in
the figure).
The poles of the celestial equator are the points (P and P 1 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 l 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 TRIGOJSTOMETKY
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 MT (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
we shall now take up
the study of the more
important of these.
29. The horizon and
meridian system. In
this case the two fixed
and mutually perpen-
dicular great circles of
reference are the hori-
zon of the observer
(SHWNE) and his
meridian (SM$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 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
( Sunset)
* At sea the altitude is usually measured by the sextant, while on land 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.
From 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 M 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,
(a)
0>)
(c)
(d)
(e)
(0
(g)
(h)
(i)
Azimuth
45
60
90
120
180
225
300
315
178
Altitude
45
30
60
76
55
60
15
82
0)
(k)
0)
(m)
(n)
(o)
(P)
(q)
Azimuth
180
90
270
360
330
76
90
Altitude
90
46
75
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 1 on the eastern
horizon (altitude zero), mounts higher and higher in the sky, on a
circle (MiM 2 M z ) parallel to the celestial equator, until it reaches the
observer's meridian M z (at noon, when its altitude is a maximum),
then sinks downward to M 8 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.
1 That is, azimuth is measured from to 360 clockwise.
240
SPHERICAL TEIGOKOMETEY
( North
pole)
30. The equator and meridian system. In this case the two fixed
and mutually perpendicular great circles of reference are the celestial
equator (EQDWQ') 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 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 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 Ephemeria, published by the United
States government.
APPLICATIONS OF SPHEEICAL TRIGONOMETKY 241
to 24, one hour being equal to 15.* When the star is at M l (on
the observer's meridian) its hour angle is zero. Then the hour
angle increases until it becomes the angle MPM (when the star is
at M). When the star sets on the western horizon its hour angle
becomes M^M* Twelve hours after the star is at M 1 it will be at
If a, when its hour angle will be 180 (= 12 hours). Continuing on
its circuit, the star rises at M and finally reaches M ly when its hour
angle has become 360 (= 24 hours), or 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 (j) 00 S, 45
(b) 60 N. 60 (k)
(c) 90 S. 45 (1) 180
(d) 120 S. 30 (m) 90 N. 90
(e) 180 N. 60 (n) 270
(f ) 6 hr. N. 75 (o) 12 hr. S. 10
(g) 15 hr. -25 (p) Shr. +80
(h) 6 hr. + 79 (q) 9 hr. - 45
(i) hr. - 90 (r) 20 hr. -f 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 about 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
(To Zenith)
z
Q
Earth
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
0P 2 ) of the celestial north pole will be parallel to CP ly since the
celestial north pole is at an unlimited distance from the earth (see
27, p. 235). The angle NOP 2 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^ ; hence the
angles NOP Z 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 SPHEEICAL TEIGONOMETRY 243
Horizon
( Observer)
horizon on the plane of the observer's celestial meridian, the horizon
will be projected into a line (as NS), and the tipper 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-J 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 circumpolar
$tars. 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
64 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' -f 40 24' = 94 40'.
Therefore
94 40'
= 47 20' = altitude of north pole
= north latitude of place of observation.
244
SPHERICAL TRIGONOMETBY
^M(Star)
Horizon Q
( 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
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 -f- declination.*
If the star is on the meridian between the zenith and the pole (as
at M" |), we will have
North latitude = NP = ZQ = QM"~ ZM"
= declination zenith distance.
M(Star)
O
( Observer)
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 16' N. What was the latitude of the observer ?
* If the star is south of the celestial equator (as at M'}, the same rule 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 SPHERICAL TRIGONOMETRY 246
Solution. Draw the semicircle NZSO. Lay off the arc SM = altitude = 68 40',
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 16' 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.) -f M Q (dec.)
= 26 20' -f 21 16' = 47 36'.
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 (dee.)
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 M Z = co-alt. = 90 - 49 40' = 40 20 7 ,
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 61' = north latitude of place.
The angle MZP is found to be 27 68'; hence the azimuth of the star
(angle SZH) is 180 - 27 63' = 152 7'.
846
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
60 46'
30 40'
Ana. 40 43'
(b) Boston
44 22'
40 20'
42 21'
(c) New Haven
68 24'
24 KX
41 17'
(d) Greenwich
64 36'
38 22'
61 29'
(e) San Francisco
66 6'
20 30'
87 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'
48 27'
38 6'
50
28 46'
67
4r26'
72
22 18'
Declination
Body is
Latitude
N. 20
S. of zenith
Ans. 60 N.
N. 32 13'
S. of zenith
46 33' N.
S. 10 52'
S. of zenith
36 41' N.
S. 44 26'
S. of zenith
7 28' N.
N. 62
N. of zenith
22 N.
N. 73 1C'
N. of zenith
12 2' N.
S. 59
S. of zenith
36 S.
S. 81 48'
S. of zenith
87 14' S.
S. 8
N. of zenith
26 S.
N. 46 25'
N. of zenith
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
60
Ans. 27 2' N.
(b) Northern
16
S. 8
66
3638'N.
(c) Northern
62
N. 19
2hr.
48 16' N.
(d) Northern
64 42'
N. 24 20'
345
834'N.
or 46 86' N.
(e) Northern
8.6
6hr.
7122'N.
(f) Northern
26
21 hr.
5318'N.
(g) Northern
N. 11 14'
68 64'
No solution
(h) Northern
9 26'
72 22'
6714'N.
(i) Southern
88
S. 12
62
8366'S.
or48'S.
(J) Southern
19*
N. 7
8hr.
62 66' S.
(k) Southern
46 18'
S. 16 28'
326
49 14' S.
(1) Southern
N. 14
38
7226'S.
(m) Southern
67 86'
2hr.
1260'S.
APPLICATIONS OF SPHEEICAL TRIGONOMETEY 247
34. To determine the time of day. A very simple relation exists
between the hour angle 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
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
pole)
( North) tf
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 Almanaa
. * . 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
P.M.
When the observation is made in the forenoon the time of day will be
- 12\ A.M.
(hour angle
IS
Ex. 1. In New York (lat. 4P 43' N.) the sun's altitude is observed to be
80 4<X. Having given that the sun's declination is 10 N. and that the observa-
Zenith 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' = 69 20'.
Side 6 = co-dec, a 90 - 10 = 80.
>-
Pole Side c = co-lat. = 90 - 40 48' = 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'
/sin sin (a a)
c= 4917' - 1 ~ l
sin b sin c
2 5 == 188 37'
log sin ^ a = J [log sin 5 -f- log sin (s a) {log sin b + log sin c}]
8 /4 1U
- a = 34 59'.
log sin 8 = 9.9988 log sin 6= 9.9934
logsin(s - a) = 9.7684 log sin c= 9.8797
log numerator = 19. 7672 log denominator = 19. 8781
log denominator = 19.8731
9.8841
2 1 19. 8841
= 9.9421
a = 122 8'.
.-. A = 180 a = 67 62' = hour angle of sun.
Third step. Time of day = our an 8 PtM . = 3 nr . 51 m i nt PtM
15
EXAMPLES
1. In Milan (lat. 4530'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
80 40' as the altitude. What is the time of day, the sun's decimation 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 36 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. 46 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. 61 31' N.) at an afternoon observation the sun's altitude
is 16 40'. Find the time of day, given that the sun's declination is 12 S.
Ans. 2 hr. 69 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 ? Ans. 4 hr. 67 min. P.M.
8. The captain of a steamship observes that the altitude of the sun is 26 80'. Jf
he is in latitude 46 80' N. 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
SPHERICAL TRIGONOMETRY
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. 46 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 67' N.), if its declination
at sunrise is 23 2' S. ? Ans. 6 hr. 16 min. A.M.
3. About the first of April of each year the declination of the sun is 4 SO' N.
I ind the time of sunrise on that date at the following places :
(a) New York (lat. 40 43' N.). Ans. 5 hr. 46 min. A.M.
(b) London (lat. 61 31' N.). 6 hr. 37 min. A.M.
(c) St. Petersburg (lat. 60 N.). 6 hr. 29 min. A.M.
(d) New Orleans (lat. 29 58' N.). 6 hr. 60 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 1 is the me-
ridian (hour circle) of
Greenwich and PDP f 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 OF 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 longitude 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 tw'o 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 4 hr. 30 min. P.M. What is the longitude of the ship ?
Solution. Greenwich having the later time,
4 hr. 30 min.
9 2 hr. 26 min.
Subtracting, 2 hr. 6 min. = west longitude of the ship.
Reducing this to degrees and minutes of arc,
2 hr. 6 min.
16
Multiplying, 81 16' = west longitude of ship.
Second method. Find Greenwich time from a Greenwich 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 be more than 12 hours, we subtract it from 24 hour*
and use the remainder instead as the difference.
252 SPHEEICAL TRIGONOMETRY
Ex. 2 An exploring party have calculated their local time to be 10 hr. A.M.
The Greenwich chronometer which they carry gives the time as Shr. 80 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. 8 hr. 30 min. P.M. Ana. 5230 / W.
(b) Noon. 7 hr. 20 min. A.M. 70 E.
(c) Midnight. 10 hr. 15 min. P.M. 2615'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 min. P.M. Midnight. 85 W.
(g) 2 hr. 15 inin. P.M. 11 hr. 20 min. A.M. 4345 / E.
(h) 10 hr. 26 min. A.M. 5 hr. 16 min. A.M. 7780'E.
(i) 1 hr. 30 min. P.M. 7 hr. 45 min. P.M. 93 45' W.
(j) Noon. Midnight. 180 W. or E.
(k) 6 hr. P.M. 6 hr. A.M. ' 180 E. or W.
(1) 6 hr. 45 min. A.M. 7 hr. 80 min. P.M. 153 45' E.
(m) 10 hr. 55 min. P.M. 8 hr. 85 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., Januaiy 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 ?
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. Find the longitude of the observer in each case.
Observer's Corresponding time and Longitude
local time longitude of the other place of observer
(a) 2 hr. P.M. 6hr. P.M. at Havana (long. 8228 / W.) Ans. 127 28 X W.
(b) 10 hr. A.M. 8 hr. P.M. at Yokohama (long. 189 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 Vera Cruz (long. 969'W.) 6889'W.
(e) 9hr. 45 min. P.M. Midnight atBatavia (long. 10652'E.) 737'E.
(f ) 7 hr. 40 min. p. M. Noon at Gibraltar (long. 6 21' W. ) 109 39' E.
(g) 4 hr. 60min. P.M. Noon at Auckland (long. 17450 / E.) 112 4(K W.
APPLICATIONS OF SPHERICAL TRIGONOMETRY 258
5. What is the longitude of each place mentioned in the examples on p. 249,
the Greenwich time for the same instant being given below ?
Example, p. 249
(a) Ex. 3
(b) Ex. 4
(c) Ex. 6
(d) Ex. 7
(e) Ex. 8
Greenwich time
2 hr. 12 min. P.M.
4 hr. 62 min. P.M.
6 hr. 9 min. A.M.
10 hr. 33 min. P.M.
6 hr. 26 min. P.M.
Longitude of place
Ans. 67 W. long, (vessel)
3 1 46' W. long, (vessel)
60 E. long, (observer)
84 W. long, (surveyor)
26 W. long, (ship)
37. The ecliptic and the equinoxes. The earth makes a complete
irciiit around the sun in one year. To us, however, it appears as if
he 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
K
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 veiv
nal equinox (PVP 1 ), also called the equinoctial colure ; and the spher-
ical coordinates of a
heavenly body are its
declination and right
ascension.
The decliuation 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
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 between the
hour circle of the body and the hour circle of the vernal equinox
measured eastward from the latter circle from to 360, or in hours
from to 24. In the figure, the angle VPD (or the arc FD) 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 J).
Ex. I. In each of the following examples draw a figure of the celestial sphere
and locate the body from the given spherical coordinates.
Right ascension
Declination
(a)0
(b) 180
o-
(c) 90
N. 90
(d) 46
N. 46
(e) 60
N. 60
(f) 120
+ 30
(g) 300
-60
(h) 12 hr.
-f 46
(i) 20 hr.
Eight ascension
(J) 90*
(k) 270
(1) 90
(m) 46
(n) 90
(o) 240
(p) 330
(q) 6 hr.
(r) 9hr.
Declination
S. 90
8.46
8.30
+ 60
-46
+ 16
-76
* The right ascensions of the sun, moon, and planets may be found in the Nautical
Almanac for any time of the year.
APPLICATIONS OF SPHEEICAL 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 declination 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 6 = 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
Star
(North celestial pole)
P
(Pole
ecliptic.
of the
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
denned 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 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 J (= arc QL) on the longest day of the
year in the northern hemisphere (June 21), the sun being then the
highest in the sky (at Z), to S. 23^ (arc Q'L 1 ) on the shortest day
of the year (December 22), the sun being then the lowest in the sky
(at L 1 ). 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) (j) 90
(b) 90 N. 90 (k) 180
(c) 180 N. 45 (1) S. 60
(d) 270 (m) 60 N. 30
(e) 46 S. 30 (n) 120 N. 46
(f) 186 + 16 (o) 270 - 76
(g) 316 -f 60 (p) 80 - 60
(h) 6 hr. - 46* (q) 9 hr.
(i) 16 hr. -f 45 (r) 18 hr. -f 80
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 JKPM on the next page.
Angle KPM = /. Q'PV + Z VPD
= 90 + right ascension
= 90+2hr. 40 min.
= 90+ 40 ^
Side PM = co-declension
= 90 -24 20'
and
APPLICATIONS OF SPHERICAL TRIGONOMETRY 257
Side KP = LQ = e = 23 80'.
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,
a,nd 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. Aw. 90.
2. Find the distance between Regulus and Antares, the right ascensions
being 10 hr. and 16 hr. 20 inin.. and the polar distances 77 19' and 116 6'.
Ans. 99 66'.
3. Find the distance in degrees between the sun and the moon when their
right ascensions are respectively 15 hr. 12 min., 4 hr. 46 min., and their decli-
aations are 21 30' S., 6 30' N. Ans. 162 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 4 hr. 27 min., and his decimation
is 16 12' N. Find the angular distance between the stars. -4ns. 46 8',
5. Given the right ascension of a star 10 hr. 50 min., and its decimation
12 30' N., find its latitude and longitude. Take e = 23 80'.
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 68".
258
SPHERICAL TRIGONOMETRY
7. The sun's longitude was 69 40'. What was its right ascension and decli-
nation? Take e = 23 27'.
Ana. Right ascension = 3 hr. 50 min., declination = 20 6' N.
Hint. The latitude of the sun is always zero, since it moves in the ecliptic. Hence in the
triangle KPM (figure, p. 257), KM = 90, and it is a quadrantal triangle. This triangle may
iheu be solved by the method explained on p. 204.
8. Given the sun's declination 16 V N., find the sun's right ascension and
ongitude. Take e = 23 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-
,ion. Take e = 23 27'. Ans. Longitude = 214 16', declination = 12 66' S.
10. Find the length of the longest day of the year in latitude 42 17' N.
16 hr. 6 min.
Hint. This will be the time from sunrise to sunset when the sun is the highest in the sky,
,hat is, when its declination is 23 27' N.
11. Find the length of the shortest day in lat. 42 17' N. Ans. 8 hr. 64 min.
Hint. The sun will then be the lowest in the sky, that is, its declination will be 23 27' S.
12. Find the length of the longest day in New Haven (lat. 41 19' N.). Take
( = 23 27'. Ans. 16 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. 69 21' N.). Take
i = 23 27'. Ans. 18 hr. 16 min.
15. Find the length of the shortest day in Stockholm. -4ns. 6 hr. 48 min.
40. The astronomical triangle. We have seen that many of our
nost important astronomical problems depend on the solution of
he astronomical triangle PZM. In any such problem the first
hing 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
^ 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.
' Latitude = 51 32' K
.-. side PZ = 90- 51 32'= 38 28 '.
Altitude =35 15'.
/. side MZ = 90- 35 15' = 54 45'.
Decimation = 21 27' K
.-. 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.
t In 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.
4. 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 semidiarneter 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 Field 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 13,338,000 sq. mi,
2. The distance between Paris (lat. 48 50 7 N.) and Berlin (lat. 62 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 ? Ana. 44 min. 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. Ana. 60.
APPLICATIONS OF SPHERICAL TRIGONOMETRY 261
4. What will be the altitude of the sun at 9 A. M. in Mexico City (lat. 19 25' N. ),
if its declination at that time is 8 23' N. ? Ans. 46 6'.
6. 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. 69 60' N.).
Ans. 6 hr. 68 min. A.M., 6 hr. 2 min. P.M.
7. What is the direction of a wall in lat. 62 30' N. which casts no shadow at
6 A.M. on the longest day of the year ?
Ans. 76 11', 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. 66 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 62 30' N., the decimation
of a star 38, its hour angle 28 1 V. Find the altitude of the star.
Ans. Altitude = 65 33'.
11. Given the latitude of the place of observation 61 19' N., the polar dis-
tance of a star 67 69', its hour angle 16 8'. Find the altitude and azimuth of
the star. Ans. Altitude = 68 23', azimuth = 27 SO'.
12. Given the declination of a star 7 64' N., its altitude 22 46', its azimuth
60 14'. Find the hour angle of the star and the latitude of the observer.
Ans. Hour angle = 46 41', latitude = 67 69' N.
13. The latitude of a star is 61 N., and its longitude 316. Find its declina-
tion. Take e = 23 27'. Ans. Declination = 32 23' N.
14. Given the latitude of the observer 44 60' 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 61 32' N. , the altitude of
the sun west of the meridian 36 16', its declination 21 27' N. Find the local
time. Ans. 3 hr. 69 min. P.M.
CHAPTEK 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 o sin B,
(4) cos .4 = cos a sin ;
(5) cos B cos b sin ^4,
(6) cos A = tan b cot c,
(7) cos B = tan a cot o,
(8) sin b = tan a cot A,
(9) sin a = tan /; cot B,
(10) cos c = cot .4 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-^, /3 = 18Q B, y = 180 C.
s = J (a -f b -f 0), or = (a + ft -f- y).
c? = diameter of inscribed circle.
B = 180 diameter of circumscribed circle.
Law of sines, p. 207.
sin a sin b sin c
or,
sin ^4 sin B sin C
sin a sin & sin c
sn or sn ft sn y
Law of cosines for the sides, p. 209.
(12) cos a cos b cos c sin 6 sin c cos a.
262
RECAPITULATION OF FORMULAS
Law of cosines for the angles, p. 209.
(15) cos a = cos ft cos y sin /3 sin y cos a.
Functions of a, & y in terms of the sides, pp. 211-213
,-nx . , /sin s sin (* a)
(18) / sm i a = -vl ; jA '
v ' J > sin b sin c
263
(19)
(20)
(27)
(28)
(29)
(30)
cos
__ I sin (s b) sin (s c)
r > sin b sin c
_ I sin & sin (.9 a)
* >l sin (s 6) sin (s c)
sin (s a) sin (s b) sin (s <
tan 1 c? = A L ^ L *
* X Bin s
__ sin (s a)
tan \ a =
sin (s
tan ^8 = -^
^ ^ tan
tan
sin (s c)
= v
tan J- oJ
Functions of the half sides in terms of a, fi, y, p. 214.
(31)
(32)
(33)
(40)
(41)
(42)
(43)
fsino- sin(cr a:)
"***=\ ii^sh
L fi sin y
cos
r x
; -N ri
sin /? sin y
f sin o- sin (o- cr)
ta.nl a = A -r QX \ , *-
' \ sin (cr p) sin (<r
y)
f sin (cr a) sin (cr /?) sin (o- y]
sincr
sin (cr a)
ffi= ''
" *" tan J 8
Sin (cr y)
tan c = - - -^ LL -
* tan 18
264 SPHERICAL TEIGONOMETBY
Napier's Analogies, p. 215.
(44)
(46)
(47) tan t ( + ff COS f < ~ *' tan t r .
v ' * v ^' cos i (a + &) * '
44. General directions for the solution of oblique spherical triangles,
pp. 216-227.
CASE I. (a) Given the three sides, p. 217.
(b) Given the three angles, p. 218.
CASE II. (a) Given two sides and their included angle, p. 219.
(&) Given two angles and their included side, p. 222.
CASE III. (a) Given two sides and the angle opposite one of them }
p. 224.
(&) Given two angles and the side opposite one of them,
p. 226.
45. Length of an arc of a circle in linear units, p. 228.
~
N = number of degrees in angle.
46. Area of a spherical triangle, p. 229.
, KA .
(54)
= A +B + C -180.
(65) tan E = Vtan \ s ta-n \ (s a) tan (s b) tan (s 0)
FOUR-PLACE TABLES OF
LOGARITHMS
COMPILED BY
WILLIAM ANTHONY GRANVILLE, Pn.D., LL.D.
FORMERLY PRESIDENT OF PENNSYLVANIA COLLEGE
GINN AND COMPANY
BOSTON" NEW YORK - CHICAGO LONDON
ATLANTA DALLAS COLUMBUS SAN FRANCISCO
ENTERED AT STATIONERS 1 HALL
COPYRIGHT, 1908, BY
WILLIAM ANTHONY GRANVILLE
ALL RIGHTS RESERVED
PRINTED IN THE UNITED STATES OF AMERICA
327.4
gfre
GINN AND COMPANY . PRO-
PRIETORS BOSTON U.S.A.
CONTENTS
PAGES
TABLE I. LOGARITHMS OF NUMBERS 1-5
K.ULES FOR FINDING THE LOGARITHMS OF THE TRIGONO-
METRIC FUNCTIONS OF ANGLES NEAR 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
ill
TABLE I
FOUR-PLACE LOGARITHMS OF 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.
1
0004
2
3
4
5
6
7
8
9
Prop. Parts
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
0000
0009
0013
0017
0022
0026
0030
0035
0039
i
i
a_
i
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9 ,
|
~5~~
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
"T"
0.4
0.8
1.2
1.6
2.0
2.4
2.8
3.2
3.6
"IT
0.3
0.6
0.9
1.2
1.5
1.8
2.1
2.4
2.7
IT
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
0043
0086
0128
0170
0212
0253
0294
0334
0374
0048
0090
0133
0175
0216
0257
0298
0338
0378
0052
0095
0137
0179
0220
0261
0302
0342
0382
0056
0099
0141
0183
0224
0265
0306
0346
0386
0060
0103
0145
0187
0228
0269
0310
0350
0390
0065
0107
0149
0191
0233
0273
0314
0354
0394
0069
0111
0154
0195
0237
0278
0318
0358
0398
0073
0116
0158
0199
0241
0282
0322
0362
0402
0077
0120
0162
0204
0245
0286
0326
0366
0406
0082
0124
0166
0208
0249
0290
0330
0370
0410
0414
0418
0422
0426
0430
0434
0438
0441
0445
0449
0453
0492
0531
0569
0607
0645
0682
0719
0755
0457
0496
0535
0573
0611
0648
0686
0722
0759
0461
0500
0538
0577
0615
0652
0689
0726
0763
0465
0504
0542
0580
0618
0656
0693
0730
0766
0469
0508
0546
0584
0622
0660
0697
0734
0770
0473
0512
0550
0588
0626
0663
0700
0737
0774
0477
0515
0554
0592
0630
0667
0704
0741
0777
0481
0519
0558
0596
0633
0671
0708
0745
0781
0484
0523
0561
0599
0637
0674
0711
0748
0785
0488
0527
0565
0603
0641
0678
0715
0752
0788
0792
0795
0799
0803
0806
0810
0813
0817
0821
0824
0828
0864
0899
0934
0969
1004
1038
1072
1106
0831
0867
0903
0938
0973
1007
1041
1075
1109
0835
0871
0906
0941
0976
1011
1045
1079
1113
0839
0874
0910
0945
0980
1014
1048
1082
1116
0842
0878
0913
0948
0983
1017
1052
1086
1119
0846
0881
0917
0952
0986
1021
1055
1089
1123
0849
0885
0920
0955
0990
1024
1059
1093
1126
0853
0888
0924
0959
0993
1028
1062
1096
1129
0856
0892
0927
0962
0997
1031
1065
1099
1133
0860
0896
0931
0966
1000
1035
1069
1103
1136
1139
1143
1146
1149
1153
1156
1159
1163
1166
1169
1173
1206
1239
1271
1303
1335
1367
1399
1430
1176
1209
1242
1274
1307
1339
1370
1402
1433
1179
1212
1245
1278
1310
1342
1374
1405
1436
1183
1216
1248
1281
1313
1345
1377
1408
1440
1186
1219
1252
1284
1316
1348
1380
1411
1443
1189
1222
1255
1287
1319
1351
1383
1414
1446
1193
1225
1258
1290
1323
1355
1386
1418
1449
1196
1229
1261
1294
1326
1358
1389
1421
1452
1199
1232
1265
1297
1329
1361
1392
1424
1455
1202
1235
1268
1300
1332
1364
1396
1427
1458
1461
1464
1467
1471
1474
1477
1480
1483
I486
1489
1492
1523
1553
1584
1614
1644
1673
1703
1732
1495
1526
1556
1587
1617
1647
1676
1706
1735
1498
1529
1559
1590
1620
1649
1679
1708
1738
1501
1532
1562
1593
1623
1652
1682
1711
1741
15(#
1535
1565
1596
1626
1655
1685
1714
1744
1508
1538
1569
1599
1629
1658
1688
1717
1746
1511
1541
1572
1602
1632
1661
1691
1720
1749
1514
1544
1575
1605
1635
1664
1694
1723
1752
1517
1547
1578
1608
1638
1667
1697
1726
1755
1520
1550
1581
1611
1641
1670
1700
1729
1758
1761
1764
1
1767
1770
1772
1775
1778
1781
1784
1787
Ho.
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
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
1761
1764
1767
1770
1772
1775
1778
1781
1784
1787
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
3
0.3
0.6
0.9
1.2
1.5
1.8
2.1
2.4
2.7
"IT
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
1790
1818
1847
1875
1903
1931
1959
1987
2014
1793
1821
1850
1878
1906
1934
1962
1989
2017
17%
1824
1853
1881
1909
1937
1965
1992
2019
1798
1827
1855
1884
1912
1940
1967
1995
2022
1801
1830
1858
1886
1915
1942
1970
1998
2025
1804
1833
1861
1889
1917
1945
1973
2000
2028
1807
1836
1864
1892
1920
1948
1976
2003
2030
1810
1838
1867
1895
1923
1951
1978
2006
2033
1813
1841
1870
1898
1926
1953
1981
2009
2036
1816
1844
1872
1901
1928
1956
1984
2011
2038
2041
2044
2047
2049
2052
2055
2057
2060
2063
2066
2068
2095
2122
2148
2175
2201
2227
2253
2279
2071
2098
2125
2151
2177
2204
2230
2256
2281
2074
2101
2127
2154
2180
2206
2232
2258
2284
2076
2103
2130
2156
2183
2209
2235
2261
2287
2079
2106
2133
2159
2185
2212
2238
2263
2289
2082
2109
2135
2162
2188
2214
2240
2266
2292
2084
2111
2138
2164
2191
2217
2243
2269
2294
2087
2114
2140
2167
2193
2219
2245
2271
2297
2090
2117
2143
2170
2196
2222
2248
2274
2299
2092
2119
2146
2172
2198
2225
2251
2276
2302
2304
2307
2310
2312
2315
2317
2320
2322
2325
2327
2330
2355
2380
2405
2430
2455
2480
2504
2529
2333
2358
2383
2408
2433
2458
2482
2507
2531
2335
2360
2385
2410
2435
2460
2485
2509
2533
2338
2363
2388
2413
2438
2463
2487
2512
2536
2340
2365
2390
2415
2440
2465
2490
2514
2538
2343
2368
2393
2418
2443
2467
2492
2516
2541
2345
2370
2395
2420
2445
2470
2494
2519
2543
2348
2373
2398
2423
2448
2472
2497
2521
2545
2350
2375
2400
2425
2450
2475
2499
2524
2548
2353
2378
2403
2428
2453
2477
2502
2526
2550
2553
2555
2558
2560
2562
2565
2567
2591
2615
2639
2662
2686
2709
2732
2755
2778
2570
2572
2574
2577
2601
2625
2648
2672
2695
2718
2742
2765
2579
2603
2627
2651
2674
2697
2721
2744
2767
2582
2605
2629
2653
2676
2700
2723
2746
2769
2584
2608
2632
2655
2679
2702
2725
2749
2772
2586
2610
2634
2658
2681
2704
2728
2751
2774
2589
2613
2636
2660
2683
2707
2730
2753
2776
2799
2594
2617
2641
2665
2688
2711
2735
2758
2781
2596
2620
2643
2667
2690
2714
2737
2760
2783
2598
2622
2646
2669
2693
2716
2739
2762
2785
190
191
192
193
194
195
196
197
198
199
2788
2790
2792
2794
2797
2801
2804
2806
2808
2810
2833
3856
2878
2900
2923
2945
2967
2989
2813
2835
2858
2880
2903
2925
2947
2969
2991
2815
2838
2860
2883
2905
2927
2949
2971
2993
2817
2840
2862
2885
2907
2929
2951
2973
2995
2819
2842
2865
2887
2909
2931
2953
2975
2997
2822
2844
2867
2889
2911
2934
2956
2978
2999
2824
2847
2869
2891
2914
2936
2958
2980
3002
2826
2849
2871
2894
2916
2938
2960
2982
3004
2828
2851
2874
2896
2918
2940
2962
2984
3006
2831
2853
2876
2898
2920
2942
2964
2986
3008
200
No.
3010
3012
3015
3017
3019
3021
3023
3025
3028
3030
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. 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
58
59
60
61
62
63
64
65
66
67
68
69
70
3010
3032
3054
3075
3096
3118
3139
3160
3181
3201
i
I
W
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
477J
4914"
5051
5185
5315
5441
5563
5682
5798
5911
4786
4800
4814
4829
4843
4857
4871
4886
4900
4928
5065
5198
5328
5453
5575
5694
5809
5922
4942
5079
5211
5340
5465
5587
5705
5821
5933
4955
5092
5224
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
55]4
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
9.5
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
17
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
48
6.4
8.0
96
11.2
12.8
14.4
15
1.5
3.0
4.5
6.0
7.5
9.0
105
12.0
13.5
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
42
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
8000
8069
8136
8202
8267
8331
8395
7868
7938
8007
8075
8142
8209
8274
8338
8401
7875
7945
8014
8082
8149
8215
8280
8344
8407
78^2
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
845]
8457
8463
8470
8476
8482
8488
8494
8500
8506
No.
1
2
3
4
5
6
7
8
9
TABLE I. LOGAEITHMS OF KUMBEES
Ho,
1
2
3
4
5
6
7
8
9
Prop. Farts
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
8567
8627
8686
8745
8802
8859
8915
8971
9025
Difference
8513
8573
8633
8692
8751
8808
8865
8921
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
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
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
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
1
2
3
4
5
6
7
8
9
6
0.6
1.2
1.8
2.4
3.0
3.6
4.2
4.8
5.4
5
0.5
1.0
1.5
2.0
2.5
3.0
3.5
40
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
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
0022
0026
0030
0035
0039
Wo.
1
2
3
4
5
6
7
8
9
RULES FOR FINDING THE LOGARITHMS OF THE
TRIGONOMETRIC FUNCTIONS OF ANGLES
NEAR 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 + log x.
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. f
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 and 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 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 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. LOGAEITHMIC SINES
Angle
log sin
diff.1'
log tan
com.
diff.r
log oot
log COB
0'
10.0000
90 00'
1'
2'
3'
4'
5'
6'
7'
8'
9'
6.4637
6.7648
6.9408
7.0658
7.1627
7.2419
7.3088
7.3668
7.4180
3
?
2-2<o'
.1*
g
B^S
lg|
6.4637
6.7648
6.9408
7.0658
7.1627
7.2419
7.3088
7.3668
7.4180
.5
I
<
Hi
3.5363
3.2352
3.0592
2.9342
2.8373
2.7581
2.6912
2.6332
2.5820
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
89 59'
89 58'
89 57'
89 56'
89 55'
89 54'
89 53'
89 52'
89 51'
10'
7.4637
1*8,0
7.4637
52
2.5363
10.0000
89 50'
11'
12'
13'
14'
15'
16'
17'
18'
19'
7.5051
7.5429
7.5777
7.6099
7.6398
7.6678
7.6942
7.7190
7.7425
a jji|
*P
235
7.5051
7.5429
7.5777
7.6099
7.6398
7.6678
7.6942
7.7190
7.7425
g- H 9
o> 2
>,(* 3
III
^&3
"I 1
235
2.4949
2.4571
2.4223
2.3901
2.3602
23322
2.3058
2.2810
2.2575
10.0000
100000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
89 49'
89 48'
89 47'
89 46'
89 45'
89 44'
89 43'
89 42'
89 41'
020'
7.7648
S23
7.7648
2.2352
100000
89 40'
21'
22'
23'
24'
25'
26'
27'
28'
29'
7.7859
7.8061
7.8255
7.8439
7.8617
7.87S7
7.8951
7.9109
7.9261
202
194
184
178
170
164
158
152
7.7860
7.8062
7.8255
7.8439
7.8617
78787
7.8951
7.9109
7.9261
209
193
184
178
170
164
158
152
2.2140
2.1938
2.1745
2.1561
2.1383
2.1213
2.1049
2.0891
2.0739
10.0000
10.0000
10.0000
10.0000
10.0000
100000
10.0000
10.0000
10.0000
89 39'
89 38'
89 37'
89 36'
89 35'
89 34'
89 33'
89 32'
89 31'
030'
7.9408
147
7.9409
2.0591
10.0000
89 30'
31'
32'
33'
34'
35'
36'
37'
38'
39'
7.9551
7.9689
7.9822
7.9952
8.007S
8.0200
8.0319
8.0435
8.0548
138
133
130
126
122
119
116
113
7.9551
7.9689
7.9823
7.9952
8.0078
8.0200
8.0319
8.0435
8.0548
138
134
129
126
122
119
116
113
2.0449
2.0311
2.0177
2.0048
1.9922
1.9800
1.9681
1.9565
1.9452
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
100000
10.0000
89 2^
89 28'
89 27'
89 26'
89 25'
89 24'
89 23'
89 22'
89 21'
040'
8.0658
8.0658
1.9342
100000
89 20'
41'
42'
43'
44'
45'
46'
47'
48'
49'
8.0765
8.0870
8.0972
8.1072
8.1169
8.1265
8.1358
8.1450
8.1539
105
102
100
97
96
93
92
89
8.0765
8.0870
8.0972
8.1072
8.1170
8.1265
8.1359
8.1450
8.1540
105*
102
100
98
95
94
91
90
1.9235
1.9130
1.9028
1.8928
1.8830
1.8735
1.8641
1.8550
1.8460
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
89 19'
89 18'
89 17'
89 16'
89 15'
89 14'
89 13'
89 12'
89 11'
50'
8.1627
8.1627
1.8373
10.0000
89 10'
51'
52'
53'
54'
55'
56'
57'
58'
59'
8.1713
8.1797
8.1880
8.1961
8.2041
8.2119
8.2196
8.2271
8.2346
84
83
81
80
78
77
75
75
73
8.1713
8.1798
8.1880
8.1962
8.2041
8.2120
8.2196
8.2272
8.2346
85
82
82
79
79
76
76
74
73
1.8287
1.8202
1.8120
1.8038
1.7959
1.7880
1.7804
1.7728
1.7654
10.0000
10.0000
9.9999
9.9999
9.9999
9.9999
9.9999
9.9999
9.9999
89 9'
89 8'
89 7'
89 6'
89 5'
89 4'
89 3'
89 2'
89 1'
060'
8.2419
8.2419
1.7581
9.9999
89 0'
log cos
diff.l'
log oot
dlFi-
log tan
log sin
'Anglo
8
9
COSINES, TANGENTS, AND COTANGENTS
1
Angle
log sin
difi.1'
logtan
log oot
log COB
1 0'
8.2419
71
8.2419
72
1.7581
9.9999
88 60'
1 r
1 2'
1 3'
1 4'
1 5'
1 6'
1 7'
1 8'
1 9'
8.2490
8.2561
8.2630
8.2699
8.2766
8.2832
8.2898
8.2962
8.3025
71
69
69
67
66
66
64
63
63
8.2491
8.2562
8.2631
8.2700
8.2767
8.2833
8.2899
8.2963
8.3026
71
69
69
67
66
,66
64
63
63
1.7509
1.7438
1.7369
1.7300
1.7233
1.7167
1.7101
1.7037
1.6974
9.9999
9.9999
99999
99999
9.9999
9.9999
9.9999
9.9999
9.9999
88 59'
88 58'
88 57'
88 56'
88 55'
88 54'
88 53'
88 52'
88 51'
1 10'
8.3088
62
8.3089
61
1.6911
9.9999
88 50'
1 ii'
1 12'
1 13'
1 14'
1 15'
1 16'
1 17'
1 18'
1 19'
8.3150
8.3210
83270
8.3329
8.3388
8.3445
8.3502
8.3558
8.3613
60
60
59
59
57
57
56
55
55
8.3150
8.3211
83271
8 3330
8.3389
8.3446
8.3503
8.3559
83614
61
60
59
59
57
56
56
55
55
1.6850
1.6789
1.6729
1.6670
1.6611
1.6554
1.6497
1.6441
1.6386
9.9999
9.9999
9.9999
9.9999
9.9999
9.9999
9.9999
99999
9.9999
88 49 /
88 48'
88 47'
88 46'
88 45'
88 44'
88 43'
88 42'
88 41'
120'
8.3668
54
8.3669
54
1.6331
9.9999
88 40'
1 21'
1 22'
1 23'
1 24'
1 35'
1 26'
1 27'
1 28'
1 29'
8.3722
8.3775
8.3828
8.3880
8.3931
8.3982
8.4032
8.4082
84131
53
53
52
51
51
50
50
49
49
8.3723
8.3776
8.3829
8.3881
8.3932
8.3983
8.4033
8.40S3
8.4132
53
53
52
51
5L
50
50
49
49
1.6277
1.6224
1.6171
1.6119
1.6068
1.6017
1.5967
1.5917
1.5868
9.9999
9.9999
9.9999
9.9999
99999
9.9999
99999
9.9999
9.9999
88 39'
88 38'
88 37'
88 36'
88 35'
88 34'
88 33'
88 32'
88 31'
130'
8.4179
48
8.4181
48
1.5819
9.9999
88 30'
1 31'
1 32'
1 33'
1 34'
1 35'
1 36'
1 37'
1 38'
1 39'
8.4227
8.4275
8.4322
8.4368
84414
8.4459
84504
8.4549
8.4593
48
47
46
46
45
45
45
44
44
8.4229
8.4276
84323
8.4370
84416
8.4461
8.4506
8.4551
8.4595
47
47
47
46
45
45
45
44
43
1.5771
1.5724
1.5677
1.5630
1.5584
1.5539
1.5494
1.5449
1.5405
9 9998
9.9998
9.9998
99998
9.9998
9.9998
9.9998
9.9998
9.9998
88 29'
88 28'
88 27'
88 26'
88 25'
88 24'
88 23'
88 22'
88 21'
140'
8.4637
43
8463S
44
1.5362
9.9998
88 20'
1 41'
1 42'
1 43'
1 44'
1 45'
1 46'
1 47'
1 48'
1 49'
8.4680
8.4723
8.4765
8.4807
8.4848
8.4890
8.4930
8.4971
8.5011
43
42
42
41
42
40
41
40
39
8.4682
8.4725
8.4767
8.4809
8.4851
84892
8.4933
8.4973
8.5013
43
42
42
42
41
41
40
40
40
1.5318
1.5275
1.5233
1.5191
1.5149
1.5108
1.5067
1.5027
1.4987
9.9998
9 9998
9.9998
9.9998
9.9998
9.9998
9.9998
9.9998
9.9998
88 19'
88 18'
88 17'
88 16'
88 15'
88 14'
88 13'
88 12'
88 11'
150'
8.5050
40
8.5053
39
14947
9.9998
88 10'
1 51'
1 52'
1 53'
1 54'
1 55'
1 56'
1 57'
1 58'
1 59'
8.5090
8.5129
8.5167
8.5206
8.5243
8.5281
8.5318
8.5355
8.5392
39
38
39
37
38
37
37
37
8.5092
8.5131
8.5170
8.5208
8.5246
8.52S3
8.5321
8.5358
8.5394
39
39
38
38
37
38
37
36
1.4908
1.4869
1.4830
1.4792
1.4754
1.4717
1.4679
1.4642
1.4606
9.9998
9.9998
9.9998
9.9998
9.9998
9.9998
9.9997
9.9997
9.9997
88 9'
88 8'
88 T
88 6'
88 5'
88 4'
88 3'
88 2'
88 1"
160'
8.5428
8.5431
1.4569
9.9997
88 0'
log 008
diff.1'
Jogoot
log tan
log sin
Angle
v
8
8
10
TABLE II. LOGARITHMIC SINES
a
Angle
log- sin
diff.1
log tan
di& m i'
log cot
log cos
2 0'
8.5428
36
85431
36
1.4569
9.9997
87 60'
2 r
8.5464
8.5467
1.4533
9.9997
87 59'
2 2'
8.5500
36
8.5503
36
1.4497
9.9997
87 58'
2 3'
8.5535
35
8.5538
35
1.4462
9.9997
87 57'
2 4'
8 5571
36
8.5573
35
1.4427
9.9997
87 56'
2 5'
8.5605
34
8.5608
35
1.4392
9.9997
87 55'
2 6'
8.5640
35
8.5643
35
1.4357
9.9997
87 54'
2 7'
85674
34
8.5677
34
1.4323
9.9997
87 53'
2 8'
8.5708
34
S.5711
34
1.4289
9.9997
87 52'
2 9"
8.5742
34
34
85745
34
34
1.4255
9 9997
87 51'
2 10'
8.5776
33
8.5779
33
1.4221
9.9997
87 50'
2 11'
8.5809
8.5812
1 4188
9.9997
87 49'
2 12'
8 5842
33
8.5845
33
1.4155
9.9997
87 48'
2 13'
8.5875
33
8.5878
33
1.4122
9.9997
87 47'
2 14'
8.5907
32
85911
33
1 4089
9.9997
87 46'
2 15'
8.5939
32
8.5943
32
1.4057
9.9997
87 45'
2 16'
85972
3.1
8.5975
32
1 4025
9 9997
87 44'
2 17'
86003
8.6007
1 .3993
9.9997
87 43'
2 18'
8 6035
8.6038
1.3962
9.9997
87 42'
2 19'
8.6066
31
31
8.6070
32
31
1.3930
9 9996
87 41'
2 20'
8 6097
31
8.6101
31
1.3899
9.9996
87 40'
2 21'
8 6128
8.6132
1 3868
9.9996
87 39'
2 22'
8.6159
31
8.61 63
31
1.3837
9.9996
87 38'
2 23'
8.6189
30
8 6193
30
1.3807
9 9996
87 37'
2 24'
8 6220
31
8.6223
30
1.3777
9.9996
87 36'
2 25'
8 6250
8.6254
31
13746
9.9996
87 35'
2 26'
8 6279
2<>
8 628 5
29
1.3717
9.9996
87 34'
2 27'
8 6309
8 6313
1.3687
9 9996
87 33'
2 28'
8.6339
8.6343
30
1.3657
9.9996
87 32'
2 29'
8 6368
86372
1.3628
9.9996
87 31'
2 30'
8 6397"
29
86401
29
1.3599
9 9996
87 30'
2 31'
8 6426
8.6430
1.3570
9.9996
87 29'
2 32'
8 6454
28
8.6459
29
1.3541
9 9996
87 28'
2 33'
8.6483
2<)
8.6487
28
1.3513
9.9996
87 27'
2 34'
86511
28
86515
28
1 3485
9 9996
87 26'
2 35'
8.6539
28
8 6544
29
1.3456
9.9996
87 25'
2 36'
8 6567
28
8.6571
27
1 3429
9.9996
87 24'
2 37'
8.6595
28
86599
28
1.3401
9.9995
87 23'
2 38'
8.6622
27
8.6627
28
1.3373
9.9995
87 22'
2 39'
8 6650
28
7
8.6654
27
28
1.3346
9.9995
87 21'
2 40'
8 6677
27
8.6682
27
1.3318
9.9995
87 20'
2 41'
8.6704
8.6709
1.3291
9.9995
87 19'
2 42'
8 6731
27
8.6736
27
1 3264
9.9995
87 18'
43'
8.6758
27
8.6762
26
1.3238
9.9995
87 17'
o 44'
8.6784
26
"8 6789
27
1.3211
9.9995
87 16'
o 4=;'
8.6810
26
8.6815
26
1.3185
9 9995
87 15'
46'
86S37
27
8 6842
27
1.3158
9.9995
87 14'
o 47'
8.6863
26
86868
26
1.3132
9.9995
87 13'
o 48 '
8.6889
8.6894
20
1.3106
9.9995
87 12'
o 49 x
8 6914
25
26
8.6920
26
25
1.3080
9 9995
87 11'
2 50'
8.6940
25
8.6945
26
1 3055
9.9995
87 10'
2 51'
8.6965
8.6971
1.3029
9.9995
87 9'
2 52'
8.6991
8.6996
25
13004
9.9995
87 8'
2 53'
8.7016
8.7021
1.2979
9.9995
87 7'
2 54'
8.7041
25
8.7046
25
1.2954
9 9994
87 6'
2 55'
8.7066
25
8.7071
25
1.2929
9.9994
87 5'
2 56'
8.7090
24
8 7096
1.2904
9.9994
87 4'
2 57'
8.7H?
25
8.7121
25
1.2879
9.9994
87 3'
2 58'
87140
25
8.7145
24
1.2855
9.9994
87 2'
2 59'
8.7164
24
8.7170
25
1.2830
9.9994
87 1'
2 60'
8.7188
8.7194
12806
9.9994
87 0'
log COB
diff.l'
log cot
com.
diff.l'
log tan
log sin
Angle
8'
r
COSINES, TANGENTS, AND COTANGENTS
11
3
Angle
log Bin
diff.l'
log tan
log cot
log COB
3 0'
8.7188
24
8.7194
24
1.2806
9.9994
86 60'
3 r
8.7212
8.7218
1.2782
9.9994"
86 59'
3 2'
8.7236
24
8.7242
24
1.2758
9.9994
86 58'
3 3'
8.7260
24
87266
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
99994
86 55'
3 6'
8.7330
23
8.7337
24
1.2663
9.9994
86 54'
3 r
8.7354
24
87360
23
1.2640
9.9994
86 53'
3 8'
8.7377
S3
87383
23
1 2617
9 9994
86 52'
3 9'
8.7400
23
23
87406
23
23
1.2594
9.9993
86 51'
3 10'
8 7423
22
8 7429
23
1 2571
9.9993
86 50'
3 11'
8 7445
8.7452
1.2548
9.9993
86 49'
3 12'
8 7468
23
8.7475
23
1 2525
9.9993
86 48'
3 13'
87491
23
8 7497
22
1.2503
9.9993
86 47'
3 14'
87513
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'
87557
22
8 756 S
23
i.2435
9 9993
86 44'
3 17'
87580
23
8 7587
22
1 2413
9 9993
86 43'
3 18'
8 7602
22
8 7609
22
12391
9 9993
86 42'
3 19'
8.7623
21
8 7631
22
1.2369
9.9993
86 41'
3 20'
87645"
22
22
8.7652
21
22
T.2348"
"99993
86 40'
3 21'
87667
8 7671
T72326
~9993
86 39'
3 22'
8 7688
21
8 7696
22
1 2304
9 9993
86 38'
3 23'
8 7710
22
87717
21
1.2283
9 9992
86 37'
3 24'
877U
21
8 7739
i>2
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'
87815
21
8 7823
21
1.2177
9.9992
86 32'
3 29'
8 7836
21
8.784-1
21
1 2156
9.9992
86 31'
3 30'
8 7857"
Bi
20
8 7865
21
21
T2135
"99992
86 30'
3 31'
8 7877
8 7886
1 2114
9.9992
86 29'
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 J 26'
3 3S'
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'
88019
20
8 8028
20
1 1972
9.9991
86 22'
3 39'
8 8039
20
20
88048
20
20
1.1952
9 9991
86 21'
3 40'
8.8059
19
"878067
20
LI933
~O999T
86 20'
3 41'
IT 8078"
8 "8087"
1.1913"
-9-9991"
86 19'
3 42'
S 8098
20
88107
20
1.1893
9.9991
86 18'
3 43'
88117
19
88126
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
8 816 S
J9
1.1835
9.9991
86 15'
3 46'
88175
19
88185
20
1 1815
9 9991
86 14'
3 47'
8 8194
19
8.8204
19
1.1796
99991
86 13'
3 48'
8 8213
19
8 8223
19
1 1777
9 9990
86 12'
3 49'
8 8232
19
19
8 8242
19
19
1 1758
9.9990
86 11'
3 50'
8.8251
19
8.8261
19
1.1739
9 9990
86 10'
3 51'
88270
8.8280
1.1720
9.9990
86 9'
3 52'
8 8289
19
8.8299
19
1.1701
9.9990
86 8'
3 53'
88307
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'
88363
18
8.8373
]8
1.1627
99990
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'
8.8418
18
8.8428
18
1.1572
9.9990
86 Y
3 60'
8.8436
18
8.8446
18
1.1554
9.9989
86 0'
log cos
diff.1'
log cot
m.'
log tan
log sin
Angle
86
12
TABLE II. LOGARITHMIC SINES
4
Angl
log- Bin
diff.
logta
aiS m r
' log co
log oo
4
8.8436
18
8.8446
19
1.1554
9.9989
85 60'
4 1
88454
8.8465
1.153
9.9989
85 59'
4 2
88472
18
8.848o
18
1.1517
9.9989
85 58'
4 3
8.8490
18
8.8501
18
1.149<
9.9989
85 57'
4 4
8.8508
JS
S 8518
17
1.1482
9.9989
85 56'
4 5
88525
17
S.853f
18
1.1464
9.9989
85 55'
4 6
88543
18
8.8554
18
1.144f
9.9989
85 54'
4 7
8.8560
17
885 72
18
1.1428
9.9989
85 53'
4 8
88=578
18
8.SSS9
17
1 1411
9.9989
85 52'
4 9
8.8595
17
8.8607
18
1.1393
99989
85 51'
4 10
8.86'lT
18
17
~8~862T
17
18
~l7l37r
9.9989
85 50'
4 11
^88630"
8.8642~
T1358
~9~998b~
85 49'
4 12
88647
17
8.8659
17
1.1341
99988
85 48'
4 13
8 8665
18
8.8676
17
1.1324
99988
85 47'
4 14
88682
17
8.8694
18
i.isor
99988
85 46'
4 15'
8 8699
17
88711
17
1.1289
9.998S
85 45'
4 16'
8 8716
17
8.8728
17
1 1272
9 9988
85 44'
4 17'
8.8733
17
8.8745
17
1.1255
9.9988
85 43'
4 18'
8.8749
16
8.8762
17
11238
9.9988
85 42'
4 19'
8.8766
17
17
8 8778
16
17
1.1222
9.9988
85 41'
4 20
8.8783
16
"8.8795"
17
1.1205
9.9988
85 40'
4 21'
8.8799
8,8812
1.1188
9.9987
85 39'
4 22'
8.8816
17
8.8829
17
1.1171
9.9987
85 38'
4 23'
8.8833
17
8 8845
16
1.1155
9.9987
85 37'
4 24'
8.8849
J6
88862
17
1.1138
9.9987
85 36'
4 25'
8.8865
1G
8.8878
16
1.1122
9.9987
85 35'
4 26'
8.8882
17
8.8895
17
1.1105
9 9987
85 34'
4 27'
88898
16
8.8911
16
1.1089
99987
85 33'
4 28'
8.8914
16
8.8927
16
1.1073
9.9987
85 32'
4 29'
8.8930
10
16
8.8944
17
16
1 1056
99987
85 31'
4 30'
8.8946
16
8.8960
16
1 1040
9 9987
85 30'
4 31'
8.8962
8 8976
1.1024
9.9986
85 29'
4 32'
8.8978
16
8.8992
16
1.1008
9.9986
85 28'
4 33'
8.8994
16
8.9008
16
1 0992
9 9986
85 27'
4 34'
8.9010
16
8 9024
16
1 .0976
9.9986
85 26''
4 35'
8.9026
16
8.9040
16
1.0960
9.9986
85 25'
4 36'
8.9042
16
8 9056
16
1 0944
9.9986
85 24'
4 37'
8 9057
15
89071
15
1.0929
9.9986
85 23'
4 38'
89073
16
8.9087
16
1.0913
9 9986
85 22'
4 39'
8 9089
16
15
8.9103
16
15
1.0897
9.9986
85 21'
4 40'
89104
15
8.9118
16
1.0882
9.9986
85 20'
4 41'
89119
8.9134"
"1.0866
99985"
85 19 X
4 42'
8 9135
16
8.9150
16
1.0850
9.9985
85 18'
4 43'
89150
ir>
8.9165
15
1.0835
9.9985
85 17'
40 44/
89166
16
9180
15
1.0820
9 9985
85 16'
4 45'
8.9181
15
8 9196
16
1.0804
9.9985
85 15'
4 46'
8.9196
15
8.9211
15
1 .0789
9.9985
85 14'
4 47'
8.9211
15
8.9226
15
1.0774
9.9985
85 13'
4 48'
8.9226
15
89241
15
1.0759
9.9985
S5 12'
4 49'
8.9241
15
15
8.9256
15
16
1.0744
A9985
85 11'
4 50'
8.9256
15
8.9272
15
1.0728
9.9985
85 10'
4 51'
8.9271
8.9287
1.0713
^.9984
85 9'
4 52'
8.9286
15
8.9302
15
1.0698
9.9984
85 8'
4 53'
.9301
15
.9316
14
1.0684
9.9984
85 7'
4 54'
.9315
14
8.9331
15
1.0669
9.9984
85 6'
4 55'
.9330
15
.9346
15
1.0654
9.9984
85 5'
4 56'
.9345
15
.9361
15
1.0639
9.9984
5 4'
4 57'
.9359
14
.9376
15
1.0624
.9984
5 3'
4 58'
9374
15
.9390
14
1.0610
.9984
5 2'
4 59'
.9388
15
.9405
15
1.0595
.9984
5 V
4 60'
.9403
15
.9420
15
1.0580
.9983
5 0'
ogcos
iff.!'
ogcot
com.
iff.l'
ogtan
ogsin
Angle
85
COSINES, TANGENTS, AND COTANGENTS
13
5-15
Angle
log sin
difi.l'
log tan
com.
diff.l'
log cot
log cos
diff.l'
5 0'
5 10'
5 20'
5 30'
5 40'
5 50'
6 0'
6 10'
6 20'
6 30'
6 40'
6 50'
7 0'
7 10'
7 20'
7 30'
7 40'
7 50'
8 0'
8 10'
8 20'
8 30'
8 -10'
8 50'
9 0'
9 10'
9 20'
9 30'
9 40'
9 50'
10 0'
10 ]0'
10 20'
10 30'
10 -10'
10 50'
11" 0'
11 10'
11 20'
11 30'
11 40'
11 50'
12 0'
12 10'
12 20'
12 30'
12 40'
12 50'
13 0'
13 10'
13 20'
13 30'
13 40'
13 50'
14 0'
14 10'
14 20'
14 30'
14 40'
14 50'
15 0'
8.9403
8.954T
8.9682
8.9816
8.9945
9.0070
14.2
13.7
13.4
12.9
12.3
11.9
11.5
11.3
10.9
10.7
10.4
10.2
99
9.7
9.5
93
9.1
8.9
8 7
8.5
84
H
7.9
7 R
7
7 3
7.3
7 1
7
8
6 8
d
h 6
64
fi.4
G .1
b 1
6 1
60
59
5 H
5.7
5 7
5.6
5 5
54
5-4
53
5.2
5.2
5.1
50
50
4 9
4.9
4 8
4.7
8.9420
14.3
13.8
135
13.0
1S.7
12.3
12.0
11 7
11.4
11.1
10.8
105
104
10.1
9.8
97
94
93
91
89
87
86
8 4
8 1
80
78
7 7
76
74
7.3
73
7 1
70
6 9
G.8
60
07
65
64
63
6.3
6.1
61
6.1
5.9
5.9
5.8
5.7
5.7
5.6
5.5
55
5.4
5.3
5.3
5.3
5.1
5.2
5.1
1 0580
9.9983
.2
.2
.2
.1
.2
2
.1
.2
.2
.1
.2
.2
.2
.2
.2
.2
.2
.2
.2
.2
.2
.2
.3
.2
2
.3
.2
.3
.2
.3
2
.3
.2
.3
.3
.2
.3
.3
.3
.3
.3
.3
.3
.3
.3
.3
.3
.3
.4
.3
.3
.4
85 0'
84 50'
84 40'
84 30'
84 20'
84 10'
84 0'
83 50'
83 40'
83 30'
83 20'
83 10'
83 0'
82 50'
82 40'
82 30'
82 20'
82 10'
82 0'
81 50'
81 40'
81 30'
81 20'
81 10'
81 0'
80 5<X
80 40'
80 30'
80 20'
80 10'
80 0'
79 50'
79 40'
79 30'
79 20'
79 10'
79 0'
78 50'
78 4(X
78 30'
78 20'
78 10'
78 0'
77 50'
77 40'
77 30'
77 20'
77 10'
770 o/
76 50'
76 40'
76 30'
76 20'
76 ICX
76 0'
75 50'
75 40'
75 3(X
75 20'
75 10 /
75 O'
8.9563
8.9701
8.9836
8.9966
9.0093
170437
1.0299
1.0164
1.0034
0.9907
9.9982
9.9981
9.9980
9.9979
9 9977
9 0192
90216
9784
9 9976
9.0311
9.0426
9 0539
9.0648
9.0755
9.0336
9.0453
9.0567
9 0678
90786
9664
0.9547
0.9433
0.9322
0.9214
9.9975
9.9973
9 9972
9 9971
9.9969
9.0859
9 0961"
9.1060
9.1157
9.1252
9 13-15
91436
91525
9.1612
9 1697
91781
9 1863
9 1943
9 2022
92JOU
92176
9 2251
9.2324
"9 2397
90891
9.0995"
91096
9.1194
9 1291
9.1385
0.9109
9 9968
9005
0.8904
0.8806
0.3709
8615
"0.8522
9 9966
9.996 +
9 9963
9 9961
9 9959
9.1478
9 9958
9.1569
9.1658
9 1 745
91S31
9 191 S
9 1997
9 1J078
9.2158
9 2236
92313
9.2389
<j247xT
8431
0.8342
08255
08169
0.8085
OS003
0.7922
07S42
7764
7687
0.7611
7537
9 9956
9 9954
99952
9.9950
9.9948
"99946
9 9944
9 9942
9 9940
9 9938
99936
9 9934
9 2468
9 2538
9 2606
9.2674
92740
9 2806
92S.-U)
9 2609
9 2680
9 2750
9.2819
"92887
0.7464
7391
7320
7250
0.7181
0.7TiT
9.9931
9.9929
9.9927
9.9924
9.9922
9.9919"
9 2870
9 2934
9 2997
93058
93119
9 29 S3
9.3020
9.3085
9.3149
9^212
0.7047
0.6980
0.6915
0685]
06788
9.9917
9.9914
9.9912
9 9909
9.9907
93179
9 3238
9.3296
9.3353
9.3410
9.3466
93521"
9 357T
9 3629
9 3682
9.3734
9.3786
9 J275
"91336"
9 3397
9.3458
93517
9.3576
9l6H
9~3691
9 3748
93804
9 3859
9 3914
6725
U666F
6603
06542
0.6483
06424
0~63~66~
"06301;"
0.6252
0.6196
6141
0.60S6
9 9904
9.9901
9 9899
9 9896
9 9893
9 9890
99887
9~9S84 r
9.9881
9.9878
9.9875
9.9872
9.3837
9.3887
9.3937
9.3986
9.4035
94083
9 3968
94021
9.4074
94127
94178
9.4230
9.428T
0.6~032
"0:5979
05926
5873
0.5822
0.5770
9 9869
"99866
9.9863
9.9859
9.9856
9.9853
9 4130
5719
9.9849
log cos
difi.l'
log cot
com.
diff . 1'
log tan
log sin
diff.l'
Angle
75-85
14
TABLE II. LOGARITHMIC SINES
15-25
Angle
log sin
diff.1'
log tan
JSffi'
log cot
log cos
diff.l'
15 0'
15 10'
15 20'
15 30'
15 40'
15 50'
16 0'
16 10'
16 20'
16 30'
16 40'
16 50'
17 0'
17 10'
17 20'
17 3CT
17 40'
17 50'
18 0'
18 10'
18 20'
18 30'
18 40'
18 50'
19 0'
19 10'
19 20'
19 30'
19 40'
19 50'
20 0'
20 10'
20 20'
20 30'
20 40'
20 50'
21 0'
21 10'
21 20'
21 30'
21 40'
21 50'
22 0'
22 10'
22 20'
22 30'
22 40'
22 50'
23 0'
23 10'
23 20'
23 30'
23 40'
23 50'
24 0'
24 10'
24 20'
24 30'
24 40'
24 50'
25 0'
9.4130
4.7
4.6
4.6
4.5
4.5
4.4
4.4
4 4
4 2
4:i
42
4.1
4 1
4 1
40
4
4
39
39
38
38
3.7
3 8
3.6
37
36
3 6
3.J
3.6
35
34
3.4
34
3.4
3 3
33
3.3
33
32
32
3.1
3.2
3 1
3 1
3.0
3 1
30
3.0
2.9
3
2.9
29
2.9
28
28
28
28
28
27
2.7
9.4281
5.0
5.0
4.9
4.9
4.8
4.8
4.7
4.7
4.7
46
46
4.5
4.5
4.5
4 4
4.4
4 4
4 3
4 3
4 2
4 2
4.2
4 2
4 1
4 1
4.0
4.0
4.0
4.0
4.0
3.9
3.9
3 8
3 9
3 8
3 8
3.7
3 8
3 7
3 7
3.7
36
36
3 6
3.6
3 6
3.5
3 6
3.5
3.4
3 5
3 4
3.5
3.4
3.4
3.3
3.4
3.3
3.4
3 3
0.5719
9.9849
.3
.3
.4
.3
.4
.4
.3
.4
.4
.3
.4
.4
.4
.4
.4
.4
.4
.4
.4
.4
.4
.5
.4
.4
.5
.4
.5
.4
.4
.4
.5
.5
4
.5
.5
.5
.5
.5
.5
.6
.5
.5
.5
.6
.5
.6
.5
.6
5
.6
.5
.6
.6
.6
.5
.6
75 0'
74 50'
74 40'
74 30'
74 20'
74 10'
74 0'
73 50'
73 40'
73 30'
73 20'
73 10'
73 0'
72 50'
72 40'
72 30'
72 20'
72 10'
72 0'
71 50'
71 40'
71 30'
71 20'
71 10'
71 0'
70 50'
70 40'
70 30'
70 20'
70 10'
70 0'
69 50'
69 40'
69 30'
69 20'
69 10'
69 0'
68 50'
68 40'
68 30'
68 20'
68 10'
68 0'
67 50'
67 40'
67 30'
67 20'
67 10'
67 0'
66 50'
66 40'
66 30'
66 20'
66 ICX
66 0'
65 50'
65 40'
65 30'
65 20'
65 10'
65 0'
9.4177
9.4223
9.4269
94314
9.4359
9.4403
9.4331
9.4381
9.4430
9.4479
9.4527
"94575
0.5669
0.5619
0.5570
0.5521
0.5473
0.5425
9.9846
9.9843
9 9839
9.9836
9 9832
9 9828
9.4447
94491
9.4533
94576
9.4618
9.4622
9.4669
9.4716
94762
9.4808
0.5378
0.5331
0.5284
0.5238
0.5192
9.9825
99821
9.9817
9.9814
9.9810
"99806"
9.4659
9.4853
0.5147
9.4700
9.4741
9.4781
9.4821
94861
9 4898
9.4943
9 4987
95031
9.5075
95118
9."5T6i
9 5203
9 5245
9 5287
9.5329
9 57370
9.5411
95451
9.5491
95531
9.5571
9.5611
05102
0.5057
5013
04969
4925
9.9802
9.979S
9 9794
9.9790
99786
9.4900
9.4939
9.4977
95015
9.5052
9.5090
93126"
9.5163
9.5199
9.5235
95270
9.5306
"9534T
0.4882
0.4839
0.4797
04755
04713
0.4671
04630
074589
0.4549
4509
0.4469
04429
4^89
9.9782
9 9778
9.9774
9 9770
9 9765
9 9761
"^9757"
9 9752
9.9748
9.9743
9 9739
9 9734
9 9730"
95375
9.5409
9.5443
9.5477
9.5510
"975543
9.5576
9.5609
9.5641
9.5673
9.5704
9 5650
9.5689
95727
9.5766
9.5804
9 5842
0.4350
0.4311
04273
0.4234
04196
4 1 58"
9 9725
99721
9 9716
9.9711
9 9706
9 9702
9.5879
9.591 7
9.5954
9.5991
9 6028
0.4121
0.40S3
0.4046
0.4009
0.3972
9 9697
9 9692
9 9687
9 96S2
9 9677
9 5736
9.5767"
9.5798
9 5828
9.5859
9 5889
"9.5919"
9.6064
9 6100
96136
9.6172
96208
9 6213
9.6279
0.3936
0~3900
0.3864
0.3828
0.3792
0.3757
113721"
9.9672
9 9667
9 9661
9.9656
99051
9 9646
99640
"9.9635"
9 9629
9.9624
9.9618
9.9613
99607"
9.5948
9.5978
9.6007
9 6036
9.6065
"9.609T
9.6314
9.6348
9 6383
96417
9.6452
9.6486
0.3686
3652
0.3617
0.3583
3548
0.3514"
0.3480"
0.3447
0.3413
0.3380
3346
OT373T3"
9.6121
9.6149
9.6177
9 6205
9.6232
9.6520
9.6553
96587
9 6620
9 6654
96687
9 9602
9.9596
9.9590
9.9584
9.9579
979577T
9 6259
log COB
diff.l'
log cot
com.
diff . 1'
log tan
log sin
difi. 1'
Angle
65-75
COSINES, TANGENTS, AND COTANGENTS 15
25-35
Angle
log sin
difi.l'
log tan
com.
diff. I/
log cot
log cos
difi.r
25 0'
9.6259
9.6687
0.3313
99573
65 0'
25 10'
9.6286
2.7
9^72CT
3.3
0.328CT
9.9567
.6
64 50'
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'
9 6392
2.6
9.6850
3.3
03150
9.9543
6
64 10'
26 0'
964TS"
2.6
9.6882
3.2
0.3118
~93TJ7~
.6
64 0'
26 10'
96444
2.6
"9.6914
3.2
0.3086
"9.9530"
.7
63 50'
26 20'
96470
26
9.6946
3.2
3054
99524
.6
63 40'
26 30'
96495
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 50'
9.6546
2 5
9.7040
3.1
2960
9.9505
.7
63 10'
27 0'
9.6570
2.4
9.7072
3.2
0.2928
"979499
.6
63 0'
27 10'
9.6595~
2.5
9.7103
3 1
0.2897
9 9492""
.7
62 50'
27 20'
9.6620
2.5
9.7134
3.1
2866
9.94S6
.6
62 40'
27 30'
9.6644
2.4
9.7165
3.1
0.2835
9.9479
.7
62 30'
27 40'
96668
2 4
9.7196
3 1
0.2804
9.9473
.6
62 20'
27 50'
9 6692
2 4
9.7226
3
0.2774
9 9466
.7
62 10'
28 0'
9 6716
2.4
9.7257
3.1
0.2743"
9.9459
.7
62 0'
28 10'
~9.6740~
2.4
9.7287
3.0
0.2713
9.9453
.6
61 50'
28 20'
9 6763
2.3
9.7317
3.0
2683
9.9446
.7
61 40'
28 30'
9 6787
2.4
9.7348
3 1
0.2652
9.9439
.7
61 30'
28 40'
9.6810
2 3
9.7378
3.0
2622
9.9432
7
61 20'
28 50'
9.6833
2.3
9.7408
3
2592
9.9425
.7
61 10'
29 0'
9.6856~
2.3
2 2
~9."7438
3
2562
9~94 18
.7
_Y
61 0'
29 10'
"96878"
9.7467
**
2533
99411
60 SO'
29 20'
9.6901
2 3
9.7497
3.0
2503
99404
.7
60 40'
29 30'
9.6923
2 2
9.7526
2.9
0.2474
9.9397
.7
60 30'
29 40'
9.6946
2 3
9.7556
3
02444
9.9390
.7
60 20'
29 50'
9.6968
2.2
9.7585
2 9
02415
9.9383
.7
.8
60 10'
30 0'
9~699()
*"~
9 7614
*,.
2386
99375
60 0'
30 10'
~9.7012
2.2
9.7644"
2 9
02356
"9 9368
.7
59 50'
30 20'
9.7033
2.1
9 7673
2.9
02327
9 9361
.7
59 40'
30 30'
9.7055
2 2
9 7701
2 8
2299
9 9353
.8
59 30'
30 40'
9.7076
2.1
9.7730
2 9
0.2270
9 9346
.7
59 20'
30 50'
9.7097
2 1
9.7759
2.9
02241
9 9338
.8
59 10'
31 0'
9T118
2.1
9 778S
2 9
~0.2212~
99331
.7
59 0'
31 10'
9.7139
2 1
97816
2.8
2184"
9.9323
.8
58 50'
31 20'
9.7160
2.1
9.7845
2 9
02155
9.9315
.8
58 40'
31 30'
9.7181
2 1
9 7873
2 8
02127
9 9308
.7
58 30'
31 40'
9.7201
2.0
9.7902
2 9
0.2098
9 9300
.8
58 20'
31 50'
9.7222
y.i
9 7930
2 8
0.2070
9 9292
.8
58 10'
32 0'
9.7242
2.0
9.7958"
2.8
2042
9 9284
8
58 0'
32 10'
9.7262
2.0
97986"
2.8
"0*20 14"
9^9276
.8
57 50'
32 20'
9.7282
2.0
9 8014
2 8
0.1986
9 9268
.8
57 40'
32 30'
9.7302
2
9 8042
2 8
0.1958
9 9260
8
57 30'
32 40'
9.7322
2.0
9.S070
2.8
1930
9 9252
.8
57 20'
32 50'
9.7342
2
9.8097
2 7
0.1903
9 9244
.8
57 10'
33 0'
9.7361
1.9
"9.^125"
2.8
0.1875"
99236
.8
57 0'
33 10'
9.7380
J 9
"9.8T53
2 8
0.1847
"9.9228"
.8
56 50'
33 20'
9.7400
2
9 8180
2.7
01820
99219
.9
56 40'
33 30'
9 7419
1 9
9 8208
2 8
0.1792
99211
.8
56 30'
33 40'
97438
] 9
9 8235
2 7
1765
9 9203
.8
56 20'
33 50'
9.7457
1.9
9 8263
2.8
0.1737
9.9194
.9
56 10'
34 0'
97476
1 9
1.8
98290
2 7
Q.7
"0717 ICT
9.9186
.8
.9
56 0'
34 10'
97494
9.8317
0.1683
9.9177
55 50'
34 20'
9.7513
1.9
9.8344
2 7
0.1656
9.9169
.8
55 40'
34 30'
9.7531
1.8
9.8371
2 7
0.1629
99160
.9
55 30'
34 40'
9.7550
1.9
9.8398
2 7
0.1602
9.9151
.9
55 20'
34 50'
9.7568
1.8
9.8425
2.7
01575
9.9142
.9
55 10'
35 0'
~9.7586
1.8
~9 8452~
2.7
0.1548
9.9134
.8
55 0'
log cos
diff. 1'
log cot
com.
diff.l'
log tan
log sin
diff.r
Angle
55-65
16
TABLE II. LOGARITHMIC SINES
5
*5-4,
5
Angle
log sin
dig.l
log tan
com.
diff.l'
log cot
log COB
diff . If
35
9.7586
1 8
9.8452
2 7
0.1548
9.9134
g
55 0'
35 10'
35 20'
35 30'
35 40'
35 50'
9.7604
9 7622
9.7640
9.7657
9 7675
J.8
1.8
1.7
1.8
1 7
9.8479
9 8506
9 8533
9.8559
9 8586
2.7
2 7
2.6
2.7
o 7
1521
1494
01467
0.1441
01414
9.9125
9.9116
9.9107
9.9098
9.9089
.9
.9
.9
.9
9
54 50'
54 40'
54 30'
54 20'
54 10'
36
9.7692
1 8
98613
2 6
01387
9.9080
1
54 0'
36 10'
36 20'
36 30'
36 40'
36 50'
9.7710
9.7727
97744
9 7761
9 7778
1.7
1.7
1.7
1.7
1 7
9 8639
9 8666
9.8692
98718
9.8745
2.7
2.6
2 6
27
t> Q
1361
0.1334
0.1308
0.1282
1255
9.9070
9.9061
9.9052
9 9042
9 9033
.9
.9
1.0
.9
1
53 50'
53 40'
53 30'
53 20'
53 10'
37 0'
9 7795
1 6
9 8771
'* 6
1229
9.9023
q
53 0'
37 10'
37 20'
37 30'
37 40'
37 50'
38 0'
977811
9 7828
9 7844
97861
97877
9 7893
1.7
1.6
1 7
1 6
1 6
1 7
9 8797
9.8824
9 8850
9 8876
9 8902
9 8928
26
2 6
'J.6
ac>
O (J
1203
0.1176
01150
01124
1098
1072
99014
9 9004
98995
9 8985
9S97S
9 8965
1.0
.9
.0
.0
.0
o
52 50'
52 40'
52 30'
52 20'
52 10'
52 0'
38 10'
38 20'
38 30'
38 40'
38 50'
9.7910
9 7926
9.7941
9.7957
9 7973
l.G
i.r>
1 G
1 6
1 6
9 8954
9 8980
9-9006
99032
9 90S 8
26
2 (>
2
2 b
O (J
1046
1020
0994
0.0968
0.0942
98955
98945
9 8935
9 8925
98915
.0
51 50'
51 40'
51 30'
51 20'
51 10'
39 0'
39 10'
39 20'
39 30'
39 40'
39 .SO'
40 0'
40 10'
40 20'
40 30'
40 40'
40 50'
9.7989
9 8004"
9.8020
9 8035
9 8050
9 8066
9cS08F
T A8097T
9.8111
9.81 25
9.8140
98155
1 5
1.6
I ft
1 ft
l.G
1 ft
1 ft
1 5
1.4
1 ft
1 5
9 9084
99110
9.9135
9.9161
9 9187
9.9212
ij923S-
9.926-f
9 9289
99315
9.9341
9 9366
2 (>
2 r>
26
2 6
2.5
26
26
2 ft
2 6
2 6
25
0916
0890
0086S
0.0839
00813
0788
0.0762"
0.0736
0.0711
00685
00659
0.0634
9 8905
"98895
9 8884
9 8874
9 8864
9 8853
9" 8843"
9 8832"
9 8821
98810
9 8800
9 8789
1
.0
1
.1
I
I
1
J
51 0'
50 SO'
50 40'
50 30'
50 20'
50 10'
50 V
49 50'
49 40'
49 30'
49 20'
49 10'
41 0'
41 10'
41 20'
41 30'
41 40'
41 50'
42 0'
42 10'
42 20'
42 30'
42 40'
42 50'
98169
98TM"
9.8198
9.8213
9 8227
9 8241
9.8255
9.8269
98283
9.8297
98311
9 8324
1 ft
J.4
1 ft
1 4
1.4
1.4
1.4
1.4
1.4
1.4
1.3
1 4
9 9392
9.9417"
9 9443
9 9468
9.9494
99S19
9 9544"
9" 95 70
9.9595
9.9(521
9 9646
9.9671
2.5
26
2 G
2.">
2f>
2.6
2.5
25
2 6
0608
0.0583
00557
0532
0.0506
004S1
0.04~5~6~
00430
00405
0.0379
0354
00329
9 8778
"98767
9 8756
98745
9 8733
9.8722
9787IT
9.8699
9 8688
9 8676
98665
9.8653
.1
.1
1.1
1.2
1 1
1.1
1 2
1.1
J.2
1.1
1 2
1.2
49 0'
48 50'
48 40'
48 30'
48 20'
48 10'
48 0'
47 50'
47 40'
47 30'
47 20'
47 10'
43 0'
9 8338
1 *3
9.9697
0.0303
9 8641
1
47 0'
43 10'
43 20'
43 30'
43 40'
43 50'
44 0'
98351
9 8365
9 8378
9 8391
9 8405
98413"
1.4
1.3
1.3
1.4
1.3
1 3
9 9722
9.974 7
9.9772
9.9798
9.9823
9.9848
2.5
25
26
2 5
25
2 6
00278
0.0253
0.0228
00202
0.0177
00152
9 8629
9.8618
9 8606
9 8594
9.8582
9.8569
1.1
1.2
1.2
1 2
1.3
1 2
46 50'
46 40'
46 30'
46 20'
46 10'
46 0'
44 10'
44 20'
44 30'
44 40'
44 50'
98431
9.8444
9.8457
98469
9.8482
1.3
1.3
1.2
1.3
9.9874
9 9899
9.9924
9 9949
9.9975
2.5
25
25
26
60126"
00101
0.0076
0.0051
0025
9.8557
9.8545
9.8532
9.S520
9 8507
1.2
1 3
1.2
1.3
45 50'
45 40'
45 30'
45 20'
45 10'
45 0'
9.8495
0.0000
00000
9 8495
45 0'
log cos
diff.l'
log cot
log tan
log sin
diff.l'
Angle
4
5-55
CONVEBSIOK TABLES FOE ANGLES
17
TO CHANGE FROM MlNUTES AND SECONDS INTO THE DECIMAL
PAKTS OF A DEGREE OR INTO RADIANS
From seconds ^\^
From minutes
From degrees into
radians
1" = 0.00028 = 0.0000048 Rad.
l'=0017=O.OOQ29Rad
1 = 0.01745 Rad.
2" = 0.00056 = 0000097
2'=0033=0 00058 tl
2 = 0.03491
3" = 0.00083 = 000014 5
3' =0050 = 0.00087 "
3 =005 236
4" = 0.001 1 1 = 0.0000194
4' = 0067 = 0.001 16 "
4=0 06981
5" = 001 39 = 0.0000242
S'=OOS3 = 00145 "
5=0 08727
6"=0.00167 = 0.0000291
6 / = 0'lOO=000175 "
6 =0.1 0472
7"= 0.00194 = 0.0000339
7'=0.117=0 00204 "
7=0.12217
8" = 0.00222 = 0.0000388
8' = 01 33 =000233 "
8 = 13963
9"= 00250 = 0000436
9' =0.1 SO = 000262 "
9=0 15708
10"=000278=00000485
10 / =0167 = 00291 "
10 = 017453
20" =0 005 56 =0 0000970
20'=0333=0 00582 "
20 = 034907
30" = 00833 = 00014 54
30' = 500= 000873 "
30 = 052360
40" = 001 111 = 00001 939
40' = 0667=0 01164 "
40 = 0.6081 3
50" = 01389= 0002424
50' =0833 = 001454 "
50 = 087 266
TO CHANGE FROM DECIMAL PARTS OF A DEGREE INTO MlNUTES
AND SECONDS
0000 = 000' = 0"
"00001 =~~0~006' =~0 36'~
0.0002= 0012'= 072"
0.0003 = 018' = 1 08"
00004- 0.024'= 1 44"
00005= 0.030'= 180"
00006= 0036'= 2.16"
0.0007= 0.042'= 2.52"
0.0008= 0.048'= 2.88"
00009= 0.054'= 324"
020= J2()'= 12'
21~1~2 6'~=T2' 36 7r
022= 132'= 13' 12"
025=^ 13S' = 13' 48"
021 = 144' = 14' 24"
025= 150'= 15'
026= 156'= 15' 36"
0.27 - 16 2' = 16' 12"
028= 168'= 16' 48"
0.29= 174' = 17' 24"
0.60 = 36 0' = 36'
~0 61~=~36 6 7 ^~36'"36 7? ~
062 = 372' = 37' 12"
063 = 378' =r 3 7' 48"
0.64 = 38.4' = 38' 24"
0.65 = 39 0' = 39'
0.66 = 39 6' = 39' 36"
0.67 = 40 2' = 40' 12"
0.68 = 40.8' = 40' 48"
069 = 41.4' = 41' 24"
0.0010= 0.060'= 3.60"
030= 18.0' = 18'
70 = 42 0' = 42'
0001 = 006' = 3.6"
0.002 = 0.12' = 7.2"
0.003 = 018' = 108"
0004 = 024' = 144"
0.005 = 0.30' = 180"
0006 = 036' = 216"
0007 = 042' = 252"
0008 = 0.48' = 288"
0009 = 054' = 324"
031= 186' = 18' 36"
0.32 = 192'= 19' 12"
033= 198' = 19' 48"
34 = 20 4' = 20' 24"
035 = 210' =21'
036 = 216' = 21' 36"
037 = 22 2' = 22' 12"
38 = 22.8' = 22' 48"
39 = 23 4' = 23' 24"
071 = 426' = 42 X 36"
72 = 43.2' = 43' 12"
73 = 43 8' = 43' 48"
74 = 44 4' = 44' 24"
75 = 45.0' = 45'
0.76 = 45.6' = 45' 36"
77 = 46.2' = 46' 12"
78 = 46 8' = 46' 48"
79 = 47.4' = 47' 24"
0010 = 060' = 36.0"
40 = 24 0' = 24'
80 = 48 0' = 48'
0.01 =06' = 36"
0.02 = 1.2' = 1' 12"
03 =18' = V 48"
0.04 = 2.4' = 2' 24"
005 = 3.0' = 3'
0.06 =36' = 3' 36"
007 = 4.2' = 4' 12"
0.08 = 4.8' = 4' 48"
09 =54' = y 24"
041 = 246' = 24' 36"
0.42 = 25 2' = 25' 12"
043 = 258' = 25' 48"
044 264' = 26' 24"
045 = 270' = 27'
46 = 27 6' = 27' 36"
047 = 28 2' = 28' 12"
48 = 28.8' = 28' 48"
49 = 29.4' = 29' 24"
81 = 48 6' = 48' 36"
82 = 49 2' = 49' 12"
083 = 498' = 4 9' 48"
84 = 50.4' = 50' 24"
0.85 = 510'= 51'
0.86 = 51.6' = 51' 36"
87 = 52 2' = 52' 12"
088 = 528' = 52' 48"
0.89 = 53 4' = 53' 24"
010 = 6.0' = 6'
50 = 30 0' = 30'
0.90 = 54 0' = 54'
0.11 = 6.6' = 6' 36"
0.12 =72' = r 12"
0.13 =78' = 7' 48"
0.14 = 8.4' = 8' 24"
0.15 = 9.0' = 9'
0.16 = 9.6' = 9' 36"
0.17 = 10.2' = 10' 12"
0.18 = 10.8' = 10' 48"
0.19 = 11 4' = 11' 24"
0.51- 30 6' = 30' 36"
052 = 31. 2' = 31' 12"
053 = 31. 8' = 31' 48"
0.54 = 32 4' = 32' 24"
0.5 5 = 330' = 33'
0.56 = 33.6' = 33' 36"
0.57 = 34.2' -. 34' 12"
0.58 = 34.8' = 34' 48"
0.59 = 35.4' = 35' 24"
91 = 54 6' = 54' 36"
0.92 = 55 2' = 55' 12"
0.93 = 55.8' = 55' 48"
94 = 56 4' = 56' 24"
095 = 57.0'= 57'
0.96 = 57.6' = 57' 36"
097 = 58.2' = 58' 12"
0.98 = 58.8' = 58' 48"
0.99 = 59.4' = 59' 24"
0.20 = 12.<y = 12'
060 = 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 10) of the sines,
cosines, tangents, and cotangents of all angles from 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 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
TABLE III. LOGARITHMIC SINES
Angle
log sin
diff.
log tan
com.
difi.
log cot
log COB
Prop. Farts
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
OJ1
0.12
0.13
0.14
015
016
0.17
018
0.19
0.20
021
022
0.23
0.24
0.25
0.26
0.27
0.2cS
0.29
0.30
031
0.32
0.33
034
0.35
0.36
0.37
0.38
0.39
0.40
0.41
0.42
0.43
0.44
045
0.46
0.47
0.48
0.49
0.50
672419"
6.5429
6.7190
6.8439
6.9108
7.0200
7.0870
7.1450
7.1961
oo <o <o <c <o o 5 o S " "J3 5 5 G Ordinary interpolation here will in general give inaccurate results.
-10- *.w o ta ,u oo o u> o> o e* o * <x u> Instead use formulas on p. 6.
a, <o <o <o o w to S 5 S - C ~ E ^ JS 5 - Ordinary interpolation here will in general give inaccurate results.
ooo~ 4.0,00 ww^occ^o . 3 o * S w Instead use formulas on p. 6.
10.0000
90.00
89.99
89.98
89.97
89.96
89.95
89.94
89.93
89.92
89.91
89.90
89.89
89.88
89.87
89.86
89.85
89.84
89.83
89.82
89.81
89.80
89.79
89.78
89.77
89.76
89.75
89.74
89.73
89.72
89.71
89.70
89.69
89.68
89.67
89.66
89 65
89.64
89 63
89.62
89.61
8960
89.59
89.58
89.57
89.56
89.55
89.54
89.53
89.52
89.51
89.50
Extra digit
Difference
6.2419
6.5429
6.7190
6.8439
69408
70200
7.0870
7.1450
71961
3.7581
3.4571
3.2810
3.1561
3.0592
2 9800
29130
28550
2.8039
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
100000
100000
10.0000
i
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
1
2
3
4
6
7
8
9
1
2
3
4
6
7
8
9
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
79
7.9
158
'23 7
31 6
39.5
474
55.3
63.2
71.1
76
7.6
15 2
22 !s
304
380
45.6
532
608
684
73
7.3
14.6
21 9
29.2
365
438
51 1
58 4
657
69
6.9
138
207
27.6
345
41 4
48.3
55.2
62 1
66
6.6
13.2
19.8
204
330
396
462
52.8
59.4
63
63
126
189
25.2
315
378
441
504
56.7
78
7.8
15.6
23.4
31.2
390
468
54.6
62.4
70.2
75
7.5
15.0
22.5
30.0
375
450
52.5
600
675
72
14.4
21.6
2H8
36.0
43.2
50.4
57.6
648
68
68
136
5*0.4
27.2
MO
408
47.6
544
(il 2
65
6.5
130
19.5
260
32 It
39.0
45.5
52.0
58.5
62
6.2
124
186
248
310
37.2
43.4
49.6
55.8
77
7.7
154
23.1
308
38.5
46.2
53.9
61.6
69.3
74
7.4
148
222
29.6
37.0
444
51 8
592
66.6
71
7.1
14.2
21.3
284
355
42.6
49.7
56.8
639
67
67
134
20.1
26.8
33.5
402
469
53.6
603
64
6.4
128
19.2
25.0
32.0
38.4
44.8
51 2
57.6
61
6.1
12.2
183
24.4
30.5
366
42.7
48.8
54.9
7.2419
7.2419
2.7581
10 0000
7 2833
7.3211
7.3558
7.3880
7.4180
7.4460
7.4723
7.4971
7.5206
775429
7.2833
7.3211
7.3558
7.3880
7.4180
7.4460
74723
7.4972
7.5206
2.7167
2.6789
2.6442
26120
2.5820
2 5540
2 5277
2.5028
2.4794
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10 0000
10 0000
10.0000
7.5429
2 4571
10.0000
7.5641
7.5843
7.6036
7.6221
7.6398
7 6568
7.6732
7.6890
7 7043
7.7190
75641
75843
7 6036
7.6221
7.6398
7.6569
7.6732
7.6890
7.7043
7.7190
2.4359
2.4157
2.3964
23779
23602
2.3431
2 3268
2.3110
2 2957
Y.2810'
10.0000
10 0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
7.73 52
77470
77604
7 7734
7.7859
7.7982
78101
78217
7.8329
7.8439~
7.7332
7.7470
7.7604
7.7734
7.7860
7.7982
78101
78217
78329
7.8439
2.2668
2.2530
2.2396
2.2266
22140
2.2018
2.1899
2.1783
21671
"27156T
10.0000
10 0000
10.0000
10.0000
10 0000
10.0000
100000
10 0000
10 0000
laoooo
10.0000
10.0000
10.0000
10.0000
100000
10.0000
100000
10.0000
10.0000
7.8547
7.8651
7.87S3
7.8853
7.8951
7.9046
7.9140
7.9231
7.9321
7.SS47
7.S6S1
7.8754
7 8853
7.8951
7.9046
7.9140
79231
7.9321
2.1453
2.1349
2.1246
21147
2.1049
2.0954
2.0860
2.0769
2.0678
7.9408
7.9409
2.0591
10.0000
log 008
diff.
log cot
com.
diS,
log tan
log sin
Angle
89
(JOSINES, TANGENTS, AND COTANGENTS
21
Angle
log sin
diff,
log tan
com.
diff.
log cot
log cos
Prop. Farts
0.50
79408
86
7.9409
86
2.0591
100000
89.50
^
051
7.9494
7.9495
2.0505
10.0000
89-49
bD
0.52
0.53
7-9579
7.9661
82
82
7.9579
7.9662
84
83
81
2.0421
2.0338
100000
10.0000
89.48
89.47
Td
1
Difference
0.54
7.9743
79743
20257
10.0000
89.46
H
0.55
7 9822
7.9823
80
20177
10.0000
89.45
0.56
7.9901
79
76
7.9901
7$
77
2.0099
10.0000
89.44
60
59
58
0.57
7.9977
79978
2.0022
100000
89.43
1
6.0
1
5.9
11 8
5.8
11 6
0.58
8.0053
8.0053
1.9947
10.0000
89.42
3
180
177
174
0.59
8.0127
73
8.0127
73
1.9873
10.0000
89.41
4
g
24.0
30
23 6
29 'i
232
29
0.60
8.0200
8.0200
1.9800
10.0000
89.40
6
3C.O
354
348
0.61
0.62
S~0272
8.0343
71
1T0272
8 0343
71
1.9728
1.9657
100000
10.0000
89.39
89.38
7
8
9
42.0
4S.O
54.0
41.3
47.2
53 1
40.6
46.4
52.2
063
80412
8.0412
1.9588
10.0000
89.37
K7
EC
KK
0.64
65
80480
80S48
68
8.0481
8 0548
67
19519
1 9452
10.0000
10 0000
89 36
89.35
1
2
57
11 4
5.6
11 2
5.5
11
066
80614
80614
1.9386
10.0000
8^.34
3
17.1
108
16.5
66
4
22.8
224
22.0
0,67
8 0679
80680
1.9320
10 0000
89.33
5
28.5
280
275
0.68
069
80744
80807
63
63
8.0744
8.0807
64
63
1.9256
1.9193
10.0000
100000
89.32
89.31
b
7
8
34 2
39.9
456
336
3,9.2
44.8
33.0
38.5
440
0.70
8.0870
fit
80870
1 9130
10.0000
89.30
9
51 3
50.4
49.5
071
80931
8 0932
1.9068
100000
89.29
54
f>3
5?
072
8.0992
8.0992
1.9008
10.0000
89.28
1
54
5 3
5.2
0.73
8.1052
8.1052
1.8948
100000
89.27
2
108
10.6
10.4
59
3
16.2
159
156
074
81111
8.1111
18889
10.0000
89.26
4
21.6
212
20.8
075
8.1169
58
r )B
8.1170
59
1 8830
10.0000
89.25
i>
27.0
32 4
265
31 8
2(5.0
31 Q
076
8.1227
8.1227
1.8773
10.0000
89.24
7
37.8
37.1
3G.4
0.77
8.1284
8.1284
57
1 8716
100000
89.23
8
<)
43.2
48.C
4'24
47.7
41 6
46.8
0.78
81340
8.1340
56
1 8660
10 0000
89.22
0.79
8 1395
53
8.1395
55
1.8605
10.0000
89.21
51
50
49
0.80
8~1450
55
53
~8"1450"
55
54
T8550"
Taoooo
89.20
1
2
51
103
50
100
4.9
9.8
0.81
8 1503
8 1504
1 8496
10.0000
89.19
3
153
15.0
14.7
0.82
0.83
8.1557
8.1609
54
52
81557
81610
53
53
1.8443
1.8390
10 0000
10.0000
89.18
89.17
4
5
C>
204
255
30.G
200
25.0
30.0
19 6
245
29.4
0.84
8 1661
81662
52
1.8338
10.0000
89.16;
7
8
357
408
350
400
343
39.2
0.85
8.1713
8.1713
51
1.8287
10.0000
89.15
9
45.9
45.0
44.1
086
8.1764
50
8.1764
51
50
1.8236
10.0000
89.14
48
47
46
087
8.1814
8.1814
1 8186
9.9999
89.13
1
48
4.7
4.6
88
8.1863
8.1864
1.8136
9.9999
89.12
2
9.6
9.4
9.2
0.89
8.1912
49
49
8 1913
49
49
1.8087
9 9999
89.11
3
4
144
192
14.1
18.8
13.8
18.4
090
8.1961
48
8.1962
48
1.8038
9.9999
89.10
5
g
240
1/> S S
23.5
28 2
23.0
27 6
0.91
8 2009
82010
1.7990
9.9999
8909
7
33 6
32.9
322
092
8.2056
8.20S7
1.7943
99999
89 08
8
38.4
37.8
36.8
0.93
8.2103
47
8.2104
47
46
1.7896
9.9999
89.07
9
43.2
42.3
414
094
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
i
4.5
4.4
4.3
0.96
8.2241
45
8.2242
46
45
1.7758
9.9999
89.04
3
9.0
135
88
132
8.6
129
0.97
8.2286
82287
1.7713
9.9999
89.03
4
5
180
225
176
22
17.2
21 5
0.98
82331
8.2331
1 7669
9 9999
89.02
27.0
20.4
25.8
0.99
S.2375
8.2376
1 7624
9.9999
89.01
7
g
31.5
36
30.8
35 2
30.1
34 4
1.00
8.2419
8.2419
1.7581
9.9999
89.00
9
40.5
39.6
38.7
log COB
diff.
log cot
com.
diff,
log tan
log sin
Angle
89
22
TABLE III. LOGARITHMIC SINES
1
Angle
log ain
diff.
log tan
com.
diff.
log cot
log cos
Prop. Parts
1.00
1.01
1.02
1.03
1.04
1.05
1.06
1.07
1.08
1.09
1.10
.11
.12
.13
.14
.15
.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
8.2419
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
8.2419
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
3J
31
31
30
30
30
30
30
29
30
29
1.7581
9.9999
89.00
88.99
88.98
88.97
88.96
88.95
88.94
8893
88.92
8891
88.90
88.89
8888
88.87
88.86
88.85
88.84
88.83
88.82
88.81
88.80
88.79
88.78
88.77
88.76
88.75
88.74
88.73
88.72
88.71
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
8857
88.56
88 55
88.54
88 53
88.52
88 51
88.50
1
'
d
&
Difference
8.2462
8.2505
8.2547
8.2589
8.2630
8.2672
8.2712
8.2753
8.2793
8.2462
8.2505
8.2548
8.2590
8.2631
8.2672
8.2713
8.2754
82794
1.7538
1.7495
1.7452
1.7410
1.7369
1.7328
1.7287
1.7246
1.7206
9.9999
9.9999
9.9999
9.9999
9.9999
9.9999
9.9999
9.9999
9.9999
1
2
3
4
5
6
7
8
9
1
3
4
5
6
7
8
9
1
2
3
4
5
6
7
s
9
1
3
4
5
6
7
8
9
1
2
3
4
8
9
1
2
3
4
5
6
7
8
9
43
4.3
8.6
12.9
17.2
215
25.8
30.1
344
387
41
4.1
12*3
16.4
20.5
246
28.7
32.8
36.9
39
39
7 8
11.7
15.6
19.5
23.4
27.3
31.2
35 1
37
37
7.4
11.1
148
18 5
259
296
33.3
34
3.4
6.8
10.2
136
170
20.4
238
27.2
306
31
31
6.2
9.3
12.4
155
186
21.7
248
27.9
42
4.2
8.4
12.6
168
21.0
25.2
29.4
33.6
37.8
40
40
8.0
12.0
16.0
20.0
240
280
320
36.0
38
3.8
7.6
11.4
15.2
190
22 8
26.6
30.4
34.2
36
36
io!s
14.4
180
21.6
25.2
288
324
35
35
7.0
10.5
140
17.5
21.0
24.5
280
315
82832
8.2833
1.7167
9.9999
8.2872
8.2911
8.2949
8.2988
8.3025
8.3063
8.3100
8.3137
83174
82873
8.2912
8.2950
8.2988
8.3026
8.3064
8.3101
83138
83175
1.7127
1.7088
1.7050
1.7012
1.6974
1.6936
1.6899
1.6862
1.6825
9.9999
9.9999
9.9999
9.9999
9.9999
9.9999
9.9999
9.9999
9 9999
8.3210
8.3211
1.6789
9.9999
8.3246
8.3282
8.3317
8.3353
8.3388
8.3422
8.3456
8.3491
8.3524
83247
8.3283
8.3318
8.3354
8.3389
8 3423
8 3458
8.3492
8.3525
1.6753
1.6717
1.6682
1.6646
1.661 1
1.6577
1.6542
1.6508
1.6475
9.9999
9.9999
9.9999
9.9999
9.9999
9.9999
9.9999
9.9999
9.9999
8.3558
8.3559
1.6441
L6408"
1.6375
1.6342
1.6309
1.6277
1.6245
1 6213
1.6182
1.6150
9.9999
8.3591
8.3621
8.3657
8.3689
8.3722
8.3754
8.3786
83817
83848
8.3592
8.3625
8.3658
8.3691
8.3723
8.3755
83787
8.3818
8.3850
9.9999
9.9999
9.9999
9.9999
99999
9.9999
9.9999
9.9999
9.9999
33
3.3
66
99
132
165
19.8
23.1
264
29.7
30
30
6.0
90
12.0
15.0
18.0
21.0
24.0
270
32
3.2
6.4
9.6
12.8
160
192
J24
25.6
288
29
H's
8.7
11 6
14.5
17.4
20.3
23.2
26.1
83880
8.3881
1.6119
9.9999
83911
8.3941
8.3972
8.4002
84032
8.4062
8.4091
8.4121
84150
83912
83943
8.3973
84003
8.4033
8.4063
8.4093
8.4122
8.4152
1.6088
1.6057
1.6027
1 5997
1.5967
1.5937
1.5907
1.5878
1.5848
9.9999
A9999
9.9999
9.9999
9 9999
9.9999
99999
9.9999
9.9999
8.4179
8.4181
1.5819
9.9999
OgCOB
diff.
ogcot
com.
diff.
ogtan
og sin
Angle
88
COSINES, TANGENTS, AND COTANGENTS
1
Angle
log sin
diff.
log tan
com.
diff.
log cot
log COB
Prop. Farts
1.50
1.51
1.52
1.53
1.54
1 55
1.56
1.57
1.58
1.59
1.60
1.61
162
1.63
1.64
1.65
1.66
1.67
1.68
1.69
1.70
1.71
1.72
1.73
1.74
1.75
1.76
1.77
1.78
1.79
1.80
1.81
182
1.83
1.84
1 85
1.86
1.87
1.88
1.89
1.90
1.91
1.92
1.93
1.94
1.95
1.96
1.97
1.98
1.99
2.00
8.4179
29
29
28
28
29
27
28
28
27
27
27
27
27
27
26
26
26
26
26
26
25
25
26
25
24
2o
24
25
24
24
24
24
23
24
23
23
24
23
23
22
23
23
22
22
22
23
22
21
22
8.4181
29
28
29
S8
28
28
28
28
27
27
27
27
20
27
26
26
26
26
26
25
25
26
25
25
24
25
24
<2:>
24
24
24
23
24
23
23
23
23
23
23
23
23
22
22
22
22
22
1.5819
9.9999
88.50
8849
88.48
88.47
88.46
88.45
88.44
88.43
88.42
88.41
88.40
88 39
88.38
88.37
88.36
88.35
88.34
8833
88.32
8831
88.30
88.29
8828
8827
88.26
88.25
8824
88.23
8822
8821
88.20
88 19
88.18
88.17
8816
88.15
8814
88 13
8812
88.11
8810
8809
8808
88.07
8806
88.05
88.04
88.03
8802
8801
8800
i
i
A
Difference
8.4208
8.4237
8.4265
8 4293
8.4322
8.4349
84377
8.4405
8.4432
8.4459
8.4210
8.4238
8.4267
8 4295
8.4323
8.4351
84379
8.4406
8.4434
~&4461~
1.5790
1.5762
1.5733
1.5705
1.5677
1.5649
1.5621
1.5594
1.5566
175539"
9.9998
99998
9.9998
9 9998
9 9998
9.9998
9.9998
9.9998
9.9998
1T9998~
1
2
3
4
5
6
7
8
9
29
2.9
58
8.7
11.6
145
174
20.3
23.2
26.1
28
2.8
5.6
84
11.2
14.0
168
19.6
224
25.2
8.4486
8.4S13
8.4540
8 4S67
8 4593
8.4619
8.4645
8.4671
8.4697
"874723"
1T4748
8.4773
8.4799
8.4824
8.4848
8.4873
8.4898
8 4922
8 4947
84971
8.4488
84515
8.4542
8 4568
8 4595
S.4621
84647
8.4673
8.4699
84725"
874750'
8 4775
8.4801
S.4826
8.4851
8 4875
S4900
84921-
8.4949
"8.497T
1.5512
1.5485
1.5458
] 5432
1.5405
1.5379
1.5353
1.5327
1.5301
"1 5275"
1 5250"
1 5225
1.5199
1 5174
1 5149
15125
1.5100
1.5076
1 5051
'175027
9 9998
9.9998
9.9998
9 9998
9 9998
9.9998
9.9998
9.9998
9 9998
"99998"
9" 9998"
9.9998
9 9998
9 9998
9 9998
9 9998
9 9998
9 9998
9.9998
"99998
1
3
4
5
6
7
8
9
1
2
3
4
5
G
7
8
1)
1
a
4
5
G
7
8
9
1
2
3
4
6
7
8
9
27
27
54
8.1
108
13 5
162
189
21.6
243
26
26
5.2
78
10.4
13.0
156
182
20.8
234
25
25
50
75
100
12.5
15.0
17.5
20.0
24
2.4
4.8
9!e
12.0
14.4
168
192
21 6
22
2.2
4.4
6.6
88
11.0
13.2
154
17.6
19.8
8.4995
8.5019
8.5043
8.5066
8.5090
8.5113
8.5136
85160
8.5183
8.4997
8 5021
8.5045
8 5068
8.5092
85115
85139
85162
85185
1 5003
1.4979
1.4955
1 4932
1 4908
1.4885
1 4861
1.4838
14815
T4792"
1 4769
1.4747
1.4724
1.4702
1 4679
1 4657
1.4635
1 4613
1.4591
174569
9 9998
9.9998
9.9998
9 9998
9 9998
9.9998
9 9998
99998
9.9998
~9 9998
"99998"
9 9998
9.9998
99998
9 9997
9 9997
9 9997
9 9997
99997
23
2.3
46
69
9.2
11.5
138
161
184
20.7
8.5206
875228"
8.5251
8.5274
8.5296
8.5318
8.5340
8.5363
8.5385
8.5406
8.5208
87523 1
8.5253
8.5276
85298
8 5321
8.5343
8.5365
8.5387
8.5409
21
42
(13
84
10.5
12 G
147
168
189
8.5428
8.5431
9.9997
log cos
difi.
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 COB
Prop. Parts
200
201
2.02
2.03
2(M
2 OS
206
2.07
2.08
2.09
2.10
2.11
2.12
2.13
2.14
2.1S
2.16
2.17
2.18
2.19
2.20
2.21
222
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
237
2.38
2.39
2.40
241
2.42
2.43
2.44
2.45
2.46
2.47
2.48
2.49
2.50
8. =5428
23
21
22
21
21
22
21
20-
21
21
21
20
21
20
20
20
20
20
20
20
20
19
20
19
19
20
19
19
19
19
19
18
19
19
18
19
18
18
18
19
18
18
18
17
18
18
17
18
17
18
8.5431
22
21
22
21
21
21
21
21
21
21
20
21
20
20
20
20
20
20
20
20
19
20
19
20
19
19
19
19
19
19
19
18
19
18
19
18
18
~18
19
18
17
18
18
18
17
18
18
17
1.4569
9.9997
88.00
87.99
87.98
87.97
87.96
87.95
87.94
87.93
87.92
87.91
87.90
87.89
87.88
87.87
8786
87.85
87.84
87.83
87.82
87.81
87.80
87.79
87.78
87.77
87.76
87.75
87.74
87.73
87.72
87.71
87.70
87.69
87.68
87.67
87.66
87.65
87.64
87.63
87.62
87.61
87.60
87 59
87.58
87.57
87.56
87.55
87.54
87.53
87.52
87.51
87.50
1
I
&
P
8.S450
8.5471
8 5493
8.5514
8 5535
85557
85578
8.5598
8.5619
8.5453
8.5474
8.5496
8.5517
8.5538
8.5559
8.5580
8.5601
8 5622
1.4547
1.4526
1.4504
1.4483
1.4462
1.4441
14420
1 4399
1.4378
9.9997
9.9997
9.9997
9.9997
9.9997
9.9997
9.9997
9.9997
9.9997
1
2
3
4
5
6
7
8
9
22
22
4.4
66
88
11.0
13.2
15.4
17.6
19.8
8 5640
8.5643
1 4357
9.9997
8 5661
8 5681
8.5702
8.5722
8 5742
8.5762
85782
8.5802
8 5822
8 5664
8 5684
85705
8.5725
8.5745
8.5765
8.5785
8 5805
8.5825
1.4336
1.4316
1.4295
1.4275
1 4255
1.4235
1 4215
1.4195
1 4175
9 9997
9.9997
9.9997
9.9997
9.9997
9.9997
9.9997
9.9997
9.9997
1
2
3
4
5
6
7
8
9
21
2.1
4.2
63
8.4
10.5
12.6
14.7
1C 8
189
8.5842
8 5845
1.415S
9.99-97
8.5862
8.5881
8.5901
8.5920
8.5939
8.5959
8 5978
8.5997
8601 6
8 5865
8 5884
8 5904
8 5923
8 5943
8 5962
8.5981
8.6000
8.6019
1.4135
1.4116
1.4096
1.4077
1 4057
1.4038
14019
1.4000
1.3981
9.9997
9.9997
9.9997
9.9997
9 9997
9.9997
99997
9.9997
9 9997
1
2
3
4
5
6
7
8
9
20
20
4.0
6.0
80
10.0
120
140
16.0
18.0
1
2
3
4
5
6
7
8
9
19
1 9
3.8
6.7
7.6
9.5
114
13.3
15.2
171
8 6035
86038
"8.6057"
8.6076
86095
86113
8 6132
8.6150
8.6169
8.6187
8.6205
1 3962
9 9996
860S4
86072
8.6091
86110
8.6128
86147
8 6165
8.6183
8.6201
1.3943
1.3924
1.3905
1.3887
1.3868
1.3850
1.3831
1.3813
1.3795
9.9996
99996
9.9996
99996
9.9996
9.9996
9.9996
99996
9.9996
1
2
3
4
5
6
7
8
9
18
1 8
36
5.4
7.2
90
108
12.6
144
162
86220
8 6223
1.3777
9 9996
86238
8.6256
8.6274
86291
8.6309
86327
8.6344
8.6362
8.6379
S6242
8 6260
8.6277
8.6295
8 6313
8.6331
8.6348
8 6366
8.6384
1.3758
1.3740
1.3723
1.3705
1.3687
1.3669
1.3652
1.3634
1.3616
9.9996
9 9996
9.9996
9.9996
99996
9.9996
9.9996
9.9996
99996
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
8.6397
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
An-gle
log Bin
diff.
log tan
com.
diff.
log cot
log cos
Prop. Parts
2.50
8.6397
17
8.6401
17
1.3599
9.9996
87.50
2.51
8.6414
8.6418
1.3582
9 9996
87.49
to
8
2.52
86431
17
8.6436
18
1.3564
9.9996
87.48
3
2.53
8.6449
18
17
8.6453
17
17
1.3547
9.9996
87.47
j
2.54
86466
8.6470
1.3530
9.9996
87.46
A
5
2.55
8 6483
17
8.6487
17
1 3513
9 9996
87.45
2.56
8.6500
17
17
8.6504
17
17
1.3496
99996
87.44
18
2.57
8.6517
8 6521
13479
9.9996
87.43
i
2
1 8
2.58
8 6534
17
8 6538
17
1.3462
9.9996
87.42
3
fK4
2.59
86550
16
8 6555
17
1 3445
9.9996
87.41
4
7.2
2.60
"8.6567
17
"865 7F
16
L3429"
~9J6~
87.40
5
6
9.0
10.8
2.61
"8.6584
J7
8 6588
17
1.3412
~O9995~
87.39
7
g
12.6
14 4
2.62
8.6600
16
17
86605
16
1.3395
9.9995
87.38
9
16.2
2.63
8.6617
16
86621
17
1.3379
9.9995
87.37
2.64
8.6633
8.6638
1 3362
9.9995
87.36
2.65
8.6650
17
8 6654
16
1 3346
9.9995
87 35
17
2.66
8.6666
16
8.6671
17
1.3329
9 9995
87.34
1
1.7
16
16
2
34
267
8.6682
86687
1.3313
9.9995
87.33
3
5.1
2.68
8.6699
17
8.6703
16
1.3297
9.999 S
87.32
4
6.8
2.69
8.6715
16
8.6719
16
1.3281
9 9995
87.31
5
6
8.5
10.2
2.70
8.6731
16
16
8.6736
17
16
1.3264"
"9 9995
8730
7
8
11.9
136
2.71
87>747~
"S//752"
~] 3248
"99995
87 29
9
15.3
272
86763
J6
8.6768
16
1.3232
9 9995
87 28
2.73
8.6779
16
16
86784
ir>
16
1.3216
9 9995
87.27
2.74
8 6795
8.6800
1.3200
9 9995
87.26
16
275
86810
]5
8 6815
15
1.3185
9 999 S
87.25
1
2
1 6
32
2.76
8.6S26
16
8.6831
10
1.3169
9 9995
87.24
3
4
4.8
64
2.77
8.6842
8.6847
1 3153
9.9995
87.23
5
80
2.78
8.6858
16
8 6863
16
1 3137
9.9995
87.22
6
9.6
2.79
8.6873
15
16
8 6878
J5
16
1 3122
9999S
87.21
7
8
11 2
128
2.80
"8.6889"
8.6894
T3106"
99995"
87.20
9
144
2.81
8.6904
J
8 6909
15
1.3091
~9995~
87 19
2.82
8.6920
16
8 6925
16
1 3075
99995
87.18
2.83
8 6935
15
S.6940
15
1.3060
9.9995
87 17
15
2.84
8 6950
15
8 6956
16
1.3044
99995
87 16
1
1.5
3.0
285
8 6965
15
8 6971
15
1 3029
9 9995
87.15
3
45
2.86
8 6981
16
J5
8.6986
15
15
1.3014
9 9995
87.14
4
6.0
7.5
287
S 6996
87001
1 .2999
9 9995
87.13
(>
90
10 5
288
87011
15
8.7016
15
1.2 ( )S4
9 9995
87.12
8
120
289
87026
15
8.7031
1 5
1 2969
9 9094
87.11
9
13 5
290
8~7041
15
15
8.7046
15
15
]~2954
~9 9994
87.10
2.91
"8.7056
~8706f
L2939~
T999T
87.09
2.92
87071
15
8 7076
15
1.2924
9.9994
87.08
14
2.93
8.7086
15
14
8.7091
15
15
12909
9 9994
87.07
1
1 4
28
294
8.7100
8.7106
12894
99994
8706
3
4
4.2
5 6
2.95
87115
15
8.7121
15
1 2879
9 9994
87.05
5
7^0
296
87130
14
8.7136
15
14
1.2864
9 9994
87.04
6
7
8.4
9.8
2.97
8.7144
8.7150
1.2850
9.9994
8703
8
11.2
298
87159
15
8.7165
15
1.2835
9 9994
87.02
9
12 6
2.99
8.7174
15
87179
14
1 2821
9.9994
87.01
3.00
8.7188"
14
87194
15
172806
"9.9994"
87.00
log COB
difi.
log cot
com.
difi.
log tan
log sin
Angle
87
26
TABLE III. LOGARITHMIC SINES
3
Angle
log sin
diff.
log tan
com.
diff.
log cot
log cos
Prop.
Parts
3.00
8.7188
14
8.7194
14
1.2806
9.9994
87.00
4>
3.01
8.7202
8.7208
1.2792
9.9994
86.99
.5?
3.02
8.7217
15
8.7223
15
1.2777
9.9994
86.98
td
8
3.03
8.7231
14
14
8.7237
14
15
1.2763
9.9994
86.97
1
3.04
S 7245
8.7252
1.2748
9.9994
86.96
W
p
3.05
8 7260
15
87266
14
1 2734
9.9994
86.95
3.06
87274
14
14
8.7280
14
12720
9.9994
86.94
15
3.07
8.72S8
8.7294
1.2706
9.9994
86.93
1
2
1.5
30
3.08
87302
87308
1.2692
9 9994
86.92
3
45
309
8.7316
14
14
8.7323
15
14
1.2677
9.9994
86 91
4
5
G.O
7 5
3.10
8.7330
14
8.7337
14
1.2663
9.9994
86.90
G
90
3.1i
8.7344
8 7351
1.2649
9.9994
86.89
8
120
3.12
8.7358
8 7365
1.2635
9.9994
86.88
9
13.6
3.13
8 7372
14
14
87379
14
13
1.2621
9.9994
86.87
3.14
8.7386
8.7392
1.2608
9.9993
86.86
315
87400
14
8.7406
14
1.2594
9 9993
86 85
3.16
87413
13
8.7420
14
1.2580
9.9993
S6.SJ
14
14
3.17
87427
8 7434
1 2566
9.9993
86.83
14
3.18
87441
14
8 7448
14
1.2552
9.9993
86.82
1
1.4 -
3.19
8.7454
13
14
8.7461
13
14
1.2539
99993
86.81
3
2.8
4.2
3.20
8 7468
14
8 7475
n
1.2525
9.9993
8680
4
5
5.6
7 o
321
8.7482
87488
1.2512
9 9993
86.79
G
8.4
3.22
8.7495
13
87502
J4
1.2498
9.9993
86.78
7
9.8
3.23
8.7508
13
14
8 7515
13
14
1.2485
9 9993
86.77
9
12.6
3.24
8 7522
87=529
1.2471
99993
8676
3.25
8.7535
13
8.7542
13
1.2458
9.9993
86.75
3.26
87549
14
8.7556
14
1.2444
9.9993
86.74
13
13
3.27
8.7562
8.7569
1.2431
9.9993
86.73
3.28
8.7575
8.7582
13
1.2418
9.9993
86.72
13
3.29
8.7588
13
14
87596'
I 4
1.2404
99993
8671
1
1.3
3.30
8.7602
13
8.76gJ
iki&a~
mm
1.2391
9.9993
86.70
3
3.9
3 31
8 7615
8.|M
12378
9 9993
86.69
4
5.2
3 32
8 7628
13
S
1.2365
99993
86.68
<;
7.8
3.33
87641
8.764?
1.2352
9.9993
86.67
7
9.1
13
13
8
10.4
3.34
87654
8.7661
1.2339
9.9993
86.66
9
11.7
3.35
8.7667
13
8.7674
13
1 2326
9.9993
86.65
3.36
8.7680
13
13
87687
13
13
1.2313
99993
86.64
3.37
8.7693
87700
12300
9 9992
86.63
3.38
8.7705
8.7713
13
12287
9.9992
86.62
3.39
8.7718
13
13
87726
13
13
1.2274
9.9992
86.61
"TsT
3,40
8.7731
13
8.7739
JO
1.2261
9.9992
8660
1
1.2
3.41
8.7744
8 7751
12249
9.9992
86.59
3
3.6
3.42
8.7756
8.7764
1.2236
9 9992
86.58
4
4.8
3.43
8.7769
13
13
87777
13
13
1.2223
9.9992
86.57
5
6
G.O
7.2
3.44
8.7782
8.7790
.2210
99992
86.56
7
8.4
3.45
8.7794
12
8.7802
12
.2198
9.9992
86.55
9
10.8
3.46
8.7807
12
8.7815
12
.2185
9.9992
86.54
3.47
8.7819
8.7827
.2173
9.9992
86.53
348
8 7832
8.7840
13
.2160
9.9992
86.52
3.49
8.7844
12
8.7852
12
.2148
9.9992
86.51
3.50
8.7857
87865
1.2135
9.9992
86.50
log 008
difi.
log oot
com.
diff.
log tan
log sin
Angle
86
COSINES, TANGENTS, AND COTANGENTS
27
3
Angle
log sin
diff.
log tan
com.
diff.
log cot
log cos
Prop. Farts
3.50
8.7857
12
8.7865
12
1.2135
9.9992
86.50
+5
3.51
8.7869
8.7877
1.2123
9.9992
86.49
So
8
352
8.7881
12
8.7890
13
1.2110
9.9992
86.48
^3
3.53
8.7894
13
12
.7902
12
12
1.2098
9.9992
86.47
3.54
8.7906
8.7914
1.2086
9.9992
86.46
w
P
3.55
8.7918
12
8.7927
13
1'.2073
9.9992
86.45
3.56
8.7930
12
13
8.7939
12
12
1.2061
9.9992
86.44
13
3.57
87943
87951
1.2049
9.9992
86 43
1
1.3
3.58
8.7955
12
8.7963
12
1.2037
9.9992
86.42
3
3.9
3.59
8.7967
12
8.7975
12
13
1.2025
9.9991
86.41
4
E
5.2
6 5
3.60
8.7979
12
8.7988
12
1.2012
9.9991
86.40
6
7.8
3.61
8.7991
8.8000
1.2000
9 9991
86.39
7
9.1
3.62
8.8003
12
8.8012
12
1.1988
9.9991
8638
9
11.7
3.63
8.8015
12
8.8024
12
1.1976
V.9991
86.37
3.64
8.8027
8.8036
1.1964
9.9991
86.36
3.65
8 8039
8.8048
12
1.1952
9.9991
8635
o o.
3.66
8.8051
12
11
8.8059
11
12
1.1941
9.9991
86.34
3.67
8.8062
8.8071
1.1929
9.9991
86.33
i
3.68
8.8074
12
8.8083
12
1.1917
9.9991
86.32
1
3.69
8.8086
12
8.8095
12
]0
1.1905
9.9991
86.31
2
3
2.4
3.6
3.70
8.8098
11
8.8107
12
1.1893
9.9991
86.30
4
4.8
3.71
8.8109
8.8119
1.1881
9.9991
86.29
6
7.2
3.72
8.8121
8.8130
1.1870
9.9991
86.28
7
8.4
3.73
8.8133
11
88142
12
1.1858
9.9991
86 27
8
9
9.6
10.8
3.74
88144
88154
1.1846
9.9991
86.26
3.75
8.8156
8.8165
11
1.1835
9.9991
86.25
3.76
8.8168
12
11
88177
12
11
1.1823
9.9991
86.24
3.77
88179
8.8188
1.1812
9.9991
86.23
3.78
8.8191
8.8200
12
1.1800
9.9991
86.22
3.79
88202
11
8.8212
12
11788 9.9990
86.21
1
1.1
3.80
8.8213
12
8.8223
11
1 Mll^fjUO
86.20
2
3
2.2
3 3
3.81
8.8225
8.8234
1.176(^^Bo
86.19
4
4.4
3.82
88236
8.8246
1 1 7 C >41P9!9990
86.18
5
5.5
3.83
8.8248
12
11
8.8257
11
12
1 1743|C990
86.17
6
7
6.6
7.7
3.84
8.8259
8.8269
1.173179.9990
86.16
8
9
8.8
9 9
3.85
8.8270
88280
1.1720 9.9990
86.15
3.86
8.8281
12
8.8291
11
11
1.1709
99990
86.14
3.87
8.8293
88302
11698
99990
86.13
3.88
8.8304
11
8.8314
12
11
1.1686
9.9990
86.12
3.89
3.90
8.8315
11
11
8.8325
11
11
1.1675
9.9990
86.11
86.10
1
10
1.0
8.8326
8.8336
1.1664
9.9990
3.91
8.8337
88347
1.1653
9.9990
86.09
2
2.0
3.92
8.8348
11
8.8358
11
1.1642
9.9990
86.08
3
3.0
3.93
8.8359
11
li
8.8370
12
11
1.1630
9.9990
86.07
5
6.0
6
3.94
8.8370
88381
1.1619
9.9990
86.06
7
7.0
3.95
8.8381
8.8392
11
1.1608
9.9990
86.05
8
8.0
3.96
8.8392
11
8.8403
11
11
1.1597
9.9990
86.04
9
9.0
3.97
8.8403
8.8414
1.1586
9.9990
86.03
3.98
8.8414
8 8425
11
1.1575
9.9990
86.02
3.99
8.8425
8.8436
11
1.1564
9.9989
86.01
4.00
8.8436
8.8446
1.1554
9.9989
86.00
log COB
diff.
log cot
com.
diff.
log tan
log sin
Angle
86
28
TABLE III. LOGARITHMIC SINES
4
Angle
log sin
diff.
log tan
com.
diff.
log cot
log cos
Prop. Parts
4.00
4.01
4.02
4.03
4.04
405
4.06
4.07
408
409
4.10
4.11
4.12
4.13
4.14
415
416
4.17
4.' ,'
r '9
4.20
4.21
4.22
4.23
424
4.25
4.26
4.27
4.28
429
4.30
431
432
4.33
4.34
435
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
11
10
11
11
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
11
n
11
11
ll
10
ll
11
10
11
11
10
11
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
jo
10
10
10
9
10
10
10
10
9
10
10
1.1554
9.9989
86.00
85.99
85.98
85.97
85.96
85 95
85 94
85.93
85.92
85.91
85.90
8589
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
8567
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
Extra digit
Difference
88447
8.8457
8.8468
8.8479
8.8490
8.8500
8.8511
8.8522
8.8532
8.84S7
8.8468
8.8479
88490
8.8501
8.8511
8 8522
8.8533
8.8543
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
i
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
11
11
2.2
33
44
55
66
7.7
8.8
9.9
10
1.0
2.0
30
4.0
50
BO
70
8.0
9.0
~9~
0.9
1.8
2.7
3.6
4.5
5.4
63
7.2
8.1
8.8543
8 8554
1.1446
9.9989
8.8553
885 \
8.85 5
8.P ,85
8 595
S606
8 8616
8.8627
8 8637
8 8565
8.8575
8.8586
8.8596
88607
8.8617
8.8628
8.8638
8.8649
1.1435
1.1425
1.1414
1 1404
1 1393
1.1383
1.1372
1.1362
11351
9 9989
9.9989
9.9989
9.9989
9.9989
9.9989
9.9988
9.9988
9.9988
88647
1T8658
8 8668
8 8678
8.8688
8 8699
88709
8.8719
88729
8.8739
8.8659
"8.~8669
8.8680
8.8690
8.8700
8.8711
8.8721
8.8731
8.8741
8.8751
1.1341
~iri33T
1 1320
1.1310
1.1300
1 1289
1.1279
1.1269
1.1259
1.1249
1 1238"
9.9988
9.9988
9.9988
9.9988
9.9988
9.9988
9.9988
9.9988
9.9988
9.9988
8 8749
88762
9 9988
8.8759
8 8769
8 8780
8 8790
8 8799
88809
88819
88829
8.8839
Ii8849~
88772
8.8782
8.8792
8.8S02
8.8S12
8.8822
8.8832
8.8842
8.8852
'8.8862
1 1228
1 3218
1.1208
1.1198
1.118S
1.1178
1 1168
1.1158
1.1148
T.1138
9.9988
9.9988
9.9988
9.99S8
9.9987
9.9987
9.9987
9 9987
9.9987
9:9987
8 8859
8 8869
8.8878
8.8888
S.889S
8.8908
88917
8.8927
8.8937
8.8872
88882
8.8891
8.8901
8.8911
8.8921
8.8931
8 8940
8.8950
1.1128
1.1118
1.1109
1.1099
1.1089
1.1079
1.1069
1.1060
1.1050
9.9987
9.99S7
9.9987
9.9987
9.9987
9.9987
9.9987
9.9987
9.9987
8.8946
8.8960
1.1040
9.9987
log COB
dig.
log cot
com.
diff.
log tan
log Bin
Angle
85
COSINES, TANGENTS, AND COTANGENTS
29
4
Angle
log sin
diff.
log tan
com.
diff.
log cot
log cos
Prop, Parts
4.50
8.8946
10
8.8960
10
11040
9.9987
85.50
+*
4.51
8.8956
8.8970
1.1030
9.9987
85.49
5
4.52
8.8966
10
8.8979
9
1.1021
9.9986
8^.48
-3
I
4.53
8.8975
9
10
8.8989
10
9
1.1011
99986
85.47
1
4.54
8.8985
8 8998
1,1002
9.9986
85.46
w
p
4.55
8.8994
9
89008
10
1.0992
9.9986
85.45
4.56
8.9004
10
9
8.9018
10
9
1.0982
9.9986
85.44
10
4.57
8.9013
89027
1.0973
9.9986
85.43
i
1.0
458
8.9023
10
8.9037
10
1 0963
9.9986
85.42
2
3
2.0
30
4.59
8.9032
9
10
8.9046
9
]0
1.0954
9.9986
85.41
4
5
4.0
5
4.60
8.9042
9
8 9056
9
1.0944
9.9986"
85.40
G
e'o
461
8.9051
8 9065
1.0935
9.9986
85 39
7
g
7.0
8.0
462
8.9060
9
89075
10
1.0925
9.9986
8S.38
9
9.0
4.63
8 9070
10
9
8 9084
9
9
1.0916
9.9986
85.37
4.64
8.9079
8.9093
10907
9.9986
85 36
4 65
8.9089
10
8 9103
10
1.0897
9.99S6
85 35
466
8.9098
9
9
8.9112
9
Hi
1.0888
9 9986
85 34
4.67
89107
8.9122
1.0878
9.9986
85 33
4.68
89116
9
8.9131
9
1.0869
9.9985
85.32
4.69
8.9126
10
9
S9HO
9
JO
1 0860
9.9985
85 31
4.70
8 9135"
9
8 9l5~0
9
10850"
"99985"
8530
471
8 r 9l44
8.9159'
10S41
"J7998 1 ;
85 29
4.72
8 91 53
8.9168
1.0832
9998 s ?
85.28
4.73
8.9162
9
10
8.9177
9
9
1.0823
9//AS5
8527
1
9
09
474
89172
8 9186
10814
9 998 S
85 26
2
1 8
4.75
89181
<)
8.9196
10
1.0804
9 9985
85.25
3
4
3(5
4.76
8.9190
9
8 9205
9
1.0795
9.9985
8524
5
45
y
9
G
5.4
477
89199
89214
1 07S6
yyysi
85 23
7
6.3
478
8 9208
9
8 9223
9
10777
9998S
85.22
8
72
4.79
8 921 7
9
S.9232
9
1.0768
9998S
85 21
9
8.1
4.80
"89226
*
1T9241"
J
"1.07 S9
"99985
8520
9
9
4.81
8/J235"
"89250
1.0750"
"99985
85.19
4.82
8 9244
9
8.9260
JO
1.0740
9.9985
8518
4.83
8 9253
9
9
8 9269
)
9
1.0731
9 9985
85.17
4.84
8.9262
8 9278
1.0722
9.9984
85.16
485
8.9271
9
8.9287
9
1 0713
9.9984
85 15
486
8 9280
9
9
8.9296
9
9
1.0704
9.9984
85.14
4.87
8.9289
8.9305
1.069 s ?
9.9984
85.13
488
8 9298
9
8.9313
8
1,0687
9.9984
85.12
489
89307
9
g
8.9322
9
9
1 0678
9.9984
85.11
8
4.90
8.9315
g
8.9331
9
1.0669
9.9984
85.10
1
2
0.8
1.6
4.91
8.9324
8 9340
1.0660
9 9984
8509
3
24
4.92
8.9333
9
8.93-19
9
1.0651
9.9984
85.08
4
5
3 2
4.0
4.93
8 9342
9
8.9358
9
1.0642
9.9984
85 07
G
4.8
9
9
7
5.6
494
8.9351
8 9367
1 0633
9.99S4
85.06
8
6.4
4.95
8 9359
8
89376
9
1 0624
9.99S4
85 05
9
7.2
4.96
8 9368
9
9
8.9384
8
9
1.0616
9.9984
85.04
4.97
8.9377
8.9393
1.0607
9.9984
85 03
4.98
8 9386
9
8.9402
9
1.0598
9 9984
85.02
4.99
8.9394
8
8.9411
9
1 0589
9.9984
85.01
5.00
"89403"
9
~8i9420"
9
1.0580
9.9983"
85.00
log 008
diff.
log cot
com.
diff.
log tan
log sin
Angle
85
TABLE III. LOGAEITHMIC SINES
5-10
Angle
log sin
diS.
log tan
com.
die.
log cot
log cos
diff.
Prop. Parts
5.0
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
6.0
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
7.0
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
8.0
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
9.0
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
10.0
8.9403
86
84
82
81
80
78
76
76
74
72
72
70
69
69
67
66
65
64
63
62
61
61
59
59
58
57
57
55
55
55
53
53
52
52
51
50
50
50
48
48
48
47
47
46
45
45
45
44
43
44
8.9420
86
85
83
82
80
79
77
76
75
73
73
71
70
69
68
66
66
65
64
63
63
61
61
59
59
58
58
57
56
55
55
54
53
53
52
52
51
50
50-
49
49
48
48
47
47
46
46
46
45
44
1.0580
9.9983
1
i
i
i
i
i
i
i
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
2
1
1
1
1
1
1
1
2
1
1
1
1
2
1
1
2
1
1
1
85.0
84.9
84.8
84.7
84.6
84.5
84.4
84.3
84.2
84.1
84.0
83.9
83.8
83.7
83.6
83.5
83.4
83.3
83.2
83.1
83.0
82.9
82.8
82.7
82.6
82.5
82.4
82.3
82.2
82.1
820
81.9
81.8
81.7
816
81.5
81.4
81.3
81.2
81.1
81.0
80.9
80.8
80.7
80.6
80.5
80.4
80.3
80.2
80.]
80.0
l
2
3
4
5
G
7
8
9
1
2
3
4
5
G
7
8
9
Difference
8.94S9
8.9573
8.9655
8.9736
8.9816
8.9894
89970
9.0046
9.0120
8.9506
8.9591
8.9674
8.9756
8.9836
8.9915
8.9992
9.0068
9.0143
1.0494
1.0409
1.0326
1.0244
1.0164
1.0085
1.0008
0.9932
0.9857
9.9983
9.9982
9.9981
9.99S1
9.9980
9.9979
9.9978
9.9978
9.9977
62
6.2
12.4
18.6
24.8
31.0
87.2
43.4
49.G
55.8
61
6.1
12.2
183
244
30.5
36.6
42.7
48.8
54.9
60
60
12.0
18.0
24.0
30.0
36.0
42.0
48.0
54.0
9.0192
9.0216
0.9784
9.9976
90264
9.0334
9.0403
9.0472
9.0539
9.0605
9.0670
9.0734
9.0797
9.0859"
9.0289
9.0360
9.0430
9.0499
9.0567
9.0633
9.0699
90761
9.0828
0.9711
0.9640
0.9570
0.9501
0.9433
0.9367
0.9301
0.9236
0.9172
9.9975
9.9975
9.9974
9.9973
9.9972
9.9971
9.9970
9.9969
9.9968
59
59
118
17.7
23.6
29.5
35.4
41 3
47.2
531
58
5.8
11.6
17.4
23.2
29.0
34.8
40.6
464
522
57
5.7
11.4
17.1
22.8
28.5
34.2
39.9
45.6
51.3
9.0891
0.9109
9.9968
1
2
3
4
5
6
7
8
9
56
5.6
11.2
16.8
22.4
28.0
83 G
39.2
448
50.4
55
55
11.0
16.5
220
27.5
330
385
440
49 5
52
52
104
15 G
208
260
31.2
364
41.G
46.8
49
49
9.8
147
19.G
24.5
294
34.3
39.2
44.1
54
5.4
10.8
16.2
21.6
27.0
324
37.8
43.2
48.6
9.0920
9.0981
9.1040
9.1099
9.1157
9.1214
9.1271
9.1326
9.1381
9 0954
9.1015
9.1076
9.1135
9.1194
9.1252
91310
9.1367
9.1423
0.9046
8985
0.8924
0.8865
0.8806
0.8748
08690
0.8633
08S77
9.9967
9.9966
9.9965
9.9964
9 9963
99962
9.9961
9.9960
9.9959
1
2
8
4
5
6
7
8
9
1
2
3
4
5
G
7
8
9
53
5.3
10.6
15.9
21.2
265
31.8
37 1
42.4
47.7
50
5.0
10.0
15.0
200
25.0
30.0
35.0
40.0
45.0
51
5.1
102
15.3
20.4
255
306
357
40.8
45.9
9.1436
9.147S
8522
9.9958
9.1489
9.1542
9.1594
91646
9 1697
9.1747
9.1797
91847
9.1895
9.1533
9.1587
9.1640
9 1693
91745
91797
9.184S
9.1898
9 1948
8467
08413
0.8360
08307
8255
0.8203
0.8152
08102
0.8052
9 9956
9.9955
9.9954
9.9953
99952
99951
9.9950
9 9949
9.9947
48
4.8
9.6
14.4
19.2
24.0
28.8
33.6
38.4
432
9.1943
9.1997
08003
9.9946
9.1991
9.2038
9.2085
9.2131
9.2176
9.2221
9.2266
9.2310
9.2353
9.2046
92094
9.2142
92189
9 2236
9.2282
9.2328
9.2374
9.2419
7954
0.7906
0.7858
0.7811
7764
0.7718
0.7672
0.7626
0.7581
9.9945
9.9944
9.9943
9.9941
99940
9.9939
9.9937
9.9936
9.9935
1
3
4
5
6
7
8
9
47
47
94
14.1
18.8
23.5
28.2
32.9
37.6
42.3
46
4.6
9.2
13.8
184
230
27.6
32.2
36.8
41.4
45
4.5
90
13.5
18.0
22.6
27.0
31.5
36.0
40.5
9.2397
9.2463
0.7537
9.9934
lOg COB
difi.
log cot
oom.
diff.
log tan
log sin
diff.
Angle
80- 85
COSINES, TANGENTS, AND COTANGENTS
31
10-15
Angle
log sin
diS.
log tan
com.
diff.
log cot
log COB
diff.
Prop. Farts
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
129
13.0
13.1
13.2
13.3
13.4
13.5
13.6
137
138
13.9
14.0
14.1
14.2
14.3
14.4
145
14.6
14.7
14.8
14.9
15.0
9.2397
42
43
42
41
41
41
40
40
40
39
39
38
38
38
38
37
36
37
36
36
35
36
34
35
34
34
34
34
33
33
33
32
32
32
32
31
32
30
31
31
30
30
30
30
29
29
29
29
29
28
9.2463
44
44
43
43
43
42
42
41
41
41
40
40
39
40
39
38
39
38
37
38
37
37
36
37
36
35
36
35
35
35
34
34
34
34
34
33
33
33
32
33
32
32
32
31
32
31
31
31
30
31
0.7537
9.9934
2
i
2
1
1
2
1
2
2
1
2
1
2
1
2
1
2
1
2
2
1
2
2
1
2
2
2
2
1
2
2
2
2
1
o
2
2
2
2
2
2
2
2
2
2
2
2
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
763
76.2
76.1
76.0
759
75.8
75.7
756
75.5
75.4
75.3
75.2
75.1
75.0
%
1
3
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
Difference
9.2439
9.2482
9.2524
9.2565
9.2606
9.2647
9.2687
9.2727
9.2767
9.2507
9.2551
9.2594
9.2637
9.2680
2.2722
9.2764
9.2805
9.2846
0.7493
0.7449
07406
0.7363
07320
0.7278
0.7236
0.7195
07154
9.9932
9.9931
9.9929
9.9928
9.9927
9.9925
9.9924
9.9922
9.9921
44
4.4
88
13.2
17.6
220
26.4
308
352
396
41
4.1
8.2
12.3
16.4
20.5
24.6
28.7
328
36.9
38
38
7.6
11.4
152
190
228
26.6
30.4
34.2
35
3.5
70
10.5
140
17.5
21
24.5
280
31 5
32
3.2
64
9.6
12.8
160
19.2
22.4
25.6
28.8
29
2.9
58
8.7
11.6
14.5
17.4
20.3
23.2
26.1
43
4.3
8.6
12.9
17.2
21 5
258
301
34.5
38.8
40
4.0
8.0
12.0
160
200
240
28.0
320
36.0
37
37
74
11.1
14.8
185
222
25.9
29.6
333
34
34
6.8
102
136
170
20.4
238
27 2
306
31
31
6.2
93
12.4
155
18.6
217
248
27.9
28
28
5.6
8.4
11.2
140
16.8
196
22.4
25.2
42
42
8.4
12.6
16.8
21.0
25.2
294
33.6
37.8
39
3.9
7.8
11.7
156
195
23.4
27.3
31.2
35.1
36
36
72
108
144
180
21.6'
25.2
288
32.4
33
3.3
66
9.9
132
1(5.5
19.8
23.1
26.4
29.7
30
3.0
6.0
9.0
12.0
150
180
210
24.0
27.0
2
0.2
0.4
0.6
0.8
1.0
12
1 4
1.6
18
9.2806
9.2887
07113
99919
9.2845
9.2883
9.2921
9.2959
9.2997
9.3034
9.3070
9.3107
9.3143
9.2927
9.2967
9.3006
9.3046
9 3085
9.3123
9.3162
9.3200
9.3237
0.7073
0.7033
0.6994
0.6954
06915
0.6877
6838
0.6800
0.6763
9.9918
9.9916
9.9915
9.9913
9.9912
9.9910
9.9909
9.9907
9.9906
9.3179
9.3275
6725
OT6688"
0.6651
0.6615
06S78
6542
6507
6471
0.6436
06401
9.990J
9.3214
9.3250
9.3284
9.3319
9.3353
9 3387
9.3421
9.3455
9.3488
9.352f
9.3554
9.3586
9.3618
9.3650
9.3682
9.3713
9.3745
9.3775
9.3806
9.3312
9.3349
9.3385
93422
9.3458
9.3493
9.3529
9.3564
9.3599
9.9902
9-9901
9 9899
99897
9 9896
9.9894
9.9892
9 9891
9.9S89
9.3634
"93668"
9.3702
9.3736
9.3770
9.3S04
9.3837
9.3870
9.3903
9.3935
0.6366
"0.6332
06298
0.6264
0.6230
0.6196
0.6163
06130
0.6097
0.6065
9.9887
99885
9.9884
9.9882
9.9880
9.9878
9.9876
9.9875
9.9873
9.9871
9.3837
9.3968
6032
9.9869
9.3867
9.3897
9.3927
9.3957
9.3986
9.4015
9.4044
9.4073
9.4102
9.4000
94032
9.4064
9.4095
9.4127
9.4158
9.4189
9.4220
9.4250
0.6000
05968
05936
0.5905
0.5873
0.5842
0.5811
0.5780
0.5750
9.9867
9.9865
9.9863
99S61
9.9859
9.9857
9.9855
9.9853
9.9851
9.4130
9.4281
0.5719
9.9849
log COB
diff.
log cot
com.
diff.
log tan
log sin
diff.
Angle
75-80
32
TABLE III. LOGARITHMIC SINES
15- 20
Angle
TsTo^
15.1
IS 2
15.3
15.4
155
15.6
15 7
IS. 8
15.9
16.0
16.1
16.2
16.3
16.4
16.5
166
167
16.8
16.9
17.0
17.1
17.2
17.3
17 T
17.5
17.6
17.7
17.8
17.9
18.0
18.1
J82
18.3
18.4
18.5
18.6
18.7
188
18.9
19.0
19.1
19.2
193
19.4
19.5
19.6
19.7
198
19.9
20.0
log sin
diff.
log tan
com.
diff.
log cot
log cos
diff.
Prop. Farts
9.4130
28
28
28
28
27
27
27
27
27
26
27
26
26
26
25
26
25
25
25
25
25
25
24
24
24
24
24
21
23
24
23
23
23
23
23
22
23
22
22
22
22
22
22
21
22
21
22
21
21
9.4281
30
30
30
29
30
29
29
29
29
29
28
2<)
28
28
28
as
28
27
27
27
07
07
20
27
26
26
L>6
20
25
20
26
25
*- >e
25
25
25*
25
24
24
24
23
24
24
24
0.5719
9.9849
2
2
2
2
2
2
2
Q
2
3
2
2
2
2
3
2
2
2
3
2
3
2
3
2
3
2
2
3
2
3
3
2
3
3
2
3
2
3
3
2
3
3
2
3
3
2
3
75.0
74.9
748
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
734
73 3
732
73.1
73.0
729
728
72.7
726
72.5
72. 1
723
722
72.1
72.0
71.9
71 8
71.7
716
71.5
71.4
71.3
71.2
71.1
71.0
709
708
70.7
70.6
70.5
704
70.3
70.2
70.1
70.0
I
5
N
w
Difference
9.4158
9.4186
94214
9.4242
9 4269
9 4296
9 4323
9 43 SO
94377
9^4403
9.4430"
9.4456
9 4482
94508
9 4533
9.4559
9.4584
9.4609
94634
94311
9.4341
9.4371
9.4400
9.4430
9.44 S9
9.4488
9.4S17
94S46
S689
5659
0.5629
0.5600
0.5570
0.5541
0.5512
0.5483
0.54S4
9.9847
9.9845
9.9843
9.9841
9.9S39
9.9837
9 9835
9.9833
9.9831
1
2
3
4
r>
t;
8
9
1
3
4
5
6
8
9
30
30
6.0
90
12.0
150
18.0
21
240
270
29
2.9
58
8.7
11.6
145
174
203
23.2
26 1
94575
0.5425
9.9828
9.4603
9.4632
9 4660
9 46SS
9.4716
9.4744
94771
9.4799
9 4826
5397
5368
0.5340
05312
S284
5256
5229
O.S201
05174
OS147
"0.512(T
S093
5066
0.5039
SO 1 3
0.4986
0.4960
4934
4908
9.9826
9.9824
9.9822
9.9820
9.9817
9.9815
9 9813
9.9811
9 9808
9 9806
9 9804
9 9801
9 9799
99797
9 9794
9.9792
9 97S9
9.9787
9 9785
9 9782
28
28
5.6
8.4
11.2
140
108
196
252
27
27
54.
8.1
108
135
162
189
21 6
24.3
9.4659
9.485 5
9 4684
9.4709
94733
9.47 S 7
94781
9.4805
94829
9.4853
9 4876
9 4880
9 4907
9 4934
9.4961
9 4987
9.5014
9 5040
9.50(>6
9.5092
1
2
3
4
5
6
7
8
9
1
3
4
6
8
9
1
2
3
4
5
6
8
9
26
2.6
r>.2
78
104
130
15 (5
182
20 8
23 4
25
2 K
50
7.5
100
125
150
175
200
22 5
9 4900
9.51 IS
0.4882
4857"
0.4831
0.4805
04780
0.4755
0.4730
4705
04680
4655
9.4923
9.4946
9.4969
9 4992
9 5015
9 5037
9 5060
9 5082
9 SH)4
9 S143
95169
9.5195
95220
9 S24S
9.5270
9.5295
9 5320
9.534S
975370
9 9780
99777
9.9775
9.9772
9.9770
9 9767
9.9764
9 9762
9.9759
24
24
48
72
96
12.0
144
168
19.2
21.6
23
2.3
46
69
92
115
138
101
184
207
9. SI 26
04630
9.9757
9S148
9 S170
95192
9 S213
9.5235
9.5256
9.5278
9.5299
9 5320
9.5394
9.5419
9.5443
9.5467
9 5491
9.5516
9.SS39
95S63
9.5 S87
04606
0.4581
04SS7
4533
0.4S09
04484
0.4461
04437
0.4413
9.9754
99751
9.9749
9.9746
9 9743
9 9741
9.9738
9.9735
9 9733
22
22
44
6.6
88
110
132
154
176
19.8
21
21
4.2
63
84
10.5
126
147
168
18.9
9.5341
9.5611
04389
9 9730
log COB
diff.
log cot
com.
difi.
log tan
log sin
diff.
Angle
70- 75
COSINES, TANGENTS, AND COTANGENTS
33
20-25
Angle
log sin
diff.
log tan
com.
difi.
log cot
log cos
difi.
Prop. Farts
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. S
21.6
21.7
21.8
21.9
220
22.1
22.2
22.3
22.4
225
22.6
227
228
22.9
230
23.1
23.2
23.3
23.4
23 S
23.6
23.7
238
23 9
240
24 1
24.2
24.3
24.4
24.5
24.6
24.7
24.8
24.9
25.0
9.5341
20
21
20
21
20
20
21
20
1<)
20
20
20
19
19
20
19
19
19
19
19
18
19
10
18
18
J9
18
18
18
18
18
17
JK
18
17
17
18
17
17
M
17
17
17
17
16
17
16
17
16
16
9.5611
23
24
23
23
23
23
23
23
23
23
22
23
22
23
22
22
22
22
22
22
22
21
22
21
22
21
21
21
22
21
21
20
21
21
21
20
21
20
21
20
21
20
20
20
20
20
20
20
20
0.4389
9.9730
3
3
2
3
3
3
3
3
3
2
3
3
3
3
3
3
3
3
3
3
3
3
4
3
3
3
3
3
4
3
3
3
3
4
3
3
4
3
3
4
3
3
4
3
4
3
4
3
4
3
70.0
69.9
69 8
69.7
696
69.5
69.4
693
69.2
69.1
69.0
68.9
68.8
687
68.6
68.5
68.4
68.3
68.2
68.1
68.0
67.9
678
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
662
66.1
66.0
65 9
65 8
65.7
65 6
65 S
65.4
653
65.2
65.1
65.0
i
*
f-<
X
w
Difference
9.5361
9.5382
9.5402
9.5423
9 5443
9 5463
95484
9.5504
9.5523
9.5634
9.5658
9.5681
9.5704
9.5727
9.5750
9.5773
9.5796
9.5819
0.4366
0.4342
0.4319
0.4296
0.4273
0.4250
04227
0.4201
0.4181
9.9727
9.9724
9.9722
9.9719
9.9716
9.9713
9.9710
9.9707
9 9704
1
2
3
4
5
6
8
9
1
3
4
5
6
7
8
9
23
2.3
4.6
6.9
9.2
11 5
13.8
16.1
18.4
207
21
2.1
42
6.3
84
10 5
126
147
168
180
22
2.2
4.4
60
8.8
11
13.2
15.4
17.0
19 8
20
2.0
40
60
80
100
120
140
160
18
9 5543
9JS563~
9.5583
95602
9.5621
9.5641
9.5660
9.5679
9.5698
9.5717
9.5736"
^5754
9.5773
95792
9.5810
9.5828
9.5847
9586S
9 5883
9.5901
9.5919
^9.593^
9.5954
9.5972
9.5990
9.6007
9.6024
9.6042
9.6059
9.6076
9.5842
~9~58M
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 61 72
9.6194
9.6215
9.6236
9.6257
9.6279
9 6300
96321
9.6341
9.6362
9 6383
9.6404
9.6424
9.6445
9.6465
0.4158
"04136
0.4113
0.4091
0.4068
0.4046
0.4024
04002
3980
039S8
3936
~03914
3892
0.3871
03849
3828
0.3806
3785
3764
0.3743
0372T
"0.3700
0.3679
0.3659
3638
0.3617
0.3596
3576
0.3555
3535
9.9702
"9.9699
9 9696
9 9693
9.9690
9.9687
9.9684
9.9681
9.9678
9 9675
~9%72
"99669
9 9666
9.9662
9.96S9
9.96 S 6
9.9653
9.9650
99647
9.9643
T9640
"99637
9.9634
9.9631
9.9627
99624
9.9621
9.9617
9.9614
9.9611
1
3
4
5
6
7
8
9
1
3
4
5
6
7
8
<)
19
1 9
38
5.7
76
95
11 4
133
15 2
171
17
1 7
34
51
6.8
85
102
11 9
136
15 3
18
1.8
36
54
72
90
108
12.6
144
162
16
1 6
4.8
6.4
80
9.G
11.2
12.8
14.4
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
2
02
04
06
0.8
10
12
1.4
1.6
1 8
9 6093
9 6486
0.3 S 14
9.9607
9.6110
9.6127
9.6144
9.6161
9.6177
9.6194
9.6210
9.6227
9.6243
9.6506
9.6527
9.6547
9.6567
9.6587
9.6607
96627
9.6647
9.6667
0.3494
0.3473
0.3453
0.3433
0.3413
0.3393
0.3373
0.3353
3333
99604
9.9601
9.9597
9.9594
99W
9.9587
99S83
9.9580
9.9576
3
03
06
09
12
1.5
18
2.1
2.4
2.7
4
0.4
0.8
1.2
l.G
2.0
24
2
3.6
9.6259
9.6687
0.3313
9.9573
log cos
diff.
log cot
com.
diff.
log tan
log sin
diff.
Angle
65-70
34
TABLE III. LOGARITHMIC SINES
25-30
Angle
log sin
diff.
log tan
com.
diff.
log cot
log cos
diff.
Prop. Fax to
25.0
25 1
25.2
25.3
25.4
25.5
25.6
25.7
25.8
25.9
26.0
26.1
26.2
26.3
26.4
26.5
26.6
26.7
26.8
26.9
27.0
27.1
27.2
27.3
27.4
27.5
27.6
27.7
27.8
27.9
28.0
28.1
28.2
28.3
28.4
28.5
28.6
28.7
28.8
28.9
29.0
29.1
29.2
29.3
29.4
29.5
29.6
29.7
29.8
29.9
30.0
9.6259
17
16
16
16
16
16
15
16
16
15
16
15
16
15
15
15
16
15
15
14
15
15
15
14
15
15
14
14
15
14
14
14
15
14
14
14
13
14
14
14
13
14
13
14
13
14
13
13
14
13
9.6687
19
20
20
19
20
19
20
19
20
19
19
19
19
19
19
19
19
19
19
19
18
19
19
18
19
18
19
18
18
19
18
18
18
19
18
18
18
18
18"
18
17
18
18
18
17
18
18
17
18
17
0.3313
9.9573
4
3
4
4
3
4
3
4
4
3
4
4
4
3
4
4
4
4
3
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
5
4
4
4
4
5
4
4
5
4
4
5
65.0
64.9
64.8
64.7
64.6
645
64.4
64.3
64.2
64.1
64.0
63.9
63.8
63.7
63.6
63 5
63.4
633
63.2
63.1
63.0 C
62.9
62.8
62.7
62.6
62.5
62.4
62.3
62.2
62.1
62.0
61.9
61.8
61.7
61.6
61.5
61.4
61.3
61.2
61.1
61.0
60.9
60.8
60.7
60.6
60.5
60.4
60.3
60.2
60.1
60.0
1
1
X
W
1
3
4
5
6
7
8
9
Difference
9.6276
9.6292
9.6308
9.6324
9.6340
9.6356
9.6371
9.6387
9.6403
9 6706
9 6726
9 6746
9.6765
9.6785
9.6804
9.6824
9.6843
9.6863
0.3294
0.3274
0.3254
0.3235
0.3215
0.3196
0.3176
0.3157
0.3137
9.9569
9.9566
9.9562
9.9558
9.9555
9.9551
9.9548
9.9544
9.9540
20
20
40
60
8.0
10.0
120
14.0
160
18.0
19
1.9
3.8
5.7
7.6
9.6
11.4
133
15.2
17.1
9.6418
9.6882
0.3118
9.9537
9.6434
9.6449
9.6465
9.6480
9.6495
9.6510
9.6526
9 6541
9.6556
9.6901
9 6920
9.6939
9.6958
9.6977
9.6996
9.7015
9.7034
9 7053
"9.7072"
0.3099
0.3080
03061
0.3042
0.3023
0.3004
2985
0.2966
0.2947
9.9533
9.9529
9.9525
9.9522
9.9518
9.9514
9.9510
9.9506
9.9503
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
18
1.8
3.6
54
72
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
9.6570
2928
~0"2910~
0.2891
0.2872
0.2854
0.2835
0.2817
2798
0.2780
0.2762
9.9499
9.6585
9.6600
9.6615
9.6629
9.6644
9.6659
9.6673
9.6687
9.6702
9.7090
9.7109
9.7128
9.7146
9.7165
9.7183
9.7202
9.7220
9.7238
9.9495
9.9491
9.9487
9.9483
9.9479
9.9475
9.9471
9.9467
9 9463
16
1.8
3.2
48
64
80
96
11.2
12.8
144
15
1.5
3.0
4.5
60
75
90
10.5
120
13.5
9.6716
9 7257
02743
9 9459
9.6730
9.6744
9.6759
9.6773
9.6787
9.6801
9.6814
9.6828
9.6842
9 7275
9.7293
9.7311
9.7330
9.7348
9.7366
9.7384
9.7402
9.7420
0.2725
0.2707
0.2689
0.2670
0.2652
2634
0.2616
2598
0.2580
9 9455
9.9451
9.9447
9.9443
9 9439
9 9435
9.9431
9.9427
99422
1
2
3
4
5
6
7
8
9
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
52
6.5
78
91
104
11.7
9.6856
9.7438
0.2562
9.9418
9.6869
9.6883
9.6896
9.6910
9.6923
9.6937
9.6950
9.6963
9.6977
9.7455
9.7473
9.7491
9.7509
9.7526
9.7544
9.7562
9.7579
9.7597
0.2545
0.2527
0.2509
02491
02474
0.2456
0.2438
0.2421
0.2403
~0.23S6
9.9414
9.9410'
9.9406
9.9401
9.9397
9.9393
9.9388
9.9384
9.9380
3
0.3
0.6
0.9
1.2
1 5
1.8
2.1
2.4
2.7
4
0.4
08
1.2
1.6
2.0
2.4
28
3.2
3.6
9.6990
9.7614
9.9375
log COB
diff.
log oot
com.
diff.
log tan
log sin
diff.
Angle
60-65
COSINES, TANGENTS, AND COTANGENTS
35
30-35
Angle
log Bin
diff.
log tan
com.
diff.
18
17
18
17
17
18
17
17
18
17
17
17
17
17
17
17
17
17
17
17
17
17
16
17
17
17
16
17
17
16
17
16
17
16
17
16
17
16
17
16
16
17
1C
16
16
17
1C
16
16
16
log cot
log cos
diff.
Prop. Farts
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
315 s
31.6
31.7
31.8
31.9
32.0
32.1
32.2
32.3
32.4
32.5
32.6
32.7
328
32.9
330
33.1
33 2
33.3
33.4
33.5
33.6
33.7
33.8
33.9
34.0
341
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
IS
13
13
13
12
13
12
12
13
12
12
12
12
12
12
12
12
12
12
11
12
12
11
12
11
12
11
12
11
11
12
11
11
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
599
59.8
59.7
59.6
59.5
59.4
59.3
592
591
59.0
58.9
58.8
58.7
58 6
585
584
58.3
58.2
58.1
580
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
559
55.8
55.7
55.6
55.5
55.4
55.3
55.2
55.1
55.0
fc
a
s
M
w
Difference
9.7003
9.7016
9.7029
9.7042
9.7055
9.7068
97080
9.7093
9.7106
9.7632
9.7649
9.7667
9.7684
97701
9.7719
97736
9.7753
9.7771
0.2368
0.2351
0.2333
0.2316
022^)9
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
1
2
3
4
5
6
7
8
9
18
1.8
3.6
5.4
7.2
9.0
108
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
9.7788
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
97839
9.7856
9 7873
9.7890
9.7907
9.7924
9.7941
2195
0.2178
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
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
8
9
16
1.6
3.2
4.8
64
8.0
9.6
112
12.8
14.4
9.7242
9.7958
0.2042
9.9284
9.7254
9.7266
9.7278
9 7290
9.7302
9.7314
9.7326
9 7338
9-7349
9.736T
9.7373"
9.7384
9.7396
9.7407
9.7419
9.7430
9.7442
97453
9.7464
9.7975
9.7992
9.8008
9 8025
9.8042
9.8059
9.8075
9.8092
9 8109
"9.8125"
"9.8142"
9.8158
9.8175
9.8191
9.8208
9.8224
9.8241
9.8257
9.8274
0.2025
02008
0.1992
01975
1958
0.1941
1925
0.1908
0.1891
"0.1875
0.1858
01S42
0.1825
0.1809
0.1792
0.1776
0.1759
0.1743
0.1726
9.9279
9.9275
9.9270
9.9265
9.9260
9.9255
9.92 SI
9.9246
9.9241
"9^2"3T
"9.9231
9.9226
9.9221
99216
99211
9.9206
9.9201
9.9196
9.9191
13
1.8
2.6
3.9
52
65
78
9.1
10.4
11.7
12
1.2
2.4
3.6
4.8
60
7.2
8.4
9.6
10.8
11
l.l
2.2
3.3
4.4
5.5
6.6
7.7
8.8
99
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
9.8323
9.8339
9.8355
9.8371
9.8388
9.8404
9.8420
9.8436
0.1694
0.1677
0.1661
0.1645
0.1629
0.1612
0.1596
0.1580
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
06
1.0
1.5
2.0
2.5
3.0
3.6
4.0
4.5
6
0.6
1.2
1.8
24
30
36
42
4.8
5.4
9.7586
9.8452
0.1548
9.9134
lOg COB
diff.
log cot
com.
diff.
log tan
log sin
diff.
Angle
55-60
36
TABLE III. LOGARITHMIC SINES
35-40
Angle
log sin
diff.
log tan
com.
diff.
log cot
log cos
9 9134
diff,
Prop. Parts
350
35.1
35 2
35.3
35 4
35.5
35.6
35.7
35.8
35.9
36.0
361
36.2
363
36.4
36.5
36.6
36.7
368
36.9
37.0
37.1
37.2
37.3
37.4
37.5
37.6
37.7
37.8
37.9
38.0
38.1
382
38.3
384
38.5
38.6
38.7
38.8
389
39.0
39.1
39.2
39.3
394
39.5
39.6
39.7
39.8
39.9
40.0
9.7586
11
jo
11
11
11
10
11
JO
11
10
1]
10
10
11
10
10
10
10
11
10
10
10
10
10
9
10
10
10
10
9
10
10
9
10
9
10
9
10
9
10
9
9
10
9
9
9
9
10
9
9
9.8452
16
16
17
16
16
16
16
16
16
16
16
15
16
16
16
10
16
16
15
16
16
16
15
16
16
15
16
16
]. r >
16
16
16
15
16
16
15
16
15
It
15
IB
15
16
15
15
16
15
16
15
0.1548
6
5
5
6
5
6
5
5
6
5
6
5
6
6
5
6
5
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
(>
6
(
6
6
d
6
6
6
6
6
7
6
6
55.0
54.9
54.8
54.7
54.6
54.5
54.4
54.3
54.2
54.1
54.0
S3.9
53.8
53.7
53.6
53.5
53.4
53.3
532
53.1
53.0
52.9
52.8
52.7
52 6
52. S
52 -^
52.3
52.2
52.P
52.0
51.9
51.8
51.7
51.6
51.5
51.4
51.3
51.2
51.1
51.0
50.9
50.8
50.7
50.6
50.5
50.4
50.3
50.2
50.1
50.0
occvic5ci*-wtot- Extra digit
Difie:
~IT~
1 7
3.4
51
68
85
10.2
11.9
1.3.6
15.3
rence
16
1 6
3.2
48
6.4
8
9.6
11 2
12.8
14.4
9.7597
9.7607
9.7618
9.7629
9.7640
9.7650
9.7661
9.7671
9.7682
9.8468
9.8484
9.8501
9.8517
9.8533
9.8549
9.8565
9.8581
9.8597
0.1532
0.1516
0.1499
01483
0.1467
0.1451
1435
0.1419
01403
99128
9.9123
99118
9.9112
9.9107
9.9101
9.9096
9.9091
9.9085
9.7692
"9.7703"
9.7713
9.7723
9.7734
9.7744
9.7754
9.7764
9.7774
9 7785
9 8613
9?8629~
9.8644
98660
9 8676
9.8692
9.8708
9.8724
9 8740
98755
01387
01371
01356
01340
0.1324
0.1308
0.1292
1276
01260
0.1245
0.122~9~
9 ( , JSO
9 9074"
9.9069
9.9063
9 9057
9.9052
9.9046
9.9041
9.9035
99029
1
2
3
4
r>
G
7
8
()
15
1.5
30
4.5
(10
75
90
105
120
13.5
9.7795
9.8771
9.9023
9.7805
9.7815
9.7825
9.7835
97844
9 7854
9 7864
9.7874
9 7884
9 7893~
9.8787
9.8803
9.8818
9.8834
9.8850
9.8865
9.8881
9 8897
9.8912
0.1213
1 197
01182
0.1166
01150
01135
01119
) 1 103
0.1088
9.9018
9.9012
9.9006
99000
9.8995
9 8989
9 8983
9.8977
9.8971
1
2
3
4
7
8
9
1
2
3
4
5
C
8
<)
11
1 1
33
44
r>r>
8*8
99
10
10
20
30
40
50
<;o
70
80
90
9.8928
9.8944
9 8959
9 8975
9.8990
9.9006
9 9022
9.9037
9 9053
9.9068
"9908^
0.1072
9 896 5
9.7903
9.7913
9.7922
9.7932
9 7941
9.7951
9.7960
9.7970
9.7979
97989"
1056
01041
0.1025
01010
0.0994
00978
0.0963
0.0947
00932
0091(3
989S9
9 89x3
9 8947
9 8941
9.8935
9.8929
9.8923
98917
98911
9.8905"
9
0.9
1 8
2.7
36
4.5
54
6.3
7.2
8.1
9.7998
9.8007
9.8017
9.8026
9.8035
9.8044
9.8053
9.8063
9.8072
9.9099
99115
9.9130
9.9146
9.9161
9.9176
9.9192
99207
9.9223
00901
0.0885
0.0870
0.0854
00839
0.0824
0.0808
0.0793
0.0777
9 8899
9.8893
9.8887
9.8880
9.8874
9.8868
9.8862
98855
9.8849
9.8843"
I
2
3
4
5
6
7
8
9
5
0.5
1.0
i r>
2.0
2.5
30
35
40
4.5
6
0.6
12
18
24
3.0
36
4.2
4.8
5.4
9.8081
9.9238
0.0762
log COB
diff,
log oot
com.
diff.
log tan
log sin
diff,
Angle
50-55
COSINES, TANGENTS, AND COTANGENTS
37
40-45
Angle
log sin
difi.
log tan
com.
diff.
log cot
log cos
diff.
jprop. Parts
40.0
40.1
40.2
40.3
40.4
405
40.6
40.7
40.8
40.9
41.0
41.1
41.2
41.3
41.4
41.5
41.6
417
41.8
41.9
42.0
42.1
42.2
42.3
42.4
42.5
42.6
42.7
428
42.9
43.0
431
432
43.3
43.4
43 S
43.6
437
438
43.9
44.0
44.1
44.2
44.3
44.4
44.5
446
44.7
44.8
449
45.0
9.8081
9.8090"
9.8099
9.8108
9.8117
9.8125
9.8134
9.8143
9.8152
98161
9
9
9
9
8
9
9
9
9
8
9
9
8
9
9
8
9
8
9
8
9
8
8
9
8
8
8
9
8
8
8
8
8
8
8
8
8
8
8
8
8
7
8
8
8
7
8
8
7
8
9.9238
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
J6
15
15
15
15
15
16
15
15
15
15
16
15
15
15
15
15
16
15
15
15
0.0762
9.8843
7
6
7
6
7
6
7
6
7
6
7
6
7
7
6
7
7
7
6
7
7
7
7
7
7
7
7
7
7
7
7
7
V
7
7
8
7
7
7
8
7
7
8
7
8
7
8
7
8
7
50.0
49.9
49.8
49.7
496
49.5
49.4
49.3
49.2
49.1
49.0
48.9
48.8
48.7
486
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
468
46.7
466
46.5
46.4
463
46.2
46.1
46.0
45.9
45 8
45.7
456
45.5
45.4
45.3
45.2
45.1
45.0
+3
5
2
M
W
1
9.9254
9.9269
9.9284
9.9300
9.9315
9.9330
9.9346
9.9361
9.9376
0.0746
0.0731
0.0716
0.0700
00685
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
1
2
3
4
5
6
7
8
9
16
1.6
3.2
48
64
8.0
9.6
11.2
12.8
14.4
9.8169
9.9392
0.0608
0.0593
0.0578
0.0562
0.0547
0.0532
0.0517
0.0501
0.0486
0.0471
9.8778
9.8178
9.8187
9.8195
9.8204
98213
9.8221
9.8230
9.8238
9.8247
9.9407
9.9422
9.9438
9.9453
9.9468
9.9483
9.9499
99514
9.9529
9.8771
9.8765
9.875S
9.8751
9.8745
9.S738
9.8731
9.8724
9.8718
1
2
3
4
5
6
7
8
9
15
1.5
3.0
45
(5.0
7.5
90
10.5
120
135
9.8255
9.9544
0.0456
9.8711
1
2
3
4
5
6
7
8
9
9
0.9
18
2.7
36
4.5
54
63
7.2
8.1
9.8264
98272
9.8280
9.8289
9.8297
9.8305
98313
9.8322
9.8330
9.9560
9.9575
9.9590
9.9605
9.9621
9.9636
9.9651
9.9666
9.9681
0.0440
0.0425
0.0410
0.0395
0.0379
00364
00349
0.0334
00319
9 8704
9 8697
9.8690
9.8683
9.8676
9.8669
9.8662
9 3655
98648
1
2
3
4
5
6
7
8
9
8
0.8
16
2.4
3.2
40
48
56
64
72
9.8338
9.9697
0303
9.3641
9.8346
9.8354
9.8362
9.8370
9.8378
9.8386
9.8394
9.8402
9.8410
9.9712
99727
9.9742
99757
9.9772
9.9788
9 9803
99818
9.9833
0.02S8
0.0273
0.0258
0.0243
0.0228
0.0212
0.0197
00182
0167
9.8634
9.8627
9.8620
9.8613
9.8606
9.8598
9.8591
9.8584
98577
1
2
3
4
5
6
7
8
9
7
0.7
1.4
21
2.8
3.5
42
49
5.6
(53
9 8418
9.9848
00152
9.8569
9.8426
9.8433
9.8441
9.8449
9.8457
9.8464
98472
9 8480
9.8487
9 9864
9.9879
9.9894
9.9909
9 9924
9.9939
9.9955
99970
9.9985
0.0136
00121
0.0106
0.0091
0.0076
0.0061
0.004 S
0.0030
0.0015
9.8562
9.8555
9.8547
9.8540
9.8532
9.S525
9.8517
9.8510
9 8502
1
2
3
4
5
6
7
8
9
6
0.6
1.2
1.8
2.4
30
3.6
42
4.8
5.4
9.8495
10.0000
0.0000
9.8495
log COB
difi.
log cot
com.
diff.
log tan
log sin
diff.
Angle
45-50
ss
TABLE OF TRIGONOMETRIC FUNCTIONS
TABLE OF NATURAL VALUES OF THE TRIGO-
NOMETRIC FUNCTIONS
Angle
Bin
008
tan
cot
860
080
.0000
1.0000
.0000
oo
1.0000
00
90
1
.0175
.9998
.0175
57.290
1.0002
57.299
89 .
2
.0349
.9994
.0349
28636
1.0006
28.654
88
3
.0523
.9986
.0524
19.081
1.0014
19.107
87
40
.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
.1765
5.6713
1.0154
5.758S
80
11
.1908
.9816
.1944
5.1446
1.01S7
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
.2867
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
.42 1 5
2.3559
1.0864
25593
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
.8572
.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
.7071
1.0000
1.0000
1.4142
1.4142
45
OOB
sin
cot
tan
CSC
860
Angle
Live Music Archive
Librivox Free Audio
Metropolitan Museum
Cleveland Museum of Art
Internet Arcade
Console Living Room
Books to Borrow
Open Library
TV News
Understanding 9/11