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CMlNo. 514- Accession No.
Author
Title
- . ..>. ,
Thislaook should be returned on orbeftwclhe datelajjt toarked below
ESSENTIALS OF
TRIGONOMETRY
WITH APPLICATIONS
BY
DAVID RAYMOND CURTISS
AND
ELTON JAMES MOULTON
PROFESSORS OF MATHEMATICS
NORTHWESTERN UNIVERSITY
D. C. HEATH AND COMPANY
BOSTON NEW YORK CHICAGO
ATLANTA SAN FRANCISCO DALLAS
LONDON
COPYRIGHT, 1942,
BY D. C. HEATH AND COMPANY
No part of the material covered by this
copyright may be reproduced in any form
without written permission of the publisher.
PRINTED IN THE UNITED STATES OF AMERICA
PREFACE
The distinguishing features of this book are (1) a concise presenta-
tion of the fundamental part of plane trigonometry up to the solution
of triangles; (2) the use of five-place logarithms in the solution of
triangles arid in oth^r applications, with due allowance for the ac-
curacy of data; (3) a chapter on spherical trigonometry; (4) a
chapter on applications, which has been written to give essential
preparation to those who may enter upon military or naval, service.
This text, therefore, should not only meet the needs of students for
whom trigonometry is mainly useful as an introduction to further
courses in mathematics, but it should be especially valuable for those
who are interested in applications.
The first five chapters arc taken from the authors' text A Brief
Course in Trigonometry with minor revision. In the next three
chapters part of the material has been adapted from the same text
and from the authors' Trigonometry, Plane and Spherical. Most of
the last chapter is new.
Use of Coordinates. As in their other texts, the authors have
here made use of coordinates, almost from the start. In preparatory
courses in mathematics, students have become familiar with rec-
tangular coordinates. The briefest and best introduction to trigo-
nometry, we believe, is based on the utilization of this knowledge,
together with the use of polar coordinates. This makes it relatively
simple to start with the general angle and to give proofs that apply
to all cases. It also lays a firm foundation for the study of analytic
geometry and calculus.
Applications. Throughout the first eight chapters both text and
exercises present many applications of trigonometry to problems of
engineering and of other sciences. The ninth chapter is devoted to
the use of trigonometry in surveying, in problems connected with
maps and with artillery, and in navigation. Under the last head will
be found sections on plane sailing, parallel sailing, dead reckoning,
great circle sailing, and celestial navigation. In connection with the
use of artillery such terms as the mil are explained, and problems of
aiming are briefly treated.
iii
iv PREFACE
Exercises and Answers. An unusually large number of exercises
have been included. Those in Chapter IX should be especially useful
as illustrations of the applications of trigonometry which are not
found in most textbooks on the subject.
Answers are given to odd-numbered problems where giving an
answer would not spoil the exercise. Where the proper number of
significant figures is a part of the exercise, attention is paid to this
point in the answer. Except where the contrary is indicated, the
five-place tables given in this book were used in computing answers
for the last four chapters.
Four- and Five-Place Tables. Both four-place and five-place
tables are available. The tables have been set in large, clear type
and arc noteworthy for their legibility.
Number of Lessons. A brief course in the essentials of plane
trigonometry is contained in the first seven chapters and may be
covered in from 30 to 35 lessons. By omission of starred sections and
by a reduction in the number of exercises which are assigned, the
course may be shortened by 5 lessons. By inclusion of Chapter VIII
on Spherical Trigonometry, Chapter IX on Applications, and the
starred sections, a course of 50 to 60 lessons is available.
D. R. CURTISS
E. J. MOULTON
CONTENTS
CHAPTER I
TRIGONOMETRIC FUNCTIONS
SECTION PAGE
1. The general angle in trigonometry 1
2. Measurement of angles 2
3. Coordinates. Standard position of an angle. Triangle of refer-
ence 4
4. Quadrants 6
5. The six trigonometric functions 6
6. Signs of the functions 8
7. Trigonometric tables obtained by measurements 9
8. Functions of 45 and related angles 12
9. Functions of 30, 60, and related angles 13
10. Functions of quadrantal angles 14
11. Problems in which a function is given 16
*12. Projection on coordinate axes 17
*13. Vectors. Components. Resultants 18
CHAPTER II
RIGHT TRIANGLES
14. Relations between the sides and acute angles of a right triangle . . 20
15. Relations between functions of the acute angles of a right tri-
angle 20
16. Solution of right triangles 21
17. Tables of values of functions 21
18. Interpolation 22
19. The table of squares of numbers. Square roots 24
20. Significant figures 26
21. Typical solutions of right triangles 27
22. Geometrical applications 28
23. Heights and distances. Bearings 30
CHAPTER III
REDUCTION FORMULAS. LINE VALUES. GRAPHS
24. Functions of a general angle expressed as functions of the corre-
sponding acute angle 35
v
vi CONTENTS
SECTION PAGE
25. Functions of 180 - 37
26. Functions of n 180 =fc 39
27. Functions of n 90 6, where n is odd 40
28. Line values 41
29. Properties of the sine function 43
30. Graph of the sine function in rectangular coordinates 44
31. Properties and graph of the cosine 45
32. Properties and graphs of the tangent and secant 46
CHAPTER IV
TRIGONOMETRIC IDENTITIES
33. Fundamental identities involving one angle 48
34. Other identities involving one angle 50
35. Addition formulas 52
36. Addition formulas for sine and cosine functions 53
37. Formulas for sin (a ft) and cos (a ft) 55
38. Formulas for tan (a + ft) and tan (ex - ft) 58
39. Functions of 2<x 59
40. Functions of 60
&
41. Formulas for sums and differences of two sines or two cosines ... 64
CHAPTER V
RADIAN MEASURE. INVERSE FUNCTIONS.
TRIGONOMETRIC EQUATIONS
42. Radian measure 68
43. Inverse functions and their principal values 71
44. Trigonometric equations 75
CHAPTER VI
LOGARITHMS. FIVE-PLACE TABLES
45. The definition of a logarithm 78
46. The table of five-place logarithms of numbers 78
47. Logarithms of positive numbers 80
48. Laws of exponents and of logarithms 82
49. Computation with the use of logarithms to base 10 83
*50. Cologarithms 84
51. Five-place logarithms of trigonometric functions 85
CONTENTS vii
CHAPTER VII
SOLUTION OF TRIANGLES. FIVE-PLACE LOGARITHMS
SECTION PAGE
52. Right triangles 89
53. The problem of solving an oblique triangle 91
54. The law of sines 92
55. Check formulas involving all six parts 93
56. Case I. Given two angles and one side 94
57. Case II. Given two sides and an angle opposite one of them . 95
58. The law of tangents 98
59. Case III. Given two sides and the included angle 99
60. The law of cosines 100
61. The half-angle formulas 101
62. Case IV. Given three sides 103
63. Two formulas for the area of a triangle 104
CHAPTER VIII
SPHERICAL TRIGONOMETRY
64. Spherical triangles 109
65. Polar triangles 110
66. Formulas for right triangles Ill
67. Napier's rules Ill
68. Proofs of the formulas for right triangles 112
69. Rules regarding the relative magnitude of parts 115
70. Solution of right spherical triangles 116
71. Isosceles triangles. Quadrantal triangles 119
72. The law of sines 120
73. The laws of cosines 121
74. Formulas for half-angles and half-sides 122
75. Napier's analogies 124
76. Solution of oblique spherical triangles 126
77. Case I. Given three sides 127
78. Case II. Given three angles 128
79. Case III. Given two sides and the included angle 129
80. Case IV. Given two angles and the included side 130
81. Case V. Given two sides and an angle opposite one of them (am-
biguous case) 131
82. Case VI. Given two angles and a side opposite one of them (am-
biguous case) 133
83. Area of a spherical triangle 134
viii CONTENTS
CHAPTER IX
APPLICATIONS
SECTION PAGE
84. Introduction 136
85. Plane surveying 136
86. Use of maps in artillery problems 140
87. Measurement of angles. The mil 142
88. Plane sailing 146
89. Parallel sailing 147
90. Middle latitude sailing 148
91. Dead reckoning 150
92. Great circle sailing 153
93. Positions on the celestial sphere 156
94. The astronomical triangle, and problems connected with its solu-
tion 158
ANSWERS TO ODD-NUMBERED PROBLEMS 161
ESSENTIALS OF
TRIGONOMETRY
CHAPTER I
TRIGONOMETRIC FUNCTIONS
In trigonometry we are concerned with relations involving angles,
which are best expressed by means of six trigonometric functions.
Trigonometry may be described as a study of these functions and
their uses.
1. The general angle in trigonometry. An angle as defined in
plane geometry is a figure, such as BAG in Figure 1, formed by
two rays, AB and AC, with common end-point at A. The word ray
here denotes a portion of a straight line extending indefinitely hi
one direction from a point A. The rays
A B and AC are said to be the sides of the
angle BAG.
In trigonometry we add to this defini-
tion by considering an angle BAG as gen-
erated by rotating a ray from the position
AB until it coincides with AC. We then
call AB the initial side of the angle, and AC the terminal side.
Rotation may be in either direction, and the amount of rotation
Initial Side
FIG. 2
Initial Side
FIG. 3
may include one or more complete revolutions. In each of the
Figures 2, 3, 4, 5 the curved arrow indicates the direction and amount
l
2 TRIGONOMETRY
of rotation for the angle BAG as thus defined. Figure 3, for example,
indicates an angle of more than three complete revolutions, while
the angle in Figure 5 has more than two, in the opposite direction.
In order to designate the direction of rotation, we call an angle
positive if it is generated by counterclockwise rotation (that is, rota-
tion in the direction opposite to that in which the hands of a clock
move; see the angles in Figures 2 and 3) and negative if it is gener-
ated by clockwise rotation, as in Figures 4 and 5.
Terminal Side
FIG. 4
Initial Side
FIG. 5
B
2. Measurement of angles. A familiar system for the measure-
ment of angles employs degrees, minutes, and seconds. An angle of
one degree is the ninetieth part of a right angle, while a sixtieth of a
degree is one minute, and a sixtieth of a minute is one second. Thus,
if we employ the usual notation,
1 right angle = 90, 1 = 60', 1' = 60".
In designating the measure of an angle we prefix a minus sign if
the angle is negative, but if the angle is positive we either prefix
a plus sign, or no sign at all. Thus in Figure 6 the measure of the
first angle is 90, of the second angle is 585, and of the third is
- 225.
FIG. 6
In another system of measurement a right angle is divided into 100 equal parts
called grades, a grade into 100 minutes, and a minute into 100 seconds. Still
another system, that of radian measure, is discussed in Chapter V of this
book.
TRIGONOMETRIC FUNCTIONS 3
A useful instrument for drawing or measuring angles is the pro-
tractor, whose use is illustrated in Figure 7. Here the instrument
FIG. 7
is so placed as to measure the angle AOB; we see that it is an angle
of 27. From the same figure we infer how to draw a ray OB making
an angle of 27 with OA, or to draw OC making an angle of - 153
with OA.
EXERCISES
Draw the angles whose magnitudes are the following:
1.270, 135, -225, -630, 2.5 right angles.
2. 180, 225, 1170, -300, 1.5 right angles.
3. - 270, 315, 1215, - 855, - 3.5 right angles.
4. - 90, 240, - 315, - 1215, - 2.5 right angles.
With a protractor draw the angles given in one of the following sets:
6. 10, 210, - 83, 527, - 933.
6. 70, 160, 318, - 482, - 872.
7. 40, 110, 262, - 198, - 753.
8. 80, 200, 478, - 413, - 1291.
4 TRIGONOMETRY
Estimate the measure in degrees of the angles A, B, C, of the following triangles,
then measure them with a protractor:
11.
13. Through an angle of how many degrees does the minute hand of a watch
turn in 140 minutes? The second hand? The hour hand?
14. Through an angle of how many degrees does the minute hand of a watch
turn in a week? The second hand? The hour hand?
15. An auto wheel whose circumference is 8.4 ft. rolls forward 14 ft. Through
an angle of how many degrees does a spoke turn as viewed from the left-hand side
of the car? As viewed from the right-hand side?
16. A bicycle wheel whose circumference is 7.5 ft. rolls forward 17.5 ft. Through
an angle of how many degrees does a spoke turn as viewed from the left-hand side
of the bicycle? As viewed from the right-hand side?
3. Coordinates. Standard position of an angle. Triangle of refer-
ence. The student is familiar with the use of coordinates in drawing
graphs. Axes X'OX and Y'OY are drawn perpendicular to each
other, intersecting at an origin as in Figure 8. To locate a point
Y P
y-b]
1
M
P :
r ]
7"
,\ y ~ b
x = a
V v'
M
A A IM
'
b
r/ -P :
FIG. 8
M]
-4
Y' P
P we drop a perpendicular PM to the X'OX axis (more briefly, the
x-axis) and, choosing a unit of length, measure OM and MP in terms
TRIGONOMETRIC FUNCTIONS 5
of this unit. If the length of OM = a units, and the length of M P - b
units, then the rectangular coordinates x, y, of P with respect to the
axes X'OX, TOY are defined as follows:
If M is to the right of 0, that is, on the ray OX, then
x = a;
if M is to the left of 0, that is, on the ray OX', then
x = a.
If P is above the z-axis, then
y = &;
if P is below the z-axis, then
y = - b.
If P is on the ar-axis, then
y - 0;
if P is on the ?/-axis, then
x = 0.
If the coordinates of a point P are # = a and y = 6, we refer to it
as the point (a, 6).
In trigonometry the standard position of an angle is that in which
its initial side is OX. On the terminal side take any point P (not 0),
and drop the perpendicular PM on the o>axis. If P is not on one of
the coordinate axes, then OMP forms a triangle called a triangle of
reference for the angle.
FIG. 9
If B (Greek letter "theta") is a measure of the angle XOP, and if
length of OP = r units,
then r and are called polar coordinates of the point P (see Figure 9).
6
TRIGONOMETRY
4. Quadrants. The axes X'OX and Y'OY divide a plane into
four parts called quadrants. If a point is in the first quadrant, both
of its rectangular coordinates are positive; if it is in the second
quadrant, its z-coordinate is negative, its ^-coordinate positive; if it
II
III
IV
Fio. 10
FIG. 11
is in the third quadrant, the coordinates are both negative; if it is
in the fourth quadrant, its ^-coordinate is positive, its ^-coordinate
negative.
An angle is said to terminate in the first quadrant when, if it is
in standard position, its terminal side is in the first quadrant. Fig-
ure 11 illustrates such an angle, and also angles that terminate in
each of the other quadrants.
An angle whose terminal side is OX, OF, OX', or OY' when the
initial side is OX is called a quadrantal angle.
Thus an angle of is a quadrantal angle, as well as any positive
or negative angle whose measure is a multiple of 90. For such an
angle there is, strictly speaking, no triangle of reference, but the
rectangular coordinates (x, y) and the
polar coordinates (r, 6) of a point P on
the terminal side of the angle still have a
definite meaning.
5. The six trigonometric functions.
Construct a triangle of reference, OMP,
for any angle 6. In doing this the point P
on the terminal side is taken at an arbi-
trary distance from 0. If we take a sec-
ond point P' we obtain a second triangle
of reference, OAf'P', as in Figure 12.
The two triangles are similar, since their corresponding angles are
equal. Hence the ratios of corresponding sides are equal.
FIG. 12
TRIGONOMETRIC FUNCTIONS 7
Moreover, if the coordinates of P are (x, y) or (r, 0), and of P' are
(x f , y') or (r', 0), the corresponding coordinates have the same
signs. Thus, in Figure 12 re and x' are negative, y and y' are positive,
and r and r' are positive.
It follows
that
y =
r
y'
7''
X
r
r
r'
r
y ~~
y''
x
7'
r'
U
x
x
y
These six ratios, whose values thus depend upon 6 but not on
the position of P on the terminal side, are called the trigonometric
functions (or ratios) of the angle 0. For convenience special names
are given to them, as listed below, where an abbreviation for each
is shown (the student may refer to Figures 9 and 13 for typical
angles 0) :
sine of = sin = -;
(1) cosine of = cos 6 = -;
tangent of 6 = tan 6 = -;
cosecant of = esc = -;
9
secant of 6 = sec = -;
cotangent of 6 = cot 6 = *
These six definitions should be memorized, since they are of con-
stant use in trigonometry.
In the case of quadrantal angles 0, 90, 180, . . ., there is no
triangle of reference, but a point P on the terminal side has CO-
FIG. 13
ordinates, x, y, r, and the equations (1) are used for such angles also,
defining values of the functions in all cases which do not involve
division by zero. Where the definition would involve division by
zero, the function is not defined.
From the definitions we observe that sin 0, for example, has one
value for each angle 0. But there are many angles which have the
8
TRIGONOMETRY
same sine; in fact, there are infinitely many (co-terminal) angles
differing by multiples of 360, for which the one triangle of reference
may be used, and sin 6 would have the same value for all of these
angles. The problem of finding all angles for which a given trigono-
metric function has a particular value will be studied later.
Besides the six trigonometric functions defined by (1), there are others some-
times encountered. For example:
versed sine of 6 = vers 0=1 cos 6,
coversed sine of 6 = covers = 1 sin 0,
haversine of 6 = havers = J vers 6.
6. Signs of the functions. The signs of x and y may be either
positive or negative, depending on the quadrant in which 6 termi-
nates, but r is always positive. Hence the functions are positive for
some angles and negative for others.
For example, by reference to Figure 13,
we observe that when 6 terminates in the
second quadrant
sin -f
csc +
others
cos = - = = ,
r +
tan -f-
cot +
others
tan 6
all
cos
sec
others
x
FIG. 14
and so on.
It is readily seen, by a similar discussion, that the signs of the func-
tions for the various quadrants are those indicated in Figure 14.
EXERCISES
Locate points whose polar coordinates are given as follows:
1. A (5, 30), B(10, 135), 0(6, 300), D(8, 270), E(4, - 110).
2. A (6, 70), B(4, 170), C(4, 200), Z>(8, 280), E(6, - 180).
3. A (8, 50), B(5, 145), C(4, - 150), D(8, - 400), J0(4, - 270).
4. A (7, 420), B(5, 210), C(4, - 90), Z>(5, 460), E(8, - 630).
Locate by means of a protractor and ruler points whose polar coordinates are given
is follows; and write down approxitnate rectangular coordinates for each point with
"espect to the axes OX and OY:
5. A (9, 38), B(6, 249), C(5, - 81), D(7, - 585).
6. 4(8, 42), B(6, 116), C(12, 283), D(10, - 407).
7. 4(5, 91), B(9, 171), C(12, - 140), D(8, - 758).
8. 4(6, 67). B(8, - 222), C(5, 263), I>(4, 317).
TRIGONOMETRIC FUNCTIONS 9
Plot the points whose rectangular coordinates are given in one of the following sets.
Find for each point by computation the value of r, sin 0, cos 6, and tan 0, where r, 6
are polar coordinates of the point.
9. A(4, 3), B(6, - 8), C(- 9, - 12), Z>(5, - 12), (0, - 3).
10. A(3, 4), B(- 8, 6), C(- 12, - 9), D(12, - 5), #(0, 5).
11. A(5, 12), B(- 12, 5), C(- 8, - 6), D(9, - 12), #(5, 0).
12. A(20, 21), B(- 5, 12), C(- 12, - 5), D(21, - 20), E(- 5, 0).
For each of the following points r 10. Compute the rectangular coordinates of
each point in one of the following sets where sin 6 and cos 6 have the values given t and
locate each point in a suitable figure:
13. Point
A
B
C
Z)
E
sin0
&
-i
H
-A
1
cos 6
A
f
-A
-tt
14. Point
A
B
c
Z)
E
sin0
J
if
-ft
-tt
- 1
COS0
f
-ft
12
tt
16. Point
A
B
C
D
E
sin 6
ft
-ft
A
-*
cos 6
tt
Si
-tt
-t
- 1
16. Point
A
B
c
D
E
sin 6
11
f
-ft
-tt
cos 6
ft
-*
tt
tt
1
For the following sets of points, find for each point the six trigonometric functions
of when the terminal line passes through the point designated:
17. A (9, 12), B(7, - 24), C(- 21,20), D(- 8, - 15).
18. A (4, 3), B(20, - 21), C(- 7, 24), D(- 6, - 8).
19. A (8, 15), B(- 4, 3), C(21, - 20), D(- 16, - 30).
20. A (20, 21), B(- 30, 16), C(- 8, - 6), D(3, - 4).
Ftnd in what quadrants 6 may terminate:
21. If sin 6 = - . 25. If sec = - 3.
22. If cos = f 26. If esc = 2.
23. If cot = - 2. 27. If sin = J.
24. If tan = - i 28. If cos = - J.
i/ie signs of the six trigonometric functions of each angle in the following
exercises: ^ L
29. 78, 213, - 310, - 601, 1111.
*- 30. 13, 471, - 500, 1000, - 1500.
^'8L 25, 301, - 140, - 800, 2000.
/^32. 107, 265, - 513, 640, 3000.
7. Trigonometric tables obtained by measurements. In using the
trigonometric functions it is convenient to have at hand a table of
approximate values of the functions for a large number of angles.
10
TRIGONOMETRY
We shall now show how such a table may be made by use of measure-
ments.*
In Figure 15 a circle of radius 50 millimeters is graduated to 5 intervals; the
cross-rulings are 2 millimeters apart. To find approximate values of the trigono-
FIG. 15
metric functions of an angle whose measure is 20, for instance, we read off for a
point P at the 20 mark on the circle the approximate values of z, y t and r t correct
to two significant figures, x = 47, y = 17, r = 50.
Hence, approximately, f
sin 20 - J5 = .34; cos 20 = $ J = .94; tan 20 = # - .36;
esc 20 = & - 2.9; sec 20 = J = 1.1; J cot 20 = ff = 2.8.
* At the end of the book is a more extensive and accurate set of tables, calcu-
lated by other methods, which will be used in the next chapter. The methods of
calculation are too difficult to explain in this book.
t By sin 20 we mean "the sine of an angle whose measure is 20"; this form
of abbreviation is constantly used in trigonometry.
TRIGONOMETRIC FUNCTIONS
11
Similarly for 250 we have
x = - 17, y - - 47, r = 50;
and hence
sin 250 = ^ - = - .94; cos 250 =
- .34; tan 250 -
= 2.8;
and so on.
By the method indicated we may make a table of values such as that shown
below.
Angle
Sin
Cos
Tan
Cot
Sec
Csc
.00
1.00
.00
1.00
5
.09
1.00
.09
ii.4
1.00
ii.5
10
.17
.98
.18
5.67
1.02
5.76
15
.26
.97
.27
3.73
1.04
3.86
20
.34
.94
.36
2.75
1.06
2.92
25
.42
.91
.47
2.14
1.10
2.37
30
.50
.87
.58
1.73
1.15
2.00
35
.57
.82
.70
1.43
1.22
1.74
40
.64
.77
.84
1.19
1.31
1.56
45
.71
.71
1.00
1.00
1.41
1.41
50
.77
.64
1.19
.84
1.56
1.31
55
.82
.57
1.43
.70
1.74
1.22
60
.87
.50
1.73
.58
2.00
1.15
65
.91
.42
2.14
.47
2.37
1.10
70
.94
.34
2.75
.36
2.92
1.06
75
.97
.26
3.73
.27
3.86
1.04
80
.98
.17
5.67
.18
5.76
1.02
85
1.00
.09
11.4
.09
11.5
1.00
90
1.00
.00
.00
1.00
95
1.00
- .09
- ii.4
- .09
- ii.5
1.00
100
.98
- .17
- 5.67
- .18
- 5.76
1.02
By use of the table, approximate solutions of certain important
problems are readily obtained, as illustrated in the following examples.
Example 1. If the rectangular coordinates of a point are (11, 60), what are
its polar coordinates?
Solution. We have x = 11, y = 60. And since
r 2 - x* + y\
we find that r = 61. Also
tan = 2 = 5.45.
x
Since the point (11, 60) is in the first quadrant, the angle must terminate there.
In the table we find that tan 75 = 3.73, tan 80 = 5.67. Hence, approximately,
6 = 80. The polar coordinates as thus determined are (61, 80).
12
TRIGONOMETRY
Example 2. If the polar coordinates of a point are (70, 100), what are the
rectangular coordinates?
Solution. Since cos = - sin = - we have the often used formulas,
x = r cos O t y = r sin 0.
From the table, we have, approximately, cos 100 = .17, sin 100 = .98. Since
r = 70, we therefore have x - 70 X - .17 = - 11.9, y 70 X .98 = 68.6.
Thus to two significant figures the rectangular coordinates are ( 12, 69).
EXERCISES
Extend the table on page II to 200 in the column indicated:
I. Sin. 2. Cos. 3. Tan. 4. Cot. 6. Sec. 6. Csc.
By use of the table on page 1 1 compute approximate polar coordinates (nearest 5
for 0} for points whose rectangular coordinates are given, or vice versa:
7. A(5, 12). 8. B(12, 5). 9. C(- 2, 11). 10. D(6, 10).
II. A (8, 25). 12. J5(25, 65). 13. C(31, 95). 14. Z>(40, 100).
8. Functions of 45 and related angles. There are special angles
the exact values of whose trigonometric functions can be found from
geometric properties of their triangles of reference.
Thus for an angle of 45 the triangle of
reference (see Figure 16) is isosceles. We
may take P on the terminal line so that
x = 1, y = 1. Then, by a law of right tri-
angles, x* + y* = r 2 , so that r = V2. Hence
\/2 , CO 1 V2
_, cos45 -^5-2-.
tan 45 = 1, cot 45 = 1,
sec 45 = \/2, esc 45 = \/2.
We give below figures for angles of 135,
225, 315.
. ,,
sin 45 = 7= =
V2
=l M X
FIG. 16
In each triangle of reference the acute angle MOP is an
y = -3
FIG. 17
angle of 45, and the lengths of OM, MP, and OP are 1, 1, V2. The
signs of x and y are as shown.
TRIGONOMETRIC FUNCTIONS
13
Thus the functions of 135 have the following values:
1 V2 _, -1 -V2
V2
sin 135 = -4= = =,
cos 135
V2 ~ 2 ' - ~ V2 ~ 2
tan 135 = -1, cot 135 = -1,
sec 135 = - V2, esc 135 = V2.
Note that we could have obtained these values by taking the
corresponding values for 45 and prefixing the minus sign where a
function of an angle terminating in the second quadrant should be
minus (see 6).
We could similarly write down values for the functions of 225,
315, 405, and so on.
9. Functions of 30, 60, and related angles. Let ABC, Figure 18,
be an equilateral triangle each of whose sides is of length 2 units.
Each angle is then an angle of 60. If we drop a perpendicular from.
Y
C
MX
Fio. 19
C to AB it bisects the angle C and the base AB. Hence, in the right
triangle A DC thus formed we have
A = 60, D = 90, C = 30,
AD = 1, AC = 2.
Since AC 2 = ID 2 + DC 2 , it follows that DC = \/3.
A triangle of reference whose angles are 30, 60, 90 and whose
sides are 1, \/3, 2 is used in each of Figures 19 and 20.
We readily obtain from Figure 19
sin 30 = ^,
tan 30 = -4= -
V3
sec 30 = -?= =
A/3
cos 30 = ^
V3
2
V3
cot 30 = V,
esc 30 = 2.
14
TRIGONOMETRY
From Figure 20 we have
V3
sin 60 = -^,
i
tan 60 = V3,
cos 60 - L
sec 60 = 2,
csc 60 = -
3 '
and
cos 120 = - i,
cot 120 = -
esc 120 - -4= =
O
FIG. 20
In a similar manner we obtain exact values of functions of 150,
210, 240, 300, 330, and angles differing from these by multiples
of 360.
Y
10. Functions of quadrantal angles.
When = 0, the point P is on OX. If
we choose P so that r = 1, as in Figure 21,
we shall have
x == 1, y = 0, r = 1.
Hence
sin
tan
sec
= 0,
= 0,
- 1,
cos =
cot =
FIG. 21
- 1,
impossible,
esc = (J) impossible.
TRIGONOMETRIC FUNCTIONS
15
We have indicated that the ratios for cot and esc lead, in this case,
to a division by 0, which is impossible;* these f unctions, have no
value for 6 = 0.
Similarly, from Figure 22, we have
and
sin 90 1,
tan 90 impossible,
sec 90 impossible,
sin 180 - 0,
tan 180 = 0,
sec 180 = -1,
cos 90 = 0,
cot 90 = 0,
esc 90 = 1;
cos 180 = -1,
cot 180 impossible,
esc 180 impossible.
The student should write down the values of the functions of
M
270, 360, 450, 540, and other
angles differing from these by
multiples of 360.
As a memorizing device it is inter-
esting to note that the values of the
sines of the angles 0, 30, 45, 60,
90 may be written as one half the
square roots of the successive in-
tegers:
sin - ^; sin 30 - -^; sin 45 - -^; sin 60
::$
180
Fro. 22
and the cosines of these angles are found by taking these numbers in the reverse
order. Values of all six functions of these angles are given in the following table:
Angle
Sin
Cos
Tan
Cot
Sec
Csc
1
1
30
1
V3
Vs
V3
2v/3
2
2
2
3
3
45
V2
V2
1
1
V2
V2
2
2
60
V3
2
1
2
V5
V3
3
2
2V3
3
90
1
1
* By definition of division as the inverse of multiplication, i, if it has a value,
must be such a number that its product by the denominator is the numerator 1.
But there is no such number.
16
TRIGONOMETRY
EXERCISES
For the following Bets find by measurement and calculation the values of the
sine, cosine, and tangent of each angle, using Figure 15.
1. 70, 80, 90, 100, 260.
2. 40, 60, 80, 110, 280.
3. 120,
4. 140,
240
250,
260
280.
300.
270.
- 30.
- 150.
- 45.
-*.
i
270,
240, 260, 280,
For the following sets find, by use of figures, approximate values of and 0':
5. If sin - i cos = - f. If cos 0' - - 1.
6. If cos - A, tan " - V- If tan 0' = i cos 0' positive.
7. If tan - Y, sin - }J. If sin 0' - - 1.
8. If cot - - ft, cos - - ft. If cot 0' = 0, sin 0' negative.
For the following sets find from suitable figures the exact values of the six functions
of each of the angles given:
9. 315, 210. 13. - 60,
10. 300, 135. 14. - 90,
11. 150, 225. 15. 180,
12. 240, 330. 16. 540,
Prove the following statements:
17. cos 60 sin 330 - cos 30 sin 300
18. cos 30 cos 330 + sin 45 cos 225
19. sin 45 cos 300 - cos 60 sin 225
20. sin 30 cos 300 - sin 60 cos 210 - 1.
11. Problems in which a function is given.
Example 1. Given sin * f ; construct possible angles and find values of
the other functions of 0.
Solution. Our construction depends on finding
a point (or points) P for which - = - We can
T 5
take y - 3, r 5 (or we could take y = 6, r 10).
Since & 2 + y 2 * r 2 , we see that x 2 r 2 j/ 2 , so
that in the present example, if y 3, r 5, we
have x 2 = 25 9 = 16, and x may be either
-|- 4 or 4. In Figure 23 we have drawn the
two corresponding triangles of reference, and have
indicated two angles 6\, 02, for each of which the
value of the sine is $. Of course, all co-terminal
angles have the same sine.
FIG. 23
From the figure we read off the values of the other functions of the anglea
We have, for example,
cos 0i - f, tan 0! = },
cos 02 - - i, tan a - - }.
The student should write the values of the other functions.
TRIGONOMETRIC FUNCTIONS
17
Example 2. Given tan 6
the other functions of 0.
Solution. Since - 75*
X L&
construct possible angles and find values of
r
we
may take x = 12, y =* 5, or x =
12, y = - 5. The relation
r SB & -f y* gives r 1 = 144 +
25 = 169, so that r = 13 (we do
not allow r to be negative). Fig-
ure 24 shows how to construct
two angles ft, ft, and their tri-
angles of reference. By using the IG *
data given in this figure, the values of the functions of ft and ft are found.
EXERCISES
Find the values of the other Junctions of the angle 0, when it is given that:
1. esc - Y- 7. sec - if 13. cos - - f.
2. sin - - Jf 8. tan - Y- !* tan = 2.
3. tan = - ^. 9. tan = - i 15. sin - - f .
4. cos - J. 10. sin - - J. 16. cot - - }.
6. sin = J. 11. cot - - 1. 17. sec = }f -
6. cos = - % . 12. sec = - ii 18. esc - - |$.
Express att six of the trigonometric functions in terms of the following:
19. sin 0. 21. tan 0. 23. sec 0.
20. cos 0. 22. cot 0. 24. esc 0.
*12. Projection on coordinate
axes. Consider a directed line AB
which makes an angle with the
z-axis of a system of rectangular
coordinates, and let CD be a seg-
ment of AB (Figure 25). On a
directed line through making
the angle 6 with the #-axis, take
OP = CD. Then the projection
C'D' of CD on the x-axis equals
>/
X
/Y
^T
M
/ C"
>'X
^
/
i
FIG. 25
OAf, the projection of OP on the z-axis. This may be written
ProjxCD = Proj*OP = OM.
Since cos
(1)
Similarly
(2)
t we have OM = OP cos 0; hence
Proj x CD OP cos 6,
Proj, CZ) = CD cos 0.
Projy CZ) = CZ) sin 9.
18
TRIGONOMETRY
The student should verify these formulas not only for the angle
of Figure 25 but also for that of Figure 26 and other figures.
FIG. 26
. Vectors. Components.
Resultants. A quantity which
may be represented by a directed
line segment CD is often called a
vector quantity. Thus force, ve-
locity, and acceleration are vector
quantities. The projections of a
vector quantity on the x- and
2/-axes are called components of
the vector.
If F is the magnitude of a force
which makes an angle with the z-axis, and if F x and F y are the
components of the force, then, by formulas (1) and (2) of 12,
F x = F cos 0, F v = F sin 6.
Similar formulas hold for velocity and acceleration.
If the components F x and F v are given, the vector F is called the
resultant. It is seen that the magnitude of F is VF X 2 + F y 2 . The
F
direction of F is given by the angle 0, where tan = -=r
f x
If two forces, represented in magnitude and direction by AB and
AC, act on a particle at A, they are equivalent to a single force, called
the resultant force, acting on the particle. The magnitude and direc-
O
FIG. 27
FIG. 28
tion of this resultant are represented by the diagonal AD of the
parallelogram of which AB and AC are two sides. This principle is
known as the Parallelogram Law of Forces. A similar law holds for
velocities and accelerations.
Example. A force of 20 Ib. acts at an angle of 40 with the horizontal. What
two forces, one horizontal, the other vertical, would be- equivalent?
TRIGONOMETRIC FUNCTIONS
19
Hence
Solution. In the vertical plane of the force, let the #-axis be horizontal, the
y-axis vertical. Then
F* - 20 cos 40, F v = 20 sin 40.
Using the table on page 11, we have
cos 40 - .77, sin 40 - .64.
F x - 15.4 lb., F y - 12.8 Ib.
These values are of course approximations.
EXERCISES
1. Draw a figure similar to those in 5, making an angle terminating in the
second quadrant, and verify formulas (1) and (2) of 12. Note that the signs
as well as the magnitudes are correct.
2. Proceed as in Exercise 1, making 6 an angle terminating in the fourth
quadrant.
3. If a boat is traveling NE with a speed of 15 mi. per hr.,what is the com-
ponent of its velocity in the eastward direction? In the northward direction?
4. If a boat is making 30 knots per hour on a course N 40 E, what are its
components of velocity in the eastward and the north-
ward directions respectively?
6. A swimmer in crossing a stream puts forth efforts
which in still water would carry him directly across at
2 mi. per hr. If the current is 5 mi. per hr., what are
the actual direction and speed of the swimmer?
6. An airplane heads west, running so that in still
air it would have a speed of 120 mi. per hr. There is a
wind from the south blowing with a speed of 30 mi. per hr. What are the actual
direction and speed of the airplane?
7. A force of 15 lb. acts vertically upward and another of 25 lb. acts hori-
zontally on a particle. What are the magnitude and the direction of the single
force equivalent to the two?
8. A force of 3 lb. acts horizontally, another of 4 lb. acts vertically on a
particle. What are the magnitude and the direction of the resultant force?
9. A boat sails on a course S 20 W (compare Exercise 4) with a speed of 14
knots per hour. What are the westward and southward components of its velocity?
10. A surveyor runs a line 600 yd. N 10 W (compare Exercise 4) from A to B.
How far west and how far north is B from A?
N
W
,N40E
40-
E
CHAPTER II
RIGHT TRIANGLES
14. Relations between the sides and acute angles of a right triangle.
Let ABC be a right triangle whose acute angles are at A and B. We
designate the angles as A, B, C y and the sides
as a, b, c (see Figure 30). It is convenient to
call a the opposite side to angle A, and b the
adjacent side; c is the hypotenuse.
If Figure 30 were placed suitably with re-
spect to rectangular axes, the triangle shown
would be a triangle of reference for the angle A ,
with x = b, y = a, r = c. The definitions of the functions of A give
us the following relations between a, 6, c, and A*
A a opposite side . , b adjacent side
Sin A = - = r - - > COt A = - = - - - rr - TJ~~
c hypotenuse a opposite side
cos A - - - ad * acent sid * sec A = - = hypotenuse
cos A - c - hypotenuse ' sec A b adjacent side '
A _ a _ opposite side , _ c _ hypotenuse
tan 4 - fc - adj - acent gide esc A - - - opposite side '
There is a corresponding set of formulas for the angle B. These
can be written by analogy with the above set if we recall that 6 is
opposite to B y and a is adjacent.
TI b n a , n b
sm B = -; cos B = -7 tan B = ->
c c a
Ti c n c , D a
esc B = r> sec B = -> cot 1* = -r*
o a 6
15. Relations between functions of the acute angles of a right
triangle. By comparing the two sets of formulas of 14, we obtain
the following relations between functions of B and functions of A :
sin B = cos A y cos B = sin A y tan B = cot A,
esc B = sec A, sec -B = esc A f cot B = tan A.
* The reader may note that we are using the same letter for a line segment,
for its measure in some unit of length, and for the number of units in that measure;
similarly for angles. The ambiguity need cause no difficulty, since it will be clear
from the context which designation is meant.
20
RIGHT TRIANGLES 21
A convenient way of remembering this last set of formulas is sug-
gested by the prefix "co" in the names of some of the functions. We
shall accordingly call cosine the cofunction of sine, and we shall also
call sine the cofunction of cosine; similarly for the pair tangent,
cotangent, and the pair secant, cosecant.
We may then summarize relations between A and B as follows:
Each function of one of the angles A, B is equal to the corresponding
cofunction of the other angle.
Since B = 90 - A, the angles A and B are said to be comple-
mentary (note that this word also begins with "co")> and the above
rule can take the following form:
Each function of the complement of an acute angle is equal to the
corresponding cofunction of the angle.
For example, 60 is the complement of 30, hence cos 60 = sin 30.
16. Solution of right triangles. Each formula of 14 gives us an
equation connecting three numbers, one of which is a value of a
function of an acute angle, while the other two refer to sides of a
right triangle.
If two sides are given, the value of a function of A is thereby de-
termined, and by the use of tables we can find A. To find the third
side we may use another of the equations, or we may use the equation
a 2 + 6 2 = c 2 . Finally, we find B from the relation B = 90 - A.
When we have thus found the sides and angles which were not given,
we are said to have solved the triangle.
If a side and an acute angle are given, we can also solve the triangle.
In solving a triangle we use tables of trigonometric functions such
as are to be found at the end of this book. Their use is explained in
the following section.
17. Tables of values of functions. On pages 4-8 of the "Logarith-
mic and Trigonometric Tables" at the end of this book, values of the
six trigonometric functions are given for angles at intervals of 10'
from to 90. The following illustrative examples will serve to
explain how the table is used.
Example 1. To find sin 21 30', turn to page 6, go down the column headed
"Degrees" to find 21 30', go across to the column headed "Sin," and find
sin 21 30' = .3665,
the value being correct to four decimal places. It is understood that the value
to five decimal places is between .36645 and .36655, the tabulated value being
as close as may be given by four-place tables.
22
TRIGONOMETRY
Example 2. Similarly we find sin 21 40'
down from above in the table.
.3692, the first digit being carried
Example 3. To find tan 63 20', we must go to the last column, with " Degrees"
at the bottom of the page, to find 63 20' (on page 6). At the bottom find the
column labeled "Tan," and opposite 63 20' find the value, tan 63 20' = 1.991.
It is observed that angles in the first column range from to 45;
for functions of these angles find the name of the function at the top
of the pages. Angles in the last column range from 45 to 90; for
functions of these angles use the names at the bottom.
Example 4. To find A when cot A = 4.061, we look in the "Cot" column for
4.061, and find, on page 5, that A = 13 50'.
Example 5. To find A when cos A = .4975, we note that we must use a column
with "Cos" at the bottom, on page 7, and read the angle in the right-hand
column. We find that cos 60 10' = .4975, and hence A - 60 10'.
EXERCISES
Complete the following equations, using four-place tables:
1. sin
2. tan
3. sec
4. sin
6. tan
6. sec
7. sin
8. tan
9. sec
10. sin
11. tan
12. sec
18 30'
55 50'
17 40'
28 0' :
15 40 ; :
63 30'
37 40'
48 0'
52 30'
47 20'
18 40'
75 10'
cos 7 0'
cot 42 30'
esc 69 40'
cos 82 30'
cot 18 50'
esc 49 40'
cos 10 30' :
cot 19 50'
esc 25 40' :
cos 5 50'
cot 87 30'
esc 24 40' '
Find the angle A in each of the following equations, using four-place tables:
13. sin A - .6361, cos A - .9750, tan A - 7.115, cot A - .3249.
14. sin A - .3746, cos A - .5225, tan A = 8.144, cot A = 3.305.
16. sin A - .5299, cos A = .9990, tan A - .1554, cot A - .4314.
16. sin A = .9903, cos A = .9781, tan A - .2309, cot A = 1.653.
18. Interpolation. In finding the value of a function of an angle,
such as 17 23', which is not given in the table but lies between two
angles that appear, we use the method of interpolation, as illustrated
in the following Examples 1 and 2. In Examples 3 and 4 the method
is applied in finding the angle when the value of one of its functions
is given.
RIGHT TRIANGLES 23
Example 1. Find sin 17 23'.
Solution. The given angle, 17 23', is three tenths of the way from 17 20' to
17 30'. We assume that sin 17 23' is three tenths of the way from sin 17 20'
to sin 17 30'. The sine of 17 23' will then be obtained by taking A of the
amount by which sin 17 30' exceeds sin 17 20', and adding this correction to
sin 17 20'. Hence
sin 17 23' = sin 17 20' + &(sin 17 30' - sin 17 20')
= .2979 + -ft(.0028) = .2979 + .00084
= .2987 approximately.
Since the Tables give values to only four places, we give only four places in our
value of sin 17 23'. This amounts to calling the correction .0008 instead of
.00084. We would have used .0008 for any correction greater than .00075 and
less than .00085. It is customary to disregard the decimal point in the tabulated
values and call the tabular difference 28 instead of .0028, and the correction
8 instead of .0008.
Another way to explain the preceding interpolation is to state that we have
assumed that when an angle increases, its sine increases proportionally; or, in
other words, that differences between angles are proportional to differences
between their sines. For the example just solved the accompanying small table
indicates these differences. We thus have
Angle
I" ri720' .29791 -)
10 3 L1723' \ x 28
L 17 30' .3007 J
Sin
28 10
Then x = 8.4 = 8 approximately; and
sin 17 23' - .2979 + .0008 - .2987.
The assumption just made that differences between angles are proportional to
differences between the values of a function of those angles is not exactly true, but
it gives rise to errors which are negligible when the differences involved are small.
Example 2. Find cot 17 15'. Angle
Solution. From the little table at the right f C f 17 10/
we have
x = & X 33 = 16.5.
[ri7 10' 3.237H -1
6 [17 15' \ X 33
17 20' 3.204 J
Cot
The correction x can be called either 16 or 17. Since the cotangent decreases
when we go from 17 10' to 17 20', the correction, which should take us ffr of the
way from cot 17 10' to cot 17 20', must be subtracted from the former, giving
either 3.237 - .016 = 3.221 or 3.237 - .017 = 3.220. In such cases we shall
always use the result which ends in an even digit; thus, cot 17 15' = 3.220.
Example 3. Given tan A - .4361. Find A.
Solution. We find that the angle A lies between 23 30' and 23 40', as shown
at the right. By the principle of proportional
differences we have An S le Tan
s = X10 = W = 3.7.
Hence
A = 23 30' + 4' - 23 34'.
r r23'30' .4348-1-1
10 *| A .4361 J 13 35
L 23 40' .4383 J
24 TRIGONOMETRY
Example 4. Given cos A = .4100. Find A.
Solution. Proceeding as before, we have Angle
x jj X 10 = 8.
Hence ^"L ""65 50'
A - 65 48'.
Cos
.4094
EXERCISES
By interpolation find the values of the following, using four-place tables:
1. sin 41 51'; sin 48 32'. 7. cos 8 15'; sin 51 35'.
2. tan 26 16'; tan 57 6'. 8. sin 37 8'; cot 68 18'.
3. cos 11 37'; esc 52 21'. 9. tan 42 24'; sec 73 11'.
4. cot 21 17'; tan 80 25'. 10. sin 20 52'; cot 79 13'.
6. sec 15 21'; sec 63 17'. 11. cos 35 14'; sec 48 5'.
6. sin 35 12'; cos 70 12'. 12. tan 14 43'; cos 84 54'.
By interpolation find the values of A to the nearest minute:
13. sin A - .2320. 21. cot A = 1.300.
14. cot A - 3.245. 22. tan A = .8446.
16. tan A = .1115. 23. cos A - .9940.
16. sin A - .5207. 24. cos A - .9105.
17. sin A = .8100. 26. cos A = .5105.
18. cos A - .4035. 26. tan A = .4050.
19. tan A - 1.127. 27. cot A = .6845.
20. cot A = .1540. 28. sin A - .9500.
19. The table of squares of numbers. Square roots. In solving
right triangles we use such formulas as
a 2 = b* + c 2 , a = Vb 2 + c 2 .
Table I, at the end of this book, shortens the work of finding squares
and square roots.
Example 1. Find (3.54) 2 .
Solution. On page 2, go down the column headed N to 3.5, then across to
the column headed 4. The entry is 12.53; hence (3.54) 2 = 12.53, approximately.
Example 2. Find (3.S43) 2 .
Solution. We interpolate, using a principle of proportional parts similar to
that employed in 18. The tabular difference, (3.55) 2 - (3.54) 2 , is found to be
.07. The correction is & X .07 = .02; hence (3.543) a - 12.53 + .02 = 12.55,
approximately.
Example 3. Find (3S.43) 2 and (.03543) 2 .
Solution. (3S.43) 2 - (10 X 3.543) 2 - 100 X (3.543)* - 1255.
(.03543) 2 - - - 0.001255.
RIGHT TRIANGLES 25
Example 3 illustrates the rule : If the number N whose square is
to be found is not between 1 and 10, move the decimal point to the
right (or left) k places so as to obtain a number N\ between 1 and 10.
Find NI* from the Table. Then N 2 can be obtained from Ni z by
moving the decimal point 2k places to the left (or right). Thus if
N - 35.43, we have Ni = 3.543, k - 1.
The square root of a number n in the interior of Table I is obtained
by reading off the corresponding number N from Table I. The first
two digits of N are at the left of the row in which n lies, and the third
digit is at the top of the column where n is located.
Example 4. Find Vl.800.*
Solution. By use of Table I we find Vl.796 - 1.34, Vl.823 = 1.35. By
interpolation, we thus obtain
Vl.800 - 1.34 + & X (.01) = 1.341, approximately.
Example 5. Find VlS.OO.
Solution. We note that VlS.OO is located between the entries 17.98 and
18.06 in the lower part of page 2. By interpolation, VlS.OO = 4.242, approxi-
mately (we chose the final digit as the even digit 2; see 18, Example 2).
Example 6. Find VlSO.O and V.001800.
Solution. The numbers under the radical sign are to be expressed as numbers
between 1 and 100, multiplied or divided by 100 (or powers of 100). Thus
VlSO.O - VlOO X 1.800 = 10 X Vl.800 = 13.41.
V.ooisoo = VTTTST^ * T^n x ^^oo - .04242.
* ^-LUvFy JLUU
Example 6 illustrates this rule : if the number n whose square root
is to be found is not between 1 and 100, move the decimal point to
the right (or left) an even number of places, say 2k, so as to obtain
a number n\ between 1 and 100. Find Vni from the Table. Then
Vn can be obtained from Vni by moving the decimal point k places
to the left (or right).
EXERCISES
Find the squares of the foUowng numbers by use of a Table of Squares:
1. 3.67. 6. 0.0882. 9. 67.2.
2. 28.4. 6. 63100. 10. 0.666.
3. 319. 7. 8.19. 11. 87600.
4. 0.218. 8. 385. 12. 0.00827.
* We write 1.800 to indicate that we wish an answer to four significant figures
(see 20).
26 TRIGONOMETRY
By interpolation, find the squares of the following to four places by use of a Table
of Squares:
13. 82.46. 17. 0.08218. 21. 6232.
14. 217.9. 18. 0.7526. 22. 7219.
15. 8227. 19. 876500. 23. 982.4.
16. 1.234. 20. 23.43. 24. 6.666.
Find the square roots of the following numbers to three places by use of a Table of
Squares:
26. 9.797. 29. 1648. 33. 537300.
26. 19.98. 30. 2007. 34. 156800.
27. 47.89. 31. 0.9293. 35. 0.002401.
28. 85.38. 32. 0.4914. 36. 0.009409.
By interpolation find the square roots of the following numbers to four places by
use of a Table of Squares:
37. 11.72. 41. 625.9. 45. 51870.
38. 21.98. 42. 7890. 46. 87550.
39. 68.91. 43. 2194. 47. 0.007750.
40. 22.08. 44. 9202. 48. 0.05842.
20. Significant figures. Tables I and II do not give exact values
of squares or of trigonometric functions; they give the four-place
numbers that are nearest to the true values. The latter are usually
unending decimals. We should not expect that computations based
on these tables will give better than four-place accuracy. We now
make these ideas more precise.
Let us define the significant figures (or digits) of a number as its
digits beginning with the first that is not zero and ending with the
last that is not zero, unless some other convention is explicitly stated.
An example of such a convention is to say that ftie number 32000 is
given only to four significant figures; we are then counting only the
first two of the final three zeros. We shall always count final zeros
that are written as part of a number following a decimal point. Thus
in .0100, the final two zeros are significant (the zero just after the
decimal point is not significant).
In general, if we multiply or divide numbers each of which is given
approximately, we should not retain more significant figures in the
result than the smallest number of significant figures in the data.
When we are using four-place tables, our results will be given to four
significant figures unless data contain numbers of less than four sig-
nificant figures. In this last case we may compute as if all data were
RIGHT TRIANGLES 27
given to four significant figures, but indicate that our final result is
reliable to fewer places.
Since we are to compute with measurements of angles as well as of
line segments, we need a definition of significant figures in degree
measure. By referring to the tables, one may note that in general
the sine and cosine change their second significant figure by about 1
when the angle changes by 1. Similar observations have led to this
general rule for angles:
An angle measurement to the nearest degree is taken to correspond to a
number of two significant figures; a measurement to the nearest 10' means
three significant figures , one to the nearest I 1 four significant figures, one
to the nearest 5" five significant figures.
21. Typical solutions of right triangles. In the examples which
follow, some of the data are given to less than four significant figures.
Our computations are all carried out to four or more figures and the
answers are first thus given, and then rounded off to the proper
number of significant figures.
An answer is checked by substitution in a formula that has not
been used in the solution and which involves at least two parts of the
answer. This tests the accuracy of the computation; if a check fails,
the answer is probably wrong.
Example 1. Given A = 30 46', c = 1.50. Find B t a, and b.
Solution. The work may be conveniently arranged as follows:
Formulas
a . A b A
- = sm A, - = cos A,
c c
B = 90 A, a = c sin A, b = c cos A.
Computation
90 = 89 60' sin 30 46' = .5115 cos 30 46' - .8593
A = 30 46' c - 1.5 c = 1.5
B - 59 14' 25575 42965
5115 8593
a - .76725 6 - 1.28895
Answer to three significant figures. B = 59 10'; a .767; 6 1.29.
Check to four significant figures, b = a tan B.
b = 1.289 a tan B - .7672 X 1.679 - 1.288
The check is as close as should be expected.
28
TRIGONOMETRY
Example 2. Given b = 100, c = 232 (to three significant figures). Find A t
B, and a.
Solution.
Formulas
sec A = T
o
sec A - ?g8
- 2.320
A - 64 28'
B - 90 - A,
Computation
B = 90 - 64 28'
- 25 32'
a = V53820 - 10000
= V43820
a - 209.3
Answer to three significant figures. A - 64 30'; # = 25 30'; a - 209.
Check to four significant figures, a = b tan A.
a = 209.3 6 tan A = 100 X 2.094 = 209.4
EXERCISES
In these exercises, two parts of a right triangle are given, C = 90, and the other
three parts are to be found. Check all solutions to four significant figures.
In Exercises 1-12, assume that the data were correct to four significant figures,
and give answers to four figures.
1. A = 27 16',
2. A - 18 34',
3. B - 42 20',
4. B = 35 15',
6. A = 42 17',
6. A - 50 42',
8.100.
44.00.
27.00.
12.00.
: 18.00.
.4800.
7. a - 1.66,
8. a - 54.1,
9. a - 11.2,
10. a - 1.35,
11. a = 4.02,
12. o - 46.3,
6 = 2.48.
6 - 81.2.
6 = 12.4.
6 = 1.27.
c - 3.04.
c = 57.8.
7n Exercises 13-24, compute to four or more significant figures, but round off
answers to the appropriate number of figures.
13. B - 80 20', a - .27. 19. b = 508,
c
= 817.
14. B - 68 34', o - 6.6. 20. b = 25.4,
c
- 41.6.
16. A - 62 54', a - 9.2. 21. a = 40.55,
6
= 35.62.
16. A = 56 39', a - .67. 22. a - 5.725,
b
- 4.125.
17. B = 48 45', b = 83. 23. a - 26.61,
b
= 15.07.
18. - 34 18', b - .44. 24. a = 4.027,
b
= 6.153.
C
22. Geometrical applications. Isosceles tri-
/
angles. If 45(7 is an isosceles triangle, let-
b/.
tered as in Figure 31, with a = fc, then the
/ H
perpendicular dropped from C to AB divides
/
D
,
the triangle ABC into two equal right tri-
c
angles. This makes it possible to solve an
FIQ. 31
isosceles triangle if a side and an angle are given, or if the three
sides are given.
RIGHT TRIANGLES
29
Example. Solve the isosceles triangle, where the base is 21.25 ft. and the
angle at the vertex is 37 26'.
Solution. In the triangle ADC we have
21.25
D = 90,
AD
37 26'
= 10.62,
18 43'.
Then
AC - AD csc /.ACD
- 10.62 X 3.116 - 33.09 - BC,
A = 90 - Z.ACD = 71 17' = B.
Regular polygons. Lines drawn from the center of a regular poly-
gon to the vertices are equal in length, so that a triangle such as
ABC in Figure 32 is isosceles. Here, if n is the
number of sides of the polygon, we easily see that
360
angle C is , and angles A and B may then be com-
puted. If, in addition, a side AB is given, or the ra-
dius CD of the inscribed circle (CD is also called the
apothem), or the radius AC of the circumscribed circle,
the other parts may be computed.
Oblique triangles. In Chapter VII the solution of oblique triangles
is discussed. A different, but usually more laborious method consists
in computing with two auxiliary right
triangles, formed by dropping a per-
pendicular from a vertex to an oppo-
site side, as illustrated by Figure 33.
Thus if 6, c, and A were given, we
could solve the right triangle ACD, A*
thus obtaining AD and DC. Since
DB = c AD, we would now know
two sides of the right triangle DCB and could solve for a and 5, while
C = 180 - (A + B).
EXERCISES
The data below are parts of isosceles triangles ABC, A and B being the equal
base angles and C the vertex angle. Solve and check your solution.
1. A - 46 22', c = 1262; find C, a, and 6.
2. C - 62 18', c - 1072; find A, a, and 6.
3. a = 1504, c = 1240; find A, C, and b.
4. A - 51 15', a - 10.51; find C, c, and 6.
6. A - 62 12', a - 88.66; find C, c, and b.
6. C - 135 8', 6 - 1.347; find A, a, and c.
D
FIG. 33
30
TRIGONOMETRY
The following problems 7-10 refer to regular pentagons. Find the parts asked for
by means of those given.
7. The side is 4.56 in.; find the radii of the inscribed and circumscribed
circles.
8. The apothem is 8.44 in.; find the perimeter and the radius of the cir-
cumscribed circle.
9. The radius of the circumscribed circle is 6.63 in.; find the length of a
eide and the radius of the inscribed circle.
10. The perimeter is 25.50 in.; find the apothem and the radius of the cir-
cumscribed circle.
11. In a regular octagon the apothem is 1.76 in.; find the perimeter and the
radius of the circumscribed circle.
12. For a regular decagon the radius of the inscribed circle is 2.72 in.; find
its perimeter and the radius of the circumscribed circle.
An oblique triangle may sometimes be solved by solving right triangles formed by
dropping a perpendicular from a vertex to the opposite side. Solve the following in
this way, the angles being designated by capital letters, the opposite sides by cor-
responding small letters.
13. A = 13 27', b - 37.21, c = 47.26; find B, C, and a.
14. B = 46 16', a = 81.22, c = 92.17; find A, C, and 6.
15. C = 98 12', a - 4.176, 6 = 2.873; find B, A, and c.
16. A 59 21', 6 - 0.1222, c = 1.242; find B, C, and a.
23. Heights and distances. Bearings. The applications of trig-
onometry in all the sciences are so numerous and important that
no attempt can be made to list them here. As examples, we shall
consider a few problems in surveying and navigation.
Suppose a surveyor wishes to find the distance between two trees A
and B on opposite sides of a stream. He can measure on one shore
*'* Angle of
elevation of C
FIG. 34
Horizontal
FIG. 35
along a line perpendicular to AB a convenient distance AC (Figure 34),
measure the angle ACB, and find the required distance by solving
the right triangle ACB.
Suppose he wishes to find the distance from a position A to a flag-
pole BC of known height (Figure 35) without leaving the position A.
An leof
RIGHT TRIANGLES 31
Assuming that A and B are in the same horizontal plane, and that
EC is vertical, he may measure the angle BA C, which is called the
angle of elevation of C for the observer at A, and solve the right
triangle ABC for the required distance AB.
Suppose a navigator on board ship wishes to find how far he is
from a certain rock R on shore at the water's edge. If he sights with
the appropriate instrument from A and
observes that the line AR (Figure 36) is ^ f
depressed below the horizontal line AH
by a certain amount, called the angle of
depression of R as observed from A, and
if he knows the height AB of his instru- p
ment above the water, he may solve the
right triangle ABR and find the required distance. (We observe
that the angle of depression of R for an observer at A equals the angle
ARB, which is the angle of elevation of A for an observer at R.)
In surveying and navigation it is a common
practice to describe the direction of a line from a
point P to a point Q by using the north-south
line through P as a line of reference and indi-
cating the angle not greater than 90 which PQ
makes with the north or south line through P.
^30
Q'
The bearing of Q from P states first whether the
north or south half line was used, the number of
FIG 37 degrees in the angle from this ray to PQ, and
whether Q is to the east or west of NS. In Fig-
ure 37, the bearing of Q from P is north 60 east, written in ab-
breviated form N 60 E. The bearing of Q' from P is S 30 W.
At the end of this section we shall give a number of exercises more
or less like those we have just presented. It will be helpful for the
student to adopt the following method of procedure:
(1) Read the problem carefully, then draw a figure to some con-
venient scale which will show those lines and angles which are given
and those to be found.
(2) Draw auxiliary lines if necessary, and decide on the simplest
plan for solving the problem.
(3) Write down all necessary formulas.
(4) Carry out the numerical calculations, retaining the appropriate
number of significant figures in each answer.
(5) Check the results.
32 TRIGONOMETRY
EXERCISES
Solve the following problems:
1. The angle of elevation of the top of a flagpole measured at a point on the
same level as the base of the flagpole and 153 ft. from it is 22 30'. How tall is the
flagpole?
2. A vertical marker on a sundial is exactly 3 in. high. What is the elevation
of the sun when the shadow cast is 5.8 in. long?
3. In a lighthouse, at 58.2 ft. above water level, the angle of depression of a
small boat is 10 30'. How far is the observer from the boat?
4. Two trees stand at points A and B, separated by a deep ravine. In order
to find the distance between them a line AC, at right angles to AB, is measured.
If AC is 102 ft. long and the angle ACB is found to be 42 20', how far apart
are the trees?
5. A surveyor wishes to measure the distance between two points A and B at
opposite ends of a lake. To do this he runs a line AC, 171 yd. long, and a line
CB at right angles to AC, 205 yd. long. What is the distance AB?
6. A ladder 15 ft. long is leaned against a wall. How high on the wall will it
reach when the foot is 8.5 ft. from the base of the wall, if the ground is level?
What angle will the ladder make with the ground?
7. Two towns are connected by a right-angled system of roads. To go from
one town to the other one must go 8 miles east and 23 miles south. How far
apart are the towns?
8. A guy wire of a telephone pole is 25 ft. long; one end is attached to the
pole 17 ft. above the ground. Find the distance of the other end of the wire
from the base of the pole, if this end is attached to a post at the level of the
ground, the ground being level and the pole vertical.
9. The top of a ladder is placed at a window 21 ft. above the ground. What
angle does the ladder make with the ground if the foot is 11 ft. from the house
(we assume the ground to be level)? How long is the ladder if 1 ft. 6 in. projects
above the point of contact with the window sill, in this position?
10. An observer at the same level as the bottom of a flagpole finds the angle
of elevation of the top to be 27 20'. If the flagpole is known to be 65.2 ft. tall,
how far is the observer from the base of the pole? From the top?
11. A shaft is designed to descend 35 ft. for each 100 ft. measured along the
shaft. What angle will it make with the horizontal? When the shaft is 152 ft.
long, how far is its end below the starting level? How much horizontal distance
has been traversed?
12. A right triangle with sides 4.32 and 2.41 inches long respectively is inscribed
in a circle. What is the diameter of the circle? What are the acute angles of the
triangle?
13. A and B stand on a level plain and sight at the same instant a balloon
directly above a point on a line joining them. The angle of elevation of the
balloon is 35 20' at A's station and 53 40' at B's station. If the two stations
are exactly a mile apart, how high was the balloon at the time of observation?
14. From one point of observation, on a level with the base of a hill, the angle
of elevation of the top is 44. From a point 253 yards farther away on the same
level surface the elevation of the top is 35. How high is the hill?
RIGHT TRIANGLES 33
15. A flagpole is mounted on the top of a wall. At a point level with the base
of the wall and 25.5 feet away, the angle of elevation of the bottom of the pole
is 37 40', and of the top is 54 30'. What is the height of the pole?
16. A power plant has built a new smokestack. At a point 252 feet from the
new stack and 176 feet from the old, the angle of elevation of the top of each is
38 30'. If the point of observation is in the same horizontal plane as the base of
each stack, how much taller is the new stack than the old one?
17. A ship sails N 13 20' E for 26.3 miles, then turns 90 to the right and
travels 8.20 miles. What is its position then with respect to the starting point?
18. An aeroplane flies N 26 35' W for 137.2 miles, then S 53 25' W for 62.4
miles. In what direction should it then fly to return to the starting point in a
straight line, and how far must it go?
19. Two observers at A and B in a horizontal plane observe a captive balloon C.
The points A, B, and C lie in a vertical plane, with C above a point between
A and B. The distance AB is 1570 yards. At A the angle of elevation of C is
25 20', at B it is 34 30'. How high is the balloon above the plane of the ob-
servers?
20. From a ship running on a course N 5 E along a shore the bearing of a
rock is observed to be N 32 E. When the ship has run 350 yards the bearing of
the rock is N 51 E. If the ship continues on its course, how close will it come
to the rock?
21. The angle of elevation of the top of a spire from a point A in a horizontal
plane is 22 23'; from a point B which is 120 feet nearer it in the same plane the
angle of elevation is 35 12'. How high is the top of the spire above the plane?
22. A boat is traveling on a course N 30 E.
When it is at a point A a rock 72 is due north.
After the boat has gone 500 feet to B, the bear-
ing of the rock is N 25 W. How close will
the boat get to the rock if it continues on its
course?
23. As a boat travels a certain course the _ ___
bearing of a rock R when the boat is at A is Boat's course
a. (alpha) to the left of the boat's course; when p IG 35
it is at B the bearing is ft (beta) to the left; and
the distance AB = d is measured. Find how much farther the boat is to travel
before it is at its closest proximity to the rock, and find how close it will then be.
A , d cot ft
An8. K = 7 - ^Tar
cot a cot ft
cot a cot ft
Also find two angles a. and ft such that the answers will be
h - d; k = 2d.
24. The planet Venus goes around the sun in an orbit which is practically
circular, its distance from the sun being about 67 X 10 6 miles. The earth's
orbit is also nearly circular, the distance from the earth to the sun being about
93 X 10 8 miles. What is the largest possible value of the angle SEV formed
34 TRIGONOMETRY
by the line through the earth and sun with the line through the earth and Venus?
Will Venus ever be seen in the east in the evening?
25. In latitude 41 on the earth's surface the sun gets to an altitude of 72 30*
above the horizon when it is at its highest point for the year. What is the shortest
possible shadow which may be cast on a horizontal plane by a vertical pole
75 feet high? And if a pole of unknown height casts a shadow which is at least
40 feet long in a horizontal plane, how tail is the pole?
CHAPTER III
REDUCTION FORMULAS. LINE VALUES. GRAPHS
24. Functions of a general angle expressed as functions of the
corresponding acute angle. Trigonometric tables extend only from
to 90. To find the value of a function of an angle 6' that is
not between and 90 we
reduce the problem to one
concerning an acute angle.
One such method of re-
duction uses the correspond-
ing acute angle 0, which is
the acute angle MOP of the
triangle of reference for the
angle 6'. By moving the triangle OMP into the first quadrant we
could use it as a triangle of reference for 0, but its sides would then
be considered positive, whereas, as in Figure 39, x' and y' may be
negative. We thus obtain the rule :
Each function of an angle 6' is equal either to the same-named
function of the corresponding acute angle 0, or to the negative of that
function.
We observe that when 0' is given we find 6 by taking the numerical
difference between the numerical value of 0' and the nearest multiple
of 180.
Thus
if 0' = 160, then = 180 - 160 = 20;
if 6' = - 570, then 6 = 570 - 540 = 30.
We conclude that
sin 160 = sin 20, cos (- 570) = =t cos 30.
It remains for us to determine whether to use the plus or the minus sign. Since
functions of acute angles are positive, so that sin 20 and cos 30 are both positive,
we can answer this question by finding whether sin 160 and cos ( 570) are
positive or negative. Since 160 is an angle terminating in the second quadrant,
its sine must be positive, and hence sin 160 = + sin 20; since - 570 termi-
nates in the second quadrant, its cosine is negative, and hence cos ( 570)
COS 30.
35
36 TRIGONOMETRY
We may now complete our rule as follows:
In the equation
function of 0' = =b same-named function of 9
we take the + sign if 0' terminates in a quadrant where the function
of 0' is positive, the - sign if the function of 6' is negative.
The method of reduction given above is not the only possible one,
since a function of 6 is equal to the cofunction of 90 - 0, and this
latter angle will also be acute. Thus
sin 160 = sin 20 = cos 70;
cos (- 570) = - cos 30 = - sin 60.
It may be observed that 90 - may be computed by taking the
numerical difference between the numerical value of 0' and the nearest
odd multiple of 90.
Example 1. Find the value of sin (- 497).
Solution. The corresponding acute angle is 3 X 180 - 497 = 43; also
497 terminates in the third quadrant, where the sine is negative. Hence
sin (- 497) = - sin 43 = - .6820.
Also
sin (- 497) = sin (- 5 X 90 - 47) - - cos 47 - - .6820.
It may aid the student if he draws a suitable figure.
Example 2. Find an angle 6' terminating in the fourth quadrant, and such
that cot 8' = - .5000.
Solution. The corresponding acute angle 6 must be such that cot 6 = + .5000.
From the tables we find 6 63 26'. We can obtain a value of 6' by subtracting
6 from such a multiple of 180 that 8' will terminate in the fourth quadrant.
Thus one solution is
6' - 360 - 63 26' - 296 34'.
We may add to 6' any positive or negative multiple of 360 and thus obtain
additional solutions.
EXERCISES
With the aid of a suitable figure, express each of the following in terms of a function
of an acute angle, and find its value by use of Tables:
1. sin 119. 11. cos 162 10'. 21. tan 100.
2. sin 222. 12. cos 216 20'. 22. tan 200.
3. sin 303. 13. cos 312 40'. 23. tan 310.
4. sin 456. 14. cos 471 50'. 24. tan 490.
5. sin 682. 15. cos 569 20'. 25. tan (- 80).
6. sin (- 38). 16. cos (- 27 20'). 26. tan (- 195).
7. sin (- 138). 17. cos (- 127 30'). 27. cot 230.
8. sin (- 216). 18. cos (- 210 10'). 28. cot (- 70).
9. esc 625. 19. sec 530 40'. 29. cot 850.
10. esc (- 888). 20. sec (- 1500 10'). 30. cot (- 311).
REDUCTION FORMULAS. LINE VALUES. GRAPHS 37
Find an angle terminating in the quadrant indicated which satisfies the given
equation:
31. II; sin - 0.8624. 43. II; tan - - 2.345.
32. Ill; sin 6 - - 0.2144. 44. Ill; tan 6 = 5.432.
33. IV; sin 6 - - 0.8188. 45. II; cot = - 0.4321.
34. Ill; sin0 - - 0.1111. 46. Ill; cot 6 = 0.4321.
35. II; cos = - 0.1234. 47. IV; cot 6 = - 0.4321.
36. Ill; cos 6 - - 0.4321. 48. II; cot - - 4.321.
37. IV; cos B = 0.4321. 49. II; sec 6 - - 1.234.
38. II; cos = - 0.4321. 50. Ill; sec - - 1.234.
39. II; tan = - 0.1234. 61. IV; sec 6 = 1.234.
40. Ill; tan 6 = 0.1234. 52. Ill; sec 6 = - 4.321.
41. IV; tan 6 - - 0.1234. 53. Ill; esc 6 = - 4.321.
42. II; tan 6 - - 1.234. 54. IV; esc 6 = - 4.321.
Find all of the angles, if there are any, which satisfy the given equation:
66. sin 6 - 0.2345. 63. tan 6 - 0.2345. 71. sec - 0.2345.
66. sin 8 - - 0.2345. 64. tan 6 = - 0.2345. 72. sec - - 0.2345.
67. sin 6 = 2.345. 66. tan = 2.345. 73. sec = 2.345.
68. sin 6 - - 2.345. 66. tan = - 2.345. 74. sec = - 2.345.
69. cos 6 = 0.2345. 67. cot 6 = 0.2345. 75. esc - 0.2345.
60. cos S = - 0.2345. 68. cot 6 = - 0.2345. 76. esc = - 0.2345.
61. cos 6 = 2.345. 69. cot = 2.345. 77. esc = 2.345.
62. cos 6 = - 2.345. 70. cot = - 2.345. 78. esc 6 = - 2.345.
Find rectangular coordinates of the point P whose polar coordinates are as given:
79. P(4, 119). 81. P(8, 303). 83. P(12, - 38).
80. P(6, - 222). 82. P(10, 456). 84. P(2, - 138).
Find polar coordinates of the point P whose rectangular coordinates are as given:
86. P(- 7, 24). 87. P(7, - 24). 89. P(- 12, - 5).
86. P(- 7, - 24). 88. P(- 12, 5). 90. P(12, - 5).
Prove that if 6 is a positive acute angle the following relations hold:
91. sin (180 - 6) = sin 6. 97. sin (90 + 0) = cos 0.
92. sin (360 - 0) = - sin 6. 98. sin (270 + 6) = - cos 0.
93. sin (540 - 0) = sin 0. 99. sin (450 + 0) = cos 0.
94. sin (720 - 0) - - sin 0. 100. sin (630 + 0) = - cos 0.
95. sin (- 180 - 0) = sin 0. 101. sin (- 90 + 0) = - cos 0.
96. sin (- 360 - 0) - - sin 0. 102. sin (- 270 + 0) = cos 0.
25. Functions of 180 - 0. If 0' = 180 - 0, then the angle 0,
if acute, must be the corresponding acute angle for 0', arid 6' must
terminate in the second quadrant. Hence if 6 is acute we obtain
the following formulas from the rules of 24:
sin (180 - 0) = sin 0, cos (180 - 0) = - cos 0,
tan (180 - 0) = - tan 0, cot (180 - 0) = - cot 0,
sec (180 - 0) = - sec 0, esc (180 - 0) = esc 0.
38
TRIGONOMETRY
It is an interesting property of many formulas in trigonometry that
each of them holds, not merely for acute angles, but for all permissible
angks; that is, for angles such that the formula does not imply a divi-
sion by zero. The above formulas are a good example; they hold
for all permissible values of 6, and not merely for values between
and 90.
We shall prove the statement by comparing the triangle of reference
for 180 with that for 0, in the four cases which arise when
terminates in the first quadrant, the second quadrant, the third
quadrant, or the fourth quadrant.
Figure 40a shows the triangle of reference OMP for an angle
terminating in the first quadrant, and the triangle of reference
180'-
M' *' O
FIG. 40a
FIG. 40b
FIG. 40c
FIG. 40d
OM'P' for 180 6. The lengths of corresponding sides arc equal.
In Figure 40b the two triangles are drawn for an angle 6 terminating
in the second quadrant; in Figures 40c, 40d, the angle 6 terminates
in the third and fourth quadrants respectively. In each case the tri-
angle OMP is geometrically equal to OM'P'; we insure this by
taking OP f equal to OP, and we then observe that the angles at
are equal. The two right triangles have two sides and an acute angle
of one equal to corresponding parts of the other, and hence the tri-
angles are equal. It follows that r f = r; but we note that x' and x,
REDUCTION FORMULAS. LINE VALUES, GRAPHS 39
while numerically equal, are of opposite sign in each of the four cases,
while y' and y are of the same sign. That is, in all cases,
x r = - x, y' = y, r f = r.
We complete the proof of the formulas for three of the functions
of 180 - as follows:
sin (180 _0)-^--sin 0,
cos (180 - 0) = ^ = ^ = - cos 0,
tan (180 - 0) = ^- f = -^- - - tan 0.
3J X
The student should write out the proof for the other three formulas.
Quadrantal angles do not possess triangles of reference, but even
for them the x, y, r relations hold, and our proof is valid except when
the angle and function are such that the function of that angle has
no value (tan 90, for example) ; that is, the proofs are valid for all
permissible angles.
26. Functions of n-180 0. If is an acute angle, it is the cor-
responding acute angle for the angle n-180 0, where n is an in-
teger, positive or negative or zero. Thus we have the rule :
7/0 is an acute angle, any function of 0' = n- 180 is equal to
plus or minus the same-named function of 0, according as the function of
0' is itself plus or minus.
We obtain particular formulas for particular values of n; and it
is true, as in the preceding section, that these formulas, including
the + or sign, hold not only when is acute, but also for all
permissible values of 0. We reproduce a few of these formulas:
sin (180 + 0) = - sin 0,
cos (180 + 0) = - cos 0,
tan (180 + 0) = tan 0,
cot (180 + 0) = cot 0,
sin (- 0) = sin (360 - 0) = - sin 0,
cos (- 0) = cos (360 - 0) = cos 0,
tan (- 0) = tan (360 - 0) = - tan 0,
cot (- 0) = cot (360 - 0) - - cot 0.
Proofs should be supplied by the student, who should draw tri-
angles of reference and compare them as was done in 25.
40
TRIGONOMETRY
27. Functions of n*90 0, where n is odd. If is an acute angle
and n is odd, then the corresponding acute angle for 0' = n-90
is 90 0, and the rules of 24 enable us to express a function of
0' as plus or minus the same-named function of 90 0, and finally
as plus or minus the cofunction of 0. We have, for example,
sin (90 - 0) = cos 0, cos (90 - 0) = sin 0, tan (90 - 0) = cot 0;
sin (90+ 0) = cos 0, cos (90+ 0) = -sin 0, tan (90 + 0) = -cot 0;
sin (270 - 0) = - cos 0, cos (270 - 0) = -sin 0, tan (270 - 0) = cot 0;
sin (270 +0) = -cos 0, cos (270 +0) = sin 0, tan (270 +0) = -cot 0.
These formulas all hold for all permissible values of 0. This may
be proved by comparing related triangles
of reference.
As an illustration we give in Figure 41 a
drawing for a case where terminates in the
second quadrant, so that 90 6 terminates in
the fourth. We note here that
sin (90 - 6) = = -
and similarly for other functions of 90 0.
We may summarize all of these for-
mulas in the following rule :
// is an acute angle and n is an odd integer , a function of 0' =
w-90 is equal to plus or minus the corresponding cofunction of 0,
according as the function of 0' itself is plus or minus. Each formula
thus obtained holds not only when is acute, but for all permissible val-
ues of 0.
For example, sin (270 + 0) = cos 6 whether 6 is acute or not, according
to the rule just stated, since this formula is correct when 6 is acute.
EXERCISES
Use the general rules of 26, 27 to express each of the following as a function of
an acute angle in two ways:
1. sin 237.
9. tan 386.
17. sec 281.
2. sin 1237.
10. tan 1386.
18. sec (- 281).
3. sin (- 237).
11. tan (- 386).
19. sec 1281.
4. sin (- 1237).
12. tan (- 1386).
20. sec (- 1281).
6. cos 156.
13. cot 123.
21. esc 482.
6. cos 256.
14. cot (- 123).
22. esc (- 482).
7. cos 356.
15. cot 1123.
23. esc 1482.
8. cos 1356,
16. cot (- 1123).
24. esc (- 1482).
REDUCTION FORMULAS. LINE VALUES. GRAPHS 41
By the method used in 25 (which requires drawing triangles of reference and com-
paring them) prove that if 6 is an angle terminating in the second quadrant each of
the following formulas holds:
25. sin (180 + 6) - - sin 0. 33. sin (270 - 6) - - cos 6.
26. cos (180 + 0) - - cos 0. 34. cos (270 - 0) - - sin 0.
27. tan (180 + 6) = tan 6. 35. tan (270 - 0) - cot 0.
28. cot (180 + 0) = cot 0. 36. cot (270 - 0) = tan 0.
29. sin (- 0) - sin 0. 37. sin (270 + 0) - - cos 0.
30. cos (- 0) = cos 0. 38. cos (270 + 0) = sin 0.
31. tan (- 0) = - tan 0. 39. tan (270 + 0) - - cot 0.
32. cot (- 0) = - cot 0. 40. cot (270 + 0) - - tan 0.
By drawing triangles of reference and comparing them t prove that if is an angle
terminating in the third quadrant each of the following formulas holds:
41. sin (180 + 0) = - sin 0. 44. cot (90 + 0) = - tan 0.
42. cos (180 + 0) = - cos 0. 46. sec (270 - 0) - - esc 0.
43. tan (180 + 0) - tan 0. 46. esc (450 + 0) - sec 0.
By drawing triangles of reference and comparing them, prove that if is an angle
terminating in the fourth quadrant each of the following formulas holds:
47. sin (- 0) = - sin 0. 60. cot (270 - 0) - tan 0.
48. cos (90 + 0) - - sin 0. 51. sec (270 + 0) - esc 0.
49. tan (180 + 0) - tan 0. 52. esc (180 + 0) - - esc 0.
28. Line values. In the triangle of reference OM P for an angle 0,
the length of OP is at our choice. If, for example, 6 is an acute
angle, and OP = 1, we have
sin Q
cos
MP
MP
MP,
OM.
OP
OM
OP
OM
1
These lengths, MP, OM, are called line values of sin 6, cos re-
spectively.*
The other four functions of an acute angle are also assigned
line values in a figure such as Figure 43 on page 42. We first con-
struct the auxiliary Figure 42 by drawing the coordinate axes OX, OY,
locating on them the points A whose coordinates are (1,0) and B
whose coordinates are (0, 1), then drawing the circle of unit radius
and center 0, together with tangents at A and J5. In Figure 43 we
add to Figure 42 by placing the angle 6 in standard position, draw-
ing its triangle of reference OMP so that OP = 1, and prolonging
OP until it meets at T the tangent line through A } and at I" the tan-
* It was once customary to use these as first definitions of the sine and cosine.
42
TRIGONOMETRY
gent line through B. Since triangles OMP, OAT, and OBT' are
similar, we have
sin = MP;
cos OM;
AT
ra-T-"*
(1)
OM
OP
CSC
OP
MP
_ BT
"OB
= 9L
= *L = ^L. = or.
*r
or
We have thus found six line segments whose lengths are the line values
of the six functions of an acute angle.
FIG. 42
FIG. 43
How may these definitions be adapted so as to apply to angles
that are not acute? We answer this question by drawing a figure
according to the same verbal specifica-
tions as for an acute angle. We have
done this in Figure 44 for an angle
terminating in the third quadrant; that is,
we have taken the fundamental Figure 42
and inserted the triangle of reference OMP
with OP * 1, and have prolonged OP
(in this case backwards) until it inter-
sects at T and T' the tangent lines
through A and B. It is clear that equa-
tions (1) now hold numerically, but they
do not exhibit the minus signs that sin 0, cos 6, sec 6, esc should
have as functions of an angle terminating in the third quadrant.
To remedy this we interpret MP, OM, AT, BT', OT, OT not
FIG. 44
REDUCTION FORMULAS. LINE VALUES. GRAPHS 43
merely as lengths, but as directed lengths according to the follow-
ing rule:
By MP, OM, AT, BT' we now mean, respectively, the ordinate
of P, the abscissa of P, the ordinate of T, and the abscissa of T' 9
while OT is to be positive if P is between and T } otherwise OT
is negative; for OT' the same rule holds as for OT.
With this rule as to signs, formulas (1) hold for any permissible
angle 6. The student should draw a figure for an angle 6 termi-
nating in the second quadrant, and also one for an angle terminat-
ing in the fourth quadrant; he should then verify formulas (1) as
above interpreted.
29. Properties of the sine function. The interpretation of sin 6
as the line value MP exhibits clearly the following properties of the
sine function:
(1) Sin 6 is never greater than 1 or less than 1.
(2) As increases steadily from one quadrantal angle to the next
greater one, sin 6 either increases steadily or decreases steadily.
Thus as 6 increases from to 90, MP = sin 6 increases steadily
from to 1 ; as 6 increases from 90 to 180, MP decreases steadily
from 1 to 0; as 6 increases from 180 to 270, MP decreases stead-
ily from to 1 ; as increases from 270 to 360, MP increases
steadily from 1 to 0.
(3) Sin has the period 360. By this we mean that for every
angle we have sin (0 + 360) = sin 0.
(4) Sin (- 0) = - sin 0.
Two more properties can be directly deduced from (2) and its
explanatory paragraph:
(5) If and 0' are two different angles terminating in the same
quadrant, then sin cannot be equal to sin 0'.
(6) If a is a number between 1 and + 1, there are at most two
angles between and 360 for which sin = a.
Property (5) follows from (2) because, according to (2), if 0' is
greater than 0, then sin 0' is either greater than or less than, but
not equal to, sin 0. To prove the statement of (6), note that the dis-
cussion of (2) shows that if a is positive, and we let increase steadily
from to 90, sin will increase steadily from to 1 and hence
will have the value a for at least one value of (but not two 0's,
by (5)). Similarly, for positive a there will be one between 90
and 180 for which sin = a; but there will be no between 180
44
TRIGONOMETRY
and 360 for which sin = a. The student should supply the proof
for a negative. If a = 0, the equation sin = a has two solutions,
= and 6 = 180, that are not less than and not as large as
360. We must interpret the word " between" in (6) accordingly.
With this understanding, we can make (6) more precise by saying
that if a = 1, or if a = + 1, there is just one value of between
and 360 for which sin 6 = a, but for all other values of a be-
tween 1 and + 1 there are just two values of 6 between and
360 for which sin a.
30. Graph of the sine function in rectangular coordinates. The
properties given in 29 are visualized if we draw a graph of the sine
function as given in the equation y = sin x. In this equation x and
y are no longer coordinates of a vertex P of a triangle of reference;
instead, x now stands for the number of degrees in an angle and y
is the positive or negative number that gives the value of sin x.
We take axes OX, OF, as in Figure 45, and for a sufficient number
of angles plot points (x, y) for which x is the number of degrees in
FIG. 45. y = sin x
the angle, and y is the value of sin x. For acute angles, we may
use a table of trigonometric functions to determine y. When x is
between 90 and 180, we may make use of the reduction formula
sin x = sin (180 - x) and obtain values for 180 - x from the
tables. In Figure 45, the black dots on the curve at and near it
to the right, indicate values of x, y that correspond to the following
tabulation:
X
10
20
30
60
90
y
0.1736
0.3420
0.5000
0.8660
1.000
When enough points have been plotted we draw through them the
smooth curve that forms the first right-hand arch in Figure 45, from
REDUCTION FORMULAS. LINE VALUES. GRAPHS 45
x = to x * 180. Property (4) of 29 enables us then to sketch
the curve from x = to x = 180; here an ordinate for x = 6
is of the same length as for x = + 0, but is drawn downwards in-
stead of upwards. Property (3) of 29 then tells us that the remainder
of the curve is a repetition of the part that we have now drawn;
the part from 180 to 180 + 360, for example, reproduces the part
from - 180 to + 180, since if x is the number of degrees in an
angle 9, and x' the number in 360 + 0, then sin x f = sin x.
The graph in Figure 45 gives a geometrical picture from which we
readily infer the properties listed in 29. The curve never rises above
+ 1 or falls below 1; this statement implies property (1). It
rises steadily from x = 90 to x = 90, falls steadily from x = 90
to x = 270, and so on; this implies property (2). Properties (3)
and (4) are inherent in the construction of the curve. Property (5)
is a statement of the noncquality of two ordinates in certain arcs
of the curve, and (6) tells how many ordinates may be equal from
x = o to x = 360.
31. Properties and graph of the cosine. The statements (1 ), (2), (3)
of 29 are true for cos 6 as well as for sin 0, but the explanatory
>/450
Fio. 46. y = cos x
paragraph under (2) must be altered; cos decreases as increases
from to 180, and increases as increases from 180 to 360. State-
ment (4) must be replaced by the following:
(4') Cos (- 0) = cos0.
Statements (5) and (6) hold for the cosine as well as for the sine.
In Figure 46 a graph of the equation
y = cos x
has been drawn. The student should observe how it exhibits prop*
erties (1), (2), (3), (4'), (5), (6). An inspection of the figure shows that
the cosine curve would coincide with the sine curve if it were moved
46
to the right so that x
of the formula cos =
TRIGONOMETRY
became x - 90.
sin (90 + 0).
This is a consequence
32. Properties and graphs of the tangent and secant. Figures 47
and 48 represent graphs of the equations y = tan x, y = sec x, re-
spectively. A study of these graphs, or a reference to the defini-
tions by line values, shows that the statements (2), (3), (5) of 29
hold for tan and sec also, but the discussion of (2) must be modi-
fied. It remains to see how (1), (4), (6), and the discussion of (2)
are to be changed.
FIG. 47.
The function tan is not limited; it can in particular be greater
than + 1 or less than - 1, but when, for example, is between
- 45 and + 45, tan is between - 1 and 1. The function sec
can take on any value greater than + 1 or less than 1, but its
value cannot lie between land+ 1. These statements replace (1).
The formulas analogous to (4) are
tan (- 0) - - tan 0, sec (- 0) = sec 0.
As for (6), the equation tan = a has two and only two solutions
for between and 360 (including but not 360), and there is
no restriction on a. The student should supply the corresponding*
statement for the equation sec = a.
REDUCTION FORMULAS. LINE VALUES. GRAPHS 47
The graph of y = tan x shows clearly that as x increases from
to 90, tan x becomes steadily and unlimitedly larger. We say that
tan x becomes infinite as x thus approaches 90; to write tan 90 = oo
(the symbol for " infinity") must be regarded as an abbreviation of
this statement, and not as an assertion that "infinity" is a number
or that tan 90 has a value. The graph, in fact, shows that tan x
always increases with x, has no value at x = 90, 270, , 90,
, but becomes infinite at those points.
The student should supply a similar discussion for sec x.
EXERCISES
Draw a graph of each of the following equations and discuss its properties:
1. y = cot x. 4. y = J sin 3x. 7. y tan 3x.
2. y csc x. 6. y = cos 2x. 8. y = sec -
i
3. y = sin 2x. 6. y i cos 3x. 9. */ 2 -f sin x.
10. Draw a graph of y sin x + cos x. Does the function sin x -f- cos x
have a period? If so, what is it?
11. Draw a graph of y = J sin x + I sin 2x. Does the function J sin +
J sin 2x have a period? If so, what is it?
12. Draw a graph of y = sin x % sin 2x. Does the function sin x i sin 2x
have a period? If so, what is it?
13. Draw a graph of y = cos x \ cos 2x. Does the function cos x \ cos 2a?
have a period? If so, what is it?
14. Draw a graph of y = sin 2 x. Does the function sin 2 x have a period?
If so, what is it?
In the same figure draw graphs of the two equations, for each of the following pairs:
16. y == sin x, y = sin (90 + x). 19. y = 2 cos x, y = cos 2x.
16. y = cos x, y = cos (180 + x). 20. y = cos x, y = cos (90 - x).
17. i/ = 2 sin x, y = sin 2x. 21. i/ = sin x, y = 1 -f- sin x.
18. j/ = cos x, 2; = 2 cos a:. 22. y = cos x, y cos x 1,
CHAPTER IV
TRIGONOMETRIC IDENTITIES
33. Fundamental identities involving one angle. By an identity
in an angle we mean an equation involving 6 which is true for all
permissible values of 6. Here we change slightly a meaning pre-
viously given to the term permissible values by making it include
all values except those for which a function in the identity is unde-
fined, or a denominator is zero.
The reduction formulas of Chapter III are examples of trigono-
metric identities in one angle; they hold for all permissible values
of 6.
From the definitions of the functions in 5, formulas (1), we see
at once that those written below in the same column are reciprocals
of each other:
The following two identities also follow directly from the defini-
tions of the functions:
(2)
v '
cos sin $
We prove the first of these identities, for example, as follows:
y
SEj.I.K.tantf.
cos u x x
r
We obtain three more identities from the relation x 2 + y 2 = r 2 ,
which holds because, in the triangle of reference, the sum of the
squares of two sides is equal to the square of the hypotenuse; the
sides are x, d= y, but their squares are always x 2 , y 2 . We write
48
TRIGONOMETRIC IDENTITIES 49
these identities in a customary notation where sin 2 8, for example,
means (sin 0) 2 :
sin 2 B + cos 2 0=1,
(3) 1 + tan 2 = sec 2 0,
i + cot 2 e = esc 2 e.
To prove any one of these three we start with
JC* + y* = r 2
and divide by one of the three terms. Thus, dividing by r 2 , we get
(sin0) 2 + (cos0) 2 = 1,
which is another way of writing the first of formulas (3). Similarly,
if we divide by x 2 we have
and if we divide by y* we have, with a change of order of terms,
These last two formulas reduce to the last two of identities (3).
Formulas (1), (2), (3) are fundamental in trigonometry, and must
be memorized.
Example 1. By means of identities (1), (2), and (3), find the values of the
other functions of B if tan = f , and B terminates in the second quadrant.
Solution. In examples of this sort it is necessary to use one of formulas (3);
except for this, (1) and (2) are sufficient. Thus, first using the second of formulas
(3), we have
sec 2 B = 1 + tan 2 - 1 + (- })* = f J;
sec B = (negative since B is in the second quadrant) ;
cos B = - ^ = (by the last of formulas (1));
sec o
sin B = cos B tan B = - f X (- J) = f (by the first of (2));
cot B = --i-^ - -^ (by a formula of (1));
tan u o
csc = - ^ = (by a formula of (1)).
sin v <3
If the quadrant in which B terminates had not been specified, we would have
written sec B = d= f , and similarly for cos 6, sin 6, csc B.
50 TRIGONOMETRY
Example 2. Express esc 8 + cos + tan 2 6 in terms of sin only.
Solution. We have from the fundamental identities (1), (2), (3),
esc 6 = -. -i
sm0
cos = i Vi - S in 2 0,
, , a sin 2 sin 2 6
tan 2
cos 2 1 - sin 2
Hence
esc + cos + tan 2 = -r =b v l _ s in' +
-r .,
sm 1 sin 2
where the d= sign indicates that cos is positive or negative according to the
quadrant where terminates.
34. Other identities involving one angle. An unlimited number
of other identities can be formed, and can be proved by means of
the fundamental identities (1), (2), (3) of 33. The following ex-
amples illustrate three methods of proof.
Example 1. Prove the identity
cos a
by transforming the left side only.
Solution. By means of the fundamental formulas and elementary algebra we
have
cos cos 01
. - - - -- - =s
1 - sin 2 cos 2 cos
Example 2. Prove the identity
cos - cot = cot (sin 0-
by reducing both sides to the same expression.
Solution. We arrange the work in parallel columns:
Q
u.
cos cot
COS
sin <
cot (sin - 1)
COS0
sin0
(sin 0-
n COS
COS
sin0
Example 3. Prove the identity of Example 2 by deducing from it, by algebraic
and trigonometric transformations, a chain of identities such that the given one
is true if the second is, the second is true if the third is, and so on until we obtain
an identity which states that an expression is equal to itself.
Solution, Such a chain is
(A) cos - cot - cot (sin - 1);
(B) cos - cot = cot sin - cot 0;
(C) cos = ^-! sin 0;
sm
(D) cos - cos 0.
TRIGONOMETRIC IDENTITIES 51
If we merely write down such a chafri as constituting a proof of (A), we must
understand that it is to be interpreted as saying that (A) is true if (B) is true,
(B) is true if (C) is true, (C) is true if (D) is true.
In proving identities it is usually best to avoid using radicals which
may have plus or minus signs. A method often effective is to re-
duce all other functions to sines and cosines, as was done in the pre-
ceding Examples 2, 3.
EXERCISES
By means of the fundamental identities (1), (2), (3) of 33, find the values ofaU
the functions of B from the given conditions:
1. sin 6 = f . 6. tan = if-. . 9. sec 6 - - J.
2. sin 6 = - ffc. 6. tan 6 = - V- 10. sec - f .
3. cos 8 = - f. 7. cot e = - |. 11. esc 6 = *.
4. cos = - |. 8. cot B - A- 12 - csc = - ^.
By means of the fundamental identities (1), (2), (3) of 33, express the following
in terms of sin 6 and cos only:
13. tan 4 sec 0. 20. cot 2 -- -
14. cot + esc 0. A cs 2 * A
16. sin 2 + sec 2 0. 21 - cos 6 tan e + sec * '
16. cos 2 4 esc 2 0. 22 - sin * cot + csc *
17. tan 2 + sec 2 6. 23 - sin * cos e tan * cot *
18. cot 2 + csc 2 0. 24 - tan * cot sec * csc *
19. tan 2 0-f
-
sec 2
o/ the fundamental identities (1), (2), (3) o/ 33, express the following
in terms of sin 6 only:
26. sec 2 4- tan 2 0. 29. cos cot 4- csc 0.
26. 2 cos 2 0-1. 30. sin tan 4 sec 0.
27. sec tan 4 csc 0. 31. sec 2 csc 2 - tan 2 0.
28. csc cot - sec 0. 32. cot 2 - sin tan 0.
Bi/ means of the fundamental identities (1), (2), (3) of 33, prove eoc/i of the
following identities by transforming the left side only:
33. sin cot = cos 0. A1 csc 2 - cot 2 _ >,
34. cos tan - sin 0. tt ' sec 2 " C S
36. sec cot = csc 0. _ sin 2 _ , fi
36. csc tan - sec 0. 42 ' COS 4 + COS 2 sin 2 ~ tan "'
37. (1 4- tan 2 0) sin 2 = tan 2 0. 43. cos csc = cot 0.
38. (1 + cot 2 0) cos 2 - cot 2 0. 44. s i n sec _ tan 0.
39. sin 2 4 cos 2 _ 45. tan 4 C ot a = sec a csc a.
cos 2 46. sin a tan a = tan a (cos a 1),
40 8ec8 tan 2 _ cgc2 ^ 47. cot a cos a csc a sin a.
sin 2 ' 48. sec 2 a 4- csc 2 a = sec* a csc 1 a.
52 TRIGONOMETRY
40 sin 4 a cos 4 a _ . . ... tan a 1 __ 1 cot a
sin a cos a ~ 8m " " tan a + 1 "" 1 + cot a'
1 -f tan a no sin a cos a __ tan a 1
W - sec a = sin a + cos a< sin a + cos a " tana + 1*
By means of the fundamental identities of (1), (2), (3) of 33, and a chain of
identities, prove the following:
53. sec a cos a = sin a tan a. 56. cot 2 a cos 2 a = cot 2 a cos 2 a.
54. tan a -f- cot a = sec a esc a. __ , . N , 1 cos a
66. sin' a sec' a + 1 = sec' a. B7 ' (csc a ~ cot a) = 1+cosa'
58. cot a cos a = sin a -f esc a (1 2 sin 2 a).
_ Q /seca 4- esc a\ 2 tan a 4- cot a
0. ( ;; ; T I
\ 1 -H tan a / tan a
60. sec 4 a 2 tan 2 a = 1 -f tan 4 a.
cos 2 a
61. r 1 -^ h 1 = tan 2 a 4- cot 2 a sin 2 a. tan 2 a.
tan 2 oc
62. sin a (1 -f- tan a) sec a = esc a cos a (1 -f cot a).
63. (sin a -f- cos a I) 2 = 2(sin a l)(cos a 1).
aM (tan cos a cot a) cot a . cos a cot a
64. = sm a
esc a sec a
35. Addition formulas. Important formulas involve functions
of two angles, such as sin (a + /3) where a, /? are the Greek let-
ters " alpha/ ' "beta." If one were careless he might assume that
sin (a + /3) = sin a + sin /3, which is usually false; for example,
sin (30 + 60) = sin 90 - 1,
sin 30 + sin 60 - +
and hence sin (30 + 60) does not equal sin 30 + sin 60.
In sections which follow we shall prove the following formulas:
(1) sin (a + ft) = sin a cos + cos a sin 0;
(2) sin (a 0) = sin a cos ft cos a sin 0;
(3) cos (a + 0) = cos a cos ft - sin a sin 0;
(4) cos (a - 0) = cos a cos + sin a sin 0;
/-x A / i /\ tan a + tan ft
(5) tan < + - t.^m,;
/\ x / *v tan a tan ft
(6) tan (a - A - 1+tangtan V
These are called the addition formulas for the sine, cosine, and
tangent functions. They hold for all permissible values of the
angles.
TRIGONOMETRIC IDENTITIES
53
N
36. Addition formulas for sine and cosine functions. We shall
now give two proofs of formulas (1) and (3) of 35. Figures 49 and
50 illustrate the case in which a, j8, and a + ]8 are positive acute
angles. The first proof to be given in this section would require
modification if one or more of the angles were not acute, but the
second proof applies without change of wording when the angles
are not thus restricted.
First proof, for a, 0, and a + ft acute. Draw a in the standard
position with respect to .XT-axes, with origin at 0. Then take
XiFi-axes with the same origin
such that OXi is the terminal
side of a, and draw /3 in the
standard position with respect to
the XiFi-axes. From P, a point
on the terminal side of /3, drop
perpendiculars PQ to OXi, and
PM to OX; then from Q drop
a perpendicular QN to OX.
The construction of Figure 49
is completed by dropping the per-
pendicular QR from the point Q
to the line MP. We observe that angle RPQ will then be equal to
a, since its sides RP and QP are respectively perpendicular to the
sides OX, OX\ of angle XOXi.
We are to prove that
(1) sin (a + ]8) = sin a cos /3 + cos a sin |8.
If we use the triangle of reference OMP for a + ]8, the triangle
OQP for )8, and the triangles ONQ and RPQ for a, formula (1) can
be replaced by
MP NQ OQ RP QP
OP OQ' OP* QP' OP 9
which is equivalent to
MP _ NQ RP
OP OP * OP'
or
(2) MP = NQ + RP.
Formula (2) is, however, obviously true, and we have thus proved
formula (1).
FIG. 49
54
TRIGONOMETRY
Similarly, the formula
(3) cos (a + jS) = cos a cos /3 - sin a sin j8
can be replaced by
OM _ GW OQ _RQ QP
OP OQ ' OP QP' OP'
which is equivalent to
OM ON RQ
OP OP OP'
or
(4) OM = ON - RQ.
Since (4) is obviously true, formula (3) must hold.
Second proof, with no restrictions on the angles. In Figure 50
we have altered Figure 49
as follows: We have, for
simplicity, omitted the letters
M, N, R. We have drawn
X 2 F 2 -axes, with Q as origin, par-
allel to the -XT-axes. We have
omitted the letter a for the angle
RPQ and have added 90 4- a
for the angle X%QP. We have
indicated coordinates thus:
FIG. 50
The coordinates of P with re-
spect to the XY-bxcs are (x, y), to the
X 2 F 2 -axes are (# 2 , t/ 2 ).
The coordinates of Q with respect to the X F-axes are (A, /c).
Finally, the length of OP is designated as r.
Following exactly the directions given above, we could construct
a figure corresponding to Figure 50 for any angles a, /3, whether posi-
tive or negative, without restriction as to their magnitude. In case
yi were negative, however, angle X^QP would be 270 + a (or a
coterminal angle).
Formulas (1) and (3) of 35 will now be proved by comparing
coordinates for the three sets of axes and using the familiar formulas
x = r cos 0, y = r sin
for the various sets of axes, with appropriate interpretation of x, t/, r, 9.
are (x\, 3/1), to the
TRIGONOMETRIC IDENTITIES 55
Taking into account the signs of the quantities represented by the
letters, we see that
(5) y = k + 2/2,
(6) x = h + x 2 (in Figure 50, z 2 is negative).
In all cases we have *
y = r sin (a + /3),
k = x\ sin a. = (r cos ]8) sin a,
2/2 = 2/i sin (90 + a) = (r sin ]8) cos a.
When these expressions are substituted in (5), and the resulting
equation is divided by r, we obtain
sin (a + j8) = cos ]8 sin a + sin /3 cos a,
which gives formula (1), 35.
Similarly, we have
x = r cos (a + /3),
h = x\ cos a = (r cos /8) cos or,
#2 = 2/i cos (90 + a) = r sin )8 ( sin a).
By substituting these expressions in (6) and dividing by r, we obtain
cos (a + /3) = cos j8 cos a sin /3 sin a,
which gives formula (3), 35.
The wording of these proofs of formulas (1) and (3) as well as the
directions for constructing Figure 50, apply to all angles a, /3, whether
positive or negative, without restriction as to their magnitude.
The student should draw the figure and verify the proof for cases
where a, j8, and a + /3 are not all positive acute angles.
The formulas should be verified when a and j8, one or both, are quad-
rantal angles, by use of the rules of 26 and 27.
37. Formulas for sin (a - j9) and cos (a - j8). In 36 we have
shown that the formulas (1) and (3) of 35, for sin (a + /3) and
cos (a + jS), are true even when one of the angles is negative, or
when both are negative. Hence we have
sin (a - )8) = sin (a + (- j3))
= sin a cos ( - j8) + cos a sin ( - /3)
= sin a cos ]8 cos a sin )8,
and we thus have proved formula (2) of 35.
* The equation for yt holds even when y\ is negative, for then
j/2 = - 2/1 sin (270 + a) - y t sin (90 + a).
56 TRIGONOMETRY
To prove formula (4) of 35 we proceed similarly:
cos (a - /3) = cos (a + (- /?))
= cos a cos ( j8) sin a sin ( /3)
= cos a cos /3 + sin a sin j8.
Example. Find the exact value of cos 15 by use of an addition formula.
Solution. We have
cos 15 = cos (45_- 30) - cos 45 cos 30 + sin 45 sin 30
= .^? 2^5 -L. ^1 I ^4- v^
'2 ' 2 "*" 2 ' 2 4
EXERCISES
Ity ws o/ /ta addition formulas, find the exact values of the following :
1. sin 15. 2. sin 75. 3. cos 75. 4. sin(- 15).
Apply the addition formulas to the following to obtain exact values , and then
check by obtaining the exact values in another way:
5. sin (60 + 30). 10. cos (60 - 30).
6. cos (60 + 30). 11. sin (90 + 60).
7. sin (45 + 45). 12. cos (90 + 30).
8. cos (45 + 45). 13. sin (270 + 60).
9. sin (00 - 30). 14. cos (270 - 45).
By use of addition formulas prove the following:
15. sin (90 + 0) = cos 6. 20. cos (180 + 0) = - cos 0.
16. cos (90 + 0) = - sin 0. 21. sin (270 - 0) = - cos 0.
17. sin (180 - 0) - sin 0. 2?. cos (270 - 0) = - sin 0.
18. cos (180 - 0) - - cos 0. 23. sin (270 + 0) = - cos 0.
19. sin (180 + 0) = - sin 0. 24. cos (270 + 0) - sin 0.
By use of Tables find approximate values of the following:
25. sin (47 + 32) - (sin 47 Jfc sin 32).
26. cos (18 + 28) - (cos 183> cos 28).
27. sin (25 - 10) - (sin 25 ^ sin 10).
28. cos (35 - 10) - (cos 35 -V cos 10).
By use of addition formulas find exact values of the following, assuming that
a. and ft are angles terminating in the first quadrant:
29. sin (a + ft) if sin ot = $, cos ft = ^j.
30. cos (a -|- ft) if cos a = g , cos ft = -5.
31. sin (a 4- ft) if sin a ^, cos ft = .
32. cos (a + 0) if sin a = f J, sin ft - f.
33. sin (a #) if sin a = f , sin =* A-
34. cos (a - ft) if cos a = f , cos = ffr.
35. sin (a - ft) if cos a = , cos /3 = ffc.
36. cos (a - ft) if sin a = J, cos = ^j.
37. sin (a - )8) if sin a = ^ tan ft = &.
38. cos (a - 0) if sin a - H, tan - Y-
TRIGONOMETRIC IDENTITIES 67
>rmi
39. sin (45 + 6)
Use addition formulas to prove the following identities:
sin + cos 6
40. sin (45 - 0) -
41. sin (60 + 6) =
42. sin (30 - 6) -
43. cos (45 + 0) =
V2
cos sin 6
V2
V3 cos 6 + sin
2
cos - V3 sin
2
cos sin
44. cos (60 _0 ) = *0+ 2 ^ Sing -
A/3 cos sin
46. cos (30 + 6)
46. cos (60 + 0)
2
cos - A/3 sin !
2
47. sin (4 + B) cos B - cos (A + B) sin B = sin A.
Hint. Let A + B = <x t B = .
48. cos (A B) cos B sin (A B) sin B = cos A.
49. sin (x + y + ) = sin a: cos y cos 2 sin a; sin y sin z + cos a? sin y cos 2
+ cos x cos y sin 2.
60. cos (a: -f y -f- 2) = cos x cos y cos 2 cos x sin y sin 2 sin a? sin y cos z
sin a; cos y sin 2.
61. sin (a; y + 2) = sin a; cos y cos 2 -f sin a; sin y sin 2 cos x sin y cos z
-h cos a; cos y sin 2.
62. cos (x y 2) = cos a; cos y cos 2 cos a; sin y sin 2 -f sin x sin y cos z
H- sin a; cos y sin 2.
63. Prove the formula for sin (a + ft) directly from a figure in which a and
ft are positive acute angles but a + ft > 90.
64. Prove the formula for sin (a -f ft) directly from a figure in which a and
ft are positive angles, a terminating in the first quadrant, ft in the second quadrant,
and a + ft in the third quadrant.
66. Prove the formula for cos (a -f- ft) directly from a figure in which a and
ft are positive angles terminating in the second quadrant and a + ft an angle
terminating in the third quadrant.
66. Prove the formula for cos (a -f ft) directly from a figure in which a and
ft are positive angles, <x terminating in the first quadrant, ft in the fourth quadrant,
and a + ft in the first quadrant.
67. Prove the formula for sin (a -f ft) directly from a figure in which a and
ft are positive angles, a terminating in the first quadrant, ft in the third quadrant,
and a+ ft in the fourth quadrant.
68. Prove the formula for sin (a ft) directly from a figure in which a, ft,
a ft all terminate in the first quadrant.
69. Prove the formula for cos (a ft) directly from the figure of Exercise 58.
68 TRIGONOMETRY
38. Formulas for tan (a + ft) and tan (a - 0). From formula (2)
of 33 and formulas (1) and (3) of 35, we have
f _i_ /9\ - sin (a + /3) sin o: cos ff + cos a sin ff
tan (Of "r (j) - ; - ; TJT ^j - : - : ^*
cos (a + p) cos a cos p sin a sm p
We can express the last fraction in terms of tan a and tan /8 if we
divide both numerator and denominator by cos a cos /3. We thus
obtain
sin a cos j(3 cos a sin ff
/ , a\ cos a cos /? cos a cos /8
tan ( + /?) = - ^ - - - K
sm a; sin p
cos a cos ]8
sin a sin j8
cos o: cos /8
sn a sin j
cos a cos |8
From this identity we at once derive formula (5) of 33,
/i\ * f , o\ tan a 4- tan j8
(1) tan (a + p) = - - - - - ~
1 - tan a tan |8
If we treat in the same way the identity
, ox sin (a 8) sin a cos 8 cos a sin 8
tan (a - |8) = - ) - ^ = - ^ . - r c,
cos (a p) cos a cos p + sin a sin p
we obtain formula (6) of 35,
/ rt \ A / o\ tan a tan j3
(2) tan (a - p) = ^
1 + tan a tan p
EXERCISES
Find exact values of the following by use of addition formulas:
1. tan 15. 2. tan 75. 3. tan 105. 4. tan 195.
By use of addition formulas prove the following identities:
7. tan (60 -
1 - tan 6 ' l 4- V3 tan
TRIGONOMETRIC IDENTITIES 59
Find exact values of the following by using addition formulas, assuming thai
a and ft are positive acute angles:
9. tan (a + ft) if sin a = f , sin ft - ft.
10. tan (a + ft) if sin a = A, cos ft - ^.
11. tan (a - 0) if tan a = i sin - Jf
12. tan (a - ft) if sec a - Y, esc ft - Y-
By use o/ addition formulas prove the identities:
ton ( + )- tony _ ten (ffi ^ ^ = = fr)
1 -f tan (x + y) tan y
tan(--y)+tany . tftn x ^ cot (a + # _ cot a cot -1
1 - tan (x - y) tan y K cot a + cot ft
tan (s 4- y) tan 2 x - tan 2 y . , _ ~ cot a cot ff + 1
"' cot (a: - y) 1 - tan 2 * tan 8 y' lf ' C * ^ P) cot /3 - cot a '
39. Functions of 2a. By taking ]8 equal to a in the formulas
for the sine, cosine, and tangent of a -f /?, we obtain formulas for
the corresponding functions of 2a; these are sometimes called double-
angle formulas. For example, we have
sin 2<x = sin (a + a) = sin a cos a + cos a sin a
= 2 sin a cos a;
and similarly
cos 2a = cos 2 a sin 2 a.
This last formula may be transformed as follows:
cos 2a = cos 2 a (1 cos 2 a) = 2 cos 2 a 1;
also
cos 2a = (1 - sin 2 a) - sin 2 a = 1 - 2 sin 2 a.
We list below the formulas for sin 2a, cos 2a, tan 2a, including
the three we have just obtained for cos 2cr.
(1) sin 2a = 2 sin a cos a;
(2a) cos 2a = cos 2 a sin 2 a
(2b) = 2 cos 2 a - 1
(2c) =1-2 sin 2 a;
/o\ A 2 tan a
(3) tan 2 = ^^^
Example. Find the sine, cosine, and tangent of 120 by means of the double-
angle formulas.
Solution. We have sin 120 = sin (2 X 60) = 2 sin 60 cos 60
V3 1 V
~ 2 ' T ' 2 - "2"
cos 120 = cos 2 60 - sin 2 60 = J - } - - }
, 10AO 2 tan 60 2>/3 /r
ten 12 ^ 1- tan 2 60 - STTJ - - ^
60 TRIGONOMETRY
40. Functions of Just as the algebraic identity
x(l - x) = x x*
remains an identity if we substitute ^ for a:, obtaining
2V 2/ 2
so likewise in a trigonometric identity in an angle a, we may sub-
stitute = for a if we do it in every place where a occurs.* If we do
this in formulas (1), (2b), and (2c) of 39, we have
(1) sin a = 2 sin ^ cos ^J
(2) cos a = 2cos 2 ^ - 1;
(3) cos a = 1 ~ 2 sin 2 ^-
We obtain two half-angle formulas by solving (3) for sin ^j>
and (2) for cos -^ Thus, from (3) we have
2 sin 2 "o = 1 "~ cos a >
and hence
(4) sin | = z
Similarly
/eN a , /I + cos a
(5) cos ^ = =t \ 5
4 V A
We obtain a formula for tan -5 by dividing (4) by (5):
/* \ *~ a _i_ /I - cos a
(6a) tan -5 = \/-j--|
2 \ 1 + cos a
If a is between and 180 (inclusive, but 180 is not permissible
in (6a)), the + sign is to be used in the above formulas; for other
angles we decide whether to use the + or - sign by the rules for the
signs of functions of -= according to the quadrant in which -5 lies.
* We could, in fact, substitute for a any expression involving a provided we
make the same substitution wherever a occurs.
tA\ cos a
(4)
TRIGONOMETRIC IDENTITIES 61
Two additional formulas for tan -= which avoid the sign of
t
(6a) are these:
. a 1 cos a
,. x a sin a
(6c > ""-
We prove (6b) by reducing the right side to the left side with the
aid of formulas (3) and (1) of the present section:
- cos a
sin a o <* a o a a
2 sm -^ cos -^ 2 sin -^ cos -^
. a
sin 7:
2 . a.
= = tan =
a 2
cos TJ
The identity formed by equating the right side of (6b) to the right
side of (6c) is easily proved.
Example 1. Find the sine, cosine, and tangent of a/2 if a is an angle between
180 and 270 for which tan a = 3.
Solution. Here a terminates in the third quadrant and a/2 in the second.
Hence sin a and cos a are negative; sin (a/2) is positive, cos (a/2) and tan (a/2)
are negative. We have
1 _ - 1 1
cos a / : 7 == - 1
sec a vT-f tan 2 a vlo
o
sin a = cos a tan a = ^=t
a /I + cos a _ /10 - \/10
008 2" "\ - 2 --- \ 20 '
a 1 - cos a _ X + v^To _ VlO 4-
2
62
TRIGONOMETRY
Example 2. Prove the identity
A
cos 2 A - 4 cos A -f 3 = 8 sin 4 -jr
&
Solution. We reduce both sides to the same expression by means of formulas
(2b)of 39 and (4) of 40.
2 cos 2 A - 1 - 4 cos A + 3
2 cos 2 A - 4 cos A + 2
2(1 - cos A) 2
8
8
cos A\ 4
2(1 - cos
EXERCISES
By use of the formulas of 39 find exact values of sin 2a t cos 2a, and tan 2a in
the following cases:
1. sin a = f, if < a < 90.
2. cos ex - - f, if 90 < a < 180.
3. tan a - *, if 180 < a < 270.
4. sec a = H, if 270 < a < 360.
6. cot a - i if 360 < a < 450.
6. esc a - - Hi if 540 < < 6 30.
7. sin a - - , if - 90 < a < 0.
8. cos a = - A, if - 180 < a < - 90.
By use of the formulas of 40 find exact values of sin ^, cos
f ii J
following cases:
9. sin a = j}, if < a < 90.
10. cos a = - f, if 90 < a < 180.
11. tan a - Y, if 180 < a < 270.
12. sec a = |J, if 270 < a < 360.
13. cot a = i if 360 < a < 450.
14. esc a - - H, if 540 < a < 630.
15. sin a - Jf , if - 90 < a < 0.
16. cos a - - A, if - 180 < a < - 90.
By use of formulas of 40, find exact values of the following:
17. sin 15, cos 15, tan 15.
18. sin 22J, cos 22*, tan 22J.
19. sin 75, cos 75, tan 75.
20. sin 105, cos 105, tan 105.
, and tan % in the
Prove the following identities:
21. (sin + cos 0)* - 1 + sin 26.
22. 1 + cos 2a = 2 cos 2 a.
2 tan a
23. sin 2a
sec 2 a
24. cot 2a
25. tan A
cot* a 1
2 cot a
1 - cos 2A
sin 2 A
sin 2A '
-f cos2A*
TRIGONOMETRIC IDENTITIES 63
26. tan 6 sin 26 = 2 sin 2 0.
27. cos 8 x sin 3 x = (cos x sin x)(l -f- i sin 2x).
28. cos 4 3 sin 4 a? = cos 2x.
29. sec 2a = 1 + tan 2a tan a.
30. sin a + cot a sin 2a = 1 -f cos 2a + sin a.
1 sec 2 a
31. sec 2a
32.
2 cos 2 a - 1 2 - sec 2 a
cos 2a cos 2 a sin 2 a 1 tan
1 -f- sin 2a (cos a -f sin a) 2 1 -f- tan a
33. 2 cot 2A = cot A - tan A.
34. 2 cot a = cot ~ - tan ^-
^2 4v
36. sin 4A = 2 sin 2A cos 2A = 4 sin A cos 5 A 4 sin 8 A cos A.
(7/m/. Let a = 2A.)
36. cos 4A = 1 - 2 sin 2 2A = 1 8 sin 2 A cos 2 A.
37. cos 4A = 8 cos 4 A - 8 cos 2 A + 1.
38. cos 4 A =5 cos 4 A 6 sin 2 A cos 2 A + sin 4 A.
*a A^ s ^ n 4^ _ 4 tan A 4 tan 3 A
<S3. tan 4^ - ^-jj - ! -6 tan 2 A -f tan 4 A'
40. sin 4A = 4 sin A cos A 8 sin 3 A cos A.
41. (1 + cos 2a)(l + tan ^ tan a) 2 = 2.
&
42. sin 3a = sin (2a + ) = 3 sin a. 4 sin 3 a.
43. sin 3a = 3 sin a cos 2 a sin 3 a.
44. cos 3a 4 cos 3 a 3 cos a.
45. cos 3a = cos OL 4 cos a sin 2 a.
- , 3 tan a. tan 3 a
46. tan3=
3 tan 2 a.
47. sin 2x -+- cos 2x + 1 = 2 cos x (sin a; + cos x).
fi I fi\ f (
48. cos 6 sin - = [ 1 2 sin - ) ( 1 + sin ;
t \ &l \ t
cos - cos
49. '
9. (l- cos |)
1 + 2 cos
60.
61.
1-tanf
62. sin a
64 TRIGONOMETRY
41. Formulas for sums and differences of two sines or two cosines.
From the formulas of 36 we obtain, by addition and subtraction,
(1) sin (a. + )S) + sin (a )8) = 2 sin a cos /J,
(2) sin (a + /3) - sin (a - /3) 2 cos a sin ]8,
(3) cos (a + j8) + cos (a - j8) = 2 cos a cos ]S,
(4) cos (a + |8) cos (a |3) = 2 sin a sin ]8.
If we read these formulas from right to left they express products
of a sine or cosine of one angle by the sine or cosine of another as
equal to one half of sums or differences of sines or cosines.
For purposes of computation it is often more convenient to deal
with products of functions than with their sums. The four formulas
express sums as products, but a change of notation is advantageous.
Let us make the substitutions
<* + /? = 4, a- = B.
By adding and subtracting these equations we obtain
2a = A + B, 20 - A - 5,
A + B a A - B
<* = -2~> 0--aT~
When a and ]8 are replaced by these values, formulas (1) to (4)
become
/~\. At V* ** A ~| ** A "~~ O
(5) sm A + sin B = 2 sm 5 cos r i
(6) sin A - sin B = 2 cos ^-~-? sin ^-y-^
(7) cos A + cos B = 2 cos ~ cos ^ i
x _ v _ A . ^n "T" D , A /*
(8) cos 4 - cos B = - 2 sm 5 sm r
A good way to memorize these identities is to put them in words.
Thus formula (5) is equivalent to the statement: The sum of the
sines of two angles is equal to twice the sine of half the sum of the angles^
multiplied by the cosine of half the difference.
Example 1. Prove that sin 40 -f sin 20 = sin 80.
Solution. By formula (5)
sin 40 + sin 200 = 2 sin 151^! cos 4 - 20
2 w 2
2 sin 30 cos 10 - 2 } cos 10
cos 10 - sin 80.
TRIGONOMETRIC IDENTITIES 66
_ ft T J.L i. 8m A sin B , A + B
&Mil* 2. Prove that CO8 A _ coa B - - * 3
Solution. By formulas (6) and (7),
A 4-B . A - B
. A . D 2 COS jr Sin S
sin A sin B 2 2
cos A - cos B . A -f B . A - B
- 2 sm g sm 3
A + B
., C08
A 4-B
'
EXERCISES
By use of formulas (1), (2), (3), (4) of 41, express the following products CM
sums or differences of sines or of cosines:
1. 2 cos sin 30. 6. 2 sin a sin 5a.
2. 2 sin cos 80. 7. cos a cos 2a.
3. 2 sin 2a cos 5a. 8. sin 5a sin 2a.
4. 2 cos 3a cos 4a. 9. sin 5a cos 6a.
6. 2 sin 3a sin 2a. 10. cos Set sin 2a.
By use of formulas (5), (6), (7), (8) of 41, prove the following identities:
11. sin a; + sin 3o; = 2 sin 2o? cos x. 14. cos 8# -f cos 4x = 2 cos 6s cos 2#.
12. sin 5x sin 3# = 2 cos 4x sin x. 15. sin 3a sin a + 2 cos 2a sin a.
13. cos 3z = cos x 2 sin 2z sin x. 16. cos 7a = 2 cos 6a cos a cos 5a,
... sin A + sin B . A + B
17. j~ 5 - te n s
cos A -h cos B 2
<0 sin A - sin B A - B A -f B
18. -: 5 : : 5 tan ^ co ^ ?J *
sin A + sin B 2 2
,, A cos 3a cos a . rt AA sin a sin 2a .3
19. -: = 5 ~ tan 2a. 20. = cot ^
sm a - sm 3a cos 2a - cos a 2
21. sin a (sin a + sin 3a) = cos a (cos a cos 3a).
22. sin x cos 2x sin 3a; = cos 2x (2 sin x + 1).
23. cos (45 - a) - cos (45 + a) - V<2 sin a.
24. sin (45 + a) + sin (45 - a) - V cos a.
25. sin (60 + 0) - sin (60 - 0) - sin 0.
26. cos (60 - 0) + cos (60 + 0) - cos 0.
Express each sum or difference as a product:
27. sin 36 + sin 44. 35. sin 2a + sin 4a.
28. sin 54 + sin 26. 36. sin 4a + sin 6a.
29. sin 40 - sin 50. 37. sin 3a - sin a.
30. sin 70 - sin 10. 38. sin 2a - sin 5a.
31. cos 40 + cos 50. 39. cos a + cos 4a.
32. cos 30 + cos 70. 40. cos 3a 4- cos 2a.
33. cos 20 cos 30. 41. cos &a - cos a.
34. cos 50 - cos 20. 42. cos 2a - cos 3a.
66 TRIGONOMETRY
Express as a product involving only tangents and cotangents:
43 sin 20 + sin 30 ._ cos a + cos 3a
cos 20 + cos 30' sin a + sin 3a*
AA sin 50 + sin 70 .- cos 5a + cos a
cos 50 + cos 70 " sin 5a + sin a
AK sin 40 sin 20 4ft cos 2a cos a
sin 40 + sin 20' cos 2a + cos a
dft sin 70 sin 50 _ fl cos 3a cos a
" sin 70 + sin 50' sin 3a - sin a*
MISCELLANEOUS EXERCISES
Prove the following identities:
1. (sin a -f cos a) (tan a + cot a) = sec a + esc a.
2. 1 + sec 4 a = 2 sec 2 a -f tan 4 a.
1 sec 2 2 A
3.
1 -f- cos 2 2A 2 + tan 2 2A
tan a, tan jS _ cot ft cot ot
1 4- tan a tan j9 cot a cot + 1
6. sin 2x esc x *= 2 cos x. 7. sin 2rc cot re = cos 2x + 1.
6. esc 2x sin x = J sec x. 8. cot 2x sin 4# = 1 + cos 4x.
9. sin 3x cos x cos 3z sin x = sin 2x.
10. cos x cos 3x sin x sin 3x = cos 4x.
4 ., sin 2x cos 2# .. cos 2x sin 2z
11. --- = sin x. 12. ---- = cos 3x.
sec x esc x sec x esc 3
13. sin 7x cos 5z + cos 7x sin 5x = sin I2x.
14. cos 7x cos 5x + sin 7x sin 5z = cos 2x.
16. 2 cot 4a - cot 2a = - tan 2a.
16. sin 2a sin 4a + sin 6oc = 4 sin a cos 2a cos 3a.
17. 2 sin 2 A = sin 4 A -f 8 sin 3 A cos A.
cos (Q{ + ft) __ 1 tan a tan ff
cos (a - j8) "" 1 + tan a tan 0"
19. cot ~ sin a = 2 cot a cos 2 7=
J ^
20. tan a = esc 2a cot 4a esc 4a.
21. cos 4a 4 sin 2a = 3 2 (sin a + cos a) 4 .
22. sin 3a sin 2a + sin a = 4 cos fa cos a sin a.
ftw 1 cos a. + sin a , a
23. T~I - = - tan ^
1 + cos a -j- sin a 2
24. cot | - tan ~ = cos a (cot | + tan
A, J3, C are on^Zcs o/ a triangle we have A + B + C = 180. Prove tfwtf
t's case the following four identities hold:
ABC
25. sin A -f sin B + sin C = 4 cos -5- cos -5 cos 75-
& 4U
26. sin A + sin B - sin C - 4 sin ~ sin ^ cos -
TRIGONOMETRIC IDENTITIES
67
ABC
27. cos A + cos B + cos C = 1 + 4 sin -=r sin -~ sin -5-
28. tan A + tan B -f tan C = tan A tan B tan C.
Other proofs than those given in 36 for the addition formulas for sine and cosine
are proposed in the following exercises. It is assumed that A and B are acute angles.
29. Let A and B be angles of a triangle ABC as shown
in the adjoining figure, and R the radius of the circum-
scribed circle. Show that:
(1) c = a cos B + & cos A;
(2) a = 2R sin A, b = 2R sin B, c = 2R sin C;
(3) sin C = sin (A + B).
From (1), (2) and (3), show that
sin (A -f B) = sin A cos B + cos A sin B.
30. For Figure 52, show that:
Area OPQ = %xy sin A ;
Area OQR = Jzt/ sin B;
Area OPR - Jo* sin (A + B);
and hence that
FIG. 51
FIG. 52
sin (A + B) = % sin A -f 2 s i n B
z x
= sin A cos B + cos A sin B.
31. For Figure 52, show that:
Area ORS = \wz cos (A + B);
Area OQ$ = %wy cos A;
Area OQR = Ji/z sin B.
Hence deduce that
cos (A + B) = cos A cos B sin A sin B.
32. In Figure 53, a is an acute angle which is inscribed in a semicircle of
radius a. Show that:
AP = 2a cos a;
MP = a sin 2a;
BP = 2a sin a;
and then derive the formulas
sin 2a = 2 sin a cos a;
2 cos 2 a = 1 -f cos 2a;
2 sin 2 a = 1 cos 2a;
, 1 cos 2a sin 2a
tan a = : ^ = r- ; R-
sin 2a 1 + cos 2a
In these formulas substitute a = | , and derive the half-angle formulas.
CHAPTER V
RADIAN MEASURE. INVERSE FUNCTIONS.
TRIGONOMETRIC EQUATIONS
42. Radian measure. So far we have been using the angle of 1
as our unit of measurement for angles. Another unit, especially
useful in higher mathematics, is the radian.
An angle of 1 radian is defined as the angle at the center of a circle
of radius r that subtends an arc of length r.
From a theorem of geometry, changing r would not change the
angle thus defined.
Another theorem of geometry states that measures of angles at
the center of a circle are proportional to the arcs which they sub-
tend. If we apply this to the angle and the
angle of 1 radian shown in Figure 54, we have
radian measure of 6 s
. - 1 =
1 radian r
Hence
,1^ .. - A s subtended arc
(1) radian measure of & = - = - -r-. -
N ' r radius
We can compare measurements in radians
and degrees by substituting for 6, in (I), an
angle of 180. Since such an angle subtends a semicircumference,
we have s = TIT, and (1) reduces to
*jrr
radian measurement of an angle of 180 = = TT.
We write this, in abbreviated form,
(2) 180 = TT radians.
Recalling that TT = 3.141593 approximately, we find from (2)
(3) 1 - radians = 0.0174533 radians;
(4) 1 radian = - 57.29578 = 57 17' 45'
7T
to the indicated number of significant figures.
68
RADIAN MEASURE 69
Our Tables enable us to change the measurement of an angle in
degrees and minutes to its measurement in radians and vice versa.
The first and second columns of Table II serve this purpose for acute
angles. Other angles are expressed as multiples of an angle of 90
plus or minus an acute angle, and the radian measure of these two
terms is combined, if we are changing from degrees to radians.
Similarly, for the reverse process.
Example 1. Express 110 23' in radians.
Solution. We shall first obtain the result as a fraction times IT radians, a
form that is often convenient; then as a decimal number of radians.
(a) 110 23' - (110 + )
6623 ^ TT 6623
X
-60" 180 10806
(b) 110 23' - 110 X 0.0174533 + f$ X 0.0174533
- 1.9199 + 0.0067
= 1.9266 radians (to four decimal places).
We could have obtained result (b) from Table II as follows:
110 23' = 90 + 20 23'.
From the Table, 90 = 1.5708 radians. The radian value of 20 23' is found
by interpolation to be 0.3549 + ^(.0029) = 0.3558. Hence
110 23' - (1.5708 + 0.3558) radians - 1.9266 radians.
Example 2. Express 5 radians in degrees and minutes.
Solution. 5 radians = 5 X (57 17' 45")
= 286 29' to the nearest minute.
Example 3. Express -g radians in degrees.
Solution. radians - ~ X - 30.
O D 7T
Example 4. If the radius of a circle is 10 ft., find in terms of TT the length of
arc subtended by an angle of 15 with vertex at the center of the circle.
Solution. According to formula (1), s = r X number of radians in 0; hence,
for this Example,
Example 5. Find both the radian and the degree measure of an angle sub-
tended at the center of a circle of radius 6 inches by an arc of length 10 inches.
Solution. According to formula (1),
radian measure of 6 = * = 1.6667.
To change this to degrees and minutes, note that 90 = 1.5708 radians, and write
1.6667 - 1.5708 + 0.0959;
then from Table II
6 - 90 + 5 30' - 95 30'.
70 TRIGONOMETRY
EXERCISES
Give the radian measures in terms of IT for the following angles:
1. 45; 60; 135; 300. 3. 30; 225; - 45; - 270.
2. 22i; 75; 195; 540. 4. 15; 67J; 330; 465.
Express in radians in decimal form, using Tables:
5. 35; 50; 60; 100. 9. 15 27'; 132 15'.
6. 30; 40; 80; 110. 10. 61 13'; 111 11'.
7. 20; 45; 70; 130. 11. 43 12'; 141 40'.
8. 10; 75; 90; 135. 12. 67 17'; 188 17'.
Without using Tables express the following angles, whose radian measure is given,
in terms of degrees and minutes:
1Q 7T 2?T 77T 1R 7T 37T 7lT .
13 ' 8' T ; T ; *' 10 * 5 ; T ; T ; *'
. 7T. 37T. 47T. 7T. 27T. 57T.
Using Tables express the following angles, whose radian measures are given, in
terms of degrees and minutes:
17. 0.9427; 1.0378. 21. 0.8623; 2.8883.
18. 0.8127; 1.0836. 22. 0.7979; 3.6542.
19. 0.5432; 1.5432. 23. 1.0114; 3.1104.
20. 0.7171; 1.7171. 24. 0.9876; 2.6789.
In each of the following problems the radius of a circle and an arc on that circle
are given; find the angle at the center subtended by the arc. Express the result first
in radians, then in degrees and minutes, assuming the measurements of radius and
arc exact.
25. Radius = 10 ft., arc = 15 ft. 29. Radius = 16 in., arc = 2 ft.
26. Radius = 12 ft., arc 18 ft. 30. Radius = 8 in., arc = 4 ft.
27. Radius = 20 in., arc 24 in. 31. Radius = 5 mi., arc = 11 mi.
28. Radius 18 in., arc == 15 in. 32. Radius = TT ft., arc = 2?r ft.
In each of the following problems an angle at the center of a circle^ and the arc it
intercepts are given; find the radius of the circle. (Assume that given measurements
are exact.)
33. Angle = 2 radians, arc = 9 in.
34. Angle = 1.5 radians, arc = 12 in.
35. Angle = 3.333 radians, arc = 20 in.
36. Angle = 1.111 radians, arc = 27.9 in.
37. Angle = 1.4286 radians, arc = 14 in.
38. Angle - 1.25 radians, arc = 8 in.
39. Angle - 20, arc - 6 in. 43. Angle - 152 13', arc = 25.5 in.
40. Angle = 30, arc - 7 in. 44. Angle - 307 10', arc = 68.8 ft.
41. Angle = 46 11', arc - 201 ft. 45. Angle = 300, arc = 70.7 ft.
42. Angle - 58 11', arc = 103 in. 46. Angle = 9, arc - 15.708 ft.
INVERSE FUNCTIONS 71
43. Inverse functions and their principal values. The statement
that sin is equal to a is equivalent to saying that 6 is an angle
whose sine is a, or, more briefly, that 6 is the inverse sine of a. Two
notations are in use to express this relation,
= arcsin a, and = sin" 1 a.
When the latter notation is used, we must be careful not to inter-
pret shr 4 a as the 1 power of sin a.
The three equations
sin = a, = arcsin a, = sin" 1 a
mean exactly the same thing.
We define similarly the other inverse functions arccos a, arctan a,
and so on; the alternative notation is cos" 1 a, tan" 1 a, and so on.
The inverse functions are many-valued. For example, we have
arcsin J = 30, 150, 390,
In order that = arcsin a have a value, it is necessary that a be not
greater than 1 or less than 1, since a = sin 0. If a has a value
so chosen, arcsin a has infinitely many values, of which one and only
one lies between 90 and + 90 (inclusive). This is called the
principal value of arcsin a, and is written Arcsin a (with the initial
letter A capitalized); it may also be written Sin" 1 a (with capital /S).
For Arcsin a we will say that the range of this principal value and
the interval of definition are as follows:
- 90 ^ Arcsin a ^ 90, - 1 ^ a g 1.
For principal values of all six inverse functions, the ranges and
intervals of definition are given in the following table:
- 90 ^ Arcsin a 90, - 1 ^ a ^ 1 ;
^ Arccos a ^ 180, 1 a - 1;
90 < Arctan a < 90, a may have any value;
< Arccot a < 180, a may have any value;
^ Arcsec a ^ 180, a^lora^-1;
- 90 g Arccsc a ^ 90, a - 1 or a 1.
In each case, when a is in the interval of definition, the inverse
function has one and only one value, the principal value, in the range
indicated.
The problem of finding further values of = arcsin a is that of
finding solutions of the equation sin = a other than the principal
value, 61. We easily solve this by observing that there is always a
72 TRIGONOMETRY
secondary solution 180 - 0i, since sin (180 - 0i) sin Bi =a (the
principal and the secondary solution here coincide if 0i 90). The
two solutions, 0i and 180 - 0i, both lie in the range - 90 ^
< 270, and are the only ones in that range. All other values of
= arcsin a are obtained by adding positive or negative multiples of
360 to the principal and secondary values of 0.
For the six inverse functions, if 0i is a principal value, secondary
values are as follows:
180 - 0i for arcsin and arccsc 0;
0i for arccos and arcsec 0;
180 + 0i for arctan and arccot 0.
In each case the principal and secondary values are the only ones
on a range of 360, and all other values are obtained from the prin-
cipal and secondary values by adding positive or negative multiples
of 360.
Example 1. Find all values of arccos f-r)i giving results in degrees.
Solution. The principal value is
Arccos ~) - 30,
and the secondary value is 30. Hence
arccos ( -^) = 30 =t n 360, - 30 n 360.
\ & I
Example 2. Find all values of arctan (- 3.000), giving results in radians.
Solution. Let 0' = arctan ( 3.000); then we are to solve the equation
tan 6' = - 3.000.
As a first step we obtain from the Tables the corresponding acute angle 0, for
which tan - 3.000; in radians, - 1.2490. A value of 0' will therefore be
1.2490 radians, and this will be the principal value, since it lies between
-^ and -f -= radians (equivalent to 90 and + 90). Thus for our Example
the principal value is
Arctan (- 3.000) - - 1.2490,
the secondary value is 1.2490 + K> and the general solution of our problem is
arctan (- 3.000) - - 1.2490 2nw, - 1.2490 + ?r d= 2nv.
Example 3. Find the values of tan (arcsin |).
Solution. We could, with the aid of the Tables, find all angles which are
values of arcsin ($) and then use the Tables and reduction formulas to find the
value of the tangent of each of these angles. It is simpler, however, to observe
INVERSE FUNCTIONS 73
that if we write a arcsin (f ), we are to find tan a, given sin a f . We have
solved problems of this sort on pages 16, 17, and 49. The solution is
tan a. tan arcsin (f ) =t $,
the positive sign to be taken for the principal value Arcsin ( j) and coterminal
angles n 360 + Arcsin (f ), the negative sign for the secondary value and
coterminal angles.
Example 4. Simplify the expressions sin sin~ l x, sin" 1 sin x, and Sin"" 1 sin x.
Solution. We are here using the alternative notation for inverse functions;
for example, Sin' 1 a means the principal value Arcsin a.
Obviously, the phrase, "the sine of an angle whose sine is x" can have but
one meaning, sin sin -i x = Xt
On the other hand, sin" 1 sin x (an angle whose sine is equal to sin x} has many
values. It is clear that x is an angle whose sine is equal to sin x, but it is also
true (in radian notation) that IT x is an angle whose sine is equal to sin x.
All other values are given by coterminal angles, hence
sin" 1 sin x x =t 2nir, w x =t 2nir.
7T 7T
Finally, Sin" 1 sin x is equal to x if x is between -x and + -= ; but if x is not '
4 t
between -^and -f i> then Sin " x sin x is an angle in this interval that is coter-
i
minal either with x, or else with TT z.
-k Example 5. Prove that if x and y arc positive, and each is less than 1, then
Arctan x + Arctan y = Arctan * __
Solution. This is an example of so-called identities in terms of inverse func-
tions, which are true only with restrictions on the variables, as x and y here.
In this problem the two sides of the identity are acute angles and our identity
will be true if
tan [Arctan x + Arctan y] = tan Arctan = t
L A ~" *2/J
tan (Arctan x) + tan (Arctan y)
1 tan (Arctan x) tan (Arctan y) "
The last equation reduces to
x + y = x + y
I xy \ xy
and the original identity is thus proved.
Remark. If x and t/ are positive, then Arctan x and Arctan y lie between
x -\- y
and 90, as noted above. If, however, 1 xy is negative, then Arctan ^ __
lies between - 90 and 0. Hence the stated formula is not valid in this case.
But since the tangent of the left member equals the tangent of the right member,
the two members differ by a multiple of 180. It follows that when x is positive,
y is positive, and 1 xy is negative,
Arctan x + Arctan y Arctan ^ J" y + 180,
74
TRIGONOMETRY
EXERCISES
Express the following angles in degrees and minutes:
11. Arccos 0.8988.
12. Arccos 0.4067.
A - /V3
13. Arcsin I -=-
14. Arccos
1. Arcsin
2. Arccos ( s~
3. Arcsin 0.4848.
4. Arcsin 0.8949.
5. Arcsin ( ).
( -- =- j
6. Arcsin ( ~
7. Arcsin (- 0.8572).
8. Arcsin (- 0.9205).
9. Arccos (i).
10. Arccos
16. Arccos (- 0.4067).
16. Arccos (- 0.8141).
17. Arctan 1.
18. Arctan VS.
19. Arctan 0.6873.
20. Arctan 1.492.
21. Arctan [ =-
22. Arctan
23. Arctan
24. Arctan
25. Arccot
26. Arccot
27. Arccot
28. Arccot
29. Arcsec
30. Arcsec
31. Arcsec
32. Arcsec
33. Arccsc
34. Arccsc
35. Arccsc
36. Arccsc
(-1).
(- 0.4245).
(- 2.300).
V3.
(-VI).
1.483.
(- 2.006).
1.206.
*
V2.
(-2).
1.
(-2).
1.729.
(- 2.215).
Express the following angles in radians, by use of tables:
37. Arcsin 0.0669. 40. Arctan 343.8. 43. Arcsec 68.76.
38. Arcsin 0.9890. 41. Arccos 0.9986. 44. Arcsec (- 1.010).
39. Arctan 0.0670. 42. Arccos (- 0.9986). 45. Arccsc (- 68.76).
Express in radians, and also in degrees and minutes, the following:
46. arrtan 1. 60. arccot 4.011. 64. arccsc 3.
47. arccos (- 1). 61. arccsc (- 6.392). 66. arctan (f).
48. arcsec (- 2). 62. arcsec ( V2). 66. arccot ( 1).
49. arcsin 0.9613. 63. arcsin (). 67. arccos (J).
Find values of the following:
68. sin Arctan (J). 63. cos arccos (f).
69. tan Arcsin (f ).
60. cos Arcsin (J).
61. sin Arcsin ().
62. cot arctan (}).
64. sin arccos (1).
65. cos arcsin JVl}.
66. tan arccos ( f).
67. arcsin sin 20.
73. sin (Arcsin (f) + Arctan (J)). 76. sin (2 Arccos ()).
74. cos (Arccos (i) 4- Arctan (J)). 77. cos (2 Arctan (A))-
76. tan (Arctan (i) -f Arctan (f)). 78. sin (Arctan (A) + 2 Arcsin (f)).
68. arctan tan 15.
69. Arctan cot 220.
70. Arcsin cos ( 100).
71. Arccos cos 550.
72. Arccot tan 100.
7T
4*
Prove the following:
*79. Arctan (J) + Arctan (J)
*80. Arcsin (f) + Arcsin (#) = Arccos (- tf).
*81. Arccos x Arccos y = Arccos (xy =t Vl
tions which should be placed on x and y.)
(State restric-
TRIGONOMETRIC EQUATIONS 75
44. Trigonometric equations. In 43 we have discussed the solu-
tion of a very simple type of equations in trigonometric functions;
the equation sin = a, in which 6 is the unknown, has the infinitely
many solutions 6 = arcsin a. In the following Examples we show
how to solve equations of other types.
Example 1. Find in degrees all solutions of the equation 3 tan 2 1=0.
Solution. We have
tan 2 = J,
tan 6 = --. t or tan 9 ;=
V'3 \/3
Hence all solutions are given by
6 = arctan => 9 = arctan ( ^V
v3 \ v3/
The principal value of arctan (1/V3) is 30, and that of arctan ( 1/V3)
is 30. In degrees, the above solutions are therefore
6 - 30 zfc n 360, = - 30 dh n 360,
210 n 360, 150 db n 360.
Example 2. Find in radians all solutions of 2 cos 3x 1 that lie between
and TT.
Solution. We have cos 3x = J, for which the general solution is
3x = -= 2n?r, db 2n7T.
The values of x that lie between and TT are
TT 5?r TTT
* " 9' T' T"
Example 3. Find in radians all solutions of sin 2x cos x = that lie be-
tween and 27T.
Solution. We have
sin 2x cos a; = 0,
2 sin x cos a: cos x = 0,
cos x (2 sin s 1) = 0,
hence the solutions are found by solving the two equations (not simultaneous)
cos x = 0; sin x = J.
The solutions are
_ TT 3?r TT 5?r
*~ 2' T* 6 1 T*
76 TRIGONOMETRY
Example 4. Find in degrees all solutions of sin x + sin 3x such that
x 180.
Solution. We have, by 41,
sin x -f- sin 3x ** 2 sin 2z cos re,
hence solutions of our equation are found by solving the two equations
sin 2x = 0; cos x 0.
Hence we have
x - 0, 90, 180.
Example 5. Find in degrees all solutions of cos -f- 2 sin = 2 such that
- 90 g 90.
Solution. We reduce this to a quadratic equation in sin 6 as follows:
cos -f 2 sin - 2,
cos = 2 2 sin 0,
cos* - (2 - 2 sin 0) 2 ,
(I - sin 2 0) = 4 - 8 sin -f 4 sin 2 0,
5 sin 2 - 8 sin -f 3 = 0,
(sin0- l)(5sin0-3) =0.
Hence all solutions of our equation are found by solving the two equations
sin0 - 1; sin0 - J;
and the solutions of these equations in the range proposed are
_ 90; - Arcsin (f) = 36 52'.
These are the only possible solutions of our problem, but we might doubt
whether they are actual solutions, since at one stage we squared both sides of
the equation cos = 2 2 sin 0, and it does not always follow that all solutions
of a squared equation are solutions of the equation itself. We should finish
by substituting our possible solutions. In the present Example this is easily done,
and we find that both the above values of are actual solutions.
It is interesting to note that if x cos 0, y = sin 0, we have x 2 + y 2 = 1,
and our problem reduces to this: Solve algebraically for y the simultaneous
equations * + *-!, * + 2y - 2;
if the solutions are y\, y-i, solve the equations
sin = yi, sin = y 2 .
Many trigonometric equations are similarly related to systems of algebraic
equations.
Another method is suggested by the observation that if, in an expression
a cos + 6 sin 0, we replace a by r cos a and b by r sin a, we have
a cos -H & sin r cos a cos -f r sin a sin = r cos (a 0) = r cos (0 a).
Thus the equation cos -f 2 sin * 2 can be replaced by
1 = r cos a, 2 * r sin a, r cos (0 a) = 2.
These equations are equivalent to the following:
1 Q
r = \/6, a arccos ;=. arccos --fa.
V6 r
The last equation gives the solutions when we use appropriate values of the
inverse functions.
TRIGONOMETRIC EQUATIONS 77
EXERCISES
Find in degrees all solutions of the following equations:
1. 3 cos 6 =2. 9. sin 2z = 2 sin x.
2. 4 tan 0=5. 10. cos 2x 2 cos x.
3. 3 sin B - 2. 11. tan 2s - 2 tan x.
4. 4 sec B 5. 12. sec 2z = 2 sec x.
6. 4 sin 2 B - 3. 13. sin 3a; - sin 3.
6. 2 cos 2 1. 14. cos 3x cos #.
7. 3 cot 2 0-1. 15. sin* 2x - cos 2 2a? = 0.
8. 3 sec 2 0=4. 16. cos 2 2x - cos 2 x - 0.
Find in radians att solutions of the following equations:
17. tan 2 x - 2. 21. sec 2 = 2.
18. 2 sin 2 as-1. 22. esc - 2 sec 0.
19. sin x 2 cos x. 23. sin 20 = 2 cos 0.
20. cos x 2 sin x. 24. cos 30 = cos 0.
in degrees all solutions of the following equations such that ^ x ** 180:
25. sin x sin 2x 0. 29. sin 2z + 2 cos 2 35 1.
26. cos 2# cos 2 x 0. 30. sin 5a; sin a; 0.
27. tan 2 x -f- sec 2 x 7. 31. cos 5a; + cos 3x 0.
28. 2 cos 35 + sin x = 2. 32. 3 cos x + 4 sin x = 5.
Find tn degrees and in radians all solutions, if there are any, of the following
equations such that ^ x ^ 360:
33. cos 2x cos x 0. 42. 8 sin a; + cos x 7.
34. cos 4a: cos 2x 0. 43. 4 sin a; 7 cos x 1.
35. tan 2 x + 2 = 3 tan x. 44. cos 3x 4 cos 2 a:.
36. 2 cos 2 a: + 3 5 cos x. 46. 6 esc x 6 sin x 5.
37. cos 2a; -f cos a; -f 1 0. 46. cos 2 x sin 2 351.
38. sec x + 2 cos a; 3. 47. sec 2 a; tan 2 35 -f 2.
39. 2 cos 2x - 4 sin a; -f 1 = 0. 48. 3 esc 2 x - 2 esc x - 1.
40. sin 4x sin 2x = cos 3x. 49. cos 2 x cos a; = 6.
41. sin x + sin 3a; cos 3; + cos 3x. 50. tan 4 a; + 3 tan* x + 2 = 0.
CHAPTER VI
LOGARITHMS. FIVE-PLACE TABLES
45. The definition of a logarithm. It is a consequence of the
definitions of real numbers that every positive real number N can
be expressed as a power of 10,
(1) N = 10*,
where rr is a real number. For a given N there is only one x } and to
increase N is to increase x.
Definition. // N = 10*, the exponent x is called the logarithm of
N to the base 10.
In symbols we write
log N = x,
or, the base being taken for granted,
log N = x.
Equation (1) can then be written
(2) 10iog^=M
We can generalize the definition of a logarithm so as to use another base than
10. If a is any positive real number other than 1, every positive real number N
can be expressed in one and only one way in the form
(3) N = a*.
We then say that x is the logarithm of N to base a, and write
(4) log# = z, a><*x = N.
46. The table of five-place logarithms of numbers. Logarithms
are, in general, unending decimals and only approximate values can
be given in tables. When we say that .30103 is the five-place loga-
rithm of 2 to base 10, we mean that .30103 is the five-place number
which is nearest to the true solution x of the equation
10* = 2.*
* We recall that 10 30103 means 10 to the power 30103/100000, that is, the
hundred-thousandth root of 10 30103 . Needless to say, tables of logarithms are not
actually computed by taking hundred-thousandth roots.
78
LOGARITHMS. FIVE-PLACE TABLES 79
On pages 74-91 of the Tables at the end of this book, five-place
logarithms to base 10 are given for numbers N ranging from 1.0000 to
9.9990. The first three digits of N are given in the left-hand column,
under N, the fourth digit in line with N near the top of the page, and
the fifth digit is zero. A decimal point is to be placed after the first
digit of N and before the first digit of each logarithm. The last three
digits of log N are to be found in the line which contains the first
three digits of N and in the column headed by the fourth digit of N.
We find the first two digits of log N 9 if the last three digits of log N
are not preceded by a star, by going back to the first column of
logarithms and going up to the first logarithm given to five places;
the first two digits of that logarithm are the same as the first two
digits of the logarithm we are seeking. If the last three digits are
preceded by a star, use for the first two digits those to be found in the
first logarithm of the line below. Thus, for example, we read from
page 76 of the Table
log 2.3360 = .36847,
log 2.3460 = .37033.
The following two Examples illustrate interpolation by the prin-
ciple of proportional parts.
Example 1. Find log 2.5432 to five places.
Solution. We have, from page 77 of the Table,
log 2.5432 - Iog2.5430+ A (log 2. 5440 -log 2. 5430) N
.40535 + A X (.40552 - .40535)
- .40535 + A X (.00017)
- .40535 + .00003 10 |
= .40538.
log AT
t T2.5430 .405351 "]
J L2.5432 ? J* 17
2.5440 .40552 J
At the right of page 77, under Prop. Pts. (Proportional Parts) for the black
type entry 17, it can be read off that A X 17 = 3.4, from which we could have
inferred that the correction .00003 was to be added to .40535.
Example 2. If log N - .54991, find N to five places.
Solution. The logarithms in the Table that are nearest to .54991 are
N
log 3.5470 - .54986,
log 3.5480 = .54998. 10 |
log AT
[T3.5470 .54986"! .1
X l ? .54991 J 5 12
3.5480 .54998 J
The given logarithm .54991 is A of the way from log 3.5470 to log 3.5480;
hence N is to be taken A of the way from 3.5470 to 3.5480, so that
N - 3.5470 + A(-OOIO)
- 3.5470 + .0004 = 3.5474.
80 TRIGONOMETRY
In the column of Proportional Parts for the black number 12 we would have found
A. S
that - = .4, from which we could see that the interpolation is .0004.
In the next section we shall explain how to find logarithms of
numbers greater than 10 or less than 1.
47. Logarithms of positive numbers. We have now explained how
to find the logarithm to base 10 of a number between 1 and 10.
Logarithms of positive numbers without such restriction are obtained
with the aid of the following fundamental property of numbers:
I. Every positive number N can be expressed in the form
(1) N - 10'+"
where c is or a positive or negative integer, and m is a positive number
kss than I (or m may be 0).
We are here merely rewriting formula (1) of 45, with x = c + m.
To say that x has this form is merely to say that a given x always
lies between two integers c and c + 1, that is, c ^ x < c + 1.
Another way of stating I is as follows:
II. Every positive number N has a logarithm to base 10 of the form
(2) log N = c + m,
where c, called the characteristic, is an integer (positive, negative, or
zero) and m, called the mantissa, is positive (or zero) and less than 1.
To find the logarithm of a number N to base 10 we must deter-
mine its characteristic and its mantissa. The solution is clearer if
we rewrite (1), by using a law of exponents (see the following 48),
N = 10 C X 10".
Since m is between and 1, we see that 10 m is a number M between
1 and 10, so that
(3) N = 10 C M, 1 M < 10.
(4) M - 10-.
We thus have the following rule for finding the characteristic and
mantissa of log N:
III. Express the number N in form (3); then the characteristic of
log N is c and the mantissa of log N is log M, which can be found by
using tables.
LOGARITHMS. FIVE-PLACE TABLES 81
Example 1. Find log 65.520.
Solution. We have
65.520 - 10 l X 6.5520.
Hence
c-1,
and from the Table
ro = log 6.5520 - .81637,
so that
log 65.520 = 1 + .81637 - 1.81637.
Example 2. Find log 0.065520.
Solution. We have
0.065520 - 10-* X 6.5520,
log N - - 2 + .81637.
Note that in Example 2 it would have been incorrect to write
- 2.81637 for log N, since - 2.81637 means - 2 - .81637. We do not
wish to change 2 + .81637 into 1.18363, so we use the notation
log N = 8.81637 - 10.
The above Examples illustrate the fact that, in formula (3), M
has the same digits as N, but in M the decimal point is just after
the first digit. If c is positive, multiplying M by 10 c merely moves
the decimal point so that in the product, N, the first significant figure
will be just c places to the left of units' place, the place just before
the decimal point; if c is zero, the first significant figure of N is in
units' place; if c is negative, the first significant figure of N is c
places to the right of units' place. Note that in Example 2 we kept
a zero in units' place to make sure that we should count correctly.
The converse of each statement in the preceding paragraph is also
true.
We state the rule for the characteristic as follows:
IV. To find the characteristic of log N, first find how far it is from
the first significant figure of N to the units' place. If units' place is
k places to the right, the characteristic is k\
k places to the left, the characteristic is k.
The student should write the converse of IV which states where
the units' place must be in N for a given characteristic of log N.
EXERCISES
Find the characteristics for the following logarithms:
1. log 34.210; log 58370. 6. log 0.32723; log 0.00723.
2. log 54760; log 3.4023. 6. log 0.01525; log 0.000742.
3. log 325.47; log 824300. 7. log 0.00273; log 0.000039.
4. log 5.0732; log 4217000. 8. log 0.00071; log 0.002423.
82 TRIGONOMETRY
Find the following logarithms t using five-place tables:
9. log 3752.0; log 2.8350. 19. log 371.30.
10. log 36.840; log 49990. 20. log 289520.
11. log 218300; log 0.01284. 21. log 48437.
12. log 6172000; log 0.6496. 22. log 0.62148.
13. log 0.8984; log 2982000. 23. log 0.0071782.
14. log 0.007263; log 92,920,000. 24. log 5.0425.
15. log 43434. 25. log 0.63787.
16. log 5284.2. 26. log 0.031814.
17. log 6.7316. 27. log 0.00025678.
18. log 82.477. 28. log 0.079784.
Find to five significant figures the numbers N whose logarithms are given as follows:
29. log N = 0.50664. 41. log N - 3.70030.
30. log N = 0.60438. 42. log N - 5.50042.
31. log N = 0.82000. 43. log N = 6.41143.
32. log N = 0.90020. 44. log N - 0.90029.
33. log N - 2.47712. 46. log N - 7.43100.
34. log N = 3.62180. 46. log N - 3.81020.
35. log N = - 2 + .71383. 47. log N = 6.44444 - 10.
36. log N = - 3 + .84634. 48. log N - 7.69920 - 10.
37. log AT = 8.90428 - 10. 49. log N - 9.87358 - 10.
38. log N = 7.94507 - 10. 60. log N = 8.78996 - 10.
39. log N = 2.41670. 61. log N - 4.70067 - 10.
40. log N = 4.61940. 62. log N - 3.49131 - 10.
48. Laws of exponents and of logarithms. If a is a positive num-
ber, and m and n are real numbers, the following laws of exponents
hold:
(I) a m X a n = a m+n ,
fin a _ n m-n
V **/ rT '
(III) (a m ) n = a mn .
In books on algebra these laws are discussed; in accordance with
these formulas, we have
(IV) a = 1,
___ 1
(V)
a n
Now let us turn these laws of exponents into corresponding laws
of logarithms. For this purpose we write
M = a m , m = logo M ;
N = a n , n = loga N.
LOGARITHMS. FIVE-PLACE TABLES 83
Thus formula (I) gives
MN - a+,
and we have
log(M2V) = m + n = log a M + log fl N.
Similarly, formula (II) gives
Iog j7 = m - n = loga M - log a N,
and formula (III) gives
logo M n = mn = n(log Af).
Thus our three laws of exponents have been expressed as the fol-
lowing laws of logarithms for all positive real numbers M, N, and
positive base a (a ^ 1) :
(1) log MN = log M + log N,
(2) log^=logM-logtf,
(3) log M" = n log M.
Since
VM = M ',
we may write, as a corollary of (3),
(3') log VM = J log M.
Formulas (IV) and (V) become
(4) log 1 = 0,
(5) lo *jjj = -logM.
49. Computation with the use of logarithms to base 10. From
law (1) of 48 we see that we can compute a product by finding the
logarithms of the factors M and N, adding them to get log MN,
and then finding the number MN from the Tables. Products of
more than two factors are handled by an extension of law (1) which
states that the logarithm of a product of two or more factors is
the sum of the logarithms of the factors. For example,
log MNP - log (MN)P = log MN + log P
- log M + log N + log P.
84 TRIGONOMETRY
Example 1. Find N - 2.8255 X .0074232.
Solution. log 2.8255 - 0.45110
log .0074232 - 7.87059 - 10
sum = log N - 8.32169 - 10
N - 0.0020974.
Example 2. Find AT = ||j!~
Solution. log 42.734 - 11.63077 - 10
log 5402.7 - 3.73261
difference = log N - 7.89816 - 10
N - 0.0079097.
-, , o T,. , M 0.38283 X 0.048296
Example 3. Find x = ^ = 0>06 2 1 92 X 8348.4 '
log 0.38283 - 9.58301 - 10 log 0.062192 - 8.79373 - 10
log 0.048296 - 8.68391 - 10 log 8348.4 = 3.92160
sum = log M = 18.26692 - 20 sum - log N = 12.71533 - 10
log N - 12.71533 - 10
difference = log x = 5.55159 - 10
x - 0.000035612.
Example 4. Find ^0.37285.
Solution. log 0.37285 = 29.57154 - 30
log v^O.37285 - J log 0.37285 - 9.85718 - 10
v/0.37285 - 0.71975.
Note that the characteristic of log 0.37285 was taken as 29 30 in order to facili-
tate division by 3 at the next step.
Negative numbers have no real logarithms. A computation in-
volving them, where logarithmic tables are to be used, must be
made to depend on one involving only positive numbers. Thus, to
find the product of - M and N, where M and N are positive, we
compute MN by using logarithms, and prefix a minus sign to the
result.
The student must remember that log (a + 6) is not equal to
log a + log b (unless a + b - a&). The logarithms defined in this
chapter are not useful in obtaining sums and differences of numbers.
*60. Cologarithms. In finding log ( -j* 1 we subtracted log N from
log M. This is the same as adding log M and ( log N). When
- log N is so written that the decimal part is positive (as we would
LOGARITHMS. FIVE-PLACE TABLES 85
write log (-^A which is equal to log JV), it is called the cologarithm
otN. Thus
colog N = log -^ - log N,
and the law for division becomes
log -^ = log M + colog JV.
In complicated computations, this device is often useful. We shall
not employ it in this text.
EXERCISES
Use a five-place table of logarithms to find the number N to five significant figure*
in the following:
1. N - 32.34 X 2.185. 11. N = (3.1624) 3 .
2. N = 7.588 X 87.62. 12. N - (71.346) 2 .
3. N = 625.44 X 3.8717. 13. N = (2.2333) 10 .
4. N - 14923 X 0.38761. 14. N = (0.31724) 2 X (4.1557) 8 .
6. N - 1066.1 X 0.077244. 15. # = VO4940.
6. N = 0.087319 X 0.11146. 16. N - ^87^7.
28.746 17 AT - ~ 61.872 X 23.750
7 ' " "" 37.254* "" 7279.3 X 6.071637 '
0.025183 1 ft AT = ^- 6254 X VQ.88888
0.0035677' - 37.247
-, _ 0.25764 lft 3/(3.5723) 2 X (- 1.1236) 3
' 1.3579 ' \ - 0.37162 X (88.714)"" "
96.172 on AT - /37.611 X OQ827Y/
1U * ^ " 0.042755' " \458.34 X 0.071714*
51. Five-place logarithms of trigonometric functions. Table VI
gives five-place logarithms of four trigonometric functions for angles
from to 90. For angles less than 45, we find the number of de-
grees at the top of the page, and the number of minutes in the left-
hand column ; for angles greater than 45, the number of degrees is
given at the bottom of the page, the number of minutes in the column
next to the Prop. Pts. column, and the column designations to be used
are those at the bottom of each column. After each logarithm as
given in this Table we must supply " - 10"; for example,
log sin 18 (X - 9.48998 - 10.
86 TRIGONOMETRY
We interpolate, usually to the nearest second,* by the principle of
proportional parts as illustrated in the following Examples.
Example 1. Find log sin 18 54' 25*.
Solution. On page 45 of the Tables (the page headed 18) we go down the first
column to 54', and opposite this entry and the next, in the L Sin column, we find
Angle
LSin
log sin 18 54' = 9.51043 - 10, f ri854' 0" 9.51043H "I
log sin 18 55' = 9.51080 - 10. eo 25 1_ 18 54' 25" ? J X \
L 18 55' 0" 9.51080 J
The angle 18 54' 25* is }$ of the way from 18 54' to 18 55', and we conclude that
(approximately) its log sin is f$ of the way from log sin 18 54' to log sin 18 55'.
Thus
log sin 18 54' 25* - 9.51043 - 10 + J X .00037
= 9.51043 - 10 + .00015
= 9.51058 - 10, to five decimal places.
By making use of other columns on page 45 we could compute the interpolation
as follows:
From the column next to the right of the L Sin column we could have read off
the difference between log sin 18 54' and log sin 18 55' (the tabular difference)
as 37 in the last two of the five decimal places, and in the Prop. Pts. (Proportional
Parts) column, under the black number entry 37 we are told how much an in-
crease of an angle by the indicated number of seconds increases the last two of the
five decimal places of the logarithm where the tabular difference is 37. Since to
increase 18 54' by 25* is to increase it by 20* + one tenth of 50", we read off from
the L Sin column and from the Prop. Pts. column
log sin 18 54' 25* - 9.51043 - 10 + .000123 + .0000308
= 9.51058 10, to five places of decimals.
Example 2. Find log tan 54 17' 42".
Solution. On page 62 of the Tables we find
log tan 54 17' - 10.14326 - 10.
The cd column to the left gives the tabular difference as 27, and in the Prop. Pts.
column under entry 27 we find that to the last two places of log tan 54 17' we
must add, when the angle is increased by 42", the correction 18.0 -f- 0.9 = 19 (to
the nearest whole number). Thus
log tan 54 17' 42" = 10.14326 - 10 + .00019
= 10.14345 - 10.
Example 3. Find log cos 73 23' 36".
Solution. On page 43 of the Tables we find 73 at the bottom of the page.
* Interpolation to the nearest five seconds would be more consistent with the
statements of page 27.
LOGARITHMS. FIVE-PLACE TABLES 87
In the column at whose foot we find L Cos we use the entry opposite 23 in the
right-hand column of minutes. We thus see that
log cos 73 23' = 9.45632 - 10.
We compute the correction for 36* by using d and Prop. Pts. columns as in
preceding examples, but note that the correction is now to be subtracted, since
L Cos decreases as we go from 73 23' to 73 24'. Thus
log cos 73 23' 36" = 9.45632 - 10 - .00026
= 9.45606 - 10.
Example 4. Find the acute angle A, given
log cot A = 8.97123 - 10.
Solution. By running through L Cot entries in the Table we find that those of
about the right magnitude are on page 32, for the column with L Cot at the foot.
We see that
LCot
log cot 84 39' = 8.97150 - 10, p p84 39' 0"
log cot 84 40' - 8.97013 - 10. 50
tr8439'0" 8.971 50"] -1
*[ ? 8.97123J 27
8440'0" 8.97013 J
Since 8.97123 - 10 is AT of the way from 8.97150 - 10 to 8.97013 - 10 we infer
that the angle A should be AV of the way from 84 39' to 84 40', that is,
A = 84 39' + fr x 60"
= 84 39' 12", to the nearest second.
To find that 12" is AV X 60" we could have used the Prop. Pts. table for 137.
In the column under that entry, the number next less than 27 is 22.8, which
corresponds to 10"; the other 4.2 in 27 corresponds to about one tenth of 20";
hence we infer that the total correction is 12".
Pages 21-24 of the Tables give special methods of interpolation
for angles near and 90.
EXERCISES
Find by use of five-place tables the values of the following:
1. log sin 22 37' 36". 7. log tan 71 19' 48".
2. log cos 32 51' 12". 8. log cot 79 17' 54".
3. log tan 42 36' 20". 9. log sin 85 25' 42".
4. log cot 51 22' 40". 10. log cos 87 21' 36".
6. log sin 55 18' 35". 11. log sin 113 12 ; 20".
6. log cos 62 47' 25". 12. log tan 209 5' 40".
By use of five-place tables find acute angles A satisfying the following equations:
13. log sin A = 9.55793 - 10. 17. log sin A = 9.90197 - 10.
14. log cos A = 9.95798 - 10. 18. log cos A - 9.76021 - 10.
16. log tan 4 = 9.52031 - 10. 19. log tan A = 0.14406.
16. log cot A = 0.46880. 20. log cot A = 8.82620 - 10.
88 TRIGONOMETRY
21. log sin A = 9.80923 - 10. 27. log tan A 0.27541.
22. log cos A = 9.86359 - 10. 28. log cot A - 9.57519 - 10.
23. log tan A - 9.11514 10. 29. log sec A = 0.04634.
24. log cot A - 0.75740. 30. log esc A - 0.32839.
25. log sin A = 9.99069 - 10. 31. log sec A = 1.15604.
26. log cos A = 9.54572 - 10. 32. log esc A - 0.14142.
CHAPTER VII
SOLUTION OF TRIANGLES. FIVE-PLACE LOGARITHMS
52. Right triangles. In Chapter II we solved right triangles by
making computations which involved multiplication and division.
We shall now illustrate how the computations may be carried out
by using logarithms.
It saves time, and tends to greater accuracy in computing, to out-
line the solution completely before referring to the Tables. The fol-
lowing steps are desirable:
I. Draw a figure from the data and estimate the unknown parts
of the triangle.
II. Write down all of the formulas which will be used in the solu-
tion and in the check.
III. Lay out the plan of computation, arranging for the placing of
each number which must be written in a position which insures
easy and accurate calculation. This must include the labeling
of each step so that a reader may follow without difficulty.
IV. Use the Tables to carry out the plan of Step III. If the check
is not satisfactory, go over each step to locate errors. If the
check is satisfactory, write down the computed values (of the
parts which were unknown) at the end of your computation.
We note that in Step III different plans are possible. Complete-
ness and neatness are important in avoiding errors and resultant re-
computations.
In the following Examples, the work of Step III is indicated in
bold type, and is done before any of the numbers are supplied.
Example 1. Solve the triangle, given C = 90, A - 49 45 ; 20", a = 734.40.
Solution.
Step I. /I Estimates
B - 40
6 = 600
b
FIG. 55 (1 cm. - 400)
90
TRIGONOMETRY
Step II.
B - 90
- A
Formulas
a cot A
a
log b = log a -f- log cot A
log c = log a log sin A
Check: 6* - c 2 - a 2
sin A
2 log b = log (c - a) + log (c + a)
Steps III and IV. Computation
90 = 89 59' 60" log a - 2.86593 log a = 12.86593 - 10
A = 49 45' 20" (+) log cot A = 9.92757 - 10 (-) log sin A = 9.88269 - 10
log b = 12.79350 - 10 log c = 2.98324
b - 621.59 c = 962.14
B - 40 14' 40"
Check
c - 962.14 log (c - a) = 2.35744
a = 734.40 log (c + a) = 3.22956
c - a - 227.74 sum = 5.58700 ]
c + a = 1696.54 2 log 6 - 5.58700 j
Answers: B - 40 14' 40", 6 = 621.59, c = 962.14.
Example 2. Given a = 0.23402, 6 = 0.31968, C = 90. Find A, B, c.
Solution. B
> Satisfactory
6 = .32
FIG. 56 (1 cm.
a = .23
C=90
0.15)
Estimates
A = 36
= 54
c = 0.39
B - 90 -
tan A = ^
q
sin A
Check: b = c sin J5
Formulas
log tan A = log a log 5
log c = log a log sin A
log b = log c 4- log sin B
log a
(-)logb
log tan A
A
B
Computation
19.36926 - 20 log a
9.50471 - 10
(-)logsinA
19.36926 - 20
9.77137 - 10
9.86455 - 10
36 12' 23"
53 47' 37"
Check
log c - 9.59789 - 10
(+ ) log sin B = 9.90682 - 10
sum = 19.50471 - 20
log c = 9.59789 - 10
c - 0.39618
log& - 9.50471 - luj
Answers (angles to nearest multiple of 5") :
A - 36 12' 25", B - 53 47' 35",
> Satisfactory check
0.39618.
SOLUTION OF TRIANGLES 91
Hereafter we shall give angles in answers to the nearest multiple
of 5".
A rule generally observed in such computation is to use formulas
involving the given data as far as possible, in preference to formulas
which express a part of the solution in terms of another part of the
solution. If we had followed this rule in Example 2, we should not
have used the formula c = - ~r which expresses the part c in terms
of the solution for A ; for, if there had been a mistake in obtaining A,
the solution c would then almost certainly be wrong. The formula
a 2 + f>2 = C 2 W ould have been the one to use, according to the rule;
this formula is not well adapted to logarithmic computation, how-
ever, and we have followed a method of solution here that uses
only logarithmic tables.
In both Examples we have followed the rule that a check formula
should be one not used in computing the solution, and should in-
volve at least two parts of the solution.
EXERCISES
Make an outline similar to those of pages 89-90 for the logarithmic solution of the
right triangles where the following parts are given (C = 90) :
I. A and a. 2. B and c. 3. 6 and c.
Solve, by use of five-place logarithms, the right triangles whose given parts are as
follows. Check the solutions. For each problem, prepare a complete outline, as
illustrated on pages 89-90, before using the Tables. In all these exercises, C = 90.
Write answers to the nearest multiple of 5".
4. A = 39 30' 10", a = 2670.5. 12. B = 18 38' 25", c = 0.073462.
6. B = 61 30' 50", 6 = 4320.4. 13. A = 72 14' 45", c = 7.2975.
6. A = 43 23' 15", a = 0.023570. 14. A = 84 22' 20", b = 0.027325.
7. B = 19 27' 35", 6 - 0.35244. 16. B 79 44' 40", c = 27.198.
8. a = 241.35, b = 415.05. 16. a = 1.7234, 6 = 3.2196.
9. c = 0.89236, b = 0.70254. 17. b = 271.42, c = 704.76.
10. a 0.023413, c = 0.035672. 18. a = 0.17239, 6 = 0.092460.
II. a = 0.042347, c = 0.062309. 19. c = 27335, 6 = 20298.
63. The problem of solving an oblique triangle. A triangle has
six parts, its three sides a, 6, c, and the three angles A y 5, C respec-
tively opposite those sides. In plane geometry we find that, if three
parts of a triangle are given, the triangle can be constructed in each
of the following Cases:
Case I. One side and two angles are given.
Case II. Two sides and an angle opposite one of them are given.
92 TRIGONOMETRY
Case III. Two sides and the included angle are given.
Case IV. Three sides are given.
We shall see later that in Case II there are sometimes two tri-
angles having given parts, but otherwise there is just one triangle.*
The problem of solving a triangle is that of finding by computation
the unknown parts. One well-known formula which we may use is
A + B + C = ISO .
Others will be derived in the following paragraphs as we prepare to
solve in succession Cases I, II, III, IV.
54. The law of sines. In Figure 57 we have drawn three triangles
j in each of which h is the perpendicular distance from C to
B A
FIG. 57
AB (produced if necessary). These illustrate respectively the cases
where angles A and B are both acute, A is acute and B obtuse, A
is obtuse and B acute. In all three cases we have
h . - h . D
r = sin A, - = sin B.
b ' a
It is readily verified that these formulas hold also when A or B is
a right angle.
From the above equations it follows by division that
fi\ a _ si* 1 A
b ~~ sin B
By dropping a perpendicular from B to AC we could prove that
(9^ a s * n ^ .
c sin C
* It is possible, however, to give values to three of the letters a, 6, c, A, B, C
which would be impossible for parts of a triangle (for example, a value of a greater
than the sum of 6 and c would be impossible for parts of a triangle since one side
cannot exceed the sum of the other two).
SOLUTION OF TRIANGLES 93
the proof would be as before, with an interchange of letters. Simi-
larly
** sin s
These three equations constitute the law of sines. They can be
expressed as the one continued equation
^ ' sin A = sin 5 = sin C*
We shall use the law of sines to solve triangles in Cases I and II,
but before doing so we shall derive check formulas which involve
all six parts of a triangle and can be used in all the Cases.
55. Check formulas involving all six parts. From formulas (2)
and (3) of the preceding section we obtain, by addition,
m a + b _ sin A + sin B
() c sinC
The right-hand side may be transformed by means of the identities
(2) sin A + sin B = 2 sin J(A + 5) cos i(A - 5), (by (5), p. 64),
= 2 sin 4(180 - C) cos 4(A - 5)
= 2 sin (90 - 4C) cos 4(A - B)
= 2 cos JC cos 4(A - B),
(3) sin C = 2 sin JC cos JC, (by (1), p. 60).
By substitution from (2) and (3), formula (1) becomes
a + b _ cos }(A - B)
w c sin 4C
Similarly
, , & + q _ cos (B - A)
W c " siniC
All six parts appear in formulas (4) * and (5); hence these formulas
are useful for checking solutions.
EXERCISES
Prove the following identities:
a -b = 2 sin $C sin j(A - B) 2 & = sin B
~ ' ' "
^
sin B ' ' "c 2 sin JC cos JC*
* Formula (4) is sometimes called Newton's formula. The pair consisting of
this formula and that of Exercise 3 of the following set are called Mollweide's
equations.
94
TRIGONOMETRY
a - b
sin
- B)
w * c cos JC
6. a = 6 cos C + c cos B.
a : b : c = sin A : sin ,
4.
sin (A
sin C
: sin C; A = 180 - (B + C).
66. Case I. Given two angles and one side. The following Ex-
ample illustrates the solution of a triangle in this Case. The Steps I,
II, III, IV described in 52 are again important. Step III is in-
dicated below by boldface type. Details of the computation were
carried out only after Step III was completed. It is recommended
that the student use a similar plan for exercises which follow.
32 24' 0", C = 63 17' 0*, c = 32.345. Find , a, b.
Example. Given A
Solution.
Estimates
B e 84 a == 19
Formulas
B = 180 - (A + C)
c sin A ,
Check:
sin C
fr + q
36
c sin B
sin C
>-A)
A = 32 24'
C = 6317'
C = 9541'
= 84 19'
A = 32 24'
logc= 1.50980
(+) log sin A - 9.72902-10
sum =11. 23882 -10
(-) log sin C = 9.95097-10
loga= 1.28785
a = 19. 402
FIG. 58 (1 cm. = 10)
logc= 1.50980
(+) log sin B = 9.99786-10
sum = 11. 50766 -10
( - ) log sin C = 9.95097 - 10
log 6= 1.55669
b = 36.032
Check
log cos KB- A) =9.95381 -10
D =0.23397
log(6+a)= 1.74377
(-)logc= 1.50980
diff . = />'= 0.23397
4C = 3138'30'
Since D = Z>', the work checks. We cannot always expect exact agreement.
Answers: B = 84 19' 0", a = 19.402, b = 36.032.
EXERCISES
Solve and check the triangles whose given parts are as follows,
logarithms. Write answers to the nearest multiple of 5".
Use five-place
1. A - 27 16' 10',
2. A = 44 20' 12*,
3. B = 62 27' 35",
4. B - 29 42' 25',
5. A - 82 24' 36',
B = 74 29' 40',
C = 55 42' 23',
C = 42 54' 45',
C = 63 18' 50',
C - 60 5' 24',
a = 16000.
c = 0.17285.
a - 24.182.
6 = 70.796.
a - 0.023478.
SOLUTION OF TRIANGLES
95
6. B = 79 33' 23', C = 55 31 ; 12", b - 0.072924.
7. A - 66 15' 35*, B - 52 17' 45*, a - 192.41.
8. A - 16 44' 50*, C - 82 3' 20*, 6 - 4.1292.
9. A = 112 32' 48*, B = 33 41' 12*, 6 - 527.37.
10. B = 45 37' 20*, C = 99 14' 20*, c - 129.62.
11. An observer at C on a hillside measures the angles of depression of two
points A and B in a horizontal plane below him. A and B are in the same direc-
tion from the observer and C, A, B are in the same vertical plane. The angle of
depression of A is 36 28' 30*, and that of B is 22 16' 0*. If the distance from A
to B is 4125.0 feet, find to five significant figures the distances of C from A and B.
12. For three points A, B, C, in a horizontal plane, the bearing of C from A is
N 34 27' 20* E, the bearing of C from B is S 72 40' 40* E, and the bearing of B
from A is N 15 24' 30* E. The distance from A to B is 2450.5 yards. Find to
five significant figures the distance from A to C.
67. Case II. Given two sides and an angle opposite one of them.
The following six figures illustrate the possibilities when values are
given to a, b, A as indicated:
,/-
C
FIG. 59
A < 90, a < b
a < b sin A
FIG. 60
A < 90, a < b
a = b sin A
FIG. 61
A < 90, a < b
a > b sin A
FIG. 62 FIG. 63 FIG. 64
A < 90, a ^ b A* 90, a ^ b A^ 90, a > 6
We see that solutions are determined as follows:
(a) No solution when the data are as stated under Figures 59, 63.
(b) One solution only when the data are as stated under Figures
60, 62, 64.
(c) Two solutions when the data are as stated under Figure 61.
On account of the possibility (c), Case II is sometimes called the
ambiguous case.
96
TRIGONOMETRY
When the data correspond to Figure 63, the decision that there
is no solution is immediate. If they correspond to Figure 62 or
Figure 64 we solve as illustrated in Example 3 of the Examples which
follow this paragraph. The subcases (where the data correspond to
Figure 59 or Figure 61) are illustrated by Examples 1 and 2. Note
that in these Examples we rely on the logarithmic computation to
determine whether conclusion (a), (b), or (c) holds. In Figure 60,
ABC is a right triangle.
Example 1. Given A - 47 13' 50', a - 0.20633, b - 0.70812. Find B, C, c.
Solution.
Estimate
No solution
Formula " J
. n b sin A
slnB = ___
(Only formula needed to show
that there is no solution)
'4714
FIG. 65 (1 cm. - 0.25)
Computation
log 6 = 9.85010 - 10
(+) log sin A = 9.86575 - 10
sum = 19.71585 - 20
(-) log a - 9.31456 - 10
log sin B - 10.40129 - 10
No such angle.
Example 2. Given A - 47 13' 35", a = 0.60631 ,
b - 0.70815. Find B, C, c.
Solution.
Two solutions:
Bi = 60
B 2 = 120
Estimates
6=0.
a-0.61
sin B =
b sin A
B 2 - 180 -
CTiecfc:
73
12
cos \
ci = 0.8
cj - 0.17
Formulas
Ci - 180 - (A
C 2 - 180 - (A
Kft - A) ft
FIG. 66 (1 cm. * 0.25)
p x a sin Ci
4 Dl} CL " sin A
, D x a sin C 2
+ ^ ft " isnr
> 4- a cos i(B 2 A)
sn
sin
SOLUTION OF TRIANGLES
97
Computation
A + Bi =
Ci =
log a =
(+ ) log sin Ci -
sum =
( ) log sin A =
log Ci =
log cos J(#i A) =
( ) log sin $Ci =
diff. = Di -
log 6 =
(+) log sin A =
sum =
(-) log en-
log sin B =
59 1' 8"
47 13' 35"
9.85013 - 10
9.86572 - 10
120 58' 52"
47 13' 35"
19.71585 - 20
9.78270 - 10
9.93315 - 10
(+) A -
A + J5 2 =
log a =
(+) log sin C 2 =
sum =
( ) log sin A =
106 14' 43"
73 45' 17"
9.78270 - 10
9.98230 - 10
168 12' 27"
11 47' 33"
9.78270 - 10
9.31041 - 10
19.76500 - 20
9.86572 - 10
19.09311 - 20
9.86572 - 10
9.89928 - 10 log & =
0.79302 cj -
Check for triangle A Bid
11 47' 33" &-
5 53' 46" a -
36 52' 38" 6 + a -
9.99769 - 10 log (b + a) -
9.77823 - 10 (-)logd -
9.22739 - 10
0.16881
0.70815
0.60631
1.31446
10.11875 - 10
9.89928 - 10
0.21946
diff. = Di' =
0.21947
Check for triangle AB 2 C 2
2 - A = 73 45' 17"
J(B 2 - A) = 36 52' 38"
iC 2 = 5 53' 46"
log cos 4( 2 - A) - 9.90305 - 10 log (6 + a) =* 10.11875 - 10
(-) log sin iC 2 - 9.01168 - 10 (-) log * 9.22739- 10
diff. - D 2 - 0.89137 diff. - Z> 2 ' = 0.89136
The checks are satisfactory.
Answers: B, - 59 1' 10", Ci = 73 45' 15",
B 2 = 120 58' 50", C 2 - 11 47' 35",
0.79302;
0.16881.
Example 3. Given A = 132 47' 20", a - 0.90635, b - 0.70810. Find C.
Solution. Since A > 90 and a > b there is but one solution (Figure 64).
We shall see how this fact presents itself in the computation of Ci and C 2 , following
the plan of Example 2.
9.85009 - 10 Bi - 34 59' 0" B 2 = 145 1' 0"
9.86561 - 10 A - 132 47' 20" A - 132 47' 20"
19.71570 - 20 A + B l - 167 46' 20" A + B 2 - 277 48' 20"
9.95729 - 10 Ci * 12 13' 40" C 2 is impossible.
logb
(f) log sin A
sum
(-)loga
log sin B
9.75841 - 10
98
TRIGONOMETRY
= 42 37' 0*,
a - 24.135,
6 - 20.270.
- 78 23' 30",
a - 0.37190,
6 - 0.34215.
- 95 41' 18",
a = 4594.3,
b = 7234.4.
= 107 55' 36",
a = 0.023566,
c = 0.045978.
= 40 5' 10",
a - 7003.2,
c = 8104.7.
- 42 11' 40",
a = 443.25,
6 = 502.76.
- 54 13' 54",
a = 0.50079,
6 = 0.19950.
24 25' 12",
6 = 125.34,
c - 267.85.
- 73 32' 20",
a = 2.3196,
6 = 3.7214.
62 18' 40",
b = 0.072381,
c = 0.10495.
- 65 23' 36",
6 - 32.504,
c - 30.333.
- 53 17' 25",
b = 7231.4,
c = 7952.7.
109 20' 20",
a = 11.305,
6 = 17.402.
= 98 15' 40",
6 = 2725.7,
c = 3010.3.
= 121 27' 24",
a = 0.49236,
c = 0.32584.
= 97 44' 48",
b = 6.2375,
c = 4.2191.
EXERCISES
Solve and check the triangles whose given parts are as fottows. Use five-place
tables. Write answers to the nearest multiple of 5".
1. A
2. A
3. B
4. C
6. A
6. A
7. A
8.
9. B
10. C
11. C
12. B
13. A
14. B
16. A
16. B
17. Three points A, B, C in a horizontal plane are so situated that the bearing
of B from A is N 17 30' 0" E and the bearing of C from A is S 24 20' 20" E. If
-the length of AB is 3210.5 yards and that of BC is 4715.0 yards, find the bearing
of B from C.
58. The law of tangents. In solving triangles under Case III we
shall need a formula derived from the law of sines
a _ sin A
b ~~ sin B
If we subtract 1 from each side of (1) we obtain
/rfcx a b sin A sin B
(2 > iSTB '
and if we add 1 to each side of (1) we have
(*\ a + b __ sin A + sin B
^ ' b sin B
If we divide the sides of equation (2) by the corresponding sides
of (3) we obtain
a b ___ sin A sin B
a + b sin A + sin B
From formulas (6) and (5) of 41 (p. 64), it follows that
a - b 2 cos %(A + B) sin (4 - B)
(1)
2 sin
+
cos
SOLUTION OF TRIANGLES
99
and hence
(4)
a - b
a+b
- B)
Formula (4) is known as the law of tangents. Since a and b are
any two sides of the triangle, we may state the law of tangents thus:
In any triangle the difference of any two sides is to their sum as the
tangent of one half the difference of the opposite angles is to the tangent
of one half their sum.
There are thus five other formulas similar to (4) ; for example,
b - a tan
- A)
b - c tan
- C)
'
b + a tan %(B + Ay b + c tan (J3 + C)
59. Case III. Given two sides and the included angle.
method of solving triangles in Case III by
using the law of tangents is illustrated in B
the following Example.
The
Example. Given b = 0.60733, c
A = 43 12' 30". Find a, B, C.
0.49188,
Solution.
a = 0.42
Estimates
B = 83
54
6.61
FIG. 67 (1 cm. = 0.2)
6 = 0.60733
c - 0.49188
6 - c = 0.11545
6 + c = 1.09921
A = 43 12
B + C = 136 47
B + C) = 68 23
B - C) = 14 51
7? = 83 15
C = 53 32
tun KB r\ b ~ C
tan 4(5 + (7)
h C) - 4(5 - (
9.06240 - 10
0.04108
o ~r c
KB - C) C = 4( -
Chrrk- & + a - cos KB - A)
c sin JC
Computation
J log (b - c) =
; (-)log (b + c) =
> diff. -
(+) log tan 4(5 + C) =
!' 30" log tan i(B - C) -
r/ 30^ 4(^ C)
I' 45" log b =
/ 15" (+) log sin A =
i' 0^ sum -
{'30" (-)logsin^ =
log a =
a =
9.02132 -
10.40229 -
10
10
9.42361 -
14 51' 15"
9.78342 -
9.83547 -
10
10
10
19.61889 -
9.99698 -
20
10
9.62191 -
0.41871
10
100
TRIGONOMETRY
log (b -fa) =
(-)logc
diff. - D -
0.60733
0.41871
1.02604
10.01117-10
9.69186-10
0.31931
B - 83 15' 0*
(-)A = 43 12' 30*
C = 53 32' 30*
iC = 2646'15*
B - A = 40 2' 30*
* - A) = 20 1' 15*
We note that D = /)'.
Answers: a - 0.41871,
logcosJ(B - A) - 9.97293-10
(-)logsinJC =9.65362-10
diff. - D' - 0.31931
= 83 15' 0*, C = 53 32' 30*.
EXERCISES
Solve and check the triangles whose given parts are as follows,
tables. Write answers to the nearest multiple of 5*.
Use five-place
1. a - 32.705,
2. a - 0.45235,
3. 6 - 1.7294,
4. 6 = 78542,
6. a - 5.0237,
6. a - 60235,
7. b - 0.071462,
8. 6 - 72.234,
9. a - 7.0252,
10. a - 782.36,
b = 54.803,
6 - 0.67413,
c - 2.7148,
c - 30314,
c = 4.7923,
c = 45921,
c - 0.042314,
c - 23.275,
6 - 2.3414,
b - 924.27,
C - 41 56' 20*.
C - 52 42' 40*.
A = 108 29' 30*.
A - 141 34' 50*.
J5 - 29 23' 48*.
B = 72 34' 24*.
A - 123 24' 15*.
A = 164 5 ; 20*.
C - 75 28' 40*.
C - 21 27' 50*.
11. Two sides of a parallelogram are 11.055 feet long, the other two sides are
13.267 feet long, and one interior angle is 72 15' 30*. Find the lengths of the
diagonals.
12. A surveyor runs a line N 35 30' 30* E from A to B, the length of AB
being 1246.5 feet. From B he runs a line S 25 14' 0* E to C, and measures
EC as 1729.6 feet long. How long is ACt
(D
(2)
(3)
60. The law of cosines. The law of cosines is given by the formulas
a* = 52 + c 2 - 26c cos A,
b 2 = a* + c* - 2ac cos B,
c 2 = a 2 + 6 2 - 2ab cos C.
We shall prove (1); an interchange of lettering in this proof would
establish (2) and (3).
C(x,y)
C(x,y) C(x,y)
FIG. 68
SOLUTION OF TRIANGLES 101
We draw coordinate axes so that the origin is at A, and AB is
on the positive x-axis. Three cases are illustrated in Figure 68.
In all cases, if (x, y) are the coordinates of C, we have
a 2 - i/ 2 + DB\
y* - fe 2 - * 2 ,
and hence
(4) a 2 - 6 2 - a* + DB\
The length of DB is either c - x or z - c, so that
Thus equation (4) becomes
(5) a 2 = 6 2 + c 2 -
Since in every case
x = 6 cos A,
formula (1) follows.
Note that, if A is a right angle, (1) becomes a 2 = 6 2 + c 2 .
When the three sides a, 6, c of a triangle are given (Case IV),
formulas (1), (2), and (3) enable us to compute the angles. How-
ever, these formulas are poorly adapted for logarithmic computation;
in the next section we derive other formulas better suited for this
purpose.
61. The half-angle formulas. Formula (1) of 60 may be written
+ c 2 - a 2
(1) cos A
e
1 cos A
2bc
Hence we have
2bc - b 2 - c 2 + a 2
26c
fl 2 - (b - c) 2
2bc
(a + 6 - c)(a - b + c)
2bc
Similarly
If we substitute these expressions in the formula ((6a), p. 60)
COS
102
we obtain
(2)
TRIGONOMETRY
tan^
(a - b + c)(a + b - c)
'
Let 2s be the perimeter of the triangle; then
a + b + c = 2s, a - b + c = 2s - 26,
b + c - a - 20 - 2a 9 a + 6 - c = 2s - 2c.
Formula (2) can now be written
tan
1^ _ ^2(s - 6) 2(8 - c)
2(s - o) 2s
a)s
-a
(s - a)(s - 6)(s - c)
or
(3)
where
(4)
Similarly
(5)
r =
- a)(* - >)( - c)
a+ 6+ c
We now show that r is the radius of the inscribed circle of the
triangle ABC. In Figure 69, the center of this circle, 0, is shown
as the intersection of the bisectors of the angles. We have
*C
tan
AF'
Since AF = AE, CE = CD, and
BF = BD, the perimeter will be
2s = 2AF + 2CD + 2BD
= 2AF + 2(CD + BD)
- = 2AF + 2a.
Hence AF = s - a, and therefore
OF = AFtm^A = (s - a)-
r.
SOLUTION OF TRIANGLES 103
62. Case IV. Given three sides. We carry out the logarithmic
computation with the aid of formulas (3), (4), and (5) of 61. We
check the final results by the formula
A + B + C = 180.
A check on a part of the computation comes from the relation
(s - a) + (s - b) + (s - c) = 3s - (a + b + c) = s.
Example. Given a = 250.32, b = 316.06, c = 287.42. Find A, B t C.
r*
Estimates
B = 72 C = 59
Formulas
2 _ (s - a)(s - 6)Qt - c) b
Solution.
A = 49
2s = a + 6 + c
tan^JB
s -b
a=250
tan^C
2 s c
Check: (s - a) + (s - 6) + (s - c) = *
Computation
c=290
FIG. 70 (1 cm. = 100)
A + B + C = 180
a = 250.32
b = 316.06
c = 287.42
2s = 853.80
s = 426.90
s a
s-b :
S C '
sum '
= 176.58
110.84
139.48
426.90
(check)
log (s - a) = 2.24694
(+) log (s - b) - 2.04470
(-f-)log(s-c) - 2.14451
sum = 6.43615
(-) logs =2.63033
logr
(-)log(s-a)
11.90291 - 10
2.24694
= 3.80582
logr = 11.90291-10
log tan \A = 9.65597- 10
JA = 2421'5r
logr = 11.90291 - 10
(-) log (s - 6) = 2.04470
logtanj= 9.85821-10
%B = 35 48' 31"
Answers: A = 48 43' 40", B
(-)log(s-c) -
log tan iC -
4C =
A = 48 43' 42"
B = 71 37' 2"
C= 59 39' 12"
Check: sum = 179 59' 56"
7137'0", C = 59 39' 10".
2.14451
9.75840-10
29 49' 36*
EXERCISES
Solve and check the triangles whose given parts are as follows. Use five-place
tables. Write answers to the nearest multiple of 5".
1. a - 144.82, b = 132.65, c = 163.87.
2. a = 0.15790, b = 0.14462, c = 0.08213.
3. a = 9.2407, b - 15.065, c = 12.847.
104
TRIGONOMETRY
b - 36758,
c - 43825.
b - 56.859,
c - 131.32.
b - 300.20,
c - 250.77.
6 - 0.23105,
c = 0.10097.
6 - 1629.9,
c * 1842.0.
b - 62.767,
c - 70.923.
& - 1007.4,
c = 1241.6.
6 = 0.061232,
c - 0.078944.
b - 0.091477,
c - 0.065231.
4. a = 31194,
6. a - 96.551,
6. a - 360.45,
7. a - 0.17042,
8. a - 3007.4,
9. a - 79.126,
10. a - 871.42,
11. a - 0.043256,
12. a - 0.079413,
For the triangles whose given sides are as follows find the three angles by using
the law of cosines ( 60) and the Table of Four-Place Values of Functions and
Radians (Tables, pages 4-8). Give solutions to as many significant figures as are
used for the given sides.
13. a - 20, 6 = 30, c = 40.
14. a = 50, 6 - 20, c = 60.
16. a - 6.20, b = 7.00, c 9.10.
16. a - 34.00, b - 65.00, c - 76.00.
63. Two formulas for the area of a triangle. Let S be the area
of triangle ABC (Fig. 71). Then, since
S = -5- > /i = 6 sin <A,
we have
(1) S = i&c sin 4.
This formula gives the area in terms of two sides and the included
angle.
Another formula, often proved in texts on plane geometry, ex-
presses S in terms of three sides as follows:
(2)
where
= Vs(s - a) (s - b)(s - c),
s - 4(0 + b + c).
Formula (2) is easily derived if we observe, from Figure 69, that S
is the sum of the areas of three triangles, each of which has altitude
r, and whose respective bases are a, 6, c. Hence
S = Jm + Jr6 + Jrc = Jr(a + b + c) = rs.
If we now substitute the value of r given by formula (4) of 61, we
have
S
(s a)(s b)(s c)
s
Vs(s - o)( - 6)( - c).
SOLUTION OF TRIANGLES
105
1. Prove the formula
<? - q2 si** B sin C
* " 2 sin A
EXERCISES
fr 2 sm C sin A
2 sin 5
c 2 sin A sin B
2 sin C
2. Prove that, if R is the radius of the circumscribed circle for triangle ABC,
then
= " A*
hence
sm A
3. Prove that, if S is the area of triangle ABC, and R is the radius of the
circumscribed circle,
~ abc
Find the areas, to five significant figures, of the triangles which have the following
given parts. Use the formula of Exercise 1 for triangles where one side and two
t are given.
3.1974,
4512.7,
0.40396,
90.265,
29 14' 10",
120 20' 20"
90.727,
4.2378,
4.
a
=
4.5625,
b
6.
a
=
1729.3,
b
6.
b
=
0.21230,
c
7.
a
=
70.827,
c
8.
b
=
342.50,
A
9.
c
=
47.231,
B
10.
a
=
75.040,
b
11.
b
=
6.3754,
c
c - 3.7286.
c - 3621.9.
72 24' 5".
42 32' 36".
C = 109 21' 45".
C = 40 25' 30".
11520'0".
97 44' 20".
A
B
B
B
MISCELLANEOUS EXERCISES
Solve and check solutions for the following
and writing answers to five significant figures;
given parts:
b - 431.01,
c = 6.3327,
A = 90,
B - 90,
c - 408.5,
c = 0.376,
c = 7.6604,
c = 2912.0,
B - 7 40' 15",
B - 112 40' 35",
B - 79 10',
A - 88 16' 10",
b - 415.75,
6 - 0.31783,
B - 62 13',
l.o- 327.04,
2. 6 = 3.6312,
3. 6 = 1.7325,
4. c = 245.03,
5. a - 607.4,
6. b - 0.4035,
7. b = 3.7125,
8. a - 724.02,
9. b - 0.2714,
10. c - 0.08535,
11. a - 476,
12. b = 4.0259,
13. a = 315.25,
14. a = 0.62451,
16. A - 74 25',
triangles, using five-place logarithms
or show that no triangle can have the
C = 27 30' 40".
A - 77 20' 35".
B - 60.
A - 30.
C - 41 13'.
C - 58 23' 20".
- 29 51' 5".
A - 15 20'.
A - 74 20' 45".
C - 42 10' 10".
C - 112 30'.
B - 95 42' 20".
c - 615.12.
c - 0.49116.
C - 51 14'.
106
TRIGONOMETRY
16.
A
=
25 10',
B
=
75 16',
C
=
75 16'.
17.
a
=
82435,
c
=
91257,
C
=
123 27'
15*.
18.
c
=
4243.2,
b
=
6012.4,
B
- 99 14' 20".
19.
a
=
237.01,
b
=
373.05,
c
=
625.18.
20.
a
=
4.0234,
b
=
7.9165,
c
=
3.80021.
21.
a
=
0.62455,
b
s=
0.31783,
c
s=
0.49112.
22.
a
=
0.71202,
b
=
0.41526,
c
=
0.42510.
23.
a
=
71.773,
c
=
68.442,
B
= 120 40' 20".
24.
b
=
28.414,
c
=
40.510,
A
=
112 16'
25".
25.
a
as
4.0727,
B
=
134 25',
C
=
22 17'.
26.
c
=
7.0909,
B
=
114 26',
C
=
31 5'.
Find to five significant figures the areas of the triangles having given parts as
follows:
27. 6 - 324.03,
28. 6 - 8.0845,
29. a = 16.241,
30. A = 18 20' 20*,
31. B = 75 35' 25*,
32. a = 0.08888,
c = 215.60,
c = 7.1727,
6 - 13.212,
B = 110 45' 25",
C = 62 12' 40*,
6 = 0.09999,
A = 129 15' 20"
B - 106 25' 10"
c - 27.427.
c = 36.202.
6 = 0.88250.
c = 0.06666.
Find the radii of the inscribed and of the circumscribed circles of the triangles
having given parts as follows:
33. a - 315.12,
34. a = 33.340,
35. b - 0.27150,
36. c = 0.085320,
37. b = 32.315,
38. a - 124.56,
39. a = 36.214,
40. b - 41972,
6 - 415.04, c
b - 29.816, c
A = 74 20' 10*, B
B - 11240'0*, C
c - 42.449,
c - 74.812,
c - 62.195,
c - 67537,
615.12.
45.728.
7 40' 40".
42 10' 55".
B = 47 8' 15".
A = 122 24' 35".
B = 44 37' 0".
A - 83 51' 20".
In the following problems, use jive-place logarithms to get answers to five significant
figures. Then write the answers to the number of significant figures appropriate for
the implied accuracy of the data.
41. The radius of a circle is 3.15 in. Find the angle subtended at the center
by a chord 1.74 in. long.
42. In a circle of radius 3.21 in., an angle of 29 30' is drawn with vertex
at the center. What is the length of the chord for the intercepted arc?
43. In a circle a chord of length 3.4 in. subtends an angle of 31 10' at the
center. Find the radius of the circle.
44. Two angles of a triangle are 71 10' and 65 30'; the radius of the in-
scribed circle is 5.17 in. Find the lengths of the sides of the triangle.
45. The sides of a triangle are in the proportion 3:4:5; the area of the tri-
angle is 108 sq. in. Find the radius of the inscribed circle.
46. An army officer observes the angle of elevation of an airplane to be 67 50',
its distance to be 2197 yards. How high is it above the horizontal plane of the
observer?
SOLUTION OF TRIANGLES 107
47. In tunneling under a river, a tunnel AB was first made at an angle of
depression of 12 30', then a horizontal tunnel BC 610 ft. long, then a tunnel CD
rising at an inclination of 12 30' , the points A and D lying in a horizontal plane.
If the maximum depth of the tunnel is 55 ft., how long is the tunnel and how
far apart are A and D? Assume that A, B, C, D lie in a vertical plane.
48. In going down in a mine, a man goes 576 ft. from A to B at a dip of 29 17',
then 481 ft. from B to C at a dip of 68 29', and finally 359 ft. from C to D at a
dip of 12 25'. How far is D below Al If the descent was always toward the
south, how far south of A is D? Assume that A, B, C, D lie in a vertical plane.
49. The latitude of the Dearborn Observatory is 42 3' 27.2" N. Assuming
that the earth is a sphere of radius 3959 miles, find the length of the circle of
latitude through this observatory. How far does the observatory move in a
minute due to the earth's rotation on its axis once in 24 hours?
60. A sailor travels from A to B going 13.25 miles N 28 E, and then from
B to C going 12.55 miles N 82 E, then from C to D going 14.35 miles S 15 E.
He wishes then to go from D to A by the shortest route. What is the distance
and the direction?
61. A flagpole 35 ft. tall stands on top of a corner of a building 54 ft. high.
What angle does the flagpole subtend at a point lying in the horizontal plane
at the foot of the building at a distance of 92 ft. from the building?
62. From a boat a tree on shore has a bearing of N 41 30' E. The boat travels
due north for 10 minutes at 9 mi. per hour; then the bearing of the tree is
N 72 40' E. How far east of the boat is the tree?
63. At a distance of 117 ft. from the foot of a building 63 ft. high a flagpole
on top of the front wall of the building subtends an angle of 23 30'. How tall is
the flagpole?
64. Two observers A and B are 1.834 miles apart, B being due north from A.
At a certain instant they observe an airplane in the northern sky; for A the
angle of elevation is 67 31', and for B it is 82 16'. How high is the airplane
above the plane of the observers A and B?
66. What is the radius of the largest gas tank that could be placed on a tri-
angular lot whose sides are 84.027 ft., 77.526 ft., and 102.473 ft. long respectively?
66. Two circles whose radii are 3.8224 in. and 2.1916 in. long intersect at an
angle of 18 35' 25". Find the distance between their centers and the length
of their common chord.
67. Two chords from a point A on a circle are of length 3.4115 in. and 2.1897 in.,
and the angle between them is 96 10' 35". Find the radius of the circle.
68. The angles of a triangle are 36 20' 20", 79 30' 40", and 64 10' 0"; the
radius of the circumscribed circle is 2.2534 in. long. Find the lengths of the
sides and the area of the triangle.
69. The hands of a clock are 2.250 ft. and 1.725 ft. long respectively. How
far apart are their tips when the time is 2 : 35?
60. Two sides of a^ triangle are 187.3 and 218.4, and the angle between them
is 151 18'. Find the lengths of the segments into which the opposite side is
divided by the bisector of this angle.
61. Two forces of 36.27 Ib. and 18.83 Ib. act at an angle of 32 30' with each
other. What is the resultant force in magnitude and direction?
62. A boat sailed 2317 yards N 30 E, from A to B, then turned to the right
108 TRIGONOMETRY
and sailed 1680 yards from B to C, then to the right again and sailed 1380 yards
from C to A, its starting point. What was the bearing of BC and of CA?
63. The distance from the sun, S, to the planet Venus, V, is 67,000,000 miles;
and from S to the earth, E t is 92,830,000 miles. At a certain time an observer
measured the angle SEV and found it to be 41 35'. How far was Venus from
the earth at that time?
64. The distance from the sun, S, to the earth, E, is 92,830,000 miles; and
from the sun to the planet Mars, M, is 141,500,000 miles. An astronomer
observes at a certain time that the angle SEM is 74 18'. How far is Mars from
the earth at that time?
65. A submarine is sailing N 48 20' E at the rate of 21 miles per hour from a
point A. A chaser is sailing N 31 30' E at the rate of 32 miles per hour from
a point B. The bearing of A from B is N 38 25' W and the distance AB is 9.35
miles. How far apart will they be after 18 minutes and what will then be the
bearing of the submarine from the chaser?
66. To find the height of a mountain peak, P t above a horizontal plane, two
observers A and /?, who were 2184 yards apart, made measurements as follows:
At A the angle of elevation of P was 4 18', the bearing of P was N 32 14' E,
and the bearing of B was S 82 10' E. At B the bearing of P was N 22 12' E.
How high was the mountain?
67. Two buoys, A and B, on a lake are known to be 1210 yards apart, and one
is due north of the other. An observer on a hilltop due north of the buoys ob-
serves with a range finder that the distance to A is 3240 yards and that the
distance to B is 4350 yards. What is the elevation of the observer, to the nearest
ten yards?
68. Two astronomers in the same longitude observe the zenith distance of
the center of the moon when it crosses their meridian. Their difference of lati-
tude is 92 14' 12"; the observed zenith distances are: A 44 54' 21." and
B - 48 42' 57*. Taking the earth's radius to be 3959 mi., find the distance
from earth to moon.
CHAPTER VIII
SPHERICAL TRIGONOMETRY
64. Spherical triangles. A great circle on the surface of a sphere
has certain properties that correspond to those of a straight line in a
plane. A great circle, it will be recalled, is the intersection of the
sphere with a plane through the center of the sphere. The poles of a
great circle are the points in which a line through the center, per-
pendicular to the plane of the great circle, cuts the sphere.
Two great circles intersect in two points that are diametrically
opposite. They divide the spherical surface into four parts, each of
which is called a lune.
The angle CAB between two arcs AC and AB of great circles is, by
definition, the angle T'AT between the tangents to these arcs at A
(Fig. 72). This angle also measures the
corresponding dihedral angle between the
planes of the great circles.
A spherical triangle ABC is a figure on
the spherical surface bounded by three arcs
of great circles. The planes of these great
circles intersect at 0, the center of the
sphere. The dihedral angles of the trihe-
dral angle 0-ABC (Fig. 72) are equal to
the corresponding angles of the spherical
triangle ABC, while the face angles A OB,
BOC, CO A measure the sides AB, BC, CA.
Consequently every theorem regarding spherical triangles corresponds
to a theorem concerning trihedral angles.
It has been noted that the angle -405 measures the side AB of the
spherical triangle ABC. The length of AB may be obtained by
writing formula (1) of 42 (p. 68) in the form s = rd (0 measured in
radians) when the angle AOB and the radius OA are known. It is
customary, instead of giving the actual length of a side AB of a
spherical triangle, to state the number of degrees, n, in angle AOB,
and to write AB = n.
If a spherical triangle has a side or an angle greater than 180, the
109
110 TRIGONOMETRY
great circles on which its sides lie will bound another spherical tri-
angle whose sides and angles are all less than 180. The sides and
angles of the former triangle are easily determined when those of the
latter are known. In case a side or an angle is equal to 180 the
triangle degenerates into a lune. Hence we shall hereafter consider
only spherical triangles each of whose angles and sides is less than 180.
For spherical triangles thus defined, the following two propositions
are proved in books on solid geometry:
The sum of the sides is less than 360.
The sum of the angles is greater than 180 (and, of course, less than
540 since each angle is less than 180).
66. Polar triangles. The vertices A, B, C of a spherical triangle
are the poles of three great circles which bound various other spherical
FIG. 73
triangles. Of these there is one, A'B'C ', where A is a pole of the circle
on which B'C' lies, or more briefly A is a pole of B'C' , B is a pole of
C'A', C is a pole of A'B' ', such that A and A' are on the same side of
BCj B and B' are on the same side of CM, and C and C" are on the same
side of AB. This triangle A'B'C' is the polar triangle of ABC
(Fig. 73).
It is proved in solid geometry that if A'B'C' is the polar triangle
of ABC, then ABC is the polar triangle of A'B'C'. Further, if the
sides opposite angles -A, 5, C of the triangle ABC are designated
a, 6, c, and those opposite angles A', 5', C' in triangle A'B'C' are
designated a', 6', c', respectively, then the following important rela-
tions are true:
A = 180 -a', B = 180 - &', C = 180 - c>,
a = 180 - A't b= 180 -5', c = 180 - C'.
SPHERICAL TRIGONOMETRY 111
Each angle of a spherical triangle is the supplement of the correspond*
ing side of the polar triangle; and each side is the supplement of the cor-
responding angle of the polar triangle.
EXERCISES
1. On a sphere of radius 2 ft. an arc AB 5 ft. long is drawn on a great circle,
What is the angular measure of the arc?
2. Under the general definition of a spherical triangle, we may have one whose
sides are a = 120, b = 120, c = 300. What are the lengths <n, h, Ci of the sides
of the smallest spherical triangle bounded by arcs of the same great circles?
3. Under the general definition of a spherical triangle we may have one whose
angles are A = 90, B = 90, C = 270. How does its area compare with that
of the whole spherical surface?
4. A man travels 791.8 miles on the earth's surface due north from a point
on the equator. What would his latitude be at the end of the journey if the
earth were a sphere of radius 3959 miles?
RIGHT SPHERICAL TRIANGLES
66. Formulas for right triangles. In Chapter II we obtained cer-
tain relations between the sides and angles of a plane right triangle
ABC with angle C equal to 90. We now use a similar notation for
a spherical right triangle ABC, where side a is opposite to angle A, b
is the side opposite to angle , and c is the side opposite to the right
angle C. Since A + B is greater than 90 for the spherical right
triangle, instead of being equal to 90 as in the plane, the situation
here is more complicated. We shall now list ten formulas, each of
which contains a different three of the five parts a, 6, c, A, B. The
following section gives a convenient memorizing device, and 68
contains proofs.
(1) sin a = sin c sin A y (2) sin a = tan b cot B,
(3) sin b = sin c sin 2?, (4) sin b = tan a cot A,
(5) cos A = cos a sin , (6) cos A = tan b cot c,
(7) cos B = cos b sin A, (8) cos B = tan a cot c,
(9) cos c = cos a cos 6, (10) cos c = cot A cot B.
67. Napier's rules. A device which makes it unnecessary to mem-
orize the ten preceding formulas is due to Napier, who also in-
vented logarithms. In Napier's two rules of circular parts for a
spherical triangle ABC with right angle at C, we call a, 6, 90 A,
90 - Cj 90 B the circular parts. The last three, which are com-
plements of A, Cj B, we designate A c , c c , B c . If the five circular parts
are marked on the triangle, or in the same order around a circle, as
112 TRIGONOMETRY
in Figure 74, each part has two others opposite to it and two adjacent
to it. Napier's rules are these:
1. The sine of any circular part is equal to the product of the cosines
of the opposite parts. (The "cos-opp" rule.)
2. The sine of any circular part is equal to the product of the tangents
of the adjacent parts. (The "tan-adj" rule.)
Of the ten formulas of the preceding section, those in the first
column can be written down by using the "cos-opp" rule, those in
FIG. 74
the second column by using the " tan-adj" rule. These rules do not
prove the formulas; they are merely a substitute for memorizing them.
To illustrate the use of the rules, let us obtain a formula which
expresses a function of A in terms of functions of b and c. The
corresponding circular parts are A c and the adjacent parts b and c c .
Rule 2 for these parts gives sin A c = tan b tan c c , and hence
sin (90 - A) = tan b tan (90 - c),
cos A = tan b cot c (formula (6), 66).
68. Proofs of the formulas for right triangles. In Figures 75a
and 75b the triangle ABC has a right angle at C, and OA, OB, OC
are radii of the sphere on which ABC lies. The point P may be taken
anywhere between and B on OB; the plane PRQ is perpendicular
to the line OA and intersects the planes OAB f OBC, OCA in lines
RP, QP, RQ respectively.
We shall now prove that each of the four faces of the pyramid
OPQR is a right triangle, the right angles being ORQ, ORP, OQP,
PQR. We recall that the plane PRQ is perpendicular to line OA at
point R, hence all lines in that plane that pass through R are per-
pendicular to OA, and angles ORQ and ORP are right angles. Fur-
ther, the plane PQR is perpendicular to the plane OA C since it is
perpendicular to the line OA of OAC, and the plane OBC is also
SPHERICAL TRIGONOMETRY
113
perpendicular to the plane OA C since C is a right angle. Hence
planes PQR and OBC have their line of intersection PQ perpendicular
to the plane OA C, and therefore to lines OC and QR\ in other words,
angles OQP and PQR are right angles. We have thus shown that
(1) Z.ORQ = LORP = ZOQP = /PQR - 90.
If the sides of triangle ABC and the angles they subtend at
are measured in degrees (or any other common unit), and if we ab-
B
Q
J' //~
R
A
FIG. 75a FIG. 75b
breviate by writing the symbol for an angle or side instead of the
measure of that angle or side, we have
(2) /.COB = a, /AOC = 6, LAOB = c.
Since RP and J?Q are perpendicular to OA, the angle QRP measures
a dihedral angle between the planes OAB and OAC\ hence it is equal
either to angle A, as in Figure 75a, or to 180 A, as in Figure 75b.
Let us first consider the case of triangles for which a < 90,
b < 90, as in Figure 75a. Here Q must fall between and C, and
R between and A. We have
(3) /.QRP = A, ZQOP = a,
Hence, in view of equations (1),
(4)
ZBOQ = 6, ZflOP - c.
sin A
QP
QP .
RP
sin a
RP
OP '
OP
sin c
RQ
RQ
RP
tan b
cos A
RP
~ OR '
OR
tan c*
tan A
QP
QP .
RQ
tana
RQ
OQ '
OQ
sin b'
OR
OR v
OQ
cose
~ OP
~ OQ X
OP
= cos b cos a.
These four formulas are equivalent to (1), (6), (4), and (9) of page 111.
114 TRIGONOMETRY
If, instead of drawing a plane through a point of OB perpendicular
to OA, we had drawn one through a point of OA perpendicular to
OB, we should have obtained formulas similar to the above, but in
which the pair A, a is interchanged with the pair B, b. Thus from the
first three equations (4) of this section we would obtain three equiv-
alent to (3), (8), (2) of page 111.
It remains to prove formulas (5), (7), (10) of page 111. If we
multiply the left sides of (3), (9), (6) of that set and equate the result
to the product of the right sides, we have
sin 6 cos c cos A = sin c sin B cos a cos 6 tan 6 cot c
= sin b cos c cos a sin ,
from which (5) follows (cos c is not zero when a < 90, b < 90).
We obtain (7) from (5) by interchanging A, a with B, b. Finally, we
derive (10) by multiplying the left sides of (2), (4), (9) and equating
the result to the product of the right sides.
Figure 75b illustrates the case where a > 90, b < 90. Here, in
equations (3) and (4) of the present section, we must replace A, a, c
by 180 - A, 180 - a, 180 - c, but the last four of the resulting
equations are equivalent to (4) ; hence the ten formulas of 66 still
hold. Similar remarks apply to cases where b > 90.
It remains only to consider cases where a or 6 is 90. Here the
angle opposite a 90 side is a right angle, arid the triangle ABC is
birectangular or trirectangular. The odd-numbered formulas of 66
are still true; others may be meaningless because tan 90 is impossible.
The formulas of 66 hold in all cases where the functions in them
have a meaning.
EXERCISES
1. In a birectangular spherical triangle, B = 90, C = 90, A ^ 90. Which
of the formulas of 66 are valid? Prove by means of these formulas that
6 _ c = 90, and a = A.
2. Prove by means of the formulas of 66 that in a trirectangular spherical
triangle the sides are each equal to 90. Prove the converse statement.
3. In a spherical triangle, b = 90, C = 90. Prove from the formulas of 66
that the triangle is birectangular or trirectangular.
4. In a spherical triangle, c = 90, C = 90. What conclusion follows from
each formula of 66 that is not meaningless?
6. Prove formulas (4) of 68 for the case a > 90, b < 90, illustrated in
Figure 75b.
6. Draw a figure and prove formulas (4) of 68 for the case a > 90, b > 90.
7. By using the relations of 65 between the parts of a spherical right triangle
SPHERICAL TRIGONOMETRY 115
ABC and its polar triangle A'B'C', show that c' = 90, and obtain from the formu-
las of 66 ten formulas for a spherical triangle A'B'C' in which c' = 90.
8. In a right spherical triangle on the earth's surface, A *= 30, and the
length of c is 100 miles. If the earth's radius, to two significant figures, is 4000
miles, find the length of a, in miles, from formula (1) of 66 and compare it with
the length obtained by treating the triangle as a plane triangle. Make a similar
comparison when c = 1000 miles.
9. By use of Napier's rules, obtain formula (7) of 66.
10. By use of Napier's rules, obtain formula (10) of 66.
69. Rules regarding the relative magnitude of parts. In solving
right spherical triangles certain propositions are convenient for the
purpose of checking results. We quote first from solid geometry
two theorems that are true for all spherical triangles, whether right-
angled or not:
1. The sum of any two sides of a spherical triangle is greater than
the third side.
2. // two angles of a spherical triangle are equal, the opposite sides
are equal, and conversely. If two angles are unequal, the sides opposite
are unequal, and the greater side is opposite the greater angle. The
converse is also true.
In addition we have the two so-called rules for species, their name
arising from the fact that they are concerned with the question of
whether sides or angles are of the same or different species, that is,
whether they terminate in the same quadrant or in different quad-
rants. We shall first state these rules, then prove them.
3. In a right spherical triangle ABC with right angle at C, the angle A
is of the same species as side a (both terminate in the same quadrant) ;
and B is of the same species as b.
4. In a right spherical triangle ABC with right angle at C, if c < 90,
then a and b are of the same species; if c > 90, then a and b are of
different species.
The proof of rule 3 follows from formulas (5) and (7) of 66. Thus
by formula (5), since sin B is positive, cos A and cos a cannot be of
opposite sign, hence A and a cannot terminate in different quadrants.
Rule 4 follows from formula (9) of 66; for if c < 90, then cos c is
positive and cos a and cos b are of like sign, so that a and b are of the
same species; if c > 90, then cos c is negative, cos a and cos & are of
unlike sign, and a and b are of different species.
116 TRIGONOMETRY
70. Solution of right spherical triangles. If values for any two
parts other than the right angle are given, we use equations (1)-(10)
of 66 to compute the three unknown parts. There may be no solu-
tion (for instance, when a rule of 69 is violated), just one set of
solutions, or just two sets of solutions (this can occur only when the
given parts are a and A, or b and B). To each possible set of solu-
tions of the equations an actual triangle corresponds, and the only
possible triangles with the given parts correspond to solutions of the
equations.
The student should prepare a complete outline of the computation
before turning to the Tables.
To check solutions we use a formula involving the three parts
that were not given.
In each of the following Examples, we shall, by use of Napier's
rules, find the three formulas needed, each of which contains one un-
known and two known parts. If a formula contains a part greater
than 90, we shall modify the formula so that it applies to the supple-
ment of that part. This is done because our tables do not extend
beyond 90 and because we must avoid minus quantities when using
logarithms (negative numbers have no real logarithms). If 6 is a part,
a convenient notation for its supplement is 6,;
6, = 180 - 9
(just as, with Napier's rules, we used 6 C for the complement of 6).
From the formulas of page 39, we have
sin 8 = sin 0, cos 0, = cos 8, tan 6 8 = tan 0,
cot 0, = - cot 6.
As an example, suppose given parts are 6 and A, both greater than
90, so that we are to solve for a, c, B by using formulas (4), (6), and
(7) of 66, so modified as to contain b a and A 8 and to avoid minus
signs. Formula (4) is
sin b = tan a cot A y
and, in terms of & -4,, this becomes
sin b a = tan a (- cot A s ).
To avoid the minus sign, we substitute tan a, for tan a; this gives
us
sin b a = tan a s cot A,.
SPHERICAL TRIGONOMETRY 117
Similarly, we obtain from (6) and (7)
cos A 8 = tan b a cot c,
cos B 9 cos 6, sin A 8 .
A useful rule is the following:
In any of the formulas of 66 we can without change of sign replace
the sine of a part by the sine of the supplement of that part; and we can
replace by their supplements any two parts for each of which the cosine,
tangent, or cotangent appears in the formula.
In the following examples, the 10 that should follow logarithms
of functions will be omitted, for brevity.
Example 1. Given c - 72 10' 45", B - 30 43' 0". Find a, b, A.
Solution.
Formulas
sin B c "* tan a tan c c sin 6 = cos c c cos B e sin c c = tan A c tan B t
cos B = tan a cot c sin 6 = sin c sin B cos c = cot A cot B
tan a = tan c cos B cot A = cos c tan B
Check: sin 6 = tan a tan A c
sin b = tan a cot A
Computation
log tan c = 10.49286 log sin c = 9.97865 log cos c = 9.48578
(+) log cos ^9.93435 (+) log sin B - 9.70824 (+) log tan B = 9.77390
log tan a = 10.42721 log sin 6 = 9.68689 log cot A = 9.25968
o = 69 29' 52" b = 29 5' 47" A - 79 41 ; 39"
C/iecfc
log tan a = 10.42721
(+) log cot A = 9.25968
log sin 6 = 9.68689 sum = 9.68689
Answers (to the nearest multiple of 5") :
a = 69 29' 50", b = 29 5' 45", A = 79 41' 40".
Here the equation for log sin b would also admit the additional solution
6 = 180 29 5' 47", but this must be rejected, since it is not of the same
species as B (Rule 3, 69).
The check we have used tests only the correctness of the logarithms involved.
We might have used the following modified form of our check formula:
sin b cot a tan A 1.
118
TRIGONOMETRY
Example 2. Given a = 156 3' 0", A = 154 44' 20" (an example of an
"ambiguous case " where there are two sets of solutions). Find 6, c, B.
Solution.
Formulas
sin a = cos c e cos A e
sin a = sin c sin A
sin c 1 __ sin a
sin c
sin & = tan a tan A,
sin b tan a cot A
sin
sin
sin &j \ _
sin 64 /
, . ,
tan a, cot A
sin A 8
Check: sin 6 = cos c c cos .
sin b = sin c sin
sin A c = cos a cos B c
cos A = cos a sin J5
sm B cos
sin
sin /? 1
sin B j
cos a,
Cora/wJato'on
a. - 23 57' 0* A, = 25 15' 40"
log tan a. - 9.64756 log sin a, = 9.60846 log cos A . = 9.95635
(+) log cot A, - 10.32618 ( - ) log sin A, - 9.63017 ( - ) log cos a, = 9.96090
log sin 6 = 9.97374 log sin c = 9.97829 log sin B = 9.99545"
6 = 70 16' 30" c= 72 2' 0" B = Sl^S'O"
6, = 109 43' 30" c = 107 58' 0" B 8 = 98 17' 0"
The rules for species allow two sets of solutions. We shall not check here.
Answers: &i = 70 16' 30", Ci 107 58' 0", Bi = 81 43' 0";
62 - 109 43' 30", c 2 = 72 2' 0", 2 - 98 17' 0".
EXERCISES
Obtain by using Napier's rules formulas for the solution of the right spherical
triangles of which the following are given parts, using supplements of parts greater
than 90; write a complete outline for logarithmic computation, but do not solve:
1.
2.
3.
4.
5.
a - 40,
a = 40,
6 = 70,
a =160,
a - 100,
b - 25.
c - 120
c 106
A = 150
B = 50.
6.
7.
8.
9.
6 = 70,
b = 75,
c = 80,
c = 120,
10. A = 100
A = 135.
B = 85.
A = 110.
B = 135.
B = 100.
Solve the right spherical triangles of which
each set of answers. Find both sets of answers
to the nearest multiple of 5".
11. c - 56 55' 0", A = 18 43' 30".
c - 102 2' 30", B = 6551'0".
35 40' 15", A - 35 40' 15".
127 17' 45", B 123 40' 15".
143 50' 20", B = 73 7' 40".
134 31' 40", 6 = 40 42' 20".
43 18' 30", c = 47 34' 30".
44 29' 0", c = 109 52' 30".
113 22' 15", A = 1929'0".
20. a - 130 28' 30", B 165 I'D*.
the following parts are given. Check
in case there are two. Write answers
12.
13. a
14. b
15. A
16. a
17. a
18. 6
19. b
21.
22.
23.
24.
26.
26.
27.
28.
29.
30.
6
a
b
a
a
A
b
a
c
c
= 14 59' 35",
102 35' 25",
= 47 18' 45",
= 48 54' 55",
= 72 49' 50",
= 77 21' 50",
= 140 16' 5",
- 67 2' 50",
= 36 21 '35",
A
B
c
c
b
B
B
A
B
A
=5
S=S
SB
75
112
95
50
121
125
83
101
110
58
21 '55".
13' 50"
48' 30".
2' 55".
23'35".
13'25"
56' 40".
59' 20".
39' 40".
59' 40".
SPHERICAL TRIGONOMETRY 119
71. Isosceles triangles. Quadrantal triangles. Rule 2 of 69
states that the base angles of an isosceles triangle are equal. The
great circle joining the vertex to the mid-point of the base bisects the
angle at the vertex and divides the isosceles triangle into two sym-
metrical right triangles. The solution of isosceles triangles can there-
fore be made to depend upon that of right triangles. The given parts
may be any two of the following four: base, base angle, side, angle
at vertex.
A quadrantal triangle is one of which a side is equal to 90. If this
side is lettered c then the polar triangle will be right-angled at C',
since C' = 180 c = c 8j according to a relation given in 65
(p. 110). If two parts other than c are given, their supplements are
parts of the polar triangle. When we have solved the latter triangle,
the supplements of sides or angles found will be angles or sides of the
original triangle. Thus if c, a, B are given parts, with c = 90, we
obtain for the polar triangle
C' = 90, A' = a,, 6' = B 9 .
We should then solve for c', a', 5', and finally obtain C, A, b from
the relations
C = c/, A = a/, 6 = B. 9 .
EXERCISES
Solve the oblique spherical triangles of which the following are given parts. Note
that Exercises 4, 8, 11, 15 have two sets of solutions. Write answers to the nearest
multiple of 5".
1. a = b = 47 32' 0", c = 52 50' 0".
2. a = 6 = 69 54' 0", C = 54 24' 0".
3. c = 86 22' 30", A = B = 167 22' 30".
4. c = 153 32' 30", C = 166 30' 30", a = 6.
5. c = 144 43' 20", A = B = 28 49' 55".
6. a = b = 117 3' 0", A = 106 34' 10".
7. A = B = 100 10' 55", C = 60 46' 10".
8. c = 44 30' 20", C = 54 57' 15", a = 6.
9. c = 90, a = 53 7' 0", 6 = 132 26' 0".
10. c = 90, a = 40 56' 0", C = 116 13' 0".
11. c = 90, b = 96 45' 30", B = 103 14' 30".
12. c = 90, a = 142 4' 40", B = 22 48' 20".
13. c = 90, a = 149 8' 45", b = 108 24' 0".
14. c = 90, A = 53 14' 5", C = 125 47' 50".
15. c = 90, a = 96 3' 20", A = 102 38' 20".
16. c = 90, A = 48 39' 15", B = 120 31' 35".
120 TRIGONOMETRY
OBLIQUE SPHERICAL TRIANGLES
72. The law of sines. There are laws of sines and of cosines for
spherical triangles; the reader should compare their form and their
derivation with those of the corresponding laws for plane triangles
(see 54, pp. 92-93, and 60, pp. 100-101).
The law of sines may be stated as follows:
In every spherical triangle the sines of the sides are proportional to
the sines of the corresponding opposite angles. This may be written
m sin a _ sin b _ sin c
U; sin A ~ sin B ~" sin C
The proof is analogous to that of 54. We here draw the great
circle arc CD, whose angular measure is h, perpendicular to AB.
Figure 76a illustrates the case where D is between A and B, Figure
76b a case where D is outside AB.
In both cases, formula (1) of 66 applied to the right triangles
ADC and BDC gives *
(2) sin h = sin 6 sin A, sin h = sin a sin B.
Hence
sin a sin B = sin 6 sin A,
or
(t ^ sin a _ sin b
sin A sin B
If D coincides with A or B, the triangle ABC is a right triangle, and
formula (3) here corresponds to either formula (1) or (3) of 66.
We can show similarly the equality of the last two expressions
in (1), and thus complete the proof of that formula.
* For Figure 76b, the angle at B in BDC is not B, but B,. However, since
sin B, = sin B, the second of formulas (2) is still true.
SPHERICAL TRIGONOMETRY 121
73. The laws of cosines. The first law of cosines is given by the
formulas
(1) cos a = cos b cos c + sin b sin c cos A,
(2) cos b = cos c cos a + sin c sin a cos B,
(3) cos c = cos a cos b + sin a sin b cos C.
The second law of cosines is given by the formulas
(4) cos A = cos B cos C + sin B sin C cos a,
(5) cos B = cos C cos 4 + sin C sin A cos 6,
(6) cos C = cos A cos B + sin 4 sin B cos c.
To prove formula (1) we again make use of Figures 76a, 76b, and
begin by applying formula (9) of 66 to the two right triangles
BDC&ndADC. This gives
(7) cos a = cos h cos DB,
(8) cos 6 cos h cos AD.
In Figure 76a, DB is c - AZ); and in Figure 76b, DB is AD - c.
Since cos 6 = cos ( 0), we have in both cases
(9) cos DB = cos (c - AD}.
If we solve (8) for cos h and substitute the result in (7), we have
cos b nD
cos a = - -TK cos DB'
cos AD '
and if we use (9) this becomes
r cos (c AZ>) , cos c cos -4Z) + sin c sin
cos a = cos 6 - - -pp: - = cos b
- -pp: - - j^r - >
cos -AD cos AD
hence
(10) cos a = cos b cos c + cos b sin c tan AD.
Formula (6) of 66, applied to the right triangle ADC, gives
cos A = tan AD cot b
or
tan AD = tan 6 cos A ;
hence for the last term in (10) we have
cos b sin c tan AD = cos b sin c tan 6 cos A = sin b sin c cos A,
and with this substitution (10) reduces to (1).
The proof of formulas (2) and (3) is similar. The proofs of the
three formulas must be modified if ABC is a right triangle or if
AD ** 90, but the formulas remain true in all cases.
122 TRIGONOMETRY
Formula (4) is proved by reference to the polar triangle A'B'C'
and use of the relations
a! = 180 - A = A 8 , b f = B 9J c' = C,, A 1 - a,.
Thus we have
cos a! = cos V cos c' + sin V sin c' cos A',
cos A, = cos B 9 cos C 9 + sin fi, sin C 8 cos a,,
- cos A = (- cos B)(- cos C) + sin B sin C (- cos a),
cos A = cos 5 cos C + sin B sin C cos a.
The proofs for formulas (5) and (6) are similar.
EXERCISES
1. Show that if havers 6 has the meaning given it on page 8, the relation
havers a = havers (6 c) + sin 6 sin c havers A
is equivalent to formula (1) of 73.
2. Show that if havers has the meaning given it on page 8, the relation
havers A, = havers (B + C) sin B sin C havers a
is equivalent to formula (4) of 73.
Solve the following problems without using logarithms:
3.
Given
a =
40,
A
=
30,
c =
25,
find
C.
4.
Given
6 =
50,
B
=
60,
a
45,
find A.
5.
Given
A -
30,
C
=
130,
c =
70,
find
a.
6.
Given
B =
135,
C
=
65,
b =
150,
find
c.
7.
Given
b -
45,
c
=
55,
A =
50,
find
a.
8.
Given
a =s
120,
b
=
70,
C -
40,
find
c.
9.
Given
5
85,
C
=
65,
a =
60,
find
A.
10.
Given
4 =
125,
C
=
110,
6 =
120,
find
B.
11.
Given
a =
45,
b
=
55,
C -
55,
find
c, A,
B.
12.
Given
A =
60,
B
=
95,
c =
130,
find
C, a,
b.
74. Formulas for half-angles and half-sides. For the half-angle
formulas of spherical trigonometry we shall use notations similar to
those of 61 for plane triangles. Thus we shall write
(1) 2s = a + b + c,
sn 5
Then the half-angle formulas are
/o\ A 4 r B r C r
(3) ten 2 = sin (-,)' tan 2 = S in(s-ft)' tan 2 = sin (5 - c)'
SPHERICAL TRIGONOMETRY 123
It will suffice to prove the first of these formulas. We begin by
solving formula (1) of the preceding section for cos A as follows:
A cos a cos b cos c
COS A = : T :
sin 6 sin c
We shall substitute this value of cos A in formula (6a) of 40, using
the positive sign before the radical, since A < 180. Thus we have
_ /I cos A
where
- A n cos a cos b cos c
I cos A = 1 : r :
sin b sm c
_ cos fr CQS c + sin b sin c cos a
sin 6 sin c
- cos (b c) cos a
sin 6 sin c
Formula (8) of page 64 gives
(by formula (4), p. 52).
J "
,, ^ . c + a.> c a
cos (o c) cos a = 2 sin - ~ - sin - ~ -
4 Z
rt . a + Z> c . a b + c
= 2 sm - s - sin - -
And from formula (1) of this section we have
a + b c a b + c
- 2 - = S ~ C ' - 2 - = S ~
Hence we may write
/r\ - A 2 sin (s c) sin (s b)
(5) 1 cos A = -- - . ~. - - -
sin b sin c
Similarly
//1N . . A 2 sin s sin (s a)
(6) 1 + COS A = - : - r TT - -
sin 6 sin c
From (4), (5), and (6) we have
(7) tan 4 = i/ sin (S ~ 6) f n (S 7 c)
2 V sin s sin (s a)
sin (s a) sin (s b) sin (s c)
sin s sin 2 (s a)
which reduces to the first of formulas (3).
124 TRIGONOMETRY
In the half-side formulas we use the notations
(8) 2S = A + B + C,
R _ /cos (S
\
- A) cos (S - B) cos (S - "
COS S
The half-side formulas are
a R h R c R
(10) C0t 2 = cos(S-X)' C0t 2 = coB(S-*)' C0t 2 = cos(S- C)'
We prove the first of formulas (10), for example, by applying the
first of formulas (3) to the polar triangle A'B'C', which gives us the
formula
A' r f
(11) tan^-^-y-f TT-
v ' 2 sin (s' - a )
We must now write this in terms of the sides and angles of triangle
ABC.
We have
(12) s' = i(o' + V + c') = 4(A. + B 9 + C,)
= |[540 - (A + B + C)] - 270 - S.
Similarly
s' - c' = 90 - (5 - C).
By definition,
'
4 / s ^
V
sin s
formulas (12) and (13) reduce (14) to the form
(15) r' = R.
We complete the reduction of (11) to the first of formulas (10) by
noting that
sin (s' - a') - sin [90 - (S - A)] = cos (S - A),
tan y = tan ~ = tan ^90 - ^J = cot |-
75. Napier's analogies.* In this section we shall obtain formulas
involving five of the six parts of a spherical triangle. The following
four Napier's analogies are typical; others may be obtained from
them by interchange of letters.
* Analogy is here used in an obsolete sense, meaning proportion.
SPHERICAL TRIGONOMETRY 125
sinl(A-B) _ tanj(g- b)
sin i(4 + B) " tan c
cos K* - 1?) _
cos }(4 + B) ~ tan *c
sin j( tf - _ ta
sin J( + *) ~ cot iC
... cos j(g - 6) _ tanjQi + g).
w cos (<* + W " cot K
We shall prove formula (1) as a consequence of the first two of
formulas (3) of 74. From these we have
tan %A tan \B ___ sin (s a) sin (s b)
tan \A + tan %B ~ r r
sin (s a) sin (5 6)
sn sn
_
cos |A cos ^g = sin (s b) - sin (s - a)
sin ^4 sin ff ~~ sin (s - 6) + sin (s - a) '
cos cos
sin %A cos ^j5 cos %A sin ^g _ 2 cos %(2s a 6) sin ^(a b)
sin iA cos JB + cos J4 sin JS ~ 2 sin J(2s - a - 6) cos (a - 6) '
sn __ cos c sn a
sin (%A + %E) " sin \c cos i(a - 6) '
sin \(A - B) ^ tan ^(o - 6)
sin \(A + B) ~ tan ^c
We obtain the second of Napier's analogies by transforming the
first. This is done by means of an auxiliary formula as follows:
From the law of sines ( 72) we easily deduce the relation
sin A sin B _ sin a sin b
sin A + sin B ~~ sin a + sin b
To both sides of this equation we apply formulas of page 64, thus
obtaining
2 cos %(A + B) sin %(A - B) = 2 cos (a + b) sin (a - b)
2 sin J(A + B) cos i(A - J5) 2 sin i(a + 6) cos i(a - 6)
tan %(a b)
~" tan f (a + 6) "
126 TRIGONOMETRY
From this we deduce the desired auxiliary formula
sin %(A - E) cos %(A - E) tan (g - b)
'
sin (A +B)~ cos (A + E) tan (a + b) '
If we replace the left member of (1) by the right member of this last
equation, the result is easily reduced to formula (2).
We obtain formulas (3) and (4) by applying (1) and (2) to the
polar triangle A'B'C'. Thus formula (1) for A'B'C' is
sin Q4' - B') tan %(a' - b 1 )
sin i(A' + B') tan %c'
from which we deduce
sin JKa, - b 8 ) _ tan %(A S - B s )
sin J(a. + b 8 ) ~~ tan \C
sin ^(6 - a) = tan \ (B - A)
sin J(a + b) cot JC
This may be at once reduced to formula (3), since
sin ^(6 a) = sin J(a 6),
tan l(B - A) = - tan (A - E).
Formula (4) may be obtained in a similar manner.
76. Solution of oblique spherical triangles. Cases may be listed
according to the given parts as follows:
I. Given three sides.
II. Given three angles.
III. Given two sides and the included angle.
IV. Given two angles and the included side.
V. Given two sides and an angle opposite one of them.
VI. Given two angles and a side opposite one of them.
In the following sections we shall explain the solution under each
case by logarithmic computation. For Cases I and II we shall use
the half-angle and half-side formulas ( 74) ; for Cases III and IV,
Napier's analogies ( 75) ; for Cases V and VI, the law of sines ( 72)
and Napier's analogies.
Here, as in solving right triangles, we omit the number 10 that
should follow certain logarithms. We keep results to the nearest
second in computations, but give answers to the nearest multiple of
five seconds.
SPHERICAL TRIGONOMETRY
127
The student should prepare a complete outline of the computation
before using the Tables.
77. Case I. Given three sides. We here use the half-angle
formulas of 74, and check by the law of sines, or one of Napier's
analogies that involves all three angles. We replace sin by sin 0,
when 6 is greater than 90.
125 10' 0", 6 - 110 22' 15*, c - 30 48' 45*. Find
Example. Given a
A,B,C.
Solution.
tan -
2s = a + 6 + c
r
Formulas
sin (s a) sin (s 6) sin (s c)
sin (s a)
r 2 =
. B r
tan 2 = sin (s - 6)
tan
sin (s c)
Check:
sin A sin B sin C
sin a
sin 6
sine
a = 125 10' 0"
6 = 110 22' 15"
c = 30 48' 45"
2s - 266 21' 0"
s = 133 10' 30"
s.= 46 49' 30"
Computation
s - a = 8 0' 30"
s - 6 = 22 48' 15"
s -c = 102 21' 45"
sum
133 10' 30"
(Check: sum = s)
(s - c). = 77 38' 15"
log r = 9.42964
(-) log sin (s - a) = 9.14401
10.28563
log tan
^ = 62 36' 48"
A = 125 13' 36"
logr
(-) log sin (s - c),
C
log sin (s - a) = 9.14400
(+) log sin (s - b) = 9.58837
(+) log sin (s - c). = 9.98981
sum = 28.72218
(-) log sins. - 9.86289
log r 2 = 2 log r - 18.85929
log r - 9.42964
(-) log sin (s - 6) = 9.58836
log tan -s = 9.84128
^ - 34 45' 20"
B - 69 30' 40"
= 9.42964
= 9.98981
log tan ^ = 9.43983
~ - 15 23' 35"
C = 30 47' 10"
Check
log sin A. = 9.91216 log sin B = 9.97162 log sin C - 9.70913
(-) log sin a, = 9.91248 (-) log sin 5, = 9.97195 (-) log sin c = 9.70947
9.99968 9.99967 9.99966
Answers: A = 125 13' 35", B = 69 30' 40", C = 30 47' 10".
128 TRIGONOMETRY
78. Case II. Given three angles. We here use the half-side
formulas of 74 and check by the law of sines or one of Napier's
analogies that involves all three sides. It can be shown that S 8 ,
S - A, S - B, S - C are all between - 90 and + 90; hence we
write
P2 _ cos (S - 4) cos (S - B) cos (S - C)
K c* >
COS O,
and in using this formula remember that cos ( 6) = cos 0.
Example. Given A - 144 20' 20*, B - 47 24' 35*, C - 56 35' 45*. Find
a, b, c.
Solution.
Formulas
D* cs (S - A) cos (S - B) cos (S - C)
- A + B + i
cos S,
, a ft 6 ft c ft
Cot 2-cosOS-A) Cot 2 " cos (S - B) Cot 2 " cos (S - C)
r , ,. cos j(A ~ B) tan [$(a -f 6)1
CA16C/C. r -, , . ; s^-- =! 7 :
cos [i (A + B)j. tan Jc
A - 144 20' 20" S - A = - 20 10' 0* log cos (S - A) - 9.97252
B - 47 24' 35" S - B - 76 45' 45* (+) log cos (S - B) - 9.35982
C - 56 35' 45* S - C - 67 34' 35* (+) log cos (S - C) - 9.58144
2S - 248 20' 40* sum - 124 10' 20* sum - 28.91378
S - 124 10' 20* (CTiedfe: sum -5) (-) log cos S, = 9.74949
& - 55 49' 40* log /? - 2 log ft - 19.16429
log ft - 9.58214 log ft - 9.58214
(-) log cos OS - A) - 9.97252 (-) log cos (S - B) - 9.35982
log cot Ja = 9.60962 log cot Jb - 10.22232
ia - 67 51' 8* #> = 30 56' 11*
a - 135 42' 16* b = 61 52' 22*
log ft = 9.58214
(-) log cos (S - C) = 9.58144
log cot Jc - 10.00070
Jc - 44 57' 14*
c = 89 54' 28*
Check
A - B - 96 55' 45* a + b 197 34' 38"
i(4 - B) - 48 27' 52* f(a + 6) - 98 47' 19"
A + B - 191 44' 55" [i(a + 6)1 - 81 12' 41"
+ B) - 95 52' 28" c - 44 57' 14*
- 84 T 32"
SPHERICAL TRIGONOMETRY
129
log cos
(-) log cos
- B)
- 9.82157 log tan [}(o + Wl - 10.81078
- 9.01008 (-) log tan \c - 9.99930
0.81149 0.81148
This gives a better check than the law of sines would yield, since c is almost 90.
Answers: a - 135 42' 15', 6 = 61 52' 20', c - 89 54' 30*.
EXERCISES
Solve the spherical triangles of which the following are given parts. Write answers
to the nearest multiple of 5*.
1. a - 108 0' 30*,
2. a - 150 20' 40',
3. a - 1160'15*,
4. a = 153 38' 45*,
5. a - 47 42' 0*,
6. a - 82 3' 5*,
7. a - 64 23' 15*,
8. a - 55 42' 10',
9. A - 122 40' 10',
10. A 145 46' 30',
11. A - 22 20' 20',
12. A - 25 10' 15',
13. A - 98 52' 40',
14. A - 116 55' 25',
15. A - 2551'0',
16. A - 70 20' 40*,
79. Case III. Given two sides and the included angle. If the
given parts are a, 6, C, we use Napier's analogies (3) and (4) of 75
to determine (A - B) and J (A + B) respectively, and from these
values we find A and B. The side c is then found by using either
(1) or (2) of 75. We check by the law of sines, or by a Napier's
analogy which we did not use in our solution. We choose our formu-
las so that each expression in them is positive. Parts greater than
90 are replaced by their supplements.
. Example. Given a - 65 40' 0*, c - 124 7' 20*, B - 159 44' 40*. Find A,
b - 96 42' 0*,
c - 40 34' 10*.
6 - 43 3' 20',
c = 129 8' 20*.
b - 57 57' 50',
c = 137 20' 45*.
6 - 400'20',
c= 118 20' 15*.
b - 50 1' 40',
c = 63 15' 10*.
6 - 82 1' 30',
c - 70 14' 10*.
6 - 100 49' 30',
c - 99 40' 50*.
b - 150 57' 5*,
c - 134 15' 55*.
B = 31 20' 20',
C - 37 30' 0*.
B - 125 12' 0',
C - 108 12' 30*.
B = 47 21' 10*,
C = 146 40' 10*.
B - 36 42' 45',
C - 160 32' 20*.
B = 60 44' 25',
C - 27 52' 55*.
B - 40 10' 30*,
C - 54 6' 20*.
B - 155 14' 25',
C - 11 34' 10*.
B - 11010'0',
C - 133 18' 0*.
Solution.
Formulas
cos
- A)
tan
+ q)l
cos
tan
130
TRIGONOMETRY
c - 124
o= 65 40' 0"
c - a = 58 27' 20"
c + a = 189 47' 20"
i(c - a) - 29 13' 40"
J(c + a) - 94 53' 40"
:*<* + a)l- 85 6' 20"
i# - 79 52' 20"
J(C - A) - 5 0' 11"*-
A) - 11841'36"<-
C = 123 41' 47"
A = 113 41' 25"
Computation
7' 20'
log sin J(c a)
(4-) log cot JB
sum
(-)logsin[J(c-f a)].
logtani(C- A)
-A)-
log cos J(c a)
sum
(-)Iogcos[4(c + a)l
We omit the computation for 6. The result is 6
Check
B = 159 44' 40"
A 113 41' 25"
B - A - 46 3' 15*
B + A - 273 26' 5"
i(# - A) = 23 1' 38"
i(B + A) - 136 43' 2"
159 50' 56".
5 =
a
b + a
a)'
Jc
9.68867
9.25195
18.94062
9.99841
8.94221
' 9.94085
: 9.25195
' 9.19280
8.93105
10.26175
61 18' 24"
159 50' 56"
65 40' 0"
225 30' 56"
112 45' 28"
67 14' 32"
62 3' 40"
(-)logcos[i(B + A)],
diff.
Answers: A
43 16' 58"
9.96394
9.86211
log tan [J(6 + a)l = 10.37727
(-)logtan^c = 10.27545
10.10183 diff. = 10.10182
113 41' 25", C = 123 41' 45", b = 159 50' 55".
80. Case IV. Given two angles and the included side. The pro-
cedure corresponds to that in Case III. We use Napier's analogies
to find the other two sides and the other angle. For the check we
may use a Napier's analogy or the law of sines.
EXERCISES
Solve the spherical triangles of which the following are given parts. Write answers
to the nearest multiple of 5".
1.
a
=
99
40' 0",
b
=
64 20' 0",
C
=
96
10' 0".
2.
a
=
60
6'0",
b
=
100 8' 0",
C
=
31
10' 0".
3.
a
= 138 20' 30",
c
=
117 15' 45",
B
=
22
20' 15".
4.
b
=
60
6' 20",
c
=
100 5' 40",
A
=
31
ll'O".
5.
a
=
56
19' 40",
b
=
20 16' 40",
C
=
114
20' 20".
6.
a
=
53
49' 25",
b
SB8
97 44' 20",
C
=
102
14' 10".
7.
a
- 146 37' 15",
c
ss
135 49' 20",
B
=
105
8' 10".
8.
b
- 102 16' 10",
c
-
130 46' 0",
A
=
136
19' 10".
SPHERICAL TRIGONOMETRY 131
9. A = 122 20' 0', B = 37 40' 0", c = 34 3' 0'.
10. A = 58 18' 0",
11. A = 152 14' 30",
12. B - 92 38' 20*,
13. A - 80 20' 10",
14. A - 59 4' 30",
15. A = 127 22' 0",
B - 76 22' 0",
C = 122 41' 15",
C - 124 54' 40",
5 = 54 8' 30",
B = 94 23' 10",
C = 107 33' 20",
c ~ 103 4' 0".
6 = 48 20' 45".
a - 61 20' 0".
c = 61 47' 25".
c - 129 11' 40".
b = 125 41 '45".
16. 5 = 133 28' 35", C = 169 38' 10", a = 85 50' 20".
81. Case V. Given two sides and an angle opposite one of them
(ambiguous case). Here, as in the corresponding case for plane
triangles (p. 95), there may be no solution, one solution, or two solu-
tions.
If the given parts are a, 6, A, we may use the law of sines to com-
pute B:
/*\ r> s * n k A
(I) sin B = sin A.
v ' sin a
If this gives log sin B a positive value, there is no solution. But if
log sin B has an admissible value, and B\ is the corresponding acute
angle given by the Tables, then
B 2 = 180 - B, = (SO,
may be also a solution. We now use the theorem that the greater
side is opposite the greater angle to determine the values of B that
can be used.
If a > b, and A > J5 2 (and therefore A > J?i) there are two solu-
tions, one with parts a, 6, A, Bi, the other with parts a, b, A, B*\ if
a > 6, and B\ < A ^ # 2 , there is a solution corresponding to B\, but
not one corresponding to B%\ and if a > 6, and A ^ j?i, there is no
solution.
If a = b the triangle is isosceles, and in this case there is but one
solution.
The student should discuss for himself the case a < 6, using the
principle that an angle B = BI or B = 5 2 gives a solution when and
only when A < B.
With a, 6, A, and the admissible value or values of J5, we calculate c
and C by Napier's analogies, and check by another Napier's analogy
or by the law of sines.
In preparing an outline of the work, one should first proceed as far
as the determination of B, and make out the remainder of the outline
only after the number of solutions has been found.
132 TRIGONOMETRY
Example 1. Given a = 30 0' 0', b - 60 (X 0", A - 75 0' 0'. Find B, C, c.
Solution.
Formula
sin B - 5|2_- sin A (Only formula needed.)
Computation
log sin fe = 9.93753
(+) log sin A =- 9.98494
sum - 19.92247
(-) log sin a - 9.69897
log sin B - 10.22350, impossible.
Answer: No triangle exists with the given parts.
Example 2. Show that there is but one triangle for which b =* 60, c * 150,
B - 45.
We find, by computing with the law of sines,
log sin C - 9.61093, d - 24 5' 43", C 2 = [d], = 155 54' 17".
We here have c > b, hence C > B. Thus C 2 is admissible, but Ci is not.
Example 3. Given a - 70 30' 20", c - 125 15' 35", C = 145 0' 0". Find
A, B, 6.
Solution.
Formulas
, sin a . ^ , ,, sinJ(C-hA). ,, N
sin A = -. sin C. tan 6 = . ;* /tan i(c - a)
sine, * smi(C A)
. , sin J(c -f a) . *. A . , , sin Ai sin Bi sin
cot JB - -; B ! ( tan J(C - A) Check: . - . v .
a sin J(c a) a sin a sin6i sin
Computation
log sin a = 9.97436
(+) log sin C. - 9.75859
sum - 19.73295
(-) log sine. 9.91198
log sin A = 9.82097
Ai - 41 27' 56" A 2 = [Ail - 138 32' 4"
c = 125 15' 35"
a - 7030 / 20"
c - a - 54 45' 15" i(c - a) - 27 22' 38"
c + a - 195 45' 55" J(c + a) - 97 52 ; 58"
H(c + )1 - 82 7' 2"
C - 145 0' 0" C - 145 0' 0"
Ai - 41 27 ; 56" A 2 - 138 32 ; 4"
C + Ai = 186 27' 56" C + A, - 283 32' 4"
i(C + A,) - 93 13' 58" J(C + A.) - 141 46' 2"
DKC 4- A01 - 86 46' 2" [J(C + A,)l - 38 13' 58"
C - Ai = 103 32' 4" C - A 2 - 6 27' 56"
J(C - Ai) - 51 46' 2" i(C - A 2 ) - 3 13' 58"
SPHERICAL TRIGONOMETRY
133
log sin [J(C + Ai)J
(+) log tan J(c - a)
sum
= 9.99931
- 9.71420
- 19.71351
(-) log sin J(C - Ai) - 9.89514
log tan i&i - 9.81837
i&i - 33 21' 13'
66 42' 26*
6i
log sin [i(c + a) J
(-) log sin i(c - a)
diff.
(+) log tan i(C - Ai)
log cot JBi
log sin Ai - 9.82097
(-) log sin a =9.97436
9.84661
9.99588
9.66261
0.33327
10.10356
10.43683
20 5' 22*
40 10' 44*
log sin [i(C 4- A,)]. - 9.79159
(+ ) log tan J(c - a) - 9.71420
sum = 19.50579
(-) log sin \(C - A t ) - 8.75123
log tan ik - 10.75456
i&a - 80 1' 11*
62 - 160 2' 22*
0.33327
(+) log tan 4(C - Ai) - 8.75192
log cot i# 2 9.08519
i# 2 - 83 3' 46*
Bt - 166 7' 32*
C/iecfc
log sin B l = 9.80968 log sin [B 2 ] 8
(-) Iogsin6i - 9.96308 (-) log sin M.
9.84660
9.37984
9.53323
9.84661
Answers: Ai - 41 27' 55*, B l
A t = 138 32' 5*, ft
40 10' 45*, 61 - 66 42' 25*;
166 7' 30*, bt = 160 2' 20*.
82. Case VI. Given two angles and a side opposite one of them
(ambiguous case). The procedure corresponds to that in Case V.
If the given parts are A, B, a, we use the law of sines to compute b.
We allow for two solutions, 6 = fei and b = [61], = 6 2 , in case log sin b
has a value less than zero (if log sin 6 > there is no solution).
We retain both 61 and 62 if they obey the rule that a is greater than,
equal to, or less than b, according as A is greater than, equal to, or
less than B. If only one of the pair bi, 62 obeys this rule, we retain
that one; if neither obeys the rule, there is no solution. With the
parts retained' we compute c and (7, and check, as in Case V.
EXERCISES
Find the number of solutions for the following given parts:
1. a = 110, 6 - 80, A - 95. 4. B - 130, C - 140, 6 = 125.
2. a - 40, A - 100, B - 130. 5. c - 105, A = 110, C - 100.
3. b - 100, c - 120, C = 130. 6. a - 114, c - 47, A - 100.
Solve the spherical triangles of which the following are given parts. If there are
two solutions, find both. Write answers to the nearest multiple of 5*.
7. a - 32 59' 30*, 6 = 47 45' 10*, A = 43 50' 40*.
8. a = 47 29' 20*, b - 50 6' 20*, A - 34 29' 30*.
9. 6 - 138 20' 15*, c = 117 9' 30*, C * 47 21' 45*.
10. 6 - 107 33' 20*, c - 52 38' 0*, C - 55 47' 40*.
134 TRIGONOMETRY
11. a - 96 34' 40', c = 55 0' 20", C - 42 30' 30".
12. a - 121 28' 10", c - 154 46' 50", A - 39 45' 10".
13. A - 111 42' 20", - 77 5' 0", a - 108 30' 30".
14. A - 105 6' 40", = 58 40' 15", 6 = 61 33' 5".
15. B = 31 26' 15", C = 130 36' 35", c - 71 15' 20".
16. B - 39 36' 55", C - 95 50' 0", c - 30 17' 35".
17. A = 133 18' 20", C - 70 16' 10", a - 155 5' 40".
18. A - 111 51' 45", C = 126 18' 40", c = 120 22' 40".
83. Area of a spherical triangle. We have seen that the sum of the
angles of a spherical triangle ABC is greater than 180. Following the
usage of books on solid geometry we write
E - A+B + C - 180,
and call E the spherical excess. It is shown in such books that the
area of the spherical triangle ABC is equal to the area of a lune whose
E
angle is 75-; that the area of a lune is to the area of the entire spheri-
&
cal surface as the angle of the lune is to 360; and that the surface
area of a sphere of radius R is equal to 4 irR 2 . If we put these state-
ments together we have
Area of AABC
~360'
(1) Area of AABC =
It is here understood that E is given in degrees. Since the number
__. 77T
of radians in E degrees is TOO> formula (1) states that the area
of A ABC is equal to the product of R 2 and the radian measure of E.
If the three angles of a spherical triangle are given, we compute
its area directly from (1). When three other parts are given we may
solve for the angles and then use (1). In more extended works on
spherical trigonometry other formulas for area are developed; we
give one here, called L'Huilier's Theorem, which enables us to com-
pute E when the three sides are known:
E
(2) tan ^ = Vtan }j tan }(* - a) tan }($ - b) tan }(* - c).
The proof is omitted.
SPHERICAL TRIGONOMETRY 135
EXERCISES
Find the area of each triangle for which the following parts are given, when R has
exactly the length indicated. Give answers to four significant figures for Exercises 1,
3, 5, 7, 9, 11, and to five significant figures for Exercises 2, 4, 6, 8, 10.
1. A = B C = 70 0'; R = 1 ft.
2. A = 50 14' 25", B = 63 20' 40", C = 134 37' 10"; R = 3959 mi.
3. a = 36 27', c = 54 20', C = 90 (exactly); R = 10 in.
4. b = 32 9' 15", A - 50 19' 20", C - 90 (exactly); 72-100 ft.
6. A = 143 5', C = 120 54', c - 90 (exactly); 72-50 ft.
6. a = 51 59' 55", b - 121 33' 0", c = 90 (exactly); 72 - 3959 mi.
7. a = 154 45', c - 34 9', B = 108 40'; 72 = 10 ft.
8. B - 104 49' 50", C = 84 51' 40", a = 96 53' 10"; 72 = 3959 mi.
9. a = 76 36', 6 = 57 8', c = 110 26'; 72 = 1000 ft.
10. a = 54 36' 50", b = 147 37' 0", c = 126 50' 50"; 72 = 47281 ft.
11. Find the number of square miles in the triangle whose vertices are Baltu
more (lat. 39 17' N, long. 76 37' W), New York (lat. 40 43' N, long. 74 0' W),
and Chicago (lat. 41 53' N, long. 87 38' W), assuming the earth to be a sphere
of radius 3959 mi.
CHAPTER IX
APPLICATIONS
84. Introduction. In this chapter we present briefly a few impor-
tant applications of trigonometry. First we shall present problems
which use plane trigonometry, (1) in plane surveying, (2) in artillery,
and (3) in navigation (of the water or of the air) ; later we shall give
applications of spherical trigonometry to problems of navigation on
the earth regarded as a sphere.
In our brief discussion we must omit all description of instruments
of measurement, of methods of handling them, and of corrections to
be made to observations. For these important topics, we must refer
to special books on surveying, artillery, and navigation.*
In computations we shall use five-place tables, and at the end of
the work round off the answers to four places or less according to data.
PLANE SURVEYING
85. Plane surveying. In plane surveying we consider small por-
tions of the earth's surface for which no important error is made if
we assume that the earth is flat, i.e., that such a portion lies in a
horizontal plane. A map of such a region is a drawing which pictures
the relative distances and directions of points in the region. In the
simplest map a coordinate system (or grid) is used, points being lo-
cated by their coordinates.
A surveyor locates points with respect to some particular object or
point, frequently a bench mark. Through this point B passes a
* E.g., Davis, R. E., and Foote, F. S., Surveying: Theory and Practice,
McGraw-Hill, New York.
Breed, C. B., and Hosmer, G. L., Principles and Practice of Surveying, Wiley,
New York.
Button, Capt. Benjamin, Navigation and Nautical Astronomy, U. S. Naval
Institute, 1939.
Lyon, Thoburn C., Practical Air Navigation, U. S. Department of Commerce.
Bowditch, N., American Practical Navigator, U. S. Hydrographic Office, Navy
Department, Washington, D. C. This contains extensive tables, which simplify
calculations.
136
APPLICATIONS
137
N
A
meridian line SN from south to north, and a line WE from west to
east (Fig. 77). A point P is located by its distance CP north or south
of WE, and its distance AP east or west of the meridian SN. In the
surveyor's language, the latitude (or
difference in latitude) of P is I units
north or south, and the departure of
P is p units east or west, where the
length of CP is I units and of A P is
p units (the unit usually being a foot,
a yard, or a rod).
The point P is also located by its
distance BP, and the bearing of BP
(which is the angle NBP if P is north
of WE, and the angle SBP if P is south of WE), written as in Exer-
cises 4, 9, 10, page 19. If the distance is d units and the bearing is
B, then we have (Fig. 77)
W
B
C E
FIG. 77
I = d cos B, p = d sin B,
and
(2)
tan B = >
P
I
sin 5 cos B
By use of (1) or (2) we may compute two of the four quantities Z, p,
J5, d when the others are known.
When the line BP is measured it is called the course.
A surveyor frequently makes a series of measurements, going from
B to Pi to P 2 to P 3 , and so on. A problem is to locate the final point
with respect to the starting point, as illustrated in the following
Example.
Example. A surveyor's notebook shows the following measurements (shown
in heavy type), calculations, arid sketch.
Course
Distance
Bearing
Latitude
Departure
N
S
E
W
BP l
PiP 2
P 2 Pa
200 yd.
300 yd.
200 yd.
N15E
S80E
S5W
193.19
52.09
199.24
51.76
295.44
17.43
Totals
193.19
251.33
347.20
17.43
Diff.
BP,
335 yd.
S80E
58.14
329.77
138
TRIGONOMETRY
Computation
I = 58.14
p - 329.77
log p - 2.51821
log I = 1.76448
log tan B - 0.75373
B = 80 0.1'
log p = 2.51821
log sin # - 9.99335
log d = 2.52486
d = 334.85
FIG. 78
In the preceding Example, the surveyor wished to know the length
and bearing of Z?P 3 , and a lake and a forest caused the scries of meas-
urements. An alternative method of locating Pa would have been
to measure the bearing and distance for BP& (see Fig. 78), the bearings
of BPt and PsP 4 and the bearing and distance for P 4 P 3 . By this
method the distance 5P 4 across the lake is found by solving the
triangle J3P 6 P 4 . We then should have sufficient data to solve the
triangle BP 4 Ps and thus find the distance and bearing of BP 3 .
To make a map of a region, we may first find latitude and departure
for a number of points, and bearing and distance for a number of
lines, and then make a drawing to scale. Angles on the drawing
should be the same as corresponding angles in the field. Lengths on
the drawing are reduced by a constant scale from those in the field.
EXERCISES
Use five-place tables in solving the following Exercises:
1. The bearing of BP is N 38 W and the distance is 324 yd. Find the latitude
and departure of P from B.
2. The bearing of BP is S 52 E and the distance is 324 rods. Find the latitude
and departure of P from B.
In a series of measurements a surveyor made the following tables. In each case
find the latitude, departure, bearing, and distance for BP&. Draw a map to some
selected scale (and give the scale).
3.
Course
Dist.
Bearing
BP,
PiP 2
P 2 ft
150 yd.
210 yd.
320 yd.
N16 W
S72W
S65E
Course
Dist.
Bearing
BP 1
PiP 2
P 2 P 3
280 rd.
302 rd.
401 rd.
N35W
N15E
S27E
APPLICATIONS
139
Course
Dist.
Bearing
PiP 2
P 2 Ps
328.1 ft.
218.7 ft.
301.6 ft.
N 3 25' E
N 88 12' E
S 4 5' W
Course
Dist.
Bearing
PiP 2
P 2 Ps
288.4 yd.
129.3 yd.
307.2 yd.
S211'E
S 59 4' W
S 6 34' E
FIG. 79
7. A surveyor found for BPi the distance 248.1 yd., the bearing S 87 15' W.
The bearing for P 2 was S 27 14' E and for PiP 2 was S 53 47' E. For P 2 P 3
the distance was 587.7 yd. and the bearing
N 76 14' E. Find the distance and bearing
of BP 3 (see Fig. 79).
8. A surveyor found for BPi the distance
124.7 rd. and the bearing S 84 18' W. The
bearing for BP 2 was S 23 13' E and for P,P 2
was S 63 44' E. For P 2 P 3 the distance was
310.6 rd. and the bearing N 82 4' E. Find the
distance and bearing of BPa.
9. A surveyor made the following measure-
ments: For BPi the distance was 220.1 yd., the bearing N 15 1' E. For P t P 2
the distance was 329.9 yd., the bearing S 80 2' E. For P 2 P 3 the distance was
219.8 yd., the bearing S 5 2' W. For P 3 B' the distance was 368.5 yd., the bear-
ing S 79 58' E. What was the calculated latitude and departure of B f from B?
If all measurements had been exact, B f and B would have coincided, since they
were the same point of observation.
10. A surveyor made a traverse A BCD A
by making measurements shown in the ad-
joining table. By calculating the latitude
and departure for each course find the re-
sultant error in latitude and departure for
the closed traverse.
According to a map, referred to a bench mark B several points (see Fig. 80) are lo-
cated as shown in the following table. Using these data, solve Exercises 11 and 12.
Course
Distance
Bearing
AB
123.4 yd.
S 20 15' E
BC
129.2 yd.
N7115'W
CD
66.8 yd.
N 8 3' W
DA
90.4 yd.
N 85 1' E
Diff. Lat.
Departure
N
S
E
W
C
427.31
631.75
D
815.23
182.14
E
128.72
581.56
F
132.68
622.28
11. Railroad surveyors at C can observe D but not
E which lies over a ridge through which a tunnel is to
be constructed. They calculate the angle DCE and
determine the direction in which the tunnel is to be
dug. Find this angle and the distance CE.
FIG. 80
\
140
TRIGONOMETRY
FIG. 81
12. Surveyors at E proceed similarly (see Ex. 11) and calculate the angle FEC,
and determine the direction EC of the other end of the tunnel. Calculate angle
FFC and the distance EC.
In the following Exercises the surveyors used observation points at different levels
as stated. In making a map the points are projected on a horizontal plane. Find
the required things and draw a map to scale (indicate the scale on the map) in Ex-
ercises 13 and 14.
13. In a mountainous country there were two points C and D which were
situated so that an observer at one could not see the other, but both were visible
from two points A and B which lay in a
horizontal plane MM'. Surveyors at A ob-
served that the bearing of AB was S 6 20' E,
of AC was N 62 34' E, and of AD was
S 84 47' E. They measured AB to be 1538
yd., and observed at B that the bearing of BC
was N 55 47' E, of BD was N 79 13' E.
The angle of elevation of BC was 5 11' and
of BD was 4 28'. Letting C' and D' be the
points in the plane MM' directly under C
and D, find the distance and bearing of C'D' and the elevation of C and D above
the plane MM'.
14. Surveyors measured AB to be 235 yd. along a hillside whose inclination
to the horizontal was 8 21'. At A the angle of elevation of a point C on a distant
hill was 6 13', the bearing of AB was S 4 11' W,
of AC was N 38 57' E. At B the bearing of
BC was N 35 11' E. What is the distance BC
and how much higher is C than B?
15. A line AB on a hillside is d yd. long, its
inclination is a. The inclination of AC is 0.
The bearing of 47? is S b E, of AC is N B E, and
of BC is N E. Find the heights h and k of B
and C above the horizontal plane MM' through
A , and find the horizontal distance from B to C.
16. Two points C and D are invisible from
each other. At a point A the bearing of AC is N ot E and of AD is N (3 E. At
another point B which lies in a horizontal plane MM' with A, the bearing of BC
is N a' E and of BD is N 0' E. The angle of elevation of A C is 0, and the length
of AB is d yd. Let C' and D' be points in the plane MM' which are directly be-
low C and D. Find the distance C'D' and the angle of elevation of CD.
APPLICATIONS 141
ARTILLERY PROBLEMS
86. Use of maps. Maps arc of great importance in military prob-
lems. Some of the principles of their construction have been briefly
described in the preceding pages. By use of surveying instruments,
which measure directions, angles, and distances, relative positions of
points of interest are found and a drawing or map is made showing
these points in their relative positions. And conversely when such
a map is available we may determine from it directions, angles, and
distances for use in the field.
Suppose that a map is at hand, and that a gun which is located at a
mapped point G is to be fired at a mapped target T. The distance
GT is scaled off from the map, and the angle
of elevation of the gun is found from ''firing
tables " which give this angle for various dis-
tances under standard conditions. For very
large guns and long ranges corrections are
made for weather conditions and other factors F
which produce predictable results. If the point
T is visible from (?, the direction to point the axis of the gun may be
determined by direct aiming, which is called "direct firing." If the
target T is not visible from the gun G, "indirect firing" is used. One
method is to use an aiming point A, which is mapped and which is
visible from G. From the map the magnitude of the angle AGT is
found with a suitable protractor, and then the gun is set so that its
axis makes this angle with the observable direction GA.
More frequently, however, the map does not show G and T, and T
is not visible from G. In this case, these points may be mapped by
observing their directions from mapped
points A, B, C . . ., and drawing lines on
the maps through these points with these di-
rections. Suppose three points A, B, C are
used. Because of inaccuracies of observa-
tion and of drawing, these sets of lines will
not exactly coincide at G or at T, but the
FIG. 84 ge j. Q intersecting lines will appear to locate
G in a triangle of position and T in another such triangle (Fig. 84).
Assuming that the points are at the centers of the triangles (per-
haps roughly determined), the gunner may proceed with "indirect
fire." In some cases more elaborate methods are used.
142 TRIGONOMETRY
EXERCISES
1. By use of a range finder an officer at observes that the distance to a
gun G is 3127 yd., and that the distance to a target T is 2463 yd. The angle TOG
is observed to be 88 21'. The gunner wishes to know the angle OGT and the
distance GT. What are they?
2. By use of a range finder an officer at observes that the distance to a gun G
is 1628 yd. and to a target T is 3181 yd. The angle TOG is observed to be
123 14'. The gunner wishes to know the angle OGT and the distance GT.
What are they?
The coordinates of points A, B, C, D on a map are as follows: A (2.3, 1.3),
#(2.4, 6.6), C(6.5, 3.4), Z>(- 2.1, 3.1). Use these points in solving Exercises 3-6.
3. Plot the points A t B, C on a sketch. From a gun emplacement G the
following bearings are taken: For GA the bearing is S 27 E; for GB the bearing
is N 27 E; for GC the bearing is N 84 E. On the sketch draw GA, GB, and
GC, and estimate the coordinates of G in the "triangle of position."
4. A target T is observed from A, B, C. The bearing of AT is N 56 E, of
BT is S 80 E, and of CT is N 72 E. On a sketch locate A, B, and C, draw AT,
BT t and CT, and estimate the coordinates of T from the " triangle of position."
5. From a gun G the bearing of B is N 38 E, of C is S 86 E, and of D is
S 74 W. Make a sketch of B, C, and D, and estimate the coordinates of G from
the "triangle of position."
6. The bearing of a target T from D is S 82 E, from B is S 61 E, and from
A is N 78 E. From a sketch estimate the coordinates of T in the "triangle of
position."
7. Write down all of the formulas that
would be used in a solution of the following
problem. An observer at A (Fig. 85) finds
angle a. and distances a and 6; and an ob-
server at .B finds angle ft and distance c. Find
angle y and distance d for the gunner at G
who wishes to fire at target T.
8. From a gun G the target, is hidden by a f p IG gg
forest. An officer at A observes that GA = 582
yd., AB = 629 yd., and angle GAB = 121 13'. An officer at B observes that
BT = 1122 yd., and angle ABT = 98 15'. Calculate the angle AGT and the
range GT for the gunner.
87. Measurement of angles. The mil. In various branches of
military service several systems of measuring angles are used. In
giving bearings of a course or of an object from a given point, (a) in
the navy angles are measured from the north clockwise from to
360; (b) in the coast artillery angles are measured from the south
clockwise from to 360; and (c) in field artillery angles may be
measured to the right or left of some observable directing point.
APPLICATIONS
143
The angles are measured (a) in degrees and minutes, or (b) in de-
grees and hundredths of degrees, or (c) in mils, a unit which we
shall now briefly discuss.
A mil is an angle which may be defined by the relation
1600 mils
It follows that
1 mil = 0.05625 = 3.37500';
1 degree = 17.77778 mils;
1 minute = 0.296296 mils.
90.
1 mil = 0.0009817 radians;
1 radian = 1018.6 mils;
The usefulness of the mil arises principally from the fact that a
vertical yardstick (BD in Fig. 86) at a horizontal distance of a thou-
sand yards from A subtends almost exactly an angle of one mil.
More precisely, for an angle of A mils
(Fig. 86),
< x > Z = IS (error about L9%) ' A
and FIG. 86
(error less than 2% for angles less than 340
mils, about 19).
The latter equation is equivalent to
B
BD A
AB = TOGO
(2) tan A =
1000
(error less than 2% when A < 340).
The adjoining table shows that
this approximation (2) is very close
for small angles, being exact when
.4=0 and also when A is about
235 mils (an angle of about 1313');
but for angles as large as 600 mils
the error is more than 10%. For
example, when A = 400,
tan A 0.41421,
4=0.40000,
and the error is 0.01421.
The following Examples illustrate how the mil is used.
A mils
tan A
Angle in
degrees
mils
0.00000
50
0.04913
2 48.75'
100
0.09849
5 37.5'
150
0.14834
8 26.25'
200
0.19891
11 15 ;
250
0.25048
14 3.75'
300
0.30335
16 52.5'
350
0.35780
19 41.25'
400
0.41421
22 30'
500
0.53451
28 7.5'
600
0.66818
33 45'
144
TRIGONOMETRY
Example 1. At a distance of d yd. from an observer at A, a line of unknown
length x yd. subtends an angle of A mils, as shown in Figure 87. Find x.
Solution. From the approximation (2) we have
T
and hence
(3)
x
d
A
1000
Ad
1000*
A
FIG. 87
This involves an error of less than 2% if A is less
than 340 mils.
Example 2. From a gun G (Fig. 87) a target T is invisible, but a directing
point D is visible such that angle ADT is approximately a right angle. An
observer at A on the line GD finds that AG = a yd., AD =* d yd., and the
angle DAT = A mils. What is the angle DOT, which gives the deflection of
the gun from the visible direction GD?
Solution. If DT * x we have, by (2),
and
Hence
(4)
x
d
-d
G
A
'' 1000'
G
" 1000'
Ad
a + d'
1000s - Ad,
1000s = (a + d).
This involves an error due to the assumption that angle ADT is approximately a
right angle. With a range finder, the observer at A can estimate the accuracy of
the assumption; for practical purposes the approximation might suffice for giving
the direction of a trial shot. If angle G is small the range GT is approximately
a-}- AT.
Example 3. A gun G (Fig. 88) fires at a target T. An observer at B on the
line GT observes that the shell S falls B mils to the right of BT, that GB - b yd.
and BS = s yd. An observer at A, where AT is perpendicular to GT, notes
that AT = a yd. and that S is A mils to the right of T. What changes should
the gunner make in setting his gun for another shot?
Solution. When S is close to T compared to distances b and s, we get a good
approximation to angle TGS by use
of (4), namely,
and the gun should be deflected to the left G mils.
If AS intersects GT at S' t then, by (3),
TS' = ^- FIG. 88 U
The elevation of the gun should be decreased to shorten the range by this amount.
APPLICATIONS 145
EXERCISES
Express the following angles in mils:
1. 10, 25, 45. 5. 7 19', 17 31'.
2. 8, 30, 55. 6. 5 23', 32 28'.
3. 4, 18, 35. 7. 2 35', 24 25'.
4. 2, 16, 40. 8. 6 28', 19 19'.
Express the following angles in degrees and minutes:
9. 125 mils, 260 mils. 11. 84 mils, 400 mils.
10. 60 mils, 441 mils. 12. 71 mils, 313 mils.
Express the following angles in radians:
13. 25 mils, 225 mils. 15. 18 mils, 455 mils.
14. 40 mils, 375 mils. 16. 44 mils, 360 mils.
Express the following angles in mils:
17. 0.25 radians, 1.2 radians. 19. 0.40 radians, 0.66 radians.
18. 0.12 radians, 0.85 radians. 20. 0.18 radians, 0.35 radians.
Find a dose approximation in mils for the following angles:
21. Tan- 1 0.0985. 26. Tan" 1 0.0777.
22. Tan- 1 0.4142. 26. Tan- 1 0.1234.
23. Tan- 1 0.3578. 27. Tan"- 1 0.2333.
24. Tan- 1 0.6682. 28. Tan" 1 0.3456.
Draw angles of the following magnitudes:
29. 100 mils. 33. 50 mils.
30. 200 mils. 34. 250 mils.
31. 300 mils. 35. 450 mils.
32. 400 mils. 36. 800 mils.
In the following when answers are measures of angles express them in mils.
Assume that in each of Exercises 37-42 the angle and the line concerned are parts
of a right triangle.
37. At a distance of 1200 yd. a vertical stick 3 yd. long subtends an angle A.
How large is A?
38. How large is an angle that is subtended by a stick 7 ft. long at 350 yd.?
39. How tall is a building which subtends an angle of 60 mils at a distance of
175 yd.?
40. A target T is 120 yd. to the left of a directing point D as seen from a gun G.
What is the angle DOT if DG = 720 yd.?
41. A target T appears 40 mils to the right of a directing point D which is
800 yd. from an observer at a gun G. How far is it from D to T?
42. A chimney C which is 160 ft. tall subtends an angle of 50 mils at an observa-
tion post P. How far is it from P to C?
43. From a gun G a target T is invisible, but a directing point D is visible such
that angle GDT is approximately 90 (Fig. 89). From a point A to the rear of
G and in line with GD, an observer finds that AG = 237 yd., AD - 1872 yd.,
146 TRIGONOMETRY
and angle GAT 135 mils. Find the angle of deflection of GT from GD, which
enables the gunner to direct his gun on the target.
44. An officer at A observes that a target T and a point D are so situated that
angle ADT is approximately a right angle, angle DAT = 85 mils, and AD is
2856 yd. A gun G is on the line AD and is 382 yd. from A as shown in Fig-
ure 89. Find the angle AGT t which may be used by the gunner in training his
gun on the target T.
B G
A G D
FIG. 89 FIG. 90
46. A shell from a gun G (Fig. 90) fired at a target T falls at S.
An observer at B observes that BS is 8 mils to the right of the line
BGT, and that BG = 152 yd., BT = 1892 yd. An observer at A observes that
AS is 21 mils to the left of AT and that AT = 692 yd.
Assuming that AT is perpendicular to GT, what
changes should be made in setting the gun for the next
shot?
46. A target T is due east from a gun G. An ob-
server A in an airplane at a height of 3000 yd. directly Q __
over T observes that a shell falls at S due southeast
from 7\ and that the angle TAS is 16 mils. If GT is
1239 yd., what adjustments in bearing and range of the
gun should be made for the next shot? FIG. 91
NAVIGATION
88. Plane sailing. In navigation of the sea or air, it frequently
suffices to consider the surface of the earth as a plane surface when
only short distances are involved. This is the method of plane sailing.
We first define a few words which will be used. The student may
note that some of these words have slightly different meanings in
surveying and in various branches of navigation.
In plane sailing the line which runs north and south through a
point on the earth is called the meridian of the point. The direction
of a line is described by giving the angle it makes with a meridian.
The course C is this angle, and it is measured from the north clock-
wise, through the east, south, and west, running from to 360.
In giving a ship's position with reference to some given place, five
terms are employed the course, the distance (d), the difference of
latitude (Z>L), the difference of longitude (DLo), and the departure
(dep). All of these terms except difference of longitude are shown in
the plane triangle of Figure 92. Here TN is a segment of the merid-
APPLICATIONS
147
N
ian through T, and QN is the perpendicular to TN from the ship Q.
The distances TN and NQ are the difference of latitude and the de-
parture; the distance TQ is called the distance; and the angle C is
the course.
We consider departures to the cast as positive,
and to the west as negative; and differences in lati-
tude to the north as positive, and to the south as
negative. We have the relations
(1) dep = d sin C;
(2) DL = d cos C;
r< dep
~ DL
DL
(3)
FIG. 92
Navigators use Traverse Tables which give values of d sin C and
d cos C for integral values of d from to 600 and angles at intervals
of 1' from to 90. Without
such tables at hand, we may
compute these products by use
of trigonometric tables.
89. Parallel sailing. Here
we no longer treat parallels of
latitude and meridians as
straight lines, but consider them
as circles on a sphere. If Q is
a point on the earth's surface,
its parallel of latitude is the
circle cut from the sphere by a
plane through Q parallel to the equatorial plane, and its meridian
is the half great circle on the sphere that passes through Q and ends
at the north and south poles.
In Figure 93, G is Greenwich, England, P is the north pole, the
center of the earth, FDE the equator, PQE a part of the meridian
through Q, and PGF part of the meridian of Greenwich; then we
define the latitude and longitude of Q as follows :
longitude ofQ = angle GPQ
= angle FOE, measured to 180 E or 180 W;
latitude ofQ = angle EOQ, measured to 90 N or 90 S.
Thus the position of Q in Figure 93 is described as follows:
Q, lat. 58 50' N, long. 110 20' W.
FIG. 93
148 TRIGONOMETRY
When a course is due east or due west, the path is along a parallel
of latitude and we have parallel sailing. In this case the distance
equals the departure, and DL = 0. The only problem is to de-
termine the difference of longitude from the departure, or vice versa.
Referring again to Figure 93, we say that TQ represents the de-
parture in parallel sailing from Q to T. The difference of longitude
of T and Q is the angle DOE or TCQ, where TQ is an arc of a parallel
of latitude with radius r and center at C. If R is the radius of the
earth, we have OD OT = R. The latitude L of T and Q is the
angle DOT. We then have
T r TQ r
hence, since TQ = dep,
(1) dep = DE cos L,
(2) DE = -^-
y cos L
A nautical mile is defined as the length of a 1' arc on a great circle
of the earth. Hence a nautical mile on the equator corresponds to a
difference of longitude of 1', and, if differences of longitude and de-
partures are measured in nautical miles, we have
(3) dep * DLo cos L,
dep
(4) DLo =
cos L
90. Middle latitude sailing. In parallel sailing the latitude L is
constant. In plane sailing the points T and Q have different latitudes,
Li and LZ. The formulas (3) and (4) of parallel sailing will give good
approximations if for L we take the middle latitude ^ - If we
call this middle latitude L m , we then have
(1) dep = DLo cos I/ m ,
dep
(2) DLo
cos L m
Thus, in general, to solve a problem in navigation (using only plane
trigonometry) it is necessary to use both the plane sailing formulas
of 88 and the middle latitude sailing formula.
Example 1. A ship sails a course of 134 25' for a distance of 135 nautical
miles. If the starting point T has a latitude of 38 15' N and a longitude of
70 20' W, what are the latitude and longitude of the ship Q?
APPLICATIONS 149
Solution. Using plane sailing formulas, we have
dep = 135 sin 134 25',
DL - 135 cos 134 25'.
Since sin 134 25' - cos 44 25', and cos 134 25' = - sin 44 25', we have
log 135 - 2.13033
log cos 44 25' = 9.85386
log dep = 1.98419
dep = 96.425
log 135 -
log sin 44 25' =
log (- DL)
2.13033
9.84502
1.97535
DL - - 94.482
FIG. 94
Thus the departure is 96.425 nautical miles and the difference of latitude is
- 94.482 nautical miles or - 94.482' or - 1 34.5'. Since DL is negative the
latitude L of Q is less than that of T, and we find that the latitude of Q is
38 15' - 1 34.5' - 36 40.5'.
The middle latitude L m is
L m -
38 15' + 36 40.5'
37 27.8'.
Hence by the formula of middle latitude sailing, we get
DLo
96.425
cos 37 27.8'
- 121.48'
= 2 1.5'.
log 96.425 - 1.98419
log cos 37 27.8' = 9.89968
log DLo = 2.08451
Since the position of Q is in west longitude, and since the ship sails in an easterly
direction, we subtract DLo from the longitude of T to get the longitude of Q.
The result is 68 18.5' W longitude.
Hence the position of Q is lat. 36 40.5' N, and long. 68 18.5' W.
Example 2. If a ship is to sail from a point
T whose longitude is 46 25' W and whose lati-
tude is 44 14' N to a point Q whose longitude
is 51 35' W and latitude is 45 10' N, what are
the distance and the course?
Solution. We find the middle latitude
L m = 44 42', and the difference of longitude
DLo - 5 10' - - 310', using the negative
sign since S is west of T ( 88). FIG. 95
160
TRIGONOMETRY
By formula (1) above, we then have
dep - - 310 cos 44 42'
= - 220.35;
and by use of formula (3), 88, since DL
or 56 nautical miles, we have
, ~ dep
tan C = ,
56'
Computation
log 310 = 2.49136
log cos 44 42' = 9.85175
log (- dep) = 2.34311
log DL = 1.74819
log tan C = 6.59492
from which we find, since the course is northwesterly,
C = 360 - 75 44.4' = 284 15.6'.
Then by formula (1), 88,
-
sin C
= 227.35.
From T to Q the distance is 227.35 nautical miles and the course is 284 15.6'.
91. Dead reckoning. Dead reckoning is the process of determin-
ing the position of a ship by use of the course and distance of its
run from a position previously determined by other methods. The
course is determined by use of a compass, and the distance by use of
an instrument known as a log or by the number of revolutions of the
ship's propeller. If there is a current affecting the ship's motion,
this will not be observed on the compass or the log, and the velocity
of the current must be added to the ship's observed velocity in de-
termining the actual velocity relative to the ground, which is required
in finding differences in latitude and in longitude.
The dead reckoning may be done by plotting on maps or by com-
putation by the method of middle latitude sailing. In the following
Examples the dead reckoning is done by computation.
Example 1. A ship starts at a point T whose latitude is 36 25' N and whose
longitude is 74 36' W. As determined by compass and log, it sails 212 nautical
miles on a course of 132 45' in 10 hours. Assuming that there is a current of
6.2 knots (nautical miles per hour), its set being 65, find the ship's position at
the end of the ten hours.
Solution. We first find the velocity over the
ocean floor by vector addition of the two velocities.
In one hour the motion relative to the water is 21.2
nautical miles on course 132 45', and of the water
relative to the floor is 6.2 nautical miles on course
65. From Figure 96 we see how to find the result-
ant DL and departure for one hour:
DL = 21.2 cos 132 45' + 6.2 cos 65 - - 11.771,
dep = 21.2 sin 132 45 ; + 6.2 sin 65 = 21.187. FIG. 96
APPLICATION'S
Then for the resultant course and distance
tan C = ^jr C 119 3.3', d =
151
dep
sin C
24.237.
The distance made good in ten hours is 242.37 nautical miles on course 119 3.3'.
In ten hours the DL - 117.71' and the dep = 211.87 nautical miles. Hence
the required latitude is 36 25' - 1 57.7' = 34 27.3'. The middle latitude is
found to be L m = 35 26.2'. Hence by (2), 90, and a brief calculation,
DLo
dep
= 260.04' = 4 20.0',
cos L m
to the nearest tenth of a minute, and therefore the required longitude is
74 36' - 4 20.0' = 70 16.0'.
Example 2. A ship is to sail from a point T whose latitude is 36 25' N and
longitude 74 36' W to a point Q whose latitude is 35 42' N and longitude is
73 12' W, its speed being 26.4 knots. There is a current of 4.6 knots on course
65. What should be the compass course of the ship, and how long will it take to
reach Q?
Solution. Here we have DL = 43',
middle latitude L m = 36 3.5', and DLo = 84'.
Hence, by (1), 90, omitting details of the
calculation, we have
dep = DLo cos L m = 67.908;
therefore, by (3) and (1), 88,
C
d
122 20.5';
^ = 80.377.
sin C
Let C" be the ship's compass course, TQ' the
velocity relative to the water, and Q'Q" the
velocity of the current. Then angle TQ' f Q' = C - 65
and Q'Q" = 4.6.
Relettering the triangle TQ"Q' as shown in Figure 97, we have
a . 4.6, 6 = 26.4, B = 57 20.5'.
By the law of sines, we get
sin A =L^L^, A =8 26.1'.
FIG. 97
57 20.5', TQ' = 26.4,
Then Ci
a sin C
sin A
28.597. Hence the ship's compass course is
Of = 122 20.5' + 8 26.1' = 130 47'.
80 377
The time required is 98597 = 2.81 hr.
152 TRIGONOMETRY
EXERCISES
In each of Exercises 1-8 find the departure and the difference in latitude for the
given distance and course (plane sailing) :
1. dist. - 125, C - 57 28'. 6. dist. = 219, C = 213 15 ; .
2. dist. = 201, C - 182 13'. 6. dist. - 371, C - 241 17'.
3. dist. = 178, C = 151 52'. 7. dist. = 175, C = 303 15'.
4. dist. = 82, C = 265 13'. 8. dist. = 144, C - 327 50'.
In each of Exercises 9-16 find the distance and course for the given departure and
difference in latitude (plane sailing) :
9. dep = 73.7, DL = 23.9. 13. dep - 64.4, DL - - 23.7,
10. dep - 86.8, DL = - 31.4. 14. dep = - 29.9, DL - - 71.1.
11. dep = - 70.7, DL - 87.1. 16. dep - 72.3, DL = - 31.6.
12. dep = - 81.1, DL = 27.7. 16. dep = 67.7, DL = - 55.5.
17. A ship sailed from lat. 44 19' N, long. 40 42' W due west for 121.4
nautical miles. What were then its latitude and longitude?
18. A ship sailed from lat. 13 12' N, long. 145 42' W due east for 212.5
nautical miles. What were then its latitude and longitude?
19. A ship sailed from lat. 55 15' N, long. 10 39' W to lat. 55 15' N, long.
13 27' W. How far did it go?
20. A ship sailed from lat. 63 12' S, long. 2 5' E to lat. 63 12' S, long. 4 15' W.
How far did it go?
21. A ship sailed from lat. 58 27' S, long. 14 33' W for 147.2 nautical miles
on course 78 15'. What were then its latitude and longitude?
22. A ship sailed from lat. 23 16' N, long. 5 10' W for 167.1 nautical miles
on course 247 10'. What were then its latitude and longitude?
23. A ship sailed from lat. 12 4' S, long. 4 14' W for 155.4 nautical miles on
course 151 25'. What were then its latitude and longitude?
24. An airplane flew from lat. 38 40' N, long. 74 13' W for 218.3 nautical
miles on course 321 32'. What were then its latitude and longitude?
25. A ship is to sail from lat. 17 32' N, long. 15 12' W to lat. 19 19' N, long.
13 7' W. What are the distance and course?
26. A ship is to sail from lat. 14 15' S, long. 120 15' W to lat. 16 6' S, long.
122 6' W. What are the distance and course?
27. A ship sailed a compass course 123 10' for 162.3 nautical miles, requiring
7.4 hr. There was a current of 3.7 knots with set 41 30'. What were the differ-
ence of latitude and the departure over the ocean floor?
28. An airplane flew a compass course 62 30' at 97.3 mi./hr. for 2 hr. 15 min.
There was a wind of velocity 35 mi./hr. with course 80 40'. What difference of
latitude and departure did it make with respect to the ground?
29. A ship sailed from lat. 37 43' N, long. 66 18' W for 132.2 nautical
miles on compass course 67 30', requiring 6 hours. There was a current of
7.2 knots with set 38. What were then the latitude and longitude by dead
reckoning?
APPLICATIONS
153
30. An airplane is to go from A to B, traveling at 135 miles per hour with
respect to the air. There is a wind of 47 miles per hour on course 125. If the
position of A is lat. 33 14' N, long. 102 7' W, and if the position of B is lat.
37 25' N, long. 93 5' W, what should be the compass course of the airplane, and
how long will it take to fly from A to B?
92. Great circle sailing. In the preceding sections, we have solved
certain problems of navigation by means of plane trigonometry.
We shall now apply spherical trigonometry to problems in naviga-
tion. While the earth is not exactly spherical, it is approximately so.
In this section we shall take the
earth to be a sphere whose radius
is R = 3959 miles.
The line of shortest distance be-
tween two points on a sphere is
the great circle on which the points
lie, the distance being measured
on the arc which is not greater
than a semicircumference. In
great circle sailing a ship (or air-
plane considered as traveling ap-
proximately on the earth's surface)
has a great circle arc as its track
from A to B.
Two principal problems of navigation are to determine the distance
from A to B along the great circle arc AB and to determine the
direction of that arc at any one of its points.
In 89 we have defined the latitude and longitude of a point on
the earth's surface, and the difference of longitude of two points.
We now add two other definitions, referring to Figure 98.
Colatitude of A = angle NO A.
Bearing of A from B = the acute angle which the great circle BA
makes with the meridian through B.
For bearings we use the notation of 85. Thus, if A were west of B
and angle NBA were 50, the bearing of A from B would be N 50 W;
if B were east of A, and angle SAB were 75, the bearing of B from
A would be S 75 E.
If the latitude and longitude of A and B are given, we can find the
distance from A to B along the great circle passing through them, and
the bearing of each point from the other. We obtain these by solving
154 TRIGONOMETRY
the spherical triangle NAB, where
AN = colatitude of A,
BN = colatitude of B,
AB = great circle distance from A to B,
(1) Z BNA = difference of longitude of A and ,
Z AM = bearing of B from 4, or supplement of that bearing,
Z W2M = bearing of A from 5, or supplement of that bearing.
When any three of these parts are given, we can solve for the other
three.
In great circle navigation the term course is used to specify direc-
tion on a great circle track. Thus, if A is in the north hemisphere,
the angle NAB taken as a positive angle between and 180 is
called the course angle of the track from A to B. The initial course,
commonly designated by the symbol C n , is the angle between and
360 which gives the value of the angle NAB when measured clock-
wise from AN to AB.
To express distances on the earth's surface in linear units when
they have been given in degrees, minutes, and seconds, we may use
the geographical or nautical mile whose length is that of a great
circle arc of 1', and which can be reduced to statute miles by means
of the formulas:
1 nautical mile = 1'.
1 nautical mile = 6080 ft.
(2) 1 statute mile = 5280 ft.
1 nautical mile = 1.1515 statute miles.
1 statute mile = 0.8684 nautical miles.
To these formulas we add, for reference,
R = earth's radius = 3959 statute miles.
Example. For the great circle track from New York, lat. 40 43' N, long.
74 0' W, to Liverpool, lat. 53 24' N, long. 3 4' W, find the distance, the bearing
of each city from the other, the course angle, and the initial
course.
Solution. If we reletter the ANB triangle as in Fig-
ure 99, with A at New York, B at Liverpool, and C at the
north pole, we have
a - 90 - 53 24' = 36 36',
b - 90 - 40 43' = 49 17', p
C = 74 0' - 3 4' = 70 56'.
The solution conies under Case III for spherical triangles ( 79, p. 129). If we
keep four significant figures in our answers, we obtain
A - 49 30', B = 75 8', c - 47 50'.
APPLICATIONS 155
Hence
distance is 47 50', or 2870 nautical miles,
bearing of Liverpool from New York is N 49 30' E,
bearing of New York from Liverpool is N 75 8' W,
course angle (New York to Liverpool) is 49 30',
initial course C (New York to Liverpool) is 49 30'.
We note that the initial course for the track from Liverpool to New York is
360 - 75 8' = 284 52'; also that, although New York is farther south than
Liverpool, an air pilot would start in a northerly direction (N 75 8' W) in
flying from Liverpool to New York by the shortest path.
EXERCISES
For the great circle track from the first to the second place specified in each of Ex-
ercises 1 to 6, find the initial course, the distance in nautical miles, and the bearing of
each place from the other:
1. Boston, lat. 42 21' N, long. 71 4' W, and San Diego, lat. 32 43' N,
long. 11710'W.
2. New York, lat. 40 43' N, long. 74 0' W, arid Cape Town, lat. 33 56' S,
long. 1829'E.
3. Valparaiso, lat. 33 2' S, long. 71 39' W, and Rio de Janeiro, lat. 22 54' S,
long. 43 10' W.
4. Seattle, lat. 47 36' N, long. 122 20' W, and Manila, lat. 14 36' N, long.
120 58' E.
6. San Francisco, lat. 37 47' 30 ; ' N, long. 122 27' 50" W, and Sydney, lat.
33 51' 40" S, long. 151 12' 25" E.
6. Bombay, lat. 18 53' 45" N, long. 72 48' 55" E, and Greenwich, lat.
51 28' 40" N.
Find the shortest distance in statute miles between the places specified in Ex-
ercises 7, 8 (for latitudes and longitudes see Exercises 1 to 6) :
7. New York and San Diego. 8. Bombay and Sydney.
9. A great circle track starts from New York (see Exercise 2) bearing
N 29 35' E. At what angle will it cross the meridian of 30' W longitude, and
how far will this point be from the north pole?
10. Determine the latitude and longitude of the point farthest north on the
great circle specified in Exercise 9. Hint. If C is this point, A is at New York,
and N at the north pole, then ACN is a right angle.
11. If <f> is the latitude of A, and R is the earth's radius, then the parallel of
latitude through A is a circle (but not a great circle) of radius R cos $. What
is the distance from Providence (lat. 4150'N, long. 7124'W) to Chicago
(lat. 41 50' N, long. 87 38' W), (a) along their common parallel of latitude,
(b) along the great circle through them?
12. From San Francisco (see Exercise 5) a ship has sailed just 2000 nautical
miles on a great circle track that started due west. What is the ship's latitude
and longitude?
13. Find the latitude and longitude of the point midway between Liverpool
(lat. 53 24' N, long. 3 4' W) and San Francisco on the shortest air route.
156
TRIGONOMETRY
93. Positions on the celestial sphere. In the following sections
we shall consider the solution of certain problems of especial interest
to navigators; for example, the problem of computing an observer's
position, or the time of observation, from the position of the sun or
the positions of stars.
We see the heavenly bodies as though they were situated on a vast
spherical surface whose center is at the observer's eye. If the diam-
eter of this sphere is taken so large that the length of the earth's
radius is negligible by comparison, we call the center of the earth
Z (Zenith)
Q
p (North
k Pole)
P' ,
(South
Pole)
Z' (Nadir)
FIG. 100
the center of the celestial sphere. By the position of a heavenly
body we mean its apparent position, that is, the position on the
celestial sphere of its projection from the observer's eye.
We shall describe two coordinate systems on the celestial sphere
comparable to the latitude-longitude system on the earth's surface.
These are the declination-hour angle system, and the altitude-
azimuth system.
The following are preliminary definitions, referring to Figure 100:
The zenith, Z, is the point vertically above the observer; the line
OZ from the center of the earth to Z passes through the observer.
The opposite point, Z', is called the nadir.
The horizon of the observer is the great circle of which his zenith,
APPLICATIONS 157
Z, is a pole; its plane is tangent to the earth at the observer's position.
It is the circle SWNE of Figure 100.
The celestial poles, P and P', are the intersections of the earth's
axis with the celestial sphere. In Figure 100, P is the north pole and
P' the south pole.
The celestial equator is the great circle QEQ'W of which P and P'
are poles. It is the intersection of the earth's equatorial plane with
the celestial sphere.
The celestial meridian of the observer is the great circle through
Z and P. Its intersections with the horizon determine the north
point, N, and the south point, S. If the observer faces north, the
east point, E, is to his right, and the west point, W, to his left.
In the declination-hour angle system a position A on the celestial
sphere is described as follows:
The hour circle of A is the great circle through P and A.
The declination of A is the distance of A (in degrees) from the
celestial equator along the hour circle of A. In Figure 100, it is the
measure of arc LA. It is analogous to latitude on the earth's surface.
If A is 20 north of the equator its declination is given as 20 N;
if it were 20 south its decimation would be 20 S. The codeclination
is 90 minus the declination (south declination taken as negative).
The hour angle of A is the angle ZPA between the celestial meridian
and the hour circle of A. It is thus analogous to the longitude angle,
but, instead of measuring it in degrees east or west of the celestial
meridian, we use hours as a unit with the connecting formula
1 hour = 15 degrees.
This relation is suggested by the apparent revolution of the celestial
sphere through 360 in 24 hours, and therefore through 15 in one
hour. The hour angle is measured westward from the celestial merid-
ian in hours, minutes, and seconds (time units) from hr. to 24 hr.
For some purposes right ascension is used instead of hour angle; in the dec-
lination-right ascension system the zero meridian is the great circle joining the
poles PP* and passing through the vernal equinox. The latter point is the inter-
section of the equator with the annual path of the sun in March.
In the altitude-azimuth system, the zenith takes the place of the
north pole, and the horizon that of the equator in the declination-
hour angle system. The corresponding definitions are:
The vertical circle of A is the great circle through Z and A.
The altitude of A is the distance of A (in degrees) above the horizon.
158 TRIGONOMETRY
It is measured along the vertical circle of A ; in Figure 100 it is the
measure of arc MA. Altitude is positive for positions above the
horizon, and negative for positions below the horizon. The coaltitude
is 90 minus the altitude.
The azimuth of A is the angle PZA between the celestial meridian
and the vertical circle of A. It is measured from arc ZPN east or
west up to 180, and is designated as so many degrees, minutes, and
seconds E or W.
Owing to the apparent daily revolution of the celestial sphere, the
point Z is not a fixed point on the sphere over any interval of time;
also at a given instant the zenith point is different for observers in
different places. Thus hour angle, altitude, and azimuth are in-
stantaneous and local, while declination is not.* As we shall see in
the next section, these contrasting qualities enable us to solve prob-
lems of time and position when both sets of coordinates of a heavenly
body are given.
94. The astronomical triangle, and problems connected with its
solution. The triangle AZP is called the astronomical triangle cor-
responding to A. Various problems of time and position can be
referred to the problem of solving the astronomical triangle.
The part ZP, which is independent of the position of A, is measured
by the angle ZOP\ this angle is the colatitude, on the earth's surface,
of the observer, since the direction of P'OP is that of the earth's
axis and OZ passes through the earth's center and the observer.
Four of the other parts of triangle AZP are expressible in terms of
the coordinates of A in the two systems. Thus we have the relations
AZ = 90 - MA = coaltitude of A,
AP = 90 LA = codeclination of A,
ZP = 90 - QZ = codeclination of the zenith,
= 90 - NP = coaltitude of the pole,
= colatitude of the observer,
Z.ZPA = hour angle of A, if this hour angle is less than 12 hr.,
= 24 hr. hour angle of A, if this hour angle is greater
than 12 hr.,
Z.PZA = azimuth of A.
* We assume here that the polar axis of the earth has an unvarying direction
and that the celestial sphere is so large that its intersections with the axial line
appear as the same point for all positions of the earth in its orbit.
APPLICATIONS 159
In these equations we are to take MA as negative if A is below the
horizon, and LA as negative if A is south of the equator.
We will now discuss some problems solvable in terms of the as-
tronomical triangle.
(a) To find the local solar time for an observer at a moment when
the sun's declination is known and its altitude is measured. The
observer's latitude is supposed to be known.
Here A is the sun's position. The observer's colatitude gives ZP
and the given data determine AP and AZ. We can therefore solve
the triangle AZPj since three sides are known, and in particular we
can find angle ZPA in degrees. We reduce this to hours, minutes,
and seconds, and the result will be the time before noon, or after
noon, as the case may be.
(b) To find the time of sunrise or sunset.
This is a special case of the preceding problem. The coaltitude of
the sun is 90 at sunrise or sunset. The triangle APZ is quadrantal
and is to be solved as in 71 (p. 119).
(c) To determine the observer's latitude after finding the altitude and
hour angle of a heavenly body whose declination is known.
This problem is at once reducible to that of finding ZP, given
AZ, AP, and angle ZPA.
(d) To find the azimuth of a heavenly body, given its declination and
hour angle and the observer's latitude.
This is the problem of determining angle PZA, given ZP, AP, and
angle ZPA. The result is used to find compass corrections.
EXERCISES
In these Exercises, carry out the computation as though the data were exact:
1. An observer in latitude 30 N takes a morning observation and finds the
sun's altitude to be 35, its declination being 15 S. Find the time of observation.
2. At the U. S. Naval Observatory in Washington (lat. 38 55' 14" N), the
altitude of the sun in the west was observed to be 60 20' 10" when its declination
was 20 31' 30" N. Find the time of observation.
3. Find the time of sunrise at New York (lat. 40 43' N) when the sun's dec-
lination is 18 25' N.
4. At San Francisco (lat. 37 47' 30" N) what is the time of sunset when the
sun's declination is 8 45' S?
6. Find the length of the longest day of the year (sun's declination 23 30' N)
in Chicago (lat. 4l50'N). What angle does the sun's path make with the
horizon at sunset?
6. Find the length of the shortest day of the year (sun's declination 23 30' S)
in Edinburgh (lat. 55 57' N). What is the bearing of the sun at sunset?
160 TRIGONOMETRY
7. On the longest day of the year in Chicago (see Exercise 5) at what time of
day is the sun due west of the observer?
8. The compass bearing of a star is N 47 E. The observer's latitude is
28 26' 55 N, the hour angle of the star is 16 hr. .59 min. 30 sec., and its declination
is 37 54' 5* N. What is the error of the compass?
9. At 3 : 35 P.M., local solar time, the sun's altitude is 27 23', and its declina-
tion is 8 27' N. Find the observer's latitude.
ANSWERS TO ODD-NUMBERED PROBLEMS
Answers have been omitted in cases where little would be left for the stu-
dent to do if the answer were given. Where approximate results were required,
four-place tables were used in Chapters T-V. In Chapters VI-IX, five-place
tables were used. Answers obtained by using other tables may be slightly
different from those given. Occasionally a problem may be solved in several
ways, and answers when approximate may differ slightly according to the method
used.
While the authors have attempted to check the answers with great care, they
can hardly hope that all errors have been avoided. They will appreciate notice
of errors which are discovered.
CHAPTER I
Art. 2, Pages 3-4
13. - 840; - 50,400; - 70. 15. 600; - 600.
9.
r
sin0
cos 6
tan0
A
B
C
D
E
5
10
15
13
3
*
-*
-*
-tt
- 1
t
i
-i
ft
3
-4
1
-
Art. 6, Pages 8-9
11.
r
sin B
cos 6>
tan 6
A
13
H
ft
B
13
A
-H
-A
C
10
a
5
~~ 5
4
D
15
-I
-J
E
5
i
13. A (8, 6); B (8, - 6); C (-
); D (- , -
(0, 10).
15. A
17.
); #(,-); C (- W, M); D (- 6, - 8); E (- 10, 0).
19.
sin0
COS0
ttaaO
cot
sec
csc B
A
B
C
D
f
-It
-H
*
ft
-H
-A
1
-V
-1?
i
-ft
-i*
A
*
-
_ 17
\
-
t&
-H
sin I?
cos B
tan B
cotfl
seed
cse B
A
B
C
D
H
*
-
-H
A
-I
It
-A
-1
-M
A
-4
-
A
-4
-
H
f
-
-H
161
162 . TRIGONOMETRY
21. Ill; IV. 23. II; IV. 26. II; III. 27. I; II.
29.
31.
6
sin
cos
tan0
cot 6
sec0
CSC
78
+
+
+
+
+
+
213
+
+
- 310
etc.
(to be
complet
ed by the
student)
- 601
1111
sin0
cos 6
tan0
cot
sec0
esc
25
+
+
etc.
301
+
etc.
- 140
etc.
(to be
complet
ed by the
student)
- 800
2000
7. r = 13, = 65.
11. x - 7.28, y - 3.36.
Art. 7, Page 12
9. r = 11.2, - 100.
13. x = - 2.79, y = 31.0.
Art. 10, Page 16
3.
70
80
90
100
260
sin (9
cos
fan ft
.94
.34
20
.98
.17
R Q
1.00
.98
- .17
C
- .98
- .17
5Q
120
240
250
270
280
sin0
COS0
fart D
.86
- .50
1 79
- .86
- .50
1 79
- .94
- .34
20
- 1
- .98
.17
e o
tan (/
6.
9.
11.
143; 180.
7. 247; 270.
sin0
cos
tan0
cot
sec
CSC
315
210
- JV2
-*
iV2
- iV
- 1
jVs
- 1
V3
V2
- *V
- v
- 2
sin0
COS0
tan0
C0t0
sec (9
CSC
150
225
- JV2
- iV3
- JV2
- *V3
1
-VI
1
-fVI
- V2
2
- V
ANSWERS
163
13.
15.
e
sinB
cos0
tan0
cot0
sec0
C3C0
- 60
270
-iV3
- 1
4
- V
- 4^3
2 '
-V3
- 1
e
sin6
cos e
tan0
C0t0
sec0
CSC
180
- 150
-*
- 1
-iVa
iV3
vif
- 1
-SVs
2
Art. 11, Page 17
Exercise
sin
COS
tan0
cot e
seed
CSC
1.
A
H
A
H
3.
ft
IT?
-A
If
6.
=t \/3
jVs
Vs
=t Vs
2
7.
fa
H
A
^
i V
9.
1
t
-1
1
t
11.
iv 2
|"S/2
- i
=fc v2
rb V2
13.
*
1
1
-*
1
15.
t
1
t
*
~s
17.
ft
H
A
V
19. cos = Vl - sin 2 0; tan
in
/)
sec =
/i
. _ ; esc - -
VI- sin 2 sm
21. sin ~-
V^
sec =
23. sin =
_
- tan 2
cos
_
Vl + tan 2
COt
C0t0
Vl - sin 2 0.
sin '
d=tan0
tan
cot0 =
-
v sec 2 0-1
_.
Vsec 2 0-1
Art. 13, Page 19
3. 10.6 mi. per hr. ; 10.6 mi. per hr.
5. 68 downstream from a line straight across, at 5.4 mi. per hr.
7. 29.2 lb., at 31 angle with horizontal.
9. Westward 4.8 knots per hour; southward 13.2 knots per hour.
164 TRIGONOMETRY
CHAPTER II
Art. 17, Page 22
1. .3173; .9925. 3. 1.049; 1.066. 6. .2805; 2.932.
7. .6111; .9833. 9. 1.643; 2.309. 11. .3378; .0437.
13. 39 30'; 12 50'; 82 0'; 72 0'. 16. 32 0'; 2 30'; 8 50'; 66 40'.
Art. 18, Page 24
1. 0.6672; 0.7494. 3. 0.9795; 1.263. 6. 1.037; 2.224.
7. 0.9896; 0.7835. 9. 0.9131; 3.456. 11. 0.8168; 1.496.
13. 13 25'. 15. 6 22'. 17. 54 6'. 19. 48 25'.
21. 37 34'. 23. 6 17'. 26. 59 18'. 27. 55 37'.
Art. 19, Pages 26-26
1. 13.47. 3. 101,800. 6. 0.007779. 7. 67.08. 9. 4516.
11. 7,674,000,000. 13. 6800. 16. 67,680,000. 17. 0.006754.
19. 768,200,000,000. 21. 38,840,000. 23. 965,100. 26. 3.13.
27. 6.92. 29. 40.6. 31. 0.964. 33. 733. 36. 0.0490. 37. 3.423.
39. 8.301. 41. 25.02. 43. 46.84. 45. 227.8. 47. 0.08803.
Art. 21, Page 28
1. a = 3.711; b = 7.200; B = 62 44'.
3. a - 19.96; 6 * 18.18; A 47 40'.
6. a = 16.37; c = 24.34; B = 47 43'.
7. A - 33 48'; B = 56 12'; c = 2.984.
9. A - 42 5'; B = 47 55'; c = 16.71. 11. Impossible.
13. A - 9 40'; b = 1.585; c = 1.608. (A - 10; b = 1.6; c = 1.6.)
15. B = 27 6'; b = 4.708; c = 10.33. (B = 27; b = 4.7; c = 10.3.)
17. A = 41 15'; a = 72.79; c = 110.4. (A = 41; a = 73; c = 110.)
19. A - 51 33'; B = 38 27'; a = 639.9. (A = 51 30'; B = 38' 30'; a = 640.)
21. A - 48 42'; B = 41 18'; c = 53.97.
23. A = 60 29'; B = 29 31'; c = 30.58.
Art. 22, Pages 29-30
1. C = 87 16'; = 6 = 914.3. 3. A = 65 39'; C - 48 42'; b - 1504.
6. C - 55 36'; c - 82.70; 6 = 88.66.
7. Inscribed radius = 3. 137 in. (or 3. 14 in. to indicated accuracy); circumscribed
radius = 3.878 in. (or 3.88 in.).
9. Side = 7.794 in. (or 7.79 in. to indicated accuracy) ; inscribed radius == 5.364
in. (or 5.36 in.).
ANSWERS
165
11. Perimeter = 11.66 in. (or 11.7 in. to indicated accuracy); circumscribed
radius =- 1.904 in. (or 1.90 in.).
13. B 38 1'; C = 128 32'; a = 14.05.
15. A = 49 59'; B = 31 49'; c = 5.394.
Art. 23, Pages 32-34
1. 63.4 ft. 3. 319 ft. 6. 267 yd. 7. 24.4 mi.
9. 62 20'; 25.2ft. 11. 20 30'; 53.2ft.; 142.4ft.
13. 2460 ft. 16. 16.1 ft. 17. N 30 40' E; 27.6 mi.
19. 440 yd. 21. 118.7 ft. 26. 23.6 ft.; at least 126.9 ft.
CHAPTER III
1. .8746.
9. - 1.004.
17. - .6088.
25. - 5.671.
33. 305 2'.
41. 352 58'.
49. 144 7'.
Art. 24, Pages 36-37
3. - .8387. 6. - .6157.
13. .6777.
21. - 5.671.
29. - .8391.
37. 295 36'.
46. 11 3 22'.
53. 193 23'.
7. - .6691.
15. - .8718.
23. - 1.192.
31. 120 25'.
39. 172 58'.
47. 293 22'.
11. - .9520.
19. - 1.013.
27. .8391.
35. 97 5'.
43. 113 6'.
61. 324 7'.
65. 13 34' + n 360; 166 26' + n 360.
67. None. 69. 76 26' + n 360.
61. None. 63. 13 12' + n 180. 66. 66 54' + n ISO .
67. 76 48' + n 180. 69. 23 6' + n 180. 71. None.
73. 64 45' + n 360. 76. None.
77. 25 15' + n 360; 154 45' + n 360. 79. x = - 1.9392, y = 3.4984.
81. x - 4.3568, y = - 6.6696. 83. x = 9.4560, y = - 7.3884.
86. r = 25, = 106 15'. 87. r = 25, 6 - 286 15'.
89. r = 13, 6 = 202 13'.
1. - sin 57; - cos 33.
6. - cos 24; - sin 66.
9. tan 26; cot 64.
13. - tan 33; - cot 57.
17. esc 11; sec 79.
21. sec 32; esc 58.
11. Period = 360.
Art. 27, Pages 40-41
3. sin 57; cos 33.
7. cos 4; sin 86.
11. - tan 26; - cot 64.
16. cot 43; tan 47.
19. - sec 21; - esc 69.
23. esc 42; sec 48.
Art. 32, Page 47
13. Period - 360.
166 TRIGONOMETRY
CHAPTER IV
Art 34, Pages 61-62
13 sinl+1. 15. sin'
cos 9
cos 2
21. sin 6 H -- - 23. sin 6 cos 0.
cos &
25. an. 27.
1 -sin 2 " smO (1 - sin 2 0)
2-sin 2 31 1 + sin 2
sin * sin 2
Art. 37, Page 56
3. - . 25. _ o.2797. 27. 0.0098.
. __ .
29. i|. 3V If-.. 33. ft. 35. - |f. 37. *&.
Art. 38, Pages 58-69
1. ^"Vl or 2 - V3. 3. - 2 - Vs. 9. ft. 11. -
Art. 40, Pages 62-63
1- ; ft; . 3. H; - Ht; -tt*.
6. ft; ft; 7. -; - HI ; t-
9 A_- 3 . i.
J VI6' 3
1. sin 40 -f sin 20. 3. sin 7a sin 3a. 6. cos a cos 5a.
7. i cos 3a -f i cos a. 9. i sin lla - i sin a. 27. 2 sin 40 cos 4.
29. - 2 cos 45 sin 5. 31. 2 cos 45 cos 5. 33. 2 sin 25 sin 5.
35. 2 sin 3a cos a. 37. 2 cos 2a sin a. 39. 2 cos - cos -7^-
4 4
41. - 2 sin 2a sin a. 43. 1 tan 25. 46. cot 30 tan 10.
47. 1 - cot 2a. 49. - tan ^ tan -
Art. 43, Page 74
1. 45.
3. 29.
6. - 30.
11. 26.
13. 60.
16. 114.
21. - 30.
23. - 23.
25. 30.
31. 45.
33. 90.
36. 35 20'.
ANSWERS 167
CHAPTER V
Art. 42, Page 70
1 E E ?E- ~ q E- ^E- _ E _ ?E
4' 3' 4 ' 3 ' 6' 4 ' 4' ~2~'
6. 0.6109; 0.8727; 1.0472; 2.0594. 7. 0.3491; 0.7854; 1.2217; 2.2690.
9. 0.2696; 2.3072. 11. 0.7540; 2.4376. 13. 22 30'; 120; 630; 114 35.5'.
15. 36; 77 8.6'; 420; 229 11'. 17. 54 1'; 59 28'. 19. 31 7'; 88 25'.
21. 49 24'; 165 49'. 23. 57 57'; 172 32'.
26. 1.5 radians; 85 57'. 27. 1.2 radians; 68 45'.
29. 1.5 radians; 85 57'. 31. 2.2 radians; 126 3'.
33. 4.5 in. 36. 6.000 in. 37. 9.800 in. 39. 17.19 in.
41. 249.4 ft. 43. 9.579 in. 46. 13.31 ft.
7. - 59. 9. 60.
17. 45. 19. 34 30 r .
27. 34. 29. 34.
37. 0.0669 radians.
39. 0.0669 radians. 41. 0.0524 radians. 43. 1.5563 radians.
46. - 0.0145 radians. 47. TT 2nir radians; 180 n 360.
49. 1.2915 2mr radians, 1.8501 =b 2mr radians; 74 n 360, 106 db n 360.
61. - 0.1571 2nir radians, - 2.9845 2nir radians; - 9 n 360, - 171
n 360.
63. 0.6431 =b 2nir radians, 2.4985 2mr radians; 36 52' =b n360, 143 8'
n 360.
65. 0.8760 nir radians; 50 11' n 180.
67. 1.2310 2mr radians, - 1.2310 2mr radians; 70 32' n 360, - 70 32'
n 360.
69. j. 61. . 63. f. 66. $.
i
67. 20 =b n 360, 160 n 360. 69. 50. 71. 170.
73. 1. 75. 2. 77. H* -
Art. 44, Page 77
1. 48 11' n 360, - 48 11' n 360.
3. 41 49' n 360, 138 11' n 360.
6. 60 n 180, 120 n 180. 7. 60 n 180, 120* =fc n 180.
9. db n 180. 11. n 180.
13. =fc n 180, 45 n 90. 15. 22 30' db n 45.
168
TRIGONOMETRY
17. 0.9552 nir, - 0.9552 db nir.
tt.5. 23. ^nw.
27. 60, 120. 29. 67 30', 157 30'.
31. 22 30', 67 30', 90, 112 30', 157 30'.
33. 0, 120, 240, 360; 0, ~^, ^, 2ir.
36. 45, 63 26', 225, 243 26'; ^ 1.1071, ^, 4.2487.
37. 90, 120, 240, 270; | ^ ^ ^-
39. 30, 150; |, ^-
41. 22 30', 90, 112 30', 202 30', 270, 292 30'; |, ~ t
43. 67 23', 232 53'; 1.1761,4.0646.
46. 41 49', 138 11';, 0.7298, 2.4118.
47. None. 49. None.
CHAPTER VI
Art. 47, Pages 81-82
19. 1.1071 nir.
26. 0, 60, 180.
1. 1; 4. 6. - 1; - 3.
11. 5.33905; 8.10857 - 10.
9. 3.57426; 0.45255.
13. 9.95347 - 10; 6.47451.
16. 4.63783.
23. 7.85601 - 10.
31. 6.6070.
39. 261.04.
47. 0.00027825.
17. 0.82812.
26. 9.80473 - 10.
33. 300.00.
41. 5015.3.
49. 0.74745.
19. 2.56983.
27. 6.40957 - 10.
36. 0.051740.
43. 2578900.
61. 0.0000050196.
21. 4.68517.
29. 3.2110.
37. 0.080220.
45. 26978000.
Art. 60, Page 86
1. 70.663. 3. 2421.6. 6. 82.350. 7. 0.77163. 9. 0.18973.
11. 31.627. 13. 3086.7. 16. 0.70285. 17. - 2.8179. 19. 0.041168.
1. 9.58515 - 10.
7. 10.47122 - 10.
13. 21 11' 0".
21. 40 7' 44".
29. 26 0' 0".
Art. 61, Pages 87-88
3. 9.96366 - 10.
9. 9.99862 - 10.
16. 18 20' 0". 17. 52 56' 0*.
23. 7 25' 38". 26. 78 10' 40".
31. 85 59' 47".
6. 9.91500 - 10.
11. 9.96336 - 10.
19. 54 20' 0".
27. 62 3' 33".
ANSWERS 169
CHAPTER VII
Art. 62, Page 91
6. A - 28 29' 10', a = 2344.4, c - 4915.4.
7. A = 70 32' 25*, a = 0.99750, c - 1.0579.
9. A = 38 4' 5", = 51 55' 55", a - 0.55022.
11. A = 42 48' 50", B = 47 11' 10", 6 - 0.045708.
13. B = 17 45' 15", a = 6.9500, 6 = 2.2252.
16. A = 10 15'J20", a = 4.8423, 6 = 26.763.
17. A - 67 20' 55", B = 22 39' 5", a - 650.40.
19. A = 42 2' 55", 5 = 47 57' 5", a = 18308.
Art. 66, Pages 94-95
1. C = 78 14' 10", b - 33650, c = 34188.
3. A 74 37' 40", 6 - 22.238, c - 17.076.
5. B = 37 30' 0", 6 = 0.014419, c = 0.020530.
7. C - 61 26' 40", 6 = 166.31, c - 184.63.
9. C =- 33 46' 0", a = 878.16, c * 528.49.
11. AC - 6368.1 ft., BC - 9990.8 ft.
Art. 67, Page 98
1. B = 34 39' 25", C - 102 43' 35", c = 34.770.
3. A = 39 11' 40", C = 45 7' 5", c = 5151.3.
6. Bi = 91 44' 5", Ci = 48 10' 45", 61 - 10,870.
B 2 = 8 5' 35", C 2 = 131 49' 15", 6 2 = 1531.0.
7. B = 18 51' 30", C = 106 54' 35", c = 0.59051.
9. A = 36 42' 30", C = 69 45' 10", c = 3.6408.
11. Ai = 37 37' 55", Bi = 76 58' 30", a,"= 20.370.
At = 11 34' 55", B 2 = 103 1' 30", a 2 - 6.6980.
13. No solution.
16. B = 24 10' 25", C = 34 22' 15", 6 = 0.23635.
17. N 2 40' 30" E.
Art. 69, Page 100
1. A = 35 39' 0", B = 102 24' 40", c = 37.504.
3. B - 26 41' 0", C - 44 49' 30", a - 3.6523.
6. A = 80 26' 15", C - 70 10' 0", 6 = 2.5006.
7. B = 36 9' 5", C - 20 26' 40", a = 0.10113.
9. A = 85 7' 40", B - 19 23' 40", c - 6.8254.
11. 14.421ft.; 19.688ft.
170 TRIGONOMETRY
Art. 62, Pages 103-104
1. A = 57 19' 0*, B - 50 26' 20", C = 72 14' 50*.
3. A - 37 37' 5", B = 84 19' 30*, C - 58 3' 30".
6. A - 41 39' 55", B = 23 2' 50", C - 115 17' 15".
7. A 42 15' 10", B = 114 16' 15", C = 23 28' 35".
9. A = 72 17' 15", B = 49 4' 55", C = 58 37' 55".
11. A = 32 58' 30", B = 50 23' 40", C = 96 37' 55".
13. A = 30, B = 50, C = 100 (to the nearest 10).
16. A = 43, B = 50, C = 87 (to the nearest degree).
Art. 63, Page 105
6. 2,945,600. 7. 2161.4. 9. 489.09. 11. 8.8740.
Miscellaneous, Pages 106-108
1. A = 46 59' 0", B = 105 30' 20", c = 206.61.
3. C = 30, a - 2.0005, c = 1.0003.
6. A! - 78 26' 20", Bi = 60 20' 40", 61 = 538.76.
A 2 = 101 33' 40", 5 2 = 37 13' 20", h = 375.02. 7. No triangle.
9. C = 97 59' 0", a = 1.9578, c = 2.0135. 11. No triangle.
13. A = 27 56' 40", B = 38 10' 10", C = 113 53' 10". 15. No triangle.
17. A = 48 54' 45", 5 = 7 38' 0", b = 14529. 19. No triangle.
21. A = 98 49' 0", B = 30 11' 30", C = 50 59' 30".
23. A = 30 26' 20", C = 28 53' 20", b - 121.85.
26. A = 23 18' 0", b = 7.3543, c = 3.9042.
27. S = 27048. 29. S = 73.157. 31. S = 0.23890.
33. r = 88.744, R = 336.93. 35. r - 0.12410, fl = 1.0161.
37. n = 10.412, Ri = 22.043.
r z = 6.6063, R z = 22.043.
39. r = 11.075, /? = 31.622.
41. 32 4' 0". (To three significant figures, 32 0'.)
43. r = 6.3281 in. (To two significant figures, r = 6.3 in.)
46. r = 3\/2 in. (By five-place logarithms, r = 4.2427 in.; to three significant
figures, r = 4.24 in.)
47. Length = 1118.22 ft., AD = 1106.16 ft. (To three significant figures,
length = 1120 ft., AD = 1110 ft.)
49. 18469 mi., 12.826 mi. (To four significant figures, 18470 mi., 12.83 mi.)
61. 13 38' 20". (To two significant figures, 14.)
63. 85.69 ft. (To two significant figures, 86 ft.)
66. 24.188ft. 67. 2.1361 in.
69. 3.6445 ft. (To three significant figures, 3.64 ft.)
ANSWERS 171
61. Angle made with greater force = 10 58' 45", magnitude = 53.123 Ib. (To
four significant figures, 10 59', 53.12 Ib.)
63. 95,758,000 mi., or 43, 1 13,000 mi. (To four significant figures, 95,760,000 mi.,
or 43, 110,000 mi.)
65. 6.9663 mi., N 61 27' 25" W. (To two significant figures, 7.0 mi., N 61 W.)
67. To the required accuracy, 1490 yd.
CHAPTER VIII
Art. 65, Page 111
1. 143 14' (to the nearest minute). 3. As 3 to 8.
Art. 70, Page 118
11. a = 15 36' 15 ;/ , 6 = 55 28' 35", B = 79 31' 0".
13. b = 90, c = 90, B = 90.
15. a = 147 31' 50", b = 60 32' 5", c = 114 31' 5".
17. b = 22 0' 50", A = 68 19' 0", B = 30 31' 0".
19. a = 17 59' 30", c = 112 9' 55", B = 97 36' 10".
21. a = 44 44' 0", c = 46 40' 5", B = 20 49' 50".
23. a - 98 35' 5", A = 96 19' 55", B = 47 38' 5".
26. c = 117 3' 5", A = 33 11' 45", B = 106 34' 15".
27. ai = 28 14' 30", d = 78 53' 20", AI = 28 49' 55".
a 2 = 151 45' 30", c 2 = 101 6' 40", A 2 = 151 10' 5".
29. a = 140 12' 0", b = 120 30' 10", A = 135 57' 40".
Art. 71, Page 119
1. A = B = 62 57' 20", C = 74 11' 10".
3. a = 6 = 136 6' 45", C = 161 39' 35".
5. a = b = 74 26' 5", C = 163 11' 40".
7. a = b = 107 50' 15", c = 57 34' 0".
9. A = 35 35' 20", B = 147 31' 10", C = 46 41' 15".
11. a, = 30 54' 50", A l = 30 14' 20", Ci = 78 35' 20".
02 = 149 5' 10", A 2 = 149 45' 40", C 2 = 101 24' 40".
13. A = 154 47' 10", B = 127 59' 10", C = 123 50' 20".
15. &i = 28 49' 30", B, = 28 14' 5", d = 78 53' 20".
& 2 = 151 10' 30", B 2 = 151 45' 55", C 2 = 101 6' 40".
Art. 73, Page 122
3. C = 19 11'. 5. a = 37 50'. 7. a = 38 55'.
9. A = 65 30'. 11. c = 42 27', A = 59 8', B - 84 0'.
172 TRIGONOMETRY
Art 78, Page 129
1. A - 109 57' 45", B - 78 59' 10", C = 39 59' 55".
3. A - 94 49' 25", B 70 1' 40", C = 131 18' 15".
6. A - 55 52' 40", = 59 4' 0", C = 88 12' 35".
7. A = 65 33' 5", B = 97 26' 30", C = 95 38' 5".
9. a - 64 11' 50", 6 = 33 47' 45", c = 40 37' 15".
11. a - 27 22' 20", 6 = 117 9' 35", c - 138 20' 40".
13. a - 47 6' 15", b = 40 18' 20", c - 20 17' 20".
15. a - 82 56' 45", 6 - 107 35' 10", c = 27 9' 40".
Art. 80, Pages 130-131
1. A = 96 8' 35", B = 65 22' 35", c = 99 40' 50".
3. A - 146 33' 25", C = 47 28' 40", b - 27 17' 0".
5. A - 55 52' 25", B - 20 9' 55", c - 66 20' 50".
7. A - 142 12' 45", C = 129 5' 30", 6 = 60 4' 55".
9. a - 64 20' 5", 6 = 40 40' 50", C - 31 39' 40".
11. a - 159 6' 50", c - 139 53' 30", B - 102 33' 20".
13. a - 64 47' 15", 6 = 48 3' 25", C = 73 46' 50".
15. a = 124 12' 30", c = 82 47' 40", B = 128 41' 50".
Art. 82, Pages 133-134
1. No solution. 3. One solution. 6. One solution.
7. 1 - 70 20' 30", Ci = 80 13' 0", ci - 50 46' 10".
Bz - 109 39' 30", C 2 = 34 4' 40", c 2 - 26 7' 55".
9. A = 22 20' 25", B = 146 39' 30", a = 27 22' 10".
11. Ai = 55 1' 25", Bi = 146 39' 0", bi = 138 12' 5".
A 2 = 124 58' 35", B* = 45 20' 20", & 2 = 59 34' 45".
13. 61 = 84 10 7 0", ci = 71 42' 35", d = 68 28' 40".
62 - 95 50' 0", c 2 - 40 49' 30", C 2 - 39 49' 55".
15. b = 40 35' 10", a = 39 10' 5", A = 30 25' 10".
17. ci - 33 0' 20", 6! = 147 4' 40", #1 = 110 5' 10".
C2 a. 146 59' 40", 6 2 = 15 6' 20", B 2 = 26 45' 55".
Art. 83, Page 135
1. 0.5236 sq. ft. 3. 26.15 sq. in.
5. 9335 sq. ft. 7. 232.6 sq. ft.
9. 873,000 sq. ft. 11. 43,310 sq. mi.
ANSWERS 173
CHAPTER IX
Art. 85, Pages 138-140
NOTE. Five-place tables were used in obtaining the answers for this chapter,
except in Art. 92 and 94. These answers are rounded off to the appropriate
number of places and given in parentheses.
1.255.32yd.; 199.48yd. (255yd.; 199yd.)
3. 55.943yd.; 48.955yd.; S4111'20"E; 74.338yd. (56*yd.; 49yd.; S41E;
74 yd.)
5.33.550ft.; 216.68ft.; N 81 11' 55" E; 219.25ft. (33.55ft.; 216.7ft.;
N8112'E; 219.2ft.)
7. 750.18yd.; S 76 51' 45" E. (750.2yd.; S7652'E)
9.0.734yd.; -0.20yd. 11. 51 23' 10"; 1303.7. (51 23'; 1304)
13. 7455.8 yd., S 38 49' 40" E; 1102.0 yd., 427.0 yd. (7456 yd., S 38 50' E;
1102 yd., 427 yd.)
., _ , , . , d cos a sin (8 -f <b) tan ft d cos a sin (6 + 0)
15. h = a sm a, K = : 75 TT \ : 7^ -77
sm (0 - </>) sm (6 - <)
Art. 86, Page 142
1. 38 52' 3"; 3923.3yd. (38 52'; 3923yd.) 3. (0.3,2.6). 5. (0.4,4.3).
Art. 87, Pages 145-146
Answers which follow when expressed in mils are given to the nearest mil,
and when expressed in degrees and minutes are given to the nearest minute.
1. 178 mils; 444 mils; 800 mils. 3. 71 mils; 320 mils; 622 mils.
6. 130 mils; 311 mils. 7. 46 mils; 434 mils.
9. 7 2'; 14 38'. 11. 4 44'; 22 30'.
13. 0.025 radians; 0.221 radians. 15. 0.018 radians; 0.447 radians.
17. 255 mils; 1222 mils. 19. 407 mils; 672 mils.
21. 100 mils. 23. 350 mils. 25. 79 mils. 27. 233 mils.
37. 2 (or 3) mils. 39. 10.5 yd. 41. 32 yd. 43. 155 mils.
45. Increase range 1 yd., deflect gun 9 mils to left.
Art. 91, Page 152
1. 105.4 east; 67.2 north. 3. 83.9 east; 157.0 south.
5. 120.1 west; 183.1 south. 7. 146.4 west; 96.0 north.
174 TRIGONOMETRY
9. 77.5; 72 2'. 11. 112.2; 320 56'.
13. 68.6; 110 12'. 16. 78.9; 246 24'.
17. 44 19' N; 43 32' W. 19. 95.8 mi.
21. 57 57' S; 10 0' W. 23. 14 20' S; 2 58' W.
25. 165.3 mi.; 48 31'. 27. 68.3 mi. south; 154.0 mi. east.
29. 39 7.6' N; 63 8' W.
Art. 92, Page 165
Four-place tables were used in computing the answers for this article, except
in Exercise 5.
1. Distance = 2239 nautical miles. Rearing of San Diego from Boston,
N 89 23' W. Bearing of Boston from San Diego, N 61 27' E.
3. Distance = 1621 nautical miles. Bearing of Valparaiso from Rio de Ja-
neiro, S 61 41' W. Bearing of Rio de Janeiro from Valparaiso, N 75 19' E.
5. Distance = 6445.5 nautical miles. Bearing of Sydney from San Francisco,
S 60 17' 20" W. Bearing of San Francisco from Sydney, N 55 44' 44" E.
7. Distance = 2428 statute miles.
9. At an angle of 86 28', 1321 nautical miles from the north pole.
11. (a) 835.8 statute miles; (b) 834.6 statute miles.
13. Lat. 63 6' N, long. 76 11' W.
Art. 94, Pages 169-160
Four-place tables were used in obtaining the answers for this article.
1. 9 hr. 49 min. A.M. 3. 4 hr. 54 min. A.M.
6. 15 hr. 3 min. Angle between sun's path and horizon = 43 21'.
7. 4 hr. 4 min. P.M. 9. 54 26' N or 26 10' S.
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