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AX
ELEMENTS
OF
PLANE AND SPHERICAL
TRIGONOMETRY
BY
C. W. CROCKETT
PROFESSOR OF MATHEMATICS AND ASTRONOMY
RENSSELAER POLYTECHNIC INSTITUTE
>>Hc
NEW YORK :• CINCINNATI : • CHICAGO
AMERICAN BOOK COMPANY
y\ /<;-
Copyright, 1896, by
AMERICAN BOOK COMPANY.
OBOCKXTT. PLANE AND SPHES. TBIGONOM.
W. P. 2
PREFACE.
This work has been prepared for the use of beginners in
the study of trigonometry. Assuming that a high degree of
proficiency cannot be expected from such students, the author
has limited himself to the selection of simple proofs of the
formulas, not striving after original demonstrations. Geo-
metrical proofs have been added in many cases, experience
having shown that the student is assisted by them to a clearer
understanding of the subject.
The student is expected, in technical institutions, to acquire
facility in the use of the tables. All of the numerical exam-
ples have been computed by the author, with special attention
to correctness in the last decimal place, and the arrangement
of the computations has been carefully considered. Five-place
tables have been adopted, and the angles in the examples are
given to the nearest tenth of a minute, because the instruments
ordinarily used by engineers are read by the vernier only to
the nearest minute of arc, while the angle corresponding to a
computed function may be found usually to the nearest tenth
of a minute by the use of five-place tables.
Credit is due particularly to the works of Chauvenet, Snow-
ball, Beasley, Woodhouse, Newcomb, and Todhunter, although
many others have been consulted. A number of the illustra-
tive examples in Art. Ill were taken from Gillespie's "Land
Surveying," the numerical values being assigned by the author
of this work.
The author cannot hope that among so many examples
there are no errors ; he therefore requests those finding such
to kindly notify him.
Rensselaer Polytechnic Institutb,
Troy, N. Y.
8
1 ^:>i)45
GREEK ALPHABET.
A, a,
B, A
r, y,
A, S
E, 6
z, ?
H, V
@, «
I, '.
K, K
A, \
M, M
a, a Alpha
/3, p Beta
7, 7 Gamma
Delta
. - Epsilon
Zeta
, Eta
Theta
Iota
Kappa
........ Lambda
Mu
N, 1/ . . . . • Nu
H, ? Xi
O, Omicron
n, TT ........ Pi
V, p Rho
2, cr, 9 Sigma
T, T Tau
T, V Upsilon
^, (^ PJd
X, X ^A*'
^, >/r Psi
12, o) Omega
CONTENTS.
PART ONE. PLANE AND ANALYTICAL TRIGONOMETRY.
CHAP. I. MEA.SURBMKNT OF ANGLES: TRIGONOMETRIC FUNCTIONS OF
Angles less than 90°.
PAOB
Trigonometry defined 7
Directed lines ; angles 7
Measurement of angles 8
Trigonometric ratios in right triangles 10
Tables of the ratios 12
Ratios for 30°, 45°, 60°' . . . . v 12
Computation of the ratios when one is given 13
Measurement of angles in the field 14
Illustrations of the applications of the ratios 16
CHAP. 11. Right Plane Triangles.
Facts derived from geometry 17
Solution defined 17
Formulas employed in the solution of right triangles 17
Relations between the sides and the angles ...*... 18
Methods of solution 19
Isosceles triangles 23
Special methods 23
CHAP. III. Trigonometric Functions of Ant Angle.
Generation of angles 30
General measure of an angle 31
Coordinates ............ 53
General definitions of the trigonometric functions 35
Geometrical representation of the functions 37
Changes in the values of the functions 40
Limiting values of the functions ......... 42
Graphical representation of the functions 42
Two angles correspond to any given function 43
CHAP, IV. Relations between the Functions of One Angle.
Relations between the functions of one angle, and their applications . . 46
Solution of trigonometric equations containing one an^ile .... 52
Functions of angles greater than 360° 54
Functions of 90° ± x, 270° ± x, 180° ± ?/, 300° - y, and - // . . .54
The trigonometric tables 60
Transformations 61
CHAP. V. Relations between Functions of Several Angles.
Functions oi x + y and of x — y 64
Functions of 2 X in terms of functions of a; 71
Functions of x in terms of functions of 2 ic 72
Multiple angles 74
To change the product of functions into the sum of functions ... 74
To change the sum of functions into the product of functions ... 76
Circular functions 77
To prove that tan aj > x > sin oj, when a; < ^ tt 79
To prove that sin x, tan x, and x approach equality as x approaches zero . 79
6
I
CONTENTS.
PAGE
Development of sin x, cos x, and tan x 80
Computation of the trigonometric functions 82
Approximate assumptions 84
Transformations 85
CHAP. VI. Trigonometric Equations.
Equations containing multiple angles 88
Special cases 89
CHAP. VII. Oblique Plane Triangles.
Facts derived from geometry 97
The sine proportion 97
c^ = a^ -}- b"^ - 2 ab cosy 98
Methods of solution . 99
Metliods of solution, using right triangles 107
Areas 110
Illustrative examples 112
PART TWO. SPHERICAL TRIGONOMETRY.
CHAP. VIII. Definitions and Constructions.
Spherical trigonometry defined " . . 123
Representation of trihedral angles . . 123
Limitation of values 124
Definitions, relations, and constructions 124
The polar triangle 126
Facts derived from geometry 127
Construction of triangles 127
CHAP. IX. General Formulas.
cos a = cos b cos c + sin 6 sin c cos a 130
cosa =— COS/3C0S7 + sin^sin7Cosa 132
The sine proportion ........... 133
Additional formulas ........... 134
CHAP. X. Right Spherical Triangles.
Formulas
Napier's rules
Species .
135
136
.138
Methods of solution 139
Isosceles triangles ........... 143
Quadrantal triangles 144
CHAP. XI. Oblique Spherical Triangles.
To find an angle, given the three sides 146
To find a side, given the three angles 148
Napier's analogies 150
Gauss's equations 152
Species 153
Methods of solution . 155
Methods of solution, using right triangles ', ' . 168
CHAP. XII. Applications of Spherical Trigonometry.
Distance between two points on the earth's surface
Miscellaneous applications
Spherical excess .
Legendre's theorem
Astronomical definitions
Astronomical applications
174
175
181
184
187
189
PART ONE.
PLAJ{E AKB AJ^ALYTICAL TRIGOKOMETRY.
CHAPTER I.
MEASUREMENT OF ANGLES; TRIGONOMETRIC FUNCTIONS OF
ANGLES LESS THAN NINETY DEGREES.
1. Analytical Trigonometry treats of the relations of lines
and angles by algebraic methods. In Plane and Spherical
Trigonometry, these relations are applied to the solution of
plane and spherical triangles.
2. Directed Lines; Angles. — A directed line is one whose
beginning, direction, and length are known. The direction of
the line is indicated by the order of the letters in its symbol \
for instance, the line AB is drawn from A to B. If one direc-
tion along the line is considered positive, the opposite direction
will be negative ; thus, if the line AB is positive, the line
BA will be negative, their numerical measures being equal, or
line AB = — line BA.
An angle is the figure formed by two intersecting lines, the
point of intersection being the vertex.
The angle between any two given lines, whether intersect-
ing or not intersecting,* is defined to be the same as the angle
formed by two lines drawn through any point parallel to and
in the same direction as the given lines. Hence an angle may
be defined as the difference in direction of two directed lines.
* That is, parallel or in space.
7
8 PLANE AND ANALYTICAL TRIGONOMETRY.
3. Measurement of Angles. — Two methods of measuring
angles are in common use, — the sexagesimal and the circular
or natural methods.
4. Sexagesimal Measure.* — The circumference of a circle
described about the vertex of the angle as a center is divided
into 360 equal parts, and the angle at the center subtended by
one of these parts is taken as the unit. The length of one of
these divisions of the circle will depend upon its radius ; but
the corresponding angle at the center will be independent
of the radius, since it is -jj^ of four right angles. This unit
angle, called a degree^ is divided into 60 parts called minutes^
each of which is subdivided into 60 parts called seconds.
These are marked °, ', " ; thus 43° 14' 35". 2 is read, "43 de-
grees, 14 minutes, and 35.2 seconds."
How many degrees are there in
1. Two tbirds of four riglit angles ? Ans. 240°.
2. Two fifths of three right angles ? Ans. 108°.
3. Five sixths of two right angles ? Ans. 150°.
5. The Circular or Natural Measure. — From geometry we
know that in any two concentric
circles the arcs intercepted by any
angle at the center are to each other
as- the radii of the circles. Therefore,
if ACB be any central angle, we have
arc ^5 'dvcA'B'
CA CA'
(1)
Hence the length of the intercepted
arc divided by the radius is a number
that is always the same for the same
angle, no matter what the radius may be.
We also know that in any circle any two central angles are
to each other as their intercepted arcs, and therefore as the
quotients of their intercepted arcs divided by the radius. We
can, then, use these quotients to measure the angles.
V
* From sezagesimus, sixtieth.
MEASUREMENT OF ANGLES. 9
The circular measure of an angle is the quotient obtained
by dividing the length of its intercepted arc, in a circle whose
center is at the vertex of the angle, by the radius of the circle.
Thus, if c is the circular measure of the angle, I its intercepted
arc, and r the radius, we have
«=- (2)
If the radius of the circle is unity,
c = «. (3)
Hence the circular measure is represented by the length of the
intercepted arc in the circle whose radius is unity.
The angle whose circular measure is one, that is, whose
intercepted arc is equal to the radius, is called the radian.
1. The length of the intercepted arc of a central angle is 4 feet in a circle
whose radius is 2 feet; the length of the intercepted arc of another central
angle is 20 meters in a circle whose radius is 5 meters. Show that the second
angle is twice as large as the first.
2. In a circle with a radius of 10 inches, the intercepted arc of a central
angle is 5 inches, and that of an angle whose vertex is on the circumference is
10 inches. Find their circular measures. Ans. \.
6. Relation between the Two Measures. — Two right angles
are measured by 180°, and also by 7rr* ^ r = tt, since irr is the
semicircumference of a circle whose radius is r. Hence, using
the equality sign to represent " corresponds to," we have
180° = 1x1^11 circular measure 5 (1)
. •. 1° = -^ in circular measure. (2)
180
Again, tt in circular measure = 180° ; (3)
. •• 1 in circular measure = i^^. C4)
.-. 1 in circular measure = 57°. 29577 95 + (5)
.-. 1 in circular measure = 206 264''.806. (6)
1. What is the circular measure of 120° ? Ans. 120 x -^ = I
2. What is the circular measure of 10° 10' 10" ?
180 '
The circular measure of 1° is -— , and that of 1" is But
180 180 X 60 X 60
10° 10' 10" = 36610". . •. Circular measure of 10° 10' 10" = ^^^^^ "^
180 X QO X 60
♦ IT denotes the ratio of the circumference of a circle to its diameter, and is
the number 3.14159 265+.
10 PLANE AND ANALYTICAL TRIGONOMETRY.
3. What is the sexagesimal measure of the angle whose circular measure
^^^'^^ w = 180°; .-. ^7r=60°.
4. What is the sexagesimal measure of the angle whose circular measure
^^^ TT .. • • 1 180° , - , 120°
Unity in circular measure = ; . •. f corresponds to
TT IT
6. What are the sexagesimal and circular measures corresponding to f of
three right angles ? Ans. 60° ; \ ir.
6. The sexagesimal measures of two angles are 22° 30' and 43° 14' 3".
1350 IT 155643 tt
Show that their circular measures are and
180 X 60 180 X 60 X 60
7. The circular measures of three angles are y^j tt, | tt, and J^ v. Show
that their sexagesimal measures are 15°, 40°, and 3° 36'.
8. The circular measures of three angles are i, f , and |. Show that their
. , 45° 300° , 40°
sexagesimal measures are — , , and _ .
TT IT TT
9. Find the sexagesimal and circular measures corresponding to
(a) Seven tenths of four right angles. Ans. 252° ; |ir.
(6) Five fourths of two right angles. Ans. 225° ; | t.
(c) Two thirds of one right angle. Ans. 60° ; | tt.
7. Centesimal Measure. — In this system, proposed by the
French, the right angle is divided into 100 parts called grades,
each of which is subdivided into 100 parts called minutes,
each minute being divided into 100 parts called seconds;
marked ^ \ '\
8. Trigonometric Ratios. — Let the sides of a right-angled
triangle be denoted as shown in Fig. 2. The trigonometric
ratios may be defined as follows:
The sine of an angle = -^ P- ; written sin -4 = -
hypotenuse h
n^^ . o I side adiacent .^^ .a
ihe cosine oi an angle = -.; ~ ; written cos A = ~
nypotenuse h
The tangent of an angle = -r-^ ^ -; written tan A = -
side adjacent «
The cotangent of an angle = ^^ — t— : written cot ^ = -
^ side opposite o
The secant of an angle = . ^^ ,. -; written sec^ = -
side adjacent «
The cosecant of an angle = -^^ r— ; written cosec A = -
side opposite o
These fundamental equations should be thoroughly memorized.
O)
MEASUREMENT OF ANGLES.
11
Fig. 2.
Fig. 8.
9. The Ratios are Constant for Any One Angle. — In Fig. 3
let BAO and BAF be two angles differing by a quantity as
sin^ll as Ave please. At any two points B and I) on AB, draw
BF and BG- perpendicular to AB ; with ^ as a center, and
radius A (7, describe the arc OIT, and draw Lff perpendicular
to AB. The triangles BAO and BAF are similar.
BC^BF
AC AF
BO BF
a constant
sin a;.
= ---- = a constant
AB AB
side opposite
hypotenuse
side opposite ,
— — = tan Xm
side adjacent
AC AF , , hypotenuse
— -— = — — = a constant = ,•: -, . = sec x,
AB AB side adjacent
10. The Values of the Ratios differ for Different Angles. —
From Fig. 3 we have, since AH = AO^
BO . • LH LH
sin. = — andsin^ = -^= — ;
BO . ^ BF
tan X = — — ; and tan y = ;
AB ^ AB
AO . AF
sec :c = — and sec ^ = — .
11. The Angle may he constructed
when One of the Ratios is known. —
Let sin x = ^. With any convenient
radius AO, describe a circle about A
as a center. Draw AB perpendic-
ular to AB, and on it lay off AB=
^AO; draw i> (7 parallel to AB until
FiQ. 4.
12 PLANE AND ANALYTICAL TRIGONOMETRY.
it intersects the circle at ; join A and (7, and BA C will be
the required angle, since
. ^,ri BC AD 1
sin BA C = -— - = -TT^ = - .
AC AC 2
Let tan a; = J. Lay off any convenient distance AB ; at B
draw BC perpendicular to AB, and lay off J9(7= ^ AB ; join
A and (7, and ^J. C will be the required angle, since
tani?^(7=4^=f.
Let sec x = 2. Lay off any convenient distance AB ; erect
the perpendicular line BC; with a radius J. (7= 2 J.^ describe
an arc cutting BC at C; join ^ and C, and jB^Cwill be the
required angle, since
secj&A(7=4^=2.
Let the student construct the angle whose cosine is |, the angle whose
cotangent is 5, and the angle whose cosecant is 4.
12. We therefore conclude that to any one angle there
will correspond a special value of each of these ratios, that the
value of each ratio will differ for different angles, and that, if
any one of these ratios is given, the angle may be constructed.
13. Tables of Sines, Cosin*es, etc. — The values of these ratios
for angles between 0° and 90° have been computed, and are
given in tables so arranged that the values corresponding to
any angle may be readily found. The tables of natural sines.,
etc., contain the actual values of these ratios ; while the tables
of logarithmic sines, etc., contain their logarithms.
14. Ratios for 30% 45°, 60°.
(rt) Ratios for 45°. — In Fig. 5 let the angle A = 45° ; then
J5=90°-^ = 45°.
.-. AC — CB, since they are opposite equal angles.
Let AC= a; then CB = a^ and AB = Va^ + a^ = a V2.
MEASUREMENT OF ANGLES.
13
^°^'''°=3l=^' cot45'=^|=l; cosec46o=^=V-2.
(5) Ratios for 30° and 60°. — In the equilateral triangle
ABC (Fig. 6), let .A.S = a ; draw D^ perpendicular to AO ;
AC will be bisected at i>, making J.i> = Ja, and the angle
^^i)=angle i>^(7=30°.
Also DB = Va2 - 1 ^2 = 1 a Vs.
sin ABi> = sin 30^^
^^ 2
tan 30° =
cos 30° = ^=—; cot 30°
AB 2
DB V3
DB V3
^ = V3
siriZU^=sin60° = :^=^; tan60° = — = >/3
^^2 ^Z>
2)^
AD
cos60° = — = i; *cot60° =
^^2 Z)^ V3
Note that the sines of 30°, 45°, and 60°, are
spectively.
sec 30° = ^ =—
DB V3
cosec 30° = ^ = 2.
AD
sec 60° = :^^ = 2 :
cosec 60° = :^= A.
DB V3
^(v/l,/^V2, and^VS re-
15. The Ratios are not Independent of Each Other ; for we
have from Fig;. 2,
so that if two of the three quantities A, o, and a, are given, the
third can be found. Hence if we know one of the ratios, that
is, the relative values of two of the three elements, we can de-
termine the relative value of the third element, and from it
the other ratios.
14
PLANE AND ANALYTICAL TRIGONOMETRY.
Thus if tan a: = |, and the other ratios are required, we have
tana:
; let = 3, a = 4 ; then A = 5.
a
4'
'
•••
sin 2: =
3
5'
COS X
a
4
^5'
cot 2:
4
^3'
sec 2:
_h_
a
5
cosec X -
5
3*
Having
given the ratic
on
the left,
find the ratios on the
right :
sin a;.
cos a?.
tan a.
cot 85.
seca;.
cosec 85.
o.-^ ^ 8
15
8
15
17
17
1.
sin X = —
—
17
17
15
8
15
8
5
12
12
5
13
13
2.
cos x = —
—
13
13
5
12
5
12
8.
tan a; = —
7
24
24
25
25
24
25
25
7
24
7
4.
cot a = 2
1
V5
2
V5
1
2
, —
IVB
V5
29
21
20
21
20
29
b.
seca; = —
20
29
29
20
21
21
6.
cosec x = 3
1
3
1--.
Iv^
2V2
\^^
—
16. Measurement of Angles in the Field. — In Fig. 7,
FG-HK represents a fixed graduated circle, and ABDE a
circle resting on the plate FGrHK^ and capable of moving
about a pivot at O'; J and are two small rods fixed to
ABBE^ and perpendicular to~the planes
of the circles ; and il[f is a mark on the
circle ABBE in the same line with /,
(7, and 0. If we wish to measure the
horizontal angle between two distant
objects, two church towers, for ex-
ample, we proceed as follows : first
place the circles in a horizontal posi-
tion ; revolve the circle ABBE, look-
ing along the line 10, until the line of
sight passes through one of the objects, and note the reading
MEASUREMENT OF ANGLES.
16
of the circle opposite the mark M; then revolve the circle
ABBE, being careful not to move FGHK, until the line of
sight passes through the second object, and note the new read-
ing of the circle opposite the mark M. The difference between
the two readings will be the angular distance required.
17. The Engineers' Transit, shown in Fig. 8, is used in
measuring horizontal and vertical angles. The lower circle is
provided with two levels, by which its horizontality is tested.
Fig. 8.
The rods I and are replaced by the telescope with a system
of intersecting wires in the common focus of the object glass
and eyepiece, the telescope being capable of rotation about
an axis parallel to the horizontal circle. The circle fixed to
the axis of the telescope is vertical when the plate bearing the
upright supports is horizontal.
16
PLANE AND ANALYTICAL TRIGONOMETRY.
18. Illustrations of the Application of the Ratios.*
1. A rope fastened to the top of a vertical pole 60 feet
high, and to a stake driven in the ground, is inclined at an
angle of 30°. How far is the stake from the bottom of the
pole ? How long is the rope ?
-^ = tan 30° = —.
AC V3
••• ^^ = V3a^ = 60V3feet.
Fig. 9.
OB . ono 1
— — = sm 30° = -.
AB 2
.-. ^j5 = 2 (7^ = 120 feet.
2. The angle at the vertex of a right circular cone is 60°,
and the slant height is 10 inches. What is the altitude and
the radius of the base of the cone ?
Fig. 11.
OB _„ ono V3
_ = cos30 =— .
OB = ^AB
5 V3 inches.
^=sin 30^ = 1
. •. AC= ^ AB = 5 inches.
3. The top of a ladder 30 feet
long rests on the upper edge of a
wall 15 feet high. What is the
inclination of the ladder ?
sin CAB =
05^15^1.
AB 30 2'
but sin30° = V. .'. (7^^ = 30°.
In these cases the ratios corresponding to the angles were
known from Art. 14. Usually it will be necessary to refer
to the tables in solving problems involving the ratios.
It is assumed that the ground is horizontal.
CHAPTER II.
RIGHT TLANE TRIANGLES.
19. It has been shown in Geometry that a right-angled tri-
angle can be constructed when two elements * besides the right
angle are known, one of the known elements being a side. We
also know that
(1) The hypotenuse is greater than either of the other '
two sides.
(2) The hypotenuse is less than the sum of the other two -^
sides.
(3) The sum of the two acute angles must be 90°. '^
(4) The greater side is opposite the greater angle. '
(5) The square on the hypotenuse is equal to the sum of "^
the squares on the other two sides.
20. A triangle is said to be solved when, having some of the
elements given, the others have been found by some process.
21. The Solution of a Right Triangle is effected by means of
the trigonometric ratios. Each equa-
tion, as sin ^ = -, contains three
c
quantities ; and two of them must be
known in order that the third may
be found. Hence in any particular
case we use the equations that con-
tain the two given elements ; thus, if ^^
a and h are given, we use tan J. =y via-^ii
to find A, and then e may be found from either sin A =- or
COS A = -,
c
* The elements of a triangle are the three sides and the three angles.
CROCK. TRIG. — 2 17
18 PLANE AND ANALYTICAL TRIGONOMETRY.
The equations used in the solution of right triangles are
Fia. 18
sin A = - = cos B,
c
cos A= - = sin B,
c
tan J. = - = cot-B.
cot A = - — tan B.
a
A + B = 90°.
^ = ^2 + ^2.
(1)
22. From the Trigonometric Ratios we have
tanJ.= ; .*. a = 5tanJ.,
cot B = -; .'. a = b cot B^
(1)
or, any side of a right triangle is equal to the other side multiplied
by the tangent of the angle opposite^ or by the cotangent of the
angle adjacent^ to the side itself.
sin^ = — ; .*. a = (? sin ^,
c
eosB = -; .*. a = c cos B,
c
(2)
or, any side is equal to the hypotenuse multiplied by the sine of
the opposite angle^ or by the cosine of the adjacent angle.
sec J. = -;
'=h sec^,
cosec^=-; .-. c=5cosecJ5,
b
(3)
or, the hypotenuse is equal to a side multiplied by the secant of
the adjacent angle^ or by the cosecant of the opposite angle.
Note. — The secant of an angle is the reciprocal of its cosine, and the cose-
cant is the reciprocal of its sine ; hence the logarithm of the secant is the arith-
metical complement of that of the cosine, and the logarithm of the cosecant is
the A. C. of that of the sine, or
log sec X = colog cos x, and log cosec x = colog sin x.
RIGHT PLANE TRIANGLES.
23. Case I. Given c and A,
Formulas:
a = (? sin A,
h = c cos A,
^=90° -A
19
1. Solve the triangle when c = 1.0034, and A = 42^ lO'.S.
.-. B = 90° - ^ = 47°49'.7.
(a) By natural functions.
a = c sin ^ = 1.0034 x 0.67136 = 0.67364.
6 = ccos^ = 1.0034 X 0.74114 = 0.74366.
(b) By the use of logarithms.
a = c sin ^ ; . •. log a = log c + log sin A.
6 = c cos ^ ; . •. log 6 = log c + log cos A.
Always write first all the formulas that will be used in the
problem; then write them in a form adapted to logarithmic
computation ; then refer to the tables and write the logarithms
in their proper places. Thus in this case we arrange the work
as follows:
logc= logc =
+ log sin ^ = + log cos A =
.-. loga= .-. log 6 =
.-. a= .'. 6 =
The positive signs preceding log sin A and log cos A indicate
that they are to be added to log c.
We now find the angle A in the table of logarithmic func-
tions and take from the table both log sin A and log cos A,
writing them in their proper places. Then we refer to the
table of logarithms of numbers and find log c, writing it oppo-
site log c. Then we add the proper quantities to find log a and
log 5, finally looking in the table of the logarithms of numbers
for the numbers corresponding to the computed values of log a
and log b.
The arrangement on the right is preferable, since it saves
log sin A =
(1)
logc =
(3)
log cos A =
(2)
.-. loga = (l) + (3)
(4)
a =
(6)
.'. log 6 =(2)4- (3)
(5)
b =
(7)
20
PLANE AND ANALYTICAL TRIGONOMETRY.
the writing of one line. The numbers in the parentheses indi-
cate the order in which the quantities should be found.
0.00147
9.82695
logc:
10 + log cos ^
log6 = 9.87137
6 = 0.74365
0.00147 or log sin A = 9.82G95 - 10
9.86990-10 logc = 0.00147
log cos ^ = 9.86990 - 10
10
Check :
c + 6 = 1.74705
c-b = 0.25975
b)
loga = 9.82842 - 10
a = 0.67363
log6 = 9.87137 -10
6 = 0.74365
log (c+ 6) = 0.24230
log (c - &) = 9.41456 ~ 10
.-. loga2 = 9.65686
loga = 9.82843
Exact agreement is not expected, since the tables give the values of the
functions only to the nearest unit in the fifth decimal place. The — lO is usually
omitted, and sin A is written for log sin J., when there is no danger of confusion.
2. Solve the triangle when c = 34.687, and B = 49° 8'.4.
Ans. A = 40° 51'.6 ; b = 26.234 ; a = 22.6925.
3. Solve the triangle when c = 305, and A = 63° 31'.14, using the natural
functions. ^^^^ ^ ^ 273.00 ; h = 136.00.
4. Solve the triangle when c = 205, and B = 49°33'.01, using the natural
Ans. a = 133.00 ; b = 156.00.
functions.
24. Case II. Given c and a.
Formulas
sin A — -.
c
b = a cot A = c cos A.
1. Solve the triangle when c = 8.7982, and a = 3.1292.
.*. logsin^ = loga — logc; logft = loga + logcot^ = logc + logcos A
log a = 0.49544 log a = 0.49544 log c = 0.94439
-logc = 0.94439 +logcot^ = 0.41958 +logcos^ = 9.97063 - 10
log sin ^ = 9.55105 -
A = 20° 50M
^ = 69° 9'.9
10
log 6 = 0.91502
b = 8.2228
log6 = 0.91502
b = 8.2228
RIGHT PLANE TRIANGLES.
21
log cot ^ = 0.41958 (5)
loga = 0.49544 (1)
-logc = 0.94439 (2)
log cos ^ = 9.97063 (6)
logsin^ = 9.55105 (l)-(2)
A = 20° 50'.1 (4)
i? = 69° 9'.9
(l) + (5)
(2) + (6)
b = 8.2228
Check : b^=(c- a){c-\- a)
c-a= 5.6690, log (c - a)
c-ha = 11.9274
0.75361
log (c + a) =1.07656
log 62
1.83006
log 6 =0.91603
log6 = 0.91602 {
2. Solve the triangle when c = 369.27, and b = 235.64.
Ans. A = 50° 20'.9 ; 5 = 39° 39M ; a = 284.31.
3. Solve the triangle when c = 281, and a = 160, using the natural functions.
Ans. A = 34° 42'.5 ; b = 231.00 or 231.01.
4. Solve the triangle when c = 365, and b = 76, using the natural functions.
Ans. A = 77° 58'.93 ; a = 357.00.
25. Case III. Given a and b.
Formulas:
tan^ =
c =
sin A cos A
5 = 90° - A.
1. Solve the triangle when a = 169.03, and b = 203.44.
. •. log tan ^ = log a - log 6 ; log c = log a - logsin A = \ogb — log cos^.
log a = 2.22796
-log 6 = 2.30843
log tan A
log a = 2.22796
log sin ^ = 9. 80555
10
log 6 = 2.30843
log cos A = 9.88602 - 10
9.91953-10
^ = 39°43'.3
J5 = 50°16'.7
logc = 2.42241
c = 264.49
logc = 2.42241
c = 264.49
or
*loga = 2.22796
(1)
logsin^ = 9.80555
(5)
log cos ^ = 9. 88602
(6)
log6 = 2.30843
(2)
log tan ^ = 9.91953
(3)
al = 39°43'.3
(4)
J5 = 50°16'.7
logc = 2.42241
c = 264.49
1(2)-
-(5)
-(6)
Check : a2 = c2 - 62
c + 6 = 467.93
c-6= 61.06
log(c + 6)= 2.67018
log(c- 6)= 1.78569
.-. loga^ =4.45587
log a =2.22794
* This form is preferable.
22
PLANE AND ANALYTICAL TRIGONOMETRY.
2. Solve the triangle when a = 4.8199, and b = 2.6492.
Ans. A = 61° 12'.8 , B = 28" il'.l ; c = 5.4999.
3. Solve the triangle when a = 60, and 6 = 91, using the natural functions.
Ans. A = 33° 23'.9 ; c = 109.00.
4. Solve the triangle when a = 72, and b = 65, using the natural functions.
Ans. A = 47° 55'.5 ; c = 97.000.
B
26. Case IV. Given a and A,
h = a cot A.
Formulas:
c =
B
sin A cos A
90° - A,
1. Solve the triangle when a = 613.35, and A = 40° 12'.6.
.-. 5 = 90° -^=49° 47 '.4.
log & = log a + log cot A.
log c = log a — log sin A = log b — log cos A.
loga = 2.78770
cot^ = 0.07295
log& = 2.86065
b = 725.52
loga = 2.78770
logsin^ = 9.80996 -10
logc = 2.97774
c = 950.04
log6 = 2.86065
log cos^ = 9.88291 - 10
logc = 2.97774
c = 950.04
or log sin ^ = 9.80996 (1)
log a = 2.78770 (3)
log cot ^ = 0.07295 (2)
logc = 2.97774 (3)-(l)
c = 950.04
log6 = 2.86065 (3) + (2)
b = 725.52
Check: a2=(c + &)(c-fe)
c + b = 1675.56, log (c + b)= 3.22416
c-b= 224.52, log (c - 6) = 2.35126
log«''^ = 5.57542
loga = 2.78771
2. Solve the triangle when a = 3.6378, and B = 69° 23'. 5.
Ans. A = 20° 36'.5 ; b = 9.6738 ; c = 10.335.
3. Solve the triangle when b = 160, and A = 55° 17'.48, using the natural
functions.
Ans. c = 281.00 ; a = 231.00.
4. Solve the triangle when a = 340, and A = 60° 55'. 85, using the natural
functions.
Ans. c = 389.00 ; b = 189.00.
RIGHT PLANE TRIANGLES.
23
27. Isosceles Triangles. — If a perpendicular to the base is
drawn from the vertex, it will bisect the base and the angle at
the vertex, forming two equal right tri-
angles.
ZABI)=ZI)BC=l^; AB = BO;
1. Solve the triangle when h = 2.1452, and
j8 = 121° 14'.6.
.-. ^i?= 1.0726; ^i52) = 60°37'.3;
o = 90°-^)8=29°22'.7.
^"^sinT^' •■• loga = logi6-logsinJ)8.
log ^6 ==0.03044
log sin i)a = 9.94022 -10
loga = 0.09022
a = 1.2309
\ogp = log J 6 + log cot J 0.
log^& = 0.03044
+ log cot ^ /8 = 9. 75049 - 10
logp = 9.78093- 10
p = 0.60385
2. Solve the triangle when a = 52° 10'. 2, and a = 600.2.
Ans. /8 = 75°39'.6; ^6=368.12; ;> = 474.07.
28. Given c and b (Special Method). — When b nearly equals
c, the angle found from the formula cos J. = - is uncertain, the
c
tabular difference for the cosine
being so small that a small error
in cos^ would produce a large
error in A.
In the figure, AB bisects the
angle A^ and BU is perpen-
dicular to AB; .'. BE= CJ).
Let OB = x=BB;
.-. tan4a = -
2 b
Also, CB=a= CB + BB= OB-^ BUsec a
a
(1)
a = X -{- X sec a ;
a ab
X =
1 H- sec a '
X
1 +
+ 4'
(2)
24
PLANE AND ANALYTICAL TRIGONOMETRY.
From (1) and (2),
tan -|- u
c -^ b c
h)(c-b}.
. •. tan J
+ 6 ^l ic + bf ■
^ C -\-
Suppose that we wish to find the greatest distance at sea at which a moun-
tain 4.3 miles high can be seen, the earth being considered as a sphere witli
a radius of 3963.3 miles, and the distance being measured as a
-A chord.
Let 5^=4.3, and CB=CB = 'dmS.S ; BD being the distance
required. Then cos DCA = — , giving log cos DCA = 9.9m52 ;
CA
and DCA as found from the tables might have any value
between 2° 40'.5 and 2° 42'.5.
Using (3), we have
CA-CD= 4.3; log = 0.63347
CA + CD = 7930.9 ; log = 3.89932
2 )6.73415 - 10
log tan IDCA = 8.36708 - 10
bpl. r= 3.53620
log (I Z>C^)'= 1.90328
.-. I DCA = 80'm5; .-. DCA =2° iO'.Ol.
Then BD = 2CDsmhDCA will give the chord BD. The arc BD is found
from the proportion :
360° : DCA = 2Tr X 3963.3 : arc BD.
Note. — Eq. (3) follows directly from (4), Art. 69 :
tan i a = \ 52ifi; where cos a
^ ^1+COSa'
29. Given a and b (Special Method). —
^ When a and b are nearly equal, the angle
a may be determined more accurately, as
follows :
Draw AI), making CAD = 45°, and
DU perpendicular to AB. Then
tan BAU = tan(a - 4o°) =
AH
But DU = DB cos a = COB- CD') cos a
(a — 5) cos a
(a - b)b
RIGHT PLANE TRIANGLES. 26
and
AE = AB — EB = AB — BB sin a = c — ^ ^- = — —
c c
^ b^ + ab
c
BE ^ (a-b)b ^ a-b
AE ab + b'^ a + h
.-. tan(«-45°) = ^. (1)
a -{•
If b were greater than a, the formula would be
tan(45°-«)=^^^. (2)
f
Note. — Eq. (1) may be found from the relation proved in Art. 100 :
a - 6 _ tan Ha - ^)^ ^j^^-^ i ^^ ^ ^)= 45°, and K^ - /3)= a - 46°.
a + 6 tanHa + /3)
EXAMPLES.
Note. — The angle between the line of sight and a horizontal plane is called
an angle of elevation when the point sighted on is above the horizontal plane,
and an angle of depression when it is below the horizontal plane.
1. The shadow of a vertical pole 30 feet high is 40 feet long. Find the
elevation of the sun above the horizon. Ans. 36° 52 '.2.
2. The vertical central pole of a circular tent is 20 feet high, and its top is
fastened by ropes 40 feet long to stakes set in the ground, the ground being
horizontal. How far are the stakes from the foot of the pole, and what is the
inclination of the ropes to the ground ? Ans. 34.641 feet ; 30°.
8. The top of a lighthouse is 200 feet above the sea level, and the angle of
depression to a buoy is 9° 52 '.8. Find the horizontal distance of the buoy from
the lighthouse. Ans. 1148.3 feet.
4. The horizontal distance from a point to the vertical wall of a tower is
1000 feet, and the angle of elevation of the top is 4° 15'.2. Find the height of
the top of the wall above the point. Ans. 74.370 feet.
5. Two points A and B are on the opposite banks of a stream*. A line AG
at right angles to ^ J5 is measured 300 feet long, and the angle ACB is found by
measurement to be 62° 30'.4. What is the distance from Ato B?
Ans. 576.45 feet.
6. From the top of a lighthouse, 150 feet above the sea level, the angle of
depression to a buoy was 12° 10'.2, and that to the shore, measured in the same
vertical plane with the buoy, was 62° 14 '.8. Find the distance in feet of the buoy
from the shore. Ans. Log. Tables, 616.60 ; Nat. Tables, 616.61
26 PLANE AND ANALYTICAL TRIGONOMETRY.
7. The angle of elevation to the top of the vertical wall of a tower is
20° 10 .4, and the angle of depression to the bottom is 10^ 11 '.0, the horizontal
distance from the observer to the wall being 250 feet. Find the height of the
wall. Ans. 136.802 feet.
8. We wish to make a ladder that would reach from a point 20 feet in front
of a building to the fourth story, a height of 45 feet. Find the length of the
ladder and the angle it would make with the ground in this position.
Ans. 49.244 feet; 66° 2'.2.
9. The ridgepole of a roof is 15 feet above the center of the garret floor,
and the garret is 40 feet wide. What is the inclination of the roof to a horizon-
tal plane ? Ans. 36° 52 '.2.
10. A chord of a circle is 20 feet long, and the angle at the center subtended
by it is 46° 43'.6. Find the radius of the circle. Ans. 25.217 feet.
11. The angle between two lines is 40° r2'.4, and a circle whose radius is
5730 feet is tangent to both lines. Find the distance from the point of tangency
to the point of intersection of the two lines when the circle is in the smaller
angle, and when it is in the larger angle formed by producing one of the lines.
Ans. 15055 and 2097.2 feet.
12. The legs of a pair of dividers are set so that the angle between them is
80° 24'.4. What is the distance between the points, the legs being 6 inches
long? Ans. 7.7460 inches.
13. An equilateral triangle is circumscribed about a circle whose radius is
10 inches. Find the perimeter of the triangle. Ans. GOVS inches.
14. A wedge measures 12 inches along the side, and its base is 2 inches
wide. Find the angle at its vertex. Ans. 9^ 33'. 6.
15. The side of a regular decagon is 2.4304 feet. Find tlie radii of the
inscribed and circumscribed circles. Ans. 3.7400 feet; 3.9325 feet.
16. The area of a regular octagon is 24 square feet. Find the radius of the
inscribed circle and the length of one of the sides. Ans. 2.6912 feet ; 2.2295 feet.
17. The radius of the circumscribing circle of a regular dodecagon is 10
feet. Find the area of the dodecagon. Ans. 300.00 square feet.
18. A cord is stretched around two wheels with radii of 7 feet and 1 foot
respectively, and with their centers 12 feet apart. Prove that the length of the
cord is 12\/3 + 10 tt feet.
19. A cord is stretched around, and crossed between, two wheels whose radii
are 5 feet and 1 foot respectively, their centers being 12 feet apart. Prove that
the length of the cord is 12\/3 + 87r feet.
20. Find the radius and the length of an arc of 1° of the parallel of latitude
at a place whose latitude is 42° 43'. 9, the earth being regarded as a sphere whose
radius is 3963.3 miles. Ans. 2911.1 miles; 50.809 miles.
21. The altitude of a right circular cone is 4. 1436 feet, and the angle at its
vertex is 20° 14'.2. Find its convex surface. Ans. 9.7780 square feet.
RIGHT PLANE TRIANGLES.
27
22. The altitude of a right pyramid with a square base is 14.463 feet, and
the sides of the base are each 4.703C feet. Find its slant height, its lateral edge,
and the angle between a face of the pyramid and its base.
Ans. 14.643 feet; 14.831 feet; 80° 45'. 5.
23. The base of a trapezoid measured 600.430 feet, and the angles at the
ends of the base were found to be 62° 14'. 3 and 74° 18'. 6. Find the length of
the other base, the altitude being 40
feet. Ans. 568.138 feet.
24. Find the length of the perpen-
dicular from the vertex of the right
angle of a triangle to the hypotenuse,
the hypotenuse being 6.4603 inches
long, and one of the angles of the tri-
angle being 40° 40'. 4.
Ans. 3.1934 inches.
25. A street-railway track is 10 feet
from the curbstone (FB = HD = 10), Fig. 20.
and in passing a corner where the
street is deflected through an angle of 60°, the rail must be 4 feet from the corner
(GC = i). Find the radius of the circular curve.
Ans. 0C =
20-4v/3
2-^3
26. Before paying for a pavement, it was necessary to find the area shaded
28750
in Fig. 21. Prove that it is 1- 7500 square feet, the streets being
50 feet wide. ^
Fig. 21.
27. In the egg-shaped sewer (Fig. 22), C is the center of the arc ADB with
a radius a ; / and J, of AF and BG respectively with the radii 3 a ; and IT, of
FEG with the radius | a. Prove that its area is
a2fj+ ^tan-i ^ + 9 tan-i?- 3^ = a2f^^ + §^tan-i?- 3^ = 4.59413 a2,
\i5 4 6 4/\8 4 4/
where tan-i - is the angle whose tangent is ^
3 3
28
PLANE AND ANALYTICAL TRIGONOMETRY.
28. A hill rises 1 foot vertically in a horizontal distance of 30 feet. What
is the difference of elevation of two points that are 1000 feet apart, the distance
being measured on the ground ?
log tan a = 8.62288 - 10
CpL T' = 3.53611
log a' = 2.05899
S' = 6.46365 - 10
log sin o=: 8. 52264 -10
log 1000 = 3.
log diff. of elev. = 1.52264
difE. of elev. = 33.315 feet.
29. The horizontal distance between the two extreme positions of the end
of a pendulum 40 inches long is 4 inches. Through what angle does it swing ?
Half-angle =2° 51 '.96.
Ans. 5°43'.92.
30. The angular diameter of the moon
is 31'.12, and its distance is 238840 miles.
Find its diameter in miles.
BAD = SV. 12,
and ^C = 238 840.
Ans. 2162.0 miles.
31. The equatorial horizontal parallax
of the sun is 8". 8, and the radius of the
earth is 3963.3 miles. Find the distance
of the sun from the earth.
BAC = 8". 8, and BC = 3963.3.
Ans. 92 896 000 miles.
32. A circular chimney 100 feet high is 10 feet in diameter at the base, and
8 feet at the top. Find the angle at the vertex of the cone of which it is a
frustum.
Half-angle = 34'.376. Ans. 1°8'.752.
Solve the following triangles, the first two elements being given
33. c = 0.02934, A = 31° 14'.2.
34. c = 4.6136, B = 47° 15'.6.
35. c = 436.53, A = 74° 10'.6.
/ 36. = 0.96724, B = 40° 40'.2.
37. = 110.97, a = 67.291.
38. = 1843.7, & = 618.42.
39. c = 8226.5: a = 814.33.
40. = 0.03672, 6=0.01296.
./ 41. = 4.8293, b = 0.31435.
^ = 58°45'.8; a = 0.015215; & = 0.025086.
^ = 42°44'.4; a = 3.1311 ; 6 = 3.3885.
B=15°49'Ai a = 419.98; 6 = 119.03.
A = 49° 19'.8 ; a = 0.73363 ; b = 0.63036.
A = 37° 19'.8 ; ^ = 52° 40'.2 ; b = 88.236.
A = 70° 24M ; 5 = 19° 35'.9 ; a = 1736.9.
^ = 81°50'.5; .5= 8° 9'.5 ; 6 = 116.74.
A = 69° 19'.9 ; B = 20° 40M ; a = 0.034357.
A = 86° 16'.1 ; ^ = 3° 43'.9 ; a = 4.8191.
RKJHT PLANE TRIANGLES.
29
42. a = 43.148, 6 = 84.107.
43. a = 769.28, ft = 61.86.
44. a = 7642.5, ft =864.7.
45. a = 0.04326, ft = 0.54318.
46. a = 903.64, A = 22° lO'.S.
47. ft =0.47922, A = 62° 16' A.
48. a = 8.4642, i? = 30° 16'.4.
49. ft = 18.430, B = 65° 15'.6.
.^ = 27° 9'.5; i? = 62°50'.5; c = 94.630.
. ^ = 86° 6'.6; B= 3°54'.4; c = 761.06.
. ^ = 83°32'.7; B= 6°27'.3; c = 7691.3.
. A= 4°33'.2; I? = 85° 26'.8 ; c = 054489.
•. 5 = 67°49'.7; ft = 2217.4 ; c = 2394.6.
•. i? = 27°43'.6; a = 0.91176 ; c = 1.0300.
: A = 59° 43'.6 ; ft = 4.9409 ; c = 9.80075.
. A =2A° 44'.4 ; a = 8.4954 ; c = 20.299.
Solve the isosceles triangles (Fig. 16) in the following examples, the first
two elements being given :
50. a =57.906, ft =62.736.
51. a=3.4782, a=20°20'.6.
52. rt =99.674, /3=40°30'.4.
53. 6=0.96042, a = 70°10'.4.
54. ft = 1146.48, i3=80°36'.4.
55. a=87.904, j9=46.812.
56. 6=6.9044, p= 5.7806.
57. J9 = 18.478, a= 37° 19'.8.
58. i)=0.46424, /3=100°36'.8.
.-. a= 67°12'.05; /8=66° 35'.9 ; p=48.673.
.-. ^ = 139°18'.8; 6=6.5224; ;) = 1.209L
.-. a= 69°44'.8; 6 = 69.008;
.-. /3= 39°39'.2; a = 1.4158;
a = 886.24;
i3=115°38'.8
.-. a= 49°41'.8
.-. a= 32° 10 '.6
.-. a= 59° 9'.2; ^3= 61°41'.6
.-. ^ = 105°20'.4; a=30.471;
.-. a= 39°41'.6; a = 0.72690;
p = 93.610.
p= 1.3319.
jr)=675.87.
6 = 148.806.
a =6.7330.
6=48.458.
6 = 1.11865.
4
CHAPTER III.
TRIGONOMETRIC FUNCTIONS OF ANY ANGLE.
30. Generation of Angles. — An angle may be considered as
generated by a line revolving about a fixed point, the vertex ;
thus OA revolving about in the
direction a, to the position OB^ de-
scribes the angle A OB. The side of
the angle /rom which the revolution
takes place is called the initial side,
and that to ivhich the describing line
moves is called the terminal side.
The letters describing the initial side are Avritten first in the
symbol of the angle, so that the angle A OB is one in which
the motion is from OA to OB.
J
Fig. 24.
31. Direction of Measurement. — The revolving line can
move from OA to OB either in the direction marked a or in
that marked h. The former motion, contrary to that of the
hands of a watch, is arbitrarily considered positive and the
latter negative. Thus if the angle a;, between OA and OB, is
30°, the angle AOB is either + 30° or - 330°.
Any angle has two measures less than 360°, one positive and
the other negative, their numerical sum
90° being 360°.
32. Quadrants. — For convenience
the measuring circle is divided into
four parts called quadrants., as in the
figure. An angle is in the first quad-
rant when its value lies between 0°
and 90°; in the second, between 90°
and 180°; in the third, between 180°
30
TRIGONOMETRIC FUNCTIONS OF ANY ANGLE. 81
and 270°; in the fourth, between 270° and 360°. Angles
between 0° and — 90° are in the fourth quadrant ; between
- 90° and - 180°, in the third ; between - 180° and - 270°,
in the second ; between — 270° and — 360°, in the first.
Also, an angle between zero and ^tt is in the first quad-
rant ; between J ir and tt, in the second ; between ir and | tt,
in the third ; and between | tt and 2 tt, in the fourth.
33. Complement and Supplement. — Two angles are said to
be complementary Avhen their algebraic sum is 90°, as 60° and
30°, 120° and -30°, 260° and -170°; and supplementary
when their algebraic sum is 180°, as 120° and 60°, 230° and
- 50°, 300° and - 120°.
Note. — In Fig. 2, ^ is the sine of B ; that is, it is the sine of the comple-
h
ment of A, and hence it is called the cosine of A.
Since -|-7r corresponds to 90°, and tt to 180°, two angles are
complementary when the algebraic sum of their circular meas-
ures is ^ TT,. and supplementary when it is tt.
1. The complement of 200° is 90° - 200° = - 110°.
2. The complement of 90° + x is 90° - (90° + x) = -x.
3. The supplement of 2u0° is 180° - 200° = - 20°.
4. The supplement of 270° + x is 180° - (270° + x) = - 90° - x.
5. The complement of -^^ tt is i tt — y^^ tt = — | tt.
6. The supplement of | tt is tt — | tt = — |7r.
Show that the complement of the first angle of each of the following pairs is
equal to the second angle :
7. 145° and -55° ; 300° and -210° ; -70° and +160° ; -200° and +290°.
8. 180° - X and - 90° + x ; 270° - x and - 180° + x ; 360° - x and
- 270° + X.
9. \ IT and \ir ; f tt and — tt ; rr — x and x — \ir ', lir + x and — ^ tt — x.
Show that the supplement of the first angle of each of the following pairs is
equal to the second angle :
10. 145° and 35° ; 225° and - 45° ; - 160° and 340° ; - 70° and 250°.
11. 270° - X and - 90° + x ; 90° + x and 90° - x ; x - 90° and 270° - x.
12. \ IT and | tt ; § tt and - | tt ; x — tt and 27r— x; |7r + x and — \-ir — x.
134. General Measure of an Angle. — The line OA may be
brought into the position OB by revolving either through the
32 PLANE AXD ANALYTICAL TRIGONOMETRY.
number of complete revolutions in either direction. The
general measure of the angle A OB
is then not x, but x+nZGO°^ where
n is any whole number, positive or
negative.
The general circular measure of
""'- — '^ the angle whose circular measure
^'°' ^^* less than 2 tt is a: would be a; + 2 titt,
since 2 tt corresponds to a complete revolution.
1. Show that 1000° is in the fourth quadrant.*
1000° = 720° + 280° = 2 X 360° + 280°, two complete revolutions and 280°
beyond ; 280° lies in the fourth quadrant.
2. Show that — 3000° is in the third quadrant.
- 3000° = - 2880° - 120° = 8(- 360°) - 120°, eight complete revolutions
and 120° beyond in the negative direction ; — 120° lies in the third quadrant.
3. Show that - (8 w + |) is in the first quadrant.
A
-(8n + f)=2nx27r + j'^7r, 2n complete revolutions and -^^ ir beyond ;
2
-j3jj TT is in the first quadrant.
4. Show that 1500° is in the first quadrant, 2690° in the second, 2720° in
the third, 2100° in the fourth.
5. Show that — 010° is in the second quadrant, - 1100° in the fourth,
- 1400° in the first, - 1920° in the third.
6. Show that -(10n4-6^ is in the third quadrant, -(12n + 2^ in the
IT ^ 3
second, t (8 w 4- 7) in the fourth, | tt (3 n + 2) in the third,
7. Show that f tt (10 n — i) is in the fourth quadrant, f tt (15 n — f ) in the
third, f TT ( - 9 n - f ) in the third, ^ tt (10 w — 9) in the first.
8. Show that ^ (9 n + 1") will lie in the third or in the first quadrant,
o
according as n is odd or even.
9. Show that the general circular measure of 0° is 2 nir, and not wr.
10. Show that the general circular measure of 90° is (2 w + ^) ir ; of 180°,
(2n + l)ir; of 270°, (2 n + |)7r.
11. If a; = 60°, show that one third of the general measure of x will be 20°,
140°, and 260°, the terminal side of the angle for all values of i x greater than
260° falling in one of these positions.
We have, using the general measure, « + w 360°,
a: = 60°, 420°, 780°, 1140°, 1500°, 1860°,-.
.-. ia; = 20°, 140°, 260°, 380°, 500°, 620°,..-
or I a; = 20°, 140°, 260°, 20°, 140°, 260°,...
if we reduce the values oi ^x that are greater than 360° to others less than 360°
by subtracting some multiple of 360°.
* That is, show that when the angle is 1000° the terminal side will lie in the
fourth quadrant.
TRIGONOMETRIC FUNCTIONS OF ANY ANGLE.
33
12. Jf X = 45°, show that ] x will be 15", 136°, 255°, three values.
13. If X = 20°, show that I x will be 5°, 95°, 185°, 275°, four values.
14. If X = 60°, show that I x will be 10°, 70°, 130°, 190°, 250°, 310°, six
values.
1 r>i°
15. If X = m°, show that -x will have n values less than 3G0°, as —
n n
m° 360'^
n n
wt° 720°
n n
rn^ (n- 1)360°
n n
35. The definitions of the trigonometric ratios in Art. 8 are
applicable only to angles less than 90°. We shall now con-
sider the more general definitions, of which those in Art. 8
are special cases.
36. Map Drawing by Coordinates.* — Let ABCD be a field
whose map is wanted. From any point in the field, measure
the distances Oa, Oh^ Oc, and Od^ and also measure the dis-
tances aA, bB, cO, and dD, at right angles to X'OX, Lay off
on the paper a line X' X of
indefinite length, and take on j
it some point to represent I
the point in the field. Lay
off Oa according to some con-
venient scale ; thus if Oa were _x
200 feet, and the scale were 20
feet to 1 inch, we would on the
map make Oa 10 inches long.
Then draw the line aA perpen-
dicular to OX on the proper
side of OX^ and lay off on it
the distance corresponding to
aA according to the same scale, thus locating the point A.
The other points would be located in a similar manner.
Since Oa and Oc are measured from in contrary direc-
tions, and a A and cC are measured on opposite sides of X' X^
there is danger of laying them off in the wrong direction ;
hence their directions must be carefully distinguished.
37. Coordinates. — The distance Oa, measured along X'OX,
is called the abscissa of the point A ; aA^ measured parallel to
* This is called the method of offsets.
CBOCK. TRIG. — 3
Fig. 27.
84
PLANE AND ANALYTICAL TRIGONOMETRY.
Fig. 28.
Y'OY^ the ordinate of A; and the two distances Oa and aA^
the coordinates of A. The line X'OX is called the axis of
abscissas ; the line Y'OY, the axis of ordinates; and the point
0, the origin of coordinates.
The abscissa of a point is its
distance from the axis of ordi-
nates measured on a line parallel
to the axis of abscissas.
The ordinate of a point is
its distance from the axis of
abscissas measured on a line
parallel to the axis of ordinates.
The abscissa is positive when
the point is on the right of the
axis of ordinates, and negative
when it is on the left; the ordi-
nate is positive when the point is above the axis of abscissas, and
negative when it is below. If we consider the abscissas as
measured from Y'OY, and the ordinates from X'OX, they will
be positive when measured to the right and upward respec-
tively.
Using the customary notation for directed lines,* Oc will
represent a line measured from to , and cO will be measured
from c to 0. The line cO measured to the right is positive,
and Oc to the left is negative. Hence the coordinates of
are Oc and cC, or — cO and —Co. For brevity, the coordinates
of a point are written in a parenthesis with a comma between
them, the abscissa being written first ; thus the point D is
called the point (^Od, dD).
The ordinate of any point on X'OX is zero, the abscissa of
any point on Y'OY is zero, and both coordinates of the origin
are zero.
The signs of the numerical coordinates of points in the
different quadrants are as follows :
Quadrant
Abscissa
Ordinate
I. 11.
+ -
III.
IV.
4-
* See Art. 2.
TRIGONOMETRIC FUNCTIONS OF ANY ANGLE.
35
38. Distance of a Point from the Origin. — Represent the
abscissa of the point by a, its ordinate by o, and its distance
from the origin by h. Then
h = Va2 + o\
since h is the hypotenuse of a right triangle whose sides are a
and 0. Although h may be either positive or negative, it will
be sufficient for our purposes to treat it as being always
positive.
Fig. 29.
39. Trigonometric Ratios. — Take the origin of coordinates
at the vertex of the angle and the initial side as the axis of
abscissas. From any point B on the terminal side of the angle,
draw AB perpendicular to the initi^ side ; denote the abscissa
OA of the point by a, its ordinate AB by o, and its distance OB
from the origin by A. The general definitions of the trigono-
metric ratios are :
The sine of the angle
The cosine of the angle
The tangent of the angle
The cotangent of the angle
The secant of the angle
The cosecant of the angle
ordinate
o
Yi
a
~ h
o
~ a
a
~ o
h
~ a
h
~ o
distance ~
abscissa
distance '
ordinate
abscissa
abscissa
ordinate ~
distance
abscissa ~
distance
ordinate ~
(1)
36 PLANE AND ANALYTICAL TRIGONOMETRY.
Note. — The origin is always at the vertex of the angle; the axis of
abscissas always coincides with the initial side ; and the positive direction of the
axis of ordinates is along the line that makes an angle of + 90° with the initial
side.
Prove that the following equations are true, using Eqs. (1):
1. ^^^ ^ = cosec X.
V sec* X — 1
sec a; = - ;
a
h
-1 J?
\/sec2 X - 1 ^ fi^ 1 Vh^
- = cosec X.
a"
2. sec X cos X = 1. 7. tan x cot x = 1.
3. cosec X sin X = 1. 8. sin^x + cos^x = 1.
4. cosec* X = 1 + cot2 x. 9. sec* x = 1 + tan-^ x.
6. ,*^!i^„ = si»r«. 10. vlli2t!^=seca..
V 1 + tan"'^ X cot x
6. Vcosec-^^-l^eosx. 11. J l + cot*x ^^^^^^
cosec X ' cosec "-^ x — 1
12. (tan X — cot x) (tan x + cot x) = — ^ -^ = sec* x — cosec* x.
13. (tanx + cotx) sinx cosx = 1.
Construct geometrically the angles, and compute the corresponding ratios in
the following examples : *
Quadrant. Sin.
Cos.
Tan.
Cot.
Sec.
Cosec.
14.
sinx= + |.
I.
—
+ f
+ |.
+ |.
+ 1-
+ f.
' II.
—
-1-
-f.
-f
-|.
+ f-
15..
sinx==-i.
III.
—
-fV2.
+ iV2.
+ 2\/2.
-|V2.
-3.
IV.
—
+ f\/2.
-iV2.
-2\/2.
+fv^.
-3.
16.
cosx= + |.
I.
+ hVs.
+ V3.
+ iV3.
+ 2.
+ |V3.
IV.
-W3.
—
-V3.
-iV3.
+ 2.
-|>/3.
17.
cosx=-}.
II.
+ fV2.
—
-2V2.
-i\^.
-3.
+ IV2.
III.
-f\/2.
—
+ 2\/2.
+ iV2.
-3.
-|V2.
18.
tanx=+^.
I.
+ iV5.
+ f\/5.
—
+ 2.
+ IV5.
+ V5.
III.
-^V5.
-fV5.
—
+ 2.
-^>/5.
-V5.
19.
tan x= -2.
II.
+ |V5.
-|\/5.
-h
-V5.
+ 1V5.
IV.
-f\/5.
+ iV5.
—
-h
+ V5.
-|Vg.
* See Arts. 11 and 15. If sin x is positive, must be positive, since h is
always positive, and the angle lies in quadrants I. and II.
TRIGONOMETRIC FUNCTIONS OF ANY ANGLE. 37
Quadrant, Sin.
Cos.
Tan.
Cot.
Sec.
Cosec.
cotx=+|.
I.
+ f
fl.
+ |.
—
+ f-
+ !•
III.
-f.
I.
+ |.
—
-i-
_5
3*
cot 2= -3.
IT.
+tVv1o.
-^VTo.
-f
—
-jVTo.
+ Vio.
IV.
-T^.VIo.
+^vlo.
-h
—
+ ^VIo.
->/io.
sec a; =+3.
I.
4f>/2.
+f
+ 2\/2.
+ iV2.
—
+IV2.
IV.
-fV2.
+1.
-2V2.
-\V2.
—
-|v^.
seca:= — ^.
II.
+ !•
-h
-|.
-I
—
+1-
III.
-f.
-f.
+ t.
+ 1-
—
-l.
cosecx=+-U.
I.
+ tV
+ H-
+ tV
+v.
+f|.
—
II.
+A.
-H-
-A-
-V-.
-ii-
—
coseca;=— y-.
III.
-^T.
-if
+iV
+¥.
-If.
—
IV.
-/5-
+ lf
-^i.
-¥•
+ l|.
—
21.
22.
23.
24.
25.
40. Trigonometric Functions. — One quc^ntity is said to be a
function of anotlier when it depends upon the latter for its
value. Thus, if 2/ = sin a;, ?/ is a functioii of x, since it depends
upon X for its value, any change in the value of x producing a
change in the value of y.
The trigonometric functions are the sine, cosine, tangent,
cotangent, secant, cosecant, versed sine, coversed sine, and
suversed sine. The last three are defined by the eq.uations :
The versed sine is vers a; = 1 - cos a?
The coversed sine is covers a; = 1 - sin a;
The suversed sine is suvers a; = 1 + cos a;. ,
(1)
41. Geometrical Representation of the Functions. — In Fig. 30
let the radius OB, of the circle described about the vertex
of the angle AOB as a center, be unity, and let the angle AOY
be equal to 90°. NM and FB are tangent to the circle at
X and F respectively ; the triangles OAB, OXM, and OYB,
are right-angled ; and the angle YB is equal to the given
angle AOB. Then the trigonometric functions of the angle
AOB are represented by the lines shown in the figure. For,
in Figs. 2 and 29, B is any point on the terminal side OB
of the angle AOB, and therefore we may choose the position
of B so that OB, or A, shall be equal to unity. Comparing
Fig. 30 with Figs. 2 and 29, and using the definitions in
Arts. 8, 39, and 40, we see that
38
PLANE AND ANALYTICAL TRIGONOMETRY.
sin AOB
cos AOB
tan AOB
cot AOB
sec AOB
cosec AOB
vers AOB
covers AOB
suvers AOB
AB
h OB
• AB,
a OA
h OB
-.OA,
AB
a OA
XM
OX
= XM,
a OA
CB
YD
AB
00
or
h OB
a OA
OM
OX
--0M,
h OB
OB
OB
YD.
AB^ 00 ^ OY ^ ^^'
1 - cos AOB = OX -0A = AX.
1 - sin AOB = OY -00= CY.
1 + cos AOB = XO + 0A = X'A.
The trigonometric functions are ratios^ — pure numbers, —
and are represented by these lines in the circle whose radius is
unity ; that is, they are actu-
ally equal to the ratios of these
lines to the radius.
If, with a radius of unity
and the vertex of the angle as
the center, a circle be described
and two tangents be drawn, one
Avhere the initial side OA cuts
the circle, and the other at a
distance of + 90° from this
point (at X and Y respec-
tively), the trigonometric functions will be represented as
follows :
The sine of an angle will he the perpendicular distance from
the point where the terminal side of thi» angle cuts the circle, to
the initial side, produced if necessary ; positive when it is above,
and negative when below, the initial side. Thus sin A OB = AB,
sin AOJI = GH, sin A OK = aK, sin A OL = AL. AB and
6rl£, above X' OX, are positive, while GK and AL are nega-
tive, being below X' OX. The sine is therefore positive when
TRIGONOMETRIC FUNCTIONS OF ANY ANGLE. 39
the angle is in the first or second quadrant, and negative when
it is in the third or fourth.
The cosine will be the distance from the center to the foot * of
the sine ; positive when measured to the rights and negative to the
left, of the center. Thus cosAOB:=OA, cos AOI£= OG,
cos A OK = 0G-, cos A 0L= OA. OA, measured to the right
of the center, is positive, while 0(r, measured to the left, is
negative. The cosine is therefore positive when the angle is
in the first or fourth quadrant, and negative when it is in the
second or third.
The tajigent will be the distance along the line tangent to the
circle at the point where the initial side cuts the circle, from this
point to the point where this tangent is cut by the terminal side of
the angle, produced if necessary ; positive when measured above,
and negative when beloiv, the initial side. Thus tan ^0^ = XM,
tan A 011= XJSr, tan A 0K= X^, tan AOL = XK XM, above
X' OX, is positive, and XJSf, below X' OX, is negative. There-
fore the tangent is positive when the angle is in the first or
third quadrant,. and negative when it is in the second or fourth.
^ The cotangent will be the distance along the second tangent
(FYD) from the point of tangency to the point ivhere this line is
cut by the terminal side of the angle, produced if necessary ; posi-
tive when measured to the right, and negative to the left, of the
point of tangency. Thus cot A OB = YD, cot A OH = YF,
cot AOK= YD, cot AOL = YF. YD, measured to the right,
is positive, and YF, measured to the left, is negative. There-
fore the cotangent is positive when the angle is in the first or
third quadrant, and negative when it is in the second or fourth.
Note, — The positive directions of measurement are above X'OX and to
the right of TOY, and the negative are below X'OX and to the left of Y'OY.
The secant ivill be the distance from the center along the ter-
minal side of the angle, produced if necessary, to its point of
intersection with the taiigent at the point of intersection of the
initial side with the circle; positive when measured along the
side itself, and negative when along the side produced. Thus
sec AOB= OM, sec A 0H= OX, sec A 0K= OM, sec AOL= ON.
* The foot of the sine is the point where the perpendicular line representin ;
the sine cuts the initial side, produced if necessary. | %
40 PLANE AND ANALYTICAL TRIGONOMETRY.
Since sec A OB and sec AOL are measured along the terminal
side itself, they are positive. The terminal sides (^OH and
OK} of the angles A OH and A OK must be produced in order
that they may intersect the tangent line iV7l[f, and therefore
^ec A Off and sec A OK are negative. Hence the secant is
positive when the angle is in the first or fourth quadrant, and
nes^ative when it is in the second or third.
The cosecant will he the distance from the center along the
terminal side^ produced if necessary^ to its intersection with
the second tangent^ FYD ; positive when measured along the
side itself^ and negative when along the side produced. Thus
GosQcAOB= OB, cosec ^Oir= OF, cosec A OK = OB,
cosec A OL = OF, Since cosec A OB and cosec A OH are meas-
ured along the terminal side itself, they are positive, while
cosQc AOK and cosqgAOL, measured along the side pro-
duced, are negative. Therefore the cosecant is positive when
the angle is in the first or second quadrant, and negative when
it is in the third or fourth.
The versed sine (1 — cos x) will he the distance from the foot
of the sine to the point where the initial side cuts the circle;
always positive, hecause cosx can never he greater than the'
radius, or unity. Thus vers A OB = AX, vers A Off = GX,
Yeis AOK= ax, Yers AOL = AX.
The coversed sine (1 — sin x) ivill he the distance from the
point C or P, where a line drawn through the point of intersection
of the terminal side and the circle parallel to the initial side cuts
Y'OY, to the point Y; always positive, since sinx can never he
greater than the radius, or unity. Thus covers A OB = CY,
covers AOff= CY, covers AOK= PY, covers AOL = FY.
The suversed sine (1 + oos x) will he the distance from the
point yj , where the initial side produced cuts the circle, to the
foot of the sine ; always positive, since cos x can never he alge-
hraically less than minus unity. Thus suvers AOB = X^A,
suvers A Off= Xa, suvers A 0K= X' a, suvers AOL = X'A.
Note. — These lines represent th^ trigonometric functions, only when the
radius of the circle is unity. If the radius differs from unity, the functions are
equal to the lengths of these lines divided by the radius.
42. Changes in the Values of the Functions. — Let OX be
the initial side of the angle, and let the terminal side first
TRIGONOMETRIC FUNCTIONS OF ANY ANGLE.
41
r
Y cot
D
\
^
. 1
C E\
A
f
/
/ cos
^
1
X
\
y^
OV ^
ven
1
^
■-^
P 1/
H
coincide with OX^ and then, in revolving about 0, come into
the positions OM, OY, OH, OX', OK, (9^^ ON, and OX, and
let us consider the resulting changes in the values of the sine
and of the tangent.
The sine of 0°, the terminal side coinciding with OX, is
zero. As the angle increases,
the sine, being positive, also in-
creases (sin A OB = AB}, until
at 90° it is equal to the radius,
or +l(sinJL07= OF). The
sine then decreases {ain A Oil
= G-H^, still being positive ;
and at 180° it is zero, the ter-
minal side coinciding with OX' .
The sine then becomes negative,
and decreases algebraically, in-
creasing numerically (p\nAOK= (tK), until at 270° it is
equal to the radius, or — 1 (sin ^ OF' = OF'). It then in-
creases algebraically, decreasing numerically (sin^OX = ^X);
and at 360° it again becomes zero.
The tangent of 0° is zero ; the tangent then becomes posi-
tive, and at 90° it is infinite, the terminal side being parallel
to XM; then negative, and at 180° it is zero ; then positive,
and at 270° it is infinite ; then negative, and at 360° it is zero.
Just before the terminal side reaches the position F, the
tangent is positive, and just after, it is negative ; therefore
the tangent of 90° is ±qo, the upper sign being that of the
function of an angle a little less than 90°.
The table gives the values of the functions of 0°, 90°,
180°, 270°, and 360°, and their signs in quadrants I., II., III.,
and IV. :
Fig. 31.
0°.
I.
90°.
II.
lS()o.
III.
270°.
IV.
360°.
sin.
+
+ 1
p
+
_
-1
_
COS.
+ 1
+
9-
—
-1.,
—
+
+ 1
tan.
cot.
CO
+
CO
_
CO
+
+
CO
—
CO
sec.
+ 1
+
CO..
- 1
—
00
+
+ 1
cosec.
00
+
+ 1
+
QO
-
- 1
-
QO
^'
v^V
42 PLANE AND ANALYTICAL TRIGONOMETRY.
43. Limiting Values of the Functions. — The sine and cosine
may have any value between +1 and —1, but they cannot have
a value numerically greater than unity.
Tlie tangent and cotangent may have any value between
+ c» and —00 ; that is, no matter what a number may be, there
will always be some angle that will have that number as
the value of its tangent, and another having it as its co-
tangent.
The secant and cosecant may have any value between -|-1
and + X), or — 1 and — oo ; but they cannot have a value numeri-
cally less than unity.
The versed sine, coversed sine, and suversed sine may have
any value between zero and -f 2.
Note. — In the first quadrant, all the functions are positive, and the sine,
tangent, and secant increase as the angle increases ; while the cosine, cotangent,
and cosecant decrease as the angle increases.
Note. — The functions change signs only when they pass through the values
zero and infinity.
44. Graphical Representation of the Functions. — Let the dis-
tance OL represent 360°, so that 1° is represented hj -^\-qOL.
At (7, such that OQ =^0L^ draw a line perpendicular to OL^ and
D E \F 226' 270°
0° SO" 46" Wf 136" 160° IW
Fig. 82.
lay off on it any convenient distance 0, to represent the sine
of 90°, above the line OL^ since sin 90°= +1. At J., such that
OA = -^^OL, lay off Aa — ^Oc, since sin 30°= + J ; ^t B, such
that 0B = \ OL. lay off Bh = (7c VJ, since sin 45° = + a^J ; at H,
such that Off=^OL, lay off JIh = OcV^, below OX, since
sin 225° = — VJ ; and so on, locating as many points a, 6, c, A,
etc., as may be necessary. Draw a smooth curve through 0, a,
ft, c, (?, e, F, A, I, y, L, and we have the sinusoid, in which the
TlUGONOxMETRlC FUNCTIONS OF ANY ANGLE.
43
abscissas correspond to the angles, and the ordinates to their
sines.
We might have taken OL
equal to the circumference of
the circle whose radius is unity,
and Ce equal to this radius.
The scale would then have been
the same for both the ordinates
and the abscissas.
The graphical representa-
tions of the other functions
may be constructed in a similar
manner.
cot
o^
c
-^
/
f
(
\
\
sin \
^
1
\
/
cos
]
X
\
G y
V
A
p
y\
^^
_^
N-
FiQ. 38.
45. Two Angles correspond to Any Given Function. — In
Fig. 33 let the arcs YB, TIT, Y'K, and Y'Lhe equal; there-
fore the arcs XB, X'H^ X' K^ and XL are equal. Hence
AB = aR=OC; AL = aK=OF; OM=ON; OD = OF.
00 is not equal to OP since they have contrary signs, 00
being positive and OP negative on account of their directions.
AB = ^mXOB', aff=smXOH:;
.-. sin XO^ = sin XOir.
ax = sin XOK ; AL = sin XOL ;
.-. sin XOK = sin XOL.
Therefore two angles that differ by equal amounts from 90°, or
from 270°, will have the same sine ; thus sin (90° + 2°) =
sin (90° - 2°), aCT~smt270°TF7^in(270° - 3°).
Note. — The two angles corresponding to a given function may be identical ;
thus, if sin X =+ 1, the only value of x is ^0°, or 90° - 0° and 90° + 0°.
Again OA = cos XOB = co^ XOL ;
and
0(7 = cos XOH = cos XOK.
Therefore two angles differing by equal amounts from 0°, from
180°, or from 360°, will have the same cosine ; thus cos ( — 5°) =
cos 5°, cos (180° -f 5°) = cos (180° -5°), and cos (360° -10°) =
cos 10°.
Also XM = tan XOB = tan XOK;
and XN = tan XOH = tan XOL,
44 PLANE AND ANALYTICAL TRIGONOMETRY.
Therefore two angles differing from each other by 180° will
have the same tangent ; thus tan 140° = tan 320°.
Again YD = cot XOB = cot XOK ;
and YF = cot XOff = cot XOL.
Therefore two angles differing from each other by 180° will
have the same cotangent ; thus cot 200° = cot 20°.
Also +0M= sec XOB; + ON = sec XOL ;
.-. sec XOB = sec XOL.
-0M= sec XOK ; - ON = sec XOH ;
.-. sec XOK = sec XOH.
Therefore two angles differing by equal amounts from 0°, from
180°, or from 360°, will have the same secant ; thus sec (— 5°) =
sec 5°, sec (180° - 3°) = sec (180° + 3°), and sec (360° - 5°) =
sec 5°.
Again -h OD = cosec XOB ; -{- OF = cosec X OK ;
. \ cosec XOB = cosec XOK.
-0D= cosec XOK; - OF = cosec XOL ;
. • . cosec XOK = cosec XOL.
Therefore two angles differing by equal amounts from 90°, or
from 270°, will have the same cosecant ; thus cosec (90° + 10°) =
cosec (90° - 10°), and cosec (270° - 60°)- cosec (270° + 60°).
The four angles XOB, XOH, XOK, and XOL, have the same functions
numerically. Thus if sin x = ± |, x will be 30°, 150^, 210°, and 330° ; the tirst
EXAMPLES.
1. What angle has the same sine as 140°?
2. What angle has the same sine as 220° ?
3. What angle has the same cosine as 330°?
4. What angle has the same cosine as 220° ?
5. What angle has the same tangent as 230° ?
6. What angle has the same tangent as 300° ?
7. What angle has the same cotangent as 240° ?
8. What angle has the same cotangent as 110° ?
9. What angle has the same secant as 315° ?
10. What angle has the same secant as 160° ?
11. What angle has the same cosecant as 110°?
12. What angle has the same cosecant as 300° ?
Ans.
40°.
Ans.
320°.
Ans.
30°.
Ajis.
140°.
Ans.
50°.
Ans.
120°.
Ans.
60°.
Ans.
290°.
Ans.
45°.
Ans.
200°.
Ans.
70°.
Ans.
240°.
6i
TRIGONOMETRIC FUNCTIONS OF ANY ANGLK. 45
Find the values of less than 360° in Exs. (13-24) : ♦
13. sin 5 = - sin 200°. Ans. 20°. 19. cot ^ =- cot 106°. An8. 76°.
14. sin ^ =- sin 100°. ^ns. 260°. 20. cot « = - cot 205°. ylns. 166°.
16. cos(? = -cosl50°. Ans. 30°. 21. sec ^ = - secl40°. Ans. 40°.
16. cos=-cos300°. Ans. 120°. 22. sec ^ = - sec 326°. Ans. 146°.
17. tan ^= -tan 350°. A71S. 10°. 23. cosec^ =- cosecl20°. ^ns. 240°.
18. tan 0= -tan 230°. Ans. 130°. 24. cosec ^ = - cosec 355°. Ans. 6°.
25. cos 3 ^ = + i y/S. Find three values of less than 180°.
3 may he 30°, or 330°, or these values plus any number of circumferences;
.-. 3^ = 30°, 390°, 760°, ••., 330°, 690°, 1050°, -.
.-. = 10°, 130°, 250°, ..., 110°, 230°, 360°, ...
Ans. = 10°, 110°, 130°.
26. sin 2 ^ = - h Find four values of less than 360°.
Ans. 106°, 285°, 165°, 345°.
27. tan 3 ^ = — 1. Find six values of less than 360°.
Ans. 45°, 165°, 285°, 105°, 226°, 345°.
28. sec 6 ^ = -2. Find five values of less than 180°.
Ans. 24°, 96°, 168°, 48°, 120^
29. cot 5 ^ = + 1. Find five values of less than 180°.
Ans. 9°, 81", 153°, 45°, 117°.
30. cos4 = — i. Find four values of less than 180°.
Ans. 30°, 120°, 60°, 150°.
31. sin = h. Show that the general measure of is (2 n + ^)ir ± i tt.
= 30° and" 150°, or 90° - 60° and 90° + 60°, or 90° db 60°, or '^ tt ± A tt.
But the general measures of are these values increased by any number (;i) of
circumferences. . •. = '^ nir -{- ^ ir ±.^ ir = (2n + ^) tt ± i tt.
32. sin = + hV2, tan = — 1 ; the general measure of ^ is 2 wtt + l-v.
Note that is in the second quadrant, since its sine is positive and its tan-
gent is negative.
Q
33. cos ^ = — ^ cosec = -\ ::; the general measure of is 2 nrr + 6',
2V2
where 6' is the value of that lies between ^ tt and w.
34. cos (9 = — I ; the general measure of ^ is (2 w + 1)t ± ^ tt.
35. sin 2 ^ = + i ; the general measures of are (2 w + |) tt ± | tt, and
(2 n + f ) ^ ± i TT.
36. cos 3 ^ = — i ; the general measures of are (2 n + |) tt ± ^ tt,
(2 n + 1) TT ± i TT, and (2 n + f ) tt i ^- ir.
Construct geometrically (Art. 11) the two angles when
37. sinx=+i. 41. tanx = + 2. 45. sec a; = +3.
38. sin a; = — ^. 42. tan x=—\. 46. sec x=—\.
39. cos a; = + |. 43. cot sc = + |. 47. cosec re = + 6.
40. cos a; = — |. 44. cot r = — f . 48. cosec a: = — |.
♦ Only one of the two answers is given.
CHAPTER IV.
RELATIONS BETWEEN THE FUNCTIONS OF ONE ANGLE.
46. Relations between the Functions of One Angle.
o2 4.^2 = ;^2.
or.
A2^A2
sin'^ dc + cos^ oc — \^
h sma?
tan X = - = - = ;
a a cos a;
I
sino?
tan 05
cos 05'
cot 05=- = 7 — -;
tan 05
cot 05
COS 05
\
sin 05
A2 = a2 + o2;
. *. sec^ 05 = 1 + tan'^ x,
7,2 = 02 + ^2;
. •. COSeC^ 05 = 1 + COt^ 05.
h 1
sec 05 = -
=1+^
7(2
^^^
a cos 05
h 1
cosec ic = - = —. — •
sinx
Ters a! = 1 - COS x.
corers a; = 1 - sin as.
snyers a; = 1 + cos a;.
46
(1)
(2)
(S)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
FUNCTIONS OF ONE ANGLE. 47
Note. — These formulas may be easily remembered by the use of Fig. 30,
where
AB^-\- OA^ = 0B\ or sin2 a; + cos2 x=l.
.„„ ^ XM AB ^^ .^ ^ sin x
tan X = — — - = — — , or tan x =
OX OA cos a;
cotar=— — = — , or cotx =
OY OC sinx
0.1/2 = 0X2 + xJlf2, or sec2 x = 1 + tan2 x.
OD'- = Or2 + YD\ or cosec2 a; = 1 + cot2 x.
\ 47. To express One Function in Terms of Each of the
I Others. — Suppose that we wish to find expressions for sin a:
[ that shall contain only cos a;, tana:, cot a;, sec a:, and cosec^
I respectively. From the preceding article we have:
^ sin^ X + cos^ a: = 1, and cosec x =
sin a;'
and sin x =
sin X = ± Vl — cos^ a^
1
cosec X
The other expressions are derived from these as follows :
^ = ± ^ from (6).
cosec X Vl + cot2 X
1 , 1 , tanx , ,„.
•. sin X = ± — = ± — - = ± — , from (3).
V 1 + cot2 X L 1 Vl + tan2 x
\ tan^x
r,i^ ^ I tanx , \/sec2x — 1 i?^^^ /c^
. •. sm X = ± •= ± , from (5).
Vl + tan2 X sec x
The double signs are due to the fact that there are two
angles corresponding to any given function ; thus if cos a; = |^,
the angle might be either^ in the first or in the fourth quadrant,
and the sine would be positive in the first case and negative
in the second. It will be seen that if any one of the functions
' is given, all the others found from it will have the double sign,
except its reciprocaL
48
PLAN"£ AND ANALYTICAL TRIGONOMETRY.
In the same way it may be shown that*
1 cot X 1 Vcosec-^ X — 1
cos X — v'l — sin^ X —
Vl + tan^ X Vl + cof-^ x sec x cosec x
•
tan X- ^^^^
Vl-cos''x_ 1 _^/sec2a. i_ 1
Vl - sin2 X
cos X cot X Vcosec^ x — 1
^^^^ - ^ - ^ • -Vcoscc^x 1
sinx
Vl - cos2 X i^aii a: Vsec2 x - 1
aan n> — —
^ - Vl + tan2 X - ^^ + ^*^'' ^ - ^^^^^ * .
Vl — sin^ X
cosx cotx Vcosec2x-l
1 1 Vl 4- tan2 X / i . ^^♦•2^ sec x
cosec X = = — — = ■ = V 1 -f cot-* X = — - ■
sinx Vl -cos2x ^anx Vsec2 x - 1
If any one of the functions is given, the others may be
found from these formulas. It is easier in general to find first
the sine and cosine, and then to find the others.
48. Find the Unknown Functions in the Following :
1. tana;=— -, 2: being in the fourth quadrant. Compute
the numerical values of the ratios
by the^ method of Art. 15, and then
select the proper signs for the func-
tions in the fourth quadrant. Thus let
= 3,
sin 2:= — -,
5
4, . A = 5
cos x — -\-
5'
cot 2; = --,
sec a: = H- -,
4
cosec a: = — -•
o
2. tan 2: = 2, a; being in the third quadrant. Then
sma:
tana:
cos a:
_1_
V5'
Vl + tan2 X V5' Vi + tan2 x
These convenient formulas may be easily remembered from
Fig. 35. Knowing sin x and cos a:, we have
cota: = - = 4-^; seca; = = _V5:
tan X A cos x
cosec X
sma; z
The radicals should be taken with the double sign.
FUNCTIONS OF ONE ANGLE. 49
8. cot X = — % X being in the second quadrant.
•. coseca; = ± Vl + cot2a;= + V5; sin 2: = =H — ;
cosec X V6
cos x = ± Vl — sin^a; = ^ ; tan x = == — - ;
V5 cot X 2
sec a; = =— -V5.
cos X I
4. sec x = — -y-, X being in the third quadrant.
. •. cos X = = — -— ; sin a: = ± V 1 — cos'^^ = ;
sec a: 17 17
, sin a: , 15 . ,8 17
tana; = =-|-— -; cota; = +-— ; cosec a; = .
cos a; 8 15 15
5. sinx = — I, X being in the third quadrant.
. •• cos X = — f ; tan x = + | ; cot x = + | ; sec x = — | ; cosec id = — | .
6. cosx = + I, X being in the fourth quadrant.
.-. sinx = — |V6; tanx = — ^VS; cotx = — |\/5; secx = +f;
cosecx = — |V5.
7. tanx = — y\, X being in the second quadrant.
. sin X = + t\ ; cos X = — {| ; cot x = — -^ ; sec x = — ^| ; cosec x = + ^?-.
8. cotx = + j\, X being in the third quadrant.
. '. sin X = — II ; cos x = — ^^.
9. sec X = — II, X being in the second quadrant.
. •. cos X = — If ; sin X = + y^y ; tan x = — -^-^.
10. cosec X = — V , X being in the fourth quadrant.
If sin J^:
If tan J^;
•. sma
^ = -/t;
; CO
.sx = +r_
where s =
cos J^=-
where s =
cosi5=
; tan %■=-
a^b^-c
2 '
■^.
11.
W^
-6)(s-
6c
_^
show that
Ms-a)_
> 6c
a + 6 4- c
2 '
12.
=#
-a)(s-
s(s - c)
^
show that
ks-c),
^ ^ ab
13. If sec 6 = a, show that sin 6 is imaginary if a is numerically less than
unity.
d = Vl - COS20 =a/1 ^ rrJl - 1
\ sec2^ ^ a2
Smt/=V1 -COS''£'=A/l ^r-^-v/i -— = — i.
»2 /7
CROCK. TRIG.— 4 Vf^^^^^^'^^V
'^ Of THE
50 PLANE AND ANALYTICAL TRIGONOMETRY.
14. If tan 6 = a, show that cosec 6 is real for all values of a.
16. If cos d = a, show that cosec 6 is imaginary when a is numerically
greater than unity.
49. The Signs of the Functions are given by the formulas
of Art. 46, so that it is necessary to remember only that the
sine is positive in the first and second quadrants and the cosine
in the first and fourth. Thus, in the second quadrant,
sin a; -h , cos a: —
tan X = = — = — ; cot X = = — = — ;
cos a; — sma; +
1 + 1 + , .
sec X = = — = — ; cosec x = - — = — = + .
cos a; — sin a; +
50. Find the Values of the Following Expressions :
1. versa:tan3T-l ^^^^ tan a: = 4, x being in the third
secic
quadrant. Find the numerical values of vers x and sec x, and
substitute.
.-. cosa; = ^, seca; = — VTf, versa;=lH — .
VlT + l .^ ^
Vr! 4VT7 + 4-V17 3VT7 + 4
-VI7 -17 17 •
2. SIP ^ sec X ^Yien versx = |, x in the fourth quadrant. Ans. + 16.
cos X cosec X
8. tan a; - cot X ^^^^^ ^^^^^ x = - V5, x in the third quadrant. Ans. - f .
tan X + cot X
when cot a; = — ^, a; in the second quadrant. Ans. — 2.
cosec X 4- cos x
5. sm a; + tan a; ^^^^^^ sec x = - f , x in the third quadrant. Ans. + ^%.
cos X + vers x
6. sec X - vers x ^^^^ cot x = - 2, x in the second quadrant,
sec X 4- vers x
Ans 9 + 2V5 ^ 29 -^ 20V5
7. su^y^ + t^"'^^ when secx = - |, x in the second quadrant. Ans. ^Wz-
cos-^ X + vers2 x
8. secx + sin X ^^^^ tan x = 2, x in the third quadrant. Ans. - ^ V5.
1 — cotx
FUNCTIONS OF ONE ANGLE. 51
9. cosecx + seca ; ^^^^ sec a; = + VIO, x in the fourth quadrant.
^°^^^°«^ Ans. -20.
10. secx-coseca; ^^^^ cotx = - 2, x in the second quadrant. Ana. - 3.
sec X + cosec x
11. versx- covers X ^^^^ sinx = - |, x in the fourth quadrant.
sec x- cosec X ^,^^ _ ^^
51. Change the Given Expression to Another containing only
One Function :
, 2 sec? X 4- sec^ x tan^ x — sec* x
sec^ a; — 1
to contain only cosec x.
It is best generally to change the expression to another con-
taining only sin x and cos x^ and then to change this into one
containing the proper function.
2 sin2 X 1
cos^ X cos* X cos* X _2 cos^ X + sin^ x — 1
1 -| cos2:c(l— cos^a;)
COS^ 2J
2 — 2 sin^ 2; + sin''^ a; — 1 1 — sin^ x 1 o
; ! ; = ; ^ = — = COSCC^ X.
(1 — sin^ x^ sin^ x (1 — sin^ x^ sin^ x sin^ x
to contain only tan x.
vers 2: — covers x
sin^ a; — cos^ x . , , tan 2: . 1
= sm X + cos x= ±
l-cosa;-l + sina: Vl + tan^a; VT+tan^^
where the signs used will depend upon the quadrant of x.
The true result is ± — where the positive sign
Vl + tan^ X
corresponds to x in the first or fourth quadrant, and the nega-
tive to X in the second or third.
Use radicals as little as possible.
3. 1 — 2(1 — covers x)2 -\ ^~-^ — - to contain only cosx. Ans. cos*x.
^ (1 + t,an2 x)2 ^
4 sec xcosecx-4sinx cosx ^^ ^^^^.^ ^^^^ ^.^ ^ ^,^^ (l-2sin2x)2^
sin X sec x sin^ x
5. (^ - covers x)^ cosec* x ^^ contain only tan x. ^ns. tan2 x + tan* x.
(cosec- X — 1) cot-x
« sec2 X — sec2 x sin* x(l + cot^ x) . . . ,
6. — ; 7^—^ ^ to contain only cosec x.
sni2 x cos2 X .
Ans. *^^^^^*^
cosec2 X — 1
52 PLANE AND ANALYTICAL TRIGONOMETRY.
7. tan2^sec2^-sin2^cos2^ to contain only cot ^. Ans. l+'^cot2g+3cot*g ^
COt4^(l+COt2 0)2
8. ^ ~ ^" ^^' (cos* X — sin* x) to contain only sin x. Ans. (1—2 sin2 x)*.
(l + tan'-ix)--^^ ^ ^ \
9. — sec2 a sin2 a — ^^ contain only cosec a. ^ns.
(tan a + 2 cot a)2 (2 cosec2 o - 1)2
- 10. ^'^'^^^"'^ to contain only sec 6. Ans. C^^^^'^--^)'.
sin- e — cos-^ d sec2 ^ - 2
- 11. sec2^cosec2^+sec2^-cosec2^-l ^^ ^^^^.^ ^^^ ^ ^^^^ cot2^+2
tan2 d - cosec2 ^ + 1 1 -cot* d
52. Solution of Trigonometric Equations. — Transform the
given equation into one containing only a single function
(usually the sine or cosine), because in a single equation we
must have only one unknown quantity. Then solve the equa-
tion algebraically for this function as the unknown quantity.
The corresponding angle may then be found from the tables.
Test the angles by substitution in the given equation.
I. sin ^ cos ^ = + J.
.-. sin eVl - sin2 (9 = + J ; .-. sin2 (9(1 - sin2 (9) = J ;
.-. sin*(9-sin2(9 + i = 0; .-. sin2(9 - 1 = ; .-. sin^ = ±Vj.
.-. e might be 45°, 135°, 225°, or 315°. But the given equa-
tion shows that the product of the sine and cosine must be
positive, and hence that they must have the same sign. Both
the sine and cosine are positive in the first quadrant, and nega-
tive in the third, but they have contrary signs in the second
and fourth quadrants. Hence the only admissible values of 6
are 45° and 225°.
2. tan ^ sec ^ = - \/2. .-. 5 = 225°, 315°.
3. cosec 5 = I tan 5. .-. 5 = 60^ 300°.
4. tan 5 + cot 5 = 2. .-. = 45°, 225°.
6, sec2 d + cosec2 6 = 4. .: d = 45°, 135°, 225°, 315°.
6. sin5=±\/3vers5. .-. 5 = 0°, 60°, 300°.
7. see + tan 5 = ± V3. .-. d = 30°, 150°. [300°, 330°.
8. sec2 + cot2 5 = J/ . .-. 5 = 30°, 60°, 120°, 150°, 210°, 240°,
9. sinx= + V3cosx. . •. x = 60°, 240°.
' 10. tan X = - 2\/3 cos X. .'. x = 240°, 300^
II. sin X cos X = - 1 V3. . •. x = 120°, 150°, 300°, 330°
12. sin 6 + cosec 9 = -^. .: $ = 210°, 330°.
13. 3 sin X = 2 cos2 x. .-. x = 30°, 150°.
14. sec X tan X = + 2 V3. . •. x = 60°, 120°.
FUNCTIONS OF ONE ANGLE.
68
16. sec 6 vers 6 = \ — tan 6.
1 (l-C08^)=l-''"^
sin ^ = 2 cos ^ — 1 ;
cos 6^ ^ cos^
.-. sin2^ = 4cos2^-4cos^ + l; .-. l-cos2^ = 4cos2^-4cos^ + l;
. •. 5 cos2 ^ - 4 cos ^ = ; . •. cos ^ (5 cos ^ - 4) = ;
. •. cos ^ = and 5 cos ^ — 4 = 0.
(a) cos ^ = gives 6 = 90° or 270°. These values are re-
jected for reasons involving the methods of the Differential
Calculus.
(J) cos ^=f gives sin ^= ± |, since this value of the cosine
will allow the angle to lie either in the first or in the fourth
quadrant. Transposing in the original equation, we have
sec 6 vers + tan ^ — 1 = 0,
and we test by substitution. For in the first quadrant, we
have
showing that 6 has a value in the first quadrant. For 6 in
the fourth quadrant, we have
not zero; and hence 6 does not have a value in the fourth
quadrant.
16. sin X tan x = — t%.
17. vers x = 2 covers x.
18. sin X tan x = 2 cos x.
19. sec X cosec x = — 2.
20. cos X cot X = — f .
21. sin X cos X = — ^|.
22. tan x = — ^20 cos x.
23. sec X + tan x = 2.
smx:
5»
cos X = — i ; quadrants II. and III.
•. cos X = I or — 1 ; first quadrant, and 180°.
•. sin X = ± Vf , cos X = ± V-J ; four quadrants.
•. sin x = ± ^\/2, cos X = T 1 V2 ; 135° and 316°.
•. sin X = — f ; quadrants III. and IV.
•. sin X = ± I or ± f ; quadrants II. and IV.
2
•. sin X = :: ; quadrants III. and IV.
V5
•. tan X = + f ; first quadrant.
24. sec X tan 2 a: (1 — 2 cos x) = 0.
The values of x are found by placing each factor equal to
zero, and solving the resulting equations. Hence we have
sec X = 0, tan 2 a; = 0, 1 — 2 cos x = 0,
64
PLANE AND ANALYTICAL TRIGONOMETRY.
But sec a; = is impossible; tan 2 a; = gives 2 rr = 0° or 180°,
and, using the general measures of the angles, x = 0°, 90°,
180°, 270°, the second and the last values being inadmissible.
1 — 2 cos a; = gives cos x = -J, and x = 60° or 300°.
25. tan ^ a; = 0.
26. vers 3 a; = 0.
27. sinxcosx(l+2 cosx)=0.
28. cos 2 X (3 - 4 cos^ x) = 0.
29. (l+tana;)(l-2sinic)=0.
30. tan X = — 2 sin X.
31. sin 2x vers3x = 0.
x = 0°.
X = 0°, 120°, 240°.
X = 0°, 90°, 180°, 270°, 120°, 240°. [330°.
X = 45°, 225°, 135°, 315°, 30°, 150°, 210°,
X = 45°, 225°, 30°, 150°.
X = 0°, 120°, 180°, 240°.
X = 0°, 90°, 180°, 120°, 240°, 270°.
53. The Functions of an Angle Greater than 360° are the
same as those of the angle less than 360°, found by increasing
or diminishing the given angle by some multiple of 360° ; for
the position of the terminal side would not be changed by
these operations. Thus
tan 1010° = tan (1010° - 720°) = tan 290° ;
cos ( - 835°) = cos ( - 835° + 720°) = cos ( - 115°),
or cos (- 835°)= cos (- 835° + 1080°)= cos 245°.
54. The Functions of 90° ± oc and of 270° ± a? are numeri-
cally equal to the cof unctions of x, but may differ from them
in signs. Let the arcs UB, EB, KJ, KM, and NP be equal,
E
T)^^
B
/
\
. /
\
/
\
\ r
^^.
\
H
i
n
r \
/
V
>
/,
J
^■--^
_^^M
JiV
Fig. 87.
K
Fitt. 36.
the radii CB and RP each being unity. Then the right tri-
angles PCB, FOB, LCJ, LCM, and SRP are equal, having
FUNCTIONS OF ONE ANGLE. 55
the same hypotenuse (unity) and the angle x the same in
each. Therefore
FB = CI= LM= DF=nC= JL = SP,
and CF=IB = HD = LQ^ MI = JH = RS.
.'. sin( ^()°-x) = IB = OF=RS = + Go^x\
] (4)
cos( W-x)=CI =FB=SP = + &mx. ) ^^^
cos( dO''+x)=CH:=FI) = -I)F*=-SP=-8inx. ] ^ ^
sin (210°-x) = IfJ= CL = -L0*= -RS= - cos a;; ) .^.
008(270'' -x)= Cir=LJ = -JL* =-SP=-sinx.] ^ ^
sin (210° +x} = lM=OL = -LO* = -RS= -cosx;
cos (270° + X) = CI = LM= SP = + sin x.
Thus sin 100°= sin (90° + 10°)= + cos 10° ;
cos 100°= cos (90° + 10°) =- sin 10°.
sin 200° = sin (270° - 70°) = - cos 70° ;
cos 200° = cos (270° - 70°)= - sin 70°.
sin 300° = sin (270° + 30°) = - cos 30° ;
cos 300° = cos (270° + 30°) = + sin 30°.
55. The Functions of i8o° ± y and of 360° — y are numeri-
cally equal to the same functions of y, but may differ from
them in signs. From Fig. 36,
sin (180° - z/)= ^D = Zg = + sin y ;
cos(180° -y)=CH =- HO* = -CI = - cos ?/. * ^^^
8m(im° +y)=HJ = -JH* =-Z5 = -siny;) .^^
cos(180°-hi/)=(7^ =-^(7* = -6Y =-cos^/. ) ^^
sin(360°-j/) = 7ilZ=-if/* =-Zg = -siny;) ,„>.
cos (360° -I/) =07" = + cos2/. ) ^^
Thus sin 100° = sin (180° - 80°) = + sin 80° ;
cos 100° = cos (180° - 80°) = - cos 80°.
sin 200° = sin (180° + 20°)= - sin 20° ;
cos 200° = cos (180° + 20°) = - cos 20°.
sin 300° = sin (360° - 60°)= - sin 60° ;
cos 300° = cos (360° - 60°) = + cos 60
* See Art. 2.
\o
66
PLANE AND ANALYTICAL TRIGONOMETRY.
56. The Functions of a Negative Angle are numerically
equal to the same functions of an equal positive angle, but may
differ from them in signs.
B
^-^
^
?
^
B
\
F
/
\
H
\
r
^
t
/
I
/
\^
\
)
^
L
\
J
J
^--^
—
M
K
Fig. 38.
sin (-?/) = IM= - MI"" = -IB\
= - sin «/ ;
cos ( — 1/) = CT= + cosy. J
Thus
sin (a:-180°) = sin [-(180°-a:)]
= - sin (180° -J)=- sin x.
cos(a:-180°) = cos[-(180°-2;)]
= + cos (180° — x)=— cos X,
(1)
57. Summary. — Using the equations of Art. 46,
, sin a; , cos a;
tan X = , cot X = ,
cos X ^ sin X
sec re
cos a;
cosec X
sin a;
and the results of Arts. T>4:^35, and 5Q, we have
sin (90° — a;) = + cosx;
tan (90° -x) = + cot x ;
sec (90° — x) = + cosec x ;
sin (90° + x) = + cos a; ;
tan (90° + a:) = - cot X ;
sec (90° + x) = — cosec x ;
sin (180° - x) = +smx;/\
QannSIE^xJ
sec (180° — x) = — secx;
sin (180° + a;) = — sinx ;
tan(180° + x) = +tanx;
sec (180° + x) = — sec X ;
sin (270° — x) = — cos X ;
tan (270° - x) = + cot x ;
sec (270° — x) = — cosec J ;
sin (270° + x) = — cos X :
tan(270° + x) = -cotx;
sec (270° + x) = + cosec x ;
sin (360° - x) = - sin x ;
tan(360°-x) = -tanx;
sec (360° — X) = + sec X ;
sin ( — x) = — sin x ;
tan ( — x) = — tan x ;
sec(— x) = + secx;
cos (90° — x) = + sin X ;
cot (90° - x) = + tan X ;
cosec (90° — x) = + sec x.
cos (90° + x) = — sin X ;
cot (90° + x) = - tanx;
cosec (90° + x) = + sec x.
cos (180° — x) = — cos X ;
cot(180°-x) = -cotx;
cosec (180° — x) = 4- cosec x.
cos (180° + x) = — cosx ;
cot(180° + x) = + cotx;
cosec (180° + x) = — cosec x.
cos (270° — x) =,^_sinjc ;
cot (270° - X) = + tan x ;
cosec (270° — x) = — sec x.
cos (270° + x) = + sin x ;
cot (270° + x) = - tan X ;
cosec (270° -|- x) = — sec x.
cos (360° - x) = + cos X ;
cot(360°-x) = -cotx;
cosec (360° — x) = — cosec x.
cos ( — x) = + cos X ;
cot (— x) = — cotx ;
cosec ( — X ) — — cosec x.
* See Art. 2.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
FUNCTIONS OF ONE ANGLE.
67
These formulas may be remembered from the three facts :
(a) Whenever the angle is 90° ± x, or 270° ± x, the func-
tions of the angle are numerically equal to the corresponding
(7of unctions of x.
(b) Whenever the angle is 180° ± x, 360° - a;, or - x, the
functions of the angle are numerically equal to the same
functions of x.
((?) The sign to be placed before the function of x is that of
the original function when x is less than 90°. Thus
sin (270° + x) =^-. cos x,
since, when x < 90°, 270° + x will be in the fourth quadrant,
and sin (270° + x) will therefore be negative.
58. General Method of Proof. — In Arts. 54, 55, and 56,
both x and y were less than 90°, but the formulas in Art. 57
are true for all values of x. Sup-
pose, for example, tbat we wish to
prove the formulas for 270° 4- x
when X is in the fourth quadrant,
that is, when x is between 270°
and 360°. Let KAUaJ=x; then
AEaJKAEaJ= 270° + x. Let
AEGKJ'^x. Then in the right
triangles JCL and J' CL' the angles
JCL and J' CL' are equal, each be-
ing 360° — X ; therefore the triangles
are equal, and CL = CL' and LJ= L'J' numerically,
braically CL = - CL' and LJ= + L'J'.
.-. sin(270° + 2:)=5'J^= CX = - OX' =-cosa:;
cos (270° ■\-x)= CH= LJ = + L'J' = + sin a;.
EXAMPLES.
1. From the preceding equations prove that
Alge-
(a) tan (- 1200°) = cot 30°.
(&) sec 1000° = cosec 10°.
(c) cos ( - 890°) = - cos 10°.
{d) cot 1700° = cot 80°.
(e) cosec ( - 1235°) = - sec 66°.
(/) sin 1340° = - cos 10°.
(gr) sin ( - 3000°) = - cos 30°.
(/i) cos 1300° = -cos 40°.
(0 tan 3200° = -tan 40°.
( j) cot ( - 1300°) = - cot 40°.
(k) sec (- 2900°) = + sec 20°.
(0 cosec 2420° = - sec 10°.
58 PLANE AND ANALYTICAL TRIGONOMETRY.
2. If tan ^ = - cot 140°, find the two values of 6 less than 360°
. •. tan 5 = - cot (90° + 50°) - + tan 50°. .-. = 50^ 230°.
8. Find the values of 6 in the following equations :
(a) sin ^ = + cos 220°. . •. 230°, 310°.
(6) sin^ = + cos310°. .-. 40°, 140°.
(c) sin^ = -cos210°. .-. 60°, 120°.
(d) sin2 d = + cos2 200°. . •. 70°, 110°, 250°, 290°.
(e) cos ^ = + sin 150°. . •. 60°, 300°.
(/) cos ^ = + sin 250°. . •. 160°, 200°.
{g) QOsd = - sin 170°. .-. 100°, 260°.
Qi) cos ^ = - sin 275°. .-. 5°, 355°.
(0 cos2 ^ :=: + sin2 100°. .'. 10°, 350°, 170°, 190°.
(j) tan ^ = + cot 100°. . •. 170°, 350°.
(A:) tan ^ = + cot 200°. . •. 70°, 250°.
(0 tan^ = -cot230°. .-. 140°, 320°
(?n) cot = + tan 260°. . •. 10°, 190°.
(n) cot^ii:+tan345°. .-. 105°, 285°.
(o) cot (? = - tan 245°. . •. 155°, 335°
{p) cot ^ = - tan 305°. .-. 35°, 216°.
(g) sec ^ = - cosec 100°. .-. 170°, 190°.
(r) sec = + cosec 130°. . •. 40°, 320°.
(s) sec ^ = + cosec 310° . •. 140°, 220°.
(0 cosec ^ = + sec 315°. . •. 45°, 135°.
(It) cosec ^ =: + sec 2.30°. . •. 220°, 320°.
(v) cosec ^ = - sec 185°. . •. 85°, 95°.
{w) cosec ^ = - sec 335°. . •. 245°, 295°.
(x) cosec2 zr + sec2 250°. .-. 20°, 160°, 200°, 340°.
{y) SQCd = - cosec 290°. . •. 20°, 340°.
{z) sin ^ = - cos 300°. . •. 210°, 330°.
4. cos = sin 2 6. Show that one value of is 30°.
5. tan n = - cot 120°. Show that one value of 6 is 30° ^ n.
6. sec3e = cosec (n- 1)^. Show that one value of ^ is 90° -^(n + 2).
7. If cot 309° = - j%, find sin 219°.
sin 219° = sin (180° + 39°) = - sin 39°.
But cot 309° = cot (270° + 39°) = - tan 39° ; .-. tan 39° = + j%.
... sin39° = -^:^|?L==-4= = -^; .'. sin219° =-- 4<
Vl +tan2 39° Vl64 Vil V41
8. If sin 21 7° = - j%, prove that tan 127° = - f-
9. If cos 125° = - a, prove that tan 325° = _^ -
vl — a2
FUNCTIONS OF ONE ANGLE. 59
10. If cot 260° = + a, prove that cos 350° = + ^
vTTa2
11. If sec 340° = + a, prove that sin 110° = - , and tan 110° = —
12. If cos 300° = + «, prove that cot 120° = ^
13. If sin 116^ = + a, prove that tan 205° sec 245° _ . VTIT^^
cosec 385° a
14. If cos 200° = - m, prove that tan 110° cosec 250° cot 290° =
m
15. If cosec 185°= -m, prove that tan 355° tan 275° cos 175° = - ^^ — -.
m
16. Show that cot ^ ( - x - 540°) = tan J x.
cot K- X - 540°) = cot (- ^ X - 90°)
= cot [ - (90° + 1 X)] = - cot (90° + i sc) = + tan ^ SB.
17. Show that sin (y — 90°) = — cos y.
18. Show that sin (y - 180°) = - sin y. -
19. Show that cos (y — 270°) = — sin y.
20. Show that sec (— x — 540°) = — sec x.
21. Show that tan (y - 360°) = + tan y.
22. Show that cos |(x - 270°) = + sin | x. [Note that 270° in the parenthe-
sis is to be multiplied by i.]
23. Show that cos |( - 810° + a - 6) = - sin ^(a - 6).
24. Show that cosec |(x - 360°) = - sec ^ x.
25. Show that sec ^ ( - 900° - x) = - sec ^ x.
26. Show that tan |(360° + a - &) = + tan ^(a - 6).
27. Show that cos (180° — x) is equal to the sine of the complementary
angle.
Complement = 90° - (180° - x) = - (90° - x) ; sin [- (90° - x)] =
- sin (90° -x)= - cos x. But cos (180° - x) = - cos x. . •. cos (180 - x) =
sin [90°- (180°- X)]. q. e. d.
28. Show that cosec (270° — x) equals the secant of the complementary
angle.
29. Show that tan (180° + x) equals the cotangent of the complementary
angle.
30. Show that sec (270° + x) equals the cosecant of the complementary
angle.
31. Show that cos (90° +x) equals the sine of the complementary angle.
32. Show that cot (360°— x) equals the tangent of the complementary angle.
33. Show that tan (270°+ a;) is equal to the negative of the tangent of the
supplementary angle.
Supplement = 180° - (270° + x) = - (90° + x) ; tan [- (90° + x)] =
- tan (90° + x) = + cot x. But tan (270° + x) = - cot x. . •. tan (270° + x) =
- tan [180° - (270° + x)]. q. e. d.
34. Show that cosec (180° + x) is equal to the cosecant of the supple-
mentary angle.
60 PLANE AND ANALYTICAL TRIGONOMETRY.
35. Show that sin (360° — x) is equal to the sine of the supplementary
angle.
36. Show that sec (90° + x) is equal to the negative of the secant of the
supplementary angle.
37. Show that cos (270°— cc) is equal to the negative of the cosine of the
supplementary angle.
38. Show that cot (270° + x) is equal to the negative of the cotangent of
the supplementary angle.
59. The Trigonometric Tables. — The relations shown in
Arts. 53 and 51 enable us to find the functions of any angle,
although the tables contain only the sines, cosines, tangents,
and cotangents of angles less than 45°. For, since
sin (90° — x) = cos x, cos (90° — x^ = sin x,
tan (90° -x) = cot x, cot (90° - x) = tan x,
the tables are immediately extended to 90° by writing the
proper degrees and minutes at the bottom and on the right
of the page respectively.
Then, since the value of any function of an angle greater
than 90° can be found in terms of a function of an angle less
than 90°, we can find the numerical value of the function from
the tables.
1. Find from the tables the logarithmic functions of 580° 42'.4.
580° 42' A = 300° + 220° 42'.4.
.-. sin 580° 42'.4 = sin 220° 42'.4 = sin (180° + 40° 42'.4) = - sin 40° 42'.4 ;
.-. log sin 580° 42'.4 = 9.81437 n.
cos 580° 42'.4 = - cos 40° 42'.4
tan580°42'.4 = + tan 40° 42'.4
cot 580° 42'.4 = + cot 40° 42'.4
.-. log cos 580° 42 '.4 = 9.87971 n.
.'. log tan 580° 42'.4 = 9.93467.
.-. los cot 580° 42'.4 = 0.06533.
2. Find from the tables the logarithmic functions of the following angles :
Angle.
log sin.
log COS.
log tan.
log cot.
499° 29'.7.
9.81258.
9.88102 w.
9.93158 w.
0.06842 n.
597° S'.B.
9.92427 w.
9. 73449 «.
0.18978.
9.81022.
689° 27'.6.
9. 70598 w.
9.93514.
9.77084 w.
0.22916 n.
3. sin b = tan 250° 15'.5 cot 278° 17'.3 ; find 6 = 203° 57'.0 or 336° 3'.0.
4. cos /3 = cos 149° 27'.6 sin 216° 44'.0 ; find 8 = 58° 59^7 or 301° 0'.3.
6. tana = sin 319° 52'.0 -- cot 254° 30'.2 ; find a = 113° 16'.5 or 293° 16'.5.
6. cotc = cos216°44'.0-v-tan329°27'.6; find c = 36°21'.6 or 216°21'.6.
FUNCTIONS OF ONE ANGLE. 61
60. Transform the First Member into the Second in the
following examples. Usually it is best to change the given
expression into one containing the sine and cosine, and then to
change this into the required form. Any operation is admis-
sible that does not change the value of the expression. Use
radicals only when unavoidable. If the expression is factored,
it is often advantageous to reduce each factor separately, not
multiplying until it becomes necessary.
tan X — sin x sec a?
sin^ X 1 -f cos X
sin a;
— sin X
tan X — ^\nx cos x _ sin a; (1 — cos a;) _ 1 — cos x
sin^ x ~~ sin^ x cos x sin^ x cos x sin^ x
1 — cos X 1 sec X
cos x(l — cos^ x^ cos x(l -\- cos a;) 1 + cos x
2. COS X tan a; + sin x cot x = sin oj + cos x.
3. (2 — vers x) vers x = sin^ x.
. COSX f„„ «
4. = tan X.
sin X cot2 X
5. (tan X + cot x) sin x cos x = 1.
6. (sec^x- l)(cosec2x — 1) = 1.
7. sec X cosec x (cos^ x — sin^ x) = cot x — tan x.
8. (sin X + cos x) (tan x + cot x) = sec x + cosec x.
9. cot X + ^^^^ = cosec x.
1 + cos X
10. sin X (sec x + cosec x) — cos x (sec x — cosec x) = sec x cosec x.
11. (cosec X- cot x)2 = ^ -cosx ^
1 + cos X
12. (1 + tan2 x) (1 - cot'2 x) = sec^ x — cosec^ x.
13. tan X - cot X _ — 2 ^ [First change to an expression containing
tan X + cot X cosec^ x
only sin x, the reciprocal of cosec x.]
14. sec2 X cosec2 x — 2 = tan^ x + cot^ x. [Substitute for sec x and cosec x
their values in terms of tan x and cot x respectively. ]
,K tan a + tan /3 .„„ ♦„„ o
15. ' — tan a tan p.
cot a + cot j3
16. cot X - sec X cosec x (1 - 2 sin^ x) = tan x. [The expression reduces to
sill X H- cos X.]
62 PLANE AND ANALYTICAL TRFGONOMETRY.
17. cosec X (sec a; — 1) — cot x (1 — cos x) = tan x — sin x. [Factor a^ sood
as possible, and reduce each factor separately. ]
18. vers x (sec x + 1 ) + covers x (cosec x + 1) = sin x tan x + cos x cot x.
j3^ vers x (1 + sec x) ^ covers x (1 + cosec x) ^ ^^^ ^ ^^^^^ ^
sin X cos X
20. sin2 X (tan2 x - 1) + cos^ x (cof-^ x - 1) = - 2 cos'^ x)^ seC^ x^
tan'^ X
21. sec X cosec x [vers x(vers x — 2) —covers x(covers x— 2)] =cot x— tan x.
22. cos* X — sin* x = cos x ( 1 — tan x) (sin x + cos x) .
^„ vers X (1 + cos x)— covers X (1 + sin x) tan*x — tan^x ^^,
23. ^^ r 5 ^ - — [Change to
sec-* X cosec-^ x sec ^ x to
an expression containing only sin x and cos x, and then substitute their values
in terms of tan x. ]
g, sec'^ X sin^^ x — cosec'^ x + cosec'^ x cos^ x _ . o ^
sec2 X sin2 x — cosec^ x cos^ x
„c * o -9 - (sec2x + l)(sec2x- 1)2
25. tan2 x - sin2 x cos^ x = -^^ ^-^^ —-
sec* X
2g cosxcotx-sinxtanx ^l_^3.^^^^3^
cosec X — sec X
2„ (sec X 4- cosec x)2 _ (1 + tan x)^
tan X + cot X tan x
2g_ 2 + 5il^i^il£2!i£ = sec2x + cosec2x.
sin2 X cos2 X
gg sec X + cosec x _ tan x + 1 _ 1 + cot x
sec X — cosec x tan x — 1 1 — cot x
„Q sin X — tan2 x covers x sin* x
cosec X cot2 X (1 + sin x) cos2 x
81. sec2 X cosec2 x [vers x (vers x — 2) — covers x (covers x — 2) ]
= cot2 X — tan2 X.
82. tan x + cot x = ^^^^ ^ "^ cosec2 j; ^ ^^^ .^ admissible to divide both
sec X cosec x
numerator and denominator by sin2 x cos2 x.]
88. tan2 a tan2 /3 - 1 = ?^5!^^^^ = ^H^^f=^.
cos2 o cos2 /3 C0S2 a cos2 /3
g^ 1 - tan2 g tan2 /3 _ cos2 a - sin2 j3 _ cos2 ^ - sin2 g
tan2 g tan2 /3 ~ sin2 ^ sin2 /3 ~ sin2 g sin2 ^
85. sin2 X tan2 x + cos2 x cot^ x = tan2 x + cot2 x — 1.
36. sin2 X tan x + cos2 x cot x + 2 sin x cos x = sec x cosec x.
37. sec* X + tan* x = 1 + 2 sec2 x tan2 x. [It is admissible to add and sub-
tract 2 sec2 X tan2 x. ]
38. (r cos 0)2 + (r sin sin 0)2 + (r sin > cos 0)2 = r2.
. •. r2 cos2 + r2 sin2 (sin2 e + cos2 0) = r2 (cos2 + sin2 0) =r2,
since sin2 x + cos^ x = 1.
FUNCTIONS OF ONE ANGLE. 63
89. (2 r sin a cos o)* + r^ (cos2 a - sin^ o)2 = r^.
40. (a sin 7)2 + (a cos 7 sin 5)^ -|- (a cos 7 cos 5)2 = a^.
41. (cos a cos & — sin a sin 6)2 + (sin a cos 6 + cos a sin 6)2 = 1.
42. (cos a cos 6 + sin a sin 6)2 + (sin a cos b — cos a sin 6)2 = l.
43. (x cos ^ - y sin ^)2 + (x sin ^ + ?/ cos dy = x2 + y2.
^ 1 4 tan2 X _ ,
(cos2 X - sin2 a;)2 (1 - tan2 x)2 ~ *
45. 1 (l-tan2x)2 ^^
4 sin2 X cos2 x 4 tan2 x
46. (3 sin a cos2 a - sin^ a)2 + (cos^ a - 3 sin2 a cos a)2 = 1.
47. x^-hy^ + s^ = ?-2 when
X = r cos a cos /3 + r cos i sin a sin /3,
y = r cos I cos a sin /3 — r sin a cos /3,
2 = r sin t sin /S.
CHAPTER V.
RELATIONS BETWEEN FUNCTIONS OF SEVERAL ANGLES.
61. Sine and Cosine of the Sum of Two Angles. — Let x and
y be the angles, each, as well as their sum, being less than 90°.
QQ \s> perpendicular to OP, BC and
BQ are perpendicular to OA^ and
EC is parallel to OA^ the radius of
the circle being unity. Then
x-\-y = AOQ,
the angle EQC=x^ 06'= cos i/, and
CQ = sin y.
sin (x-^y')=DQ = BO-\- EQ
= OC^mBOC+ OQ cos EQC
= cos y sin x + sin y cos x,
or sin (a? + y) = sin a? cos ?/ + cos a? sin y. (1)
cos (a: + 2/)= ^J^= OB-EO= OOcosBOO- CQsmEQO
= cos y COS X — sin ^ sin x,
cos (ac + y) = cos a? cos y - sin a? sin y. (2)
or
1. sin 90° = sin (60° + 80°) = sin 60° cos 30° + cos 60" sin 30°
^\/8 V3 , 1. 1^1
2*2 2*2
2. cos 90° = cos (60° + 30°) = cos 60° cos 30° - sin 60° sin 30°
= 1. V3 V3 , 1 ^
2*2 2*2
3. If sin a = f , and sin /3 = y\, find sin (a + /3) and cos (a + /3) when
a < 90°, and /3 < 90°
A71S. sin (a + iS) = |f , cos (a + /S) = |f
64
J'UNCTIONS OF SEVERAL ANGLES.
66
4. If tana = f, and tan/3 = ,'j, find sin (o + /3) and cos(o + /3) when
a < 90°, and /3 < 90°.
Ans. sin (a + P)= *, cos (o + /3) = g.
Note. — At a point A the angle of elevation DAB to the top of a vertical
wall is o, and the angle of depression
CAD to its base is /3. Find the height
CB of the wrall, the horizontal distance
AD being a feet.
CB= CD -\- DB = a tan /3 + a tan a
= a(tana + tan/3). (3)
^^/sina^sin^X
\cosa COSjS/
_ sin g cos /3 + cos a sin /3
cos a cos /3
^ sin (g + /3)
cos g cos /3
Fig. 41.
(4)
Eq. (3) w^ould be solved by the use of the natural functions, while (4) is
adapted to logarithmic computation.
62. Sine and Cosine of the Difference of Two Angles. — Let
X and y be the angles, each being less than 90° and x being
greater than t/. QC is perpendicu-
lar to OP, BO and DQ are perpen-
dicular to OA, and EQ is parallel
to OA, the radius of the circle
being unity. Then x — y = AOQ,
EOQ=x, 0(7=cos?/,..and C§=siny.
sin (x-y)=DQ= BO- EO
= 0Os'mB0O- OQcosEOQ
= cos ?/ sin X — sin «/ cos x,
or siii(x-2/) = siu£ccos2/-cosa5sin2/. (1)
cos (x - y)= OB = OB + EQ = 00 cos BOO + OQ sin EOQ
= cos 1/ cos X + sin z/ sin x,
or cos (pc-y)= cos a? cos y + sin x sin y, (2)
In this proof we have assumed that x is the greater angle,
but (1) and (2) are true when y is greater than x. To prove
this, let y8 be greater than «. Then
sin (a — /3) = sin [ — (/8 — «)] = — sin (yS — a),
CROCK. TRIG. — 5
66 PLANE AND ANALYTICAL TRIGONOMETRY,
and, developing sin (^ — a) by (1),
sin (rt — yS) = — (siny8 cos a — cos ^ sin a)
ss sin a cos /S — cos a sin ^.
Also cos (a — /3) = cos [ — (/S — «)] = cos (/S — a)
= cos )Q cos a H- sin /3 sin a.
Q.E.D.
Q.E.D.
1. sin 30° = sin (90° - 60°) = sin 90° cos 60° - cos 90° sin 60°
1.1-0.^
2. cos 30° = cos (90° - 60°) = cos 90° cos 60° + sin 90° sin 60°
= 0.1+1.^ = ^.
2 2 2
3. If sin a = j^j, and sin /3 = :f\, find sin (a - /3) and cos (a - /3) when
o<90°,and/3<90°.
Ans. sin (a - /S) = ^^i \ cos (a - /3) = |f f .
4. If tan a = f , and tan/3 = f, find sin (a — /3) and cos (a — /3) when
a < 90°, and /3 < 90.°
Ans. sin (a - /3) = ^V J cos (a - /3) = f f .
Note. — At a pointy^ on a horizontal plane, the angle CAD to the top of
a crag is 7, and a feet farther away in the same
vertical plane (at 5), the angle CBD is 7'.
^,,„, .,, Find^C = a;.
Ci> = ^C tan 7 = a; tan 7.
CD = BCid^ny' =(a + a;) tan 7'.
. •. X tan 7 = (a + a;) tan 7'.
Fig. 43.
^ _ q tan 7^
tan 7 — tan 7'
osiny C0S7
g sin 7^ cos 7
sin 7 cos 7' — cos 7 sin 7' sin (7 — 7')
(3)
(4)
Eq. (3) would be solved by the use of the natural functions, while (4)
is adapted to logarithmic computation.
63. General Proof of the Addition Formulas. — These formulas were
shown in Art. 61 to be true when a;, y^ and x + y were each less than 90°. That
they are true for all values of these angles may be shown by proving the special
cases separately. Let us consider first the case when
»< 90°, y < 90°, X + 2/ > 90° and < 180^^
FUNCTIONS OF SEVERAL ANGLES. 67
Let x = 90°-a, y = 90°-p; .-. x + y = 180° -(a + /3).
.-. a = 90° - x, )8 = 90'' - y, a + i8 = 180° - (x + j/).
.-. a < 90°, /3 < 90°, a + /3 < 90°, since a; + y > 90°.
Then sin (a + /3) may be developed by (1), Art. 61, since the conditions of that
article are satisfied. But
sin (x + y) = sin [180° — (a + /3)] = sin (a -f |3) = sin a cos /3 + cos a sin /3
= sin (90° - x) cos (90° - y) + cos (90° - x) sin (90° - y)
= cos X sin 2/ + sin 05 cos 2/. q.e.d.
Also cos (x + y)~ cos [180° - (a + /3)] = - cos (a + ^) = - cos a cos j8 + sin a sin |3
= - cos (90° - x) cos (90° -y) + sin (90° - x) sin (90° - y)
= — sin X sin y 4- cos X cos y. q.e.d.
Hence the formulas are true for x < 90°, y < 90°, x + y < 180°.
To illustrate the proof for any special case, let us take x in the second and
y in the fourth quadrant. Place x = 90° + a, and y = 270° + /3, so that a and /3
are each less than 90°. Then
sin (x + y)= sin [360° + (a + /3)] = sin (a + jS) = sin a cos /3 + cos a sin /3
=; sin (x - 90°) cos (y - 270°) + cos (x - 90°) sin (y - 270°)
= (— cos x)(— sin ?/) + sin X cos y
= cos X sin y + sin x cos y. q.e.d.
Let the student prove the addition formulas in the following cases :
1. X in the first, and y in the third quadrant.
2. X in the second, and y in the second quadrant.
3. X in the second, and y in the third quadrant.
4. X in the third, and y in the third quadrant.
6. X in the third, and y in the fourth quadrant.
6. X in the fourth, and y in the fourth quadrant.
64. General Proof of the Subtraction Formulas. — These formulas were
shown in Art. 62 to be true when x and y were each less than 90°, both for
x>y and for x < y. That they are true 'for all values of the angles may be
shown by proving the special cases separately. For illustration, let x be in the
second, and y in the third quadrant. Place x = 90° + a, and y = 180° + /3, so
that a < 90°, and ^ < 90°. Then
sin(x-y)=sin[90°+a-(180°+/3)]=sin[-90° + (a-i3)] = -sin[90°-(a-i3)]
= — cos (a — /3) = — COS a cos iS — sin a sin
= - cos (X - 90°) cos (y - 180°) - sin (x - 90°) sin (y - 180°)
= — sin X ( — cos 2/) — ( — cos x) ( — sin y)
= sin X cos y — cos X sin ?/. q.e.d.
Also cos {x-y)= cos [- 90° + (a - 0)] = cos [90° - (a - ^8)] = sin (a - 0)
= sin a cos $ — cos a sin
= sin (x - 90°) cos (y - 180°) - cos (x - 90°) sin (y - 180°)
= (— cos x) (— cos y) — sin X (— sin y)
= cos X cos y + sin X sin y. q.e.d.
68 PLANE AND ANALYTICAL TRIGONOMETRY.
Let the student prove the subtraction formulas in the following cases ;
1. X in the fourth, and y in the first quadrant.
2. X in the fourth, and ?/ in the second quadrant.
3. X in the fourth, and y in the third quadrant.
4. X in the third, and y in the third quadrant.
5. a; in the third", and y in the fourth quadrant.
6. X in the second, and y in the fourth quadrant.
65. Tangent of the Sum and of the Difference of Two Angles.
sin (x + ?/) sin x cos y + cos x sin y
tan (a: + ?/) = t^~^ = ^ —• -^•
^ '^ cos {x + y) COS X COS ^ — sm a: sin i/
Divide both numerator and denominator by cos x cos y.
sin X cos ^/ cos a: sin y sin a: sin y
, , cos a: cos v cos a: cos y cos a; cos ?/
.-. tan (2:+?/)= ^^ ^ ^ = : -^;
cos X COS y sin a; sin «/ ^ sin a; sm z/
cos X cos ?/ cos X cos y cos a; cos y
In the same way, we may show that
. , . tan 05 - tan y ^o\
1. tan75°=tan(45°+30°) = ;^"^^°+\^"-l": =-^^^g±J^2+V3.
^ ' 1 — tan 45° tan SO"' i VS — 1
V3
2. tan 15-tan (45°-30°) = J^li5;;i5a^„= ^ =2^ = 2- V3.
^ ^ 1 + tan 4d° tan 30° 1 ^_|_ ^
V3
3. If sina = |f and sinj3 = |, find tan (a + 5) and tan (a-;8), when a<90°
and B < 90°. Ans. tan (a + ;8) = - f | ; tan (a - )3) = f |.
66. Geometrical Proof. — In Fig. 44, let 0^1 = 1, AOB = x, BOC = y.
Draw .5C perpendicular to OJ?, and CD parallel to OA; :. DBC = x^
DCE = x + y. Then
tan (x + y)= AE = AB + BD + BE.
FUNCTIONS OF SEVERAL ANGLES.
69
But BD = BC cos X = OB tan y cos x — sec x tan y cos x — tan y,
and DE = CD tan (x + y) = BC sin x tan (x + y) = OZ? tan y sin x tan (x -f y)
= sec X tan y sin x tan (x + y) = tan x tan y tan (x + y).
. •. tan (x -f y) = tan x + tan y + tan x tan y tan (x + y).
tan X + tan y
/. tan (x + y) =
1 — tan X tan y
Fig. 45.
In Fig. 45, let 0.4 = 1, AOB = x, COB = y. Draw 50 perpendicular to
OB, and Z>(7 parallel to OA ; .-. DBC = x, DCE=x-y. Then
tan (.X - ?/) = ^E- =r ^i^ - i>5 - ^D.
Rut DB = BC cos X — OB tan y cos x = sec x tan y cos x = tan y,
and ^i) — DC tan (x — y) = BC sin x tan (x — y) = Oi? tan y sin x tan (x — y)
= sec X tan y sin x tan (x — y) = tan x tan y tan (x — y).
.-. tan (x — y) = tan x — tan y — tan x tan y tan (x — y).
tan X — tan y
tan (x — y)
1 + tan X tan y
EXAMPLES.
Find by inspection one value of x in Exs, (1-6) :
1. sin (w — l)a cos a + cos (n — l)a sin a = sin x. ^ws. x = wa.
2. cos (10^= + a) cos (10° - a) + sin (10° + a) sin (10°- a) = cos X. J.>iS. x = 2a.
3. sin(a-)8 + 10°)cos(i8-a + 10°)-cos(a-i3 + 10°)sin ()3-a + 10°)= sin x.
Ans. X = 2(a — i8).
4. cos 45° cos (90° - a) - sin 45° sin (90° - a)= cos x. .4ns. x = 135° - o.
5. sin (90° + h, a) cos (90° - i a) + cos (90°+ i a) sin (90° - i a) = sin x.
Ans. x= 180°.
6. cos(45°-a)cos(45°+a)-sin(45°-a)sin(45°+a)=cosx. ^ns.x = 90°.
7. Given the functions of 30° and 45°, find those of 75°.
Ans. sin 75° = ^^+ ^ ; cos 75° = ^~^ ; tan 75° = ->^J- = 2 + V3.
2V2
2V2
Vs-i
70 PLANE AND ANALYTICAL TRIGONOMETRY.
8. Given the functions of 30° and 45°, find those of 15°.
Ans. sin 15° = ^~ ^ ; cos 15° = ^^ ; tan 15° = 2 - VS.
2V^ 2\/2
9. If tan a = I and sin ;8 = }|, find the functions oi a + when a is in
the third, and j8 in the second quadrant.
Ans, sin (a + j8) = - f | ; cos (a + j8) = f f ; tan (a + 3) = - f f .
10. If cos a = — If and sin )8 = - y\-, find the functions of o — /3 when a
is in the third, and j8 in the fourth quadrant.
Ans. sin (a -)3) = -!§!; cos (a - )8) = - ||| ; tan (a - )8) = + f f f .
11. If cos o = f and sin ;8 = — |, find the functions of o + ;8 and ot a —
when a is in the fourth, and in the third quadrant.
Ans. sin (a + )8) = + /^ ; cos (a + 3) = - If ; tan (a + ;8) = - /^ ;
sin (a - 3) = + 1 ; cos (a - i8) = ; tan (a - fi) = oo.
Transform the first member into the secpnd (or last) in Exs. (12-32):
12. sin (a + ;8) sin (a ~ $)= sin2 a — sin^ $ = cos^ & — cos^ a.
13. cos (a + ;8) cos (a — )8) = cos^ a — sin^ $ = cos^ — sin^ o.
14. sin (60° + a) - sin a = sin (60° - o).
15. (r' cos v' — r cos vy + (r' sin v' — r sin u)2 = r^ + r'^ — 2 rr' cos (v' — v).
16. cos^ a + cos2 /3 — 2 cos o COS i8 cos w = sin2 w, when w = o + j8. [Place
a = w — iS.]
tPf 4- , tan sec a *„„ / , .\
17. tan o H ■^-—. = tan (a + 0).
cos a — tan
a
— cos2 2a:
(1 + cos 22:)
2'
1 + cos 2 a5
Also
_ /l— cos 2 2: 1 — cos 22:_ /(I — cos 22:)2_
^ 1 + cos 2x 1 — cos 2x ^ 1 — cos^ 2 x
.*• tan 0?
1 - cos 2 05
•
sin 2 a?
(8)
(4)
(5)
(fi)
FUNCTIONS OF SEVERAL ANGLES.
73
Note. — The double sign is not used in (5) and (6), for
sin 2 X 2 sin x cos x
and
1 + cos 2 a; 2 cos-' x
1 — cos 2 X 2 sin^ x
sin 2 X 2 sin x cos x
= tan X,
r= tan X.
1. To express the functions of 20° in terms of those of 40°, we have
2sin2 20° = 1 -cos 40°;
2cos2 20° = 1 + cos 40°;
1 -cos 40°
tan2 20^
tan 20° =
1 + cos 40°
sin 40° _ 1 - cos 40^
1 + cos40°~
sin 40=
2. tan 2 ^ = — 2, 2 ^ being in the second quadrant. Find the functions of B.
1 o
.-. cos 2 ^ = —, and sin 2 ^ = -f -=-.
cos ^ = ± Y" ( 1 - -^) , from (2) ;
.*. sin
tan 5 =
2
V6
1 _ J_ Vs - 1
V5
from (5).
Since 2 ^ is in the second quadrant, B may be either in the first or in the
third quadrant ; hence sin 6 and cos 6 have the double sign, and tan B is positive.
3. Given the functions of 30°, find those of 15°.
Ans. sin 15°^ ^- ^
cos 15=
^^ll±l- tan 15° = 2 -V3.
2V2 ■ 2V2
4. Given the functions of 45°, find those of 22J°.
Ans. sin22i° = ^V2-V2; cos22|° =i V2 + \/2 ; tan22|° = V2 - 1.
70. Geometrical Proof. — Let the radius CA of the circle be unity.
e^ ^P_ ^~6B'Ba _ABa ^Jba
OP VOBToA ^ OA
= J g-^ - CB ^ j \ -cos2x
OB
OB
OP y/OB.OA ^0.\
=M-r-^
lOC+CB J\ +cos2x
= \— ^ =\ 1
74 PLANE AND ANALYTICAL TRIGONOMETRY.
BP BP sin 2 a;
tana; =
tana:
OB 0C+ CB 1 + cos2a:
BA ^ CA- CB ^ 1 -cos2x
BP BP sin2x
71. Multiple Angles. — Suppose that we wish to express
sin 3 X in terms of powers of sin x.
sin 3 a: = sin (2 a; + ^) = sin 2 x cos x + cos 2 a: sin a:
= 2 sin X cos^ a: + (1 — 2 sin^ x} sin x
C^'^'Z ^ / j^ ^OyUji^J ^^ = 2 sin a; — 2 sin3 a: + sin a; — 2 sin^ a:
^ = 3«ina; — 4sin3,a^. Q.E.I.
1. Show that cos 3 x = 4 cos^ x — 3 cos x.
2. Show that tan3x=^^^^^-^^"'^.
1 -3tan2a;
3. Show that sin 4 x = 4 sin x cos a; — 8 sin^ x cos aj. [Use 4x = i2x + 2x.]
4. Show that cos 4 x = 1 — 8 sin^ x + 8 sin* x.
E ou *u * * A 4tanx(l - tan2x)
6. Show that tan 4 x = -.^ \ — ^ •
1—6 tan2 X + tan* x
6. Show that sin 6 x = 5 sin x — 20 sin^ x + 16 sin^ x. [Use 5 x = 3 x + 2 x.]
7. Show that cos 5 x = 5 cos x — 20 cos^ x + 16 cos^ x.
8. Find the functions of 18°, of 36°, and of 72°.
Place X = 18° ; then, since cos 54° = sin 36°, we have
cos 3 X = sin 2 x.
.•. 4 cos-5 X — 3 cos X = 2 sin X cos x.
.'. cos X (4 cos2 X — 3 — 2 sin x) = 0.
/. 1—4 sin2 X — 2 sin X = 0.
.-. sinx = ^(— 1 ± Vo).
.-. sin 18° = cos72° = K>/5 - 1) ; cos 18° = sin 72° = jVlO + 2V5.
Hence sin 36° = 2 sin 18° cos 18° = ^VlO-2V5 ;
cos36° = 1 -2sin218° = i(^^+ 1).
72. To change the Product of Functions of Angles into the
Sum of Functions. — From Arts. 61 and 62,
sin (a; + ^) = sin x cos y + cos x sin i/; /^
sin (x — y^= sin x cos y — cos x sin y. ^^^
.'. sin (a; H- «/) + sin (a; — ?/) = 2 sin a: cos «/, (1)
and sin (x -\- y}— sin (a; — «/) = 2 cos x sin y. (2)
FUNCTIONS OF SEVERAL ANGLES. 75
Also cos (a: + y) = cos x cos y — sin a: sin y\ ^ —
cos (x — y)= cos X cos y + sin x sin y. -^
.*. cos (a; 4- y ) + cos (a; — ?/) = 2 cos a; cos y, (3)
and cos (x + y)— cos (a: — ?/)= — 2 sin a; sin y . (4)
Reversing (1), (2), (3), and (4), we have
sina3COS2/ = ^sin(a!; + 2/)+^sitt(a5-2/), \ (5)
cos ic sin 2/ = I sin (05 + 2/)-^ sin (05-1/). [ (6)
cosxcosy = |cos(a5 + 2/)+ |cos(a5 -y). (7)
sina5sin2/ = -|cos(a5 + i/) +|cos(a5- J/). (8)
In applying these formulas, let x represent the larger angle.
1. sin 4 ^ cos 2 = ^ sin (4 ^ 4- 2 ^) + I sin (4 ^ - 2 ^), from (5),
= 1 sin 6 ^ + 1 sin 2 d.
2. cos 6 d sin 2 ^ = ^ sin (6 ^ + 2 ^) - ^ sin (6 ^ - 2 6), from (6),
= ^ sin 8 ^ - ^ sin 4 ^.
3. cos8^ cos2^ = ^coslO^ + ^cos6^, from (7).
4. sin 6 ^ sin 4 ^ = - I cos 10 ^ + ^ cos 2 6, from (8).
5. cos 2 sin 4 = i sin 6 ^ - i sin ( - 2 ^), from (6),
= i sin 6 ^ + ^ sin 2 , as in Ex. 1.
6. sin2^cos6^= isin8^ + |sin (- 4^), from (5),
= i sin 8 — I sin 4 ^, as in Ex. 2.
7. cos2^ cos 8^= 1208 10^+ I cos (- 6^), from (7),
= ^ cos 10 ^ + I cos 6 ^, as in Ex. 3.
8. sin 4 e sin 6 ^ = - ^ cos 10 ^ + ^ cos ( - 2 ^), from (8),
= — ^ cos 10 ^ + I cos 2 d, as in Ex. 4.
9. sin2 ^ cos ^ = sin d [sin 6 cos d] = sin d [^ sin {6 -{- 6)+ \ sin (^ — ^)]
= sin ^ [^ sin 2 ^ + I sin 0°] = | sin sin 2 d
= K - i cos 3 ^ + ^ cos ^] = - I cos 3 ^ + i cos ^.
10. Reduce sin^ a cos a to ^ sin 2 a — ^ sin 4 a.
sin3 a cos o = sin2 a • sin a cos a ;
using (8) and (5), or the relations in Arts. 69 and 67, we have
sin3 a cos o = I (1 — cos 2 a) . ^ sin 2 o = ^ sin 2 a — ;^ sin 2 a COS 2 a
= ^ sin 2 a — ^ sin 4 o.
11. Reduce sin^ d cos2 ^ to ^ (1 — cos 4 d).
12. Reduce sin2 d cos^ ^ to | (cos ^ - ^ cos 3 ^ — ^ cos 5 ^).
13. Reduce sin^ cos^ 6 to ^^5 (3 sin 2 ^ - sin 6 ^).
14. Reduce cos^ d to ^-^ (10 cos ^ + 5 cos 3 ^ + cos 5 d).
15. Reduce cos^ 6 to \ (cos 3 ^ + 3 cos 0).
16. Reduce sin^ cos^ to -^^ (3 sin 2 ^ - sin 4 ^ - sin 6 ^ + ^ sin 8 6).
76
PLANE AND ANALYTICAL TRIGONOMETRY.
73. To change the Algebraic Sum of Functions of Angles into
the Product of Functions. — Let x + y = u and (2; — ^) = v,
.'. X = l(^u -{- v~) and y = ^(u — v).
Substituting in (1), (2), (3), and (4), Art. 72, we have
sinw + sinv = 2siii|(w + v)cos^(m- v). (1)
sin 1* - sin V = 2 cos| {u + v) sin | (w - v), (2)
COSW + cosv = 2cos^(w + i^)cos|(w -V). (3)
cosM - cosv = - 2sin|(w + v)sin|(M — v).
In applying the formulas, let u represent the larger angle.
1. Reduce sin 3 ^ + sin ^ to 2 sin 2 ^ cos 6.
Let u = ZO and v = ^ in (1).
2. Reduce cos ^ — cos 3 ^ to 4 sin^ d cos d.
cos d — cos 3 ^ = — (cos •') — cos d). Let ?( = 3 ^ and v = ^ in (4).
. •. - (cos 3 ^ - cos d) = - ( - 2 sin 2 ^ sin ^) = + 2 sin 2 ^ sin
= 4 sin ^ cos ^ sin <^ = 4 sin^ d cos d.
3. Reduce sin 3 ^ + cos ^ to a product.
sin 3 ^ + cos e = sin 3 ^ + sin (90° -6) =2 sin (45° + d) cos {2 6 - 45°)
= 2 sin (45° + 6) cos (45° -26).
74. Geometrical Proof. — In the figure, OD bisects the angle QOP, and
is therefore perpendicular to QF. Using the notation there shown, we have
QOP=u-v\ .: qOD^DOP=\{u-v);
AOD = AOq+ qOD = v+\ {u-v) = \{u^v);
FPQ=GDQ = AOD=] (u+v). Then, if the
radius = 1,
sin w+sin v = BP+ CQ = 2 ED=2 OD sin AOD
= 2 OP cos DOP sin AOD
= 2sm \{u-\-v)cosl(u — v). (1)
sin u-sinv=BP- CQ = 2 GD=2 DQ cos GDQ
= 2 OQ sin QOD cos GDQ
= 2 cos I (u-^v) sin | (u—v). (2)
cos M + cos r=OZ?+OC= 2 0E=2 OD cos A0D=2 OP cos DOP cos AOD
= 2 cos I (m + u) cos I (w — r). (3)
cosM-cosv= OS- 0C= -2 GQ=-2DQs\n GDQ=-2 OQ sin QOD sin GDQ
= — 2 sin I (u + v) sin | (u — v) . (4)
Fig. 49.
FUNCTIONS OF SEVERAL ANGLES.
7T
EXAMPLES.
Show that the first member of the equation may be reduced to the second
(or last) in Exs. (1-7):
1. sin (45° + a-) + sin (45° — x) = 2 sin 45° cos x = V2 cos x.
3. sin (90° + X) - sin (180° + re) = 2 cos_(135° + a;) sin ( - 46°)
= -\/2cos(135° + a;).
3. cos (180° + x) + cos (180° - a-) = 2 cos 180° cos x = - 2 cos x.
4. cos (270° + X) - cos (270° - x) = - 2 sin 270° sin x = + 2 sin x.
6. sin 3 X + 2 sin 5 X + sin 7 x = 4 sin 5 x cos^ x.
6. cos 3 X + 2 cos 5 X + cos 7 X = 4 cos 5 x cos^ x.
7. cos (6 — c) — cos a =3 + 2 sin i (a + & — c) sin |(a — 6 + c) .
8. Show tliat sin (\"— X') — sin (X"— X) + sin (X'— X) may be reduced to
4 sin \ (X' — X) sin \ (\" — \') sin ^ (X"— X). [The formula sin x=2 sin | x cos | x
is iised in the process.]
If a 4-j8 + 7 = 180°, reduce the first member to the second in Exs. (9-14):
9. sin o + sin /3 + sin 7 = 4 sin J (a + j8) cos J a cos ^ 0.
7 = 180° -(a + B); .: sin y = sin (a + $). Then
sino+sin /3 + sin (a4-)8)=2sin^ (a + i3) cos J (a — )3)4-2 sin J (a + )8) cos^(a + ;8)
= 2sin^(a + i8)[cos^(a - /3)+cos J(a+ 3)]
= 2 sin ^ (a + )8) (2 cos J a cos J i3).
10. cos a + cos )8 + cos 7 = 4 cos ^ (a + j8) sin J a sin ^ )8 + 1. [Note that
cos 7 = — COC (a + )8) = — 2 cos2 ^ (a + i8) + 1.]
11. cos 2 a + cos 2 )8 + cos 2 7 = — 4 cos a cos B cos 7 — 1.
12. sin 2 a + sin 2 j8 + sin 27 = 4 sin a sin $ sin 7.
13. 2 sin2 a + 2 sin^ 3 + 2 sin^ 7 = 4 + 4 cos a cos $ cos 7.
14. sin 3 a + sin 3 )3 + sin 3 7 = — 4 cos | a cos | ^3 cos 1 7.
15. If a + 3 + 7 = 360°, sin a + sin ^ + sin 7 = 4 sin J a sin J /3 sin J 7.
16. If a + ;8 + 7 = 360°, sina + sin)3 + 2 sin -^-7=4 sin |(a + )8) cos2 1 (a-/3).
75. Circular, or Inverse Trigonometric, Functions. — If y
is the sine of the angle or arc x, then x is the arc whose
sine is y. This is written x = sin~^ y, read
" X is the arc whose sine is y." So also if
tan X = m, then '•'■ x is the arc whose tangent is
w," written x = tan~^ m.
In consequence of this notation, if we have
-: and wish to bring sin x into the numer-
sm a; o .
ator, we must write it in a parenthesis with the exponent — 1 ;
Fig. 50.
78 PLANE AND ANALYTICAL TRIGONOMETRY.
- = (smx)~^. All other exponents may be written above
sin a; -.
the name of the functions ; . ., = sin"^ x = (sin x)~^.
sin'* X
1. y = tan~i m + tan-i n. Find tan y.
Let tan-i m = a and tan-i n = b ; .-. tan a = m, tan b = n.
tan a + tan b m -{■ n
.-. y = a + b; .-. tan 1/
1 — tan a tan 6 1 — win
2. tan-i — = tan-i — \- tan"! x. Find x.
w w 4- w
. •. tan~i X = tan~i tan~i Let a = tan-i — , 6 = tan-i — - —
m wi + n m m -\- n
1 1
/ , * / 1.N tan a — tan 6 m m + n
.*. tan-i x = a — 6; .-. x = tan (a — 6) =
1 + tan a tan 6 , ,
•. X
m{m + n)
m'^ + ?»n 4- 1
S.* y = sin-i ^ + tan-i |. Find sin y. Aris. sin y = J^ (4 + 3V3).
4. tan-i — = tan-i — ^ tan"! x. Find x= ^
m m — n m'^ — mn + 1
6. tan-i a = tan-i \ + tan-i ^i- Find a = \.
6.* y = sin~i m + sin~i n. Find sin ?/ = m Vl — n'-^ + n Vl — m^.
7.* y = cos-i 771 + cos~i n. Find sin y = nVl — w'^ -f my/ 1 — n^.
8.* y = cos~i wi — sin"i n. Find cos y = wVl — n'^ 4- ?iVl — rri^.
' 9. tan-i a = tan~i J — tan-i ^. Find a = ^
10.* m = tan-i J + tan-i i Find m = 45^.
11.* m = tan-i J + tan-i | + tan-i y^- ^ii^d m = 45°.
12.* w = 2 tan-i ^ - tan-i f Find m = 45°.
Let tan-i ^ = a, tan-i 4^ = 6; .-. m = 2a + 6; .-. tan wi = , etc.
13.* m = 2 tan-i | + tan-i j. Find m = 45°.
14.* Show that tan-i |(1 - w) = sec- 1 1 Vs - 2 m + wi'-^.
Let tan-i l(\ — m) =x ; . •. tan x = ^ (1 — w) ; sec x= Vl +tan2 x ;
. •. sec X = IV 5 — 2m-\- m'^. . •. x = sec-i I Vo — 2m -\- m^.
15. Show that tan-i m = i tan-J ^^ .
' 1 - Wi2
Let X = tan-i ,jj^ or m = tan x. If the equation is true, we must have
X = i tan-i -1^, or 2 X = tan-i ^^^"^ , or tan 2 x = Al^^l^,
1 -m'^ 1 - tan2 x 1 - tan2 x
a formula proved in Art. 67.
16. Show that cos-i m = ^ cos-i (2 m^ — 1).
17.* Show that sin-i ^V^ ^ tan- 1 ?^^.
a + & a — 6
* When the angles are less than 90°.
FUNCTIONS OF SEVERAL ANGLES. 79
18. Showthatsinf--2tan-iA/l-^^ = a;.
19.* Show that | vers ^ a^ — sin-^ i a is constant for all possible values
of a.
Let 6 =\ vers-i \a^ — sin-i \ a, and let m = vers-^ ^0:^^71 = sin-^ ^ a.
cos 2 ^=cos m cos 2 n+sin wi sin 2 n.
But
i; .'. 2^=m-2ni
sin n = ^ a ; . •• cos n = J \/4 — a^
. •. sin 2 71 = I a V4 — a^ ; cos 2 w = i (2 — a^).
Also
cos m = 1 — vers wi = 1 — | a^ ; . •. sin m = - V4 — a*.
•. cos 2^ =
2 - a2 2 - a2 . a
2 2 2
2^ = 0°, or d = (f
20.* Show that taii-i ^4-sin-i ^ jg constant for all possible values of a.
a
21.* Show that vers-i a — 2 cot-^A/ ~ ^ is constant for all possible
values of a.
X
a
22.* Show that vers-i — — 2 sin-i-v/— is constant for all possible values
12 >'24 ^
of z.
76. To prove that tan a? > £c> sin « when a; < — , a? being ex-
2
pressed in Circular Measure. — Let AOB = BOC = x^ the radius
being unity. Evidently AT> SB, or tan x > sin x.
Also, since the shortest distance
from a point to a line is perpen-
dicular to the line, SB < AB, or
sin x&vcABO, since ABO is
a convex polygon lying in the tri- At^
angle formed by a chord A with fig. 6i.
the tangent lines TA and TO. Then
2 AT > arc ABO, or AT>a>TGAB, or tanic>a;.
Hence tan x>x, and x > sin x. q.e.d.
77. To prove that sin a?, tan a?, and a? approach Equality as
the Angle a? approaches Zero. — As the angle AOT decreases,
When the angles are less than 90*^
80 PLANE AXD ANALYTICAL TRIGONOMETRY.
the points B and T approach A^ and hence approach each
other. But
SB sin X
AT i'Awx
cos a;.
When the angle x approaches zero as its limit, cos x approaches
unity as its limit. Hence -j-^, or , approaches unity as its
limit, or sin x and tan x approach equality.
The arc x is intermediate in value between sin x and tan x ;
hence the three quantities approach equality as the angle be-
comes smaller. That is, the three ratios
sin X sin x tan x
tan XX X
approach unity as the angle approaches zero.
Hence we may say that when the angle is small, its sine and
its tangent are equal to the arc itself, and its cosine is equal
to unity. The smaller the angle, the more nearly correct will
be the assumption.
78. Development of sin a?, of cos «, and of tan a?. — Let us
assume that
sin X = a + hx -\- cx^ -\- ds? + es^ +f^ + ••• (1)
is true for all values of x. Then it is true when x has the
values 4- y ^^^ ~ V j hence
sin^ = a-\-hy^-cf'-\- dy^ + ey^ -{-fy^ + ••• (2)
and sm(— y}=a — by -{- cy^ — dy^ + ey^ —fy^ -\ (3)
But sin y = — sin ( — ^), or sin ?/ + sin (—?/)= 0. Adding
(2) and (3),
2 a + 2 £?/ + 2 e/ + ... = 0. (4)
But (4) is true for all values of y, since (1) is true for all
values of x. In order that all values of y may reduce the left
member of (4) to zero, we must have a = 0, c = 0, e = 0, ..-•
Hence (1) becomes
sin x = hx -\- dx^ -\-fx^ + ••• (5)
or ?HL£ = 6 + ^a;2+A*+- (6)
X
FUNCTIONS Or SEVERAL ANGLES. 81
But as X approaches zero, ^^^ approaches unity, and h + dx^
X
-f/r*+ ... approaches h. Hence
1 = J, (7)
and (5) becomes
sin £c = a? + doc^ + fx^ + ••• (8)
Again, let
cos a: = ^ 4- J5a: + Cx^ ^ Bt? -\- Ex"^ + Fx^ + -.• (9)
Since cos x = cos ( — x')^ we have
A + Bx + Cx'^ + Dj^ + E3^ + F3^-[- ...
= A-Bx-\-Cx'^- B^ H- J^;^;* - i<:r5 ^ ... (IQ)
or 2j52: + 2i>ar^ + 2jP2:5+ ... = 0. (11)
In order that this may be true for all values of x^ we must have
^ = 0, i> = 0, ^=0 ..., and (9) becomes
cos a: = ^ + Cx^ + Ed" + - (12)
But when a: = 0, (12) reduces to
1 = A, (IS)
and hence (12) becomes
cos a^ = 1 + Cic- + Bqc^ + ... (14)
Substituting from (14) and (8) in the formula
cos 2 a: = cos^ a: — sin^ a:,
we have 1 + 4 Ca:2 _j_ iq ^^4 _^ ... ^ 1 ^(9 (7_ i)^2
+ (2 ^ + 6^2 _ 2 d):»^ 4- ••• (15)
Equating the coefficients of like powers of x^
4(7=2(7-1, or 2(7+1=0. (16)
lG^=2^+(72_2^, or WE-C^^ld^^. (17)
Substituting from (14) and (8) in the formula
sin 2 a: = 2 sin x cos a:,
we have 2 a: + 8 cZa;3 4. 32 /.^;5 + ... ^ 2 a: + 2 ((7+ d^T?
+ 2(iE + Cd+f^7^+... (18)
CROCK. TRIG. 6
82 PLANE AND ANALYTICAL TRIGONOMETRY.
Equating the coefficients of like powers of a;,
4.d = C-\-d\ or 3c^-(7=0. (19)
lQf=E + Cd+f, or 15f-U-Cd = 0. (20)
From (16), 0=-^. (21)
From (19), ^ = -1=-^. (22)
From (17), U=+^ = +^ (23)
From (20), / = +^=+^. (24)
These values, substituted in (8) and (14), give
^inx = a^-^ + ^^.-^ ij^-^ (25)
i_n
cosaj= 1
g?^ . ag^
L2. LI
Dividing (25) by (26),
tan i» = 05 + 4 ^3 ^ ^ ar;5 _j. ...
15
(26)
(27)
In (25), (26), and (27), which are the required develop-
ments, X must be' expressed in circular measure.
79. Computation of the Trigonometric Functions (First
Method). — The functions may be computed by (25), (26), and
(27), Art. 78. Thus, to find sin 20°, we place x=\'ir, the
circular measure of 20°.
log 7r8= 1.49145
col93 = 7.13727 -10
col6 = 9.22185 -10
log 7r6 = 2.4857
col96 = 5.2288 -10
C0II2O = 7.9208 - 10
log*^ - 5.6353 - 10
li
.-.^=0.0000432 .
X
: sin
= - = 0.34906 59
9
^.= 0.00708 88
|3
0.34197 71
log ^ = 7.85057 - 10
.-. ^ = 0.0070888
^=0.00004 32
20° = 0.34202 03
li
In the tables, sin 20° = 0.34202.
FUNCTIONS OF SEVERAL ANGLES. 83
80. Computation of the Trigonometric Functions (Second
Method). — From (25), Art. 78, it may be shown that
sin 1" = 0.00000 48481 36811 07637,
while arc 1'' = 0.00000 48481 36811 09536.
. •. arc 1'' - sin 1'' = 0.00000 00000 00000 02.
Again, sin V = 0. 00029 08882 04563 42460,
while arc 1' = 0.00029 08882 08665 72160.
.-. arc r- sin 1^ = 0.00000 00000 04.
Again, sin 1° = 0.01745 24064 37283 51282,
while arc 1° = 0.01745 32925 19943 29577.
.-. arc r- sin 1° = 0.00000 09.
' Also, from (26), Art. 78,
cos V = 0.99999 99999 88 = 1 - 0.00000 00000 12.
cos 1' = 0.99999 99576 92 = 1 - 0.00000 00423 08.
cos 1° = 0.99984 76952 = 1 - 0.00015.
In computing a set of five-place tables, we may assume
sin 1' = arc V = 0.00029 08882 with an error of 5 x lO-^^,
and cos 1' = 1 with an error of 4 x 10"^.
Then sin 2^ = 2 sin 1' cos 1' ; cos 2' = cos^ 1^ — sin^ l^
sin 3' = sin 2' cos 1' + cos 2' sin 1' ;
cos 3' = cos 2' cos 1' — sin 2' sin V.
sin 4' = sin (3' + 1^ ; cos 4' = cos (3' + 1^,
or sin 4' = 2 sin 2' cos 2' ; cos 4' = cos2 2' - sin2 2'.
And so on.
This method would be employed until the functions of all
angles less than 30° had been computed. Then, since
sin (30° + a:) = cos a; - sin (30° - rr),
and cos (30° -h x)= cos (30° — x)— sin x,
the functions of angles between 30° and 45° would be found by
combining the functions already found. Thus, if 2; = 10°, we
have sin 40° = cos 10° - sin 20°,
and cos 40° = cos 20° - sin 10°.
84 PLANE AND ANALYTICAL TRIGONOMETRY.
It is possible to compute independently the sine and cosine
of 3°, 6°, 9°, ..., 39°, 42°, 45°. We have found in this chapter*
the sine and cosine of 15°, of 18°, and of 36°, and we have
3° = 18°-15°, 6° = 3G°-30°, 9° = 45°~36°, 12° = 30°-18°,
21° = 36°-15°, 24° = 45°-21°, 2T° = 45°-18°,
33° = 18° + 15°, 39° = 45°- 6°, 42° = 45°- 3°.
The values found from these relations would serve as checks
upon the computation.
The computations may also be checked by Euler's and
Legendre's verification formulas :
sin (36° + A)- sin (36° -A)- sin (72° + A)+ sin (72° - A)
= sin A.
cos (36° + ^) + cos (36° -A}- cos (72° + ^) - cos (72° - A)
= cos^.
81. Approximate Assumptions. — It can be shown that
tanV' - arc 1" = 0.00000 00000 00000 04 ;
arc 1'' - sin 1" = 0.00000 00000 00000 02 ;
tan 1" - sin 1" = 0.00000 00000 00000 06.
Hence we may assume that
siiil'' = taiil" = arcl". (1)
In the whole circumference of a circle there are 1296000",
so that the error due to placing arc V = sin 1" in finding the
circumference of a circle with a radius of unity will be only 2^
units in the eleventh decimal place.
In the computation of elliptic orbits there occurs the
equation M= U — esinU,
where M and U are expressed in circular measure. If M" is
the number of seconds in the angle, M= M" SircV\ and ap-
proximately M= M" sin 1^' and U = U" sin 1".
Hence the equation may be written
M" = E" ^sinJS'.
sm 1"
* Ex. 3, Art. 69, and Ex. 8, Art. 71.
FUNCTIONS OF SEVERAL ANGLES. 86
Another assumption that is often made is that for small
angles Sinn" = n sin 1". (2)
The error introduced is
for 1', w" = 60", error = + 0.00000 00000 04 ;
for 1°, n" = 3600'S error = + 0.00000 09.
Thus, if sin a = 0.4 sin 2°, we should have, since a must be
small, a" sin 1" = 0.4 sin 2° or a" = ^liiHl^!.
sm 1'^
82. Transform the First Member into the Second (or last)
in the following examples ;
. cos a — sec a a <> ^ /^ t ^ IN
1. = 4 cos^ 1 a (cos2 i a — 1).
sec a 2 V 2 J .
The first member contains the angle a and the second J a ;
hence we must change the angle.
1_
= cos2 a - 1 = (2 cos2 1 a _ 1)2 _ 1
cos a —
cos a
cos a ^4 gQg4 1 « - 4 cos2 i « = 4 cos2 J a (cos^ |^ « - 1).
2. cosec2a + cot2a = cota. 8. cosec 2 a - cot 2 a ^ ^^^, ^
cosec 2 a 4- cot 2 a
4. cot a — tan a = 2 cot 2 a.
We may either reduce the expression as far as possible before
changing the angle, or change the angle and then reduce.
^ ^ cos a sin a cos^ a — sin^ a cos 2 a o ^ 4- o
(a) = — : = - — : — -— = 2 cot 2 a.
sin a cos a sin « cos a ^ sin 2 «
,j. 1 + cos 2 « 1 - cos 2 « _ 2 cos 2 « _ g ^^^ g ^
sin 2 a sin 2 a sin 2 a
Note. — Avoid radicals if possible.
6. sec a cosec o = 2 cosec 2 a. 8. cot J ^ 4- tan J ^ = 2 cosec 6.
6. (sin \d -\- cos J 0)2 = 1 -i- sin d. 9. sin x — 2 sin^ x = sin x cos 2 x.
l-tan^^^^ 1^, ^(secg + sec2g)= l+tan^^ .
l + tan2jv ^^ ^ (l-tan2je)2
S6 PLANE AND ANALYTICAL TRIGONOMETRY.
. - 2 tan ^ V _ . 13. 1 + tan x tan J x = sec x.
^ 14. ^ (1 + tan i o)2 = i±^HLi^.
opf»2 fl 1 + COS a
12. _§^1_^ = sec 2 ^.
2 — sec2 ^ 15. tan J a 4- 2 sin^ J o cot a = sin a.
16. «i"^a-tan^^)/ L_^_^ 1 \^sin2x.
sec^x Vcosx — sinx cosxf sinx/
17. (1 - tan2 d) sin cos ^ = cos 2 e-J-
— cos 2 g
+ cos 2 ^
18. l + ^^I^=sec2a. 20. ^-^^l±22l2J:l = cot^ 6.
1 — tan^ a sec ^ 4- cos ^ — 2
-g cosg— sing _ l— sin2g _ cos2g 21 * tan g ^ "^ ^^ ^ ^ — ^^" ^
cos^+sin^"" cos2g "~l4-sin2g* * 1 - tan J g~ 1 - sin^
22. sec 2 a + tan 2 a + 1 = ^
1 — tan o
23. (vTTsIna - Vl -sina)2 = 4sin2 Jo.
t-
24. (Vl + sin a + VT— sino)2 =:4cos2 Ja.
25. 2 sin ^ - sin (^ - 5) - 4 sin A sin2 J 5 = sin (^ 4- -B).
26.t cos (36° + ^) + cos (36° - ^) - cos (72° + ^) - cos (72° -A)= cos ^.
27.t sin (36° + A)- sin (36° -A)- sin (72° + ^4) + sin (72° -A)z= sin ^.
28. ?ilL^±iHLl^ = cotJx.
cos X — cos 2 X
29. 1 + cot2 ^v =
sin V tan J v
30. tannv(l + cot2Av)8=-§ —
sin^t)
Q J sin a cos ^ g — 2 cos a sin ^ a _ „ ^^ga i
2 sin J a — sin a ^
32 tan2 ^ x + cot2 j^ x __ _ 1 + cos2 x
tan2 1 X — cot2 J X 2 cos x
33. Given tan J v =a/ "^ ^ tan J ^, show that
'1 — e
1 — e cos ^
(1 + e) cos2 J V + (1 _ e) sin2 J v 1 - e*
34. tan (45° + ^) - tan (45° -A) =2 tan 2^.
tan 45° + tan A tan 45° — tan A
(a)
1 — tan 45° tan ^ 1 + tan 45° tan A
_ 1 + tan A _ 1 — tan A _ 4 tan J. _ 2 tan 2 ^
1 — tan A 1 + tan A 1 — tan^ A
* After substituting, multiply both numerator and denominator by the
quantity sin 0— 1 + cos d.
t cos 36° = Kl + v^).
FUNCTIONS OF SEVERAL ANGLES. 87
.,. 1 - cos(90° -f 2 A) 1 - co8(9Q° -2 A)
^ ^ sin (90° + 2 A} sin (90° - 2 ^)
1 + sin 2 ^ 1 - sin 2 ^ 2 sin 2 A o .. o a
COS 2 A cos 2 ^ cos 2 A
35 tan (45° + M) + tan (45° - ^ A) _
^^ tan (45° + M) - tan (45° - M) ~
36. tan (45° + ^) - cot (46° + ^) = 2 tan 2 ^.
37. tan2 (45° + 6)+ cot2 (45° + ^) = 2 + 4 tan^ 2 $,
38. tan2 (450 + „) - cot2 (45° + o) = 4 tan 2 o sec 2 a.
jg tan (45° + ^6) _ l + 8md
tan (45° - ^ ^) 1 - sin tf'
^^40. tan^tan(45° + J^^)= ®^^^
1 — sm ^
41. cot (45° - J a) - tan (45° - J a) = 2 tan o.
42. tang(45"4-^a) = ^ + ^"^" .
1 — sin a
43. tan (45° + 6)+ tan (45° - ^) = 2 sec 2 d,
44. l-tan2(45°-^)^^.^^,^
iiK ^ />ico . 1 .l + tanArc l+sino;
45. tan (45° + J x) ^^ —
1 + tan2 (45° - 6)
IJ
1 — tan ^x 1 — sin a;
tan (45° + ^x) _ 1 + sin a;
***• 1 + cot2 (45° + ^x) - *°°^^ 1 - sinx*
47. sin (45° - J ^) + cos (45° - J e) = V2 cos J ^ = ^^^^ .
VI - cos e
CHAPTER VI.
TRIGONOMETRIC EQUATIONS.
83. One Equation Containing Multiple Angles.* — Change
the equation so that it shall contain a single angle, and then
proceed as in Art. 52.
1. cos 3 a; = sin 1x ; find x. (See Ex. 8, Art. 71.)
4 cos^ x—K> cos a; = 2 sin x cos x .
. * . cos 2: (1 — 4 sin^ 2: — 2 sin rr) - = 0.
. • . cos a; = 0, giving x = 90° and 270° ;
and 1 — 4 sin^ x — 2 sin x = 0, giving sin x = ^ ( V5 — 1)
and sin a; = - ^ ( V5 + 1), or a: = 18°, 162°, 234°, 306°.
2. cos2 e + cos^ = - 1 ; find 6. Ans. 90°, 270°, 120°, 240°.
3. cot2 + tan = - f V3 ; find d. Ans. 150°, 330°, 120°, 300°.
4. C0S2.X + sinx ^ + 1 ; find x. Ans. 0°, 30°, 150°, 180°.
5. sin 3 X + sin 2 X = sin x ; find x. Ans. 0°, 180°, 60°, 300°.
6. tan 2 X = - 2 sin X ; find x. Ans. 0°, 60°, 180°, 300°.
7. tan2xtanx = + 1 ; find x. Ans. 30°, 150°, 210°, 330°.
8. tan2 X tan 2 X + 2 tan X = + V3 ; find x. Ans. 30°, 120°, 210°, 300°.
9. sin 4 2r - 2 sin 2 = ; find z. Ans. 0°, 90°, 180°, 270°.
The equation may sometimes be solved by the use of the
equations of Art. 73.
10. cos 3 a; — sin 2 a; = ; find x.
cos 3 a; — sin 2 a; = sin (90° + 3 a;) — sin 2 a;
= 2 cos (45° + f a;) sin (45° + 1 a:) = 0.
cos (45°+ fa;) = gives 45°+ fa: = 90°, 270°, 450°, 630°, 810°,
or x= 18°, 90°, 162°, 234°, 306°.
sin (45° + ia:)= gives 45° + la; = 0°, 180°,
or a; = -90° and 270'' .
* See Art. 52 for the solution of equations wlien only one angle is involved.
88
TRIGONOMETRIC EQUATIONS. 89
11. cos 9 — cos 3 e = sin 2 fl ; find 6 by both methods.
Ans. 0°, 30^ 90°, 150°, 180°, 270°.
12. sin 3 ^ + sin 2 ^ -f sin ^ = ; find by both methods.
Ans. 0°, 90°, 120°, 180°, 240°, 270°.
13. cos2 = sin^; find d by both methods. Ans. 30°, 160°, 270°.
14. cos6^-cos3 + sin^ = O; findfl. ^ns. 0°, 180°, (2 n + J ± ^) -.
4
16. sin 5 + sin 3 ^ + 2 cos = ; find 6. Ans. 90°, 270°, (2 n + f ) -.
4
16. sin (60° -X)- sin (60° + x) = + J \/3 ; find x. Ans. 240°, 300°.
17. sin (30° + X) - cos (60° + a;) = - ^ V3 ; find x. Ans. 210°, 330°.
18. cos 4;? - cos 2 ;? = ; find 2. Ans. 0°, 60°, 120°, 180°, 240°, 300°.
84. Find r and from the Equations
a and b being known.
r sin «(» = a,
r cos = &,
(1)
(2)
(1)^(2) gives
tan<^=:^.
(3)
From (1) and (2)
r- ^ -
h
(4)
s'm(f)
1. Find r and when loga=0.47141, and log 6=0.63927 n, r being positive.
log (r sin 0)= log a = 0.47 141 (1)
logsin0 = 9.74972 (5)
logcos0 = 9.91758n (6)
log (r cos , affect the
logarithmic cosine less than the logarithmic sine. The angle > is placed in the
second quadrant, since r cos (f> is negative and r sin positive, r being considered
positive.
2. Find r and = 201° 30'.0 ; r = 79.497.
90 PLANE AND ANALYTICAL TRIGONOMETRY.
85. Find r, <|>, and 6 from the Equations
r cos cos 9 = a, ]
(1)
r
sin 4> cos = 6,
(2)
r
a, h, and o being known
sine =c,
(3)
(2)-^(l)gives
tan cos ^ sin
1. Given loga = 0.46472, log6 = 0.72413 n, logc = 0.62817, find r, 0, and
df d being numerically less than 90"^, and r being positive.
log (r cos «^ cos e) = log a = 0.46472 (1)
log cos (f) = (9.68314) Only as a check. (5)
log sin 0= 9.94256 w (5)
log (r sin cos d) =\ogb= 0.72413 n (2)
(2) - (1) = log tan = 0.25941 n (3)
= 298° 49'.4 (4)
(2)-(5) = (l)-(5)=log(rcos^)= 0.78157 (6)
log cos ^= 9.91291 (10)
log sin d = (9.75951) Only as a check. (10)
logc = log(rsin^)= 0.62817 (7)
(7) - (6) = log tan ^ = 9.84660 (8)
0= 35° 5'. 1 (9)
(6) - (10) = (7) -(10)= log r= 0.86866 (11)
r= 7.3903 (12)
The angle is placed in the fourth quadrant, since r cos 6 is positive, and
therefore cos must be positive and sin negative, r cos cos being positive
and r sin cos negative.
2. Given log a = 0.26903 n, log & = 0.32426, log c = 0.36903 w, find r, 0,
and 0, r being positive and numerically less than 90°.
Ans. = 131° 22'.0 ; = - 39° 45'. 6 ; r = 3.6572.
3. Given log a = 9.43942 w, log b = 9.40403 w, log c = 9.56700 w, find r,
0, and ^, r being positive and numerically less than 90°.
Ans. = 222° 40'.1 ; d = - 44« 36'.4 ; r = 0.525425 or 0.52544.
TRIGONOMETRIC JJQUATIONS. 91
86. Find ^ from the Equation
a sin<|) + 6 cos 4» = c (1)
by formulas adapted to logarithmic computation, a, J, and c
being known.
Let M be an auxiliary angle and m a positive constant,
so that
m cos M=b. J ^ -^
The angle M is always possible, for we have, by division,
tanif=i (3)
and since the tangent may have any value between + oo and
— 00, there will always be some angle whose tangent is equal
to Y* Also, squaring and adding Eqs. (2), we have
m^ sin^ il[f + m^ cos^ M= m^= a^ -\- h\
or m = Va^ + b'^. (4)
Therefore the assumptions in (2) are always possible, since M
and m will be real quantities if a and h are real.
Substituting (2) in (1), we have
m sin M sin = c^
\ or m cos ((/> — Jf ) = c. (5)
Hence, from (2) find M and m by the method of Art. 84 ;
from (5) find <\) — iJf (two values < 360°), and thence find >.
1. Find when 2 sin ^ — 3 cos = 1.
Ans. M= 146° 18'.6 ;- = 220° 12'. 5, or 72° 24'. 7.
2. Find when 2 sin + 4 cos = — 3.
Ans. ilf=26°33'.9; = 158°41'.8, or 254°26'.0.
87. Find <() from the Equation
atan<|> + &cot<|> = c
by formulas adapted to logarithmic computation, a, 5, and c
being known.
92 PLANE AND ANALYTICAL TRIGONOMETRY.
Substituting for tan and cot > in terms of sin (f> and cos <^,
we have, after reducing,
(a — 6) cos 2(l> -\- c sin 2(f) = a -{- b.
Let msinM=a — b^
m cosilff = c.
\ .-. wsin (il[f 4- 2(/>) = a4- ^.
1. Find when 2 tan — cot = — 3.
^ns. ilf=135°; = 15°41'.O, 119M9'.0, 195°41'.0, 299° 19'.0.
2. Find when tan + 3 cot = — 2 VS.
^ns. M = 210° ; = 120° or 300°.
88. Find <}> from the Following Equations, a and a being
known:
(a) sin (<^ -f- «) = « sin >. (1)
Expanding, sin <^ cos a + cos <^ sin a = a sin <^.
. • . sin (^ (a — cos a) = cos <^ sin a,
, , sin a ,o\
.-. tan = (2)
a — cos a
Eq. (2) is not adapted to logarithmic computation. But
from (1) we have
sin (0 -f «) _ a
sin<^ i
and, by composition and division,
sin (<^ + a) + sin <^ _ a + 1
sin ((/) + «)— sin <^ a — V
and this, from the equations of Art. 73, becomes
tan (<^ -f 1^ a) _ a 4- 1
tan^^a a — l'
or tan (<^ + 1 a) = ^ "^ tan \ a. (3)
Let tan/3 = a, and note that tan 45° = 1.
^ ^^ . ^ tan /3 + tan 45° ^ ,
... tan(c/> + J«)=^-^^^^-^^^^^tanla
sin(y8 + 45°)^ ,
= . \ri TF^ tan i a.
sm (/S — 45°) 2
.-. tan()-fla)=cot(/3-45°)tanla. (4)
TRIGONOMETRIC EQUATIONS. 93
(6) COS (<^ -f a) = <« cos .
. • . tan (<^ -f J «) = tan (45° — /S) cot J a, if tan /8 = a.
() sin (a — (f)^= a sin <^.
. • . tan (<^ — I «) = tan (45° — y8) tan J a, if tan y9 = a.
(^) sin (> + a) =3 a cos >.
.-. sin(<^ + a)=«sin(90° +<^);
.-. tan(4'5° + ) + |«) = cot(45°-)8)tan(45°-^a), if tan)S=a.
(g) cos () + a) == a sin <^.
.-. cos(> + «)=acos(90° - <^);
. • . tan (<^ 4- i a - 45°) = tan (45° - ff) cot (45° + J a), if tan = a.
Note. — The equation a sin (0 + a) = a' sin (0 + a') and similar equations
may be solved by expansion, the solution of the given equation being
,„ , a' sin a' — a sin a
tan = -.
a cos a — a' cos a'
A solution adapted to logarithmic computation may be found by the method
of this article, giving
tan [0 + ^ (o + a')] = cot (j8 - 45°) tan K« - «')» i^ tan & = —-
89. Find from the Following Equations, a and a being
known :
(a) sin (<^ + a) sin = a.
From (8), Art. 72,
cos a — cos (2 <^ + a) = 2 a.
. • . cos (2 <^ + a) = cos a — 2a. (1)
Let tan /3 = ^-^. (2)
sin a
^n , , ^ ' . a COS a COS j8 — sin a sin y3
.*. cos(2 + a) = cosa — sinatan)^= — ^•
cos)8
■ •■ coa^24> + a)= ^°^("+/\ (3)
cosp ^ ^
(5) sin (a — <^) sin (\> = a.
. • . cos (a — 2 <^) — cos a = 2 a ;
. • . cos (a — 2 <^) = cos a + 2 a.
^ o JL\ cos(a — /3) ./. . o 2 a
• o cos (« — 2 0) = ^^ — — ^-^ , if tan p =
cos p sin a
94 PLANE AND ANALYTICAL TRIGONOMETRY.
(c) sin (0 + a) cos cf) = a.
. • . sin (2 > + a) -t- sin « = 2 = a.
. • . cos (2 (^ + a) + cos a = 2 a.
, • . cos (2 + «)=« tan (f>.
tan ( a — l'
sin (2 <^ + ct) _ g + 1
sin a a — 1
Let tan/8 = a; (2)
.-. ^ = cot(yS-45°),
and sin (2 <^ + «) = cot (/3 - 45°) sin a. (3)
Find /3 from (2) and 2 <^ + « from (3).
(V) tan ((^ + a) = a cot .
. • . cos (2 <^ + «) = tan (45° — yS) cos a, if tan = a.
(c) cot (a — <^^= a cot <^.
. • . sin (2 (/) — a) = tan (/3 — 45°) sin a, if tan /3 = a.
(c?) cot () + a) = a cot (<^ — a).
. • . sin 2 (/> = cot (45° — yS) sin 2 a, if tan fi = a.
(e) tan ((^ + a) = a tan (> + «').
. • . sin (2 <^ + a + «') = cot (y8 - 45°) sin (a - «'), if tan yS = a.
(/) cot ((/> + a) = a cot (<^ + a')-
.• . sin (2 > + a 4- "0 = cot (/S - 45°) sin (a' - a), if tan /S = a.
(^) cot (<^ + a) = a tan (> + «')•
. • . cos (2 (^ + a + a') = tan (y3 — 45°) cos (a — a'), if tan yS = a.
(1)
TRIGONOMETRIC EQUATIONS. 95
91. Find ^ from the Following Equations, a, a, and a' being
known : *
(a) tan (<^ -|- «) tan -}- a) sin > = a cos (> + a) cos ^.
From the equations of Art. 72, we have
— cos (2 > + a) + cos a = a cos (2 (f) -{• a) -\- a cos a;
.-. cos (2).H- a) = --^^^— cosa.
1 + ^
Let tan fi = a.
. • . cos (2 > + «) = tan (45° — ;8) cos a.
(V) tan (<^ + a) cot > = «.
. • . sin (2 ) + «)= cot (y8 — 45°) sin a, if tan ^ = a.
(c) tan (<^ + a) tan ((f> — a)= a.
. • . cos 2 <^ = tan (45° — y8) cos 2 a, if tan P=a.
(d) tan (^ + a) cot (<^ + «') = a.
. • . sin (2 > + « + «0 = cot (y8 — 45°) sin (a — a'), if tan ^ = a.
92. Find r and from the Following Equations, a, h, a, and
/9 being known :
r sin (<^ + «) = a, I (1)
rcos(<^ + /3)= J. J (2)
sin (d) -f a) <* i • ^ j , \ ^ i , o\
. • . 5 sin <^ cos a-\-h cos <^ sin a = a cos <\> cos ^ — a sin <^ sin yS.
. • . 5 sin (\> cos a 4- a sin <^ sin p = a cos <^ cos /S — b cos <^ sin a.
. • . sin (f) (b cos a-{- a sin y8) = cos > (a cos /3 — 5 sin a).
. • . tan 6 = — r, (3)
b cos a -\- a sin p
and r = = (4)
sin(<^ + «) cos(> + y8)
The quadrant of (j) will be determined by the sign assigned
to r.
* The method of Art. 90 may be used, since tan x = and cot x = •
cot X tan X
96
PLANE AND ANALYTICAL TRIGONOMETRY.
1. If r sin (0 + a) = a, and r sin (0 + /3) = 6, show that
6 cos a — a cos/3
2. If r cos (0 + a) = a, and r cos (0 + /3) = 6, show that
♦«^ -J. rt cos /3 — ?> cos a
tan =
as>'\n3 — b sin a
;:i
93. Find r and <|) from the Following Equations, a, 5, a, and
/3 being known, and the formulas derived being adapted to
logarithmic computation :
r sin () + «) = a,
r sin {(f) + /3)=b
(1) 4- (2) = r [sin (<^ + «) + sin (<^ + yg)] = « + 5 ;
.-. 2rsin[<^ + l(a + y3)] cos | (a - y8)= a + J.
(1) - (2) = r [sin ((^ + «) - sin ((^ -f ^)] = « - ^ ;
. • . 2 r cos [(^ + ^ (a + /S)] sin |^ (a - y8) = a - ^>.
From (3) and (4), we have
a + b
0)
(2)
r sin [<^ + K« + ^)]
rcos[<^4-K" + ^)] =
2cos-|-(te-y3)'
2
a — h
2sini(«-y8) J
from which r and > 4- | (« + y8) are found by the method of
Art. 84.
1. If r cos (0 4- a) = a, and r cos (0 + /3) = &, show that
a -\-h
rcos[0+i(« + ^)] =
rsin[0 4-Ka + '8)] =
2 cos I (a — ii)*
h — a
2 sin \{a- fi)
2. If r sin (0 4- a) = a, and r cos (0 + y8) = &, show that by placing
cos (0 + /3) = sin (90"^ + + 3) we may obtain
a + &
rsin[0 + 45° + Ha + )8)] =
rcos[0 + 45° + Ka + i8)]
2 cos [45°- Ka-^)]
h — a
2 sin [45°- K«-^)]
3. Find r and when rsin (0 + 100°) = 2, and r sin (0 + 200°) = 3,
r being positive.
Ans. = 290° 28'. 4 ; r = 3.9436.
CHAPTER VII.
OBLIQUE PLANE TRIANGLES.
94. It has been shown in Geometry that a triangle can be
constructed when three elements, one being a side, are known.
If the three angles only are given, there will be an infinite
number of triangles satisfying the conditions of the problem,
since the data determine the shape and not the size of the
triangle.
We also know that in any triangle
(1) The sum of the three angles is 180°.
(2) If one angle is 90°, the sum of the other two is 90°.
(3) The greater side is opposite the greater angle, and
conversely.
(4) Any side is less than the sum of the other two.
95. The Sine Proportion. — The sides of a triangle are to each
other as the sines of the opposite angles.
In Fig. 52, p = a sin 7 ; p = c sin a.
.*. a sin 7 = b,
1. Find a when c = 1.4345, b = 2.3671, and o = 112°43'.4.
log tan x = I (log b + log c) + log 2 + log sin ^ a + col (b — c\.
log a = log (6 — c) — log cos X.
log b = 0.37422
logc = 0.15070
log (6 -c) = 9.96970
-logcosic = 9.46361
log 6c = 0.53092
log a = 0.50609
log a/6c = 0.26546
log 2 = 0.30103
logsin|a = 9.92041
col (&-c)= 0.03030
log tan ic = 0.51720
X = 73° 5'.6
a = 3.2069
2. Find 6 when a = 101.47, c = 99.367, jS = 47°48'.2.
.-. x = 88° 31'. 17; 6=81.396.
3. Find a when 6 = 19.937, c = 62.475, a = 130° 9'.4.
.-. x = 56°23'.7; a = 76.858.
OBLIQUE TRIANGLES SOLVED BY RIGHT TRIANGLES.
103. Case I. Given a, a, y, — In Figs. 63 and 64, on the
next page, draw DB perpendicular to AC Considering the
first figure, in the triangle BDO we know a and 7, and we
compute DB and BQ\ then in the triangle BDA we know
DB and a, and we compute AD and ; then b = AD -{- DC,
108
PLANE AND ANALYTICAL TRIGONOMETRY.
completing the solution. In the second figure, where 7 is
obtuse, we know, in the triangle BDC^ a and DCB = 180° — 7,
and we compute DB and CD\ then in the triangle BDA we
know DB and a, and we compute c and AD\ then h — AD
— QD^ completing the solution.
Fig. 63.
1. Solve the triangle when a = 3.4356, o = 17° 43'.4, 7 = 60° 35'.7.
.-. j8 = 101° 40'.9 ; X>C= 1.6868; i)J5 = 2.9929 ; ^D = 9.3650 ;
c = 9.8315; & = 11.0518.
2. Solve the triangle when a = 54.376, 7 = 103° 3'.2, /3 = 40° 10'. 3.
Ans. o = 36° 46'.5 ; c = 88.478 ; h = 58.592.
3. Solve the triangle when c = 230.47, a = 21° 32'.2, /3 = 36'^ 24'.4.
Ans. 7 = 122° 3'.4 ; a = 99.825 ; h = 161.3975.
104. Case II. Given a, c, a. — In the right triangle ABB
we know c and «, and we compute AD and DB; then in the
triangle CBD we know DB and a, and we find DQ and 7;
then
6 = ^2) + DC; /3 = 180°-(« + 7);
b' = AD-DC; y = 180°-7; ^8' = 180° -(« + 7').
Two solutions are possible only when a is acute and a is less
than and greater than DB,
OBLIQUE PLANE TRIANGLES.
109
If a is obtuse, as in Fig. 66, we solve first the triangle
BAD, then the triangle BCD, and find b = DC- DA.
1. Solve the triangle when c = 23.647, a = 14.135, a = 33° 17'.3.
. •. AD = 19.767 ; DB = 12.979 ; 7 = 66° 40'.0 ; DC = 5.5986 ;
r 7= 66°40'.0; j8 =80° 2'.7 ; b =25.3656;
\7' = 113°20'.0; ^' = 33°22'.7; 6' = 14.1684.
2. Solve the triangle when a = 2.4741, c = 1.0003, a = 6t)° 14'.8.
•• 7 = 22° 12'.8 ; /3 = 88° 32'.4 ; AD = 0.35445 ; DC = 2.2905 ; b = 2.64495.
3. Solve the triangle when a = 10.473, b = 12.987, a = 44° 11 '.3.
^ /3 = 59°48'.5 ; y = 76° 0'.2 ; c = 14.5793 ;
^ns.
\(j' = 120° 11'.5 ; 7' = 15° 37'.2 ;
4.0455.
4. Solve the triangle when a = 0.43477, b = 0.40031, a = 94° 17'.6.
Ans. $ = 66° 39'.6 ; 7 = 19° 2'.8 ; c = 0.142282.
105. Case III. Given a, &, c. — In Fig. 63,
p^ = c^- AD^; jt?2 = a2 _ j)02^
... (^-AD^=a'^-DC\
,'. AD^-DC^ = (^-a^.
(c + g) (g - a) _ (g + a^ (c - a')
• • ^^~^^=~~AD^fDC~- l '
from which AD —DC may be computed. Then
AD=i[b+CAD-DC)],
and DC = ilb-(AD-DC)^,
If either J.i) or DC is negative, it is exterior to the tri-
angle; that is, the point D is on the line AC produced.
Having found AD and DC, the angles are found from ths
right triangles DBA and DBC.
1. Solve the triangle when a = 27.103, b = 16.432, c = 12.511.
... c-a = -14.592; ^i> - i)C = - 35.178 ; .41) = -9.373; 2)0 = 25.805.
^n.«?. a = 138° 31'.2 ; 7 = 17° 48'.6 ; )8 = 23° 40'.3.
In this example D lies to the left of A.
110
PLANE AND ANALYTICAL TRIGONOMETRY.
2. Solve the triangle when a = 32.456, h - 41.724, c = 53.987.
.-. AD- DC = 44.(507 ; AD = 43.1655 , DC = - 1.4415.
Alls, a - 36° 54'.7 ; 7 = 92° 32'. 7 ; & - 50° 32'.6.
3. Solve the triangle when a = 0.14679, h - 0.10433, c = 0.04796.
.-. ^2> - i>C = - 0.18448 ; ^Z) = - 0.040075 ; DC = + 0.144405.
Ans. a = 146° 40'. 75 ; 7 = 10° 2l'.0 ; )8 = 22° 58'.25.
106. Case IV. Given 6, c, a. — In the triangle J.i>^, know-
ing c and «, find AT) and BB. Then in the triangle BBC we
know BB and BC =h — AB, so that we can compute a and 7.
1. Solve the triangle when b = 1143.7, c = 1822.4, a = 15°6'.4.
.-. ylZ>= 1759.5; Z>^ = 474.96; DC = -615.8.
Ans. 7 = 142° 21'.5 ; a = 777.68 ; 3 = 22° 32'. 1.
The negative value of DC shows that D is to the right of C.
2. Solve the triangle when b = 19.937, c = 62.475, a = 130° 9'.4.
.-. ^D = - 40.288; DC = 60.225.
Ans. 7 = 38° 24'.5 ; i8 = 11° 26M ; a = 73.857, or 76.858.
Note that a is obtuse.
8. Solve the triangle when a = 101.47, c = 99.367, j8 = 47°48'.2.
Ans. 7 = 64° 44'.6 ; a = 67° 27'.2 : b = 81.394.
AREAS OF TRIANGLES.
107. Given Two Sides and the Included Angle (6, c, a).—
Represent the area by A. From geometry, in Fig. 67,
A = \pb.
But p = c sin a.
.'. A = ^ 6c sin a, (1)
or, the area of a triangle is equal to half the product of the two
sides multiplied by the sine of their included angle.
OBLIQUE PLANE TRIANGLES. HI
108. Given One Side and the Three Angles («», a, p, -y). —
Substitute in (1), Art. 107, the value of c found from the sine
proportion, _ ^ sin 7
sin P
giving ^=f-^-^w|P- (1)
109. Given the Three Sides («, h, c).— We have
A = \hc mna = hc sin J a cos J a.
From (2) and (3), Art. 99, we have
110. Given Two Sides and the Angle Opposite One of them
(6, c, P). — First find 7 by the formula
c .
sin 7 = - sin
^.
Then a = 180°
and A = ^bc
sin a.
EXAMPLES.
1. Find the area when b = 0.14367,
c = 0.11412, = 42° 14'.6.
2. Find the area when a = 3.4356,
o=17°43'.4, 7 = 60°36'.7.
log 6 = 9.15737
log c = 9.05737
log sin a = 9.82755
col 2 = 9.69897
log ^=7.74126
A = 0.0055114
.-. 3 = 101° 40'.9.
loga2 = 21oga= 1.07200
col 2= 9.69897
logsini8= 9.99091
log sin y = 9.94010
col sin a = 0.51652
log^= 1.21850
A = 16.539
3. Find the area when a = 0.0093146, b = 0.0176530, c = 0.0095768.
2s = 0.0365444 logs= 8.26179
5=0.0182722 \og(s-a)= 7.95219
s - a = 0.0089576 log (s-b)= 6.79183
s- 6 = 0.0006192 log(s-c)= 7.93929
s-c= 0.0086954 2 )10.94510-20
sum = 0.0365444 ^°S^ ^ 5.47255-10
a check. ■ ^= Q-QQQQ29686
112 PLANE AND ANALYTICAL TRIGONOMETRY.
4. Find the area when a = 9.4672, c = 14.433, o = 11° 14'.3.
log c = 1. 15936 log a = 0.97622 log a = 0.97622
log sin = 9.28979 logc= 1.15936 logc = 1.15936
col a = 9.02378 col 2 = 9.69897 col 2 = 9.69897
log sin 7 = 9^47293 log sin ^= 9.67899 log sin /3^ = 9.02259
7= 17°17M log^= 1.51354 log ^1' = 0.85714
7' = 162° 42'.9 A = 32.624 A' = 7.1968
.-. /3= 151°28'.6 ■
/3' = 6° 2'.8
Note that log A and log A' can be found by adding log sin p and log sin /3'
respectively to log a + log c -f col 2, a shorter method than that given in this
example.
6. Find the area when a = 0.013456, b = 0.023678, a = 40° 31 '.4.
Ans. 0.00010351.
6. Find the area when c = 43.145, o = 40° 40'.3, = 60° 30'.3. Ans. 538.19.
7. Find the area when a = 1.4142, b = 1.6735, c = 2.8533. A7is. 0.83826.
8. Find the area when a = 14.135, c = 23.647, a = 33° 17'.3.
Ans. 164.61 or 91.948.
111. Illustrative Examples. — The hearing of a line is the
angle it makes with the magnetic meridian, shown by the mag-
netic needle. The letter indicating whether the line is meas-
ured north or south of the point of beginning is written, then
the number of degrees and minutes in the angle, and then the
letter indicating whether the line lies to the east or to the west
of the magnetic meridian. Thus, if the bearing of the line AB
is S. 60° W., the line is measured from A to the west of south
by an angle of 60°.
The distances and the angles given in the examples are
horizontal unless otherwise specified.
1. From a point on a horizontal plane the
angle of elevation to the top of a crag is 40° 28'.6,
and 4163.2 feet farther away in the same vertical
plane the angle is 28° 50 '.4. Find the distances
from the points to the top of the crag, and its
height above the horizontal plane.
BD = 13399 feet ; AD = 9956.2 feet ; CD = 6463.0 feet j
BC = 11737 feet : AC = 7573.2 feet.
OBLTQW: PLANK TRTANOLES.
113
3. A tower 160.43 feet high is situated at tlie top of a hill (Fig. C9) ; 600
feet down the hill the angle between the surface of the hill and a Jine to the
top of the tower is 8° 40'.4. Find the distance to the top of the tower, and the
inclination of the ground to a horizontal plane.
.-. ^BC = 136°59'.7; ^C :- 726.60 feet ; DAB = 46° 59^.7.
'>B
Fig. 70.
3. To find the horizontal distance from a point A to an inaccessible point
B (Fig. 70), the horizontal distance AC and the angles a and y were measured
and found to be 1042.3 feet, 72°9'.4, and 14° 13'.7, respectively.
• .-. AB = 256.69 feet ; CB =z 994.15 feet.
4. To find the distance between two points A and B not visible from each
other (Fig. 71). — Select a third pohit C from which A and B are visible, and
measure the distances (7^ = 444.38 feet, C5 = 222.76 feet, and the angle
^C^=17'' 17'.6. Ans. AB = 240.97 feet.
--~i>B
Fig. 71.
Fig. 72.
5. To find the distance from a point A to another point JB, the latter being
inaccessible and invisible from A (Fig. 72). — Select two points C and D so
that C, J., and D shall be in the same straight line, A and B being visible both
from C and from D. From measurement it is found that CA — 456.72 feet,
AD = 490.74 feet, y = 71° 22'. 7, 5 = 36° 19'.4.
.-. CB = 589.10 feet ; DB = 942.475 feet ; AB = 619.51, or 619.53 feet.
CROCK. TRIG. 8
114
PLANE AND ANALYTICAL TRIGONOMETRY.
6. To find the elevation of the top of a church steeple D (Fig. 78) above
the horizontal plane ACB, and the distances of the steeple from A and B. — Let
the horizontal distance AB = 435.53 feet, the horizontal angles a = 140° 40'.2 and
/3 = 10° 7'.6, and the vertical angles 7 = 32°45'.6 and 7' = 10° 7'.3.
. •. AC= 156.95 feet ; BC = 565.74 feet ; CD = 100.99, or 101.00 feet.
The agreement of the values of CD is a check upon the observed angles and
upon the computations.
7. To find the elevation of the top of a church steeple D (Fig. 74) above
the iYfo points A and B, not in the same horizontal plane, the inclined distance
from A to B, and its angle of inclination 5 to a horizontal plane being measured,
as well as the angles a, ^, 7, and 7', shown in the preceding example. — Let
^5=134.70 feet, 5=3°2'.7, a=43°14'.8, /3=63°17'.5, 7 = 56°36'.6, 7' = 62°17'.3.
[First find the horizontal distance AF and the vertical distance FB in the
right triangle AFB ; then solve the horizontal triangle AFC; and then find CD
and FD from the right triangles ACD and BFD respectively.]
. •. AF= 134.51 feet ; FB = 7.1553 feet ; FC = BE = 96.135 feet ;
AC = 125.34 feet ; CD = 190.17 feet ; ED = 183.02 feet.
Check: CD = FB + ED.
Fig. 75.
8. To find the distance between
two inaccessible points A and B. —
Select two points C and D from which
both A and B can be seen, and measure
CD = 456.82 feet, a = 30° 40'.6,
)8 =r 40° 14'.8, 7=35°16'.4,
5 = 56°47'.4.
.-. .42) =449. 09 feet; ^C=274.41 feet ;
jBJ9 = 398.66 feet ; 5C=616.66 feet ;
AB = 405.57, or 405.58 feet.
OBLIQUE PLANE TRIANGLES.
115
9. To find the distance between two inaccessible points A and
being visible from only one accessible point
C. — Select a point D from which A and C
are visible, and another point E from which
B and C are visible. From measurement
CZ>=943.37 feet, C^=673.33 feet,
a = 72°9'.3, /3 = 60°17'.9,
7 = 32° 14'.6, 8 = 67° 33'.9,
e = 19° W.I.
. '. CA = 1217.0 feet ; CB = 222.28 feet ;
AB = 1035.8 feet.
B, both
10. To find the distance between two inaccessible points A and B^ there
being no accessible point from which both A and B are visible (Fig. 77). —
Select the points C, 2>, E, and F so that A, (7, and E shall be visible from i>,
and D, F, and B from E. Measure the angles a, /3, y, 5, e, and d, and the dis-
tances CD, DE, and EF, Show how AB may be found from the data thus
obtained.
Ar,
Fig. 77.
Fig. 78.
11. Two points A and B, 8763.6 feet apart (Fig. 78), are situated at the
sea level in the same north and south line ; a vessel is seen at C, and an
hour later at D. The required quantities are AC, BC, AD, BD, CD, and
the angle that CD makes with the north and south line, having measured
BAC=: 120° 30'.6, BAD = 30° 14'.4, ABC = 40° 18'.8, ABD = 140° 28'.2.
.-. AC= 17260 feet ; BC = 22985 feet ; AD = 34552.5 feet ; BD = 27340 feet ;
ACD = 63° 14'.5 ; ADC = 26° 29'.3 ; BCD = 44° 3'.8 ; BDC = 35° 46'.8 ;
CD=SSed6, 38697, or 38699 feet ;
e=S60°-BAC-ACB-BCD=nG° 15'.0,
or =ABD+BDA-{-ADC=176°U'.9.
12. In measuring the line from A to B, whose direction was known, it was
necessary to pass an obstacle at F (Fig. 79). A distance CD =144.31 feet was
measured, making an angle y = 19° 53'.4 with AB, and the angle 5 = 140° 10'. 3
116
PLANE AND ANALYTICAL TRIGONOMETRY.
was laid off with the transit. It is required to find the distance DE to the line,
the distance CE, and the angle e, in order that tlie line AC may be prolonged.
Arts. CE = 27L06 feet ; DE = 143.98 feet ; e = 160° 3'.7.
x:;;'^a
G B
Fig. 80.
13. Ill passing an obstacle at F it was necessary to use the broken line
CDEG (Fig. 80). The distances CD and DE and the angles 7, 5, and e were
measured. It is required to find the distance EG to the line AB, the dis-
tance CG, and the angle 6, when CD = 100.37 feet, DE = 94.367 feet, 7 = 80°,
5 = 101°19'.8, ande = 110°.
.-. DCE=S7°[>3'.3; DEC = iO° AG'.O ;
CE = 150.67 feet ; EG = 108.46 feet ;
CG' = 151.22 feet ; = 111°19'.8.
14. From the top of a lighthouse ABy
200 feet above the sea level, the angle of
depression to a ship was 7 = 10° 14'.3 : an hour
later it was 7' = 11° lO'.O; the horizontal angle
between the directions of the ship at the two
instants was a = 127° 14'.4. Find the distance
sailed by the ship.
.-. AC = 1107.3 feet ; AD = 1012.2 feet ;
CD = 1899.3 feet.
15. A ladder 52 feet long is set 20 feet in
front of an inclined buttress,' and reaches 46
feet up its face. Find the inclination of the
face of the buttress.
Ans. ABC = do°51'.S, or 95°51'.9.
t^
^ 16. The sides of a city block measured AB = 423.24, BC=
162.36, CD = 420.81, and DA = 160.62 feet, the first two sides
being perpendicular to each other. Find the angles between
the other sides.
.-. ^0-^453.31 feet; BCA = C0° 0'.8;
^^C = 20°59'.2; ACD = 20° ^b'.O ;
CAD = G8° 8'.8'; CZ>yl = 91° 6'.4 ;
BCD = 89° 45'.S; BAD = S9° 8'.0.
Fig. 83.
OBLIQUE PLANE TRIANGLES.
117
17. A ship B is 12 miles S. 45° W. of a lighthouse A,
and sails S. 50° E. to C, a distance of 15 miles. Find its
distance from the lighthouse. B
Ans. AC= 18.374, or 18.375 miles. ''^"
Fig. 84.
18. In surveying a field a thick wood prevents the measurement of the
angle ABD and of the distance BD. The
angle ABC = 70° 14'.6 is measured, a line
^O is run 743.86 feet, the angle BCD is
found to be 62° 14'.4, and the distance CD
to be 912.82 feet.
.-. CjB2) = 68°28'.1; CDB = i9°lT.5;
BD = 868.34, 868.30, or 868.38 feet ;
^i?Z)z=138°42'.7.
Fig. 85.
19. The distance OE and its bearing E'OE are required, the engineer
having measured the distances a, b, c, d, and e and their respective bearings,
N. 30° W., S. 60° E., N. 20° E., N. 40° W., and
N. 50° E.
OE' = OA' - B'A' JrB'C'+C>D>-{- D'E'
= a cos 30° - h cos 60° + c cos 20°
+ d cos 40° -f e cos 50°.
E'E = -AA'-\- B"B + C"C-DD"^E"E
= - a sin 30° + b sin 60° + c sin 20°
- d sin 40° + e sin 50°.
Then OE cos E'OE = OE',
OEsmE'OE = E'E;
whence OE and E'OE can be found. Then
the quadrant of E'OE fixes the direction of
the line OE ; thus, if E'OE = 40°, the bearing is N. 40° E. ; if ^^0^"= 110°,
the bearing is S. 70° E. ; if E'OE = 230°, the bearing is S. 50° W. ; if E'OE =
310°, the bearing is N. 50° W.
20. At a certain point the angles of elevation of the base of a vertical
tower and of its top are a and /3 respectively, the height of the tower being
h feet. Prove that the horizontal distance from the point to the tower is
Fig. 86.
118
PLANE AND ANALYTICAL TRIGONOMETRY.
j^cosa cos/3 cosec(i3 — a), and that the elevation of its top above the point
is h cos a sin /3 cosec (/3 — a) .
21. At the top of a vertical tower whose height is h, the angles of depression
to two points M and N in the same vertical plane with the tower were o and /3
respectively (/3 > a), the points being in the same horizontal plane with the base
of the tower. Prove that the distance MN is h sin (/3 — a) cosec o cosec /3.
22. Two points M and iV in a horizontal plane are in the same vertical plane
with a tower. The angle of elevation of the
top of the tower from 31 is a, and from N it'
is iS, /3 being greater than a. Prove that the
horizontal distance of the tower from iV" is
MN sin a cosjS cosec {^ — a).
23. Three points, A, B, and C, are in the
same horizontal line, the distances AB and
BG being a and b feet respectively (Fig. 87).
The angles of elevation of the top of a tower
measured at A, B, and C were a, /3, and y
respectively. Find the elevation of the top
of the tower above the horizontal plane
through the points, and the horizontal dis-
tances of the tower from the three points.
Fig. 87.
m
= h cot a ;
n = h cot ^; p =
/i cot 7 ;
m^
= a^-\-n^
-2 an cos ABD ;
P'
= b-2 + W2
+ 2bn cos ABD ;
a2
+ n2-
-m2
p^- 62
- w2
2 an
2bn
>
.
h^
ab(a + b)
a (cot2 7 - cot2 /3) + 6 (cot2 a - cot2 /3)
24. In Fig. 88 the distances a and & and the angles a, /3, and y are known,
and the distance BG = x is required, ABGD
being an inaccessible straight line.
^<^r;||v:
FB
a + X
FG
Fig. 88.
sin a sin J. '
sin (a + jS) sin A '
FB a
sin (a + /S)
FG a + x
sin a
b FG .
b + x _ FB
sin 7 sin Z) '
sin (^ + 7) sin D
FG b
sin(j8 + 7).
FB 6 + x
sin 7
Multiplying (1)
and (2), we have
(a + x) (6 + x) sin a sin 7 = ab sin (a + jS) sin (^ + 7),
from which x may be found, since the equation is a quadratic in x.
(1)
(2)
OBLIQUE PLANE TRIANGLES.
119
a tower
25. Two points A and B in the same vertical plane with the top of
are on a sidehill whose angle of inclination to a
horizontal plane is 5, the inclined distance AB
being a feet. The angles of elevation of the top
of the tower were measured at A and B, and
found to be a and )3. Prove that the horizontal
distance of the top of the tower from B is
a (cos S tan a — sin 5) cos a cos cosec (3 — a),
and that the elevation of the top above B is
a (cos 5 tan a — sin 5) cos a sin |3 cosec (/3 — a).
26. In a hydrographical survey, the distances between three points, A, Bj
and C, on the shore having been determined, the observer in the boat P measures
the angles 5 and e subtended by AB and BC. It is required to find the dis-
tances of the boat from the three points.
(1) Graphical Solution. — Construct on AB the segment of a circle APB
that shall contain the measured angle 5, and
on BC the segment of a circle BPC that
shall contain the angle e. Their point of in-
tersection P will be the position of the boat.
There are four possible solutions, only one
being shown in the figure. .'
(2) Analytical Solution. — Let J.Z>CP »
be the circle through A, C, and P. Then
DAC= e, and DC A = 8. Hence in the tri-
angle ADC we know one side AC and the
three angles ; find AD and CD. In the tri-
angle ABC we know the three sides; find ^
the three angles. In the triangle DAB we '^^^- ^^^
know two sides and the angle DAB = CAB — CAD ; find ABD. Then in the
triangle ABP we know one side and the three angles ; find AP and BP. Also,
compute DBC from the triangle DCB, and then BP and CP from the triangle
BPC. The values of BP should agree.
In the following examples find the last three elements, the first three being
given:
27. a = 1.0431, /3 = 4°4'.4, 7 = 22°3'.6.
.'. a = 153° 62'.0 ; b = 0.16822 ; c = 0.88942.
28. a = 103.37, a = 10° 11 '..3, /3 = 83°43'.6.
7-.
M: & = 580.89: c = 583.02.
29. c = 74.344, a = 105° 6'. 7, /S = 60° 14'.4.
.-. 7 = 14°38'.9; a = 283.82 ; 6 = 255.21.
30. c = 0.047365, p = 40° 7'. 7 y = 39° 41'. 9.
.'. a = 100° 10'.4 ; a = 0.072990 ; b = 0.047792.
31. c = 4.4479, a = 11° 11'.3, y = 57° 37'.4.
. •. /3 = 111° 11'.3 ; a = 1.0219 ; b = 4.9106.
120 PLANE AND ANALYTICAL TRIGONOMETRY.
82. b = 143.97, /3 = 30° 36'.8, y = 107° 15'.5. .•'.
.-. a =42°7'.7; a = 189.64; c = 269.98.
33.6 = 10.467, c = 1.4321, /S = 114° 10'.3.
.-. 7 = 7°10'.2; a = 58°39'.5; a = 9.79875.
34. a = 0.67375, b = 0.43213, a = 147° 11'.3.
.-. i3 = 20°20'.2; 7 = 12°28'.5; c = 0.26858.
35. a = 1.4742, c = 0.97674, o = 25° 19'.9.
. •. 7 = 16° 28'. 1 ; j3 = 138° 12'. ; 6 = 2. 2966.
36. a = 943.42, b = 647.15, a = 104° 6'.9,
.-. /3 =41°42'.0; 7 = 34° 11'. 1; c = 646.59.
37. a = 0.10321, c = 0.047323, a = 45°9'.7.
.-. 7 = 18°58'.4; /3 = 115°51'.9; 6 = 0.13097.
38. a = 4.4321, c = 5.4763, 7=100°11'.9.
' .-. o =52°48'.l; /3 = 27°0'.0; 6=2.5261.
39.0=23.111, 6 = 19.476, 7 = 47°16'.7.
.-. ^=38°15'.0; a = 94°28'.3; a = 31.363.
40. a = 0.11111, = 0.12767, a = 23° 15'.6.
. •. 7 = 26° 59'. 1 ; /3 = 129° 45'. 3 ; 6 = 0.21630 ;
7' = 153° 0'.9 ; /3' = 3° 43'.5 ; 6' = 0.018279.
4L 6 = 1.4326, c= 1.3671, 7 = 44° 17'.3.
.-. /3= 47° 1'.9; a = 88°40'.8; a = 1.9574 ;
/3' = 132° 58'. 1 ; a' = 2° 44'. 6 : a' = 0.093706.
42. a = 46.703, 6 = 57.147, a = 19° 17'. 7.
.-. /3 = 23°50'.9; 7 = 136°51'.4; c = 96.652 ;
/3' = 156° 9'.1 ; 7' = 4°33'.2 ; c' = 11.221.
43. a = 9.4327, c = 10.4751, a = 63° 17'.3.
.-. 7 =82°45'.0; ^ = 33°57'.7; 6 = 5.8990;
7' = 97° 15'.0 ; ^' = 19° 27'.7 ; 6' = 3.5182.
44. a = 0.034337, c = 0.062774, a=9°6'.7.
.: y= 16° 49'. 7 ; ^ = 154° 3'.6 ; 6 = 0.094846 ;
7' = 163°10'.3; /3' = 7°43'.0; 6' = 0.029115.
46. a = 0.79797, 6 = 0.46731, jS = 23° 19'.6.
.-. a= 42°32'.5; 7 = 114° 7'.9 ; r.= 1.07705;
a' = 137°27'.5; 7'= 19° 12'.9 ; c' = 0.38841.
46. a = 37.456, 6 = 43.987, c = 13.498.
. •. i a = 26° 31'.0 ; ^ jS = 55° 7'.0 ; i 7 = 8° 22'.0.
47. a = 2.4568, 6 = 2.4743, c = 1.0047.
.-. ia = 38°38'.0; i^ = 39°36'.7; i7 = ll°45'.3.
48. a = 47.474, 6=100.980, c = 93.929.
.-. ia = 13°56'.8; |^ = 42°10'.2; i7 = 33°53'.0.
49. a =14.567, 6 = 9.4769, c = 11.113.
.-. ia = 44°50'.9; i /3 = 20° 17'.5 ; i7 = 24°51'.5.
OBLIQUE PLANE TRIANGLES. 121
50. rt = 2.1476, 6 = 1.9397, c = 3.4345.
.-. ^0=17^22^8; i/3=15°29'.8; ^y = b7°T.3.
61. a = 115.03, b = 129.15, c = 112.06.
.-. ia = 28°12'.9; J /3 = 34° 39'.2 ; J 7 = 27° 7'.9.
52. b = 113.47, c = 227.79, a = 19° 43'.4.
.-. /3= 17°33'.8; 7 = 142°42'.8; a = 126.90 ;
or log tan x = 9.68278 ; a = 126.89.
53. a = 99.416, c = 90.432, /3 = ll°7'.8.
.-. a = 110°20'.4; 7 = 58°31'.8; 6 = 20.467;
or log tan x = 0.31110 ; b = 20.467.
54. a = 1.4342, b = 9.7672 ; 7 = 109° 19'. 0.
.-. o=7°31'.7; ^ = 63°8'.7; c = 10.330, or 10.331 ;
or log tan x = 9. 86498 ; c = 10.331.
55. a = 1003.7, b = 943.67, 7 = 101° 19'.8.
.-. a = 40°46'.9; i8 = 37°53'.3; c = 1506.7 ;
or log tan x = 1.39930 ; c = 1506.7.
56. a = 222.76, 6 = 444.38, 7 = 17° 17'.6.
.-. a = 15°57'.0; /S = 146°45'.4 ; c = 240.97;
or log tan x = 9.63029 ; c = 240.97.
67. a = 363.24, 6 = 146.18, 7 = 68° 14'.4.
.-. a = 88°2'.6 ; /3 = 23°43'.0 ; c = 337.55, or 337.56;
or log tan x = 0.07590 ; c = 337.55.
PART TWO.
SPHERICAL TBIGOJfOMETBY.
^-if^Oo
CHAPTER YIII.
DEFINITIONS AND CONSTRUCTIONS.
112. Spherical Trigonometry treats of the relations between
the face angles and the edge angles of a trihedral angle.
An edge angle is the angle between
two of the three planes forming the ^--^t^'^ rV
trihedral angle ; it is measured by ct<<^^^^^ --^^^^^ ?l A
the angle between the lines cut from >^>i5^^^ \ < j ry
the two planes by a plane perpen- ^^^^^'^^' J^
dicular to the edge in which the two b^^^-J^
planes intersect.
A face angle is the angle between two of the edges.
113. Representation of Trihedral Angles. — The relations
between the elements of a trihedral angle are discussed by
means of the spherical triangle formed by the intersections of
the faces with a sphere described with any radius about the
vertex as a center. The faces will cut arcs of great circles
from the surface of the sphere, their angular measures being
the same as those of the face angles; and the angles of the
spherical triangle will correspond to the edge angles, each
being measured by the angle between two lines lying in the
planes of the faces and perpendicular to the line of intersection
of the faces.
1^
124 SPHERICAL TRIGONOMETRY.
Hence, in the spherical triangle the sides correspond to the
face angles, and the angles to the edge angles of the trihedral
angle.
The lengths of the sides in linear measure will depend upon
the radius of the sphere, and are computed, when the radius is
known, by the proportion
360°:a = 2Trr:«, (1)
where a is the number of degrees in the arc, and I is its length.
114. Limitation of Values. — We shall consider only those
triangles in which each element is less than 180°. In the gen-
eral spherical triangle the sides and angles may have values
greater than 180°, but in such a case it is always possible to
substitute for the triangle, in the computations, another in
which each element shall be less than 180°.
115. Definitions and Relations. — A great circle is cut from
the surface of a sphere by a plane passing through its center;
its radius is equal to the radius of the sphere.
A %mall circle is cat from the surface by a plane not passing
through the center; its radius is always less than the radius of
the sphere.
Two planes passing through the center will intersect in a
diameter of the sphere, and the two corresponding great circles
will intersect at the ends of this diame-
ter. Hence any two great circles will
intersect at two points 180° apart.
To describe a great circle on a
sphere, separate the points of a pair
of compasses by a distance equal to
the chord of 90°, or rV2, and describe
an arc about any point. If any other
^ _ distance is used, a small circle will be
Fig. 92.
described. The point used as the
center is called the fole of the great circle; its distance from
all points on the great circle is evidently 90°.
Any great circle passing through the pole of another great
circle will be perpendicular to that great circle
DEFINITIONS AND CONSTRUCTIONS. 125
Any two great circles drawn perpendicular to a third great
circle will intersect in its pole.
A great circle perpendicular to two great circles will pass
through the poles of both, and its plane will be perpendicular
to the diameter joining the points of intersection of the two
great circles.
The angle between two arcs of great circles is measured
by the arc of a great circle described about the vertex as a
pole, and limited by the sides, produced if necessary
The shortest distance between two points on a sphere is the
arc of the great circle passing through the points.
116. Constructions. — To find the pole of a given great cir-
cle: from any two points on the circle as poles, describe arcs of
great circles, and their intersection will be the point required.
To draw a great circle through two points: find the pole as
before, and describe the great circle.
To draw a great circle through a given point perpendicular
to a given great circle: from the point as a pole describe an
arc of a great circle; its point of intersection with the given
circle will be the pole of the required circle. Or, find the pole
of the given circle, and then draw the great circle through this
pole and the given point.
To cut from a great circle an arc n° long: separate the
points of the compasses by a distance equal to the chord of
n^^ or 2 r sin ^ 7i°, place the points on the great circle, and the
arc intercepted will be the one re-
quired.
To construct a great circle pass-
ing through a given point and mak-
ing a given angle with a given great
circle: in Fig. 93, let ACB be the
given great circle,* P its pole, F the
given point, and a the given angle.
With P as a pole, draw the small
circle P'P" such that the angular FilTaT
* The planes of the great circles ACB and CF, and of the small circle
P'P"^ are perpendicular to the paper.
126 SPHERICAL TRIGONOMETRY.
distance PP' = a; then the pole of the required great circle
must be on this small circle. With jP as a pole, describe an
arc of a great circle cutting the small circle P'P" in two points;
these points will be the poles of two great circles through F,
both of which satisfy the given conditions. Only the great
circle OF, whose pole is P', is shown in the figure.
To construct a great circle making a given angle with a
given great circle, the point of intersection being given: from
the given point as a pole describe a great circle, lay off on it
from the given circle a distance equal in angular measure to
the given angle, and pass a great circle through the point thus
found and the given point of intersection.
117. Definitions. — A right spherical triangle is one which
has one angle equal to 90°; a birectangular triangle lias two
angles each equal to 90°; a trirectangular triangle has three
angles each equal to 90°.
A quadrantal triangle has one side equal to a quadrant, or
90°; a biquadrantal triangle has two sides each equal to a
quadrant; a triquadrantal triangle has three sides each equal
to a quadrant.
A birectangular triangle is also biquadrantal, and a tri-
rectangular triangle is also triquadrantal; and vice versd.
118. The Polar Triangle of any triangle is constructed by
describing arcs of great cireles about the vertices of the origi-
nal triangle as poles. Thus, about A,
B, apd C as poles, describe the arcs
B' C\ A!'Q\ and A! B\ respectively; that
triangle is called the polar in which
the vertices. ^ and A! , B and B\ C and
C are on th'^ same side of BO, AO, and
AB, respectively.
The vertices of the polar triangle
will be the poles of the sides of the
original triangle, so that either triangle will be the polar of
the other.
The sides of a triangle are the supplements of the opposite
angles of the polar, and the angles are the supplements of the
DEFINITIONS AND CONSTRUCTIONS.
127
Fio. 95.
opposite sides of the polar; a'= 180°— a, aJ — 180"— a. Thus,
if the angles of a triangle be 120°,
80°, and 60°, the opposite sides of
the polar will be 60°, 100°, and 120°.
The polar of a quadrantal tri-
angle is a right triangle, the angle
in the polar opposite the quadrant
being equal to the supplement of
90°; the polar of a biquadrantal tri-
angle is birectangular; the polar of
a triquadrantal triangle is trirectan-
gular; and vice versd.
The triquadrantal triangle is its own polar, each vertex
being the pole of the opposite side.
119. In Any Spherical Triangle :
(1) Each side must be an arc of a great circle.
(2) Each side must be less than the sum of the other two.
(3) The greater side is opposite the greater angle, and
conversely. Equal sides are opposite equal angles.
(4) The sum of the sides must be less than 360°.
(5) The sum of the angles must be greater than 180° and
(ess than 540°.
120. Construction of Triangles. — (1) Given the three sides,
a, 5, c. — Draw an arc of a great circle and lay off on it an arc
equal to one of the sides, as a. From the extremities of this
arc as poles, with radii equal to the chords of h and c respec-
tively,, describe arcs of small circles with the compasses, and find
their point of intersection. Join this point and the extremities
of a by arcs of great circles, and the triangle will be constructed.
(2) Given the three angles, a, /3, 7. — Find the sides of the
polar triangle, construct it, and then construct the given tri-
angle by using the vertices of the polar as poles.
(3) Given two sides and the included angle, a, 5, 7. — Draw
an arc of a great circle, and lay off on it an arc equal to one
of the sides, as a. Pass an arc of a great circle through one
extremity of a, making the angle 7 with a, and lay off on it an
arc equal to h. Join the extremities of a and h by an arc of a
great circle, and the triangle will be constructed.
v'^V^^'
ouw^^^
s^
t>f
CAi \rcS
128
SPHERICAL TRIGONOMETRY.
(4) Given two angles and their included side, a, /3, o. — In
the polar we know two sides and the included angle, and hence
we can construct it by the method just given. Having the
polar, we can then construct the required triangle.
Or, draw a great circle and lay off on it an arc equal to e ;
at the extremities of this arc, construct arcs of great circles
making the angles a and y3 with c ; their point of intersection
will be the third vertex.
(5) Given two sides and the angie opposite one of them,
a, 5, a. — Draw any great circle ADA\ and through any point
on it, as A^ draw a great circle
making the angle DA C= a with
it. On this circle l:iy off from
A the distance AC =h. With
-^ ^ ^ ^ ^6^ as a pole, describe a small
Fig. 96. • ^ ^ ^
Circle whose radius is equal to
the chord of a, using the compasses ; pass arcs of great circles
through C^ and the points B and B' where this small circle
intersects the first great circle ADA'^ and the triangle will be
constructed.
There will be, in general, two points of intersection, and
there may therefore be two triangles that will satisfy the con-
ditions of the problem. Only those triangles can be taken in
which each side is less than 180°, i.e. both B and B' must lie
on the arc ADA' between A and A\ these points being 180°
apart.
If a is acute, as in Fig. 96, a must be greater than p and
less than the shorter of the two
distances CA and CA' (h and
180° - 6) in order that there
may be two solutions.
If a is obtuse, as in Fig. 97,
CD' is the least and CD the
greatest distance of C from
ADA'D', DCD' being perpen-
dicular to ADA'D'. Therefore
a must be less than jt?, in ordei
that the small circle may cui
■piQ, 97. ADA'D' ; a must also be greatei
DEFINITIONS AND CONSTRUCTIONS. 129
than the longer of the two distances CA and CA' (6 and
180° — 6) in order that the two points of intersection may fall
on the arc ABA'.
The conditions, therefore, for two solutions are :
a acute : a > p^ a < b^ a < 180° — b.
a obtuse : a b^ a> 180° — b.
Or, a must be intermediate in value between p and both b and
180° - b.
If a is intermediate in value only between p and either b or
180° — 6, there will be one solution.
If a is not intermediate in value between p and either b or
180° — 6, no solution will be possible, but ii p = a, there will
be one solution — a right triangle.
(6) Given -two angles and the side opposite one of them,
«, /3, a. — In the polar triangle we know two sides and the
angle opposite one of them, and we can construct it ; having
the polar we can construct the required triangle.
As the polar triangle may admit of two solutions, there may
be two solutions of the problem.
CROCK. TRIG. — 9
CHAPTER IX.
GENERAL FORMULAS.
Fia. 98.
121. The Cosine of Any Side of a Spherical Triangle is equal
to the product of the cosines of the other two sides, increased hy the
product of the sines of these two sides
multiplied hy the cosine of their included
angle. — Let the phxne BAC be per-
pendicular to OA at any point A, and
let J. (7, BC, and BA be its intersections
with the faces of the trihedral angle.
Then BA C — a, and AB and A are
perpendicular to OA, i.e. OAB and OAO are triangles right-
angled at A.
In the triangle BA we have
BC^ = AB^ -{- AC^- 2 AB' AC cos a.
In the triangle BOO,
BC'^= 0&-{- OC^-I OB' 00 cos a.
Equating the values of BC^, and transposing,
2 OB ' O0cosa = OB^ - AB^ + 00^ - AO^ + 2AB' AOcosa.
In the right triangles OAB and OAO,
OB"- - AB"^ = 0A\ and 00^ - AO^ = OA^.
.'. 2 OB' 00 cos a = OA^ + OA'^ + 2 AB ■ AC cos a;
OB ' 00 cos a = OA^ + AB ■ AC cos a.
or
or
cos a
cos a
OA
00
OA AC
OB 00
AB
OB
cos a
cos b cos c + sin 6 sin c cos a.
130
(1)
GENERAL FORMULAS.
131
In this proof b and c are assumed to be less than 90°, while a and a may
have any values less than 180°. The formula is true, however, when either b
or c, or both b and c, exceed 90°.
If, in the triangle represented by the full lines (Fig. 99), b is greater than
90°, then in the dotted triangle formed by completing the arcs of great circles,
the two sides are 180° — 6, and c, both less than 90°, and the other side and its
opposite angle are 180° — a, and 180° — a. Hence we can apply (1) to the
dotted triangle, giving
Fkj. 100.
cos (180° - a)= cos (180° - b) cose + sin (180° - 6) sine cos (180° - o).
.-. — cosa = — cos6 cose — sin6 sinecoso.
.*. cos a = cos 6 cose + sin & sine cos a. q.e.d.
If both b and c are greater than 90°, as in Fig. 100, then in the dotted
triangle the two sides are 180° — b, and 180° — c, and the other side and its
opposite angle are a and a.
.-. cos a = cos (180° - b) cos (180° - c) + sin (180° - b) sin (180° - e) cos a
= ( — cos &) ( — cos c) + sin b sin e cos a.
.-. cos a = cos 6 cos c -f sin 6 sin c cos o. q.e.d.
Therefore the formula is always true when each of the elements of the
triangle is less than 180°.
No assumption, then, has been made concerning any element
that is not true for all the others. We may therefore change
any angle to another, as a to yS, if at the same time we change
the sides opposite, as a to 6, making also the reverse changes,
5 to a and ^ to a, in the formula ; for this is equivalent to
changing the names arbitrarily assigned to the sides and angles.
Thus, to permute (1) to find cos , we change a to , « to 7, e to
a, and 7 to a, if they occur in the formula, while b and yS will
not be affected.
. •. cos c = cos b cos a H- sin 5 sin a cos 7.
If we assume that our triangle is right-angled, 7 being
equal to 90°, we can permute between a and 6, and d and yS,
since no assumption is made concerning a and a that is not
equally true concerning b and y8. But we cannot permute
132
SPHERICAL TRIGONOMETRY.
between a and 7, because 7 is assumed to be equal to 90°, while
no such assumption is made concerning a.
Permuting (1) in the oblique-angled triangle, we have
cos a = COS 6 cos c + sin 6 sin c cos a,
cos h — cos a cos c + sin a sin c cos P,
cos c = cos a cos 6 + sin a sin b cos 7.
(2)
>
Eq. (1) is called the fundamental equation of spherical trig-
onometry, since all the other formulas may be derived from it.
122. The Cosine of Any Angle of a Spherical Triangle is equal
to the product of the sines of the other two angles multiplied by the
cosine of their included side, diminished
by the product of the cosines of the qther
two angles. — We have
cos a = cos b cos c -\- sin b sin c cos a. (1)
Since the angles of the polar triangle
are the supplements of the sides oppo-
site in the original triangle, and vice
versd, we have
a = 180° -a', b = 180° - yS', c^ 180° - 7', a = 180° - a'.
Substituting in (1),
cos (180° - a') = cos (180° - yS') cos (180° - 7')
+ sin (180° - 13') sin (180° - 7') cos (180° - a'),
(— cosy8')(— cos 7') 4- sinyS' sin 7' (~ cos a').
— cos yS' cos 7' -1- sin /3' sin Q = (9^ sin b sin 7.
sin yS _ sin 7
sin
sin /9 _ sin 7
sin b sin c
Fig. 103.
Permuting,
sin 5
sin«
sin a
O)
124. Additional Formulas. — We have
cos b = cos a cos c + sin a sin cos /S.
sin a sin c? cos /S = cos 6 — (cos b cos -f sin b sin c cos «) cos
= cos b — cos 6 cos^ c — sin 5 sin c cos c cos a
= cos b sin^ c — sin b sin cos c cos a.
.•. sin a cos p = cos 6 sin c - sin ft cos c cos a, (1)
Applying (1) to the polar triangle and dropping the primes,
sin a cos 6 = cos p sin 7 + sin p cos 7 cos a, (2)
Dividing (1), member for member, by the equation
sin a sin yS = sin b sin «,
we have
COtyS
cot b sin c — cos c cos a
sin a
.*. sin a cot p = cot 6 sin c - cos c cos a.
Transposing,
sin c cot b — sin « cot yS + cos c cos a.
Permuting,
sin a cot 6 = cot p sin v 4- cos a cos -y.
Other formulas may be found by permuting (1), (2), and
(3). Among these are the following :
from (1), sin a cos 7 = cos c sin b — sin c cos b cos «; (5)
from (3), sin a cot 7 = cot c sin 5 — cos 5 cos a, (6)
and sin 7 cot a = cot a sin b — cos 6 cos 7. (7)
(3)
(4)
CHAPTER X.
RIGHT SPHERICAL TRIANGLES.
125. Formulas for Right Spherical Triangles. — The follow-
ing equations have been shown in Chap. IX to be true for
all spherical triangles :
cos c = cos a cos h -f sin a sin h cos 7, (a)
cos a = — cos /3 cos 7 + sin y8 sin 7 cos «, (]))
cos 7 = — cos a cos ^S + sin a sin /S cos c, ()
sin a sin 5 sine ^7^
-: — = — — -z — — — 1 \.^)
sin a sinyS sin 7
sin a "(308 7 = cose sin 6 — sine cos 6 cos a, (^)
sin 7 cot a = cot a sin b — cos 6 cos 7. (/)
By making 7 = 90° we get seven formulas applicable to right
triangles, and by permuting these three others are found.
From (a),
cos c = cos a cos 6. '
^
(1)
From (c),
cos c = cot a CQt p.
(2)
From (^),
cos « = sin yS cos a.
cos yS = sin « cos h. .
(3)
Permuting,
From (e),
cos a = tan h cot c.
cos y8 = tan a cot c. .
1
(4)
Permuting,
From ((^),
sin a = sin sin a.
i
(5)
From ((?),
sin 5 = sin c sin /3. .
^''''■'
From (/),
sin b = tan a col a.
sin a = tan b cot yQ. .
(6)
Permuting,
135
136
SPHERICAL TRIGONOMETRY.
126. Formulas for Right Spherical Triangles. Geometrical
Proof. — Let OB be unity. From B pass the plane BAC per-
pendicular to OA, Then AB and A Q are perpendicular to OA,
and OB is perpendicular to 00.
.'. OB = iima, 00= cos a, AB = smc,
OA = cos c, OAB = a.
.*. AO = OB cot a = sin a cot a, (a)
AO = AB cos a = sin c cos a, (6)
AO = 00 sin h = cos a sinh^ ■ ()
^ (7 = OJ. tan 5 = cos tf tan h. (d)
^Equating these values of A 0, we obtain the following for-
mulas :
From (a) and (^), sin a = sin c sin a.
Permuting, sin i = sin c sin /S.
From (a) and (t-), sin 5 = tan
Permuting, sin a = tan
a cot a. 1
h cot yS. 1
sin a
From (a) and (^?), cos c = cot a ;
tan 6
cos c = cot a cot /3.
sin 5
.-. from (6),
From (i) and ((?),
cos a = cos a
sni d?
.-. from the sine proportion or from (5)
cos a = cos a sin y8. "[
cos yS = cos b sin a. J
Permuting,
From (^) and (rZ),
Permuting,
From ((7) and (cZ),
cos a = tan 5 cot c.
cos /3 = tan a cot e.
cos (? = cos a cos 5.
(5)
(6)
(2)
(3)
0)
127. Napier's Rules. — Napier, the celebrated Scotch mathe-
matician, devised two rules by which the ten formulas connect-
ing the elements of a right spherical triangle may be easily
written.
\^
RIGHT SPHERICAL TRIANGLES.
137
Fig. 106.
He called the sides a and h about the right angle, and the
complements of the two oblique angles
and of the hypotenuse, the parts of the
triangle, not considering the right angle
as a part ; the parts, then, are a, 5,
90° - c, 90° - «, 90° - /9, which we
shall call a, 5, c\ «', /S'. By reference
to the circular figure, in which the parts
are arranged in their order in going
around the triangle, it will be seen that
if any three parts are considered, either
one will lie between the two others,
being adjacent to both, or one will be
separated from the other two by inter-
mediate parts. Thus h lies between a'
and a, being adjacent to both, and /3^ is
separated from both a' and h. The part
which lies between two others adjacent
to it or is separated from both the others
by intervening parts, is called the middle part; the two others,
if adjacent to it, are called adjacent parts^ and if separated,
opposite parts. Thus, if c\ a'<, and h are considered, a' is the
middle part, and c' and b are the adjacent parts ; if c\ ff^ and b
are considered, b is the middle part, and c' and ^' are the oppo-
site parts.
Napier's rules are :
1. The sine of the middle part is equal to the product of the
tangents of the adjacent parts.
2. The sine of the middle part is equal to the product of the
cosines of the opposite parts.
The rules may be easily remembered by the a in the words
tangent and adjacent and the o in cosine and opposite.
If, in a right triangle, any two elements besides the right
angle are given, the other elements may always be expressed in
terms of these two by Napier's rules. Thus, let the given ele-
ments be a and c.
(1) To find a ; of the three parts a, ', and «', a is the
middle part, and c' and a' are the opposite parts.
138
SPHERICAL TRIGONOMETRY
. •. sin a = cos c^ cos a' = cos (90° — c) cos (90° — «) = sin c sin «.
sin a
.*. sina =
sine
(2) To find yS ; of the three parts «, c\ and /3', /8' is the
middle part, and a and ' are the adjacent parts.
. •• sin yS' = tan a tan c'.
. • . sin (90° - yS) = tan a tan (90° - c) .
.-. cosyS = tana cote.
(3) To find h ; of the three parts a, c\ and 5, ' is the
middle part, and a and h are the opposite parts.
.-. sin e' = cos a cos 6.
. •. sin (90° — c) = cos a cos h.
.'. cos c = cos a cos 6.
cose
.-. COS 5 =
COS a
128. Species. — Two angular quantities are said to be of
the same species when both are less or both greater than 90°,
i.e. when they are in the same quadrant ; and of different
species when they are in different quadrants.
129. Rules for Species in Right Spherical Triangles.
(1) An oblique angle and its opposite side are always of the
same species. From Napier's rules,
sin h = tan a cot a.
But sin h is always positive, and therefore tan a and cot a must
have the same sign ; if they are both positive a and a will be in
the first quadrant, and if both are nega-
tive a and a will be in the second quad-
rant.
(2) If the hypotenuse is less than
90°, the tivo oblique angles (and therefore
the two sides') are of the saine species;
if it is greater than 90°, the two angles
(and therefore the two sides) are of dif-
ferent species. From Napier's rules,
cos c = cot a cot yS = cos a cos b.
RIGHT SPHERICAL TRIANGLES. 139
If c is less than 90° its cosine will be positive ; cot a and cot/8
must therefore have the same sign, and hence a and y8 must be
in the same quadrant. If c is greater than 90° its cosine will
be negative ; cot a and cot fi must therefore have different
signs, and hence a and /9 must be in different quadrants.
Thus, if a = 40° and ^ = 60°, a and b must be, from the
lirst rule, in the first quadrant ; and, since a and ff (or a
and h') are in the same quadrant, c must be, from the second
rule, in the first quadrant. If a = 70° and c = 110°, from the
second rule we see that yS must be in the second quadrant, and
from the first rule that a is in the first and h in the second
quadrant.
130. Solution of Right Spherical Triangles. — There are six
possible cases, all of which may be solved by Napier's rules :
I. Given the hypotenuse and an angle.
II. Given the hypotenuse and a side.
III. Given the two angles.
IV. Given the two sides.
V. Given an angle and the adjacent side.
VI. Given an angle and the opposite side.
The required elements should always be determined directly
from the given elements.
First write the three formulas, each containing the two
given elements and one required element ; arrange the three
formulas for logarithmic computation ; and then write the
values of the functions in their proper places, being very care-
ful about writing n after the logarithms of the negative func-
tions. If the number of negative factors is even, the result
will be positive ; if it is odd, the result will be negative and
n should be written after the resulting logarithm.
If the sine of a quantity is found by the computation to
be positive, the quantity may be either in the first or in the
second quadrant, the proper quadrant being determined by the
rules for species ; if the sine is negative the triangle is impos-
sible, since the elements of the triangle are each less than 180°.
If the cosine, tangent, or cotangent is found to be positive, the
quantity lies in the first quadrant ; if negative, the quantity
lies in the second quadrant.
140
SPHERICAL TRIGONOMETRY.
A check formula in each case is found by applying Napier's
rules to the three unknown elements ; thus, if a and h are
given, a, /S, and c will be computed, and the check formula is
cos e = cot a cot ff,
131. Case I. Given the Hypotenuse and an Angle.
1. c? = 129M4'.6,
43° 15'. 7.
To find ff ; cos c = cot a cot ^ ;
.-. cotyS
cose
cot a
To find b ; cos a = tan b cot c ;
.-. tan 6
cos a
cot c
To find a ; sin a = sin a sin c.
Check : sin a = tan b cot /S.
logcosc = 9.80114/1
logcot a = 0.02637
logcot/3 = 9.77477 ?t
.-. /3 = 120° 46'.03
log cos a = 9.86227
logcotc = 9.91214 n
log tan & = 9.95013 n
b = 138° 16'.96
log sin o = 9.83590
+ log sine = 9.88900
log sin rt = 9.72490
a =32° 3'. 4
By the rules for species j8 must be in the second, a in the first, and b in the
second quadrant.
\y 2. c = 110'
/3 = 48°28'.6.
. a = 118°46M; 6 = 44° 42'.
111°7'.2.
132. Case II. Given the Hypotenuse and a Side.
V 1. c= 75° 0'.4, a = 32°56'. .-. b= 72° 2'.8 ; a = 34° 15'.0 ; /3 = 80° 0'.6.
.r^. c = 100°12', 6 = 40°30'.3. .-. a = 103°28M ; a = 08° 50'.5 ; ^3 = 41° 17'.7.
133. Case III. Given the Two Angles.
1. a= 30°51'.2, /3= 71° 36'. .-. a = 25"12'.8;6= 52° 0'.75;c = 56° 9'.6.
2. = 130° 20', /3 = 100°10'.9. .-. a = 131° 7'.0 ; 6 = 103°24'.5^ c=8in3'.7.
134. Case IV. Given the Two Sides.
1. a= 43° 20', 6 = 74° 13'. .-. c = 78° 35'.3 ; a = 44° 26'.0 ; iS = 79° 1'.4.
2. a = 100°, 6 = 98° 20. .-. c = 88° 33'.5 ; a=99°53'.8; /3 = 98° 12'.5.
135. Case V. Given One Side and the Adjacent Angle.
1. 6 = 66° 29', a = 50° 17'.
2. a = 24° 41', /3= 140° 34'. 7.
a=47°40'.5; c= 74° 27'.6; /3 = 72° 7'.5.
b = 161°3'.2
149°15'.0: a = 54°45'.6.
RIGHT SPHERICAL TRIANGLES.
141
136. Case VI. Given One Side and the Angle Opposite.
Let a and a be the given elements.
To find b ; sinh = tan a cot a.
To find c ; sina = sin (? sina ; ' .*. sine
sma
To find /3 ; cos « = cos asin/S ; . *. sin yS =
sma
cos a
cos a
All three quantities are found by their sines ; each of the
three quantities may therefore be in either the first or the
second quadrant, and there will be two solutions.*
If P is the pole of AC, PC will be perpendicular to AG^
and the distances BO will be equal if m
is equal to n. Either of the triangles
AB will satisfy the conditions of the
problem, being right-angled at O and
having A = a and B0= a.
The rules for species must be applied
in determining which values belong to
each solution. Thus, if a is greater
than 90°, and /Q in the first solution is
less than 90°, c must be greater than
90° in that triangle ; in the second
triangle, where y3 is greater than 90°, c must be less than 90°.
Each side, of course, is of the same species as its opposite
angle. These results may be written :
First solution : « > 90° ; /3 < 90°; > 90° ; a > 90° ; h< 90°.
Second solution : « > 90° ; yS > 90° ; c<90°; a > 90° ; b> 90°.
1. a = 160^ 12'.2, a = 150° 37'.
log sin a = 9.52979
log sin a = 9.69077
log sin c
Fig. 109.
log tan a = 9.55625 n
+ log cot a = 0.24942 n
log sin b = 9.80567
b= 39''44M
b' = 140° 15'.9
log cos a = 9.94020 n
log cos a = 9.97354/1
9.83902
c'= 48°39M
c = 136° 20' .9
log
sin|3 =
9.96666
/3 =
67° 50'
2
/3' =
112° 9'
8
b = 40° 50', /3 = 62° 14'. .: a = 27° 3'.9 ; a = 38° 0'.4 ; c = 47° 38'.6 ;
or a' = 152° 56'. 1 ; a' = 141° 59'.6 ; c' = 132°21'.4.
• The triangle is supposed to be possible. The two solutions are identical
when a = o.
142
SPHERICAL TRIGONOMETRY.
137. Special Cases.
1. C = 90°, a = 90^
2. c = 90°, a = 90^.
3. a = 90°, /3 = 90°.
4. « = 90°, & = 90°.
6. a = 90°, i3 = 90°.
6. a = 20°, a = 20°.
. a = 90° ; b and /3 indeterminate.
. a = 90° ; b and /3 indeterminate.
•. rt := 90° ; b = 90° ; c = 90°.
. c = 90° ; a = 90° ; /3 = 90°.
. c = 90° ; 6 = 90° ; a = 90°.
■. c = 90° ; & = 90° ; /3 = 90°.
138. Additional Examples.
1. rt = 40°42'.4, c = 63°20'.
.-. 6=53°41'.9; a = 46° 52'.25 ; /3 = 64°24'.0.
2. a = 70°15'.5, a = 81°42'.7.
.: b= 23°57'.0; ^= 25° 15'.7 ; c = 72° 1'.25 ;
or &' = 156° 3'.0 ; ^' = 154° 44'.3 ; c' = 107° 58'.75.
3. 6 = 30°32'.4, a = 36° 44'.
.-. a=20°46'.0; c = 36°21'.6; /3 = 58°59'.7.
4. c = 72° 10', a = 30° 43'.
.'. a = 29° 5'.0 ; & = 69° 29'.0 ; j8 = 79° 41'.25.
6. = 106° 34- .2, /3 = 33°11'.7.
.-. a = 121°23'.6; 6 = 29° ll'.O ; c = 117° 3'.0.
6. a = 28° 47', & = 110°27'.3.
.-. c = 107°50'.2; a = 30° 23'.1 ; /3 = 100° 10'.9.
7. c = 54°12'.2, )S = 164°50'.4.
.-. a = 99°0'.3; 6 = 167°45'.2; a = 126°45'.9.
8. a = 40° 8', i3 = 74°30'.2.
.-. 6 = 66°43'.5; c = 72°25'.0; a = 42°32'.7.
W 9. c = 102°30', a = 125°13'.4.
.-. « = 127°8'.l; 6 = 68°49'.0; /3=72°49'.8.
10. a = 40°42'.4, ^ = 67°51'.0.
.-. a = 35°4'.4; ?> = 54°42'.0; c = 61°46'.6.
11. 6 = 163°14'.2, c = 112°41'.8.
. •. rt = m° 14'.1 ; a = 82° 45'.75 ; /3 = 161° 46'.9.
12. a = 120° 30'.2, b = 140° 12'.
.-. c = 67°2'.8; a = 110° 39'.7 ; /3 = 135° 57'.7.
13. c = 50° 20'.2, /3 = 101° 29'.4.
.-. rt = 166°29'.5; ?> = 131° 1'.7 ; a = 162''20'.1.
14. a = 82°4'.4, /3 = 8°22'.3.
.-. a = 18°42'.2: c = 18° 53'.25 ; Zy = 2° 43' or 2° 44'.
RIGHT SPHERICAL TRIANGLES.
143
15. a = 130°40'.7, c= 76° 31 '.6.
.-. 6 = 112°33'.0; a = 128°26'.6; /3 = 107" 28'. 76.
16. 6 = 10°10'.2, /3= 16°40'.6.
.: a= 39° 43'.9 ; c = 40° 48'. 1 ; o = 78° 0'.7 ;
or a' = 140° 16M ; c' = 139° 11'.9 ; a' = 101° 59'.3.
17. 6 = 67°8'.3, a = 104°16'.2.
.-. a = 106° 50'.8 ; c = 99° 2'.8 ; ^ = 58° 16'.4.
18. a = 20° 64', b - 6^° 26'.7.
.-. c = m° 14'. 1 ; a = 22° 66'.5 ; /3 = 80° 19'. 2.
139. Isosceles Triangles. — If an arc of a great circle be
drawn from the vertex perpendicular to the base, it will bisect
both the base and the angle at the ver-
tex, dividing the triangle into two equal
right triangles that may be solved by
Napier's rules.
1. a = 110°47'.3, y8=92°14'.6.
.-. -l-y8 = 46°7^3.
To find a: cosa= cot « cot JyS ;
, cos«
.-. cot a =
Fig. 110.
COt-l-yS
To find J h : sin \h = sin a sin ^ ff.
To find jt?: cos-J/S = tanj? cota; .-. tanjo = -^ —
Check : sin J5 = tanjt? cot a.
log cos a = 9.55013 n log sin a = 9.97076 log cos ^ /3 = 9.84081
log cot I /S = 9.98299 f log sin ^ /3 = 9.85783 - log cot a = 9.57936 n
log cot a = 9.56714 n log sin ^b = 9.82859 log tanp = 0.26145 n
110° 15'.54 I b = 42° 22'.1 p = 118° 42'.6
■ — • b = 84° 44'.2
2. a = 82° 26', /3 = 64° 42'.
3. & = 56°41', /3 = 112°44'.6.
.-. a= 77°53'.6; \b= 31°32'.75.
.-. a= 38°69'.6; a= 34°45'.6;
or a' = 141° 0'.4 ; a' = 146° 14'.4.
144
SPHERICAL TRIGONOMETRY.
140. Quadrantal Triangles. — The polar of a quadrantal tri-
angle is a right triangle whose angles
are the supplements of the sides, and
whose sides are the supplements of
the angles, of the original triangle.
^90° We may therefore solve the polar by
Napier's rules, and then find the ele-
ments of the original triangle by
taking the supplements of the ele-
ments of the polar.
A b'
Fig. 111.
1. ^=90°, a= 23°14'.7, b=: 27°14'.6.
.-. y = 90°, a' = 156°45'.3, yS' = 152° 45'.4
are the elements of the polar triangle.*
To find c' : cos jS' = tan a' cot c' ;
To find b' : sin a' = tan b' cot /3' ;
To find a' : cos a' = cos a' sin ff.
Check: cos a' = cot c' tan b'.
cot c'
tan b'
^ cos 13'
tan a'
_ sin a'
~cot/3'"
log cote' = 0.31594
log tan 6' = 9.30795 n
log cos a' ■= 9.62389 n
c'= 64°12'.8
6'=168°30'.8
a' = 114°52'.4
.-. 7 = 115°47'.2
.: p= 11°29'.2
.: a= 65° 7'.6
log cos /3' = 9.94894 n
log tan a' = 9.63300 n
log sin a' = 9.59623 log cos a' = 9.96324 n
log cot /3' = 0.28828 n + log sin /3' = 9.66065
c= 90°, 7
22'.7,
I = 150° 47
150°26'.2
90'
90^
90^
a = 121° 30', /3 = 112°16'.2.
.: b = 108°51'.l: -
94°43'.5; jS = 99°36'.6.
123°30'.75; y= 102° 4'.7.
a = 138°47'.8, 6 = 107°54'.9.
.-. a = 142°15'.2; ^ = 117°50'.25; 7 = 111° 40'. 1.
a = 112° 6'. 5, 7= 74° 30'.
. •. 6 =._ 56° 39'.6 ; a = 116°46'.4; /3 = 53°36'.9.
6. c = 90°, a = 83° 20'.6, /3 = 77° 14'.3.
.-. a= 83°30'.3; &= 77°19'.3; 7= 91°28'.0.
7. c= 90°, a= 94°22'.2, a = 108° 13'.3.
.-. b= 14° 6'.2; /3= 13°25'.3; 7= 72° 17'.5;
or 6' = 165° 53'.8 ; ^' = 166° 34'.7 ; 7' = 107° 42'.5.
* Note that a', /3', and c' are not the parts of the right triangle, but their
complements.
RIGHT SPHERICAL TRIANGLES.
145
141. Quadrantal Triangles may also be solved by the use of
Fig. 112, in which B^ one of the vertices adjacent to the
quadrantal side, is the pole of the great circle MDGN.
If the triangle has one side less than 90°, as BO in the tri-
angle ABC, produce that side to B.
In the triangle ACB, ABC =90",
BAC = 90° - «, ACB = 180° - 7,
AC=b, (7i>=90°-a, SindAB = l3
since AB = ABB. Therefore, if any
two elements of ACB besides the
quadrant are given, we know two
elements of the right triangle ACB
in addition to the right angle. Hence
we could solve it by Napier's rules,
thence obtaining the elements of ABC.
If one side of the triangle is greater than 90°, as in BUF,
then in the triangle GBF we have GF = a - 90°, GF = /3,
FF = 6, FGF = 90°, GFF =a- 90°, and GFE = 7. If any
two elements of BFF besides the quadrantal side are given,
we then know two elements of the triangle GFF in addition
to the right angle. Hence we could solve it by Napier's rules,
thence finding the elements of BFF.
Fig. 112.
CROCK. TRIG.
10
CHAPTER XI.
OBLIQUE SPHERICAL TRIANGLES.
142. To find an Angle, having given the Three Sides.
(a) cos a = cos b cos c
4- sin 5 sin cos a;
, „^^ cos a - cos b cos c
• • LOs CI — ; 9
sin 6 sin c
which may be solved by the use of the
natural functions.
(5)* To adapt (1) to logarithmic
computation, subtract each member from
unity,
-i -j cos g — cos 5 cos c _ sin b sine — cos a -f cos 6 cosg
sin b sin c sin b sin e
sin b sin c
Applying (4) of Art. 73,
cos u — COS v= —2 sin J (u + v) sin J (w — v)^
we have cos (b — c) — cos a = — 2 sin -J (5 — e + a) sin | (b
a)
since sin ( — a;) = — sin x.
Let a-\-b-\-c=^2s',
+ 2 sin J (a + 6 — f) sin J (a — 5 + ),
cos (6
a + 6-e = 2s-2c = 2(s-e);
a-6 + e=2s-25 = 2(s-^.).
) — cos a — 2 sin (s — e) sin (^s — b').
c)
Permuting,
. o 1 sin (s - b) sin {s
2l
sin^^p
8in«|Y =
sin 6 sin c
sin (s - a) sin (s
-c)
sin a sin c
sin (s - a) sin {s -
-b)
sin a sin 6
Compare with Art. 99.
146
(2)
OBLIQUE SPHERICAL TRIANGLES.
147
((?)* Add each member of (1) to unity.
cos a — cos h cos c
1 -h cos « = 1 +
sin h sin c
__ cos a — (cos h cos g — sin h sin g)
sin h sin g
o o 1 cos a — cos (h-\-c)
.'. 2cos2Ja = : — ; — H^ — ■ — ^•
sin sin g
Applying (4) of Art. 73, we have
cos a — cos (6 H-g)= —2 sin J (a -}- ^ + ) sin J(a — 6 — g)
= +2 sin-Ka + J + Osin J(5 + g — a).
Let a + b + c = 2s; .-. ^»4-g-a = 2s-2a=2(«-a).
.-. cosa — cos(6 + g) = 2sins sin(s — a).
Permuting,
^ Sin 6 sin c
cos^ip = ^'»^^^"<^^-^\
^ sin a sin c
f^a^i _ sing sin(8-c)
MIS nj — ; z 7 •
* sm a sm &
(3)
(c?) Dividing sin^ J '^ by cos^ ^ a, we have
Permuting,
We may write
. 2 1 Sin (s - b) sin (s - c)
2 sm « sin (s - a)
2 sm s sm (s - o)
tan« 1 ^ ^ sinCs-CT)sin(g-6) ^
2 ^ sin s sin (« - c)
(4)
Let
1 /sin (s — a') sin (s — 6) sin (s — g")
tan i a = -r— 7 ^\/ ^= ^^ ^ --
^ sin (s — a) ^ sin s
_ / sin (8 - a ) sin (s - 6) sin (s — c)
Permuting,
sins
••*^"«* = 8in(5-a)-
**"|P=sin(r-6)'
* Compare with Art, 99.
(5)
(6)
148
SPHERICAL TRIGONOMETRY.
Note. — The center of the inscribed circle of a spherical triangle is the
point of intersection of the arcs of great circles bisecting the angles of the
triangle. From this point draw the arcs
OZ, OM, and ON perpendicular to the
sides of the triangle.
.-. MA = AN\ NB = BL) CM=LC.
.'. MA + NB-h CM= s.
.: b + NB^s, since MA + CM = b.
M
Fig. 114. . •. NB = s - b.
In the right triangle OBN, by Napier's rules,
sin NB = tan O.Vcot NBO.
,-. tanO-V = -^^Hi^=sin(s-6)tan^)8
cotiV^O
= sin (s — b)
4
sin (s — a) sin (s — c)
sin s sin (s — b)
4
sin (s — a) sin (s — 6) sin (s — c)
sms
.-. tan ON=r.
Hence r is the tangent of the radius of the inscribed circle.
143. To find a Side, having given the Three Angles,
(a) cos a = — cos yS cos 7
+ sin /3 sin 7 cos a.
.•.cosa = ?^^^«5ii5^, (1)
sinpsln-y ^ ^
which may be solved by the use of the
natural functions.
^ (5)* To adapt (1) to logarithmic
computation, subtract each member from
unity.
- ^ cos a + cos 8 cos 7
.-. 1 — cosa = l -. — p5— T^-
sm p sin 7
_ cos y8 cos 7 — sin ff sin 7 + cos a
sin yS sin 7
o . 1 cos (/3 + 7) + cos a
.'. 2sm2Aa= ^^^ — y^.
^ sm p sin 7
Applying the equation (Art. 73)
cosw + cos V = 2 cos 1^(2* + t>)cos J(w — v"),
we have cos(y8+7)+cosa=2cos J (a + /3+7) cos J(y8 + 7 — '0
♦ Compare with Art.
OBLIQUE SPHERICAL TRIANGLES.
149
Let a-|-;9 + 7 = 2AS'; .-. /9 + 7 - « = 2aS' - 2a = ^(.S'- a).
. *. cos (/S + 7) + COS a = 2 cos JS cos (aS' — a).
Permuting,
^ sin p sin Y
gin2 15= -cosScos(5-p)
^ sin a sin y
sin'' - c = ~ *®'' '^ ^®* C-S^ - 7).
(2)
sin a sin P
(c?)* Add each member of (1) to unity.
^ ^ cos a + cos yS cos 7
.-. l + cosa = lH : — ri—'
sin p sin 7
_ cos a 4- cos /Q cos 7 + sin ^ sin 7
"" sin j3 sin 7
cos a -\- cos (yS — 7)
sin /3 sin 7
2cos^^a =
Applying the equation (Art. 73)
cos u + cos V = 2 cos 1^ (t^ 4- v) cos J (m — v),
we have cosa + cos(^ — 7) = 2 cos J(a + /Q — 7) cos J(a — /3 + 7).
Let a + yS + 7 = 2AS'; .-. a + /3 - 7 = 2aS'- 27 = 2 (aS^ - 7);
« - yg + 7 = 2aS'- 2/3 = 2(aS'- /3).
. •. cos a + COS (yS — 7) = 2 cos (^S — /3) cos (aS' — 7).
.. cos ^a- sinpsinv
•r> , . o 1 , COS (5> - a) cos (S - y)
Permuting, cos« ^ 6 = sinasinY '
«i cos (S - g) cos {S - P)
^^^^^= sinasinp
(cT) Dividing sin^Ja by cos^i-a, we have
-cosScos(S -g)
cos(«-P)cos(;S-y)'
(3)
tan»^a
Tj .• X oi^ - cos aS cos (*S - p)
Permuting, tan«-6 = c^c^S - g) cos (^ - y) ^
tan^i^^ -cos^^cos(^-y) ^
2 cos(-S'-g;cos(S-p) J
(4)
♦ Compare with Art. 99.
150 SPHERICAL TRIGONOMETRY.
We may write
tan i a = cos {S - a} J -cosS'
COS (S — a) cos (aS' — /3) cos (^S— 7)'
>'C0S(^-a)C0S(>S-P)C0SC^'-Y)* ^^
.•. tan I a = JB cos (S - a).
Permuting, tan 1 6 = U cos (5 - P), (6)
tan I c = iJ cos (S - 7).
Note. —Since the sum of the angles 2 S must be between 180° and 540°, S
must be between 90° and 270°, so that COSTS' is always negative and hence
— cos S is always positive.
Note. — The center of the circumscribed circle of a spherical triangle is
the point of intersection of the arcs of great circles perpendicular to the sides of
the triangle at their middle points.
.-. AN = NB; BL=LC; CM = MA.
.-. 0AM = OCM; OANz= OBN; OCL = OBL.
.'. 0AM + OAN-\- OCL= S,
.'. OCL = S-{OAM-\-OAN) = S-a.
In the right triangle OCL^ by Napier*s rules,
cos OCL = tan LCcot OC.
-.^ tanLC tania
.-. tan 0C = 7vFrF = ttt — r
cos OCL cos(/S^— a)
S-a)\i
COS /S cos (^ — a)
cos (S — a) \cos («S'— /3)cos (S —7)
4
- cosS
'cos (6'-a)cos (S-p) C9S (0
cot ^ y' ~ — cos ^ (a' + ^')'
Changing the signs and dropping the primes,
tan I (tt + p) cos l(a-b)
— ^-1 — = —i (4)
cot ^7 COS ^ (a + 6)
Eqs. (1), (2), (3), and (4) are called Napier's Analogies.
145. Gauss's Equations. — From (2) and (3), Art. 142, we
have
. - - -, sin (s — 6)^ /sin s sin (s — (?) sin (s — h') ,
sin i a cos i )8 = ^^ ^\ ■. ^. , ^ = ^ ^ cos i 7 ;
^ ^ sine ^ sin asm 6 sm^ ^
- .10 sin Cs — a')^ /sin s sin (s — ) sin(s — a) .
cos i a sin 1 /8= ^ ^\ : ^ — j-^= ^ ^cos i 7 ;
^ • ^ sin ^ sm a sin sin c ^
, , ^ sin s^ /sin (s — a) sin (s — 6) sin s . ,
cos i a cos A /S= ^ — \ ^ ■ ■ \ -=- — sin i- 7 ;
^ ^ s\nc ^ sin asm 6 sine ^
^_ sin (s — c) /sin (s — a) sin (s — 5)
~" sin (? ^ sin a sin b
sin (s — c) . ,
= ^^ sm 1 7.
sin (? "^
(1) sinK«+ /g) = ^^nr^'^'''''^^ ~ ^)+ sin {s - a)]
_ cos ^ 7 sin [g — j- (g + ^)] cos |(a — 6)
sin ^ c cos ^ tf
_ cos J 7 cos 1^ (a — J)
~~ QOS^C .
••. cos|csin|(a + p) =cos|7COs|(a-6). (1)
OBLIQUE SPHERICAL TRIANGLES. 163
(2) COS K« + /3) = ^^ [sin > - sin (»-<;)]
_ sin 1 7 cos (g — ^ g) sin j- c
"" sin ^ (? cos J g
__ sin J 7 cos ^ (« + ft)
~~ cos ^ g
.•. cos I c COS ^ (a + p) = sin |-y cos ^ (a + 6). (2)
(3) sin K« - ^) = ^^ [sin (s - b) - sin (« - a)]
_ cos 1 7 cos [8 — j^ (g + ^)] sin j (a — h')
"~ sin 1^ c cos ^ g
_ cos ^ 7 sin i(ct — h)
~" sin J (?
.•. sm^csiii|(a-p) = cos|7siii|(a-6). (3)
(4) cos^ (a — yS) = ^ . ^ ^ [sin s + sin (s — )]
_ sin J 7 sin (s — |^ (?) cos ^ c
sin ^ c cos I g
_ sin^7sin j (a + ^)
"~ sin^^c
.«. sin|ccos|(a-p) =sm^7siii|(a + 6). (4)
Eqs. (1), (2), (3), and (4) are known as Gauss's Equations^
or Delambre's Analogies.
146. Rules for Species in Oblique Spherical Triangles.
(1) If a side (^or angle') differs more than another side (^or
angle) from 90°, it is of the same species as its opposite angle
(or side).
We wish to show that cos a and cos a will have the same
sign when the difference between a and 90° is numerically
greater than the difference between b and 90°. In the formula
cos a — cos b cos c ^i v
cos a = : — ; — : (1)
sm 6 sni c
the denominator is always positive, so that the sign of the frac-
tion, and hence that of cos a, is the same as that of the numera-
154 SPHERICAL TRIGONOMETRY.
tor. But if a differs more than h from 90°, cos a is numerically
greater than cos 5, and hence greater than cos h cos c, since cos c
cannot exceed unity. Therefore the numerator has the same
sign as cos a; i.e. cos a and cos a have the same sign, so that a
and a are in the same quadrant.
By a similar process, using the formula
cos a -f cos B cos 7 ,«.
cos a = 7^ .^ 1, (2)
sinpsin7
we can show that, when a differs more than yS or 7 from 90°,
a and a are of the same species.
Since two sides will in general differ more than the third
from 90°, two angles will in general be of the same species as
their opposite sides. Thus, if a = 140°, h = 50°, and c = 110°,
we see that a and b differ more from 90° than c does; therefore
a will lie in the second, and /3 in the first quadrant, while the
quadrant of 7 is not determined by this rule.
2. Half the sum of two sides must be of the same species as
half the sum of the two opposite angles. — From (3), Art. 144,
tan i(a + ft) = tan J c ^^"f^'^"^^ - (3)
^ ^ ^ ^ cos J (« + p)
But c must be less than 180°; hence ^c must be less than 90°,
so that tan ^c is positive. Also, a — /3 must be numerically
less than 180°; hence J(« — /S) must be numerically less than
90°, so that cos|(a — /8) is always positive. Hence tan|(a + 6)
and cos^(«H-/3) must have the same sign. But a -{- b and
«-i-/S must each be less than 360°; hence ^(« + ft) and ^(« + yS)
must each be less than 180°, so that they must be in the same
quadrant in order that tan ^(a + 5) and cos J(a + yS) may have
the same sign. Thus, if |(a + yS) is in the first quadrant, its
cosine will be positive; the second member of (2) will be posi-
tive, and therefore J(a + 5) must be in the first quadrant.
If cos J(a + /3) is negative, tanj(a + ft) will be negative;
therefore J(« + yS) and ^(« + ft) must both be in the second
quadrant.
In the example under the first rule, after a and yS have been
computed, the quadrant in which 7 will lie may be determined
by the second rule.
OBLIQUE SPHERICAL TRIANGLES.
156
147. Solution of Oblique Spherical Triangles. — Any spherical
triangle may be solved by the use of the following formulas:
sin a _ sinb _ sinc^
sin a sin p sin y
**''2''=Sill(S-«)'
tan|a = Bcos(S-a)5 R=yj—^^-—^
/sin(s-
-a)sin(s -ft) sin(«-
-c)
r-^^
sins
„_J
— cohS
eosiS-^)co8iS-y)
a ^ ■'
2 ^ *^^
tan|c
~8in|(a + P)y
tan^(a + 6)
_COs|(a-P) ]
ten|c
~COs|(a + P)'
ten|(a-P)
sin|(a-6)-
C0t|7
~sin|(a + 6)
tan|(a + p)
cos|(a-5)
cotl-/
COS -(a + 6)
0)
(2)
(3)
(4)
(5)
(6)
(7)
iM<'
There are six possible cases:
I. Given the three sides.
II. Given the three angles.
III. Given two sides and the included angle.
IV. Given two angles and the included side.
V. Given two sides and the angle opposite one of them.
VI. Given two angles and the side opposite one of them.
148. Case I. Given the Three Sides
(a, bf c) . — Find the angles by the for-
mulas
_ /sin (s — a) sin (s — 5) sin (s — c')
tan J a =
sin«
r
sin (s — a)
Check by the sine proportion.
156
SPHERICAL TRIGONOMETRY.
1. Solve the triangle when a = 114° 43'.3, 6 = 136° 19'. 6, c = 43° 18'.5.
8 = 147° 10'. 7 col sin s = 0.26598 .-. log tan ha = 9.89910
5 - a = 32° 27'.4 log sin {s- a)= 9.72970 | o = 38° 24'.2
8-h= 10°51'.l log sin (s- 6)= 9.27478 log tan ^ /3 = 0.35402
s-c = 103° 52'.2 log sin (s - c) = 9.98714 ^ /3 = 66° 7'.6
2s = 294°21'.4 logr2 = 29.25760 -30 log tan ^ 7 = 9.64166
a check. logr= 9.62880 i7- 23°39 '.7
In finding log tan ^ a write log r on the margin of a slip of paper, place it
above log sin (s — a), and write the difference opposite log tan ^ o ; then find
log tan ^ j8 and log tan | 7 in the same manner.
2. a = 76°40'.4, 6 = 54°21'.3, c = 36°8'.7.
. •. I o = 60° 1'.8 ; ^ ^ = 23° 8'.6 ; i 7 = 15° 49'.3.
8. a = 124° 34'.9, b = 6G° 7'.2, c = 109° 43'.5.
.-. Aa = 60°l'.3; |/3 = 37°0'.8; ^7 = 49°6'.8.
4. a = 30°17'.6, 6 = 22°14'.4, c = 18°51'.8.
.-. |o = 47°55'.0; ^i3 = 24°8'.5; ^ 7 = 19° 48'.45.
6. a = 130°46'.0, b = 113°21'.4, c = 102° 16'.2.
•. ^a = 72°38'.0; ii3 = 68°9'.6; ^7 = 66°20'.5.
149. Case II. Given the Three Angles (a, p, -y). — Find the
sides by the formulas
B
-4
— GOSS
COS (S - «) COS iS - yS) cos CS-ryy
tan J a = J? cos (^S — a).
Check by the sine proportion.
b
Fig. 118.
1. Solve the triangle when a = 116° 19'.4, /3 =
: 83° 19'. 2, 7 = 106°10'.6.
/S'=152°54'.6
col (-COSTS') = 0.05047
.-. log tan ^ a = 0.23789
S-a= 36°35'.2
logcos(5'-a)=9.90469
^a = 59° 57'.7
S-p= 69°35'.4
log cos (>S'-^)= 9.54249
log tan i& =9.87569
S-y= 46°44'.0
log COS C^*- 7) =9-83594
J & = 36° 54'.6
3 .5= 305° 49'. 2
a check.
col i?2 = 9.33359
logi22 = 0.66641
log i2 = 0.33320
log tan ^ c = 0. 16914
Jc =55°63'.l
OBLIQUE SPHERICAL TRIANGLES. 157
In finding log tan J a, write log B on the margin of a slip of paper, place
it above log cos (/i? — a), and write the sum opposite log tan J a; then find
log tan J b and log tan J c in a similar manner.
2. a = 110° 36'. 4, /3 = 122° 8'. 7, y = 140° 20'.3.
.-. Ja = 4r66'.3; J6 = 57°57'.5; Jc = 68°39'.4.
3. a = 120°60'.6, /3 = 78° 6'. 1, 7= 81° 12'.3.
.-. Ja = 59°65'.2; i6=40°40M; Jc = 43°23'.4.
4. 0= 80° 20'. 2, /3= 73° 46'. 7, 7= 54° 8'. 5.
.-. Ja = 32°23'.6; J6 = 30°63'.7; Jc = 24°l'.7.
6. a = 100°61'.3, /3= 80°47'.6, 7= 74°3'.3.
.-. ia = 49°22'.4; J&=41°42'.5; ic = 37°41'.6.
150. Case III. Given Two Sides and the Included Angle
(6, c, a). — By permuting (6) and (7), Art. 147, we have
tanK^-7)=cotJ„?i^i|^, (1)
tanK^ + 7)=cotJ-«5^jl|^. (2)
Then /3 = K^ + 7)+K/8-7).
Note that the larger angle must be opposite the larger side.
To obtain a, we permute (4) and (5), Art. 147 :
tan^a^tanK^-0;i:fg!g. (3)
tan.a = tanK* + 0^^|ig±g. (4)
The agreement of the values oi ^a found from (3) and (4) is
a check upon the computation. The sine proportion may also
be used as a check.
Note. — In using these formulas, the larger side and the larger angle should
be written first in the expressions b — c and /3 — 7. Thus for c > 6, (1) would
be written
* 1 ^ ON * 1 sin i (c - 6)
tan i (7 - /3) = cot i a -^-i — — ix*
^^' ^^ ^ sm ^ (c + &)
Eq. (1) may be read : " The tangent of half the difference of the required
angles is equal to the cotangent of half the given angle, multiplied by the sine of
half the difference of the given sides, ancf divided by the sine of half their sum."
158
SPPIERICAL TRIGONOMETRY.
1. Solve the triangle when h = 105" 14'.8, c = 43° 17'.2, a = 112°
6=105° 14'. 8 logcot^a = 9.82251 log cot J a =
c= 43°17'.2 log sin i(6-c) =9.71159
col sin J (6 + c) =0.01658
J(6 + c)= 74°1C'.0
i(6-c)= 30°58'.8
Ja= 56°23'.7
log tan i (i3-7) =9.55068
K^-7)=19°33'.8
logtan|(&-c) =9.77843
log sin I (/3 + 7) =9.95566*
col sin ^(P-y) =0.47514*
log tan I a =0.20923
|a=58°17'.8
log tan 1(6+ c)= 0.55019
log cos ^ (^ + 7) =9.63322*
col cos I (/3-7) =0.02582*
logtan|a=0.20923
|a=58°17'.8
log cos J^ (6 — C):
col cos ^ (6 + C):
logtani(/3+7):
H/3 + 7):
K/3-7):
47'.4.
: 9. 82251
9.93316
:0.56677
: 0.32244
-64°
32'
.9
19°
33'
.8
84°
6'
.7
:44°
69'
.1
2. a = 103° 44'. 7, 6 = 64° 12'.3, y = 98° 33'. 8.
.-. K« + /S) = 82° 37'.0 ; K« - /S) = 16° 19'.0 ; a = 98° 56'.0 ;
/3 = 66°18'.0; ^c = 51°45'.3.
S. a = 156°12'.2, 6 = 112°48'.6, 7 = 76°32'.4.
.-. K« + ^)=120°45'.6; K« - /3)= 33° 18'.5 ; a = 154°4M;
)8 = 87°27'.l; Ac = 31° 54'. 4.
151. Case III. Second Method. Given 6, c, a, to find One
Element only.
(1) To find a only.
Let
cos a = cos b cos c + sin ^ sin c cos a.
w sin M= sin c cos a, '
m cos M= cose.
cos a = m (cos 5 cos ilff -f sin 5 sin JkT),
cos a — m cos (5 — i^f ).
(1)
(2)
(2) To find one angle only, /8 or 7. — From (6), Art. 124,
sin a cot 7 = cot ^ sin 6 — cos b cos a.
cot c sin b — cos ^ cos a cos sin ^ — cos b sin cos a
.'. cot 7;
sma
sm a sin c
♦ The functions of \{^ - 7) and of K/3+7) should be found by using the
fraction from which the decimal of a minute is found. Thus,
log sin 1 (^ + 7) = 9.95561 + f | x 6 = 9.95561 + 5 = 9.95666.
OBLIQUE SPHERICAL TRIANGLES.
159
Let
?n siniHf = sine
m cos M = cos c
cot 7
cos a, 1
_ m sin (b — M)
sin a sin c
(3)
(4)
The formula for cot yS may be found by permuting h and c
in (3) and (4).
1. 6 = 105°14'.8,
To find a.
log sine =9. 83611
log cos a= 9. 588 11 n
log (to sin i)/)
=9.42422 n
log sin Jlf = 9.53499 n
log cos ilf= 9. 97286
log cose = 9. 86209
log tan i¥ =9.56213 71
ilf=-20°2'.7
6 = 105° 14'.8
ft -itf = 125° 17^5
logcos(& — ilf)
=9.76173 n
log TO = 9.88923
log cos a =9. 65096 w
a=116°35'.7
c = 43°17'.2, o = 112°47'.4.
To find 7.
(1) log sin c =
(3) log cos o=
log(TOsinJJf)
(4)
(7) log sin ilf=
(8) log cos ilf =
(2) log cos c=
(5) log tan ilf =9.56213 n
(6)
(10)
(11)
(12)
(9)
(13)
(14)
(1)
(3)
(5)
(8)
(9)
(2)
(6)
Jf = -20°2'.7 (7)
6 = 105° 14'.8 (11)
6-3/ = 125°17^5 (12)
log sin ( 6 — ilf)
=9.91180
log TO =9. 88923
col sin a =0.03530
col sine = 0.1 6389
log cot 7 =0.00022
7=44° 59M
(13)
(10)
(4)
(14)
(15)
(16)
To find /3.
log sin 6 = 9.98444
log cos g = 9.58811 n
log(TO sin M)
=9.57265 n
log sin iW =9.91 266 n
log cos Jf= 9.76002 n
log cos 6 = 9.41992 n
log tan i¥ =0.15263
JW=234°52'.0
e= 43°17'.2
c-ilf= - 191° 34\8
log sin (c—ilf)
=9.30263
log TO = 9.65989
col sin 0=0.03530
col sin 6= 0.01556
log cot i3= 9.01338
i8=84°6V7
2. a = 103°44'.7, 6 = 64° 12'.3, 7 = 98°33'.8.
.-. ilf = 211° 19'.8, c = 103°30'.6, a = 98°56'.0; Jf=-17°7'.4, /3 = 66°18'.0.
3. a = 156°12'.2, 6 = 112° 48' .6, 7 = 76°32'.4.
.-. iW=174°8'.4, c = 63°48'.9, a = 154°4'.2; ilf = 151°2'.3, /3 = 87°27M.
152. Case IV. Given Two Angles and the Included Side
(a, p, c). — From (4) and (5), Art. 147, we have
Then
4. 1 ^ 7N ^ 1 Sin ^(« — /3)
tani (a - ^i) = tan ^ c . f ) — -^,
tan 1 (a + ^) = tan J «?
cosj-(«-y9)
cos J (a + yS)
« = K^ + ^) + K« - ^)'
(1)
(2)
160
SPHERICAL TRIGONOMETRY.
To obtain 7, use (6) and (7), Art. 147 :
cot J 7 = tan -|- (<^ + /3)
cos -|- (a + h')
(3)
COS 1^ (« — 6)
The agreement of the values of ^y
*^ found from (3) and (4) is a check upon
the computation. The sine proportion
may also be used as a check.
See the note, Art. 150.
1. Solve the triangle when o = 104° 30'.7, /3 = 62° 52M, c = 66° 6'.4.
a = 104° 30'. 7
i8= 62°52M
Ka + ^)^
}°41'.4
logtan^c = 9.72665
logsini(a- /3) = 9.55079
col sin i(a + ^) = 0.00264
^(a - ^) = 20° 49'. 3 log tan |(a - &) = 9.28008
^c= 28° 3'.2
logtani(a-i3) = 9.58012
logsinK«+ &)= 9.98908
col sin |(a- 6) = 0.72768
log cot^7 = 0.29748
^7=20°45'.2
^(a-6) = 10°47'.4
log tan I (a + ^) = 0.95633
log cos -|(a + &) = 9.33339
col cos Ka -&) = 0. 00775
logcot 17 = 0.29747
|7=26°45'.2
logtanic = 9.72665
logcosi(o-/3)= 9.97066
col cos ^(a + py= 0.95897
log tan l(a +b)= 0.65628
i(a + 6)=77°33'.4
K«-&) = 10°47'.4
a=88°20'.8
6=66°46'.0
60°43'.6.
1
2. a = 140°43'.2, /3 = 100°4'.6, c
.-. |(a + 6)=132°38'.88
6 = 119° 22'. 56 ; ^7 = 40° 7'. 42.
3. a = 140° 24'.6, ^ = 12° 18'.6, c = 28° 7'.7.
... ^(rt ^i))= 24° 55'.9 ; 1 (a - &) = 13° 3'.0 ; a = 37° 58'.9 ;
6 = 11°52'.9; |7 = 14°36'.7.
153. Case IV. Second Method. Given p, -y, a, to find One
Element only.
(1) To find a only.
cos a = — cos y3 cos 7 + sin /3 sin 7 cos a.
Let m sin M = sin 7 cos .^ ^
sin = sin yS (1)
sin a
Find c by (4) and (5), Art. 147,
tan J c = tan K« - ^) "^^ f ^^^ + ^> (2)
tan J ^ = tan J (« + *) ^^'f^^ + ^> (3)
2 2 V ^^ COS ^ (« — yS) "" ^
Find 7 by (6) and (7), Art. 147,
cot 1 7 = tan 1 (a - /3) ^^^f^^ + ^> (4)
2' 2v ^^sinj(a — 6) ^^
^1 ^ 1^ , ON cos A (a +^) ,rx
coti7-tanK« + ^) ^^3|^^_^j - (5)
The agreement of the values of J and of J 7 is a check
upon the computation.
Since b is found by means of its sine, it may be either in
the first or in the second quadrant; hence there may be two
solutions. If /3 differs more than « from 90°, ff and b must be
of the same species, and the quadrant in which b lies is fixed.
But if ff does not differ more than a from 90°, we cannot deter-
mine by the first rule for species the quadrant in which b must
lie, and both values of b may be admissible. Hence, inspect for
two solutions when the angle opposite the required side differs less
from 90° than the angle opposite the given side. After finding 5,
the second rule of Art. 146 will show whether both values are
admissible.
1. Solve the triangle when a = 143° 17'.4, jS = 70° 18'.4, a = 160°40'.6.
Since /3 differs less than a from 90°, there may be two solutions.
logsin a = 9.51969 i (« + /3)= 106°47'.9 ^(a-j8)= 36°29'.5
log sin ^ = 9.97383 l(a + b)= 96° 2'.65 ^"'(rt + 6')= 154°37'.95
col sin g = 0.22347 ^(a-6)= 64°37'.95 i(a-6')= 6° 2'.65
log sin 6 = 9.71699
6= 31°24'.7
6' = 148°35'.3
166
SPHERICAL TRIGONOMETRY.
The second rule for species is satisfied for both b and h' ; hence there are
two solutions.
First.
log tan ^ (a - 6) = 0.32409
log sin i (o + /3)= 9.98106
colsin ^ (a -/3) = 0.22570
log tan^c = 0.53085
lc= 73°35'.28
c = 147° 10'. 56
log tan ^ (a + 6) = 0.97517 n
log cos ^ (o + /S) = 9.46091 n
col cos i (o - /3) = 0.09478
logtan^c = 0.53086
lc= 73°35'.30
c = 147° lO'.OO
Second.
9.02483
9.98106
0.22570
9.23159
9° 40'. 38
19^20'.76
9.67591 n
9.46091 n
0.09478
9.23160
9° 40'. 39
c'=19°20'.78
log tan ^ (a - /3) = 9.86908
log sin J (a + 6) = 9.99758
col sin i (a - 6) = 0.04403
log cot ^7 = 9.91069
. ^7= 50° 51'. 00
7 = 101°42'.00
log tan ^ (a -f /3) = 0.52016 n
log cos ^ (a + 6) = 9.02241 n
col cos ^ (a -6) = 0.36813
log cot ^7 = 9.91070
^7= 50°50'.96
7 = 101°41'.92
9.86908
9.63187
0.97759
0.47854
18° 22'.74
^' = 36°45'.48
0.52016 n
9.95597 n
0.00242
0.47855
18° 22'.71
7' = 36°45'.42
2. a = 117°54'.4, /3 = 45°8'.6, a = 76°37'.5.
6 = 51°17'.9; ^ c = 20° 32'.3 or 20° 32'.4 ; ^7 = 18°19'.4.
8. a = 104° 40'.0, iS = 80° 13'.6, a = 126° 50'.4.
.-. b= 54°36'.8; |c=73°48'.4 or 73°48'.5; ^7=69°49'.5 or
and 6' = 125°23'.2; ^c'= 3°25'.6 or 3°25'.5; 17'= 4° 8'.8.
49'.6 ;
157. Case VI. Second Method. Given a, p, a, to find One
Element only.
(1) To find b only.
1 sin S . ^-1 V,
sm = — sm a. (1)
sin a
(2) To find c owZ?/.— Permuting (3), Art. 124,
cot a sin c — cos c cos yS = sin y8 cot a.
. •. cos a sin — sin a cos c cos yS = sin a sin ^ cot a.
Let
OBLIQUE SPHERICAL TRIANGLES.
m sin M = sin a cos yS, )
mcosM=cosa. j
. •. m sin ((? — M) = sin a sin /8 cot a.
• ^ 7i.r\ sin a sin 8 cot a
.-. sin ((? — i)[f) = ^
m
167
(2)
(3)
c — M may be either in the first and second quadrants, or in
the third and fourth ; if there are two solutions, both values of
c-M\vi\\ give c<180°.
(3) To find 7 onli/,
— cos yS cos 7 + sin jS sin 7 cos a = cos a.
Let m smM= cos a sin j3, |
mcosiHf = cosyS. J
. '. m cos (iHf H- 7) = — cos a.
cos a
(4)
cos(M-{- 7)= —
m
(5)
ilf + 7 may be either in the first and fourth quadrants, or in
the second and third ; if there are two solutions, both values of
if + 7 will give 7 < 180°.
: = 143° 17'. 4, /3
To find c.
log sin a = 9.51969
logcosiS = 9.52761
log (wi sin ilf)= 9.04730
log sin 3/ = 9.06947
log cos iW= 9.99699 w
log cos a = 9.97482 n
log tan i¥= 9.07248 n
Jf=173°15'.7
(1)
(3)
(5)
(8)
(9)
(2)
(6)
(7)
cologm = 0.02217 (10)
logsina = 9.51969 (1)
log sin /3 = 9.97383 (4)
log cot = 0. 12746 w (11)
log sin {c- M)= 9.64315 n (12)
c-M.= 20Q° 5M (13)
Jf=173°15'.7 (15)
c' - i»f = 333° 54'.9 (14)
c = 19°20'.8 (16)
c' = 147°10'.6 (17)
Two values.
70°18'.4, a = 160°40'.6.
To find 7.
log cos rt = 9.97482 n
log sin /3 = 9.97383
log (m smM)= 9.94865 n
log sin il/'= 9.97082 «
logcositf= 9.54977
log cos j8 = 9.52761
log tan iJf= 0.42104 n
M = 290° 46'.2
(1)
(2)
(4)
(7)
(8)
(3)
(5)
(6)
colog(- w)= 0.02217 w (9)
log cos a = 9.90400 n (10)
logcos(ilf+ 7)= 9.92617 (11)
M-\-y= 32°28'.2 (12)
ilf=290°46'.2 (14)
i»f+7'=327°31'.8 (13)
7 =101°42'.0 (15)
7'= 36°45'.6 (16)
Two values.
168
SPHERICAL TRIGONOMETRY.
2. a = 117°54'.4, ^ = 45°8'.G,
a
'6'' 37'
31=71" 22'.3, c = 41° 4'.9; M = 13° 5'.3, 7 = 36° 38'.8.
:80°13'.6, a = 126°50'.4.
167° 14'.0, c = 147° 36'.9 or 6° 51'.1 :
3. a = 104°40'.0, )S
.-. 31 =
3f=~'JS° 58'.3, 7 = 139° 39'.0 or 8° 17'.6.
OBLIQUE TRIANGLES SOLVED BY RIGHT TRIANGLES.
158. General Method. — From any vertex O of the triangle
draw an arc p of a great circle perpendicular to the opposite
z)^-:-
Fig. 122.
side, dividing the triangle into two right triangles. Denote
the segments of the side by m and n, and the corresponding
segments of the angle by M and N.
The opposite side must in some cases be produced to meet
the perpendicular arc, as in Fig. 123. The segments of the
side are AD and DB, and their signs are so taken that tlieir
algebraic sum shall be equal to the side ; that is, if a segment is
entirely/ exterior to the triangle, it is negative.
The perpendicular p may have either of two supplemental
values ; we shall always place it in the same quadrant as its
opposite angle in the triangle first used in the solution, in
accordance with the rule for species.
159. Special Formula. — To prove
tan I {tn + n) tan | {m - n) = tan | (a + 6) tan | (a - 6).
In both Fig. 122 and Fig. 123, by Napier's rules,
cos a = cos m cosjo, and cos b = cos n cos p.
(1)
OBLIQUE SPHERICAL TRIANGLES. 169
cos a cos h
• *
cosjt?
= =
cosm
cosw
cos w
cos a
cosw
cos 6
cosm -
- cosw
cos a -
- COS 6
cos m + cos n cos a + cos 6'
which becomes, from (4) and (3), Art. 73,
tan \(m + ii) tan J (jn — n) = tan J (« + ^) tan | (a — 5). q.e.d.
160. Case I. Given a, 6, c. — From (1), Art. 159, we have
tan ^ (jn — n)= tan ^ (a -\- b^ tan |^ (a — ft) cot ^ e, (1)
since m +n = c. We shall consider |^ (ra — n) as being numeri-
cally less than 90°, so that it will be a negative angle when its
tangent is negative. After | (m — n) has been found, we have
n = ^c — l. (jn — n)
A negative value of m ov oi n indicates that the segment, and
hence the corresponding triangle, is exterior to the given tri-
angle. Note that m is always measured from the side that is
called «, and n from h.
In the triangles A CD and I) CB we now know the two sides,
so that the other elements can be found by Napier's rules.
The example shows the method of finding the elements of the
original triangle from the results of the computation.
1. a = 114° 43'.3, h = 136° lO'.G, c = 43° 18'.5.
From (1), ^ (m - n)= 33°oG'.81, whence m = 55°36'.06, w =- 12° 17'.56.
The negative value of n shows that ACD is exterior to the triangle.
From BCD we find DB€ = ^ = 132° 15'.3, DCB = iJf = 65° 17'. 0.
From ^CZ> we find D.4C= 180°- «= 103° 11'.6, ^CZ) = iV= - 17° 57'.5,
giving iY the negative sign since it is exterior to the triangle. Hence
a = 76° 48'.4 ; 7 = 3/-I- iV= 47° 19'.5.
2. a = 76°40'.4, 5 = 54°21'.3, c = 36°8'.7.
.-. J(»w-w)=53°0'.38: w = 71°4'.73: n = - 34° 66'.03 ; /3 = 46°17'.3;
Jlf=76°27'.0; iV = - 44° 48'.2 : 7 = 31°38'.8; a = 120°3'.6.
170 SPHERICAL TRIGONOMETRY.
3. a = 124° 34'.0, h = 6Q° 7'.2, c = 109° 43'.5.
.-. J(m-w) = -76°37'.32; m = - 21°45'.57 ; n = + 131°29'.07 ; /3 = 74° 1'.7 ;
ilf = - 26° 45'.6 ; a = 120° 2'.7 ; iV = + 124° 69'.2 ; 7 = 98° 13'.6.
4. a = 30°17'.6, 6=22°14'.4, c = 18°51'.8.
.-. ?>i = 21° 14'.6 ; n = ~2°22'.8; /3 = 48°17M; J/=45°54'.8;
a = 95° 50'.0 ; iV^ = - 6° 17'.9 ; 7 = 39° 36'.9.
5. a = 130°46'.0, b = 113°21'.4, c = 102° 16'.2.
. •. i (m - «) = - 11° 8'.6 ; m = 39° 59'.5 ; n = 62° 16'.7 ; /3 = 136° 19'.25 ;
Jf = 58° 3'.4 ; a = 145° 15'.9 ; i\r = 74° 37'.75 ; 7 = 132° 41'.2.
161. Case II. Given a, p, y, — ^Pply the method of Case I
to the polar triangle, and thence find the elements of the
original triangle.
1. a = 116° 19'.4, /3 = 83° 19'.2, 7 = 106° 10'.6.
In the polar triangle,
a' = 63° 40'.6, b' = 96° 40'.8, c' = 73° 49'.4.
.-. J (m' - n') = - 66° 18M, m' =- 29° 23'.4, n' = + 103° 12'.8.
The negative value of m' shows that B'C'D' is exterior to the triangle.
From B'C'D' we find
D'B'C = 180° - /3' = 73° 49'.2, D'C'B' = 3I'=- 33° 11'.8,
giving M' the negative sign since it is exterior to the triangle.
From A' CD' we find
D'A'C = a' = 60° 4'.7, N' = + 101° 25'. 5.
. •. p' = 106° lO'.S, 7' = M' + N'= 68° 13'.7.
Passing from the polar to the original triangle,
a = 119° 55'.3 ; 6 = 73° 49'.2 ; c = 111° 46'.3.
2. a = 110° 36'.4, /3 = 122° 8'.7, 7 = 140° 20'.3.
. •. J (m' - n') = 29° 27'.90 ; m' = 49° 17'. 75 ; w' = - 9° 38'.05 ;
/3' = 64° 4'.9 ; M' = 54° 5'.4 ; a' = 96° 7'.4 ; iY' = - 11° 24'.0 ; 7' = 42° 41'.4 ;
.-. a = 83° 52'.6, b = 115° 55M, c = 137° 18'.6.
3. o = 120° 50'.6, /3 = 78° 6'.1, 7 = 81° 12'.3.
. •. i (m' - w') = - 63° 33'.19 ; m' = - 14° 9'.34 ; n' = 112° 67'.04 ;
/3' = 98° 39'. 7 ; M'=- 1G° 33'.0 ; a> = 60° 9'.6 ;
N' = 109° 46'.0 ; 7' = 93° IS'.O ;
.: a = 119° 50'.4, b = 81° 20'.3, c = 86° 47'.0.
4. a = 80° 20'.2, /3 = 73° 46'.7, 7 = 54° 8'.5.
.-. i (m' - n') = 7° 15'.69 ; m' = 70° 11'.44 ; n' = 55° 40'.06 ;
/3' = 118° 12'.7 ; M' = 72° 37'.5 ; a' = 115° 12'.8 ;
N' =59°19'.l; 7' = 131° 56'.6 ;
.'. a= 64° 47'.2, b = 61° 47'.3, c = 48° 3'.4.
OBLIQUE SPHERICAL TRIANGLES.
171
6. a = 100° 61'.3, /3 = 80° 47'.C, 7 = 74° 3'.3.
. •. J (m' - n') = - 83° 60'.76 ; m' = - 30° 62'.41 ; n' = 130° 49'.11 ;
/3' = 96° 35'.0 ; 3P = - 31° 30'.0 ; a' = 81° 16M ;
N' = 136° 6'.8 J 7' = 104° 36'.8 ;
.: a = 98° 44'.9, b = 83° 25'.0, c = 76° 23'.2.
162. Case III. Given a, 6, -y. — From the end of one of the
sides, as 5, let fall an arc of a great circle perpendicular to the
other side. In the triangle J) AC we
know b and 7 ; hence we find w, iV,
and p by Napier's rules, considering
p as of the same species as 7. Then
m = a — n, being negative when n>a,
showing that the triangle BAD is then
exterior to the triangle BA C.
Now in the triangle BAD we know
DB and AD^ and we find , iHf, and
ABD by Napier's r«les.
1. a = 105° 14'.8, 6 = 43° 17'.2, 7 = 112° 47'.4.
.;. n = 159° 57'.3, N= 150° 0'.4, p = 140° 47'.53.
. '. w = — 54° 42'.5, showing that BAD is exterior to BAC.
In the triangle BAD we find
ABD = 180° - j8 = 135° 0'.8, c = 116° 35'. 6, ilf = -65° 63'.7,
giving M the negative sign since it is exterior to the triangle.
Hence ^ = 44° 59'.2, a = iV^+ ilf= 84° 6'. 7.
2. a = 103° 44'.7, b = 64° 12'.3, 7 = 98° 33'.8.
.-. iY=100°54'.7; ?i = 162° 52'.6 ; ;)= 117° 5'.1 ; m=-69°7'.9;
c = 103° 30'.6 ; ^ = m° 18'.0 ; 31 = - 61° 58'. 7 ; a = 98° 56'.0.
3. a = 156° 12'.2, b = 112° 48'.6, 7 = 76° 32'.4.
.: 3f= 148° 18'.6 ; n = 151° 2'.3 ; p = 63° 41'.8 ; ?» = 5° 9'.9 ;
c = 63" 48'.8 ; /3 = 87° 27'.1 ; M = 5° 45'.5 ; a = 154° 4M.
163. Case IV. Given a, p, c. — Let fall from the vertex of
one of the angles, as a = BAC (Fig. 124), an arc of a great
circle perpendicular to the opposite side. In the triangle ABD
we know e and yS, and we find m, M, and p by Napier's rules,
considering p as of the same species as ^. Then iV = a — iltf, a
negative value of JS^ showing that the point D lies on BC pro-
duced, the triangle A CD being then exterior to the given
triangle.
172 SPHERICAL TRIGONOMETRY.
In the triangle ACD we now know p and CAD^ and we
find 6, 7, and n by Napier's rules.
\. a = 140° 43'.2, ^ = 100° 4'.6, c = 60° 43'.6.
. , rr> = 162° 39'.9, j? = 120° 48'.86, 31 =:. 160° VJ.
Then iV = a - i¥ = - 19° 18^5.
. •. & = 119° 22'.5, ACD = 180° - 7 = 99° 45M, n = - 16° 44'.8,
giving n the negative sign since it is exterior to the triangle.
.-. 7 = 80° 14'.9, a = m-\-n = 145° 55M.
2. a = 104° 30'.7, /3 = 62° 52M, c = 56° 6'.4.
. •. M = 42° 34'.8 ; iNr= 61° 55'.9 ; m = 34° 10'.2 ; p = 47° 37'. 5 ;
b = 66° 46'.0 ; 7 = 53°30'.4 ; 7i = 54° 10'.7 ; a = 88° 20'.9.
8. a = 140° 24'.6, /3 = 12° 18'.6, c = 28° 7'.7.
. •. ilf = 79° 6'.4 ; iNT = 61° 18'.2 ; w = 27° 34'.7 ; p = 5° 46M ;
& = 11° 52'.9 ; 7 = 29° 13'.3 ; n = 10° 24'.3 ; a = 37° 59'.0.
>
164. Case V. Given a, b, a. — Let fall an arc of a great
circle from the intersection of a and J,
^^ perpendicular to c. In this case there
^^^^Vi\ will be two solutions if a is inter-
y^ / 1 \a mediate in value between p and both
X 7 'f \ b and 180° - h (Art. 120).
''^^O-— - ^/ \d ^ r ^^^ ^^® triangle ACD, knowing b
"'^'^ ^ and «, find 771, M, and p by Napier's
rules. Then in the triangle DOB^
knowing p and a, find DB^ DCB^ and BBC. Then in the tri-
angle A CB we have
c = AB = m-\-BB, y = ACB = lM-\- BOB, ^ = BBO;
and in the triangle ACB' ^
c'=AB' = m-BB, y' =ACB' = M-BOB, j3' = 180'' -BBC.
1. a = 148° 34'.4, h = 142° 11'.6, a = 153° 17'.6.
.-. p = 164° 0'.52, and there are two solutions.
->. = 34° 43'.5, M = 68° 19'.4.
Also, DB = 27° 25M, DBC = 148° 6'.3, DCB = 62° 1'.8.
.-. c = 62° 8'.6, 7 = 130° 21'.2, ^ = 148° 6'.3,
and c' = 7° 18'.4, 7' = 6° 17'.6, j8' = 31° 53'.7.
OBLIQUE SPHERICAL TRIANGLES. 173
2. a = 40° 20'.4, b = 20° 18'.2, a = 60° 44'.4.
.'. p = 17° 37'.3 ; m = 10° IG'.O ; ^ = 30° 61 '.2 ; /3 = 27° 52'.9 ;
DB - 36° 53 .7 ; DCB = 68° 2'.3 ; c = 47° 8'.7 ; 7 = 98° 63'.6.
3. a = 98° IC, & = 74° 38', a = 78° 40'.
.: p = 70° 59'.25 ; m = 35° 34'.0 ; iW= 37° 6'.1 ; /3 = 72° 49'.25 ;
DB = 116° 11'.4 ; DCB = 114° 66'.4 ; c = 151° 45'.4 ; 7 = 152° 2'.5.
165. Case VI. Given a, p, a. — Pass to the polar triangle,
in which we shall know a', Z>^ and a', and solve by the method
of Art. 164. There may be two solutions of the polar triangle,
and therefore of the triangle itself.
1. a = 143° 17'.4, /3 = 70° 18'.4, a = 160° 40'.6.
.-. a' = 36° 42'.6, h> = 109° 41'.6, a' = 19° 19'.4.
• •. p' =z 18° 9'. 13, and there will be two solutions.
M' = 96° 44'.3, m' = 110° 46'.3.
Also D'B' = 32° 28'.25, D'C'B' = 63° 54'.9, D'B'C = 31° 24'.7.
.'. ci' = 143° 14'.55, 7i' = 160° 39'.2, /3i' = 31° 24'.7,
and ci" = 78° 18'.05, 71" = 32° 49'.4, jSi" = 148° 35'.3.
Taking the supplements to obtain the elements of the original triangle,
7 = 36° 45'.45, c = 19° 20'.8, b = 148° 35'.3,
and 7' = 101° 41'.95, c' = 147° 10'.6, b' = 31° 24'.7.
2. o = 117° 54'.4, p = 45° 8'.0, a = 76° 37'.5.
.: p' = 136° 23'.8 ; 31' = 18° 37'.7 ; m' = 13° 5'.3 ;
D'C'B' = 120° 17'.5 ; D'B'C = 128° 42'.1 ; D'B' = 130° 15'.9 ;
7' = 138° 55'.2 ; c' = 143° 21'.2 ;
. •. 6 = 51° 17'.9 ; c = 41° 4'.8 ; 7 = 36° 38'.8.
3. a = 104° 40'.0, /3 = 80° 13'.6, a = 126° 50'.4.
.-. p' = 52° 3'.8 ; 31' = 102° 46'.0 ; m' = 106° 1'.7 ;
D'C'B' = 70° 22'. 9 ; D'B'C = 54° 36'.8 ; D'B' = 65° 40'.7 ;
7i' = 172° 8'.0 ; 71" = 32° 23'.1 ; d' = 171° 42'.4 ;
ci" = 40° 21'.0 ; /3i' = 54° 36'.8 ; /3i" = 125° 23'.2.
. •• c = 7° 51'.1 ; 7 = 8° 17'.6 ; b = 125° 23'.2 ;
and c' = 147° 36'.9 ; 7' = 139° 39'.0 ; h' = 54° 36'. 8.
CHAPTER XII.
APPLICATIONS OF SPHERICAL TRIGONOMETRY.
166. To find the Shortest Distance between Two Points on
the Surface of the Earth,* the earth being treated as a sphere. —
North latitudes and west longitudes are considered positive.
Let QQ' be the equator, P the north pole, A and B the two
points, and PM the meridian
from which the longitudes are
measured. The longitude of
A is MPC and that of B is
MPD^ both being positive
since they are measured west-
ward. The latitudes are CA
and DB^ the former being
negative since it is measured
southward.
In the triangle APB the
sides AP and BP are found
by algebraically subtracting
the latitudes from 90°, and the angle APB is the algebraic
difference of the longitudes. Hence we know two sides and
their included angle, so that we can solve the triangle, using
the method of Art. 151 when the distance only is required,
and that of Art. 150 when we wish to find all the elements.
. 1. Find the shortest distance between New York, 40° 45'.4 N., 73° 58'. 4 W.,
and Rio Janeiro, 22° 54'.4 S., 43° 10'.4 W.
.-. I?P=49° 14'.6, ^P=112° 54'.4, ^PS=30° 48'.0. Ans. .45=69° 48'. 2.
2. Find the shortest distance between New York, 40° 45'. 4 N., 73° 58'.4 W.,
and Paris, 48° 50'.2 N., 2°.20'.2 E. Ans. AB = 52° 2C'.8.
* The shortest distance between two points on a sphere is the arc of the
great circle passing through the points.
174
Fig. 126.
APPLICATIONS OF SPHERICAL TRIGONOMETRY. 175
If the bearings of the great circle AB at A and B are required, it -will be
necessary to find the angles PAB and PBA.
3. A ship sailed from Calcutta, 22° 34'.8 N., 88°27'.3E., on an arc of a
great circle to Melbourne, 37° 48'.0 S., 144° 58'.0 E. Find the distance sailed
and the bearings * at both points.
Ans. At Calcutta, S. 41° 56'. 61 E. ; at Melbourne, S. 61° 21 '.47 E. ; dis-
tance, 80°22'.4or80°22'.6.
4. A ship sailed from the Cape of Good Hope, 34° 22' S., 18° 29' E., on an
arc of a great circle to Cape St. Roque, 5° 28' S., 36° 16' W. Find the distance
sailed and the bearings * at both points.
Ans. At G. H., N. 72° 28'.0 W. ; at S. R., N. 52° 16'.0 W. ; distance, 57° 20'.4.
6. A ship sailed from Bombay, I&° 56' N., 72° 53' E., on an arc of a great
circle to the Cape of Good Hope, 34° 22' S., 18° 29' E. Find the distance sailed
and the bearings * at both points.
Ans. At Bombay, S. 44° 12'.8 W. ; at G. H., S. 53° 2'.6 W. ; distance,
74° 15'.2 or 74° 15'.4.
6. A ship sailed from Bombay, 18° 66' N., 72° 53' E., on an arc of a great
circle for the Cape of Good Hope, 34° 22' S., 18° 29' E. Find the distance to the
equator and the bearing* and longitude at the equator. [Use the triangle
BDE; the angle PBA - 135° 47'.2 was found in Ex. 6.]
Ans. S. 41° 16'. 1 W. ; distance, 25° 34'.5 ; longitude, 65° 21'.8 E.
7. From a point whose latitude is 17° N. and longitude 130° W. a ship
sailed an arc of a great circle over a distance of 4150 miles, starting S. 64° 20' W.
Find its latitude and longitude if the length of 1° is 69^ miles.
Ans. Lat., 19° 40'.62 or 19° 40'.60 S. ; Long., 178° 20'.9 W.
167. Given the Lengths of the Three Edges of a Parallelo-
piped that meet in a Point, and the Angles between them, to
find the Surface and the Volume of the Parallelopiped. — Let
OGr he the solid, AD the perpendicular from A to BOO, and
hence AOD a plane per-
pendicular to BOO. Let
the angles and edges be
BOC=:a, AOC=h,
AOB = c, OA = l,
0B =
m,
00 =
n.
Describe a sphere with a
radius of unity about as a
center, its intersections with
the planes forming the figure marked by the primed letters.
Fig. 127.
The course of the ship.
176
SPHERICAL TRIGONOMETRY.
Then the surface is
^=2 OBEC^- 2 OAFC+ 2 OBHA
= 2 {mn sill a + In sin b + Im sin c). (1)
In the triangle A'B'B', right-angled at D', we have
sin B'A' = sin B'A' sin A'B'I)' ;
. •. sin D'A' = sin 6> sin A'B'L'.
But in the triangle A' B' C we know the three sides a, b^ c ;
hence
sin A'B'B' = 2 sin l A'B'B' cos ^ J.'^'i>'
2
sin a sin
.-. fiuiD'A' = sin BOA
2
Vsiii s sin (s — «) sin (s — ft) sin (.s- — c).
sin a
i>A= OA sin BOA
21
Vsin s sin (s — «) sin (s — b) sin (s — ).
sin a
Hence the volume is
F= OB EC X D^
Vsin s sin (s — a) sin (s — 6) sin (s — ).
(2)
= 2 Zww Vsin 8 sin (s — a) sin (s — ft) sin (s — c).
168. To find the Volume of a Regular Polyhedron. — Let
AB be the edge in which two adjacent faces intersect, B its
middle point, C and B the centers of the
polygonal faces, and the center of the sphere
inscribed in the polyhedron, the faces being
tangent to the sjjhere at and B. Then
BC=BB; BA = BB;
CDA = CBB = BBA = EBB = 90°;
i>(7O = 90°; BEO==90\
Let a = length of an edge AB^
s = number of sides of each polygonal
face,
n = number of faces meeting at a vertex
of the polyhedron,
APPLICATIONS OF SPHERICAL TRIGONOMETRY. 177
iV^= number of faces of the polyhedron,
E = edge angle CDE of the polyhedron.
i Then CD = AD Got ACD = \a cot ^-^^'
CO = CD tan CDO = CD tan -^ E,
.'. CO = I acot^^ tan IE, (1)
About as a center, with a unit radius, describe a sphere,
and let its intersections with the three planes form the triangle
A' CD'. Then
A'C'D' = ACD = ^^; A'i>'(7' = 90°; C'^'D' = lM2!.
s n
By Napier's rules,
cos C'A'D' = cos CD' sin A^ CD',
180° ^,j.f . 180°
or cos = cos C^'-D' sin
n s
But
cos CD' = cos COD = cos (90° - CDO} = sin CD0= sin J J57.
180° . 1 rr . 180°
. •. cos = sm J E sin
• 1 XT 180° 180° ,oN
. •. sm -^ E — cos cosec (2)
7h 8
Then, if A is the area of a face, the volume is
V=lCOxAxN= ^\Ma^ cot2 1?^ tan ^ E. (3)
Find -1 -^from (2) and then Ffrom (3).
1. Dodecahedron, formed by 12 regular pentagons, 3 meeting at a vertex.
. •. s = 5, n = 3, iV = 12. log cos 60° = 9.G9897 lo^ ^ = 39-Q4
log cosec 36° = 0.23078 ^24 " ''
log sin I E = 9.92975 ^'^S cof^ 36° = 0.27748
log tan ^E=z 0.20896
0.88438
.-. F= 7.663 a*.
2. Tetrahedron, formed by 4 equilateral triangles, 3 meeting at a vertex.
■.-. s = 3, 71 = 3, iY=4. Ans. F= 0.1179 a^.
3. Cube, formed by 6 squares, 3 meeting at a vertex.
.-. s = 4. n = 3, iV = 6. Ans. V=a\
CROCK. TRIG. — 12
178
SPHERICAL TRIGONOMETRY.
4. Octahedron, formed by 8 equilateral triangles, 4 meeting at a vertex
.-. s = 3, « = 4, N=S. Ans. V=0A7Ua\
6. Icosaliedron, formed by 20 equilateral triangles, 5 meeting at a vertex.
.-. s = 3, n = 5, iV=20. Ans. F= 2.182 a^.
169. If from Any Point in a Trirectangular Triangle Arcs of
Great Circles are drawn to the Vertices,
COS'^ a + COS^ p + COS^ 7 = 1»
where «, /3, and 7 are the arcs. — In Fig. 129, produce YP and
ZP to D and U. In the right triangle PDX,
sin PD = sin a sin PXD ; .-. cos /3 = sin ct sin PXi>. (1)
In the right triangle PJSX,
sinPU = sinasinPXU ; .-. cos 7 = sin a cos PXD. (2)
Squaring (1) and (2), and adding, we have
cos^ yS + cos^ 7 = sin^ «.
. •. cos^ a + cos^ /3 + cos^ 7 = 1. Q.E.i).
Fig. 130.
170. If from Any Two Points P and I*' in a Trirectangular
Triangle Arcs of Great Circles are drawn to the Three Vertices,
and if v is the Length of the Arc rr', prove that
cos V = cos a cos a' + COS p cos p' + cos 7 cos y'.
In the triangle PYP^ (Fig. 130),
cos V = cos yS cos /3^ + sin yQ sin yS' cos PYP'. (1)
But
cos PZP^ = cos CZYPf - ZYP^.
.-. cos P rP' = cos ZrP^ cos ZrP + sin ZZP' sin ZrP. (2)
APPLICATIONS OF SPHERICAL TRIGONOMETRY. 179
(3)
In ZYP, cos 7 = sin y3 cos ZYF.*
In ZYF^, cosy = sinks' cos ZYF'.*
In XYF, cos « = sin /9 cos XYF * = sin /3 sin ZYF.
In XrP', cos a' = sin /S' cos XYF' * = sin /3' sin ZYF'.
Substituting in (1) the values found from (2) and (3),
cos V = cos /3 cos fi' + cos 7 cos y' -\- cos a cos a', q.e.d.
This is the formula for the cosine of the angle between two
lines in space, the angles made by them with tliree lines at right
angles to each other being «, yS, 7, and a'^ 13', 7', respectively.
171. To find the Angle a' between the Chords of Two Sides
of a Spherical Triangle, having given the Two Sides b and c, and
the Angle a between them. — Let AB = c, AC = b^ the spherical
angle BA C = «, and the plane angle
BAC = «', being the center of the
sphere. About ^ as a center de-
scribe a sphere, and let its intersec-
tions with the planes GAB, OAC,
and BAG form the triangle BUF.
Then
I)F= GAB =90°
FDE = « ; FF= BAC=a'. ^^^ ,31
. •. cosFF= cos BF cos BF + sin BF sin BF cos FBF.
.-. cos «' = sin -|- 6 sin I" c? -f- cos I 5 cos -|- e cos «. (1)
This formula is true for all values of 5, c, and «. When
b and e are small, the correction that must be applied to a to
obtain a' may be found from (1) as follows :
Let p = b -\- c^ and q = b — c. Then, from Art. 72,
cos a' = l cos J q — ^ cos -|-|) + -|- (cos \p + cos J q) cos a
= — sin2 \q -{- sin2 |-p + (1 — sin^ -^ j; — sin^ ^ ^) cos a
= (sin^ I j;> — sin^ -^ (^) (sin^ -i- « -f cos^ 1 a) -|- cos a
— (Qi\i?\p + sin2 ^ g) (cos2 -|- « — sin^ | «) .
. ♦. cos a' = cos a — 2 sin^ J e^' cos^ 1 a 4- 2 sin^ ^jo sin^ |- a. (2)
*Eq. (2), Art. 12L
l"
180
SPHERICAL TRIGONOMETRY.
Let «' = «-{- 6^ where 6 is so small that we may place
sin 6 = 6^ and cos 6 = 1.
. '.' cos a' = cos a cos — sin a sin 0.
.'. cos a' = cos a — sin a. (3)
Comparing (2) and (3),
2 6 sin 1^ a cos |- « = 2 sin^ ^ ^ cos^ i « — 2 sin^ ^ jt? sin^ |^ «.
. •. ^ = sin2 ^ ^ cot -|- a — sin^ ^p tan -J a.
' '' ^" ^ ^hTl^ ^'''' * ^ ''''^ 2 « - ^i^,- sinH^ tan ^ a, (4)
since 6? = 6"' sin 1'^ (Art. 81).
172. The Angles of Elevation of Two Points, in the Direc-
tions OA and OB, above a Horizontal Plane, and the Inclined
Angle AOB, were measured with a Sextant. Find the Hori-
zontal Angle between the Points, as seen
from O. — Let OZ be the vertical line, Oah
the horizontal plane; aOA = h^ and hOB=k
the measured altitudes; and AOB=:c the
inclined angle. Describe a sphere about
as a center. Then in the triangle AZB,
AZ = 90° -h, BZ= 90° - 7c, AB = c,
and hence the required angle aOb = AZB
may be computed, since we know the three
sides of the triangle.
When h and k are small, the correction
to be applied to the measured value c to obtain a Ob may be
found as follows :*
From (2), Art. 121,
Fig. 132.
cos^Z^=
c — sin h sin k
cose
hJc
cos h cos k
cos c — hk
(Art. 78)
1 - K^' + ^')
. cos AZB= cos c + 1(^2 4- F) cos c
(l-J-A2)(l_l/,2)
= (cos c - hk) [1 + i Qi^ + A;2)] .
A^. (1)
* Neglecting powers of h and k above the second.
APPLICATIONS OF SPHERICAL TRIGONOMETRY. 181
Let 6 be the correction to c so that AZB = c + ^.
. *. cos AZB = cos c cos 6 — sin c sin 6.
. •. cos AZB = cos c — ^ sin c. (2)
Comparing (1) and (2),
(/,2 ^ ^2) (cos^ -|- ff - sin2 lc)-2hk (cos^ ^ g + sin^ i g)
~~ 4 sin ^ g cos ^ g
.-. ^= i(^ + A:)2tanJg-|(A-/c)2cotl(7, (3)
where ^, A, and A: are expressed in circular measure. To find
e in seconds, let 6 = 6" sin 1", ^ = h" sin l'^ /c = k" sin 1^'.
... B'f ^\Qi'' +y'y&mV' t2in^c-\(h'' -h''y^mV' Qot I c. (4)
SPHERICAL EXCESS.
173. Area of a Spherical Triangle. — From geometry we
know that the areas of any two triangles are to each other as
their spherical excesses, the spherical excess being the amount
by which- the sum of the three angles exceeds 180°. We also
knoAV that the area of the trirectangular triangle is -|- Trr^, and
that its spherical excess is 90°. If A is the area of any triangle,
and E its spherical excess expressed in degrees, we have
Ai\'nr'^=Ei^{i\ (1)
.-. A = E^^, (2)
180° ^ r
and E = a'^' ' (3)
174. Lhuillier's Theorem. — We have ^V
tan 1 xT_s^^i (« + ^ + 7-'^) 2cos^(ot + ^ + 7r-7)
^ - ^(jg 1 (^^ + ^ + ^ _ ^) 2 cos|(a + /S 4- TT - 7)
_ sin -|- (« + /3) — sin j- (tt — 7)
"" cos I (a + yS) + cos ^ (tt ~ 7)'
from (6) and (7), Art. 72.
^ cos i (« + y8) 4- sm J 7
182 SPHERICAL TRIGOXOMETRY.
Hence, from (1) and (2), Art. 145, substituting for sin^(a-f-/3)
and cos ^ (a 4- yS), we have
- ^ cos i (a — h)— cos i c cos X y
tan i J5;= \^ — —j^ f ^-f-i-
* cos ^ (a + 6) + cos J sin J 7
^ sinK^'^-^ + ^)si"i(^ + g-a) ^^^ 1
cos :^ (a + 6 + cos ^ (a + 6 — c) 2" v»
from (4) and (3), Art. 73.
, _ sin J (s— 5)sinJCs — «)^ / sin s sin (s — f?)
.-. tani^= ^ "^-T-/— — ^\-^— 7 ^ r
* cos ^ s COS ^Qs — c) ^ sm (« — a) sin (s —
4
sin2 j- (g - ^>) sin^ 1 (g - a) ^^
COS^ -1- 8 COS^ J (S — t?)
sin J s cos J s sin J (s — c) cos ^ (s —
sin ^ (s — a) cos J (s — «) sin |^ (s — ^) cos
f) 1.
. tan ^ -E'= Vtan |^ s tan |^ (s — a) tan |^ (s — ^) tan ^ (s — c?). Q.E.i.
175. Spherical Excess in Terms of Two Sides and their
Included Angle.
tan ^ J? =r ^""^ 2 (« + /^ + 7 - tt) ^ - cos-|- (« + ^ + 7)
2 cos 1 (a 4- ^ + 7 — tt) sin |- (a + /3 + 7)
_ sin I (a + /3) sin j^ 7 — cos j- (a 4- /5) cos j- 7
"" siji ^ (a + 13} cos-|- 7 + cos ^ (« + fi) sin J 7
Substituting for sin |- (« + /3) and cos J (a + /3) from (1) and
(2), Art. 145,
^ ^ _ sin ^ 7 cos I 7[cos J (a — h) — cos |^ (« + ^)]
2 "~ cos|^(rt — 6) cos2^7 4- cos J(a + 6) sin2l7
sin -|- 7 cos -|- 7 f 4- 2 sin J a sin ^ />!
~^[cos^(a — 6)4-cos^(a4-^)J + -2-[cos^(^a— 6) — cos-^(«4-^)]cos7
_ sin I" a sin J 5 sin 7
cos ^ a cos ^b -{- sin ^ a sin J 6 cos 7
... tanA^=, tan^atan^ismy . ^.^.l.
"* 1 4- tan J a tan J 6 cos 7
176. Approximate Value of the Spherical Excess, neglecting
Powers above the Second. — Let the sides of the triangle be so
APPLICATIONS OF SPHERICAL TRIGONOMETRY. 183
small that the powers of their circular measures higher than
the second may be neglected. We have, from Art. 78,
tana; = a; -I- Ja;3 4- •••, (1)
where x is expressed in circular measure.
Let the lengths of the sides be a, 6, and c when expressed
in circular measure, and a\ h\ and c' in linear measure, r being
the radius of the sphere. Then
« = p ^ = p ^=7 (2)
Placing these values of a, 6, and c for x in (1), and substituting
in Lhuillier's theorem, we have, neglecting powers above the
second,
for, 1 7? ^V^^ s' — a' &' — V s' — c' ,oN
where s' = ^(a' + 5' + c'). (4)
.-. tani^ = -!-Vs'(s' - a')(«' - ^-OC^' - O- C^)
4 H
Since \Il\s, small, we place its tangent equal to its arc.
.-. \E=^ A^«'(«' - ^')(s' - ^')0' - O (6)
4H
^ = 1^, (7)
where ^ is the area of the plane triangle whose sides are a', V ^
and c\ E being expressed in circular measure.
To find the value of E in seconds of arc, divide both sides
by sin 1''.
= En = —A_. (8)
E
sin 1 f f r^ sin 1 "
Hence, whenever the third powers of the circular measures
of the sides can be neglected, the spherical excess is found by
computing the area of the triangle, considering it as a plane
triangle, and dividing the area by r^miV .
184 SPHERICAL TRIGONOMETRY.
177. Approximate Value of the Spherical Excess, neglecting
Powers above the Fourth. — From Lliuillier's theorem,
* L2r 24?-aJL 2r ^ 24r3 J
fs' - b' (s' - hy ~\[ s' - c' (js' - gpn
L 2r 24^3 JL 2r 24?-3 J
where A^ = s'(.s' - a')(^' - ^00' - ^')-
.4^ A^
16 r* 192 r"
4r2V 24?-2 /
••• i^'s^^^l' =4^(,^-^ 247^ ^j-
r2sinl"V 24 r2 J ^
This value exceeds that found in Art. 176 by
A a^2 ^ yi 4. ^/2
r2sinl''* 24 r2
If a' = b' = c' = 100 miles, and r = 3963.3 miles, we obtain
-^^— - = 56^^863; ^!-^tJ!l±^ ^qmOOS;
?-2 Sin 1'^ 24 H
so that the correction to the value of U'' given by (8), Art.
176, is only
56'^863 X 0.00008 = 0'^003.
178. Legendre*s Theorem. — If the sides of a spherical tri-
angle are veri/ small compared ivith the radius of the sphere, the
angles of the plane triangle whose sides are of the same length as
APPLICATIONS OF SPHERICAL TRIGONOMETRY. 185
those of the spherical triangle, are equal to the corresponding
angles of the spherical triangle diminished hy one third of the
spherical excels. — Let a', h\ and c' be the lengths of the sides
of the spherical triangle expressed in linear measure, and a, 6,
and c the lengths in circular measure.
a' J h' c' .^^
r r r
Let a be an angle of the spherical triangle and a' the corre-
sponding angle of the plane triangle. We have
cos a — cos b cos c
cos a =
(2)
cosa = -V[(^2 + c2_^2)
2 be
sin b sin c
From Art. 78,
cos a = 1 — I a2 -f 2^ a* — ••• sinb = b — -J- b^ + ...
cos6 = 1 - -J62 + ^\b^ smc = c - i6'3 + ...
cos (? = 1 — -J c2 + 2f ^ — •••
X (1,2 _!_ ^2 _ ^2) + 1 (^4 _ 54 _ .4 _ e J2^2)
•*• cosa = ^^ j-jr — ^t-77 , 0.-1 ^' (^)
the terms of orders higher than the fourth being neglected.
[(52 + ^2 _ ^2)
+ J^ (a4 _ J4 _ ^4 _ e 52^2)-| [-1 _ ^ (^2 + ^2)-]-l
+ _i_ (a4 _ /,4 _ ^4 _ 6 62^2)] [1 4- K^' + ^') + •••] •
52 4. ^2 _ ^2 ^4 + 54 _f_ ^4 _ 2a2^2 _ 2a\^ - llP-c^
.-. cos « = \ + — ^- ^ -— ,
2bc 24: be /'4\
the terms of orders higher than the fourth being neglected, as
before.
In the plane triangle,
^ ^'2 + ^r2_^/2 J2 4.^,2_^2
^^^^= 2 6V = 2bc ' ^^^
from (1)
, «4 + ^4 + ^4 _2a2^»2_ 2^2-2^2^2
.-. cosa = cosa'H ■ — . (6)
24 oc
, 1 a'4 + 5'4 4-c'4_2a'26'2_2a'2e'2-2 6V2
... cos« = cos«+- ^j^,-^
1S6 SPHERICAL TRIGONOMETRY.
Let 8' = Ka' + b' -\- c') ; then
= - iV («'^ + ^" + ^'^ - 2 a'25^2 _ 2 a'V2 - 2 5'V2). (8)
But the area of the plane triangle is
Vs'(s'-a')(s'-6')(s'-60 = ^ 6'tf' sin a^ (9)
Hence (7) becomes, from (8) and (9),
cos a = cos a' — r—7,h'c' sin^a'. nO")
6 r^ ^
Let a = a' + ^. (11)
.-. cos a = cos a' cos ^ — sin a' sin ^. (12)
Since 6 is small, we may place cos ^ = 1, and sin 6 = d.
. *. cos a = cos a' — sin a^ (13)
Comparing (10) and (13),
a ^ ti I • / 11 b'c' sin a' ^^.^
e = — 6Vsm«'=-.-._^-. (14)
Hence, from (7), Art. 176,
and, from (11), a' = a — ^ K Q.B.D.
179. Application of Legendre*s Theorem. — In the New York
State Survey the angles of the spherical triangle, whose ver-
tices were at Howlett, Gilbertsville, and Eagle, were measured,
the distance from Howlett to Gilbertsville having been already
computed. The measured values were
At Howlett, a = 85° 18' 57". 71 logh = 4.54227 32
At Eagle, /3 = 51° 35' 41''.61 logr= 6.80459 32
At Gilbertsville, 7= 43° 5' 24". 24
... « -1-/3 + 7 = 180° 0' 3". 56
The formula for the spherical excess is (Art. 176)
■pn _ ^ _ 1 52 sin a sin 7 1
r^sinl" 2 sinyS r'^sinV
APPLICATIONS OF SPHERICAL TRIGONOMETRY. 187
log52 = 9.08455
colog 2 = 9.69897 -10
logsina = 9.99855 -10
log sin7 = 9.83451 -10
colsin/3 = 0.10588
cologr2= 6.39081 -20
colsinr' = 5.31443
logU'f = 0.42770 .-. E" = 2''. 6773;
.-. 1^^' = 0'^8924.
The errors due to observation therefore amounted to 3''. 56
— 2''. 677 = 0''.883. This discrepancy was distributed among
the three angles according to the method of least squares,*
gfving the following results :
Observed
Angles.
Correction.
Spherical
Angles.
i^".
Plane Angles.
o= 85° 18'
57".71
-0''.747
56".963
0".893
56".070 = a'
jS = 51° 35'
41".61
+ 1".355
42".965
0".892
42".073 = ^'
7= 43° 5'
24".24
- 1".491
22". 749
0".892
2l".8J7 = 7'
Suni=180° 0' 3".56 -0".883 2".677 2".677 0".000
Using the plane triangle, we find by the sine proportion :
log 5
col sin /S
log sin a
log a
a
= 4.542 2732
= 0.105 8837
= 9.998 5468
10
4.646 7037
44330.61 meters.
log5' = 4.542 2732
colsinyQ' = 0.105 8837
log sin y = 9.834 5089-10
log(?' = 4.482 6658
c' = 30385.46 meters.
These are the distances between the points measured on the
great circles joining them.
ASTRONOMICAL APPLICATIONS.
180. Definitions. — Let us consider the earth as a point
(Fig. 134), and let a sphere be described about as a center,
with a radius indefinitely great, so that all the stars shall be
within the sphere. The figure represents the sphere as seen
from the outside.
Eleven angles were involved in the adjustment.
188
SPHERICAL TRIGONOMETRY.
The zenith Z is the point where a vertical line — the plumb
line — pierces the sphere.
The horizon HWNE is the great circle cut from the sphere by
a plane through perpendicular to the plumb line. iV, E, H^
and TF^are the north, east, south, and west points of the horizon.
^"7- ' ^
FiQ. 134.
Vertical circles are great circles whose planes pass through
the plumb line OZ, as ZST in the plane OZT.
The meridian HZN is the vertical circle passing through
the north and south points of the horizon.
The altitude TS of a star or point is its angular distance
above the horizon, measured on a vertical circle.
The zenith distance ZS is the complement of the altitude.
The azimuth of a star or point is the arc NT or the angle
NZT between the meridian and the vertical circle through the
star * or point. It is usually measured from the south point of
the horizon through the west.
The poles P and P' are the intersections of the axis of the
earth with the sphere. P is here the north pole. In conse-
quence of the earth's rotation about its axis the stars appear to
* That is, whose plane passes through the star.
APPLICATIONS OF SPHERICAL TRIGONOMETRY. 189
describe small circles about P as the pole, apparently moving
in the clirection EQWQ',
The equator EQ WQ' is the great circle cut from the sphere
by a plane through perpendicular to the axis of the earth.
The latitude of the observer is the angular distance QZ from
the equator to the zenith. Since PQ = 90° and ZN = 90°, we
liave NP = QZ^ i.e. the elevation of the pole above the horizon
is equal to the latitude of the place.
The hour circle of a star is the great circle PSD through the
star * and the pole. All the hour circles are perpendicular to
the equator.
The hour angle of a star is the angle at the pole between the
meridian and the hour circle of the star, measured from the
meridian to the west. Thus the hour angle of S is — ZPS^
negative since it is measured to the east. It is so named
because, if the angle ZPS is 15°, one hour will elapse before
PS coincides with PZ ; for 15° = 360° -- 24, and the star
appears to make a complete revolution about P in 24 hours of
sidereal (^i.e. star) time.
The declination DS of a star is its angular distance from the
equator, measured on its hour circle, and positive when the star
is north of the equator.
The right ascension of a star is the angular distance along
the equator from a certain point on the equator, called the ver-
nal equinox^ to the foot of the hour circle through the star,
measured towards the east ; or it is the angle at the pole be-
tween the hour circle of the vernal equinox and that of the
star.
Hence the angle between the hour circles of two stars is
equal to the difference between their right ascensions.
181. At a Place in Latitude 42° N. the Altitude of a Star,
whose Declination is + 60°, was measured and found to be
50°, the Star being East of the Meridian. At what Time did
the Star reach the Meridian? — In the triangle ZPS, ZP = 48°,
ZS = 40°, PS = 30° ; . •. by Art. 148, -|- ZPS = 29° 55^9 ;
.-. ^^.9= 59° 51'. 8 or 3* 59^5. Hence the star reached the
meridian 3*59™ .5 after the observation was made. f
• That is, whose plane passes through the star. t Sidereal time.
1^0
SPHERICAL TRIGONOMETRr.
182. The Latitude of the Place being 42° N., find the Interval
of Time between the Rising of a Star above the Horizon and its
Passage across the Meridian, its Declination being + 10°. — In
the triangle ZPS, S will be on the horizon NEH at the instant
of rising, so that ZS =90°. -
. •. cos ZaS' = = cos ZP cos SP + sin ZP sin SP cos ZPS.
. •. cos ZPS = - cot ZP cot SP ^- cot 48° cot 80°.
.-. ZPS=m° 8'.ror 6* 36'".5.*
Hence the star will be abcute_the horizon 13* 13"*.0.*
183. The Latitude of the Place being 42° N., and the Declina-
tion of the Star + 20°, find the Interval between the Instant
when it is due East and that when it is due West. — In the
triangle ZPS, PZS =90°.
. •. cos ZPS = tan ZP cot SP = tan 48° cot 70°.
.-. 2^P^=6G° 9^4. .-. 2Zi^AS'=132° 18'.8 = 8*49'».3.
Hence the interval required is 8* 49'".3.*
184. The Latitude being 42° N. and the Declination of the
Star + 80°, find the Azimuth of the Star when it is at its
Greatest Western Elongation ; that is, when the Star has reached
its Farthest Distance towards
the West, afterwards moving
East. — In the figure the
ZPS triangle is projected
upon the plane of the hori-
zon, so that Z is the zenith,
P the pole, S the star,
MSM' the apparent diurnal
path of the star about the
pole, ZP the meridian, ZS
the vertical circle of the star, and PZS the angle required, the
angle ZSP being a right angle.
Fig. 135.
* Sidereal lime.
APPLICATIONS OP SPHERICAL TRIGONOMETRY. 191
. •. sin SF = sin ZP sin PZS. . •. sin FZS = sin 10° cosec 48° ;
.-. PZaS'=13°30'.8.
Note. — This is the method ordinarily used by the engineer to determine
the north and south line.
185. The Right Ascensions of Two Stars
are a and a', and their Declinations 6 and 8' ;
find the Angular Distance between the Two
Stars.
SP=90°-B, S'P=:90°-S', SPS' = a'-a.
Hence we know two sides and the in-
cluded angle, and we find the third side SS' ^ fig. m.
by Art. 151 or by Art. 150.
186. If a' and a'^ are the Right Ascensions, and h' and 8'' the
Declinations of Two Stars, find the Inclination to the Equator of
the Great Circle passing through the
Stars, and also the Right Ascension of ^J^
the Point where it cuts the Equator. — ^^
Let B and D be the two stars, UQ v^e 'a
the equator, V the vernal equinox, fiq. ist.
E the intersection of the great circle BD with the equator,
VE = ay In the right triangle EAB,
sinEA = tiinAB cot AEB. .-. cot^=sin(a^ — «i)cot S'. (1)
In the right triangle ECD^
sin EO=tiiii CI) cot CED. .-. cot z = sin(«''-«i)cot 8'^ (2)
sin («" — wj) _ cot 8'
sin ((x' — cii) ~ cot h"
. ^ sin (f/^ - «^) + sin {a' - a^ _ cot h' + cot h^'
sin («" — «i) — sin («' — «j) ~~ cot 8' — cot 3''
,. tan I (ja' ^ + r/ - 2 a^^ _ sin (h" + 8^)
tan !-(«"-«') ~sin(8''-S')
... tanK«'^ + «'-2«0=|^§^^tanlC«''-«0. (3)
From (3) find \(^a'^ + «' — 2 Wj), thence finding a^i then i
may be found from either (1) or (2).
192
SPHERICAL TRIGONOMETRY.
187. The Right Ascension and Declination of a Star are a and
8, and those of Another Star are a' and 5' ; find the Hour Angle
Ox the First Star and their Common Azimuth when the Stars are
^g in the Same Vertical Circle, the Lati-
tude of the Place being <(). — There
are two positions, one when both
stars are west, and the other when
they are both east, of the meridian.
(1) S'F = dO°-S'; SF = 90°-^;
SPS' = a-a'\ ZP = ^T-(i>. In the
triangle SPS\ find P^S^^S' and PSS'.
Then in the triangle S'PZ we know
S^P, ZP, and PS'Z, and Ave find
PZS' and ZPS'. In the triangle
SPZ we know SP, ZP, and PSZ =
180° - PSS', and Ave find PZS and ZPS,
The checks are PZS' = PZS, and S' PZ-SPZ=:a-«! .
(2) S^P = 90° - S, S^P = 90° - a', S^PS^ = ar-a'; find
PS^S^' and PS^'S-^^, these angles being the same as those at
S and S' in the first case. Then from the tAvo triangles PS^Z
and PS^'Z Ave find the angles PZS^ and PZS^', Avhich should
bs identical, and also the angles S^PZ and S-^'PZ, Avhose dif-
ference should be a — a'.
"-•^^l Ir^
LOGARITHMItr AND iRIGONOiMETRIC
TABLES
FIVE DECIMAL PLACES
EDITED BY
C. W. CROCKETT
PROFESSOR OF MATHEMATICS AND ASTRONOMY
RENSSELAER POLYTECHNIC INSTITUTE
J'i9^<^
NEW YORK •:• CINCINNATI .:• CHICAGO
AMERICAN BOOK COM^^amv
:92
CONTENTS.
PAGH
3
Table I. Logarithms of Numbers
S\T',S'\T",ioT2r-s' ...... 24
II. Logarithms of Trigonometric Functions . . 25
III. Natural Trigonometric Functions .... 71
IV. Lengths of Circular Arcs 9-
V. Conversion of Logarithms
Formulas q-
Constants j^-
Explanation of the Tables 10-
NoTE. — The well-known tables of Gauss, Becker, and
/■il^.Tr^t have been taken as the standards, the proof sheets
have been read with gTCat care, and it is believed that the
numl)er of errors cannot be iar°^e. The arrangement of
the figures on the page is in accordance v'ith that adopted
in the standard six and seven place tables.
'I'he natural tables were reduced from seven-place tables
and compared with published five-place tables.
For convenience in using the tables, the explanation
' ten placed after them instead of before them.
\KIGHT, 1896, BY AMERIC-VN 3oOK
■
3
\
f'^ >\
(
. UNIVERSITY ]
I.
COMMON
LOGARITHMS OF
NUMBERS
'
• '
FROM I TO I 1000.
N.
Log.
N.
Log.
N.
Log.
N.
Log.
N. Log.
2
3 *
—
20
21
22
23
1.30 103
1.32 222
1.34242
1-36173
40
41
42
43
1.60206
60
61
62
63
1-77815
80
1.90309
O.OO CXXD
0.30 103
0.47712
1.61 278
1.62 325
1-63347
1-78533
1.79 239
1-79 934
81
82
1.90849
1.91 381
1.91 908
4-,
0.77815
24
25
26
1. 38 021
1-39 794
1.41 497
44
45
46
.1-64345
1.65321
1.66276
64
65
66
1.80 618
1.81 291
1.81 954
!4
8^^
1.92428
1.92942
1.93450
7
8
9
10
i t
0.84510
0.90309
0,95 424
1. 00 000
r.04 139
1.07 918
1.11 394
27
28
29
30
31
11
1-43 136
1.44716
1.46 240
49
50
51
52
53
1.67 210
1.68 124
1.69020
67
68
69
70
71
72
73
1.82607
1-83251
1.83885
1.84 510
89
90
91
92
93
1-93952
,1-54448
1-94539
M77I2
1.49-
1.50 i .
1.51851
1.69897
1.95 424
1.70757
1. 7 1 600
1.72428
1.85 126
1.85 733
1.86332
1.95904
1.96379
1.96848
H
IS
i6
1.14613
1.17609
1.20412
34
11
1-53 148
1.54407
1-55630
54
11
1-73239
1.74036
1.74 819
74
75
76
1.86923
1.87506
1.88081
94
96
1-97313
1.97772
1.98227
17
i8
1 20
1.23045
1-25 527
1.27875
1.30 103
37
38
39
40
1.56820
»-57 978
1.59 106
57
58
59
60
1-75 587
1-76343
1.77085
77
78
79
80
1.88649
1.89 209
1.89763
1.90309
97
98
99
100
1.98677
1.99 123
1.99564
1.60 206
1.77 815
2.00000
o
I
S'.
' 6.46 371
373
T.
37:^
37 ^
0° 0' - .
I -
i — 1
0"
60
20
S". T".
4-68 557 557
557 557
.^57 557
S'. T'.J N.
L.
1 2
3
4
5
6
7
8 ' 9
P. P. ^
6 "'■' f
.. ,^^_
366'
3.=
iDO
00000
043
087
130
^73
217
260
303
346
389
44 43 42
366
38s j 'Ol
432
475
578
561
604
/647
♦072
689
732
775
817
366
38, i02
860
903
945
988
♦030
*ii5|*i57
*i99
*242
J ^ . . ^ ,. „
^•4 4- j 4--^ '
8.8 8.6 8.4
13.2 12.9 12.6
17.6 17.216.8
366
• 366
3&'' 1 03
01284
703
326
745
3^8
787
410
828
452
870
494
912
536
^11
578
995
620
♦036
662
*078
2
3
4
366
X^i- \ 05
02 119
I bo
202
243
284
325.
366
407
449
490
5
6
22.0 21.5 2 f.o
26.425.825.2
366
3«n
io6
531
572
612
653
694
735
776
816
857
898
366
3S7
',07
938
979
♦019
*o6o
*IOO
*i4i
*i8i
♦222
*262
*302
7
30.8 30.1 2C.4
365
387
108
03342
383
423
463
503
543
583,
623
663
703
8
35.2 34.4 33.6
365
'365
I365
387
^7
3S8
T09
III
743
J^2^
822
862
902
941
9^1
*02I
*o6o
*IOO
9
39-6 38.7 37-8
04i39_
J79_
218
258
"297^
336
~37^
415
454
493
41 40 39
532
571
610
650
689
727
766
805
844
883
365
388
112
922
961
999
*038
*o77
*ii5
*^54
*192
♦231
*269
^
4.1 4.0 3.0
8.2 8x3 7.:,,
365
38S
"3
05 308
346
385
423
461
500
538
576
614
652
2
■'
36S
389
114
690
729
767
805
843
881
918
956
994
♦032
3
4
12.3 12.0 11.'^
16.4 16.0 15.61
20.5 20.0 19.5
24.6 ? 1 n 23,4 1
365
389
115
06070
108
145
185
221
258
296
333
371
408
364
389
116
446
483
521
558
595
633
670
707
744
781
5
6
364
389
ri7
819
856
893
930
967
*oo4
♦041
*o78
*ii5
*i5i
7
28.7 28.0 27.3
364
390
118
07 188
225
262
298
335
372
408
445
482
518
8
32.832.031.2
364
- 364
390
'I9
120
555_
918
59t
954
628
6b4
700
737
*099
773
*i35
809
846
882
9
36.9 36.0 3^.1
990
♦027
*o63
*i7i
*207
*243
38 37 36
364
391
121
08 279
3H
350
386
422
458
493
529
565
600
363
39'
J 22
636
672
707
743
778
8.4
849
884
920
955
I
3.» 3-7 3-^\
363
39 ■
123
991
*026
*o6i
♦096
*I32
*i67
♦202
*237
*272
*307
2
7.6 7.4 7-2'
363
39 r
124
09342
377
412
447
482
517
552
587
621
656
3
11.4 11. 1 10.^'
15.2 14.8 14./
19X) 18.5 18.0
22.8 22.2 2I.(
3<^3
392
125
691
726
760
795
830
864
899
934
368
♦003
4
5
6
,363
352
126
10037
072
106
140
ns
209
243
278
312
346
,36,
392
127
380
415
449
483
517
551
585
619
653
687
7
26.6 25.9 25.2
I363
393
128
721
755
789
823
857
890
924
958
992
♦025
8
^0.4 29.6 28.8
362
362
393
393
394
129
I30
131,
II 059
093
126
160
193
227
261
294
327
361
9
34-2 33-3 32.4'
394
727
428
461
494
528
561
594
628
661
694
35 34 33
760
793
826
860
893
926
9t;Q
992
♦024
\lt..
394
132
120^7
090
123
156
189
222
254
287
320
352
I
3-5 3-4 3-3
362
3;>4
^^2,
385.
4j8
450
483
516
548
581
613
646
678
2
7.0 6.8 6.6
362
-
395
134
710
743
775
808
840
872
905
937
969
*OOI
3
10.5 10.2 9,9
14.0 13.6 13.2
17.5 17.0 16.5
21.0 204 19.8
24.5 23.8 23.1
28.0 27.2 26.4
^i c -in 6 00-7
361
393
J 35
13033
066
oq8
130
162
194
226
258
290
322
4
361
361
395
396
136
U7
* 354
672
386
704
418
735
450
767
481
799
513
830
545
862
577
893
609
925
640
956
5
6
361
396
138
988
*oi9
*oqi
♦082
*ii4
*I45
*i76
*2o8
*239
♦270
361
396
139
14301
333
364
^1
426
457
489
520
551
582
Q
361
360
397.
397
140
141
613
922
644
953
675
983"
706
"737"
*045
768
*076
799
*io6
829
*'37
860 ] 891
32 31 30
*oi4
*i68
*i98
i;6o
397
142
11; 229
259
290
320
351
381
412
442
473
503
I
3-2 3-t 3-0
\yo
393
143
■^534
564
594
625
655
685
715
746
776
806
2
6.4 6.2 6.0
J360
398
144
836
866
897
927
957
987
*oi7
*047
*077
*io7
3
9.6 9.3 9.0
I360
398
^45
16137
167
197
227
-2-56-
286
316
346
376
406
4
12.8 12.4 12.0
: 360-
399
146
435
465
49?
524
554
584
613
643
673
702
5
'^•° "il Tn
359
399
147
732
761
791
820
850
879
909
938
967
997
6
19.2 18.6 18.0
359
399
148
17026
056
085
114
143
^73
202
231
260
289
7
8
22.4 21.7*21.0
25.6 24,8 24.0
359
359
400
400
149
ISO
' 3- 348
609 538
377
406
435
464
493
522
81 r
551
580
667
696
725
754
782
840
869
9 ^^'"^ -^i'y ^/-^
N.
L. 1 1
2
3
4
5
6
7
8
9
P.P.
S.' T.'
II S." T."
S." T."
I' 646 373 373
0° i'= 60" 4.68557 557
0° 19'= 1 140" 4-68557 558,.
2 373 373
2 = 120 557 557
20=1200 557 558
21 =1260 5f7 50
22=1320 557 5ju
' '■' "^' :■ / J
3 = 180 ^K-j 557
i73
16 = 960 557 558
: " N>73
17 =IC20 557 558
c 23=1380 ■ • 557 558
^3 t::\
11 18 "=:I0S0 S"^/ 558
:r! =1440 557 558
11 ,
,r. —
f 1 »r»
c^7
<;•,?
i
5 =^
500
557 558 1
^
.(_
_,_
b . T'.'
N~
L. 1.
2 1 3 1 4
5 1 6
7
8
9
p. p. 1
1 ■■'■
400
40t
150
15'
I7OO9
898
9:^13
667
OoOi 725
754
♦041
*070
811
*«S99~
840
*I27
869
*I56
29 9.^
i| 2.9
2! 5.8
3i 8.7
4!ii.6 1 i
955
984 1*013
401
IS2
18184
2K3
241
j/O , 298
327
355
384
412
441
401
^53
469
498
526
55-J 1 583
611
<539
667
696
724
402
154
752
780
808 1 8:57 ! S65
893
921
, 949
977
*cx)5
358
402
155
19033
061
089
ii7 H5
173 j 201
229
257
28S
5 M-5 '4^'
617.4 r.,X
il
1358
402
156
312
340
368
390 1 424
45' 1 479
507
535
562
i
358
403
157
590
ii%
645
673! 700
728
756
783
811
838
720.3 I
i
357
403
i';S
86^.
893
921
948 976
*oo3
♦030
*o5il
♦085
*II2
8 23.2 J
926.1 ^ ,.
27 iH
,
357
151
357
404
40.
404
159
160
161
20 140
it>7
194
222 249
^6
303
330
385
H'2
439
466
493 520
548
575
602
629
656
' 683
710
'737
7631 790
817
844
'871
898
925
357
405
162
952
978
♦005
*032 1*059
♦085
*iU
*i39
♦165
*I92
I
^•7 -•'-■ 1
356
405
163
2Jl, 219
245
272
299
325
352
378
405
431
458
2
11 11 i
356
400
164
"484
5"
537
564
Sm
617
643
6(69
696
722
3 o» /" :
4 lp.8 ,io.4 1
513-5 J 30 ,
6 16.2 i;.6 '
7li3.9 18.2 !
356
406
167
.748
775
801
^-^^
880
906
932
958
985
356
356
406
407
22 0ir
272
037
298
063
324
0S9
350
"5
376
141
401
}^
'94
220
479
246
505
427
453
355
407
168
531
557
il
608
634
660
686
712
737
763
821.6 20.8 i
r
255
355
408
'403
408
169
I70
171
789
814
866
891
917
943
968
994
*oi9
9124.3 23.4 !
25
23?4|_
070.
096
121
147
172
198
223
249
274
'300
325
350
_
376
401
426
452
477
502
52^8
\
409
172
553
578
603
629
654
679
704
729
754
779
I
2.
409
^n
805
830
851
880
905
930
9^
980
*oo5
♦030
2
.5-
410
174
24055
080
103
130
155
180
204
229
254
279
3
4
i
7
Tn
TO
4
410
175
304
329
353
378
403
428
45^
477
502
527
12
15... ;
1354
411
176
551
576
601
625
659
674
699
724
748
773
'353
411
177
797
822
846
895
920
944
969
993
♦oi8
7
353
411
178
25042
666
091
115
139
164
188
212
237
261
8
20' \
353
353
412
179
180
285
527
310
55J_
.314_
575
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2° 8'= 7680" 4-68547 578
8
373
373
2 9=7740 547 578
2 10=7800 547 578
"211 =7860 547 579
75
80
369
369
380
380
2
5=7500 /S48 577
2
6=7560 ./ 548 577
7 =7620 / 548 577
2 12=7920 547 579
2
2 13=7980 547 579
2
8. .7680 . 547 578
2 14=8040 546 579
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2
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18 = 8280
546 580
546 581
80
81
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369
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368
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380
381
381
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2 13^=7980 -\47 579
2 14 = 8040 5:,^ 579
2 15 =8100 54t 580
2 i6 = 8i'o 546] 5 So
19=8
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300
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9
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512
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596
601
606
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109
114
119
124
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134
139
143
148
153
158
163
168
173
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182
187
192
197
202
207
211
216
221
226
896
231
236
240
245
250
255
260
265
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274
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279
284
289
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328
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4
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8'
6.46 373 373
0° 14'= h
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2°25' = 87
00" .
^68545 583
9
373 373
15 - 9
00
557 558
2 26 =87
60
20
544 584
544 584
8S
368 381
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545 582
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86
368 382
2 22 =81:
20
545 582
2 28=8S
80
544 584
89
368 382
2 23 =8c
80
545 583
2 29=89
40
544 585
r
368 383
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2 25-=8;
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700
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545 583
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00
544 585
20
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2
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900
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497
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516
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521
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530
535
540
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554
559
564
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574
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588
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617
622
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670
674
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718
722
727
732
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751
756
907
761
766
770
775
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918
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933
938
942
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961
966
971
976
980
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104
109
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142
147
152
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166
171
175
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190
194
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213
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270
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298
303
308
313
317
322
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341
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350
355
360
365
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440
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544
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558
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600
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633
638
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656
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661
666
670
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179
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188
192
197
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206
211
216
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220
225
230
234
239
243
248
253
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8
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940
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271
276
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470
474
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483
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497
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516
520
525
529
534
539
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548
552
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571
575
580
585
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594
598
603
607
612
617
621
626
630
947
635
640
644
649
653
6s8
663
667
672
676
948
681
685
690
695
699
704
708
713
717
722
949
950
727
731
736
740
745
749
754
759
763
768
772
777
782
786
791
795
800
804
809
813
N.
L.
1 1
2 1
3 1
4
5 1 6 1 7 1
8 1
9 1 P.P.
S/
T.'
S." T."
S." T."
9'
6.46 373
373
0° I
5'- S
00" 4-68557 558
2° 3^
^'=92
^0" 4.68543 587
10
373
373
I
6= c
)6o 557 558
2 3
2 3
5=93
3 = 93
30 543 587
So 543 587
90
368
383
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= 9C
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91
368
383
2 -i
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2 3
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20 542 588
92
367
383
2 3
2=91
20 543 586
2 3
^ = 94
So 542 588
4.0 542 588
94
367
383
2 3
3 = 91
80 543 586
2 3
:^=95
95
367
384
2 3
4 = 9:
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p.pr
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100
105
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150
155
159
164
168
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195
200
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214
218
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5
227
272
232
236
241
245
250
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304
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308
268
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277
281
286
290
295
299
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322
327
331
336
340
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349
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408
367
412
372
417
376
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381
426
385
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2
3
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5
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453
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462
466
471
475
480
484
489
493
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498
502
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516
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547
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556
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565
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632
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109
114
118
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123
127
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136
140
145
149
154
158
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167
171
176
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193
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216
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326
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568
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594
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616
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629
634
638
642
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656
660
664
669
673
677
682
686
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695
699
704
708
712
717
721
726
730
734
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739
743
747
752
756
760
765
769
774
778
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782
787
791
795
800
804
808
813
817
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870
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900
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930
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961
965
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978
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004
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013
017
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026
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035
039
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1 I
2 1 3
4 1 5 1 6 1 7 1 8
1 9 1 p.p.
S.' T.'
S." T.'
S/' T."
9' 6.46373 373
0° 15'= 900" 4.68557 55^'
2° 41'= c
)66o" 4.68542 589
10 373 373
16 = 960 557 55^
2 42= c
> 2 43= c
)720 541 590
?78o 541 590
?840 541 590
95 367 384
98 367 384
17=1020 557 55^
2 38=9480 542 58^
\ 2 44 = <
99 367 385
2 39 =9540 542 58^
\ 2 45 = <
J906 541 591
100 366 385
2 40 =9600 542 58c
J 2 46= <
^960 541 591
2 41 =9660 542 58c
) 2 47=i(
D020 540 592
N.
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2
3 1 4
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1
2
3
4
5
6
7
8
9
S."
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S." T."
2° 46' = 9960
' 4.
58541
591
2°5i' = 10260"
4.68 540 593
2 47 = 10020
540
592
2 52 = 10320
539 594
2 48 = 10080
540
592
2 53 = 10380
539 594
2 49 = 10140
540
592
2 54 = 10440
539 595
2 50 = 10200
540
593
2 55 = 10500
539 595
23
N.
L. 1 1
2
3 1 4
5
6 1 7 1 8
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1051
1052
1053
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N.
L.
1
2
3
4
5
6
7 8
9
S."
T."
S." T."
2° 55' = 10500" 4.(
38539
595
3°o'= 10800" 4.68538 597
2 56 = 10560
539
.595
3 I = 10860 537 598
2 57 = 10620
538
596
32= 10920 537 598
2 58 = 10680
538
596
3 3= 10980 537 599
2 59 = 10740
538
597
34=1 1040 537 599
24
3°
/
M.
S'. T'.
Sec. 1 S". T". ll
6.46 1
4.68 II
I
180
353
412
10800
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181
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2
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13860
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239
240
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337
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14400
522 1 628 II
' M. 1 S'. T'. 1
Sec. 1 S". T". 1
1
6.46 1
4.68
240
337
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522 ! 628
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2
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14880
520
633
9
10
II
249 335 i 449
14940
520
634
250 i 334 1 449
251 i 334 ! 450
15000
519
519
634
635
15060
12
252 1 334 1 450
15120 1 519
63s
13
253 333 451
15180 j 518
636
H
254
333 452
15240 i 518
637
15
255
333 1 452
15300 518
637
16
256
332
453
15360 1 517
638
17
257
332
454
15420 1 517
638
18
258
332
454
15480 1 517
639
19
20
21
259
260
261
332
455
15540 1 516
640
331
456
15600 1 516
640
331
456
15660
516
641
22
262
331
457
15720
515
642
23
263
330
457
15780
515
642
24
264
330
458
15840
515
643
25
26s
330
459
15900
514
644
26
266
329
459
15960
5H
644
27
267
329
460
16020
514
645
28
268
329
461
16080
513
646
29
30
31
269
328
461
16140
513
646
270
328 1 462
16200 1 513
647
271
328
463
16260 1 512
648
32
272
327
463
16320 1 512
648
33
273
327
464
16380 1 512
649
34
274
327
465
16440 511
650
35
275
326
465
16500 511
650
36
276
326 ! 466
16560 1 511
651
37
277
326 i 467
16620 1 510
652
38
278
325 467
16680 510
652
39
40
41
279
280
325
325
468
16740 510
653
469
16800 i 509
654
281
324 ' 469
16860 1 509
654
42
282
324 470
16920 509 1 655 1
43
283
324 1 47»
16980
508
656
44
284
323 ! 472
17040
508
656
45
28s
323 j 472
17100
508
657
46
286
323
473
17160
507
658
47
287 i 322
474
17220
507
659
48
288
322 i 474
17280
507
659
49
50
51
289
321 ! 475
17340
506
b6o
290
321 j 476
1740G 506 i 661
291 1 321 1 477
17460
506 1 661
52
292 1 320 i 477
17520
505 j 662
53
293
320 478
17580
505
663
54
294
320 ! 479
17640
505
664
55
29 s
319
479
17700
504
bb4
56
296
319
480
17760
504
665
57
297
319
481
17820
503
666
58
298
318
482
17880
503
666
59
60
299
318
482
17940
503
667
300
317
483
18000
502
668
25
r
1
II.
1
THE LOGARITHMS
•
OF THE
TRIGONOMETRIC FUNCTIONS
FOR EACH MINUTE.
Formulas for the Use of the Auxiliaries 5 and T.
1. When a is in the Hrst five degrees of the quadrant :
log sin a = log a' + S.'
log a' = log sin a + cpl SJ
log tan a = log a' + 7'.'
= log tan a + cpl 7'.'
log cot a = cpl log tan a.
= cpl log cot a 4- cpl 7'.'
log sin a = log a" + S."
log a" = log sin a + cpl 5."
log tan a = log a" + TJ'
= log tan c + cpl 7"."
log cot a = cpl log tan a.
= cpl log cot a -1- cpl 7."
2. When a is in the last five degrees of the quadrant :
log cos a = log(90° - a)' + SJ
log(90° — a)' = log cos a + cpl SJ
log cot a = log(90° - a)' + TJ
= log cot a + cpl y\'
log tan a = cpl log cot a.
= cpl log tan a + cpl T.'
log cos a 3= log(90° - a)" + 6'."
log(90°-a)"= log cos a -f cpl 5."
log cota = log(90° - a)" + 7V'
= log cot a + cpl r."
log tan a = cpl log cot a.
= cpl log tan a + cpl 7'."
a = 90° -(90° -a).
/
26
0°
//
/
L. Sin.
1 d.
iCpl. S'. 1 Cpl. T'.
L. Tan. jc. d.
L. Cot.
1 L. Cos.
o
646 373
'^0101
—
—
—
30103
17609
12494
—
0.00000
60
59
60
3-53 627
3.53627
6.46 373
3-53 627
0.00000
120
2
6.76476
17609
12494
3-53627
3-53627
6.76476
323 524
0.00 000
S8
1 80
3
6.94 085
353627
3-53627
6.94 085
3-05915
0.00 000
57
240
4
7.06579
9691
3-53 627
3.53 627
7.06579
9691
7918
66q4
2.93421
0.00 000
S6
300
5
7.16 270
3-53 627
3-53627
7.16 270
2.83 730
0.00 000
55
360
6
7.24 188
6694
3-53 627
353627
7.24 188
2.75812
^ 0.00 000
54
420
7
7.30882
5800
3-53 627'
3-53627
7.30 882
5800
2.69 118
1 0.00 000
53
480
8
7.36682
3.53627
353627
7.36 682
2.63318
1 0.00 000
52
540
600
9
10
7.41 797
4576
4139
3779
3476
3218
2997
3-53 627
353627
7.41 797
4576
4139
3779
3476
3219
2oa6
2.58 203
0.00 000
51
50
49
746373
3-53627
3-53627
7-46 373
2-53 627
0.00000
660
750 5' 2
3-53627
3-53627
7.50512
2.49 488
0.00000
720
12
7.54291
•3-53 627
3-53627
7-54 291
2.45 709
o.coooo
48
780
13
7-,57 767
3-53627
3.53 627
7-57767
2.42 233
0.00 000
47
840
14
7.60 9S5
3-53628
3-53627
7.60 986
2.39014
0.00 000
46
900
15
7.63 982
2802
3-53 628
3-53627
7.63 982
2803
2633
2.36018
o.co 000
45
960
16
7.66 784
2633
353628
3-53 627
7.66 785
2.33215
0.00 000
44
1020
7
7.69417
2483
3-53 628
3-53 627
7.69418
2.30 582
9-99 999
43
1080
18
7.71 900
2348
2227
2119
3-53 628
3-53627
7.71 900
2348
2228
2119
2.28 100
9.99 999
42
1 140
19
20
21
7.74 248
3-53 628
353627
7-74 248
2.25 752
9-99 999
41
40
39
1200
1260
7-76 475
3.53628
353627
7.76476
2.23 524
9-99 999
7-78594
353628
3-53 627
7-78 595
2.21 405
9-99 999
1320
22
7.80615
3.53628
3.53627
7.80615
2.19385
9.99 999
38
1380
23
7-82 545
1930
1848
353628
3-53627
7.82546
1931
1848
2.17454
9.99 999
37
1440
24
7-84 393
3-53 628
3-53627
7-84 394
2.15 606
9 99 999
36
1500
2S
7.86 166
353628
3-53627
7.86 167
2.13833
9-99 999
35
1560
26
7.87 870
1639
353628
3-53627
7.87871
1639
2.12 129
9.99 999
34
1620
27
7.89 509
1579
1524
1472
1424
3.53 628
3-53626
7.89510
1579
1524
1473
1424
2.10490
9 99 999
V^
1680
28
7.91 088
3-53 628
3-53.626
7.91 089
2.08911
9-99 999
32
1740
29
30
31
7.92 612
3-53 628
3-53626
7.92613
207387
9.99 998
31
30
29
1800
i860
7.94084
3-53 628
3-53626
7.94 086
2.05 914
9-99 998
7-95 508
3.53628
3-53626
7.95510
2.04 490
9-99 998
1920
32.
7.96887
1336
1297
3.53 628
3-53 626
7.96 889
1336
1297
2.03 1 1 1
9-99 998
28
1980
33
7.98 223
3.53 628
3-53 626
7.98 225
2.01 775
9-99 998
27
2040
34
7.99 520
3.53628
3-53 626
7.99 522
2.00 478
9-99 998
26
2100
3S
8.00 779
353628
3-53626
8.00 781
1.99 219
9.99 998
25
2160
36
8.02 002
1 190
1158
3-53 628
353626
8.02 004
1190
"59
1 128
1.97996
9-99 998
24
2220
37
8.03 192
3.53 628
3-53 626
803194
1.96806
9.99 997
23
2280
3«
8.04 350
3-53 628
3-53 626
804 353
1.95 647
9.99 997
22
2340
2400
39
40
41
S.05 478
1 100
1072
1046
353628
3.53 626
8.05 481
IIOO
1072
1.94 5 19
9-99 997
21
20
19
8.06 578
3-53 628
353625
8.06581
1.93419
9-99 997
2460
8.07 650
3-53 628
3-53 625
8.07 653
1-92347
9-99 997
2520
42
8.08 696
3-53 628
353625
8.08 700
1.91 300
9.99 997
18
2580
43
8.09718
999
976
3-53 629
353625
8.09 722
998
976
1.90 278
9.99 997
17
2640
44
8.10717
3-53 629
3-53625
8.10 720
1.89280
9.99 996
16
2700
4S
8. 1 1 693
3.53 629
3-53625
8. 1 1 696
1.88304
9-99 996
15
2760
46
8.12647
954
934
3.53629
3-53625
8.12651
955
934
1-87349
9 99 996
14
2820
47
8.13 581
3-53 629
3-53 625
8.13585
1.86 415
9.99 996
13
2880
48
8.14495
914
896
877
860
843
812
3-53 629
3-53625
8.14500
H'5
895
878
860
843
828
812
1.85 500
9.99 996
2940
49
50
SI
8.15391
3-53629
353624
8.15395
1.84605
9.99 996
11
10
9
3000
3060
8.16268
3-53 629
3-53 624 I
8.16273
1.83727
9.99 995
8.17128
3-53 629
3.53 624
8.17 133
1.82867
9-99 995
3120
S2
8.17971
3-53629
3-53 624
8.17976
1.82024
9-99 995
8
3180
53
8.18798
3-53629
3-53 624
8.18804
1.81 196
9-99 995
7
3240
S4
8.19 610
782
769
353629
3-53 624
8.19616
797
782
769
756
742
730
1.80384
9-99 995
6
3300
ss
8.20407
3-53 629
3-53 624
8.20413
1-79587
9-99 994
5
3360
56
8.21 189
3-53 629
3 53 624
8.21 195
1.78 805
9.99 994
4
3420
57
8.21 958
TCC
353629
3-53623
8.21 964
1.78036
9.99 994
3
3480
58
8.22713 ;r, 1
3-53 629
3-53623
8.22 720
1.77 280
9.99 994
2
3540
59
60
8.23456
730
3 53 630
3-53623
8.23 462
1.76538
9.99 994
I
3600
8.24 186
3.53 630
3-53 623
8.24 192
1.75 808
9-99 993
L. Cos.
d. 1
1
L. Cot.
c. d.
L. Tan.
L. Sin.
/
89'
27
I L. Sin.
d. I Cpi. s^ i cpl.T^
L. Tan. c d. L. Cot.
L. Cos.
3600
3660
3720
3780
3840
3900
3960
4020
4080
4 140
4200-
4260
4320
4380
4440
4500
4560
4620
4680
4740
4800
"486^
4920
4980
5040
5100
5160
5220
5280
5340
5400
5460
5520
5580
5640
5700
5760
5820
5880
5940
6000
6060
6120
6180
6240
6300
6360
6420
6480
6540
6600
~66fe
6720
6780
6840
6900
6960
7020
7080
7140
7200
o
I
2
3
4
5
6
7
8
9
iO
II
12
13
H
»5
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
8.24 186
8.24 903
8.25 609
8.26 304
8.26 988
8.27661
8.28324
8.28977
8.29 621
8.30 25 ^
8.30 879
8.31 495
8.32 103
8.32 702
8.33 292
8.33875
8.34 450
8.35018
8.35 578
8.36 131
8.36 678
8.37217
8.37 750
8.38 276
8.38 796
8.39310
8.39818
8.40 320
8.40816
8.41 307
8.41 792
8.42 272
8.42 746
8.43 216
8.43 680
8 44 139
8.44 594
8.45 044
8.45 489
8.45 930
.46 306
8.46 799
8.47 226
8.47 650
8.48 069
8.48 485
8.48 896
8 49 304
8.49 708
8.50 108
■50 504
8.50 897
8.51 287
8.51673
852055
8.52434
8.52810
8.53 183
8-53 552
8.53919
■54 282
717
706
695
684
673
663
653
644
634
624
616
608
599
590
583
575
568
560
553
547
539
533
526
520
5H
508
502
496
491
485
480
474
470
464
459
455
450
445
441
436
433
427
424
419
416
411
408
404
400
396
393
390
386
382
379
376
373
369
367
363
3:53 ^30
3-53 630
3-53 630
3-53 630
3.53 630
3-53 630
3-53 630
3-53 630
3-53 630
3-53 630
3.53 630
3-53 630
353631
3-53631
353631
353631
353631
353631
3-5363'
3-53631
3.53631
353631
3-53 632
3-53 632
3-53 632
353632
3-53632
3-53 632
353632
353632
3-53 632
3-53 632
3-53 633
3-53 633
3-53 633
353633
3-53633
3-53 633
3-53633
3-53 633
3-53 634
3-53 634
3-53 634
3-53 634
3-53 634
3.53 634
3-53 634
3-53 634
3-53 635
3-53 63 5
3-53 635
3-53 635
3-53 635
3-53635
3-53635
3-53 635
3.53 636
3-53 636
3 53 636
3_5 3 6^6
3-53636
3i53 623
353623
3-53623
3-53 623
3.53 622
353622
353622
353622
353622
353622
3-53621
3-53621
3-53621
3-53621
353621
3.53 620
3.53 620
3.53 620
3-53 620
3-53 620
3-53620
3-53619
3-53619
3-53619
3-53619
3-53619
3-53618
3-53618
3-53618
3-53618
3-53617
3-53617
3-53617
3-53617
3-53617
3-53616
353616
3-5:
616
3-53616
3-53615
3-53615
3-53615
3-53615
3-53614
3-53614
3-53614
3-53614
3-53613
3-53613
3-53613
3-53613
3-53612
3-53612
3-53612
3-53611
3-53611
353611
3-53611
3-53610
3-53 610
•53610
8.24 192
8.24910
8.25616
8.26312
8.26 996
8.27 669
8.28 332
8.28 986
8.29 629
8.30 263
8:30888
8.31 505
8.32112
8.32 7"
8.33 302
8.33 886
8-34461
8.35 029
8.35 590
8.36 143
8.36689
8.37 229
8.37 762
8.38 289
8.38 809
8-39 323
8.39 832
8.40 334
8.40 830
8.41 321
8.41 807
8.42 287
8.42 762
8.43 232
8.43 696
8.44 156
8.44 61 1
8.45 061
8.45 507
8.45 948
8.46 385
8.46817
8.47 245
8.47 669
8.48 089
8.48 505
8.48917
8.49 325
8.49 729
8.50 130
8-50527
8.50 920
8.51 310
8.51 696
8.52079
8.52459
8.52831
8.53 208
8.53578
8.53 94 5
8.54 308
718
706
696
684
673
663
654
643
634
625
617
607
599
591
584
575
568
561
553
546
540
533
527
520
514
509
502
496
491
486
480
475
470
464
460
455
450
446
441
437
432
428
424
420
416
412
408
404
401
397
393
390
386
383
.380
376
373
370
367
363
75808
75090
74384
73688
73004
72331
71 668
71 014
70371
69737
9-99 993
69 112
68T95
67888
67289
66698
66 1 14
65539
64971
64 410
63857
9-99 993
9.99 993
9 99 993
9.99 992
9 99 992
9 99 992
9.99 992
9-99 992
9i99 99i
9.-99 99 y
9.99991
9.99 990
9.99 990
9.99 990
9.99 990
9.99 989
9.99 989
9 99 989
9-99 989
63311
62771
6:2238
61 711
61 191
60 677
60 168
59 666
59170
58679
9.99-9 88
9.99 988
9-99 988
9-99 987
9 99 987
9.99 987
9-99 986
999986
9.99 986
9-99 985
58193
57713
57238
56768
56304
55844
55389
54 939
54 493
54052
9-99 98 5
9-99 985
9-99 984
9.99 984
9.99 984
9.99 983
9-99 983
9.99 983
9.99 982
9.99 982
53615
9.99 982
53183
52755
52331
51 911
5M95
51083
50675
50271
49870
9.99 981
9.99 981
9.99981
9-99 980
9.99 980
9.99979
9.99979
9.99 979
9-99978
49 473
49080
48 690
48304
47921
47541
47 165
46 792
46 422
46055
9.99 9 78
9.99977
9.99977
9.99977
9-99 976
9-99 976
9-99 975
9-99 97l
9.99 974
9-99 974
45692
9-99 974
L. Cos.
L. Cot. |c. d. I L. Tan.
L. Sin.
QQ°
28
2^
n
/
L. Sin. j d.
LCpl. S'.
Cpl. T'.
1 L.Tan.
c. d. 1 L
.. Cot.
L. Cos.
7200
7260
I
8.54 282
360
357
355
351
1 3-53 636
3.53610
8.54 308
361 '•
45692
9-99 974
60
S9
8.54 642
3.53 636
3.53 609
8.54 669
301
45331
9-99 973
7320
2
8.54 999
3-53 637
3-53 609
8.55 027 :^^r i I
44 973
9.99 973
58
• 7380
3
8-55 354
3-53 637
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,09 920
.09 760
.09 601
•09 443
.09 285
.09 128
.08 971
9.99 865
9 99 864
9.99 863
9.99 862
9.99 861
9.99 860
9-99 859
9.99 858
9-99 857
08815
9-99 856
08660
oS 505
08350
08 197
08043
07 890
07738
07586
07435
07 284
07 134
,06 984
06835
06 687
06 538
06391
06 244
06097
05951
05 805
9-99 855
9.99 854
9-99 853
9.99 852
9.99 851
9 99 850
9.99 84S
9 99 847
9.99 846
9-99 845
9.99 844
9.99 843
9 99 842
9.99841
9.99 840
9-99 839
9.99 838
9-99 837
9.99 836
60
59
58
57
56
55
54
53
52
51
50
49
48
47
46
45
44
43
42
41
40
39
3^
37
36
35
34
33
32
31
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
15+T
9-99 834
182
181
ISO
1
18.2
18.1
18.0
2
36.4
36.2
36.0
3
54-b
54-3
540
4
72.8
72.4
72.0
5
91.0
QO.s
1,0.0
b
109.2
108 6
108.0
7
1274
126.7
126.0
8
145.6
144.8
144-0
9
163.8
162.9
162.0
177
176
175
I
17.7
176
17-5
2
35 4
35-2
35-0
3
53-1
52.8
52.. s
4
70.8
70.4
70
5
88. s
88.0
«7.S
6
106.2
105.6
105.0
7
123.9
1232
122.5
b
141.0
140.8
140.0
9
159-3
158.4
157.5
179
17.9
35-3
53-7
71.6
^9-5
107.4
125.3
143-2
161. 1
17.4
34.8
52.2
69^6
87.0
104.4
121.8
139.2
is6.6
178
17.8
35-6
53-4
71.2
89.0
106.8
124.6
142.4
160.2
17-3
34-6
5'-9
69.2
86.5
103.8
121. 1
138.4
155-7
172 171 170
17.2
34-4
51-6
68.8
86.0
103.2
120.4
137-6
1548
17. 1
34-2
51-3
68.4
85-5
102 6
119.7
136.8
170
34-0
51.0
68.0
85 o
169
16.9
168
16.8
33-8 33-6
50.7 50.4
67.6 67.2
84.5 84.0
02.0 101.4 100.8
19.0 118.3 117.6
36.0 135.2 134.4
530 152.1 151.2
167 166 165 164 163
16.7 16.6 16.5 16 4 16.3
33-4 33-2 33.0 32.8 32.6
50.1 49.8 49.5 49.2 48.9
66.8 66.4 66.0 65.6 65.2
83.5 830 82.5 82.0 81.5
100.2 99.6 99.0 98.4 97.8
116.9 u6.2 115.5 114.8 114. 1
133.6 132.8 132.0 131 2 130.4
150.3 149.4 148.5 147.6 146.7
162 161 160 159 158
16.2 16.1 16.0 15.9 15.8
32 4 32 2 32.0 31.8 31.6
48 6 48.3 48 o 47.7 47.4
64 8 64.4 64.0 63.6 63.2
81.0 80.5 80.0 79.5 79.0
97.2 96.6 96.0 95.4 94 8
113. 4 1127 112.0 III. 3 ito.6
129.6 128.8 128.0 127.2 126.4
145.8 144.9 ^44-o 143-1 142.2
157
15.7
31-4
47-1
62.8
785
94 2
109.9
125-6
141-3
156
15-6
31.2
46.8
62.4
78.0
936
[09
155 154 153
15-5
31.0
46.5
62.0
77-5
93 o
0S.5 107
124.8 124.0 123 2
140.4 139.5 1386
15-4
30.8
46.2
61.6
77.0
92.4
149
149
29.8
44 7
59-6
74-5
894
106 4 105.7 105.0 104.3
121 6 120.8 120.0 119. 2
136.8 135.9 135-0 134. 1
152
151
15.2
I5-I
30.4
30.2
45 b
45-3
bo.8
bo.4
76.0
75-5
91.2
90.6
150
15.0
30.0
45-0
60 o
750
90.0
15-3
30.6
45-9
61.2
76.5
91.8
107.1
122.4
137-7
148
14.8
29.6
44-4
59-2
74.0
88 8
103.6
118.4
133 2
I L. Cos.
L. Cot. 0. d
L. Tan.
L. Sin.
P. P.
fifi°
31
I L. Sin.
d. I L. Tan. |c.d. j L. Cot.
L. Cos.
P.P.
O
I
2
3
4
5
b
7
8
9
iO
II
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
3«
39
40
41
42
43
44
45
46
47
48
49
50
5'
52
53
54
55
56
57
5«
59
60
8.94 030
8.94 '74
J^.94317
8.94 461
8.94 603
8.94 746
8.94 887
8.95 029
8.95 170
8^5310
8-95_4|o_
8.95 5^9
8.95 728
8.95 867
8.96 005
8.96 143
8.96 280
8.96417
^•96553
8.96 689
8.96 825
8.9b 960
8.97 095
8.97 229
8.97 363
8.97 49b
8.97 629
8.97 762
8.97 894
8.98 026
8.98157
8.98 288
8.98419
8.98 549
8.98 679
8.98 808
8.98937
8.99 066
8.99 194
8.99 322
8.99 450
8.99 577
8.99 704
8.99 830
8.99956
9.00 082
9.00 207
9.00 332
9.00456
9 00 581
9.00 704
9.00 828
9.00951
9.01 074
901 196
9.01 318
9.01 440
9 01 561
9.01 682
901 803
9.01 923
144
H3
144
142
143
141
142
141
140
140
139
139
139
138
138
^37
137
136
136
136
135
135
134
134
133
133
133
132
132
131
131
131
130
130
129
129
129
128
128
128
127
127
126
126
126
125
125
124
125
123
124
123
123
122
122
122
121
121
121
120
8.94 195
8.94 340
8.94 485
8.94 030
8.94 773
8.94917
8.95 000
8.95 202
8.95 344
8.95 4«6
8-95 ^27
8.95 767
8.95 908
8.96 047
8.96 187
8.96 325
8.96 464
8.96 602
8.96 739
8.96 877
8.97013
8.97 150
8.97 285
8.97421
8-97 556
8.97 691
8.97825
8.97 959
8.98 092
8.98 225
8.98 358
8.98 490
8.98 622
8.98 753
8.98 884
8-99015
8.99 145
8.99 275
8.99 405
8.99 534
8.99 662
8.99 791
8.99919
9.00 046
9.C0 174
9.C0301
9.00427
9-00553
9.C0 b79
9.00 805
9.00 930
9.0 J 055
9.01 179
9.01 303
9.01 427
9 01 550
9.01 673
9.01 796
9.01 918
g. 02 040
9.02 ib2
M5
•45
145
H3
144
M3
142
142
142
141
140
141
139
140
138
139
138
137
^3^
136
137
135
136
135
135
134
134
133
^33
133
132
132
131
131
131
130
130
130
129
128
129
128
127
128
127
126
126
126
126
125
125
124
124
124
123
123
123
122
122
122
1.05805
1 .05 660
105 5»?
1.05 370
1.05 227
1.05083
1 .04 940
1 .04 798
1.04 656
104373
9-99 834 60
9.99 ii33
9.99 832
9.99831
9.99 830
9.99 829
9.99 828
9.99 827
9.99 825
9.99 824
9 -99 823
1.04233
1 .04 092
1-03953
1.03 813
1-03673
I;03 53^
1.03398
1.03 261
1.03 123
9.99 822
9.99 821
9.99 820
9.99819
9.99817
9.99 816
9.99815
9.99814
9-99813
1.02 987
9.99 812
1.02 850
1.02 715
1.02579
1.02444
1.02 309
1-02 175
1.02 041
1. 01 908
i.oi 775
9.99 810
9-99 809
9.99 808
9-99 807
9.99 806
9.99 804
9.99 803
9.99 802
9.99 801
I.OI 642
9.99 8CO
I.OI 510
I.OI 378
I.OI 247
I.OI 116
1.00985
1.00855
1. 00 725
1. 00 595
1 .00 466
.00338
1 .00 209
1 .00 08 1
0.99 954
0.99 826
0.99 699
0.99 573
0.99 447
0.99321
0.99 195
0.99 070
0.98 945
0.98821
0.98 697
0.98 573
0.98 450
0.98 327
0.98 204
0.98 082
0.97 960
0.97 838
L. Cos. I d. I L. Cot, led. L. Tan. | L. Sin. |
■99 798
■99 797
•99 796
..99795
9-99 793
9.99 792
9.99 791
9.99 790
9-99 788
9-99 787
, -99 786
9-99 78I
999 783
,-99782
9.99781
9.99 780
9-99 778
9.99 777
9-99 776
999 775
9.99 773
9-99 772
9.99771
9-99 769
9.99 768
9-99 767
9-99 765
9.99 764
•99 763
9.99 761
59
58
57
56
55
54
53
52
51
50
49
48
47
46
45
44
43
42
41
40
39
38
37
36
35
34
33
32
31
30
29
28
27
26
25
24
23
22
21
20
147
147
294
44.1
588
73 5
88.2
102.9
117. 6
132.3
146
146
29 2
438
58.4
73 o
87.6
102. 2
I16.8
I3I4
I
143
14.2
2
28.6
284
S
42.9
426
4
57-2
56.8
t^
71-5
71.0
6
85.8
85.2
7
100. 1
99.4
8
"4 4
113. 6
9
128.7
127.8
13-9
•27.8
41.7
55.6
69.5
834
97-3
III. 2
125.1
135
13-5
27.0
40-5
54-0
67.5
94.5
108.0
131
I3-I
26.2
39-3
52.4
655
78.6
91 7
104.8
117.9
127
12.7
25-4
38.1
50.8
635
76 2
88.9
loi 6
114.3
123
24.6
369
49.2
615
73-8
86. 1
"•4
138
27 6
41.4
552
69 o
82.8
966
1 10.4
124 2
145
39.0
43 5
58.0
725
870
101.5
1160
130.5
14 I
28.2
423
564
70-5
84.6
98.7
112.8
126.9
137
137
27.4
41. 1
54.8
68.5
82.2
95 9
109.6
123 3
14-4
288
576
72 o
864
X00.8
115.3
129.6
140
14.0
280
42.0
56 o
70.0
84.0
980
112.0
126.0
136
13-6
27 2
40.8
54 4
68.0
81 6
95-2
108.8
134 133 132
13-4
26.8
40.2
. 53.6
67 o
80.4
93-8
107 2
120.6
130
130
26.0
390
52.0
65.0
78.0
91. o
104.0
117.0
126
12.6
25 2
37.8
50-4
63 o
75-6
88.2
100.8
II3-4
12.2
24.4
36.6
48.8
61.0
73-2
854
97.6
133
26.6
39 9
53-2
665
798
93 I
106 4
119.7
129
12.9
258
387
516
64.5
77-4
90-3
103 2
116.1
125
12.5
25 o
37 5
50 o
62 5
750
87.5
100 o
112.5
12. t
24 2
363
484
605
72.6
847
96.8
13-2
26.4
396
52.8
660
79.2
92 4
105 6
128
128
25 6
38.4
51 2
64.0
76.8
89.6
102 4
115-2
124
12.4
24.8
37 2
496
62 o
74-4
86 8
99.2
111.6
12 o
24.0
36 o
48.0
60.0
72.0
840
96.0
110.7 109.8 108 9 108.0
P. P.
Qy1°
32
■^
2
3
4
5
6
7
8
9
10
II
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
2>7
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
L. Sin.
9.01 923
9.02 043
9.02 163
9.02 283
9.02 402
9.02 520
9.02639
9.02 757
9.02 874
9.02 992
903 109
9.03 226
9-03 342
9-03 458
903574
9.03 690
9.03 805
9.03 920
9.04 034
9.04 149
9.04 262
9.04 376
9.04 490
9.04 603
9.04715
9.04 828
9.04 940
9.05 052
9.05 164
9-05 27?
9-05 386
9-05 497
9.05 607
9.05 717
9.05 827
905 937
9.06 046
9.06155
9.06 264
9.06 372
9.06481
9.06 589
9.06 696
9.06 804
9.06 911
9.07018
9.07 124
9.07 231
907 337
9.07 442
9-07 548
9-07 653
9.07 758
9.07 863
9.07 968
9.08 072
9.08 176
9.08 280
9.08 383
9.08 486
9.08 589
L. Cos.
i«^1
L. Tan.
9.02 162
9.02 283
9.02 404
9.02 525
9.02 645
9.02 766
9.02 885
9.03 005
9.03 124
9.03 242
903 361
9.03 479
9-03 597
9.03 714
9.03 832
9.03 948
9.04 065
9.04 181
9.04 297
9.04413
9.04 528
9.04643^
9.04 758
9.04 873
9.04 987
9.05 lOI
9.05 214
9.05 328
9.05 441
9-05 553
9.05 666
9-05 778
9.05 890
9.06 002
9.06 113
9.06 224
906 335
1.06 445
9.06 556
9.06 666
9-o6 775
9.06 885
9.06 994
9.07 103
9.07 211
9.07 320
9.07 428
9.07 536
9.07 643
9.07751
9.07 858
9.07 964
9.08071
9.08177
9.08 283
9.08 389
9.08 495
9.08 600
9.08 705
9.08 810
9.08914
L. Cot.
c.d.
21
21
21
20
21
19
20
19
i8
19
18
18
17
18
16
17
16
16
16
15
15
15
15
14
14
13
H
13
12
13
12
12
12
II
II
II
10
II
10
09
10
09
09
08
09
08
08
07
08
07
06
07
06
06
06
06
05
05
05
04
cTd!
L. Cot.
0.97 838
0.97717
o 97 596
0.97 475
0-97 355
0.97 234
0.97115
0.96 995
0.96 876
0.96 758
0.96 639
0.96 521
0.96 403
o 96 286
0.96 168
0.96052
0-95 935
0.95 819
0.95 703
0-95 587
095472
0.95 357
0.95 242
0.95 127
0.95013
0.94 899
0.94 786
0.94 672
0.94 559
0.94 447
0-94 334
0.94 222
0.94 no
0.93 998
0.93 887
0.93 776
0.93 665
0.93 555
0.93 444
0.93 334
0.93 225
0.93115
0.93 006
0.92 897
0.92 789
0.92 680
0.92572
0.92 464
0.92 357
0.92 249
0.92 142
0.92 036
0.91 929
0.91 823
0.91 717
0.91 611
0.91 505
0.91 400
0.91 295
0.91 190
0.91 086
L. Tan.
L. Cos.
9»9" 761
9.99 760
9-99 759
9.99 757
9-99 755
9-99 755
9-99 753
9-99 752
9-99751
9.99 749
9-99 748
9-99 747
9-99 745
9.99 744
9.99 742
9.99 741
9.99 740
9-99 738
9-99 737
9.99 73(>
9-99 734
9-99 1Z2>
9-99 73'
9-99 730
9.99 728
9.99 727
9.99 726
9.99 724
9-99 723
9 -99 72'
9.99 720
9.99718
9.99717
9.99 716
9.99714
9-99713
9.99711
9.99 710
9.99 708
9.99 707
■99 705
9.99 704
9.99 702
9.99 701
9.99 699
9.99 698
9.99 696
9.99 695
9-99 693
9-99 692
9.99 690
9.99 689
9.99 687
9.99 686
9.99 684
9.99 683
9.99 681
9.99 680
9.99 678
9.99 677
^99 675
L. Sin.
60
59
58
57
56
55
54
53
52
51
50
49
48
47
46
45
44
43
42
41
40
39
^^
37
36
35
34
Z2>
32
31
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
15
14
13
12
10
9
8
7
6
5
4
3
2
I
O
P.P.
121
120
119
118
I
12. 1
12.0
j;:i
11.8
2
24.2
24.0
2^.6
3
36.3
36.0
35.7
^S-4
4
4«.4
48.0
47-b
47.2
5
bo.5
60.0
59-5
59-0
b
72.6
72.0
71.4
70.8
7
«4-7
84.0
«S.^
82.6
«
96.8
96.0
95.2
94.4
9
108.9
108.0
107. 1
106.2
117
116
115
114
I
11.7
11.6
"•5
II. 4
2
23.4
23.2
23.0
22.8
3
4
J?:s
34.8
46.4
34.5
46.0
34-2
4,S.6
5
6
7
58.5
70.2
81.9
58.0
69.6
81.2
69.0
80.5
79.8
8
93-6
92.8
92.0
91.2
9
105.3
104.4
103-5
102.6
"•3
22.6
33-9
45-2
56.5
67.8
79.1
90.4
101.7
109
10.9
21.8
32.7
43-6
54-5
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9.48 998
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42
43
42
42
42
42
42
42
42
42
42
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0.48 437
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4
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48
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4
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45
39 38 37
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44
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38
38
38
38
38
38
38
38
37
38
38
37
38
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37
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37
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38
37
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9-52 115
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9.49 882
9-52157
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37
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9.49 920
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0.47 800
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36
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949 958
9-52 242
0.47 758
9.97717
4
35
26
949 996
9.52 284
0.47 716
9-97713
4
5
4
4
4
5
4
4
4
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4
4
4
5
4
4
4
5
4
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27
9-50034
9-52 326
0.47 674
9-97 708
33
36 5 4
28
9.50072
9-52 368
0.47 632
9.97 704
32
29
30
31
9.50 no
9.52410
0.47 590
9-97 700
31
30
29
2
3
4
3.0 0.5 0.4 1
7.2 i.o 0.8
10.8 1.5 1.2
14.4 2.0 1.6
18.0 2.5 2.0
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9.50 148
9.52452
0.47 548
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9.50 185
9-52494
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9-97 691
32
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28
33
9.50 261
9-52578
0.47 422
9.97 683
27
34
9.50 298
9.52 620
0.47 380
9.97 679
26
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9-50336
9.52 661
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9-97 674
25
8
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9-50 374
9.52 703
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9-97 670
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9.50411
9-52 745
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9.97 666
23
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39
40
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9.50449
9-52787
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22
21
20
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9.50486
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9-97 653
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37
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36
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37
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37
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41
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42
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41
41
41
42
41
41
41
41
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0.46 839
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9-97 623
9.97619
5
4
4
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4
4
5
4
4
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4
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L. Cos.
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54
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9-97 506
4
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9.51 811
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0.45 691
9-97 501
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lb
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9.54 390
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43
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9.51 919
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0.45 569
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42
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20
9.51 955
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4
4
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21
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9.97475
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36
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35
36
36
36
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9.52099
9-54 633
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33-3 32.4 31-5
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9-52135
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41
40
40
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9-54 794
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9.52 278
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9.54 875
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9-97 430
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36
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9-54 995
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9-97426
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9-52456
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9-52527
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9-52 598
9-55035
9.55075
9-55 "5
9.55155
9.55 195
0.44 965
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0.44 8-45
0.44 805
9.97421
9.97417
9.97412
9-97 408
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27
26
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9-55 235
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22
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1 L. Cos. 1 d. 1 L. Cot. |c. d.| L. Tan. | L. Sin.
1 d.
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1 P.P.
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46
20'
L. Sin.
L. Tan. c. d
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L. Cos.
P.P.
9
10
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12
13
14
15
16
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25
26
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28
29
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9-54 263
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9-54 534
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9.58 115
9-58153
9.58 191
9-58 229
9.58 267
9-58 304
958342
9.58 380
9.58418
39
39
39
40
39
39
39
39
39
39
39
39
39
39
39
39
39
39
39
38
39
39
39
38
39
39
38
39
38
39
38
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38
39
38
38
39
38
38
39
38
38
38
38
39
38
38
38
38
38
38
38
38
38
38
38
37
38
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0.43 893
9-97 299
0.43 854
9.97 294
0.43815
9.97 289
0.43 776
9-97 285
0-43 736
9.97 280
0.43 697
9 97 276
0.43 658
9.97 271
0.43 619
9.97 266
0.43 580
9.97 262
0.43 541
9-97 257
0.43 502
9-97 252
0.43 463
9.97 248
0.43 424
9.97 243
0.43 385
9.97 238
0.43 346
9-97 234
0.43 307
9.97 229
0.43 268
9.97 224
0.43 229
9.97 220
0.43 190
9.97215
0.43 151
9.97 210
0.43 I '3
9.97 206
0.43 074
0.43 035
0.42 996
0.42958
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0.42 880
0.42 842
0.42 803
0.42 765
9.97 201
9-97 196
9-97 192
9.97 187
9.97 182
9-97 178
9-97 173
9.97 168
9-97 ^63
0.42 726
9-97 159
0.42 688
0.42 649
0.42 611
042572
0.42 534
0.42 496
0.42 457
0.42419
0.42 381
9-97 154
9-97 149
9 97 145
9-97 140
9-97 135
9.97 130
9.97 126
9.97 121
9.97 116
0.42 342
9 97"
0.42 304
0.42 266
0.42 228
0.42 190
0.42 151
0.42 113
0.42 075
0.42037
0.41 999
9.97 107
9 97 102
9.97 097
9.97 092
9-97 087
9-97 083
9.97 078
997073
9.97 068
0.41 961
9.97 063
0.41 923
0.41 885
0.41 847
0.41 809
0.41 771
0.41 733
0.41 696
0.41 658
0.41 620
9-97 059
997054
9.97 049
9-97 044
9-97 039
9-97035
9-97 030
9-97 025
9.97 020
0.41 582 I 9.97015
L. Tan. [ L. Sin.
fift°
60
59
58
57
56
55
54
53
52
51
50
49
48
47
46
45
44
43
42
41'
40
39
3^
37
36
35
34
33
32
31
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
15
14
13
12
II
10
9
8
7
6
5
4
3
2
40 39
4.0
8.0
12.0
16.0
20.0
7.6
II.4
15.2
19.0
3-9
78
11.7
15.6
19-5
24.0 23.4 22.8
28.0 27.3 26.6
32.0 31.2 30.4
36.0 35.1 34.2
37 35 34
3-7
7-4
II. I
14.8
18.5
3.4
6.8
10.2
13.6
17.0
3.5
7-0
10.5
14.0
17-5
22.2 21.0 20.4
25.9 24.5 23.8
29.6 28.0 27.2
33-3 31.5 30.6
33
I
3.3
0.5
0.4
2
6.6
I.O
0.8
3
9-9
1.5
1.2
4
13.2
2.0
1.6
5
16.S
2.5
2.0
6
19.8
30
2.4
7
23.1
.3-5
2.8
8
26.4
4.0
3-2
9
29.7
4.5
3-6
AAA
40 39 38
4.0 3-9 3-8
12.0 II. 7 11.4
20.0 19.5 19.0
28.0 27.3 26.6
36.0 35.1 34.2
AAA
37 39 38
3-7 4-9
[I.I 146
4-8
[4.2
18.5 24.4 23.«
25-9 34-1 33.2
33-3 — —
L. Cos.
d.
L. Cot.
c. d.
P. P.
47
O
I
2
3
4
5
6
7
8
9
10
II
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
L. Sin.
9-55 43:
9-55 4^^
9.55 499
9.55 532
9-55 564
9-S5 597
9-55 630
9-55 663
9-55 695
9-55 728
9-55 761
9-55 793
9.55 826
9-55 858
9-55891
955923
9-55 956
9 55 988
9.56021
956053
9-56 085
9.56 118
9.56 150
9.56 182
9.56215
9.56 247
9.56 279
9563"
9-56 343
9.56375
9.56 408
9-56 440
9.56472
9.56 504
9-56536
9.56 568
9-56 599
9.56631
9.56 663
9-56 695
9-56 727
956 759
9-56 790
9.56822
9.56854
9.56 886
9.56917
9.56 949
9.56 980
9.57012
9-57 044
9-57075
9.57 107
9-57 138
9.57 169
9.57 201
957232
9.57 264
9-57 295'
9-57 326
9-57 358
L. Cos. I d.
L. Tan. c. d. L. Cot.
9.58418
958455
958493
9.58531
9.58 569
9.58606
9:58 644
9.58681
9.58719
9-58757
9. 58 794
9.58 832
9-58 869
9.58907
958 944
9-58981
9.59019
9.59056
9-59 094
9-59 131
9.59 168
9.59 205
9-59 243
9.59 280
9.59317
9-59 354
9-59 391
9-59 429
9.59466
9-59 503
9-59 540
9-59 577
9.59614
9.59651
9.59 688
9-59 725
9-59 762
9-59 799
9-59 835
9.59872
9-59 909
9-59 946
9.59 983
9.60019
9.60056
9.60 093
9.60 130
9.60 166
9.60 203
9.60 240
9.60 276
9.60313
9.60349
9.60 386
9.60 422
9.60459
9-60 495
9.60 532
9.60 568
9.60 605
9.60 641
L. Cot.
37
38
38
38
37
3^
37
38
38
37
38
37
38
37
37
38
37
38
37
37
37
38
37
37
37
37
38
37
37
37
37
37
37
37
37
37
37
36
37
37
37
37
36
37
37
37
36
37
37
36
37
36
37
36
37
36
37
36
37
36
Q.41 582
0.41 545
0.41 507
0.41 469
0.41 431
0.41 394
0.41 356
0.41 319
0.41 281
0.41 243
0.41 206
0.41 168
0.41 131
0.41 093
0.41 056
0.41 019
0.40 981
0.40 944
0.40 906
0.40 869
0.40 832
0.40 795
0.40 757
0.40 720
0.40 683
0.40 646
0.40 609
0.40571
0.40 534
0.40 497
0.40 460
0.40 423
0.40 386
0.40 349
0.40312
0.40 275
0,40 238
0.40 201
0.40 165
0.40 1 28
0.40091
0.40 054
0.40017
0.39 981
0.39 944
0.39 907
0.39 870
0.39 834
0.39 797
0.39 760
0.39 724
0.39 687
0.39651
0.39614
0.39 578
0.39 541
0.39 50?
0.39 468
039 432
0.39 395
0.39 359
c. d.| L. Tan.
L. Cos.
9.97015
9.97 010
9-97 005
9.97 001
9.96 996
9.96991
9.96 986
9.96981
9.96 976
9.96971
9.96 966
9.96 962
9-96957
9.96952
9-96 947
9.96 942
9-96 937
9-96 932
9.96927
9.96922
9.96917
9.96 912
9-96 907
9.96 903
9.96 898
9.96 893
9.96 888
9.96 883
9.96 878
996 873
9.96 868
9.96 863
9.96 858
996853
9.96 848
9.96 843
9.96 838
9.96 833
9.96 828
9.96823
9.96818
9.96813
9.96 808
9.96 803
9.96 798
996 793
9.96 788
9.96 783
9.96 778
9.96 772
.96 767
9.96 762
996757
9-96 752
9.96 747
9.96 742
996 737
9.96 732
9.96 727
9 96 722
996 7^7
L. Sin.
vr
60
59
58
57
56
55
54
53
52
51
50
49
48
47
46
45
44
43
42
41
40
39
38
37
36
35
34
33
32
31
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
15
14
13
12
10
9
P. P.
38 37
3-8
7.6
11.4
15.2
190
3.7 3-6
7-4 7-2
II. I 10.8
14.8 14.4
18.5 18.0
22.8 22.2 21.6
26.6 25.9 25.2
30.4 29.6 28.8
34.2 33-3 32.4
33 32 31
3-3
6.6
9-9
13.2
16.5
19.8
23.1
3.2
6.4
9.6
12.8
16.0
19.2
22.4 21.7
26.4 25.6 24.8
29.7 28.8 27.9
9-3
12.4
J5-5
18.6
I
0.6
0-5
0.4
2
1.2
I.O
0.8
3
1.8
1-5
1.2
4
2.4
2.0
1.6
5
30
2-5
2.0
6
3-6
30
2.4
7
4.2
3-5
2.8
8
4.8
4.0
3-2
9
5-4
4.5
3.6
7
6
2
5
3
4
4
3
5
2
I
6
37
3.1
9-2
15-4
21.6
27.8
33-9
3.6
10.8
18.0
25.2
32-4
5
38
3.8
U.4
19.0
26.6
34.2
38
4.8
14.2
23-8
33-2
37
3-7
II. I
18.5
25-9
33-3
37
4.6
139
23.1
32.4
P. P.
68°
48
1
L. Sin.
d.
L. Tan.
c. d.
L. Cot.
L. Cos.
d.
P.P.
I
9-57358
31
31
31
31
32
31
31
9.60.641
36
36
I
36
36
36
36
3.6
36
^i
36
36
^i
36
36
36
36
36
36
35
36
36
36
35
36
^%
36
35
36
35
36
36
35
36
35
36
35
36
35
36
35
35
36
35
35
36
35
0-39 359
9.96717
6
5
60
59
9-57 389
9.60 677
0.39 323
9.96711'
2
9.57420
9.60714
0.39 286
9.96 706
58
37 36 35
3
9-57451
9.60 750
0.39 250
9.96 701
5
5
5
5
5
5
6
5
5
57
I
3-7 3-6 3-5
4
9.57 482
9.60 786
0.39214
9.96 696
56
2
7-4 7-2 7.0
S
9-57 5H
9.60 823
0.39177
9.96 691
55
3
II. I 10.8 10.5
6
9-57 545
9.60 859
0.39 141
9.96 686
54
4
14.8 14.4 14.0
7
9-57576
9.60 895
0.39 105
9.96 681
S3
5
18.5 18.0 17.5
8
9.57 607
31
9.60931
0.39 069
9.96 676
52
6
22.2 21.6 21.0
9
10
II
9-57 638
31
31
31
31
9.60 967
0.39 033
9.96 670
51
50
49
7
8
9
25.9 25.2 24.5
29.6 28.8 28.0
33-3 32.4 31-5
9.57 669
9.61 004
0.38 996
9.96 665
9-57 700
9.61 040
0.38 960
9.96 660
12
9-57 731
9.61 076
0.38 924
9.96655
5
48
13
9-57 762
31
31
31
31
30
31
31
31
30
9.61 112
0.38 888
9.96 650
5
5
5
6
5
5
5
5
6
47
H
9-57 793
9.61 148
0.38 852
9.96 645
46
32 31 30
15
16
9-57 824
9-57855
9.61 184
9.61 220
0.38816
0.38 780
9.96 640
9.96 634
45
44
I
2
3.2 3-1 30
6.4 6.2 6.0
17
9-57 §85
9.61 256
0.38 744
9.96 629
43
3
9.6 9-3 9.0
18
9.57916
9.61 292
0.38 708
9.96 624
42
4
12.8 12.4 12.0
19
20
21
9-57 947
9.61 328
0.38 672
9.96 619
41
40
39
5
6
7
16.0 15.5 15.0
19.2 18.6 18.0
22.4 21.7 21.0
9-57978
9.61 364
0.38 636
9.96 614
9.58008
9.61 400
0.38 600
9.96 608
22
9.58 039
31
9.61 436
0.38 564
9.96 603
5
38
8
25.6 24.8 24.0
23
9.58070
31
31
30
31
9.61 472
0.38 528
9.96 598
5
37-
9
28.8 27.9 27.0
24
9.58 lOI
9.61 508
0.38 492
9-96 593
5
36
2S
9-58 131
9.61 544
0.38 456
9.96 588
\
5
5
5
5
6
5
5
35
■
25
9.58 162
9.61 579
0.38421
9.96 582
34
29 6 5
27
9.58 192
30
31
9.61 615
0.38385
9.96577
33
2.9 0.6 0.3
5.8 1.2 I.O
8.7 1.8 1.5
1 1.6 2.4 2.0
14.5 3.0 2.5
28
9.58223
9.61 651
0.38 349
9.96572
32
2
29
30
31
9.58253
.5°
31
30
31
9.61 687
0.38313
9.96 567
31
30
29
3
4
9.58 284
9.61 722
0.38 278
9.96 562
9.58 3'4
9.61 758
0.38 242
9-96 556
32
958 345
9.61 794
0.38 206
9-96551
28
6
17.4 3.6 3.0
Zl
9-58375
S^
9.61 830
0.38 170
9-96 546
27
7
20.3 4-2 3-5
34
9.58 406
31
9.61 865
0.38135
9.96 541
5
6
5
5
5
6
5
5
6
5
5
\
26
8
23.2 4.8 4.0
3S
9-58 436
30
31
30
30
9.61 901
0.38 099
9.96 535
25
9
26.1 5.4 4.5
36
9.58467
9.61 936
0.38 064
9.96 530
24
^^
9-58497
9-58527
9.61 972
9.62 008
0.38 028
0.37 992
9.96 525
9.96 520
23
22
39
40
41
9.58 557
30
31
30
9.62 043
0.37957
9.96514
21
20
19
6 6
36 35
9-58 588
9.62079
0.37921
9-96 509
9.58618
9.62 114'
0.37 886
9.96 504
42
9.58 648
30
9.62 150
0.37 850
9.96498
18
3.0 2.9
9.0 8.8
15.0 14.6
43
44
9-58678
9.58 709
30
31
9.62 185
9.62 221
0.37815
0.37 779
9-96 493
9.96 488
17
16
I
2
4S
9-58 739
30
30
9.62 256
0.37 744
9.96483
15
3
21.0 20.4
46
9.58 769
9.62 292
0.37 708
9.96477
5
14
4
27.0 26.2
47
9-58 799
30
9.62 327
0.37 673
9.96472
13
330 32.1
48
9.58829
30
9.62 362
0.37 638
9.96467
12
49
50
5'i
9.58859
30
30
30
9.62 398
0.37 602
9.96 461
5
5
II
10
9
5 5 5
9.58889
9.62 433
0.37 567
9-96456
9.58919
9.62 468
0.37 532
9.96451
S2
9-58 949
30
30
9.62 504
0.37 496
9-96 445
5
8
37 36 35
53
9-58979
9.62 539
0.37461
9.96 440
7
54
55
9-59 009
9-59 039
3*^
30
9.62574
9.62 609
3b
%
35
35
35
35
0.37 426
0.37391
9-96 435
9.96429
5
6
5
5
6
5
5
6
5
I
2
3-7 3.6 3.5
I I.I 10.8 10.5
18.5 18.0 17.5 1
25.9 25.2 24.5
33-3 32.4 31.5
5^
57
58
9.59 069
9-59 098
9.59 128
29
30
9.62 645
9.62 680
9.62715
0.37 355
0.37 320
0.37 285
9.96 424
9.96419
9.96413
4
3
2
3
4
5
59
60
9-59 158
30
30
9.62 750
0.37 250
9.96 408
I
9-59 188
9.62 785
0.37215
9-96 403
L. Cos.
d.
L. Cot. |c. d.
L. Tan.
L. Sin.
d.
/
P.P.
67°
49
L. Sin.
L. Tan. c. d
L. Cot.
L. Cos.
d.
P.P.
2
3
4
5
6
7
8
9
10
II
12
13
H
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
3B
39
40
41
42
43
44
45
46
47
48
49
50
51
52
5.3
54
55
56
57
58
59
60
9.59 188
9.59218
9.59 247
9-59277
9-59 307
9-59 33i^
9-59 366
9-59 396
9-59425
9.59 451
9-5 9 484
9-59 5 H
9-59 543
9-59 573
9.59 602
9-59 ^32
9.59651
9.59 690
9.59 720
9.59 749
9-59 778
9.59 808
9-59 837
9-59 866
9-59 895
9-59 924
9-59 954
9-59983
9.60012
9.60041
9.60070
9.60 099
9.60 128
9.60 157
9.60 186
9.60 215
9.60 244
9.60 273
9.60 302
9.60331
9.60359
9.60 388
9.60417
9.60 446
9.60 474
9.60 503
9.60 532
9.60 561
9.60 589
9.60618
9.60 646
9.60 675
9.60 704
9.60 732
9.60 761
9.60 789
9.60818
9.60 846
9.60 875
9.60 903
9.60931
29
30
29
30
29
30
29
30
29
29
30
29
29
30
29
29
29
29
30
29
29
29
29
29
29
29
29
29
29
29
29
29
28
29
29
29
28
29
29
29
28
29
28
29
29
28
29
28
29
28
29
28
28
9.62 785
9.62 820
9.62855
9.62 890
9.62 926
9.62 961
9.62 996
9.63 031
9.63 066
9.63 lOI
9»fa3 135
9.63 170
9.63 205
9.63 240
9.63 275
9.63310
9-63 34l
9-63 379
9.63414
9-63 449
9.63 484
9.63519
9.63 553
9.63 588
9.63 623
9.63 657
9.63 692
9.63 726
9.63 761
9.63 796
9.63 830
9.63 865
9.63 899
9-63 934
9.63 968
9.64 003
9.64 037
9.64 072
9.64 106
9.64 140
9-64175
9.64 209
9.64 243
9.64 278
9.64312
9.64 346
9.64 381
9.64415
9.64 449
9.64 483
9.64517
9.64552
9.64 586
9.64 620
9.64 654
9.64 688
9.64 722
9.64 756
9.64 790
9.64 824
9.64858
35
35
35
36
35
35
35
35
35
34
35
35
35
35
35
35
34
35
35
35
35
34
35
35
34
35
34
35
35
34
35
34
35
34
35
34
35
34
34
35
34
34
35
34
34
35
34
34
34
34
35
34
34
34
34
34
34
34
34
34
0.37215
9.96 403
0.37 180
0.37 M5
0.37 no
0.37 074
0.37 039
0.37 004
0.36 969
0.36 934
0.36 899
9.96 397
9.96 3^2
9.96 387
9.96 381
9.96 376
9.96 370
9-96 365
9.96 360
9-96 354
o.36't>65
9.96 349
0.36 830
0.36 795
0.36 760
0.36 725
0.36 690
0.36655
0.36 621
0.36 586
0-36551
9-96 343
9.96 33^
9-96 333
9.96327
9.96 322
9.96316
9.96 311
9.96 305
9.96 300
036510
0.36481
0.36 447
0.36412
0.36 377
0.36 343
0.36 308
0.36 274
0.36 239
0.36 204
0.36 170
0.36135
0.36 lOI
0.36 066
0.36032
0-35 997
0.35 963
0.35 928
0.35 894
0.35 860
0.35 825
0.35 791
0.35 757
0.35 722
0.35 688
0.35 654
0.35 619
0.35 585
0-35551
0-35517
0-35 483
0.35 448
0.35414
0.35 380
0.35 346
0.35312
0.35 278
0.35 244
0.35 210
0.35 176
0.35 H2
9- 96 294
9.96 289
9.96 284
9.96 278
9.96 273
9.96 267
9.96 262
9.96 256
9.96 251
9.96 245
9.96 240
9.96 234
9.96 229
9.96 223
9.96 2l8
9.96 212
9.96 207
9.96 201
9.96 196
9.96 190
9-96 185
9.96179
9.96174
9.96 168
9.96 162
9.96157
9-96151
9.96 146
9.96 140
9-9 6 135
9.96 129
9.96 123
9.96 118
9.96 112
9.96 107
9.96 lOI
9.96 095
9.96 090
9.96 084
9^96 079
9-96 073
60
59
58
57
56
55
54
53
52
51
50
49
48
47
46
45
44
43
42
41
40
39
3S
37
36
35
34
33
32
31
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
15
14
13
12
II
10
9
8
7
6
5
4
3
2
I
O
35 34
2
3.6 3.5
7.2 7.0
3-4
6.8
3
10.8 10.5
10.2
4
5
6
144 M-o
18.0 17.5
21.6 21.0
13.6
17.0
204
7
8
25.2 24.5
28.8 28.0
23.8
27.2
9
324 3' -5
30.6
I
2
3-0
6.0-
2.9
S.8
3
4
9.0
12.0
. 8.7
11.6
5
6
7
15.0
18.0
21.0
14-5
17.4
20.3
8
9
24.0
27.0
23.2
26.1
6
0.6
1.2
1.8
24
3-0
3-6
4.2
4.8
5.4
29 28
2.8
5.6
84
II. 2
14.0
16.8
19.6
22.4
25.2
0.5
i.o
1-5
2.0
2.5
3-0
3.5
4.0
4-5
6
36
35
2.9
8.8
3.0
9.0 5.5 5.5
15.0 14.6 14.2
21.0 204 19.8
27.0 26.2 25.5
33.0 32.1 31.2
35
3.5
10.5
17-5
24-5
31-5
34
34
10.2
17.0
23.8
30.6
I L. Cos. i d. I L. Cot, c. d. L. Tan.
L. Sin.
d. I
P. P.
50
L. Sin. d.
L. Tan. |c. d.
L. Cot.
L. Cos.
d.
P.P.
O
I
2
3
4
5
6
7
8
9
10
II
12
13
H
^5
i6
17
18
19
20
21
22
^3
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
9.60931
9.60 960
9.60 988
9.61 016
9.61 045
9.61 073
9,61 lOI
9.61 129
9.61 158
9.61 186
9.61 214
9.61 242
9.61 270
9.61 298
9.61 326
9.61 354
9.61 382
9.61 411
9.61 438
9.61 466
9.61 494
9.61 522
9.61 550
9.61 578
9.61 606
9.61 634
9,61 662
9.61 689
9.61 717
9.61 745
9.61 773
9.61 800
9.61 828
9.61 856
9.61 883
9.61 911
9.61 939
9.61 966
9.61 994
9.62021
9.62 049
9.62076
9.62 104
9.62 131
9.62 159
9.62 186
9.62 214
9.62 241
9.62 268
9.62 296
9.62 323
9.62 350
9.62 377
9.62 405
9.62 432
9.62459
9.62 486
9.62513
9.62 541
9.62 568
9-62 595
L. Cos.
9.64
9.64 892
9.64 926
9.64 960
9.64 994
9.65 028
9.65 062
9.65 096
9.65 130
9.65 164
9.65 197
9.65 231
9.65 265
9.65 299
9.65 333
9.65 366
9.65 400
9-65 434
9.65 467
9.65501
9-65 535
9.65 568
9.65 602
9.65 636
9.65 669
9.65 703
9.65 736
9.65 770
9.65 803
9-65 837
9.65 870
9.65 904
9-65 937
9.65 971
9.66 004
9.66038
9.66071
9.66 104
9.66138
9.66 171
(.66 204
9.66 238
9.66 271
9.66 304
9-66 337
9.66371
9.66 404
9.66437
9.66 470
9-66 503
9-66 537
9.66 570
9.66 603
9.66 636
9.66 669
9.66 702
9-66 735
9.66 768
9.66 801
9.66 834
9.66 867
L. Cot.
34
34
34
34
34
34
34
34
34
33
34
34
34
34
33
34
34
33
34
34
33
34
34
33
34
33
34
33
34
33
34
33
34
33
34
33
33
34
33
33
34
33
33
33
34
33
33
33
33
34
33
33
33
33
33
33
33
33
33
33
cTd
0.35 142
9.96073
0.35 108
0.35 074
0.35 040
0.35 006
0.34972
0.34 938
0.34 904
0.34 870
0.34 836
9.96067
9.96062
9.96056
9.96050
9.96 045
9.96039
9.96 034
-9.96 028
9.96 022
0.34 803
9.96017
0.34 769
0-34 735
0.34 70 i
0.34 667
0.34 634
0.34 600
0.34 566
0.34 533
0.34 499
9.96 01 1
9.96 005
9.96 000
9.95 994
9.95 988
9.95 982
9.95 977
9-95971
9-95 965
0.34 465
9.95 960
0.34 432
0.34 398
0.34 364
0.34331
0.34 297
0.34 264
0.34 230
0.34 197
0.34 163
9-95 954
9.95 948
9-95 942
9-95 937
9-95 931
9-95 925
9.95 920
9-95 9H
9.95 908
0.34 130
9.95 902
0.34 096
0.34 063
0.34 029
0.33 996
0.33 962
0.33 929
0.33 896
0.33 862
0.33 829
9-95 897
9.95 891
9.95 885
9.95 879
•9-95 873
9.95 868
9.95 862
9-95 856
9-95 850
0.33 796
9-95 844
0.33 762
0.33 729
0.33 696
0.33 663
0.33 629
0.33 596
0.33 563
0.33 530
0.33 497
9-95 839
9-95 833
9.95 827
9.95821
9-95 815
9.95 810
9.95 804
9-95 798
9-95 792
0.33 46:
•95 786
0.33 430
0.33 397
0.33 364
0-33 33^
0.33 298
0.33 265
0.33 232
0.33 199
0.33 166
9.95 780
9-95 775
9-95 769
9-95 763
9-95 757
9-95751
9-95 745
9-95 739
9-95 733
0-33 ^33
L. Tan.
9-95 728
L. Sin.
60
59
58
57
56
55
54
53
52
51
50
49
48
47
46
45
44
43
42
41
40
39
38
37
36
35
34
33
32
31
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
15
14
13
12
II
10
34
3-4
6.8
10.2
13-6
17.0
20.4
23-8
27.2
30.6
33
3-3
6.6
9.9
13.2
16.5
19.8
23.1
26.4
29.7
29 28 27
I
2.9
2.8
2.7
2
5.8
5.6
S-4
3
8.7
8.4
8.1
4
11.6
II. 2
10.8
s
14.5
14.0
1.^.5
6
17.4
16.8
16.2
7
20.3
19.6
18.9
8
23.2
22.4
21.6
9
26.1
25.2
24.3
6
0.6
1.2
1.8
2.4
30
3-6
4.2
4.8
5-4
0-5
i.o
1-5
2.0
2-5
30
3-5
4.0
4-5
6^
34
2.8
8.5
33
2.5
8.2
14.2 13.8
19.8 19.2
_5^
34
3-4
10.2
17.0
2^.s
25.5 24.8 30.6
31.2 30.2 —
p.p.
a^°
5'
O
I
2
3
4
5
6
7
8
9
10
II
12
13
H
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
31
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
L. Sin.
9.62 595
9.62 622
9.62 649
9.62 676
9.62 703
9.62 730
9.62 757
9.62 784
9.62 81 1
9.62 838
9.62 86g
9.62 892
9.62918
9.62 945
9.62972
9.62 999
9.63 026
9.63052
9.63 079
9.63 106
9-63 133
9.63 159
9.63 186
9.63213
9.63 239
9.63 266
9.63 292
963319
9-63 345
9-63 372
9-63 398
9.63 425
9.63451
9.63 478
9-63 504
9-63 531
9-63557
963 583
9.63 610
9.63 636
9.63 662
9.63 689
9-63715
9-63 741
9.63 767
9-63 794
9.63 820
9.63 846
9.63 872
9.63 898
•63 924
9-63 950
9.63 976
9.64 002
9.64028
9.64054
9.64 080
9.64 106
9.64 132
9.64 158
9.64 184
L. Tan.
9.66 867
9.66900
9-66 933
9.66 966
9.66 999
9.67 032
9.67 065
9.67 098
9.67 131
9.67 163
9.67 196
9.67 229
9.67 262
9.67 295
9.67 327
9-67 360
9-67 393
9.67 426
9.67 458
9.67 491
9.67 524
9.67 556
9.67 589
9.67 622
9.67 654
9.67 687
9.67 719
9.67 752
9.67 785
9.67 817
9.67 850
9.67 882
9.67915
9-67 947
9.67 980
9,68012
9.68 044
9.68077
9,68 109
9.68 142
9.68 174
9.68 206
9-68 239
9.68 271
9.68 303
9.68336
9.68 368
9.68 400
9.68 432
9.68 465
9.68 497
9.68 529
9.68 561
9-68 593
9.68 626
9.68 658
9.68 690
9.68 722
9.68 754
9.68 786
9.68818
c. d
32,
33
33
33
33
33
33
33
32
33
33
33
33
32
33
33
33
32
33
33
32
33
33
32
33
32
33
33
32
33
32
33
32
33
32
32
33
32
33
32
32
33
32
32
33
32
32
32
33
32
32
32
32
33
32
32
32
32
32
32
L. Cot.
^13J33^
0.33 too
0.33067
0.33 034
0.33001
0.32 968
0.32 935
0.32 902
0.32 869
0.32837
0.32 804
0.32771
0.32 738
0.32 705
0.32 673
0.32 640
0.32 607
0.32 574
0.32 542
0-32 509
0.32 476
0.32 444
0.32 411
0.32 378
0.32 346
0.32313
0.32 281
0.32 248
0.32 215
0.32 183
0.32 150
0.32 118
0.32085
0.32053
0.32 020
0^1 988
0.31 956
0.31 923
0.31 891
0.31 858
0.31 826
0.31 794
0.31 761
0.31 729
0.31 697
0.31 664
0.31 632
0.31 600
0.31 568
0-31 535
0.31 503
0.31 471
0.31 439
0.31 407
0.31 374
0.31 342
0.31 310
0.31 278
0.31 246
0.31 214
0.31 182
L. Cos.
995 728
9-95 722
9-95 716
9-95 710
9.95 704
9.95 698
9.95 692
9.95 686
9.95 680
9-95 674
9-95 668
9.95 663
9.95 657
9-95 651
9-95 64I
9-95 639
9-95 633
9-95 627
9.95 621
995615
9.95 609
9.95 603
9-95 597
9-95 591
9.95 585
9-95 579
9-95 573
9-95 567
9-95 561
9-95 555
9-95 549
9-95 543
9-95 537
9-95 531
9-95 525
9-95519
9-95513
9-95 507
9-95 500
9-95 494
9.95 488
9.95 482
9-95 476
9-95 470
9-95 464
9-95 458
9-95 452
9-95 446
9.95 440
9-95 434
9-95 427
9-95 421
9-95415
9.95 409
9-95 403
9-95 397
9-95 391
9.95 384
9-95 378
9-95 372
9-95 366
60
59
58
57
56
55
54
53
52
51
50
49
48
47
46
45
44
43
42
41
40
39
3^
37
36
35
34
33
32
31
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
15
14
13
12
II
10
9
8
7
6
5
4
3
2
I
O
P. P.
33 32
3-2
6.4
9.6
12.8
16.0
19.2
22.4
25.6
28.8
27 26
I
.3.3
2
6.6
3
9-9
4
13.2
5
16.5
6
19.8
7
23.1
8
26.4
9
29.7
2-7
5-4
8.1
10.8
13-5
16.2
18.9
21.6
243
2.6
5-2
7-8
10.4
13.0
15.6
18.2
20.8
23-4
I
0.7
0.6
0.5
2
1.4
1.2
I.O
3
2.1
1.8
1.5
4
2.8
2.4
2.0
5
3.5
3-0
2-5
6
4.2
3.6
3-0
7
4-9
4.2
3-5
8
5.6
4-8
4.0
9
6.3
5-4
4-5
32
2-3
6.9
11.4
16.0
20.^
25.1
29.7
32
2.7
8.0
13-3
18.7
24.0
29-3
33
3-3
9-9
16.5
23.1
29.7
L. Cos.
L. Cot.
c. d.
L. Tan-
fid."
L. Sin. I d.
P.P.
/
L. Sin.
d.
L.Tan. c. d.
L. Cot. 1 L. Cos.
d.|
P. P. 1
I
9.64 184
26
26
26
9.68818
32
0.31 182
9.95 366
6
6
60
59
9.64 210
9.68 850
0.31 150
9.95 360
2
9.64 236
9.68 882
32
32
0.31 118
9-95 354
58
3
9.64 2b2
26
25
26
9.68914
0.31 086
9-95 348
57
4
9.64 288
9.68946
32.
32
10
0.31 054
9-95 341
7
5
56
32 31
5
964313
9.68978
0.31022
9-95 335
6
55
I
3-2 3.1 1
6
9-64 339
26
26
26
25
26
26
9.69 OIO
32
32
32
32
32
32
32
32
32
31
0.30 990
9-95 329
6
6
7
6
6
6
54
2
6.4 6.2
7
9.64 365
9.69 042
0.30 958
9-95 323
53
3
9-6 9-3
8
9.64391
9.69 074
0.30 926
9-95317
52
4
12.8 12.4
9
10
II
9.64 4P7
9.69 106
0.30 894
9-95 310
51
50
49
' 5
6
7
8
16.0 15.5
19.2 18.6
22.4 21.7
25.6 24.8
9.64 442
9.69 1.38
0.30 862
9-95 304
9.64 468
9.69 170
0.30 830
9.95 298
12
9.64 494
9.69 202
0.30 798
9.95 292
A
48
9
28.8 2-7.0
13
9.64519
25
26
26
9-69 234
0.30 766
9.95 286
47
14
9-64 545
9.69 266
0.30 734
9.95 279
7
46
IS
9-64571
9.69 298
0.30 702
9-95 273
6
6
45
i6
9.64 596
25
26
11
25
26
9.69 329
0.30671
9.95 267
44
17
9.64622
9.69 361
32
0.30 639
9-95 261
43
26 25 24
18
9.64 647
9-69 393 C
0.30 607
9-95 254
7
t
6
42
I
2.6 2.5 2.4 1
19
20
21
9.64 673
9.69 425
32
31
0-30 575
9-95 248
41
40
39
2
3
4
5.2 5.0 4.8
7-8 7-5 7-2
10.4 lO.O 9.6
9.64 698
9-69 457
'^.30 543
9.95 242
9.64 724
9.69 488
0.30 5 1 2
9.95 236
22
9.64 749
25
26
25
26
9.69 520
32
32
32
31
32
32
31
32
32
31
32
31
32
32
31
32
31
•32
31
32
31
32
0.30 480
9.95 229
7
6
6
6
3^
7
8
13.0 12.5 12.0 1
15.6 15.0 14.4
18.2 17.5 16.8
20.8 20.0 19.2 ]
23
24
9-64 775
9.64 800
9.69 552
9.69 584
0.30 448
0.30416
9.95 223
9-95217
37
36
2.S
9.64826
9.69615
0.30385
9.95 211
7
6
35
g
22.A 22. C 21.6 1
25
9.64851
26
9.69 647
0.30 353
9-95 204
34
27
9.64877
9.69 679
0.30321
9.95 198
33
28
9.64 902
25
9.69710
0.30 290
9.95 192
32
29
9.64927
^5
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47
9.84 006
^3
9-98 155
O.OI 843
9.85851
13
12
13
2
5-0 A
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48
9.84020
M
9.98 180
11
25
25
O.OI 820
9-85 839
12
3
7.0 6
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49
50
51
984033
13
13
9.9S 206
O.OI 794
9-85 827
12
12
II
10
9
4
I
7
9.0 8
1 1.0 IC
13.0 12
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-5 13.5
9.84046
9.98231
O.OI 769
9.85 815
9.84059
9.98 256
O.OI 744
9.85 803
52
9.84072
13
9.98281
26
O.OI 719
9-85 791
8
8
15.0 14.4 15.0
S3
9.84085
M
9.98 307
O.OI 693
985 779
7
Q
17.0 16.3 17.7
19.0 18.3 19.8
21.0 20.2 21.9
23.0 22.1 24.0
54
55
9.84098
9.84 112
14
13
9-98 332
^9.98357
^5
25
26
O.OI 668
O.OI 643
9.85 766
9-85 754
12
1 2
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5
10
II
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9.84125
9-98383
25
O.OI 617
9.85 742
4
12
25.0 2^
l-.o —
57
9.84 138
13
9.98 408
O.OI 592
9.85 730
3
13
58
9.84151
13
998433
25
O.OI 567
9.85718
I-
2
59
60
9.84 164
13
13
998458
il
O.OI 542
9.85 706
13
I
9.841/7
9-98 484
O.OI 516
9.85 693
L. Cos. 1 d.
L. Cot. |c. d.
1 L. Tan.
L. Sin. 1 d.
t
P.P.
Aa^
70
'
L. Sin.
d.
L. Tan.
c. d.
L. Cot.
L. Cos.
1 d.
P. P.
I
9.84177
13
13
13
13
13
13
9.98484
25
25
26
o.oi 516
9.85 693
12
60
59
9,84 190
9.98 509
0.0 1 491
9.85 681
2
9.84 203
9-98 534
O.OI 466
9.85 669
S8
26 25 14
3
9.84216
9.98 560
25
25
25
26
O.OI 440
9.85 657
T
57
I
2.6 2.5 14
4
9.84 229
9.98 585
O.OI 415
9.85 645
13
56
2
5.2 5.0 2.8
S
9.84 242
9.98 010
O.OI 390
9.85 632
55
3
7.8 7.5 42
6
9-^4 255
998 635
O.OI 365
9.85 620
54
4
0.4 lo.o 5.6
7
9.84 269
14
9.98661
O.OI 339
9.85 608
5S
5
3-0 12.5 7.0
8
9.84 282
13
9.98 686
25
O.OI 314
9-85 596
52
6
5.6 15.0 8.4
9
10
11
9.84 295
9.64 308
9.84321
13
13
13
13
9.98 711
9-98 737
2^
25
25
25
26
O.OI 289
9.85 583
^3
12
12
12
13
51
50
49
7 15.2 17.5 9.5
8 20.8 20.0 II. 2
9 23.4 22.5 I2.b
O.OI 263
9-85 571
9.98 762
O.OI 238
9-85 559
12
9-^4 334
9.98 787
O.OI 213
9.85 547
48
13
9-^4 347
9.98 812
O.OI 188
9-85 534
47
1.^ 19.
H
9.84 360
9.98 838
25
25
25
26
O.OI 162
9.85 522
12
13
12
1 2
46
IS
9.^4 373
12
I -1
9.98 863
O.OI 137
9.85510
45
1.3 1.2
2.6 2.4
3-9 3.6
5.2 4.8
i6
9.84 3«5
9.98 888
O.OI 112
9-85 497
44
3
I?
9.84 398
I ■>
9.98913
O.OI 087
9.85 485
43
i8
9.84 41 1
13
13
9-98 939
25
25
26
25
25
25
26
O.OI 061
9-85 473
42
5
6
I
(x.S 6.0
7.8 7.2
9.1 8.4
10.4 9.6
19
20
21
9.84 424
9-^4 437
9.84450
9.98 964
9.98 989
O.OI 036
9.85 460
13
12
12
13
12
12
41
40
39
O.OI on
9.85 448
9.99015
0.00985
9-85 436
22
9.84 463
13
13
9.99 040
0.00 960
9.85 423
38
9
11.7 10.8
23
9.84476
9.99 065
0.00 935
9-85411
37
24
9.84489
9.84 502
9.84515
9.99 090
9.99 116
9.99 141
0.00910
0.00 884
0.00 859
9-85 399
9-85 i86-.
9:35-374
36
25
26
13
25
%
26
25
25
«-
4
27
9.84 528
^j
9.99 166
0.00 834
9.85361
13
ZZ
13 13
28
9.84 540
13
13
13
13
13
13
12
13
9.99 191
0.00 809
9-85 349
32
29
30
31
9-84 553
9.99217
0.00 783
9-85 337
13
12
13
12
13
31
30
29
26 25
9.84 566
9.99 242
0.00 758
9-85 324
I
I.O I.O
3.0 2.9
9M 579
9.99 267
0.00 733
9.85312
32
9.84 592
9.99 293
25
25
25
26
0.00 707
9.85 299
28
2
5.0 4.8
Zi
9.84 605
9.99318
0.00 682
9.85287
27
3
7.0 6.7
34
9.84618
9-99 343
0.00 657
9-85 274
26
4
9.0 8.7
3S
9.84 630
9.99 368
0.00 632
9.85 262
12
25
I I.O 10.6
36
9.84 643
9-99 394
0.00 606
9-85 250
24
I
13.0 12.5
37
9.84 656
13
9.99419
25
0.00 581
9-85 237
13
23
15.0 14.4
38
9.84 669
13
9.99 444
25
0.00 556
9.85 225
22
9
17.0 16.3
19.0 18.3
21.0 20.2
23.0 22.1
25.0 24.0
39
40
41
42
9.84 682
13
12
13
13
13
9.99 469
26
25
25
25
26
0.00 531
9.85212
13
12
13
12
13
21
20
18
10
II
12
13
9.84 694
9.99 495
0.00 505
9.85 200
9.84 707
9.84 720
9.99 520
9-99 545
0.00 480
0.00 455
9.85 187
9.85 175
43
9-84 733
9-99 570
0.00 430
9.85 162
17
44
9-84 745
13
13
13
12
9.99 596
25
25
26
25
25
25
26
25
25
0.00 404
9.85 150
13
12
13
12
16
45
9.84 758
9.99 521
0.00 379
985 137
15
12 12
46
9.84771
9.99 646
0.00 354
9-85 125
14
26 25
47
9.84 784
9.99 672
0.00 328
9.85 112
13
I.I I.O
48
9.84 796
9.99 697
0.00 303
9.85 100
12
I
3-2 3-1
49
50
51
9.84 809
13
13
12
13
9.99 722
0.00 278
9.85 087
13
12
13
1 2
II
10
9
2
3
4
5
6
5-4 5-2
7-6 7-3
9-8 9-4
9.84 822
9-99 747
0.00 253
9.85 074
9.84835
9-99 773
0.00 227
9.85 062
52
9.84847
9.99 798
0.00 202
9.85 049
8
11.9 11.5
53
9.84 860
9.99 823
0.00 177
9.85 037
7
14.1 13-5
54
9.84873
13
12
9.99848
25
26
0.00 152
9.85 024
13
12
6
16.2 15.6
18.4 17.7
20.6 19.8
22.8 21.9
24.9 24.0
55
9.84 885
9.99 874
0.00 126
9.85 012
5
9
56
9.84898
9.84 91 1
13
13
9.99 899
9.99 924
25
25
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0.00 10 1
0.00 076
9.84 999
9.84 986
13
4
3
10
II
5«
9.84 923
13
13
9.99 949
0.00051
9.84 974
13
12
2
12
59
60
9.84 936
9-99 975
25
0.00 025
9.84961
I
9.84 949
0.00 000
0.00 000
9.84 949
L. Cos.
d.
L. Cot. |c. d.
L. Tan.
L. Sin.
d. '
P.P. ;
46'
n
III.
NATURAL
TRIGONOMETRIC FUNCTIONS
FOR EACH MINUTE.
72
,
0°
1
^
1
t
N.Sin.
|N.Tan.|N.Cot.
N.Cos.l
I
.00000
.00000
00
1. 0000
60
59
029
029
3437-7
000
2
058
058
1718.9
000
58
3
087
087
"45-9
000
57
4
116
116
85944
000
56
5
.00145
.00145
687.55
1. 0000
55
6
175
175
572.96
000
54
7
204
204
491. II
000
53
8
233
233
429.72
000
52
9
10
II
262
262
381.97
000
51
50
49
.00291
.00291
343.77
1. 0000
320
320
312.52
-99999
12
349
349
286.48
999
48
13
37«
378
264.44
999
47
14
407
407
245-55
999
46
IS
.00436
.00436
229.18
.99999
4S
16
465
465
214.86
999
44
■
17
495
495
202.22
999
43
18
524
524
190.98
999
42
19
20
21
553
553
180.93
998
41
40
39
.00582
.00582
171.89
.99998
611
611
163.70
998
22
640
640
156.26
998
38
23
669
669
149.47
998
37
24
698
698
143.24
998
36
25
.00727
.00727
137-51
■99997
3S
26
756
756
132.22
997
34
27
785
785
127.32
997
997
33
28
814
81S
122.77
32
29
30
31
844
844
118.54
996
31
30
29
.00873
.00873
"4-59
.99996
902
902
110.89
996
32
931
931
107.43
996
28
33
9bo
960
104.17
995
27
34
.00989
.00989
lOI.II
995
26
35
.01018
.01018
98.218
.99995
25
36
047
047
95.489
995
24
37
076
076
92.908
994
23
38
105
105
90.463
994
22
39
40
41
134
135
88.144
994
21
20
19
.01164
.01164
85.940
.99993
193
193
83.844
993
42
222
222
81.847
993
18
43
251
251
79.943
992
17
44
280
280
78,126
992
16
4S
.01309
.01309
76.390
.99991
15
46
33^
338
74.729
991
11,4
47
367
367
73-139
991
13
48
396
396
71.615
990
12
49
50
SI
425
425
70.153
990
II
10
9
.01454
•01455
68.750
.99989
483
484
67.402
989
S2
513
513
66.105
989
8
53
542
542
64.858
988
7
S4
571
571
63-657
988
6
ss
.01600
.01600
62.499
-99987-
-^
56
629
629
61.383
987
4
S7
658
658
60.306
986
3
sB
687
687
S9.266
986
2
59
60
716
716
58.261
985
I
•01745
.01746
57.290
.99985
N. Cos.
N.Cot.lN.Tan.
N.Sin.j ' i
^
1
' iN.Sin.
N.Tan.|N.Cot.
N.Cos
I
•01745
.01746
57-290
.99985
60
S9
774
775
56.351
984
2
803
804
55^442
984
S8
3
832
833
54^561
983
57
4
862
862
53-709
983
S6
5
.01891
.01S91
52.882
.99982
SS
6
920
920
52.081
982
54
7
949
949
51^303
981
S3
8
.01978
.01978
50.549
980
S2.
9
10
II
.02007
.02007
49.816
980
51
50
49
.02036
.02036
49.104
.99979
065
066
48.412
979
12
094
095
47.740
978
48
13
123
124
47-085
977
47
14
152
153
46.449
977
46
15
.02181
.02182
45-829
.99976
4S
16
211
211
45.226
976
44
17
240
240
44.639
97S
43
18
269
269
44.066
974
42
19
20
21
298
298
43-508
974
41
40
39
-02327
.02328
42.964
•99973
356
357
42.433
972
22
385
386
41.916
972
38
23
414
415
41.411
971
37
24
443
444
40.917
970
36
25
.02472
-02473
40.436
.99969
3S
26
501
502
39.965
969
34 ■
27
530
531
39-506
968
33
28
560
560
39.057
967
32
29
30
31
589
589
38.618
966
31
30
29
.02618
.02619
38.188
.99966
647
648
37.769
965
32
676
677
37.358
964
28
33
705
706
36.956
963
27
34
734
735
36.563
963
26
35
.02763
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36.178
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2S
36
792
793
35.801
961
24
37
821
822
35-431
960
23
3^
850
851
35-070
959
22
39
40
41
879
881
34.715
959
21
20
19
.02908
.02910
34.368
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938
939
34.027
957
42
967
968
33.694
956
18
43
.02996
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33.366
955
17
44
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33.045
954
16
45
.03054
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32.730
•99953-
IS
46
083
084
32.421
952
14
47
112
114
32.118
952
13
48
141
143
31.821
951
12
49
50
SI
170
172
31.528
950
II
10
9
.03199
.03201
31.242
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228
230
30.960
948
S2
257
259
.30.683
947
8
53
286
288
30.412
946
7
S4
316
317
30.145
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6
SS
■03345
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29.882
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5
56
374
376
29.024
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4
57
403
405
29.371
942
3
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432
434
29.122
941
2
59
60
461
463
28.877
940
.03490
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28.636
•99939
1
N.Cos.|N.Cot.
N.Tan.
N. Sin.
'
89°
88'
2°
A
1
N. Sin.
N. Tan. In. Cot.
N.Cos.
I
.03490
519
.03492
28.636
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.99939
938
60
59
5ii
2
54Ji
550
28.166
937
58
3
577
579
27-93r
936
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4
606
609
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935
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5
•03635
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27,490
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55
bb4
667
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933
54
7
693
696
27.057
932
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8
723
725
26.845
931
52
9
10
11
752
754
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930
49
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26.432
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927
810
812
12
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26.031
926
48
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25-835
925
47
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900
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46
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45
lb
955
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44
17
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25.080
921
43
18
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24.898
919
42
19
20
21
042
046
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918
41
40
39
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24.542
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100
104
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916
22
129
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915
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23
159
162
24.02b
913
37
I24
188
191
23.859
912
36
25
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35
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246
250
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34
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275
279
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304
308
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32
29
30
31
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23.058
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31
30
29
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22,904
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391
395
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904
32
420
424
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28
33
449
454
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901
27
34
478
483
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900
26
35
.04507
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22.164
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2S
3b
536
541
22.022
897
24
37
565
570
21.881
896-
23
3«
594
599
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894
22
39
40
41
623
628
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21
20
19
.04653
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21.470
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682
687
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890
42
711
716
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889
18
43
749
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21.075
888
17
44
769
774
20.946
886
16
45
.04798
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20.819
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15
4b
827
833
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883
H
47
856
862
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882
13
48
885
891
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881
12
49
50
51
914
920
•325 ! 879
II
10
9
•04943
.04949
.04978
20.206 1 .99878
.04972
20.087
87b
52
.05001
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19.970
87.5
8
53
030
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873
7
54
059
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6
55
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i9-627
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5
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117.
124
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4
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146 1 153
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867
3
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175
182
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866
2
59
60
205
212
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864
ii
.05234 1 .05241
19.081
.99863
.N.Cos.|n. Cot.|N.Tan.| N. Sin.) ' J\\
8
T
)
^
3°
73
/
N. Sin. N. Tan.
N. Cot. N.Cos.
.05234
.05241
19.081
.99863
60
I
263
270
18.976
861
59
2
292
299
.871
860
S8
3
321
328
.768
858
57
4
350
357
.666
857
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.0099
059
17
44
381
073
.0094
039
16
45
.70401
•991 31
1.0088
.71019
15
46
422
189
.0082
.70998
14
47
443
247
.0076
978
13
48
463
304
.0070
957
12
49
50
51
484
362 .0064
937
II
10
9
•70505
.99420
1.0058
.70916
525
478
.0052
896
52
546
536
.0047
875
8
53
567
594
.0041
855
7
54
587
652
•0035
83.4
6
55
.70608
.99710
1.0029
.70813
5
56
628
768
.0023
793
4
57
649
826
.0017
772
3
58
670
884
.0012
752
2
59
60
690
.99942
.0006
731
I
.70711
1. 0000 j 1. 0000
.70711
|n Cos.|N.Cot.|N.Tan.! N. Sin.| '
A.*i'
IV. CIRCULAR ARCS
WITH
R/
\I)IUS
UNITY. 95
DEGREES.
1 MINUTES.
SECONDS.
O^
0.0000000
60
61
1. 047 1 9 76
120^
121
2.09439 5 »
2.1118484
0' , o.ooocx) 00
O", 0.0000000
0.0174533
1 .06465 08
I
0.0002909
I
0.0000048
2
0.03490 66
62
1.0821041
122
2.12930 17
2
0.00058 18
2
0.0000097
3
0.05235 99
63
1-09955 74
123
2.1467550
3
0.00087 27
3
0.0000 1 45
4
0.06981 32
64
1.11701 07
124
2.1642083
4
0.001 16 36
4
0.00001 94
5
0.08726 65
65
1. 1 3446 40
125
2.18166 16
5
0.0014544
5
0.00002 42
6
0.1047 1 9^
66
1.15191 73
126
2.1991149
b
0.0017453
6
0.00002 91
7
0.1 22 1 7 30
67
1. 1 6937 06
127
2.2165682
7
0.00203 62
7
0.00003 39
8
0.13962 63
68
1. 1 8682 39
128
2.23402 14
8
0.00232 71
8
0.00003 88
9
10
II
0.1570796
69
70
71
1.20427 72
1.2217305
129
130
131
2.2514747
9
10
11
0.00261 80
10
11
0.00004 36
0.1745329
2.26892 80
0.00290 89
0.00004 85
0.00005 33
0.1919862
1.23918 38
2.2863813
0.0031998
12
0.20943 95
72
1.25663 71
132
2.30383 46
12
0.00349 07
12
0.00005 82
13
0.22689 28
73
1.2740904
^33
2.3212879
13
0.00378 15
13
0.00006 30
H
0.2443461
74
1.2915436
1.34
2.3387412
14
0.00407 24
M
0.00006 79
15
0.2617994
75
1.3089969
135
2.356194?
15
0.00436 3i
15
0.00007 27
16
0.27925 27
76
1.3264502
136
2.37364 78
16
0.00465 42
16
0.00007 76
17
0.2967060
77
1-3439035
1.37
2.391 10 11
17
0.00494 5 1
17
0.00008 24
18
0.3141593
78
1.36135 68
138
2.40855 44
18
0.00523 60
18
0.00008 73
19
20
21
0.33161 26
79
80
81
1.37881 01
139
140
141
2.42600 77
19
20
21
0.00552 69
19
20
21
0,00009 21
0.34906 59
1.3962634
2.44346 10
0.00581 78
0.00009 70
0.36651 91
1.41371 67
2.46091 42
0.00610 87
o.oooio 18
22
0.38397 24
82
1.4311700
142
2.47836 75
22 1 0.00639 95
22
o.oooio 67
23
0.40142 57
83
1.4486233
143
2.49582 08
23
0.00669 04
23
0.0001 1 15
24
0.4188790
84
1.46607 66
144
2.5132741
24
0.00698 13
24
0.000 11 64
25
0.43633 23
85
1.4835299
HS
2.53072 74
25
0.00727 22
25
0.00012 12
26
0.45378 56
86
1.50098 32
146
2.5481807
26
0.0075631
26
0.00012 61
27
0.4712389
87
1. 5 1 843 64
H7
2.56563 40
27
0.00785 40
27
0.0001309
28
0.48869 22
88
1-5358897
148
2-58308 73
28
0.0081449
28
0.0001357
29
30
31
0.50614 51
89
90
91
1.5533430
149
150
151
2.6005406
29
30
31
0.00843 58
29
30
31
0.0001406
0.52359 88
1-5707963
2-6179939
0.00872 66
0.00014 54
0.54105 21
1.5882496
2.63544 72
0.00901 75
0.00015 03
32
0.5585054
92
1.60570 29
152
2.6529005
32
0.00930 84
32
0.00015 5^
33
0.57595 87
93
1.6231562
153
2.67035 38
33
0.00959 93
33
0.0001600
34
0.59341 19
94
1.6406095
154
2.68780 70
34
0.00989 02
34
0.0001648
35
0.61086 52
95
1.6580628
iSS
2.7052603
35
0.01018 u
35
0.0001697
36
0.62831 85
96
1.67551 61
156
2.72271 36
36
0.01047 20
36
0.00017 45
37
0.64577 18
97
1.6929694
157
2.74016 69
37
0.01076 29
37
0.00017 94
3«
0.6632251
98
1.7104227
158
2.7576202
38
0.01105 38
38
0.0001842
39
40
41
0.68067 84
99
100
lOI
1.72787 60
159
160
161
2.7750735
39
40
41
0.0113446
39
40
0.0001891
0.69813 17
i.''4532 93
2.7925268
0.0116355
0.00019 39
0-7155850
1.7627825
2.8099801
0.01 192 64
41
0.00019 88
42
0.73303 ^3
102
1.78023 58
162
2.82743 34
42
0.01221 73
42 1 0.00020 36
43
0.75049 16
103
1.79768 91
163
2.8448867
43
0.0125082
43 0.00020 85
44
0.7679449
104
1.81514 24
164
2.86234 00
44
0.0127991
44 0.00021 33
45
0.78539 82
105
1-8325957
i6s
2.87979 33
45
0.0130900
45 0.00021 82
45
0.80285 15
106
1.8500490
166
2.89724 66
46
0.0133809
46 i 0.00022 30
47
0.82030 47
107
1.8675023
167
2.91469 99
47
0.01367 17
47 1 0.00022 79
48
0.83775 80
108
1.88495 56
168
2.93215 31
48
0.0139626
48 1 0.00023 27
49
50
51
0.85521 13
109
110
m
1.9024089
169
170
171
2.94960 64
49
50
51
0.01425 35
49
50
0.00023 76
0.8726646
1.91986 22
2.96705 97
0.0145444
0.00024 24
0.89011 79
I-9373I 55
2.98451 30
0.01483 53
51
0.00024 73
52
0.90757 12
112
1.9547688
172
3.0019663
52
0.01512 62
52
0.00025 21
53
0.92502 45
"3
1.97222 21
^73
3.01941 96
53
0.01541 71
53
0.00025 70
54
0.94247 78
114
1-9896753
174
3.03687 29
54
0.0157080
54
0.00026 18
55
0.95993"
115
2.00712 86
17s
3.05432 62
55
0.0159989
55 0.00026 66 II
5^
0.9773844
116
2.02458 19
176
30717795
56
0.0162897
56
0.00027 15
57
0.99483 77
117
2.0420352
177
3.08923 28
57
0.0165806
57
0.00027 63
5«
1.01229 10
118
2.05948 85
178
3.10668 61
58
0.01687 '5
58
0.00028 12
59
60
1.0297443
119
120
2.07694 18
179
180
3-1241394
59
60
0.01716 24
59
60
0.00028 60
1. 047 1 9 76
2.0943951
3.1415927
0.0174533
0.00029 09
DE
IGREES.
MINUTES.
SECONDS. i|
96
V.
CONVERSION
OF
LOGARITHMS.
Base of common logarithms
= 10. 1
Base of Naperian logarithms () = 2.71828 18284 59045 23536
Com. Log. e = Jlf (Modulus of Com. Logs.) = 0.43429 44819 03251 82765
Nap. Log. 10 = —
= 2.30258 50929 94045 68402
Com. Log. N = jV X Nap. Log. A'. ""
Nap. Log. JV =— X Com. Log. N.
► where iV denotes any number.
Multiples of M.
Multiples of J .
M
2
3
0.00000 000
50
51
52
53
21.71472 410
22.14901 858
22.58331 306
23.01760754
I
2
3
0.00000 000
2.30258 509
4.60517019
6.90775J28
50
51
52
53
115.12925465
0.43429 448
0.86858 896
1.30288345
117.43183974
119-73442484
122.03700993
4
I
1-73717793
2.17147241
2.60576 689
54
11
23.45190202
23.88619 6^0
24.32049 099
4
5
6
9.21034037
1 1. 5 1 292 546
13.81551056
54
55
56
124.33959502
126.64218011
128.94476521
9
10
II
12
13
3.04006 137
3-47435 586
3.90865 034
57
58
59
60
61
62
63
24-75478 547
25.18907995
25.62337 443
I
9
10
II
12
13
1 6. 1 1 809 565
18.42068074
20.72326 584
11
59
60
61
62
63
131.24735030
133-54993539
135-85252049
4.34294 482
26.05766891
23.02585 093
138.15510558
140.45769067
142.76027577
145.06286086
477723 930
5-21153378
5.64582 826
26.49196340
26.92625 788
27.36055 236
25.32843 602
27.63102 112
29.93360621
6.08012275
6.51441 723
6.94871 171
64
65
66
27.79484 684
28.22914 132
28.66343 581
14
32.236J9 130
34-53877 639
36.84136 149
64
147-36544 595
149.66803 104
151.97061 614
19
20
21
22
23
7.38300619
7.81730067
8.25159 516
67
68
69
70
71
72
73
29.09773 029
29.53202 477
29.96631 925
17
18
19
20
21
22
23
39.14394658
41.44653 167
43.74911677
%
69
70
71
72
73
154.27320123
15657578632
158.87837 142
8.68588 964
J040o6]tj73_
30.83490 822
31.26920270
31.70349718
46.05170 186
161. 18095 651
9.12018412
9.55447 860
9.98877 308
48.35428 695
50.65687 205
52.95945 714
163.48354 160
165.78612670
168.08871 179
24
^1
10.42306 757
10.85736 205
11.29165653
74
75
76
32.13779 166
32.57208 614
33.00638 062
24
25
26
55.26204 223
57.56462 732
59.86721 242
74
75
76
170.39129688
172.69388 197
174.99646707
27
28
29
30
3^
32
33
11.72595 lOI
12.16024 549
12.59453998
77
78
79
80
81
82
83
33.44067511
33.87496 959
34.30926 407
_3474355 855^
35- '7785 303
35.61214752
36.04644 200
27
28
29
1 30
31
32
33
62.16979751
64.47238 260
66.77496 770
W
79
80
81
82
83
177.29905216
179.60163725
181.90422235
13.02883446
69.07755 279
^71.38013788
73.68272 298
75.98530 807
184.20680744
13.46312894
13.89742342
H-33I7I 790
186.50939253
188.81197763
191. 11456 272
34
35
36
14.76601 238
15.20030687
15.63460 135
84
85
86
36.48073 648
36.91503096
3734932 544
34
35
36
78.28789316
80.59047821;
82.89306 ss's
84
11
193.41714781
i95-7'973 290
198.02231 800
37
3«
39
40
41
42
43
16.06889583
16.50319 031
16.93748479
17.37177928
17.80607376
18.24036 824
18.67466 272
87
88
89
90
91
92
93
3778361 993
38.21791 441
38.65220 889
37
38
39
40
41
42
43
85.19564844
87.49823 3 S3
89.800S1 863
87
88
89
90
91
92
93
200.32490309
202.62748818
204.93007 328
39.08650 337
39.52079 785
39-95509 234
40.38938 682
92.10340372
207.23265 837
94.40598 881
9670857 391
99.01 u 5 900
209.53524346
211.83782 8s6
214.14041 365
44
45
46
19.10895 720
19.54325 169
19-97754617
94
95
96
40.82368 1 30
41.25797578
41.69227026
44
IOI.3I374409
103.61632 918
I05.9I89I 428
94
11
216.44299874
218.74558383
221.04816893
49
50
20.41184065
20.84613513
21.28042 961
97
98
99
100
42.12656474
42.56085 923
42.99515371
ti
49
50
108.22149937
110.52408446
112.82666956
99
100
22335075 402
225.65333911
227.95592421
21.71472410
43-42944819
115. 12925 465
230.25850930
97
TRIGONOMETRIC FORMULAS.
sm^a + cos^a = l.
sec'-^a = I +tan2a.
cosec^a = I + cot^a.
sino = db
QOta
Vi + tan2,
cos a
_ cos^
sin a
cos a —^±
sec a
i_
COSO
I
sin a
sin {a ± (3) = sin a cos j3 ± cos a sin /3.
cos ( a ± /y) = cos « cos /3 =F sin a sin ^^
» / , «\ f^" <"■ ± tan /3
, tan (a ± p) ■= ■ — — ^—^
I =F tan a tan /3
sin a + sin /3 =2 sin ^ (a-J-/3) cos^ (a — ^).
sin a — sin /3 =2 cos J (^4-/3) sin ^ (a — /3).
cosa + c«>sj3 = 2cos J (a4-/3)cos^ (a — /3).
cosa — ct)S/3=— 2sin^(a + /3)sini(a — /3).
Fig.
sin a sin /3 = ^ cos (a — /3) — ^ cos (a + ^) .
cos a cos /3 =r J cos (a — j3) + ^ cos (a + /3) .
sin a cos ^ = ^ sin (a + P) + ^ sin (a — /3).
sin2 a - sin2 /3 = cos^jS - cos^a = sin (a + /3) sin (a - jS).
cos^ a — sin2 j3 = cos'^ /3 — sin^ a = cos (a + /3) cos (a — /3).
sin 2 a = 2 sin o cos a.
cos 2 o = cos- o — sin^ a.
2 sin2 ^ a = I — cos o
tan
tan 2 a =
2 tan g
I - tan2a
i«=±Vr
cos a
2 C0S2 ^ o = I -f cos O.
sin a I — COS a
+ COS a I + cos a sin a
sin a + sin (o + ;r) + sin (a -f 2 jc) -f ••• + sin (a + nx)
_ sin ^ (« 4- I ) jT sin (o + ^ nx)
sin i^x
cos a f cos {a + x)+ cos (a + 2^)+ ••• + cos (a -f «jir)
_sin^ (« + .rcos(a -f ^ «^)
~ sin ^ jr
■\/ — I. ^* = cos X + «■ sin jr.
cos X = - (^» -f ^-**) .
2
^ •" = cosjT — t smx.
sin jr = — («■ — <»-**).
2?
^»«» = (cos X + i sin x)" = cos nx +
.f^ O- THf
%
^l^^gSi
or
98
PLANE TRIANGLES.
a + d _tain^ (A -\- B)
V^^b ~ Xzxi\\A - B) '^"
a — h _ 2 sin ^ r ,—r
c = , if tan X = Vao.
a _ b _ c
sin A sin B sin C .
a = <5cos C+ ccosB.
a^ = IP- + c' — 2 be cos A.
a sin \{B - C)-(b-c) cos J A.
a cos ^(B- C) = {b-\- c) sin ^ A.
asm B
c — a cos B
\{ s=\{a -\- b -^ c)\
smM = A/^— 4^ ^•
/(^-^)(^-0
.s^A=^jf^. tanM = V^^^^
J (s-a)0-i)(s -^_ ,a„M=— • tan J 5=-^. ta,>JC=-il-.
'J ^ s—a ^ s — b ^ s—c
. 1 JL • ^ f2 sin /4 sin i5
Area — \ab%yi\ C = — '
2 sin C
Radius of inscribed circle = r.
Diameter of circumscribed circle
= \/ s{s- a)(^s-b) {s- c).
sin A
DIFFERENTIAL FORMULAS FOR PLANE TRIANGLES.
i/A + dB + dC = o.
— - cot A dA ='^ - cotBc/B = - - cot C dC.
a b c
da - cos Cdb + cos B dc -{- b sin CdA.
a dB = sin Cdb — s'm B dc — b cos CdA.
RIGHT SPHERICAL TRIANGLES (C=90°).
sin « = sin ^ sin c.
sin a = cotB ta.nb.
a cos ^ = sin -5 cos a.
cos A = tan b cot c.
sin ^ = sin ^ sin c.
sin <^ = cot .4 tana,
cos i9 = sin ^ cos b.
cos B = tan a cot r.
cos c = cos a cos b = cot -r4 cot B.
Fig. 3.
99
OBLIQUE SPHERICAL TRIANGLES.
sin a _ sin^ _ sing
sin /^ sin .5 sin C
cosrt = cos(^ cosr + sin <^sinr cos/^.
cos A =— cos ^ cos C 4- sin i9 sin C cos a.
sin a cos i9 = cos b sin f — sin b cos r cos /4.
sin A cos ?cos(5-^).
sin \c s\n^{A — B)— cos ^ C sin i (« — b).
sin I i
r = tangent of the angular radius of the inscribed small circle.
R = tangent of the angular radius of the circumscribed small circle.
SPHERICAL EXCESS.
E = A + B-\-C- i8o°.
sm\a sin^b .
s\nlt E = — ; — sin C.
cos ^ c
tan^^
_ tan ^rt tan^^sin C
~ I + tan ^ rt tan ^ b cos C
tan2 J ^ = tan ^ J tan ^ {s - a) tan ^ (j - {z);
BINOMIAL THEOREM.
I I • 2 I • 2 • 3
EXPONENTIAL THEOREM.*
M \ M I 2\ \ M I 3\ \ A/ I 4\
^ = I + ;c + -^2 + 7^ + — ^ H ^ + ^ + •-.
2 6 24 120 720
* n ! denotes " factorial «," or the product i • 2 • 3 • 4 ••• «.
LOGARITHMIC SERIES.*
,o.(...=,o....[.(i)--Q>;(^7-'(,7....|.
log(i + x)= A/ {x - ^ x-i + -X^ --Ji-* + -x^ ).
23 4 5
l0g(l - X)=: - M {X -^ ^~X^ -\- ^ X^ + ^X* + - X^ -^ •••)•
iVj// 2!^^// 3!V^'^/ 4!\^/
+ -.
TRIGONOMETRIC SERIES.* t
I 3! 5! 7!
;r'^ , x^ x^ ,
cos ;r=l -\ — h""-
2! 4! 6!
3 15 315 2835 155925
cot X = X x^ x^
•^ 3 45 945 4725 93555
sec^ = I + i;c2 + -^x^ + ^x^ + ^x^ + ....
2 24 720 8064
cosec ;r = - + -;i:+ -^ x^ -\ ^ x^-{- ^x'' + •••.
X 6 360 15120 604800
sin-ij = v + -y + -^ vS ^ _5_ y + _35_ y + ....
6 40 112 1152
-^2-^6-^ 40-^ 112-^ 1152-^
tan-ijj/ = J j3 4- _y jj,7 _|_ _y _ ....
3 5 7 9
>' 3r 5r 7r 9r
log sin .r= log X - M ( ^-x^ + -^x^ -\- -^x^ + —^ x^ -i- "\
V6 180 2835 37800 /
\ogcos X = - A/ ( ~x^ -j- — x^ + —x^ + -^x» + "•].
\2 12 45 2520 /
log tan or = log.r + Afl-x^ + ^x^ + -^x^ + -^^ + -V
V3 90 2835 18900 I
logsin-Jj = log;/ + AI f iy^ + -^-y + Jli j^ + ...).
\.6 180 5070 /
logtan-i/ = log7-il/f-jj/2-i3y + ^^/ V
\3 90 2835 J
logsin^ = logtanjc- Afl-tan'^x - - tan* ;ir + i tan^ j; --^tan^jr + ... ).
V2 4 6 8 /
log tan X = log sin x + A/( - sin^ ;r + - sin* x + - sin^ x + - sin^ j: + •• • ) .
\2 4 6 8 /
* n ! denotes " factorial «," or the product I • 2 • 3 • 4 ... «.
t The angles are expressed in circular measure.
DIFFERENTIATION.
d {ax + /J) = where
« is even.
J^ ^ ^ 3 180 1512 '
where P = Lx [^'0 + ;'„ + 2 {yo + _j/4 + . . . + j„_2) ] ,
^=: 2Ax[>'i + V3 + ••• +J>'«-l],
Fn'" = f{x') when x = abscissa of jn*
2. Simpson's rule :
^ = ^[JO + Jn + 2 (^2 +^4 +••• +JJ'„_2)+ 4(>'l + J3 + - +/«-!)].
3
3. Weddle's rule (for seven ordinates) :
A - ^— ^ [jj'o + 72 + ^'4 + jJ'6 + jJ^3 + 5 (/I 4-/3+ J5)].
10
4. Prismoidal formula : F = — [^ + /4' + 4 /^m] = - [^ + ^' + 4 ^«].
3 6
I03
Constants.
Base of Naperian logs : ^ = . .
Modulus of common logs : log e •=. M ■■
Degrees in arc = radius: 180'^ h- tt =
Minutes in arc = radius :
Seconds in arc = radius :
360^^ expressed in minutes of arc : . . .
360° expressed in seconds of arc : . . .
24 hours expressed in minutes of time :
24 hours expressed in seconds of time:
TT = 314159 26535 89793 23846
logTT =0.49714 98726 94133 85435
sin i" =0.00000 48481 3681 1 07637
arc i" =0.00000 48481 3681 1 09536
. . . . 2.71828 183 .
. . . . 0.43429448 .
• • . 57°-29577 95i •
• .3 437'-74677- • • •
2o6 264".8o6 . . . .
21 600' .
I 296000" .
I 440"* .
86400* .
Eng. inch 0.02540
Eng. foot 0.30480
Eng. yard 0.91440
Eng. statute mile 1.60935
meter 39-3700
meter 3.28083
meter i. 09361
kilometer O.62137
sq. foot 9.29034
sq. inch 6.45163
sq. meter 10.7639
sq. centimeter 0.15500
cubic foot 0.02831
cubic inch 16.3872
cubic meter 35-3145
cubic decimeter (liter) . . . 61.0234
avoirdupois pound 453.59242
avoirdupois ounce 28.34953
Troy ounce 31.10348
grain 64.79892
kilogram 2.20462
kilogram 35.2740
kilogram 32.1507
gram 15-43235
foot-pound 0.13825
kilogram-meter 7.23300
pound per sq. in 70.3067
gram per sq. cm 0.01422
pound per cu. ft 0.0 1 601
grain per cu. in 0.00395
gram per cu. cm 62.4283
gram per cu. cm 252.8925
70
77
639
5
meters
meters
meters
kilometers ....
Eng. inches . . .
Eng. feet
Eng. yards ....
Eng. statute miles
sq. decimeters . .
sq. centimeters .
sq. feet
sq. inches ....
cubic meters . . .
cubic centimeters
cubic feet ....
cubic inches . . .
grams . . . .
grams . . . .
grams . . . .
milligrams . .
avdp. pounds
avdp. ounces
Troy ounces .
grains . . . .
kilogram-meters
foot-pounds . .
grams per sq. cm.
34 lbs. per sq. in. . .
84 grams per cu. cm.
425 grams per cu. cm.
lbs. per cu. ft. . .
grains per cu. in.
Logarithms.
0.43429 448
9.6377^ 431
1. 75812 263
3.53627 38«
5.31442 513
4-33445 375
6. 1 1 260 500
3.15836249
4-93651 374
0.49714987
.68557487 — 10
-68557 487 - 10
Logarithms.
8.40483 5 — 10
9.48401 6 — 10
9.961137- 10
0.20665 O
I-595165
0.515984
0.03886 3
9-79335 o - 10
0.96803 2
0.80966 9
1.031968
9.19033 I - 10
8.45204 7—10
1.214504
1-54795 3
1.785496
2.65666 6
1.452546
1 .49280 9
1.811568
0-34333 4
1.54745 4
1.50719 I
1. 18843 2
9.14068 2—10
0.85931 8
1.84699 7
8.153003 - 10
8.20461 8 — 10
7.597064- lO-
1.795382
2.40293 6
Logarithms.
dynes (^jf in meters).
dynes (^ in meters) 0.811568
ergs {g'\v\ centimeters) .... 4.140682
ergs (^ in centimeters),
ergs per sec.
watts (^ in meters) 1.881044
watts (approximately) .... 2.87273 9
32.086 528 + 0.171 293 sin2 — 0.000003 h. in feet (Harkness).
= 9-779 886 + 0.052 210 sin^t^ — 0.000003 f^- i" meters (Harkness).
/ = 39.012 540 -f 0.208 268 sin2 — 0.000000 3 //. in inches (Harkness)
= 0.990 910 -f 0.005 290 sin2 — 0.000 000 3 h. in meters (Harkness).
Wt. of mass of 1 gram . . loo^
Wt. of mass of I grain . . 6.47989 2g
1 foot-pound 1 3825.5 i"
1 kilogram-meter looooo^
I watt lo''
I horse-power 76.0404^
I horse-power 746
g
EXPLANATION OF THE TABLES.
INTRODUCTORY.
1. When we have a number with six or more decimal places, and we
wish to use only five :
{a) If the sixth and following figures of the decimal are less than
5 in the sixth place, they are dropped ; thus, 0.46437 4999 is called
0.46437.
(^) If the sixth and following figures of the decimal are greater than
5 in the sixth place, the fifth place is increased by unity and the sixth
and following places are dropped ; thus, 0.46437 5001 is called 0.46438.
(c) If the sixth figure of the decimal is 5, and if it is followed only by
zeros, make the fifth figure the nearest even figure ; thus, 0.46437 500 is
called 0.46438, while 0.46438 500 is also called 0.46438. The number
is thus increased when the fifth figure is odd, and decreased when it is
even, the two operations tending to neutralize each other in a series of
computations, and hence to diminish the resultant error.
2. Hence any number obtained according to Art. i may be in error
by half a unit in the fifth decimal place.
3. When the last figure of a number in these tables is 5, the number
printed is too large, the 5 having been obtained according to Art. i (^) ;
if the 5 is without the minus sign, the number printed is too small,
the figures following the 5 having been dropped according to Art. i (a).
4. The marginal tables contain the products of the numbers at the
top of the columns by i, 2, 3, •••9 tenths, and may be used in multiply-
ing and dividing in interpolation.
{a) To multiply 38 by .746 :
38 X .7 = = 26.6
38 X .4= 15-2; ••• 38 X .04 = 1.52
38 X .6 = 22.8 ; .-. 38 X .006 = .228
.-. 38 X .746 = 28.348
In multiplying by the second figure (hundredths), the decimal point
in the table is moved one place to the left ; in multiplying by the third
(thousandths), two to the left ; and so on.
88
I
3-8
2
7.6
3
1 1.4
4
15.2
s
19.0
6
22.8
7
26.6
8
30.4
9
34.2
(105)
EXPLANATION OF THE TABLES.
{b) To divide 28 by t,^ :
Dividend,
28
38
Next less,
26.6
corresponding to
•7
I
2
3.8
76
Remainder,
I 4
3
11.4
Next less,
I 1.4
corresponding to
•03
4
5
15.2
19.0
Remainder,
26
6
7
22.8
26.6
Nearest,
. Quotient,
26.6
corresponding to
.007
.737
8
9
304
34-2
to the nearest third decimal place. The decimal point is moved one
place to the right in each remainder, since the next figure in the quotient
will be one place farther to the right.
To divide 23 by 38 :
Dividend, 23
22.8 corresponding to .6
0.0 corresponding to .00
2 o.
Nearest,
.-. Quotient,
I 9.0 corresponding to .005
.605
The computer should use the marginal tables mentally.
LOGARITHMS.
5. The logarithm of a number is the exponent of the power to which
a given number called the dase must be raised to produce the first
number. U A = ", a is called the logarithm of the number A to the
base i 1 tan a
^ =log — T'
a'
where «' is the number of minutes in the angle. Their values from
0° to 1° 40' (=100') are given at the bottom of pages 3-21 ; from
i°4o'to 3° 20' at the left margin of pages 4 and 5, the first three
figures being found at the top ; and from 3° to 5° on page 24. Thus,
for i'= i' (pages), ^' = 6.46 373, ^' = 6.46 373 ;
for 15'=: 15' (pages), ^' = 6.46 372, r' = 6.46 373 ;
for 2° 40'= 160' (pages), ^'=6.46357, r'= 6.46404;
for 4° 20' =260' (page 24), 6*' = 6.46 331, 7"' = 6.46 456.
Each of these numbers should have —10 written after it (Art. 11).
X EXPLANATION OF THE TABLES.
Note. — The logarithmic cosine of a small angle is found by the ordinary method.
The cotangent of an angle is the reciprocal of the tangent, and hence the logarithmic
cotangent is the arithmetical complement of the logarithmic tangent. The formulas
for finding the logarithmic cosine, tangent, and cotangent of angles near 90° are
given on page 25.
37. To find the logarithmic sine or tangent of a small angle. — From
Art. 36, we have
log sin « = 6"' + log «',
log tan « = 7"' + log «'.
Hence, to find the logarithmic sine or tangent of an angle less than
5°, find the value of the 6"' or T^ corresponding to the angle, interpolat-
ing if necessary, and add it to the logarithm of the number of minutes
in the angle.
Find log sin o°42'.6. Since the angle is nearer 43' than 42', we take
6"' =6.46 371
log 42.6 = 1.62 941
.*. log sin 0° 42 '.6 = 8.09 312
Find log tan i°53'.2. Since the angle is nearer i°53' (= 113') than
114', we take
r' = 6.46 388
log 113.2 = 2.05 385
/. log tani°53'.2 = 8.51 773
Note. — When the angle is given in seconds, either reduce the seconds to deci-
mals of a minute, or use the values of 6"" and 7^" given at the bottom of pages
3-23 and on page 24. They are defined by the equations
5"=log^-i^, and 7^" = log^,
a" a"
where a" is the number of seconds in the angle. Hence
log sin a = .S" + log a", and log tan a— T" -\- log a".
38. To find the S7?iall angle corresponding to a given logarithmic sine
or tangent. — From Art. 36,
log «' = log sin « — .S', '
log a' = log tan a — T\ .
or log a' = log sin a + cpl S ',
log «' = log tan a + cpl T\
When the angle is less than 3°, find on pages 26-28 the value of
cpl 6"' (or cpl 7") corresponding to the function, interpolating if neces-
sary, and add it to log sin « (or log tan «) ; the sum will be the loga-
rithm of the number of minutes in the angle.
In finding the angle whose logarithmic sine is 8.09006, we see from
EXPLANATION OF THE TABLES.
XI
theZ. Sin column (page 26) that the angle is between 0° 42' and o°43',
and that the value of cpl S* must be either 3.53628 or 3.53629. The
given logarithmic sine is nearer that of 42' than that of 43'; hence we
take
cpl ^' = 3-53628
log sin a = 8.09006
log a' = 1.62634 .*. «' = 42'.300.
When the angle is between 3° and 5°, we may find S' and T' from
page 24 after finding the angle approximately from pages 29 and 30.
Thus in finding the angle whose logarithmic tangent is 8.77237 we
find from page 29 that the angle is between 3° 23' and 3° 24', being
nearer 3° 23'. Then on page 24 we have
r'=: 6.46423
log tan ot = 8.77237
/ .*. log tan a — 7"' = log «'= 2.30814 /. a'= 203'.30 = 3° 23'.3o.
^ -CJL-v/yv -4 \^ Angles Greater than 90°.^* M^ O*"^ '
39. To find the logarithmic sine, cosine, tangent, or cotangent of an
angle greater than 90°, subtract from the given angle the largest multi-
ple of 90° contained therein. If this multiple is even, find from the
tables the logarithmic sine, cosine, tangent, or cotangent of the remain-
ing acute angle. If the multiple is odd, the logarithmic cosine, sine,
cotangent, or tangent, respectively, of the remaining acute angle will be
the function required ; thus, sin 120° = sin (90° + 30°) = cos 30°.
x =
I. Quadrant.
a
II. Quadrant.
90°+a
III. Quadrant.
iSo'+a
IV. Quadrant.
270'+ a
%\X\X —
+ sin a
+ cosa
— sin a
— cosa
cos.r =
+ cosa
— sin a
— cosa
+ sin a
tan X =
+ tana
— cot a
+ tana
— cot a
cot X =
-f cot a
— tana
+ cot a
- tana
Or we could find the difference between the angle and 180° or 360°,
and find from the tables the same function of the remaining acute
angle ; thus, cos 300° = cos (360° — 60°) = cos 60°, etc.
x =
I. Quadrant.
a
II. Quadrant.
180° -a
III. Quadrant.
i8o'+tt
IV. Quadrant.
360*— a
or -a
sin X =
COSJf =
tan X =
cot X =
+ sin a
+ cosa
+ tana
+ cot a
+ sin a
— cosa
— tan a
— cot a
— sin a
— cosa
4- tana
+ cot a
— sin a
+ cosa
— tan a
— coto
To indicate that the trigonometric function is negative, n is written
after its logarithm.
xii EXPLANATION OF THE TABLES.
40. To find the angle corresponding to a given function, find the
acute angle a corresponding thereto, and the required angle will be «,
1 80° ± «, or 360° — a, according to the quadrant in which the angle
should be placed.
41. There are always two angles less than 360° corresponding to any
given function. Hence there will be ambiguity in the result unless
some condition is known that will fix the angle ; thus, if the sine is
positive, the angle may be in either of the first two quadrants, but if we
also know that the cosine is negative, the angle must be in the second
quadrant.
Given One Function of an Angle, to fijid Another without Jindifig
the Angle.
42. Suppose log tan a = 9.79361, and log cos a is sought. On page
57 the tabular difference for log tan « is 28, and that for log cos «
is 8, the given logarithmic tangent exceeding 9.79354 by 7, Hence
28 : 7 = 8 : a: ; .*• jc = 2^ X 7 = 2 = correction to 9.92905, giving
log cos a = 9.92903.
In the margin are tables to facilitate the process. In the column
headed ^V, the numerator is the tabular difference for the column
headed Z. Cos, and the denominator that for the logarithmic tangents.
The numbers in the marginal column are the values of A, — the excess
of log tan a over the next smaller logarithmic tangent, found in the
tables, — such that ^^^A shall be 0.5, 1.5, 2.5, etc. ; and the numbers
on the left are the corrections x to be appHed to the numbers in the
column headed L. Cos. Thus, if A is between 1.8 and 5.2, x is between
0.5 and 1.5, and is called i. Note that i is opposite the space between
1.8 and 5.2. In the example above, the excess A is 7, which lies
between 5.2 and 8.8 ; hence x is 2, the number on the left opposite the
space between 5.2 and 8.S.
For example, if we have given the logarithms of the sides of a
right-angled triangle, \oga = 2.98227 and log /^ = 2.90255, to find the
hypotenuse, we use the formulas
a /T />
tan a = -, and e =
b sin a cos «
The value of log tan a being found in
\oga = 2.98227 (t) the column marked Z. Tan at the bot-
logsin« =9.88571 (4) torn, the right column will contain the
log/!' = 2.90255 (2) logarithmic sine of the corresponding
log tan « = 0.07972 (3) angle. Also, the correction to 9.88563
.-. logm'^-//
s^.
EXPLANATION OF THE TABLES.
— ^j; J^-\]
TABLE V.
45. Conversion of common logarithms into Naperian^ and vice versd
(page 95) . — We have
\Q%y^N=M\o%,N, and \og, N = ^\og,^ N .
Table V. contains the multiples of M and — by numbers from
I to 100.
Find the common logarithm of 2, its Naperian logarithm being
0.69314 718056.
J/ X .69 =0.2996631925
J/ X .0031 = .001346312894
J/ X .000047 = .000020411841
J/ X .00000018 = .000000078173
M X .0000000005 = .00000 00002 1 7
M X .00000000006 = .oooQo 00000 26
/. Iogio2 =0.301029995651
(True value = 0.30102 99957)
Find the Naperian logarithm of 0.2, its common logarithm being
9.30102 99957 — 10. Hence the true logarithm is
logio 0.2 = — I H- .30102 9995 -] = — 0.69897 00043.
= 1.58878 37142
= .02049 30073 28
X .000070 = .00016 1 1809 57
X .0000000043 = .00000 00099 01
/. log^ 0.2 = — 1.60943 79123 86
(True value = — 1.60943 79124)
— X .0080
M
M
I
'm.
Typography by J, S. Gushing & Co., Norwood, Mass.
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UNIVERSITY OF CALIFORNIA LIBRARY
This book is DUE on the last date stamped below.
Fine
a'
o
OECJ 19521.1'
3Har'55JP
fiiho{,Jil^^lM
h\
10ct'55KC
SfP 17 1955 LU
REC'D LD
MAY 3 195a
REC'D LD
JAN 16 1959
2Feb'59MJ
RECD LD
JUL 1 6 1959
5sl6)4120
YC 76228