Trigonometry
AND
Stereographic Projections
Trigonometry
AND
Stereographic Projections
(REVISED)
PREPARED FOR THE USE OF THE
MIDSHIPMEN AT THE UNITED STATES
NAVAL ACADEMY
BY
STIMSON J. BROWN
Professor of Mathematics, United States Navy
1913
Copyright, 1913, by
RALPH EARLE
Secy, and Treas. U. S. Naval Institute
(gafttmore (p
BALTIMORE, MD., U. S. A.
PKEFACE.
I have attempted in the work here presented to give a thor-
oughly practical treatment of the subject of trigonometry, adapted
to the requirements of the course of study of the Naval Academy.
The chapter on stereographic projections represents largely
the methods of Professor W. W. Hendrickson, U. S. Navy, by
whose courtesy some of the articles and plates have been taken
directly from his work on that subject.
S. J. B.
JANUABY 18, 1913.
3G0378
CONTENTS.
PLANE TRIGONOMETRY.
CHAPTER I. PAGE
Angles and their Measure 1
Sexagesimal Units 2
Circular Units 3
Compass Units 5
CHAPTER II.
Definitions of Trigonometric Functions . 8
Complementary Functions 11
Fundamental Formulae 12
Line Representation of Functions 15
Reduction of Functions to the First Quadrant 16
Periodicity of Trigonometric Functions 20
Inverse Functions 20
CHAPTER III.
Explanation of the Use of Trigonometric Tables 23
Graphs of Trigonometric Functions 31
CHAPTER IV.
Solution of Plane Right Triangles 35
Polar Coordinates 38
Graphs, Polar Coordinates 40
CHAPTER V.
Functions of the Sum or Difference of Two Angles, Double-angle,
and Half-angle 43
Sum of Two Inverse Functions.. 48
viii CONTENTS.
CHAPTER VI. PAGE
Formulae for Oblique Triangles 50
The Sine Formula 50
The Cosine Formula 51
The Tangent Formula 51
Angles in Terms of the Sides 52
Area of a Triangle 53
CHAPTER VII.
Solution of Oblique Triangles 55
Radius of Circumscribed and Inscribed Circles 61
CHAPTER VIII.
Graphic Representation of Complex Numbers 63
De Moivre's Theorem 65
Functions of Multiple Angles 66
Series for Sine and Cosine 67
Roots of Imaginaries 70
SPHERICAL TRIGONOMETRY.
CHAPTER IX.
General Formulae 71
The Three Sides and an Angle 72
The Three Angles and a Side 73
The Sine Formula 74
The Half-angles in Terms of the Sides 76
The Half-sides in Terms of the Angles 77
Gauss's Equations 78
Napier's Analogies 79
CHAPTER X.
Solution of Spherical Right Triangles 80
Napier's Rules 81
The Law of Quadrants.. 82
CONTENTS. ix
CHAPTER X. Continued. PAGE
The Double Solution 83
Triangles with a Small Part 85
Quadrantal Triangles 86
CHAPTER XI.
Solution of Oblique Spherical Triangles 87
Solution by Perpendicular and Napier's Rules 88
Solution by Cosine Formula 90
Solution by Napier's Analogies 92
Double Solution 93
Solution by Tangent Formula (the Three Sides Given) 98
STEREOGRAPHIC PROJECTIONS.
CHAPTER XII.
Definitions and Fundamental Properties 100
To Project a Circle with Given Pole and Polar Distance 107
Line of Centers of Circles Through a Given Point 109
To Draw a Great Circle at a Given Angle with NS 110
To Find the Pole of a Given Circle Ill
True Length of a Projected Arc. Ill
Terrestrial Spherical Triangles 112
Great Circle Distances 112
The Celestial Sphere 115
Celestial Coordinates 115
The Astronomical Triangle 117
Given L, d and h, to Project Meridian and Horizon and Solve 121
Given t, d and I/, to Project on Horizon and Meridian and Solve. 126
Method of St. Hilaire 130
Given t. d, h to Project on Equator and Solve 130
CORRECTIONS TO LATEST EDITION OF BROWN'S
TRIGONOMETRY.
(DECEMBER 10, 1916.)
1. Date of edition not correct.
2. p. ix, 4th line from bottom of page, omit the words " Meridian
and Horizon."
3. p. ix, 3d line from bottom of page, omit the words " on Hori-
zon and Meridian/'
4. p. 6, 9th line, read"/! ~~Y instead of
5. p. 13, 6th line, omit " or " at left-hand edge of page.
6. p. 16, next to last line, read " side " instead of " sides."
7. p. 18, 3d line, read " sin (180 + 0) = -sin 0" instead of
"sin (180+0) = -0."
8. p. 20, last line, read "cos" 1 VT^r 2 " instead of "cos^Vl-z 2 ."
9. p. 21, 4th line from bottom, read " terms of sin x " instead of
" terms sin x." _
10. p. 26, example, middle of page, read " 5 59' 50" " instead of
"5 59' #0"."
11. p. 28, 15th line, read " (68 16') " for " (68 15')."
12. p. 35, 9th line from bottom, read "a cotan A" for "a cotan 5."
13. p. 41, 5th line, read " Note " for " Notes."
14. p. 41, 2d line from bottom, read " 135 " for " 105."
15. p. 47, end of 2d line, read '7~ + 0Y for '7-~
16. p. 48, near middle of page, read " tan' 1 " for
2 CORRECTIONS TO BROWN'S TRIGONOMETRY.
17. p. 48, next line of work, read
" 6 - ff = tan" 1 x - tan- 1 y = tan' 1 %=*-
1 + xy
18. p. 50, 4th line from bottom, read " a sin B " for " a sin A"
19. p. 51, in Fig. 38, read " D" placed directly under C, for " <Z."
20. p. 51, 5th line, omit " and," reading in one line,
"in EDO, BD = aco$B=c bco$A." And begin next line,
" Equating
21. p. 59, 5th line, read "Arts. 59 and 60 " for "Arts. 48 and 49"
22. p. 61, 6th line from bottom, read "(Fig. 42)" for "(Fig. 44) ."
23. p. 66, last line of Art. 77, read " formula " for " formula."
24. p. 66, last line, read " Art. 49 for " Art. 32."
25. p. 66, next to last line, read "79" for "75" (number of Art.) .
26. p. 68, 3d line, read " 35) : " for " 35)."
27. p. 69, 6th line from bottom, read
28. p. 70, 4th line from bottom, read " i sin" for
29. p. 75, 7th line from bottom, read " sin b " for " sid 6."
30. p. 77, 2d line, read " cos 2 $A " for " cas 2 %A."
31. p. 78, 9th line from bottom, read "(Art. 53)" for "(Art. 42) ."
32. p. 86, 4th line, read " -45 " for " =45."
33. p. 87, 5th line from bottom, read " A, b, c or A, B, c " for
" AbcoTABc"
34. p. 88, 6th line, read " solutions " for " solution."
35. p. 90, 16th line, read " and </> is " for " and is."
36. p. 91, 6th line from bottom, read " JA 22 40' 12" " for
"A22 50' 12"."
CORRECTIONS TO BROWN'S TRIGONOMETRY. 3
37. p. 92, lower right-hand corner of page, read " tan 9.79928 "
for " tan 0.79928."
38. p. 94, 15th line, read "6 = 93 36' 2"" for & = 93 35' 2"."
39. p. 98, first two lines of Art. 115, read "This problem" for
" The solution of this problem."
40. p. 98, 3d line of Art. 115, read " (Art. 90) " for " (Art. 77) ."
41. p. 99, 10th line from bottom, read " cosec 0.06320 " for
" cosec 0.36320."
42. p. 100, 1st line of small capitals, read " ARTICLES " for
" CHAPTERS."
43. p. 103, 1st line, read " as " for " at."
44. p. 109, 2d line from bottom, read " point " for " line."
45. p. 115, 8th line, read " A 2 = 8 h 3 m 50 s E." for " A^etc."
46. p. 115, 5th line of Art. 132, read " objects " for " subjects."
47. p. 118, 10th line from bottom, read " PZM = Azimuth of M "
for "Plf^ = etc."
48. p. 123, 3d line, read " PZ = 9Q -L " for " PZ + 90 -L"
49. p. 124, last line, read " =70 " for " -70."
50. p. 128, 1st line, read " ZM " for " Zm"
PLANE TRIGONOMETRY.
CHAPTER I.
1. Trigonometry is the branch of mathematics which treats of
the methods of subjecting angles and triangles to numerical com-
putation.
Plane trigonometry treats of plane angles and triangles;
spherical trigonometry, of the angles and triangles formed by arcs
of great circles on the surface of a sphere.
2. Comparison of Geometric and Trigonometric Methods. By
the methods of plane geometry, if any three of the six parts of a
plane triangle are given, except the three angles, the triangle may
be constructed and the unknown parts measured, a process neces-
sarily inconvenient and inaccurate.
By the methods of trigonometry the unknown parts may be
found by numerical computation to any desired degree of accu-
racy; and this computation is called the solution of the triangle.
Before such a solution can be made we shall have to express
the parts in numbers by the adoption of suitable units of meas-
ure, and define certain functions of angles which depend upon
their magnitude.
3. Angles. Although the solution of plane triangles involves
the consideration of angles not greater than 180, many problems
of mechanics and higher mathematics require the consideration
of angles of unlimited magnitude.
To represent such an angle, we consider it to be generated by
the revolution of a line about a fixed point. Assume such a line
2
2 PLANE TRIGONOMETRY.
to start in coincidence with OA and to revolve into the position
OB, generating the angle AOB, in the direction of the arrow
(Fig. 1). The revolution may be continued, in the same direc-
tion, until the generating line comes again into coincidence with
OB, and so on indefinitely, the angle always having the same
boundary lines, but with each revolution greater by 4, 8 or any
number, 4n, of right angles. The side OA is called the initial,
and OB the terminal, side of the angle.
FIG. 1. FIG. 2.
Angles generated in this direction, to the left, or counter-
clockwise, are usually regarded in mathematical works as positive,
in distinction from angles formed by a revolution of the generat-
ing line in the opposite direction, to the right, or clockwise.
Thus the angle A OB, with the same initial and terminal sides, but
generated in the direction of the arrow (Fig. 2), is regarded as
negative. The positive direction of revolution is, however, a ques-
tion of convenience, as, for instance, in the determination of the
errors of the compass in ships, where angles are regarded as
positive when generated clockwise.
We shall always regard positive angles, unless otherwise stated,
as generated in the counter-clockwise direction.
4. Units of Measurement of Angles. Angles occurring in the
solution of triangles are expressed in the well-known sexagesimal
units of angular measure : degrees, minutes and seconds
PLANE TRIGONOMETRY. 3
the degree, or -fa of a right angle ;
the minute, or -fa of a degree ;
the second, or -fa of a minute
which are designated by the symbols 1, 1', 1", respectively.
5. Measure of Arcs. Since arcs of a circle .are proportional
to the angles they subtend at the center, a degree of arc is -g^-g-
of the circumference, and is subdivided into minutes and seconds
just as the degree of angle. The actual length of a degree of arc
depends upon the size of the circle, and is -^ r- = ^-^ in a circle
obU loU
of radius r.
6. The Circular Unit. For certain investigations of angles and
arcs a circular unit of measure is used. The circular unit of
arc is an arc of the same length as the radius ; the circular unit of
angle, called the radian, is the angle at the center subtended by
this arc.
In any circle the semicircumference, whose length is TT times
the radius, subtends an angle at the center TT times as great as a
radian; thus TT radians and 180 are the same, and the equation
connecting the two units is
TT radians =180.
In writing an angle in radians no symbol is used ; thus, the angle
J means an angle of a radian; the angle ~ means ^- or 1.5708
radians, the equivalent of 90.
7. Table for Converting Circular into Sexagesimal Units:
TT radians = 180
1 " = 57.29578+ Its log is 1.758123
1 " = 3437.7468' " 3.536274
1 " =206264.8" " 5.314425
4 PLANE TRIGONOMETRY.
For Converting Sexagesimal into Circular Units :
180 = 3.14159265 radians Its log is 0.497150
1 = 0.01745329 " " 8.241877-10
1' = 0.00029089 " " 6.463726 - 10
1" = 0.00000485 " " 4.685575-10
8. In numerical work the minute makes a convenient unit; and
an angle 6', expressed in minutes, is changed into circular units
by the formula
0=0.000290890',
or the circular value of 1' multiplied by the number of minutes
in the angle.
For the reverse process,
Example 1: Express 35 42' 30" in circular measure.
Angle 0' = 2142.5' log =3.33092
log 1' = 6.46373 -10
= 0.62323 log = 9.79465 - 10
Example 2: Express 0= 1.3786 in angular measure.
0= 1.3786 log =0.13944
log 1' = 6.46373 -10
0' = 4739.2' log =3.67571
0=78 59.2'
Examples.
1. Give the value in angular units of
7T 7T 2-JT 3-7T 3?T 9?T 17?T
T' T ? T~ : ~4~ .-8 J 4- J ~13~'
2. Change the following angles into circular measure :
15, 18, 75, 150, 240, 330, 420.
3. What is the angle at the center of a circle of 2 ft. radius
subtended by an arc 12 inches in length ? Also 28 inches ?
Ans. 28 38' 52V4; 66 50' 42C'4.
PLANE TRIGONOMETRY.
4. Find the length of V of arc on the earth's equator, assuming
3962 miles as the radius. This distance is the geographical mile;
what is its value in feet? Ans. 6085.2 ft.
5. Find the length in feet of 1" of arc on the earth's equator.
Ans. 101.4 ft.
FIG. 3.
9. Angular Units of the Mariner's Compass. Angles indicated
by the mariner's compass are frequently expressed in points, each
quadrant being divided into eight points. The value in degrees
of one point is therefore 11J, and further subdivisions are ex-
pressed in half- and quarter- points. The names of the points are
indicated in Fig. 3.
6 PLANE TRIGONOMETRY.
10. Designation of Angles. Angles are often designated by
the letters of the Greek alphabet,* but they may be expressed by
any other algebraic notation according to convenience. Their
magnitude may be expressed numerically in either unit, although
aliquot parts of the circumference are conveniently expressed in
circular measure, as 2-n- for 360, (IT 6} instead of (180 0),
7T
9 for 90
The complement of an angle 6 is expressed as either (90 6) or
X
The supplement of the same angle is (ISO 0) or (IT 6).
The angular space about the
vertex of an angle may be
divided into four quadrants of
1 90 by a line coinciding with
X the initial side of the angle,
and another line at right
angles, as in Fig. 4, the angle
XOY, from OX to OY, being
the first quadrant. Angles are
PIG. 4. said to be :
In the first quadrant, when less than 90.
In the second quadrant, when between 90 and 180
In the third quadrant, when between 180 and 270
In the fourth quadrant, when between 270 and 360
*For convenience the Greek Alphabet is given:
A a Alpha I i Iota P p Rho
B j8 Beta K K Kappa 2 a Sigma
T 7 Gamma A X Lambda T T Tau
A 5 Delta M p Mu T v Upsilon
E e Epsilon N v Nu $ Phi
Z f Zeta S Xi X x Chi
H tj Eta O o Oinicron ^ i// Psi
6 Theta II it Pi ft o Omega
PLANE TRIGONOMETRY. 7
It is important to note that this designation of angles refers
JJonly to their angular magnitude.
The modern navy compass, however, is graduated in degrees
from the North point through East, South and West from to
360. Positive angles in this system are thus generated right-
handed, and the angular magnitude increases through the four
quadrants in the opposite direction to those given in Fig. 4.
s.i
PLANE TRIGONOMETRY.
CHAPTER II.
THE TRIGONOMETRIC OR CIRCULAR FUNCTIONS.
11. Positive and Negative Lines. Let the vertex of an angle,
0, be taken as the origin of a system of rectangular coordinates;
X
y'
PIG. 7.
y
FIG. 8.
the initial side as the positive axis of X; and the terminal side of
the first quadrant as the positive axis of Y. Figs. 5, 6, 7 and 8
show the angle in each of the four quadrants.
PLANE TRIGONOMETRY.
9
The position of any point P of the terminal side of the angle
is determined by its perpendicular distance from the axes X'X
and YY' ; distances from X'X are taken as positive when P is
above X'X, negative when below ; distances from the axis YY' are
considered positive when P is to the right, negative when to the
left.
The line PM, measured parallel to YY', is the ^-coordinate of
P, or simply the ordinate; the line OM, or a line parallel to X'X
from P to YY', is the ^-coordinate, or the abscissa.
The terminal side of the angle, OP, is positive in whatever
quadrant, and PO produced back through the vertex is negative;
thus, the line OP' (Fig'. 5) is a negative line, relatively to OP.
12. Definitions of the Trigonometric or Circular Functions of
Angles. If from any point in one side of an angle a perpendicular
be drawn to the other side, produced if necessary, a right triangle
is formed, as in Fig. 5, 6, 7 and 8, in which PM, the side opposite
the angle, is the perpendicular, OM the base, and OP the hy-
pothenuse.
The six ratios which may be formed between the three sides of
this triangle are called the trigonometric ratios, or the trigo-
nometric or circular functions, of the angle, and are denned as
follows :
The sine of
The cosine of 6
hypothenuse
abbreviated sin 9=
hypothenuse '
The tangent of = Pedicular , tan 0=
The cosecant of 6 =
jr r-, " *cosec0=^.
perpendicular y
* $ome mathematicians and navigators prefer to abbreviate " cose-
cant " as " esc," thus making formulas symmetrical, all functions
having three letters.
10 PLANE TRIGONOMETRY.
The secant of = h yP then ^ se abbreviated, sec 0= -^
base x
The cotangent of 9 = - , , " cot = - .
perpendicular y
The versine of 0= 1 cos 0, abbreviated vers 0= - - , is fre-
quently used as one of the functions.
It is important to remember that the three principal functions
are the sine, cosine and tangent, and that the cosecant, secant and
cotangent are the reciprocals of the first three, respectively.
13. These functions are always the same for the same angle;
for if the perpendicular be drawn from any other point, P 1} or P 1
in the side produced, the right triangles thus formed are similar,
and the ratios of the homologous sides are the same, not only in
numerical magnitude, but also in sign. (Note that all the lines
in P'OM' (Fig. 5) are negative, in accordance with the assump-
tions of Art. 11.)
The ratios are in general different for different angles ; for two
right triangles in which the acute angles of the one are not equal
to the acute angles of the other are not similar, and the ratios
of their homologous sides are not equal.
The ratios thus depend on the magnitude of the angle, and upon
this alone, and are therefore functions of the angle.
14. Signs and Numerical Value of the Trigonometric Func-
tions for Different Quadrants. Since the terminal side, r, of the
angle is always positive, the sign of the trigonometric function
will depend only on the sign of x or y, or both x and y, in the
triangle of reference (Figs. 5, 6, 7 and 8).
Thus the sine is positive only where y is positive, in the first and
second quadrants ; the cosine is positive where only x is positive, in
the first and fourth quadrants ; the tangent will be positive where
both x and y have like signs, in the first and third quadrants.
PLANE TRIGONOMETRY.
11
From these simple relations, the following table is easily made,
showing the signs of the functions for the four quadrants :
Quadrant I
II
Ill
IV
Sine and cosecant
+
Cosine and secant
+
+
Tangent and cotangent
+
+
Since the perpendicular and base are never greater than the
hypothenuse of the right triangle, neither the sine nor the cosine
of an angle is ever greater than unity; the cosecant and the secant
are therefore never less than unity. The limits of the numerical
values of all the functions are given in the following table. They
may be readily found by following a point P in the terminal side
as it revolves around as a center, and noting the signs of the
coordinates of P, and the value of the ratio.
6
..
.90
90...
.180
180.
...270
270.
...360
Sine
+0 ..
+ 1
+ 1 ..
.+0
.. .1
1 .
...
Cosine
+ 1 ..
.+0
..
. 1
1
...0
+0 .
...+1
Tangent
+0 ..
+
00 . .
.0
+0
...+*
GO .
...
Cosecant
+..
+ 1
+1 ..
.+00
GO
. ... 1
1 .
. . . GO
Secant
+1 ..
.+
QC . .
.1
1
. . . QO
+00.
...+1
Cotangent
+ 05..
.+0
. .
. 00
+ 00
....+0
.
. . . GO
15. Complementary Functions. Let ABC (Fig. 9) be a right
triangle, with the right angle at C. Since
Z? = 90 ,4, we may deduce from this rela-
tion, and the definitions of the trigonometric
ratios, the functions of the complement of an
angle. We have :
PIG. 9.
12 PLANE TRIGONOMETRY.
= :=cos=cos(90 -A),
c
tsMA = ~-
cosecA = = sec B= sec (90 -A).
a
On account of this relation, the sine and cosine, tangent and
cotangent, and secant and cosecant are called complementary
functions.
16. Fundamental Formulae. In any of the right triangles of
Figs. 5, 6, 7 and 8 we have always
r 2 r 2
From the definition of the sine and cosine, this becomes
sin 2 + cos 2 (9=1. (1)
If we divide the same equation, x 2 + y 2 = r 2 , by x 2 , we obtain
+ 'x 2 ~~ 'x 2 '
or
l + tan 2 0=sec 2 0. (2)
If we divide by y 2 , we obtain
l + cot 2 0=cosec 2 0. (3)
These equations are graphically represented in the following
figures, the designation of the sides of the right triangle being so
made that the ratio involved in the definition gives at once the
function ; thus, sin 6 S11 ^ .
PLANE TRIGONOMETRY.
13
sin
tan
j
FIG. 10.
cor
From the definitions of the functions we have the following
important equations:
111
cosec 6 = -. - ,
sin 6
Since we always have
cot 6 =
tan 6 '
sin 6 _ y _._ x_ _ y
cos 6 r r "" a;
tan =
sin
cos 6
(4)
17. The values of all the functions may be expressed in terms
of any one function by means of the formulae of Art. 16. For in-
stance,
sin 0=sin 9,
cos0= Yl sin 2 0,
sin sin
tan0=
cos 6 VI -sin 2 (9
cos0~ Vl-sin 2
, etc.
The same result may always be more readily obtained by mak-
ing the given function the appropriate side of a right triangle.
Thus, for the preceding example, the right triangle is given in
Fig. 11, where by definition sin 0= - . The value of any func-
tion in terms of sin 6 is then derived directly from its definition.
14
PLANE TRIGONOMETRY.
FIG. 11.
4
FIG. 12.
If the numerical value of one function is given, the graphic
method furnishes the simplest means of finding the numerical
value of all the other functions. As an example, if the given
function is tan 0=}, the right triangle whose perpendicular is 3
and whose base is 4 (Fig. 12) gives:
tan 0= J, sin 0=f, cos 0=f, etc.
18. Examples.
1. Find the remaining functions from the following data (Arts.
16-17) :
(a) sina= T 5 7 . (b) seca=-i-J-.
(d)
(b)
(e)
m* n'
(c)
(f)
COS a =
COSa=
2. Is the equation sec 2 B
a possible equation ?
from the
3. Derive the values of all the functions of 45
isosceles right triangle whose two equal sides are a.
Ans'. sin 45 = -i=- , tan 45 = 1.
4. Derive all the functions of 30 and 60 by dividing the
equilateral triangle, of side a, into two equal right triangles.
Ans. sin 30 =4, cos 30=iV3, tan 30=\/3.
5. Prove the following identities :
(a) (tan0 + cot0)sin0cos0=l.
(b) (sin0+cos0) 2 -f (sin 0-cos0) 2 = 2.
(e) (tan 2 0-sin 2 0)=tan 2 0sin 2 0.
(d) sec 2 0+csc 2 0-sec 2 esc 2 0.
r l sin _ 1 sin _ cos
COS0
/
V
PLANE TRIGONOMETRY.
15
6. From the relations of complementary functions find a value
of the angle involved in the following equations :
(a) sina=cos2a (note that sinea = sin (90 2a), and thus
a = 90-2a, a = 30.
(b) tan2a=cot3a.
(c) CSC I a = SCO a.
(d) sin 60 = cos 39.
(e) sin(0-0)=J, cos(0+<j&)=0.
19. Trigonometric Functions Represented in Numerical Value
by the Length of Lines. Since the numerical values of the trigo-
nometric ratios are independent of the length of the terminal side,
FIG. 14.
we may suppose the point P to generate in its revolution a circle
of radius unity, as in Fig. 13. Then, from the figure, x and y are
always \ess than unity ; and
~
cos 0= T ,
or
i '
and the lengths of these lines represent graphically the numerical
values of the functions in terms of r=l. And since the angle
POY=9Q -6, by construction
= cot 6, and OT' = cosec 6.
16
PLANE TRIGONOMETRY.
When the angle is in the second quadrant (Fig. 14), the termi-
nal side, produced, meets the tangent at A, produced, in T and the
values of the secant and tangent are negative ; i. e.,
sec B- OT, tan 6= AT;
The construction of the functions for the other two quadrants
is readily made (Figs. 15 and 16).
PIG. 15.
3d Quadrant.
AT =tan 0, positive.
OT = sec 0, negative.
Fr'=cot0, positive.
= cosec 6, negative.
20. Reduction of the Functions of Any Angle to the First
Quadrant. The functions of any angle may be expressed in terms
of functions of an angle in the first quadrant. The necessary
equations are easily derived by means of the coordinates of a point
in the terminal sidefc of the angle, as used in the general definition
of the functions.
FIG. 16.
4th Quadrant.
AT = tan 0, negative.
OT =sec 6, positive.
YT' = cot 0, negative.
6, negative.
PLANE TRIGONOMETRY.
17
21. Functions of (90 + <9). Let
X0P=90 + be an angle in the
second quadrant, and draw OP' per-
pendicular to OP, thus making the
angle XOP' = B (Fig. 17). The tri-
angles P'OM' and POM are equal,
since the angle POX' = 90 -0; and
hence y = x' and x= y'.
We have, from the figure :
sin (90
cosec(90
COS i
= cos 0,
sec (90 + 0) = -^-= --p
"
tan(90 + 0) = = ^
- sec0;
= sin
= cosec ;
= cot
cot(90+0)=- = F =
i/ a;
tan0.
Expressed in words, these formulae state that the function of an
angle in the second quadrant is the complementary function of its
excess above 90 , with the proper sign of the desired function for
the second quadrant.
22. Functions of (180 -0). If X'OP (Fig. 18) is laid off
equal to 0, an angle XOP' in the first quadrant, then XOP= 180
6 ; hence, from the figure :
sin(180-0) = sin0, cosec ( 180 -B)= cosec 6;
cos(180 -6) = - cos 0, sec(180-0) = -sec 0;
tan(180-0) = -tan 0, cot(180-0) = -cot 0.
3
18
PLANE TRIGONOMETRY.
P'
FIG. 19.
23. Functions of (180 +0). Let XOP be an angle (Fig.
19), and XOP '= (180 +0) ; then by definition :
<9
tan (180 -Ml) = tanfl.
The value and signs of the reciprocal functions are obvious.
24. Functions of (270db0). Let TOP and TOP (Fig. 20)
each equal an angle in the first quadrant ; then
PLANE TRIGONOMETRY. 19
whence, by definition :
sin (270 -0)= sin (270 +0)= -- = -^ = - cos 0,
cos (270 -0) = - cos (270 + 0) = ^ = -^ = -sin 0,
7/ T
tan (270 0) = tan (270 + 0) = -*?-= = +cot 0,
x y
the other functions being obvious.
25. Functions of (360 -0) or (-0). In Fig. 21 let XOP
and XOP' be equal angles, 0, measured from the initial side, OX,
in opposite directions. Then XOP' may be regarded either as a
negative angle ( 0) or as an angle in the ' fourth quadrant
(360 0), and we have, in either case,
sin(-0) = sin(360-0) = -sin 0,
cos( -0) = cos(360 -0) = +cos 0,
tan( -0) =tan(360 -0) = -tan 0.
26. All the relations of functions of angles in the various quad-
rants to the functions of an angle in the first quadrant may be
summarized in the two formulae
f(n~6) =co-f(0), when n is odd;
and
f(n r 6) = f(0), when n is even,
to
where /(0) is any desired function of 0, and co-/(0) the comple-
mentary function of 0, with the appropriate sign of the function
for the quadrant of (^- 0) .
20 PLANE TRIGONOMETRY.
27. Periodicity of the Trigonometric Functions. We have
seen, in Art. 3, that the initial and terminal sides of a given angle
may indicate an angle 0, or the same angle plus one or more revo-
lutions of the terminal side through 360, or 2?r. Hence, from
the definitions of the functions of angles, it is evident that they
remain the same when the angle is increased or diminished by any
multiple of 360, or, expressed as an equation,
sin ( 2/iTT + 0) = sin 0, cos ( 2mr + 0)=cos6,
and, generally,
f (%**+$) =f (6),
where / denotes a circular function.
Exercises.
1. Make a table of the value and sign of all the functions of
0, 30, 60, 90, 120, etc., to 360.
2. Make a table of the functions of 0, 45, 90, 135, etc., to
360.
28. Inverse Functions. In the equation z = sin 6, x is said to
be an explicit function of 0. To express 6 as an explicit function
of x, the equation is written
which is read " 6 is the angle or arc whose sine is x " and sin' 1 x
is called an inverse function of x.
To express 9 by any other inverse function most conveniently,
we may graphically represent 9 and its given function as parts of
a right triangle, as in Fig. 22. From the parts
of the right triangle,
PLANE TRIGONOMETRY. 21
29. Angles Corresponding to Given Functions. We have seen
that for a given angle there is but one value to each of its func-
tions ; but if we have to find the angle corresponding to a given
function, there is an indefinite number of angles which have the
same function.
If we have, corresponding to a given sine, ~b, an angle (3, then we
have, also (from Art. 22), TT , and (from Art. 27) an indefinite
number of angles formed by adding multiples of 2-n- to these; i. e.,
A 27T+/J, 47T+ p, etc.;
TT-/J, 3*& 57r-/?,etc.,
which may be reduced to the formula
-l) n p.
Corresponding to a given cosine, b, if ft be one value, then
will also be another, since cos (3 = cos ( /?), and, generally,
- 1
cos- = nir
Corresponding to a given tangent, 6, if /? be one value, then
TT+/J is another, since tan /?=:tan(7r+/?), and, generally,
tan- 1 & =
Equations frequently occur which are expressed in terms of the
functions of an angle. They are solved by the elementary proc-
esses of algebra, after first transforming the equation into one
which contains only a single trigonometric function.
As an illustration, let it be required to solve the equation sin x
= tan 2 x. u
Transforming to terms.sin x, it becomes
DUA .*/ _ ; ^ ,
cos 2 a; 1 sm 2 x
which reduces to
sina;(l sin a; sin 2 x) = 0.
22 PLANE TRIGONOMETRY.
The factor sin x=Q gives
X=UTT(\. e., 0, TT, 2^r, etc.).
The other factor,
sin 2 a;+sin a; 1 = 0,
solved as a quadratic in terms of sin x as the unknown, gives
sina; = ~ l v 5 =0.61803 or -1.61803.
2
Since sin x cannot be numerically greater than unity, the second
value is obviously impossible, although it satisfies the original
equation.
30. Examples.
1. Find all the angles less than 2-ir corresponding to the fol-
lowing functions :
sin- 1 ^ tan^VS, co^TTp cot'^-l), sec-^-D.
2. What is the sin (cos- 1 4) ? tan (cot' 1 x) ?
3. Solve the following equations :
(a) 2 cos 0= sec 0. Ans. ^TT -j-.
(b) 4 sin ^=3 esc (9. Ans. n7r47r.
(c) 3 tana; cot a: =0.
Ans. n 7 r~ > (30, 150, 210, 330, etc.) .
(d) tan0+cot0=2.
(e) 3 tan 2 0-4 sin 2 0=1. Ans. sin 0= 7 ~.
V-4
(f) 2sin0-tan0=0.
PLANE TRIGONOMETRY. 23
CHAPTER III.
TRIGONOMETRIC TABLES.
31. Tables of Natural Sines, Cosines and Tangents. The
numerical values of sines, cosines and tangents, usually called the
natural functions, have been computed and collected in many
different tables, distinguished by names indicating the degree of
approximation to which the values are given ; thus there are four-,
five-, six- or seven-place tables, giving the values of the functions
correct to that number of decimal places.
In Bow ditch's Useful Tables, Table 41 gives only the natural
sines and cosines to five places, for each minute from to 90.
The functions of intermediate values may be found by inter-
polating for the seconds, on the assumption that the algebraic
increase of the function is proportional to the increase in the
angle. Thus, for example, if we wish the sine of 30 2 6' 35" the
sine of 30 26' is 0.50654, and the tabular difference for one
minute between 26' and 27', is .00025, and for 35", ff of .00025
or .00015 and the interpolated value is 0.50669.
Auxiliary tables of proportional parts are given in the first and
last column of each page, marked " Prop, parts 25 " in the ex-
ample cited above, the number 25 being the tabular difference for
1'. The interpolation for the 35" is found in this column opposite
the 35 in the column of minutes immediately adjoining.
The cosine of the same angle 30 26' is 0.86222, the tabular
difference for 1' is 15, and the interpolation for 35" is f- X 15
= 9; the interpolated value is thus 0.86213, since the cosine is
a decreasing function.
The auxiliary table for this interpolation is found in the last
column, marked " Prop, parts 16," and the proportional part for
24 PLANE TRIGONOMETRY.
35", opposite the 35 in the column of minutes immediately ad-
joining, is 9.
The columns of minutes, marked " M," thus serve the double
purpose of giving the minutes of the tabulated angle and also in-
dicating the seconds for which the value of the "proportional
part " is given in the same line.
The tabulated " proportional part " for a given page is made on
the assumption that the tabular difference for 1' is constant for the
page ; while the actual difference for this page varies from 26 to
23. When greater accuracy is desired, the actual computation is
easily made.
Since every angle between and 45 is the complement of
another angle between 45 and 90, every function of an angle
less than 45 is the complementary function of another angle
greater than 45. Therefore it is necessary to extend the table
only to 45, the degrees at the top of the page extending from to
45, those at the bottom from 45 to 90. Each number, there-
fore, in each column, stands for the function named at the top of
the column and the complemental function at the bottom of the
same column. Thus, in looking for a function of an angle less
than 45, the degrees are found at the top of the page, and the
minutes in the left-hand column, marked " M " ; for an angle
greater than 45, the degrees are found at the bottom, and the
minutes in the right-hand column, marked " M."
32. Tables of Logarithmic Functions. The logarithms of the
trigonometric functions, which are of far greater use in computa-
tions, are tabulated on the same general plan as the natural func-
tions ; and the tables are likewise distinguished by the number of
decimal places to which the mantissas are carried.
Table 44, of Bowditch's Useful Tables, gives the logarithms of
all six functions to five places. Since the sine and cosine of all
angles and the tangents of angles less than 45 are less than unity,
PLANE TRIGONOMETRY. 25
their logarithms are negative. To avoid the use of such loga-
rithms, the tabular logarithms of all the functions of Table 44 are
increased by 10, which must be subtracted from the tabular loga-
rithm to give the correct logarithm for use in computing. For
instance, the tabular log sin 30 is 9.69897, from which 10 must
be subtracted, making the correct log sin 30, 9.69897 10.
The arrangement of the degrees from to 45, and the corre-
sponding log function, at the top of the page, and from 45 to 90
at the bottom, is the same as for Table 41 ; but in addition there
is also given, at the opposite end of each column of minutes, an
angle 90 greater.
Thus the log functions of any angle from to 180 may be
taken directly from Table 44, without resorting to the use of
complemental functions, as explained in Arts. 21 to 28. An in-
spection of the table shows that in all cases the minutes of a given
angle are found' in the " M " column immediately below the
degrees, when these are at the top, and immediately above, when
at the bottom, of the page.
33. Tables of Proportional Parts, Table 44. For angles inter-
mediate between the minutes of the table, the interpolation for the
given seconds may be made on the usual assumption, that the
change in the logarithm is directly proportional to the change in
the angle.
For the five degrees at each end of the first and second quad-
rants, the tabular difference for 1' is given for each function in
the column marked " Diff. 1' " ; and for angles found on these
pages the interpolation for the seconds must be computed.
It should be noted, however, since the sine and cosecant, tangent
and cotangent, cosine and secant, are respectively reciprocal func-
tions, that the log of one of these pairs is the colog of the other,
and their tabular difference is the same numerically, but opposite
in sign.
26 PLANE TRIGONOMETRY.
For angles outside of these five-degree limits, the proportional
part for each second of angle is contained in the column marked
" Diff./' and the corresponding seconds are always found in the
left-hand column of minutes, no matter where the degrees of the
angle may be found.
These auxiliary tables for interpolation depend upon the as-
sumption that the tabular difference for 1' for a given function
is the same for the whole column. For angles near 5 or 85, this
assumption leads to an error which may be as great as 12 units in
the last figures of the logarithm. It is therefore advisable to com-
pute the interpolation, when the tabular differences are large.
As an example of this error, the log sine of 5 59' 50" is found
below :
5 59' log sin = 9.01803-10
Tab. Diff. for 50" = + 110
5 59' Jb" log sin = 9.01913-10
The tabular difference for 1' between 59' and 60' is 120, and the
proportional part for 50" is 120 xf = 100; and the more nearly
correct logarithm is 9.09103 - 10.
If the interpolation is made backwards, however, for 10" from
5 60', the log sine is 9.01901 10, a much closer approximation
than the first.
Example 1 : Find the log sine and log cosine of 28 59' 24".
28 59' sin 9.68534 10 cos 9.9418910
" Diff." 2T, "A" + 9 Diff." C " 3
28 59' 24" sin 9.6854310 cos 9.9418610
The tabular difference for 1' at 59' is 23, and the interpolation
for 24" is 0.4x23 = 9.2, or, to the nearest fifth place, 9, which is
found in the column marked "Diff.," opposite 24 in the first
column.
PLANE TRIGONOMETRY. 27
Example 2: Find the log tangent and log cosecant of 59 54'
48".
At the bottom of the page for 59 (opposite 54' in the right-
hand column), is found
59 54' tan 0.23681 cosec 0.06291
"Diff." 48", "B" + 23 "Diff.""C" 6
59 54' 48" tan 0.23704 cosec 0.06285
The interpolation for 48" is taken from the column marked
" Diff.," between the tangent and cotangent and opposite 48" in
the left-hand column.
Example 3: Find the log cosine and log tangent of 111 25'
35". The degrees 111 are found at the bottom of the page under
the first column, and the minutes must be taken in the same
column ; thus :
111 25' cos 9.56247 10n tan 0.40646w
"Diff." 35", "A" + 19 Diff.""B" 22
111 25' 35" cos 9.56266 Wn tan 0.40624n
Note that the minutes increase from the top of the page, and
the direction in which the interpolation is to be applied is found
by noting whether the log function for the next higher minute
is greater or less. As both the required functions in the second
quadrant are negative, this is indicated by the " n " affixed to the
logarithms.
34. The Angle Corresponding to a Given log Function.
Except for the sine and cosecant, the sign of the function indi-
cates whether the corresponding angle is in the first or second
quadrant. If the given function is found in the tables, take out
the degree and minute corresponding to it. Otherwise, find the
two tabular functions between which the given function lies, and
note the difference between the given function and the tabular
function corresponding to the smaller minute; find the number of
28 PLANE TRIGONOMETRY.
seconds in the column " M " on tine left side of page, corresponding
to this difference, and add to the degrees and minutes already
written down. When greater accuracy is required, the seconds
may be computed from the proportion which the difference bears
to the tabular difference for 1' (or 60").
If the function is the sine or cosecant, there are always two
corresponding angles, one in the first quadrant, and the other its
supplement.
Example 1 : Find the angle whose log cosine is 9.56843 10.
The given log cosine lies between
9.56854, corresponding to 68 16',
and
9.56822, corresponding to 68 17';
and the difference between it and the tabular function for the
smaller minute (68 Iff) is 11, corresponding to which, in the
first column, " M," is 20" or 21".
The difference for 11 corresponds'to ^ x60 = 21"; and the
angle is 68 16' 21".
Example 2 : Find the angle whose log tangent is n9. 62456 10.
The angle is in the second quadrant ; and the given log tangent
lies between
9.62468-10, for 157 9',
and
9.62433-10, for 157 10'.
Corresponding to the difference 12, in " B," we find 20" in the
left-hand column, " M." The angle is 157 9' 20".
35. Functions of Small Angles. Let 6 (Fig. 23) be an acute
angle, expressed in circular units, and ABC a circular arc of
radius r, the point C determined by producing the perpendicular
AM to C. The angle AOC then equals 20. Draw the two tangents
AT and CT.
PLANE TRIGONOMETRY.
or
or
FIG. 23.
From plane geometry,
chord AC<a,TcABC<AT+CT,
2rsm0<2rO<2rtsin0,
sin0<0<tan0,
or
sm 6
(2)
Now let the angle decrease indefinitely, approaching zero as a
limit ; sec 6 approaches unity, and when 0= 0, sec 0= 1.
n
Hence the limit of - z , which lies between unity and sec 6,
sin 6'
is unity, or
= 1.
0=0
30 PLANE TRIGONOMETRY.
Also
tan sin /]
TT- X sec =
When is a small angle, we may then write the equations
sin 0=0, tan 0=0
as approximate formulae, the use of which will depend upon the
degree of accuracy required in a given problem.
As an illustration,
log sin 1 = 8.24186 -10,
log ^=8.24188 -10.
The tabular difference for 1' for 1 is 717 ; hence the use of log
for log sin would cause an error in an angle of 60' of T f T X 60"
=". For 2,
log sin = 8.54282 -10,
log ^=8.54291 -10.
This error is -gfa x 60" = 1.5".
The error involved will be shown more clearly in the accurate
expression of sin in terms of 0.
In practice with five-place tables, the formula
sin0=0 / sinl /
gives the last place of five-figure logarithms correctly to 23' ;
tan 0=0' tan 1',
to only 16'.
Example: Find the log sin 21' 21".
21.35' log 1.32940
1' log sin 6.46373-10
21.35' log sin 7.79313-10
PLANE TRIGONOMETRY. 31
36. Graphs of Trigonometric Functions. If x be considered
as a variable angle and y the corresponding trigonometric func-
tion, the curves of the trigonometric functions may be plotted or
traced in the same general way as the curves of algebraic functions.
In order that the linear or numerical value of the function in
terms of r=l (Art. 19) may be represented by the graph, the
unit of values must be the same for the angle and the function ;
hence the angle must be expressed in circular measure in terms
of 7r=3.1416, while the function corresponding to the given angle
is expressed in terms of the radius unity.
The angle increases indefinitely, while the functions repeat
themselves in the same order from every multiple of TT or 360 ;
it follows that all such curves, like the functions, will be periodic.
As we desire to show principally the limiting values of the func-
tions, we shall trace the curves instead of plotting them.
In the curves of sines, the sinusoid,
y = sinx, (1)
the limiting value of y is 1 for -^- , and 1 for ~ , repeating these
values in succession for every addition of 2ir to these angles.
The curve crosses the axis of x f i. e., y = Q, when x = Q, TT/^TT, 3ir
and so on indefinitely for every multiple of TT.
To find the direction of the curve for these points, we have,
from (1) and Art. 35,
y _ sin x ~] _ i
x x J x=0 ~
To find the direction of the curve at XTT, when y = 0, we have
to resort to the equation
y=sin(7r+x) = sin x,
or
y_ _ sin (IT + a;) _ ^ sinafl _ _ ^
x x x \ x=0 ~
32
PLANE TRIGONOMETRY.
Hence the curve cuts the axis of x at the angles of 45 and 135,
for x and X = TT respectively, and is traced in Fig. 24.
2.TT
FIG. 24.
The curve y = cosx, since
cos #=sin f~ x\,
is a succession of similar curves, except that the phase is advanced
90, or^- (i. e., for z-0, y = l, etc.).
/v
The curve y = tan x crosses the axis of x at the origin, at an
angle of 45, since
y_ _ tan afl _ -,
~z~ " * J ^=o ~
When a; approaches -^- , or 90, since tan $=s the tangent,
2 cos a;
or #, approaches oo as a limit ; when x passes through ^ , the cosine
becomes negative and the ratio approaches oo as a limit. Hence
the value of the ordinate y changes from + oo to oo as x passes
through the value -^- into the second quadrant.
PLANE TRIGONOMETRY.
FIG. 25.
Since tan (TT + X) =tan x, the direction of the curve in passing
through TT becomes
JL
x ~
tana
_ -
and the curve consists of a series of similar curves always crossing
the axis of x at an angle of 45, for every value of x which is a
multiple of ?r.
The curve i sec x
cosx
: The limiting values of y are +1
for every value 2mr of x, 1 for every value (2n 1) ?r of x; and
for every odd multiple of -^- the value of y changes from + oo to
- oo, or the reverse, as x passes through the value (2n 1) - .
34
PLANE TRIGONOMETRY.
Y i
PIG. 26.
The logarithmic curve y = logo x t on account of its showing the
limits of the logarithms of the trigonometric function, is given
below.
[o=10].
y= when x 1
y y=~l " x=.l
y=-2 " z=.01
y= oo " x
y= 1
O
x = a
Y'
FIG. 27.
PLANE TRIGONOMETRY.
35
a
CHAPTEE IV.
SOLUTION OF PLANE RIGHT TRIANGLES POLAR COORDINATES.
37. The tabulated values of the trigonometric functions and
their logarithms enable us, from their definitions alone, to effect
the numerical solution of right triangles. Omitting the right
angle, if two of the remaining five parts of the triangle are given,
one of them being a side, each of the other parts may be found
from a simple equation which expresses the D
definition of the trigonometric function in-
volved, in terms of the two given parts and
the required part.
Let ABC (Fig. 28) be a plane right tri-
angle, C the right angle, and a, &, and c the
sides opposite the angles A, B and C, re-
spectively. The equations necessary for the ** 6 ^
solution, depending on the parts given, are : FIG. 28.
a= c sin A = c cos B= I tan A I cotan B= Vc 2 6 2 ,
b = c sin B= c cos A a tan B= a cotanf^ Vc 2 a 2 ,
c = a cosec A = a sec B = I sec A = b cosec B
It is not necessary to memorize the formulae, as they are readily
derived for any given case from the definitions of the trigo-
nometric ratios.
It is always advisable to make a rough sketch of the triangle,
indicating the given parts, and also a form for the entire computa-
tion, before taking from the tables any of the desired functions.
Example: Given A = 35 16' 25", c=672.34.
36 PLANE TRIGONOMETRY.
The formulae are :
B=9QA, a=csinA, b = ccosA.
The solution is :
A = 35 16' 25" sin 9.76153-10 cos 9.91190-10
c = 672.34 log 2.82759 log 2.82759
a = 388.25 log 2.58912
5 = 548.90 log 2.73949
When there is no liability of confusing the natural and loga-
rithmic functions, the abbreviations log sin, log cos, etc., are
usually still further abbreviated into sin, cos, etc., as in the above
example.
38. Examples.
1. A = 44 10' 38", c = 24.896.
Ans. a= 17.350, 6 = 17.854.
2. a= 396.25, 6 = 645.32. Ans. 1 = 31 33' 4", c = 757.25.
3. c = 98.245, a=95.573.
Ans. 5=13 23' 38", 6 = 22.757.
4. 5 = 27 9' 14", a = 35.421. Ans. 6 = 18.168, c = 39.81.
39. The examples given below involve the solution of right tri-
angles, the known parts of which are found by the measurement of
horizontal distances or vertical heights, and vertical or horizontal
angles.
A vertical line is a line determined by a plumb-bob.
A vertical plane is a plane determined by a vertical line.
A horizontal plane is a plane perpendicular to a vertical line,
and for small distances it may be regarded as coinciding with a
body of water or other liquid at rest.
An angle of elevation is the vertical angle of an object above a
horizontal plane.
An angle of depression is the vertical angle below a horizontal
plane.
PLANE TRIGONOMETRY.
37
40. Examples.
(To be worked without logarithms.)
1. At a distance of 100 feet from the foot of a flag-staff, the
angle of elevation of its top is 30; what is the height of the
staff? Ans. 57.735 ft.
2. The angle of elevation of the top of a flag-staff is found at
a certain point to be 60 ; 150 feet further away, in the same
vertical plane, the angle is found to be 30 ; find its height above
the horizontal plane. Ans. 129.9 ft.
3. The angles of elevation of the top and bottom of a flag-
staff 50 feet high, standing on the top of a building, from a point
in the plane are found to be 60 and 45 respectively; what are
the height of the building and the distance of the point of ob-
servation? Ans' 68.3 ft.
4. The angle of elevation of a balloon bearing N.W. from A is
45 ; from B its angle of elevation is the same, and its bearing
N.E.; A and B are one mile apart on a line running E. and W.
Find the height of the balloon. Ans. 3733 ft.
5. Later the balloon is due north of A, and its elevation is 60 ;
at the same instant its elevation from B is 45 ; what is its height?
6. A ship steaming N.E. by E. observes a lighthouse bearing
N". by E. ; after steaming 18 miles, the lighthouse bears N.W. by
N". ; what is the distance of the light at the time of the last bearing ?
7. A ship steaming S.W. observes a lighthouse bearing W.S.W.,
and after steaming 16 miles observes the light bearing W. How
much further will she steam, on the same course, to pass the light-
house at the least distance ?
41. Projections. The orthogonal projection, P, of a point A,
upon a given line OX, is determined by the perpendicular AP
38
PLANE TRIGONOMETRY.
P(r0>
o
upon the line OX. The projection of the line AB upon the given
line is obviously PQ; its projection upon a line OF at right angles
to OX is RS. The projections of a given line of length r upon
two axes at right angles are usually called the horizontal and
vertical projections of the line, and from the right triangle in-
volved in the construction we have at once, calling these pro-
jections x and y respectively,
xr cos0,
y r sin 0,
where 6 is the angle of the line with OX
42. Polar Coordinates. Besides the familiar method of fixing
the position of a point in a plane
by two rectangular coordinates, we
may employ the method of polar
coordinates in which the position
of the point is determined by its
distance r=OP from a fixed point
in the line OA, with which OP
makes the angle 6, as shown in
Fig. 30. The point is called the pole or origin, and the line OA
the initial line; the distance r is the radius vector and 6 the
vectorial angle of the point P, designated by the symbol (r, 0).
The angle is measured counter-clockwise if positive ; clockwise if
negative ; the distance r is measured along the terminal side, of
the angle if r is positive, in the opposite direction if r is negative.
43. Transformation of Coordinates. The relation between
polar and rectangular coordinates is
simple where they have the same origin
and the initial line coincides with the
re-axis, as shown in Fig. 31. The fol-
x lowing equations connecting the two
systems are readily derived from the
FIG. 31. figure:
P(-r,0)
FIG. 30.
PLANE TRIGONOMETRY. 39
x=r cos 0,~]
In the following examples, where we have given x and y to find r
and B y we may regard the process as the solution of two simul-
taneous trigonometric equations
rcos x,
rsin Q,
or as finding the polar coordinates of a given point from its
rectangular, an extension of the ideas involved in the solution of
a right triangle to angles of any magnitude.
Dividing the given equations, we have
from which we determine 0, and thence get
r=x sec Qy cosec 6.
Example: Given x= 17.366, y 20.462, find r and 0.
check
# = 29.462 log 1.46926 log 1.46926.
x- 17.366 log 1.23970 log 1.23970.
0= 59 29.0' I tan 0.22956 I sec 0.29432 I esc 0.06475.
r=34.199 log 1.53402 log 1.53401.
Examples.
Find the polar or spherical coordinates of each of the fol-
lowing points, giving the logarithmic solution and check :
(a) (3, 4), (b) (-2.67, 3.48), (c) (2.72, -6.76), (d)
(_3.478, -7.286), (e) (*V3, 1), (f) (-2.7183, -.36777),
(g) (6.765,0), (h) (0, -7.2143).
40 PLANE TRIGONOMETRY.
44. Equations of Curves in Polar Coordinates. The equation
of a curve in polar coordinates may be
found by expressing its geometric defini-
tion in polar coordinates. For instance,
if the pole is on the circumfer-
ence, and the initial line is the diam-
eter of the circle through 0, as in Fig.
32, we have at once, since APO is a
right angle,
32. r=2acos0
as the polar equation of the circle with the given origin and initial
line.
The polar equation thus found is transformed into rectangular
coordinates by substituting the values of r and cos 6 given above,
W~2_|_ 2_Q ^_ _ 2flff
which reduces to
x 2 + y 2 2ax.
45. Tracing or Plotting a Curve in Polar Coordinates. When
the equation of a curve is given in polar coordinates, we may either
plot a sufficient number of points, by computing the value of r for
the corresponding angle and connecting the points with a smooth
curve, or we may trace the curve, as in rectangular coordinates,
by finding its limits and critical points and sketching in the curve.
The following examples illustrate both methods.
Plot the locus of the equation
r = 2a cos
by computing the value of r for every 15.
PLANE TRIGONOMETRY.
41
Taking a equal to one inch, the values of r are found from a
table of natural trigonometric functions as follows :
15 30 45
r 2 1.93 1.73 1.41
60 75
1.00 0.52
90'
N"ote^f that the values of r from 90 to 180 are numerically the
same, but opposite in sign, as from 90 to in reverse order. The
FIG. 33.
first half of the curve is thus plotted in Fig. 33, and the smooth
curve drawn in.
Trace the curve whose polar equation is
r=acos20.
The limits of r are seen to be a for the angle and 180, and
-a for 90 and 27Q. The curve passes through the pole (r=0)
for values of 45, 1?5, 225, and 315, making the same angles
in each case with the initial line as the radius vector.
PLANE TRIGONOMETRY.
FIG. 34.
The form at the pole is first drawn ; then starting from A the
curve is sketched in through the pole to B f from B through the
pole to G, from C through the pole to D, from D through the pole
back to A.
46. Examples.
1. Trace the curves given by the following polar equations :
(a) rsm6=a.
(b) rcos0=a.
r=a(l cos 0) (cardioid).
(c)
(d)
(e)
(f)
= a 2 cos 20(lemniscate).
r=asin30.
2. Transform the following curves from polar to rectangular
coordinates :
(a)
(b) r=a.
(c) r=asecQ.
(d) r=2aBin
PLANE TRIGONOMETRY.
43
CHAPTEE V.
FUNCTIONS OF THE SUM OR DIFFERENCE OF Two OR MORE
ANGLES. OF MULTIPLE ANGLES.
47. The Sine and Cosine of the Sum or Difference of Two
Angles. We have the trigonometric functions of two angles,
6 and 0', and wish to find the sine and cosine of (0+0') and
(00'), in terms of the sine and cosine of and 0'.
Let POM (Fig. 35) be the
sum of the two given angles.
From P, any point in the side
OP (which is the common
terminal side of (0+0') and
of ff=NOP), draw PM per-
pendicular to OM and PN per-
pendicular to ON; then, from
plane geometry, the angle NPM
= 0. Also draw NQ perpen-
dicular to PM and designate PIG 35
the magnitudes of the various
lines as in the figure. From the definitions of the trigonometric
functions, we have
sin(0+0') =
PM
e
c
OP c - b c
= sin cos tf + cos sin 0' ;
~d
also
OM
a
c
" OP
= cos cos 0' sin sin ;
A A JL . A
b c ~ d c
44 PLANE TRIGONOMETRY.
In the applications of the general definitions of the trigo-
nometric functions, we have seen that the definitions apply to
angles of any magnitude if we consider the signs of the auxiliary
lines involved in the definitions. It follows that trigonometric
formulae deduced, with due regard to signs, from geometric figures
are of general application. We may therefore assume that ff may
be replaced by ( 6'), in the above formula?, giving thus the sine
and cosine of (6 9') (see Art. 25) :
sin(0 ff) =sin 0cos & cos 6 sin 0',
cos(0 0') =cos 6 cos ff -\- sin sin 6' '.
These four formulae are fundamental in the consideration of
functions of two angles and should be carefully memorized:
sin(0+0') =sin 6 cos 0' + cos sin 0', (5)
cos(0+0') =cos cos 0'-sin 0sin 0', (6)
sin (0-0') =sin cos 0' - cos sin 0', (7)
cos(0-0') =cos cos 0' + sin 0sin 0'. (8)
As an exercise, prove by use of these formulae the results derived
geometrically in Arts. 21-26; thus, from (6),
cos(90+0)=cos90 cos0-sin90 sin 0= -sin 0.
48. Tangent of (00'). Dividing the first of the formulae of
Art. 47 by the second, and the third by the fourth, we get
+ ta \ a\ sm cos 0'4-cos sin 0'
V > ~ eos 6 cos ff- sm0 sin ff '
, ( fi ff\_ sin cos 6' cos sin ff
} ~ cos cos 0' + sin sin 0"
If each term in each of these fractions be divided by cos cos 0',
PLANE TRIGONOMETRY. 45
49. Functions of Double Angles. If, in the functions of
(0 + 0') (Equations 5-10), we make 0=0', there results:
sin 20 =2 sin B cos 0, (11)
cos 20= cos 2 B -sin 2 0, (12)
If we replace cos 2 6 by 1 sin 2 6, and sin 2 6 by 1 cos 2 0, in the
value of cos 20,
cos 20 =1-2 sin 2 0, (14)
cos 20=2 cos 2 0-1. (15)
50. Functions of the Half -angle. Since J-s a variable angle,
we may replace 20 by <f>, and by J<, in formulae of Art. 49;
getting from (14) and (15), after transposition,
2 sin 2 40 = 1 -cos <, (16)
2cos 2 44>=l + cos</>, (17)
and, from (16) -(17),
_
2
These relations are three of the most useful formulae of trigo-
nometry. If we rationalize first the numerator and then the
denominator, (18) is freed from radicals:
tan - < _ -< M9)
2 "
51. Graphic Derivation of Functions of the Half -angle. Let
be an angle at the center of the circle ; then J0 is the inscribed
angle subtended by the same arc, as in Fig. 36.
Calling the radius unity, we have :
46 PLANE TRIGONOMETRY.
From these data and the definition of the trigonometric functions,
we derive:
whence
tan e=
sin 6
1 + COS0
1 cos0
1 cos0
sin 6 '
tan 2 U=
1 + COS0 *
FIG. 36.
52. Examples.
1. Find the value of sine, cosine and tangent of 15 from the
functions of (45 -30).
2. Find the functions of 15 and 22J from the formula of
Art. 50.
3. Show that the value of the functions of 15 derived in
Example 2 are the same as those derived in Example 1.
4. Let & ^Q in formulae (5) and (6) and thus derive the
sine and cosine of 36.
Ans. sin 30 = 3 sin 6 4 sin 3 0, cos 30=4 cos 3 6 3 cos 6.
PLANE TRIGONOMETRY. 47
5. From Art. 50, tan -- = cosec <f> cot <, show that
T .~~
tan f ~ + -- J = sec B + tan 0. Assume <f> = (~ + j .
6. Show that sine 2x= . 2 * anx and cos 2x= '
. 2 .
1 + tan 2 x 1 + tan 2 a;
7. Show that tan(45 + z) =cot(45 -a;) = l + i ^ x .
1 tan x
8. From formula (18) show that
I it x\ /I sin :r 1 sin #
tan - zzA/i bm ^ - - =sec a; tana:.
U 2y Vl + sino; cos a:
53. Formulae for Transforming Sums into Products. From
the four formulae of Art. 47, we get by addition and subtraction :
sin(0+0') +sin(0-0 / ) - 2 sin 6 cos 0',
sin(0 + / ) -sin(0-0') = 2 cos sin ff,
006(0+?) +cos((9-(9') = 2 cos cos 0',
008(0+?) -cos(0-^) = -2 sin sin 0'.
If we take
0+0' = $,
e-e f =<j>' f
from which
and substitute these values in the preceding equations, they
become :
sin< + sin</>'= 2 sin i(</> + <#>') cos (< <'), (20)
sin^> sin<'= 2 cos i(^> + ^>') sin |(< 0'), (21)
cos </> + cos <'= 2cos J(<^> + ^') cos i(<^> </>'), (22)
cos> cos <>'= 2 si
48 PLANE TRIGONOMETRY.
From the quotient of the first of these by the second, and of the
third by the fourth, we obtain :
sin<4-sin<' tan
sin ^ ~tan^(< <f>')
54. Sum of Two Inverse Functions. If we have two angles
given as inverse functions, the formulae of Art. 48 may be em-
ployed to express the sum or difference of the two angles as inverse
functions.
If, for instance, # = tan 6, and ?/ = tan &, formulae (9) and (10)
become, since 6 tan' 1 x and 0'^tan' 1 y,
0+6' = tan' 1 x + tan" 1 y = tan
l-xy
and
0-j-0' = tan" 1 x tan" 1 y = tan' 1 .f ^ .
1 + zy
If the numerical values of the functions are given, as 0=
and tf = tan- 1 J, we derive
' = tan- 1 % + tan" 1 J =
and
lrf-t
55. Examples.
1. Prove the following identities:
/ x sin :r-fsin y _
\ a .
/T x sin a: sin y
(b) - =
y cos x + cosy
2. Simplify
PLANE TRIGONOMETRY. 49
sin 3a + sin 5a
COS 3a COS 5a
c, . T - sin x + sin 2x + sin 3x
3. Simplify- =-.
cos z-f-cos 2a; + cos ox
4. Show that tan (2 tan' 1 a) 1 2 .
x a
5. If tan' 1 J + tan- 1 J + tan' 1 -J + tan- 1 f = a, find a.
6. Solve sin 2a; = 2 cos x.
7. Solve cos 3x cos 5z = sin a:.
8. 4 tan- 1 J-tan' 1 ^ = ^ .
9. sin' 1 f + sin- 1 -fr + sin' 1 1| = ^ .
w
12. sin" 1 (cos x) H-cos" 1 (sin y) +x + y = 7r.
2
x
13. tan-' +tan" _ +t an- , =,.
50 PLANE TRIGONOMETRY.
CHAPTER VI.
FORMULA FOR THE SOLUTION or OBLIQUE TRIANGLES.
56. The Sine Formula. Let ABC (Fig. 37) be any oblique
triangle, and from the vertex C drop a perpendicular to the oppo-
site side c. Let A, B and C be the values of the angles, a, ~b and c
the values of the opposite sides, respectively, and p the value of
the perpendicular.
C
From the two right triangles BCD and ACD, we have
p = asinB = b sin A.
By dropping a perpendicular from B on h, we obtain
p' = a sin C=c sin A.
From these two equations,
sin A "~ sin B ~~ sin C '
If the angle A is obtuse, the perpendicular falls on BA pro-
duced, and then
r>
p l sin(180 A) =b sin A = a sin 2^
as before. Stated in words:
The sides of a plane triangle are proportional to the sines of
their opposite angles.
PLANE TRIGONOMETRY.
51
57. The Cosine Formula. In the oblique triangle ABC (Fig.
38), draw the perpendicular p from
C to the side c; then in the right
triangle ACD the side AD = b cos A,
&JL, in BDC, BD = a cos B=c-b
cos A. Equating the value of p 2
from each of the two right triangles,
from which
= b 2 + c 2 -2bccosA.
Similar formulas for b 2 and c 2 are obtained by interchanging
letters. Hence
The square of any side of a triangle is equal to the sum of the
squares of the other two sides diminished by twice the rectangle
of these sides multiplied by the cosine of the included angle.
58. The Tangent Formulae. From Art. 56,
a_ _ sin A
T ~~ sin B '
and, by composition and division,
a b sin A sin B
From Art. 53,
Hence
a + b sin A + sin B'
sin A sin B _ tan^(A B)
sin A + sin B ~ ta
a-b _tan|(A-B)
a + b . tan|(A-f B) '
Similar formulae for b and c, and a and c, are readily derived by
interchanging letters.
52 PLANE TRIGONOMETRY.
59. Functions of Half -angles in Terms of the Sides. If in Art.
57 we put cos A - 1 - 2 sin 2 \A,
. 2 i A _a 2 -(b-c) 2 _ (a-b + c)(a+b-c)
~fc~ ~^fc~
If we put 2s = (a+b + c), then
(a-b + c)=2(s-b), (a+b-c)=2(s-c) ;
and, substituting these values in the above equation, we obtain
ac
the last two from interchange of letters.
60. If in the same formula we put cos A = 2 cos 2 \A 1, we get
or
^2 -LA-
COS 2-i-
Put a+6 + c=25; then (b + c a) =2(s a), whence
ac
The last two are written by interchange of letters.
PLANE TRIGONOMETRY. 53
61. Dividing the formulae of Art. 59 by those of Art. 60 :
s(s-c)
62. From Art. 56, we may obtain
a -f b _ sin A + sin B
c sin C
and
a b _ sin A sin B
c sin C
hence by 49 and 53,
a + b sin
c sin |(7 cos
a-b c
c sin ^( cos (
But A+B=180-C, and J(A+) =90-- |C; and thus
sin (A +B) =cos%C, and cos %(A+B) =sin|(7.
Substituting these values, we obtain:
a + b _ cosl(A-B) _ cos^(A-ff)
c sin^(7 ~ cos %(A +B)'
a-b sinl(A-B) _ am$(A-B)
c cosJC ~ sin
63. Area of an Oblique Triangle in Terms of Its Three Sides.
We have
sin A = 2 sin \A cos \A,
and, by Arts. 59 and 60, this becomes
sin A ^r- ys(s-a) (s b) (sc).
be
54 PLANE TRIGONOMETRY.
Hence, K, the area of the triangle (Fig. 37), is
K=^pc=^bc sin A,
whence
K= Vs(s-a) (s l) (s-c).
We also have
p= \/s(s-a)(s-b)(s-c).
PLANE TRIGONOMETRY. 55
CHAPTER VII.
SOLUTION OP PLANE OBLIQUE TRIANGLES.
64. Case I: Given Two Angles and One Side, A, B and a.
The third angle C is found from the formula
C=1SO-(A+B).
The sides are found by the sine formula,
I sin B . c sin C
~ T and = 7 j
a smA a smA
or
. _ . n FIG. 39.
o = a sin # cosec A, c= a sin C cosec A.
Example: Given A = 50 38' 50", 5=60 7' 25", a = 412.67.
(7 = 180 -(110 46' 15") =69 13' 45".
a=412.67 log 2.61560 log 2.61560
A = 50 38' 50" cosec 0.11168 cosec 0.11168
J5 = 60 r 25" sin 9.93807-10
(7 = 69 13' 45" sin 9.97082-10
= 462.76 log 2.66535
c = 499.00 log 2.69810
65. Second Method. When a and & are nearly equal, and
greater accuracy is desired, we may write :
a sin B sin A sin B
: r =0,' - = -
sm A sin A
But
whence
,
a o
sin A
or
sin A
56
PLANE TRIGONOMETRY.
Thus in the preceding example,
i(5 + A)=55 23' 7.5"
$(B-A)= 4 44' 17.5"
4 = 50 38' 50"
b-a = 50.083
a=412.67
6 = 462.753
cos 9.75439-10
sin 8.91699-10
cosec 0.11168
log 2.91663
log 1.69969
66. Case II: Given Two Sides and an Angle Opposite One of
Them, a, b and A. By the sine formulae
Having found B, we have
and
c=asin C cosec A.
The angle B corresponding to the com-
puted sine may be either an angle in the
first quadrant or its supplement, and hence
two solutions may be possible, as indicated
in the geometrical construction of Fig. 40.
If the side a is greater than b, the second
point of intersection, B 2 , falls on the side
c produced, and only one of the solutions
fills the given conditions, as the triangle AB 2 C contains 180 A.
If the computed value of sin B is unity, the two triangles are
two coincident right triangles; while if the computed value of
sin B is greater than unity, the triangle is impossible.
Example: Given A = 27 47' 45", a = 219.91, 6 = 251.33.
PIG. 40.
PLANE TRIGONOMETRY. 57
6 = 251.33 log 2.40024
a = 219.91 colog 7.6577510 log 2.34225 log 2.34225
A =27 47' 45" sin 9.6686910 cosec 0.33131 cosec 0.33131
B,= 32 12' 15" sin 9.7266810
# 2 =147 47' 45"
O l =l2Q 0' 0" sin 9.9375310
Cf= 4 24' 30" sin 8.8857210
c,= 408.40 log 2.61109
C 2 = 36.248 log 1.55928
Second Method. Having found B from its sine, we may com-
pute AP= b cos A = x, and PB^ = a cos B = y, the negative value
resulting from the value of B in the second quadrant, whence
67. Case III: Given Two Sides and the Included Angle, or
a, b and C. From the tangent formulae (Art 58),
B a-b .
and since! (A +B) =4(180 -C) = 90 -\C,
tani(A+)=
Substituting this value, we derive
- 6 co
The value of c may now be found from the sine formula, or
more conveniently from the formulae of Art. 62, from which
-(a + M c
'c
or
c-(a-b) s _
b) s
Example: Given a = 6.2387, & = 2.3475, (7=110 32'.
58 PLANE TRIGONOMETRY.
a b 3.3912 log 0.59008 log 0.59008
a+6 8.5862 colog 9.06620 10 log 0.93380
l(A+B)=34 44' 0" tan 9.84092 10 cos 9.9147710 gin 9.7556910
I (A B)=17 26' 33" tan 9.49720 10 sec 0.02044 cosec 0.52324
c = 7.3962 log 0.86901
c 7.3962 log 0.86901
A =52 10' 33"
B =17 IT 27"
The two values of c are computed here for a check.
68. Second Solution. If we wish to find c only, we have, by
the law of cosines (Art. 57),
c 2 = a 2 + b 2 -2abcosC,
which, however, is not adapted to logarithmic computation. If
we put cos C = T- 2 sin 2 %C,
Let
tan 0=
a b
and the radical in the above equation becomes
2 0=sec0,
from which
c= (a &)sec 6.
Examples.
1. Given a=. 062387, 6 = . 023475, 0=110 32', find
10' 33", 5 = 17 ir 27", c=. 073962.
2. Given a= 35.237, 6 = 18.482, 0=110 40' 30", find
49' 58", B = 22 29' 32", c = 45.198.
3. Find c in both the above examples by Art. 68.
PLANE TRIGONOMETRY. 59
69, Case IV: Given the Three Sides, a, b and c. When all
three angles are required, the formula of Art. 61 for the tangent
of the half -angle is the most convenient, and is accurate for all
values of the angles. When only one angle is required, the
formulae of Arts. 4Sand <WTiiay be used, although the tangent
formula is always more accurate in practice.
The formula for tan \A. may be written
tan p = -L.
The other angles, by permuting letters, are readily obtained.
Note that the radical expression is the same for all three angles.
Put _
p_ / (s-a)(s-b)(s-c)^
* s
and the formulae become :
fcniA = , tan^=-, tan *(7=-.
Example: Given a = 6.8235, 6 = 5.2063, c= 3.1628.
a= 6.8235
b= 5.2063
c- 3.1628
25=15.1926
s= 7.5963 colog 9.11940-10
8-a = 0.7728 log 9.88807-10
s-b = 2.3900 log 0.37840
s-c = 4.4335 log 0.64675
P 2 log 0.03262
P log 0.01631
iA = 53 20' 20" tan 0.12824
lB = 23 28' 51" tan 9.63791-10
10=13 10' 49" tan 9.36956-10
Check: 90 0' 0"
60
PLANE TRIGONOMETRY.
D
FIG. 41.
70. Second Method. When a, b
and c are given, we may solve as
follows :
Draw the perpendicular p from
C to the side c, and call the seg-
ments of the base x and y, as in Fig.
41. Then
whence
But
and thus
whence
and
Examples.
1. Given a = 25.783, 6 = 37.686, c = 50.401; solve, by Art. 59;
4 = 29 49' 22", = 46 37' 43", (7-103 32' 55".
2. Find the angles in Example 1 by Article 70.
71. Solutions of Eight Triangles by Using Functions of Half-
angles. The formulas involving sines and cosines may be con-
verted into formulae involving functions of the half -angles. Thus
in the case of the right triangle (Art. 37) when the hypothenuse
and one leg are nearly equal and accuracy is desired, use the
following method :
PLANE TRIGONOMETRY.
ttl
whence
C -^=2 sin 2
c
c + a _ _ o
c a
c+a
Since b = Vc 2 a?= V(c a) (c+a), these formulae furnish
a convenient and accurate solution :
Given
a= 382.4
c=383.2
c-a= 0.8
log 9.90309
log 2.88400
-10
5 = l 51' 5"
5=3 42' 10"
2
6 = 24.748
2)7.01909-10
tan 8.50954-10
log 2.78709
log 1.39354
72. The Radius of the Circumscribed
Circle. The center of the circumscribed
circle (Fig. 4$) lies in the perpendicu-
lars erected at the middle points of the
sides. From the figure we have
sin CO^sin A = =
whence
FIG. 42.
R =
2 sin A '
62 PLANE TRIGONOMETRY.
From Art. 63,
2K
smA = -=.
be
Hence
p _ ale
~ 4K'
73. Radius of the Inscribed Circle. The area of the triangle
ABC (Fig. 43) is equal to the area of the three triangles having
FIG. 43.
the sides of the triangle as bases and the radius of the inscribed
circle as altitudes. Hence
Examples.
1. Find the radii of the inscribed and circumscribed circles of
the triangle of Art. 89.
2. Find the same for the triangle of Example 1, Art. 70.
3. Find the area of both the triangles of Examples 1 and 2.
PLANE TRIGONOMETRY. 63
CHAPTER VIII.
COMPLEX NUMBERS, DE MOIVRE'S THEOREM, DEVELOPMENT OF
SINE AND COSINE IN TERMS OF THE ANGLE.
74. Complex Numbers. In the Algebra the subject of complex
numbers was briefly treated, where expressions, like a + bi were
subjected to certain operations, a and b representing real num-
bers, positive or negative, i the so-called imaginary unit, V 1.
We shall here show how such numbers may be graphically repre-
sented and the results of operations performed by or upon them
interpreted. They are much used in the practical computations of
rotating machinery, such as steam turbines, dynamos and motors.
We recall first, from the Algebra (Arts. 3-5), that from the
definition of i V 1, the powers of i are :
i=i f * 2 =-l, i 3 =-i, t 4 =+l,
i 5 =i, i e = -l, i*= -i, i 8 = + 1, etc.
Secondly, that if we have an equation between two complex
quantities like
a+bi=c-\-di,
then must a c and b = d; in other words, an equation containing
real and imaginary quantities may be broken up into two equa-
tions, one containing only the real quantities, the other only the
imaginary quantities.
75. Graphical Representation of Complex Numbers. If we
multiply a real number a by i, and this product again by i, the
result is a, or, multiplying a twice by i reverses its sign. This is
graphically shown in Fig. 44, where the number OP =3 units is
changed to OP^ 3 units, by a change of 180 in its direction.
In other words, by using the operator i twice in succession as a
64
PLANE TRIGONOMETRY.
multiplier, the line OP has been turned through two right angles.
Multiplying a number by i may be interpreted as turning the
direction of the line which represents it through one right angle,
-X-
-3
25 <?| 3 7>
. -3-i
Q.
V
FIG. 44.
or 90 . Thus the quantity represented by the line OP, becomes
the quantity OQ, when multiplied once by i; and for this reason
we take the axis of y as the axis of imaginary numbers, just as we
have taken the #-axis as the axis of real numbers. To plot a com-
plex number like 3 + 4t, for instance (Fig. 45), lay off OQ = 3
PIG. 46.
units horizontally on the axis of real numbers, and the 4t units,
QP, at right angles (parallel to the axis of imaginary numbers,
which are therefore more properly called quadrature numbers).
PLANE TRIGONOMETRY. 65
The general complex number x + yi is represented in Fig. 46,
from which we illustrate the following terms used in connection
with such numbers. The vector OP rVx 2 + y 2 is called the
modulus or the absolute value of x + yi. The angle XOP 6 is
called the amplitude, or the argument, or the arc of x-\-yi.
Modulus of (x+yi) =r=
Arc of (x + yi) = 0=tsn\- 1
x
76. Polar Coordinates of Complex Numbers. From Fig. 46 we
have
x=rcos0, y
r 2 (cos 2 0+sin 2 6) =
and
x + yi=r(cos B+i sin 0).
This trigonometric expression shows more clearly the vector char-
acter of complex numbers and leads to simpler methods in their
use.
Examples.
1. Plot the complex number 3 + 4i, and show geometrically and
trigonometrically that the angle is turned through 90 by multi-
plying the number by i.
Ans. ^=tan- 1 (t), ^ / = tan- 1 (-f), i. e., 0' = 90 +6.
2. Find the modulus and amplitude of the following complex
numbers, and show that the amplitude of each is increased by 90
by multiplying by i.
3 + 3t, - , n , - . -
77. De Moivre's Theorem. The theorem of De Moivre is stated
by the equation
(cos 0+tsin 0) n = cos n0 + isin n0,
which may be derived in the following way :
66 PLANE TRIGONOMETRY.
If we square the complex number
x + yi=z = r(cos 0+i sin 0),
we get
= r 2 (cos 2 0-sin 2 0+2i sin cos 6)
= r 2 (cos20+tsin20).
Multiplying z 2 by 2 = r(cos0+tsin 6),
z z = i*\ (cos 20 cos sin 20 sin 0) + 1 (sin 20 cos + cos 26 sin 0) [
= r 3 (cos30+isin30).
If this process be continued to n factors, we shall get
2 n = r n (cos 0+i sin 0) n = r"(cos nQ+isinnO).
Dropping the common factor s r n ,
(cos 0+i sin 0) n = cos n0+i sin nO.
The theorem may also be developed by multiplying together two
expressions like (cosa+tsina) and (cos /? + tsin/?) and noting
that the product is <( cos (a +/?) +t sin (a +(3) [-, and so on, for any
number n of angles; finally making a =p=y, etc., we derive the
formula^ as given.
78. Functions of Multiple Angles. If we write De Moivre's
theorem,
cos n0 + t sin nO= (cos + t sin 0) n ,
we may expand the second member by the binomial theorem, and
equate the real and imaginary terms of the identity (Art. 74).
For example,
cos 20+i sin 20= (cos + i sin 0) 2
= cos 2 sin 2 + i2 sin cos 0,
whence
cos 20- cos 2 0- sin 2 0,
isin20 = i2 sin cos 0,
thus establishing the formulae of Art. 8&r U (\
PLANE TRIGONOMETRY. 67
Ifw=3,
cos 30 + i sin 30
= cos 3 + t3 cos 2 sin 3 cos sin 2 tsin 3 .
Equating real and imaginary terms,
cos 30= cos 3 0-3 cos sin 2 0,
tsin30 = i(3cos 2 0sin0 sin 3 0),
which may be reduced to
cos 30 =4 cos 3 3 cos 0,
sin 30=3 sin 4 sin 3 0.
Expanding the second member of the identity and equating real
and imaginary terms, we express cos nO and sin nO in a power-
series of sine and cosine 0. Noting that even powers of i are,
real, we may write :
cos n0= cos n 0- 2^=11 cos"- 2 sin 2
+ n(n-l)(n-2)(n-3) cos n
sin n0= cos"- 1 sin 0- rc(n-l)(K,-2) cos n-3 # sin 3 e
L5
+ tt(n-l)(tt-2)(n-3)(n-4) cos n- 5 e sin 5 # +
l
Express the following in trigonometric form and find the values
of the first, second, third and fourth powers :
7$. Cosine a and Sine a as Power-series of a. In the expres-
sions for cos nO and sin nO of the preceding article, let n0=a, and
68 PLANE TRIGONOMETRY.
let n increase without limit, while a remains unchanged. Then
0= diminishes without limit; and at the limit we have (Art.
35).
sin 0= sin = , cos = cos =1.
n] n=M n' nj n=x
Introducing these values into the expansion of cos nO and sin nO
(Art. 78), they become:
n(nl) a 2
cosa=l f * 7
2 n 2
n(n-l)(n-2)(n-3)
|4
-
n [3 n 3
, n(n-l)(n-2)(n-3)(n-4) a 5
~[5~ "5F"
When n increases without limit, the factors containing n all
vanish, except the highest power of n in each term; thus, for
example,
n(n-l)(n-2) , 3 _21 ,
n 3 " n + n 2 J n=00
The expansions then become :
COSa = -
a 2 a 4 a
/T \
a a a /TT x
Sina-a- -||- + -jr"-]j; + .....
Dividing (II) by (I) (synthetic division),
PLANE TRIGONOMETRY. 69
Examples:
QO -< i " \ J- I W \ -L
cos 3 = 1 - -T-- +
a =.052360.
a 2 = . 002742^- 2 = 0.001371.
a s = - 000143^- 6 = 0.000024.
24= 0.
With these values, we obtain, to five places :
sina= .052360 -.000024 = 0.05234.
cos a = 1 - .001371 = 0.99863,
tana= .052360 + .000048 = 0.05241.
2. Expand e i0 by the exponential theorem (Alg. Art. 161) and
by putting iO for x in the expansion
show that
= cos#+isin 0.
3. Show also that e~ i9 =cos i sin 0, and hence that
cos 0=
A
and
70 PLANE TRIGONOMETRY.
80. De Moivre's Theorem with Fractional Exponents. If we
r\
replace 6 by , in
fi
(cos 0+i sin 6) n = cos nO+isin n0,
we get
(cos + tsin ) = c
V n n /
whence
n i\ j
cos +i sin = (cos 0+i sin 0) n.
IV it
Since the values of cos 6 and sin remain unchanged when 6 is
increased by any multiple of 2?r ?
i
(cos 0+i sin 6)^ =-{cos(2
We may therefore derive the n values of (cos 0+i sin 6) T by mak-
ing k successively equal to 0, 1, 2, 3, . . . . , (n 1) .
Examples.
1. Let x+yi=8 = 8 (cos + t sin 0) =8 (cos 2for + isin 2&7r) ;
then (x + yi)* = 2 (cos ~p + i sin ^) . Thence show that the
three cube roots of 8 are 2, l + tV% and 1 tVs".
2. Find the three roots of ( 8) and plot them; verify them by
multiplying the three trigonometric expressions together.
SPHERICAL TRIGONOMETRY.
71
SPHERICAL TRIGONOMETRY.
CHAPTEE IX.
GENERAL FORMULAE.
81. The object of Spherical Trigonometry is the investigation
of the relations of the circular functions of the sides and angles
of a spherical triangle, and thus finding formulas for the solution
of the triangle.
The sides and angles are expressed in the sexagesimal units of
angular measure, and their numerical values are independent of
the size of the sphere. For this reason, we shall consider the
radius of the sphere as unity.
PIG. 47.
82. Let ABC (Fig. 47) be a spherical triangle, formed on the
surface of the sphere by the intersection of three planes through
72 SPHERICAL TRIGONOMETRY.
the center of the sphere, 0, forming the triedral angle ABC.
The sides of the triangle a, b and c, measure the face angles of the
triedral angle, BO C, AOC and AOB, respectively, while the angles
between the three planes are equal to the angles of the spherical
triangle.
A plane tangent to the sphere at A, and thus perpendicular to
OA, intersects the three planes in the plane triangle ARC', and
the angle at A measures the angle between the faces AOB and
A 00, and thus also the spherical angle BAC A.
In the right triangle OAB' let OA \\ then OB' = sec c, and
AB' = i&n c; in the right triangle OAC', 0.4 = 1, 0C" = sec &, and
AC' = ta,n I; let x-ffC'.
83. Relation between the Sides and an Angle of a Spherical
Triangle. We have from plane trigonometry, in the triangles
OVVvn&AVV:
x 2 = sec 2 b + sec 2 c 2 sec b sec c cos a,
# 2 = tan 2 & + tan 2 c2 tan b tan ccos A.
Subtracting the second equation from the first, and noting that
sec 2 B tan 2 0=1, we obtain
= 2 + 2 tan b tan c cos A 2 sec b sec c cos a.
Transposing the last term, and multiplying the equation by
J cos b cos c, there results
cos a=cos b cos c + sin b sin c cos A.
Writing similar formulas for cos b and cos c, we have the funda-
mental formulae for spherical trigonometry:
cos a=cos b cos c + sin b sin c cos A, "1
cos b = cos a cos c + sin a sin c cos B, I (A)
cos c = cos a cos b + sin a sin b cos C. J
84. We snail assume, in accordance with the principles of gen-
erality involved in the definition of the circular functions of plane
SPHERICAL TRIGONOMETRY.
73
trigonometry, that these formulae are general and apply to spheri-
cal triangles whose sides or angles may exceed 90 ; in the practical
solution of spherical triangles, however, we shall consider those
only whose parts are less than 180.
85. Relations Existing between a Side and the Three Angles
of a Spherical Triangle. Let A'B'C'
be the polar triangle of the triangle &'
ABC (Fig. 48).
From the geometric principles of
polar triangles, we have :
-6, 6'= 180 -B,
cr=i8o-c, c'=i8o-a * V
From Art. 83, in the triangle A'B'C',
, F IG. 4o.
we have * tj (L^Q . fa* (L -t (*
cos a' = cos V cos c' + sin V sin c' cos A'.
Replacing the values of a', b', c' and A' by the values given above,
^-cosA= (cosB) (cos (7) sin B sin C cos a.
Changing signs of all the terms, and interchanging the different
letters, we get the three formulae :
cos A cos B cos C+ sin B sin C cos a^
cos B= cos A cos (7 + sin A sin C cos 6, L (B)
cos C = cos A cos B + sin A sin B cos c. J
86. The Sine Formulae. From the first of the formulae of Art.
83, we get
A cos a cos b cos c
cos A = ; - : .
sin b sin c
Squaring both sides, and noting that
1 cos 2 A = sin 2 /!,
74 SPHERICAL TRIGONOMETRY.
2 A sin 2 sin 2 c cos 2 a cos 2 cos 2 c + 2 cos a cos cos c
bin A ; r-^ ; = .
sin * sin c
Putting for sin 2 b sin 2 c its equal, (1 cos 2 0) (1 cos 2 c), we get
sin 2 b sin 2 c=l cos 2 cos 2 c + cos 2 b cos 2 c,
whence
. 2 , 1 cos 2 a cos 2 cos 2 c+2 cosacos cos c
sin 2 A = - . . - .
sin 2 sin 2 c
Dividing both sides by sin 2 a,
sin 2 ^. _ 1 cos 2 a cos 2 b cos 2 c + 2 cos a cos & cos c _ -,
sin 2 a ~ sin 2 a sin 2 b sin 2 c
a constant, since the right-hand side of the equation remains un-
changed, when A and a are replaced by either B and b, or C and c.
Hence,
sin 2 A _ sin 2 B _ sin 2 C
sin 2 a sin 2 b ~ sin 2 c '
or
sin ^4. sin B _ sin (7
sin a " " sin " " sin c '
or
sin a sin B= sin & sin A,!
sin a sin (7 = sin c sin A, i- ( C)
sin b sin C = sin c sin B. J
Stated in words, these formulae read :
In any spherical triangle, the sines of the angles are propor-
tional to the sines of the opposite sides.
87. The sine formula may be derived from a geometrical figure
(Fig. 49) . From any point P in OC, draw PM perpendicular to
the plane OAB, and draw MA' and MB' perpendicular to OA and
OB, respectively.
SPHERICAL TRIGONOMETRY.
75
Then the angle PA'M = the angle A, and PB'M=ihe angle B.
PM PM
hence
BinA = Pl"
sin A _ PB f
sin B ~ PA' '
(1)
FIG. 49.
In the triangles POA' and FOR, the angle POA' = l t and
POB' = a; and
sma=
PA '
and
FO '
sin a PB'
sin b ~ PA'
From (1) and (2),
or
sin A _ sin a
sin J5 sin &
sin A _ sin B
sin a sin ft
76
SPHERICAL TRIGONOMETRY.
88. Formulae for the Half -angles. In the first of the formulae
(A), put
cos A = 1 2 sin 2 \A. ;
then
cos a cos b cos c + sin b sin c 2 sin b sin c sin 2 \A, (I)
whence, since
cos b cos c + sin b sin c=cos(& c),
(2)
..
2 sm b sin c
From Art. 53, Plane Trigonometry,
cos (6 c) cosa= 2 sin i(6 c + a)sin(& c a)
Let
then
and
J(a+c &) =s b.
Substituting the values of (3) and (4) in
Similarly,
and
sin 2 1? A _ sin (s-fr) sin (s-c)
sin & sin c
sin 2 ig- sin (s-ft) sin (s-c)
sin a sin c
. n2
sin a sm
89. Substituting in the same formula of Art. 83,
cos A = 2 cos 2 \A. 1,
cos a cos b cos c sin b sin c + 2 sin b sin c cos 2 \A.
= cos ( & -f c) + 2 sin b sin c cos 2 JA,
(3)
(4)
SPHERICAL TRIGONOMETRY.
77
whence
C^S ifxl : ; : ~ y
2 sin b sin c
cos a cos ( b + c) = 2 sin
= 2 sin
Introducing, as in Art. 88, s =
Similarly,
a I c)
a).
sin 6 sm c
sn a sin c
(E)
sin a sin b
90. The quotient of (D) divided by (E) gives:
tan a *A =
sin (5 &) sin (s c}
sin s sin (s a)
sin s sin ( s b )
tan
sin 5 sin (s c)
91. Formulae for the Half -sides. To find similar formulae for
the half -sides, apply the formulae (D), (E) and (F) to the polar
triangle, thus :
= 90
c=180-C"
= 270-^
Whence we have
s-a=
5-6=
sn = cos
cos
= sin 2 Ja', tan 2 $A = cot 2 Ja'.
78
SPHERICAL TRIGONOMETRY.
Substituting the above values in the formulae (D), (E) and (F),
and dropping primes, we obtain
sm B sin G
tan 2 la
1 fl -
sin B sin C
-*
92. Formulae of Gauss or De Lambre, From (D) and (E),
sn cos =
sm c
/sm
Y si
sin a sin 6
or
sm c
cos4(7.
Similarly,
cos \A. sin 45 =
cos if A cos 45 =
sin 4 sin 5=
sm c
sm c
cos
sn
(1)
(2)
( 3 )
(4)
From Plane Trigonometry (Art. 4^), (1) + (2)
gives
sine
Reducing,
Similarly,
cos 4c sin
sm \c cos %c
^ cos a
(5)
sm c
2 sm -|c cos ^c
sin Jc sin 4 (A 5) =0034^ sin 4 (a &).
(6)
SPHERICAL TRIGONOMETRY. 79
In the same way,
1 / A , T>\ in f -sins sin (s c) \
cos4(A+)=sm4(7| -- ^ -- )
= |^l(^+&)smi C |
I 2 sin Jc cos Jc J
or
and, similarly,
sinico>si(A--#)=siiil<7sin4(a+&). (8)
Collecting (5), (7), (6), and (8), we have Gauss' Equations:
cos c sn
cos Jc cos ^ (A + B) =8in4(7co8j(o-f-J),
sin Jc sin J (A 5) =cos-J(7 sinj(a 6), |
sin \c cos ^ (A 5) ^sinJC' sini(a+&).j
93. Napier's Analogies (or Eatios) .Dividing (6) by (5)
(Art. 92), and transposing,
tan^c
From (8) -(7),
_ cos ^ (A -B)
tan|c c
From (6) -7- (8),
From (5) -(7),
tan \ (A +B] _
cot 4(7 ~
80 SPHERICAL TRIGONOMETRY.
CHAPTER X.
SOLUTION OF SPHERICAL RIGHT TRIANGLES.
94. When one of the angles of a spherical triangle is a right
angle, the general formulas of Chapter IX assume forms which
may be generalized under two simple rules.
Assume the angle = 90, and the formulae of (A), (B) and
(C) become:
FromA(3), cos c=cosacos b.- (1)
" B(3), cosc=cot^cot#. (2)
" B(l), cos A = cos a sin B. (3)
" B(2), cos B= coat sin A.- (4)
" C(2), sin a = sin c sin A. (5)
" 0(3), sin 6 = sine sin B. (6)
From the triangle AB'C' (Fig. 47),
cos A = ^^ = tan b cot c, (7)
tan c
cos B= tan a cot c, ( 8 )
by interchanging a and b and A and B in (7).
We have tan A = sm , and substituting the values of sin A
cos A
and cos A from (5) and (7),
tan A
tan b sin c cos c tan b 3
and putting cos c= cos a cos I, from (1), we obtain, finally,
cos a sm sin
or
sin b = tan a cot A ; (9)
and by interchange of letters,
sin a = tan b cot B. (10)
SPHERICAL TRIGONOMETRY. 81
These ten formulae are sufficient for the solution of any spherical
right triangle of which any two parts are given (excluding the
right angle).
95. Napier's Rules. The ten formula? necessary for the solu-
tion of any right triangle from two given parts, may be readily
derived from two artificial rules, devised by Lord Napier.
Omitting the right angle, and taking the complements of the
hypothenuse and of the two adjacent angles, the five parts are
called the circular parts of the right tri-
angle, and are lettered as in Fig. 50.
If we consider any three of these circular
parts, one of them is the middle part and
the other two either adjacent or opposite
parts. For example, of the parts b, a and C0i
co-B, a is the middle part and b and co-B \
adjacent parts, the right angle being
omitted; while of the three parts co-A, co-B and b, co-B is the
middle part and b and co-A opposite parts, since each is separated
from co-B by an intervening part.
Napier's two 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.
Example: Given a and b, to find the formulae for the other
parts of the triangle in terms of functions of the given parts alone.
To find the angle A : The three circular parts are a, b and co-A;
b is the middle part; and the other two adjacent parts; hence
sin b = tan a cot A, or
cot A = sin b cot a. (1)
To find B: a is the middle part and b and co-B adjacent parts ;
hence sin a tan b cot B, or
cot B= sin a cot b. (2)
82 SPHERICAL TRIGONOMETRY.
To find c: co-c is the middle part, a and b opposite parts;
hence
cos c cos a cos 6. (3)
A check formula involving the three required parts, co-c, co-A
and co-B, is found by taking co-c as the middle part, co-A and
co-B as adjacent parts, giving
cos c cot A cot B. (4)
Note that the check formula is always expressed in terms of the
functions l>y which the parts are obtained.
96. Solution of Spherical Right Triangles. When two parts
of a spherical right triangle are given (not including the right
angle), each of the other parts is to be determined in terms of the
functions of the two given parts only, by means of Napier's rules,
and the computation checked by a " check formula " involving
the three computed parts.
It is customary to divide the solution into five cases, according
to the parts given; but this is not necessary, as Napier's rules
furnish an obvious method for any case that may arise.
97. To Determine the Quadrant in Which a Computed Part
Belongs. If the computed part is found from its cosine, tangent
or cotangent, the sign of the function determines the quadrant;
but if it is found from its sine, it may be in either the first or
second quadrant. Except in the particular case of a double solu-
tion, the proper quadrant for a part found from its sine may be
determined by one of the two following principles.
98. 1. In a right spherical triangle an angle and its opposite
side are always in the same quadrant.
For we have
cos A = sin B cos a;
and since sin B is always positive (#<180), cos A and cos a
must have the same sign; that is, A and a must be either both
less or both greater than 90.
SPHERICAL TRIGONOMETRY. 83
99. 2. When the two sides including the right angle are both
in the same quadrant, the hypothenuse is less than 90 ; when the
two sides are in different quadrants,, the hypothenuse is greater
than 90.
This follows from the relations between the three sides,
cos c^cos a cos b.
For if a and b are in the same quadrant, cos a and cos b have like
signs and cos-c is positive, so that c is in the first quadrant; i. e.,
less than 90. If a and b are in different quadrants, their cosines
have different signs ; cos c is therefore negative, and c is in the
second quadrant ; i. e., >9^j^
100. Double Solutions. When the two given parts are a side
and its opposite angle, there are always two solutions, as in such a
case the required parts are all found from their sines, and may be
therefore in either the first or second quadrant. In arranging
and marking the parts which go together, attention must be paid
to the principle of Art. 99.
101. Examples.
Example 1: Given a = 37 45' 20",
b = 113 10' 30". Taking the two parts
a and b with each of the other three in
turn as the third part, we find, Fig. 51 : Fio.^51.
cos c= cos a cos b; (1)
sin b = tan a cot A, or cot A = sin b cot a; (2)
sin a = tan b cot 5, or cot B sin a cot b; (3)
Check: cos c = cot .4 cot #. (4)
Note that the check formula for a right triangle always' con-
tains the functions from which the required parts should be found.
84 SPHERICAL TRIGONOMETRY.
= 37 45 20 cot 0.11101 sin 9.78696 cos 9.89798
=113 10 30 sin 9.96340 cot 9.63153n cos 9.59499w
A 40 6 42 cot 0.07447
41 15 . - cot 9.41849n
c =108 7 43 cos 9.49297?!
Check: (cot A cot B)=cos 9.49296n
Example 2: Given & = 154 7', #=152 23'.
Formulae : ( 1 ) sin a = tan b cot 5,
(2) sin A = sec & cos B f
(3) sin c=sin b cosec 5,
(4) sin a= sine sin A (check).
o / "
6 =154 7 tan 9.68593ft sec 0.04591ft sin 9.64002
B =152 23 cot 0.28127w cos 9.94747ft cosec 0.33390
a x = 68 02 36 sin 9.96730
a 2 =lll 57 24
A,= 80 1 20 sin 9.99338
Ao= 99 58 40
c 2 = 70 20 30 sin A 9.99338
Check: sin a 9.96730
For the proper grouping of the parts (Kules 1 and 2), a and A^
are put in the same quadrant, the first; and c^ must be in the
second quadrant, since # and b are in different quadrants.
Derive the formulae and check, and solve the following right
triangles :
o / " o ' " o ' "
3. c=109 41 50 4. o= 67 6 53 5. A= 55 18 13
6 = 25 52 34 A= 80 10 30 c = 75 20 30
6. A=138 27 18 7. a =116 24 25 8. 6=154 032
B 80 55 27 6= 16 50 30 5=126 57 37
9. 6=60 10. a= 90 11. a= 90
A= 90 6 = 60 6 = 90
SPHERICAL TRIGONOMETRY. 85
102. Additional Formulae for the Solution of Spherical Right
Triangles. In cases where a part is found from its computed sine
or cosine, the formulae of the preceding articles may have to be
transformed into formulae involving functions of the half-angle
to insure sufficiently accurate results.
Thus, if we have given c and a, we have
. A _ sin a
8111 /t - ,
sin c
from which
1 sin A __ sin c sin a
1 + sin A ~ sin c + sin a '
sin c sin a=2 cos -J(c+a)sin -J(c a),
sin c + sina=2 sin J('c+a)cos(c a).
Hence
tnn(45"-iA)=
From cos c=cos a cos &, or cos &= cos c , we derive
cos a
1 cos & _ cos a cos c
1 + cos 6 ~~ cos a + cos c '
or
From cos B=
tan c '
1 cos B _ tan c tan a
1 + cos L ~ tan c + tan a '
tan 2 iB=^ r/ " ^
86 SPHERICAL TRIGONOMETRY.
Examples.
1. Derive, from cos c cot A cot B, the formula
2. From cosa=? 3 derive
103. Quadrantal Triangles. If one side of a spherical triangle
is 90, it is called a quadrantal triangle. In its polar triangle, the
angle opposite the quadrantal side is 90 ; and the formulae for the
right triangle thus formed are easily transformed into those for
the given triangle. Generally but one or two parts of the triangle
are required and formulae (A), (B), (C) are readily converted
by putting the given side of 90 in the proper formulae.
SPHERICAL TRIGONOMETRY.
87
CHAPTER XI.
SOLUTION OF OBLIQUE SPHERICAL TRIANGLES.
104. Any oblique spherical triangle may be divided into two
right triangles by a perpendicular from an angle to its opposite
side, and the complete solution of the triangle obtained in terms
of the given parts and certain parts of the right triangles as
auxiliaries, thus reducing the solution of all spherical triangles
to the application of Napier's rules for right triangles.
When the three given parts are the three sides (a, b, c) or the
three angles (A, E, C], however, the
solution by the formulae for the half-
angle is much more expeditious and
accurate.
In solving other cases, draw a plane
triangle to represent the spherical tri-
angle, and letter the segments of the
base <f> and <', and the parts of the
opposite angle 6 and #', as in Fig. 52. The perpendicular p is
eliminated in the solution and for that reason is not lettered. The
other parts of the triangle are to be so lettered that (1) two of the
given parts shall be in the first triangle and (2), whenever pos-
sible, two of the required parts in the second triangle.
It will be found, that when Ajbp^OT AJ3f ,(two parts including
a third part) are given, there is a choice of two ways in which the
diagram may be lettered; and that one should be taken which
brings the two required parts, when only two are required, in the
second triangle.
PIG. 52.
88
SPHERICAL TRIGONOMETRY.
When two parts and a part opposite one of them are given, the
perpendicular can be drawn in but one way.
There is no ambiguity about the quadrant in which the com-
puted parts belong, as the two rules concerning right triangles
(Arts. 98 and 99) apply to both the right triangles.
Note that in double solutions is in the same quadrant as $, &
in the same quadrant as <f>'.
105. Case I: Given A, b, c, required E, a Let two of the
parts of the first triangle be A
and b, as in Fig. 53, making the
two required parts fall undivided
in the second triangle.
From the first triangle, by
Napier's rules,
tan <f> = cos A tan ~b, (1)
*'=c-$. (2)
To find a: Associate a and <j>' with b and <f> :
cosa=cos<f>' cosp,
cos & = cos <f> cos p.
Eliminating p by division,
cos a = cos b cos </>' sec </>.
To find B: Associate B and </>' with A and <f> :
sin <' = cot B tan /?/
sin 4> = cot A tan p.
(3)
Eliminate tanp:
cot B = cot A sin <f>' cosec <,
cos B = cot a tan <f>' (check).
(4)
(5)
SPHERICAL TRIGONOMETRY. 89
If the angle C is also required, we find 6 and tf, using the given
parts and the two auxiliary parts <f> and <f>'.
cot 0= cos & tan A. (6)
sin p = cot & tan <',
sin p = cot 0tan<.
Eliminating sinp,
cot tf^cot 6 tan < cot <', (7)
0=0+0*. (8)
The whole computation may be checked by the sine formula,
sin A _ sin B _ sin C /o^
sin a sin 6 sin c *
106. If tan (/> from (1) is negative, and A is obtuse, the per-
pendicular will fall to the left of A on the side BA produced, in
which case < will be negative and less than 90 ; this case is shown
in Pig. 54.
M fl A C tf fa
FIG. 54. FIG. 55.
If tan (f> from ( 1 ) is negative, and A is acute, the perpendicular
will fall to the right of A (either upon AB or AB produced), in
which case <f> will be positive and lie in the second quadrant ; this
case is illustrated in Fig. 55.
The same result will be reached, however, if, in all cases, the
auxiliary <f> is taken in the first or second quadrant according to
90 SPHERICAL TRIGONOMETRY.
the sign of tan </>, and the proper signs of the functions of the
auxiliary arcs <f> and c (/> are observed.
107. Case I : Second Solution, Given A, 6 and c. This prob-
lem is of importance in navigation, where, however, only the side
a and the angle B are required.
The angle B is then usually taken from " azimuth tables," com-
puted for that purpose, and the side a is computed from the
formula
cos a = cos & cos c + sin & sin c cos A,
after substituting 1 2 sin 2 ^A for cos A, giving
cosa=cos(& c) 2 sin & sin csin 2 |A; (10)
or else by putting cos A = 1 vers A., giving
cosa=cos(& c) sin 6 sin c vers A. (11)
The following example is solved by both methods, as well as by
the formulae of Art. 93. cfc'
Note, in the first solution, tan <f> is negative an<Lis taken in the
second quadrant, where in reality it belongs from the given parts.
If, however, <f> is taken as a negative angle, its value would be
- (59 28' 39") ; <' becomes 124 57' 9". Following these parts
through the computation, we get A and B from their functions,
in agreement with the results already obtained by taking </> in the
second quadrant. Cot 9 will be negative as before, and will have
to be taken as a negative angle, to be in the same quadrant as <f> ;
cot & will still be negative and must be taken in the second quad-
rant with (j> f ; thus
=-(68 49' 23")
&- 114 40' 9"
e+tf= 45 50' 46"
as before.
SPHERICAL TRIGONOMETRY.
91
o
)0
co
o
CO
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ft ft
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05
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SPHERICAL TRIGONOMETRY.
CO T
i^ a
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to rH C
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SPHERICAL TRIGONOMETRY.
93
1. Given A = 50'
6 = 59
c=109
2. Given 5=102'
a= 98
c= 99
Examples.
10' 10"
29 30
39 40 find a, B, and C.
Ans.
55' 4"
8 18
9 48 find &, A, and (7.
Ans.
a= 69 34' 56"
#= 44 54 31
(7 = 129 29 52
find c, A, and B.
Ans.
17' 58"
16 48
6 20
3. Given C = 99 44' 46"
a=81 2 30
6 = 99 58 38
3' 0"
43 8
30 24
111. Case II: Given A, B and c, to Find C and a. Letter the
diagram (Fig. 56) so as to bring the required parts (7 and a in the
second triangle. Here one of the given angles, B, is divided, and
we may first deduce 6 and ff.
A = 100
(7=101
c=101 (
A= 82
B= 98
cot = cos c tan A,
^ = -0,
cot a=cot c cos 0' sec 0,
cos (7 = cos A sin 0' cosec 0,
cos a = cot 6' cot (7 (check).
(3)
(3)
(4)
(5)
94 SPHERICAL TRIGONOMETRY.
If b is required,
tan < = cos A tan c, (6)
tan <' = tan < cot tan 0', ( 7 )
&:=< + </>'. (8)
The whole computation is to be checked by the sine formula.
The problem, however, never comes in navigation, and very sel-
dom, if ever, in any practical way.
Examples.
1. Given 1 = 115 56' 23"
B= 50 51 43
c= 52 29 24 find a, b, and C.
Ans. a=89 29' 30"
6 = 59 35 50
$' (7 = 45 30 35
2. Given b = 93 -96' 2"
A = 97 17 38
= 100 10 54 find 5, a, and c.
Ans. 5 = 94 48' 16"
a = 96 33 58
c = 99 40 34
3. Given a= 99 40' 48"
5=114 26 50
C= 82 33 31 find A, b, and c.
Ans. A= 95 38' 4"
6 = 115 36 45
c= 79 10 30
112. Case III: Given a, b and A, to Find B and c. The per-
pendicular can be drawn in only one way, as in Fig. 57.
tan < = cos A tan b. (1)
cos a= cos <' cos p,~]
cos & = cos < cos p. J
SPHERICAL TRIGONOMETRY. 95
Eliminating cos p,
cos </>' = cos (j> cos a sec &, (2 )
c = **', (3)
sin <'=cot B tan p,'
sin </> cot A tan>.
Eliminating tan p,
cot J5 cot A sin <' cosec </>. (4)
It is important to note that <j>', being found from its cosine, may
be either a positive or a negative arc, and hence there are two solu-
tions possible. This is always the case in spherical triangles
where two of the given parts are an angle and a side opposite.
This problem is of frequent occurrence in navigation, where,
however, only the parts B and c are required.
If the angle C were required, it would be computed from the
formulae
cot = cos b tan^l, (6)
cot e f = cot tan <f> cot <', (7)
(8)
Check by the sine formula.
113. The following example of Case III is solved completely
and checked :
96
SPHERICAL TRIGONOMETRY.
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g
SPHERICAL TRIGONOMETRY.
97
I. Given a=99 c
6 = 64
4=95
40' 48"
23 15
38 4
Examples.
find E, c, and (7.
Ans.
B= 65
c=100
33'
49
9"
28
C= 97 26 26
2. Given a=
6 =
A=
40
29
find B, c, and C.
Ans.
37' 20"
38 42
27
40
40
1
22
53
114. Case IV:
5' 26"
22 7
42 34
5= 42 C
c = 153
(7=160
# = 137
c'= 90
<7' = 50
Given A, B and a, to Find c and &.
tan <f> =cosB tan a,
sin $'= sin < cot A tan #,
cos b = cos a cos <' sec <,
c = <p -\- <$>' and
is determined by its
sine, it may be in either the first
or the second quadrant; and c,
therefore, has two values, as noted.
The corresponding two values of
b are determined from the two
values of 4,'.
The angle, C, if required, is determined from the formulae :
cot 6 = cos a tan B, (5)
cot & - cot 6 tan <f> cot </>', ( 6 )
C=(0+tf) or 0+(180-0'). (7)
This problem does not occur in navigation, and is of rare occur-
rence elsewhere.
8
Since
98 SPHERICAL TRIGONOMETRY.
Examples.
1. Given 4 = 115 36' 45"
B= 80 19 12
6= 84 21 56 find a, c, and (7.
Ans. 0=114 26' 50"
c= 82 33 31
C= 79 10 30
2. Given a=53 18' 20"
5 = 46 15 15
4 = 79 30 45 find &, c, and (7.
Ans. c = 50 24' 57"
(7=70 55 35
& = 36 5 34
3. Given B = 76 41' 13"
4 = 117 44 36
a=126 17 22 find I, c, and C.
Ans. 6 = 117 35' 35"
c= 24 16 50
C= 26 50 24
&' = 62 24 25
c' = 120 53 50
(7 = 109 34 34
115. Case V: Given the Three Sides, a, b and c
4t this problem is most conveniently and accurately solved by
means of the tangents of the half -angles (Art. W-), especially
when all three angles are required.
It is one of the most frequent problems of navigation (where,
however, only the angles 4 and B are required) and, for reasons
peculiar to the problem, 4 is found from sin 2 4 (Art. 88), and
B from cos 2 # (Art. 89).
The tangent formulae may be put in a more convenient form by
multiplying the expressions of Art. 90 by
sin(s a] sin(s b) -, sin(s c)
sin(s a) ' sin(s b) sin(s c} '
SPHERICAL TRIGONOMETRY. 99
respectively, giving
tan B=
, ~
sm(s-a) sm(s-b)
P
sin (5 c) 9
where
p /sin (s a] sin (s b ) sin (s c)
V sins
Example: Given a=56 37', 6 = 108 14', c=75 29'.
a = 56
& =108
c =75
37
14
29
0,
cosec 0.30933 . 6 33.0
sin 9.95198
sin 9.31549
sin 9.84707
s =120
10
56=
63
11
44
33
56
41
P . .
2)9.17774
9.58887
U =
i* =
io =
23
61
28
25
56
53
55
54
28
tan
tan
tan
9.63689
0.27338
9.74180
These results should be checked by the sine formula.
100 STEREOGRAPHIC PROJECTIONS.
STEREOGRAPHIC PROJECTIONS.
THE FOLLOWING CHAPTERS, TOGETHER WITH THE PRECEDING METHODS OF SPHER-
ICAL TRIGONOMETRY ARE ESSENTIAL TO THE STUDY OF NAVIGATION; AND,
AS A PROFESSIONAL SUBJECT, THEY SHOULD BE MOST THOROUGHLY UNDER-
STOOD BY ALL MIDSHIPMEN.
CHAPTEE XII.
116. Definitions. The projection of the sphere on the plane of
one of its great circles, when the point of sight is taken at one of
the poles of the great circle on which the projection is to be made,
is called a stereo graphic projection.
The plane on which the projection is made is called the primi-
tive plane, and the circle the primitive circle.
The polar distance of a point on the sphere is its angular dis-
tance on the surface of the sphere from one of the poles of the
primitive circle.
The polar distance of a circle is the angular distance of any
point of its circumference from either of its own poles.
117. The inclination of a circle is the angle between its plane
and the primitive plane.
Let WBED (Fig. 59) be a great circle cut from the sphere by
the primitive plane, N and 8 its poles, the pole 8 being taken as
the point of sight.
Let P! be any point of the sphere, and NP^BS a great circle
cut from the sphere by a plane through P intersecting the primi-
tive plane in the straight line BOD. Then NP t is the polar dis-
tance of P 1 ; and p ly the point in which SP 1 pierces the primitive,
is the stereographic projection of the point P x .
STEREOGRAPHIC PROJECTIONS',
101
The angle NSP^ is measured by one-half the polar distance of
P l ; hence Op^ the distance of p from the center of the primitive
circle, is
Op^-rtan^p,
where p is the polar distance, and r the radius of the primitive
circle.
N
^ .
\ 1 S~~ -~
! \ !//' "^
\ i 7 ^/
A. j^:
<
.\ /<|0
X. ^
\l
118. Let Cd-R (Fig. 59) be a small circle of the sphere, its
plane making an angle <j> with the primitive plane. Its pole P
is the extremity of the diameter of the sphere, OP, passing
through the center of the circle perpendicular to its plane.
The -polar distance of the circle is PQ PR. Its inclination,
(f>, to the primitive WEED is measured by the arc NP, which
measures the angle between their poles, since the angle between
the planes is equal to the angle between the normals ON and OP
drawn from to the two planes.
102
STEREOGPA^HIC PROJECTIONS.
The line WE, the intersection of the plane passed through NO
and OP with the primitive plane, is called the line of measures of
the circle QQ^R.
119. Let NSWE (Fig. 60) be a circle cut from the sphere by a
plane through NS and OP, the axes of the primitive circle and the
small circle QR, respectively, and WE the straight line in which
it intersects the primitive plane. Then WE is the line of measures
of the circle QR.
N
P being the pole of the circle QR, PQ = PR is its polar distance,
and PN its inclination (since the angle between two circles is
equal to the distance between their poles) . 8 is the point of sight,
and q and r the projections of the extreme or principal elements
of the oblique circular cone SQR, which is formed by the pro-
jecting lines of all points of the circle QR.
Denoting the polar distance of the circle QR by p, and its
inclination by <, we have, in the two right triangles SOq and SOr
STEEEOGRAPHIC PROJECTIONS.
103
-at" the distances on the line of measures from the center of the
primitive circle to the projected extremities of the diameter of QR.
120. The Stereographic Projection of a Circle is a Circle.
Let QR (Fig. 61) be any circle, P the vertex of a cone tangent to
the sphere along QR, and Q any point in QR. Pass a plane
through SQP, cutting from the primitive a line, pq, which is the
projection of the element PQ; cutting a tangent plane at 8 in ST,
which is parallel to pq; and from the sphere a circle (not shown
in the figure) to which PQ and ST are tangents, 8Q being the
chord of contact. Fig. 62 represents the circle omitted in Fig. 61,
rotated into the plane of the paper.
104 STEREOGRAPHIC PROJECTIONS.
Draw PK parallel to pq (Fig. 62) ; it is also parallel to TS,
and the alternate interior angles TSQ and QKP are equal. But
as T8 and TQ are tangents from a common point T, they are
equal, and the angle TSQ = angle TQS Bangle KQP. Hence
PK=PQ.
K-, .,?
FIG. 62.
The triangles Sqp and SKP are similar, hence
or pq = PQ - , which is equal to a constant, since PQ,
pS and PS are all fixed lengths.
Since pq is constant, the point p being fixed, it follows that the
locus of q is a circle whose center is the projection of the vertex
of the tangent cone.
STEREOGRAPH: c PROJECTIONS.
105
We shall see hereafter that this gives us a simple construction,
to find the center of the projected circle.
121. Angles on the Sphere are Unaltered in the Stereographic
Projection. Let MR and MT (Fig. 63) be tangents to two
curves on the sphere at the common point M. (These curves are
S "*--..
FIG. 63. T
not shown in the figure.) Let these tangents be projected on the
primitive plane by the planes RIMS and TMS, respectively, in the
lines mr and mi, and let MaS and Mis be the circles cut from the
sphere by these planes; also let the lines cut from the tangent
plane at 8 be SR and ST. Since the tangent plane is parallel to
the primitive, the lines SR and ST will be parallel to mr and mt,
and the angle RST-rmt.
106
STEREOGRAPHIC PROJECTIONS.
Join RT. Now, since R8 and RM are tangents to the same
circle, they are equal; and for the same reason TM=TS; hence
the triangles RMT and R8T are equal, and the angle RMT=
RST=rmt. That is, the angle between two lines drawn tangent
to the sphere at a common point is equal to the angle between their
projections.
Y
FIG. 64.
/
As the angle between two curves on the sphere is measured by
the angle between the tangents at the common point, it follows
that angles on the sphere are unaltered by 'the stereographic pro-
jection.
122. The principal properties of the stereographic projection
are those just proved; viz.:
1. That all circles are projected as circles.
2. That angles are not altered.
STKREOGRAPHIC PROJECTIONS.
107
These properties enable us to construct the projections of
spherical triangles by convenient and simple methods.
123. To Project Any Circle of the Sphere, Given its Polar Dis-
tance and the Projection of its Pole. First Method. Let NESW
be the primitive circle (Fig. 64), and Z the projection of the
point diametrically opposite the point of sight, the latter lying
in the axis of the primitive circle at a perpendicular distance r
from Z, below the primitive plane.
Let P be the projection of the pole of the given circle ; draw the
diameters NS through P, and WE at right angles. The diameter
NPS is the projection of the great circle through P; and we re-
volve it about the diameter N8 into the plane of the primitive,
bringing tlie point of sight to W. NS is the line of measures, and
108
STEREOGRAPH: c PROJECTIONS.
Q } the intersection of WP (produced] with NQE, is the revolved
position of the pole P.
Lay off from Q, Qr and Qt equal to the given polar distance,
and draw Wr and Wt, cutting the line of measures in r and t 19
the extremities of the diameter of the required projection (Art.
119), and c, the middle point of r 1? is its center.
f
Second Method. Find Q, as in the first method, take Qr (Fig.
64) equal to the given polar distance, and draw Wrjr, giving one
extremity of the diameter at r v . Draw rV tangent at r, meeting
ZQV in V, the vertex of the tangent cone, and join VW, cutting
NS in c, the center of the required circle (see Art. 120, Fig. 61).
Draw the circle with radius c/v
This method is most convenient when the pole is on the circum-
ference of the primitive circle, as in Fig. 65. Here we simply
draw the diameter NS through P and WE at right angles, laying
off the polar distance Pr, and drawing the tangent at r to intersect
NS in the center c.
STEREOGRAPHIC PROJECTIONS. 109
124. To Project a Great Circle. When the polar distance is
90, the circle is a great circle and passes through W and E (Fig.
66), which are each 90 from P.
Proceed as for a small circle to find Q, lay off 0r = 90, and
draw Wr t r. The tangent cone to the sphere becomes in this case
a tangent cylinder, of which ZQ is the axis ; and the line from W,
parallel to ZQ, cuts the line of measures at c, the required center.
Draw the circle with radius cr^ cW.
Since W, r and E are three points of the required circle, a per-
pendicular to the middle point of Wr^ will intersect N8 in c, a
much less convenient way of getting c.
125. To Find the Locus of the Centers of the Projections of
All Great Circles Passing Through a Given Point. Let P (Fig.
67) be the given point through which the projections of great
circles are to pass; draw the diameters NFS and WE at right
angles, and draw WPQ to Q.
1. The projections of all great circles through P must also pass
through a point 180 from P; hence draw the diameter Qr, and
draw Wr, cutting NS, the line of measures, in T; then T is the
projection of the point 180 from P.
Since all the required circles pass through P and T f their center
must lie on the straight line perpendicular to PT at its middle
point c; the line is called the line of centers.
2. Since a great circle may always be drawn through the points
W, P and E (Art. 124), the point c may be found by drawing a
perpendicular to WP at its middle point, intersecting N8 in c.
3. The triangle WcP is isosceles, and the angle PWs=ihe angle
WPS, which is measured by J(90 +QN) =%QNW; i. e., the arc
QEs = QNW. Hence lay off QEs=QNW, and draw Wcs. This
is equivalent to laying off a polar distance QNW from Q; and
thus the line of centers is the projection of a small circle passing
through the iwe of sight and having the polar distance QNW
= 180 <, where <f> denotes its inclination.
110
STEREOGRAPHIC PROJECTIONS.
4. From the figure, Wr = QE, and rSs=lSO -QEs = 180
-QNW=QE. Hence lay oft*WSs=2QE f and draw Wcs.
5. Since the angle WEs=QZE=WZr, a line joining E and s
is parallel to Qr, which gives a fifth method.
Of these methods the most convenient are :
( 3 ) Lay off QEs =QNW and draw Ws f
(4) l&joftWS8=2QE.
126. To Draw a Great Circle Through P Making a Given Angle
with NS. The tangent to the required circle at P (Fig. 67)
makes the required angle (x) with PZS (Art. 121) ; the per-
pendicular to the tangent (i. e., the radius), makes with PZS the
angle 90 x. Hence construct ZPC = 9Qx, intersecting the
line of centers in C, the center of the required circle.
STEREOGRAPHIC PROJECTIONS.
Ill
Note that the projection of a great circle always meets the
primitive circle at the extremities of a diameter, LK, of the
primitive circle.
127. To Find the Pole of a Given Circle. 1. Let Wr^E (see
Fig. 66) be a given great circle. Draw the diameters WE and NS
at right angles, and draw Wr r. Lay off rQ = 9Q and join WQ,
cutting NS in P, the required pole.
2. Let r^ (Fig. 64) be a given small circle; through its center
c draw the diameter NS, and WE at right angles. Draw Wr^ and
Wt 19 intersecting the primitive circle in r and t. Bisect the arc
tQr in Q, and draw WQ intersecting NS in P, the required pole.
128. Given the Projected Arc of a Great Circle, to Find its
True Length. Let NM (Fig. 68) be the arc to be measured.
Fing the pole X of this arc, and draw the straight line XMm;
then Nm is the true length required. For XMm is the projection
112 STEREOGRAPHIC PROJECTIONS.
of a small circle which passes through the point of sight; i. e.,
through a pole of the primitive ; and since it passes through the
poles of two circles, NM and the primitive, it makes equal angles
with them, the triangle NMm is isosceles, and NM=Nm.
The distance Nm might also be found by drawing the tangent
Me to meet NS at c; then a tangent to the primitive through c
meets the primitive at ra. In other words, c is the vertex of a
tangent cone of which a small circle passing through M and m
(pole at N) is the base. But this method would generally be in-
convenient.
The solution of the reverse problem is obvious, to lay off the pro-
jected length NM on a given circle Nr^S, from its true length Nm,
by simply drawing from the pole X, Xm to m, cutting the given
circle in M .
129. Spherical Triangles on the Earth Regarded as a Sphere.
Definitions. The axis of the earth is the diameter about which
the earth revolves.
The north and south poles are the extremities of the earth's axis.
The equator is a great circle of the earth perpendicular to its
axis.
Meridians are great circles of the earth passing through its
poles, and are therefore perpendicular to the equator.
The latitude of a point on the earth's surface is its angular dis-
tance north or south of the equator (measured on the meridian
passing through the point) .
The longitude of a point on the earth's surface is the angle at
the poles between the meridian of the point, and a fixed meridian
taken as the origin of longitudes, as the meridian of Greenwich,
or of Washington Longitude is reckoned from the zero meridian,
positive to the west, negative to the east, from O h to 12 h , or from
to 180.
130. To Find the Great Circle Distance between Two Given
Points. The angle between the meridians of the two planes is
STEREOGRAPHIC PROJECTIONS.
113
obviously the algebraic difference of their longitudes. The two
adjacent sides are the co-latitudes of the two places ; and we have
two sides and the included angle (Fig. 69) to project the triangle,
measure, and compute the desired parts.
FIG. 69.
Example : Find the great circle distance from Cape Flattery,
Long. (AJ =8 h 20 m W., Lat. (LJ =48 N., to Java, Long. (X 2 )
= 7 h 28 m E., Lat. L 2 = 9 S.
Take the meridian of Cape Flattery as the primitive circle (Fig.
69) and P as the North Pole of the earth. Lay off PM 1 = 90 L^
9
114
STEREOGRAPHIC PROJECTIONS.
= 42. Draw through P a great circle making the angle 123
with PMi (or 33 with PP') . See Art. 126. This is conveniently
done by laying off P'Q'c 114, twice the angle cfP', then draw-
ing PCC-L to c the center of the required circle PM 2 P'. Find its
pole p by Art. 127. Make PQs=99 (the north polar distance
of Java, 9 S.), and draw p s cutting off PM 2 , the projected length
of 99. Draw the diameter M^m^ and the diameter perpendicular
to it; then find the center of the great circle through MJl z by
perpendicular bisector of chord M^M Z) meeting line of centers at
c 2 , and draw MJtt z m^ The triangle PM^M 2 is the projection of
the given triangle.
After finding the pole, p 2 , of M^M 2 m^ the true length of M^M.,
is given on the primitive circle (Art. 128) by the arc D intercepted
between p^M^ and p 2 M 2 (produced), about 118 30' . The angle
PM^M 2 is measured by the projected arc mp of the great circle
90 from M 19 the true length of which is marked M^ on the primi-
tive circle, about 70.
131. Numerical Computation. From Art. 105, Spherical
Trigonometry, calling
\ 2 -Xi=A, 6 = 90-L 2 , c=90-L 1 ,
we get
tan (f> = cos ( A 2 Aj ) cot L 2 ,
$' = (90 -LJ -$ =
cosZ> = sinL 2 sec < sin(
cot(A 2 Ajcosec
L! 48 00' 00"
X 2 \,=123 00' 00"
^2
4
_9 00' 00"
= 73 47' 8"
cos
cot
tan
9.7361 In
0.80029w
cot 9.81252n
D
M,
D
0.53640
47' 8"
=118 30' 35"
= 70 23' 30"
=7110.6 geographical miles.
sin 9.19433w
sec 0.55404
sin 9.92943
cos 9.67880w
esc
cos
0.01763
9.72160n
cot 9.55175
STBREOGRAPHIC PROJECTIONS. 115
The values found from the projection agree as closely with the
computed values as could be expected from the scale of the pro-
jections (Fig. 69).
Examples.
Project the triangle and compute the great circle course
between :
(1) San Francisco, L 1 = 37 47' 48" N., A 1 = 8 h 9 ra 43 s W. ;
Manila, L 2 = 14 35' 25" N., A^=8 h 3 m 50 s E.
Ans. = 100 '50' 55".
M^N. 61 45' 34" W.
(2) New York, L=40 40' N., A= 4 h 55 m 54 s AY.;
Cape Good Hope, L = 33 56' S., X= l h 13 m 55 s E.
(3) San Francisco, L = 37 47' N., X= 8 h 09 m 43 s W.;
Sydney, L = 33 52' S., A=10 h 04 m 50 s E.
132. The spherical coordinates of a point on the earth, latitude
and longitude, as defined in Art. 129, can be determined only by
astronomical observations of the heavenly bodies, the sun, moon,
planets and fixed stars. ^ \ (fc^ftf
^In such observations these 01103 cotes are seen projected on the
background of the sky, and appear to have a revolution about an
axis from east to west. This apparent revolution is due to the
rotation of the earth on its axis in the opposite direction, but we
shall regard the apparent revolution of the heavenly bodies as real.
133. The Celestial Sphere. The celestial sphere is an imagi-
nary spherical surface of indefinite radius, having the eye of the
observer at its center, and upon which the heavenly bodies are
projected.
The poles of the sphere are the points in which the earth's axis
produced meets the celestial sphere, and about which the celestial
sphere apparently revolves once in 24 hours.
The equinoctial (or celestial equator) is the great circle 90
from the poles obviously the same as the great circle formed by
the indefinite extension of the earth's equator.
116 STEREOGRAPHIC PROJECTIONS.
Hour circles are great circles through the poles or the great
circles produced by the indefinite extension of the earth's
meridians.
The zenith and nadir are the points of the celestial sphere
vertically overhead and vertically underneath the observer. They
may be conceived to be the points in which the celestial sphere is
pierced by the indefinite extension of the radius of the terrestrial
sphere -through the position of the observer.
The celestial horizon is the great circle of which the zenith and
nadir are the poles ; or it is the great circle cut from the celestial
sphere by the tangent plane to the earth at the observer's position.
Since the radius of the celestial sphere is infinite, we may sub-
stitute for the tangent plane, a parallel plane through the center
of the earth, for these two planes will meet the celestial sphere
in the same great circle, which is the celestial horizon. (See Fig.
71, Art. 135.)
Vertical circles are great circles passing through the zenith and
nadir.
The celestial meridian (of a place) is the hour circle passing
through the zenith and nadir ; it may also be defined as the vertical
circle passing through the poles. The celestial meridian is ob-
viously the great circle formed by the indefinite extension of the
plane of the terrestrial meridian of a given place.
The prime vertical is the vertical circle passing through the east
and west points of the horizon, and is perpendicular to the
meridian and to the horizon.
The declination (d) of a point of the celestial sphere is its
angular distance north or south of the equinoctial, measured on
the hour circle passing through the point. It is marked north or
south like latitude on the earth.
The polar distance (p) of a point of the celestial sphere is its
angular distance from the pole; p = 9Q d if the declination is of
STEREOGRAPHIC PROJECTIONS. 117
the same name as the pole from which it is reckoned, otherwise
p = 90 + d.
The altitude (h) of a body is its angular distance above the
horizon.
The zenith distance (z) of a body is its angular distance from
the zenith; z = 90 -h.
The hour angle (t) of a point of the celestial sphere is the angle
at the pole between the meridian of a place and the hour circle
passing through the point.
The azimuth (Z) of a body is the angle at the zenith between
the meridian of a place and the vertical circle passing through the
body.
The amplitude of a body is the arc of the horizon between the
east or west point and the body when in the horizon.
The ecliptic is the great circle of the celestial sphere which is
the apparent path of the sun, due to the real motion of the earth
around the sun.
The vernal equinox, or the first point of Aries, is the point in
which the ecliptic cuts the equinoctial, as the sun passes from
south tojnorth declination.
The right ascension of a heavenly body is the angle at the pole
between the hour circle passing through the body and the hour
circle through the vernal equinox.
134. The Astronomical Triangle. The spherical triangle
whose vertices are the pole, the zenith and any point M on the
celestial sphere is called the astronomical triangle. The angle at
the pole P is the hour angle (t), the angle at Z is the azimuth
(Z), the angle at the point M is the position angle. The side
PZ is the co-latitude (90 -L) ; the side PM is the polar distance
from the elevated pole, and is equal to 90 d when the latitude
and declination have the same name, to 90 +d when they have
different names ; the side ZM is the zenith distance of the point,
or the co-altitude, 90 h.
118
STEREOGRAPH ic PROJECTIONS.
The relation of the astronomical triangle to the definitions of
Art. 133 is shown by the diagram of Fig. 70 :
NWS = Horizon.
Mrrii Vertical circle of M.
Mm, = Altitude of M.
= Azimuth of M.
PQP'Q' = Meridian.
(Q'WQ = Equator.
I PMP' = Hour circle of M .
mM Declination of M .
[ MPZ = Hom angle of M.
Either of these two groups forms &' system of spherical co-
ordinates, by which the position of a body on the celestial sphere
is known with reference to the observer's position :
1. When its hour angle and declination (or polar distance) are
known.
2. When its azimuth and altitude (or zenith distance) are
known.
135. The latitude of a place on the earth is equal to the declina-
tion of its zenith or to the altitude of the elevated pole.
STEREOGRAPHIC PROJECTIONS.
119
Let pzq (Fig. 71) be the terrestrial meridian of a place z, pp^
the axis of the earth, and q^Oq the equator.
Extend the axis of the earth, the equator, and the radius (or
vertical) of the place, to meet the celestial sphere in P, Q and Z.
The plane NOS, parallel to the tangent plane nzs at z, is the
celestial horizon.
The arc qz, which is by definition the latitude of z, is obviously
equal to QZ, which by definition is the declination of the point Z,
the zenith.
FIG. 71.
The elevated pole (P) is the pole visible above the observer's
horizon (the north pole for places in north latitude, the south pole
for places in south latitude) .
The arc NP, the altitude of the elevated pole P, is equal to QZ,
the latitude of z, since each is equal to 90 less the arc PZ. The
arc PZ, included between the zenith and the pole, is therefore
90 L, where L is the latitude of a place.
136. To Find the Latitude of a Given Place. The latitude of
a given place (i. e., the declination of its zenith) is most easily
found from the observation of the altitude of the sun or other
120
STEREOGRAPHIC PROJECTIONS.
celestial body when it is just on the meridian of the place. Sup-
pose the meridian altitude of such a body to have been measured
and reduced by various small corrections to give the true altitude
above the celestial horizon. The declinations of the brighter
celestial bodies are tabulated in the American Nautical Almanac.
With these two data the method of finding the latitude is shown
in Fig. 72, where the celestial sphere is projected on the plane of
the meridian of the given place. ZSNaN is the meridian, Z the
zenith, NWS the horizon, P the elevated pole, Q^Q the equator,
and QZ the latitude. If m is the observed body, then QM is its
declination and m^Z its zenith distance = 90 h : . We have from
the figure
If the body is in south declination, as at ra 2 , then
L=ZM z -QM 2 =Z 2 +(-d 2 ),
STEREOGRAPHIC PROJECTIONS.
121
south declination in this case being negative (the latitude being
north).
If the body is north of the zenith, as at M 3 , then
FIG. 73.
137. Projection of the Astronomical Triangle. Case I : Given
L = 20 N., d=32 N., h = 4:Q, body bearing W., to Project the
Triangle on the Plane of the Horizon. The three sides of the
given triangle are PZ = 70, PM=58, ZM = 50.
Let NE8W be the horizon, then Z will be the center of the
primitive circle (Fig. 73). Draw the diameters NZ8 and WZE
122 STEREOGRAPHIC PROJECTIONS.
at right angles, and take NZ8 as the line of measures. The point
of sight (Na) lies below Z perpendicular to the primitive plane
and at a distance r. The projected pole P lies between Z and N f
and for the purpose of finding this, the point of sight is revolved
about NS into the position W.
Lay off EA = 90 - L, and draw WA cutting N8 in P, the pro-
jected pole. Construct the small circle of 58 polar distance from
P by laying off Ar and Ar' equal to the polar distance 58, and
joining Wr f intersecting NS in r one extremity of the diameter
(Art. 123). The center c is found by the second method of Art.
123, the tangent at / intersecting ZA (produced) in V, and join-
ing WV, cutting NS in c; cr^ is the radius of the required circle.
Lay off EH=9Q -h = 5Q, and draw WH cutting N8 in h;
then Zh is the polar distance 50 from Z. With Zh as a radius
(and Z as center) draw a circle cutting Mr in M. This point
being at the given distance from both P and Z respectively, and
lying west of the meridian NS, is the required point.
Lay off As=ANW, and join Ws> cutting NS in c ; draw the
line of centers of great circles through P, perpendicular to NS
at c . Bisect the chord PM, and draw a perpendicular at its
middle point meeting the line of centers in c, the center of the
required great circle PM.
Draw the great circle ZM, passing through the point of sight,
as a straight line, completing the construction of PZM. The
values of the required quantities t and Z are indicated on the
projection ; the measure of the angle Z by the arc of the primitive
circle included between ZPN and ZM produced, approximately
64 ; the value of the hour angle ZPM=t is measured by the arc
of the primitive circle marked t, approximately 53 45', which
is the true length of Qm the projected arc of the equator, WQE,
intercepted between the two hour circles PZQ and PMm. The
angle PcC l is also a measure of the angle ZPN. (See Art. 126.)
STEREOGRAPHIC PROJECTIONS.
123
138. To Project the Same Triangle on the Meridian. Let
PZSN be the meridian (Fig. 74), and draw the diameters ZWNa
and NWS at right angles. Lay off PZ&3Q-L and draw the
diameters PP' and QQ' at right angles. Since the object is west
of the meridian, the center of the primitive circle is W.
FIG. 74.
From Z, lay off ZH = 90 h, and draw a tangent at H inter-
secting WZ in c, the center of the small circle having a polar dis-
tance of 90 -h (Art. 123, Second Method); with radius cH
draw the small circle HM.
From P lay off Pr=90 -d = 58, and draw a tangent at r,
intersecting PP^ in c x , the center of the small circle of 58.
With radius c^r draw the small circle rM, intersecting HM in M,
which has the given polar distance from both Z and P, re-
spectively.
124
STEREOGRAPHIC PROJECTIONS.
The perpendicular at the middle point of the straight line ZM
intersects NS, the line of centers of Z, in the center for the great
circle ZM. Construct the great circle PM in a similar way, giving
PZM, the projection of the given triangle.
The value of the angle PZM is given by the arc of the primitive
marked Z (about 64), which measures the projected arc of the
horizon Nm, which is in turn the measure of the azimuth. The
hour angle t is measured by the arc of the primitive marked t, the
true length of the projected arc Q'm^ of the equator, about 53 50'.
139. To Project the Same Triangle on the Equator. In the
projection (Fig. 75) the primitive circle is the celestial equator or
equinoctial, and the pole P is its center. " Call the vertical diam-
eter QQ' the projection of the meridian and lay off WA = 9Q L
^70, drawing EA cutting QQ' in Z, the projected zenith. Find
STEREOGRAPHIC PROJECTIONS. 125
c l9 the center of the small circle of (90 ft) =50 polar dis-
tance, and construct the circle; find Pb, the projected length of
the polar distance PM = 58, and draw the small circle inter-
secting the first one in M. Find the line of centers of all great
circles through Z, and the center c of that one which passes
through M. The construction is obvious. The value of t, 54 30',
is measured by the arc of the primitive circle as indicated. The
azimuth Z is measured by the true length of the intercepted arc
Nm of the horizon ENW, indicated by Z on the primitive circle,
about 64.
140. Numerical Solution of the Triangle. This problem is the
time sight, of almost daily occurrence in the navigation of a ship
at sea. The altitude of the observed body is measured with the
sextant, the latitude is assumed as known, and the declination
of the observed body is taken from the tables of the Nautical
Almanac.
To derive the formulae, used in the solution of the triangle, let
t = A, Z = B, in the formulae of Arts. 88 and 89 (Sph. Trig.),
sin o sin c
sin a sin c
Then
c=90-L
s c = %(L + p h)=s f h
where s 1 ' = %(h + L + p) ; substituting these values in the above
formulae,
sin 2 \t cos s' sin (s' h) sec L cosec p.
cos 2 \Z = cos s' cos (s' p)secL sec h.
126 STEREOGRAPHIC PROJECTIONS.
Note that p, the polar distance, is used, and not d.
In making the solution, always use the form given below.
h= 40 00' sec 0.11575
L= 20 00' sec 0.02701 sec .02701
p= 58 00' esc .07158
2s' = 118 00'
s'= 59 00' cos 9.71184 cos 9.71184
s'-h= 19 00' sin 9.51264
s'-p= 1 00' cos 9.99993
2)9.32307
*= 27 18' 12" sin 9.66153
t= 54 36' 24" 2(9.85453
3 h 38 m 25.6 s \Z 32 14' 34" cos 9.92726
Z 64 29' 8"
141. Case II : Given t, d and L, or Two Sides and the Included
Angle. Take L = 39 K, =2 h 20 m =35, d=S K
1. On the Horizon : Let NESW be the horizon (Fig. 76), and
draw NZS and WZE at right angles.
Find ZP, the projected length of 90 -L=EA = 51 ; construct
the projection of the equator, Ws'E, by drawing We parallel to
ZA (Art. 124), intersecting NS at c, the center, giving We, the
radius.
The pole p of the hour circle PM, making an angle of 35 with
PZ, must lie on the equator 35 from the pole of PZ or E; hence
lay off Ee = 35 and draw Pe, cutting the equator at p. Since the
poles and center of any great circle are in the same straight line
as the center of the primitive, draw Zpc^ meeting the line of
centers at c lf the center of the required circle, and construct the
great circle PM. From its pole p draw pm, making the true
length of PM, measured by p'm = 90 -d = 82.
STEREOGRAPHIC PROJECTIONS,
127
Draw ZMriL i the great circle through Z and M, completing the
construction of PZM, and find its pole p 19 laying off m 1 p 1 = 90.
To find the altitude h, draw pji^ giving mji l} the true length
of the projected h (m 1 7lf)=45 50'. The azimuth Z is given
directly by the arc NWm 1 = l2Q.
2. The same triangle projected on the meridian in Fig. 77,
where PZ 90 L as before. Draw the equator Q'WQ, which is
FIG. 76.
the line of centers, as also the locus of the poles, of all great circles
through P. Find the pole of PM making an angle of 35 with
the meridian by taking P'a=35, giving Wp the projected length
of the distance between the poles of the two circles ; find its center
c by taking P'& = 70, twice the angle P'Pc=35. From the
center c, draw PtP' and lay off Pm = 9Q -d = 82. Draw pm
cutting the great circle at M; PM projected length of Pm = 82.
128
STEREOGRAPH ic PROJECTIONS.
Draw a perpendicular to middle point of the chord Z'fiQ inter-
secting the line of centers of great circles through Z at c 1? the
center of ZMm^Na. Find its pole p i by taking ,sz t 90. Draw
FIG. 77.
p^M, giving the true length of the projected altitude Mm^ as
indicated approximately 45 50'. The azimuth PZM is measured
by the true length of the projected arc of the horizon N Wm^
that is, by the arc NQ'z, about 125.
3. The same triangle is projected in the plane of the equator in
Fig. 78, where SPN' is taken as the projection of the meridian,
and the projection of the zenith found by making PZ = 9Q
-L. Draw PMt, making the hour angle ZPM = 35, and find
the point M by drawing pM from pole p of the hour circle PM,
cutting off Mt = S projective length.
Bisect chord ZM as before by a perpendicular meeting the line
of centers at c; then construct ZM meeting horizon produced at
m, and find its pole p by joining cP.
The altitude Mm is measured by A-^ according to usual
STEREOGRAPHIC PROJECTIONS.
method, and is about 45 45'. The azimuth, MZP, is measured by
the true length of the arc of the horizon N Wm, or N'Wz
nearly.
142. Numerical Solution of Case II. By Napier's rules, drop-
ping the perpendicular from M on PZ :
tan <f> = cos t cot d,
sin/t=:sin^ sin(Z/ + <)sec </>,
cot Z = cot t cos (L + <f>) cosec <.
Note that when d has a different name from L, d is negative.
L = 39 00' 00"
cos 9.91336 cot 0.15477
cot 0.85220 sin 9.14356
= 35 00' 00"
= 8 00' 00"
= 80 15 53"
=119 15' 53"
= 45 53' 00"
=125 18' 40"
10
tan 0.76556
sec 0.77178
sin 9.94070
esc 0.00630
cos 9.68917n
sin 9.85604
cot 9.85024w
130 STEREOGRAPHIO PROJECTIONS.
143. Method of St. Hilaire. In the practice of navigation the
value of Z is taken from Azimuth Tables, and h is derived from
the formula of Art. 107 (Sph. Trig.),
sin h = sin d sin L+ cos d cos L cos i,
=cos(Ld) 2 cos d cos L sin 2 1/.
log cos L 9.89050 cos (L-d) 0.85717 (Table 41)
log cos d 9.99575 No. 0.13917
log 2 0.30103
0.18728 sinfc 0.71800 (Table 41)
log sin 2 $t 8.95628 A- 45 53' 23"
No. 0.13917 log 9.14356
The computed value of h is compared with the observed value,
thus furnishing a method of establishing the position of the ship,
instead of computing t by Art. 140.
Log sin 2 ^ may be taken directly from Table 45 of latest
edition of Bowditch's Useful Tables.
144. Case III : Given t, d and h, to Find L. We have given
two sides and an angle (t) opposite co-h, thus furnishing a double
solution. Assume t=32, d=21 N., ^ = 40 W. The projection
is made on the equator, on account of the simplicity of the con-
struction.
Let EQ'WQ be the primitive circle, the equator, which has the
pole P at the center. Draw the diameter QPQ', and EW at right
angles (Fig. 79). Construct the angle QPM =2=32, and PM
as shown in the figure equal to 90 d=89. Prom M as a pole
construct the small circle having a polar distance (90 h) =50,
on which lie all points having the given zenith distance of 50.
Then the two points Z^ and Z 2 , in which this circle of equal alti-
tudes intersects PQ, satisfy the conditions of the problem. Find
the line of centers for M and construct the great circles Z^M and
Z 2 M, completing the two triangles.
STEEEOGRAPHIC PROJECTIONS.
131
Since P is the projection of the North Pole, Z 1? lying between
it and the equator, is the projected zenith of a place in north lati-
tude, while Z 2 , lying at a greater polar distance than 90, is the
projected zenith of a place in south latitude. The true length of
Z^Q is measured by the arc marked L 1 = 66 50' N.; the true
length of QZ 2 , by the arc marked L 2 = 18 S.
FIG. 79.
145. Numerical Solution of Case III. Draw the perpendicular
from M to the side PZ Z 2) and by Napier's rules derive :
tan (j> = cos t cot d,
cos <' = cos <j> sin h coseo d,
/, = 90 -(<#>: </'),
cos Z = tan <f>' tan h.
132
STEREOGRAPHIC PROJECTIONS.
Note that <f>', being found by its cosine, may be either a positive
a negative arc.
or
h = 40 00' 00"
t = 32 00' 00"
d = 21 00' 00" N.
= 65 38' 46"
0'= (42 17' 55")
Z l= 40 13' 38"
Zj= 139 46' 22"
L 1= : 17 56' 41" S.
L,= 66 39' 9" N.
cos 9.92842
cot 0.41582
tan 0.34424
sin 9.80807 tan
esc 0.44567
cos 9.61529
9.92381
cos 9.86903 tan () 9.95899
cos () 9.88280
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