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Eau^I \feq.03,-sT^
HARVARD
COLLEGE
LIBRARY
THE GIFT OF
Miss Ellen Lang Wentworth
of Exeter J Nifw Hampshire
3 2044 097 047 112
LA'^tX-.
'Xa^
--CiT
Cic
\amV\M —
\
/
I
PLANE
TRIGONOMETRY
BY
G. A. WENTWORTH
AuTHOB OF ▲ Sebies jdf Text-Books in Mathematics
SECOND REVISED EDITION
BOSTON, U.S.A.
GINN & COMPANY, PUBLISHERS
1903
HARVARD COLLEGE LIBRARY
GIFT OF
MISS ELLEN L. WENTWORTH
MAY tt 1^39
Copyright, 1882, 1895, 1902, bt
G. A. WENTWORTH
ALL BIGHTS RE8EBYED
PREFACE
In preparing this work the aim has been to furnish just so much
of Trigonometry as is actually taught in our best schools and
colleges. Consequently, all investigations that are important only
for the special student have been omitted, except the development
of functions in series. The principles have been unfolded with the
utmost brevity consistent with simplicity and clearness, and inter-
esting problems have been selected with a view to awaken a real
love for the study. Much time and labor have been spent in devising
the simplest proofs for the propositions, and in exhibiting the best
methods of arranging the logarithmic work.
The author acknowledges his obligation to G. A. Hill, A.M., of
Cambridge, Mass., to Dr. F. N. Cole, of New York, N.Y., to Pro-
fessor S. F. Norris, of Baltimore, Md., and to Professor B. F. Yanney,
of Alliance, Ohio.
G. A. WENTWORTH.
Exeter, N.H.,
January, 1903.
ui
CONTENTS
[The numbers refer to the pages.}
PLAITS TRIGONOMETRY
CHAPTER I. Trigonometric Fdnctions of Acute Angles:
Angular measure, 1 ; trigonometric functions, 3 ; representation
of the functions by lines, 7 ; changes in the functions as the angle
changes, 10; functions of complementary angles, 11; relations of
the functions of an angle, 13; formulas for finding all the other
functions of an angle when one function of the angle is given, 16 ;
functions of 46°, 17 ; functions of 30° and 60°, 18.
CHAPTER n. The Right Triangle :
Given parts of a right triangle, 20 ; solutions withdut logarithms,
20 ; Case I, when an acute angle and the hypotenuse are given, 20
Case II, when an acute angle and the opposite leg are given, 21
Case III, when an acute angle and an adjacent leg are given, 21
Case IV, when the hypotenuse and a leg are given, 22; Case V,
when the two legs are given, 22 ; general method of solving the right
triangle, 23 ; solutions by logarithms, 26 ; area of the right triangle,
27 ; the isosceles triangle, 32; the regular polygon, 34.
CHAPTER in. Goniombtry:
Definition of goniometry, 36 ; positive and negative quantities, 36 ;
co-ordinates of a point in a plane, 37 ; angles of any magnitude, 38 ;
functions of ahy angle, 40; functions of a variable angle, 42 ; func-
tions of angles larger than 360°, 44 ; extension of formulas for acute
angles to angles of any magnitude, 44 ; reduction of the functions of
all angles to the functions of angles in the first quadrant, 47 ; func-
tions of angles that differ by 90°, 60 ; functions of a negative angle,
61 ; functions of the sum of two angles, 63 ; functions of the differ-
ence of two angles, 66 ; functions of twice an angle, 68 ; functions
of half an angle, 68 ; sums and differences of functions, 69 ; anti-
trigonometric functions, 61.
V
vi CONTENTS
CHAPTER IV. The Oblique Triangle :
Law of sines, 64 ; law of cosines, 66 ; law of tangents, 67. Solu-
tions : Case I, when one side and two angles are given, 69 ; Case II,
whdn two sides and the angle opposite one of them are given, 71 ;
Casfe III, when two sides and the included angle are given, 76;
Cas^ rv, when the three sides are given, 80 ; area of a triangle, 85.
CHAPTER y. Miscellaneous Examples:
Problems in Plane Trigonometry : right triangles, 90 ; oblique tri-
angles, 93 ; areas, 98 ; plane sailing, 101 ; parallel and middle lati-
tude sailing, 103; traverse sailing, 106; problems in goniometry,
107 ; solution of single equations, 112 ; systems of equations, 116.
CHAPTER VI. Construction of Tables:
Logarithms, 119 ; exponential and logarithmic series, 122 ; trigo-
nometric functions of small angles, 127 ; Simpson ^s method of con-
structing a trigonometric table, 129 ; De Moivre's theorem, 131 ;
expansion of sin x, cos x, and tan x in infinite series, 135.
FORMULAS 139
PLANE TRIGONOMETRY
CHAPTER I
TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES
SECTION I
ANGULAR MEASURE
As lengths are measured in terms of various conventional
units, the foot, the meter, etc., so different units for measur-
ing angles are employed, or have been proposed.
In the common or sexagesimal system the circumference of
a circle is divided into 360 equal parts. The angle at the
i^ntre subtended by each of these pai'ts is taken as the unit
angle and is called a degree. The degree is subdivided into
60 mimUes, and the minute into 60 seconds. Degrees, minutes,
and seconds are denoted by symbols. Thus, 6 degrees 5
minutes 7 seconds is written 6** 5' 7".
Note. The sexagesimal system was employed by the early Babylonian
astronomers to conform with their year of 360 days.
In the circular system an arc of a circle is laid off equal in
length to the radius. The angle at the centre subtended by
this arc is taken as the unit angle and is called a radian.
The number of radians in 360® is equal to the number of
times the length of the radius is contained in the length of the
circumference. It is proved in Geometry that this number is
1
2 PLANE TRIGONOMETRY
2 TT for all circles, tt being equal to 3.1416, nearly. Therefore
the radian is the same angle in all circles.
The circumference of a circle is 2 tt times the radius.
Hence, 2 tt radians = 360**, and tt radians = 180**.
iQAO
Therefore, 1 radian = ^^^ = 57^ 17' 45",
TT
TT
and 1 degree = t^ radian = 0.017453 radian.
By the last two equations the measure of an angle can be
changed from radians to degrees or from degrees to radians.
iQAO
Thus, 2 radians = 2 x =^^ = 2 x (57^ 17' 45") = 114^ 35' 30".
Note. The circular system came into use early in the eighteenth
century. It is found more convenient in the higher mathematics, where
the radians are expressed simply as numbers. Thus, the angle x means
T radians, and the angle 3 means 3 radians.
Oki the introduction of the metric system of weights and measures at
the dlose of the eighteenth century, it was proposed to divide the right angle
into 100 equal parts called grades, which were to be taken as units. The
grade was subdivided into 100 minutes and the minute into 100 seconds.
This French or centesimaX system, however, never came into actual use.
EXERCISE I
[Assume Tt = 3.1416.]
1. Reduce the following angles to circular measure, express-
ing the results as fractions of tt :
6o^ 45^ l5o^ 195°, n^ 15', 123^ 45', 37^ 30'.
2. How many degrees are there in f tt radians ? J tt radians ?
I TT radians ? 1| ^ radians ? i^y tt radians ?
3. What decimal part of a radian is 1** ? 1' ?
4. How manv seconds in a radian? *
TRIGONOMETRIC FUNCTIONS 3
5. Express in radians one of the interior angles of a regular
octagon ; of a regular dodecagon.
6. On the circumference of a circle of 50 feet radius an
arc of 10 feet is laid off. How many degrees in the angle at
the centre subtended by this arc ?
7. The earth's equatorial radius is approximately 3963
miles. If two points on the equator are 1000 miles apart^
what is their difference in longitude?
8. If the difference in longitude of two points on the
equator is 1°, what is the distance between them in miles ?
9. What is the radius of a circle, if an arc of 1 foot sub-
tends an angle of V at the centre ?
10. In how many hours is a point on the equator carried
by the rotation of the earth on its axis through a distance
equal to the earth's radius?
11. The minute hand of a clock is 3^ feet long. How far
does its extremity move in 25 minutes ? (Take tt = ^.)
12. A wheel makes 15 revolutions a second. How long does
it take to turn through 4 radians ? (Take tt = ^ .)
SECTION II
THE TRIGONOMETRIC FUNCTIONS
The sides and angles of a plane triangle are so related that
any three given parts, provided at least one of them is a side,
determine the shape and the size of the triangle.
Geometry shows how, from three such parts, to construct
the triangle.
Trigonometry shows how to compute the unknown parts of a
triangle from the numerical values of the given parts.
Geometry shows in a general way that the sides and angles
of a triangle are mutually dependent. Trigonometry begins
PLANE TRIGONOMETRY
by showing the exact nature of this dependence in the. right
trianglC; and for this purpose employs the ratios of the sides.
Let MAN (Fig. 1) be an acute
angle. If from any points B, D,
F in one of its sides perpendicu-
lars BC, DE, FG are let fall to
the other side, then the right tri-
angles ACB, A ED, AGF thus
formed have the angle A com-
mon, and are therefore mutually
equiangular and similar. Hence,
the ratios of their corresponding
sides, pair by pair, are equal. That is,
AC AE AG A£ AE AG
BC
BC DE FG
AB AD AF' BC DE FG' AB AD AF
These ratios, therefore, remain unchanged so long as the
angle A remains unchanged.
Hence, for every value of an acute angle A there are certain
numbers that express the values of the ratios of the sides in
all right triangles that have this acute angle A,
There are all together six different ratios :
I. The ratio of the opposite leg to the hypotenuse is called
the Sine of A, and is written sin A.
it. The ratio of the adjacent leg to the hypotenuse is
called the Cosine of A, and is written cos A.
III. The ratio of the opposite leg to the adjacent leg is
called the Tangent of A, and is written tan A.
IV. The ratio of the adjacent leg to the opposite leg is
called the Cotangent of A, and is written cot A.
V. The ratio of the hypotenuse to the adjacent leg is called
the Secant of A, and is written sec A.
VI. The ratio of the hypotenuse to the opposite leg is called
the Cosecant of A, and is written esc A.
TRIGONOMETRIC FUNCTIONS
These six ratios are called the Trigonometric Functions of
the angle A,
To these six ratios are often added the two following func-
tions, which also depend only on the angle A :
VII. The Versed Sine of A is 1 — cos ^, and is written vers A.
VIII. The Coversed Sine of A is 1 — sin ^, and is written
covers A.
In the right triangle A CB (Fig. 2)
let a, b, c denote the lengths of the
sides opposite the acute angles A , B,
and the right angle C, respectively,
these lengths being all expressed in
terms of a common unit. Then,
sin A = - =
tan A=T- =
sec -4 = 7 =
vers -4 = 1 —
_a _ opposite leg
hypotenuse
opposite leg
adjacent leg'
hypotenuse
adjacent leg
b c-b
a
b
cos .4 = - =
cot ^ = - =
CSC -4 = - =
_ ft _ adjacent leg
hypotenuse
adjacent leg
opposite leg
hypotenuse
opposite leg
c
a
. a c — a
covers -4=1 =
c c
EXERCISE n
1. What are the functions of the other acute angle B of
the triangle A CB (Fig. 2) ?
2. Compare the functions of A and B, and show that
sin A = cos B, sec A = esc B,
cos A = sin B, esc A = sec B,
tan A = cot B, vers A = covers J5,
cot A = tan B, covers A = vers B,
6 PLANE TRIGONOMETRY
3. Find the values of the functions of ^, if a, J, c, respec-
tively, have the following values :
(i) 3, 4, 5. (rii) 8, 15, 17. (v) 3.9, 8, 8.9.
(ii) 5, 12, 13. (iv) 9, 40, 41. (vi) 1.19, 1.20, 1.69.
4. What condition must be fulfilled by the lengths of the
three lines a, ft, c (Fig. 2) in order to make them the sides of
a right triangle ? Is this condition fulfilled in Example 3 ?
5. Find the values of the functions of ^, if a, ft, c, respec-
tively, have the following values :
(i) 2 mn, rnP' — n^y m^ + n^, (iii) pqr, qrs, rsp,
,... 2xy ic^ -f- y^ X. V wi?i mv nr
(u) — ^> ^-\-y, -^^' (iv) — y — > —
^ ^ X — y X — y ^ ^ pq sq ps
6. Prove that the values of a, ft, c, in (i) and (ii).
Example 5, satisfy the condition necessary to make them
the sides of a right triangle.
7. What equations of condition must be satisfied by the
values of a, ft, c in (iii) and (iv). Example 5, in order that
the values may represent the sides of a right triangle ?
Given a^ + ft^ = c^ ; find the functions of A and B when :
8. a = 24, ft = 143. 11. a = Vp^ + q^^ h = V2^.
9. a = 0.264, c = 0.265. 12. a = -slp^ + pq^ c=p-\-q.
10. ft = 9.5, c = 19.3. 13. b = 2-\/pq,c=p + q.
Given a^ -\- b^ = c^', find the functions of A when :
14. a = 2 ft. 16. a -f ft = I c.
15. a = I c. 17, a — b = ^ c.
18. Find a if sin A = ^, and c = 20.5.
19. Find ft if cos A = 0.44, and c = 3.5.
20. Find a if tan A = ^^, and ft = 2^.
21. Find ft if cot ^ = 4, and a = 17.
TRIGONOMETRIC FUNCTIONS 7
22. Find c if sec ^ = 2, and h = 20.
23. Find c if esc A = 6.45, and a = 35.6.
Construct a right triangle, given ;
24. c «= 6, tan ^ = |. 26. ft = 2, sin ^ = 0.6.
26. a = 3.5, cos -4 = J. 27. ft = 4, esc -4=4.*
28. In a right triangle c = 2.5 miles, sin A = 0.6, cos -4 ::= 0.8 ;
compute the legs.
29. Construct with a protractor the angles 20*, 40*, and
70*; determine their functions by measuring the necessary
lines, and compare the values obtained in this way with the
more nearly correct values given in the following table :
20'
40"
70"
sin
COB
tan
cot
sec
qsc
0.342
0.643
0.940
0.940
0.766
0.342
0.364
0.839
2.747
2.747
1.192
0.364
1.064
1.305
2.924
2.924
1.556
1.064
30. Find, by means of the above table, the legs- of a right
triangle if ^ = 20*, c = 1 ; also if ^ = 20*, c = 4.
31. By dividing the length of a vertical rod by the length
of its horizontal shadow, the tangent of the angle of elevation
of the sun at the time of observation was found to be 0.82.
How high is a tower, if the length of its horizontal shadow at
the same time is 174.3 yards ?
SECTION m
REPRESENTATION OF THE FUNCTIONS BY LINES
The functions of an angle, being ratios, are numbers ; but
we may represent them by lines if we first choose a unit of
length, and then construct right triangles, such that the
denominators of the ratios shall be equal to this unit.
8
PLANE TRIGONOMETRY
The most convenient way is the following :
About a point O (Fig*. 3)
a centre, with a radius equal
one unit of length, describe
circle, and diaw the liorizont
diameter AA^ and the dia]iiet<
BB^ perpendicular to A A'.
A circle with radius equal t
1 is called a unit circle.
Let A OP be an acute a«ngl€
and let its value (in degrees
etc.) be denoted by x. We maj
regard the angle x as generated by a line OP that revolves
about from the initial position OA to the terminal position OP.
Draw PM ± to OA, PN ± to OB,
In the rt. A OMP the hypotenuse OP = 1.
MP
Therefore, sin a; = -— = MP ; cos x
OM
OP
= OM.
Through A and B draw tangents to the circle meeting OP
produced in T and S, respectively ; then, in the rt. A OAT
and OBS, OA = 1, the leg OB = 1, and the Z OSB = the Z x.
Therefore,
tana; = —— = AT-,
OT ^^
secaj =7-7 = OT]
cotic = — = 55;
OS
cscar = — = 05;
vers x = l — OM = MA ; covers x — 1 — ON = NB,
These eight line values of the functions are all expressed in
terms of the radius of the circle as a unit ; and it is clear that
as the angle varies in value the line values of the functions
will always remain equal numerically to the ratio values.
Hence, in studying the changes in the functions as the angle
TRIGONOMETRIC FUNCTIONS 9
is supposed to vary in value, we may employ the simpler line
^^ ^ (?• values instead of the ratio values.
ndmt
^^^^^ EXERCISE ra
^hi' he.
, . 1. Represent by lines the functions of an acute angle
. "" larger than that shown in Fig. 3.
liiii e If x is an acute angle^ show that :
Tclt 2. sin X is less than tan x.
i^'utei 3. sec x is greater than tan x.
'^ ^f" 4. CSC X is greater than cot x.
Construct the angle x, if :
.. 6. tan a; = 3. 7. cos aj = ^. 9. sin x = 2 cos x,
6. CSC 05 = 2. 8. sin x = cos x. 10. 4 sin x = tan «.
11. Show that the sine of an angle is equal to one-half the
chord of twice the angle.
12. Find x if sin x is equal to one-half the side of a regular
inscribed decagon.
/
^ Given X and y, x + y being less than 90 ; construct :
a'
, 13. The value of sin (a? + y) — sin x,
14. The value of tan (a? + y) — sin (x -\-y)-\- tan x — sin x.
Given an angle x ; construct an angle y such that :
15. sin y = 2 sin sc. 17. tan y = 3 tan x,
16. cos y = i cos a. 18. sec y = esc x,
19. Show by construction that 2 sin ^ > sin 2 ^.
20. Given two angles A and B^ A '\- B being less than 90° ;
show that sin (-4 + 5) < sin A + sin J5.
21. Given sin 05 in a unit circle^. ,find the length of a line
corresponding in position to sin 05 in a circle whose radius is r.
22. In a right triangle, given the hypotenuse c, and also
sin A =m, cos -4 = w ; find the legs.
10
PLANE TRIGONOMETRY
SECTION IV
CHAHGES or THE FUNCTIONS AS THE ANGLE CHANGES
If we suppose the Z-40P, or x (Fig. 4), to increase gradu-
ally to 90* by the revolution of the moving radius OP about
0, the point P moves along the arc
AB towards B, T moves sdong the
tangent A T Bwaj from A, S moves
along the tangent BS towards B,
and M moves along the radius OA
towards 0.
Hence, the lines MP, AT, OT
gradually increase in length, and
the lines OM, BS, OS gradually
decrease. That is.
As an acute angle increases to
90*, its sine, tangent, and secant
also increase, tohile its cosine, co-
tangent, and cosecant decrease.
On the other hand, if we suppose x to decrease gradually,
the reverse changes in its functions occur.
If we suppose x to decrease to 0", OP coincides with OA
and is parallel to BS, Therefore, MP and AT vanish, OM
becomes equal to OA, while BS and OS are each infinitely
long and are represented in value by the symbol oo.
And if we suppose x to increase to 90*, OP coincides with
OB and is parallel to AT Therefore, MP and OS are^each
equal to OB, OM and BS vanish, while AT and Or are each
infinite in length.
Hence, as the angle x increases from 0* to 90*,
sin X increases from to 1,
cos X decreases from 1 to 0,
TRIGONOMETRIC FUNCTIONS
11
tan X increases from to oo,
cot X decreases from oo to 0,
sec X increases from 1 to oo,
CSC X decreases from oo to 1.
The values of the functions of 0" and of 90° are the limiting
values of the functions of an acute angle. It is evident that
for acute angles,
Sines and cosines are always less than 1;
Secants and cosecants ai*e always greater than 1 ;
Tangents and cotangents have all values between and oo.
Remark. "We are now able to understand why the sine, cosine, etc.,
of an angle are called functions of the angle. By a function of any mag-
nitude is meant another magnitude which remains constant so long as the
first magnitude remains constant, but changes in value for every change
in the value of the first magnitude. This, as we now see, is the relation
in which the sine, cosine, etc., of an angle stand to the angle.
SECTION V
FUNCTIONS OF COMPLEMENTARY ANGLES
The general form of two complementary angles is A and
In the rt. AACB (Fig. 5),
A +B:=90''; hence 5 = 90° - A,
Hence, putting 90° — A for B in the formulas on p. 5,
sin A = cos B == cos (90° — A),
cos A = sin B = sin (90* — A),
tan A = cot 5 = cot (90° — A),
cot A = tan 5 = tan(90° — A),
sec A = CSC 5 = CSC (90° — A),
CSC A = sec 5 = sec (90° — A).
12 PLANE TRIGONOMETRY
Therefore,
Each function of an aetUe angle is equal to the co-named
function of the complementary angle.
Note. Cosine, cotangent, and cosecant are sometimes called co-
functions; the words are simply abbreviated forms of complement^ 8
sine, complement's tangent, and complement's secanL
Hence, also,
Any function of an angle between 45* and 90* m^y be found
by taking the co-named function of the complementary angle
between 0® and 45*.
EXERCISE IV
1. Express as functions of the complementaxy angle :
sin 30*. tan 89*. esc 18* 10'. cot 82* 19'.
cos 45*. cot 15*. cos 37* 24'. esc 54* 46'.
2. Express as functions of an angle less than 45* :
sin 60*. tan 57*. esc 69* 2'. cot 89* 59'.
cos 75*. cot 84*. cos 85* 39'. csC 45* 1'.
3. Given tan 30* = J Vs ; find cot 60*.
4. Given tan A = cot A ; find A,
5. Given cos A = sin 2 A ; find A,
6. Given sin A = cos 2 A ; find A.
7. Given cos A = sin (45* — ^ ^) ; find A.
8. Given cot i A = tan A ; find A,
9. Given tan (45* -\- A) = cot A ; find A,
10. Find A if sin A = cos 4 A.
11. Find ^ if cot A = tan 8 A.
12. Find A if cot A = tan nA.
TRIGONOMETRIC FUNCTIONS
13
SECTION VI
RELATIONS OF THE FUNCTIONS OF AN ANGLE
Since (Fig. 6) a^ + ft^ = c\ therefore,
I'-l-. » (0"<0"--
But — = sin ^, and - = cos A ;
c c
therefore, (sin Ay + (cos Ay = 1 ; a
or, as it is usually written for con-
venience,
8in*A + C08*A = 1. [1]
That is : The sum of the squares of the sine and the cosine
of an angle is equal to unity.
Formula [1] enables us to find the cosine of an angle when
the sine is known, and the sine when the cosine is known.
The values of sin A and of cos A deduced from [1] are :
in ^ = Vl — cos^^.
sin
cos
A = Vl -sinM.
Since
abaca
C C C
and since - = sin ^, - = cos A, and - = tan A^
c c ^
therefore.
tan A =
sin A
COS A
[2]
That is : The tangent of an angle is equal to the sine
divided by the cosine.
Formula [2] enables us to find the tangent of an angle when
the sine and the cosine are known.
14
PLANE TRIGONOMETRY
Now
Therefore,
each ha
and
d , h a
- = sm Af - = cos A, -7 = taJi -4>
ft c c
-==cot^, 7 = 8ec^, - = csc-4.
O a h a
sin A X CSC A = 1
cos A X sec A = 1
tan A X cot A = 1
[3]
That is : The sine and the cosecant of an angle, the cosine
and the secant, and the tangent and the cotangent, pair by
pair, are reciprocals.
The equations in [3] enable us to find an unknown func-
tion contained in any pair of these reciprocals when the other
function in this pair is known.
EXERCISE V
1. Prove Formulas [1], [2], [3], using for the functions
the line values in the unit circle given in Sect. Ill, page 8.
Prove that :
2. 1 + tan* A = sec* A. 3. 1 + cot' A = esc* A.
Note. The equations in Examples 2 and 3 should be remembered.
4. cot A =
cos^
sin A
5. siuil sec^ = tan^.
6. sin ' ^os.4.
7. cos A CSC A = cot A,
8. tan A cos A = sin A,
9. sin A sec A cot A =1.
TRIGONOMETRIC FUNCTIONS 15
10. cos A CSC A tan -4=1.
11. (1 - sin^ A) tan* A = sin* A.
12. Vl — cos* -4 cot A = cos A,
13. (1 + tan* A) sin* A = tan* ^.
14. (1 — sin* -4) CSC* -4 = cot*^.
15. tan*^ cos* A + cos* -4 =1. ^
16. (sin* A - cos* ^)* = 1 - 4 sin* A cos* A,
17. (l-tan*^)* = secM -4tan*^.
^ sin ^ . cos -4
18. 7 + -: 7 = sec ^ CSC A,
cos ^ sin ^
19. sin* A — cos* A = sin* ^ — cos* A.
20. sec -4 — cos ^ = sin -4 tan -4.
21. CSC ^ — sin -4 = cos A cot A.
cos ^ 1 + sin ^
22.
1 — sin -4 cos A
SECTION VII
APPLICATION OF FORMULAS [1], [2], [8]
Formulas [1], [2], and [3] enable us, when any one func-
tion of an angle is given, to find all the others. A given value
of any one function, therefore, determines all the others.
Example 1. Given sin ^ = f ; fiiid the other f imctions.
By [1], p. 13, cos A = Vl-J = V| = J VH.
By [2], p. 13, taji^ = S-5-jV5 = |X-^ = -^=fV5.
By [3], p. 14, cot ^ = J V5, sec -4 = J V5, esc -4 = J.
16 PLANE TRIGONOMETRY
Example. 2. Given tan ^ = 3 ; find the other functions.
And by [1], p. 13,
sin'^ -I- cos*^ =1.
If we solve these equations (regarding sin A and cos A as
two unknowni quantities), we find
sin^ = ^^ViO, cos^=tV'^-
Then, by [3], p. 14, cot ^ = i, sec ^ = VlO, csc ^ = i VlO.
Example 3. Given sec -4 = m ; find the other functions.
By [3], p. 14, cos ^ = —
HI
By [1], p. 13, smA=Jl^^ = J^^=:-V^;j^.
% [2], p. 13, tan ^ = Vw^ - 1.
By [3], p. 14,
Cot^ = -^i— Vi;^^^; CSC ^ = -^-r Vm« - 1.
EXERCISE VI
Find the values of the other functions, when :
1. Sin A = If. 5. tan ^ = J. 9. esc ^ = V2.
2. sin A = 0.8. 6. cot ^ = 1. 10. sin ^ = m.
2m
l + m«
2 Trill
3. cos^ = ff 7. cot ^ = 0.5. 11. sin A =
4. cos ^ = 0.28. 8. sec ^ = 2. 12. cos ^ = „
m^ -|- 71*
13. Given tan 45** = 1 ; find the other functions of 45°.
14. Given sin 30° = ^ ; find the other functions of 30**.
TRIGONOMETRIC FUNCTIONS 17
15. Given esc 60° = f VS ; find the other functions of 60^
16. Given tan 15° = 2 — V3 ; find the other functions of 15°.
17. Given cot 22° 30' = V2 + 1 ; find the other functions
of 22* 30'.
18. Given sin 0° = ; find the other functions of 0°.
19. Given sin 90° = 1 ; find the other functions of 90°.
20. Given tan 90° = oo ; find the other functions of 90°.
Express the values of all the other functions in terms of :
21. sin -4. 22. cos ^. 23. tan -4. 24. cot^.
25. Given 2 sin -4 = cos A ; find sin A and cos A.
26. Given 4 sin A = tan A ; find sin A and tan A,
27. If sin ^ : cos ^ = 9 : 40, find sin A and cos A.
28. Transform the quantity tan^-4 + cot^-4 — sin*^ — cosM
into a form containing only cos A. .
29. Prove that sin ^ + cos ^ = (1 + tan A) cos A,
30. Prove that tan A -{- cot ^ = sec -4 x esc A.
SECTION VIII
FUNCTIONS OF 45''
Let A CB (Fig. 8) be an isosceles right triangle, in which
the length of the hypotenuse AB is
equal to 1 ; then -4 C is equal to BC,
and the angle A is equal to 45°.
Since AC^ + BC^ = 1, therefore
2AC^= 1, and ^C = V^ = I- V2.
Therefore, by Sect. II, p. 5,
8in45° = cos45° = iv^;
tan 45° = cot 45° = 1 ; "** H Vi
sec 45° = CSC 45° = V2. »»-* «
18
PLANE TRIGONOMETRY
SECTION IX
FUNCTIONS OF 80"* AND 60"*
Let ABC (Fig. 9) be an equilateral triangle, in which the
length of each side is equal to 1 ; and let CD bisect the angle C
Then CD is perpendicular to -45 and bisects AB.
Hence, AD = i, and CD = Vl - i = Vf = ^ VS.
In the right triangle ADC, the angle ACD = 30*, and the
angle CAD = 60^ Whence, by Sect. II, p. 5,
sin 30° = cos 60° = ^ ;
cos 30° = sin 60° = ^ V3 ;
. tan30° = cot60° = -7= = iV3;
V3 ^
cot30° = tan60°= V3;
sec 30° = CSC 60° = -p = f VS;
esc 30° = sec 60° = 2.
The results for sine and cosine of 30°, 45°, and 60° may be
easily remembered by arranging them in the following form :
Angle
30°
45"
60°
iVl = 0.5
Sine
iVI
iV2
iV3
i V2 = 0.70711
Cosine
iV3
iV2
iVI
i V3 = 0.86603
EXERCISE Vn
Solve the following ejq[uations :
1. 2 cos X = sec X,
2. 4 sij) '" — ""'* «?.
3. tan X = 2 sin x,
4, sec X = V2 tan x.
TRIGONOMETRIC FUNCTIONS 10
5. sin^ aj = 3 cos* a;. 9. sin'o; — cosoj = J.
6. 2 sin* X + cos* a; = |. 10. tan* x — sec x = 1,
7. 3tan*a; — sec*aj = 1, 11, sin a; + V3 cos a; = 2.
8. tan x + cot » = 2. 12. tan* x + c8C*x = 3.
13. 2 cos X + sec x = S.
14. cos* X — sin* aj = sin x.
15. 2 sin X -f cot aj = 1 + 2 cos «.
16. sin* a; + tan*aj = 3 cos* a:.
17. tan X •\-2 cot a: = J esc x.
Note. Wentworth & HilPs Five-place Logarithmic and Trigonometric
Tables have full explanations, and directions for using them. Before pro-
ceeding to Chapter II the student should learn how to use these tables.
Table VI is to be used in solutions withovJt logarithms. This four-
place table contains the natural functions of angles at intervals of V,
The decimal point must be inserted before each value given, except when
it appears in the values of the table.
CHAPTER II
THE RIGHT TRIANGLE
SECTION X
THE GIVEN PARTS
In order to solve a right triangle, two parts besides the
right angle must be given, one of them at least being a side.
The two given parts may be :
I. An acute angle and the hypotenuse.
II. An acute angle and the opposite leg.
III. An acute angle and the adjacent leg.
IV. The hypotenuse and a leg.
V. The two legs.
SECTION XI
SOLUTION WITHOUT LOGARITHMS
The following examples illustrate the process of solution
when logarithms are not employed.
Case I
^ Given A = 43^ 17', c = 26; find B, a, b.
1. B = 90^ - ^ = 46M3'.
2. - = sin^; ,', a = csin.A.
c
3. - = cos^; .*. 6 = ccos-4.
c
20
THE RIGHT TRIANGLE
21
sinid =
0.6866
c= 26
41136
13712
a = 17.8256
cos A = 0.7280
c= 26
43680
14560
b = 18.9280
Case II
Given A = 13^68', a = 15.2 ; find J5, 6, c.
1. B = 90** - ^ = 76** 2'.
2. - = cot -4 ; .'. 6 = a cot -4.
o. - = sin A ; .*.c = -: — -•
c sm -4
cot^l
a
4.0207
15.2
80414
201035
40207
b = 61.11464
a =
15.2, sin ^ = 0.2414.
0.2414)15.200(62.9
14 484
c = 62.9
7160
4828
2332
Case III
Given A = 27** 12', 6 = 31 ; find B, a, c.
1. B = 90*» - ^ = 62^ 48'.
a
2. -r = tan A ; .•.a = 6tan^.
6
6. - = cos A ; .'. c = 7
c cos ^
22
PLANE TRIGONOMETRY
tan^= 0.5139
i= 31
5139
15 417
a = 15.9309
31, cos A = 0.8894.
0.8894)31.000(34.9
26 682
c = 34.9
4 3180
3 5576
7604
a =
Case IV
Given a = 47, c = 63 ; find A, B, b,
1. sin A = -*
c
2. 5 = 90^-^.
3. 5 = Vc2-a2=V(c + a)(c-a).
47, c = 63.
63)47.0(0.7460
441
2 90
2 52
sin ^ = 0.7460 380
.-. A = 48^ 15' 378
B = 41^ 45' 2
c — a
110
16
660
110
P = 1760
5:
Vl760
41.95
Case V
Given a = 40, 5 = 27 ; find A, B, c.
1. tan A =-'
2. 5 = 90^-^.
3. c = v^2*:r^.
THE RIGHT TRIANGLE
23
a = 40, 5 = 27.
J^ = 1.4815
tan .4 = 1.4815
.-. A = 55^ 69'
B = 34^ 1'
a
5«
» - 1600
729
2329
V23a9
= 48.26
SECTION XII
GENERAL METHOD OF SOLVING THE RIGHT TRIANGLE
From these five cases it appears that the general method of
finding an unknown part in a right triangle is as follows :
Choose from the equation A + B = 90", and the equations
that define the functions of the angles, an equation in which
the required part only is unknown; solve this equation, if
necessary, to find the value of the unknown part ; then com-
pute the value.
Note. In Case IV, if the given sides (here a and c) are nearly alike
in value, then A is near 90°, and its value cannot be accurately found
from lihe tables, because the sines of large angles differ little in value
(as is evident from Fig. 4). In this case it is better to find B first, by
means of the formula given on page 69, namely.
tan^B
-4
c — a
c-\- a
Example.
c
c
a
a
Given a = 49, c = 50 ; find A, B, b.
1, c + a = 99.
c -\- a
4
€ — a
c-}- a
tmiB
.-. iB
B
A
= 0.01010
= 0.1005
= 0.1005
= 5° 44'
= 11° 28'
= 78° 32'
c — a
c -f a
1
99
b
= 99
= 99
= V99
= 9.95
24
PLANE TRIGONOMETRY
EXERCISE Vm
1. In Case II give another way of finding e, after b has
been found.
2. In Case III give another way of finding e, after a has
been found.
3. In Case IV give another way of finding b, after the
angles have been found.
4. In Case Y give another way of finding c, after the
angles have been found.
6. Given B and c \ find A, a, b,
6. Given B and b ; find A, a, c.
7. Given B and a ; find A, b, c.
8. Given b and c ; find A, B, a.
Solve the following right triangles ;
9
Given
Bbquirrd
a = 3, 6 = 4.
A = 36° 62', B = 63° 8', c = 6.
10
a = 7, c = 13.
A = 32° 36', 5 = 57° 26', 6 = 10.964.
11
a = 6.3, A = 12° 17'.
B = 77°43', 6 = 24.342, c = 24.918.
12
a = 10.4, B = 43° 18'.
A = 46° 42', 6 = 9.800, c = 14.290.
13
c = 26, A = 37° 42'.
B = 62° 18', a = 16.900, 6 = 20.672.
14
c=140, B = 24°12'.
A = 66° 48', a = 127.694, 6 = 67.386.
15
6 = 19, c = 23.
A = 34° 18', B = 65° 42', a = 12.961.
16
6 = 98, c = 136.2..
A = 43° 33', B = 46° 27', a = 93. 139.
17
6 = 42.4, A = 32° 14'.
B = 57°46', a = 26.733, c = 60.124.
18
6 = 200, B = 46°ll'.
A = 43° 49', a = 191.900, c = 277.160.
19
a = 96, 6 = 37.
A = 68° 43', B = 21° 17', c = 101.961.
20
a = 6, c = 103.
^ = 3° 21', J? = 86° 39', 6 = 102.826.
21
a = 8.12, B = 6°8'.
A = 84° 52', 6 = 0.280, c = 3.133.
22
a = 17, c = 18.
A = 70° 48', J? = 19° 12', 6 = 6.916.
23
c = 67, A= 38° 29'.
5= 51° 31', a = 35.471, 6 = 44.620.
24
a + c = 18, 6 = 12.
A = 22° 37', B = 67° 23', a = 6, c=13.
26
a 4- 6 = 9, c = 8.
A = 82° 18', B = 7° 42', 1 ^ "= ^'^^^^
* \6 = 1.072.
THE RIGHT TRIANGLE
25
SECTION XIII
SOLUTION BY LOGARITHMS
Case I
Given A = 34^ 28', c = 18.75 ; find B, a, h,
1. 5 = 90^-^ = 55*32'.
2. -=sin^; .•.a = csin^.
c
3. -=cos^; .'.6 = 0008^1.
c
log a
logc
log sin A
log a
a
log c + log sin A
1.27300
9.75276 - 10
1.02576
10.611
log 6 = log c 4- log cos A
log c = 1.27300
log cos A = 9.91617-10
log i = 1.18917
^=15^9
Case II
Given A = 62** 10', a = 78 ; find B, ft, c.
1. 5 = 90* - ^ = 27* 50'.
2. -=cotA; .•.ft = acot-4.
3. - = sm ^.
c
a
.', a = sin yl, and e = -: ,
sin^
log b = log a + log cot A
log a = 1.89209
log cot A = 9.72262 - 10
. log ft = 1.61471
ft = 41.182
logc =loga+cologsin^
log a =1.89209
colog sin ^= 0.05340
logc =1.94549
c =88.204
PLANE TRIGONOMETRY
Case
Ill
/l^
Given A = 50° 2', 6 = 88 ; find B,
/
1. B = 90° - X = 39" 58'.
/
a
2. Y=ta.iiA; .■.a = b^aA.
,/V.v
3. - =coa^.
"" 1,-8. -
Tia. 17
b
.■.6 = cco8^, andc = — — ■
log.-
logi 4-logtan ,4 ! logc =log6+colog
logS =
1.94448
log6 = 1.94448
log tan ^ =
log..-
10.07670
2.02118
-10
cologcos,!- 0.19223
log c = 2.13671
105.00
1 c = 137.00
Case IV
J.B
Given e = 58.40, a = 47.55 ; fin
/
fi,b.
/
/
n
1. sin A =-■
/
c
2. B=90''-.4.
3. - = cotyl; .■.b = acotA.
\- log cot vl
5
0-10
THE RIGHT TRIANGLE
27
Case Y
Given a = 40, ft = 27 ; find A, B, c.
a
1. tan A = -'
2. B = 90^-.4.
o. - = sm A.
c
,\ a = c sin -4, and c = —
a
sin^
log tan A
log a
colog h
log tan ^
B
log a 4- colog h
1.60206
8.56864 - 10
10.17070 - 10
BB"" 59'
34° 1'
logc
log a
colog sin ^
logc
c =
loga + cologsin^
1.60206
0.0815 1
1.68357
48.258 .
Note. In Cases IV and V the unknown side may also be found from
the equations
(for Case IV) h = Vc2 _ a^ = V(c -\.a)(c- a) ;
(for Case V) c= Va2 + &2.
These equations express the values of b and c directly in terms of the
two given sides ; and if the values of the sides are simple numbers {e.g., 5,
12, 13), it is often easier to find 6 or c in this way. But this value of c is
not adapted to logarithms, and this value of h is not so readily found by
logarithms as the value of b given under Case IV. See also p. 23.
SECTION XIV
AREA OF THE RIGHT TRIANGLE
The area of a triangle is equal to one-half the product of
the base by the altitude ; therefore, if a and b denote the legs
of a right triangle, and F the area, F = ^ab.
Hence, the area may be found when a and b are known.
28
PLANE TRIGONOMETRY
For example : Find the area,
Case I (Sect. XIII, p. 25).
A = 34* 28', c = 18.75.
First find (as in Sect. XIII,
p. 25) log a and log b,
log F = log a + log b 4- colog 2
log a = 1.02576
log b = 1.18917
colog 2 = 9.69897 - 10
log F= 1.91390
F = 82.016
having given :
Case IV (Sect. XIII, p. 26).
a = 47.55, c = 58.40.
First find (as in Sect. XIII,
p. 26) log a and log b.
log F = log a -f log b 4- colog 2
log a = 1.67715
log i = 1.58015
colog 2 = 9.69897 - 10
log F = 2.90627
F = 805.88
EXERCISE IX
Solve the following triangles by logarithms, finding the
angles to the nearest minute:
1
Given
Required
a=6,
c=12.
>1=30°,
B=60°,
6=10.392.
2
>1=00°,
6=4.
B=30°,
c=8,
a=6.9282.
3
^=30°,
a=3.
B=60°,
c=6.
6=6.1961.
4
a=4,
6=4.
^ = B=45°,
c=6.6568.
6
a=2,
c=2.82843.
^=B=45°,
6=2.
6
c=627,
^=2.3° 30'.
B=66° 30',
a =260. 02,
6=676.0.
7
c=2280,
^ = 28° 6'.
B=ei° 65',
a=1073.3.
6=2011.6.
8
c=72.15,
^=39° 34'.
B=50° 26',
a =45. 968,
6=56.620.
9
c = l,
>1=36«.
B=54°,
a=0.68779.
6=0.80902.
10
c=200,
B=2l° ir.
^=68° 13',
a=186.72.
6=74.22.
11
c=93.4,
5=76° 25'.
^ = 13° 36',
a=21.936.
6=90.788.
12
a=637,
A= 4° 36'.
B=85° 25',
6=7946,
c=7971.5.
13
a =48. 632,
^=36^44'.
B=53°16',
6=65.031,
c=81.144.
14
a=0.0008,
^=86°.
B=4°,
6=0.0000669,
c=0.000802.
15
6=60.937,
B=43° 48'.
^ = 46° 12',
a =63. 116,
c=73.69.
16
6=2,
B= 3^38'.
^=86^22',
a=31.496.
c=31.669.
THE RIGHT TRIANGLE
29
17
GnricK
Kequirrd
a =992,
5=76° 19^.
^ = 13° 41',
6=4074.5,
c =4193. 6.
18
a =73,
B=6S° 62'.
A =21° 8',
6=188.86,
c=202.47.
19
a=2.189.
1^=45° 25'.
^=44° 36',
6=2.2211,
c =3. 1185.
20
6=4,
^=37° 66'.
B=62°4',
a=3.1176,
c=6.0714.
21
c=8590,
a=4476.
^=31° 24',
5=68° 30',
6=7332.8.
22
c=86.53,
a=71.78.
^ = 56<'3',
J? =33° 67',
6=48.324.
23
c=9.35.
a=8.49.
^=65° 14',
5=24° 46',
6=3.917.
24
c=2194.
&=1312.7.
^ = 63^16',
5=36° 45',
a =1758.
26
c=30.69,
6=18.256.
>1 = 53°30',
5=36° 30',
a=24.07.
26
a=38.313.
6=19.622.
^=63°,
5=27°,
c=43.
27
a=1.2291,
6=14.960.
A= 4° 42',
5=85° 18',
c=16.
28
a=416.38,
6=62.080.
^=81° 30',
5= 8° 30',
c=420.
29
a =13.690,
6=16.926.
^=38° 68',
5=61° 2',
c=21.769.
30
c=91.92,
a=2.19.
A= P22',
5=88° 38',
6=91.894.
Compute the unknown parts and also the area, having given :
36. c = 68,
37. c = 27,
38. a = 47,
39. b = 9,
A = 69^ 54'.
B = 44^ 4'.
B = 48^ 49'.
5 = 34*» 44'.
40. c = 8.462, 5=:86M'.
31. a = 5, b = 6.
32. a = 0.616, c = 70.
33. i = V^, c = V3.
34. a = 7, A= 18* 14'.
35. 6 = 12, A= 29* 8'.
41. Find the value of F in terms of c and A,
42. Find the value of F in terms of a and A .
43. Find the value of F in terms of b and A,
44. Find the value of F in terms of a and c.
45. Given F= 58, a = 10; solve the triangle.
46. Given jP = 18, b = 5; solve the triangle.
47. Given jP = 12, A= 29° ; solve the triangle.
48. Given F — 100, c = 22; solve the triangle.
49. Find the angles of a right triangle if the hypotenuse is
equal to three times one of the legs.
30 PLANE TRIGONOMETRY
50. Find the l^s of a tight triangle if the hypotenuse is 6,
and one angle is twice the other.
61. In a right triangle given c, and A — nB; find a and fi.
52. In a right triangle the difference between the hypote-
nuse and the greater leg ia equal to the difference between
the two legs. Find the angles.
js^*as4;-
The ant/le of elevatwn of an object, or the an^le of depres-
sion, is the angle which a line from the eye to the object makes
with a horizontal line in the same vertical plane.
Thus, if the obsei-ver is at (Fig. 20), x is the angle of
elevation of B, and j/ is the angle of depression of C.
53. At a horizontal distance of 120 feet from the foot of a
steeple, the angle of elevation of the top was found to be
60° 30'. Find the height of the steeple.
. 54. From the top of a rock that rises vertically 326 feet out
of the water, the angle of depression of a boat was found to be
24°. Find the distance of the boat from the foot of the rock.
65, How far is a monument, in a level plain, from the eye,
if the height of the monument is 200 feet and the angle of
elevation of the top 3° 30' ?
6G. A distance AB is measured 96 feet along the bank of a
river from a point A opposite a tree C on the other bank.
The angle A"' "'" "**'. Find the breadth of the river.
THE RIGHT TRIANGLE 31
57. What is the angle of elevation of an inclined plane if it
rises 1 foot in a horizontal distance of 40 feet ?
58. Find the angle of elevation of the sun when a tower
120 feet high casts a horizontal shadow 70 feet long.
59. How high is a tree that casts a horizontal shadow 80 feet
in length when the angle of elevation of the sun is 50** ?
60. A ship is sailing due northeast at a rate of 10 miles
an hour. Find the rate at which she is moving due north,
and also due east.
61. In front of a window 20 feet high is a flower-bed 6 feet
wide. How long is a ladder that will just reach frojn the
edge of the bed to the window ?
62. A ladder 40 feet long may be so placed that it will reach
a window 33 feet high on one side of the street, and by turn-
ing it over without moving its foot it will reach a window 21
feet high on the other side. • Find the breadth o:^ the street.
63. From the top of a hill the angles of depression of two
successive milestones, on a straight level road leading to the
hill, are observed to be 5** and 15°. Find the height of the hill.
w
64. A fort stands on a horizontal plain. The angle of
elevation at a certain point on the plain is 30**, and at a point
100 feet nearer the fort it is 45°. How high is the fort ?
65. From a certain point on the ground the angles of eleva-
tion of the belfry of a church and of the top of the steeple
were found to be 40° and 51°, respectively. From a point 300
feet farther off, on a horizontal line, the angle of elevation of
the top of the steeple is found to be 33° 45'. Find the distance
from the belfry to the top of the steeple.
66. The angle of elevation of the top C of an inaccessible
fort observed from a point A is 12°. At a point B, 219 feet
from A and on a line AB perpendicular to AC, the angle ABC
is 61° 45'. Find the height of the fort:
32
PLANE TRIGONOMETRY
SECTIOli XV
THE ISOSCELES TRIANGLE
An isosceles triangle is divided by the perpendicular from
the vertex to the base into two equal right triangles.
Therefore, an isosceles triangle is determined by any two
parts that determine one of these right triangles.
Let the parts of an isosceles tri-
angle CAB (Fig. 21), among which
the altitude CD is to be included,
be denoted as follows:
a = one of the equal sides,
c = the base,
h = the altitude,
A = one of the equal angles,
C = the angle at the vertex.
For example : Given ft and c ; required A, Cy h,
1. cos ^ = ^— =
a 2 a
2. C-f 2^ = 180^; .-. C = 180^-2^ =2(90^-^).
3. h may be found by any one of the equations :
/.a
whence
Also,
whence
A2 + - = a2;
4
A = V(a + ic)(a-ic).
- = sin A, and — = tan A ;
a \c
h=^ a sin A, and h = \c tan A .
When c and h are known, the area can be found by the
formula „ , ,
THE RIGHT TRIANGLE
33
EZERaSE X
Solve the following isosceles triangles, finding the angles
to the nearest second :
1.
Given a and A
; ^x\(\ C,
c, h.
2.
Given a and C
\ find Ay
Cy h.
3.
Given c and A
; find C,
ay h.
4.
Given c and C \
, find Ay
ay h.
5.
Given h and A ;
find C,
ay c.
6.
Given h and C ;
find Ay
a, c.
7.
Given a and h \
find Ay
CyC.
8.
Given c and h ;
find i4,
Cy a.
9. Given a = 14.3, c = 11 ; find Ay C, h.
10. Given a = 0.295, ^ = 68^ 10' ; find c, h, F.
11. Given c = 2.352, C = 69* 49'; find a, A, F.
12. Given A = 7.4847, A = 76° 14' ; find a, c, F.
13. Given a = 6.71, h = 6.6 ; find Ay Cy c.
14. Given c = 9, A = 20 ; find ^, C, a.
15. Given c = 147, F = 2572,5 ; find ^, C, a, A.
16. Given A = 16.8, F = 43.68 ; find Ay Cy a, e.
17. Find the value of F in terms of a and c.
18. Find the value of F in terms of a and C
19. Find the value of F in terms of a and A,
20. Find the value of F in terms of h and C
21. A barn is 40 x 80 feet, the pitch of the roof is 45° ;
find the length of the rafters and the area of the whole roof.
22. In a unit circle what is the length of the chord corre-
sponding to the angle 45° at the centre ?
23. If the radius of a circle is 30, and the length of a
chord 1 . 44, find the angle subtended at the centre.
34 PLANE TRIGONOMETRY
24. Find the radius of a circle if a chord whose length
is 5 subtends at the centre an angle of 133°.
25. What is the angle at the centre of a circle if the corre-
sponding chord is equal to f of the radius ?
26. Find the area of a circular sector if the radius of the
circle is 12, and the angle of the sector is 30°.
SECTION XVI
THE REGULAR POLYGON
Lines drawn from the centre of a regular polygon (I^ig. 22)
to the vertices are radii of the circumscribed circle ; and lines
drawn from the centre to the middle points of the sides are
radii of the inscribed circle. These lines divide the polygon
into equal right triangles. Therefore, a regular polygon is
determined by a right triangle whose sides are the radius of
the 6ircumscribed circle, the radius of the inscribed circle, and
half of one side of the polygon.
If the polygon has n sides, the angle of this right triangle
at the centre of the polygon is equal to - f j , or ;
and the triangle may be solved when a side of the polygon or
one of the radii is given.
Let
n = number of sides,
c = length of one side,
r = radius of circumscribed circle,
h = radius of inscribed circle,
p = the perimeter,
F = the area.
Then, by Geometry,
^<*- 22 F = i hp.
THE RIGHT TRIANGLE 36
EXERCISE XI
Pind the remaining parts of a regular polygon, given :
1. 71 = 10, c = l. 3. 71 = 20, 7- = 20. 5. 7^ = 11, F= 20.
2. 71 = 18, r = 1. 4. 71 = 8, A = 1. 6. 7i = 7, F = 7.
Pind the side of :
7. A regular decagon inscribed in a unit circle.
8. A regular decagon circumscribed about a unit circle.
9. If the side of an inscribed regular hexagon is 1, find the
side of an inscribed regular dodecagon.
10. Given n and c, and let b denote the side of the inscribed
regular polygon having 2 n sides ; find b in terms of n and c.
11. Compute the difference between the areas of a regular
octagon and a regular nonagon if the perimeter of each is 16.
1 2. Compute the difference between the perimeters of a regu-
lar pentagon and a regular hexagon if the area of each is 12.
Fina the area of :
13. The regular octagon formed by cutting away the corners
of a square whose side is 1.
• 14. A regular pentagon if its diagonals are each equal to 12.
15. A regular polygon of. 11 sides inscribed in a circle, if
the area of an inscribed regular pentagon is 331.8.
16. A circle inscribed in an equilateral triangle whose perim-
eter is 20.
17. A regular polygon of 15 sides inscribed in a circle, if
the area of a regular inscribed polygon of 16 sides is 100.
18. Find the perimeter of a regidar dodecagon circum-
scribed about a circle the circumference of which is 1.
19. The area of a regular polygon of 25 sides is 40 ; find
the area of the ring comprised between the circumferences of
the inscribed and circumscribed circles.
CHAPTER III
GONIOMETRT
SECTION xvn
DEFINITION OF GONIOMETRT
To prepare the way for the solution of the oblique triangle,
we now proceed to extend the definitions of the trigonometric
functions to angles of all magnitudes, and to deduce certain
useful relations of the functions of different angles.
That branch of Trigonometry which- treats of trigono-
metric functions in general, and of their relations, is called
Goniometry. •
SECTION XVIII
POSITIVE AND NEGATIVE QUANTITIES
In measurements it is convenient to mark the distinction
y between two magnitudes that are measured in
opposite directions, by calling one of them posi-
tive and the other negative.
Jji
i^
O Thus, if OX (Fig, 23) is considered to be
positive, then OZ' is considered to be negative ;
y' and if Oy is considered to be positive, then
Fio. 23 oy is considered to be negative.
When this distinction is applied to angles, an angle is
considered to be positive, if the rotating line that describes
it moves counter-clockwise, that is, in the direction opposite
36
GONIOMETRY 37
to the hands of a clock, and to be negative, if the rotating
line moves clockwise, that is, in the same direction as the
hands of a clock.
Arcs corresponding to positive angles are
considered j905i^4ve, and arcs corresponding to
negative angles are considered negative. \ -s. /
Thus, the angle A OB (Fig. 24) described by \^ ^/^'
a line rotating about from OA to OB is pos-
itive, and the arc AB \^ positive; the angle
AOB^ described by the line rotating about from OA to OB' is
negative, and the arc AB^ is negative.
SECTION XIX
COORDINATES OF A POINT IN A PLANE
Let XX^ (Fig. 25) be a horizontal line and let 77' be a line
y perpendicular to XX^ at the point 0.
Then the plane determined by the
lines XX* and YY* is divided into
four quadrants which are numbered
S- I, II, III, IV.
Any point in thb plane is deter-
n
in
IV
mined by its distance and direction
from each of the perpendiculars XX*
and YY\ Its distance from YY\ meas-
^* * ured on XX\ is called the abscissa of
the point; its distance from XX* y measured on YY\ is called
the ordinate of the point.
The abscissa and the ordinate of a point are called the
co-ordinates of the point ; and the lines XX* and YY* are called
the axes of co-ordinates. XX* is called the axis of abscissas or
the axis of x ; YT is called the axis of ordinates or the axis of y ;
and the point O is called the origin.
38
PLANE TRIGONOMETRY
Pf
A
^iBt
Bz
f-4
O
A,
^.
4— f
^1
B4
Bi
^s
Fio. 26
In Fig. 26 the co-ordinates Pi, Pg,
Pj, P4 are as follows :
The abscissa of Pi is OBi,
and the ordinate of Pi is OAi;
the abscissa of Pj is OB^,
and the ordinate of Pa is OA2 ;
the abscissa of Pj is O-Bg,
and the ordinate of Pg is O^
the abscissa of P4 is O-B4,
and the ordinate of P4 is OA^.
89
Abscissas to the rigiht of YY' are positive.
Abscissas to the left of YY' are negative.
Ordinates above XX' are positive.
Ordinates below XX' are negative.
Therefore,
in Quadrant I,
the abscissa is positive, the ordinate is positive ;
in Quadrant II,
the abscissa is negative, the ordinate is positive ;
in Quadrant III,
the abscissa is negative, the ordinate is negative ;
in Quadrant IV,
the abscissa is positive, the ordinate is negative.
SECTION XX
ANGLES OF ANT MAGNITUDE
If the line OP (Figs. 27-30) is revolved about from OX
as its initial position counter-clockwise, as shown by the curved
arrows, the line during one revolution will form with OX all
angles from 0** to 360®.
GONIOMETRY
39
Any particular angle is said to be an angle of that quadrant
in which its tenninal side lies.
F^
"^^^^
AtS^
^M
V III
/
ly /
Fig. 27
Pig. 28
Fig. 29
Fig. 90
Angles between 0° and 90® are angles of Quadrant I.
Angles between 90** and 180° are angles of Quadrant II.
Angles between 180** and 270** are angles of Quadrant III.
Angles between 270° and 360° are angles of Quadrant IV.
Fig. 31
Fio. 32
Fig. 33
Fig. 34
If the revolving line makes another revolution (Figs. 31-34),
it will describe all angles from 360° to 720° ; and so on.
Fig. 35
Fig. 36
Fig. 37
Fig. 38
If the line OP is revolved from OX clockwise (Pigs. 35-38),
it will describe all negative angles.
Thus we arrive at the conception of an angle of any magni-
tude, positive or negative.
I
40
PLANE TRIGONOMETRY
SECTION XXI
FUNCTIONS OF ANT ANGLE
Figs. 39-42 show the functions in a unit circle drawn for
an angle A OP in each quadrant, taken in order. The tangents
to the circle are always drawn through A and B.
Let the angle A OP formed with OA by the moving radius
OP be denoted by x; then, in each quadrant,
sin X = MP,
cos X = OM,
tana: = ^7;
cot X = BS,
/
.
5
/^
^\ !]
f
L
f C
A
\
y
P
\^
J
s
B'
FlQ, 41
Fia. 42
GONIOMETRY
41
If the terminal line of any angle x extends tlirough the
vertex indefinitely both ways, and if the circumference of a
unit circle cuts the terminal line at P, the axis of abscissas at
Af and the axis of ordinates at B, then
sin X = the ordinate of P ;
cos X = the abscissa of P ;
tan X = the tangent from A to meet the terminal line ;
cot X = the tangent from B to meet the terminal line ;
sec X = the segment of the terminal line between the vertex
and the tangent ;
CSC x = the segment of the terminal line between the vertex
and the cotangent.
Sines and tangents extending from the axis of abscissas
upwards are positive ; downwards^ negative.
Cosines and cotangents extending from the axis of ordinates
towards the right are positive ; towards the left, negative.
The signs of the secant and cosecant are determined by
the signs of the cosine and sine, respectively. Therefore,
secants and cosecants extending from the centre, in the direc-
tion of the terminal line, are considered poi^itive ; in the oppo-
site direction, negative. Hence,
Qua nit ANT
I
II
III
IV
sin and esc
cos and sec
tan and cot
+
+
+
In Quadrant I all the functions are positive.
In Quadrant II the sine and cosecant only are positive.
In Quadrant III the tangent and cotangent only are positive.
In Quadrant IV the cosine and secant only are positive.
42
PLANE TRIGONOMETRY
The signs of all the functions of any quadrant are known
when the signs of the sine and cosine are known.
If the sine and cosine have like signs, the tangent and cotan-
gent are positive; if unlike signs, negative. The sine and
cosecant have like signs ; the cosine and secant have like signs.
SECTION XXII
FUNCTIONS OF A VARIABLE ANGLE
Let the angle A OP (Fig. 43) increase continuously from 0®
to 360®. The values of its functions change as follows :
1. The Sine. In the first quad-
rant the sine MP increases from
to 1 ; in the second it remains
positive, and decreases from 1 to
; in the third it is negative, and
increases in absolute value from
to 1; in the fourth it is nega-
tive, and decreases in absolute
value from 1 to 0.
2. The Cosine. In the first
quadrant the cosine OM decreases
from 1 to ; in the second it be-
comes negative, and increases in
absolute value from to 1 ; in
T' the third it is negative, and de-
creases in absolute value from 1
to ; in the fourth it is positive, and increases from to 1.
3. The Tangent. In the first quadrant the tangent AT
increases from to oo ; in the second it becomes negative,
and decreases in absolute value from oo to ; in the third it
is positive, and increases from to oo ; in the fourth it is
negative, and decreases in absolute value from oo to 0.
Fig. 43
GONIOMETRY
43
4. The Cotangent. In the first quadrant, the cotangent
BS decreases from oo to ; in the second it is negative, and
increases in absolute value from to oo; in the third and
fourth quadrants, it has the same sign, and undergoes the
same changes as in the first and second quadrants, respectively.
5. The Secant, In the first quadrant, the secant OT
increases from 1 to oo; in the second it is negative, and
decreases in absolute value from oo to 1 ; in the third it is
negative, and increases in absolute value from 1 to oo ; in the
fourth it is positive, and decreases from oo to 1.
6. The Cosecant. In the first quadrant, the cosecant OS
decreases from oo to 1 ; in the second it is positive, and
increases from 1 to oo; in the third it is negative, and
decreases in absolute value from oo to 1 ; in the fourth it is
negative, and increases in absolute value from 1 to oo.
The limiting values of the functions are as follows :
Sine
0»
90*»
1S0*»
270<>
360<>
±0
+ 1
±0
1
±0
Cosine
+ 1
±0
-1
±0
+ 1
Tangent
±0
±00
±0
•± 00
±0
Cotangent
±00
±0
±00
±0
±00
Secant
+ 1
±00
-1
± 00
4-1
Cosecant
±00
+ 1
±00
1
±00
Sines and cosines vary in value from 4-1 to — 1 ; tangents
and cotangents, from 4- oo to — oo ; secants and cosecants, from
± 00 to ± 1, and from — 1 to — oo.
In the table given above the double sign ± is placed before and oo .
From the preceding investigation it appears that the functions always
change sign in pacing through and oo ; and the sign + or — prefixed
to or 00 simply shows the direction from which the value is reached.
44
PLANE TRIGONOMETRY
SECTION xxni
FUNCTIONS OF ANGLES LARGER THAN ZeO"*
The functions of 360® + x are the same in sign and in
absolute value as those of x ; for the moving radius has the
same position in both cases. If w is a positive integer,
The functions of (n x 360® + x) are the same as those of x.
For example : The functions of 2200® (6 x 360® + 40®) are
equal to the functions of 40®.
SECTION XXIV
EXTENSION OF FORMULAS
The Formulas [1], [2], [3] established for acute angles on
pp. 13, 14 hold true for all angles. Thus, in each quadrant
;2
Fig. 44
MP -{- OM = OP ,
Therefore,
sin* X -j- cos* X = 1. [1]
We have in each quadrant from
the similar triangles OMP, OAT,
OBS the proportions
AT:OA=MP: OM,
or tan x :!== sin x : cos x ;
MP:OP= OB : 05,
or sin a: : 1 = 1 ; esc x ;
OM: 0P= OA: OT,
or cos a^ : 1 = 1 : sec x ;
AT:OA = OB: BS,
or tan x : 1 = 1 : cot x.
GONIOMETRY 45
That is, tanx = ?5J?. r2-|
C08X ^ -^
sin X X C8C X = 1 1
cos X X sec X = 1
tan X X cot X = 1
> •
[3]
Formulas [l]-[3] enable us, from a given value of one
function, to find the absolute values of the other five func-
tions, and also the sign of the reciprocal function. But in
order to determine the proper signs to be placed before the
other four functions, we must know the quadrant to which
the angle in question belongs, or the sign of any one of these
four functions ; for, by Sect. XXI, p. 40, it will be seen that the
signs of any two functions that are not reciprocals determine
the quadrant to which the angle belongs.
Example. Given sin x = -|- |, and tan x negative ; find
the values of the other functions.
Since sin x is positive, x is an angle in Quadrant I or II ;
but, since tan x is negative. Quadrant I is inadmissible.
By[l], cosa:=±Vl-iJ=±f.
Since the angle is in Quadrant II, the minus sign must be
taken, and we have
■ cos a; = — f .
By [2] and [3],
tan X = — J, cot 05 = — J, sec a; = — |, esc a; = |.
EXERCISE Xn
1. Construct the functions of an angle in Quadrant II.
What are their signs ?
2. Construct the functions of an angle in Quadrant III.
What are their signs ?
46 PLANE TRIGONOMETRY
3. Construct the functions of an angle in Quadrant IV.
What are their signs ?
4. What are the signs of the functions of the following
angles : 340°, 239^ 145^ 400°, 700°, 1200°, 3800° ?
5. How many angles less than 360° have the value of the
sine equal to -f- f , and in what quadrants do they lie ?
6. How many values less than 720° can the angle x have
if cos a; = -f f , and in what quadrants do they lie ?
7. If we take into account only angles less than 180°, how
many values can x have if sin a: = f ? if cos x = J ? if cos x =
-f? iftanaj = §? ifcotic=-7?
8. Within what limits must the angle x lie if cos x = — f ?
if cotaj = 4? if sec a; = 80? if esc a; =-3? (Ifaj<360°.)
9. In what quadrant does an angle lie if sine and cosine
are both negative ? if cosine and tangent are both negative ?
if cotangent is positive and sine negative ?
10. Between 0° and 3600° how many angles are there whose
sines have the absolute value f ? Of these sines how many
are positive and how many negative ?
11. In finding cos x by means of the equation cos x =
± Vl — sin^a;, when must we choose the positive sign and
when the negative sign?
12. Given cos x = — V^; find the other functions when x
is an angle in Quadrant II.
13. Given tan x = VS ; find the other functions when x is
an angle in Quadrant III.
14. Given sec x ^-\-l, and tan x negative ; find the other
functions of x,
15. Given cot a; = — 3 ; find all the possible values of the
other functions.
16. What functions of an angle of a triangle may be nega-
tive ? In what case are they negative ?
GONIOMETRY
47
17. Why may cot 360° be considered equal either to -f oo
or to — 00 ?
18. Obtain by means of Formulas [l]-[3] the other func-
tions of the angles, given :
(i) tan 90° = 00. (iii) cot 270° = 0.
(ii) cos 180° = - 1. (iv) esc 360° = - oo.
19. Find the values of sin 450°, tan 640°, cos 630°, cot 720°,
sin 810°, CSC 900°.
Compute the values of the following expressions :
20. a sin 0° -f J cos 90° - c tan 180°.
21. a cos 90° - h tan 180° + c cot 90°.
22. a sm 90° - h cos 360° -f (a - J) cos 180°.
23. (a* - h^) cos 360° - 4 aJ sin 270°.
SECTION XXV
REDUCTION OF FUNCTIONS TO THE FIRST QUADRANT
In a unit circle (Fig. 45) draw two diameters PR and QS
equally inclined to the horizontal diameter AA\ or so that
the angles AOP^ A^OQ, A' OR, and
A OS shall be equal. From the
points P, Q, R, S let fall perpen-
diculars to AA' ', the four right
triangles thus formed, with a
common vertex at 0, are equal;
because they have equal hypote-
nuses (radii of the circle) and
equal acute angles at 0, Ther^
fore, the perpendiculars PM, QN,
RN, SM are equal, and are the
sines of the angles A OP, AOQ, A OR, A OS, respectively.
48
PLANE TRIGONOMETRY
. Therefore, in absolute valtie,
sin AOP = sin ^OQ = sin ^012 = sin AOS.
And from Sect. XXIV, p. 44, it follows that in absolute
value the cosines of these angles are also equal ; and likewise
the tangents, the cotangents, the
secants, and the cosecants.*
Hence, For every acute angle
there is an angle in each of the
higher quadrants whose functions,
in absolute value, are equal to
those of this acute angle.
Let AAOP^x, APOB=^y)
then X -\- y = 90®, and the func-
tions of X are equal to the co-
named functions of y (Sect. V, p. 11) ; and
Z^OQ (in Quadrant II) = 180** - x = 90**-!-^,
/LA OR (in Quadrant III) = 180° -h aj = 270® - y,
/LAOS (in Quadrant IV) = 360® -x = 270® + y,
HenOe, prefixing the proi)er sign (Sect. XXI, p. 40), we have :
Angle in Quadrant II
8in(180® — x)= sinx.
sin (90® + y) — cos y.
cos (180® — x) = — cos X.
cos (90® -J- y)= — sin y.
tan(180® — x)= — tan X.
tan(90® + y)= — cot y .
cot(180® — x)= —cot X.
cot (90® + y)= — tany.
* In future, secants, cosecants, versed sines, and coversed sines will be
disregarded. Secants and cosecants may be found by Formula [3],
versed sines and coversed sines by VII and YIII, p. 5, if wanted, but
they are seldom used in computations.
GONIOMETRY 49
Angle in Quadrant III
sin (180® + aj) = — sin x. sin (270® — y) = — cos y,
cos (180° -f- x) = — cos X. cos (270° — y) = — sin y,
tan (180° + aj) = tana;. tan(270°-y)= cot y.
cot (180° + aj) = cot X. cot (270° -y)= tan y.
Angle in Quadrant IV
sin (360° — aj) = — sin x, sin (270° + y) = — cos y,
cos (360° — a;) = cos x, cos (270° + y) = sin y,
tan (360° - aj) = - tan x. tan (270° -f- y) = - cot y.
cot (360° - a;) = - cot x. cot (270° -f- y) = - tan y.
Remark. The tangents and cotangents may be found directly from
the figure, or by Formula [2].
It is evident, from these formulas,
1. The functions of all angles can he reduced to the functions
of angles in the first quadrant, and therefore to functions of
angles not greater than 45° (Sect. V, p. 11).
2. If an acute angle is added to or subtracted from 180° or
360**, the functions of the resulting angle are equal in absolute
value to the like-named functions of the acute angle ; but if an
acute angle is added to or subtracted from 90° or 270°, the
functions of the resulting angle are equal in absolute value to
the co-named functions of the acute angle,
3. A given value of a sine or cosecant determines two supple-
mentary angles, one acute, the other obtuse ; a given value of
any other function determines only one angle : acute if the
value is positive, obtuse if the value is negative. [See fiinc-
tions of (180° - a;).]
50
PLANE TRIGONOMETRY
SECTION XXVI
FUNCTIONS OF ANGLES THAT DIFFER BT 90""
The general form of two angles whose difference is 90® is x
and 90® + x, and they must lie in adjoining quadrants. The
relations between their functions were found in Sect. XXV,
p. 48, but only for the case when x is acute. These relations,
however, may be shown to hold true for all values of x.
In a unit circle (Fig. 47) draw
two diameters PR and QS per-
pendicular to each other, and let
fall to A A* the perpendiculars
PM, QH, RK, and SN, The
right triangles OMP, QHO, OKR,
and SNO are equal, because they
have equal hypotenuses and equal
acute angles POM, OQH, ROK, and
OSN,
Therefore,
and
OM=QH=OK = NS,
PM =OH=RK= ON,
Hence, taking into account the algebraic sign.
sin AOQ= cos A OP
QosAOQ = — sin^OP
sin A OR = cos A OQ
cos AOR= — sin AOQ
sin AOS = cos A OR ;
cos AOS = — sin A OR ;
sin (360® ■i-AOP)= cos A OS ;
cos (360® + ^OP) = - sin ^O^.
In all these equations, if x denotes the angle on the right-
hand side, the angle on the left-hand side is 90® + x.
Therefore, if a? is an angle in any one of the four quadrants,
sin (90® + x) = cos x, tan (90® -|- x) = — cot x.
cos (90® -f- aj) = — sin x, cot (90® + x) = — tan x.
GONIOMETRY
51
In like manner, it can be shown that all the formulas of
Sect. XXV, p. 48, hold true, whatever the values of x and y.
Hence, In every case the algebraic sign of the function
of the resulting angle is the same as when x and y are both
acute,
SECTION XXVII
FUNCTIONS OF A NEGATIVE ANGLE
If the angle x is generated by the radius moving from the
initial position OA to the terminal position OS, it will have the
sign — , and its terminal side will
be identical with its position for
the angle 360® — x. Therefore,
the functions of the angle — x
are the same as those of the angle
360** - a;; or (Sect. XXV, p. 49),
sin (— a;) = — sin x,
cos (— x) = cos X,
tan (—«) = — tan x,
cot (— a;) = — cot x.
EXERCISE Xm
1. Express sin 250® in terms of the functions of an acute
angle less than 45®.
Solution, sin 260° = sin (270° - 20°) = - cos 20°.
Express the following functions in terms of the functions
of angles less than 45® :
2. sin 172®. 6. cot 91®. 8. sin 204®.
3. cos 100®. 6. sec 110®. 9. cos 359®.
4. tan 125®. 7. esc 157®. 10. tan 300®.
62 PLANE TRIGONOMETRY
11. cot 264°. 14. sinl63M9'. 17. cot 139M7'.
12. sec 244^ 15. cos 195° 33'. 18. sec 299° 45'.
13. CSC 271°. 16. tan 269° 15'. 19. esc 92° 25'.
Express all the functions of the following negative angles
in terms of the functions of positive angles less than 45° :
20. - 75°. 22. - 200°. 24. - 52° 37'.
21. - 127°. 23. - 345°. 26. - 196° 54'.
26. Find the functions of 120°.
Hint. 120° = 180° - 60°, or 90° + 30° ; then apply Sect. XXV, p. 48.
Find the functions of the following angles :
27. 135°. 29. 210°. 31. 240°. 33. -30°.
•28. 150°. 30. 225''. 32. 300°. 34. -225°.
35. Given sin x = — ^ V2, and cos x negative; find the other
functions of x, and the value of x,
36. Given cot a? = — VS, and x in Quadrant II ; find the
other functions of x, and the value of x.
37. Find the functions of 3540°.
38. What angles less than 360° have a sine equal to — ^ ?
a tangent equal to — Vs ?
39. Which of the angles mentioned in Examples 27-34
have a cosine equal to — J V2 ? a cotangent equal to — V3 ?
40. What values of x between 0° and 720° will satisfy the
equation sin a; = + i ?
41. Find the other angle between 0° and 360° for which the
corresponding function (sign included) has the same value as
sin 12°, cos 26°, tan 45°, cot 72°, sin 191°, cos 120°, tan 244°,
cot 357°.
GONIOMETRY
53
42. Given tan 238** = 1.6 ; find sin 122**.
43. Given cos 333^ = 0.89 ; find tan 117^
Simplify the following expressions :
44. a cos (90** - x) + 6 cos (90** + x).
45. m cos (90** - X) sin (90* - x),
46. (a - h) tan (90** - a:) + (a + *) cot (90** + x).
47. a^ + *2 _ 2 aft cos (ISO*' - x),
48. sin (90* + x) sin (180* + «) + cos (90* + x) cos (180* - x).
49. cos (180*+ x) cos (270*- y) - sin (180*+ x) sin (270*- y).
50. tan X + tan (— y) — tan (180* — y).
51. For what values of x is the expression sin x + cos x
positive, and for what values negative ?
52. Answer the questions of Example 51 for sin x — cos x.
53. Find the functions of a; — 90* in functions of x.
64. Find the functions oi x — 180* in functions of x.
SECTION XXVIII
FUNCTIONS OF THE SUM OF TWO ANGLES
In a unit circle (Fig. 49) let the angle AOB = x, the angle
BOC = y ; then the angle AOC =^
x-\ry.
In order to express sin (x + y)
and cos {x + y) in terms of the
sines and cosines of x and y, draw
CF i. OA, CD ± OB, DE ± OA,
DG i. CF; then CD = sin y, OD =
cos y, and the angle DCG = x.
Also,
sin(aj + y) =CF = DE-^ CG.
54
PLANE TRIGONOMETRY
DE
Now — - = sin X ; hence, DE = sin x X OD = sin x cos y.
And — - = cos X ; hence,
CG = cos X X CD = cos X sin y.
Therefore,
8in(z-|-y) = 8inzc087-|-C08Z8iny. [4]
Again,
cos (x + y)= OF = OE — DG,
OE
OD
= cos X ; hence,
OE = cos X X OZ) = cos X cos y.
DG
-— = sin aj ; hence, DG = sin cc X CD = sin x sin y.
Therefore,
cos (z -|- y) r= cos z cos 7 — sin z sin y. [5]
In this proof x and y, and also the sum aj + y, are assumed
to be acute angles. If the sum
aj .+ y of the acute angles x and
y is obtuse, as in Fig. 51, the
proof remains, word for word,
the same as above, the only dif-
ference being that the sign of
OF will be negative, as DG is
now greater than OE. The above formulas, therefore, hold
true for all acute angles x and y.
If these formulas hold true for any two acute angles x and
y, they hold true when one of the angles is increased by 90®.
Thus, if for x we write a:' = 90° + x, then, by Sect. XXV,
p. 48,
sin (x' -{-y)= sin (90° -\- x -^y)= cos (x + y),
cos (x' + 2/) = cos (90° -|- X 4- y) = — sin (x + y).
GONIOMETRY 55
Hence, by [5], sin (x' + y)= cos x cos y — sin x sin y,
by [4], cos (x' -\- y) = — sin X cos y — cos x sin y,
Now, by Sect. XXV, p. 48,
cos X = sin (90° + a;) = sin x',
sin a; = — cos (90° -{- x) = — cos aj'.
Substitute these values of cos x and sin aj, then
sin (x' -{- y)= sin x' cos y + cos x' sin «/,
cos (x' -{- y) = cos a' cos y — sin a;' sin y.
It follows that Formulas [4] and [5] hold true if either
angle is repeatedly increased by 90° ; therefore they apply to
'all angles whatever.
By Sect. XXIV, p. 45, Formula [2],
sin (x + y) __ sin x cos y + cos x sin y
tan (x ~r y) — z ; r — : : *
^ ^ cos (x + y) cos X cos y — sm x sm «/
If we divide each term of the numerator and denominator
of the last fraction by cos x cos y, we have
sin X sin y
+
, . . cos X COS y
tan (x -^y) = —
sin X sin y
COS aj cos y
r.^1 . . ^ V tan X + tan y ^^_
That IS, . tan (x + y) = ^^ ^ ^ - [6]
^ ' -'^ 1— tanxtany ^ -^
cos X cos y — sin a; sin y
A 1 ^ / . V cos (a? 4- 2/) c(
Also, cot (x + y)= . ) i = -
' ^ ^^/ sin(a; + y) si
sm a; cos y — cos x sm y
Divide each term of the numerator and denominator by
cos X cos 1/
sin X sin y, remembering that — = cot x and -: — - = cot y:
, sin X sm y
we have
, . cot X cot y — 1 --,
cot(x + y)= ^ [7]
^ ^ ^^ coty + cotx ■- -^
56
PLANE TRIGONOMETRY
SECTION XXIX
FUNCTIONS OF THE DIFFERENCE OF TWO ANGLES
In a unit circle (Fig. 52) let the angle AOB =:x, COB = y ;
then the angle AOC = x ^y.
In order to express sin (x — y)
and cos (x — y) in terms of the
sines and cosines of x and y, draw
CF ± OA, CD ± OB, DE ± OA,
DG ± FC prolonged ; then CD =
sin y, OD = cos y, and the angle
DCG = X,
Now sin (x — y) = CF= DE—CG.
Fig. 62
DE
OD
CG
= sin X ; hence, DE = sin a x OD = sin aj cos y.
CD
Therefore,
Again,
OE
OD
DG
= cos X ; hence, CG = cos x x CD = cos x sin y.
sin (z — y) = sin z cos y — cos z sin y.
cos (a? — y) = OF = 0^ -h DG.
= cos X ; hence, 0^ = cos x x OD = cos x cos y.
[8]
CD
Therefore,
= sin aj ; hence, DG = sin a; x CD = sin x sin y,
cos (z — y) = cos z cos y + sin z sin y. [9]
In this proof, both x and y are assumed to be acute angles ;
but, whatever be the values of x and y, the same method of
proof will always lead to Formulas [8] and [9], when due
regard is paid to the algebraic signs.
The general application of these formulas may be at once
shown by deducing them from the general formulas estab-
lished in Sect. XXVIII, p. 64, as follows :
GOmOMETRY 57
It is obvious that (a; — y) + y = aj. If we apply Formulas
[4] aad [6] to (a; — y) -f- y, then
Qin.\(x — y)+y\ or sino; = sin (x — y) cosy + cos (x — y) siny,
cos \(pi^ — y)+y\ or cosa? = cos (x — y) cosy — sin (a? — y) sin y.
Multiply the first equation by cos y, the second by sin y,
sin X cos y = sin (x — y) cos^ + cos (x — y) sin y cos y,
cos X sin y = — sin (a; — y) sin^ + cos (x — y) sin y cos y ;
whence, by subtraction,
sin X cos y — cos aj sin y = sin (a; — y) (sin^ + cos'y).
But sinV + cosV = 1 (Sect. XXIV, p. 44).
Therefore, by substitution and transposition,
sin (x — y) = sin x cos y — cos x sin y.
Again, if we multiply the first equation by sin y, the second
equation by cos y, and add the results, we obtain, by reducing,
cos (x — y) =z cos X cos y -f- sin x sin y.
Therefore, Formulas [8] and [9], like [4] and [5], from
which they have been derived, are universally true.
From [8] and [9], by proceeding as in Sect. XXVIII, p. 56,
we obtain
^ . V tan X — tan y _ . ^_
tan(x — y) = £-. [10]
^ "^l+tanxtany '--'
^/ vCOtxcoty+1 -.^T
cot (x — y) = i-i— . [11]
^ ^ coty — cotx •- -^
Formulas [4]-[ll] may be combined as follows:
sin (x±y) =^ sin x cos y ± cos x sin y,
cos (x±y) = cos a; cos y ip sin a; sin y,
, , . tan X ± tan y
tan (x±y) = — '—f
^ ^^ 1 ip tan X tan y
. , , . cot a; cot y t 1
cot (x±y)=^ — 7 — X 7 —
^ ^^ cot y ± cot a;
68 PLANE TRIGONOMETRY
SECTION XXX
FUNCTIONS OF TWICE AN ANGLE
liy — Xj Formulas [4]-[7] become
sin 2 X = 2 sin x cos x. [12]
cos 2 X = cos*x — sin*x. [13]
^ „ 2 tan X ^^ ^ ^
tan2x = 5-. [14]
1 — tan*x •- -J
. -, cot*x — 1 _ . ^_
cot 2 X = [15]
2 cot X ^ -*
By these formulas the fimctions of twice an angle may be
found when the functions of the angle are given.
SECTION XXXI
FUNCTIONS OF HALF AN ANGLE
Formula [1] is cos^o; + sin'a: = 1.
Formula [13] is cos^a — sin^a; = cos 2 x.
Subtract, 2 sin^a = 1 — cos 2 x.
Add, 2 cos^aj = 1 + cos 2 x.
Whence,
, /l — cos 2 aj , /I + cos 2 x
smaj = ±'Y 9 ' cos a; = ±'Y o
These values, if « is put for 2 x, and hence ^ z for x, become
s 1 j_ /l — cos z ^^^-
sin^zrri-y' [16]
1 I /l + cos Z ^.-,_
cos i z = ± \-^I- [17]
aONIOMETRY 69
Hence, by division (Sect. XXIV, p. 46),
taniz=±JL=~iI. [18]
^1 +C08Z L -'
^ 1 j_ /l + COS Z _.^-
cot^z=±^_L [19]
By these formulas the functions of half an angle may be
computed when the cosine of the entire angle is given.
The proper sign to be placed before the root in each case
depends on the quadrant in which the angle i z lies (Sect.
XXII, p. 42).
Let the student show from Formula [18] that
tan i 5 = V^ — - • (See p. 23, Note.)
SECTION XXXII
SUMS AND DIFFERENCES OF FUNCTIONS
From [4], [6], [8], and [9], by addition and subtraction,
sin (aj -f y) + sin (x — y) = 2 sin a? cos y,
sin (aJ + y) — sin (a; — y) = 2 cos x sin y,
cos (« + y) + cos (a; — y) = 2 cos x cos y,
cos (x-^-y)— cos (x — y) = — 2 sin x sin y ;
or, by making a; -f y = ^, and x — y = B, and, therefore,
a? = i(^ + ^), and y = ^(^ ~ 5),
8iiiA + 8iiiB= 2 8in^(A + B)co8^(A-B). [20]
8inA-8inB=: 2 C08^(A + B)8in^(A- B). [21]
C08A + C08B= 2 co8|(A + B)co8|(A — B). [22]
C08A-c08B = -2 8in^A + B)8in|(A-B). [23]
60 PLANE TRIGONOMETRY
From [20] and [21], by division, we obtain
sin ^ -h sin ^ , . , . , ^. ^ , , ^ „v
or, since ootHA - B) = :^^^^^-^y
sin A + sin B _ tan ^ (A + B)
sin A — sin B ~ tan ^(A — B)
EXERCISE XIV
[24]
1. Find the value of sin (x + y) and cos (x -f y) when
sin aj = J, cos a; = f , sin y = t^, cos y = |f .
2. Find sin (90** - y) and cos (90** - y) by making a; = 90^
in Formulas [8] and [9].
Find, by Formulas [4]-[ll], the first four functions of:
3. 90** + y. 8. 360* -y. 13. - y.
4. 180** -y. 9. 360** + y. 14. 45** -y.
5. 180** + y. 10. aj-90^ 15. 45** + y.
6. 270** -y. 11. x- 180**. 16. 30** + y.
7. 270** 4- y. 12. 05-270**. 17. 60** - y.
18. Find sin 3 a; in terms of sin x,
19. Find cos 3 a; in terms of cos x.
20. Given tan ^ a; = 1 ; find cos x.
21. Given cot ^ a; = V3 ; find sin x,
22. Given sin a; = 0.2 ; find sin ^ x and cos ^ x,
23. Given cos aj = 0.5 ; find cos 2 x and tan 2 x.
24. Given tan 45** = 1 ; find the functions of 22** 30'.
25. Given sin 30** = 0.5 ; find the functions of 15**.
sin 33** 4- sin 3**
26. Prove that tan 18** =
cos 33** + cos 3*
»-■
GONIOMETRY 61
Prove the following formulas :
• o 2tanaj ^^ ^ , sinaj
27. sin 2 g = — - — 29. tan^aj =
1 + tan^x ' 1 4- cos x
o 1 — tan^a; ^^ ^ _ sin x
28. co8 2a; = :j — - — 30. cotia; =
1 + tan^a; 1 — cos x
31. sin -J- a; ± cos ix = Vl ± sin x.
tana? i tan v
32. — ; r^ = ± tan x tan y,
cot a; ± cot y ^
33. tan(45^-a;) = f "^^^ ^
^ "^ 1 + tanaj
If ^, 5, C are the angles of a triangle, prove that :
34. sin A + sin B + sin C = 4 cos i A cos i B cos ^ C
35. cos A 4- cos B + cos C = 1 -f- 4 sin ^ ^ sin ^ 5 sin ^ C.
36. tan A ■+■ tan 5 + tan C = tan A x tan 5 x tan C.
37. cot ^ ^ + cot ^ B + cot ^ C = cot ^ ^ X cot 1^ B X cot ^ C
Change to a form more convenient for logarithmic compu-
tation :
38. cot X + tan X. 43. 1 + tan x tan y.
39. cot X — tan X. 44. 1 — tan x tan y.
40. cot X -f tan y. 46. cot x cot y -f 1.
41. cot X — tan y. 46. cot x cot y — 1.
,^ 1 — cos 2 aj tan x 4- tan «/
*2. ri JT"* 47. — - — -y — 7^-
1 + cos 2 a; cot x + cot y
SECTION XXXIII
ANTI-TRIGONOMETRIC FUNCTIONS
If y is any trigonometric function of an angle x, then x is
said to be the corresponding anti-trig onometric function of y.
Thus, if y = sin x, x is the anti-sine or inverse sine of y.
62 PLANE TRIGONOMETRY
The anti-trigonometric functions of y are written
sin~^y, tan"^y, sec~^y, vers~^y,
cos"^y, cot~^y, csc~^y, covers" ^y.
These are read, the angle whose sine is y, and so on.
For example, sin 30® = ^ ; hence, 30® = sin~^^. Similarly,
90® = Oos-^0 = sin-^l, and 45® = tan-^l = sin-^^ ■y/2, etc.
The symbol —^ must not be confused with the exponent — 1. Thus,
sm-ix is a very different expression from , which would be written
smx
(sin z)^K On the continent of Europe mathematical writers employ the
notation arcsin, arccos, etc., for sin— i, cos-i, etc.
There is an important difference between the trigonometric
and the anti-trigonometric functions. When an angle is given,
its functions are all completely determined ; but when one of
the functions is given, the angle may have any one of an indefi-
nite number of values. Thus, if sin y = ^, y may be 30®, or
150®, or either of these increased or diminished by any integral
multiple of 360® or 2 7r, but cannot take any other values.
Accordingly, sin"^^ = 30® ± 2 nir, or 150® ± 2 wtt, where n is
any positive integer. Similarly, tan^^l = 45® ±2mr or
225® ± 2 nTT ; t\e., tan-^1 = 45® ± mr.
Since one of the angles whose sine is x and one of the angles
whose cosine is x together make 90®, and. since similar relar
tions hold for the tangent and cotangent, for the secant and
cosecant, and for the versed sine and coversed sine, we have
sin~^a5 -h cos~^a5 = -^^ sec~^a; -h csc^a? = -^j
tan~^a; + cot~^a; = — > vers~^aj + covers~^a; = — j
where it must be understood that each equation is true only
for a particular choice of the various possible values of the
functions. Thus, if x is positive, and if the angles are always
4-airAn in thc first quadraut, the equations are correct.
GONIOMETRY 63
EXERCISE ZV
1. Find ail the values of the following functions :
sin-i|V3, tan-^iV3, vers- 4, cos-^(-iV2),
cso""^ V2, tan"^ oo, sec~^ 2, co8~^ (— i VS).
2. Prove that
8in~^(— a;) = — sin~^a;; cos~^(— a;)= tt — cos~^ic.
3. If sin~^a; + 8in~^y = tt, prove that x = y.
4. If y = sin~"^-J-, find tan y. \
5. Prove that cos (sin~^a;) = Vl — x\
6. Prove that cos (2 sin~^a;) =l — 2x\
7. Prove that tan (tan~^a; + tan~^y) = -j ^•
8. If ic = V^, find all the values of sin~^a; 4- cos^^aj.
9. Prove that tan"^ ( , i = sin~-^
V Vl - xy
X,
10. Find the value of sin (tan"" ^ ^).
11. Find the value of cot (2 sin" ^ ^ ) .
12. Find the value of sin (tan~^^ + tan~^^).
13. If sin~^x = 2 cos~^a;, find x.
2x
14. Prove that tan (2 tan~ ^ x) = •
^ 1 — x^
2x
16. Prove that sin (2 tan~^a:) = -'
CHAPTER IV
THE OBLIQUE TRIANGLE
SECTION XXXIV
LAW OF SINES
Let Ay By C denote the angles of a triangle ABC (Figs. 53
and 54), and a, 6, c, respectively, the lengths of the opposite
sides.
Draw CD JLABy and meeting AB (Fig. 53) or AB produced
(Fig. 54) at D, Let CD = A.
C
C >f\
D
In either figure,
In Fig. 53,
In Fig. 54,
- = sin^.
- = sm B.
a
h
- = sin (180'' -B)= sin B,
a
Therefore, whether h lies within or without the triangle, we
obtain, by division,
a sin A ^^kt
64
THE OBLIQUE TRIANGLE
65
By. drawing perpendiculars from the vertices A and B to
the opposite sides, we may obtain, in the same way,
b
c
sin^
sinC
a
c
sin A
• sin C
Hence the Law of Sines :
The sides of a triangle are lyrajportional to the sines of the
opposite angles.
If we regard these three equations as proportions, and take
them by alternation, it is evident that they may be written in
the symmetrical form
a h c
sin A sin B sin C
Each of these equal ratios has a simple geometrical meaning
which will appear if the Law of Sines is proved as follows :
Circumscribe a circle about the
triangle ABC (Fig. 66)j and draw
the radii OB, OC, Let R denote
the radius. Draw OM _L BC\ By
Geometry, the angle BOC = 2 A ;
hence, the angle BOM=: A, then
BM = R sin BOM = i2 sin ^.
.'. BC OT a = 2 R sm A,
In like manner,
b = 2 R sin B, and c = 2R sin C.
Whence we obtain
a b
TiQ. 55
2R =
c
sin A sin B sin C
That is : The ratio of any side of a triangle to the sine of
the opposite angle is numerically equal to the diameter of the
circumscribed circle.
66
PLANE TRIGONOMETRY
SECTION XXXV
LAW OF COSINES
This law gives the value of one side of a triangle in terms
of the other two sides and the. angle included between them.
BD = c — AD,
BD^AD — c\
In Figs. 56 and 57,
In Fig. m.
In Fig. 57,
In either case, BI>^ = Iff — 2 c x ^4/> + c\
Therefore, in all cases, a^ = h^ -{- AD^ -^ c^ — 2 c x AD,
Now, h^ + aH" = b\
and AD = b(io^ A,
Therefore, a" = b* + c» - 2 be cos A. [26]
In like manner it may be proved that
h^ = a^ -{■ c^ — 2 ac cos -B,
c^ = a^-irb^-2ah cos C.
The three formulas have precisely the same form, and the
Law of Cosines may be stated as follows :
The square of any side of a tria7i(/le is equal to the sum of
the squares of the other two sides diminished by twice their
product into the cosine of the included angle.
THE OBLIQUE TRIANGLE 67
SECTION XXXVI
LAW OF TANGENTS
By Sect. XXXIV, p. 64, a-, 6 = sin ^ ; sin 5 ;
whence, by the Theory of Proportion,
a — b sin A — sin B
■ ■ ■ I ; , , ^ .1 ■.■ , ■■■■■ ■ ■ ■■■ ■ «
a -\-b sin A + sin B
But by [24], p. 60,
sin A — sin B _ tan ^{A — B)
sin ^ 4- sin B tan ^(A -\- B)
rw^i o a — b tan i(A — B) -^„^
Therefore, r = t — ttt sr ' [27]
' a + b tan-i-(A + B) ^ -^
By merely changing the letters,
g — c _ tan^(^-C) 6 ~ c _ tan |(jg - C)
a + c""tani(.4 + C)' ft + <? "" tan^ (5 + C)
Hence the Law of Tangents :
The difference of two sides of a triangle is to their sum as
the tangent of half the difference of the opposite angles is to
the tangent of half their sum.
Note. If in [27] 6 > a, then B>A, The formula is still true, but to
avoid negative numbers'the formula in this case should be written
6-a _ tani(j? -A)
h + a ~ tan i (B + ^)
EXERCISE XVI
1. What do the formulas of Sect. XXXIV, p. 64, become
when one of the angles is a right angle ?
2. Prove by means of the Law of Sines that the bisector
of an angle of a triangle divides the opposite side into parts
proportional to the adjacent sides.
68 PLANE TRIGONOMETRY
3. What does Formula [26'] become when A = 90" ? when
^ = 0" ? when A = 180" ? What does the triangle become in
each of these cases ?
Note.' The case when A = 90° eKplaiius why the theorem of Sect.
XXXV, p. 06, is sometimes called the Generalized Theorem of PythagoroA.
4. Prove (Figs. 66 and 57) that whether the angle B is
acute or obtuse c = a cos B-\-b cos A, What are the two sym-
metrical formulas obtained by changing the letters? What
does the formula become when B = 90" ? -
6. From the three following equations (found in the last
example) prove the theorem of Sect. XXXV, p. 66 :
c = a cos B -^b cos A,
b = a cos C -\- c cos -4,
a =: b cos C -{- c cos B,
Hint. Multiply the first equation by c, the second by 6, the third
by a ; then from the first subtract the 8um of the second and third.
6. In Formula [27] what is the maximum value of i (A —B) ?
7. Find the form to which Formula [27] reduces, and
describe the nature of the triangle, when
(i) C = 90"; (ii) A-B = 90", and 5 = C.
SECTION XXXVII
THE GIVEN PARTS
The formulas established in Sects. XXXIV-XXXVI, pp.
64-67, together with the equation
A+B + C = 180", are sufficient
for solving every case of an
oblique triangle. The three parts
that determine an oblique triangle
may be :
THE OBLIQUE TRIANGLE
69
I. One side and two angles ;
II. Two sides and the angle opposite one of these sides ;
III. Two sides and the included angle ;
IV. The three sides.
SECTION XXXVIII
SOLUTION OF AN OBLIQUE TRIANGLE
Case I
Griven one side a and two
angles A and B; find the re-
maining parts C, b, and c.
1. C = 180*-(^ + 5).
a sm A
^ e sin C .
a sin^
a sin 5
sin A
a sin C
a
sin^
a
sin A sin A
X sin B,
X sinC.
Example, a = 24.31, A = 45'' 18', B = 22"" 11'.
The work may be arranged as follows :
a = 24.31
log a = 1.38578
= 1.38578
A = 46^ 18'
colog sin A = 0.14825
= 0.14825
JB = 22M1'
log sin B = 9.57700
log sin C = 9.96556
^+5 = 67° 29'
log b = 1.11103
log c = 1.49959
C = 112^31'
b = 12.913
c = 31.593
Note. When — 10 is omitted after a logarithm or cologarithm, it
must be remembered that the log or colog is 10 too large.
EXERCISE XVn
1. Given a = 500,
A = 10^ 12',
5 = 46*36';
find C = 123^ 12',
b = 2051.5,
c = 2362.6.
2. Given a = 795,
A = 79* 59',
5-44*41';
find C = BB"" 20',
b = 567.69,
c = 663.99.
70 PLANE TRIGONOMETRY
3. Given a = 804, ^=99*55', 5 = 45M';
findC = 35M', 6 = 577.31, c = 468.93.
4. Given a = 820, ^ = 12M9', 5 = 141*69';
find C = 25* 12', h = 2276.6, c = 1573.9.
5. Given e = 1005, A = 78* 19', . 5 = 54* 27' ;
find C = 47* 14', a = 1340.6, h = 1113.8.
6. Given b = 13.57, 5 = 13* 57', C = 57* 13' ;
find\4 = 108* 50', a = 53.276, c = 47.324.
7. Given a = 6412, A = 70* 55', C = 52* 9' ;
find 5 = 56* 56', 6 = 5685.9, c = 5357.5.
8. Given b = 999, A = 37* 58', C = 65* 2' ;
find 5 = 77*, a = 630.77, c = 929.48.
9. In order to determine the distance of a hostile fort A
from a place B, a line BC and the angles ABC and 5C^ were
measured and found to be 322.55 yards, 60* 34', and 56* 10',
respectively. Find the distance AB,
10. The angles B and C of a triangle ABC are 50* 30' and
122* 9', respectively, and BC is 9 miles. Find AB and AC.
11. Two observers 5 miles apart on a plain, and facing each
other, find that the angles of elevation of a balloon in the
same vertical plane with themselves are 55* and 58*, respec-
tively. Find the distance from the balloon to each observer,
and also the height of the balloon above the plain.
12. In a parallelogram given a diagonal d and the angles
X and y which this diagonal makes with the sides; find the
sides. Find the sides ii d = 11.237, x = 19* 1', and y = 42* 54'.
13. A lighthouse was observed from a ship to bear K. 34* E. ;
after the ship sailed due south 3 miles it bore N. 23* E. Find
the distance from the lighthouse to the ship in each position.
Note. The phrase to bear N. S49 E. means that the line of sight to
the lighthouse is in the northeast quarter of the horizon and makes,
with a line due north, an angle of 34°.
THE OBLIQUE TRIANGLE 71
14. In a trapezoid given the parallel sides a and h, and the
angles x and y at the ends of one of the parallel sides ; find
the non-parallel sides. Compute the results when a = 15,
6 = 7, aj = 70^ y = 40^
Solve the following examples without using logarithms :
16. Given h = 7.07107, A = 30*, C = 105* ; find a and c.
16. Given c = 9.562, A = 45*, 5 = 60* ; find a and h
17. The base of a triangle is 600 feet and the angles at the
base are 30* and 120*. Find the other sides and the altitude.
18. Two angles of a triangle are, the one 20*, the other 40*.
Find the ratio of the opposite sides.
19. The angles of a triangle are as 5 : 10 : 21, and the side
opposite the smallest angle is 3. Find the other sides.
20. Given one side of a triangle equal to 27, the adjacent
angles equal each to 30* ; find the radius of the circumscribed
circle. (See Sect. XXXIV, p. 65.)
SECTION XXXIX
Case II
Given two aides a and b and the angle A opposite the
side a ; find the remaining parts B, C, c.
This case, like the preceding case, is solved by means of
the Law of Sines.
„. sin 5 ft ,, « . _ hsmA ,
Since -: — 7 = -> therefore sm B = >
sm A a a
C = 180* -(A+ B).
... c sin C ,, . a sin C
And since - = -: — - > therefore c = —. — — •
a sm A sm A
72
PLANE TRIGONOMETRY
When an angle is determined by its sine it admits of two
values which are supplements of each other (Sect. XXV,
p. 48) ; hence, either value of B may be taken unless excluded
by the conditions of the problem.
If a>b, then by Geometry A > B, and B must be acute
whatever be the value of A ; for
a triangle can have only one obtuse
angle. Hence, there is one, and
only one, triangle that will satisfy
the given conditions.
If a = b, then by Geometry
-4=5; both A and B must be
acute, and the required triangle is
isosceles.
Ji a <b, then by Geometry A < B, and A must be acute in
order that the triangle may be possible. If -4 is acute, it
is evident from Fig. 61,
wheie ZBAC = A, AC = b,
CB = CB' = a, that the two
triangles ACB and ACB'
will satisfy the given condi-
tions, provided a is greater
than the perpendicular CP;
that is, provided a is greater
than b sin A (Sect. XI, p. 20). The angles ABC and AB'C are
supplementary (since Z.ABC=:Z. BB'C) ; they are, in fact,
the supplementary angles obtained from the formula
.^B
PiQ. 61
sin 5 =
b sin A
a
If, however, a = b sin A = CP (Fig. 61), then sin JB = 1,
B = 90®, and the triangle required is a right triangle.
If a<bs\nA, that is, < CP, then sin 5 > 1, and the tri-
angle is impossible.
THE OBLIQUE TRIANGLE 73
These results, for convenience, may be thus stated ;
Two solutions / if ^ is acute and th6 valu6 of a lies between
h and b sin A.
No solution ; if ^ is acute and aKh sin A \
or if ^ is obtuse and a < 6, or a = J.
One solution ; in all other cases.
The number of solutions can often be determined by inspec-
tion. In case of doubt, find the value of ^ sin ^.
Or we may proceed to compute log sin B, If log sin 5 = 0,
the triangle required is a right triangle. If log sin 5 > 0, the
triangle is impossible. If log sin -B < 0, there is one solution
when a > 6 ; there are two solutions when a < ft.
When there are two solutions, let B\ C\ c\ denote the
unknown parts of the second triangle; then,
JB' = 180** - 5, C = 180* -{A'\-B^ = B-A,
a sin C
c' =
sin^
Example 1. Given a = 16, ft = 20, ^=106''; find the
remaining parts.
In this case a<h and A > 90® ; therefore, the triangle is impossible.
Example 2. Given a = 36, ft = 80, A= SO** ; find the
remaining parts.
Here we have ft sin ^ = 80 x i = 40 ; so that a < 6 sin ^ and the
triangle is impossible.
Example 3. Given a = 72,630, ft = 117,480, A = 80* 0' 50" ;
find By Cy c.
a = 72,630
b = 117,480
^ = 80° 0' 50''
cologa = 5.13888
log 6 = 5.06997
log sin A = 9.99337
log sin 5 = 0.20222
Here log sin J? > 0.
.'. no aolviion.
74
PLAKE TRIGONOMETRY
Example 4. Given a = 13.2, b = 15.7, A = 57' 13' 15";
find B, C, c.
a = 13.2
6 = 15.7
^ = 57<'13'15''
Here log sin B^Q.
.'. a ri^^ triangle.
Example 5.
find -B, C, c.
a = 767
6 = 242
A = m"* 5.r 2''
cologa
log 6
log sin A
log sin £
B
.-. C
8.87943
1.19590
9.92467
o.oouoo
32<» 46' 45''
c
log 6
log 006 ^
logc
c
6co8^
1.19590
9.73352
0.92942
8.5
Given a = 767, b = 242, ^=36** 53' 2";
Here a > 6,
and log sin B<0.
.'. one solution.
colog a
log 6
log sin A
log sin B
B
.-. C
7.11520
2.38382
9.77830
9.27732
10° 54' 58"
132° 12' 0"
log a
log sin C
colog sin A
logc
c
2.88480
9.86970
0.22170
2.97620
946.68
Example 6. Given a = 177.01, b = 216.45, A = 35' 36' 20" ;
find the other parts.
a = 177.01
colog a = 7.75200
log a = 2.24800
2.24800
6 = 216.45
log 6 = 2.33536
logsinC =9.99462
9.23035
A = 35° 36' 20"
logsin^ = 9.76507
log sin J? = 9.85243
colog sin ^ = 0.23493
log c = 2.47755
0.23493
Here a<b,
1.71328
and log sin B<0.
B = 45° 23' 28"
c =300.29 or 51.675
.'. two solutions.
or 134° 36' 32"
.-. C = 99°0'12"
or 9° 47' 8"
EXERCISE XVm
1. Find the number of solutions of the following :
(i) a = 80, b = 100, A = 30^
(ii) a = 50, b = 100, A = 30^
(iii) a = 40, b = 100, A= 30°.
(iv) a = 13.4, • b = 11.46, A = 77° 20'.
THE OBLIQUE TRIANGLE 75
(v) a = 70, b = 75, A = 60^
(vi) a = 134.16, b = 84.54, B = 52** 9' 11".
(vii) a = 200, b = 100, A = 30^
2. Given a = 840, b = 485, ^ = 21* 31' ;
find 5 = 12M3' 34", C = 146* 15' 26", <? = 1272L1.
3. Given a = 9.399, b = 9.197, A = 120* 35' ;
find B = 57* 23' 40", C = 2* 1' 20", c = 0.38525.
4. Given a = 91.06, ft = 77.04, ^ = 51* 9' 6";
find 5 = 41* 13', C = 87* 37' 54", c = 116.82.
5. Given a = 55.55, b == 66M, 5 = 77*44' 40";
find ^=54* 31' 13", C = 47*44'7", c = 50.481.
6. Given a = 309, ft = 360, ^ = 21* 14' 25";
find B = 24* 57' 54", C = 133* 47' 41", c = 615.67,
B' = 155* 2' 6", C = 3* 43' 29", c' = 55.41.
7. Given a = 8.716, -ft = 9.787, ^ = 38* 14' 12";
find B = 44* 1' 28", C = 97* 44' 20", c = 13.954,
B' = 135* 58' 32", C = 5* 47' 16", c' = 1.4202.
8. Given a = 4.4, ft = 5.21, ^ = 57* 37' 17";
find 5 = 90*, C = 32* 22' 43", c = 2.7901.
9. Given a = 34, ft = 22, 5 = 30*20';
find A = 51* 18' 27", C = 98* 21' 33", c = 43.098,
A' = 128* 41' 33",C' = 20* 58' 27", c' = 15.593.
10. Given ft = 19, c = 18, C = 15* 49';
find 5 = 16* 43' 13", A = 147* 27' 47", a = 35.519,
B' = 163* 16' 47",'^ = 0* 54' 13", a'= 1.0415.
11. Given a = 75, ft = 29, 5 = 16* 15' 36" ; find the differ-
ence between the areas of the two corresponding triangles.
12. Given in a parallelogram the side a, a diagonal d, and
the angle A made by the two diagonals ; find the other diagonal.
Special case : a = 35, <Z = 63, A = 21* 36' 30".
76 PLANE TRIGONOMETRY
SECTION XL
m
Case III
Given two sides a and b and the included angle C ; find the
remaining parts A, B, and c.
Solution I. The angles A and B may both be found by-
means of Fonnula [27], Sect. XXXVI, p. 67, which may be
written ,
tan |(^ - 5) = ——7 X tan i(^ + JB).
Since i(^ +5) = i(180* - C), the value of i(A-\-B) is
known, go that this equation enables us to find the value of
i(A — 5). We then have
i(A-\-B)^i(A-B) = A
and i(A '{-B)-i(A-B) = B.
After A and B are known, the side c may be found by the
Law of Sines, which gives its value in two ways, as follows :
a sin C b sin C
c = — : — --9 or c = — : — — ■•
sin A sm B
Solution II. The third side c may be found directly from
the equation (Sect. XXXV, p. 66)
c = Va^ + 6^ - 2 a6 cos C ;
and then, by the Law of Sines, the following equations for
computing the values of the angles A and B are obtained :
sin C . „ , sin C
sin A = a X. > sin B =^ o X
THE OBLIQUE TRIANGLE
77
Solution III. If, in the triangle ABC (Fig. 62), BD is
drawn perpendicular to the side A C, then
B
BD
tan -4 = — — =
BD
Now
and
'.tan^ =
AD AC -DC
BDzzza sin C
DC = a cos C
a sin C
b — a cos C
By merely changing the letters,
b sin C
D
FlQ. G2
C
tan 5 =
a — b cos C
It is not necessary, however, to use both formulas. When
one angle, as A, has been found, the other, B, may be found
from the relation ^ + 5 + C = 180^
When the angles are known, the third side is found by the
Law of Sines, as in Solution I.
Note. When all three unknown parts are required, Solution I is the
most convenient in practice. When only the third side, c, is desired, Solu-
tion II may be used to advantage, provided the values of a^ and 6^ can
be obtained readily without the aid of logarithms. But Solutions II and
III are not adapted to logarithmic work.
Example 1. Given a = 748, b = 375, C = 63'' 35' 30" ; find
-4, B, and c.
a + 6 =
1123
a — 6 =
373
^+5 =
116°
24' 30^'
\(A + B) =
bS""
12' 15''
H^-B) =
28<»
10' 64"
A =
86°
23' 9'
B =
30°
1'21"
log(a-- 6)= 2.57171
colog(a + 6) = 6.94962
log tan i (^ + J?) = 020766
log t!ini(A-B) = 9. 72899
i (^ - ^ = 28° 10' 64"
log 6
log sin C
colog sin B
logc
c
2.67403
9.96214
0.. 30073
2.82690
671.27
Note. In the above example we use the angle B in finding the side
c rather than the angle A, because A is near 90°, and therefore the use
of its sine should be avoided. See Note, p. 23.
78 PLANE TRIGONOMETRY
Example 2. Given a = 4t,c = 6, B = 60®; find the third side b.
Here Solution II may be used to advantage. We have
b = Va2 + c2 - 2 ac cos B = Vl6 + 30 - 24 = V28;
log 28 = 1.44716, log V28 = 0.72358, V28 = 5.2915 ;
that is, b = 5.2915.
EXERCISE XIX
1. Given a = 77.99, 5 = 83.39, C = 72°15';
find A = 5r 15', B = 56° 30', c = 95.24.
2. Given b = 872.5, c = 632.7, ^ = 80® ;
find 5 = 60® 45' 2", C = 39® 14' 58", a = 984.83.
3. Given a = 17, ft = 12, C = 59®17';
find ^ = 77® 12' 53", B = 43® 30' 7", c = 14.987.
4. Givenft = V5, <j = V3, ^=35® 53';
find B = 93® 28' 36", C = 50® 38' 24", a = 1.3131.
6. Given a = 0.917, ft = 0.312, C = 33®7'9";
find A = 132® 18' 27", B = 14® 34' 24", c = 0.6775.
6. Given a = 13.715, c = 11.214, 5 = 15®22'36";
find^ = 118®55'49", C = 45®41'35", ft = 4.1554.
7. Given ft = 3000.9, c = 1587.2, A = 86® 4' 4" ;
find B = 65® 13' 51", C = 28® 42' 5", a = 3297.2.
8. Given a = 4527, ft = 3465, C = 66®6'27";
find ^=68® 29' 15", 5 = 45® 24' 18", c = 4449.
9. Given a = 55.14, ft = 33.09, C = 30®24';
find^ = 117®24'32", 5 = 32®11'28", c = 31.431.
10. Given a = 47.99, ft = 33.14, C = 175®19'10";
find A =^® 46' 8", B = 1® 54' 42", c = 81.066.
11. If two sides of a triangle are each equal to 6, and the
included angle is 60®, find the third side.
THE OBLIQUE TRIANGLE 79
12. If two sides of a triangle are each equal to 6, and the
included angle is 120°, find the third side.
13. Apply Solution I to the case in which a is equal to h ;
that is, the case in which the triangle is isosceles.
14. If two sides of a triangle are 10 and 11, and the included
angle is 50°, find the third side.
15. If two sides of a triangle are 43.301 and 25, and the
included angle is 30°, find the third side.
16. In order to find the distance between two objects, A
and B, separated by a swamp, a station C was chosen, and the
distances CA = 3825 yards, CB = 3475.6 yards, together with
the angle ^C^ = Q2^ 31', were measured. Find the distance
from A to B,
17. Two inaccessible objects, A and By are each viewed from
two stations, C and D, on the same side oi AB and 562 yards
apart. The angle ACB is ^2"" 12', BCD 41° 8', ADB 60° 49',
and ADC 34° 51'; required the distance AB.
18. Two trains start at the same time from the same station
and move along straight tracks that form an angle of 30°, one
train at the rate of 30 miles an hour, the other at the rate of
40 miles an hour. How far apart are the trains at the end of
half an hour ?
19. In a parallelogram given the two diagonals 5 and 6,
and the angle that they form 49° 18' ; find the sides.
20. In a triangle one angle is 139° 54', and the sides forming
the angle have the ratio 5 : 9. Find the other two angles.
.21. In order to find' the distance between two objects, A
and B, separated by a pond, a station C was chosen, and the
distances CA = 426 yards, CB = 322.4 yards, together with
the angle ACB = 68° 42', were measured. Find the distance
from A to B.
80
PLANE TRIGONOMETRY
SECTION XLI
Cask IV
Given the three sides a, b, c ; find the angles A, B, C.
The angles may be found directly from the fonnulas estab-
lished in Sect. XXXY, p. 66. Thus, from the formula
a* = b^'\-c^ — 2becosAy
cos A =
2 be
Erom this equation formulas adapted to logarithmic work
are deduced as follows :
For the sake of brevity, let
a '\- b '\- c = 2s;
then 6 -f- c — a = 2 (5 — a),
a — b-{-c = 2(s — b),
and a-^b — c = 2(s — c).
Then the value of 1 — cos A is
P -{- c"^ - a^ _ 2bc - b^ - c^ -\- a^ _ a^ - (b - cY
2bc ~ 2 be ~ 2 be
— (^ + ^ — c) (fl^ — ^ + g)
■" 2 be
^ 2(s-b)(s-e)
be
and the value of 1 -f- cos A is
b^-\-c^-a^ ^ 2be-^b^-\-e^-a^ _ (b + ey-a^
2be " 2 bo ~ 2 be
^ (b -\- e -\- a) {b -\- e - a) ^ 2s{s-a)
2 be " be '
THE OBLIQUE TRIANGLE 81
But from Formulas [16] and [17], p. 58, it follows that
1 — cos A =2 sin^ ^ A, and 1 -\- cos A =z2 cos^ i A,
o • 01 ^ 2(s — b)(s — c) , _ ,_ ^ 28(8 — 0)
.'.2sm^iA=-^ f^ ^> and 2cosH^= — h ^>
be be
whence sin ^ A = ^iill^Hillii , [28]
■cosiA = ^II, [29]
and by [2], tan ^ A= J(8 - b) («» - c) . .g^,
' 8 (s — a)
By merely changing the letters,
ae ao
eosifi = V^^3, cos i C = V*-^^-
^ ^ ae ^ ^ ab
* ^ 8(s — b) ^ ^ s(s — c)
There is then a choice of three different formulas for finding
the value of each angle. If half the angle is very near 0®,
the formula for the cosine will not give a very accurate result,
because the cosines of angles near 0° differ little in value ; and
the same holds true of the formula for the sine when half the
angle is very near 90®. Hence, in the first case the formula
for the sine, in the second that for the cosine, should be used.
But, in general, the formulas for the tangent are to be
preferred.
It is not necessary to compute by the formulas more than
two angles ; for the third may then be found from the equation
A-\-B + C = 180°.
S2
PLANE TRIGONOMETRY
There is this advantage, however, in computing all three
angles by the formulas, that we may then use the sum of the
angles as a test of the accuracy of the results.
In case it is desired to compute all the angles, the formulas
for the tangent may be put in a more convenient form.
The value of tan i A may be written
(^-c)
or
s
^ s(s ^ a)
1 l (s-a)(s-b)(s-c)
— a^ s
Hence, if we put
1 (8 - a) (8 - b) (73^ ^ ^^
[31]
we have
tan^ A =
s — a
T T
Likewise, tan \B = r > tan \C =
s-b
s — c
[32]
Example 1. Given a = 3.41, b = 2.60, c = 1.58 ; find the
angles.
Using Formula [30] and the corresponding formula for tan^B, we
may arrange the work as follows :
a
b
c
2s
s
8 — a
8-b
8 — C
3.41
2.60
1.68
7.59
3.795
0.385
1.195
2.215
colog 8 = 9.42079
colog (s — a) = 0.41454
log (8 -&) = 0.07737
log (8 - c) = 0.34637
2 )0.25807
log tan i^ =0.12903
i^= 53° 23' 20"
A = 106° 46' 40''
colog 8= 9.42079- 10
log (8- a) = 9.58546- 10
colog (8 -b)= 9.92263 - 10
log (8 - c) = 0.34637
2 )19.27425-20
log tan i B = 9.63713 - 10
i B = 23° 26' 37"
B = 46° 53' 14"
,'.A + B = 153° 39' 54", and C = 26° 20' 6".
THE OBLIQUE TRIANGLE
83
Example 2. Solve Example 1 by finding all three angles
by the use of Formulas [31] and [32],
H^re the work may be compactly arranged as follows, if we find
log tan^^, etc., by subtracting log(s — a), etc., from logr instead of
adding the cologarithm:
d = 3.41
6 = 2.60
c = 1.68
2« = 7.69
« = 3.795
8 - a = 0.385
«- 6 = 1.195
«-c = 2.215
log(«~ a) = 9.58546
log (s-b) = 0.07737
log(s-c) = 0.34537
colog 8 = 9.42079
log r2 = 9.42899
logr =9.71450
2« = 7.590 (check).
log tan i -4 = 10.12903
log tan iB = 9.63713
log tan i C = 9.36912
iA= 53° 23' 20
iB= 23°26'37
iC= 13°10' 3
//
//
//
A = 106° 46' 40^'
B= 46° 53' 14''
C= 26° 20' 6"
Check, A-\-B-\-C= 180° 0' 0"
Note. Even if no mistakes are made in the work, the sum of the
three angles found as above may differ very slightly from 180° in conse-
quence of the fact that logarithmic computation is at best only a method
of close approximation. When a difference of this kind exists, it should
be divided among the angles according to the probable amount of error
for each angle.
EXERaSE XX
Solve the following triangles, taking the three sides as the
given parts :
1
a
b
c
A
B
C
51
65
20
38° 52^ 48"
126° 52' 12"
14° 15'
2
78
101
29
32° 10' 55"
136° 23' 50"
11° 26' 16"
3
111
145
40
27° 20^ 32"
143° 7' 48"
9° 31' 40"
4
21
26
31
42° 6' 13"
56° 6' 36"
81° 47' 11"
5
19
84
49
16° 25' 36"
30° 24'
133° 10' 24"
6
43
50
57
46° 49' 35"
57° 59^ 44"
75° 10' 41"
7
37
58
79
26° 0'29"
43° 25' 20"
110° 34' 11"
8
73
82
91
49° 34' 58"
58° 46' 58"
71° 38' 4"
9
14.493
55.4363
66.9129
8° 20' 1"
33° 40' 5"
137° 69' 54"
10
V6
V6
V7
61° 53' 12"
59° 31' 48"
68° 35'
84 PLANE TRIGONOMETRY
11. Given a = 6, b = S, c = 10; find the angles.
12. Given a = 6, ft = 6, c = 10 ; find the angles.
13. Given a = 6, 6 = 6, c = 6; find the angles.
14. Given a = 6, 6 = 9, c = 12 ; find the angles.
15. Given a = 2, b = VG, c = VS — 1 ; find the angles.
16. Given a = 2, b = V6, c = Vs + 1 ; find the angles.
17. The distances between three cities, A, J5, and C, are as
follows : AB = 165 miles, AC = 72 miles, and J5C = 185 miles.
B is due east from A, In what direction is C from A ? What
two answers are admissible ?
18. Under what visual angle is an object 7 feet long seen
by an observer whose eye is 5 feet from one end of the object
and 8 feet from the other end ?
19. When Formula [28] is used for finding the value of an
angle, why does the ambiguity that occurs in Case II not exist ?
20. If the sides of. a triangle are 3, 4, and 6, find the sine of
the largest angle.
21. Of three towns, A, B, and C, A is 200 miles from B and
184 miles from C, B is 150 miles due north from C. How far
is A north of C ?
22. The sides of a triangle are 78.9, 65.4, 97.3, respectively.
Find the largest angle.
23. The sides of a triangle are 487.25, 512.33, 544.37,
respectively. Find the smallest angle.
24. Find the angles of a triangle whose sides are — ^ )
V3-1 V3 \. , 2V2
-^) -^> respectively.
25. The sides of a triangle are 14.6 inches, 16.7 inches, and
18.8 inches, respectively. Find the length of the perpendic-
ular from the vertex of the largest angle upon the opposite side.
THE OBLIQUE TRIANGLE
85
SECTION XLH
AREA OF A TRIANGLE
Case I
When two sides and the included angle are given^
I^ the triangle ABC (Fig. 64 or 65),
F = ^cxCD.
Now, CD = a sin B,
Therefore, F = ^ ac sin B.
Also, F = ^ a6 sin C, and F = ^bc sin A,
[33]
Case II
When a side and the two adjacent angles are given,
sin A : sin C = a\c, (Sect. XXXIV, p. 65.)
a sin C
Therefore,
sin^l
Putting this value of c in Formula [33],
„ a* sin B sin C
F =z
2 sin A
But sin (5 -f- C) = sin (180° - ^ ) = sin A . (Sect. XXV, p. 48.)
a^ sin B sin C
Hence,
F =
2sin(B + C)
[34]
S6
PLANE TRIGONOMETRY
Case III
When the three sides of a triangle are given.
By Formula [12], p. 58,
sin 5 = 2 sin ^ 5 X cos i B,
Now, by Formula [28], p. 81,
• 1 ^ ^ /(s — a)(s — c)
smiB = \^ ^ ^9
and by Formula [29], p. 81,
COS
ac
By substituting these values of sin i B and cos ^ -B in the
above equation, we have
sm
mB = — V5 (s — a) (« — b) (s — c),
etc
By putting this value of sin B in [33], we have
F = V8(8 — a) (s — b) (s — c).
[36]
Case IV
When the three sides and the radius of the circumscribed
circle or the radius of the inscribed circle are given.
If R denotes the radius of the
circumscribed circle, we have, from
Sect. XXXIV, p. 65,
b
8inB
2R
By putting this value of sinB
in [33], we have
F =
abc
4R
[36]
Fig. 66
THE OBLIQUE TRIANGLE 87
If r denotes the radius of the inscribed circle, divide the
triangle into three triangles by lines from the centre of this
circle to the vertices ; then the altitude of each of the three
triangles is equal to r. Therefore,
F = ^r(a + b + c) = rs; [37]
By putting in this formula the value of F given in [35],
whence r, in [31], Sect. XLI, p. 82, is equal to the radius of
the inscribed circle.
EXERCISE XXI
Find the area :
1. Given a = 4474.5, 6 = 2164.5, C = 116° 30' 20".
2. Given b = 21.66, c = 36.94, A = 66° 4' 19".
3. Given a = 510, c = 173, B = 162° 30' 28".
4. Given a = 408, b = 41, c = 401.
5. Given a = 40, 6 = 13, c = 37.
6. Given a = 624, 6 = 205, c = 445.
7. Given b = 149, A = 70° 42' 30", B = 39° 18' 28".
8. Given a = 215.9, c = 307.7, A = 25° 9' 31".
9. Given 6 = 8, c = 5, A = 60°.
10. Given a = 7, c = 3, A = 60°.
11. Given a = 60, 5 = 40° 35' 12", area = 12 ; find the
radius of the inscribed circle.
12. Obtain a formula for the area of a parallelogram in
terms of two adjacent sides and the included angle.
13. Obtain a formula for the area of an isosceles trapezoid
in terms of the two parallel sides and an acute angle.
88 PLANE TRIGONOMETRY
14. Two sides and included angle of a triangle are 2416,
1712, and 30° ; and two sides and included angle of another
triangle are 1948, 2848, and 150°. Find the sum of their areas.
15. The base of an isosceles triangle is 20, and its area is
100 -i- Vs ; find its angles.
16. Show that the area of a quadrilateral is equal to one-
half the product of its diagonals into the sine of their included
angle.
EXERCISE XXn
1. From a ship sailing down the English Channel the Eddy-
stone was observed to bear N. 33° 45' W., and after the ship
had sailed 18 miles S. 67° 30' W. it bore N. 11° 15' E. Find
its distance from each position of the ship.
2. Two objects, A and B, were observed from a ship to be
at the same instant in a line bearing N. 15° E. The ship then
sailed northwest 5 miles, when it was found that A bore due
east and B bore northeast. Find the distance from A to B,
3. A castle and a monument stand on the same horizontal
plane. The angles of depression of the top and the bottom of
the monument viewed from the top of the castle are 40° and
80° ; the height of the castle is 140 feet. Find the height of
the monument.
4. If the sun's altitude is 60°, what angle must a stick make
with the horizon in order that its shadow in a horizontal plane
may be the longest possible ?
5. If the sun's altitude is 30°, find the length of the longest
shadow cast on a horizontal plane by a stick 10 feet in length.
6. In a circle with the radius 3 find the area of the part
comprised between parallel chords whose lengths are 4 and 5.
(Two solutions.)
CHAPTER V
MISCELLANEOUS EXAMPLES
Problems in Plane Trigonometry
Cr.
1^^^
i
-/.•
• ■• - — •
Pig. 67
If two objects are not in the same horizontal plane with
each other or with the point of observation, we may suppose
vertical lines to be passed through the two objects and to
meet the horizontal plane of the point of observation in two
points. The angular distance of these two points is the bear-
ing of either of the objects from the other. Thus, the angle
COD' (Fig. 67) is the bearing of C from 2).
Note. "Problems in Plane Trigonometry" are selected from those
published by Mr. Charles W. Seaver, Cambridge, Msuss. The full set can
be obtained from him in pamphlet form.
89
90 PLANE TRIGONOMETRY
EXERCISE XXm
RIGHT TRIANGLES
1. The angle of elevation of a tower is 48** 19' 14", and
the distance of the base from the point of observation is
95 feet. Find the height of the tower and the distance of
its top from the point of observation.
2. From a mountain 1000 feet high, the angle of depression
of a ship is 77** 35' 11". Find the distance of the ship from
the summit of the mountain.
3. A flagstaff 90 feet high, on a horizontal plane, casts a
shadow of 117 feet. Find the altitude of the sun.
4. When the moon is setting at any place, the angle at the
moon subtended by the earth's radius passing through that
place is 57' 3". If the earth's radius is 3956.2 miles, what is
the moon's distance from the earth's centre ?
5. The angle at the earth's centre subtended by the sun's
radius is 16' 2" and the sun's distance is 92,400,000 miles.
Find the sun's diameter in miles.
6. The latitude of Cambridge, Mass., is 42° 22' 49". What
is the length of the radius of that parallel of latitude ?
7. At what latitude is the circumference of the parallel of
latitude half of that of the equator ?
8. In a circle with a radius of 6.7 is inscribed a regular
polygon of thirteen sides. Find the length of one of its sides.
9. A regular heptagon, one side of which is 5.73, is
inscribed in a circle. Find the radius of the circle.
10. A tower 93.97 feet high is situated on the bank of a
river. The angle of depression of an object on the opposite
bank is 25° 12' 54". Find the breadth of the river.
MISCELLANEOUS EXAMPLES 91
11. From a tower 58 feet high the angles of depression of
two objects situated in the same horizontal line with the
base of the tower, and on the same side, are 30® 13' 18" and
45° 46' 14". Find the distance between these two objects.
12. Standing directly in front of one corner of a flat-roofed
house, which is 150 feet in length, I observe that the hori-
zontal angle which the length subtends has for its cosine V|^,
and that the vertical angle subtended by its height has for its
3
sine — ;=• What is the height of the house ?
V34
13. A regular pyramid, with a square base, has a lateral
edge 150 feet long, and a side of its base is 200 feet. Find
the inclination of the face of the pyramid to the base.
14. From one edge of a ditch 36 feet wide, the angle of
elevation of a wall on the opposite edge is 62° 39' 10". Find
the length of a ladder that will just reach from the point
of observation to the top of the wall.
15. The top of a flagstaff has been partly broken off and
touches the ground at a distance of 15 feet from the foot of
the staff. If the length of the broken part is 39 feet, find the
length of the whole staff.
16. From a balloon, which is directly above one town, is
observed the angle of depression of another town, 10° 14' 9".
The towns being 8 miles apart, find the height of the balloon.
17. From the top of a mountain 3 miles high the angle of
depression of the most distant object which is visible on the
earth's surface is found to be 2° 13' 50". Find the diameter
of the earth.
18. A ladder 40 feet long reaches a window 33 feet high,
on one side o£ a street. Being turned over upon its foot, it
reaches another window 21 feet high, on the opposite side of
the street. Find the width of the street.
92 PLANE TRIGONOMETRY
19. The height of a house subtends a right angle at a
window on the other side of the street; and the angle of
elevation of the top of the house, from the same point, is 60**.
The street is 30 feet wide. How high is the house ?
20. A lighthouse 54 feet high is situated on a rock. The
angle of elevation of the top of the lighthouse, as observed
from a ship, is 4^ 52', and the angle of elevation of the top of
the rock is 4^ 2', Find the height of the rock and its distance
from the ship.
21. A man in a balloon observes the angle of depression of
an object on the ground, bearing south, to be 35® 30'; the
balloon drifts 2^ miles east at the same height, when the angle
of depression of the same object is 23® 14'. Find the height
of the balloon.
22. A man standing south of a tower, on the same horizon-
tal plane, observes its angle of elevation to be 54® 16' ; he goes
east 100 yards, and then finds its angle of elevation is 50® 8'.
Find the height of the tower.
23. The angle of elevation of a tower at a place A, south of
it, is 30® ; and at a place B, west of A, and at a distance of a from
it, the angle of elevation is 18®. Show that the height of the
rz __ ^
tower is = ; the tangent of 18® being
V2-f-2V5' VlO -f- 2 V5
24. A pole is fixed on the top of a mound, and the angles
of elevation of the top and the bottom of the pole are 60® and
30®, respectively. Prove that the length of the pole is twice
the height of the mound.
26. At a distance a from the foot of a tower, the angle
of elevation A of the top of the tower is tha complement of
the angle of elevation of a flagstaff on top of it. Show that
the leugth of the staff is 2 a cot 2 ^.
MISCELLANEOUS EXAMPLES 93
26. A line of true level is a line every point of which is
equally distant from the centre of the earth. A line drawn
tangent to a line of true level at any point is a line of
apparent level. If at any point both these lines are drawn,
and extended one mile, find the distance they are then apart.
27. In Problem 1, page 90, determine the effect upon the
computed height of the tower of an error in either the angle
of elevation or the measured distance.
OBLIQUE TRIANGLES
28. To determine the height of an inaccessible object
situated on a horizontal plane, by observing its angles of
elevation at two points in the same line with its base, and
measuring the distance between these two points.
29. The angle of elevation of an inaccessible tower situated
on a horizontal plane is 63° 26'; at a point 500 feet farther
from the base of the tower the angle of elevation of its top is
32* 14'. Pind the height of the tower.
30. A tower is situated on the bank of a river. Prom the
opposite bank the angle of elevation of the tower is 60° 13',
and from a point 40 feet more distant the angle of elevation
is 50° 19'. Pind the breadth of the river.
31. A ship sailing north sees two lighthouses 8 miles apart,
in a line due west ; after an hour's sailing, one lighthouse
bears S.W., and the other S.S.W. Pind the ship's rate.
32. To determine the height of an accessible object situated
on an inclined plane.
33. At the distance of 40 feet from the foot of a tower on
an inclined plane, the tower subtends an angle of 41° 19' ; at
a point 60 feet farther away, the angle subtended by the tower
is 23° 45'. Pind the height of the tower.
94 PLANE TRIGONOMETRY
34. A tower makes an angle of 113" 12' with the inclined
plane on which it stands ; and at a distance of 89 feet from its
base, measured down the plane, the angle subtended by the
tower Is 23'' 27'. Find the height of the tower.
35. From the top of a house 42 feet high the angle of
elevation of the top of a pole is 14** 13'; at the bottom of
the house it is 23** 19'. Find the height of the pole.
36. The sides of a triangle are 17, 21, 28. Prove that the
length of a line bisecting the greatest side and drawn from
the opposite angle is 13.
37. A privateer, 10 miles S.W. of a harbor, sees a ship sail
from it in a direction S. 80" E., at a rate of 9 miles an hour.
In what direction, and at what rate, must the privateer sail
in order to come up with the ship in 1^ hours ?
38. A person goes 70 yards up a slope of 1 in 3^ from the
edge of a river and observes the angle of depression of an
object on the opposite bank to be 2^^". Find the breadth of
the river.
39. The length of a lake subtends, at a certain point, an
angle of 46" 24', and the distances from this point to the two
extremities of the lake are 346 and 290 feet. Find the length
of the lake.
40. Two ships are a mile apart. The angular distance of
the first ship from a fort on shore, as observed from the second
ship, is 36" 14' 10" ; the angular distance of the second ship
from the fort, observed from the first ship, is 42" 11' 53".
Find the distance in feet from each ship to the fort.
41. Along the bank of a river is drawn a base line of 500
feet. The angular distance of one end of this line from an
objection the opposite side of the river, as observed from the
other end of the line, is 53" ; that of the second extremity
MISCELLANEOUS EXAMPLES 95
from the same object, observed at the first, is 79** 12'. Find
the breadth of the river.
42. A vertical tower stands on a declivity inclined 15® to
the horizon. A man ascends the declivity 80 feet ^from the
base of the tower, and finds the angle then subtended by the
tower to be 30°. Find the height of the tower.
43. The angle subtended by a tower on an inclined plane is;
at a certain point, 42° 17'; 325 feet farther down it is 21° 47'.
The inclination of the plane is 8° 53'. Find the height of the
tower. •
44. A cape bears north by east, as seen from a ship. The
ship sails northwest 30 miles, and then the cape bears east.
How far is it from the second point of observation ?
45. Two observers, stationed on opposite sides of a cloud,
observe its angles of elevation to be 44° 56' and 36° 4'. Their
distance from each other is 700 feet. What is the height of
the cloud ?
46. From a point B at the foot of a mountain, the angle of
elevation of the top A is 60°. After ascending the mountain
one mile, at an inclination of 30° to the horizon, and reaching
a point C, the angle A CB is found to be 135°. Find the height
of the mountain in feet.
47. From a ship two rocks are seen in the same right line
with the ship, bearing !N". 15° E. After the ship has sailed
northwest 5 miles, the first rock bears east, and the second
northeast. Find the distance between the rocks.
48. From a window on a level with the bottom of a steeple
the angle of elevation of the steeple is 40°, and from a second
window 18 feet higjier the angle of elevation is 37° 30'. Find
the height of the steeple.
96 PLANE TRIGONOMETRY
49. To determine the distance between two inaccessible
objects by observing angles at the extremities of a line of
known length.
60. Wishing to determine the distance between a church A
and a tower Bj on the opposite side of a river, I measure a
line CD along the river (C being nearly opposite A), and
observe the angles A CB, 58** 20' ; A CD, 95** 20' ; ADB, 53** 30';
BDC, 98** 46'. CD is 600 feet. What is the distance required?
51. Wishing to find the height of a summit A, I measure a
horizontal base line CD, 440 .yards. At C, the angle of eleva-
tion of A is 37** 18', and the horizontal angle between D and
the summit is 76** 18' ; at D, the horizontal angle between C
and the summit is 67** 14'. Find the height.
52. A balloon is observed from two stations 3000 feet apart.
At the first station the horizontal angle of the balloon and the
other station is 75** 35', and the angle of elevation of the balloon
is 18**. The horizontal angle of the first station and the bal-
loon, measured at the second station, is 64** 30'. Find the
height of the balloon.
53. Two forces, one of 410 pounds, and the other of 320
pounds, make an angle of 51** 37'. Find the intensity and the
direction of their resultant.
54. An unknown force, combined with one of 128 pounds,
produces a resultant of 200 pounds, and this resultant makes
an angle of 18** 24' with the known force. Find the intensity
and direction of the unknown force.
55. At two stations, the height of a kite subtends the same
angle A, The angle which the line joining one station and
the kite subtends at the other station is B ; and the distance
between the two stations is a. Show that the height of the
kite is ^ a sin A sec B,
MISCELLANEOUS EXAMPLES 97
66. Two towers on a horizontal plane are 120 feet apart. A
person standing successively at their bases observes that the
angle of elevation of one is double that of the other; but,
when he is half-way between them, the angles of elevation are
complementary. Prove that the heights of the towers are 90
and 40 feet.
67. To find the distance of an inaccessible point C from
either of two points A and B, having no instruments to meas-
ure angles. Prolong CA to a, and CB to h, and join AB^ Ah,
and Ba, Measure AB, 500 ; a A, 100 ; aBy 660 ; bB, 100 ; and
Ab, 550. Compute the distances AC and BC,
68. Two inaccessible points A and B are visible from D,
but no other point can be found whence both are visible.
Take some point C, whence A and D can be seen, and meas-
ure CD, 200 feet; ADC, 89**; ACD, 60° 30'. Then take some
point E, whence D and J5 are visible, and measure DE, 200
feet; 5Z)^, 64** 30' ; 5£:Z), 88** 30'. At Z) measure ^Z)5, 72** 30'.
Compute the distance AB,
69. To compute the horizontal distance between two inac-
cessible points A and B, when no point can be found whence
both can be seen. Take two points C and D, distant 200 yards,
so that A can be seen from C, and B from Z). From C meas-
ure CF, 200 yards to F, whence A can be seen ; and from D
measure DE, 200 yards to E, whence B can be seen. Measure
AFC, 83**; ACD, 53** 30'; ACF, 54** 31'; BDE, 54** 30'; BDC,
166** 25'; DEB, 88** 30'.
60. A column in the north temperate zone is east-southeast
of an observer, and at noon the extremity of its shadow is
northeast of him. The shadow is 80 feel in length, and the
elevation of the column, at the observer's station, is 45**.
Find the height of the column.
98 PLANE TRIGONOMETRY
61. From the top of a hill the angles of depression of two
objects" situated in the horizontal plane of the base of the hill
are 45® and 30**; and the horizontal angle between the two
objects is 30°. Show that the height of the hill is equal to
the distance between the objects.
62. Wishing to know the breadth of a river from A to B, 1
take AC, 100 yards in the prolongation of BA, and then take
CDy 200 yards at right angles to AC. The angle BDA is
37° 18' 30". Find AB,
63. The sum of the sides of a triangle is 100. The angle at
A is double that at By and the angle at B is double that at C
Determine the sides.
64. If sinM + 5 cos^^ = 3, find A.
65. If sin*^ = m cos A — n, find cos A.
66. Given sin ^ = m sin B, and tan A =n tan B ; find sin A
and cos B,
67. If tanM + 4 sin^^ = 6, find A,
68. If sin A = sin 2Aj find A.
69. If tan 2 ^ = 3 tan ^, find A.
70. Prove that tan 50° + cot 50° = 2 sec 10°.
71. Given a regular polygon of n sides, and calling one of
them a, find expressions for the radii of the inscribed and the
circumscribed circles in terms of n and a.
If P, H, D are the sides of a regular inscribed pentagon,
hexagon, and decagon, prove P^ = H^ -\- D\
AREAS
72. Obtain the formula for the area of a triangle, given two
sides h, c, and the included angle A.
73. Obtain the formula for the area of a triangle, given two
angles A and B, and included side c.
MISCELLANEOUS EXAMPLES 99
74. Obtain the formula for the area of a triangle, given the
three sides.
75. If a is the side of an equilateral triangle, show that
its area is — - —
4
76. Two consecutive sides of a rectangle are 52.25 chains
and 38.24 chains. Eind the area.
77. Two sides of a parallelogram are 59.8 chains and 37.05
chains, and the included angle is 72® 10'. Find the area.
78. Two sides of a parallelogram are 15.36 chains and 11.46
chains, and the included angle is 47® 30'. Find the area.
79. Two sides of a triangle are 12.38 chains and 6.78 chains,
feind the included angle is 46® 24'. Find the area.
80. Two sides of a triangle are 18.37 chains and 13.44
chains, and they form a right angle. Find the area.
81. Two angles of a triangle are 76® 54' and 57® 33' 12",
and the included side is 9 chains. Find the area.
82. Two sides of a triangle are 19.74 chains and 17.34
chains. The first bears N. 82® 30' W. ; the second S. 24® 15' E.
Find the area.
83. The three sides of a triangle are 49 chains, 50.25 chains,
and 25.69 chains. Find the area.
84. The three sides of a triangle are 10.64 chains, 12.28
chains, and 9 chains. Find the area.
86. The sides of a triangular field, of which the area is 14 *
acres, are in the ratio of 3, 5, 7. Find the sides.
86. In the quadrilateral ABCD we have AB, 17.22 chains ;
AD, 7.45 chains ; CD, 14.10 chains ; EC, 5.25 chains ; and the
diagonal A C, 15.04 chains. Find the area.
100 PLANE TRIGONOMETRY
87. The diagonals of a quadrilateral are a and ft, and they
intersect at an angle D. Show that the area of the quadri-
lateral is ^ ah sin Z>.
88. The diagonals of a quadrilateral are 34 and 56, inter-
secting at an angle of 67**. Find the area.
89. The diagonals of a quadrilateral are 75 and 49, inter-
secting at an angle of 42®. Find the area.
90. , Show that the area of a regular polygon of n sides, of
, no" 180°
which one is a, is — ^ cot
4 n
91. One side of a regular pentagon is 25. Find the area.
92. One side of a regular hexagon is 32. Find the area.
93. One side of a regular decagon is 46. Find the area.
94. Find the area of a circle whose circumference is 74 feet.
95. Find the area of a circle whose radius is 125 feet.
96. In a circle with a diameter of 125 feet find the area of
a sector with an arc of 22°.
97. In a circle with a radius of 44 feet find the area of a
sector with an arc of 25°.
98. In a circle with a diameter of 50 feet find the area of
a segment with an arc of 280°.
99. Find the area of a segment (less than a semicircle) of
which the chord is 20, and the distance of the chord from the
middle point of the smaller arc is 2.
100. If r is the radius of a circle, the area of a regular
180°
circumscribed polygon of n sides is nr^ tan
n
The area of a regular inscribed polygon is - r^ sin
MISCELLANEOUS EXAMPLES 101
101. If a is a side of a regular polygon of n sides, the area
of the inscribed circle is —r- cot^
^ ira^ 180"
The area of the circumscribed circle is — r- esc*
4 n
102. The area of a regular polygon inscribed in a circle is
to that of the circumscribed regular polygon of the same
number of sides as 3 to 4. Find the number of sides.
103. The area of a regular polygon inscribed in a circle is
the geometric mean between the areas of an inscribed and a
circimiscribed regular polygon of half the number of sides.
104. The area of a circumscribed regular polygon is the
harmonic mean between the areas of an inscribed regular
polygon of the same number of sides and of a circumscribed
regular polygon of half that number.
106. The perimeter of a circumscribed regular triangle is
double that of the inscribed regular triangle.
106. The square described about a circle is four-thirds the
inscribed regular dodecagon.
107. Two sides of a triangle are 3 and 12, and the included
angle is 30**. Find the hypotenuse of an isosceles right tri-
angle of equal area.
PLANE SAILING
Plane Sailing is that branch of Navigation in which the
surface of the earth is considered a plane. The problems
which arise are therefore solved by the methods of Plane
Trigonometry.
The difference of latitude of two places is the arc of a
meridian comprehended between the parallels of latitude
passing through those places.
102 PLANE TRIGONOMETRY
The departure between two meridians is the arc of a parallel
of latitude comprehended between those meridians. It dimin-
ishes as the distance from the equator increases.
When a ship sails in such a manner as to cross successive
meridians at the same angle, it is said to sail on a rhumb-line.
This angle is called the course, and the distance between two
places is measured on a rhumb-line.
If we consider the distance, departure, and difference of
latitude of two places to be straight lines, lying in one plane,
they form a right triangle, called the triangle of plane sailing.
If ABC is a plane triangle, right-angled at B, and BC repre-
sents the difference of latitude of B and C, A CB will be the
course from C to ^, C^ the distance, and D the departure,
measured from B, between the meridian of A and that of B.
108. Taking the earth's equatorial diameter to be 7925.6
miles, find the length in feet of the arc of one minute of a
great circle.*
109. A ship sails from latitude 43° 45' S., on a course
N. by E. 2345 miles. Find the latitude reached, and the
departure made.
110. A ship sails from latitude V 45' N., on a course S.E.
by E., and reaches latitude 2** 31' S. Find the distance, and
the departure.
111. A ship sails from latitude 13** 17' S., on a course N.E.
by E. i E., until the departure is 207 miles. Find the distance,
and the latitude reached.
112. A ship sails on a course between S. and E. 244 miles,
leaving latitude 2** 52' S., and reaching latitude 5° 8' S. Find
the course, and the departure.
* The length of the arc of one minute of a great circle of the earth is
called a geographical mile or a knot. In the following problems, this is
the distance meant by the term **mile,'^ unless otherwise stated.
MISCELLANEOUS EXAMPLES
103
113. A ship sails from latitude 32° 18' N., on a course
between If. and W., a distaace of 344 miles, and a departure
of 103 miles. Find the course, and the latitude reached.
114. A ship sails on a course between 8. and E., making a
difference of latitude 136 miles, and a departure 203 miles. .
Find the distance, and the course.
116. A ship sails due north 15 statute miles an hour, for
one day. What is the distance, in a, straight line, from the
point left to the point reached ? (Take earth's radius, 3962.8
statute miles.)
PABALLEL AND MIDDLE LATITUDE SAILING
The difference of longitude of two places is the angle at the
pole made by the meridians of these two places ; or, it is the
arc of the equator comprehended between these two meridians.
In Parallel Sailing a vessel is supposed to sail due east or
due west. The distance sailed is the departure made; and
the difference of longitude is found as follows:
116. Given the departure between any two meridians at
any latitude ; find the difference of longitude of any point on
one meridian from any point on the other.
PLANE TRIGONOMETRT
In rt. A ODA, Z ^ OZ) = 90° - lat.
Hence, — = sin (90° — lat.) = coa lat.
The A DAB and OEQ are Bimilar.
nj AD DA AD
Therefore, -
Hence, I
' OE
slat.
EQ
AB
Therefore, E<1 =
EQ
AB
cos lat
OA EQ
= AS X sec lat.
no. -m That is, Difi. long. = depart, x sec lat.
117. A ship in latitude 42° 16' N., longitude 72° 16' W.,
sails due east a distance of 149 miles. What is the position
of the point reached?
118. A ship in latitude 44° 49' S., longitude 119° 42' E.,
sails due west until it reaches longitude 117° 16' E. Find
the distance made.
In Middle Latitude Sailing the departure between two places
is measured on that parallel of latitude which lies midway
between the, parallels of the two places. Except in very high
latitudes or excessive runs, this assumption produces no great
error. Hence, in middle latitude sailing,
Diff. long. = depart, x sec mid. lat.
MISCELLANEOUS EXAMPLES 105
119. A ship leaves latitude 31** 14' K, longitude 42^ 19' W.,
and sails E.N.E. 325 miles. Find the position reached.
120. Pind the bearing and distance of Cape Cod from
Havana. (Cape Cod 42** 2' N., 70** 3' W. ; Havana, 23'' 9' N.,
S2'' 22' W.)
121. Leaving latitude 49* 57' K, longitude 16** 16' W., a
ship sails between S. and W. till the departure is 194 miles,
and the latitude is 47® 18' N. Find the course, distance, and
longitude reached.
122. Leaving latitude 42" 30' K, longitude 58** 51' W., a
ship sails S.E. by S. 300 miles. Find the position reached.
123. Leaving latitude 49* 57' K, longitude 30** W., a ship
sails S. 39'' W., and reaches latitude 47** 44' N. Find the
distance, and longitude reached.
124. Leaving latitude 37** N., longitude 32** 16' W., a ship
sails between N. and W. 300 miles, and reaches latitude 41** N.
Find the course, and longitude reached. *
125. Leaving latitude 50** 10' S., longitude 30** E., a ship
sails E.S.E., making a departure of 160 miles. Find the dis-
tance, and position reached.
126. Leaving latitude 49** 30' K, longitude 25** W., a ship
sails between S. and E. 215 miles, making a departure of 167
miles. Find the course, and position reached.
127. Leaving latitude 43** S., longitude 21** W., a ship sails
273*Tniles, and reaches latitude 40** 17' S. What are the two
courses and longitudes which will satisfy the data ?
128. Leaving latitude 17** N., longitude 119** E., a ship sails
219 miles, making a departure of 162 miles. What four sets
of answers do we get ?
129. A ship in latitude 30** sails due east 360 statute miles.
What is the shortest distance from the point left to the point
reached ? Solve the same problem for latitude 45**, 60**.
106
PLANE TRIGONOMETRY
FlO. 73
TRAVERSE SAILING
Traverse Sailing is the application of
the principles of Plane and Middle Lati-
tude Sailing to cases when the ship sails
from one point to another on two or
more different courses. Each course is
worked by itself, and these independent
results are combined, as may be seen in
the solution of the following examples.
130. Leaving latitude ST** 16' S., longitude 18** 42' W., a
ship sails N.E. 104 miles, then KN.W. 60 miles, then W. by
S. 216 miles. Find the position reached, aild its bearing and
distance from the point left.
We have, for the first course, difference of latitude 73.5 N".,
departure 73.6 E. ; for the second course, difference of latitude
55.4 N., departure 23 W. ; for the third couree, difference of
latitude 42.1 S., departure 211.8 W.
On the whole, then, the ship has made 128.9 miles of north
latitude, and 42.1 miles of south latitude. The place reached
is therefore on a parallel of latitude 86.8 miles to the north of
the parallel left, that is, in latitude 35° 49.2' S.
The departure is, in the same way, found to be 161.3 miles
W. ; and the middle latitude is 36** 32.6'. With these data,
and the formula after Example 118, we find the difference
of longitude to be 201', or 3° 21' W. Hence, the longitude
reached is 22° 3' W.
With the difference of latitude 86.8 miles, and the departure
161.3 miles, we find the course to be N. 61° 43' W., and the
distance 183.2 miles. The ship has reached the same point
that it would have reached if it had sailed directly on a
course N. 61° 43' W. for a distance of 183.2 miles.
MISCELLANEOUS EXAMPLES 107
131. A ship leaves Cape Cod (Example 120), and ^ails S.E.
by S. 114 miles, K. by E. 94 miles, W.N.W. 42 mile^. Solve
as in Example 130.
132. A ship leaves Cape of Good Hope (latitude 34'' 22' S.,
longitude IS** 30' E.), and sails K.W. 126 miles, N. by E. 84
miles, W.S.W. 217 miles. Solve as in Example 130.
EXERCISE XXIV
PROBLEMS IN GONIOMETRY
Prove that :
1. sin X -\- cos X = V2 cos (« — J tt).
2. sin X — cos a; = — V2 cos (« + J tt).
3. sin X + Vs cos a; = 2 sin (a; -|- J tt).
4. sin (« + J tt) + sin (a; — J tt) =;= sin x,
6. cos (ic + i tt) H- cos (a; — J tt) = V3 cos x.
6. tan X + sec x = tan (i ic + i tt).
7. tan 05 -h sec a; =
8.
sec X — tan aj
1 — tan X cot a; — 1
1 + tan X cot a; + 1
sm X 1 -h cos X ^
9. T— 1 : = 2 CSC a;.
1 + cos X sm X
10. tan X + cot X = 2 esc 2 ar.
11. cot a; — tan x = 2 cot 2 x,
12. 1 + tan x tan 2 aj = sec 2 x.
13. sec 2 aj =
sec^a;
2 — sec^ a;
14. 2 sec 2 a; = sec (x -\- 45**) sec (x - 45°).
108
PLANE TRIGONOMETRY
16. tan 2x + sec 2x =
cos X + sin X
cos X — sin X
16. sin 2x =
2 tana;
1 + tan^x
17. 2 sin a; + sin 2 x =
2 sin'* a;
18. sin Sx =
19. tan 3 a; =
1 — cos X
sin^2a5 — sin* a;
sin X
3 tan a; — tan'a;
1-3 tan^a;
tan 2 a; -|- tan x sin 3 x
tan 2x — tan aj sin a:
sin (aJ 4- y) + cos (x — y) = 2 sin (a; + J tt) sin (2^ + J ^).
sin (x + y) — cos (aj — y) = — 2 sin (a; — J tt) sin (y — J tt).
sin (a; + y)
cos aj cos y
sin 2 aj -f- sin 2 y
20.
21.
22.
23. tan a; + tan y =
24. tan (aj -h y) = o . o
^ ^'^ cos 2 a; + cos 2 y
sin a; + cos V tan li(x -{-y) -\- 45**^
sin a; — cos y tan | J (a; — y) — 45®|
26. sin 2x -{- sin 4 aj = 2 sin 3 a; cos aj.
27. sin 4 a; = 4 sin a; cos a; — 8 sin*a; cos x
= 8 cos' a; sin aj — 4 cos x sin x,
28. cos 4 aj = 1 — 8 cos* a; + 8 cos*a; = 1—8 sin* a; -h 8 sin* ax
29. cos 2x -{- cos 4 aj = 2 cos 3 x cos x,
30. sin 3 aj — sin x = 2 cos 2 a; sin x.
31. sin'aj sin 3 aj + cos'a; cos 3 a; = cos' 2 x.
32. cos* a; — sin* a; = cos 2 a;.
MISCELLANEOUS EXAMPLES 109
33. cos*a5 + sin*a; = 1 — J sin^ 2 x.
34. cos^a; — sin* a; = (1 — sin^x cos^oj) cos 2 x.
36. cos* a; + sin*aj = 1 — 3 sin^a5 cos^oj.
sin 3 a5 + sin 5 aj
36. ' z— = cot X.
COS 3 a; — cos 5 x
^^ sin 3 a; 4- sin 5 a; ^ ^
37. —. : — - — = 2 cos 2a;.
sin X -f sin 3 x
38. CSC X — 2 cot 2 X cos a; = 2 sin x.
39. (sin 2x — sm2y) tan (x + y) = 2 (sin^a; — sin^y). •
y- . ^ . ^ . , . . sec a; esc a;
40. (1 + cot a; + tan x) (sm a; — cos x) = — 5 r--
^ '^ ^ ^ csc^a; sec^a;
. • o . • (? sin23a;
41. sin a; 4- sin 3 a; 4- sm 5 a; = — :
sma;
3 cos X 4- cos 3 a; ^,
42. TT-' ^~"F~ = cot' a:.
3 sm a; — sm 3 x
43. sin 3 a; = 4 sin x sin (60® 4- x) sin (60** — x),
44. sin 4 a; = 2 sin a; cos 3 a; 4- sin 2 x.
46. sina;4- sin(a; — f 7r)4-sin(i7r — a;)= 0.
46. cos X sin (y — «)4- cos y sin (« — a:)4- cos z sin (a: — y)= 0.
47. cos (a; 4- y) sin y — cos (x 4- «) sin z
= sin (a; 4- y) cos 1/ — sin (x 4- «) cos «.
48. cos (a; 4- 2/ + «) 4- cos (a; 4- 2/ — ^) + cos (a; — y -j- z)
4- cos (y 4- « — a;) = 4 cos a; cos y cos «.
49. sin (x 4- y) cos (a; — y) 4- sin (y 4- «) cos (y — z)
4- sin (« 4- x) cos (« — a;) = sin 2 a; 4- sin 2 y 4- sin 2 «.
^^ sin 75** 4- sin 15° , ^^o
^0. -r-w^ r-^r^ = tan 60 .
sm 75 — sm 15
51. cos 20'' 4- cos 100° 4- cos 140° = 0.
110 PLANE TRIGONOMETRY
62. cos 36* 4- sin 36* = V2 cos 9°.
53. tan 11* 15' + 2 tan 22* 30' + 4 tan 45* = cot 11* 15'.
Ji A, Bf C are the angles of a plane triangle, prove that :
64. sin2^ + sin2^ + sin2 C =4sin^ sin^sinC.
65. cos 2 ^ + cos 2 JB + cos 2 C = — 1 — 4 cos A cos B cos C.
QA Qd 3(7
56. sin 3 -4 4- sin 3 J5 4- sin 3 C = — 4 cos -^ cos —^ cos -^«
Zi A At
67. cos"^ 4- cos'-B 4- cos^ C = 1 -- 2 cos A cos B cos C.
If ^ 4- ^ 4- C' = 90*, prove that :
68. tan A tan B 4- tan J5 tan C 4- tan C tan A =1.
59. sin"^ 4- sin»5 + sin^C = 1 - 2 sin ^ sin :B sin C.
60. sin 2 ^ 4- sin 2 J5 4- sin 2 C = 4 cos A cos -B cos C.
Prove that :
61. sin (sin" ^aj 4- sin~^y) = x Vl — y" + y Vl — a;*.
62. tan (tan~^a5 4- tan""^v) = :: — '
^ ^ 1 — XV
xy
2x
63. 2 tan"^a; = tan"^ ^
1 — a:^
64. 2 sin-^a; = sin-^ (2 cc Vl - x^.
65. 2 cos-^aj = cos-^ (2 a;" — 1).
Sx — x^
66. 3tan"^a;=»= tan~^:j r— t:*
1 — 3a;^
67. sin-i\/- = tan-i\/— ^— •
68. sin-i\P^ = tan-i\/^^^.
MISCELLANEOUS EXAMPLES 111
69. siii~^a; = sec~^
VT
X'
70. 2 sec~^a; = taii~^ — -^ ^r— •
2 — x^
71. tan-ii4-tan-ii = 45^
72. tan-i J + tan-^J = tan-^f
73. sin-^f 4-sin-^Jf = siii-^ff.
1 4
74. sin"^ , — 4- sin~^ —7= = 45°.
V82 .V41
75. sec-i|-f sec-^jf = 75° 45'.
76. tan-i (2 + V3) - tan-^ (2 - VS) i= sec-i2.
77. tan-^J 4-tan-^i4-taii-^| 4-tan-iJ = 45°.
78. tan-^q i^ — r^T"^ + ^an" ^ q— -7^ — — p^ = tan" ^ 75-^
1 — 2a; + 4a;^ l-f2a;4-4a;2 2a;^
79. Given cos a; = J ; find sin ^ x and cos ^ cc.
80. Given tan a; = J ; find tan J x.
81. Given sin a; -f cos x = V^ ; find cos 2 aj.
82. Given tan 2 a; = ^^i 5 find sin x.
83. Given cos 3 a; = f f ; find tan x,
84. Given 2 esc a; — cot x = Vs ; find sin J x.
85. Find sin 18° and cos 36°.
Find the value of :
86. a sec x -i-b esc aj, when tan x = \/- •
87. sin 3 x, when sin 2 x = Vl — m^.
88. sin X, when tan^x + 3 cot^x = 4.
112 PLANE TRIGONOMETRY
csc^x — sec^aj , ^ rr
89. — z z" ' when tan x = V 1.
csc'aj + sec'aj
90. cos X, when 6 tan a: + sec a5 = 5.
91. sec Xf when tan x =
V2a + 1
Simplify the following expressions :
(cos a; -f cos y)* + (sin x + sin y)'
cos^^(x — y)
sin (a; 4- 2 y) — 2 sin (g.+ y) -f sin x
cos (a; + 2 y) — 2 cos (aj + y) 4- cos x
gin (a; —• g) 4- 2 sin a; 4- sin (a; -f g)
sin (y — «) 4- 2 sin y 4- sin (y 4- »)
cos 6 ar — cos 4 x
93.
94.
Sin 6 a; 4- sm 4 aj
96. tan-i(2x4- l) + tan-^(2a;-l).
1 1 1.1
1 4- sin" a; 1 + cos" a; 1 + sec'aj 1 + csc*aj
98. 2 sec"aj — sec* a; — 2 esc" a; 4- csc*aj.
SOLUTION OF SINGLE EQUATIONS
To solve a single equation that involves different functions
of the same angle, or the same or different functions of related
angles, first transform the equation, if necessary, into an
equivalent equation that involves a single function of the
same angle.
Employ the method of factoring, if possible, in the algebraic
part of the solution.
Completely solve each equation, and check the results by
substitution in the given equation.
MISCELLANEOUS EXAMPLES
113
Solve cos aj = sin 2 x.
By [12], p. 58, sin 2 X = 2 sin X cos z.
.'. cos X = 2 sin X cos x.
.-. (1 — 2 sin x) cos x = 0.
.*. cos X = 0, or 1 — 2 sin X = 0.
.-. X = 90° or 270°, or 30° or 150°.
Each of these values satisfies the given equation.
Solve the following equations
99. sinaj = 2sin(^7r + ic).
•100. sin 2x = 2 cos x.
101. cos 2 aj = 2 sin X.
102. sin aj + cos a: = 1.
103. sin 05 4- cos 2 aj = 4 sin^a.
104. 4 cos 2 a; + 3 cos x = 1.
106. sin X -f sin 2x = sin 3 x.
106. sin 2 aj = 3 sin^x — cos* a;.
107. cot^ = Jtan^.
108. 2 sin ^ = cos 0.
109. 2 sin*aj -f 5 sin a; = 3.
110. tan X sec x — V2.
111. sin a; = cos 2 a;.
112. tan X tan 2x — 2,
113. sec aj = 4 CSC a;.
114. cos ^ + cos 2^ = 0.
115. cot J ^4- CSC ^ = 2.
116. cot X tan 2 a; = 3.
117. sin X sec 2 a; = 1.
118. sin^aj + sin 2 a; = 1.
119. cosa; sin2aj cscaj = 1.
120. cot a; tan 2 a; = sec 2 a;.
121. sin 2 a; = cos 4 a;.
122. sin2«cot» — sin*« = i.
123. tan x + tan 2x = tan 3 x,
124. cot X — tan x = sin x -f cos x.
125. tan* a; = sin 2 x,
126. tan x -f cot x = tan 2 a;.
1 — tan X
127.
1 -f tanx
= cos 2 ar.
114 PLANE TRIGONOMETRY
128. sin X -f sin 2 aj = 1 — cos 2 x,
129. sec 2 a; -f 1 = 2 cos x.
130. tan 2a; 4- tan 3x = 0.
131. tan(i7r 4- a;)-f tan(i7r — aj) = 4.
132. Vl -f sin X — Vl — sin x = 2 cos x,
133. tan X tan 3 x = — |.
134. sin (45** -\-x)-{- cos (45° — aj) = 1.
135. tan X 4- sec x — a,
136. cos 2 a; = a (1 — cos x),
137. (1 — tan x) cos 2 a; = a (1 4- tan x),
138. sin^x 4- cos* a; = ^^ sin^ 2 x.
139. cos 3 a; 4- 8 cos'aj = 0.
140. sec (x 4- 120**) + sec (x — 120*^) = 2 cos x.
141. CSC X = cot X 4- Vs.
142. 4 cos 2x4-6 sin x = 5
143. cos X — cos 2x = 1.
144. sin 4 X — sin 2 x = sin x.
145. 2sin2x4-sin22x = 2.
146. cos 5x4- cos 3x4- cos x = 0.
147. sec X — cot X = CSC x — tan x.
148. tan^x 4- cot^a? = V-
149. sin4x — cos 3x = sin 2x.
150. sin X -f cos x = sec x.
151. 2 cos X cos 3 X 4- 1 = 0.
152. cos 3 X — 2 cos 2x4- cos x = 0.
MISCELLANEOUS EXAMPLES j 116
163. tan 2 a tan a; = 1. y
154. sin (x 4- 12°) + sin {x 4t 8**) = sin 20°.
165. tan (60° -f «) tan (60° '-x)^-2.
156. sin (x 4- 120°) + sin {x + 60°) = J.
157. sin (a; + 30°) sin (x - 30°) = J.
158. sin'*a5 + cos^cc = |.
159. sin*aj — cos*aj = /y.
160. tan (a; 4- 30°) = 2 cos «.
161. sec x=^2 tan aj + J.
162. sin 11 aj sin 4 a; 4- sin 5 aj sin 2 aj = 0.
163. cos X 4- cos 3 a; 4- cos 6 a; 4- cos 7 aj = 0.
164. sin {x 4- 12°) cos (x - 12°) = cos 33° sin 67°.
165. sin-^a; 4- sin-^J aj = 120°.
166. tan-^a;4-tan-^2a; = tan-i3 Vs.
167. sin~^a; 4- 2 cos"^aj = § tt.
168. sin-ia;4-3cos-^a; = 210°.
169. tan~^a; 4- 2 cot-^x = 136°.
170. tan-i(a;4-l)4-tan-^(a; — l)=tan-^2aj.
4- tan-i
171. tan
_j3c_+2 . . ,x — 2
172. tan
— 1
aj4-l
2a;
7 = JTT.
05 — 1 *
= 60°.
1-x^
173. cos 2^ sec ^4- sec ^4- 1 = 0.
174. sin X cos 2 x tan x cot 2 x sec aj esc 2 a; = 1.
175. sin J X (cos 2 a; — 2) (1 — tan^x) = 0.
Hint. Equate to each factor except the second. The second factor
cannot equal 0.
116 PLANE TRIGONOMETRY
176. sin3a; = cos2aj — 1. 178. sin 2 ^ = cos 3 ^.
177. tanic-f taii2aj = 0. 179. (3— 4cos^a;)siii2a; = 0.
180. sin X 4- sin 2 aj -f sin 3 a; = 0.
181. sin ^ + 2 sin 2 ^ 4- 3 sin 3 ^ = 0.
182. sin^a; cos^aj — cos^a; — sin^a; -f 1 = 0.
183. sin X + sin 3 a; = cos x — cos 3 x.
184. (1 — Vl — tan^a;) cos 2 x vers 3 a; = 0.
185. tan (^ -f 45**) = 8 tan ^.
186. sin (a; - 30°) = J V3 sin a;.
187. tan (^ -f 45°) tan ^ = 2.
188. sin-^Jaj = 30°.
SYSTEMS OF EQUATIONS
189. Solve for x and y the system
X sin a -f y sin ^ = a, (1)
x cos a -f- y cos P = b, (2)
(1) X cos a, X sin a cos a + y sin /3 cos a = a cos a. (3)
(2) X sin a, x sin a cos a + y cos /3 sin a = 6 sin a, (4)
(3) — (4), y (sin /3 cos a — cos /3 sin a) = a cos a — 6 sin a. (6)
a cos a — 6 sin a
o. .1 1 6 sin fl — a cos fl
Similarly, x =
sin (j8 — a)
in /3 — a cot
sin 03 — a)
190. Solve for x and 2/ the system
sin X 4- sin y ^ a^ (1)
cos a; 4- cos y = b, (2)
MISCELLANEOUS EXAMPLES IIT
*
Transform (1) and (2), by Sect. XXXII,
by [20], p. 59, 2 sin i (x + y) cos ^{x-y) = a, (3)
by [22], p. 59, 2co8i(x + y)cosi{x -y) = h, (4)
(3)-^{4), tani(x + y) = ^. (5)
.-. sini(x -\-y)= . ^ (6)
Va^ + 62
Substitute value of sin i(x + y) in (3),
cosi{x-y) = iVaM^. (7)
From (5), x + y = 2 tan-i ? . (8)
From (7), x - y = 2 cos-^i Va^ 4- 6^. (9)
Whence x = tan-i \ + cos-^i Va^ + 62,
and y = tan-i - - cos-H Va^ + 62.
6
191. Solve for r and ^ the system
r sin ^ = a, (1)
r cos ^ = h, (2)
(3)
(4)
(1) ^ (2),
tan^:
"6*
From (3),
Q:
= tan-
-1?.
5
Square (1)
and
(2)
and add,
•
r2
(sin2^ + cos2^)
= a2 + 62.
.-. r
= V^
+ 6«.
192. Solve for r and ^ the system
r sin (^ + a) = a, (1)
r cos (^ + iS) = ft. (2)
Expand (1) and (2),
r sin ^ cos a-\-r cos ^ sin a = a. (3)
r cos e cos /3 — r sin ^ sin /3 = 6. (4)
Now solve (3) and (4) for r sin Q and r cos ^, as in Example 189. Then
solve for r and d, as in Example 191.
LIS PLANE TRIGONOMETRY
193. Solve for r, 0, and .^ the system
r COS ^ sin = a,
r COS tb cos S = b,
{l)-i-(2), tan# = -- .:$■■
SquEira (1) and (2) and add.
(1)
(2)
(3)
e (3) and add to (5),
Solve the following ayatema for r, 9, <ft, x, and y :
. 194. X sin 21° + y cos 44° = 179.70,
I cos 21° + y sin 44° = 232.30.
196.
C08 3!-
sin y = 0.7038,
cosy = -0.7245.
196.
rsintf
j-cosfl
= 92.344,
= 205.309.
CHAPTER VI
CONSTRUCTION OF TABLES
SECTION XLIII
LOGARITHMS
Properties of Logarithms. Any positive number except unity-
being selected as a hasBj the index or exponent which the base
must have to produce a given number is the logarithm of that
number to the given base.
Thus, if a" = N, then n = log^N,
n = log^iV is r6ad, n is equal to log N to the base a.
Let a be the base, M and N any positive numbers, m and n
their logarithms to the base a ; so that
a™ = M, a" = N,
m = log„3f, n = log^iNT.
Then, in any system of logarithms :
1. The logarithm of 1 is 0.
For, a° = 1. .-. = log^l.
2. The logarithm of the base itself is 1.
For, a^ = a, .'. 1 = log„a.
3. The logarithm of the reciprocal of a positive number is
the negative of the logarithm, of the number.
For, if a» = N, then — = — = »-».
' N a""
••• log« i-^J = -n = - log^N.
119
120 PLANE TRIGONOMETRY
4. The logarithm of the product of two or more positive
numbers is found by adding together the logarithms of the
several factors.
For, M X I^ = a"* X a"* = »"•+".
/. log^(M xN) — m + n = log^M + log^iV:
Similarly for the product of three pr more factors.
5. The logarithm of the quotient of two positive numbers
is found by subtracting the logarithm of the divisor from the
logarithm of the dividend.
^ • Ma""
For, — = — = a"-».
•Mog,(f)-
m — n = log„ 3f — log^N,
6. The logarithm of a power of a positive number is found
by multiplying the logarithm of the number by the exponent
of the power. •
For, NP = (a'^y = a"".
.'. log„ (N^) = np=p log^N.
7. The logarithm, of the real positive value of a root of a
positive number is found by dividing the logarithm of the
number by the index of the root.
For, Vn = Va" = a\
r
Change of System. liogarithms to any base a may be con-
verted into logarithms to any other base b as follows :
Let N be any number, and let
n = log^N and m = log^N.
Then, N = a*" and N = b"^.
,'. a" = J"*.
CONSTRUCTION OF TABLES 121
Taking logarithms to any base whatever,
n log a =^ m log 6,
or, log a X log„iV = log ft x log^iV,
from which log^iV may be found when log a, log ft, and log^iV
are given; and conversely, log^iV may be found when log a,
log b, and log^iV are given.
Two Important Systems. Although the number of different
systems of logarithms is unlimited, there are but two systems
which are in common use. These are :
1. The common system, also called the Briggs, denary, or
decimal system, of which the base is 10.
2. The natural system of which the base is the fixed value
which the sum of the series
^ 1 ^ 1.2 ^ 1.2.3 ^ 1.2.3.4 ^
approaches as the number of terms is indefinitely increased.
This fixed value, correct to seven places of decimals, is
2.7182818, and is denoted by the letter e.
The common system is used in actual calculation; the
natural system is used in the higher mathematics.
EXERCISE XXV
1. Given logio 2 =0.30103, logio3 =0.47712, logio7 =0.84510;
find logio6, logiol4, logio21, logio4, logiol2, logio5, logjoi,
logioj, logioj, logiofi-
2. With the data of Example 1, find logglO, logs 5, logs 5,
3. Given logioC = 0.43429 ; find log, 2, log, 3, log, 5, log, 7,
log,8, log,9, log,§, log,J, log,J^, log,/^.
4. Find x from the equations 5' = 12, 16^ = 10, 27* = 4.
122
PLANE TRIGONOMETRY
SECTION XLIV
EXPONENTIAL ANB LOGARITHMIC SERIES
Exponential Series. By the binomial theorem^
(. , l\^ . , 1 , nx(nx-l) 1
\ nj n 1-2 rr
nx (nx — 1) (nx — 2) 1
■^ 12. 3 ^n'^
X
= l + a; +
G-0,!B)H)
+
(1)
2 ■ [3
This equation is true for all real values of x, since the bino-
mial theorem may be extended to the case of incommensurable
exponents (Wentworth's College Algebra, § 299) ; it is, how-
ever, true only for values of n numerically greater than 1, since
- must be numerically less than 1 (College Algebra, § 418).
As (1) is true for all values of x, it is true when x = 1.
-(-o-=
1-
1 + 1 +
w I \ nj\ ^ nJ
[3
Hence, from (1) and (2),
+ •••(2)
1 + 1 + i+\ ^ZV V
' + '+ 12 + \3
+
= l+x +
X[ X I 07(05 ][ X 1
[2 + [3
-f
J[
CONSTRUCTION OF TABLES 123
This last equation is true for all values of n numerically
greater than 1. Taking the limits of the two members as n
increases without limit, we obtain
and this is true for all values of x. It is easily seen that both
series are conyergent for all values of x.
The sum of the infinite series in parenthesis is the natural
base e.
Hence, by (3), «* = H-a:-f i^ + il^-f- • (4)
To calculate the value of e, we proceed as follows :
Adding,
To ten places,
1.000000
2
1.000000
3
0.500000
4
41
0.166667
6
0.041667
6
0.008333
7
0.001388
8
0.000198
9
0.000025
c =
e =
0.000003
= 2.71828.
= 2.7182818284.
2 ^ n'
Limit of ( 1 + - J . By the binomial theorem,
(, . x\^ ^ . X ^ nin -
14--) =14- nX--f-V
nj n 1 •
^ 1.2.3 n*^
= ^+"+— "+-^ — Y — ^" +
x^ . x^
l + aj4-r77 + ^ + ---
124 PLANE TRIGONOMETRY-
This equation is true for all values of n greater than x
{College Algehray § 418). Take the limit as n increases with-
out limit, X remaining finite ; then
limit A ^ ^Y^
= ^ = n ^« ("l + -T-
Logarithmic Series.
Let y = log^(H-iK);
then l + - = «' = «'!^':(l+|)"-
If n is merely a large number, but not infinite,
(5)
(-0-=
1 + ic + c,
where c is a variable number which. approaches the limit 0,
when n increases without limit. Hence,
y =^n Vl + 05 + c — n.
If n increases without limit, and consequently c approaches
as a limit, we have
y = « i^« [« ^^1^ - «]•
If X is less than 1, we may expand the right-hand member
of this equation by the binomial theorem. The result is
=,'?i[-a-Oi<=-oa-)i-]
CONSTRUCTION OF TABLES 126
X^ . 05* iC*
.-. log,(l + a;) = a; - ^ + — - - + . . .
This series is known as the logarithmic series. It is con-
vergent only if X lies between — 1 and -f 1, or is equal to -f 1.
Even within these limits it converges rather slowly, and for
these reasons it is not well adapted to the computation of loga-
rithms. A more convenient series is obtained as follows.
Calculation of Logarithms. The equation
iog.(i + y) = y-f + f-^+-- (1)
holds true for all values of y numerically less than 1 ; there-
fore, if it holds true for any particular value of y less than 1,
it will hold true when we put — y for y ; this gives
log.(l-y) = -y-|-§-^-... (2)
Subtracting (2) from (1), since
loge (1 + y) - log, (1 - y) = log, (jZ^y
we find iog^^l±|) = 2(y-ff-ff + ...)•
Put
then
2^ = 27Tl'
. = ,
1 — y z
and log, f ^-J- J = loge (« + !)- ^og,z
= 2^-^-4- ^ + ^ + ...\
This series is convergent for all positive values of z.
126
PLANE TRIGONOMETRY
Logarithms to any base a can be calculated bj the series
loga(« + l)-log««
1
log.a\2« + l
+
i +
3(2» + l)» 6(2 «
Example. Calculate log«2 to five places of decimals.
Let « = 1; then »-f- 1 = 2, 2« + l=3,
and
_2 2 2 2
+
The work may be arranged as follows :
3
9
9
9
9
9
2.000000
0.666667
0.074074
0.008230
1 = 0.666667
3 = 0.024601
6 = 0.001646
0.000914 -f- 7 = 0.000131
0.000102 -r- 9 = 0.000011
0.000011 ^ 11 = 0.000001
' loge2 = 0.693147
Note. In calculating logarithms the accuracy of the work may be
tested ev^ry time we come to a composite number by adding the logarithms
of the several factors. In fact, the logarithms of composite numbers are
best found by addition, and then only the logarithms of prime numbers
need be computed by the series.
V
EXERCISE XjCVI
1. Calculate to five places of decimals lbg«3.
2. Calculate to five places of decimals log«5.
3. Calculate to five places of decimals log^7.
4. Calculate to ten places of decimals log^lO.
5. Calculate to five places of decimals logio2, logioe, logioll.
CONSTRUCTION OF TABLES
127
SECTION XLV
TRIGONOMETRIC FUNCTIONS OF SMALL ANGLES
Let AOP (Fig. 74) be any angle less than 90° and x its cir-
cular measure. Describe a
circle of unit radius about
O as a centre and take
AAOP' = — /LAOP, Draw
the tangents to the circle at
P and P'y meeting OA in T,
Then, from Geometry,
chord PP' < arc PP'
<PT-\- P'T,
or, by dividing by 2,
MP<SiXGAP<PT,
or sin x<x< tan x.
Fig. 74
Hence, dividing by sin x,
x
1 < —. — < sec X,
sm x
sm X
1 > > cos X,
X
Then — - — lies between cos x and 1.
X
(1)
If now the angle x is constantly diminished, cos x approaches
the value 1.
sin X
Accordingly, the limit of ) as x approaches 0, is 1.
X
sin a;
differs
X
In other words, if a; is a very small angle, then
from 1 by a small value c ; and this small value c approaches
as a; approaches 0.
128 PLANE TRIGONOMETRY
Example. To find the sine and cosine of 1'.
If X is the circular measure of 1',
the next figure in x being 8.
Now sina;>0 but<a;; hence, sin 1' lies between and
0.000290889.
Again, cos 1' = VI - sin^' > Vl - (0.0003)^ > 0.9999999.
Hence, cos 1 ' = 0. 9999999 + .
But, from (1), sin x > a; cos x,
.-.sin l'> 0.000290888 x 0.9999999
> 0.000290888 (1 - 0.0000001)
> 0.000290888 - 0.0000000000290888
> 0.000290887.
Hence, sin 1' lies between 0.000290887 and 0.000290889;
that is, to eight places of decimals
sin 1' = 0.00029088 +,
the next figure being 7 or 8.
»
EXERCISE XXVn
Given it = 3.141592653589 :
■
1. Compute sin 1', cos 1', and tan 1' to eleven places of
decimals.
2. Compute sin 2' by the same method, and also by the
formula sin 2 a; = 2 sin x cos x. Carry the operations to nine
places of decimals. Do the two results agree ?
3. Compute sin 1° to four places of decimals.
CONSTRUCTION OF TABLES 129
X
4. From the formula cos cc = 1 — 2 sin* -? show that
x^
cos a > 1 — ^ •
5. Show by aid of a table of natural sines that sin x and x
agree to four places of decimals for all angles less than 4° 40'.
6. If the values of log x and log sin x agree to five decimal
places, find from a table the greatest value x can have.
SECTION XLVI
SIMPSON'S METHOD OF CONSTRUCTING A TRIGONOMETRIC TABLE
By Sect. XXXII, p. 59,
sin {A -f 5) + sin (^ — 5) = 2 sin A cos B.
If we put A =x -\-2y, B = y,
we have sin (a + 3 y) -f- sin (x + y) = 2 sin (x -\- 2y) cos y, -
or sin (a; -f- 3 y) = 2 sin (x '\- 2y) cos y — sin (x -f- y).
Similarly, cos (a; -f 3 y) = 2 cos (x -\-2y) cos y — cos (x + y). (1)
If y = 1', the last two equations become
sin(aj + 3') = 2 sin (x + 2') cos 1' - sin (x + 1'),
cos (x + 3') = 2 cos {x -f 2') cos 1' — cos {x + 1').
Hence, taking x successively equal to — 1', 0', 1', 2 ',•••, we
obtain
sin 2' = 2 sin 1' cos 1',
sin 3' = 2 sin 2' cos 1' — sin 1',
sin 4' = 2 sin 3' cos 1' — sin 2',
• ..♦.♦
cos2' = 2cosn'-l,
cos 3' = 2 cos 2' cos V — cos V,
cos 4' = 2 cos 3' cos 1' — cos 2',
130 PLANE TRIGONOMETRY
Since sin 1' and cos V are known, these equations enable us
to compute step by step the sine and cosine of any angle.
The tangent may then be f ouncT in each case as the quotient
of the sine divided by the cosine.
This process need be carried only as far as 30°. For
sin (30* + a) + sin (30° - x) = 2 sin 30° cos x -= cos x,
cos (30° + a) - cos (30° - aj) = - 2 sin 30° sin a = - sin x.
.'.sin (30° -\- x)= cos x — sin (30° — x),
cos (30° -f- aj) = — sin a; -h cos (30° — x).
Moreover, the sines and cosines need be calculated only to
46°, since
sin (45° + x) = cos (45° — x),
cos (45° -{- x) = sin (45° — x).
In using this method, the multiplication by cos V, which
occurs at each step, can be simplified by noting that
cos 1' = 0.9999999 = 1 - 0.0000001.
Note. Simpson^s method is superseded in actual practice by much
more rapid and convenient processes in which we employ the expansions
of the trigonometric functions in infinite series.
EXERCISE XXVm
1. Compute the sine and cosine of 6' to seven decimal places.
In Formula (1) let y = 1°. Assuming
sin 1° = 0.01 7454+, cos 1° = 0.999848+ :
2. Compute the sine and cosine of two degrees.
3. Compute the sine and cosine of three degrees.
4. Compute the sine and cosine of four degrees.
6. Compute the sine and cosine of five degrees.
CONSTRUCTION OF TABLES 131
SECTION XLVn
DE MOIVRE'S THEOREM
Expressions of the form
cos X -^ i sin x,
when i = V— 1, play an important part in modem analysis.
Given two such expressions,
cos X -{-i sin x, cos y -{-i sin y
their product is
(cos X -^ i sin x) (cos y -{-i sin y)
= cos x COS y — sin a; sin y -\- i (cos x sin y -f- sin x cos y)
= cos (aj -f- y) + *' sin (a; -f- y)'
Hence, the product of two expressions of the form
cos a; -f i sin a?, cos y + i sin y
is an expression of the same form in which x or y is replaced
by aj + y- In other words, the angle which enters into such a
product is the sum of the angles of the factors.
If X and y are equal, we have at once, from the preceding,
(cos a? -f t sin xy = cos 2x -\- i sin 2 x ;
and again,
(cos X -^ i sin »)• = (cos a; + t sin xy (cos a; -f- * sin a;)
= (cos 2 a; -f t sin 2 a;) (cos a? -f i sin x)
= cos 3 a; + t sin 3 x.
Similarly,
(cos aj -f t sin xy = cos 4.a; -f- i sin 4 a;,
and in general, if n is a positive integer,
(cos X -\- isin xy = cos nx + i sin nx, (1)
132 PLANE TRIGONOMETRY
Hence,
To raise the expression cos x -h i sin x to the nth power when
nis a positive integer , we have only to multiply the angle x by n.
Again, if 71 is a positive integer as before,
(cos - 4- i sin - ) = cos x -^-i sin x,
n nj
— X X ■
.'. (cos X -{- i sin xY = cos - -{- i sin —
^ ^ n n
Since, however, x may be increased by any integral multiple
of 2 TT without changing cos a; + * sin x, it follows that all the
n expressions,
X ^ , . X X + 27r ^ , . X -\-2ir
cos - + i sin -} cos h i sin ;
n n n n
x + 47r , , . X -\- 47r
cos h I sin > • • ♦,
n. n
x-\-(n — l)2ir . . . ic + (n- l)27r
cos ^^ ^ — - + t sm ^ — -9
n n
are nth roots of cos x -{-i sin x. There are no other roots, since
x-^n27r , . . x-^n2ir
cos — h I sin
n n
X , , X
cos- + tsm-)
n n
= cosf- + 27rj + isinf-4-27rj =
, a; + (yi + l)27r , . . x -{- (n -{- 1) 2 ir
and cos ^^ h i sm — ^-^^ — - — ^
n n
(x±2iT , . \ _^ . . (x±2ir ^^ \
= cos ( h 2 TT 1 + t sm ( h 2 TT 1
X + 2tt . . . aj + 2iT
= cos h i sin }
n n
and so on.
CONSTRUCTION OF TABLES 133
Hence, if ti is a positive integer,
(cos a; -f t sin a;)» -*
= cos h t sin (A; = 0, 1, 2, • • n — 1). (2)
n n
From (1) and (2) it follows at once that if m and n are
positive integers,
m 1
(cos 05 -f t sin a;)" = I (cos x -\- i sin a;)" j**
AM AM
= COS — (a; + 2 kir) + i sin — (aj + 2 kir)
n ^ ' n^ ^
(A: = 0, 1, 2, ...» - 1). (3)
AM
Finally, if is a negative fraction.
(cos x-\- i sin a?) * =
(cos x-^-i sin aj)
m
n
^ . 1 COS a; — t sin x
But
COS a; + t sin x (cos a; -f * sin x) (cos a; — i sin a)
__ cos X — i sin aj
cos^a; -f sin^aj
= cos X — i sin a?
= cos (— x) + 1 sin (— aj).
m nt
Hence, (cos aj -h t sin x) " = | cos (— a) + 1 sin (— x) \ *
= cos --(—x-V^ kif) + I sin — (— a; + 2 Ajtt),
(Aj = 0, 1, 2, . . . w - 1)
= cos -{ {x-\-2 kii) 1 4- 1 sin -I (a; + 2 kir) \ >
(Aj = 0, 1, 2, . . . w - 1). (4)
134 PLANE TRIGONOMETRY
Consequently, if w is a positive or negative integer or
fraction,
(cos x-\-i8mxy = oos(n (x + 2 kir) ] + 1 sin [n (aj -h 2 kir) ],
(Aj = 0, 1, 2, . . . w - 1). (5)
Example. Find the three cube roots of — 1.
We have - 1 = cos ISO** + i sin ISO**.
.-. (-1)4 = cos ^ +t8in ^ (& = 0, 1, 2).
For the three cube roots of — 1 we find, therefore,
cos 60® + I sin 60% cos 180® + i sin 180®, cos 300® + i sin 300®,
1 + £V3 , l-iV3
or , —1,
2 ' 2
By aid of De Moivre's Theorem, we may express sin nO and
cos nBj when n is an integer, in terms of sin 6 and cos B,
Thus, cos nO + i sin n$ = (cos ^ -f t sin ^)*
= cos"^ + in cos»-^^ sin $ + i^ ^^^i7 ^ cos—*^ sin^^
. „ n (n — 1) (n — 2) , ^ . . >» .
+ t* -^^ f^^^ ^ cos»-«^ sin»^ -f . . •
Or, since i* = — 1, i* = — i, i* = -f 1, . . .,
cos w^ -f i sin 71^ = cos"^ + in cos""^^ sin
^Hr^ — ^ cos"^^ sm^^ — t —^ r^^^ ^ cos"-^^ sin»^ H
Equating now the real parts and the imaginary parts sepa-
rately, we obtain
cos nO = cos»^ — ^ \ 7 cos--^^ sin^^
. n(n-l)(n-2)(n-3) ^^ . ^^
+ -^ ^^-S-i cos"-*^ sm*^ ,
[4
CONSTRUCTION OF TABLES 135
smn$ = n cos"-^^ sin ^ - ^ |3 cos»-*^ sin»^
, n(7i-l)(w-2)(/i-3)(w-4f n 5^ ' sn
-I — i^ L^ Z^ ^ ^ cos""*^ sin*^
[5
EXERCISE XXIX
1. Find the six 6th roots of — 1 ; of + 1.
2. Find the three cube roots of i.
3. Find the four 4th roots of — i.
4. Express sin 4 $ and cos 4 ^ in terms of sin and cos 0.
SECTION XLVIII
EXPANSION OF SIN X, COS X, AND TAN X IN INFINITE SERIES
Let one radian be denoted simply by 1, and let
cos 1 -f i sin 1 = k.
Then cos a; -f t sin a; = (cos 1 -f t sin 1)*= /c*,
and, putting — x for x,
cos (— x) + i sin (— x) = cos a? — i sin x = k"*.
That is, cos x -{- i sin x = k"",
and cos x — i sin x = A;""*.
By taking the sum and difference of these two equations,
and dividing the sum by 2 and the difference by 2 1, we have
1 1
cos a; = - (k' -\- k-'), sin x = — (k^ — k-").
But k' = (e^'^^y = e*»o«*, k-' = 6-**'**,
and e*^*«* = 1 + ajlogAj H nr 1 1 ° ^ H ,
136 PLANE TRIGONOMETRY
, ^ ^ , , . x^fiogkY a:*(logA;)« .
g-xiog* = 1 - ajlog A; + \^" ^ Ss~ ^'"
If, , , x' (log ky ^ x^logky ,
It only remains to Iind the value of k, and this can be
obtained by dividing the last equation through by x and
letting X approach indefinitely.
Then we have
• S. •
But limitJ?ilL^>i==i.
« = 0\^ X )
,'. log k = i.
Therefore, we have
From the last two series we obtain, by division, '*■ ^
sin a; , x^ 2x^ , 17 a;"^
By the aid of these series the trigonometric functions of any
angle are readily calculated.
In the computation it must be remembered that x is the^'
circular measure of the given angle. / ^
«
CONSTRUCTION OF TABLES 137
EXERCISE XXX
Verify by the series just obtained ttat :
1. sin^a? + cos^x = 1.
2. sin (— a:) = — sin x and cos (— x) = cos x,
3. sin 2 a; = 2 sin x cos x.
4. cos 2 03 = 1 — 2 sin^a:. *
5. Find the series for sec x as far as the term containing
the 6th power of x,
6. Find the series for x cot x^ noting that
X cot X = —, — cos X,
sin a;
7. Calculate sin 10° and cos 10° to five places of decimals.
8. Calculate tan 15° to five places of decimals.
9. From the exponential value of cos x show that
cos 3 a; = 4 cos'aj — 3 cos x,
10. From the exponential value of sin x show that
sin 3 aj = 3 sin aj — 4 sin'a;.
FORMULAS
PLANE TRIGONOMETRY
1. sin'^ 4- cos* -4 = 1.
sin A
2. tan A =
cos -4
fsin^ X csc^ = 1.
3. •< cos A X sec ^ = 1.
I tan^ X cot ^ = 1.
4. sin (a; 4- y) = sin x cos y 4- cos x sin y,
5. cos (a; 4- y) = cos x cos y — sin x sin y.
. tan a; 4- tan y
6. tan (x 4- y) = i^ -r^- •
^ ^^ 1 — tan aj tan-^^
. cot X cot y — 1
7. cot (a; 4- 2/) = — z T"^-:
^ ^^ cot y 4- cot a;
8. sin (a; — y) = sin a; cos y — cos a: sin y,
9. cos (a; — y) = cos a; cos y 4- sin a; sin y.
. ^ ^ , s tan a; — tan y
10. tan(a;— y)= ^
11. cot (aj — y) =
1 4- tan X tan i/
cot X cot y 4- 1
cot y — cot X
12. sin 2 a; = 2 sin x cos a;.
13. cos 2 aj = cos' x — sin* x,
2 tan a;
14. tan 2 a; =
1 — tan*a;
139
140 PLANE TRIGONOMETRY
cot^cc — 1
15. cot 2 a: =
2 cot a;
16. sin ^ «
/I — cos z
=±\-^ —
Awf , I /I + cos z
17. cosi« = ±\ — ^
^r> . , . /I — cos z
18. taiij« = ±\3—
'' ^ 1 H- cos «
1Q 4.1 _L ^ /I + C08"g
19. coti« = ±\/:;
'^ ' 1 — cos z
20. sin ^ + sin 5 = 2 sin i(^ + 5)co8^(^ - B).
21. sin^ -sinJ5 = 2cosi(^ + 5)sini(^ -5).
2% cos^ 4- cos5 = 2cos i(^ + B)cosi(^ — 5).
23. cos^ -cosB = - 2sini(^ +-B)sin J(^ - B).
sin A + sin -B tan \{A -\' B)
24.
sin ^ — sin J5 tan \{A — B)
a sm A
Zo, — = — : — — •
sm J5
26. a2 = 52^c2- 25CCOS A
2^ a-5 ^ tan^(^ — B)
a + 5 tan i (^ + B) '
28. smM=V^^4^^
29. cosi^=A/^5^
'' ^ DC
FORMULAS Ul
30. ^^^=.J(Lzm^.
^ ^ s{s — a)
32. tani^ =— ^^ —
33, F = J ac sin B,
a?" sin -B sin C
34. iJ' =
2 sin {B + C)
35. F = Vs (s - a) (s - h) (s-c),
86. .-1^.
37. F=ir(a + 5 + c) = r«.
ANSWERS
PLANE TRIGONOMETRY
Exercise I. Page 2
1. i^; i^; i^; {i^; A.^; ii^; !i\^-
2. 12(y>; 136°; 112° SCT; 168° 45'; 84°.
3. 0.017453; 0.0002900. 8. 69.166 miles.
4. 206,265''. 9. 67 feet 3.66 inches.
5. I ;r ; I ;r. 10. 3 hours 49 minutes 11 seconds.
6. 11° 27' 33". 11. 9 feet 2 inches.
7. 14° 27' 28". 12. yj^ seconds.
Exercise II. Page 5
1. sinB = -: cos B = - : tan B = -; cot B = =- ; sec J? = - ; esc J? = v .
c c a 6 a b
8. (i) sin = f, cos = J, tan = }, cot = |, sec = }, esc = f ;
(ii) sin = -i^j, cos = ||, tan = y\, cot = -»/, sec = j|, esc = -^ ;
(iii) sin = ^y, cos = jf , tan = ,85, cot = V-, sec = {i, esc = V ;
(iv) sin = y\, cos = jf , tan = :f%, cot = V, sec = } J, esc = V- ;
(v) sin = If, cos = f §, tan = |g, cot = |g, sec = |§, esc = f | ;
(vi) sin = iif, cos = ||g, tan = |ig, cot '= ^
sec = if J, CSC = i}|.
4. The required condition is that a^ -{- b^ = c^. It is.
. • . 2mn m^-n^ ^ 2mn
5. (1) sm = » cos = > tan =
^ ' /7i2 + n2 r7i2 + n2
, m2-n2 m2 + n2
cot = » sec = 1 esc = — ^
2 mn m2 — n2
,..v . 2xy iC2_y2
(11) sm = 1 cos = , tan =
X2 + 2/2 X^ + 2/2
X2 - 2/2 X2 + 2/2
cot = — 1 sec = - — , CSC =
2zy x2 _ y2 2 X2/
1
m2-n2
m2 + n2
2mn
2xy
X2 _ ?/2 '
x2 + 2/2 .
2
PLANE TRIGONOMETRY
111) Bin
(iv) sin
cot
7. In (iii)
8. sin A -
cot -4 =
9. sin ^ :
cot -A :
10. sin A =
cot A =
11. sin ^ :
tan A :
sec -4 :
= - , cos = - »
8 P
_ ma
~ qr
__ pv
mpv
cos = -^^—
nqr
nqr
sec = —2—:
mpv
p 8
tan = - , cot = - 1
8 p
ns
tan = — ,
pv
P
sec = — »
8
esc = -;
CSC =
-^
ms
p2g2 _|. g282 -. p2g2 . in (iy) m^n^as + m^v^ = n^^,
= |}| =sinJ?; tanil =^ = cot5;
= J}f = CSC 5 ; CSC -4 = ^ = sec B.
= ^ = sinBj tan^=\«^ = cotB;
= ^y. = CSC 5 ; CSC -4 = }|| = sec B.
= ^g\ = sin B ; tan -4 = J^*^ = cot B ;
= -Y^ = CSC B ; CSC -4 = }§| = sec B.
^ = C08B
J^^* = tan B
m = cos B
^ = tanB
if} = cos B
JV^y = tan B
cos A
sec .4
cos^
sec^
cos A
sec A
= cotB
12. sin^
tan .4
sec A
13. sin A
tan^
sec A
14. sin A
sec^
15. sin A
sec A
Vp2 + gr2
^ ^ =C0SB;
V2pg
^ = cscB;
V2pg
P + 9
V2pg
cos -4 = ^-^
= sin-B;
V2pa ^
cot A = — ^^ = tan B ;
esc -4 = '^
yp2 _|_ g2
= sec B.
=n/I
«
P + g
= cot B ;
cos A = ^ / ^^ = sin 5 ;
p + q
cot A = -v/- = tan B ;
\p >
Vg2 + pq
p-q
= C8CB
CSC
A = -IM= =
P-^Q
= cos\B ;
Vp2 _^pg
2Vp^
= sec jB.
cos A = — = sin B :
P + g
2Vpg
2 Vpa
cot -4 = ^-^ = tan 5 ;
p-q
esc -4 =
_P + g
P-9
= sec B.
cot -4 = J ;
2Vpq
f Vs ; cos ^ = J Vs ; tan -4 = 2
Vs ; CSC -4 = ^ Vs.
f; cos-4=iV5; tanA = |V5; cot-4 = jV5;
I VS; esc^ = J.
ANSWERS 3
16. ain^ = i(6+ V?); cos A = I (b - Vl) ;
tan^ = i (16 + 5 V7); cot ^ = i(16 - 6 V?) ;
8ec-4 = J(6+ V?); esc -4 = J (6 - V?).
17. 8m^ = i(V8T+l); cos^ = J(^^ - 1);
tSLnA = ^{lQ-{- Vsi); cot^=^(16- VsT);
sec^ = ^(V81 + l); csc^ = ^(Vsi - 1).
18. a = 12.3. 20. a = 9. 22. c = 40.
19. 6 = 1.54. 21. 6 = 68. 23. c = 220.62.
24. Construct a rt. A with legs equal to 3 and 2, respectively ; then con-
struct a similar A with hypotenuse equal to 6.
28. a = 1.6 miles ; 6 = 2 miles.
SO. a = 0.342, b = 0.940 ; a = 1.368, b^ 3.760. 81. 142.926 yards.
Exercise in. Page 9
6. Through A (Fig. 3) draw a tangent, and take AT equal to 3; the
angle ^OT is the required angle.
6. From O (Fig. 3) as a centre, with a radius equal to 2, describe an
arc cutting at 8 the tangent drawn through B; the angle A OS is
the required angle.
7. In Fig. 3, take OM equal to i, and erect MP JL OA^ intersecting the
circumference at P ; the angle POM is the required angle.
8. Since sinic = cosx, OM = PM (Fig. 3), and x = 45°; hence, con-
struct X equal to 46°.
9. Construct a rt. A with one leg equal to twice the other ; the angle
opposite the longer leg is the required angle.
10. Divide OA (Fig. 3) into four equal parts ; at the first point of divi-
sion from O erect a perpendicular meeting the circumference at
some point P. Draw OP ; the angle A OP is the required angle.
12. X = 18°. 21. r sin x. 22. a = mc ; 6 = nc.
Exercise IV. Page 12
60°; sin 46°; cotl°; tan 75°;
71° 60' ; sin 52° 36' ; tan 7° 41' ; sec 35° 14'.
30°; sin 16°; cot 33°; tan 6°;
20° 58' ; sin 4° 21' ; tan 0° 1' ; sec 44° 59'.
1. COB
sec
2. cos
sec 20° 58' ; sin
PLANE TRIGONOMETRY
8. iVs.
6. 30°.
9. 22° 80'.
4. 46°.
7. 90°
10. 18°.
5. 80°.
8. 00°.
11. 10°.
1«. «^
n-f 1
Exercise VI. Page 16
1. 008-4=^^^; tan^=J^; cot-4=^; 8QcA=i^; cscA=\^,
2. cofl-4=0.6;tan-4=1.3333;cot-4=0.76; 860-4 = 1.6667; cscA = 1.25.
8. Bin-4=J|; tan-4=JJ; cot-4=fJ; 8ec-4 = |J; C8C-4=f|.
4. sin -4 =0.06; tan-4 =3.4286; cot-4 =0.2917; sec-4 =3.5714;
csc-4 = 1.0417.
5. siii-4=0.8; 008-4=0.6; cot -4 =0.76; 860-4 = 1.6667; C80-4 = 1.25.
6. 8in-4 = jV2; oos-4=jV2; tan-4 = l; 860-4 = V^; 080-4 = V2.
7. 8in-4=0.90; oo8-4=0.46; tan-4=2; 860-4=2.22; 080-4=1.11.
8. 8in-4 = iV3; 008-4 = J; tan-4=V3; oot-4=jV3; 080-4 = fV3.
9. 8in-4=jV2; 0O8-4 = iV2; tan-4=l; cot-4 = l; sec-4=V2.
"* Vl -m2; oot-4 = -Vl - m^ ;
m
1 1
860-4 =
11. 0O8-4 =
860-4 =
Vl - m2
1
Vl-m^
l-m2
•
H-m2'
1 +ma
l-m2'
•
m2 + n2 *
m^ + n^
l-m2
- 1
; 080-4 = —
m
tan-4 = ;
cot-4 =
2m
osc-4 =
2m
12. 8in-4=-— -; tan-4 = ; oot-4 =
860-4= : 080-4 =
2 mn * m2 -- n^ *
2 ?nn m2 — >i2
18. sin = J V^ ; cos = J V2 ; cot = 1 ; sec = V2 ; esc = V2.
14. cos = jV3; tan = jV3; cot = VS; seo = fV3; esc = 2.
15. sin = iV3; co8 = i; tan=V3; cot = ^V3; sec = 2.
16. sin = i V2 - VS ; cos = i V2 + Vs ; cot=2+V3;
= 2(2-V8)V2 + V3; esc = 2 (2 + Vs) ^2 - Vs.
1
860
ANSWERS • i
17. Bin = 5 V2 ~ V^; cos = J V2 -f v^; tan = V2~l;
sec = (4 ~V2)V2+V2; esc = (2 + ^2 ) V2 - V2.
18. cos = 1 ; tan = ; cot = go ; sec = 1 j esc = go.
19. cos = ; tan = go ; cot = ; sec = go ; esc = 1.
SO. sin =: 1 ; cos = ; cot = ; sec = go ; esc = 1.
A« A /I . o A J. A sin -4 ^ . Vl — sin^ A
21. cos-4 = vl — sin*-4; tan A^ - ; cot-4 =
Vl--8inM. sm^
A '^ A ^
aecA= — ; esc A = •
Vl-sin2 2l sin^ .
S2. sinA=Vl-cos2^; tan ^ = ^^^Zf^^lZ ; cot^ = -^^^
cos -4 Vl - cos2^
sec ^ = ; esc ^ = — •
cos-d Vl -cos^^
AA 1 .1 tan^ . 1 ,1
28. sin ^ = — ] cos A = — ; cot-4. =
Vl + tan2^ Vl + tanM. tan^
A ^/ i . ^ A A Vl + tana a
sec -4. = V 1 -f tan^^ ; esc -4. = .
tan^
AM • A 1 A cot A ^ . 1
24. sin ^ = — ; cos A = , — ; tan^ =
Vl + cot2^ Vl +eot2^ cot^
. Vl +eot2A . n — '• — -r-r
see -4 = ; esc -4. = V 1 H- cot^ A,
cot^
25. sin^ = jV5; cos^ = i>^. 8''- sin^ = :j\; cos^ = Jf.
«« . ^ t /7T ^ J /77 A<» 1 — 3 cos2 ^ H- 3 cos* -4.
26. sin-4 = i V16: tan-4.=Vl5. 28.
eos2 A — cos* A
Exercise VII. Page 18
1. a; = 46^ 6. x = 45°. 11. x = 30°. 16. x = 46^.
2. x = 30° 7. x = 46°. 12. x = 45°. 17. x = 60°.
8. X = 0°, or 60°. 8. X = 45°. 18. x = 0°, or 60°.
4. x = 45°. 9. x = 60°. 14. x = 80°.
5. x = 60°. 10. x = 60°. 16. x = 30° or 45°.
PLANE TRIGONOMETRY
Exercise Vm. Page 24
a
1.
c = -.
cos A
2. c = -^
sin A
8. b = c cos A.
4. c = -^—-
BlliA
5.
^ = 90*-
B; a -.
= c cos B; 6 =
CBinB,
6.
A = 90° -
B; a =
= 6 cot J5 ; c =
b
ainB
7.
8.
^ = 90°-
COS -4. = -;
c
B; b =
Bz
= atajiB; c =
= 90°-^; a =
a
cos 5
V(c + 6) (c - b).
Exercise IZ. Page 28
81.
c = 7.8112 ;
^ = 39° 48' ,
; B = 50°12';
F=16.
82.
h = 69.997 ;
A = 9(y 12"
; B = 89° 29' 48" ;
F= 21.525.
88.
a =1.1886;
A = 43° 20' ,
; B = 46° 40' ;
F = 0.74876.
f
84.
d = 21.249 ;
c = 22.372 ;
B=71°46';
F= 74.372.
85.
A = 6.6882 ;
c = 13.738 ;
B = 60° 52' ;
F= 40.129.
86.
A = 63.859 ;
b = 23.369 ;
B = 20° 6' ;
F= 746.15.
87.
d = 19.40 ;
b = 18.778 ;
A = 45° 56' ;
F= 182.15.
88.
b = 53.719 ;
c = 71.377 ;
A = 41° 11' ;
F = 1262.4.
89.
a = 12.981 ;
c = 16.796 ;
A = 55° 16' ;
F= 58.416.
40.
a = 0.58046 ;
b = 8.442 ; .
^ = 3° 56' ;
F = 2.4501.
41.
F= ic^ sin A cos A.
48. F = i62tan^.
42.
F=ia^ cot A.
44. F=iaVc2-
-a2.
45.
h = 11.6 ; c
= 15.315 ; A :
= 40° 45' 48" ; B = 49°
14' 12".
46.
a = 7.2; c
= 8.7658 ; A -.
= 55° 13' 20" ; B = 34°
46' 40".
47.
a = 3.6474 ; b
= 6.58 ; c :
= 7.6233 ; B = 61°
•
48.
a = 10.283 ; b
= 19.449 ; A =
= 27° 62' ; B = 62°
8'.
49.
19° 28' 17" and 70° 31' 43".
52. 36° 52' 12" and 53° 7' 48".
50.
3 and 5.1961.
58. 212.1 feet.
51.
90°
a = c cos
n + 1
. 90°
b = CBin
n+1
•
1
* •
54. 732.22 feet.
55. 3270 feet.
56. 37.3 feet.
ANSWERS
57. P?6'56''. 58. 69° 44' 36".
60. 7.0712 miles in each direction.
61. 20.88 feet. 63. 686.9 feet.
62. 66.66 feet. 64. 136.6 feet.
59. 96.34 feet.
65. 140 feet.
66. 84.74 feet.
1.
2.
8.
4.
Szercise Z. Page 88
C = 2(90P-A); c = 2acosA; h=za8mA.
-4 = J (180° - C) ; c = 2 a cos -4 ; ^ = a sin -4.
c
C = 2(90°-^); a =
^=i(180°-C);
2 cos A *
c
h = a sin .A.
a =
2 cos A '
h
A = a sin A,
5.
C = 2(90°-^);
smA
c = 2 a cos A.
6.
^=i.(180°-C);
h
a = -T--7;
sin^
c = 2 a cos A,
7.
'A ^
Sin i4 = - ;
a
C = 2(90°-^);
c=z2acosA.
8.
* A 2^
tan^ = — ;
c
C = 2 (90° - ^) ;
h
Bin .4
9.
v4 =67°22'60";
C = 45° 14' 20" ;
h = 13.2.
10.
c = 0.21943;
^ = 0.27384;
F= 0.03004.
11.
a = 2.066 ;
h = 1.6862 ;
F = 1.9819.
12.
a = 7.706;
c = 3.6676 ;
F= 13.726.
18.
^ = 79° 36' 30^';
C = 20° 47' ;
c = 2.4206.
14.
^ = 77° 19' 11";
C = 25° 21' 38" ;
a = 20.6.
15.
A = 26° 27' 47" ;
C = 129° 4' 26" ;
a = 81.41 ; h
16.
A = 81° 12' 9" ;
C = 17° 35' 42" ;
a = 17 ; c
17.
F = J c V4 a2 - c2.
22. 0.76636.
18.
F=a2siniCcos JC.
23. 94° 20'.
19.
F = a^ sin A cos A,
24. 2.7261.
20.
F-h^UnlC,
■
25. 38° 56' 33".
21.
28.284 feet; 4626.44
square feet.
26. 37.699.
= 36.
= 6.2.
8
PLANE TRIGONOMETRY
Exercise XI. Page 86
1.
r= 1.618;
h
= 1.5388; F= 7.694.
2.
^ = 0.0848,
; P
= 6.2514; i^= 3.0782.
8.
^= 19.754;
c
= 6.257; F=1236.
4.
r = 1.0824 i
f c
= 0.82842; F= 3.3137.
5.
r = 2.6933 ;
; h
= 2.4882 ; c = 1.4615.
6.
r= 1.5994;
h
= 1.441; p = 9.716.
7.
0.61803.
11. 0.2238.
16.
11.636.
8.
0.64984.
12. 0.310.
17.
99.640.
9.
0.51764.
18. 0.82842.
18.
1.0286.
Q
6- ""
o'
14. 94.63.
15. 414.97.
19.
0.636.
v«
2cos —
n
Exercise XII. Page 46
5. Two angles ; one in Quadrant I, one in Quadrant II.
6. Four values ; two in Quadrant I, two in Quadrant IV.
7. X may have two values in the first case, and one value iu each of the
other cases.
8. If cos X = — I, X is between 90® and 270° ; if cot x = 4, x is between
0° and 90° or between 180° and 270° ; if sec x = 80, x is between 0°
and 90° or between 270° and 360°; if cscx = - 3, x is between
180° and 360°.
9. In Quadrant III ; in Quadrant II ; in Quadrant III.
10. 40 angles ; 20 positive and 20 negative.
11. + , when X is known to be in Quadrant I or IV ; — , when x is known
to be in Quadrant II or III.
12. sinx =+ i">/2; tanx = — 1; cotx = — 1; secx= — V2;
esc X = + V2.
18. sinx=— jVS; cosx = — J; cotx = +J">/3; secx = — 2;
cscx=-|V3.
14. sinx=— ^VS; cosx = |; tanx = — 4^3; cotx = — ^J^VS;
CSC X =; — ^5 V3.
15. sinx=:± i^ff>^; cosx = =Ft»|jViO; tanx = - J; 8ecx = T jVlO;
esc X = i: vio.
ANSWERS
16. The cosine, the tangent, the cotangent, and the
when the angle is obtuse.
18. sin90° = l, cos 90° =0, cot 90° =0, sec 9
secant are negatiye
sin90° = l, cos 90° =0, cot 90° =0, sec 90° =
CSC 90°= 1;
sin 180° = 0, tan 180° = 0, cot 180° = ao, sec 180° = - 1,
CSC 180° = 00 ;
sin 270° = - 1, COS 270° = 0, tan 270° = oo, sec 270° = a>,
CSC 270° = - 1 ;
sin 360° = 0, t*rM afto^ = i . tjtn R(V)° = n ont .<wno = ««.
MOM AI\J — — ± J
sin 360° = 0, cos 360° = 1, tan 360° = 0, cot 860° =
sec 360°= 1.
19. sin450°=l; tan 640° =0; cos630° = 0; cot720°=a>;
sin 810° = 1 ; CSC 900° = oo.
20. 0. 21. 0. 22. 0.
23. a2- 524.405.
Exercise XIII. Page 61
1.
- cos 20°
8.
— sin
24°.
16. -cos 15° 33'.
2.
sin 8°.
9.
cos
1°.
16. cot 0° 46'.
8.
- sin 10°.
10.
- cot 30°.
17. -cot 40° 43'.
4.
- cot 36°.
11.
tan 6°.
18. CSC 29° 45'.
5.
- tan 1°.
12.
— CSC
26°.
19. sec 2° 25'.
6.
- CSC 20°
18.
-sec
1°.
7.
CSC 23°.
14.
sin
16° 11
•
20.
sin (-75°)=-
- cos 15° ;
28.
sin(
[-346°) = sin 16°;
cos (- 76°) =
sin 15° ;
COSI
[-345°) = cos 16°;
tan (-75°)= -
-cot 15°;
tan<
;- 346°) = tan 15°;
cot (-75°) = -
tan 15°.
cot|
;- 345°) = cot 15°.
21.
sin (-127°) =
-cos 37°;
24.
sin(
[-62° 37')= -cos 37° 23';
cos (-127°) =
- sin 37° ;
cos(
-62° 37') = sin 37° 23';
tan (-127°) =
cot 37°;
tan<
;- 62° 37')= -cot 87^23';
cot (-127°) =
tan 37°.
cot I
[-52° 37')= -tan 37° 23'.
22.
sin (- 200°) =
sin 20° ;
«
26.
sin 1
[-196° 64')= sin 16° 54';
cos (-200°) =
-cos 20°;
cos<
"-196° 54')= -cos 16° 64';
tAn(-200°)-
-tan 20°;
tan(
[-196° 64') = -tan 16° 64';
cot (- 200°) =
- cot 20°.
COt(
[-196° 64')= -cot 16° 64'.
10 PLANE TRIGONOMETRY
26. sinl20°=+iV3; co8l20°=-i; tanl20^=-V3; cot 120°=- J Vs.
27. sin 136°=+ J V2; oosl36°=-i V^; tan 136°= -1; cot 135°= -1.
28. siiil50°=i; C08l60°=-} VS; tanl50°=-iV3; cotl60°=-V3.
29. sin210°=-i; cos210°= -J VS; tan 210°= + J Vs ; cot210°= + V3.
80. sin226°=-iV2; cos225°=-i V2; tan 225°= 1; cot 226°= 1.
81. 8m240°=--jV3; cos240°=-J; tan240°= + V3; cot240°= +J Vs.
82. sin300°=-jV3; cos300°=i; tan300°=-V8; cot 300°= -J Vs.
88. 8in(-30°)=- J; cp9,( - 30°) = + J Vs ; tan(- 30°)*=- J V^;
cot{-30°)=-V3. ^. '. c
84. sin (- 225°) = + J Vi ;, cos(- 225°) = - J v^ ; tan (- 226°) = - 1 ;
cot(-226°) = -l.
85. cosx = — i V5; tanx = 1 :. cotx = 1 ; x = 226°.
86. 8inx = J; cosx=- J V3;'v iix = - J V3; x = 160°.
87. Bm3540° = -iV3; cos3540° = }; tan3640° = -V3;
cot 3640° = - i V3.
88. 210° and 330° ; 120° and 300°.
89. 136°, 225°, and - 226° ; 150° and - 30°.
40. 30°, 150°, 390°, and 510°.
41 . sin 168° ; cos 334° ; tan 226° ; cot 262° ;
sin 349° ; cos 240° ; tan 64° j cpt 177°.
42. 0.8480. 48. -1.9522. 44. (a -6) sin x.
45. msinxcosx. 48. 0.
46. (a — 6) cotx — (a + 6) tanx. 49. cos x sin y — sin x cos y.
47. a2 + 6a + 2a6cosx. 50. tanx.
51. Positive between x = 0° and x = 135°, and between x = 316° and
X = 360° ; negative between x = 136° and x = 316°.
52. Positive between x = 46° and x = 225° ; negative between x = 0° and
X = 45°, and between x = 225° and x = 360°.
58. sin (x — 90°) = — cos x ; cos (x — 90°) = sin x ;
tan (x - 90°) = - cot x ; cot (x - 90°) = - tan x.
54. sin (X - 180°) = - sin x ; cos (x - 180°) = - cosx ;
tan(x - 180°) = tanx; cot (x - 180°) = cotx.
ANSWERS
11
Exercise ZIV. Page 60
1. sin (X + y) = Jf ; cos {x + y) = f f .
3. sin ( 90° - y) = cos y
8. sin ( 90° + y) = cos y
tan( 90° + 2/) = -coty
4. sin (180° — y)= sin y
tan(180°-y)=-tany
6. sin (180° + y) = - sin y
tan(180°+y)= tany
6. sin (270° - y) = - cosy
tan(270°-y)= coty
7. sin (270° + y)=- cosy
tau(270° + y) = -coty
8. sin (360° - y) = - sin y
tan (360° - y) = - tan y
9. sin (360° + y) = siny
tan (360° + y ) = tan y
10. sin (x - 90°) = - cosx
tan(x- 90°)=-cotx
11. sin (X- 180°)=- sin X
tan (x - 180>°) = tan x
12. sin (x-270°)= cosx
tan (X - 270°) = - cot x
18. sin (— y) = — siny; cos(— y)= cosy;
tan (— y) = — tany ; cot (— y) = — cot y.
14. sin (45° — y)= jV2(cosy — siny); cos(46°— y) = J V2(cosy+siny);
cos ( 90° — y) = sin y.
cos ( 90° + y) = — sin y ;
cot ( 90° 4- y) = - tan y.
cos (180° — y) = — cos y ;
cot (180° -y)=- cot y.
cos (180° 4 7/^ - — cos y ;
cot (180*" = coty.
cos (270° — y) = — sin y ;
cot (270° - y) = tan y.
cos " + y) = sin y ;
cot V ■* + 1/) = — tan y.
cos (360° — y) = cos y ;
cot (360° - y) = - cot y.
cos (360° + y) = cos y ;
cot (360° + y)= coty.
cos (x — .' ~^°) = sin X ;
cot (x -"? ^0°) = - tan x.
cos (x — 180°) = — cos X ;
cot (x - 180°) = cot X.
cos (x — 270°) = — sin X ;
cot (x - 270°) = - tan x.
tan(45°-y) =
1 — tan y
cot(45°— y) =
cot y 4- 1
1 + tany' ' *" coty — 1
16. sin (46° + y)= jV2 (cosy + siny) ; cos(45°+y) = | v^ (cosy— siny);
tan(46° + y) =
1 + tan y
cot(45°+y) =
cot y — 1
1 — tan y' ' ^ ' *" cot y + 1
16. sin(30° + y)=ncosy + V3siny); cos (30°+ y) = i(V3 cosy-sin y);
tan(30° + y) =
^ Vs + tan y
1 _ 1 Vs tan y '
cot(30°+y) =
V3cot2/-l
cot y + V3
12 PLANE TRIGONOMETRY
17. sin {60° - y)= i{VScosy - siny); cos(60°-y) = J(cosy+ V^siny);
tan(60O-v)=:%i?5J^; cot (60°-y) = ^ ^""^^ +J.
l+v3tany coty-jVs
18. Ssinx -4sin8x. 19. 4cos8x - 3cosx. 20. 0. 21. jVs.
22. sin Jx = V ~ ^'^ = 0.10061 ; cos ix = -^i±M:^ = 0.99493.
23. cos2aJ = - i, tan2x = - V3.
24. sin 22i° = J V2 - V2 = 0.3827 ; cos 22^° = J V2 + V2 = 0.9239 ; .
tan 22^° = V2 - 1 = 0.4142 ; cot 22^° =V2 -f 1 = 2.4142.
26. sin 15° = J V2-V3 = 0.2588 ; cos 16° = J V2 + V3 = 0.9659 ;
tan 15° = 2 - V3 = 0.2679 ; cot 15° = 2 + Vs = 3.7321.
84. sin ^ + sin JB + sin C = sin A + sin B + sin [180° - (A -\- B)]
= sin ^ + sin B + sin (^ + E)
By [20] and [12],
= 2sin|(^ 4- B) cos|(^ -JB) + 2sin^(^ -\- B)cos^{A -\- B)
= 2 sin i (A + B) [cos ^(A-B)-\- cos | (A + B)]
By [22],
= 2 sin 5 (^ 4- B) (2 cos J ^ cos J J5)
= 4 sin J (-4 4- B) cos J -4 cos I B.
But cos ^ C = cos [90° -l(A-^ B)] = sin J (A + B).
.-. sin ^ 4- sin B 4- sin C = 4 cos J ^ cos J B cos ^ C.
86. Proof similar to that for 34.
38 ^ *2' tan2x. ^^ cos (x 4- y)
sin 2 X cos (x — y) sin x sin y
48. •
89. .2 cot 2 X. cos x cos y 47. tan x tan y.
^^ cos (x - y) ^^ co s(x4 -y)
sin X cos y cos x cos y
^j cos( x4-y) ^g cos (X - y)
sin X cos y sin x sin y
Exercise XV. Page 63
1. sin-H^ = 60°4-2n;r orl20°4-2n;r;
tan-H >^= 30° 4- 2 n;r or 210° -\-2n7t;
vers-H = 00° 4- 2n7r or 300° -\-2n7t;
ANSWERS 13
CDS-^i (- 1 y/2) = 135° + 2 n;r or 226° •\-2nft;
csc-i V2 = 45° + 2 n;r or 135° + 2 n;r :
tan-i 00 = 90° + 2 nTT or 270° + 2 iitt ;
sec-i 2 = 60° + 2 nTT or 300° + 2 n;r ;
cos-i (- 1 Vs) = 160° + 2 n;r or 210° + 2 n;r.
4. iV^. 10. ±^j. 12. iiV2.
8. 0°, 90°, 180°. 11. ± A. is. X = or ± i Vi.
Exercise XVI. Page 67
1. If, for instance, C = 90°, [25] becomes - = sin -4.
8. a2 = 62 + c2 ; a2 = 62 -I- c2 - 2 6c ; a2 = 62 4- c2 + 2 6c; a right triangle;
a straight line ; a straight line.
4. b = a cos C + c cos A\ a=h cos C •}- c cos B; c = b cos -4.
6. 90°.
7. (i) = tan (-4 — 45°) ; a right triangle.
a + 6
(ii) a + 6 = (a — 6) (2 + V3); an isosceles triangle with the angles 30°,
30°, 120°.
Exercise XVII. Page 69
9. 300 yards. 16. o = 5 ; c = 9.6693.
10. ^JB = 69.664 miles; 16. a = 7; 6 = 8.573.
AC = b4. 285 miles. 1 7 . sides, 600 feet and 1039. 2 feet ;
11. 4.6064 miles; 4.4494 miles; altitude, 519.6 feet.
3.7733 miles. 18. 855:1607.
12. 4.1501 and 8.67. 19. 5.433 and 6.857.
18. 6.1433 miles and 8.7918 miles. 20. 15.588.
14. 8 and 6.4723.
Exercise XVIII. Page 74
1. Two ; one ; no solution ; one ; two ; no solution ; one.
11. 420. 12. 124.617.
14 PLANE TRIGONOMETRY
Exercise XIX. Page 78
11. 6. 15. 25. 18. 10.266 miles.
12. 10.392. 16. 3800 yards. 19. 5.0032 and 2.3385.
14. 8.9212. 17. 729.67 yards. 20. 26® 0' 10^' and 14° 5' 60''
21. 430.85 yards.
Exercise XX. Page 83
11. ^ = 36° 52' 12"; 5 = 63° 7' 48"; C = 90°
12. ^ = 5 = 33° 33' 27"; C = 112° 63' 6".
13. ^=:JB=C = 60°.
14. ^ = 28° 67' 18"; JB = 46°34'6"; C = 104° 28' 36".
16. ^ = 46°; 5 = 120°; C = 16°. . 21. 54.616 miles.
16. ^ = 46° ; 5 = 60° ; C = 75°. 22. 84° 14' 34".
17. 4° 23' 2" W. of N., or W. of S. 28. 54° 48' 54".
18. 60°. 24. 106°; 16°; 60°.
2%. 0.88877. 25. 12.434 inches.
Exercise XXI. Page 87
1. 4,333,600. 6. 26,208. 11. 0.19975.
2. 365.68. 7. 15,540. 12. F=a6sin^.
8. 13,2^0. 8. 29,450 or 6982.8. 18. F=\(a^-l^)iaknA.
4. 8160, 9. 17.3206. 14. 2,421,000.
6. 240. 10. 10.392. 16. 30°; 30°; 120°.
Exercise XXII. Page 88
1. 21.166 miles; 24.966 miles. 4. 30°.
2. 6.3399 miles. 6. 20 feet.
8. 119.29 feet. 6. 2.6247 or 21.4587.
Exercise XXIII. Page 90
1. 106.70 feet; 8. 37°34'5". 6. 2922.4 miles.
142.86 feet. 4. 238,410 miles. 7. 60°.
2. 1023.9 feet. 6. 861,860 miles. 8. 3.2068.
ANSWERS
9.
6.6031.
81.
13.657 miles per
10.
199.56 feet.
hour.
11.
43.107 feet.
88.
56.564 feet.
12.
45 feet.
84.
51.595 feet.
18.
26° 34'.
85.
101.892 feet.
14.
78.367 feet.
87.
N. 76«56'B.;
15.
75 feet.
13.938 miles per
16.
1.4446 miles.
hour.
17.
7912.8 miles.
88.
442.11 yards.
18.
56.649 feet.
89.
255.78 feet.
19.
69.282 feet.
40.
3121. 1 feet;
20.
260.21 feet;
3633.5 feet
3690.3 feet.
41.
529.49 feet.
21.
1.3438 miles.
42.
41.411 feet.
22.
235.81 yards.
48.
234.51 feet.
26.
8.0076 inches.
44.
25.433 miles.
29.
460.46 feet.
45.
294.69 feet.
80.
88.936 feet.
A — Wl ±
46.
12,492.6 feet.
flfi
Vr/ja 4- 4 (^ 4. 1)
15
47. 6.3397 miles.
48. 210.44 feet.
50. -757.50 feet.
51. 520.01 yards.
52. 1366.4 feet.
58. 658.36 poimds;
22«>23'47''with
first force.
54. 88.326 pounds;
45« 37' 16" with
known force.
57. 536.28; 500.16.
58. 345.48 feet.
59. 345.46 yards.
60. 61.23 feet.
62. 307.77 yards.
68. 19.8; 35.7; 44.5.
64. 45^ 135^ 225^
or 315^
66. sin A
Vm2-n2 _ n \\ -ma
67.
68.
69.
71.
72.
78.
74.
76.
77.
78.
79.
80.
81.
82.
60°, 120°, 240°, or 300°.
0°, 60°, 180°, or 300°.
0°, 30°, 150°, 180°, 210°, 330°.
a , 180°
r = - cot
2 n
„ a 180°
B. = -esc
2 n
F= ^hcsinA.
F=ic^smA8ijiBc8c(A-\-B).
F= V5(«-o)(«-6)(«-c).
199 acres 8 square chains.
210 acres 9.1 square chains.
12 acres 9.78 square chains.
3 acres 0.392 square chains.
12 acres 3.45 square chains.
4 acres 6.634 square chains.
14 acres 5.54 square chains.
83. 61 acres 4.97 square chains.
84. 4 acres 6.633 square chains.
85. 13.93 chains; 23.21 chains;
32.50 chains.
86. 9 acres 0.055 square chains.
88. 876.34.
89. 1229.5.
91. 1075.3.
92. 2660.4.
98. 16,281.
94. 435.76 square feet.
95. 49,088 square feet.
96. 749.95 square feet^
97. 422.38 square feet.
16 PLANE TRIGONOMETRY
98. 1834.96 square 108. 6086.4 feet. 111. 228.98 miles;
feet. 109. 6°25'6"S.; IP 39' 6'' S.
99. 26.88. • 467.49 miles. 112. S. 66° 7' 82" E. ;
102. 6. 110. 460.79 miles; ^02.68 miles.
107. 6. 383.13 miles.
118. N. 17° 25' 22" W. ; 119. 33° 18' 22" N. ; 36° 23' 63" W.
37° 46' 13" N. 120. N. 28°47' 26" E. ; 1292.8 miles.
114. 244.36 miles ;S. 66° 10' 49" E. 121. S. 50° 39' 44" W. ;
116. 369.87 miles. . 250.84 mUes ; 20° 9' 30" W.
117. Long. 68° 54' 39" W. 122. 38° 20' 34" N.; 56° 12' 4" W.
118. 103.67 miles. 128. 171.14 miles; 32° 43' 38" W.
, 124. N. 36° 62' 12" W. ; 36° T 37" W.
126. 173.18 miles ; 61° 16' 16" S. ; 34° 12' 43" E.
126. S. 60° 67' 48" E. ; 47° 14' 36" N. ; 20° 48' 37" W.
127. N. 63° 20' 21" E., 16° 6' 67" W. ; <yr N. 63° 20' 21" W., 25° 63' 3" W.
128. N.47°42'33"E.,19°27'22"N.,121°50'34"E.; orN. 47°42'33" W.,
19° 27' 22" N., 116° 9' 26" E. ; or S. 47° 42' 33" E., 14° 32' 38" N.,
121° 48' 20" E. ; w S. 47° 42' 33" W., 14° 32' 38" N., 116° 11' 40" E.
129. 369.82 miles; 359.73 miles; 359.50 miles.
180. 35° 49' 10" S., 22° 2' 44" W. ; N. 61° 42' W. ; 183.16 miles.
131. 42° 15' 29" N., 69° 5' 11" W.; N. 72° 32' 40" E.; 44.939 miles.
182. 32° 63''34" S., 13° 1' 63" E.; N. 72° 3' 43" W.; 287.16 miles.
Exercise XXIV. Page 107
{The solutions here given are for angles less than 360^.)
79. siniiC = ±} V6; cosiz=± f VS. 81. iiVs.
80. ± Vs - 2. 82. ± I or ± |.
88. ± 2 V2, ± ^\(9 V3 + 8 V2), or ± ^^^^(O V3 - 8 V2).
84. ±i. 86. i(V6-l); i(V5 + l).
86. (a^ + &*)*. 91. 4-- ^^- . 96. tan-i^^ —
^ V2a + 1 1-2x2
87. (1^^\ (lT27nV 92. 4. 97. 2.
88. ±^V2or±^V3. 93. tan(aj + y). 98. - tan2x -f cot^aj.
sill dC
89. f. 94. 99. x = i7t OT ^7t.
sin y
90. J or - f 96. - tan x. 100. x = 90° or 270°.
ANSWERS 17
101. a; = 21°28'orl68°32'.
102. x = 0«or90°.
108. « = 30°, 160°, 199° 28', or 340° 32'.
104. X = 51° lO', 180°, or 308° 41'.
106. X = 0°, 120°, 180°, or 240°.
106. X = 45°, 161° 34', 225°, or 341° 34',
107. 6 = 60°, 120°, 240° or 300°.
108. ^ = 26° 34' or 206° 34'. 110. x = 45° or 135°.
109. X = 30° or 150°. 111. x = 30°, 150°, or 270°.
112. X = 35° 16", 144° 44', 215° 16', or 324° 44'.
118. X = 75° 58' or 255° 68'.
114. e = 60°, 180°, or 300°. 116. x = 30°, 150°, 210°, or 330°.
116. ^ = 90° or 143° 8'. 117. x = 30°, 160°, or 270°.
118. X = 26° 34', 90°, 206° 34', or 270°.
119. X = 45°, 135°, 226°, or 316°.
120. X = 45°, 135°, 225°, or 315°.
121. X = 16°, 75°, 135°, 195°, 265°, or 315°.
122. 2 = 45°, 135°, 225°, or 316°.
128. X = 0°, 60°, 120°, 180°, 240°, or 300°.
124. X = 27° 58', 135°, 242° 2', or 315°.
126. X = 0°, 46°, 180°, or 225°.
126. X = 32° 46', 147° 14', 212° 46', or 327° 14'.
127. X = 0°, 45°, 90°, 180°, 226°, or 270°.
128. X = 0^, 66° 42', 180°, or 204° 18'.
129. X = 0°, 90°, 120°, 240°, or 270°.
130. X = 0°, 36°, 72°, 108°, 144°, 180°, 216°, 252°, 288°, or 324°.
181. X = 80°, 150°, 210°, or 330°.
182. x = 60° or 240°.
188. X = 54° 44', 126° 16', 234° 44', or 305° 16'.
184. X = 105° or 345°.
a2-l
186. x = tan-i
186. X = cos
2a
, -a± Va2 + 8 a + 8
4
18 PLANE TRIGONOMETRY
187. X = 136^ 316°, or i sin-i (1 - a).
188. X = 30°, 60°, 120°, 160°, 210°, 240°, 300°, or 330°.
189. X = 60°, 90°, 120°, 240°, 270°, or 300°.
140. X = 60°, 90°, 120°, 240°, 270°, or 300°.
141. x = 120°. 143. X = 14° 29^, 30°, 160°, or 166° Sr.
148. X = 60°, 90°, 270°, or 300°.
144. X = 0°, 20°, 100°, 140°, 180°, 220°, 260°, or 340°.
146. X = 46°, 90°, 136°, 226°, 270°, or 316°.
146. X = 30°, 60°, 90°, 120°, 160°, 210°, 240°, 270°, 300°, or 330°.
147. X = 0°, 46°, 90°, 180°, 226°, or 270°.
148. X = 30°, 60°, 120°, 160°, 210°, 240°, 300°, or 330°.
149. X = 30°, 90°, 160°, 210°, 270°, or 330°.
160. X = 0°, 46°, 180°, or 226°.
161. X = 46°, 60°, 120°, 136°, 226°, 240°, 300°, or 316°.
162. X = 0°, 46°, 136°, 226°, or 316°.
168. X = 30°, 90°, 160°, 210°, 270°, or 330°.
164. x = 8° or 168°.
166. X = 40° 12', 139° 48', 220° 12', or 319° 48'.
166. x = 30° or 330°.
167. X = 60°, 120°, 240°, or 300°.
168. X = 30°, 60°, 120°, 160°, 210°, 240°, 300°, or 330°.
169. X = 63° 8', 126° 62', 233° 8', or 306° 62'.
160. x = 30°. 161. X = 22° 37' or 143° 8'.
162. X = 0°, 20°, 40°, 60°, 80°, 100°, 120°, 140°, 160°, 180°, 200°, 220°, 240°,
260°, 280°, 300°, 320°, or 340°.
168. X = 22° SO', 46°, 67° SO', 90°, 112° 30', 136°, 157° SO', 202° 30', 226°,
247° SO', 270°, 292° 30', 316°, or 337° 30'.
164.
X = 46° or 226°.
169.
x = l.
166.
X = ± 1 or ± 1 V2I.
170.
X = or ± 1.
166.
X = J Vs or - J Vs.
171.
X = ± i V2 .
167.
X = i Vs.
172.
x = iVs.
168.
X = i.
178.
e = 120° or 2i
174.
X = 60°, 120°, 240°, or 300°.
176.
X = 0°, 46°, 136°, 225°, or 315°.
ANSWERS 19
176. X = 0°, 180®, 220® 39', or 310° 21'.
177. X = 0®, 60°, 120°, 180°, 240°, or 300°.
178. e = 18°, 90°, 162°, 234°, 270°, or 306°.
179. X = 0°, 30°, 90°, 160°, 180°, 210°, 270°, or 330°.
180. X = 0°, 90°, 120°, 180°, 240°, or 270°.
18i. <? = 0°, 74° 5', 127° 25', 180°, 232° 86', or 286° 55'.
182. X = 0°, 90°, 180°, or 270°.
188. X = 0°, 45°, 90°, 180°, 225°, or 270°.
184. X = 0°, 46°, 120°, 135°, 180°, 226°, 240°, or 315°.
186. <? = 10° 12', 34° 48', 190° 12', 187. ^ = 29° 19', 105° 41', 209° 19',
or 214° 48'. or 285° 41'.
186. x = 90° or 270°. 188. x = 1.
, _ - 6 sin fl — a cos B a cos a — 6 sin a
l5w. x = J y = : •
sin (/3 - a) sin (/3 — a)
190. X = tan-i - + cos-i \ Va^ + h^ j y = tan-i^ - cos-i J Va^ + 62.
h
191. ^ = tan-J?; r=V^T^.
192. r =
(,acos/3— 6sina \ /^ .acosfl— ftsir^a . ^\
tan-J ^- h a ) cosi tan-i ^- -1-/3 )
asin/3+&cosa / \ asin/S+ocosa /
^ , , a cos /3 -- 6 sin «
= tan-* — ; — - — ;
asinp + cos a
193. r =Vo*^ + 62_}.c2; ^rztan-i^j = tan-i ^
^ Vaa + 62
194. X = 100 ; y = 200. 196. x = 76° 10' j y = 15° 30'.
196. r = 225.12, = 24° 13' j or r = - 225.12, = 204° 13'.
197. r = 151, ^ = 42° 28' ; or r =r - 151, = 222° 28'.
198. r = 108, <f> = 120°, = 330° ; r =108, = 300°, = 210°;
• r=-108, = 120°, ^=150°; orr = -108, = 300°, ^ = 30°.
199. X = r vers-i^-^ ^F V2 ry - y^.
Exercise XXV. Page 121
1. logio 6 = 0.77815 ; logio 14 = 1.14613 ; logio 21 = 1.32222
logio4 =0.60206; logio 12 =1.07918; logio 6 =0.69897
logio i =1.69897; logio i =1.39794; logio J =1.89086
logio ii = 0.02119.
20
PLANE TRIGONOMETRY
2.
logs 10
= 3.3219 ;
log2 5
= 2.3219 ;
logs 6
= 1.4660;
l0g7i
= - 0.3662 ;
logs 7f T
= - 2.2620.
8.
l0ge2
= 0.69316 ;
10ge3
= 1.09862 ;
loge5
= 1.60945 ;
10ge7
= 1.94593;
logc8
= 2.07943 ;
log«9
= 2.19724;
log. J
= -0.40547;
loget
= - 0.22316 ;
lOgef^
= 0.26962 ;
logeA
= - 2. 14846.
4. X = 1.64396 ; x - 0.83048 ; x = 0.42062.
1. loge3
4. logelO
5. Iogio2
Exercise XXVI. Page 126
1.09861. 2. loge6 =1.60944. 8. loge7 =1.94691.
2.3025860930. .
0.30103 ; logio 6 = 0.43429 ; logio 11 = 1.04139.
1. sin V
tanr
2. sin 2'
Exercise XXVII. Page 128
0.00029088820; cos V = 0.99999996769 ;
0.000290888212.
0.000681776. 8. sin 1° = 0.0176.
6. 0°40'9^
Exercise XXVIII. Page 180
1. sin 6' = 0.0017463 ; cos 6' = 0.9999986.
2. sin 2° = 0.034902 ; cos 2° = 0,999392.
8. sin 3° = 0.052340 ; cos 3° = 0.998632.
4. sin 4^ = 0.069762 ; cos 4° = 0.997568.
5. sin 5° = 0.087163 ; cos 6° = 0.996201.
Exercise X^IX. Page 135
1. The 6 sixth roots of — 1 are :
V3 + i . -V3 + i -V3-i
> «■»
2 ' ' 2 2
The 6 sixth roots of + 1 are :
-I,
v^-i
2.
^'2-2
V3 + i -Vs + i
, -1,
_1 -yTTz 1 -VITs
— I.
ANSWERS 21
8. cos 67i° + i sin 67i^ cos 157i^ + i sin 157i^, <»s 247^° + i sin 247^°,
co8 337J^+isin337i°.
4. sin4^ = 4cos8^sin^ — 4cos^sin3tf;
cos 4 ^ = cos* — Q C082 e sin2 $ + sin* ^.
Exercise XXX. Page 137
5. secx = lH h-..
2 24 720
6. » cot X = 1
3 45 945
7. sin 10° = 0.173648; cos 10° = 0. 984808.
8. tan 15° = 0.267958.
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