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ELEMENTARY PLANE TRIGONOMETRY. With
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PLANE AND SPHERICAL TRIGONOMETRY. In
Three Parts, comprising those portions of the
subjects, theoretical and practical, which are
required in the Final Examination for Rank of
Lieutenant at Greenwich. 8vo. 8*. 6d.
LONGMANS, GREEN, & CO., 39 Paternoster Row,
London, New York, Bombay, and Calcutta.
PLANE AND SPHEEICAL
TEIGONOMETEY
IN THREE PARTS
BY
H. B. GOODWIN, M.A.
NAVAL INSTRUCTOR, ROYAL NAVY
tPUBLISHED, UNDER THE SANCTION OF THE LORDS COMMISSIONERS OF
THE ADMIRALTY, FOR USE IN THE ROYAL NAVY>
EIGHTH IMPRESSION
LONGMANS, G E E E N, AND CO.
39 PATERNOSTER ROW, LONDON
NEW YORK, BOMBAY, AND CALCUTTA
1907
All rights reserved.
PBEFACE
TO
THE FOURTH EDITION.
As indicated in the original Preface, this . treatise in the first
instance was intended to • serve as an introduction to the study
of Navigation and Nautical Astronomy for the junior officers
under training in H.M. Fleet.
Since, however, it has had the good fortune to secure a
somewhat more extended circulation, the Author takes advan-
tage of the production of the fourth edition to largely supple-
ment the number of examples, both theoretical and practical,
so that while more fully meeting the requirements of naval
students, the work may at the same time be rendered more
complete in itself, and therefore more available for general
purposes.
The new examples, to the number of nearly three hundred
and fifty, will be found in an Appendix at the end of the
volume. They have been selected from the papers set in
examinations held under the direction of the Royal Naval
College during recent years, and will, it is hoped, afford a
sufficient field of exercise for the student in all branches of the
subject.
ROYAL NAVAL COLLEGE:
September 1893.
357856
PBEFACE.
THE following pages have been compiled chiefly for the use of
the junior officers of H.M. Fleet, in whose studies the subjects
of Plane and Spherical Trigonometry, forming, as they do, the
basis of the sciences of Navigation and Nautical Astronomy,
must necessarily occupy a very important place.
Since the establishment of the Royal Naval College at
Greenwich a considerable advance has been made in the standard
of mathematical knowledge attained by the junior officers of the
Fleet, and for some time the need of a suitable treatise upon
Plane and Spherical Trigonometry has been making itself more
and more apparent.
The text-books in Trigonometry commonly used in the
Service of late years have been four in number, viz. Hamblin
Smith's Plane Trigonometry, Todhunter's Spherical Trigono-
metry, Johnson's Trigonometry (used on board H.M.S. Bri-
tannia), and Jeans' Trigonometry (used chiefly afloat).
The inconvenience which must attend the use of so varied a
list of text-books is obvious, and, to remedy this drawback, in
the year 1884 the Lords Commissioners of the Admiralty were
pleased to give their approval to the preparation of the present
work.
The Author has endeavoured to include within the compass
of a single volume as much of the more theoretical portions of
Plane and Spherical Trigonometry as is required in the final
Vlll PEEFACE.
examination of acting sub-lieutenants at Greenwich, and at
the same time not to lose sight of the special character which
must belong to a work intended for naval students, in whose case
the practical application of the logarithmic formulae must neces-
sarily be of paramount importance.
The book is divided into three parts, the third of which is
devoted to the practical application of the various formulae
established in Parts I. and II.
In Part I., dealing with the theoretical portion of Plane
Trigonometry, the ground covered is practically identical with
the subject matter of the well-known manual of Hamblin
Smith, — a work which has during the last ten years proved of
great value as an elementary text-book.
Part II. contains as much of the theory of Spherical
Trigonometry as is necessary to establish the various relations
required in the solution of spherical triangles. This is a subject
which has generally been found to present special difficulties
to the young officer, because, on account of the early age at
which he is compelled to give it his attention, he enters upon
its study with a much smaller amount of mathematical know-
ledge than is possessed by those who take it up simply as
a branch of their general education. An effort has therefore
been made to exhibit the subject in its simplest form, and the
chief purpose of its study by naval officers, viz. to serve as an
introduction to the subject of Nautical Astronomy, has been
kept steadily in view.
Part III., the practical portion of the work, consists, to a
great extent, of the examples in the use of logarithms and in the
solution of plane and spherical triangles, compiled by the late
Mr. H. W. Jeans, formerly Mathematical Master at the Royal
Naval College at Portsmouth. Jeans' Trigonometry has been
in constant use in the Royal Navy for many years, and there
seems reason to believe that the collection of examples given in
that book has been found to answer satisfactorily the purposes
for which it was intended.
PREFACE. IX
The miscellaneous examples given at the end of Part III..
as exercises in Practical Spherical Trigonometry, may perhaps
be considered to belong rather to the sciences of Navigation and
Nautical Astronomy ; but, as no collection of such examples is
to be found in any of the text-books in ordinary use afloat, the
importance of these problems appears to justify their introduc-
tion here.
It will be observed that in the practical solution of triangles
the cumbrous verbal rules which, in the darker days of naval
education, were considered necessary, have been discarded, and
the particular process of computation has been deduced directly
from the appropriate formula in each case.
In obtaining the answers to the various practical problems
the ordinary custom has been followed of looking out logarithms
for the value given in the Tables which is nearest to the given
angle, that is, in general, to the nearest 15".
The Author wishes to take this opportunity of thanking the
several friends who have been good enough to assist him with
criticism and suggestions. To the Rev. J. B. Harbord, Chaplain
of the Fleet, the Rev. J. C. P. Aldous and the Rev. S. Kenah,
of H.M.S. Britannia, and the Rev. J. L. Robinson, of the
Royal Naval College, his acknowledgments are especially due.
ROYAL NAVAL COLLEGE, GEEENWICH :
March 1886.
CONTENTS,
PART I.
PLANE TRIGONOMETRY.
CHAPTER
I. ON MEASUREMENT, UNIT, KATIO
II. ON THE MEASUREMENT OF ANGLES .
III. ON THE APPLICATION OF ALGEBRAICAL SIGNS
PAGB
3
C
18
IV. ON THE TRIGONOMETRICAL EATIOS 15
V. ON THE CHANGES IN VALUE OF THE TRIGONOMETRICAL
EATIOS . . 18
VI. ON THE EATIOS OF ANGLES IN THE FIRST QUADRANT . 22-
VII. ON THE EATIOS OF THE COMPLEMENT AND SUPPLEMENT . 25
VIII. ON THE EELATIONS BETWEEN THE TRIGONOMETRICAL
EATIOS FOR THE SAME ANGLE 28
IX. ON THE EATIOS OF ANGLES UNLIMITED IN MAGNITUDE . 85
X. ON THE EATIOS OF THE SUM AND DIFFERENCE OF ANGLES 39
XI. ON THE EATIOS FOR MULTIPLE AND SUBMULTIPLE ANGLES 47
XII. ON THE SOLUTION OF TRIGONOMETRICAL EQUATIONS . 56
XIII. ON THE INVERSE NOTATION ... . . 59
XIV. ON LOGARITHMS 61
XV. ON THE ARRANGEMENT OF LOGARITHMIC TABLES . . 70
XVI. ON THE FORMULAE FOR THE SOLUTION OF TRIANGLES . 82
XVII. ON THE SOLUTION OF EIGHT-ANGLED TRIANGLES . . 93
XVIII. ON THE SOLUTION OF TRIANGLES OTHER THAN EIGHT-
ANGLED . . . . . . . . . .95
XIX. PROBLEMS ON THE SOLUTION OF TRIANGLES . . . 102
XX. OF TRIANGLES AND POLYGONS INSCRIBED IN CIRCLES, &c. 108
ANSWERS TO THE EXAMPLES
117
Xll CONTENTS.
PART II.
SPHEEICAL TEIGONOMETEY.
CHAPTER PAGE
I. THE GEOMETRY OF THE SPHERE ...... 123
II. ON CERTAIN PROPERTIES OF SPHERICAL TRIANGLES . . 133
III. ON FORMULA CONNECTING FUNCTIONS OF THE SIDES AND
ANGLES OF A SPHERICAL TRIANGLE 141
IV. ON THE SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES 151
V. ON THE SOLUTION OF EIGHT-ANGLED SPHERICAL TRIANGLES 157
VI. ON THE SOLUTION OF QUADRANTAL SPHERICAL TRIANGLES . 164
MISCELLANEOUS EXAMPLES .169
PAET III.
PRACTICAL TEIGONOMETEY.
PLANE AND SPHERICAL.
I. ON THE METHOD OF USING TABLES OF LOGARITHMS . . 177
II. THE SOLUTION OF RIGHT-ANGLED PLANE TRIANGLES . .189
III. THE SOLUTION OF OBLIQUE-ANGLED PLANE TKIANGLES . 192
IV. AREAS OF PLANE TRIANGLES 200
V. THE SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES . 203
VI. THE SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES . 213
VII. THE SOLUTION OF QUADRANTAL SPHERICAL TRIANGLES . 216
MISCELLANEOUS EXAMPLES IN PLANE TRIGONOMETRY . . .219
MISCELLANEOUS EXAMPLES IN SPHERICAL TRIGONOMETRY . . 225
ANSWERS TO THE EXAMPLES 233
APPENDIX.
A COLLECTION OF EXAMPLES SELECTED FROM EXAMINATION PAPERS
SET AT THE ROYAL NAVAL COLLEGE BETWEEN THE YEARS
1880-1893 241
ANSWERS TO THE EXAMPLES . . 269
PART I.
PLANE TRIGONOMETRY
CHAPTER I.
ON MEASUREMENT, UNIT, RATIO.
1. IN a subject dealing with concrete quantities, it is neces-
sary to fix upon standards of measurement, by reference to
which we may form definite conceptions of the magnitude of
those quantities. Thus, in such expressions as l a journey of
fifty miles,' c a reign of thirty years,' ' a legacy of one thousand
pounds,' our notions of the distance, time, and value involved are
derived from the several standards — one mile, one year, and one
pound respectively. We consider how many times each of the
fixed measures is contained in the aggregate quantity which
we have in view, and thus arrive at its numerical measure.
2. And since each standard measure contains itself once, its
measure is therefore represented by unity, and with reference to
that particular system of measurement it is termed a unit.
3. In trigonometry the systems of measurement with which
we have to deal relate, firstly, to lines, and secondly, to angles.
4. To measure a line AB, we fix upon a line of given
length as a standard of linear measurement. Thus, if AB con-
tain the line p times, p is called the measure of AB, and AB
is represented algebraically by the symbol p.
5. Two lines are called commensurable when such a line, or
unit of length, can be found that it is contained an exact num-
ber of times in each. When this is not the case, the lines are
said to be incommensurable, and the comparison of their lengths
can only be expressed approximately by figures.
Example 1. — What is the measure of 1^ miles, when a
length of 4 feet is taken as the unit ?
1J miles = | x 1760 x 3 feet = 7920 feet = 1980 x 4 feet.
B 2
4 PLANE TEIGONOMETEY.
Therefore the measure of 1£ miles is 1980.
Example 2. — If 360 square feet be represented by the
number 160, what is the unit of linear measurement?
The unit of area, that is, the square area which has for its
side the unit of length, = — - square feet = - square feet.
/9 3
Therefore the side of this area = \ / - linear feet = -
V 4 2
linear feet, or 18 inches.
Example 3.— If 4| inches be the unit of length, find the
volume of a block of wood which is represented by the number
Of)
OZ.
The unit of volume, viz., the cube which has for its edge
999 729
the unit of length = - x ~ x - = — - cubic inches.
a «. 2 o
729
Therefore the volume of the cube represented by 32 = - —
8
x 32 cubic inches =2916 cubic inches.
EXAMPLES. — I.
1 . If 5 inches be the unit of length, by what number will
6 yards 4 inches be represented ?
2. If 3 furlongs be represented by the number 11, what is
the unit ?
3. Find the measure of an acre when 11 yards is the unit of
linear measurement.
4. Find the measure of a yards when the unit is b feet.
5. A line referred to different units has measures 7 and 3 ;
the first unit is 9 inches : what is the other ?
6. A block of stone containing 3430 cubic inches is repre-
sented by the number 270 ; what is the unit of length ?
6. Thus it will be seen that in order to obtain the measure of
a given straight line, with reference to a given unit, we in reality
form a fraction, which -has for its numerator the length of the
given straight line, and for its denominator the length of the
given unit ; a fraction which, when reduced to its lowest terms,
is not necessarily a proper fraction, but is sometimes a whole
number, and sometimes an improper fraction.
MEASUREMENT, UNIT, KATIO. 5
The fraction so formed is said to represent the ratio of the
given straight line to the given unit.
7. To ask, therefore, ' What is the ratio of 3 yards to 7
inches ? ' is the same thing as to ask, ' What is the measure of 3
yards when 7 inches is taken as the unit ? ' or, ' What fraction
is 3 yards of 7 inches ? '
Generally, then, to express the ratio of one straight line to
another, we have but to reduce the lengths of the lines to the
same denomination ; then taking one of these values for nume-
rator, and the other for denominator, we obtain a fraction which
expresses the ratio of the lengths of the two lines.
Thus, if we have two lines, AB 3 inches in length, and CD
Q
1£ feet, the ratio of AB to CD is expressed by the fraction—,
1
or -. Similarly the ratio of CD to AB is represented by the
D
number 5.
8. Let us suppose that, in accordance with the system of
measurement which has been described, the values of the three
sides of a right-angled triangle are represented when referred to
a common unit of measurement by the letters p, q, r, the last of
these being opposite to the right angle.
The theorem established by Euclid, I. 47, furnishes us with
an equation —
p2 -f q2 = r2,
from which, if any two of these quantities be known, we may
determine the third. Thus, if p = 6, and r = 10, we have —
36 + q*= 100;
.-. £2 = 64, and £ = 8.
EXAMPLES. — II.
1. The ratio of the heights of two walls is as 7 to 5 ; the
height of the second one is 8 ft. 4 in. : what is the height of
the first ?
2. A field containing one acre is represented in one system
of measurement by 10m2 ; in another system a furlong is de-
6 PLANE TRIGONOMETRY.
noted by 20m ; find the ratio of the linear units made use of in
the two cases.
3. The length of the hypothenuse of a right-angled triangle
is 80 feet, and one of the sides including the right angle is 64
feet ; find the third side.
4. Find approximately the diagonal of a rectangle, the sides
of which are 7 feet and 5 feet respectively.
5. ABC is an isosceles triangle, each of the equal sides
being 20 feet ; the length of the line which bisects the vertical
angle is 16 feet ; find the third side of the triangle.
6. The area of a square is 12,769 yards ; find approxi-
mately the length of its diagonal.
7. If in an equilateral triangle the length of the perpen-
dicular let fall from an angular point upon the opposite side be
a feet, what is the length of the side of the triangle ?
8. A square is inscribed in a circle ; find the ratio of the
side of the square to the radius of the circle.
9. The length of the chord of a circle is 8 feet, and its
distance from the centre is 3 feet ; find the length of the
diameter.
10. The radius of a circle is 1 foot ; find approximately
the length of a chord which is 2 inches distant from the centre.
CHAPTER II.
ON THE MEASUREMENT OF ANGLES.
9, THE student has been accustomed in geometry to the
definition of a plane rectilineal angle as c the inclination of two
straight lines to one another, which meet, but are not in the
same straight line ' ; a definition which restricts the magnitude of
the angle to angles less than two right angles. In trigonometry
this definition is extended, and the conception of an angle is
derived from the revolution of a straight line round one end of
it, which remains fixed.
Thus, let AOB, COD be two straight lines intersecting at O,
ON THE MEASUREMENT OF ANGLES.
and at right angles to each other, and let OP be a line free to
revolve round the point 0.
The trigonometrical angle is measured by the angle moved
through by this revolving line.
Thus let the line be supposed at first to coincide with OA,
and to move from right to left, or in a direction contrary to that?
of the hands of a watch. c
We have in the figure four
positions of the revolving line,
and the angle traced out, AOP,
lies in the several cases within
the following limits : — B •
(1) AOP is less than one
right angle.
(2) AOP is less than two
right angles, but greater than one
right angle.
(3) AOP is less than three, but greater than two right
angles.
(4) AOP is less than four, but greater than three right
angles.
Since, after reaching the initial position OA, the revolution
of the line OP may be continued indefinitely, there are evi-
dently no limits to the magnitude of the trigonometrical
angle.
10. It becomes necessary, as in the 'case of straight lines, to
adopt suitable units of measurement, which will enable us to
form an idea of the magnitude of a given angle.
11. The most obvious method which suggests itself is to
take for our unit a right angle, or some fraction of a right
angle. Accordingly in what is known as the sexagesimal system
a right angle is divided into 90 parts, each of which is called a
degree, and the degree is subdivided into 60 smaller parts called
minutes, while the minute also is divided into 60 seconds.
Degrees, minutes, and seconds are then represented respectively
by the symbols °, ', " ', so that an angle is written as follows :
76° 18' 25".
12. We need not, however, confine ourselves to the right
8
PLANE TRIGONOMETRY.
angle as the primary unit. Any angle who3e value is invariable
will answer the purpose.
The method of measuring angles now to be described is
known as the system of ' circular measurement,' and the unit to
which all angles are referred is the angle which at the centre of
any circle is subtended by an arc equal in length to the radius
of the circle.
If this angle is to be taken as the unit, its magnitude must
be shown to be invariable, whatever the size of the circle.
13. We shall first show that the circumferences of circles vary
as their radii, that is, that if the radius of a circle be given, the
length of the circumference may be found by multiplying the
radius by a certain constant quantity.
Let 0 and o be the centres of two circles.
Let AB and db be the sides of regular polygons of n sides
inscribed in the circles. Let P be the perimeter of the polygon
inscribed in the first circle, p the perimeter of that inscribed in
the second circle, and let C, c be the circumferences of the two
circles respectively.
Then the angles AOB, aob, being each equal to -th part of
71
four right angles, are equal to one another.
Hence it is evident that the two triangles are equiangular.
Therefore their sides about the equal angles are proportionals.
Therefore OA : oa:: AB : ah ;
: : n . AB : n . ab ;
ON THE MEASUREMENT OF ANGLES. 9
But when n is very large, the perimeters of the polygons
may be regarded as equal to the circumferences of the circles.
Therefore OA : oa : : C : c ;
or alternately 0 A : C : : oa : c.
14. It is thus established that in any circle the circumference
bears a constant ratio to the radius of the circle. The fraction,
i . ! ,T , . circumference ->
therefore, which represents the ratio, viz. ^ and
radius
,T /, ,. circumference . ,, P „
therefore also the fraction — — - ± , is the same for all
diameter
circles. The numerical value of this ratio is an incommensurable
quantity, but it has been determined approximately by various
methods. The fraction £-4s a rough approximation ; it is more
355
nearly expressed by IrjoJ an(^ ^s value to five places of deci-
mals is 3-14159. The numerical value of this ratio is universally
expressed by the symbol TT.
The value of TT may be roughly tested by actual measure-
ment. Thus, if we take a circular hoop of radius 3'5 inches, and
a piece of string be drawn round the hoop, its length will be
found to be very nearly 22 inches.
„-, o,. circumference ., ~ «
15. Since — Tr— J- = 77% it follows that
diameter
circumference ~ 2 TIT.
Therefore the arc of a quadrant = j 2 TIT =~o"
EXAMPLES.— III.
22
(In these examples the value of IT may ~be taken as -— .)
1. The diameter of a coin is 1^ inches ; find the length of its
circumference.
2. The wheel of a bicycle makes 110 revolutions in a quarter
of a mile ; find the diameter of the wheel.
3. The minute hand of a watch is '6 inch in length; in
what time will the extremity travel one inch ?
4. How many inches of wire would be necessary to make a
10
PLAXE TRIGONOMETRY.
figure consisting of a circle of diameter one inch, with a square
inscribed in it ?
5. The barrel of a rifle is 1-75 inches in circumference;
find the radius of a spherical bullet which will just fit the
barrel.
6. A. person who can run a mile in 5 minutes runs round a
circular field in lm 25s ; find the diameter of the field.
16. To slww that the angle subtended at the centre of a circle
by an arc equal in length to the radius is an invariable anqle.
Let 0 be the centre of the circle whose radius OA=/\
At 0 in the straight line AO make the angle AOC a right
angle, and let AB be an arc equal in length to the radius of
the circle.
Then since, by Euclid, VI. 33,
angles at the centre of a circle
have to one another the same ratio
as the arcs upon which they stand,
A we have —
angle AOB
angle AOC
arc AB /•
: ^
arc AC irr
2
Therefore the angle
A ^-n. 2 * r\n two risrht angles
AOB =- angle AOC, or= ?
7T 7T
Now TT, as we have seen, is a quantity whose value does not
change.
Therefore the angle AOB has always the same magnitude,
whatever be the value of the radius OA, so that this angle
may properly be adopted as the unit in a system of measure-
ment.
17. To show that the circular measure of an angle is the ratio
of the arc on which it stands to the radius of the circle.
Thus, let AB be an arc equal to r, the radius of the circle.
Let AC be an arc, of length /, subtending an angle AOC at
the centre.
OX THE MEASUKEMENT OF ANGLES.
Then, by Euclid, VI. 33,
angle AOC_arc AC I
angle AOB arc AB r
Therefore AOC =~ AOB = - ,
r r
since AOB is unity.
The fraction - is called the
r
' circular measure ' of the angle
AOC, so that l circular measure ' may be defined as follows : —
The circular measure of an angle is the fraction which has
for its numerator the length of the arc subtended by the angle at
the centre of any circle, and for its denominator' the radius of the
same circle.
In the case of a right angle, since, as has been shown, the
7TT
arc of a quadrant is — , the fraction - becomes — ,or ^,
The circular measure of a right angle is therefore ^ ; so
2
that the circular measure of two right angles, or 180°, is TT.
18. We shall proceed to investigate methods of converting
the measure of an angle from one system to the other.
Thus let D be the number of degrees in an angle; it is
required to find its circular measure x.
Since TT is the circular measure of 180°, and since, whatever
be the system of measurement, the values which represent two
angles must have the same ratio to one another in each system,
we obtain the equation —
D therefore x = -^r (1).
^ 180
Again, when #, the circular measure, is given, and it is re-
quired to find the number of degrees.
Let y be the number required ; then
^-J; therefore ,=§180 (2).
To obtain the number of degrees in the unit of circular
12 PLANE TEIGONOMETEY.
measure, i.e. in the angle which at the centre of any circle sub-
tends an arc equal in length to the radius, we have only to
substitute unity for 6 in equation (2).
1 £0°
Thus, y= _ = 57°-29577 . . .
If we remember this value, we can at once realise in degrees,
&c., the value of an angle given in circular measure.
Thus, suppose the circular measure to be - ; its value in
2
degrees is i(57°-29577 . . .), or 28°-64788 . . . and so on.
2
EXAMPLES. — IV.
1. Express in degrees, &c., the angles whose circular
measures are
(1)1; (2) if; (3)11; (4)j. (g) . (0) 17^
2. Express in circular measure the following angles :—
(1) 30° (2j 75° (3) 420°
(4) the angle of an equilateral triangle.
3. If the unit of angular measurement be one-sixth part
of a right angle, by what number would an angle of 100° be
expressed ?
4. What must be the unit of measurement when three-
fourths of a right angle is expressed by the number 9 ?
5. An isosceles triangle has each of the angles at the base
four times the vertical angle ; express each of the angles in
circular measure.
6. If a right angle be taken as the unit of measurement,
what is the measure of the angle which, at the centre of a circle,
subtends an arc equal in length to the radius ?
7. The angles of a triangle are in arithmetical progression;
show that one of them must be 60°.
8. If the unit be the angle subtended at the centre of a
circle by an arc equal in length to twice the radius, by what
number would three right angles be represented ?
9. One angle of a triangle is 45°, and the circular measure
of another is 1*5 ; find the circular measure of the third angle.
10. The angles of a triangle, in ascending order of magni-
ON THE MEASUREMENT OF ANGLES. 13
tude, if expressed 'in terms of 1°, 100', and 200' respectively as
units, are numerically equal ; find the angles.
1 1 . The angles of a plane triangle are in arithmetical pro-
gression; if the circular measure of the smallest be equal to
-j-J-oth part of the number of the degrees of the angle next in
size, find the circular measure of the greatest angle.
12. Express in circular measure the angle of a regular
hexagon.
13. If ABCD ... be a regular decagon, and the sides AB,
DC be produced to meet, express in degrees and in circular
measure the angle contained by the pair of sides so produced.
14. In a certain circle an angle of 30° at the centre sub-
tends an arc three feet in length ; find the length of the radius.
15. The measure of a certain angle in degrees exceeds its
circular measure by unity ; find the value of the angle in degrees.
16. The angle subtended by the moon's diameter is 32' 30" ;
find approximately its length in miles, if the distance from the
observer be 240,000 miles.
*V ~\ ^ •^^"^
fr, • ** ^\>0 £ 3* ' L '- t'V
•*l CHAPTER III. &
b A • /& n^
ON THE APPLICATION OF ALGEBRAICAL SIGNS. , f^P
19. LET MN be a straight line, and 0 a fixed point in it.
*T— -s— 3— — i 5
The position of any point in the line will be determined if
we know the distance of the point from 0, and if we know also
upon which side of 0 the point lies. It is, therefore, convenient
to bring into requisition the algebraical signs + and — , so that
if distances measured along the straight line from the point 0 in
one direction be considered positive, distances measured along the
straight line in the opposite direction will be considered negative.
For instance, let A be a point to the right of O, distant from
it a linear units. Then if B be a point to the left of O, and
equally distant from it, the distance of B from 0 is properly
expressed by — a.
It should be observed that in the first instance either direc-
14
PLANE TRIGONOMETRY.
tion may be selected as the positive direction, provided that
when once made the selection is rigidly adhered to afterwards.
20. Let AOA', BOB' be two straight lines intersecting at
right angles in 0.
Then if lines measured along OA,
OB, be regarded as positive, lines
measured along OA', OB', must be
considered negative.
This convention, as it is called,
•A is extended to lines parallel to
AA' and BB' as follows :—
Lines drawn parallel to AA'
are considered positive when mea-
sured to the right of BB', and ne-
'e' gative when to the left of BB'.
Lines drawn parallel to BB'
are reckoned positive when measured upwards from A A', and
negative when measured downwards from AA'.
21. We have seen in art. 9 that the angle in trigonometry
is generated by the revolution of a straight line round one
B extremity, which remains fixed.
Let OP be the revolving line,
and let us suppose that it is in
the initial position, coinciding
with OA.
A- It is obvious that the straight
line may trace out angles in two
ways, according as it moves in
the direction of, or contrary to,
the hands of a watch.
Let us therefore suppose the
B'
revolving line to have left the initial position, and, moving con-
trary to the hands of a watch , 'to have reached the position OP.
It is the general custom to call this the positive direction.
But if the revolving line be supposed to move downwards
from OA until it reaches the position OP', moving in the direc-
tion of the hands of a watch, the angle AOP' and all other
angles so traced out are considered negative angles.
ON THE APPLICATION OF ALGEBRAICAL SIGNS.
15
22. Let the revolving line, moving in the positive direction,
come to rest in the position OP, having traced an angle of A°.
Afterwards let us suppose the revolving line to move through
another angle of B°, first in the positive, then in the negative direc-
tion, finally arriving at the positions OP'., OP" in the two cases.
0 A
Then the angles AOP', AOP" are respectively (A + B)°, and
(A— B)°, these being the angles moved through by the revolving
line in the two cases.
CHAPTER IV.
ON THE TRIGONOMETRICAL RATIOS.
23. LET the line OP, leaving the initial position OA, and re-
volving round 0 in the positive direction, describe an angle AOP.
(l) (2) (3) (4)
ziK
MA
M A
N
From P drop a perpendicular upon OA, the initial line, or
upon OA produced.
In fig. (1) the angle described is an acute angle, and is less
than one right angle.
In fig. (2) it is greater than one right angle, but less than
two right angles.
In fig. (3) it is greater than two, but less than three right
angles.
16 PLANE TRIGONOMETRY.
In fig. (4) it is greater than three, but less than four right
angles.
In each case our construction gives us a right-angled tri-
angle POM, which is called the triangle of reference.
And in this triangle the side PM is the perpendicular, OM
the base, and OP the hypothenuse.
The ratios which these parts of the triangle POM bear to
one another are of great importance in dealing with the angle
AOP, and to each of these ratios accordingly a separate name
has been given.
24. The principal ratios are six in number, as follows : —
In the triangle POM
PM perpendicular . , , . n * /-\T-»
/vrr, or £— - » is the sine of AOP.
OP hypothenuse
, or _ - -- , is the cosine of AOP.
hypothenuse
PM perpendicular . ,, , c Arrn
==-0 or v F. -- , is the tangent of AOP.
OM base
OP hypothenuse . ,,
or -££ — ^ — *—. is the cosecant of AOP.
PM perpendicular
OP hypothenuse . , , n * -^
-— r.., or -^- -- , is the secant of AOP.
OM base
or . -, is the cotangent of AOP.
PM perpendicular
To these may be added another function of the angle, viz.
the versine, which is the defect of the cosine from unity. Thus
versine AOP= 1 — cosine AOP.
25. The words sine, cosine, &c. are abbreviated in practice,
the several ratios being written —
Sin. A, cos. A, tan. A, cosec. A, sec. A, cot. A, vers. A.
26. The powers of the various ratios are expressed in the
following way : —
(sin. A)2 is written sin.2 A ;
and so on for the other ratios.
[The reader should carefully guard against confusing this
symbol with sin. 2 A, the meaning of which will appear further on.]
ON THE TRIGONOMETRICAL RATIOS.
17
27. It should be noticed that the second three ratios defined
above are respectively the reciprocals of the three first given.
Thus—
1
cosec. A =
sec. A =
cot. A =
sin. A '
1
cos. A J
1
tan. A'
28. The trigonometrical ratios remain the same so long as the
angle is unchanged.
It has been explained that in Trigonometry an angle is
supposed to be traced by the re-
volution of a straight line round
one of its extremities.
Let AOB be any angle.
Take points P, P' in OB,
such that OP' = m . OP.
Then we may suppose that
either OP or OP', originally co-
inciding with OA, has traced the angle AOB.
From P, P' let fall perpendiculars PM, P'M' upon OA.
Then the triangles POM, P'OM' are equiangular, and there-
fore similar.
Hence the sides of the triangles about the equal angles are
proportional.
And
So
P'M' m.PM.
OF = m~TVP
OM1
OP'
PM
OP
And similarly for the other functions of the angle AOB. So
that it is a matter of indifference whether we consider the angle
AOB to belong to the triangle POM or to the triangle P'OM'.
Inreach case the values obtained for the several ratios will be
the same.
c
18
PLANE TRIGONOMETRY.
CHAPTER V.
TO TRACE THE CHANGES, IN SIGN AND MAGNITUDE, OF THE
DIFFERENT TRIGONOMETRICAL RATIOS OF AN
ANGLE INCREASES FROM 0° TO 360°.
ANGLE, AS THE
29. LET AA', BB' bisect each other at right angles at the
point 0, and let a line OP revolve in the positive direction round
the fixed point 0, so that its ex-
tremity P traces the circumference
of a circle.
Thus, as in art. 23, if P, P15
P2, P3 be the positions of P in
the first, second, third, and fourth
quadrants respectively, by drop-
ping perpendiculars PN, P^,
P2N2, P3N3 upon the initial line,
we obtain in each case a triangle
of reference, from the sides of
which the geometrical representa-
tions of the several ratios _, _, &c. may be derived.
And in accordance with the convention of the preceding
chapter, the lines PN, P^, &c., will be considered positive so
long as they are above the line AA', and negative when below
that line.
Similarly the lines ON, ON1? &c. will be regarded as posi-
tive when drawn to the right of BB', and negative when to the
left of that line.
The line OPl is to be regarded as uniformly positive.
30. To trace the changes in the sine of an angle as the angle
increases from 0° to 360°.
ON CHANGES IN VALUE OF THE TRIGONOMETRICAL RATIOS. 19
Let A be the angle traced by the revolving line.
By art. 24, sin. A = Pedicular = PN
hypothenuse OP
Now when A=0°, OP coincides with OA, and the perpen-
dicular PN vanishes.
Therefore sin. A= —— = Q.
When the angle therefore is zero, so also is its sine. As
A increases from 0° to 90° PN is positive, and continually
increases from 0 to OP.
Thus when A =90°
*- -§£•
Hence in the first quadrant sin. A is positive, and continually
increases from 0 to 1.
As A increases from 90° to 180° the perpendicular PN
remains positive, but decreases continually until when A=180°
PN again vanishes, and sin. 180°=:0. In the second quadrant
therefore sin. A is positive, and decreases from 1 to 0.
As A increases from 180° to 270° the perpendicular PN
becomes negative in sign, but increases until on reaching 270°
it coincides with OB', so that sin. 270°= — 1.
Hence in the third quadrant sin. A is negative, and increases
numerically from 0 to— 1.
As A increases from 270° to 360° PN remains negative in
sign, and decreases, till on reaching the initial line it again
vanishes.
Thus in the fourth quadrant sin. A is negative, and de-
creases numerically from — 1 to 0.
31. To trace the changes in the sign and magnitude of the cosine
of an angle as the angle increases from 0° to 360°,
The same construction being made, we have
base ON
L/OS. A = , -J " = 7TFT'
. hypothenuse OP
c 2
20 PLANE TKIGONOMETRY.
Proceeding in the manner of the preceding article, we have,
when the angle is 0°, OP coinciding with OA.
,o OA
Hence cos. 0° = — =1.
As the angle increases from 0° to 90°, ON is positive, and con-
tinually decreases until, when OP reaches OB, ON vanishes alto-
gether. Hence cos. 90° = 0.
Therefore in the first quadrant the cosine is positive, and
decreases from 1 to 0.
By the same process the following results may easily be
established : —
In the second quadrant cos. A is negative, and increases
numerically from 0 to — 1 .
In the third quadrant cos. A is negative, and decreases
numerically from — 1 to 0.
In the fourth quadrant cos. A is positive, and increases from
0 to 1.
32. To trace the changes in the sign and magnitude of the
tangent as the angle increases from 0° to 360°.
It should be observed that in selecting the proper sign to be
affixed to the sine or cosine of a particular angle, we have only
to consider what sign is due to the numerator of the fraction in
each instance. With regard to the tangent and cotangent the
case is otherwise, for with these ratios the sign belonging to both
the numerator and denominator has to be taken into account.
When the angle is 0°, PN vanishes as before, while ON is
equal to OA.
Therefore tan. A = Jp[= J*
ON OA
Therefore tan. 0°=0.
As A increases PN, the numerator, is positive in sign, and
increases until it ultimately coincides with OB. But ON, which
was originally equal to OA, decreases continually, and eventually
vanishes ; so that since the numerator continually increases, while
the denominator decreases, the value of the fraction continues to
increase until upon the angle reaching 90°
ON CHANGES IN VALUE OF THE TRIGONOMETEICAL KATIOS. 21
Hence by taking the angle sufficiently near 90° the tangent
may be made greater than any assigned value.
This is expressed by saying that tan. 90° is infinity, which
is denoted by the symbol oo.i
And between 0° and 90°, ON and PN bein£ both positive,
PN
tan. A, or-——, is also positive.
Thus in the first quadrant tan. A is positive, and increases
from 0 to oo.
As A increases from 90° to 180°, PN continues positive, and
decreases, until at 180° it vanishes. ON is negative, and in-
creases until it coincides with OA'.
Therefore in the second quadrant tan. A is negative, and
decreases from <=o to 0.
As A increases from 180° to 270°, PN is negative, and in-
creases until ultimately it coincides with OB1 ; ON also is nega-
tive, and decreases until at 270° it vanishes.
The tangent is therefore positive in sign, and increases from.
0 to oo.
Between 270° and 360°, PN is negative, and decreases; ON
is positive, and increases.
The tangent is therefore negative, and decreases numerically
from oo to 0.
33. Since by art. 27 cosec. A = — , sec. A =
sin. A cos. A
cot. A = -, the changes in sign and magnitude of the
tan. A
other ratios may be deduced from those already investigated,
or they may be obtained directly from the figure. To trace
them for himself will be found a useful exercise for the
student.
1 This result may be illustrated as follows. Suppose ON = - OB, then
OB
tan. A = 1 QB = n. Now as ON diminishes the quantity n increases, so that
n
when ON becomes indefinitely small, tan. A or n becomes indefinitely large.
That is, an angle may be found approximately equal to 90° whose tangent is
greater than any assignable quantity.
22
PLANE TRIGONOMETRY
34. The following table exhibits the changes in the values
of the several ratios in a convenient form.
The intermediate columns show the sign possessed by the
particular ratio as the angle increases from one value to the
next higher in magnitude :- —
Value of A
0°
90°
180°
270°
360°
Sine . . .
0
+
1
+
0
-1
0
Cosine . .
1
+
0
—
-1
—
0
-t-
1
Tangent .
0
+
00
—
0
+
00
—
0
Cosecant
00
+
1
4-
oo
—
-1
—
00
Secant . .
1
+
00
—
-1
—
oo
+
1
Cotangent .
00
+
0
—
00
+
0
—
00
35. The following points may be noticed in connection with
the values assumed by the different ratios : —
The sine and cosine are never greater than unity.
The cosecant and secant are never less than unity.
The tangent and cotangent may have any values whatever
from zero to infinity.
The trigonometrical ratios change sign in passing through
zero or infinity, and through no other values.
CHAPTER VI.
ON THE RATIOS OF CERTAIN ANGLES IN THE FIRST QUADRANT.
36. WE have seen that the trigonometrical ratios, the sine,
cosine, &c. are simply numerical quantities. They can be found
approximately for all angles by methods which cannot be at
present explained. There are, however, certain angles the
ratios of which can be determined in a simple manner. Among
these are the angles 30°, 45°, and 60°.
37. To find the trigonometrical ratios for an angle of 30°.
ON KATIOS OF CERTAIN ANGLES IN THE FIRST QUADRANT. 23
Let OP, revolving from the position OA, describe an angle
AOP, equal to one-third of a
right angle, that is, an angle
of 30°.
From P draw PM perpendi-
cular to OA, and produce PM to
meet the circle in P'. Join OP'.
Then the two triangles 0PM,
OP'M are equal in all respects.
And the angle OP'M = 0PM
= 90° - AOP = 60°.
Thus the triangle OPP' is equilateral.
Therefore PM = 1 PF = 1 OP.
Z 4
Let 2m be the measure of OP. Then m is the measure of
PM. And OM =
PM m 1
Then sin. 30 = _=_=-,
c-.8Q-.gf-
tan. 30° = ™ =
whence also cosec. 30° = 2, sec. 30° = — -, cot. 30° =
38. To find the trigonometrical ratios for an angle of 45°.
Let OP, revolving from the position OA, describe an angle
AOP, equal to half a right angle,
that is, an angle of 45°.
Draw PM perpendicular to 0 A.
Then, since POM, OPM are
together equal to a right angle,
and POM is half a right angle,
therefore OPM also is half a right
angle.
Thus OPM is equal to POM,
and PM equal to OM.
Let the measure of OM or PM be m.
24
PLANE TKIGONOMETKY.
Then the measure of OP is \/m* + m'2 =
Thus sin. 45'
PM
OP
m
1
.,0 OM m 1
cos. 45 = -—^ = — — = — — ;
O P ^/2m /y/2
mnm
IIU t
tan. 4D- = ?rT? = - •= 1 ;
OM m
and cosec. 45° = \X2, sec. 45° = \/2, cot. 45° = 1.
89. To find the trigonometrical ratios for an angle o/60°.
Let OP, revolving from the
position OA, describe an angle
equal to two-thirds of a right
angle, that is, an angle of 60°.
Draw PM perpendicular to
OA, and join AP.
Because OP is equal to OA,
therefore the angle OAP is equal
to the angle OPA.
But these two angles are to-
gether equal to 120° ; therefore each is equal to 60°, and the
triangle AOP is equilateral.
Hence OA is bisected in M.
Let the measure of OM be m.
Then the measure of OP is 2m.
And the measure of PM is V'4m2—m2 =
Then sin. 60° = =
r.ro OM
COS. bO = -- = - = —
m
2m
1
;
2 '
OP
tan. 60'==™ = ^ =
OM m
and cosec. 60° = sec. 60° = 2, cot. 60° = -=•
ON EATIOS OF CERTAIN ANGLES IN THE FIRST QUADRANT. 25
EXAMPLES. — V.
If A = 90°, B= 60°, C = 45°, D = 30°, prove the following
relations : —
(1) Sin.2 D + cos.2 D =-• 1.
(2) Cos.2 B - sin.2 B = 1 - 2 sin.2 B.
(3) Sec.2 B = 1 + tan.2 B.
(4) Sin. B tan. D + tan. C sin. D = 1.
(5) 2 cos.2 C + tan.2 B = cot.2 D sec.2 C - cosec.2 C.
Sin. Q - sin. D = __
V J Sin. C + sin. D
(7) Sin. A sin. D — cos. A cos. 0 = 2 vers.2 B.
(8) Tan.2 B - tan.2 D ^
sin. A sin. P
cos.2 B cos.2 D'
CHAPTER VII.
ON THE RATIOS OF THE COMPLEMENT AND SUPPLEMENT.
40. Def. Two angles are said to be the complements of each
other when their sum amounts to 90°.
Thus 60° is said to be the complement of 30°, and vice versa ;
— 30° is the complement of 120°, ^ of J, and so on.
6 o
41. To compare the trigonometrical ratios of an angle and its
complement.
Let AA', BB' be two diameters of a circle intersecting at
right angles in 0.
Let a radius OP revolving
from OA trace out the angle
AOP = A.
Next let the radius revolve
from OA to OB and back again
through an angle BOP' equal to A.
Then the angle AOP = 90° - A.
Draw PM, P'M' perpendicular
to OA.
Then the angle OP'M' = EOF
= A = MOP.
26 PLANE TRIGONOMETRY.
Therefore the triangles OPM, OP'M', having one side and
two angles equal, are equal in all respects ; so that OM£ = PM,
and P'M' = OM.
Then f;.
Sin. (90° - A) = sin. POM' =
= cos. A.
Cos. (90° - A) = cos. POM' =
= sin. A. *»
= = cos. MOP
= ^ = sin. MOP'
And similarly it may be shown that tan. (90°— A) = cot. A,
cosec. (90° - A) = sec. A, sec. (90° - A) = cosec. A,
cot. (90° - A) = tan. A.
42. JDef.* Two angles are said to be the supplements of each
other when their sum amounts to
180.°
Thus 150° is the supplement of
is the supplement of 5.
o 6
To compare the trigonometrical
ratios of an angle and its supple-
ment.
Let AA', BB' be two diame-
ters of a circle intersecting at
right angles in O.
Let the line OP, revolving from OA, trace out an angle A.
Next let OP revolve from OA to OA' and back through an
angle equal to A, coming to rest in the position OP', so that
AOP' = 180° - A.
Drop perpendiculars PM, P'M' upon A A'.
The two triangles POM, P'OM' are equal in all respects.
Thus PM = P'M', OM = OM'.
Therefore sin. (180° - A) =
™' = ~ =sin. A ;
OP'
OP
Again tan. (180° — A) = — tan. A ; and in the same manner
the other ratios may be compared.
ON THE EATIOS OF THE COMPLEMENT AND SUPPLEMENT. 27
43. To compare the trigonometrical ratios of the angle
(90° + A) with those of A.
Let AA', BB' be two diameters of a circle intersecting at
right angles in 0.
Let OP, revolving from OA,
move to OP, describing an
angle A.
Let OP then continue to re-
volve until after passing OB the
angle BOP = A.
Then AOF = 90° + A.
Drop perpendiculars PM,
PM' upon AA'.
In the triangle POM' the
angle 0PM' = FOB = A, and
the two triangles POM, P'OM' are equal in all respects.
Thus OM'= PM, P'M' = OM.
Therefore sin. (90° + A) =
PM'
= =
OP'
OM' -PM
OP
OM A
— = cos.A;
= —sin. A.
Thus tan. (90° + A) = — cot. A, and similarly the other
ratios may be compared.
44. By processes similar to those already given the ratios of
the angles (180° -I- A) and (—A) may be compared with those
of the angle A, and will be found to be as follows :—
Sin. (180° + A) =•- - sin. A ; sin. (- A) = -sin. A ;
cos. (180° + A) = - cos. A ; cos. (- A) = cos. A ;
tan. (180° + A) = tan. A ; tan. (- A) = -tan. A.
EXAMPLES. — VI.
1. Write down the complements of the following angles : —
(1) 27° 37' 48". (2) 50° 16' 38". (3) 105°.
(4) -37°. (5) " (6)-~
28 PLANE TRIGONOMETRY.
2. Write down the supplements of the following angles
(1) 79° 36' 15". (2) 101° 19' 43". (3) 200°.
(4) - 70°. (5) *£. (6) - *
o i
3. Reduce to simpler forms the following equations : —
/-, \ fir \ cos. (TT — A)
(1) cos. ( - — x ) = 2 lm
\2 ) sin. (- A)
(2) sin. f j? + x J = cos. (TT — A) cosec. (TT + A),
(3) sin. (TT — x) cos. (TT — B)
cos. (~ + A j sec. (TT + B)
cosec. I - —
CHAPTER VIII.
ON THE RELATIONS BETWEEN THE TRIGONOMETRICAL RATIOS
FOR THE SAME ANGLE.
45. THE principal trigonometrical ratios, as defined in art.
24, are six in number, viz. the sine, cosine, tangent, cosecant,
secant, and cotangent.
Three of these, as already pointed out, are the reciprocals of
the other three ; thus —
Cosec. A = _ — - , sec. A = - , cot. A =
sin. A cos. A tan. A
By means of these relations, and others now to be established,
any one of the ratios may be expressed in terms of each of the
others.
46. Let AOP represent any angle A traced out by the
revolution of OP round one extremity 0, and let a perpendicular
ON RELATIONS BETWEEN THE RATIOS FOE THE SAME ANGLE. 29
PM be let fall upon OA, or OA produced, so that in each case
POM is our triangle of reference.
We can now prove the following relations :~—
r ; cot. A = LU&'A
cos. A sin. A
For tan. A =
PM
PM
OP
sin. A
OM OM cos. A'
OP
OM
PM
OP
II. Sin.2 A+ cos.2A= 1.
2 A PM2 OM2 PM2 + OM2
For sm.2A + cos.2A= _ + ^ = - — _
OP2
OP2
III. Sec.2 A = 1 + tan.2 A ; cosec.2 A = 1 + cot.2 A.
2 A - 2?? - QM2 + PM2 ._ l PM2
~ OM2 ~ OM2
OP2 PM2 4- OM2 , OM2
= 1.
n 2 A
Cosec.2 A =
PM2
PM2
PM2
, ,
cot.2 A.
Collecting our results, we know now that if A denote any
angle, the following statements are true : —
(1) cosec. A = —
(2) tan. A =
sin.A '
sin. A
sec. A =
1
cos.A '
cos. A
cot. A =
tan. A
, cot. A = - .
cos. A sin. A
(3) sin.2 A + cos.2 A = 1 ; sec.2 A = 1 4- tan.2 A ; cosec.2 A
=n 1 + COt.2 A.
30 PLANE TRIGONOMETRY.
46a. We shall now give a few examples of what are known
as ' identities.' An identity is an expression which states in
the form of an equation certain relations between the functions
of angles, these relations being true whatever magnitudes are
assigned to the angles involved.
Example 1 .—Prove that cot.2 d — cos.2 6 = cot.2 0 cos2 6.
Since cot.20
cot' 0 - cos.* 9 = ^ - cos.* e = «*.'tf-cos.'flain.'g
sm.2 0 sin.2 0
cos.2 6(1 - sin.2 0) cos.2 0
= -- . 9 „ - -= . 9 a cos.2 0 = cot.2 0 cos.2 0.
sm.2 0 sm.2 0
Here our first step was to express cot.2 0 in terms of the sine
and cosine. This process of substituting for the tangent and
other functions of an angle, their value in sines and cosines will
be found of frequent advantage in such cases.
Example 2. — Prove that sin.2 0 tan. 0 + cos.2 0 cot. 0
-f 2 sin. 0 cos. 0 = sec. 0 cosec. 0.
Since ten.* = ^if, cot. <9 = ^ -f.
cos. 0 sin. 0
sin.2 0 4n.-#-+ cos.2 0 eoir.^ + 2sin. 0cos. 0= ™^% + cos-3 ?
^U cos. 0 sin. 0
+ 2 sin. 0 cos. 0
= sin.4 0 + cos.4 0 -4- 2sin.2 0 cos.2 0
. sin. 0 cos. 0
(sin.2 0 + cos.2 0)2 ]
* - - - — ~ - -
a a a
sin. 0 cos. 0 sin. 0 cos. 0
= sec' cosec
EXAMPLES. — VII.
Prove the identities—
(1) Sin. A sec. A = tan. A.
(2) (tan. A + cot. A) sin. A cos. A = 1.
(3) (tan. A — cot. A) sin. A cos. A = sin.2 A — cos.2 A.
(4) (cos.4 A - sin.4 A) = 1 - 2sin.2 A.
(5) cos.3 A— sin.3 A = (cos. A— sin. A)(l + sin. A cos. A).
(6) cos.6 A + sin.6 A = 1 — 3cos.2 A sin.2 A.
(7) 2 (sin.6 A + cos.6 A) -3 (sin.4 A + cos.4 A) + 1 = 0.
ON RELATIONS BETWEEN THE RATIOS FOR THE SAME ANGLE. 31
(8) (1-2 cos.2 A) (tan. A + cot. A) =
(sin. A — cos. A) (sec. A + cosec. A).
(9) sec.2 d — cos.2 6 — cos.2 9 tan.2 6 + sin.2 6 sec.2 0.
(10) (cosec.2 0—1) (2vers. 0 - vers.2 0) = cos.2 0.
(11) sin. 0 (cot. 0 + 2) (2 cot. 0 + 1) = 2 cosec. 0
+ 5 cos. 0.
(12) vers.2 A + 2cos. A — sin.2 A = 2cos.2 A.
(13) cosec.2 0 — vers. 0 = vers. 0 cot.2 0 + cos. 0 cosec.2 0.
(14) cos.6 A + 2cos.4 A sin.2 A + cos.2 A sin.4 A + sin.2 A
= 1.
n KN tan. A + tan. B A .
(15) — — = tan. A tan. B .
cot. A -f cot. B
/-, /?N cot. A -f' tan. B
(16) — — = cot. A tan. B.
tan. A + cot. B
(17) * + cos- A = (Cosec. A + cot. A)2.
1 — cos. A
/I Q\ 2 A 2 -D COS'2 B — COS.2 A
(18) tan.2 A — tan.2B = - .
cos.2 A cos.2 B
(19) sec. 0 4- cosec. 0 tan.3 0 (1 -f cosec.2 0) = 2 sec.3 0.
(20) 2vers. 0 - vers.2 0 = sin.2 0.
47. To express all the other ratios in terms of the sine : —
Since sin.2 A + cos.2 A = 1
cos.2 A = 1 — sin.2 A ;
.-. cos. A = ± Vl — sin.2 A '
A . A sin. A sin. A
Again tan. A = = =. ;
cos. A ± VI — sin.2 A
cosec. A =
sec. A =
sin. A '
1 1
cos. A ± Vl _ sin.2 A '
cot.A^CQS- = ±l -sin.2 A
sin. A sin. A
It will be observed that corresponding to a given value pf the
sine, there will be two values of the cosine, tangent, secant, and
cotangent, since on tho right-hand side of the equation either
the positive or negative sign may be taken. This arises from
32 PLANE TEIGONOMETEY.
the fact that when the sine of an angle is given, more than one
angle may be found which possesses this sine, although the co-
sine, tangent, &c. may not be the same for the two angles.
As a simple case, let sin. A = -.
a
From art. 37 we know that the sine given is that of 30°.
But since, by art. 42, sin. (180° — A) = sin. A, the angle
150° also has this sine.
Thus there are two angles possessing the given sine, viz. 30°
and 150°.
To find the cosines of these angles we have the formula —
cos. A = ± Vl — sin.2 A = + -
u
For 30°, since the angle lies in the first quadrant, the posi-
tive sign must be taken ; for 150° the negative.
For the present the ambiguity in sign may, however, be
neglected.
48. In a similar manner, by means of the formulae given
above, the other ratios may be expressed in terms of the cosine,
tangent, &c. There is, however, another method of arriving at
these expressions which is worthy of notice.
It will first be necessary to show how to construct an angle
having a given sine, cosine, or tangent.
49. To construct an angle having a given sine or cosine.
Let it be required to construct the angle which has for its
sine a given ratio a, a being less than unity.
Take AB equal to the unit
of length, and upon AB as dia-
meter describe a circle.
With centre B and radius BC,
equal to the fraction a of the unit
of length, describe a circle.
Let C be one of the points
where the circumference of this
circle intersects the first circle,
and join AC, BC.
ON RELATIONS BETWEEN THE RATIOS FOR THE SAME ANGLE. 33
Then ACB, the angle in a semicircle, is a right angle
(Enc.III.31).
Thus sin.
Therefore BAG is the angle required.
If the cosine of the required angle is to be a, the same con-
struction may be made, and ABC will be the angle required.
50. To construct an angle having a given tangent or cotangent.
Let it be required to construct an angle of which the tan-
gent shall be a given quantity a.
Draw a straight line AB equal to the unit c
of length. At B draw BC at right angles to
AB, and equal in length to a times the unit,
and join AC.
Then tan. BAG =^-=a.
AB A" _B
Therefore BAC is such an angle as is required.
If the cotangent of the required angle is to be a, then the
same construction may be made, and ACB will be such an angle
as is required.
51. To express the other trigonometrical ratios in terms of
the sine.
Let s be the sine of the given angle.
As in art. 49, construct a right-angled triangle having its
hypothenuse equal to the unit of length, and the side BC
s times the unit.
It follows that AC = Vl-s
AC
Then we have cos. A =-—=l—
BC
tan. A = -;-= =
AC Vl-s2 " A tJT^r c
Similar expressions will easily be found for the other ratios.
52. To express the other trigonometrical ratios in terms of the
tangent.
Let t be the tangent of the angle.
D
34 PLANE TRIGONOMETRY.
As in art. 50, construct a right-angled triangle having the
side AC, one of those including the right
angle, equal to the unit of length, and the
other side BC t times the unit, so that BAG
is the angle which has the given tangent.
Then AB^
BC
and sin. A — T^ = ->-=
AB VI
And so for the other ratios.
53. In the case of the tangent the above method is the
simplest for deducing expressions for the other ratios. We
might, however, deduce the value of the sine, &c. in terms of
the tangent by proceeding as follows : —
sin. A
Tan. A=
cos. A '
sin. 2A sin. 2A
Therefore tan. 2 A =
cos. 2A 1 — sin. 2A*
/. tan.2 A — tan.2 A sin. 2 A = sin.2 A.
tan. A
Whence sin. A= -7=
Vl+tan.2A
In the same way the expression for the cosine might be
obtained.
EXAMPLES. — VIII.
1. Express the sine of an angle in terms of the cotangent.
2. Express the sine and cotangent of an angle in terms of
the secant.
3. Express the principal trigonometrical ratios in terms of
the versine.
4. Given that cos. A = — , find tan. A.
o
5. Given that sec. A=2, find cot. A.
o
5. If tan. 6 = -^- , find sin. 6 and sec. 6.
V5
7. If sec. A = L find cot. A.
5
ON RELATIONS BETWEEN THE RATIOS FOR THE SAME ANGLE. 35
8. If sin. A= -, find tan. A.
o
9. If tan. A — - , find cosec. A.
o
10. If cos. 0 = a, and tan. 0=fr, find the equation con-
necting a and b.
1 1 . Construct the angles whose sines are - and -_ .
12. Construct the angle whose tangent is \/2 — 1.
CHAPTER IX.
ON THE RATIOS OF ANGLES UNLIMITED IN MAGNITUDE.
54, IT is shown in art. 41 that sin. (180° — A) = sin. A. It
follows from this that if an angle be constructed in the first
quadrant having its sine equal to a given positive quantity a,
another angle may always be obtained in the second quadrant
having its sine equal to the same quantity.
If the condition were that cos. A should be equal to a} since
cos. A = cos. ( — A), or cos. (360° — A), there would also be two
values for the angle A, one in the first and the other in the
fourth quadrant.
If tan. A=a, then the angle must be in the first or third
quadrant.
55. Thus'let sin. A=i. Here A may be either 30° or 150°, and
the positions of the revolving line
will be OP and OP' respectively.
Moreover, since the trigonome-
trical angle is unlimited in mag-
nitude, let us suppose the revolv-
ing line on reaching the initial
position to continue its revolu-
tion ; as often as it takes up the
position OP or OP' we shall ob-
tain an angle having its sine equal to the assigned value, viz. -'
Ll
D 2
36 PLANE TRIGONOMETRY.
And OP of course is free to move either in the positive or
negative direction.
Thus, if OP revolve from OA in the positive direction, we
have a series of angles as follows: 30°, 150°, 390°, 510,° . . .
all having the given value of the sine.
But if the negative direction be taken, we obtain a series
of negative angles, viz. — 210°, — 330°, — 570,°. . . also having
the given sine.
Thus generally if a be the circular measure of the smallest
angle having the given sine, by the addition or subtraction of
2?r, 4-7T, 6?r . . . 2w7r, we shall obtain a series of angles having
their sine equal to the value given. And the same statement
applies to the other ratios.
We shall proceed to find expressions which will include all
angles having a given sine, cosine, or tangent.
56. To find an expression which will include all angles having
a given sine.
Let a be the angle having a
given sine.
The other position of the re-'
volving line which gives the same
value for the sine is that for the
angle IT— a.
All other angles positive and
negative which have the same sine
may be derived by adding or sub-
tracting some multiple of 2?r from these two values.
Thus if we suppose the revolving line to move in the posi-
tive direction we shall obtain the two series of angles
a, 2-7T -I- a, 4?r + a 2m?r + a
TT — a, 3-7T— a, 57T— a (2m+l)7T — a
If in the negative direction, we have
— TT — a, — STT— a, — 5?r— a .... — (2w-f 1) TT— a
— 2-TT + a, — 47r-|-a, — 67T-4-a .... — 2m7T-f a
where m may have any integral value.
ON THE KATIOS OF ANGLES UNLIMITED IN MAGNITUDE. 37
Comparing these four series we see that
(1) TT may have any integral coefficient, odd or even, posi-
tive or negative.
(2) When the coefficient of TT is even the sign of a is
+ , when odd the sign of a is — .
Now both the conditions of (2) are satisfied by the expression
mr + (- 1)» a
where n is any integer, positive or negative.
For if n be even (— l)n must be positive, but if n be odd
( — 1)TC will be negative.
By assigning, therefore, in succession the numbers 1, 2, 3,
4, ... and — 1,— 2,— 3,— 4, . . .to the letter rc, we shall
obtain the two series of angles which have the given sine.
57, To find an expression which will include all angles having
a given cosine.
Let a be the angle having the
given cosine.
The only other position of the
revolving line which will give the
same value for the cosine is that
for the angle 2?r — a.
All other angles having the
given cosine may be obtained by
adding or subtracting 2-Tr, and
multiples of 2?r, from each of
these values.
The four series so obtained will be
a, 2?r + a, 4-yr + a . . . 2m?r + a
2^7- _ a? 4-7T — a, GTT — a . . . 2m?r — a,
— 27r + a, —477 + 0-5 — 6V + a . . . — 2m?r + a
— a, -- 2-7T — a, — 4-7T — a ... — 2m?r — a
where m may have any integral value.
Here we observe
(1) That the coefficient of TT is always even, but either
positive or negative in sign.
(2) That a may have either a positive or negative sign.
38
PLANE TKIGONOMETKY.
Thus the expression 2ri7r ± a, where n is any integer,
positive or negative, will furnish all values of the angle having
the given cosine.
58. To find an expression which ivill include all angles having
a given tangent.
Let a be the angle whose
tangent is given.
The only other position of the
revolving line for which the tan-
L gent will have the given value is
that for the angle IT + a. All
other values of the angle having
the given tangent will be obtained
by adding or subtracting 2?r, and
multiples of 2?r, from one of these angles.
Thus we obtain the series
27T + a, 4-7T + a . . . 2m7r + a
3?r + a, 5-Tr + a . . . (2m + 1) TT -f a
— 4-7T + a, — 6?r + a . . . — 2m?r + a
— STT -fa, — 5?r -fa... — (2m -f 1) TT -f a
where m is any integer.
:
Thus we find that
(1) The coefficient of TT may be either odd or even, positive
or negative.
(2) The sign of a is always positive.
So that the expression mr + a, where n has any integral
value, positive or negative, contains all the angles required.
EXAMPLES. — IX.
1. Write down the values of the following ratios : —
(1) tan. 225° (2) cos. (- 60°) (3) tan. 780°
(4) cot. 1035° (5) sec. 240° (6) cot. 210°
(7) cosec. 570° (8) sin. (-210°) (9) cos. (- 120°)
(10) tan. !2p (11) cot. -^ (12) tan. 6-rr
a,
TT + a,
- 2-rr + a,
— TT + a,
ON THE EATIOS OF ANGLES UNLIMITED IN MAGNITUDE. 39
2. Find the general value of 0 under the following circum-
stances : —
(1) When tan. 0=1.
(2) When sin. 0 = \.
a
(3) When cos. 0 = - |.
(4) When sec. 0 = - 1.
(5) When sec.2 (9 = 2.
(6) When tan.2 0 = 3.
CHAPTER X.
ON THE TRIGONOMETRICAL RATIOS OF THE SUM OR DIFFERENCE
OF ANGLES.
59, THE object of the present chapter is to establish formula
for expressing the ratios of angles made up of the sum or
difference of other angles in terms of functions of these angles
themselves.
60. The formulae first to be established are as follows : —
Sin. (A + B) = sin. A cos. B + cos. A sin. B
Cos. (A -f B) = cos. A cos. B — sin. A sin. B
Sin. (A — B) = sin. A cos. B — cos. A sin. B
Cos. (A — B) = cos. A cos. B + sin. A sin. B
61. To show that
Sin. (A + B) = sin. A cos. B + cos. A
Cos. (A + B) = cos. A cos. B — sin. A
Let the angle BAG be repre-
sented by A, and the angle CAD
byB.
Then the angle BAD will be
represented by A -f B.
From P any point in the line
AD, draw PM at right angles to
AB, and PN at right angles to AC.
sin. B
sin. B
D
40
PLANE TEIGONOMETEY.
From N draw NO at right angles to PM, and NQ at right
angles to AB.
Then the angle OPN = 90° - PNO = ONA = NAQ = A.
_NQ . OP__NQ AN OP PN
" AP "*" AP AN' AP + PN ' AP
= sin. A . cos. B + cos. A sin. B ;
AQ AN_ON PN
AN ' AP PN ' AP
= cos. A . cos. B — sin. A sin. B.
62. To show that
Sin. (A — B) = sin. A cos. B — cos. A sin. B
Cos. (A — B) = cos. A cos. B + sin. A sin. B.
Let the angle BAG be re-
presented by A, and the angle
CAD by B.
Then the angle BAD will re-
present A — B.
Take any point P in AD, and
from P draw PM at right angles
to AC, and PN at right angles
toAB. A Q N B
From M draw MO at right angles to NP produced, and MQ
at right angles to AB.
Then the angle MPO = 90° - OMP = CMO = A.
PN ON - OP MQ - OP
Now sm. (A - B) = — = ___=_-bL__
_MQ_OP_MQ AM_ OP PM
= AP AP ~ AM * AP PM ' AP
= sin. A cos. B — cos. A sin. B ;
8L
ON THE EATIOS OF COMPOUND ANGLES. 41
and cos. (A_B)= =
Al Ax AP
= AQ OM
AP AP
AM OM MP
AM ' AP MP ' AP
= cos. A cos. B + sin. A . sin. B.
In the construction of the figure of this and the preceding
article, it may assist the student to notice that P, the point from
which perpendiculars are let fall, lies in the line which bounds
the compound angle A + B or A — B.
63. Expressions for tan. (A 4- B) and tan. (A — B), in
terms of tan. A and tan. B, may be established independently
by means of the figures given in arts. 61 and 62. It is, how-
ever, simpler to deduce them from the formulae already esta-
blished.
sin. A cos^B^h cos. A sin. B
cos. A cos. B — sin. A sin. B'
and this by dividing each term of numerator and denominator
by cos. A cos. B becomes
sin. A cos. B cos. A sin. B
cos. A cos. B cos. A cos. B _ tan. A + tan. B
cos. A cos. B sin. A sin. B 1 — tan. A tan. B
cos. A cos. B cos. A cos. B
/A T>\ sin. (A — B)
Again, tan. (A - B) = ^ _ ^
sin. A cos, B — cos. A sin. B
cos. A. cos. B 4- sin. A sin. B
sin. A cos. B _ cos. A sin. B
cos. A cos. B cos. A cos. B
cos. A cos. B sin. A sin."B
cos. A cos. B cos. A cos. B
tan. A — tan. B
" 1 -f tan. A tan. B'
PLANE TKIGONOMETEY.
In the same way may be deduced the following results : —
T-J, cot. A cot. B — 1
cot. (A + B) = - -— — —5- ;
cot. A 4- cot. B
/ A T>\ cot. A cot. B -f 1
cot. (A — B) = — —5— ~TT •
cot. B — cot. A
64. The formulae now established will enable us to find the
values of the sines and cosines of 75° and 15°.
Thus, to find the value of sin. 75°.
Sin. 75° = sin. (45° + 30°) = sin. 45° cos. 30° + cos. 45° sin. 30°
, /3 + *
2 V2 2 2 V2
Again, to find the value of sin. 15°, we have
sin. 15° = sin. (45° - 30°) = sin. 45° cos. 30° -cos. 45° sin. 30°.
65. The diagrams of arts. 61, 62 are constructed for the
simplest cases ; thus, in art. 61, A, B are together less than 90°,
while in art. 62 A is less than 90°, and B less than A.
The formulas obtained, however, are universally true, as may
be shown in any particular case.
Thus, for example., let A, B each lie between 45° and 90°,
so that A + B lies between 90° and 180°.
Let A' =90° -A.
B/ = 90°-B.
Then sin. (A' + B') = sin. (180°-A + B) = sin. (A + B).
Since A, B are each greater than 45°, it follows that A' + By
is less than 90°.
Therefore, by making use of the diagram of art. 61, we have
sin. (A' + BO = sin. A' cos. B' + cos. A' sin. B'
=cos. A sin. B + sin. A cos. B.
And from above, sin. (A' 4- Br) = sin. (A + B).
Therefore sin. (A + B) = sin. A cos. B + cos. A sin. B.
ON THE RATIOS OF COMPOUND ANGLES. 43
Andcos.(A'-fB')=cos.(180°-A + B) = - cos. (A+B).
But by art. 61
cos. (Ax + B') = cos. A' cos. B' - sin. A' sin. B'
= sin. A sin. B— cos. A cos. B.
Therefore
cos. (A -f B) = — cos. (A' + B') = cos. A cos. B — sin. A sin. B.
In this particular instance the value of A — B is clearly less
than 90°, and the values of sin. (A— B), cos. (A— B) may there-
fore be deduced directly from the diagram of art. 62.
By processes similar to the above the expansions for A -f B
and A— B may be shown to be true for all values of A and B.
EXAMPLES. — X .
Show that the following relations are true : —
1. Sin. (A + B) cos. A — cos. (A + B) sin. A = sin. B.
2. Cos. (A + B) cos. A + sin. (A + B) sin. A = cos. B.
3. Sin" <A + B> = tan. A + tan. B.
cos. A cos. B
4 sin. (A + B) sin. (A - B) = ^ A _ ^ R
cos.2 A cos.2 B
5. Cos. (A + B) cos. (A - B) = cos.2 A - sin.2 B.
6. Sin. (A + B) sin. (A - B) = cos.2 B - cos.2 A.
7 tan. A cot. B + 1 = tan> (A + B)>
cot. B — tan. A
8 tan. (A + B) - tan. A _ ^ R
1 + tan. (A + B) tan. A
q tan. (A - B) + tan. B = ^ A
' 1 -tan. (A-B) tan. B
,Q cot. A + cot. B _ sin. (A + B)
cot. B — cot. A ~~ sin. (A — B)'
11. (sin. A- sin. B)2 4- (cos. A - cos. B)2 = 2 vers. (A-B).
12. Cot.Acot.Bcos.(A + B) = cos,Acos.B(cot.Acot.B — 1).
IQ Q /A , -m sec. A sec. B
13. Sec. (A + B) = - — — -j— - —^5.
1 — tan. A tan. B
44 PLANE TKIGONOMETKY.
14. Cos. A - sin. A = V2 cos. (A + 45°).
- - 1 -tan. (45° - A) •
15' 1+ tan. (45° -A) = ^ A'
16. Cos. 80° + 2 sin. 10° sin. 70° = -,
Q K
17. If sin. A = — , and cos. B » JL, find the value of
17 lo
sin. (A + B).
18. If sin. A = — L, and cos. B = ?, find the value of
V 10 £
tan. (A + B).
19. Find the value of cos. 105°, of tan. 75°, and of tan. 15°.
20. If tan. A= , and tan. B = , show that A + B = 45°.
EXAMPLES. -XI.
Apply the formulge established in this chapter to prove the
following relations :—
1. Cos. (180° + A) = - cos. A.
2. Sin. (90° + A) = cos. A.
3. Cos. (360° - A) = cos. A.
4. Cos. (180° + A) = cos. (180° - A).
5. Cos. (270° + A) = sin. A.
6. Tan. (225° + A) = cofc. (225° - A).
66. In the reduction of trigonometrical expressions, it is
often an advantage when we meet with an expression involving
the sum or difference of functions of angles to transform it into
another containing only a product.
The values obtained for sin. (A + B), cos. (A + B), &c. in
terms of sines and cosines of A and B may usefully b'e employed
for this purpose.
Thus
Sin. A = sin. {£ (A + B) + £ (A - B)} =
sin. i (A + B) cos. \ (A-B)+ cos. \ (A + B) sin. \ (A-B)
Sin. B = sin. {\ (A + B) -£ (A - B)} =
sin. i (A + B) cos. \ (A-B)- cos. i (A + B) sin. \ (A-B)
ON THE EATIOS OF COMPOUND ANGLES. 45
Cos. A = cos. Ji (A + B) + 4 (A - B)} =
cos. i (A + B) cos. | (A-B)- sin. 1 (A + B) sin. ± (A — B)
Cos. B = cos. {i (A + B) - i (A - B)} =
cos. J (A + B) cos. J (A-B)+ sin. i'(A + B) sin. 4 (A-B).
Adding and subtracting these four expressions, two by two,
we arrive at the following results : —
Sin. A + sin. B = 2 sin. x (A + B) cos. £ (A - B)
Sin. A - sin. B = 2 cos. £ (A + B) sin. £ (A - B)
Cos. A + cos. B = 2 cos. 4 (A + B) cos. J (A - B)
Cos. B - cos. A = 2 sin. 4 (A + B) sin. £ (A — B)
67. These results are of great importance, and should be
committed to memory. Expressed in words they may be stated
as follows : —
(1) The sum of the sines of two angles is equal to twice the
sine of half their sum multiplied by the cosine of half their dif-
ference.
(2) The difference between the sines of two angles is equal to
tivice the cosine of half their sum multiplied by the sine of half
their difference.
(3) The sum of the cosines of tiuo angles is equal to twice the
cosine of half their sum multiplied by the cosine of half their
difference.
(4) The cosine of the lesser of two angles minus the cosine of
the greater is equal to twice the sine of half their sum multiplied
by the sine of half their difference.
Example. — :Express sin. 9 A -f sin. 3 A in the form of a pro-
duct.
Here half the sum is 6A, half the difference 3A.
Hence by rule (1) we have
" sin. 9A + sin. 3A = 2 sin. 6A cos. 3A.
If the fundamental formula of art. 60 are added and sub-
tracted, two by two, we shall obtain the expressions
sin. (A + B) + sin. (A — B) = 2 sin. A cos. B
sin. (A + B) — sin. (A — B) = 2 cos. A sin. B
cos. (A -f B) -f cos. (A — B) = 2 cos. A cos. B
cos. (A — B) — cos. (A + B) = 2 sin. A sin. B
46 PLANE TRIGONOMETRY.
The formulae in this shape are convenient for converting an
expression given as a product of sines and cosines into the form
of a sum or difference.
Thus
2 sin. 5A sin. 2A = cos. (5A ^ 2A) - cos. (5A + 2A)
= cos. 3A — cos. 7A.
EXAMPLES. — XII.
Prove the following relations : —
1. Sin. 8 A + sin. 2 A = 2 sin. 5 A cos. 3 A.
2. Cos. A — cos. 5A = 2 sin. 3A sin. 2A.
3. Cos. A + cos. 4A = 2 cos. -— cos. -—- .
2 2
4. Sin.A + rin.2Ag CQt_ A
cos. A — cos. 2A 2
Sin- 3A + sin-
cos. 3A + cos. A
= tan. 2A.
6 Sin. A - sin. B = cot A 4- B
cos. B — cos. A 2
7. Sin. (30° 4- A) 4- sin. (30° - A) = cos. A
8. Cos. (30° - A) - cos. (30° + A) = sin. A.
9. Sin. (45° + A) - sin. (45° - A) = ^2 sin. A.
10. Cos. (45° 4- A) 4- cos. (45° - A) = Vifcos. A.
11. Cos. A 4- cos. (A + 2B) = 2 cos. (A + B) cos. B.
12. Cos. (A 4- B) sin. B — cos. (A -f C) sin. C
= sin. (A + B) cos. B — sin. (A 4- C) cos. C.
13. Sin. (A 4- B - C) 4- sin. (A 4- .0 - B)
4- sin. (B 4- C — A) — sin. (A4-B4-C) = 4 sin. A sin. B sin. C.
14. Cos. (A 4- B-C) 4- cos. (A + C— B) 4- cos. (B + C — A)
4- cos. (A + B 4- C) = 4 cos. A cos. B cos. C.
15. Cosec. 18° - cosec. 54° = 2.
l6' cos. 18° - sin. 18° =
17. Sin. 70° - sin. 10° = cos. 40°.
18. Sin. 10° 4- sin. 20° 4- sin. 40° 4- sin. 50° = sin. 70°
4- sin. 80°.
ON THE RATIOS OF COMPOUND ANGLES. 47
19. Cos. 181° 4- cos. 127° + cos. 113° + cos. 61° + cos. 59C
-f cos. 7° = 0.
20. Cos. 4° 30' + sin. 7° 30' + sin. 37° 30' - cos. 40° 30'
^ 4 sin. 22° 30' cos. 28° 30' cos. 43° 30'.
EXAMPLES. — XIII.
Express as the sum or difference of two trigonometrical
ratios the following expressions : —
1. 2 cos. A cos. B.
2. 2 sin. 4 A cos. A.
3. 2 sin. 3A sin. 4A.
9A . 5A
4. 2 cos. — sin. — .
L- L
5. 2 cos. (A + B) sin. A.
6. 2 co, ^3? A
66 .
7. 2 sin. 50° cos. 20°.
8. 2 sin. 25° sin. 10°.
9. Showthat sin.4Acos. A— sin. Acos.2A=sin. 3Acos. 2A.
10. Show that
5A 3A . 3A . A QA A
cos. — cos. — — sin. — - sin. — = cos. 3A cos. A.
a ft a a
CHAPTER XI.
ON THE TRIGONOMETRICAL RATIOS FOR MULTIPLE AND SUB-
MULTIPLE ANGLES.
68. IF in the formula sin. (A -f B) = sin. A cos. B + cos. A sin.B
we make A equal to B, we obtain
sin. 2A=sin. (A-f A) = sin. A cos. A + cos. A sin. A
= 2 sin. A cos. A.
Similarly cos. 2A=cos. (A + A) = cos. A cos. A— sin. A sin. A.
= cos.2 A— sin.2 A = 2 cos.2 A— 1,
or =1-2 sin.2 A.
A . OA /A , AN tan- A+tan. A 2 tan. A
Again, tan. 2A=tan. (A + A)= — -— — — = —
1— tan. A. tan. A 1— tan 2 A
48 PLANE TRIGONOMETRY.
If for 2A we substitute A, and consequently for A write — ,
2i
we shall have
A A
sin. A = 2 sin. — cos. — ;
A A A
cos.A=cos.2 — — sin.2—- = 2 cos.2 — — 1, or 1 — 2
m' ft '—
2 tan. ~
tan. A =
l-tan'f
EXAMPLES.— XIV.
Prove the following identities : —
1— tan.2 A
1. Cos. 2 A =
2. Tan. 2 A =
1+tan.2 A-
2 cot. A
cot.2 A — 1'
„ Sec. 2A-1 , .
3. - -=tan.2 A.
sec. 2A+1
A A\2
A = l+sin. A.
4. I Sin. —
5. Cos.2 A ^l+tan.|y=l+sin. A.
6. Cos.4 A -sin.4 A = cos. 2 A.
,. Sec.2 A
7. TT-860- 2A-
1— tan.2 A
Q Cosec.2 A
8. - — - — -=sec. 2A.
cot.2 A— 1
0 1— cos. A 2A
9. — =tan.2 — .
1 + cos. A 2
10. Tan.<45° + 4) = — ^
\ 27 1— sin. A'
.,-. Cos. A + sin. A 0 A
11. - -»•..* =tan. 2A + sec. 2A.
cos. A— sm. A
12 CQS-3 A 4- sin.3 A 2 — sin. 2 A
cos. A + sin. A 2
1Q 1 + sin. A— cos. A A
lo. : -. r =tan. — .
1 4 sin. A H- cos. A
ON MULTIPLE AND SUBMULTIPLE ANGLES. 49
14. 2 cos.8 A - 2 sin.8 A = cos. 2A (1 + cos.2 2A).
an'
A
Sin. A + sin. 2 A
15. - £ - 5T=tan' A>
1 4- cos. A 4- cos. 2 A
1 -f cos. A 2
Cos. (A 4- 45°)
• - il — J^
cos. (A— 45 )
+t y
2 ) -
= sec- 2A— tan. 2A.
A
1 4- tan. A tan. — ..
18.- __ 2-itan.A.
tan. -+ cot. -
19. Tan.180°-2A + tan. 180°+2A=2 sec. A.
4 4
20. Cos.6 A - sin.6 A = cos. 2 A (cos.2 2 A + 1 sin.2 2 A).
69. In the same manner as in art. 68 it is easy to deduce
the values of sin. 3A, cos. 3A, and tan. 3A in terms of sines,
cosines, and tangents of A respectively.
Thus sin. 3A = sin. (2 A + A)
= sin. 2 A cos. A + cos. 2 A sin. A
= 2 sin. A cos. A cos. A + (1 — 2 sin.2 A) sin. A
= 2 sin. A cos.2 A + sin. A — 2 sin.3 A
= 2 sin. A— 2 sin.3 A + sin. A— 2 sin.3 A
= 3 sin. A— 4 sin.3 A.
70. Cos. 3A=cos. (2A + A)
=cos. 2 A. cos. A — sin. 2 A sin. A.
= (2 cos.2 A — 1) cos. A — (2 sin. A cos. A) sin. A
= 2 cos.3 A— cos. A— 2 sin.2 A cos. A
= 2 cos.3 A— cos. A — 2 cos. A + 2 cos.3 A
= 4 cos.3 A— 3 cos. A.
71. Tan. 3A = tan. (2 A + A)
tan. 2A + tan. A
" 1 —tan. 2A tan. A
2 tan. A
1— tan.2 A
4- tan. A
'-,2 tan. A . tan. A
1-tan.2 A
PLANE TRIGONOMETRY.
2 tan. A + tan. A — tan.3 A
1— tan.2 A
1— tan.2 A— 2 tan.2 A
~~1— tan.2 A
3 tan. A— tan.3 A
1-3 tan.2 A
EXAMPLES. — XV.
Prove the identities
= 2 cos. 2A— 1
cos. A
3. 3 sin. A-sin.
cos. 3A + 3 cos. A
Sin. 3A-cos. 3 A
— — — T—
sm. A + cos. A
= 2 sm. 2A— 1.
5 tan73A-tan. A + cot. A- cot. 3A=COt" 2A*
A 180° - A . 180° + A - A
6. 4 sm. — sm. - — - -- sin. - - - = sin. A.
o o o
7 Tan. (3A-450) __2 sin. 2A-1
tan. (A 4- 45°) ~2 sin. 2A+1 '
8. 4 cos.3 0 sin. 30 + 4 sin.3 Q cos. 30 = 3 sin. 40.
9. Cos. 4A = 8 cos.4 A— 8 cos.2 A+ 1.
-i A m A i 4 tan. A — 4 tan.3 A
10' Tan' 4A= 1 -6 tan.* A -f tan.* A'
sec. cosec. *
12. Tan.2 A + cot.2 A = 2 + 4 cot.2 2 A.
13. Cot. A-tan. A-2 tan. 2A-4 cot. 4A=0.
14. Tan. A tan. 2 A = sec. 2 A - 1.
15. Tan. A tan. (A — B) (cot. A + tan. B) = tan. A — tan. B.
16. Sin.2 B + sin.2 (A-B) + 2 sin. B sin. (A-B) cos. A
sin.2 A.
'
ON MULTIPLE AND SUBMULTIPLE ANGLES. 5]
in.32A-4 sin.* A
18. Cos. (A + B) + cos. A = cos. (A - B) + cog. 3A
/ "Dx r "R\
+ 4 sin. f A -- -J sin. A cos. I A + — Y.
19. Sin. 2A + sin. 4A + sin. 6A=4 cos. A cos. 2 A sin. 3 A.
20. Vers. (A — B) vers. {180°-(A + B)} = (sin. A— sin. B)2.
9, Cos. nA— cos. (Ti + 2) A
«1. -= — 7 — , Ox A - : -- r=tan. (n+ 1) A.
sm. (n+ 2) A — sin. nA
00 0 cot. (w — 2) A cot. n A+ 1 A A
22. 2 • --rS1' ON A - -V- = cot. A — tan. A.
cot. (n — 2) A— cot. nA
23. Tan. (30°- A) tan. (30° + A) = 2 cos' 2A~1>
2 cos. 2A+1
24. Sin.2 (30° + A) + sin.2 (30°-A)=1-^ cos. 2 A.
L
25. Tan. (45°- A) + tan. (45° + A) = 2 sec. 2A.
26. Cos. (A- 45°) cos. (B - 45°) + cos.2 ( ^~ + 45
2A-B
= cos.2 — - — .
27. Sin. 6A=4 sin. 2A sin. (60° + 2A) sin. (60°- 2A).
28. Cot. A + cot. (60° + A)+.cot. (120°+A) = 3 cot. 3A.
29. Tan. 50° + cot. 50° = 2 sec. 10°.
30. Tan. 20° + tan. 20° tan. 25° + tan. 25°= 1.
31. Cos. 20° cos. 40° cos. 80°=-.
o
32. Cot. 22° 30x-tan. 22° 30' = 2.
33. Cos. 40° + cos. 50° + cos. 60° -f cos. 150°
= 4 cos. 45° cos. 50° cos. 55°.
34. Cos.(A + B + C)
= cos. A cos.B cos.C (1 — tan.B tan.C — tan.C tan. A — tan A tan. B).
35. Sin. 2 (A-B) + sin. 2 (A-C)-f sin. 2 (C-B)
= 4 cos. (B-C) cos. (C-A) sin. (A-B).
36. Sin. A + sin. B -sin. C- sin. (A + B — C)
. A-C . B-C . A+B
-4 sm. — — sm. — ^— sin.— ^~
M 2
52 PLANE TRIGONOMETRY.
72. We now proceed to obtain formulas for expressing the
sine, cosine, and tangent of half an angle in terms of functions
of the angle.
To prove that sin. — = ± A/ - -(
— cos. A
Since cos. A = 1 - 2 sin.2 ~ (art. 68),
2
.;. 2 sin.2-^ = 1 - cos. A;
. 9A 1 — cos. A
sm-v= —2 — ;
73. To prove that cos. ~ = ± /I
4 V
cos.
Since cos. A = 2 cos.2 ^ - 1 (art. 68),
2 cos.2 ~ = 1 + cos. A ;
2l
2 A 1 + cos. A
cos-2-2- = 2 ;
cos. * ± -• cos' A
From the ambiguity of sign it is clear that each value of
cos. A (nothing else being known of the angle A) gives two
A A
values of sin. — and two values of cos. — . If besides cos. A
- 2
we have given the magnitude of A, so that we can tell in which
quadrant the revolving line will fall, after describing the angle
^
—5 we are in a position to affix the proper signs to the root
ft
A A
symbol, both for sin. — and cos. — .
Thus, let it be given that cos. A = -.
Then sin. ^ = ± A
a v
cos.^ = ± /I
2 'V
ON MULTIPLE AND SUBMULTIPLE ANGLES. 53
Now the positive angles less than 360° which have the given
cosine are 60° and 300°, the negative angles are —60° and — 300°.
A
If the angle be 300°, since — - = 150°, we must take the
2
A A
positive sign for sin. — , and the negative for cos. — , and so on.
2 2
It is sufficient, therefore, to know in which quadrant
A
the angle — lies ; we can then affix the appropriate signs of
2
sin. -,-, cos. - accordingly.
2 ij
74. To prove that
^
2 cos. - = + Vl + sin. A ± VI — sin- A 5
^ *
2 sin. - = ± Vl + sin. A + Vl — sin. A.
o -^- • o -^X •*
Since cos.2— 4- sin.2 — = 1,
A A
and 2 cos. — sin. — . = sin. A,
2 2
... cos.2 4 + 2 cos- ^ sin- ^ -»- sin.2 4 = 1+ sin. A,
2 22 2
A A A A
and cos.2 — — 2 cos. — sin. - + sin.2 — == 1 — sin. A.
2 a a £
Hence, taking the square root of each side of the equations,
cos. - + sin. — = ± Vl -r sin. A,
and cos. — — sin. - = ± Vl — sin. A.
Therefore adding, we have
A
2 cos. - = ± Vl + sin. A + VI — sin- A5
a
and subtracting,
2 sin. - = ± Vl 4- sin. A ip V^—'sm. A-
A A
Hence cos. — and sin. — have each four values, corresponding to
2 2
a given value of sin. A.
54 PLANE TEIGONOMETKY.
If the value of A be given we may remove the ambiguity by
assigning the proper signs to the root symbols in the interme-
diate equations ; thus
(1) cos. — + sin. -5 = ± A/1 + sin. A.
(2) cos. ^ - sin. - = ± VI - sin. A.
Let us suppose that A = 210°, so that sin. A = — -.
Then, since — = 105°, we know that sin. -- is positive, and
i
cos. - negative.
2
Moreover sin. — is numerically greater than cos. — .
2 u
Hence we must take the positive sign in equation (1), the
negative in equation (2).
Thus
cos. 105° -f sin. 105° = /y/ 1 - | = >y/ i ;
cos. 105° - sin. 105° = - I + = --
Hence
cos. 105° = (1 -V3)5
sin. 105° = - (1 +
Here, also, it is not necessary to know the actual value of
A, but only the limits between which A lies, the essential points
A A
being, in the first place, the signs belonging to cos. — and sin. -
2* A '
A A
and in the second, the relative magnitude of cos. — and sin. -
75. Thus let it be given that A lies between 90° and — 90°.
Then ~ lies between 45° and - 45°.
ON MULTIPLE AND SUBMULTIPLE ANGLES. 55
A A
Thus cos. — is positive, and greater than sin. — .
Therefore cos. — -f sin.— = Vl 4- sin. A ;
cos. ~ - sin. _ = Vl - sin. A ;
Hence 2 cos. — = Vl + sin. A + Vl — sin. A ;
a
2 sin. - = Vl + sin. A — A/1 — sin. A.
ft
76. To find the sine and cosine of an angle of 18°.
Let A denote the angle which contains 18°; then 2 A contains
36°, and 3A contains 54°.
Therefore 2A + 3A=90°.
Hence sin. 2A = cos. 3A.
Therefore 2 sin. A cos. A = 4 cos.3 A — 3 cos. A.
Dividing both sides by cos. A,
2 sin. A = 4 cos.2 A - 3 = 4 - 4 sin.2 A - 3 ;
therefore 4 sin.- A + 2 sin. A — 1 = 0.
Solving this as a quadratic equation, in which sin. A is the
unknown quantity, we obtain
- 1 ± A/5
sin. A = .
4
Since sin. 18° must be a positive quantity, the positive sign
only is available for /v/5 5
therefore sin. A = . ;
cos. 18= = Vl - sin.' 18° =
77. To find the sine and cosine of an angle of 36°.
Since cos. 36° = 1 - 2 sin.2 18° ;
0/V5 — IV 6 — 2 V5
therefore cos. 36° = 1 — 2f - - -- 1 = 1 — - - -
56 PLANE TKIGONOMETRY.
... cos. .36° = ?JlJ^ = 5/Lt
and sin. 36° = Vl - cos.2 36° = z
4
(A A\ /A A
cos. - + sin. -_) and ( cos. - — sin. - j
EXAMPLES. — XVI.
1. State the signs of
A)
when A has the following values : —
(1) 50°; (2) 160°; (3) 220°;
(4) 320°; (5) - 24°; (6) - 460°.
2. If A lies between 180° and 270°, show that
A _
2 sin. - = ^/l + sin. A 4 A/1 — sin. A.
3. If A lies between 450° and 630°, prove that
2 cos. - = ^Y~- sin. A - - VI + sin. A,
4. Find the limits between which A must lie when
2 cos. 2 = - A/I + sin. A - yT^~smTA.
. 5. Find the values of
(1) sin. 72°; (2) cos. 54°; (3) sin. 9°; (4) cos. 9°; (5) cos. 81
CHAPTER XII.
ON THE SOLUTION OF TRIGONOMETRICAL EQUATIONS.
78. A TRIGONOMETRICAL EQUATION is a relation other than an
identity between one or more functions of one or more angles.
In solving the equation we obtain the value of one or other of
the functions, whence we may determine, exactly or approxi-
mately, the magnitude of the angle with respect to which the
given conditions are true.
ON THE SOLUTION OF TRIGONOMETEICAL EQUATIONS. 57
As an example let us consider the equation
Cos.2 0+2 sin. 0 = 7-..
Our first step will be to express cos.20 in terms of sin. 6.
The equation may then be treated as an ordinary algebraical
equation, in which sin. 6 takes the place of x.
Thus 1— sin.8 0 + 2 sin. 0=^;
4
.-. sin.2 0 — 2 sin. 0=—?;
.-. sin.2 0-2 sin. 0+1— i;
.-.sin. 0-1= ±1;
' . a 3 1
and sin. 0 = - or -.
2 '2
The greater of these two values must be rejected, since the
sine can never exceed unity.
The remaining value is the sine of 30°. Therefore one value
of 0 which satisfies the equation is ^.
6
But by art. 56 all angles furnished by the formula nir + ( — 1)"0
will have the same sine as 0.
Hence any one of these angles will satisfy the equation.
In the present instance 0= ^; nir +(— l)n7^ is therefore
6 o
the general solution of the equation proposed.
79. As a second example take the equation
Sin. 30 + sin. 20 + sin. 0 = 0.
Since sin. 3$ + sin. 0=2 sin. 20 cos. 0,
therefore 2 sin. 20 cos. 0 + sin. 20 = 0,
or sin. 20 (2 cos. 0 + 1) = 0.
Thus the equation is satisfied if either
sin. 20=0, or 2 cos. 0+1 = 0.
If sin. 20=0, we have 20=0, and the general value is
given by 20 = ?i7r.
If 2 cos. 0 + 1 = 0, cos. 0= -i
a
58 PLANE TEIGONOMETRY.
The general value is therefore, by art. 57,
80. It will sometimes be convenient to proceed as in the
following example.
Thus, in the equation
v/3 sin. 0-cos. 0= A/ 2",
dividing both sides of the equation by 2 we obtain
V3 . 1 1
_sm.0--c<*, 0=;
therefore sin. 0 cos. _— cos. 0 sin. - = — ^,
6 6 A/ 2
or sin. I 0 — ^ ) = sin. ^ .
6/ 4
Hence 0-^:
EXAMPLES.— XVII.
Find values of 0 not greater than two right angles which
will satisfy the following equations : —
1. Sin. 0=cosec. 6.
2. 4 cos. 0 = 3 sec. 6.
3. Sin. 6. -f cos. 0=0.
4. 2 tan. 0 = sec.2 d.
5. Sin. 0 + cos. 0 cot. 0=2.
6. Cot. 0=tan. °
£
7. Cosec. 0— sec. 0 tan. 0 = 0.
8. Cot.2 0= V2 cosec. 0 — 1.
' 9. Sec. 0 cosec. 0 = 4 tan. 0.
10. 1+cos. 0=? sec. 0.
4
11. Cot. 0 = 2 v/3 sin. 0.
12. Cot.2 0 + 4 cos.2 0=3.
13. 3 cot. 0-sec. 0 cosec. 0=1.
14. 4-x/ITcot. 0=7 cosec. 0—4 sin. 0.
ON THE SOLUTION OF TKIGONOMETRICAL EQUATIONS. 59
15. Sin. 0 (cos. 40°-cos. 20°)=cos. 0 (-sin. 40° + sin. 20°).
16. Cos. 40-cos. 20= -1.
17. Cos. 80 + 2=3 cos. 40.
EXAMPLES.— XVIII.
Find all the values of 0 which satisfy the equations —
1. Sin. 0=1.
2. Cot. 0=1.
3. Sec. 0= <S2.
4. Tan.2 0=1.
o
5. Cos. 0 + sin. 0 = 4s-
V2
6. \/2 sin. 0 + A/2 cos. 0 = \/3.
7.Tan.Q+0)cot.g--0)=3.
8. 2 sec.2 0 + (2 V2 + 1) sec. 0 + A/2 = 0.
9. Cos. 0 + cos. 10 = cos. | 0.
2 4
10. Cos. 0 + cos. 20 + cos. 30 = 0.
CHAPTER XIII.
ON THE INVERSE NOTATION.
81. IN the equation sin. 0 = a, we state that the sine of a
certain angle 0 is a.
This equation is sometimes written in another form, viz.
0 = sin.~~1&.
Thus sin.-1^ is an angle, and a mode of expressing the
particular angle whose sine is a. For instance, sin. 30°=-;
therefore 30° = sin.'1 i
Z
82. The following example will illustrate the method of
using this notation : —
60 PLANE TKIGONOMETKY.
Show that cos.-1 + cosr1 = 90°.
5 o
4
Let the angle cos."1 p- be denoted by 0, and the angle
cos."1 - by <£.
5
We have to show that 0-}- 0 = 90°.
That this may be the case sin. <£ must be equal to cos. 0.
9 4
Now sin. <f>= VI— cos.2 <f>— A / 1 — ^- =-=cos. 6.
V Zo 5
Therefore 6 is complementary to <£.
It must be remembered that in statements like the above
Q
the value of the angle indicated by cos."1 - is its least positive
5
value.
83. Again, let it be required to show that
Cos.-1 a i=2 sin.'1
As before, let cos.-1 a=6, sin.-1 A/ - -=</>.
v 2
We have to show that 0 — 2(f>.
If this be the case cos. 6 will be equal to cos. 2</>.
Now cos. 20=1-2 sin.2 0=1 - 2. 1~a=l-l + a = a
2
= cos. 6.
EXAMPLES. — XIX.
Prove the following relations : —
2. Tan-1 2 + tan.-1 3 = *£.
4
3. Tan.- 1- tan.- U tan.- ^.
4. Tan.-' | = tan.-1 | + tan.-' i - tan.-' jjl
ON THE INVERSE NOTATION. 61
6. Tan.-1 + 4 tan.-1 + 2 tan.-1 =
7. Tan. ("(tan.-1 | + tan."1 |V = 1.
V O O'
8. Sin.- + 2 tan.-1 = .
5 o Z
9. Cot. (cosec."1 a) = *Ja? — 1.
x ^\
10. a cos. ( sin."1 _ ) = \fa? — b*.
\ as
11. Sin. cof1 a = tan. cos.'1
12. Tan.-1 (2 + V3)-tan.-1 (2- ^S^sec.-
13. Cot.-1 ^/3- cot.-1
14. {Tan. (sin.-1 a) + cot. (cos.'1 a)}2=2a tan. (2 tan."1 a).
1 the value of x in the follow:
15. Tan.-1 VaT+l = 2 tan.'1
Find the value of x in the following equations : —
1
Vo+t"
16. Sin.-1 1=4 sin.'1-.
55 3 JC
17. Cotr1 0 + cot.-1 (aj_l)=^.
18. Tan.-1 — — — tan.'1 — L_=tan.-1 a.
x—l x+l
19. Cos.-1 0 + cos.-1 (1— a}) = cos."1 (— x).
20. Tan.-1 0 + tan.-1 2^-tan.-
CHAPTER XIV.
ON LOGARITHMS.
84, THE logarithm of a number to a given base is the index
of the power to which the base must be raised that it may pro-
duce the number.
62 PLANE TRIGONOMETRY.
Thus, if m = a*, x is said to be the logarithm of m to the
base a.
For example, 64 = 43 ; then 3 is the logarithm of 64 to the
base 4.
85. The logarithm of a number m to the base a is written
thus : — x — log.am.
Hence, since m = ax, m =. a log-am-
86. It follows from the definition that to construct a table
of logarithms of a series of numbers, 1, 2, 3, . . . to a given
base, as, for example, 10, we have to solve a series of equations :
10* = 1, 10* = 2, 10* = 3, . . .
These equations can in general be solved only approximately.
Thus, for example, we cannot find a value of x which will make
10X = 3 ; we may, however, find such a value as will make 1035
differ from 3 by as small a quantity as we please.
87. We shall now establish some of the properties which
render the use of logarithms advantageous in practical calcu-
lations.
88. The logarithm of unity is 0, whatever the base may be.
For we know by algebra that a° = 1, whatever may be the
value of a.
89. The logarithm of the base itself is unity.
For ax = a when x = 1.
90. The logarithm of a product is equal to the sum of the
logarithms of its factors.
Let ax = m, ay = n.
Then x = log.am, y = log.an.
And ax ay = ax+v = mn.
Thus log.amn = x + y — log.am + log.aw.
91. The logarithm of a quotient is equal to the logarithm of
the dividend diminished by the logarithm of the divisor.
Let ax = m, av = n.
Then x = log.am, y = log.a?i.
ON LOGARITHMS. 63
Therefore = a*-» = -
Thus log.0™ = x-y = log.am - log.aw.
71
92. The logarithm of any power, integral or fractional, of a
number is equal to the product of the logarithm of the number by
the index of the power.
For let m = ax, so that x = log.am ;
Therefore mr = (ax)r = arx ;
Thus log.amr = rx = r log.am.
93. The following results flow from these important theo-
rems : —
By the use of logarithms
Multiplication is changed into Addition.
Division ,, „ Subtraction.
Involution ,, ,, Multiplication.
Evolution „ „ Division.
In practical calculations the only base that is used is 10,
logarithms to the base 10 being known as common logarithms.
We shall proceed to point out certain advantages which result
from the employment of 10 as a base.
94. The logarithms of numbers in general consist of two
parts, an integral part and a decimal part. The integral part is
known as the characteristic, the decimal part as the mantissa.
95. In the common system of logarithms, if the logarithm of
any number be known, we can determine the logarithm of the pro-
duct or quotient of that number by any power of 10.
For log.10 (10* x N) = log.10 N + log.10 10n = log.10 N+ n,
So that if we know the logarithm of a given number we
may determine that of any number which has the same sequence
of figures, and differs only in the position of the decimal point.
96. Thus, having given that log.lo 5'2502 is 720176,
64 PLANE TRIGONOMETRY.
let it be required to find the logarithm of log.10 5250*2 and
log.10 -0052502.
Log.10 5250-2 = log. (5-2502 x 1000) = log. 5-2502 + 3
= 3720176.
Log.10 -0052502 = log. = log. 5-2502 - 3
= __ 3 + -720176, or, as it is generally written, 3720176.
97. The form in which the logarithm of -0052502 is given
should be very carefully noticed. This logarithm is really a
negative quantity, and is equal to — 2*279824.
In order, however, that all numbers expressed by the same
series of digits may have the same mantissa in their logarithms,
it is usual to agree that the mantissa shall in all cases remain
positive.
This result is attained by simply adding unity to the man-
tissa^ and subtracting it from the characteristic, the value of the
whole logarithm being unaltered.
Thus in the present instance the mantissa becomes
- -279824 + 1, or 4- -720176.
The characteristic becomes — 2 — 1, that is — 3; and, as we
have said, the whole logarithm, is written 3-720176.
98. In the common system of logarithms the characteristic
of the logarithm of any number may be determined by inspection.
For suppose the number to be greater than unity, and to lie-
between 10* and 10n+1.
Its logarithm must be greater than ?i, and less than n -f 1.
Therefore the characteristic, or integral part of the logarithm,
is n. Thus in the example given above the characteristic of
log.10 5250-2 is 3, for 5250-2 lies between 1000, or 103, and
10000, or 104.
Next suppose the number to be less than unity, and to lie
between — - and — — , that is, between 10~n and 10~("+1).
The integral part of the logarithm would be — n if both
characteristic and mantissa were to retain the negative sign.
But since, to avoid the inconvenience arising from a negative
ON LOGARITHMS. 65
mantissa, unity has to be subtracted from the characteristic, the
latter will become — (n+ 1) instead of — n.
Thus, since -0052502 lies between '01 and '001, or between
(10)~2 and (10)~3, we know that —3 must in this case be taken
for the characteristic.
99. The rules for determining at once the characteristic of
the logarithm of a number may be stated as follows :
(1) If the quantity be greater than unity the characteristic is
positive, and is one less than the number of figures which form the
integral part of the number.
(2) If the quantity be less than unity the characteristic is
negative, and, when the quantity is expressed as a decimal frac-
tion, is one more than the number of cyphers between the decimal
point and the first significant figure to the right of the decimal
point.
100. As has been said, tables of logarithms for practical use
are calculated to the base 10. The actual processes by which
the numerical values of logarithms are determined do not fall
within the scope of this treatise ; it may, however, be stated
that the logarithms which are first calculated have for their base
an incommensurable quantity, known by the symbol e, which is
the sum of the series —
2 + l + 2i3 + 2-f^+---ad^'
and this series may be taken as approximately equal to
2-7182818.
Such logarithms are known as Napierian logarithms, from
Napier, the inventor of logarithms; and sometimes naturae
logarithms, as being those which are first determined.
We shall proceed to show how, having given the logarithm
of a number to the base e, we may deduce the logarithm of the
number to the base 10.
101. Having given the logarithm of a number to a particular
base, to find the logarithm of the number to a different base.
Let a be the given base, b the other base, m the number.
66 PLANE TKIGONOMETRY.
Let x be the logarithm to base a, y the logarithm to the base b.
Then m = ax = by.
Therefore at = b,
x
y
qn
and - = log.a&.
Therefore x = y log.ab, and y = - - — ;
Therefore log.6 m = - log. m.
log-«6
Thus, if we have given the logarithms of a series of num-
bers calculated to a given base a, we may find the logarithms of
these numbers to the base b by multiplying each of the given
logarithms by the constant factor- - -.
log.06
102. In the particular case before us the number repre-
sented by e corresponds to a, while 10 may be written for b.
Thus
If; therefore, we assume that the logarithms of numbers may
be calculated to the base e, it is clear that the logarithms of the
same numbers to the base 10 may be deduced from them.
The value of the constant factor - - — is expressed ap-
Iog.e10
proximately by the figures '43429448 : it is known as the
modulus of the common system.
103. The following easy examples will illustrate the principles
explained in this chapter. When no other base is indicated the
number 10 is to be understood as the base.
Example 1.— Given log. 2 = -301030.
log. 3 = -477121,
find log. 12 and log. -— .
A/3
Log. 12 = log. (4x3) = log. 4 + log. 3 = 2 log. 2 + log. 3
= -602060 + -477121 = 1-079181.
ON LOGAKITHMS. 67
Again, log. 3- = log. */2 — log. \X3
\/o
= \ log. 2 - I log. 3
= 1 (-301030) - 1 (-477121)
A O
= -150515 - -159040
= - -008525 =1-991475.
Example 2. — Given log.10 7 = a, find log.7 490.
Let iog.7 490 = x.
Then T = 490 ;
Therefore x log.10 7 = log.10 490
= log.10 (49 x 10) = log.10 49 + log.10 10
= 2 log.lo 7 + 1.
Hence (x — 2) log.10 7 = 1;
1
therefore x — 2 =
i
and x = 2
log.,0 ? '
2a + I
log.
10
EXAMPLES . — XX ,
1. Given log. 2 = -301030, and log. 3 = -477121, find the
logarithms of 30,000, -0002, 2'25, and -04.
2. Given log.10 7 = '845098, and log.IO 6-3 = '799341, find
27
**i.5.
3. Givenlog.3 = '477121, find log. {(2- 7)3 x ('81)f -=- (90)1).
4. Given log. 24 = 1-380211, and log. 36 = 1-556303, find
the values of log.10 54, and log.9 8.
5. Find the logarithms of
(a) -00001 to base 10.
(6) — = to base 8.
y2
(c) 81 ^a to base3f.
6. Given log.10 2 = -301030, find log.10 -05, log.100 2, log.2 1 00.
68 PLANE TKIGONOMETKY.
7. Having given log. 28 = 1-447158, and log. 35 = 1-544068,
find log.056 7 to 3 places of decimals.
8. Given log. 2 = -301030, and log. 3 = -477121, find x from
the equation 3X = 32.
9. Given log. 2- 7 = -431364, and log. 5-172818 = -713727,
find the value of 3~T-
10. Find the logarithm of 9 to base 3 V'3, and of 125 to
base V& ^.
11. If | log. (2*) = 30-103, and log. 2 = -301030, find x.
o
12. Assuming that log. 250 and log. 256 differ by -0103,
find log. 2.
13. Given log.6 a = c, find log.a am bm.
14. Prove that log.a&. log.6a= l,and that log.a&. log.6 c. log.c a
= 1.
15. If the logarithms of the numbers a, 6, and c be p, q, r
respectively, prove that
aq-r frr-p tf-g _. 1
16. If a2 + b* = 7ab, prove that
log. I (a + 6) = 1 (log. a + log. 5).
17. Having given log. 3 = -477121, and log. 7 = '845098,
solve the equation
G)'- «•
18. How many positive integers are there whose logarithms
to base 3 have 5 for a characteristic ?
19. Show how a system of logarithms having 2 for its base
may be transformed into the system which has 8 for its base.
20. Prove that log.a x : log.b x :: log.c b : log.c a.
FOKMUL^E OF KEFEKENCE. 69
FORMULA OF REFERENCE (I.).
(1) sin. (90°-A) = cos. A, cos. (90°-A) = sin. A (art. 41)
sin. (lSO°~A)=:sin. A, cos. (180°-A) = -cos. A (art. 42)
sin. (90° + A)=cos. A, cos. (90° -j- A) = -sin. A (art. 43)
sin.(1800 + A)=-sin.A.,cos.(180° + A)=-cos.A
/ A \ 'A /A\
sin. ( — A)= — sin. A, cos. ( — A) = cos. A
/ftN . A sin. A A cos. A
(2) tan. A= - — , cot. A = - — -
cos. A sm. A ,
sin.2A 4- cos.2A = 1 , sec.2A = 1 + tan.2A, cosec.2A ( *
= l+cot.2A
(3) sin. (A + B) = sin. A cos. B + cos. A sin. B ) r f fi n
cos. (A 4- B) = cos. A cos. B — sin. A sin. B )
sin. (A— B) = sin. A cos. B — cos. A sin. B ) , ^
cos. (A — B) — cos. A cos. B + sin. A sin. B )
-T.X tan. A H- tan. B -\
tan. (A4-B)=-— -~—
y 1— tan. A tan. B I ,
^ > ..... (art. 63)
/A -r,v tan. A— tan. B I
tan. (A — B)=— - — 5 \
1-t-tan. A tan. B )
(4) sin. A + sin. B = 2 sin. i (A + B) cos. J (A - B) ^
sin. A -sin. B = 2 cos. £ (A-f B) sin. 1 (A-B) I ( 6fi,
cos. A + cos. B = 2 cos. J (A + B) cos. £ (A - B) |
cos. B-cos. A=2 sin. ± (A + B) sin. J (A— B) )
(5) sin. 2 A=2 sin. A cos. A -\
cos. 2 A=cosM-sin.2A I (art 6g)
„ A 2 tan. A (
tan. 2 A= —
1 - tan.2A
(6) sin. 3 A= 3 sin. A- 4 sin.3A (art. 69)
cos. 3 A=4 cos.3A— 3 cos. A (art. 70)
Q v 3 tan. A— tan.3A
tan.8A=
70 PLANE TKIGONOMETBY.
(7) sin.A = ± A/l=^l
cos. — + sin. - = ± /v/1 + sin. A
. . . (art. 73)
(art. 74)
_
cos. — - sin. — = ± Vl —sin. A
Z J
(8) Log.artt,n=log.am + log.a™ (art. 90)
log.a— =log.0m — log.an (art. 91)
log.amr=rlog.am (art. 92)
log.6ra =, - ;-log.ara (art. 101)
log. Jb
CHAPTER XV.
ON THE ARRANGEMENT OF LOGARITHMIC TABLES.
104. THE object of the present chapter is to give an account
of the arrangement of the principal logarithmic and trigono-
metrical tables, and the method of using them. The tables
which will be kept specially in view are those compiled by the
late Dr. Inman, which are in general use in the Royal Navy.
The various sets of tables, however, differ from one another
chiefly in matters of detail, so that it is hoped that the explana-
tions here given will enable the student to avail himself of any
collection that may be at hand.
The logarithms in all cases are taken to the base 10.
105. The tables with which we are at present concerned are
the following : —
(1.) The table of the logarithms of numbers from 1 to 10,000.
(2.) Tables of logarithms of the six principal trigonometrical
ON THE ARRANGEMENT OF LOGARITHMIC TABLES. 71
ratios, viz. the sine, cosecant, tangent, cotangent, secant, and
cosine from 0° to 90°.
(3.) The table of the natural versines from 0° to 180°.
106. In the table which we have called (1) the mantissas of
the logarithms from 1 to 10,000 are given as far as six places of
decimals. It should be noticed that in all approximate calcula-
tions it is usual to take for the last figure which is retained the
figure which gives the nearest approach to the true value. Thus
if we have the fraction -7536, and it is desired to retain only
three places of decimals, WB should write '754, and not '753,
the rule being that when a number of figures are struck off, the
last remaining figure is to be increased by 1 if the first figure
removed be not less than 5.
It is unnecessary to print the characteristic as well as the
mantissa, because, as has been already pointed out, the charac-
teristic may be always determined by inspection in the case of
logarithms to the base 10.
107. To find from the tables the logarithm of a given number.
If the number be contained in the tables, we have only to
take out the mantissa and prefix the characteristic.
Thus, to find the logarithm of -527 and of -00527. The
table gives '721811 as the mantissa of 527, and since the series
of digits is the same in each case, the mantissa of the two loga-
rithms will be the same. Thus we have
log. 527 = 2-721811. log. -00527 = 3-721811.
If, however, the number be not exactly contained in the
table, which, as we have said, goes only as far as 10,000, we
must proceed as follows : Let us suppose that the logarithm
required is that of 800076. The number in question lies
between 800000 and 800100, both of which may be taken at
once from the tables. Thus —
log. 800100 = 5-903144
log. 800000 = 5-903090
difference = -000054
The required logarithm must of course lie between the two
72
PLANE TEIGONOMETEY.
which we have taken out, and if we may assume that the in-
crease in the logarithm will be proportional to the increase
in the number, we obtain the following proportion (where x is
the increase required corresponding to an increase of 76 in the
number) : —
x : -000054 :: 76 : 100;
.-.»=— -000054 = -000041.
Then log. 800076 = 5-903090 + -000041 = 5-903131.
Thus we have taken three numbers, 800000, 800076, and
800100, and have proceeded upon the assumption that the in-
crements in the logarithms of the two higher numbers will be
proportional to the increments of the numbers themselves.
This is a case of what is known as the principle of propor-
tional parts, a principle which though not strictly true is
sufficiently accurate for practical purposes in cases where the
differences in the three numbers are small in comparison with the
numbers themselves.
108. The process of finding the corresponding logarithm of
a given number not contained in the tables is facilitated by the
use of a small table known as the table of proportional parts.
Thus in Inman's tables the logarithms of 8000 and the nine
numbers next following are printed as follows :—
No.
Log.
Part.
8000
•903090
00
8001
•903144
05
8002
•903198
11
8003
•903253
16
8004
•903307
22
8005
••903361
27
8006
•903416
32
8007
•903470
38
8008
•903524
43
8009
•903578
49
To obtain the small table headed ' Part,' proceed as follows
ON THE ARRANGEMENT OF LOGARITHMIC TABLES. 73
Take the difference between two consecutive logarithms of those
given and divide by ten. Then multiply the resulting decimal
by two, three, &c. up to nine, and retaining only six places of
decimals write down the last three figures of these as in the
table.
Thus in the preceding case
log. 8001 = 3-903144
log. 8000 = 3-903090
difference = -000054
Therefore one-tenth of this difference is -0000054, and mul-
tiplying by two, three, four, &c. we obtain -0000108, -0000162,
•0000216, &c. or, striking off the last figure in each case,
-000011, -000016, -000022, &c.
These are then arranged as in the table.
109. The use of this table may be illustrated by applying it
in the example given above, viz. in finding the logarithm of
800076.
As we saw, when x is the excess of the logarithm required
over that of 800000, we have
= -000054 = 1 + A) .000054.
That is, we have to multiply '000054 by — - and by — - -, and
10 100
add the two products together.
The table of proportional parts gives us the yalue of the
several tenth parts, and therefore also of the several hundredth
parts of the decimal in question ; we may therefore proceed as
follows : —
log. 800000 = 5-903090
add for Jg- -000038
add for f|^ '0000032
5-9031812
Therefore, retaining only six places of decimals, we have
log. 800076 = 5-903131.
110. The example given was the case of an integral number.
If a decimal fraction or a mixed quantity, made up of a whole
74 PLANE TRIGONOMETRY.
number and • a decimal fraction, were given, the process would
be the same. We have only to disregard the decimal point,
calculate the mantissa of the whole number so obtained as
before, and then prefix the proper characteristic.
Thus, suppose the logarithm of 14-9037 to be required.
Kejecting the decimal point, we find, proceeding as before,
the logarithm of 149037 to be 5-173294.
The logarithm of 14*9037 will only differ from this logarithm
in its characteristic, which will be 1 instead of 5. The required
logarithm, therefore, will be 1-173294.
111, To find from the tables the number 'which corresponds
to a given logarithm.
If the mantissa be found in the table we have only to write
down the corresponding number, and place the decimal point
in the position indicated by the characteristic.
Thus, let the given logarithm be 4*155336. On reference
to the table we find the mantissa given to be that of the logarithm
of 143.
The required number is therefore 14300.
Next suppose that the mantissa is not contained exactly in
the table, as, for instance, if the given logarithm be 5-348390.
Here we must again have recourse to the 'principle of propor-
tional parts' Turning to the tables, we find that the given
logarithm lies between those of 223000 and 223100. Thus
log. 223100 = 5*348500 log. given = 5-348390
log. 223000 = 5-348305 log. 223000 = 5-348305
•000195 -000085
Thus, if x be the excess of the required number over 223000,
we have x . 1QO .. .000085 . .QQ0195 ;
•'••-&«»-"
Thus the number required is 223044.
112. This process also may be somewhat abbreviated by
recourse to the table of proportional parts.
Set down in full, the operation will be as follows : —
ON THE ARRANGEMENT OF LOGARITHMIC TABLEb. 75
195) 8500 (43-589
780
700
585
1150
975
1750
1560
1900
1755
The products 780, 585, 975, &c., are furnished approximately
for us in the table of proportional parts. Thus, in the table
which commences opposite to the number 2230, we find opposite
to the figure 4, 078, that is, four-tenths of 195 = 78, or four
times 195 = 780.
Using this table, therefore, we may proceed as follows : —
8500
4 780
6
Thus we have 43*6 as the value for », which is sufficiently
near to the true value to be accurate enough for ordinary
purposes. In cases where great exactness is required the actual
division should, of course, be performed.
113. The table denoted by (2), next to be noticed, contains
the logarithms of the six principal trigonometrical ratios, viz. the
sine, cosecant, tangent, cotangent, secant, and cosine for all angles
from 0° to 90°, calculated for each change of 15" in the angle.
Since, however large an angle may be, by means of the
expressions for sin. (180° — A), sin. (180° + A), &c. an angle
may always be found in the first quadrant having for one of
its functions the same numerical value as the required function
of the given angle, it is unnecessary to take into account angles
greater than 90°.
Thus cos. 125° = — cos. 55°, tan. 260° = tan. 80°, and
so on.
76 PLANE TKIGONOMETRY.
Again, since sin. (90° — A) = cos. A, tan. (90°— A) = cot. A,
cosec. (90°— A) = sec. A, it is obvious that logarithms need not
be calculated for angles higher than 45°. The logarithm of
sin. 46°, for instance, is the same as that of cos. 44°. It will be
seen on reference to the tables that the figures which denote
the values of angles less than 45° are printed at the top and
left-hand side of each page, and the figures for angles greater
than 45° at the bottom and right-hand side ; while each column
of logarithms has a denomination both at the top and bottom,
the column which is headed sine having cosine printed at the
bottom, and so on.
114. The trigonometrical ratios being, in a majority of cases,
less than unity, the logarithms of these ratios would have a
negative characteristic. To avoid the frequent recurrence of
the negative sign in the tables, it is customary to add 10 to the
logarithms of the angles. Logarithms so increased are called
tabular logarithms, and are denoted by the letter L.
Thus L sin. A = log. sin. A + 10.
115. To find the tabular logarithm of the sine of a given
angle.
If the angle given be one of those contained in the tables
the required sine may be at once taken out. If the angle be
not given exactly we must have recourse to the principle of
proportional parts.
Thus let it be required to find L sin. 40° 30' 24".
We find from the tables that
L sin. 40° 30' 30" = 9-812618
L sin. 40 30 15 = 9-812581
•000037.
Assuming that the increase of the logarithm is proportional
to the increase in the angle, we have, taking x as the required
increase in the logarithm —
x : -000037 : : 9" : 15" ;
Therefore x = ~ -000037 = -000022.
15
Thus L sin. 40° 30' 24"= 9-812581 +-000022 = 9-812603.
ON THE ARRANGEMENT OF LOGARITHMIC TABLES. 77
116. To find the angle which corresponds to a given tabular
logarithmic sine.
Thus let it be required to find the angle which has for the
tabular logarithm of its sine 9*688723.
Reference to the tables shows that the given tabular
logarithm lies between those of the sines of 29° 13' 45" and
29° 14'.
Thus L sin. 29° 14' 0"= 9-688747
L sin. 29 13 45 = 9-688690
difference = -000057
L sine of angle required = 9-688723
L sin. 29° 13' 45" . .= 9-688690
difference ="^000033
Hence if x be the excess of the required angle above
29° 13' 45", we have
x : 15" : : -000033 : -000057.
qq
Therefore x = ~ 15"= 9" nearly.
57
And the angle required = 29° 13' 45"+ 9" = 29° 13' 54".
117. To find the tabular logarithmic cosine of a given angle.
Let it be required to find the tabular logarithmic cosine of
the angle 47° 44' 21".
We have from the tables
L cos. 47° 44' 15" =9-827711
L cos. 47 44 30 = 9-827676
difference = -000035
Since in the first quadrant the cosine decreases as the angle
increases, let x be the required decrease in the tabular logarithm,
corresponding to the increase of 6" in the angle, from 47° 44' 15",
namely, to 47° 44' 21".
Then, making use, as before, of the principle of proportion,
we have
»: -000035:: 6": 15";
therefore x = | -000035 = -000014.
5
78 PLANE TRIGONOMETRY.
Therefore the tabular logarithm required is 9-827711
-•000014 = 9-827697.
118. To find the angle corresponding to a given tabular
logarithmic cosine.
Suppose the given tabular logarithm to be 9*864532.
The tables give us
L cos. 42° 56' 30" . . = 9-864539
L cos. 42 5645 . . = 9-864510
difference . = -000029
L cos. 42° 56' 30" . . = 9-864539
L cosine of angle required = 9*864532
difference . = -000007.
Thus we have to find a?, the increase in the angle correspond-
ing to -000007, the decrease in the tabular logarithm.
Therefore x = ~ 15"= 4" nearly.
^y
And the angle sought for is 42° 56' 30"+ 4"= 42° 56' 34".
119. It is unnecessary to give further examples. For any
of the six functions the tabular logarithms of angles inter-
mediate in value between those given in the tables may be
found by applying the principle of proportion, and if we have
given the logarithm we may determine the value of the angle
to which it belongs by the same method. It must, of course,
be remembered that in the first quadrant the tangent and secant
increase with the increase of the angle, while the cosecant and
cotangent decrease within these limits.
120. It should be observed in connection with these tables
that the tabular logarithmic sines are given for each second of
angle up to 50'. The reason for this is to be found in the fact
that the sines of small angles vary with great rapidity, so
that the ordinary method of proportioning for seconds would
lead to erroneous results.
121. Of a similar character to those lately described is the
table containing what are called the logarithmic haversines,
which occupies a prominent position in Inman's tables. It is
ON THE ARRANGEMENT OF LOGARITHMIC TABLES. 79
of considerable importance in connection with what is known
as the ' solution of triangles,' as will appear in the next chapter.
The word haversine is a contraction of half-versine. The
versine has been already defined to be the defect of the cosine
from unity. ^
A 1 A 2 Sin'2 9~ A
* A
Thus hav. A =
== = _
Z <B . »
The haversine of an angle is therefore the square of the
sine of half the angle, and the logarithmic haversines might be
deduced from the table of logarithmic sines. They are, how-
ever, of constant use in practical calculations, and it is con-
venient that they should be given separately.
Since the value of the versine increases continually from 0°
to 180°, the tabular logarithmic haversines are given for all
angles up to this limit. To 135° the logarithms are set down
for each 15"; as the angle approaches 180°, however, the haver-
sine changes very slowly, and it is found sufficient to give the
logarithm for a change of each minute of angle.
The tabular logarithmic haversines for values of the angle
intermediate in value between those given in the tables may
be found by applying the principle of proportion, as in the
cases of the sine, cosine, &c.
122. On account of the great simplification which the use
of logarithms introduces into the several processes of calcula-
tion, it is in general more important to have the tables of the
logarithms of the sine, cosine, &c. of an angle than the actual
values of the functions themselves. In cases where we have
need of the actual value of one of the ratios, .as, for instance,
of the sine, or, as it is called, the natural sine, we may easily
deduce its value from the tabular logarithm by making use of
the table of logarithms of numbers.
Thus L sin. 30°= 9-698970, or log. sin. 30°= 1-698970.
Referring to the table of logarithms of numbers, we find
this to be the value of '5, or -.
2
123. One case, however, occurs in which the value of the
function itself is frequently required. It is that of the versine.
80 PLANE TKIGONOMETRY.
The values of the natural versme are therefore recorded in a
table for every minute of angle from 0° to 180°.
By means of a column of ' proportional parts for seconds,'
constructed upon the same principle as those given with the
logarithms of numbers, the natural versine of an angle given
to the nearest second may be readily determined with great
exactness.
Since the value of the versine in general takes a fractional
form, it is found advantageous for convenience in printing the
tables to multiply each versine by 1000000 ; in other words, to
omit the decimal point.
In order, therefore, to obtain the real value of the versine
from that given in the tables, which is called the tabular versine,
we must first divide by that quantity.
Thus the tabular versine of 60° is given as 500000.
124. The following example will serve to explain the method
of using the table of natural versines :—
Let it be required to find the tabular versine of 78° 16' 27".
The value of the tabular versine of 78° 16' is found directly
to be 796643. To find the quantity to be added for 27", we
carry the eye along the horizontal line of figures set on the
right-hand page against 27" until the column headed 78° 0' is
reached, whence is taken the number 128, being the part
for 27".
Then we have
tabular versine of 78° 16' - 796643
part for 27" = 128
tabular versine of 78° 16' 27" = 796771.
Had the number of minutes in the given angle been 30, or
greater than 30, the parts for seconds would have been taken
from the vertical column headed 78° 30'.
125. Again, suppose we have given the tabular versine
439672, and that it is required to find the corresponding angle.
ON THE ARRANGEMENT OF LOGARITHMIC TABLES. 81
We find in the tables that 439602, the tabular versine of
55r 55', is the next in value below that given. Thus
tabular versine given = 489672
tabular versine of 55° 55' = 439602
difference = 70
This difference must be due to the parts for seconds.
Looking out 70 in the column headed 55° 30', we find that 70
lies midway between the parts for 17 and for 18 seconds. The
angle 55° 55' 18" may therefore be taken for the value required.
126. A few examples are added here to assist the student in
familiarising himself with the arrangement of the several tables.
In Part III., the practical portion of this treatise, ample
opportunity will be afforded of acquiring proficiency in the
ready and accurate practical use of logarithms and logarithmic
tables.
EXAMPLES. — XXI.
The following two logarithms will be required in solving some
of the examples : —
Log. 2 = • 301030 ; log. 3 = • 477121.
1. Deduce from the table of logarithms of numbers the
tabular logarithms of sin. 45°, tan. 45°, sin, 60°, cos. 60°, cot. 30°,
and tan. 210°.
2. Prove that L tan. A + L cot. A = 20, and that L cos. A
4- L tan. A + L cosec. A = 30.
3. If L sin. A = 9700280, and L cos. A = 9-937092, find
L tan. A.
4. If L tan. A exceeds L sin. A by '062762, find the value
of L cos. A.
5. Given L sin. ~ = 9-741889, and L cos. ^ = 9-921107,
2 A
find L cosec. A.
6. If L sec. A + L cosec. A = 20-316086, find L sin. 2A.
7. If L hav. A = 9-301030, find L cos. A.
8. If L cos. A = 9-477121, find the tabular versine
o* A.
82 PLANE TRIGONOMETRY.
9. If L hav. 2A = 9-847183, find L sin. A.
10. If the tabular versine of A = 1500000, find L hav. A.
11. If L hav. A = 9-000000, find the tabular versine
of A.
12. The difference between the tabular versines of two angles
is 300000, and the sum of their tabular logarithmic cosines is
19 ; find the natural cosines of the angles.
CHAPTER XVI.
ON THE FORMULA FOR THE SOLUTION OF TRIANGLES.
127. WE have now reached a point where the actual value
of trigonometry, considered as a practical science, becomes
manifest. One of its principal objects, as its name implies, was
to establish certain relations between the sides and angles of
triangles, so that when some of these are known the rest may
be determined. Certain relations of this kind we have already
from geometry. We shall proceed to determine others.
128. Every triangle has six parts, as they are termed, viz.
three sides and three angles, some of which being given we are
able to determine the remainder.
The simplest cases which occur are those which deal with
right-angled triangles. These will therefore be first treated.
129. Let ABC be a triangle having the angle at C a right
angle. Let the letters a, 6, c be the measures of the sides
opposite to the angles A, B, C respectively.
130. In any right-angled triangle each side is equal to the
FOKMUL^E FOB THE SOLUTION OF TKIANGLES.
83
product of the hypothenuse into the sine of the opposite angle, or
is equal to the product of the hypothenuse inio the cosine of the
adjacent angle.
By definition, — — = - = sin. A ;
a = c sin. A.
AC b
Again, by definition, — = - = cos. A;
C
b = c cos. A.
131. In the same way it may be shown that a = b tan. A,
or = b cot. B ; or, as it may be stated in words : In any right-
angled triangle each side is equal to the product of the tangent
of the opposite angle into the other side, or is equal to the product
of the cotangent of the adjacent angle into the other side.
Generally it will be found that one side of a right-angled
triangle may always be expressed in the form of a product, in
which one -of the other sides is multiplied by some function of
one of the angles of the triangle.
Collecting these results, we have
a = c sin. A or = c cos. B or = b tan. A or = b cot. B ;
b = c cos. A or = c sin. B or = a tan. B or = a cot. A ;
c = a cosec. A or = a sec. B or = b cosec. B or = b sec. A.
132. The formula now to be established are true for all
plane triangles, right-angled or otherwise.
133. In any triangle the sides are proportional to the sines of
the opposite angles.
Let ABC be any triangle, and from A draw AD perpen-
dicular to the opposite side, meeting that side, or side produced,
G 2
84 PLANE TKIGONOMETKY,
in D. If B and 0 be both acute angles, we have from the left-
hand figure
AD = AB sin. B, and AD = AC sin. C.
Therefore AB sin. B = AC sin. C.
c _ sin. C
b sin. B
If one of the angles, as C, be obtuse, we have from the
right-hand figure
AD = AB sin. B., and AD = AC sin. (180° - C) = AC sin. C.
Therefore AB sin. B = AC sin. C.
c sin. C
b ~ sin. B '
If the angle C be a right angle, we have from the figure of
art. 129
AC = AB sin. B ;
c _ 1 _ sin. C
b sin. B sin. B
Similarly it may be shown that
a sin. A -.a sin. A
7 = - — r, , and - = - —7= .
0 sm. B c sin. C
These results may be written thus :
sin. A _ sin. B _ sin. C
a b c
134. To express the cosine of an angle of a triangle in terms
of the sides.
A A
B. 3>
Let ABC be a triangle, and suppose C an acute angle, as in
the left-hand figure. Then by Euclid, II. 13,
AB2 = BC2 + AC2 - 2BC . CD,
and CD = AC cos. C ;
FORMULAE FOR THE SOLUTION OF TRIANGLES. 85
Next suppose C an obtuse angle, as in the right-hand figure.
Then by Euclid, II. 12,
AB2 == BC2 + AC2 + 2BC . CD,
and CD = AC cos. (180° - C) = - AC cos. C ;
c2 = a2 + b2 — Zabcos. C.
Thus in both cases
r< a2 + b2 — c2
cos. C = — — .
2ab
If C be a right angle, cos. C = 0, so that the formula
c2 = a2 + b2 — 2ab, cos. C becomes c2 = a2 + b2, which is true
by Euclid, I. 47.
The relation, therefore, holds, whatever may be the value of
the angle C.
Similarly it may be shown that
7i2 i_ 2 *
cos- A = - -2fc-~J
cos. B = --^ — .
135. In every triangle each side is equal to the sum of the
product of each of the other sides into the cosine of the angle
which it makes with the first side.
If the angle C be acute, we have from the left-hand figure
BC = BD + DC = AB cos. B + AC cos. C.
Therefore a — c cos. B + b cos. C.
If the angle C be obtuse, we have from the right-hand
figure
A A
B D C B C
BC = BD - DC = AB cos. B - AC cos. (180°- C)
= AB cos. B + AC cos. C.
Therefore a = c cos. B + b cos. C.
86 PLANE TRIGONOMETRY.
Similarly we shall have
1) = a cos. 0 + c cos. A, and c = b cos. A + a cos. B.
136. The value of the cosine of an angle in terms of the
sides may be made to depend on this property. Thus
b = a cos. C + c cos. A .-. b2 = ab cos. 0 + be cos. A
c = a cos. B + b cos. A .'. c2 — ac cos. B + be cos. A
a = b cos. C + c cos. B .*. a2 = ab cos. C + ac cos. B.
By adding each side of the first two equations, and from the
sum subtracting the third equation, we obtain
6> + c2 - a2 = 2bc cos. A ;
...„* A = * + * = *;
2bc
and in the same way the values for cos. B and cos. C may
be derived.
137. To express the sine, the cosine, and the tangent of half of
an angle of a triangle in terms of the sides.
To show that
ei-n A _ /(s _ M (s _ c) , a + b -f- c
sm. - — A / \ - ZJ, -- / where s = — -- — — .
^ V be 2
Since cos. A = 1 — 2 sin.2 f",
,•. 2sin.2J| = l -cos. A
<L
7,2 1-2 _ n
— — c2+ a*
2bc
a?- ffl_ ^bc +c2)
_ ' (a + & — c) (a — b + c)
2bo
Let a + b + c = 2s.
Then a + & — c = 2s — 2c, and a — b + c = 2s — 26;
FORMULAE FOR THE SOLUTION OF TRIANGLES. 87
therefore 2 sin.* = i-<«Zl_«) 2j (t=l) .
2
and sm.
And the sign to be affixed to the root symbol must be always
positive, since A, being an angle of a triangle, must be less than
180°, and therefore - less than 90°.
Next to show that
A / s (s — a) ! a + b + c
cos- 9 = V S— 'where S = - — .
A
Since cos. A = 2 cos.2 - — 1,
A
therefore 2 cos.2 — = cos. A + 1
t?B + 26c + c2 — a2
(6 + c + <:t) (6 + c — a)
But if a + 6 + c = 2,9, fo + c - a = 2s — 2a ;
,T f r» o A 2s . 2 Cs — a)
therefore 2 cos.2 — = — — '.
2 . 2bc
Therefore cos. ^ = A. (.9 -a),
2 V - 60
A A
From the values of sin. — , and cos. — " , we obtain
2 2
A
And, as in the case of the sine, since — is always less than 90°,
2t
A A
the values of cos. — and tan. -- must always be positive.
88 PLANE TBIGONOMETKY.
138, To express the haversine of an angle in terms of the sides.
Since COS.A=b-+c*-a
I—COS. A=l
2bc
2bc
2bc
But 1 —cos. A= vers. A= 2 hav. A ;
_
46c 46c fcc
A
Since, as pointed out in art. 121, hav. A = sin.2 — , the value
2
here obtained for hav. A might have been deduced from that
j^
found for sin. g- in the preceding article. The expression for the
haversine, however, is of very great importance in the solution
of triangles, and the proof is therefore exhibited separately.
139. To express the sine of an angle of a triangle in terms of
the sides.
Sin. A = 2 sin. ^ cos. ~
2 2
-fc V* (*-*)(*- *) (*-^
140, To S/IOM; tfwtf
A-B a_£> 0
tan. — - — = cot. — .
2 a + b 2
a sin. A
Since r= -. — -5'
b sin. 13
therefore
a — b _ sin. A — sin. B
a + 6 - sin. A + sin. B
FOKMUL.E FOE THE SOLUTION OF TKI ANGLES.
89
cos.
B . A -B
- sin. --
A + B A -B
2 sin. — |— cos. — —
A- B , A + B
= tan. — - — cot.
= tan.
2
A-B
.,
Therefore
141, To show that the area of a triangle is equal to one-half
the product of two sides multiplied by the sine of the angle in-
cluded between those sic
Let ABC be any triangle, and AD the perpendicular upon
the side BC.
Since a triangle is equal to one-half the rectangle upon the
same base, and of the same altitude.
The area of ABC = ^BC.AD=^ a.c sin. B, or ~ a.b sin. C.
u 2t 2
In the same way the area may be shown to = - b.c sin. A.
142. To express the area of a triangle in terms of the sides.
Since the area of triangle ABC = - be sin. A (art. 141).
2
But, by art. 139, sin. A = " Vs (s — a) (s — b) (s — c).
Therefore the area of ABC = Vs (s — a) (s — b) (s — c).
This is denoted sometimes by the symbol S, so that
S = Vs 8-af8^V8^c.
90 PLANE TKIGONOMETKY.
143. Since the three angles of a triangle are equal to 180°,
the following results will hold : —
(1) Sin. (A + B) = sin. C.
(2) Cos. (A + B) = - cos. C.
A 4- B C
(3) Sin.— — = cos.-.
, ,N n A + B .0
(4) Cos. _ = sin.-.
144. By the employment of suitable artifices a large number
of relations may be established between the angles and sides of
a triangle in addition to those already given.
Thus, let it be required to show that, if A, B, C be the
angles of a triangle,
ABC1
cos. A + cos. B — cos. 0 = 4 cos. - cos. — sin. -—1.
a' 2 2
Since cos. C = cos. (180° — A — B) = — cos. (A -f B),
Therefore
cos. A + cos. B — cos. C = cos. A -f cos. B -f cos. (A + B)
= 2 cos. ^p cos. A-^ + 2 CQS.2 A + B _ l
0- A + B
= 2 cos. — -
/0 A B\
(2 cos. - cos. -J - 1
A . C A B
4 sm. — cos. _ cos. — — 1.
- _
'EXAMPLES.— XXII.
1. In any plane triangle ABC establish the following rela-
tions : —
(1) Sin. A -f sin. B -f sin. C = 4 cos. ^ cos. ? cos. ~ .
222
(2) Sin. A - sin. B + sin. C = 4 sin. ^ cos. ? sin. ~.
2i 2i &
(3) Sin.2 A -f sin.2 B + sin.2 C-2 cos. A cos. B cos. C = 2.
FOEMUL^E FOE THE SOLUTION OF TEIANGLES. 91
(4) Cot. + cot. + cot. = cot. cot. cot. .
(5) Tan. £ tan. 5 + tan..-? tan. 5 + tan. 2 tan. ~ = 1.
4 a a & a a
sin. A + sin, B - sin. G = A B
k ; sin. A + sin. B + sin. C 22'
(7) Cot. A + cot. B -h cot. C = cot. A cot. B cot. C
-I- cosec. A cosec. B cosec. C.
. ...
222 44 4
(9) Sin. 2A + sin. 2B + sin. 2C = 4 sin. A sin. B sin. C.
(10) Cos.2 A + cos. 2B + cos. 2C + 4 -cos. A cos.B cos.C + 1=0.
2. Show that if in the triangle ABC the angle C is a right
angle, the following relations hold : —
(3)
2
Cos. 2B - cos. 2A
sm. 2A
(4) Sec. 2A=r
— a
3. In any plane triangle ABC prove the following relations: —
(1) a sin. - (B - C) = (b - c) cos. -*
v } 2 v 2
(2) a cos.2? + b cos.2 ^ = 6 cos.2| + c cos.2f
2 2 - ..2
= c cos.2 — + a cos.2—.
(3) a sin. (B - C) + b sin. (C - A) + c sin. (A - B) = 0.
(A,\ s*n> A _ cos. A cos. C + cos. B
sin. B cos. B cos. C + cos. A*
(5) (- + -1 cos. A + (G- + a] cos. B +f? + ^Vcos. C = 3.
\c o/ \a c/ \o a/
92 PLANE TKIGONOMETEY.
(6) Sin. 2A + sin. 2B + sin. 20
= 4 cos. A cos. B cos. C (tan. A + tan. B + tan. C).
/7\ pnt A _ cos. C - cos. B.
Ot' 2 ~ sin. B - sin. C
(8) (8 -a) tan. ^ = («-&) tan. ?= (s-c) tan. 2.
(9) If p, q, r be the perpendiculars from A, B, C upon the
opposite sides, then
p sin. A = 2 sin. B = r sin. C.
(10) (a2-62) sin. A=ac sin. (A-B).
nv\ l+cos. (A-B) cos. C = a
> ' l + cos. (A-C)cos. B~a2
(12) If A:B:C::2 : 3 : 4, then
(13) 46csin.2 + casin.2~ + a&sin.2 + 2 (a2 + 62 + c2)
(14) If 2a=b + c, tan. tan. =.
Z Z o
4. Show that the area of a triangle is given by each of the
following expressions : —
(1) be Vhav.Ahav. (B + C).
(2) 1 (a + 6 + c)2 tan. ~ tan. 5 tan. P.
(3) o A (cos- B cos- C + cos. A).
2 sm. A
(4) 9Sin'2A. (c-a cos. B) (6-acos.C).
2 cos.2 A v
5. If two triangles have angles A, B, C, and 90°— ^ 90° — ?,
- LJ
C
90° — — respectively, and a common side a opposite to the first-
a
FORMULAE FOE THE SOLUTION OF TRIANGLES. 93
named angle in each, show that their areas will be in the
B C A
proportion of 2 sin. -- sin.— to sin. — .
22 a
6. Show that the length of the straight line joining the
point A with the middle point of the side BC of the triangle
ABC is equal to - V b2 + c2 + 2bc cos. A.
a
7. AD is the perpendicular from the vertex A on the base
BC of a plane triangle ABC, and E, F are the middle points of
AD, BC respectively ; prove that
(1) Cot. EFC = cot. B-cot. C.
(2) If EF = |, then either a2 = 3 (&2-c2), or the triangle is
Z
right-angled.
8. Two plane triangles with a common angle A, and the
sides containing it &, c, and I', cf respectively, have equal areas ;
show that two other triangles, which have each an angle A, and
have sides containing it 7>, b' and c, c' respectively, are equiangular.
9. If the sum of the squares of the cosines of the angles be
equal to unity, show that the triangle is right-angled.
CHAPTER XVII.
ON THE SOLUTION OF RIGHT-ANGLED TRIANGLES.
145. THE solution of triangles is the process by which when
a sufficient number of the six parts of any triangle are given
the rest may be determined.
It will be seen that of these six parts three must always be
given, and of these at least one must be a side. This is obvious
from the consideration that if two sides of a triangle be produced,
and lines be drawn, parallel to the base, to meet the sides so
produced, an indefinite number of triangles will be obtained
having for their angles the same values as in the original tri-
angle. In this case therefore we cannot determine the lengths
of the sides, but only their ratios to one another.
94
PLANE TRIGONOMETRY.
146. Since the three angles of a triangle are together equal
to two right angles, if two angles A, B of the triangle ABC be
given, the value of the third angle may be
obtained by subtracting the sum of A and
B from 180°.
147. Let ABC be a right-angled tri-
angle, having the angle C a right angle.
Since in the right angle one part is
always given, it will be necessary that only
c two other parts should be given. The
several cases that may occur divide themselves into two heads,
according as we have given —
i. Two sides as a, b.
ii. One side and one angle as a, A.
148. To solve a right-angled triangle, having given two sides.
Let us suppose that the two sides a, b are given.
The third side might at once be found from the well-known
A relation between the sides of a right-
angled triangle c2 = a2 + b2.
The employment of this formula,
giving, as it does, the value of c2 in the
form of the sum of the squares of the
two known quantities, would deprive us
of the advantage to be derived from the
use of logarithms : such a form should
therefore in general be avoided.
It will commonly be more advantageous first to determine
one of the unknown angles. Thus
tan. A = —
therefore log. tan. A = log. a — log. b,
andL tan. A = 10 + log. a — log. b.
Having found the value of A from the tables, since A + B
= 90°? to obtain B we have only to subtract A from 90°.
To find c we have, by art. 131,
c—a cosec. A.
log. c=log. a + log. cosec. A = log. cH-L cosec. A— 10.
ON THE SOLUTION OF EIGHT-ANGLED TRIANGLES. 95
149. If the parts given be one side and one angle, as a, A, we
may at once find the value of the third angle B by subtracting the
sum of A, 0 from 180°. To find the sides &, c we have, by art. 131,
b = a cot. A;
c = a cosec. A;
whence the values of the two sides may be determined as
already explained.
CHAPTER XVIII.
ON THE SOLUTION OF TRIANGLES OTHER THAN RIGHT-ANGLED.
150. IN the solution of triangles other than right-angled,
four cases present themselves, according as we have given —
(I.) Three sides a, &, c.
(II.) Two angles and one side, as A, B, a or A, B, c.
(III.) Two sides, and the angle opposite to one of those
sides, as a, 6, A.
(IV.) Two sides and the included angle, as &, &, C.
The several cases will be considered in order.
CASE I.
151. Given the three sides a, 6, c to solve the triangle,
The parts required are the three angles of the triangle.
In art. 134 we have established the formula
&2 + c2_a2
cos. A= — — .
2bc
This formula may sometimes be adopted in simple cases.
Thus let a = 5, 6 = 4, c = 7.
16 + 49-25 40 5
Then we nave cos. A= — = ^- = — .
5o oo 7
Therefore L cos. A=10 + log. 5— log. 7;
10 + log. 5 = 10-698970
log. 7= -845098
L cos. A= 9-853872
whence A =44° 24' 55".
96 PLANE TEIGONOMETE1'.
152. In practice, however, the custom of resorting at once
to logarithms is almost universal ; we must therefore have re-
course to one of the logarithmic expressions, deduced from the
fundamental formula, viz. —
COS
2
A /(s-b)(s-c).
tan.^=V s(s-a)^
(s_&) (g-c)
hav. A= v « *.
foo
153. If the student is obliged to confine himself to the use
of the ordinary tables of logarithmic sines, cosines, tangents, &c.
A A
he may make use of the expression for sin. — or cos. — . It is
2 2
of little importance whether one or other of these is employed,
but when the value of the angle is required to the nearest
A /A
second, it may be observed that the logarithm of sin. — f — being
always less than 90°) will increase as the angle increases,
whereas the logarithm of the cosine decreases as the angle in-
creases. The process of obtaining by proportion the exact num-
^
ber of seconds is therefore slightly less troublesome for sin. — .
In the practical solution of triangles it is in general sufficient to
take out angles only to the nearest 15", so that it is immaterial
which formula is employed.
154. If, however, as in the case of Inman's collection, a table
of logarithmic haversines is at hand, the solution of the triangle
is somewhat simplified by the adoption of the haver sine formula,
since in the first place we avoid the root symbol on the right-
hand side of the equation, and in the second place we obtain at
once the value of the angle A, instead of that of half the angle,
as in the other cases.
Reduced to logarithms the formula will appear —
L hav. A = 10 4 log. (s — fr) + log. (s — c)~ log. b — log. c.
SOLUTION OF TEIANGLES OTHER THAN EIGHT-ANGLED. 97
Thus we obtain the angle A. To obtain B we may, if we
please, employ the formula
^
Sin. B=- sin. At
a
In that case, however, any error which may have arisen in
the calculation of the angle A will vitiate our results for B also.
It is better, therefore, to calculate B independently by the same
formula already employed for A. Thus —
L hav. B = 10 + log. (s — a) + log. (s — c) — log. a — log. c.
Having now the values of A and B, by subtracting the sum
from 180° we obtain C. When, however, accuracy is of great
importance, it may be advisable, by a third application of the
haversine formula, to determine C also independently.
We may then test the accuracy of our results by adding the
three angles together, when the sum should, of course, be equal
to 180°.
CASE II.
155. Given two angles, and one side, to solve the triangle.
Let the angles given be A, B. Then since A + B + C =
180° ; C also is known.
First let the side a be given.
mi . b sin. B c sin. C ,
Then, since - = -, - = - — -, we have
a sin. A a sin. A
log. b = log. a + L sin. B — L sin. A.
log. c = log. a + L sin. C — L sin. A.
If the parts given be A, B, c, the process will be similar to
that given above. In each case we have all the angles and one
side given, and to find the remaining sides we have only to
apply the sine formula as before.
156. It may be observed that the equation b = a . sm' —
sin. A
may be written b = a sin. B cosec. A, so that
log. b = log. a + L sin. B + L cosec. A — 20 ;
a formula which is perhaps slightly less troublesome than the
other, the process of subtraction being changed into addition.
H
98 PLANE TEIGONOMETEY.
CASE III.
157. Given two sides and the angle opposite to one of them.
Let the parts given be the sides a,, &, and the angle A.
The consideration of this case divides itself into two parts
according as the side a, which is opposite to the angle A, is
greater or less than b.
First let the side a be greater than b. For instance, let us
suppose the parts given to be a = 72, b = 64, A = 61°.
We have, from the sine formula,
L sin. B = L sin. A + log. b — log. a
Lsin. A= 9-941819
log. I = 1-806180
sum. = 11-747999
log. a = 1-857333
difference = L sin. B = 9-890666
Now, when all that we know of an angle is its sine, since
sin. A = sin. (180° — A), there will be two angles less than
180° possessing this sine, and we may be in doubt which of
these to select.
Thus, in the present instance, 9*890666 is the tabular
logarithmic sine corresponding to each of the angles 51° V 36",
and 128° 58' 24".
Since, however, the greater angle of every triangle must be
opposite to the greater side, the value 128° 58' 24" is inadmis-
sible, being greater than 61°, which is the value of the angle
given as subtending the greater of the two sides.
There is, therefore, no ambiguity, and the value 51° V 36"
is clearly that which we seek.
158. Secondly, let the same parts be given, viz. 'a, b, and A,
but let a be less than b.
As in the previous case, we have
L sin. B = L sin. A + log. b — log. a.
From this formula, as before, we may obtain two values of
B, each less than 180°, and each greater than A.
Each of these values may therefore be retained consistently
SOLUTION OF TRIANGLES OTHER THAN RIGHT-ANGLED. 99
with the necessary condition that the greater angle shall be
opposite to the greater side.
This is an instance of what is known as ' the ambiguous
case' in the solution of triangles, and, as will be seen, two
triangles may sometimes be found which satisfy the given
conditions.
As a simple case, let a = 1, b = \^3, A = 30°.
Take any straight line AC = \/3 units of length.
At A make the angle CAD = 30°.
With centre C, and distance equal to the unit of length,
describe a circle.
This will cut AD in B, B'.
We have now two triangles, ABC, AB'C, which satisfy the
given conditions, viz. that a = 1, b = \/3, A = 30°.
To obtain the angles B, B' we have
sin. B=- sin. 30° = ^.
a 2
Thus B may be either 120°, as in the triangle ABC, or 60°,
as in AB'C.
159. The geometrical construction given in the preceding
article may be extended to illustrate generally the several cases
involved in the solution of a triangle, in which the parts given
are the two sides, and the angle opposite to one of these
sides.
Thus, having given A, a, 6, to construct the triangle ABC.
As before, draw the straight line AC equal to 5, and at the
point A in CA make the angle CAB equal to A.
With centre C and radius CB, equal to a, describe an arc of
a circle.
In order that a triangle may be constructed having the
H 2
100
PLANE TRIGONOMETKY.
assigned parts, AB must either touch or cut the circumference
of this circle.
c
First let AB touch the circumference of the circle.
Then ABC is the triangle required, and it has one angle B
a right angle.
Next let the radius CB, or a, be greater than b. Then CB
will cut AB twice, but one point of intersection will lie upon
the left of A, as B'. Thus there will be one triangle only, viz.
ABC, which possesses the given parts A, a, b.
Thirdly, let CB be less than CA or b. Then, as shown in
the previous article, the circumference of the circle will inter-
sect AB in two points B, B7, upon the same side of A, and two
triangles will be formed, ABC, AB'C, each of which has the
given values for the angle A and the sides a, b.
160. It should be observed that a must not be less than
b sin. A, for in that case CB would be less than CD, so that the
construction would fail, and no triangle could be formed having
the values given for its several parts.
CASE IV.
161. Given two sides, and the included angle, to solve the
triangle.
Let the parts given be the two sides a, b and the angle C.
We have to determine the two remaining angles A, B, and
the third side c.
We may sometimes in simple cases make use of the formula
c2 = a2 + b* — 2ab cos. C.
Then, having the three sides, we may proceed to find the
angles as already explained.
In practice, however, it rarely happens that the use of this
formula is advantageous, and though by suitable artifices the
SOLUTION OF TEIANGLES OTHER THAN HIGHT-'ANdT'FD, 101
expression may be transformed into a logarithmic shape, the
process of determining the third side independently by this
method is a cumbrous one, and has no advantage to compensate
for the amount of trouble involved.
It is better, therefore, to first determine A and B ; having
then a, A and C, we may easily find c.
It should be observed that A is supposed to be the greater
of the two angles.
To obtain A, B.
The sum of these angles is already known, for it is the
supplement of the angle C. If, therefore, we can find the
difference, we have two simultaneous equations, whence the
values of the two quantities A, B may be obtained.
Now by art. 140
tan A~B = a-1 cot -
2 ~ a + b C( ' 2 *
^ -Q Q
Therefore Ltan. — - — = log. (a— b) + L cot. - —log. (a + fy.
L ->
This equation gives the value of — ^ — , and — i — is
— 2
already known.
The sum of the two values is therefore equal to A, the
difference to B.
To find c we have, by art. 133, - = . ' . , from which
a sm. A
we obtain
log. c = L sin. C + L cosec. A + log. a — 20.
162. To find the area of a triangle., having given two sides
and the included angle.
Let the parts given be a, 5, C. It is required to find the
area of the triangle.
By art. 141 area = - ab sin. C.
Therefore log. area = log. a +log. b + L sin. C — log. 2 — 10.
163. To find the area of a triangle, having given the three
By art. 142 area=<v/6- (s — a) (s — b) (s — c).
102 PLANE TRIGONOMETRY.
Therefore
log. area=-^ {log. s + log. (s— a)+log. (s— 6) + log. (« — c)}.
z
Thus in every case where we have sufficient data to solve
the triangle we are in a position to find its area.
CHAPTER XIX.
PROBLEMS ON THE SOLUTION OF TRIANGLES.
164. WE shall show in the present chapter the practical
application of the processes explained in previous chapters in
determining the heights and distances of visible but inaccessible
objects.
The following definitions will be required : —
The angle which the line joining the eye of the observer with a
distant object makes with the horizontal plane is called the angle
of elevation when the object is above the observer.
The same angle is called the angle of depression when the object
is below the observer.
By means of a theodolite, angles between objects situated in
the same horizontal or vertical plane may be observed; by
means of a sextant the angle between objects situated in any
plane may be determined.
165. To find the height of a column standing on a horizontal
plane, the base of the column being attainable.
Let AB be the vertical column.
From the base B measure the horizontal line BC in any
direction. . A
At C observe the angle of elevation
ACB.
Then AB = BC tan. ACB.
The line BC is called a base line.
The employment of such a base line
is very general in problems of this
nature, and its careful measurement is essential to accuracy ia
the final result.
PKOBLEMS ON THE SOLUTION: OF TEIANGLES.
103
166. To find the height of a flag -staff on the top of a tower.
Let AB be the flag-staff standing on the tower BC.
From C measure the base line CD.
At D observe the angles of elevation ADC, BDC.
Then AB = AC - BC = CD tan. ADC - CD tan. BDC.
167. To find the height of a column standing on a horizontal
plane, when the base of the column is inaccessible-
Let AB be the column.
Measure a distance CD in the same horizontal line with B.
At C, D observe the angles of elevation ACB, ADB.
Then in the triangle ACD we know the angle ACD, since it
is the supplement of the angle ACB. Moreover the angle ADC
is known, and also the side CD.
Therefore AC may be determined.
Then in the triangle ACB
AB = AC sin. ACB.
168. To find the distance between two inaccessible oojects in
the same horizontal plane.
Let A, B be the objects.
104 PLANE TKIGONOMETEY.
Measure a base line CD in the same horizontal plane as A, B.
At C observe the angles ACB, BCD.
At D observe the angles ADC and BDA.
Then in the triangle ADC we know the angles ACD and
ADC, and the side CD.
A B
D
Hence AC may be determined.
In the triangle BDC we have the angles BDC, BCD and
the side CD.
Hence BC may be determined.
Then in the triangle ABC, having the two sides AC, BC antl
the included angle ACB, AB may be obtained.
169. — The practical portion of this work will furnish the
student with abundant exercise in the solution of practical
problems, involving heights and distances, by the aid of loga-
rithmic tables. In the collection of examples which follows the
results required may be established without the aid of logarithms.
Many of the examples here given have been proposed in the
periodical examinations for the rank of lieutenant held at the
Royal Naval College during the past five years.
EXAMPLES.— XXIII.
1. From the top of a perpendicular cliff two rocks are
observed in a straight line with the base, which are known to
be a yards apart ; their respective depressions are a and 3a :
show that the height of the cliff is '-,
2 cos. a
2. A man walking due east along a straight road observes
that a certain church tower bears E.N.E. ; a mile farther on
the bearing of the tower is N.N.E. : show that the shortest
distance of the tower from the road is half a mile.
PROBLEMS ON THE SOLUTION OF TRIANGLES. 105
3. The altitudes of the top of a flag-staff, at points due south
and due west of it, are observed to be 45° and 30° respectively ;
show that the height of the flag-staff is half of the distance
between the points of observation.
4. The elevation of two clouds to a person in the same line
with them is a ; when standing on the shadow of one of them
its elevation is 2 a, and the other is vertically over him : show
that the heights of the clouds are as 2cos.2a to 1.
5. A person wishes to find the distance between two places
A and B on opposite sides of a brook. He walks from B to a
bridge ,two miles off. Crossing this he continues his walk 6
miles in the same direction to C, which he knows to be 3 miles
from A. If A be 4 miles from the bridge, show that AB = 5-86
miles nearly.
6. The elevation of a tower standing on a horizontal plane is
jserved ; a feet nearer it is found to be 45° ; b feet nearer still
j is the complement of the first angle observed : show that the
height of the tower is — — feet.
a — b
7. A column stands in a field which has the form of an
equilateral triangle, and subtends angles tan."1 — ^ , tan."1-— ,
V3 V7
tan."1 - at the three corners; show that the height of the
column is equal to the length of a side of the field.
8. A person at the top of a mountain observes the angle of
depression of an object in the horizontal plane beneath to be
45° ; turning through an angle of 30° he finds the depression of
another object in the plane to be 30° : show that the distance
between the objects is equal to the height of the mountain.
9. A flag-staff a feet high, on the top of a tower, is seen
from a certain point in the horizontal plane on which the tower
stands to subtend at the eye of the observer an equal angle 0
with that subtended by the tower itself; show that the height
of the tower = a cos. 20.
10. A statue which stands upon level ground, on a pedestal
27 feet high, subtends an angle tan."1 -J-4- at a point 36 feet
106 PLANE TKIGONOMETKY.
from the foot of the pedestal ; show that the height of the
statue is 21 feet.
11. A fort is seen from a ship bearing E.N.E. ; when the
ship has sailed due east 4 miles it bears N.N.E. : show that the
distance of the fort is now \/16 — 8\/2 m^es-
12. The elevation of a tower standing on a horizontal plane
is observed to be 6; at a station m feet nearer it is 90° — 0,
and at a point n feet nearer still it is 20 : show that the height
of the tower is A / (m + n)2 — -— •
V 4
13. On the bank of a river a column 186 feet high supports
a statue 31 feet high ; the statue, to an observer on the
opposite bank, subtends the same angle as a man 6 feet high
standing at the foot of the column : show that the breadth of
the river is approximately 98 '5 feet.
14. An observer at the sea-level notices that the elevation
of a certain mountain is a ; after walking directly towards it
for a distance d along a road inclined at an angle 7 to the
horizontal, lie finds the elevation of the mountain to be ft :
show that the height of the mountain is d sin. a . ' y — ^f .
sm. (a— ft)
15. From the top of a tower, which stands near a river, it
is observed that the distance of the foot of the tower from the
river subtends the same angle as the breadth of the river ; the
height of the tower being /i, and the distance of the tower from
the river d : show that the breadth of the river is l, - d.
hr — d
16. A tower standing on a horizontal plane is surrounded
by a moat, which is equal in width to the height of the tower ;
a person at the top of another tower, whose height is h, and
distance from the moat c, observes that the first tower subtends
an angle of 45°: show that the height of the first tower is
k — c
17. A person wishing to determine the length of an in-
accessible wall places himself due south of one end, and due
west of the other, at such distances that the angle subtended by
PKOBLEMS ON THE SOLUTION OF TEIANGLES. 107
the wall is in each case equal to a. If I be the distance be-
tween the two stations, show that the length of the wall is
I tan. a.
18. A and B are two points in a level field, which is adjacent
to a lake, in which are moored two buoys, C and D ; A is
distant d yards from B and both at A and B the distance CD
subtends an angle a : show that if the straight lines AD and
BC are inclined at an angle 0, the distance between C and D is
either
d sin. a ' d sin. a
or
sin. (Q + a) sin. (0—a)
19. A flag-staff a feet long stands on the top of a tower b
feet high ; an observer, whose eye is c feet above the level of
the foot of the tower, finds that the staff and the tower subtend
equal angles at his eye : prove that the length of the straight
line joining his eye to the foot of the tower is
— 2c c
feet.
a — b
20. The top of a ladder placed against a vertical wall, the
foot of which makes an angle of a with the horizon, rests on a
window-sill ; when its foot is moved m feet farther away from
the wall, it makes an angle /3 with the horizon, and rests on a
second sill : show that the distance between the two sills is
,
m cot. — -—
u
21. A column on a pedestal 20 feet high subtends an angle
of 45° to a person standing on the level ground upon which
the column stands; on approaching 20 feet the observer finds
that the angle subtended is again 45° : show that the height of
the column is 100 feet.
108 PLANE TRIGONOMETRY.
CHAPTER XX.
OF TRIANGLES AND POLYGONS INSCRIBED IN CIRCLES, AND
CIRCUMSCRIBED ABOUT THEM.
170. To find the radius of the circle inscribed in a triangle.
A
D C
Let ABC be a triangle, 0 the centre of the inscribed circle,
which touches the sides BC, CA, AB in D, E, F respectively.
Let r denote the radius of the circle.
Then
area of triangle BOC = ~ BC . OD = ^ ;
area of triangle AOC =~ AC . OE = ~;
2 2
area of triangle AOB = i AB . OF = - .
2 2 *
Therefore, by addition,
- (a + b + c) r=area of triangle ABC = S (art. 142) ;
therefore r = - .
s
Thus the radius of the inscribed circle is equal to the area
of the triangle divided by one-half the sum of the sides.
171. A circle which touches one side of a triangle, and the
other two sides produced, is called an escribed circle.
172. To find the radius of an escribed circle.
Let ABC be a triangle, and let 0 be the centre of the circle
which touches the side BC in D, and AC, AB produced in E, F
respectively.
PROPERTIES OF TRIANGLES. 109
Let rl be the radius of this circle.
It will be seen from the figure that the quadrilateral figure
OBAC may be regarded as the sum of the two triangles OAB,
OAC.
Therefore the quadrilateral figure OBAC = -^ + -^- .
Again, the same figure may be divided into the two triangles
OBC and ABC.
Therefore the figure OBAC = -1 + S.
Hence ^± + ^J- = El + S;
therefore — fc -f b — a*} rx = S ;
therefore rl = 5 .
s — a
Similarly if r2 be the radius of the circle which touches AC
and BA, BC produced, and r3 the radius of the circle which
touches AB, and CA, CB produced, it may be shown that
S S
— c
173. To find the radius of the circle described about a triangle.
Let ABC be a triangle, and 0 the centre of the circle de-
scribed about it.
Draw OD perpendicular to BC. Then BC is bisected at D.
Let R denote the radius of the circle.
Since the angle at the centre of a circle is double of the
110
PLANE TKIGONOMETKY.
angle at the circumference upon the same base, the angle BOO
is double of the angle BAG. (Euc. III. 20.)
But the angle BOO is double of the angle BOD.
Therefore the angle BOD = the angle BAG.
In the triangle BOD, BD = BO sin. A.
Therefore - = E sin. A, and R =
a
2 sin. A
Similarly it may be shown that E =
2 sin. B
or
174. To find the radii of the inscribed and circumscribed
circles of a regular polygon.
Let AB be the side of a regular polygon, that is, of a polygon
which has all its sides and angles equal, and let n be the number
of sides.
Let 0 be the centre of the circles, OD the radius of the in-
scribed, and OA that of the circumscribed circle.
Let AB = a, OA = E, OD = r.
The angle AOB is the nth part of four right angles ;
therefore AOB = -2>7r, and AOD = - ,
n . n
PROPERTIES OF TRIANGLES.
Ill
and AD = - = R sin. - = r tan. ? .
Therefore R =
2 sin. -
n
r =
n
a
2 tan. 2T
The area of the polygon may be expressed by means of the
radius of the inscribed circle, or by that of the circumscribed
circle.
Thus, the area of the polygon = n times the area of the
triangle AOB
= n . AD . OD = n ~ r = nr* tan. - .
2 n
7T
7T
Again, since AD = R sin.- . and OD = R cos.- :
n n
therefore the area of the polygon = n R2 sin. - cos. —
n n
175. To prove that if 6 be the circular measure of a positive
angle less than a right angley 6 is greater than sin. 0, but less than
tan. 6.
Let AOB be any angle less than a right angle, and let OB
= OA.
From B draw BD perpendicular to OA, and produce it to
C, so that DC = DB, and join 00.
Draw BE at right angles to OB, meeting OA produced in
E, and join EC.
112 PLANE TRIGONOMETRY.
Then the triangles BOD, COD are equal in all respects, so
that the angle BOD is equal to the angle COD.
Therefore the triangles BOB, COE are equal in all respects,
the angle ECO is a right angle, and EC = EB.
With centre 0, and radius OB, describe an arc of a circle BAG.
This will touch EB at B, and EC at C.
Then if we may assume provisionally that the straight line
BC is less than the arc BAG, we have BD, the half of the
straight line, less than BA, the half of the arc.
Therefore •=-== is less than 77^-; that is, the sine of the
O-D U.D
angle AOB is less than its circular measure.
Again, if we assume that the arc BAG is less than the sum
of the two straight lines EB and EC, we have AB less than EB.
EB
Hence -^=- is less than - — ; that is, the circular measure
OB OB
of the angle AOB is less than its tangent.
176. With regard to the two assumptions that have been.
made, the first is sufficiently obvious to be readily admitted.
The second one, namely, that the two straight lines EB, EC are
together greater than the arc BAG, requires more consideration.
At the point A in the figure of the preceding article, draw the
straight line EG, touching the arc, and meeting EB, EC in F, G.
Then because two sides of a triangle are together greater
than the third, it follows that BE, EC are together greater than
BF, FG, GC.
Now if the angle BOC = — , where n is a positive integer,
BE, EC will be together equal to the side of a regular polygon
of n sides described about the circle of which BAG is an arc.
Again, since FB, FA are tangents drawn to the circle from
the same point F.
Therefore FB is equal to FA.
For a similar reason GC is equal to GA.
Therefore BF, FG, GC are together equal to twice FG, that
is, they are together equal to two sides of a regular polygon of
2n sides described about the circle.
PROPERTIES OF TRIANGLES. 113
Hence the perimeter of the polygon of n sides described
about a circle is greater than the perimeter of a polygon of 2n
sides described about the same circle.
Similarly it may be shown that the perimeter of a polygon
of 2n sides is greater than the perimeter of a polygon of 4<n sides,
and that generally the perimeter decreases as the number of
sides increases.
But since when the number of sides is indefinitely increased
the circumscribed polygon approaches the form of a circle, it
is clear that the circumference of a circle must be less than
the perimeter of any polygon that may be described abort it.
And therefore any arc BAG, which is - th part of the Cir-
ri,
cumference, is less than the sum of EB, EC, which are together
- th part of the perimeter of the circumscribed polygon.
71
177. To show that the limit of 6^ — , when 6 is indefinitely
u
diminished, is unity.
Since sin. 0, 6, and tan. 6 are in ascending order of magni-
6 1
tude, dividing by sin. 0, we have 1, - — - , — in ascendinef
sin. a cos. u
order of magnitude.
As 9 is diminished, cos. 6 approaches unity, and the smaller
cos. 6 becomes, the more nearly does approach unity.
cos. c/
Q
Therefore by diminishing 0 sufficiently we can make
sin. 0
differ from unity by less than any assignable quantity, however
Q
small, that is to say, - — - will approach the limit unity,
sin. (7
Therefore also - — ~ approaches the limit unity.
6
tan. 0 sin. 9 1 , ,. . f
Moreover, since — -^- = — ^ — x . , the limit of
a u cos. a
— — — , when 9 is indefinitely diminished3 is also unity.
114 PLANE TRIGONOMETRY.
178. It follows from the preceding article that the limit of
m sin.— , when m increases indefinitely, is a.
m
For m sin.— = - — , and when m becomes indefinitely
CL
sin._
great - — is unity.
a
m
Similarly the limit of m tan. — , when m increases inde-
m
finitely, is a. ,
179. To find the area of a circle.
By art. 174 the area of a regular polygon of n sides de-
scribed about a given circle of radius r
= nr2 tan. ^.
n
Now when n is increased without limit, the figure of the
polygon approaches the form of the circle, and therefore the
area of the circle will be equal to the limit of the above
expression.
But when n is indefinitely great Titan. — = TT, by art. 178.
Ifb
Therefore the area of the circle = irr2.
EXAMPLES.— XXIV.
1. A square and an equilateral triangle are inscribed in the
same circle. Find the ratio between (1) their sides, and (2)
their areas.
2. Compare the areas of regular pentagons inscribed within,
and described about, a given circle.
3. Find the value of the interior angle of any regular
polygon of n sides.
4. Show that the square described about a circle is equal to
4
— of the inscribed duodecagon.
o
PEOPEETIES OF TKI ANGLES. 115
5. Show that the area of a regular polygon inscribed in a
circle is a mean proportional between the areas of an inscribed
and circumscribed polygon of half the number of sides.
6. Find the radii of the inscribed, and each of the escribed
circles of the triangle ABC, when a = 5, b = 7, c — 9.
7. With the usual notation, prove the following relations :—
(1) B =
(2) Vr.rl.r2r3 = S.
(3) rj + r2 + r3 - r = 4E.
rl-r r2 + r3 o
„, 111 1 o? + ft2 + c2
(5) -3 + -, + -, + -,= -gr--;
(6) E (sin. A + sin. B + sin. C) = ,<?.
(7) 4 E2 (cos. A + cos. B cos. C) = be.
8. Show that the distance between the centre of the in-
scribed circle, and that of the escribed circle touching the side
BC, is A
a sec. -.
9. If C is a right angle, show that R + r = a ^ .
A
10. If A is a right angle, show that r2 + r3 = a.
11. If 0 is the centre of the circle described about the
triangle ABC, and AO is produced to meet BC at D, show
that
DO cos. (B - C) = AO cos. A.
12. Find the angle of the sector in which the chord of the
arc is three times the radius of the circle inscribed in the
sector.
13. In any triangle show that the area of the inscribed circle
ABC
is to the area of the triangle as IT to cot. — cot. — . cot. -.
ft ' 2 2
14. Show that in any triangle the radius of the inscribed
12
116
PLANE TRIGONOMETRY.
circle bears to the radius of the circumscribed circle the ratio of
A B C
4sm.~ sin.- sin,- : 1.
a • & &
15. If DEF be the triangle formed by the lines joining the
feet of the perpendiculars from A, B, C upon the opposite sides,
then the radius of the circle inscribed within DEF is to the
radius of the circle inscribed within ABC as
Cos. A cos. B cos. C : 2 sin.- A sin.-- B sin.- C.
- _
FORMULA OF REFERENCE (II.).
1. In any plane triangle, ABC,
(2)
(3) a = c cos. B + b cos. C
(4)
* be
(5) Hav.A=(*~5)(g~c)
oc
(6) Tan.^=^|cot.|
(7) S=itcsin. A
(8) S=V«(«— o)(»— &)(*— c)
(9) r=-
(art. 133)
(art. 134)
(art. 135)
(art. 137)
(art. 138)
(art. 140)
(art. 141)
(art. 142)
(art. 170)
. 172)
2. The area of a circle of radius r-—7rr*
(art. 179)
ANSWEBS TO THE EXAMPLES.
117
ANSWERS TO THE EXAMPLES.
1. 44.
4 T
1. lift. 8in.
5. 24 feet.
8. As 1 to </2.
I. (page 4).
2. 60 yards.
5. 1ft. 9in.
8. 40.
6. 2£ inches.
II. (page 5).
2. As 2 to 1. 3. 48 feet. 4. 8-6 feet nearly.
6. 159-8 yards nearly.
1. 3-93 inches.
4. 5-971 inches.
7. 4=..
9. 10 feet.
III. (page 9).
2. 45£ inches.
6. -278 inch.
10. 23-6 inches nearly.
3. 15mm. 54-5 sec.
6. 158-7 yards.
1. (1) 19° & 55".
(4) 60°.
2- (1) !•
3. 6f. 4. 7° 30'.
8. 2-356 nearly.
11. 1-494 nearly.
14. 5£ feet.
IV. (page 12).
(2) 100° 16' 3".
(5) 22° 30'.
(2) ^TT. (3) ?7T.
e 47T- 4?r TT
9"' "9 ' 9* 77
9. -856 nearly. 10. 30°, 50°, 100°*
12. ??.
(3) 39° 23' 27".
(6) 191° 15'.
6. ?.
13, 108°, or .
5
15. 1°-018 nearly. 16. 2,269 miles nearly.
VI. (page, 27).
(2) 39° 43X 22".
(5) I'
(2) 78° 40' 17".
1. (1) 62° 22' 12".
(4) 127°.
2. (1) 100° 23' 45".
(4) 250°-
3. (1) Sin. x = cos. A cosec. A.
(2) Cos. x - cos. A cosec. A.
(3) Sin. x = — sin. A cos. A sec.2 B.
(5) |.
, (3) -15°.
(6) 7f.
(3) -20°.
(6) ^.,
118 PLANE TRIGONOMETBY.
VIII. (page 34).
1. Sin. 8 = - l
cot.2 6
2. Sin. 6 = sec- ~ , cot. 0 =
sec. 0 \/sec.2 6 - 1
8. Sin. 6 = N/ 2 vers. 0 - vers.2 0, cos. 6 = 1 - vers. 6, &c.
4. Tan. A = - . 5. Cot. A - -L .
4 V3
V. OUIt V = ~ j Ot
;u. f = ~ . i • ^y^u. ii. — — — .
-s/6 2^6
a Ton A -
9f!nqo/> A — -
3\
/7 «
10. «262 + a2 = 1.
IX. (page 38).
1. (1) 1.
(2) J. (3) VS. (4) -I-
(6) -2.
(6) V3. (7) -2. (8) \
w.-}
(10) V3. (11) 0. (12) 0.
2. (1) 7l7T+£.
(2) wr + (-!)«' (3) 2W7r±?f,
D o
(4) (2w + l)r
(6) mr±£. (6) n^ + J.
X. (page 43).
17. ||. 18. 3. 19. -
XIII. (page 47).
1. Cos. (A + B) + cos. (A - B). 2. Sin. 6A + sin. 3A.
3. Cos. A- cos. 7A. 4. Sin. 7A - sin. 2 A.
6. Sin. (2 A + B) - sin. B. 6. Cos. (A + ??) + cos. — .
V 2i / 2
7. Sin. 70°+ sin. 30°. 8. Cos. 15°- cos. 35°.
XVI. (page 56).
1. (1) ++. (2) +-. (3) +-. (4) --.
(5) ++. (6) +_.
4. Between 270° and 450°.
6. (1) Sin. 72°= cos. 18°. (2) Cos. 54° = sin. 36°.
(3) Sin. 9° = 1 (x/s + V5 - ^6 -
(4) Cos.9° =
(5) Cos. 81°= sin. 9°.
ANSWEKS TO THE EXAMPLES. 119
XVli, (page 58)'.
T.
e
-|. 2.
B
JT
" 6*
3.
0 -.
07T
"4*
4.
(9
Tf
" 4*
6.
B
= ?. 6.
B
7T
" 3*
7.
^ =
7T
4'
8,
(9
= 4'
9.
e
-1 10'
B
7T
" 3*
11.
B =
7T
6'
12.
(9
7T
~ 4*
13.
6
= |. 14.
B
7T
= 6'
15.
(9 =
67T
9'
16.
B
-5°
'?
17.
B
=T?or°-
XVIII.
(page
69).
1.
B
= 2W7T + £.
2. i
0 = W7T
•*• T.
3.
7. ^ + =W7r+ . a e
9. ^ = W7r+5or = 2w7r±r. 10.
4 2 4 o
XIX. (page 60).
16. a? = 2. 16. a: -2. 17. a: = 3.
10 x = A /2 in r ._ 1 20 # =s — -
J.O. "* A/ — • J£7. *t — — « ^V/. «* — _
XX. (page 67).
1. 4-477121, 4-301030, -352182, 2-647818.
2. 1-741167. 3. 2-778074.
4. 1-732395, -94639. 5. (a) -5, (ft) -^, (c) ^.
6. 2-698970, -150515, 6-64386. 7. - -6751.
8. ^ = 3-1547. 9. -5172818.
10. *, i§. 11. ^-76. 12. -301030.
3 5
13. mc + 7n. 17. ^ = 5-435. 18. 486.
c
19. By dividing each, logarithm by 3.
XXI (page 81).
1. L sin. 45° = 9-849485, L tan. 45° = 10-000000, L sin. 60° = 9-937531,
L cos. 60° = 9-698970, L cot. 30° = 10-238561, L tan. 210° = 9-761439.
120
PLANE TRIGONOMETRY.
3. 9-763188.
6. 9-984944.
9. 9-923592.
12. land*.
4. 9-937238.
7. 9-778151.
10. 9-875061.
5. 10-035974.
8. 700000.
11. 200000.
XXIV. (page 114).
1. The sides as »J2 to V3 ; the areas as 8 to
2. As 3 + A/5 to 8. 3. IT (l - -
6. r =
12. 60°.
PART II.
SPHEEICAL TEIGONOMBTEY
123
CHAPTER I.
THE GEOMETRY OF THE SPHERE.
1. A SPHERE is a solid bounded by a surface every point of
which is equally distant from a certain fixed point, which is
called the centre of the sphere.
The straight line which joins any point of the surface with
the centre is called a radius.
A straight line drawn through the centre, and terminated
both ways by the surface, is called a diameter.
2. In order to establish the fundamental theorems upon
which the science of spherical trigonometry is based, it will be
necessary to take for granted certain definitions and propositions
of the eleventh book of Euclid. It will be convenient to give
here the definitions which will be required, and the enunciations
of the propositions, the truth of which we shall have afterwards
to assume.
DEFINITIONS.
3. A straight line is perpendicular, or at right angles, to a
plane when it makes right angles with every straight line
meeting it in that plane (Euc. XI.
def. 3).
Thus if AB be at right angles
to the plane which contains the
straight lines BC, BD, BE, then the
angles ABC, ABD, ABE are each
of them a right angle.
A plane is perpendicular to a
plane, when the straight lines drawn
in one of the planes, perpendicular to
the common section of the two planes, are perpendicular to the
other plane (Euc. XI. def. 4).
124
SPHERICAL TRIGONOMETRY.
Thus if the plane ABCD be at right angles to the plane
ABEF, and GH be drawn in the plane ABCD at right angles to
AB, the common section,
the angles which GH
makes with HE, HF, &c.,
any straight lines drawn
in the plane ABEF, are
all right angles.
The inclination of a
straight line to a plane is
the acute angle contained by that straight line, and another drawn
from the point in which the first line meets the plane, to the point
in which a perpendicular to the plane drawn from any point of the
first line above the plane meets the same plane (Euc. XI. def. 5).
Thus the inclination of the straight line EF to the plane
ABCD is found by dropping EG perpendicular to the plane from
any point E in EF, and joining
FG. The angle EFG is the in-
B clination of EF to the plane
ABCD.
The inclination of a plane to a
plane is the acute angle contained
by two straight lines drawn from
any the same point of their
common section, at right angles to it, one in one plane, and the
other in the other plane (Euc. XI. def. 6).
Let ABCD, ABEF be two planes, of which the common
section is AB, and let HG, HK be drawn in the two planes at
right angles to AB, then the angle GHK measures
clination of the two planes.
the
in-
ON SPHERICAL GEOMETRY.
125
A solid angle is that which is made by more than two plane
angles , which are not in the same plane, meeting at one point
(Euc. XI. def. 9).
B C
Thus in the figure the three plane angles ADB, BDC, CDA
contain a solid angle at D.
THEOREMS.
4. // a straight line stand at right angles to each of two straight
lines at the point of their intersection, it shall also be cut right
angles to the plane which passes through them ; that is, to the
plane in which they are (Euc. XI. 4).
In the figure attached in the preceding article to def. 3, if
ABD, ABE be both right angles, AB will be at right angles
to the plane containing BD, BE.
If two planes which cut one
another be each of them per-
pendicular to a third plane,
their common section shall be
perpendicular to the same plane
(Euc. XI. 19).
If the two planes ABCD,
EBCF be each of them perpen-
dicular to the plane ABE, then
BC, the common section of the
first two planes, is perpendicular to the plane ABE.
If a solid angle be contained by three plane angles, any two of
them are together greater than the third (Euc. XI. 20).
Every solid angle is contained by plane angles, which are
together less than four right angles (Euc. XI. 21).
Thus in the figure attached to def. 9, any two of the three
126
SPHERICAL TRIGONOMETRY.
plane angles ADB, BDC, CDA are greater than the third angle,
and the sum of the three angles is together less than four right
angles.
5. The section of the surface of a sphere, made by any plane,
is a circle.
Let AB be the section of a sphere
made by any plane, 0 the centre of
the sphere.
Draw 00 perpendicular to the
plane, so that 00 produced both
ways is a diameter of the sphere.
Take any point D in the section,
and join OD, DC.
Since 00 is perpendicular to the
plane, the angle OCD is a right angle
(Euc. XI. def. 3).
Therefore CD2=OD2-OC2.
But 0 and 0 being fixed points, 00 is constant, or has a
fixed value.
And OD is constant, for it is a radius of the sphere.
Therefore CD also is constant ; that is, it has the same value
wherever the point D is taken in the section.
Thus all the points in the plane section are equally distant
from the fixed point 0, so that the section is a circle, of which 0
is the centre.
6. If the cutting plane pass through the centre of the sphere
00 vanishes, and CD becomes equal to OD, the radius of the
sphere.
7. The section of the surface of a sphere is called a great
circle if the plane passes through the centre of the sphere, and a
small circle if the plane does not pass through the centre of the
sphere. The radius of a great circle is therefore equal to the
radius of the sphere.
8. Through the centre of a sphere and any two points on its
surface only one plane can in general be drawn ; but if the two
points are situated at the extremities of a diameter, an infinite
number of such planes can be drawn.
ON SPHEEICAL GEOMETRY.
127
If therefore two points are not at the extremity of a dia-
meter, only one great circle can be drawn through them, which
is unequally divided at the two points. The shorter of the two
arcs is commonly spoken of as the arc of a great circle joining
the two points.
9. The axis of any circle of a sphere is that diameter of
the sphere which is perpendicular to the plane of the circle, and
the extremities of the axis are known as the poles of the circle.
Thus in the figure to art. 5, EF is the axis of the circle ABD,
and E, F are the poles of the circle,
In the case of a great circle the poles are equally distant
from the plane of the circle.
With a small circle this is not so, and the two poles are
called respectively the nearer and the further pole. Thus in the
figure, E is the nearer pole and F the further pole, of which the
nearer is commonly known as the pole.
10. A pole of a circle is equally distant from any point in the
circumference.
Let 0 be the centre of the sphere, and AB be any circle of
the sphere. Let PP' be the axis of
the circle. Then C, the point in
which the axis meets the plane of
the circle, is the centre of the circle
(art. 5), P,P' being the poles.
Take any point D in the circum-
ference of the circle, and join CD,
PD.
ThenPD2-=PC2-fCD2.
But PC, CD are the same in
length wherever D may be.
Therefore PD is constant.
Hence the length of the straight line joining the pole P
with any point in the circumference of the circle is invariable.
Similarly it may be shown that the pole P' is equally distant
from every point in the circumference of the circle.
Again, let a great circle be described through the points
P and D, when D is any point in the circle AB.
Since the length of the chord PD is the same for all points
128
SPHEKICAL TEIGONOMETEY.
in the circumference AB, and since in equal circles the arcs
which are cut off by equal straight lines are equal (Euc. III. 28).
Therefore the arc of a great circle intercepted between P and
D is constant for all positions of D on the circle AB.
Thus the distance of a pole of a circle from every point of
the circle is constant, whether the distance be measured by
the straight line joining the points, or by the arc of a great
circle intercepted between the points.
11. The arc of a great circle luhich is drawn from a pole of a
great circle to any point in its circumference is a quadrant.
Let ACBD be a great circle, and
P its pole, 0 being the centre of the
sphere.
Join OP, and draw any diameters
of the circle AOB, COD.
Because OP is at right angles to
the plane of the circle, therefore
POA, POC, POB, POD are all right
angles (Euc. XL def. 3).
Therefore the arcs PA, PC, PB, PD are all quadrants (Euc.
VI. 33).
12. Two great circles bisect each other.
Let BAD, BFD be two arcs
of great circles, intersecting one
another in the points B, D.
Then BD, the common sec-
tion of the planes of the two
great circles, is a straight line
(Euc. XL 3).
Join BD.
And because the centre of
the sphere lies in each of the
planes of the circles, it must
therefore be a point in their common section, as 0.
Therefore BD must be a diameter of the sphere, and of each
of the two circles BAD, BFD.
Therefore the two circles BAD, BFD bisect each other.
13. If the arcs of great circles joining a point on the surface of
ON SPHERICAL GEOMETRY.
129
a sphere ivith two other points on the surface of the sphere, which
are not at opposite extremities of the same diameter, be each of
them quadrants, then the first point is a pole of the great circle
through the last two points.
Let O be the centre of the sphere, P a point on the surface,
and PA, PB be two arcs, each equal to a quadrant.
Through the points A, B let
a great circle be described, and
join'OA, OB, OP.
Then, because PA, PB are
quadrants, therefore POA, POB
are each right angles.
Therefore PO stands at right
angles to each of the straight
lines OA, OB at the point of their
intersection.
Therefore PO is at right
angles to the plane AOB in which they are (Euc. XI. 4).
Therefore OP is a portion of the axis of the circle AB (art. 9),
and P, being an extremity of the axis, is a pole of the circle AB.
14. If from a point on the surface of a sphere there can be
draivn two arcs of great circles, not parts of the same areat circle,
the planes of ivhich are at right angles to the plane of a qiven circle,
that point is a pole of the given circle.
Let 0 be the centre of the
sphere, AB a great circle of the
sphere, and PA, PB two great
circles whose planes are at right
angles to the plane of AB.
Then, since the planes of the
circles PA, PB are at right
angles to the circle AB, therefore
PO, their common section, is at
right angles to AB (Euc. XI. 19).
Therefore P is a pole of the great circle AB.
Similarly P will be the pole of any small circle which has
its plane parallel to that of AB, so that the planes of PA, PB
are at right angles to it.
130
SPHEKICAL TEIGONOMETKY.
COR. Hence it follows that if in the circumference of a
given circle any two points be taken, and arcs drawn through the
two points at right angles to the given circle, the point of inter-
section of these two arcs is a pole of the given circle.
15. Great circles which pass through the poles of another
great circle are said to be secondaries to that circle.
Thus, in the figure of art. 14, PA, PB are secondary circles
to AB.
16. Def. A spherical angle is the inclination of two arcs of
great circles at the point where they meet.
Thus let ABD, ACB be two
great circles intersecting in A, D,
and having the straight line AD
for their common section.
And let Ab be the tangent at
the point A to the circle ABD,
and Ac the tangent at the point A
to the circle ACD.
Then BAG is a spherical angle.
And since at the point A the
two arcs AB, AC have the same
directions as their tangents, the spherical angle BAG is
measured by the angle bA.c.
But since Afr, Ac are straight lines drawn in the planes ABD,
ACD at right angles to AD, the common section of the two
planes, the angle 6 Ac measures the inclination of the two planes
(Euc. XI. def. 6).
Hence the spherical angle BAG is equal to the inclination
of the two planes ABD, ACD.
17. The angle between any two great circles is measured by the
arc intercepted by the circles on the great circle to wliicli they are
secondaries.
Let AB be the arc of a great circle of which P is the pole,
so that PA, PB are arcs of great circles, secondaries to AB, and
let 0 be the centre of the sphere.
Then the spherical angle APB will be measured by the arc AB,
ON SPHEEICAL GEOMETRY.
131
Because P is the pole of the circle AB,
Therefore the angles POA, POB are each of them right
angles.
Therefore the angle AOB represents the inclination of the
planes of the circles PA, PB (Euc. XL def. 6).
But the angle AOB is measured by the arc AB.
Therefore the arc AB measures the inclinations of the planes
of the circles PA, PB.
18. The angle subtended at the centre of a sphere by the arc of
a great circle which joins the poles of tivo great circles is equal to
the inclinations of the planes of the great circles.
Let 0 be the centre of the sphere, CD, CE the two great
circles intersecting in the point C,
A and B the poles of CD and CE ^ --B
respectively.
Let a great circle be drawn
through A, B, meeting CD, CE in
F, G respectively.
Join OA, OB, OC, OF, OG.
Because A is the pole of the
circle CDF, therefore the angle AOC c
is a right angle.
And because B is the pole of the circle CEG, therefore the
angle BOC is a right angle.
Therefore OC stands at right angles to the two straight lines
OA, OB at the point of their intersection.
Therefore it is at right angles to the plane AOB (Euc. XI. 4).
K 2
132 SPHERICAL TKIGONOMETRY.
Therefore OC is at right angles to OF, OG, two straight
lines in the plane AOB.
Hence, FOG represents the inclination of the two planes
OOD, OCE (Euc. XI. def. 6).
And since the angles AOF, BOG are both right angles,
therefore the angle AOB = AOF - EOF = BOG- EOF = FOG.
That is, the angle subtended at the centre of the sphere by
the arc of a great circle joining A, B, the poles of the two circles
CD, CE, is equal to the inclination of the planes of the circles
CD? CE.
19. To compare the arc of a small circle, subtending any angle
at the centre of that circle, with the arc of a great circle subtend-
ing the same angle at its centre.
Let ab be the arc of a small circle, c the centre of the circle,
P the pole of the circle, 0 the centre of the sphere.
Through P draw the great circles
PaA, P6B, meeting the great circle of
which P is a pole at A and B respec-
tively.
Join ca, cb, OA, OB.
Because P is the pole of ab,
Therefore the angles Pea, Pcb are right
angles.
Therefore the angle acb represents
the inclination of the planes PoA, PfrB
(Euc. XI. def. 6).
And because P is the pole of AB ;
therefore the angles POA, POB are right angles.
Therefore the angle AOB represents the inclination of the
planes PaA, PbE (Euc. XI. def. 6).
Therefore the angle acb is equal to the angle AOB.
* -• arc ab arc AB
And - _ - — - = — _ __ for these fractions represent
radius ca radius OA
the circular measure of the angle acb or AOB (Part I. art. 17).
Therefore arc ab radius ca
arc AB radius OA'
ab ca
arc AB ~~ OA
TT arc ab ca ca . -r,^
Hence — = _ _ = _ _ = sm. P0«=cos. &OA = cos. aA.
OX SPHERICAL GEOMETRY.
133
Therefore the length of the arc ab is equal to the length of
the arc AB multiplied by the cosine of the arc <zA.
This result is of very great importance, and is of frequent
use in the problems of navigation and nautical astronomy.
CHAPTER II.
ON CERTAIN PROPERTIES OF SPHERICAL TRIANGLES.
20. A SPHERICAL triangle is the figure formed on the surface
of a sphere by three arcs of great circles intersecting one another.
Thus in the figure, if O be the centre of the sphere, the
great circles ABG, ACG, CBD intersecting at A, B and C will
form on the surface of the sphere
a figure ABC, which is called a
spherical triangle.
The arcs AB, BC, CA are the
sides of the triangle, and the
angles formed by the arcs at the
points A, B, C are called the
angles of the triangle.
And, as in plane trigono-
metry, it is usual to denote the
angles by the letters A, B, C,
and the sides respectively opposite to them by the letters &, ft, c. .
21. The sides of a spherical triangle are therefore arcs of
great circles, and are proportional to the three angles which
at the centre of the sphere form the solid angle 0.
In the various relations which will hereafter be established
between the six parts of a spherical triangle, we shall in general
only have to consider the angles subtended at the centre of the
sphere by the arcs of great circles which form the sides of the
triangle, and not the actual lengths of the sides themselves ;
and in the several formulae functions of the angles so subtended,
such as the sine and cosine of these angles, will appear.
Thus by cos. a is to be understood the cosine of the angle
subtended at the centre by the arc BC, or the angle BOC, in
the figure of art. 20.
It may however be observed that if the radius of a particular
134 SPHERICAL TRIGONOMETRY.
sphere be given we may always obtain the actual length of an
arc which subtends a given angle at its centre, by means of the
circular measure of the angle.
Thus, let us suppose that the sphere depicted in art. 20
represents the earth, and that B, C are two places on the earth's
surface. Let us also suppose that the arc BC joining the two
places subtends an angle of 45°, and that the radius of the earth
is known to be 4,000 miles. Then, by Part I. art. 17, the cir-
•p>p
cular measure of the angle BOG is — ^.
and ' BC = - 4000, or 3146 miles nearly.
22. As shown in art. 16, the spherical angle formed at A by
the arcs AB, AC is equal to the inclination of the planes which
contain the two arcs.
The angles of a spherical triangle are therefore the inclina-
tions of the plane faces which form the solid angle.
23. It is found convenient in spherical trigonometry to agree
that each side of a triangle shall be less than a semicircle.
Thus in the figure let ADEF be a great circle, divided by A
and B into two unequal arcs ADEB, AFB, of which ADEB is
greater than a semicircle.
Thus the triangle ABC may be regarded as formed by the
arcs CA, CB, AFB, or by the arcs CA, CB, ADEB.
It is to be understood, when the triangle ABC is spoken of,
T
that it is the first of these, namely, the triangle formed by the
arcs CA, CB, AFB which is meant.
ON CERTAIN PROPEETIES OF SPHERICAL TRIANGLES. 135
24. And if each side of a spherical triangle is less than a
semicircle, it will follow also that each angle of a spherical
triangle will be less than two right angles.
J>
For let a triangle be formed by BC, CA, and BDEA, having
the angle BCA greater than two right angles.
Let E be the point where BC if produced will meet AE.
Then BDE is a semicircle (art. 12).
Therefore BDEA is greater than a semicircle ; that is, the
proposed triangle is one which we have agreed to exclude from
consideration.
25. Polar triangle.
Let ABC be a spherical triangle, and let the points A', B', C'
be those poles of BC, CA, AB which lie on the same sides of those
arcs as the opposite angles A, B, C.
Then the triangle A'B'C' is said
to be the polar triangle of the
triangle ABC.
Since each side of a spherical
triangle has two poles, eight tri-
angles in all may be formed having
for their angular points poles of
the sides of the given triangle, but
in only one of these do the poles
A', B', C' lie towards the same parts
with the corresponding angles A, B, C, and this is the one
which is known as the polar triangle.
The triangle ABC is known as the primitive triangle with
respect to the triangle A'B'C'.
136
SPHERICAL TRIGONOMETRY.
A'
If one triangle be the polar triangle of another, the latter will
be the polar triangle of the former.
Let ABC be a triangle, and A'B'C' its polar triangle.
It is required to show that the
triangle ABC is the polar triangle
of A'B'C'.
First to prove that A is a pole
of B'C'.
Since B' is a pole of AC,
Therefore B'A is a quadrant.
And since C' is a pole of AB,
Therefore C'A is a quadrant.
Hence since AB', AC' are both quadrants, A is a pole of
B'C' (art. 13).
Next to show that A, A' lie upon the same side of B'C'.
Since A' is a pole of BC, and A, A' lie upon the same side of
BC,
Therefore A' A is less than a quadrant.
And since the arc AA', drawn from A, a pole of B'C', to A',
is less than a quadrant, the two points A, A' must lie upon the
same side of B'C'.
Similarly it may be shown that B is a pole of A'C', and
that B, B' are upon the same side of A'C'. Also that C is a
pole of A'B', and that C and C' are on the same side of A'B'.
Thus ABC is the polar triangle of A'B'C'.
26. The sides and angles of the polar triangle are respectively
the supplements of the angles and sides of the primitive triangle.
Let the side B'C' of the polar triangle, produced if neces-
sary, meet the sides AB, AC, pro-
duced if necessary, in D, E.
Then since A is the pole of
B'C', the arc DE measures the
angle A (art. 17).
Because B' is a pole of AC,
Therefore B'E is a quadrant.
And because C' is a pole of AB,
Therefore C'D is a quadrant.
Therefore B'E, C'D are together equal to a semicircle.
A'
ON CERTAIN PROPERTIES OF SPHERICAL TRIANGLES. 137
But the arcs B'E, C'D together make up the arcs B'C' and
DE, that is, the arc B'C' and the angle A.
Therefore the side B'C' and the angle A are supplementary
to each other.
Similarly it may be shown that A'C', A'B' are supplementary
respectively to B, C.
And since ABC is the polar triangle of A'B'C', it follows
that BC, CA, AB are respectively the supplements of the angles
A7, B', C'.
27. From these properties a primitive triangle and its polar
triangle are sometimes spoken of as supplemental triangles.
Thus if A, B, C, a, &, c denote the angles and the sides of
the primitive triangle, and A', B', C', a1 ', &', c' those of the polar
triangle, all expressed in circular measure, we have
A' = 7r-a, B' = jr — fe, C' = 7r-c
28. If a general equation be established between the sides and
the angles o/ a spherical triangle, the supplements of the sides and
angles respectively may be Substituted for the angles and sides
'which are involved in the equation.
For since the equation is general it is true for any triangle,
and it holds, therefore, when a', &', c', A', B', C' (the sides and
angles of the polar triangle), are substituted for a, 6, c, A, B, C
respectively. In the equation as it then stands we may substi-
tute for the sides and angles of the polar triangle their equiva-
lents drawn from the primitive triangle ; thus,
a' = 7r- A, ?/ = 7r-B, &c.
If, then, a general theorem be established connecting the
sides and angles of a spherical triangle, we may substitute for
each side the supplement of the corresponding angle, and for
each angle the supplement of the corresponding side, and the
relation between the sides and angles so obtained will be also
true for all triangles. The great importance of this result will
appear later.
138 SPHERICAL TRIGONOMETRY,
29. Any two sides of a spherical triangle are together greater
than the third.
For any two of the three plane angles
AOB, AOC, BOO, which form the solid
angle at O, are together greater than the
third (Euc. XI. 20).
Therefore any two of the arcs AB, AC,
BC, which measure these angles respec-
tively, are together greater than the third.
From this proposition it is clear that
any side of a spherical triangle is greater
than the difference of the other two.
30. The sum of the three sides of a spherical triangle is less
than the circumference of a great circle.
The sum of the three plane angles which form the solid
angle at 0 is less than four right angles (Euc. XI. 21).
Therefore the sum of the circular measures of these angles is
less than the circular measure of four right angles.
Therefore -=—• + pr-r + ~-r is less than 2?r.
Therefore AB + BC + AC is less than 2-rr x OA ; that is,
the sum of the three arcs is less than the circumference of a
great circle.
31. The three angles of a spherical triangle are together greater
than two and less than six right angles.
Let A, B, C be the angles of a spherical triangle, and let a',
bf, c' be the sides of the polar triangle.
Then by art. 30, a' + V -f c' is less than 2-Tr.
Therefore TT — A + TT — B + TT — Cis less than 2?r ;
Or, TT - (A + B + C) is less than 0.
That is, A -F B + C is greater than TT.
And since each of the angles A, B, C is less than TT, the sum
of A, B, C is less than STT.
32. The angles at the base of an isosceles triangle are equal
to one another.
Let ABC be a spherical triangle having the side AC equal
to the side BC, and let 0 be the centre of the sphere.
ON CERTAIN PEOPERTIES OF SPHERICAL TRIANGLES. 139
At the point A draw a tangent to the arc AC, and let this
meet 00 produced in D.
Join BD.
Then in the triangles AOD,
BOD, the side AO is equal to BO,
OD is common, and the angle AOD C
is equal to the angle BOD, since
these two angles are measured re-
spectively by the equal arcs AC,
CB. Therefore BD = AD, and
OBD = the right angle OAD.
Hence BD touches the arc BC.
At the points A, B draw tan-
gents AT, BT to the arc AB, and join TD.
Then in the two triangles ATD, BTD we have AT = BT,
being tangents drawn to a circle from the same point T, and
AD is equal to BD, and TD common to both triangles.
Therefore the angle TAD is equal to the angle TBD.
And TAD measures the angle BAC, and TBD measures
ABC (art. 16).
33. If two angles of a spherical triangle are equal, the opposite
sides are equal.
In the spherical triangle ABC let the angles A, B be equal.
Therefore in the polar triangle the sides a' ', V are equal.
Hence in the same triangle the angles A', B' are equal (art. 32).
Therefore in the primitive triangle the sides a, b are equal.
34. If one angle of a spherical triangle be greater than another,
the side opposite to the greater angle is greater than the side oppo-
site to the less angle.
Let ABC be a spherical tri-
angle, and let the angle A be
greater than the angle B.
At A make the angle BAD
equal to the angle B.
Then because the angle BAD B'
is equal to the angle ABD, therefore the arc DB is equal to the
arc DA (art. 33).
140 SPHERICAL TRIGONOMETRY.
But the two arcs DA, DC are together greater than AC.
Therefore the two arcs DB, DC are together greater than
the arc AC; that is, the side BC is greater than AC.
35. If one side of a spherical triangle be greater than another ,
the angle opposite to the greater side is greater than the angle
opposite to the less side.
In the spherical triangle ABC let the side BC be greater
than the side AC.
Then shall the angle BAG be greater than the angle ABC.
For if not, the angle BAG must be either equal to or less
than the angle ABC.
If the angle BAG were equal to the angle ABC, then would
the side BC be equal to the side AC (art. 33).
But by hypothesis BC is not equal to AC.
Similarly if the angle BAG were less than the angle ABC,
the side BC would be less than the side AC (art. 34).
But by hypothesis BC is not less than AC.
Therefore the angle BAG must be greater than the angle
ABC.
From these two propositions it follows that in any spherical
triangle A— B, a — b must have the same sign.
36. The following is a summary of the more important pro-
perties of spherical triangles established in this chapter.
1. Each side of a spherical triangle must be less than a
semicircle (art. 23).
2. Each angle must be less than two right angles (art. 24).
3. Any two sides of a spherical triangle are together greater
than the third (art. 29).
4. If two sides of a spherical triangle are equal, the angles
opposite to them are equal, and conversely (arts. 32, 33).
5. The greater side is opposite to the greater angle, and con-
versely (arts. 34, 35).
6. A — B and a — b have the same sign (art. 35).
7. The three sides of a spherical triangle are together less
than four right angles.
8. The three angles of a spherical triangle are greater than
two right angles and less than six right angles (art. 31).
141
CHAPTER III.
ON FORMULAE CONNECTING FUNCTIONS OF THE SIDES AND ANGLES
OF A SPHERICAL TRIANGLE.
37. The sines of the angles of a spherical triangle are pro-
portional to the sines of the opposite sides.
Let ABC be a spherical triangle, and 0 the centre of the
sphere.
Take any point D in OA, and
from D let fall DE perpendicular to
the plane BOC.
From E draw EF, EG perpen-
dicular to OB, OC respectively, and
join DF, DG, OE.
Because DE is perpendicular
to the p]ane BOC, it is at right
angles to every straight line meet-
ing it in that plane (Euc. XI.
def. 3).
Therefore the angles DEG, DEF, DEO are all right angles.
Therefore DF2 = DE2 + EF2 = OD2 - OE2 + OE2 - OF2
= OD2-OF2.
Thus the angle DFO is a right angle.
Similarly it may be shown that the angle DGO is a right
angle.
Now since DF, FE are each at right angles to OB, the angle
DFE represents the inclinations of the planes AOB, BOC, that
is, the angle B of the triangle ABC (art. 16).
Similarly it may be shown that DGE represents the angle C.
Thus DE = DF sin. B = OD sin. c sin. B.
Again, DE = DG sin. C = OD sin. b sin. C.
Therefore sin. c sin. B = sin. b sin. C ;
sin. B sin. C
sin. b sin. c '
Or,
142
SPHERICAL TRIGONOMETRY.
Similarly it may be shown that
sin. B sin. A
sin. b sin. a
sin. A sin. B sin. C
sin. a sin. b sin. c '
Thus
38. To express the cosine of an angle of a spherical triangle in
terms of sines and cosines of the sides.
Let ABC be a spherical triangle, O the centre of the sphere.
Let the tangent at A to the arc AB meet OB produced in D,
A
and let the tangent at A to the arc AC meet OC produced in E.
Join DE.
Then the angle DAE measures the angle A of the triangle,
and the angle DOE measures the side a.
And in the triangles ADE, ODE
DE2=AD2 + AE2-2AD AE cos. A.
DE2=OD2 + OE2-20D . OE cos. a.
Therefore, subtracting each side of the first equation from
the second,
OD2-AD2 + OE2-AE2 + 2AD . AE cos. A
-20D . OE cos. a=0.
But since the angles OAD, OAE are right angles,
0 A2 = OD2 - AD2 = OE2 - AE2 ,
therefore 20A2 + 2AD . AE cos. A— 20D . OE cos. a=0.
Therefore OD . OE cos. a=OA2 + AD . AE cos. A ;
and
cos. a=
OA
OE
OA AE
OD OE
AD
OD
cos. A.
FORMULAE CONNECTING SIDES AND ANGLES. 143
Therefore cos. a=cos. b cos. c-fsin. b sin. c cos. A;
. cos. a — cos. b cos. c
and cos. A= : — - — : .
sin. b sin. c
o- -I i -o cos. b — cos. c cos. a
Similarly cos. B = : :
sin. c sin. a
r* cos. c — cos. a cos. b
and cos. C = r— — ; — = .
sin. a sin. b
Note. — In the construction of the figure given above, it has
been assumed that the tangents AD, AE will meet the radii
OB, 00 respectively, when produced.
That this may be the case it is evident that each of the arcs
AB, AC, which include the angle A, must be less than a quadrant.
It is easy, however, to show that if tne theorem is true when
AB, AC are each less than a quadrant, it is true also when
either one, or both of these arcs, are greater than quadrants.
(1) Let one of the sides which contain the angle A, as AB,
be greater than a quadrant.
B'<"
Produce BA, BC to meet at B'.
Then BAB', BOB' are semicircles (art. 12).
Let AB'=c', CB'^a'.
In the triangle AB'C, since AB', AC are each less than a
quadrant, we have
cos. a' = cos. b cos. c' + sin. b sin. c' cos. B'AC.
And a' = w—a, C' = TT-C, B'AC = 7r-A;
therefore cos. a=cos. b cos. c-t-sin. b sin. c cos. A.
(2) Let both of the sides AB, AC, which contain the angle
A, be greater than quadrants.
-B_—
•c
Produce AB, AC to meet at A', and let A'B=c', A'C=fc'.
144 SPHERICAL TRIGONOMETRY.
Then in the triangle A'BC we have
cos. a=cos. V cos. c' + sin. V sin. c' cos. A'.
And b' — ir — b, C' = TT — c, A'^A;
therefore cos. &=cos. b cos. c + sin. b sin. c cos. A.
39. To express the cosine of a side of a spherical triangle in
terms of functions of the angles.
Let ABC be any spherical triangle, and A'B'C' its polar
triangle, so that A/ = TT — a, a/ = ?r — A, &c.
In the triangle A'B'C' we have
cos. A'=cos- ft'-cos. ?/ cos^' (art a8)<
sin. b sm. c
But A'=7r — a, a'=7r-A, &'=7r-B, c' = 7r- C.
Substituting for A', a', &c. we have
f N cos. (TT — A) — cos. (TT — B) cos. (IT — C) e
COS* ( vr — *" * Ct i ^^ .---.-- . -— L- . • . . — - -i --•- -i— *
sin. (IT— B) sin. (TT— Cj
therefore -cos. a= r003' A-COS' B cos' °
or, cos. a =
sin. B sin. C
cos. A + cos. B cos. C
sin. B sin. C
c. M i 7 cos. B + cos. C cos. A
bimilarly cos. 6 = -
sm. C sm. A
cos. C + cos. A cos. B
cos. c —
sin. A sin. B
40, To show that in a spherical triangle ABC,
cot. a sin. & = cot. A sin. C + cos. b cos. C.
Let ABC be a spherical triangle.
Produce the side CA to D, making AD equal to a quadrant,
and ioin BD.
FOKMUUE CONNECTING SIDES AND ANGLES. 145
Then in the triangle BCD we have, by art. 38,
p^cos. BP— cos. (90° + 6) cos, a
sin. (90° + 6) sin. a
_cos. BD + sin. b cos, a >
cos. b sin. a
therefore cos. BD=cos. b sin. a cos. C— sin. b cos. a.
Again, in the triangle ABD we have
/ Ax cos. BD— cos. 90° cos. c
cos. (TT— A)= : — — — : — ;
sin. 90 sin. c
therefore cos. BD = — cos. A sin. c.
Equating the two values of cos. BD we obtain
. cos. b sin. a cos. C — sin. b cos. a = —cos. A sin. c;
therefore cos. a sin. &=cos. A sin. c + cos. b sin. a cos. C.
Dividing both sides of the equation by sin. a,
Q-J-*-| /»
cot. a sin. 6= cos. A -r-1 — [-cos. b cos. C.
sm. a
But — : — = —7—^- (art. 37), and may therefore be sub-
sin. A sm. a ^
stituted for it on the right hand side of the equation.
Therefore cot. a sin. & = cot. A sin. C + cos. b cos. C.
41. In the formula established in the preceding article it
will be noticed that the four parts involved, namely, a, C, &, A,
j^ are adjacent parts of the triangle ABO,
occurring in the order given as we go
round the triangle. The relation obtained
may be stated in words as follows, and in
this form, perhaps, it is easier to retain
the formula in the memory.
Cotangent of extreme side x sine of other side = cotangent of
extreme angle x sine of other angle + product of cosines of the
two middle parts.
Thus in the formula of art. 40, a is the extreme side, b fall-
ing between the angles A and C ; A is the extreme angle, since
L
146 SPHEEICAL TEIGONOMETKY.
C is included between a and b. By taking in turn each side as
the extreme side, we shall obtain six formulas in all, each in-
volving four parts of the triangle, adjacent to one another.
These formulae are as follows : —
Cot. a sin. ~b = cot. A sin. C + cos. b cos. C,
Cot. a sin, c = cot. A sin. B + cos. c cos. B.
Cot. b sin. a=cos. B sin. C + cos. a cos. C.
Cot. b sin. c=cot. B sin. A + cos. c cos. A.
Cot. c sin. a=cot. C sin. B + cos. a cos. B.
Cot. c sin. 6 = cot. C sin. A + cos. b cos. A.
42. To express the sine, cosine, and tangent of half an angle of
a spherical triangle in terms of the sides.
Since in the triangle ABC
A cos. a— cos. b cos. c , , 00>.
cos. A= : — - — : (art. 38).
sin. b sin. c
f i A i cos. a — cos. b cos. c
therefore 1 — cos . A = 1 — . : — - — :
sin. b sm. c
sin. b sin. c — cos. a + cos. b cos. c
sin. b sin. c
_cos. (b— c) — cos, a m
sin. b sin. c
e • 2A sin. \ (a + b — c) sin. i (a— b + c)
therefore sm.2 — = *-i — - — : — ( — : — •-* ^
2 sm. b sin. c
Let 2s = a + b + c, so that s is half the sum of the sides of the
triangle.
Then a + b — c=2s — 2c=2 (s— c),
. 0 A sm. (s — ?>) sm. (s — c)
therefore sm.2 — = ^ — f—. — ^ +,
Z sm. b sm. c
and sin. |= ^/sin. (8-6) sin, (s-c)
Again,
* -, , cos. a — cos. b cos. c cos. a— cos. (b + c)
1 + COS. A=l+ : — =. : — -^ ^;
sm, b sm. c sin. 6 sm. c
FORMULA CONNECTING SIDES AND ANGLES. 147
therefore ' n.
sm. 6 sin. c
_sin. s sin. (s — a)
sin. 6 sin. c
A /sin. s sin. (s — a)
and cos.— =A/ - : - i - /.
2 V sin-. 5 sin. c
A A
From the expressions for sin. —• and cos. — we obtain by
- 2
division
4.nT1 A_ . /sin, (g-6) sin, (g-c)
77 — \/ • ~s
J V sin. s sin. (s — a)
The positive sign must in each case be given to the radicals
in these equations, because, A being less than two right angles,
j^
-• must be less than 90°. Consequently the sine, cosine, and
tangent of that angle will all be positive.
Proceeding in the same manner, it may be shown that
B /sin. (s — a) sin. (s — c)
m. — = A/ : : £,
v sm. a sm. c
sin
cos
B_ /sin, s sin, (s — b)
v sin. a sin. c
C /sin. (s — a) sin. (s — b)
in. - = A/ -^ ^-; ^ '-•>
2 V sin. & sin. ft
C_ /sin, g sin, (s — cj
2 v sin. a sin. b
cos. —
43. To express the haversine of an angle of a spherical triangle
in terms of functions of the sides.
Let ABC be a spherical triangle.
Then
A cos. a — cos. b cos c • , Qo\
cos. A = — — : — - — : art. 38)o
sin. b sin. c
Therefore 1 - cos. A = 1 - COB, a- COB. 6 cos, c ;
sin. b sin. c
A sin. b sin. c — cos. a + cos. b cos. c .
vers. A= - ;
sin. b sin. c
L 2
148 SPHERICAL TRIGONOMETRY.
sin. b sin. c
2 sin. ^ (a + 6 — c) sin. ^ (a — b + c).
sin. b sin. c
But sin. -J (a + & — c) = Vhav. (& + &— c)
sin. J (&— o + c) = Vhav. (a— 6 + c) /
Vhav. (a + 6 — c) Vhav. (a — b + c)
therefore hav. A = - — ; — \ — —
sm. b sm. c
= \/hav. (a + 6 — c) Vhav. (a— 6 + c) cosec. 6 cosec. c.
44. To express the versine of any side of a spherical triangle
in terms of functions of the other two sides, and of the angle in-
cluded by those sides.
Since in the triangle ABC we have
cos. a— cos. b cos. c
A
cos. A=
therefore 1— cos. A=l —
/
vers. A=
: — - — : -
sin. b sin. c
cos. a — cos. b cos. c
sin. o sin. c
sin. b sin. c — cos. a + cos. b cos. c
sin. b sin. c
cos. (fc~c)^cos. a=sin. b sin. c vers. A,
and —cos. a = —cos. (&~c) + sin. b sin. c vers. A.
Adding unity to both sides we obtain
1 — cos. c&=l — cos. (6~c) + sin. b sin. c vers. A,
or, vers. a = vers. (&~c) + sin. b sin. c vers. A.
45. To express the sine, cosine, and tangent of half a side of a
spherical triangle in terms of functions of the angles.
In the spherical triangle ABC we have
cos. A + cos. B cos. C , , QOA
cos. a= : — ^—. — ^ (art. 6\)) ;
sin. B sin. C
sin. B sin. C— cos. A— cos. B cos. C
therefore I— cos. a= — : — ^ — : — ^ ?
sm. B sm. C
and 2 sin.2 ? = _ cos. A + cos. (B + C)
2 sin. B sin. C
2 cos, i (A + B + C) cos, j (B + C-A).
sin. B sin. C
FORMULAE CONNECTING- SIDES AND ANGLES. 149
Let A + B + C = 2S, so that B + C-A=2 (S— A).
Then sin.' 2 = _COB. S cos (S-A)
sin. B sm. C
and sin. £ = /cos. S cos. (S-A).
2 V gin. B sin. C
In a similar manner it may be shown that
cos. ?. = A /cos. (S-B) cos. (S-0)
2 V sin. B sin. C
tan - = A /_ cos. S cos. (S-A)
V cos. (S-B) cos. (S-C)'
Similar expressing may be obtained for - and -•
u 2i
46. It should be observed that the values obtained in the
preceding article are always real.
For by art. 31 the sum of the three angles of a spherical
triangle lies always between TT and STT, so that cos. S is always
a negative quantity.
Again, since two sides of a spherical triangle are always
greater than the third, we have by the properties of the polar
triangle
7T — B+7T— C > 7T — A.
Therefore B +C - A < TT.
Hence ^ (B + C — A) is less than ^, and consequently
2i
cos. (S— A) is a positive quantity, as also are cos. (S — B),
cos. (S-C).
47. To demonstrate Napier's Analogies.
By art. 37 sm> A = sm' f . Let each of these ratios be
sin. a sin. o
equal to m.
Then we have, by algebra,
sin. A + sin. B /1N
m = — — : — - ...... (1)
sin. a + sin. 0
sin. A — sin. B /ON
m = — : — - (2)
sm. a — sm. b
150 SPHERICAL TRIGONOMETRY.
Now by art. 39
cos. A-r cos. B cos. C = sin. B sin. C cos. a=m sin. C sin. b cos, a.
And cos. B + cos. A cos. C = sin. A sin. C cos. b
= m sin. C sin. a cos. b ;
therefore, by addition,
(cos. A + cos. B) (1 -4- cos. C) = m sin. C sin. (a + b) . . (3)
Therefore by (1) and (3), when the two equations are divided
one by the other, we obtain
sin. A + sin. B __ sin. a + sin. b 1 + cos. C e
cos. A + cos. B sin. (a + 6) sin. C
therefore 2 sin, j (A + B) cos, j (A- B)
2 cos. i (A + B) cos. i (A-B)
/~1
= 2 sin. IQ + fr)cos.iO-?A 2 COS'2 2~
"" 2 sin. i (a + 6) cos. £ (a + 6) Q cQg C_'
and ta,HA + B) = -^|^co,C ... (4)
Similarly from (2) and (3) we obtain
sin. A — sin. B _ sin. a — sin. b 1 + cos. C
cos. A + cos. B sin. (a + b) sin. C
By substituting TT — A for a, &c. in formulae (4) (5), as ex-
plained in art. 28, these formulas become
These four equations (4) (5) (6) (7; are called from their
discoverer Napier's Analogies. Equations (6) (7) may be esta-
blished independently by commencing with the formulas of art. 38,
viz,
cos. a— cos. b cos. c = sin. b sin. c cos. A
cos. b — cos. a cos. c = sin. a sin. c cos. B.
To so deduce them for himself will be a useful exercise for
the student.
FOEMUL^E CONNECTING SIDES AND ANGLES. 151
48, In the relation (4) of the previous article cos. ^ (a — 6)
C
and cot. -- must be positive quantities. It follows, therefore,
that tan. ^ (A + B), cos. ^ (a + b) must necessarily have the
same sign, that is, J (A + B), ^ (a + b) are either both less or
both greater than a right angle. This is expressed by saying
that ^ (A + B) and -J (a + b) are of the same affection.
CHAPTER IV.
ON THE SOLUTION OF OBLIQUE ANGLED SPHERICAL TRIANGLES.
49. THE formulae established in the preceding chapter will
enable us in all cases when three of the six parts of a spherical
triangle are given to determine the other parts. The three parts
given may be either sides or angles. The several cases will be
as follows : —
CASE I.
Three sides of a spherical triangle being given, to solve the
triangle.
By art. 43 we have
Hav. A = cosec. b cosec. c Vhav. (a + b — c) hav. (a — b + c) ;
therefore L hav. A = L cosec. b + L cosec. c
+ i L hav. (a + b ~ c) + J L hav. (a — b ~ c) — 20.
In practice it is unnecessary to write down more than the
mantissa of L cosec. b and L cosec. c. We need not then sub-
tract 20 from the sum of the logarithms.
In the later editions of Inman's i Nautical Tables ' a table
giving the values of half the logarithmic haversines is included,
so that the trouble of dividing the tabular logarithms is avoided.
Having obtained one of the angles, as A, to determine B and
C3 we may if we please make use of the formulae
TT, sin. b . • n sin. c . A
sin. B = ™ sin. A; sin. C = sm. A.
sm. a sin. a
On account, however, of the ambiguities which attend the
use of these formulas (which will be considered later in arts. 52,
152 SPHERICAL TEIGONOMETKY.
53), it is better to determine B and C by the haversine formula,
as in the case of A.
CASE II.
50. Having given two sides of a spherical triangle, and the angle
included by those sides, to determine the other parts.
Let 6, c, A be the parts given.
We shall first determine the side a.
By art. 44
vers. a = vers. (b ~ c) + sin. b sin. c vers. A.
Since sin. b sin. c vers. A is never greater than vers. A, for
sin. b, sin. c cannot either of them be greater than unity, an
angle 9 may always be found such that
vers. 9 = sin. b sin. c vers. A.
So that vers. a = vers. (fr~ c) + vers. 9.
To find 9.
Since vers. 9 = sin. b sin. c vers. A,
therefore hav. 9 = sin. b sin. c hav. A.
By means of the table of logarithmic haversines we may then
find 9.
Thus L hav. 9 = L sin. b + L sin. c + L hav. A — 20.
Then since vers. a = vers. (b ~ c) + vers. 0,
therefore tab, vers. a __ tab, vers. (b ~ c) tab, vers. 9 §
1,000,000 : 1,000,000 1,000,000" ;
and tab. vers. a = tab. vers. (b ~ c) + tab. vers. 9.
Having now the three sides of the triangle, we may proceed
to determine the remaining angles by the haversine method, as
already explained.
51. When two sides and the included angle are given, we
may if we please determine the remaining angles directly from
the data. Thus, if &, c, A are given, we have by Napier's
Analogies (art. 47)
tan.i (B + C) = <™.*(&-e) ^ A;
cos. i (6 + c) 2 '
tan. i (B - C) = Sin-^~c) cot. ^.
sm. i (b + c) 2
SOLUTION OF OBLIQUE ANGLED SPHERICAL TRIANGLES. 153
From these formulas we obtain the values of -J- (B + C) and
^ (B — C), whence by addition and subtraction the values of B,
C will be found.
In practical calculations it is probably better to proceed as.
in art. 50, and first determine the third side, although in solving
examples it will be found a useful exercise to solve the same
triangle by each method.
CASE III.
52. Having given in a spherical triangle two sides and the
angle opposite to one of them, to determine the other parts.
Let a, b, A be the parts given.
n- T-> sin. b . A
Since sin. B = — - sin. A,
sin. a
therefore L sin. B = L sin. 6 -t- L sin. A — L sin. a.
Since the angle B is determined from its sine there will be
two angles less than 180°, corresponding to the tabular logarithm
L sin. B, and it will sometimes happen that there are two tri-
angles having the given parts a, 6, A.
It becomes necessary, therefore, whenever the above formula
is used in obtaining an angle of a spherical triangle, to ascertain
whether the value of B which is required is greater or less than
90°, or whether both values are admissible, so that two triangles
are possible which satisfy the data of the problem.
In the decision of this point we must be guide'd by the con-
sideration that in any spherical triangle the greater angle must be
subtended by the greater side (arts. 34, 35). From this property it
follows that if a be greater than 6, A must be greater than B ;
so that a — 6, A — B must always have the same sign.
For suppose that we have given a = 80°, b = 70°, A = 40° ;
we find from the tables that L sin. B = 9-787702, so that
B = 37° 50' or 142° 10'. Since b is less than a, B must be less
than A, and the second of these values must therefore be re-
jected. Again, let a = 70°, b = 100°, A = 50°. Here L sin. B
= 9-904619, and B = 53° 24' or 126* 36'. Since b is greater
than a, Bmust be greater than A ; a condition which is satisfied
154 SPHERICAL TRIGONOMETRY,
whichever value of B be taken. The case is therefore an am-
biguous one, and two triangles may be constructed having the
given parts.
As will be shown later, the cases in which only one solution
is to be expected are those in which a, the side opposite to the
angle given, lies between b and TT — b.
Having now two sides, and the angles opposite to these sides,
we may proceed to determine the remaining parts c, C by
Napier's Analogies.
Thus, to find c we have, by art. 47,
, £ __ cos. HA + B) fl
while C is given by the formula
tan C cj*y_(^)
2 cos. | (a +6)
53. To show that in a spherical triangle, in which a, &, A are
given, there will be only one solution if d lies between b and TT — b.
Let ADE be a great circle, and at the point A let a second
great circle ACE cut ADE at the given angle A, supposed less
than a right angle, and let the two circles intersect again
inE.
(1) (2)
B I)
Then AE is a semicircle (art. 12).
From AE cut off AC equal to the given side 6, supposed less
than -.
2
Therefore CE is equal to TT— b.
SOLUTION OF OBLIQUE ANGLED SPHERICAL TKI ANGLES. 155
With C as centre, and radius equal to &, describe the arc of
a small circle, cutting ADE in D, so that CD = CA=b.
First let a, the side opposite to the given angle, be less than
&, and consequently less than TT — b also.
With centre C, and radius equal to the given value a,
describe the arc of a small circle.
This will cut the circle ADE in two points, as B, B', between
A and D, as shown in figure (1).
Two triangles will then be formed, viz. ABC, AB'C, each
of which has the given parts a, 5, A, and there is therefore a
double solution.
Next, let a lie between b and TT— b in value.
The same construction being made, the small circle described
with radius equal to a will cut the circle ADE in two points B,
B' as before, but these points will lie on opposite sides of the
point E, as shown in figure (2).
There will therefore be only one triangle, viz. ABC, having
the given parts, since in the triangle AB'C, the value of the
angle at A will be TT— A instead of A.
Note. — In the above investigation we have taken as granted
that the value of any arc of a great circle drawn through C, and
cutting the circle ADE between D and E, is intermediate in
value between CD and CE. The point may be for the present
assumed. At a later stage, when the student has made himself
acquainted with the relations between the sides of right angled
spherical triangles, by dropping a perpendicular from C upon
the circle ADE he may easily establish it for himself.
Moreover, although we have selected for illustration the
particular case in which the side b and the angle A are each less
than — , the statement of limitations is true generally, and may
be shown to hold for all values of b and A.
CASE IV.
54. Having given two angles, and the side opposite to one of
them, to determine the other parts.
Let a, A, B be the parts given.
156 SPHERICAL TRIGONOMETRY.
Then sin. I = ^^ sin. a.
sin. A
The same ambiguities will arise in this case as in the pre-
ceding one, and we must be guided by the same considerations.
And it is to be remembered that when a, A, B are given,
one solution only is possible whenever A lies between B and
7T- B.
To find c and C we make use of Napier's Analogies, as in
art. 52.
CASE V.
55. Having given two angles, and the side included between
them, to find the other parts.
Let A, B, c be the parts given.
We may, if we please, make use of Napier's Analogies to
find a. b.
-.»<«»>- -|.
It is, however, not unusual in this case to resort to the
polar triangle.
Thus, since in the primitive triangle A, B, c are known, we
have in the polar triangle a', b', C', two sides and the included
angle.
From these data we may obtain c', as shown in art. 50 ;
then, having three sides, we may determine A', B' by the
haversine formula used in art. 49.
Thus the six parts of the polar triangle are completely de-
termined, and the supplements of these parts, which are the
elements of the primitive triangle, are therefore known also.
CASE VI.
56. Having given the three angles of a spherical triangle, to
find the three sides.
In art. 45 are established certain expressions for sin. -,
&c., in terms of functions of the angles, which may, if we please
SOLUTION OF OBLIQUE ANGLED SPHERICAL TEIANGLES. 157
be made available. Here again, however, it is preferable to
resort to the polar triangle. Thus A, B, C being known in the
triangle ABC, a', &', c', the three sides of the polar triangle are
known also.
Hence by art. 49 we may determine the three angles
A', B', C', and consequently the supplements of these angles,
the three sides a, &, c.
CHAPTER V.
ON THE SOLUTION OF RIGHT ANGLED SPHERICAL TRIANGLES.
57. IF one angle of a spherical triangle be a right angle,
the triangle may be solved by processes simpler than those given
in the preceding chapter.
In Chapter III. we have established a series of fundamental
formulas connecting functions of the sides and angles of a
spherical triangle, each formula involving four parts of the
triangle.
Let one of the angles, as C, be a right angle.
Then, if we select all those equations which contain C, and
substitute for the sine and other functions of C their actual
values, we shall obtain a series of equations connecting three
parts of the given triangle.
And from these equations, as will be shown, when two parts
of a triangle are given, any other may be determined by the
addition or subtraction of two logarithms.
If, then, C be taken as the right angle, the formulas which
we shall require will be as follows :
Sin. a sin. C = sin. c sin. A ) , ,
W art-
7 . n • -D
sin. b sm. C = sin. c sm. B
Cos. c = cos. a cos. b + sin. a sin. b cos. C (b) art. 38.
Cot. a sin. b = cot. A sin. C + cos. b cos. C
cot. b sin. a = cot. B sin. C + cos. a cos. C
~ . „ -T. > (C) art.
cot. c sm. a = cot. C sm. B + cos. a cos. B
cot. c sin. b = cot. C sin. A + cos. b cos. A
158 SPHERICAL TRIGONOMETRY.
Cos. c sin. A sin. B = cos. C + cos. A cos. B -\
cos. a sin. B sin. C = cos. A + cos. B cos. C >- (d) art. 39.
cos. b sin. A sin. C = cos. B + cos. A cos. C)
If C = 90°, we shall derive the following relations from
those given :
From (a) sin. a = sin. A sin. c
sin. b = sin. B sin. c
(b) cos. c = cos. a cos. b
(c) sin. I = cot. A tan. a
sin. a = cot. B tan. I
cos. B = tan. a cot. c
cos. A = tan. b cot. c
(d) cos. c = cot. A cot. B
cos.
cos.
3. c = cot. A cot. B ^
5. A = cos. a sin. B >
i. B = cos. & sin. A J
Thus we have in all ten formulas, each involving three of
the five parts &, &, c, A, B. And since ten is the total number
of combinations which can be formed by five quantities taken
three together, it follows that when two of these quantities are
given we may always determine the third by one or other of the
above equations.
58. The formulae for right angled triangles may be esta-
blished independently as follows :
Let ABC be a spherical triangle, having a right angle at C,
and let 0 be the centre of the sphere. From any point D in
OA draw DE perpendicular to
OC, and from E draw EF per-
pendicular to OB, and join DF.
Then DE is perpendicular to
EF, because the plane AOC is
perpendicular to the plane BOC
(Euc. XI. def. 4).
And
E2 + OE2 - OF2 = OD2 - OF2.
Therefore the angle OFD is a right angle, so that the angle
SOLUTION OF RIGHT ANGLED SPHERICAL TRIANGLES. 159
EFD represents the inclination of the planes AOB, BOC} that
is, it represents the angle B (art. 16).
DE DE DF
OD = DFOD'°] m.Bsm.c.
By a similar construction it may be shown that
sin. a = sin. A sin. c.
. OF OF OE
Agam' OD = OE OD' °r C°S- ° = C°S- * COS' b' (2>
, EF EFDE ^
OE = DE OE' °r Sm* a = C0t' B tan' b' I (3)
Similarly it may be shown that sin. b = cot. A tan. a. )
A -, EF EFOF
DF = OFDF'°rCOS<I 1(4)
Similarly it may be shown that cos. A = tan. b cot. c.)
Multiplying together the two formulae in (3) we obtain
cot. A cot. B tan. a tan b = sin. a sin. b •
therefore cot. A cot. B = cos. a cos. b = cos c. (5)
Again, by (4)
cos. B = tan. a cot. c
_sin. a cos. c
cos. a sin. c *
But by the second equation of (1) sin, A = Sm' a .
sin. c
sin. A cos. c
Therefore cos. B =
cos. a
and by (2) cos. b = ^-? ;
cos. a
therefore cos. B = sin. A cos. b. *j
In the same way we may show that > (6)
cos. A = sin. B cos. a. . J
59. From the two formulae (3) of art. 58
sin. a = cot. B tan. b ;
sin. b = cot. A tan. a.
Since, a, b are each less than 180°, it follows that
the expressions cot. B tan. b and cot. A tan. a must both be
positive.
160
SPHERICAL TRIGONOMETRY.
And since A, a are each less than 180°, A, a must be either
both less or both greater than 90°; that is, A, a are of the same
affection.
Similarly, B, b are of the same affection.
60. The ten formulae which we have given for right angled
triangles are comprehended under two rules, called, after their
inventor, Napier's Rules of Circular Parts.
These rules may be explained as follows :
Let ABC be a spherical triangle in which 0 is the right
angle.
Then, excluding C, the circular parts are the two sides
including the right angle, and the
complements of the hypothenuse and
of the angles A and B.
These five parts may be ranged
round a circle in the order in which
they occur with respect to the
triangle.
Any one of these parts may be
selected, and may be called the
middle part ; then the two parts next
to it are called the adjacent parts, and the other two parts are
called the opposite parts.
Napier's rules are two in number.
Sine of the middle part = product of tangents of adjacent
parts.
Sine of the middle part = product of cosines of opposite parts.
The occurrence of the vowel i in the words ' sine,' ' middle,'
of the vowel a in ' tangent ' and : adjacent,' and of the vowel o
in the words ( cosine ' and ' opposite ' renders these rules easy to
remember.
61. By taking in detail each of the five parts as the middle
part, and writing down the equations furnished by these two
rules, we shall obtain the same ten equations established in
art. 57.
SOLUTION OF EIGHT ANGLED SPHEEICAL TKIANGLES. 161
Thus, commencing with c,
,Sin.(f-e)=tan.(|-A)tan.(|-B)
.'. cos. c = cot. A cot. B.
Sin.f — — cj = cos. a cos. fr .-. cos. c = cos. a cos. ~b.
Sin/^ — Bj = tan.a tan.f^— cj . • . <jos. B = tan. a cot. c.
Sin.(|-B) = cos. 6 cos. (| -A) .-. cos. B = cos. 6 sin. A.
Sin. a = tan.f - — BJ tan.fr /. sin. a = cot. B tan. fr.
Sin. a = cos.n^ — Aj cos.f^— cj .-. sin. a = sin. A sin. c.
Sin.fr = tan. (^- — A J tan. a .-. sin. 6 = cot. A tan. a.
Sin.fr = cosY^— Bj cos/^— cj .-. sin. fr = sin. B sin. c.
SinY— — AJ = tan. frtan.T^— c 1 .-. cos. A = tan. fr cot. c.
inV^— A J = cos.a cosY — — B J .-. cos. A = cos. asin. B.
62. The method of applying Napier's Rules is shown in the
following example.
Let ABC be a spherical triangle, having the angle C a right
angle, and let the parts given be fr, A. It is required to find
the other parts. Let A be greater than 90°, and fr less than 90°.
First to find a.
Referring to the circle of art. 60, we see that of the three
parts a, fr,^"— A, fr is the middle part, and that it has a, — - — A
2 <->
for its adjacent parts.
f t
Therefore sin. fr=tan. a tan. f 5- — A J, or tan. a = sin. fr tan. A.
Thus L tan. a = L sin. fr + L tan. A — 10.
M
162 SPHERICAL TRIGONOMETRY:
To determine whether a is greater or less than 90°, we
must be guided by the algebraical sign of tan. a, which depends,
of course, upon the sign due to the product which forms the
right-hand side of the equation. We must therefore be careful
to write over each function the appropriate sign, as shown above.
To find c, we may if we please make use of the formulas
connecting a, b, c, now that the value of a is determined.
It is better, however, to solve the triangle completely by
means of the two parts first given, since by making use of the
side a, any error which has been made in calculating a will
vitiate the results for c and B also. Moreover, the labour of
calculating is slightly simplified when the same two parts are
made use of.
7T TT
We have then to consider next the three parts £, — — c, — — A.
LJ 4
7T A • ii • i 11 L 7 7T
Here ——A is the middle part, 6, - — c are the adjacent parts.
2 2t
Hence sinY ^— Aj = tan. 6 tan.f ^ — cj, or
— +
cot. c = cos. A cot. b.
Therefore L cot. c = L cos. A -f L cot. b — 10, and cot. c
being negative, c is greater than 90°, so that the value required
is the supplement of the angle found in the tables.
To find B we have _ — B as the middle part, A and b as
ij 2
the opposite parts.
Then an.Qp- B) = cos.b cos.(~-A\
+ + +
whence cos. B = cos. b sin. A,
and L cos. B = L cos. b + L sin. A — 10.
63. When, as in the example of the preceding article, the
unknown parts of the triangle are determined by means of the
tangent, cotangent, or cosine, there can be no ambiguity, as the
algebraic sign of the function from which the part required is
determined indicates whether the latter is greater or less than 90°.
SOLUTION OF RIGHT ANGLED SPHEEICAL TRIANGLES. 163
But if the required part is determined from its sine, there
will be two values, each less than 180°, which will satisfy the
equation, so that sometimes two triangles may be found possess-
ing the given data.
The five equations in which the unknown part is found from
the sine may be divided into two classes, as follows : —
sin. a . 7 tan. a . -r, cos. A
(1). sm. c = - — r , sin. b = - ~ , sin. B = — ,
sm. A tan. A cos. a
in each of which the two given parts are a side and its opposite
angle.
(2). sin. a = sin. c sin. A, sin. A = Sm' a .
sm. c
In the first class the solution is a double one, as may be
shown.
Thus, let ABC be a spherical triangle in which C is a right
angle.
Then if the sides AB, AC are produced to A', a second
triangle A'BC is obtained, having the side BC in common with
the triangle ABC, and the angle A' equal to the angle A ; but
the side A'B=180° -AB, A'C = 180°-AC, and the angle A'BC
= 180°- ABC.
The equations in the first class afford a double solution, since
in each case the parts given are a, A, which are common to both
triangles ABC, A'BC.
In the second class, however, there is but one solution, and,
as shown in art. 59, a will be greater or less than 90°, according
as A is greater or less than 90°, and conversely.
M 2
164 SPHERICAL TRIGONOMETRY.
CHAPTER VI.
ON THE SOLUTION OF QUADRANTAL SPHERICAL TRIANGLES.
64. In the spherical triangle ABC let one side be a quadrant.
In this case also the fifteen fundamental formulae, established in
Chapter III., may be simplified so as to furnish at once loga-
rithmic expressions for the solution of the triangle similar to
those already obtained for right-angled triangles.
Thus, let us suppose that the side c = 90°.
Then, as before, there will be ten formulas containing the
side c, as follows :
sin. A sin. c = sin. C sin. a 1 ftpfc>
in. a 1
in. b J
sin. B sin. c = sin. C sin.
cos. a = cos. b cos. c + sin. l> sin. c cos. A \
cos. b = cos. a cos. c + sin. a sin. o cos. B > (b) art. 38.
cos. c = cos. a cos. b + sin. a sin. b cos. C )
cot. a sin. c = cot. A sin. B + cos. c cos. B
cot. c sin. a = cot. C sin. B + cos. a cos. B
cot. b sin. c = cot. B sin. A + cos. c cos. A
cot. c sin. b = cot. C sin. A + cos. b cos. A
cos. C = sin. A sin. B cos. c — cos. A cos. B
In these formulas, if we substitute for sin. c and cos. c the
values of sin. 90° and cos. 90°, we shall obtain ten formulas as
follows : —
sin. A = sin. a sin. C ~| / \
sin. B = sin. b sin. C j
cos. a = sin. b cos. A ~j .
cos. b = sin. a cos. B >- (b)
cos. C = — cot. a cot. b )
cot. a = cot. A sin B
cos. a = — tan. B cot C
cot. b = cot. B sin. A
cos. b = — tan. A cot. C
cos. C = — cos. A cos. B
SOLUTION OF QUADEANTAL SPHEEICAL TEIANGLES. 165
65. The ten formulas for quadrantal triangles may, if we
please, be obtained from those established in art. 57 for right-
angled triangles, by making use of the polar triangle.
Thus, let ABC be a spherical triangle having the angle C a
right angle. Then if A'B'C' be the polar triangle to ABC, it
has the side cf a quadrant.
In the triangle ABC, if we take the first three formulae
established in art. 57, we have
sin. a = sin. A sin. c,
sin. b = sin. B sin. c,
cos. c = cos. a cos. b.
And since a = TT — A', b =*. TT — B', &c., the formulas may
be written
sin. (?r — A') = sin. (TT — of) sin. (TT — C') ;
sin. (TT - B') = sin. (TT - &') sin. (TT — C') ;
cos. (TT - C') = cos. (TT - A') cos. (TT - B').
Therefore sin. A' = sin. a! sin. C' ;
sin. B' = sin. V sin. C7 ;
cos. C7 = — cos. A7 cos. B' ;
results which agree with those obtained by other methods in
art. 64.
Proceeding in the same manner, we might deduce the
remainder of the formulas for quadrantal triangles from those
established for right-angled triangles in art. 57.
66. The rules of circular parts explained in the previous
chapter for right-angled triangles may be utilised for quadrantal
triangles under certain modifications.
The rules for quadrantal triangles differ from those for right-
angled triangles in two particulars : —
(1). The elements of the triangle of which we have to take the
complements are the sides a, 6, and the angle C, C being the qua-
drant.
(2). Whenever the two adjacent, or the two opposite parts, are
both sides, or both angles, the sign — must be attached to the
product.
166 SPHEKICAL TRIGONOMETEY.
The circular parts for a triangle ABC when c is a quadrant,
will therefore be arranged as follows :
And by taking each in turn as the middle part, and applying the
two principles —
sine of the middle part = the product of the tangents of the adja-
cent parts ;
sine of the middle part = the product of , the cosines of the opposite
parts,
the ten formulas of art. 64 will be obtained.
67. In the quadrantal, as in the right-angled triangle, an
ambiguity arises when the two parts given are a side and the
opposite angle, so that two solutions are then possible.
In all other cases in which an apparent ambiguity presents
itself, the proper value of the required part may be determined
from the consideration that in a quadrantal triangle an angle and
the side opposite to it must be either both greater or both less
than 90°.
That in a triangle ABC which has the side c a quadrant, A, a,
and B, b must be respectively of the same affection will easily
appear from a consideration of the two formulae (c) of art. 64, viz.
cot. a = cot. A sin. B
cot. 6 = cot. B sin
:!
68. Appended is a collection of exercises, of varying degrees
of difficulty, upon the subject matter of the foregoing chapters.
They have for the most part been taken from the papers of
questions set in the examinations for rank of lieutenant during
the last five vears.
FORMULA OF REFERENCE
167
FORMULAE OF REFERENCE (III.).
In any spherical triangle
sin. A sin. B sin. C
(1) —. - = — : — T-= — -
v J sin. a sin. b sm. c
cos. a —cos. b cos. c
(2)C08'A=
sin. 6 sin. c
cos. A 4- cos. B cos. C
' = sin. B sin. C
(4) cot. a sin. 6=cot. A sin. C-f-cos. b cos. C
(5) sin.| =
(art. o7)
(art'38>
(art. 40)
sm. sin. c
A /sin. « sin. (s — a
(«)«*• 2= V sin. b sin. c
(7) tan. A=
sin. 8 Bin. (a — «)
01
(8; hav. A= Vhav. (a-f b— c) Vhav. (a— 6-fc) cosec. & cosec c
(art. 43)
(9) vers. a=vers. (fc^c)H-sin. b sin. c vers. A (art. 44)
(10) sin.
^
^
2
j*±B?£ <8-A),where2S=A + B +C
sm. B sin. C
cos.(S~B)cos.(S-C)
sin. B sin. 0
a /~~ cos. S cos. (8 -A)
(12) tan. ^ = A/ -- /0 ^^^^ „
2 V cos. S — Bcos.S— C
.sin.
tan. -
. . (art. 47)
168
SPHERICAL TRIG-ONOMETKY
In any spherical triangle having C a right angle
(15) sin. «=sin. A sin. c
sin. fr=sin. B sin. c
cos. c=cos. a cos. b
sin. 6= cot. A tan. a
sin. a=cot. B tan. b
cos. B = tan. a cot. c
cos. A = tan. b cot. c
cos. c=cot. A. cot. B
cos. A=cos. a sin. B
cos. B = cos. b sin. A
In any spherical triangle having c a quadrant
(16) sin. A = sin. a sin. C
sin. B = sin. b sin. C
cos. a=sin. b cos. A
cos. 6= sin. a cos. B
cos. C = — cot. a cot. b
cot. a — cot. A sin. B
cos. a= —tan. B cot. C
cot. 6= cot. B sin. A
cos. b= —tan. A cot. C
cos. C= —cos. A cos. B /
(art. 57)
(art. 64)
169
MISCELLANEOUS EXAMPLES.
1. Show that in any equilateral spherical triangle ABC the
following relations hold :
cos. a
(a) -Cos. A =
1 + cos. a
/? \ c\ Cb • A- i
(b) 2 cos.- sm.-=l.
(c) tan.2-| = l-2 cos. A.
2
(d) Sec. A— sec. a=l.
(e) 1 + 2 cos. a = cot.2-£.
2. Show that in a spherical triangle which has each of its
sides equal to 60° the cosines of each of the angles are equal
to £.
3. If D is the middle point of BC, and AD be joined,
prove that
cos. & + cos. c=2 cos. AD cos. — ,
A
4. If 6 + c=90°, prove that
^
(a) Cos. a=sin. 2 c cos.2 -- .
(6) (Cos. c + sin. c) sin. A = 2 cos.2 ~ sin. (B + C).
2
5. Show that the perpendicular from any vertex of a spheri-
cal triangle upon the opposite side divides the angle and that
side into two parts whose tangents have the same ratio.
170 SPHEKICAL TRIGONOMETRY.
6, If A be one of the base angles of an isosceles spherical
triangle, whose vertical angle is 90£, and a the side opposite,
show that cos. a = cot. A, and determine the limits between
which A must lie.
7. If p. q, r be the perpendiculars from the vertices on the
opposite sides a, b, c respectively, prove that sin. p sin. a=sin. q
sin. 6 = sin. r sin. c
= 2 v' sin. s sm (s — a) sin. (s — 6) sin. (s — c).
8. The sides of a spherical triangle are all quadrants, and
x, y, z the arcs joining any point within the triangle with the
angular points, prove that
cos.2 a? + cos.2 2/ + COS.2 0=1.
9. In a spherical triangle ABC, if the angle B be equal to
the side c, show that
sin. (A — a) = sin. a sin. A cos. B cot. B.
10. If the sum of two sides of a spherical triangle exceeds
the third side by the semicircumference of a great circle, show
that the sine of half the angle contained by these two sides is a
mean proportional between their cotangents.
11. If in a spherical triangle ABC, the angle C is a right
angle, prove that the following relations will hold :
(a) Cos. (a + ft) + cos. (a — b) = 2 cos. c.
(b) If tan. a=2 and tan. 6=1, then tan. c = 3.
(c)
sm.
/ 7\ m o^ c + a , c — a
(d) Tan.2 Q =tan. — ^— tan. — — .
ft U 61
(e) Sin. (c + a) sin. (c — a) = sin.2 ~b cos.2 a.
(/) Sin.' |-«m.»|ooB.»|+co8>-|rin.s|.
MISCELLANEOUS EXAMPLES. 171
B B
(g) Sin. (c — <x)=tan. b cos. c tan. — = sin. 6 cos. a tan. —
2 2
Sin. (a — 6) = sin. a tan.— — sin. 6 tan.-—.
2 2
(0
.
sm.B — sm.
12. In a spherical triangle ABC, if & = c, prove that
(a) cos.— = cot. b sin.- tan. B.
(b) sin.2 B cos.2 ~ sin.2 6 = sin.2 5-sin.2-|.
2i 2
13. In a spherical triangle ABC, having a right angle at C,
if two arcs, a?, ?/, be drawn from C to c, of which x is perpendicu-
lar to the side c, and y bisects it, prove that
(a) cot. x=\/ cot.2 a + cot.2 b.
cos. a + cos. b
(b) cot. y =
A/sin.2 a + sin.2 b
,^ c sin. y
(c) sin. -= — — .
V / O /-« o
« Vl + siri. aj
14. Two great circles of a sphere Pa A, P/?B intersect two
other great circles QBA, Qba in points A, B and a, b : prove that
sin. AQ_sin. Aa sin. Pb
sin. BQ sin. B6 sin. Pa'
15. If E, F are the middle points of the sides AC, AB of a
spherical triangle ABC, and EF produced meets BC produced at
D, prove that
b c
sin. DE cos.- = sin. DF cos.-.
2 2
16. The middle points of the sides AB, AC of a spherical
triangle are joined by the arc of a great circle, which cuts the
base produced towards C at D. Prove that BD + CD=180°,
and that
. ^ . b + c . c — 1) a
cos. AD = sin. -^— sm. — — cosec.-,
AB being the greater of the two sides.
172 SPHERICAL TKIGONOMETEY.
17. If D, E are the middle points of the sides AB, AC of an
equilateral spherical triangle, prove that
BC DE
tan. — - = 1 sin. .
— _
18. If in the spherical triangle ABC, D is the middle point of
BC, and AE is drawn perpendicular to BC, prove that
tan. DE = tan. ~ — 6' tan. L— ^ cot. .
2 a &
19. If in the spherical triangle ABC, AD is drawn perpen-
dicular to BC, and DE perpendicular to AC, cutting AB in F,
prove that
tan. EF
— —— =cos. b tan. A tan. G.
tan. ED
I
20. In the spherical triangle ABC the angle C = 120°, show
that the arc of a great circle drawn through C to meet AB at
/o"
right angles is tan."1^— , if
a
cot.2 a + cot. a cot. 6 + cot.2 &=1.
21. Show that in any spherical triangle ABC,
COB. A + COB. B=i^^> «i».4.
sin. G 2
22. In a spherical triangle, if b -f- c = 60°, show that
Sa
COS. ~n~ A
cos. (b — c) = cos. a + - - tan.2 — .
2 cos. I
23. If in a spherical triangle ABC, a -f b + c = 180°, prove
that
AT) /"I
• 9 J3 , • 9 O -|
Sm'* 2 + Sm" 2 + Sm' 2 = l-
24. If D is the middle point of ^.B, prove that
i. -onT\ A r(T\ sin-2 A — sm-2 B
cot. BCD — cot. ACD = -
sin. A sin. B sin. C*
25. If ABC be a spherical triangle, of which the sides are
MISCELLANEOUS EXAMPLES. 173
each equal to a quadrant, and if R be the pole of the great circle
passing through any two points P and Q, prove that
cos. AR sin. PQ = cos. PB cos. QC — cos. PC cos. QB.
26. If the sides a, b, c of a spherical triangle ABC are in
arithmetical progression, show that
Sin.fsin.C
27, If P be any point in the base AB of an isosceles spherical
triangle ACB, prove that
. COS-OA'
f4
28. If arcs be drawn from the angles of a spherical triangle
to the middle points of the opposite sides, and if a, /3 be the
parts of the arc which bisects the side a, show that
.
sin./3 2'
29. If APB, DPE be two arcs of great circles meeting the
same small circle of the sphere in A, B and D, E, and each
other in P, then
30. A spherical triangle ABC has the angle C = 120°, and
b = 3a. Show that the length of an arc d bisecting C and ter-
minated by the opposite side is given by the equation
tan.c? = - (tan. a + tan. 2a).
PART III.
PEACTICAL TKIGONOMETBY
PLANE AND SPHEBICAL
CHAPTER I.
ON THE METHOD OF USING TABLES OF LOGARITHMS.
1. To find from the tables the logarithm of a given number.
(See Part I., Chapters XIV. and XV.)
Example 1. — Find the logarithms of 7963 and -02795.
Since 7963 is less than 10,000 the mantissa may be taken
at once from the tables, and is found to be -901077.
The characteristic is found by inspection to be 3 (Part L,
art. 99).
Thus the complete logarithm is 3-901077.
To find the logarithm of the decimal fraction -02795 we
treat it as a whole number (Part I., art. 110) and look out the
mantissa of 2795.
This is found to be -446382.
The characteristic is seen to be 2.
Thus log. -02795 is 2-446382.
Example 2.— Find the logarithm of 109872-5.
As before, we first treat 109872-5 as a whole number.
The mantissa of the logarithm of 1098000 is -040602
The proportional parts for 700 are 278
20 „ 079
11 11 i» & ,, lyy
(sum) -04088988
Rejecting the last two figures, and increasing the last figure
retained by unity (Part L, art. 106), we have for the mantissa
of the required logarithm -040890 ; and the characteristic is 5.
Thus
log. 109872-5 is 5-040890.
178 PRACTICAL PLANE TRIGONOMETRY.
Example 3.— Find the logarithm of -00384757.
Regarding the decimal given as a whole number,
The mantissa of 384700 is -585122
The proportional parts for 50 are 056
7 „ 079
(sum) -5851859
and the appropriate characteristic is 3.
We have therefore
log. -00384757 is 3-585186.
Should the number be given in the form of a vulgar fraction
it may be converted into a decimal fraction, and its logarithm
obtained as before.
Or the logarithm of the denominator may be taken from that
of the numerator, and the result will be the logarithm of the
given fraction. The method first given is, however, generally
preferable.
EXAMPLES. — I.
Required the logarithms of the following numbers : —
1. 248. 7. -fr.
2. 1476. 8. ^W.
3. 14-06. 9. -75234.
4. 4196-534. 10. 90^.
5. 121004-15. 11. 1000-
S2 5
6. 49£. 12. -00732596.
2. To find the number corresponding to a given logarithm.
(See Part I., arts. Ill, 112.)
Example 1. — Let the given logarithm be 2-619406 ; it is
required to find the corresponding natural number.
Turning to the tables we find that the given mantissa is
that set down against the number 4163.
We know, therefore, from the characteristic given, viz. 2,
that the number required is 416-3.
Example 2. — Let the given logarithm be 6-150822.
The mantissa given in the tables next below the value given
is -150756, corresponding to the number 1415. Thus we have :
METHOD OF USING TABLES OF LOGAEITHMS. 179
The mantissa of the given
logarithm . . -150822
The mantissa next lower
in the tables . . -150756 corresponding to 1415
(difference) 66^
Prop, parts next lower . . 61 „ „ 2
~~50
3> >J • _\ » »
190
. . 184 „ „ 6
(sum) 1415216
The process might be carried further if necessary.
Since the characteristic is 6, the natural number required
will be 1415216 (Part I., art. 99>
EXAMPLES. — II.
Required the natural numbers corresponding to the following
logarithms : —
1. 2-394452. 7. 2-380211.
2. 6-394452. 8. 5-544068.
3. 2-415671. 9. T841672.
4. -217845. 10. 7-875061.
5. 2-310101. 11. 3-602060.
6. 5-082800. 12. 2;394452.
3. To find the tabular logarithmic sine, cosine, &c. of an angle
given to the nearest second. (See Part I., arts. 115, 117.)
Example 1. — Let it be required to find the tabular loga-
rithmic sine of 31° ir 12".
We have from the tables
L sin. 31° 11' 15" . . . 9-714196
L sin. 31 11 0 ... 9-714144
(difference) . . -QQQQ52
N2
180 PEACTICAL PLANE TRIGONOMETRY.
Let x represent the excess of
L sin. 31° 11' 12" over L sin. 31° II7 0".
Then x : '000052 :: 12" : 15".
Therefore x = -(-000052), or -000042 nearly.
5
Hence L sin. 31° 11' 12" = 9-714144 + -000042
= 9-714186.
If the angle given be in the second quadrant, since
sin. (180° - A) = sin. A, cos. (180° - A) = - cos. A, &c. we
must look out the logarithm of the same function of the supple-
ment of the angle. Thus L sin. 137° = L sin. 43°.
EXAMPLES. — III.
1. Required the tabular logarithmic sines, tangents, and
cosecants of the following angles : —
(1) 10° 10' 6". (3) 48° 35' 35".
(2) 19 10 40. (4) 61 24 40.
2. Find L hav. 61° 9' 53" and L hav, 135° 21' 37".
3. Find L sin. 102° 37' 44" and L hav. 215° 17' 33".
4. To find to the nearest second the angle corresponding to
a given tabular logarithmic sine, cosine, or haversine. (See
Part L, arts. 116, 118.)
Let it be required to find the angle which has for its tabular
logarithmic sine 9-469003.
From the tables
L sin. 17° 7' 30" . . . 9-469022
Lsin. 17 7 15 ... 9-468920
(difference) -000102
Tabular logarithm given 9-469003
L sin. 17° 1' 15" 9-468920
(difference) ~ -000083"
METHOD OF USING TABLES OF LOGAEITHMS. 181
Let x represent the excess of the given angle over
17° r 15".
Then x : 15" :: -000083 : -000102.
Thus x = —15" = 12".
Therefore the required angle is 17° 1' 27".
EXAMPLES. — IV.
1. Find to the nearest second the value of the angle A in
the following cases : —
(1) L sin. A = 9-641452. (3) L sec. A = 10-723465.
(2) L cot. A = 10-200970. (4) L tan. A = 9-488763.
(5). L hav. A = 9-757633.
2. What values of A, less than 360°, have the values given
below for the several tabular logarithms ? —
(1) L sin. A = 9-763520. (3) L tan. A = 10-790490.
(2) L cos. A = 9-840750. (4) L hav. A = 9-632720.
5. To find from the tables the tabular versine of an angle, and
to find the angle corresponding to a given tabular versine. (See
Part L, arts. 124, 125.)
Let it be required to find the tabular versine of 48° 29' 47".
We have from the tables,
Tabular versine of 48° 29' . . . 0337162
Parts for 47" .... 169
Tabular versine of 48° 29' 47" . . 0337331
Example r2, — Find the angle corresponding to the tabular
versine of 0077895.
The given tabular versine is ... 0077895
From the tables tab. vers. 22° 45' is . . 0077799
(difference) ... 96
Looking in the column of c parts for seconds ' headed by
22° 30', we find that 96 corresponds to 52".
The complete value of the angle required is therefore
22° 45' 52".
182 PRACTICAL PLANE TRIGONOMETRY.
EXAMPLES. — V.
1. Find the tabular versines of the following angles : —
(1) 37° 56' 42". (2) 70° 18' 10". (3) 101° 27' 32".
2. Find the angles corresponding to the tabular versines
(1) 0072653. (2) 0987321. (3) 0003721.
Multiplication by logarithms.
6. By Part I., art. 90 the logarithm of a product is equal
to the sum of the logarithms of the several factors ; we have
therefore only to take the logarithm of each number involved,
and add these logarithms together. The sum will be the
logarithm of the product.
EXAMPLES. — VI.
Find by logarithms the products of the following quan-
tities : —
1. 84 x 96.
2. 6 x 4 x 12 x 32.
3. 36 x 48 x 62 x -4.
4. 72 x 96 x 124 x -05.
5. 2-4 x -007 x -54 x -1.
6. 784 x -000079 x -0000036.
Division by logarithms.
7. Since, by Part I., art. 91, the logarithm of a quotient is
equal to the logarithm of the dividend diminished by the
logarithm of the divisor, we have only to write down the
logarithm of the dividend and subtract from it the logarithm of
the divisor. The difference will be the logarithm of the quotient.
Example 1. — Divide 472 by 32-2.
From the tables log. 472 ... 2-673942
„ „ log. 32-2 . . . 1-507856
(difference) log. quotient . . 1-166086
Whence the quotient is 14-658.
METHOD OF USING TABLES OF LOGAKITHMS. 183
Example 2.— Divide -0472 by 3-22.
log. -0472 . . . 2.673942
log. 3-22 . . . -507856
(difference) log. quotient iJ- 166086
Therefore the quotient is -014658.
EXAMPLES. — VII.
1. Find by logarithms the value of
(1) 2004-64 divided by 34.
(2) 19 „ 72.
(3) 1 „ -45.
(4) 1 „ -0004572.
2. Find the value of the fractions
(1) 242 x 559 x 63
781 x 432
(2) 84 x -00769 x -683
598 x -0000146 x -039
Involution and Evolution by Logarithms.
8. By Part I., art. 92 the logarithm of any power, integral
or fractional, of a number is equal to the logarithm of the
number multiplied by the index of the power.
Example 1. — Required the value of (1-05)16.
From the tables log. 1-05 . . . -021189
4
•084756
4
log. (1-05)16 -339024
Thus (1-05)16 = 2-18285.
In the next example the quantity to be raised to a power is
a decimal fraction, so that its logarithm will have the character-
istic negative and the mantissa positive.
The multiplication of the two parts must therefore be
184 PRACTICAL PLANE TRIGONOMETRY.
performed separately, and the required logarithm is the alge-
braical sum of the two products.
Example 2.— Find the value of (-2)10.
log. -2 ... 1-301030, or 1 + '301030
10 10
TO + 3.010300
The algebraical sum of the two is 7-010300.
Thus (-2)10 = -0000001024.
If the index of the power be fractional, the process of
division is substituted for that of multiplication.
Example 3. — Required the value of (1234)*.
From the tables log. 1234 is 3-091315.
3)3-091315
1-030438
Thus the value of (1234)* is 10-726.
Example 4.— Required the value of (-00005214)*.
From the tables log. -00005214 is 5-717171 ;
5)5-717171
1-143434
Therefore (-00005214)* = -13913.
A slight complication arises when the characteristic of the
logarithm to be divided is negative, and does not contain the
divisor an exact number of times. In this case it is customary to
increase the negative characteristic by the least number of units
which will render it divisible by the divisor without remainder,
adding to the positive portion of the logarithm, that is, to the
mantissa, the same number of units, so that the value of the whole
logarithm remains unaltered.
Thus to divide 4-681241 by 3. By adding and subtracting
2 we may write the logarithm in the form
6 + 2-681241, which when divided by 3 is 2-893747.
Example 5.— Find the value of (-00345)*.
Log. -00345 is 3-537819, orlO + 7-537819.
Therefore log. (-00345)* is T-753782,
and the value of (-00345)^ is -56726.
METHOD OF USING TABLES OF LOGARITHMS. 185
If the index of the power be negative it will in general be
easiest to reduce it in the first place to an equivalent expression
with a positive index, by substituting for the given quantity its
reciprocal with the sign changed.
Thus a- = -l;«f'=^ ;?!"=?;
an' tf'b-*> an'
and so on.
Example 6. — Find the value of ('4)~5.
We have by algebra
(1 \5
I — loo*
Ino- 1
- 5 log. -4.
•4 / ~ ° '
O4)5
From the tables
log. 1 . -000000
log. -4 .
. 1-602060
5 log. -4 . 2-010300
5
(difference) 1-989700
5 log. -4
. 2-010300
Thus the value of (-4)-5 =
97-656.
EXAMPLES. — VIII.
Find by logarithms the values of the following quantities : —
1. (12-5)3. 16. (-0125)*.
2. (4-7215)6. 17. V-0093".
3. (1-05)150. 18. (-000048)i
4. (1-0125)200. 19. (19)*.
5. (1-0125)1000. 20. (-096)*.
6. (2)5. 21. (472)*.
7. (-09163)4. (466871)? x fr(3576)16
8. (-975)200. 996003 x v/70077
9. (784)*. 23. (-042)8'3.
10. -v/365". 24. (-00563)'07.
11. ^12345. 25. 3-7.
12. (2)*. 26. 3-».
13. V-093". 27. (-045)-*.
14. (7-0825)*. 1
15. 000125)*. 28. (-2)-4'
186 PKACTICAL PLANE TKIGONOMETKY.
9. To adapt an expression to logarithmic computation.
Let it be required to find by logarithms the value of x in the
equation _ \/ab &c
By making use of the properties of logarithms established in
Part I., Chapter XIV., we may adapt the equation to logarithmic
computation as follows : —
- log. x = - log. a + log. 6 + -- log. c — 2 log. d.
& & 4
EXAMPLES.— IX.
1. Adapt the following equations to logarithmic calculation : — -
(1) x = a*b c d\ (3) x -
2. Find the value of the following expressions :
_2
2
(5)'
3. Find an approximate value of a? in the following equa-
tions : —
(1) x^ = 14. (5) x = (-02445)*.
(2) 32* = 20. (6) x = 200^4769T.
(3) tf =- 004. W^Viis'
(4) x -» = 4 J. (8) £ = A 3/ ?—
V 3-14159
METHOD OF USING TABLES OF LOGAEITHMS. 187
4. Show how x may be determined by means of logarithms
in the equations
(!)«* = 6. V)~=c. (3)a*=
5. Find x in the equations
(1) 43* = -005. (2) (-04)3* = -001.
(3) (-04)^=5.
10. The expression of a trigonometrical formula in logarithms.
In the reduction of an expression involving the trigono-
metrical ratios to logarithms, it must be remembered that the
logarithms of the several ratios given in the tables are increased
by 10. In the final result, therefore, due allowance must be made
for the several tens belonging to the tabular logarithms employed.
Attention must also be given to the algebraic signs of the
different ratios.
Example 1. — Find x in the equation
7-8 sin. 49° cosec. 112°
= tan. 52 .
x2
Multiplying both sides by £2, and transposing, we have
x2 tan. 52° = 7*8 sin. 49° cosec 112°,
+ + +
.-. x2 = 7-8 sin. 49° cosec. 68° cot. 52° ;
that is, 2 log. x = log. 7'8 + L sin. 49° + L cosec. 68°
-f L cot. 52°— 30.
log. 7-8 ...
L sin. 49° ...
L cosec. 68° . .
L cot. 52° ...
(sum)
(difference) 2 log. x .
•892095
. 9-877780
. 10-032834
. 9-892810
30-695519
30-000000
•695519
Therefore log. x = i (-695519) = -347759, and x = 2-227
a
nearly.
188 PEACTICAL PLANE TKIGONOMETEY.
Example 2. — Find x in the equation
120-5cos-1300=73sec.70°tan.*210°.
Therefore 73 x= 120-5 cos. 130° cos. 70° cot.2 210° ;
log. x = log. 120-5 + L cos. 130° + L cos. 70° + 2 L cot. 210
-40 -log. 73.
log. 120-5 . .
L cos. 130°. . .
. 2-080987
. 9-808067
L cos. 70° ...
. 9-534052
2 L cot. 210°. . .
. 20-477122
41-900228
(difference),
log. 73 .
(difference).
40-000000
. 1-900228
. 1-863323
•036905
In this case, as appears from the equation, since cos. 130°
is negative, the value of x must be negative also.
Therefore x = -1-089 nearly.
EXAMPLES. — X.
Find x in the equations
1. x sin. 68° = 117 tan. 48° sec.2 10°.
o cosec. 100° 8-5 sin.2 50°
°
3. v? cot. 109° = 129-6 sin. 73° cosec. 119°.
4 41-3 tan. 200° _ ^^inTBO5
Vx (-75)-2 "
SOLUTION OF EIGHT-ANGLED PLANE TRIANGLES. 189
CHAPTER II.
THE SOLUTION OF RIGHT-ANGLED PLANE TRIANGLES.
(See Part I., Chapter XVII.)
11. Example 1. — In the plane triangle ABC let a-= 117,
b = 201, and C = 90°, required the other parts.
B
(1) To find the angle A.
Since tan. A= -, L tan. A = 10 + log. a— log. b.
10-000000
log. 117 2-068186
(sum) 12-068186
log. 201 2-303196
(difference) L tan. A. 9-764990
Therefore A = 30° 12' 15".
(2) To find the angle B.
Since A + B = 90°, B = 90° -30° 12' 15"= 59° 47' 45".
(3) To find the side c.
Since c = a cosec. A,
log. c = log. a -f L cosec. A — 10.
log. 117 2-068186
L cosec. 30° 12' 15". . 10-298361
(sum)
(difference)
12-366547
10-000000
2-366547
Therefore c = 232-5.
190
PRACTICAL PLANE TRIGONOMETRY,
Example 2. — In the plane triangle ABC, having given
a =-02, c = •!, and B = 90°, find the other
c
a
(1) To find the angle A.
Since tan. A = — ,
c
therefore L tan. A = 10 + log. a — log. c.
10-000000
log. -02 2-301030
(sum) 8-301030
log. -1 1-000000
(difference) L tan. A. 9-301030
Therefore A = 11° 18' 30".
(2) To find the angle C.
Since A + C = 90°, C = 90° -11° 18' 30" = 78° 41' 30".
(3) To find the side b.
Since b = c sec. A,
log. b = log. c + L sec. A — 10.
log. -1 . . .
L sec. 11° 18' 30"
(sum)
1-000000
10-008514
9-008514
10-000000
(difference) log. b . . 1-008514
Therefore b = -102.
Example 3. — In the plane triangle ABC, having given
h = 33, A = 37° 40', and C = 90°, required the other parts.
B
(1) To find the angle B.
Since A + B = 90°, B = 90° -37° 40' = 52° 20'.
SOLUTION OF EIGHT-ANGLED PLANE TRIANGLES. 191
(2) To find the side a.
Since a = b tan. A,
.-. log. a = log. b + L tan. A — 10.
log. 33 1-518514
L tan. 37° 40' ... 9-887594
(sum) 11-406108
10-000000
(difference) log. a . . 1-406113
Therefore a = 25-5.
(3) To find the side c.
Since c = b sec. A,
log. c = log. b + L sec. A — 10.
log. 33 1-518514
L sec. 37° 40' ... 10-101506
(sum) 11-620020
10-000000
(difference) log. c . . 1-620020
and c = 41-7.
EXAMPLES. — XI.
1 . In the plane triangle ABC find the other parts, having given
(1) a = 35-76, I = 45, B = 90°.
(2) a = 384, c = 331, B = 90°.
(3) a = 3555, b = 2354, C = 90°.
(4) b = -2 A = 40°, B = 90°.
(5) c = -04, 0=40°, B =90°.
(6) b = 1777-5, c = 1177, A = 90°.
2. The sides of a rectangular field are 156 feet and 117 feet
respectively ; find the distance between two opposite corners.
3. At a point 153 feet from the foot of a tower the angle of
elevation (Part L, art. 164) of the top of the tower is 57° 19' ;
find the height of the tower.
4. From the top of a cliff, known to be 100 feet above the
sea-level, the angle of depression (Part L, art. 164) of a boat at
sea is 11° 19'. How far is the boat from the foot of the cliff?
5. A ladder 52 feet in length reaches a window 37 feet
from the ground. When the ladder is turned over about its foot
192 PKACTICAL PLANE TRIGONOMETRY.
it reaches a window-sill on the other side of the street 43 feet
from the ground. Find the width of the street.
6. The length of the shadow of a perpendicular stick is 6-25
feet, when the altitude of the sun is 31° 17'. What will be the
length of the shadow of the same stick when the altitude of the
sun is 50° 26' ?
7. A lighthouse bore N.N.B. from a ship at anchor.
After the ship had sailed E.N.E. 17 miles the lighthouse bore
N.N.W. What was the distance of the anchorage from the light-
house ?
8. A column 100 feet high, standing in the middle of a
square court, subtended an angle of 21° at a corner of the
court. Find the length of a side of the court.
9. A certain port, B, is due south of another port, A, dis-
tant 40 miles. At noon a ship left A, steaming S. x° W., 3 knots ;
and at the same time a second ship left B, steaming N. (90 — x)°
W., 4 knots. The two ships met the same evening. At what
time did this occur ?
CHAPTER III.
THE SOLUTION OF OBLIQUE-ANGLED PLANE TRIANGLES.
CASE I.
12. Having given the three sides of a plane triangle, to find
the angles. (See Part. L, arts. 152-154.)
In the plane triangle ABC having given a = 20, b = 30,
c = 40 : find the angles A, B, C.
(1) To find the angle A.
A (s — b) (s — G)
Since hav. A = ^- — ^,
be
therefore L hav. A = 10 + log. (s — b) + log. (s — c) — log. b
— log. c.
a . . 20 s . . 45 s . . 45
b . . 30 b . . 30 c . . 40
c . . 40 ,9 — 6.15 s — c . 5
(sum) 90
— (sum)=s = 45.
SOLUTION OF OBLIQUE-ANGLED PLANE TEIANGLES. 193
10-000000
1-176091
•698970
log. 30 . .
log. 40 . .
(sum)
1-477121
1-602060
11-875061
3-079181
3-079181
log. 15
log. 5
(sum)
(difference) L hav. A 8*795880
.-.. A = 28° 57' 15".
(2) To find the angle B.
As before,
L hav. B = 10 + log. (s — a) -f log. (s — c) — log. a — log. c.
s
. 45
s . .
. 45
a .
. 20
c
. 40
s — a .
. 25
S — G .
. 5
10-000000
. . .
. 1-397940
. -698970
log. 20 .
log. 40 .
(sum)
. 1-301030
. 1-602060
12-096910
2-903090
2-903090
log. 25
log. 5
(sum)
(difference) L hav. B 9-193820
.-.B = 46° 34' 0".
.(3) To find the angle C.
The angle C may now be obtained by subtracting the sum
of A and B from 180°.
Thus C = 180° - 28° 57' 15" - 46° 34' 0" = 104° 28' 45".
Or C may be obtained by a third application of the haver-
sine formula.
Thus
L hav. C = 10 -f log. (s — a) + log. (s — 6) — log. a — log. I
s ... 45 s . . '. 45
a 20 b 30
s — a
25
s-b .
15
10-000000
1-397940
1-176091
log. 20 .
log. 30 .
(sum)
. 1-301030
. 1-477121
12-574031
2-778151
. 2-778151
194 PRACTICAL PLANE TRIGONOMETRY.
log. 25
log. 15
(sum)
(difference) L hav. C 9795880
.-. C = 104° 28' 45".
Note. — In finding the second angle B, we may, if we please,
make use of the relation
T> b • A
sm. n = - sin. A.
a
In that case, however, an error made in calculating the angle A
will affect the result of B also. It is preferable, therefore, at
the expense of slightly increased labour in computation, to
obtain B directly from the original data.
EXAMPLES. — XII.
In the plane triangle ABC find the angles A, B, C, having
given
1. a = 798, l> = 460, c = 654.
2. a = 512, b =r 627, c = 430.
3. a = 649, b = 586, c = 757.
4. a = 627, b = 1140, c= 718-9.
5. a =-025, 6 = -125, c= '115.
6. a= -8, 6 = -672, c= "275.
7. a == -5, ft = -75, c = 1-013.
8. a = -25, ft = -541, c = -674.
CASE II.
13. In a plane triangle, having given tivo angles and one side,
to find the other parts. (See Part. I., arts. 155, 156.)
In the plane triangle ABC, having given A = 70° 36',
B = 57° 19', and c = 37, required the other parts.
SOLUTION OF OBLIQUE-ANGLED PLANE TKIANGLES. 195
(1) To find the angle C.
Since A + B 4- C = 180°, C = 180° - 70° 36' - 57° 19'
= 52° 5'.
(2) To find the side a.
Since - = _ _L_ , log. a — log. c + L sin A + L cosec. C — 20.
log. 37 ..... 1-568202
L sin. 70° 36' . . . 9-974614
L cosec. 52° 5' 10-102975
(sum) 21-645791
20-000000
(difference) log. a 1-645791
a =45-2.
(3) To find the side b.
Since - = - -~-, log. b = log. c -f- L sin. B 4- L cosec. C — 20.
c sin. O
log. 37 1-568202
L sin. 57° 19' . . . 9-925141
L cosec. 52° 5' 10-102975
(sum) 21-596318
20-000000
(difference) log. b 1-596318
.-. 6=39-5.
CASE III.
14. In a plane triangle, having given two sides and the angle
opposite to one of them, to find the other parts. (See Part L,
arts. 157-160).
Example 1. — In the plane triangle ABC, having given
b = 56, c = 38, and B = 101° 20', required the other parts.
02
196 PRACTICAL PLANE TRIGONOMETRY.
(1) To find the angle C.
Q. sin. C c
sin. B 6'
therefore
L sin. C = L sin. B + log. c — log. b
L sin. 101° 20' . . . 9-991448
log. 38 1-579784
(sum) 11-571232
log. 56 1-748188
(difference) L sin C . . 9-823044
.-. C =41°42/30//.
(2) To find the angle A.
As before,
A = 180° - 101° 20' - 41° 42' 30" = 363 57'' 30".
(3) To find the side a.
c,. a sin A
Since - = -7— — ,
b sm B
therefore log. a = log. b -f L sin. A + L cosec. B — 20.
log. 56 1-748188
L sin. 36° 57r 30" . 9-779044
L cosec. 101° 20' . 10-008552
(sum) 21-535784
20-000000
(difference) log. a . . 1-535734
.-. a = 34-3.
In the preceding example, since the angle given was that
opposite to the greater side, the parts which remained to be
found were determined without ambiguity.
The example which follows will illustrate the ambiguity
discussed in Part. I., arts. 158-160.
SOLUTION OF OBLIQUE-ANGLED PLANE TRIANGLES. 197
Example 2. — In the plane triangle ABC, having given
b = 73, c = 101, and B = 39° 33', required the angle C.
L sin. C = L sin. B + log. c — log. b.
L sin. 39° 33' . . 9-803970
log 101 .... 2-004321
(sum) 11-808291
log. 73 .... 1-863323
(difference) L sin. C 9-944968
.-. 0 = 61° 45' 45", as in the
triangle ACB ; or 118° 14' 15", as -
in the triangle AC'B.
The remaining angles A, A' and the remaining sides a, of of
the triangles ABC, A'BC may be found as in Example 1.
EXAMPLES. — XIII.
1. In the plane triangle ABC, find the other parts, having
given
(1) a = 214, b = 191, A = 41° 19' 15".
(2) a = 17-25, c = 10-75, A = 47° 0' 30".
(3) a = 96, c = 48, A = 101° 41'.
(4) a = 2-75, A = 43° 24' 15'.', B = 48° 33' 15".
(5) a = -5, I = -75, B = 45° 10'.
(6) c = 376, A = 48° 3', B = 40° 14'.
(7) a = 242, A = 60°, B = 72°.
(8) a = 178-3, b = 145, B = 41° 10'.
(9) a = 2597-84, b = 3084-33, A = 56° 12' 45".
2. A fort bears from a ship at anchor S. 75° E., and is 2-35
miles distant ; and a lighthouse bears S. 5° W. from the same
ship. If the distance of the fort from the lighthouse be 3-25
miles, find the bearing of the one from the other.
3. From a ship steering W. by S. a beacon bore N.N.W.,
and after the ship had sailed 1 2 miles farther the bearing of the
beacon was N.B. by E. : at what distance had the ship passed
the beacon ?
198 PEACTICAL PLANE TKIGONOMETKY.
4. A boat is sent out from a ship with orders to row S. by E.
until a rock situated 12 miles E.S.E. of the ship bears N.E. by N.
How long will the boat take to reach the required position,
rowing at the rate of 5 miles an hour ?
5. A ship which can sail within seven points of the wind
wishes to reach the mouth of a river 25 miles N.E. of her. The
wind being N.N.E. she starts on the port tack, sailing 7'5
knots : after what interval should she go about ?
6. Two forts, A and B, were four miles apart, and A bore
from B, W. by N. A ship ordered to run in with A bearing
due north until B bore E.N.E., anchored by mistake when B
bore N.E. : how far was she then from her required position ?
CASE IV.
15. In a plane triangle, having given two sides and the included
angle, to find the other parts. (See Part I., art. 161.)
In the plane triangle ABC let a = 798, b = 460, and C
= 55° 2' 1 5", it is required to find the remaining parts A, B,
and c. ,
(1) To find the angles A and B.
By Part I, art. 140,
A-B a-b . C
A _ 13 n
therefore L tan. - - = log. (a— b) + L cot. -—log. (a + b).
2i 2
a . . 798 A + B + C. . 180° 0' 0"
b . 460 C. 55 2 30
a + b .1258 (difference) A + B 12457 30
a-b
.-. = 2731 15
a
SOLUTION OF OBLIQUE-ANGLED PLANE TRIANGLES. 199
log. 338 2-528917
L cot. 27° 31' 15" . . 10-283138
(sum) 12-812055
log. 1258 .... 3-099681
(difference) L tan. Az^ 9-712374
• A^? - 27° 16' 45"
and = 62 28 45
(sum) A = 89 45 30
(difference) B = 35 12 0
(2) To find the side c.
0- c sin. C
Since -= -, - -,
a sin. A
therefore log. c = log. & + L sin. C + L cosec. A— 20.
log, a ..... 2-902003
L sin. C ..... 9-913563
L cosec. A ..... 10-000004
(sum) 22-815570
20-000000
(difference) log. c . . 2-815570
c = 654.
EXAMPLES. — XIV.
1. In the plane triangle ABC find the other parts, having
given
(1) b = 64, c = 70, A - 66° 20r 30".
(2) a = 512, b = 627, C = 42° 537 45".
(3) b = 54, c = 79, A = 105° 27X 30".
(4) a =-036, b =-027, C = 75° 16r 30".
2. Two ships leave a harbour at noon ; one sails N.E. by N.,
7 knots, the other E. by S., 9 knots. How far will they be apart
at midnight?
200 PRACTICAL PLANE TRIGONOMETRY.
3. From a certain station a fort, A, bore N., and a second
fort, B, N.E. by E. Guns are fired simultaneously from the two
forts, and are heard at the station in 1-5 seconds and 2 seconds
respectively. Assuming that sound travels at the rate of 1,100
feet per second, find the distance of the two forts apart.
4. A boat ds anchored half a mile from one end of a break-
water, and three-quarters of a mile from the other end, and an
observer with a sextant finds that the breakwater subtends an
angle of 50°. Find the length of the breakwater.
5. A ship was moored 7 miles N.N.E. of a lighthouse, and a
boat, breaking adrift from the ship, was picked up four hours
after 4 miles due east of the lighthouse. Find the direction
and rate of the current bv which the boat was set.
CHAPTER IV.
AREAS OF PLANE TRIANGLES.
CASE I.
16. In a plane triangle , having given two sides and the included
angle, to find the area. (See Part I., Art. 162.)
In the plane triangle ABC let a — 798 feet, b = 460 feet,
and C = 55° 2' 15", it is required to find the area.
Since area = - al) sin. C,
therefore log. area = log. a + log. fr + L sin. C — log. 2 — 10.
log. 798 ... 2-902003 log. 2 • -301030
log. 460 ... 2-662758 10-000000
Lsin. C. 9-913563
(sum) 10-301030
(sum) 15-478324
10-301030
(difference) log. area 5-177294
Therefore area = 150416 square feet nearly.
The work may be slightly simplified by dividing one of the
sides by 2 in the first instance.
AREAS OF PLANE TKIANGLES 201
ThUS,
area = \ 798 x 460 sin. 55° 2' 15".
or, = 399 x 460 sin. 55° 2' 15".
So that
log. area = log. 399 + log. 460 + L sin. 55° 2' 15" - 10.
EXAMPLES. — XV.
1. In the plane triangle ABC find the area, having given
(1) a = 245 yards, b = 760 yards, C = 60°.
(2) b = 53 feet, c = 91 feet, A = 71° 36' 30".
(3) a = 78 feet, b = 101 feet, C = 109 27' 30".
(4) a =1-23 feet, b =-97 feet, 0= 81° 40' 0".
(5) a = 103 feet, c = 76 feet, A = 95° 37' 15".
2. An enclosure has the form of a parallelogram, the sides of
which are 1 £ miles and 7 furlongs respectively, and the smaller
of the angles included by the sides is 50°. Calculate its area in
acres.
3. An isosceles triangle, the vertical angle of which is 78°,
contains 100 square yards. Find the lengths of the equal sides
of the triangle.
4. The area of a triangle is 763 square feet, and the sides
which contain the smallest angle are 51 feet and 65 feet re-
spectively. Find the angles of the triangle.
5. The base of an isosceles triangle is 150 feet, and the
vertical angle 100°. Find the side of the equilateral triangle
which has an area equal to that of the given triangle.
CASE II.
17. Three sides of a plane triangle being given, to find the
area. (See Part I. art. 163.)
In the plane triangle ABC let a = 711 feet, b = 681 feet,
o = 327 feet, to find the area.
By Part L, art. 142
area = \A' (s — a) (s — b) (s — c).
202 PEACTICAL PLANE TRIGONOMETRY.
Therefore
log. area = J {log. s + log. (s — a) -flog, (s — 6)4- log. (s — c)},
a = 711 s . 859-5 s . 859-5 s . 859-5
I = 681 a. 711-0 b . 681-0 c . 327-0
C = 327 s-a . 148-5 *-6 . 178-5 s-c
1719
s = 859-5
log. 859-5 .... 2-934246
log. 148-5 .... 2-171726
log. 178-5 .... 2-251638
log. 532-5 .... 2-726320
(sum) 2)10-083930
5-041965
Therefore area =110145 square feet.
EXAMPLES. — XVI.
1. In the plane triangle ABC find the area, having given
(1) a = 101-4 feet, b = 76-5- feet, c = 91-3 feet.
(2) a*= 1721 feet, b = 1946 feet, c = 2030 feet,
2. Having given a = -08 feet, b = -12 feet, c = -15 feet;
find the number of square inches which the triangle contains.
3. Express in acres the area of a triangular field, the
sides of which are 125 yards, 143 yards, and 159 yards respec-
tively.
4. In a quadrilateral figure ABCD, AB = 90 yards, BC
= 100 yards, CD = 110 yards, DA=120 yards, and BD = 178'8
yards. Find the area of the figure.
SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. 203
CHAPTER V.
THE SOLUTION OF OBLIQUE-ANGLED SPHERICAL
TRIANGLES.
CASE I.
18. Three sides of a spherical triangle being given, to find the
angles. (See Part II., art. 49.)
In the spherical triangle ABC let a = 124° 10', b =
89° 0' 15", c = 108° 40', it is required to find the angle A.
To find the angle A.
By Part II., art. 43
hav. A = cosec. b cosec. c \/ hav. (a + b ~ c) hav. (a— b ~ c).
It has been explained in Part. II. that by taking the loga-
rithms of cosec. b and cosec. c, instead of the tabular logarithms,
we avoid the necessity of subtracting 20 in the final result.
Therefore
L hav. A = log. cosec. b -f log. cosec. c -f ^ L hav. (a -f b ~ c)
hav. (a — b ~ c).
c . . 108° 40' 0" log. cosec. c . . -023468
b . . 89 0 15 log. cosec. b . . -000066
c-b . . ~19 39 45 i L nav. (a + c-V) . 4-978001
a . . 124 10 0 i L hav. (a— c — b) . 4-898006
143 49~45~ (sum) L hav. A . 9-899541
104 30 15
Therefore A = 125° 56' 30".
By the same process we may find the angles B and C from
the formulas.
L hav. B = log. cosec. a + log. cosec. c + -J L hav. (b + a ~ c)
+ J L hav. (b — a~ c).
L hav. C = log. cosec. a+log. cosec. 6 + -J L hav. (c + a ~ b)
-f -J L hav. (c — a ~ b).
204 PRACTICAL SPHERICAL TRIGONOMETRY.
Note. — The formula
-o sin. b .
sm. B = . sin. A
sm. a
might be used in finding B. Its use, however, is open to two
objections : the first, that it involves the angle A, which is not
one of the original data of the question ; the second, that, as
the angle is determined from its sine, we shall have to consider
whether the angle first taken from the tables, or the supplement
of that angle, is the value required, whereas by the haversine
formula B is determined without ambiguity. In this case,
therefore, as in similar instances which will be met with later,
it is recommended that the use of the sine formula should in
general be avoided.
EXAMPLES.— XVII.
In the spherical triangle ABC find the angles, having
given
1. a = 49° 10' 0", b = 58° 25' 0", c = 56° 42' 0".
2. a = 119° 42' 30", ft = 108° 4' 15", c = 68° 53' 45".
3. a = 87° 10'' 15", ft = 62° 36' 45", c = 100° 10' 15".
4. a •= 156° 10' 30", ft = 137° 3' 45", c = 47° 57' 15".
5. a = 40° 31' 15", ft = 50° 30' 30", c = 61° 5' 0".
CASE II.
19. Having given two sides of a spherical triangle, and the
angle included by those sides, to find the other parts. (See Part II.,
art. 50.)
In the spherical triangle ABC, having given ft = 108° 4' 15",
c = 119° 42' 15", and A = 75° 31' 30", it is required to find
the other parts.
To find the side a
Since vers. a — vers. (ft ~ c) + sin. ft sin. c. vers. A,
tab. vers. a = tab. vers. (ft ~ c) -f tab. vers. #, where
Lhav. 0 = L sin. ft + L sin. c + L hav. A — 20.
SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. 205
To find 0.
L sin. b . . .
L sin. c . . .
L hav. A. . .
(sum)
9-978031
9-938818
9-574056
c . 119°
I .108
42' 15''
4 15
11
38 0
29-490905
20-000000
(difference) L hav. 6 9-490905
and 6 = 67° 37' 30".
tab. vers. 67° 37' . . . 619199
parts for 30" .... 134
tab. vers. 11° 38' . 20542
(sum) tab. vers. a 639875
Therefore a = 68° 53' 32".
Having now the three sides a, b, c, we may find the angles
B, C by the method given under case I.
Or the angles B and C may be found directly from the data
by means of Napier's Analogies. (Part II., art. 51.)
Since by Part II., art. 47
sin.
therefore
L tan. i (C + B) = L cos. J (c — 1) + L sec. i (c + &)
+ Lcot. £-20.
6
L tan. i (C — B) = L sin. £ (c — &) + L cosec. J (c + fc)
+ Lcot. ^-20.
Thus, in the triangle given,
c ...... 119° 42' 15"
& 108 4 15
r + 6 ..... 227 46 30
4(c + &) • • • • H3 53 15
c-b 11 38 0
ifc-^ . 5 49 0
206 PEACTICAL SPHERICAL TRIGONOMETRY.
L
cos. -J (c — b) .
9-
997758
L
sin.
i (c-&) •
9-
005805
L
sec. i (c + fr) .
10-
392607
L
cosec. ^ (c + fr)
10-
038891
L
cotf . . .
10-
110905
L
cot.
A
2
10-
110905
(sum)
30-
501270
(sum)
29-
155601
20-
000000
20-
000000
(difference) ) 10.501270 (difference) )
L tan. J (0+.B).} * Ltan. 1 (C-B) |j
therefore J (0 + B) = 107° 30' 0" . 1 (C-B) = 8° 8' 30".
£(C + B) = 107° 30' 0"
Q-B = 8 8 30
C = 115 38 30
B = 99 21 30
It will be noticed that 107° 30', and not 72° 30', is taken as
the value of ^ (C + B). For we know that -J (C + B) must be
greater than 90°, because, as explained in Part II. art. 48,
J (C + B) and ^ (c 4- V) must be of the same affection, that is,
must be both less or both greater than 90° ; we therefore
select the greater of the two values.
EXAMPLES.— XVIII.
1. In the spherical triangle ABC find the third side,
having given
(1) A = 96° 32' 0", b = 76° 42' 0", c = 89° 10' 30".
(2) A = 50° 0' 0", I = 70° 45r 15", c = 62° 10r 15".
(3) a = 100° 8r 45", b = 98° 10' 0", C = 88° 24r 30".
(4) 6 = 118° 2r 15", c = 120° 18X 30", A = 27° 22X 30".
(5) a = 87° 10' 15", b = 62° 36X 45", C = 102° 58X 30".
(6) a= 69° 19r 15", 5 = 78° 59X 15", C = 110° 48r 45".
2. In the spherical triangle ABC find directly by Napier's
Analogies the remaining angles of the triangle, having given
(1) A = 85° 31X 15", b = 49° 36' 0", c = 100° 17r 30".
(2) a = 109° 15r 30", b = 93° 267 30", C = 53° 21r 30".
SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. 207
CASE III.
20, Having given two sides of' a spherical triangle, and the
angle opposite to one of them, to find the other parts. (See Part
II., arts. 52, 53.)
In the spherical triangle ABC let a = 80° 5' 0", b = 70° 10' 30",
and A = 33° 15'; required the other parts.
(1) To find the angle B.
oc -r> sin. b • A
Since sin. r> = - - sin. A.
sin. a
therefore L sin. B = L sin. A + L sin. b + L cosec. a — 20.
L sin. 33° 15' 0" . . . 9-739013
Lsin. 70° 10' 30" ... 9-973466
L cosec. 80° 5' 0" 10-006538
(sum) 29-719017
20-000000
(difference) L sin. B . . 9-719017
There are two angles less than 180°, which correspond to
the given logarithm, viz. 31° 34' 30" and 148° 25' 30", and we
have to determine which of the two is the angle sought, or
whether both values are admissible.
Since b is less than a, the angle B must be less than A
(Part II., art. 35), that is, less than 33° 15'.
The value 148° 25' 30" is therefore plainly inadmissible,
and 31° 34' 30" is the only value of B.
(2) To find the side c.
We have now the two sides a, b, and the angles opposite
to these sides, A, B. Hence, by Napier's Analogies,
c cos. i (A + B") .
therefore L tan. 1 = L cos. ^ (A -f B) + L sec. f (A — B)
2
+ L tan. J (a -f 6) - 20.
208
PEACTICAL SPHEKICAL TRIGONOMETRY.
A
B
A
i<
A
*(
+ B . .
;A + B) .
33° 15
31 34
0"
30
a
b . .
80°
70
5'
10
0"
30
64 49
32 24
30
45
a + b . 150
i (a 4- b) 75
. . 9-926451
.. . 10-000046
. . 10-575879
15
7
30
45
-B. .
A-B).
L cos.
L sec.
L tan.
1 40
0 50
KA +
HA-
(sum)
30
15
B) .
B) .
b) .
30-
20-
502376
000000
(difference) L tan. -
2i
10-502376
Therefore
and
1 = 72° 32' 30",
c =145° 5' 0".
(3) To find the angle C.
Having now the three sides a, &, c, we may find C as in
Case I., art. 18.
Thus L hav. C = log. cosec. a + log. cosec. b +
J L hav. (c + a — b) + \
a ... 80° 5' 0"
b ... 70 10 30
a — b . .
c .
c+a — b .
9 54 30
145 5 0
L hav. (c — a — b).
log. cosec. a.
log. cosec. b .
i L hav.
c — a — b
154° 59' 30"
135° 10' 30"
therefore 0 = 161° 2
^L hav. (c — a — b)
(sum) L hav. C .
•006538
•026534
4-989581
^965902
9-988555
0".
21. In the preceding example, as we have seen, one triangle
only was possible for the given data. That which follows
supplies an illustration of the ambiguities discussed in articles
52 and 53 of Part. II.
In the spherical triangle ABC having given a = 50° 45X 15",
b = 69° 12' 45", A = 44° 22r 15", find the angle B.
SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. 209
As before,
L sin. B = L sin. A + L sin. b + L cosec. a — 20.
L sin. 44° 22' 15" ... 9-844633
L sm. 69 12 45 ... 9-970767
L cosec. 50 45 15 . , . 10-111013
(sum) 29-926413
20-000000
(difference) L sin. B . . 9-926413
The two angles less than 180° which correspond to the
tabular logarithm 9-926413 are 57° 34' 45" and 122° 25' 15".
From the data we see that the angle B must be greater than the
angle A, a condition which is satisfied by either of the values
obtained, and two triangles are therefore possible.
In the example given b is 69° 12' 45", and 180°— b is there-
fore 110° 47' 15." Since a, the side opposite to the given
angle, is in this case 50° 45' 15", and therefore does not lie
between those values, it might have been anticipated that two
solutions would be obtained. (See Part II., art. 53.)
Having now in each of the two triangles two sides, and the
angles respectively opposite to those sides, the remaining parts
in each triangle may be determined by the processes of the
preceding article.
EXAMPLES. — XIX.
1. In the spherical triangle ABC find the angle B, having
given
(1) a = 119° 21' 0", b = 50° 26' 0", A = 108° 35' 30".
(2) a = 58° 27' 30", b = 117° 46' 30", A = 35° 17' 15".
2. In the spherical triangle ABC find the other parts, having
given
a = 87° 36' 0", b = 45° 31' 30", A = 75° 27' 30".
CASE IV.
22. Having given two angles of a spherical triangle, and a
side opposite to one of them, to find the other parts. (See Part II.,
art., 54.)
In the spherical triangle ABC let A = 75° 31' 0",
B = 110° 16' 45", a = 83° 13' 30".
210 PEACTICAL SPHERICAL TRIGONOMETRY.
To find the side b.
o- • 7 sin. B .
bmce sin. b = sin. a,
sin. A
therefore L sin. b = L sin. B + L sin. a + L cosec. A — 20
L sin. 110° 16' 45" . . . 9-972210
Lsin. 83 13 30 ... 9-996957
L cosec. 75 31 0 ... 1Q-Q14Q26
(sum) 29-983193
20-000000
(difference) L sin. 6 . . 9-983193
Therefore b = 74° 9' 45", or 105° 50' 15". And since b must
in this case be greater than a, which is 83° 13' 30", the second
of these values is clearly the only one admissible; that is,
I = 105° 50' 15".
By a test similar to that applied in Case III. we may ascer-
tain whether the given conditions may be expected to furnish
one result or two by considering whether the angle A, opposite
to the given side, falls between B and 180°— B. In the present
instance these angles are 110° 16' 45" and 69° 43' 15". The
value of A, 75° 31' 0", lies between these limits, and consequently
only one value of a was to be looked for.
We have now in the triangle ABC two sides and the angles
opposite to those sides ; the remaining parts of the triangle may
therefore be determined by the methods explained under Case III.
EXAMPLES. — XX.
In the spherical triangle ABC find the side fr, having given
(1) a = 81° 37' 30", A = 78° 29' 0", B = 75° 49' 30''.
(2) a = 105° 30' 0", A = 99° 36' 30", B = 86° 25' 15".
CASE V.
23. Having given two angles of a spherical triangle and the
included side, to find the other parts. (See Part II., art. 55.)
In the spherical triangle ABC let A = 113° 33' 30", B
=-. 51C 30' 30", c = 60° 18' 0"; it is required to find the other
parts.
SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. 211
To find the angle C.
Let A'B'C' be the polar triangle to ABC.
Then A + a! = 180°, B + V = 180°, c + V = 180° (Part II.,
art. 26).
A . 113° 33' 30" B . 51° 30' 30" c . 60° 18' 0"
180° 180° 180°_
(difference) af . 66° 26' 30" V . 128° 29' 30" C' . 119° 42' 0"
Hence in the triangle A'B'C', having given two sides and
the included angle, we may find the third side, as in Case II.
L sin. a! . . . 9-962205 V . . . 128° 29' 30"
L sin. V . . . 9-893595 a' . . . 66 26 30
Lhav.C'. . . 9-873744 fe'_a' . 62 3 0
29-729544
20-000000
(difference) L hav. 6 . . 9-729544
therefore 0 = 94° 11' 0"
tab. vers. (9 ... 1072948
tab. vers. (&' - a') . . . _531_299
tab. vers. c' . . . 1604247
therefore c' . . . 127° 10' 29"
180 0 0
(difference) C . . . 52° 49' 31"
To obtain the remaining two sides, a, 6, of the triangle ABC,
we first determine the angles A', B' of the polar triangle, as in
Case I. Thus :
. . 128 29 30 log. cosec. c' . -098654
. . 127° 10' 30" log. cosec. j/_ . -106405
1 19 0 jLhav. (a' + 6'^V) 4-746248
66 26 30 jLhav. (a' -~FW) 4-731009
67 45 30 L hav. A' ... 9-682316
. . 65 7 30 .-. A' . . 87° 50' 45"
180
(difference) a = 92° 9' 15"
And in the same way the remaining side b may be determined.
p2
212 PEACTICAL SPHERICAL TRIGONOMETRY.
CASE VI.
24. Having given the three angles of a spherical triangle, to
find the three sides. (See Part II., art. 56.)
In the spherical triangle ABC let A = 101° 17' 30", B
= 109° 52' 0", C = 102° 31' 0"; it is required to find the sides.
Then if A'B'C' be the polar triangle, we shall have A + a'
= 180°, B + V = 180°, C + c' = 180°.
A . . 101° 17' 30" B . . 109° 52' 0" C . . 102° 31' 0"
180 180 180
a' . . 78° 42' 30" V . . 70° 8' 0" c' . . 77° 29' 0"
To find the side a.
c' . .
c' - V .
IT
70
29'
8
0"
0
log.
log.
rL hav.
cosec.
cosec.
c'
V ,
. . .
4
4
•010447
•026648
•834054
•765896
7
78
21
42
0 \
30 \
(a! +c
f
7
b').
V).
86° 3' 30" (sum) L hav. A' ... 9-637045
71° 21' 30" V. A'= 82° 21' 45"
180
(difference) a = 97° 38' 15"
And in the same manner the other sides of the triangle may
be determined.
EXAMPLES. — XXI.
In the spherical triangle ABC find the other parts, having
given
(1) A = 81° 24' 45", B = 61° 31' 45", C = 102° 59'.
(2) A = 78° 36' 30", B = 83° 16' 15", C = 98° 34' 30".
(3) A = 69° 37' 45", B = 88° 35' 45", C = 121° 37' 30".
SOLUTION OF EIGHT-ANGLED SPHERICAL TKI ANGLES. 213
CHAPTER VI.
THE SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES.
(See Part II., Chapter V.)
25. As lias been explained in Part II., arts. 57 — 63, when in
a right-angled spherical triangle any two parts beside the right
angle are given, each of the remaining parts may be obtained
by the addition or subtraction of two logarithms.
The requisite formulae are furnished by the system of rules
called Napier's Rules of Circular Parts.
Let ABC be a spherical triangle, having
i , the angle C a right angle.
The two sides a, b and the complements of
the side c and of the angles A, B are known
~C as the circular parts, and are usually arranged
round a circle, as described in Part II., art. 60.
Since there are five parts in all, it must happen that the
three involved in a particular for-
mula either all lie together, or else,
two being together, that the third is
separated from each of them.
When all three lie together the
part which is between the other two
is the middle part, the parts on either
side being called the adjacent parts.
When one of the three is separated
from the other two it is still called the
middle part, and the other two parts are called the opposite parts.
All the formulas necessary for the complete solution of the
triangle are now furnished by the rules.
Sine middle part = product of tangents of adjacent parts.
Sine middle part = product of cosines of opposite parts.
The student should be careful to write over each function of
214
PRACTICAL SPHERICAL TRIGONOMETRY.
a known part its appropriate sign, in order that he may determine
the sign belonging to the ratio from which the unknown part is
to be obtained.
Example. — In the spherical triangle ABC, having given
c = 118° 21', A = 23° 40', B = 90°, find the remaining parts.
It is convenient to mark the two parts given, as shown in
the figure.
(1) To find the side a.
The three parts involved are a, c, ^ — A, of which c is the
middle part, a and - — A being the adjacent parts.
Hence
sin. c = tan. a tan. (^ — Aj,
or, sin. c = tan. a cot. A.
Therefore
tan. a = sin. c tan. A,
and L tan. a = L sin. c + L tan. A — 10.
And the signs of sin. c and tan. A being in each case -f , the sign
of tan. a also is -f .
(2) To find the side b.
The three parts in tt
~ — A is the middle part, and c, ".— b are the adjacent parts
" '—:
The three parts in this case are |" — 6, ? — A, c, of which
7T
7T
SOLUTION OF EIGHT-ANGLED SPHERICAL TRIANGLES. 215
Hence /V A \ /TT
sin. (- — AJ = tan. c tan. (~ — 1) 1,
or, cos. A = tan. c cot. b.
Therefore + —
cot. b = cos. A cot. GJ
and L cot. & = L cos. A + L cot. c — 10.
In this case cos. A is + , and cot. c is — . Therefore cot. b
also is — , which shows that b is greater than 90°.
(3) To find the angle C.
Here f- — CJ is the middle part, e, ^ — A being the opposite
parts. Therefore
sin. (? - G) = cos. c cos. (| - A),
or, +
cos. C = cos. c sin. A,
and L cos. C = L cos. c + L sin. A — 10.
Here again cos. c is — , so that cos. C also is — , and C must
be greater than 90°.
The necessary logarithms may now be set down in order ;
then, adding each pair and rejecting 10 from the characteristic
in each sum, we shall obtain the logarithms of the parts required.
(1) (2)
L sin. c . . 9-944514 L cot. c . . 9-732043
L tan. A . 9-641747 L cos. A . . 9-961846
(sum) L tan. a . . 9-586261 (sum) L cot. b . . 9-693889
(3)
L cos. c . . 9-676562
L sin. A . . 9-603594
(sum) L cos. C . . 9-280156
therefore a=21° 5; 30", & = 180°-63° 42r 15"=116° 17y 45",
C = 180°-79° 0' 45" = 100° 59' 15".
216 PEACTICAL SPHERICAL TRIGONOMETRY.
EXAMPLES.— XXII.
In the right-angled triangle ABC find the other parts,
naving given
(1) b = 60° 10' 0", c = 100° 0' 0", A = 90°.
(2) B = 100° 0' 0", C = 87° 10' 0", A = 90°.
(3) c = 46° IS' 30", B = 34° 27' 30", A = 90°.
(4) a = 85° 17' 0", b = 102° 26' 15", A = 90°.
(5) a = 100° 42' 0", B = 78° 10' 0", A = 90°.
(6) c = 53° 14' 15", A = 91° 26' 0", B = 90°.
(7) a = 120° 18' 45", b = 101° 9' 0", C = 90°.
(8) B = 72° 19' 0", b = 50° 50' 0", A = 90°.
Note. — Example (8) affords an illustration of the ambiguity
discussed in Part II., art. 63. A diagram may easily be con-
structed similar to the one given in that place, from which it
will at once be seen how it is that a double set of results may be
found to satisfy the data of the question.
CHAPTER VII.
THE SOLUTION OF QUADRANTAL SPHERICAL TRIANGLES.
(See Part II., Chapter VI.)
26. If one side of the spherical triangle ABC, as c, be a
quadrant, the rules of Circular Parts, given in art. 25, may be used
in its solution, the five circular parts in this instance being A, B,
and the complements of a, 6, and C.
It must be remembered, as pointed out in Part II., art. 66,
that whenever the two adjacent parts, or the two opposite parts,
are both sides or both angles, the sign — must be attached to
their product.
SOLUTION OF QUADKANTAL SPHEEICAL TRIANGLES. 217
Example. — In the spherical triangle ABC let a = 90°,
B = 80° 10', and C = 48°'50 ; it is required to find the other
parts.
(1) To find the angle A.
(\
^ — AJ is the middle part; B, C are the opposite
parts.
Thus we have
sin. ( ^ — A J = — cos. B cos. C,
or, + +
cos. A = — cos. B cos. C ;
therefore L cos. A = L cos. B -f L cos. C — 10.
And the sign of cos. B and of cos. C being in each case -f , it
follows that the sign of cos. A must be — , and A therefore
greater than 90°.
(2) To find the side b.
In this case C is the middle part, B and ( - — b J the ad-
jacent parts.
Therefore
Cx
— b) tan. B,
— /
or, cot. b — sin. C cot. B ;
and L cot. b = L sin. C + L cot. B — 10 ;
and the sign of cot. b being + , b is less than 90°.
(3) To find the side c.
Here B is the middle part, C and (- — c J the adjacent
parts.
218 PRACTICAL SPHERICAL TRIGONOMETRY.
Hence
sin. B = tan. C tan. -
or, cot. c = sin. B cot. C ;
therefore L cot. c = L sin. B + L cot. C— 10,
and the sign of cot. c is + .
We have now only to add the several pairs of logarithms,
and reject 10 from the characteristic in each case.
(1) (2)
L cos.
L cos.
L cos.
B .
C .
A.
. 9-232444
. 9-818392
L cot. B .
L sin. C .
L cot. b .
(3)
. . 9-993572
. . 9-941713
. 9-238872
. 9-876678
. 9-050836
. 9-115550
L sin. B
L cot. C
L cot. c
. 9-935285
therefore A = 180°- 83° 32' 45" = 96° 27' 15'
c = 49° 15' 15".
= 82 34.
EXAMPLES.— XXIII.
In the spherical triangle ABC find the other parts, having
given
(1) A = 100°, c = 50° 10', a = 90 .
(2) B =45°, c = 72°, a =90°.
(3) B = 80° 10', C = 50° 2', a = 90°.
(4) A = 72° 49' 45", b = 47° 44' 30" a = 90°.
(5) c = 49° 23' 45", b = 76° 41', a = 90°.
(6) a = 60° 10' 15", b = 80° 20' 30", c = 90°.
Note. — In the solution of right-angled and quadrantal
spherical triangles it will be well for the beginner to arrange
the several parts of the triangle in the five compartments of a
circle, as has been done in this and in the previous chapter.
As, however, he acquires familiarity with the Rules, he will
probably find that he is able to dispense with the circle, and
deduce the appropriate relation between the three parts involved
directly from inspection of the triangle.
219
MISCELLANEOUS EXAMPLES IN PLANE
TRIGONOMETRY.
1. An observer on the bank of a river finds that the elevation
of the top of a tower on the opposite bank, known to be 216
feet high, is 47° 56', the height of his eye from the ground being
5 feet. Find the distance of the observer from the foot of the
tower.
2. A straight line, AD, 100 feet in height, stands at right
angles to another straight line BDC at the point D. At B the
angle subtended by AD is 36° 48', at C the angle is 54° 30'.
Find the length of BDC.
3. A field is in the form of a right-angled triangle. The
base is 200 feet, and the angle adjacent to the base 67°. How
long will a man take to walk round it at the rate of four miles
an hour ?
4. Two points, B and C, are 100 feet apart, and a third
point, A, is equally distant from B and C. What must be the
distance of A from each of the other two points in order that the
angle BAC may be 150° ?
5. A May-pole being broken off by the wind, its top
struck the ground at a distance of 15 feet from the foot of the
pole. Find the original height of the pole, if the length of the
broken portion is 39 feet.
6. A ladder 36 feet long reaches a window 30-7 feet from
the ground on one side of a street, and when turned over about
its foot just reaches another window 18'9 feet high on the other
side of the street. Find the breadth of the street.
7. From the bottom of a tower a distance, AB, is measured
in the horizontal plane, and found to be 50 yards, and at A
the angle BAC is observed to be 25° 17'. Required the height
of the tower BC.
220 MISCELLANEOUS EXAMPLES IN PLANE TRIGONOMETRY.
8. To determine the distance of a ship at anchor at C a
distance, AB, of 1,000 yards was measured on the shore, and
the angles CAB, CBA were found to be 32° 10' and 83° 18'
respectively. Find the distance of the ship from A.
9. Two ships, A, B, are anchored two miles apart. At A
the angle between the other ship and an object, C, on shore is
found to be 85° 10', at B the angle between C and the other
ship is 82° 45'. Find the distances of the two ships from C.
10. The two sides, AB, BC, of the right-angled triangle
ABC are 18 and 24. Find the length of the perpendicular let
fall from the right angle upon the hypothenuse.
11. To determine the height, AB, of a tower inaccessible at
the base, two stations, C, D, are chosen in a horizontal plane,
so that A, C, and D are in the same vertical plane. The dis-
tance CD is 100 yards, the angle ACB is 46° 15', and BDA
31° 20'. Find the height AB.
12. From the decks of two ships at C and D, 880 yards
apart, the angle of elevation of a cloud at A, in the same vertical
plane as C, D, is observed, and at C found to be 35°, at D 64°.
Find the height of the cloud above the surface of the sea, the
height of eye in each case being 21 feet.
13. A tower subtended 39° to an observer stationed 200 feet
from the base. Find its height, and also the angle which it
will subtend to an observer at 350 feet from its bas.e.
14. To determine the distance between two ships at sea an
observer noted the interval between the flash and report of a
gun fired on board each ship, and measured the angle which the
two ships subtended. The intervals were 4 seconds and 6 seconds
respectively, and the angle 48° 42'. Find the distance of the
ships from each other, having given the velocity of sound 1,142
feet per second.
15. From the top of a ship's mast, 80 feet above the water,
the angle of depression of another ship's hull was observed, and
found to be 20°. Eequired the distance between the two ships.
16. Two monuments are 50 feet and 100 feet in height re-
spectively, and the line joining their tops makes with the
horizontal plane an angle of 37°. Find their distance apart.
17. To determine the distance between two ships at anchor,
MISCELLANEOUS EXAMPLES IN PLANE TRIGONOMETRY. 221
0, D, a base AB is measured on the beach, and found to be 670
yards. The following angles were then observed at the ex-
tremities of the base : at A the angle BAD 40° 16', BAG 97° 56',
at B, ABC 42° 22', and ABD 113° 29'. Find the distance of
the two ships apart.
18. To determine the distance of two forts, C, D, at the
mouth of a harbour, a boat is placed at A, with its bow towards
a distant object E, and the angles CAD, DAE, are observed and
found to be 22° 17' and 48° V respectively. The boat is then
rowed to B, a distance of 1,000 yards, directly towards E, and
the angles CBD, DBE are observed to be 53° 15' and 75° 43'
respectively. Find the distance CD.
19. To determine the height of an object, EB, on the top of
an inaccessible hill, the angle of elevation, ACE, of the top of
the hill was observed, and found to be 40°, and that of the top
of the object, ACB, was found to be 51°. After walking a dis-
tance of 100 yards in a horizontal line directly away from the
object, the observer found the angle of elevation of the top of
the object, ADB, to be 33° 45'. Find the height of the object.
20. To determine the distance from an inaccessible object at 0,
without observing any angles, a straight line, AB, of 500 yards,
was measured, so that O was visible from each extremity. From
A, B, in a direct line from 0, AC and BD were measured, each
equal to 100 yards. Finally the distances AD and BC were
measured. The former was 550 yards, the latter 560 yards.
Find the distances of the object from A and B.
21. The angle of elevation of a tower 100 feet high due
north of an observer was 50°. What will be its elevation when
the observer has walked due east 300 feet ?
22. The elevation of a balloon was observed at a certain
station to be 20°, and its bearing was N.E. At a second station
4,000 yards due south of the former one its bearing was N. by E.
Find its height.
23. From a window which seemed to be on a level with the
bottom of a steeple the angle of elevation of the top of the
steeple was 40°. At another window, 18 feet vertically above
the former, the angle of elevation was 37° 30'. Find the height
of the steeple.
222 MISCELLANEOUS EXAMPLES IN PLANE TRIGONOMETRY,
24. At B, the top of a castle which stood on a hill near the
seashore, the angle of depression, HBS, of a ship at anchor was
4° 52', and at R, the bottom of the castle, its depression, NRS,
was 4° 2'. Find the height of the top of the building above
the level of the sea, the height of the castle itself being 54 feet.
25. In order to find the breadth of a river a base line of
500 yards was measured in a straight line close to one side of
it, and at each extremity of the base the angle subtended by the
other end and a tree upon the opposite bank were measured.
These angles were 53° and 79° 12' respectively; find the
breadth of the river.
26. The elevation of the top of a spire at one station, A, was
23° 50' 15", and the horizontal angle at this station between
the spire and another station, B, was 93° 4' 15". The horizontal
angle at B was 54° 28' 30", and the distance between the sta-
tions 416 feet ; what was the height of the spire ?
27. In order to find the distance of a battery at B from a
fort at F, distances BA, AC were measured to points A, C,
from which both the fort and battery were visible, the former
distance being 2,000, and the latter 3,000 yards. The follow-
ing angles were then observed : BAF = 34° 10', FAC = 14° 42',
and FCA = 80° 10'. From these data find the distance of the
fort from the battery.
28. From a ship sailing along a coast a headland, C, was ob-
served to bear N.E. by N. After the ship had sailed E. by N. 15
miles the headland bore W.N.W. Find the distance of the head-
land at each observation.
29. A cape, C, bore from a ship N.W., and a headland, H,
bore N.N.E. ^ E. After the ship had sailed E. by N. J- N. 23 miles
the cape bore W.N.W. and the headland N. by W. £ W. Find
the bearing and distance of the cape from the headland.
30. From a ship sailing N.W. two islands appeared in sight,
one bearing W.N.W. , the other N. When the ship had sailed
six miles farther they bore W. by S. and N.E. respectively. Find
their bearing and distance from each other.
31. A ship was 2,640 yards due south of a lighthouse. After
the ship had sailed N.W. by N. 800 yards the angle of elevation
of the top of the lighthouse was 5° 25'. Find its height.
MISCELLANEOUS EXAMPLES IN PLANE TKiaONOMETKY. 223
32. A church, C, bears from a battery, B E.N.E., 960 yards
distant. How must the church bear from a ship at sea which
runs in until the battery is due north, 2,000 yards distant ?
33. What angle will a tower subtend at a distance equal to
six times the height of the tower ? Where must the observer
station himself that the angle of elevation may be double the
former angle ?
34. A May-pole was broken by the wind, and its top struck
the ground 20 feet from the base. Being again fixed it was
broken a second time 5 feet lower down, and its top reached the
ground at a point 10 feet farther than before. Find the height
of the pole.
35. The summit, A, of a hill bore east from a spectator at
B, and E.N.B. from a spectator at C, a point due south of B.
The angle of elevation of the point A at B being 20°, find its
elevation at C.
36. An observer finds the angle of elevation of a tower at a
point B to be 23° 18'. After walking from B 300 feet, in a
direction at right angles to the line joining B with the foot of
the tower, he found the elevation to be 21° 16'. Required the
distance of the tower from B.
37. The distance between two objects, C and D, is known to
be 6,594 yards. On one side of the line CD there are two
stations, A, B, at which angles are observed. The angle CAD
is 85° 46', DAB 23° 56', CBD 68° 2', and CBA 31° 48'. From
these observations find the distance between A and B.
38. The area of a triangle is 6 square feet, and two of its
sides are 3 feet and 5 feet. Find the third side.
39. Find the area of a regular octagon, the side of which is
16 yards.
40. The area of a regular decagon is 3233*5 square yards.
Find a side.
41. If at the top of a mountain the true depression of the
horizon is 1° 31', find the height of the mountain, supposing the
earth to be a sphere of diameter 8,000 miles.
42. A flagstaff, 12 feet high, on the top of a tower, sub-
tended an angle of 48' 20" to an observer at a distance of 100
yards from the foot of the tower. Find the height of the tower.
224 MISCELLANEOUS EXAMPLES IN PLANE TRIGONOMETRY.
43. Walking along a road I observed the elevation of a
tower to be 20°, and the angular distance of its top from an
object in the road was 30°. The shortest distance from the
tower to the road being 200 feet, find the height of the former.
44. The sides of a triangle were 6, and the angles were to
each other :: 1 : 2 : 3. Find the sides.
45. The perimeter of a triangle is 100 yards, and the
angles are to each other in the proportions of 1, 2, 4. Find
the sides of the triangle.
46. The perimeter of a right-angled triangle is 24 yards, and
one of its angles is 30°. Find the sides.
47. In the plane triangle ABC the side a is 400 feet, and
the sum of b and c is 600 feet, the angle A being 80°. Find
b and c.
48. The perimeter of a right-angled triangle is 24 feet, and
'its'base is 8 feet. Find the other sides.
49. At the distance of 80 feet from a steeple the angle made
by a line drawn from its top to the platfe of observation was
double that made by a line -drawn fr\«fr the top to a point 250
feet from the steeple on the same level. Find the height of the
steeple.
50. In the plane triangle ABO, the side c is 70 feet, a — b
= 13 feet, and A — B = 20°. Solve the triangle.
225
MISCELLANEOUS EXAMPLES IN SPHERICAL
TRIGONOMETRY.
1. Having given the Sun's meridian altitude 70° (zenith
north of the Sun) and its declination 20° N., required the lati-
tude of the place.
2. Having given the Sun's meridian altitude 70° (zenith
north) and its declination 5° S., required the latitude.
3. Having given a star's meridian altitude 70° (zenith south)
and decimation 25° N., required the latitude.
4. Having given the Sun's meridian altitude 30° (zenith
south) and declination 10° N., required the latitude.
5. Find the maximum altitude attained by a body of de-
clination 20° N. in latitude 40° N.
6. Having given the Sun's meridian altitude 30° (zenith
south) at a place in latitude 50° S., find its declination.
7. The meridian altitude of a star at a place on the Equator
is 57°; find its declination (zenith north of the star).
8. The meridian altitudes of a circumpolar star at its superior
and inferior transits were 70° and 20° respectively ; required the
latitude.
9. A circumpolar star passes the zenith of a place, and its
altitude at the inferior transit is 20°; required the latitude.
10. If the altitude of a circumpolar star at its inferior transit
is equal to its zenith distance at its superior transit, required
the latitude.
11. In latitude 60° N. find the altitudes of a star at the
inferior and superior transits, the declination being 40° N.
12. What is the declination of a star that passes the zenith
of a place in latitude 50° 48' N., and what will be its altitude
at the inferior transit?
226 MISCELLANEOUS EXAMPLES IN SPHEEICAL TEIGONOMETKY.
13. Find the latitude of a place at which the Sun's centre just
touches the horizon without setting on the longest day.
14. In latitude 50° 48' N. the altitude of the Sun was 46° 20'
(west of the meridian) and its decimation was 23° 27' 45" N. ;
find the azimuth and the apparent time.
15. The azimuth of a heavenly body was N. 111° 51' W.,
its altitude at the same time was 46° 20', and declination
23° 27' 45" N. ; find the apparent time.
16. Find the altitude of a star, whose hour-angle is 2h. 32m.
and declination 16° N., at a place in latitude 50° 48' N.
17. In latitude 50° 48' N., the Sun's declination being
12° 29' N., find his azimuth at 2h. 53m. Is. A.M., apparent time.
18. Having given the Sun's altitude 42° 30', declination
22° 10' N., and azimuth S. 57° 45' W., find the latitude.
19. Having given the Sun's altitude 37° 20', hour-angle
2h. 15m., and declination 22° 30' N., find the latitude (zenith
north of the Sun).
20. Having given the Sun's altitude 30°, when due west,
and its declination 20° N., find the latitude.
21. Having given the Sun's declination 23° 27' 45", and the
latitude of the place 50° 48' N., find the time when he will be
on the prime vertical, and the altitude at the time.
22. Two stars are due east at the same time at a place in
latitude 50° 48' N. ; their altitudes are 20° and 40° ; find the
difference of their hour-angles.
23. Having given the Sun's altitude at six o'clock 18° 45',
and declination 20° 4' N., find the latitude.
24. Having given the latitude of the place 50° 48' N., and
the Sun's declination 23° 27' 45" N., find the altitude and
azimuth at six o'clock.
25. Find the apparent time of sun-rise at a place in latitude
50° 48' N. when the amplitude is E. 10° S., neglecting the
effects of refraction.
26. Having given the Sun's amplitude W. 37° 30' N., and
decimation 15° 12' N., required the latitude.
27. Having given the latitude of the place 50° 48' N., and
the Sun's decimation 18° 28' N. ; find the amplitude and
the length of the day.
MISCELLANEOUS EXAMPLES IN SPHERICAL TRIGONOMETRY. 227
28. Where will the Sun rise in latitude 50° 48' N. when the
day is 14 hours long ?
29. Having given the Sun's altitude 22° 56', the hour-angle
3h., and the declination 0°, find the latitude.
30. The right ascension of the Sun is 4h. 10m. 20s., and the
obliquity of the ecliptic 23° 27' 45"; find the declination and
longitude.
31. The right ascension of a heavenly body is 2h. 59m. 37s.,
the declination 21° 27' 45" N. ; find its latitude and longitude.
32. The latitude of a heavenly body was 46° 6' 15" K, and
the longitude 234° 36' 30"; find the declination and right
ascension.
33. Two places have the same latitude, 45° N., and their
difference of longitude is 10° 36' ; find their distance apart,
measured on the parallel of latitude passing through them.
34. Find the distance between Portsmouth and Buenos
Ayres, measured upon the arc of the great circle passing through
these places, having given
Lat. Portsmouth. ... 50° 48' N.
„ Buenos Ayres ... 34 37 S.
Long. Portsmouth ... 16 W.
,, Buenos Ayres . . 58 24 W.
35. A ship from latitude 50° 10' N. starts on a great circle,
sailing S. 45° W. What course will she be steering after fol-
lowing the arc of the great circle for 100 miles ?
36. What is the highest latitude attained by a ship sailing
on the arc of a great circle from Port Jackson to Cape Horn,
their latitudes being 33° 51' S. and 55° 58' S. respectively, and
the difference of longitude 140° 27' ?
37. Required the Sun's depression below the horizon at 7h.
P.M., when the declination is 10° 15' S., and the latitude of
the place 50° 4$' N.
38. Determine the azimuth of the two stars Aldebaran and
Pollux when on the same vertical circle, the latitude of the place
being 25° 1ST. ; the R.A. and declination of the former star being
4h. 26m. 46s. and 16° 11' N.; of the latter 7h. 35m. 13s. and
28° 24' 30" N.
Q2
228 MISCELLANEOUS EXAMPLES IN SPHERICAL TRIGONOMETRY,
39. In a certain latitude (zenith N.) the Moon's true altitude
was 18° 2' 30" (east of meridian) when upon the same vertical
circle as a star whose R.A. and declination were 9h. 59m. 42s.
and 12° 45' 45" N. The Moon's R.A. and decimation being
12h. 35m. 54s. and 1° 42' 30" S., required the latitude.
The problems hitherto given may be regarded as exercises in
the practical solution of spherical triangles. Those which
follow require for the most part a greater degree of mathematical
skill, and are therefore distinguished by an asterisk :—
* 40. The altitude of a star when due east was 20°, and it
rose E. by N. Find the latitude.
* 41. The altitude of a star when due east was 10°, and when
due south 40° ; find the latitude.
* 42. Having given the altitude of the Sun when due west,
and at six o'clock, to find the latitude and declination.
Example. — Altitude when west 27° 24', at six o'clock
14° 43' 30".
* 43. Having given the Sun's altitude at six o'clock, and his
amplitude, to find the latitude of the place and declination of
the sun.
Example. — Altitude at six o'clock 14° 43' 30", amplitude
W. 30° 44' 30" N.
* 44. Having given the Sun's altitude at six o'clock, and the
hour-angle at setting, to find the latitude and declination.
Example. — Altitude at six o'clock 14° 43' 30" ; hour-angle
at setting 7h. 35m. 22s.
* 45. Having given the times at which the Sun sets, and is
west, on the same day, at a particular place, to find the latitude
of the place and the declination.
Example. — Hour-angle when west 4h. 43m. 28s., hour-angle
at setting 7h. 35m. 22s.
* 46. Having given the Sun's declination, and the interval
between the times at which he is west, and sets, to find the
latitude of the place.
Example. — Declination 20° N., interval 2h. 51m. 54s.
MISCELLANEOUS EXAMPLES IN SPHEEICAL TRIGONOMETRY. 229
* 47. Having given the amplitude of the Sun, and the azi-
muth at six o'clock, to find the latitude of the place and de-
clination of the Sun.
Example. - - Amplitude W. 30° 44' 30" N., azimutb
N. 76° 18' 45" W.
*48. Having given the Sun's meridian altitude, and his
altitude at six o'clock, to find the latitude of the place, and
declination of the Sun.
Example. — Meridian altitude 62°, altitude at six o'clock
14° 43' 30".
*49. Having given the Sun's meridian altitude, and the
hour-angle when rising, to find the latitude of the place.
Example. — Meridian altitude 56°, hour-angle at rising 7h.
* 50. Having given the interval between the times at which
the Sun bears west, and sets, at a place whose latitude is known,
to find the declination of the Sun.
Example. — Latitude 48° N., interval 2h. 51m. 54s.
* 51. At a given place, to find the greatest azimuth of a
heavenly body whose declination is greater than the latitude of
the place ; to find also the time and altitude on a given day
when the heavenly body will have the greatest azimuth, and
when, consequently, it will appear to move perpendicularly to the
horizon.
Example. — In latitude 20° N. when the Sun's declination is
23° 28' N., required the time and altitude when its azimuth is
the greatest, and also its greatest azimuth.
*52. In latitude 20° N. when the decimation is 23° 28' N.,
required the time when the Sun will appear stationary in
azimuth, the period during which the shadow will move in a
contrary direction, and the number of degrees through which it
will appear to go back.
*53. When the Sun's declination was 10° 15' N., and that
of the Moon was 12° 46' S., they were observed to rise at the
same time ; required the latitude of the place and the time of
the observation, the right ascension of the Sun being Ih. 53m.
42s. greater than that of the Moon.
230 MISCELLANEOUS EXAMPLES IN SPHERICAL TRIGONOMETRY.
*54. Find the decimation of the Sun when he is in the
horizon of Dublin and of Pernambuco at the same instant, the
respective latitudes being 53° 21' N. and 8° 13' S., and the
longitudes 6° 19' W. and 35° 5' W.
*55. ABCD is a square field, each of whose sides is 100
yards. In the middle of the field stands an obelisk 60 feet high ;
find the altitude of the Sun when the shadow of the obelisk just
reaches the corner of the square.
* 56. At noon on the shortest day the shadow of a perpen-
dicular stick was seven times as long as its shadow at noon on
the longest day ; required the latitude, the declination being
23° 28'.
*57. Compare the lengths of the shadow of a perpendicular
stick at noon in latitude 45° N. on the two days when the Sun's
declination is 15° N. and 15° S. respectively.
* 58. In latitude 33° 30' N. I observed that my shadow bore
to my height the proportion of 5 : 3 ; required the altitude and
hour-angle of the Sun, having given the declination 10° 15' N.
* 59. The length of the shadow of a perpendicular object
was 4 feet, and its longest when sloping was 5 feet ; required
the Sun's altitude.
*60. The elevation of a cloud was observed to be 20°, and
at the same time the Sun's altitude was 22°, the Sun and cloud
being in the same vertical plane with the observer, whose
distance from the shadow was 400 yards. Find the height of
the cloud.
* 61. In latitude 45° N., the meridian altitude of the Sun
was 30° ; show that the tangent of one quarter of the length of
the day was
^3
* 62. At a certain place the Sun rose at 7h. A.M., apparent
time, and its meridian zenith distance was twice the latitude ;
required the latitude.
* 63. In latitude 45° N. the Sun rose at 4h. A.M., apparent
time ; show that the tangent of the meridian altitude was 3.
*64. In latitude 50° N. when the Sun's decimation is
5° 38' N., required the time it will take to rise out of the hori-
zon, its semidiameter being 16 ,
MISCELLANEOUS EXAMPLES IN SPHERICAL TRIGONOMETRY. 23 L
* 65. Required the time the Sun's semidiameter will take
to pass the meridian, the declination being 23° 4' and semi-
diameter 16' 17".
* 66. In latitude 45°, required the difference in the lengths
of the longest and shortest days.
* 67. In what latitude will the difference between the longest
and shortest days be just 6 hours ?
*68. At a certain place, when the Sun's declination was
10° N. it rose an hour later than when it was 20° N. ; required
the latitude.
* 69. In what latitude will the shortest day be just one-
third the longest ?
* 70. At a certain place, when the Sun's declination was d,
the length of day was 13h. 38m., and when the declination
was 2d the length of day was 15h. 26m. ; find the latitude of
the place and the declination of the Sun.
233
ANSWERS TO THE EXAMPLES.
I. (page 178).
1. 2-394452. 2. 3-169080. 3. 1-147985. 4. 3-622891.
5. .5-082800. 6. 1-692406. 7. 2-602060. 8. 4*698970.
9. T-876414. 10. 1-954435. 11. 3-000003. 12. "864866.
II. (page 179).
1.
248.
2.
248COOO. 3.
260-418.
5.
204-221
6.
121004-19. 7.
•024.
9.
•6945.
10
•00000075. 11.
•004.
III. (page 180).
1.
L sin.
Ltan.
(1)
9-246845
9-253720
(2)
9-516536
9-541332
]
(8)
9-875079
10-054613
(4)
9-943532
10-263631
2.
9-413054 and 9-932357. 3. 9-9;
1.
2.
(1)
(2)
(3)
(4)
4. 1-6514.
8. -000035.
12. -0248.
L cosec.
10-753155
10-483464
10-124921
10-056468
9-989363 and 9-958137.
IV. (page 181).
(1) 25° 58' 30". (2) 32011'33".
(3) 79° 6' 18". (4) 17° 7' 36".
(5) 98° 18' 56".
35° 27' 33", 144° 32' 27", 215° 27' 33", 324° 32' 27".
46° 7' 47", 133° 52' 13", 226° 7' 47", 313° 52' 13".
80°47/53", 99° 12' 7", 260° 47' 53", 279° 12' 7".
81° 51' 59", 278° 8' 1".
(1) 0211397.
(1) 21° 58' 31".
V. (page 182).
(2) 0662951.
(2) 89° 16' 25'
1. 8064.
4. 42854.
VI. (page 182).
2. 9216.
5. -0009072.
(3) 1198665.
(3) 4° 56' 42".
3. 42854.
6. -0000002229.
234
ANSWEES TO EXAMPLES.
VII. (page 183).
1. (1) 58-96. (2) -26389. (3) 2-222.
2. (1) 25-26. (2) 1295-71.
(4) 2387-23.
VIII. (page 185).
1.
1953-127.
2.
11078-5.
3.
1507-82
4.
11-989.
5.
247742-3.
6.
•00032.
7.
•000070494. 8.
•0063241.
9.
3-79195.
10.
19-105.
11.
231116.
12.
1-071776.
13.
•304959.
14.
1-47923.
15.
•10772.
16.
•2321.
17.
•09644.
18.
•036342.
19.
10544.
20.
•272016.
21.
30-586.
22.
1717740.
23.
•000000000003741.
24.
•695883.
25.
•0004572.
26.
•85475.
27.
1-8593.
28.
•0016.
IX. (page 186).
1.
(1)
log
.# = 2
log.
a + log. b + log. c + 2
log. d.
(2)
log
,# = 2
log.
a + log. 6 + £log. c-
1.
(3)
log. x = $ (log. a + log. b) - log. c - 2 log. d.
(4)
log.*. I
(log. a + log. £ + 4 log. c)
-4 log.
d.
2.
(1)
110592.
(2) -07509.
(3) -00000675.
(4)
107-124.
(5) 54-95.
(6) 1747-6.
(7) 94-795.
o
(1)
x =
2-4101
.
(2) a- = 1-8634.
(3)
£ _ . J5g'
(4)
x =
•6057.
(5) x = -06183.
(6)
# = 1-00<
<7)
x =
•1275.
(8) # = -6827.
4.
(1)
*-
iogTa'
(2) r-
log. c- log. b
n
&log. a — nlog.b'
(3)
X =
n
log. b
V /
wlog.
b — log. a — r log. c*
5.
(1)
-1
•274.
(2) 7153.
(3) A/J?
"=-•5.
X. (page 188).
1. # = 144-5. 2. or = -074764. 3. x= -7-4382.
XL (page 191).
4. o?» 19878-98.
1.
(1) c = 27-31, A = 52° 37r 30", C = 37° 22' 30".
(2) b = 506-9, A = 49° 14' 16", C = 40° 45X 45".
(3) c = 4264, A = 56° 29' 15", B = 33° 30' 45".
(4) a= -1286, c =-1532, 0 = 50°.
(5) a = -04767, b = -06223, A = 50°.
(6) « = 2132-1, B = 56°29/15//, 0 = 33° 30' 45"
2.
195 feet. 3. 238-5 feet. 4. 499-7 feet. 5.
65-8 feet.
6.
3-14 feet. 7. 24 miles. 8. 368-4 feet. 9.
At 8h p.m.
ANSWEKS TO EXAMPLES. 235
XII. (page 194).
1. A= 89° 45' 45'', B = 35° 12' 0", C- 55° 2' 15 '.
2. A= 54° 8' 15", B = 82° 58' 0", 0 = 42° 53' 45".
3. A« 56° 4' 0", B= 48° 31' 0", C= 75° 25' 0".
4. A = 29°44/ 0", B = 115° 36' 30", C = 34° 39' 30".
5! A= 10° 68' 0", B = 107° 58' 45", 0= 61° 3' 15".
6. A = 107° 46' 15", B= 53° 7' 30", C= 19° 6' 15",
7. A = 28° 14' 15", B = 45° 12' 30", C = 106° 33' 15".
8. A = 20U 11' 30", B = 48° 19' 15", C = 111° 29' 15".
XIII. (page 197).
L (1) c = 316-3, B= 36° 6' 30", C = 102° 34' 15".
(2) 6=2268, B = 105° 52' 30", C= 27° 7r.
(3) b = 73-98, B - 49°, C - 29° 19'.
(4) b = 3, c = 4, 0 = 88° 2' 30".
(5) c = 1-013, A = 28° 13', C = 106°37'.
(6) a = 279-7, 6 = 243, 0=91° 43'.
(7) 6=265-8, c = 207-6, C = 48°.
(8) A = 54° 2' 15" or 125° 57X 45", C = 84° 47X 45" or 12° 62' 15";
c- 219-3 or 49-06.
(9) B = 80°39/ 45" or 99° 20' 15", 0 = 43° 7' 30" or 24° 27".
c = 2136-7 or 12937.
2. Bearing of lighthouse from fort, S. 60° 24' W.
3. 4-59 miles. 4- 3h. 20m. nearly.
5. 8h. 32m. nearly. 6. 2-298 miles.
XIV. (page 199).
1. (1) B = 52° 54' 45", C = 60° 44' 45", a = 73'5.
(2) A = 54° 8' 15", B = 82° 67' 45", c = 430.
(3) B = 29° 7' 45", C = 45° 24' 45", a = 106-92.
(4) A = 62° 51' 30", B - 41° 52', c = -0391.
2. 108-52 miles. 3. 1,879 feet. 4. 1,012 yards.
5. S. 11° 32' E., 1-65 mile per hour.
XV. (page 201).
1. (1) 80627 square yards. (2) 2288-3 square feet.
(3) 3714 square feet. (4) -5902 square feet,
(5) 2362-5 square feet.
2 536-23 acres. 3. 42-9 feet.
4. 27° 24' 30", 102° 37' 45", 49° 57' 45". 5. 104-4 feet.
XVI. (page 202).
1. (1) 3353 square feet. (2) 1540280 square feet,
2! -688 square inch. 3. 1 '7604 acres.
4. 97687 square yards.
236 ANSWERS TO EXAMPLES.
XVII. (page 204).
1. A= 59° 2' 15", B- 74°54/ 0", 0= 71° 18' 30".
2. A = 115°39/ 0", B = 99° 2V 30", 0 = 75° 31' 45".
3. A= 81° 24' 45", B= 61° 31' 45", 0 = 102° 59' 30".
4. A = 147° 3' 0", B = 113°28' 0", C= 90° 0' 15".
5. A = 47° 55' 45", B = 61° 60' 30", C = 89° 59' 30".
XVIII. (page 206).
1. (1) «= 96° 10' 2". (2) a= 46° 19' 38".
(3) c = 87° 0' 48". (4) a = 23° 57' 10".
(5) c=100° 9' 33". (6) c = 105° 0' II".
2. (1) B = 49° 30' 15", C = 100° 43' 45".
(2) A = 111° 18' 15", B = 80° 5' 45".
XIX. (page 209).
1. (1) B = 56° 57' 15". (2) B = 36° 51' 16" or 143° 8' 45".
2. B = 43° 44', c = 101° 1' 30", C = 108° V 30".
XX. (page 210).
(1) * = 78° 13' 0". (2) b = 77° 16' 15" or 102° 43' 45".
XXI. (page 212).
(1) a = 87° 10' 30", b = 62° 36' 45", c = 100° 10' 15".
(2) « = 79° 26' 15", b= 84° 49' 0", c = 97° 26' 0".
(3) a = 66° 48r 0", b = 101° 25X 0", c = 123° 23r 30".
XXII. (page 216).
(1) a = 94° 57' 15", B= 60° 32' 45", C = 98° 41' 45'*.
(2) a = 90° 30' 0", b = 100° Or 45", c = 87° 1' 15".
(3) « = 51° 46' 15", b = 26° 23' 30", C = 66° 59' 30".
(4) B = 301° 31' 15", C = 111° 58' 0", c = 112° 26' 45'".
(5) b = 74° 5' 45", c = 132° 39' 30", C - 131° 32' 45".
(6) a = 91° 47' 15", b = 91° 4' 15", 0= 53° 15' 0".
(7) c= 84° 24' 0", A = 119° 50' 15", B= 99° 39' 30".
(8) a = 54° 28' 0" or 125° 32' 0", c- 23° 2' 30" or 156° 57' 30",
C = 28° 44' 45" or 151° 15' 15".
XXIII. (page 218).
(1) b ** 78° 14' 30", B= 74° 367 30'', C= 49° 8r 0".
(2) b = 47° 44' 30", A = 107° 10' 15", C = 65° 19' 15".
(3) b= 82° 26' 0", c= 50° 27' 0", A= 96° 17' 45".
(4) c ='108° 0' 0", B= 45° 0' 0", C = 114° 40' 45".
(5) A = 101° 42' 30", B = 72° 20' 15", 0 = 48° 1' 30".
(6) A= 59° 41' 45", B- 78° 51' 0", C = c,-5° 36' 0".
ANSWERS TO EXAMPLES.
237
MISCELLANEOUS EXAMPLES IN PLANE TKIGONOMETKY.
1.
190-4 feet.
2.
3.
3m- 21 -6s-
4.
5.
75 feet.
6.
7.
23-6 yards.
8.
9.
9-478 miles and 9-52 miles.
10.
11.
145-9 yards.
12.
13.
162 feet ; 24° 60'.
14.
15.
219-8 feet.
16.
17.
1174-4 yards.
18.
19.
46-67 yards.
20.
21.
17° 47' 45".
22.
23.
210-4 feet.
24.
25.
529-5 yards.
26.
27.
5422 yards.
28.
29.
S. 87° 40' W. 42-3 miles.
30.
31.
192 yards.
32.
33.
9° 28' ; 2-9 times the height.
34.
35.
18° 35' 15".
36.
37.
4694 yards.
38.
39.
1236-1 yards.
40.
41.
1-402 miles.
42.
43.
187-5 feet.
44.
45.
19-8, 35-69, 44-51 yards.
46.
47.
369 feet and 231 feet.
48.
49.
150 feet.
50.
205 feet.
51-76 feet.
49-44 feet.
1100 1 yards.
14-4 feet.
942-7 yards.
5147-9 feet.
66-4 feet.
1290 yards.
From A 536 yards, from B
500 yards.
511-3 yards.
314-2 feet.
278-7 feet.
8-5 and 10-8 miles.
S. 58° 40' W. 9-71 miles.
N. 20° 32' E.
50 feet.
633-4 feet.
4 or A/52 feet.
20-5 yards.
401-4 feet.
1-268, 2-196, 2-536.
5-072, 8-784, 10-144 yards.
10 feet and 6 feet.
« = 103-7 feet, 6 = 907 feet,
A = 79° 14', B = 69°14'.
MISCELLANEOUS EXAMPLES IN SPHERICAL TRIGONOMETRY.
1. 40° N.
3. 6°N.
5. 70°.
7. 33° S.
9. 55° N. or S.
11. 70° arid 10°.
13. 66° 32' N. or S.
15. 2h. 57m, 16s,
17. N. 44° 12' E.
19. 71°31'N.
21. At 4h. 37m. 4s. P.M. ; 30° 55'
23. 69° 32' N.
25. 6h. 31m. 7s.
27. E.30°4'N.; 15h. 13m. 22s.
2. 15° N.
4. 50° S.
6. 10° N.
8. 45° N. or S., according as the
declination is N. or S.
10. 45° N. or S.
12. Dec. 50° 48' N. ; alt. 11° 36'.
14. N. 111° 51' W. ; 2k. 57m. 16s.
16. 43° 49'.
18. 59° 4' N.
20. 43° 9' N.
22. 59m. 56s.
24. Alt. 17° 58'; Az. N. 74° 39' E.
26. 64°29'N.
28. E. 19° 5' N.
238
ANSWERS TO EXAMPLES.
29.
31.
32.
33.
35.
37.
39.
41.
43.
44.
46.
48.
50.
51.
52,
53.
54.
56.
58.
59.
62.
65.
67.
69.
56° 33'. 30.
Lat. 4° 14' 45" N. ; long. 48° 37' 30'
Dec. 25° 50' N. ; R. A. 16h. 14m. 3s.
449-7 miles. 34.
S. 43° 38' W. 36.
17° 24'. 38.
19° 55' 30" N. 40.
58° 31' N. 42.
Lat. 42° or 48° N. ;
Lat. 48° ; Dec. 20°.
42° or 48°.
Dec. 21° 4' N. ; long. 64° 33'.
5,950 miles.
72°41/S.
N. 75° 5' W.
29°42'N.
Lat. 48° ; Dec. 20°.
Dec. 22° 20' or 20° N.
45. Lat. 48°; Dec. 20°.
47. Lat. 48° N. ; Dec. 20° N.
Dec. 20°.
49. 47° 24'.
Lat. 48° 0' 15"
20" N.
Azimuth N. 77° 28' E.; time 9h. 47m. 53s. A.M. ; altitude, 59° 11' 30".
Tima 9h. 47m. 53s. A.M.; period 4h. 24m. 15s.; shadow went back
through 12° 32' 30".
Lat. 50° 18' 30" N.; time 5h. 9m. 40s. A.M.
55. 15° 47' 30".
57.
hour angle 3h. 58m. 3s.
60.
64.
66.
68.
70.
18° 6'.
38° 27' 45"
Alt. 30° 58'
36° 52' 15';
26° 58'.
1m. 11s.
41° 24'.
68° 27'.
As 1 to 3.
1,468 yards.
3m. 21s.
6h. 51m. 40s.
52° 27r N.
Lat. 54°17X
= 8° 40' 30".
APPENDIX
A COLLECTION OF EXAMPLES
SELECTED FEOM
EXAMINATION PAPERS SET AT THE ROYAL NAVAL COLLEGE
BETWEEN THE YEAKS 1880—1893
WITH ANSWEES
APPENDIX.
A.— PART I. CAP. I. II.
N.B. — In the following examples, unless otherwise stated, TT is
22
taken as — -.
1. If on a map a square inch represents 10 acres, how many
yards are represented by the diagonal of a square inch ?
2. Find the distance at which a person looking towards the
Sun must hold a coin whose diameter is half an inch in order
that it may just hide the Sun, assuming that the angle subtended
by the Sun's diameter is 32'.
3. An angle measuring 1*25 is subtended by an arc of 16
feet : calculate the radius.
4. Determine the number of degrees in the angle subtended
at the centre of a circle, the radius of which is 10 feet, by an
arc the length of which is 9 inches.
5. Find the measure, both in degrees and circular measure, of
the angle between two consecutive spokes of a wheel of 14
spokes.
6. If D, 9 represent the same angle, as measured in degrees
and circular measure respectively, and TT be the circular measure
of two right angles, show that
r> 20
90 ar '
7. (a) Find the radius of a planet on which the arc joining
two places that have the same longitude, but latitudes differing
by 60°, is 2,000 miles long.
(6) Find the diameter of a globe on which an arc of 21° of
a meridian measures 5ft. Gin.
K
242 APPENDIX.
8. How large a mark on a target 1,000 yards off will
subtend an angle of one second at the eye ?
9. It has been found from the transit of Venus in 1882 that
the Earth's semidiameter subtends an angle of 8"'82 at the Sun.
Find the Sun's distance, the Earth's radius being 3,963
miles.
10. Assuming the Moon's distance from the Earth to be
equal to 60 times the Earth's radius, express in degrees and in
circular measure the approximate value of the angle subtended
at the Moon by the Earth's radius.
11. The measure of the same angle is given by one person as
15, and by another as *16. Taking it for granted that the first
person means 15 degrees, find the unit employed by the second
person.
12. Find the number of degrees in the unit of angular
A
measurement when the right angle is measured by — .
7T
13. The perimeter of a certain sector of a circle is equal to
the length of the arc of a semicircle of the same radius. Find
the number of degrees, minutes, and seconds in the angle of the
sector.
14. If an angle subtended by an arc equal to TT times the
radius be taken as the unit, what number will measure an
angle of 45° ?
15. One angle of a triangle is 30°, and the circular measure
of another is - : find the circular measure of the third angle.
16. Find the angle whose circular measure is equal to IT
times the square root of the number of right angles in the
angle.
17. In sailing three-quarters of a mile a ship changes the
direction of her course from S.E. to E.S.E. : find the radius of
the circular arc she describes.
18. The arc of a certain sector of a circle is equal in length
to the sum of its bounding radii : find the number of degrees,
minutes, and seconds in the angle between these radii.
19. In a circle of radius unity a certain arc subtends at the
centre an angle whose circular measure is A. In a second
MISCELLANEOUS EXAMPLES. 243
circle an arc of the same length subtends an angle of A° : find
the radius of the latter.
20. Show that the supplement of half the complement of an
angle is greater than the complement of half its supplement by
the angle of a regular octagon.
21. The circular measure of the difference between the
vertical angle and either of the angles at the base of an isosceles
triangle is ^", the vertical angle being the smaller. Express the
angles in degrees.
22. The four angles of a quadrilateral figure are in arith-
metical progression, and the greatest is to the least as 17 to 3.
Express each angle in degrees.
23. The angle subtended at the centre by a certain chord is
such that the square root of its cosine is equal to the ratio of
the chord to the radius of the circle. Express the angle both
in degrees and circular measure.
24. Show with the aid of logarithmic tables that log.
cos. 6 = log. (circular measure of &) very nearly, when
0 = 42° 18' 5" and TT = 3-14159.
B.— PART L CAP. VIII.-XI.
Prove the identities : —
25. (sin. A + cos. A) (sin.3 A + cos.3 A)
— (sin. A — cos. A) (sin.3 A — cos.3 A) = 2 sin. A cos. A.
26. 1 + tan.6 A = sec.4 A (sec.2 A - 3 sin.2 A).
27. sec.2 Avers. (90° — A) vers. (90° + A) = 1.
28. (tan. A + cot. A — 1) (sin. A + cos. A) = tan. A sin. A
4- cot. A cos. A.
29. 1 + cos.4 A cosec.2 A + sin.4 A sec.2 A = cot.2 A + tan.2 A.
30. (1 + cos. A cosec.2 A + cos.2 A cosec.2 A) vers. A = 1.
31. {(cos. A + sin. A)2 - 1} (1 — tan.2 A)
= 2 (cos.2 A — sin.2 A) tan. A.
32. (sin. A — cosec. A)2 + (cos. A + sec. A)2
= 1 4- sec.2 A cosec.2 A.
33 1 — sin. A cos. A sin.2 A — cos.2 A _ . .
cos. A (sec. A — cosec. A) sin.3 A + cos.3 A ~
B 2
244 APPENDIX.
34. sin. 4A = 4 sin. A cos.3 A — 4 cos. A sin.3 A.
35. sin. 5 A = 16 sin.5 A - 20 sin.3 A + 5 sin. A.
«fi 1 + 2 cos. A A
oo. - - : - T-T- — sec. a*.
1 + cos. A + cos. 2A
37. 2 sin. 2A + sin. 4A = 4 sin. 2A cos.2A.
38. (cos. 2 A — cos. 4A) cos. 3A = (sin. 4A — sin. 2 A) sin. 3A.
39. tan. 2A - tan, A = - 2A sm' A Q A .
cos. A -t- cos. 3A
40. sin. 3A + sin. 5A = 8 sin. A cos.2 A cos. 2 A.
41. sin. A + sin. 3A + sin. 5A 4- sin. 7A
= 16 sin. A cos.2 A cos.2 2A.
42. sin. 2 A + sin. 4 A + sin. 6 A — sin. 12A
= 4 sin. 3A sin. 4A sin. 5A.
-„ cot. A -f cot. 4A -- 1 9A
-~-sec. 4 A..
. —
cot. 2A + cot. 3A 2
44. g (cos.6 A + sin.6 A) - i (cos.4 A- sin.4 A)2 = *-.
45. (cos. A + sin. A) (cos. 2 A + sin. 2 A) = cos. A + sin. 3A.
gj-Q 8 f |_
46 tan. A + cot. A 4- 2
in 2 (—
\4
tan. A + cot. A - 2 gin ,
- A)
47. sin. A sin. 2 A + 2 cos. A cos. 2 A = 2 cos.3 A.
48. sec.2 A - tan.2 A(l + 2 cos. 2A)2 = (1 - 2 cos. 2A)2.
49. 1 + (cos.2 A - sin.2 A)3 = 2 cos.2 A (cos.4 A + 3 sin.4 A).
50. cos. 6A r= 16 (cos.6 A - sin.6 A) - 15 cos. 2A.
51. tan. 3A — tan. 2 A — tan. A = tan. A tan. 2 A tan. 3A.
o -I
52. sin. A 4- sin. 2A + sin. 3A = 4 sin.— -A cos. A cos. - A.
"<u A
53. tan. ^A + tan- 7L=T^ = 2 sec' ^
44^
54. (sin. 2A -f 2 sin. A)2 + (cos. 2A + 2 cos. A + I)2 =
16 cos.4 ~
^i
55. cos.2^ (1 - 2 cos. A)2 -f sin.2 ^ (1 + 2 cos. A)2 = 1.
U t-l
2A
cosec.2-
56. tan. A + cot. ~ =
2 A A
cot. - - tan. g
MISCELLANEOUS EXAMPLES, 245
57. cos. A — tan. ±= sin. A = cos. 2A + tan. — sin. 2A.
'2 £
58. cos. A + cos. 2A + cos. 3A + cos. 4A
A A 5A
= 4 cos. — cos. A cos. — .
4 4
59. log. (cos. A) + log. (cos. 2A) 4- log. (cos. 4A)
= log. (sin. 8A) - 3 log. 2 - log. (sin. A).
sin.2 A — sin2 B
60. tan. (A + B) = : — ~ =5-.
sin. A cos. A — sm. B cos. B
61. sin. (A - B) cos. (A + B) + sin. (B - C) cos. (B + C)
+ sin. (C - D) cos. (C + D) + sin (D - A) cos. (D + A) = 0.
62. 4 sin. 5 A sin. 5B sin. 5 (A + B) = sin. 10 A + sin. 10 B
- sin. 10 (A + B).
63. sin. 2A + sin. 2B + sin. 20 = sin. 2 (A + B + C) +
4 sin. (A + B) sin. (B -f C) sin. (C + A).
64. sin. (A + B) sin. (B + C + D) => sin. A sin. (C + D)
+ sin. B sin. (A + B + C + D).
2 (sin. A + sin. B)
65.
( A + B A - B)2
JCOS. -+-+COB.—
= (tan.f+tan.|) (l + tan.f tan. f ).
66. cot. (A + B) + cot. (A - B) = ^'^ A>
67. 2 {1 — cos. (A — B) cos. A cos. B} = sin.2 A + sin.2 B
sin.2 (A - B).
/sin. 45- - sm. 8g = ^ ^0 _ ^
V sin. 45° H- sin. 30°
69. cos. (A + 15°) + sin. (A - 15°) = sin. (A + 45°).
70. sin. + sin. + sin. + sin.
7T 7T 7T
= 4 cos. - cos. - cos. —.
4 8 16
71. cos. 75° x cos. 15° = .
72. tan. 60° - tan. 165° = 2.
73. sin.4 15° + cos.4 15° = |.
246 APPENDIX.
74. tan.2 72° + tan.2 36° = 10
75. cos.2 72° + cos.2 36° = ?.
4
76. cos.2 18° + cos.2 30° + cos.2 54° = 2.
77. sin. 87° — sin. 59° — sin. 93° + sin. 61° = sin. 1°.
78. tan. 36° - A/5 tan. 18° = 0.
79. (a) 4 sin. 9° = A/3 + V5 - V5 - ^6.
(b) 2 sin. 11° 15' = V{2 — A/2 + A/2} .
f)A A
80. If sin. A = ~, find the values of sin. -.
81. If A lies between 270° and 360°, show that 2 cos. —
2
= — VI — sin- A — A/1 + sin. A.
82. If cot. A = 2>v/2, find sin. A.
83. If sin. A = ~, sin. B = jj., find tan. (A - B).
84. If tan. A = A} tan. 2 = |, find sin. (A + B).
1 /S
85. If cos. A = 4 (A/6~ + A/2), and cos. B = ~-, find
cos. (A + B).
86. If tan. A = -, prove that
a
_ 2 cos. A
a + b A/COS. 2A*
87. If sin. (A - B) = ± 3 tan. A = 5 tan. B, find the value
b
of sin. (A + B).
88. If cot. A = 5 cos. B = 3, show that 2A + B = 90°.
O
89. If the sine of an angle in the third quadrant is , find
5
the cosine of half the angle.
O
90. The cosine of an angle in the second quadrant is :
find the values of the sine and tangent.
91. If tan. 6 = -, show that a cos. 26 + b sin. 20 = a.
a
92. If cot. A = 2 tan. (A - B), show that tan. B = 1~3cos-2A.
3 sin. 2A
MISCELLANEOUS EXAMPLES. 247
C._ PART I. CAP. XII.-XIV.
93. If sin. 0 + 3 cos. 0=1, find tan. 6.
94. If 5 sin. 0+12 cos. (9=13, find tan. 6.
95. If tan. 0 tan. 30 = - |, find tan. 0 and tan. 30.
5
96. If sec. A — cosec. A = 2\/2, find sin. 2A.
97. Having given that (sin. 6 + cos. 0)2 = * + V% +
2v 2
find cos. 40.
98. Solve the equations : —
(a) sin. 0 + cos. 20 = 1.
(6) 11 cos. 0 + 6 sin.20= 10.
(c) tan. (45° + 0) = 1 + tan. 0.
(d) 2 cos. 0 = 3 + 2 sec. 0.
99. Having given that a cos. 0 + 6 sin. 0 = c, and that
= c^7£ = ^ + fe2"' Prove that ° = 2n7r + a ± /3-
100. Solve the equation :
sin. 2x + sin. 30° = */2 sin. (» + 45°).
Prove the following relations : —
101. cot.-1 I = cot.-1 3 + cot.-1-.
o 4
102. cot.-1 3 + cosec.-1 ^5 = -.
4
. . .
104. tan."1 A + 2 tan.-1 i = tan.-1 1.
107. tan.- + tan.
248 APPENDIX.
108. cot.
. (cot.-1 ~ + cot.-1 *\ = - 1.
109. 2 tan.-1. A^| tan. ? = cos.-1 b + * COS' *
V a + o 2 a + & cos. a?
110. Find x from the equation :
tan."1 (x— l) + tan.-J (2 — a?) = 2 tan. ~ 1 V3 as — ^ - 2.
111. Having given log.102 = -301030, log.103 = -477121,
find log.23.
112. Using the values of the logarithms given above, solve
the equations :
4* = 500; (16-2)^ = 900.
113. Solve also the equation :
114. Prove that (1 + 2 log.a r) log.(a,.2) a = 1.
115. Prove that if log. a, log.y b log.z c are in arithmetic
progression, with a common difference unity, then x z = ym,
wherem =
log. b — log. y log. b + log. y
116. Show that log.r {log.a br log.b cr2 log.c af3\ = 6.
D.— PART I. CAP. XVI.-XIX.
In any plane triangle prove the following relations: —
117. cot. B - cot. A = ^-V Cosec. C.
ao
118. cos.2 - — cos.2 (A + — ) = sin. A sin. B.
<s \ 2tJ
ABC
cos. — cos. - cos. -
119. 222
120. sin. 2A sin. 2B (tan. A tan. B tan. C — tan. C) =
4 sin. A sin. B sin. C.
1 2B , 1 2C (a + b + cV
— r»ns 2 — 4- — r.ns 2 — = 1 ! — - — I — L .
101 2 , 2 , 2
121. - cos.2 - + - cos.2 - + - cos.2 - =
a 2 ' b 2 ' c 2 4abc
a sin. A + b sin. B + c sin. G _ 2 sin. A
ab cos. C + ac cos. B + be cos. A a
MISCELLANEOUS EXAMPLES. 249
b cos. B — c cos. C c cos. C — a cos. A
1 —
be cos. A ca cos. B
a cos. A — b cos. B
ab cos. C
124, Show that in any plane triangle
a2 sin. B sin. 0
= 0.
(\ » U/ OA1A.
a) Area = T - —
2 sin.
(B + C)
(6) cot.scot.0
(c) Area = *(&' + c')-"5^5 sin-.°p
2 v b sm. B + c sm. C
Area -
"
4 (cot. A 4- cot. B + cot. C)'
125. In a plane triangle ABC, in which A = 90°, prove that
1 + tan. ?L=-2 : 1 - tan.5-=I! :: b : o.
2 A
126c If C be a right angle, prove that
„ A c — b
tan.2- = - -.
2 c + b
127. Show also, if C be a right angle, that
128. A plane triangle ABC is such that
cos. 2 A - cos. 2 B = ten. B _ ^ A
sin. 2 A
show that it is either isosceles or right-angled.
129. In any triangle prove that
O QJ2
a cos. A -f b cos. B + c cos. C = — -— .
aoc
130. The bisector of the angle A of a triangle ABC meets
the opposite side in D. Prove that
tan.
--.~.
b — c 2
131. If x, y be the lengths of the two diagonals of a quadri-
lateral figure, and 0 the angle between them, show that
Area = % x y sin. 6.
250 APPENDIX.
132. In any plane triangle ABC, if
x + - = cos. A, y + - • = 2 cos. B,
x y
prove that one of the values of bx + - is c.
y
133. If the sines of the angles of a triangle are in the ratios
of 13 : 14 : 15, prove that the cosines are in the ratios of
39 : 33 : 25.
134. The sides of a triangle are 2xy + a?2, a?2 + xy -f y2,
x't _ y*m Show that the angles opposite are in arithmetical pro-
gression, the common difference being
2 tan.-'
135. If from any angle of an equilateral triangle any straight
line be drawn, and from the other two angles perpendiculars
p, q, be drawn to the straight line, prove that the area of the
, . p* 4 pq -4- q'2
triangle is £ £•* *~
^ o.
136. If the cosines of the angles A, B, C of a plane triangle
are in arithmetical progression, show that s — a, s — 6, s — c
are in harmonical progression.
137. If D and E are the points of trisection of the side BO
of a plane triangle ABC, prove that
9 cot. ADB cot. AEB = 2 (cot.2 B + cot.2 C) — 5 cot. B cot. C.
138. If from the angular points of a plane triangle ABC
lines be drawn outwards, inclined at an angle of 30° to the sides,
show that the vertices of the isosceles triangles so formed will
be at equal distances from each other.
139. If AD, BE, CF are the perpendiculars drawn from the
angular points of a plane triangle ABC upon the opposite sides,
prove that the area of the triangle DEF to that of ABC is as
2 cos. A. cos. B. cos. C to 1.
140. If a straight line of length p bisect the angle A of a
triangle ABC, and divide the base into two parts of lengths m
and n, prove that p2 = be — mn.
141. If ABC be a plane triangle, and another triangle be
MISCELLANEOUS EXAMPLES. 251
c c
constructed having c, (a + b) sin. -, (a — b) cos. - for the re-
iL <L
spective lengths of its sides, show that one of its angles is a right
angle.
142. An equilateral triangle is described having its angular
points on three given parallel straight lines, of which the outer
ones are at distances a and b from the middle one.
Show that the side of the triangle
= V 3 (a2 + ab + 62).
143. In the triangle ABC, D is a point in BO such that BD
2 CD ; show that
AD = >/ 662 + 3c2 _ 2aa.
144. The sides of a triangle are in arithmetical progression,
and its area is four-fifths that of an equilateral triangle of the
same perimeter ; show that the sides of the triangle are as
7 : 10 : 13.
145. The sides of a plane triangle are in the proportion of a,
b, \/ a2 + b2 -f ab ; show that the greatest angle = 120°.
146. Standing on a horizontal plane, I observe that the ele-
vation of a hill due north is a, and after walking a yards due east,
its elevation is ft. Show that its height above the horizontal
plane is
a sin. a sin. ft
V sin. (a + ft) sin. (a — ft)
147. At the top P of a tower of height h, the angles of
depression of two objects in the horizontal plane upon which
the tower stands are 45° — a and 45° -f a respectively, P, A
and B being in the same vertical plane. Show that AB =
2h tan. 2a.
148. A flag- staff standing in the centre of a circular pond,
whose radius is equal to the height of the staff, is observed to
subtend an angle of 45°, from the top of a column 20 feet high,
whose base is 12 feet from the edge of the pond. Show that
the height of the staff is 68 feet.
149. Standing on an eminence a feet above the level of a
lake, I observe the elevations of the top and bottom of a tower
252 APPENDIX.
on a hill opposite (a, /3), and the depression of the reflection of
the bottom of the tower in the water (7).
.« '•• i . :* . » V* • n cos. w sin. (a — 8}
Show that the height of the tower is 2a - L _ — : - 'I-L.
cos. a sin. (7 — p)
150, The shadows of two vertical walls, which are at right
angles to each other, and are a feet and a' feet in height respec-
tively, are observed when the sun is due south to be b feet and
b' feet in breadth ; show that if a be the sun's altitude, and ft
the inclination of the first wall to the meridian,
151. A flag-staff of known height k feet stands on a hori-
zontal plane. The angle of elevation of its top is observed at
three points, 0, A, B, situated in a horizontal line through D,
the foot of the flag-staff. If the increase in the angle of eleva-
tion in passing from O to A is equal to the increase in passing
from A to B, and if OA = a feet, AB = b feet, and OD = x feet,
show that
(a - b) («.» + #») = a* (2x - a - b).
152. A man walking due north along a straight road observes
that at a certain milestone two distant objects bear N.E. and
S.W., and that at the next milestone the straight line drawn to
each object makes an angle of 30° with the direction of the road.
Show that the distance to each object is A/ 6 miles.
153. From a certain point on a level platform a flag-staff has
the same bearing, S.W., and the same elevation, 45°, as a lamp-
post of height of 14 feet. From another point, due west of the
former point, and due north of the lamp-post, the elevation of
the lamp-post is tan."1 1-24. Show that the height of the flag-
staff is nearly 31 feet.
154. A man finds that he walks 40 feet in going straight
down the slope of the embankment of a railway which runs due
east and west, and when he has walked 20 feet along the foot
of the embankment he finds that he is exactly N.E. of the point
from which he started at the top of the bank. Show that the
inclination of the bank to the horizon is 60°.
155. Three points, A, B, C, are situated in the same hori-
MISCELLANEOUS EXAMPLES. 253
zontal line, and are such that AB = 150 feet, BC = 50 feet,
and the angles of elevation of the top of a tower (which lies off
the straight line) from the three points A, B, C are respectively
60°, 45°, 30°. Show that the height of the tower is 75 feet.
156. From a house on one side of a street observations (a, /3)
are made of the angle subtended by the house opposite, first
from the level of the street, next from a room window at a
known height, c, above the street. Prove that the height, A, of
the opposite house is given by
= cot. /3 cot. a — cot.2 a.
h2 h
157. The hypothenuse of a right-angled triangle is a, and
one of the acute angles is a. Prove that the distance between
the centres of the squares described on the sides of the triangle,
and external to it, is a sin. (45° + a).
158. From any point in the circumference of a circle three
chords are drawn, such that the two outer ones are equally
inclined to the inner one, and are together equal to it. Show
that the outer extremities of the three chords are the angular
points of an equilateral triangle.
159. Two triangles, ABC, A'BC, equal in every respect, are
similarly situated 011 opposite sides of BC. Show that if A be
30°, and AA' equal to BC, the values of B and C will be
75° + -, when cos. a = —
~" a 2
160. Two parallel chords of a circle, lying on the same side
of a circle, subtend angles of 72° and 144° at the centre. Show
that the distance between the chords is equal to half the radius
of the circle.
161. A vessel starts from halfway between two points, B, C,
and steams at uniform speed on a due north course at right
angles to the line BC. An observer at B, at a certain moment,
perceives that she is in the same straight line with a buoy A ;
five minutes later he sees that she bears N.E. ; and five
minutes later still an observer at C sees her to be in the same
straight line with the buoy A. Prove that in the triangle ABC
tan. A (tan. C - tan. B)2 + 8 = 0.
162. The isosceles triangle ABC has each of its angles B, C
254 APPENDIX.
double of the angle A. If BD be drawn perpendicular to AC,
and then DE be drawn perpendicular to EC, prove that
CE : BE :: tan.2 18° : 1.
163. The angles at the base of a triangle are 22° 30' and
112° 30' respectively. Show that the area of the triangle is
equal to the square on half the base.
164. In the ambiguous case for plane triangles, show that
the rectangle contained by the third sides of the two triangles
which satisfy the given conditions is equal to the difference of
the squares of the two given sides.
165. If in the ambiguous case the area of the larger triangle
is double that of the smaller, show that the tangent of one of the
angles at the base is three times that of the other.
E.— PART I. CAP. XX.
166. A square and a regular hexagon are inscribed in the
same circle. Show that their areas are as 4 to 3\/3.
167 A regular octagon has a side of 2 inches. Show that
the radius of the inscribed circle is (V2 + 1) inches.
168. Show that in any plane triangle
fa) 2-+-1 + _^ = JL
' be ca ab Kr '
A B C
(b) ra cos. — = a cos. — cos. - .
(c) a cos. A + b cos. B + c cos. C = 4 R sin. A sin. B sin C.
169. In any plane triangle show that the sum of the pro-
ducts, taken two at a time, of the radii of the inscribed and
escribed circles is ab + be + ca.
170. Show that the radius of the circle which touches the
sides AB, AC of the triangle ABC, and also touches the
inscribed circle (of radius p). is
MISCELLANEOUS EXAMPLES. 255
171. A circle is inscribed in the triangle ABC, and a, /3, 7
are the angles subtended at the centre by the sides of the tri-
angle. Prove that
4 sin. a sin. ft sm. 7 = sin. A + sin. B + sin. C.
172. In any plane triangle show that
r 4- rb + rc — ra = 4R cos. A.
173. If three circles, of radii r, /, r", touch one another,
prove that their common tangents at the points of contact meet
at a point, and that the distance of this point from each point
of contact = : , where A is the area of the triangle
r + r ' + r"
whose vertices are at the centres of the three circles.
174. If the three bisectors of the angles of a triangle meet
in O, show that the squares on OA, OB, 00 are respectively
proportional to
s — a s — b s — c
•, - , .
a b c
175. Show that the sides of the triangle formed by the bisectors
of the exterior angles of a triangle are a cosec. — , b cosec. — ,
2 2
r1
c cosec. -, and that its area is equal to 2.s.R.
£
176. If the tangents at A, B, C fco the circumscribed circle
of a triangle ABC meet the opposite sides produced in the
points D, E, F respectively, prove that the reciprocal of one of
the distances AD, BE, CF is equal to the sum of the reciprocals
of the other two.
177. If O is the centre of the circle described round an acute-
angled triangle, and AO is produced to meet BO in D, show
that ,
QT-X _ R cos. A
" cos. (B — C)'
178. If the inscribed circle of a triangle ABC touch the
sides BC, CA, AB in D, E, F, prove that tan. ADB = 7-^-'
b — C
where rl is the radius of that escribed circle which touches
BC
S56 APPENDIX.
179. Prove the following expression for the area of a
triangle : —
| R2 {sin.3 A cos. (B - C) + sin.3 B cos. (C - A)
o
+ sin.3C cos. (A - B)}.
180. The triangle ABC has a right angle at C ; E is the
point at which the inscribed circle touches BC, and F the point
at which the circle drawn to touch AB and the sides AC, BC
produced meets CA. Show that if EF be joined, the triangle
FEC is half of the triangle ABC.
181. Show that the sum of the (n) perpendiculars from any
point within a regular polygon upon the sides is constant, and
equal to
a , TT
n - cot. — .
2 n
182. The radius of the circle inscribed in an isosceles
triangle is 16 inches, and of the circle described about the
triangle 50 inches. Show that the ratio of the base to one of
the equal sides is as 8 to 5.
183. If the bisectors of the angles A, B, C of the triangle
ABC meet the opposite sides in D, E, F, prove that
4 (area of ABC) x (area of DEF) = AD . BE . CF . r.
184. Show that the sum of the squares of the distances of
the angular points of an equilaterial triangle from any point on
the circumference of the inscribed circle is equal to five times
the square on half the side of the triangle.
F. — PART I. MISCELLANEOUS.
185. If sin. A cos. B = -, show that sec. 2A = 1 + sec. 2B.
a
186. If tan. C = tan. A tan. B, show that
tan. 2C = {sec. (A + B) + sec. (A - B)} sin. A sin. B.
187. If 2 cos. 0 = x + -, show that 2 cos.3 d = x3 + i
x %
188. If sin. 3x = — sin. 2x, prove that one value of cos. ss
is sin. 18°.
MISCELLANEOUS EXAMPLES. 257
189. If A + B + C = 90°, then
1 — tan. —
cos. A + sin. B + sm. C _ 2
sin. A + cos. B + sin. 0 B
- tan' "2
190. If A + B + C = 90°, then
sin. 2A + sin. 2B -f sin. 2C = 4 cos. A cos. B cos. C.
191. If cos.3 20 -f cos.2 26 + ^ cos. 20 = ^2, show that
tan.3 0 -f tan.2 0 + p tan. 0=1.
192. Show that if cos. 0 = cos- u ~ e then
1 — e cos. u
TT
193. If cos. A = a>, show that
cos. 5 A = 5& —
194. If x cos. fB + y sin. a = z, and a? sin. /3 — y sin. a = 0,
prove that
x y _ z
sin. a sin. £ ~~ sin. (a + /:?)'
195. The sines of three acute angles in arithmetical pro-
gression are as the numbers 2, \/6, 1 + ^3. Show that the
angles are 45°, 60°, 75°.
196. If cos. (f) — cos. 0 = m, and sin. <j> — sin. 0 — n, show
that the value of sin. (0 + <b) is — .
m2 ± n2 — 1
197. Show that
tan .-i * COS- * - tan.-' x ~ sin' * = *.
1 — ai sin. <£ cos. (ft
198. Show that sin. 63° is \/% times the arithmetic mean
between sin. 18° and cos. 18°.
199. Show that a cos. 0 + b sin. 0 will have its greatest
value when 0 = tan.~l -.
a
200. A tower h feet high subtends an angle a at a person's
eye ; if the height of eye is a, show that x, the distance of the
tower from the observer, is given by the equation
a?2 — h cot. a . x 4 a2 — ah = 0.
258 APPENDIX.
G. — PART II. MISCELLANEOUS.
201. Prove that in any spherical triangle ABC,
T>
cos. b = cos. (a + c) 4- 2 sin. a sin. c cos.2 - '.
z
202. If the sum of the sides of a spherical triangle is 180°,
show that the haversine of each angle is equal to the product of
the cotangents of the two adjacent sides.
203. The equal sides, AB, AC, of an isosceles spherical
triangle are bisected in P, Q respectively. Prove that
. BC PQ AB
sin.— = 2 sin. -^ cos. — .
204. If a be the side of an equilateral spherical triangle, and
of that of its polar triangle, show that
cos. -J a cos. ^ a' = J.
205. If A be the angle of an equilateral spherical triangle,
and A' that of its polar triangle, show that
hav. A . hav. A' = hav. 60°.
206. Any point P is taken within a right-angled spherical
triangle, and perpendiculars PM, PN are drawn to its sides CA,
CB, which contain the right angle. Show that, if CM = a?,
CN = y, PN = x', PM = y',
tan.2 x + tan.2 y — tan.2 CP (i)
sin. x' x..v
(11)
(iii)
207. In a spherical triangle, right-angled at C, show that
(i) If a + b + c = 90°, tan. A = cosec. b — sec. I.
(ii) If a 4- b + c = 180°, cot. A = ! (sec. b - cos. 6).
208. If D be the middle point of the base BC of a spherical
triangle ABC, and AB, AD, AC be in arithmetical progression,
show that AB is a quadrant.
209. A spherical triangle ABC is right-angled at C. If p
be the perpendicular from C upon AB, prove that
sin.2 p sin.2 c = sin.2 a + sin.2 b — sin.2 c.
MISCELLANEOUS EXAMPLES. 259
210, In a spherical triangle ABC the arc AB is a quadrant,
and CD is drawn perpendicular to AB. Prove that
cot.2 CD = cot.2 A + cot.2 B.
211, In any spherical triangle prove that
. a + b + c . A B C
sin. sin. — = cos. — cos. — sin. a.
-j A 2 2
212, Through the vertical angle A of an isosceles spherical
triangle there is drawn an arc of a great circle meeting the base
in D. Show that
tan. I BD tan. \ CD = tan. ~ (BA + AD) tan. \ (BA - AD).
a & a 2i
213, A triangle has two sides quadrants, and the difference
between the sum of its angles and that of the angles of its polar
triangle is a sixth of^the sum of all six angles. Show that the
angles of the triangle are 90°, 90°, 135°.
214, If the cosine of each of the equal sides of an isosceles
spherical triangle is — — , and the base a quadrant, show that the
v 3
angles of the triangle are 45°, 45°, 120°.
215, Show that the cosine of the angle between the chords
of the two sides b, c of a spherical triangle is equal to
b . c be
sin. - sin. - -f cos. -- cos. - cos. A.
& a A 2>
216, If a, /3, 7 be the angles made with the sides BC, CA,
AB of a spherical triangle by the connectors of their middle
points with A, B, C respectively, then
tan. a — —
s 2
260 APPENDIX.
217. The diagonals, AC, BD, of a spherical quadrilateral
ABCD meet at right angles in 0. If a, /3, 7, 8 are the arcs of
great circles drawn from O perpendicular to the sides of ABCD,
taken in order, prove that
cot.2 a + cot.2 7 = cot.2 /3 + cot.2 S.
218. The square ABCD is inscribed in a sphere whose
centre is 0. If each side of the square is er, and the diameter
of the sphere is d, prove that the cosine of the acute angle
a2
between the planes OAB and OBC is — — — -.
d? — a2
219. If in the spherical triangle ABC the arc which bisects
the angle A meet the opposite side in D, and E be the middle
point of that side, show that
tan. DE . tan. ^ (b + c) = tan. £ tan. I (b - c).
u ft , a
220. A pyramid stands on an irregular polygonal base. If
one face be an equilateral triangle, inclined at an angle of 45°
to this base, show that the inclination of the next face, which is
a right-angled isosceles triangle, is 60°.
221. If a, 6, c, d be the sides of a spherical quadrilateral,
taken in order, 8, S' the diagonals, and <f> the arc joining the
middle points of the diagonals, show that
g &'
4 cos. - cos. — cos. <f) — cos. a + cos. b -f cos. c 4- cos. d.
2 u
222. Prove that in a spherical triangle
tan. b cos. A + tan. a cos. B
tan. c =
1 — cos. A cos. B tan. a tan. b
223. A spherical triangle ABC has all its angles right
angles, and DEF is a great circle which meets BC in D, CA
in E, and AB produced in F. Prove that
8 cos. DE . cos. EF . cos. FD = sin. 2BD . sin. 2CE . sin. 2AF.
224. The circumference of a great circle is trisected in
A, B, C, and three equal small circles are described having
A, B, C for their poles. Show that they will intersect one
another if the area of each small circle is greater than three-
fourths of the area of a great circle.
MISCELLANEOUS EXAMPLES. 261
225. In a spherical triangle, right-angled at C, show that
a . b a b
1 Sm'2Sm'2 1 cos. -cos. -
sin. o E = , cos. o E = ,
« c * c
cos. - cos. -
where E = A + B
H.— PART III. CAP. I.-IV.
226. Find to five places of decimals the value of
Find the value of x in the following equations : — •
vers. 100° a/ _
227. - T - = ^vers. 50°.
vx
228. x3 cot. 108° = 128 sin. 72° cos. 18°.
•
23°- - 7?r — N = (2-71828)*.
cos. (22° + x)
3 27-2968 sin.3 101° 101-78 cos.* 320°
Vx ~~ (-009)* tan. 215° '
232. (sin. 8° + cos. 8°)2* = 2 sin. 16° (tan. 32°)*.
233. log. y? + log. 2» + 1 = 0.
Solve the following plane triangles : —
234. a = 32, b = 26, A = 46° 20'.
235. a = 35, c = 20, A = 107° 25'.
236. A = 84° 52' 30", B ^ 52° 25' 30", a = 235.
237. a = 56, 6 = 73, A = 41° 10'.
238. b = 230, c = 327, A = 78° 30'.
239. & = 3000, c = 3266, A = 49° 28' 15".
240. a = 93-4, b = 70-4, C = 116° 12'.
24].. In the plane triangle ABC, a = 152 feet, b = 188 feet,
c = 142 feet ; find the angles A, B.
242. The sides of a triangle are 35, 43, 48. Show that one
of the angles is 60°.
243. The sides of a plane triangle are '75, -93, 1-23; find
the greatest angle.
262 APPENDIX.
244. In a plane triangle the sum of two sides is 160 feet,
their difference is 35 feet, and the difference between the angles
opposite these sides is 10°. Solve the triangle.
In the following triangles find the area : —
245. a = 101-5 feet, b = 167-5 feet, C = 79° 33'.
246. b = 65-3 feet, c = 89-4 feet, C = 88° 30'.
247. a = 77 feet, b = 75 feet, c = 68 feet.
248. In the triangle ABC, a = 241 yards, b = 169 yards,
C = 104° 3' 45" ; find the side of a square which has the
same area.
249. Show that the area of a plane triangle, of which the
sides are 98'29 yards, 105*72 yards, and 115-25 yards, is very
nearly an acre.
250. Express in acres and decimals of an acre the area of a
triangular field whose sides are respectively 316, 558, and 726
yards.
251. In a plane triangle ABC, AB = 37, BC = 45, AC = 74 ;
find the length of the straight line drawn from. C to the middle
point of AB.
252. A quadrilateral figure ABPQ has one side, AB, 300
yards. The angle QAB = 58° 20', PAQ = 37°, ABQ = 53° 30',
and PBQ = 45° 15' Calculate the side PQ.
253. From a station, A, an object, 0, bore S.E. by S., but
after the observer had walked 1,200 yards S.W. the object bore
E. by S. Find AO.
254. The angle of elevation of a tower 100 feet high from a
place N.N.W. of it is 45°, and from another place E. by S. of it
is 60° ; find the bearing of the second place from the first. '
255. What angle will a flag-staff 18 feet high, on the top
of a tower 200 feet high, subtend to an observer on the same
level with the foot of the tower, and 100 yards distant from it ?
256. A quadrilateral has its sides AB, BC, CD, DA and its
diagonal BD, respectively equal to 25, 39, 52, 60; and 65 inches.
Show that the angles at A and C are right angles, and find the
other two angles of the quadrilateral.
257. The angle of elevation of the top of a hillock being
2° 33' 15", a man proceeds to walk up it (by a direct course)9
MISCELLANEOUS EXAMPLES. 268
the angle of elevation of his path for 100 yards being 2°, and
afterwards 3°. Find the vertical height of the hillock.
258. A flag-staff, 27 feet in height, standing on the edge of
a cliff, subtends an angle of 0° 40 ' at a ship at sea, the angle of
elevation of the cliff being 15°. Find the distance of the cliff
from the ship (the point of observation being considered to be
in the same horizontal plane with the foot of the cliff).
259. Two objects, A and B, lie in the same straight line
and the same horizontal plane with the base, D, of a tower CD,
whose height is 98 feet. If the angles ACB, ABO are observed
to be 7° 21' and 8° 12' respectively, find the distance AB.
260. At a station due south of a circular fort a man observes
the horizontal angle subtended by the fort to be 2° 11' 30"-
He then walks E.N.E. to a station a quarter of a mile distant,
and finds the angle subtended to be the same as before. Find
the diameter and distance of the fort.
261. A flag-staff which leans over to the east is found to
cast shadows of 198 feet and 202 feet, when the sun is due
east and due west respectively, and his altitude is 7°. Find the
length of the flag-staff and its inclination to the vertical.
262. At two points in a straight sea-wall, 440 yards apart,
the lines drawn to a ship are found to be inclined respectively
at angles 60° and 66° to the direction of the wall. Find the
distance of the ship from the nearest point of the wall. (N.B.—
The stations lie on opposite sides of the ship.)
263. The elevations of a tower from two points in the same
straight line with its foot, at a distance of 41 yards from one
another, are 58° 23' 30" and 27° 26' 15". Find the height of
the tower.
264. A privateer is lying 10 miles W.S.W. of a harbour,
when a merchantman leaves it steering S.E. 8 miles an hour.
If the privateer overtakes the merchantman in 2 hours, find hei
course and rate of sailing.
265. At the point C is a light-ship. The directions CA, CB
are known to include an angle of 30° 20', and A bears N.E. by
E. from C. The distance CA being 3 miles, a steamer is seen
to start from A at a speed of 9 knots. In what direction must
she steer so that in 40 minutes she may appear to the light-ship
in thf> line CB ?
264 APPENDIX.
266. The legs of a pair of compasses are 1O5 inches and
6 inches respectively. At what angle must they be placed to
mark off a distance of 12 '5 inches ?
267. An observer, situated due south of a landmark A,
notices that the bearing of another landmark, B, is N. 15° 22' W.
He then walks one statute mile N. 45° W., and finds that B is
due east while A bears S. 84° 15' E. Find the distance AB.
268. A man walking along a road which runs N.E. observes
that a distant object bears N. 60° E. After walking 1,500
yards further the object bears N. 80° E. Find the distance of
the object from the first point of observation, and its shortest
distance from the road.
269. From the junction of two lines of railway two trains
start together at different speeds. The angle between the
directions of the lines is 108°, and the trains reach their re-
spective destinations, which are 70 miles apart, in lh 30m. If
the speed of one train be 20 miles an hour, find the speed of
the other.
270. At 4h 30°* P.M. on a certain day a vessel sailing S. by
W. at a uniform rate of 7*5 knots passed another vessel sailing
E. by S.-JS. 9 knots. Assuming the courses and rates to have
been uniform, find the bearing and distance of one ship from the
other at the preceding noon.
271. A column stands at the top of a hill whose inclination
is 10°, and at two stations 40 feet apart on the slope of the hill,
and in a direct line with the foot of the column, the angles of
elevation (above the horizon) of the top of the column are 40°
and 46°. Find its height.
272. A man standing at a certain point in a straight road
observes that the straight lines drawn from that point to two
landmarks on the same side of the road are each inclined at an
angle of 45° to the direction of the road. He walks to a point
250 yards further along the road, and there finds that the former
angles of inclination are changed to 25° and 65° respectively.
Find the distance between the landmarks.
273. A hill rises from a horizontal plain. On the hill is a
tower. From a point C in the plain the angles of elevation of
the top, A, and the bottom, B, of the tower are respectively
'MISCELLANEOUS EXAMPLES.. 265
72° 15' and 54° 45'. A distance CE of 200 feet is then
measured horizontally, and the angle of elevation of A as seen
from E is then 44° 12'. The straight lines CA, EA, CB being
in the same vertical plane, find the height AB.
274. A spectator observes the explosion of a meteor due
south of him at an altitude of 28° 45'. To another spectator
11 miles S.S.W. of the former it appears at the same instant to
have an altitude of 42° 15' 30". Show that there are two
possible heights above the earth's surface at which it may have
exploded, and find these heights.
275. A person is standing on a cliff looking north (200 feet
above the sea), and observes the elevation of a cloud to be
53° 7' 45", and the depression of its shadow on the sea to be
7° 7' 30". The sun being due south, and the line joining sun
and cloud being inclined at an angle of 45° to the horizon, find
the height of the cloud above the sea.
276. A tower 120 feet high stands in the middle of a
horizontal field whose shape is that of an equilateral triangle.
From the top of the tower each side of the field subtends an
angle of 100°. Find the length of a side of the field.
277. Two towers on a hill-side are 70 feet and 130 feet high
respectively. The line connecting their bases, which are 80 feet
apart, makes an angle of 10° with the horizon. Find the
inclination to the horizon of the line joining their tops.
278. A ship sailing at a uniform rate was observed to bear
N. 30° 57' 30" E. After 20 minutes she bore N. 35° 32' 15"
E., and after 10 minutes more N. 37° 52' 30" E. Find the
direction in which she was sailing.
K. —PART III. CAP. V.-VII.
In the following spherical triangles find the angles : —
279. a = 158° 54' 15", b = 118° 21', c = 63° 42'.
280. a = 95° 18' 15", b = 50° 45' 15", c = 69° 12' 45".
281. a = 58° 34' 30", b = 60° 18', c = 92° 10'.
In the following spherical triangles find the third side : —
282. a ^ 112° 30', b = 53° 45' 30", C = 23° ; find c.
266 APPENDIX.
283. b = 62° 36' 45", c = 100° 10' 15", A = 81° 24' ;
find a.
284. a =107° 15', 6 = 94° 12' 45", 0 = 91° 14' 30";
find c.
285. b = 70° 14' 15", c = 38° 46' 15", A = 48° 56' ;
find a.
286. b = 58° 25', c = 49° 10', A = 71° 18' 30" ; find a.
In the following spherical triangles find the other angles :— -
287. a = 65° 20' 15", b = 50° 30' 15", C = 118° 10' 45".
288. a = 49° 10', b = 58° 25', C = 71° 18' 30".
289. a = 117° 30', b = 57° 45', C =84° 15'.
290. a = 45° 30', b = 75° 45', C = 80°. i
291. In the spherical triangle ABC, given A = 124° 13',
B = 49° 7', a =115° 6'; find I.
292. Having given b = 64° 30' 45", c = 95° 7' 45",
C = 100° 48', 15"; find a.
293. Having given A = 126° 37', B = 48° 30', a = 115° 20';
solve the triangle.
294. Having given a = 101° 37', A = 98° 13' 30", B =
85° 17'; find b.
295. Having given a = 52° 13', b = 70° 21' A == 46° 18' •
findB.
296. Having given A = 67°, B = 120°, c = 43°; find C.
297. Having given A = 71°, B = 59°, c = 38°; find C.
298. Having given A = 112° 25', B = 73° 15', c = 24° 38';
find a, &, C.
299. Having given B = 62° 20' 15", C = 60° 13' 45"
a = 150° 10' 30"; find A.
300. Having given A = 130° 50', B = 121° 35' c =
108° 41' 30"; find C.
301. Having given A = 85°, B = 65°, c = 130°; find a, b.
302. Having given A = 110°, B = 54° 30' c= 22° 30' 30"-
findC.
303. Having given A = 105° 15' 30", B - 65° 30' C =
85° 20' 45" ; find the sides.
304. Having given A = 121° 36' 30", B = 42° 15' 15"
C =34° 15'; find the sides.
MISCELLANEOUS EXAMPLES. 267
305. Having given A = 94° 30', B = 104° 33' 45", C =
144° 9' 30"; find a.
306. Find the side of a spherical triangle each angle of which
is 70° 46' 15".
Solve the following right-angled spherical triangles : —
307. C = 90°, a = 49° 17' 30", b = 95° 36' 15".
308. C = 90°, a = 72° 27', c = 91° 18'.
309. C = 90°, a = 54° 41' 45", B = 101° 48'.
310. C = 90°, a = 18° 25' 45", b = 72° 15' 30".
311. C = 90°, a = 24° 15', c = 145° 30'.
312. C = 90°, a = 132° 39' 30", B = 78° 10'.
313. A = 90°, a = 98° 14' 30", b = 54° 41' 45".
314. C = 90°, a = 57° 16', b = 96° 24' 15".
315. C = 90°, a = 22° 30', A = 30°.
316. C = 90°, a = 38° 47' 15", A = 42° 50' 45".
317. C = 90°, a = 37° 25' 30", B = 40° 4' 15".
318. A = 90°, B = 101° 50', C = 48° 27' 15".
319. c = 90°, A = 131° 30', B = 120° 32'.
320. a = 90°, A = 100°, B = 74° 36' 30".
321. a = 90°, A = 80°, B = 75°.
322. a = 90°, A = 85°, b = 97°.
323. a = 90°, B = 79° 54' 15", C = 97° 56' 45"
324. a = 90°, b = 78° 14' 30", c = 50° 10'.
325. In the spherical triangle ABC, a = 90°, b = 90°, and
c is one half the side of an equilateral triangle in which each
angle is 75°. Find A, B and 0.
269
ANSWERS TO APPENDIX.
1. 220 72 yards. 2. 8 feet 11-4 inches.
3. 12 feet 9-6 inches. 4. 4° 17' 50".
5 25° 42' 51" ; '449. 7- («) 191° miles nearly » ^ 30
8. '1744 inch. ' 9. 92,678,723 miles.
10. - ; 57' 18" nearly 11. 90°.
12. 141°-43 nearly. *>. 65° 28' 51".
14 1 15. 1-37 nearly.
16. 360°. "' l^1-1—^'
-IQ -*-"U
18. 114° 35' 30". w '
21. 75°, 30°. 22. 27°, 69°, 111°, 153°.
23. 48° 11' 30" ; '841.
B
1 _§§§
fifl " nr 82. -r. ^ RAJ.
Ov» "3 "-^ p^* VJrx^t
5 o
84 — - 85- -^' 87' 12'
a4' 325 72
1 -j
/_L °0. sine — — , tangent - -5-
• ~ V 10* >/iao
93. - |.
96. I or - 1.
98. («) 0 = ™
(c) d = «•
(e) d = w;
C.
94. ^. 95.
r»T <v o
97. ^-.
r or mr + (— l)ns«
TT or w?r — 2"%
7T
rr or WTT ± -£
; tan.
± -
270
ANSWERS TO APPENDIX.
100. 2x = WTT ± 5.
o
110. |.
112. 4-48289 nearly ; 2-443 nearly.
111. 1-58496.
113. 1-062 nearly.
H.
226.
229.
232.
234.
235.
236.
237.
238.
239.
240.
241.
244.
245.
247.
250.
252.
254.
256.
257.
259.
260.
261.
262.
264.
266.
268.
270.
271.
273.
275.
277.
279.
280.
281.
282.
284.
286.
227. 2-73605.
230. 61° 59'.
228. - 7-0895.
231. -00016736.
233. -3684.
43-84.
23-36.
1-47687.
41411.
- -8345 nearly.
B = 35° 59' 45", C = 97° 40' 15", c
B = 39° 32' 30", C = 33° 2' 30", b -
C = 42° 42', b = 187, c = 160.
B = 59° 6' or 120° 54', 0 - 79° 44' or 17° 56', c
B = 38° 43', C = 62° 47', a = 360-3.
B = 60°, C = 70° 31' 45", a = 2633.
A = 36° 53' 45", B = 26° 54' 15", c - i39-6.
A = 52° 38', B = 79° 25' 15". 243. 93° 30' 45".
Angles 26° 48', 16° 48', 136° 24'. Sides 97-5, 62-5, 149-1.
246. 2049-5 square feet.
83-71 or 26-2.
59°29'
248.
251.
253.
255.
30".
258.
140-5 yards.
58-38.
1411 yards.
2° 18' 45".
2159-2 feet.
8359-6 square feet.
2310 square feet.
17-2046 acres.
206 yards nearly.
S. 42°29'E.
B = 120° 30' 30", I
30 feet nearly.
327-9 feet.
Diameter 22 yards ; distance 575 yards.
24-6 feet ; inclination to vertical 4° 39'.
430-3 yards. 263. 31-28 yards.
S. 70° E. 10-9 knots. 265. S. 78° 48' E.
94° 33'. 267.
2515-6 yards ; 651-6 yards. 269.
S. 64° 48' W. 60 miles nearly.
114-2 feet. , 272.
154-5 feet. 274.
1000 feet. 276.
43° 9' 45". 278.
364 yards nearly.
36*43 miles per hour.
731 yards.
4-33 miles or 13-21 miles,
394 feet.
S. 44° 38' E.
K.
A = 156° 19', B = 100° 59' 45", C --= 90°.
A = 115° 58' 15", B = 44° 22' 15", C = 57°- 35'.
A = 51° 31' 15", B = 52° 50', C = 113° 33' 30".
c = 62° 38'. 283. a = 87° 10' 15".
c = 89° 56' 15". 285. a = 49° 24' 15"
a = 56° 42'. 287. A = 63° 24', B = 42° 59'.
ANSWERS TO APPENDIX. 271
288. A = 59° 2' 15", B = 74° 53' 45".
289. A = 116° 23' 30", B = 58° 39' 30".
290. A = 47° 16' 15", B = 86° 32' 45".
291. b •= 55° 53' 1,5". 292. a = 79° 41'.
293. b = 57° 30', c = 82° 26' 30", C = 61° 41' 30 '.
294. b = 80° 31' 15" or 99° 28' 45".
295. B = 59° 29' or 120° 31'. 296. C = 38° 53' 15".
297. C = 61° 54' 15".
298. a = 107° 42' 15", b = 99° 19' 15", 0 = 23° 51' 30".
299. A = 153° 50'. 300. C = 123° 18'.
301. a = 104° 5' 45", b = 61° 55' 45".
302. 0 = 25° 8'.
303. a = 104° 39' 30", b = 65° 51' 15", c = 91 49' 15".
304. a = 76° 36', b = 50° 11' 15", c = 40°.
305. a = 77° 13' 30". 306. 60° 34' 30".
307. c = 93° 39' 15", B = 94° 15' 15", A = 49° 25' 45".
308. A = 72° 29' 45", B = 94° 7', b = 94° 19'.
309. A = 55° 23', b = 104° 21' 30", c = 98° 14' 30".
310. A = 19° 17', B = 84° 13' 30", c = 73° 11' 45-'.
311. A = 46° 28' 45", b = 154° 40' 30", B = 130° 57'.
312. A = 131° 32' 45", b = 74° 5' 45", c = 100° 42'.
313. B = 55° 33', C = 101° 48' 15", c = 104° 21' 45".
314. A = 57° 2 45", B = 95° 23' 30", c = 93° 27' 30".
315. b = 45° 50' 45" or 134° 9' 15", B - 69° 37' or 110° 23', c = 49° 56' 30"
or 130° 3' 30".
316. b = 60° 3' 15" or 119° 56' 45", B = 70° 9' 15" or 109° 50' 45",
c = 67° 6' or 112° 54'.
317. b = 27° 4' 30", c - 45°, A = 59° 15' 15".
318. a = 100° 42', b = 105° 54' 15", c = 47° 20' 30".
319. a = 127° 18' 30", b = 113° 50', 0 = 109° 40'.
320. b = 78° 14' 15", c = 50° 10' 15", C = 49° 8' 15 .
321. b - 78° 45' 45", c = 131° 9' 15", C = 132° 8' 15' .
322. c = 54° 38', B = 98° 35' 45", C = 54° 19' 30".
323. b = 80°, c = 97° 49' 30", A = 88° 36' 45".
324. A - 100°, B = 74° 36' 15", 0 = 49° 8'.
325. A = B = 90° ; C = c = 34° 47'.
Spottiswoocle & Co, Ltd,, Printers, New-street Square, London.
UNIVERSITY OF CALIFORNIA LIBRARY
BERKELEY
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