i
u-
^
ELEMENTARY
TRIGONOMETRY
BY
W. E. PATERSOX, M.A., B.Sc.
MATHEMATICAL MASTER, SIERCERs' SCHOOL
AUTHOR OF ' SCHOOL ALGEBRA
OXFORD
AT THE CLARENDON PRESS
LONDON, EDINBURGH, NEW YORK, TORONTO
AND MELBOURNE
HENRY FROWDE
1911
BY THE SAME AUTHOR
SCHOOL ALGEBRA
THIRD EDITION
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PREFACE
The common examination syllabus ' Trigonometry up
to and including solution of triangles ' has caused most
textbooks to treat the subject as if the sole use of
Trigonometry were to solve triangles, and the practical
examples deal almost exclusively with various forms of
triangle-solving under the heading ' Heights and Dis-
tances '. Further it is customary to define the trigono-
metrical ratios by means of a right-angled triangle;
this encourages the mistaken idea that the ratios are
fundamentally attached to a triangle, and does not
impress upon the pupil the fact that they are the pro-
perty of an angle and of an angle only.
In this book the trigonometrical ratios are introduced
as functions of the angle. The trigonometrical proper-
ties of the single angle are treated fully in the early
chapters, and from the beginning the examples apply
Trigonometry wherever it may be useful, to Geometry,
Mensuration, Analytical Geometry, Physical formulae,
&c. The right-angled triangle definitions are given in
Chapter Y. This chapter contains, in addition to the
usual matter, a short treatment of Plane Sailing in
Navigation. It is hoped that the examples in Navigation
will provide practice in the use of tables, and at the
same time be of interest to the pupil. Other examples
260011
4 PREFACE
in this chapter lead up to the formulae dealing with the
ordinary triangle.
The formulae for ^ A are proved by Geometry, indepen-
dently of tlie addition formulae; thus the triangle is
treated fully without breaking the sequence with a dis-
cussion of the A + B and allied formulae.
A chapter is devoted to Projection. This includes
a discussion of Vector Quantities, their composition and
resolution, and finishes with a geometrical treatment
of Demoivre's Theorem. In the last chapter, the addi-
tion formulae and the allied formulae are treated fully ;
the projection proofs are used and recommended but the
old-fashioned proofs are also given
Throughout the book the student is given every oppor-
tunity of developing the subject for himself. A large
portion of the bookwork first occurs among the examples
of earlier chapters. Also, when a formula has been
proved, the proofs of others of the same kind are left
for the student to supply. Thus, when siii^A + cos^^ = 1
has been proved, the student should have no difficulty
in proving the connexion between sec^A and tsui^A;
when sin(^ + ^) has been found, the student should
himself find the expanded form of cos (J. + B), Sec.
The sets of examples in the body of a chapter arc num-
bered IV. a, IV. b, &c. ; these deal only with the matter
immediately preceding them. The last set of examples in
a chapter has no distinguishing letter and serves for
revision of the whole chapter. There are also three
sets of Revision Examples, l^ookwork is frequentl}^ set
as an example, both in the Revision Sets and elsewhere ;
only by constant repetition, oral or written, can the
PEEFACE 5
bookwork be learnt. There are a few sets of oral
examples; these are intended to fill up spare minutes
at the end of a lesson and often bring out the weak
points in a pupil's knowledge. The book contains nearly
1,000 examples ; it is not intended that any one should
attempt all these, but it is hoped that they include
a sufficient variety of types and a sufficient number of
each type to meet all requirements.
Many examples are taken from Examination Papers
by kind permission of the following authorities : —
The Controller of His Majesty's Stationery Office.
The University of Cambridge.
The Joint Matriculation Board of the Scottish Univer-
sities.
The Intermediate Education Board for Ireland.
The Oxford and Cambridge Schools Examination
Board.
The Delegacy for Oxford Local Examinations.
The Syndicate for Cambridge Local Examinations.
The College of Preceptors.
I am indebted to Mr. Norman Chignell, B.A., of
Charterhouse, for many suggestions and for assistance
in correcting the proof-sheets. It is too much to hope
that the answers are wholly free from mistakes, and
I shall be grateful to receive early intimation of any
corrections that may be found necessary.
VV. E. P.
April, 1911.
CONTENTS
CHAP. PAGE
Preliminary. Propositions in Geometry. Graphs.
Logarithms. Greek Alphabet ... 9
I. Angles and their Measurement. The Degree.
Positive and Negative Angles. The
Grade. Connexion between Arc and
Angle. Circular Measure. The Points
OF THE Compass. Gradient .... 25
II, Definitions of Trigonometrical Ratios. Their
Graphs. Inverse Notation. Construc-
tion OF Angle with given Ratio . . 37
IIT. Elementary Formulae. Identities. Ratios
OF Complementary and Supplementary
Angles. Ratios of 0^ 90", 30°, 60°, 45^
The Small Angle 49
Revision Examples A .65
TV. Use of Tables. Harder Graphs. General
Solutions of Equations. Use of Loga-
rithms 70
V. The Right-angled Triangle Definitions of the
Trigonometrical Ratios. Solution of
Right-angled Triangles. Navigation:
Plane Sailing. Ratios of the Double
Angle and Half Angle 85
VI. Relations between the Sides and Angles of a
Triangle. Area of Triangle. The Circles
of the Triangle 107
Vn. Solution of Triangles 119
Revision Examples B 134
3,.^., CONTENTS
CHAP. PAGE
VIII. Projection. Vector Quantities. Composition
AND Resolution of Vectors. Projection
OF AN Area on a Plane. Geometrical
Representation of Imaginary Quantities.
Demoivre's Theorem 139
IX. Ratio of the Sum or Difference of two Angles.
Sum or Difference of Two Sines or Co-
sines. Product of Sines and Cosines.
The Double Angle and Half-Angle . .155
Revision Examples C 175
Miscellaneous Problems (taken from recent
Army Entrance and Civil Service Exami-
nation Papers) 184
Examination Papers 190
INDEX 201
The following course of reading is suggested for l3eginners : —
Chapter I, §§ 1-4, 9-11.
Chapter II.
Chapter III, §§ 21-31.
Chapter IV, §§ 34-7, 40-2.
Chapter V, §§ 43-5, 51-3.
Chapter VI, §§ 54-8.
Chapter VII.
PliELIMINAEY CHAPTEE
GEOMETRY
A KNOWLEDGE of the following geometrical facts is required.*
In this book these propositions are referred to by the numbers
given below.
Angles.
Prop. 1. If a straight line meets another straight line, the
adjacent angles are together equal to two right angles.
Prop. 2. If two straight lines cut, the vertically opposite angles
are equal.
Prop. 3. The angle at the centre of a circle is double an angle
at the circumference standing on the same arc.
Prop. 4. Angles in the same segment of a circle are equal.
Prop. 5. Angles at the centre of a circle standing on different
arcs are in the same ratio as the lengths of the arcs.
Triangles.
Prop. 6. [a] The three angles of a triangle are together equal
to two right angles.
(&) If one side be produced the exterior angle equals the sum of
the two interior opposite angles.
Prop. 7. Any two sides of a triangle are together greater than
the third.
Proj). 8. Two triangles are congruent (i. e. are equal in every
respect) if they have —
(a) two sides of the one equal to two sides of the other, each
to each, and the angle contained by the two sides of the one equal
to the angle contained by the two corresponding sides of the other ;
or (h) three sides of the one equal to three sides of the other, each
to each ;
or (c) two angles of the one equal to two angles of the other,
* For proofs see Warren's Experimental and Theoretical Geometry
(Clarendon Press), or any standard textbook.
10 GEOMETRY
each to each, and a side of the one equal to the corresponding side
of the other.
Prop. 9. If two triangles have an angle of the one equal to
an angle of the other, and the sides about another pair of angles
equal, each to each, then the third angles are either equal or
supplementary.
Prop. 10. (a) If two sides of a triangle are equal, the opposite
angles are equal.
(&) If two sides are unequal, the greater side is opposite a greater
angle.
(c) If all the sides of a triangle are equal, all the angles are equal.
Prop. 11. (a) If two angles of a triangle are equal, the opposite
sides are equal.
{b) If two angles are unequal, the greater angle is opposite a
greater side.
(c) If all the angles of a triangle are equal, all the sides are equal.
Prop. 12. Two triangles are similar (i. e. their angles are equal,
each to each, and the ratio of pairs of sides opposite equal angles
is the same for all three angles) if they have —
{a) their angles equal each to each ;
(6) their sides in the same ratio ;
(c) an angle of the one equal to an angle of the other, and the
sides about the equal angles in the same ratio.
Prop. 13. (Pythagoras' Theorem.) In a right-angled triangle the
square on the hypotenuse is equal to the sum of the squares on the
other two sides.
Parallel Lines.
Prop. 14. (a) If a line is drawn to cut two parallel lines, it
makes (i) the alternate angles equal, (ii) the interior angles on
the same side of it together equal to two right angles, (iii) the
exterior angle equal to the interior opposite angle.
(b) The opposite sides and angles of a parallelogram are equal.
Area.
The unit of area is the area of a square whose side is of unit
length.
Prop. 15. The number of units of area in a rectangle is equal to
the product of the number of units of length in one side multi-
plied by the number of units of length in the other.
GEOMETRY 11
Or, more shortly : Area of rectangle = length x breadth.
Proio. 16. The area of a triangle = \ base x altitude.
The Concurrencies of the Triangle.
Prop. 17. The lines bisecting the sides of a triangle at right
angles are concurrent (i. e. meet at a point).
The point in which they meet is the centre of the circle passing
through the three vertices and is called the circumcentre.
Prop. 18. The lines drawn from the vertices to bisect the oppo-
site sides are concurrent.
These lines are called the medians and the point of concurrency
is called the centroid.
Prop. 19. (a) The lines bisecting the angles are concurrent.
The point of concurrency is the centre of the circle that touches
all the sides, and is called the incentre.
(&) If two of the sides be produced, the lines bisecting the ex-
terior angles so formed and the line bisecting the interior angle
contained by the produced sides are concurrent.
The point of concurrency is the centre of the circle that touches
the two sides when produced and the third side (not produced) ;
it is called an e-centre.
Prop. 20. The perpendiculars let fall from the vertices on the
opposite sides are concurrent.
The point of concurrency is called orthocentre.
The Circle.
Prop. 21. The straight line passing through the centre, at right
angles to a chord, bisects the chord.
Prop. 22. [a) The angle at the centre of a circle is twice the
angle at the circumference on the same arc. (&) Angles in the
same segment are equal, (c) The opposite angles of a quadri-
lateral inscribed in a circle are together equal to two right angles.
Prop. 23. The tangent at any point is at right angles to the
radius drawn to that point.
Prop. 24. (a) Two tangents can be drawn to a circle from any
external point. (&) The parts of these tangents between the
external point and the points of contact are equal, (c) The line
joining the external point to the centre bisects the angle between
the tangrents.
12
GEOMETRY
Prop. 25. The ratio of the circumference of any circle to its
diameter is the same for ail circles.
This ratio is denoted by the symbol tv : its value is 3*1416 correct
to five significant figures.
Prop. 26. The area of a circle equals the area of the rectangle
contained by the radius and a straight line equal to half the
circumference.
This is usually expressed in the formula : Area = tt r^.
GRAPHS*
Geometrical. If two straight lines are drawn in a plane, the
position of any point in the plane can be determined by means of
its distances from those lines.
Y
Fig. 1.
It is usual to draw one of the lines horizontal and the other
perpendicular to it. The customary notation is shown in Fig. 1.
X'OX is called the axis of x ;
Y'OY i% called the axis of y;
0 is called the origin ;
ON is called the abscissa of the point P;
NP is called the ordinate of the point P.
The abscissa and ordinate are called the co-ordinates t of the
point P.
* For a fuller treatment of Graphs see School Algebra published by
the Clarendon Press.
t Tliese co-ordinates are called Cartesian co-ordinates because they
were first used by the French mathematician, Descartes.
GRAPHS
13
The abscissa is said to be positive if drawn to the right, negative
if drawn to the left. Similarly, the ordinate is positive if drawn
upwards from N, negative if drawn downwards. The number of
units of length in ON, preceded by the proper sign, is usually
denoted by x, and the number of units of length in NP, preceded
by the proper sign, is denoted by tj. In each case the sign + is
often omitted.
Thus, in Fig. 2, the co-ordinates of A are x = -4, y = 2, of
C, x = 4:, y = ^, oi L, x = (), y = -7.
1
1 1
1 1
1
Y
io
1
B
5
1
ID
1
c
X
A
X
O
r~
5
K
1
1
1
Gl
E
H
5
.
i 1
1
1
V
1
LI
1
1
1
^°,
1
1
1
i
1
1
Y| 1
1
1 1
i
Fig. 2.
Very often a point is described by writing the values of the
co-ordinates in brackets ; e. g. the point H might be described as
the point ( — 6, —4).
Exercise. Write down the co-ordinates of all the points in
Fig.- 2.
Graphs of Statistics. The magnitude of any quantity may be
represented by a straight line which contains as many units of
length as the quantity contains units of its own kind.
If two quantities are changing their values at the same time,
14
GRAPHS
the simultaneous values may be represented in the same figure
by taking horizontal lengths to represent one magnitude and
vertical lengths to represent the other.
1
~
5
0
1
'
1
1
i
/
4
5
/
/
'
\
4
0
1
\
\
/
\
/
/
3
5
1
\
/
1
i3
/
«
3>
/
/
c
i
1
1
O
/
V
/
j 1
\
1
1
S.\M.
T.
W. 1 Th.
F.
Sot.
1
i i 1
1
1 1 1
1
Fig. 3.
In Fig, 3 the changing quantities are time and temperature ;
and the dots show that at noon on Sunday the temperature was
29°, on Monday the temperature was 35, &c. In fact the figure
conveys the same information as the following table : —
Sun. Mon. Tues. Wed. Thurs. Fri. Sat.
Temp. 29° 85° 42° 31° 27° 43° 50°
If there is no information about intermediate temperatures,
the points are joined by a series of straight lines. The figure
now forms a graph.
In describing such a graph we should say that the abscissae
represent time and the ordinates temperature.
Graphs of functions. If two quantities x and // are such that
a change of value in the one causes a change of value in tiie
other, then either of them is said to be a function of the other.
GRAPHS
15
This is expressed thus : y =f{x), or x =f{y) where f[x) means
a function of x. A graph can be drawn in which the abscissae
are proportional to the values of x and the ordinates to the values
of y. This graph is called the graph of the function f{x) or of
the equation y=f{x). This maybe more easily understood by
considering a few algebraical functions.
Example I. Draw the graph when y = \x-%
(Choose values of x which will make y a whole number.)
y = \{^x-2)
X -6-149
Zx-1 -20 -5 10 25
y _ 4 -1 2 5
-
1
0
-
f^
5
y
^
y
!
/
r^
1 I>^1 1
^ ! '
0
5
u
"51
"l
O
^
f"
-
!
1/
1 ' ' ' 1
y
1 !
: I 1
/^
y
1
1
k-
5
i
1
1
Fig. 4.
When the points corresponding to these values of x and y are
plotted, it is found that they lie on the straight line shown in
Fig. 4, It is also found
(i) That any simultaneous values of x and y connected by the
given equation are the co-ordinates of some point on this straight
line ;
(ii) That the co-oidinates of any point on the straight line
satisfy the equation.
It is found by experience (and can be proved from the geo-
16
GRAPHS
metrical propositions on proportion) that, when x and y are
connected by an equation of the first degree, the graph is always
a straight line.
Example II. In the same figure draw the graphs of
y = x^-^x-V^ and r» = 2f/H3.
Neither of these- equations is of the first degree, therefore
neither of the graphs is a straight line. At least six points must
be found on each.
yr=x^-Zx\'l. (i)
X
-3
— 2
-1
0
1
2
3
4
5
x^
9
4
1
0
1
4
9
16
25
-3a;
9
6
3
6
0
2
-3
-6
-9
-12
6
-15
12
y
20
12
0
0
2
JT 20
f I
t -1
X II
4 3f
4 ^ 4t
1
:: ri i
-L \ J
i f:
u 3
l" 7
4 A =+
r _ — — =-^
j^^-;^"
-- -^ S^Z
5 ^ i ^ 5 lO 5
^ -- — -___
Fig. 5.
GRAPHS
17
(ii)
y
-3
-2
-1
0
1
2
3
4
2j/2
18
8
2
0
-3
2
8
18
32
X
15
5
-1
-1
5
15
29
The co-ordinates of every point on graph i satisfy the first equa-
tion, and of every point on graph ii satisfy the second equation.
Hence the co-ordinates of any points which are on both graphs,
that is, the co-ordinates of the points of intersection, satisfy both
equations.
Fig. 5 shows, therefore, that the values x = '3, t/ = I'S, and
X = 2'8, y = 1*6, are the solutions of the two equations.
This graphical method of solving equations is very useful,
but is, of course, only approximate. If more accurate answers are
required, the graphs must be drawn on a larger scale in the
neighbourhood of their points of intersection.
LOGARITHMS *
Fractional and Negative Indices. It is shown in Algebra
that p _ 1
where p and g are any positive integers and x is any positive
quantity, integral or fractional.
A fractional index may be expressed as a decimal ; thus such
expressions as 4'^^ 10*'°^ have a definite value. This value could
in theory be found by reducing the decimal to a vulgar fraction
and then replacing the power with a fractional index by a root,
e.g.
10-301
This is obviously not practical. The value can be found by a
graphical method which is easy but only approximate.
Di-aw the graph of x = lO*'.
* For a fuller treatment of Indices and Logarithms see School Algebra,
Chapters XXI and XXII.
_301_ ,
IQIOOO _ 1000 /j^QSOl
\
18
LOGARITHMS
y
0
•5
•25
•125
'lb
•625
•875
1
X
1
3-16
1-76
1-33
5-62
426
7-49
10
The values of x are obtained as follows : —
10° = 1 by definition given above.
10-» = 10^ = VIo.
10-25 = 10^= V^\/ 10; similarly W^'-^^^ ^W\
W^ = 10-5+-25 = 10-5 X 10*25, &c.
The graph is shown on a small scale in Fig. 6.
It is seen that 10'^° is almost exactly 2 ; any other power of 10
can be found approximately from this graph when the index is
between 0 and 10.
Definition of a logarithm. The logarithm of a number to a
given base is the index of the power to which the base must be
raised to equal the number. Thus 3^'= 9, therefore the logarithm
of 9 to base 3 equals 2 ; this is written log3 9 = 2.
In dealing with numbers the base is 10. In the remainder of
this chapter it is assumed that the base is always 10, so that
log 731 means logarithm of 731 to base 10.
The equation x = 10^ may be written t/ = log x.
Hence Fig. 6 provides an approximate means of finding the
logarithm of any number between 1 and 10.
Characteristic and mantissa. Consider a number, such as
4878. It means
4xlO.H3x 102 + 7x10 + 8.
Also a decimal number, such as '0376, means
A A _I_ _?_
10 "^ 1Q2 "^ 10^ "^10^*
If we use negative indices, this may be written
•0376 = 3 X 10-2 + 7 X 10-^ + 6 x 10"*.
Similarly
537-13 = 5 X 102 + 3 x 10^ + 7 X 10'' + 1 X 10-i + 3 x lO"''.
It follows that
4378 > 103 but < 10\
•0376 > 10-2 but < 10-^
587-13 > 102 but < 10»,
4-37 > 10° but < 10\
•. log 4378 = 3 + a decimal ;
•. log "0376 = - 2 + a decimal ;
•. log 537-13 = 2 + a decimal ;
•. log 4'37 = 0 + a decimal.
LOGARITHMS
19
We now see (i) that the logarithm of any number consists of an
integer (which may be positive, zero, or negative) and a positive
Fig. 6.
decimal, (ii) that the integer is the index of the highest power of
10 contained in the number.
The integral part of a logarithm is called the characteristic ;
the decimal part the mantissa.
Index
5
,
3
2
1
0
-1
-2
-3
-4
-5
-6
Number
3
7
8
9
0
0
)j
4
3
7
6-
5
2
?>
3-
4
6
7
8
»>
0-
0
0
7
3
4
8
if
2
0
0
0-
0
0
0
20
LOGAEITHMS
Consideration of the preceding table shows that the character-
istic (i. e. the highest index) may always be found by the following
rule : Count from the unit place to the first significant figure
(i. e. the first figure which is not 0), the unit place being counted
as nothing. The characteristic is positive or zero if the number
is greater than one, negative if it is less than one.
The mantissa is independent of the position of the de-
cimal point. An example will make this clear.
Given that log 4-376 = -6411, find log 4376 and log -004376.
4376 = 1000 X 4-376
but 4-376 = 10-"" since log 4*376 = '6411.
.-. 4376 = 10^ X 10-<'^^i = 10s-6"S i. e. log 4376 = 3-6411
•004376 = 1 J-oo X 4*376 = 10-^ x lO-^^i = io-3+-64n
.-. log -004376 = -3 + -6411.
The negative sign of a characteristic is always placed on top
and the + before the decimal is omitted. Thus log '004376 =
3-6411.
To find the logarithm of any number.
(a) Four-figure tables. The mantissa is found from tables, of
which a specimen is given below.
Logarithms.
0
1
2
3
4
5
6
7
8
9
I 2
3 4
5
6 7 8
51
53
54
7076
7160
7243
7324
7084
7168
7251
7332
7093
7177
7259
7340
7101
7185
7267
7348
7110
7193
7275
7356
7118
7202
7284
7364
7126
7210
7292*
7372.
7135
7218
7300
7380
7143
7226
7308
7388
7152
723s
7316
7396
I 2
I 2
I 2
I 2
3 3
2 3
2 3
2 3
' i 7
5 ^ I
566
5 <i 6
55
7404
74r2j
7419
7427
7435
7443
7451
7459
7466 7474
I 2
2 3
5 5 6
Consider the logarithms of 5467 and '05467.
By counting, the characteristic of log 5467 is found to be 3, and
that of log '05467 to be 2.
Both logarithms have the same mantissa. Look for 54 in t)ie
extreme left-hand column. In the same line with 54 and under 6
we find 7372 ; this is the mantissa of log 546. Under the 7 in the
small columns to the right, and in a line with 54, we find 6 ; this
must be added to the last digit of the mantissa already found.
Hence the mantissa is '7378.
Therefore log 5467 = 3'7378 and log '05467 = 2*7378.
LOGARITHMS
21
(b) Five-figure tables. Find the logarithm of 346'73.
Proceeding as -with four-figure tables, we find that the mantissa
of log 346 is '53908. ,Under 7 in the side columns, we find 88 ;
this must be added to the last two digits already found. For a 3
in the fourth place we should add 38, for a 3 in the fifth place we
add, therefore, ^^ x 38, i. e. 4 to nearest integer. Hence
log 3467-3 = 2-53908 + 88 + 4 = 2-54000.
To find the number corresponding to any logarithm.
Method I. Reverse the process for finding a logarithm. Sup-
pose the logarithm is 3*7271.
Look in the logarithms for the mantissa nearest to 7271, but
less than it. We find 7267, level with 53 and under 3 ; the first
three figures of the number are 533. This leaves 7271 - 7267 = 4 ;
in the right-hand columns 4 is found under 5. Hence the first four
figures are 5335.
The characteristic is 3, therefore the left-hand digit 5 represents
5 X 10^ ; hence the number is bd\p.
The number is called the antilogarithm of the logarithm.
Method II. If tables of antilogarithms are available, they are
used in the same way as kcarithm tables.
AxmLOGARITHMS.
0
1
2
3
4
5
6
7
8
0
12 3 4
5
6 7 8 9
-012
5023
5035
5047
5164
5284
5408
5534
5662
3058
3176
5297
5420
5546
3675
5070
508^
5200
3321
3445
5572
5702
5093
5212
5333
3458
5585
5715
5105
3224
3346
5470
5598
5728
5117
5236
3358
5483
5610
5741
1245
6
7 8 9 n
5370
5493
5140
5260
5383
5508
5152
5272
5393
3521
3188
3309
3433
3559
1245
1245
1345
1345
6
6
6
6
7 8 10 II
7 9 10 II
8 9 lo II
8 9 10 13
5623
5636
3649
5689
1343
7
8 9 10 12
Look for *72 in the left-hand column ; level with *72 and under 7
we find 5333 ; in the small columns we find 1 under 1. Hence the
first four figures are 5334.
The characteristic is 3 ; as before, the number is 5334.
Note. — The two methods give results differing by 1 in the last
figure ; this shows that the number is between the two results. On
using five-figure tables, it is found that the antilogarithm of 3'7271
is 5334-5.
22 LOGARITHMS
Use of Logarithms.
By the definition of logaritlim a = lO^oga, & = lO^os^,
.-. db = 10log«x IQlos^ = iQlogrt+logft.
Hence log (a6) = loga + log &.
Similarly log ajh = log a — log &,
log a'" = wi log a,
log Vi/a = — log a.
Thus, instead of
multiplying, we may use logarithms and add ;
dividing, ,, „ subtract;
raising to a power, „ ,, multiply ;
taking a root, „ „ divide.
Note. — There is no process with logarithms to correspond with
addition or subtraction with ordinary numbers.
1 T 17- ^ M 1 P 516-5 X -852
Example I. Find the value of —^- — •
log of fraction = log SIG'S + log '852 -log 36500
= 2*7130
+ 1-9304 -4-5623
= 2-6434
- 4-5623
= 2-0811
.-. fraction = -01205.
Example II. Find the cube root of -1765.
log of cube root = ^ log '1765
= ^ of 1-2467
= ^3 + 2-2467)
= 1-7489.
Notice carefully this method of division when the characteristic
is negative.
Hence V-1765 = -5610.
LOGARITHMS
Exercises. Find the value
of
(1)
(2)
(3)
(4)
(5)
^^319-2 X r756.
•03056 X 0-4105.
3-142 x(71-43)^
^x 3-142 X (9-67)3.
254-3 -r 0-09027.
Ans.
Ans.
Ans.
Ans.
Ans,
31-37.
•01254.
16030.
3787.
2817.
(6)
(-1136)^x^/81-86
^x V2000
An?.
•1874.
23
THE GREEK ALPHABET
Greek letters are used so frequently in Trigonometry and other
branches of Mathematics that it is useful to have the complete
alphabet for reference.
Name.
Small.
Capital.
alpha
a
A
beta
^
B
gamma
y
r
delta
d
A
epsilon
e
E
zeta
c
Z
eta
V
H
theta
e
e
iota
L
I
kappa
K
^ :
lambda
X
mu
/*
M
nu
V
N
xi
i
S .
omicron
o
O
pi
IT
n
rho
P
p
sigma
a-
2
tau
T
T
upsilon
V
Y
phi
0
*
chi
X
X
psi
^
^
omega
<o
Q«
CHAPTER I
ANGLES AXD THEIR MEASUREMENT
1. Any angle such as BAC (Fig. I) may be thought of as
having been formed by rotating the line AC about the point A
from the position of coincidence with AB to its final position AC.
Fig. I.
This way of regarding an angle shows clearly the intimate
connexion between angles and arcs of circles and this connexion
leads to the usual method of measuring angles.
2. The Degree.* From very early times it has been the custom
to divide the circumference of a circle into 360 equal parts or
* 'The current sexagesimal division of angles is derived from the
Babylonians through the Greeks. The Babylonian unit angle was the
angle of an equilateral triangle; following their usual practice this
was divided into 60 equal parts or degrees, a degree was subdivided
into 60 equal parts or minutes, and so on ; it is said that 60 was
assumed as the base of the system in order that the number of degrees
corresponding to the circumference of a circle should be the same as
the number of days in a year which it is alleged was taken, at any rate
in practice, to be 360.' (From A Short Account of the Hisfory of Mathe-
matics, by W. W. Rouse Ball.)
26 ANGLES AND THEIR MEASUREMENT
degrees, each degree into 60 parts or minutes,* each minute into
60 seconds.^
The angle at the centre of a circle, subtended by an arc
of 1 degree, is taken as the unit angle, and it, too, is called
a degree ; it is divided into minutes and seconds in the
same way as the arc degree.
The notation used is shown in the following example : —
47° 15' 37" is read 47 degrees 15 minutes 37 seconds.
If the line makes a complete rotation, thus returning to its
original position, it has turned through an angle of 360°.
A right angle is produced by one-quarter of a complete rotation,
and is, therefore, equal to 90°.
If two angles together equal a right angle, either of them is
called the complement of the other. When the sum equals two
right angles, either angle is the supplement of the other.
3. Positive and Negative Angles. In discussing the
properties of a single angle it is usual to draw the initial
line so that it is horizontal and to name it OA. If the
rotating line moves in a direction opposite to that of the
hands of a clock, the angle is said to be positive ; if in
the same direction as the hands of a clock, the angle is
negative.
Fi- III.
* Minutes derived from the Latin jmrfcs minutae ; seconds from the
Latin partes fuinntae secundac.
ANGLES AND THEIR MEASUREMENT 27
In Fig. II the line OP has made i of a complete turn, hence
the angle ^OP = i x 360° = 45° ; in Fig. Ill the angle ^ OP is
reflex * and is equal to § x 360° = 225°. If, in Fig. II, the line
OP reached its position b}-- turning in the negative direction it
would have made | of a complete turn so that the reflex angle
^OP in Fig. II = -315°. Similarly in Fig. Ill the obtuse angle
AOP= -135°.
4. Angles unlimited in size. In Fig. II the line OP might
have made one, two, or any number of complete turns, either
positive or negative, and then have moved on to its final position :
hence the angle AOP may represent 405°, or 765°, or —675°. All
possible values are included in the general formula
^0P= 360 n + 45,
where n is any whole number, positive, zero, or negative.
Unless the problem under discussion allows the possibility of
the angle being greater than 360°, it is always assumed that the
angle is less than 360°.
5. The Grade. When the metric system was invented, the
French Mathematicians introduced a new unit, the Grade, such that
100 grades = 1 right angle,
100 minutes = 1 grade,
100 seconds = 1 minute.
This system never came into general use, even in France, and now
exists only in old-fashioned examination papers.
Examples I a.
1. Find the complement of each of the following angles:
32°, 47° 23', 75° 13' 14", 68°0'13", 27° 42' 18-6".
2. Write down the supplements of 75°, 68° 14', 115° 17' 48"
90°, 78° 24' 36".
3. The angles of a triangle are found to be 42° 13' 17", 73° 47' 5",
64° 0' 38". Is this correct ?
4. Two angles of a triangle are 17° 43', 92° 16'; calculate the
third angle.
5. In a triangle ABC, 1{A + B) = 77° 29' and \{A-B) = 16° 25' ;
find all the angles.
* A reflex angle is an angle greater than two right angles, but less
than four right angles.
28 ANGLES AND THEIR MEASUREMENT
6. Express in degrees, minutes, and seconds the angle of
(a) a square, {h) a regular pentagon, (c) a regular heptagon.
7. Express each of the angles of question 6 in grades.
8. The magnitude of an angle may be expressed either as
D degrees or G grades ; find the equation connecting D and G.
9. Draw the angles A and ^A in each of the following cases:
(«) A = 54°, (b) A = 414°, (c) A = 774°, (d) A = 1134°, (e) 234^
(/) -126°.
10. Through what angles do the hour, minute, and second
hands of a watch respectively turn between 12^ 30' a.m. and
5h 3' a.m. ?
6. The ratio of the length of the circumference of a circle
to the length of its diameter is the same for all circles.
This constant value is denoted by the Greek letter -n (pro-
nounced pi), so that if
the circumference = c units of length,
and diameter = cl units of length,
then - = 77.
d
The value of tt can be found, correct to two or three significant
figures, by actual measurement. By geometrical and trigono-
metrical calculations its value can be calculated to any desired
number of places.
Correct to 5 significant figures, n = 31416.
Correct to 6 significant figures, n = 314159.
For mental calculations tt may be taken as 31.
7. By using Prop. 5, p. 9, problems dealing with the lengths of
circular arcs may often be solved.
Example. Find the length of an arc which
subtends an angle of 49° at the centre of a
circle whose 7-adius is 5 feet.
arcylP _ angled OP
semi-circumference 2 right angles
ai'c AP __ 49°
Fig. IV. i-^- "^5^ ~180°*
The calculation is easily completed.
ANGLES AND THEIR MEASUREMENT 29
,,. ., , area of sector ^OP an^le ^OP ,t^ o^\
Similarly r^l = , .^, . 1— , (Pi'op. 26)
"^ area of circle 4 rierht angles
1. e.
area of sector AOP _ 49
2blT 360
8. Circular measure. By the method of the last section
it is easily shown that the length of an arc of a circle,
Att
radius r, subtending an angle A° at the centre is r -— - .
loO
In many other formulae the fraction -— occurs in con-
nexion with the angle A°. In theoretical work it has,
therefore, been found convenient to use another unit angle,
which simplifies formulae considerably.
The radian is the angle subtended at the centre of any
circle by an arc equal in length to the radius.
Let x° equal 1 radian
arc equal to radius _ angle of 1 radian _ x°^
semi-circumference 2 right angles 180°
'•^* 7rr~180°'
Since n is the same for all circles, it follows that the radian is
the same for all circles and may, therefore, be taken as a unit
of measurement.
The number of radians in an angle is often called the
circular measure of the angle. For this reason the
symbol <^ is used to show that the angle is measured in
radians, e. g. 2^^ means 2 radians.
When the radian is the unit angle, it is customary to use
Greek letters to denote the number of radians, and the
symbol '^ is then often omitted. When capital English
letters are used, it is usually understood that the angle is
measured in degrees.
30 ANGLES AND THEIR MEASUREMENT
Examples I b.
1. How many times is an arc equal to the radius contained in
the semi-circumference ? Reduce 180°, 90°, 60°, 30° to radians.
(Do not substitute for tt.)
2. Show by simple geometry that the radian is less than 60°.
3. How many radians are there in 10°, 75°, 138°, respectively ?
Give the answers correct to 2 decimal places.
4. Express the angle of (i) an isosceles right-angled triangle,
(ii) a regular nonagon, in circular measure. Give the answers in
terms of 73-.
5. One angle of a triangle is ^ tt, another is -J- n ; what is the
circular measure of the third angle ?
6. Find the length of an arc of a circle which subtends an
angle 78° at the centre, the radius being 18 feet.
7. An arc of length 5 feet subtends an angle of 132° at the
centre ; what is the radius of the circle ?
8. Find the area of the sector of a circle if the radius is 12 feet
and the angle 40°.
9. What time does the minute hand of a watch take to turn
through (i) 3000°, (ii) 3000 grades, (iii) 3000 radians ?
10. Fill in the missing values in the following table, which gives
data about circular arcs.
Length.
413 feet
220 yards
1 kilometre
half a mile
11. Express in radians the angle of a sector of a circle, being
given that the radius is 7 inches and the area of the sector 100 sq.
inches.
12. Show that the length of an arc subtending an angle 0^ at
the centre of a circle, radius r, is r6. What is the area of the
corresponding sector ?
13. Find the circular measure of V and of \'\ correct to 5 signi-
ficant figures.
Kadius.
Angle.
(1)
5 inches
2 radians
(2)
7-6 centimetres
74*6 grades
(3)
314 degrees
(4)
100 yards
radians
(5)
320 metres
degrees
(6)
yards
5 radians
ANGLES AND THEIR MEASUREMENT
31
9. The points of the compass. The card of the Mariner's
Compass is divided into four quadrants by two diameters pointing
North and South, East and AVest respectively. These are the
Cardinal Points. Two other diameters bisecting the angles
between the previous diameters give four other points, viz. NE.,
NW., SW., SE. The eight angles so formed are bisected and
eight more points are thus obtained. These are named by com-
-biningthe names of the points between which they lie, beginning
with the cardinal point. Thus the point midway between E. and
SE. is ESE. (East South-East).
The sixteen angles now formed are bisected so that the circum-
ference is finally divided into thirty two equal divisions. From
their names the last sixteen points are called by-points. The
point midway between N. and NNE. is called N. by E. ; that mid-
way between SW. and SSW. is called SW. by S., &c.
32
ANGLES AND THEIR MEASUREMENT
The angle between two consecutive points of the compass is also
called a point, thus N. 2 points E. is the same as NNE. ; WSW. \ W.
means | a point W. of WSW.
The ordinary degree is sometimes used in defining a direction,
for instance ENE. can be referred to as 221° N. of E. Similarly
we may have 32° W. of N., 40° S. of W., &c.
10. Latitude and Longitude. The position of a point on
a sphere can be defined by two angles, which may be compared
with the abscissa and ordinate of plane geometry. These angles
are easily understood by considering the special case of Longitude
and Latitude.
In Fig. V the meridian through Greenwich cuts the equator
at A ; the meridian through P cuts at B. 0 is the centre of the
Earth.
The Longitude of Pis the angle AOB and may be either East
or West of the Greenwich meridian.
The Latitude of Pis the angle FOB and may be either North
or South of the Equator.
Note. — A geographical or nautical mile is the lengtli of an arc of a
meridian (or of the equator) subtending an angle of 1' at the centre of
the earth.
ANGLES AND THEIR MEASUREMENT 33
A ship travelling at the rate of I nautioal mile per hour is said to
have a speed of one knot.
11. Gradient. It is usual to estimate the inclination to
the horizontal of a road or hill by the distance risen veitically for
a certain horizontal distance. Thus a hill might be said to rise
3 in 5 ; this would mean that if a horizontal line were drawn
through a point B on the hill to meet the vertical line through
a lower point A at C, then AC/BC would equal ?. The hill is said
to have a gradient or slope of 3 in 5.
Fig. VI.
It is clear that in many cases it is easier to measure AB than BC;
and some books take a gradient of 3 in o to mean a rise of 3 vertically
for a distance 5 measured along the incline ; so that in the figiire
AC/AB would be f. This latter interpretation of gradient is very
common in books on Theoretical Mechanics.
If the inclination is small, it makes no practical difference whicli
interpretation of gradient is taken.
It should be noticed that the angle is the same whatever units
be used ; that is whether we consider a rise of 3 inches in 5 inches,
3 furlongs in 5 furlongs, 3 miles in '5 miles. This follows from
Prop. 12 0.
Examples I c.
1. Express in degrees the angle between
(a) NNE. and E. by N. ; (/) S. 2 points W. and W. 2 points S. ;
(&) W. by S. and SE. by N. ; (r/) 40° N. of W. and 30° E. of S. ;
(c) ESE. and NE. by N. ; {h) NE. by E. and 1 point W. of N. ;
(d) NNW. and SSE. ; (/) 30° S. of W. and ESE. ;
(e) N. by W. and SW. ; (k) S. 2 points W. and W. 2 points N.
In the following questions take the radius of the Earth to be
4000 miles.
34 ANGLES AND THEIR MEASUREMENT
2. Two places on the Equator are 300 miles apart, find the
ditt'erence of their Longitudes.
3. Quito (Longitude 79' W.) and Macapa (Longitude SIJt W.)
are both on the Equator, find the distance between them. What
time is it at Macapa when it is noon at Quito ?
4. Find the distance between Poole (Lat. 50° 43' N., Long.
1°59' W.) and Berwick (Lat. 55° 46', Long. 1°59'W.).
5. Find the distance between Cape Breton Island (Lat. 45° 50' N.,
Long. 60° W.) and the Falkland Isles (Lat. 51° 32' S., Long. 60° W.).
Oral Questions.
1. What is a degree ? How many degrees are there in the angle
of a regular pentagon ?
2. How big is each acute angle of an isosceles right-angled
triangle ?
3. One angle of a triangle is A°, another 30°, how big is the
third angle ?
4. What is meant by a negative angle ? When screwing an
ordinary screw in, is the turning in the positive or negative
direction ?
5. Does the earth rotate in the positive or negative direction ?
In which direction does the sun appear to move ?
6. Do you usually draw a circle in the positive or negative
direction ?
7. The needle of a mariner's compass is deflected from its normal
position through a positive angle 33i| degrees, to what point of the
compass does it then point ?
8. Express the following angles in circular measure : 90°, 00°,
180°, 45°, 30°. (Give the answers in terms of n.)
9. What is the locus of all places having latitude 35° N. V
10. What is the locus of all places having longitude 15° W. V
11. It is noon at the same time at two different places, what do
you know about their longitudes or latitudes ?
12. Give the latitude and longitude of the N. pole.
Examples I.
1. In a triangle ABC, A = 43° 15', B = 07° 38', calculate the
number of degrees in (i) the angle C, (ii) the angle subtended at
ANGLES AND THEIR MEASUREMENT 35
the centre of the c ire urn circle by the side BC, (iiij the angle sub-
tended at the centre of the inscribed circle by the side BC.
2. Express in circular measure, correct to 3 significant figures,
(a) the supplement of I'S? radians, (/;) 74°, [c) the angle of a regular
octagon.
8. Define a radian and a grade. If an angle, containing D
degrees, may be expressed as either 6 radians or G grades, prove
that 2)/180 = 6/tt = G/200.
4. The hands of a clock are coincident at noon, through what
angle does the hour hand turn before they next coincide ?
5. Prove that whatever be the radius of a circle the size of the
angle at the centre, which subtends an arc ec^ual to the radius, is
constant. What is this angle called ? Show, by a geometrical
construction, that it is a little less than 60°.
6. A wheel of a cart is 4 feet in diameter, through what angle
does it turn when the cart moves forward 10 feet?
7. Explain how to find the length of a circular arc being
given the number of degrees in the angle subtended at the centre
and the length of the radius.
^'. Two places on the p]quator differ in longitude by 87° 16', find
the distance between them, correct to three significant figures.
(Radius = 4000 miles.)
9. Find the distance between a place, longitude 45° 17' E., lati-
tude 0°, and another place, longitude 88°43'W., latitude 0°.
10. Through what angle does the Earth turn between 9.30 a.m.
and 4 p.m. ?
11. When it is noon at Greenwich what time is it at (a) Calcutta
(88° 15' E.j, (&) New York (74° W.), ic) Hawaii (156° W.) ?
12. The co-ordinates of two points P and Q are (7, 8), (9, 11)
respectively, find the gradient of the line PQ.
13. Draw an angle AOP = 85^ in OP take 8 points P, Q, li such
that OP = 1 inch, OQ = VI inch, OR = 2d inches. From P draw
PH Sit right angles to OA, at Q draw QK at right angles to OQ,
and from R let fall RL perpendicular to OA. -Measure OH, OK,
OL and calculate, to 3 decimal places, the ratios OH/ OP, OQ/OK,
OLjOR. Justify the result.
14. Explain what is meant by a radian, and find how many
degrees and minutes it contains.
Express in degrees, and also in radians, the angle of a regular
polygon of 100 sides.
c2
S(j ANGLES AND THEIK MEASUREMENT
15. An explorer reaches a latitude of 87° 28' 4«". Find how
many miles he is distant from the pole, assuming the earth to be
a sphere whose circumference is 25000 miles.
16. Find the gradient of a straight line joining two points
whose co-ordinates are {x', y'} and (x", y"). Hence find the equation
of the straisrbt line.
CHAPTER II
THE TRIGONOMETRICAL RATIOS
12. Definitions. Let OA the initial line be taken as axis
of X, the axis of// being perpendicuhir to it at 0 ; in the final
P
./I
N A
\
Fi?. VI r.
Figure VII
position of the rotating line take any point P.
shows four possible types of positions of OP.
Let fall PX perpendicular to OA or OA produced, so that
ON is the abscissa of P and PN the ordinate. Then
as THE TRIGONOMETRICAL RATIOS
NP ordinate y
sine of AGP =
cosine of AGP =
GP radius r
GN abscissa x
GP radius r
. . . ^^ NP ordinate y *\ a
tangent of AGP = -—- = , — ^ = - • 7. a.
GN abscissa x ^
These are the most important ratios ; the others are their
reciprocals, viz. :
G P radius r
cosecant of AGP —
secant of AGP —
cotangent of AGP =
NP ordinate y
GP radius r
GN abscissa x
GN abscissa x
NP ordinate y
TWfollowing abbreviations are usually used:
sin A instead of sine of AOP,
cos A ,, ,. cosine of ^ OP,
tan^ „ „ tangent of .4 OP,
cosec A „ ,, cosecant of AOP.
sec A ,, „ secant of J OP,
cotan ^ ,, ,, cotangent of. f OP.
Similarly, if ^OPismeasured in radians, sin 6, cos f/^ cosec \//, &c.,
are used.
13. Trigonometry was developed by Arabian and Greek astrono-
mers who based their work on the circular arc and not on the
angle. In the Middle Ages this early mathematical work was
translated into Latin, and so the present names of the nitio were
derived. The following section shows the reasons for these names.
14. Draw a circle with centre 0 cutting the initial line at .1 and
the perpendicular to it at B.
Take a point P on the circumference of the circle.
Draw the tangent at yl and product' OP to meet it at T.
Draw PN perpendicular to ()A.
AT wiiH called the tangent of the arc AP.
OT, which cuts the circle, was called the secant of the arc AP.
THE TRIGONOMETRICAL RATIOS
39
. NP was called the sine * of the arc AP.
Clearly the lengths o^ AT, 02] AP change when the radius OA
changes, even if the angle .lOP remain constant.
But from similar triangles it is seen that, if the angle is constant,
the ratios oi AT, OT, NP to the radius are also constant. Hence,
as Trigonometry developed, it was seen to he advisable to divide hy
* The word ' sine' is derived from the Latin sinus. If in Fig, VIII
P^y be produced to meet the circumference at P', then PAP' resembles
a bow (Latin arcus) of which PXP' is the string or chord (Latin chorda}.
To use the bow, the string is pulled till JV touclies the bosom (Latin
sinus^ ; hence PX is called the sine. NA is often called the sagitta of
the arc.
40 THE TRIGONOMETRICAL RATIOS
the radius and to treat the subject as depending on the angle
AOF rather than on the arc A P. Thus we have
angle AOP = '—^. — (when the angle is measured in radians),
" radius
,^, sine of arc ^P iYP
sin AOP =
tan AOP =
sec AOP =-
radius OP
tangent of arc AP _ AT _ yP
radius OA ON
secant of arc AP OT OP
radius OA ON
Now make a similar construction for the complementary arc BP.
Then
sine of the complement of the angle AOP
__ sine of the complementary arc BP
radius
^ XP
OP
_ ON
~ OP ' ''
' Sine of the complement of ' was shortened into co-sine. Possibly
' complementary sine ' was an intermediate stage. [Similarly, co-
tangent and cosecant were derived.
Since the values of the ratios depend on the values of the angle,
the term Trigonometrical Functions is often used instead of
Trigonometrical Ratios. Frequently the ratios are referred to as
Circular Functions.
15. Ratios rarely used, in Fig. VIII
N A is called the vorsine (i. e. versed sine) of the are AV.
X'B is called t])e coversine of the arc AP.
AP (not joined in the figure) is the chord of the arc AP.
If we divide each of these by the radius we get the corresponding
ratios of the angles AOP. These ratios are very rarely used. Another
function that is now rarely used is tlie haversino, i. c. half the versed
sine.
16. Projection Formulae. It is useful to remember that
ON (i. e. the projection of the radius on the initial line)
= r cos 6
and iYi^(i. e. the projection ot the radius on a line perpen-
dicular to tlie initial line) = ;• sin 0.
THE TRIGONOMETRICAL RATIOS 11
17. Polar co-ordinates. The position of a point P is
determined if the distance OF from a fixed point is known
and also the angle this distance makes with a fixed line OX
through 0. The length is usually denoted by y and the
angle by 6 ; these are the polar co-ordinates of P. In this
connexion 0 is called the pole.
18. Graphs of Trigonometrical Functions.
The definitions of the last suction lead to an easy method of
drawing the graphs. On page -12 the sine graph (i.e. the graph of
the equation «/ = sin x) is given. It is obtained as follows :
Step 1. On the extreme left of the paper (which should be ruled
in squares) draw a circle with its centre at the intersection of two
lines. Take the horizontal radius CA as initial line.
Step 2. The perpendicular B'CB gives the angles 90° and 270°.
The diagonals through C give the angles 45°, 135°, 225°, 315°.
By stepping off chords equal to the radius, starting from A, the
angles 60", 120°, 240°, &c., are obtained ; and, by starting at B,
•the angles 30°, 150°, &c., are obtained.
Only the points P, P' , . . ., on the circumference need be obtained
as is shown in the third quadrant ; the radii are not needed for
drawing the sine grajih.
Step 3. Take a point 0 as origin, some distance along the initial
line, and, with a convenient scale, mark off abscissae to represent
the angles 30°, 45°, 60°, &c., and, as far as space allows, mark off
the negative abscissae.
Stej) 4. Through the points on the circumference draw parallels
to the initial line to cut the corresponding ordinates. These
points of intersection are points on the graph.
The ordinates of this graph* are proportional to the sines; if
we divide by the radius, the actual values of the sines are found.
The sine graph is shown on a larger scale in Fig. X.
19. The tangent graph. To obtain the ordinates for the
tangent graph the radii must be produced to meet the tangent to
* The giaph of the sine is a wavy or sinuous curve. Tlie name sine
is therefore appropriate, although it is improbable that the originators
of the name ever drew the graph.
-.^ — __ _ —
7^ — 1 —
_. /
?
yt
y
7^
y-
' t
S
I— *
^ i i
\
:is.
\,
^g
S
\
\
\
\
\
p
J5
~t
/
-»- -/-
y
i
± -7- it
y
I
o
r <5>
\^ '
v
. ^
- -- ^o
^
N
^
\
:;:: "::""\
-— o ._. ,^
0> 4
7
---- -
^ r- - -. - ^ ^
^^ Ij;-- <-v:'^ ^^
^^^ %-\ / A 7
, ;2;..^_^^_._/7 r<
t..:-s\//.i:.\
ffl_-_ ^^^^ 5,
ffl ^'lo ffi
^- V ^' / . :'7 "
\: 7- - ,^r /
-^ . i Z-_. A ^i
--- - ±.-^^' _ ...
- - -- ... ^.-.
o^
fl>
1 1^ ; t
i 1 i
; V ■- ! ^
\ i ""
i\i i
t4^ -^ i ^
1 1 \ 1 . o
1 M . •*!
' ' \ i !
i ^1 i
( , , |v^ , 1 ■ ; : . j .
! j \ 1
\i ^
i '^ ill
It- ^v^ mitx
-1 — ^ 1 — ^ — r
\ ' o
1 ■ \ ' 1 ' , fn,
! ; \ 1
i 1
1 \
i \ in
1 It it ^ -X--
"^ 1
1 : \ '
\ ^
a> do ri- <© i> ^- fo fu_\
... . • . . . ; i 1 . . o
1
^^— -" I <
^•^ !
.^
y^
/
y
/
"^ X 7 it
' / '
(
■ 1 / i
'/i 1
/ 1 1
/ '
/ i ■
' 1 1
' -l-i-
\ !
I 1
iffl o
44 THE TRIGONOMETRICAL RATIOS
the circle drawn at A. This is done in Fig. TX, and a few points of
the graph are marked, but the graph is not drawn.
The cosine graph. Since the cosine of x is the sine of the
complement of x, the student should be able to modify the method
for the sine graph so as to obtain the cosine graph.
The secant graph. This is obtained b}^ marking off along the
respective ordinates the corresponding values of CT (see Fig. IX).
Examples II a.
(Answers should be given correct to 2 significant figures.)
By drawing to scale find the trigonometrical ratios of the
following angles :
1. 30°. 2. 49°. 3. 79°. 4. 100°.
5. 78°. 6. 170°. 7. 250°. 8. 25°.
9. 300°. 10. 156°. 11. -80°. 12. 415°.
Find the ratios of the angle ^40/' when the coordinates of P are
13.(4,3). 14.(4,-3). 15. (-4, -3). 16. (-4, 3).
17. (3,2). 18. (-7, -3). 19. (-5,4). 20. (63, -16).
The following graphs should be drawn carefully and kept for use :
21-26. A graph on a large scale for each function, for angles
from 0° to 90°.
27-32. A graph on a smaller scale for each function for angles
from -360° to +360°.
33. The blanks in the following table are to be filled with the
sign ( + or — ) of the respective ratios :
Angle 0"-90° 90° -180° 180° -270° 270" -360°
sine
cosine
tangent
34. If the gradient of a hill, inclined at A° to the horizon, is
known, what trigonometrical ratio of the angle is known ?
35. Construct an angle whose (i) tangent is 1*45, (ii) sine is '75,
(iiij cotangent is 1*45, (iv) secant is 2*7, (v) cosecant is 2"7,
(vi) cosine is *75. Measure each angle in degrees.
20. Powers of the Trigonometrical Fimctions. The
sc^uare of sin^ is written sin^^l ; and a similar notation is
used for other powers and ratios ; thus, in general,
sin" ^4 means (sin ^4)".
THE TRIGONOMETRICAL RATIOS
45
Inverse notation. There is one exception to the above
statement. Suppose sin A — a. then A is an angle whose
sine is a. This is written A = sin ' ffv Similar!}^, tan"'«,
means an angle whose tangent is a ; and so for the other
ratios.
If w^e wish to express -^ — - as a power of sin J., we must
write (sin A)~'^.
sin A
Note. Continental mathematicians denote tlie angle whose sin is x
by arc sin oc. Tlxis notation sometimes occurs in English books.
Example. Determine, hy draivlng, the angle siir'^ 5.
Step 1. Draw axes OA, OB. [
Step 2. Draw circle centre 0, radius 3 units.
Step 3. Along OB mark off 07v equal 2 units.
Step 4. Through K draw n parallel to OA cutting circle at
FOP. Join OF, OP'.
We now have two angles AOP, AOP' each of which has its sine
equal to |. AOPifir.^AOP' 139°.
Fig. XL
46 THE TRIGONOMETRICAL RATIOS
It is always under.stood that siii^^rt means the smallest
positive angle that has the sine equal to a ; and similarly
for the other ratios.
Examples II.
1. Find, by drawing to scale, the sine, cosine, and tangent of 30°,
45", 60°. Verify the results by calcidation.
2. The sine of an acute angle is ^-* ; find the cosine, tangent, and
secant.
3. The sine of an angle, not acute, is f ; find the cosine and
tangent.
4. The cosine of an angle is j%; find the sine and tangent.
5. Draw as many angles as possible having the tangent equal
to -8.
6. Given that sin 63'= '89 find cos 63^ and cos 27°.
7. Find the value of sinM + cosM, it being known that
sin A = -3907, cos A = -9205.
Also find the values of tan A and sec A.
8. Given tan d = ^fi, find cot 6 and sec 6.
9. Draw and measure an angle A such that (i) sin ^ = —'5,
(ii) cos J. = -"5. (iii) tan.i = -'5, (iv) sec ^ = - *5.
10. Find the value of sec- 0 - tun^ 6 , when sec^ = r22l and
tan 6 ■= '7002. Justify the answer by geometrical reasoning.
11. Are any of the following data inconsistent or impossible?
Give reasons for your answers.
(rt) sin A = I; [b] sec ^ = f ;
(c) sin A = i, cos A = j ; {d} sin A = •4, cos^l = '6 ;
(e) sin .4 = '6, cos A = 'S, tan A = '9 ;
(/)sec^l = "35, tanyl = 1*35;
ig) tan A — 1 ; [h) sin ^1 = 1; (/) cosec .4=1.
12. Prove, by means of the definitions in § 12, that
cosyl = sin (90-^) and tan (^77 -0) = cot ^.
13. Find, by drawing to scale, (a) sin 36^ and sin 144' ; (h) cos 42°
and cos 138° ; (c) cos 246° and cos 66 .
14. By means of graphs (or otherwise) test the following state-
ments : («) sin (180-^) = sin yl ; (h) cos (180 + ^) = -cos^ ;
(f) sin (90 + ^) = sin^.
15. By means of graphs find values for sin-^ *6, tan~^ 2"5
C0S~^ '34, C08~^ I'b.
THE TRIGONOMETRICAL RATIOS 47
IG. Given sin36'' = -587;^, find cos 54", sin 144', sin 21G^ sin 324'^
17. Given cos 53' = -6018, find sin 37% cos 127 , cos 233", cos 413^,
cos 307".
18. Prove that sin 117" = cos 21\
19. Is it possible to find angles to satisfy the following equations ?
Give reasons.
(i) tan^=l; (ii) cos ^ = :|^g ;
(iii) sin ^ + cos ^ = 1 ; (iv) sin^ d + cos^ 6=1;
(v) sec ^ = 31416; (vi) cosec ^ = i ;
(vii) sin ^ = 0 ; (viii) tan ^ = 100 ;
(ix) cos <9 = 1 ; (x) sec <9 = 78.
20. Show that (i) sine and cosine cannot be numerically greater
than 1 ; (ii) tangent and cotangent may be either greater or less
than 1 ; (iii) secant and cosecant cannot be numerically less than 1.
Why is the word numerically inserted ?
21. Find all the trigonometrical functions of 0"^ and 90\
22. (i) Show that the straight line whose equation is )/ — mx
makes an angle tan"^ m with the axis of x.
(ii) What is the tangent of the angle made with tlie axis of x
by the stmight line joining the two points {x^, yj and {x^, y^) ?
(iii) Show that the equation of the line joining the two points
, - . ?/ — Vi X- —x.
(ivj If the equation of a straight line is y = mx + c, give the
geometrical meanings of m and c.
23. Show that, if x be any numerical quantity, positive or
negative, an angle can be found whose tangent is equal to x.
Show what limitations in value, if any, exist in the case of each
of the other trigonometrical ratios.
24. State concisely the changes in the sign and magnitude of
sin ^ as ^ increases from 0" to 360^.
25. Define the cosine of an angle of any magnitude, explaining
the conventions regarding the signs of the lines referred to in your
definitions. Draw the graph of cos 6 from ^ = 0 to ^ = .^ tt.
26. Define the sine of an angle and find by geometrical reason-
ing the values of sin 45°, sin 90^ sin 135=.
48 THE TRIGONOMETEICAL RATIOS
27. Define the tangent and the versed sine of an angle ; and
find the greatest and least values which each can have.
28. With ruler and compasses construct an angle whose cosine
is ^; also an angle whose cosine is — ^. Calculate the sine of the
latter angle to three places of decimals.
29. ABC is a triangle in which AN is the perpendicular from
A to BC. U AB = 2'9 inches, AC = 2-5 inches, AN = 2 inches,
find the values of sin 5, cos C, tan 5, cosec C Calculate the
length of ^C correct to one decimal place.
30. If A, B, C are the angles of a triangle, express sin ^ (A + B),
cos ?; (A + B), tan \ {A + B in terms of ratios of \ C.
CHAPTER III
ELEMENTARY FORMULAE
21. Reciprocal Relations.
By definitions, sin A = -, cosec A =^-^ '
. '. sin A cosec A = 1 :
1 1
1. e. sm A = T- . cosec A = - — - •
cosec A sm A
In a similar way it can be proved that
cos A sec A = 1. &c.
tan A cot A = 1, &c.
22. Relations deduced from Pythagoras' Theorem
(Prop. 13, p. 10).
In Fig. YII, § 12, we have in all cases
i. e. X- + 7/' = r-.
Three sets of formulae are obtained by dividing in turn
by r, x\ \f.
f^ ' x^ iP"
'^Divide by n, - + — = 1 ;
but - = cos .1. - = sin A.
r r
Substitute, cos- A + sin- A = 1.
The equivalent formulae must also be learnt, viz. :
sin- A = 1 — cos^ J., sin^ = + Vl — cos'- A ;
cos- ^4 = 1 — sin' J., cos^l = -j- Vl ~ sin"- -4.
In a similar way the student should prove that
tan^ A -f 1 = sec- A.
and cot- A4 1 = cosec- A,
1216 D
50 ELEMENTARY FORMULAE
23. Relation between sine, cosine, and tangent.
Substitute,
tan A =
5;
X
^.
t
r
—
X
r
tan A —
sin A
cos A
is proved that
cot A=:
cos A
sin A
24. Identities. By means of the relations proved in the
preceding sections, any expression containing trigonometrical
functions can be put into a number of forms. It is a useful
exercise to prove that two expressions, apparently different,
are identical ; such exercises serve to fix the relations in the
memory and lead to facility in dealing with trigonometrical
expressions.
Example. Prove that scc^ A + cosec- A ^ sec- A coser A.
[Express all ratios in terms of sine and cosine.]
by § 21
Bthod I.
L.H.S.
_ 1 1
cos2 A sin2 A
sin' A + cos'^ A
%\\\^ A cos^J
1
sin'^ A cos' A
1 1
cosM ' sin'^l
using Formula of J^ 22
= scc-yi co?L'C-yl. l>v § 21
Q.KD.
ELEMENTARY FORMULAE 51
Method
II.
sec^ A + cosec^
•A
=
1
cos^
A
+
1
A
sin^ A + COS'
'■A
sin'^
siii'^ A cos^
A
1
1
sec^A cose c^ A =
sin^ A cos^ A '
1
cos'^ A sin^ A sin^ A cos'^ A '
.•. sec^^ + cosec^^ = sec^^ cosec^^. Q.E.D.
Method III. This method is clumsy, and should be used
only if Methods I and II have been tried unsuccessfully.
sec'^ J. + cosec'^^ = sec^^ cosec*^,
j^ _1_ ^ 1 _ ^ 1
cos'^-4 sin^^ cos^^ sin'^^ '
i.e. if sin^^ + cos'^J. = 1.
But sin'^^ + cos'^^ does equal 1 ;
sec"^4-cosec-^ = 860^^-4 cosec^^. ,
Note. The introductory * if, or some similar conjunction, is vital
to the logical statement of the work and must not be omitted.
25. Elimination. If two equations are satisfied by the
same value of a single variable, there must be a relation
connecting the constants of the equations ; this is also the
case when n equations are satisfied by the same values of
n—1 variables. In order to find this relation we eliminate
the variable or variables.
Example. Eliminate 6 from the equations si7i d = a, tan 6 = h.
By formulae tan^(9= — sTi
sin2^
Substitute h^
1 1
l-8in2<9
a'
l~a' '
I.e. - - — = 1.
a^ Tt'
The result is called the diminant of the original equations.
D 2
52 ELEMENTARY FORMULAE
Examples III a.
1 If sin A = ^§ use formulae to find the remaining ratios.
Draw a figure to explain why some of the ratios may be either
positive or negative.
2. Given that tan 6 = V find cot 6 and sin 6.
3. Find sec S in the following cases :
(i) cos 6 = "7921 ; (iij tan d = 1'352 ; (iiij cosec 6 = 2'583.
4. Show how all the ratios may be found when (i) the cosine,
(ii) the tangent is known.
5. Prove the following identities :
(i) sin A cot A + cos A tan A = sin A + cos A :
(ii) tan A + cot ^ = sec ^ cosec A ;
(iii) sin 8 tan ^ + cos ^ cot ^ = sec ^ + cosec ^ - sin ^ — cos 0 ;
(iv) sec2 6 - cosec^ 6 = tan^ S - cot^ 0 ;
(v) l-2sinM = 2cosM-l;
. . tan a + tan (:i _ siiux cos /^ + cos a sin 3 ,
'■^^^ 1 — tan a tan (:i ~~ cos a cos /3 — sin a sin j3 '
(vii) (1 -tanM) -r (1 ^tan^^) = cosM -sin^4 ;
(viii) (sin A + cos A)"^ = 1 + 2 sin ^ cos A ;
(ix) sin^^-cos^^ = (sin ^ — cos^) (1 + sin A cob A) ;
(x) sin A cot A cosec A + cos A tan A sec A = sec A cosec u4. ;
(xi) tan X -tan Y — (sin X cos Y—cobX sin Y) -^ cos JV cos Y;
(xii) (tan J[ - tan B) -r (cot A-cot B) = - tan ^ tan 5 ;
(xiii) cos^ 3 - sin* 6 = 2 cos^ ^ - 1 ;
(xiv) (3-4 sinM) ^ cosM = 3-tanM.
6. Prove that versin ^ = 1 -cos ^, coversin ^ = 1 —sin A.
7. Show that the numerical value of sin^^ -^ (1 — cos^) di-
minishes from 2 to 0 as A increases from 0° to 180°, and illustrate
your answer by a diagram.
8. Which is greater, the acute angle whose cotangent is 4, or
the acute angle whose cosecant is | ?
9. Prove that, if 6 is an angle less than 180" for which
1 4 sin 6 = 7c cos 6, then cos 6 ~ 2k -i- (1 + l^) : and express tan 0 in
terms of k.
10. Eliminate 9 from the following :
(i) sin d ■■= a, cos 6 ~ h;
(ii) sin d-\ cos 6 = a, sin 8 - cos 6 — h ;
(iii) sec 6 - tan B = a, sec 9 + tan 0 = },;
ELEMENTARY FORMULAE
53
(iv) a sin d + h cosS = p, a sin d-b C08 d = q;
(V) a sin 6 + b cos 6 = p, «' sin 6 + 1/ cos 6 = j^'.
11. If rf (1— sin^) = h cos 6, prove that & (1 + sin ^j = « cos ^.
12. If a (sec ^+ 1 j = b tan 6, prove that 6 (sec d-l) = a tan ^.
13. If a; = a cos ^ cos (f), y = a cos ^ sin ^, z = a sin ^, eliminate
^ and 0.
26. Ratios of complementary angles.
Let XOF^ A° I Fig, XII) and XOQ - 90-^° ; make OQ = OP,
and let fall QK, PN perpendicular to OX.
K NX
Fig. XII.
Then the triangles QOK, POX are congruent (Prop. 8c); so
that KQ = OX and OK = XP.
Hence sin XOQ = ^
~ OP
= cos XOP.
i. e. sin (90 — A) = cos A.
In a similar way it is proved that '^
cos (90 -A) =
tan (90 -A) =
Compare these results with § 11.
What are the corresponding results when angles nro measured
in radians ?
27. Ratios of supplementary angles.
Mako XOP = .1^, and XOQ = 180 ->!'', so that QOK = A°.
* The student is expected to complete thee formulae.
54
ELEMENTAKY FORMULAE
Make OQ = OP and let fall the perpendiculars PiS', QK.
Then the triangles QOK, PON are congruent (Prop. 8 c) so that
0K= Oy (in magnitude) and KQ = NP. But OK and ON are of
opposite sign.
Y
Q
^\
p
\
V )
N
y
X
\
y
1
y^
Y
^
r~
< o
¥-
W X
Fig. XIII.
Hence cos,XO^ ■=--
OQ
- z9^
OP
= - cos XOP,
i.e. cos (180- A) = -cos A.
In a similar way it is proved that
sin (180 -A) =
tan (180 -A) -
[OK) * (where [OK) denotes the magnitude
of OK with the proper sign prefixed)
28. Ratios of negative angles.
Make XOP = + ^" and XOQ = - A".
Then XOP = XOQ in magnitude.
Make OQ = OP.
Join PQ cutting OX at N.
Then in the triangles PON, QON
OP = OQ,
ON is common,
included angle NOI' = included angle NOQ.
Nl' = NQ in magnitude,
and ONP= ONQ, so that PQ is perpendicular to OX.
* In writing it is usual to use tlu; symbnl OK to denote length
preceded by correct sign ; it is more convenient to print {OK).
ELEMENTARY FORMULAE
{OX}
DO
Hence cos XOQ
e. cos (-A) = cos A.
Y
OQ
_ (ON)
OP
= cos A' OP,
0\
Q
Fic^. XIV.
In a similar way it is proved that
sin ( — A) =
tan ( — A) =
The student should also work out the ratio of 90 + ^4, 180 + ^,
270-^, &c.
29. By means of the hist three sections the ratios of any
angle can be expressed in terms of the ratios of an acute
angle not greater than 45°. For example
cos 139' = cos (180" - 41' j = - cos 4r,
cos 246' = cos(-114'j
= cos(114='j
-cos(180'-6G")
= -cos 66°
= -cos (90' -24^)
= -sin24^
56 ELEMENTARY FORMULAE
It is usually easy to work directly from the figure; thus in
cos 246° =
Fig. XV, where XOP = 246° and XOQ = 66°,
(ON)
OP
OQ
— cos 66°
-sin 24°.
Y
N
1
Q
K
P
0/
X
Fig. XY.
30. Ratios of 0° and 90°.
If XOV =0,7' and A coincide ; so thnt XP = 0, OX = OP.
Hence sin 0° = ^ - 0.
If XOP ^ 90 , then PX fulls along the y axis and X coincides
with the origin 0. In this case NP= OP and OX = 0.
Hence
ELEMENTARY FORMULAE
NP
57
sin 90 "" =
cos 90'
OP
ON
6F
1,
^ „ NP NP
tan 90^=^^= -^ = 00
31. Ratios of 30°, 45°, 60°.
Make XOP equal to 30", Fig. XVI.
Let fall PX perpendicular to OX.
Make XOQ equal to 30"" in magnitude, and produce PA" to meet
OQ in Q.
Fio. XVI.
Then, by Prop. Sc, the triangles PON, QOX are congruent. It
follows that the triangle OPQ is equilateral.
* The syml>nl x means 'infinity', i.e. a nnml;er greater than any
number we can imagine.
Consider the value of 1/x as x gets smaller and smaller,
1=10. 4 = 1000, .^j -_. 1000000,
As X diminishes, 1/x increases, and, by making x suificiently small,
We can make 1/x exceed any assigned value however great. This is
expressed thus: when x = 0, 1/x = co . Or more generally, if a is a
constant, then a/x = oo when x = 0.
58
ELEMENTARY FORMULAE
Hence
also
0^
Hence sin 30*^
cos 30° = '
tan 30° =
Similarly sin 60° =
OP.
PX = IPQ since PX = QN
= \0P since PQ = OP ;
0.Y-' = OP'-PN^ = OP'- 1 OP'- = ^^OP"".
VB
2
NP
OP 2
Ni' 1
f = -866,
OP
ON
•866,
tan 60" = -v/3 = 1*732.
As an exercise the stndent should find the values of the ratios
of45^
sin 45^ =
cos 45° =
tan 45"" =
Fig. XVIL
32. The very small angle.
In Fig. XVll let the circular measure ol" the angle AOP be $.
ELEMENTARY FORMULAE
59
Then arc AP - rO, NP - r sin 0, AT = r tan ^.
Hence Area of triangle AOP =^ lOA . NP
= |7-2sin^;
Ai-ea of sector AOP = \r^6;
Area of triangle^ or =\OA.AT
But, if ^OP is an acute angle,
Triangle AOP < sector ^ OP < Triangle AOT,
i e. Ir^sin^ <\r'e
i. e. sin S <6
This relation is true for
throughout by ;• we have
rBin^< rS < rtan^,
< * ;^ tan 6,
< tan d.
any acute angle
(Prop. 16)
(§7)
(Prop. 16)
if we multiply
NP < arc AP< AT.
Fi-. XVIII.
But, as the angle diminishes, these three lengths more and
more nearly coincide ; and are practically indistinguishable when
the angle is very small. This is shown in Fig. XVIII, which also
shows that ON is indistinguishable from OA.
Hence, when 6 is very small, there is very slight error in
saying NP = arc AP = AT, and OiY = OA.
60 ELEMENTARY FORMULAE
Substituting the trigonometrical values for the lengths of these
lines, we have
sin S = 6 = tan 6 and cos ^ = 1 ,
when 6, the circular measure of the angle, is small.
This may also be expressed thus : The limit of or of ^^^ >
when 6 is zero, is 1 ; or in symbols
T sin ^ = . «i T tan <9 , ^
L —X- 1 an Jj -.- = 1.*
33. Error involved. Whatever be the value of ^, it has been
shown that cos^^ + sin'^^ = 1.
Using the above approximations, we have
This last statement is true only when 6"^ is so small as to be
negligible. Hence
If ^ is so small that $- maybe neglected, we may say that
sin S = e, cos ^ = 1, tan 6 = 0.
It is shown in Higher Trigonometr}'- that sin i^' = ^ gives correct
results if }r6^ is negligible.
Example. // accuracy /.s required to four decimal places, find
the sine of 1 degree.
1° = ToTj 77 radians = -01745 radian
■017452= -000295. (We are not, therefore, justified in saying
cos r = 1.)
•01745' = '000005. (This does not aft'ect the first four places so
we may use the approximation sin 6 = 6.)
Hence sin 1° = '0175, correct to four decimal places.
Examples III b.
1. Write down the sine, cosine, and tangent of
(i) 150°, 240°, 330°, 840^
(ii) 60°, 800°, 135°, 225"^;
(iii) 180°, 270-^, 405°, 210°.
2. Find the secant and cosecant of G0°, 45°, 120°, 225^
3. Use Ihe definitions of § 14 to find the ratios of 180-^1.
* Fur explanation of the wonl ' limit " seo School Algtlra, I'art ii, p. 440.
ELEMENTARY FORMULAE 61
4. Correct, if necessary, the following statements :
Bin (180 -.4) = cos^ ; cos (270+ ^) = -cos^ ;
tan(180 + ^)=tan^; sec (90-^) = sec ^ ;
cot (90 + ^) = cot ^.
5. In a right-angled triangle the hypotenuse is 5 feet long and
one of the angles is 60° ; find the lengths of the other two sides.
6. A ladder 25 feet long is leaning against a wall and is in-
clined 45° to the horizontal ; how far up the wall does it reach ?
7. Find sin 1', correct to 3 significant figures.
8. Find sin 10', cos 10', tan 10' correct to 5 decimal places.
9. What angle does a halfpenny (diameter 1 inch) subtend at
the eye when at a distance of 10 feet ?
10. A p3st 25 feet high subtends an angle of 30' at a certain
point on the ground. How far from the post is the point ?
11. Find approximately the distance of a tower which is 51 feet
high and subtends at the eye an angle Sfy'.
12. Prove that
tan^ 60° - 2 tan^ 45^ - cot- 80° -2 sin- 80° - f cosec- 45°.
13. Find approximately the number of minutes denoting the
inclination to the horizon of an incline which rises 5^ feet in
420 yards.
14. In any triangle show that
cos(^ -\~B)=- cos C, sin (B + C) = sin J, tan {B +C)= - tan A .
Write down the other similar relations.
Oral Examples.
Fill in the right-hand sides of the following equalities:
1. (i) sin2^= 2. (ij sec^J-tan^^ =
(ii) sin 45' = (ii) cos 60' =
(iii) cos 135° = (iii) tan d =
(iv) tan^TT = (iv) sin (180 -^» =
(v) sinyl cot^ (v) sec (90-5) =
3. (i; cos2 60° + sin^60" = 4. (i) cos (9 tan ^ =
(ii) cosec'^ C = (ii) 1 - sin' x =
(iii) cot^7r= (iii) tan 210' =-
(iv) cos (180-^) = (iv) cos-^A =
I v) cot^ d = i v) cos'^ i TT + sin'^ i T
62 ELEMENTARY FORMULAE
5. (i) 1 + cotM = 6. (i) cos-i ^ =
(ii) sin ^ cot ^ = (ii) cos 225° =
(iii) sin(180 + ^j = (iii) cos (90 + A) =
(iv) cos^ 63° + sin^ 63 = (iv) cos(-^) =
(v) tan 330° = (v) cos (180 -Z^j =
7. (i) tanirr - 8. (i) sin (360-^) -
(ii) sec 60' - (ii) sin-i 2 =
(iii) sin^ J^ + cosH^ = (iii) sec^ i tt — tan^ ^ rr =
(iv) tan (|7r + ^) = (iv) cos 0° =
(v) tan 135° = (v) cosec 120° =
9. (i) tan 150° = 10. (i) sec 150° =
(ii) cos ^/ sin 6 = (ii) cos (360°-^) =
(iii ) cos 90° = (iii) cos-^ ^Z =
(iv) sec 240° = (iv) sin 77° cot 77" =
(v) cot (180-^)= (v)tant7r =
11. (i) tan 1200°= 12. (i) 008^23° + cos'
(ii) tan (180° + ^)= (ii) cos (270° + 5)
(iii) tan|7r= (iii) sin-^ •4 + cos^(^=
(iv) tan 15° cot 15° = . (iv) sin.(-(/)) =
(v) tan-M-l)= (v) sin 225° =
Examples III.
1. Prove, from first principles, that sin (90 + ^) = cos ^,
cos(180 + ^) = -cos^, tan(360-^) = -tan^.
2. Show that sin (180-^) = sin ^, when A is (i) obtuse, (ii)
between 180° and 270°, (iii) between 270° and 360°.
3. Show that cos (90 - ^) = sin J., when A is (i) obtuse, (ii)
between 180° and 270°, (iii) between 270° and 360°.
4. Show that tan (180 + ^) = tan ^, when A is (i) obtuse, (ii)
between 180° and 270°, (iii) between 270°, and 360°.
5. Give 6 different solutions of each of the following equations :
(i) sin A=\\ (ii) sin A= \. \ (iii) cos Q = ^^ ;
(iv) tan ^ = 1 ; (v) cos ^ = - ^ ; (vi) sin ^ = - ^ •
6. Show that all angles having the same sine as A are included
in one or other of the forms: 180/? + ^, if n is an even integer,
ELEMENTARY FORMULAE 63
180 71- A, it' n is an odd integer ; and that these are included in
the single form 180 « + ( — !)" A where n is any integer, positive
or negative.
7. Show that all angles having the same cosine as A are in-
cluded in the form 360 n ± A, where n is any integer.
8. Show that all angles having the same tangent as A are
included in the form 180 n± A, where n is any integer.
9. What do the forms of the three previous examples become
when the angle is measured in radians ?
10. If a small angle equals A°, what is the value of sin A ?
11. Show that using the approximation sin ^ = ^ is equivalent
to regarding a circle as a polygon with a large number of
sides.
12. What do the following equalities become when the angle S
is so small that 6^ is negligible ?
(i) sin2^ = 2sin^cos^; "^'
(ii) cos 2 ^ = 1 - 2 sin2 d ;
(iii) sin (S + cf)) = sin 6 cos ^ + cos 6 sin 'p ;
(iv) C = G tan 0 ;
(v) t'2 = 4^,.sin2,i^.
13. Two strings are tied to two pegs A and B in the same hori-
zontal line, and knotted together at C; when the strings are
pulled tight, it is found that ^C is 18 inches long and that the
angles CAB^ CBA are 30^ and 60" respectively ; how far apart are
the pegs and how far is C from AB ?
14. An inclined plane, length 4 feet, is inclined at 30" to the
horizontal, what is the length of the base ?
15. A pendulum is held so as to make an angle of 30" with the
vertical, what is then the distance of the end of the pendulum
from the vertical line through point of support ?
16. Prove the following identities:
(i) cot^ A cos^ A = cot^ A — cos'^ A ;
(ii) sec"^ A — sm"^ A = ta.n'^ A + coi"^ A ;
(iii) sin^ (cosec ^ — sin 6) = cos^^;
(iv) (cos A + cosec A) (sin A + sec A)
= 2 -f sin yl cos ^ -F sec A cosec A :
(v) (cos J. -f sec A) (sin.l -^ cosec .^j
= sin A cos .1 4 2 sec A cosec A :
(vi) sec A - sin A tan A = cos A ;
(vii) (sec A — cosec ^) (sin A + cos ^Ij 4 sec^ ^-1 cot .4 = 2 tan A.
64 ELEMENTARY FORMULAE
17. (i) If 6 and cf) dift'er by ^ n, prove that tan 6 tan 0 = — 1 ;
(ii) Show that the lines whose equations are, respectively,
1/ = mx and y = inx, are at right angles if mm' = — 1 ;
(iii) Show that the graphs of the equations ax + hy + c = 0,
ax + b'y + c = 0 are at right angles if aa+hb' = 0, and are
parallel if a/a' = b/V.
18. If tan 6 = b/a, find the value of a cos d + h sin 6.
19. If tan^ B = b/a, show that «/cos 6 + b/sin 6 = a^ + &3) 2.
20. Give a general formula for all values of A which satisfy the
equation cos J. = — 1 .
21. If a sin^ 6 + b cos^ 6 ^ c and a cos'^ /9 + & sin'^ (9 = d, prove that
« + & = c + rf.
22. From the vertex Cof an equilateral triangle ABC a perpen-
dicular CD is let fall on AB ; DC is produced to E so that CE
equals CA, and JLJ? is drawn. From the resulting figure find the
sine, cosine, and tangent of 15° and 75°.
23. A is an angle between 180° and 270°, also cos ^4 ^ -'i\ find
the value of cosec A + tan A.
24. Define the cosine of an angle of any magnitude and express
the cosine of an angle between 180° and 270° in terms of each of
the other trigonometrical ratios.
If cos ^ = — 65? fin^^ sin^, sec B, coiB, and explain any double
signs which occur in your answer.
25. Prove the following identities :
(i) (sin Jl cos J5 + cos A sin B)"^ + (cos A cos B — ^m A sin BY= 1 ;
(ii) sin^ B + cos*^ ^ = 1 - 3 sin^ B cos^ B ;
(iii) cot .4— tan A = sec A cosec ^ (1 — 2 sin"^) ;
(iv) (1— sin ^-cos A'f = 2 (1-sin A) (1 —cos A) ;
(v) (2 cos J. — sec -4)-r (cos yl-sin ^)= 1 + tan A.
(vi) (3 sin B cos^ B - sin^ B)^ + (cos" B-Sco&B sin^ Bf = 1 ;
(vii) sec' B - tan' ^ = 1 + 3 tan^ B sec^ B ;
(viii) versin (270° + ^) . versin (270°-^) = cos^ J.
26. Prove that
cos (180° -yl) = -cos^, and cos (90° + .4) = -sin.l.
For what values of A is tan A = v/8 and sec A = -2 ?
27. Solve for x the equations :
(i) .rM 2 .;>• sec o 4 1 = 0;
(ii) x"^ + 2 X cos a = sin^ a ;
(iii) x"^ + (tan a + cot a) a; + 1 = 0.
ELEMEXTAKY FORMULAE 65
28. Prove that the number of ^^ecolKl^^ in an angle whose circular
measure is unity is 206,265.
The moon subtends at the eye of an observer an angle of 30',
its distance is 240,000 miles, find its radius.
29. If tan-^ = |, find versin 0, and explain the double result.
30. Eliminate 0 from
( i) a tan 0 + bcotd = c, a' tan ^ -f /^' cot ^ = c' ;
(ii) a tan ^ + & sin ^ = r, a tan 6 + h' sin 6 = >:'.
Revision E:samples A.
i. Define the tangent of an angle. From your definition find
tan 45^ and tan 135^, and prove that tan iItt-O) = cot 6.
2. A surveyor goes 10 chains in a direction 35^ S. of E., then
7'8 chains U^ E. of S. ; then 5'6 chains 10' N. of W. Find by
drawing how far he is now from his starting-point.
3. Prove the relation 1 -sin-^ = cos- .4 for the case where A
lies between 90 ~ and ISC.
Show that (sin ^ + cos^j* = l -f 4 sin.4 cos .4 + 4sin^^-4sinM.
4. The gi-adient of a railway is 1 in 270 ; find the inclination
to the horizontal to the nearest second.
5. When the sun's altitude is 60', find the length of the shadow
cast by a vertical rod whose length is 10 feet.
6. Draw the graph of cos x between x = 15" and x — 135' with-
out usinor tables.
7. Explain how to find the length of the arc of a circle of given
radius, when the angle subtended at the centre is given in degrees.
A wheel, radius 4| feet, rolls along the gi-ound ; what hori-
zontal distance does the centre travel when the wheel turns
through 157' ?
8. Why is the secant so called ? Prove that the secant is the
reciprocal of the cosine.
Given sec A = 2^, find tan A and sin A.
9. Show that the graph of the straight line y = 2x-Tj is in-
clined to the axis of x at an angle tan"^ 2. Verify this by a care-
ful drawing.
10. Trace the changes in sin ^ as 6 changes from 0" to 360' and
exhibit these changes by means of a graph.
121C E
6Q ELEMENTARY FORMULAE
11. Find the smallest angle which satisfies the equation
3 cos^ + 2sin2^ = 0.
Give also four other solutions.
12, If sin A = f, prove that sec A + 1/cotyl = 2.
13. What is a radian ? Prove the formula
arc = rx 6.
Show that if 6 is small, sin ^ = ^ api^roximately.
14. Show that tan (180 n + A) = tan^ where n is any integer.
If tan 3^ =^3, state three possible values for A that do not
differ by 360°.
15. Find the value of the expression cosec yl — j: cot yl, if
sin A = ff (i) when A is acute, (ii) when A is obtuse.
16. Prove the identity 2 sin ^4 cos ^ = (2 tan J) H- (1 + tan^ A).
17. In a triangle ABC, C == 90°, AB = 15, sin ^4 = '37 ; find the
length of^C and 5(7.
18. Criticize the following statements:
(«) sin2 (9=4; (h) sin 6 tan ^ = 1 ;
(c) sin-^ ( - -3) = 170° ; {d) sin ^ + cos^ ^/sin 6 = tan 9.
19. Explain clearly what is meant by latitude.
A place has latitude 30° N., what is its distance from (i) the
earth's axis, (ii) the Equator, measured along the surface ? (Radius
of earth = 4000 miles.)
20. Give a definition of cosine that applies to angles of any size.
Prove that cos (180 — ^1) = — cos^.
If sin A = l^ and A is obtuse, find cos A.
21.. Prove that
(cos A cos 5 + sin A sin i?/ + (sin ^ cos 5 - cos A sin Bf = 1.
22. Draw the graph of y = seer from ;r = 0° to .r = 180°.
23. What is meant by the statement that tan 90° = oo ?
Is sec 90° equal to tan 90° ? Give reasons.
24. Construct an angle A such that cos J = -5 and tan^l is
positive.
25. Name the points of the compass between West and South.
How many degrees are there in the angle between SW. by S.
and S. by E. ?
ELEMENTARY FORMULAE 07
26. Find the values of sin45% cts^^ J. tan \7t.
27. Prove by means of a figure that
sin- A + tan- A = sec- A - coa- A.
Is this true when the angle is measured in radians ? Give
reasons.
28. Construct an angle such that its tangent = ^ and its versine
is greater than unity.
29. Prove the identity cos^ J. - sin-^l = (coL-^-1) -h cosecM.
30. Find the value of tan 0 from the equation
3 tan-^' = 2-v/3tan(9-l.
Hence find three different values of ^ that satisfy the equation.
3L Write down six positive angles which have the same cosine
as the angle a ; and find the positive values of 6 less than two
right angles which satisfy the equation
sin4^ = cos 5^,
32. Show how to find by calculation the value of sin 30' correct
to four decimal places.
Verify, by substitution,
(i) sin 60' = 2 sin 30' cos 30' ;
lii; sin 120' -sin 60' = 2 cos 90' sin 30' ;
liiij cos 60' - cos 120' = 2 sin 30' sin 90^
33. Prove the identities :
(i) cosec- J. — cotan- .4 = 1 ;
.. 5jfl3^nirf , 12 -13 cos 6i _
* "^ r2 + 13'cos e^ 5 - 13 sin ^' ~
34. A steamer travels along the equator from longitude o7V W.
to longitude 5' 30' E. in 4 days. ^Yhat is the distance travelled
in nautical miles ? What was her average rate in knots ?
35. What is meant by the chord of an angle ?
For which angle is the chord equal to unity ?
Explain how to draw an angle when a table of chords is
given.
36. Express the following ratios as ratios of angles not greater
than 45' :
sin 172', cos 412', tan 246', sec 76', cosec 147', sec 236',
cot 138=, cosine 150°, sin 67', tan 102°.
E 2
68 ELEMEI^TAEY FORMULAE
87. If the circumt'ercii'es ot* the quadrants of two circles be
divided similarly to the right angles they subtend, what would be
the radius of a circle divided according to the French scale, in
which the length of the arc of one grade would be equal to the
length of the arc of one degree on a circle whose radius was
18 feet?
38. Point out which of the trigonometrical functions are never
numerically less than unity, and which may be either less or
greater than unity.
Express the numerical values of sin 135° and tan 150° with their
proper signs.
39. If n be a positive v/hole number, show that the angles
(2??. 180° + ^) and {(2;? 4 1)180°-^}
have the same sine as y1.
Express these in a single formula.
40. Distinguish carefully between (sin A)~^ and sin"^^.
Show that cos-i |. + 2 sin"' i = 120°.
41. Trace the changes in sign and magnitude of the expression
cos re -sin a; as .r increases from 0 to 27r. Illustrate your answer
by a graph.
42. A church spire, whose height is known to be 45 feet, subtends
an angle of 9' at the eye ; find its distance approximately.
43. What is meant by tan"'' m ?
If y= mx + c represents a straight line, state the geometrical
interpretation of the coefficients m and r?
What is the angle between the lines whose equations are
ij = x-4:, y = a/Sx + 2'?
44. Show that the equation of the line joining the points
45. Find the equation of a line passing through the origin and
(i) parallel to, (ii) perpendicular to, the line whose equation is
y = mx + c.
Deduce the conditions that the two lines whose equations are
ax + hy + c = 0, a'x + h'y + / = 0, should be (i) parallel, (ii) per-
pendicular.
ELEMENTARY FORMULAE 69
46. Find the eijuatioii of the line joining the origin to the point
F whose co-ordinates are {x', y').
Find the equation of the line perpendicular to OP and passing
through P.
Hence show that the equation of the tangent to a circle at the
point x\ y' is xx ■\- yy' = v"-, the equation of the circle being
X- + ?/- = ;•-.
47. If (;•, 6) are the polar co-ordinates of a point, what locus is
represented by
(i) r = 3, (ii) ^ = ^ tt, (iii) r cos 0 = 5, (ivi r = 5 cos 6?
48. If (x, y) are the Cartesian co-ordinates, (r, 6) the polar
co-ordinates, of the same point, what relations connect them ?
Express the equations of the previous example in Cartesian
co-ordinates.
Express (i) x^ + y'^ — 4:X + hy — 1, (ii) 3x'-f4// = 5 in polar co-
ordinates.
49. The sum of two angles is 3 radians, their difference is
10 degrees. Find each angle in degrees, assuming that
4377 = 135.
50. A ring, 10 inches in diameter, is suspended from a point
one foot above its centre by six equal strings attached to its
circumference at equal intervals. Find the angle between two
consecutive strings.
CHAPTER IV
USE OF TABLES
34. It has been shown in the previous chapters that the
trigonometrical ratios of any angle may be foancl roughly by
drawing to scale or by means of graphs. By methods which are
explained in more advanced books on Trigonometry, the ratios
can be calculated to any required degree of accuracy. There are
many collections of tables published, containing not only the
actual trigonometrical ratios (the natural functions as they are
called) but also the logarithms of these ratios. These collections
differ slightly in their arrangement, but the following general
remarks apply to most of them.
35. Since any ratio of any angle is equal in magnitude to the
same ratio of some angle less than 90°, it is necessary to tabulate
the ratios only for angles between 0° and 90°. Thus
sin 156° = sin (180° -24°) = sin 24°,
cos 215° = cos (1.^0° + 35°) = - cos 35°.
But the tables may be made even shorter, for any function of
an angle between 45° and 90° is equal to the complementary
function of an angle less than 45°. Thus
sin 76° = sin (90° -14°) = cos 14°,
^ tan69°= tan(90°-21°) = cot21°.
This fact is used in two different ways. Some tables give all
the ratios for angles from 0° to 45° ; so that if, for instance,
sin 72° is required, it must be looked up as cos 18°. Other tables
give the values of sine, tangent, and secant for angles from 0°to 90° ;
in this case, cosine, cotangent, and cosecant must be looked for as
the si]ic, tangent, and secant respectively of the complementary
angle.
The slight mental work involved is avoided by giving each
column a " footing " as well as a heading. Thus '26892 is, in some
tables, found on a page headed Natural Sines, on a level with 15° in
the extreme left hanl column and under 36', i.e. '26892 = sin 15°36'.
USE OF TABLES 71
Bat the same page has Natural Cosines at the bottom, "26892 is on
same level as 74° in the extreme right-hand column and above 24',
i.e. -26892 = cos 74°24'.
A few minutes' inspection will make the arrangement of any set
of tables quite clear.
36. Logarithmic Functions. Since the sine and cosine
cannot be greater than unity, their logarithms cannot be greater
than zero ; hence these logarithms have a negative characteristic.
In order to avoid difficulties of printing it has been the custom to
add 10 to all these logarithms, and to the other logarithmic func-
tions. The values thus tabulated are called Tabular Logarithms
and are denoted in writing by L, thus L tan 75^ = log tan 75'' + 10.
Some of the modern tables give the ordinary logarithms with
the negative characteristics.
When tabular logarithms are used it is advisable to sub-
tract 10 mentally and to work with the correct logarithm.
37. Interpolation. It is impossible to give the ratios for all
angles. Four-figure tables usually give values for every 6', seven-
figure tables for every 1'. Intermediate values may, in some tables,
be found from side columns giving the differences^ as in the case
of ordinary logarithms. If these side columns are not given, the
method of proportional parts * must be used. This method is
equivalent to assuming that the graph of the tabulated function
may be treated as a straight line for portions lying between the
points corresponding to two consecutive tabulated values. The
practical use is easily followed from an example or two.
Example i. Given that
sm28°9'.= -4717815, ami sin 28^ 10' = -4720380,
find sm 28=^9' 43".
sin 28^ 10' = -4720380, "000004275 x 43
sin28'' 9' = -4717815. '00017100
Increase for 60" = '0002565 ; 1282
Increase for 43" = |^ x '0002565 '00018382
= '0001838;
sin 28" 9' 43" =-4719653.
^ For a fuller treatment see School Algclm, Part IT, p. 376.
72 USE OF TABLES
In practice the zeros are omitted as in the following example.
Example ii. Given that Zo^ cos 73° 15' = 1-4058617, and
log cos 73^16' = 1*4053816, ^/j72cl the angle ivhen the log cosine is
1-4056348.
Denote the angle hy 73° 15' x'\
- log cos 73° 15' = T-4058617. log cos 73= 15' - 1-4058617.
log cos 73° 15' x" = T-4056348. log cos 73° 16' - 1-4053816.
Decrease for £c" = 2269. Decrease for 60"- 4801.
X 2269 2269
^^"^^ 60^4801- ^1
.r-28. 4801)136140
4012
171
.". required angle = 73° 15' 28" to the nearest second.
Note. It is important to recollect that cosine, cotangent,
cosecant, and their logarithms decrease as the angle in-
creases ; consequently proportional differences must be suh-
tr acted, not added.
If the graphs of the functions are carefully drawn, it is seen that
in some parts they approach much more nearly to straight lines
than in others. It follows that the method of proportional parts
is more accurate for some angles than for others. For a complete
discussion of Proportional parts see Nixon's Elementary Plane
Tngonometnj (Clarendon Press) or any advanced textbook.
Examples IV a.
Find, from tables, the natural function of the following angles,
find the logarithm of the number found, and then look up the
logarithmic function in the tables. There may be a slight dis-
crepancy in the fourth decimal place.
1.
sin 17= 15'.
2.
cos 73° 47'.
3.
tan 16° 39'.
4.
cos 23° 19'.
5.
sec 67° 15'.
G.
cotan44°5'.
7.
tan 78° 53'.
8.
sin 83° 43'.
9.
cos 63° 28'.
10.
sin 156° 17'.
11.
tanl7G°16'.
12.
cot 100° 10'.
13.
cos 137° 42'.
14.
sin 126° 37'.
15.
tan 173° 14'.
Explain carefully the difficulty that arises in connexion with
some of the angles.
USE OF TABLES 73
16. Find the Cartesian co-ordinates of a point whose polar co-
ordinates are (i) 17, 16°; (ii) 25, 114°; (iii) 49, 227^
Find the angles less than 180' which are determined by the
following data :
17. sin e = '8732. 18. cos A = _'8469.
19. sin 5= -9340. 20. logtan .4 = 1*7932.
21. Lcos^= 9-7432. 22. sec B = 2'5732.
23. logsin^ = 1-3465. 24. L tan .4 = 10-4385.
25. L cote = 10-7386.
Find the sine, cosine, and tangent of the following angles,
which are measured in radians :
26. ^n. 27. iV-- 28. 1-2. 29. ^;r.
30. Verify that
sin 112^ = sin 70' cos 42' -f cos 70° sin 42'. ' .
31. Find from the tables the values of ■ I
sinjV77 and sin 27' 18'/cos 32"' 45'. " ' — •
32. Employ the tables to verify the formula
cot 24° 45' - cot 49= 30' = cosec 49' 30'.
33. Find the values of cos 110", cot 160°, sin 250".
A quantity fi is such that /n = sin //sin ;• ; complete the following
tables :
I
r
H-
34.
W
12"
35.
26° 18'
1-427.
36.
31° 52'
1-467.
37.
53° 49'
1-5.
38. Find the polar co-ordinates of points w^hose Cartesian co-
ordinates are (i) (3, 7) ; (iii (-3, 7) ; (iii) (-3, -7) ; (iv) (3, -7).
39. The angle of friction e and the coefficient of friction fx are
connected by the relation /^t = tane. Determine the missing
quantity in the following cases :
6 40'' 15' i 17=39' I 47' 8' : |
,1 I -67 I -37 ' -50
40. In a circle of radius 17 find the lengths of chords subtending
angles (i) 37°, (ii) 73", (iii) 143° at the centre. What are the
areas of the corresponding segments V
74
GEAPHS
38. Graphs.
Example. Draw the graph of
3 sin (^+30°)- 2 cos (a;-30°) from x =^ 0° to x =^ 120°.
In other words, draw the gra2)h of
^j = S sin {x + 30°) - 2 cos (ic- 30°).
X
0
•500
15°
30^
45°
60°
75°
90°
•866
105°
120^
•500
sin (x + 30^) .
•707
•866
-966
1-000
•966
.0.
cos (x-30°) .
•866
•966
1-000
-966
•866
•707
•500
•259
0-000
3 sin (x + 30°)
1^500
2-121
2^598
3-000
2^898
2^598
2-121
1-500
2 cos (.x-30°)
1^732
1-932
2-000
1-732
1^414
•1-000
•518
•000
y
-•232
•189
-598
-966
1-268
1-484
1-598
1-603
1-500
The graph is shown in Fig. XIX.
Use of Graph.
Interpolation. The value of the function can be found for any
intermediate value of the angle. From the graph it is seen that
v/ = r07 when x = 50°, and y = 1'54 when x = 81°. Calculation
shows that the correct values are 1'075 and 1*542 respectively.
This is a useful method of testing the accuracy of a graph.
Maximum and Minimum. When, as x increases, ij continually
increases to a certain value and then decreases, that value is said
to be a miaximum ; similarly, when j/ first decreases and then
increases there is a minimum value. These maximum and
minimum values are clearly shown on the graphs ; the corre-
sponding points are called turning-points.
From the graph the maximum value of
3 sin (.r + 30°) - 2 cos {x - 30°)
is found to be 1*62, and the correi^ponding angle is OO"".
GRAPHS
75
x
S
1 - it
I 1
t ^
\f\
o
1 ^
^\r -^^
\^ J
JV
^ Ul
4^5
I^^ 1
X-^
^^di i
v5 °
\^ ^
- ^x
^!s
^^ \^^ "^
^s: ^ '
T ^^
^
oK
^, o
\ (0
^
\
\
^^ ^
\ -
\.
\
V
h ib ^
•- -■ o n
-__L
76 GRAPHS
Kate of change of the function. The graph shows that, when
X changes from 15° to 30°, the increase in y is more than when x
changes from 60° to 75°; consequently the curve is steeper
between 15° and 30° than between 60° and 75°. Thus the rate at
which y changes compared with x is shown by the steepness of
the curve.
Join two points P and Q on the graph, and draw PK, QK
parallel to the axes to meet in K. Then
increase in v/ KP , ■ n , i i nr^r.-
^ — i' = = tangent of the angle PQK
increase in x QK
= tangent of the angle PQ makes with the axis of x.
This is called the slope of the line PQ.
When Q approaches indefinitely near to P, the chord PQ
becomes a tangent. Hence
The rate of increase of y at the point P is measured by the
slope of the tangent at the point P.
Notice that the slope diminishes in the neighbourhood of a
turning-point and is zero at the turning-point itself.
39. Solution of equations. By finding where two graphs
intersect or where one graph intersects the axis of x or a line
pai'allel to the axis, equations can be solved just as in Algebra.
Example. Solve the equation
3 sm (a;-}- 30°) -2 cm- (a; -30°) - r5.
It is seen in Fig. XIX that the graph cuts the line whose
equation is ?/ = I'S where x = 77° and x = 120°. These are,
therefore, the solutions within the range of the graph.
Examples IV b.
(The graphs should be verified in the way that the example
of § 38 is verified.)
Draw the graphs and find the turning-points of:
1. sin^ X from it; == 0° to a; = 90°.
2. cos 1 X from x= -90° to x = 90°.
3. sin ^ ic -f cos ^J- a; from x = 15° to x — 135°.
4. I tan {x-m°) from x = 0" to x - 90°.
5. ^-sin e from ^ = 0 to ^ - ^tt.
6. sec X — tan x from O'' to 90°.
GRAPHS 77
7. cos- i ar-f sin- ] x from 0 to 360\
8. Draw the graph of sin .r + cos.r between o-- = 0 and x = 360°.
Solve sin 3-+ cos .r = '89, and find the slope of the graph at the
points corresponding to these values of ,r.
9. Draw the graph of cos.r between the values of 0 and 2 tt for x\
Show that an acute angle can be found to satisfy the equation
X = cos X.
10. Draw the graphs from x = -1 to j^^ = -f 1 of (i) sin-^r,
(ii) cos-^rr, (iii) tan-^a;. How are they related to the graphs of
sin.r, cos.r, tana? respectively?
11. Draw the graphs whose polar equations are
(i) r sin 6 = 11 ; (ii) r =10 sin 0 ;
(iii) ;• = 10 cos 6 : (iv) tan 6 = 2-45.
12. Find from your tables the values of cos 2 x for the values
0^ 10", 20% 30^ 40^ 50\ 60^ of .r.
Draw the graph of cos2.r-cosa; as x increases from 0°
to 60°.
13. Find, by drawing graphs of 2 sin .4 and sin 2.4, for what
values of A, less than 90', 2 sin ^ — sin 2.4 = 1,
14. Find, by the aid of the tables, the values of sin ,r — tan2a: for
the values 0=, 10^ 20", 45°, 60' of x.
Make a graph to give the values of sin.r — tan2.r from ,r = 0 to
X = 60°.
15. Make a table giving the values of cos^ at intervals of one-
fifth of a radian from ^ = 0 to ^ = two radians, taking the radian
as 57^30'.
From your table plot the graph of t^cos^; and hence find for
what value of 6, between the limits 0 and 2, ^cos^ is greatest.
16. Plot the function ^ {sin ^-f. sin 2 (^ + 20°)} between <9 = 0^
and 6 = 180°, and find the maximum and minimum values of the
function which occur within this range, and the corresponding
values of d.
17. Draw, in the same diagram, the graphs of sin a- and 2 cos a;
between x = 0° and x = ISO". Show how to find from your
diagram an angle whose tangent is 2.
18. Taking tt as 8*1416 and using your tables, find the values of
^-sinf? when 3 = ^7r, ^tt, ^ tt, ^tt, f^rr, and -^ tt ; and hence
make a graph to give ^ — sin ^ from 6 = 0 to 6 = Itt.
19. Draw the graph of (i) sin~^a'-l-cos~^ic ; (ii) sin-'(l/a:).
78 SOLUTTON OF EQUATIONS
Solution of Equations.
40. To solve a trigonometrical equation,
(i) express all the ratios involved in terms of one ratio,
(ii) find the value of tliis ratio by ordinary algebraical
methods,
(iii) find the angle from the tables,
(iv) give the general solution.
Example i. Solve 2 sin .r-f 3 cos a? = 2.
Express in terms of sine, -^
+ 3 V*! - sin^iP =2 — 2 sin x.
Square 9 — 9 si n^ a; = 4 — 8 sin x + 4sin^ x.
Transpose ISsin^aj-S sina7-5 = 0.
Factorize (13 sin x + h) (sin x-\) = 0.
.'. sin ^ = — j5 or 1.
Substituting in the original equation, we find that :
(i) If sin. r = — j%, cos^r = \'i = '9281.
Hence the bounding line is in the fourth quadrant.
From the tables it is found that cos 22° 37' = -9231.
Hence the smallest positive angle satisfying the equation is
360"-22"37' = 337''2r.
But we may add or subtract any multiple of 360° without
altering the position of the bounding line ; hence any angle
satisfies the equation whose value is 360° n + 337° 23', where n is
any integer positive or negative. This is the general solution.
(ii) If sin a; = 1, cos.^ = 0.
The smallest solution is x = 90°.
The general solution is 360°n + 90° or (4n + l)90°.
Note. The same difficulty has arisen here tliat arises in Algebra
when the original equation contains surds. After we have squared,
the i-esulting equation is exactly the same as if we liad started with
the equation 2 sin x — 3 cos x = 2. For this reason, after we found the
value of sin 9, it was necessary to substitute in tlie original equation
to find the corresponding value of cos x.
SOLUTION OF EQUATIONS
Example ii. Solve tan- ^ + 4 sin'^ A = 5.
Express in terms of tan ^.
79
tanM + 4
tan^^
1 + tan-^l
tan^ ^ = 5.
4 log tan ^ = -6990.
log tan ^ = -1747.
= log tan 56° 13'.
^ = 56° 13'.
Consideration of the fundamental figure shows tliat the general
solutionis ^ = 180>i + 56° 13'.
Multiply by l + tan^^
Take logarithms
Use tables
Hence a solution is
41.
General Solutions.
General solutions can always be obtained by mentally
considering the possible positions of the radius vector that
give angles having the same function as some angle already
found. This is what has been done in the two preceding
examples. It is, however, useful to know the formulae that
give these general solutions.
Find an expression for all angles that have a sine
equal to sin oc.
We have to solve sin 6 = sin a.
Fig. XIX a.
The bounding line may have either of the positions shown
in Fig. XIX a.
en
80 SOLUTION OF EQUATIONS
Thus the line may revolve through n.r, where ii is even, and then
go on CX, or may revolve through mr, where )i is odd, and th
come back Oi. Hence ^ = n rr + 3c if n is even,
or nrr — y if 71 is odd.
These are included in the one formula
e= 1177 + {-iyo(.
If sin X = sin A, then x = ISO n 4 ( - 1 j" A.
Exercises. In a similar way prove that
i. 6 = 2mr±0(, when cos d = cos 0( ;
X = S60 u ± A, when cos.r = cos^.
ii. 0=1177 +CX, when tan^=tanCX;
;r ='180 n + ^, when tan .-c = tan ^ .
Example. Solve sin 3^ = cosb^.
This is the same as
sin 36 = sin (i/r-5^);
3^ = n7r + (-l)"(lrr-5^).
If n is odd Sd = mr-}>7T + 5e;
20 = 1
- n TV.
Put 11 = - 2 i? + 1, then 0 = ^; 77 - 1 rr, where ^; is any integer.
If n is even 3 ^ = ??. tt + (In - 5 ^) ;
Put n = 22) ^ = P^T^ + it^ ^' where p is any integer.
The complete solution is
d =2)rr- ^ 77 or jj Itt + ^tt.
Examples IV c.
Solve :
1. 2cos2(9 = 3(l-sin^). 2. sin^ + cos^=l.
3. sin 0 + cos 0 = 72. 4. 12 tan^ .4 - 13 tan A f 3 = 0.
5. 2 008^0-- 1 = l-sin^.r. 6. sin 3^ = sin 4^.
7. 3cot^<9-10cot2^+3 = 0. 8. 2sin .4 = tanvl.
9. tan^ + 3cot^ = 4. 10. sinj.r + ^) = cos(a;-^).
11. tanM + 4sinM = 6. 12. v'3tan2^ + 1 = (UV3)tan^.
13. cosec^ = cot ^+73.
14. cos(135° + ^)+sin(135°-^) = 0.
15. cos^ ^ - cos ^ sin A - sin^ yl = 1.
16. cos3^ + sin(9 = 0. 17. 3tan2 2^ = l.
USE OF TABLES
81
18. tan2.r = tan2/;r.
20. 2cos2^-f3cos^-l =0.
22. l-7sin^--73 = 0.
24. 2tan2^ + 7tan^-f3 = 0.
19. 2sin2^-3sin^-2 = 0.
21. tan2^ + sec-^-2.
23. 3 sin (9 + 2 cos 0 = 2.
25. tan^-2cotr' = 17.
42. Examples of the use of logarithms.
Example i. Given that
a ^ 250, h = 240. A = 72' o\
sin .4
a
sin A
n)i B
find B wJicn
We have
1. e.
sin 5
~V
sin 5
a
b sin A
Take logs. log sin B = log h + log sin .4 — log r/
= 2*3802 -2-3979
+ r*9784
= 2-3586
-2-3979
= f-9607 ;
[.-. Lsin5= 9-9607]:
.'. sinjB = sin66^
Hence jB = 180n' + (-1)» 66^
After a little practice the work may be arranged so that the
logarithms are kept quite distinct from the remainder of the work.
This same example is worked below to show the shorter method
and the use of five- figure tables.
Logarithms.
sinB
sin-B
smA
a"'
240 sin 72"^ 5' ^250
sin 65^ 59'.
2-38021
+ 1-97841
2-35862
2-39794
1-96068
Example 11. The sides and angles of a irianglc are con-
nected hij the relation tan \ [A —B) — — -, cot \ C: find A a)>d
•^ ^ ^ a-\-h
B irhen a
1216
242o, 6 = 164-3, C= 54^36'
F
82
USE OF TABLES
-^-^ '^-2 cot 27=18'
2 - 406-8 """"
r 78-2 , ,
= tan 20° 26'
.'.
^? = 20-26',
by question
^^^ = 62-42-.
Hence
A = 83" 8',
5 = 42" 16'.
Logarithms.
1-89321
+ -28723
2-18044
2-60938
1-57106
The step in brackets is required if the tables do not give the
cotangents. Since A and B are angles of a triangle, 1{A-B)
cannot ec^ual any of the angles 180 n + 20° 26' (except when n = 0).
so that there is no need to give the general solution.
Example iii. // a, h, c are the sides of a triangle,
(g y\ [g (A
fan lA—^ ^ — -^ — —-^ iclierc s is half the sum of the
V s(s—a)
sides. Find A when a = 1762, h = 893, c = 1386.
Logarithms.
s = (1762 + 893 + 1386)-f-2 =
4041-f-2
= 2020-5,
305212
s-a= 258*5,
280243
s-h = 1127-5,
+ 5-85455
s-c = 634-5 ;
3-30546
^ A 71127-5x634-5.
' * "^^^2 ~V 2020-5x258-5'
2-41246
-5-71792
.-. ^ = 49° 29',
2)0-13663
006831
A = 98° 58'.
Examples I"V.
1. Use the tables to find the values of sin 52'', cos 140^, tan 220".
cos 340", sm340^
2. Divide sin52 ])y cos 52'; verify your answer liy finding the
value of tan 52 ' from the tables.
USE OF TABLES 83
3. Write down by using tables the values of sin 140', cos 160",
cos 220°, tan 320°.
4. Find the smallest positive value of B which satisfies
cos 0 = sin {(4 m + 3) | tt + 0(}.
5. Find all the values of 6 which satisfy the equation
4cos6J-3sec^ = 2tan^.
6. Find the inclination to the horizon of an incline which rises
5i feet in 420 yards.
7. Solve the equation tan-t^ - (1 + ^/S) tan (9 + ^3 = 0.
8. Given that tsui^C = \/(s-a) {s-h)-=r s{s-c), find C when
a = 32, b = 40, c = 66.
9. Solve the equations cos(2x + 3 //; = |, cos(3.r + 2 //) = ^^3/2.
10. Given log 2 = '30103 and log 3 = -47712, find (without the
tables) L sin 60= and L tan 30° .
11. Find the acute angle whose cosine equals its tangent.
12. The current C in a circuit, as determined by a tangent
galvanometer, equals G tan 6, where 6^ is a constant depending on
the galvanometer only and 6 is the deflexion of the needle.
Determine the ratio of two currents which give deflexions of
27' 14', 35' 23' respectively.
13. The length of a degree of latitude in latitude (p is
(11 11-317 - 5-688 cos 0) 10* centimetres.
Find the length at London (latitude 5r3rN.) and Melbourne
(latitude 37"50'S.).
14. The length of the seconds pendulum in centimetres, at
a place whose latitude is A, is 99-3563 --2536 cos 2A. Find the
length of the seconds pendulum at Paris (lat. 48° 50' N.) and
Calcutta flat. 22' 33' N.).
15. The acceleration of a falling body at a place whose latitude
is X, when measured in centimetres per second per second, is
980-6056-2-5028 cos 2 X.
Find the acceleration at Montreal (lat. 45° 30' N . i and Cape Town
(lat. 33° 40' S.).
16. A quantity A is determined by the relation A = ^ ah sin C.
Complete the following table :
A
a
b
C
i.
17
43
IT 14
ii.
342-6
21-3
38-19
iii.
984-2
43-82
f2
43 21'
84 USE OF TABLES
17. Draw the graph of
tan 6-0 from ^ i= 0 to ^ = ^•
Hence solve tan 6 = 6 + S.
18. Given that A and B, the angles of a triangle, are connected
by the relation asm B = fesin^, find B when a = 181, b = 217,
A = 34° 15'.
19. If 2i? = a'sin^, find the value of A when i? = 179*4 and
a = 300.
20. Verify that
cos 146' 43' - cos 56" 51' = - 2 sin 44° 56' sin lOP 47'.
21. Find the length of (i) the chord, (ii) the arc, subtending an
angle 70° at the centre of a circle of radius 25 cm. Find also the
area of the segment.
22. Find the length of the side of a regular decagon (i) inscribed
in, (iij described about, a circle of radius 2 "7 inches.
CHAPTER V
THE RIGHT-ANGLED TRIANGLE
43. In the previous chapters we have had to deal with only
one angle at a time, and have been able to draw one of the lines
containing that angle horizontal. In applications of Trigonometry
we often have to deal with several angles in the same example,
and the lines containing them are drawn in various directions ; in
such examples it would be difficult to apply the definitions of
§ 12. But it has been shown that the ratios of any angle can
be expressed in terms of the ratios of an acute angle. In practice,
therefore, it will often be found advisable to use the following
definitions, which apply only to acute angles.
In a right-angled triangle an acute angle is contained by
the hypotenuse and one of the other sides which is called
the side adjacent to that angle. The remaining side is
called the side opposite. Then in Fig. XX
86
THE KIGHT-ANGLED TRIANGLE
sin BAG =
cos BAG =
tan BAG =
opposite
hypotenuse '
adjacent
hypotenuse '
opposite ^
adjacent '
cosec BAG =
sec BAG
cot BAG =
hypotenuse
opposite '
hypotenuse
adjacent '
adjacent
opposite
These are clearly the same definitions as in § 12, the triangle
BAC taking the place of the triangle PON; and the various
formulae proved in Chap. Ill can be proved directly from the
definitions of this section.
44. It is usual to denote the angles of any triangle ABC by
the capital letters A, B, C; the lengths of the sides opposite the
angles A, B, C are denoted by a, b, c respectively.
Hence, in a triangle ABC, right-angled at C,
a
sin^
I.e. a = c sin A
cos A = -,
c
tan^
&'
i.e. b = ccos^
e. a = &tan^.
Examples Va.
1. Prove, from the definitions of § 43, that
(i) cos ^ = sin 2? = sin (90 — ^) ;
(ii) sin A = cos B = cos (90 — ^) ;
(iii) tan A = cotB = cot (90 - A.).
2.
P
N M
Fig. XXI.
THE EIGHT-ANGLED TRIANGLE
87
In the figure PNO, QMO, QKP are right angles.
If 0N=o, NP=1, OM=G, MQ = o, find the values of
sinPO.V, inn KPQ, tan KQP, sec QOM, cos KQO, cosec XRO.
3. If, in Fig. XXI, OP = 8, POQ = 30", QON = 45°, PQO = 90^,
find the lengths of OQ, PQ, PK, QM, OM.
4. A circle is described on a horizontal diameter ^5 of length
10 inches; a point C is taken on the circumference, such that
BC = 7, and CD is let fall at right angles to AB. Find the size
of the angle BAC and the length of CD.
5. In a triangle, right-angled at C, a perpendicular is let fall
from C to the hypotenuse ; prove, by Trigonometry, that this
perpendicular is a mean proportional between the sides containing
the ris^ht angle.
D
C
B
^\
A
Fig. XXII.
In the above figure (which is not drawn to scale) AO is at
right angles to DE, OC is at right angles to AG, OG is at right
angles to AO and EF', also G is the middle point of AB.
Use this figure in the following examples.
6. UAC= 10, CAD = 40°, find, if possible, the lengths of all the
other lines.
7. If CD = S, AB = 24, find sin CAD.
8. If GF= 18, AE=b, OC = 5, find cos ACD and the length
of^a
88
THE RIGHT-ANGLED TRIANGLE
9. If AB = I, CAD = 0, find CD and AO.
10. If CG = a, CGO = 6, find AD and AO.
Fig. XXIII.
With the ordinary notation for the sides and angles of a
triangle, find in the above figure :
11. The length of AD when c = 70, i? = 49°.
12. The length of AD when 1= 42, C = 72°.
13. The length of BD when c = 76, B = 39°.
14. The length of CD when h = 114, C = 114°.
(What geometrical fact does the negative sign in the result
show ?)
15. Prove that the area of the triangle = i«& sin C; give the
proof, also, when C is obtuse.
Solution of Right-angled Triangles.
45. The angles and sides of a triangle are sometimes
called the six parts of a triangle. The determination of all
the parts, when only some of the parts are known, is called
solving the triangle. If the triangle is known to be right-
angled, the triangle can be solved if one side and one other
part are known.
Example i. A man, standinci 100 feet from the foot of
a clmrelt stee2)le, finds that the inu/le* of elevation of the top
* If a i)er.son is looking upwards, the angle Ins line of sight makes
with the horizontal is the angle of elevation ; similar!}', if lie is look-
ing downwards, the angle his line of sight makes with the horizontal
is the angle of depression.
THE RIGHT-ANGLED TRIANGLE
89
Ls 50°. If Ms cue is 5^ feet from ilie gyoiind, what is the
height of the steeple ?
[The figure should be drawn neatly but need not be drawn to scale.]
In Fig. XXIY AE represents the steeple, BC the man; CD is
drawn parallel to BA.
[Mental. In the right-angled triangle CDE we know that
CD = BA= 100 ft., angle DCE - 50^
and w"e wish to find DE.
unknown side
Here
some ratio of known anglt
= tan DCE,
known side
DE
. BC
i.e. /)i^ = 100 tan 50° feet
= 100 xl'1918 feet
= 119-18 feet.
Therefore height of steeple = AD + DE= 124'68 feet.
Example ii. The shadoir, cast by the sun on a horizontal
plane, of a vertical pjole 10 feet high, is ohscrred to he 14 feet
long ; find the altitude of the sun [i. e. the angle of elevation
of the sun).
90
THE RIGHT-ANGLED TRIANGLE
In Fig. XXV AB represents the pole, AC the shadow; so
that CB is the direction of one of the sun's rays.
loF^
[Mental. In the right-angled triangle BAC ^ve know 5^ and AC,
and wish to find the angle ACB.
ratio of known sides = some ratio of required angle.]
— = tan ACB ; ^^^ ^^, ^^, ^ .^^^^9.
.-. tan ACB = ]^ = -71429. 440 is diff. for 10'.
.-. ^CJ5 = 35°32'. .-. 100
Sun's altitude = 35° 32' to nearest minute.
10_Q' _ 0'
44
Here
sec BAC:
Example ill. A ship C is observed at the same time from
two coastguard stations A and B, 1459 yards apart. The
angle ABC is found to he 90°, and the angle
BAC to he 67° 14', ivhat is the distance of
the ship from station A ?
AC
AB
.'. log^O = logl459-+logsecG7°14'
= 3-16406
+ -41111
+ 120
= 3-57637,
i.e. ^0=3770-3.
Distance of ship from A = 3770 yards to
nearest yard.
Fig. XXVI.
^ If tlie tables do not contain the secants, the working must be made
to depend on the cosine.
THE RIGHT-ANGLED TRIANGLE
91
AC
1
AB~
cos BAC
log AC =
log 1459-
-log
cos
67°
14
=
3-1641
_
1-5877
=
3-5764,
.e. AC =
3770.
Example iv. Ttvo men, A and B, 1370 yards axmrt,
observe an aeropJane C at the same instant and find the
respective angles of elevation to he 40° and 67°. If the plane
ABC is vertical, ralcidate the height of the aeroplane.
Fig. XXVII.
Let h feet be height of aeroplane.
From triangle ADC, AD = h cot 40°.
From triangle BDC, BD = h cot 67' ;
but AD + BD = AB;
.-. hcoti0° + hcot6r= 137.
1370
cot 40° + cot 67"
1370
1-61622'
.-. h = 847-62 ;
Height of aeroplane = 848 yards to nearest yard
cot40°= 1-19175
cot 67°= -42447
Logarithms
8-13672
- -20852
2-92820
Examples V b.
1; The string of a kite is known to be 500 feet long, and it is
observed to make an angle of 55° with the horizontal ; find the
height of the kite.
92 THE RIGHT-ANGLED TRIANGLE
2. From the top of a cliff, 215 feet high, the angle of depression
of a ship is observed to be 23° 20' ; what is the distance of the ship
from the foot of the cliff?
3. From a point 56 feet from the foot of a tree the angle of
elevation of the top is 73'' ; find the height of the tree.
4. The top of a conical tent is 9 feet above the ground ; the
radius of the base is 5 feet ; what is the inclination of the side of
the tent to the horizontal ?
5. The shadow thrown by a flagstaff is found to be 55^ feet long
when the sun's altitude is 53° 15'; what is the height of the
flagstaff?
6. I know that a certain tower is 144 feet high. I find that its
elevation observed from a certain point on the same level as the
base of the tower is 37° 16'. Find the distance of that point from
the base of the tower.
7. A sphere of radius 4 inches is suspended from a point A in
a vertical wall so that it rests against the wall. The string is
11 inches long and is in the same straight line as a radius of the
sphere. Find the inclination of the string to the vertical.
8. From the top of a cliff, 254 feet high, the angle of depression
of a ship was found to be 9° 28', and that of the edge of the sea
72° 40'; how far distant was the ship from the edge of the sea?
9. Two observers on the same side of a balloon and in the same
vertical plane with it, a mile apart, find its angles of elevation to
be 15° and 65° 30' at the same moment. Find the height of the
balloon.
10. From the top of a tower, 108 feet high, the angles of de-
pression of the top and bottom of a vertical column are found to
be 30° and 60° respectively. What is the height of the column ?
11. A flagstafF, 80 feet high, is fixed in the centre of a circular
tower 40 feet in diameter. From a point on the same horizontal
plane as the foot of the tower the elevations of the top of the
flagstaff and the top of the tower are observed to be 35° and 30"^
respectively. Find the height of the tower,
12. A river, the breadth of which is 2C0 feet, Hows at the foot of
a tower, which subtends an angle 25° 10' at a point on the further
bank exactly opposite. Find the height of the tower.
THE RIGHT-ANGLED TRIANGLE 98
13. A person standing at the edge of a river finds that the eleva-
tion of the top of a to'»ver on the edge of the opposite bank is 60'' ;
on going back 80 feet he finds the elevation to be 45'^; find the
breadth of the river,
14. From the top of a tower. 50 feet high, the angle of depression
of a man, walking towards the tower, is noticed to be 30' ; a few
moments after it was 45'. How far had the man walked between
the two observations ?
15. Two p6.=!ts, 400 yards apart, at the sides of a straight road
running E. and W., are observed to bear N. 20° E. and E. 20° N.
respectively. Find the distance of the observer from the road.
16. Two points A and B and the foot D of a tower CD are in
a horizontal straight line, and the angles of elevation of C, the top
of the tower, as seen from A and B respectively, are 25" 46' and
35=25'. If the distance AB is 200 feet, find the height of the
tower.
17. A vertical post casts a shadow 15 feet long when the altitude
of the sun is 50= ; calculate the length of the shadow when the
altitude of the sun is 32°.
!>:. A vertical mast, having its base at A, is set up on a horizontal
plane. B and C are points in the plane in a line with A, and such
that the angular elevations of the top of the mast, when observed
at these points, are respectively y and (3. If tan ^ = f , tan ,3 = f
and the length of BC is 105 feet, find the height of the mast.
19. A man standing on a tower at a height of 80 feet from the
ground observes that the angles of depression of two objects on
a straight level road running close to the foot of the tower are 60°
and 30°. If the objects are on the same side of the tower, how far
are they apart ?
20. A, B, C are three points in succession on a straight level
road, and P is another point so situated that the angles PAB,
PBA, PCA are respectively 90°, 60°, and 45°. If a man walks at
a uniform rate from ^ to JB in 25 seconds find, to the nearest
second, how long it will take him, at the same rate, to walk from
Bio C.
21. A ray of light passes through a hole A in a graduated
horizontal scale AB in a direction perpendicular to the scale and
is reflected by a vertical mirror which is distant 30 inches from the
scale and makes an angle x'' with the incident (i.e. approaching)
94 THE RIGHT-ANGLED TRIANGLE
ray. After reflection the ray makes the same angle with the
mirror as before and shines on the scale at a distance 8 inches from
A. Find the value of x.
If the mirror now swings through an angle 1°, how far will the
spot of light on the scale move ?
Elementary Navigation.
{The student sliould revise §§9 and 10 dealing with latitude and
longitude and the points of the compass. [
46. When a ship is sailing, the angle between its direction of
sailing and the meridian the ship is crossing is called the course.
If the course is constant, the ship
is said to sail on a rhumb-line.
The distance between two positions
of the ship is then measured along
the rhumb-line. The difference of
latitude of two places is the arc of
a meridian intercepted between the
parallels of latitude passing through
the two places. The departure
' between two meridians is the dis-
Fig. XXVIII. tance between the two meridians
measured along a parallel of lati-
tude ; thus the departure between any two given meridians is not
a constant but diminishes from the equator to the poles.*
47. A small portion of the earth's surface may be regarded as
a plane ; for distances small, compared with the earth's radius, we
may therefore use the formulae of Plane Trigonometry.
Plane Sailing is the name given to that part of navigation
which treats the surface of the eaith as a plane. On this assump-
tion the meridians become parallel straight lines, the rhumb-line
becomes the hypotenuse of a right-angled triangle of which the
departure is the side opposite to the course, and the difference of
latitude is the side adjacent. Thus problems on Plane Sailing are
merely examples in the solution of right-angled triangles.
* In navigation distances are usually measured in nautical miles ;
a nautical mile is the length of an arc of a meridian (or the equator)
which subtends an angle of 1' at the centre of the eartli ; thus a
distance of 75 nautical miles is usually written 75'.
THE RIGHT-AXCxLED TRIANGLE
95
Examples V c.
(The distances are given in nautical miles.)
1. A ship sails SE. by S., a distance 81 miles; what is her de-
parture and difference of latitude ?
2. A ship sails N. 49"^ 41' W., a distance 73 miles ; what is the
departure and difference of latitude ?
3. A ship sails SSW. until its departure is 198 miles , what is the
distance sailed and the difference of latitude?
4. If the course is 3^ points W, of N., and the difference of
latitude 149 miles, what is the distance ?
5. A ship sails between North and West, maldng a difference of
latitude 157^ miles and departure 79 miles ; what is the course ?
6. A ship sails westward 247 miles along the equator from
meridian 16" E. ; what is now the longitude ?
7. A ship sails 247 miles eastward along the parallel 40" N. :
what is the change in longitude ?
8. When a ship sails any distance (great or small) along a
parallel of latitude, show that
difference of longitude in minutes = departure x secant of latitude.
9. A ship, from latitude 54' 22' 10" N., sails 195^ miles I of a
point S. of SE. ; what is now the latitude ?
10. Leaving latitude 49' 37' N., longitude 15' 22' W., a ship sails
SW. by W. 150 miles ; find the new latitude and longitude.
Fig. XXIX.
48. Parallel Sailing. If X' is the latitude, then the radius
of the parallel of latitude (AT in Fig. XXIX) is cos X x radius of the
96 THE RIGHT-ANGLED TRIANGLE
earth. If B is the radian measure of the difference of longitude of
two places on the same parallel, the length of the arc between
them is 6 cos X x radius of the earth. The radius of the earth is
21600 ,. , .,
-^^ — nautical miles.
a TV
iience departure = - ?, — 6 cos X.
When d is reduced to minutes, this relation iDecomes
departure = difference of longitude x cosine of la,titude.
49. Middle Latitude Sailing. In Middle Latitude Sailing,
the departure between two places, whose latitudes are X and X', is
taken to be the departure between their meridians, measured at
the latitude |(X + X'). On this assumption,
departure = diff. of longitude x cos |(A + X').
50. Traverse Sailing, If a ship sails on different courses,
from A to B, from B to C, from C to Z>, &c., then, by the methods
of Plane Sailing, the total changes in latitude and longitude can
be worked out. This is called the method of Traverse Sailing.
This method can only be used when the whole area traversed can be
regarded as plane without introducing a great amount of error.
Example. A ship left a j^osition in which Oporto Light
(lat. 41° 9' N., long. 8° 38' T^.) lore W. hy N., 15 miles
distant. Afterwards she sailed as under :
Courses.
Distances
N.W.
70'
S. by W. ^W.
55'
E.
35'
N.N.W.
42'
S.E.
51'.
Find her bearing and distance from the Light in her last
position.
We liave a sories of right-angled triangles to solvo, the hypotenuso
and an acute angle being given in each case. In practical navigation
special tables are userl, calU'd Traverse Tables.
THE RIGHT-ANGLED TRIANGLE
97
The angle the hypotenuse makes with the meridian is taken iu
each case.
Oto A.
Hypotenuse 15', angle 7 points = 18^ 45'. ri761
Diff. of latitude = 15' x cos 78= 45' 1-2902
-1 2-926' S. ^63
Departure = 15' x sin 78M5' 1-1761
= 14-71' E. r-9916
ri677
C
Fig. XXX.
Note that 0 bears W. by N. from A, but A bears E. by S. from 0.
A to B.
Hypotenuse 70', angle 45=.
Diff. of latitude = 70 x cos 45=
= 49-497' N.
Departure = 49-497' E.
Bio a
Hypotenuse 55', angle 1| points = 16= 52|'.
Diff. of latitude = 55 cos 16= 52 V
= 52-64' S.
Departure = 55 sin 16° 52i'
= 15-97' W.
The other triangles are worked in the same way.
•7071
1-7404
1-9809
1-7213
r7404
r4628
1-2032
98 THE RIGHT-ANGLED TRIANGLE
Tabulate the results thus :
Distance.
Diff. of Latitude.
Depai
ture.
N.
S.
E.
W.
E. byS
15'
2-93
14-71
NW.
70'
49-50
49-50
S.byW.iW.
55'
52-64
15-97
E.
35'
35
NNW.
42'
S8-80
16-07
SE.
51'
36-06
36-06
88-30
91-63
88-30
99-21
68-10
68-10
3-33
31-11
We see now that the final difference of latitude from the light is
3-33' S., and departure 31*11 E. ; so that we have to solve a right-
angled triangle given the two sides.
8*33 -5224
31*11' 1-4929
F0N=6°r. 1^95
^^_ 31-11 1*4929
In Fig. XXX tan FON =
cos7^(
:>N 1*9975
= 31*29
1*4954
In her final position the ship bore G'^?' S. of E., 3r3 miles
distant from the Light.
To find thf lo7igitude of the ship.
Latitude of 0
41° 9'N.
Diff. of latitude for F
3*3' S.
Latitude of F
4V 5-7' N.
Middle latitude
41" 7'.
Difference of longitude in minutes =
departure
cosine of middle latitude
31*11 1*4929
~
cos 41° 7' -1-8770
=
41*30'. 1-6159
Longitude of i''= 8°38'-41*30'
= 7" 57' W.
THE RIGHT-ANGLED TRIANGLE 99
Examples V d.
1. Find the distance on the parallel between Cape Agulhas
(lat. 34°50'S., long. 20M' E.) and Monte Video (lat. 34° 50' S.,
long. 56° 9' W.).
2. A ship steamed at the rate of 12 knots from Albany
(lat. 35° 3' S., long. 118° 2' E.) to Cape Catastrophe (lat. 35° 3' S.,
long. 135° 58' E.}. How long did she take on the voyage ?
3. A ship sailed from Port Elizabeth (lat. 34° T S., long. 25°40' E.)
SE. i^ S., until her departure was 397' ; find her final position.
4. Find the course and distance from Syracuse (lat. 37° 3' N.,
long. 15° 15' E.) to Fano (lat. 39° 52' N., long. 19° 19' E.).
5. A ship left a position from which Cape Clear (lat. 51° 26' N.,
long. 9° 29' W.) bore NE. by E. 12*5 miles distant and sailed
South 150' and then West 290 miles. Find the bearing and
distance of Cape Clear from the ship in her last position.
6. Find, by Middle Latitude Sailing, the departure between
two places whose positions are 13° S., 50° E. and 20° S., 60° E.
7. A ship sails from 50° N., 50° W. to latitude 48° N., the
distance being 157' ; find the new longitude.
8. Cape Ortegal (lat. 43° 45' N., long. 7° 6' W.) bore SW. JW.
12 miles distant. Afterwards sailed as under :
Prue Courses.
Distances.
NNW.iW.
70'
ESE.
85'
NNE.fE.
lor
S.
50'
wsw.
92'
Find the final latitude and longitude.
9. A ship left the Texel (latitude 52° 58' N.) and then sailed W.
by N. 34', S. by E. 45', W. by S. 35', SSE. 44', WSW. iW. 42'.
Find the course and distance to Dungeness which lies 139' West of
the Texel in latitude 50° 55' N. '
10. A ship, latitude 17° 10' N., is making for a harbour, latitude
13°10'N., and 180' W. of the ship. She sails SW. by W. 27',
WSW.|W. 30', W. by S. 25', W. by N. 18', SSE. 32', SSE.f E., 27',
S. by E. 25', S. 31', SSE. 39'. Find the course and distance to the
harbour.
g2
100
THE RIGHT-ANGLED TRIANGLE
11. A ship left a position in which Heligoland bore ENE. 12',
and then sailed NW. 24', S. by W. 20', NW. by W. 32', S. by E. 36',
WNW.iW. 42', SSE. AE. 16', W.fN.45'. What is then the
position of the ship ? Heligoland lies 54° 12' N., 7° 54' E.
• 12. A ship sailed from Barcelona (41°25'N., 2"10'E.) SE. by
E.|E. until she reached latitude 36°2rN. What was then her
longitude ?
13. A ship left a position in which Sable Island (43°24'N.,
65° 36' W.) bore NW. | W., distant 12 miles.
Afterwards sailed as under :
Courses. Distances.
ESE. 72'
SW.iW. 37'
NNE. 42'
E. 25'
Required the latitude and longitude reached.
51. The Double Angle.
In Fig. XXXI, the Sing\eBAC = A° ; onAB [i semicircle
is described with centre 0, so that angle BOC =2A.
Let fall CN perpendicular to AB.
N B
COS 2 ^ =
ON
OC
_AN-AO
~ OC ~
_AN__
"" OC
m
:hej
— 1 hypotenuse of the triangle of '
2 OC which AN is a side. J
THE RIGHT-ANGLED TRTANC^I^
n A ^J Fill in the vacant places with the
_ AN AG__
~ ag'ab
= 2cosM- 1.
Exercises. In a similar way prove
i. sin 2 ^ = 2 sin A cos A.
ii. 003 2.4= l-2sin2^.
Deduce
iii. cos 2 ^ = cos^ A - sin^ A.
. ^ ^ , 2tan^
IV. tan 2A =
l-tan^^
V. sin ^ = 2 sin | ^ cos ^ ^ ; cos A = cos^ \A- sin^ \ A.
vi. 2 cos*^ i ^ = 1 + cos A,
vii. 2sin^^yl = 1-cos^.
sin J. 1— cos^
viii. ■ ■
... , , , /I— COS^ SI
111. tcin IA= . I, = ^ —
'NJl + cos^ 1 +
cos>4 sin J.
ix. Prove the formulae for sin 2^ and cos 2^ when 2yi is
obtuse.
X. Do these proofs apply to angles of any size? If not,
between what limits do they apply ? Why is the ambiguous sign
omitted in viii ?
52. Geometrical questions may often be solved by using
Trigonometry. For example :
If from a point outside a circle a secant and a tangent be drawn, the
rectangle contained by the ivhole secant and the part outside the circle
is equal to the square on the tangent.
102 I'HE EIGHT-ANGLED TRIANGLE
In Fig. XXXII it is required to prove that rect. PA . PB = eq.
on PT.
r>
Fig. XXXII.
Let radius = r, OP = c, and angle OPB = ^, angle OAC = (/).
PA = PC -AC
= ccos6 — rcos(j).
PB = PC+CB
= PC+AC (Prop. 21)
= ccos^ + rcos0
PA.PB = c' cos- e-7^ COS^ 0
= c^ — i^ — c^ sin'^ 6 + 7-^ sin^ (p.
But c sin ^ = OC from triangle OPC
= rsincp from triangle OAC.
Hence PA . PB = c" - r""
= OP^-OT^
= PT2 since OTP is a right angle.
53. Known results in Geometry are useful for proving
Trigonometrical relations.
Show that, in any triangle,
tan^jA-B) _a-h
tan^{A+B) ~ r7T7>*
With centre C and radius CA (i.e. ?>), describe a circle cutting
CB in E and CB produced in D.
THE RIGHT-ANGLED TRIANGLE
103
Then BE = a-h, and BD= a + h.
Join AD and AE.
Through E draw i?i^ parallel to DA and meeting AB at F.
Then the angle DCA at the centre = 180 - C = ^ + 5.
So that the angle DEA at the circumference = \{A^-B).
Fig. XXXIII.
Also the angle BAE = BAC-EAC= A-\{A+B) ==\{A-B).
Also the angle EAD, being in a semicircle, is a right angle.
tani(^ + £) = ^,
tan ^ (^ -P) = — - , since AEF= EAD = a right angle.
AE
Hence
tan^(^-^)^^i?^
tan|(J. + J5) ^Z)
= ^— since -&i^is parallel to AD. (Prop. 12 «.
a-b
~ a + b'
Corollary. ^ + B+C=180^ .-. ^(A + B) = 90-^C.
Hence the above result may be written
tan4(A-B) = ?— ^cot-|C.
^^ ^ a + b ^
This formula will be used in a later chapter.
104 THE KIGHT-ANGLED TRIANGLE
Examples V.
In the following examples :
A, B, C are the angles of a triangle ABC.
a, b, c are the sides, s = half the sum of the sides ; R is the
radius of the circumcircle.
r is the radius of the inscribed circle.
f\ is the radius of the escribed circle touching the side BC.
A is the area of the triangle.
D, E, F are the middle points of the sides BC, CA, AB,
respectively.
X, Y, Zare the feet of the perpendiculars let fall from A, B, C
respectively on the opposite sides.
0 is the centre of the circumcircle.
/ is the centre of the inscribed circle.
K is the orthocentre.
1. Express in terms of the sides and angles the lengths of AX,
BX, CX, AK, BK, CK.
2. Express the length of ^i) in terms of (i) a,b,C, (ii) a,b,B,
(iii) a, b, c.
3. Show that a/sin A — b/sinB = c/sin C = 2R. Deduce that
R = abc/iA.
4. Prove that r (cot IB + cot^C) = a. Write down the two
similar formulae.
5. Prove that r = A/s. (No trigonometiy required.)
Deduce that tan|^ = A-r {s(s-f/)}.
6. Show that BX = a — ?>cos c ; hence prove that
c^ = a"^ + b^ — 2 ab cos c.
7. Prove that (i) A = ^ab sin C, (ii) A = rs, (iii) A = abc-i-iR,
(iv) A = -v/s(s-«)(s-&)(s-c).
8. Prove that
(i) sin^A= ^/{s~b) (s-c) -^bc, (ii) cos^^= \/s{s-a} -r-bc,
(iii) tan ^A = y/{s -b){s — c)-TS{s — a).
THE RIGHT-ANGLED TRIANGLE 105
9. Show that the triangles ABC and A ZZare equiangular ; hence
prove that YZ = a cos A.
10. Two tangents are drawn from a point P to a circle of
radius 10 cm. ; the tangents contain an angle of 43°. Find the
lengths of the tangents and the distance of P from the centre.
11. A sheet of iron is shaped so that it can be rolled up to form
a conical funnel 6 feet high with open circular ends 2 feet and
6 feet diameter respectively. Draw a plan of the sheet before
rolling. What is the inclination of the edge of the funnel to the
line joining the centres of the ends ?
12. A circle rolls without slipping along a straight line : prove
that the co-ordinates of a point fixed to the circumference are such
that x = a (^ — sin 6), i/ = a (1 -cos 6) ; the origin being taken at
the point where the fixed point meets the straight line, and 6 being
the angle turned through by the circle.
13. One of the angles of a right-angled triangle is the acute
angle whose sine is §, and the length of the shortest side of the
triangle is 10 feet. Find the lengths of the other two sides.
14. ^ is the highest point of a sphere with centre 0 ; a particle
slides from a position P, where the angle AOP = 6, to the position
Q where the angle AOQ is 0. How much lower is Q than P and
how much further from OA ?
15. The time t of sliding from rest down a length s inclined at
6 to the horizon is given by s = ^^^^sin^ where g is a constant.
A circle is held with a diameter AB vertical ; prove that the time
of sliding along a chord from the highest point A to the circum-
ference is the same whatever be the inclination of the chord, and
that the time of sliding from the circumference along a chord to
B is also independent of the inclination of the path.
16. A plane, inclined at 20" to the horizon, is placed with the line
of greatest slope pointing north. A line is drawn on the plane,
pointing NNE. ; find the inclination of this line to the horizontal.
17. A man 6 feet high walks along a straight line which passes
3 feet from a lamp-post. If the light is 9 feet from the ground,
find the length of the man's shadow when his distance from the
point on his path nearest to the lamp is 10 feet. What is the locus
traced out by the extremity of his shadow as he walks along the
line ?
106 THE RIGHT-ANGLED TRIANGLE
18. If, in the previous question, there is a vertical wall parallel
to the man's path and distant 2 feet from it on the side remote
from the lamp, what is then the length of the shadow and the
locus traced by its extremity ?
19. Draw the graph of 6 /sin 6 from ^ = 0 to ^ = i tt.
Use the graph to solve the following problem.
A string 30 inches long is tied to the ends of a cane 35 inches
long, thus forcing the cane into a circular arc. Find the radius of
the arc correct to the nearest inch.
20. Find the length of a strap which passes tightly round two
pulleys of radii 2 feet and 3 feet, their centres being 6 feet apart.
CHAPTER VI
THE TKIANGLE
Several formulae connecting the sides and angles of a triangle
have been proved in the examples of the preceding chapters.
They are here gathered together for reference and proofs are
given. Care should be taken that the proof applies when
the triangle is obtuse-angled ; if it does not, a separate proof
must be given.
Relations between the sides and angles.
54. The angle formula. A + B + C = 2 right angles.
a ^ ^ ^
sin A ~" sin B "~ sin C
A
The sine formula.
(=211).
Fig. XXXIV.
Let 0 be the centre of the circumcircle, and D the middle point
of^a
Join OB, OC, OD.
Then, in the left-hand circle of Fig. XXXIV,
angle 300 = 2'' angle BAC
= 2A.
Triangles BOD and COD are congruent ; (Prop. 8 a.)
.'. BOD = COD = A.
Also BD==^BC = ^ a.
108
THE TRIANGLE
In the right-angled triangle BOD,
BD = OB sin BOD,
i.e. \a = R^mA\
Sin A
In a similar way it may be proved that
-X^ = 2E and ^ = 2R.
sin B sm C
sin J. sin^ sm C
Hence
Exercise. Supply the proof when the angle A is obtuse.
Note. In using this formula the following algebraic result is often
useful :
If - =z - ^ - then each fi*action equals ; :..
55. The cosine formula cos A = — ^r^. , and its
2 be
equivalent a^ = b^ + c^— 2 be cos A.
This can be proved very shortly by assuming Euclid II, 13 and 14 ;
but it is better to base the proof on the theorem of Pythagoras.
Let CZ be the perpendicular from C on AB, Fig. XXXV.
ThenZC= &sin^, AZ=b cos A, and BZ=c-b cos A.
BC^ = BZ^ + ZC\
a' = {c-h cos Af + (b sin Af
= c'^-2bccosA + b^cos^A + b'^sm'^A,
i.e. a'' = bUc'' -2 be cos A,
h-i + c^-a'
or
cos^ =
2 he
THE TKIANGLE 109
If ^ is obtuse, then in Fig. XXXV a,
ZC = & sin (180-^) = &sin^,
^Z=Z>cos(180-^) = -&cos^,
BZ= BA + AZ= C + {-bcosA} = c-hcosA.
The proof is now the same as before.
Exercise. Write down the corresponding formulae for cos B
and cos C.
56. The Projection formulae
c = b cos A + a cos B.
In Fig. XXXV, BZ is the projection of BC on BA ; and AZ is
the projection of ^C.
AB = AZ+BZ,
i.e. c = hcosA + a cosB.
Exercises. Supply the proof when A is obtuse.
Write down the other two corresponding formulae.
57. Area formulae
The symbol A is used to denote area of triangle.
(i) A = ^ any side x perpendicular from opposite angle.
(Prop. 16.)
(ii) A = ^ AB X ZC = ^ c . b sin A = ^ be sin A.
(iii) A = ^/3 (s - a) (s - b ) (s - c).
In Fis. XXXV. BZ ^ acosB =a X '-±!! "- .
2ca
.'. (2c.Z(7)2
= {a-\-h + c){a — b + c){a + h — c) -h + c — o).
Let 2s= a + b + c, then h-^c — a = 2(s — a) &c. ; so that
2c . ZC= ^/2s.2(s-a). 2 (s - &)T2 (s-c) ;
A =iAB.ZC
= >/s(s-a)(s-b)(s-c).
Exercise. Show that
16 A2 = 2 (&2c2 + c-a^ + a' h") - (a* + h' + c').
no
THE TRIANGLE
58. From these formulae others may be deduced.
Example i. To show that in any triangle
cos (A + B) = cos AcosB — sin A sin B.
From sine formula «sin ^ — &siii A = 0. (i)
From projection formula a cos B + h cos A = c. (ii)
Square and add, a"^ + &" + 2 cch (cos A cos -B — sin A sin B) = c^
From cosine formula a'^ + h'^ — 2 ah cos C = c^
It follows that
cos C = — (cos A cos ^ — sin ^ sin B).
From the angle formula C = 180 -(^ + -5),
i.e. cos C= — cos (^ + jB).
Hence cos (A + B) = cos AcosB-~ sin A sin B.
Example ii. In any triangle
sin {A — B) = silt A cos B — cos A sin B.
Multiply together equations (i) and (ii) above.
a^ sin ^ cos 5 - h"^ sin ^ cos yl - ah (sin ^ cos 5 - cos A sin B) = 0.
From Fig. XXXVI it is seen that
«2 sin B cos B = BZ.ZC= 2'' triangle BZC,
and h"^ sin ^ cos ^ = 2*^^ triangle AZC
= 2'Uriangle .4'ZC,
{ZA'= ZA, so that triangles CZA, CZA' are congruent).
.-. ci^ sm B cos B -W sm A cos A = 2 ''triangle 5C/1'
^BC.CA' sin BCA'
= ahsui{A — B).
Comparing this with the result above, we see that
sin {A-B) = sin AcosB- cos A sin B.
This result can, however, be obtained more quickly.
THE TRIANGLE
111
For
I.e.
sin BC A'
~ZrB~
sinjA-B)
a cos B — b cos ^1
sin CBA'
sin^
„ • / < T,N « sin 5 cos 5 < • D
Hence sin (^ — 5) = cos^smJ?
= sin A cos B — cos A sin B
since a sin B = h sin A
Example iii. To show that the area of a quadrilateial inscribed
in a circle is \/{s — a) {s -b){s — c){s — d) where s — lUi + b + c + d).
In Fig. XXXVII
Area of ABCD = sum of triangles ABD and BCD
= I ad sin A + ^lc sin {180 -A)
= I (ad + bc) sin A.
From triangle ABD,
BD"' = a"^ -\- d"^ -2 ad cos A.
Fig. XXXVII.
From triangle BCD,
BD'' = b'' + c'' -2 be cos {180 -A).
Hence a^ + d'^-2 ad cos A = h"^ + c"- + 2 be cos A,
i.e. 2{ad + bc)cosA = a'' + (P-{b''-hc'');
2{ad + bc){l + cosA) = {a + dy-{b-c)'',
and 2(((d + bc){l-cosA) = {b + c)''-{a-d)\
112 THE TRIANGLE
Hence i {ad + bc)^ {1 - cos"^ A)
= {-a + b + c + d){a-h + c + cl)(a + h-c + d){a + b + c-d),
i.e. {i(af? + &c)siii^}2
= I {- a + b + c + d) I {a-b + c + d) l{a + b - c + d) ^{a + b + c- d)
.'. Area of ABCD = ^/{s-a){s-b){s-c){s-d}.
Examples Via.
1. From the three projection formulae deduce the three cosine
formulae.
2. Prove that sin^ = sinPcos C+ cos^sin 0; and deduce that
sin (B + C) = sin BcosC + cos B sin C.
3. Prove that cos(J.-5) = cosvl cos5 + sin^sin5.
4. Show that A = |- (&^ sin CcosC + c^ sin B cos B).
5. Show that A = i c^ {sin AsinB-^ sin (^ + ^)} .
6. Prove that sin A + sinB> sin C.
7. Prove that cot yl + cot5 = ccosec-B-^a.
What third expression are these equal to ?
8. Show that
R (i.e. the radius of the circumcircle) = s -^ (sin ^ + sin ^ + sin C).
9. Use the formula cos^ = l-2sin'^|^ to prove that
sin 1^ = \/(s -b){s-c)-r- be.
Write down the similar formulae for sin J 5 and sin | C.
10. In a similar way to that suggested in the previous example,
prove that cos |^ = ^/s {s-a) -f be. Write down the formulae for
cos 1 5 and cos I C. What is the formula for tan^yl ?
11. Given cf = 17, 5 = 12, 5^= 37° 15', find A.
12. Given a = 14, b = 13, c = 12, find the greatest angle.
13. Given a = 45, A = 45°, B = 60°, find b.
14. Given b=ll, c = 42, A = 72°, find a.
15. Given a = 176, b = 291, c = 352, find all the angles.
(Choose a formula adapted for logarithms.)
16. Given « = 7, fc = 5, C = 49°, find e.
17. Given b = 9,c=10,C= 57°, find a.
18. By considering two forms for the area of an isosceles triangle,
prove that sin ^ = 2 sin J A cos | A.
t
THE TRIANGLE 113
19. Two sides of a triangle are 3 and 12 and the contained angle
is 30^ ; find the hypotenuse of an isosceles right-angled triangle of
equal area.
20. Two adjacent sides of a parallelogram, 5 inches and 8 inches
long respectively, include an angle of 60°. Find the length of the
two diagonals and the area of the figure.
21. If in a triangle C = 60^, prove that
l/(a + c) + l/{b + c) = 3/(rt + b + c).
22. On a straight line AB, 4 inches long, describe a semicircle,
and on the arc of the semicircle find points P, Q, i?, S such that
the areas of the triangles APB, AQB, ARB, ASB are 1 square inch,
2 square inches, 3 square inches, and 4 square inches respectively.
If C is the centre of the circle, determine the sines of the angles
ACP, ACQ, ACR, and ACS, and hence find, from the tables, the
values of these angles.
23. If a quadi-ilateral can be inscribed in one circle and circum-
scribed about another, show that its area is ^/ctbcd, where a, h, c, d
are the lengths of the sides.
The circles of the triangle.
59. It is shown in any Geometry textbook that
(i) the centre of the circumcircle is the point of concurrence of
the perpendicular drawn at the middle points of the sides ;
(ii) the centre of the inscribed circle is the point of concurrence
of the three lines bisecting the three angles ;
(iii) the centre of an escribed circle is the point of concurrence
of the bisector of the opposite interior angle with the bisectors of
the two adjacent exterior angles.
In Fig. XXXVIII, we have
AQ = AR, (Prop. 24.)
BP = BR,
CP=CQ;
.'. AQ-\-BP+CP=^. sum of sides = s.
Hence AQ = s-a.
Exercise. In a similar way, prove that
BP = CF =
CQ = QQ' =
AQ' = PF =
1216 H
114
THE TRIANGLE
Fig. XXXVIII.
Examples VI b.
Prove the following formulae :
\. K = a^2&mA. 2. 7? = rt/;c-r4 A.
3. r = A/s. (Considerthesum of the triangles P/C, (7/.4,yi/B.)
4. r = rt^(cot^i? + cot|C). 5. ;-j = A/(.s-cf).
6. ri = rt^(tanJi?-t tan ^ C).
THE TEIANGLE 115
Using the above formulae, prove the following relations ;
7. In a right-angled triangle R + r = ^{a + h).
8. l//'j + 1/^2 +1/^3 = 1/r. 9. l/r^+l/r^ = 2 -=-6 sine.
10. r7\ror2,= ^". 11- r r-^ = {s-h){s-c).
12. (aZ^c-fsin J.sin^sinC)^. 13. 27?/- = «&c-f(a + & + c).
14. 4i? sin yl sin B sin C = a cos ^ + 6 cos B + c cos C
15. tanf^ = V'(s-6j(s- c)-rs(s- a).
16. s2= AcotMcotii^cotiC.
17. If ABC is a triangle such that 2h = a-\-c, and ^ is the
length of the perpendicular from B upon AC, show that tan | J.
and tan \ C are equal to the roots of the equation
18. Show that the sum of the radii of the escribed circles of
a triangle is equal to the radius of the inscribed circle together
with four times the radius of the circumscribing circle.
19. Show that the area of the triangle formed by joining the
centres of the escribed circles is
Si^^cos-l^cosi^cos^O.
20. The sides of a triangle are 3, 5, 6 ; find the radii of the inscribed
and circumscribed circles.
21. In an isosceles triangle the base is 100 cm. and the perpen-
dicular from the vertex is 70 cm. ; find the radii of the inscribed
and circumscribed circles.
22. A triangle is described with base BC = 5 inches and angle
A = 70°. What is the radius of the circumcircle ? Find the dis-
tance of the centre of the circumcircle from BC.
23. Find the radius of the circumcircle of the triangle ABC being
given that BC = 7, CA = 6, and C = 60°.
24. If a = 32, & = 16, C = 42°, find R and r.
25. The area of a parallelogram having base 5*8 cm. and angle
123° is 37*7 sq. cm. Find the other sides and angles. Find the radii
of the circles which pass through three of the corners of this
parallelogram.
26. Two of the sides of a triangle are 7'5 cm. and 9'3 cm., the
included angle is 37'. Find the radius of the circle which touches
these sides produced and the third side.
h2
116 THE TRIANGLE
Oral Revision Examples.
Complete the following identities and equations :
1. sin (270-^)- 2. cos2(9 =
3. 2 tan ^ cot ^ = 4. If sin 6 = 1,0 =
5. In any triangle &^ = 6. In any triangle R =
1. sm2A= 8. tan 225° =
9. If cos ^ = #, tan 6 = 10. A in terms of the sides =
11. tan-il = 12. secM-1 =
13. length of arc = radius x 14. sin'^5 + sin^(90 — J5) =
15. Definition of tangent. 16. In any triangle cosC'
17. In any triangle h cos C+c cos B =
18. In any triangle &csin^ = 19. In any triangle r =
20. tan I TT =
21. Definition of sine. 22. cos (360° -jB) =
23. In any triangle r^ = 24. In any triangle cos^
25. What formula connects a, b, and B?
26. tan-'(--s/3) = 27. If cos/?= -^6 =
28. tan2 73l° + l= 29. abc =
30. 37° = ? radians.
31. cos2(^-45°) + sin2(^-45°) =
32. &sinC =
33. Express R in terms of the sides.
34. If sin (9 = sin 0^, then 6= 35. cos^= (in terms of sin |^),
36. Area of triangle = 37. aH c^ - 2 «c cos B =
38. cos 1200° =
39. Maximum value of 2 sin ot cos a =
40. cos^^-sin*^ = (in its simplest form).
41, A -r (.<;-«)= 42. tan (180-5) =
43. acos C + CC0S.4 = 4t4:. he sin A =
45. sui^ {A + B) + cos"^ {A + B) = 46. If cos^ = cosyl, then x =
47. In any triangle cos A = 48. tan 60° =
49. cos 2 ^ = 50. How many radians = >4° ?
THE TKI ANGLE 117
Examples VI.
1. Prove that (a cos A - h cos B) -r (a^ - h') + cos C/c — 0.
2. Prove that c- = {a + hf sin- 1 C + {a- h)- cos- }, C.
3. In a triangle ^i?Cthe lines drawn from A and C, perpen-
dicular to the opposite sides, intersect in 0. If the angle A is
acute, show that OA == b cos A/sin B.
Also draw a diagram in which A is an obtuse angle, and establish
the corresponding expression for OA in that case.
4. Show that in any triangle the product of a side and the sines
of the two adjacent angles is the same, whichever side be taken.
5. Find the area of a regular polygon of n sides circumscribed
about a circle of radius r.
6. Regular polygons of 1 5 sides are inscribed in and circumscribed
about a circle whose radius is one foot ; show that the difference
of their areas is nearly 20 square inches*
7. ABCD are four points on a circle such that the angles BAC
and BCA each equal 6. Show that AD +CD = 2BD cos 6.
8. If 2 cos B = sin A/sin C, prove that the triangle is isosceles.
9. If tan.-l/tan^ = sin-.-1/sin-i?, show that the triangle is
isosceles or right-angled.
10. Express the sides of a triangle in terms of the angles and the
semi-perimeter.
11. In a triangle ABC perpendiculars AD and BE are let fall on
the opposite sides ; prove that the radius of the circle circum-
scribing the triangle CDE equals R cos C.
12. If in a triangle the median bisecting the base AB is perpen-
dicular to the side AC, prove that 2 tan A -f tan C = 0.
13. If ^; and q are the lengths of the perpendiculars from A, B on
any arbitrary line drawn through the vertex 0 of a triangle, prove
that a'p"" -f &' 5' - 2«Z> pq cos C = w^ ^2 ^^^2 q
14. An isosceles triangle, vertical angle 35°, is inscribed in a circle
whose radius is r65 inches. Find the lengths of the sides.
15. Show that in any triangle
cos A cos B cos C ^ a^ ■\-}r A- (?
a b c 2abc
118 THE TRIANGLE
16. If R is the radius of the circumcircle of any triangle and
X, y, z are the lengths of the perpendiculars let fall from its centre
on the sides, prove that
R^-{x^ + y'^ + z')R-2xyz = 0.
17. The rectangular co-ordinates of the angular points of a triangle
are (4, 5), (6, 7), (8, 6) ; determine the sum of the two smaller
angles.
18. A rod AB, length 2 a, can turn about a hinge fixed to the
wall at ^ ; it is supported by a string BC, length /, fastened to
a point C on the wall at a height h above A.
(i) If BC is horizontal, what is the inclination of the rod to
the vertical ?
(ii) If BC is horizontal, what is the inclination to the vertical
of the line joining the hinge to the middle point of the string ?
(iii) If the string and rod are inclined at 6 and 0 to the vertical
respectively, prove that (i) 2 a sin 0 = Z sin 6, (ii) J cos 6 — 2 a cos (^ = /i.
(iv) In the general case, what is the angle between the string
and the rod ? Give the answer in terms of li, a, 6 or li, I, 0.
(v) In the general case, what is the inclination to the vertical
of the line joining the hinge to the middle point of the string?
Give the answer in terms of h, a, (p.
19. Three equal spheres of radius 7 centimetres are fixed in
a horizontal plane so as to touch each other ; a sphere of radius
6 cm. rests upon these three. Find the height of the centre of the
fourth sphere above the horizontal plane, and the inclination to
the vertical of the line joining the fourth centre to one of the
lower centres.
20. Three equal rods of length 54 inches are fixed so as to fonii
a tripod. If their feet are at the corners of an equilateral triangle,
side 18 inches, find the inclination of each rod to the vertical.
21. In any triangle prove that the centroid trisects the line
joining the circumcentre to the orthocentre.
22. Find the lengths of the sides of the pedal triangle of the
triangle ABC. Find also the radii of the inscribed and circum-
scribed circles of that triangle.
(The pedal triangle is formed by joining the feet of the perpen-
diculars let fall from the vertices on the opposite sides.)
23. If rt = 5 and ft = 4, draw a graph to show the value of c as
C varies from 0° to 180°. Hence find the value of c when C = 40^
CHAPTER VII •
SOLUTION OF TRIANGLES
60. It is known from Geometry that, if three parts of a
triangle are given, the remaining parts can in some cases be
found; and that, in other cases, relations between the missing
parts may be found even though their exact values cannot be
determined. When actual numbers are given, results can be
obtained to a greater degree of accuracy by Trigonometrical
methods than by drawing to scale. In all cases a formula is
sought which shall contain the three given letters and one
unknown letter.
61. Case I. Three angles given.
The angle formula shows that A + B+C must be 180°. No
formula contains the three angles and one side only ; but from
the sine formula, viz. -. — - = -. — ., = -. — -, we can find the ratios
sm ^-1 sin B sm C
of the sides.
62. Case II. Two angles and one side given.
The third angle can be found immediately since
A + B+C= 180".
Suppose a is the given side ; and it is required to find b. The
formula must contain a, h, and two of the angles ; hence we use
b _ a
sin B sin A
This is adapted for the use of logarithms as it involves no
addition or subtraction. If the tables in use give the logarithms of
the cosecant, it may be advisable to use the following logarithmic
form log b = log a + log sin B + log cosec A.
63. Case III. One angle and the two sides containing
the angle are given.
Suppose a, b, C are the given parts. Then the cosine formula
c^ = ci^ + b'^ — 2 ab cos C enables us to determine c. When c is
120 SOLUTION OF TRIANGLES
determined, the remaining angles can be found by the sine
formula.
This method is of practical use only when the numbers involved
are small ; the cosine formula is not adapted for the use of
logarithms. It is usual, therefore, to use the formula proved in
§ 53,* viz. tan 1{A-B) = ^^ cot I C.
This determines h{A—B)', also ^{A + B) equals the complement
of I C; hence A and B are found by adding and subtracting.
The value of c is then calculated by the sine formula.
64. Case IV. One angle and the two sides not
containing the angle are given.
Suppose a, h, A are given. Then we can determine c from the
formula a^ = b'^ + c'^ — 2bc cos A.
This is a quadratic equation to determine c, and it is seen that
there is the possibility of two distinct values for c. This is also
seen from the geometrical construction. On this account this
case is usually known as the Ambiguous Case.
If there are two values of c, there will be two values for B and
for C. This is seen independently if the sine formula is used
(as it usually is, on account of its adaptability for logarithms) :
sin B _ sin A
b a
Suppose that this leads to the result
sin B = sin x.
Then B = x or 180 -ic.
This shows that, if there are two solutions, those two solutions
are supplementary. Hence one of the solutions will be obtuse.
Preliminary geometrical considerations often show that there can
be only one solution.
(i) If the given angle A is not acute, then B must be acute and
the obtuse-angled solution must be rejected.
(ii) If a>b or = h, then A> B or = B; consequently B
cannot be obtuse.
Exercises. When a, b, A are given, show (i) from the geo-
* Another proof is given on p. 103.
SOLUTION OF TKIANGLES 121
metrical solution, (ii) from the cosine formula, (iii) from the
sine formula, that
(a) there is no solution, if a<h sin A ;
(6) there is one solution only, if a = & sin ^ ;
(c) there are two solutions, if a>68in^ but < &;
[d) there is one solution only, if « > &.
Point out the difference in nature of the one solution in {h)
and [d).
65. Case V. Three sides given.
Here again the cosine formula may be used, if the numbers
involved are not inconveniently large. For logarithmic calcula-
tion the formula for sin^^, cos|^, or tan|^l is used. These
half-angle formulae are derived from the cosine formula.
2bin2M = l-co8^1* (§51)
~~ 2bc
2 he
{a — b + c){a + b-c)
ainiA- j(s-b)(B-c)
srn.A-^
2
Similarly cos ^ A = \ ~^ — '
Divide tan^A=J^^:^l^
^ \ s(s-a)
Of these three formulae it is best to use the tangent formula ; for
the logarithms used in finding tan 4 A are the same as those required
for finding tan i B or tan I C. If only one angle lias to be found, it is
indifferent which formula is used.
There is a simple geometrical proof for tan \ A.
* In old books on Trigonometry the ' haversinc ' was used for
solving triangles, and the values of log haversinc were tabulated in
mathematical tables. The haversinc equals half the versed sine ;
hence haversin -4 = J versin A = (1 — cos^)-^2 = sin^ \ A. The for-
mula for solution of the triangle then becomes
havei'sin A = (s—b) (s — c)-T-bc.
122 SOLUTION OF TRIANGLES
In Fig. XXXVIII / is the centre of the inscribed circle, £" is the
point of contact of the circle with AC.
Then tan ^ = 42
2 AQ
r
s~a
A
(§59)
(s-rt)
ls-h){s-c)
\ s{s — a)
Examples Vila. (See p. 81 for arrangement of work.)
In the following triangles when
Case I.
1. A = 79° 20', B = 64" 10', find the ratios of the sides.
Case II.
2. Ar= 58° 12', B = 64° 33', a = 385, find b.
3. ^ = 38° 24', C = 95° 5', c = 7-832, find a and b.
4. 5 = 63° 55', C = 48°27', c = b'16, find 6.
Case III.
5. a = 409, b = 381, C = 58° 12', find A and B.
6. B = 23° 46', c = 9-72, a = 8*88, find A and 0.
7. a = -532, c = '259, B = 39° 33', find A and C.
8. ^ = 73° 15', b = 7315, c = 8013, find B and 0.
Case IV.
9. A = 38° 14', a = '33, 6 = '44, find C.
10. « = 409, b = 385, A = 64° 32', find B and C.
11. 6 = 6-901, c = 5-749, 0=48° 27', findi?.
12. A = 73° 15', a = 7315, c = 8013, find B and C.
Case V.
13. a = 17, fe = 13, c = 12, find the least angle.
14. a = 793, b = 937, c = 379, find all the angles.
15. s = 1410, a = 1437, b = 811, find all the angles.
16. s = 1437, a = 1410, b = 811, find all the angles.
SOLUTION OF TRIANGLES 123
17. ft = 13, h = l, C= 60^ find A and B.
18. a = 32, h = 40, c = 66, find C.
19. a = 250, Z) = 240, A = 72° 4', find B and C.
20. a = 2 &, O = 120°, find ^, 5 and the ratio of c to a.
21. rt = 86, & = 63, c = 81, find the smallest angle.
22. & = 5, c = 3, ^ = 42°, find B and C.
Oral Examples.
State the formula to be used in the following cases :
1. Given a, h, C, find c. 2. Given a, h, C, find A and B.
3. Given h, c, C, find B. 4. Given b, c, C, find «.
5. Given c, a, C, find A. 6. Given c, a, C, find 5.
7. Given c, A, B, find C. 8. Given c. «, A, find &.
9. Given a, h, B, find C. 10. Given a, 5, ^, find c.
11. Given «, b, c, find C. 12. Given A, B, C, find a.
13. Given ^, C, b, find a. 14. Given a, b, B, find ^.
15. Given a, c, B, find C. 16. Given c. A, B, find &.
17. Given «, fe, r, find B. 18. Given &, c, A, find £.
19. What is the ambiguous case?
20. When a, c, A are given, what are the conditions that there
should be no ambiguity?
Examples VII b.
Solve the following triangles :
1. a = h,b = l, C=30°.
3. a = 65, & = 68, c = 16.
5. a = 7, ^ = 120°, A = 45°.
7. 6 = 926*7, ^ = 48° 24', B
S. a = 407-4, c = 115'9, A = 127° 45'
9. rt = 1263, b = 1359, c = 1468.
10. a = 53-94, b = 156-5, C = 15° 13'.
11. b = 457-2, c = 342-6, A = 73° 45'.
12. rt = 246-7, Z; = 342-5, B = 32° 17'.
13. c = 79-48, A = 54° 16', B = 85° 6'.
14. rt = 7-956, b = 10-35, c = 9-412.
15. b = 9463, c = 7590, C = 43° 47'.
16. a = 739, c = 937, £ = 146° 12'.
2.
b =
4,
c =
= 3,
C =
= 60°
4.
b =
8,
c =
= 9,
c =
:45°,
6.
a =
= 6,
& =
= 7,
c =
5.
31'
= 13'
124
SOLUTION OF TKIANGLES
17. c = 79-5, A = 35° 14', C = 117° 35'.
18. A = 89°, B = 18° 47', C = 72° 13'.
19. a = 87-6, b = 57'4, c = 46*8. 20. a = 79, c = 97, A = 2437.
21. A = 79°, C = 97°, R = 17-2. 22. b = 73-6, R = 57, a = 48*9.
23. ft2 + &2^34i^ sinC= 1, tanJ5 = fQ.
24. ^ = 42° 35', a = 83, b = 74.
25. a = 2-740, b = '7401, C = 59° 27'.
Heights and Distances.
66. First a figure must be drawn, not necessarily to
scale ; the known lengths and angles should be indicated
in the figure. It may be necessary to solve, or partly solve,
more than one triangle before the required measurement is
found. The scheme for working should be carefully thought
out before the work is actually begun.
Example i. Wishing to find the height of a house standing
on the summit of a hill of uniform slope, I descended the hill
for 40 feet, and then found the height subtended an angle of
34° 18'. On descending a further distance of 60 feet, I found
the subtended angle to be 19° 15'. Find the height of the
house.
Fig. XXXIX.
Scheme.— In triangle ADC we know one side CD and all the angles ;
so AC can bo found. Then in the triangle A CB two sides AC, CB aro
known, and the included angle, hence AB can be found.
HEIGHTS AND DISTANCES
125
From triangle ACD,
AC
CD
sin ADC sin DAC
i.e. ylC = 60sin 19^ 15' cosec IS*"
= 76-182.
From triangle ABC,
h — a
3'
tan ^(B-A) = , cot i C
36-182
Again,
116-182
^{B-A) =45° 16',
i{B+A) = 12°bV-
A = 27° 35'.
AB _ ^inACB
'CB ~ sin cab'
40 sin 34=
cot 17^ 9'
Logarithms.
1-77815
1-51811
-58559
1-88185
1-55849
+ -51061
AB =
18'
2-06910
2-06514
-00396
1-60206
+ 1-75091
35'
1-35397
1-66562
1-68835
sin 27'
= 48-792.
Height of house = 48-8 feet.
Example ii. Wanting to Icnoiv the height of a castle on
a rock, I measured a base line of 100 yards, and at one
extremity found the angle of elevation of the castle's top to he
45° 15', and the angle subtended hy the castle's height to he
34° 30'; also the angle subtended by the top of the castle
and the oilier extremity of the base line was 73° 14'. At
the other extremity the angle between the first extremity and
tJie top of the castle teas 73° 18'. Find the height of the castle.
This requires a rough perspective figure of the whole, and sub-
sidiary plane figures.
A
Fig. XL.
126
HEIGHTS AND DISTANCES
AB represents the castle.
C is the point in the same vertical as AB, and in the same
horizontal plane as DE, the base line.
The following magnitudes are known :
DE = 100 yards.
ACD and ACE are each right angles.
ABC, ABB are known, therefore BBC is known.
ABE, AEB are known.
Scheme. In triangle ADE, DE and the adjacent angles are known ;
hence AD can be found. AB can now be found from triangle ABD.
A
From triangle ABE,
AB BE
I.e.
sin 73° 18'
AB
2-
+ 1-9813
- 17415
log^Z) = 2-2398
Fig. XLII.
From triangle ABB,
AB
.ID sin 34° 30'
sin 79° 15'
Height of castle = 100 yards.
2-2398
+ 1-7531
1-9929
-1-9923
2-0006
HEIGHTS AND DISTANCES
127
Example iii. From the top of the Feak of Teneriffe the
dip of the horizon is found to he 1° 58'. If the radius of
the earth be 4000 miles, ivhat is the height of the mountain?
In Fig. XLIII C is the centre of the earth, AB is Teneriffe ;
BH is the tangent drawn from B to the earth's surface, so that H
is the farthest point seen from B ; in other words, H is on the
horizon. The angle between BH and BD (the perpendicular to
the vertical) is called the dip of the horizon.
Fi;?. XLIII,
From triangle BCH,
BC^
CH
sec BCH,
angle BCH = complement of CBH = HBD ;
i?C = 4000 sec 1°58'
= 4000x1-00059
= 4002*36 miles.
Height of mountain is 2*36 miles.
Note. Fig. XLIII is drawn much out of scale ; for small heights BH
and BD are practically identical. Even for mountains the dip is very
small, as in this example ; in fact, so small that we may use the
approximation sine of dip = tan of dip = circular measure of dip.
128 HEIGHTS AND DISTANCES
If E be the other extremity of the diameter through B, we have,
from § 52,
BA. BE = Bir\
i.e. h{2r + h)=- ct",
whore r is radius of earth, h is height of place of observation, d is
the distance of the horizon.
Hence d = \/2 rh-\-h'^',
= -s/2 rh ( \ + — — — ^... ) by the Binomial Theorem.
\ 4 r 32 r' /
So far the work is accurate ; usually h/r is so small that it may be
neglected. Hence for ordinary heights
Distance of horizon = v 2 rh.
Exercise, (i) In the formula just obtained r, /;, and the distance
are all expressed in the same units. By taking r = 3960 miles, prove
that
Distance of horizon in miles
= •v/f X height of place of observation in feet.
(ii) Show also that
Dip in minutes = '9784 V height in feet.
Examples VII.
1. Standing at a horizontal distance 100 yards from the foot of
a monument, a man observes the elevation of its top to be 25° 35'.
Assuming the man's eye to be 5 feet from the ground, find the
elevation of the top when the man stands 50 yards from the foot.
2. OA and OB are two straight roads intersecting at 0 and
making with each other an angle of 85° 12'. vl is a house
1572 yards from 0, and 5 is a house 1129 yards from 0. Find the
direct distance between A and B.
3. A man observes the angles subtended by the base of a round
tower at three points A, B, and C, in the same horizontal straight
line with the centre of the circular base, to be 2 a, 2 /3, 2 y respec-
tively. Find the ratio of AB to BC. and find the diameter of the
tower in terms of ^IC.
SOLUTION OF TRIANGLES 129
4. A man observes that the elevation of the top of a tower is
37^ 40', and that the elevation of the top of a flagstaff on the tower
is 43= 59'; show that the height of the flagstaff is one-fourth of the
height of the tower very nearly.
5. Having given that the least side of a triangle is 17'3 inches,
and that two of the angles are 63=20' and 72M0', find the
greatest side.
6. If two sides of a triangle are 7235 feet and 4635 feet respec-
tively, and if the included angle is 78=26', find the remaining
angles of the triangle.
7. The base of a triangle being 7 feet, and the base angles
129=23' and 38=36', find the length of the shortest side.
8. Explain the ambiguous case of the solution of triangles.
When a, h, A are given and the question is asked whether, from
these data, two triangles, one triangle, or no triangle can be
constructed, show that the question can be answered from a
consideration of the roots of the equation
x'^ — 2hx cos A-Irl? = c<}.
9. From each of two ships, a mile apart, the angle is observed
which is subtended by the other ship and a beacon on shore ;
these angles are found to be 52° 25' and 75° 10' respectively.
Find the distances of the beacon from each of the ships.
10. A ship sailing due north observes two lighthouses bearing
respectively NE. and NNE. After the ship has sailed 20 miles
the lighthouses are seen to be in a line due east. Find the
distance in miles between the lighthouses.
11. The angles A, J5, C of a triangle ABC are 40°, 60=, and 80°
respectively, and CD is drawn from C to the base bisecting the
angle ACB', if AB equals 100 inches, find the length of CD.
12. A man standing at a certain station on a straight sea-wall
observes that the straight lines drawn from that station to two
boats lying at anchor are each inclined at 45° to the direction
of the wall, and when he walks 400 yards along the wall to
another station he finds that the angles of inclination are 15° and
75° respectively. Find the distance between the boats and the
perpendicular distance of each from the sea-wall.
13. From a house on one side of a street observations are made
of the angle subtended by the height of the opposite house, first
121C I
130 SOLUTION OF TKIANGLES
from the level of the street, in which case the angle is tan~^ (3),
and afterwards from two windows, one above the other, from each
of which the angle is found to be tan-^( — 3). The height of the
opposite house being 60 feet, find the height of each of the two
windows above the street.
14. A segment of a circle stands on a chord AB 10 cm. long and
contains an angle of 40°. A point C travels along the arc ; for what
value of the angle ABC is the chord CA three times the chord CB ?
Verify by drawing a graph showing the chord CA as a function of
the chord CB.
15. If the sides of a triangle are 1011 and 525 feet, and the
difference of the angles opposite to them is 24°, find (correct to the
nearest degree) the smallest angle of the triangle.
16. A ladder is placed against the wall of a room and is inclined
at an angle Oc to the floor. If the foot of the ladder slips outwards
from the wall a distance of a feet, and the inclination of the ladder
to the floor is then /3, show that the distance which the top of the
ladder will slide down the wall is a cot|(a + /3).
17. A man travelling due west along a straight road observes
that when he is due south of a certain windmill the straight line
drawn to a distant tower makes an angle of 30° with the direction
of the road. A mile further on the bearings of the windmill and
tower are NE. and NW. respectively. Find the distances of the
tower from the windmill, and from the nearest point of the road.
18. A statue 10 feet high, standing on a column 100 feet high,
subtends at the eye of an observer in the horizontal plane from
which the column springs the same angle as a man 6 feet high
standing at the foot of the column ; find the distance of the
observer from the column.
19. It is found that two points, each 10 feet from the earth's
surface, cease to be visible from each other over a level plain at
a distance of 8 miles ; find the earth's diameter.
20. A plane, inclined at 33° to the horizontal, meets a horizontal
plane in the line BC. From B a line BD is drawn on the inclined
plane making an angle 27° with the horizontal plane. If BD is
18 inches long, find the height of D above the horizontal plane,
and its distance from BC. Also find the angle BD makes
with BC.
SOLUTION OF TRIANGLES 131
2L A lighthouse was observed from a ship to be N. 23" E. ;
after the ship had sailed due south for 3 miles, the same light-
house bore N. 12" E. Find the distance of the lighthouse from
the latter position of the ship.
22. Two streets meet at an acute angle; the one lies N. 51° W.,
and the other S. 48" \V. The distance from the corner to a
chemist's door in the first street is 315 yards; and the distance
from the corner to a doctor's door in the other street is 406 yards.
Find the length of a telephone wire going direct from the doctor's
house to the chemist's.
23. From a vessel at anchor two rocks are observed to the
westward, the one (A) bearing WNW., and the other (B) W. by S.
from the vessel. From the chart it is found that A bears NNE.
from B and is distant 645 yards from it. What are the distances
of the rocks from tlie vessel ?
24. Three objects A, B, and C forming a triangle are visible
from a station D at which the sides subtend equal angles. Find
AD, it being known that
AB = 12 miles, AC = ^ miles, CAB = 46" 34'.
25. A tower on the bank of a river, whose breadth is 100 feet,
subtends angles 22 1" and 67^" at two points A and B on the
opposite bank of the river, whose distance apart is 6C0 feet, on
a level with the base of the tower. Find the height of the tower.
*26. A, B, C are three given stations, so that the triangle ABC
is completely known. Show how to determine, by means of
angles measured at a fourth station P, the distances PA, PB, PC,
the four stations being all in one plane, the case for considera-
tion being that in which P is within the angle A, and the points
P and A on opposite sides of BC.
If ABC is equilateral, and the angle BPC equals 60', show
that 2 cos {(^0^ + BAP) + cos (ABP- BPA) =0.
27. A tower stands on the edge of a circular lake ABCD. The
foot of the tower is at D, and the angles of elevation of the top of
the tower from A, B, C, are 0(, ,3, y respectively. If the angles
BCA, BAC be each equal to 6, show that
cotan y. + cotan 7 = 2 cotan 3 cos 9.
* This example is best solved by using the formulae of §§ 83 and 84.
*i2
132 SOLUTION OF TRIANGLES
28. A mountain is observed from a place A to have elevation
15° 17' and to bear N. 24°29' W. From another place B which is
2347 yards north of A its bearing is N. 37° 2' W. Deduce the
elevation from B.
29. The extremity of the shadow of a flagstaff 6 feet high,
standing on the top of a regular j^yramid on a square base, just
reaches a side of the base and is distant 56 feet and 8 feet from the
extremities of that side. If the height of the pyramid be 34 feet,
find the sun's altitude.
30. A man observes that when" he has walked c feet up an
inclined plane the angular depression of an object in the horizontal
plane through the foot of the slope is (X ; and that, when he has
walked a further distance of c feet, the angular depression of the
object is ^. Show that the inclination of the slope to the horizon
is cot-^ (2 cot /3 — cot 0() ; and determine the distance of the object
observed from the foot of the slope.
31. A straight flagstaff, leaning due east, is found to subtend an
angle 0( at a point in the plain upon which it stands, a yards west
of the base. At a point h yards east of the base, the flagstaff sub-
tends an angle /3. Find at what angle it leans.
32. Four rods are loosely jointed at their extremities to form
a parallelogram with sides 4 and 5 inches long. Two of the
opposite corners are connected by an elastic string of length
7 inches. Find the angle between the string and the shorter side.
If the length of the other diagonal be diminished by 1 inch,
what does the angle become ?
*33. Three posts on the border of a lake are at known distances
from each other, namely 63 yards, 44 yards, and 76 yards. At
a boat on the lake it is found that the two posts, whose distance is
63 yards, subtend an angle 89° 15', and the two posts, whose
distance is 76 yards, subtend an angle 130° 45'. Find the distances
of the boat from the three posts.
34. A base line AB is drawn 2 chains in length on a plane
in the same horizontal plane as C the foot of a tree. The angles
ABC, BAC are found to be 79° 56' and 78° 18' respectively; the
angle of elevation of the top of the tree is found to be 19° 46' at A.
Find the height of the tree to the nearest foot.
* This example is best solved by using the formulae of §§ 83 and 84,
SOLUTION OF TRIANGLES 133
35. A base line AB, 2527 links long, is measured on the sea-
shore along the high water mark. 0 is a point where a distant
rock meets the sea ; the angles BAC, ABC are found to be 89" 15',
86" 21' respectively. The angle of elevation of the highest point
of the rock, which is vertically above C, as observed at A, is 1°48'.
Neglecting the curvature of the earth, find the height of the rock
and its distance from A.
36. A hill slopes upwards towards the North at an inclination
14" to the horizontal. The sun is 15° W. of S., at an altitude
of 47° ; find the length of the shadow cast on the hill by a vertical
post 39 feet high.
37. If, in the previous question, the post is perpendicular to the
surface of the hill, what is the lenorth of the shadow ?
134 REVISION EXAMPLES
Revision Examples B.
1. Define the tangent of any angle, and prove from the defini-
tion that (i) tan(90 + ^j= -cot^; (ii) tan(180-.4)= -tan^.
Express the other trigonometrical ratios in terms of the tangent.
2. Show by substitution that
sin 45" + sin 30° > sin 60°,
and cos 30° - cos 45° < cos 60°.
3. Find the value of sin 45° without using tables.
Solve the equation 4 sin ^ cos ^ -f 1 =2 (sin 6 -h cos S).
Give the general solutions.
4. A man walks directly across the deck of a ship, which is
sailing due North at 4 miles an hour, in 12 seconds, and finds that
he has moved in a direction 30° East of North. How wide is
the deck?
5. Show that in any triangle ABC,
(i) sin A /a = sin B/b = sin C/c ;
(ii) sin C {a cos B-b cos ^) = (a + b) (sin A — sin B).
6. Prove geometrically that
cos 2^ = l-2sinM.
Hence find the value of sin 15°.
7. The angle of elevation of the top of a spire seen from A
is 30°, and it is found that at a point B, 115| feet nearer the foot
of the spire, it is 60°. Find the height of the spire to the nearest
foot.
8. Plot a curve giving the sum of 4sin^ and 3 sin 2^ from
^ = 0° to ^ = 180° ; and read off the angles at which the greatest
and least values respectively of this sum occur.
Estimate the slope of the curve when 6 = 90° and when ^ = 135°.
9. Define a radian. Express in degrees and minutes an angle
of 1'36 radians.
Find the number of radians in the angle of a regular decagon.
10. Prove
(i) sin'^^ -f cos'^^ = 1 ;
(ii 1 tan ^ -i- (1 - cot ^) + cot ^ -^ (1 - tan A) = sec A cosec ^ + 1.
REVISION EXAMPLES 135
11. Draw the sine and cosine graphs, in the same figure, from
^ = 10° to <9 = 20°.
From the graph find the angle which satisfies
sin ^ + cos ^= V2.
12. Find an expression which will include all angles having
a given tangent. Write down the values of tan 225^, tan 780",
cot 1035% cot210\
Construct an angle, having given the cotangent.
VS. Find a/cos A-^b/coi B + c/cos C in a form adapted to
logarithmic calculation.
14. In any triangle prove that (i) a = 6cos 6'+ ccos ^ ; (ii)
«(6cos C-ccosi?) = ^'^-c^; (iii) rcoalA = as\n^Bsin\C.
15. If the sides of a parallelogram are a, b, and the angle
between them co, prove that the product of the diagonals is
'^/a*-2a-b- cos CO + ¥.
16. A vessel is steaming towards the East at 10 miles an hour.
The beanng of a lighthouse as seen from the vessel is 42^24' North
of East at noon, and 25° 12' East of North 25 minutes later. Find
how far the vessel was from the lighthouse at noon, and find also
at what time the bearing of the lighthouse will be due North.
17. Assuming that a circle may be treated as a regular polygon
with an infinite number of sides, show that the ratio of the
circumference of a circle to its diameter is constant.
What is the circular measure of the least angle whose sine is |,
and what is the measure in degrees, &c., of the angle whose
circular measure is '15708?
18. Prove by a geometrical construction that
cos 2^ = cos^^ — sin^x4.
Solve the equation cos 2 A = (cos^ + sin Af.
19. For what data will the solution of a triangle become
ambiguous ? Explain this.
Given J5 = 30°, c = 150, Z> = 50 ^/3, show that of the two
triangles that satisfy the data one will be isosceles and the other
right-angled. Find the third side in the greater of these triangles.
Would the solution be ambiguous if ^ = 30°, c = 150, h = 75 ?
20. AB is a horizontal line whose length is 400 yards; from
a point in the line between A and B a balloon ascends vertically,
136 KEVISION EXAMPLES
and after a certain time its altitude is taken simultaneously from
^ and jB ; at ^ it is observed to be 64° 15' ; at B 48° 20' ; find the
height of the balloon.
21. Find the radius of the circle circumscribing a triangle, in
terms of its sides. If c^ = ci^ + W, show that this radius equals \ c.
22. Define the trigonometrical ratios of A involved in the
equation cot ^ + tan A = sec A cosec A ; and establish its tmth by
a geometrical constiiiction.
23. Prove that
cos
' ^/{a -x)-^{a-h) = sin ^ ^y{x -b)-r- {a - h)
= cot""^ \/{a — x)-7-{x — b).
24. Prove that sin 6 = tan 6-^ ^/l + tan2<9.
Having given tan ^ = |, find sin 6, cos 9, and versin 0.
25. If 6 is an acute angle whose sine is j%, calculate the value
of tan ^ + sec ^.
What would the value be if 6 were obtuse ?
26. What is the angle between the diagonal of a cube and one
of the edges at its extremity ?
27. Obtain an expression for all the angles which have a given
tangent.
Find all the angles lying between —360° and +360° which
satisfy the equation
2
tan'^x = tan re — 1 = 0.
28. A circular wire of 3 inches radius is cut and then bent so as
to lie along the circumference of a hoop whose radius is 4 feet.
Find the angle which it subtends at the centre of the hoop.
29. A triangle ABC has angle A = 34°, a = ll'O cm., c = 7*8 cm.
Calculate the perpendicular from B on h, and the remaining
angles and side of the triangle.
30. In a triangle a = 14:, & = 37, c = 97 ; find the value of
(i) acos^ + tcos^, (ii) cisinB — bBinA.
31. If ABC be a triangle, and 6 an angle such that
sin 0 = 2 \/ah cos J C -f (a + h),
find c in terms of a, h, and 0.
If a = 11, h=.2ry, and C= 106° 15^', find c.
REVISION EXAMPLES 137
32. Find the area of a regular quindecagon inscribed in a circle
of one foot radius.
33. Find an expression for all angles having the same sine as
the angle Oc.
Solve the equation sin {0(. + x) + sin {3 + x) = 0.
34:. An angle 0( is determined by the equations v^ = 2(/h,
-b = tv sin a -Igf, tv cos 'X = a. Show that
«2 tan^ Ot - 4 ha tan y + a'^-i hh = 0.
35. Criticize the proposition that three measurements are suffi-
cient and necessary to determine a triangle uniquely in shape
and size.
36. A square house, measuring 30 feet each way, has a roof
sloping up from all four walls at 35° to the horizontal. Find the
area of the roof.
37. Draw up a table showing in three columns the values of
10sin<9, 10 cos (9, and 8 sin ^ + 6 cos ^ for each 30° from 0° to
360°. From the table draw, in the same figure, the graphs of
?/ = 10sin^ and ?/ = 8 sin^ + 6cos^ ; and from the curves deter-
mine approximately a value of 6 for which tan ^ = 3.
38. Taking the earth as a sphere of radius 4000 miles, find the
distance London travels in an hour in consequence of the rotation
of the earth. (Latitude of London 51° 30' N.)
39. ABCD is a quadrilateral in which AB and DC are parallel
and 40 feet apart, and AB is 100 feet long. The angle DAB is
72° 30', and the angle CBA is 38° 15'. Find the lengths of AD,
DC, and CB, and the area of the quadrilateral.
40. State the local time at the following places when it is noon
at Greenwich.
Cape Town 33° 56' S., 18° 25' E. Fiji 18° 0' S., 178° 0' E.
Edinburgh 55°57'N., 3° 10' W. Singapore 1° 17' N., 103° 50' E.
41. Define the cosine and the tangent of an angle, and show
how to express the tangent in terms of the cosine.
Having given that cos^l = '8, and that A is Jess than 90°, find
the value of tan .4; and by means of the tables find the value
of Af both from its cosine and from its tangent.
138 REVISION EXAMPLES
42. Prove that, in any triangle ABC, sin B: sin C = b:c. In
the triangle ABC the angle CAB is 50°, the angle ABC is 65°,
and the side BC is 4 inches long. Find the length of the side AB.
43. Show how to find the height of a tree by means of a chain
for measuring lengths and of an instrument for measuring angles.
44 Find an expression for all the angles which have (i) a given
tangent, (iij a given sine.
45. Explain how it is that, tan ^ being given, tan 2^ is known ;
but that, sin ^ being given, sin 2^ may have either of two values.
46. Prove that the area of a triangle is \/.'i {s-a) {s- b) {s — c).
Show also that the area is ^ c^ -f (cot J. + cot B .
47. Find the radius of the circumscribing circle of the triangle
for which A = 66° 30', B = IV 30', c = 200 feet.
48. A ship is sailing due East at a uniform rate : a man on
a lighthouse observes that it is due South at 1 p.m. and 16^30'
East of South at 1.20 p.m. In what direction will he see it
at 2 p.m.?
CHAPTEE VIII
PROJECTION. VECTOES
Fi-. XLIV.
67. If from the extremities of u line FQ, of definite
length, perpendiculars PK, QL are let fall on a line AB,
which may be produced if necessary, then KL is called the
Projection of PQ on the line AB.
Projections are subject to the same
convention of sign as are abscissae
and ordinates. Thus, in the above
figure, KL is positive, but LK is
negative. It follows that the projec-
tion of PQ is not the same as the ^
projection of QP, so that the order
of the letters in naming a line is
of great importance when we are
dealing with projection. When the direction of the line is to be
taken into account as well as its length, it is called a directed
length ; and we shall, in future, use the symbol {PQ) * to denote
the directed length of the line from Pto Q. The number of unit> of
length in that line we shall continue to denote by the symbol I'Q.
Thus, in Fig. XLIY, the projection of (PQ) is (KL),
and the projection of [QP) is {LK).
Note. When we speak of the sum of directed lengths in the same
straight line, tlie algebraical sum is always meant. Geometrically
this means that we require the directed length between the starting-
point and final point, and not the length of the actual path traversed.
68. If the length of PQ is I, and if 6 is the angle
between PQ and the line AB, then
projection of {PQ) on AB = I cos 0.
This is usually written PQ.
140 PROJECTION. VECTORS
Some care is necessary in applying this formula ; the safest
plan is to keep ? and 6 both positive.
Consider, for instance, the projection of {QP) in Fig. XLV.
Imagine a line drawn from the initial point Q parallel to the
line AB. Then it is seen that the angle between {QP) and AB
is 6 + IT, while the length ^Pis I.
Hence projection of {QP) on AB = lcos{d + 7r) = -I cos 6,
Two other methods of treatment give the same result.
A B
Fig. XLV.
In Fig. XLV the line QX is actually drawn parallel to AB ; but
it is usually sufficient to imagine it. Then we may take the angle
between (QP) and QX to be the negative angle XQP, i. e.
— (n — 6); the length ^Pis positive so that
projection of QP = lcos{ — Tr — 6) = — Zcos d.
Or we may regard 6 as being the angle between {QP) and QX ;
but this requires that the length of {QP) should be taken as —I,
and so the projection of (QP) on AB = -/cos d.
It will be found that, in all cases, I cos 0 gives both the
magnitude and sign of the projection of (PQ) on AB.
Similarly,
the projection of (PQ) on a line perpendicular to AB
= I sin 6,
69. Proposition A. The sum of the projections on any
PROJECTION. VECTORS
141
Jhie of two sides {AB), (BC) of (a triangle is equal to the
project ioti of the third side {AC).
Fig. XLVI.
In either of tbe above figures (or in any other figure)
projection of (.45) + projection of (BC) = {ah) + {hc)
= {ac)
= projection of {AC).
Proposition B. The sum of the projections on any line of
the three sides {AB), {BC), {CA) of a triangle is zero.
Sum of projections of iAB), iBC), {CA) = {ah) + (be) + (ca).
Hence on the line of i^rojection we start at the point a and
finish at the same point, so that the distance between the initial
and final points is zero. That is, the sum of the projections
Proposition C. In any closed figure ABC ... HK, the stm
of the projectioyis of the sides {AB), {BC) ... {HE) equals the
projection of(AK).
Proposition D. In any closed figure the sum of the pro-
jections of all the sides taJcen in order in the same direction
is zero.
Propositions C and D are proved exactly in the same way as Pro-
positions A and B.
142 PROJECTION. VECTORS
Example. Prove that
cos A + cos{120 + A)-\- cos (120 ~ ^) = 0.
Fig. XLVII.
Draw an equilateral triangle PQR, side a units.
Draw a line OX inclined at an angle A to (QR).
Then {RP) is inclined at .4 + 120 degrees to OX; and {PQ} is
inclined at ^1 + 240 degrees.
Project on OX; then, by Proposition B,
a cos A + a cos {A + 120) + a cos (A + 240) = 0 ;
but cos (.4 + 240) = cos {360 -(120 -.4)} = cos (120-^);
.-. cos A + cos (120 + A) + cos (120 -A) = 0.
Examples VIII a.
(These examples should be verified by drawing a figure to scale.)
1. Show that the projection of a line on a line parallel to itself
is equal to the projected line, and that the projection of a line on
a line perpendicular to itself is zero.
2. A line of length r, making an angle 6 with OX is projected
on OX and at right angles to OX ; calculate the lengths of the
projections in the following cases :
(i) r = 5, ^ = 60°; (ii) r=-5, ^ = 120°;
(iii) ;• = 5, ^ = 248° ; (iv) r = 5, d = 300° ;
(v) r = -5, ^ = 330°.
3. Two rods AB, BC, of lengths 5 feet and 10 feet respectively,
are joined together at an angle of 135°. The rods are fixed in
PROJECTION. VECTORS 143
a vertical plane so that CB is inclined at 60'' to the horizontal,
and the angle ABC is beneath the rods ; by projecting horizon-
tally and vertically, find the inclination of the line ^C to the
horizontal.
4. By projecting a diagonal and two sides of a square on a line
making an angle .4' with one of the sides, prove that
cos (.4 + 45'') = (cos A - sin A) -r >/2.
Find a similar value for sinu-1 + 45").
5. PQR is a triangle right-angled at Q, having the angle at P
equal to A' ; FQ is inclined to OX at an angle B^.
Prove by projection that
cos {A + B) = cos A cos B — sin A sin B,
and sin (A + B) = sin A cos B + cos A sin B.
70. If the projections of a line on two lines at right angles are
given, the length and direction of the projected line can be found,
but not its actual position.
Let r be the length of the line and 0 the angle it makes with
one of the lines of projection. Then r cos S and rsin 6 are known ;
suppose these values are x and y respectively, so that >'cos^ = x
and rsin 6 = ij.
Then r = a/x- + i/- and tan 6 = y/x.
The projected line has therefore a definite length and a definite
direction ; it is the simplest example of a group of quantities
called vector quantities or vectors.
71. A quantity which possesses a direction as well as
magnitude is called a vector. Yelocities and forces are
examples of such quantities. The magnitude and direction
can be represented by the length and direction of a directed
straight line ; hence the properties of a directed straight
line that depend only on its length and direction represent
properties common to all vectors.
144 PROJECTION. VECTORS
72. Vector addition or Composition of Vectors.
A displacement from A to B followed by a displacement from
B to C produces the same result as a single displacement from
^to a
B
Fig. XLVIir.
Or we may regard the displacements as being simultaneous.
Suppose a point to start from A and move along AB, and while
this point is moving, suppose the line AB to move parallel to
itself, the point B moving to C while the point travels from
A to B. The result of the two simultaneous displacements is that
the point has travelled from A to C.
Hence the vector (AC) is called the resultant of the vectors
(AB) and (BC).
Finding one quantity equivalent to two or more of the same
kind is equivalent to the process of addition in Arithmetic.
If we use the sign + to denote this process, we have
{AC) = {AB) + {BC).
If P, Q, and R are the respective magnitudes of the vectors
represented by {AB\ {BC), and {AC), and if 6 is the angle
between the directions of {AB) and {BC) (in Fig. XL VIII the
angle ABC is the supplement of 6) ;
then 112 ^ P2 + Q2 + 2 PQ cos 6.
Similarly, if a number of vectors are represented by the directed
lengths {AB), {BC), {CD)...{HK), then their resultant is repre-
sented by the directed length {AK).
73. Besolution of vectors.
In Fig. XLVIII the vector {AG) may be replaced by the
two vectors (AB) and {BC). Viewed in this light they are
called the components of the vector (AC).
PROJECTION. VECTORS
145
When we talk of the component of a vector in a given direction,
and no mention is made of the direction of the other component,
it is understood that the other component is at right angles to the
first.
P
Fig. XLIX.
If (OP) in Fig. XLIX represents a vector of magnitude E inclined
at an angle 6 to OX, then its projection {OX) represents the
component along OX, and the projection (XP) represents the
component perpendicular to OX.
The vector is now said to be resolved along and perpendicular
to OX.
Resolving along OX, we find that the component is Rcosd.
Resolving peqDendicular to OX, we find that the component is
Psin^.
74. All the work of § 69 on projections can be applied to
vectors and their components. For instance, Proposition C gives
the following proposition :
The sum of the components of any number of vectors in a given
direction is equal to the component of their resultant in that
direction.
Examples VIII b.
[In the following examples the letters P, Q, R imply that the
vectors are forces ; the letters u, v, w imply that the vectors are
velocities. When possible, figures should be drawn to scale to
check the calculation.]
1. Find the resultant R in the following cases :
(i) P= 17, <?= 13, e= 40";
fii) P= 17, (2= 13, 6^ 140';
(iii) P- 114. Q= 75, 6 = 65';
(iv) P= 123, (,^ = 496, e^nV.
145 PROJECTION. VECTORS
2. Find P when Q the other vector, 6 the angle between them,
and R their resultant have the following values :
(i) ^ = 176, i? = 249, e^ 72°;
(ii) Q= 73, i^ = 193, 6= 110°;
(iii) Q = 245, i? = 92, e= 130° ;
(iv) Q= 36, 7? = 84, ^ = 20°.
8. Show that, if the resultant of three forces is zero, the sum of
their components in any direction is zero.
4. Show that if three forces produce equilibrium (their resultant
is, therefore, zero) they are parallel and proportional to the sides
of a triangle.
5. A boat is being rowed due E. at a speed of 6 miles an hour ;
at the same time a current carries it due S. with a speed of 3 miles
an hour ; find the magnitude and direction of the actual velocity.
6. Find the resultant of velocities u and v inclined at an angle By
when
(i) 11 = 14, f = 16, 8= 180°;
(ii) « = 14, i^ = 16, 0= 65°;
(iii) u = U, v=lQ, e = 135°.
7. Vectors of magnitudes 7, 8, 9 respectively are parallel to
three consecutive sides of a regular hexagon. Find the sum of
their components (i) parallel to, (iij perpendicular to, the middle
one of these sides. Hence find the magnitude and direction
of their resultant.
8. Find the magnitude and direction of the resultant of four
forces of magnitudes 5, 10, 15, 20 respectively, which act along
the sides of a square.
9. A stream flows at the rate of 2 miles an hour. In what
direction must a man swim in order that he may actually go
straight across the river, his rate of swimming being 3 miles
an hour?
10. A rod 5 feet long is hung by a string, attached to its two
endS; over a smooth peg ; it rests, at an angle of 20° to the
horizontal, so that the two portion^^ of the string are each inclined
35" to the vertical. Find the length of the string.
PKOJECTION. VECTORS
147
Projection on a Plane.
75. If from every point in a line, atniight or curved, a perpen-
dicular be let fall on a plane, the locuj; of the feet of the
perpendiculars is called the projection of the line on the plane.
If from every point in the boundary of a surface a perpendicular
be let fall on a plane, the area bounded by the locus of the feet
of the perpendiculars is called the projection of the surface on
the plane.
76. The angle between a straight line and its projection on
a plane is called the angle between the straight line and the
plane. It follows that the projection on a plane of a straight line
of length I, making an angle a with the plane, is I cos (X.
Any two planes, not parallel, intersect in a straight line. If
from any point P in this line two perpendiculars FA, PB are
drawn to it, one in each plane, then the angle APB measures the
angle between the planes.
77. If any plane surface, of area A, is projected on a
plane making an angle a with its own plane ; then the area
of the projection is A cos ol.
*Step I. Consider a rectangle ABCD, having the side AB
parallel to the plane of projection, and the side BC making an
angle 0( with that plane ; then y is the angle between the plane
of the rectangle and the plane of projection.
Then, in Fig. L, abed is the projection of ABCD.
D
Fig. L.
* A slight knowledge of solid geometry is assumed lu this proof.
k2
148
PROJECTION. VECTORS
Now Bb is perpendicular to the plane abed, and therefore to the
line ab ;
.-. Bb is perpendicular to AB ;
but BC is perpendicular to AB ;
.". AB is perpendicular to plane BCcb ;
.•. ab is perpendicular to plane BCcb ;
.-. ab is at right angles to he,
i.e. abed is a rectangle.
Hence art-a of abed = abxbc
= ABxBC cos Oi
= area of ABCD x cos CX.
Step II. Consider a plane area with curved or rectilinear
boundary. In the plane of the figure draw any line PQ parallel
to the plane of projection. Then in the area we can inscribe
a number of rectangles having the short sides parallel to PQ and
the longer sides perpendicular to PQ.
Fig. LI.
The sum of these rectangles is less^than the original area, but may
be made to differ from that area by as small a quantity as we please by
making their width small enough ; and then the sum of their projec-
tions will differ from tlie projection of the area by an even smaller
quantity. Hence in the limit, when the width is indefinitely small,
the sum of each set of rectangle;s will equal the area of the corre-
sponding circumscribing figure.
But the sum of projections of rectangles -- sum of rectangles x cos A ;
.*. the area of projected figure =■ area of original figure x cos A.
PROJECTION. VECTORS 149
Examples VIII c.
1. A pyramid VABCD has a square ba?e A BCD, side a, and the
faces VAB, &c., are equilateral triangles. Find the length of the
projection of VA on the base.
Verify that the sura of the areas of the projections of the four
faces is equal to a^.
2. A square house, whose side is 28 feet long, has a roof sloping
up from all four walls at 40° to the horizontal, find the area
of the roof.
3. Find, by projection, the area of the curved surface of a right
circular cone, having height h, and semi-veitical angle 2 y.
4. From a cone 6 feet high a smaller cone 2 feet high is cut off.
If the radius of the base of the small cone is TG feet, find the area
of the curved surface of the remainder of the large cone.
Verify your answer by projecting this surface on the base.
5. A circle with radius a is projected into an ellipse with semi-
axes a and h ; show by projection that the area of the ellipse
is TTCib.
6. The vertical angle of a conical tent is 67", and the radius of
the base is 5| feet; find (i) the slant height, (ii) the area of
canvas used, (iii) the content of the tent.
7. A pyramid on a square base is such that each of the other
faces is an isosceles right-angled triangle, find by projection the
angle between a triangular face and the base.
Geometrical representation of imaginary quantities.
78. In Fig. LII OA is of length r.
By the usual convention a line OA drawn to the right is
B
A' 0 A
Fig. LII.
considered positive, so that {OA) represents -fr. If now [OA)
is turned through two right angles, it takes up the position {OA')
and, by the usual convention, {OA') represents - r. Hence the
150
PROJECTION. VECTORS
geometrical operation of turning through two right angles repre-
sents the algebraical operation of multiplying by -1. Let us
consider what the operation of turning through one right angle
represents.
This is an operation which, if performed twice in succession,
turns through two right angles, which represents multiplica-
tion by — 1.
But the algebraical operation of multiplying by aZ-I, if
performed twice in succession, multiplies by —1.
Hence it seems reasonable that the operation of turning a vector
line through a right angle represents the algebraical operation
of multiplying by \/-l ; that is, (OB) at right angles to (OA)
represents r x V^— 1, i.e. -%/ — Ir.
In future we shall denote \/—l by /.
79. With the interpretation of i suggested by the last section,
cc + ii/ is represented by a vector line of length or followed by
a vector line of length y at right angles to the first vector.
X + iy ^ {0N)+ (NP) (Fig. LIII.)
= (OP). (By vector addition, § 72.)
Or, in words, or + iy is represented by the vector (OP), that is
by a vector line of length >v/.r^ + f/^,
with the positive direction.
laking an angle tan"'
80. For our purjioses this statement is more useful if reversed,
VIZ. (OP) =^ x + iy
= }'cos6+ ir Bin 6
= (cos ^ + 1 sin B) r.
PEOJECTIO^\ VECTORS 151
Or, in words, the vector line of length r, in direction ^, repre-
sents the magnitude r multiplied by cos ^ + /sin ^. This gives the
important result that
turning through an angle 6 represents multiplication by
cos^ + j sin^.
Hence
turning twice in succession through d represents multipli-
cation by cos ^ + /sin B repeated twice ;
i.e. turning through 26 represents multiplication ly
(cos 0 + / sin 6f ;
but turning through 2 d represents multiplication by
(cos2^ + /sin2^).
Hence the suggested interpretation of -y/— 1 or /, leads to the
identity fcos 26+ i sin 2 ^) = (cos ^ + j sin 6)-.
If this is verified by algebraic multiplication and by the use of
the ordinary formulae for cos 2^ and sin 2^, it will be found
correct.
Carrying on the argument in the same way, we deduce that
(cos 7id + i sin n 6) = (cos 6 + i sin 6)'\
where n is any positive integer.
Again, turning through a half 6 is an operation which, if re-
peated, turns through 6, and, therefore, represents a multiplication
which, if repeated, multiplies by cos 6 + / ?in 6 ;
• Le. (cos 16 + { sin 16)=> (cos 6 + i sin 6} '^.
/ 6 6 \ -
Similarly. ( cos - + / sin - j = ( cos ^ + / sin 6)" ;
and ( cos - ^ + / sin - ^ ) = ( cos ^ -f / sin 6) i •
a '1 /
Lastly, turning through -6 cancels turning through 6, and,
therefore, represents an operation which cancels multiplication
by (cos (9+ i%\n6) ;
i.e. {co8(-^) + /sin(-^)y = .cos^ + /sin^;-'.
r!fimilarly,
CCS { — n6) + i sin { — n6) = (cos 6 + / sin ^j~",
where n is any positive quantity.
152 PROJECTION. VECTORS
We have now clediiced from the geometrical interpretation
of \/-l. that
(cosn^ + isinn^) = (cos ^ + i sin ^)"
for all real values of n.
This is known as De Moivre's Theorem.
Example. Use De Mo iv re s Theorem to find \^\.
cos 2 nTT = 1, sin 2 « 77 = 0,
where n is zero or any integer.
Hence 1 = cos2«7r-f /sin 2;?7r ;
.'. ^T = (cos2n7r + isin2«7r)i
= cosf ^?7r + isin§n7r.
If >?-0, >v/r= cosO + /sinO = 1.
If n = 1, ^T ^ cos Itt + / sin §77 = -^ ( - 1 + / yS).
If n = 2, ^r=cos^7r + /sin^n- = H-1-?V3).
If ;^ = 3, v^l = cos27r + » sin2n- =1.
For other values of n it is seen that the three roots are repeated,
Hence De Moivre's Theorem shows that there are three different
cube roots of unity. They are, of course, the three roots of the
equation ic^ — 1 = 0. The student should verify that the same
roots are obtained hy Algebra.
Examples VIII d.
Represent graphically and by imaginary quantities the follow-
ing vectors :
1. (i) Magnitude, r = 25 ; direction 6 = y where a = tan-^g^^.
(ii) „ r = 25; „ 6 = tt-OL;
(iii) „ r = 25; „ 6 = tt + 0L]
(iv) „ r=25; „ ^ = 27r-a;
(v) ,. r = 25; „ 6=-0(.
2. Show graphically that
[x + iy') + {x" - iy") = (.r' + x") + i (//' 4- y").
3. Express the following in the form y (cos 6 -f / sin 6) :
(i) 3+4/; (ii) 5 + 6/; (iii) 7-8/;
(iv) -5-12/; (v) -5 + 12/; (vi) 8/.
4. Interpret geometrically (cos a + /sinOK) (cos 0(-t sina) r; and
justify your interpretation.
PKOJECTION. VECTORS 153
5. Show graphicall}^ that
(cos 0( + -/ sin CX) (cos jS + / sin (3) = cos {0C + (3) + i sin {OC + /3).
6. Verify De Moivre's Theorem by calculation, when n = 2, 3,
-1, -2.
7. Assuming De Moivre's Theorem, prove that v^ — l has three
values, viz. -1, ^(1 + ^-3) and 1(1-7-3).
8. («) Prove De Moivre's Theorem by induction when n is a posi-
tive integer.
(b) Deduce the proof when n is not a positive integer, by
methods similar to those used for the Binomial Theorem in
Algebra."^
Examples VIII.
1. A man walks one kilometre in a direction 16 degrees North of
East ; he then turns to the left, through an angle of 110 degrees,
and walks one kilometre in the new direction. How far is he North
and how far East of his starting-point ?
2. Show that «cos^ + 6sin^ can be expressed in the form
r cos {d - Oc). Illustrate by a figure.
3. A number of rods are jointed together, and the two free ends
are secured to two points A and B in the same horizontal line
and distant c inches. If the length of the r^^ rod is «,. , and its
inclination to the horizontal is 0,. (all the angles being measured in
the same sense), prove that (i) 2 (ctr cos ^,.) = c ; (ii) 2 («,. sin 6,.) = 0.
(See § 89, Example ii.)
4. Prove by projection that
sin(90 + ^) = cos^ and sin (270-^) = -cos ^.
5. In what respects can a vector quantity be represented by
a straight line?
If three forces P, Q, R, acting at a point 0, are such that
P/B\nQR= Q/smFP= R/sinPQ (where sin P^ denotes the sine
of the angle between P and Q), show that the three forces produce
equilibrium.
6. A man walks one kilometre in a direction A"" North of
East, one kilometre in a direction making 120" with the first
direction, and one kilometre at an angle 240° with thefirst direction.
Draw a figure showing that he has now returned to the starting-
* See School Algebra, pp. 407, 435, 4G5.
154 PROJECTION. VECTORS
point ; and by considering the distances he has gone to the East
and North write down two trigonometrical identities concerning
the sines and cosines of A, 120 + -4, 240 + ^.
7. Suggest a geometrical construction which may help to sum
the series :
(i) cos a + cos (a + /3) + cos (CX + 2 /3) + . . .n terms ;
(ii) sin (X + sin {0( + ^)+ sin (a + 2^) + . . .w terms.
Deduce that both these sums become zero if «,3 = 2it.
8. A body which weighs 12 lb. is kept at rest by means of two
cords, one being horizontal and the other inclined to the horizontal
at an angle whose tangent is | ; find the forces exerted by the cords.
9. A mine shaft is 1650 feet in length. It slopes downwards at
an angle of 45° to the horizon for a certain part of its total
length, say x feet, and at an angle of 35° for the rest of its length.
If the total depth reached is 1000 feet, obtain an equation for x,
and hence calculate x.
10. A man playing five holes of a golf course first walks 260 yards
due East, then 140 yards 20° South of East, then 300 yards due
South, then 200 yards 40° West of North, then 220 yards 30° West
of South, thus arriving at the fifth hole. Find how far the fifth
hole is from the first tee.
11. The perpendicular from the origin on a straight line equals
2) and makes an angle a with the axis o( x; by projecting the
co-ordinates of any point on the line show that the equation of
the straight line may be pat in the form a; cos a + 1/ sin Oc = p.
(This is known as the perpendicular form of the equation of
a straight line.)
Hence find the length of the perpendiculars from the origin on
the lines whose equations are (i) Sx + 4ij = 7 ; (ii) 5x-l2ij = 2;
(iii) x + 2t/ = 6. Verify by drawing to scale.
12. The co-ordinates of a point referred to rectangular axis
OX, OY are x, y; referred to two rectangular axes OF, OW
through the same point 0 the co-ordinates are |, //. Prove by
projection that ^ = xco^oc + ijs'm OC, where Oi is the angle between
OX and or.
Find three other similar relations connecting ^, »/, x, //,
CHAPTER IX
FORMULAE FOR (i) THE FUNCTIONS OF THE
SUM OR DIFFERENCE OF TWO ANGLES,
(ii) THE SUM OR DIFFERENCE OF THE
FUNCTIONS OF TWO ANGLES, (iii) THE FUNC-
TIONS OF THE DOUBLE ANGLE AND THE
HALF-ANGLE
81. To express cos (A + B) in terms of the sines and
cosines of A and B.
Fig. LV.
Let OX be the initial line ; and let the revolving line first turn,
through the angle A to the position OA and then continue to
turn through an additional angle B to the position OB. Then OB is
the bounding line of the angle A + B. Along OB measure a length
OP=r units.
Project OP on the initial line, produced backwards if necessary
also project OP on OA, produced backwards if necessary.
Figures LIV and LV show two of the many possible cases.
156 FORMULAE FOR FUNCTIONS OF THE SUM
In all cases
the projection of (OP) on OX
= sum of the projections of (OK) and (KP) on OX ;
i.e. rcos{A + B) = (OK) cos A + {KP) cos (^ + 90)
= r cos B cos A + r sin B { — sin A).
. Hence cos (A + B) = cos ^ cos £ — sin A sin B. *
Several proofs of this have already been given, but the earlier
proofs have implied that A and B are together less than two right
angles ; this proof applies whatever be the values of A and B.
Exercises, (i) Deduce the formula for sin (A + B) by substitu-
ting 90 -v4 in place of A, and -B m place of B.
(ii) By similar substitutions deduce the formulae for cos {A — B)
and sin(.4 — J5).
(iii) By projecting perpendicular to OX, find the expanded
form of sin {A -f B).
(iv) Modify the construction so as to prove directly, by pro-
jection, the formula for sin (A-B) and cos (A~B).
(v) Complete the following formulae :
cos {A -¥B) =
cosU-5) =
nn{A + B) =
sin(^-^) =
(vi) Learn these formulae in woids, as :
cos sum = cos cos - sin sin.
* When Fig. LV is being used it must be recollected that {OK) is
negative, and tliat its inclination to OX is XOA not A'OK, see § 68,
If (07v) is regarded as positive, its actual length is — r cos B ; but the
angle is tlien JCOK, the cosine of wliich is —cos A. Whatever way it
is taken, the projection of {OK) on OA' is found to be r cos £ cos A.
OK DIFFEEENCE OF TWO ANGLES
15'
82. The followinrr proof does not involve any knowledge of
projection ; its chief drawback is that it applies only to the case
when A + B is less than a right angle. It is easily modified to
suit any other given case.
Let XOA = A and AOB ^ B ;
then XOB = A + B, and in Fig. L VI is less than 90°.
Take a point P in the bounding line oi A + B;
let fall PN perpendicular to OX-,
PK ,. „ OA;
„ KL ,, ,, PX and. therefore, parallel to OX
KM „ „ OX.
ON OM-MN
goh{A+B) =
OP
OP
OM
OP
LK -\
OPj
OM OK
OK OP
LKPK
PK OP
The spaces in the bracketed line (which does not appear in the
completed work) are filled in with the hypotenuses of the triangles in
which the respective numerators occur.
Now angle LPK = 90° - LKP = LKO = A ; therefore
LK PK = sin A.
Hence cos (A + B) = cos A cos B - sin A sin B.
From the same figure, prove that
s'm(A + B) =
To find the functions of A—B, the angle AOB must be made on
the negative side of OA. The point P must be taken in the
158 FORMULAE FOR FUNCTIONS OF THE SUM
Dounding line of A — B, and the construction and proof proceed as
before. It is found that
cos{A-B} =
sin (.4 -5) =
Exercises, (i) Prove the four formulae when A and B are each
less than 90^ but ui + B is greater than 90^.
(Make the same construction as when ^ + i>' is less than 90^, and
pay careful attention to the signs of the lines.)
(ii) Prove the four formulae when A and B are each obtuse
and together greater than 270°.
Examples IX a.
1. By using the formulae of § 81, verify that sin (90 — -4) = cos^
cos(90 + ^)= -sin^, sin (180-^) = sin .4, cos (180-^1 j= -cos^
sin(270 + ^)= - cos ^, cos (360-^) = cos^.
2. Express cosTO"" in terms of the functions of (i) 40° and SO"*
(ii) 45° and 25° ; (iii) 95° and 15° ; (iv) 35°.
3. Express sin 40° in terms of (i) 30° and 10° ; (ii) 25° and 15°
(iii) 70° and 30° ; (iv) 20°.
4. From the expansions of sin(^ + i^) and coii{A + B) deduce
the expansion of tan(^ + J5) in terms of tan^ and tani?.
5. From the expansions of sin(^ — ^) and cos (^-^) deduce
the expansion of tan {A - B) in terms of tan A and tan B.
6. Verify that sinO° = 0 and cosO° = 1 by using the formulae
for^-^.
7. Show that (i) sin (^ + -B) cos -B - cos (^ + 5) sin 5 = sin 5 ;
(ii) cos {A + B) cos B + sin {A + B) sin B = cos B.
8. From the formulae for A + B deduce that
sin 2^ = 2 sin -4 cos ^ and cos2^ = cos^vl — sin'-^.
What is the value of tan 2^ ?
9. Find the values of
(i) sin(^ + jB)-} sin(^-^) ; (ii) cos ( J + B) -f cos (.4 - £) ;
(iii) sin {A 4- B) - sin {A - B) ; (iv) cos {A + B)- cos {A - B).
Account for the signs of (iii) and (iv) from first principles.
OR DIFFERENCE OF TWO ANGLES 159
10. Prove thiit
sin A = sin }, (A + B) cos \ {A-B) + cos \ [A + B) sin 1{A- B).
Prove similar results for cos A, sin B, and cos B.
11. From the results of 10 deduce that
sin ^ + sin J5 = 2 sin \{A + B) cos \{A- B),
and three similar results.
12. Prove that
(i) cos- 6 -f cos- (p — 2 cos 6 cos (p cos ($ + (p) = sin- (d + cf));
(ii) sin- 6 + cos- 0 — 2 sin ^ cos 0 sin (^ + </>) = cos^ {B + </>).
83. Sums and difi'erences of sines or cosines ex-
pressed as products.
These formulae are most easily derived from the formulae of
§ 81, as suggested in Examples IX «. They can be proved inde-
l)endently by projection.
Make the angle XOA = .1 and the angle XOB = B.*
On OA and OB take lengths OP, OQ respectively, each equal to
r units. Join PQ.
B
Fig. LVII. Fig. LVIII.
Bisect the angle QOP by a line cutting PQ at R.
Then the angles BOP, ROQ each equal 1{A-B) ; and the angle
XOP = l{A + B).
From congruent triangles EP = RQ, and PRO is a right angle.
The projection of (OP) = sum of projections of (OR) and (RP),
and the projection of (0^; = sum of projections of (02?) and (RQ).
* Notice the difference between tbi> construction and tbc construc-
tion of § 81.
160 FORMULAE FOR SUM OR DIFFERENCE OF
.•. projection of ( OP j + projection of (0^)^*2'^ projection of
(OR), since projections of {HP) and (BQ) are equal but opposite.
Projecting on a line perpendicular to OX, we have
r sin A + r sin B = 2'^'' the projection of {OB).
But {OR) is the projection of {OP) on the direction OR,
i.e. {OR) = rco&\{A-B).
rsin^ + rsin J5 = 2rco&\{A-B)^m\{A + B).
Hence sin ^ + sin £ = 2 sin ^ (^ + B) cos \{A-B).
From the same figure, by projection on OX, we have
cos^ + cosjB = 2cos|(^ + jB)cos^(^-jB).
Again, projection of (OP) -projection of (0$) = 2^^ projection
of {RP).
Hence sin^ — sin P = 2 cos | (A + P) sin \{A — B),
cos^-cosP= -2sin|(^-P)sin J(^ + P).
The proofs apply to all cases whatever be the magnitudes of
A and P.
The reason for the negative sign in the last of these formulae is
obvious, for, if A> B, then cos A < cos P.
It is useful to learn the formulae in words, it being understood
in all cases that the greater angle is put first.
sine + sine = 2 sine half sum cos half difference.
84. Products of sines and cosines expressed as sums
or differences.
In the formulae of the last section put
i(^ + P) = X, \{A-B)=Y',
so that A^X+Y, B = X-Y.
Then sin(X+ r) + sin {X- Y) = 2 sin X cos Y,
i. e. 2 sin X cos Y = sin {X+ F) + sin(X- Y).
Similar results are obtained from the other formulae. If we
replace X by ^ and Y by P, the formulae become
2 sin ^ cos P = sin (.1 + P) + sin (.1 - P),
2 cos .1 sin B = sin (.1 + P) - sin (.1 - P),
2 cos.l cos P = cos {A + P) -f cos {A - B),
2 sin A sin P ^ cos (.1 - P) - cos {A + P).
These are more easily proved direct from the A-vB and A—B
formulae.
FUNCTIONS OF TWO ANGLES 161
In using these formulae it is usual (but not necessary) to put the
greater angle first ; this shows why there are distinct formulae for
2 sin .4 cos i? and 2 cos .4 sin i?.
Express the four formulae in words :
Twice sine cos = sin sum + sin difference.
Examples IX b.
1. Apply the formulae of §§ 83, 84 to the following cases and
verify from the tables :
(i) A = 70°, B = 30° ; (iii) A = 72°, 5 = 18°;
(ii) A = 110°, 5 = 75° ; (iv) A = 78°, B = 46°.
2. Prove, from the formula for sin A + sin B, that
sin 2 ^ = 2 sin 6 cos d,
and, in a similar way, show that
1 + cos ^ = 2 cosH ^, 1 - cos JL = 2 sinH ^.
3. Prove that
(i) sin ^ + cos ^ = ^/2 cos {A — 45) ;
(n) sm A — cosB= - 2 sm ( 45 7,— 1 sm ( 4o ^ — j ;
/•••x . • -^ ^ /.K A-B\ { ,^ A + B
(ill) cos A + smB = 2 cos f 45 ^ — 1 cos ( 45 +
4. Prove that (i) sin 50° + sin 130° = 2 cos 40° ;
(ii) cos 50° - cos 130° = 2 sin 40°.
Verify these by squaring and adding.
5. Prove that (i) 2 cos 40° sin 50° = 1 - sin 10° ;
(ii) 2 cos 40° sin 40° = sin 80°;
(iii) 2 sin 64° sin 26° = cos 38°.
Verify this last result from the tables.
6. Fill iu the right-hand side of the following :
(i) sin /0° + sin50° = (ii) cos 30° -cos 110° =
(iii) 2 sin 75° cos 10° = (iv) sin 37° + cos 24° =
(v) 2 cos 84° cos 72° = (vi) cos 79° -cos 52° =
(vii) sin 75° -sin 116°= (viii) 2 cos 80° cos 35° =
(ix) cos 24° -sin 76° = (x) 2 sin 17° sin 48° =
1216 L
162 FORMULAE FOR FUNCTIONS OF
(xi) 2 cos 73° sin 15° = (xii) 2cos U°cos 166° =
(xiii) cos |7r-cos|rr = (xiv) sin 1 7r + COS J tt =
(xv) 2 sin 43° cos 47° = (xvi) 2 cos 97° sin 46° =
(xvii) sin 81° + sin 10° = (xviii) sin 49° -sin 53° =
(xix) 2sin79°sinl5° = (xx) cos 43° -cos 216° =
7. Prove that 4 cos (75° + A) sin (75° - ^) = 1 - 2 sin 2 ^.
Formulae for the double angle and half- angle.
85. It lias already been shown in § 51 that
sin 2 ^ = 2 sin A cos A;
cos 2A = co&'^A - sin^^
= l-2sin2^
= 2jiQsM-l. ,
The proof there given assumed that 2 J. is less than 180°.
If we put A instead of B in the A + B formulae the same results
are obtained ; thus they are true for all values of ^.
The results can easily be proved independently by projection ;
the proofs are the same as in § 81, A taking the place of B.
86. From the last section, by putting -^^ in place of A, we
have cos'* ^A — sin'^ -| J. = cos A ;
also cos'^^^ + sin^^ J. = 1.
Add 2cos^i^ = l + cosX
Subtract 2 sin^ ^A = 1- cos A.
Hence cos^^ = + Kv^l + cos^) ;
sin|^= +|(-v/l -cos^).
If the value of A is given, there is no ambiguity of sign. If, for
instance, A = 140°, then ^A = 70°, and the sine and cosine are
both positive; if ^ = 264°, then -^.4 = 132°, and the sine is
positive, the cosine negative.
If the value of cos^ is given but not the value of A, the
ambiguity cannot be removed. Suppose cos .4 = ^, then A may
have any value of the form 360° n ± 60°. Hence I A may have any
value given by 180° « + 30°. If we tabulate these values, we have
r igle cosine sine
30° +|-v/3 +^
150° -^^/S +1
210° -},^/3 -I
330° +^^3 -I
DOUBLE ANGLE AND HALF-ANGLE 163
87. Tangent formulae.
From the sine and cosine formulae the following tangent formulae
are derived ; the proof of the first only is given :
, , „, sinU+5)
tan \A-{-B) = — — =r(
' cos (^ + 5)
sin A cos B + cos A sin B
cos A cos B — sin A sin B
_ ^n A + tan B (By dividing throughout by
~l-tanAtanB cos^eo3J5.)
Similarly,
tan (^-5) =
tan 2.^ =
1 A /I -COS A
tan I A = + A z
^ - A' 1 + cos A
1-cosA sin A ,„ ,. ... ,
= + — -. — - — or + _ . (By rationaliznig. )
- sin A -1 + cosA ^ ^ °'
88. These may all be proved directly from the figures used for
the sine and cosine formulae ; e.sr. in Fier. LVI.
tan (^ + 5) = '^,
MK^LP
OM-
MK
OM
LK
LP
"^ OM
(By dividing so as to make the first
~ LK PK' term in the denominator to be L)
PKOJl
The triangles LPK, OKM are similar ;
LP PK , ^
TT L / i T,\ tan A + tan B
Hence tan(^ + ^)=:, — 7 tt ^•
1-tan^tan^
Exercises. Prove that
tanyl + tan5 + tan(7 — tan^tan^tanC
(i) tan(^4-5+C) =
l-tanjBtanC-tanCtan^-tan^tan^
(ii) tan.i + ta„£ = '^^^^^';
cosJ-cos-B
(iii) tan X- tan 5
sin(^--B)
cos^ cos5
l2
164 FORMULAE FOR FUNCTIONS OF
(iv) cot A + cot B = -. — -—. — =, ;
sill A sm B
{,)cotA-ootB =-^i'^Mz^'.
sm A sm A
Example. To prove that tan'^ g\ + tan ~ ^ ^ J y = tan~ ' ^^q •
Let A = tan-^ ^g, B = tan-^ ^tf ;
so that tan A = ^^, tan B — o|y .
tan A + tan B
tan (^ + 5) =
1 — tan A tan ^
■9^ + -
1 ]L V _1-
■■■ 99 "^ 2'59
239 + 99 338
23900-240 23660
i. e. tan ^ t^^^ + tan"^ 239 = tan"
-1 1
7(J'
Examples IX c.
1. Prove that (sin A + cos Af = 1 + sin 2^,
a nd (sin A - cos Af = 1 - sin 2 ^.
2. Assuming the values of sin 45°, cos 45°, tan 45°, deduce
sin 90°, cos 90°, tan 90°.
3. Find a formula for cot 2^ in terms of cot^.
4. Show that sin ^A = 3sin^ -4 sin^ JL. Explain how it is that
there are three values of sin J. when sin 3 J. is given.
5. Find the values of tan 22^°, tan OTf, tan 157i°.
6. Prove that 2sin-|^ = + -y/l + sin A ± ^/l-smA.
Find sin ^.4 when sin^ = |. Illustrate by a figure.
7. Find cosl^ when sin^ = ^. Illustrate by a figure.
8. Prove that (i) sin 2^ = 2 tan^^(l +tan2^) ;
(ii) cos2^ = (l-tanM)~(l+tanM);
(iii) tan 2^ = 2 tan ^ ^ (1 - tanM).
9. Prove that sin ^A = 3 sin^l — 4 sin^^ ;
cos 3^ = 4cos^u4-3cos^ ;
tan 3^ = (3 tan^- tan^^) ^(1 - 3 tanM).
DOUBLE ANGLE AND HALF-ANGLE 165
10. Show that
(cos A + sin Af + (cos A - sin Af = 3 cos ^ — cos 3 A.
11 (rt). Show that sin^^ + cos^^ = + \/l + sin^,
and sin ^A — cos ^ J. = + v^l — sin A.
(b) Having given 4 sin 54°= ^/b + l, apply the formulae in
(a) to find sin 27° and cos 27°, explaining how the ambiguities of
sign are cleared up.
(c) Show that 8(sinU2°-cos2 78°) = -/S + l.
12. Prove that
fA R r — ^^'^^^tan^ + tan C— tan^tan5tan C
'^ ~ 1-tan^tan C — tan C7tan^— tan^tan^
Deduce the formula for tan dA-
What can be deduced ifA + B + C equals (i) 180°, (ii) 90° ?
13. If tan ^ = If and tan B = 2%, show that
A + B = {4:n + l)l7r.
14. Show that
cos ^ + cos 3 ^ + cos 5 ^ + cos 7 ^ = 4 cos ^ cos 2 ^ cos 4 S.
15. Find all the solutions of the equation
sin 6 sin 3 ^ = sin 5 6 sin 7 6.
16. If tan^ = I, tan 5 = §, tan C = fj, and each angle is acute,
prove that A + B + C = ^7r,
17. If tan^ = tanjcxtani^, show that
tan 2 ^ = (sin OC sin ^) -r (cos y. + cos /S).
18. (i) If ^ = tan-4, find tan 2^.
(ii) Show that 2 tan"^ | + tan-^ } = in.
19. Prove that
cos2^-cos2^ = 2(cos-^-cos2 5) = 2(sin-5-sin^^).
20. Prove that
.. _J sec4^
^^^ a + bcosd~ {a + b) + {a-b)ta.n'^^B''
1 1+tan^l^ ^
^"^ acos^ + fesin^ ~ a + 26tan|^-atan2i^ *
21. Solve the equations
(i) x2-^/2sin(i7r + Oi)a; + isin20( = 0;
(ii) «2_2cot2/3.a;-l = 0.''
22. Solve for OC and Fthe following equations :
2ag = V- sin 2CX, 2bg = V^ sin^ y.
166 FORMULAE FOR FUNCTIONS OF
23. A hemispherical shell of radius 16 inches rests with its rim
on a horizontal table ; a rod is hinged to a vertical wall, 25 inches
from the centre of the shell, at a point 5 feet above the table.
The rod is in the same vertical plane as the hinge and centre of
the shell, and touches the shell. Find its inclination to the
vertical.
Oral Examples.
(a) (i) sin(P-^) = (ii) cosX + cos r =
(iii) cos (90- 1 J+^) = (iv) sin 270°
(v) 2 sin a cos /3 = (vi) cos^ 6 - sin^ 6 =
(vii) tan (A-B) = (viii) sin ^ - sin C =
(ix) cos245°-sin2 45°= (x) cos 2.1 =
(6) (i) cos(C+.4) = (ii) sini? + sinC
(iii).2sini(5-f Cji(P-C)= (iv) cos^ + cos(/) =
(v) 2cos2 1(7-1 = (vi) tan25
(vii) sin (180-5 + C) = (viii) cos^TS^ + sin^TS^ =
(ix) cos{^A + B + lA-B)= (x) sin (360-2 Cj =
(c) (i) sin 2 5 = (ii)sin(P+^)
(iii) cosin-oc + li) = (iv) 2siniCcosJC =
(v) 1-2 sin^ B = (vi) cos^ | C- sin^ i C =
(vii) cos2 1 C + sin^ A C = (viii) tan (B-C) =
(ix) cosC-cos^ = (x) (sin j5 + cos 5)^ =
(d) (i) sin^cosC-sin^sin C = (ii) cos(X-r) =
(iii) sin 3^ = (iv) sin-^i? =
(v) cos 5 + cos (7 = (vi) 2 cos 5 cos C =
(vii) sin^ {B + C) + cos- {B + C) = (viii) 2 cos- ^ ^ - 1 =
(ix) (cos 1^- sin 1 .1)2 = (x) tan (90 -C) =
89. The preceding formulae load to a number of useful identities
in the cases where A + B + C=90° or 180". The method of dealing
with these is shown in the following illustrative examples.
DOUBLE ANGLE AND HALF-ANGLE 167
Example i. In any triangle tan H^"" ^) = r — ^^^ \ ^-
[Here - — - gives the clue to the proof.]
0 + c
b c
By the sine formula, -. — ^ = ~ — ;;,'
*' ' sm j5 sm C
sin B b
I.e. . ^ = -,
sm C c
^ , i T -i T sin 5 — sin 0 b — c
Componendo et dividendo, -; — ^^ -. — i^ = , ;
2cosl(B+C)sm^iB-C) _ b-c^
28in^{B+C)cos^{B-C)~ b + c'
tan^(g-CO _ b-c ^
^•®* tani(.B+(7j ~ b + c'
hut A + B+C=180''; .'. IA + 1{B+ C) = 90.
Hence tan |(5+ C) = tan (90 -M) = cotj^.
b — c
Substituting above, tan ^(-B - C) = cot ^A.
This formula has been proved geometrically in § 53 ; it is usually
proved by the method given above.
Example ii. In any triangle
^^~~cUS^-\-€o^^ + cosC = 1 + 4 sin i A sin J B sin i C.
L.H.S. = cos J. + cos^ + cos C
= 2 cos * (^ + jB) cos 1 U - -B) + 1 - 2 sin^ -i C
= 2sin|Ccosi(^-5) + l-2sin|Ccos^(^ + J5),
since ^C = %-l{A + B)
= l + 2sin|C(cosi^-^-cos^^ + ^)
= l+4sin|^sin^5sin"?C.
The symbol 2 cos A is sometimes used to denote
cos^ + cos5 + cos C;
and nsin^ to denote sin ^ sin 5 sin C. The above result can be
written: 2 cos^ = 1 4-4nsin^^.
168 VARIOUS ILLUSTRATIVE EXAMPLES
Example iii. In any triangle ^cos'^A = l — 2Yicos A.
(Questions involving the sum of the squares of sines or cosines are
usually solved by expressing these squares in terms of the cosine of
the double angle.)
2 2 cosM = 2 cos2 A + 2 cos'^B + 2 cos^ C
= 1 + cos 2^ + 1 + cos 25 + 2 cos^ C. (Note that one angle
is left unchanged.)
.-. 2 cos^^ = 1 + cos (^ + B) cos (A-B) + cos^ C
= 1 - cos C cos (^-5) -cos Ccos (A + B),
since C= 180 -(^ + 5)
= 1 - cos C [cos {A ~ B)-cos(A + B)]
= 1—2 cos Acqs B cos C.
Example iv. Solve the eqimtion
sin 0 + sin 2e + sinS0 + sin 4:6 = 0.
Rearrange sin ^ + sin 4 ^ + sin 2 ^ + sin 3 ^ = 0.
Use formula for sum of two sines
2sin#^cos|^ + 2sin|^cos|^ = 0
.*. either sin f ^ = 0 or cosp+cos^^ = 0
i.e. -^6 = nrr or 2 cos^cos ^^ = 0 :
i.e. 6 = ^nrr, or cos ^ = 0 or cos^^ = 0
i.e. ^=(2n + l)*7r or |^ = (2/i4- l)i7r.
Hence the complete solution is
^=(2w + l)7r, (2n + l)i7r or fnTT.
Example v. To prove that
r = 4:B sin J A sin I B sin i C.
From the figure of § 59,
rcot^5 + rcot^C= a,
/coshB cosiCX ^ „ . ,
r . l^ + -7-^-y, ) = 2i? sin ./,
\sm|5 sin|C/ '
siniBcos^C+cos^^sin-JC ^^ .
I.e. r —^ . ', p . , V, ^ = 27?sm^,
sinf5sin|(7 '
sini(J5 + C) ^^ . ,
I.e. r . {1. . '^ = 27?sin^,
sm|5sin^C
i. e. r _._;o^i;^ = 4 7^ sin I A cos I A ;
sinl^sin^C
''sincei(P+C) = 90-|^;
.-. >• = 4 7? sin I ^ sin ^5 sin J C.
Exercise. Prove that r^ = 4 7i? sin J ^ cos ^ 5 cos ^ C.
I.e.
VAKIOUS ILLUSTEATIVE EXAMPLES 169
Example vi. To show that the distance between the circum-
centre and in-centre = \^{Ii- — 2Br).
A
In Fig. LIX, with the usual notation,
BD = E sin A, DO = R cos A,
BP= 1' cot IB, PI=r.
Or- = {BP-BDf + {IP-ODf
= {rcoi\B-R%\nAf + {i'-Rco^Af
= i?2_2i;-(sin^coti5 + cos^) + »^(l + cot45)
sin|^ sin^l^
^0 nr. sinU+*5) ,^ sini^sinJ^sinlO (Substitut-
smi^ sm'^i^ mgforr.)
^o or> cos^{A-C)-2smlAsm^C
= U' — 6iir -. — , ,
sin ^5
since A + },B = ^0 + lA-lC
siniB
= R!^-2Rr. since i(^ + Cj = 90 -i5.
This is more shortly proved by Pure Geometry ; but the method
used here is a general method to find the lengths of lines connected
with the triangle.
Example vii. To p'ove that
sin A + sin {A-\-B)-\- sin {A + 2B)-{- ... to n terms
sin h nB sin {A + ^n—l B)
sin I B
170 VARIOUS ILLUSTRATIVE EXAMPLES
Let S* = sin ^ + sin (^ + i?) + . . . + sin (A + n - 1 B).
Multiply by 2sinjP.
Then 2smlB.S
= 2sin^sini^ + 2sin(^ + ^)sini5+ ... +2sin(^ + n-l^)sini5.
Use the formula for the product of two sines.
2sini^.5'= cos(^-ij5)-cos(^+|5)
+ cos (A + IB)- cos [A + IB)
+ .... -
+ cos (.4 + i2n- 35) -cos (^+12^1-1^)
= cos(^-i5)-cos(^ + i2M-15);
^^m\)iB%m{A + \n-lB)
sin^?i5
Note, Compare tliis with the formula for the sum of n terms of an
Arithmetic Progression. Notice that A + \n — \ B = half the sum of
the first angle {A) and the last angle (^4-n — 1 B),
Examples IX.
1. Prove the following identities :
(i) sin 3 ^ = 4 sin A sin (60^ + A) sin (60'' - A) ;
(ii) sin 3 A sin^^ + cos ZA cos^ A = cos' 2 A ;
(iii) (l-2sin2^)-f(l + sin2^) = (1 -tan^) f (1 + tan^) ;
,. . tan (45° + ^)+ tan (45°-^) ^ ,
(IV) r~ y-Ar^o-^Tx , 7TE5 7\ = COSeC 2 ^ ;
^ ' tan (45° + ^) -tan (45°-^)
(v) sin {y->iZ-x) + sin {z + x-y) + dn{x + y-z)-^\n{x + y + z)
— 4 sin a: sin y sine';
(vi) cot J^ — cot^ = cosec^ + cosec-|^;
(vii) cos 4 ^ + 2 (cos ^ + sin J.)* = 3 + 4 sin 2 A ;
(viii) sin ^ + sin 5 = sin (^ + -B) + 4 sin \ AQm\B sin \{A + B)',
(ix) sin ^ - 3 sin 3 ^ + 3 sin 5 ^ - sin 7 ^ = 8 sinM cos 4 ^ ;
(x) cos(^ + J?+C)
= cos A cos B cos C — cos A sin 5 sin C — sin A cos B sin C
— sin^ sin 5 cos C\
(xi) cos \ A (2 sin A - sin 2 A) = sin^ I A (2 sin ^ + sin 2 ^) ;
(xii) cos A-vco%B + cos C + cos (^ + i? + C)
= 4 cos ^ {A + B) cos I {B + C) cos },{C-\- A).
SUM AND DIFFERENCE FORMULAE 171
2. If A, B, C be the angles of a triangle, show that
(i) tan A + tan B + tan C = tan A . tan B . tan C ;
(ii) sin 2 ^ + sin 2 5 + sin 2 C = 4 sin ^ sin 5 sin C ;
(iii) sinH^ + sin2i5 + sin2iC + 2sin*^sin*5sin^ C= 1;
(iv) sin ^ + sin jB + sin C = 4 cos \ A cos \ B cos \ C ;
(v) cot A cot 5 + cot A cot C + cot 5 cot C = 1 :
(vi) cot A + cot B + cot C
= cot ^ cot B cot C + cosec .4 cosec B cosec C ;
(vii) tan B tan C + tan Ctan J. + tan ^ tan B
= 1 + sec A sec 5 sec C ;
(viii) cos ^ sin (5 -C'j + cos 5 sin (6'-^j + cos Csin(^-5) = 0 ;
(ix) (tan A + tan B) (tan A - cot C) = sec^^ ;
(x) tani5tan^C + taniCtanM + tani^tani5 = 1.
3. Show geometrically that sin(^ + 5) = sin^cos^ + cos^sini?
when each of the angles A and B is between \tt and it, and A + B
is less than frr.
4. Solve the equation cos 3^ + cos 2^ + cos ^1 = 0.
5. Find all the values of 6 which satisfy
(i) cos(9 + cos2^ + cos3^ + cos4^ = 0;
(ii) sin3<9 + sin4^ + sin5<9 = 0.
6. Solve (i) sin {A + 30°j = 1 -^ ^2 ;
(ii) V'-3 sin A + cos A = ^/2 ;
(iii) sin J. + cos ^ = 1 ;
(iv) sin A + ^/2^ cos ^ = 2 ;
(v) ^/2 (cos OX + sin 3a^) = 1 ;
(vi) a cos6 + b sin 6 = c (put a = r cos 0(, h = r sin 0().
7. Prove that (i) 2 sin-i ^ ^/2 = 90° ;
(ii) 2tan-^| = tan-4.
8. In any triangle show that
i? (sin 2^ + sin 25 + sin 2 Cj = 2r (sin^ + sin J5 + sin C).
9. In any triangle show that
a^ COS. 2 B + b'^ cos2 A = a- + &'^-4rt& sin^sin 5.
10. Prove the formula (& + c) tan 1{B-C) = {b-c) cot hA.
Write down two corresponding formulae.
11. Using the fact that 3 x 18° = 90°-2 x 18°, find the values of
sin 18° and cos 18°.
Give a geometrical method for determining sin 18°.
172 SUM AND DIFFERENCE FORMULAE
12. Simplify
., /sin 4^ cos4vl\ , , , .^ ,,
V sm J. cos^ /
sin 5^ -sin 3^ 2 sin 5 (9 + sin 3^
(") .^.^^ , ...o^ +
cos5^ + cos3^ sin2^ cos5^ — cos3^
13. D, E, F are the feet of the perpendiculars from A, B, C on
the opposite sides ; P is the orthocentre. Prove that
(i) ^P=2i?cos^; (ii) PZ) = 2PcosP cos C;
(iii) perimeter of triangle DEF = 4P sin ^ sin B sin C.
14. State the general formula for all angles having a given
cosine.
Solve sin 3 ^ + sin 5 ^ + sin 7 ^ = 0.
15. Find sec {A + B) in temis of the secant and cosecant of .4 and
5, and prove secl05° = 72 (1 + ^3).
16. Prove that
sinl8° = |(v^-l) ; and that sin^ 30° = sin 18° sin 54°.
Show that in any circle the chord of an arc of 108° is equal to
the sum of the chords of arcs of 36° and 60°.
17. Given cos ^='28, determine the value of tan ^.4, and
explain fully the reason of the ambiguity which presents itself in
your result.
18. Prove that
cos~'ii; + cos~^y = sin"^ {x \/\ —if^-y^/X —x^),
and solve the equation
tan-i{(a:+l)-f(a^-l)}+tan-i{(a^-l)-^a;} = tan-'(-7).
19. Express sin3^ -r (sin2^ — sin^) in terms of cos ^.
20. Prove the identities :
(i) (l + cos^)tan2i.4= 1-cos^;
(ii) (sec ^ + 2 sin A) (cosec ^ - 2 cos ^) = 2 cos 2 A cot 2 .-1.
21. In any triangle prove that [h — c) cos ^A = a sin I {B — C).
Ii A = 80°, a = 10, h-c = 5, find B and" C.
22. Prove the identity cos 2^; sin 3a7 = sin x cos ix + cos .r sin 2.r.
23. Solve the equations cos 2^ = cos(^-rt) ; cos 3^ = sin (6-^).
24. (i) If the equation of a straight line is put in the form
ij = mx + c, what is the geometrical interpretation of m ?
(ii) Show how to find the angle between two lines whose
equations are y = mx + c, y = rnx + c\
SUM AND DIFFEKENCE FORMULAE 173
(iii) Deduce that the lines are at right angles if mm' = - 1 ;
and parallel if m — m'= 0.
(iv) Prove that the lines whose equations are ax + hi/ + c = 0,
ax + Vy + c^O, are perpendicular if aa +W =0y and parallel
if a/ a' = h/h'.
25. Find the angle between the lines whose equations are
(i) Sx-4:y=b, 4x-2ij = 7;
(ii) 4a; + 3?/ = 6, Sx-4i/ = 9',
(iii) 2x— y =3, 4a7 + 5?/=l;
(iv) 2x-y =3, 4x + 2?/ = 5;
(v) 2a; + 4i/ = 5, a; + 2t/=3.
In each case verify by drawing to scale.
26. Find the equations of the straight lines drawn through the
point (3, 5), and respectively parallel and perpendicular to the
line whose equation is 3ic — 4f/ = 5.
27. Find the equation of the straight line, parallel to the line
whose equation is iccos a+i/sin a =_p, and passing through the
point [x, y). Deduce that the length of the perpendicular from
[x', y') to the line x cos 0( + y sin oc = p i& x' cos Oi + ?/'sin Oi —p.
28. Find in its simplest form the equation of the line joining
the points {acos(CX + ^), &sin(CX + ^)}, {a cos (a -/3), &sin(a-/3)}.
29. Prove that sin 55° sin 15° - sin 50° sin 10° - sin 65° sin 5° = 0.
30. Show that in any triangle
a^sin(J9-C) &^sin(C-^) c^sin(J. -B)
h + c c + a a + b
31. If 2cos^ = ic + l/a; and 2 cos 0 = 2/ + 1/^/, prove that
2 cos {d + (f)) = xy + 1/xy and 2 cos {d — (p) = x/y + y/x.
82. If d + (f) = 240°, and versing = 4 versing, find the values of
6 and 0.
33. Draw a curve to represent the variations in sign and magni-
tude of (sin^->v/3cos^)-7-(-/3sin^ + cos^), from <9 = 0 to e = n.
34. If oc and /3 are the roots of « sin ^ + & cos ^ + c = 0, prove that
cos-|((X4-3) _ COS |(CX-^) _ sin^(0( + /3)
b —c a
35. Eliminate 6 and 0 from
a sin ^ + & sin 0 = h, (i)
rt cos ^ — & cos (p = k, (ii)
, cos {d + (f)) = I. (iii)
174 SUM AND DIFFERENCE FOEMULAE
36. Eliminate 6 and 0 when two equations are the same as
(i) and (ii) in Ex. 35, and the third equation is (i) sin(^ + 0) = I,
(ii) tan(^ + 0) = /.
37. Eliminate B and 0 from the equations
a cos{(b-hOC) sine/) . ,. x , • /,
h cos(^-a) sm ^ ^ ■ v-r /
38. Expand sin 5^ in terms of sin^, and cos 6^ in terms of
cos^.
39. If sinJ5 is the arithmetic mean between sin^ and cosyl,
prove that cos2^ = cos2(^ + 45°).
40. If a cos ^ + & sin B = c, show that
B = tan-^ h/a + cos-i cj{ Va" + ¥).
41. Find the maximum and minimum values of
acos^ + Z>sin^ = c.
Verify your answer when a = 3, Z> = 5, by drawing a graph.
42. Prove that
(i) sin B + sin 2B + sin 3 ^ + ... to n terms
sin|«^sin|«4- 1 B ^
^ sinT^ " '
(ii) cos ^ + cos (^ + 5) + cos (^ + 2 i?) + . . . to n terms
s,va.\nB C0& {A + \n~ 1 B)
~ sin \B
(iii) cos a + cos (a /3) + cos(a + 23)+ ...to nterms = 0, if
n^ = 2n.
43. Find the sum of n terms in the following series :
(i) sin^ A + sin2 {A + B) + sin^ {A-v2B)+ ...;
(ii) cosM + cos2(^ + jB) + cos2(^ + 2^)+...;
(iii) sin^ sin 2^ + sin 2^ sin 3^4 + sin 3 J. sin 4-1 + ....
REVISION EXAMPLES 175
Revision Examples C.
(All the following examples are taken from recent Examination
Papers.)
1. Find, without reference to the tables, the values of (i) sin 45° ;
(ii) cos 150^; (iii) the tangent of the obtuse angle whose sine is
1/v/lO.
2. Trace the graph of the function cos^ + 2sin^ between the
values 0 and 180° of 6, and determine from your figure the value
of 6 for which the function (ij is greatest, (ii) is decreasing most
rapidly.
3. Express tan 6 in terms of sec 6.
Show that (sin 6 - cos d) (sec 6 + cosec 6) = tan 6 - cot 0.
4. Prove that the sines of the angles of a triangle are in the
ratios of the sides opposite them.
5. Solve the equation 2 cos a; + sin a? = 2.
6. In a right-angled triangle ACB, C being the right angle,
the angle A is 35°, the side AB is 10 inches ; find the other sides.
7. If cos(^ + -Bj = cos^cos5-sin.4sin5, calculate cos(A + B)
when A = 50° and B = 50°.
8. If OC is measured in radians,
sin a = a-aV,3 + aV|5_-a7l2+ ...
where j 5 means 1x2x^3x4x5. Find sin a correct to four
significant figures when oc = 0'3. What is the angle OC in degrees ?
9. Define the tangent of an angle in such a way that your
definition is true for all angles.
If 6 be an acute angle, prove that cos (90 + ^) = -sin 0.
10. Arrange in. order of magnitude the angles
2 sin~^ -51, i cos-i '32, tan'^ 8-9.
11. Draw the graph of cos a; for values of x lying between
0° and 90°.
Use your figure to solve roughly the equation .r = 100cos:r°,
and verify your solution by the tables.
176 REVISION EXAMPLES
12. Given that sin 20° = O'Si and cos 20° = 0*94, write down
the values of sin 160° + cos 160°, of sin 250° + cos 250°, and of
sin 340° + cos 340°.
13. In any triangle ABC, show that
(i) c = acos^ + &cos^ ;
(ii) c2 = a2 + &2-2a&cosC.
Find c when a = 5, & = 6, and 0=155° 31', having given
cos24°29' = 0*91. Verify your result by a diagram drawn to
scale.
14. Find to the nearest degree the angle subtended at a man's
eye by a tower 50 feet high, when the man has stepped back
30 feet from the tower, assuming the height of his eye above the
ground to be 5 feet 6 inches.
15. Write down a formula for sin^J. in terms of the sides of
the triangle ABC and explain the notation. How is the formula
modified when h — c^
Given that the sides are 100, 200, 160 units in length, calculate
the smallest angle.
16. A and B are two acute angles but A + B \^ obtuse; prove
that cos {A + B) = cos A cos i? — sin ^ sin B.
Solve completely cos x + ^mx = cos Oi — sin OL.
17. Define the tangent of an angle, and show geometrically that
tan^tan(90° + ^) + l = 0.
18. Draw a circle of diameter 1 inch. Draw a diameter AB
and the tangent to the circle at B, divide either of the semi-
circumferences between A and B into 8 equal parts, join A to
the points of section, and produce the joining lines to meet the
tangent at B. Measure the distances of the points so found from
B, and use the results obtained for drawing the graph of tan ^4
from ^ = 0° to ^ = 90^.
19. Prove that the area of the triangle ABC is
\a^ sin B sin C/sin A.
Use this expression to find the area of the triangle when
a = 106-5 yards, A = 56° 37', B = 75° 46.'
REVISION EXAMPLES 177
20. A person walking along a straight level road running clue
East and West observes that two objects P and Q are in a line
bearing North-West, and after walking a further distance d he
observes that P bears due North and that the direction of Q
makes an angle A with the direction in which he is walking.
Prove that the distance PQ is cZcos^/sin (^-45°). Find PQ
when d = 1372 yards, and the angle A = 56° 31'.
21. (i) Show that (sin^ + cos^)2 + (sin^-cos^)2 = 2.
(ii) Considering only values of A between 0° and 90°, find the
value of A when sin^ cos J. has its greatest value, and show that
the same value of A gives the greatest value of sin ^ + cos ^.
22. Let AD bisect the angle ^ of a triangle ABC, and let it
meet BCin D -, show that BD sin B = CD sin C.
Hence show that BD.AC= DC. AB.
23. (i) Show geometrically that
sin {A + B) = sin ^ cos ^ + cos A sin B,
when A, B, and A + B are each less than 90°.
(ii) By means of this formula, and in view of the restrictions
under which it has been obtained, show that
sin 464° = sin 153° cos 311° + cos 153° sin 311°.
24. Find tan 6 and x in terms of a and b from the equations
a sin ^ + & cos 6 = ^x,
« cos ^ — 2 & sin ^ = 2 37.
25. An angle is made to increase gradually from 0° to 360° ;
state briefly how the values of its sine and of its cosine change
during the increase of the angle.
26. Calculate the values of A between 0° and 360° for which
tan J. -2 cot ^ = 1.
27. A and B are two milestones on a straight road running due
East across a horizontal plane, C an object on the plane. The
bearings of C as viewed from A and B are 35° North of East,
and 55° North of West respectively. Find, to the nearest foot,
(1) the distance of C from A, (2) the distance of C from the
nearest point of the road.
1916 M
178 REVISION EXAMPLES
28. Plot in relation to the same axis and origin the values of
tana; and 2 sin a; for the values 0°, 12=^30', 37° 30', 50°, 62° 30',
75° of X, draw the graphs of tan x and 2 sin x, and find from them
the values of x for which tana? = 2 sin a;. Give the general solu-
tion of the equation tan a? = 2 sin x.
29. Prove that (cos A + sin A) -h- (cos A - sin A) = tan {A + 45°).
30. Prove for a triangle in which the angle B is obtuse the
relation sin Bjh = sin C/c, and deduce the relation
tan ^{B-C} = {b- c)/{b + c) cot I A.
If & = 27*3 yards, c = IS'S yards, A = 48° 36'", find B and C.
31. Prove that in a triangle ?• = 4 J? sin ^ J. sin 1 5 sin | C
ABC is a triangle ; B'C is drawn through A parallel to BC, A'C
through C perpendicular to AC, and A'B' through B perpendicular
to AB. Prove that the area of the triangle A'B'C is
\ or cos- {B — C)-ir cos B cos C sin A.
32. (i) Find sin ^ + sin 5 in terms of functions of half the sum
and of half the difference of the angles A and B.
(ii) If ^ + 5 is between 90° and 180°, find under what circum-
stances tan J. + tan 5 will be negative.
33. Find to the nearest minute the angle of a regular polygon of
17 sides.
What angles less than 360° satisfy the equation
2cos2^-f llsin^-7 = 0?
34. Prove the identity
tan^^ coi^A _ 1 — 2 sin-.l cos-.l
1 + tan^^ 1 + cot^^ sin J. cos ^
35. Assuming the formula a^ = h'^ + c^ — 2 he co^ A, establish a
formula for tan^^ in terms of the sides of the triangle, and find
the greatest angle of the triangle whose sides are 13, 14, 15.
36. Prove that for any triangle ABC
rt/sin A = &/sin B = r/sin C.
U B = 39° 17', a = 4*2, and b = 3-5, solve the triangle fully ;
draw a figure to illustrate your solution.
37. The angles of elevation of a vertical pole from two points on
a horizontal line passing through its liase and 6 feet apart are
a and /3 ; prove that the height of the pole is b/[cot 0( — cot^) feet.
KEVISION EXAMPLES 179
38. From a point on a horizontal plane passing through the foot
of a tower the angles of elevation of the top and bottom of
a flagstaff 20 feet high, placed vertically at the summit of the
tovrer, are 51*2^ and 47"3''. Find the height of the tower.
39. Prove that (i) sin (A + B) = sin AcosB + cos A sin B ;
(ii) cos2^{l + tan2^)= l-tan-.4.
Use (ii) to find the value of tan 15°.
40. Reduce the fraction a -r (cos^^ — sin-^; to a form suitable
for logarithmic calculation, and perform the calculation when
« = 10, A = 29' 55', and B = 15" 5'.
41. Prove that sin*^ .1 + cos'^ .^ = 1 for all values of A less
than 180°.
A and B are each less than 180^ sin ^ = '3900, smB= -9208,
find four possible values of A + B.
42. Find from your tables the value to two decimal places of
the expression sin^ + sin2^, when B is 10^ 20°, 30°, ... 90', and
from these draw a graph of the expression on a suitable scale.
43. In a triangle ABC prove that
(i) 2bc cos A = b- + c^- a-;
(ii) cosM + cos2-B + cos^co35 = f, if C=60°.
44. In a triangle a = 12*76, h = 10-87, c = 8-37, find C.
45. Show how to find the distance between two visible but
inaccessible objects.
46. In any triangle ABC show that four times the area equals
{(("^ + b- + c^) -^ (cotan A + cotan B + cotan C).
Show also that when C is a right angle this expression reduces
to 2a&.
47. Prove the identities :
(ij 1/sin 2.4 = 1/tan A - 1/tan 2.4 = tan A + 1/tan 2^ ;
,.., . ^ , sin ^^ + sin 4 J.
(ii) tan ?>A = ^^-p- T-, •
^ ' co8 2J. + cos4^
48. What is the meaning of tan-* x ?
Prove that tan"' a? + tan"'?/ = id,n-'^[{x + y) -f {\-xy)].
Prove that 45° is one value of tan~^ -| + tan"' |- + tan"* j^jj.
49. Prove that (i) secM = 1 4- tan^ ^ ;
(ii) cosec^-cot^ = tan^^.
M 2
180 REVISION EXAMPLES
50. Construct an angle whose sine is 0'76. From your figure
obtain the value of the cosine of the angle.
51. On squared paper draw graphs of tan^ and cot ^ between
^ = 10° and 6 = 80". From the graph, or otherwise, find angles
which satisfy the equation tan 6 + cot ^ = 3.
52. Let D be the point in which one of the escribed circles
touches the side BC of a triangle ABC. If the sides a, h, c of
the triangle are given, find expressions for the radius of that
circle and for BD and CD.
53. A tree which grows at a point A on the north bank of
a river is observed from the points B and C on the south bank.
The distance BC is 200 metres, the angle ABC is 46° 80', and the
angle ACB is 58° 20'. Calculate the distance of A from the
straight line BC.
54. Prove the formula sin|J. = \/{s — h) (s — c) -r he.
If, in a triangle ABC, 2b = a + c, prove that
sin|5 = 2 sin 1^ sin I C
55. Find the angles B and C and the radius of the circum-
scribed circle of a triangle ABC in which A = 32° 42', a = 36,
6 = 44.
56. State De Moivre's Theorem, and, assuming it for integral
indices, prove it for fractional indices.
Write down all the values of (>/— l)o.
57. U A is an obtuse angle whose sine is i%, find the values of
cos^ and tan^.
58. (i) Show, by drawing graphs of the two expressions sinu;
and cos (.r + 90°), that sina;= —cos (a; + 90°).
(ii) If sin a? = ^ \/2, find a formula which gives all the values
of .r which satisfy the equation.
59. Prove that in a triangle
(i) tan ^B = ^{s -c}{s-a}-^s {s — b) ;
(ii) b cos B + c cos C = a cos {B — C).
EEVISION EXAMPLES 181
60. If two sides of a triangle and the angle opposite one of them
are given, show how to solve the triangle, and discuss by the aid of
a figure all the cases that can arise.
One side of a triangle is 20 inches long, the opposite angle is
34° 42' ; another side is 30"41 inches. Find the sides and angles of
the two possible triangles.
61. Assuming the formulae for the sine and cosine of half an
angle of a triangle in terms of the sides, prove that
(i) r = ^/{s — a){s — b) ( s — c) -r s ;
(ii) Z?= a/2 sin ^.
62. I observe the altitude of an airship to be 35°, and that of the
sun, which is in the same vertical plane as my eye and the airship,
to be 40^ The shadow of the airship falls on a tree on the same
level as my eye and 500 feet in front of me. Find the height of
the airship.
63. In any triangle prove that
sin ^ - sin ^ + sin C = 4sin ^vi cosi^sin^C
Assuming the formula for expanding ti\.n{A + B), find expressions
for tan 2^ and tan 3 J. in terms of tan J..
64. Make an angle AOC and bisect it by the line OB. From
any point A in OA draw ABC perpendicular to OB, meeting
OB, OC in the points B and C respectively, and draw AX perpen-
dicular to OC. Use this figure to prove that
(i) sin2J.<2sin^; (ii) tan 2 xt > 2 tan .4.
65. Prove that sin^^ + cosM = 1.
Having given that the sine of an angle is '56, calculate its cosine.
66. Show how to construct an angle whose sine is "6.
Find a value of x which satisfies the equation
4 sin ic + 3 cos a; = 1.
67. Given two sides of a triangle and the included angle, show
how to find the remaining side and the other angles. Prove such
formulae as you require.
If a = 1097 feet, b = 781 feet, C = 31° 30', find c to the nearest
foot.
68. A ship is sailing at the rate of 7 miles an hour. A man
walks forward across the deck at the rate of 4 miles an hour
182 KEVISION EXAMPLES
relative to the deck, in a direction inclined to the keel at an angle
of 60°. Find the direction of his actual motion in space.
69. Prove the formula cos {A-B) = cos AcosB + sin A sin B.
Show that if xy = a'^ + 1 then
cot~^ {a + x) + cot"' (« + y) = cot"' a.
70. Find an expression for cos{Oi. + ^ + y) in terms of sines and
cosines of Of, 0, and y.
Prove the identity
cos OC cos 0 + 7) + COS /3 cos {y + OC) + COS y COS {01. + ^)
= COS (ex + /3 + -y) -h 2 COS OC COS /3 cos y.
71. At what angle must forces of 4 dynes and 5 dynes act so
that their resultant may be a force of 6 dynes ?
72. If B be the circular measure of an angle, prove that, as 6 is
indefinitely diminished, the ratios 6 : sin d, 6 : tan 6 approach to
the limit unity.
A man standing beside one milestone on a straight road observes
that the foot of the next milestone is on a level with his eyes, and
that its height subtends an angle of 2' 55". Find the approximate
height of that milestone.
73. Write down the values of sin 36° and cos 36° as given by
your tables. Calculate the sum of the squares of these numbers
to six decimal places, and explain why the result differs from unity.
74. Give definitions of the tangent and cotangent of an angle
which is greater than 90° and less than 180°.
Prove that (i) tan (180 - ^) = - tan ^ ;
(ii) tan(90 + <9)=-cot^.
75. In any triangle prove that a/sin A = hj&m B = c/sin C.
If BC be 25 inches, and CA be 30 inches, and if the angle ABC
be twice the angle CAB, find the angles of the triangle ABC, and
show that the length of the third side is 11 inches.
76. F, Q, R are three villages. P lies 7 miles to the North-East
of Q, and Q lies 11^ miles to the North-West of B. Find the
distance and bearing of 7' from B.
11. A point is moving with velocity 50 feet per second in
a direction 60° North of East. Find the resolved parts of the
velocity in directions East and North.
EEVISION EXAMPLES 183
78. A man has before him on a level plane a conical hill of
vertical angle 90^. Stationing himself at some distance from its
foot he observes the angle of elevation Oc of an object which he
knows to be half-way up to the summit. Show that the part
of the hill above the object subtends at his eye an angle
, tan0c(l-tan3()
tan~ — •
1 + tan DC (1 + 2 tan y)
79. The latitude of London is 51' N., and the radius of the
Earth 4000 miles. How far is London from the Equator measured
along the Earth's surface, and how far from the Earth's axis ?
80. Prove that sin A + sinB = 2 sin l (A + B) cos I {A - B).
Show that sin 10' + sin 20' + sin 40' + sin 50' = sin 70' + sin 80°.
MISCELLANEOUS PKOBLEMS
(The following examples are taken from recent Army Entrance
and Civil Service Papers.)
1. I take measurements to determine the air space of a rect-
angular hall : length 18*4 metres, breadth 11*8 metres, inclination
to floor of diagonal of side wall 31 "8°, of diagonal of end wall 44°.
Calculate the air space.
More measurements were taken than were necessary. Check
the measurements by deducing one of them from the other three.
2. The ancient Greeks measured the latitude of a place by
setting up a vertical rod and comparing its length with the
length of its shadow. Supposing observations taken at mid-day
at the equinox (when the sun is vertical at the equator) to give
^ as the ratio of the rod to shadow at Alexandria, and ^ as the
ratio at Carthage, find the latitude of each place.
3. The following method of determining the horizontal distance
PR, and the difference of level QR between two points P and Q,
is often used. A rod with fixed marks A, B on it is held vertical
at Q, and the elevations of these points, viz. ACD = (X, BCD = jS,
are read by a telescope and divided circle at C, the axis of the
telescope being a distance CP=a above the ground at F. If
QA = h, and AB = s, write down expressions for PB and QB.
Find PB and QB when 0( = 6° 10', 8 = 1° 36', the values of a, h,
and s being 5 feet, 2} feet, and 5 feet respectively.
4. Three balls, 5 cm. in diameter, lie on a floor in contact, and
a fourth equal ball is placed on them. Find the height of the
centre of the fourth ball above the i)lane of the other three
centres. Find also the inclination to the vertical of any line
that touches both the top ball and one of the lower balls.
5. The curved surface of a right circular cone whose semi-
vertical angle is 45° is made by cutting out a sector from a circular
sheet of copper, the diameter of the sheet being 5C cm. Deter-
mine the angle of the required sector.
MISCELLANEOUS PROBLEMS 185
6. If tangents be drawn to the inscribed circle of a tiiangle
parallel to tbe sides of the triangle, show that the areas of the
triangles cut off by these tangents are inversely proportional
to the areas of the corresponding escribed circles.
7. A rod BC, of length 5*8 cm., rotates about B. Another rod CA,
of length 8*6 cm,, has one end C hinged to the first rod, while the
other end A slides along the line BO. By drawing the rods in
various positions, find how the length o^ BA varies as the angle B
increases ; and show BA as a function of angle 5 in a graph for
one revolution of BC, showing the actual length of BA and repre-
senting 30° by 1 cm.
Write down an equation connecting the angle B and the lengths
of the three sides of the triangle ABC. Solve the equation to find
the length of BA when angle B is 35"".
8. The extreme range of the guns of a fort is 8000 metres.
A ship, 14000 metres distant, sailing due East at 24 kilometres an
hour, notices the bearings of the fort to be 20° 30' North of East.
Find, to the nearest minute, when the ship will first come within
range of the guns.
9. The face of a building is 136 feet long. A photographer
wants to take the building from a point at which the face subtends
an angle of 37°, and for this purpose he starts off from one corner
of the building in a direction making an angle of 127° with the
face in question. Find by calculation the distance from the corner
at which he must take the photograph. Calculate the area of
ground in the triangular space between his position and the face of
the building.
10. From the top of a telephone pole three wires radiate in
a horizontal plane. One wire, A, exerts a tension of 100 lb.
weight ; the next, B, makes an angle of 90° with A and exerts
a tension of 80 lb. weight ; the third, C, makes an angle of 35°
with B and an angle of 125° with A, and exerts a tension of
90 lb. weight. It is required to equilibrate the three tensions by
means of a fourth wire. Find its direction and tension.
11. A man passing along a straight road measures the angle
between the direction of his advance and a line drawn to a house
on his left. At a certain moment the angle is 36° 21'. He walks
on 1500 yards and finds that the angle between the same direction
186 MISCELLANEOUS PROBLEMS
and the line to the house is now 125° 36'. Find the distance of
the house from the road.
12. Plot a curve giving the sum of 4sin^ and 3 sin 2^ from
S = 0° to 6 = 180°, and read off the angles at which the greatest
and the least values respectively of this sum occur. For the angle
use 1 cm. to represent 10 degrees, and for 4 sin ^ + 3 sin 2^ use
1 cm. to represent unity. Also estimate the slope of the curve
when 6 = 90° and when d = 135°.
13. A, B, and C are three buoys marking the corners of a
triangular yacht racecourse round an island. The angles A, B,
and C of the triangle ABC are found to be 75°, 63°, and 42°
respectively. P is a flagstaff on the island, from which A and B
can be seen, and the distances of P from A and B are found by
a range-finder to be 650 yards and 585 yards resi3ectively, and the
angle APB to be 187°. Calculate the length of one lap of the
course.
14. Draw an angle XOP of 30°, making OP 2" long : through P
draw PQ parallel to OX and in the same direction : produce XO to
X\ making OX' = OP, and join X'P: cut off PQ = PX. Join OQ
and measure the angle XOQ carefully. Now denote XOQ by 0,
XOP by 6, and OP by e, and write down an expression for the
length PQ. Deduce an equation for 6 and 0, and solve it for tan (p.
Use your tables to evaluate (p when 6 = 30°, and compare your
result with the measured value. It is said that the given con-
struction trisects an angle. What is the percentage error for 30° ?
15. In running a survey the lengths of a series of lines are
measured, and the angle each line makes with the direction of
magnetic Noiih is measured by a theodolite. The data booked are
given in the table below : —
Line. Length in feet. Bearing.
AB 433 29° 15'
BC 521 89° 12'
CD 352 182° 38'
DE 417 233° 25'
The angles are measured clockwise from the magnetic North
direction.
By an error the measured length of the closing line LA of the
survey was not recorded, nor its bearing; from the data given in
the table calculate these missing data.
MISCELLANEOUS PROBLEMS 187
16. AOB and COD are two straight roads crossing one another
at an angle of 57". A motor-car, travelling at the rate of 18 miles
an hour along AOB, is 1500 yards from 0, when a man, walking at
the rate of three miles an hour along COD, is a quarter of a mile
from 0; car and man are both approaching 0. Find graphically
the motion of the car relative to the man. Hence find the least
distance between the car and the man, and when they are at this
distance from one another.
17. In a triangle a = 10 cm., b = l cm., one angle is 95°.
There being no restriction as to which angle of the triangle is 95",
discuss how many distinct triangles can be made. Select any one
case, and for this case calculate the remaining sides and angles.
18. X and Y are two fixed points in a straight line, P a point
which so moves that cosPA^Z+cos PYX = Jc (a constant). Prove
the accuracy of the following construction for obtaining the locus
of P: With X and Y as centres describe circles of radius XY/k.
From any point X in XY draw XAB perpendicular to XF cutting
the former circle in A and the latter in B. Draw XA and YB,
intersecting in P. Then P is a point on the locus.
19. A candle, C, is placed on the floor at a distance r from
a point 0 on a wall, and at the same level as the candle-flame,
and the angle which OC makes with a perpendicular to the wall
at 0, is 6. The illumination received on the wall at 0 from the
candle is known to be equal to Acosd/f- where ^ is a constant.
If the candle be moved about on the floor in such a way that this
illumination remains constant, plot on a diagram the curve
described by the candle-flame.
20. Two small islands are 5 miles apart, and there is known to
be a rock distant 3 miles from each. A ship is in such a position
that the islands subtend an angle of 66^ at the ship. Calculate,
to the nearest hundredth of a mile, her least possible distance
from the rock.
21. Find by means of a graph two acute angles 6 for which
5sin2^ = 3sin(9 + 2-5.
Find also the greatest value of 5 sin 2^ -3 sin ^ when 6 is an
acute angle, and the angle to which this value corresponds.
22. The elevation of an aeroplane which is flying horizontally
on a flxed course at a height of 150 feet is taken at two instants
188 MISCELLANEOUS PROBLEMS
at an interval of 20 sees. At the first observation the elevation
is 10° and the bearing is clue North, and at the second the
elevation is 6|° and the bearing is N. 35° E. Find the course and
speed of the aeroplane.
23. The strength of an electric current C is obtained from the
formula C = k tan 6 where 6 is the angle read off in degrees on
an instrument, and k is n constant. If an observer makes an error
of 8^ in reading the angle ^, prove that the value of C thus
obtained will be wrong by an amount equal to J^tt Ccosec2^S^.
Hence find the error per cent, in C produced by making a mistake
of -j^j degree when 6 is 60°.
What value of B is likely to produce the smallest error in the
value of C ?
24. If P denote the pressure of wind in lb. per square foot on
a plane surface at right angles to the direction of the wind, and
2) denote the normal pressure of wind in lb. per square foot on
a plane surface inclined at an angle 6 to the direction of the
wind, the following formulae are used to determine the ratio p : P.
(i)iV-P=(sin^)^'^^'-''''^-^;
(ii) p/P = 2 sin 6/(1 + sin^ S).
Compare the values of p/P given by these formulae for the
values 10° and 50° of d.
25. A man walks due W. from a point ^ up a straight path
inclined at 10° to the horizon. After walking 2 miles he reaches
B, and turns up another straight path to the NE., sloping 15°
upwards. He reaches C after walking one mile from B. What is
the distance in a straight line from C to ^ ? What is the height
of C above the level of A ? Taking the face of the hill ABC
as a plane surface, what is the greatest slope ?
26. A flagstaff stands vertically on horizontal ground. Four
ropes, each 56 feet long, are stretched from a point in the flagstaff',
50 feet above the ground, to four pegs in the ground, arranged
at the corners of a square. Calculate the angle between two
adjoining ropes.
27. Q is the centre of a circle of radius 10 cm., and QO is
a radius. The seven points ABC ... lie on the circumference and
the angles OQA, OQB, OQC... have the values 10", 20", 30°... 70°.
Find by drawing or calculation the lengths of the chords OA,
OB, OC..., and tabulate the results.
MISCELLANEOUS PROBLEMS 189
Draw a graph to give the length of chord of the circle in terms
of the angle which it subtends at the centre (for angles up to 70").
Show the chord's actual size, and represent 4 degrees by 1 cm.
From jour graph find the length of the chord which subtends
an angle of 48°. Make a triangle having one side of this length,
and the other two sides 10 cm. long, and therefore having an
angle of 48''.
Check the accuracy of your drawing by measuring this angle.
28. A square made of jointed rods each 4 inches long is
deformed into a rhombus having half the area of the square-
Calculate the lengths of the diagonals of the resulting figure and
check by drawing. If it is part of a lattice-work, the original
height of which is 6 times the diagonal of one of these squares,
find by calculation how much the height of the lattice-work could
be increased if each square were reduced to half its area.
29. A straight rod AB, 3 feet 9 inches long, is held under water,
A being 2 feet 6 inches and B 9 inches below the surface. Calcu-
late (a) the distance below the surface of a point C on the stick
which is 12 inches from A, (h) the angle which the stick makes
with the surface of the water.
If a parallelogram is held under water, show that in every
position the sum of the depths of the 4 corners is 4 times the
depth of the point of intersection of the diagonals.
30. If a closed loop of thread is placed on a soap-film that
covers a ring of wire, and the film within the loop is joierced, the
film outside takes up as small an area as possible and thus pulls
the thread at A into a circle. Calculate the diameter and the
area of the circle formed by the thread if length of thread forming
the loop is 6 cm.
If the ends BC of the thread are attached to the ring, and the
film on one side of the thread is pierced, the thread again becomes
a circular arc. If the thread BC is 6 cm. long, and the angle
it subtends at the centre of the circle of which it forms an arc
is 120°, calculate the length of the chord BC,
EXAMINATION PAPERS
OXFOKD AND CAMBRIDGE SCHOOLS' EXAMINA-
TION BOARD.
School Certificate, 1910.
1. Define the tangent of an angle.
Construct an acute angle whose sine is "6, and find its cosine and
cotangent.
2. Prove that cos (180 -a) = -cos Of.
Arrange the angles Oi, /3, y in order of magnitude, if
sin 3^ = -8211, cosi3 = -7738, tany= -O'GlOi,
the angles being positive and each less than 180°.
3. What is the length of the shadow of a man, 5 feet 8 inches
high, cast by the sun when its altitude is 55° 30' ?
4. Draw the graph of 10 + 10 cos 2a; for values of x between
0° and 60°. Find a value of x to satisfy the equation
a; = 10 + 10 cos 2 0^°.
[Take one-tenth of an inch as unit along both axes.]
5. Prove that in any triangle sin A/a = sin B/b.
U A = 63°, B = 49°, a = 50 inches, find b to the nearest tenth
of an inch.
6. Prove that
... cos ^ + sin ^ cos ^- sin ^_ 2
^^ cos ^ -sin ^ cos(^ + sin^ ~ l-2sin-^'
(ii) (sec 6 + tan 3) (cosec 6 ~ cot 6) = (cosec ^ + 1) (sec ^ - 1).
7. If 2 sin ^ + 5 cos ^ = 5, prove that tan^ = 0 or 20/21.
8. Prove that sin {A — B) = sin A cos B — cos A sin B, where A
and B are both acute angles and A is greater than B.
^ , , , sin 5 ^ + sin ^ , _ » „ ^
Prove that . ^ , ; — - = 1 + 2 cos 2 -4. .
sind^ —sin J.
9. Show that in any triangle ABC
b + c _ coslJB-C)
a sin h A
If fc + c = 24'8 cm., a = ir89 cm.,''yl = 39°, find B and C.
EXAMINATION PAPERS 191
10. A lighthouse is observed from a ship which is steaming due
N. to bear 62= W. of N. ; after the ship has sailed 10 miles the
lighthouse is observed to bear 40° W. of S. Calculate the distance
of the ship from the lighthouse when it was nearest to it.
Higher Certificate, 1910.
Part I.
1. Give a definition of cos^ that holds for all angles from 0' to
180°. Show that cos ( 180 -d)= - cos 0.
2. Show that secM = 1 + tan2^.
Draw the graph of 1 +sin dx°, where x lies between 0° and 60°.
3. Construct an acute angle whose cotangent is 2, an obtuse
angle whose sine is 3, and an obtuse angle whose secant is —3*5.
Measure these angles as accurately as you can with the protractor,
and verify your results by means of tables.
4. (i) Verify that 30'', 45°, and 60'' are solutions of the equation
sin3a; + cos3rr = 2 cos 2a;.
(ii) Show that
(cosec^ + sec^j^ + (cosec A - sec A)^ = 2 cosec'^ (3 sec^A — 2).
5. Show that in an obtuse-angled triangle
sin A/a = sin B/h = sin C/c.
A man observes that the angular elevation of the foot of a
tower on a distant hillside is OC, and that the angular elevation
of the top of the tower is /3, and he knows that the height of
the tower is h feet. Show that his horizontal distance from the
tower is ^ cos (X cos /3 cosec (/3 — 3ii).
Part II.
6. Draw the graph of cot.r between the values —180 and + 180
of X, taking the unit of x to be -^q inch and the unit of y to
be one inch.
Find an acute angle to satisfy the equation x = 60 cot a;''.
7. Show that sin (^-5) = sin .4 cos ^- sin 5 cos. 1, taking A
and B to be acute angles of which A is the greater.
If tana; = A; tan (^ -a;), show that
{k-l)smA= {k-hl)3in{2x-A).
Use this result and tables to solve the equation
tana: = 2 tan (50° -a;).
192 EXAMINATION PAPERS
8. In the triangle in which a = 72 feet, B = 40°, and C = 55°,
find c.
9. Find in terms of a, b, and c the radius of the circle escribed to
the side BC of the triangle ABC.
If Jj is the centre of this circle, show that
aAI,''-hBI,^-cCI,' = abc.
10. AB is a diameter of a circle whose centre is 0 ; on AB an
equilateral triangle ABC is described, and a point D is taken in AB
such that 1BD = 2AB ; CD is produced beyond D to meet the circle
at E. Show that tan ADC = 7/^/3 and that sin OED = 3/^52.
Hence, or otherwise, show that the error made in taking the arc
BE to be one-seventh of the circumference of the circle is less than
'2 per cent.
Part III was beyond the scope of this book.
OXFORD LOCAL EXAMINATIONS.
JUNIOE. 1910.
1. (i) Find the sine of 60° ;
(ii) If -4 is an acute angle, and cos^ = ^, find the value of
4 tan ^ + 5 sin A.
2. P and Q are points on a straight stretch of a river bank and R
is a point on the other bank. If cot PQR = '32, cot ^Pi^ = '43,
and the length of PQ is 15 yards, find the breadth of the river.
3. Draw the graph of sin (45° + 2a:) between x = 0 and 180^
4. U A, B, A — B are all positive acute angles, prove that
cos {A — B) — cos AcosB + sin A sin B.
5. (i) A, B, C are the angles of a triangle; if tan^ = ^ and
tan ^ = ^, find the angle C.
,.., „ ,- , cos5u4 + cos3^ , .
(u) Prove that - — p— : — ir—r = cot A.
^ ^ sin 5 ^ - sm 3 ^
6. Solve the equation cos2^ + sin^ =?= 1,
EXAMINxVTlON PAPERS 193
7. Prove for any triangle that
(i) rt/sin A = b/tiin B = c/sin C ;
( ii) {b + c) cos A + {c + a) cos ^ + (« + 6j cos C = a + 6 -I- c.
8. Find the angles A and i? of a triangle ABC in which « = 13,
/> = 14, c = 15, having given:
Iog2 = -o010, log? = -8151,
Z tan 26= 34' = 9-6990,
Z tan 29' 44' = 9-7569.
Senior. 1910.
1. Find the tangent of 30"".
Using the values of tan 30° and tan 45'', prove that
tan 75° = 2 + -\/3.
2. A man on a straight level road observed two objects Pand Q
{P being the nearer) in a horizontal straight line inclined to the
direction of the road at an angle 0(. If tan Oi = -75, FQ = 400 yards,
and the shortest distance of F from the road is 180 yards, what is
the shortest distance of Q from the road ?
3. Prove that cos3^ = 4cos^.4-3cos.l. Find sin 18°.
4. If ABC is a triangle in which 6 = c = 5 inches and
a = 8 inches, find the values of tan A and tan B.
5. Prove that
cos2 j^ ^ cos2 B = sin2 {A + B) + 2 cos A cos B cos {A + B).
6. Prove that in any triangle c = (« + &) sin^, where
cos ^ = 2 ^/ab cos | C/{a + &).
In a triangle ABC, a = 36 feet, & = 4 feet, C = 55°. Using
the above formula, find the third side, having given
log 6= -7782, Z cos 57° 51' =9-7261,
Zcos27°30' = 9-9479, sin 57° 51'= '8467.
7. Find the radius of the circle inscribed in the triangle ABC.
C is the centre of a circle of diameter d, and A, B are two
points on the circumference of the circle. If I is the length of the
chord AB and S is the diameter of the circle which touches
CA, CB and also the arc AB at its middle point, prove that
1/5 = 1/(^+1//.
1216 N
194 EXAMINATION PAPERS
CAMBRIDGE LOCAL EXAMINATION.
Junior. 1909.
1. Define the sine of an angle. What are the greatest and least
values which the sine of an angle can have ?
Prove that sin^ = cos A x tan J., and that
sin A sin B cot B = cos A cos B tan A.
2. Construct an angle vrhose tangent is 1*45, and measure it
with a protractor. Verify your results with the help of the tables.
3. Prove that
(i) sin A = tan A/{ -/l + tanMj ; (ii) cos (90° + A)= - sin A.
4. Find by drawing graphs of sin^ and sin 2^ for what value
of A, less than 90°, 2 sin ^ - sin 2 ^ = 1.
5. A vertical post casts a shadow 15 feet long when the altitude
of the sun is 50° ; calculate the length of the shadow when the
altitude of the sun is 32°,
6. Prove that sin ^ + sin i? = 2 sin 1{A + B) cos ^{A- B), and
that tan 2 J = 2 tan A/{1 - tan-^).
Show that sin ^ - 3 sin 3 ^ + 3 sin 5 ^ - sin 7 ^ = 8 sin^^ cos 4 A.
7. Prove that, in any triangle ABC, acos,B+ h cos A = c.
Show also that ftan A + tan B) (tan A — cot C) = sec"^.
8. Show how to solve a triangle when three sides are given.
Find the greatest angle of the triangle whose sides are 5'2 inches,
77 inches, and 9*1 inches.
Senior. 1909.
1. Show that the ratio of the circumference to the diameter of
a circle is an invariable quantity.
Find to an inch the diameter of a wheel which makes 400
revolutions in rolling along a track one mile long.
2. Any positive proper fraction being given, show that there are
two angles, one acute and the other obtuse, such that the sine of
either is equal to this fraction.
If the fraction is |, use the tables to find the angles, and the
cosine and tangent of each.
3. Find by aid of the tables the values of sin it- -tan 2x for the
values 0°, 10°, 20°, 30°, 45°, 60° of x.
Make a graph to give the values of sin it' — tan 2 jc from ^=0 to
X = 60°.
EXAMINATION PAPERS 195
4. Show that sin A + sin 5 = 2 sin i (.1 + B) cos I (A - B).
Prove also that
(i) tan2^ = (l-cos2.4)^(l + cos2^);
(ii) sin 55° sin 15° - sin 50° sin 10° - sin 65° sin 5° = 0.
5. Find the greatest angle of a triangle whose sides are 15, 21,
28 inches in length.
Show that in any triangle
a'^sin(^-C) &-sin(C-^) c=^sin(.l-^)
h + c c-^a a + b
6. Find an expression for the radius of the inscribed circle of
a given triangle.
Determine to one place of decimals the length of the radii of the
inscribed circle, and of the escribed circle opposite the greatest
angle of the triangle referred to in Question 5.
Questions 7 and 8 were outside the scope of this book.
(The two following questions may be taken instead of 7 and 8, but
considerably lower marks will be assigned to them.)
A. Show that if .4, B, Care the angles of a triangle,
tan A + tan B + tan C = tan A tan B tan C.
Show also that
tanA5tan^C + tan*CtanA.i + tani^tan|^= 1.
B. Solve the equation a cos d + h sin 6 = c.
Find all the solutions of sin ^ sin 3 ^ = sin 5 6 sin 7 0.
COLLEGE OF PRECEPTORS.
Cheistmas, 1910.
H Hours.
[Four-place tables of logarithms and of natural functions and
square-ruled paper are provided. All diagrams should be
drawn as accurately as possible.]
Part I.
1. Define a radian, and find its magnitude in degrees to two
places of decimals (tt = -->').
If the angle of an equilateral triangle were taken to be the unit
angle, what would be the measure of a radian to two places of
decimals ?
196 EXAMINATION PAPERS
2. Define the sine and tangent of an acute angle. Prove that
sin'^^ + cos'^^ = 1.
If tan^ = i\^, find the value of cos ^-8 sin ^.
3. Find, geometrically, tan 30°.
If ^ = 30°, B = 45°, C = 60°, D = 90°, find the value of:
(i) sin ^ cos 5 — sin BcosA;
(ii) (tan^^ - cosec^^) / (cot C + cos D).
4. Use logarithms to find as nearly as possible the values of:
(i) 3-142 X •9342/-00532 ; (ii) >/562'3/-00^984.
5. Solve, using the tables, the triangle in which C = 90°,
« = 654, ^ = 38°45'.
Part II.
6. Find all the positive values of 0, less than 360°, which satisfy
the equations :
(i) cos2^-sin"^ = 0;
(ii) 4sin2^cos^^-sin^<9= |.
Which of the following statements are possible ?
(i) tan^= -2; (ii) sin^ = f.
7. Write down, without proof, the expansions of sin (.1 - B),
cos{A-B).
Find the value of ianA — B in terms of tan J, tan 2?.
If tan^l = ^, tan J5 = -/, find tan (.4 + ^).
8. Prove that, in a triangle, a^ = &- + r-26fcos.4 when the
angle A is (i) acute, (ii) obtuse.
Deduce that tan lA = \/ (s -h) {s - c) -^ s [s - a).
Find the greatest angle in the triangle whose sides arc 256,
389, 401.
9. AB is a horizontal straight line. A vertical straight line is
drawn from B upwards, and in it two points P, Q are taken, such
that BQ is five times BP. If the angle BAP is 30'\ calculate
UmPAQ.
EXAMINATION PAPKRS 107
LEAVING CERTIFICATE EXAMINATION
(SCOTLAND). 1910.
1. Explain the circular measurement of angles.
Express 30°, 50°, 166° 40' in radians.
Express '0187 radian in degrees, minutes, and seconds, taking
77 = 3-1416.
2. Taking a horizontal inch to represent 10° and 5 vertical inches
to represent the unit of length, plot, with the help of your tables,
the values of tan 6 when ^ = 0, 10°, 20°, 30°, 40°, 50°.
Plot also the values of sin 6 for the same angles, join both series
of points by smooth curves, and thus find a graphic solution of the
equation 5 (tan ^- sin ^) = 1.
3. State the relation which exists between the sine and cosine of
any angle.
Use this relation to find, and express in a diagram, all the values
of 0^, less than 180°, which satisfy the equation
5 sin a + 6 cos'' 3^ = 7.
Either, 4 a. A man walked 5 miles due North and then walked
6 miles in a direction 27° East of North. Find by a figure drawn
to scale how far he now is from his starting-point, and in what
direction he should have originally started in order to go straight
to his final position. Verify your results by calculation.
Or, 4 b. The sides of a parallelogram are 2 inches and 3 inches
in length, and its area is 3? square inches. Find by a diagram the
sizes of its angles and the length of its longer diagonal. Verif}-
your results by calculation.
Either, 6 a. Draw a circle of radius 2 inches, and inscribe in it
a triangle ABC, such that ZB = .34°, ZC= 73°.
Measure the lengths of the sides as nearly as possible.
Calculate with the help of the tables the lengths of the sides to
the nearest hundredth of an inch, and thus test the correctness of
your di awing.
Or, 5 b. State and prove the formula which gives tan(yl + i?) in
terms of tan A and tan B.
Apply this formula to find expressions for tan 2^, tan3yl, and
tan 5 ^ in terms of tan A.
198 EXAMINATION PAPERS
INTERMEDIATE EXAMINATION (IRELAND).
Middle Grade (Pass). 1910.
1. Prove that sin^^ + cos^^ = 1, where A is an obtuse angle.
2. Find the value of the expression cosec^-gcot^, if sinyl = ^^,
when A is acute, and when A is obtuse.
3. Prove the identity (1 - tanM) -r- (2 cosM - 1) = secM.
4. In a triangle C = 90°, c = 65°, tan A = '28. Find a and h
each to two decimal places.
5. In a triangle a = b \/3, /; = 11, C = 150°. Find c and cos^.
6. In a triangle B = 45°, fc = 20, c = 4. Find sin C, and prove
that the perpendicular from A on BC divides BC into two
segments one of which is seven times the other.
7. Prove that the length of the perpendicular from the vertex A
of a triangle on the opposite side BC is equal to «/(cot-B + cotC),
considering the cases when both angles^ are acute, when one is
right, and when one is obtuse.
8. Find the angles between 0° and 360° which satisfy the equation
6sin^-4 cosec^ + cot6^ = 0, being given cos 48° IT 23" = §.
Middle Grade (Honours). 1910.
1. Show by a graph the values of cosec A for values of A between
-90° and 360°.
2. If yl is an angle in the first quadrant, prove that
sin A + cos A 4- tan A + cot ^ > sec ^ + cosec A.
3. Prove the identity
3 (sin ^ - cos ^)H 0 (sin A + cos Af + 4 (sinM + cos^i) = 13.
4. The sides of a triangle are 37, 7, and 40. Find all the angles,
being given that eo? 69° 25' 48" = ?; ; .
5. In a triangle a = \/5, h = -y/lS, C = 45°. Find c, and prove
that cotyl = 2-v/f-l.
6. Find a solution between 180° and 270° of the equation
5(1+ sin .r) = — 3 cos .r,
being given cos 28° 4' 21" = 1;.
EXAMINATION PAPEES 199
7. Prove by drawing a line through B, making an angle x with
the side BC, or otherwifse, that in a triangle ABC,
c cos {B - x) + h COS {C + x) = acosx.
P is a point on the hypotenuse AB of a right-angled triangle
ABC. AP = x, PB = ij, PC = z. Find cos CPB interms of jr, y,
and z. Find the sides of the triangle when x = S — ^/d, y = ^/3 + 1 ,
z= ye.
Senior Grade (Pass). 1910.
1. Find the distance from the earth to the moon, assuming that
the moon's diameter, 2165 miles, subtends an angle of 31' lU" at
the earth.
2. Prove that tan \A = {\- cos ^)/sin A.
Find tan 15° and tan 22i" without using the tables.
3. Find x if cos-^;r-f cot"' 2 = ^tt.
4. Assuming the formulae for the sines of the sum and difference
of two angles, prove that
sin A-sinB = 2cosl{A + B] sin \ {A - B).
Find the corresponding expressions in factors for cos^ -cos B.
5. Find the solutions between 0' and 360^ of the equation
cot 2. r- 3 tan re = 3.
6. In a triangle a = 183, h = 247, C = Sr 40'. Find A and B.
7. In a triangle A ^ 54° 80', B = 69° 20', a = 341. Find h and c.
8. Prove that in a triangle
a cos B-bcosA= {a-- h'^)/c.
9. Prove that in a triangle rcot|^=s — a, where r is the
radius of the inscribed circle, and s the perimeter.
Senior Grade (Honours). 1910.
1. An arc 40 feet in length is taken on a circle whose radius
is 35 feet. Find, to the nearest inch, the length of the perpen-
dicular from the centre on the chord of this arc.
2. Prove the identity
cos5^/sin^ + sin5^/cos^ = 2 cosec2^-4 sin 2 .4.
3. If cos x + cos y + cos z + cos x cos y cos 2- = 0, prove that
tan I X tan J y tan Iz = ± 1 .
4. If X = cot~^ V'cos y — tan~' -y/cosy, prove that
y = 2tan~^ -y/sin.r.
5. In a triangle A = 35° 20', a = 127, h = 104. Fin<l B, C,
and c.
INDEX
INDEX
A.bscis.sa, 12.
Addition formulae, 155,
Altitude of sun, 89.
Ambiguous case, 120.
Angle, 25.
circular measure of, 29.
measurement of, 25.
negative, 26.
.»^f elevation, 88.
>of depression, 88.
Antilogarithm, 21.
Arc, length of, 28.
functions of, 38.
Area of triangle, 109.
Axes, 12.
Characteristic, 18.
Chord of angle, 15.
ircular functi
measure, 29.
Complementary angle, ratios of,
40.
Co-ordinates, 12.
cartesian, 12.
polar, 41.
Cosecant, 38.
Cosine, 38.
formula, 108,
cos (A±B), 155.
cos A ± cos B, 159.
Cotangent, 38.
Course, 94.
Coversine, 40.
Degree, 25.
De Moivre's theorem, 152.
Departure, 94.
Difference of Latitude, 94.
of two angles, ratios of, 15G.
of two sines, 159.
of two cosines, 159.
Dip of the horizon, 127.
Double angle, ratios of, 100, 162.
Elementary formulae, 49.
Eliminant, 51.
Elimination, 51.
Equations, solution of, 78.
solution by graphs, 17, 76.
general solution of, 79.
Functions, algebraical, 14.
circular, 40,
trigonometrical, 40.
Geometry enunciations, 9.
Grade, 27.
Gradient, 33.
Graphs, general treatment, 12.
of trigonometrical functions, 41,
163.
harder trigonometrical, 74.
Greek alphabet, 24.
Half-angle, ratios of, 101, 104, 121,
162.
Haversine, 40, 121.
Heights and distances, 88, 124.
-identities, 50, 64, 167, 168, 170.
Imaginary quantities, 149.
Indices, fractional, 17.
negative, 17.
Interpolation, 37.
Inverse functions, 45, 164.
Latitude, 32.
Logarithm, 18.
base of, 18.
use of, 22, 81.
tabular, 36.
Longitude, 32.
Mantissa, 19.
Mariner's compass, 31.
Minimum value, 74.
Navigation :
plane sailing, 94.
parallel sailing, 95,
middle latitude sailing, 96,
traverse sailing, 96.
204
INDEX
Negative angle, 26.
ratios of, 54.
direction, 13.
Oral examples, 34, Gl, 116, 123,
166.
Ordinate, 12.
Origin, 12.
Point, 32.
Points of the compass, 31.
Powers of trigonometrical ratios,
44.
Product of sines and cosines, 160.
Projection, definition, 139.
of area, 147.
propositions, 140.
length of, 40.
formulae, 40, 109.
Proportional parts, 71.
Quadrilateral, area of cyclic, 111.
Radian, 29.
Radius of inscribed circle, 114,168.
of circumcircle, 114.
of escribed circle, 114, 168.
Rate of change of function, 76.
Ratios, trigonometrical, 37.
of 90-^, 53.
of 180-^, 53.
of -^,54.
of 0° and 90^, 56.
of 30°, 45°, 60°, 57.
of 2 A, 100, 162.
of i A, 101, 162.
of 3 A, 164.
ofA + B, 154.
of A -B, 156.
ofSA, 164.
Rhumb line, 94.
Secant, 38.
Sector, area of, 29.
Sine, 38.
formula, 107.
graph, 41.
sin {A±B), 154.
sin A ± sin B, 159.
Slope of curve, 76.
Small angle, 59.
sine and tangent of, 60.
>* Solution of right-angled triangle,
88.
of triangles, 119.
of equations, 78, 168.
of equations by graphs, 17, 76.
Sum of two angles, ratios of,
154.
of two sines, 159.
of two cosines, 159.
Summation of a series, 169,
Supplementary angle, ratios of,
53.
Tabular logarithm, 36.
Tangent, 38.
graph, 41.
tan iA±B), 163.
Triangle formulae, 107.
solution of, 119.
Trigonometrical ratios :
general definition of, 38.
right-angled triangle, definition
of, 86.
Turning points, 74.
Vectors, 143.
addition of, 144.
resolution
Versine, 40.
Five-figure Logarithmic
and
Trigonometrical Tables
ARRANGED BV
W. E. PATERSON, M.A., B.Sc.
MATHEMATICAL MASTER, MERCERs' SCHOOL
AUTHOR OF 'SCHOOL ALGEBRa/ 'ELEMENTARY TRIGONO.METRY '
OXFORD: AT THE CLARENDON PRESS
LONDON: HENRY FROWDE, AMEN CORNER, E.C.
AXD AT
EDINBURGH, GLASGOW, NEW YORK, TORONTO
AND MELBOURNE
OXFORD : HOKACE HAET
TRINTER TO THE UNIVERSITY
These five-figure tables are intended to give results
correct to four figures ; tlie fifth figure in tlie answer may
be inaccurate.
The decimal point is printed before all the logarithms
of numbers ; it is hoped that this will obviate the common
mistake of reading off logarithms instead of antilogarithms,
and vice -versa.
The trigonometrical tables are arranged so that, at one
opening of the tables, all the functions of an angle may
be found on the left-hand page and their logarithms on
the right-hand page ; here again confusion is avoided.
The characteristics of the logarithmic functions are the
true characteristics ; no useful purpose is served by
increasing them by 10.
It should be noticed that, instead of dividing by a sine,
one may multiply by the cosecant, &c., and, similarly,
instead of subtracting the logarithm of a sine, one may add
the logarithm of the cosecant, &c. In many cases this
shortens calculation.
For quick reference the last page may be used, which
gives the trigonometrical functioas, to four figures only,
for every whole degree up to 90° and the corresponding
circular measure to five figures.
R
1-0025
1-005
1-0075
i-oi .
1-0125
I -015
I -0175
I -02
1-0225
1-025
1-0275
I -03 .
Logarithms of R for
log R
00108438
00216606
00324505
00432137
00539503
00646604
00753442
00860017
00966332
01072387
01178183
01283722
Compound Interest
R logR
1-0325 . , . . . -01389006
I -035 -01494035
I-0375 -01598811
I-04 . . ' . . . . -01703334
1-0425 -01807606
1-045 •01911629
I -0475 -02015403
1-05 . . . . . , -02118930
1-0525 -02222210
1-055 -02325246
1-0575 -02428038
I -06 -02530587
Constants used in Mensuration and their Logarithms
7r= 3-14159265
in = 1-57079633
iyr =052359878
•A 77 = 4-18879020
^/n= 1-77245385
7f2 = 9-86960440
^n= I -46459 1 89
7r/l8o = 0-01745329
logarithm
0-497150
0-196120
1-718999
0-622089
0-248575
0-994300
0-165717
2-241877
I -=- 77
1 -^ 477
-</6-r~^
\/3 -^ 4 TT
\/l 4- 77
= 0-31830989
= 0-07957747
= 1-24070098
= 0-62035049
= 0-56418958
1 -^ 77^= 0-10132118
^^= 2-14502940
180/77= 57-29577951
logarithm
1-502850
2-900790
0-093667
1-792637
1751425
1-005700
0-331433
1-758123
Naperian (or Natural) Logarithms
e= 2-7182182 logjoe = -43429448 logg 10 = 2-30258509
logio -V = logg A" X logio e. log^ A' = logio ^V x logg 10
LOGARITHMS OF NUMBERS
Mean Differences
^1
10
II
12
13
14
IS
j'i) i6
7-^_i7
i8
J9
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
S3
54
0
•00000
•04139
•07918
•II394
•I46I3
P7609
•20412
•23045
•25527
•27875
•30103
32222
■34242
36173
38021
•39794
•41497
•43136
•44716
•46240
•47712
•49136
•50515
•5I85I
•53148
S4407
55630
56820
57978
59106
1
•00432
•04532
•08279
•II727
1-14922
•17898
•20683
•23300
•25768
•28103
•60206
•61278
•62325
■^^3347
■64345
65321
66276
67210
68124
69020
•30320
•32428
•34439
•36361
•38202
•39967
•41664
•43297
•44871
•46389
•47857
•49276
•50651
•51983
•53275
•54531
'55751
56937
58093
59218
l
69897
•70757
71600
72428 I -72509
3239 I -73320
•60314
•61384
•62428
•63448
•64444
•65418
•66370
•67302
•6S215
•69108
69984
■70S42
■71684
•00860
-04922
-08636
•12057
•15229
•18184
20952
'23S53
•26007
•28330
•30535
•32634
•34635
•36549
'3^3^2
•40140
•41830
•43457
•45025
•46538
•48001
•49415
•50786
•52114
•53403
•S4654
•55871
•57054*
•58206
•59329
3
•01284
•05308
0899 r
•12385
•15534
-18469
-21219
-23805
•26245
-28556
•30750
-32838
•34830
■36736
•38561
•40312
■41996
■43616
45179
46687
•01703
•05690
-09342
•12710
•15836
•18752
-21484
•24055
-26482
•28780
•30963
•33041
•35025
•36922
•38739
40483
42160
43775
45332
46835
•60423
•61490
•62531
•63548
•64542
•65514
■66464
■67394
68305
69197
•48144
•49554
•50920
•52244
■53529
54777
55991
57171.
58320
59439
60531
•61595
•62634
•63649
•64640
•65610
•66558
•67486
■68395
■69285
•70070 -70157 70243
-70927 -71012 •71096
•71767 ^71850 ^71933
•72591 •72673-72754
-73400 (-7^480 1 -731; 60
•48287
•49693
•51055
•52375
•53656
•54900
•56110
•57287
•58433
•59550
60638
61700
•62737
•63749
•64738
•65706
•66652
•67578
-68485
•69373
42
39
35
33
•o
28
27
25
24
23
21
20
19
18
17
17
16
15
15
14
14
13
13
13
12
12
12
II
II
II
I I
10
10
TO
10
9
9
9
9
9
8
8
8
85
77
71
66
61
57
S3
50
48
45
43
41
39
37
36
34
33
32
31
30
127
116
106
98
91
85
80
76
71
68
64
6
59
56
54
52
50
48
46
45
170
155
142
131
122
114
107
lOI
95
90
86
82
7^
75
72
69
66
64
61
59
43 57
33 44
32
32
31
30
29
29
28
28
27
26
26
25
25
43
42
41
40
39
38
38
37
36
35
35
34
33
212
193
177
164
152
142
134
126
119
113
107
102
98
94
90
86
S3
80
77
74
72
70
67
65
63
62
60
58
57
55
254
232
213
197
183
171
160
151
'•$^
i'35
297
270
248
229
213
199
187
176
167
i^r
1281150
1231143
ii7|i37
112 131
108
103
99
96
92
89
8 16
24 33
24 1 32
54
53
51
50
49
48
47
46
45
44
43
42
42
S3
81
78
76
74
72
70
68
66
65
63
62
60
59
58
56
55
54
53
52
51
50
125
120
116
112
108
104
339
309
284
262
244
228
21,4
201
190
180
86 10 1
411 49
40
97
94
91
89
86
84
82
80
78
76
74
72
70
69
67
66
64
63
62
60
59
58
57
48 I 56
172
164
158
149
143
138
132
128
123
119
IIS
111
108
105
101
99
96
93
91
89
86
84
82
80
79
77
75
73
72
71
31
3^
3'
2C
2;
21
24
22
21
2C
19
18
16
16:
15'
14.
14.
13;
13^
I2(
I.
121
11^
114
II]
108
105
102
100;
57
95
93
90
88
86
84
83
81
79
69 78
68
67
65
64
76
75
73-^
72
LOGARITHMS OF NUMBERS
5
6
7
8
9
I 2
^
4
'
6
7
8
9
lO
02119 02531
•02938
03342
•03743
40 sT
121
162
202
242
283
323
364
II
•06070 -06446
-06819
-07188
•07555
37|74
III
148
185
222
259
296
III
12
•09691 -10037
-10380
-10721
•11059
34|68
102
^37
170
204
238
272
307
13
•13033 -13354
m
-13988
-14301
32:63
i
126
1^8
147
190
221
253
284
14
•16137 -16435
-17026
•17319
29 59
118
177
206
236
265
15
•19033 19312
•19590
•19866
•20140
28 55
83
no
138
165
193
221
248
i6
•217481 -22011
■22272
•22531
-227S9
26 52
is
104
130
156
182
208
233
17
•24304-24551
•24797
•25042
•25285
24 49
73
98
123
147
171
196
220
i8
•26717! -269^1
•27184
•27416
•27646
23
46
70
93
116
139
162
185
208
19
•29003
-29236-
"•29447
•29667
-29885
22
44
66
88
no
132
154
176
198
20
•31175
•31387
•31597
-31806
•32015
21
42
63
84
los
126
147
168
188
21
•33^44
•33445
•33646
•33846"
•34044
20 40
60
80
100
120
140
160
180
22
■35218 -35411
•35603
•35793
•35984
19 38
57
77
96
H15
134
153
172
23
■37107 -37291
•37475
•37658
•37840
18,37
55
73
91
no
128
146
165
24
•38917 -39094
-39270
•39445
•39620
i8!35
53
70
88
105
123
140
158
25
40654 1 -40824
•40993
•41 162
•41330
17 34
51
67
84
lOI
118
135
152
26
•423251-42488
-42651
-42813
•42975
16 I 32
49
65
81
97
114
130
146
27
•43933
•44091
-44248
•44404
-44560
16 31
47
63
78
94
no
125
141
28
•45484
•45637
-45788
•45939
-46090
i5j3o
45
61
76
91
io6
121
136
29
•46982
•47129
•47276
•47422
•47567
14; 29
1
44
58
73
87
102
117
131
30
48430
•48572
•48714
-48855
•48996
1 ^
14 28
42
56
71
85
99
113
127
31
■49831 P49969
-50106
-50243-50379
14 27
41
55
68
82
96
109
^23
32
•51188 -51322
•51455
-51587 -51720
^1,27
40
53
66
80
93
106
119
33
•52504 -52634
■52763
-52892 -53020
13 26
39
51
64
77
90
103
116
34
•53782! -53908 .54033
-54158 1-54283
13 25
3^
50
63
75
88
100
113
35
•S5023 1 55145
•55267 55388 -55509
12 24
36
49
61
73
85
97
109
36
■56229 1-56348
•56467 1 -56585 1-56703
12 24
31
47
59
71
S3
95
106
37
■57403.
•57519
•57634
•57749
•57864
12,23
35
46
58
69
81
92
104
38
•58546
•58659
•58771
•58883
•58995
11,22
34
45
56
67
78
90
lOI
39
•596601-59770
•59879
•59988
-60097
II 22
33
44
55
66
76
87
98
40
1 1
•60746 -60853 ! -609591 -61066
-61172
II 21
32
43
53
64
74
8S
96
41
•61805 -^'909
-62014 •62118
•62221
10 21
31
42
52
62
73
83
94
42
•62839
-62941
•63043 ^63144
•63246
10 20
30
41
51
61
71
81
91
43
•63849
-63949
•64048 -64147 j -64246
10 20
30
40
50
60
70
79
89
44
•64836
•64933
•65031 -65128-65225
10 19
29
39
49
58
68
78
87
45
65801
-65986
•65992 66087 66181
10 19
29
38
48
57
67
76
86
46
•66745
-66839
-66932
-670251-67117
9 19
28
37
47
56
65
74
84
47
•67669
•67761
-67852
•67943
-68034
9! 18
27
36
46
55
64
73
82
48
•68574
•68664
•68753
-68842
•68931
9!i8
27
36
45
54
62
71
80
49
•69461
-69548
-69636
-69723
-69810
9;i7
1
26
35
44
52
61
70
78
SO
70329
•70415
•70501
•70586 ^70672
9 17
26
34
43
SI
60
68
77
51
■71181
-71265
•71349
•7i433i^7i5i7
8,1/
25
34
42
50
59
<^7
75
52
•72016
-72099
-72181
■72263
•72346
8 16
■^5
33
41
49
58
66
74
S3
•72835
•72916
-72997
•7307s
•73159
8!i6
24
12
40
48
57
65
73
54
-73640
•73719
•73799
■73878
•73957
8|i6
24
32
40
48
55
63
71
Mean Differences
LOGARITHMS OF NUMBERS
Mean Differences
1 0 1 1
2
3
4
I
2
3
4
5
6
7
8
9
55
•74036
•741 15
•74194
•74273
•74351
8
^
24
3!
39
47
55
6j
71^
56
74819
•74896
•74974
75051
•75128
8
15
^l
31
39
46
54
62
69
57
75587
75664
•75740
75815
75891
8
15
23
30
38
46
53
61
68
58
76343
•76418
•76492
•76567
•76641
7
15
22
30
1>7
45
52
60
67
59
77085
•77150
77232
•77305
'77119
7
15
22
29
17
44
51
59
66
6o
•77?i5
•77887
•77960
•78032
•78104
7
14
22
29
36
43
51
58
65
6i
78533
•78604
•78675
•78746
78817
7
14
21
28
36
43
50
57
64
62
79239
•79309
•79379
•79449
•79518
7
14
21
28
35
42
49
56
63
63
79934
•80003
•80072
•80140
•80209
7
14
21
27
34
41
48
55
62
64
•80618
•80686
•80754
•80821
•80889
7
14
20
27
34
41
47
54
61
65
•81291
•81358
•81425
•8149 1
•81558
7
13
20
27
33
40
47
53
60
66
•81954
•82020
•82086
•82151
•82217
7
13
20
26
Zl
39
46
52
59
67
•82607
•82672
'^^7Z7
•82802
•82866
6
13
19
26
32
39
45
52
58
68
•83251
•83315
•83378
•83442
•83506
6
13
19
25
Z^
Z^
45
51
57
69
•83885
•83948
•8401 1
•84073
•84136
6
13
19
25
31
38
44
50
56
70
•84510
•84572
•84634
•84696
•84757
6
12
19
25
31
37
43
49
S6
71
•85126
•85187
•85248
•85309
•85370
6
12
18
24
31
17
43
49
55
72
•85733
•85794
•85854
•85914
•85974
6
12
18
24
30
36
42
48
54
73
•86332
•86392
•86451
•86510
•86570
6
12
18
24
30
36
42
48
53
74
•86923
•86982
•87040
•87099
•87157
6
12
18
23
29
35
41
47
53
75
•87506
•87564
•87622
•87679
•87737
6
II
17
23
29
35
40
46
52
76
•88081
•88138
•88195
•88252
•88309
6
II
17
23
29
34
40
46
51
77
•88649
•88705
•88762
•88818
•88874
6
II
17
22
28
34
39
45
51
78
•89209
•89265
•89321
•89376
•89432
6
II
17
22
2\
ZZ
39
44
50
79
•89763
•89818
•89873
•89927
•89982
5
II
16
22
27
Zl
38
44
49
80
•90309
•90363
•90417
•90472
•90526
5
II
16
22
27
33
38
43
49
81
•90849
•90902
•90956
•91009
•91062
1 1
16
21
27
32
Z7
43
48
82
•91381
•91434
•91487
•91540
•91593
II
16
21
27
Z^
Z7
42
48
83
•91908
•91960
•92012
•92065
•92117
10
16
21
26
31
Z7
42
47
84
•92428
•92480
•92531
•92583
•92634
10
15
21
26
31
36
41
46
85
•92942
•92993
•93044
•9309s
•93146
10
15
20
26
31
36
41
46
86
•93450
•93500
•93551
•93601
•93651
10
15
20
25
30
35
40
45
87
•93952
•94002
•94052
•94101
•941 5 1
10
15
20
25
30
35
40
45
88
•94448
•94498
•94547
•94596
•94645
10
15
20
25
30
34
39
44
89
•94939
•94988
•95036
•95085
•95134
10
15
19
24
29
34
39
44
90
•95424
•95472
•95521
•95569
95617
10
14
19
24
29
34
39
43
91
•95904
•95952
•99599
•96047
•96095
10
14
19
24
29
ZZ
Z'^
43
92
•96379
•96426
•96473
•96520
•96 5 Cy
•970/5
9
14
1 9-.
24
2%
zz
38
42
93
•96H48
•96895
•96942
•96988
9
14
19
23
28
ZZ
Z7
42
94
•97313
•97359
•97405
•97451
•97497
9
14
18
23
2"^
z-
Z7
41
95
•97772
•97818
•97864
•97909
•97955
9
14
18
23
27
32
36
41
96
•98227
•98272
•98318
•983^>3
•9S408
9
14
18
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-7
32
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41
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•9H677
•98722
•98767
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9
13
18
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31
36
40
98
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•99167
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•99255
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9
13
18
22
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31
35
40
.99
•99564
•99607
•9965 1
•99695
•99739
4
9
13
17
22
26
31
35
39
LOGARITHMS OF NUMBERS
5 6 7
8 1 9
I
8
2
16
3
23
4 5
3TI39
1'.
47
7
55
8
62
9
55
•74429 -74507 •74586
•74663
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70
56
•75205
•75282 1 -75358
•75435
-75511
8
15
23
31
38
46
53
61
69
57
•75967
•76042. -76118
•76193
•76268
8
15
23
30
38
45
53
60
68
58
76716
•767901-76864
•76938
-77012
7
15
22
30
37
44
52
59
66
59
•77452
•77525 -77597
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7
15
22
29
36
44
51
58
65
6o
•78176
•78247 78319
•78390
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7
14
21
29
36
43
50
57
64
6i
•ySiSS
•78958^79029
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•79169
7
14
21
28
35
42
49
56
63
62
•79588
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7
14
21
28
35
42
48
55
62
. 63
•^ozjj j -803461 -So^l^^
•80482
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7
14
20
27
34
41
48
55
61
64
•80956! -81023 -81090
•81158
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7
13
20
27
34
40
47
54
60
65
•81624 81690 81757
•81823
•81889
7
13
20
26
33
40
46
53
59
66
•82282
•82347 ,•82413
•82478
-82543
7
13
20
26
33
39
46
52
59
67
•82930
-82995
-83059
-83123
-831^
6
13
19
26
32
39
45
51
58
68
•83569
-83632
•83696
-83759
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6
13
19
25I32
38
44
51
57
69
•84198
•84261
■^AZ27,
-84386
•84448
6
12
19
25
31
37
44
50
56
70
•S4819
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•85065
6
12
18
25
31
37
43
49
55
71
•^'5431
•85491
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6
12
18
24
30
36
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•^6034
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6
12
18
24
30
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54
73
•86629
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6
12
18
24
29
35
41
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53
74
•87216
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6
12
17
23
29
35
41
46
52
75
•87795
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6
II
17
23
29
34
40
46
51
76
•88366
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•88480
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6
II
17
23
28
34
40
45
51
77
•88930
•88986
•89042
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6
II
17
22
28
33
39
45
50
78
•89487
-89542
-89597
•■89653
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6
II
17
22
28
33
39
44
50
79
■90037
•90091
•90146
•90200
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5
II
16
22
27
33
38
44
49
80
•90580
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s
II
16
22
27
32
38
43
48
81
•91 1 16
•9II69
•91222
•91275
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5
II
16
21
27
32
37
42
48
82
•91646
•91698
•91751
-91803
•91855
5
10
16
21
26
31
37
42
47
83
•92169
-92221
.92273
•92324
•92376
5
10
16
21
26
31
36
41
47
84
•92686
•92737
-92788
•92840
•92891
5
10
15
20
26
31
36
41
46
85
•93197 93247
-93298
-93349 ! 93399
5
10
15
20
25
30
35
40
46
86
•93702-93752
•93802
•93852 ^93902
5
10
15
20
25
30
35
40
45
87
•94201 1-94250
•94300
•94349
•94399
5
10
15
20
25
30
35
40
44
8a
:24694
•94743
-94792
•94841
•94890
5
10
15
20
25
29
34
39
44
89
•95182
•95231
•95279
•95328
-95376
5
10
15
19
24
29
34
39
44
90
•9566s
•95713
-95761
-95809
•95856
5
10
14
19
24
29
33
38
43
91
•96142
•96190
-96237
•96284
-96332
5
9
14
19 24
28
33
38
43-
92
•96614
-96661
-£6708
-96755
-96802
5
9
14
19
23
2S
33
37
42
93
•97081
•97128
•97174
•97220
-97267
5
9
14
19
23
28
32
37
42
94
•975431-97589
•97635
•97681
.97727
5
9
14
18
23
27
32
37
41
95
•98000 98046 98091
•98137 i 98182
5
9
14
18
23
27
32
36
41
96
•98453-98498 -98543
•98588
•98632
4
9
13
18
22
27
31
36
40
97
•98900 -98945
-98989
-99034
•99078
4
9
13
18
22
27
31
36
40
98
•99344, -99388
•99432 -99476
•99520
4
9
13
18
22
26
31
35
40
J 99_
•99782 -99826
-99870 '-99913
•99957
4
9
13
17
22
26
31
3<>
39
Mean Differences
ANTILOGARITHMS
Mean Differences
0
1
2
3
4
2
2 5
3
~7
4
~9
5
12
6
7
i6
8
19
9
•oo
1 0000
10023
10046
10069
10093 -
21
•01
10233
10257
10280
10304
10328 2 5
7
9
12
14
17
19
21
•02
1047 1
10495
10520
10544
10568 2 5
7
10
12
15
U
20
22
•03
10715
10740
10765
10789
10814 2 5
7
10
12
15
17
20
22
•04
10965
10990
1 1015
11041
1 1066 3 5
8
10
13
15
18
20
23
•OS
11220
1 1 246
1 1272
1 1298
I 1324 :
J 5
8
10
13
16
18
21
23
•06
1 1482
1 1 508
"535
11561
11588 3| 5
8
II
13
16
18
21
24
•07
1 1 749
1 1776
1 1803
11830
11858 :
5 5
8
II
14
16
19
22
24
•08
12023
12050
12078
12106
12134 .
5 6
8
11
14
U
19
22
25
•09
12303
12331
12359
12388
12417 .
5 6
9
11
14
17
20
23
26
•10
12589
12618
12647
12677
12706 i
( 6
9
12
15
18
20
23
26
•II
12882
12912
12942
12972
13002 :
5 6
9
12
15
18
20
24
27
•12
13183
13213
13243
13274
13305 :
5 6
9
12
15
18
21
24
27
•13
13490
13521
13552
13583
1 3614 :
6
9
12
16
19
22
25
28
•14
13804
13836
13868
13900
13932 :
6
10
13
16
19
22
26
29
•15
14125
141S8
14191
14223
14256 3
i 7
10
13
16
20
23
26
30
•16
14454
14488
14521
14555
14588 :
7
10
13
17
20
24
27
30
•17
14791
14825
14859
14894
14928 :
7
10
14
17
21
24
27
31
•18
15136
15171
15205
15241
15276 ^
^ 7
II
14
18
21
25
28
32
•19
15488
15524
15560
15596
15631 A
^ 7
11
14
18
22
25
29
32
•20
15849
15885
15922
15959
15996 A
^ 7
II
15
18
22
26
29
33
•21
16218
16255
16293
16331
16368 A
8
1 1
15
19
23
26
30
34
•22
16596
16634
16672
16711
16749 A
^ 8
II
15
19
23
27
31
34
•23
16982
17022
1 706 1
1 7 100
17140 A
^ 8
12
16
20
24
28
32
35
•24
17378
17418
17458
17498
17539 A
8
12
16
20
24
28
32
36
•25
17783
17824
17865
17906
I'jgA'j 4
[ 8
12
16
21
25
29
33
37
•26
18197
18239
18281
18323
18365 4
8
13
17
21
25
30
34
38
•27
1 862 1
18664
18707
18750
18793 A
9
13
17
22
26
30
34
39
•28
19055
19099
19143
19187
19231 A
9
13
18
22
26
31
35
40
•29
19498
19543
19588
19634
19679 5
9
14
18
23
27
32
36
41
•30
19953
19999
2004s
20091
20137 5
9
14
18
23
28
32
37
42
•31
20417
20464
20512
20559
20606 5
9
14
19
24
28
33
3»
43
•32
20893
20941
20989
21038
21086 5
10
15
19
24
29
34
39
44
•33
21380
21429
21478
21528
21577 5
10
15
20
25
30
35
40
44
•34
21878
21928
21979
22029
22080 5
10
15
20
2S
30
35
40
46
•35
22387
22439
22491
22542
22594 5
10
16
21
26
31
36
41
47
•36
22909
22961
23014
23067
23121 5
1 1
16
21
27
3^
37
42
48
•37
23442
23496
23550
23605
23659 5
11
16
22
27
33
3»
44
49
•38
23988
24044
24099
24155
24210 6
1 1
17
7 T
28
33
39
44
50
•39
24547
24604
24660
24717
24774 6
1 1
17
23
28
34
40
45
51
•40
25119
25177
2523s
25293
25351 6
12
17
23
29
35
41
47
52
•41
25704
25763
25823
25882
25942 6
12
18
24
30
36
42
48
54
•42
26303
26363
26424
26485
26546 6
12
18
24
30
36
43
49
55
•43
26915
26977
27040
27102
27164 6
12
19
25
31
37
44
50
56
•44
27542
27606
27669
277Z1>
27797 6
13
19
26
32
38
45
51
57
•45
28184
28249
28314
28379
2844s 7
13
20
26
33
39
46
52
59
•46
28840
28907
2^97i
29040
29107 7
13
20
27
35
40
47
53
60
•47
29512
29580
29648
29717
29785 7
14
21
27
34
41
48
55
62
•48
30200
30269
30339
30409
30479 7
14:211
28
35
42
49
56
63
•49_
30903
30974
31046
31117
31189 7
14I21I
29
36
43
50
57
64
ANTILOGARITHMS
5
6
7
8
9
I
2
2
5
3
"7
4
9
5
12
6
14
7
16
8
19
9
•00
10116
IOI39
10162
10186
10209
21
•01
10351
10375
10399
10423
10447
2
5
7
10
12
14
17
19
22
•02
10593
I06I7
10641
10666
1 069 1
2
5
7
10
12
15
17
20
22
•03
10839
10864
10889
10914
10940
3
5
8
10
13
15
18
20
23
•04
1 1092
I II 17
11143
11169
11194
3
5
8
10
13
15
18
20
n
•OS
1 1350
1 1376
1 1402
1 1429
1 1455
3
5
8
II
13
16
18
21
24
•o6
11614
II64I
11668
11695
11722
3
5
8
1 1
14
16
19
22
24
•07
11885
II9I2
11940
11967
11995
3
6
8
11
14
17
19
22
25
•o8
12162
12190
12218
12246
12274
3
6
8
11
14
17
20
23
25
•09
12445
12474
12503
12531
12560
3
6
9
12
14
17
20
23
26
•10
12735
12764
12794
12823
12853
3
6
9
12
15
18
21
24
26
•11
13032
13062
13092
13122
13152
3
6
9
12
15
18
21
24
^7
•12
13335.
13366
13397
13428
13459
3
6
9
12
16
19
22
25
28
•13
13646
^1^71
13709
13740
13772
3
6
9
13
16
19
22
25
28
•14
13964
13996
14028
14060
14093
3
6
10
13
16
19
^l
26
29
•15
14289
14322
14355
14388
14421
3
7
10
13
17
20
23
26
30
•i6
14622
14655
14689
14723
14757
3
7
10
14
17
20
24
27
30
•17
14962
14997
15031
15066
15101
3
7
10
14
17
21
24
28
31
•i8
15311
15346
15382
15417
15453
4
7
11
14
18
21
25
28
32
•19
15668
15704
15740
15776
15812
4
7
11
14
18
22
25
29
11
•20
16032
16069
16106
16144
16181
4
7
II
15
19
22
26
30
33
•21
16406
16444
16482
16520
16558
4
8
11
15
19
23
27
30
34
•22
16788
16827
16866
16904
16943
4
8
12
16
19
'-I
27
31
35
•23
17179
17219
17258
17298
17338
4
8
12
16
20
24
28
32
36
•24
17579
17620
17660
17701
17742
4
8
12
16
20
24
29
33
17
•25
17989
18030
18072
18113
18155
4
8
12
17
21
25
29
33
37
•26
18408
18450
18493
18535
18578
4
9
13
17
21
26
30
34
38
•27
18836
18880
18923
18967
19011
4
9
13
18
22
26
31
35
39
•28
19275
19320
19364
19409
19454
4
9
13
18
22
27
31
36
40
•29
19724
19770
19815
19861
19907
5
9
14
18
23
27
32
Z7
41
•30
20184
20230
20277
20324
20370
5
9
14
19
23
28
33
37
42
•31
20654
20701
20749
20797
20845
5
10
14
19
24
29
IZ
38
43
•32
21135
21184
21232
21281
21330
5
10
15
20
25
29
34
39
44
•33
21627
21677
21727
^-^m
21827
5
10
15
20
25
30
35
40
45
•34
22131
22182
22233
22284
22336
5
10
15
20
26
31
36
41
46
•35
22646
22699
22751
22803
22856
5
II
16
21
26
32
37
42
47
•36
23174
23227
23281
^ms
23388
5
11
16
21
27
32
38
43
48
•37
23714
23768
2ZZ21
23878
23933
5
II
16
22
27
^^
38
44
49
•38
24266
24322
24378
24434
24491
6
1 1
17
22
28
34
39
45
51
•39
24831
24889
24946
25003
25061
6
12
17
23
28
35
40
46
52
•40
2S4IO
25468
25527
25586
25645
6
12
18
24
29
35
41
47
53
•41
26002
26062
26122
26182
26242
6
12
18
24
30
36
42
48
54
•42
26607
26669
26730
26792
26853
6
12
18
25
31
17
43
49
55
•43
27227
27290
27353
27416
27479
6
13
19
25
32
38
44
50
57
•44
27861
27925
27990
28054
28119
6
13
19
26
32
39
45
52
S8
•45
28510
28576
28642
28708
28774
7
13
20
26
33
40
46
53
59
•46
29174
29242
29309
29376.
29444
7
14
20
^7
34
41
47
54
61
•47
29854
29923
29992
30061
30130
7
14
21
28
35
42
48
55
62
•48
30549
30620
30690
30761
30832
7
14
21
28
35
42
50
57
64
•49
31261
31333
314OS
31477
31550
7
14
21
29
36
43
51
58
65
Mean Differences
II
Antilogarithms
Mean Differences
0
1
2
3
4
I 2
3
4
5
6
7
8
9
•SO
31623
3T696
31769
31842
31915
~7
IS
^
29
37
44
SI
59
66
•SI
32359
32434
32509
32584
32659
8
15
^3
30
3S
45
53
60
68
•S2
33113
33189
33266
33343
33420
8
15
23
31
38
46
53
61
69
'S3
33884
33963
34041
341 19
34198
8
16
24
31
39
47
55
63
71
•54
34674
34754
34834
34914
34995
8
16
24
32
40
48
56
64
72
•ss
35481
35563
35645
35727
35810
8
16
25
53
41
49
S8
66
74
•S6
36308
36392
36475
36559
36644
8
17
2q
34
42
50
59
67
76
•S7
37154
37239
37325
37411
37497
9
17
26
34
43
52
60
69
78
•S8
38019
38107
38194
38282
38371
9
18
26
35
44
53
62
70
79
•S9
38905
38994
39084
39174
39264
9
18
27
36
45
54
63
72
81
•60
3981 1
39902
39994
40087
40179
9
18
28
37
46
55
65
74
83
•61
40738
40832
40926
41020
41115
9
19
28
38
47
57
66
76
85
•62
41687
41783
41879
41976
42073
10
19
29
39
48
58
67
77
87
•63
42658
42756
42855
42954
43053
10
20
30
40
49
59
69
79
89
•&4
43652
43752
43853
43954
44055
10
20
30
40
51
61
71
81
91
•6s
44668
44771
44875
44978
45082
10
21
31
41
52
62
73
83
93
•66
45709
45814
45920
46026
46132
1 1
21
32
42
53
63
74
85
95
•67
46774
46881
46989
47098
47206
1 1
22
32
43
54
65
76
86
97
•68
47863
47973
48084
48195
48306
1 1
22
33
44
55
66
78
89
100
.69
48978
49091
49204
49317
49431
1 1
23
34
45
57
68
79
91
102
•70
SOI 19
50234
50350
50466
50582
12
23
35
46
58
70
81
93
104
•71
51286
51404
51523
51642
51761
12
24
36
48
59
71
83
95
107
•72
52481
52602
52723
52845
52966
12
24
36
49
61
73
85
97
109
•73
53703
53827
53951
54075
54200
12
25
37
50
62
75
87
100
1 12
•74
54954
55081
55208
55335
55463
13
25
38
51
64
76
89
102
114
•7S
56234
56364
56494
56624
56754
13
26
39
52
65
78
91
104
117
.76
57544
57677
57810
57943
58076
13
27
40
53
67
80
93
107
120
•77
58884
59020
59156
59293
59429
14
27
41
55
68
82
95
109
123
•78
60256
60395
60534
60674
60814
14
28
42
56
70
84
98
112
126
•79
61660
61802
61944
62087
62230
14
29
43
57
71
86
100
114
128
•80
63096
63241
63387
63533
63680
15
29
44
58
73
88
102
117
131
•81
64565
64714
64863
65013
65163
15
30
45
60
75
90
105
120
135
•82
66069
66222
<^^^374
66527
66681
15
31
46
61
77
92
107
122
138
•83
67608
67764
67920
68077
68234
16
31
47
63
78
94
no
125
141
•84
69183
69343
69502
69663
69823
16
32
48
64
80
96
112
128
144
•85
70795
70958
71121
71285
71450
16
33
49
66
82
98
IIS
131
147
•86
72444
7261 1
72778
72946
73114
17
34
50
(^7
84
101
117
134
151
•87
74131
74302
74473
74645
74817
17
34
Si
69
86
103
120
137
154
•88
75858
76033
76208
76384
76560
18
35
S3
70
88
105
123
140
158
•89
77625
77804
77983
78163
7S343
18
36
54
7~
90
108
126
144
162
•90
79433
79616
79799
79983
80168
18
37
55
74
92
no
129
147
166
•91
81283
81470
81658
81846
82035
19
38
56
75
94
113
132
151
169
•92
83176
83368
83560
83753
83946
19
39
58
77
96
116
135
154
174
•93
85114
85310
85507
85704
85901
20
39
59
79
99
118
138
158
177
•94
87096
87297
87498
87700
87902
20
40
61
81
101
121
141
161
182
•95
8912s
89331
89536
89743
89950
21
41
62
83
103
124
144
165
186
•96
91201
91411
91622
91833
92045
21
42
63
84
106
127
148
169
190
•97
93325
93541
93756
93972
94189
22
43
65
80
108
130
151
173
195
•98
95499
95719
95940
96161
96383
22
44
6()
88
1 1 1
133
155
177
199
M
97724
97949
98175
98401
98628
23
45
68
90
113
136
158
181
204
12
ANTILOGARITHMS
5
6
7
8
9
^\l
3
4
5
6
7
\'
9
•so
31989
32063
32137
32211
32285
7
15
22
30
37
44
52
59
67
•51
32734
32810
32885
32961
33037
8
15
23
30
38
45
53
61
68
•52
33497
33574
33651
33729
33806
8
15
23
31
39
46
54
62
70
'53
34277
34356
34435
34514
34594
8
16
24
32
40
48
55
C^3
71
•54
35075
35156
35237
35318
35400
8
16
24
32
41
49
57
65
73
•55
35892
35975
36058
36141
36224
8
17
25
33
42
50
S8
67
75
•S6
36728
36813
36898
36983
37068
9
17
26
34
43
51
60
68
77
•57
37584
17670
Z77S7
37844
37931
9
17
26
35
44
52
61
70
78
•58
38459
38548
38637
38726
38815
9
18
27
36
45
54
62
71
80
•59
39355
39446
39537
39628
39719
9
18
27
36
46
55
64
73
Sjj
•6o
40272
40365
40458
40551
40644
9
19
28
37
47
56
6S
75
84
•6i
41210
41305
41400
41495
41591
10
19
29
38
48
57
67
76
86
•62
42170
42267
42364
42462
42560
10
20
29
39
49
59
68
7S
88
•63
43152
43251
43351
43451
43551
10
20
30
40
50
60
70
80
90
•64
44157
44259
44361
44463
44566
10
20
31
41
51
61
72
82
92
•65
45186
45290
45394
45499
45604
10
21
31
42
52
63
73
84
94
•66
46238
46345
46452
46559
46666
II
21
32
43
54
64
75
86
96
•67
47315
47424
47534
47643
47753
II
22
33
44
55
66
77
88
99
•68
48417
48529
48641
48753
48865
II
22
34
45
56
(^7
79
90
lOI
.69
49545
49659
49774
49888
50003
II
23
34
46
57
69
80
92
103
•70
50699
50816
50933
51051
51 168
12
23
35
47
59
70
82
94
106
.71
51880
52000
52119
52240
52360
12
24
36
48
60
72
84
96
108
•72
53088
53211
^ZZIZ
53456
53580
12
25
37
49
62
74
86
98
III
•73
54325
54450
54576
54702
54828
13
25
38
50
63
75
88
100
113
•74
55590
55719
55847
55976
56105
13
26
39
52
64
77
90
103
116
•75
56885
57016
57148
57280
57412
13
26
40
S3
66
79
92
105
119
.76
58210
58345
58479
58614
58749
13
27
40
54
67
81
94
108
121
•77
59566
59704
59841
59979
601 17
14
28
41
55
69
Si
97
no
124
•78
60954
61094
61235
6\Z7^
61518
14
28
42
56
71
85
99
113
127
•79
^^Z7l
62517
62661
62806
62951
14
29
43
58
72
87
lOI
116
130
.80
63826
63973
641 2 1
64269
64417
15
30
44
59
74
88
104
118
133
•81
^sm
65464
65615
65766
65917
15
30
45
60
76
91
106
121
136
•82
66834
66988
67143
67298
67453
15
31
46
62
77
93
108
124
139
•83
68391
68549
68707
68865
69024
16
32
48
63
79
95
III
127
143
•84
69984
70146
70307
70469
70632
16
32
49
65
81
97
114
130
146
•85
71614
71779
71945
721 1 1
72277
17
33
50
66
83
100
116
133
149
•86
73282
73451
73621
73790
73961
17
34
51
6S
84
102
119
136
153
•87
74989
75162
75336
75509
75683
17
35
52
70
87
104
122
139
156
•88
76736
76913
77090
77268
77446
18
36
53
71
89
107
124
142
160
.89
78524
78705
78886
79068
79250
18
36
55
73
91
109
127
145
164
•90
80353
80538
80724
80910
81096
19
37
56
74
93
112
130
149
167
•91
82224
82414
82604
82794
82985
19
3S
57
76
95
114
133
1^2
171
.92
84140
Hill
84528
84723
84918
19
39
58
78
97
117
136
i=;6
175
•93
86099
86298
86497
86696
86896
20
40
60
80
100
120
140
160
179
•94
88105
88308
88512
88716
88920
20
41
61
82
102
122
143
163
184
•95
90157
90365
90573
90782
90991
21
42
63
84
104
125
146
167
188
.96
922^ M70
-^ -24
92683
92897
93111
21
43
64
85
107
128
150
171
192
.n-it
94842
95060
95280
22
44
66
87
109
131
153
175
197
W > ^ '
97051
97275
97499
22
45
67
90
112
134
157
179
201
^-
_.j993i2|
99541
99770
13_
46
69
9^
115
137
160
183_
206
M
ean ]
Diffe
renc
2S
0° NATURAL FUNCTIONS
Differences are given for every lo'. Intermediate values can be found by n
method of proportional parts ; e. g. : —
To find tan 43° 56' and cos 37° 34'
tan 43° 5o'= -96008 cos 37° 3°'= 79335 1
+ diff. for 6'= 337 -diff. for4'=- 71 f
.-. tan 43° 5^'= -96345 ••• cos 37° 34'= -79264
When there is no entry in the difference column, the value of the function char
too rapidly for correct interpolation by proportional parts. C4reater accuracy is t
obtained by expressing the function in terms of the sine and cosine.
To find tan 67° 23'
tan 67° 20'= 2-39449 Diff. for 10' = 1972
by proportional parts, diff. for 3'= 592
This gives tan 67° 23'= 2-40041. (The correct value is 2-40038.)
Subtract differences when dealing with co-functions
0°
10'
20'
30'
40'
';o^
i''
10'
20'
30'
40'
50'
2°
10'
20'
30'
40'
so'
3°
•00000
•00291
•00582
•00873
•o I 1 64
•01454
•01745
•02036
•02327
•02618
•02908
•03199
•0349Q
•03781
•0407 1
•04362
•04653
■04943
05234
cosine
291
291
291
291
290
291
291
291
291
290
291
291
291
290
291
291
290
291
D
00
34378
171-89
114-59
85-946
68-757
57-299
49-114
42-976
38-202
34-382
'31-258
28-654
26-451
24-562
22-926
2 1 -494
20-230
19-107
secant
tangent
-00000
•00291
•00582
•00873
•01 164
•01455
•01746
•02037
•02328
•02619
•02910
■03201
03492
■03783
■04075
•04366
■04658
04949
05241
D cotangent
291
291
291
291
291
291
291
291
291
291
291
291
291
292
291
292
291
292
^^
cotangent
00
343-77
171-89
114-59
85-940
68-750
57290
49-104
42-964
38^188
34-368
31-242
28636
26-432
24-542
2 2 -904
21-470
20-206
I9081
tang(;nt
secant
I -00000
I -ooooo
I -00002
I -00004
I -00007
I -000 1 I
I 000 I 5
I ^0002 1
1^00027
I -00034
I -00042
I -0005 I
I 0006 1
I -00072
1-00083
I -00095
I -00108
I -OO I 2 2
I -00137
cosecant
000
002
002
003
004
004
005
006
007
008
009
010
Oil
01 1
012
013
014
015
I -ooooo
•99998
-99996
-99993
-99989
-99985
•99979
•99973
-99966
•99958
-99949
-99939
•99929
•99917
•99905
-99892
-99878
-99863
000
002
002
003
004
004
006
006
007
008
009
010
GIG
012
012
013
014
015
|9C
5<
4c
3C
2C
IC
8S
5t
4C
3c
2C
IC
50
4oi
30
20
10
87
0^ LOGARITHMIC FUNCTIONS
The values given here are the true logarithms ; the characteristic is not
increased by lo as in many tables.
Differences are given for every lo'. Intermediate values can be found by
the method of proportional parts.
The differences for the logarithm of a function and of the reciprocal of the
function are the same in magnitude but opposite in sign.
When there is no entry in the difference column, the rate of change of the
logarithm changes too rapidly for correct interpolation by proportional parts.
The following rules may be used when the angle is small : —
Log sine. Add &6S$S7 to the log of the angle expressed in seconds and
subtract ^ of^tlie log secant.
Log tan. Add 6-68557 to the log of the angle expressed in seconds and
add ^ of the log secant.
When the log sine is given, the angle is found in seconds by adding
5-31443 to the log sine and ^ of the corresponding log secant (found
in the ordinary way).
When the log tan is given, the angle is found in seconds by adding
5*31443 to the log tan and subtracting | of the corresponding log
secant (found in the ordinary way).
Subtract differences when dealing with co
-functions
log sin
D
log cosec
log tan
D |log cotan
log sec
D ] log COS
d^
— 00
00
— 00
00
0-00000
0-00000
90°
10'
3-46373
2-53627
3-46373
2-53627
0-00000
001
o-ooooo
50'
20'
376475
2-23525
3-76476;
2-23524
0-0000 1
001
1-99999
40'
30'
3-94084
2-05916
3-94086
2-05914
0 -00002
^; 11-99998
30'
40'
2-06578
1-93422
2-06581
jl-93419
0-00003
002
1-99997
20'
50'
2-16268
1-83732
2-16273
1-83727
0-00005
002
1-99995
10'
r
2-24186
1-75814
2-24192
1-75808
0-00007
002
1-99993
89"^
10'
2-30879 1
I -69 1 2 1
2-30888
i I -691 12
0-00009
00 i I -9999 1
50'
20'
2-366781
1-63322
2-36689
I-633II
0-00012
003
00 ■?
1-99988
40'
30'
2-41792,
1-58208
2-41807
1-58193
0-00015
1-99985
30'
40'
2-463661
1-53634
2-46385
1-53615
0-00018
^l ■•9»«-'|
20'
50'
2-50504
I -49496
2-50527
1-49473
0-00022
004
004
1-99978
10'
2^
2-S4282I
1-45718
2-54308
1-45692
000026
1-99974
88^
10'
2-57757
1-42243
2-57788
1-42212
0-00031
1-99969
50'
20'
2-60973
1-39027
2-61009
I -38991
0-00036
005 ^-99964
40'
30'
2-63968.
1-36032
2-64009
1-35991
0-00041
o^!i-99959
30'
40'
2-66769!
1-33231
2-66816
1-33184
o-o(J047
0^^99953
20'
50'
2-69400'
1-30600
2-69453
j I -30547
o-(;<KJ53
006 ^-99947
10'
3"
2-71880'
1-28120
2-71940
11-28060
000060
i -99940
ST
log COS j D
log sec
log cotan
D 1 log tan
log cosec
D
log sin
87=
3° NATURAL FUNCTIONS
05234
05524
05814
06105
06395
06685
06976
07266
07556
07846
08136
08426
08716
09005
09295
09585
09874
0164
0453
0742
1031
1320
1609
1898
2187
2476
2764
3052
3341
3629
3917
4205
4493
4781
5069
5356
5643
5931
6218
6505
6792
7078
736s
290
290
291
290
290
291
290
290
290
290
290
290
289
290
290
289
290
289
289
289
289
289
289
289
289
288
288
289
288
288
288
288
288
288
287
287
288
287
287
287
286
287
D
19-1073
18-1026
17-1984
16-3804
15-6368
14-9579
14-3356
13-7631
13-2347
127455
12-2913
11-8684
11-4737
11-1046
10-7585
IO-4334
10-1275
9-83912
956677
9-30917
9-06515
8-83367
8-61379
8-40466
8-20SSI
8-01565
7-83443
7-66130
7-49571
7-18530
7-03962
6-89979
6-76547
6-63633
6-51208
639245
6-27719
6-16607
6-05886
5-95536
5-85539
5 75877
tangent
D
-05241
•05533
-05824
•061 16
•06408
•06700
-06993
•07285
•07578
-07870
-08163
•08456
-08749
•09042
-09335
•09629
•09923
-IO216
-IO51O
•10805
•I 1099
-II394
-I1688
-II983
-12278
-12574
•12869
•13165
•I 3461
•13758
•14054
-1435 I
• 1 4648
-14945
•15243
•15540
•15838
•I 61 37
•16435
•16734
•17033
'^7Zr:s
•17633
292
291
292
292
292
293
292
293
292
293
293
293
293
293
294
294
293
294
295
294
295
294
295
295
296
295
296
296
297
296
297
297
297
298
297
298
299
298
299
299
300
300
cotangent] D
cotangent
I9081I
18^0750
17-1693
16-3499
I 5 -6048
14-9244
14-3007
13-7267
13-1969
12-7062
12-2505
11-8262
Iit430i
11-0594
10-7119
10-3854
10-0780
978817
9-S1436
9-25530
9-00983
8-77689
8-55554
8 -34496
8-14435
7-95302
7-77035
7-59575
7-42871
7-26873
7^ii537
6-96823
6-82694
6-691 16
6-56055
6^43484
631375
6-19703
6-08444
5-97576
5-87080
5-76937
567128
tangent
I 00137
I -00153
I -00 1 69
I -001 87
1-00205
1-00224
1-00244
1-00265
1-00287
1-00309
I -00333
1-00357
I 00382
I -00408
I -00435
I -00463
I -0049 1
■I -005 2 I
I -00551
1-00582
I -006 1 4
I -00647
I -0068 I
I -007 I 5
I 0075 1
1-00788
1-00825
1-00863
I -00902
I -00942
I 00983
1-01024
I -01067
I -O I I I I
I-OII55
I -01 200
I 01247
1-01294
I -01 342
I -01 39 1
I -01 440
I -01 49 1
I 01543
D
016
016
018
018
019
020
021
022
022
024
024
025
^f026
027
028
028
030
030
031
032
033
034
034
036
037
037
038
039
040
041
041
043
044
044
045
047
047
048
049
049
051
052
99863
99847
99831
99813
99795
99776
99756
99736
■99714
■99692
■99668
■99644
99619
■99594
■99567
•99540
■995 1 1
•99482
•99452
-99421
-99390
-99357
•99324
-99290
•99255
•99219
•99182
-99144
-99106
•99067
•99027
•98986
•98944
•98902
■98858
■98814
98769
■98723
■98676
•98629
■98580
98531
98481
80^
LOGARITHMIC FUNCTIONS
loo^ sin
D
log cosec
log tan
log cotan
1-28060
lo^ sec
D
log cos
3"
271880
1-28120
271940
0-00060
006
i -99940
87^
lO'
2-74^26
1-25774
2-74292
1-25708
■ J -00066
008
1-99934
50'
20'
276541
1-23549
2-76525
1-2 347 S
0-00074
007
Of 18
1-99926
40'
30'
278568
1-21432
2-78649
1-21351
0-0008 1
1-99919
30'
4'''
2-80585
1-19415
2-80674
1-19326
0-00089
LHJO
008
T -999 11
20'
;o'
2-S2SI3
1-17487
2-82610
1-17390
0-00097
009
009
009
010
1-99903
10'
4^
10'
284358
2-86128
1-15642
1-13872
2 84464
5-60243
1-15536
1-13757
000106
0-001 15
199894
1-99S85
86^
50'
20'
2-87829
1-12171
2-87953
I -12047
0-00124
1-99876
40'
3<J'
2-89464
1-10536
2-89598
1-10402
0-00134
0 T r>
1-99866
30;
40'
2-91040
I -08960
2-91185
1-08815
0-00144
(J lU
Oil
1-99856
20'
Sf'j'
5-92561
1-07439
2-92716
1-07284
0-00155
Oil
1-99S45
10'
5"
2 94030
105970
2-94195
1-05805
000166
Oil
1-99834
85^
10'
^•95450
1-04550
2-95627
1-04373
0-00177
r> T T
1-99823
50'
20'
2-96825
1-03175
2-97013
1-02987
0-00188
L) 1 1
012
013
012
1-99812
40'
30'
2-98157
1-01843
2-98358
1-01642
0-00200
1-99800
30'
40'
2-99450
1-00550
2-99662
1-00338
0-00213
1-99787
20'
;o'
I -00704
0-99296
1-00930
0-99070
0-00225
014
013
1-99775
10'
6^~
101923
098077
i-02162
097838
000239
i -99761
84°
10'
T -03 109
0-96891
1-03361
0-96639
0-00252
1-99748
50'
20'
1-04262
0-95738
1-04528
0-95472
0-00266
014
014
015
r\T -
1-99734
40'
30'
1-05386
0-94614
1-05666
0-94334
0-00280
1-99720
30'
40'
I -0648 1
0-93519
1-06775
0-93225
0-00295
1-99705
20'
50'
1-07548
0-92452
1-07858
0-92142
0-00310
oio
0 T ;
1 -99690
10'
7=^
1-08589
0-91411
108914
0-91086
0-00325
Ul3
016
199675
83^
10'
I -09606
0-90394
I -09947
0-90053
0-00341
016
016
1-99659
50'
20'
1-10599
993
0-89401
I -10956
987
966
945
926
908
889
873
856
840
825
811
0-89044
0-00357
1-99643
40'
30'
1-11570
971
0-88430
I-II943
o'88o57
0-00373
1-99627
30'
40'
8=
10'
1-12519
1-13447
I 14356
1-15245
949
928
909
889
871
854
837
821
805
790
0-87481
0-86553
085644
0-84755
I -12909
1-13854
i- 14780
1-15688
0-87091
0-86146
085220
0-84312
0-00390
0-00407
000425
0-00443
017
017
018
018
018
019
019
1-99610
1-99593
I -99575
1-99557
20'
10'
82°
50'
20'
i-i6ii6
0-83884
T-16577
0:83423
0-00461
1-99539
40'
3^'
1-16970
0-83030
1-17450
0-82550
0-00480
1-99520
30'
40'
1-17807
0-82193
1-18306
0-81694
0 -00499
1-99501
20'
50'
9~
1-18628
1-19433
0-81372
080567
1-19146
1-19971
0-80854
0 80029
0-00518
0-00538
019
020
020
1-99482
1-99462
10'
81°
10'
20'
1-20223
1-20999
776
762
748
0-79777
0-79001
1-20782
I -2 1 578
796
783
769
0-79218
0-78422
0-00558
0-00579
.021
021
021
022
022
D
I -99442
1-99421
50'
40'
30'
40'
1-21761
1-22509
078239
077491
1-22361
I -23 1 30
0-77639
0-76870
0-00600
0-00621
1 -99400
1-99379
30'
20'
^0'
10""
1-23244
1-23967
735
723
0-76756
0-76033
1-23887
1-24632
log cotan
757
745
0-76113
0-75368
0-00643
0 00665
log cosec
1-99357
1-99335
log sin
10'
80
log COS
D
log sec-
D
log tan
80=
17
10'' NATURAL FUNCTIONS
M
17365
■ I 765 1
■17937
•18224
•18509
■18795
19081
■19366
•19652
•19937
•20222
•20507
•20791
•21076
•21360
•21644
•21928
•22212
•22495
•22778
•23062
•23345
•23627
•23910
•24192
•24474
•24756
•2503S
•25320
•25601
•25882
•26163
•26443
•26724
•27004
•27284
•27564
•27843
•28123
•28402
•28680
•28959
•29237
286
286
287
285
286
286
285
286
285
285
285
284
285
284
284
284
284
283
283
284
283
282
283
282
282
282
282
282
281
281
281
280
281
380
280
280
279
280
279
278
279
278
D
5-75877
5-66533
5-57493
5-48740
5-40263
5-32049
524084
5-16359
5-08863
5-01585
4-94517
4-87649
480973
4-74482
4^68167
4-62023
4^56o4i
4^50216
4-44541
4-39012
4-33622
4^28366
4-23239
4-18238
4-13357
4-08591
4-03938
3-99393
3^94952
3^90613
386370
3-82223
3^78166
3-74198
3-70315
3-66515
3 62796
3-59154
3^55587
3-52094
3-48671
3-45317
342030
tangent
17633
17933
18233
18534
18835
19136
19438
19740
20042
20345
20648
20952
21256
21560
21864
22169
22475
22781
23087
23393
23700
24008
24316
24624
24933
25242
25552
25862
26172
26483
26795
27107
27419
28046
28360
28675
28990
29305
29621
29938
30255
30573
otangont
300
300
301
301
301
302
302
302
303
303
304
304
304
304
305
306
306
306
306
307
308
308
308
309
309
310
310
310
311
312
312
312
313
314
314
315
315
315
316
317
317
318
cotangent D
567128
5^57638
5-48451
5-39552
5-30928
5^22566
5-14455
5-06584
4-98940
4-91516
4-84300
4-77286
4-70463
4-63825
4-57363
4-51071
4-44942
4-38969
433148
4-27471
4-21933
4-16530
4-11256
4-06107
401078
3-96165
3^91364
3 •86671
3-82083
3-77595'
373205
3-68909
3-64705
3 60588
3-56557
3-52609
3-48741
3-44951
3-41236
3-37594
3-34023
3-30521
327085
tangent
I-0I543
i^oi595
1-01649
1-01703
I •01758
1-01815
I 01872
1-01930
1-01989
I -02049
I -021 10
I -02 1 7 1
I 02234
1-02298
1-02362
I ^02428
I -02494
1-02562
I 02630
1-02700
1-02770
1-02841
1-02914
1-02987
I 03061
1-03137
I -032 1 3
1-03290
1-03368
I -03447
I 03528
I ^03609
1-03691
I -03774
1-03858
I -03944
I 04030
1-04117
I -04206
1-04295
1-04385
I -04477
I 04569
052
054
054
055
057
057
058
059
060
061
061
063
064
064
066
066
068
068
070
070
071
073
073
074
076
076
077
078
079
081
081
082
083
084
086
086
087
089
089
090
092
092
I)
98481
98430
98378
98325
98272
98218
98163
98107
98050
97992
^7934
);875
9781S
97754
97692
97630
97566
97502
97437
97371
97304
97237
97169
97100
97030
96959
96887
96815
96742
96667
96593
965^7
96440
.96363
96285
96206
96126
96046
95964
95882
95799
95715
95630
73=
18
10=
LOGARITHMIC FUNCTIONS
log sin D
71]
23967
•24677
•25376
•26063
•26739
•27405
•28060
I
698
687
676
666
655
64s
635
625
616
= 3/S 582
-960 ^^^
■33534 ,66
■5ATOO. ^
•29340
•29966
■305.8
•31
•3
^-rlOO
•3465
•34100!^ g
. -J - -7 - T
537
•35752
•36289 'J'
•37341 'i I
4/:>
■41300 ^g
•^'^^' 464
■-^''^'1458
•43591 :|^3
•44034 438
■^^-^'"^ 433
•44905 li'
■45334/: 2^
-. . -_.o 424
log coscc
076033
075323
0-74624
073937
0-73261
0-72595
071940
0-71295
0-70660
0-70034
0-69418
0^688 1 1
0^682I2
0^67622
0-67040
0-66466
0-65900
0-65342
0-64791
0^64248
0-637II
0-63181
0-62659
; 0-62 1 42
061632
0-61129
0^60631
0^60140
0-59654
0-59175
0-58700
0-58232
;0-57768
io-57310
0-56857
0-56409
0-55966
0-55528
0-55095
0-54666
0-54242
0-53822
053406
46594
log COS D log sec
log tan D log cotan
•24632
•26086
•26797
•27496
•28186
•28865
■-9S3S
•30195
•30846
•31489
•32122
•32747
•33365
•33974
•34576
•35170
■3S7S7
-36336
•36909
•37476
•38035
•38589
•39136
•39677
•40212
•40742
•41266
•41784
•42297
•42805
•43308
•43806
•44299
•44787
•45271
•4S7SO
•46224
•46694
•47160
•47622
•48080
48534
733
721
711
699
690
679
670
660
651
643
633
62
075368
0-74635
0-73914
0-73203
0-72504
0-71814
0-71135
0-70465
0-69805
0-69154
0-68511
0-67878
579
573
567
559
6,8 067253
^ ^ 0-6663 ^
5^9 0.66026
601 ,
0-65424
'f 0-64830
0-64243
o 63664
0-63091
0-62524
^, 0-61965
^^4 0-61411
:)47
541
535
530
524
518
513
508
503
498
493
488
484
479
474
0-60864
060323
0-59788
0^59258
0-58734
io-58216
'0-57703
0-57195
0^56692
0^56194
0-55701
0-55213
0-54729
054250
0-^3776
^66 ^-53306
^ 0^52840
0-52378
0-51920
051466
462
458
454
log cotan
log sec D log cos
o 00665
0^00687
0^00710
0-00733
0-00757
0-00781
000805
0-00830
0-00855
©•00881
©•00907
o -009 3 3
o 00960
©•00987
0-01014
©•01042
©•01070
©•01099
001128
©•01157
0-01187
©-01217
0-01247
©•01278
001310
©•©1341
©•©1373
©•01406
©•01439
©•©1472
001506
©•©154©
<J-oi574
©•©i6©9
©•©1644
©•0168©
001716
©•©1752
©•01789
©•01826
©•01864
©•©1902
00 1940
log cosic
022
023
023
024
©24
©24
025
025
©26
©26
026
027
027
027
©28
©28
©29
©29
029
03©
©3©
03©
031
032
031
032
033
033
033
034
034
034
035
035
036
©36
036
037
037
038
038
038
u
1-99335
1-99313
1-99290
1-99267
1^99243
r992i9
199195
1-99170
1-99145
1-991 19
1-99093
I •99©67
i 99040
I -990 1 3
1^98986
T^98958
1-98930
i^989©i
198872
1-98843
1-98813
1-98783
1-98753
1^98722
I 98690
T^98659
i^98627
1-98594
i^9856i
i^98528
I 98494
I •9846©
1-98426
1-98391
1-98356
1-98320
I 98284
i^98248
i^982i I
1-98174
I -98 1 36
I -98098
I 98060
log
50'
40'
30'
20'
10'
79^
50^
40'
30'
20'
I©'
78^
50'
40'
30'
20'
10'
77°
50'
40'
30'
20'
I©'
76'
50'
40'
30'
20'
10'
75'
50'
40'
30'
20'
10'
740
50'
40'
30'
20'
10'
73°
73''
19
17° NATURAL FUNCTIONS
•29237
•29515
•29793
•30071
•30348
•30625
•30902
•31178
•31454
•31730
•32006
•32282
•32557
•32832
•33106
•33381
•33655
33929
•34202
34475
34748
•35021
•35293
•35565
•35837
•36108
■36379
•36650
•36921
•37191
37461
37730
37999
38268
38537
38805
39073
39341
39608
39875
40142
40408
40674
278
278
278
277
277
277
276
276
276
276
276
275
275
274
275
274
274
273
273
^71
273
272
272
272
271
271
271
271
270
270
269
269
269
269
268
268
268
267
267
267
266
266
3 42030
3-38808
3-35649
3-32551
3-29512
3-26531
3-23607
3-17920
3-I5I55
3-12440
3-09774
3-07155
3-04584
3-02057
3-99574
2-97135
2-94737
292380
2^90063
2^87785
2-85545
2-83342
2-81175
279043
2-76945
2^7488i
2^72850
2^70851
2^68884
2-66947
2^65040
63162
61313
59491
2-57698
55930
2-54190
2-52474
2-50784
491 19
2-47477
2-45859
"I
tangent
-30573
-30891
-31210
-31530
•31850
-32171
•32492
•32814
•33136
•33460
•34108
•34433
-34758
-35085
-35412
-35740
•36068
-36397
•36727
-37057
•37388
•37720
■38053
•38386
-38721
-39055
•39391
•39727
•40065
-40403
-40741
•4108 1
•4 1 42 1
•41763
•42105
42447
•42791
•43136
■43481
•43828
•44175
•44523
secant D cotangent
D
318
319
320
320
321
321
322
122
324
323
325
325
325
1^7
2,27
328
328
329
330
330
331
332
33Z
335
334
336
338
338
338
340
340
342
342
342
344
345
345
347
347
348
cotangent D
3 27085
3-23714
3-20406
3-17159
3-13972
3-10842
3 07768
3-04749
3-01783
2-98869
2-96004
2-93189
2 9042 I
87700
2-85023
2-82391
79802
77254
274748
2^72281
853
2^67462
2^65109
2^62791
2 60509
2-58261
2-56046
2-53865
2-51715
2-49597
2-47509
2^45451
2-43422
2-41421
2^39449
2^37504
235585
2^33693
31826
2-29984
2'28i67
2-26374
2 24604
tangent
I 04569
I ^04663
I -04757
1-04853
1-04950
1-05047
I 05146!
1-05246
1-05347
1-05449
1-05552
1-05657
1-05762
1-05869
1-05976
1-06085
1-06195
I -06306
I -06418
I -065 3 1
I -06645
I -06761
1-06878
I -06995
IO7115
1-07235
1-07356
I -07479
I -07602
1-07727
107853
I -0798 1
I -08 109
1-08239
1-08370
1-08503
I 08636
I -0877 1
I -08907
I -09044
1-09183
1-09323
I -09464
094
094
096
097
097
099
100
lOI
102
103
105
105
107
107
109
no
1 1 1
1 12
113
114
116
117
117
120
120
121
123
123
125
126
128
128
130
131
133
133
135
136
-^2,7
139
140
141
cosecant D
•95630
•95545
•95459
•95372
•95284
•95195
•95106
•95015
•94924
•94832
•94740
•94646
•94552
-94457
-94361
-94264
-94167
-94068
-93969
-93869
■93769
•93667
•93565
■93462
93358
93253
93148
93042
•92935
-92827
•92718
•92609
•92499
•92388'
•922761
•92164'
-92050 1
-91930!
•91822
•91706
•91590
•91472
91355
085
086
,087
I'
1089
091
1091
092
092
J094
1 094
095
096
097
097
099
099
100
100
102
102
103
104
105
105
106
107
108
109
109
no
111
1 12
112
114
114
114
116
116
118
117
5^
4C
3<
2C
IC
6£
5c
4c
3C
20
10
68
50
40
30
20
iO
67
50
40
30
20
10'
66'
D
66^
20
IT
LOGARITHMIC FUNCTIONS
loj; sin I D I log coscc
•46594
•47005
•4741 1
•47814
•48213
•48607
401
406
•48998
■49385
•49768
403
399
394
391
387
3S3
r^8i^^°
'f'368
f 4 36s
''^'^ 362
;i99i
;235o
2705
jOI
•505
■5
SI
)0^^3
50896'
•53057 ^^8
53405 346
359
355
352
5375^
•54093
•54433
•54769
•55102
•55433338
■55761 ...
•56085 ^^^
•56408
■•56727
342
340
336
333
33
323
319
317
317
57044 1/
57978
58284
•58588
306
304
301
-■ ' -^ 299
■S9I88 ^H
.9/78
•60070 ~
•60359
•60646
•60931
]o^ COS
289
287
285
053406
0-52995
0-52589
0-52186
0-51787
0-51393
0SI002
0-50615
0-50232
0-49852
0-49477
0-49104
048736
0-48371
0-48009
0-47650
0-47295
0-46944
0-46595
0-46249
0-45907
0-45567
0-45231
0-44898
044567
0-44239
0-43915
0-43592
0-43273
0-42956
042642
0-42331
0-42022
O-41716
0-41412
0-41 1 1 1
0-40812
0-405 16
0-40222
0-39930
0-39641
0-39354
0-39069
log SLC
log tan D log cotan
•48534
•489S4
•49430
-49872
•5031 1
•50746
•51178
•51606
-52031
•52452
•52870
•53285
•53697
•54106
•54512
•54915
•55315
•55712
•56107
•56498
-56887
•57274
•57658
•58039
•58418
•58794
•59168
•59540
•59909
•60276
•60641
•61004
-61364
-61722
-62079
•62433
•62785
•63135
•63484
•63830
•64175
•64517
•64858
439
435
432
428
425
421
418
415
412
409
406
403
400
397
395
391
389
3S7
384
381
379
376
372
369
367
365
360
358
357
354
352
350
349
346
345
342
341
^^^0.51466
446 ^•5^"^^
442 °'5°570
^^ 0-50128
0-49689
0-49254
o 48822
0-48394
0-47969
0-47548
1 0-47 1 30
0-46715
o 46303
0^45894
o^45488
0^45085
0-44685
0-44288
043893
0-43502
0-43113
0-42726
0-42342
0-41961
041582
^_^ 0-41206
^^4 1 0.40832
0-40460
0-40091
0-39724
363:^-^9359
0-38996
0-38636
0-38278
0-37921
0-37567
0-37215
0-36865
0-36516
0-36170
0-35825
0-35483
i 0-35142
og cotan j D
log tan
log sec
001940
0-01979
0-02018
0-02058
0-02098
0-02139
002179
0-0222I
0-02262
0-02304
0^02347
0-02390
039
039
040
040
041
040
042
041
042
043
043
043
log cos
0-024 J 7 '
0-02521
0-02565
0-02610
0-02656
002701
0-02748
0-02794!
0-02841
0-02889
0-02937!
002985
0^03034
0-03083 I
0-031321
0-03182 I
0-03233!
0-03283
0-03335
0-033861
0-03438!
0-03491 I
0-03544 1
003597
0-03651 I
0-03706
0-03760
0-03815
0-03871
003927 '
log coscc D
044
044
045
046
045
047 I
0461
047!
048:
048'
048!
049
049
049
050
051
050
052
0511
052
053
053
053
054
055'
054
055
056
o;6
-98060
•98021
-97982
•97942
-97902
-97861
-97821
-97779
•97738
•97696
•97653
-97610
■97567
•97523
•97479
•97435
-97390
•97344
•97299
-97252
-97206
•97159
•971 1 1
•97063
•97015
•96966
-96917
-96868
•96818
-96767
•96717
-96665
•96614
•96562
•96509
•96456
96403
•96349
•96294
-96240
-96185
•96129
•96073
log sin
66^
21
24° NATURAL FUNCTIONS
40674
■40939
■41204
■41469
•41734
•41998
42262
•42525
■42788
•43051
•43313
■43575 j
•43837 i
•44098 i
•44359!
•44620 1
•44880
•45140
•45399
•45658
•45917
•46175 j
•46433
•46690
•46947
•47204
•47460
•47716
•47971
•48226
•48481
•48735
•48989
•49242
•49495
•49748
•Soooo
•50252
•50503
•50754
•51004
•51254
5 1 504
26s
265
265
265
264
264
263
263
263
262
262
262
261
261
261
260
260
259
259
259
258
258
257
257
257
256
256
255
255
255
254
254
253
253
253
252
252
cosecant
2^45859
2-44264
2-42692
2-41142
39614
38107
36620
35154
33708
32282
30875
29487
28117
26766
2-25432
2-24116
2-22817
2-21535
2-20269
2-19019
2-17786
2-16568
2-15366
2^i4i78
2^I3005
2-1 1847
2-10704
2-09574
2-08458
2-07356
2-06267
2-05191
2-04128
2-03077
2-02039
2-OIOI4
2-00000
1-98998
1-98008
1-97029
I -96062
1-95106
1-94160
■^ secant
990
979
967
956
946
1)
tangent
44523
44872
45222
45573
45924
46277
46631
46985
47341
47698
48055
48414
48773
49134
49495
49858
50222
50587
S0953
51320
51688
52057
52427
52798
53171
53545
53920
54296
54673
55051
5543 1
55812
56194
56577
56962
57348
57735
58124
58513
58905
59297
59691
60086
Dtangent
349
350
351
351
353
354
354
356
357
357
359
359
361
361
363
364
365
366
367
368
369
370
Z7Z
374
375
376
378
380
381
382
383
385
386
387
389
389
392
392
394
395
D
cotangent
2-24604
2-22857
2-21132
2-19430
2-17749
2-16090
2-14451
2-12832
2-11233
2-09654
2-08094
2-06553
2-05030
2-03526
2-02039
2-00569
1-99116
1-97681
I -96261
1-94858
1-93470
1-92098
1-90741
I -89400
1-88073
1-86760
1-85462
1-84177
1-82906
1-81649
1-80405
1-79174
1-77955
1-76749
1-75556
1-74375
1-73205
1-72047
1-70901
I ^69766
1-68643
1-67530
I 66428
tangent
09464
09606
09750
09895
10041
10189
10338
10488
10640
10793
10947
11103
1 1 260
11419
11579
11740
11903
12067
12233
12400
12568
12738
12910
13083
13257
13433
13610
13789
13970
14152
14335
14521
14707
14896
15085
15277
15470
15665
1 5861
16059
16259
16460
16663
osecant D
142
144
145
146
148
149
150
152
153
154
156
157
159
160
161
163
164
166
167
1<
170
172
173
174
176
177
179
181
182
183
186
186
II
189
192
193
195
196
198
200
201
203
22
24°
LOGARITHMIC FUNCTIONS
24°
lO
20
30
40
50'
25°
10'
20'
30'
40
50'
26°
10'
40
50'
27°
10'
20'
30'
40'
_Jo'
28°
10
20
30
40
50
29°
10'
20'
30
40'
50'
30°
10'
20
30'
40'
31°
loR
D log cosec
1-60931 2g3 039069
I-6I2I4 ,,g 0-38786
I -61494 ^-^0*38506
1-61773 276 ^'^"^^^7
1-62049 2-4 ^•'"'^'^
1-62323
1-62595
1-62865
£•63133
1-633981
0-37951
0-37677
037405
272
270
268°-^7i35
265
264
0-36867
0-36602
I -63662 i^gj 0-36338
1-63924; 260'°'^^°^^
1-64184^ g 0-35816
"64442|^^^iO-35558
•64698
256
1-64953!^^^
1-65205 1
I -65456 1
1-65705
1-65952!
1-66197
r-66441
1-66682
1-66923
1-67161
1-67398!
1-67633
1-67866
I 68098
1-68328
i 68557
1-68784
I -690 10
1-69234
1-69456
1-69677
1-69897
1-70115
1-70332
1-70547
I -70761
1-70973
1-71184
2
252
251
249
247
245
244
241
241
238
237
1235
233
232
log; cos
0-35302
0-35047
0-34795
0-34544
0-34295
0-34048
0-33803
0-33559
0-33318
'0-33078
0-32839
10*32602
'0-32367
[0-32134
0-31902
230 "^ ^
^on 0-31672
0-31443
0-3I2I6
0-30990
0-30766
0-30544
0-30323
0-30103
0-29885
0-29668
0-29453
0-29239
0-29027
0-28816
log sec
log tan D log cotan
339
338
229
227
;226
'224
'222
221
220
218
'217
j2I5
:2i4
212
211
1-64858
I -65197
1-65535 ,,,
1-65870:^^5
1-66204 ^^^
1-66537 ^^-^
1-66867
I -67196
1-67524
1-678501
330
329
328
326
f-68174'3^4
r-68497 \~,\
i-688i8 •^"'
- r o 320
1-69138 I
£-69457;^,^
£•69774 "f,;
I -700891^
I -70404 '^j^
1-70717
I -71028 1
1-71339'^"
i-7i648'^°9
T4:955i307
1-72262
i-72567
"72872
73175
log sec
log cos
I
T
£-73476
\-7l777
1-74077
1-74375
£-74673
£-74969
1-75264
£•75558
£•75852
1-76144
£•76435
1-76726
£-77015
\j77Z^Z
i-7759i|
i -77877!
307
305
305
303
!30i!
'300 1
298
298
296
295
294!
294
292
291
I 290
;29ol
288^
288,
286
03927 os6
-03983:057
0-04040:^^^
-04098.^
•^4156!^^
0-04214^58
°'«4272 ^^
0-04332
0-0439I 060
0^04451 i 061
0-04512 i^gj
°-°4573'o6i!
004634062'
°-°4696|^62
^•°4758o63
0-04821 I ^
0-04884 i°4
°'°4948,^6^
°°5oi2|o6s
°-°5077|o65
0-05142^6^
-^•^^ 207 1066
•°5 273 1 067
•°5340^6;
°°S407!o67
°-°5474lo68
^•°5 542 1^68
0-05679^6^
0-05748 I
005818
0-05888!
^ 071
0-05959
0-06030,^^^
0-06102 ^^^
o-o6i74'^;3
o 06247 0^3
0-06320,
r 074
0-06394! '^
o.o6^68|^;4
0-06543 l;^
0-06618 r^5
'075
o 06693
log cotaii
0-35142
0-34803
0-34465
0-34130
0-33796
0-33463
0-33133
0-32804
0-32476
0-32150
0-31826
0-31503
0-31182
0-30862
0-30543
0-30226
0-2991 1
0-29596
029283
0-28972
0-28661
0-28352
0-28045
0-27738
0-27433
0-27128
I0-26825
0-26524
0-26223
0-25923
025625
0-25327
0-25031
0-24736
0-24442
0-24148
023856
0-23565
0-23275
0-22985
0-22697
0-22409
0-22123 f
log tan I log cosec D log sin
i -96073
T-96017
1 1-95960
,£•95902
[£•95845
'1-95786
195728
: £-95668
1-95609
1-95549
1-95488
1-95427
I 95366
£-95304
£-95242
£•95179
1-95110
1-95052
i -94988
1-94923
1-94858
£-94793
£•94727
I -94660
1-94593
£-94526
£-94458
£-94390
£-94321
1-94252
1-94182
T^94ii2
1-94041
£-93970
£•93898
1-93826
i 93753
1-93680
1-93606
£•93532
£•93457
1-93382
i 93307
66°
50'
40'
30'
20'
10'
65°
50'
40'
30'
20'
10'
64°
50'
40'
30'
20'
\o'
63°
50'
40'
30'
20'
10'
62°
7o^
40'
30'
20'
10'
61°
50'
40'
30'
20'
10'
60°
50'
40'
30'
20'
10'
59°
59=
23
31° NATURAL FUNCTIONS
5 1 504
51753
52002
52250
52498
52745
52992
53484
53730
53975
54220
S4464
54708
54951
55194
55436
55678
55919
56160
56401
56641
56880
57119
57358
57596
S7^U
58070
58307
58543
S8779
59014
59248
59482
59716
59949
60182
60414
60645
60876
61 107
61337
61566
D
249
249
248
248
247
247
246
246
246
245
245
244
244
243
243
242
242
241
241
241
240
239
239
239
238
2Z7
2Z7
^17
236
236
235
234
234
234
233
232
231
231
231
230
229
cosecant D
94160
93226
92302
91388
90485
89591
88708
87834
86970
86116
85271
84435
83608
82790
81981
81180
79604
78829
78062
77?>o?>
76552
75808
75073
7434S
73624
72911
72205
71506
70815
70130
69452
68782
68117
67460
66809
66164
65526
64894
64268
63648
63035
62427
934
924
914
903
894
883
874
864
854
845
836
827
818
809
801
792
784
775
767
759
751
744
735
728
721
713
706
699
691
685
678
670
665
657
651
645
638
632
626
620
613
608
n
tangent D
"60086
60483
60881
61280
61681
62083
62487
62892
63299
63707
641 17
64528
64941
65355
65771
66189
66608
67028
67451
67875
68301
68728
69157
69588
70021
70455
70891
71329
71769
72211
72654
73100
73547
73996
74447
74900
75355
75812
7(^7 i?>
77196
78129
i)tanj)«nt I)
397
398
399
401
402
404
405
407
408
410
411
413
414
416
418
419
420
423
424
426
427
429
431
433
434
436
438
440
442
443
446
447
449
451
453
455
457
460
461
463
465
468
cotangent
1-66428
1-65337
1-64256
1-63185
1-62125
I -61074
I 60033
1-59002
1-57981
1-56969
1-55966
1-54972
I 53987
I-53010
1-52043
I -5 1084
1-50133
1-49190
1-48256
1-47330
I -4641 1
I-45501
1-44598
1-43703
I-42815
I -41934
I-41061
I -40195
1-39336
1-38484
137638
1-36800
1-35968
1-35142
1-34323
1-33511
I 32704
I -3 1904
I-31110
1-30323
1-29541
1-28764
1-27994
tangent
994
985
977
967
959
951
943
934
926
919
910
903
895
888
881
873
866
859
852
846
838
832
826
819
812
807
800
794
7^7
782
777
770
D
•16C63
-16868
•17075
-17283
-17493
-17704
•17918
•18133
-18350
-18569
-18790
-19012
•19236
•19463
•19691
-19920
-20152
-20386
-20622
•20859
-21099
-21341
-21584
-21830
•22077
-22327
-22579
•23089
■2i2>A7
•23607
•23869
-24134
•24400
•24669
-24940
-25214
-25489
•25767
•26047
-26330
-26615
I 26902
20s
207
208
210
211
214
215
217
219
221
222
224
227
22i
229
232
234
236
237
240
242
243
246
247
250
252
254
256
251
260
262
265
266
269
271
274
275
278
280
285
287
24
LOGARITHMIC FUNCTIONS
log sin
1-71184
£•71393
171602
I71809
I72OI4
I72218
172421
T72622
172823
173022
I73219
I73416
1-73611
173805
173997
1-74189
174379
1-74568
1-74756
174943
I-75128
I75313
1-75496
1-75678
1-75859
176039
1-76218
176395
176572
_ '6747
i -76922
1-77095
1-77268
177439
1-77609
177778
1-77946
1-78113
T-78280
1-78445
1-78609
1-78772
I 78934
log COS
D log cosec
™ 0-28816
200
l^^ 0-28398
207 o
205 °''^'?3
,4,0.27986
30], 0-27782
201 1 0-27579
,01 0-27378
197
197
195
194
192
192
0-26978
jO-26781
1 10-26584
026389
0-26195
0-26003
0-2581 1
;i9-362.
^8^10-25432
0-25244
0-25057
0-24872
0-24687
0-24504
0-24322
0-24141
0-23961
0-23782
0-23605
0-23428
0-23253
0-23078
0-22905
0-22732
0-22561
187
185
185
183
182
181
180
179
177
177
175
175
173,
173
1711
17O'
169 °-^-39i
^^^0-22222
0-22054
0-21887
0-21720
0-21555
0-21391
0-21228
167
167
165
164
163'
162
0-21066
i log sec
log tan
177877
178163
1-78448
1-78732
I-79015
1-79297
i -79579
1-79860
r -80 1 40
1-80419
1-80697
1-80975
1-81252
1-81528
1-81803
1-82078
1-82352
1-82626
i'-^2899
1-83171
1-83442
1-83713
1-83984
1-84254
1-84523
1-84791
1-85059
£•85327
1-85594
1-85860
1-86126
1-86392
1-86656
1-86921
1-87185
1-87448
1-87711
1-87974
1-88236
r-88498
r-88759
1-89020
i 89281
log cotan
286
285
284
283
282
282
281
280
279
278
278
2771
276,
275 I
275 I
274 I
274'
273
272^
271
271
271
270
269
2681
268
2681
267 I
266
2661
266
264
265
264
263
263
263
262
262
261
261
261
log cotan
0-22123
0-21837
0-21552
0-21268
0-20985
0-20703
0-20421
0-20140
0-19860
O-19581
0-19303
0-19025
0-18748
0-18472
0-18197
0-17922
0-17648
0-I7374
0-17101
0-16829
0-16558
0-16287
0-16016
0-15746
0-15477
0-15209
0-14941
0-14673
0-14406
0-14140
0-13874
0-13608
0-I3344
0-13079
0-12815
0-12552
0-12289
0-12026
0-1 1764
o-i 1502
0-11241
0-10980
0-10719
log tan
log sec
0-06693
0-06770
0-06846
0-06923
0-07001
0-07079
007158
0-07237
0-07317
0-07397
0-07478
0-07559
007641
0-07723
0-07806
0-07889
0-07973
0-08058
008143
0-08228
0-08314
0-08401
0-08488
0-08575
log cos
077
076
077
078
078
079
079
080
080
081
081
082
082
083!
083!
0841
085 1
085:
085'
086 1
087!
087!
087!
o,. 089
008664 ^gg
0-08752
0-08842
0-08931
0-09022
0-09113
o 09204
0-09296
0-09389
0-09482
0-09576
0-09670!
009765
U-0986I
0-09957
0-10053
0-IOI5I
0-10248
0-10347
log cosec
090
089
091
091
091 I
092
093
093;
094 I
094'
095
096
096
096
098
097
099
93307
93230
93154
93077
92999
92921
92842
92763
92683
92603
92522
92441
92359
92277
92194
92111
92027
91942
91857
91772
91686
91599
91512
91425
91336
91248
91158
91069
90978
90887
90796
90704
9061 1
90518
90424
90330
90235
90139
90043
89947
89849
89752
89653
log sin
52*
25
J" NATURAL FUNCTIONS
•6is66
•61795
•62024
•62251
•62479
•62706
•62932
•63158
•63383
•63608
•63832
■64056
•64279
•64501
•64723
-64945
■65166
•65386
65606
•65825
■66044
■66262
■66480
•66697
66913
67129
67344
67559
67771
679^7
68200
68412
68624
68835
69046
69256
69466
69675
69883
70091
70298
70505
7071 1
D
229
229
227
228
227
226
226
225
225
224
224
223
222
222
222
221
220
220
219
219
218
218
217
216
216
215
215
214
214
213
212
212
21 I.
211
210
210
206
208
208
207
207
206
I 62427
i^6i825
1-61229
I -60639
I ^600 5 4
1-59475
I •58902
1-58333
I-5777I
1-57213
i^5666i
i^56ii4
I 55572
1-55036
1-54504
1-53977
1-53455
1-52938
I 52425
1-51918
1-51415
1-50916
1-50422
1-49933
1-49448
1-48967
I -48491
I -48019
I -475 5 1
1-47087
146628
1-46173
I -4572 1
1-45274
1-44831
I -44391
I 43956
1-43524
I -43096
1-42672
I -4225 1
1-41835
1-41421
1602
596
590
585
579
573
569
562
I558
552
547
542
536
'532
1527
522
517
:5i3
507
i503
499
494
489
485
481
476
472
468
464
459
455
452
447
443
440
435
432
428
424
421
416
414
tangent
•78129
•78598
-79070
-79544
-80020
-80498
-80978
-81461
-81946
-82434
-82923
-83415
•83910
•84407
•84906
•85408
•85912
•86419
•86929
•87441
-87955
•88473
•88992
-89515
•90040
•90569
•91099
•91633
-92170
-92709
•93252
•93797
-94345
-94896
-95451
-96008
•96569
-97133
-97700
-98270
-98843
-99420
00000
J I -00000
cotangent
D
469
472
474
476
478
480
483
485
488
489
492
495
497
499
502
504
507
510
512
514
518
519
523
525
529
530
534
537
539
543
545
548
551
555
557
561
564
567
570
573
577
580
D
cotangent
1^27994
1-27230
1-26471
I-25717
I -24969
1-24227
1-23490
1-22758
I -2203 I
I-21310
1-20593
I-19882
I-I9175
I -18474
I-I7777
1-17085
1-16398
1-15715
1-15037
1-14363
1-13694
I -1 3029
1-12369
1-11713
1-11061
1-10414
1-09770
1-09131
I -08496
1-07864
I 07237
I -066 1 3
1-05994
1-05378
I ^04766
I ^04 1 5 8
I 03553
1-02952
I-02355
I •o 1 76 1
I •o 1 170
1^005 8 3
I 00000
tangent j
26
D
764
759
754
748
742
72>7
732
727
72T
717
711
707
701
697
692
687
67S
674
669
665
660
656
652
647
644
639
635
632
627
624
619
616
612
608
605
601
597
594
591
587
583
289
1-2777S
1-2807 s
1-28374
I 28676
1-28980
1^29287
1-29597
1-29909
1-30223
1-30541
I -30861
1-31183
I -3 1 509
1-31837
1-32168
I 32501
1-32838
1-33177
1-33519
1-33864
1-34212
1-34563
1-34917
1-35274
1-35634
1-35997
1-36363
1-36733
1-37105
1-37481
i^3786o
1^38242
1^38628
1-39016
1-39409
1-39804
I ^40203
I ^40606
I^4IOI2
I 4 142 1
1-26902
1-27483 295
297
299
302
304
307
310
312
314
318
320
322
326
328
331
333
337
339
342
345
348
351
354
357
360
363
366
370
372
376
379
382
386
388
393
395
399
403
406
409
D
179-
180
181
I 18.
7
)
•78801
•78622
•78442
•78261
•78°79^8,
-778971^82
•777151,84
•77531 ,8
'77347
•77162
-76977
-767911,8; 1
-766041,8^ 5
•7^417 ji88 '
•76229,^88
•76041
•75851
185
185
186
190
190
90
•7566
■7S47I
-75280. ^^
-75088 9;
-74896 ;9-
-74509 1 ^^ K
•74314 1 4}
•741201
-73924
•7372s
-73531
•73333
196
196
197
198
- 198
•73135 8
•72937 200
•72737 200
-72537 200
-72137
/ 00/ 201
-72136 2^_.
•7^934 202
■71732
' ' ^ 20\
.71529 -^
•71325
-7II2I
•70916 -^
-7071 1
5(
4^
3<
2C
4]
5C
4^'
3C
2L
IC
46
5<'-
204 40
J04 -^^
:o5
10
45
D
41
I
LOGARITHMIC FUNCTIONS
log sin 1 D log cosec
178934 ,61
1-79256
1-79415
1795731
I -7973 1
159
158
158
156
IS6
154
154
153
1-79887
1-80043
1-80197
£•80351
^•80504,
1-806561^5-
1-80807'
1-80957:
I -8 1 106 j
1-81254I
T-81402 [
T-8i549|
1-816941
1-81839^
T-81983
1-82126
1-82269
1-82410
1-82551
1-82691 •
1-82830I
1-82968 I
1-83106^
1-83242^
183378
1-83513
1-83648
1-83781:
1-83914
1-84046
1-84177
T-84308
1-84437
1-84566
1-84694
1-84822
1-84949
151
ISO
149
148
148
147
145
14s
144
143
143
141
! 141
140
139
138
138
136
136
135
135
133
133
132
131
131
129
129
128
128
127
0-21066
0-20905
0-20744
0-20585
0-20427
0-20269
0-20II3
0-19957
0-19803
0-19649
0-19496
0-I9344
0-19193
0-19043
|o-i8894
0-18746
10-18598
:o-i845
0-18306
0-18161
0-18017
0-17874
0-17731
0-17590
0-17449
0-17309
0-17170
|o-i703
10-1689
'0-16758
0-16622
:o-i6487
|o-i6352
]0-i62i9
jO-i6o86
10-15954
0-15823
15692
15563
15434
15306
15178
15052
log COS I D I log sec
log tan D
1-89281
I-89541
I-89801
T- 9006 1
T-90320
1-90578
1-90837
1-91095
1-91353
1-91610
T-91868
1-92125
1-92381
1-92638
1-92894
1-93150
1-93406
1-93661
1-93916
1-94171
1-94426
I -9468 1
1-94935
I -95 190
1-95444
1-95698
1-95952
1-96205
1-96459
I -967 1 2
1-96966
1-97219
1-97472
1-97725
1-97978
1-98231
1-98484
T-98737
260
260
260
259
258
259
2S8
258
257
258
'log cotan
O-IO719
■ 0-10459
{0-10199
! 0-09939
j 0-09680
0-09422
0-09163
0-08905
1 0-08647
0-08390
0-08132
:% 0-07875
^^^0-07619
I -99242 j
1-99495 '
1-99747'
0-00000
0-07362
0-07106
^^0-06850
i ^5 0.06594
12^. .0-06339
0-06084
0-05829
0-05574
0-05319
0-05065
0-04810
o 04556
0-04302
0-04048
0-03795
0-03541
0-032
0-03034
0-02781
0-02528
0-02275
0-02022
0-01769
O-OI516
0-01263
2 5 2'
•^ 10-0101 I
253 L.,
256
256
255
1255
i255
I254
I255
'254
254
254
!253
'254
253
'254
253
,253
253
1253
i2S3
■253
253
253
252
253
10-00758
0-00505
0-00253
0-00000
log cotan I D ' log tan
log sec
0-10347
0-10446
0-10545'
0-10646
0-10746
10848
0-10950
O-IIO52
0-11156
O-II259
O-II364
o- 1 1469
0-II57S
0-11681
0-11788
0-11895
0-12004
0-12113
0-12222
0-12332
0-12443
0-12554
0-12666
0-12779
0-12893
0-13007
0-13121
0-13237
0-I3353
0-13470
0-13587
0-13705
0-13824
0-13944
0-14064
0-14185
0-14307
0-14429
0-14552
0-14676
0-14800
0-14926
0-15052
log cosec 1 D
099
099,
loi !
100 1
102
102
102
104
103
105
105
106
106
I 107
;.o7i
109 I
! 1091
1 109
1 10
III
'hi;
: 112 !
I 114!
"4
iii6l
;ii6|
117;
117
118
119
120
120
121
122
122
123
124
1241
126
126
log cos
89653
89554
89455
89354
89254
89152
89050
8S948
88844
88741
88636
88531
88425
88319
88212
88105
87996
87887
87778
87668
87557
87446
87334
87221
87107
86993
86879
86763
86647
86530
86413
86295
86176
86056
85936
85815
85693
85571
85448
85324
85200
85074
84949
45^
27
FOUR-FIGURE TRIGONOMETRICAL. TABLES
Radians
De-
grees
Sine
Cosec,
Tangent
Cotan.
Secant
Cosine
•OOOOO
0
•0000
CO
•0000
00
I -0000
I -OOOO
90
1-57080
•01745
I
-0175
57-2986
-0175
57-2899
I -0002
•9998
1-55334
•03491
2
-0349
28-6537
-0349
28-6362
I -0006
-9994
88
1-53589
•05236
3
•0523
19-1073
•0524
19-08 1 1
1-0014
•9986
87
i-5i''44
•06981
4
•0698
14-3356
-0699
14-3OJ6
I -0024
•9976
86
1-50-98
•08727-
5
•0872
11-4737
-0875
11-4301
I -0038
-9962
85
I -48 - -
•10472
6
•1045
9-5668
•IO51
9-5144
1-0055
•9945
84
i-46(.;'
•12217
7
•1219
8-2055
•1228
8-1443
1-0075
•9925
83
1-44^^:.
•13963
8
•1392
7-1853
•1405
7-1154
I -0098
•9903
82
1-431^7
•15708
9
•1564
6-3925
•1584
6-3138
1-0125
•9877
81
1-41372
•17453
10
•1736
5-7588
•1763
5-6713
I-OI54
•9848
80
1-39626
•19199
II
•1908
5 '2408
•1944
5-1446
I-O187
.9816
79
1-77881
•20944
12
•2079
4-8097
•2126
4-7046
1-0223
•97 I
78
I--6i::5
•22689
13
14
•2250
•2419
4-4454
4-1336
•2309
-2493
4-3315
1-0263
•972
77^
76
J -34:90
•24435
4-0108
I -0306
-9703
1-3264
•26180
IS
•2588
3-8637
-2679
Vll^-i^
1-0353
-965^
75
i'3(^<-.
•27925
16
-2756
3-6280
•2867
3-4874
I -0403
-961
4
1-291--
•29671
17
•2924
3-4203
•3057
3-2709
1-0457
-9563
73
"r274c
•31416
18
•3090
3-2361
•3249
yo777
I-0515
-95 II
72
I..56C
•33161
19
.3256
3-0716
•3443
2-9042
1-0576
•9455
71
|.239ifc
•34907
20
-3420
2-9238
•3640
2-7475
I -0642
-9397
10
1-22173
•36652
21
•3584
2-7904
•3839
2-6051
I -07 1 1
-9336
*/%-
1-20428
•38397
22
-3746
2-6695
•4040
2-4751
1-0785
•9272
^9
r-18682
•40143
23
•3907
2-5593
-4245
2-3559
I -0864
•9205
ft?
/-16937
•41888
24
•4067
2-4586
-4452
2-2460
I -0946
-9135
0^
/•15192
•43633
25
•4226
2-3662
•4663
2-1445
I-IO34
-9063
65
1-13446
•45379
26
•4384
• 2-2812
•4877
2-0503
I-II26
•8988
64
1-11701
•47124
27
-4540
2-2027
-5095
1-9626
I-1223
•8910
63
1-09956
•48869
28
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