UC-NRLF
IN MEMORIAM
FLOR1AN CAJORI
x~\
C^y^>^
PEACTICAL TKIGONOMETKY
PRACTICAL TRIGONOMETRY
BY
H. C. PLAYNE, M.A.
HEADMASTER OF BANCROFT'S SCHOOL AND FORMERLY
ASSISTANT MASTER AT CLIFTON COLLEGE
AND
R. C. FAWDRY, M.A.
ASSISTANT MASTER AT CLIFTON COLLEGE
THIRD IMPRESSION
NEW YORK:
LONGMANS, GREEN & CO
LONDON: EDWARD ARNOLD
[All Rights reserved]
PREFACE.
DURING the last few years a great change has come
over the teaching of Elementary Mathematics.
The laborious months hitherto spent in acquiring skill
in the manipulation of elaborate Algebraical and
Trigonometrical transformations have often given the
beginner a dislike for Mathematics and have retarded
his progress.
It has been shown that it is quite possible to
arrange (for the average student) a course of Mathe-
matics which is both interesting and educational, by
constantly keeping before him the practical application
of the subject, and omitting as much as possible those
parts of Mathematics which are purely academical.
The object of this book is to give the reader such a
working knowledge of elementary Trigonometry, with-
out avoiding the difficulties or sacrificing thoroughness.
Much that has hitherto been found in the text-books
has been omitted, and the examples throughout will be
seen to be more practical than is usually the case.
The book contains many and varied examples to be
worked out by the student, but we have avoided the
grouping together of batches of examples of the same
type, believing that such a system is the cause of
much mechanical and unintelligent work. Collections
VI PREFACE
of miscellaneous examples occur frequently, so that the
student may be constantly revising what he has learnt
in the earlier chapters. We have avoided those artificial
questions which have gradually been evolved by the
ingenuity of examiners, but are never met with in the
practical application of Mathematics, and have intro-
duced as many examples as possible to illustrate the
use of Trigonometry in Mechanics, Physics and Analy-
tical Geometry. In numerical work we have indicated
the degree of accuracy to which the results are reliable.
Enough examples are worked out in the text to
show how each new principle may be applied, and to
show the best way of arranging the work — which is of
especial importance when logarithms are used ; but we
have endeavoured to leave the student as much as
possible to his own intelligence.
Another special feature of the book is Chapter X,
which deals with solid figures and angles which are not
in one plane. We have also added an introduction to
Trigonometrical Surveying.
We believe that the book will be of value to those
who are preparing for Army and Civil Service Examin-
ations, to Technical Students, and to all who require
Trigonometry for practical purposes.
Our best thanks are due to several friends and
colleagues for much kind help, and in particular to
Mr G. W. Palmer of Clifton College.
December, 1906.
H. C. P.
R. C. F.
CONTENTS.
PAGE
Chapter I. Measurement of Angles .... 1
Chapter II. Trigonometrical Functions ... 5
Miscellaneous Examples. A. ... 16
Chapter III. The use of four figure tables of natural
functions 18
Miscellaneous Examples. B. . 33
Chapter IV. Functions of angles greater than a right
angle 36
Chapter V. Relations between the sides and angles
of a triangle 55
Miscellaneous Examples. C. 61
Chapter VI. Projection, and Formulae for Compound
Angles 64
Miscellaneous Examples. D. . 84
Chapter VII. Logarithms 86
Miscellaneous Examples. E. . . . 100
Chapter VIII. Solution of Triangles : Circumsciibed,
Inscribed and Escribed Circles . . . 103
Miscellaneous Examples. F. . . 117
Chapter IX. Radian, or Circular Measure of Angles. 120
Miscellaneous Examples. G. . . . 132
Chapter X. Angles which are not in one plane :
Trigonometrical Surveying . . . . 135
Miscellaneous Examples. H. . . , 153
Answers
CHAPTER I.
ANGLES.
1. LET OX be a fixed straight line, and let a straight
line OP, initially coincident with OX, turn about the point O
in one plane; then, as it turns, it is said to describe the
angle XOP. The magnitude of the angle depends on the
amount of revolution which OP has undergone.
Fig. 1.
OX is called the initial line.
In Trigonometry there is no limit to the magnitude of
the angles considered.
When OP reaches the position OX', i.e. when X'OX is
a straight line, it has turned through an angle equal to two
X O X
Fig. 2.
right angles; and when it again becomes coincident with OX
it has turned through four right angles.
p. F. 1
A.
2 PRACTICAL TRIGONOMETRY
2. Sexagesimal Measure.
Since all right angles are equal, a right angle might be
chosen as the unit of measurement of angles but it is too
large to be convenient. The unit selected is one-ninetieth
part of a right angle and is called a degree (1°).
A degree is subdivided into 60 equal parts, each of
which is called a minute (!'), and a minute into 60 equal
parts, each of which is called a second (I").
Thus 15° 42' 21" is read 15 degrees, 42 minutes, 27
seconds.
This system of measurement of angles is called the
Sexagesimal measure.
Another unit, called a Radian, is used especially in
theoretical work and will be discussed in Chap. ix.
Example (i).
The angle subtended at the centre of a circle by the side
of an inscribed regular figure may readily be expressed in
Sexagesimal Measure.
Let the regular figure be a Pentagon.
Then at the centre O we have five equal angles whose sum is
four right angles;
Fig. 3.
O£»A°
,'. the angle subtended by each side= — — —72°.
ANGLES
Example (ii).
The angle of a regular figure, e.g. an octagon, may be found
thus : —
Join any angular point A to the other angular points.
Six triangles are formed, the sum of all their angles being
12 right angles.
But these angles make up the eight angles of the figure ;
.*. each angle of the figure =
1080°
= 135°.
Or we may make use of the geometrical theorem that all the
interior angles of any rectilineal figure, together with four right
angles, are equal to twice as many right angles as the figure has
sides.
Thus if a figure has n sides, the interior angles make up
2ft — 4 right angles.
If the figure is regular, each angle is
i — 4
— right angles.
1—2
4 PRACTICAL TRIGONOMETRY
Examples. I.
1. Express in degrees the angles of an equilateral triangle.
2. One angle of a right-angled triangle is 37° 15' 20", find
the other acute angle.
3. Two angles of a triangle are 42° 14' and 100° 12', find
the other angle.
4. What are the angles between the two hands of a clock at
5 o'clock?
5. Express in degrees the angles between the two hands of
a clock at 6.15.
6. Through how many degrees does the minute hand of a
clock turn between 3.10 and 7.25 ?
7. Express 27° 14' 5" in seconds.
8. Find the sexagesimal measure of -486 of a right angle.
9. Find to the nearest second the angle of (i) a regular
hexagon, (ii) a regular heptagon, (iii) a regular pentagon.
10. Express 51° 17' 45" as a decimal of a right angle to
5 places of decimals.
CHAPTER II.
TRIGONOMETRICAL FUNCTIONS.
3. Note on Similar Triangles.
Two equiangular triangles are proved in Geometry to
have their corresponding sides proportional, and the
triangles are called Similar.
That is to say if ABC, A'B'C' are two triangles in
which the angles at A, B, C respectively equal those at
A', B', C', then
AB _ BC _ CA
A'B' ~~ B'C'~ C'A'*
€'
B C
Fig. 5.
Conversely, if
AB _ BC _ CA
A7? ~~ &Cf ~ C/A"' '
the two triangles ABC, A'B'C' are equiangular, having those
angles equal which are opposite corresponding sides.
The student who is unfamiliar with the properties of
similar triangles should carefully work through the follow-
ing Exercise.
Draw an angle XOR equal to 50°. Take any three points
P> PI> ^2) on OR. From these points drop perpendiculars
PN, PiNj, P2N2, on OX. Measure these perpendiculars and
PRACTICAL TRIGONOMETRY
the lengths ON, ON1? ON2. Then write down the values of
the following ratios correct to 2 decimal places :
NP NiPi N2P2> ON ONj ON2 NP N^ N2P2
OP' O
OP2 OP'
OP2' ON' ONX
R
on OR.
N
Fig. 6.
Now take a point P3 on OX and drop a perpendicular P3N3
Measure OP3, P3N3, ON3 and find the values of
N3P3 ON3 NgP3
OP3' OP3' ON3*
State what conclusions you draw from your results.
4. Trigonometrical Functions.
Let XOR "be any angle 0. From any point P in one
of the boundary lines of the angle draw PN perpendicular
P X.
Kg. 7.
to the other boundary line. From the properties of similar
TRIGONOMETRICAL FUNCTIONS 7
triangles, or by actual measurement, it may be shown that
the ratios
NP ON NP
OP' OIP' ON
are constant for all positions of P so long as the magnitude
of the angle remains unchanged.
These ratios, which depend only on the magnitude of 0
are called respectively the sine, cosine, tangent of 0, and
their reciprocals are called respectively the cosecant, secant,
cotangent of 0.
They are thus abbreviated :
NP 1 OP
sin^op, sin0-NP>
ON 1 OP
OP' COS 0 ON3
-^ cot0_J^_™
ON' ~tan<9~NP'
Note. In view of a distinction in Sign which will be
made in Chap. iv. between the direction NP and the
direction PN, it is preferable here to write NP and not PN
in the expressions for sin 0 and tan 0.
Calling NP the side opposite the angle 0, ON the side
adjacent to 0, and OP the hypotenuse, we may write them
. . _ side opposite to 0 ^
hypotenuse
„ side adjacent to 0
cos 0 = — , — *-- ;
hypotenuse
n side opposite to 0
tan 0 = -TJ §£ , ;
side adjacent to 0
and similarly for cosec 0, sec 0 and cot 0.
These ratios are called the trigonometrical functions or
ratios of the angle 0.
Note, (sin A)2 is written sin2 A; i.e. if
sin A = | sin2A = |.
8
PRACTICAL TRIGONOMETRY
5. The definitions of the trigonometrical functions
still hold good for an angle greater than 90°.
If from a point P in one of the boundary lines of the
ir o x
Fig. 8.
angle 0, PN be drawn perpendicular to the other boundary
line produced if necessary, then
NP
/i ON
cosfl = — ;
NP
tan 0 = — etc.
N
Fig. 9.
For the present we shall confine our attention to acute
angles, and it will be explained in Chap. iv. that there are
certain conventions of sign to be adopted in treating of the
ratios of angles greater than a right angle.
TRIGONOMETRICAL FUNCTIONS
9
6. Variation in the value of the ratios as the
angle increases.
In order to compare the values of fractions in Arith-
metic it is convenient to express them with the same
denominator, so in Trigonometry we can compare the
values of various ratios by keeping OP (called the radius
vector) of constant length.
-^
Fig. 10.
As the angle XOP increases from 0° to 90°, does sin A
increase in value or diminish ?
Discuss what happens to the other trigonometrical
ratios.
Why is sin A not greater than 1 ? What is the greatest
value of cos A ? Can tan A exceed 1 ?
Note. The angles of a triangle ABC are conveniently
denoted A, B, C, and the sides opposite these angles re-
spectively a, b, c.
Examples. II a.
1. ABC is a triangle, B being a right angle, AC = 5", AB=4".
Calculate the length of BC and write down sin A, cosC, tan A,
sec A, cosecC. Find the value of sin2 A + cos2 A,
sin A
cos A
, tan A,
l + tan2C, sec2C.
2. In a triangle, c=17, a = 8, £ = 15; prove that B = 90° - A.
Write down the values of sin A, sin B, tan B, cos A, cot A, cosec B,
cos (90° - A), sin (90° - A), tan (90° - A). What ratio of A is equal
to (i) cos (90° - A), (ii) sin (90° - A), (iii) tan (90° - A), (iv)
10 PRACTICAL TRIGONOMETRY
3. The hypotenuse c of a right-angled triangle is 15" and
the side a =12". Calculate the length of b. Find the values
of l+tan2A, sec2 A, l+tan2B, sec2 B, l+cot2A, cosec2A,
sin2 A + cos2 A, sin2 B + cos2 B.
4. In the triangle ABC, A = 90°, and AD is drawn per-
pendicular to BC. From the triangles ABD and ABC write
down two values of sin B, and two values of cos B. Hence find
AD and BD if a=41, c=40, £ = 9.
5. A point A on the circumference of a circle is joined to
BC the extremities of the diameter. AD is drawn perpendicular
to BC. Prove that L BAD = C, and L DAC = B.
From the triangles ABC and ABD write down two values of
sin C. Hence prove AB2= BC . BD.
Prove in a similar way that AC2= BC . CD.
6. In the same figure from the triangles ABD, ADC write
down two values for cot B. Hence prove AD2=BD.DC.
7. ABC is any triangle, AD, BE, CF are the perpendiculars
drawn from the angular points to the opposite sides.
...AD FC FC BE ,...N CD EC
Prove W - = — , (n) ___, (m) _ = — .
8. AB is the diameter of a circle, C a point on the circum-
ference. The tangent at B meets AC produced at D.
Prove 2lCBD=^CAB.
From the triangles ACB and BCD write down two values of
tan A. Hence prove BC2 = CA.CD.
DB BC
9. In the same figure prove : (i) — = — ,
10. ABCD is a rectangle, AD = 12", AB = 5". Draw AE
perpendicular to BD. Write down two values for sinADE.
Hence find the length of AE.
7. Geometrical constructions for trigonome-
trical ratios with, given angles.
It will be found useful to employ squared paper for
these examples, and generally to write the ratio in the
form of a fraction with 10 as its denominator.
TRIGONOMETRICAL FUNCTIONS
11
Example (i).
Draw an angle of 49° and find from measurements the value
of sin 49°.
O NX.
Fig. 11.
Draw the angle XOP by means of a protractor: since the
hypotenuse is to be the denominator, mark off OP = 10 units and
draw PN perpendicular to OX.
Then
Example (ii).
Draw an angle of 54°, and find from the drawing sec 54°.
Draw the angle XOP = 54°.
O N
Fig. 12.
Mark off ON = 10 units.
Erect the perpendicular N P.
Then 80064--^ -g-1-7.
12
PRACTICAL TRIGONOMETRY
8. Geometrical construction for angles with
given ratios.
Inverse Notation. The angle whose sine is x is written
sin~lx.
Thus cos"1 '86 is read "the angle whose cosine is *86."
Example.
To construct an angle whose sine is *72, that is sin"1 '72.
7*2
Since "72=^, draw two lines PN and ON at right angles.
Mark off PN = 7*2 units and with centre P, radius 10 units, strike
an arc PO.
Then L PON is sin"1 -72 = 46°.
9. The trigonometrical ratios of 60°, 30°, 45°.
These ratios may be found by Geometrical reasoning
without accurate drawing.
/GO
M
Fig. 14.
TRIGONOMETRICAL FUNCTIONS
(i) If L PON = 60°, then L. OPN -30°.
Complete the equilateral triangle OPM.
Then if OP - 2 units,
ON = 1 unit;
and OP2=PN2 + NO2;
.-. PN-^/3 or 1-732;
.'. sin 60° - ^ or -866 ; cos 60° = | or '5 ;
tan 60°- V3 or 1'732.
(ii) From the same triangle, since L OPN - 30°,
sin 30° = - or '5 ; cos 30° = ^- or '866,
J3 _ J3 1-732
13
0_ 1 1
""
(iii) If L PON - 45°, then L OPN = 45° ;
/. if PN = ON = 1 unit,
snce
Fig. 15.
1 = 2; .'. OP=,/2.
J2 1-414
cos 45° - s = '
V^
tan 45° = 1.
14 PRACTICAL TRIGONOMETRY
10. Relations between the Trigonometrical
Ratios.
N
(1) To prove ~ = tan A.
NP
SJnA_ OP_ NP
COSA~ON~ON~
OP
Similarly
cos_A_ 1
sin A tan A
= COt A.
(2) To prove sin2 A + cos2 A = 1.
NP2 ON2 NP2 + ON2
sm2 A + cos2 A = — 2 + _ = Qp2
_OP2
~OP2
(3) Prove in a similar manner
sec2A = 1 + tan2 A,
cosec2A = 1 + cot2 A.
Relations such as these which are true for all angles are
called Identities,
TRIGONOMETRICAL FUNCTIONS
15
11. Given one ratio of an angle to find the other
ratios.
If it is not required to find the angle, the ratios may be
calculated without accurate drawing.
Example.
Given sin A=^7, to find the other trigonometrical ratios of A.
If PN =8 units and OP= 17 units, then
172=32+ ON2,
Fig. 17.
= 172-82 = (17 + 8)(17-8)
=(25) (9),
=T85, etc.
Or using the result of Article 10 (2), we have
We shall disregard the negative sign until Chap. iv.
Examples. II b.
1. Draw an angle of 37°. Find its ratios by measurement
to two decimal places.
2. Draw an angle of 49°. Find by measurement sin 49°,
cos 49°, sec 49°, tan 49° ; with your results test the following,
sin2 49+ cos2 49 = 1, sec2 49 -tan2 49=1.
3. Construct the angle whose cosine is *52 ; measure it, and
find its sine and tangent.
4. Given that sinA = f, calculate the value of sec A and
tan A to two decimal places. Using your results, find by how
much sec2 A differs from l+tan2A.
16 PRACTICAL TRIGONOMETRY
5. Given that cosecA = |, find the values of (sin A + cos A)2
and sin2 A -h cos2 A.
6. Draw the angle A whose tangent is 6. Bisect this angle
A
and find by measurement tan—. By how much does it differ
from £tanA?
7. Construct the angle whose cosecant is 2*14. Measure it
and find its cosine and secant to one decimal place.
8. Draw an angle of 40°. Find its tangent. Bisect the
angle and from measurements find tan 20°. From the same
diagram find cot 70°.
9. If sin d = -5, find the value of 1+ tan2 6.
10. If sin 6 =•£ . prove that cos 6 = V^ ~^ .
2 ?
11. The diagonal of a rectangle is twice one of the sides:
prove that the ratio of the sides is \/3 : 1.
12. ABC is a right-angled triangle with BA = BC. BD is
BD 1
drawn perpendicular to AC. Prove that -— = -j- and that
BC *J 2i
BD = DC.
13. ABCDEF is a regular hexagon. If AB = 4", find the
lengths of BE and BF.
Miscellaneous Examples. A.
1. Draw with your protractor an angle of 142°, also one
of 210°.
2. Draw an angle of 48°. From measurements of your
drawing find tan 48°.
3. Draw a triangle ABC having B a right angle, 5 = 15,
c=12. Write down sin A, cosC. What relation is there be-
tween the angles A and C ?
4. Find the number of degrees in the angle of a regular
hexagon. Prove that the side of a regular hexagon equals the
radius of the circumscribing circle.
5. Express *2145 of a right angle in degrees, minutes and
seconds.
TRIGONOMETRICAL FUNCTIONS 17
6. Construct the angle whose tangent is 2, and prove that
.. . . ,
its sine is — f— .
5
7. The angle subtended by a side of a regular figure at the
centre of its inscribed circle is 36°. How many sides has the
figure ?
8. Draw carefully the angle whose cosine is *37. From
measurements find the cotangent of the angle.
9. What decimal of a right angle is 52° 12' ?
10. An isosceles triangle has each of its equal sides double
the base ; find the cosine and cotangent of the base angles.
11. Find the angle of a regular figure of 12 sides.
X*1 ?/2
12. If x=a cos <p and y = b sin <£, prove that —2 + ^ = 1.
13. Draw accurately the angle cosec~12'4; also the angle
!^. Measure the angles and find their difference in
degrees.
14. A diameter AB of a circle bisects the chord CD at O.
If sinABC=f and AC = 10", find AO.
15. Given secA=J^, calculate tan A. Show that for this
angle sin2 A = 1 — cos2 A.
16. Two tangents OA, OB are drawn to a circle of radius 5"
from a point 12'' from the centre C. Prove that sin CAB =^
and hence that the distance of C from AB = 2iy.
17. The three angles of a right-angled triangle are such that
2B=A + C ; find them in degrees.
18. Prove that sin 60° = 2 sin 30° cos 30°,
and that cos 60° = cos2 30° - sin2 30°.
p. p.
CHAPTEE III.
THE USE OF FOUR FIGURE TABLES.
12. THE values of the Trigonometrical Ratios will be
found in Bottomley's 4-figure Tables, pp. 32 — 43.
The ratios are given at intervals of 6 minutes with
difference columns for variations of 1, 2, 3, 4, 5 minutes.
Since all the sines and cosines are ^>1 the values of
these ratios are entirely decimal, and the decimal points
are not printed ; but in all other ratios the decimal point
and any integral part is printed in the first column only.
Note that as the angle increases from 0° to 90° the
cosine, cotangent, and cosecant diminish (see Chap. n. § 6).
Example (i).
To find the value of sin 31° 47'.
The following is an extract from the table of Natural Sines
on p. 32 of Bottomley's tables.
a
6'
12'
18'
24'
3V
36'
42'
48'
54'
123
4 5
31
5150
5165
5180
5195
5210
5225
5240
5255
5270
5284
257
1012
In the row opposite 31° and in the column under 42' we find
5255.
The difference for 5' is given in the same row in the last
column under 5 : we find 12.
Thus sin 31° 42' =-5255,
difference for 5' = -0012 ;
.-. sin 31° 47' = '5267.
The difference is added since the sine increases if the angle
increases.
THE USE OF FOUR FIGURE TABLES 19
Example (ii).
To find the value of cos 49° 21'.
From the tables
cos 49° 18'= -6521,
difference for 3' = '0007 ;
.'. cos 49° 21' = -6514.
The difference is subtracted since tho cosine diminishes as
the angle increases.
Note. The correct value will not be found by taking cos 49° 24'
from the tables and using the difference table.
Example (iii).
To find the angle whose cotangent is 4 '8 142.
Since the difference column is to be subtracted we find the
nearest angle with a cotangent greater than 4*8142.
The bar over the figures in the tables denotes that the whole
number has changed in the row and in this case is no longer 5
but 4.
Thus cot 1 1° 42' = 4-8288,
difference for 2'=- -0148;
.-. cot 11° 44' = 4-8140,
i.e. the angle 00^4-8142 is 11° 44' to the degree of accuracy
given by the tables.
Example (iv).
By using the tables we can find angles to satisfy given
equations. The identities in Chap. n. § 10 will be found useful in
throwing the equation into a form suitable for solving.
Find the acute angles which satisfy the equation
3cosec2<9-llcot0 + 7 = 0.
By using the identity cosec20 = l + cot2# the equation can be
written in terms of one unknown,
3(l+cot2<9)-llcot<9 + 7 = 0,
3cot20~ 11 cot (9 + 10=0,
(3cot0~5)(cot<9-2)=0;
.-. cot 6 = 1-6667 or cot (9 = 2;
.-. 0=30° 58' or 6 = 26° 34' from the tables.
2—2
20
PRACTICAL TRIGONOMETRY
1. sin 19°.
2.
4. cos 18° 5'.
5.
7. tan 16° 50'.
8.
10. cosec 9° 42'.
11.
13. sin-1 -8867.
14.
16. tan"1 5-0577.
17.
Examples. Ill a.
Look up in the tables,
sin 33° 22'. 3. cos 65° 4'.
cot 30° 21'. 6. sin 63° 50'.
cosec!4°31'. 9. sec 70° 10'.
cot 11° 37'. 12. tan 80° 48'.
tan-12-0248J 15. cos"1 '4830.
cot ~ ! -2600. 18. sec ~ 1 4'0855.
Find the acute angles which satisfy the following equations : —
19. 10 sin2 6 -7 sin 6 + 1=0. 20. 15 cos 0 + 8 sec 6 = 22.
21. 9 cos2 6 + 18 sin 0 = 17. 22. 4sec2<9- 17 tan<9 + ll=0.
13. Right-angled triangles.
It is very important to be able to write down at once
the sides of a right-angled triangle in terms of a side and
the ratios of a given angle.
Example (i).
Given the side OP =x and the angle PON=0, PNO being
a right angle.
Fig 18.
We have
THE USE OF FOUR FIGURE TABLES
Example (ii).
Given BC = 10" and Z.ABC=40°, ^ACB = 90°,
B
21
A C
Fig. 19.
AB = 10x
10 tan 40°
10 x -8391 =8-391.
AB
10
= 10 sec 40°
= 10x1-3054
= 13-054.
Exercise.
Practice writing down the other sides of the following right-
angled triangles in terms of the ratios of the given angle and the
given side.
22
PRACTICAL TRIGONOMETRY
A
*> Ob
Fig. 21.
Look up the ratios in the tables and write down the lengths
of the other sides of the triangle ABC (Fig. 21) from the following
data : —
(vii) B = 36°, 6 = 10. (viii) A = 24°, 6 = 10.
(ix) B = 28°ll', a = 20. (x) A = 41° 19', a =10.
(xi) B = 38°14', c=25. (xii) A = 33°17/, c = 20'25.
14. Angles of Elevation and Depression.
The angle which a line joining the eye of an observer
and a distant object makes with the horizontal plane is
. A ngle of Elei'atLon
{Angle of 'Depression
Fig. 22.
THE USE OF FOUR FIGURE TABLES
23
called the Angle of Elevation if the object be above the
observer, and the Angle of De^ession if the object be below
the observer.
Thus in fig. 1 if A B be the horizontal line through A, to
the observer at A the angle BAG is the angle of elevation of
the point C.
In fig. 2 the angle BAG is the angle of depression of the
point C.
Example.
Find the angle of elevation of the sun if the shadow cast by
a stick 6 ft. high is 4 ft. 4 in.
4ft 4in.
6ft.
Fig. 23.
Let B be the angle required ; then
tan i9 = |f = l'4;
. - . from the tables 6 = 54° 3(X.
Examples. Ill b.
1. Find the altitude of an equilateral triangle whose sides
are 4".
2. In the triangle ABC, A = 90°, C = 50°, 6 = 10". Draw AD
perpendicular to BC and find the lengths of AD, CD, AB, BD.
3. I observe the angle of elevation of the top of a tower
240 feet high to be 37°. What is my horizontal distance from
the foot of the tower?
24? PRACTICAL TRIGONOMETRY
4. Find the angle of elevation of the sun if a tower 212 feet
high casts a shadow 327 feet long.
5. The steps of a staircase are 10" wide and 7" high. How
many degrees are there in the slope of the staircase?
6. AD is the perpendicular from A on the side BC of a
triangle ABC. If B = 32°, BD = 7 ft., DC = 5 ft., find AD, AB and
the angle C.
7. The angle of depression of a boat from the top of a cliff
200 ft. high is 36° 13'. Find the distance of the boat from the
foot of the cliff.
8. The sides of a parallelogram are 4 ft. and 5 ft. and the
acute angle between them is 50°. Find the lengths of the
perpendicular distances between the parallel sides.
9. Find the lengths of the three perpendiculars from the
angular points to the opposite sides of an isosceles triangle whose
equal sides are 12 cms. and the included angle 40°.
10. From the top of a spire the angle of depression of an
object 100 feet from its base is 56° ; find the height of the spire.
11. In a triangle ABC, B = 70°, C = 50°, c = 20". Draw AE
perpendicular to BC and BD perpendicular to AC. Find the
lengths of BD, BE, AE, AC.
12. From a point 500 feet from its base the angle of
elevation of a tower is 26° 11'. Find the height of the tower.
13. ABCD is a quadrilateral inscribed in a circle of 10 ft.
radius. If AC is a diameter and Z_ABD = 15°, Z_ACB = 40°, find
the lengths of the sides of the quadrilateral.
15. Illustrative Examples.
In the following examples the angles are assumed to be
acute, but it will be shown in Chap. Y. that the theorems
are true also when the angles are obtuse.
Example (i).
Prove that the area of a triangle —\ product of two
sides x sine of included angle.
We have, area of triangle (A) = Jap, when p is the perpen-
dicular on the side a from the opposite angular point.
THE USE OF FOUR FIGURE TABLES 25
But
= b sin C ;
Exercise.
(1) Prove also that
(2) Find a formula for the area of a parallelogram in terms
of two adjacent sides and the included angle.
(3) Show that the sides of a triangle are proportional to the
sines of the opposite angles, i.e.
a b c
sin A ~~ sin B ~ sin C *
(4) If two triangles ABC, DEF have B= E, prove that
AABC_AB.BC
ADEF~DE.EF*
Example (ii).
To find the area of a regular figure, e.g. a pentagon inscribed
in a given circle.
Let O be the centre of the circumscribing circle and AB a
side of the figure.
26
PRACTICAL TRIGONOMETRY
We can find the angle AOB and we thus know two sides and
the included angle of the triangle. Five times its area gives the
area of the pentagon.
O
Fig. 25.
Exercise.
(1) Find the area of a regular pentagon inscribed in a circle
of radius 10 in.
(2) Find also the perimeter of the pentagon.
(3) Find the area and perimeter of a regular pentagon
circumscribed about a circle of 10 in. radius.
Example (iii).
Show that in a triangle
of the circumscribing circle.
a
sin A
= 2R where R is the radius
Fig. 26.
THE USE OF FOUR FIGURE TABLES
27
Let O be the centre of the circle and D the middle point
of BC.
Show that LBOC =
easily follows.
Exercise.
(1) Show that
and hence Z.BOD = A. The result
= 2R.
sin A sin B sin C
(2) Prove this also by producing BO to meet the circum-
ference at E and joining EC.
16. Example (i).
To an observer on a tower the angles of depression of two
points due S. known to be 100 ft. apart are 54° 11' and 33° 17'.
Find the height of the tower above the horizontal plane on which
these points lie.
Let x be the required height in feet, AB the tower and C, D
the points observed.
Then BD =x cot 33° 17',
BC =.27 cot 54° 11'.
.-. 100 =x (cot 33° 17'- cot 54° 11')
=# (1-5234- -7217)
v- 100-125ft
-•8017"
A more convenient method of solving problems of this nature
by the aid of logarithms is given in Chap. v. Art. 31.
28
PRACTICAL TRIGONOMETRY
Example (ii).
To an observer at A the angle of elevation of the top of a
tower 220 feet away is 25°, and the angle subtended by the spire
above it is 14°. Find the height of the spire.
Let BC represent the tower and CD the spire.
We have L DAB = 39° (this
is the angle of elevation of the
top of the spire).
DB = 220xtan39J,
CB = 220xtan25°;
.-. CD = 220 (tan 39° -tan 25°)
= 220(0-8098-0-4663)
= 220 (-3435)
JSt.
= 75-6 ft. Fig. 28.
17. The Compass.
For purposes of indicating direction the compass is used.
In all there are 32 points of the compass, that is, 32
differently named directions from any one point.
THE USE OF FOUR FIGURE TABLES 29
Hence the angle between any two consecutive points
In the figure we have shown the points in one quadrant.
As an Exercise the student should fill in the points in the
other quadrants by analogy.
Directions are also often given in degrees. Thus
N. 30° E., or 30° East of North, is the direction to the East
of North making 30° with the direction North, i.e. be-
tween N.N.E. and N.E. by N.
Example.
A man observes a spire in a direction E. 10° N. He walks
500 yards to the S.E. and observes that the bearing of the spire
is N.E. How far is he now from the spire?
Let A be his position when he first observes the spire B in
the direction AB where /_EAB = 10°.
He walks in the direction AC, 500 yards where /.EAC = 45°.
At C the angle BCN=45° where N is the direction of North.
Pig. 30.
2LACB being 90° we have
= 500tan55°
= 714 yds approx.
30 PRACTICAL TRIGONOMETRY
Examples. Ill c.
1. Find the area of a triangle, given a = 5", 5 = 6", C = 43°.
2. The side of a regular octagon inscribed in a circle is 4".
Find the radius of the circle.
3. A small weight swings at one end of a string 5 ft. long,
the other end being fixed. How far is the weight above its
lowest position when the string is inclined at 10° to the vertical?
4. From the top of a cliff 200 feet high the angles of
depression of two boats due S. were observed to be 37° and 52°.
How far apart were the boats?
5. Find the area and perimeter of a regular hexagon inscribed
in a circle of 6" radius.
6. From a point A on the ground, the angle of elevation
of the top of a tower 60 feet high is 43° 13'. How far is the
observer from the foot of the tower and what is the elevation
of the tower from a point 10 yards nearer?
7. By how many feet does the shadow cast by a spire 150 ft.
high lengthen as the sun sinks from an elevation of 67° 14? to an
elevation of 37° 20'?
8. From a point 8 in. from the centre of a circle of radius
4 in. two tangents are drawn to the circle. Find the angle
between them. What is the angle between the radius at the
point of contact and the chord of contact? Find the length
of the chord of contact.
9. Find the area of a parallelogram whose sides are 4 ft. and
5 ft., the acute angle between them being 47° 17'.
10. A triangle is inscribed in a circle of radius 4*5 cms. with
base angles 44° and 56°. Find the lengths of its sides.
11. The sides of a rectangle are 4" and 7". Find the angle
between the diagonals.
12. At a point 100 yards from the foot of a cliff the angle of
elevation of the top of the cliff is 35° 11', and the angle subtended
by a tower on its edge is 11° 53'. Find the height of the tower.
THE USE OF FOUR FIGURE TABLES 31
13. A man at a point A observes the angle of elevation of
the top of a flagstaff to be 35°. He then walks past the flagstaff
to a place B on the other side where he observes the angle of
elevation to be 63°. From A to B is 120 feet. Find the height
of the flagstaff.
14 One side of a triangle inscribed in a circle is 4 in. and
the angle opposite it is 27° 11'. Find the diameter of the circle.
15. The road to the top of a hill runs for J mile inclined at
10° to the horizon, then for 500 yards at 12° : then for 200 yards
at 15°. Find the height of the hill in feet.
Show that the compass directions of the three parts of the
road are not required.
16. If a ship sails 4 points off the wind (i.e. in a direction
making 45° with the direction of the wind), how far will she have
to sail in order to reach a point 30 miles to windward ?
17. The shadow of a tower is 55 ft. longer when the sun's
elevation is 28° than when it is 42°. Find the height of the
tower and the length of the shorter shadow.
18. Find the height of a hill if the angles of elevation taken
from two points due North of it and 1000 feet apart are 51° 13'
and 67° 5'.
19. A man in a balloon at a height of 500 ft. observes the
angle of depression of a place to be 41°. He ascends vertically
and then finds the angle of depression of the same place to be
62°. How far is he now above the ground?
20. A man surveying a mine measures a length AB of
16 chains due E. with a dip of 5° to the horizon ; then a length
BC of 10 chains due E. with a dip of 3°. How much deeper
vertically is C than A? Answer in feet.
21. A building 100 feet long and 50 feet wide has a roof
inclined at 35° to the horizon. Find the area of the roof and
show that the result will be the same whether the roof has a
ridge or not.
22. A man travels 5 miles from A to B in a direction 20°
N. of E., then 3 miles to C in a direction N. 25° E. Find the
distance of C (1) North of A, (2) East of A, (3) from A. Verify
by a figure drawn to scale.
32 PRACTICAL TRIGONOMETRY
23. The angle of elevation of the top of a house 100 feet
high observed from the opposite side of the street is 65°, and the
elevation of a window of the house from the same point is 40°.
Find the height of the window from the ground.
24. A regular polygon of 10 sides is inscribed in a circle
of radius 5 feet. Find the area and perimeter of the polygon and
of a circumscribed polygon of the same number of sides.
25. From one end of a viaduct 250 feet long a man observes
the angle of depression of a point on the ground beneath to be
37°, and from the other end the angle of depression of this point
is 71°. Find the height of the viaduct.
26. The top C of a tower 80 feet high is observed from the
top and from the foot of a higher tower AB. From A the angle
of depression of C is 18° 11', and from B the angle of elevation is
23° 31'. Find the height of AB and its distance from the other
tower.
27. From a ship the direction of a lighthouse is observed
to be N. 25° E., and after the ship has sailed 10 miles North-East,
the bearing of the lighthouse is North- West. If the ship now
changes her course and sails in direction W. 25° N., how near
will she approach the lighthouse?
28. A man standing at a point A on the bank of a river
wishes to find the distance of a point B directly opposite him
on the other bank. He noticed a point C also on the other bank
and found LBA.C to be 55°; he walked directly away from the
river for 100 yards to a point D and found the angle ADC to be
35°. Find the distance AB.
29. From a steamer moving in a straight line with a uniform
velocity of 10 miles per hour the direction of a lighthouse is
observed to be N.W. at midnight, W. at 1 a.m., S. at 3 a.m. Show
that the direction of the steamer's course makes an angle cot ~ 1 3
with the N. Find the least distance of the steamer from the
lighthouse.
30. B is 50 yards from A in a direction E. 20° S., C is
100 yards from B in a direction E. 32° 15' N., D is 80 yards from
C in a direction W. 46° 10' N. Find how far D is from A and in
what direction.
THE USE OF FOUR FIGURE TABLES 33
Miscellaneous Examples. B.
1. Draw two straight lines OB, OC at right angles and OA
between them making 39° with OB. With centre O and radius
10 cms. draw a circle cutting OB in Q. and OA in P. From P let
fall perpendiculars PS on OB and PR on OC. At Q draw a
tangent QT cutting OA in T. Measure PR, PS, QT to the
nearest millimetre and write down their lengths. Hence find
sin 39°, cos 39°, tan 39° and compare with the values given in
the tables.
2. The diagonal of a rectangle is 12 cms. long and makes
an angle of 34° with one of the sides. Find the length of the
sides.
3. Prove that (sin A + cos A)2 = 1 + 2 sin A cos A ; and hence
evaluate \fl+ 2 sin 53° cos 53°.
4. Find the values of
(i) sin 47° sec 47° ;
(ii) tan 74° cosec 74°.
5. The base of an isosceles triangle is 8 cms. and the
diameter of its circumscribing circle is 12 cms. Find its vertical
angle and its altitude.
6. AB is a diameter of a circle, centre O, and OC is a radius.
If OC = a and L COB = a, show that AC = 2acos ^ and the length
A
of the perpendicular from O on AC = a sin - .
2i
7. Draw accurately a triangle with base BC = 5 cms.,
BA = 8cms., B = 40°. Calculate the length of the perpendicular
from A on BC. Find the area of the triangle and from measure-
ments of your diagram find cos 40°.
8. A man 5 ft. 9 in. high standing 134*2 ft. from the foot of
a tower observes the elevation of the tower to be 30° 14'. Find
the height of the tower.
9. Prove that if cos A = a then tan A == .
a
10. P, Q, R are three villages. P lies 7 miles to the N.E. of
Q and Q lies 11 J miles to the N.W. of R. Find the distance and
bearing of P from R.
p. F. 3
34 PRACTICAL TRIGONOMETRY
11. Two adjacent sides of a parallelogram are AB=6cms.,
BC = 7 cms., the included angle being 52°. Find the angles
between the diagonal BD and the sides AB and BC. Verify
by an accurate drawing.
12. A ladder 20ft. long rests against a vertical wall and
makes an angle of 50° with the ground. What will be its
inclination to the ground when the foot of the ladder is 5 ft.
farther from the wall?
13. Express the equation 2 cos2 6 + sin 0 = 2, in terms of
sin$, solve it and find from the tables two values of 6 to
satisfy it.
14. Two equal forces P making an angle a with one another
act at a point O. Their resultant R is represented by the
diagonal passing through O of the parallelogram in which the
lines representing the forces form two adjacent sides. Prove
15. Show from a figure that cot 53°= tan 37° and hence find
a value of 6 which satisfies the equation cot (Q-\- 16°)= tan 6.
16. In a triangle ABC, a =2", c = 3", B = 37°; calculate the
length of the perpendicular drawn from A to BC. Also if PBC
be an isosceles triangle on BC as base and of the same altitude
as the triangle ABC, find its angles.
17. Express 16sin0 + 3cosec0 = 16 as a quadratic in sin#
and find two values of 6 to satisfy it.
18. On a tower 85ft. high stands a pole of length 10ft.
What angle does this pole subtend at a point on the horizontal
plane on which the tower stands, at a point 40 ft. from its
base?
19. Find the area of a regular pentagon inscribed in a circle
of 4" radius.
20. O the mid-point of AC is the centre of the circle
circumscribing the right-angled triangle ABC. If 6 = 13, c=12,
find a. Prove that L BOC = 2A. Find sin2A, sin A, cos A, and
verify the relation sin 2 A = 2 sin A cos A.
THE USE OF FOUR FIGURE TABLES 35
21. A man at a point B observes an object at C and walks
200 yards in a direction making an angle of 68° with BC, to a
point A where the angle CAB also equals 68°. Find the distance
from B to C.
22. A set square has its hypotenuse 12" long and the shorter
side 4". The hypotenuse slides along a scale which is held fixed,
and an arrowhead on the hypotenuse is placed in succession
against marks at intervals of 0*15 of an inch on the scale. In
each position a line is ruled along the longer side of the set
square. How far apart are these lines ? If an error of O'Ol of
an inch was made in placing the set square, what error in the
position of the line would result ?
23. If a ship after sailing 25 miles is 12 miles to windward
of her starting point, what angle does her course make with the
direction of the wind ?
24. Construct the angle whose cotangent is 1*62. Measure
it and compare with the angle given in the tables.
25. Find two values of 6 to satisfy the equation
26. In the side of a hill which slopes at an angle of 20° to
the horizontal, a tunnel is bored sloping downwards at an angle
of 10° with the horizontal. How far is a point 40 ft. along the
tunnel vertically below the surface of the hill ?
27. Find two values of Q to satisfy the equation
6cos20 + 7sin<9-8=0.
28. A surveyor finds two points A, B on a hillside to be 3 chains
43 links apart, and finds the line AB to be inclined at 17° 30' to
the horizontal. On his plan these points must be shown at their
horizontal distance apart. What is this to the nearest link?
Given 1 chain = 100 links.
29. If .27=asec#, y = &tan$, prove that —^ — j- =1.
ci/ o
30. C is the right angle of a right-angled triangle ABC. AD
and BD are drawn perpendicular to AC and AB respectively.
Prove that AD = BCcosec2A.
3—2
CHAPTER IV.
FUNCTIONS OF ANGLES GREATER THAN A RIGHT
ANGLE.
18. Note on the Convention of Sign.
If a line OX of indefinite length be drawn from a point O,
and any length such as OM be taken as a unit, we may
O M B1 A 5 i
Fig. 31.
represent any integral number by the length of a segment
containing this number of units, e.g. OA, which contains OM
six times, represents the number 6, and AB the number 2.
If we wish to add the two numbers represented by
OA, AB we may place AB at the end of OA and we ha,ve their
sum represented by OB.
If we wish to subtract AB from OA we have only to
mark off AB' equal to AB but in the opposite direction, and
we have OB' their difference.
If AB is longer than OA, B' falls to the left of O and
the difference is represented by OB', measured from O from
right to left and not from left to right.
B' 0 A
Fig. 32,
ANGLES GREATER THAN A RIGHT ANGLE 37
It will thus be seen that lengths measured along a line
X' A' 0 A X
Fig. 33.
XX' from a point O will be conveniently regarded as positive
if taken in the direction OA to the right of O but as
negative if drawn to the left.
This difference in sign may also be represented by the
order of the letters; thus OA may be considered as — AO.
OA and AO are said to denote the same segment taken
in opposite senses.
Similarly, for lengths measured along a line OY at right
X
Y'
Pig. 34.
angles to XX', the direction OY is considered positive and
OY' negative.
This convention is applied, not only to lengths measured
along XX' and YY', but also to lines drawn parallel to these.
It will be found that, by the adoption of these conven-
tions, trigonometrical formulae are considerably simplified
and that instead of requiring different formulae for cases in
which the angle involved is acute or obtuse, positive or
negative, we are able to use the same formula for all cases.
38
PRACTICAL TRIGONOMETRY
19. As in Article 1 we will suppose a straight line, called
the radius vector, to turn about O from an initial position
OX; then the amount of revolution it has undergone in
coming to the final position OP measures the angle XOP.
Also it will he remembered that if PN be the perpendicular
drawn from P to the initial line OX, then
sin XOP =,
ON
NP
— ,
whatever position OP may have.
It is important to notice that all angles are supposed to
be described by revolution/rom the position OX.
This revolution maybe in the opposite direction to that
of the hands of a clock, called the positive direction ; or in
the same direction as the hands of a clock, called the
negative direction.
Also the radius vector may make any number of com-
plete revolutions before coming to rest.
From our definition it follows that all angles which have
the same boundary line OP have the same trigonometrical
functions. Such angles are called coterminal angles.
ANGLES GREATER THAN A RIGHT ANGLE ' 39
20. In the figure, Art. 19, YOY' is drawn perpendicular
to XOX', so that any circle described with O as centre is
divided into four quadrants. The quadrants XOY, YOX',
X'OY', Y'OX are called the first, second, third, and fourth
quadrants respectively.
Now if the lines PiOP3, P2OP4 are equally inclined to XX',
we have four congruent triangles
Hence it follows that the trigonometrical functions of
the angles XOP1? XOP2, XOP3, XOP4 are numerically the
same; also that there are four and only four positions
which the boundary line may have in order that any one
trigonometrical function of the angle may have a given
numerical value.
If 0 be the acute angle XOPX we see from the figure that
sin<9, sin (180° -0), sin (180°+ (9), sin (360° -0}
are numerically equal ; and so for the other functions.
Here it is convenient to adopt the convention of sign
which we have mentioned already.
The convention of sign is as follows:
The radius vector OP is always considered positive.
ON is considered positive if measured along OX, and
negative if measured along OX'.
NP is considered positive if measured in the direction
OY, and negative if measured in the direction OY'.
Hence if OP lies in the first quadrant,
ON and NP are positive;
.'. all the functions are positive.
If OP lies in the second quadrant,
ON is negative, and NP is positive;
.". the sine and cosecant are positive, but all the other
functions are negative.
40 ' PRACTICAL TRIGONOMETRY
If OP lies in the third quadrant,
ON and NP are negative;
.*. the tangent and cotangent are positive, but all the other
functions are negative.
If OP lies in the fourth quadrant,
ON is positive, and NP is negative;
.'. the cosine and secant are positive, but all the other
functions are negative.
Thus if in the figure of Art. 19, PN : ON : OP = 3 : 4 : 5,
sin XOP1 = sin XOP2 = f , sin XOP3 = sin XOP4 = — |;
COS XOPi = COS XOP4 = |, COSXOP2=:COSXOP3=: — -|;
tan XOPX = tan XOP3 = f , tan XOP2 = tan XOP4 = - -J.
Now, having regard to the sign of the function, we see
that there are two positions which the boundary line may
have when we are given the value of any one function.
21. The point of chief importance for us is that we
may be able to obtain at once any trigonometrical function
of any angle a with the help of tables which give the
functions of acute angles only.
The most convenient method is to notice in which
quadrant the boundary line of a lies, and then to obtain
from the tables the required functions of the acute angle 0,
where
a = 180° — 0 for the second quadrant,
a = 180° + 0 for the third quadrant,
a = 360° - 0 for the fourth quadrant.
We then only have to prefix the proper sign, which can
be done by drawing a figure, or mentally after a little
practice.
ANGLES GREATER THAN A RIGHT ANGLE 41
Example (i).
Find the functions of 140°.
The boundary line of the angle is in the second quadrant,
and the corresponding acute angle is 40°, since 140° = 180° — 40°.
Also in the second quadrant the sine is positive, and the cosine
and tangent are negative ;
.'. sin 140°= sin 40°= '6428 ;
cos 140° = - cos 40° = - -7660 ;
tan 140° = - tan 40° == - -8391 .
In a similar way we have
cos 200° = cos (180° + 20°)= -cos 20°= - '9397 ;
tan 313° = tan (360° - 47°) = - tan 47° = - 1 '0724 ;
cosec 127° = cosec (180° - 53°)= +cosec 53° = 1-2521 ;
cot 197° 24'=cot (180° + 17° 24')= +cot 17° 24' = 3*1910.
Example (ii).
Find the positive angles less than 360° which satisfy
(1) tan 6 = '4734; (2) cos 6 =- -4360.
(1) Since the tangent is positive the boundary lines of the
angles must be in the first and third quadrants.
From the tables, -4734 = tan 25° 20' ;
,\ the angle in the first quadrant is 25° 20' ;
and the angle in the third quadrant is
180° + 25° 20', i.e. 205° 20'.
(2) Since the cosine is negative the boundary lines of the
angles must be in the second and third quadrants.
From the tables, -4360 = cos 64° 9' ;
.-. the angle in the second quadrant is
180° -64° 9', i.e. 115° 51';
and the angle in the third quadrant is
180° + 64° 9', i.e. 244° 9'.
42 PRACTICAL TRIGONOMETRY
22. If we are given sin 0 - f , we have
The meaning of the double sign, which we disregarded
in Art. 11, can now be explained.
There are two positions which the boundary line of 0
may have in order that sin 0 may be f , one in the first
quadrant and one in the second.
The cosines of angles which have one of these two
boundary lines are numerically ^ ; but if the boundary line
is in the first quadrant the cosine is positive, and if the
boundary line is in the second quadrant the cosine is"
negative.
23. We can state our results more generally as
follows :
sin (180° -0) = sin 0,
cos (180° - 0) = - cos 0,
tan (180° -0) = - tan 0,
sin (180° + 0) = - sin (9,
cos (180° + 0} = - cos (9,
tan (180° + (9) = tan (9.
Also since the angles — 0, 360°— 0 are coteriniiial, we
have
sin (- 0} = sin (360° - 0} = - sin (9,
cos (- 0) = cos (360° - 0) = cos 0.
The student who wishes to acquire skill in trigono-
metrical transformations should make himself familiar with
the results in the above form, and with the functions of
90°- 0 and 90° + 0 which we discuss in Articles 24, 25.
ANGLES GREATER THAN A RIGHT ANGLE
43
24. To prove that
sin (90°- 0) = cos Ot
and cos (90°- 0) = sin 0.
Let a radius vector start from OX and revolve until it
has described an angle 0, taking up the position OP.
Fig. 36.
Then let the radius vector start from OX and revolve
through 90° to the position OY and back through an angle 0
to the position OP'. Then XOP' is the angle 90°- 0.
If we draw perpendiculars PN, P'N' to OX, we have two
congruent triangles PON, ON'P'.
sin (90°- 0) = —7- = — = cos 0,
Hence
Thus we have important relations between the functions
of complementary angles.
The sine of an angle is the cosine of its complement.
The tangent of an angle is the cotangent of its com-
plement.
The secant of an angle is the cosecant of its comple-
ment.
44 PRACTICAL TRIGONOMETRY
Example.
Find a value of 6 to satisfy sin 60 = cos 40.
The equation is satisfied if 6$ and 40 are complementary
angles ; that is if 6$ + 4$ = 90° ; hence 6 = 9° is a solution of the
equation.
25. To prove that
sin (90°+ 0) = cos 0,
and cos (90°+ 0) = -sin 0.
Let a radius vector start from OX and revolve until it
has described an angle 0, taking up the position OP.
Then let the radius vector start from OX and revolve
through 90° to the position OY and then on through the
angle 0 to the position OP'.
Then XOP' is the angle 90° + #.
If we draw the perpendiculars PN, P'N' to OX, we have
two congruent triangles PON, ON'P'; and hence
ON' = NP in magnitude but is of opposite sign,
N'P'= ON in magnitude and is of the same sign ;
rig. 37.
ANGLES GREATER THAN A RIGHT ANGLE
26. The results of Articles 23, 24, 25 have been
obtained for the case in which 0 is an acute angle, but
the importance of the sign convention will be realised by
noticing that we obtain the same formulae whatever the
magnitude of 0 may be. The following figures illustrate
the relations between the ratios of 0 and 180° — 0.
(i) (ii>
T
(iii) (iv)
Fig. 38.
In each figure XOP represents the angle 0, taken in turn,
in each of the four quadrants.
XOP' represents the angle 180° — 0 formed by OP' turn-
ing from OX through 180° and then backwards in the
negative direction through an angle equal to 0.
It will be seen that in each case NP = N'P', ON = — ON'
and OP is always considered positive ;
/. sin (180° - 0) = sin (9, cos (180° - 0) - - cos 0,
tan (180° -0) = - tan 0.
Exercise. Draw figures to prove the relations between the
ratios of the angles (9, 180° + 0, 90° -(9, 90° + 0 when
(i) (9 = 150°, (ii) (9 = 215°, (iii) 0=-30°.
46 PRACTICAL TRIGONOMETRY
Examples. IV a.
1. With the help of the tables, find the following :
(1) sin 115°. (2) cos 130°. (3) sec 175°.
(4) tan 142°. (6) cos 312°. (6) cot 127°.
(7) sin 125° 37'. (8) cos 98° 14'. (9) sin 216°.
(10) tan 243° 15'. (11) cosec 164°. (12) cot 192° 33'.
2. Find in each of the following cases two positive values of
6 less than 360° :
(1) tan 6 =-2-1426. (2) tan0='3466. (3) sin0='8916.
(4) cos 6 =-'3870. (5) cosec 6 =-1*1432. (6) cot0 = 2'9515.
3. Draw the boundary lines of all the angles whose tangent
is '7. Measure the two smallest positive angles with a protractor,
and verify your results with the tables.
4. Draw a figure to show that if sin 6 = -f% , then tan 0 = ± -f%>
5. When A = 130°, draw figures to show that
sin (90° + A) = cos A, sin (180° + A) = - sin A,
tan (180° - A) =- tan A.
6. In a triangle ABC, 6 = 5, c = 3; show that the area is the
same whether A = 50° or 130°.
7. If A is an angle of a triangle, find its magnitude from the
following equations :
(1) 3sinA = l-7; (2) 4cosA = 2'5; (3) 5cosA + 2 = 0.
8. Show that no root of 5 sin #4-4 = 0 can be an angle of a
triangle.
9. Find all the positive angles between 0° and 360° which
satisfy the equations
(1) 2 cos2 (9 = 3 sin (9; (2) 10 sin2 <9- 3 sin 0-4 = 0 ;
(3) 10 tan e -5 cot d = 23.
10. By making use of the relations which exist between the
functions of complementary angles, find a value of 6 to satisfy
the equations
(1) sin 30= cos 2<9 ; (2) tan 50 = cot 40.
11. By using the relations which exist between functions of
6 and 180°±0, find a value of 6 to satisfy the equations
(1) sin 40 = sin 0; (2) sin 40 = - sin 0 ; (3) cos 40=- cos 0.
12. If A, B, C are the angles of a triangle, show that
(1) sin(B + C) = sinA; (2) cos(B + C) = -cos A ;
. A + B C
(3) sin__ = cos-.
ANGLES GREATER THAN A RIGHT ANGLE 47
27. Limiting Values.
If the denominator of a fraction remains constant while
the numerator decreases, it is clear that the fraction de-
creases ; and by decreasing the numerator sufficiently the
fraction can be made as small as we please.
sc
Thus in the fraction - , if a remains constant while x
a
decreases, the fraction also decreases and approaches zero ;
and zero is called the limiting value of -, when #=0.
QJ
A convenient notation to express this is
This is read " when # = 0, the limiting value of the
n . • oc • jt
fraction - is zero.
If the numerator of a fraction remains constant while
the denominator decreases, the fraction increases.
= 1000000a.
By making the denominator sufficiently small we can
make the fraction as large as we please ; and in this case
the value of the fraction eventually becomes infinitely
great, and is denoted by the symbol oo .
A convenient notation to express this is
T a
L — = oo .
3=0 #
Similarly it will be seen that
L «=0.
PRACTICAL TRIGONOMETRY
28. Functions of 0° and 90°.
Let XOP be any small angle.
o
Fig. 39.
With our usual notation we have
sin XOP =£?.
Now as the radius vector approaches the position OX,
NP decreases while OP remains constant.
Hence as the angle XOP decreases, sin XOP also de-
creases; and in the limit, when OP lies along OX, we have
Also as OP approaches OX, ON becomes more nearly
equal to OP, and in the limit we have
OP
cosO°=-=l.
OP
Again, if we consider the ratio — , we see that as the
angle XOP decreases OP remains constant while NP de-
OP
creases; and therefore the ratio — increases. In the
limit when NP vanishes, the ratio becomes infinitely great;
and hence we have
cosec 0° = QO .
In a similar way it can be shown that
oc, secO° = l,
ANGLES GREATER THAN A RIGHT ANGLE 49
Now let us suppose OP to approach the line OY. In
this case NP approaches OP and coincides with it when
L XOP = 90°,
and ON decreases and becomes zero when
L XOP = 90°.
OP
Hence sin 90° = — = 1,
cos 90° = -=<),
OP
tan 90° = -TT- = co .
When the angle XOP becomes slightly greater than 90°,
ON becomes negative and the tangent of the angle is infi-
nitely great and of negative sign. The tangent is said to
change its sign when passing through the value infinity.
It will be noticed that 0° and 90° are complementary
angles and consequently their functions obey the laws of
Art. 24.
Exercise. Write down the values of :
(i) cosec 90°, sec 90°, cot 90°.
(ii) sin 180°, cos 180°, tan 180°.
(iii) cosec 180°, sec 180°, cot 180°.
(iv) sin 270°, cos 270°, tan 270°.
(v) cosec 270°, sec 270°, cot 270°.
p. r.
50
PRACTICAL TRIGONOMETRY
29. To trace the changes in the functions as
the angle changes from. 0° to 360°.
Fig. 40.
"With the same figure as before, let L XOP = 0.
NP
Then sin 0 — — .
Now OP remains constant in magnitude and sign, so
the changes in sin 0 are due to the changes in N P only.
When 0 = 0° we have sin (9 = 0 [Art. 28].
As 0 increases from 0° to 90°,
N P increases and is positive ;
.'. sin 0 increases and is positive.
When 0 = 90°, sin 0 = 1 [Art. 28].
As 0 increases from 90° to 180°,
N P decreases and is positive ;
.'. sin# decreases and is positive.
When (9=180°, sin 0=^ = 0.
ANGLES GREATER THAN A RIGHT ANGLE
51
As 0 increases from 180° to 270°,
NP increases and is negative;
.'. sin 0 increases and is negative.
When
= 270°,
OP
sin0 = ---=-l.
As 0 increases from 270° to 360°,
NP decreases and is negative;
.'. sin0 decreases and is negative.
When
(9 = 360°,
The changes in the value of a function can be shown
conveniently by means of a curve drawn on squared paper.
Draw two axes OX, OY at right angles to one another.
Along OX take a length ON to represent the magnitude of
an angle, and erect a perpendicular NP to represent the
value of the function. The locus of P will be a curve which
is called the graph of the function.
We have given below the graph of the sine.
1
1
•16
•6
•26
0
/
^
x
\
/
\
/
\
/
\
3
0° 6
Foil
0' 9
Les t
0' -L
>f 0.
IO' -k
O' 1
v
o- &
0' 2
0' 3(
«.
30' 3
w
-16
\
/
\
Z
s
^
^
/
Tig. 41.
4—2
52 PRACTICAL TRIGONOMETRY
NP
30. To trace the changes in tan 0, we have tan 0 — — ,
and both NP and ON change with 0.
When 0 = 0°, tan 0 = ~=0.
As 0 changes from 0° to 90°,
NP increases and is positive,
ON decreases and is positive;
.'. tan 0 increases and is positive.
OP
When 0 = 90°, tan0= — =o>.
As 0 changes from 90° to 180°,
NP decreases and is positive,
ON increases and is negative;
.'. tan 0 decreases and is negative.
When (9=180°, tan 0 = ^=0.
As 0 changes from 180° to 270°,
NP increases and is negative,
ON decreases and is negative;
.*. tan 0 increases and is positive.
OP
When 0-270°, tan0 = — = 00.
As 0 changes from 270° to 360°,
NP decreases and is negative,
ON increases and is positive;
.'. tan0 decreases and is negative.
When 0 = 360°,
ANGLES GREATER THAN A RIGHT ANGLE
53
The graph of tan 0 is given below. Note that since
tan (180° + 0)= tan 6,
the curve for values of 0 from 0° to 180° is repeated for
values of 0 from 180° to 360°.
In drawing graphs of the functions the student should
note that the function changes sign only after passing
through the values zero or infinity.
-3
-4
-5
-6
-7
Values of
e.
Fig. 42.
Examples. IV b.
1. Discuss the changes in the following functions as 6
changes from 0° to 360°, and illustrate by a graph in each case :
(1) cos B. (2) cot 6. (3) cosectf. (4) sectf.
2. Draw, with the same axes of reference, graphs of sin0
and cos$; and from your figure obtain values of 6 between
0° and 360° for which (1) sin0 = cos<9; (2) sin B=- cos 6.
Also with the help of your figure draw the graph of
sin B + cos B.
3. Trace the changes in sign and magnitude of tan# as B
decreases from 180° to 0°.
4. Draw a curve on squared paper to show the length of
the shadow cast by a tree 100 ft. high for all elevations of the
sun up to 50°.
5. N is the foot of the perpendicular from a moving point P
on the fixed straight line OX. If all positions of P are obtained
by giving different values to B in the equations
ON = 5 cos 0, PN=4sin6>,
find for what values of 6, P is (1) nearest to O, (2) farthest
from O, and obtain the distance of P from O in each case.
Draw on squared paper a curve showing the positions of P
for values of B between 0° and 180°.
6. A particle projected with a velocity of 100 feet per second
in a direction making an angle a with the horizontal plane
10000 sin 2a A
strikes the horizontal plane again at a distance — ft.
o'2t
from the point of projection. For what value of a is this
distance greatest, and what is the greatest distance?
Also find two values of a for which the range of the particle
would be 100ft.
CHAPTER Y.
KELATIONS BETWEEN THE SIDES AND ANGLES OF
A TRIANGLE.
31. To prove that in any triangle
a b e
sin A ~~ sin B ~~ sine "
180- C
D
Fig. 43.
C
Fig. 44.
If p be the length of the perpendicular AD drawn from A
to the side BC, we have
p = c sin B = b sin C ;
. b c
sin B sin C '
and in a similar way each of these ratios may be shown to
be equal to — — .
sin A
If one of the angles be obtuse, such as C in Fig. 44, the
same result holds, for p = b sin (180° - C) = b sin C as before.
Note. Prove that the formula \ab sin C for the area
of a triangle (Art. 15), holds good when C is an obtuse
angle.
56 PRACTICAL TRIGONOMETRY
32. To prove that in any triangle
c2 = a2 + b2 - 2ab cos c.
In Fig. 43, where C is acute, we have, by a well-known
theorem in Geometry,
AB2 = BC2 + CA2 - 2BC . CD
= BC2 + CA2 - 2BC . CA COS C;
i.e. c2= a? + b2 -2abcosC.
In Fig. 44, where C is obtuse, we have, by Geometry,
AB2 = BC2 + CA2 + 2BC . CD
= BC2 + CA2 + 2BC . CA COS (180° - C)
= BC2 + CA2 - 2BC . CA COS C,
i.e. <? = a2 + b2 - 2ab cos C.
Note that the sign convention enables us to have one
formula for both cases, (i) C acute, (ii) C obtuse.
Thus we can obtain the third side of a triangle when
we are given two sides and the included angle.
And since the above formula can be written
COSC=
and similarly
COSA=
2bc > 2'
we can obtain any angle of a triangle of which the sides are
known.
33. When any three independent parts of a triangle
are given, the formulae proved above are sufficient to
determine the remaining parts, but the complete solution of
triangles without the use of logarithms generally involves
clumsy work, and we shall therefore postpone it until
Chapter vin.
There are however many occasions on which logarithmic
work is not required, and it is well that the student should
become familiar with the formulae at this stage.
SIDES AND ANGLES OF A TRIANGLE
57
Example (i).
Find the largest angle of the triangle whose sides are 3, 4, 6.
If a=3, 6 = 4, c=6, we have
cosC
2ab
= 180°-62°43/
=117° ir.
94-16-36
~24 =
-g—4583;
Example (ii).
A man observes the elevation of a tower to be a; after
walking a distance c towards the tower he observes the elevation
to be j8. Find the height of the tower.
Let A, B denote the points of observation, arid CD the tower.
Then CD = BDsin£.
Now from the A ABD, we have
BD c
sin a sin (/3 — a) J
and the height of the tower is
c sin a sin /3
sinO-a)
This method is usually more convenient than that given in
Article 16, as the result is suitable for logarithmic work.
58
PRACTICAL TRIGONOMETRY
34. If we are given two sides of a triangle and the
angle opposite one of them, say a, b, A, we may proceed to
find the remaining parts in two ways.
We may find the angle B from the relation
b sin A
smB = - - ..................... (1),
a
or we may find the side c by considering the relation
............... (2)
as a quadratic equation in c.
Now from (1) we get two values for B, which are supple-
mentary angles [Art. 23], and from the quadratic equation
(2) we get two values for c.
There may consequently be ambiguity concerning the
solution of the triangle, which we will now discuss geo-
metrically.
35. To construct the triangle, draw the angle XAC
equal to A, and make AC equal to b. With centre C and
radius a describe a circle, which (if the data are possible)
will meet AX at the required point B.
If a = the perpendicular drawn from C to AX = b sin A,
the circle touches AX at B [see Fig. 46].
B
Fig. 46. Fig. 47.
If a > b sin A < b, the circle cuts AX at two points B, B',
and we have ambiguity; for both triangles CAB, CAB' have
the given parts [see Fig. 47].
SIDES AND ANGLES OF A TRIANGLE 59
If a = b, the point B' coincides with A, and we have one
triangle only.
If a > b, the points B, B' are on opposite sides of A, and
we only have one triangle with the given parts; for L CAB'
is the supplement of the given angle A [see Fig. 48].
Fig. 48.
Thus we see that ambiguity can only arise when the
side opposite the given angle is less than the other side.
Examples. V.
1. Find the largest angle of the triangle whose sides are
6, 7, 8 feet.
2. Given B = 114°, a = 2, c = 3, find b.
3. Find the vertical angle of an isosceles triangle whose
equal sides are 3 ft. and base 5 ft. ; (1) by using the fact that
the bisector of the vertical angle is perpendicular to the base
and bisects it; (2) by using the formula giving the cosine of an
angle of a triangle in terms of the sides.
4. Find the lengths of the diagonals of a parallelogram of
which two sides are 2, 5 metres and are inclined at 50°.
5. Show that the parts B = 40°, 6 = 5, c = 20 cannot form a
triangle.
6. If a = 4, 6 = 5, c = 6, find the angles.
7. The diagonals of a parallelogram are 4, 6 ft. and intersect
at 28° ; find the sides.
8. If b = 10, c = 8, A = 47°, solve the triangle.
60 PRACTICAL TRIGONOMETRY
9. Show that there are two triangles having Z>=3, c=4,
B = 40°, and find the angle A in each case.
10. The sides of a parallelogram are 4 ft., 5 ft., and the
shorter diagonal is 2 ft. ; find the other diagonal.
11. Given c=10, a = 12, B = 35°, find the length of the
median which bisects BC.
12. Find the obtuse angle in the triangle whose sides are
as 2 : 5 : 6.
13. Prove with help of figures (1) when A, B are acute,
(2) when A is obtuse, that c=acos B-f&cos A.
14. In a triangle A = 115°, a = 3, c = 2; find the other angles.
15. OX, OY are two straight roads inclined at 60°. A man
A walks along OX at 4 miles an hour, and B starts along OY at
the same time. If B is 7J miles from A at the end of 2 hours,
obtain a quadratic equation for the distance B has walked in
that time and solve it.
16. a, 6, c, d are the sides of a quadrilateral inscribed in a
circle, and 6 is the angle contained by a, b ; by writing down two
expressions for the diagonal opposite $, prove that
cos$=
17. If x be the length of a diagonal of a parallelogram which
makes angles a, j8 with the sides, show that the sides are
x sin a , x sin £
and —
flin(a+j3) sin (a + /3)'
18. A straight line AD divides the angle A of a triangle
ABC into two parts a, ft and meets BC at D : show that
BD _csina
DC~~ frsin/S*
19. The parts a, c, A of a triangle are given. Write down a
quadratic equation for the remaining side b. If 5l9 b2 are the
lengths of the third side in the two triangles which have the
given parts, show that bi + b% — 2c cos A and bib% = c2 — a2.
Also prove that the sum of the areas of the two triangles is
c2 sin A cos A, and consequently independent of a.
20. In the A ABC, if the line joining A to the mid -point of
BC is perpendicular to AC, prove
2(c2-a2)
cos A cos C = — ^r • .
3ac
SIDES AND ANGLES OF A TRIANGLE 61
Miscellaneous Examples. C.
1. Draw two straight lines OX, OY at right angles, OX to
the right, OY up, and find a point P 4" from OX and I" from OY.
Now imagine P to remain fixed while YOX is revolved counter-
clockwise about O. Determine both by drawing and calculation,
(1) what amount of revolution would make OX pass through P;
(2) the distance of P from OX when OX has turned through 60°.
2. Being given cos 41° 24' = j, find two values of 6 less than
180° which satisfy 4 cos 20 +3=0.
3. In a triangle ABC prove that
a sin B
tan A =
c — a cos B *
4. A man surveying a mine measures a length AB of 12
chains due E. with a dip of 8° to the horizon ; then a length BC
of 20 chains due E. with a dip of 5°. How much deeper vertically
is C than A? A chain =66 ft. Give the answer in feet.
5. Two lines OA, OB of length rly r2 respectively make
angles of 6l and $2 with a third line OX. Prove
AB2 = /y5 + r22 - 2?v'2 cos (<92 - ^).
6. In a triangle sm2C = siii2A-f siri2B. Prove that the
triangle is right-angled.
7. A BCD is a parallelogram :
AB = 2'5", BC = 4", and L ABC = 65°.
Calculate the area of the parallelogram and the length of the
diagonals.
8. A is 200 yards from B in the same horizontal plane.
The angular elevation at A of a kite vertically above B is 55° 30'.
How far must the kite descend before its angular elevation as
seen from A is half that angle?
9. If D be the mid-point of the side BC of an equilateral
triangle ABC, and O the point of intersection of the medians,
prove by finding the lengths of AD and AO in terms of a side
of the triangle that AO = 2.OD.
62 PRACTICAL TRIGONOMETRY
10. A mast is secured by 3 equal stays connecting its
highest point with 3 pegs on the ground at the corners of an
equilateral triangle. If the length of each stay is 45 feet and
the distance between 2 pegs is 30 ft., find the height of the mast.
11. Find from the definitions a formula which will give
cos 6 when tan 6 is known.
Taking 0=34° 43', find its tangent from the tables: then find
its cosine from your formula and compare the result with that
given in the tables.
12. A balloon is vertically over a point which lies in a direct
line between two observers who are 2000 ft. apart and who note
the angles of elevation of the balloon to be 35° 30' and 61° 20':
find its height.
13. The half ABC of a rectangular sheet of paper ABCD,
AB = 5", BC = 7", is folded about the diagonal AC. Find by
using tables the angle between CD and the new position of CB.
Find also the length of the line joining the old and the new
positions of B.
14. Two circles whose radii are 5 cms. and 3 cms. have their
centres 10 cms. apart ; prove that the common tangents make
angles sin"1 -2, or sin"1 -8 with the line joining the centres.
15. A line of length x is drawn from A to any point in the
side BC of a triangle ABC and makes angles of 0, </> respectively
with AB, AC : prove by using the formula for the area of a
triangle that
sin 6 sin </> _ sin
~~ ~~
16. Take a line OA, length 5 cms., near the lower edge of
the page and draw perpendicular to it a line AB of unlimited
length. Find with your instruments eight points P, Q, R . . . on
AB such that AOP = POQ=QOR= ... =10°. From the figure find
the average increase of the tangent of the angle for each degree
between 0° and 10°, 10° and 20°, etc.
What happens to the tangent as the angle increases from
80° to 90°?
SIDES AND ANGLES OF A TRIANGLE 03
17. Two adjacent sides of a parallelogram inclined at an
angle a are P and Q. The diagonal passing through their point
of intersection is R. Prove
18. Find the positive values of 6 between 0° and 3«0° which
satisfy the equation
6ootl+I
19. OX, OY are two straight lines intersecting at an angle 6.
A point A is taken on OY such that OA = a, and then AB is drawn
perpendicular to OY meeting OX in B ; BC is drawn perpen-
dicular to OX meeting OY in C and CD is drawn perpendicular
to OY meeting OX in D. Prove that
CD=atan<9(l+tan2<9).
20. From a point O three straight lines OA, OB, OC are
drawn in the same plane of lengths 1, 2, 3 respectively arid with
the angles AOB, BOG each equal to 60°. Find the angle ABC.
CHAPTER VI.
PROJECTION. FORMULAE FOR COMPOUND ANGLES.
36. Projection.
If from the extremities of a line OP, perpendiculars OM,
PN be dropped to another line AB, then MN is said to be
the projection of OP on the line AB.
4 ft AT B
Fig. 49.
Length of projection.
Let the angle between OP and AB be 0.
Then from the diagram
2V B
-R,
Fig. 50.
i.e. the length of the projection of a line OP on another
line equals OP x cosine of angle OP makes with the line on
which it is projected.
. If 0 = 90°, then the projection of OP
- OP cos 90° = 0.
If 0-0° then the projection of OP
= OP cos 0° = OP.
FORMULAE FOR COMPOUND ANGLES 65
Exercise, (i) Take two fixed points P and Q and any
straight line AB. Let R be any other point. Project PQ, PR,
RQ on AB. Show by a diagram that by adopting the sign con-
vention we have, for all positions of R :
Projection of PR + Projection of RQ= Projection of PQ.
(ii) Show by a diagram that the sum of the projections on
any straight line, of the sides, taken in order, of any closed
polygon is zero.
37. Trigonometrical ratios of compound angles.
It is frequently useful to express the trigonometrical
ratios of compound angles such as A + B, or A — B, in terms
of the ratios of A and B.
The beginner is apt to think that sin (A + B) is
= sin A + sin B,
a statement which can at once be shown to be incorrect by
the help of tables.
For instance, 74° = 40° + 34° ;
but sin 74° -'9613,
and sin 40° + sin 34° = '6428 + '5592 = 1 '2020.
In the following articles we shall prove that
sin (A + B) = sin A cos B + cos A sin B,
cos (A + B) = cos A cos B — sin A sin B,
sin (A — B) = sin A cos B — cos A sin B,
cos (A - B) = cos A cos B + sin A sin B
Exercise.
Which is the greater, cos (A-|- B) or cos A?
Why is the statement cos (A + B) = cos A+cos B obviously
absurd ?
p. F. 5
66
PRACTICAL TRIGONOMETRY
38. To prove cos (A + B) = cos A cos B — sin A sin B.
Let XOQ be the angle A, and POQ the angle B.
In the line OP bounding the compound angle A + B take
a point P and let fall a perpendicular PQ, on OQ.
O N M X.
Fig. 51.
Project OQ, OP on OX.
Now OM=ON + NM,
i.e. projection of OQ = projection of OP + projection of PQ,
or OQ cos A = OP cos (A + B) 4- PQ cos (90° — A)
(by producing PQ we see that the angle PQ makes with OX
is (90° - A) since OQP is a right angle),
or
i.e. cos (A + B) = cos A cos B - sin A sin B.
OQ , , PQ .
— cos A = cos (A + B) 4- — sm A,
Note. If in this formula we write — B for B, we get
cos (A — B) = cos A cos (— B) — sin A sin (— B)
= COS A COS B 4: sill A SHI B,
since cos (- B) = cos B and sin (- B) = - sin B.
FORMULAE FOR COMPOUND ANGLES
67
39. To prove sin (A + B) = sin A cos B+cos A sin B.
With the same construction as before, but projecting on
OY at right angles to OX,
ON = OM+MN,
Fig. 52.
i.e. projection of OP = projection of OQ + projection of QP,
OP cos [90° - (A + B)] = OQ cos (90° - A) + QP cos A.
By producing QP we see that the angle QP makes with
OY is A, since it is the complement of QOY; .'. we have
OP sin (A + B) = OQ sin A + QP cos A,
OQ QP
or sin (A + B) = — sin A + — cos A
= sin A cos B + cos A sin B.
Note. If for B we write — B, we get
sin (A — B) = sin A cos B - cos A sin B.
5—2
68
PRACTICAL TRIGONOMETRY
40, Independent proofs that
cos (A — B) = cos A cos B + sin A sin B.
sin (A — B) = sin A cos B — cos A sin B.
Let OP describe the angle A in the positive direction,
and then the angle B in a negative direction.
As before, take a point P in the line OP bounding the
compound angle A — B and drop a perpendicular on OQ.
Project on OX for cos (A - B), on OY for sin (A - B),
ON =OM + MN;
projection of OP = projection of OQ + projection of QP,
OP cos (A - B) = OQ cos A + QP cos (90° - A),
from which
cos (A - B) = cos A cos B + sin A sin B.
Taking projections on OY, we have
OM'=ON'+ N'M',
Y
M1
O M N X
Fig. 53.
i.e. projection of OQ = projection of OP + projection of PQ ;
.', OQ cos (90° - A) - OP cos (90° - A - B) + PQcosA;
.'. OQ sin A = OP sin (A — B) + PQ cos A ;
whence sin (A - B) = sin A cos B - cos A sin B.
FORMULAE FOR COMPOUND ANGLES 69
Note. We have seen that the formula for cos (A — B)
may be deduced from that for cos (A + B).
If we write (90 - A) for A in the formula for cos (A + B),
we get
cos (90 - A 4- B) = cos (90 - A) cos B - sin (90- A) sin B,
i.e. cos [90 - (A - B)]
= sin (A — B) = sin A cos B — cos A sin B.
Similarly we can obtain the formula for
sin (A + B).
Example (i).
Prove sin (A + B) sin (A - B) = sin2 A - sin2 B.
sin (A + B) sin (A— B) = (sin A cos B -fcos A sin B)
(sin A cos B — cos A sin B
= sin2A cos2 B - cos2 A sin2 B
= sm2A(l-sm2B)-(l-sm2A)sin2B
= sin2 A- sin2 B.
Example (ii).
Expand sin ( A -f B -f C).
Treating (B + C) as a single angle we have
sin [A -}- ( B + C)] = sin A cos ( B + C) -f cos A sin ( B + C)
= sin A (cos B cos C — sin B sin C)
+cos A (sin B cos C + cos B sin C)
= siii A cos B cos C +sin B cos A cos C
-f sin C cos A cos B — sin A sin B sin C.
Example (iii).
Find the value of
cos 34° cos 42° - sin 34° sin 42°.
By comparing with the formula for cos(A + B) we see that
this expression equals cos (34° +42°)= cos 76° = '2419 (from the
Tables).
70
PRACTICAL TRIGONOMETRY
Example (iv).
The following example shows the use made of Projection
in Statical problems.
A weighted rod AB, 4 ft. long, is suspended by a string, fastened
to its two ends, which passes over a pulley at O so that each portion
is inclined at an angle of 35° to the vertical. The rod makes an
angle of 20° with the horizon. Find the length of the string.
Let x and y be the lengths of the two portions of the string.
Project on AC the horizontal line through A.
Projection of AO + projection of OB = projection of AB ;
.-. x sin 35° +y sin 35° = 4 cos 20°;
4 cos 20°
.*. x+y = — -. — ^rr=6*5;
sin3o
. •. length of string =6-5 ft. approximately.
FORMULAE FOR COMPOUND ANGLES 71
Examples. VI a.
1. If cos a= f, and cos /3 = |-, calculate the values of
sin a, sin/3, sin(a-{-/3), cos(a + /3), sin(a-/3), cos(a-/3).
2. If sina = J, and sinj3 = J, calculate the values of cos a,
cos ft sin (a + ft). Verify by finding the angles a, ft by the help
of the tables.
3. If cosa = *2 and cos £='5, find cos (a — /3) and verify from
the tables.
4. Expand cos (90° — A), and show that it equals sin A.
Expand also cos (180° + A), and sin (90° + A).
5. Find the values of
(i) sin 47° cos 16° — cos 47° sin 16°,
(ii) sin 52° sin 27° - cos 52° cos 27°.
6. By writing cos 75° as cos (45° + 30°) and expanding, prove
V6-V2
cos 75 = - — , .
4
7. Prove that cos 15°=- . , and find sin 15°.
8. Prove that cos (A + B) cos (A - B) = cos2A - sin2 B.
9. Show that V2 sin (A + 45°) = sin A + cos A.
10. Prove that S1P (AA + ^ = tan A + tan B.
cos A cos B
11. Show that cos A - sin A = ^2 cos (A + 45°).
12. Find the values of
(i) cos 1 8° cos 36° - sin 1 8° sin 36°,
(ii) sin 18° cos 36° + cos 18° sin 36°.
13. Prove that
sin (A -f B) +cos (A - B) = (sin A + cos A) (sin B + cos B).
14. Factorise sin(A-B)+cos(A+B).
72 PRACTICAL TRIGONOMETRY
15. Expand cos(A + B+C).
16. If B and d> are both less than 180°, and sm ^ sin *= - 1,
cos 6 cos cf)
show that 6 and <f> differ by a right angle.
17. A sphere of radius r rolls down an inclined plane which
makes an angle a with the horizon. Prove that the height of the
centre of the sphere above the horizontal plane when the point
of contact of the sphere is at a distance I from the foot of the
inclined plane is r cos a + 1 sin a.
18. OX, OY are two straight lines at right angles. P is a
point 4" from OX and 3" from OY. Through O a straight line
is drawn making an angle 6 with OX. Prove by projection that
the length of the perpendicular from P on this line is 4cos0 — 3sin0.
41. To prove
tan A + tan B
tan (A + B) = _- - .
' 1-tanAtanB
/A v sin(A+B)
tan (A + B) = - — )- — -(
COS (A + B)
___ sin A cos B + cos A sin B
cos A cos B - sin A sin B
sin A cos B cos A sin B
_ cos A cos B cos A cos B
cos A cos B sin A sin B
COS A COS B COS A COS B
(dividing numerator and denominator by cos A cos B)
tan A + tan B
~ 1 — tan A tan B '
Prove in a similar way
, ^ tan A - tan B
tan (A - B) = -
' 1 + taiiAtanB
FORMULAE FOR COMPOUND ANGLES 73
Examples. VI b.
1. Prove that tan (45° + A) = - .
1 - tan A
P tan 47° -tan 20°
2. Find the value of
tan 20° tan 47° + 1 '
3. Prove that tan 75° = 2 + ^/3, and find tan 1 5°.
4. Expand tan (90° 4- A) and show that it equals - cot A.
5. In a similar way prove that
tan (180° + A) = tan A,
and tan (180° - A) = - tan A.
6. By writing cot (A+ B) as - — ~r — I and expanding, prove
sin ^/\ ~p &)
, , , . , , cot A cot B - 1
that it equals — ,-- — .
cot A + cot B
7. Express cot (A — B) in terms of cot A and cot B.
8. Given tan a = 1 and tan (a + /3) = 2, find tan /3.
9. If tan A = £ and tan B = J , show that A -I- B = 45°, supposing1
A and B to be acute angles.
10. The perpendicular from the vertex of a triangle is 6"
long and it divides the base into segments which are 2" and 3"
respectively. Find the tangent of the vertical angle.
11. ABC is an isosceles triangle, right angled at C, and D is
the middle point of AC. Prove that DB dividas the angle B into
two parts whose cotangents are in the ratio 2 : 3.
12. If two straight lines make with a third straight line angles
6 and & such that tan 6 = m and tan & — m', prove that the angle
,. . , , m~mr
between the two lines is tan ~ l — — ; .
1 + mm
13. Expand tan(a+/3-fy) first in terms of tana and
tan (/3 + y) and hence in terms of tan a, tan /3, tan y. Use your
result to show that (i) if a + /3 + y=180°, then
tan a + tan /3 + tan y = tan a tan /3 tan y.
(ii) if a+/3 + y=90°, then
tan (3 tan y + tan y tan a + tan a tan /3 = 1.
74 PRACTICAL TRIGONOMETRY
14. A vertical pole more than 100 ft. high consists of two
parts, the lower being J of the whole. At a point in the
horizontal plane through the foot of the pole and 40 ft. from
it, the upper part subtends an angle whose tangent is J. Find
the height of the pole.
42. To express sin2A, cos2A and tan2A as
functions of A,
We have
sin 2A - sin ( A + A) = sin A cos A + cos A sin A ;
.'. sin 2A = 2 sin A cos A (1).
Also
cos 2A = cos (A + A) = cos A cos A — sin A sin A ;
/. cos 2 A = cos2 A - sin2 A (2).
Writing 1 - cos2 A for sin2 A, we get
cos2A = 2cos2A-l (3).
Writing 1 — siii2A for cos2 A, we get
cos2A = l- 2 sin2 A (4).
The results (3) and (4) may be written
l + cos2A = 2cos2A ...(5),
l-cos2A=2sin2A (6),
and in this form are of much importance.
From (5) and (6) we have
1 - cos 2A
-A = tan2 A.
1 + cos 2A
Again,
, . tan A + tan A
tan 2A = tan (A + A) = — - — — — ;
1 - tan A tan A
2 tan A
/. tan2A = 1— —s— (7).
1 - tan2 A
It is important to notice that the above formulae enable
us to express functions of an angle in terms of the functions
of half the angle.
FORMULAE FOR COMPOUND ANGLES 75
f\ A
Thus sin 0 = 2 sin - cos - ;
2 2
A A
cos 0 — cos2 - — sin2 -
2 2
.30 30
sin 30 = 2 sin — cos — ;
~> 2
e
0 2tan4
tan- =
2 l-tan^'
Note. The expression 1 — cos S is of frequent occurrence in
Nautical computations and is called versine 6. Half-versine is
A
contracted to Haversine and from the formula cos$ = l - 2 sin2-,
. vers 6 1 — cos B . „ 6
we see, hav 6 = — - — = — = sin2 - .
22 2
Example (i).
Prove
A A
We have cos A = cos2 - — sin2 —
2 2
A
9 " i
= cos2— ( 1 -
76 PRACTICAL TRIGONOMETRY
Example (ii).
Prove sin 3A = 3 sin A — 4 sin3 A.
We have
sin3A = sin(2A + A)
= sin 2 A cos A + cos 2 A sin A
= 2 sin A cos2 A + (1 — 2 sin2 A) sin A
= 2sinA(l-sin2A) + (l-2sin2A)sinA
=3 sin A — 4 sin3 A.
Exercise.
Prove in a similar way that
(1) sinA= ' ; (2) cos3A = 4cos3A-3cosA ;
/ox o. OA 3 tan A -tan3 A
(3)tan3A=-T-3— 2— .
Examples. VI c.
1. If sin a = i, calculate cos a, sin 2a, cos 2a.
2. Given cos a = '4, find sin 2a, cos 2a, tan 2a.
3. Find the value of 2 sin 25° cos 25°, 1-2 sin2 25.
4. Prove that (sin 6 - cos (9)2 = 1 - sin 2<9.
5. Find tan 2A when tan A = -5.
6. Factorise cos4 A — sin4 A and prove it equal to cos 2A.
/5
7. If cos 2a = f, prove tan a= ~ .
o
8. If l + cos2a = ff, find cosa.
9. If 1 - cos 2a=|, find sin a.
10. Given that tan a = J, prove cos 2a = ± 4.
FORMULAE FOR COMPOUND ANGLES 77
11. Find the values of \/ and
1- cos 56° __j /I + cos 56°
2
12. Find the value of \/
V
1+ cos 40
13. Find tan — , given cos a = f .
x
14. Find the value of 2 cos 20 + 3 sin 20 when tan 6 = f .
15. Prove that 1 — cos a cos ft — sin a sin /3= 2 sin2 ^— .
2s
16. Find the positive values of A between 0° and 360° which
satisfy the equations
(i) cos2A + sin2A=--=f ; (ii) tan 2 A = 3 tan A.
17. Express cos 4a in terms of cos a.
18. Find the value of a cos 20 + 6 sin 20 when tan 0 == - .
a
19. If cot 0 = 1~C°^ , prove that f =90° - 0.
sm <p ' 2
20. Express cos2 a — sin2 /3 as half the sum of two cosines
and hence evaluate cos2 63° — sin2 47°.
21. If tan 0 = - , simplify tan 20 + sec 20.
Cd
22. If cot2 0- cot 0 = 1, prove cot 20=1.
23. AB is the diameter of a circle of radius r, whose centre
is at C. P is a point on the circumference where Z.BCP = 0.
A
Prove that the projection of AP on the diameter equals 2rcos2-.
2i
Shew that this result is true whether 0 is acute or obtuse.
24. A point P moves round the circumference of a wheel of
radius r, centre O, placed in a vertical plane. If A is the lowest
position of P show that the vertical height of P above A at any
A
time is 2rsin2 - where L AOP = 0.
a
25. Two radii OP, OQ of a circle of radius r are inclined at
an angle 0. The perpendicular from O on PQ, cuts the chord at
n
A and the arc at B. Prove AB = 2r sin2 - .
4
78 PRACTICAL TRIGONOMETRY
43. The formulae of Article 37 are useful for obtaining
solutions of equations of the form
a sin 0 + b cos 0 = c.
Example.
Find a solution of the equation 3 sin 6 — 2 cos B = 2.
Let a be an acute angle such that tan a = § ; then
2 3
cos a
The equation can now be written
\/13 (sin B cos a — cos B sin a) = 2 ;
2
whence sin (0 - a) =
_2y/13
~T3~~
_ 2 x 3-606
13
= '5548
= sin 33° 42'.
Also a = tan-1|-tan-1 -6667 = 33° 41';
.-. a solution of the equation is given by
<9-33° 41' = 33° 42';
whence 0 = 67° 23'.
The angle a which has been introduced in the work is called
a subsidiary angle. Other occasions when a subsidiary angle
is of use will be found in Articles 56, 57.
Beginners sometimes solve equations of the form
a cos B + b sin B = c
by substituting v 1 — sin2 6 for cos B and squaring : but this
method is not satisfactory, as in consequence of squaring we
obtain some values of B which are not roots of the given
equation,
FORMULAE FOR COMPOUND ANGLES 79
Examples. VI d.
1. Show that 3 sin 6 + 4 cos 6 = 5 sin (6 + a), where a = tan ~ * £ ;
and hence prove that the greatest value of 3 sin 0+4 cos 0, when
6 may have any value, is 5. What is the value of 6 in this case 1
2. Find a solution of 3 sin 6 + 4 cos 6 = 2.
3. Find a value of # which satisfies cos x+ sin #='5.
4. Find a solution of 4 cos # — 3 sin # = 3.
5. If - = tan 0, prove that
jo cos a — q sin a = Jp2+q2 sin (0 — a),
and find the greatest value of p cos a — q sin a if a varies.
44. We have proved
(i) sin A cos B + cos A sin B = sin (A + B),
(ii) sin A cos B — cos A sin B = sin (A — B),
(iii) cos A cos B - sin A sin B = cos (A + B),
(iv) cos A cos B + sin A sin B = cos (A - B).
Adding (i) and (ii), we get
(a) 2 sin A cos B = sin (A + B) + sin (A — B)
= sin (sum) + sin (difference).
Subtracting (i) and (ii)
(/3) 2 cos A sin B = sin (A + B) - sin (A - B)
= sin (sum) — sin (difference).
Adding (iii) and (iv)
(y) 2 COS A COS B = COS (A + B) 4- COS (A — B)
= cos (sum) -f cos (difference).
Subtracting (iii) from (iv)
{since (A + B) > (A - B) ; .'. cos (A + B) < cos (A - B)}.
(3) 2 sin A sin B = cos (A — B) - cos (A + B)
= cos (difference) - cos (sum).
80 PRACTICAL TRIGONOMETRY
These formulae enable us to express products of sines
and cosines as sums or differences, and should be learnt in
the verbal form.
It will be noticed that in both (a) and (/?) we have the
product of a sine and a cosine; but either formula gives
the same result, as will be seen from the following example.
Example (i).
2 sin 50 cos 20 = sin (sum) + sin (difference)
= sin (50 + 20) + sin (50 - 20)
= sin 70+ sin 30.
If however we apply formula (/3) which also gives the product
of a sine and a cosine, we have
2 cos 20 sin 50 = sin (sum) — sin (difference)
= sin (20+50) - sin (20 - 50)
= sin 70 -sin (-30)
= sin 70+ sin 30,
for sin ( - 30) = - sin 30.
Example (ii).
cos 20 cos 50 = J [cos (sum) + cos (difference)]
= i[cos (20+50)+cos (20-50)]
=J[cos70+cos(-30)]
_ cos 70+ cos 30
2
since cos ( - 30) = cos 30.
Example (iii).
2 sin 50 sin 20 = cos (difference) — cos (sum)
= cos (50 - 20) - cos (50 + 20)
;= COS 30 -COS 70.
FORMULAE FOR COMPOUND ANGLES 81
Examples. VI e.
Express as the sum or difference of two Trigonometrical
ratios : verify approximately the numerical examples by help
of the tables.
1. 2 sin 30 cos 6. 2. 2 cos 3(9 cos <9. 3. sin 30 sin 0.
4. 2 cos 30 sin 6. 5. sin A cos 2A. 6. sin A cos B.
7. cos2Acos2B. 8. sin 50 sin 6. 9. 2 sin 20° cos 70°.
10. 2 cos 40° cos 30°. 11. 2 sin 10° sin 20°. 12. cos 50° cos 30°.
13. 2 sin (A + B) cos (A - B). 14. 2cos(A+2B)cos(2A+B).
A A
15. 2 cos — sin--. 16. sin 3a sin a.
'2i 2i
45. The formulae of Article 44 give us sums and
differences expressed as products, but it is more convenient
to put the formulae in a different form, as follows.
"Writing X for (A + B), and Y for (A - B), we have
A + B=X,
A-B = Y;
or A = — — ;
X-Y
and 2B = X - Y, or B - — — .
A
Substituting in (a), (/?), (y), (8), of Article 44, we get
X _i_ Y X _ Y
from (a) sin X + sin Y = 2 sin — - cos — — ,
2 A
i.e. sum of sines = 2 sin (half sum) cos (half difference) ;
X + Y . (X-Y)
from (p) sm X - sin Y = 2 cos — — sin ^-— — - ,
2 £
i.e. difference of sines = 2 cos (half sum) sin (half difference);
X + Y X — Y
from (y) cos X + cos Y = 2 cos - — cos — - — ,
"Z A
sum of cosines = 2 cos (half sum) cos (half difference) ;
X + Y X _ Y
from (8) cos Y - cos X = 2 sin —5— sin - - ;
- *-i
difference of cosines = 2 sin (half sum) sin (half difference
reversed).
p. F. 6
82 PRACTICAL TRIGONOMETRY
Example (i).
Express as a product sin 30+ sin 20,
sin 3d + sin 20= 2 sin (half sum) cos (half difference)
. 50 0
= 2 sm — cos ^r .
Example (ii).
cos 30 - cos 50 = 2 sin (half sum) sin (half difference reversed)
. 30 + 50 . 50-30
= 2 sin— — sm— —
= 2 sin 40 sin 0.
Example (iii).
Prove that
sin a — sin 2a + sin 3a =4 sin - cos a cos — ,
'2i 2»
sin a — sin 2a + sin 3a = sin a + sin 3a — sin 2a
= 2 sin 2a cos a — 2 sin a cos a
= 2 cos a (sin 2a — sin a)
3a . a
=2 cos a 2 cos — sin -
2i A
. . a 3a
= 4 sin - cos a cos — .
Examples. VI f.
Express as products :
1. sin 3A -f sin A. 2. sin 3A — sin A. 3. cos 3A + cos A.
4. cos A- cos 3A. 5. sin 20- sin 0. 6. cos 30 -cos 20.
7. sin B + sin A. 8. cos 2a + cos 2/3. 9. cos 2a - cos 2£.
10. sin 23° + sin 14°. 11. cos 32° -cos 41°.
12. sin41° + cosl20. 13. cos 18° + cos 43°.
14. Prove that
,-
_ — -- _ =taii — s-2- cot —zr
sin 0 - sm (f) 2 2
cos 0 - cos 6 0 + <b , 0
15. Prove --- ^— -?- = - tan — TT— tan
cos 0 + cos <£ 2
FORMULAE FOR COMPOUND ANGLES 83
. „ T. sin 5° + sin 47°
16. Prove -- = tan 69 .
cos 5 —cos 47
sin 10° + sin 26°
17. Prove 0,.0=cot72°.
cos 10 -f cos 26
18. If
x (sin 6 — sin <£) +y (cos </> — cos ff) + cos 6 sin <£ - sin 0 cos </> = 0,
0 + (t> . 0 + 6 0-d>
show that x cos — g-2- -fy sin — — — =cos — 5-*- .
2i 2i 2>
19. Prove cos 3a sin 2a — cos 4a sin a = cos 2a sin a.
20. If ^7 cos a-fy sin a — c=0,
•
and ,#cos/3+2/ sin/3 — c=0,
c cos G sin — — -
prove that x = — . ?/ = — .
a — p a — p
cos -
2
21. In any triangle prove that
A-B
cos •
q + ft _ 8
c A+B*
cos-g-
22. If ^7COSj8+y cosa=/> and #sin/3 — y sina=0,
p sin a io sin 8
prove x—-:—-r — -. and y= -.--- r — ^-rr.
sm(a + 0) y sm(a-f^)
23. If
A cosu—e 6 /l+e , u
cos 6 = - ---- ~ , prove tan - = \/ - -- . tan - .
1-ecos^' r 2 V i -e' 3
24. From the equations
T! cos 6 + T2 cos 0 = 100,
T! sin 0 - T2 sin </> = 0,
100 sin (/> 100 sin B
show that T! = -I — 7^ — ~r and T2= - — ^ — rr.
0,« /a . ^\ sm(0-f <^>)
OK ,. cos 40° -f cos 12°
25. Prove Tf^ --» = tan 116 .cot 14 .
cos 40 —cos 12
6—2
84 PRACTICAL TRIGONOMETRY
Miscellaneous Examples. D.
1. Two straight lines make with another line angles,
measured in the same direction, whose tangents are m and in'.
If these two lines are at right angles, prove 1-f mm' = 0. What
is the relation between m and m' if the lines are parallel ?
2. If a, /3 are the angles which satisfy the equation
4 tan2 6 — 3 tan 0 — 2 = 0, find the value of tan a + tan ft tan a tan ft
tan(a+£).
3. AB is a diameter of a circle of radius 5 '6 ft. At A a line
AC is drawn meeting the circle at C and the tangent at B in D.
If BAG = 32° 45', find the length of CD. Also if O be the centre
and OD cuts the circumference in E, find the length of DE.
4. One mast of a ship is 12 feet longer than the other and
both slope towards the stern at an angle of 10° to the vertical.
The line joining their tops is inclined at 40° to the horizon.
Find the horizontal distance between the masts.
5. If - cos <i> -h T sin <b = l and -sin <f>— ycos<f> = — 1.
a o a b
6. The angles a and /3 are acute, sina=4 and sm/3=553-.
Calculate the value of sin(a+/3) and of a-fft
Construct a A ABC in which AD the perpendicular from
A on BC is 6 cms. long and the angles DAB, DAC are the angles
a, ft Measure the angle BAG and compare it with the value
already found for a+ft
7. Solve a2=b2 + c2 — 2bccos A as a quadratic equation in
which b is the unknown quantity. And hence, or otherwise,
calculate the positive value of b when a = llcms., c=9cms.,
A = 40°. Check your result by an accurate drawing.
8. A hemispherical bowl, centre C, radius r, rests with its
lowest point O on a horizontal plane. It is tilted until the line
CO makes an angle 6 with the vertical. Prove that the height
A
of O above the plane is now 2r sin2 - .
FORMULAE FOR COMPOUND ANGLES 85
9. A ladder 30 ft. long just reaches the top of a house and
makes an angle of 67° with the ground. It is let down until it
rests on a sill and then makes an angle of 48° with the ground.
How far is the sill vertically below the point where the ladder
first rested ?
10. The angles which satisfy the equation
tan2 B- 3 tan 6- 1=0
are a and £. Prove that the difference between a and /3 is 90°.
11. Solve the equations
x sin j8 +y cos /3 = 1,
x cos a +y sin a = 0.
12. The sides of a parallelogram are a, 5, and the angle
between them is 6. Prove that (1) the sum of the squares on
the diagonals is 2(a2-f&2); (2) the difference of the squares
on the diagonals is 4ab cos 6.
13. Three lines OA, OB, OC of length r^ r2, r3 are drawn
making angles $!, $2? $3 with the horizontal through O, prove
that the area of the triangle ABC is
tf\ sin (Bl -
14. ABC is a triangle, B = 90°, BA = 2, BC = 3, CD is the
median joining C to the mid-point of AB. Prove that
15. The sights of a gun are 2 ft. apart and the back sight is
raised till it is 2" above the front sight when the barrel of the
gun is pointing horizontally. I raise the gun till the line of
sights points directly towards the top of a tower 100ft. high
and 500 yards distant. Find the tangent of the angle of
elevation at which the barrel points and hence calculate the
angla
CHAPTER VII.
LOGARITHMS.
46. Definition.
The logarithm of a number to a given base is the index
of the power to which the base must be raised in order to
equal the number.
Thus if x = ey, then y is the logarithm of x to the base e.
This is written
Example.
Find logsx/27.
Let x= Iog3 \/27,
then 3a:=x/27
For practical purposes the base to which logarithms are
calculated is 10; such logarithms are called common
logarithms, and we shall confine ourselves to them.
Thus log 1 7 denotes the logarithm of 1 7 to the base 10.
LOGARITHMS 87
47. Iii the first place we must prove certain funda-
mental laws of logarithms, on which the utility of logarithms
depends.
I. log ab = log a + log b.
Let log a = x, and log b = y.
Then a = 10% and b = 1.0*;
.'.by definition log ab =
= log a + log b.
II. log ?J = log a- log b.
We have =
= log a — log 6.
III. log an = n log a.
We have an = (lO*)* = 10"*;
= n log a.
Example*.
log (35 x 47) = log 35 + log 4*7,
log ||f = log 213 -log 42 1,
log v/57 = log 57* = J log 57,
log —~- =- log 34 + i log 29 - log 53.
88 PRACTICAL TRIGONOMETRY
48. An inspection of the following table will enable us
to formulate rules for writing down at sight the integral
part of the logarithm of a number.
104 = 10,000, .'. log 10,000 = 4 ;
103 =1,000, .'. log 1,000 -3;
102 =100, /. log 100 = 2;
101 - 10, .'. log 10 = 1 ;
10° =1, .'. logl =0;
10-^V =•!, /. log'1 -1;
10-a=Tfo ='01, .-. log -01 = -2;
10-8=Tnftnr=-001, .'. log -001 --3.
It will be noticed that the only numbers whose logarithms
are whole numbers are those which are integral powers
of 10.
The logarithms of numbers which lie between these
various powers of 10 will be partly integral and partly
decimal : thus, since 126*4 lies between 100 and 1000 its
logarithm will lie between 2 and 3,
i.e. log 126*4 = 2 + a decimal.
The integral part of the logarithm is called the charac-
teristic.
The decimal part is called the mantissa, and it is always
arranged that the mantissa is positive. The mantissa is
obtained from Tables, as will shortly be explained, and the
characteristic is found as follows.
All numbers with only one digit in the integral part
have 0 as the characteristic of their logarithm ; hence the
characteristic for any number is the index of the power of
ten by which the number must be divided in order that it
may have one digit in the integral part, thus :
261-3 = 2-613 xlO2;
.'. log 261'3 = log 2-613 + log 102
= 0-4171 + 2
(the mantissa being taken from the tables)
= 2-4171.
LOGARITHMS 89
Again '002613 = 2*613 x 10~3;
.'. log '002613 = log 2'613-f log 10~3
= 0-4171-3
= 3-4171.
The negative sign is written over the 3 since the charac-
teristic only is negative, the mantissa remaining positive.
We write the logarithm in this form, and not —2*5829,
since by this device the mantissa will remain unaltered for
all numbers having the same significant figures.
Various other mnemonics are often given for writing
down characteristics, and are here stated for the benefit of
those who prefer to use them.
1. The characteristic of the logarithm of a number
which is greater than one is one less than the number of
digits before the decimal point.
The characteristic of the logarithm of a number less
than one is negative, and is one more than the number of
zeros that follow the decimal point or is the same as the
number of the place in which the first significant figure
occurs.
2. Begin at the first significant figure and count the
digits to the unit figure (not including the unit figure),
this rule applying whether the number is greater or less
than one.
Example.
Given log 2933 = 3'4673,
we have log 29-33 = 1 '4673 ;
for the characteristic is 1, since there are 2 digits in the integral
part, and the mantissa remains unaltered.
Similarly log -002933 = 3'4673.
Again, we have -4673= log 2 '933; for there can only be one
digit in the integral part, since the characteristic is zero, and
•4673 is the mantissa corresponding to the digits 2933.
Similarly 2'4673 = log -02933.
90
PRACTICAL TRIGONOMETRY
Examples. VII a.
1. Write down the characteristics of the logarithms of the
following numbers.
12-8, 161-4, -3279, '061, 1538,
2-749, -0006, 13864, -002, -87.
2. Given that log 4023 = 3 '6045, write down
log 4'023, log 402-3, log '4023, log '004023, log 40230.
3. Given that log 21 74 = 3*3373, write down the numbers
whose logarithms are
1-3373, 2-3373, -3373, 4'3373, 3'3373, 2*3373, T'3373.
4. Given log 2 = '3010 and log 3 = '4771, find the logarithms
of: 4, 5, 6, 8, 9, 12, 15, 16, 18, 20.
Also since approximately 74=2400, 112 = 120, 192=360, find
roughly log 7, log 11, log 14, log 19.
Taking difference of logs proportional to small difference
in the numbers, find log 13 since log 130 lies between log 128
and log 132.
Now since 17 x 10=169 (approximately), find log 17.
49. To obtain the logarithm of any number we write
down the characteristic by rule, and obtain the mantissa
from the tables as follows.
For purposes of explanation we give the following extract
from Bottomley's Four Figure Tables :
LOGARITHMS.
57
0
1
2
3
4
5
6
7
8
9
123
456
789
7559
7566
7574
7582
7589
7597
7604.
7612
7619
7627
122
345
567
From this portion of a page we read that the mantissa
corresponding to 574 is *7589 (note that the decimal point
is not printed in the tables), and so we have
log 574 =2*7589,
log 57400 = 4-7589,
log '0574 =2-7589.
LOGARITHMS
91
If we require the mantissa corresponding to 4 digits, we
must add on the difference obtained from the right hand of
the page.
Thus mantissa for 574 is "7589,
diff. for 6 is 5;
.'. mantissa for 5746 is *7594.
After a little practice the student will have no difficulty
in adding the difference mentally.
50. The reverse operation, to find the digits corre-
sponding to a given mantissa, can be easily performed with
the same tables; but more quickly with tables of anti-
logarithms, as shown below.
ANTILOGARITHMS.
0
1
2
3
4
5
6
7
8
9
123
456
789
75
5623
5636
5649
5662
5675
5689
5702
5715
5728
5741
134
578
91012
Example.
Find x, being given log x— 2*7594.
From the extract of the tables given above, we have
*759 is the mantissa for 5741
4 is the difference for 5 ;
. \ *7594 is the mantissa for 5746.
Since the characteristic is 2, we must have 3 digits in the
integral part.
.-. #=574-6.
Examples. VII b.
1. Write down the logarithms of
473, 4-735, -2864, 456000, 87'67, -003724.
2. Write down the numbers whose logarithms are
•4726, 3-7458, T'8642, 4*2175, 3*6847.
92 PRACTICAL TRIGONOMETRY
51. Examples to illustrate the use of logarithms.
Example (i).
Find the value of
3-562 x -06875
(7843)2
If we denote the fraction by #, we have
log #=log 3-562 + log -06875 - 2 log *7843,
log 3-562= -5516,
log -06875 = 2-8373
1-3889 (by addition),
2 log -7843 = 2 x 1 '8945 =1-7890 (since -2 + 1 '7890 = 1 '7890),
log #=1-5999 (by subtraction) ;
.*. x= -398(0) (from antilogarithm tables).
KB. (i) For addition and subtraction arrange logarithms
in columns.
(ii) The result is only correct to 3 significant figures but
the fourth figure gives an approximation to the correct value
which is '3981 to four significant figures.
Example (ii).
Evaluate ^-0276.
Let ^=^-0276,
then log x = i log '0276
= J of 2-4409
=i of ( -3 + 1-4409) (see note)
= _ i + -4803
= 1-4803;
.-. #= -3022.
Note. Since the negative characteristic is not exactly divisible
by the divisor 3, it is increased iintil it is a multiple of the
divisor, proper correction being made.
LOGARITHMS 93
Example (iii).
Find the reciprocal of 275*4.
Let ^ = ^4 = (275*4)"1'
Then log#= -Iog275'4
= - 2-4399 (both integral and decimal part being
negative)
=» - 3 + 1 - '4399= - 3 + (l - '4399)
== 3-5601 (making the mantissa positive) ;
.-. #='003632.
Example (iv).
Solve 575 x (1-03)*= 847.
We have, by taking logarithms of both sides.
log 575 + x log 1 -03 = log 847 ;
log 847 -log 575
logl-03
•1682
2-9279
2-7597
•1682
128)1682(13
402
18
•0128
= 13'(1).
Note. We cannot obtain x to a greater degree of accuracy
without using tables which give more than 4 figures.
The above equation gives the number of years in which
£575 would amount to £847 at 3% compound interest.
For the interest on £1 for 1 year=£*03;
.'. in 1 year £1 amounts to £1'03.
During the second year each £1 in this amounts to £1*03 ;
1 'O'?
.-. £1-03 amounts to — -x £1'03 = £(1'03)2, and so on.
.*. after x years £1 amounts to £(l'03)x and £575 to
£575 x (1-03)*.
94
PRACTICAL TRIGONOMETRY
52. Change of base.
If the logarithms of numbers to any base are known it
is easy to obtain the logarithms to any other base.
Suppose logarithms to any base a are known and we
wish to obtain the logarithm of any number n to the
.'. \ogan = logabx
Let
then
Hence to transform logarithms calculated to base a to
logarithms calculated to base b, we only have to multiply
. logo?
. • u/ — -I
This multiplier is commonly called the modulus.
Examples. VII c.
Evaluate to three significant figures. State the fourth
significant figure obtained although it cannot be relied upon
as correct.
1. 23-61 X -0324x1 -384.
23-68
2-174*
-0264x123-6x18-41
•00326 x 106-4
1
23-68*
1 -274 x -0623 x -001
•0362
•004671 "
r 21-63 x\/12-18
"361-8
1
"xMTsr
2-7 18 x -000526
9. 4/2174. 10. (31-76)1
12. 1 13. /«.
v/6'783 V 5'8
15. 483x('04172)5. 16. ^'0176.
17.
LOGARITHMS 95
18. (-00268)§x ('0246)1 19. ('01001)1
20. Find the number of digits in 917.
21. Find the number of ciphers before the first significant
digit in (ft)io. _
22. Obtain the square root of - -- - .
'UUUol
23. Solve (A)-- A-
24. Find approximately the amount of £317 in 10 years
at 3% compound interest.
25. If the population of a town increases at the rate of 4 %
each year, in how many years will the population be doubled ?
26. Find the mean proportional between 14*76 and 35-82.
27. Calculate the surface and volume of a sphere of radius
13'27 ft., given that the surface is 47rr2, and volume is f Trr3.
(TT = 3-142.)
28. Evaluate \/28'65 x 14-35 x 11*05 x 3'25.
29. Find the product of 4-177, 0-04177, 0*0004177, 4177000,
and find the square root of (0'07346):{.
30. Knowing the number of pounds in a cubic inch of a
substance, you can find the number of kilograms in cubic cm.
by multiplying by 0-4536 x (2-54) ~3. Express this multiplier
as a decimal to 3 places.
If steel weighs 488 Ibs. per cubic foot, how many kilograms
per cubic centimetre does it weigh?
31. Without using the tables find the characteristics of
(1) Iog7 15914, (2) Iog8 0-00187.
32. Obtain the value of
327*4 x
33. Calculate, as accurately as your tables permit, the value
of the fraction
1234 x (2345)2 x(345-l)3
^45-12x^5-123
34. Solve asx-i^2**!.
35. Find the values of Iog12 432, logao "2164.
96 PRACTICAL TRIGONOMETRY
36. The time of oscillation of a pendulum in sees, is given by
find this if 7r = 3'142, Z=126'2 cms., # = 981.
37. The reduction factor of a galvanometer is given by
t=*L.
2ir*'
find k when ?r = 3'142, r= 16'2 cms., H = -18, ?i=5.
38. Find the critical temperature of a gas given by the
formula
T = J-^, when R = l-^j^, a=-00874, 6=-0023.
A i r\0 2t i «3
39. Calculate the velocity of sound in cms. per sec. from
the formula
'=>/?•
. when 7 = 1-40, p = 76x13-6x981, p = -001293.
40. Find the temperature of a gas expanding adiabatically
according to the formula T = 273x2'y~1, where y=l'40.
41. Find the wave-length of sodium light from the formula
. — , AJ. M,— ^wic, ^01., #='089 cms., D = 358 cms.
42. Calculate (ri) the modulus of torsion in a wire, given
tt= — ^ — , where I — 144*1 cms., # = 4'10 sees., a = *0625cms.,
_ 6079 x (4-325)2
2
43. Find M, the viscosity of water, given M = ~o~fw" > when
P=39'25x 981, R2=-00788, #=47 sees., L = 23 -3 cms., V = 102'5c.c.
44. Find the ratio of L to I2, given r = A — ^> where
I2 t2 — *
#! = 3-81 sees., #2= 5'19 secs-> ^=3-26 sees.
45. Find C, the capacity of a condenser from the formulae
C=-, Q=* ' (l + -p), given that D = 1'3, logK = 7'3432,
T= 6'333 sees., X = -425, E=1'08.
46. Evaluate Y= ™^ 3 (Young's modulus), when
w = 20 grams, £=38'2, # = '32 cms., ^=981, 6 = 1'287 cms.,
^=•00656 cms.
LOGARITHMS 97
53. Logarithms of Trigonometrical Functions.
The logarithms of the trigonometrical functions of acute
angles are to be obtained from tables. As the character-
istic cannot be seen by inspection it is printed as well as
the mantissa. Also, to save confusion with regard to the
sign of the characteristic, the number 1 0 is added in each
case. The result is called the Tabular logarithm. In
order to obtain the logarithm we mentally subtract 10 as
we read the tables.
Thus in the table of Logarithmic Sines, we have the
tabular logarithm of sine 68° 18' is 9 '9681, and hence
log sin 68° 18' -1*9681.
Note that the characteristic is printed once only, at the
beginning of each line.
The same rules concerning the subtraction of differences
for cosines, cotangent, cosecant hold good as in the tables
of natural functions.
Examples. VII d.
1. "Write down from the tables : —
log sin 56° 40', log tan 27° 13', log sec 56° 47', log cos 43° 26',
log cot 19° 44', log sin 123° 15'.
2. Find 6 in each of the following cases : —
(1) log sin 0=T'4762. (2) logcos0=T'6254.
(3) log tan 0= *5843. (4) log sec 0= -8765.
(5) logtan0=T'5843.
3. Find the values of
(1) sin 43° 12' x cos 28° 17'. (2) sin 130° 15' x cos 120° 3'.
tan27°_ll'
w cosec56°23''
15-4 sin 47° 13' A , . ,
4. If smA= Z-Q-S , find two values of A less
lo*7
than 180°.
P. F. 7
98 PRACTICAL TRIGONOMETRY
/SS'GSxU^
5. If cos *=v/-_-5S-rf find A
356 sin 37° 16' .
a GlVena= Bin 63° 27- >find*
. - sin 25° cos 37°
7. Obtain the value of --^-0 — .
tan 130
8. The area of a A being \ab sin C, find the area where
a=798ft., 6 = 460 ft., C = 55° 2'.
9. If tan 6 = fflfo cot 28° 54', find 6.
10. In a A, sin E
<j=59-21 ft., 0=27° 22'.
11. Find the value of , the coefficient of diurnal
aberration where a — radius of earth = 3960 mis., V = velocity of
light =186,000 mis. per sec., ^observer's latitude = 51° 7'.
12. The electric current in a wire is given by C= — — - ;
2iTT
find its value when H = *18, r= 16*01 cms., tan 2^> = '1723,
TT- 3-142.
13. The refractive index for glass is given by
10. In a A, sinB= — — ; find B when &= 127*3 ft.,
c
Find fj. when S=43° 51', (9 = 64° 54'.
14. The coefficient of mutual induction being given by
M = ~, where Q=^x3'6xlO~9 and C = Ktana, find M
L» t
when R = 400ohms, Z = 4'5 sees., K==1'3, 5=11°, 7r=3'142.
15. The strength of a magnetic field is given by the formula
H = nn \/:57~2 • Evaluate H when n = 3*142, n = -42, < = 274*6,
r=26, <9 = 59° 7'.
LOGARITHMS 99
16. In solving a triangle it was necessary to use the
A
2 v be cos --
o
formulae cos<£ = 7 , a=(b+c) sin <£. Find a when
o ~f~ c
6=13-2 cms., c=15-6cms., A =48° 28'.
17. Given that the force required to prevent a body slipping
down a rough inclined plane of angle a is — ^ , where X
cosX
is the angle of friction. Find this force if W = 52 grams weight,
a = 32° 14', X = 15° 20'.
2 *Jbc sin —
18. If a—(b — c*)sec</> where tan<£=— T , find a
when 6= 11-64 cms., c= 9'38 cms., A = 52° 14'.
19. Find H from the formula H tan 6 = !!L_. } where
10 (a2 +^8)*
rc=25, a=13'97cms., 0=20°, C = '62 amperes, #=36'1 cms.
20. In a conical pendulum the angle the string makes with
the vertical is given by cos#= ~-$— ^\ find 6 if <7=32, n — -8.
4% 7T &
7r = 3-142, £ = 11-86.
21. The number of minutes in the angle of deviation of the
plumb line due to the earth's rotation being
180 x 60 rfasm X cos X
" TT ~ g
find the angle if «=oZ-|^j^, «=4000x 1760x 3, g=
X = 52° 4'.
7—2
100 PRACTICAL TRIGONOMETRY
Miscellaneous Examples. E.
1. Find the angle of elevation of the sun when the shadow
cast by a tower 200 ft. high is 12 J ft. less than it was when the
elevation of the sun was 27°.
^
2. Given cos A= *34, find the value of tan ^ and explain the
2i
double answer.
3. If you had no book of tables and had to find out whether
the following were approximately correct, state how you would
do so, giving your working and reasoning :
(i) log 3 = '5, (ii) the no. whose log is - \ is '56,
(iii) log -12 = 2 log -35.
4. Find four angles between 0° and 360° which satisfy the
equation
5. Two sides of a triangle are 13'6 cms. and 15-4 cms. and
the included angle is 46°. What would be the increase in area
if each of the two sides was lengthened by 0'3 cm. ?
6. I have two tables containing the logarithms of all
numbers and the tabular logarithms of sines of all angles
from 0° to 90° but I have no tabular logarithms of cosines or
tangents. I want to find the tabular logarithm of the cosine
and tangent of a certain angle, say 34° 27'. How am I to
do so?
7. Evaluate .--^2-— ^ , where 7r=3-142, K=074, ^=
log^-logV
= 1-25, r2=l'55.
8. Two adjacent sides AB, AD of a parallelogram are 4"
and 5" respectively. The diagonal AC is 7". Calculate the
angle BAD.
LOGARITHMS 101
9. The line OC joining a point O on the circumference of a
circle of radius a to the centre C, makes with OX, any line
through O, an angle a. If r be the distance of any other point
P on the circumference from O and 6 the angle OP makes with
OX, prove r— 2a cos (6 ~ a).
10. A regular pentagon is inscribed within a circle of
radius r ; show that its perimeter is lOr sin 36° and its area
5r2 sin 36° cos 36°, and find its perimeter and area as nearly as
the tables allow when r=5".
11. One side of a right-angled triangle is 6*432 ft. long and
the angle opposite to it is 37° 27'. Find (i) the area of the
triangle, (ii) the length of the perpendicular from the right angle
on the hypotenuse.
12. If a body is projected up an inclined plane of angle /3,
with a velocity V ft. per sec. making an angle a with the horizon,
. 2V2 cos a sin (a -8) „. , ,.
its range is - ------ -y-^ - — . Find the range when V=56*4,
a = 64° Iff, /3 = 28° 16', # = 32-2.
13. The angle between two tangents of length a, from an
external point to a circle of radius r, is 6. Prove by projection
a-cos rcos TJ. 71_ ,, ,.,
that r= -1-.— a — -, a= — — r— ; — '- . If d be the distance from
sin 6 sin 6
the external point to the centre of the circle, prove
B
•=.
2t
14. Find to the nearest tenth the positive value of x which
satisfies ^— =tan 12°.
1 ~~ oc
15. A ray of light after reflexion at a plane mirror makes
with the perpendicular to the mirror at the point of incidence
an angle equal to the angle it makes with this perpendicular at
incidence. Prove that if the mirror is turned through an angle
a the reflected ray will be turned through an angle 2a.
102 PRACTICAL TRIGONOMETRY
16. XB is the projection of AB on MN, the angle AXB being
a right angle. Find the length of XB when AB = 5 inches and
the angle ABX (a) is equal to 33°. If AB and BC are the sides of
a square, and XB, BY their projections on MN, how must the
square be placed for XY to have (i) the least, (ii) the greatest
possible length, consistently with the conditions that B is always
to be on M N and the square is to be above M N and in the same
plane with it ?
Fig. 55.
17. If a and /3 are two different angles which satisfy the
equation 3 + 2 tan x— sec #, prove that tan(a+/3) = :1^.
18. An error of 1*5% excess is made in measuring the side
a of a triangle and of 1'8 % defect in measuring b. What is the
resulting percentage error in the area as calculated from the
formula i
19. Find in acres the area of a triangular field, two of
whose sides measure 576 and 430 yards, and meet at an angle
of 54°.
20. A chord AB of a circle cuts a diameter CD at right
angles at O. A line OE at right angles to the plane of the circle
subtends at the points C, B, D angles of 6, a, <p respectively.
Prove cot <f> = cot2 a . tan 6.
CHAPTER VIII.
THE SOLUTION OF TRIANGLES.
£2 _j_ 02 _ g2
54. The formula cos A = — —r --- , proved in Art. 32,
2 be
is not suitable for logarithmic work, but we can obtain from,
it formulae that are.
Thus we have
+ e2 - a?
Now let a + b + c = 25,
then b + c - a = 2s - 2a = 2 (s - a) •
j 2^ . 2 (s — a)
and . . 1 4- cos A = -- - - J- •
s —
.'. COS - =
104 PRACTICAL TRIGONOMETRY
Similarly it can be shown that
, . 2A 2(*-6)(s-c)
1 — cos A = 2 sin2 -r = ~ —^ -'- :
2 be
(s - b) (s - c)
Prom (1) and (2) we have
Any one of these three formulae can be conveniently
used for finding the angles of a triangle when the sides are
given.
Example.
Find the angles of the triangle if a = 243'4, 6 = 147 '6, c= 185*2.
a =243-4
6 = 147-6
c= 185-2
2)576-2
5=288-1
s-a= 44-7
5-6 = 140-5
s - c = 102-9
[A convenient test of accuracy (s — a) + (s — b) + (s — c) = s.]
A_ /140-5x 102-9
In2~ V 288-1x44-7 ;
.-. logtan £ =J{log 140-5+ log 102-9 -log 288-1 -log 44-7}
log 140-5 = 2-1476
= •0250;
^=46° 39';
log 102-9 =2-0123
4-1599
log 288-1 =^4596
A=93°18'. log 44-7 = 1-6503
2) -0500
•0250
THE SOLUTION OF TRIANGLES
105
Also
44-7 x 102-9
288*1 x 140-5 '
log tan - = 1-5277;
/. |= 18° 38';
m
/. B= 37° 16';
/. A+B =130° 34';
.-. C= 49° 26'.
log 44-7 = 1-6503
log 102 -9 = 2-0123
3-6626
log 288-1 = 2T4596
log 140-5 = 2-1476
2)1-0554
T-5277
Note. We use the formula for the tangent here because we
then only require to obtain four logarithms from the tables, viz.
log 5, log (s - a), log (s - b\ log (s - c).
Q
To test accuracy we can find — by the same method.
2i
55. To solve a triangle when, two sides and the
included angle are given.
Let a, b9 C be the given parts.
We have
sin A _ a
sin B ~ b ;
sin A - sin B a — b
sin A + sin B
. A-B A+B
2 sin —7;^ — cos — - —
rt . A+B A-B
2 sin —^-- cos
a — b
. ,
. . tan
2 2
A-B a-l
+
tan
A + B
/. tan
since
2 a + b
A- B_q -6
2
A + B
106 PRACTICAL TRIGONOMETRY
The above formula is suitable for logarithmic work,
and from it we obtain the value of — - — .
And hence, since is known, we get the values of
A and B. The side c can then be found, since
_ a sin C
sin A
Example (i).
Given 6=253, c=189, A=72° 14', solve the triangle.
First method.
™- , B-C6-C.A
We nave tan — = — = , — cot —
= ?% cot 36° 7';
/. log tan ^-^- = log 64 - log 442 + log cot 36° 7'
= 1-2976;
log 64 = 1-8062
B-C no13,. log 442 = 2-6454
1-1608
log cot 36° 7'= -1368
we have — ~- =53° 53'. T2976
By addition B=65° 6'.
By subtraction C = 42° 40'.
csinA 189 sin 72°
Also
sin42°4(X
.-. Ioga = logl89 + logsin72° 14' - log sin 42° 40'
log 189=2-2765
=2-4242- Iogsin72°14'=r9788
2*2553
.-. a=265-6. loggin 42o 40/^1.831!
2-4242
Second method.
The following method does not involve the use of any special
formula, and may sometimes be of use, but the results are likely
to be less accurate than those obtained by the first method.
THE SOLUTION OF TRIANGLES
107
Let BD be perpendicular to AC.
A. D
Fig. 56.
Then AD = 189 cos 72° 14';
.-. log AD = 1-7610;
/. AD = 57-68;
.-. CD = 195-32.
Also BD = 189 sin 72° 14' ;
.'. logBD = 2'2553;
.'. BD = 180-0.
r. BD
tanC= — =
180
log 189 = 2-2765
log cos 72° 14' =1-4845
1-7610
log 189 =2-2765
log sin 72° 14' =1-9788
2;2553
" 195-3 '
log tan C = 1-9646;
.-. C = 42° 40'.
The rest of the solution is the same as in the first method.
log 180 =2-2553
log 195-3 = 2-2907
1-9646
Example (ii).
Given a =324, 5=287, B=34° 17', solve the triangle.
a sinB 324 sin 34° 17' .
b 287 ;
We have sinA=-
or
Since
-. log sin A=log 324 - log 287 + log sin 34° 17'
= T-8033;
A = 39° 28';
140° 32'.
both values of A
are possible, and we have an
ambiguous case. [Art. 35.]
log 324 = 2-5105
log 287 = 2-4579
"•0526
log sin 34° 17' =1-7507
V8033
108
PRACTICAL TRIGONOMETRY
(1) When
and
A= 39° 28';
A-hB= 73° 45';
.-. C = 106° 15';
287 sin 106° 15' 287 sin 73° 45'
sin B
.-. logc=2-6895;
(2) When
sin 34° 17' sin 34° 17'
I log 287 = 2-4579
log sin 34° 17' = 17507
¥7072
log sin 73° 45' =1-9823
2-6895
A = 140° 32';
= 174° 49';
C= 5° 11';
and c =
287 sin 5° 11'
sin 34° 17'
.-. log c= 1-6626;
.-. c=45-98.
log 287 -log sin 34° 17' = 27072
log sin 5° 11' = 2-9554
1-6626
Examples. VIII a.
Solve the following triangles :
1. a = 56-4, 5=75-7, c= 107*5.
2. A = 37° 14', B = 65° 15', c=83.
3. B = 75° 27', C = 43° 12', 6 = 27'8.
4. a = 264, 6 = 435, C = 81° 25'.
5. 6 = 14-76, c= 28-47, C = 46°30'.
6. a = 28, c=33, A =36° 24'.
7. A = 107°, a =456, 6=312.
8. a =345-2, 6=281*7, c=261'5.
9. B = 41° 15', A=103° 7', c=3«47.
10. B = 122°, a = 43-56, c = 51«45.
11. A = 57° 14', B=83°35', 6 = 3147.
12. In a triangle ABC, a=35, 6=43 and C = 75° 11', find the
angles A and B.
THE SOLUTION OF TRIANGLES 109
13. Given A=42°, a=141, 5=172-5, find all solutions of the
triangle ABC.
14. If a=447, c = 341, C = 37° 22', find the two values of B ;
and draw a figure showing the two triangles obtained.
15. A, B are two points on one bank of a straight river,
distant from one another 649 yards ; C is on the other bank, and
the angles CAB, CBA are respectively 48° 31' and 75° 25'; find
the width of the river.
16. The angles A, B of a triangle are respectively 40° 30' and
45° 45', and the intervening side is 6 feet ; find the smaller of the
remaining sides.
17. Find the greatest angle of the triangle whose sides are
184, 425 and 541.
18. In a triangle ABC the angles B and C are found to be
49° 30' and 70° 30' respectively, and the side a is found to be
4 '375 inches. Find A, b and c as accurately as the tables
permit.
19. If a =1000 inches, b =353 inches, B=20° 35', find the
angles A and C, taking A to be obtuse.
20. From Bristol to Richmond is 99 miles. From Richmond
to Nottingham is 112 miles. From Nottingham to Bristol is
122 miles. If Richmond is due E. of Bristol, find the bearing of
Nottingham from Bristol to the nearest degree.
21. A man walking along a road due E. sees a fort 4 miles
away in a direction E. 32° N. If the guns have a range of
3 miles, how far must he go before he is (i) within range, (ii) out
of range again ?
22. OABC is a quadrilateral in which OA = 12'5 ft., OC = 11 ft.,
L AO B == 27° 40', L BOC = 35° 25'. Find the angle O AC, and hence
the distance of the intersection of the diagonals from O.
23. A rock slope is inclined at 40° to a horizontal plane.
A man stands 30 yards from the foot of the slope, on the
horizontal plane through it, and notices that the slope subtends
20° at his eye. If his eye is 5 ft. above the horizontal plane, find
the length of the slope.
110 PRACTICAL TRIGONOMETRY
56. Frequently by the use of a subsidiary angle ex-
pressions may be thrown into a form suitable for logarithmic
work.
Thus
a sin 0 + b cos 0 =•= a ( sin 0 + - cos 0 }
\ a J
= a (sin 0 + tan a cos 0), where tan a = -
- (sin 0 cos a + cos 0 sin a)
COS a
= a sin (0 + a) sec a.
Here, by the use of the subsidiary angle a, we have
thrown the expression a sin 0 + b cos 0 into a form suitable
for logarithmic work.
57. Again the formula c2 = a2 + b'2— 2ab cos C can be
put in various forms suitable for logarithmic work with the
help of subsidiary angles ; so that when two sides and the
included angle of a triangle are given the third side can be
found without first finding the other angles.
We proceed to give an example of this.
We have
<? = a? + b2 — 2ab cos C
= a2 + b2 — 2ab ( 2 cos2 - -
>»\
= (a, -f &)2 — 4ab cos2 -
Now since 4a6 < (a + b)2 and cos — < 1, we can find an
acute angle 0 such that
. , 2 Jab C
sin 0 = r cos - .
Ot + b 2
THE SOLUTION OF TRIANGLES
We then have
111
.'. c = (a + fy cos 0.
Example.
The sides of a triangle are 237 and 158, and the contained
angle is 58° 40'. Find the value of the base, without previously
determining the other angles.
If a =237, 6 = 158, C = 58°40/,
we have c = (a + b) cos 0,
2 slab C 2\/237>Tl58
where ism 6 = T cos — = — — cos 29 20 .
a + b 2 395
To find 0, we have
log sin 6 = log 2 + £ (log 237 + log 158) - log 395 + log cos 29° 20',
log 237 = 2-3747
log 158 = 2-1987
2)4-5734
2-2867
-3010
= 1-9316;
0 = 58° 41'.
c= 395 cos 58° 41';
c = 2-3124;
0=205-3.
log 2 =
log cos 29° 20' =1-9405
2T5282
log 395 = 2-5966
^9316
log 395 = 2-5966
log cos 58° 41' =1-7158
2-3124
112 PRACTICAL TRIGONOMETRY
Examples. VIII b.
1. Show that Va2 -f 62 can be thrown into the form a sec 0,
where ^=tan~1-.
a
Give a geometrical interpretation to this by supposing a, 6
to be sides of a right-angled triangle.
5 sin 0 + 3-584 cos 6 .
2. Throw the expression — — — ^ — ----- — ^ into a form
5 sin (9 -3-584 cos Q
suitable to logarithmic calculation when different values of B
are introduced, and use your form to evaluate the expression
when (9 = 71° 59'.
3. In any triangle if tan <f) = — —7- cot — , prove that
f\
c—(a + b) sin — sec </>.
2
Hence find c if a = 423, 5 = 387, C = 46°.
4. Prove the formula
Apply it to find the side a of a triangle when 5 = 132'5feet,
c= 97 *32 feet, A = 37° 46', as accurately as the tables permit.
5. If ABC be a triangle, and B such an angle that
C
find c in terms of a, 6 and 6.
If a = 11, 6 = 25 and C = 106°16', find c.
58. The area of a triangle in terms of the sides.
In Article 15 it was shown that A = \bc sin A.
Hence we have
A = - be . 2 sin - cos -
2 22
= *Js (s - a) (s — b)(s- c).
THE SOLUTION OF TRIANGLES
59. Radius of circumscribed circle.
From Article 15 Ex. (iii) we have
abc
whence
' sin A sin B sin C '
abc
' 2bc sin A
abc
113
60. Radius of inscribed circle.
D
Fig. 57.
Let I be the centre of the inscribed circle of the triangle
ABC, and D, E, F the points of contact with the sides. Let
r be the radius. Then
AABC=ABIC
.'. A = \ra + \rl) + \
= \r (a + b + c)
__A_
P. F.
114 PRACTICAL TRIGONOMETRY
61. Radii of escribed circles.
Let E be the centre of the escribed circle which touches
BC and the other two sides produced.
Let P, Q, R be the points of contact, and r± the radius.
Then
Fig. 58.
AABC= AEAC+AEAB-AEBC;
.*. A = ^rj> + \r& — \r^a
= ^Ti(b + c- a)
s — a
Similarly the radii of the other escribed circles are
*-b>
THE SOLUTION OF TRIANGLES 115
62. There are many forms in which the radius of the
inscribed circle may be expressed. Another form which is
sometimes convenient can be obtained as follows.
Since tangents drawn from a point to a circle are equal,
we have (Fig. 57)
BD = BF, CE=CD, AF = AE;
= half the perimeter of the triangle
Similarly CD =s-c, and AE = s-&.
Hence we have
B B
r — BD tan - = (s — H] tan -
J J
C A
= (s - c) tan - = (s - a) tan - similarly.
2 2i
By combining this formula with r = — prove the for-
s
^
mulae expressing tan- etc. in terms of the sides of the
2
triangle.
63. We can also obtain i\ as follows, since (Fig. 58)
BR=BP,
CQ=CP,
and AR = AQ;
/. AR = i(AR + AQ) = J(AB + BP + AC + CP)
= ^(a + b + c)
A A
/. TI = AR tan - = s tan - .
B C
Similarly rz = s tan - , r% — s tan - .
Jt A.
8—2
116 PRACTICAL TRIGONOMETRY
Examples. VIII c.
1. Find correct to the tenth of a sq. inch the area of a
triangle whose sides are 2 -45, 3*17, 2*21 inches.
2. Find the radius of the inscribed circle of a triangle whose
sides are 27'6, 13'8, 20'5.
3. A circle is circumscribed about a triangle whose sides
are 17, 32, 43; find its radius.
4. A chord of a circle is 15 '7 cm. in length, and the angle
in one of the segments is 47°.; what is the radius of the circle ?
5. Find the radius of the largest circle which can be cut out
of a triangle whose sides are 423, 375, 216 ft. Also calculate the
area of the circle.
6. The lengths of the sides of a triangle are 375 links,
452 links, and 547 links. Find the length of the perpendicular
upon the shortest side from the opposite corner, and the radius
of the inscribed circle.
7. If the sides of a triangle are 17, 23, 30 inches in length,
in what ratios do the points of contact of the inscribed circle
divide them?
8. Prove that in an equilateral triangle the radii of the
inscribed, circumscribed and escribed circles are as 1 : 2 : 3.
9. The sides of a triangle are 17, 25, 36; show that the
radii of the escribed circles are as 21 : 33 : 154
10. Prove that the radii of the inscribed and escribed circles
can be expressed as
, C . A C A
b sin — sin — b cos — cos —
, and respectively.
B B
COS2 ^2
11. Express the area of a triangle in terms of one side and
the angles.
12. Prove that the distances between the centre of the
inscribed circle and the centres of the escribed circles are
ABC
, t>sec-, csec-.
THE SOLUTION OF TRIANGLES 117
Miscellaneous Examples. P.
1. The distances of a point P from two other points Q and R
are wanted and cannot be directly measured. The distance
between Q and R is found to be 1370 yds. PQR = 33°40',
PRQ=96°25'. Find the distances of P from Q and R, both by
calculation and drawing.
2. If in the triangle ABC, C = 90°, prove
, A b+c
cot - = .
2 a
AC 73
3. Calculate Young's Modulus from the formula Y= ' ,
bk3x '
where F=500x981, £=70, 6 = 2'22, A=1'28, #=2.
4. Two adjacent sides of a parallelogram are 6" and 5".
Find the angle between them if the diagonal passing through
their point of intersection is 9".
5. Given that the diagonals of any quadrilateral are of
length #, and y, and intersect at an angle #, prove that the area
of the figure is \xy sin 6.
6. The corner-post C of a property was fixed as being
87*6 chains from a tree and in the direction S. 56° 50' E. This
post having now been moved to a point C' 25 chains due N. of C,
the distance and direction of C' from the tree must be determined.
Find them by calculation.
7. A point P lies 3 miles from a point O in a direction
31° north of East; another point Q lies 5J miles from O in a
direction E. 57° N. Calculate the distance between P and Q
to the nearest tenth of a mile.
8. An isosceles triangle of vertical angle a is suspended
by a string tied to its vertex and to an extremity of the base
and rests so that the lower of the equal sides is horizontal.
The angle made with the vertical by each portion of the string
h cos n
is 6 and I is the length of the string, prove 1= , where
h is the altitude of the triangle. 2
118 PRACTICAL TRIGONOMETRY
9. Find 6 from the formula cos 6 — — J—-. . where <7 = 32,
4tt27T26
ra=-84, TT = 3'142, Z = 11'8.
10. Find the radius of a sphere of volume 320 c.c., given
that volume=|7rr3. (*-»3*142.)
11. In a triangle ABC, BC = 93 yards, ABC = 59°19',
AC B = 43° 15'. Calculate the length of AB.
Also find what error is made in the length of AB if the angle
ACB is through a wrong measurement taken as 43° IT.
12. Find the number of years in which £320 will amount
to £450 at 4°/0 Compound Interest.
13. A person on a cliff observes that the angles of depression
of the light of a lightship 500 yds. away and its image by reflexion
in water (which is the same distance vertically below the surface
as the light is above) are Dl and D2, prove that the height of the
cliff is 250 (tan DI+ tan D2) yards.
14. In a triangle ABC, a = 25", 6 = 30", and B = 2A, find the
angles of the triangle and the third side.
15. Solve the equation 2*2= 16*-1.
Find the number of digits in 1933.
Find the number of zeros following the decimal point in the
value of (TV)*3.
16. P and Q are two forts on the same side of a straight
entrenchment. A base line XY of 1000 yards is measured along
the entrenchment and the following angles are observed : —
YXP = 95°, XYP = 43°, XYQ=105°, QXY = 27°.
Find the distance between the forts and check your result
by drawing a plan to a scale of 6" to a mile.
You mav find useful the formula tan = ^ — • cot — or
2 b+c 2
the formula a=(b+c) cos <£, where <f> is given by
(b + c) sin 0 = 2 \/6c cos — .
THE SOLUTION OF TRIANGLES 119
IV. An obtuse angled triangle has a = 15*3 cms., 6 = 97 cms.,
and B = 31°45'. Calculate the remaining angles and draw the
triangle accurately.
18. ABCD is a rectangular piece of paper having AB = 14",
BC = 10". The paper is folded so that the corner C lies on AB
and the crease makes 26° with the original position of the side
CD. Calculate the length of the crease.
19. Prove that in any triangle
B± C s-a
20. Find the volume of a regular tetrahedron (a pyramid,
each face being an equilateral triangle) whose edge is 12" long.
Given, vol. of pyramid = J area of basex altitude.
21. A rod AB, 3 feet long, is suspended by a string fastened
to its two ends, which passes over a pulley at O, so that both
portions of the string, OA and OB, make an angle of 20° with the
vertical. If AB is inclined at 15° to the horizontal find the
length of the string.
22. Express cos 6+ sin 6 as the product of two cosines and
hence find for what positive values of 0, less than 90°, the
expression is (i) a maximum, (ii) a minimum.
23. If an error of 2 °/0 excess is made in measuring the sides
a and b of a triangle, find the percentage error in the area calcu-
lated from the formula \ab sin G.
24. When the sun is vertically overhead at the equator, an
upright pole, 10 feet high, casts a shadow of 12 feet at a certain
place. Find approximately the latitude of the place.
25. A is a point in the line XY. B and C are two points on
the same side of XY. AB = 4", AC = 6", YAB = 40°, BAC = 60°.
Calculate BC and find, by projecting on XY, the angle it makes
with XY.
CHAPTER IX.
RADIAN OR CIRCULAR MEASURE OF ANGLES.
64. IT may be either proved theoretically or verified
by actual measurements that the circumference of a circle
bears a constant ratio to the diameter.
This constant ratio is represented by the Greek letter TT,
., . circumference
so that — -T. 7 = 7T,
diameter
or circumference of a circle = 2-n-R where R is the radius.
The value of TT has been calculated to some 707 decimal
places. For accurate results it may be taken as 3'14159
or 3*1416 ; but for rougher approximations -2T2"(= 3*143),
which is correct to two places, will be more useful :
ffl gives 3*14159.
In working examples TT is taken to be -2T2- or 3*142 or
3*14159 according to the degree of accuracy required, and
the answer must be given up as correct only to the number
of significant figures justified by the data.
65. In theoretical investigations angles are not
measured in degrees but in terms of a much more con-
venient unit called a Radian.
RADIAN OR CIRCULAR MEASURE OF ANGLES 121
A Radian is the angle subtended at the centre of a circle
by an ARC equal in length to the Radius.
It will be noticed that the angle subtended by a chord
equal to the radius is 60°, so that a radian will be slightly
less than 60°.
It will shortly be seen that the angle is of constant
magnitude and in no way varies with the dimensions of
the circle, otherwise of course it could not be used as a unit
of measurement.
66. To measure any angle in terms of a Radian.
Let AOP be the angle.
Fig. 60.
With centre O and any radius (r) draw a circle APB
and suppose the arc AB = r, AP =#. Then
/. AOB=: 1 radian.
Since angles at the centre of a circle are proportional to
the arcs on which they stand, we have
x
1 radian r '
.'. the number of radians in L AOP is - .
r
Hence if 0 be the number of radians in an angle which
is subtended at the centre of a circle of radius r by an arc
of length x, we have
122 PRACTICAL TRIGONOMETRY
67. If the angle at the centre of the circle is 180°, we
have
180° _ semicircumference
1 radian r
_irr
r
.'. 180° = 7r radians;
180°
. . 1 radian = -
7T
= 57° 17' 44" approximately,
and is therefore of constant magnitude.
It is important to remember that TT denotes a number,
namely, the ratio of the circumference of a circle to its
diameter, which is approximately 3*1416 ; but it is usual to
speak of "the angle ir? meaning an angle of ?r radians,
which is 180°.
Similarly "the angle -T" means an angle of - radians,
O O
which is 60°.
Example (i).
Express 20° 14' in radian measure.
We have 20° 14' = 20^°,
acwL
^--^TT radians
607 T
= -3532 radians.
Example (ii).
Assuming the earth to be a sphere of 4000 miles radius, find
the distance measured on the earth's surface between two places
on the same meridian whose latitudes are 55° 16' and 37° 40'.
RADIAN OR CIRCULAR MEASURE OF ANGLES 123
Let A, B represent the two places, and C the point where the
meridian through A and B meets the equator.
Fig. 61.
Then L AOC = latitude of A = 55° 16',
and L BOC = latitude of B = 37° 40' ;
« j|g «• radians;
. •. arc AB = — ~0*— x 4000 miles
loO
= 1230 miles approx.
Or thus, from first principles
_arc^B _17f°
2ir~x40bO "3605*
Examples. IX a.
1. Express in radian measure as a fraction of IT the angles
30°, 150°, 65°, 74° 35'.
2. Express in sexagesimal measure the angles whose radian
TT 2?r 5ir 5?r
measures are - , — , — , — .
3. Find, to 2 places of decimals, the radian measures of
72° 15', 47° 24', 134° 13'.
124 PRACTICAL TRIGONOMETRY
4. Express in sexagesimal measure, to the nearest minute,
the angles 1*24, *63 radians.
5. Find the length of the arc of a circle of 12 cm. radius,
which subtends an angle of 40° at the centre. Answer to the
nearest millimetre.
6. Find the number of radians in the angle subtended at
the centre of a circle of radius 5 ft. by an arc 3 inches long.
7. Express in radians the angle turned through by the
minute hand of a clock in 20 minutes.
8. An angle whose radian measure is *45 is subtended at
the centre of a circle by an arc 4 inches long ; find the radius
of the circle.
9. Find the number of degrees in the angle subtended at
the centre of a circle of 10 cms. diameter by an arc of length
4 cms.
10. Express in degrees and in radians the angle of a regular
figure of 8 sides.
11. The length of a degree of latitude on the earth's surface
being 69 J miles, find the radius of the earth.
12. A wheel makes 20 revolutions per second; how long
will it take to turn through 5 radians?
13. The circumference of a circle is found by measurement
to be 21*43 cms. with a possible error of 1 mm. ; find its radius
as accurately as this measurement justifies.
14. The distance between two places on the equator is
150 miles; find their difference in longitude. Take the radius
of the earth to be 4000 miles, correct to two significant figures.
15. The driving wheel of a locomotive engine 6 ft. in
diameter makes 3 revolutions in a second. Find approximately
the number of miles the train passes over in an hour.
16. By considering regular hexagons inscribed in, and
circumscribed about a circle, show that the ratio of the
circumference of a circle to its diameter lies between 3 : 1
and 2^3 : 1.
RADIAN OR CIRCULAR MEASURE OF ANGLES 125
17. Find the distance on the earth's surface between two
places on the same meridian whose latitudes are 23° N. and
14° S. respectively; assuming the earth to be a sphere of
4000 miles radius, correct to 2 significant figures.
18. Two circles whose centres are A and B and radii 1*8 in.
and 0*6 in. respectively are placed so as to touch one another
externally at C. A line is drawn to touch the first circle at P
and the second circle at Q. Calculate the lengths of the common
tangent PQ and of the arcs PC, CQ.
19. A band is stretched tightly round two wheels of radii
1 ft. and 4 ft. respectively whose centres are 10 ft. apart. Find
the total length of the band to the nearest inch.
68. Limiting values.
Let an arc BB' of a circle subtend an angle of 20 radians
at the centre O.
\B
Fig. 62.
Draw BT, B'T the tangents at B and B'. Join BB'
and OT.
We shall assume that chord BB' <arc BAB' < BT + TB'.
(Note. A rigid proof that arc BAB'< BT + TB' is diffi-
cult and is beyond the scope of this book.)
Hence we have
BC arc BA BT
OB < OB < OB ;
i.e. sin 0, 0, tan 0
are in ascending order of magnitude.
126 PRACTICAL TRIGONOMETRY
Dividing by sin 0, we have
0
1, -^— £, sec0
sin 0
are in ascending order of magnitude.
Now as 0 approaches the value zero, sec0 approaches
unity ; .'. since - — ^ lies between 1 and sec 0, we have that
A
the limiting value of — — ^ , when 0 = 0 is 1 .
sm0
Using the notation of Art. 27, we have
Lt J^ = l (1).
0=0 sin 0
Again, by dividing sin 0, 0, tan 0 by tan 0, we have
cos<?' SE3' !
in ascending order of magnitude.
And as 0 approaches zero, cos 0 approaches unity ;
.% Lt~ = l (2).
0=0 tan 0
From the results (1) and (2) we see that, if the angle is
small, we may use its radian measure in place of its sine or
tangent.
We may verify this by means of the tables.
Thus
radian measure of 3° = "0524,
sin 3° = -0523,
tan 3° = '0524.
For still smaller angles the degree of accuracy may be
estimated from the following extract from 7 -figure tables :
sin 10' = '0029089,
tan 10'= '0029089,
radian measure of 10' = '0029089 ;
sin 19'- '0055268,
tan 19' = '0055269,
radian measure of 19'= -0055269.
RADIAN OR CIRCULAR MEASURE OF ANGLES 127
69. To find the area of a circle.
Suppose a regular polygon ABC of n sides to be inscribed
in the O, and one of n sides A'B'C' to be described about
the O.
Pig. 63.
Then area of inscribed polygon
= n . ^OA . OB sin AOB
n a . 27T
= - r2 sin — .
2 n
Area of circumscribed polygon
= n. JA'B'.OA
= tt. AA'. OA
= n . AO tan- . OA
n
= nr* tan - .
n
The area of the circle lies between these values however
great the number of sides may be.
Now when n is made infinitely great
27T
T, n „ . ZTT T nr
Lt — r2 sin — = Lt —
n=°o £ fl n=°o &
sm-
n
~2^~
2?T T
. -- = LtTT/*2.
sin —
ft
IF
n
= 7T/-2, by Art. 68,
since
128 PRACTICAL TRIGONOMETRY
Also
tan -
Lt nr~ tan - = Lt nr^ . - . — —
Vt tl 1C
71 = 00 iv n=oo w
n
tan-
- LtTT^. - = 7rr>;
?l = oo ^
.'. area of circle = Trr2.
70. Area of a sector of a circle.
If a sector of a circle contain an angle of 0 radians at
the centre, since sectors are proportional to the angles they
contain, we have :
area of sector _ 0 radians t
area of circle ~ 2?r radians '
, 0 Or
. . area of sector = — . -n-r2 = — .
%7T 2
This result may be written \r(0r) = \rx where x is the
length of arc subtended by 0.
71. If a distant object subtend a small angle at the
point of observation, we can find a formula connecting the
radian measure of the angle, and the approximate length
and distance of the object.
Let I, d be respectively the approximate length and
distance of the object, and let 0 be the radian measure of
the object subtended. Then the relation between these
three quantities is
If we consider the length of the object as the length of
]l
RADIAN OR CIRCULAR MEASURE OF ANGLES 129
an arc of a circle of radius d, we have the above formula
at once from Article 66.
Tig. 65.
If we consider the length of the object as the base of
an isosceles triangle of which the altitude is d, we have
since 0 is small, Art. 68 ;
.'. l=dO.
Example.
Given that the sun subtends an angle of 32' at a point on the
earth's surface, and that the distance of the sun is 92 x 106 miles ;
find the sun's diameter.
327T
The radian measure of 32' =
.-. the diameter of the sun
60 x 180 '
327T
: 60 x 180
- -8563 x 106
.= 856000 miles.
x 92 x 106 miles approximately
log 32 = 1-5051
logTr = '4972
log 92 = 1-9638
3^661
log 60 = 17782
log 180 = 2-2553
F9326 = log -8563
Note. Since the distance of the sun is only correct to two
significant figures, we cannot rely on the above answer to
more than two figures. Hence the result should be given as
860,000 miles. Also it should be remembered that results
obtained by means of four figure tables cannot be expected to
be accurate to more than three figures.
p. P. 9
130
PRACTICAL TRIGONOMETRY
72. Dip of the Horizon.
Let ATB represent the earth, and O the position of an
observer; then if tangents be drawn from O to the earth's
surface they will touch the earth in a circle, called the
Visible Horizon.
If OH be the horizontal plane through O the angle HOT
is called the Dip of the Horizon.
Ex. Find the dip of the horizon from a point 200 feet above
sea-level, assuming the earth a sphere of radius 4000 miles.
From the figure L HOT = ,L TOO and OT2 = OA. OB, where B
is the other extremity of the diameter. If r be the radius of the
earth and h the length of O A in miles,
but since Ti is very small compared with r, A2 is so small that it
may be neglected ;
This is called the Distance of the Horizon.
Also since 6 is a very small angle,
6 radians = tan 6 = — — =
/. the number of minutes in 6 = \ I — x x 60
v r TT
2x200
4000x1760x3 3142
180 x 60 , . ._,
X -^r-^r- =14-96'.
RADIAN OR CIRCULAR MEASURE OF ANGLES 131
Examples. IX b.
1. Find the area of a circle of 10 inches radius.
2. Find the radius of a circle whose area is 426*24 sq. cms.
3. What is the area of a sector of a circle of radius 4 ft.
which is bounded by two radii inclined at an angle of 60° ?
Also find the area of the segment bounded by the chord joining
the extremities of these radii.
4. The mean angular diameter of the moon being 31' when
it is 240,000 miles away, find the diameter in miles.
5. If the sun is 93 x 106 miles distant, and subtends at the
earth an angle of '0093 radians, find its diameter.
6. Find the dip of the horizon from the top of a lighthouse
250ft. high.
7. What is the distance of the visible horizon from the top
of a cliff 300 ft. high?
8. Two lighthouses, each 200 ft. high, are so placed that the
light of each is just visible from the other ; what is the distance
between the lighthouses 1
A
9. From the formula cos 6 — 1 - 2 sin2 - , prove that if 6 be
2
01
an acute angle cos 6 lies between 1 and 1 — — .
10. Deduce from the above result that sin 6 lies between 6
andtf™.
2i
11. Taking sin 0=0 (in radians) for a small angle, find
sin 20' correct to three significant figures.
12. If d be very small, prove that approximately
sin (a + #) = sin a + #cosa,
cos (a + 0} = cos a — 6 sin a,
tan (a + 6) = tan a + 6 sec2 a.
13. Prove that approximately the height of an object in feet
, , distance in yards x elevation in degrees
is equal to * =^ — .
14. Taking the diameter of a halfpenny to be 1 inch, find at
what distance it will subtend 1° at the eye.
15. Find the perimeter and area of the crescent-shaped
figure bounded by the arcs of two equal circles of radius
5 inches whose centres are 4 inches apart.
9—2
132 PRACTICAL TRIGONOMETRY
Miscellaneous Examples. G.
1. 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?
2. A man standing beside one milestone 011 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.
3. A rod ABC of length 7 ft. is held vertically at a point C
on the side of a hill. From a point E at the foot the angle of
elevation of A, the top of the rod, is 8° 12' and of B a point on
the rod 3 ft. from the bottom the angle of elevation is 7° ] 8'.
Find the vertical height of C above E.
4. If D be the mid-point of BC in the triangle ABC, prove
that
cot CDA=^(cot B-cot C).
5. Two tangents are drawn to a circle of radius 4" from a
point 10" from its centre. Find the lengths of the two arcs
between the points of contact.
6. XAY is a straight line, AO a line 3 cms. long perpen-
dicular to XAY, P is a point in XA, and the angle OPA is
6 radians. With centre P and radius PO the circular arc OB
is drawn to the line XAY and the tangent OC to this arc meets
XAY in C. Suppose P to move continually away from A along
AX and show what values the angle #, the arc OB, the straight
,. __
line OC, — ^— , — — , approach as P moves away.
o 6
Express 5° in radians, and compare it with the values of
sin 5° and tan 5 given by the tables.
7. How many miles an hour does London move in con-
sequence of the rotation of the earth ? Take the earth as a
sphere of radius 3960 mis. London is in latitude 51° 30' N.
RADIAN OR CIRCULAR MEASURE OF ANGLES 133
8. A man is on the perimeter of a circular space, and
wishing to know its diameter, he selects two points in the
boundary a furlong apart, which at a third point also in the
boundary, subtend an angle of 164° 43'. Find the diameter to
the nearest foot.
9. Find the radius of a sphere whose volume is 216'8 c.c.,
given volume=|7rr3, 7r = 3'142.
10. What is the distance of the visible horizon from the
mast of a ship 80 feet high ?
11. From a quadrant AB of a circle an arc AP is marked off
subtending an angle of XQ at the centre. A circle with centre A
passes through P and cuts the chord AB in P'. Express AP' in
terms of x. Suppose the chord graduated so that every point P'
corresponding to an integral value of x is marked x. How could
you from a ruler graduated like this chord construct an angle of
given magnitude ?
12. Show that if an object of height A at a distance d from
the observer subtends a small angle of A degrees at his position,
A<^
then roughly h =— — . Use this to find the height of a tower
o7o
which subtends an angle of 9° at a point 170 yards away.
13. A girder to carry a bridge is in the form of a circular
arc : the length of the span is 120 ft. and the rise of the arch
(i.e. the height of the middle above the ends) is 25 ft. Find the
angle subtended by the arc at the centre of the circle and the
radius of the circle.
14. Find the value of (-03642)* x cos 61° 23'.
15. If the light from a lighthouse 250 ft. high can just be
seen from the top of a mast 80 ft. high, find the approximate
distance of the ship from the lighthouse, assuming the earth a
sphere of 4000 mis. radius.
16. Taking sin 6=6 (in radians) for small angles, find sin 25'
correct to four significant figures.
134
PRACTICAL TRIGONOMETRY
17. Two places A and B on the earth's surface are on the
same parallel of latitude 52° 30'. The difference of their longi-
tudes is 32° 15'. Take the earth as a sphere of such size that a
mile on the surface subtends an angle of 1' at the centre, and
find (i) the radius of the parallel of latitude on which A and B
lie, (ii) the distance in a straight line between A and B, and
(iii) the distance between A and B along a great circle, i.e. along
a circle which passes through these points and has its centre at
the centre of the earth.
18. A circle of radius r rolls on a horizontal straight line.
A point P on the circle coincides with a point O on the straight
line and after the circle has rotated through an angle 6 the
horizontal and vertical distances of P from O are x and y.
Prove x =r6 — r sin #, y=r — r cos 6.
19. The figure is a rough sketch of a railway from A to B,
which is made up of three straight pieces and two circular arcs.
Calculate the length of the railway from A to B.
ZG-5 Chains
-s*...
$&•
Fig. 67.
20. A chasm in level ground is bounded by parallel vertical
sides. The depth AB of the chasm at A is wanted, and, it being
impossible to take measurements from C, the point opposite A,
a point D 50 yards along the side from C is chosen. The angle
ADB is 43° and the angle ADC is 52°. Find the depth of AB.
CHAPTER X.
ANGLES WHICH ARE NOT IN ONE PLANE.
73. WE will begin by reminding the reader of some of
the definitions and theorems of Solid Geometry.
(1) The intersection of two planes is a straight line.
(2) The angle between two planes is the angle be-
tween two straight lines drawn from any point in the line
of intersection of the planes and perpendicular to it, one
being in each plane.
Thus in the figure, XY is the line of intersection of the
two planes AXY, BXY.
136
PRACTICAL TRIGONOMETRY
Also if PQ, PR are both perpendicular to XY, and one of
them lies in the plane AXY and the other in the plane BXY,
then L QPR is the angle between the planes.
(3) The angle a straight line makes with a plane is the
angle between the straight line and its projection on the
plane.
(4) If a straight line is perpendicular to each of two
intersecting straight lines it is perpendicular to the plane
which contains them ; that is, it is perpendicular to every
straight line in that plane which meets it.
(5) If N be the foot of the perpendicular from a point P
to a plane, and Q be the foot of the perpendicular drawn
from N to any straight line XY on the plane, then XY is
perpendicular to the plane PNQ.
Fig. 69.
Thus in the figure, PN is perpendicular to every straight
line which lies in the plane NXY and passes through N.
NQ is the projection of PQ on the plane, and PQN is the
angle of inclination of PQ to the plane. XY is perpendicular
to the plane PNQ.
ANGLES WHICH ARE NOT IN ONE PLANE 137
Example (i).
Suppose OX to be the intersection of a vertical with a
horizontal plane.
Fig. 70.
Let OA be in the horizontal plane making the angle a with
OX ; and let OB be in the vertical plane making the angle ft
with OX.
To find
(1) The angle AOB.
(2) The inclination of the plane AOB to the horizon.
From any point P in OB draw PN perpendicular to OX, and
draw NO. perpendicular to OA.
Then PQ is perpendicular to OA. [Art. 73 (5).]
Now OQ = ONcosa
= OP cos (3 cos a;
OQ
.'. cos L AOB = — — = cosacos/3.
Again
PN=ONtanft
QN=ON sin a.
Now the inclination of AOB to the horizon
-Z.PQN, [Art. 73(2)]
PN
and we have
tan L PQN = -
tan/3
138
PRACTICAL TRIGONOMETRY
Example (ii)
A desk slopes at 15° to the horizon ; find the inclination to
the horizon of a line on the desk which makes 40° with the line
of greatest slope.
Fig. 71.
Let AB be the intersection of the plane of the desk with a
horizontal plane. Also let AC be a line of greatest slope, and
AD the line on the desk making 40° with AC. Take any point
D in AD.
Draw DE parallel to AC, and DF perpendicular to the
horizontal plane ; then 6 is the angle required.
Now Z.DEF = 15°,
and we have DF = D E sin 15°
= DA cos 40° sin 15°,
since D E A is a right angle.
loS cos 40° = i'8843
Iogsinl5° =
= cos 40° sin 15° ;
logsm^l'2973;
.'. 0 = 11° 26'
approximately.
ANGLES WHICH ARE NOT IN ONE PLANE 139
Example (iii).
Two set squares, whose sides are 3, 4, 5 inches, are placed so
that their shortest sides coincide, and the angle between the set
squares is 40°. Find the angle between the longest sides.
Let ABC, ABD denote the set squares. We require the
angle DAC.
Now /.CBD=40° ; and if E be the middle point of CD, we
have
CE = 4siii20°;
.*. sinCAE
and
L. CAD = 31° 46' nearly.
140
PRACTICAL TRIGONOMETRY
Example (iv).
The figure represents a rectangular box of which the sides
are 3, 4, 5 feet.
5
Fig. 73.
(1) The angle made by the plane ABH E with the plane ABG F
= 0 = tan-1|=tan-1l-6667 = 59° 2'.
(2) To find the angle between the planes AEC and ADEF ;
draw DN perpendicular to AE ; then CN is also perpendicular to
AE. Then </> is the angle required.
We have DN = DEsin DEA = 3x
tan DC = 4v/34>
... </> = 57° 15'.
(3) To find the angle CAE, we have
15
, since AE = /s/34;
log 4= -6021
\ log 34= -7657
1-3678
log 15= 1-1761
•1917
CN =
and
CN
/ 769
log 769 = 2-8859
•'• SmCAE-CA
/7fiQ
/ • Ml
log 34 = 1-5315
log 41 = 1-6128
2)1-7416
V 34 ' V4J
V 34x41'
8'.
ANGLES WHICH ARE NOT IN ONE PLANE 141
Example (v).
To find the angle between two faces of a regular tetrahedron
(i.e. a figure enclosed by four equal equilateral triangles).
Let D, A, B, C be the vertices of the figure, and let a be the
length of the side of each triangle.
Let E be the middle point of BC and N the foot of the
perpendicular from D on the plane ABC.
Since DE is perpendicular to BC, and DN is perpendicular to
the plane ABC,
.*. EN is perpendicular to BC at E the mid-point, and bisects
the angle BAG. Similarly BN bisects the angle ABC.
.*. we have
and
EN_EBtan30°
> = EBtan60°=-
—
= 008-! -3333
= 70° 32'.
And this is the angle between two faces.
142 PRACTICAL TRIGONOMETRY
Examples. X a.
1. Find the angle between a diagonal of a cube and a
diagonal of one of the faces which meets it.
2. Find the angle between the diagonals of any two adjacent
faces of a cube.
3. The edges of a rectangular box are 4, 3, 6 inches ; find
the length of a diagonal of the box, and the angle it makes with
the longest side.
4. A triangle whose sides are as 3 : 4 : 5 is inclined to the
horizon at an angle of 35°, and the longest side is horizontal.
What are the inclinations of the other sides to the horizon ?
5. A rectangle 6 ft. by 4 ft. is turned about the shorter side
through an angle of 40° ; find the angle between the two positions
of one of the diagonals.
6. A desk slopes at 15° to the horizon and AB, the lower
edge of it, is horizontal. A straight line AC is drawn on the
desk making 35° with the lower edge and of length 20 inches.
(1) How far is C from AB ? (2) How far is C above the hori-
zontal plane through AB ? (3) What is the inclination of AC to
the horizontal plane?
7. All the edges of a pyramid are of length a and its base
is a square. Find the angle between one of the slant edges and
the diagonal of the base which meets it. Find also the altitude
of the figure.
8. A square of side 5" rests on one edge and is inclined at
an angle of 35° to the horizontal plane. Find the angle between
a diagonal and its projection on the plane.
9. A rectangle 5 ft. by 4 ft. rests with its longer edge on a
horizontal plane and is inclined at an angle of 52° to this plane.
Find the length of the projection of a diagonal of this rectangle
on the plane and the angle between the diagonal and its pro-
jection.
10. An isosceles triangle, base BC, 8", equal sides AB, AC,
12", rests with its base on a horizontal plane and is tilted over
until it makes an angle of 40° with the plane. Find the height
of the vertex above the plane and the angle between AC and its
projection on the plane.
ANGLES WHICH ARE NOT IN ONE PLANE 143
11. Two equal 45° set squares ABC, ABD are placed at right
angles to one another and at right angles to a horizontal plane
so that the edges AB coincide and B is on the plane. Find the
angle the plane ACD makes with the horizontal plane, and the
perpendicular distance of B from the plane ACD, if the shorter
sides of the set squares are 5".
12. Two vertical planes ZOX, ZOY inclined to one another
at an angle of 20° intersect the horizontal plane in OX and OY.
In the plane ZOY a point P is taken 8" from OZ and 10" from
OY. Find the angle between the line OP and the plane ZOX.
13. Up a hillside sloping at 26° to the horizontal plane runs
a zigzag path which makes an angle of 60° to the line of greatest
slope. What is the length of the path to the top of the hill
which is 1200 feet high and what angle does the path make with
the horizontal plane ?
14. O is a corner of a rectangular solid, and A, B, C are
points on the three edges which meet at O. If OA, OB, OC
are respectively 1, 2, 3 inches, find the angles the plane ABC
makes with the faces of the solid.
15. Three straight lines OA, OB, OC are mutually at right
angles, and their lengths are a, b, c. Show that the tangent of
the angle between the planes OAB, ABC is =- — , and hence
that the area of A ABC is \ Jb2c2 + c2a?+a2b*.
16. A roof of a porch is built out at right angles to a
vertical wall. The ridge AF is horizontal and of length 10 ft.
The front face is an isosceles triangle FDE, whose edges FD, FE
slope at 45° to the horizon, and the edge DE is 6ft. The lower-
edges parallel to AF are each 14 ft. in length. Calculate the area
of the roof.
17. XOY is the floor of a room; ZOX, ZOY are two vertical
walls at right angles to one another. A stick AB rests with its
end A on the floor 6 ft. from OX, and 3 ft. from OY. The other
end B is fastened to the wall ZOY, 2 ft. from OY and 1 ft.
from OZ, Find the length of the stick and of its projections
on the walls ZOY, ZOX.
144 PRACTICAL TRIGONOMETRY
74. To find the height of a distant object.
Let AB denote the object, and let its height be h feet.
A
Prom a point C measure a straight line CD in any
direction on a horizontal plane, and let its length be a feet.
Let the angles ACB, ACD, ADC be observed to be a, fi, y
respectively. Then we have
AC = li cosec a.
Also from the triangle ACD, we have
AC CD
whence
sin(180°-/2
L a sin y
fl COSeC a = -. — -N ;
/. h =
a sin a sin y
If the observations were made with a theodolite, the
angles BCD, CDB would be observed instead of ft and y,
a tan a sin CDB
In this case prove Ji =
sin (BCD + CDB)'
ANGLES WHICH ARE NOT IN ONE PLANE 145
Example.
A man at A observes the angle of elevation of the top of a
tower BC to be a.
Fig. 76.
He walks x yards towards the tower up a road inclined at y
to the horizon and then observes the angle, of elevation of B to
be p. Find BC.
From the triangle BDC we have
BD h
sin (90 + y) ~~ sin (/3 - y) '
In the triangle ABD the angle ABD=/3 — a,
BD x
and
sin(a-y) sin(/3-a)'
=
sin 03 -a) '
, _#sin(a — y) sin (/3 - y)
~~ sin (/3 — a) cos y
Note that BD forms a connecting link between x and h.
In Art. 74 AC formed the connecting link between CD and h.
P. F.
10
146
PRACTICAL TRIGONOMETRY
75. Projection of an area.
Let A BCD be a rectangle inclined at an angle 6 to the
horizon and having the side BC horizontal. Then if a, d
are the projections of A and D on the horizontal plane, the
Fig. 77.
rectangle Bade is the projection of the rectangle ABCD;
and the area of Bade is the area of ABCD multiplied by
COS0.
For Cd=CDeos0;
.". .area of Bade = BC x cd
= BC x CD COS 0
= area of ABCD x cos 0.
It follows that if we have any figure of area A on a plane
inclined to another plane XY at an angle 0, the area of the
projection of the figure on the plane XY is equal to Acos0.
Fig. 78.
For the figure A may be considered to be composed of small
rectangles having one side parallel to the line of section of
the planes.
ANGLES WHICH ARE NOT IN ONE PLANE 147
Examples. X b.
1. The elevation of a tower was observed at a certain station
to be 25° and its bearing N.E. At a second station 1000 feet
due S. of the former its bearing was N. by E. Find its height.
2. From a point A an observer finds that the angle of
elevation of a peak B is 37°. He walks 1000 yards to a point
C on the same horizontal plane as A and observes the angles
BAG = 65°, ACB = 70°. Find the height of the peak.
3. BC is a tower standing on a horizontal plane. From
A and D two points in the plane 500 feet apart the angles
of elevation of B, the top of the tower, are observed to be 20° 5'
and 27° 17' respectively. The angle CAD =40°. Find the height
of the tower.
4. A ship was 2 miles due S. of a lighthouse. After sailing
1 mile W. 30° N. the angle of elevation of the top of the lighthouse
was 2°. Find the height of the lighthouse above sea-level.
5. The angle of elevation of A the top of an inaccessible
tower AB is observed from a point C to be 24°. A base line
400 ft. long is drawn from C to a point D and the angles BCD,
CDB are observed to be 95°, 54° respectively. Find the height
of the tower.
6. A lighthouse is seen N. 20° E. from a vessel sailing
S. 30° E., and a mile further on it appears due N. Find its
distance at the last observation.
7. A man at sea-level observes that the elevation of a
mountain is 32° 11': after walking directly towards it for a mile
along a road inclined at an angle of 10° to the horizontal, he finds
the elevation of the mountain to be 47° 23'. Find the height of
the mountain.
8. From the top of a hill the depression of a point on the
plain below is 40°, and from a place f of the way down the
depression of the same point is 20°. Find the inclination of the
hill.
10—2
148 PRACTICAL TRIGONOMETRY
9. To find the distance of a battery B from a fort F,
distances BA, AC were measured on the ground to points A
and C, BA being 1000 yards and AC 1500 yards. The following
angles were observed: BAF = 33°41', FAC = 73° 35', FCA = 81°4/.
Find the distance BF.
10. From a certain station the angular elevation of a peak
in the N.E. is observed to be 32°. A hill in the E.S.E. whose
height above the station is known to be 1200ft. is then ascended
and the peak is now seen in the N. at an elevation of 20°. Find
the height of its summit above the first station.
11. A balloon was observed in the N.E. at an elevation of
51° 50' : 10 minutes afterwards it was found to be due N. at an
elevation of 31°. The rate at which the balloon was descending
was afterwards found to be 6 miles per hour. Find the velocity
of its horizontal motion (supposed uniform), the wind at the
time being in the East.
12. A rectangular vertical target standing on a horizontal
plane faces due S. Compare the area of the target with that
of its shadow when the sun is S. 20° E. and at an altitude of 53°.
13. Find the height of a mountain whose summit is A, given
that the length of a horizontal base line BC is 1500 yards,
Z.ABC = 61°10', Z_ACB = 52°11', and the angle which AB makes
with the vertical = 57° 18'.
14. A hill which slopes to the N. is observed from two
points on the plane due S. at distances of 200 and 500 yards.
If the angles of elevation of the top of the hill from these
points are 32° and 25° respectively, find the inclination of the
hill to the vertical.
15. From the top of a hill 1000 ft. above a lake the angle
of elevation of a cloud is 21° 11', and the angle of depression
of its reflexion in the lake is 46° 3'. Find the height of the cloud.
16. A and B are two places 10 miles apart, B bearing
E. 18° N. of A. A man is at P which bears S. 18° 36' W. of A,
and S. 52° 17' W. of B. Find in what direction he must move to
walk straight to a place Q 7 miles away from both A and B
to the South of AB. Calculate also the distance from P to Q.
ANGLES WHICH ARE NOT IN ONE PLANE 149
17. A seam of coal, 10 ft. thick, is inclined at 20° to the
horizon. Find the volume of coal under an acre of land.
18. The area of the cross-section of a cylinder is 147 sq. ins.
What is the area of a section making an angle of 10° with the
cross-section ?
19. A district in which the surface of the ground may be
regarded as a sloping plane has an area of 5*8 sq. mis. It is
shown on the map as an area of 4*6 sq. mis. At what angle
is the plane inclined to the horizon?
20. A vertical wall 40 ft. long and 10 ft. high runs east and
west ; calculate the area of the shadow cast by it on the ground
when the sun is S.S. W. at an elevation of 20°.
150 PRACTICAL TRIGONOMETRY
TRIGONOMETRICAL SURVEYING.
76. Triangulation.
A district or country is surveyed by constructing
a series of triangles, the sides of which are calculated from
measurements of the various angles and the known length
of one side of the initial triangle called the Base Line.
Angles in a horizontal or vertical plane are measured by
an instrument called a Theodolite.
A survey which extends over a country large enough to
necessitate the application of Spherical Trigonometry to
allow for the curvature of the earth's surface is called
a Geodetic Survey.
The Base Line for such a survey may be as much as
14 miles in length and is measured with great accuracy by
a nickel-steel wire which has no coefficient of expansion for
variations of temperature. Since the base line is not
horizontal, the differences of level have to be measured and
the observations reduced to sea-level.
For smaller triangulations the base line is measured
with sufficient accuracy by a surveyor's chain, 22 yards
long, consisting of 100 links.
The triangles observed should be as nearly equilateral
as possible and small angles should be avoided as any error
in their measurement would considerably affect the accuracy
of the calculations.
If the angles of the triangle do not add up to 180° the
difference between their sum and 180° is divided equally
among them.
TRIGONOMETRICAL SURVEYING
151
Example (i).
The base line CD was 8*895 chains.
At C the angles ECD, DCF were measured, also ECF and the
re-entrant angle ECF, to check the observations.
Fig. 79.
At D similar angles were observed. At E observations of
C, F and D were made, and at F observations of C, E and D.
From the triangle ECD find EC, and from the triangle ECF
find EF.
We have :
EC
sin 55° 13' " sin 63° 41' '
EF EC
CD
EC = 8'895sin55°13'
sin 63° 41'
8-895 sin 55° 13' sin 34° 28'
sin 145° 32' sin 15° 44"
sin 63° 41' sin 15° 44'
= 17*01 chains.
Show that the same result is obtained by working with the
triangle CDF to find DF and then with the triangle EFD to
find EF.
Example (ii).
The diagram shows part of the triangulation of a river.
When the principal triangulation is completed other points
are fixed by using the sides of these triangles as base lines
and the course of the river is determined by measurements
of offsets from known points and lines.
152
PRACTICAL TRIGONOMETRY
Work with the triangles ABD and ADC to obtain
DC = 294-45 ft.
460 ft.
Then check by working with triangles ABC and BCD
DC = 294-39 ft.
Taking DC = 294-4 ft. work out the lengths of DE, FE, FG, GE.
In practice the length EG would be measured as a check base
to confirm the accuracy of the observations and calculations.
Exercise.
The corners of a triangular field PQR are determined with
reference to a base line AB by the dimensions PAB = 57°,
PBA = 84°, QAB = 64°, QBA = 101°, RAB = 115°, RBA = 47°, AB
is 50 feet long. Calculate the sides of the triangle PQR to the
nearest foot.
TRIGONOMETRICAL SURVEYING 153
Miscellaneous Examples. H.
1. Calculate the following by logarithms, and show how
you would roughly check your results:
(1) prn, where p = 9375, r=lO3, w = 4;
(2) ^Trr3, where ir = ?f£, r= 5-875.
2. A man surveying a road from A to B, goes first 7 chains
in a direction S. 63° E., then 8*3 chains S. 80° E., then 12 chains
N. 46° E., and then 5-7 chains N. 16° W. to B. Find (1) how
far B is east of A; (2) how far B is north of A; (3) the distance
AB; (4) the bearing of B from A. Verify by a figure drawn
to scale.
3. The sides of a quadrilateral taken in order are 4, 5, 8, 9 ft.,
and one diagonal is 9 ft. ; find its angles and area.
4. ABCD is the rectangular floor of a room, the length BA
being 48 ft. The height at C subtends at A an angle of 18°,
and at B an angle of 30°. Find the height of the room.
5. In any triangle, prove
(1 ) sin 2 A + sin 2 B + sin 2C = 4 sin A sin B sin C ;
ABC
(2) sin A + sin B+sinC = 4cos- cos- cos — .
6. Calculate as accurately as the tables permit
52-45 x 378-4 x '02086
87-32 x '5844
(2) (1-246)4195.
7. A ship sailing north sees two lighthouses which are
4 miles apart in a line due West. After sailing for an hour one
of these bears S.W. and the other S.S.W. Find the ship's rate.
8. AB and DE are two chords of a circle at right angles to
each other intersecting in C: AC = 40 ft., DC = 30 ft., and the
radius of the circle is 100 ft. Find the sides and angles of the
quadrilateral AD BE and determine its area.
154 PRACTICAL TRIGONOMETRY
rt T ,. sin (C - $) cos A ,
9. If V--;.—- = , where A, B, C are the angles of a
sin 6 cos B '
triangle, prove that cot 6 = tan B.
10. The plane side of a hill running E. to "W. is inclined
to the horizon at an angle of 20° : it is required to construct
a straight railroad upon it inclined at 5° to the horizon.
Determine the point of the compass to which it must be
directed.
11. A bed of coal 14 ft. thick is inclined at 23° to the
surface. Calculate the number of tons of coal that lie under
an acre of surface. A ton of coal occupies 28 c.ft. The 14 ft.
is to be regarded as a measurement at right angles to the surface
of the coal bed.
12. A pyramid of height 57 in. stands on a triangular base,
one side of which is 25 in., the angles at the extremities of that
side being 45° and 57° 30'. Find the volume to 2 significant
figures.
13. The area of a triangle is 96 sq. ft. and the radii of the
three escribed circles are 8, 12, 24 ft. respectively. Find the
sides.
14. The angle of elevation of a tower 100 ft. high, arid due
N. of an observer is 50°. What will be its elevation to the
observer when he has walked 300 ft. due E. of his former position?
15. If a, b, c are three consecutive integers, prove that
log b - log a > log c — log b.
16. If the sides of a triangle are 51, 68, 85 ft., show that the
shortest side is divided by the point of contact of the inscribed
circle into two segments, one of which is double of the other.
17. In a triangle ABC the side BC is 200 ft. long, and the
angles at B and C are 79° and 75° respectively. B and C are
observation stations and it is impossible to approach nearer to A.
A body in the air h ft. immediately above A is observed to have
an elevation of 40° at C. Calculate h.
TRIGONOMETRICAL SURVEYING 155
18. OX, OY are two straight lines at right angles. On OX
take a point P such that OP = 10 cm. Now imagine OP to revolve
to the position OY, and to vary in length in such a way that its
length at any moment is equal to its original length multiplied
by the cosine of the angle it has revolved through. Thus at 60°
its length will be 5 cm. If x, y are the distances of P at any
moment from OX, OY, show that #2+y2- 10#=0.
19. In any triangle, prove that the area is equal to
Rr (sin A 4- sin B-f sin C), where R, r are the radii of the circum-
scribed and inscribed circles.
20. An upright pole 10 ft. high casts a shadow 12 -6 ft. long
at midday on a certain day. Another upright pole of the same
height 100 miles further north casts a shadow 13*2 ft. long at the
same time. Deduce the Earth's perimeter, supposing the Earth
a sphere.
21. A man whose eye is 5 ft. above the ground stands 20 ft.
from the wall of a room, and observes the angle of elevation of
one of the corners of the ceiling to be 30°. After walking 16 ft.
directly towards the wall he finds the angle of elevation of the
same corner to be now 60°. Find the height of the room.
22. A pole 15 ft. long leans against a wall with one end on
the ground 9 ft. from the foot of the wall. This end is pulled
away until the angle the pole makes with the ground is half
what it was originally. Prove, without the use of tables, that
the end is now 6>J5 ft. from the wall.
23. The side AB of a triangle ABC is divided at P in the
ratio of m : n. The angles PC A, PCB, CPB are a, /3, 6 respec-
tively. Prove that
m cot a — n cot fi=n cot A — m cot B = (m+n) cot 6.
24. Given x — a cos B + b cos 20,
y = a sin 0 + b sin 20 ;
., , #2+^_a2_&2
prove that cos0 = 97. — ~~ •
25. It was known to early Hindu mathematicians that if
x, y and z are three angles such that x — y—y — z=^ then
sintf — siny=sin?/— sins + £ sin 3/, and they used this formula
to check tables of sines. Express k in terms of the angle A, and
check your tables for the case #=52° 24', y=48° 42', z = 45°.
156 PRACTICAL TRIGONOMETRY
26. A man has before him on a level plain a conical hill of
vertical angle 90°. Stationing himself at some distance from its
foot he observes the angle of elevation « 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
_t tan a (1— tana)
l+tana(l-f 2 tan a)*
27. A rectangle ABCD in which AB = 6, BC = a is placed so
that its diagonal AC, of length d, makes an acute angle <£ with
AX, a line passing through A. If AB makes an angle 6 with AX,
prove that
dcos <f> = b cos 6 — a sin $,
A b tan <t> — a
and tan0=,—
b + a tan <p
28. A straight bar 2 ft. long is suspended horizontally by
two strings, each 2 ft. long, attached to its ends. The bar is
twisted round its centre, the strings being kept tight, and the
bar horizontal, till the centre is raised a foot. Through what
angle is the bar twisted?
29. Two planes inclined at angles 0, cf> to the horizon slope
in opposite directions. A rod of length 2a making an angle a
with the horizon rests with one end on each plane so that its
mid-point is vertically over the line of intersection of the planes.
Assuming that the line of intersection of the planes is horizontal,
and that the rod lies in a vertical plane at right angles to this
line, prove that tan 6 ~ tan $ = 2 tan a.
30. An observer wishing to determine the length of an
object in the horizontal plane through his eye, finds that the
object subtends an angle a at his eye when he is in a certain
position A. He then finds two other positions B, C where the
object subtends the same angle a. Show that the length of the
object is — , where a, 6, c are the sides, and A the area
Saa
of the triangle ABC.
1. 60°.
4. 150°, 210°.
7. 98045.
ANSWERS.
2.
5.
8.
I. p. 4.
52° 44' 40".
97° 30', 262° 30'.
43° 44' 24".
3. 37° 34'.
6. 1530°.
9. (i) 120°, (ii) 128° 34' 17", (iii) 108°. 10. '56995.
II a. p. 9.
BC = 3, I, l,f, |, f; 1, f, f, y, y.
£•> if, ¥, Hi ¥, H, A, if, ¥•
(i) sin A, (ii) cos A, (iii) cot A, (iv) tan A.
6 = 9; %5-, y, ff, ff, ff, ff, 1, 1.
AD AC ,60 BA
AB » BC ' BA ' BC 5
AB BD BD AD
CB' BA* DA' DC'
(i) sinABD, (ii) tan BAC, (iii) cosACD.
BC CD
AC' CB*
(i) tan A, (ii) cos A, (iii) From sin A. 10. 4^.
lib. p. 15.
1. sin 37° = -60, cos 37° = '80, tan 37° = '75,
cosec 37° = 1 '66, sec 37° = 1 '25, cot 37° = 1 '33.
2. sin 49° = -75, cos 49° = '66, sec 49° = 1 '52, tan 49° =1 '15.
3. 58° 40', sin 58° 4(X = -85, tan 58° 4(X = 1 '6.
P. F.
11 PRACTICAL TRIGONOMETRY
4. sec A = 1-94, tan A = 1'66. 5. $8, 1.
6. A = 80° 36', tan ^ = '85, 2'15.
7. 28°, cos28° = '9, sec 28° = 1-1.
8. tan 40° = -84, tan 20°= cot 70° = -38. 9.
13. BE = 8", BF=6-9".
MISCELLANEOUS EXAMPLES A. p. 16.
2. tan 48° = 1-11. 3. sinA=-(
4. 120°. 5. 19° 18' 18". 6. 63° 30'.
7. 10. 8. 68°, -40. 9. -58. 10. -25, '26. 11. 150°.
13. 32° nearly. 14. 8". 15. 2-4. 17. 30°, 60°, 90°.
Ill a. p. 20.
1. -3256. 2. -5500. 3. '4215. 4. '9506.
5. 1*7079. 6. -8976. 7. '3025. 8. 3-9894.
9. 2-9478. 10. 5-9351. 11. 4-8642. 12. 6-1742.
13. 62° 28'. 14. 63° 43'. 15. 61° 7'. 16. 78° 49'.
17. 75° 26'. 18. 75° 50'. 19. 11° 32', 30°.
20. 36° 52', 48° 11'. 21. 41° 49'. 22. 51° 20', 71° 34'.
EXERCISE.
p. 21.
W
AB=#cosec#,
BC = #cot 0.
(ii)
AB=#sec<£,
BC=ytan<£.
(iii)
BC=#tan 0,
AC=#sec0.
(iv)
AB = ?/cos<£,
BC = ?/sin</>.
(v)
AC = x cot 0,
B C — x cosec 0.
(vi)
AB = ?/cos<£,
AC=y sin(/>.
(vii)
c= 17'013,
a =13-764.
(viii)
c = 10-946,
a = 4-452.
(ix)
c= 22-69,
6=10-718.
(x)
c = 15-146,
6=11-376.
(xi)
6 = 15-4725,
a =19-6375.
(xii)
6=16-929,
« = 11-1132.
ANSWERS 111
Illb. p. 23.
1. 3-464in. 2. 7 '66, 6-43, 11-92, 9-13 in. 3. 318-5 ft.
4. 33°. 5. 35°. 6. 4-37 ft, 8-25 ft., 41° 11'.
7. 273ft. 8. 3-06 ft., 3-83 ft. 9. 11-28 cm. 7 '71 cm.
10. 148-26 ft. 11. 17-32, 6-84, 18-79, 24'53in.
12. 246ft. 13. 12-86, 15*32, 19-32, 5*18 ft.
EXERCISE, p. 26.
1. 237'8 sq. in. approx. 2. 58-8 in. approx.
3. 363 sq. in. approx., 72-6(5) in. approx.
IIIc. p. 30.
1. 10-23 sq. in. 2. 5-23 in. 3. '076 ft. 4. 109 ft.
5. 93-53 sq. in., 36 in. 6. 63'86 ft., 60° 34'.
7. 133-7 ft. " 8. 60°, 30°, 6'93 in. 9. 14-69 sq. ft.
10. 8-86, 6-25, 7'46 cms. 11. 59° 29'. 12. 37 yds.
13. 61 -9 ft. 14. 8-76 in. 15. 696ft. 16. 42-4 mis.
17. 71-4, 79-3 ft. 18. 2630 ft. approx. 19. 1081ft.
20. 126-6 ft. 21. 6104 sq. ft.
22. (i) 4-43 mis., (ii) 5'97 mis., (iii) 7'4 mis. 23. 39 ft.
24. 73-5 sq. ft., 30-90 sq. ft. ; 81-2 sq. in., 32'49 in.
25. 149-6 ft. 26. 140, 184 ft. 27. 1-245 mis.
28. 96-2 yds. 29. 6 miles. 30. 121 yds., E. 51° N.
MISCELLANEOUS EXAMPLES B. p. 33.
1. 7*8 cms., 6*3 cms., 8*1 cms. 2. 9*95 cms., 6*71 cms.
3. 1-40. 4. (i) 1-0724, (ii) 3'6280.
5. 41° 49', 10'47 cms. 7. 5-14 cms., 12'86 sq. cms.
8. 84ft. 10. 13jmls., N. 13° 7' W. 11. 23° 51', 28° 9'.
12. 26° 47'. 13. 0°, 30°. 15. 37°.
16. 1-805 in., 61° 1', 61° 1', 57° 58'. 17. 48° 35', 14° 29'
18. 2° 22'. 19. 38-04 sq. in.
20. a = 5, sin2A = '71, sinA = T%, cosA = }H-.
21. 266-95 yds. 22. -05 ins., -0033 ins. 23. 61° 19'.
24. 31° 41'. 25. 56° 19', 53° 8'. 26. 21 -3 ft.
27. 30°. 41° 49'. 28. 3 ch. 27 links.
IV PRACTICAL TRIGONOMETRY
IV a. p. 46.
1. (1) -9063.
(4) -'7813.
(7) '8129.
(10) 1-9841.
(2) -'6428.
(5) '6691.
(8) --1432.
(11) 3-6280.
(3) -1-0038.
(6) --7536.
(9) -'5878.
(12) 4-4919.
2. (1) 115° I', 295° r. (2) 19° 7', 199°
(3) 63° 5', 116° 55'. (4) 112° 46', 247° 14'.
(5) 241° 1', 298° 59'. (6) 18° 43', 198° 43'.
3. 35°, 215°.
7. (1) 34° 31' or 145° 29'. (2) 51° 19'. (3) 113° 35'.
9. (1) 30°, 150°. (2) 53° 8', 126° 52', 210°, 330°.
(3) 168° 41', 348° 41', 68° 12', 248° 12'.
10. (1) 18°. (2) 10°.
11. (1) 36°. (2) 60°. (3) 36° or 60°.
EXERCISE, p. 49.
(i) 1, oo, 0. (ii) 0, -1, 0. (iii) oo, -1, -oo.
(iv) -1, 0, oo. (v) -1, -GO, 0.
IV b. p. 54.
2. (1) 45°, 225°. (2) 135°, 315°.
5. (1) 0=90° or 270°. OP = 4. (2) <9 = 0° or 180°, OP = 5.
6. 45°, 312-5 ft., 9° 20', 80° 40'.
V. p. 59.
1. 75° 31'. 2. 4-23. 3. 112° 53'. 4. 6 '47 m., 4 '02 m.
6. A = 41°24', B = 55°47', C = 82°49'. 7. 4 '86 ft., 1-55 ft.
8. a = 7'41, B = 80°49', C = 52°ll'. 9. 81°, 19°.
10. 8-83 ft. 11. 6-14. 12. 110° 29'.
14. B = 27°50', C = 37°10'.
15. 4.^2-32^ + 31=0, 6-87, 1-13 miles.
ANSWERS
MISCELLANEOUS EXAMPLES C. p. 61.
1. (1) 75° 58'. (2) 1-134". 2. 69° 18', 110° 42'.
4. 225-3 ft. 7. 9-06 sq. in., 3«71, 5-54 in. 8. 185-8 yds.
10. 41-5 ft. 11. cos 6= — 7= 12. 1026ft.
±Vtan20 + l
13. 18° 56', 8-14 in.
16. -018, -019, -021, -026, '035, -054, -102, -292. Increases
from 5*67 to oo .
18. 36° 52', 146° 19', 216° 52', 326° 19'. 20. 109° 6'.
Via. p. 71.
1- i, *, 1, 0, .ft, ft. 2. -9428, '9683, '5585.
3. -9484. 5. (i) -5150. (ii) -'1908. 7. ^6~ .
12. (i) -5878. (ii) -8090. 14. (cos A + sin A) (cos B- sin B).
15. cos A]cos B cos C — cos A sin B sin C — cos B sin A sin C
— cos C sin A sin B.
2. '5095.
10. 1.
3. 2-V&
14. 120 ft.
VI b. p. 73.
cotAcotB + 1
7.
cot B - cot A
8. J.
VI c. p. 76.
L 5, Ii -A- 2. ±-7333, --6800, ±1*078.
3. -7660, -6428. 5. |. 8. ±|. 9. ±j.
11. -4695, -8829. 12. '3640. 13. ±J. 14. 2
16. (1) 30°, 150°, 210°, 330°.
(2) 0°, 30°, 150°, 180°, 210°, 330°.
17. 8cos4a-8cos2a + L 18. a.
20. i(cos2a+cos2/3), -%3288. 21. ±-*.
23. Projection equals r+r cos 6.
24, Height equals r-r cos 0.
A. 3
VI PRACTICAL TRIGONOMETRY
VI d. p. 79.
1.
36° 52'. 2. 103° 17'. 3. 114° 18'. 4. 16°
16'.
5.
&+f-
Vie. p. 81.
1.
4,
6.
7.
9.
sin 40 + sin 20. 2. cos 40+ cos 20. 3. £ (cos 20 - cos 40).
sin 40 — sin 20. 5. J (sin 3 A — sin A).
\ {sin (A + B) + sin (A — B)}.
\ {cos 2 (A + B) + cos 2 (A - B)} . 8. \ (cos 40 - cos 60).
1- sin 50°. 10. cos 70° + cos 10°.
11.
13.
15.
cos 10° - cos 30°. 12. \ {cos 80°+cos 20°} .
sin2A+sin2B. 14. cos3(A+B) + cos (A- B).
sin A. 16. \ (cos 2a - cos 4a).
VI f. p. 82.
1.
2 sin 2A cos A. 2. 2 cos 2 A sin A. 3. 2 cos 2 A cos A.
4.
O/1 /I
2 sin 2A sin A. 5. 2 cos — sin - .
6.
. 50 . 0 7o- A+B ,A~B
2i sin _ sin ~. , t . 2i sin „ cos .
8.
10.
2cos(a + 0)cos(a-£). 9. 2sin(a + j8)sin(j3-a).
2 sin 18° 30' cos 4° 30'. 11. 2 sin 36° 307 sin 4° 30'.
12.
sin 41° + sin 78° = 2 sin 59° 30' cos 18° 30'.
13.
2 cos 30° 30' cos 12° 307.
21.
a+b sin A + sin B
— . and in a ti lanorle
c sinC
C = 180- (A + B), .'. sin C= sin (A + B).
MISCELLANEOUS EXAMPLES D. p. 84.
1.
4.
7.
m=m'. 2. f, -i, J. 3. 3-90 ft. 3-52 ft.
16-16 ft. 5. Square and add. 6. ff, 75° 45'.
c cos A + V«2 - ^ sin2 A, 16 '25 cms. 9. 5 '32 ft.
10.
tan 0j tan 6*= - 1. See Qu. 1.
11
sin a cos a - _
/y — 9/ — - 1 r» - ' X Xzt'
ANSWERS
Vll
VII a. p. 90.
1. 1, 2, -1, -2,_3; 0,_-4, 4, -3, -1.
2. -6045, 2-6045, 1-6045, 3'6045, 4-6045.
3. 2174, '02174, 2-174, 21740, -002174, 217'4, '2174.
4. -6020, -6990, '7781, -9030, -9542, 1-0791, 1-1761, 1-2040,
1-2552, 1-3010.
•845, 10395, 1-146, 1-2781. 1-113. 1'226.
VII b. p. 91.
1. 2-6749, -6754, 1*4570, 5*6590, 1*9428, 3-5710.
2. 2-969, 5569, '7314, 16500, "004839.
1.
2.
3.
4.
7.
10.
13.
16.
19.
VII c. p. 94.
1. 1-059.
5. -2086.
9. 12-95.
13. -8555.
17. 1*975.
21. 6.
25. 18.
28. 121-5.
2. 10-89.
6. -04223.
10. 13-38.
14. 4-108.
3. 7-750.
7. 127-8.
11. -8950.
15. -00006101.
4. 173-2.
8. -05551.
12. -3840.
16. -2601.
20. 14.
18. -005610. 19. -3163.
22. 39-98. 23. *95. 24. £425. 15s.
26. 22-99. 27. 2214 sq. ft., 9790 cu. ft.
29. 304-2, -01991. 30. -028, -00782.
31. (1) 4, (2) -4. 32. 3-484. 33. 7757 x 1013.
34. -938. 35. 2'442, --511. 36. 2-254. 37. "09281.
38. 305-5. 39. 33130. 40. 360-2. 41. -00005903.
42. 8028 xlO8. 43. -01848. 44. -2384
45. -0000003243. 46. 9888 x 10s.
VH d. p. 97.
(1) -6029, (2) --3822,
1-9219, 1-7112, -2614, 1-8611, '4453, 1-9224.
(1) 17°25', (2) 65° 2', (3) 75° 24', (4) 82° 22', (5)21°.
(3) -4276.
16° 28'. 6. 241.
150400 sq.ft. 9. 22° 16'.
•2004. 12. -0393.
7-958x10-°. 15. -1803.
15-68 grams wt. 18. 9*475 cms.
37° IT, 142° 49'.
-•2831.
81° 12'.
1*518.
12*03.
•01289.
5.
8.
11.
14.
17.
20.
83° 53'.
5-780.
Vlil PRACTICAL TRIGONOMETRY
MISCELLANEOUS EXAMPLES E. p. 100.
1. 27° 45'. 2. ±-7018.
3. (i) 10'5 = V10 = 3 approx., (ii) 10-1 = ^^ = '56 approx.,
(iii) (*35)2 = -12 approx.
4. 14° 2', 45°, 194° 2', 225°. 5. 3-16 sq. cms.
6. log cos 6 = log sin (90° - 0) ; log tan 6 = log sin 0 - log cos 6.
7. 642-2. 8. 78° 28'. 10. 29*4 in., 59'4 sq. in.
11. (i) 27-01 sq. ft, (ii) 5-106 ft.
12. 65-1 ft. 14. ^= tan 6° = -1.
16. 4-193 in. ; XY = 5 (cos a + cos (90° - a)} = lOcos 45°. cos(45° - a),
.-. XY least when a=0, greatest when a =45°.
18. -3%. 19. 20-7.
VIII a. p. 108.
1. A =29° 56', B=42°3', C = 108°l'.
2. C = 77°31', a=51-4, 6=77'2.
3. A = 61°21', a=25-2, c=19«7.
4. A = 33°26', B = 65°10', c=474.
5. A = lll°24', B=22°6', a=36'55.
6. B = 99°13', C = 44°23', 6=46'6.
or B = 7°59', C = 135°37', 6=6'56.
7. B = 40°52', C = 32°8', c=254.
8. A = 78° 48', B = 53°10', C = 48°2'.
9. C = 35°38', a = 5'80, 6=3'93.
10. A = 26°22', C=31°38', 6 = 83'18.
11. C = 39°ll', a=2663, c=2001.
12. A = 44° 49', B=60°.
13. B = 54°56', C = 83°4', c = 209.
or B = 125°4', C = 12°56', c=47'2.
14 89° 55' or 15° 21'. 15. 567 yds. 16. 3'9 ft.
17. 120°. 18. A = 60°, 6 = 3-84 in., c= 4*76 in.
19. A = 95° 12', 0 = 64° 13'. 20. N. 30° E.
21. (i) 1-27 miles, (ii) 5-51 miles.
22. 10-1 ft. 23. 26-1 yards.
VHIb. p. 112.
2. 1*61. 3. 318. 4. 81ft. 5. 30.
ANSWERS ix
VIII c. p. 116.
1. 2-7 sq. in. 2. 4-403. 3. 24'7. 4. 10'7.
5. 79-8 ft., 20,000 sq. ft. 6. 448, 122 links.
7 IQ.K K ift •*•< 11 a2 sift B sin C
7. 12:5, 5:18, 3:2. 11. — . .
2 sin A
MISCELLANEOUS EXAMPLES F. p. 117.
1. 1779 yds., 992'6 yds. 3. 7'228 x 1010.
4. 70° 32'. 6. 76-9 chains ; S. 72° 38' E.
7. 31 miles. 8. Project on the horizontal side.
9. 84° 25'. 10. 4-243 cms.
11. 65-28, -06yds. 12. 87 yrs.
14. A = 53° 8', B = 106°16', C = 20°36', c = 10-99.
15. 2, 43, 42. 16. 1317 yards.
17. A = 56° 5', C = 92° 10' ; or A = 1 23° 55', C = 24° 20'.
18. 14-12". 20. 203-6 c. in. 21. 8-47 ft.
22. 2 cos 45° cos (45° - 6). Max. when 6 = 45°. Min. when 6 = 0°.
23. 4%. 24. 50° 12'. 25. 5-29", 39° 6'.
IX a. p. 123.
7T 57T 137T 1797T
6 ' T' 36 ' 432 '
2. 45°, 120°, 128° 34' 17}", 300°. 3. 1'26, -83, 2'34.
4. 71° 3', 36° 6'. 5. 8-4 cms. 6. -05.
7. ~. 8. 8 Jin. 9. 45° 50'. 10. 135°, ~.
11. 3960 miles approx. 12. -0398 sees. 13. 3*4 cms.
14. 2° nearly. 15. 38-5. 17. 2600 miles.
18. 2-08 in., 1-88 in., 1'26 in. 19. 36'62 ft.
IX b. p. 131.
1. 314-16 sq. in. 2. 11*65 cms.
3. 8-38 sq. ft., 1'45 sq. ft. 4. 2165 miles.
5. 860000 miles. 6. 16'7'. 7. 2 1-3 miles.
8. 34-8 miles. 11. O0582. 14. 57'3 in.
15. 31-416 in., 38-9 sq. in.
PRACTICAL TRIGONOMETRY
MISCELLANEOUS EXAMPLES G. p. 132.
1. 3560, 2517 miles. 2. 54 ins. 3. 29 ft.
5. 9-27, 15-86 ins. 6. 0, 3 cm., 3 cm., 1, 1, -0874 radians,
•0872= sin 5°, -0875 = tan 5°. 7. 645 miles per hr.
8. 2504ft. 9. 3-73 cms. 10. 11 miles.
11. AP' = 2r sin'-. Describe a circle whose radius is the dis-
tance from A to graduation 60°. An angle of x° is subtended
at the centre of the circle by a chord whose length is the
distance from A to the graduation x.
12. 26-7 yds. 13. 90° 29', 84-5 ft. 14. -1587.
15. 30-5 miles. 16. -007272.
17. (i) 2092, (ii) 1162, (iii) 1168.
' 19. 78-4 chains. 20. 75-75 yds.
X a. p. 142.
1. 35° 16'. 2. 60°. 3. 7-810 in., 39° 48'.
4. 27° 19', 20° 8'. 5. 33° 4'.
6. (1) 11-47 in. (2) 2'97 in. (3) 8° 32'.
7. 45°,^?. 8. 23° 56'. 9. 5-57 ft., 29° 30'.
10. 7-27 in., 37° 18'. 11. 54° 44', 2'89 in. 12. 12° 20'.
13. 5475ft., 12° 40'. 14. 31°, 64° 37', 73° 24'.
16. 159-15 sq. ft. 17. x/38, ^29, v/13 ft.
Xb. p. 147.
1. 164ft. 2. 800yds. 3. 172 ft. or 391-5 ft.
4. 106yds. 5. 280 feet. 6. 2-24 miles.
7. 6510ft. 8. 56°approx. 9. 2690 yds. nearly.
10. 2874 ft. nearly. 11. 5 miles per hr. 12. 1-412:1.
13. 2092ft. 14. 51° nearly. 15. 2193ft.
16. 8-63 miles, N. 17° 54' E. 17. 17169 cu. yds.
18. 14-9 sq. in. 19. 37° 30'. 20. 1014 sq. ft.
TRIANGULATION. p. 152.
DE-366ft., FE = 412ft., FG = 274ft., EG=308ft.
ANSWERS XI
EXERCISE, p. 152.
PQ = 112ft., QR = 148ft., RP = 102ft.
MISCELLANEOUS EXAMPLES H. p. 153.
1. (1) 105-5, (2) 849-4.
2. (1) 21-5 chains E., (2) 9'2 chains N., (3) 23-4 chains,
(4) N. 66° 45' E.
3. 77° 10', 139° 21', 84° 16', 59° 13', 37 '45 sq. ft. 4. 18-9 ft.
6. (1)8-116, (2)2-516. 7. 6 -83 miles per hr.
8. DA-50ft., AE = 160ft., EB = 194ft., BD = 120ft.;
112° 23', 67° 37', 128° 39', 51° 21', 14440 sq. ft.
10. 13° 54' with a line going E. and W. 11. 23660 tons.
12. 3600 cu. in. 13. 12,16,20ft. 14. 17° 48' nearly.
17. 376ft. 20. 28,000 miles. 21. 17ft.
25. -4sin2^. 28. 120°.
CAMBRIDGE : PRINTED BY JOHN CLAY, M.A. AT THE UNIVERSITY PRESS.
UNIVERSITY OF CALIFORNIA LIBRARY
BERKELEY
Return to desk from which borrowed.
This book is DUE on the last date stamped below.
MAY 18 1954 Ul
2lNov57KJ'
REC'D LD
MOV 2 6 1957
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LD 21-100m-9,'48(B399sl6)476
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