(navigation image)
Home American Libraries | Canadian Libraries | Universal Library | Community Texts | Project Gutenberg | Children's Library | Biodiversity Heritage Library | Additional Collections
Search: Advanced Search
Anonymous User (login or join us)
Upload
See other formats

Full text of "Practical trigonometry"

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. 

OA 

,'. 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. 



12 



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 
A 7 ? ~~ &C f ~ 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, P2N 2 , on OX. Measure these perpendiculars and 



PRACTICAL TRIGONOMETRY 



the lengths ON, ON 1? ON 2 . Then write down the values of 
the following ratios correct to 2 decimal places : 

NP NiPi N 2 P 2> ON ONj ON 2 NP N^ N 2 P 2 

OP' O 



OP 2 OP' 



OP 2 ' ON' ON X 
R 




on OR. 



N 



Fig. 6. 

Now take a point P 3 on OX and drop a perpendicular P 3 N 3 
Measure OP 3 , P 3 N 3 , ON 3 and find the values of 
N 3 P 3 ON 3 NgP 3 
OP 3 ' OP 3 ' ON 3 * 
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 
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 , sin 0- NP > 

ON 1 OP 



OP' COS ON 3 

-^ cot 0_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 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 ^ 
hypotenuse 

side adjacent to 

cos = , *-- ; 

hypotenuse 

n side opposite to 

tan = -TJ , ; 

side adjacent to 

and similarly for cosec 0, sec and cot 0. 

These ratios are called the trigonometrical functions or 
ratios of the angle 0. 

Note, (sin A) 2 is written sin 2 A; i.e. if 
sin A = | sin 2 A = |. 



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 = 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 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 sin 2 A + cos 2 A, 



sin A 
cos A 



, tan A, 



l + tan 2 C, sec 2 C. 

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+tan 2 A, sec 2 A, l+tan 2 B, sec 2 B, l+cot 2 A, cosec 2 A, 
sin 2 A + cos 2 A, sin 2 B + cos 2 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 AB 2 = BC . BD. 

Prove in a similar way that AC 2 = BC . CD. 

6. In the same figure from the triangles ABD, ADC write 
down two values for cot B. Hence prove AD 2 =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 BC 2 = 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~ l x. 
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 OP 2 =PN 2 + NO 2 ; 

.-. 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 



_ 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 sin 2 A + cos 2 A = 1. 

NP 2 ON 2 NP 2 + ON 2 
sm 2 A + cos 2 A = 2 + _ = Qp2 

_OP 2 
~OP 2 

(3) Prove in a similar manner 

sec 2 A = 1 + tan 2 A, 
cosec 2 A = 1 + cot 2 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 
17 2 =3 2 + ON 2 , 




Fig. 17. 

= 17 2 -8 2 = (17 + 8)(17-8) 
=(25) (9), 



= T 8 5, 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, 
sin 2 49+ cos 2 49 = 1, sec 2 49 -tan 2 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 sec 2 A differs from l+tan 2 A. 



16 PRACTICAL TRIGONOMETRY 

5. Given that cosecA = |, find the values of (sin A + cos A) 2 
and sin 2 A -h cos 2 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+ tan 2 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~ 1 2'4; 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 sin 2 A = 1 cos 2 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 = 2 i y. 

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 = cos 2 30 - sin 2 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 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 
3cosec 2 <9-llcot0 + 7 = 0. 

By using the identity cosec 2 = l + cot 2 # the equation can be 
written in terms of one unknown, 

3(l+cot 2 <9)-llcot<9 + 7 = 0, 

3cot 2 0~ 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. 

22 



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!431'. 9. sec 70 10'. 

cot 11 37'. 12. tan 80 48'. 

tan- 1 2-0248 J 15. cos" 1 '4830. 

cot ~ ! -2600. 18. sec ~ 1 4'0855. 

Find the acute angles which satisfy the following equations : 
19. 10 sin 2 6 -7 sin 6 + 1=0. 20. 15 cos + 8 sec 6 = 22. 

21. 9 cos 2 6 + 18 sin = 17. 22. 4sec 2 <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 = 28ll', a = 20. (x) A = 41 19', a =10. 

(xi) B = 3814', c=25. (xii) A = 3317 / , 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 = 220xtan39 J , 
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 cos 2 6 + sin = 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 

6cos 2 + 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 = BCcosec 2 A. 

32 



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 B 1 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' A 

Fig. 32, 



ANGLES GREATER THAN A RIGHT ANGLE 37 

It will thus be seen that lengths measured along a line 



X' A' 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 PiOP 3 , P 2 OP 4 are equally inclined to XX', 
we have four congruent triangles 



Hence it follows that the trigonometrical functions of 
the angles XOP 1? XOP 2 , XOP 3 , XOP 4 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 be the acute angle XOP X 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 XOP 1 = sin XOP 2 = f , sin XOP 3 = sin XOP 4 = |; 

COS XOPi = COS XOP 4 = |, COSXOP 2 =:COSXOP 3 =: -|; 

tan XOP X = tan XOP 3 = f , tan XOP 2 = tan XOP 4 = - -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 for the second quadrant, 

a = 180 + for the third quadrant, 
a = 360 - 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 - 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 
may have in order that sin 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 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- and 90 + which we discuss in Articles 24, 25. 



ANGLES GREATER THAN A RIGHT ANGLE 



43 



24. To prove that 

sin (90- 0) = cos O t 
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 
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 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 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 may be. The following figures illustrate 
the relations between the ratios of 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 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 + 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 = -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 = can be an angle of a 
triangle. 

9. Find all the positive angles between and 360 which 
satisfy the equations 

(1) 2 cos 2 (9 = 3 sin (9; (2) 10 sin 2 <9- 3 sin 0-4 = ; 
(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 1800, find a value of 6 to satisfy the equations 

(1) sin 40 = sin 0; (2) sin 40 = - sin ; (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. 

= 100 0000a. 



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 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 = 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 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. to 360. 




Fig. 40. 

"With the same figure as before, let L XOP = 0. 

NP 
Then sin . 

Now OP remains constant in magnitude and sign, so 
the changes in sin are due to the changes in N P only. 

When = we have sin (9 = [Art. 28]. 
As increases from to 90, 

N P increases and is positive ; 
.'. sin increases and is positive. 
When = 90, sin = 1 [Art. 28]. 
As 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 increases from 180 to 270, 

NP increases and is negative; 
.'. sin increases and is negative. 

When 



= 270, 



OP 
sin0 = ---=-l. 



As 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 































/ 


^ 


x 


\ 




















/ 






\ 


















/ 










\ 
















/ 










\ 
















3 


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. 



42 



52 PRACTICAL TRIGONOMETRY 

NP 

30. To trace the changes in tan 0, we have tan , 
and both NP and ON change with 0. 

When = 0, tan = ~=0. 

As changes from to 90, 

NP increases and is positive, 

ON decreases and is positive; 

.'. tan increases and is positive. 

OP 
When = 90, tan0= =o>. 

As changes from 90 to 180, 

NP decreases and is positive, 
ON increases and is negative; 
.'. tan decreases and is negative. 

When (9=180, tan = ^=0. 

As changes from 180 to 270, 

NP increases and is negative, 
ON decreases and is negative; 
.*. tan increases and is positive. 

OP 

When 0-270, tan0 = = 00. 

As changes from 270 to 360, 

NP decreases and is negative, 

ON increases and is positive; 

.'. tan0 decreases and is negative. 

When = 360, 



ANGLES GREATER THAN A RIGHT ANGLE 



53 



The graph of tan is given below. Note that since 
tan (180 + 0)= tan 6, 

the curve for values of from to 180 is repeated for 
values of 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 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 
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 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 
c 2 = a 2 + b 2 - 2ab cos c. 

In Fig. 43, where C is acute, we have, by a well-known 
theorem in Geometry, 

AB 2 = BC 2 + CA 2 - 2BC . CD 

= BC 2 + CA 2 - 2BC . CA COS C ; 

i.e. c 2 = a? + b 2 -2abcosC. 

In Fig. 44, where C is obtuse, we have, by Geometry, 
AB 2 = BC 2 + CA 2 + 2BC . CD 

= BC 2 + CA 2 + 2BC . CA COS (180 - C) 
= BC 2 + CA 2 - 2BC . CA COS C, 

i.e. <? = a 2 + b 2 - 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 
= 180-6243 / 

=117 ir. 



94-16-36 
~24 = 



-g4583; 



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 5 l9 b 2 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% = c 2 a 2 . 

Also prove that the sum of the areas of the two triangles is 
c 2 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(c 2 -a 2 ) 

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 r ly r 2 respectively make 
angles of 6 l and $ 2 w ith a third line OX. Prove 

AB 2 = /y 5 + r 2 2 - 2?v' 2 cos (<9 2 - ^). 



6. In a triangle sm 2 C = siii 2 A-f siri 2 B. 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 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 and 30 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+tan 2 <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 = 90, then the projection of OP 

- OP cos 90 = 0. 

If 0-0 then the projection of OP 
= OP cos = 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. 



52 



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 
M 1 




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) = sin 2 A - sin 2 B. 

sin (A + B) sin (A B) = (sin A cos B -fcos A sin B) 

(sin A cos B cos A sin B 
= sin 2 A cos 2 B - cos 2 A sin 2 B 
= sm 2 A(l-sm 2 B)-(l-sm 2 A)sin 2 B 
= sin 2 A- sin 2 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) = cos 2 A - sin 2 B. 

9. Show that V 2 sin (A + 45) = sin A + cos A. 

10. Prove that S1P ( A A + ^ = 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, supposing 1 
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~m r 
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 = cos 2 A - sin 2 A (2). 

Writing 1 - cos 2 A for sin 2 A, we get 

cos2A = 2cos 2 A-l (3). 

Writing 1 siii 2 A for cos 2 A, we get 

cos2A = l- 2 sin 2 A (4). 

The results (3) and (4) may be written 

l + cos2A = 2cos 2 A ...(5), 

l-cos2A=2sin 2 A (6), 

and in this form are of much importance. 
From (5) and (6) we have 

1 - cos 2A 

- A = tan 2 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 - tan 2 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 = 2 sin - cos - ; 

2 2 

A A 

cos cos 2 - sin 2 - 
2 2 



.30 30 
sin 30 = 2 sin cos ; 

~> 2 

e 

2tan 4 
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 sin 2 -, 

. vers 6 1 cos B . 6 

we see, hav 6 = - = = sin 2 - . 

22 2 

Example (i). 
Prove 




A A 

We have cos A = cos 2 - sin 2 

2 2 

A 

9 " i 

= cos 2 ( 1 - 




76 PRACTICAL TRIGONOMETRY 

Example (ii). 

Prove sin 3A = 3 sin A 4 sin 3 A. 

We have 

sin3A = sin(2A + A) 

= sin 2 A cos A + cos 2 A sin A 

= 2 sin A cos 2 A + (1 2 sin 2 A) sin A 

= 2sinA(l-sin 2 A) + (l-2sin 2 A)sinA 

=3 sin A 4 sin 3 A. 

Exercise. 

Prove in a similar way that 



(1) sinA= ' ; (2) cos3A = 4cos 3 A-3cosA ; 



/ox o. OA 3 tan A -tan 3 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 sin 2 25. 

4. Prove that (sin 6 - cos (9) 2 = 1 - sin 2<9. 

5. Find tan 2A when tan A = -5. 

6. Factorise cos 4 A sin 4 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 sin 2 ^ . 

2s 

16. Find the positive values of A between and 360 which 
satisfy the equations 

(i) cos2A + sin 2 A=--=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 == - . 

a 

19. If cot = 1 ~ C ^ , prove that f =90 - 0. 

sm <p ' 2 

20. Express cos 2 a sin 2 /3 as half the sum of two cosines 
and hence evaluate cos 2 63 sin 2 47. 

21. If tan = - , simplify tan 20 + sec 20. 

Cd 

22. If cot 2 0- cot = 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 2rcos 2 -. 

2i 

Shew that this result is true whether 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 2rsin 2 - 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 sin 2 - . 

4 



78 PRACTICAL TRIGONOMETRY 

43. The formulae of Article 37 are useful for obtaining 
solutions of equations of the form 

a sin + b cos = 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 = 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 sin 2 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 = Jp 2 +q 2 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 
= 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 + cosl2 . 13. cos 18 + cos 43. 

14. Prove that 



,- 
_ -- _ = taii s- 2 - cot zr 

sin - sm (f) 2 2 

cos - cos 6 + <b , 

15. Prove --- ^ -?- = - tan TT tan 
cos + 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 ,. =cot72. 

cos 10 -f cos 26 

18. If 

x (sin 6 sin <) +y (cos </> cos ff) + cos 6 sin < - sin cos </> = 0, 

+ (t> . + 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 cosue 6 /l+e , u 

cos 6 = - ---- ~ , prove tan - = \/ - -- . tan - . 
1-ecos^' r 2 V i - e ' 3 



24. From the equations 

T! cos 6 + T 2 cos = 100, 
T! sin - T 2 sin </> = 0, 

100 sin (/> 100 sin B 

show that T! = -I 7^ ~r and T 2 = - ^ rr. 
, /a . ^\ sm(0-f <^>) 



OK ,. cos 40 -f cos 12 

25. Prove Tf ^ -- = tan 116 .cot 14 . 

cos 40 cos 12 



62 



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 tan 2 6 3 tan 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= 5 5 3 -. 
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 a 2 =b 2 + c 2 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 sin 2 - . 



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 

tan 2 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(a 2 -f& 2 ); (2) the difference of the squares 
on the diagonals is 4ab cos 6. 

13. Three lines OA, OB, OC of length r^ r 2 , r 3 are drawn 
making angles $!, $2? $3 with the horizontal through O, prove 
that the area of the triangle ABC is 

tf\ sin (B l - 



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 = e y , then y is the logarithm of x to the base e. 
This is written 



Example. 

Find logsx/27. 

Let x= Iog 3 \/27, 

then 3 a: = 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 a n = n log a. 

We have a n = (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. 

10 4 = 10,000, .'. log 10,000 = 4 ; 

10 3 =1,000, .'. log 1,000 -3; 

10 2 =100, /. log 100 = 2; 

10 1 - 10, .'. log 10 = 1 ; 

10 =1, .'. logl =0; 

10-^V =!, /. log'1 -1; 

10- a =Tfo ='01, .-. log -01 = -2; 
10- 8 =Tnft nr =-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 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 xlO 2 ; 
.'. log 261'3 = log 2-613 + log 10 2 

= 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 7 4 =2400, 11 2 = 120, 19 2 =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 






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. 








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 



.'. \og a n = log a b x 



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 91 7 . 

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 47rr 2 , and volume is f Trr 3 . 
(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) Iog 7 15914, (2) Iog 8 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 Iog 12 432, 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, R 2 =-00788, #=47 sees., L = 23 -3 cms., V = 102'5c.c. 

44. Find the ratio of L to I 2 , given r = A ^> where 

I 2 t 2 * 
#! = 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 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 cosec5623'' 

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 --^- . 

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 = K tana, 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 (a 2 +^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'. 



72 



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 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 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, r 2 =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 
5r 2 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, 

. 2V 2 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> = cot 2 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 






+ e 2 - 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 

, . 2 A 2(*-6)(s-c) 

1 cos A = 2 sin 2 -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 
In 2~ 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=9318'. 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 = 2 T 4596 

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, b 9 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 n o 13 ,. 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 



sin424(X 
.-. Ioga = logl89 + logsin72 14' - log sin 42 40' 

log 189=2-2765 

=2 -4242- Io g sin7214'=r9788 

2*2553 
.-. a=265-6. loggin 42 o 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 = 4630'. 

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=347. 

10. B = 122, a = 43-56, c = 5145. 

11. A = 57 14', B=8335', 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 + b cos == a ( sin + - cos } 
\ a J 

= a (sin + tan a cos 0), where tan a = - 
- (sin cos a + cos 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 + b cos into a form suitable 
for logarithmic work. 

57. Again the formula c 2 = a 2 + 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? + b 2 2ab cos C 

= a 2 + b 2 2ab ( 2 cos 2 - - 



>\ 

= (a, -f &) 2 4ab cos 2 - 



Now since 4a6 < (a + b) 2 and cos < 1, we can find an 
acute angle such that 

. , 2 Jab C 

sin = 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 = 5840 / , 

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; 
= 58 41'. 



c= 395 cos 58 41'; 
c = 2-3124; 
0=205-3. 



log 2 = 
log cos 29 20' =1-9405 
2 T 5282 

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 Va 2 -f 6 2 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 + 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 = 10616', 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 r z = s tan - , r% s tan - . 

Jt A. 



82 



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 

COS 2 ^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 = 3340', 

PRQ=9625'. 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= ' , 

bk 3 x ' 
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, 

4tt 2 7T 2 6 

ra=-84, TT = 3'142, Z = 11'8. 

10. Find the radius of a sphere of volume 320 c.c., given 
that volume=|7rr 3 . (*-3*142.) 



11. In a triangle ABC, BC = 93 yards, ABC = 5919', 
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/ 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 D l and D 2 , prove that the height of the 
cliff is 250 (tan DI+ tan D 2 ) 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 19 33 . 

Find the number of zeros following the decimal point in the 
value of ( T V)* 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 = 2 \/6c cos . 



THE SOLUTION OF TRIANGLES 119 

IV. An obtuse angled triangle has a = 15*3 cms., 6 = 97 cms., 
and B = 3145'. 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 / 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 - 2 T 2 "( = 3*143), 
which is correct to two places, will be more useful : 

ffl g ives 3*14159. 

In working examples TT is taken to be - 2 T 2 - 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 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 = ~ * x 4000 miles 
loO 

= 1230 miles approx. 
Or thus, from first principles 



_arc^B _17f 
2ir~x40bO "360 5 * 



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 
are in ascending order of magnitude. 



126 PRACTICAL TRIGONOMETRY 

Dividing by sin 0, we have 



1, -^ , sec0 
sin 

are in ascending order of magnitude. 

Now as approaches the value zero, sec0 approaches 

unity ; .'. since - ^ lies between 1 and sec 0, we have that 

A 

the limiting value of ^ , when = is 1 . 
sm0 

Using the notation of Art. 27, we have 

Lt J^ = l (1). 

0=0 sin 

Again, by dividing sin 0, 0, tan by tan 0, we have 

cos<? ' SE3' ! 

in ascending order of magnitude. 

And as approaches zero, cos approaches unity ; 

.% Lt~ = l (2). 

0=0 tan 

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 
= - r 2 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 r 2 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 = Trr 2 . 

70. Area of a sector of a circle. 

If a sector of a circle contain an angle of radians at 
the centre, since sectors are proportional to the angles they 
contain, we have : 

area of sector _ radians t 

area of circle ~ 2?r radians ' 

, Or 

. . area of sector = . -n-r 2 = . 

%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 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 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 10 6 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 10 6 
.= 856000 miles. 



x 92 x 10 6 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 OT 2 = 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, A 2 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 10 6 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 sin 2 - , 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 sec 2 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. 

92 



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=|7rr 3 , 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 X Q 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. 



lo S cos 40 = i' 8843 
Iogsinl5 = 



= cos 40 sin 15 ; 
logsm^l'2973; 

.'. = 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 

= = tan- 1 |=tan- 1 l-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 \ Jb 2 c 2 + c 2 a?+a 2 b*. 

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 

= 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. 

102 



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 = 3341', FAC = 73 35', FCA = 814 / . 
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 = 6110', Z_ACB = 5211', 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) pr n , where p = 9375, r=lO3, w = 4; 

(2) ^Trr 3 , 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 +y 2 - 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 + b sin 20 ; 
., , #2 + ^_ a 2_&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 yy 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 = 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 = 4124', B = 5547', C = 8249'. 7. 4 '86 ft., 1-55 ft. 

8. a = 7'41, B = 8049', C = 52ll'. 9. 81, 19. 
10. 8-83 ft. 11. 6-14. 12. 110 29'. 

14. B = 2750', C = 3710'. 

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., 371, 5-54 in. 8. 185-8 yds. 

10. 41-5 ft. 11. cos 6= 7= 12. 1026ft. 

Vtan 2 + 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. 8cos 4 a-8cos 2 a + 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 . 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 30 7 sin 4 30'. 




12. 


sin 41 + sin 78 = 2 sin 59 30' cos 18 30'. 




13. 


2 cos 30 30' cos 12 30 7 . 




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 - ^ sin 2 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 10 13 . 

34. -938. 35. 2'442, --511. 36. 2-254. 37. "09281. 
38. 305-5. 39. 33130. 40. 360-2. 41. -00005903. 
42. 8028 xlO 8 . 43. -01848. 44. -2384 

45. -0000003243. 46. 9888 x 10 s . 



VH d. p. 97. 



(1) -6029, (2) --3822, 



1-9219, 1-7112, -2614, 1-8611, '4453, 1-9224. 
(1) 1725', (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 - 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=423', C = 108l'. 

2. C = 7731', a=51-4, 6=77'2. 

3. A = 6121', a=25-2, c=197. 

4. A = 3326', B = 6510', c=474. 

5. A = lll24', B=226', a=36'55. 

6. B = 9913', C = 4423', 6=46'6. 
or B = 759', C = 13537', 6=6'56. 

7. B = 4052', C = 328', c=254. 

8. A = 78 48', B = 5310', C = 482'. 

9. C = 3538', a = 5'80, 6=3'93. 

10. A = 2622', C=3138', 6 = 83'18. 

11. C = 39ll', a=2663, c=2001. 

12. A = 44 49', B=60. 

13. B = 5456', C = 834', c = 209. 
or B = 1254', C = 1256', 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', = 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 a 2 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 10 10 . 

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 = 10616', C = 2036', 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. 56approx. 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. -4sin 2 ^. 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 
2lFebS8RS 



LD 21-100m-9,'48(B399sl6)476 



M30624S 

Q./4 53 



THE UNIVERSITY OF CALIFORNIA LIBRARY