UC-NRLF
$B 53D 7ET
LIBRARY
OF THE
University of California.
Mrs. SARAH P. WALSWORTH.
Recei^d October, i8g4.
z/Iccessions No.Q^y'/3*5~^ Class No,
3<
V'\ '■•••- \jM, • - •
"%
ELEMENTS
OF
GEOMETRY AND TRIGONOMETRY.
TRANSLATED FRO^ THE FRENCH OF
A. M. LE'GENDRE,
BY DAVID BREWSTER, LL. D.
f
REVISED AND ADAPTED TO THE COURSE OF. MATHEMATICAL INSTRUCTION*
IN THE UNITED STATES,
BY CHARLES DAVIES,
AUTHOR OF MENTAL AND PRACTICAL ARITHMETIC, ELEMENTS OF SURVEYING,
ELEMENTS OP DESCRIPTIVE AND OF ANALYTICAL GEOMETRY,
ELEMENTS OF DIFFERENTIAL AND INTEGRAL CALCULUS,
AND SHADES SHADOWS AND PERSPECTIVE.
PHILADELPHIA :
PUBLISHED BY A.S.BARNES AND CO.
21 Minor-street.
DAVIES' ^1i^
COpRSE OF MATHEMATICS.
DA VIES' FIRST LESSONS IN ARITHMETIC,
DESIGNED FOR BEGINNERS.
DA VIES' ARITHMETIC,
DESIGNED FOR THE USE OF ACADEMIES AND SCHOOCS.
KEY TO DAVIES' ARITHMETIC.
DA VIES' FIRST LESSONS IN ALGEBRA;
Being an introduction to the Science, and forming a connecting link between
Arithmetic and Algebra.
DAVIES' ELEMENTS OF GEOMETRY.
This work embraces the elementary principles of Geometry. The reasoning is plain
and concise, but at the same time strictly rigorous.
DAVIES' PRACTICAL GEOMETRY,
Embracing the facts of Geometry, with applications in Artificer's Work,
Mensuration and Mechanical Philosophy.
DAVIES' BOURDON'S ALGEBRA,
Bemg an abridgment of the work of M. Bourdon, with the addition of practical
examples.
DAVIES' LEGENDRE'S GEOMETRY and TRIGONOMETRY,
Being an abridgment of the work of M. Legendre, with the addition of a Treatise
on Mensuration of Planes and Solids, and a Table of Logarithms and
Logarithmic Sines.
DAVIES' SURVEYING,
With a description and plates of, the Theodolite,- Compass, Plane-Table, and
Level — also, Maps of the Topographical Signs adopted by the Engineer
Department — an explanation of the method of surveying
the Public Lands, and an Elementary Treatise on
Navigation.
DAVIES' ANALYTICAL GEOMETRY,
Embracing the Equations of the Point and Straight Line — of the Conic
Sections — of the Line and Pla^e in Space — also, the discussion of the
General Equation of the second degree, and of Surfaces
OF the second order.
DAVIES' DESCRIPTIVE GEOMETRY,
With its application to Spherical Projections.
DAVIES' SHADOWS and LINEAR PERSPEPTIVE.
DAVIES' DIFFERENTIAL and INTEGRAL CALCULUS.
Entered according to the Act of Congress, in the year 1834,
By CHARLES DAVIES,
in the Clerk's Office of the District Court of the United States, for the Southern District of New York.
PREFACE
TO THE AMERICAN EDITION.
The Editor, in offering to the public Dr. Brewster'a
translation of Legendre's Geometry under its present
^ form, is fully impressed with the responsibility he
assumes in making alterations in a work of such de-
served celebrity.
In the original work, as well as* in the translations
' of Dr. Brewster and Pr^essor Farrar, the proposi-
tions are not enunciated in general terms, but with
reference to, and by the aid of, the particular diagrams
used for the demonstrations. It is believed that this
departure from the method of Euclid has been gene-
rally regretted. The propositions of Geometry are
general truths, and as such, should be stated in gene-
ral terms, and without reference to particular figures.
The method of enunciating them by the aid of particu-
lar diagrams seems to have been adopted to avoid the
difficulty which beginners experience in comprehend-
ing abstract propositions. But in avoiding this diffi-
culty, and thus lessening, at first, the intellectual
labour, the faculty of abstraction, which it is one of
the primary objects of the study of Geometry to
strengthen, remains, to a certain extent, unimproved.
iv . PREFACE. ,
Besides the alterations in the enunciation of the
propositions, others of considerable importance have
also been made in the present edition. The propo-
sition in Book V., which proves that a polygon and '#
circle may be made to coincide so nearly, as to differ
from each other by less than any assignable quantity,
has been taken from the Edinburgh Encyclopedia.
It is proved in the corollaries that a polygon of ap •*.
infinite number of sides becomes a circle, and this
prmciple is made the basis of several important de-
monstrations in Book VIIL 4-
* •>♦•'
Book Il.jOn Ratios and Proportions, has been partly
adopted from the Encyclopedia Metropolitana, and
will, it is believed, supply a deficiency in the original
work.
Very considerable alterations have also been made
in the manner of treating the subjects of Plane and
Spherical Trigonometry. It has also been thought
best to publish with the present edition a table of
logarithms and logarithmic sines, and to apply the
principles of geometry to the mensuration of sur-
faces and solids.
Military Academy,
West Point, March, 1834.
1
CONTENTS
The principles,
Ratios and Proportions,
BOOK I.
BOOK II.
BOOK III.
The Circle and the Measurement of Angles,
Problems relating to the First and Third Books,
BOOK IV.
The Proportions of Figures and the Measurement of Areas,
Problems relating to the Fourth Book,
BOOK V.
Regular Polygons and the Measurement of the Circle,
BOOK VI.
Planes and Solid Angles, - -
BOOK VII.
Polyedrons,
BOOK vm.
The three round bodies, - . - - -
BOOK IX.
Of Spherical Triangles and Spherical Polygons, -
APPENDIX.
The regular Polyedrons, - . -
84 ]
41
67
68
98
10?
126
142
166
186
209
vi CONTENTS.
PLANE TRIGONOMETRY,
Division of the Circumference, - - - - - 207
General Ideas relating to the Trigonometrical Lines, - - 208
Theorems and Formulas relating to the Sines, Cosines, Tan-
gents, &c. - . - ' ' - - 215
Construction and Description of the Tables, - - - 223
Description of Table of Logarithms, - - - - - 224
Description of Table of Logarithmic Sines, ... 228
Principles for the Solution of Rectilineal Triangles, - - 231
Solution of Rectilineal Triangles by Logarithms, - - 235
Solution of Right angled Triangles, ... - 237
Solution of Triangles in general, 238
SPHERICAL TRIGONOMETRY.
First principles, -------- 246
Napier's Circular Parts, 252
Solution of Right angled Spherical Triangles by Logarithms, 255
Qiiadrantal Triangles, ------- 257
Solution of Oblique angled Triangles by Logaritlims, - - 250
MENSURATION.
Mensuration of Surfaces, - - - - - - 274
Mensuration of Solids, ------- 285
Alf INDEX
SHOWING THE PROPOSITIONS OF LEGENDRE WHICH CORRESPOND TO
THE PRINCIPAL PROPOSITIONS OF THE FIRST SIX BOOKS OF EUCLID.
Euclid.
Legendre.
Euclid.
Legendre.
Euclid.
Legendre. 1
Book I.
Book I.
Cor. 2. of 3*2
Prop. 27
Prop. 26
Prop. 15i
33
34
30
28
28
29
5:
5
Prop. 4
Prop. 5
5
Cor. of 5
11
Cor. of n
Book IV.
S Cor. 2. >iQ
} ^3. r
6
12
35
1
31
8
13
10
1
36
37
1
Cor. 2. of 2
Book IV.
14
3
38
Cor. 2. of 2
35
28
15
4
4
2
36
30'
Cor. 1.5, 5
& 2. ) ^^
16
17
Sch. of 4
5 Cor. of 25
) 25
47
11
8
5 Cor. 1. of 4
\ Cor. of 6
Book VI.
Book II.
1
4
18
13
12
13
19
20
21
24
13
13
12
2
5 15!
16l
171
18|
7
8
9
Book 111.
Book III.
3
4
Prop. 3
Prop. 6
25
9
10
Cor. of 7
5
19
26
6
11
Cor. of 14
6
20!
27
Cor. 1. of 19
12
Cor. of 14
8
22
28
Cor. 2. of 19
14
8
14
) 25
I Cor. of 15
29
\ Cor. 2. &
I 4. of 20
15
2
15
18
9
19
25!
30
22
20
18
S 26!
I 27|
'Cor.l.of32
26
21
Cor. of 18
20
i
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ELEMENTS OF GEOMETRY.
BOOK I.
THE I^INCIPLES.
Definitions.
I. Geometry is the science which has for it| object the
ij^'isurement of extension.
^^Ixtension has tiiree dimensions, length, breadth, and height,
^v thickness.
*3. A line i§ length without breadth, or thickness. ^
The extremities of a line arc called points : a point,. there-
fyre, has neither length, breadth, nor thickness, but position
only.
3. A straight line is the shortest distance from one point to
another.
4. Every line which is not straight, or composed of straight
lines, is a curved line.
Thus, AB is a straight line ; ACDB is a
broken line, or one composed o| straight A.<
lines ; and AEB is a curved line.
The word line, when used alone, will designate a scraight
line ; and the word curve, a curved line.
5. A surface is that which has length and breadth, without
height or thickness.
0. A plane is a surface, in which, if two points be assumed
at pleasure, and connected by a straight line, that line will lie
wholly in the surface.
7. Every surface, which is not a plane surface, or composed
of plane surfaces, is a curved surface.
8. A solid or body is that which has length, breadth, and
thickness ; and therefore combines the three dimensions of
extension.
10
GEOMETRY
9. When two straight lines, AB, AC, meet
each other, their inclination or opening is call-
ed an angle^ which is greater or less as the
lines are more or less inchned or opened. The
point of intersection A is the vertex of the ^,
angle, and the lines AB, AC, are its sides.
The angle is sometimes designated simply by the letter at
the vertex A ;• sometimes by the three letters BAC, or CAB,
the letter at the vertex being always placed in the middle.
Angles, like all other quantities, are susceptible of addition,
subtraction, multipllGation, and division.
Thus the angle DCE is the sum of
the two angles DCB, BCE ; and the an-
gle DCB is the difference of the two £^
angles DCE, BCE.
10. When a straight line AB meets another
straight. line CD, so as to make the adjacent
angles BAC, BAD, equal to each other, each
of these angles is called a right angle ; and the
line AB is said to be peiyendicular to CD. ;
A
D
11. Every angle BAC, less than a^>
right angle, is an acute angle ; and
every angle DEF, greater than a right
angle, is an obtuse angle.
-r
12. Two lines are said to hQ parallel, when —
being situated in the same plane, they cannot
meet, how far soever, either way, both of them
be produced.
13. A "plane figure is a plane terminated on
all sides by lines, either straight or curved.
If the lines are straight, the space they enclose
is called a rectilineal figure, or polygon, and the
lines themselves, taken together, form the contour,
or perimeter of the polygon.
14. The polygon of three sides, the simplest of all, is called
a triangle ; that of four sides, a quadrilateral; that of five, a
pentagon; that of six, a hexagon ; that of seven, a heptagon;
that of eight, an octagon ; that of nine, a nonagon; that of ten, a
decagon ; and that of twelve, a dodecagon.
^M::
BOOK I.
11
15. An equilateral iriaugle is one 'which has its three sides
equal ;'an isosceles triangle, one which h^s two of its sides
equal ; a scalene triangle, one which has its three sides unequal
16. A right-angled triangle is one which
has a right angle. The side opposite the
right angle is called the liypothenUse. Thus,
in the triangle ABC, right-angled at A, the
side BC is the hypothenuse.
17. Among the quadrilaterals, we distinguish :
The square, which has its sides equal, and its an-
gles right-angles.
The rectangle, which has its angles right an-
gles,- without having its sides equal.
The parallelogram, or rhomboid, which
has its opposite sides parallel. /
The rhombus, or lozenge, which has its sides equal,
without having its angles right angles.
And lastly, the trapezoid, only two of whose sides
are parallel.
18. A diagonal is a line which joins the ver-
tices of two angles not adjacent to each other.
Thus, AF, AE, AD, AC, are diagonals. ^^
19. An equilateral polygon is one which has all its sides
equal ; an equiangular polygon, one which has all its angles
equal.
20. Two polygons are mutually equiloieral, when they have
their sides equal each to each, an.d placed in the same order ;
'^^ Oft xm ■
12 GEOMETRY.
that is to say, when following their perimeters in the same di-
rection, the first side of the one is equal to the first side of the
other, the second of the one to the second of the other, the
third to the third, and so on. The phrase, mutually equian-
gular, has a corresponding signification, with respect to the
angles.
In both cases, the equal sides, or the equal angles, are named
homologous sides or angles.
Definitions of terms employed in Geometry,
An axiom is a self-evident proposition.
A theorem is a truth, which becomes evident by means of a
train of reasoning called a demonstration.
A problem is a question proposed, which requires a solu-
tion,
A lemma is a subsidiary truth, employed for the demonstra-
tion of a theorem, or the solution of a problem.
The common name, proposition, is applied indifferently, to
theorems, problems, and lemmas.
A corollary is an obvious consequence, deduced from one or
several propositions.
A scholium is a remark on one or several preceding propo-'
sitions, which tends to point out their connexion, their use, their
restriction, or their extension.
A hypothesis is a supposition, made either in the enunciation
of a proposition, or in the course of a demonstration.
Explanation of the symbols to be employed.
The sign = is the sign of equality; thus, the expression
A=B, signifies that A is equal to B.
To signify that A is smaller than B, the expression A<B
is used.
To signify that A is greater than B, the exp#i'ession,A>B
is used ; the smaller quantity being always at the vertex of the
angle.
The sign + is called plus : it indicates addition.
The sign — is called jninus : it indicates subtraction.
Thus, A + B, represents the sum of the quantities A and B;
A — B represents their diiference, or what remains after B is
taken from A ; and A — B + C, or A + C — B, signifies that A
and C are to be added together, and that B is to be subtracted
from their «jum.
BOOK I. 13
The sign x indicates multiplication : thus, A x B represents
tlie product of A and B. Instead of the sign x , a point is
Bometimes employed ; thus, A.B is the same thing as A x B.
The same product is also designated without any intermediate
«ign, by AB ; but this expression should not be employed, when
there is any danger of confounding it with that of the line AB,
which expresses the distance between the points A and B.
The expression A x (B + C — D) represents the product of
A by the quantity B + C — D. If A + B were to be multiplied
by A — B + C, the product would be indicated thus, (A + B)x
(A — B + C), whatever is enclosed within the curved lines, being
considered as a single quantity.
A number placed before a hne, or a quantity, serves as a
multiplier to that line or quantity ; thus, 3AB signifies that
the line AB is taken three times ; i A signifies the half of the
angle A.
The square of the line AB is designated by AB^ ; its cube
by AB^ What is meant by the square and cube of a line, will
be explained in its proper place.
The sign V indicates a root to be extracted ; thus -^2
means the square-root of 2 ; \/ A x B means the square-root of
the product of A and B.
Axioms.
1. Things which are equal to the same thing, are equal to
each other.
2. If equals be added to equals, the wholes will be equal.
3. If equals be taken from equals, the remainders will be
equal.
4. If equals be added to unequals, the wholes will be un-
equal.
5. If equals be taken from unequals, the remainders will be
unequal.
6. Things which are double of the same thing, are equal to
each other.
7. Things which are halves of the same thing, are equal to
each other.
8. The whole is greater than any of its parts.
9. The whole is equal to the sum of all its parts.
10. All right angles are equal to each other.
1 1 From one point to another only one straight line can be
drawn.
12. Through the same point, only one straight line can be
drawn Which shall be parallel to a given line.
13. Magnitudes, which being applied to each other, coincide
throughout their whole extent, are equal.
B
14
GEOMETRY.
PROPOSITION I. THEOREM.
Ij one straight line meet another straight line, the sum of the
two adjacent angles will he equal to two right angles.
Let the straight line DC meet the straight
line AB at C, then will the angle ACD +
the angle DCB, be equal to two right angles.
At the point C, erect CE perpendicular to
AB. The angle ACD is the sum of the an-v p ^
gles ACE, ECD: therefore ACD + DCB is ^ "^
the sum of the three angles ACE, ECD, DCB : but the first
of these three angles is a right angle, and the other two
make up the right angle ECB ; hence, the sum of the two an-
gles ACD and DCB, is equal to two right angles.
Cor. 1. If one of the angles ACD, DCB, is a right angle,
the other must be a right angle also.
Cor. 2. If the line DE is perpendicular
to AB, reciprocally, AB will be perpendicu-
lar to DE.
For, since DE is perpendicular to AB, the
angle ACD must be equal to its adjacent an-
gle DCB, and both of them must be right
angles (Def. 10.). But since ACD is a
right angle, its adjacent angle ACE must also be a right angle
(Cor. 1.). Hence the angle ACD is equal to the angle ACE,
(Ax. 10.) : therefore AB is perpendicular to DE.
Cor. 3. The sum of all the successive
angles, BAC, CAD, DAE, EAF, formed
on the same side of the straight line BF,
is equal to two right angles. ; for their sum
is equal to that of the two adjacent an-
gles, BAC, CAR
D
33
PROPOSITION II. THEOREM.
2\vo straight lines, which have two points common, coincide with
each other throughout their whole extent, and form one ana
the same straight line.
Let A and B be the two common
points. In the first place it is evident
that the two lines must coincide entirely
between A and B, for otherwise there
would be two straight lines between A t"— ng"
and B, which is impossible (Ax.4 1 ) . Sup-
BOOK I. 15
pose, however, that on being produced, these' lines begin to
separate at C, the one becoming CD, the other CE. From
the point C draw the hne CF, making with AC the right angle
ACF. Now, since ACD is a straight line, the angle FCD will
be a right angle (Prop. I. Cor. 1.); and since ACE is a straight
line, the angle FCE will likewise be a right angle. Hence, the
angle FCD is equal to the angel FCE (Ax. 10.); which can
only be the case when the lines CD and CE coincide : there-
fore, the straight lines which have two points A and B com-
mon, cannot separate at any point, when produced ; hence they
form one and the same straight line.
PROPOSITION III. THEOREM.
If a straight line meet two other straight lines at a common
point, making the sum of the two adjacent angles equal to two
right angles, the two straight lines which are met, will form
one and the same straight line.
Let the straight line CD meet the
two lines AC, CB, at their common
point C, making the sum of iht; two
adjacent angles DCA, DCB, equal to j^
two right angles ; then will CB be the
prolongation of AC, or AC and CB
will form one and the same straight line.
For, if CB is not the prolongation of AC, let CE be that pro-
longation: then the line ACE being straight, the sum of the
angles ACD, DCE, will be equal to two right angles (Prop. I.).
But by hypothesis, the sum of the angles ACD, DCB, is also
equal to two right angles : therefore, ACD + DCE nmst be equal
to ACD + DCB ; and taking away the angle ACO from each,
there remains the angle DCE equal to the angle "DCB, which
can only be the case when the lines CE and C !> coincide ;
hence, AC, CB, form one and the same straight li. ;.
PROPOSITION IV. THEOREM.
When two straight lines intersect each other, the ojrp 'I'e or Dcr-
tical angles, which they foam, are equciL
«♦
16
GEOMETRY.
Let AB and DE be two straight^
lines, intersecting each other at C ;
then will the angle ECB be equal to
the angle ACD, and the angle ACE to
the angle DCB. T) B"
For, since the straight line DE is met by the straight line
AC, the sum of the angles ACE, ACD, is equal to two right
angles (Prop. L) ; and since the straight line AB, is met by the
straight line EC, the sum of the angles ACE and ECB, is equal
to two right angles : hence the sum ACE + ACD is equal to
the sum ACE + ECB (Ax. 1.). Take away from both, the com-
mon angle ACE, there remains the angle ACD, equal to its
opposite or vertical angle ECB (Ax. 3.).
Scholium. The four angles formed about a point by two
straight lines, which intersect each other, are together equal to
four right angles : for the sum of the two angles ACE, ECB,
is equal to two right angles ; and the sum of the other two,
ACD, DCB, is also equal to two right angles : therefore, the
sum of the four is equal to four right angles.
In general, if any number of straight lines
CA, CB, CD, &c. meet in a point c, the >s,
sum of all the successive angles ACB,BCD,
DCE, ECF, FCA, will be equal to four
right angles : for, if four right angles were
formed about the point C, by two lines per-
pendicular to each other, the same space
would be occupied by the four right angles, as by the succes-
sive angles ACB, BCD, DCE, ECF, FCA.
PROPOSITION V. TliEORlIM.
If two triangles have two sides and the included angle of the one,
equal to two sides and the included angle of the other, each to
each, the two triangles will be equal
Let the side ED be equal
to the side BA, the side DF
to the side AC, and the an-
gle D to the angle A ; then
will the triangle EDF be
equal to the triangle BAC. ^^ . — -= — ^
For, these triangles may be so applied to each other, that they
shall exactly coincide. Let the triangle EDF, be placed upon
the triangle BAC, so that the point E shall fall upon B, and the
side ED on the equal side BA; then, since the angle D is equal
to the angle A, the side DF will take the direction AC. But
%
BOOK I. 1^
DF is equal to AC ; therefore, the point F will fall on C, and
the third side EF, will coincide with the third side BC (Ax. 11.):
therefore, the triangle EDF is equal to the triangle BAG
(Ax. 13.).
Cor, When two triangles have these three things equals
namely, the side ED=BA, the side DF=AC, and the angle
D=A, the remaining three are also respectively equal, namely,
the side EF=BC, the angle E=B, and the angle F=C
PROPOSITION VI. THEOREM.
If two triangles have two angles and the included side of the one,
equal to two angles and the included side of the other, each to
each, the two triangles will be equal.
Let the angle E be equal
to the angle B, the angle F
to the angle C, and the in-
cluded side EF to the in-
cluded side BC ; then will
the triangle EDF be equal
to the triangle BAC. ^ ^^ C
For to apply the one to the other, let the side EF be placed
on its equal BC, the point E falHng on B, and the point F on
C ; then, since the angle E is equal to the angle B, the side ED
will take the direction BA ; and hence the point D will be found
somewhere in the hne BA. In like manner, since the angle
F is equal to the angle C, the line FD will take the direction
CA, and the point D will be found somewhere in the line CA.
Hence, the point D, falhng at the same time in the two straight
lines BA and CA, must fall at their intersection A: hence, the
two triangles EDF, BAC, coincide with each other, and are
therefore equal (Ax. 13.).
Cor. Whenever, in two triangles, these three things are equal,
namely, the angle E=B, the angle F=C, and the included side
EF equal to the included side BC, it may be inferred that the
remaining three are also respectively equal, namely, the angle
D=A, the side ED=BA, and the side DF=AC.
Scholium, Two triangles are said to be €qual, when being
applied to each other, they will exactly coincide (Ax. 13.).
Hence, equal triangles have their like parts equal, each to each,
since those parts must coincide with each other. The converse
of this proposition is also true, namely, that two triangles which
have all the parts of the one equal to the parts of the other, each
B*
18
GEOMETRY.
to each, are equal ; for they may be applied to each other, and
the equal parts will mutually coincide.
PROPOSITION VII. THEOREM.
The sum of any two sides of a triangle, is greater than the
third side.
Let ABC be a triangle : then will the
sum of two of its sides, as AC, CB, be
greater than the third side AB.
For the straight line AB is the short-
est distance between the points A and
B (Def. 3.) ; hence AC + CB is greater
than AB,
PROPOSITION VIII. THEOREM.
If from any point within a triangle, two straight lines he drawn
to the extremities of either side, their sum will he less than the
sum of the two other sides of the triangle.
Let any point, as O, be taken within the trian-
gle BAC, and let the lines OB, OC, be drawn
to the extremities of either side, as BC ; then
willOB + OC<BA+AC.
Let BO be produced till it meets the side AC
in D : then the line OC is shorter than OD + DC^
(Prop. VII.): add BO to each, and we have BO + OC<BO+
OD + DC (Ax. 4.), or BO + OC<BD + DC.
Again, BD<BA+ AD: add DC to each, and we have BD +
DC<BA + AC. But it has just been found that BO + OC<
BD + DC ; therefore, still more is BO + OC<BA+AC.
*
PROPOSITION IX. THEOREM.
if two triangles have two sides of the one equal to two sides of the
other, each to each, and the included angles unequal, the third
sides*will he unequal ; and the greater side will belong to the
triangle which has the greater included angle.
Let BAC and EDF
bd two triangles, having
the sideAB=DE, AC
=DF, and the angle
A>D; then will BC>
EF.
Make the angle C AG^
= D; take AG=:DE,
and draw CG. The
BOOK I.
19
triangle GAC is equal to DEF, since, by construction, they
have an equal angle in each, contained by equal sides, (Prop.
V.) ; therefore CG is equal to EF. Now, there may be three
cases in the proposition, according as the point G falls without
the triangle ABC, or upon its base BC, or within it.
First Case. The straight line GC<GI + IC, and the straight
line AB<AI + IB; therefore, GC + AB< GI + AI + IC + IB,
or, which is the same thing, GC + AB<AG+BC. Take away
AB from the one side, and its equal AG from the other ; and
there remains GC<BC (Ax. 5.) ; but we have found GC=EF,
therefore, BC>EF.
Second Case, If the point G
fail on the side BC, it is evident
that GC, or its equal EF, will be
shorter than BC (Ax. 8.).
Third Case, Lastly, if the point G
fall within the triangle BAC, we shall
have, by the preceding theorem, AGj^l*
GC<AB + BC; and, taking AG from
tlie one, and its equal AB from the other,
ihere will remain GC<BCorBC>EF.B
Scholium. Conversely, if two sides
BA, AC, of the triangle BAC, are equal
to the two ED, DF,of the triangle EDF,
each to each, while the third side BC of
the first triangle is greater than the third
side EF of the second ; then will the an-
gle BAC of the first triangle, be greater
than the angle EDt^ of the second.
For, if not, the angle BAC must be equal to EDF, or less
than it. In the first case, the side BC would be equal to EF,
(Prop. V. Cor.) ; in the second, CB would be less than EF ; but
either of these results contradicts the hypothesis ; therefore, BAC
is greater than EDF.
PROPOSITION X. THEOREM.
If two triangles have the three sides of the one equal to the three
sides of the other, each to each, the three angles will also he
equal, each to each, and the triangles themselves will be equal.
m-
20 GEOMETRY
Let the side ED=BA,
ihe side EF=BC, and the
side DF=AC ; then will
the angle D=A, the angle
E=B, and the angle F
= C. E TB C
For, if the angle D were greater than A, while the sides
ED, DF, were equal to BA, AC, each to each, it would fol-
low, by the last proposition, that the side EF must be greater
than BC ; and if the angle D were less than A, it would follow,
that the side EF must be less than BC : but EF is equal to BC,
by hypothesis ; therefore, the angle D can neither be greater
nor less than A ; therefore it must be equal to it. In the same
manner it may be shown that the angle E is equal to B, and
the angle F to C : hence the two triangles are equal (Prop.
VI. Sch.).
Scholium. It may be observed that the e(|Ual angles lie op-
posite the equal sides : thus, the equal angles D and A, lie op-
posite the equal sides EF and BC.
PROPOSITION XI. THEOREM.
In an isosceles triangle, the angles opposite the equal sides
are equal.
Let the side BA be equal to the side AC ; then
will the angle C be equal to the angle B.
For, joife the vertex A, and D the middle point
of the base BC. Then, the triangles BAD, DAC,
will have all the sides of the one equal to those
of the other, each to each ; for BA is equal to AC,]^
by hypothesis ; AD is common, and BD is equal
to DC by construction : therefore, by the last proposition, the
angle B is equal to the angle C.
Cor. An equilateral triangle is likewise equiangular, that is
to say, has all its angles equal.
Scholium. The equality of the triangles BAD, DAC, proves
also that the angle BAD, is equal to DAC, and BDA to ADC,
hence the latter two are right angles ; therefore, the line drawn
from the vertex of an isosceles triangle to the middle point of its
base, is perpendicular to the base, and divides tlie angle at the
vertex into two equal parts.
In a triangle which is not isosceles, any side may be assumed
indifferently as the base ; and the vertex is, in that case, the
vertex of the opposite angle. In an isosceles triangle, however
BOOK I. 21
(■•
that side is generally assumed as the base, which is not equnl
to either of the other two.
PROPOSITION XII. THEOREM.
Conversely, if two angles of a triangle are equal, the sides oppo-
site them are also equal, and the triangle is isosceles.
Let the angle ABC be equal to the angle ACB ;
then will the side AC be equal to the side AB.
For, if these sides are not equal, suppose AB
to be the greater. Then, take BD equal to AC,
and draw CD. Now, in the two triangles BDC,
BAC, we have BD=:AC, by construction; the
angle B equal to the^ngle ACB, by hypothesis ;b^
and the /side BC common : therefore, the two
txidriglefe, BDC, BAC, have two sides and the included angle in
the on^, equal to two sides and the included angle in the other,
each to each : hence they are equal (Prop. V.). i But the part
«annot be equal to the whole (Ax. 8.) ; hencej'^rliej'e is^no
'nequahty ijutw^en thq sides BA, AC ; therefore, the triangle
C is isosceles.
..^
PROPOSITION XIII. THEOREM.
■g^y^The greater side of every triangle is opposite to the greater an-
' gle ; and conversely, the greater angle is opposite to the
greater side.
First, Let the angle C be greater than the angle
B ; then will the side AB, opposite C, be greater
than AC, opposite B.
For, make the angle BCD=B. Then, in the
triangle CDB, we shall haveCD-BD (Prop.XIL).
Now, the side AC < AD + CD; butAD+CD=C'
AD + DB=AB : therefore AC< AB.
Secondly, Suppose the side AB>AC; then will the angle C,
opposite to AB, be greater than the angle B, opposite to AC.
For, if the angle C<B, it follows, from what has just been
proved, that AB<AC; which is contrary to the hypothesis. It
the angle C=B, then the side AB=AC (Prop. XJI.); which is
also contrary to the supposition. Therefore, when AB>AC,
the angle C must be greater than B.
22 GEOMETRY.
PROPOSITION XIV. THEOREM.
From a given point, without a straight line, only one perpendicu-
lar can he drawn to that line.
Let A be the point, and DE the given
line.
Let us suppose that we can draw two
perpendiculars, AB, AC. Produce either
of them, as AB, till BF is equal to AB, and D-
draw FC. Then, the two triangles CAB,
CBF, will be equal: for, the angles CBA,
and CBF are right angles, the side CB is Np
common, and the side AB equal to BF, by construction ; there-
fore, the triangles are equal, and the angle ACB=:BCF (Prop.
V. Cor.). But the angle ACB is a right angle, by hypothesis ;
therefore, BCF must likewise be a right angle. But if the adja-
cent angles BCA, BCF, are together equal to two right angles,
ACF must be* a straight line (Prop. IIL) : from whence it fol-
lows, that^tween the same two points, A and F, two straight
lines can be drawn, which is impossible (Ak. 1 1,).- iience, two
perpendiculars cannot be drawn from the same point to the^^.
same straight line.
Scholium, At a given point C, in the line j; i
AB, it is equally impossible to erect two per-
pendiculars to that line. For, if CD, CE,
were those two perpendiculars, the angles
BCD, BCE, would both be right angles:—
hence they would be equal (Ax. 10.); and
the line CD would coincide withCE; otherwise, a part would
be equal to the whole, which is impossible (Ax. 8.).
PROPOSITION XV. THEOREM.
If from a point without a straight line, a perpendicular he let
fall on the line, and ohlique lines he drawn to different points :
1st, The perpendicular will he shorter than any ohlique line,
2d, Any two ohlique lines, drawn on different sides of the perpen-
dicular, cutting off equal distances on the other line, will he
equal,
Sd, Of two ohlique lines, drawn at pleasure, that which is farther
from the perpendicular will he the longer.
BOOK I. 23
Let A be the given point, DE the given
line, AB the perpendicular, and AD, AC,
AE, the oblique lines.
Produce the perpendicular AB till BF
is equal to AB, and draw FC, FD. D^
First. The triangle BCF, is equal to the
triangle BCA, for tliey have the right angle
CBF=CBA, the side CB common, and the ^F
side BF=BA ; hence the third sides, CF and CA are equal
(Prop. V. Cor.). But ABF, being a straight line, is shorter than
ACF, which is a broken line (Def. 3.) ; therefore, AB, the half
of ABF, is shorter than AC, the half of ACF ; hence, the per-
pendicular is shorter than any oblique line.
Secondly. Let us suppose BC=BE; then will the triangle
CAB be equal to the the triangle BAE ; for BC=BE,the side
AB is common, and the angle CBA=ABE ; hence the sides
AC and AE are equal (Prop. V. Cor.) : therefore, two oblique,
lines, equally distant from the perpendicular, are equal.
Thirdly. In the triangle DFA, the sum of the lines AC, CF,
is less than the sum of the sides AD, DF (Prop. VIIL) ; there-
fore, AC, the half of the line ACF, is shorter than AD, the half
of the line ADF ; therefore, the oblique line, which is farther
from the perpendicular, is longer than the one which is nearer.
Cor. L Th-e perpendicular measures the shortest distance
of a point from a line.
Cor. 2. From the same point to the same straight line, only
two equal straight lines can be drawn ; for, if there could be
more, we should have at least two equal oblique lines on the
same side of the perpendicular, which is impossible. «
PROPOSITION XVI. THEOREM.
F fj'om the middle point of a straight line, a .perpendicular he
drawn to this line ;
ist, Every point of th& perpendicular will he equally distant
from the extremities of the line.
2dy Every point, without the perpendicular, will he unequally dis'
tant from those extremities.
%
%.
24
GEOMETRY.
Let AB be the given straight line, C the
middle point, and ECF the perpendicular.
First, Since AC=CB, the two oblique lines
AD, DB, are equally distant from the perpen-
dicular, and therefore equal (Prop. XV.). So,
(ike wise, are the two oblique hues AE, EB, the 7^^
two AF, FB, and so on. Therefore every point
m the perpendicular is equally distant from the
extremities A and B.
Secondly, Let I be a point out of the perpen-
dicular. If lA and IB be drawn, one of these lines will cut
the perpendicular in D; from w^hich, drawing DB, we shall
have DB=DA. But the straight line IB is less than ID+DB,
and ID + DB^ID + DA-IA; therefore, IB<IA; therefore,
every point out of the perpendicular, is unequally distant from
the extremities A and B.
Cor. If a straight line have two points D and F, equally dis-
tant from the extremities A and B, it will be perpendicular to
AB at the middle point C.
PROPOSITION XVII. THEOREM.
If two right angled triangles have the hypothenuse and a side oj
the one, equal to the hypothenuse and a side of the other, each to
each, the remaining parts will also he equal, each to each, and
the triangles themselves will ^^ equal.
In the two right angled *
triangles BAG, EDF, let the
hypothenuse AG = DF, and
the sideBA=ED: then will
the side BC=EF, the angle ^
A=D, and the angle C=F.
If the side BC is equal to EF, the like angles of the two
triangles are equal (Prop. X.). Now, if it be possible, suppose
these two sides to be unequal, and that BC is the greater.
On BC take BG=:EF, and draw AG. Then, in the two
triangles BAG, DEF, the angles B and E are equal, being right
angles, the side BA=ED by hypothesis, and the side BG=EF
by construction : consequently, AG =DF (Prop. V. Cor.). But,
by hypothesis AC=DF; and therefore, AC=AG (Ax. 1.).
But the oblique line AC cannot be equal to AG, which lies
nearer the perpendicular AB (Prop. XV.) ; therefore, BC and
EF caDnot be unequal, and hence the angle A=D, and the
angle C=F; and I'^ereforc, the triangles are equal (Prop. VI.
Sch.).
BOOK I. • 25
PROPOSITION XVIIl. THEOREM.
If two straight lines are perpendicular to a third line, they will
be parallel to each other : in other words, they will never meet,
how far soever either way, both of them be produced.
Let the two lines AC, BD, A. C
be perpendicular to AB ; then
will they be parallel.
For, if they could meet in
a point O, on either side of
AB, there would be two per-
::::r:^o
D
pendiculars OA, OB, let fall from the same point on the same
straight line; which is impossible (Prop. XIV.).
PROPOSITION XIX. THEOREM.
If two straight lines meet a third line, maldng the sum of the
interior angles on the same side of the line me,i, equal to two
right angles, the two lines will be parallel.
Let the two lines EC, BD, meet
the, third line BA, making the an- ^ ^
gles BAC, ABD, together equal to
two right angles: then the hnes
EC, BD, will be parallel.
From G, the middle point 'of
BA, draw the straight line EGF,
perpendicular to EC. It will also
be perpendicular to BD. For, the sum BAC + ABD is equal
to two right angles, by hypothesis ; the sum BAC + BAE is
likewise equal to two right angles (Prop. I.) ; and taking away
BAC from both, there will remain the angle ABD=BAE.
Again, the angles EGA, BGF, are equal (Prop. IV.) ; there-
fore, the triangles EGA and BGF, have each a side and two
adjacent angles equal ; therefore, they are themselves equal,
and the angle GEA is equal to the angle GFB (Prop. VI. Cor.) :
but GEA is a right angle by construction ; therefore, GFB is a
right angle ; hence the two Hnes EC, BD, are perpendicular to
the same straight line, and are therefore parallel (Prop. XVIIL).
26
GEOMETRY.
Scholium. When two parallel
straight lines AB, CD, are met by a
third line FE, the angles which are
formed take particular names. .
Interior angles on the same side, are
those which lie within the parallels, ^-
and on the same side of the secant
Mine : thus, 0GB, GOD, are interior
angles on the same side ; and so also
are the the angles OGA, GOC.
Alternate angles lie within the parallels, and on different
sides of the secant line : AGO, DOG, are alternate angles ;
and so also are the angles COG, BGO.
Alternate exterior anglts lie without the parallels, and on dif-
ferent sides of the secant line : EGB, COF, are alternate exte-
rior angles ; so also, are the angles AGE, FOD.
Opposite exterior and interior angles lie on the same side of the
secant line, the one without and the other within the parallels,
but not adjacent : thus, EGB, GOD, are opposite exterior and
interior angles ; and so also, are the angles AGE, GOC.
Cor. 1. If a straight line EF, meet two straight lines CD,
AB, making the alternate angles AGO, GOD, equal to each
other, the two lines will be parallel. For, to each add the an-
gle 0GB; we shall then have, AGO + 0GB = GOD + 0GB ;
but AGO + 0GB is equal to two right angles (Prop. I.) ; hence
GOD -4- 0GB is equal to two right angles : therefore, CD, AB,
are parallel.
Cor. 2. If a straight line EF, meet two straight lines CD,
AB, making the exterior angle EGB equal to the interior and
opposite angle GOD,the two lines will be parallel. For, to each
add the angle 0GB: we. shall then have EGB + 0GB = GOD
+ 0GB : but EGB + 0GB is equal to two right angles ; hence,
GOD 4- 0GB is equal to two right angles ; therefore, CD, AB,
are parallel.
PROPOSITION XX. THEOREM.
If a straight line meet two parallel straight lines, the sum of the
interior angles on the same side will he equal to two right angles.
Let the parallels AB,CD,be
met by tlie secant line FE : then
will OGB + GOD, or OGA +
GOC, be equal to two right an-
For, if OGB + GOD be not
equal to two right angles, let
IGH be drawn, making the sum
OGH+GOD equal to two
BOOK 1. . 27
right angles ; then IH and CD will be parallel (Prop. XIX.),
and hence we shall have two lines GB, GH, drawn through
the same point G and parallel to CD, which is impossible (Ax.
12.): hence, GB and GH should coincide, and 0GB + GOD is
equal to two right angles. In the same manner it may be proved
that OGA+GOC is equal to two right angles.
Cor. 1. If 0GB is a right angle, GOD will be a right angle
also: therefore, every straight line perpendicular to one of two
parallels, is perpendicular to the other.
Cor. 2. If a straight line meet two ^
parallel lines, the alternate angles will
be equal.
Let AB, CD, be the parallels, and
FE the secant line. The sum 0GB +
GOD is equal to two right angles. But ^
the sum OGB + OGA is also equal to
two right angles (Prop. I.). Taking
from each, the angle OGB, and there
remains OGA=:GOD. In the same manner we may prove that
GOC=OGB.
Cor. 3. If a straight line meet two parallel lines, the oppo-
site exterior and interior angles will be equal. For, the sum
OGB + GOD is equal to two right angles. But the sum OGB
+ EGB is also equal to two right angles. Taking from each the
angle OGB, and there remains GOD=EGB. In the same
manner we may prove that AGE =GOC.
Cor. 4, We see that of the eight angles formed by a line
cutting two parallel lines obliquely, the four acute angles are
equal to each other, and so also are the four obtuse angles.
PROPOSITION XXI. THEOREM.
If a straight line meet two other straight lines, making the sum of
the interior angles on the same side less than two rightangles^
the two lines will meet if sufficiently produced*
Letthe line EFmeet the two
4ines CD, lil, making the sum
of the interior angles OGH,
GOD, less than two right an-
gles : then will IH and CD
meet if sufficiently produced.
For, if they do not meet they
are parallel (Def. 12.). But they
are not parallel, for if they were,
the sum of the interior angles OGH, GOD, w^ould be equal to
two right angles (Prop. XX.), whereas it is less by hypothesis :
hence, the lines IH, CD, are not parallel, and will therefore
meet if sufficiently produced.
28
GEOMETRY.
Co7\ It is evident that the two lines IH, CD, will meet on
that side of EF on which the sum f»f the two angles OGH,
GOD, is less than two right angles
PROPOSITION XXII. THEOREM.
Two straight lines which are parallel to a third line, are parallel
to each other.
.A.
H
■JS
Let CD and AB be parallel to the third line EF ; then arc
they parallel to each other.
Draw PQR perpendicular to EF, and
cutting AB, CD. Since AB is parallel to__
EF, PR will be perpendicular to AB (Prop.E"
XX. Cor. 1.) ; and since CD is parallel to
EF, PR will for a like reason be perpen-C
dicular to CD. Hence AB and CD are
perpendicular to the same straight line
hence they are parallel (Prop. XVIII.).
t
PROPOSITION XXIII. THEOREM.
Two parallels are every where equally distant.
Two parallels AB, CD, being c K
given, if through two points E
and F, assumed at pleasure, the
straight lines EG, FH, be drawn
perpendicular to AB,these straight^
lines will at the same time be
perpendicular to CD (Prop. XX. Cor. 1.) : and we are now to
show that they will be equal to each othcv.
If GF be drawn, the angles GFE, FGII, considered in refer-
ence to the parallels AB, CD, will be alternate angles, and
therefore equal to each other (Prop. XX. Cor. 2.). Also, the
straight lines EG, FH, being perpendicular to the same straight
line AB, are parallel (Prop. XVIII.) ; and the angles EGF,
GFH, considered in reference to the parallels EG, FH, will be
alternate angles, and therefore equal. Hence the two trian-
gles EFG, FGH, have a common side, and two adjacent angles
in each equal ; hence these triangles are equal (Prop. VI.) ;
therefore, the side EG, which measures the distance of the
parallels AB and CD at the point E, is equal to the side FH.
which measures the distance of the same parallels at the
point F.
BOOK I. 29
PROPOSITION XXIV. THEOREM.
If two angles have their sides parallel and lying in the same di-
rection, the two angles will he equal.
Let BAG and DEF be the two angles,
having AB parallel to ED, and AC to EF ;
then will the angles be equal.
For, produce DE, if necessary, till it
meets AC in G. Then, since EF is par-
allel to GG, the angle DEF is equal to ^
DGC (Prop. XX. Cor. 3.); and since
DG is parallel to AB, the angle DGC is equal to BAG ; hence,
the angle DEF is equal to BAG (Ax. 1.).
Scholium, The restriction of this proposition to the case
where the side EF lies in the same direction with AG, and ED
in the same direction with AB, is necessary, because if FE
were produced towards H, the angle DEH would have its sides
parallel to those of the angle BAG, but would not be equal to
it. In that case, DEH and BAG would be together equal to
two right angles. For, DEH + DEF is equal to two right angles
(Prop. I.) : but DEF is equal to BAG : hence, DEH + BAG is
equal to two right angles.
PROPOSITION XXV. THEOREM.
In every triangle the sum of the three angles is equal to two
■ right angles.
Let ABG be any triangle : then will the an-
gle G+A+B be equal to two right angles.
For, produce the side GA towards D, and at
the point A, draw AE parallel to BC. Then,
since AE, GB, are parallel, and GAD cuts them,
the exterior angle DAE will be equal to its inte- C AD
rior opposite one AGB (Prop. XX. Cor. 3.) ; in like manner,
since AE, GB, are parallel, and AB cuts them, the alternate
angles ABG, BAE, will be equal : hence the three angles of
the triangle ABG make up the same sum as the three angles
GAB, BAE, EAD ; hence, the sum of the three angles is equal
to two right angles (Prop. L).
Cor, 1. Two angles of a triangle being given, or merely
their sum, the third will be found by subtracting that sum from
two right angles.
30 GEOMETRY.
Cor. 2. If two angles of one triangle a re respectively equal
to two angles of another, the third angles will also be equal
and the two triangles will be mutually equiangular.
Cor, 3. In any triangle there can be but one right angle :
for if there were two, the third angle must be nothing. Still
less, can a triangle have more than one obtuse angle.
Cor. 4. In every right angled triangle, the sum of the two
acute angles is equal to one right angle.
Cor. 5. Since every equilateral triangle is also equiangular
(Prop. XI. Cor.), each of its angles will be equal to the third
part of two right angles ; so that, if the right angle is expressed
by unity, the angle of an equilateral triangle will be expressed
byf. , ,
Cor. 6. In every triangle ABC, the exterior angle BAD is
equal to the sum of the two interior opposite angles B and C.
For, AE being parallel to BC, the part BAE is equal to the
angle B, and the other part DAE is equal to the angle C.
PROPOSITION XXVI. THEOREM.
TJie sum of all the interior angles of a polygon, is equal to two
right angles, taken as many times less two, as the figure has
sides.
Let ABCDEFG be the proposed polygon.
If from the vertex of anyone angle A, diagonals jj^
AC, AD, AE, AF, be drawn to the vertices of
all the opposite angles, it is plain that the poly-^^
gon will be divided into five triangles, if it has
seven sides ; into six triangles, if it has eight; and, A.
in general, into as many triangles, less two, as
the polygon has sides ; for, these triangles may be considered
as having the point A for a common vertex, and for bases, the
several sides of the polygon, excepting the two sides whiqh form
the angle A. It is evident, also, that the sum of all the angles
in these triangles does not differ from the sum of all the angles
in the polygon : hence the sum of all the angles of the polygon
is equal to two right angles, taken as many times as there are
triangles in the figure ; in other words, as there are units in the
number of sides diminished by two.
Cor. 1. The sum of the angles in a quadrilateral is equal
to two right angles multiplied by 4 — 2, which amounts to four
BOOK I. 31
right angles : hence, if all the angles of a quadrilateral are
equal, each of them will be a right angle ; a concluslop which
sanctions the seventeenth Definition, where the four angles of
a quadrilateral are asserted to be right angles, in the case of the
rectangle and the square.
Cor. 2. The sum of the angles of a pentagon is equal to
two right angles multiplied by 5 — 2, which amounts to six right
angles : hence, when a pentagon is equiangular, .each angle
is equal to the fifth part of six right angles, or to | of one right
angle.
Cor. 3. The sum of the angles of a hexagon is equal to
2 X (6 — 2,) or eight right angles ; hence in the equiangular
hexagon, each angle is the sixth part of eight right angles, or |
of one.
Scholium. When this proposition is applied to
polygons which have re-emtrant angles, each re-
entrant angle must be regarded as greater than
two right angles. But to avoid all ambiguity, we
shall henceforth limit our reasoning to polygons
with salient angles, which might otherwise be named convex
polygons. Every convex polygon is such that a straight line,
drawn at pleasure, cannot meet the contour of the polygon in
more than two points.
PROPOSITION XXVII. THEOREM.
If the sides of any polygon he produced out, in the same direc-
tion , the sum of the exterior angles will he equal to four right
angles.
Let the sides of the polygon ABCD-
FG, be produced, in the same direction ;
then will the sum of the exterior angles
a + & + c -i- d-^f-\-g, be equal to four right
angles.
For, each interior angle, plus its ex-
terior angle, as A + a, is equal to two
right angles (Prop. I.). But there are
as many exterior as interior angles, and as many of each as
there are sides of the polygon : hence, the sum of all the inte-
rior and exterior angles is equal to twice as many right angles
as the polygon has sides. Again, the sum of all the interior
angles is equal to two right angles, taken as many times, less
two, as the polygon has sides (Prop. XXVI.) ; that is, equal to
twice as many right angles as the figure has sides, wanting
four right angles. Hence, the interior angles plus four right
32 GEOMETRY.
angles, is equal to twice 'as many right angles as the polygon
has oides, and consequently, equal to the sum of the interior
angles plus the exterior angles. Taking from each the sum of
the interior angles, and there remains the exterior angles, equal
to four right angles.
PROPOSITION XXVIII. THEOREM. .
In every parallelogram, the opposite sides and angles are equal.
Let ABCD be a parallelogram : then will jy c »
AB=DC, AD=BC, A=C, and ADC= ABC. "' '
For, draw the diagonal BD. The triangles
ABD, DBC, have a common side BD ; and
since AD, BC, are parallel, they have also the
angle ADB=DBC, (Prop. XX. Cor. 2.) ; and since AB, CD,
are parallel, the angle ABD==BDC : hence the two triangles
are equal (Prop. VI.) ; therefore the side AB, opposite the an-
gle ADB, is equal to the side DC, opposite the equal angle
DBC ; and the third sides AD, BC, are equal: hence the op-
posite sides of a parallelogram are equal.
Again, since the triangles are equal, it follows that the angle
A is equal to the angle C ; and also that the angle ADC com-
posed of the two ADB, BDC, is equal to ABC, composed of
the two equal angles DBC, ABD : hence the opposite apgles
of a paKallelogram are also equal.
Cor. Two parallels AB, CD, included between two other
parallels AD, BC, are equal ; and the diagonal DB divides the
parallelogram into two equal triangles.
PROPOSITION XXIX. THEOREM.
Jf the opposite sides of a quadrilateral are equal, each to each,
the equal sides will he parallel, and the figure will he a par-
allelogram.
Let ABCD be a quadrilateral, having
its opposite sides respectively equal, viz.
AB=DC, and AD=:BC ; then will these
sides be parallel, and the figure be a par-
allelogram.
For, having drawn the diagonal BD,
the triangles ABD, BDC, ha^Q all the sides of the one equal to
BOOK I. 33
m
Ihe corresponding sides of the other ; therefore they are equal,
and the angle ADB, opposite the side AB, is equal to DBC,
opposite CD (Prop. X.) ; therefore, the side AD is parallel to
BC (Prop. XIX. Cor. 1.). For a hke reason AB is parallel to
CD : therefore the quadrilateral ABCD is a parallelogram.
PROPOSITION XXX. THEOREM.
If two opposite sides of a quadrilateral are equal and parallel
the remaining sides will also he equal and parallel, and the
figure will he a parallelogram.
Let ABCD be a quadrilateral, having
the sides AB, CD, equal and parallel ;
then will the figure be a parallelogram.
For, draw the diagonal DB, dividing
the quadrilateral into two triangles. Then, J-
since AB is parallel to DC, the alternate
angles ABD, BDC, are equal (Prop. XX. Cor. 2.) ; moreover,
the side DB is common, and the side AB=DC ; hence the tri-
angle ABD is equal to the triangle DBC (Prop. V.) ; therefore,
the side AD is equal to BC, the angle ADB = DBC, and conse-
quently AD is parallel to BC ; hence the figure ABCD is a
parallelogram.
PROPOSITION XXXI. THEOREM.
The two diagonals of a parallelogram divide each other into equal
parts, or mutually bisect each other.
Let ABCD be a parallelogram, AC and
DB its diagonals, intersecting at E,then will
AE=EC, and DE=EB.
Comparing the triangles ADE, CEB, we
find the side AD = CB (Prop. XXVIIL),
the angle ADE=CBE, and the angle
DAE=ECB (Prop. XX. Cor. 2.); hence those triangles are
equal (Prop. VI.) ; hence, AE, the side opposite the angle
ADE, is equal to EC, opposite EBC ; hence also DE is equal
to EB.
Scholium. In the case of the rhombus, the sides AB, BC,
being equal, the triangles AEB, EBC, have all the sides of the
one equal to the corresponding sides of the other, and are
therefore equal: whence it follows that the angles. AEB, BEC,
are equal, and therefore, that the two diagonals of a rhombus
cut each other at right angles.
34 GEOMETRY.
BOOK 11.
OF RATIOS AND PROPORTIONS.
Definitions,
1. Ratio is the quotient arising from dividing one quantity
by another quantity of the same kind. Thus, if A and B rep-
resent quantities of the same kind, the ratio of A to B is ex-
pressed by -V-.
The ratios of magnitudes may be expressed by numbers,
either exactly or approximatively ; and in the latter case, the
approximation may be brought nearer to the true ratio than
any assignable difference.
Thus, of two magnitudes, one of them may be considered to
be divided into some number of equal parts, each of the same
kind as the whole, and one of those parts being considered as
an unit of measure, the magnitude may be expressed by the
number of units it contains. If the other magnitude contain
a certain number of those units, it also may be expressed by
the number of its units, and the two quantities are then said
to be commensurable.
If the second magnitude do not contain the measuring unit
an exact number of times, there may perhaps be a smaller unit
which will be contained an exact number of times in each of
the magnitudes. But if there is no unit of an assignable value,
which shall be contained an exact number of times in each of
the magnitudes, the magnitudes are said to be incommensurable.
It is plain, however, that the unit of measure, repeated as
many times as it is contained in the second magnitude, would
always differ from the second magnitude by a quantity less
than the unit of measure, since the remainder is always less
than the divisor. Now, since the unit of measure may be made
as small as we please, it follows, that magnitudes may be rep-
resented by numbers to any degree of exactness, or they will
differ from their numerical representatives by less than any
assignable quantity.
Therefore, of two magnitudes, A and B, we may conceive
A to be divided into M number of units, each equal to A' :
then A=M x A': let B be divided into N number of equal units,
each equal to A'; then B=N x A'; M and N being integral num-
bers. Now the ratio of A to B, will be the same as the ratio
of M X A' to N X A'; that is thasame as the ratio of M to N, since
A' is a common unit.
BOOK 11. 35
In the same manner, the ratio of any other two magnitudes
C and D may be expressed by P x C to Q x C, P and Q being
also integral numbers, and their ratio will be the same as that
ofPtoQ.
2. If there be four magnitudes A, B, C, and D, having such
values that —-is equal to— , then A is said to have the same ratio
A O
to B, that C has to D, or the ratio of A to B is equal to the ratio
of C to D. When four quantities have this relation to each
other, they are said to be in proportion.
To indicate that the ratio of A to B is equal to the ratio of
C to D, the quantities are usually written thus, A : B : : C : D,
and read, A is to B as C is to D. The quantities which are
compared together are called the terms of the proportion. The
first and last terms are called the two extremes, and the second
and third terms^ the two means.
3. Of four proportional quantities, the first and third are
called the antecedents, and the second and fourth the conse-
quents ; and the last is said to be a fourth proportional to the
other three taken in order.
4. Three quantities are in proportion, when the first has the
same ratio to the second, that the second has to the third ; and
then the middle term is said to be a mean proportional between
the other two.
5. Magnitudes are said to be in proportion by inversion, or
inversely, when the consequents are taken as antecedents, and
the antecedents as consequents.
6. Magnitudes are in proportion by alternation, or alternately,
vv^hen antecedent is compared with antecedent, and consequent
with consequent.
7. Magnitudes are in proportion by composition, when the
sum of the antecedent and consequent is compared either with
antecedent or consequent.
8. Magnitudes are said to be in proportion by division, when
the difference of the antecedent and consequent is compared
either with antecedent or consequent.
9. Equimultiples of two quantities are the products which
arise from multiplying the quantities by the same number :
thus, m X A, m X B, are equimultiples of A and B, the common
multiplier being m.
10. Two quantities A and B are said to be reciprocally
proportional, or inversely proportional, when one increases in
the same ratio as the other diminishes. In such case, either
of them is equal to a constant quantity divided by the other,
and their product is constant.
86 GEOMETRY.
PROPOSITION I. THEOREM.
k:
When four quantities are in proportion^ the product of the
two extremes is equal to the product of the two means
Let A, B, C, D, be four quantities in proportion, and M : N
: ; P : Q be their numerical representatives ; then will M x Q=
N O
Nx P; for since the quantities are in proportion — =- there-
O MP
fore N=Mx^,orNxP=MxQ.
Cor, If there are three proportional quantities (Def. 4.), the
product of the extremes will be equal to the square of the
mean.
PROPOSITION II. THEOREM.
If the product of two quantities he equal to the pr-oduct of two other
quantities^ two of them will he the extremes and the other two
the means of a proportion.
Let M X Q=N X P ; then will M : N : : P : Q.
For, if P have not to Q the ratio which M has to N, let P
have to Q', a number greater or less than Q, the same ratio
that M has to N; that is, let M : N : : P : Q' ; then MxQ'=
NxP (Prop. L) : hence, Q'= ^^^ ; but Q=Z^ ; con-
feequently, Q=Q' and the four quantities are proportional; that
is, M:N:;P:Q.
PROPOSITION III. THEOREM.
If four quantities are in proportion, they will he in proportion
when taken alternately.
Let M, N, P, Q, be the numerical representatives of four
quanties in proportion ; so that
M : N : : P : Q, then will M : P : : N : Q.
Since M ; N : : P : Q, by supposition, M x Q=N x P ; there-
fore, M and Q may be made the extremes, and N and P the
means of a proportion (Prop. IL) ; hence, M : P : : N : Q.
BOOK IT. 37
PROPOSITION IV. . THEOREM.
[f there he four proportional quantities, and four other propor-
tional quantities, having the antecedents the same in both, the
consequefits will be proportional.
Let M : N : : P : Q
and M : R : : P : S
then will N : Q : : R : S
P O
For, by alternation M;P::N:Q,or tj=5-
P S
and M : P ; : R ; S, or 5|=g"
Q S
hence :j^=:5- ; or N : Q : : R : S.
Cor. If there be two sets of proportionals, having an ante-
cedent and consequent of the first, equal to an antecedent and
consequent of the second, the remaining terms will be propor-
tional.
, PROPOSITION V. THEOREM.
If four quantities be in proportion, they will be in proportion when
taken inversely.
Let M:N::P:Q.; then will t
N : M : : Q : P.
For, from the first proportion we have M x Q=N x P, or
NxP=MxQ.
But the products N x P and M x Q are the products of the
extremes and mean^ of the four quantities N, M , Q, P, and these
products being equal,
N:M::Q:P(Prop.IL).
PROPOSITION VI. THEOREM.
If four quantities are in proportion, they will be in proportion by
composition, or division.
D
38 GEOMETRY.
Let, as before, M, N, P, Q, be the numerical representatives
of the four quantities, so that
M : N : : P : Q ; then will
M±N:M::Pd=Q:P.
For, from the first proportion, we have
MxQ=NxP, orNxP=MxQ;
Add each of the members of the last equation to, or subtract
It from M.P, and we shall have,
M.P±N.P=:M.P±M.Q;or
(M±N)xP-(P±Q)xM.
But MrbN and P, maybe considered the two extremes, and
P±Q and M, the two means of a proportion : hence,
M±N : M : : FiQ : P.
PROPOSITION VII. THEOREM.
Equimultiples of any two quantities^ have the same ratio as the
quantities themselves.
Let M and N be any two quantities, and m any integral
number ; then will
m.M.:m. N : : M : N. For
m. MxN=?w. NxM, since the quantities in
each member are the same ; therefore, the quantities are pro-
portional (Prop. II.) ; or
m. M : 771. N : : M : N.
PROPOSITION VIII. THEOREM.
Of four proportional quantities, if there he taken any equimul-
tiples of the two antecedents, and any equimultiples of the two
consequents, the four resulting quantities will be proportional
Let M, N, P, Q, be the numerical representatives of four
quantities in proportion ; and let m and n be any numbers
whatever, then will
m. M : ?i. N : : m. P : 71. Q.
For, since M : N : : P : Q, we have M x Q=N x P ; hence,
m. M X 71. Q=?i. N X 77Z. P, by multiplying both members of the
equation by ??i x n. But m, M and n. Q, may be regarded as
the two extremes, and n. N and m. P, as the means of a propor-
tion ; hence, tw. M : n. N : ; m, P : ti. Q.
BOOK II. a^
PROPOSITION IX. THEOREM.
Of four proportional quantities, if the two consequents he either
augmented or diminished hy quantities which have the same
ratio as the antecedents, the resulting quantities and the ante-
cedents will he proportional.
Let
M : N : : P : Q, and let also
M : P : : m : 71, then will
M : P : : N±»z : Qrbw.
For, since
M : N : : P : Q, MxQ=NxP.
And since
M : P : : m : 71, Mx7i=Px7?i
Therefore,
MxQ±Mx7i=NxP±Px77z
or.
Mx(Q=i=7i)=Px(N±m):
hence
M : P : : Ndbm : Q±7i (Prop. II.).
PROPOSITION X. THEOREM.
[f any numher of quantities are proportionals, any one antece-
dent will he to its consequent, as the sum of all the antecedents
to the sum of the consequents.
Let M : N : ; P : Q : : R j^ &c^then will
M : N : : M + P + R : N + Q + S
For, since M : N : : P : Q, we have MxQ=JixP
And since M : N : : R : S, we have Mx S=]?lxR
Add MxN=MxN
and we have, M.N+M.Q + M.S=M.N+N.P + N.R
or Mx(N + Q+S)^Nx (M + P + R )
therefore, M : N : : M+P + R : N + Q+S.
PROPOSITION XI. THEOREM.
If two magnitudes he each increased or diminished hy like parts
of each, the resulting quantities will have the same ratio as the
magnitudes themselves.
m
40 GEOMETRY.
Let M and N be any two magnitudes, and — and i- be like
m m
parts of each : then will
M ; N : : Md=M : N ±—
m m
For, it is obvious that Mx(N±_\ =Nx(M±_\ since
m / m /
each is equal to M.Ndb_L_. Consequently, the four quan-
m
titles are proportional (Prop. II.),
PROPOSITION XII. THEOREM.
//* four quantities are proportional, their squares or cubes will
also be proportional.
Let M : N : P : Q,
then will M2 : N2 : : P2 : Q2
and M^ : N^ : : P3 : Q3
For, MxQ=NxP, since M : N : : P : Q
or, M^ X Q^=N^ X P^ by squaring both members,
and M^ X Q^=N^ X P^ by cubing both members ;
therefore, M^ ; N^ : : P^ : Q2
and M^ : N^ : : P3 : Q3
Cor. In the same way it may be shown that like powers or
roots Jtf proportional quantities are proportionals.
PROPOSITION XIII. THEOREM.
Jf there be two sets of proportional quantities, the products ofths
corresponding terms will be proportional.
Let
and
then will
For since
and
M : N : : P : Q
. R • S : : T : V
*MxR : NxS : : PxT : QxV
MxQ=N>?P
R X V= S X T, we shall have
MxQx^xV=NxPxSxT
or
MxRxQxV=NxSxPxT
therefore,
MxR : NxS : : PxT : QxV
BOOK III. 41
BOOK III.
THE CIRCLE, AND THE MEASUREMENT OF ANGLES.
Definitions,
1. The circumference of a circle is a
curved line, all the points of which are
equally distant from a point within,
called the centre.
The circle is the space terminated by ^.j
this curved line.*
2. Every straight line, CA, CE, CD,
drawn from the centre to the circum-
ference, is called a radium or semidiam- E"
eter ; every line which, like AB, passes through the centre, and
is terminated on both sides by the circumference, is called a
diameter.
From the definition of a circle, it follows that all the radii
are equal ; that all the diameters are fqual also, and each
double of the radius.
3. A portion of the circumference, such as FHG, is called
an arc.
The chord, or subtense of an arc, is the straight line FG, which
joins its two extremities.f
4. A segment is the surface of portion of a circle, included
between an arc and its chord.
5. A sector is the part of the circle included between an
arc DE, and the two radii CD, CE, drawn to the extremities
of the arc.
6. A straight line is said to be inscribed in
a circle, when its extremities are in the cir-
cumference, as AB.
An inscribed angle is one which, like BAC,
has its vertex in the circumference, and is
formed by two chords.
* Note. In common language, the circle is sometimes confounded with its
circumference : but the correct expression may always be easily recurred to if
we bear in mind that the circle is a surface which has length and breadth,
while the circumference is but a line.
t Note. In all cases, the same chord FG belongs to two arcs, FGH, FEG,
and consequently also to two segments : but the smaller one is always meant,
# unless the contrary is expressed.
D*
42 GEOMETRY.
An inscribed triangle is one which, like BAG, has its three
angular points in the circumference.
And, generally, an inscribed figure is one, of which all the
angles have their vertices in the circumference. The circle is
then said to circumscribe such a figure.
7. A secant is a line which meets the circum-
ference in two points, and lies partly within ^-
and partly without the circle. AB is a secant.
8. A tangent is a hne which has but one
point in common with the circumference. CD
is a tangent.
The point M, where the tangent touches the c
circumference, is called the point of contact.
In like manner, two circumferences touch
each other when they have but one point in
common.
9. A polygon is circumscribed about a
circle, when ail its sides are tangents to
the circumference : in the same case, the
circle is said to be inscHbed in the po-
lygon.
PROPOSITION I. THEOREM.
Every diameter divides the circle and its circumference into two
equal parts.
Let AEDF be a circle, and AB a diameter.
Now, if the figure AEB be applied to AFB,
their common base AB retaining its position,
the curve line AEB must fall exactly on the
curve line AFB, otherwise there would, in
the one or the other, be points unequally dis-
tant from the centre, which is contrary to
tlie definition of a circle. ^
I
BOOK III. 43
PROPOSITION II. THEOREM.
Every chord is less than the diameter.
Let AD be any chord. Draw the radii
CA, CD, to its extremities. We shall then
have AD<AC + CD (Book I. Prop. VII.*); ^
or AD<Ap.
Cor, [Hence the greatest Hne which can be inscribed in a
Y circle is its diameter.^
PROPOSITION III. THEOREM.
A straight line cannot meet the circumference of a circle in more
than two points.
For, if it could meet it in three, those three points would be
equally distant from the centre ; and hence, there would be
three equal straight lines drawn from the same point to the
same straight line, which is impossible (Book I. Prop. XV.
Cor. 2.).
PROPOSITION IV. THEOREM.
In the same circle, or in equal circles, equal arcs are subtended by
equal chords ; and, conversely, equal chords subtend equal arcs.
Note. When reference is made from one pronosition to another, in the
same Book, the number of the proposition referred to is alone given; but when
the proposition is found in a different Book, the number of the Book is also
given.
44 GEOMETRY.
If the radii AC, EO, are
equal, and also the arcs
AMD, ENG; then the chord
AD will be equal to the jA
chord EG.
For, since the diameters
AB, EF, are equal, the semi-
circle AMDB maybe applied
exactly to the semicircle ENGF, and the curve line AMDB
will coincide entirely with the curve line ENGF. But the
part AMD is equal to the part ENG, by hypothesis ; hence, the
point D will fall on G ; therefore, the chord AD is equal to the
chord EG.
Conversely, supposing again the radii AC, EO, to be equal,
if the chord AD is equal to the chord EG, the arcs AMD,
ENG will also be equal.
For, if the radii CD, OG, be drawn, the triangles ACD,
EOG, will have all their sides equal, each to each, namely,
AC^EO, CD = OG, and AD=EG; hence the triangles are
themselves equal ; and, consequently, the angle ACD is equal
EOG (Book I. Prop. X.). Now, placing the semicircle ADB
on its equal EGF, since the angles ACD, EOG, are equal, it is
plain that the radius CD will fall on the radius OG, and the
point D on the point G ; therefore the arc AMD is equal to the
arc ENG.
PROPOSITION V. THEOREM.
In the same circle y or in equal circles , a greater arc is subtended
by a greater chords and conver'sely^ the greater chord subtend
the greater arc.
Let the arc AH be greater than
the arc AD ; then will the chord AH
be greater than the chord AD.
For, draw the radii CD, CH. The
two sides AC, CH, of the triangle . /
ACH are equal to the two AC, CD, -^^
of the triangle ACD, and the angle
ACH is greater than ACD ; hence, the
third side AH is greater than the third
side AD (Book I. Prop. IX.) ; there- X
fore the chord, which subtends the greater arc, is the greater.
Conversely, if the chord AH is greater than AD, it will follow,
on comparing the same triangles, that the angle ACH is
BOOK III. 4i,
greater than ACD (Bk. I. Prop. IX. Sch.) ; and hence that
the arc AH is greater than AD ; since the whole is greater
than its part.
Scholium. The arcs here treated of are each less than the
semicircumference. If they were greater, the reverse pro-
perty would have place ; for, as the arcs increase, the chords
would diminish, and conversely. Thus, the arc AKBD is
greater than AKBH, and the chord AD, of the first, is less
than the chord AH of the second.
PROPOSITION VI. THEOREM.
The radius which is perpendicular to a chord, bisects the chords
and bisects also the subtended arc of the chord.
Let AB be a chord, and CG the ra-
dius perpendicular to it : then will AD =
DB, and the arc AG^GB.
For, draw the radii CA, CB. Then
the two right angled triangles ADC,
CDB, will have AC=CB, and CD com-
mon ; hence, AD is equal to DB (Book
T. Prop. XVII.).
Again, since AD, DB, are equal, CG
is a perpendicular erected from the mid-
dle of AB ; hence every point of this perpendicular must be
equally distant from its two extremities A and B (Book I. Prop.
XVI.). Now, G is one of these points ; therefore AG, BG, are
equal. But if the chord AG is equal to the chord GB, the arc
AG will be equal to the arc GB (Prop. IV.) ; hence, the radius
CG, at right angles to the chord AB, divides the arc subtended
by that chord into two equal parts at the point G.
Scholium. The centre C, the middle point D, of the chord
AB, and the middle point G, of the arc subtended by this
chord, are three points of the same line perpendicular to the
chord. But two points are sufficient to determine the position
of a straight line ; hence every straight line which passes through
two of the points just mentioned, will necessarily pass through
the third, and be perpendicular to the chord.
It follows, likewise, that the perpendicular raised from the
middle of a chord passes through the centre of the circle y and
through the middle of the arc subtended by that chord.
For, this perpendicular is the same as the one let fall from
the centre on the same chord, since both of them pass through
the centre and middle of the chord.
46 GEOMETRY.
PROPOSITION VII. THEOREM.
fi-^<i^^jf^\
Through three given points not in the same straight line, one cir-
cumference may always he made to pass, and but one.
Let A, B, and C, be the given
points.
Draw AB, BC, and bisect these
straight Unes by the perpendiculars
DE, FG : we say first, that DE and
FG, will meet in some point O.
For, they must necessarily cut
each other, if they are not parallel.
Now, if they were parallel, the line AB, which is perpendicular
to DE, would also be perpendicular to FG, and the angle K
would be a right angle (Book I. Prop. XX. Cor. 1.). But BK,
the prolongation of BD, is a difierent line from BF, because the
three points A, B, C, are not in the same straight line ; hence
there would be two perpendiculars, BF, BK, let fall from the
same point B, on the same straight line, which is impossible
(Book I. Prop. XIV.) ; hence DE, FG, will always meet in
some point O.
And moreover, this point O, since it lies in the perpendicular
DE, is equally distant from the two points, A and B (Book I.
Prop. XVI.) ; and since the same point O lies in the perpen-
dicular FG, it is also equally distant from the two points B and
C : hence the three distances OA, OB, OC, are equal ; there-
fore the circumference described from the centre O, with the
radius OB, will pass through the three given points A, B, C.
We have now shown that one circumference can always be
made to pass through three given points, not in the same
straight line : we say farther, that but one can be described
through them.
For, if there were a second circumference passing through the
three given points A, B, C, its centre could not be out of the
line DE, for then it would be unequally distant from A and B
(Book I. Prop. XVI.); neither could it be out of the line FG, for
a like reason ; therefore, it would be in both the lines DE, FG.
But two straight lines cannot cut each other in more than one
point ; hence there is but one circumference which can pass
through three given points.
Cor. Two circumferences cannot meet in more than two
pomts ; for, if they have three common points, there would be
two circumferences passing through the same three points ;
which has been shown by the proposition to be impossible.
BOOK III.
47
\.
■\
PROPOSITION VIII. THEOREM.
Two equal chords are equally distant from the centre ; andoj two
unequal chords, the less is at the greater distance from the
centre.
First. Suppose the chord AB=
DE. Bisect these chords by the per-
pendiculars CF, CG, and draw the
radii CA, CD.
In the right angled triangles CAF,
DCG, the hypothenuses CA, CD, are
equal ; and the side AF, the h^lf of
AB, is equal to the side DG, the half
of DE : hence the triangles are equal,
and CFis eciual to CG (Book I. Prop.
XVII.) ; hence, the two equal chords
AB, DE, are equally distant from the centre.
Secondly Let the chord AH be greater than DE. The
arc AKH will be greater than DME (Prop. V.) : cut off from
the former, a part ANB, equal to DME ; draw the chord AB,
and let fall CF perpendicular to this chord, and CI perpendicu-
lar to AH. It is evident that CF is greater than CO, and CO
than CI (Book I. Prop. XV.) ; therefore, CF is still greater
than CI. But CF is equal to CG, because the chords AB,
DE, are equal : hence we have CG>CI ; hence of two unequal
chords, the less is the farther from the centre.
n^^
PROPOSITION IX. THEOREM.
A straight line perpendicular to a radius, at its extremity, is a
tangent to the circumference.
Let BD be perpendicular to the B
radius C A, at its extremity A ; then
will it be tangent to the circumfe-
rence.
For, every oblique line CE, is
longer than the perpendicular CA
(Book I. Prop. XV.); hence the
point E is without the circle ; therefore, BD has no point but
A common to it and the circumference ; consequently BD is a
tangent (Def. 8.).
48
GEOMETRY.
Scholium. At a given point A, only one tangent AD can
be drawn to the circumference ; for, if another could be drawn,
it would not be perpendicular to the radius CA (Book I. Prop.
XIV. Sch.) ; hence in reference to this new tangent, the radius
AC would be an oblique line, and the perpendicular let fall
from the centre upon this tangent would be shorter than CA ;
hence this supposed tangent would enter the circle, and be a
secant.
PROPOSITION X. THEOREM.
aTm:
Two parallels intercept equal arcs on the circumference.
There may be three cases.
First. If the two parallels are se-
cants, draw the radius CH perpendicu-
lar to the chord MP. It will, at the
same time be perpendicular to NQ
(Book I.Prop.XX.Cor. 1 .) ; therefore, the
point H will be at once the middle of
the arc MHP, and of the arc NHQ
(Prop. VI.) ; therefore, we shall have
the arc MH=HP, and the arc NH =
HQ ; and therefore MH— NH=:HP— HQ ; in other words,
MN=PQ.
Second. When, of the two paral-
lels AB, DE, one is a secant, the
other a tangent, draw the radius CH
to the point of contact H ; it will be
perpendicular to the tangent DE
(Prop. IX.), and also to its parallel
MP. But, since CH is perpendicular
to the chord MP, the point H must be
the middle of the arc MHP (Prop.
VI.) ; therefore the arcs MH, HP, in-
cluded between the parallels AB, DE, are equal.
Third. If the two parallels DE, IL, are tangents, the one
at H, the other at K, draw the parallel secant AB ; and, from
what has just been shown, we shall have MH=HP, MK=KP;
and hence the whole arc HMK=HPK. It is farther evident
that each of these arcs is a semicircumference.
BOOK III.
49
PROPOSITION XI. THEOREM.
If two circles cut each other in two points, the line which passes
through their centres, will be perpendicular to the chord which
joins the points of intersection, and will divide it into two
equal parts.
For, let the line AB join the points of intersection. It will
be a common chord to the two circles. Now if a perpendicular
be erected from the middle of this chord, it will pass through
each of the two centres C and D (Prop. VI. Sch.). But no
more than one straight line can be drawn through two points ;
hence the straight line, which passes through the centres, will
bisect the chord at right angles.
PROPOSITION Xn. THEOREM.
If the distance between the centres of two circles is less than the
sum of the radii, the greater radius being at the same time
less than the sum of the smaller and the distance between the
centres, the two circumference^'jivill cut each other.
For, to make an intersection
possible, the triangle CAD must
be possible. Hence, not only-
must we have CD < AC + AD,
but also the greater radius AD<
AC + CD (Book I. Prop. VII.).
And, whenever the triangle CAD
can be constructed, it is plain
that the circles described from the centres C and D, will cut
each other in A and B.
m
50
GEOMETRY.
PROPOSITION XIII. THEOREM.
If the distance between the centres of two circles is equal to the
sum of their radii, the two circles will touch each other exter-
nally.
Let C and D be the centres at a
distance from each other equal to
CA + AD.
The circles will evidently have the
point A common, and they v^ill have
no other; because, if they had two
points common, the distance between
their centres must be less than the sum of their radii.
PROPOSITION XIV. THEOREM.
)(
If ike distance between the centres of two circles is equal to the
difference of their radii, the two circles icill touch each other
internally.
Let C and D be the centres at a dis-
tance from each other equal to AD — CA.
It is evident, as before, that they will
have the point A common : they can have
no other; because, if they had, the greater
radius AD must be less than the sum of
the radius AC and the distanceCD between
the centres (Prop. XIL); which is contraiy
to the supposition.
Cor. Hence, if two circles touch each othef, either eiter-
nally or internally, their centres and the point of contact "S^ill
be in the same right line.
Scholium. All circles which have their centres on the right
line AD. and which pass through the point A, are tangent to
each other. For, they have only the point A common, and if
through the point A, AE be drawn perpendicular to AD, the
straight line AE will be a common tangent to all the circles.
BOOK III. 51
PROPOSITION XV. THEOREM.
KW^
In the same circle, or in equal circles, equal angles having their
vertices at the centre, intercept equal arcs on the circumference :
and conversely, if the arcs intercepted are equal, the angles
contained by the radii will also be equal, •
Let C and C be the centres of equal circles, and the angle
ACB-DCE.
First. Since the angles ACB,
DCE, are equal, they may be
placed upon each other ; and [ C \ ( C
since their sides are equal, the
point A will evidently fall on D,
and the point B on E. But, in iiN J^B 3^
that case, the arc AB must also
fall on the arc DE ; for if the arcs did not exactly coincide, there
would, in the one or the other, be points unequally distant from
the centre ; which is impossible : hence the arc AB is equal
to DE.
Secondly. If we suppose AB=DE, the angle ACB will be
equal to DCE. For, if these angles are not equal, suppose
ACB to be the greater, and let ACI be taken equal to DCE.
From what has just been shown, we shall have AI=DE : but,
by hypothesis, AB is equal to DE ; hence AI must be equal to
AB, or a part to the whole, which is absurd (Ax. 8.) : hence,
the angle ACB is equal to DCE.
PROPOSITION XVI. THEOREM.
In the same circle, or in equal circles, if two angles at the centre
are to each other in the proportion of two whole numbers, the
intercepted arcs will be to each other in the proportion of the
same numbers, and we shall have the angle to the angle, as the
corresponding arc to the corresponding arc.
62
GEOMETRY.
Suppose, for example, that the angles ACB, DCE, are to
each other as 7 is to 4; or, which is the same thing, suppose
that the angle M, which may serve as a common measure, is
contained 7 times in the angle ACB, and 4 times in DCE
C C
The seven partial angles ACm, mCUf nCp, &c., into which
ACB is divided, being each equal to any of the four partial
angles into which DCE is divided ; each of the partial arcs
Am, mn, np, <fec., will be equal to each of the partial arcs Da-,
xy, &c. (Prop. XV.). Therefore the whole arc AB will be to
the whole arc DE, as 7 is to 4. But the same reasoning would
evidently apply, if in place of 7 and 4 any numbers whatever
were employed ; hence, if the ratio of the angles ACB, DCE,
can be expressed in whole numbers, the arcs AB, DE, will be
to each other as the angles ACB, DCE.
Scholium, Conversely, if the arcs, AB, DE, are to each
other as two whole numbers, the angles ACB, DCE will be to
each other as the same whole numbers, and we shall have
ACB : DCE : : AB : DE. For the partial arcs. Am, 7W7i,&c.
and Dx, xy, &c., being equal, the partial angles ACm, mCn,
&c. and DCa;, xCy, &c. will also be equal.
PROPOSITION XVII. THEOREM.
Whatever be the ratio of two angles, they will always he to each
other as the arcs intercepted between their sides ; the arcs being
described from the vertices of the angles as centres with equal
radii.
Let ACB be the greater and
ACD the less angle.
Let the less angle be placed
on the greater. If the propo-
sition is not true, the angle
ACB will be to the angle ACD
as the arc AB is to an arc
greater *or less than AD. Suppose this arc to be greater, and
let it be represented by AO ; we shall thus have, the angle
ACB : angle ACD : : arc AB : arc AO. Next conceive the ai'c
BOOK III. 53
AB to be divided into equal parts, each of which is less than
DO ; there will be at least one point of division between D and
O ; let I be that point ; and draw CI. The arcs AB, AI, will be
to each other as two whole numbers, and by the preceding
theorem, we shall have, the angle ACB : angle ACI : : arc AB
: arc AI. Comparing these two proportions with each other,
we see that the antecedents are the same : hence, the conse-
quents are proportional (Book II. Prop. IV.) ; and thus we find
the angle ACD : angle ACI : : arc AO : arc AI. But the arc
AO is greater than the arc AI ; hence, if this proportion is true,
the angle ACD must be greater than the angle ACI : on the
contrary, however, it is less ; hence the angle ACB cannot be
to the angle ACD as the arc AB is to an arc greater than AD.
By a process of reasoning entirely similar, it may be shown
that the fourth term of the proportion cannot be less than AD ;
hence it is AD itself ; therefore we have
Angle ACB : angle ACD : : arc AB : arc AD.
Cor. Since the angle at the centre of a circle, and the arc
intercepted by its sides, have such a connexion, that if the one
be augmented or diminished in any ratio, the other will be
augmented or diminished in the same ratio, we are authorized
to establish the one of those magnitudes as the measure of the
other ; and we shall henceforth assume the arc AB as the mea-
sure of the angle ACB. It is only necessary that, in the com-
parison of angles with each other, the arcs which serve to
measure them, be described with equal radii, as is implied in
all the foregoing propositions.
Scholium 1. It appears most natural to measure a quantity
by a quantity of the same species ; and upon this principle it
would be convenient to refer all angles to the right angle ;
which, being made the unit of measure, an acute angle would
be expressed by some number between and 1 ; an obtuse an-
gle by some number between 1 and 2. This mode of express-
ing angles would not, how^ever, be the most convenient in
practice. It has been found more simple to measure them by
arcs of a circle, on account of the facility with which arcs can
be made equal to given arcs, and for various other reasons. At
all events, if the measurement of angles by arcs of a circle
is in any degree indirect, it is still equally easy to obtain the
direct and absolute measure by this method ; since, on
comparing the arc which serves as a measure to any an-
gle, with the fourth part of the circumference, we find the'
ratio of the given angle to a right angle, which is the absolute
measure.
E*
'W^:":.
54
GEOMETRY.
Scholium 2. All that has been demonstrated in the last three
propositions, concerning the comparison of angles with arcs,
holds true equally, if applied to the comparison of sectors with
arcs ; for sectors are not only equal when their angles are so,
but are in all respects proportional to their angles ; hence, two
sectors ACB, ACD, taken in the same circle, or in equal circles^
are to each other as the arcs AB, AD, the bases of those sectors.
It is hence evident that the arcs of the circle, which serve as a
measure of the different angles, are proportional to the different
sectors, in the same circle, or in equal circles.
PROPOSITION XVIII. THEOREM.
An inscribed angle is measui^ed by half the arc included between
its sides.
Let BAD be an inscribed angle, and let
us first suppose that the centre of the cir-
cle lies within the angle BAD. Draw the
diameter AE, and the radii CB, CD.
The angle BCE, being exterior to the
triangle ABC, is equal to the sum of the
two interior angles CAB, ABC (Book I.
Prop. XXV. Cor. 6.) : but the triangle BAC
being isosceles, the angle CAB is equal to
ABC ; hence the angle BCE is double of BAC. Since BCE lies
at the centre, it is measured by the arc BE ; hence BAC will be
measured by the half of BE. For a like reason, the angle CAD
will be measured by the half of ED; hence BAC + CAD, or BAD
will be measured by half of BE + ED, or of BED.
Suppose, in the second place, that the
centre C lies without the angle BAD. Then
drawing the diameter AE, the angle BAE
will be measured by the half of BE ; the
angle DAE by the half of DE : hence their
difference BAD will be measured by the
half of BE minus the half of ED, or by the
half of BD.
Hence every inscribed angle is measured
bv half of the arc included between its sides.
BOOK III.
55
Cor. 1. All the angles BAG, BDC,
BEC, inscribed in the same segment are
equal ; because they are all measured by
the half of the same arc BOC.
Cor. 2. Every angle BAD, inscribed in a
semicircle is aright angle ; because it is mea-
sured by half the semicircumference BOD,
that is, by the fourth part of the whole cir-
cumference.
Cor. 3. Every angle BAG, inscribed in a
segment greater than a semicircle, is an acute
angle ; for it is measured by half of the arc
BOC, less than a semicircumference.
And every angle BOG, inscribed in a
segment less than a semicircle, is an obtuse
angle ; for it is measuf ed by half of the arc ^
BAG, greater than a semicircumference.
Cor. 4. The opposite angles A and G, of
aa inscribed quadrilateral ABGD, are to-
gether equal to tvv^o right angles : for the an-
gle BAD is measured by half the arc BGD,
the angle BGD is measured by half the arc
BAD ; hence the tvro angles BAD, BGD, ta-
ken together, are measured by the half of the
circumference ; hence their sum is equal to two right angles.
PROPOSITION XIX. THEOREM.
The angle formed by two chords, which intersect each other, is
measured by half the sum of the arcs included between its sides
56
GEOMETRY.
Let AB, CD, be two chords intersecting
each other at E : then will the angle
AEC, or DEB, be measured by half of
AC + DB.
Draw AF parallel to DC : then will
the arc DF be equal to AC (Prop. X.) ;
and the angle FAB equal to the angle
DEB (Book I. Prop. XX. Cor. 3.). But
the angle FAB is measured by half the
arc FDB (Prop. XVIII.); therefore, DEB
is measured by half of FDB ; that is, by half of DB + DF, or
half of DB + AC. In the same manner it might be proved that
the angle AED is measured by half of AFD + BC.
PROPOSITION XX. THEOREM.
The angle formed hy two secants, is measured by half the diffe-
i^ence of the arcs included between its sides.
Let AB, AC, be two secants : then
will the angle BAC be measured by
half the difference of the arcs BEC
and DF.
Draw DE parallel to AC : then will
the arc EC be equal to DF, and the
angle BDE equal to the angle BAC.
But BDE is measured by half the arc
BE ; hence, BAC is also measured by
half the arc BE ; that is, by half the
difference of BEC and EC, or half the
difference of BEC and DF.
PROPOSITION XXI. THEOREM.
The angle formed by a tangent and a chord, is measured by half
of the arc included between its sides.
BOOK III.
57
Let BE be the tangent, and AC the chord.
From A, the point of contact, draw the
diameter AD. The angle BAD is a right
angle (Prop. IX.), and is measured by
half the semicircumference AMD ; the
angle DAC is measured by the half of
DC: hence, BAD + DAC, or PAC, is
measured by the half of AMD plus the
half of DC, or by half the whole arc
AMDC.
It might be shown, by taking the difference between the an-
gles DAE, DAC, that the angle CAE is measured by half the
arc AC, included between its sides.
-«.e@o«N-
PROBLEMS RELATING TO THE FIRST AND THIRD BOOKS
PROBLEM I.
To divide a given straight line into two equal parts, ' *
*
><*
Let AB be the given straight line.
From the points A and B as centres, with
a radius greater than the half of AB, describe
two arcs cutting each other in D ; the point
D will be equally distant from A and B. Find,
in like manner, above or beneath the line AB, -
a second point E, equally distant from the
points A and B ; through the two points D
and E, draw the line DE : it will bisect the
line AB in C.
For, the two points D and E, being each equally distant from
the extremities A and B, must both lie in the perpendicular
raised from the middle of AB (Book I. Prop. XVI. Cor.). But
only one straight line can pass through two given points ; hence
the line DE must itself be that perpendicular, which divides
AB into two equal parts at the point C.
X^
59
GEOMETRY.
PROBLEM II.
At^ given pointy in a given straight line, to erect a perpendicu-
lar to this line.
>N>
■^
Let A be the given point, and BC the
given line.
Take the points B and C at equal dis-
tances from A ; then from the points B and
C as centres, vi^ith a radius greater than
BA, describe two arcs intersecting each
other in D ; draw AD : it will be the perpendicular required.
For, the point D, being equally distant from B and C, must
be in the perpendicular raised from the middle of BC (Book I.
Prop. XVI.) ; and since two points determine a hne, AD is that
perpendicular.
Scholium, The same construction serves for making a right
angle BAD, at a given point A, on a given straight line BC.
PROBLEM III.
From a given point, without a straight line, to let fall a perpen-
dicular on this line.
Let Abe the point, and BD the straight
line.
From the point A as a centre, and with
a radius sufficiently great, describe an
arc cutting the line BD in the two points
B and D ; then mark a point E, equally
distant from the points B and D, and
draw AE : it will be the perpendicular required.
For, the two points A and E are each equally distant from
the points B and D ; hence the line AE is a perpendicular
passing through the middle of BD (Book I. Prop. XVL Cor.).
PROBLEM IV.
At a point in a given line, to make an angle equal to a given
angle.
BOOK III. 59
Let A be the given point, AB the given line, and IKL the
given angle.
From the vertex K, as a cen- t o<^
tre, with any radius, describe the ^/^1\ .^-^vv
arc IL, terminating in the two- ^^ \ ^^^ \
sides of the angle. From the '^^^ 1 .A B~
point A as a centre, with a dis-
tance AB, equal to Kl, describe the indefinite arc BO ; then
take a radius equal to the chord LI, with which, from the point
B as a centre, describe an arc cutting the indefinite arc BO, in
D ; draw KD ; and the angle. DAB will be equal to the given
angle K.
For, the two arcs BD, LI, have equal radii, and equal chords ;
hence they are equal (Prop. IV.) ; therefore the angles BAD,
IKL, measured by them, are equal.
PROBLEM V.
To divide a given arc, or a given angle, into two equal parts.
First. Let it be required to divide the
arc AEB into two equal parts. From the
points A and B, as centres, with the same
radius, describe two arcs cutting each other
in D ; through the point D and the centre
C, draw CD : it will bisect the arc AB in
the" point E.
For, the two points C and D are each
equally dptant from the extremities A and ^(^
B of the Aord AB ; hence the line CD bi-
sects the^hord at right angles (Book I. Prop. XVI. Cor.) ;
hence, it bisects the arc AB in the point E (Prop. VI.).
Secondly. Let it be required to divide the angle ACB into
two equal parts. We begin by describing, from the vertex C
as a centre, the arc AEB ; w^iich is then bisected as above. It
is plain that the line CD will divide the angle ACB into two
equal parts.
Scholium. By the same construction, each of the halves
AE, EB, may be divided into two equal parts ; and thus, by
successive subdivisions, a given angle, or a given arc may
be divided into four equal parts, into eight, into sixteen,
and so on.
ao
GEOMETRY.
PROBLEM VI.
Through a given point, to draw a parallel to a given straight
line.
Let A be the given point, and BC |;^ -p^
the given hne. JS
From the point A as a centre, with
a radius greater than the shortest dis-
tance from A to BC, describe the in- Jl i>I
definite arc EO ; from the point E as O
a centre, with the same radius, describe the arc AF ; make
ED— AF, and draw AD : this will be the parallel required.
For, drawing AE, the alternate angles AEF, EAD, are evi-
dently equal ; therefore, the lines AD, EF, are parallel (Book 1.
Prop. XIX. Cor. 1.).
PROBLEM VIL
Two angles of a triangle being given, to find the third,
Dr^wthe indefinite line DEF;
at any point as E, make the an-
gle DEC equal to one of the
given angles, and the angle
CEH equal to the other : the
remaining angle HEF will be
the third angle required ; be-
cause those three angles are
together equal to two right angles (Book I. Prop. I.
XXV).
and
PROBLEM VIII.
Two sides of a triangle, and the angle which they contain, being
given, to describe the triangle,
Ltt the lines B and C be equal to
the given sides, and A the given an-
Having drawn the indefinite line
DE, at the point D, make the angle _
EDF equal to the given angle A ; ^ ^
then take DG=B, DH=C, and draw GH ; DGH will be the
triangle required (Book I. Prop. V.).
BOOK III.
01
PROBLEM IX.
A side and two angles of a triangle being given, to describe the
triangle.
The two angles will either be both ad-
jacent to the given side, or the one adja-
cent, and the other opposite : in the lat-
ter case, find the third angle (Prob.
VII.) ; and the two adjacent angles will
thus be known : draw the straight line
DE equal to the given side : at the point D, make an angle
EDF equal to one of the adjacent angles, and at E, an angle
DEG equal to the other ; the two lines DF, EG, will cut each
other in H ; and DEH will be the triangle required (Book I.
Prop. VI.).
PROBLEM X.
The three sides of a triangle being given, to describe the triangle.
Let A, B, and C, be the sides.
Draw DE equal to the side A :
from the point E as a centre, with
a radius equal to the second side B,
describe an arc ; from D as a cen-
tre, with a radius equal to the third
side C, describe another arc inter-
secting the former in F ; draw DF,
EF ; and DEF will be the triangle
required (Book I. Prop. X.).
Scholium. If one of the sides were greater than the sum of
the other two, the arcs would not intersect each other : but the
solution will always be possible, when the sum of two sides, any
how taken, is greater than the third.
F
GEOMETRY.
PROBLEM XI.
Two sides of a triangle^ and the angle opposite one of them, being
given, to describe the triangle.
Let A and B be the given sides, and C the given angle.
There are two cases.
First. When the angle C is a right
angle, or when it is obtuse, make
the angle EDF=C; take DE=A ;
from the point E as a centre,
with a radius equal to the given
side B, describe an arc cutting DF
in F; draw EF : then DEF will be
the triangle required.
In this first case, the side B must
be greater than A ; for the angle C,
being a right angle, or an obtuse an-
gle, is the greatest angle of the tri-
angle, and the side opposite to it must, therefore, also be the
greatest (Book I. Prop. XIIL).
Ai-
Secondly, If the angle C is
acute, and B greater than A, the
^ame construction will again ap-
ply, and DEF will be the triangle
required.
But if the angle C is acute, and
the side B less than A, then the
arc described from the centre E,
with the radius EF=B, will cut
the side DF in two points F and
G, lying on the same side of D :
hence there will be two triangles
DEF, DEG, either of which will
satisfy the conditions of the pro-
blem.
Bh
SchoUum, If the arc described with E as a centre, should
be tangent to the line DG, the triangle would be right angled,
and there would be but one solution. The problem would be
impossible in all cases,. if the side B were less than the perpen-
dicular let fall from E on the line DF,
BOOK III. 6B
PROBLEM XII.
T/ie adjacent sides of a parallelogram, with the angle which they
contain, being given, to describe the parallelogram.
Let A and B be the given sides, and C the given angle.
Drav^rthe hne DE=A; at i\vi
point D, make the angle EDF—
C ; take DF=B ; describe two
arcs, the one from F as a cen-
tre, with a radius FG=DE, the
L
other from E as a centre, with
a radius EG=DF; to the point Ai — i
G, where these arcs intersect b i 1
each other, draw FG, EG ;
DEGF will be the parallelogram required.
For, the opposite sides are equal, by construction ; hence the
figure is a parallelogram (Book I. Prop. XXIX.) : and it is
formed with the given sides and the given angle.
Cor, If the given angle is a right angle, the figure will be
a rectangle ; if, in addition to this, the sides are equal, it will
be a square.
PROBLEM Xin.
To find the centre of a given circle or arc.
Take three points. A, B, C, any-
where in the circumference, or the
arc; drawAB,BC, or suppose them
to be drawn ; bisect those two lines
by the perpendiculars DE, FG :
the point O, where these perpen-
diculars meet, will be the centre
sought (Prop. VI. Sch.).
Scholium. The same construc-
tion sei-ves for making a circum-
ference pass through three given points A, B, C ; and also for
describing a circumference, in which, a given triangle ABC
shall be inscribed.
CA
GEOMETRY.
PROBLEM XIV.
Through a given pointy to draw a tangent to a given circle.
If the given point A lies in the circum-
ference, draw the radius CA, and erect
AD perpendicular to it : AD will be the
tangent required (Prop. IX.).
If the point A lies without the circle,
join A and the centre, by the straight
line CA : bisect CA in O ; from O as a
centre, with the radius OC, describe a
circumference intersecting the given cir-
cumference in B ; draw AB : this will be
the tangent required.
For, drawing CB, the angle CBA be-
ing inscribed in a semicircle is a right
angle (Prop. XVIII. Cor. 2.) ; therefore
AB is a perpendicular at the extremity
of the radius CB ; therefore it is a tan-
gent.
Scholium. When the point A lies without the circle, there
w^ill evidently be always two equal tangents AB, AD, passing
through the point A : they are equal, because the right angled
triangles CBA, CDA, have the hypothenuse CA common, and
the side CB = CD; hence they are equal (Book I. Prop. XVII.);
hence AD is equal to AB, and also the angle CAD to CAB.
And as there can be but one line bisecting the angle BAC, it
follows, that the line which bisects the angle formed by two
tangents, must pass through the centre of the circle.
PROBLEM XV.
To inscribe a circle in a. given triangle.
Let ABC be the given triangle.
Bisect the angles A and B, by
the lines AO and BO, meeting in
the point O ; from the point O,
let fall the perpendiculars OD,
OE, OF, on the three sides of the
triangle: these perpendiculars will
all be equal. For, by construe-
BOOK III. 65
lion, we have the angle DAO=OAF, the right angle ADO=
AFO ; hence the third angle AOD is equal to the third AOF
(Book I. Prop. XXV. Cor. 2.). Moreover, the side AO is com-
mon to the tvi^o triangles AOD, AOF ; and the angles adjacent
to the equal side are equal : hence the triangles themselves are
equal (Book I. Prop. VI.) ; and DO is equal to OF. In the same
manner it may be shown that the two triangles BOD, BOE,
are equal ; therefore OD is equal to OE ; therefore the three
perpendiculars OD, OE, OF, are all equal.
Now, if from the point O as a centre, with the radius OD,
a circle be described, this circle will evidently be inscribed in
the triangle ABC ; for the side AB, being perpendicular to the
radius at its extremity, is a tangent ; and the same thing is true
of the sides BC, AC,
Scholium. The three lines which bisect the angles of a tri-
angle meet in the same point.
PROBLEM XVI.
On a given straight line to describe a segment that shall contain
a given angle ; that is to say, a segment such, that all the an-
gles inscribed in it, shall be equal to the given angle.
Let AB be the given straight line, and C the given angle.
Produce AB towards D ; at the point B, make the angle
DBE=C; draw BO perpendicular to BE, and GO perpen-
dicular to AB, through the middle point G ; and from the point
O, where these perpendiculars meet, as a centre, with a dis-
tance OB, describe a circle : the required segment will be
AMB.
For, since BF is a perpendicular at the extremity of the
radius OB, it is a tangent, and the angle ABF is measured by
half the arc AKB (Prop. XXL). Also, the angle AMB, being
an inscribed angle, is measured by half the arc AKB : hence
we have AMB=ABF=:EBD=C : hence all the angles in-
scribed in the segment AMB are equal to the given angle C.
F*
66 GEOMETRY.
Scholium. If the given angle were a right angle, the required
segment would be a semicircle, described on AB as a diameter.
PROBLEM XVII.
To find the numerical ratio of two given straight lineSy these lines
being supposed to have a common measure.
Let AB and CD be the given lines. A C
From the greater AB cut off a part equal to the less
CD, as many times as possible ; for example, twice,
with the remainder BE.
From the line CD, cut off a part equal to the re-
mainder BE, as many times as possible ; once, for ex-
ample, with the remainder DF.
From the first remainder BE, cut off a part equal to
the second DF, as many times as possible ; once, for
example, with the remainder BG.
From the second remainder DF, cut off a part equal \.q.
to BG the third, as many times as possible.
Continue this process, till a remainder occurs, which ^
is contained exactly a certain number of times in the preced-
ing one.
Then this last remainder will be the common measure of the
proposed lines ; and regarding it as unity, we shall easily find
the values of the preceding remainders ; and at last, those of
the two proposed lines, and hence their ratio in numbers.
Suppose, for instance, we find GB to be contained exactly
twice in FD ; BG w^ill be the common measure of the two pro-
posed lines. Put BG=1 ; we shall have FD = 2 : but EB con-
tains FD once, plus GB ; therefore we have EB = 3 : CD con-
tains EB once, plus FD ; therefore we have CD=:5 : and,
lastly, AB contains CD twice, plus EB ; therefore we have
AB = 13 ; hence the ratio of the hues is that of 13 to 5. If the
line CD were taken for unity, the line AB would be ^^ ; if AB
were taken for unity, CD would be ^3.
Scholium, The method just explained is the same as that
employed in arithmetic to find the common divisor of two num-
bers ; it has no need, therefore, of any other demonstration.
How far soever the operation be continued, it is possible
that no remainder may ever be found, which shall be contained
an exact number of times in the preceding one. When this
happens, the two lines have no common measure, and are said
to be int)ommensurable. An instance of this will be seen after-
BOOK III. C7
wards, in the ratio of the diagonal to the side of the squaro.
In those cases, therefore, the exact ratio in numbers cannot be
found ; but, by neglecting the last remainder, an approximate
ratio will be obtained, more or less correct, according as the
operation has been continued a greater or less number of times.
PROBLEM XVIII.
Two angles being given, to find their common measure, if they
have one, and by means of it, their ratio in numbers.
Let A and B be the given an-
gles.
With equal radii describe the
arcs CD, EF, to serve as mea-
sures for the angles : proceed
afterwards in the comparison of
the arcs CD, EF, as in the last
problem, since an arc may be cut off from an arc of the same
radius, as a straight line from a straight line. We shall thus
arrive at the common measure of the arcs CD, EF, if they have
one, and thereby at their ratio in numbers. This ratio will be
the same as that of the given angles (Prop. XVII.) ; and if DO
is the common measure of the arcs, DAO will be that of the
angles.
Scholium. According to this method, the absolute value of
an angle may be found by comparing the arc which measures
it to the whole circumference. If the arc CD, for example, is
to the circumference, as 3 is to 25, the angle A will be /^ of four
right angles, or if of one right angle.
It may also happen, that the arcs compared have no com-
mon measure ; in which case, the numerical ratios of the angles
will only be found approximatively with more or less correct-
ness, according as the operation has been continued a greater
or less number of times.
68
GEOMETRY.
%■
BOOK IV.
OF THE PROPORTIONS OF FIGURES, AND THE MEASUREMI>i>{T
OF AREAS.
Definitions,
1. Similar figures are those which have the angles of the one
equal to the angles of the other, each to each, and the sides
about the equal angles proportional.
2. Any two sides, or any two angles, which have like po-
sitions in two similar figures, are called homologous sides or
angles.
3. In two different circles, similar arcs, sectors, or segments,
are those which correspond to equal angles at the centre.
Thus, if the angles A and O are equal,
the arc BC will be similar to DE, the
sector BAG to the sector DOE, and the
segment whose chord is BC, to the seg-
ment whose chord is DE.
4. The base of any rectilineal figure, is the side on which
the figure is supposed to stand.
5. The altitude of a triangle is the per- a.
pendicular let fall from the vertex of an
angle on the opposite side, taken as a
base. Thus, AD is the altitude of the
triangle BAG
6. The altitude of a parallelogram is the
perpendicular which measures the distance
between two opposite sides taken as bases.
Thus, EF is the altitude of the parallelo-
gram DB.
7. The altitude of a trapezoid is the per-
pendicular drawn between its two parallel
sides. Thus, EF is the altitude of the trape-
zoid DB.
DEC
8. The area and surface of a figure, are terms very nearly
synonymous. The area designates more particularly the super-
ficial content of the figure. The area is expressed numeri-
BOOK IV. 69
cally by the number of times wm'ch the figure contains some
other area, that is assumed for its measuring unit.
9. Figures have equal areas, when they contain the same
measuring unit an equal number of times.
10. Figures which have equal areas are called equivalent.
The term equal, when applied to figures, designates those which
are equal in every respect, and which being applied to each
other will coincide in all their parts (Ax. 13.) : the term equi-
valent implies an equality in one respect only : namely, an
equality between the measures of figures.
We may here premise, that several of the demonstrations
are grounded on some of the simpler operations of algebra,
which are themselves dependent on admitted axioms. Thus,
if we have A=B + C, and if each member is multiplied by the
same quantity M, we may infer that AxM=BxM + CxM;
in like manner, if we have, A =6 + C, and D— E — C, and if the
equal quantities are added together, then expunging the +C
and — C, which destroy each other, we infer that A + D=B +
E, and so of others. All this is evident enough (Jf itself; but
in cases of difficulty, it will be useful to consult some agebrai-
cal treatise, and thus to combine the study of the two sciences.
PROPOSITION I. THEOREM.
Parallelograms which have equal bases and equal altitudes^ are
equivalent.
Let AB be the common base of-j) cp EDPCE
the two parallelograms ABCD, V S/'
ABEF : and since they are sup- \
posed to have the same altitude, \
their upper bases DC, FE, will be A B A B
both situated in one straight line parallel to AB.
Now, from the nature of parallelograms, we have AD— BC,
and AF=BE; for the same reason, we have DC=:AB, and
FE=AB ; hence DC=rFE : hence, if DC and FE be taken
away from the same line DE, the remainders CE and DF will
be equal : hence it follows that the triangles DAF, CBE, are
mutually eqilateral, and consequently eqiial (Book I. Prop. X.).
But if from the quadrilateral ABED, we take away the tri-
angle ADF, there will remain the parallelogram ABEF ; and
if from the same quadrilateral ABED, we take away the equal
triangle CBE, there will remain the parallelogram ABCD.
70 GEOMETRY.
Hence these two parallelograms ABCD, ABEF, which have
the same base and altitude, are equivalent
Cor, Every parallelogram is equivalent to the rectangle
which has the same base and the same altitude.
PROPOSITION II. THEOREM.
Evejy triangle is half the parallelogram which has the same base
and the same altitude.
Let ABCD be a parallelo-
gram, and ABE a triangle,
having the same base AB,
and the same altitude : then
will the triangle be half the
parallelogram. FA B
For, since the triangle and the parallelogram have the same
altitude, the vertex E of the triangle, will be in the line EC, par-
allel to the base AB. Produce BA, and from E draw EF
parallel to AD. The triangle FBE is half the parallelogram
FC, and the triangle FAE half the parallelogram FD (Book I.
Prop. XXVIII. Cor.).
Now, if from the parallelogram FC, there be taken the par-
allelogram FD, there will remain the parallelogram AC : and
if from the triangle FBE, which is half the first parallelogram,
there be taken the triangle FAE, half the second, there will re-
main the triangle ABE, equal to half the parallelogram AC.
Cor 1 . Hence a triangle ABE is half of the rectangle ABGH,
which has the same base AB, and the same altitude AH : for
the rectangle ABGH is equivalent to the parallelogram ABCD
(Prop. I. Cor.).
Cor. 2. All triangles, which have equal bases and altitudes,
are equivalent, being halves of equivalent parallelograms.
PROPOSITION III. THEOREM.
Two rectangles having the same altitude^ are to each other as their
bases.
BOOK IV.
71
D
1?
(
1
J
i i
1,
=
A
E
B
Let ABCD, AEFD, be two rectan-
gles having the common altitude AD :
they are to each other as their bases
AB, AE.
Suppose, first, that the bases are
commensurable, and are to each other,
for example, as the numbers 7 and 4. If AB be divided into 7
equal parts, AE will contain 4 of those parts : at each point of
division erect a perpendicular to the base ; seven partial rect-
angles will thus be formed, all equal to each other, because all
have the same base and altitude. The rectangle ABCD will
:iontain seven partial rectangles, while AEFD will contain four:
hence the rectangle ABCD is to AEFD as 7 is to 4, or as AB
is to AE. The same reasoning may be applied to any other
ratio equally with that of 7 to 4 : hence, whatever be that ratio,
if its terms be commensurable, we shall have
ABCD : AEFD : : AB : AE.
Suppose, in the second place, that the bases
AB, AE, are incommensurable : it is to be
shown that we shall still have
ABCD : AEFD : : AB : AE.
For if not, the first three terms continuing
the same, the fourth must be greater or less
than AE. Suppose it to be greater, and that we have
ABCD : AEFD : : AB : AO.
Divide the line AB into equal parts, each less than EO.
There will be at least one point I of division between E and
O : from this point draw IK perpendicular to AI : the bases
AB, AI, will be commensurable, and thus, from what is proved
above, we shall have
ABCD : AIKD : : AB : AI.
But by hypothesis we have
ABCD : AEFD : : AB : AO.
In these two proportions the antecedents are equal ; hence
the consequents are proportional (Book II. Prop. IV.) ; and
we find
AIKD : AEFD : : AI : AO.
But AO is greater than AI ; hence, if this proportion is cor-
rect, the rectangle AEFD must be greater than AIKD : on
the contrary, however, it is less ; hence the proportion is im-
possible ; therefore ABCD cannot be to AEFD, as AB is to a
line greater than AE.
EIOB
72 GEOMETRY.
Exactly in the same manner, it may be shown that the fourth
term of the proportion cannot be less than AE ; therefore it is
equal to AE.
Hence, whatever be the ratio of the bases, two rectangles
ABCD, AEFD, of the same altitude, are to each other as their
bases AB, AE.
PROPOSITION IV. THEOREM.
Any two rectangles are to each other as the products of their bases
multiplied by their altitudes.
Let ABCD, AEGF, be two rectangles ; then will the rect-
angle,
ABCD : AEGF : : AB.AD : AF.AE.
Having placed the two rectangles, if ^ "D
so that the angles at A are vertical,
produce the sides GE, CD, till they
B
meet in H. The two rectangles ^
ABCD, AEHD, having the same al-
titude AD, are to each other as their q
bases AB, AE : in like manner the
two rectangles AEHD, AEGF, having the same altitude AE,
are to each other as their bases AD, AF : thus we have the
two proportions,
ABCD : AEHD : : AB : AE,
AEHD: AEGF : : AD : AF.
Multiplying the corresponding terms of these proportions
together, and observing that the term AEHD may be omit-
ted, since it is a multiplier of both the antecedent and the con-
sequent, we shall have
ABCD : AEGF : : ABxAD : AExAF.
Scholium. Hence the product of the base by the altitude may
be assumed as the measure of a rectangle, provided we under-
stand by this product, the product of two numbers, one of
which is the number of linear units contained in the base, the
other the number of linear units contained in the altitude. This
product will give the number of superficial units in the surface ;
because, for one unit in height, there are as many superficial
units as there are linear units in the base ; for two units in
height twice as many ; for three units in height, three times as
many, &c.
Still this measure is not absolute, but relative : it supposes
BOOK IV
7a
that the area of any other rectangle is computed m a similar
manner, by measuring its sides with the same linear unit ; a
second product is thus obtained, and the ratio of the two pro-
ducts is the same as that of the rectangles, agreeably to the
proposition just demonstrated.
For example, if the base of the rectangle A contains three
units, and its altitude ten, that rectangle will be represented
by the number 3x 10, or 30, a number which signifies nothing
while thus isolated ; but if there is a second rectangle B, the
base of which contains twelve units, and the altitude seven, this
second rectangle will be represented by the number 12x7 =
84 ; and we shall hence be entitled to Conclude that the two
rectangles are to each other as 30 is to 84 ; and therefore, if
the rectangle A were to be assumed as the unit of measurement
in surfaces, the rectangle B would then have |4 for its absolute
measure, in other words, it would be equal to ^^ of a super-
ficial unit.
It is more common and more
simple, to assume the square as
the unit of surface ; and to se-
lect that square, whose side is
the unit of length. In this case
the measurement which we have
regarded merely as relative, becomes absolute : the number 30,
for instance, by which the rectangle A was measured, now
represents 30 superficial units, or 30 of those squares, which
have each of their sides equal to unity, as the diagram exhibits.
In geometry the product of two lines frequently means the
same thing as their rectangle, and this expression has passed
into arithmetic, where it serves to designate the product of two
unequal numbers, the expression square being employed to
designate the product of a number multiplied by itself.
The arithmetical squares of 1, 2, 3,
&c. are 1, 4, 9, &c. So likewise, the
geometrical square constructed on a
double line is evidently four times
greater than the square on a single one ;
on a triple fine it is nine times great-
er, i&c.
A
>^ Of
74 GEOMETRY.
PROPOSITION V. THEOREM.
The area of any parallelogram is equal to the product of its base
by its altitude.
For, the parallelogram ABCD is equivalent 3? j)
to the rectangle ABEF, which has the same
base AB, and the same altitude BE (Prop. I.
Cor.) : but this rectangle is measured by AB
xBE (Prop. IV. Sch.); therefore, ABxBE
is equal to the area of the parallelogram ABCD.
Cor. Parallelograms of the same base are to each other as
their altitudes ; and parallelograms of the same altitude are to
each other as their bases : for, let B be the common base, and
C and D the altitudes of two parallelograms :
then, BxC:BxD::C:D, (Book II. Prop. VII.)
And if A and B be the bases, and C the common altitude,
we shall have
AxC : BxC : : A : B.
And parallelograms, generally, are to each other as the pro-
ducts of their bases and altitudes.
PROPOSITION VI. THEOREM.
The area of a triangle is equal to the product of its base by half
its altitude.
For, the triangle ABC is half of the par-
allelogram ABCE, which has the same base
BC, and the same altitude AD (Prop. II.) ;
but the area of the parallelogram is equal to
BC X AD (Prop. V.) ; hence that of the trian-
gle must be iBC x AD, or BC x ^AD.
Cor. Two triangles of the same altitude are to each other as
their bases, and two triangles of the same base are to each
other as their altitudes. And triangles generally, are to each
other, as the products of their bases and altitudes.
BOOK iV. 76
PROPOSITION VII. ' THEOREM.
The area of a trapezoid is equal to its altitude multiplied by the
half sum of its parallel bases.
Let ABCD be a trapezoid, EF its alti-
tude, AB and CD its parallel bases ; then
will its area be equal to EFx i(AB + CD).
Through I, the middle point of the side
BC, draw KL parallel to the opposite side
AD ; and produce DC till it meets KL. ^ ^
In the triangles IBL, ICK, we have the side IB=IC, by
construction; the angle LIB = CIK; and since CK and BL
are parallel, the angle IBL=ICK (BookL Prop. XX. Cor. 2.);
hence the triangles are equal (Book L Prop. VL) ; therefore,
the trapezoid ABCD is equivalent to the parallelogram ADKL,
and is measured by EF x AL.
But we have AL=DK ; and since the triangles IBL and
KCI are equal, the side BL=CK: hence, AB + CD- AL +
DK=2AL ; hence AL is the half sum of the bases AB, CD ;
hence the area of the trapezoid ABCD, is equal to the altitude
EF multiplied by the half sum of the bases AB, CD, a result
which is expressed thus: ABCD=EFx^^^— .
Scholium. If through I, the middle point of BC, the line IH
be drawn parallel to the base AB, the point H will also be the
middle of AD. For, since the figure AHIL is a parallelogram,
as also DHIK, their opposite sides being parallel, we have
AH=IL, and DII=IK; but since the triangles BIL, CIK, are
equal, we already have IL=IK; therefore, AH = DH.
It may be observed, that the line HI=AL is equal to
— ; hence the area of the trapezoid may also be ex-
pressed by EF x HI : it is therefore equal to the altitude of the
trapezoid multiplied by the line which connects the middle
points of its inclined sides.
76 GEOMETRY.
PROPOSITION VIII. THEOREM.
If aline is divided into two parts, the square described on the
whole line is equivalent to the sum of the squares described on
the parts, together with twice the rectangle contained by the
parts. I
Let AC be the line, and B the point of division ; then, is
AC2 or (AB + BC)2=ABHBC2+2ABxBC.
Construct the square ACDE ; take AF= S H J O
AB ; draw FG parallel to AC, and BH par-
allel to AE. ^ fi — O
The square ACDE is made up of four parts ;
the first ABIF is the square described on AB,
since we made AF=AB : the second IDGH is K B C
the square described on IG, or BC ; for since we have AC =
AE and AB=AF, the difference, AC — AB must be equal to
the difference AE — AF, which gives BC=EF ; but IG is equal
to BC, and DG to EF, since the lines are parallel ; therefore
IGDH is equal to a square described on BC. And those two
squares being taken away from the whole square, there re-
mains the two rectangles BCGI, EFIH, each of which is mea-
sured by AB X BC ; hence the large square is equivalent to the
two small squares, together with the two rectangles.
Cor. If the line AC were divided into two equal parts, the
two rectangles EI, IC, would become squares, and the square
described on the whole line would be equivalent to four times
the square described on half the line.
Scholium. This property is equivalent to the property de-
monstrated in algebra, in obtaining the square of a binominal ;
which is expressed thus :
(a-Vbf=a^-V2abVh\
PROPOSITION IX. THEOREM.
The square described on the difference oftwe lines, is equivalent
to the sum of the squares described on the Jhe^, r^inus twice
the rectangle contained by the lines.
BOOK IV.
77
B.
Let AB and BC be two lines, AC their difference ; then is
AC^, or (AE— BC)2=AB2+BC2— 2ABxBC.
Describe the square ABIF ; take AE Xi ] ? G- I
= AC ; draw CG parallel to to BI, HK
parallel to AB, and complete the square
£FLK.
The two rectangles CBIG, GLKD,
arc each measured by AB x BC ; take
them away from the whole figure ^
ABILKEA, which is equivalent to ^ ^
AB^-f BC^ and there will evidently remain the square ACDE;
hence the theorem is true.
Scholium, This proposition is equivalent to the algebraicai
formula, (a—by=ar-~-2ab+b\
G I
PROPOSITION X. THEOREM.
The rectangle contained by the sum and the difference of two
lines, is equivalent to the difference of the squares of those
lines.
Let AB, BC, be two lines ; then, will
(AB+BC) X (AB— BC)=AB2— BC2.
On AB and AC, describe the squares n
ABIF, ACDE ; produce AB till the pro-
duced part BK is equal to BC ; and p,
complete the rectangle AKLE.
The base AK of the rectangle EK,
is the sum of the two lines AB, BC ; its
altitude AE is the difference of the
same lines ; therefore the rectangle j
AKLE is equal to (AB + BC) x (AB—
BC). But this rectangle is composed of the two parts ABHE
+ BHLK ; and the part BHLK is equal to the rectangle EDGP,
because BH is equal to DE, and BK to EF ; hence AKLE is
equal to ABHE + EDGF. These two parts make up the square
ABIF minus the square DHIG, which latter is equal to a
square described on BC : hence wo have
(AB+BC) X (AB— BC)=AB2— BC2
Scholium. This proposition is equivalent to the algebraical
formula, (a+6) x (a — b)=a^ — b^.
H
D
C B K
G*
78
GEOMETRY.
PROPOSITION XI. THEOREM
The square described on the hypothenuse of a right angled tr> -
angle is equivalent to the sum of the squares described on th4
other two sides.
Let the triangle ABC be right
angled at A. Having described
squares on the three sides, let
fall from A, on the hypothenuse,
the perpendicular AD, which
produce to E; and draw the
diagonals AF, CH.
The angle ABF is made up
of the angle ABC, together with
the right angle CBF ; the angle
CBH is made up of the same
angle ABC, together with the
right angle ABH ; hence the - ^ e
angle ABF is equal to HBC. But we have AB
sides of the same square ; and BF==BC, for the same reason
therefore the triangles ABF, HBC, have two sides and the in-
cluded angle in each equal ; therefore they are themselves
equal (Book I. Prop. V.).
The triangle ABF is half of the rectangle BE, because they
have the same base BF, and the same altitude BD (Prop. II.
Cor. 1.). The triangle HBC is in like manner half of the
square AH : for the angles BAC, BAL, being both right angles,
AC and AL form one and the same straight line parallel to
HB (Book I. Prop. HI.) ; and consequently the triangle HBC,
and the square AH, which have the common base BH, have
also the common altitude AB ; hence the triangle is half of the
square.
The triangle ABF has already been proved equal to the tri-
BH, being
angle HBC ;
hence the rectangle BDEF, which is double of
the triangle ABF, must be equivalent to the square AH, which
is double of the triangle HBC. In the same manner it may be
proved, that the rectangle^ CDEG is equivalent to the square
AI. But the two rectangles BDEF, CDEG, taken together,
make up the square BCGF : therefore the square BCGF, de-
scribed on the hypothenuse, is equivalent to the sum of tht
squares ABHL, ACIK, described on the two other sides ; is
other words, BC^^ABHAG^.
BOOK IV.
79
Co7\ 1. Hence the square of one of the sides of a right an-
gled triangle is equivalent to the square of the hypothenuse
diminished by the square of the other side ; which is Lius ex-
pressed : AB2=BC2— AC2.
Cor. 2. It has just been shown that the square AH is equi-
valent to the rectangle BDEF ; but by reason of the common
altitude BF. the square BCGF is to the rectangle BDEF as the
base BC is to the base BD ; therefore we have
BC2 : AB2 : : BC : BD.
Hence the square of the hypothenuse is to the square of one of
the sides about the right angle, as the hypothenuse is to the seg-
ment adjacent to that side. The word segment here denotes
that part of the hypothenuse, which is cut off by the perpen-
dicular let fall from the right angle : thus BD is the segment
adjacent to the side AB ; and DC is the segment adjacent to
the side AC. We might have, in like manner,
BC2 : AC2 : : BC : CD.
Cor. 3. The rectangles BDEF, DCGE, having likewise the
same altitude, are to each other as their bases BD, CD. But
these rectangles are equivalent to the squares AH, AI ; there-
fore we have AB^ : AC^ : : BD : DC.
Hence the squares of the two sides containing the right angle,
are to each other as the segments of the hypothenuse which lie
adjacent to those sides. '
Cor. 4. Let ABCD be a square, and AC its
diagonal : the triangle ABC being right an-
gled and isosceles, we shall have AC^=AB^+
BC^=2AB^: hence the square described on the
diagonal AC, is double of the square described
on the side AB.
This property may be exhibited more plainly,
by drawing parallels to BD, through the points A and C, and
parallels to AC, through the points B and D. A new square
EFGH will thus be formed, equal to the square of AC. Now
EFGH evidently contains eight triangles each equal to ABE ;
and ABCD contains four such triangles : hence EFGH is
double of ABCD.
Since we have AC^ : AB^ : : 2 : 1 ; by extractmg the
square roots, we shall have AC : AB : : \/2 : 1 ; hence, the
diagonal of a square is incommensurable with its side ; a pro-
perty which will be explained more fully in another place.
80
GEOMETRY.
PROPOSITION XII. THEOREM.
In every triangle, the square of a side opposite an acute angle fa
less than the sum of the squares of the other two sides, by twice
the rectangle contained by the base and the distance from the
acute angle to the foot of the perpendicular let fall from the
opposite angle on the base, or on the base produced.
Let ABC be a triangle, and AD perpendicular to the base
CB ; then will AB^^ AC2+ BC^— 2BC x CD.
There are two cases.
First. When the perpendicular falls within
the triangle ABC, we have BDrrrBC— CD,
and consequently BD2=BC2+CD^— 2BC
xCD (Prop. IX.). Adding AD^ to each,
and observing that the right angled trian-
gles ABD, ADC, give ADHBD^^ AB^, and
ADHCD2=AC2, we have AB^-BC^^- _
AC^— 2BCxCD. ^
Secondly. When the perpendicular AD
falls without the triangle ABC, we have BD
= CD— BC ; and consequently BD^^CD^-j-
BC2— 2CD X BC (Prop. IX.). Adding AD^
to both, we find, as before, AB^^BCHAC^
— 2BCxCD.
PROPOSITION XIII. THEOREM.
In every obtuse angled triangle, the square of the side opposite the
obtuse angle is greater than the sum of the squares of the other
two sides by twice the rectangle contained by the base and the
distance from the obtuse angle to the foot of the peipendicular
let fall from the opposite angle on the base produced.
Let ACB be a triangle, C the obtuse angle, and AD perpen-
dicular to BC produced ; then will AB2=AC2+BC-+2BCx
CD.
The perpendicular cannot fall within the
triangle ; for, if it fell at any point such as
E, there would be in the triangle ACE, the
right angle E, and the obtuse angle C, which
is impossible (Book L Prop. XXV. Cor. 3.) :
BOOK IV. 81
hence the perpendicular falls without ; and we have BD=BC
+ CD. From this there results BD^^BCHCDHSBC x CD
(Prop. VIII.). Adding AD^ to both, and reducing the sums as
in the last theorem, we find AB^^BCH AC-H2BC x CD.
Scholium. The right angled triangle is the only one in which
the squares described on the two sides are together equivalent
to the square described on the third ; for if the angle contained
by the two sides is acute, the sum of their squares will be
greater than the square of the opposite side ; if obtuse, it will
be less.
PROPOSITION XIV. THEOREM.
In any triangle, if a straight line he drawn from the vertex to the
middle of the base, twice the square of this line, together with
twice the square of half the base, is equivalent to the sum of the
squares of the other two\sides of the triangle.
Let ABC be any triangle, and AE a line drawn to the mid-
dle of the base BC ; then will
2AEH2BE2=AB2+AC2.
On BC, let fa'l the perpendicular ADr A
Then, by Prop. XII.
AC^riAEHEC^— SEC x ED.
And by Prop. XI 11.
AB2-AEHEB2+2EB x ED. j^^ ^-^
Hence, by adding, and observing that EB and EC are equal,
we have
AB2 + AC2r=2 AE2 + 2EB2.
Cor. Hence, in every parallelogram the squares of the sides
are together equivalent to the squares of the diagonals.
For the diagonals AC, BD, bisect each q q
other (Book I. Prop. XXXI.) ; consequently
the triangle ABC gives \ ""^^
AB^-f BC2z=2AEH 2BE2.
The triangle ADC gives, in like manner.
AD2+ DC2-2AE2 ^ 2DE2.
Adding the corresponding members together, and observing that
BE and DE are equal, we shall have ^
AB2+AD2fDC2-fBC2n4AE2+4DE2. >f>
But 4AE2 is the square of 2AE, or of AC ; ^DW is the square
of BD (Prop. VIII. Cor.) : hence the squares of the sides are
together equivalent to the squares of the diagonals.
82
GEOMETRY.
PROPOSITION XV. THEOREM.
If a line he drawn parallel to the base of a triangle^ it will divide
tfie other sides proportionally,
Eet ABC be a triangle, and DE a straight line drawn par
allel to the base BC ; then will
AD : DB : : AE : EC.
Draw BE and DC. The two triangles BDE,
DEC having the same base DE, and the same
altitude, since both their vertices lie in a line
parallel to the base, are equivalent (Prop. II.
Cor. 2.).
' The triangles ADE, BDE, whose common
vertex is E, have the same altitude, and are to
each other as their bases (Prop. VI. Cor.) ;
hence we have
ADE ; BDE : : AD : DB.
The triangles ADE, DEC, whose common vertex is D, have
also the same altitude, and are to each other as their bases ;
hence
ADE : DEC : : AE : EC.
But the triangles BDE, DEC, are equivalent ; and therefore,
we have (Book II. Prop. IV. Cor.)
AD : DB : : AE : EC.
Cor. 1. Hence, by composition, we have AD + DB : AD : :
AE + EC : AE, or AB : AD : : AC : AE ; and also AB :
BD : : AC : CE.
Cor. 2. If between two straight lines AB, CD, any number
of parallels AC, EF, GH, BD, &lc. be drawn, those straight
lines will be cut proportionally, and we shall have AE : CF '>
EG : FH : GB : HD.
For, let O be the point where AB and
CD meet. In the triangle OEF, the line
AC being drawn parallel to the base EF,
we shall have OE : AE : : OF : CF, or
OE : OF : : AE : CF. In the triangle
OGH, we shall likewise have OE : EG
: : OF : FH,orOE : OF : : EG : FH.
And by reason of the common ratio OE :
OF, those two proportions give AE : CF
: : EG : FH. It may be proved in the
same manner, that EG ; FH : : GB : HD, and so on ; hence
the lines AB, CD, are cut proportionally by the parallels AC,
EF, GH, i&c.
1
BOOK IV.
8d
PROPOSITION XVI. THEOREM.
Conversely, if two sides of a triangle are cut proportionally hy a
straight line, this straight line will be parallel to the third side.
In the triangle ABC, let the line DE be drawn, making
AD : DB : : AE : EC : then will DE be parallel to BC.
For, if DE is not parallel to BC, draw DO paral-
lel to it. Then, by the preceding theorem, we shall
have AD : DB : : AO : OC. But by hypothe-
sis, we have AD : DB : : AE : EC : hence we
must have AO : OC : : AE : EC,orAO ; AE
: : OC : EC ; an impossible result, since AO, the
one antecedent, is less than its consequent AE,
and OC, the other antecedent, is greater than its
consequent EC. Hence the parallel to BC, drawn from the
point I), cannot differ from DE ; hence DE is that parallel.
Scholium. The same conclusion would be true, if the pro-
portion AB : AD : : AC : AE were the proposed one. For
this proportion would give AB — AD : AD : : AC — AE :
AE, or BD : AD : : CE : AE.
PROPOSITION XVII. THEOREM.
The line which bisects the vertical angle of a triangle, divides the
base into two segments, which are proportional to the adjacent
sides.
bisecting
the angle
AC.
E
In the triangle ACB, let AD be drawn
CAB ; then will
BD : CD : : AB :
Through the point C, draw CE
parallel to AD till it meets BA
produced.
In the triangle BCE, the line AD
is parallel to the base CE ; hence
we have the proportion (Prop.
XV.),
BD : DC : : AB : AE.
But the triangle ACE is isos-
celes : for, since AD, CE are parallel, we have the angle ACE
=DAC, and the angle AEC=BAD (Book I. Prop. XX. Cor.
2 & 3.) ; but, by hypothesis, DACinBAD ; hence the an-
gle ACE— AEC, and consequently AErrAC (Book I. Prop.
XII.). In place of AE in the above proportion, substitute AC,
and we shall have BD : DC : ; AB : AC.
84 GEOMETRY.
PROPOSITION XVIII. THEOREM.
Two equiangular triangles have their homologous sides propor-
tional, and are similar.
Let ABC, CDE be two triangles which E
have their angles equal each to each, /\
namely, BAC=CDE, ABCz=:DCE and
ACBn DEC ; then the homologous sides,
or the sides adjacent to the equal angles,
will be proportional, so that we shall
have BC ; CE : : AB : CD : : AC :
DE.
Place the homologous sides BC, CE in the same straight
line ; and produce the sides BA, ED, till they meet in F.
Since BCE is a straight line, and the angle BCA is equal to
CED, it follows that AC is parallel to DE (Book I. Prop. XIX.
Cor. 2.). In like manner, since the angle ABC is equal to
DCE, the line AB is parallel to DC. Hence the figure ACDF
is a parallelogram.
In the triangle BFE, the line AC is parallel to the base FE ;
hence we have BC : CE : : BA : AF (Prop. XV.); or put-
ting CD in the place of its equal AF,
BC : CE : : BA : CD.
In the same triangle BEF, CD is parallel to BF which may
be considered as the base ; and we have the proportion
BC : CE : : FD : DE ; or putting AC in the place of its equal FD,
BC : CE : : AC : DE.
And finally, since both these proportions contain the same
ratio BC : CE, we have
AC : DE : : BA : CD.
Thus the equiangular triangles BAC, CED, have their ho-
mologous sides proportional. But two figures are similar when
they have their angles equal, each to each, and their homolo-
gous sides proportional (Def. 1.) ; consequently the equiangu-
lar triangles BAC, CED, are two similar figures.
Cor, For the similarity of two triangles, it is enough that
they have two angles equal, each to each ; since then, the
third will also be equal in both, and the two triangles will be
equiangular.
BOOK IV. m
SchuUum. Observe, that in similar triangles, the homolo-
gous sides arc opposite to the equal angles ; thus the angle ACB
being equal to DEC, the side AB is homologous to DC ; in like
manner, AC and DE are homologous, because opposite to the
equal angles ABC, DCE. When the homologous sides are de-
termined, it is easy'to form the proportions :
AB : DC : : AC : DE : : BC : CE.
PROPOSITION XIX. THEOREM.
Two triangles^ which have their homologous sides proportional^
are equiangular and similar.
In the two triangles BAC, DEF,
suppose we have BC : EF : : AB
. DE : : AC : DF ; then wilj the
triangles ABC, DEF have their an-
gles equal, namely, A=D, B=E,
C=F. ^
At the point E, make the angle
FEG = B, and at F, the angle EFG = C ; the third G will be
equal to the third A, and the two triangles ABC, EFG will be
equiangular (Book I. Prop. XXV. Cor. 2.). Therefore, by the
last theorem, we shall have BC : EF : : AB : EG ; but, by
hypothesis, we have BC : EF : : AB : DE; hence EG =DE.
By the same theorem, we shall a!so have BC ; EF : : AC :
FG ; and by hypothesis, we have BC : EF : : AC : DF ;
hence FG=:DF. Hence the triangles EGF, DEF, having their
three sides equal, each to each, are themselves equal (Book I.
Prop. X.). But by construction, the triangles EGF and
ABC are equiangular ; hence DEF and ABC are also equian-
gular and similar.
Scholium 1. By the last two propositions, it appears that in
triangles, equahty among the angles is a consequence of pro-
portionality among the sides, and conversely ; so that either of
those conditions sufficiendy determines the similarity of two
triangles. The case is different with regard to figures of
more than three sides : even in quadrilaterals, the proportion
between the sides may be altered without altering the angles,
or the angles may be altered without altering the proportion
between the sides ; and thus proportionality among the sides
cannot be a consequence of equality among the angles of two
quadrilaterals, or vice versa. It is evident, for example, that
H
86 GEOMETRY.
by drawing EF parallel to BC, the angles of
the quadrilateral AEFD, are made equal to
those of ABCD, though the proportion be-
tween the sides is different ; and, in like man-
ner, without changing the four sides AB, BC,
CD, AD, we can make the point B approach
D or recede from it, which will change the
angles.
Scholium 2. The two preceding propositions, which in strict-
ness form but one, together with that relating to the square of
the hypothenuse, are the most important and fertile in results
of any in geometry : they are almost sufficient of themselves
for every application to subsequent reasoning, and for solving
every problem. The reason is, that all figures may be divided
into triangles, and any triangle into two right angled triangles.
Thus the general properties of triangles include, by implica-
tion, those of all figures.
PROPOSITION XX. THEOREM. ;
Two triangles, which have an angle of the one equal to an angle
of the other, and the sides containing those angles proportional,
are similar.
In the two triangles ABC, DEF, let
the angles A and D be equal ; then, if
AB : DE : : AC : DF, the two trian-
gles will be similar.
Take AG-.DE, and draw GH paral-
lel to BC. The angle AGH will be equal
to the angle ABC (Book I. Prop. XX.
Cor 3.) ; and the triangles AGH, ABC, will be equiangular : \
hence we shall have AB : AG : : AC : AH. But by hypo-
thesis, we have AB : DE : : AC : DF ; and by construction,
AG=DE : hence AH=DF. The two triangles AGH, DEF, '
have an equal angle included between equal sides ; therefore ^
they are equal : but the triangle AGH is similar to ABC : there*
fore DEF is also similar to ABC.
J
I
BOOK IV
#7
PROPOSITION XXI. THEOREM.
Two triangles, which have their homologous sides parallel, or
perpendicular to each other, are similar.
Let BAG, EDF, be two triangles.
First. If the side AB is parallel to DE, and
BC to EF, the angle ABC will be equal to
DEF (Book I. Prop. XXIV.) ; if AC is parallel
to DF, the angle ACB will be equal to DFE,
and also BAC to EDF ; hence the triangles
ABC, DEF, are equiangular; consequently
they are similar (Prop. XVIII.).
Secondly. If the side ©E is perpen-
dicular to AB, and the side DF to AC,
the two angles I and H of the quadri-
lateral AIDH will be right angles ; and
since all the four angles are together
equal to four right angles (Book I. Prop.
XXVI. Cor. l.),the remaining two I AH,
IDH, will be together equal to two right ^
angles. But the two angles EDF, IDH, are also equal to two
right angles : hence the angle EDF is equal to lAH or BAC.
In like manner, if the third side EF is perpendicular to the third
side BC, it may be shown that the angle DFE is equal to C, and
DEF to B : hence the triangles ABC, DEF, which have the
sides of the one perpendicular to the corresponding sides of the
other, are equiangular and similar.
Scholium. In the case of the sides being parallel, the homolo-
gous sides are the parallel ones : in the case of their being per-
pendicular, the homologous sides are the perpendicular ones.
Thus in the latter case DE is homologous with AB, DF with
AC, and EF with BC.
The case of the per[A3ndicular sides might present a rela-
tive position of the two triangles different from that exhibited
in the diagram. But we might always conceive a triangle
DEF to be constructed within the triangle ABC, and such that
its sides should be parallel to those of the triangle compared
with ABC ; and then the demonstration given in the text would
apply.
68
GEOMETRY.
PROPOSITION XXII. THEOREM.
In any triangle, if a line he drawn parallel to the base, then, all
lines drawn from the vertex will divide the base and the pai^-
allel into proportional parts.
Let DE be parallel to the base BC, and
the other lines drawn as in the figure ;
then will
DI : BF : : IK : FG : : KL : GH.
For, since \)\ is parallel to BF, the
triangles ADI and ABF are equiangu-
lar ; and we have DI : BF : : AI :
AF ; and since IK is parallel to FG,
we have in like manner AI : AF : :
IK : FG ; hence, the ratio AI : AF being common, we shall
have DI : BF : : IK : FG. In the same manner we shall
find IK : FG : : KL ; GH ; and so with the other segments ;
hence the line DE is divided at the points I, K, L, in the same
proportion, as the base BC, at the points F, G, H.
Cor. Therefore if BC were divided into equal parts at the
points F, G, H, the parallel DE would also be divided into equal
parts at the points 1, K, L.
PROPOSITION XXIII. THEOREM.
If from the right angle of a right angled triangle, a peiyendicu-
lar he let fall on the hypothenuse ; then,
\st. The< two partial triangles thus formed, will he similar to each
other, and to the whole triangle.
2d. Either side including the right angle will he a mean propor-
tional hetiveen the hypothenuse and the adjacent segment.
*Sd. The perpendicular will be a mean proportional between the
two segments of the hypothenuse.
Let BAC be a right angled triangle, and AD perpendicular
to the hypothenuse BC.
First. The triangles BAD and BAC
have the common angle B. the right
angle BDA=BAC, and therefore the
third angle BAD of the one, equal to
the third angle C, of the other (Book
I. Prop. XXV. Cor 2.) : hence those
two triangles are equiangular and
BOOK IV. 89
similar. In the same manner it may be shown that the trian-
gles DAC and BAG are similar ; hence all the triangles are
equiangular and similar.
Secondly. The triangles BAD, BAC, being similar, their
homologous sides are proportional. But BD in the small tri-
angle, and BA in the large one, are homologous sides, because
they lie opposite the equal angles BAD, BCA ; the hypothe-
nuse BA of the small triangle is homologous with the hypo-
thenuse BC of the large triangle : hence the proportion BD :
BA : : BA : BC. By the same reasoning, we should find
DC : AC : : AC : BC ; hence, each of the sides AB, AC, is
a mean proportional between the hypothenuse and the segment
adjacent to that side.
Thirdly. Since the triangles ABD, ADC, are similar, by
comparing their homologous sides, we have BD : AD ; : AD
: DC ; hence, the perpendicular AD is a mean proportional
between the segments BD, DC, of the hypothenuse.
Scholium. Since BD : AB : : AB : BC, the product of the
extremes will be equal to that of the means, or AB^=:BD.BC.
For the same reason we have AC^— DC.BC ; therefore AB^+
AC2=BD.BC + DC.BC= (BD + DC).BC=BC.BC=BC2; or
the square described on the hypothenuse BC is equivalent to
the squares described on the two sides AB, AC. Thus we again
arrive at the property of the square of the hypothenuse, by a
path very diflferent from that which formerly conducted us to
it : and thus it appears that, strictly speaking, the property of
the square of the hypothenuse, is a consequence of the more
general property, that the sides of equiangular triangles are
proportional. Thus the fundamental propositions of geometry
are reduced, as it were, to this single one, that equiangular tri-
angles have their homologous sides proportional.
It happens frequently, as in this instance, that by deducing
consequences from one or more propositions, we are led back
to some proposition already proved. In fact, the chief charac-
teristic of geometrical theorems, and one indubitable proof of
their certainty is, that, however we combine them together,
provided only our reasoning be correct, the results we obtain
are always perfectly accurate. The case would be different,
if any proposition were false or only approximately true : it
would frequently happen that on combining the propositions
together, the error would increase and become perceptible.
Examples of this are to be seen in all the demonstrations, in
which the reductio ad absurdum is employed. In such demon-
strations, where the object is to show that two quantities are
equal, we proceed by showing that if there existed the smallest
H*
90
GEOMETRY.
inequality between the quantities, a train of accurate reason-
ing would lead us to a manifest and palpable absurdity; from
which we are forced to conclude that the two quantities are
equal.
Cor. If from a point A, in the circumference
of a circle, two chords AB, AC, be drawn to
the extremities of a diameter BC, the triangle
BAG will be right angled at A (Book III. Prop. ^ J^ C
XVIII. Gor. 2.) ; hence, first, the perpendicular AD is a mean
proportional between the two segments BD, DG, of the diameter^
or what is the same, AD^=BD.DG.
Ilence also, in the second place, the chord AB is a mean pro-
portional between the diameter BC and the adjacent segment BD,
or, what is the same, AB^nBD.BC. In like manner, we have
AC^=GD.BG ; hence AB^ : AC^ : : BD : DC : and com-
paring AB'^ and AC^, to BC^, we have AB^ : BC^ : : BD :
BC, and AC^ : BC^ : : DC : BC. Those proportions between
the squares of the sides compared with each other, or with the
square of the hypothenuse, have already been given in the third
and fourth corollaries of Prop. XI.
PROPOSITION XXIV. THEOREM.
Two triangles having an angle in each equals are to each other
as the rectangles of the sides which contain the equal angles.
In the two triangles ABC, ADE, let the angle A be equal to
the angle A ; then will the triangle
ABC : ADE : : AB.AC : AD.AE.
Draw BE. The triangles
ABE, ADE, having the com-
mon vertex E, have the same
altitude, and consequently are
to each other as their bases
(Prop. VI. Cor.) : that is,
ABE : ADE : : AB : AD.
In like manner,
ABC
ABE
AC : AR
Multiply together the corresponding terms of these proportions,
omitting the common term ABE ; we have
ABC : ADE : AB.AC : AD.AE.
BOOK IV. 01
Cor. Jlcnce the two triangles would be equivalent, if the
rectangle AB.AC were equal to the rectangle AD.AE, or if
we had AB : AD : : AE : AC i which would happen if DC
were paraliv^.l to BE.
PROPOSITION XXV. THEOREM.
Two similar triangles are to each other as the squares described
on their homologous sides.
Let ABC, DEF, be two similar trian-
gles, having the angle A equal to D, and
the angle B=E.
Then, first, by reason of the equal an-
gles A and D, according to the last pro-
position, we shall have
ABC : DEF : : AB.AC : DE.DF.
Also, because the triangles are similar,
AB : DE : : AC : DF,
And multiplying the terms of this proportion by the corres-
ponding terms of the* identical proportion,
AC : DF : : AC : DF,
there v/ill result •
AB.AC : DE.DF : : AC^ : DP.
Consequently,
ABC : DEF : : AC^ : DF^.
Therefore, two similar triangles ABC, DEF, are to each
other as the squares described on their homologous sides AC,
DF, or as the squares of any other two homologous sides.
PROPOSITION XXVI. THEOREM.
7\vo similar polygons are composed of the same number of tri-
angles ^ similar each to each, and similarly situated.
92 GEOMETRY.
Let ABCDE, FGHIK, be two similar polygons.
From any angle A, in c
the polygon ABCDE,
draw diagonals AC, AD
to the other angles. From
the homologous angle F,
in the other polygon
FGHIK, draw diagonals
FH, FI to the other an-
gles.
These polygons being similar, the angles ABC, FGH, which
are homologous, must be equal, and the sides AB, BC, must
also be proportional to FG, GH, that is, AB : FG : : BC :
GH (Def. 1.). Wherefore the triangles ABC, FGH, have each
an equal angle, contained between proportional sides ; hence
they are similar (Prop. XX.) ; therefore the angle BCA is equal
to GHF. Take away these equal angles from the equal angles
BCD, GHI, and there remains ACD==FHI. But since the
triangles ABC, FGH, are similar, we have AC : FH : : BC :
GH ; and, since the polygons are similar, BC : GH : : CD :
HI ; hence AC : FH : : CD : HI. But the angle ACD, we
already know, is equal to FHI ; hence the triangles ACD, FHI,
have an equal angle in each, included between proportional
sides, and are consequently similar (Prop. XX.). In the same
manner it might be shown that all the remaining triangles are
similar, whatever be the number of sides in the polygons pro-
posed : therefore two similar polygons are composed of the
same number of triangles, similar, and similarly situated.
Scholium, The converse of the proposition is equally true :
If two polygons are composed cf the same number of triangles
similar and similarly situated, those two polygons will be similar.
For, the similarity of the respective triangles will give the
angles, ABC=FGH, BCA^GHF, ACD=FHI : hence BCD=
GHI, likewise CDE=HIK, &c. Moreover we shall have
AB : FG : : BC : GH : : AC : FH : : CD : HI,&c.; hence
the two polygons have their angles equal and their sides pro-
portional ; consequently they are similar.
PROPOSITION XXVII. THEOREM.
The conto^irs or perimeters of similar polygons are to each other
as the homologous sides : and the areas are to each other as
the squares described on those sides.
BOOK IV. 03
First. Since, by the
nature of similar figures,
we have AB : FG : :
BC : GH : : CD ; HI,
&c. we conclude from
this series of equal ratios
that the sum of the ante-
eecieuts AB + BC + CD,
&c., which makes up the perimeter of the first polygon, is to
the sum of the consequents FG + GH + HI, &c., which makes
up the perimeter of the second polygon, as any one antecedent
is to its consequent ; and therefcjre, as the side AB is to its cor-
responding side FG (Book II. Prop. X.).
Secondly. Since the triangles ABC, FGH are similar, we
shall have the triangle ABC : FGH : : AC^ : FH^ (Prop.
XXV.) ; and in like manner, from the similar triangles ACD,
FHI, we shall have ACD : FHI : : AC^ : FH^; therefore, by
reason of the common ratio, AC^ : FH^, we have
ABC : FGH : : ACD : FHI.
By the same mode of reasoning, we should find
ACD : FHI : : ADE : FlK;
and so on, if there were more triangles. And from this series
of equal ratios, we conclude that the sum of the antecedents
ABC + ACD + ADE, or the polygon ABCDE, is to the sum cf
the consequents FGH + FHI + FIK, or to the polygon FGHIK,
as one antecedent ABC, is to its consequent FGH, or as AB^
is to FG- (Prop. XXV.) ; hence the areas of similar poly-
gons are to each other as the squares described on the homolo-
gous sides.
Cor. If three similar figures were constructed, on the
three sides of a right angled triangle, the figure on the hypo-
thenuse would be equivalent to the sum of the other two : for
the three figures are proportional to the squares of their
homologous sides ; but the square of the hypothenuse is
equivalent to the sum of the squares of the two other sides ;
hence, &c.
PROPOSITION XXVIII. THEOREM.
The segments of two chords, which intersect each other vn a circle,
are reciprocally proportional.
u
GEOMETRY.
Let the chords AB and CD intersect at O : then will
AO : DO : : OC : OB.
Draw AC and BD. In the triangles ACO,
BOD, the angles at O are equal, being verti-
cal ; the angle A is equal to the angle D, be-
cause both are inscribed in the same segment
(Book III. Prop. XVIII. Cor. 1.) ; for the same
reason the angle C— B ; the triangles are there-
fore similar, and the homologous sides give the proportion
AO : DO : : CO : OB.
Cor. Therefore AO.OB=DO.CO: hence the rectangle
under the two segments of the one chord is equal to the rect-
angle under the two segments of the other.
PROPOSITION XXIX. THEOREM.
If from the same point without a circle, two secants he drawn
terminating in the concave arc, the whole secants will he recip'
srannJIg/ 2^r'e?po<rii^9^ty1 i-» i?ioit' eJCtcmul segments.
Let the secants OB, OC, be drawn from the point O
then will
OB : OC : : OD : OA.
For, drawing AC, BD, the triangles OAC,
OBD have the angle O common ; likewise the
angle B=C (Book IIL Prop. XYIII.Cor. 1.);
these triangles are therefore similar ; and their
homologous sides give the proportion,
OB : OC : : OD : OA.
Cor. Hence the rectangle OA.OB is equal
to the rectangle OC.OD. B
Scholium, This proposition, it may be observed, bears a
great analogy to the preceding, and differs from it only as the
two chords AB, CD, instead of intersecting each other within,
cut each other without the circle. The following proposition
may also be regarded as a particular case of the proposition
just demonstrated.
BOOK IV.
95
PROPOSITION XXX. THEOREM.
If from the same point without a circle , a tangent and a secant
be drawn, the tangent will he a mean proportional between the
secant and its external segment.
From the point O, let the tangent OA, and the secant OC be
be drawn ; then will
OC : OA : : OA : OD, or OA^^OC.OD.
For, drawing AD and AC, the triangles O
OAD, OAC, have the angle O common; also
the angle OAD, formed by a tangent and a
chord, has for its measure half of the arc AD
(Book III. Prop. XXI.) ; and the angle C has
the same measure : hence the angle OAD =
C ; therefore the two triangles are similar,
and we have the proportion OC : OA : :
AO : OD, which gives OA'-=OC.OD.
X
PROPOSITION XXXI. THEOREM.
If either angle of a triangle he bisected by a line terminating in
the opposite side, the rectangle of the sides including the bi-
sected angle, is equivalent to the square of the bisecting line
together with the rectangle contained by the segments of the
third side.
In the triangle BAC, let AD bisect the angle A ; then will
AB.AC=:ADHBD.DC.
Describe a circle through the three points
A, B, C ; produce AD till it meets the cir-
cumference, and draw CE.
The triangle BAD is similar to the trian-
gle EAC ; for, by hypothesis, the angle
BAD = EAC; also the angle iB=E,' since
they are both measured by half of the arc
AC ; hence these triangles are similar, and
the homologous sides give the proportion BA : AE : : AD :
AC ; hence BA.AC=AE.AD ; but AE=AD + DE, and multi-
plying each of these equals by AD, we have AE.AD=:AD'H
Ab.DE; now AD.DE=BD.DC (Prop. XXVIII.) ; hence,
finallv,
BA.AC==AD2+BD.DC.
m
OG
GEOMETIIY.
PROPOSITION XXXII. THEOREM.
In every triangle, the rectangle contained hij two sides is equiva-
lent to the rectangle contained by the diameter of the circum-
scribed circle^ and the perpendicular let fall upon the third
side.
In the triangle ABC, let AD be drawn perpendicular to BC ;
and let EC be the diameter of the circumscribed circle : then
will
AB.AC=AD.CE.
For, drawing AE, the triangles ABD,
AEC, are right angled, tiie one at D, the
other at A: also the angle B — E ; these tri-
angles are therefore similar, and they give
the proportion AB : CE : : AD : AC ; and
hence AB.ACrr:CE.AD.
Cor. If these equal quantities be multiplied by the same
quantity BC, there will result AB.AC.BC = CE.AD.BC ; now
AD.BC is double of the area of the triangle (Prop. VI.) ; there-
fore the product of three sides of a triangle is equal to its area
multiplied by twice the diameter of the ci?xumscribed circle.
The product of three lines is sometimes called a solid, for a
reason that shall be seen afterwards. Its value is easily con-
ceived, by imagining that the lines are reduced into numbers,
and multiplying these numbers together.
Scholium. It may also be demonstrated, that the area of
a triangle is equal to its perimeter multiplied by half the radius
of the inscribed circle.
For, the triangles AOB, BOC,
AOC, which have a common
vertex at O, have for their com-
mon altitude the radius of the
inscribed circle ; hence the sum
of these triangles will be equal
to the sum of the bases AB, BC,
AC, multiplied by half the radius
OD ; hence the*^ area of the triangle ABC is equal to the
perimeter multiplied by half the radius of the inscribed circle.
BOOK IV.
97
PROPOSITION XXXIII. THEOREM.
In every quadrilateral inscribed in a circle, the rectangle of the
two diagonals is equivalent to the sum of the rectangles of the
opposite sides.
In the quadrilateral ABCD, we shall have
AC.BD=AB.CD+AD.BC.
Take the arc CO = AD, and draw BO
meeting the diagonal AC in I.
The angle ABD=CBI, since the one
has for its measure half of the arc AD,
and the other, half of CO, equal to AD ;
the angle ADB=BCI, because they are
both inscribed in the same segment
AOB ; hence the triangle ABD is similar
to the triangle IBC, and we have the
proportion AD ; CI : : BD : BC ; hence AD.BC=:CI.BD.
Again, the triangle ABl is similar to the triangle BDC ; for the
arc AD being equal to CO, if OD be added to each of them,
we shall have the arc AO=r:DC ; hence the angle ABI is equal
to DBC ; also the angle BAI to BDC, because they are in-
scribed in the same segment ; hence the triangles ABI, DBC,
are similar, and the homologous sides give the proportion AB :
BD : : AI : CD ; hence AB.CD=AI.BD.
Adding the two results obtained, and observing that
AI.BD + CI.BD = (AI + CI).BD=:AC.BD,
we shall have
AD.BC+AB.CD=AC.BD.
OS
GEOMETRY.
^t
PROBLEMS RELATING TO THE FOURTH BOOK.
PROBLEM I.
To divide a given straight line into any number of equal parts,
or into parts proportional to given lines.
First. Let it be proposed to divide the line
AB into five equal parts. Through the ex-
tremity A, draw^ the indefinite straight line
AG ; and taking AC of any magnitude, apply
it five times upon AG ; join the last point
of division G, and the extremity B, by the
straight line GB ; then draw CI parallel to
GB : AI will be the fifth part of the line
AB ; and thus, by applying AI five times
upon AB, the line AB will be divided into
five equal parts.
For, since CI is parallel to GB, the sides AG, AB, are cut
proportionally in C and I (Prop. XV.). But AC is the fifth
part of AG, hence AI is the fifth part of AB, ^
Secondly. Let it be pro-
posed to divide the line AB
mto parts proportional to
the given lines P, Q, R.
Through A, draw the indefi-
nite hne AG ; make AC =
P, CD=Q, DE:=R; join
the extremities E and B ;
and through the points C,
D, draw CI, DF, parallel to EB ; the line AB will be divided
into parts AI, IF, FB, proportional to the given lines P,
Q, R.
For, by reason of the paraLcls CI, DF, EB, the parts AI,
IF, FB, are proportional to the parts AC, CD, DE ; and by
construction, these are equal to the given lines P, Q, R.
BOOK IV.
PROBLEM II.
To find a fourth proportional to three given lineSy A, B, C.
Draw the two indefi-
nite lines DE, DF, form-
ing any angle with each
other. Upon DE take
DA=A, and DB=B ;
upon DF take DC=C ;
draw AC ; and through
the point B, draw BX
parallel to AC ; DX will be the fourth proportional required ;
for, since BX is parallel to AC, we have the proportion
DA : DB : : DC : DX ; now the first three terms of this pro-
portion are equal to the three given lines : consequently DX is
the fourth proportional required.
Cor. A third proportional to two given lines A, B, may be
found in the same manner, for it will be the same as a fourth
proportional to the three lines A, B, B.
PROBLEM in.
To find a mean proportional between two given lines A and B.
Upon the indefinite line DF, take
DE=A, and EF— B : upon the whole
line DF, as a diameter, describe the
semicircle DGF ; at the point E,
erect upon tl}e diameter the perpen-
dicular EG meeting the circumfe-
rence in G ; EG will be the mean
proportional required.
For, the perpendicular EG, let fall from a point in the cir-
cumference upon the diameter, is a mean proportional between
DE EF, the two segments of the diameter (Prop. XXIII.
Co I .) ; and these segments are equal to the given lines A
and B.
Ai 1
PROBLEM IV.
To divide a given line into two parts, such that the greater part
shall be a mean proportional between the whole line and the
other part.
100
GEOMETRY.
Let AB be the given line.
At the extremity B of the line
AB, erect the perpendicular BC
equal to the half of AB ; from the
point C, as a centre, with the ra-
dius CB, describe a semicircle ;
draw AC cutting the circumfe-
rence in D ; and take AF=AD : A :f
the line AB will be divided at the point F in the manner re-
quired ; that is, we shall have AB : AF : : AF : FB.
For, AB being perpendicular to the radius at its extremity,
is a tangent ; and if AC be produced till it again meets the
circumference in E, we shall have AE : AB : : AB : AD
(Prop. XXX.) ; hence, by division, AE— AB : AB : : AB—
AD : AD. But since the radius is the half of AB, the diame-
ter DE is equal to AB, and consequently AE-t-AB= AD=AF ;
also, because AF=AD, we have AB — ^AD=r:FB ; hence
AF : AB : : FB : AD or AF ; whence, by exchanging the
extremes for the means, AB : AF : : AF : FB.
Scholium, This sort of division of the line AB is called di
vision in extreme and mean ratio : the use of it will be per-
ceived in a future part of the work. It may further be
observed, that the secant AE is divided in extreme and mean
ratio at the point D ; for, since AB=DE, we have AE : DE
: : DE : AD.
PROBLEM V.
Through a given point, in a given angle, to draw a line so that
the segments comprehended between the point and the two sides
of the angle, shall be equal.
Let BCD be the given angle, and A the given point.
Through the point A, draw AE paral-
lel to CD, make BE=CE, and through
the points B and A draw BAD ; this will
be the line required.
For, AE being paralV! +0 CD, we have
BE : EC : : BA : AD , W* B^.^TC ;
therefore BA=AD.
BOOK IV.
PROBLEM VI.
101
To describe a square that shall he equivalent to a given paralleh'
granif or to a given triangle.
First, Let ABCD be
the given parallelogram,
AB its base, DE its alti-
tude ; between AB and
DE find a mean propor-
tional XY ; then will the
square described upon
E
XY be equivalent to the parallelogram ABCD.
For, by construction, AB ; XY : : XY : DE ; therefore,
XY2= AB.DE ; but x\B.DE is the measure of the parallelogram,
and XY^ that of the square ; consequently, they are equiva-
lent.
Secondly, Let ABC be the
given triangle, BC its base,
AD its altitude : find a mean
proportional between BC and
the half of AD, and let XY be
that mean ; the square de-
scribed upon XY will be equi-
valent to the triangle ABC.
For, since BC : XY : : XY : lAD, it follows that XY^^
BC.iAD ; hence the square described upon XY is equivalent
to the triangle ABC.
PROBLEM VII.
Upon a given line, to describe a rectangle that shall he equiva-
lent to a given rectangle.
Let AD be the line, and ABFC the given rectangle.
Find a fourth propor
tional to the three lines
AD, AB, AC, and let AX
be that fourth propor-
tional ; a rectangle con-
structed with the lines
AD and AX will be equi-
valent to the rectangle ABFC.
For, since AD : AB : : AC : AX, it follows that AD. AX =
AB.AC ; hence the rectangle ADEX is equivalent to the rect-
angle ABFC.
102 GEOMETRY.
PROBLEM VIII.
To find two lines whose ratio shall he the same as the ratio of
two rectangles contained by given lines.
Let A.B, CD, be the rectangles contained by the given Hnes
A,B, Candl).
Find X, a fourth proportional to the three
lines B, C, D ; then will the two lines A and
X have the same ratio to each other as the
rectangles A.B and CD.
For, since B : C : : D : X, it follows that
CD=B.X; hence A.B : CD : : A.B : B.X
: : A : X.
Cor, Hence to obtain the ratio of the squares described
upon the given lines A and C, find a third proportional X to
the lines A and C, so that A : C : : C : X; you will then
have ,
A.Xrz:C2, or A2.X=A.C2 ; hence
A2 : C2 : : A : X.
PROBLEM IX.
To find a triangle that shall be equivaleik to a given polygon.
Let ABCDE be the given polygon.
Draw first the diagonal CE cutting off
the triangle CDE ; through the point
D, draw DF parallel to CE, and ipeet-
ing AE produced ; draw CF: the poly-
gon ABCDE will be equivalent to the
polygon ABCF, which has one side
less than the original polygon.
For, the triangles CDE, CFE, have the base CE common,
they have also the same altitude, since their vertices D and F,
are situated in a Hne DF parallel to the base : these triangles are
therefore equivalent (Prop. II. Cor. 2.). Add to each of them
the figure ABCE, and there will result the polygon ABCDE,
equivalent to the polygon ABCF.
The angle B may in like manner be cut off, by substituting
for the triangle AfiC the equivalent triangle AGC, and thus
the pentagon ABCDE will be changed into an equivalent tri-
angle GCF.
The same process may be applied to every other figure ;
for, by successively diminishing the number of its sides, one
being retrenched at each step of the process, the equivalent
triangle will at last be found.
BOOK IV. 108
Scholium. We have already seen that every triangle may
be changed into an equivalent square (Prob. VI.) ; and thus a
square may always be found equivalent to a given rectilineal
figure, which operation is called squaring the rectilineal fiigure,
or finding the quadrature of it.
The problem o{ the quadrature of the circle^ consists in find-
ing a square equivalent to a circle whose diameter is given.
PROBLEM X.
To find the side of a square which shall he equivalent to the sum
or the difference of two given squares.
Let A and B be the sides of the
given squares.
First. If it is required to find a
square equivalent to the sum of
these squares, draw the two indefi-
nite lines ED, EF, at right angles
to each other; take ED = A, and
EG=B; draw DG: this will be the side of the square re-
quired, f
For the triangle DEG being right angled, the square de-
scribed upon DG is equivalent to the sum of the squares upon
ED and EG.
Secondly, If it is required to find a square equivalent to the
difference of the given squares, form in the same manner the right
angle FEH ; take GE equal to the shorter of the sides A and
B ; from the point G as a centre, with a radius GH, equal to
the other side, describe an arc cutting EH in H : the square
described upon EH will be equivalent to the difference of the
squares described upon the lines A and B.
For the triangle GEH is right angled, the hypothenuse
GH=:A, and the side GE=B; hence the square described
upon EH, is equivalent to the difference of the squares A
and B.
Scholium. A square may thus be found, equivalent to the
sum of any number of squares ; for a similar construction which
reduces two of them to one, will reduce three of them to two,
and these two to one, and so of others. It would be the same,
if any of the squares were to be subtracted from the sum of
the others.
104
GEOMETRY.
PROBLEM XI.
To find a square which shall he to a given squai e as a given
line to a given line.
Let AC be the given
square, and M and N the
given lines.
Upon the indefinite
line EG, take EF=M,
and FG=N ; upon EG
as a diameter describe
a semicircle, and at the point F erect the perpendicular FH.
From the point H, dravvr the chords HG, HE, which produce
indefinitely : upon the first, take HK equal to the side AB of
the given square, and through the point K draw KI parallel to
EG ; HI will be the side of the square required.
For, by reason of the parallels KI, GE, we have HI : HK
: : HE : HG; hence, HP : HK^ : : HE^ : HG^: but in the
right angled triangle EHG, the square of HE is to the square
of HG as the segment EF is to the segment FG (Prop. XI.
Cor. 3.), or as M is to N ; hence HP : HK^ : : M : N. But
HK=AB ; therefore the square described upon HI is to the
square described upon AB as M is to N.*
PROBLEM XII.
Upon a given line, to describe a polygon similar to a given
polygon.
Let FG be the given
line, and AEDCB the
given polygon.
In the given polygon,
draw the diagonals AC,
AD; at the point F
make the angle GFH=
BAC, and at the point
G the angle FGH=ABC ; the lines FH, GH will cut each
other in H, and FGH will be a triangle similar to ABC. In
the same manner upon FH, homologous to AC, describe the
triangle FIH similar to ADC ; and upon FI, homologous to AD,
describe the triangle FIK similar to ADE. The polygon
FGHIK will be similar to ABCDE, as required.
For, these two polygons are composed of the same number
of triangles, which are similar and similarly situated (Prop.
XXVLSch.).
BOOK IV.
105
PROBLEM XIII.
Tloo similar figures being given, to describe a similar figure
which shall be equivalent to their sum or their difference.
Let A and B be two homologous sides of the given figures.
Find a square equivalent to the
sum or to the difference of the
squares described upon A and B ;
let X be the side of that square ;
then will X in the figure required,
be the side which is homologous
to the sides A and B in the given
figures. The figure itself may then
be constructed on X, by the last problem.
For, the similar figures are as the squares of their homolo-
gous sides ; now the square of the side X is equivalent to the
sum, or to the difference of the squares described upon the
homologous sides A and B ; therefore the figure described upon
the side X is equivalent to the sum, or to the difference of the
similar figures described upon the sides A and B.
PROBLEM XIV.
To describe a figure similar to a given figure, and bearing to it
the given ratio of M to N.
Let A be a side of the given figure, X
the homologous side of the figure required.
The square of X must be to the square of
A, as M is to N : hence X will be found by
(Prob. XL), and knowing X, the rest will be
accomplished by (Prob. XIL).
106
GEOMETRY.
PROBLEM XV.
To construct a figure similar to the figure P, and equivalent to
the figure Q.
Find M, the side of a square
equivalent to the figure P, and
N, the side of a square equiva-
lent to the figure Q. Let X be
a fourth proportional to the three
given lines, M, N, AB ; upon
the side X, homologous to AB,
describe a figure similar to the figure P ; it will also be equiva-
lent to the figure Q.
For, calling Y the figure described upon the side X, we have
P ; Y : : AB2 : X^ ; but by construction, AB : X : : M : N,
or AB2 : X^ : : M2 : N2 ; hence P : Y : : M'^ : W, But by
construction also, M-=P and N2=Q; therefore P : Y : : P :
Q; consequently Y=Q; hence the figure Y is similar to the
figure P, and equivalent to the figure Q.
PROBLEM XVI.
To construct a rectangle equivalent to a given square, and having
the sum of its adjacent sides equal to a given line.
Let C be the square, and AB equal to the sum of the sides
of the required rectangle.
Upon AB as a diame-
ter, describe a semicir-
cle ; draw the line DE
parallel to the diameter,
at a distance AD from it, __
equal to the side of the A 1?]3
given square C ; from the point E, where the parallel cutp the
circumference, draw EF perpendicular to the diameter ; AF
and FB will be the sides of the rectangle required.
For their sum is equal to AB ; and their rectangle AF.FB is
equivalent to the square of EF, or to the square of AD ; hence
that rectangle is equivalent to the given square C.
Scholium, To render the problem possible, the distance AD
must not exceed the radius ; that is, the side of the square C
must not exceed the half of the line AB.
BOOK IV.
107
PROBLEM XVII.
To construct a rectangle that shall he equivalent to a given
square, and the difference of whose adjacent sides shall he
equal to a given line.
Suppose C equal to the given square, and AB the difference
of the sides.
Upon the given line AB as a diame-
ter, describe a semicircle : at the ex-
tremity of the diameter drav^^ the tan-
gent AD, equal to the side of the square
C ; through the point D and the centre
O draw the secant DF ; then will DE
and DF be the adjacent sides of the
rectangle required.
For, first, the difference of these sides
is equal to the diameter EF or AB ;
secondly, the rectangle DE, DF, is
equal to AD^ (Prop. XXX.) : hence that rectangle is equivalent
to the given square C.
PROBLEM XVIII.
To find the common measure, if there is one, between the diagonal
and the side of a square.
Let ABCG be any square what-
ever, and AC its diagonal.
We must first apply CB upon
CA, as often as it may be contained
there. For this purpose, let the
semicircle DBE be described, from
the centre C, with the radius CB.
It is evident that CB is contained
once in AC, with the remainder
AD ; the result of the first operation
is therefore the quotient 1, with the remainder AD, which lat-
ter must now be compared with BC, or its equal AB.
We might here take AF=AD, and actually apply it upon
AB ; we should find it to be contained twice with a remain-
der : but as that remainder, and those which succeed it, con-
108
GEOMETRY.
tinue diminishing, and would soon
elude our comparisons by their mi-
nuteness, this would be but an imper-
fect mechanical method, from which
no conclusion could be obtained to
determine whether the lines AC, CB,
have or have not a common measure.
There is a very simple way, however,
of avoiding these decreasing lines,
and obtaining the result, by operating
only upon lines which remain always of the same magnitude.
The angle ABC being a right angle, AB is a tangent, and
4E a secant drawn from the same point ; so that AD : AB : :
AB : AE (Prop. XXX.). Hence in the second operation, when
AD is compared with AB, the ratio of AB to AE may be taken
instead of that of AD to AB ; now AB, or its equal CD, is con-
tained twice in AE, with the remainder AD r the result of the
second operation is therefore the quotient 2 with the remain-
der AD, which must be compared with AB.
Thus the third operation again consists in comparing AD'
with AB, and may be reduced in the same manner to the com-
parison of AB or its equal CD with AE ; from which there will
again be obtained 2 for the quotient, and AD for the re-
mainder.
Hence, it is evident that the process will never terminate ;
and therefore there is no common measure between the diago-
nal and the side of a square : a truth which was already known
by arithmetic, since these two lines are to each other : : V2 : 1
(Prop. XI. Cor. 4.), but which acquires a greater degree of
clearness by the g-eometrical investigation.
BOOK V. 100
BOOK V.
REGULAR POLYGONS, AND THE MEASUREMENT OF THE
CIRCLE.
Definition,
A Polygon, which is at once equilateral and equiangular, is
called a regular polygon.
Regular polygons may have any number of sides : the equi-
lateral triangle is one of three sides ; the square is one of four.
PROPOSITION I. THEOREM.
Two regular polygons of the same number of sides are similar
figures.
Suppose, for example,
that ABCDEF, ahcdefi
are two regular hexa-
gons. The sum of all the
angles is the same in both
figures,being in each equal
to eight right angles (Book I. Prop. XXVI. Cor. 3.). The angle
A is the sixth part of that sum ; so is the angle a : hence the
angles A and a are equal ; and for the same reason, the angles
B and 5, the angles C and c, &c. are equal.
Again, since the polygons are regular, the sides AB, BC, CD,
&c. are equal, and likewise the sides ah, he, cd, &c. (Def ) ; it is
plain that AB : ah : -. BC i he i : CD : cd, &c. ; hence the
two figures in question have their angles equal, and their ho-
mologous sides proportional ; consequently they are similar
(Book IV. Def 1.).
Cor. The perimeters of two regular polygons of the same
number of sides, are to each other as their homologous sides,
and their surfaces are to each other as the squares of those sides
(Book IV. Prop. XXVIL).
Scholium. The angle of a regular polygon, like the angle of
an equiangular polygon, is determined by the number of its
Bides (Book I. Prop. XXVL).
K
no GEOMETRY.
PROPOSITION II. THEOREM.
Any regular polygon may he inscribed in a circle^ and circum-
scribed about one.
Let ABCDE &c. be a regular poly-
gon : describe a circle through the three
points A, B, C, the centre being O, and
OP the perpendicular let fall from it, to
the middle point of BC : draw AO and
OD.
If the quadrilateral OPCD be placed
upon the quadrilateral OPBA, they will _
coincide ; for the side OP is common ; IP
the angle OPC=:OPB, each being a right angle ; hence the
side PC will apply to its equal PB, and the point C will fall on
B : besides, from the nature of the polygon, the angle PCD=:
PBA ; hence CD will take the direction BA ; and since CD=:
BA, the point D will fall on A, and the two quadrilaterals will
entirely coincide. The distance OD is therefore equal to AO ;
and consequently the circle which passes through the three
points A, B, C, will also pass through the point D. By the
same mode of reasoning, it might be shown, that the circle
which passes through the three points B, C, D, will also pass
through the point E ; and so of all the rest : hence the circle
which passes through the points A, B, C, passes also through
the vertices of all the angles in the polygon, which is therefore
inscribed in this circle.
Again, in reference to this circle, all the sides AB, BC, CD,
&c. are equal chords ; they are therefore equally distant from
the centre (Book III. Prop. VIII.): hence, if from the point O
with the distance OP, a circle be described, it will touch the
side BC, and all the other sides of the polygon, each in its mid-
dle point, and the circle will be inscribed in the polygon, or the
polygon described about the circle.
Scholium 1. The point O, the common centre of the in-
scribed and circumscribed circles, may also be regarded as the
centre of the polygon ; and upon this principle the anglo AOB
is called the angle at the centre, being formed by two radii
drawn to the extremities of the same side AB.
Since all the chords AB, BC, CD, &c. are equal, all the an-
gles at the centre must evidently be equal likewise ; and there-
fore the value of each will be found by dividing four right an-
gles by the number of sides of the polygon.
BOOK V.
Ill
Scholium 2. To inscribe a regu-
lar polygon of a certain number of
sides in a given circle, we have only
to divide the circumference into as
many equal parts as the polygon
has sides : for the arcs being equal,
the chords AB, BC, CD, &c. will
also be equal ; hence likewise the
triangles AOB, BOC, COD, must
be equal, because the sides are equal each to each ; hence all
the angles ABC, BCD, CDE, &c. will be equal ; hence the
figure ABCDEH, will be a regular polygon.
PROPOSITION III. PROBLEM.
To inscribe a square in a given circle.
Draw two diameters AC, BD, cut-
ting each other at right angles ; join
their extremities A, B, C, D : the figure
ABCD will be a square. For the an-
gles AOB, BOC, &c. being equal, the
chords AB, BC," &c. are also equal :
and the angles ABC, BCD, &c. being
in semicircles, are right angles.
Scholium. Since the triangle BCO is right angled and isos-
celes, we have BC : BO : : v/2 : 1 (Book IV. Prop. XI.
(Jor. 4.) ; hence the side of the inscribed square is to the radius^
as the square root of 2, is to unity.
PROPOSITION IV. PROBLEM.
In a given circle, to inscribe a regular hexagon and an equilate-
ral triangle.
112
GEOMETRY.
Suppose the problem solved,
and that AB is a side of the in-
scribed hexagon ; the radii AO,
OB being drawn, the triangle
AOB will be equilateral.
For, the angle AOB is the sixth
part of four right angles ; there-
fore, taking the right angle for
unity, we shall have AOB=| —
f : and the two other angles
ABO, BAO, of the same trian-
gle, are together equal to 2 — |
=f ; and being mutually equal, _
each of them must be equal to | ; hence the triangle ABO W
equilateral ; therefore the side of the inscribed hexagon is equal
to the radius.
Hence to inscribe a regular hexagon in a given circle, the
radius must be applied six times to the circumference ; which
will bring us round to the point of beginning.
And the hexagon ABCDEF being inscribed, the equilateral <
triangle ACE may be formed by joining the vertices of the
alternate angles.
Scholium, The figure ABCO is a parallelogram and even
a rhombus, since AB=BC=CO=AO ; hence' the sum of the
squares of the diagonals AC^+BO- is equivalent to the sum of
the squares of the sides, that is, to 4AB^ or 4B0^ (Book IV.
Prop XiV. Cor.) : and taking away BO^ from both, there will
remain AC2=3BO^ hence AC^ : BO^ : : 3 : l,orAC : BO
: : \/3 : 1 ; hence the side of the inscribed equilateral triangle
is to the radius as the square root of three is to unitu*
PROPOSITION V. PROBLEM.
In a given circle^ to inscribe a regular decagon; then apentagony
and also a regular polygon of fifteen sides.
BOOK V.
113
Divide the radius AO in
extreme and mean ratio at
the point M (Book IV. Prob.
IV.) ; take the chord AB equal
to OM the greater segment ;
AB will be the side of the
regular decagon, and will re-
quire to be applied ten times
to the circumference.
For, drawing MB, we have
by construction, AO : OM
: : OM : AM ; or, since AB
C.OM, AO : AB : : AB :
AM ; since the triangles ABO, AMB, have a common angle A,
included between proportional sides, they are similar (Book
IV. Prop. XX.). Now the U'iangle OAB being isosceles, AMB
must be isosceles also, and AB=BM ; but AB=:OM ; hence
also MB = OM ; hence the triangle BMO is isosceles.
Again, the angle AMB being exterior to the isosceles trian-
gle BMO, is double of the interior angle O (Book I. Prop.
XXV. Cor. 6,) : but the angle AMBzrrMAB ; hence the trian-
gle OAB is such, that each of the angles OAB or OBA, at its
base, is double of O, the angle at its vertex ; hence the three
angles of the triangle are together equal to five times the angle
O, which consequently is the fifth part of the two right angles,
or the tenth part of four ; hence the arc AB is the tenth part
of the circumference, and the chord AB is the side of the reg-
ular decagon.
2d. By joining the alternate corners of the regular decagon,
the pentagon ACEGI will be formed, also regular.
3d. AB being still the side of the decagon, let AL be the
side of a hexagon ; the arc BL will then, with reference to
the whole circumference, be } — j\, or yV ; hence the chord BL
will be the side of the regular polygon of fifteen sides, or pente-
decagon. It is evident also, that the arc CL is the third of CB.
Scholium. Any regular polygon being inscribed, if the arcs
subtended by its sides be severally bisected, the chords of those
semi-arcs will form a new regular polygon of double the num-
ber of sides : thus it is plain, that the square will enable us to in-
scribe successively regular polygons of 8, 16, 32, &c. sides. And
m like manner, by means of the hexagon, regular polygons of
12, 24, 48, &c. sides may be inscribed ; by means of the deca-
gon, polygons of 20, 40, 80, &c. sides ; by means of the pente-
decagon, polygons of 30, 60, 120, &c. sides.
It is further evident, that any of the inscribed polygons will
be less than the inscribed polygon of double the number of
sides, since a part is less than the whole.
K*
114
GEOMETRY.
PROPOSITION VI. PROBLEM.
A regular inscribed poly gon being gimn^ to circumscribe a sim
ilar polygon about the same circle.
Let CBAFED be a regular polygon.
At T, the middle point
of the arc AB, apply the
tangent GH, which will
be parallel to AB (Book
III. Prop. X.)-; do the
same at the middle point
of each of the arcs BC,
CD, &c. ; these tangents,
by their intersections,
will form the regular
circumscribed polygon
GHIK &c. similar to
the one inscribed. _^
ik: Q ii
Since T is the middle point of the arc BTA, and N the mid-
dle point of the equal arc BNC, it follows, that BT=:BN ; or
that the vertex B of the inscribed polygon, is at the middle
point of the arc NBT. Draw OH. The line OH will pass
through the point B.
For, the right angled triangles OTH, OHN, having the com-
mon hypothenuse OH, and the side OT=:ON, must be equal
(Book I. Prop. XVH.), and consequently the angle TOHr:r
HON, wherefore the line OH passes through the middle point
B of the arc TN. For a like reason, the point 1 is in the pro-
longation of OC ; and so with the rest.
But, since GH is parallel to AB, and HI to BC, the angle
GHI=ABC (Book I. Prop. XXIV.) ; in like manner HIKrr
BCD ; and so with all the rest : hence the angles of the cir-
cumscribed polygon are equal to those of the inscribed one.
And further, by reason of these same parallels, we have GH :
AB : : OH : OB, and HI : BC : : OH : OB ; thefefore GH :
AB : : HI : BC. But AB=BC, therefore GHrrHI. For the
same reason, HI =IK, &c.; hence the sides of the circum-
scribed polygon are all equal ; hence this polygon is regular,
and similar to the inscribed one.
Ccr. 1. Reciprocally, if the circumscribed polygon GHIK
&c. were given, and the inscribed one ABC &c. were re-
quired to be deduced from it, it would only be necessary to
BOOR V.
115
draw from the angles G, H, I, &c
lines OG, OH, &c. meeting the
A, B, C, &c. ; then to join
AB, BC, &c. ; this would form
easier solution of this problem
points of contact T, N, P, <fec.
which likewise would form an
the circumscribed one.
. of the given polygon, straight
circumference in the points
those points by the chords
the inscribed polygon. An
would be simply to join the
by the chords TN, NP, &c.
inscribed polygon similar to
Cor, 2. Hence we may circumscribe about a circle any
regular polygon, which can be inscribed within it, and con-
versely.
Cor. 3. It is plain that NH + HT=rHT + TG=HG, one of
the equal sides of the polygon.
PROPOSITION VII. PROBLEM.
A circle and regular circumscribed polygon being given, it is
required to circumscribe the circle by another regular polygon
having double the number of sides.
Let the circle whose centre is P, be circumscribed by the
square CDEG ; it is required to find a regular circumscribed
octagon.
Bisect the arcs AH, HB, BF,
FA, and through the middle
points c, d, a, b, draw tangenls to
the circle, and produce them till
they meet the sides of the square :
then will the figure ApKdB &c.
be a regular octagon.
For, having drawn Vd, Va, let
the quadrilateral P^^B, be ap-
plied to the quadrilateral PB/a,
so that PB shall fall on PB.
Then, since the angle dVB is
equai to the angle BPa, each being half a right angle, the line
Vd will fall on its equal Va, and the point d on the point a. But
the angles Vdg, Vaf, are right angles (Book HI. Prop. IX.) ;
hence the line dg will take the direction af. The angles PB^,
PB/, are also right angles ; hence B^ will take the direction
Bf ; therefore, the two quadrilaterals will coincide, and the
point ^ will fall at/; hence, B^=Bf, c?^=«/, and the angle
dgB = Bfa. By applying in a similar manner, the quadrilate-
rals BBfa, VFha, it may be shown, that af^ah, fB=Fh, and
the angle Bfa—ahF. But since the two* tangents /z, /B, are
116 GEOMETRY.
equal (Book III. Prob. XIV. Sch.), it follows that fh, which is
twice /a, is equal to^, which is twice /B.
In a similar manner it may be shown that /(/r=/ii, and the
angle Yit=Yha, or that any two sides or any two angles of the
octagon are equal : hence the octagon is a regular polygon (Def.).
The construction which has been made in the case of the square
and the octagon, is equally applicable to other polygons.
Cor It is evidentthat the circumscribed square is greater than
the circumscribed octagon by the four triangles, Cnp, kDgf
hEf, Git ; and if a regular polygon of sixteen sides be circum-
scribed about the circle, we may prove in a similar way, that
the figure having the greatest number of sides will be the least ;
and the same may be shown, whatever be the number of sides
of the polygons : hence, in general, any circumscribed regular
polygon, will he greater than a circumscribed regular polygon
having double the number of sides.
PROPOSITION VIII. THEOREM.
Two regular polygons, of the same number of sides, can always
be formed, the one circumscribed about a circle, the other in-
scribed in it, which shall differ from each other by less than
any assignable surface.
Let Q be the side of a square
less than the given surface.
Bisect AC, a fourth pare of
the circumference, and then bi
sect the half of this fourth, and
proceed in this manner, always
bisecting one of the arcs formed ^
by the last bisection, until an
arc is found whose chord AB is -
less than Q. As this arc will
be an exact part of the circum-
ference, if we apply chords AB,
BC, CI), &c. each equal to AB, the last will terminate at A,
and there will be formeji a regular polygon ABCDE &c. in
the circle.
Next, describe about the circle a similar polygon abcde (fee.
(Prop. VI.) : the difference of these two polygons will be less
than the square of Q.
For, from the points a and b, draw the lines aO, bO, to the
centre O : they will pass through the points A and B, as was
BOOK V. 117
shown in Prop. VI. Draw also OK to the point of contact
K : it will bisect AB in I, and be perpendicular to it (Book III.
Prop. VI. Sch.). Produce AO to E, and draw BE.
Let P represent the circumscribed polygon, and p the in-
scribed polygon : then, since the triangles aOb, AOB, are like
parts of P and p, we shall have
aOb -i AOB : : P : p (Book II. Prop. XI.) :
But the triangles being similar,
aOb : AOB : : Oa'' : 0A«, or OK^
Hence, P : p : : Oa^ : OK^
Again, since the triangles 0«K, EAB are similar, havmg
their sides respectively parallel,
Oa'^ : OK^ : : AE^ r EB\ hence,
F : p : : AE^ : EB^ or by division,
P : P-p : : AE=^ : AE^^-EB^ or AB^.
But P is less than the square described on the diameter AE
(Prop. VII. Cor.); therefore F—p is less than the square de-
scribed on AB ; that is, less than the given square on Q : hence
the difference between the circumscribed and inscribed poly-
gons may always be made less than a given surface.
Cor. 1. A circumscribed regular polygon, having a given
number of sides, is greater than the circle, because the circle
makes up but a part of the polygon : and for a like reason, the
inscribed polygon ie less than the circle. But by increasing
the number of sides of the circumscribed polygon, the polygon
is diminished (Prop. VII, Cor.), and therefore approaches to
an equality with the circle ; and as the number of sides of
the inscribed polygon is increased, the polygon is increased
(Prop. V. Sch.), and therefore approaches to an equality with
the circle.
Now, if the number of sides of the polygons he indefinitely in-
creased, the length of each side will be indefinitely small, and the
polygons will ultimately become equal to each other, and equal
also to the circle.
For, if they are not ultimately equal, let D represent their
smallest difference.
Now, it has been proved in the proposition, that the differ-
ence between the circumscribed and inscribed polygons, can
be made less than any assignable quantity : that is, less than
D : hence the difference between tlie oolygons is equal to D,
and less than D at the same time, which is absurd : therefore,
the polygons are ultimately equal. But when they are equal
to each other, each must also be equal to the circle, since the
circumscribed polygon cannot fair within the circle, nor the
inscribed polygon without it.
118
GEOMETRY.
Cor. 2. Since the circumscribed polygon has the same num-
ber of sides as the corresponding inscribed polygon, and since
the two polygons are regular, they will be similar (Prop, I.) ;
and therefore when they become equal, they will exactly coin-
cide, and have a common perimeter. But as the sides of the
circumscribed polygon cannot fall within the circle, nor the
sides of the inscribed polygon without it, it follows that the
perimeters of the polygons will unite on the circumference of the
circle, and become equal to it.
Cor. 3. When the number of sides of the inscribed polygon
is indefinitely increased, and the polygon coincides with the
circle, the line 01, drawn from the centre O, perpendicular to
the side of the polygon, will become a radius of the circle, and
any portion of the polygon, as ABCO, will become the sector
OAKBC, and the part of the perimeter AB + BC, will become
thearcAKBC.
PROPOSITION IX. THEOREM.
Tile area of a regular polygon is equal to its perimeter, multi-
plied by half the radius of the inscribed circle.
Let there be the regular polygon
GHIK, and ON, OT, radii of the in-
scribed circle. The triangle GOH
will be measured by GH x ^OT ; the
triangle OHI, by HIxiON: but
ON^^OT; hence the two triangles
taken together will be measured by
(GH + HI)xiOT. And, by con-
tinuing the same operation for the
other triangles, it will appear that
the sum of them all, or the whole
polygon, is measured by the sum of the bases GH, HI, &c.
or the perimeter of the polygon, multiplied into ^OT, or half
the radius of the inscribed circle.
Scholium. The radius OT of the inscribed circle is nothing
else than the perpendicular let fall from the centre on one of
the sides : it is sometimes named the apothem of the polygon.
BOOK V.
no
PROPOSITION X. THEOREM.
The perimeters of two regular polygons, having the same num-
ber of sides, are to each other as the radii of the circumscribed
circles, and also, as the radii of the inscribed circles ; and their
areas are to each other as the squares of those radii.
Let AB be the side of the one poly-
gon, O the centre, and consequently
OA the radius of the circumscribed
circle, and OD, perpendicular to AB,
the radius of the inscribed circle ; let
ah, in like manner, be a side of the
other polygon, o its centre, oa and od
the radii of the circumscribed and the
inscribed circles. The perimeters of
the two polygons are to each other as the sides AB and ah
(Book IV. Prop. XXVII.) : but the angles A and a are equal,
being each half of the angle of the polygon ; so also are the
angles B and b ; hence the triangles ABO, abo are similar, as
are likewise the right angled triangles ADO, ado ; hence
AB : ab : : AO : ao : : DO : do ; hence the perimeters of the
polygons are to each other as the radii AO, ao of the circum-
scribed circles, and also, as the radii DO, do of the inscribed
circles.
The surfaces of these polygons are to each other as the
squares of the homologous sides AB, ab ; they are therefore
likewise to each other as the squares of AO,ao,the radii of the
circumscribed circles, or as the squares of OD, oc?,the radii of
the inscribed circles.
PROPOSITION XI. THEOREM.
The circumferences of circles are to each other as their radii,
and the areas are to each other as the squares of their radii.
120
GEOMETRY.
Let us designate the circumference of the circle whose radius
is CA by arc. CA ; and its area, by area CA : it is then to be
shown that
circ, CA : circ, OB : : CA : OB, and that
area CA : area OB : : CA^ : OB^
Inscribe within the circles two regular polygons of the same
number of sides. Then, whatever be the number of sides,
their perimeters will be to each other as the radii CA and OB
(Prop. X.). Now, if the arcs subtending the sides of the poly-
gons be co/itinually bisected, until the number of sides of the
polygons shall be indefinitely increased, the perimeters of the
polygons will become equal to the circumferences of the cir-
cumscribed circles (Prop. VIII. Cor. 2.), and we shall have
circ, CA : circ, OB : : CA : OB.
Again, the areas of the inscribed polygons are to each other
as CA^ to OB^ (Prop. X.). But when the number of sides of
the polygons is indefinitely increased, the areas of the polygons
become equal to the areas of the circles, each to each, (Prop.
VIII. Cor. 1.) ; hence we shall have
area CA : area OB : : CA^ : OB^.
V
:e
Cor, The similar arcs AB,
DE are to each other as their
radii AC, DO ; and the similar
sectors ACB, DOE, are to each
other as the squares of their
radii.
For, since the arcs are simi- ^ '
lar, the angle C is equal to the angle O (Book IV. Def. 3.) ;
but C is to four right angles, as the arc AB is to the whole cir-
cumference described with the radius AC (Book III. Prop.
XVII.) ; and O is to the four right angles, as the arc DE is to
the circumference described with the radius OD : hence the
arcs AB, DE, are to each other as the circumferences of which
BOOK V. 121
they form part : but these circumferences are to each other as
their radii AC, DO ; hence
arc AB : arc DE : : AC : DO.
For a Hke reason, the sectors ACB, DOE are to each other
as the whole circles ; which again are as the squares of their
radii ; therefore
sect, ACB : sect. DOE ; : AC^ : D0\
PROPOSITION XII. THEOREM.
The area of a circle is equal to the product of its circumference by
half the radius.
Let ACDE be a circle whose
centre is O and radius OA : then
will
area OA=^OAx arc. OA.
For, inscribe in the circle any ^
regular polygon, and draw OF
perpendicular to one of its sides.
Then the area of the polygon
will be equal to ^OF, multiplied
by the perimeter (Prop. IX.).
Now, let the number of sides of the polygon be indefinitely
increased by continually bisecting the arcs which subtend the
sides : the perimeter will then become equal to the circumfe-
rence of the circle, the perpendicular OF will become equal to
OA, and the area of the polygon to the area of the circle
(Prop. VIIL Cor. 1. & 3.). But the expression for the area
will then become
area OA=^OA x circ, OA :
consequently, the area of a circle is equal to the product of
half the radius into the circumference.
Cor. 1. The area of a sector is equal
to the arc of that sector multiplied by half
its radius.
For, the sector ACE is to the whole
circle as the arc AMB is to the whole
circumference ABD (Book III. Prop.
XVII. Sch. 2.), or as AMBx^AC is to
ABDx^AC. But the whole circle is
equal to ABD x ^AC ; hence the sector
ACB is measured by AMB x i AC.
L
Iii2 GEOMETRY.
Cor. 2. Let the circumference of the
. circle whose diameter is unity, be denoted
by n: then, because circumferences are
to each other as their radii or diameters,
we shall have the diameter I to its cir-
cumference TT, as the diameter 2CA is
to the circumference whose radius is CA,
that is, 1 : 7t : : 2CA : arc. CA, tliere-
fore circ. CA=7r x 2CA. Multiply both
terms by iCA ; we have ]CJA x circ. CA
= 7tx CA^ or area CA=7i x CA^ : hence- the area of a circle is
equal to the product of the square of its radius by the constant
number ^, which represents the circumference whose diameter
is 1, or the ratio of the circumference to the diameter.
In like manner, the area of the circle, w-hose radius is OB,
will be equal to ^r x OB'^ ; but tt x CA^ : n x OB^ : : CA^ : OB^ ;
hence the areas of circles are to each other as the squares of
their radii, which agrees with the preceding theorem.
Scholium. We have already observed, that the problem of
the quadrature of the circle consists in finding a square equal
in surface to a circle, the radius cf which is known. Now it
has just been proved- that a circle is equivalent to the rectangle
contained by its circumference and half its radius ; and this
rectangle may be changed into a square, by finding a mean
proportional between its length and its breadth (Book IV.
Frob. III.). To square the circle, therefore, is to find the cir-
cumference when the radius is given ; and for effecting this, it
is enough to know the ratio of tTie circumference to its radius,
or its diameter.
Hitherto the ratio in question has never been determined
except approximatively ; but the approximation has been car-
ried so far, that a knowledge of the exact ratio would afford
no real advantage whatever beyond that of the approximate
ratio. Accordingly, this problem, which engaged geometers
so deeply, when their methods of approximation were less per-
fect, is now degraded to the rank of those idle questions, with
w^hich no one possessing the slightest tincture of geometrical
science will occupy any portion of his time.
Archimedes showed that the ratio of the circumference to
the diameter is included between 3|^ and 3|f ; hence 3^ or
\2 affords at once a pretty accurate approximation to the num-
ber above designated by ?? ; and the simplicity of this first ap-
proximation has brought it into very general use. Metius,
for the same number, found the much more accurate value ?f |.
At last the value of 7r,-developed to a certain order of decimals,
wasfound by other calculators to be 3.1415926535897932, 6ic.i
BOOK V. 123
and some have had patience enough to continue these decimals
to the hundred and twenty-seventh, or even to the hundred
and fortieth place. Such an approximation is evidently equi-
valent to perfect correctness : the root of an imperfect power
is in no case more accurately known.
Tiie following problem will exhibit one of the simplest ele-
nientary methods of obtaining those approximations.
PROPOSITION XIII. PROBLEM.
The surface of a regular inscribed poly gon^ and that of a simi-
lar polygon circumscribed, being given; to find the surfaces of
the regular inscribed and circumscribed polygons having
double the number of sides.
Let AB be a side of the given
inscribed polygon; EF, parallel to
AB, a side of the circumscribed
polygon ; C the centre of the cir-
cle. If the chord AM and the
tangents AP, BQ, be drawn, AM
will be a side of the inscribed
polygon, having twice the num-
ber of sides; and AP+PM= 2PM
or PQ, will be a side of the simi-
lar circumscribed polygon (Prop. -
VI. Cor. 3.). Now, as the same ^
construction will take place at each of the angles equal to
ACM, it will be sufficient to consider ACM by itself, the tri-
angles connected with it being evidently to each other as the
whole polygons of which they form part. Let A, then, be
the surface of the inscribed polygon whose side is AB, B that
of the similar circumscribed polygon ; A' the surface of the
polygon whose side is AM, B' that of the similar circumscribed
polygon : A and B are given ; we have to find A' and B'.
First. The triangles ACD, ACM, having the common ver-
tex A, are to each other as their bases CD, CM ; they are like-
wise to each other as the polygons A and A', of which they
form part : hence A : A' : : CD : CM. Again, the triangles
CAM, CME, having the common vertex M, are to each other
as their bases CA, CE ; they are likewise to each other as the
polygons A' and B of which they form part ; hence A' : B : :
CA : CE. But since AD and ME are parallel, we have
CD : CM : : CA : CE; hence A : A' : : A' : B ; hence the
polygon xV, one of those required, is a mean proportional between
the two given polygons A and B and consequently A' = V A x B.
121 GEOMETRY.
Secondly- The altitude CM be-
ing common, the triangle CPM is
to Jthe triangle CPE as PM is to
PE ; but since CP bisects the an-
gle MCE, we have PM : PE : :
CM : CE (Book IV. Prop.
XYII.)::CD : CA : : A : A' :
hence CPM : CPE : : A : A' ;
and consequently CPM : CPM +
CPEorCME::A:A + A'. But
CMPA, or 2CMP, and CME are
to each other as the polygons B' ^
and B, of which they form part : hence B' : B : : 2A : A + A'.
Now A' has been already determined ; this new proportion will
serve for determining B', and give us B'— - ^; and thus by
A+ A'
means of the polygons A and B it is easy to find the polygons
A' and B', which shall have double the number of sides.
PROPOSITION XIV. PROBLEM.
To find the approximate ratio of the circumference to the
diameter.
Let the radius of the circle be 1 ; the side of the inscribed
square will be \/2 (Prop. III. Sch.), that of the circumscribed
square will be equal to the diameter 2 ; hence the surface of
the inscribed square is 2, and that of the circumscribed square
is 4. Let us therefore put A:r=2, and B=4 ; by the last pro-
position we shall find the inscribed octagonA' r= V8=2.8284271,
1 r*
and the circumscribed octagon B'=^-—t^= 3.3 137085. The
mscribed and the circumscribed octagons being thus deter-
mined, we shall easily, by means of them, determine the poly-
gons having twice the number of sides. We have only in this
case to put A=:2.8284271, B = 3.3137085 ; we shall find A' =
2A B
N/A.B = 3.0614674,and B'=j— ^, = 3.1825979. These poly-
gons of 16 sides will in their turn enable us to find the polygons
of 32 ; and the process may be continued, till there remains
no longer any difference between the inscribed and the cir-
cumscribed polygon, at least so far as that place of decimals
where the computation stops, and so far as the seventh place,
in this example. Being arrived at this point, we shall infer
BOOK V. 12?
that the last result expresses the area of the circle, which,
since it must always lie between the inscribed and the circum-
scribed polygon, and since those polygons agree as far as a
certain place of decimals, must also agree with both as far as
the same place.
We have subjoined the computation of those polygons, car-
r^'ed on till they agree as far as the seventh place of decimals.
Number of sides Inscribed polygon. Circumscribed polygon.
4 2.0000000 .... 4.0000000
8 2.8284271 .... 3.3187085
IG 3.0G14674 .... 3.1825979
32 3.1214451 .... 3.1517249
64 .... . 3.1365485 .... 3.1441184
128 3.1403311 .... 3.1422236
256 3.1412772 .... 3.1417504
512 3.1415138 .... 3.1416321
1024 3.1415729 .... 3.1416025
2048 3.1415877 .... 3.1415951
4096 3.1415914 .... 3.1415933
8192 3.1415923 .... 3.1415928
16384 3.1415925 .... 3.1415927
32768 3.1415926 .... 3.1415926
The area of the circle, we infer therefore, is equal to
3.1415926. Some doubt may exist perhaps about the last de-
cimal figure, owing to errors proceeding from the parts omitted ;
but the calculation has been carried on with an additional
figure, that the final result here given might be absolutely cor-
rect even to the last decimal place.
Since the area of the circle is equal to half the circumfe-
rence multiplied by the radius, the half circumference must be
3.1415926, when the radius is 1 ; or the whole circumference
must be 3.1415926, when the diameter is 1 : hence the ratio
of the circumference to the diameter, formerly expressed by tt,
is equal to 3.1415926. The number 3.1416 is the one gene-
rally used. I^*
126 GEOMETRY.
V
BOOK VI.
PLANES AND SOUD ANGLES.
Definitions,
1. A straight line is perpendicular to a plane, when it is per-
pendicular to all the straight lines which pass through its foot
in the plane. Conversely, the plane is perpendicular to the
line.
The foot of the perpendicular is the point in which the per-
pendicular line meets the plane.
2. A line is parallel to a plane, when it cannot meet thai
plane, to whatever distance both be produced. Conversely,
the plane is parallel to the line.
3. Two planes are parallel to each other, when they cannot
meet, to whatever distance both be produced.
4. The angle or mutual inclination of two planes is the quan-
tity, greater or less, by which they separate from each other ;
this angle is measured by the angle contained between two
lines, one in each plane, and both perpendicular to the common
intersection at the same point.
This angle may be acute, obtuse, or a right angle.
If it is a right angle, the two planes are perpendicular to
each other.
5. A solid angle is the angular space in- S
eluded between several planes which meet , y/f^
at the same point. yy l \
Thus, the solid angle S, is formed by ..^ /Y / 1
the union of the planes ASB, BSC, CSD, JJ^—f-^C
^®^- .. / /
Three planes at least, are requisite to ^ ^
form a solid angle. -^ ^
BOOK VI. 127
PROPOSITION I. THEOREM.
A straight line cannot be partly in a plane, and partly out of it.
For, by the definition of a plane, when a straight li.ne has
two points common with a plane, it lies wholly in that plane.
Scholium. To discover whether a surface is plane, it is ne-
cessary to apply a straight line in different ways to that sur-
face, and ascertain if it touches the surface throughout its whole
extent.
PROPOSITION II. THEOREM.
Two straight lines, which intersect each other, lie in the same
plane, and determine its position.
Let AB, AC, be two straight lines which
intersect each other in A ; a plane may be
conceived in which the straight Hne AB is
found ; if this plane be turned round AB, until
it pass through the point C, then the line AC,
which has two of its points A and C, in this
plane, lies wholly in it ; hence the position of
the plane is determined by the single condition of containing
the two straight lines AB, AC.
Cor. 1. A triangle ABC, or three points A, B, C, not in a
straight line, determine the position of a plane.
Cor. 2. Hence also two parallels
AB, CD, determine the position of a
plane ; for, drawing the secant EF,
the plane of the two straight lines
AE, EF, is that of the parallels
AB, CD.
PROPOSITION III. THEOREM.
If two planes cut each other, their common intersection will be a
straight line.
128
GEOMETRY.
Let the two planes AB, CD, cut
each other. Draw tlie straight line
EF, joining any two points E and F in
the common section of the two planes.
This line will lie wholly in the plane
AB, and also wholly in the plane CD
(Book J. Def. 6.) : therefore it will be
in both planes at once, and conse-
quently is their common intersection.
-,P=*C1
»^..
.--B
D
■iS PROPOSITION. IV. THEOREM.
If a straight line be perpendicular to two straight lines at their
point of intersection^ it will be perpendicular to the plane oj
those lines.
Let MN be the plane of the
two lines BB, CC, and let AP
be perpendicular to them at
their point of intersection P ;
then will AP be perpendicular
to every line of the plane pass-
ing through P, and consequently
to the plane itself (Def. 1.).
Through P, draw in the plane
MN, any straight line as PQ,
and through any point of this
line, a:s Q, drawBQC, eo that BQ shall be equal to QC (Book
IV. Prob. V.) ; draw AB, AQ, AC.
The base BC being divided into two equal parts at the point
Q, the triangle BPC will give (Book IV. Prop. XIV.),
PCHPB2=2PQH2QC2.
The triangle BAC will in like manner give,
AC2+AB2=2AQ-+2QCl -
Taking the first equation from the second, and observing
that the triangles APC, APB, which are both right angled at
P, give
AC2— PC2= AP2, and AB^— PB2= AP^ ;
we shall have
AP2+AP2=2AQ2— 2PQ^.
Therefore, by taking the halves of both, we have
AP2=:AQ^— PQ2, or AQ^^AF+PQ^ ;
hence the triangle APQ is right angled at P ; hence AP is per-
pendicular to PQ.
BOOK VI.
120
Schohupi. Thus it is evident, not only tiiat a straight line
may be perpendicular to all the straight lines which pass
through iis foot in a plane, but that it always must be so, when
ever it is perpendicular to two straight lines drawn in the
plana ; which proves the first Definition to be accurate.
Cor. 1. The perpendicular AP is shorter than any oblique
line AQ ; therefore it measures the true distance from the point
A to the plane MN.
Cor. 2. At a given point P on a plane, it is impossible to
erect more than one perpendicular to that plane ; for if there
could be two perpendiculars at the same point P, draw through
these two perpendiculars a plane, whose intersection with the
plane MN is PQ ; then these two perpendiculars would be per-
pendicular to the line PQ, at the same point, and in the same
plane, which is impossible (Book I. Prop. XIV. Sch.).
It is also impossible to let fall from a given point out of a
plane two perpendiculars to that plane ; for let AP, AQ, be
these tv:o perpendiculars, then the triangle APQ would have
two right angles APQ, AQP, which is impossible.
PROPOSITION V. THEOREM.
If from a point without a plane, a perpendicular be drawn to the
plane, and oblique lines be drawn to different points,
1st. Any two oblique lines equally distant from the perpendicular
will be equal.
2d. Of any two oblique lines unequally distant from the perpen-
dicular, the more distant will be the longer.
Let AP be perpendicular to
the plane MN ; AB, AC, AD,
oblique hues equally distant
from the perpendicular, and
AE a line more remote : then
will AB-AC=AD; and AE
will be greater than AD.
For, the angles APB, APC,
APD, being right angles, if we
suppose the distances PB, PC,
PI), to be equal to each other, the triangles APB, APC, APD,
will have in each an equal angle contained by two equal sides ;
herefore they will be equal ; hence the hypothenuses, or the
oblique lines AB, AC, AD, will be equ^l to each other. In like
130
GEOMETRY.
manner, if the distance PE is greater than PD or its pqual PB,
the obhque hne AE will evidently be greater than Ali, or its
equal AD.
Cor. All the equal oblique
lines, AB, AC, AD, &c. termi-
nate in the circumference BCD,
described from Pthe foot of the
perpendicular as a centre ;
therefore a point A being given
out of a plane, the point P at
which the perpendicular let fall
from A would meet that plane,
may be found by marking upon
that plane three points B, C, D, equally distant from the pomt A,
and then finding the centre of the circle which passes through
these points ; this centre will be P, the point sought.
Scholium. The angle ABP is called the inclination of the
oblique line AB to the plane MN ; which inclination is evidently
equal with respect to all such lines AB, AC, AD, as are eqtially
distant from the perpendicular ; for all the triangles ABP, ACP,
ADP, &c. are equal to each other.
PROPOSITION VI. THEOREM.
If from a point without a plane, a perpendicular he let fall on the
plane, and from the foot of the perpendicular a perpendicular
be drawn to any line of the plane, and from the point of inter-
section a line be drawn to the first point, this latter line will he
perpendicular to the line of the plane.
Let AP be perpendicular to the
plane NM, and PD perpendicular to
BC ; then will AD be also perpen-
dicular to BC.
Take DB=DC. and draw PB, PC,
AB, AC. Since DB-DC, the ob-
lique line PB — PC: and with regard
to the perpendicular AP, since PB=:
PC, the oblique line AB=:AC (Prop.
V. Cor.) ; therefore the line AD has
two of its points A and D equally distant from the extremities
B and C ; therefore AD is a perpendicular to BC, at its middle
point D (Book I. Prop. XVI. Cor.).
Ei
BOOK VI.
131
Cor. It is evident likewise, that BC is perpendicular to the
plane APD, since BC is at once perpendicuiar to the two
straight lines AD, PD.
Scholium. The two lines AE, BC, afford an instance of two
lines which do not meet, because they are not situated in the
same plane. The shortest distance between these lines is the
straight line PD, which is at once perpendicular to the line AP
and to the line BC. The distance PD is the shortest distance
between them, because if we join any other two points, such
as A and B, we shall have AB>AD, AD>PD; therefore
AB>PD. ^
The two lines AE, CB, though not situated in the same plane,
are conceived as forming a right angle with each other, because
AE and the line drawn through one of its points parallel to
BC would make with each other a right angle. In the same
manner, the line AB and the line PD, which represent any two
straight lines not situated in the same plane, are supposed to
form with each other the same angle, which would be formed
by AB and a straight line parallel to PD drawn through one
of the points of AB.
/
PROPOSITION VII. THEOREM.
If one of two parallel lines he perpendicular to a plane, the othe?
will also be perpendicular to the same plane.
Let the lines ED, AP, be
parallel ; if AP is perpen-
dicular to the plane NM,
then will ED be also per-
pendicular to it.
Through the parallels AP,
DE, pass a plane ; its inter-
section with the plane MN
will be PD ; in the plane MN
draw BC perpendicular to PD, and draw AD.
By the Corollary of the preceding Theorem, BC is perpen-
dicular to the plane APDE ; therefore the angle BDE is a right
angle ; but the angle EDP is also a right angle, since AP'is
perpendicular to PD, and DE parallel to AP (Book I. Prop.
XX. Cor. 1.) ; therefore the line DE is perpendicular to the
two straight lines DP, DB ; consequently it is perpendicular to
their plane MN (Prop. IV.).
132
GEOMETRY.
Cor. 1. Conversely, if the
straight Hnes AP, DE, are
perpendicular to the same
plane MN, they will be par-
allel ; for if they be not so,
draw through the point D. a
, line parallel to AP, this par-
allel will be perpendicular
to the plane MN ; therefore
through the same point D more than one perpendicular might
be erected to the same plane, which is impossible (Prop. IV.
Cor. 2.).
Cor, 2. Two lines A and B, parallel to a third C, are par-
allel to each other ; for, conceive a plane perpendicular to the
line C ; the lines A and B, being parallel to C, will be perpen-
dicular to the same plane ; therefore, by the preceding Corol-
lary, they will be parallel to each other.
The three lines are supposed not to be in the same plane ;
otherwise the proposition would be already known (Book I.
Prop. XXII.).
PROPOSITION VIII. THEOREM.
If a straight line is parallel to a straight line drawn in a plane^
it will bp parallel to that plane.
Let AB be parallel to CD
of the plane NM ; then will
it be' parallel to the plane
NM.
For, if the line AB, which
lies in the plane ABDC,
could meet the plane MN,
this could only be in some
M
K
point of the line CD, the .common intersection of the two
planes : but AB cannot meet CD, since they are parallel ;
hence it will not meet the plane MN ; hence it is parallel to
that plane (Def. 2.).
PROPOSITION IX. THEOREM.
1
TiJDO planes which are perpendicular to the same straight line,
are parallel to each other.
BOOK VI.
133
• — r~
t2
^
■x:
K"
Let the planes NM, QP, be per- *••
pendiciilar to the hne AB, then will P
they be parallel.
For, if they can meet any where,
let O be one of their common
points, and draw OA, OB ; the line
AB which is perpendicular to the
plane MN, is perpendicular to the \ IQ.
straight line OA drawn through its foot in that plane ; for the
same reason AB is perpendicular to BO ; therefore OA and OB
are two perpendiculars let fall from the same point O, upon
the same straight line ; which is impossible (Book I. Prop. XIV.);
therefore the planes MN, PQ, cannot meet each other ; consf^?
quently they are parallel.
• PROPOSITION X. THEOREM.
If a plane cut two parallel planes, the lines of intersection will he
parallel.
Let the parallel planes NM,
QP, be intersected by the plane
EH ; then will the lines of inter-
section EF, GH, be parallel.
For, if the lines EF, GH, lying
in the same plane, were not par-
allel, they would meet each other
when produced ; therefore, the
planes MN, PQ, in which those
lines lie, would also meet ; and
hence the planes would not be
parallel.
M E
PROPOSITION XI. THEOREM.
If two planes are parallel, a straight line which is perpendicular
to cne, is ulso perpendicular to the other.
M
134
GEOMETRY.
M
i Sr
B
Let MN, PQ, be two parallel
planes, and let AB be perpendicu-
lar to NM ; then will it also be per-
pendicular to QP. -J
Having drawn any line BC in
the plane PQ, through the lines AB
and BC, draw a plane ABC. inter-
secting the plane MN in AD ; the
intersection AD will be parallel to BC (Prop. X.) ; but the line
AB, being perpendicular to the plane MN, is perpendicular to
the straight line AD ; therefore also, to its parallel BC (Book
I. Prop. XX. Cor. 1.): hence the line AB being perpendicular
to any line BC, drawn through its foot in the plane PQ, is con-
sequently perpendicular to that plane (Def. 1.). j
ia
PROPOSITION XII. THEOREM.
The parallels comprehended between two parallel planes are
equal.
Let MN, PQ, be two parallel
planes, and FH, GE. two paral-
lel lines ; then will EG=FH
For, through the parallels EG,
FH, draw the plane EGHF, in-
tersecting the parallel planes in
EF and GH. The intersections
EF, GH, are parallel to each
other (Prop. X.) ; so likewise are
EG, FH ; therefore the figure
.kEGHF is a parallelogram ; con-
sequently, EG =FH.
Cor. Hence it follows, that two parallel planes are every
where equidistant : for, suppose EG were perpendicular to the
plane PQ ; the parallel FH would also be perpendicular to it
(Prop. VII.), and the two parallels would likewise be perpen-
dicular to the plane MN (Prop. XL) ; and being parallel, tliey js
will be equal, as shown by the Proposition.
BOOK VI.
135
PROPOSITION XIII. THEOREM.
If two angles, not situated in the same plane, have their sides
parallel and lying in the same direction, those angles ivill le
eqhal and their planes will he parallel
Let the angles be CAE and DBF.
Make AC-BD, AE= JM
BF ; and draw CE, DF,
AB, CD, EF. Since AC
is equal and parallel to
BD, the figure ABDC is
a parallelogram ; therefore
CD is equal and parallel
to AB. For a similar rea-
son, EF is equal and par-
allel to AB ; hence also CD
is equal and parallel to
EF ; hence the figure
CEFD is a parallelogram,
and the side CE is equal
and paiallel to DF; therefore the triangles CAE, DBF, have
their corresponding sides equal; therefore the angle CAE —
DBF.
Again, the plane ACE is parallel to the plane BDF. For
suppose the plane drawn through the point A, parallel to BDF,
were to meet the lines CD, EF, in points different from C and
E, for instance in G and H ; then, the three lines AB, GD, FH,
would be equal (Prop. XII.) : but the lines AB, CD, EF, are
already known to be equal; hence CD=GD, and FH=EF,
which is absurd ; hence the plane ACE is parallel to BDF.
Cor. If two parallel planes MN, PQ are met by two other
planes CABD, EABF, the angles CAE, DBF, formed by the
mtersections of the parallel planes will be equal ; for, the inter-
section AC is parallel to BD, and AE to BF (Prop. X.) ; there-
fore the angle CAE = DBF.
PROPOSITION XIV. THEOREM.
If three straight lines, not situated in the same plane, are equal
and parallel, the opposite triangles formed by joining the ex-
tremities of these lines will be equal, and their planes will he
i
parillel
136
GEOMETRY.
Let AB, CD, EF, be the
lines.
Since AB is equal and
parallel to CD, the figure
ABDC is a parallelogram ;
hence the side AC is equal
and parallel to BD. For a
like reason the sides AE,
BF, are equal and parallel,
as also CE, DF ; therefore
the two triangles ACE, BDF,
are equa^ ; hence, by the last
Proposition, their planes are
parallel.
-M.
C
H
^
/\G^
\.\-r.
\
A^
-^.E
A
\ \
\
N
P
\ \
\?^
^\
\
B
1?
PROPOSITION XV. THEOREM.
If two straight lines he cut hy three parallel planes, they will he
divided proportionally.
Suppose the line AB to meet
the parallel planes MN, PQ,
RS, at the points A, E, B ; and
the line CD to meet the same
planes at the points C, F, D :
we are now to show that
AE : EB : : CF : FD.
Draw AD meeting the plane
PQ in G, and draw AC, EG,
GF, BD ; the intersections EG,
BD, of the parallel planes PQ,
RS, by the plane ABD, are
parallel (Prop. X.) ; therefore
AE : EB : : AG : GD ;
in like manner, the intersections AC, GF, being parallel,
AG : GD : : CF : FD ;
the ratio AG : GD is the same in both ; hence
AE : EB : : CF : FD.
1^
Y^
I
M
I^
^4^
.R
^\v^
b' — — -^_^ •
PROPOSITION XVI. THEOREM.
Ifalt^ ) is perpendicular to a plane, every plane passed through
thf zrpendicular, will also he.peipendicular tojhe plane.
BOOK VI.
137
Let AP be perpen^ular to the
plane NM ; then will every plane
passing through AP be perpendicu-
lar to NM.
Let BC be the intersection of the
planes AB, MN ; in the plane MN,
draw DE perpendicular to BP : then
the line AP, being perpendicular to
the plane MN, will be perpendicu-
lar to each of the two straight lines
BC, DE ; but the angle APD, formed by the two perpendicu-
lars PA, PD, to the common intersection BP, measures the
angle of the two planes AB, MN (Def. 4.) ; therefore, since that
angle is a right angle, the two planes are perpendicular to each
other.
Scholium, When three straight lines, such as AP, BP, DP,
are perpendicular to each other, each of those lines is perpen-
dicular to the plane of the other two, and the three planes are
perpendicular to each other.
PROPOSITION XVII. THEOREM.
If two planes are perpendicular to each other, a line drawn in
one of them perpendicular to their common intersectionj .will
be perpendicular to the other plane, '
Let the plane AB be perpen-
dicular to NM ; then if the line
AP be perpendicular to the inter-
section BC, it will also be perpen-
dicular to the plane NM.
For, in the plane MN draw PD
perpendicular to PB ; then, be-
cause the planes are perpendicu-
lar, the angle APD is a right an-
gle ; therefore, the line AP is
perpendicular to the two straight
lines PB, PD ; therefore it is perpendicular to their plane MN
(Prop. IV.).
Cor, If the plane AB is perpendicular to the plane MN, and
if at a point P of the common intersection we erect a perpen-
dicular to the plane MN, that perpendicular will be in the plane
AB ; for, if not, then, in the plane AB we might draw AP per-
M*
138
GEOMETRY.
pendicular to PB the common intersection, and this AP, at the
same time, would be perpendicular to the plane MN; therefore
at the same point P there would be two perpendiculars to the
plane MN, which is impossible (Prop. IV. Cor. 2.).
PROPOSITION XVIII. THEOREM.
Ij two planes are perpendicular to a third plane, their common
intersection will also he peipendicular to the third plane.
Let the planes AB, AD, be per-
pendicular to roi; then will their
intersection AP be perpendicular
toNM.
For, at the point P, erect a per-
pendicular to the plane MN" ; that
perpendicular must be at once in
the plane AB and in the plane AD
(Prop. XVII. Cor.) ; therefore it
is theii common intersection AP.
PROPOSITION XIX. THEOREM.
If a solid angle isfonmd by three plane angles, the sum of any
two of these angles will he greater than the third.
The proposition requires demonstra-
tion only when the plane angle, which
is compared to the sum of the other
two, is greater than either of them.
Therefore suppose the solid angle S to
be formed by three plane angles ASB,
ASC, BSC, whereof the angle ASB is
the greatest; Ave are to show that
ASB<ASC + BSC.
In the plane ASB make the angle
straight line ADB at pleasure; and having taken SC = SD,
draw AC, BC.
The two sides BS, SD, are equal to the two BS, SC ; the
angle BSD=:BSC ; therefore the triangles BSD, BSC, are
equal; therefore BD=BC. But AB<AC + BC; taking BD
from the one side, and from the other its equal BC, there re-
BSD=:BSC, draw the
139
mains AD<AC. The two sides AS, SD, are equal to the
two AS, SC ; the third side AD is less than the third si-^e AC ;
therefore the angle ASD<ASC (Book I. Prop. IX. Sch.).
Adding BSD=:=BSC, we shall have ASD + BSD or ASB<
ASC + BSC.
PROPOSITION XX. THEOREM.
The sum of the plane angles which form a solid angle is always
less than four right angles.
Cut the solid angle S by any plane S
ABCDE ; from O, a point in that plane, /A
draw to the several angles the straight // \\
lines AO, OB, OC,OD,bE. // \\
The sum of the angles of the triangles / •' / l\
ASB, BSC, &c. formed about the^vertex / ,.(M. ilry
S, is equal to the sum of the angles of an />-''' */ .a\
equal number of triangles AOB, BOC, &c. A.^- y- ■-)<o 1 1
formed about the point O. But at the >. // xll
point B the sum of the angles ABO, OBC, ^^ ^
equal to ABC, is less than the sum of the
angles ABS, SBC (Prop. XIX.) ; in the same manner at the
point C we have BCO + OCD<BCS + SCD; and so with all
the angles of the polygon ABCDE : whence it follows, that the
sum of all the angles at the bases of the triangles whose vertex
is in O, is less than the sum of the angles at the bases of the
triangles whose vertex is in S ; hence to make up the defi-
ciency, the sum of the angles formed about the point O, is
greater than the sum of the angles formed about the point S.
But the sum of the angles about the point O is equal to four
right angles (Book I. Prop. IV. Sch.) ; therefore the sum of the
plane angles, which form the solid angle S, is less than four
right angles.
Scholium. This demonstration is founded on the supposition
that the solid angle is convex, or that the plane of no one sur-
face produced can evej: meet the solid angle ; if it were other-
wise, the sum of the plane angles would no longer be limited,
and might be of any magnitude.
PROPOSITION XXI. THEOREM.
If two solid angles are contained by three plane angles which are
equal to each other, each to each, the planes of the equal angles
will be equally inclined to eachMher,
140 GEOMETRY.
Let the angle ASC=DTF,the
angle ASB=DTE, and the an-
gle BSC=ETF; then will the
mclination of the planes ASC,
ASB, be equal to that of the
planes DTF, DTE.
Having taken SB at pleasure,
draw BO perpendicular to the
plane ASC ; from the point O, at which the perpendicular
meets the plane, draw OA, OC perpendicular to SA, SC ;
draw AB, BC ; next take TE = SB ; draw EP perpendicular to
the plane DTF ; from the point P draw PD, PF, perpendicular
respectively to TD, TF ; lastly, draw DE, EF.
The triangle SAB is right angled at A, and the triangle TDE
at D (Prop. VI.) ; and since the angle ASB = DTE we have
SBA=TED. Likewise SB = TE ; therefore the triangle SAB
is equal to the triangle TDE; therefore SA = TD, and AB = DE.
In like manner, it may be shown, that SC=TF, and BC=EF.
That granted, the quadrilateral SAOC is equal to the quadri- •
lateral TDPF: for, place the angle ASC upon its equal DTF; S
because SA=:TD, and SC=TF, the point A will fall on D,.: i
and the point C on F ; and at the same time, AO, which is per-
pendicular to SA, will fall on PD which is perpendicular to
TD, and in like manner OC on PF ; wherefore the point O
will fall on the point P, and AO will be equal to DP. But the
triangles AOB, DPE, are right angled at O and P ; the hypo-
thenuse AB=I)E, and the side AO=DP: hence those trian- ^
gles are equal (Book I. Prop. XVII.) ; and consequently, the |
angle OAB = PDE. The angle OAB is the inclination of the '
two planes ASB. ASC ; and the angle PDE is that of the two
planes DTE, DTF ; hence those two inclinations are equal to
each other.
It must, how^ever, be observed, that the angle A of the right
angled triangle AOB is properly the inclination of the two
planes ASB, ASC, only when the perpendicular BO falls on
the same side of SA, with SC ; for if it fell on the other side,
the angle of the two planes would be obtuse, and the obtuse
angle together with the angle A of the triangle OAB would
make two right angles. But in the same case, tl^e angle of the
two planes TDE, TDF, would also be obtuse, and the obtuse
angle together with the angle D of the triangle DPE, would
make two right angles ; and the angle A being thus always
equal to the angle at D, it would follow in the same manner that
the inclination of the two planes ASB, ASC, must be equal to
that of the two planes TDE, TDF.
Scholium. If two solid angles are contained by three plane
BOOK VI. 141
angles, respectively equal to each other, and if at the same time
the equal or homologous angles are disposed in the same man-
ner in the two solid angles, these angles will be equal, and they
will coincide when applied the one to the other. We have
already seen that the quadrilateral SAOC may be placed upon
its equal TDPF ; thus placing SA upon TD, SC falls upon TF,
and the point O upon the point P. Btit because the triangles
AOB, DPE, are equal, OB, perpendicular to the plane ASC,
is equal to PE, perpendicular to the plane TDF ; besides, those
perdendiculars lie in the same direction ; therefore, the point
B will fall upon the point E, the line SB upon TE, and the two
solid angles will wholly coincide.
This coincidence, however, tai^es place only when we suj>-
pose that the equal plane angles are arranged in the same man-
ner in the two solid angles ; for if they were arranged in an in-
verse order, or, what is the same, if the perpendiculars OB, PE,
instead of lying in the same direction with regard to the planes
ASC, DTF, lay in opposite directions, then it would be impos-
sible to make these solid angles coincide with one another. It
would not, however, on this account, be less true, as our Theo-
rem states, that the planes containing the equal angles must
still be equally inclined to each other; so that the two solid an-
gles would be equal in all their constituent' parts, without,
however, admitting of superposition. This sort of equalit}%
which is not absolute, or such as admits of superposition, de-
serves to be distinguished by a particular name : we shall call
it equality hy symmetry.
Thus those two solid angles, which are formed by three
plane angles respectively equal to each other, but disposed in an
inverse order, will be called angles equal hy symmetry , or simply
symmetrical angles.
The same remark is applicable to solid angles, which are
formed by more than three plane angles : thus a solid' angle,
formed by the plane angles A, B, C, D, E, and another solid
angle, formed by the same angles in an inverse order A, E, D,
C, B, may be such that the planes which contain the equal an-
gles are equally inclined to each other. Those two solid angles,
are likew^ise equal, without being capable of superposition, and
are called solid angles equal by symmetry , or symmetrical solid
angles.
Among plane figures, equality by symmetry does not pro-
perly exist, all figures which might take this name being abso-
lutely equal, or equal by superposition ; the reason of which is,
that a plane figure may be inverted, and the upper part taken
indiscriminately for the under. This is not the case with solids ;
in which the third dimension may be taken in two different
directions.
142
GEOMETRY.
BOOK VII.
POLYEDRONS.
Definitions,
1. The name solid pnlyedronf or simple jjolyedron, is given
to every solid terminated by planes or plane faces; which
planes, it is evident, will themselves be terminated by straight
lines.
2. The common intersection of two adjacent faces of a
polyedron is called the side, or edge of the polyedron.
3. The prism is a solid bounded by several parallelograms,
which are terminated at both ends by equal and parallel
polygons.
IC 7c
T
To construct this solid, let ABCDE be any polygon ; then
if in a plane parallel to ABCDE, the lines FG, GH,*HI, &c. be
drawn equal and parallel to the sides AB, BC, CD, &c. thus
forming the polygon FGHIK equal to ABCDE ; if in the next
place, the vertices of the angles in the one plane be joined with
the homologous vertices in the other, by straight lines, AF, BG,
CPI, &c. the faces ABGR BCHG, &c. will be parallelograms,
and ABCDE-K, the solid so formed, will be a prism.
4. The equal and parallel polygons ABCDE, 'FGHIK, are
called the bases of the prism; the parallelograms taken together
constitute the lateral or convex surface of the prism; the equal
straight lines AF, BG, CH, &c. are called the sides, or edges of
the prism.
5. The altitude of a prism is the distance between its two
bases, or the perpendicular drawn from a point in the upper
base to the plane of the lower base.
BOOK VII.
143
6. A prism is right, when the sides AF, BG, CH, &;c. are
perpendicular to the planes of the bases ; and then each of them
is equal to the altitude of the prism. In every other case the
prism is oblique, and the altitude less than the side.
7. A prism is triangular, quadrangular, pentagonal, hex-
agonal, &c. when the base is a triangle, a quadrilateral, a
pentagon, a hexagon, &c.
8. A prism whose base is a parallelogram, and
which has all its faces parallelograms, is named a
parallelopipedon.
The parallelopipedon is rectangular when all
its faces are rectangles.
9. Among rectangular parallelopipedons, we
distinguish the cube, or regular hexaedron, bounded
by six equal squares.
10. A pyramid is a solid formed by
several triangular planes proceeding from
the same point S, and terminating in the
different sides of the same polygon
ABCDE.
The polygon ABCDE is called the
base of the pyramid, the point S the
vertex ; and the triangles ASB, BSC,
CSD, &c. form its convex or lateral sur-
11. If from the pyramid S-ABCDE,
the pyramid S-abcde be cut off by a
plane parallel to the base, the remaining
solid ABCDE-c?, is called a truncated
pyramid, or the frustum of a pyramid.
12. The altitude of a pyramid is the
perpendicular let fall from the vertex upon
base, produced if necessary.
IS. A pyramid is triangular, quadrangular, &c. according
as its base is a triangle, a quadrilateral, &c.
14. A pyramid is regular, when its base is a regular poly-
gon, and when, at the same time, the perpendicular let fall
from the vertex on the plane of the base passes through the
centre of the base. That perpendicular is then called the axis
of the pyramid.
15. Any line, as SF, drawn from the vertex S of a regular
pyramid, perpendicular to either side of the polygon w^hich
forms its base, is called the slant height of the pyramid.
16. The diagonal of a polyedron is a straight line joining
the vertices of two solid angles which are not adjacent to each
other.
A
the plane of the
144 ^
GEOMETRY.
17. Two polyedrons are similar when they are contained
by the same number of similar planes, similarly situated, and
having like inclinations with each other.
PROPOSITION I. THEOREM.
The convex surface of a right prism is equal to the perimeter oj
its base multiplied hy its altitude.
liCt ABCDE-K be a right prism : then
will its convex surface be equal to
(AB + BC + CD + DE + EA) x AF.
For, the convex surface is equal to the
sum of all the rectangles AG, BH, CI,
DK, EF, which compose it. Now, the
altitudes AF, BG, CH, &c. of the rect-
angles, are equal to the altitude of the
prism. Hence, the sum of these rectan-
gles, or the convex surface of the prism,
is equal to (AB + BC + CD + DE + EA) x
AF ; that is, to the perimeter of the base of the prism multi
plied by its altitude.
Cor. If two right prisms have the same altitude, their con-
vex surfaces will be to each other as the perimeters of theii ;
bases
PROPOSITION II. THEOREM.
In every prism, the sections formed hy parallel planes, are equal
polygons. ^
Let the prism AH be intersected by
the parallel planes NP, S'V ; then are the
polygons NOPQR, STVXY equal.
For, the sides ST, NO, are parallel,
being the intersections of two parallel
planes with a third 'plane ABGF ; these
same sides, ST, NO, are included between
the parallels NS, OT, which are sides of
the prism: hence NO is equal to ST.
For like reasons, the sides OP, PQ, QR,
&c. of the section NOPQR, are equal
to the sides TV, VX, XY, <&c. of the sec-
tion STVXY, each to each. And since
BOOK VII.
145
the equal sides are at the same time parallel, it follows that the
angles NOP, OPQ, &c. of the first section, are equal to the
angles STV,TVX, &c. of the second, each to each (Book VI.
Prop. XIIL). Hence the two sections NOPQR, STVXY, are
equal polygons.
Cor. Every section in a prism, if drawn parallel to the base,
is also equal to the base.
PROPOSITION III. THEOREM.
If a pyramid he cut by a plane parallel to its base,
1st. The edges and the altitude will be divided proportionally.
2d. The section will be a polygon similar to tlie base.
Let the pyramid S-ABCDE,
of which SO is the altitude,
be cut by the plane abcde ;
then will Sa : SA : : So : SO,
and the same for the other
edges : and the polygon abcde,
will be similar to the base
ABCDE.
First. Since the planes ABC,
abc, are parallel, their intersec-
tions AB, ab, by a third plane
SAB will also be parallel
(Book VI. 'Prop. X.) ; hence the triangles SAB, Sab are simi-
lar, and we have SA : Sa : : SB : S6 ; for a similar reason,
we have SB : S6 : : SC : Sc; and so on. Hence the edges
SA, SB, SC, &:c. are cut proportionally in a, 6, c, &c. The
altitude SO is likewise cut in the same proportion, at the point
o ; for BO and bo are parallel, therefore we have
SO : So : : SB : Sfe.
Secondly. Since ab is parallel to AB, be to BC, cd to CD, &c.
the angle abc is equal to ABC, the' angle bed to BCD, and so on
(Book VI. Prop. XHL). Also, by reason of the similar trian-
gles SAB, S«6, we have AB : ab : : SB : S6 ; and by reason
of the similar triangles SBC, Sbc, we have SB : Sb : : BC :
be ; hence AB : ab : : BC : be ; we might likewise have
BC :bc : : CD : cd, and so on. Hence the polygons ABCDE,
abcde have their angles respectively equal and their homolo-
gous sides proportional ; hence they are similar.
N
146
GEOMETRY.
Cor. 1. Let S-ABCDE,
S-XYZ be two pyramids, hav-
ing a common vertex and the
same altitude, or having their
bases situated in the same
plane ; if these pyramids are
cut by a plane parallel to the
plane of their bases, giving the
sections abcde, xyz, then will
the sections abcde, xyz, he to each
other as the bases ABCDE,
XYZ.
For, the polygons ABCDE, abcde, being similar, their sur-
faces are as the squares of the homologous sides AB, ab ; but
AB : «& : : SA : S«; hence ABCDE : abcde : : SA^ : ^a\
For the same reason, XYZ : xyz : : SX^ : Sx^. But since
abc and xyz are in one plane, we have likewise SA : Saj : :
SX : So; (Book VI. Prop. XV.) ; hence ABCDE : abcde : :
XYZ : xyz ; hence the sections abcde, xyz, are to each othei
as the bases ABCDE, XYZ.
Cor. 2. If the bases ABCDE, XYZ, are equivalent, any sec-'
tions abcde, xyz, made at equal distances from the bases, will
be equivalent likewise.
PROPOSITION IV. THEOREM.
The convex surface of a regular "pyramid is equal to the perime
ter of its base multiplied by half the slant height.
For, since the pyramid is regular, the
point O, in which the axis meets the base,
is the centre of the polygon ABCDE
(Def. 14.) ; hence thelines OA, OB, OC,
&c. drawn to the vertices of the base,
are equal.
In the right angled triangles SAO, SBO,
the bases and perpendiculars are equal :
since the bypothenuses are equal : and
it may be proved in the same way that
all the sides of the right pyramid are
equal. The triangles, therefore, which
form the convex surface of the prism are
all equal to each other. But the area of
either of these triangles, as ES A, is equal
BOOK VII.
147
to its base EA multiplied by half the perpendicular SF, which
is the slant height of the pyramid : hence the area of all the tri-
angles, or the convex surface of the pyramid, is equal to the
perimeter of the base multiplied by half the slant height.
Cor. The convex surface of the frustum of a regular pyra-
mid is equal to half the perimeters of its upper and lower bases
multiplied by its slant height.
For, since the section abcde is similar to the base (Prop. III.),
and since the base ABCDE is a regular polygon (Def. 14.), it
follows that the sides e«, ab, be, cd and de are all equal to each
other. Hence the convex surface of the frustum ABCDE-c?
is formed by the equal trapezoids EAae, AB6a, &c. and the
perpendicular distance between the parallel sides of either of
these trapezoids is equal to Ff the slant height of the frustum.
But the area of either of the trapezoids, as AEea, is equal to
|(EA + ea) xF/ (Book IV. Prop. VII.) : hence the area of all
of them, or the convex surface of the frustum, is equal to half
the perimeters of the upper and lower bases multiplied by the
slant height.
PROPOSITION V. THEOREM.
If the three planes which form a solid angle of a prism, are equal
to the three planes which form the solid angle of another prism,
each to each, and are like situated, the two prisms will be equal
to each other.
Let the base ABCDE be equal to the base abcde, the paral-
lelogram ABGF equal to the parallelogram abgf and the par-
allelogram BCHG equal to bchg-, then will the prism ABCDE-K
be equal to the prism abcde-k.
For, lay the base ABCDE upon its equal abcde ; these two
bases will coincide. But the three plane angles which form
148
GEOMETRY
the solid angle B, are respectively equal to the three plane
angles, which form the solid angle b, namely, ABCrrc/ic,
ABGrrra^^, and GBC =gbc ; they are also similarly situated .
hence the solid angles B and bare equal (Book YI. Prop. XXL
Sch.) ; and therefore the side BG will fall on its equal bg. It
is likewise evident, that by reason of the equal parallelograms
ABGF, abgff the side GF will fall on its equal gf, and in the
same manner GH on gh ; hence, the plane of the upper base,
FGHIK will coincide with the plane fghik (Book VI. Prop. II.).;
K 7c
But the two upper bases being equal to their corresponding
lower bases, are equal to each other : hence HI will coincide
with hi, IK with ik, and KF with A/; and therefore the lateral
faces of the prisms will coincide : therefore, the two prisms
coinciding throughout are equal (Ax. 13.).
Co?\ Two right prisms, which have equal bases and equal al-
titudes, are equal. For, since the side AB is equal to ab, and
the altitude BG to bg, the rectangle ABGF will be equal to
abgf; so also will the rectangle BGHC be equal to bghc ; and
thus the three planes, which form the solid angle B, will be
equal to the three, which form the solid angle b. Hence the
two prisms are equal.
PROPOSITION VI. THEOREM.
In every parallelopipedon the opposite planes are equal and
parallel.
By the definition of this solid, the bases
ABCD, EFGH, are equal parallelograms,
and their sides are parallel : it remains
only to show, that the same is true of any
two opposite lateral faces, such as AEHD,
BFGC. Now AD is equal and parallel
to BC, because tiie figure ABCD is a par-
E
H
/ r
.4/
V— ""
v^
B
c
BOOK VII. 149
allelogram ; for a like reason, AE is parallel to BF : hence the
angle DAE is equal to the angle CBF, and the planes DAE,
CBF, are parallel (Book VI. Prop. XIII.) ; hence also the par-
allelogram DAEH is equal to the parallelogram CBFG. In the
same way. it might be shown that the opposite parallelograms
ABFE, DCGH, are equal and parallel.
Cor. 1. Since the parallelopipedon is a solid bounded by six
planes, whereof those lying opposite to each other are equal
and parallel, it follows that any face and the one opposite to it,
may be assumed as the bases of the parallelopipedon.
Cor. 2. The diagonals of a parallelopipedon bisect each
other. For, suppose two diagonals EC, AG, to be drawn both
through opposite vertices : since AE is equal and parallel to
CG, the figure AEGC is a parallelogram ; hence the diagonals
EC, AG will mutually bisect each other. In the same manner,
we could show that the diagonal EC and another DF bisect
each other ; hence the four diagonals will mutually bisect each
other, in a point which may be regarded as the centre of the
parallelopipedon.
Scholium. If three straight lines AB, AE, AD, passing
through the same point A, and making given angles with each
other, are known, a parallelopipedon may be formed on those
lines. For this purpose, a plane must be passed through the
extremity of each line, and parallel to the plane of the other
two ; that is, through the point B a plane parallel to DAE,
through D a plane parallel to BAE, and through E a plane
parallel to BAD. The mutual intersections of these planes will
form the parallelopipedon required.
PROPOSITION VII. THEOREM.
The two triangular prisms into which a parallelopipedon is di'
vided by a plane passing through its opposite diagonal edges,
are equivalent.
N*
150 GEOMETRY.
Let the parallelopipedon ABCD-H be
divided by the plane BDHFpassingthrough
its diagonal edges : then will the triangular
prism ABD-H be equivalent to the trian-
gular prism BCD-H.
Through the vertices B and F, draw the
planes Bade, Fehg, at right angles to the
side BF, the former meeting AE, DH, CG,
the three other sides of the parallelopipe-
don, in the points a, d, c, the latter in e, A,
g : the sections Bade, Fehg, will be equal
parallelograms. They are equal, because
they are formed by planes perpendicular to the same straight
line, and consequently parallel (Prop. II.) ; they are parallelo-
grams, because aB, dc, two opposite sides of the same section,
are formed by the meeting of one plane with two parallel
planes ABFE, DCGH.
For a like reason, the figure B^zeF is a parallelogram ; so also
are BF^c, cd/ig, adhe, the other lateral faces of the solid Badc-g ;
hence that solid is a prism (Def 6.) ; and that prism is right,
because the side BF is perpendicular to its base.
But the right prism Badc-g is divided by the plane BH into
two equal right prisms Bad-h, Bcd-h ; for, the base^JB^j^, Bed,
of these prisms are equal, being halves of the same parallel-
ogram, and they have the common altitude BF, hence they are
equal (Prop. V. Cor.).
It is now to be proved that the oblique triangular prism
ABD-H will be equivalent to the right triangular prism Bad-h ;
and since those prisms have a common part ABD-/i, it will
only be necessary to prove that the remaining parts, namely,
the solids BaADi, FeEH/«, are equivalent.
Now, by reason of the parallelograms ABFE, «BFe, the
sides AE, ae, being equal to their .parallel BF, are equal to each
other; and taking away the common part Ae, there remains
AanrEe. In the same manner we could prove Dc?=HA.
Next, to bring about the superposition of the two solids
BflADd/, FfiEH/i, let us place the base Yeh on its equal Bad :
the point e falling on a, and the point h on d, the sides eE, AH,
will fall on their equals a A, dD, because they are perpendicu-
lar to the same plane Bad. Hence the two solids in question
will coincide exactly with each other ; hence the oblique prism
BAD-H, is equivalent to the rigf;t one Bad-h.
In the same manner might the oblique prism BCD-H, be
proved equivalent to the right prism Bcd-h, But the two right
prisms Bad-h, Bcd-h, are equal, since they have the same alti-
tude BF, and since their bases Bad, Bde, are halves of the
same parallelogram (Prop. V. Cor.). Hence the two trian-
BOOK VII. 151
gular prisms BAD-H, BDC-G, being equivalent to the equal
right prisms, are equivalent to each other.
Cor, Every triangular prism ABD-HEF is half of the paral-
lelopipedon AG described with the same solid angle A, and
the same edges AB, AD, AE.
PROPOSITION VIII. THEOREM. -
If two parallelopipedons have a common hase, and their upper
bases in the same plane and between the same parallels^ they
will be equivalent.
Let the parallelopipe-
dons AG, AL, have the
common base AC, and
their upper bases EG,
MK, in the same plane,
and between the same
parallels HL, EK ; then
will they be equivalent.
There may be three
cases, according as EI is
greater, less than, or equal to, EF ; but the demonstration is
the same for all. In the first place, then we shall show that
the triangular prism AEI-MDII, is equal to the triangular
prism BFK-LCG.
Since AE is parallel to BF, and HE to GF, the angle AEI
=BFK, HEI-GFK, and HEA=GFB. Also, since EF and
IK are each equal to AB, they are equal to each other. To
each add FI, and there will result EI equal to FK : hence the
triangle AEI is equal to the.triangle BFK (Bk. I. Prop. V), and
the paralellogram EM to the parallelogram FL. But the par-
allelogram AH is equal to the parallelogram CF (Prop. VI) :
hence, the three planes which form the solid angle at E are
respectively equal to the three which form the solid angle at
F, and being like placed, the triangular prism AEI-M is equal
to the triangular prism BFK-E.
But if the prism AEI-M is taken away from the solid AL,
there will remain the parallelopipedon BADC-L ; and if the
prism BFK-L is taken away from the same solid, there will
remain the parallelopipedon BADC-G ; hence those two paral-
lelopipedons BADC-L, BADC-G, are equivalent.
152
GEOMETRY.
PROPOSITION IX. THEOREM.
Two parallelopipedons, having the same base and the same alti-
tude, are equivalent.
Let ABCD be the com-
mon base of the two par-
allelopipedons AG, AL ;
since they have the same
altitude, their upper bases
EFGH,IKLM,willbein
the same plane. Also the
sides EF and AB will be
equal and parallel, as well
as IK and AB ; hence EF
is equal and parallel to
IK; for a like reason, GF
is equal and parallel to
LK. Let the sides EF, GH, be produced
B
and likewise
IM, till by their intersections they form the parallelogram
NOPQ ; this parallelogram will evidently be equal to either
of the bases EFGH, IKLM. Now if a third parallelopipedon
be conceived, having for its lower base the parallelogram
ABCD, and NOPQ for its upper, the third parallelopipedon
will be equivalent to the parallelopipedon AG, since with the
same lower base, their upper bases lie in the same plane
and between the same parallels, GQ, FN (Prop. VIII.).
For the same reason, this third parallelopipedon will also be
equivalent to the parallelopipedon AL ; hence the two paral-
lelopipedons AG, AL, which have the same base and the
same altitude, are equivalent.
PROPOSITION X. THEOREM.
Any parallelopipedon may be changed into an equivalent rectan
gular parallelopipedon having the same altitude and
equivalent base.
an
BOOK VII.
153
Let AG be the par-
allelopipedon proposed.
From the points A, B, C,
D,drawAI,BK,CL,DM,
perpendiculartothe plane
of the base ; you will thus
form the parallelopipe-
don AL equivalent to
AG, and having its late-
ral faces AK, BL, &c.
rectangles. Hence if the
base ABCD is a rectan-
gle, AL will be a rectan-
gular parallelopipedon equivalent to AG, and consequently,
the parallelopipedon required. But if ABCD is not a rectangle,
draw AO and BN perpendicular to CD, and MQ XP
OQ and NP perpendicular to the base ; you
will then have the solid ABNO-IKPQ, which
will be a rectangular parallelopipedon : for
by construction, the bases ABNO, and IKPQ
are rectangles ; so also are the lateral faces,
the edges AI, OQ, &c. being perpendicular
to the plane of the base ; hence the solid AP
is a rectangular parallelopipedon. But the
two parallelopipedons AP, AL may be con-
ceived as having the same base ABKl and
the same altitude AO : hence the parallelopipedon AG, which
was at first changed into an equivalent parallelopipedon AL,
is again changed into an equivalent rectangular parallelopipe-
don AP, having the same altitude AI, and a base ABNO equi-
valent to the base ABCD.
PROPOSITION XI. THEOREM.
Two rectangular parallelopipedons, wJiich have the same base,
are to each other as their altitudes.
(F
M
K
D
\
154 GEOMETRY.
Let the parallelopipedons AG, AL, have the same base ED;
then will they be to each other as their altitudes AE, AT.
•First, suppose the altitudes AE, AI, to be j; .H
to each other as two whole numbers, as 15 is
to 8, for example. Divide AE into 15 equal
parts ; whereof AI will contain 8 ; and through q
X, y, z, &c. the points of division, draw planes ^\in
parallel to the base. These planes will cut
the solid AG into 15 partial parallelopipedons,
all equal to each other, because they have 55..
equal bases and equal altitudes — equal bases, ^'.
since every section MIKL, made parallel to
the base ABCD of a prism, is equal to that
base (Prop. II.), equal altitudes, because the
altitudes are the equal divisions Ax, xy, yz,
&c. But of those 15 equal parallelopipedons, 8 are con-
tained in AL ; hence the solid AG is to the solid AL as 15 is to
8, or generally, as the altitude AE is to the altitude AI.
Again, if the ratio of AE to AI cannot bfe exactly expressed
in numbers, it is to be shown, that notwithstanding, we shall
have
solid AG : solid AL : : AE : AI.
For, if this proportion is not correct, suppose we have
sol. AG : soL AL : : AE : AO greater than AI.
Divide AE into equal parts, such that each shall be less than
01 ; there will be at least one point of division m, between O
and I. Let P be the parallelopipedon, whose base is ABCD,
and altitude Am ; since the altitudes AE, Am, are to each other
as the two whole numbers, we shall have
50/. AG : P : ; AE : Am.
But by hypothesis, w^e have
sol AG : sol AL : : AE : AO ;
therefore,
sol AL : P : : AO : Am.
But AO is greater than Am ; hence if the proportion is correct,
the solid AL must be greater than P. On the contrary, how-
ever, it is less : hence the fourth term of this proportion
sol AG : sol AL : : AE : x,
cannot possibly be a line greater than AI. By the same mode
of reasoning, it might be shown that the fourth term cannot be
less than AI ; therefore it is equal to AI ; hence rectangular
parallelopipedons having the same base are to each other as
their altitudes.
BOOK VII.
155
PROPOSITION XII. THEOREM.
Two rectangular parallelopipedons, having the same altitude
are to each other as their bases.
1
A
I
^
Let the parallelopipedons x E 11
AG, AK, have the same al-
titude AE ; then will they be \K_
to each other as their bases
AC, AN.
Having placed the tvv^o "^ |-HG
solids by the side of each
other, as ithe figure repre-
sents, produce the plane
ONKL till it meets the
plane DCGH in PQ ; you
will thus have a third par-
allelopipedon AQ, which
may be compared with each
of the parallelopipedons
AG, AK. The two solids
AG, AQ, having the same
base AEHD are to each other as their altitudes AB, AG ; in
like manner, the two solids AQ, AK, having the same base
AOLE, are to each other as their altitudes AD, AM. Hence
we have the two proportions,
sol. AG : sol AQ : : AB : AG,
sol AQ : sol AK : : AD : AM.
Multiplying together the corresponding terms of these propor-
tions, and omitting in the result the common multiplier sol AQ ;
we shall have
50/. AG : sol AK : : ABxAD : AOxAM.
But AB X AD represents the base ABCD ; and AO x AM rep-
resents the base AMNO ; hence two rectangular parallelopipe-
dons of the same altitude are to each other as their bases.
PROPOSITION XIII. THEOREM.
Any two rectangular parallelopipedons are to each other as the
products of their bases by their altitudes, that is to say, as the
products of their three dimensions.
156
GEOMETRY.
E
•%
K
V
X
Z
M
H
T?
^>.
D
For, having placed the two
soHds AG, AZ, so that their
surfaces have the common
angle BAE, produce the
planes necessary for com-
pleting tlie third parallelopi-
pedon AK having the same
altitude vt^ith the parallelopi-
pedon AG. By the last propo-
sition, v^^e shall have
sol. AG : sol AK : :
ABCD : AMNO.
But the two parallelopipedons
AK,AZ, having the same base
AMNO, are to each other as
their altitudes AE, AX ; hence _
we have
sol AK : sol AZ : : AE : AX.
Multiplying together the corresponding terms of these propor-
tions, and omitting in the result the common multiplier sol AK ;
we shall have
sol AG : 50/. AZ : : ABCDxAE : AMNO x AX.
Instead of the bases ABCD and AMNO, put ABxADand
AO X AM it will give
50/.AG : solAZ : : ABxADxAE : AOxAMxAX.
Hence any two rectangular parallelopipedons are to each
other, &c.
Scholium. We are consequently authorized to assume, as
the measure of a rectangular parallelopipedon, the product
of its, base by its altitude, in other words, the product of its
three dimensions.
In order to comprehend the nature of this measurement, it
is necessary to reflect, that the number of linear units in one
dimension of the base multiplied by the number of Hnear units
in the other dimension of the base, will give the number of
superficial units in ihe base of the parallelopipedon (Book IV.
Prop. IV. Sch.). For each unit in height there are evidently
as many solid units as there are superficial units in the base.
Therefore, the number of superficial units in the base multi-
plied by the number of linear units in the altitude, gives the
number of solid units in the parallelopipedon.
If the three dimensions of another parallelopipedon are
valued according to the same linear unit, and multiplied together
in the same maimer, the two products will be to each other as
BOOK VII. 157
the solids, and will serve to express their relative magni-
tude.
The magnitude of a sohd, its volume or extent, forms Vi^hat is
called its solidity ; and this word is exclusively employed to
designate the measure of a solid : thus we say the solidity of a
rectangular parallelopipedon is equal to the product of its base
by its altitude, or to the product of its three dimensions.
As the cube has all its three dimensions equal, if the side is
1, the solidity will be 1 x 1 x 1 = 1 : if the side is 2, the solidity
will be 2 X 2 x 2 — 8 ; if the side is 3, the solidity will be 3 x 3 x
3 = 27 ; and so on : hence, if the sides of a series of cubes are
to each other as the numbers 1, 2, 3, &c. the cubes themselves
or their solidities will be as the numbers 1, 8, 27, &c. Hence
it is, that in arithmetic, the cube of a number is the name given
to a product which results from three factors, each equal to
this number.
If it were proposed to find a cube double of a given cube,
the side of the required cube would have to be to that of the
given one, as the cube-root of 2 is to unity. Now, by a geo-
metrical construction, it is easy to find the square root of 2 ;
but the cube-root of it cannot be so found, at least not by the
simple operations of elementary geometry, which consist m
employing nothing but straight lines, two points of which are
known, and circles whose centres and radii are determined.
Owing to this difficulty the problem of the duplication of
the cube became celebrated among the ancient geometers, as
well as that of the trisection of an angle, which is nearly of the
same species. The solutions of which such problems are sus-
ceptible, have however long since been discovered ; and though
less simple than the constructions of elementary geometry, they
are not, on that account, less rigorous or less satisfactory.
PROPOSITION XIV. THEOREM.
The solidity of a parallelopipedon, and generally of any prism,
is equal to the product of its hose by its altitude.
For, in the first place, any parallelopipedon is equivalent to
a rectangular parallelopipedon, having the same altitude and
an equivalent base (Prop. X.). Now the solidity of the latter
is equal to its base multiplied by its height ; hence the solidity
of the former is, in like manner, equal to the product of its base
by its altitude.
In the second place, any triangular prism is half of the par-
allelopipedon so constructed as to have the same altitude and
a double base (Prop. VII.). But the solidity of the latter is equal
158
GEOMETRY.
to its base multiplied by its altitude ; hence that of a triangular
prism is also equal to the product of its base, which is half that
of the parallelopipedon, multiplied into its altitude.
In the third place, any prism may be divided into as many
triangular prisms of the same altitude, as there are triangles
capable of being formed in the polygon which constitutes its
base. But the solidity of each triangular prism is equal to its
base multiplied by its altitude ; and since the altitude is the
same for all, it follows that the sum of all the partial prisms
must be equal to the sum of all the partial triangles, which con-
stitute their bases, multiplied by the common altitude.
Hence the solidity of any polygonal prism, is equal to the
product of its base by its altitude.
Co7\ Comparing two prisms, which have the same altitude,
the products of their bases by their altitudes will be as the
bases simply ; hence two prisms of the same altitude are to each
other as their bases. For a like reason, two prisms of the same
base are to each other as their altitudes. And when neither their
bases nor their altitudes are equal, their solidities will be to
each other as the products of their bases and altitudes.
PROPOSITION XV. THEOREM.
TVjo triangular pyramids, having equivalent bases and equal
altitudes, are equivalent, or equal in solidity.
Let S-ABC, S-ahc, be those two pyramids ; let their equiva-
lent bases ABC, abc, be situated in the same plane, and let AT
be their common altitude. If they are not equivalent, let ^-ahe
BOOK VII. 159
be the smaller : and suppose Ka to be the altitude of a prism,
which having ABC for its base, is equal to their difference.
Divide the altitude AT into equal parts Ax, xy, yz, &c. each
less than A«, and let k be one of those parts ; through the points
of division pass planes parallel to the plane of the bases ; the
icorre spending sections formed by these planes in the two pyra-
mids will be respectively equivalent, namely DEF to def, GHI
to ghi, &c. (Prop. III. Cor. 2.).
This being granted, upon the triangles ABC, DEF, GHI, &c.
taken as bases, construct exterior prisms having for edges the
parts AD, DG, GK, &c. of the edge SA ; in like manner, on
bases dej, ghi, klm, &c. in the second pyramid, construct inte-
rior prisms, having for edges the corresponding parts of Sa.
It is plain that the sum of all the exterior prisms of the pyramid
S-ABC will be greater than this pyramid ; and also that the
sum of all the interior prisms of the pyramid S-abc will be less
than this pyramid. Hence the difference, between the sum of all
the exterior prisms and the sum of all the interior ones, must be
greater than the difference between the two pyramids them-
selves.
Now, beginning with the bases ABC, ahc, the second exte-
rior prism DEF-G is equivalent to the first interior prism def-a,
because they have the same altitude k, and their bases DEF,
def, are equivalent ; for like reasons, the third exterior prism
GHI-K and the second interior prism ghi-d are equivalent ;
the fourth exterior and the third interior ; and so on, to the last
in each series. Hence all the exterior prisms of the pyramid
S-ABC, excepting the first prism ABC-D, have equivalent cor-
responding ones in the interior prisms of the pyramid S-abc :
hence the prism ABC-D, is the difference between the sum of
all the exterior prisms of the pyramid S-ABC, and the sum of
the interior prisms of the pyramid S-abc. But the difference
between these two sets of prisms has already been proved to
be greater than that of the two pyramids ; which latter diffe-
rence we supposed to be equal to the prism a-ABC : hence the
prism ABC-D, must be greater than the prism a-ABC. But in
reality it is less ; for they have the same base ABC, and the
altitude Ax of the first is less than Aa the altitude of the second.
Hence the supposed inequality between the two pyramids can-
not exist ; hence the two pyramids S-ABC, S-abc, having equal
altitudes and equivalent bases, are themselves equivalent.
leo GEOMETRY.
PROPOSITION XVI. THEOREM.
Ev.ery triangular pyramid is a third part of the triangular prism
having the same base and the same altitude.
Let F-ABC be a triangular
pyramid, ABC-DEF a triangular
prism of the same base and the
same altitude ; the pyramid will
be equal to a third of the prism.
Cut off the pyramid F-ABC
from the prism, by the plane
FAC ; there will remain the solid
F-ACDE, which may be consi-
dered as a quadrangular pyramid,
whose vertex is F, and whose base
is the parallelogram ACDE.
Draw the diagonal CE ; and pass
the plane FCE, which will cut the B
quadrangular pyramid into two triangular ones F-ACE,F-CDE.
These two triangular pyramids have for their common altitude
the perpendicular let fall from F on the plane ACDE ; they
have equal bases, the triangles ACE, CDE being halves of the
same parallelogram ; hence the two pyramids F-ACE, F-CDE,
are equivalent (Prop. XV.). But the pyramid F-CDE and the
pyramid F-ABC have equal bases ABC, DEF; they have also the
same altitude, namely, the distance between the parallel planes
ABC, DEF ; hence the two pyramids are equivalent. Now the
pyramid F-CDE has already been proved equivalent to F-ACE ;
hence the three pyramids F-ABC, F-CDE, F-ACE, which
compose the prism ABC-DEF are all equivalent. Hence the
pyramid F-ABC is the third part of the prism ABC-DEF, which
has the same base and the same altitude.
Cor. The solidity of a triangular pyramid is equal to a third
part of the product of its base by its altitude.
PROPOSITION XVII. THEOREM.
The solidity of every pyramid is equal to the base multiplied by
a third of the altitude,.
BOOK VII. 161
Let S-ABCDE be a pyramid.
Pass the planes SEB, SEC, through the
diagonals EB, EC ; the polygonal pyraniid
S-ABCDE will be divided into several trian-
gular pyranriids all having the same altitude
SO. But each of these pyramids is measured
by multiplying its base ABE, BCE, or CDE,
by the third part of its altitude SO (Prop. XVI.
Cor.) ; hence the sum of these triangular pyra-
mids, or the polygonal pyramid S-ABCDE
will be measured by the sum of the triangles
ABE, BCE, CDE, or the polygon ABODE, ^
multiplied by one third of SO ; hence every pyramid Is mea-
sured by a third part of the product of its base by its altitude.
Cor, 1. Every pyrarnid is the third part of the prism which
has the same base and the same altitude.
Coj\ 2. Two pyramids having the same altitude are to each
other as their bases.
Cor. 3. Two pyramids havmg equivalent bases are to each
other as their altitudes.
Cor. 4. Pyramids are to each other as the products of their
bases by their altitudes.
Scholium. The solidity of any polyedral body may be com-
puted, by dividing the body into pyramids ; and this division
may be accomplished in various ways. One of the simplest
is to make all the planes of division pass through the vertex
of one solid angle ; in that case, there will be formed as many
partial pyramids as the polyedron has faces, minus those faces
which form the solid angle whence the planes of division
proceed.
PROPOSITION XVIII. THEOREM.
If a pyramid be cut hy a plane parallel to its base, the frustum
that remains when the small pyramid is taken away, is equi-
valent to the sum of three pyramids having for their common
altitude the altitude of the frustum, and for bases the lower
base of the frustum, the upper base, and a mean proportional
between the two bases,
O*
162
GEOMETRY
Let S-ABCDE be a pyra-
mid cut by the plane abcdCf
parallel to its base; let T-FGH
be a triangular pyramid hav-
ing the same altitude and an
equivalent base with the pyra-
mid S-ABCDE. The two
bases may be regarded as
situated in the same plane ; in
which case, the plane abed, if
produced, will form in the triangular pyramid a section fgk
situated at the same distance above the common plans of the
bases ; and therefore the section j/^/i will be to the section aftc^/e
as the base FGH is to the base ADD (Prop. III.), and since
the bases are equivalent, the sections will be so likewise.
Hence the pyramids S-abcde, T-fgh are equivalent, for their
altitude is the same and their bases are equivalent. The whole
pyramids S-ABCDE, T-FGH are equivalent for the same rea-
son ; hence the frustums ABD-dab, FGH-hfg are equivalent ;
hence if the proposition can be proved in the single case of
the frustum of a triangular pyramid, it will be true of every
other.
liCt FGH-hfg be the frustum of a tri-
angular pyramid, having parallel bases :
through the three points F, g, H, pass
the plane F^H ; it will cut off from the
frustum the triangular pyramid g-FGH.
This pyramid has for its base the lower
base FGH of the frustum ; its altitude
likewise is that of the frustum, because
the vertex g lies in the plane of the up-
per base fgh.
This pyramid being cut off, there will
remain the quadrangular pyramid
g-f/iKF, whose vertex is g, and base fJiHF. Pass the plane
fgH through the three points /, g, H ; it will divide the quad-
rangular pyramid into two triangular pyramids g-F/H, g-fhH.
The latter has for its base the upper base gfh of the frustum ;
and for its altitude, the altitude of the frustum, because its ver-
tex H lies in the lower base. Thus we already know two of
the three pyramids which compose the frustum.
It remains to examine the third ^-FfH. Now, if ^K be
drawn parallel to /F, and if we conceive a new pyramid
K-FfH, having K for its vertex and FfH for its base, these
two pyramids will have the same base F/"H ; they will also
have the same altitude, because their vertices g and K lie in
the line ^K, parallel to F/, and consequently parallel to the
BOOK VII. 163
plane of the base : hence these pyramids are equivalent. But
the pyramid K-F/H may be regarded as having its vertex in
/, and thus its altitude will be the same as that of the frufitum :
as to its base FKH, we are now to show that this is a mean
proportional between the bases FGH and fgh. Now, the tri-
angles YMYLjfgh, have each an equal angle F=/; hence
FHK : /^/i : : FKxFH : fgxfh (Book IV. Prop. XXIV.) ;
but because of the parallels, FK==^, hence
FHK : /^A : : FH : fh.
We have also,
FHG : FHK : : FG : FK or fg.
But the similar triangles FGH,7^/i give *
FG:/^: : FH : /A ;
hence,
FGH : FHK : : FHK :/^A;
or the base FHK is a mean proportional between the two
bases FGH, fgh. Hence the frustum of a triangular pyramid
is equivalent to three pyramids whose common altitude is that
of the frustum and whose bases are the lower base of the
frustum, the upper base, and a mean proportional between the
two bases.
PROPOSITION XIX. THEOREM.
Similar triangular prisms are to each other as ike cubes of their
homologous sides.
Let CBD-P, chd-p, be two
similar triangular prisms, of
which BC, he, are homologous
sides : then will the prism
CBD-P be to the prism chd-p,
as BC^ to hc^.
For, since the prisms are
similar, the planes which con-
tain the homologous solid an- C ^ B
gles B and h, are similar, like placed, and equally inclined to
each other (Def. 17.) : hence the solid angles B and h, are equal
(Book VI. Prop. XXI. Sch.). If these solid angles be applied
to each other, the angle c/;6?will coincide with CBD, the side ha
with B A, and the prism chd-p will take the position Bcc?-p. From A
draw AH perpendicular to the common base of the prisms : then
will the plane BAH be perpendicular to the plane of the com-
164
GEOMETRY.
mon base (Book VI. Prop. XVI.). Through a, in the plane BAH.
draw ah perpendicular to
BH : then will ah also be per-
pendicular to the base BDC
(Book VI. Prop. XVII.) ; and
AH, ah will be the altitudes
of the two prisms.
Now, because of the similar
triangles ABH,«BA, an(l of the
similar parallelograms AC, ac,
we have
AH : flf/i : : AB : ez6 : : BC : he.
But since the bases are similar, we have
base BCD : base bed : : BC^ ; he"^ (Book IV. Prop. XXV.) ;
hence,
base BCD : base bed : : AH^ : ah\
Multiplying the antecedents by AH, and the consequents by
ahf and we have
base BCD X AH : base bed x ah : : AH^ ah\
But the solidity of a prism is equal to the base multiplied by
the altitude (Prop. XIV.) ; hence, the
prism BCD-P : prism bcd-p : : AH^ : ah^ : : BC^ : bc\
or as the cubes of any other of their homologous sides.
Cor. Whatever be the bases of similar prisms, the prisms
will be to each other as the cubes of their homologous sides.
For, since the prisms are similar, their bases will be similar
polygons (Def. 17.) ; and these similar polygons may be di-
vided into an equal number of similar triangles, similarly placed
(Book IV. Prop. XXVI.) : therefore the two prisms may be
divided into an equal number of triangular prisms, having their
faces similar and like placed ; and therefore, equally inclined
(Book VI. Prop. XXI.) ; hence the prisms will be similar. But
these triangular prisms will be to each other as the cubes of
their homologous sides, which sides being proportional, the
sums of the triangular prisms, that is, the polygonal prisms, will
be to each other as the cubes of their homologous sides.
I
PROPOSITION XX. THEOREM.
Two similar pyramids are to each other as the cubes of their
homologous sides.
BOOK VII. 165
For, since the pyramids are similar, the soUd
angles at the vertices will be contained by the
same number of similar planes, like placed,
and equally inchned to each other (Def. 17.).
Hence, the solid angles at the vertices may
be made to coincide, or the tv^^o pyramids
may be so placed as to have the solid angle
S common.
In that position, the bases ABCDE, abcde,
vf'iW be parallel ; because, since the homolo-
gous faces are similar, the angle S<z& is equal
to SAB, and S6c to SBC ; hence the plane
ABC is parallel to the plane ahc (Book VI. Prop. XIII.). This
being proved, let SO be the perpendicular drawn from the
vertex S to the plane ABC, and o the point where this perpen-
dicular meets the plane ahc : from what has already been
shown, we shall have
SO : So : : SA : Sa : : AB : ah (Prop. III.) ;
and consequently,
iSO : iSo : : AB : ah.
But the bases ABCDE, abcde, being similar figures, we have
ABCDE : abode : : AB^ : aU' (Book IV. Prop. XXVIL).
Multiply the corresponding terms of these two proportions ;
there results the proportion,
ABCDE xiSO : ahcdex^So : : AB^ : ah\
Now ABCDE X iSO is the solidity of the pyramid S-ABCDE,
and abcdexjSo is that of the pyramid S-abcde (Prop. XVII.) ;
hence two similar pyramids are to each other as the cubes of
their homologous sides.
General Scholium.
The chief propositions of this Book relating to the solidity ol
polyedrons, may be exhibited in algebraical terms, and so
recapitulated in the briefest manner possible.
Let B represent the base of a prism ; H its altitude : the
solidity of the prism will be B x H, or BH.
Let B represent the base of a pyramid ; H its altitude : the
solidity of the pyramid will be B x ^H, or H x ^B, or ^BH.
Let H represent the altitude of the frustum of a pyramid,
having parallel bases A and B ; VAB will be the mean pro-
portional between those bases ; and the solidity of the frustum
willbeiHx(A + B+VAB).
In fine, let P and p represent the solidities of two similar
prisms or pyramids ; A and a, two homologous edges : then we
shall have
P : p ; : A3 : a\
166
GEOMETRY.
BOOK VIII.
THE THREE ROUND BODIES.
Definitions.
E
M
:n:
\TI>
^:::^P^^
Li
K
G
1. A cylinder is the solid generated by the revolution oif a
rectangle ABCD, conceived to turn about the immoveable
side AB.
In this movement, the sides AD, BC, con- ^
tinuing always perpendicular to AB, describe
equal circles DHP, CGQ, which are called
the bases of the cylinder, the side CD at the
same time describing the convex surface.
The immoveable line AB is called the axis
of the cylinder.
Every section KLM, made in the cylinder,
at right angles to the axis, is a circle equal to
either of the bases ; for, whilst the rectangle
ABCD turns about AB, the line KI, perpen-
dicular to AB, describes a circle, equal to the base, and this
circle is nothing else than the section made perpendicular to
the axis at the point I.
Every section PQG, made through the axis, is a rectangle
double of the generating rectangle ABCD.
2. A cone is the solid generated by the revolution of a right-
angled triangle SAB, conceived to turn about the immoveable
side SA.
In this movement, the side AB describes
a circle BDCE, named the base of the cone ;
the hypothenuse SB describes the convex
surface of the cone.
The point S is named the vertex of the
cone, SA the axis or the altitude, and SB
the side or the apothem.
Every section HKFI, at right angles to
the axis, is a circle ; every section SDE,
through the axis, is an isosceles triangle,
double of the generating triangle SAB.
3. If from the cone S-CDB, the cone S-FKH be cut off by
a plane parallel to the base, the remaining sohd CBHF is called
a truncated cone, or the frustum of a cone.
BOOK VIII.
167
We may conceive it to be generated by the revolution of a
trapezoid ABHG, whose angles A and G are right angles, about
the side AG. The immoveable line AG is called the axis or
altitude of the frustum^ the circles BDC, HEK, are its bases, and
BH is its side.
4. Two cylinders, or two cones, are similar, when their
axes are to each other as the diameters of their bases.
5. If in the circle ACD, which forms the
base of a cylinder, a polygon ABCDE be
inscribed, a right prism, constructed on this
base ABCDE, and equal hi altitude to the
cylinder, is said to be inscribed in the cylin-
der, or the cylinder to be circumscribed
about the prism.
The edges AF, BG, CH, &c. of the prism,
being perpendicular to the plane of the base,
are evidently included in the convex sur-
face of the cylinder ; hence the prism and
the cylinder touch one another along these
edges.
6. In like manner, if ABCD is a poly-
gon, circumscribed about the base of a
cylinder, a right prism, constructed on this
base ABCD, and equal in altitude to the
cylinder, is said to be circumscribed about
the cylinder, or the cylinder to be inscribed
in the prism.
Let M, N, &c. be the points of contact
in the sides AB, BC, &c. ; and through the
points M,N,&c. let MX, NY, &c. be drawn Ak
perpendicular to the plane of the base : x^
these perpendiculars will evidently lie both
in the surface of the cylinder, and in that 13
of the circumscribed prism ; hence they will be their lines of
contact.
7. If in the circle ABCDE, which forms
the base of a cone, any polygon ABCDE
be inscribed, and from the vertices A, B,
C, D, E, lines be drawn to S, the vertex
of the cone, these lines may be regarded
as the sides of a pyramid whose base is
the polygon ABCDE and vertex S. The
sides of this pyramid are in the convex
surface of the cone, and the pyramid is
said to be inscribed in the cone.
168
GEOMETRY.
8. The sphere is a solid terminated by a curved surface, all
the points of which are equally distant from a point within,
called the centre. "-
The sphere may be con-
ceived to be generated by the
revolution of a semicircle
DAE about its diameter DE :
for the surface described in
this movement, by the curve
DAE, will have all its points
equally distant from its cen-
tre C.
9. Whilst the semicircle
DAE revolving round its di-
ameter DE, describes the
sphere ; any circular sector,
as pCF or FCH, describes a
solid, which is named a spherical sector.
10. The radius of a sphere is a straight line drawn from the
centre to any point of the surface ; the diameter or axis is a .
line passing through this centre, and terminated on both sides ,
by the surface.
All the radii of a sphere are equal ; all the diameters are
equal, and each double of the radius.
11. It will be shown (Prop. VII.) that every section of the
sphere, made by a plane, is a circle : this granted, a great cir-^
cle is a section which passes through the centre ; a small circle^
is one which does not pass through the centre.
12. A plane is tangent to a sphere, when their surfaces have
but one point in common:
13. A zone is a portion of the surface of the sphere included
between two parallel planes, which form its bases. One of
these planes may be tangent to the sphere ; in which case, the
zone has only a single base.
14. A spherical segment is the portion of the solid sphere,
included between two parallel planes which form its bases.
One of these planes may be tangent to the sphere ; in which
case, the segment has only a single base.
15. The altitude of a zone or of a segment is the distance
between the tw^o parallel planes, which form the bases of the
zone or segment.
Note. The Cylinder, the Cone, and the Sphere, are the
three round bodies treated of in the Elements of Geometry.
BOOK VIII.
1G9
PROPOSITION I. THEOREM.
ne convex surface of a cylinder is equal to the circumference of
its base multiplied by its altitude.
Let CA be the radius of the
given cylinder's base, and H its
altitude : the circumference
whose radius is CA being rep-
resented by circ. CA, we are to
show that the convex surface of
the cyhnder is equal to circ. CA
xH.
Inscribe in the circle any
regular polygon, BDEFGA, and
construct on this polygon a right
prism having its altitude equal to H, the altitude of the cylin-
der : this prism will be inscribed in the cylinder. The convex
surface of the prism is equal to the perimeter of the polygon,
multiphed by the altitude H (Book VII. Prop. I.). Let now
the. arcs which subtend the sides of the polygon be continually
bisected, and the number of sides of the polygon indefinitely
increased : the perimeter of the polygon will then become equal
to circ. CA (Book V. Prop. VIII. Cor. 2.), and the convex sur-
face of the prism will coincide with the convex surface of the
cylinder. But the convex surface of the prism is equal to the
perimeter of its base multiplied by H, whatever be the number
of sides : hence, the convex surface of the cylinder is equal to
the circumference of its base multiplied by its altitude.
PROPOSITION II. THEOREM.
The solidity of a cylinder is equal to the product of its base by its
altitude.
I'TO GEOMETRY.
Let CA be the radius of the
base of the cylinder, and 11
ihe altitude. Let the circle
whose radius is CA be repre-
sented by area CA, it is to be
proved that the solidity of the
cylinder is equal to area CA x Hi
Inscribe in the circle any regu-
lar polygon BDEFGA, and con-
struct on this polygon a right
prism having its altitude equal
to H, the altitude of the cylinder : this prism will be inscribed
in the cylinder. The solidity of the prism will be equal to the
area of the polygon multiplied by the altitude H (Book VIL
Prop. XIV.). Let now the number of sides of the polygon be
indefinitely increased : the solidity of the new prism will still
be equal to its base multiplied by its altitude.
But when the number of sides of the polygon is indefinitely
increased, its area becomes equal to the area CA, and its pe-
rimeter coincides with circ. CA (Book V. Prop. VUL Cor. 1.
& 2.) ; the inscribed prism then coincides with the cylinder,
since their altitudes are equal, and their convex surfaces per-
pendicular to the common base : hence the two solids will be
equal ; therefore the solidity of a cylinder is equal to the product
of its base by its altitude.
Cor. L Cylinders of the same altitude are to each other as
their bases ; and cylinders of the same base are to each other
as their altitudes.
-i;.i4^- -:v-. ■■■■ ' ' •
^•-'•Coi\ 2. Similar cylindo -s are to each other as the cubes of
their altitudes, or as the cuoes of the diameters of their bases.
For the. bases are as the squares of their diameters ; and the
, cylinders behig similar, the diameters of their bases are to
. each other as the altitudes (Def. 4.) ; hence the bases are
as the squares of the altitudes ; hence the bases, multiplied
by the altitudes, or the cylinders themselves, are as the cubes
of the altitudes.
Scholium, Let R be the radius of a cylinder's base ; H the
altitude : the surface of the base will be tt.W (Book V. Prop.
XII. Cor. 2.) ; and the solidity of the cylinder will be jiR-xH
or^r.Rs.H.
BOOK VIII.
171
PROPOSITION III. THEOREM.
The convex surface of a cone is equal to the circumference of its
'■ haselmultipliedhy half its side.
Let the circle ABCD be the
base of a cone, S the vertex,
SO the altitude, and SA the
side : then will its convex sur-
face be equal to circ. OA x ^S A.
For, inscribe in the base of
the cone any regular polygon
ABCD, and on this polygon as
a base conceive a pyramid to
be constructed having S for its
vertex : this pyramid will be a
regular pyramid, and will be inscribed in the cone.
From S, drav/ SG perpendicular to one of the sides of the
polygon. The convex surface of the inscribed pyramid is equal
to the perimeter of the polygon which forms its base, multiplied
by half the slant height SG (Book VII. Prop. IV.). Let now
the number of sides of the inscribed polygon be indefinitely
increased; the perimeter of the inscribed polygon will then
become equal to circ, OA, the slant height SG will become
equal to the side SA of the cone, and the convex surfape of
the pyramid to the convex surface of the cone. But wfe-atever
be the number of sides of the polygon which forms 'the base,
the convex surface of the pyramid is equal to the perimeter of
the base multiplied by half the slant height: hence the. convex
.surface of a cone is equal to the circumference of the base
multiplied by half the side.
Scholium. Let L be the side of a cone, R the radius. of its
base ; the circumference of this base will be 27r.R, and the sur-
face of the cone will be 2iR x^L, or jiRL.
y?{
PROPOSITION IV. THEOREM.
The convex surface of the frustum of a cone is equal to its side
multiplied by half the sum of the circumferences of its two
bases.
172
GEOMETRY
Lei BTA-DE be a frustum of a
cone : then will its convex surface be
equal to AD x ^ circ.OA + circ.CU ^
. For, inscribe in the bases of the
frustums two regular polygons of the
same number of sides, and having
their homologous sides parallel, each
to each. The lines joining the ver-
tices of the homologous angles may
be regarded as the edges of the frus-
tum of a regular pyramid inscribed
in the frustum of the cone. The con-
vex surface of the frustum of the
pyramid is equal to half the sum of the perimeters of its bases
multiplied by the slant height fh (Book VII. Prop. IV. Cor.).
Let now the number of sides of the inscribed polygons be
^ indefinitely increased : the perimeters of the polygons will be-
come equal to the circumferences BI A, EGD ; the slant height
fh will become equal to the side AD or BE, and the surfaces
of the two frustums will coincide and become the same surface.
But the convex surface of the frustum of the pyramid will
still be equal to half the sum of the perimeters of the upper
and low^er bases multiplied by the slant height : hence the sur-
face of the frustum of a cone is equal to its side multiphed by
half the sum of the circumferences of its two bases.
Cor, Through/, the middle point of AD, draw /KL paral-
* lel to AB, and /i, Dd, parallel to CO. Then, since A/, 11), are
'^ equal, Ai, id, will also be equal (Book IV. Prop. XV. Cor. 2.) :
hence, K/ is equal to ^(OA + CD). But since the circumfe-
rences of circles are to each other as their radii (Book V.
Prop. XL), the circ. Kl=l(circ. 0A + circ. CD) ; therefore, the
convex surface of a frustum of a cone is equal to its side multi-
plied hy the circumference of a section at equal distances from
the two bases. , ,
Scholium. If a line AD, lying wholly on one side of the line
OC, and in the same plane, make a revolution 'around OC,
tlie surface, described by AD will have for its measure ADx
/circ.AQ + arc. DC\ ^^ ^j^ ^ ^.^^ ^j^. '^^^ jj^^^ ^^ ^^^ ^^
bemg perpendiculars, let fall from the extremities and from
the middle point of AD, on the axis OC.
For, if AD and OC are produced till they meet in S, the
surface described by AD is evidently the frustum of a cone
BOOK VIII. 173
•
having AO and DC for the radii of its bases, the vertex of
the whole cone being S. Hence this surface will be measured
as we have said.
This measure will always hold good, even when the point
P falls on S, and thus forms a whole cone ; and also when the
line AD is parallel to the axis, and thus forms a cylinder. In
the first case DC would be nothing ; in the second, DC would
be equal to AO and to IK,
PROPOSITION V. THEOREM.
The solidity of a cone is equal to its base multiplied hy a third of
its altitude.
Let SO be the altitude of a cone,
OA the radius of its base, and let
the area of the base be designated
by area OA : it is to be proved that
the solidity of the cone is equal to
area. OAx^SO.
Inscribe in the base of the cone
any regular polygon ABDEF, and
join the vertices A, B, C, <fec. with
the vertex S of the cone : then will
there be inscribed in the cone a
regular pyramid having the same vertex as the cone, and hav-
ing for its base the polygon ABDEF. The solidity of this
pyramid is equal to its base multiplied by one third of its ahi-
tude (Book VII. Prop. XVII.). Let now the number of sides
of the polygon be indefinitely increased : the polygon will then
become equal to the circle, and the pyramid and cone will
coincide and become equal. But the solidity of the pyramid
is equal to its base multiplied by one third of its altitude, what-
ever be the number of sides of the polygon which forms its
base : hence the solidity of the cone is equal to its base multi-
plied by a third of its altitude.
Cor. A cone is the third of a cylinder having the same base
and the same altitude ; whence it follows,
1. That cones of equal altitudes are to each other as their
bases ;
2. That cones of equal bases are to each other as their
altitudes ;
3. That similar cones are as the cubes of the diameters of
their bases, or as the cubes of their altitudes.
P*
174 GEOMETRY.
Cor. 2. The solidity of a cone is equivalent to the solidity of
a pyramid having an equivalent base and the same altitude
(Book VII. Prop. XVIL).
Scholium. Let R be the. radius of a cone's base, H its alti-
tude ; the solidity of the cone will be nW x ^H, or ^^iR^H.
PROPOSITION VI. THEOREM
The solidity of the frustum of a cone is equal to the sum of the
solidities of three cones whose common altitude is the altitude
of the frustum, and whose bases are, the upper base of the frus-
tum, the lower base of the frustum, and a mean proportional
between them.
Let AEB-CD be the frustum of a
cone, and OP its altitude ; tlien will its
solidity be equal to
^^ X OP X (A02+DP2+ AO X DP).
For, inscribe in the lower and- upper
basea two regular polygons having the
same number of sides, and having their
homologous sides parallel, each to each.
Join the vertices of the homologous
angles and there will then be inscribed
in the frustum of the cone, the frustum
of a regular pyramid. The sohdity of
the frustum of the pyramid is equivalent to three pyramids
having the common altitude of the frustum, and for bases, the
lower base of the frustum, the upper base of the frustum, and
a mean proportional between them (Book VIL Prop. XVIII.).
Let now, the number of sides of the inscribed polygons be
indefinitely increased : the bases of the frustum of the pyramid
■will then coincide with the bases of the frustum of the cone,
and the two frustums will coincide and become the same solid.
Since the area of a circle is equal to R-.tt (Book V. Prop. XII.
Cor. 2.), the expression for the solidities of the frustum will
become
for the first pyramid -\0P x OA^rr.
for the second iOP x PD^.^r
for the third J OP x AO x PD.Tt ; since
AO x PD.Ti is a mean proportional between OA^.^r and PD^.rr.
Hence the solidity of the frustum of the cone is measured by
4«0P x (OAH PDH AO X PD).
BOOK VIII. 175
PROPOSITION VII. THEOREM.
Every section of a sphere, made by a plane, is a cir.c!e.
Let AMB be a section, made by a
plane, in the sphere whose centre is C.
From the point C, draw CO perpen-
dicular to the plane AMB ; and diffe-
rent lines CM, CM, to different points
of the curve AMB, which terminates
the section.
The oblique lines CM, CM, CA, are
equal, being radii of the sphere ; hence
they are equally distant from the perpendicular CO (Book VI.
Prop. V. Cor.) ; therefore all the lines OM, OM, OB, are equal ;
consequently the section AMB is a circle, whose centre is O.
Cor 1. If the section passes through the centre of the sphere,
its radius will be the radius of the sphere; hence all great
circles are equal.
Co7\ 2. Two great circles always bisect each other ; for
their common intersection, passing through the centre, is a
diameter.
Cor. 3. Every great circle divides the sphere and its surface
into two equal parts : for, if the two hemispheres were sepa-
rated and afterwards placed on the common base, with their
convexities turned the same way, the two surfaces would
exactly coincide, no point of the one being nearer the centre
than any point of the other.
Cor. 4. The centre of a small circle, and that of the sphere,
are in the same straight line, perpendicular to the plane of the
small circle.
Cor. 5. Small circles are the less the further they lie from
the centre of the sphere ; for the greater CO is, the less is the
chord AB, the diameter of the small circle AMB.
Cor. 6. An arc of a great circle may always be made to pass
through any two given points of the surface of the sphere ; for
the two given points, and the centre of the sphere make three
points which determine the position of a plane. But if the
two given points were at the extremities of a diameter, these
two points and the centre would then lie in one straight line,
\ and an infinite number of great circles might be made to pass
s through the two given points.
176
GEOMETRY.
PROPOSITION VIII. THEOREM.
Evejy plane perpendicular to a radius at its extremity is tangent
to the. sphere.
Let FAG be a plane perpendicular
to the radius OA. at its extremity A.
Any point M in this plane being as-
sumed, and OM, AM, being drawn,
the angle 0AM will be a right angle,
and hence the distance OM will be
greater than OA. Hence the point
M lies without the sphere ; and as the
same can be shown for every other
point of the plane FAG, this plane can
have no point but A common to it and the surface of the sphere ;
hence it is a tangent plane (Def. 12.)
Scholium. In the same way it may be shown, that two
spheres have but one point in common, and therefore touch
each other, when the distance between their centres is equal to
the sum, or the difference of their radii ; in which case, the
centres and the point of contact lie in the same straight line
PROPOSITION IX. LEMMA.
If a regular semi-polygon he rev^olved about a line passing
through the centre and the vertices of two opposite angles, the
surface described by its perimeter will be equal to the axis mul-
tiplied by the circumference of the inscribed circle.
Let the regular semi-polygon ABCDEF,
be revolved about the line AF as an axis :
then will the surface described by its pe-
rimeter be equal to AF multiplied by the
circumference of the inscribed circle.
From E and D, the extremities of one of
the equal sides, let fall the perpendiculars
EH, DI, on the axis AF, and from the cen-
tre O draw ON perpendicular to the side
DE : ON will be the radius of the inscribed
circle (Book V. Prop. H.). Now, the sur-
face described in the revolution by any one
side of the regular -polygon, as DE, has
BOOK VIII.
177
been shown to be equal to DE x circ. NM (Prop. IV. Sch.).
Biit since the triangles EDK, ONM, are similar (Book IV
Prop. XXL), ED : KK or HI : : ON : NM, or as circ. ON .
circ. NM ; hence
ED X circ. NM = HI x circ. ON ;
and since the same may be shown for each of the other sides,
it is plain that the surface described by the entire perimeter i$
equal to
(FH + HH-IP + PQ+QA)xaVc ON=AFxcirc. ON.
Cor. The surface described by any portion of the perime-
ter, as EDC, is equal to the distance between the two perpen
diculars let fall from its extremities on the axis, multiplied by
the circumference of the inscribed circle. For, the surfac(
described by DE is equal to HI x circ. ON, and the surface
described by DC is equal to IPx circ. ON : hence the surface
described by ED + DC, is equal to (HI + IP) x arc. ON, o]
equal to HP x circ. ON.
PROPOSITION X. THEOREM.
The surface of a sphere is equal to the product of its diameter 5y
the circumference of a great circle.
Let ABODE be a semicircle. Inscribe in
it any regular semi-polygon, and from the
centre O draw OF perpendicular to one of
the sides.
Let the semicircle and the semi-polygon
be revolved about the axis AE : the semi-
circumference ABODE will describe the
surface of a sphere (Def. 8.) ; and the pe-
rimeter of the semi-polygon will describe
a surface which has for its measure AE x
circ. OF (Prop. IX.), and this will be true
whatever be the number of sides of the po-
lygon. But if the number of sides of the polygon be indefi-
nitely increased, its perimeter will coincide with the circumfe-
rence ABODE, the perpendicular OF will become equal to
OE, and the surface described by the perimeter of the semi-
polygon will then be the same as that described by the semi-
circumference ABODE. Hence the surface of the sphere is
equal to AE x circ, OE.
Cor. Since the area of a great circle is equal to the product
of its circumference by half the radius, or one fourth of the
178
GEOMETRY.
uiameter (Book V. Prop. XII.), it follows that the surface of a
siihere is equal to four of its great circles : that is, equal to
4.^.0A2 (Book V. Prop. XII. Cor. 2.).
Scholium 1. The surface of a zone is equal to its altitude mul-
tiplied by the circumference of a great circle.
For, the surface described by any portion
of the perimeter of the inscribed polygon, as
BC + CD, is equal to EH xcfrc. OF (Prop.
IX. Cor.). But when the number of sides
of the polygon is indefinitely increased, BC
f CD, becomes the arc BCD, OF becomes
equal to OA, and the surface described by
BC + CD, becomes the surface of the zone
described by the arc BCD : hence the sur-
*kce of the zone is equal to EH x circ. OA.
Scholium 2. When the zone has but one
base, as the zone described by the arc ABCD, its surface will
still be equal to the altitude AE multiplied by the circumference
of a great circle.
Scholium 3. Two zones, taken in the same sphere or in equal
spheres, are to each other as their altitudes ; and any zone is to
the surface of the sphere as the altitude of the zone is to the
diameter of the sphere.
PROPOSITION XL LEMMA.
If a triangle and a rectangle, having the same base and the same
altitude, tujm together about the common base,the solid described
by the triangle will be a third of the cylinder described by the
rectangle. *^'
Let ACB be the triangle, and BE the rectangle.
On the axis, let fall the perpcn- ^
dicular AD : the cone described by
jthe triangle ABD is the third part of
the cylinder described by the rectan-
gle AFBD (Prop. V. Cor.) ; also the
cone described by the triangle ADC
is the third part of the cylinder de-
sci-ibed by the rectangle ADCE ; hence the sum of the two
cones, or the solid described by ABC, is the third part of the
two cylinders taken together, or of the cylinder described by
the rectangle BCEF.
BOOK VIII.
179
If the perpendicular AD falls without
the triangle ; the solid described by ABC
will, in that case, be the difference of the
two cones described by ABD and ACD ;
but at the same time, the cylinder de- .
scribed by BCEF will be the difference ^ ^ ^
of the two cylinders described by AFBD and AECD. Hence
the solid, described by the revolution of the triangle, will still
be a third part of the cylinder described by the revolution of
the rectangle having the same base and the same altitude.
Scholium. The circle of which AD is radius, has for its
measure ^rx AD-; hence t^x AD^xBC measures the cylinder
described by BCEF, and ^ttx AD-xBC measures the solid
described by the triangle ABC.
PROPOSITION XII. LEMMA.
^^tMJ^dtL^,: .
If a triangle he revolved about a line drawn at pleasure through
its vertex, the solid described by the triangle will have for its
measure, the area of the triangle multiplied by two thirds of the
circumference traced, by the middle point of the base.
Let CAB be the triangle, and CD the line about which it
revolves.
Produce the side AB till it
meets the axis CD in D ; from the
points A and B, draw AM, BN,
perpendicular to the axis, and CP
perpendicular to DA produced.
The solid described by the tri-
angle CAD is measured by \n x
AM^x CD (Prop. XI. Sch.) ; the solid described by the triangle
CBD is measured by ^n x BN- x CD ; hence the difference of
those solids, or the solid described by ABC, will have for its
measure i7r(AM2— BN^) x CD.
To this expression another form may be given. From I, the
middle point of AB,draw IK perpendicular to CD ; and through
B, draw BO parallel to CD : we shall have AM + BN=2lK
(Book IV. Prop. VII.) ; andAM— BN=AO; hence (AM +
BN) X* ( AM— NB), or AM2_BN2=2IK x AO (Book IV. Prop.
X.). Hence the measure of the solid in question is ex-
pressed by
l^xIKxAOxCD.
MX^sr
18t/
OEOMETRf.
M:x3>r
But CP being drawn perpendicular to AB, the triangles ABO
DCP will be similar, and give the proportion
AO : CP : : AB : CD;
hence AOxCD=CPxAB;
but CP X AB is double the area of the triangle ABC ; hence
we have
A0xCD=2ABC;
hence the solid described by the
triangle ABC is also measured
by |7r X ABC X IK, or which is the
same thing, by ABC x Icirc. IK,
arc. IK being equal to S^rxIK.
Hence the solid described by the
revolution of the triangle ABC, has
for its measure the area of this triangle multiplied hy two thirds
of the circumference traced by I, the middle point of the base.
Cor, IfthesideAC-=:CB,
the line CI will be perpen-
dicular to AB, the area ABC
will be equal to ABx|CI,
and the solidity ^n x ABC x
IK will become fTixABx
IKxCI. But the triangles
ABO, CIK, are similar, and
give the proportion AB : BO
or MN : : CI : IK; hence ABxIK=MNxCI; hence the
solid described by the isosceles triangle ABC will have for its
measure fTrxCFxMN : that is, equal to two thirds of n into
the square of the perpendicular let fall on the base, into the
distance between the pwo perpendiculars let fall on the axis.
Scholium. The general solution appears to include the sup-
position that AB produced will meet the axis ; but the results
would be equally true, though AB were parallel to the axis.
Thus,the cylinder described by AMNB p /\ j^
is equal to tt.AM^.MN ; the cone descri-
bed by ACM is equal to In.KW.QM,
and the cone described by BCN to
^nkW CN. Add the first two solids and
take away the third ; we shall have the
solid described by ABC equal to tt.AM^.
(MN + iCM— iCN): and since CN— CM =MN, this expres-
sion is reducible to tt.AM-.^MN, or fTi.CP^.MN; which agreeg
with the conclusion found above.
BOOK VIII.
181
PROPOSITION XIII. LEMMA.
Ij a regular semi-polygon he revolved about a line passing
through the centre and the vertices of iivo opposite angles, the
solid described will be equivalent to a cone, having for its base
the inscribed circle, and for its altitude twice the axis about
which the semi-polygon is revolved.
Let the semi-polygon FABG be revolved
about FG : then, if 01 be the radius of the
inscribed circle, the solid described will be
measured by ^area 01 x 2FG.
For, since the polygon is regular, the
triangles OFA, OAB, OBC, &c. are equal
and isosceles, and all the perpendiculars let
fall from O on the bases FA, AB, &c. will
be equal to 01, the radius of the inscribed
circle.
Now, the solid described by OAB is mea-
sured by frr OP+MN (Prop. XII. Cor.) ;
the solid described by the triangle OFA has for its measure
fTiOP X FM, the solid described by the triangle OBC, has for
its measure f tiOP x NO, and since the same may be shown for
the solid described by each of the other triangles, it follows
that the entire solid described by the semi-polygon is mea-
sured by |7iOP.(FM+MN + NO + OQ+QG), or f^OPxFG ;
which is also equal to ^tiOP x 2FG. But yr.OP is the area of
the inscribed circle (Book V. Prop. XII. Cor. 2.) : hence the
solidity is equivalent to a cone whose base is area 01, and
altitude 2FG.
PROPOSITION XIV. THEOREM.
The solidity of a sphere is equal to its surface multiplied by a
third of its radius.
182 GEOMETRY.
Inscribe in the semicircle ABCDE a
regular semi-polygon, having any number
of sides, and let 01 be the radius of the
circle inscribed in the polygon. .
If the semicircle and semi-polj^gon be
•'evolved about EA, the semicircle will
describe a sphere, and the semi-polygon a
solid which has for its measure fyiOPx
EA (Prop. XIII.) ; and this will be true
whatever be the number of sides of the
polygon. But if the number of sides of
the polygon be indefinitely increased, the E
semi-polygon will become the semicircle, 01 will become
equal to OA, and the solid described by the semi-polygon will
become the sphere : hence the solidity of the sphere is equal
to fTiOA^xEA, or by substituting 20 A for EA, it becomes
jn.OA^ X OA, which is also equal to 47rOA2 x ^OA. But 47r.OA2
is equal to the surface of the sphere (Prop. X. Cor.) : hence
the solidity of a sphere is equal to its surface multiphed by a
third of its radius.
Scholium 1. The solidity of every spherical sector is equal to
the zone which forms its base, multiplied by a third of the radius.
For, the solid described by any portion of the regular poly-
gon, as the isosceles triangle OAB, is measured by f^rOFx AF
(Prop. XII. Cor.) ; and when the polygon becomes the circle,
the portion- OAB becomes the sector AOB, 01 becomes equal
to OA, and the solid described becomes a spherical sector. But
its measure then becomes equal to f^r. AO^ x AF, which is equal
to 27r.AO X AF X ^AO. But 27r.AO is the circumference of a
great circle of the sphere (Book V. Prop. XII. Cor. 2.), which
being multiplied by AF gives the surface of the zone which
forms the base of the sector (Prop. X. Sch. 1.) : and the proof
is equally applicable to the spherical sector described by the
circular sector BOC : hence, the solidity of the spherical sector
is equal to the zone which forms its base, multiplied by a third
of the radius.
Scholium 2. Since the surface of a sphere whose radius is
R, is expressed by 47rR2 (Prop. X. Cor.), it follows that the
surfaces of spheres are to each other as the squares of their
radii ; and since their solidities are as their surfaces multiplied
by their radii, it follows that the solidities of 'spheres are to
each other as the cubes of their radii, or as the cubes of their
diameters.
BOOK VIII.
183
Scholium 3. Let R be the radius of a sphere ; its surface
will be expressed by 47rR^, and its solidity by AnK^ x ^R, or
|7rR^. If the diameter is called D, we shall have R=iD,
and R'*'=|D^ : hence the solidity of the sphere may likewise be
expressed by
PROPOSITION XV. THEOREM.
The surface of a sphere is to the whole surface of the circum-
scribed cylinder^ including its bases, as 2 is to 3 : and the so-
lidities of these two bodies are to each other in the same ratio.
D
^ « ^
s^
^^^
/ "^
: \
/^^— -
^
^
^
Let MPNQ be a great circle of the
sphere ; ABCD the circumscribed
square : if the semicircle PMQ and
the half square PADQ are at the
same time made to revolve about the
diameter PQ, the semicircle will gene- ]M
rate the sphere, while the half square
will generate the cylinder circum-
scribed about that sphere. .
The altitude AD of the cylinder is
equal to the diameter PQ ; the base of
the cylinder is equal to the great circle, since its diameter AB
is equal to MN ; hence, the convex surface of the cylinder is
equal to the circumference of the great circle multiplied by its
diameter (Prop. 1.). This measure is the same as that of the
surface of the sphere (Prop. X.) : hence the surface of the sphere
is efiual to the convex surface of the circumscribed cylinder.
But the surface of the sphere is equal to four great circles ;
hence the convex surface of the cylinder is also equal to four
great circles : and adding the two bases, each equal to a great
circle, the total surface of the circumscribed cylinder will be
equal to six great circles; hence the surface of the sphere is to
the total surface of the circumscribed cylinder as 4 is to 6, or
as 2 is to 3 ; which was the first branch of the Proposition.
In the next place, since the base of the circumscribed cylin-
der is equal to a great circle, and its altitude to the diameter,
the solidity of the cylinder will be equal to a great circle mul-
tiplied by its diameter (Prop. II.). But the solidity of the
sphere is equal to four great circles multiplied by a third of the
radius (Prop. XIV.); in other terms, to one great circle multi-
plied by ^ of the radius, or by | of the diameter ; hence the
sphere is to the circumscribed cylinder as 2 to 3, and conse-
quently the solidities of these two bodies are as their surface*
184 GEOMETRY. M
%
Scholium. Conceive a polyedron, all of whose faces touch
the sphere ; this polyedron may be considered as formed of
pyramids, each having for its vertex the centre of the sphere,
and for its base one of the polyedron's faces. Nov*^ it is evi-
dent that all these pyramids will have the radius of the sphere
for their common altitude : so that each pyramid will be equal
to one face of the polyedron multiplied by a third of the radius ;
hence the whole polyedron will be equal to its surface multi-
plied by a third of the radius of the inscribed sphere.
It is therefore manifest, that the solidities of polyedrons cir-
cumscribed about the sphere are to each other as the surfaces
of those polyedrons. Thus the property, which we have shown
to be true with regard to the circumscribed cylinder, is also
true with regard to an infinite number of other bodies.
We might likewise have observed that the surfaces of poly-
gons, circumscribed about the circle, are to each other as their
perimeters.
PROPOSITION XVI. PROBLEM.
If a circular segment he supposed to make a revolution about a
diameter exterior to it^ required the value of the solid which it
describes, A
Let the segment BMD revolve about AC.
On the axis, let fall the perpendiculars
BE, DF ; from the centre C, draw CI
perpendicular to the chord BD ; also draw
the radii CB, CD.
The solid described by the sector BCD . C
is measured by f^ CB^.EF (Prop. XIV. Sch. 1). But the
solid diescribed by the isosceles triangle DCB has for its mea-
sure fTT.CP.EF (Prop. XII. Cor.) ; hence the sohd described
by the segment BMD=|^.EF.(CB2— CP). Now, in the right-
angled triangle CBI, we have CB^— CP^BP^iBD^ ; hence
the solid described by the segment BMD will have for its mea-
sure f TT.EF.iBD^, or itt.BDIEF: that is one sixth of n into
the square of the chord, into the distance between the two per-
pendiculars let fall from the extremities of the arc on the
axis.
Scholium. The solid described by the segment BMD is to
the sphere which has BD for its diameter, as ^^r.BD-.EF is
to i^.BD^ or as EF to BD.
BOOK VIII. 185
PHOPOSITION XVII. THEOREM.
Every segment of a sphere is measured hy the half sum of
its bases multiplied hy its altitude, plus the solidity of a
sphere whose diameter is this same altitude.
Let BE, DF, be the radii of the two
bases of the segment, EF its altitude, the
segment being described by the revolu-
tion of the circular space BMDFE about
the axis FE. The solid described by the
segment BMD is equal to ^rr.BD^.EF
(Prop. XVI.) ; and the truncated cone de-
scribed by the trapezoid BDFE is equal
to i 7r.EF. (BE2 + DF-+ BE.DF) (Prop.V I.) ;
hence the segment of the sphere, which is the sum of those two
solids, must be equal to i7r.EF.(2BE2+2DF2+2BE.DF+BD:)
But, drawing BO parallel to EF, we shall have DO=DF— BE,
hence DO'^^DF^— 2DF.BE + BE2 (Book IV. Prop. IX.) ; and
consequently BD2=zB02+ D0'^=:EF2+ DF^— 2DF . BE + BE'^.
Put this value in place of BD^ in the expression for the value
of the segment, omitting the parts which destroy each other ;
we shall obtain for the solidity of the segment,
i7rEF.(3BE2+3DF2+EF2),
an expression which may be decomposed into two parts ; the
/7t.BE2+^.DF2\
one i^.EF.(3BE2+3DF2), or EF.( ^ ) being the
rhalf sum of the bases multiplied by the altitude ; while the
other iTt.EF^ represents the sphere of which EF is the diame-
ter (Prop. XIV. Sch.) ; hence every segment of a sphere, &c.
;. Cor, If either of the bases is nothing, the segment in ques-
% tion becomes a spherical segment with a single base ; hence
jl' any spherical segment, with a single base, is equivalent to half
§ Vie cylinder having the same base and the same altitude, plus the
'■k sphere of which this altitude is the diameter..
General Scholium.
Let R be the radius of a cylinder's base, H its altitude : the
I .solidity of the cylinder will be ttR^ x H, or ttR^H. -
^ Let R be the radius of a cone's base, H its altitude: the
solidity of the cone will be nWx ^H, or i^iR^H.
Let A and B be the radii of the bases of a truncated cone,
a*
186 GEOMETRY.
H its altitude : the solidity of the truncated cone will be iTr.H.
(A^+BHAB).
Let R be the radius of a sphere ; its solidity will be "flifR^.
Let R be the radius of a spherical sector, H the altitude of
the zone, which forms its base : the solidity of the sector will
be f ^R2H.
Let P and Q be the two bases of a spherical segnient, H its
altitude : the solidity of the segment will be -_I_z.H+|7r.H^.
If the spherical segment has but one base, the other being
nothing, its solidity will be ^PH + ^tiH^.
A
BOOK IX.
OF SPHERICAL TRIANGLES AND SPHERICAL POLYGONS.
Definitions.
1. A spherical triangle is a portion of the surface of a sphere,
bounded by three arcs of great circles.
These arcs are named the sides of the triangle, and are
always supposed to be each less than a semi-circumference.
The angles, which their planes form with each other, are the
angles of the triangle.
2. A spherical triangle takes the name of right-angled,
isosceles, equilateral, in the same cases as a rectilineal triangle.^
3. A spherical polygon is a portion of the surface of a sphere %
terminfited by several arcs of great circles. "
4. A lune is that portion of the surface of a sphere, which is
included between two great sehii-circles meeting in a common
diameter.
5. A spherical wedge or ungula is that portion of the solid
sphere, which is included between the same great semi-circles,
and has the lune for its base.
6. A spherical pyramid is a portion of the solid sphere, in-
cluded between the planes of a solid angle whose vertex is
the centre. The base of the pyramid is the spherical polygon
intercepted by the same planes.
7. The fole of a circle of a sphere is a point in the surface
equally distant from all the points in the circumference of this
circle. It will be shown (Prop. V.) that every circle, great or
small, has always two poles.
BOOK IX. 187
PROPOSITION I. THEOREM.
In every spherical triangle, any side is less than the sttM of the
other two.
Let O be the centre of the sphere, and
ACB the triangle ; draw the rad i i O A, OB,
OC. Imagine the planes AOB, AOC,
COB, to be drawn ; these planes will form
a solid angle at the centre O ; and the an-
gles AOB, AOC, COB, will be measured
by AB, AC, BC, the sides of the spherical
triangle. But each of the three plane an-
gles forming a solid angle is less than the
sum of the other two (Book VI. Prop.
XIX.) ; hence any side of the triangle
ABC is less than the sum of the other two.
PROPOSITION II. THEOREM.
The shortest path from one point to another, on the surface of a
sphere, is the arc of the great circle which joins the two given
points.
Let ANB be the arc of a great circle
which joins the points A and B ; then will it
be the shortest path between them.
1st. If two points N and B, be taken on
the arc of a great circle, at unequal distan-
ces from the point A, the shortest distance
from B to A will be greater than the short-
est distance from N to A. ''b
For, about A as a pole describe a circumference CNP. Now,
the line of shortest distance from B to A must cross this circum-
ference at some point as P. But the shortest distance from P to
A whether it be the arc of a great circle or any other line, is
equal to the shortest distance from N to A; for, by passing the
arc of a great circle through P and A, and revolving it about the
diameter passing through A, the point P maybe made to coincide
with N, when the shortest distance from P to A will coincide
with the shortest distance from N to A : hence, the shortest dis-
tance from B to A, will be greater than the shortest distance
from N to A, by the shortest distance from B to P.
If the point B be taken without the arc AN, still making AB
greater than AN, it may be proved in a mannerentirely similar
to the above, that the shortest distance from B to A will be great-
er than the shortest distance from N to A.
If now, there be a shorter path between the points B and A,
than the arc BDA of a great circle, let M be a point of the short-
88 GEOMETRY.
est distance possible; then through M draw MA. MB. arcs ot
great circles, and take BD equal to BM. By the last theorem,
BDA< BM + MA; take BD = BM from each, and there will re-
main AD< AM. Now, since BM=:BD, the shortest path from B
to M is equal to the shortest path from B to D: hence if we sup-
pose two paths from B to A, one passing through M and the other
through D, they will have an equal part in each ; viz. the part
from B to M equal to the part from B to D.
. But by hypothesis, the path through M is the shortest path from
jB to A : hence the shortest path from M to A must be less than
'ftre shortest path from D to A, whereas it is greater since the
arc MA is greater than DA : hence, no point of the shortest
distance between B and A can lie out of the arc of the great
circle BDA.
I *v-
PROPOSITION Illv THEOREM.
^The sum of the three sides of a spherical triangle is less than the
circumference of a great circle.
Let ABC be any spherical trian-
gle ; produce the sides AB, AC, till
they meet again in D. The arcs ABD,
ACD, will be semicircumferences,
since two great circles always bisect
each other (Book VIII. Prop. VII.
Cor. 2.). But in the triangle BCD, we
have the side BC<BD + CD (Prop
I.); add AB+AC to both; we shall
have AB + AC + BC<ABD + ACD,
thatistosay,lessthanacircumference.
PROPOSITION IV. THEOREM
The sum of all the sides of any spherical polygon is less than the
circumference of a great circle.
Take the pentagon ABCDE, for
example. Produce the sides AB, DC,
till they meet in F; then since BC is
less than BF + CF, the perimeter of
the pentagon ABCDE will be less
than that of the quadrilateral AEDF.
Again, produce the sides AE, FD. till ^ A
they meet in G; we shall have ED<EG + DG; hence the pe-
rimeter of the quadrilateral AEDF is less than that of the tri-
angle AFG ; which last is itself less than the circumference of
a great circle ; hence, for a still stronger reason, the perimeter
of the polygon ABCDE is less than this same circumference.
L C Dt
/^
BOOK IX.
189
Scholium. This proposition is fun-
damentally the same as (Book VI.
Prop. XX.) ; for, O being the centre
of the sphere, a sohd angle may be
conceived as formed at O by the plane
angles AOB, BOC, COD,&c., and the
sum of these angles must be less than
four right angles ; which is exactly
the proposition here proved. The A
demonstration here given is different from that of Book VI.
Prop. XX. ; both, however, suppose that the polygon ABCDE
is convex, or that no side produced will cut the figure.
PROPOSITION V. THEOREM.
The poles of a great circle of a sphere, are the extremities of that
diameter of the sphere which is perpendicular to the circle ;
and these extremities are also the poles of all small circles
"parallel to it.
Let ED be perpendic-
ular to the great circle
AMB ; then will E and
D be its poles ; as also
the poles of the parallel
small circles HPI,FNG.
For, DC being per-
pendicular to the plane
AMB, is perpendicular
to all the straight lines
CA, CM, CB,&c. drawn
through its foot in this
plane ; hence all the arcs
DA, DM, DB, &c. are
quarters of the circumfe-
rence. So likewise are
all the arcs EA, EM, EB, &c. ; hence the points D and E are
each equally distant from all the points of the circumference
AMB ; hence, they are the poles of that circumference (Def. 7.).
Again, the radius DC, perpendicular to the plane AMB, is
perpendicular to its parallel FNG ; hence, \\ passes through O
the centre of the circle FNG (Book VIII. Prop. VII. Cor. 4.) ;
hence, if the oblique lines DF, DN, DG, be drawn, these ob-
lique lines will diverge equally from the perpendicular DO,
and will themselves be equal. But, the chords being equal,
190
GEOMETRY.
ihe arcs are equal ; hence the point D is the pole of the small
circle FNG ; and for like reasons, the point E is tlie other pole.
Cor. 1. Every arc DM,
drawn from a point in
the arc of a great circle
AMBto its pole, is a quar-
ter of the circumference,
which for the sake of
brevity, is usually named
a quadrant : and this
quadrant at the same
time makes a right angle
with the arc AM. For,
the line DC being per-
pendicular to the plane
AMC, every plane DME,
passing through the line
DC is perpendicular to
the plane AMC (Book VI. Prop. XVI.); hence, the angle of
these planes, or the angle AMD, is a right angle.
Cor. 2. To find the pole of a given arc AM, draw the indefi-
nite arc MD perpendicular to AM ; take MD equal to a quad-
rant ; the point D will be one of the poles of the arc AM : or
thus, at the two points A and M, draw the arcs AD and MD
perpendicular to AM ; their point of intersection D will be the
pole required.
Cor. 3. Conversely, if the distance of the point D from each
of the points A and M is equal to a quadrant, the point D will
be the pole of the arc AM, and also the angles DAM, AMD,
will be right angles.
For, let C be the centre of the sphere ; and draw the radii
CA, CD, CM. Since the angles ACD, MCD, are right angles,
the line CD is perpendicular to the two straight lines CA, CM ;
hence it is perperpendicular to their plane (Book VI. Prop.
IV,) ; hence the point D is the pole of the arc AM ; and conse-
quently the angles DAM, AMD, are right angles.
Scholium. The properties of these poles enable us to describe
arcs ^of a circle on the surface of a sphere, with the same
facility as on a plane surface. It is evident, for instance, that
by turning the arc DF, or any other line extending to the same
distance, round the point D, the extremity F will describe the
small circle FNG ; and by turning the quadrant DFA round
BOOK IX.
191
the point D, its extremity A will describe the arc of the great
circle AMB.
If the arc AM were required to be produced, and nothing
were given but the points A and M through which it was to
pass, we should first have to determine the pole D, by the
intersection of two arcs described from the points A and M as
centres, with a distance equal to a quadrant ; the pole D being
found, we might describe the arc AM and its prolongation,
from D as a centre, and with the same distance as before.
In fine, if it be required from a given point P, to let fall a
perpendicular on the given arc AM ; find, a point on the arc
AM at a quadrant's distance from the point P, which ig'done by
describing an arc with the point P as a pole, intersecting AM in S :
S will be the point required, and is the pole with which a per-
pendicular to AM may be described passing through the point P.
PROPOSITION VI. THEOREM.
The angle formed hy two arcs of great circles^ is equal to the an-
gle formed by the tangents of these arcs at their point of inter-
section, and is measured by the arc described from this point
of intersection, as a pole, and limited by the sides, produced if
necessary.
O
Let the angle BAG be formed by the two A.
arcs AB, AC ; then will it be equal to the
angle FAG formed by the tangents AF, AG,
and be measured by the arc DE, described
about A as a pole.
For the tangent AF, drawn in the plane
of the arc AB, is perpendicular to the radius
AO ; and the tangent AG, drawn in the plane
of the arc AC, is perpendicular to the same
radius AO. Hence the angle FAG is equal
to the angle contained by the planes ABO,
OAC (Book VI. Def 4.) ; which is that of Hi
the arcs AB, AC, and is called the angle BAG.
In like manner, if the arcs AD and AE are both quadrants,
the lines OD, OE, will be perpendicular to OA, and the angle
DOE will still be equal to the angle of the planes AOD, AOE :
hence the arc DE is the measure of the angle contained by
these planes, or of the angle CAB.
Cor. The angles of spherical triangles may be compared
together, by means of the arcs of great circles described from
their vertices as poles and included between their sides : hence
it is easy to make an angle of this kind equal to a given angle*
192
GEOMETRY.
'Wr^'
Scholium. Vertical angles, such
as ACO and BCN are equal ; for
either of them is still the angle
formed by the two planes ACB,
OCN.
It is farther evident, that, in the
intersection of two arcs ACB, OCN,
the two adjacent angles ACO, OCB,
taken together, are equal to two
right angles.
PROPOSITION VII. THEOREM.
If from the vertices of the three angles of a spherical triangle, as
poles, three arcs he described forming a second triangle, the
vertices of the angles of this second triangle, will he respectively
poles of thei sides' of the first.
From the vertices A, B, C,
as poles, let the arcs EF, FD,
ED, be described, forming on
the surface of the sphere, the
triangle DFE ; then will the
points D, E, and F, be respec-
tively poles of the sides BC,
AC,AB.
For, the point A being the
pole of the arc EF, the dis-
tance AE is a quadrant ; the
point C being the pole of the arc DE, the distance CE is like-
wise a. quadrant : hence the point E is removed the length of a
quadrant from each of the points A and C ; hence, it is the
pole of the arc AC (Prop. V. Cor. 3.). It might be shown, by
the same method, that D is the pole of the arc BC, and F that
of the arc AB.
Cor, Hence the triangle ABC may be descril)ed by means
of DEF, as DEF is described by means of ABC. Triangles
so described are called polar triangles, or supplemental tri-
angles
BOOK IX. 193
PROPOSITION VIII. THEOREM. ^
The same supposition continuing as in the last Proposition^ each
angle in one of the triangles, will be measured by a semicir-
cumference, minus the side lying opposite to it in the otiier
triangle.
For, produce the sides AB,
AC, if necessary, till they meet
EF, in G and H. The point A
being the pole of the arc GH,
the angle A will be measured
by that arc (Prop. VI.). But
the arc EH is a quadrant, and
likewise GF, E being the pole
of AH, and F of AG ; hence
EH + GF is equal to a semi-
circumference. Now, EH+ H
GF is the same as EF+GH ; hence the arc GH, which mea-
sures the angle A, is equal to a semicircumference minus the
side EF. In like manner, the angle B will be measured by
^^circ. — DF : the angle C, by | circ. — DE.
And this property must be reciprocal in the two triangles,
since each of them is described in a similar manner by means
of the other. Thus we shall find the angles D, E, F, of the triangle
DEFtobe measured respectivelyby^ arc. — BC, ^ circ. — AC,
^ circ. — AB. Thus the angle D, for example, is measured by
the arc MI; but MI + BC=MC + BI=a circ; hence the are
MI, the measure of D, is equal to ^ circ. — BC ; and so of all
the rest.
Scholium. It must further be observed,
that besides the triangle DEF, three others
might be formed by the intersection of
the three arcs DE, EF, DF. But the
proposition immediately before us is ap-
plicabJe only to the central triangle,
which is distinguished from the other
three by the circumstance (see the last
figure) that the two angles A and D lie
on the same side of BC, the two B and E on the same side of
AC, and the two C and F on the same side of AB.
R
104 GEOMETRY.
y PROPOSITION IX. THEOREM.
£f around the vertices of the two angles of a given spherical tri-
angle, as poles, the circumferences of two circles be described
ivhich shall pass through the third angle of the triangle; if then,
through the other point in which these circumferences intersect
and the two first angles of the triangle, the arcs of great cir-
cles be drawn, the triangle thus formed will have all its parts
equal to those of the given triangle.
Let ABC be the given triangle, CED,
DFC, the arcs described about A and B
as poles ; then will the triangle ADB have
all its parts equal to those of ABC.
For, by construction, the side AD=r
AC, DB=BC, and AB is common ; hence
these two triangles have their sides equal,
each to each. We are now to show, that
the angles opposite these equal sides are
also equal.
If the centre of the sphere is supposed to be at O, a solid
angle may be conceived as formed at O by the three plane
angles AOB, AOC, BOC ; likewise another solid angle may be
conceived as formed by the three plane angles AOB, AOl),
BOD. And because the sides of the triangle ABC are equal
to those of the triangle ADB, the plane angles forming the one
of these solid angles, must be equal to the plane angles forming
the other, each to each. But in that case we have shown that
the planes, in which the equal angles lie, are equally inclined
to each other (Book VI. Prop. XXI.) ; hence all the angles of
the spherical triangle DAB are respectively equal to those ot
the triangle CAB, namely, DAB--BAC, DBA=ABC, and
ADB = ACB; hence the sides and the angles of the triangle
ADB are equal to the sides and the angles of the triangle AC3.
Scholium. The equality of these triangles is not, however,
an absolute equality, or one of superposition ; for it would be;
impossible to apply them to each other exactly, unless theyj
were isosceles. The equality meant here is what we have!
already named an equality by symmetry ; therefore w'e shalJ
call the triangles ACB, ADB, symmetrical tnangles.
^» BOOK IX. 196
PROPOSITION X. THEOREM.
T\uo triangles on the same sphere, or on equal spheres, are equal
in all their parts, when two sides and the included angle of the
one are equal to two sides and the included angle of the other^
each to each.
Suppose the side AB=EF, the side
AC =EG, and the angle BACmFEG ;
then will the two triangles be equal
in all their parts.
For, the triangle EFG may be
placed on the triangle ABC, or on
ABD symmetrical v/ith ABC, just as
two rectilineal triangles are placed
upon each other, when they have an
equal angle included between equal sides. Hence all the parts
of the triangle EFG will be equal to all the parts of the trian-
gle ABC ; that is, besides the three parts equal by hypothesis,
we shall have the side BC=FG, the angle ABC = EFG, and
the angle ACB^EGF.
PROPOSITION XI. THEOREM.
Two triangles on the same sphere, or on equal spheres, are equal
in all their parts, when two angles and the included side of the
one are equal to two angles and the included side of the other,
each to each.
For, one of these triangles, or the triangle symmetrical with
it, may be placed on the other, as is done in the corres-
ponding case of rectilineal triangles (Book I. Prop. VT.).
PROPOSITION XII. THEOREM.
If two triangles on the same sphere, or on equal spheres, have all
their sides equal, each to each, their angles will likewise he
equal, each to each, the equal angles lying opposite the equal
sides.
lOG
GEOMETRY.
This truth is evident from Prop. IX,
where it was shown, that with three given
sides AB, AC, BC, there can only be two
triangles ACB, ABD, differing as to the
position of their parts, and equal as to the
magnitude of those parts. Hence those
two triangles, having all their sides re-
spectively equal in both, must either be
absolutely equal, or at least symmetrically
so ; in either of which cases, their corres-
ponding angles must be equal, and lie opposite to equal sides.
PROPOSITION XIII. THEOREM.
In every isosceles spherical triangle^ the angles opposite the equal
sides are equal ; and conversely, if two angles of a spherical
triangle are equal, the triangle is isosceles.
First. Suppose the side AB = AC; we shall
have the angle C=B. For, if the arc AD be
drawn from the vertex A to the middle point
D of the base, the two triangles ABD, ACD,
will have all the sides of the one respectively
equal to the corresponding sides of the other,
namely, AD common, BD=DC, and AB=:
AC : hence by the last Proposition, their an-
gles will be equal ; therefore, B = C.
Secondly. Suppose the angle B = C ; we shall have the side
AC=AB. For, if not, let AB be the greater of the two ; take
BO=:AC, and draw OC. The two sides BO, BC, are equal to
the two AC, BC ; the angle OBC, contained by the first two
is equal to ACB contained by the second tv^^o. Hence the
two triangles BOC, ACB, have all their other parts equal
(Prop. X.) ; hence the angle OCB— ABC : but by hypothesis,
the angle ABC=rACB ; hence we have OCB=ACB, which is
absurd ; hence it is absurd to suppose AB different from AC ;
hence the sides AB, AC, opposite to the equal angles B and C,
are equal.
Scholium. The same demonstration proves the angle BAD=^
DAC, and the angle BDA=ADC. Hence the two last are
right angles ; hence the arc drawn from the vertex of an isosceles
spherical triangle to the middle of the base, is at right angles to
that base, and bisects the vertical angle.
BOOK IX. 107
rROlt)SITION XIV. THEOREM.
In any spherical triangle, the greater side is opposite the greater
angle ; and conversely, the greater angle is opposite the greater
side.
Let the angle A be greater
than the angle B, then will BC
be greater than AC ; and con-
versely, if BC is greater than
AC, then will the angle A be
greater than B.
First, Suppose the angle A>B ; make the angle BAD=B ;
then we shall have AD=DB (Prop. XIII.) : but AD + DC is
greater than AC ; hence, putting DB in place of AD, we shall
haveDB + DCorBOAC.
Secondly. If we suppose BC>AC, the angle BAC will be
greater than ABC. For, if BAC were equal to ABC, we
should have BC=AC ; if BAC were less than ABC, we should
then, as has just been shown, find BC<AC. Both these con-
clusions are false : hence the angle BAC is greater than ABC.
PROPOSITION XV. THEOREM.
If two triangles on the same sphere, or on equal spheres, are
mutually equiangular, they will also he mutually equilateral.
Let A and B be the two given triangles ; P and Q their polar
triangles. Since the angles are equal in the triangles A and
B, the sides will be equal in. their polar triangles P and Q
(Prop. VIII.) : but since the triangles P and Q are nnutually
evuilateral, they must also be mutually equiangular (Prop.
XII.) ; and lastly, the angles being equal in the triangles P
and Q, it follows that the sides are equal in their polar trian-
gles A and B. Hence the mutually equiangular triangles A
and B are at the same time mutually equilateral.
Scholium. This proposition is not applicable to rectilineal
triangles ; in which equality among the angles indicates only
proportionality among the sides. Nor is it difficult to account
for the difference observable, in this respect, between spherical
and rectilineal triangles. In the Proposition now before us,
R*
J 98 GEOMETRY.
as well as in the preceding ones, which treat of the comparison
of triangles, it is expressly required that ihe arcs be traced on
the same sphere, or on equal spheres. Now similar arcs are
to each other as their radii ; hence, on equal spheres, two tri-
angles cannot be similar without being equal. Therefore it is
not strange that equality among the angles should produce
equality among the sides.
The case would be different, if the triangles were drawn
upon unequal spheres ; there, the angles being equal, the trian-
gles would be similar, and the homologous sides would be to
each other as the radii of their spheres.
PROPOSITION XVI. THEOREM.
The sum of all the angles in any spherical triangle is less than
six right angles, and greater than two.
For, in the first place, every angle of a spherical triangle is
less than two right angles : hence the sum of all the three is
less than six right angles.
Secondly, the measure of each angle of a spherical triangle
is equal to the semicircumference minus the corresponding side
of the polar triangle (Prop. VIII.) ; hence the sum of all the three,
is measured by the three semicircumferences 7ninusi\\e. sum of all
the sides of the polar triangle. Now this latter sum is less than a
circumference (Prop. III.) ; therefore, taking it away from three"
semicircumferences, the remainder will be greater than one
semicircumference, which is the measure of two right angles ;
hence, in the second place, the sum of all the angles of a sphe-
rical triangle is greater than two right angles.
Cor. 1. The sum of all the angles of a spherical triangle is
not constant, like that of all the angles of a rectilineal triangle ;
it varies between two right angles and six, without ever arriving
at either of these limits. Two given angles therefore do not
serve to determine the third.
Cor. 2. A spherical triangle may have two, or even three of
its angles right angles ; also two, or even threes of its angles
obtuse. i^
BOOK IX. 199
Cor. 3. If the triangle ABC is hi-rectangular,
in ether words, has two right angles B and C,
the vertex A will be the pole of the base BC ;
and the sides AB, AC, will be quadrants
(Prop. V. Cor. 3.).
If the angle A is also a right angle, the tri- ^
angle ABC will be iri-rectangular ; its angles ^
will all be right angles, and its sides quadrants. Two of the
tri-rectangular triangles make half a hemisphere, four make a
hemisphere, and the tri-rectangular triangle is obviously con-
tained eight times in the surface of a sphere.
Scholium. In all the preceding
observations, we have supposed, in
conformity with (Def. 1.) that sphe-
rical triangles have always each of
their sides less than a semicircum-
ference ; from which it follows that
any one of their angles is always
less than two right angles. For, if
the side AB is less than a semicir-
cumference, and AC is so likewise,
both those arcs will require to be E
produced, before they can meet in D. Now the two angles
ABC, CBD, taken together, are equal to two right angles ;
hence the angle ABC itself, is less than two right angles.
We may observe, however, that some spherical triangles do
exist, in which certain of the sides are greater than a semicir-
cumference, and certain of the angles greater than two right
angles. Thus, if the side AC is produced so as to form a whole
circumference ACE, the part which remains, after subtracting
the triangle ABC from the hemisphere, is a new triangle also
designated by ABC, and having AB, BC, AEDC for its sides.
Here, it is plain, the side AEDC is greater than the semicir-
cumferencc AED ; and at the same time, the angle B opposite
to it exceeds two right angles, by the quantity CBD.
The trianjrles whose sides and angles are so large, have been
excluded by the Definition ; but the only reason was, that the
solution of them, or the determmation of their parts, is always
reducible to the solution of such triangles as are comprehended
by the Definition. Indeed, it is evident enough, that if the sides
and angles of the triangle ABC are known, it will be easy to
discover the angles and sides of the triangle which bears the
same name, and is the difference between a hemisphere and the
former triangle.
I
200 GEOMETRY.
PROPOSITION XVII. THEOREM.
'llie surface of a lune is to the surface of the sphere^ as the angle
of this lune, is to four right angles, or as the arc which mea-
sures that angle, is to the circumference.
Let AMBN be a lune ; then will its
surface be to the surface of the sphere
as the angle NCM to four right angles,
or as the arc NM to the circumference
of a great circle.
Suppose, in the first place, the arc
MN to be to the circumference MNPQ
as some one rational number is to ano-
ther, as 5 to 48, for example. The cir-
cumference MNPQ being divided into
48 equal parts, MN will contain 5 of them ; and if the pole A
were joined with the several points of division, by as many-
quadrants, we should in the hemisphere AMNPQ have 48 tri-
angles, all equal, because all their parts are equal. Hence the
whole sphere must contain 96 of those partial triangles, the lune
AMBNA will contain 10 of them ; hence the lune is to the
sphere as 10 is to 96, or as 5 to 48, in other words, as the arc
MN is to the circumference.
If the arc MN is not commensurable with the circumference,
we may still show, by a mode of reasoning frequently exem-
plified already, that in that case also, the lune is to the sphere
as MN is to the circumference.
Cor, 1. Two lunes are to each other as their respective
angles.
Cor. 2. It was shown above, that the whole surface of the
sphere is equal to eight tri-rectangular triangles (Prop. XVI.
Cor. 3.) ; hence, if the area of one such triangle is represented
by T, the surface of the whole sphere will be expressed by 8T.
This granted, if the right angle be assumed equal to l,the sur-
face of the lune whose angle is A, will be expressed by 2AxT:
for,
4: A: : 8T : 2AxT
in which expression, A represents such a part of unity, as the
angle of the lune is of one right angle.
Scholium. The spherical ungula, bounded by the planes
AMB, ANB, IS to the whole solid sphere, as the angle A is to
BOOK IX. 201
four right angles. For, the lunes being equal, the spherical
ungulas will also be equal ; hence two spherical ungulas are to
each other, as the angles formed by the planes which bound
them.
PROPOSITION XVIII. THEOREM.
Two symmetrical spherical triangles are equivalent.
Let ABC, DEF, be two symmetri-
cal triangles, that is to say, two tri-
angles having their sides AB=DE,
AC=DF, CB=EF, and yet incapa-
ble of coinciding with each other : / I \ q p /
we are to show that the surface ABC
is equal to the surface DEF.
Let P be the pole of the small
circle passing through the three points
A, B, C ;* from this point draw the
equal arcs PA, PB, PC (Prop. V.) ; at the point F, make the
angle DFQzrACP, the arc FQ=CP ; and draw DQ, EQ.
The sides DF, FQ, are equal to the sides AC, CP ; the an-
gle DFQ=ACP : hence the two triangles DFQ, ACP are equal
in all their parts (Prop. X.) ; hence the side DQ=AP, and the
angle DQF=APC.
In the proposed triangles DFE, ABC, the angles DFE, ACB,
opposite to the equal sides DE, AB, being equal (Prop. XII.).
if the angles DFQ? ACP, which are equal by construction, be
taken ^way from them, there will remain the angle QFE, equal
to PCB. Also the sides QF, FE, are equal to the sides PC,
CB ; hence the two triangles FQE, CPB, are equal in all their
parts ; hence the side QE^PB, and the angle FQE = CPB.
Now, the triangles DFQ, ACP, which have their sides re-
spectively equal, are at the same time isosceles, and capable of
coinciding, when applied to each other; for having placed AC
on its equal DF, the equal sides will fall on each other, and
thus the two triangles will exactly coincide : hence they are
equal ; and the surface DQF— APC. For a like reason, the
surface FQE=CPB, and the surface DQE=APB ; hence we
» The circle which passes through the three points A, B, C, or which cir-
cumscribes the triangle ABC, can only be a small circle of the sphere ; for if
it were a great circle, the three sides AB, BC, AC, would lie in one plane, and
the triangle ABC would be reduced to one of its sides.
202
GEOMETRY.
have DQF+FQE— DQE=APC + CPB— APB, or DFE=:
ABC ; hence the two symmetrical triangles ABC, DEF are
equal in surface.
Scholium, The poles P and Q
might lie within triangles ABC,
DEF: in which case it would be
requisite to add the three triangles
:)0 P.^
DQF, FQE, DQE, together, m or-
der to make up the triangle DEF ;
and in like manner, to ?dd the three
triangles APC, CPB, APB, together,
in order to make up the triangle
ABC : in all other respects, the de-
monstration and the result would still be the same.
PROPOSITION XIX. THEOREM.
If the circumferences of two great circles intersect each other on
the surface of a hemisphere, the sum of the opposite triangle^y
thus fo7^med, is equivalent to the surface of a lune whose angle
is equal to the angle formed hy the circles.
Let the circumferences AOB, COD,
intersect on the hemisphere OACBD ;
then will the opposite triangles AOC,
BOD, be equal to the lune whose an-
gle is BOD.
For, producing the arcs OB, OD, on
the other hemisphere, till they meet in
N, the arc OBN will be a semi-circum-
ference, and AOB one also ; and taking
OB from each, we shall have BN= AO.
For a like reason, we have DN=CO, and BD=AC. He-nce,
the two triangles AOC, BDN, have their three sides respect-
ively equal ; they are therefore symmetrical ; hence they are
equal in surface (Prop. XVIII.) : but the sum of the triangles
BDN, BOD, is equivalent to the lune OBNDO, 'whose angle is
BOD: hence, AOC + BOD is equivalent to the lune whose
angle is BOD.
Scholium. It is likewise evident that the two spherical pyra-
mids, which have the triangles AOC, BOD, for bases, are toge-
ther equivalent to the spherical ungula whose angle is BOD.
BOOK IX. 203
PROPOSITION XX. THEOREM.
The surface of a spherical triangle is measured by the excess of
the sum of its three angles above two right angles, multiplied
by the tri-rectangular triangle.
Let ABC be the proposed triangle : pro-
duce its sides till they meet the great circle
DEFG drawn at pleasure without the trian-
gle. By the last Theorem, the two triangles
ADE, AGH, are together equivalent to the
lune whose angle is A, and which is mea-
sured by 2A.T (Prop. XVII. Cor. 2.).
Hence we have ADE + AGH=2A.T ; and
for a like reason, BGF+BID = 2B.T, and
CIH + CFE=2C.T But the sum of these
six triangles exceeds the hemisphere by twice the triangle
ABC, and the hemisphere is represented by 4T ; therefore,
twice the triangle ABC is equal to 2A.T + 2B.T + 2C.T— 4 T;
and consequently, once ABC = (A + B-j-C — 2)T; hence every
spherical triangle is measured by the sum of all its angles minus
two right angles, multiplied by the tri-rectangular triangle.
Coj\ 1. However many right angles there may be in the sum of
the three angles minus two right angles,just so many tri-rectan-
gular triangles, or eighths of the sphere, will the proposed trian-
gle contain. If the angles, for example, are each equal to f of
a right angle, the three angles will amount to 4 right angles, and
the sum of the angles minus two right angles will be represented
by 4 — 2 or 2; therefore the surface of the triangle will be equal
to two tri-rectangular triangles, or to the fourth part of the
whole surface of the sphere.
Scholium, While the spherical triangle ABC is compared
with the tri-rectangular triangle, the spherical pyramid, which
has ABC for its base, is compared with the tri-rectangular py-
ramid, and a similar proportion is found to subsist between
them. The solid angle at the vertex of the pyramid, is in like
manner compared with the solid angle at the vertex of the tri-
rectangular pyramid. These comparisons are founded on the
coincidence of the corresponding parts. If the bases of tho
JiOl GEOMETRY.
pyramids coincide, the pyramids themselves will evidently co-
incide, and likewise the solid angles at their vertices. From
tliis, some consequences are deduced.
First. Two triangular spherical pyramids are to each other
as their bases : and since a polygonal pyramid may always be
divided into a certain number of triangular ones, it follows that
any two spherical pyramids are to each other, as the polygons
which form their bases.
Second. The solid angles at the vertices of these pyramids, are
also as their bases ; hence, for comparing any two solid angles,
we have merely to place their vertices at the centres of two
equal spheres, and the solid angles will be to each other as the
spherical polygons intercepted between their planes or faces.
The^ vertical angle of the tri-rectangular pyramid is formed
by three planes at right angles to each other : this angle, which
may be called a right solid aiigle, will serve as a very natural
unit of measure for all other solid angles. If, for example, the
the area of the triangle is f of the tri-rectangular triangle,
then the corresponding solid angle will also be f of the
right solid an^le.
PROPOSITION XXI. THEOREM
The surface of a spherical polygon is measured by the sum of all
its angles,m\n\is two right angles multiplied by the number of
sides in the polygon less two, into the tri-rectangular triangle.
From one of the vertices A, let diago-
nals AC, AD be drawn to all the other ver-
tices ; the polygon ABCDE will be di-
vided into as many triangles minus two as
it has, sides. But the surface of each tri-
angle is measured by the sum of all its an-
gles minus two right angles, into the tri-
rectangular triangle ; and the sum of the angles in all the tri-
angles is evidently the same as that of all the angles of the
polygon ; hence, the surface of the polygon is equal to the sum
of all its angles, diminished by twice as many pght angles as
it has sides less two, into the tri-rectangular triangle.
Scholium. Let s be the sum of all the angles in a spherical
polygon, n the number of its sides, and T the tri-rectangular tri-
angle ; the right angle being taken for unity, the surface of the
polygon will be measured by
(s-^2 (n— 2,)) T, or (s— 2 w^4) T
APPENDIX,
THE REGULAR POLYEDRONS.
A regular polyedron is one whose faces are all equal regular
polygons, and whose solid angles are all equal to each other.
There are five such polyedrons.
First. If the faces are equilateral triangles, polyedrons may
be formed of them, having solid angles contained by three of
those triangles, by four, or by five : hence arise three regular
bodies, the tetraedron, the octaedron, the icosaedron. No other
can be formed with equilateral triangles ; for six angles of such
a triangle are equal to four right angles, and cannot form a
solid angle (Book VI. Prop. XX.).
Secondly. If the faces are squares, their angles may be ar-
ranged by threes : hence results the hexaedron or cube. Four
angles of a square are equal to four right angles, and cannot
form a solid angle.
Thirdly. In fine, if the faces are regular pentagons, their
angles likewise may be arranged by threes : the regular dode-
caedron will result.
We can proceed no farther : three angles of a regular hexa-
gon are equal to four right angles ; three of a heptagon are
greater.
Hence there can only be five regular polyedrons ; three formed
with equilateral triangles, one with squares, and one with pen-
tagons.
Construction of the Tetraedron,
Let ABC be the equilateral triangle
which is to form one face of the tetrae-
dron. At the point O, the centre of this
triangle, erect OS perpendicular to the
plane ABC ; terminate this perpendicular
in S, so that AS=AB; draw SB, SC :
the pyramid S-ABC will be the tetrae-
dron required.
For, by reason of the equal distances
OA, OB, OC, the oblique lines SA, SB, SC, are equally re-
S
206
APPENDIX.
moved from the perpendicular SO, and
consequently equal (Book VI. Prop. V.).
One of them SA=AB ; hence the four
faces of the pyramid S-ABC, are trian-
gles, equal to the given triangle ABC.
And the solid angles of this pyramid
are all equal, because each of them is
formed by three equal plane angles:
hence this pyramid is a regular tetrae-
dron.
Construction of the Hexaedron.
Let ABCD be a given square. On the
base ABCD, construct a right prism whose
altitude AE shall be equal to the side AB.
The faces of this prism will evidently be
equal squares ; and its solid angles all equal,
each being formed with three right angles : ■
hence this prism is a regular hexaedron or
cube.
E
C\
T?
13
The following propositions can be easily proved.
1." Any regular polyedron may be divided into as many
regular pyramids as the polyedron has faces ; the common
vertex of these pyramids will be the centre of the polyedron ;
and at the same time, that of the inscribed and of the circum-
scribed sphere.
2. The solidity of a regular polyedron is equal to its sur-
face multiplied by a third part of the radius of the inscribed
sphere.
3. Two regular polyedrons of the same name, are two simi-
lar sojids, and their homologous dimensions are proportional ;
hence the radii of the inscribed or the circumscribed spheres
are to each other as the sides of the polyedrons.
4. If a regular polyedron is inscribed in a sphere, the planes
drawn from the centre, through the different edges, will divide
the surface of the sphere into as many spherical polygons, all
equal and similar, as the polyedron has faces.
APPLICATION OF ALGEBRA.
TO THE SOLUTION OF
GEOMETRICAL PROBLEMS.
A problem is a question which requires a solution. A geo-
metrical problem is one, in which certain parts of a geometri-
cal figure are given or known, from which it is required to de-
termine certain other parts.
When it is proposed to solve a geometrical problem by
means of Algebra, the given parts are represented by the first
letters of the alphabet, and the required parts by the final let-
ters, and the relations which subsist between the known and
unknown parts furnish the equations of the problem. The solu-
tion of these equations, when so formed, gives the solution of
the problem.
No general rule can be given for forming the equations. The
equations must be independent of each other, and their number
equal to that of the unknown quantities introduced (Alg.
Art. 103.). Experience, and a careful examination of all the
conditions, whether explicit or implicit (Alg. Art. 94,) will
serve as guides in stating the questions ; to which may be
added the following particular directions.
1st. Draw a figure which shall represent all the given parts,
and all the required parts. Then draw such other lines as will
establish the most simple relations between them. If an angle
is given, it is generally best to let fall a perpendicular that shall
lie opposite to it; and this perpendicular, if possible, should be
drawn from the extremity of a given side.
2d. When two lines or quantities are connected in the same
way with other parts of the figure or problem, it is in general,
not best to use either of them separately; but to use their sum,
their difference, their product, their quotient, or perhaps ano-
ther line of the figure with which they are alike connected.
3d. When the area, or perimeter of a figure, is given, it is
sometimes best to assume another figure similar to the propo-
sed, having one of its sides equal to unity, or some other known
quantity. A comparison of the two figures will often give a re-
quired part. We will add the following problems.*
* The following problems are selected from Hutton's Application of Algebra
to Geometry, and the examples in Mensuration from his treatise on that subject.
208 APPLICATION OF ALGEBRA
PROBLEM I.
In a right angled triangle BAG, having given the base BA,
and the sum of the hypothenuse and perpendicular, it is re-
quired to find the hypothenuse and perpendicular.
Put BA=c=3, BC=a:;, KC=y and the sum of the hypo-
thenuse and perpendicular equal to s = 9
Then, x-^y=s=Q.
and x'=y''+c^ (Bk . IV. Prop. XL)
From 1st equ: x=s — y
and x^= s^ — 2sy -\-y^ p
By subtracting, = s^ — 2sy — c^
or 2sy=s'^ — c^
s'^—c^
hence, y=-^ =4=Ae
Therefore a; + 4 =9 or a:=5=BC.
PROBLEM II.
In a Hght angled triangle, having given the hypothenuse, and ike
sum of the base and perpendicular, to find these two sides'
Put BC=a=5, BA=a:, AC=y and the sum
of the base and perpendicular =5=7
Then x-\-y=s=l
and a;2+3/^=ar^
From first equation x=s — y
or x^=s'^—2sy-{-y'^
Hence, y^==a^-.s'2+2sy^y^
or 2?/^ — 2sy=a^—s^
or 'ir—sy-.
2
By completing the square y^ — 5?/ + i5^=|a^ — \s^
or y =is± yia^— {5-=4 or 3
Hence a:= J5=F Vid-—\s'^='^ or 4
TO GEOMETRY.
209
PROBLEM III.
In a rectangle^ having given the diagonal and perimeter, to find
the sidea.
Let ABCD be the proposed rectangle.
Put AC=^=10, the perimeter = 2a :^ 28, or
AB + BC=a=14: also put AB=a;and BC=y.
Then, x^+y'=d^
and x+y=a
From which equations we obtain,
=ia=fc Vld^—ia^=8 or 6,
and
a;=iflr=p V^d^—ia^=6 or 8.
PROBLEM IV.
Having given the base and perpendicular of a triangle^ to find
the side of an inscribed square.
Let ABC be the triangle and HEFG
the inscribed square. Put AB=b, CD=a,
and HE or GHrra; : then Cl=a — x.
We have by similar triangles
AB: CD:; GF: CI
or b: a:: x: a — x
Hence, ab — bx=ax
ab
or
a + b
the side of the inscribed square ;
which, therefore, depends only on the base and altitude of the
triangle.
PROBLEM V.
In an equilateral triangle, having given the lengths of the
three perpendiculars drawn from a point within, on the three
sides: to determine the sides of the triangle.
210 APPLICATION OF ALGEBRA
Let ABC be the equilateral triangle ;
DG, DE and DF the given perpendicu-
lars hi fall from D on the sides. Draw
DA, DB, DC to the vertices of the angles,
and let fall the perpendicular CH on
the base. Let DG=a, J)E=b, and
DF=c : put one of the equal sides AB
=2x; hence AH=x, and CH=
Now since the area of a triangle is equal to half its base
into the altitude, (Bk. IV. Prop. VL)
iAB X CH=a; x x Vs^x^ -/s^triangle ACB
iAB X DG=.r X a =ax = triangle ADB
iBCxJ)E=xxb =bx =triangle BCD
^ACxDF=a;xc =cx =triangle ACD
But the three last triangles make up, and are consequently
equal to, the first ; hence,
x^ V3=ax + bx + cx^=x{a + 6 + c) ;
or xV3=a-\-b + c
a-\-b + c
therefore",
V3
Remark. Since the perpendicular CH is equal to xy 3, it ^
is consequently equal to « + 6 + c; that is, the perpendicular let I
fall from either angle of an equilateral triangle on the oppo- •
site side, is equal to the sum of the three perpendiculars let "
fall from any point within the triangle on the sides respectively. ^
PROBLEM VI.
In a right angled triangle, having given the base and the dif-
ference between the hypothenuse and perpendicular, to find
the sides.
PROBLEM VII.
In a right angled triangle, having given the hypothenuse and
the difference between the base and perpendicular, to deter-
mine the triangle.
►
TO GEOMETRY. 211
PROBLEM VIII.
Having given the area of a rectangle inscribed in a given
triangle ; to determine the sides of the rectangle.
PROBLEM IX.
In a triangle, having given the ratio of the two sides, togeth-
er with both the segments of the base made by a perpendic-
ular from the vertical angle ; to determine the triangle.
PROBLEM X.
In a triangle, having given the base, the sum of the other two
sides, and the length of a line drawn from the vertical angle
to the middle of the base ; to find the sides of the triangle.
PROBLEM XI.
In a triangle, having given the two sid6s about the vertical
angle, together with the line bisecting that angle and terminating
in the base ; to find the base.
PROBLEM XII.
To determine a right angled triangle, having given the
lengths of two lines drawn trom the acute angles to the mid^
die of the opposite sides.
PROBLEM Xni.
I
To determine a right-angled triangle, having given the pe-
rimeter and the radius of the inscribed circle.
PROBLEM XIV.
To determine a triangle, having given the base, the per-
pendicular and the ratio of the two sides.
PROBLEM XV.
To determine a right angled triangle, having given the
hypothenuse, and the side of the inscribed square.
PROBLEM XVI.
To determine the radii of three equal circles, described
within and tangent to, a given circle, and also tangent to
each other.
212 APPLICATION OF ALGEBRA
PROBLEM XVII
In a right angled triangle, having given the perimeter and
the perpendicular let fall from the right angle on the hypothe-
nuse, to determine the triangle.
PROBLEM XVIII.
To determine a right angled triangle, having given the
hypothenuse and the difference of two lines drawn from the
two acute angles to the centre of the inscribed circle.
PROBLEM XIX.
To determine a triangle, having given the base, the perpen-
dicular, and the difference of the two other sides.
PROBLEM XX.
To determine a triangle, having given the base, the perpen-
dicular and the rectangle of the two sides.
PROBLEM XXI.
To determine a triangle, having given the lengths of three
lines drawn from the three angles to the middle of the opposite
sides.
PROBLEM XXII.
In a triangle, having given the three sides, to find the radius
of the inscribed circle.
PROBLEM XXIII.
To determine a right angled triangle, having given the side
of the inscribed square, and the radius of the inscribed circle.
PROBLEM XXIV.
To determine a right angled triangle, having given the
hypothenuse and radius of the inscribed circle.
PROBLEM XXV.
To determine a triangle, having given the base, the line
bisecting the vertical angle, and the diameter of the circum-
scribing circle.
PLANE TRIGONOMETRY. 213
PLANE TRIGONOMETRY.
In every triangle there are six parts : three sides and three
angles. These parts are so related to each other, that if a
certain number of them be known or given, the remaining
ones can be determined.
Plane Trigonometry explains the methods of finding, by cal-
culation, the unknown parts of a rectilineal triangle, when
a sufficient number of the six parts are given.
When three of the six parts are known, and one of them is a
side, the remaining parts can always be found. If the three
angles were given, it is obvious that the problem would be in-
determinate, since all similar triangles would satisfy the con-
ditions.
It has already been shown, in the problems annexed to Book
III., how rectilineal triangles are constructed by means of three
given parts. But these constructions, which are called graphic
methods, though perfectly correct in theory, would give only
a moderate approximation in practice, on account of the im-
perfection of the instruments required in constructing them.
Trigonometrical methods, on the contrary, being independent
of all mechanical operations, give solutions with the utmost
accuracy.
These methods are founded upon the properties of lines called
trigonometrical lines, which furnish a very simple mode of ex-
pressing the relations between the sides and angles of triangles.
We shall first explain the properties of those lines, and the
principal forraulas derived from them ; formulas which are of
great use in all the branches of mathematics, and which even
furnish means of improvement to algebraical analysis. We
shall next apply those results to the solution of rectilineal tri-
angles.
DIVISION OF THE CIRCUMFERENCE.
I. For the purposes of trigonometrical calculation, the cir-
cumference of the circle is divided into 360 equal parts, called
degrees ; each degree into 60 equal parts, called minutes ; and
each minute into 60 equal parts, called seconds.
The semicircumference, or the measure of two right angles,
contains 180 degrees ; the quarter of the circumference, usually
denominated the quadrant, and which measiu'es the right an-
gle, contains 90 degrees.
II. Degrees, minutes, and seconds, are respectively desig-
214
PLANE TRIGONOMETRY.
nated by the characters ; «, ', " : thus the expression 16° 6' 15"
represents an arc, or an angle, of 16 degrees, 6 minutes, and
15 seconds.
III. The complememt of an angle, or of an arc, is what re-
mains after taking that angle or that arc from 90°. Thus the
complement of 25° 40' is equal to 90°— 25° 40' r= 64° 20' ; and
the complement of 12° 4' 32" is equal to 90°— 12° 4' 32" = 77°
55' 28".
In general, A being any angle or any arc, 90° — A is the com^
plement of that angle or arc. If any arc or angle be added
to its complement, the sum will be 90°. Whence it is evident
that if the angle or arc is greater than 90°, its complement will
be negative. Thus, the complement of 160° 34' 10" is — 70°
34' 10". In this case, the complement, taken positively, would
be 9, quantity, which being subtracted from the given angle or
arc, the remainder would be equal to 90°.
The two acute angles of a right-angled triangle, are together
equal to a right angle ; they are, therefore, complements of each
other.
IV. The supplement of an angle, or of an arc, is what re-
mains after taking that angle or arc from 180°. Thus A being
any angle or arc, 180° — A is its supplement.
In any triangle, either angle is the supplement of the sum of
the two others, since the three together make 180°.
If any arc or angle be added to its supplement, the sum will
be 180°. Hence if an arc or angle be greater thaii 180°, its
supplement will be negative. Thus, the supplement of 200°
is — 20°. The supplement of any angle of a triangle, or indeed
of the sum of either two angles, is always positive,
GENERAL IDEAS RELATING TO TRIGONOMETRICAL LINES.
V. The sine of an arc is
the perpendicular let fall from
one extremity of the arc, on
the diameter which passes
through the other extremity.
Thus, MP is the^ine of the
arc AM, or of the angle ACM.
The tangent of an arc is a
line touching the arc at one
extremity/ and limited by the
prolongation of the diameter
which passes through the
other extremity.'. Thus AT is
the tangent of tne arc AM,
or of the angle ACM. -
PLANE TRIGONOMETRY. 215
The secant of an arc is the line drawn from the centre of
the circle through one extremity of the arc land limited by the
tangent drawn through the other extremity. Thus CT is the
secant of the arc AM, or of the angle ACM.
The versed sine of an arc, is the part of the diameter inter-
cepted between one extremity of the arc and the foot of the
sine. Thus, AP is the versed sine of the arc AM, or the angle
ACM.
These four lines MP, AT, CT, AP, are dependent upon the
arc AM, and are always determined by it and the radius ; they
are thus designated :
MP=sin AM, or sin ACM,
ATintangAM, or tang ACM,
CTz^secAM, or sec ACM,
APr=ver-sin AM, or ver-sin ACM.
VI. Having taken the arc AD equal to a quadrant, from the
points M and D draw the lines MQ, DS, perpendicular to the
radius CD, the one terminated by that radius, the other termi-
nated by the radius CM produced ; the lines MQ, DS, and CS,
will, in like manner, be the sine, tangent, and secant of the arc
MD, the complement of AM. For the sake of brevity, they
are called the cosine, cotangent, and cosecant, of the arc AM,
and are thus designated :
MQ=cosAM, or cos ACM,
DS=cot AM, or cot ACM,
I CS=cosec AM, or cosec ACM.
In general, A being any arc or angle, we have
cos A=sin (90°— A),
cot A = tang (90°— A),
cosec A = sec (90° — A).
The triangle MQC is, by construction, equal to the triangle
CPM ; consequently CPrrrMQ : hence in the right-angled tri-
angle CMP, whose hypothenuse is equal to the radius, the two
sides MP, CP are the sine and cosine of the arc AM : hence,
the cosine of an arc is equal to that part of the radius inter-
cepted between the centre and foot of the sine.
The triangles CAT, CDS, are similar to the equal triangles
CPM, CQM ; hence they are similar to each other. From
these principles, we shall very soon deduce the different rela-
tions which exist between the lines now defined : before doing
so, however, we must examine the changes which those lines
undergo, when the arc to which they relate increases from zero
to 180«.
The angle ACD is called the first quadrant ; the angle DCB,
the second quadrant ; the angle BCE, the third quadrant ; and
the angle EC A, the fourth quadrant.
210
PLANE TRIGONOMETRY.
B
D
N
^
Q^^
V
\
P^
/
p' \
y
I
\
k
T^'
R J
s
E
VII. Suppose one extrem-
ify of the arc remains fixed in
A, wiiile the other extremit}%
marked M, runs successively
throughout the whole extent
of the semicircumference,
from A to B in the direction
ADB.
When the point M is at A,
or when the arc AM is zero,
the three points T, M, P, are
confounded with the point A ;
whence it appears that the
sine and tangent of an arc
zero, are zero, and the cosine and secant of this same arc, are
each equal to the radius. Hence if R represents the radius ol
the circle, we have
V sin 0=0, tang — 0, cos 0=R, secO=R.
VIII. As the point M advances towards D, the sine increases,
and so likewise does the tangent and the secant; but the cosine,
the cotangent, and the cosecant, diminish.
When the point M is at the middle of AD, or when the arc
AM is 45°, in which case it is equal to its complement MD,
the sine MP is equal to the cosine MQ or CP ; and the trian-
gle CMP, having become isosceles, gives the proportion
MP : CM : : 1 : x/2,
or sin 45° : R : : 1 : V2.
Hence
sin 45°=cos45o=-7-=:iR\/2
V 2
In this same case, the triangle CAT becomes isosceles and
equal to the triangle CDS ; whence the tangent of 45° and its
cotangent, are each equal to the radius, and consequently we
have
tang 45° = cut ..15° =R.
IX. The arc AM continuing to increase, the sine increases
till M arrives at D ; at which point the si^ie is equal to the ra-
dius, and the cosine is^zero. Hence we have ,
sin90°=:R, cos 90° = 0;
and it may be observed, that these values are a consequence
of the values already found for the sine and cosine of the aiv
zero ; because the complement of 90« being zero, we have
sin yO''— cos 0°=R, and
cos 90°=rsin 0°=0.
PLANE TRIGONOMETRY. 211
As to the tangent, it increases very rapidly as the point M
approaches D ; and finally when this point reaches D, the tan-
gent properly exists no longer, because the lines AT, CD,
being parallel, cannot meet. This is expressed by saying that
the tangent of 90° is infinite ; and we write tang 90" r: ao
The complement of 90" being zero, we have
tang O=cot 90" and cot Orztang 90°.
Hence cot 90° =0, and cot 0=ao .
X. The point M continuing to advance from D towards B,
the sines diminish and the cosines increase. Thus MT' is the
sine of the arc AM', and M'Q, or CP' its cosine. But the arc
M'B is the supplement of AM', since AM' + M'B is equal to a
semicircumference ; besides, if M'M' is drawn parallel to AB,
the arcs AM, BM', which are included between parallels, will
evidently be equal, and likewise the perpendiculars or sines
MP, M'P'. Hence/i/ie sine of an arc or of an angle is equal to
the sine of the siipplement of that arc or angle^
The arc or angle A has for its supplement 180" — A: hence
generally, we have
sin A = sin (180"— A.)
The same property might also be expressed by the equation
sin (90^ + B) = sin (90°— B),
B being the arc DM or its equal DM'.
XI. The same arcs AM, AM', which are supplements of
each other, and which have equal sines, have also equal co-
sines CP, CP' ; but it must be observed, that these cosines lie
in different directions. The line CP which is the cosine of the
arc AM, has the origin of its value at the centre C, and is esti-
mated in the direction from C towards A ; while CP', the cosine
of AM' has also the origin of its value at C, but is estimated in
a contrary direction, from C towards B.
Some notation must obviously be adopted to distinguish the
one of such equal lines from the other ; and that they may both
be expressed analytically, and in the same general formula, it is
necessary to consider all lines which are estimated in one di-
rection as j)ositiv)% and those which are estimated in the con-
trary direction as negative. If, therefore, the cosines which
are estimated from C towards A be considered as positive,
those estimated from C towards B, must be regarded as nega-
tive. Hence, generally, we shall have,
cos A= — cos (180° — A)
that is,frAe cosine of an arc or angle is equal to the cosine of its
supplement taken negatively^
The necessity of changing the algebraic sign to correspond
T
218
PLANE TRIGONOMETRY
with the change of direction
in the trigonometrical line,
may be illustrated by the fol-
lowing example. The versed
sine AP is equal to the radius
CA minus CP the cosine AM :
that is,
ver-sin AM.=:R— cos AM.
Now when the arc AM be-
comes AM' the versed sine
AP, becomes AF, that is equal
to R + CP'. But this expression
cannot be derived from the
formula,
ver-sin AM=r:R — cos AMj
unless we suppose the cosine AM to become negative as soon
as the arc AM becomes greater than a quadrant.
At the point B the cosine becomes equal to — R ; that is,
cos 180^=— R.
For all arcs, such as ADBN', which terminate in the third
quadrant, the cosine is estimated from C towards B, and is
consequently negative. At E the cosine becomes zero, and for
all arcs which terminate in the fourth quadrant the cosines are
estimated from C towards A, and are consequently positive.
The sines of all the arcs which terminate in the first and
second quadrants, are estimated above the diameter BA, w^hile
the sines of those arcs which terminate in the third and fourth
quadrants are estimated below it. Hence, considering the
former as positive, we must regard the latter as negative.
XII. Let us now see what sign is to be given to the tangent
of an arc. The tangent of the arc AM falls above the line BA,
and w^e have already regarded the lines estimated in the direc-
tion At as positive : therefore the tangents of all arcs which
terminate in the first quadrant will be positive. But the tan-
gent of the arc AM', greater than 90^, is determined by the
intersection of the two lines M'C and AT. These lines, how-
ever, do not meet in the direction AT ; but they meet in the
opposite direction AV. But since the tangents estimated in the
direction AT are positive, those estimated in thfe direction AV
must be negative : therefore, the tangents of all arcs which ter-
minate in the second quadrant will be negative.
When the point M' reaches the point B the tangent AV will
become equal to zero : that is,
tang 1 80° = 0.
When the point M' passes the point B, and comes into the
position N', the tangent of the arc ADN' will be the line AT :
PLAJSE TRIGONOMETRY. 219
hence,^ the tangents of all arcs which terminate in the third quad-
rant are positive.
At E the tangent becomes infinite : that is,
tang270° = Q0.
When the point has passed along into the fourth quadrant
to N, the tangent of the arc ADN'N will be tl^e line AV : hence,
the tangents of all arcs which terminate in the fourth quadrant
are negative.
The cotangents are estimated from the line ED. Those which
lie on the side DS are regarded as positive, and those which lie
on the side DS' as negative. /"Hence, the cotangents are posi-
tive in the first quadrant, negative in the second, positive in the
third, and negative in the fourth.\ When the point M is at B
the cotangent is infinite ; when m E it is zero : hence,
cot 180°=— 00 ; cot 270° = 0.
Let q stand for a quadrant ; then the following table will show
the signs of the trigonometrical lines in the different quadrants.
Iq 2q Sq 4q
Sine + + •— —
Cosine + — — + .
Tangent + — + —
Cotangent 4- — -{- —
XIII. In trigonometry, the sines, cosines, iSz^c. of arcs or an-
gles greater than 180° do not require to be considered ; the
angles of triangles, rectilineal as well as spherical, and the
sides of the latter, being always comprehended between and
180°. But in various applications of trigonometry, there is fre-
quently occasion to reason about arcs greater than the semi-
circumference, and even about arcs containing several circum-
ferences. It will therefore be necessary to find the expression
of the sines and cosines of those arcs whatever be their
magnitude.
We generally consider the arcs as positive which are esti-
mated from A in the direction ADB, and then those arcs must
be regarded as negative which are estimated in the contrary
dii-ection AEB.
We observe, in the first place, that two equal arcs AM, AN
with contrary algebraic signs, have equal sines MP, PN, with
contrary algebraic signs ; while the cosine CP is the same for
both.
The equal tangents AT, AV, as well as the equal cotangents
DS, DS', have also contrary algebraic signs. Hence, calling
X the arc, we have in general,
sin {^x)= — sin x
' i cos ( — x) = cos X \
\ tang ( — x) = — tang x \
cot ( — x)= — cot a;
220
PLANE TRIGONOMETRY.
By considering the arc AM, and its supplement AM', and
recollecting what has been said, we readily see that,
sin (an arc) = sin (its supplement)
cos (an arc) = — cos (its supplement)
tang (an arc) :=— tang (its supplement)
cot (an arc) = — cot (its supplement).
It is no less evident, that e' D ^
if one or several circumfe-
rences were added to any
arc AM, it would still termi-
nate exactly at the point M,
and the arc thus increased
would have the same sine as
the arc AM ; hence if C rep-
resent a w^hole circumfe-
rence or 360°, we shall have
sin X = sin (C + x)= sin x = sin
(2C + x), &c.
The same observation is ap-
plicable to the cosine, tan-
gent, &c.
Hence it appears, that whatever be the magnitude of x the
proposed arc, its sine may always be expressed, with a proper
sign, by the sine of an arc less than 180°. For, in the first
place, we may subtract 360° from the arc x as often as they
are contained in it ; and y being the remainder, we shall have
sin a;=sin y. Then if?/ is greater than 180°, make y— 180°-f-2,
and we have sin y= — sin z. Thus all the cases are reduced
to that in which the proposed arc is less than 180° ; and since
we farther have sin (90° + a;) = sin (90° — x), they are likewise
ultimately reducible to the case, in which the proposed arc is
between zero and 90°.
XIV. The cosines are always reducible to sines, by means
of the formula/ cos A = sin (90° — A)J or if we require it, by
means of the fArmula cos A = sin (90° -f A) : and thus, if we can
find the value of the sines in all possible cases, we can also find
that of the cosines. Besides, as has already been shown, that
the negative cosines are separated from the posilive cosines by
the diameter DE; all the arcs whose extremities fall on the
right side of DE, having a positive cosine, while those whose
extremities fall on the left have a negative cosine.
Thus from 0° to 90° the cosines are positive ; from 90° to
270° they are negative ; from 270° to 300° they again become
positive ; and after a whole revolution they assume the same
values as in the preceding revolution, for cos (360°+a;)=cosa:.
PLANE TRIGONOMETRY.
221
From these explanations, it will evidently appear, that the
sines and cosines of the various arcs which are multiples of the
quadrant have the following values :
sin 0^ =
sin 90°=R
cos 0°=R
cos 90° =0
sin 180°=0
sin 270°=— R
cos 180°=— R
cos 270° =0
sin 360^=0
sin 450° =R
cos 360° =R
cos 450° =0
sin 540° =0
sin 630°=— R
cos 540°=— R
cos 630°=0
sin 720° =0
sin 810°=R
cos 720° =R
cos 810°=0
&c.
&c.
&c.
&c.
And generally, k designating any whole number we shall
have
sin 2A;.90°=0, cos (2A:+1) . 90° = 0,
sin (4A;+ 1) . 90°=R, cos Ak . 90°=R,
sin (4^—1) . 90°=— R, cos (4A; + 2) . 90°=— R.
What we have just said concerning the sines and cosines
renders it unnecessary for us to enter into any particular de-
tail respecting the tangents, cotangents, &c. of arcs greater
than 180° ; the value of these quantities are always easily de-
duced from those of the sines and cosines of the same arcs :
as we shall see by the formulas, which we now proceed to
explain.
THEOREMS AND FORMULAS RELATING TO SINES, COSINES,
TANGENTS, &c.
\
XV.^T/ie sine of an arc is half the chord which subtends a
^ double arc,^
'■)
For the radius CA, perpen-
dicular to the chord MN, bi-
sects this chord, and likewise
the arc MAN ; hence MP, the
sine of the arc MA, is half the
chord MN which subtends
the arc MAN, the double of
MA.
The chord which subtends
the sixth part of the circum-
ference is equal to the radius ;
hence
^^^'orsin30°=iR,)
sm
12
in other words, the sine of a third part of the right angle is
equal to the half of the radius.
T *
222
PLANE TIIIGOIMOMETRY.
X(
I v^
^ XVI. The square of the sine
hf an arc, together with the
Square of the cosine, is equal
;io the square of the radius ; so
that in general terms we have
sin^A + cos2A=:R2.
This property results im-
mediately from the right-an-
gled triangle CMP, in which
MP-+CP'^=CM2.
It follows that when the
sine of an arc is given, its co-
sine may be found, and re-
ciprocally, by means of the
formulas cos A = d= V (R^ — sin^A), and sin A = rfc \/ (R^ — cos^A).
The sign of these formulas is +, or — , because the same sine
MP answers to the two arcs AM, AM', whose cosines CP, CP',
are equal and have contrary signs ; and the same cosine CP
answers to the two arcs AM, AN, whose sines MP, PN, are
also equal, and have contrary signs.
Thus, for example, having found sin 30^=|R, we may de-
duce from itcos30^orsin60^=\/(R2— iR2) = yfR2— iRv/3,
XVII. The sine and cosine of an drc A'heing 'given, it is re-^
quired to find the tangent, secant, cotangent, and cosecant of the
same arc.
The triangles CPM, CAT, CDS, being similar, we have the
proportions :
CP : PM : : CA : AT ; or cos A : sin A : : R : tang A=
CP : CM : : CA : CT ; or cos A : R : : R : sec A =
PM : CP : : CD : DS ; or sin A : cos A : : R : cot A=
R sin A
cos A
_R2_
cos A
RcosA
PM : CM : : CD : CS ; or sin A : R : : R : cosec A=-
sin A
R^
sin A
which are the four formulas required. It may also be observed,
that the two last formulas might be deduced from the first two,
by simply putting 90° — A instead of A.
From these formulas, may be deduced the values, with their
proper signs, of the tangents, secants, &c. belonging to any
arc whose sine and cosine are known ; and since the progres-
sive law of the sines and cosines, according to the different
arcs to which they relate, has been developed already, it is
unnecessary to say more of the law which regulates the tan-
gents and secants. ,-
PLANE TRIGONOMETRY. 223
By means of these formulas, several results, which have
already been obtained concerning the trigonometrical lines,
may be confirmed. If, for example, we make A =00^, we
shall have sin A=R, cos A — ; and consequently tang 90°-=
W
— , an expression which designates an infinite quantity ; for
the quotient of radius divided by a very small quantity, is very
great, and increases as the divisor diminishes ; hence, the quo-
tient of the radius divided by zero is greater than any finite
quantity.
The tangent being equal to R ; and cotangent to R.-^- ;
cos siii.
.V
itf(
follows that tangent and cotangent will both be positive
when the sine and cosine have like algebraic signs, and both
negative, when the sine and cosine have contrary algebraic
signs. Hence, the tangent and cotangent have the same
sign in the diagonal qiiadrants : that is, positive in the 1st and
3d, and negative in the 2d and 4th ; results agreeing witii those
ofArt. Xlf.
The Algebraic signs of the secants and cosecants are readily
determined. For, the secant is equal to radius square divided
by the cosine, and since radius square is always positive, it
follows that the algebraic sign of the secant will depend on
that of the cosine: hence, it is positive in the 1st and 4th
quadrants and negative in the 2nd and 3id.
Since the cosecant is equal to radius square divided by the
sine, it follows that its sign will depend on the algebraic sign
of the sine : hence, it will be positive- in the 1st and 2nd
quadrants and negative in the 3rd and 4th.
XVIII. The formulas of the preceding Article, combined
with each other and with the, equation sin "A + cos ''^A:=R^
furnish some others worthy of attention.
First we have R^ -f- tang'^ A =: R- + "^IjI^-^ z=,
cos^ A
R- (sin*^ A + cos- A) R'' , tiq , * 2 a • o *
__1 i=: ; hence R^+tang^ A=sec- A, a
cos -A cos" A
formula which might be immediately deduced from the righi-
angled triangle CAT. By these formulas, or by the right-an-
gled triangle CDS, we have also R'-^ + cot- Arzcosec" A.
Lastly, by taking the product of the two formulas tang A=r
RsinA 1 . A RcosA , , . . * tio
--, and cot A= — — we have tang Ax cot A^R-, a
cos A sin A
R' R2
formula which gives cot A= ^.^,^^ ^ , and tang A=
We likewise have cot B=
tang A ' o ■ ■ cot A,
I
tangB
L.?-:.
224
PLANE TRIGONOMETRY
j Hence cot A : cot B : : taiig B : tang A ; that is, the cotan- '
\ gents of two arcs are reciprocally proportional to their tangents.
The formula cot Ax tang A=R'"^ might be deduced imme-
diately, by comparing the similar triangles CAT, CDS, which
give AT : CA : : CD : DS, or tang A : R : : R : cot A
XIX. The sines and cosines of two arcs, a and b, being given,
it is required to find the sine and cosine of the sum or difference
of these arcs.
Let the radius AC=R, the arc
AB=a, the arc BD=6, and con-
sequently ABD=a + h. From
the points B and D, let fall the
perpendiculars BE, DF upon AC ;
from the point D, draw DI per-
• pendicular to BC ; lastly, from
the point I draw IK perpendicu-
lar, and IL parallel to, AC. F' C FXTKE
The similar triangles BCE, ICK, give the proportions,
sin a cos /;.
CB : CI : : BE : IK, or R : cos Z) : : sin « : IK=
CB : CI : : CE : CK, or R : cos 6 : : cos a : CK=
R
co§ a cos '6.
R
The triangles DIL, CBE, having their sides perpendicular,
each to each, are similar, and give the proportions,
CB : DI : : CE : DL, or R : sin ft : : cos a : DL=
CB : DI : : BE : IL, or R : sin 6 : : sin a : IL:
cos a sin b,
R
sin « sin b.
R
But we have
IK+DL=DF:
Hence
sin (a + 6), and CK— IL-:CF=rcos (a + b).
sin a cos ft + sin i cos a
sin (a + b) =
cos {a-\-b) =
R
cos a cos ft — sin a* sin ft.
R
The values of sin (a — ft) and of cos {a — ft) might be easily
deduced from these two formulas; but they may be found
directly by the same figure. For, produce the sine DI till it
meets the circumference at M ; then we have BM— BD=ft,
and MI==ID = sin ft. Through the point M, draw MP perpen-
dicular, and MN parallel to, AC ifsince Mlrr^DI, we have MN
=IL, and IN=DD But we have IK— IN=MP = sin («— ft),
and CK+MN=CP=^cos (a—b) ; hence
PLANE TRIGONOMETRY. ' 225
. , ,. sm a cos b — sin h cos a
sm {a—h)^ ^
-. cos a cos 6 + sin a sin h
cos {a — o) = —
R
These are the formulas which it was required to find.
The preceding demonstration may seem defective in point
of generaHty, since, in the figure which we have followed, the
arcs a and b, and even a + &, are supposed to be less than 90°.
But first the demonstration is easily extended to the case in
which a and b being less than 90°, their sum a + 6 is greater
than 90°. Then the point F would fall on the prolongation of
AC, and the only change required in the demonstration would
be that of taking cos {a -\- b) = — CF' ; but as we should, at the
same time, have CF' = rL' — CK', it would still follow that cos
(a + b) = CK! — I'L', or R cos (a + b)=cos a cos b — sin a sin b.
And whatever be the values of the arcs a and b, it' is easily
shown that the formulas arc true : hence we may regard them
as established for all arcs. We will repeat and number the
formulas for the purpose of more convenient reference.
. , . ,. sin a cos ^ + sin b cos a .^ .
sm {a + b) = g (1.).
. , ,v sin a cos b — sin b cos a .^.
sin (a—b)= ^^ (2.).
, ^ cos a cos b — sin a sin b . ^
cos (a+b)=-. ^ (3.)
,, cos a cos & + sin a sin ft ..
cos (a—b) = ^ (4.)
XX. If, in the formulas of the preceding Article, we make
6=a, the first and the third will give . •■ , ^
. _ 2 sin a cos a ^ cos^ a — sin^ a 2 cos^ a — R"
sm2a= p -f cos 2a= ^ =-^=^ — ^
formulas which enable us to find the sine and cosine of the
double arc, when we know the sine and cosine of the arc itself.
To express the sin a and cos a in terms of ^a, put |a for a,
and we have
. ^ 2 sin la cos ^a cos^ ^a — sin^ la
sm a= ± £_, cos a= i =— .
R R
To find the sine and cosine of Ja in terms of a, take the
equations
cos- i^-hsin^ |«=R^ and cos-i« — sin^ J-a=R cos a,
there results by adding and subtracting
cos- |a-^|R- + iR cos a, and sin2-i,Gt=iR2— iR cos a;
whence
sin ia= V (|R2— iR cos a)=iV2R2— 2R cos «.'
cos 'a= \/(iR2+iR cos a)=iV2ilH2Rcosa.
826 PLANE TRIGONOMETRY.
If we put 2a in the place of a, we shall have,
sin (?= n/ (iR2_4R cos 2a) =i V 2R"^— 2R cos 2a.
cos a=V(iR2+iR cos 2a)=\s/2W-^ 2R cos 2a.
Making, in the two last formulas, a=45°, gives cos2a=0, and
sin 45°= \/iR2=RV| ; and also, cos 45°= VlR^R V-^.
Next, make a =22° 30', which gives cos 2a=R V^, and we have
sin 22° 30'=R V(i— ^Vi) and cos 22° 30'=RN/(^-+iVi).
XXI. If we multiply together formulas (1.) and (2.) Art.
XIX. and substitute for cos'-^ a, R^ — sin''^ a, and for cos- b,
R^ — sin^ h ; we shall obtain, after reducing and dividing by R^,
sin {a + h) sin {a — h) — sin^ a — sin^ h = (sin a + sin h) (sin a — sin h),
or, sin {a — V) : sin a — sin h : : sin a + sin 6 : sin (a + 6).
XXII. The formulas of Art. XIX. furnish a great number of
consequences ; among which it will be enough to mention those
of most frequent use. By adding and subtracting we obtain
the four which follow,
2
sm («+6) + sin (a — h)= — sin a cos 6.
R
2
sin {a-{-p) — sin {a — 6)=-_sin b cos a.
R
2
cos (a + 6) + COS (a — 6) = —- cos a cos b.
R
2
cos {a — b) — cos (a + 6)=-psin a sin b.
and which serve to change a product of several sines or co-
sines into linear sines or cosines, that is, into sines and cosines
multiplied only by constant quantities.
XXIII. If in these formulas we put a-{-b=pf a — b=q, which
p-\-q p — q
gives a=—^j 5=—^, we shall find
2
sin j9-f sin ^=-— -sin ^(p + q) cos |(jo — q) (1.)
R "
2
sin p — sin q~-:^sm -J (p — q) cos h{p + q) (2.)
R
W cosp + cosq = -^cos -^ {p + q) cos h (p — q) (3.)
cosg — cos 7?=^ sin h (p + q) sin J {p — q) (4.)
R "
PLANE TRIGONOMETRY. 227
') - 6^^"$^ /- ■ ■
If we make g=o, we shall obtain,
2 sin i» cos \p
sin p — ^-^ -—
„ 2 cos^ \ p
R + cos p= „ "
^ 2 sin^ I p
R — cos j9= : hence
R
sin j9 _ tang ^p R
wky R+cos /?~ R ""cotjjp
- si np _^cot ^p_ R _
R — cos p R tang ^p-
formulas which are often employed in trigonometrical calcula-
tions for reducing two terms to a single one.
XXIV. From the first four formulas of Art XXIII. and the first
of Art. XX.,dividinff, and considerinjsj that = — M—=
^ ^ cos a R cot a
we derive the following :
I sin ;7 + sin q _^\n \ {p -\- q ) cos ^ ( p—g^) _ ta ng -|(;? + y)
> sin j9 — sin ^ cos|^ (^-f g) sin J (jo — q) tangj(jt? — q)
' sin ;? + sin 7 _ sin j^ (;? + 9^) _ tangl (jp + y)
cosp-fcos^' Qos^{p-^q) R
i s|g_jH;ig^^_ cos^ (/?— y) _ cot I (p—q) ^
' cos q — cos p sin -J (p — q) R
sin /?— sin q _s\nl- {p—q) _ tanrri (p—q)
COSJ9 + COS5' COS J (2? — q) R
::in /> — sin q _<^os^(p + q) ^cot^ (p + q)
cos q — COS JO sin ^{p-\-q) R
( i)sp + cos ^ _ cos }f (p + q) COS | {p — 9')_cot h{p-rq)
r <s q — cos p sin h (p+q) sm |- (p — qj tang-^ (p—q)
sin /) + sin gf 2sin j (/? + ^) cos -} (p — q) cos | (jo — g)
sin (p + ^)~2sin J {p-^q) cos J- (p + g)~cosi (p + g)
s in p — si n q 2sin | (p — g^) cos ^ (p + q) sin |^ (jo — q)
sin (io + ^)~2sini (p + ^)cosi(;? + 5')'^sin| (;? + g')
Formulas which are the expression of so many theorems.
From the first, it follows that the sum of the sines of two arcs is
to the difference of these sines, as the tangent of half the sum of
\ the arcs is to the tangent of half their difference.
828 PLANE TRIGONOMETRY.
XXV. In order likewise to develop some formulas relative
to tangents, let us consider the expression
. , ,. R sin (« + &) . , . , , , . . ,
tang {a-\-o)= -^ — — i, m which by suostitutmg the values
^ cos (a^b) '' ^
of sin (a +6) and cos {a+h), we shall find
* / . i.\ R (sin a cos & + sin 6 cos a)
tang(cz + &)=— A ■ ^ . , . \
cos a cos — sm bsma
MT , . cos a tanff a , . , cos h tang h
JNow we have sm a= p , and sm &= ^ — ^-- :
substitute these values, dividing all the terms by cos a cos h ;
we shall have
R^ — tang a tang 6
which is the value of the tangent of the sum of two arcs, ex-
pressed by the tangents of each of these arcs. For the tangent
of their difference, we should in like manner find
R^ (tang ^— tang h)
tang {a^h) =R2+tang « tang 6.
Suppose 6=a ; for the duplication of the arcs, we shall have
the formula
2 B? tang a
tang 2a—rfrr, — ~ :
° H~ — tang^a
Suppose &=2a ; for their triplication, we shall have the for-
mula
tang 3«=?^J&^5g_^±*^M^ ;
* W — tang a tang 2a'
in which, substituting the value of tang 2 a, we shall have
3R^ tang a — tang ^a
tang 3 a= -_— — ^ & —
^ R'^_3 tang ^a. ■»
XXVI. Scholium, The radius R being entirely arbitrary, isl
generally taken equal to 1, in which case it does not appear in
the trigonometrical formulas. For example the expression for
the tangent of twice an arc when R=l, becomes,
2 tang a
tang2<z= ^ -
1 — tang- fl-
it we have an analytical formula calculated to the radius of 1,
and wish to apply it to another circle in which the radius is R,
we must multiply each term by such a power of R as will make
all the terms homogeneous: that is, so that each shall contain the
same number of literal factors.
PLANE TRIGONOMETRY. 229
CONSTRUCTION AND DESCRIPTION OF THE TABLES.
XXVII. If the radius of a circle is taken equal to 1, and the
lengths of the lines representing the sines, cosines, tangents,
cotangents, &c. for every minute of the quadrant be calculated,
and written in a table, this would be a table of natural sines,
..cosines, &c. sj^
W^ XXVIII. If such a table were known, it would be easy to
^ calculate a table of sines, &c. to any other radius ; since, in
ditferent circles, the sines, cosines, &c. of arcs containing the
same number of degrees, are to each other as their radii.
XXIX. If the trigonometrical lines themselves were used, it
would be necessary, in the calculations, to. perform the opera-
tions of multiplication and division. To avoid so tedious a
method of calculation, we use the logarithms of the sines, co-
• sines, &:c. ; so that the tables in common use show the values
of the logarithms of the sines, cosines, tangents, cotangents, &c.
for each degree and minute of the quadrant, calculated to a
- given radius. This radius is 10,000,000,000, and consequently
its logarithm is 10.
XXX. Let us glance for a moment at one of the methods
of calculating a table of natural sines.
The radius of a circle being 1, the semi-circumference is known
to be 3.14159265358979. This being divided successively, by
180 and 60, or at once by 10800, gives .0002908882086657,
for the arc of 1 minute. Of so small an arc the sine, chord,
and arc, differ almost imperceptibly from the ratio of equality ;
so that the first ten of the preceding figures, that is, .0002908882
may be regarded as the sine of 1' ; and in fact the sine given
in the tables which run to seven places of figures is .0002909.
By Art. XVI. we have for any arc, cos= V(l — sin^). This
theorem gives, in the present case, cos 1' = .9999999577. Then
by Art. XXII. we shall haVe
2 cos I'xsin 1'-— sin 0'=sin 2' = .00058 17764
2 cos I'xsin 2'— sin r=sin 3' =.0008726646
2 cos I'xsin 3'~-sin 2' = sin 4' = .00 11635526
2 cos I'xsin 4'— sin 3' = sin 5' = .0014544407 *
2 cos I'xsin 5'— sin 4'=sin 6' = .0017453284
&c. &c. &c.
Thus may the work be continued to any extent, the whole
difficulty consisting in the multiplication of each successive re-
sult by the quantity 2 cos 1'— .1.9999999154.
U
230 PLANE TRIGONOMETRY.
Or, the sines of T and 2' being determined, the work might
be continued thus (Art. XXI.) :
sin T: sin 2' — sin I' : : sin 2' + sin 1' : sin 3
sin 2': sin 3' — sin 1' : : sin 3' + sin 1' : sin 4'
sin 3' : sin 4' — sin 1' :: sin 4' + sin 1' : sin 5'
sin 4' : sin 5' — sin 1' :: sin 5' + sin 1' : sin 6'
&c. &;c. &c.
In like manner, the computer might proceed for the sines of
degrees, &c. thus :
sin 1° : sin 2°--sin 1° : : sin 2'' + sin 1° : sin 3°
sin 2° : sin 3°-— sin 1° : : sin 3° + sin 1° : sin 4°
sin 3° : sin 4°— sin 1° : : sin 4° + sin 1° : sin 5°
&:c. &c. &c.
Above 45° the process may be considerably simplified by
Ihe theorem for the tangents of the sums and differences of
arcs. For, when tiie radius is unity, the tangent of 45° is also
unity, and tan (a + b) will be denoted thus :
tan(45° + fe)=:l±i^.
^ ^ 1— tan b
And this, again, may be still further simplified in practice
The secants and cosecants may be found from the cosines and
sines.
TABLE OF LOGARITHMS.
XXXI. If the logarithms of all the numbers between 1 and
any given number, be calculated and arranged in a tabular form,
such table is called a table of logarithms. The table annexed
shows the logarithms of all numbers between 1 and 10,000.
The first column, on the left of each page of the table, is the
column of numbers, and is designated by the letter N ; the deci-
mal part of the logarithms of these numbers is placed directly
opposite them, and on the same horizontal line.
The characteristic of the logarithm, or the part which stands
totheleftof the decimal point, is always known, being 1 less than
the places of integer figures in the given number, and there-
fore it is not written in the table of logarithms. ' Thus, for all
numbers between 1 and 10, the characteristic is 0: for num-
bers between 10 and 100 it is 1, between 100 and 1000 it is
2, &c.
(
\
PLANE TRIGONOMETRY. 231
*- PROBLEM.
To find from the table the logarithm of any number,
CASE I.
Wlien the number is less than 100.
Look on the first page of the table of logarithms, along the
columns of numbers under N, until the number is found ; the
number directly opposite it, in the column designated Log., is
the logarithm sought.
CASE II.
IVhen the number is greater than 100, and less than 10,000,
Find, in the column of numbers, the three first figures of the
given number. Then, pass across the page, in a horizontal
line, into the columns marked 0, 1,2, 3, 4, &c., until you come
to the column which is designated by the fourth figure of the
given number : to the four figures so found, tw^o figures taken
from the cohimn marked 0, are to be prefixed. If the four
figures found, stand opposite to a row of six figures in the column
marked 0, the two figures from this column, which are to be
prefixed to the four before found, are the first two on the left
hand ; but, if the four figures stand opposite a line of only four
figures, you are then to ascend the column, till you come to the
line of six figures : the two figures at the left hand are to be
prefixed, and then the decimal part of the logarithm is obtained.
To this, the characteristic of the logarithm is to be prefixed,
which is always one less than the places of integer figures in
the given number. Thus, the logarithm of 1 122 is 3.049993.
In several of the columns, designated 0, 1, 2, 3, &c., small
dots are found. Where this occurs, a cipher must be written
for each of these dots, and the two figures which are to be prer
fixed, from the first column, are then found in the horizontal
line directly below. Thus, the log. of 2188 is 3.340047, the two
dots being changed into two ciphers, and the 34 from the
column 0, prefixed. The two figures from the colum 0, must
also be taken from the line below, if any dots shall have been
passed over, in passing along the horizontal line : thus, the loga-
rithm of 3098 is 3.491081, the 49 from the column being
taken from the line 310.
232 PLANE TRIGONOMETRY.
CASE III.
When the number exceeds 10,000, or consists of five or mort
places of figures.
Consider all the figures after the fourth from the left hand,
as ciphers. Find, from the table, the logarithm of the first four
places, and prefix a characteristic which shall be one less than
the number of places including the ciphers. Take from the last
column on the right of the page, marked D, the number on the
same horizontal line with the logarithm, and multiply this num-
ber by the numbers that have been considered as ciphers :
then, cut off from the right hand as many places for decimals
as there are figures in the multiplier, and add the product, so
obtained, to the first logarithm : this sum will be the logarithm
sought.
Let it be required to find the logarithm of 672887. The log.
of 672800 is found, on the 1 1th page of the table, to be 5.827886,
after prefixing the characteristic 5. The corresponding num-
ber in the column D is 65, which being multiplied by 87, the
figures regarded as ciphers, gives 5655 ; then, pointing ofif two
places for decimals, the number to be added is 56.55. This
number being added to 5.827886, gives 5.827942 for the loga-
rithm of 672887 ; the decimal part .55, being omitted.
This method of finding the logarithms of numbers, from the
table, supposes that the logarithms are proportional to their
respective numbers, which is not rigorously true. In the exam-
ple, the logarithm of 672800 is 5.827886 ; the logarithm of
672900, a number greater by 100, 5.827951 : the difference of
the logarithms is 65. Now, as 100, the diiference of the numbers,
IS to 65, the diflference of their logarithms, so is 87, the diffe-
rence between the given number and the least of the numbers
used, to the difference of their logarithms, which is 56.55 : this
diflference being added to 5.827886, the logarithm of the less
number, gives 5.827942 for the logarithm of 672887. The use
of the column of differences is therefore manifest.
When, however, the decimal part which is to be omitted ex-
ceeds .5, we come nearer to the true result by increasing the
next figure to the left by 1 ; and this will be done in all the
calculations which follow. Thus, the difference to be added,
was nearer 57 than 56 ; hence it would have been more exact
to have added the former juimber.
The logarithm of a vulgar fraction is equal to the loga-
rithm of the numerator, minus the logarithm of the denom-
I
PLANE TRIGONOMETRY. 233
inator. The logarithm of a decimal fraction is found, hy ccn-X
sidering it as a whole numher, and then prefixing to the decimal \
part of its logarithm, a negative characteristic, greater hy unity '
than the number of ciphers between the decimal point and the first
significant place of figures. Thus, the logarithm of .0412. is
2^614897. •
PROBLEM.
To find from the table, a number answering to a given logarithm,
XXXII Search, in the column of logarithms, for the decimal
part of the given logarithm, and if it be exactly found, set down
the corresponding number. Then, if the characteristic of the
given logarithm be positive, point otf, from the left of the number
found, one place more for whole numbers than there are units
in the characteristic of the given logarithm, and treat the other
places as decimals ; this will give the number sought.
If the characteristic of the given logarithm be 0, there will
be one place of whole numbers ; if it be — 1, the number will
be entirely decimal ; if it be — 2, there will be one cipher be-
tween the decimal point and the first significant figure ; if it be
— 3, there will be two, &c. The number whose logarithm is
1.492481 is found in page 5, and is 31.08.
But if the decimal part of the logarithm cannot be exactly
found in the table, take the number answering to the nearest
less logarithm ; take also from the table the corresponding dif-
ference in the column D ; then, subtract this less logarithm from
the given logarithm ; and having annexed a sufficient number
of ciphers to the remainder, divide it by the difference taken
from the column D, and annex the quotient to the number an-
swering to the less logarithm : this gives the required number,
nearly. This rule, like the one for finding the logarithm of a
number when the places exceed four, supposes the numbers to
be proportional to their corresponding logarithms.
Ex. I. Find the number answering to the logarithm 1.532708-
Here,
The given logarithm, is ... 1.532708
Next less logarithm of 34,09, is - - 1.532627
Their difference is - _ . . ~l gj"
And the tabular difference is 128 : hence
128) 81.00 (63
which being annexed to 34,09, gives 34.0963 for the number
answering to the logarithm 1.532708.
234 , . PLANE TRIGOKOMETRY,
^ Ex. 2. Required the ^number answering to the logarithnj
3.233568.
The given logarithm is 3.233568
The next less tabular logarithm of 1712, is 3.233504
Diff.= 64
Tab. Difr. = 253) €A.OO- (25
Hence the number sought is 1712.25, marking four places
of integers for the characteristic 3.
TABLE OF LOGARITHMIC SINES.
XXXIII. In this table are arranged the logarithms of the
numerical values of the sines, cosines, tangents, and cotangents,
of all the arcs or angles of the quadrant, divided to minutes,
and calculated for a radius of 10,000,000,000. The logarithm
of this radius is 10. In the first and last horizontal line, of each
page, are written the degrees wiiose logarithmic sines, &:c. are
expressed on the page. The vertical columns on the left and
right, are columns of minutes.
CASE L
To findy in the table, the logarithmic sine, cosine, tangent, or co-
tangent of any given arc or angle.
1. If the angle be less than 45*^, look in the first horizontal
Ime of the difterent pages, until the number of degrees be
found ; then descend along the column of minutes, on the left
of the page, till you reach the number showing the minutes ;
then pass along the horizontal line till you come into the column
designated, sine, cosine, tangent, or cotangent, as the case may
be : the number so indicated, is the logarithm sought. Thus, the
sine, cosine, tangent, and cotangent of 19"" 55', are found on
page 37, opposite 55, and are, respectively, 9.532312, 9.973215,
9.559097, 10.440903.
2. If the angle be greater than 45°, search along the bottom
fine of the different pages, till the number of degrees arc found ;
then ascend along the column of minutes, on the right hand
side of the page, till you reach the number expressing the mi-
nutes ; then pass along the horizontal line into the columns
designated tang., cotang., sine, cosine, as the case may be ; the
number so pointed out is the logarithm required.
PLANE TRIGONOMETRY. 235
It will be seen, that the column designated sine at the top of
the page, is designated cosine at the bottom ; the one desig-
nated tang., by cotang., and the one designated cotang., by
tang.
The angle found by taking the degrees at the top of the page,
and the minutes from the first vertical column on the left, is the
complement of the angle, found by taking the corresponding
degrees at the* bottom of the page, and the minutes tiaced up
in the right hand column to the same horizontal line. This
being apparent, the reason is manifest, why the columns desig-
nated sine, cosine, tang., and cotang., when the degrees are
pointed out at the top of the page, and the minutes countec*
downwards, ought to be changed, respectively, into cosine, sine,
cotang., and tang., when the degrees are shown at the bottom
of the page, and the minutes counted upwards.
it' the angle be greater than 90^, we have only to subtract it
from 180^, and take the sine, cosine, tangent, or cotangent of
the remainder.
The secants and cosecants are omitted in the table, being
easily found from the cosines and sines.
R2 .
For, sec. = ; or, taking the logarithms, log. sec.=2
COS.
log. R — log. COS.— 20 — log. COS. ; that is, the logarithmic secant
is found by substracLing the logarithmic cosine from 20. And
R^
cosec. — , or log. cosec.=2 log. R — log. sine = 20 — log.
sine
sine ; that is, the logarithmic cosecant is found by subtracting the
logarithmic sine from 20.
It has been shown that R-rrtang. x cotang. ; therefore, 2 log.
Rr=:log. tang. + log. cotang.; or 20=r:log. tang. + log. cotang.
The column of the table, next to the column of sines, and
on the right of it, is designated by the letter D. This column
is calculated in the following manner. Opening the table at
any page, as 42, the sine of 24° is found to be 9.609313 ; of
24^ 1', 9.609597 : their difference is 284 ; this being divided by
60, the number of seconds in a minute, gives 4.73, which is
entered in the column I), omitting the decimal point. Now,
supposing the increase of the logarithmic sine to be propor-
tional to the increase of the arc, and it is nearly so for 60", it
follows, that 473 (the last two places being regarded as deci-
mals) is the increase of the sine for 1". Similarly, if the arc
be 24° 20', the increase of the sine for 1", is 465, the last two
places being decimals. The same remarks are equally appli-
cable in respect of the column D, after the column cosine, and
of the column D, between the tangents and cotangents. The
column D, between the tangents and cotangents, answers
23(5 PLANE TRIGONOMETRY.
to either of these columns ; since of the same arc, the log
tang. -f log. cotangiz:20. Therefore, having two arcs, a and h,
log. tang 6 + log. cotang fe^log. tang fl + log. cotang a; or,
log. tang h — log. tang «r=log. cotang a — log. cotang h.
Now, if it were required to find the logarithmic sine of an
arc expressed in degrees, minutes, and seconds, we have only
to find the degrees and minutes as before ; then multiply the
corresponding tabular number by the seconds, cut off two places
to the right hand for decimals, and then add the product to the
number first found, for the sine of the given arc. Thus, if we
wish the sine of 40° 26' 28".
The sine 40° 26' - - - - 9.811952
Tabular difference =:= 247
Number of seconds m 28
Product=:69.16, to beadded = 69.16
Gives for the sine of 40° 26' 28" =9.812021.16
The tangent of an arc, in ,which there are seconds, is found
in a manner entirely similar. In regard to the cosine and co-
tangent, it must be remembered, that they increase while the
arcs decrease, and decrease while the arcs are increased, con-
sequently, the proportional numbers found for the seconds must
be subtracted, not added.
Ex, To find the cosine 3° 40' 40".
Cosine 3° 40' 9.999110
Tabular diflference i= 13
Number of seconds = 40
Product m 5.20, which being subtracted =: 5.20
Gives for the cosine of 3° 40' 40" 9.999104.80
CASE II.
To find the degrees, minutes, and seconds answering to any given
logarithmic sine, cosine, tangent, or cotangent.
Search in the table, and in the proper column, until the num-
ber be found ; the degrees are shown either at the top or bot-
tom of the page, and the minutes in the side columns, either at
the left or right. But if the number cannot be exactly found in
the table, take the degrees and minutes answering to the nearest
less logarithm, the logarithm itself, and also the corresponding
tabular difference. Subtract the logarithm taken, from the
PLANE TRIGONOMETRY. 237
given logarithm, annex two ciphers, and then divide the re-
mainder by the tabular difference : the quotient is seconds, and
is to be connected with the degrees and minutes before found ;
to be added for the sine and tangent, and subtracted for the
cosine and cotangent.
Ex, 1. To find the arc answering to the sine 9.880054
Sine 49° 20', next less in the table, , 9.879963
Tab. DilF. 181)9100(50"
Hence the arc 49° 20' 50" corresponds to. the given sine
9.880054.
Ex. 2. To find the arc corresponding to cotang. 10.008688.
Cotang 44° 26', next less in the table 10.008591
Tab. Diff: 421)9700(23"
Hence, 44°26'— 23"=44°25'37" is the arc corresponding
to the given cotangent 10.008688.
i
PRINCIPLES FOR THE SOLUTION OF RECTILINEAL TRI-
ANGLES.
THEOREM L
In every right angled triangle, radius is to the sine of either
of the acute angles, as the hypothenuse to the opposite side :
and radius is to the cosine of either of the acute angles, as
the hypothenuse to the adjacent side.
Let ABC be the proposed tri-
angle, right-angled at A : from
the point C as a centre, with a
radius CD equal to the radius of
the tables, describe the arc DE,
which will measure tiie angle C ;
on CD let fall the perpendicular
EF, which will be the sine of the
angle C, and CF will be its co-
sine. The triangles CBA, CEF, are similar, and give the pro-
portion,
CE : EF : : CB : BA : hence
R : sin C : : BC : BA.
238
PLANE TRIGONOMETRY
But we also have,
CE : CF :
R : cos C :
Cor.
CB
CB
CA : hence
CA.
If the radius R=l, we shall have,
AB = CB sin C, and CA=CB cos C.
Hence, in every right angled triangle, the perpendicular is equal
to the hypothenuse multiplied by the sine of the angle at the base ;
and the base is equal to the hypothenuse multiplied by the cosine
of the angle at th,e base ; the radius being equal to unity.
THEOREM II.
In every right angled triangle, radius is to the tangent of ei-
ther of the acute angles, as the side adjacent to the side op-
posite.
Let CAB be the proposed tri-
angle.
With any radius, as CD, de-
scribe the arc DE, and draw the
tangent DG.
From the similar triangles
CDG, CAB, we shall have,
CD : DG : : CA : AB : hence,
R : tang C : : CA : AB.
Cor.l. If the radius R=l,
AB=CAtangC.
Hence, the perpendicular of a right angled triangle is equal to
the base multiplied by the tangent of the angle at the base, the
radius being unity.
Cor. 2. Since the tangent of an arc is equal to the cotangent
of its complement (Art. VI.), the cotangent of B may be sub-
stituted in the proportion for tang C, which will give
R : cot B : : CA : AB.
THEOREM III.
In every rectilineal triangle, the sines of the angles are to each
other as the opposite sides.
PLANE TRIGONOMETRY.
239
Let ABC be the proposed triangle ; AD
the pcrpcnciicula^ let fall from the vertex A
on the opposite side BC : there may be two
cases.
First. If the perpendicular falls within -o
the triangle ABC, the right-angled triangles
ABD, ACD, will give,
R : sin B : : AB : AD.
R : sin C : : AC : AD.
In these two propositions, the extremes are equal ; hence,
sin C : sin E : : AB : AC.
Secondly. If the perpendicular falls
without the triangle ABC, the right-
angled triangles ABD, ACD, will still
give the proportions,
R : sin ABD : : AB : AD,
R:sinC ;:AC:AD;
Trom which we derive
sin C : sin ABD
But the angle ABD is the supplement of ABC, or B ; hence
sin ABD = sin B ; hence we still have
sinC:sinB;:AB:AC.
AB : AC.
THEOREM IV.
In every rectilineal triangle, the cosine of either of the angles is
equal to radius multiplied by the sum of the squares of the sides
adjacent to the angle, minus the square of the side opposite,
divided by twice the rectangle of the adjacent sides.
Let ABC be a triangle : then will
AB^+BC2— AC2
cos B=R-
2ABxBC.
First. If the perpendicular falls within
,'the triangle, we shall have AC^^rAB^H-
BC2— 2BCxBD(Book IV. Prop. XII.); B
. T3^, AB^+BC^— AC2 ^ , . ^ . ^ , , . ,
I hence BD=: —.^^^ . But m the right-angled triangle
jABD, we have
2BC
R : cos B : : AB : BD ;
^40 PLANE TRIGONOMETRY.
hence, cos B— ^p , or by substituting the value of BD,
AB2+BC2_AC2
cos B— R X
2AB X BC
Secondly. If the perpendicular falls
without the triangle, we shall have
AC2z=:AB2+BCH2BCxBD; hence
p^ AC^—AB^— BC2
^^== 2BC
But in the right-angled triangle BAD,
RxBD D B C
we still have cos ABD=:— -^g- ; and the angle ABD being
supplemental to ABC, or B, we have
cosB=-cosABD=-?-f|^.
AB
hence by substituting the value of BD, we shall again have
P „ AB2+BC2— AC2
""^^=^><— 2AB^BG--
Scholium. Let A, B, C, be the three angles of any triangle ;
a, h, c, the sides respectively opposite them : by the theorem,
we shall have cos B=R x — ^ . And the same principle,
when applied to each of the other two angles, will, in like man-
ner give cos A=R x — kt , anc cos C=R x — tt-t — .
*= 26c 2ab
Eithei- of these formulas may readily be reduced to one in which
the computation can be made by logarithms.
Recurring to the formula R'-^ — R cos A=2sin^ ^A (Art.
XXIII.), or 2sin^^A=R~ — RcosA, and substituting for cosA,
we shall have
2sin^^ArrR^-^R^x \^^
Wx2bc--n\h^+c^-^') a^^b''-^c^+2bc
2bc — R X 2^^
^ R.xf!:^)Wx (^^=4M£l±). , Hence
2oc 2bc
^ 46c ^
For the sake of brevity, put
J (a + /; + c) =/?, or a + 6 + c =2p ; we have a + h — c=2p — 2c,
a-\-c — 6=2p — 2h ; hence
a.nJA=RV(i^=^^)).
PLANE TRIGONOMETRY. '''"'^'^241
THEOREM V.
In every rectilineal triangle, the sum of two sides is to their diffe-
rence as the tangent of half the sum of the angles opposite those
sides, to the tangent of half their difference.
For. AB : BC : : sin C : sin A (Theo-
rem III.). Hence, AB + BC : AB— BC
; : sin C -t-sin A : sin C — sin A. But
. ^ . . . ^ . . C + A
sinC + sinA: sinC — sin A : : tang—— — :
tang — ^r— (Art. XXIV.) ; hence,
AB+BC : AB— BC : : tang ^±^ : tang "IZH^, which is
fi <«
the property we had to demonstrate.
With the aid of these five theorems we can solve all the
cases of rectilineal trigonometry.
Scholium. The required part should always he found from
the given parts ; so that if an error is made in any part of the
work, it may not affect the correctness of that which follows.
SOLUTION OF RECTILINEAL TRIANGLES BY MEANS OF
LOGARITHMS.
It has already been remarked, that in order to abridge the
calculations which are necessary to find the unknown parts of
a triangle, we use the logarithms of the parts instead of the
parts themselves.
Since /the addition of logarithms answers to the mqltiphca-
tion of tneir corresponding numbers, and their subtraction to
the division of their number^; it follows, that the logarithm of
the fourth term of a proportion will be equal to the sum of
the logarithms of the second and third terms, diminished by
the logarithm of the first term.
Instead, however, of subtracting the logarithm of the first
term from the sum of the logarithms of the second and third
terms, it is more convenient to use the arithmetical complement
of the first term.
/The arithmetical complement of a logarithm is the number
which remains after subtracting the logarithm from 10. J Thus
10 — 9.274687 = 0.725313: hence, 0.725313 is the arithmetical
complement of 9.274687. X
242 PLANE TRIGONOMLrUY.
It is now to be shown that, the difference hcfvhen rvo V/'if
r'lthms is truly founds hij adding to the first logarxihm ih-° ^'rith-
meticdl complement of the logarithm to he subtracledi end dimin-
ishing their sum by 10.
Let a =z the first logarithm.
b = the logarithm to be subtracted.
c = 10 — b=xhe arithmetical complement of 6,
Now, the difference between the two logarithms will be
expressed by a — b. But from the equation c=10 — b, we have
c — 10= — b : hence if we substitute for ^-b its value, we shall
have
a — b=a-{-c — 10,
which agrees with the enunciation.
/ When we wish the arithmetical complement of a logarithm,
/we may write it directly from the tables, by subtracting the
/ left hand figure from 9, then proceeding to the right, subtract
f each figure from 9, till we reach the last significant figure, which
\ must be taken from 10 : this will be the same as taking the
logarithm from 10.
Ex. From 3.274107 take 2.104729.
Common method. By ar.-comp.
3.274107 3.274107
2.104729 ar.-comp. 7.895271
Diff. 1.169378 sum 1.169378 after re-
jecting the 10.
We therefore have, for all the proportions of trigonometry,
tlie following
RULE.
Add together the arithmetical complement of the logarithm of the
the first term, the logarithm of the second term, and the loga-
rithm of the third term, and their sum after rejecting 10, will
he the logarithm of the fourth term.. And if any expression
occurs in which the arithmeticcd com,plement is twice used, 20
must be rejected fivm the sum.
i
c
PLANE TRIGONOMETRY. 243
SOLUTION OF RIGHT ANGLED TRIANGLES.
Let A be the right angle of the proposed
right angled triangle, B and C the other two
angles ; let a be the hypothenuse, b the side
opposite the angle B, c the side opposite the
angle C. Here we must consider that the ^
two angles C and B are complements of each other ; and that
consequently, according to the different cases, we are entitled
to assume sin C^=cos B, sin B=cos C, and likewise tang B=
coi C, tang C= cot B. This being fixed, the unknown parts
of a right angled triangle may be found by the first two theo-
rems; or if two of the sides are given, by means of the pro-
perty, that the square of the hypothenuse is equal to the sum
of the squares of the other two sides.
EXAMPLES. '^
Ex. 1. In the right angled triangle BCA, there are given t!:e
hypothenuse (2=250, and the side 6=240 ; required the other
parts.
R : sin B : : a : 6 (Theorem L).
or, « : 6 : : R : sin B.
• When logarithms are used, it is most convenient to write tlie
proportion thus.
As hyp. a - 250 ' - ar.-comp. log. - 7.602060
To side/; - 240 2.380211
So is R 10.000000
To sin B - 73° 44' 23" (after rejecting 10) 9.982271
But the angle C=90°— B=90^— 73° 44'23"=16° 15' 37"
or, C might be found by the proportion,
As hyp. a - 250 - ar.-comp. log. - 7.002060
To side /? - 240 - - 2.380211
So is R - - lO.OOOOGO
To cos C - 16° 15' 37" 9.982271
To find the side c, we say,
As R - - ar. comp. log. - 0.000000
To tang. C 16° 15' 37" - - - - 9.464889
So is side 6 240 ; - - - 2.380211
To side c 70.0003 - - - 1.845100
214 PLANE TRIGONOMETRY.
Or the side c might be found from the equation
For, c^=a^—h''={a-\-h)x{a—b):
hence, 2log. c=log. (a+t) + log. (a — ^, or
log. c=Jlog. {a-\~h) ^l\og. (a-^b)
a4-5=250 + 240=490 log. 2.690196
a— &=250— 240=10 - - 1.000000
2 ) 3.690196
Log. c 70 ---..- - 1.845098
Ex. 2. In the right angled triangle BCA, there are given,
side 6=384 yards, and the angle B=53° 8' : required the other
parts.
To find the third side c.
R : tang B : : c : b (Theorem II.)
or, tangB : R : : h : c. Hence,
As tang B 53° 8' ar.-comp. log. 9.875010
is to R ... - - 10.000000
So is side b 384 - - - - . 2.584331
To side c 287.965 - - - - "2^459341
Note. When the logarithm whose arithmetical complement
is to be used, exceeds 10, take the arithmetical complement
with reference to 20 and reject 20 from the sum.
To find the hypothenuse a.
R : sin B : : a : 6 (Theorem I.). Hence,
As sin B 53° 8' ar. comp. log. 0.096892
Is to R - - - - - 10.000000
So is side 6.384 - - - - 2.584331
To hyp. a 479.98 - - - 2.681223
Ex. 3. In the right angled triangle BAG, there are given,
side c=195, angle B=47° 55',
required the other parts.
Ans. Angle G=42° 05', a==290.953^ 6=215.937.
SOLUTION OF RECTILINEAL TRIANGLES IN GENERAL.
Let A, B, G be the three angles of a proposed rectilineal tri-
angle ; ff, 6, c, the sides which are respectively opposite them ;
the different problems which may occur in determining three of
these quantities by means of the other three, will all be redu-
cible to the four following cases.
TLANE TRIGONOMETRY. 245
CASE I.
Given a side and two angles of a triangle, to find the remaining
parts.
First, subtract the sum of the two angles from two right an-
gles, the remainder will be the third angle. The remaining
sides can then be found, by Theorem 111.
I. In the triangle ABC, there are given the angle A =58° 07',
the angle B = 22° 37', and the side 07= 408 yards: required ih^
remaining angle and the two other sides.
To the angle A =58° OT
Add the angle B - ... - =22° 37'
Their sum - - - - - - =80° 44'
taken from 180° leaves the angle C - =99° 10'.
This angle being greater than 90° its sine is found by taking
that of its supplement 80° 44'.
To find the side a.
As sine C
Is to sine A
So is side c
99° 16'
58° 07'
408 -
ar.-comp.
log.
0.005705
9.928972
2.610660
So side a
351.024
-
-
2.545337
To find the side h.
As sine C
Is to sine B
So is side c
To side h
99° 16'
22° 37'
408 -
158.976
ar.-comp.
log.
0.005705
9.584968
2.610660
2.201333
2. In a triangle ABC, there are given the angle A = 38° 25'
B = 57° 42', and the side c=400 : required the remaining
parts.
Ans, Angle C=83° 53', side a=249.974, side &=340.04.
CASE II. Y
Given two sides of a triangle, and an angle opposite one of them,
to find the third side and the two remaining angles,
X*
246
PLANE TRIGONOMETRY.
1 . In the triangle ABC, there
are given side AC = 21 6, BC=:
117, and the angle A =22° 37',
to find the remaining parts.
Describe the triangles ACB,
ACB', as in Prob. XI. Book III.
Then find the angle B by-
Theorem III.
As sideB'C or BC 117
Isioside AC 216
So is sine A 22° 37'
To sine B' 45° 13' 55" or ABC 134° 46' 05"
Add to each A 22° 37' 00 " 22° 37' 00' ^
Take their sum 67" 50 5'5 " ' 157° 23' 05"
From 180° 00' 00" 180° 00' 00"
Rem. ACB' 112° 0905" ACB=22° 36' 55"
ar.-comp. log.
7.931814
2.334454
9.584968
9.851236
To find the side AB or AB'.
As sine A 22° 37' ar.-comp.
Is to sine ACB' 112° 09' 05" -
So is side B'C 117 -
To side AB' 281.785
log.
0.415032
9.966700
2.068186
2.449918
The ambiguity in this, and similar examples, arises in con-
sequence of the first proportion being true for both the trian-
gles ACB, ACB'. As long as the two triangles exist, the am-
biguity will continue. But if the side CB, opposite the given
angle, be greater than AC, the arc BB' will cut the line ABB',
on the same side of the point A, but in one point, and then
there will be but one triangle answering the conditions.
If the side CB be equal to the perpendicular Crf, the arc
BB' wil,l be tangent to ABB', and in this case also, there will
be but one triangle. When CB is less than the perpendicular
Cd, the arc BB' will not intersect the base ABB', and in that
case there will be no triangle, or the conditions are impossible.
2. Given two sides of a triangle 50 and 40 respectively, and
the angle opposite the latter equal to 32° : required the remain-
ing parts of the triangle.
-4715. If the angle opposite the side 50 be acute, it is equal
to 41° 28' 59", the third angle is then equal to 106° 31' 01", and
the third side to 72.368. If the angle opposite the side 50 be
obtuse, it is equal to 138° 31' 01", the third angle to 9° 28' 59',
and the remaining side to 12.436.
PLANE TRIGONOMETRY. 247
CASE III.
Given two sides of a triangle^ with their included angle, to find
the third side and the two remaining angles.
Let ABC be a triangle, B the given
angle, and c and a the given sides.
Knowing the angle B, we shall like-
wise know the sum of the other two an-
gles C + A=180°— B, and their half sum
J (C + A)=90— JB. We shall next A! b *C
compute the half difference of these two angles by the propor-
tion (Theorem V.),
c + a : c — a : : tang J (C + A) or cot J B : tang J (C — A,)
in which we consider c>a and consequently C>A. Having
found the half difference, by adding it to the half sum
^ (C + A), we shall have the greater angle C ; and by subtract-
ing it from the half-sum, we shall have the smaller angle A.
For, C and A being any two quantities, we have always,
Crz:i(C + A)+^(C— A)
A::z:i(C+A)-J(C-A).
Knowing the angles C and A to find the third side h, we liave
the proportion.
sin A : sin B : : a : &
Ex. L In the triangle ABC, let a =450, c=:540, and the in-
cluded angle B=z: 80"^ : required the remaining parts.
c + a— 990,c— a = 90, 180"— B = 100°=C + A.
Asc + a 990 ar.-comp. log. 7.004365
Is toe— a 90 1.954243
So is tang J (C + A) 50° -. - - 1 0.076187
To tangi (C— A) 6° 11' - - - "9. 0;i4795
Hence, 50° + 6° ll' = 56° ll'^C; and 50^—6° ir=43'' 49'
=A.
To find the third side b.
As sine A 43" 49' ar.-comp. log. 0.159672
Is to sine ^80° 9.993351
So is side a 450 - - . - 2. 653213
To side b 640.082 - - - . 2r8"06236
Ex. 2. Given two sides of a plane triangle, 1686 and 960,
and their included angle 128° 04': required the other parts.
Ans. Angles, 33° 34' 39 ", 18° 21' 21", side 2400.
248 PLANE TRIGONOMETRY.
CASE IV.
Given the three sides of a triangle, to find the angles.
We have from Theorem IV. the formula,
.in -J A=R^(li^(Z^ in which
p represents the half sum of the three sides. Hence,
sn.^A=w (P-'Y/-') , or
2 log. sin JA=2 log. R+log. (p — Z>)-l-log. (p — c) — log. c —
log. b,
Ex. 1. In a triangle ABC, let b=40, c=34, and a=25:
required the angles.
40 + 34 + 25 , ^
Here jo= =^49.5, p — 0=9.5, and p — c=15.5.
2 Log. R - - - - - - 20.000000
log. ip—h) 9.5 - - - - - 0.977724
log. (p-^c) 15.5 - - - - - 1.190332
— log. c 34 ar.-comp. - - 8.468521
— log. h 40 ar.-comp. - - 8.S97940
2 log. sin J A 19.034517
log. sin J A 19° 12' 39'' - - - 9-517258
Angle A=38° 25' 18".
In a similar manner we find the angle B=83° 53' 18" and
the angle C=57° 41' 24".
Ex. 2. What are the angles of a plane triangle whose sides
are, a=60, ^=50, and c=40?
Ans. 41° 24' 34", 55° 46' 16" and 82° 49' 10".
APPLICATIONS.
Suppose the height of a building AB were tequired, the
foot of it being accessible.
PLANE TRIGONOMETRY.
249
On the ground which we
suppose to be horizontal or very
nearly so, measure a base AD,
neither very great nor very
small in comparison with the
altitude AB ; then at D place
the foot of the circle, or what-
ever be the instrument, with
which we are to measure the
angle BCE formed by the hori-
zontal line CE parallel to AD, A I>
and by the visual ray direct it to the summit of the building.
Suppose we find AD or CErr67.84 yards, and the angle
BCE^=41" 04' : in order to find BE, we shall have to solve
the right angled triangle BCE, in which the angle C and the
adjacent side CE are known.
To find the side EB.
AsR
ar.-comp.
0.000000
Is to tang. C 41° 04' ". . . 9.940183
So is EC 67.84 1.831486
ToEB
59.111 1.771669
Hence, EB=59.111 yards. To EB add the height of the
instrument, which we will suppose to be 1.12 yards, we shall
then have the required height AB=60.231 yards.
If, in the same triangle BCE it were required to find the
hypothenuse, form the proportion
' As cos C 41° 04' ar.-comp. - - log. 0.122660
Is to R : - - - 10.000000
So is CE 67.84 1.831486
ToCB 89.98 ---..-... 1.954146
Note, If only the summit B of the building or place whose
height is required were visible, we should determine the dis-
tance CE by the method shown in the following example ;
this distance and the given angle BCE are sufficient for solv-
ing the right angled triangle BCE, whose side, increased by
the height of the instrument, will be the height required.
250
PLANE TRIGONOMETRY.
2. To find upon the ground
the distance of the point A
from an inaccessible object
B, we must measure a base
AD, and the two adjacent
angle? BAD, ADB. Sup-
pose we have found AD=
588.45 yards, BAD = 103°
55' 55", and BD A = 36° 04';
we shall thence get the third
angle ABD=40° 05", and to
obtain AB, we shall form the
proportion
As sine ABD 40° 05"
Is to sin BDA 36° 04'
So is AD 588.45
To AB - - 538.943
ar.-comp.
log.
- 0.191920
- 9.709013
- 2.769710
2.731543
If for another inaccessible object C, we have found the an-
gles CAD=35° 15', ADC = 119° 32', we shall in like manner
find the distance AC = 1201.744 yards.
3. To find the distance between two inaccessible objects C
and C, we determine AB and AC as in the last example ; we
shall, at the same time, have the included angle BAC = BAD —
DAC. Suppose AB has been found equal to 538.818 yards,
AC = 1201.744 yards, and the angle BAC = 08° 40' 55"; to
get BC, we must resolve the triangle BAC, in which are known
two sides and the included angle.
AsAC + AB 1740.562 ar.-comp. log.- 6.759311
Is to AC— AB 662.926 2.821465
P -4- C
So is tang.— ^ . 55° 39' 32" - - - - -10.165449
To tang.
B-
2
-C
29° 08' 19" 9.74u225
Hence - - - -
But wc have - -
B— C
2
B + C
:29° OS' 19
= 55° 39' 32'
Hence - - -*- - - B =84° 47' 51"
and C =26° 31' 13"
PLANE TRIGONOMETRY. 25J
Now, to find the distance BC make the proportion,
As sine B 84^ 47' 51" ar.-comp. - log. - 0.001793
Is to sine A 08'" 40' 55 ' 9.909218
So is AC 1201.744 - - 8.079811
To BC 1124.145 - - -. 3.050822 ^
4. Wanting to know the distance between two inaccessible
objects which lie in a direct line from the bottom of a tower
of 120 feet in height, the angles of depression are measured,
and found to be, of the nearest, 57° ; of the most remote,
25^ 30' ; required the distance between them.
Ans. 173.G5G feet.
5. In order to find the distance between two trees, A and
B, which could not be directly measured because of a pool
which occupied the intermediate space, the distance of a third
poi'.it C from each, was measured, viz. CA=588 feet and CB
=072 feet, and also the contained angle ACB=55° 40': requi-
red the distance AB.
Ans. 592.907 feet.
6. Being on a horizontal plane, and wanting to ascertain
the height of a tower, standing on the top of an inaccessibje
hill, there were measured, the angle of elevation of the top of
the hill 40^ and of the top of the tower 51° : then measuring
in a direct line 180 feet farther from the hill, the angle of ele-
vation of the top of the tower was 33° 45' : required the height
of *he tower.
Ans. 83.9983 feet.
7. Wanting to know the horizontal distance between two
inaccessible objects A and B, and not finding any station from
which both of them could be seen, two points C and D, were
chosen, at a distance from each other equal to 200 yards, from
the former of which A could be seen, and from the latter B,
and at each of the points C and D a staff was set up. From
C a distance CF was measured, not in the direction DC, equal
to 200 yards, and from D, a distance DE equal to 200 yards,
and the following angles were taken, viz. AFC=83° ACF=
54° 31', ACD=:53° 30 , BDC=156° 25', BDE=54° 30', and
BED =88° 30' : required the distance AB.
Ans. 345.46 yards.
8. From a station P there can be seen three objects. A, B
and C, whose distances from each other are known, viz. AB=
800, AC=:600, and BC=400 yards. There are also measured
the horizontal angles, A?C = 33° 45', BPC=22° 30'. It is re-
quired, from these data, to determine the three distances PA»
PC and PB.
Ans. PA=710.193, PC=:1042.522, PB=934.291 yards.
>
252 SPHERICxVL TRIGONOMETRY.
SPHERICAL TRIGONOMETRY.
I. It has already been shown that a spherical triangle is
formed by the arcs of three great circles intersecting each other
on the surface of a sphere, (Book IX. Def. 1). Hence, every
spherical triangle has six parts : the sides and three angles.
Spherical Trigonometry explains the methods of determin-
ing, by calculation, the unknown sides and angles of a spheri-
cal triangle wh-en any three of the six parts are given.
II. Any two parts of a spherical triangle are said to be of
the same species when they are both less or both greater than
90*^ ; and they are of different species when one is less and the
other greater than 90°.
III. Let ABC be a spherical JV
triangle, and O the centre of the /l\^\^
sphere. Let the sides of the tri- / |\ ^"^^
angle be designated by letters / || \ ^^^^^^
cori-esponding to their opposite / lV\ \ ^ ^T^^^
angles : that is, the side opposite B [ — rtrr^ ¥ — H"t^O
the angle A by a, the side oppo- \i-i/\ \ 4i ^y^
site B by i, and the side opposite \ '<:t"- """"j'ii y^
C by c. Then the angle COB \j\'-<\'ly^
will be represented by a, the an- \\ y^
gle CO A by h and the angle ^q
BOA by c. The angles of the
spherical triangle will be equal to the angles included between
the planes which determin.3 its sides (Book IX. Prop. VI.).
From any point A, of the edge OA, draw AD perpendicular
to the plane COB. From D draw DH perpendicular to OB,
and DK perpendicular to OC ; and draw AH and AK : the
last lines will be respectively perpendicular to OB and OC,
(Book VI. Prop. VI.)
The angle DH A will be equal to the angle B of the spheri-
cal triangle, and the angle DKA to the angle C.
The two right angled triangles OICA, ADK, will give the
proportions
R : sin AOK : : OA : AK, or, R x AKrr OA sin b.
R : sin AKD : : AK : AD, or, R x AD=AK sin C.
Hence, R^ x AD=AO sin h sin C, by substituting for AK its
value taken from the first equation.
SPHERICAL TRIGONOMETRY. 253
In like manner the triangles AHO, ADH, right angled at
H and D, give
R : sin c : : AO : AH, or R x AH= AO sin c
R : sin B : : AH : AD, or R x ADr=:AH sin B.
Hence, R^ x AD= AO sin c sin B.
Equating this with the value of R^ x AD, before found, and di-
viding by AO, we have
. ^ • . T. sin C sin c
sm b sm C=:sin c sm B, or - — ^=-^ — r ( 1 )
' sin B sin 6 ^ '
or, sin B : sin C : : sin 6 : sin c that is,
The sines of the angles of a spherical triangle are to each
other as t/ie sines of their opposite sides.
IV. From K draw KE perpendicular to OB, and from D draw
DF parallel to OB. Then will the angle DKF=:COB=a,
since each is the complement of the angle EKO.
In the right angled triangle OAH, we have
R : cos c : : OA : OH ; hence *
AO cos c=RxOH=RxOE+R.DF.
In the right-angled triangle OKE
R : cos a : : OK : OE, or Rx OE=OK cos a.
But in the right angled triangle OKA
R : cos 6 : : OA : OK, or, Rx OK=OA cos b,
vy TT TJ /^T^ r^ A COS « cos b
^ Hence RxOE=OA. ^
In the right-angled triangle KFD
R : sin « : KD : DF, or R X DF=KD sin a.
But in the right angled triangles OAK, ADK, we have
R : sin 6 : : OA : AK, or Rx AK=OA sin b
R : cos K : AK : KD, or R x KD- AK cos C
, T^Tx ^A sin b cos C
hence KD = ^ , and
OA sin a sin & cos C
K X Dr = j^2 : therefore *
OA cos « cos 6 AO sin a sin b cos C
OA cos c= ^ + =^2 1 ^
R^ cos c— R cos a cos &+sin a sin b cos C.
Y
254 SPHERICAL TRIGONOMETRY.
Similar equations may be deduced for each of the other
sides. Hence, generally,
R^ cos a=R cos b cos c+sin b sin c cos A.
R2 cos b=p, cos a cos c + sin a sin c cos B.
R^ cos c==R cos b cos a + sin b sin a cos C.
That is, radius square into the cosine of either side of a spheri-
cal triangle is equal to radius into the rectangle of the cosines of
the two other sides plus the rectangle of the sines of those sides
into the cosine of their included angle.
V. Each of the formulas designated (2) involves the three
sides of the triangle together with one of the angles. These
formulas are used to determine the angles when the three sides
are known. It is necessary, however, to put them under an-
other form to adapt them to logarithmic computation.
Taking the first equation, we have
R^ cos a — R cos b cos c
cos A=-
sin b sin c
Adding R to each member, we have
R2 cos a-f R sin b sin c — R cos 6 cos c
R+cos A:
sin b sin c
But, R + cos A^ p^^ (Art. XXIII.), and
■R sin b sin c— R cos b cos c= — R^ cos (6 + c) (Art. XIX.) ;
- 2 cos^^A R2 (cos a — cos(6 + c))
hence, g— = sin b sin c =
. ^^ sin| (a + Z> + c) sin^ (b^c-a) ^^^^^
sm sm c ^
Putting s = a + 5 + c, we shall have
i5=i(a+6+c) and is— a=J (&+c— a) : hence
cos i K=Yisy^^SMWSI^\
^ ▼ cm h cin r
sin b sin c
. ^ T» /^in i
»cos
sm a sm c
)■ (3-)
cos J C=R^— ^-^5^,-55^6
SPHERICAL TRIGONOMETRY.
255
Had we subtracted each member of the first equation from
R, instead of adding, we should, by making similar reductions,
have found
sm
sm
sm
/sin ^(a + b — c) sin -|- (a + c — b) ^
sin b sin c
/ sin l(a + b — c) sin ^ (b + c — a)
2^I3=Kv sin a sine
n4.)
sin a sin 6
Putting s=a+6 + c,we shall have
^s — a=^{b+c — a), ^ — 6=J {a+c — 6), and Js — c=^{a+b — c)
hence,
sin 1A=R. An ft^-^)^ina»-ft)
sin 6 sin c
sin ^R^R^/ sm (^s— c) sin ftg=^)
M5.)
sm a sm c
sin XCr=R. / sin (1^— ^) sin (^g— a)
^ sin a sin 6 -^
VI. We may deduce the value of the side of a triangle in
terms of the three angles by applying . equations (4.), to
' the polar triangle. Thus, if a'; ft', c', A', B', CVrepresent the
I sides and angles of the polar triangle, we shall have
A==180°— ff', B = 180°— &', C=180°— c' ;
a= 180°— A', Z>= 180°— B', and c= 180°— C
(Book IX. Prop. VII.) : hence, omitting the ', since the equa-
tions are applicable to any triangle, we shall have
cos
l.^R^ A"^ h (A + B-C) cos i (A + C-B) ^
sin B sin C
cos 1 h=R^ Ao^ i (A + B-C) cos I (B + C-A)
sin A sin C
cos I c==R^/' cos i (A+C— B) cos j (B + C— A)
pin A sin B.
(6.)
256
SPHERICAL TRIGONOMETRY.
Putting S=A + B + C, we shall have
JS— A=|(C + B--A), JS— B=:J(A+C— B) and
^S— C=i(A+B--C), hence
cosi
sin B sin C
cos U=R^ A'^^ gS— C) cos qS— A)
sin A sin C
cos
,^^^^^^ /cos gS-B) cos (xS
-A)
sin A sin B
HI-)
VII. If we apply equations (2.) to the polar triangle, we
shall have
— R2 cos A'=R cos B' cos C — sin B' sin C cos a\
Or, omitting the ', since the equation is applicable to any tri-
angle, we have the three symmetrical equations,
R^.cos A=sin B sin C cos a — R cos B cos C \
R^.cos B=sin A sin C cos b — R cos A cos C > (8.)
R^.cos C=sin A sin B cos c — R cos A cos B /
That is, radius square into the cosine of either angle of a sphe-
rical triangle, is equal to the rectangle of the sines of the two other
angles into the cosine of their included side, minus radius into the
rectangle of their cosines.
VIII. All the formulas necessary for the solution of spheri-
cal triangles, may be deduced from equations marked (2.). If
we substitute for cos b in the third equation, its value taken
from the second, and substitute for cos^ a its value R^ — sin^ a,
and then divide by the common factor R.sin a, we shall have
R.COS c sin flnsin c cos a cos B + R.sin b cos C.
T, ' ,. /, X . . , sin B sin c
But equation (1.) gives sin 6= -. — — — ;
hence, by substitution,
R cos c sin a=sin c cos a cos B + R.
Dividing by sin c, we have
n cose . T> , T>
R -; — sm a=cos a cos B + R
sin c
sin B cos C sin c
sin C
sin B cos C
sin C
(
SPHERICAL TRIGONOMETRY. 257
But, ^J-^ (Art. XVII.).
Sin It
Tlierefore, cot c sin a=cos a cos B + cot C sin U.
Hence, we may write the three symmetrical equations,
cot a sin &=cos b cos C + cot A sin C n
cot b sin c=cos c cos A + cot B sin A > (9.)
cot c sin arzco's a cos B + cot C sin B /
That is, in every spherical triangle, the cotangent of one of the
sides into the sine of a second side, is equal to the cosine of the se-
cond side into the cosine of the included angle, plus the cotangent
of the angle opposite the first side into the sine of the included
angle.
IX. We shall terminate these formulas by demonstrating
J^apier's Analogies, which serve to simplify several cases in the
; solution of spherical triangles.
If from the first and third of equations (2.), cos c be elimi-
nated, there will result, after a little reduction,
R cos A sin c=R cos a sin b — cos C sin a cos b.
By a simple pernlutation, this gives
R cos B sin c=R cos b sin a — cos C sin b cos a.
Hence by adding these two equations, and reducing, we shall
have
sin c (cos A + cos B)=(R — cos C) sin (a +6)
. sin c sin a sin 6
But smce - — 7^=-^^ — t = - — 5, we shall have
sm C sm A sm B
sin c (sin A'+sin B)=sin C (sin a+sin 6), and
sin c (sin A — sin B)=sin C (sin a — sin b).
Dividing these two equations successively by the preceding
one ; we shall have
sin A + sin B_ sin C sin cK + sin b
cos A + cos B~R — cos C ' sin {a-\-b)
sin A — sin B_ sin C sin a — sin b
cos A+cos B~R — cos C * sin (a +6)*
Y*
258 V SPHERICAL TRIGONOMETRY,
And reducing these by the formulas in Articles XXIII. and
XXIV., there will result -^
ta„gJ(A+B)=cotiC.^-^ii)^ \
cos^(a+i) jifj^
tangi(A-B)=cotJC.t"4^.
-^ Sin J (a + 6)
Hence, two sides a and h with the included angle C being
given, the two other angles A and B may be found by the
analogies,
cos^-(a+i) : cosJ(a! — h) : : cot^C : tangJ(A+B)
sin J (rt + 6) : sin \ (a — b) : : cot | C : tang J (A — B).
If these same analogies are applied to the polar triangle of
ABC, we shall have toput 180°— A', 180°— B', 180°— a', 180°- 6', •
180° — c', instead of a, 6, A, B, C, respectively; and for the result,
we shall have after omitting the ', these two analogies,
cosJ(A + B) : cosJ(A — B) : : tangjc : tangJ(flr+6)
sinJ(A + B) : sin^(A — B) : : tangjc : tang J (a — J),
by means of which, when a side c and the two adjacent angles.^
A and B are given, W3 are enabled to find the two other sides
a and b. These four proportions are known by the name of
Napier^s Analogies,
X. In the case in which there are given two sides and an
angle opposite one of them, there will in general be two solu-
tions corresponding to the two results in Case II. of rectilineal
triangles. It is also plain that this ambiguity will extend itselt
to the corresponding case of the polar triangle, that is, to the
case in which there are given tvv^o angles and a side opposite
one of them. In every case we shall avoid all false solutions
by recqllecting,
1st. That every angle, and every side of a spherical triangle
is less than 180°.
2d. That the greater angle lies opposite the greater side, and
the least angle opposite the least side, and reciprocally.
NAPIER'S CIRCULAR PARTS.
XI. Besides the analogies of Napier already demonstrated,
that Geometer also invented rules for the solution of all the
cases of right angled spherical triangles.
fl^ ^ "Spherical TRIGONOMETRY. 259
In every right angled spherical
triangle BAG, there are six parts :
three sides and three angles. If
we omit the consideration of the
right angle, which is always
known, there will be five remain-
ing parts, two of which must
. be given before the others can
I be determined.
f The circular parts, as they are called, arp the two sides c and h,
i about the right angle, the complements of the oblique angles B
and G, and the complement of the hypothenuse a. Hence there
are five circular parts. The right angle A not being a circular
part, is supposed not to separate the circular parts c and 6, so
that these parts are considered as adjacent to each other.
If any two parts of the triangle be given, their corresponding
circular parts will also be known, and these together with a
required part, will make three parts under consideration. Now,
these three parts will all lie together, or one of them will he sepa-
rated from both of the others. For example, if B and c were
given, and a required, the three parts considered would lie
together. But if B and G were given, and h required, the parts
would not lie together ; for, B would be separated from C by
the part a, and from h by the part c. In either case B is the
middle part. Hence, when there are three of the circular parts
under consideration, the middle part is that one of them to which
both of the others are adjacent, or from which both of them are
separated. In the former case the parts are said to be adjacent,
*' and in the latter case the parts are said to be opposite.
This being premised, we are now to prove the following
rules for the solution of right angled spherical triangles, which
it must be remembered apply to the circular parts, as already
defined.
1st. Radius into the sine of the middle part is equal to the rect-
angle of the tangents of the adjacent paints.
2d. Radius into the sine of the middle part is equal to the rect-
angle of the cosines of the opposite parts.
These rules are proved by assuming each of the five circu-
lar parts, in succession, as the middle part, and by taking the
extremes first opposite, then adjacent. Having thus fixed the
three parts which are to be considered, take that one of the
general equatigns for oblique angled triangles, which shall con •
tain the three corresponding parts of the triangle, together with
the right angle : then make A = 90^, and after making the reduc-
tions corresponding to this supposition, the resulting equation
will prove the rule for that particular case.
260 SPHERICAL TRIGONOMETRY.
For example, let comp. a be the middle part and the ex-
tremes opposite. The equation to be applied in this case must
contain a, b, c, and A. The first of equations (2.) contains these
four quantities : hence
R^ cos a=R cos b cos c+sin b sin c cos A.
If A=90° cos AznO ; hence
Rcos fl!=cos 6cosc;
that is, radius into the sine of the middle part, (which is the
complement of a,) is equal to the rectangle of the cosines of the
opposite parts.
Suppose now that the complement
of a were the middle part and the ex-
tremes adjacent. The equation to be
applied must contain the four quan-
tities fif, B, C, and A. It is the first
of equations (8.).
c
R^ cos A:=sin B sin C cos a — R cos B cos C.
Making A =90°, we have
sin B sin C cos fl=R cos B cos C, or
R cos a=cot B cot C ;
that is, radius into the sine of the middle part is equal to the
rectangle of the tangent of the complement -of B into the tan-
gent of the complement of C, that is, to the rectangle of the
tangents of the adjacent circular parts.
Let us now take the comp. B, for the middle part and the
extremes opposite. The two other parts under consideration
will then be the perpendicular b and the angle C. The equation
to be applied must contain the four parts A, B, C, and 6 : it is the
second of equations (8.),
R" cos B=sin A sin C cos b — R cos A cos C.
Making A = 90°, we have, after dividing by R,
R cos B = sin C cos b.
Let comp. B be still the middle part and the extremes adja-
cent. The equation to be applied must then contain the four
four parts a, B, c, and A. It is similar to equations (9.).
cot a sin c=cos c cos B 4- cot A sin B
But if A =90°, cot A =0 ; hence,
cot a sin cncos c cos B ; or .
R cos B=:cot a tang c.
SPHERICAL TRIGONOMETRY. 261
And by pursuing the same method of demonstration when each
circular part is made the middle part, we obtain the five fol-
lowing equations, which embrace all the cases,
R cos a=cos b cos c=cot B cot C"
R* cos Brrcos b sin C=cot a tang c
R cos C=cos csinB=cot a tang b Y O-^')
R sin Z>=sinasinB=tangccotC
R sin c=sin<2sinC=tang6cotB>
We see from these equations that, if the middle part is required
we must begin the proportion with radius ; and when one of the
extremes is required we must begin the proportion with the other
extreme.
We also conclude, from the first of the equations, that when
the hy pothenuse is less than 90°, the sides b and c will be of the same
species, and also that the angles B and C will likewise be of the
same sjiecies. When a is greater than 90°, the sides b and c will
be of different species, and the same will be true of the angles B
and C. We also see from the two last equations that a side and
its opposite angle will always be of the same species.
These properties are proved by considering the algebraic
signs which have been attributed to the trigonometrical lines,
and by remembering that the two members of an equation must
always have the same algebraic sign.
SOfejJTION OF RIGHT ANGLED SPHERICAL TRIANGLES BY
LOGARITHMS.
It is to be observed, that when any element is discovered in
the form of its sine only, there may be two values for this ele-
ment, and consequently two triangles that will satisfy the ques-
tion ; because, the same sine which corresponds to an angle or
an arc, corresponds likewise to its supplement. This will not
take place, when the unknown quantity is determined by means
of its cosine, its tangent, or cotangent. In all these cases, the
sign will enable us to decide whether the element in question is
less or greater than 90° ; the element will be less than 90°, if its
cosine, tangent, or cotangent, has the sign + ; it will be greater
if one of these quantities has the sign — .
In order to discover the species of the required element of
the triangle, we shall annex the minus sign to the logarithms of
all the elements whose cosines, tangents, or cotangents, are
negative. Then by recollecting that the product of the two
262 SPHERICAL TRIGONOMETRY. v ,
extremes has the same sign as that of the means, we can at once
determine the sign which is to be given to the required element*
and then its species will be known.
EXAMPLES.
1. In the right angled spherical tri-
angle BAG, right angled at A, there
are given «— 64° 40' and 6=42° 12':
required the remaining parts.
First, to find the side c.
c
The hypothenuse a corresponds to the middle part, and the
extremes are opposite : hence
R cos «=cos h cos c, or
As cos h 42° 12' ar.-comp. .log. 0.130296
Is to R - - - • - - - 10.000000
So is cos a 64^' 40' - - - - 9.631326
Tocos c 54° 43' 07" - - ' -. 9.761622
To find the angle B. ' •
The side h will be the middle part and the extremes oppo-
site : hence
R sin 6=cos (comp. a) x cos (comp. B)=sin a sin B.
As sin a 64° 40' ar.-comp. log. 0.043911
I§ to sin h 42° 12' - - - - 9.827189
Sois R - 10.000000
To sin B 48° 00' 14" ... - 9.871100
To find the angle C^
The angle C is the middle part and the extremes adjacent ;
hence
R cos C=cot a tang &.
As R .-• ar.-comp. log. 0.000000
Is to cot a 64° 40' - - - - ' 9.675237
So is tang 6 .42° 12' ... - 9.957485
To cos C 64° 34' 46;' - - - - 9.632722
2. In a right angled triangle BAG, there are given the hy-
pothenuse a=105° 34', and the angle B=80° 40' : required the
remaining parts.
SPHERICAL TRIGONOMETRY. 2G3
To find the angle C.
The hypothenuse
jacent : hence,
will be the middle part
and the extremes
R cos a=cot B cot C.
As cot
Is to cos
So is
B
a
R
80
105
^ 40' ar.-comp. log.
^34' -
0.784220 +
9.428717—
10.000000 +
To cot
C
148
= 30' 54" -
10.212937—
Since the cotangent of C is negative the angle C is greater
than 90°, and is the supplement of the arc which would corres-
pond to the cotangent, if it were positive.
To find the side c.
The angle B will correspond to the middle part, and the
extremes will be adjacent : hence,
R cos B=cot a tang c.
Ascot a 105° 34' ar.-comp. log. 0.555053 —
Is to R 10.000000 +
So is cos B 80° 40' . - - . 9.209992 +
To tang c 149° 47' 36" - - - 9/765 045-^
To find the side h.
The side h will be the middle part and the extremes oppo-
site: hence,
R sin 6= sin a sin B.
As R - ar. comp. log. - 0.000000
To sin a 105° 34' - . - . 9.983770
So is sin B 80° 40' - - - . 9.994212
To sin h 71°54' 33" .... 9.977982
OF QUADRANTAL TRIANGLES.
A quadrantal spherical triangle is one which has one of its
sides equal to 90°.
Let BAG be a quadrantal triangle
in which the side a =90°. If we pass
to the corresponding polar triangle,
we shall have A' = 180°— a =90°, B' =
180°— &, G' = 180°— c, a' = 180°— A,
5'=180°— B,c' = 180°— C; from which
we see, that the polar triangle will be
264 SPHERICAL TRIGONOMETRY. .
-# ,
right angled at A', and hence every case may be referred to
a right angled triangle.
But we can solve the quadrantal triangle by means of the
right angled triangle in a manner still more simple.
In the quadrantal triangle BAG, C
in which BC = 90°, produce the side A
CA till CD is equal to 90°, and con- / \
ceive the arc of a great circle to be y \ ,
drawn through B and D. Then C ^^ \
will be the pole of the arc BD, and ^^^^^ _^ A
the angle C will be measured by B^-s^Cl^ \
BD (Book IX. Prop. VI.), and the \.., ^ \b
angles CBD and D will be right an- "i^'---.. /
gles. Now before the remaining "D
parts of the quadrantal triangle can
be found, at least two parts must be given in addition to the
side BC = 90° ; in which case two parts of the right angled tri-
angle BDA, together with the right angle, become known.
Hence the conditions which enable us to determine one of these
triangles, will enable us also to determine the other.
3. In the quadrantal triangle BCA, there are given CB=r50°,
the angle C=42° 12', and the angle A=115° 20' : required the
remaining parts.
Having produced CA to D, making CD =90° and drawn the
arc BD, there will then be given in the right angled triangle
BAD, the side «=C=42° 12', and the angle BAD=180°—
BAC = 180°— 115° 20'=64°40',to find the remaining parts.
To find the side d.
The side a will be the middle part, and the extremes oppo-
site: hence,
R sin a=sinA sin d.
As sin A
Is to R
So is sin a
64° 40'
42° 12'
ar.-comp.
log. 0.043911
10.000000
9.827189
To sin d
48° 00' 14"
■
9.871100
To find the angle B.
The angle A will correspond to the middle part, and the ex-
tremes will be opposite : hence
R cos A=sin B cos a.
As cos a
Is to R
So is cos A
42° 12' ar.-comp.
64° 40'
log.
0.130^96
10.000000
9.631326
To sin B
35° 16' 53"
.
9.761622
SPHERICAL TRIGONOMETRY. 265
To find the side b:
The side b will be the middle part, and the extremes adja-
cent : hence,
R sin 6= cot A tang a.
As R
ar.-comp. log.
0.000000
Is to cot A
64° 40' . - - .
9.675237
So is tang a
42° 12' - . .. .
9.957485
To sin 6
25° 25' 14"
9.632722
Hence, CA=90°— 6=90°— 25° 25' 14" =64° 34' 46"
CBA=:90°— ABD = 90°— 35° 16' 53" =54° 43 07"
BA=^ . . - . =z48°00'15".
4. In the right angled triangle BAC, right angled at A, there
are given <z=115° 25', and c=60° 59' : required the remaining
parts.
( Bzr:148°56'45"
Ans. ) C= 75° 30' 33"
( 6 =152° 13' 50".
5.: In the right angled spherical triangle BAC, right angled
at A, there are given c= 116° 30' 43", and 6=29° 41' 32": re-
quired the remaining parts.
( C=:103° 52' 46"
^715. ) B= 32° 30' 22"
(a =112° 48' 58".
6. In a quadrantal triangle, there are given the quadrantal
side —90°, an adjacent side =115° 09', and the included angle
= 115° 55' : required the remaining parts.
( side, 113° 18' 19"
^^^' \ or.„lo. i 117° 33' 52"
) angles, ^ j^^, ^^, ^^,,^
SOLUTION OF OBLIQUE ANGLED TRIANGLES BY LOGARITHMS
There are six cases which occur in the solution of oblique
angled spherical triangles.
1. Having given two sides, and an angle opposite one of
them.
2. Having given two angles, and a side opposite one o(
them.
3. Having given the three sides of a triangle, to &<t tS^
•angles.
266 SPHERICAL TRIGONOMETRY.
4. Having given the three angles of a triangle, to find the
sides.
5. Having given two sides and the included angle.
6. Having given two angles and the included side.
CASE I.
Given two sides, and an angle opposite one of them, to find the re-
maining parts.
For this case we employ equation (1.) ;
As sin a : sin 5 : : sin A : sin B.
Ex. 1. Given the side fl=44°
13' 45", 6=:84° 14' 29" and the
angle A=32° 26' 07" : required
the remaining parts.
To find the angle B.
As sin a 44° 13' 45" ar.-comp. log. 0.156437
Is to sin h 84° 14' 29" - - - 9.997803
So is sin A 32° 26' 07" - - - 9.720445
To sin B 49° 54' 38" or sin E' 130° 5' 22" 9.88368 5
Since the sine of an arc is the same as the sine of its supple-
ment, there will be two angles corresponding to the logarithmic
sine 9.883685 and these angles will' be supplements of each
other. It does not follow however that both of them will satisfy
all the other conditions of the question. If they do, there will
be two triangles ACB', ACB ; if not, there will be but one.
To determine the circumstances under which this ambiguity
arises, we will consider the 2d of equations (2.).
R^ cos 6=R cos a cos c+sin a sin c cos B.
from which we obtain
_ R^ cos h — R cos a cos c
C0SB = :— : ; . '
sm a sm c
Now if cos h be greater than cos a, we shall have .
R^ cos fe>R cos a cos c,
or the sign of the second member of the equation will depend
on that of cos 6. Hence cos B and cos h will have the same
SPHERICAL TRIGONOMETRY. 2G7
sign, or B and h will be of the same species, and there will be
but one triangle.
But when cos &>cos a, sin 6<sin a : hence,
If the sine of the side opposite the required angle he less than
the sine of the other given side, there will he hut one triangle.
If however, sin 6>sin a, the cos h will be less than cos a,
and it is plain that such a value may then be given to c as to
render
R^ cos & < R cos a cos c,
or the sign of the second member may be made to depend on
cos c.
We can therefore give such values to c as to satisfy the two
equations
_, R^ cos h — R cos a cos c
H-cos B= —
;os B =
sm a sm c
R^ cos h — R cos a cos c
sin a sin c
Hence, if the sine of the side opposite the required angle he
greater than the sine of the other given side, there will he two tri-
angles which will fulfil the given conditions.
Let us, however, consider the triangle ACB, in which we are
yet to find the base AB and the angle C. We can find these
parts most readily by dividing the triangle into two right angled
triangles. Draw the arc CD perpendicular to the base AB :
then in each of the triangles there will be given the hypothe-
nuse and the angle at the base. And generally, when it is
proposed to solve an oblique angled triangle by means of the
right angled triangle, we must so draw the perpendicular that
it shall pass through the extremity of a given side, and lie oppo-
site to a given angle.
To find the angle C, in the triangle ACD.
As cot A 32° 26' 07" ar.-comp. log. 9.803105
Is to R 10.000000
So is cos 6 84'^ 14' 29" - - - 9.0 01465
To cot ACD 86^ 21' 06" - - - ¥.804570
To find tlio angle C in the triangle DCB.
As cot B 49° 54' 38" ar.-comp. log. 0.074810
Is to R 10.000000
So is cos a 44° 13' 45" - - - 9.855250
To cot DCB 49° 35' 38" - - - 9.930060
t Hence ACB=135o 56' 47'^
^08 SPHERICAL TRIGONOMETRY.
To find the side AB.
As sin A 32° 26' 07" ar.-comp. log. 0.270555
Is to sin C 135° 56' 47" - - - 9.842191
So is sin a 44° 13' 45" - - - 9.843563
Tosin c 115° 16' 29" - - - 9.956309
The arc 64° 43' 31", which corresponds to sin c is not the
value of the side AB : for the side AB must be greater than 6,
since it lies opposite to a greater angle. But 6 = 84° 14' 29" :
hence the side AB must be the supplement of 64° 43' 31", or
115° 16' 29".
Ex. 2. Given &=91° 03' 25", a=40° 36' 37", and A=35° 57'
15": required the remaining parts, when the obtuse angle B is
taken.
( B = 115°35'41"
^715. ) C= 58° 30' 57"
/ c = 70° 58' 52"
CASE II.
Having given two angles and a side opposite one oftliem^ to find
the remaining parts.
For this case, we employ the equation (l.^
sin A : sin B : : sin a : sin h.
Ex. 1. In a spherical triangle ABC, there are given the angle
A=:50° 12', B = 58° 8', and the side a=62° 42' ; to find the re-
maining parts.
To find the side h.
As sin A 50° 12' ar.-comp. log. 0.114478
Is to sin B 58° 08' - - - - 9.929050
So is sin a 62° 42' - - - - 9.948715
To sin h 79° 12' 10", or 100° 47' 50" 9.992243
We see here, as in the last example, that there are two arcs
corresponding to the 4th term of the proportion, and these arcs
are supplements of each other, since they have the same sine.
It does not follow, however, that both of ih^m will satisfy all
the conditions of the question. If they do, there will be two
liiangles ; if not, there will be but one.
SPHERICAI. TRIGONOMETRY. 269
To determine when there are two triangles, and also when
there is but one, let us consider the second of equations (8.)
R^ cos B=sin A sin C cos b — R cos A cos C, which gives
. R^cosB + R cos AcosC
cos 0= = A = Ti .
sm A sin C
Now, if cos B be greater than cos A we shall have
W cos B >R cos A cos C,
and- hence the sign of the second member of the equation will
depend on that of cos B, and consequently cos b and cos B will
have the same algebraic sign, or b and B will be of the same
species. But when cos B >cos A the sin B<sin A : hence
If the sine of the angle opposite the required side be less than
the sine of the other given angle, there will be but one solution.
If, however, sin B>sin A, the cos B will be less than cos A,
and it is plain that such a value may then be given to cos C, as
to render
R^cos B<R cos A cos C,
or the sign of the second member of the equation may be made
to depend on cos C. We can therefore give such values to C
as to satisfy the two equations
R^ cos B + R cos A cos C
+COS 6= -. — 7 — -. — Pi , and
sm A sm C
R^ cos B + R cos A cos C
cos 0= : -r—. -^ .
sm A sm C
Hence, if the sine of the angle opposite the required side he
greater than the sine of the other given angle there will be two
solutions.
Let us first suppose the side b to be less than 90°, or equa.
to 79° 12' 10".
If now, we let fall from the angle C a perpendicular on the
base BA, the triangle will be divided into two right angled tri-
angles, in each of which there will be two parts known besides
the right angle.
Calculating the parts by Napier's rules we find,
C=130°54'28"
c=119°03'26".
If we take the side 6=100° 47' 60", we shall find
C=156°15'06"
c^l52° 14' 18".
270 SPHERICAL TRIGONOMETRY.
Ex. 2. In a spherical triangle ABC there are given A=103°
59' 57", B=46° 18' 7", and a=42^ 8' 48" ; required the remain-
ing parts.
There will but one triangle, since sin B<sin A.
/ b =30°
Ans. ) C=36° 7' 54'-
} c =24° 3' 56".
CASE III.
Having given the three sides of a spherical triangle to find the
angles.
For this case we use equations (3.).
/sin J 5 sin {^s — a)
cos^A=RV sin 6 sine
Ex, 1. In an oblique angled spherical triangle there are
given fl=:56° 40', 6=83° 13' and c=114° 30' ; required the
angles.
l{a + h^c)=ls =127° 11' 30"
^(6 + c— a)=(^5— a)=70° 31' 30".
Log sin ^s 127° 11' 30" - - - 9.901250
log sin (|5— a) 70° 31' 30" - - - 9.974413
—log sin h 83° 13' ar.-comp. 0.003051
—log sin c 114° 30' ar.-comp. 0.0409 77
Sum 19.919691
Half sum =log cos \k 24° 15', 39" - - 9.959845
Hence, angle A=48° 31' 18".
The addition of twice the logarithm of radius, or 20, to the
numerator of the quantity under the radical just cancels the 20
which is to be subtracted on account of the arithmetical com-
plements, to that the 20, in both cases, may be omitted.
Applying the same formulas to the angles B and C, we find,
B= 62° 55' 46"
C = 125° 19' 02".
Ex. 2. In a spherical triangle there are given arr40° 18' 29",
6=67° 14' 28", and c~89° 47' 6" : required the three angles.
( A= 34° 22' 16"
Ans. ; Br= 53° 35' 16"
' { C = 119o 13' 32"
SPHERICAL TRIGONOMETRY.
:27J
CASE IV.
Having given the three angles of a spherical triangle, to find the
three sides.
For this case we employ equations (7.)
^ ^./cos(^S-B)cos(^S-C)
cos§a=Rv T> r '
sm 15 sin O
Ex. 1. In a spherical triangle ABC there are given A=48°
30', B-=125° 20', and C = 62° 54' ; required the sides.
J(A + B + C) = |S= 118° 22'
(iS-A) .
= 69° 52'
(iS-B) -
=_ 6° 58'
(^S-C) -
= -650 28'
Log cos (iS— B) —6° 58'
...
9.996782
log cos (JS— C) 55° 28'
-
9.753495
—log sin " B 125° 20'
ar.-comp.
0.088415
—log sin C 62° 54'
ar.-comp.
0.050506
Sura - - - -
-
19.889198
Half sum=log cos iA=28° 19'
48"
9.944599
Hence, side a=56° 39' 36".
In a similar manner we find.
6 = 1140 29' 58"
c— 83° 12' 06".
Ex. 2. In a spherical triangle ABC, there are given A= 109°
65' 42", B = 116° 38' 33", and 0=120° 43' 37" ; required the
three sides.
Ans.
a= 98° 21.' 40"
&=109° 50' 22"
c = 115° 13' 26"
CASE V.
Having given in a spherical triangle, two sides and their in-
cluded angle, to find the remaining parts.
272 SPHERICAL TRIGONOMETRY.
For this case we employ the two first of Napier's Analogies,
cos ^{a-{-b) : cos i(a — b) : : cot iC : tang i(A + B)
sin ^(a-\-b) : sin ^{a—b) : : cot ^C : tang i(A— B).
Having found the half sum and the half diflference of the
angles A and B, the angles themselves become known ; for, the
greater angle is equal to the half sum plus the half difference,
and the lesser is equal to the half sum minus the half diffe.
rence.
The greater angle is then to be placed opposite the greater
side. The remaining side of the triangle can then be found by
Case II.
Ex. 1. In a spherical triangle ABC, there are given a=68°
46' 2", b=ST 10', and 0=39=^ 23' ; to find the remaining parts.
^(a + &) = 52° 58' 1", i(a—b) = l5o 48' 1", JC = 19°41' 30".
As cos !(« + &) 52° 58' 1" log. ar.-comp. 0.220210
' Is to cos L(a—h) 15° 48' 1" - - - 9.983271
So is cot iC 19° 41' 30" - - - 10.446254
Totangi{A + B) 77° 22' 25" - - - 10.649735
As sin ^(a + b) 52° 58' 1" log. ar.-comp. 0.097840
Is to sin i{a—b) 15° 48' 1" - - - 9.435016
So is cot "iC 19° 41' 30" - - - 10.446254
Totangi(A— B) 43°37'21" - - - 9.979110
Hence, A=77° 22' 25" + 43° 37' 21"=120° 59' 46"
B=77° 22' 25"— 430 37' 21"= 33° 45' 04"
side c ' - - - = 43° 37' 37".
Ex, 2. In a spherical triangle ABC, there are given 6=:83''
19' 42':, c=23° 27' 46", the contained angle A=20° 39' 48";
to find the remaining parts.
( B = 156° 30' 16"
Ans, )C= 9° 11' 48"
) a= 61° 32' 12".
CASE VI.
In a spherical triangle^ having given two angles and the included
side to find the remaining parts.
SPHERICAL TRIGONOMETRY. 273
For this case we employ the second of Napier's Analogies,
cos J(A + B) : cos J (A — B) : : tang |c : tangj(a + fc)
sin^-(A + B) : sin J (A — B) : : tang Jc : tang J (a — b).
From which a and b are found as in the last case. The re-
maining angle can then be found by Case I.
Ex. 1. In a spherical triangle ABC, there are given A= 81°
38' 20", B=70° 9' 38", c=59° 16' 28" ; to find the remaining
parts.
J(A + B)=75° 53' 59",l(A— B)=5°44'21",^c=29° 38' 11".
As cos i{A + B) 75° 53' 59" log. ar.-comp. 0.613287
Tocos KA— B) 5° 44' 21" - - 9.997818
So is tang ic 29° 38' 11" - - 9.755051
To tang i(a + 6) 66°42'52" - - 10.366156
As sin ^(A+B) 75° 53' 59" log. ar.-comp. 0.013286
To sin 4(A— B) 5° 14' 21" - - 9.000000
So is tang ^c 29° 38' 11" - - 9.755051
To tang i{a—b) 3° 21' 25" - - 8.768337
Hence «=66o 42' 52" + 3° 21' 25"=70° 04' 17'"
6=66° 42' 52"-^3° 21' 25"=63° 21' 27"
angle C - - - =64° 46' 33".
Ex, 2. In a spherical triangle ABC, there are given A=34'*
15' 3", B=42° 15' 13", and 0=76° 35' 36" ; to find the remain-
ing parts.
( a =40° 0' 10"
Ans, H =50° 10' 30"
(C =58° 23' 41".
i 274 )
MENSURATION OF SURFACES.
The area, or content of a surface, is determined by finding
how many times it contains some other surface which is as-
sumed as the unit of measure. Thus, when we say that a
square yard contains 9 square feet, we should understand that
one square foot is taken for the unit of measure, and that this
unit is contained 9 times in the square yard.
The most convenient unit of measure for a surface, is a
square whose side is the hnear unit in which the linear dimen-
sions of the figure are estimated. Thus, if the linear dimen-
sions are feet, it will be most convenient to express the area in
square feet ; if the linear dimensions are yards, it will be most
convenient to express the area in square yards, &c.
We have already seen (Book IV. Prop. IV. Sch.), that the
term, rectangle or product of two lines, designates the rectan-
gle constructed on the lines as sides ; and that the numerical
value of this product expresses the number of times which the
rectangle contains its unit of measure.
PROBLEM I.
To find the area of a square, a rectangle, or a parallelogram.
Rule. — Multiply the base by the altitude, and the product will
be the area (Book IV. Prop. V.).
1. To find the area of a parallelogram, the base being 12.25
and the altitude 8.5. Ans. 104.125.
2. What is the area of a square whose side is 204.3 feet 1
Ans. 41738.49 sq.ft.
3. What is the content, in square yards, of a rectangle whose
base is 66.3 feet, and altitude 33.3 feet? Ans. 245.31.
4. To find the area of a rectangular board, whose length is
12J- feet, and breadth 9 inches. Ans. 9f sq.ft.
5. To find the number of square yards of painting m a par-
allelogram, whose base is 37 feet, and altitude 5 feet 3 inches
Ms. 21yV
PROBLEM ir.
To find the area of a triangle. '*
CASE L
When the base and altitude are given.
Rule. — ^Multiply the base by the aliiiude, and take half the
product. Or, multiply one of these dimensions by half the
of/icr (Book IV. Prop. VI.).
MENSURATION OF SURFACES. 275
1. To find the area of a triangle, whose base is 625 and alti-
tude 520 feet. Ans. 162500 sq.ft.
2. To find the number of square yards in a triangle, whose
base is 40 and altitude 30 feet. Ans, 66|.
3. To find the number of square yards in a triangle, whose
base is 49 and altitude 25^ feet. ^715. 68.7361.
CASE II.
When two sides and their included angle are given.
Rule. — Add together the logarithms of the two sides and the
logarithmic sine of their included angle ; from this sum sub-
tract the logarithm of the radius, which is 10, and the remain-
der will he the logarithm of double the area of the triangle.
Find, from the table, the number answering to this logarithm,
and divide it by 2 ; the quotient will be the required area.
Let BAG be a triangle, in which there
are given BA, BC, and the included an-
gleB.
From the vertex A draw AD, perpen-
dicular to the base BC, and represent the
area of the triangle by Q. Then,
R : sin B : : BA : AD (Trig
hence, ^p^ BAxsinB
R
But, Q^J^GxAD (Book IV. Prop. VI.) ;
.^
hence, by substituting for AD its value, we have
Q_ BCxBAxsinB ^^ QO_ BCxBAxsin B
2R R
Taking the logarithms of both numbers, we have
log. 2Q=log. BC + log. BA + log. sin B— log. R;
which proves the rule as enunciated.
1. What is the area of a triangle whose sides are, BC=:
125.81, BA=57.65, and the included angle B = 57° 25'?
' +log. BC 125.81 .... 2.099715
T-Uor. i^„ on J +log. BA 57.65 .... 1.760799
Then, log. 2Q= ^ ^j^| ^.^ g ^^^ 25' 9.925626
—log. R - —10.
log. 2Q 3.786140
and 2Q=6111.4, or Q=: 3055.7, the required area.
276 MENSURATION OF SURFACES.
2. What is the area of a triangle whose sides are 30 and 40,
and their included angle 28° 57' '( Ans. 290.427.
3. What is the number of square yards in a triangle of which
the sides are 25 feet and 21.25 feet, and their included angle
45°? . . Ans. 20.8694.
CASE III.
When the three sides are known.
Rule. — 1. Add the three sides together , and take half their sum.
2. From this half-sum subtract each side separately.
3. Multiply together the half-sum and each of the three re-
mainders, and the product will he the square of the area oj
the triangle. Then, extract the square root of this product,
for the required area.
Or, After having obtained the three remainders, add together the
logarithm of the half -sum and the logarithms of the respective
remainders, and divide their sum by 2 : the quotient will be
the logarithm of the area.
Let ABC be the given triangle. ^C
Take CD equal to the side CB, and
draw DB; draw AE parallel to DB, ly
meeting CB produced, in E : then ,''' /x \y
CE will be equal to CA. Draw / I>/-".^: -v\B
CFG perpendicular to AE and DB, / i{y
and it will bisect them at the points '.
G and F. Draw FHI parallel to k.<'' /^ J/
AB, meeting CA in H, and EA pro- '^ ,;A yG ' "^
duced, in I. Lastly, with the cen- K'
tre H and radius HF, describe the circumference of a circle,
meeting CA produced in K: this circumference will pass
through I, because AI=FB=FD, therefore, HF=H1 ; and it
will also pass through the point G, because FGI is a right
angle.
Now, since HA=HD, CH is equal to half the sum of the
sides CA, CB ; that is, CH=iCA+iCB; and since HK is
equal to iIF=iAB, it follows that
CK=iAC + iCB + iAB=iS,
by representing the sum of the sides by S.
Again, HK=HI=iIF=iAB, or KL=AB.
Hence, CL=CK-KL=iS— AB,
and AK=CK- CA=|S— CA,
and AL=DK=CK— CD=iS— CB.
Now, AG X CG= the area of the triangle ACE,
and AG x FG= the area of the triangle ABE ;
therefore, AG x CF=: the area of the triangle ACB.
MENSURATION OF SURFACES. 277
Also, by similar triangles,
AG : CG : : I)F : CF, or AT : CF ;
therefore, AG x CF=: triangle ACB-CG x DF=CG x AI ;
consequently, AGxCFxCGx AI= square of the area ACB.
But CGxCF=CKxCL-JS(iS— AB),
and AGx AI =AKx AL=(iS— CA) x (^S— CB) ;
therefoie,AGxCFxCGxAI -iS^S — AB) x GS — CA) x
QS — CB), which is equal to the square of the area of the
triangle ACB.
1. To find the area of a triangle whose three sides are 20,
30, and 40.
20 45 45 45 half-sum.
W 20 30 40
40 -- — —
. — 25 1st rem; 15 2d rem. 5 3d rem.
2)90
45 half-sum.
Then, 45 X 25 X 15 X 5=84375.
V The square root of which is 290.4737, the required area.
2. How many square yards of plastering are there in a tri-
angle whose sides are 30, 40, and 50 feet ? Ans. 66|.
PROBLEM III.
To find the area of a trapezoid.
Rule. — Add together the two parallel sides : then multiply their
sum by the altitude of the trapezoid, and half the product will
be the required area (Book lY. Prop. VII.).
1. In a trapezoid the parallel sides are 750 and 1225, and
the perpendicular distance between them is 1540 ; what is the
area? Ans. 152075.
2. How many square feet are contained in a plank, whose
length is 12 feet 6 inches, the breadth at the greater end 15
inches, and at the less end 11 inches? Ans. 13|^| sq.ft.
3. How many square yards are there in a trapezoid, whose
parallel sides are 240 feet, 320 feet, and altitude 66 feet ?
Ans. 2053^.
PROBLEM IV.
To find the area of a quadrilateral.
Rule. — Join two of the angles by a diagonal, dividing the quad*
rilateral into two triangles. Then, from each of the other
angles let fall a 'perpendicular on the diagonal : then multiply
A a
218
MENSURATIC»N OF SURFACES.
the diagonal by half the sum of the two perpendiculars^ and
the product wid be the area.
1. What is the area of the quad-
rilateral ABCD, the diagonal AC
being 42, and the perpendiculars
D^, B&, equal to 18 and 16 feet ?
Ans. 714.
2. How many square yards of paving are there in the quad-
rilateral whose diagonal is 65 feet, and the two perpendiculars
let fall on it 28 and 33^ feet ?
Ans. 222J
PROBLEM V.
To find the area of an irregular polygon.
Rule. — Draw diagonals dividing the proposed polygon into
trapezoids and triangles. Then find the areas of these
figures separately J and add them together for the content of
the whole polygon.
1. Let it be required to determine
the content of the polygon ABCDE,
having five sides.
Let us suppose that we have mea-
sured the diagonals and perpendicu-
lars, and found AC = 36.21, EC=:
39.11, B6=4, J)d=7,26, Aa-
4.18, required the area.
Ans. 296.1292.
PROBLEM VI.
To find the area of a long and irregular figure, bounded on
one side by a right line.
Rule. — 1. At the extremities of the right line measure the per-
pendicular breadths of the figure, and do the some at several
intermediate points,- at equal distances from each other.
2. Add together the intermediate breadths and haf the sum of
the extreme ones : then multiply this sum by one of the equal
parts of the base line : the j^roduct will be the required area,
very nearly.
Let AEert be an irregular figure, hav-
ing for its base the right line AE. At
the points A, B, C, D, and E, equally
distant from each other, erect the per-
pendiculars Aa, B6, Cc, Ddy Ee, to the
MENSURATION OF SURrACES. 27D
base line AE, and designate them respectively by the letters
a, h, c, 6?, and e.
Then, tlie area of the trapezoid ABZ?a= -'x AB,
the area of the trapezoid BCc6=— I— xBC,
c-{-d
the area of the trapezoid CDdc — x CD,
d-\ e
and the area of the trapezoid DEe<i= xDE ;
hence, their sum, or the area of the whole figure, is equal to
/« + 6 h-\-c c-\-d d+e\
since AB, BC, &c. are equal to each other. But this sum is
also equal to
(l. + b + c+d+l-)xAB,
\2 2/
which corresponds with the enunciation of the rule.
1. The breadths of an irregular figure at five equidistant
places being 8.2. 7.4, 9.2, 10.2, and 8.6, and the length of the
base 40, required the area.
8.2 4)40
8.6 —
2(16.8
10 one of the equal parts.
8.4 mean of the extremes.
7.4 35.2 suni.
9.2 10
10.2
352= area.
35.2 sum.
2. The length of an irregular figure being 84, and the
breadths at six equidistant places 17.4, 20.6, 14.2, 16.5, 20.1.
and 24,4; what is the area? Ans. 1550.64.
PROBLEM VII.
To find the area of a regular polygon.
Rule I. — Multiply half the perimeter of the pohjgon hy the
apotheni, or perpendicular let fall from the centre on one of
the sides, and the product will be the area required (Book V,
Prop. IX.).
280
MENSURATION OF SURFACES.
Re.ma.rk I. — The following is the manner of detcrmmmg
the perpendicular when only one side and the number of sides
of tiie regular polygon are known : —
First, divide 3G0 degrees by the number of sides of the poly-
gon, and the (juotient will be the angle at the centre ; that is,
the angle subtended by one of the ecjual sides. Divide this
angle by 2, and half the angle at the centre will then be known.
Now, the line drawn from the centre to an angle of the
polygon, the perpendicular let fall on one of the equal sides,
and half this side, form a right-angled triangle, in which there
are knowii, the base, which is half the equal side of the poly-
gon, and the angle at the vertex. Hence, the perpendicular
can be determined.
1. To find the area of a regular hexa-
gon, whose sides are 20 feet each.
G)3G0°
eO°=:ACB,the angle at the centre.
30°=ACD, half the angle at the centre
Also, CAD=90°— ACD=60°; and AD = ]0.
Then, as sin ACD . . . 30% ar. comp 0.301030
: sin CAD ... 00^ 9.93753J
•: AD 10 1.000000
: CD . . . 17.3205 '. . . 1.238561
Perimeter =120, and half the perimeter =60.
Then, 60 X 17.3205 = 1039.23, the area.
2. What is the area of an octagon whose side is 20 ?
Ans. 1931.36886.
Remakk TI. — The area of a regular polygon of any number
of sides is easily calculated by the above rule. Let the area?
of the regular polygons whose sides are unity, or 1, be calcu-
lated and arranged in the following
MENSURATION OF SURFACEvS. 281
TABIE.
nics. Sides. Areas.
Triangle . .
3
0.4330127
S(|iiare . .
. 4
1.0000000
Pentagon . .
5
1.7204774
Hexagon . . .
. 6
2.59807G2
Heptagon . .
. 7
3.G339124
Octagon . .
. 8
4.8284271
Nonagon , .
. 9
0.1818242
Decagon . .
. 10
7.0942088
Undecagon .
. 11
9.3G50399
Dodecagon . .
. 12
11.1901524
Now, since the areas of similar polygons ai^e to each other
as tlie scjuares of their homologous sides (Book IV. Prop.
XXVII.), we shall have
1- : tabular area : : any side squared : area.
Or, to find the area of any regular polygon, we have
Rl'le II. — 1. Square the side of the polygon.
2. Then multiply that square hy the tabular area set opposite
the polygon of the same number of sides, and the producl will
be the required area.
1. What is the area of a regular hexagon whose side is 20?
20- = 400, tabular area =2.5980702.
Hence, 2.5980762x400=1039.2304800, as before.
2. To find the area of a pentagon whose side is 25.
Ans. 1075.298375.
3. To find the area of a decagon whose side is 20.
Ans. 3077.G8352.
PROBLEM Vlir.
To find the circumference of a cir<:le when the diameter is
given, or the diameter when the circumference is given.
\\v\.T.. — Multiply the diameter by 3.14 IG, and the product will
he the circumference ; or, divide the circumference by 3.1410,
and the quotient will be the diameter.
It is shown (Book V. Prop. XIV.), that the circumference
of a circle whose diameter is 1, is 3.141592G, or 3.1410. But
since the circumferences of circles are to each other as their
radii or diameters, we have, by calling the diameter of tlie
second circle d,
1 : d :: 3.141G : circumference,
or, cZx 3.1410= circumference.
TT 1 J circumference
Hence, also. a=
3.1410
Aa2
282 MENSURATION OF SURFACES.
1. What is the circumference of a circle whose diameter
is 25 ? .4715. 78.54.
2. If the diameter of the earth is 7921 miles, what is the
circumference? Ans. 24884.()136.
3. Wliat is the diameter of a circle whose circumference is
11652.1904? Ans. 37.09.
4. What is the diameter of a circle whose circumference is
6850? ' Ans. 2180.41.
PR^OBLEM IX
To find the length of an arc of a circle containing any numbei
of degrees.
Rule. — Multiply the numher of degrees in the given arc hy
0.0087266, and the product hy the diameter of the circle.
Since the circumference of a circle whose diameter is 1, is
3.1416, it follows, that if 3.1416 bo divided by 360 degrees,
the quotient will be the length of an arc of 1 degree : that is,
^•^'^^"=0.0087266=: arc of one degree to the diameter 1.
360
This being multiplied by the number of degrees in an arc, the
product will be the length of that arc in the circle whose diam-
eter is 1 ; and this product being then multiplied by the diam-
eter, will give the length of the arc for any diameter whatever.
Remark. — When the arc contains degrees and minutes, re-
duce the minutes to the decimal of a degree, which is done by
dividing them by 60.
1. To find the length of an arc of 30 degrees, the diameter
being 18 feet. Ans, 4.712364.
2. To find the length of an arc of 12° 10', or 12^°, the diam-
eter being 20 feet. Ans. 2.123472.
3. What is the length of an arc of 10° 15', or 10^°, in a cir-
cle whose diameter is 68 ? Ans. 6.082396.
PROBLEM X.
To find the area of a circle.
Rule I. — Multiply the circumference hy half the radius (Book
V. Prop. XII.).
Rule II. — Multiply the square of the radius hy S.14l^ (Book
V. Prop. XII. Cor. 2).
1. To find the area of a circle whose diameter is 10 and
circumference 31.416. Ans. 78.54.
MENSURATION OF SURFACES. 283
2. Find thp area of a circle whose diameter is 7 and cir-
cunriference 21.9912. Ans. 38.4846.
3. How many square yards in a circle whose diameter is
3^- feet? Ans. 1.069016.
4. What is the area of a circle whose circumference is 12
feet? Ans. 11.4595.
PROBLEM XI.
To find the area of the sector of a circle.
Rule T. — Multiply the arc of the sector hy half the radius (Book
V. Prop. XII. Cor. l).
Rule II. — Compute the area of the whole circle: then say^ as
360 decrees is to the degrees in the arc of the sector, so is the
area of the whole circle to the area of the sector.
1. To find the area of a circular sector whose arc contains
18 degrees, the diameter of the circle being 3 feet.
Ans. 0.35343.
2. To find the area of a sector whose arc is 20 feet, the
radius being 10. Ans. 100.
3. Required the area of a sector whose arc is 147° 29', and
radius 25 feet. * Ans. 804.3980.
PROBLEM XIL
To find the area of a segment of a circle.
Rule. — 1. Find the area of the sector having the same arc, hy
the last problem.
2. Find the area of the triangle formed hy the chord of the
segment and the two radii of the sector.
3. Then add these two together for the answer when the seg-
ment is greater than a semicircle, and subtract' them when it
is less.
1. To find the area of the segment p •
ACB, its chord AB being 12, and the ^
radius EA, 10 feet. /^
AsEA lOar. comp. . . 9.000000 -^At"
: AD 6 0.778151 [ ^^^>
:: sinD 90^ 10.000000
sin AED 30° 52' = 36.87 9.778151
2
73.74 = the degrees in the arc ACB.
jiH«>.
184 MENSURATION OF SURFACES.
Then, 0.008726G x 73.74 x 20 = 12.87 = arc ACB, nearly.
5 "
G4.35 = areaEACB.
Again, VEA'^— AD" = V 100—36= n/G4=8=:ED;
and Gx8=:48 = the area of tlie triangle EAB.
Hence, sect. EACH— EAB = 64.35— 48 = iG.35 = ACB.
2. Find the area of the segment whose height is 18, the
diameter of the circle being 50. Ans. G3G.4834.
3. Required the area of the segment whose chord is IG, the
diameter being 20. Ans. 44.7G4.
PROBLEM XIII. ♦
To find the area of a circular ring: that is, the area included
between the circumferences of two circl-es which have a
common centre.
Rule. — TaJie the difference between the areas of the two circles.
Or, subtract the square of the less radius from the square of
the greater i and multiply the remainder by 3. HI 6.
For the area of the larger is R-t
and of the smaller r-'t
Their difference, or the area of the ring, is (R- — y-j-r.
1. The diameters of two concentric circles being 10 and G,
required the area of the ring contained between their circum-
ferences. Ans. 50.2G5G.
2. What is the area of the ring when the diameters of the
circles are 10 and 20? Ans. 235.G2.
PROBLEM XIV.
To find the area of an ellipse, or oval.*
Rule. — Multiply the two semi-axes tdgether^ and their product
by 3.1416.
r
1. Required the area of an ellipse
whose semi-axes AE, EC, are 35 and 25.
Ans. 2748.9.
* Although this rule, and the one for (he following prollem, cannot be de-
monstrated without the aid of principles not yet C(wiFi(lered, ^till it was tlion{jl»t
host to insert them, as they coinplelo tlic rules necessary for the mensuration
of planes.
MENSURATION OF SOLIDS. 285
2. Required the area of an ellipse whose axes are 24 and 18.
Ans. 339.2928.
PROBLEM XV.
To find the- area of any portion of a parabola.
Rule. — Multiply the base by the perpendicular height, and take
two-thirds of the product for the required area.
C
1. To find the area of the parabola
ACB, the base AB being 20 and the al-
titude CD, 18.
^715. 240.
/ ^ A
2. Required the area of a parabola, the base being 20 and
the altitude 30. Ans, 400.
MENSURATION OF SOLIDS.
The mensuration of solids is divided into two parts.
1st. The mensuration of their surfaces ; and,
2dly. The mensuration of their solidities.
We have already seen, that the unit of measure for plane
surfaces is a square whose side is the unit of length.
A curved line which is expressed by numbers is also referred
to a unit of length, and its numerical value is the number of
times which the line contains its unit. If, then, we suppose the
linear unit to be reduced to a right line, and a square con-
structed on this line, this square will be the unit of measure
for curved surface^/
The unit of solidity is a cube, the face r»f which is equal to
the superficial unit in which the surface of the solid is estimated,
and the edge is equal to the linear unit in which the linear di-
mensions of the solid are expressed (Book VII. Prop. XIII.
Sch.).
The following is a table of solid measures : —
1728
cubic inches
= 1 cubic foot.
27
cubic feet
= 1 cubic yard.
44921
cubic feet
= 1 cubic rod.
282
cubic inches
= 1 ale gallon.
231
cubic inches
= 1 wine gallon
2150.42
cubic inches
= 1 bushel.
280 MENSURATION OF SOLIDS.
OF POLYEDRONS, OR SURFACES BOUNDED BY PLANES.
PROBLEM I.
To find the surface of a right prism.
Rule. — Multiply the perimeter of the base hi/ the altitude, and
the product will be the convex surface (Book VII. Prop. I.).
To this add the area of the two bases, when the entire surface
is I'equired.
1. To find the surface of a cube, the length of each side
being 20 feet. Ans. 2400 sq.ft.
2. To find the whole surface of a triangular prism, whose
base is an equilateral triangle, having each of its sides equal
to 18 inches, and altitude 20 feet. Ans. 91.949.
3. What must be paid for lining a rectangular cistern with
lead at 2d. a pound, the thickness of the lead being such as to
require libs, for each square foot of surface ; the inner dimen-
sions of the cistern being as follows, viz. the length 3 feet 2
inches, the breadth 2 feet 8 inches, and the depth 2 feet C inches ?
^ , Ans. 21. 3s. lO'^d.
PROBLEM IL
To find the surface of a regular pyramid.
Rule. — Multiply the perimeter of the base by haf the slant
height, and the product will be the convex surface (Book VII.
Prop. IV.) : to this add the area of the base, when the entire
surface is required.
1. To find the convex surface of a regular triangular pyra-
mid, the slant height being 20 feet, and each side of tlie base
3 feet. Ans. 90 sq.ft.
2. What is the entire surface of a regular pyramid, w'hose
slant height is 15 feet, and the base a pentagon, of which each
side is 25 feet ? Ans. 2012.798.
PROBLEM III.
To find the convex surface of the frustum of a regular
pyramid.
Rule. — Multiply the half sum of the perimeters of the two
bases by the slant hei<rht of the fn/stum. and the product will
be the convex surface (Book VlL Prop. iV. Cor.).
MENSURATION OF SOLIDS. 287
1. How many square feet are there in the convex surface of
he frustum of a square pyramid, whose slant height is 10 feet,
each side of the lower base 3 feet 4 inclies, and each side of
the upper base 2 feet 2 inches? Ans. 110 sq. ft.
2. Wiiat is the convex surface of the frustum of an hepta-
gona^ pyramid whose slant iieight is 55 feet, each side of the
lower base 8 i'eet, and each side of the upper base 4 leet ?
Ans. 2310 s^. ft.
PROBLEM IV
To find the soli-dity of a prism.
Rule. — 1. Find the area of the base.
2. Muhip'y the area of the base by the altitude, and the pro-
duct will be the soliditij of the prism (Book VII. Prop. XIV.).
1. What is the solid content of a cube whose side is 24
inches? ' Ans. 13824.
2. How many cubic feet in a block of marble, of which the
length is 3 feet 2 inches, breadth 2 feet 8 inches, and height or
thickness 2 feet 6 inches? Ans. 21^.
3. How many gallons of water, ale measure, will a cistern
contain, whose dimensions are the same as in the last example ?
Ans. 129if
4. Required the solidity of a triangular prism, whose height
is 10 feet, and the three sides of its triangular base 3, 4, and 5
feet. Ans. 60.
PROBLEM V,
To find the solidity of a pyramid.
Rule. — Multiply the area of the base by one-third of the alti-
tude, and the product will be the solidity (Book VII. Prop.
XVII.).
1. Required the solidity of a square pyramid, each side of
its base being 30, and the altitude 25. Ans. 7500.
. 2. To find the solidity of a triangular pyramid, whose alti-
tude is 30, and each side of the base 3 feet. Ans. 38.9711.
3. To find the solidity of a triangular pyramid, its altitude
being 14 feet 6 inches, and the three sides of its base 5, 6, and
7 feet. Ans. 71.0352.
4. What is the solidity of a pentagonal pyramid, its altitude
being 12 feet, and each side of its base 2 feet ?
Ans. 27.5276.
5. What is the solidity of an hexagonal pyramid, whose alti-
tude is 6.4 feet, and each side of its base 6 inches ?
Ans. 1.38564.
2S8 MENSURATION OI- SOLIDS.
PROBLEM VI.
To find the solidity of the frustum of a pyramid.
Rule. — Add together the areas of the two bases of the frustum
and a mean proportional between them, and then multiply the
sum by one-third of the altitude (Book VII. Prop. XVllI.).
1. To find the number of solid feet in a piece of timber,
whose bases are squares, each side of the lower base being 15
inches, and each side of the upper base 6 inches, the altitude
being 24 feet. Ans. 19.5.
2. Required the solidity of a pentagonal frustum, whose alti-
tude is 5 feet, each side of the lower base 18 inches, and each
side of the upper base 6 inches. Ans. 9.31925.
Definitions.
1. A wedge is a solid bounded by five c? H
pian^ps : viz. a rectangle ABCD, called
the base of the wedge ; two trapezoids
ABHG, DCHG, which are called the
sides of the wedge, and which intersect
each other in the edge GH ; and the two
triangles GDA, HCB, which are called
the ends of tlie wedge.
When AB, the length of the base, is equal to GH, the trape-
zoids ABHG, DCHG, become parallelograms, and the wedge
is then one-half the parallelopipedon described on the base
ABCD, and having the same altitude with the wedge.
The altitude of the wedge is the perpendicular let fall from
any point of the line GH, on the base ABCD.
2. A rectangular prismoid is a solid resembling the frustum
of a quadrangular pyramid. The upper and lower bases are
rectangles, having their corresponding sides parallel, and the
convex surface is made up of four trapezoids. The altitude of
the prismoid is the perpendicular distance between its bases.
PROBLEM VIL
To find the solidity of a wedge.
Rule. — To twice the length of the base add the length of the
edge. Multiply this sum by the breadth of the base, and then
by the altitude of the wedge, and take one-sixth of the product
for the solidity.
MENSURATION OF SOLIDS.
289
the
length
Let L=AB,
the base.
/=GH, the length
the edge.
b=BC, the breadth of
the base.
/i=PG, the altitude
the wedge.
Then, L— Z=AB— GH =
AM.
Suppose AB, the length of the base, to be equal to GH, the
length of the edge, the solidity will then be equal to half the
parallelopipedon having the same base and the same altitude
(Book VII. Prop. Yll.y. Hence, the solidity will be equal
to iblh (Book VII. Prop. XIV.).'
If the length of the base is greater than that of the edge,
let a section MNG be made parallel to the end BCH. The
wedge will then be divided into the triangular prism BCH-M,
and the quadrangular pyramid G-AMND.
The solidity of the prism =^bhl, the solidity of the pyramid
-=^bh(h—l); and their sum, ibhl+^bh(L—r)=}bhSl+ibh2h
— J-fe/i2/=jM(2L + /).
If the length of the base is less than the length of the edge,
the solidity of the wedge will be equal to the difference be-
tween the prism and pyramid, and we shall have for the solid-
ity of the wedge,
yblil—^bh{l—L) =}bh2l—l bh2l+l bh2L=ibh{2L + /).
1. If the base of a wedge is 40 by 20 feet, the edge 35 feet,
and the altitude 10 feet, what is the solidity ?
Ans. 3833.33.
2. The base of a wedge being 18 feet by 9, the edge 20
feet, and the altitude 6 feet, what is the solidity ?
Ans, 504.
PROBLEM VIII.
To find the solidity of a rectangular prismoid.
Rule. — Add together the areas of the two bases and four times
the area of a parallel section at equal distances from the
bases : then multiply th$ sum by one-sixth of the altitude.
Bb
290 MENSURATION OF SOLIDS.
Let L and B be the length and
breadth of the lower base, / and h the
length and breadth of the upper base,
M and m the length and breadth of the
section equidistant from the bases, and
h the altitude of the prismoid.
Through the diagonal edges L and
t let a plane be passed, and it will di-
vide the prismoid into two wedges,
having for bases, the bases of the prismoid, and for edges the
lines L and l'=L J
The solidity of these wedges, and consequently of the pris- '
moid, is
But since M is equally distant from L and /, we have
2M=L + /, and 2m=:B + b',
hence, 4Mm=(L+/) x (B + 6)=BL + BZ+6L+R
Substituting 4Mm for its value in the preceding equation,
and we have for the solidity
i/i(BL+M+4Mm).
Remark. — This rule may be applied to any prismoid what-
ever. For, whatever be the form of the bases, there may be
inscribed in each the same number of rectangles, and the num-
ber of these rectangles may be made so great that their sum
in each base will differ from that base, by less than any assign-
able quantity. Now, if on these rectangles, rectangular pris-
moids be constructed, their sum will differ from the given pris-
moid by less than any assignable quantity. Hence the rule is
general.
1. One of the bases of a rectangular prismoid is 25 feet by
20, the other 15 feet by 10, and the altitude 12 feet ; required
the solidity. Ans. 3700.
2. What is the solidity of a stick of hewn timber, whose
endg are 30 inches by 27, and 24 inches by 18, its length being
24 feet? Ans. 102 feet.
OP THE MEASURES OP THE THREE ROUND BODIES,
PROBLEM IX.
To find the surface of a cylinder.
Rule. — Multiply the circumference of the base by the altitude,
and the product will be the convex surface (Book VIII. Prop.
I.). To this add the areas of the two bases , when the entire
surface is required.
MENSURx\TION OF SOLIDS. 291
I -
1. What is the convex surface of a cylinder, the diameter
of whose base is 20, and whose altitude is 50 ?
Ans. 3141.6.
S 2. Required the entire surface of a cylinder, whose altitude
is 20 feet, and the diameter of its base 2 feet.
^715. 131.9472.
■ PROBLEM X.
' To find the convex surface of a cone.
Rule. — Multiply the circumference of the base by half the side
\ (Book VIII. Prop. III.) : to which add the area of the base,
I when the entire surface is required.
f 1. Required the convex surface of a cone, whose side is 50
feet, and the diameter of its base 8^ feet. Ans. 667.59.
2. Required the entire surface of a cone, whose side is 36
and the diameter of its base 18 feet. Ans. 1272.348.
PROBLEM XI.
To find the surface of the frustum of a cone.
Rule. — Multiply the side of the frustum by half the sum of the
circumferences of the two bases, for the convex surface (Book
VIII. Prop. IV.) : to which add the areas of the two bases,
when the entire surface is required.
1. To find the convex surface of the frustum of a cone, the
side of the frustum being 12^ feet, and the circumferences of
the bases 8.4 feet and 6 feet. Ans. 90.
2. To find the entire surface of the frustum of a cone, the
side bemg 16 feet, and the radii of the bases 3 feet and 2 feet.
Ans. 292.1688.
PROBIiEM XII.
To find the solidity of a cylinder.
Rule. — Multiply the area of the base by the altitude (Book VIII.
Prop. II.).
1. Required the solidity of a cylinder whose altitude is 12
feet, and the diameter of its base 15 feet. Ans. 2120.58.
2. Required the solidity of a cylinder whose altitude is 20
feet, and the circumference of whose base is 5 feet 6 inches.
Arts. 48.144.
292 MENSURATION OF SOLIDS.
PROBLEM XIII.
To find the solidity of a cone.
Rule. — Multiply the area of the base by the altitude^ and take
one-third of the product (Book VIII. Prop. V.).
1. Required the soHdity of a cone whose aUitude is 27 feet,
and the diameter of the base 10 feet. Ans. 706.86.
2. Required the sohdity of a cone whose altitude is 10^ feet,
and the circumference of its base 9 feet. Ans. 22.56.
PROBLEM XIV.
To find the solidity of the frustum of a cone.
Rule. — Add together the areas of the two bases and a mean
proportional between them, and then multiply the sum by one-
third of the altitude (Book VIII. Prop. VI.).
1. To find the solidity of the frustum of a cone, the altitude
being 18, the diameter of the lower base 8, and that of the
upper base 4. Ans. 527.7888.
2. What is the solidity of the frustum of a cone, the altitude
being 25, the circumference of the lower base 20, and that of
the upper base 10? Ans. 464.216.
3. If a cask, which is composed of two equal conic frustums
joined together at their larger bases, have its bung diameter 28
inches, the head diameter 20 inches, and the length 40 inches
how many gallons of wine will it contain, there being 231 cubic
inches in a gallon ? Ans. 79.0613.
PROBLEM. XV.
To find the surface of a sphere.
-iix" Rule I. — Multiply the circumference of a great circle by the
diameter (Book VIII. Prop. X.).
Rule II. — Multiply the square of the diameter^ or four times
the square of the radius, by 3.1416 (Book VIII. Prop. X.
Cor.).
1. Required the surface of a sphere whose diameter is 7.
Ans. 153.9384.
2. Required the surface of a sphere whose diameter is 24
inches. Ans. 1809.5616 in.
3. Required the area of the surface of the earth, its diam-
eter being 7921 miles. Ans. 197111024 sq. miles.
4. What is the surface of a sphere, the circumference of its
great circle being 78.54? Ans. 1963.5.
MENSURATION OF SOLIDS. 293
PROBLEM XVI.
To find the surface of a spherical zone.
Rule. — Multiply the altitude of the zone by the circumference
of a great circle of the sphere, and the product will he the
surface (Book Vlll. Prop. X. Sch. 1).
1. The diameter of a sphere being 42 inches, what is the
convex surface of a zone whose altitude is 9 inches ?
Ans. USl[.5248sq.in,
2. If the diameter of a sphere is 12^ feet, what will be the
surface of a zone whose altitude is 2 feet?
Ans, 78.54 sq, ft,
PROBLEM XVII.
To find the solidity of a sphere.
Rule I. — Multiply the surface by one-third of the radius (Book
VIII. Prop. XIV.).
Rule II. — Cube the diameter, and multiply the number thus
found by ^rt ; that is, by 0.5236 (Book VIII. Prop. XIV.
Sch. 3).
1. What is the solidity of a sphere whose diameter is 12?
Ans. 904.7808.
2. What is the solidity of the earth, if the mean diameter
be taken equal to 7918.7 miles ? Ans, 259992792083.
PROBLEM XVIII.
To find the solidity of a spherical segment.
Rule. — Find the areas of the two bases, and multiply their sum
by half the height of the segment ; to this product add the-
solidity of a sphere whose diameter is equal to the height of
the seginent (Book VIII. Prop. XVII.).
Remark. — ^When the segment has but one base, the other is
to be considered equal to (Book VIII. Def. 14).
1. What is the solidity of a spherical segment, the diameter
of the sphere being 40, and the distances from the centre to the
bases, 16 and 10. Ans. 4297.7088.
2. What is the solidity of a spherical segment with one base,
the diameter of the sphere being 8, and the altitude of the
seffment 2 feet? Ans, 41.888
Bb2
294 MENSURATION OF SOLIDS.
3. What is the solidity of a spherical segment with one base,
the diameter of the sphere being 20, and the altitude of the
segment feet ? Ans. 1781.2872.
PROBLEM XIX.
To find the surface of a spherical triangle.
Rule. — 1. Compute the surface of the sphere on which the trian-
gle is formed, and divide it by 8 ; the quotient will be the sur-
face of the tri-rectangular triangle,
2. Add the three angles together ; from their sum subtract
180"^, and divide the remainder by 90^ : then multiply the tri-
rectangular triangle by this quotient, and the product will be
the surface of the triangle (Book IX. Prop. XX.).
1. Required the surface of a triangle described on a sphere,
whose diameter is 30 feet, the angles being 140"^, 92°, and 68°.
Ans. 471.24 sq.ft.
2. Required the surface of a triangle described on a sphere
of 20 feet diameter, the angles being 120° each.
Ans. Sl4:.l6 sq.ft.
PROBLEM XX.
To find the surface of a spherical polygon.
Rule. — 1. Find the tri-rectangular triangle, as before.
2. From the sum of all the angles take the product of two
right angles by the number of sides less two. Divide the re-
mainder by 90°, and. multiply the tri-rectangular triangle by
the quotient : the product will be the surface of the polygon
(Book IX. Prop. XXL).
1. What is the surface of a polygon of seven sides, de-
scribed on a sphere whose diameter is 17 feet, the sum of the
angles being 1080° ? Ans. 226.98.
2. What is the surface of a regular polygon of eight sides,
described on a sphere whose diameter is 30, each angle of the
polygon being 140° ? ^715.157.08.
OF THE REGULAR POLYEDRONS.
In determining the solidities of the regular polyedrons, it
becomes necessary to know, for each of them, the angle con-
tained between any two of the adjacent faces. The determi-
nation of this angle involves the following property of a regu-
lar polygon, viz. —
MENSURATION OF SOLIDS.
295
Half the diagonal wliich joins the extremities of two adjacent
sides of a regular polygon, is equal to the side of the polygon
multiplied hy the cosine of the angle which is obtained by di-
viding 360° by twice the number of sides : the radius being
equal to unity.
Let ABODE be any regular poly-
gon. Draw the diagonal AC, and from
the centre F draw FG, perpendicular
to AB. Draw also AF, FB ; the lat-
ter will be perpendicular to the diag-
onal AC, and will bisect it at H (Book
in. Prop. VI. Sch.).
Let the number of sides of the poly-
gon be designated by n : then,
AFB =??2!, and AFG =. CAB
360'=
n 2n
But in the right-angled triangle ABH, we have
AH=AB cos A-AB cos —- (Trig. Th. I. Cor.)
271
Remark 1 . — ^When the polygon in question is the equilateral
triangle, the diagonal becomes a side, and consequently half
the diagonal becomes half a side of the triangle.
The perpendicular BH=AB sin frl (Trig.
Remark 8.
Th. I. Cor.).
2n
To determine the angle included between the two adjacent
faces of either of the regular polyedrons, let us suppose a plane
to be passed perpendicular to the axis of a solid angle, and
through the vertices of the solid angles which lie adjacent.
This plane will intersect the convex surface of the polyedron
in a regular polygon ; the number of sides of this polygon will
be equal to the number of planes which meet at the vertex of
either of the solid angles, and each side will be a diagonal of
one of the equal faces of the polyedron.
Let D be the vertex of a solid angle,
CD the intersection of two adjacent faces,
and ABC the section made in the convex
surface of the polyedron by a plane per-
pendicular to the axis through D.
Through AB let a plane be drawn per-
pendicular to CD, produced if necessary,
and suppose AE, BE, to be the lines in
2*J6
MENSURATION OF SOLIDS.
which this plane intersects the adjacent
faces. Then will AEB be the angle in-
cluded between the adjacent faces, and
FEB will be half that angle, which we
will represent by ^A.
Then, if we represent by n the num-
ber of faces which meet at the vertex of j^
the solid angle, and by m the number of
sides of each face, we shall have, from what has
been shown,
300°
BF=:BC cos ??2!,
and EB=BC sin
2n
But
hence,
BF
2m
— =sin FEB = sin ^A, to the radius of unity ;
£B
cos
sin ^A=.
360°
~2n'
sm
360°
42'
This formula gives, for the plane angle formed by every two
adjacent faces of the
Tetraedron 70° 31'
Hexaedron . 90°
Octaedron . . . . . . . . 109° 28'
Dodecaedron 116°
Icosaedron 138°
18"
33' 54"
ir 23'
Having thus found the angle included between the adjacent
faces, we can easily calculate the perpendicular let fall from
the centre of the polyedron on one of its faces, when the faces
themselves are known.
The following table shows the solidities and surfaces of the
regular ()olyedrons, when the edges are equal to 1.
A TABLE OP THE REGULAR POLYEDRONS WHOSE EDGES ARE 1.
Names.
No. of Faces.
Tetraedron .
. . . . 4 . . .
Hexaedron .
... 6 . . .
Octaedron. .
... 8 . . .
Dodecaedron
. . . . 12 . . .
Icosaedron .
. . . 20 . . .
Surface. Solidity.
1.7320508 .... 0.1178513
6.0000000 . . . : 1.0000000
3.4641016 .... 0.4714045
20.0457288 .... 7.6631189
8.6602540 .... 2.1816950
MENSURATION OF SOLIDS 297
PROBLEM XXI.
To find the solidity of a regular polyedron.
Rule I. — Multiply the surface hy one-third of the perpendicular
let fall from the centre on one of the faces ^ and the product
will be the solidity.
Rule- II. — Multiply the cube of one of the edges by the solidity
of a similar polyedron, whose edge is 1.
The first rule results from the division of the polyedron into
as many equal pyramids as it has faces. The second is proved
by considering that tw^o regular polyedrons having the same
number of faces may be divided into an equal number of simi-
lar pyramids, and that the sum of the pyramids which make
up one of the polyedrons will be to the sum of the pyramids
which make up the other polyedron, as a pyramid of the first
sum to a pyramid of the second (Book II. 'Prop. X.) ; that is,
as the cubes of their homologous edges (Book VII. Prop. XX.) ;
that is, as the cubes of the edges of the polyedron.
1. What is the solidity of a tetraedron whose edge is 15?
lAns. 397.75.
2. What is the solidity of a hexaedron whose edge is 12?
Ans, 1728.
3. What is the solidity of a octaedron whose edge is 20 ?
Ans. 3771.236.
4. What is the solidity of a dodecaedron whose edge is 25 ?
Ans. 119736.2328.
5. What is the solidity of an icosaedron whose side is 20 ?
Ans, 17453.56.
V ^
A TABLE
OF
LOGARITHMS OF JTUMBERS
FROM 1 TO 10,000.
N.
IjOC.
N.
Lofi.
N.
l^f'K.
N.
Loff.
1
0.000000
26
1.414973
51
1.707570
76
1.880814
2
0.301030
27
1.431364
52
1.716003
77
1.886491
3
0.477121
28
1.447158
53
1.724276
78
1.892095
4
0.602060
29
1.462398
54
1.732394
79
1.897627
5
0.698970
30
1.477121
55
1.740.363
80
1.903090
6
0.778151
31
1.491362
56
1.748188
81
1 . 908485
7
0.845098
32
1.505150
57
1.7.55875
82
1.913814
8
0.903090
33
1.518514
58
1.763428
83
1.919078
9
0.954243
34
1.531479
59
1.770852
84
1.924279
If)
1.000000
35
1.544068
60
1.778151
85
1.929419
li
1.041393
36
1.556303
61
1.785330
86
1.934498
12
1.079181
37
1.. 568202
62
1.792392
87
1.939519
13
1.113943
38
1.579784
63
1.799341
88
1.944483
14
1.146128
39
1.591065
64
1.806180
89
1.949390
15
1.176091
40
1.602060
G5
1.812913
90
1.954243
16
1.204120
41
1.612784
66
1.819544
91
1.959041
17
1.230449
42
1.623249
67
1.826075
92
1.963788
18
1.255273
43
1.633468
08
1.832509
93
1.968483
19
1.278754
44
1.643453
69
1.838849
94
1.973128
"zO
1.3010.30
45
1.653213
70
1.845098
95
1.977724
21
1.322219
46
1.662758
71
1.851258
96
1.982271
22
1.342423
47
1.672098
72
1.8.57333
97
1.986772
23
1.361728
48
1.681241'
73
1.863323
98
1.991226
24
1.380211
49
1.690196
74
1.869232
99
1.995635
25
1.397940
50
1.698970
75
1.875061
100
2.000000
N.B. In the following table, in the last nine columns of each
page, Avhere the first or leading figures change from 9's to O's,
points or dots are introduced instead of the O's through the rest
Df the line, to catch the eye, and to indicate that from thence
the annexed first two figures of the Logarithm in the second
column stand in the next lower line.
1
A TABLE OF LOGARITHMS FROM 1 -£0 10,000.
IN. 1 1 1 1 2 1 3 i 4 i 5' i 6 1 7 1 8 1 9 1 D. I
100
OOOOOOi 0434
0868
1301
1734
2166
2598 1 3029
3461
3891
432
101
4321
4751
5181
5609
6038
6466
6S94
7321
7748
8174
428
102
8G00
9026
9451
9876
.300
.724
1147
1570
1993
2416
424
103
012837
3259
3680
4100
4531
4940
5360
5779
6197
6616
419
104
7033
7451
7868
8284
8700
9116
9532
9947
.361
.775
416
105
021189
1603
2016
2428
2841
3252
3664
4075
4486
4896
412
106
5306
5715
,6125
6533
6942
7350
7757
8164
8571
8978
408
107
9384
9789
.195
.600
1004
1408
1812
2216
2619
3021
404
108
033424
3826
4227
4628
5029
5430
5830
6230
6629
7028
400
109
110
7426
7825
1787
8223
2182
8620
2576
9017
2969
9414
3362
9811
3755
.207
4148
.602
4540
.998
4932
396
393
041393
111
5323
5714
6105
6495
6885
7275
7664
8053
8442
8830
389
112
9218
9606
9993
.380
.766
1 1.53
1638
1924
2309
2694
386
113
053078
3463
3846
4230
4613
4996
5378
5700
6142
6524
382
114
6905
7286
7666
8046
8426
8805
9185
9563
9942
.320
379
115
060698
1075
1452
1829
2206
2582
2958
3333
3709
4083
376
116
4458
4832
5206
5580
5953
6326
G699
7071
7443
7815
372
117
8186
8557
8928
9298
9668
..38
.407
.776
1145
1514
369
118
071882
2250
2617
2985
3352
3718
4085
4451
4816
5182
366
119
120
5547
5912
9543
6276
9904
6640
.266
7004
.626
7368
.987
7731
13"47
8094
1707
8457
2067
8819
2426
363
3§0
079181
121
082785
3144
3503
3861
4219
4576
4934
5291
5647
6004
357
122
6360
6716
7071
7426
7781
8136
8490
8845
9198
9552
355
123
9905
.258
.611
.963
1315
1667
2018
2370
2721
3071
351
124
093422
3772
4122
4471
4820
5169
6518
5866
6215
6562
349
125
6910
7257
7G04
7951
8298
8644
8990
9335
9681
..26
346.
126
100371
0715
1059
1403
1747
2091
2434
2777
3119
3462
343
127
3804
4146
4487
4828
5169
5510
5851
6191
6531
6871
340
128
7210
7549
7888
8227
8565
8903
9241
9579
9916
.253
338
129
130
110590
0926
4277
1263
4611
1599
4944
1934
2270
2605
5943
2940
6276
3275
6608
3809
6940
335
333
113943
5278
5611
131
7271
7603
7934
8265
8595
8926
9256
9586
9915
.245
330
132
120574
0903
1231
1560
188S
2216
2544
2871
3198
3525
328
133
3852
4178
4504
4830
5156
5481
5806
6131
6456
6781
325
134
7105
74291
7753
8076
8399
8722
9045
9368
9690
..12
823
135
130334
06551
0977
1298
1619
1039
2260
2580
2900
3219
321
136
3539
3858
4177
4496
4814
5133
5451
5769
6086
6403
318
137
6721
7037
7354
7671
7987
8303
8618 8934
9249
9564
315
138
9879
.194
.508
.822
1136
1450
1763
2076
2389
2702
314
139
140
143015
3327
6438
3639
6748
3951
7058
4263
7337
4574
7676
4885
7985
5196
8294
5507
8603
5818
8911
311
309
146128
141
9219
9527
9835
.142
.449
.756
1063
1370
1676
1982
307
142
152288
2594
2900
3205
3510
3815
4120
4424
4728
5032
305
143
5336
6640.
5943
6216
6549
6852
7154
7457
7759
8061
303
144
8362
8664
8965
9266
9567
9868
.168
.469
.769
1068
.301
U6
161368
1667
1987
2266
2564
2863
3161
3460
3758
4055
299
146
4353
4650
4947
5244
5541
5838
6134
6430
6726
7022
297
147
7317
7613
7908
8203
8497
8792
9086
9.380
9674
9968
295
148
170262
0555
0848
1141
1434
1726
2019 2311
2603
2895
293
149
150
3186
3478
6381
3769
6670
4060
6959
4351
7248
4541
7536
4932
7825
5222
8113
5512
8401
5802
8689
291
289
176091
151
8977
9264
9552
9839
.126
3270
.699
.985
1272
1558
287
152
181844
2129
2415
2700
2985
3555 3839
4123
4407
285
153
4691
4975
5259
5542
5825
6108
6391 6674
6956
7239
283
154
7521
7803
8084
8366
8647
8928
9209
9490
9771
..51
281
155
190332
0612
0892
1171
1451
1730
2010
2289
2567
2846
279
156
3125
3403
3681
3959
4237
4514
4792
5069
5346
5623
278
157
5899
6176
6453
6729
7005
7281
7556
7832
8107
8382
276
158
8657
8932
9206
9481
9755
..29
.303
.577
.8.50
1124
274
159
201397
1670
1943
2216
2488
2761
3033
3305
3577
3848
272
N. 1 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 1 D. 1
A TABLE OP LOGAKITIIMS FROM 1 TO 10,000.
M. 1 1 I I 2 1 3 1 4 1 5 1 G 1 7 1 8 1 9 1 n. 1
~lW
204120
4391
4663 4934
5204
5475
5740
6016; 6286
6556
271
IGl
6826
7096
7305 7634
7904
8173
8441
87101 8979
9247
269
102
9515
9783
..51 .319
.586
.853
1121
13S8| 1654
1921
26?
163
212 188
2454! 2720i 2986
3252
3518
3T83
4049J4314
4579 i 266 |
104
4844
5109
5373
5638
5932
6100
0430
6694' 6957
7221
264
165
7484
7747
8010
8273
8538
8798
9060
9323 9535
9846
262
lOfi
220108
0370
0631
0892
1153
1414
1075
1936 2196
2456
201
107
2716
2976
3236
3490
3755
4015
4274
4533 4792
.5051
259
168
6309
5568
5826
0084
0342
6600' 6858
71151 7372; 7630
258
169
7887
8144
8400
8057
8913
9170
9420
9082 9933
.193
256
170
23U449
07041 0960
1215
1470
1724
1979
2234 2488
2742
2.54
171
2996
3250
3504
3757
4011
4264
4517
4770 .5023
5276
253
172
5528
5781
6033
6285
0537
6789
7041
7292! 7544
7/95
252
173
8046
8297
8548
8799
9049
9299
9550
9800
1 ...50
.300
250
174
240549
0799
1048
1297
1546
1795
2044
2293
2541
2790
249
175
303S
3280
3534
3782
4030
4277
4525
4772
.5019
5266
248
176
5513
5759 1 6006
6252
6499
674-5
0991
7237
7482
7728
240
177
7973
8219 8464
8709
8954
9198
9443
9687
9932
.176
245
178
250420
0664' 0908
1151
1395
1638
1881
2125
2368
2610
243
179
2853
30'96i 33381 3580
3822
4004
4300
4548
4790
5031
242
180
255273
5514 5755
5996
6237
6477
6718
6958
7198
7439
241
181
7079
7918 8158
8398
8637
8877
9116
9355
9594
9833
239
182
260071
0310 0548
0787
1025
1263
1501
1739
1976
2214
238
183
2451
2688 2925
3162
3399| .3636
3873
4109
4346
4582
237
184
4818
5054 5290
5525
5761
5995
6232
6407
6702
6937
235
185
7172
7406 7641 7875
8110
8,344
8578
8812
9046
9279
234
186
9513
9746 9980
.213
.446
.079
.912
1144
1377
1609
233
187
271842
2074
2306
2538
2770
3001
3233
3464
3696
3927
232
188
4158
4389
4620
4850
5081
5311
5542
5772
6002
6232
230
189
190
6462
6692
8982
6921
92'l f
7151
9439
7380 7609
78381 8067i 8296
8525
.806
229
228
278754
9667
98951 .1231 ..351
.578
191
281033
1261
1488
1715
1942
2109 2396
2622
2849
3075
227
192
3301
3527
3753
3979
4205
4431 4656
4882
5107
5332
220
193
5557
5782
6007
6232
6456
6681 6905
7130
7354
7578
225
194
7802
8026
8249
8473
8696
8920
9143
9366
9589
9812
223
195
290035
0257
0480
0702
0925
1147
1369
1.591
1813
2034
222
196
2256
2478
269i»
2920
3141
3363
3584
3804
4025
4246
221
197
4466
4687
4907
5127
5347
5567
5787 600-; 1 G226I
6446
220
198
6065
6884
7104
7323
7542(7761
7979 8198
8416
8635
219
199
200
8853
301030
9071
1247
9289
1464
9507
1681
9725
1898
9943
.161
2331
.378
2547
.595
2764
.813
2980
218
217
2114
201
3196
3412
3628
3844
4059
4275
4491
4706
4921
5136
216
202
5351
5566
5781
5996
6211
6.425
^64
6639
6854
7068
7282
215
203
7496
7710
7924
8137
8351
8778
8991
9204
9417
213
204
9030
9843
..56
.268
.481
.693
.906
1118
1330
1.542
212
205
311754
1986
2177
2389
2600
2812
3023
3234
3445
3656
211
200
3807
4078
4289
4499
4710
4920
5130
5340
.5551
5760
210
207
5970
6180
6390
6599
6809
7018
7227
7436
7646
7854
209
208
8063
8272
8481
8089
8898
9100
9314
9522
9730
9938
208
209
210
320146
0354
2426
0562
2633
0709
2839
0977
3046
1184
3252
1391
3458
1598
3665
1805
3871
2012
4077
207
206
322219
211
4282
4488
4694
4899
5105
5310
5516
.5721
5926
6131
205
212
6336
6541
6745
0950
7155
7359
7563
7767
7972
8176
204
213
8380
8583
8787
8991
9194
9398 9001
9805
...8
.211
203
214
330414
0017
0819
1022
1225
1427
1630
1832
2034
2236
202
215
2438
2040
2842
3044
3246
3447
3049
3850
4051
4253 202
216
4454
4655
4856
5057
.5257
.5458
5658
.5859
60.59
6260 201
217
6460
66601 6860
7060
7260
7459
7659
7858
8058
8257 200
218
8456
86561 885-5
9054
9253
9451
9650
9349 ..47!
.246 199
219
340444 0642' 0841 1 1039
1237' 1435' 1632' 1830' 2028'
2225 198
N i 1 I 1 2 1 3 1 4 1 5 1 6 1 7 1 8 i 9 j D.l
15*
CO
A TABLE OF LOGARITHMS FROM 1 TO lO.OOG.
jN.
1 1 1 2 I 3 1 4 1 5 1 6 1 7 1 8 1 9 '( D. 1
230
342423, 2620
2817! 3014 3212, 3409, 3oi)6 3802
399i4|41<vfii 197
221
4392
4589
4785
4981
5178
5374
5570
6766
69621 81571 196
222
6353
6549
6744
6939
7135
9083
7330
7525
7720
7915] 8110! 195
223
8305
8500
8694
8889
9278
9472
9666
98G0i ..64| 194
224
350248
0442
0636 0829
105i3
1216
1410
1603
1796! 1989| 193
225
2183
2375
2568 1 2761
2954
31-17
3339
3532
3724139161 193
226
4108
4301
4493
4685
4876
5068
6260
5452
661315834! 192
227
6026
6217
6408
6599
6790
6981
7172
7363
7554 77'14i 191
228
7935
8125
8316
8506
8696
8888
9076
9266
^456! 96461 190
229
23J
9835
..25
1917
.215
2105
.404
2294
.593
2482
.783
2671
.972
2859
1161
3048
1350 1539: 189
3238 3124; 188
361728
231
3612
3800
3988
4176
4363
4551
4739
4928
5113 630 1| 188
232
5488
5675
5862
6049
6236
6423
6610
0796
8983j7169| 187
233
7356
7542
7729
7915
8101
8287
8473
8659
8845 9030 i 186
234
9216
9401
9587
9772
9958
.143
.328
.513
.698 .883' 185
235
371068
1253
3098
1437
1622
1806
1991
2175
2360
2544 2728' 184
236
2912
3280
3464
3G47
3331
4015
4198
4332 456-3; 184
237
4748
4932
5115
6298
5481
5064
584G
6029
6212
6394:' 183
238
6577
6759
6942
7124
7306
7488
7670
7852
8034
82l6i 182
239
240
8398
8580
8761
0573
8943
0754
9124
0934
9306
1115
9487
1298
9668
1476
9849
1656
ll30
1837
181
181
380211
0392
241
2017
2197
2377
2557
2737
2917
3097
.3277
3456
3836
180
242
3815
3995
4174
4353
4533
4712
4891
5070
5249
5428
179
243
5606
5785
5964
6142
6321
6499
0677
6856
7034
7212
178
244
7390
7568
7746
7923
8101
8279
8456
8634
8811
89891 178
245
9166
9343
9520
9698
9875
..51
.228
.405
.582
.759! 177
246
390935
1112
1288
1464
1641
1817
1993
2189
2345
2521
176
247
2697
2873
3048
3224
3400
3575
3751
3926
4101
4277
176
248
4452
4627
4802
4977
5152
5326
5501
5876
5850
6025
175
249
6199
6374
6548
6722
6896
7071
7245
7419
7592
7766
174
250
397940
8114
8237
8481
8634
8808
8981
9154
9328
9501
173
251
9674
9847
..20
.192
.365
.538
.711
.883
1056
12281 173
252
401401
1573
1745
1917
2089
2261
2433
2605
2777
2949! 172
253
3121
3292
3464
3635
3807
3978
414S
4320
4492; 48831 171 |
254
4834
5005
5176
6346
55.7
5688
5858
6029
6199
6370
171
255
6540
6710
6881
7051
7221
7391
7661
7731
7901
8070
170
256
8240
8410
8579
8749
8918
9087
9257
9426
9595
9764
169
257
9933
.102
.271
.440
.609
.777
.946
1114
1283
1451
189
258
411620
1788
1956
2124
2293
2461
2629
2796
2964
3i32| 168
259
260
3300
3467
5140
3635
5307
3803
5474
3970
6641
4J37
6808
4305
5974
4472
6141
4639
6308
4808 1 167
6474! 167
414973
281
6641
6807
6973
7139
7306
7472
7638
7804
7970181351 166
262
8301
8467
8633
8798
8964
9129
9295
9160
9625! 979 1| 165
263
9956
.121
.286
.451
•616
.781
.945
1110
1275
14391 165
264
421604
17C8
1933 2097
■2261
2426
2590
2754
2918
3082i 164
9.65
•3246
3410
3574 3737
3901
4065
4228
4392
4555
4718! 164
266
4882
5045
5208
5371
6534
5097
5860
6023
6186
6349
163
267
6511
6674
6S36
6999
716]
7324
7488
7648
7811
7973
162
268
8135
8297
8459
8621
8783
8944
9106
9268
9429
9591
162
269
9752
9914
..75
.236
.398
.559
.720
.881
1042
1203
161
270
431364
1525
1685
1846
2007
2167
2328
2488
J>349
2809! 181 1
271
2969
3130
3290
3450
3610
3770
3930
4090
4249144091 160 f
272
4569
6163
4729
4883
5048
5207
5307
6626
5685
6844 i 6004
1 59
273
6322
6481
6640
6798
6957
7116
7275
7433' 7592
159
274
7751
7904
8087
8226
8384
8542
8701
8869
^017 9175
.694 .752
158
275
9333
9491
9648
9806
9964
.122
.279
.437
1.08
276
440909
1066
1224
1381
1538
1695
1852
2009
216612323
157
277
2480
'?537
2793
2950
3106
3263
3419
3576
373213889; 1571
278
4045
4201
4357
4513
4069
4825
4981
6137
529315449 156
279
6604
5760
59151 60711 62261 6382 6537 6692 6848' 70031 155 |
N.
1
i 1
i 2
1 3
1 4
5
6
7
8
9
D.I
A TABLE OF LOGARITHMS FROM 1 TO 10,000.
iN.
1 1 i 2 1 3 1 4 1 5 1 6 1 7 ! 8 1 9 1 D. 1
280
447158
7313
7468
7623
7778
7933
8088
8242
8397
9941
8552
155
281
8700
8861
9015
9170
9324
9478
9633
9787
..95
154
282
450249
0403
0557
0711
0865
1018
1172
1326
1479
1633
154
283
1786
1940
2093
2247
2400
2553
2706
2859
3012
3165
153
284
3318
3471
3624
3777
3930
4082
4235
4387
4540
4692
1.53
285
4845
4997
5150
5302
5454
5600
5758
5910
6062
6214
152
286
6366
6518
6670
6821
6973
7125
7276
7428
7579
7731
152
287
7882
8033
8184
8336
8487
8638
8789
8940
9091
9242
151
288
9392
9543
9694
9845
9995
.146
.296
.447
.597
.748
151
289
460S9S
1048
119.8
1348
1499
1649
1799
1948
2098
2248
150
290
462398
2548
2697
2847
2997
3146
3296
3445
3594
3744
150
291
3893
4042
4191
4340
4490
4639
4788
4936
6085
5234
149
292
5383
5532
5680
5829
5977
6126
6274
6423
6571
6719
149
293
6868
7016
7164
7312
7460
7608
7756
7904
8052
8200
148
294
8347
8495
8643
8790
8938
9085
9233
9380
9527
9675
148
295
9822
9969
.116
.263
.410
.55*
.704
.851
.998
1145
147
296
471292
1438
1585
1732
1878
2025
2171
2318
2464
2610
146
297
2756
2903
3049
3195
3341
3487
3633
3779
3926
4071
146
298
4216
4362
4508
4653
4799
4944
5090
5235
5381
5526
146
299
5671
5816
5962
6107
6252
6397
6542
6687
6832
6976
145
300
477121
7266
7411
7555
7700
7844
7989
8133
8278
8422
146
301
8566
8711
8855
899 EJ
0143
9287
9431
9575
9719
9863
144
302
480007
0151
0294
0438
0582
0725
0869
1012
1156
1299
144
303
1443
1586
1729
1872
2016
2159
2302
2445
2588
2731
143
304
2874
3016
3159
3302
3445
3587
3730
3872
4015
4157
143
305
4300
4442
4585
4727
4869
5011
5153
5295
5437
5579
142
306
5721
5863
6006
6147
6289
6430
6572
6714
6856
6997
.42
307
7138
7280
7421
7563
7704
7845
7986
8127
8269
8410
141
308
8551
8692
8833
8974
9114
9255
9396
9537
9677
9818
141
309
310
9958
..99
1502
.239
1642
.380
.520
1922
.661
2062
.601
2201
.941
2341
1081
2481
1222
2621
140
140
491362
1782
311
2760
2900
3040
3179
3319
3458
3597
3737
3876
4015
1.39
312
4155
4294
4433
4572
4711
4850
4989
5128
6267
5406
139
313
5544
5683
5822
5960
6099
6238
6376
6515
6653
6791
139
314
6930
7068
7206
7344
7483
7621
7759
7897
8035
8173
138
315
8311
8448
8586
8724
8862
8999
9137
9275
9412
9550
138
316
0687
9824
9962
..99
.236
.3/4
.511
.648
.785
.922
137
317
501059
1196
1333
1470] 1607
1744
1880
2017
2154
2291
137
318
2427
2564
2700
2837! 2973
3109
3246
3382
3518
3655
136
319
3791
.3927
4063
4199
4335
4471
4607
4743
4878
5014
136
320
505150
5286
5421
5557
5693
5828
5964
6099
6234
6370
136
321
6505
6640
6776
6911
7046
7181
7316
7451
7586
7721
135
322
7856
7991
8126
8260
8395
8530
8664
8799
8934
9068
135
323
9203
9337
9471
9606
9740
9874
...9
.143
.277
.411
134
324
510545
0679
0813
0947
1081
1215
1349
1482
1016
1750
.134
325
1883
2017
2151
2284
2418
2551
2684
2818
2951
3084
133
326
3218
3351
3484
3617
3750
3883
4016
4149
4282
4414
133
327
454S
4681
4813
4946
5079
5211
5344
5476
5609
5741
133
328
5874
6006
6139
6271 16403
6535
6668
6800
6932
7064
132
329
7196
7328
7460
7592
7724
7855
7987
8119
8251
8382
132
330
518514
8646
8777
8909
9040
9171
9303
9434
95C6
9697
131
331
9828
9959
..90
.221
.353
.484
.615
.745
.876
1007
131
332
521138
1269
1400
1530
1661
1792
1922
2053
2183
2314
131
333
2444
2575
2705
2835
2966
3096
3226
3356
3486
36 J 6
130
334
3746
3876
4006
4136
4266
4396
4526
4656
4785
4915
130
335
5045
5174
5304
5434
5563
5693
5822
5951
6081
0210
129
338
6339
6469
6598
6727| 6856
6985
7114
7243
7372
7.501
129
337
7630
7759
7888
8016.1 8145
8274
8402
8531
8660
8788' 129
338
8917
9045
9174
9302i 9430
9559
9687
9815
9943
..72| 128
339
5302001 03281 0456' 0584i 0712
084010968' 1096
1223
13511 128
1 1 1 2 1 3 ! 4 i 5 I 6 1 7 ! 8 1 9 1 i). i
^'^ww
A TABLE OF LOGARITHMS F1103I 1
TO 10,000
;^- i 1 1 1 2 1 3 1 4 1 5 1 6 i 7 1 S 1 9 1 I). 1
340
5314791 1607
1734
1 1862
1990
2117
1 22451 2372
2500
^i)27
128
341
2754
2882
3009
31.36
3264
3391
I3518
3645
3772
3899
127
342
4026
4153
4280
4407
4534
4661
4787
4914
5041
5167
127
343
5294
5421
5547
5674
5800
5927
6053
6180
6306
6432
126
344
6558
6685
6811
6937
7063
7189
7315
7441
7567
7693
126
345
7819
7945
8071
8197
8322
8448
8574
8699
8825
8951
126
34G
9076
9202
9327
9452
9578
9703
9829
9954
..79
.204
125
347
540329
0455 0580
0705
0830
0955
1080
1205
1330
14.54
125
348
1579
1704
1829
1953
2078
2203
2327
2452
2576
2701
125'
349
2825
2950
3074
3199
3323
3447
3571
3696
38ro
3944
124
350
544068
4192
4316
4440
4564
4688
4812
4936
5060
5183
124
351
5307
5431
5555
5078
5802
6925
6049
6172
629^
6419
124
352
6543
6666
6789
6913
7036
7159
7282
7405
7529
7652
123
353
7775
7898
8021
8144
8267
8389
8512
8635
87.58
8881
123
354
9003
9126
9249
9371
9494
9616
9739
9861
9984
.106
123
355
550228
0351
0473
0595
0717
0840
0962
1084
1206
13281 122 1
356
1450
1572
1694
1816
19a8
2060
2181
2303
2425
2547
122
357
2068
2790
2911
3033
31.55
3276
3398
3519
3640
3762
121
358
3883
4004
4126
4247
4368
4489
4610
4731
4852
4973
121
359
360
5094
5215
5336
6544
5457
6664
5578
6785
5699
6905
5820
7026
6940
7146
6061
7267
6182
7387
121
120
556303
6423
361
7507
7627
7748
7868
7988
8108
8228
8349
8469
8589
120
362
8709
8829
8948
9068
9188
9308
9428
9548
9667
9787
120
363
9907
..26
.146
.265
.385
.504
.624
.743
.863
.982
1!9
364
561101
1221
1.340
1459
1578
1698
1817
19.36
2055
2174
119
365
2293
2412
2531
2650
2769
2887
3006
3125
3244
3362
119
366
3481
3600
3718
.3837
3955
4074
4192
4311
4429
4548
119
367
4666
4784
4903
5021
5139
5257
5376
5494
5612
5730
118
368
6848
5966
6084
6202
6320
6437
6555
6673
6791
6909
118
369
7026
7144
7262
7379
7497
7614
7732
7849
7967
8084
118
370
568202
8319
8436
8554
8671
8788"
8905
902:^
9140!
9257
117
371
9374
9491
9608
9725
9812
9959
..76
.193 .309i
.426
117
372
570543
0660
0776
0893
1010
1126
1243
1359
1476
1592
117
373
1709
1825
1942
2058
2174
2291
2407
2523
2639
2755
116
374
2872
2988
3104
3220
3336
3452
3568
3684
3800
3915
116
375
4031
4147
4263
4379
4494
4610
4726
4841
4957
5072
116
376
5188
5303
5419
5534
5650
5765
5880
.5996
6111
6226
115
377
6341
6457
6572
6687
6802
6917
7032
7147
7262
7377
115
378
7492
7607
7722
7836
7951
8066
8181
8295
8410
8525
115
379
380
8639
579784
8754
9898
8868
..12
8983
.126
9097
.241
9212
.355
9320
.469
9441
9555
.697
9669
.811
114
114
..583
381
580925
1039
1153
1267
1381
1495
1608
1722
1836
1950
114
382
2063
2177
2291
2404
2518
2631
2745
2858
2972
3085
114
383
3199
3312
3426
3539
3652
3765
3879
3992
4105
4218
113
384
4331
4444
4557
4670
4783
4896
6009
5122
5235
5348
113
385
5461
5574
5686
5799
.5912
6024
61.37
6250
6362
0476
113
386
6587
6700
6812
6925
7037
7149
7262
7374
7486
7599
112
387
7711
7823
7935
8047
8160
8272
83S4
8496
8608
8720
112
388
8832
8944
9056
9167
9279
9391
9503
9615
9726
9838
112
3c9
9950
..61
.173
.284
..S96
.507
.619
.730
.843
.953
112
390
591065
1 176
1287
1399
1510
1621
1732
1843
1935
2066
111
391
2177
2288
2399
2510
2621
2732
2843
2954
3064 3175
111
392
3286
3397
3508
3618
3729
38401
3950
4061
4171 4282
III
393
4393
4503
4014
4724
4834
4945
.5055
5] 65
.5276 53S6
110
394
5496
5606
5717
.5827
5937
6047
61.57
626.
6377 6487
no
395
6597
6707
6817
6927
7037
7146!
7256
7366
7476 7586
110
396
7695
7805
7914
8024
8 KM
8243 i
8353
8462
8572i 8681
110
397
8791
8900
9009
9119
922S
9;w7
9446
95.56
9665, ^^774
109
3:)8
9883
9992
.101
.210
.3191
.428
.537
.646
.755i «64
109
399
600973
I0S2
119!
1299 14081
1517
1 625
L731
1843. ly;.l
109
N. 1 1 1 1 2 1 3 i 4 1 5 i 6 '[ 7 1 8 1 9 ! P. 1
A TABLE OF LOGARITHMS FROM 1 TO 10.000.
N. I
I 1 I 2
4 ! 5 I 6 I 7 i 8 I 9 I D
400
401
V)^
103
404
405
406
407
408
'100
4l0
'111
412
413
414
415
416
417
418
4_19
420
421
422
423
424
425
420
127
428
4vi9
4S0
431
432
433
431
435
436
437
438
439
440
441
442
443
444
445
446
447
44 S
.449
450
451
452
453
454
455
456
457
45S
45!)
¥7
602060
4-2169
1 2277
2386
3 I 'l>tBP*2f»3
13361
3469
422^334
4442
4550
5305
5413
5521
6628
6381
6489
6596
6704
7455
7562
'7669
7777
8526
8633
8740
8847
9594' 9701
9808
9914
610660
0767
0873
0979
1723
, 1829
' 2890
1936
2996
2042
3102
612i'84
3842
3947
4053
4159
489?
5003
6108
5213
5950
6055
6160
6265
7000
7105
7210
7315
8048
8153
8257
8362
9093
9198
9302
9106
620136
0240
0344
0448
1176
12S0
1384
1488
2214
2318
2421
3456
2525
3559
623249
3353
4282
4385
4488
4591
5312
5415
5518
6621
6340
6443
6546
6648
7366
7468
7571
7673
8389
8491
8593
8695
9410
9512
9613
9716
630428
0530
0631
0733
1444
1545
1647
1748
2457 2559
2660
2761
633468 3569
3670
3771
4^177
4578
4679
4779
5484
5584
5685
5786
6488
6588
6688
6789
7490
7590
7690
7790
8489
8589
8689
8789
9486
9586
9686
9785
640481
0581
0680
0779
1474
1573
/672
1771
2465
2563
3551
2662
3650
2761
3749
643453
4439
4537
4636
4734
5422
5521
5619
6717
6404
6502
6600
6698
7383
7481
7579
7676
8360
8458
8555
8653
9335
9432
9530
9627
650308
0405
0502
0599
1278
1375
1472
1569
2246
2343
2440
3405
2636
3502
653213
3309
4177
4273
4369
4465
5138
5235
5331
5427
6098
6194
6290
6386
7050
7152
7247
7343
8011
8107
8202
8298
8965
9060
9155
9250 1
9916
..11
.106
.201
680865
0960
10.55
1150
IS 13
1907
2002
2096
2494
1 2603
2711
3577
3686
3794
4058
4766
4874
5736
5844
5051
6811
6919
7026
^8841 7991
809S
8954! 906 1
9167
..21
.128
.234
1086
1192
1298
2148
2254
2360
3207
3313
3419
4264
4370
4475
6319
5424
5529
6370
6476
6581
7420
7525
7629
8466
8571
8676
9511
9615
9719
0552J 0666
0760
1592
1695
1799
2628
2732
2836
3663
3766
3809
4695
4798
4901
67241 5827
5929
6751
6853
69561
7776
7878
7980
8797
8900
9002
9817
9919
..2ll
0835
0936
1038
1849
1961
2052
2862
2963
3064
3872
3973
4074
4880
4981
6081
6886
6986
6087
6889
6989
7089
7890
7990
8090
8888
8988
9088
9885
9984
..84
0879
0978
1077
1871
1970
2069
2860
2959
3946
3058
3847
40441
4832
4931
60291
5816
6913
6011
6796
6894
6992
7774
7872
7969
8750
8848
8345
9724
9821
9919
0696
0793
0890
1666
1762
1859
2633
2730
2826
35981 3895
3791
4502
4658
4754
5523
6619
5715
6482
6577 66731
74.381 7534! 76291
8393
8488
8584
9346
9441
9536
.296
.391
.486
1245
1339
1434
2101
22S(>
2330
2819
3902
4982
6059
7133
8205
9274
.341
1405
2466
3525
4581
6634
6686
2928
4010
5089
6166
7241
8312
9381
.447
1611
2572
3630
4686
6740
6790
^734 7839
8884
8780
9824
0864
1903
2939
39731
50041
60321
70581
80821
9104
. 123;
1 1 39 1
2153|
3105
9928
0968
2007
3042
4175
5182
6187
7189
8190
9188
.183
1177
2168
3156
4143
5127
6110
7089
8067
9043
..16
0987
1956
2923
3888
4850
.5810
6769
7725
8879
9631
.581
1629
2475
4076
5107
6136
7161
8185
9206
.224
1241
2256
3206
4276
5283
6287
7290
8290
9287
.283
1276
2267
3255
4242
5226
6208
7187
8165
9140
.113
1084
2053
3019
3036
4118
5197
6274
7348
8419
9488
.554
1617
2678
3736
4792
6845
68951
79431
8989
..32
1072
2110
3JI46
4179
6210|
6238
72631
8287
93081
.326
1342
2330
3367
3984
4946
.5906
6864
7820
8774
9726
.6761
1623
2569 !
4376
6383,
6388
7390
8389i
9387
.382
1375!
2366
3354
4340
5324
6306
7285
8262
9237
.210
1181
2150
3116
4080
5042
60021
69601
79161
8870(
9821
.7711
17l8i
2663!
108
u.s
108
107
107
107
107
100
H)6
1(!6
106
105
106
105
106
104
101
1(>4
1(|4
\03
103
103
103
102
102
102
102
101
101
100
100
100
100
99
99
99
99
99
_99
98
98
98
98
98
97
97
97
97
97
96
96
96
96
96
95
95
96
95
96
3 I 4
6 I 7 I 8 I 9
A TAIILE OF I.'-»GARI'niMS FKOM 1 TO 10.000.
Jl-J
U i I i 2 i 3 1 4 1 5 1 6 1 7 1 8 i 9 ! I). 1
4fi0
6627581
2852 1 2947 1 304 1 , 3 1 35 3230, 3324, 34 1 8 1 35 1 2 1 3607 , 94 1
461
3701
37951
38891
3983 407814172
4266 4360 44.54
4.548!
94
462
4642
4736
48301
4924' 50181 5112
.5206 5299 5393
5487!
94
463
5581
5675
5769!
.58621 59.16 jMjSO
614:3 62371 6331
6424
94
464
6518
6612
67051
6799 6892 i 6986
7079 717317266
7360
94
165
7453]
7546
7640J
7733!
7826:7920
8013 810618199
8293
93
466
8386]
8479
8572 j
8665
8759188521
8945 9038 1 9131
9224
93
467
9317
9410
9503!
9596
9689 i 9782!
9875
99671 ..60
.1.53
93
468
670246
0339
043 1 i
0524
06171
0710
0802
0895 0988
1080
93
469
1173
1265
1358
1451
1.543,
1636
1728
1821 1913
2005
9.3
470
672098
2190
2283
23751
2467
2560
2652
2744 2836
2929
'92
471
1021
3113
3205
3297
3390
3482
C 74
3666137.58
38.50
92
472
3942
4034
4126
4218
4310
4402
4494
4.58614677
4769
•92
473
4861
4953
5045
5137
5228
5320
.5412
.55031 .5.595
5687
92
474
5778
5870
5962
6053
6145
6236
6328
6419
6511
6602
92
475
6694
6785
6876
6968
7059
7151
7242
7333
7424
7516
91
476
7607
7698
7789
7881
7972
8063
8154
8245
8336
8427
91
477
8518
8609
8700
8791
8882
8973
9064
91,55
9246
9337
91
478
9428
9519
9610
9700
9791
9882
9973
..63
.154
.245
91
479
680336
0426
0517
0607
0698
0789
0879
0970
1060
1151
91
480
681241
13{^2
1422
1.513
1603
1693
"1 784
1874
1964
2055
90
481
2145
2235
2326
2416
2506
2596
2686
2777
2867
2957
90
482
3047
3137
3227
3317
3407
3497
3.587
,3677
3767
3857
90
483
3947
4037
4127
4217
4307
4396
4486
4576
4666
4756
90
484
4845
4935
5025
5114
5204
.5294
5383
.5473 5563
5652
90
485
5742
583 1
5921
6010
6100
6189
6279
63681 64.58
6547
89
486
6636
6726
6815
6904
6994
7083
7172
7261
7351
7440
89
487
7529
7618
7707
7796
7886
7975
8064
81.53
8242
8331
89
488
8420
8509
8598
8687
8776
8865
8953
9042
9131
9220
89
489
9309
9398
9486
9575
9664
9753
9841
9930
.,19
.107
89
490
690196
0285
0373
0462
0550
0639
0728
0816
0905
0993
89
491
1081
1170
1258
i.347
14.35
1524
1612
1700
1789
1877
88
492
1965
2053
2142
2230
2318
2406
2494
2583
2671
2759
88
493
2847
2935
3023
3111
3199
.3287
33751 ,34631 3551
3639
88
494
3727
3815
3903
3991
4078
4166
42.54
4342
4430
4517
88
495
4605
4693
4781
4868
4956
5044
5i3l
5219
5307
.5394
88
496
5482
5569
5657
5744
5832
.5919
6007
6094
6182
6269
87
497
6356
6444
0531
6618
6706
6793
6880
6968
7055
7142
87
498
7229
7317
7404
7491
7578
7665
77.52
7839
7926
8014
87
499
8101
8188
8275
8362
8449
8535
8622
8709! 8796
8883
87
500
698970
9057
9144
9231
9317
9404
9^191
95781 9664
9751
87
501
9838
9924
..11
..98
.184
.271
.3.58
.444
.531
.617
87
502
700704
0790
0877
0963
10.50
1136
1222
1309
1.395
1482
86
503
1568
1654
1741
1827
1913
1999
2086
2172
2258
2344
S6
504
2431
2517
.2603
2689
2775
2861
2947
3033
3119
3205
86
605
3291
3377
3463
3549
3635
3721
3807
3895
3979
4005
86
50ff
4151
4236
4322
4408
4494
4579
4665
4751
4837
4922
86
507
5008
5094
5179
.5265
1 53.50
.5436
5522
5607
5693
5778
86
508
5864
5949
6035
6120
6206
6291
6376
6462
6547
6632
85
509
6718
6803
6888
6974
7059
7144
7229
7315
7400
7485
85
510
707570
7655
774'0
7826
7911
7996
8081
8166
8251
8330
85
511
8421
8506
8591
8676
8701
8846
8931
9015
9100
9185
85
512
9270
9355
9440
9524
9609
9694
9779
9803
9948
..33
85
513
710117
0202
0287
0371
0456
0.540
0625
0710
0794
0879
85
514
0963
1048
1132
1217
1301
1385
1470
1.554
1639
1723
84
615
1807
1892
1976
2060
2144
2229
2313
2397
2481
2566
84
516
2650
2734
2818
2902
2986
1 3070
3154
3238
3323
3407
84
517
3491
3575
36.50
3742
! 3826
13910
3994
4078
4162
4246
84
518
4330
4414
4497
4581
1 4665
4749
4833
4916
5000
5084
84
519
1 5167
5251
5335
.5418
! .5502' .'■^586
5669'.57.53' .58361 5920
84
LiL
1 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 ! 9 1 dH
A TARLF. OF LOaAKITirSIS FKOM I TO 10,000.
N.
1 1 1 1 2 1 3 1 4 1 5 1 6 i 7 1 « 1 W
!).
"520
716003. 6087. G170
6254
6337
642.1
6504
6588 6671
6754
"8:7
oZl
6838
692 i
7004
7088
7171
7254
7338
7421,7504
7587
m
522
7671
7754
7837
7920
8003
8086
8169
8253 8336
8419
t^.<
r)23
8502
8585
8668
8751
8834
8917
9000
9083 9165
9248
83
524
9331
9414
9497
9580
9663
9745
9828
9911 9994
..77
83
525
720159
0242
0325
0407
0490! 0573i 0655
0738 0821
0903
83
52r,
0986
1068
1151
1233
1316
1398 1481
1663, 1646
1728
82
527
1811
1893
1975
2058
2140
2222 2305
2.387 2469
2552
82
528
2634
2716
2798
2881
2963
3045 3127
3866 3948
3209 3291
3374
82
529
3456
3538
3620
3702
3784
4030! 4112
4194
82
530
724276
4358
4440
4522
4604
4685
4767
4849
4931
.5013
"82
531
5095
5176
5258
5340
5422
6503
5,585
5667
5748
,5830
S2
532
5^12
5993
6075
6156
6238
6320
6401
6483
6664
6646
82
533
G727
6809
6890
69'72
7053
7134
7216
7297
7379
7460
81
534
754 1
7623
7704
7785
7866
7948
8029
8110
8191
8273
81
535
8354
8435-
8516
8597
8678
8759
8841
8922
9003
9084
81
53')
9165
9246
9327
9408
9489
9570
9651
9732
9813
9893
81
537
9974
..55
.136
.217
.298
.378
.459
•540
.621
.702
81
538
730782
08*^3
0fct4
1024
1105
1186
1266
1347
1428
1508
81
539
1589
1669
1750
1830
1911
1991
2072
2152
2233
2313
81
540
732394
2474
2555
2635
271.5
2796
2876
2956
i3037
3117
"80
541
3197
3278
33.58
3438
3518
35981 3679
3759
3839
3919
80
542
3999
4079
4160
4240
4320
4400 i 4480
4560
4640
4720
80
543
4800
4880
4960
5040
5120
52001 5279
5359
6439
5519
80
644
5599
5679
5759
58.38
.5918
59981 6078
6157
6237
6317
80
545
6397
6476
6556
6635
6715
6796 1 6874
69.54
7034
7113 80 1
54(5
7193
7272
7352
7431
7511
75901 7670
7749
7829
7908
79
547
7987
8067
8146
8225
8305
8384 8463
8543
8622
8701
79
548
87y;
8860
8939
9018
9097
9177
9266
9335
9414
9493
79
549
9572
9651
97? I
9810
9889
9968
..47
.126
.205
.284
79
550
740363
0442
0.521
0800
0078
0757
0836
0915
0994
1073
79
551
1152
1230
1309
1388
1467
1546
1624
1703
1782
1860
79
552
1939
2018
2096
2175
2254
2332 2411
2489
2668
2646
79
553
2725
2804
2882
2961
3039
3118
3196
3275
3353
,3431
78
554
3510
3588
3667
3745
3823
3902
3980
4058
4136
4215
78
555
4293
4371
4449
4528
4606
4684
4762
4840
4919
4997
78
556
5075
5153
5231
6309
5387
5465
5543
5621
.5699
.5777
78
55/
5855
5933
6011
6089
6167
6245
6323
6401
6479
6666
78
558
6C34
6712
6790
6868
6945
7023
7101
7179
7256
7334
78
559
7412
7489
7567
7645
7722
7800
7878
7955
8033
8110
78
5(50
748188
8266
8343
8421
8498
85"76
8653
8731
88J)8
8886
7V
561
8963
9040
9118
9195
9272
9350
9427
9.504
9582
9659
77
562
9736
9314
9891
9968
..45
.123
.200
.277
..3.54
.431
77
563
750508
0586
0663
0740
0817
0894
0971
1048
1125
1202
77
564
1279
1356
1433
1510
1587
1664
1741
1818
1895
1972
77
565
2048
2125
2202
2279
2356
2433
2509
2586
2663
2740
77
566
2816
2893
2970
3047
3123
3200
3277
3353
34.30
3.506
77
567
3583
3660
3736
3813
3889
3986
4042
4119
4196
4272
77
558
4348
4425
4501
4578
46.54
4730
4807
4883
4960
5036! 76 1
569
570
5112
5189
5951
5265
6027
5341
6103
5417
6180
5494
6256
5570
5646
5722
6484
5799
6560
76
76
756875
6332
6408
571
6636
6712
6788
6864
6940
7016
7092
7168
7244
7320
76
572
7396
7472
7548
7624
7700
7775
7851
7927
8003
8079
76
573
8155
8230
8306
8382
8458
8533
8609
8685
8761
8836
76
574
8912
8988
9063
9139
9214
9290
9366
9^141
9517
9.592
76
575
9668
9743
9319
9894
9970
..45
.121
.196
.272
.347
75
576
769422
0498
0573
0649
0724
0799
0875
0950
1025
1101
76
577
1176
1251
1326
1402
1477
1562
1627
1702
1778 18.53
75
578
1928
2003
2078
2153
2228
2303| 2378
2463
2529 2604
75
579
2679
27541 2829
2904' v»i 8
30631 3128' 3203
3278 .3353' 75 1
N.
U 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 H i y ' J). 1
10
A TABLE OP LOGAKITirOTS FROM 1 TO 10,000.
N.
1' 1 1 f 2 1 3 i 4 1 5 1 6 1 7 1 8 i 9 ! D.
55o
7()342fc
i 350:j
3a78r 3B53, 3727 3802, 3877, 39521 4027 1 4l0l| 75
581
417C
4251
4326! 4400| 4475 4550 46241 4691
477414848! . 75 i
582
492C
; 4998
5072 5147 5221 5296, 53701 544?
55201 5594' 75 1
583
566ij
5743
5818, 5892 5966 6041 1 61 151 619'
6264
633S
74
584
6412
6487
6562 66361 6710 67851 68591 6933 7007
7082
74
585
7156
7230
7304 7379 7453 7527| 7601 7675 774'J
782?
74
586
789S
17972
8046 8120 8194 826818342 841618490
8564
74
587
8633
18712
87861 8860 8934 9008| 9082i 9 J 56
9525! 9599 9673i 97461 9820i 9894
0263 0336 0410;' 0484! 0557| 0631
9230
9303
74
588
9377
I 9451
9968
..42
74
589
770115
0189
0705
107781 74 1
5ii0
770852
!0926
09991 1073 1146
1220 1293! 1367
1440
1 i5"l4| 74 1
591
158?
i 1661
1734 1808
1881
1955i 2028' 2102
2175
2248
73
592
2322
! 2395
24681 2542
2615
26881 27621 2835
2908
2981
73
593
3055
13128
3201! 3274
3348
3421| 3494 3567
3640
3713
73
594
3786
3860
3933
4006
4079! 4152! 42251 4298
4371
4444
73
595
4517
4590
4663
4736
4809! 48821 4955! 5028
5100
5173
73
596
5246
5319
5392
5465
!5538
56101 5683. 575615829
5902
73
597
5974
6047
6120
6193
6265
63331 641116483
6556
6629
73
598
6701
6774
0848
69i9
6992
7064! 7137
7209
7282
7354
73
599
7427
i 7499
7572
7644
7717
7789| 7862
7934
8006
8079
[ 72
600
77815118224
8296
8368
8441
85l3i 8585
8658
8730
8802
"72
601
8874 8947
9019
9091
9163
92361 9308
9380
9452
9524
72
602
9596 9669
9741
9813
9885
9957! ..29
.101
.173
.245
72
603
780317 0389
0461
0533
0605
0677
10749
0821
0893
0965
72
604
1037 1109
1181
1253
1324
1396
1 1468
1540
1612
16S4
72
605
1755 1827
1899
1971
2042
2114
i2186
2258
2329
2401
72
606
2473 i 2544
2616
26SS
2759
2831
12902
2974
3046
3117
72
607
3189 3260
3332
34031 3475
3546
3618
3689
3761
3832
71
608
390413975
4046,4118! 4189
4261
4332
4403! 4475
4546
71
609
461714689
4760
4831! 4902
4974
5045
5116| 5187
5259
71
610
7853301 5401
5472
6543
5615
5686
5757
582'8
5899
5970
71
611
604116112
6183
6254
6325^ 6396
6467
6538
6609
6680! 71 1
612
67511 6822
6893 6964
7035! 7106
7177
7248
7319
7390
71
613
74601 7531
7602
7673
7744 7815
78S5
7956
&027
8098
71
614
816818239
8310
8381
8451! 8522
8593
8663
8734
8804
71
615
88751 8946
9016
9087
915719228
9299
9369
9440
9510
71
616
958 1| 9651
9722
9792
9863! 9933
...4
..74
.144
.215
70
617
7902851 0356
0426
0496
0567
0637
0707 0778
0848
0918
70
618
0988 1059
1129! 119&
12691
1340
1410
1480
1550
1620
70
619
1691 1761
183 1| 1901
1971!
2041
2111
2181
2252;
2322J 70 1
620
792392,
2462
2532! 2602| 2672!
2742
2812
2882
2952i
3022
70
621
3092
3162
3231 1 330 l!
3371
3441
3511
3581
3651
3721
70
622
3790
3860
39301 4000|
4070
4139
4209
4279
4349
4418
70
623
4488
4558
5254
5949
4627 4697
4767 4836
4906
4976
5045
5115
70
624
5185
5324! 5393
5463 5532
5602
5672
5741
5811
70
625
5880
60191 6088
6158 6227
6297
6366
6436
6505
62
626
6574!
6644
6713! 6782! 685216921
6990 7060!
7129
7198
69
627
7268 1
7337
7406! 7475! ^545^ 7614
7683 7752
7821
7890
69
628
7960;
8029
8098|8167| 8236 8305
8374 8443!
8513
8582
69
629
865l|
8720 i
8789! 8858 8927J 8990
9005i 91341
9203!
9272
69
630
7993411
9409
94781 95^^71 96 10 9685
9754! 98231
9892
9961
69
631
800029;
0098|
0167| 0236] (n305 03731 0442 051 1!
0580
06481 69 1
632
0717
0786
0854! 0923 0992 106l|ll29 11981
1266
13351
69
633
1404
1472|
15411 1609i J678 1747! 1815 18841 1952j 2021
69
634
2039
2158;
2226! 2295| 2363 2432i 2500 2568! 2637 2705
69
635
27741
2842
29l0i 2979! 3047 3116; 3184: 3252i 3321 1 3389
68
633
3457!
3525!
3594! 3662! 3730 3798! 38071 3935i 4003' 4071
68
637
41391
42081
42761 4344' 4412 4480' 4548, 46l6| 46S5| 47531
68
638
48211
4889^ 4957! 5025 5093 5161J 52291 52'J7; 53651 543,5,' 68 1
639 '
550 ll
5.^09' 56371 5705 5773' 584l' 59031 59^6' 6044' 6! 12^ 68
N. I
I 1 1 2 1 3 1 4 ! 5 1 6 1 7 1 8 1 9 i D. i
A TARLE OF LOGARITHMS FROM I TO 10,000.
11
040'
l_o_
1 '
1 2
1 3_
i 4
1 5
L6
1 7
1 8
LJL_
i^^
S06I80;624«|6316 6384
6451
6519
6.587| 6656, 6V'^3
67yi>
68
641
6858 i 6926
6994] 706 1
7129
7197
7201! 7.332
7400
7467
68
612
7535
7603
76701 7738
7806
7873
7941
8008
8076
8143
68
643
8211
8279
8346! 8414
8481
8549
8616
8684
8751
8818
f)7
644
8886
8953
9021
9098
9156
9223
9290
9358
9425
9492
67
645
9560
9627
9694
9762
9829
9896
9964
..31
..98
.165
67
646
8 1 0233 {0300
0367
0434
0501
0569
06'36
0703
0770
0837
67
647
0904 0971
1039
1106
1173
12401 1307
1374
1441
1.508
67
648
1575 1642
1709
1776
1843
19I0I 1977
2044
2111
2178
67
619
2245 2312
2379
2445
2512
2579
26461 2713
2780
2847
6/
650
812913
2980
3047
3114
3181
3247
3314
3381 .3448
.3514
67
651
3581
3648
3714
3781
3848
3914
3981
4048
4114
4181
67
652
4248
4314
4381
4447
4514
4,581
4647
4714
4780
4847
67
653
4913
4980
5046
5113
5179
5246
.5312
5378
5445
.5511
66
654
5578
5644
57 1 1
5777
5843
.5910
5976
6042
6109
6175
66
655
6241
6308
6374
6440
6.506
6573
6639
6705
6771
6838
66
656
6904
6970
7036
7102
7169
7235
7301.
7367
7433
7499
06
657
7565
7631
7698
7764
7830
7896
7962
8028
8094
8160
66
658
8226
8292
8,358
8424
8490
8.556
8622
8688
87.54
8820
66
659
8885
895 1
9017
9083
9149
9215
9281
9.346
9412
9478
66
660
819544
9610
9676
9741
9807
9873
9939
...4
..70
. 136
66
661
820201
0267
0333
0399
0464
0530
0.595
0661
0727
0'"92
66
662
0858
0924
0989
10.55
1120
1186
1251
1317
1382
1448
66
663
1514
1579
1645
1710
1775
1841
1906
1972
2037
2103
65
664
2168
2233
2299
2364
2430
2495
2560
2626
2691
27.56
65
665
2822
2887
2952
3018
3083
3148
3213
3279
3344
3409
65
666
3474
3539
3605
3670
3735
3800
3865
3930
3996
4061
65
667
412(;'
4191
4256
4321
4386
4451
4516
4581
4646
471J
65
668
4776
4841
4906
4971
5036
5101
5166
.5231
6296
.5361
65
069
5426
5491
5556
5621
5686
.5751
.5815
5880
5945
6010
65
670
82607^
6140
6204
6269
6334
6399
6464
6528
6593
6658
65
671
6723
6787
6852
6917
6981
7046
7111
7175
7240
7305
05
672
7369
7434
7499
7563
7628
7692
7757
7821
7886
7951
65
673
8015
8080
8144
8209
8273
8338
8402
8467
8.531
8595
64
674
8660
8724
8789
8853
8918
8982
9046
9111
9175
9239
64
675
9304
9368
0432
9497
9561
9625
9690
97.54
9818
9882
64
676
9947
..11
..75
.139
.204
.268
.332
•396
.460
.525
64
677
830589
0653
0717
0781
0845
0909
0973
1037
1102
1166
64
678
1230
1294
13.58
1422
1486
1,5,50
1614
1678
1742
1806
64
679
1870
1934
1998
2062
2126
2189
2253
2317
2381
2445
64
680
832509
2573
2637
2700
2764
2828
2892
29"56
3020
3083
64
681
3147
.3211
3275
3338
3402
3466
3530
3593
3657
3721
64
682
3784
3848
3912
3975
4039
4103
4166
4230
4294
4357
64
683
4421
4484
4548
4611
4675
4739
4802
48.66
4929
4993
64
684
5056
5120
5183
5247
i.310
.5373
.5437
5500
5564
5627
63
685
5691
5754
.5817
.5881
5944
6007
6071
6134
6197
6261
63
686
6324
6387
6451
6514
6577
6641
6704
6767
6830
6894
63
687
6957
7020
7083
7146
7210
7273
7336
7399
7462
7525
63
688
7588
7652
7715
7778
7841
7904
7907
8030
8093
81.56
63
689
8219
8282
8345
8408
8471
8534
8597
8660
8723
8786
.63
690
838849
8912
8975
9038
9101
9164
9227
92891 9352
9415
63
691
9478
9541
9604
9667
9729
9792
9855
991819981
..43
63
692
840106
0169
0232
0294
0357
0420
0482
05451 0608
0671
63
693
0733
0796 0859
0921
0984
1046
1109
1172! 1234
1297
63
694
1359
1422 1485
1.547
1610
1672
1735
17971 1860
2422 2484
1922
63
695
1985
2047 2110
2172
2235
2297
2360
2.547
62
696
2609
2672 2734
2796
2859
2921
2983
3046 3108
3170 62
697
3233
3295 3357
3420
3482
3.544
3606
3669 3731
3793 62
69S
3855
3918 3980
4042
4104
4166
4229
4291:43.53
4415 62
699
4477
4539 460:
4«f)I
4726
4788
48 5(
4912 4974; 5036' o2 '
1^1
Il|2!3i4|5|6|7|sl9|n. 1
16
12
A TABLE OF L0GARITH3TS FROM 1 TO 10,000.
N. 1 1 1 1 2 1 3 1 4 1 5 1 fi 1 7 1 8 1 9 1 D. 1
7t)()
845098
15160.6222
5284
5904
5346.540815170
5532
5594
5656
02
701
5718
5780
5842
5966
6028 6090
0151
6213
6275
62
702
6337
1 6399
6461
6523
6585
6646! 6708
6770
6832
6894
62
703
6955
17017
7079
7141
7202
7264
7326
7388
7449
7511
62
704
7573
7634
76 £6
7758
7819
7881
7943
8004
8066
8123
62
705
8189
8251
8312
8374
8435
8497
8559
8620
8682
8743
02
706
8805
8866
8928
8989
9051
9112
9174
9235
9297
.y358
61
707
9419
9481
9542
9604
9665
9726
9788
9849
9911
9972
61
708
850033
0095
0156
0217
0279
0340
0401
0462
0524
0585
61
709
710
0046
0707
1320
0769
1381
0830
1442
0891
1503
0952
1564
1014
1625
1075
1686
1136
1747
1197
1809
61
61
85125S
711
1870
1931
1992
2053
2114
2175
2236
2297
2358
2419
61
712
2480
2541
2602
2663
2724
2785
2846
29^7
2968
3029
61
713
3090
3150
3211
3272
3333
3^9^
3455
3516
3577
3637
61
714
3898
3759
3820
3881
3941
4002
40' 3
4124
4185
4245
61
715
4306
4367
4428
4488
4549
4610
4670
4731
4792
4852
61
716
4913
4974
5034
5095
5156
5216
5277
5337
5398
5459
61
717
5519
5580
5640
5701
5761
5822
5882
5943
6003
6064
61
718
6124
6185
6245
6306
6366
6427
6487
6548
6608
6668
60
719
6729
6789
6850
6910
6970
7031
7091
7152
7212
7272
60
720
857332
7393
7453
7513
7574
7634
7694
7755
7815
7875
60
721
7935
7995
8056
8116
8176
8236
8297
8357
8417
8477
60
722
8537
8597
8657
8718
8778
8833
8398
8958
9018
9078
60
723
9138
9198
9258
9318
9379
9439
9499
9559
9619
9679
60
724
9739
9799
9859
9918
9978
..33
..93
.1.58
.218
.278
60
725
860333
03&S
0458
0518
0578
0637
0697
0757
0817
0877
60
726
0937
0996
1056
1116
1176
1236
1295
1.355
1415
1475
60
727
1534
1594
1654
1714
1773
1833
1893
1952
2012
2072
60
728
2131
2191
2251
2310
2370
2430
2489
2549
2608
2688
60
729
730
2728
2787
3382
2847
3442
2906
3501
2956
3025
3620
3085
3680
3144
"3739
3204
3799
3263
3858
60
59
863323
356 1
731
3917
3977
4036
4096
4155
4214
4274
4333
4392
4452
59
732
4511
4570
4830
4689
4748
4808
4867
4926
4935
5045
59
733
5104
5163
5222
5282
5341
5400
5459
5519
5578
5637
59
734
5696
5755
5814
5874
5933
•5992
6051
6110
6169
6228
59
735
6287
6346
6405
6465
6524
6583
6642
6701
6760
6819
59
73b
6878
6937
6996
7055
7114
7173
7232
7291
7350
7409
59
737
7467
7526
7585
7644
7703
7762
7821
7880
7939
7998
59
738
8056
8115
8174
8233
8292
8350
8409
8468
8527
8586
59
739
740
8644
8703
9290
8762
8821
9408
8879
9466
8938
95-^5"
8GJ7
9584
9056
9642
9114
9701
9173
976;)
59
59
869232
9349
741
9818
9377
9935
9994
..53
.111
.170
.228
.287
.345
59
742
870404
0462
0521
0579
063S
0696'
U755
0813
0872
0930
58
743
0989
1047
1106
1164
1223
1281
1339
1398
1456
1515
58
744
1573
1631
i690
1748
1806
1865
1923
1981
2040
2098
58
745
2156
2215
2273
2331
2389
2448
2506
2564
2622
2681
58
746'
2739
2797
2855
2913
2972
3030
3088
3146
3204
3262
58
747
3321
3379
3437
3495
3553
3611
3669
3727
3785
3344
58
748
3902
3960
4018
4076
4134
4192
4250
4308
4366
4424
58
749
4482
4540
4598
4656
4714
4772
4830
4888
4945
5003
58
750
875061
5119
5177
5235
5293
5351
5409
5466
5524
.5582
58
751
5640
5698
5756
5313
5871
5929
5987
6045
6102
6160
58
•752
6218
6276
6333
6391
6449
6507
6564
6622
6680
6737
58
753
6795
6853
6910
6968
7026
7083
7141
7199
7256
7314
58
754
7371
7429
7487
7544
7602
7659
7717
7774
78;?2
7889
58
755
7947
8004
8062
8119
8177
82:4
8292
8349
8407
8464
57
756
8522
8579
8637
8694
3752
8309
8866
8924
8931
9039
57
757
9096
9153
9211
9268
9325
9333
9440
9497
9555
9612
57
T^Q
9669
9726
9784
9841
9398
9956
..13
.70
.127
.185
57
759
880242
0299
0356
0113
0471 ■).>2S
0585
.642
0699
0756
57
Ji.
1 I i 2. 1 3 1 4 1 5 1 6 1 7 i a 1 9 1 D. 1
A TABI.F OF LOGAItlTIIMS FR03I 1 TO 10,000.
13
N.
1 1 ! 2 1 3 1 4 1 5 1 6 i 7 1 8 i 9 1 D.
Wo'
880814,0871,
0928 09851
1042
1099
1 166
I213j 12711
1328. 57
701
1385
1442
1499
1556
1613
1670
1727
1784!
1841
1898! 57
703
1955
2012
2069
2126
2183
2240
2297
2354
2411
24681 57
763
2525
2581
2638
2695
2752
2809
2866
2923
2980130371 57
7(54
3093
3160
3207
3264
3321
3377
3434
3491
3548! 3605 57
765
3661
3718
3775
3832
3888
3945
4002
4059
4115U172' 57
766
4229
4285
4342
4399
4455
4512
4569
4625
46821 4739
57
767
4795
4852
4909
4965
5022
5078
6135
5192
52481 5305
57
768
5361
5418
5474
5531
5587
5644
5700
5757
5813, 6870
57
769
770
5926
886491
5983
6547
6039
6096 ■
6660
6152
6716
6209
6773
6265
6829
6321
6885
63781 6'i34
6942! 6998
56
56
6604
771
7054
71111
7167
7223
7280
7336
7392
7449
7505 1 7561
56
772
7617
7674
7730
7786
7842
7898
7955
8011
8067 8123
56
773
8179
8236
8292
8348
8404
8460
85 J 6
8573
8629 8685
56
774
8741
8797
8853
8909
8965
9021
9077
9134
9190 9246
56
775
9302
9358
9414
9470
9526
9582
9638
9694
97501 9806
56
776
9862
9918
9974 ..30|
..86
.141
.197
.253
.309 .365
56
777
890421
0477
0533
0589
0645
0700
0756
0812
0868 0924
56
778
0980
1035
1091
1147
1203
1259
1314
1370
1426 1482
66
779
1537
1593
1049
1705
1760
1816
1872
1928
1983 2039
56
780
892095
2150
2206
2262
2317
2373
2429
2484
2540 2595
~56
781
2651
2707
2762
2818
2873
2929
2985
3040
3096 3151
56
782
3207
3262
3318
3373
3429
3484
3540
3595
3651
3706
56
783
3762
3817
3873
3928
3984
4039
4094
4150
4205
4261
55
784
4316
4371
4427
4482
4538
4593
4648
4704
4759
4814
55
785
4870
4925
4980
5036
5091
5146
5201
5257
5312
5367
65
786
5423
5478
5533
5588
5644
5699
5754
5809
6864
6920
55
787
5975
6030
6085
6140
6195
6251
6306
6361
6416
6471
55
788
6526
6581
6636
6692
6747
6S02
6857
6912
6967
7022
55
789
790
7077
7132
7682
7187
7242
7792
7297
7847
7352
7902
7407
7957
7462
8012
7517
8067
7572
8122
55
55
897627
7737
791
8176
8231
8286
8341
8396
8451
8506
8561
8616
8670
55
792
8725
8780
8835
8890
8944
8999
9054
9109
9164
9218
65
793
9273
9328
9383
9437
9492
9547
9602
9656
9711
9766
55
794
9821
9875
9930
9985
..39
..94
.149
.203
.258
.312
65
795
900367
0422
047G
0531
0586
0640
0695
0749
0804
0869
55
79G
0913
0968
1022
1077
1131
1186
1240
1295
1349
1404
55
797
1458
1513
1567
1622
1676
1731
1785
1840
1894
1948
54
798
2003
2057
2112
2166
2221
2275
2329
2384
2438
2492
54
799
800
2547
2601
3144
2655
3199
2710
2764
3307
2818
3361
2873
3416
2927
3470
2981
3524
3036
3578
64
64
903090
3253
801
3633
3687
3741
3795
3849
3904
3958
4012
4066
4120
54
802
4174
4229
4283
4337
4391
4445
4499
4553
4607
4661
64
803
4716
4770
4824
4878
4932
4986
5040
5094
5148
6202
54
804
5256
5310
5364
5418
5472
5526
5580
56.34
5688
6742
54
805
5796
5850
5904
5958
6012
6066
6119
6173
6227
6281
54
806
G335
6389
6443
6497
6551
6604
6658
6712
6766
6820
54
807
6874
6927
6981
7035
7089
7143
7196
7250
7304
7358
64
80ri
7411
7465
7519
7573
7626
7680
7734
7787
7841
7895
54
809
810
7949
908485
8002
8539
8056
8110
8163
8217
8753
8270
8807
8324
8378
8914
8431
8967
• 64
54
8592 8646
8699
8860
811
9021
9074
9128 9181
9235
9289
9342
9396
9449
9503
54
812
9556
9610
966319716
9770
9823
9877
9930
9984
..37
53
813
910091
0144
0197
0251
0304
0358
0411
0464
0518
0571
53
814
0624
0678
0731
0784
0838
0891
0944
0998
1051
1104
63
815
1158
1211
1264
1317
1371
1424
1477
1530
1584
1637
53
816
1690
i 1743
1797
1850
1903
1956
2009
2063
2116
2169
53
817
2222
12275
2328
2381
2435
2488
2541
2594
2647
2700
53
818
2753
12806
2859
2913
2966
3019
3072
3125
3178
3231
53
82^9
3284
i 3337
3390
3443
3496
3549
1 3002
3655! 3708
376 ll 53 1
^nT
1 1 1 1 2 1 3 1 4 ! 5 1 6 1 7 i 8 i 9 1 O.
1
14
A TAnLE OF IOnAl?ITir3IS FK031 I TO 10,000.
N.
( 1 1 ! 2 I 3 1 4 1 5 1 G 1 7 1 8 i 9 1 D. 1
S2()
913314, 38871 3920| 3973, 4026
4079
(4132 41S4|4237
I 4290r'63
821
4343 4396
1 4449 j 4502
4555
4608
4660
1 4713
4766
4819: .53
822
4872 4925
1 4977! 5030
50S3
5135
5189
1 5241
52941 5347
53
823
5400 5453
55051 5558
5611
5664
5716
1 5769
5822
! 5875
! 53
824
5927 5980: G033! 6035
6138
6191
6243
16296
6349
! 6401
! 53
825
6454(6507
655916612
6664
6717
6770
16822
6875
i 692?
53
826
6980i 7033
7085 7138
7190
7243
7295
7348
7400
7453
1 53
827
7506
1 7558
761 1| 7663
7716
7768
7820
78731 7925
7978
52
828
8030
8083
813518188
8240
8293
8345
1 8397i 8450
8502
62
829
830
8555
919078
8607
9130
8659|8712
91831 9235
8764
9287
8816
8869
8921
9444
8973
9496
9026
9549
62
'.02
9340
9392
831
9601
9653
9706 j 9758
9810
9862
9911
9967
..19
..71
52
832
920123
0176
0228 0280
0332
0384
0438
0489
0541
0593
52
833
0645
0697
0749 0801
0853
0906
0958
1010
1062
1114
52
834
1166
1218
1270 1322
1374
1426
1478
J. 530
1582
1634
52
835
1686
1738
1790| 1842
1894
1946
1998
2050
2102
2154
52
836
2206
2258
231012362
2414
2466
2518
2570
2622
2671
52
837
2725
2777
2829 2881
2933
2985
3037
3089
3140
3192
62
838
3244
3296
3348 3399
3451
3503
3555
3607
3658
3710
52
839
840
3762
3814
4331
3865 3917
4383! 4434
3969
4486
4021
4533
4072
4589
4124
4641
4176
4693
4228
4744
52
52
924279
84 J
4796
4848
4899
4951
.5003
5054
5108
5157
5209
5261
62
842
5312
5364
5415
5467
5518
5570
5621
5673
5725
6776
.52
843
5828
5879
5931
5982
6034
6085
6137
6188
6240
6291
61
844
6342
6394
6445
6497
6648
6800
6661
67021 67.54
6805
51
845
6857
6908
6959} 7011
7062
7114
7165
7216J7268
7319
61
846
7370
7422
7473 7524
7576
7627
7678
7730
7781
7832
6L
847
7883
7935
7986 8037
8088
8140
8191
8242
8293
8.345
51
848
8396
8447
8498 8549
8601
8652
8703
8754
8805
8857
51
849
850
8908
929419
8959
9470
9010|9061
9112
9623
9163
9215
9725
9266
9317
9827
9368
9879
51
51
9521
9572
9674
9776
851
9930
9981
..32
..83
.1.34
.185
.236
.287
.333
.389
51
852
930440
0491
0542
0592
0643
0694
0745
0796
0847
0898
51
853
0949
1000
1051
1102
1153
1204
1254
1305
1356
1407
51
854
1458
1509
1560
1610
1661
1712
1763
1814
1865
1915
51
855
1966
2017
2068
2118
2169
2220
2271
2322
2372
2423
51
856
2474
2524
2575
2626
2677
2727
2778
2829
2879
2930
51
857
2981
3031
3082
3133
3183
3234
3285
3335
.3386
3437
51
858
3487
3538
3589
3639
3690
3740
.3791
3841
3892
3943
51
859
860
3993
4044
4549
4094
4599
4145
4650
4195
4700
4246
4751
4296
4801
4347
4852
4397
4448
49.53
51
60
934498
4902
861
5003
5064
5104
5154
5205
5255
5306
5356
.5406
.5457
50
862
5507
5558
6608
5658
5709
6759' 6809
5860
.5910
5960
60
863
6011
6061
6111
6162
6212
6262' 6313
6363
6413
C463
60
864
6514
6564
.6614
6665
6715
6765
6815
6865
6916
6966
50
865
7016
7066
7117
7167
7217
7267
7317
7367
7418
7468
60
8661
7518
7568
7618
7668
7718
77J69
7819
7869 7919
7969
60
867
8019
8069
8119
8169
8219
8269
8320
8370 8420
8470
50
868
8520
8570
8620
8670
8720
8770
8820
8870 8920
8970
50
869
9020
9070
9120
9170
9220
9270
9320
9369 9419
9469
50
870
9395-9
9569
9619
9669
9719
9769
9819
9869
9918
9968
60
871
940018
0068
0118
0168
0218
0267
0317
0367
0417
0467
50
872
0516
0566
0616
0666
0716
0765
0815
0865
0915
0964
50
873
UH4
1064
1114
1163
1213
1263
1313
1362
1412
1462
50
874
1511
1561
1611
1660
1710
1760
1809
1859
iq09
1958
50
875
2008
2058
2107
2157
2207
2256
2306
2355
2405
2455
50
876
2504
2554
2603
2653
2702
2752
2801
28511
2901
2950
50
877
3000
3049
3099
3148
3198
3247
3297
33461 33961 3445 i
49
878
3495
35-44
3593
3643
3692
3742!
3791
3841 3890i 3939]
49
879
3989
4038 4088' 4137
41861 42361 4285
43351 4334' 4433 49 |
N.
i 1 1 2 1 8 1 4 1 5 ' 6 1 7 1 8 ' 9 1 D. 1
A TABLE Cil'- LOOAKTTIIMS FROM T TO 10,000.
1^
N.
1 1 1 2 1 3 1 4 i 5 1 6 1 7 1 8 1 9 ; l>. 1
88b"
944483 4532
45 SI 46311
4680
4729
4779
48ii8i4877i49--;71 49 1
881
4976
5025
5074
5124
5173
5222
5272
5321 .5370
54191 49
882
5469
5518
5567
5616
5665
5715
5764
.581315862
59 1 21 49
883
5981
6010
6059
6108
6157
6207
6256
6305] 0354
6403! 49
884
6452
C501
6551
6600
6649
6698
6747
6796 1 6845
6894 49
88.'i
6943
6992
7041
7090
7140
7189
7238
72871 7336
7385
49
886
7434
7483
7532
7581
7630
7679
7728
7777 7826
7875
49
887
7924
7973
8022
8070
8119
8168
8217
8266
8315
8364
49
888
8413
8462
8511
8560
8609
8657
8706
8755
8804
8853
49
889
8902 8951
8999
9048
9097
9146
9195
9244
9292
9341
49
890
949390
94nyi 4488
9536
9585
9634
9683
9731
9780
9829 49 1
891
9878
99/6:9975
..24
..73
.121
.170
.219 .2671
.316
49
892
950365
04 4; 0462
0511
0560
0608
0057
0706
0754
0803
49
893
08*=:
0.>00
0949
0997
1046
1005
1143
1192
1240
1289
49
894
13J8 1386
1435
1483
1532
1.580
1629
1677
1726
1775
49
8^5
1823
18:2
1920
1969
2017
2066
2114
2163
2211
2260
48
d96
2308
2356
2405
2453
2502
2550
2599
2647
2696
2744
48
897
2792
2841
2889
2938
2986
3034
3083
3131
3180
3228
48
898
3276
3325
3373
3421
3470
3518
3566
3615
3663
3711
48
899
3760
3808
3856
3905
3953
4001
4049
4098
4146
4194
48
9U0
954243
4291
4339
4387
4435
4484
4532
4580
4628
4677
ll8
901
4725
4773
4821
4869
4918
4966
.5014
5062 5110
5158
48
902
5207
5255
5303
5351
5399
5447
5495
5543 5592
5640
48
903
5688
5736
5784
5832
5880
5928
5976
6024 6072
6120
48
904
6168
6216
6265
6313
6361
6409
6457
6505 6553
6601
48
905
6649
6697
6745
6793
6840
6888
6936
6984
7032
7080
48
go*?
7128 7176
7224
7272
7320
7368
7416
7464
7512
7559
48
907
7607 7655
7703
7751
7799
7847
7894
7942
7990
8038
48
908
8086
8134
8181
8229
8277
8325
8373
8421
8468
8516
48
909
8564
8612
8659
8707
8755
8803
8850
8898
8946
8994
48
910
959041
9089
9137
9185
9232
9280
9328
9375
9423
9471
48
911
9518
9566
9614
9661
9709
9757
9804
9852
9900
9947
48
912
9995
..42
..90
.138
.185
.2.33
.280
.328
.376
.423
48
"913
960471
0518
0566
0613
0661
0709
0756
0804
0851
0899
48
914
0946
0994
1041
1089
1136
1184
1231
1279
1326
1374
47
915
1421
1469
1516
1563
1611
1658
1706
1753
1801
1848
47
916
1895
1943
1990
2038
2085
2132
2180
2227
2275
2322
47
917
2369
2417
2464
2511
2559
2606
2053
2701
2748
2795
47
918
2843
2890
2937
2985
3032
3079
3126
3174
3221
3268
47
919
3316
3363
3410
3457
3504
3552
3.-99
3646
3693
3741
47
920
963788
3835
3882
3929
3977
4024
4071
4118
4165
^212
47
921
4260
4307
4354
4401
4448
4495
4542
4590
4637
46841 47 1
922
4731
4778
4825
4872
4919
4966
.''-013
5061
5108
51.55
47
923
5202
5249
5296
5343
5390
5437
5484
5531
55781 5625
47
924
5672
5719
5766
5813
5860
5907
6954
6001
6048
6095
47
925
6142
6189
6236
6283
6329
6376
6423
6470
6517
6564
47
926
6611
6658
6705
6752
6799
6845
6892
6939
6986
7033 47
927
7080
7127
7173
7220
7267
7314
7361
7408
7454
7501 47
928
7548
7595
7642
7688
8156
7735
7782
7829
7875
7922
7969 47
929
930
8016
8062
8530
8109
8576
8203
8249
8296
8763
8343
8810
8390
8856
8430
8903
47
47
96S483
8623! 8670i 8716
931
8950
8996
9043
9090
91.36 9183
9229
9276
9323 1 9369
47
932
9416
9463
9509
9556
9602 9649
9695
9742
978919835
47
933
9882
9928
9975
..21
..68 .114
.161
.207
.254
.300
47
934
970347
0393
0440
0486
05.33 0579
0626
0672
0719
0765
46
935
0S12
0858
0904
0951
0997 1044
109011137
1183
1229
46
936
1276
1322
1,369
1415
1461 1.508! 15.541 1601
1647
1693
40
937
1740
1786
1832
1879
1925 19711 2018
2064
2110
2157! 461
933
2203
2249
2295
2342
238812434! 2481
2527
257312619! 46 |
_939_
266612712
' 2758' 2804' 2851' 2897' 2943' 2989' 3035' 3082 46 |
N.
1 1 1 1 2 1 3
I ^•-.
1 5 ! .6 i 7 1 8 I 9 1 0. 1
'16 ~
16
A TABLE OF LlWJAKITIi:«S FROM 1
TO 10,000
1 I 1 2 1 3 I 4 1 5 1 6 1 7 1 8 1 9 1 D. 1
9i0|
97312813174132201
3266 331313359! 34051
3451 34971
35431
46
941
3590
3636! 36S2
3728
3774 3820]
3866
3913
3959
4005
46
942
4051
409714143
4189
4235
4281
4327
43 ?4
4420
4466
46
943
4512
45581 4604
4650
4696
4742
4788
4S34
4880
4926
46
944
4972
5018 5064
5110
5106
5202
5248
.5294
5340
5386
46
945
5432
5478 5524
5570
5616
5662
5707
5753
5799 58451
46
916
5891
59371 5983
6029 60751
6121
6167
6212
6258
6ii04
46
917
6350 6396! 6442|
6488
6533
6579
6625 66711
6717
6763
46
918
6808
6854 6900
6946
6992
7037
7083! 712917175
7220
46
949
7266
7312 7358
7403
7449
7495
7541
75861 7632
7678
46
9ol)
977724
7769 7815
786 r
790G
71.52
7993
8043 8089
81.35
'46
951
8181
8226 8272
8317
8363
8409
8454
8500 8540
8591
9047
46
952
8637
8683 8728
8774
8819
8865
8911
8956 9002
4a
953
9093
9138
9184
9230
9275
9321
9366
9412 9457
9503 46 1
954
9548
9594
9639
9685 9730
9776
9821
9867 9912
9958
46
955
980003 j 00491
0094
0140
0185
0231
0276
0322
0367
0412
45
956
0458
0503
0549
0594
0640
0685
0730
0776
0821
0867
45
957
0912
0957
1 003
1048
1093
1139
1184
1229
1275
1320
45
958
1366
1411
1456
1501
1547
1592
1637
1683
1728
1773
45
959
960 1
1819
1864
2316
1909
2362
1954
2407
2000
2045
2497
2090
2543
2135 2181
258^ 2633
2226
2678
45
45
982271
2452
961
2723
2769
2814
2859
2904
2949
2994
304C
3085
3130
45
962
3175
3220
3265
3310
3356
3401
3446
3491
3536
3581
45
963
3626
3671
3716
3762
3807 3852i 3897
3942
3987
4032
45
961
4077
4122
4167
4212
4257; 4302
4347
4392
4437
4-182
45
965
4527
4572
4617
4062
4707
4752
4797
4842
4887
4932
45
966
4977
5022
5067
5112
51.57
5202
5247
5292
5337
5382
45
967
5426
5471
5516
5561
5606
5651
5696
5741
5786
.5830
45
968
5875
5920
5965
6010
6055
6100
6144
6189
6234
0279
45
969
970
6324
6369
6413
6861
6458
6906
6503
6951
6548
6996
6593
7040
6637
7085
6682
7130
6727
7175
45
45
986772
6817
971
7219
7264
7309
7353
7398
7443
7488
7532
7577
7622
45
972
7666
7711
7756
7800
7845
7890
7934
7979
8024
8068
45
Q73
81 13
8157
8202
8247
8291
8336
8381
8425
8470
8514
45
074
8559
8604
8648
8693
8737
8782
8826
8871
8916
8960
45
975
9005
90491 9094
9138
9183
9227
9272
,9316
9.361
9405
45
976
9450
9494
9539
9583
9628
9672
9717
9701
9806
9850
44
977
9895
9939
9983
..28
..72
.117
.161
.206
.2.50
.294
44
978
990339
0383
0428
0472
0516
0.561
0005
0650
0694
0738
44
979
98^
0783
991226
0827
0871
0916
1359
0960
1403
1004
1448
1049
1492
1093
1536
1137
1580
1182
1625
44
44
1270
1315
981
1669
1713
1758
1802
1846
1890
1935
1979
2023
2067
44
982
2111
2156
2200
2244
2288
2333
2377
2421
2465
2509
44
983
2554
2598
2642
2686
2730
2774
2819
2863
2907
2951
44
984
2995
3039
3083
3127
3172
3216
3260
3304
3348
33921 44 1
985
3436
3480
3524
3568
3613
3657
3701
3745
3789
3833
44
986
' 3877
3921
3965
4009
4053
4097
4141
4185
4229
4273
44
987
4317
4361
4405
4449
4493
4537
4581
4625
4669
4713
44
988
4757
4801
4845
4889
4933
4977
5021
.5065
5108
51.52
44
989
990
5196
5240
5679
6284
5328
5372
.5811
5416
5854
5460
.5898
5504
5942
5547
5986
5.591
6030
44
44
995635
5723
5767
991
6074
6ir<
6161
6205
6249
6293
6337
6380
6424
64681 44 1
992
6512
6555
6599
6643
6687
6731
6774
6818
6862
6906
44
l;93
6949
6993
7037
7080
7124
7168
7212
7255
7299
7343
4-1
094
7386! 7430
7474
7517
7.561
7605! 7648
7692
77.3C
7779
44
P')5
7823
7867
7910
7954
7998
8041
8085
8129
8172
8216
44
£96
8259
8303
8347
8390
8434
8477
8521
8564
8603
8652
44
997
8695
8739
8782
882018869
8913
8956
9000
9043
9087
44
998
9131
9174
9218
90j1 9305
9348
9392
9435
9479
9522
44
939
9565
9'>09
9R5'2l 9Hnn|973!tl 9783
9826
9870
99131 99 >7
43
N.
1 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 o ' D. 1
A TABLE
OF
LOGARITHMIC
8INES AND TANGENTS
FOR EVERT
DEGREE AND MINUTE
OF THE QUADRANT.
N. B The minutes in the left-hand column of each page,
increasing downwards, helong to the degrees at the top ; and
those increasing upwards, in the right-hand column, belong to
the degrees below.
18
(0 Tegree.) a table of logarithmic
M.
1 Sine
1 1).
1 cnsi,... |n.
I Ta,.!:. 1 1).
1 <,"ni:uu. 1 1
"T
O.OihJOJO
10.000000
000000
0.000000
lllii:llll:.
"60
1
G. 463726
.501 717
00
6.463726
501717
13.. 536274
59
2
764756
293485
000000
00
764756
293483
235244
58
3
940817
208231
000000
00
940847
208231
0591.53
57
4
7.065786
161517
000000
00
7.06578H
161517
12.934214
56
f)
162696
13198S
OJOJOO
00
162696
131939
83/304
55
6
241877
111.575
9.99');)99
01
241878
1 1 1 578
7.58122 54
7
308824
966.53
999999
01
308825
996.53
691175 53
8
366816
85254
999999
01
.366817
852.54
633183 52
9
417968
76263
999999
01
417970
76263
6820301 51
10
il
463725
7.5U5118
68988
'62981
999998
01
01
463727
68988
.536273
50
9.999998"
7.505120
62981
12.494880
49
12
542906
57936
999997
01
642909
57933
4.57091
48
13
577668
53641
999997
01
577672
53642
422328
47
14
609853
49933
999996
01
609857
49939
390143
46
15
639816
46714
999996
01
639820
46715
360180
45
16
667845
43881
999995
01
667849
43882
332151
44
17
694173
41372
999995
01
694179
41373
30582 i
43
18
718997
39135
999994
01
719003
39136
280997
42
19
742477
3/127
999993
01
742484
37128
2.57:- 16
41
20
764754
3.5315
999993
01
764761
J5136
235239
40
21
7.785943
33672
9. 99999 .
01
7.785951
33673
12.214U49
39
22
806146
32175
999991
01
8061.55
.32176
19.3845
38
23
82545J
843931
30805
999990
01
825460
30806
174.540
37
24
29547
999989
02
843944
29549
1.56056
36
25
861662
28388
999988
02
861674
2S390
138326
35
26
878695
27317
999988
02
873708
27318
121292
34
27
895085
26323
999987
02
895099
26325
104901
33
28
910879
25399
999986
02
910894
2.5401
089106
32
29
926119
24538
999935
02
926134
24^^40
073S66
31
30
940842
23733
999933
02
940858
23735
059142
30
3f
7.955082
22980
9.9990vS2
02
7.955100
22981
12.044J00
29
32
968S70
22273
999981
02
968889
22275
031111
28
33
982233
21608
9'Jl» J 80
02
982253
21610
017747
27
b4
995198
20981
9& 979
02
995219
2 )983
004781
26
35
8.007787
20390
999977
02
8.007809
2 1392
11.992191
25
36
020021
19831
999976
02
020045
1)8:^3
979955
24
37
031919
19302
999975
02
031945
19305
968055
23
38
04350 i
18801
999973
02
043527
18803
956473
22
39
0.54781
18325
999972
02
054809
18.327
945 19 1
21
40
065776
17872
999971
02
065806
17874
. 934194
20
41
8.076.500
17441
9.iK)9969
02
8.076531
17444
11.923469
19
42
086966
17031
999968
02
08699?
17034
913003
18
43
097183
16639
999966
02
097217
16042
•902783
17
44
107167
16265
999964
03
107202
13268
892797
16
45
116926
15908
999963
03
116963
15910
883037
15
46
• 126471
1.5.566
999961
03
126510
l,5.56e
873490
!4
47
13.5810
1.5233
9999.59
03
13.5851
1.5241
864149
13
48
144953
14924
999958
03
144996
14^-^7
855004
12
49
1.53907
14622
999956
03
1.539.52
14327
846048
11
50
162681
14333
9999.54
03
162727
14336
837273
10
51
8.171280
140.54
9. J9yu.52
03
8.171328
1405'<
11 828672
9
52
179713
13786
9999.50
03
179763
13790
820237
8
53
187985
13.529
999948
03
188036
13532
811964
7
54
196102
13280
999946
03
1961.56
13284
, 803844
6
55
204070
13J41
999944
3
204126
13044
79.5874
5
56
211895
12810
999942
4
211953
12814
7880/. 7
4
67
219581
12.587
999940
04
219:541
12590
780359
3
58
227134
12372
999938
01
227195
12.37F
772805
2
60
231557
12164
999936
01
23462
12168
765379
1
60
241855
11 F3
999934 04
24I9>I
11)17
758079
n
rNHmT"!
Sine 1
(;..iu.jr. 1
•IW iM.|
m Degiecd.
SIXES AND TANGENTS. (1 Degree.]
10
A.
Sinu 1
l>.
Cosine 1 1).
r-.uiii.
D
<'()t;iim. 1
6
8.241855
11963
9.999^34
04
8.241921
11967
ll.V 580791 60
I
249033
11768
999932
04
249102
11772
75089^ y-y
2
256094
11580
999929
04
256185
11584
743835 .jH
3
263042
11398
999927
04
263115
1 1402
736885 57
4
269881
11221
999925
04
269956
11225
7300441 56
5
276814
11050
999922
04
276691
11054
7233091 55
6
2S3243
10883
999920
04
283323
10S87
716677
54
7
289773
10721
999918
04
2898.50
10728
710144
53
8
296207
10565
999915
04
296292
10570
703708
52
9
302546
10413
999913
04
302634
10418
697366
51
10
11
308794
8.314954
10266
999910
9.999907
04
04
308884
10270
10126
691116
50.
49
10122
8.315048
11.684954
12
321027
9982
999905
04
321122
9987
678878
4.S
13
327016
9847
999902
04
327114
9851
672888
47
14
332924
9714
999899
05
333025
9719
686975
46
15
338753
9586
999897
05
338S56
9590
681144
45
IG
344504
9460
999894
05
344610
9465
655390
44
17
.350181
9338
999891
05
350289
9343
64971 1
43
18
355783
9219
999888
05
355895
9224
644105
42
19
361315
9103
999885
05
301430
9108
638570
41
20
366777
8990
999882
05
368895
8995
633105
40
'Zl
8.372171
8880
9.999879
05
8.372292
8885
11.627708
39
22
377499
8772
999876
05
377622
8777
622378
38
23
382762
8667
999873
05
382889
8672
617111
37
24
337962
8564
999870
05
388092
8570
611908
36
25
393101
8464
999867
05
393234
8470
606766
35
20
.398179
8366
999864
05
398315
8371
601685
34
27
403199
8271
999861
05
403338
8276
596662
33
28
408161
8177
999858
05
408304
8182
691696
32
29
41.3068
8086
999854
05
413213
8091
686787
3i
30
417919
7996
999851
06
418068
8002
581932
30
31
8.422717
7909
9.999848
06
8.422869
7914
11.577131
29
32
427462
7823
999844
06
427618
7830
672332
28
33
432156
7740
999841
06
432315
7745
667685
27
34
436800
7657
999838
06
436962
7663
6630331 26 |
35
441394
7577
999834
06
441580
7583
658440
25
36
445941
7499
999831
06
448110
7505
553890
24
37
450440
7422
999827
06
450813
7428
549387
23
38
454893
7346
999823
06
455070
7352
544930
22
39
459301
7273
999820
06
459481
7279
640519
21
40
41
463665
8.467985
7200
993816
9.999812
06
06
463849
8.468172
7206
638151
20
19
7129
7135
11.531828
42
472263
7060
999809
06
472454
7066
527.546
18
43
473498
6991
999805
06
476693
8993
523307
17
44
480693
6924
999801
06
480892
6931
51910S
IG
45
484848
6859
999797
07
485050
6865
514950
15
46
489963
6794
999793
07
439170
8801
610830
14
47
493040
6731
999790
07
493250
6738
506750
13
48
497078
6869
999788
07
497293
6876
502707
12
49
501080
6608
999782
07
601298
6616
498702
11
50
505045
6548
999778
07
505267
8555
49-1733
10
5l'
8.508974
6489
9.999774
07
8.509200
6496
11.490300
9
52
512867
6431
999769
07
513098
6439
48H902
8
53
516726
6375
999765
07
516961
6382
483039
7
54
520551
6319
999761
07
520790
6326
479210
6
55
524343
6264
999757
07
524586
6272
475414
5
56
.528102
6211
999753
07
523349
6218
471651
4
57
531828
6158
999748
07
532080
6165
467920
3
58
535523
6108
999744
07
535779
6113
464221
2
59
539186
6055
999740
07
539447
8062
460553
I
CO
.542819
6004
999 73^
07
543;)8 !
6012
456916
n
Cosine
i Si.c |.
C'.;lan^.
Tiiiii? |M. 1
tib Degrees
so
(2 Derrn
3es.) A
TAHLE OF LOGAKITHMIC
M.
.-'IPK
1).
«'<!siiie 1 l».
'P.u. ir. i n.
1 r......... 1
.•3..)'i:^>iy
tkh) 1
9. 999 735
[07
8..543J>S4| 6.)'-^
ll i .45(iJiOi u.>
I
545422
5955
999731
07
.548691
5962
4533091 59
2
549J95
5998
999726
07
.550268
5914
449732 i 53
3
553539
5858
9^9722
08
553817
5866
448183; .57
4
557054
5811
999717
08
557336
5319
442664' 58
5
5fiJ540
583999
5765
999713
08
560828
5773
43917yL55
6
5719
999708
08
56429 I
5727
435709; 54
7
507431
5674
999704
08
567727
5682
432273; 53
8
570S30
5639
999899
08
571137
5638
428383 52
9
574214
5537
999894
08
• 574520
5595
425480; 5!
10
5775613
5544
999689
08
577877
5552
422123:50
IT
S.bSOSdZ
5592
9.999635
08
8.581208
5510
11.418 792 49
12
584193
5469
999680
03
.584514
5468
415486143
13
6-37469
5419
999675
08
587795
5427
412205:47
M
599721
5379
999670
03
591051
5337
408919; 46
15
593948
5339
999865
05
594283
5347
405717: 45
ffi
597152
5300
999660
03
597492
5308
402508, 44
17
699332
5261
999855
03
600877
5270
399323; 43
18
693489
5223
999850
08
603839
5232
3981611 42
19
696623
5186
999645
09
606978
5194
3930221 41
20
699734
5149
999840
09
610094
51.58
339996; 40
21
8.612S23
5112
9.999635
09
8.613189
5121
11.338311 39
22
615891
.5976
999829
09
616262
5035
383733; 33
23
618937
5941
999824
09
619313
.5050
3306371 37
24
621962
5998
999819
09
622343
5015
377857! 36
25
624965
4972
999814
ol;
625352
4931
3746481 35
2t)
627948
4938
999803
09
628340
4947
3ri680; 34
27
639911
4904
999803
09
631398
4913
388692; 33
2S
633354
4871
999597
09
634256
4880
365744 32
29
63t)776
4839
999592
09
637184
4848
382316 31
:v.)
639689
4806
999586
09
640093
4816
359907 30
31
8.642563
4775
9.999531
09
8 . 642982
4784
11.357018; 29
32
645428
4743
999575
09
645353
• 4753
354147| 28
31
648274
4712
999570
09
648704
4722
35I296i 27
31
601102
4632
999564
09
651537
4691
34S463' 26
35
653911
4652
999558
10
654352
4661
345648| 25
33
656702
4822
999553
10
057149
4631
34285 1 1 24
37
659475
4592
999547
10
G5992S
4692
3400721 23
3S
662230
4583
999541
10
662639
4573
337311122
39
664968
4535
999535
10
665433
4544
334567121
40
41
6G7639
8.6r0393
4506
4479
999529
9.999524
10
To
688160
8.670870
4526
33 1 840 1 20
11. 3291391 19
4438
42
673039
4451
999518
10
673563
4461
3284371 18
43
675751
4424
999512
10
676239
4434
323761 17
44
678405
4397
999506
10
678900
4417
3211001 16
45
> 681043
4370
999590
10
681.544
4380
3184581 15
46
683665
4344
999493
10
684172
4354
31.58281 14
47
636272
4318
999487
10
686784
4328
313216! 13
43
633363
4292
999481
10
639381
4303
3106!9| 12
49
691433
4267
99)475
10
691963
4277
308037; 1 1
5)
69399^
4242
999489
10
694529
4252
39547 ij 10
51
8.696543
4217
9.99r»4H3
ll
8.697081
4228
11.302919 9
52
699973
4192
999456
11
699617
4203
3003-53i 8
53
791589
4168
999450
11
702139
4179
29786 1 j 7
54
704099
4144
999443
11
704646
4155
' 295354! 6
55
706577
4121
999437
11
707140
4132
2928fH)| 5
SS
709949
4097
999431
11
709818
4108
29i>?32' 4
57
711507
4074
999424
11
712083
4085
237917 3
58
713952
4051
999418
11
714534
4062
235465; 2
5!)
7 1 63 -{3
40^9
999411
11
716972
4040
23:^028 j 1
fiO
_7l_SS00
40;)6
9'^M9l
11
719398
4') 17
28.n(»4'
C-^tnti
1
Smh 1
.;....-...t:. j 1
i:m.c. pr
UuKif
•^;
W
SINF.3 AND TANGE?fTS. ^3 DcgVCCS.^
21
M..,. 1
1). 1
<;n.«irie 1 It 1
T.... 1 M
<'..•,,..... 1
8.718800
4003
9.99Ji'»4(
11
8.719396
4017
ll.280«)04i hu
1
721204
3984
99939b
11
721806
3995
278l9l!.'i
2
723595
3982
999391
11
7242 J4
?..»74
275796 ..a
3
725972
3941
999384
11
726588
3952
2734 12 57
4
728337
3919
999378
LI
728959
3930
271041 .56
5
730(588
3898
999371
11
731317
3909
2686S3I .55
6
733027
3877
999364
12
733663
3889
266337 1 54
7
735354
3S57
999357
12
735996
3868
;^848
264004 53
8
7370o7
3835
999350
12
738317
261683152
9
739969
3816
999343
12
740626
3827
2.59374151
ill
742259
3796
999336
12
742922
3807
25707 8
50
li
8.744536
3776
9.999329
12
8.74.5207
3787
11.2.54793
49
12
746802
3756
999322
12
747479
3768
252521
48
la
749055
3737
999315
12
749740
3749
250260
47
14
751297
3717
999308
12
751989
3729
248011
46
15
753528
3698
999301
12
7.54227
3710
245773
45
IR
755747
3679
999294
12
756453
.3692
243547
44
17
757955
3661
999286
12
758668
3673
241332
43
18
760151
3642
999279
12
760872
36.55
239128
42
ly
762337
3624
999272
12
763965
3636
236935
41
20
764511
3606
999265
12
765246
3618
234754
4«)
2i
8.766675
3588
9.9992.57
12
8.767417
3600
11.232583
39
22
768828
3570
9992.50
13
769578
3583
23{)422
38
23
770970
3553
999242
13
771727 3.565
228273
37
24
773101
3535
999235
13
773866
3548
226134
36
25
775223
3518
999227
13
775995
3531
224005
35
2fi
777333
3501
999220
13
778114
3514
22188H
34
27
779434
3484
999212
13
780222
3497
219778
33
28
781.524
3467
999205
13
782320
34S0
217680
32
29
783605
3451
999197
13
784408
3464
21.5,592
31
:H)
785675
.3431
999189
13
786486
344 7
213514
30
31
8.787736
3418
9.999181
13
8.788.554
3131"
11.211446
29
32
789787
3402
999174
13
790613
3414
209387
28
33
791828
3386
999166
[3
792662
3399
207338
27
3t
793859
3370
9991.5S
13
7947(»l
3383
205299
26
35
795881
3354
999150
13
796731
3368
203269
25
3f)
797894
3339
999142
13
79S752
3352
201248
24
37
"799897
3323
999134
13
800763
3337
199237
23
3S
801892
3398
999126
13
802765
3322
197235
22
39
803876
3293
999118
13
804758
3307
195242
21
40
805852
3278
999110
13
80674 2 1 3292
1932.58
20
41
8.807819
3263
9.999102
13
8.808717 3278
11.191283
19
42
809777
3249
999094
14
810683 3262
189317
18
43
811726
3234
9990861 14
812641 3248
187359
17
44
813667
3219
9990771 14
814589 3233
18.5411
16
45
815599
3205
999069
14
816529
3219
183471
15
46
817522
3191
999061
14
818461
3205
181.539
14
47
819436
3177
999053
14
820384
3191
179616
13
48
821343
3163
999044
14
822298
3177
177702
12
49
823240
3149
999036
14
824205 i 3163
175795
11
50
8251.30
31.35
999027
9.999019
14
14
826103
8.827992
3150
3f36
173897 10 1
11.172008 9 1
8.827011
3122
52
828884
3108
999010
14
829874
3123
170126
8
53
830749
3095
999002
14
831748
3110
1682.52
7
54
832607
3082
998993
14
833613
3096
166.387
6
55
834456
3069
998984
14
835471
3083
164.529
5
5fi
83'^297
3056
998976
14
837321
3070
162679
4
57
838130
3043
998967
15
8J9163
3057
160837
3
58
839956
3030
998958
15
840998
3045
1590021 2
1.57175 1
59
841774
3017
99S950
15
842825
3032
no
843585
3000
9989H
15
8446441 3019
1.553561
z
(J<i.-ine
1
Si,e j
Vv.vms. 1
Ta .!■. \M
86 Uuifrcus.
22
(
4 Degrees.') a
TABLE OF LOGAKlTir.MlC
'.VI
f^iiif
1 n
Cosine 1 1).
_:!:^^_
__!'_
(Manj:.
"(T
8. y 43585
3005
J. 99894 11 15
8.844644
3019
ii. l5.5:}5Tr
fiT
1
845387
2992
998932 15
846455
3007
153545
"59
o
847183
2980
998923 15
848260
2995
151740
58
3
848971
2967
998914 15
850057
2982
149943
57
4
8507Jil
2955
998905 15
851846
2970
148154
56
5
852525
2943
998896 15
853628
2958
146372
55
i\
854291
2931
998887 15
8.5.5403
2946
14459/'
54
7
856049
2919
998378 15
857171
2935
142829
53
8
857801
2907
998869 15
858932
2923
141068
52
9
859546
289e
998860 15
860686
29 1 1
139314
51
10
861283
2884
998851 15
862433
2900
137567
50
11
8.863014
~2873
9.998841
15
8.864173
2888
11.135827
49
12
864738
2861
993832
15
865906
2877
134094
48
13
866455
2850
Q93823
16
867632
2866
132368
47
14
868165
2339
993813
16
869351
2854
130649
46
15
869368
2828
998804
16
871064
2843
128936
45
IT)
871565
2817
998795
16
872770
2832
127230
44
17
873255
2S06
998785
16
874469
2821
125.531
43
18
874938
2795
998776
16
876162
2811
123833
42
19
876615
2786
998766
16
877849
2800
122151
41
20
21
878285
8.87994'J
2773
998757
9.993747
16
16
879529
8.881202
2789
120471
40
39
2763
2779
11.118798
22
881607
2752
998733
16
882S69
2788
117131
38
23
883258
2742
998728
16
884530
2758
11.5470
37
24
884903
2731
998718
16
886185
2747
113815
3f)
25
886542
2721
998708
16
887833
2737
112167
35
20
888174
2711
993699
16
889476
2727
110524
34
27
889801
2700
998689
16
891112
2717
108388
33
28
891421
2690
998679
16
892742
2707
107258
.32
29
893035
2680
998669
17
894ii66
2697
105634
31
30
894643
2670
998659
17
895984
2687
104016
30
31
8.896246
2660
9.998649
17
8.897.596
2677
11.102404
29
32
897842
2651
998639
17
899203
2667
10(»797
28
33
899432
2641
998629
17
900803
26.58
099197
27
34
901017
2631
998619
17
902398
2648
097602
26
35
902596
2622
998609
17
903987
2638
096013
25
36
904169
2612
998599
17
905570
2629
094430
24
37
905736
2603
998589
17
907147
2620
0928.53
23
38
907297
2593
998578
17
908719
2610
091281
22
39
908353
2584
998568
17
910285
2601
039715
21
40
910404
2575
998558
17
911846
2592
088154
20
41
8.911949
2560
9.998.548
17
8.913401
2583
11.086599
19
42
913488
2150
998.537
17
914951
2574
08,5049
18
43
9150'^2
2547
998.527
17
916495
2565
083505
17
44
916.55«
2538
998516
18
918034
2556
081966
16
45
, 918073
2529
998.506
18
919568
2547
080432
15
46
919591
2520
998495
18
921096
2538
078904
14
47
921103
2512
998485
18
922619
2530
07738 1
13
48
922610
2.503
998474
18
924136
2.521
075864
12
49
924112
2494
998464
18
925649
2512
074351
H
50
925609
2486
998453
18
927156
2503
072844
10
51
8.927100
24^7
9.998442
18
8.9286.58
2495
11.071342
9
62
928.587
2469
998431
18
9301.55
2486
069845
8
53
930068
2460
998421
18
931647
2478
068353
7
54
931.544
2452
998410
18
933134
2470
' 066H66
6
55
933015
2443
908399
18
934616
2461
065384
5
56
934481
2435
998388
18
936093
2453
063907
4
57
935942
2427
998377
18
937565
2445
062435
3
58
937398
2419
998366
18
9390.32
2437
060968
2
59
93S850
2411
998355
18
940494
2430
059506
1
_60_
940296
2403
998344
18
941952
2421
058048
Co..;,e
ijiii<> 1
Ct>liin<!
Tana.
"mT'
eS DeKreen.
SINKS AND TANGENTS. (5 Degrees.)
23
M
t^iiio
I..
Cosine j 1)
i iViMu. 1 I).
1 C.au !
^
6.i)^>UJ6
■4iJ.i
9 . 90 -id k^
19
8.91:19)*
ZhZi
1 1 .0.^dsIM] 60
1
94l7tJS
2394
998333
10
943404
2413
056596' 59
2
943174
2387
993822
19
944S52
2405
055148158
a
944606
2379
90 S3 11
19
546205
^4773 1
2397
0.53705, 57
4
946034
2371
903300
19
2390
052266 .5G
f,
947456
2363
993239
19
949168
2382
050832 55
f.
948 S74
2355
993277
19
950597
2374
049403 54
V
950287
2343
993266
19
952021
2366
047979 53
b
951696
2340
998255
19
95344 1
2360
046559! 52
i
953100
2332
998243
19
954856
2351
045144151
0437331 50
10
954499
2325
998232
19
956267
2344
il'
8.935894
2317
9.99322.)
19
8.957674
2337
11.0423:iGJ4ji
Ik
957284
2310
998209
19
959075
2329
040925 48
i;
958670
2302
993197
19
960473
2323
039527 47
14
980052
2295
903186
19
961866
2314
038134 46
U
961429
2283
993174
19
963255
2307
036745 45
U>
962801
2280
998163
19
964639
2300
035361 44
17
964170
2273
908151
19
966019
2293
033981 43
IH
965534
2206
998139
20
917394
2286
032606 42
l!»
966893
2259
998128
20
963768
2279
031234 41
2;)
96 -{249
2252
998116
20
970133
2271
029867 40
2l"
8.96080)
2244
9.998104
20
8.971496
2265
11.028504 39
22
970947
2238
908092
20
972855
2257
027145 33
23
972289
2231
998080
20
974209
2251
025791 37
24
973628
2224
998068
20
975560
224-4
0244401 36
25
974962
2217
998056
20
976906
2237
023094 35
26
978293
2210
993044
20
978248
2230
021752 34
27
977619
2203
998032
20
979588
2223
020414 33
2S
978941
2197
993020
20
9309211 2217
019079 32
29
980259
2190
998008
20
932251
2210
0177491 31
3f)
9^1573
2183
997996
20
983577
2204
016423(30
31
8.982883
2177
9.997984
20
8.984899
2197
11.015101
29
3-7
984189
2170
997972
20
936217
2191
013783
28
33
985491
2163
997959
20
937532
2184
012468
27
3t
986739
2157
997947
20
938842
2178
011158
26
35
983083
2150
997935
21
990149
2171
009851
25
3^
989374
2144
997922
21
991-451
2165
008549
24
37
900660
2138
997910
21
992759
2158
007250
23
3S
931943
2131
9^7897
21
994045
2152
005955
22
33
903222
2125
997885
21
995337
2146
004663
21
40
901497
. 2119
907872
21
996624
2140
003376
20
41
8.0J5768
2112
9.907860
21
8.997903
2134
11.002002
19
42
907036
2105
997847
21
999188
2127
000812
18
43
908290
2100
997335
21
9.000465
2121
10.999535
17
44
999560
2094
937822
21
001738
2115
993262
16
45
9. 000816
2087
997800
21
093007
2109
996993 15
46
002069
2082
907797
21
004272
2103
9057281 14
47
003318
2076
99778 1 21
005534
2097
9944661 13
48
004563
2070
997771 21
008792
2091
903208 12
49
005805
2064
997758 21
003047
2085
991953 11
50
007044
2058
907745 21
009298
2080
990702 10
51
9.00S278
2052
9.997732
21
9.010546
2074
10.989454; 9
52
009510
2046
907719
21
011790
2068
988210 6
53
010737
2040
907706
21
01 3031
2062
986969; 7
54
011953
2134
907693
22
014263
2056
985732 n
55
013182
2029
907680
22,
015502
2051
98 44 OS 5
56
014400
2023
9976671 22
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2045
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974741;'
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026455 2000
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0276551 1995
972345
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026386 1967
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971 14 H
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03004 6 1 1985
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(»28744; 1957
997607
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0312371 1979
96876;]
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997493
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967575
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9.997466
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9. 03479 ll 1964
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964031
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0345821 1930
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037144 1953
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0357411 1925
997425
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038316 1948
961684
46
15
036896! 1920
997411
23
0394851 1943
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45
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1915
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040651 1938
959349
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039197
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041813! 1933
9.58187
43
18
040342
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042973 1928
957027
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19
041485
1899
997355
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0441301 1923
955870
41
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042625
1894
997341
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04.52841 1918
9.54716
40
21
9. 0437621 1889
9.997327
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9.0464341 1913
10.9.53566
39
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0448951 1884
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0,52144 1889
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053277 1884
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28
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1855
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24
054407 1879
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32
29
052749
1850
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05.55351 1874
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053859
1845
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nl
0549661 1841
9.99718.5
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9.0.57781 1865
10.942219
29
32
0560711 1836
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058900 1869
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1831
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24
060016 18.55
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1827
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0611.30 1851
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23
212291
1273
994150
35
218142
1308
781858
37
24
213055
1271
994129
35
218926
1305
781074
36
25
213818
1268
994108
35
219710
1303
780290
35
26
214579
1266
994087
35
220492
1301
779508
34
27
215338
1264
994066
35
221272
1299
778728
33
28
216097
1261
994045
35
222052
1297
777948
32
29
216854
1259
994024
'o5
222830
1294
777170
31
30
31
217609
1257
1255
994003
35
35
223606
1292
776394
30
29
9.218363
9.993981
9.224382
1290
10.775618
32
219116
1253
993960
35
225156
1288
774844
28
33
219868
1250
993939
35
225929
1286
774071
27
34
220618
1248
993918
.35
226700
1284
773300
26
35
221367
1246
993896
36
227471
1281
772529
25
36
222115
1244
993875
36
228239
1279
771761
24
37
222861
1242
993854
36
229007
1277
770993
23
38
223606
1239
993832
36
229773
1275
770227
22
39
224349
1237
993811
36
230539
1273
769461
21
40
41
225092
9.225S33
1235
1233
993789
36
36
231302
1271
768698
20
19
9.993768
9.232065
1269
10.767935
42
226573
1231
993746
36
232826
1267
767174
18
43
227.^.11
1228
993725
36
233586
1265
766414
17
44
228048
1220
993703
36
234345
1262
765655
16
45
228784
1224
993681
36
235103
1260
764897
15
46
229518
1222
993660
38
235859
1258
764141
14
47
230252
1220
993638
36
236614
1256
763386
13
48
230984
1218
993616
36
237368
1254
762632
12
49
231714
1216
993594
37
238120
1252
761880
11
50
61
232444
9.233172
1214
1212
993572
9.99.3550
37
37
238872
1250
761128
10.760378
io
9
9.239622
1248
52
233899
1209
993528
37
240371
1246
759620
8
53
234625
1207
993.506
37
241118
1244
758882
7
54
235349
1205
993484
37
241865
1242
758135
6
55
236073
1203
993462
37
242610
1240
757390
5
56
236/95
1201
993440
37
243354
1238
756646
4
57
237515
1199
993418
37
244097
1236
755903
3
58
238235
1197
993396
37
244839
1234
7.55161
2
59
238958
1195
993374
37
245579
1232
764421
1
60
239670
1193
993351
37
246319
1230
753681
i 1
Cosine i i
Siae 1 1
Cotaiig. 1
Tang. |M. 1
17*
8U
EE
28
(10 Dogr
ees.) A
TABLE OF LOGARITHMIC
M.
Situ;
D.
vvosine 1 D.
Tnus.
D.
1 Cotano. 1 1
"o"
9.239670
1193
9.993351
37
9.246319
1230
10.753081
60
1
240386
1191
993329
37
247057
1228
752943
69
2
241101
1189
993307
37
247794
1226
762206
58
3
241814
1187
993286
37
248530
1224
761470
57
4
242526
1186
993202
37
249264
1222
750736
66
6
243237
1183
993240
37
249998
1220
750002
55
6
243947
1181
993217
38
260730
1218
749270
64
7
244656
1179
993195
38
261461
1217
748539
63
8
245363
1177
993172
38
252191
1215
7478091 52 1
9
246069
1176
993149
38
252920
1213
7470801 51 1
10
11
246775
9.247478
1173
1171
993127
38
38
263648
9.264374
1211
7463.52
50
49
9.993104
1209
10.745626
12
248181
1169
993081
38
256100
1207
744900
48
13
248883
l.o7
993069
38
255824
1205
744176
47
14
249583
1165
993036
38
266.547
1203
743453
46
15
250282
1163
993013
38
257269
1201
742731
45
16
250980
1161
992990
38
257990
1200
742010
44
17
251677
1159
992967
38
258710
1198
741290
43
18
252373
1158
992944
38
269429
1196
740571
42
19
253067
1156
992921
38
260146
1194
7398.54
41
20
21
253761
1164
11,52
992898
9.992875
38
38
260863
1192
1190
739137
4_0
39
9.254453
9.261678
10.738422
22
265144
1150
992852
38
262292
1189
737708
38
23
255834
1148
992829
39
263005
1187
736995
37
24
266523
1146
992806
39
263717
1185
736283
36
26
257211
1144
992783
39
264428
1183
735572
36
26
257898
1142
992759
39
266138
1181
734862
34
27
268583
1141
992736
39
266847
1179
7341.53
33
28
269268
1139
992713
39
266565
1178
733445
32
29
259951
1137
992690
39
267261
1176
732739
31
30
31
260633
3135
992666
9.992643
39
39
267967
1174
1172
732033
10.731329
30
29
9.261314
1133
9.268671
32
261994
1131
992619
39
269376
1170
730626
28
33
262673
1130
992596
39
270077
1169
729923
27
34
263351
1128
992572
39
270779
1167
729221
26
35
264027
1126
992649
39
271479
1165
728621
25
36
264703
1124
G92626
39
272178
1164
727822
24
37
265377
1122
992601
39
272876
1162
727124
23
38
266051
1120
992478
40
273573
1160
726427
22
39
266723
1119
992464
40
274269
1158
725731
21
40
267395
1117
992430
40
274964
1157
725036
20
41
9.268065
1115
9.992406
40
9.276668
1156
10.724342
19
42
268734
1113
992382
40
276361
1153
723649
18
43
269402
1111
992359
40
277043
1151
722957
17
44
270069
1110
992336
40
277734
1150
722266
W5
45
270736
1108
992311
40
278424
1148
721576
16
46
271400
1106
992287
40
279113
1147
720887
14
47
272064
1105
992263
40
279801
1146
720199
13
48
272726
1103
992239
40
280488
1143
719512
12
49
273388
1101
992214
40
281174
1141
718826
11
60
51
274049
9.274708
1099
992190
40
40
281858
1140
1138
718142
10.717468
50
9
1098
9.992166
9.282542
52
275367
1096
992142
40
283225
1136
716776
8
53
276024
1094
992117
41
283907
1136
716093
7
54
276681
1092
992093
41
284688
1133
, 715412
6
-55
277337
1091
992069
41
286268
1131
714732
5
56
277991
1089
992044
41
285947
1130
714063
4
57
278644
1087
992020
41
286624
1128
713.376
3
58
279297
1086
991996
41
287301
1126
712699
2
59
279948
1084
991971
41
287977
1125
712023
1
60
280599
1082
991947
41
288662
1123
711348
Cosine 1
Bine
Ctang. 1
1
Tang. IM.|
97 Degrees.
siKES AND TANGENTS. (11 Degrees.)
29
M.
1 Sine
1 n.
1 Cosiiifi 1 n.
i Tgn..
1 D.
1 Ccrta.,!:. 1 \
"(T
[9.280599
1082
9.991947
41
9.288652
1123
10.711348, 60 1
1
281248
1081
991922
41
289326
1122
710674
59
2
281897
1079
991897
41
289999
112U
710001
58
3
282544
1077
991873
41
290671
1118
709329
57
4
283190
1076
991848
41
291342
1117
708658
56
5
283836
1074
991823
41
292013
1115
707987
55
6
284480
1072
991799
41
292682
1114
707318
54
7
285124
1071
991774
42
293350
1112
7066501531
8
285766
1069
991749
42
294017
1111
705983! 521
9
286408
1067
991724
42
294684
1109
705316
51
10
11
287048
9.287687
1066
991699
9.991674
42
42
295349
1107
704651
50
49
1064
9.296013
1106
10.703987
12
288326
1063
991649
42
296677
1104
703323
48
13
288964
1061
991624
42
297339
1103
702661
47
14
289600
1059
991599
42
298001
1101
701999
46
15
290236
1058
991574
42
208662
1100
701338
45
16
290870
1056
991549
42
299322
1098
700678
44
17
291.504
1054
991524
42
299980
1096
700020
43
IS
293137
1053
991498
42
300638
1095
699362
42
19
292768
1051
991473
42
301295
1093
698705
41
20
293399
1050
991448
42
301951
1092
698049
40
21
9.294029
1048
9.991422
42
9.302607
1090
10.697393
39
22
294658
1046
991397
42
303261
1089
696739
38
23
295286
1045
991372
43
303914
1087
696086
37
24
295913
1043
991346
43
.304567
1086
695433
36
25
296539
1042
991321
43
305218
1084
694782
35
26
297164
1040
991295
43
305869
1083
694131
34
27
297788
1039
991270
43
306519
1081
693481
33
28
298412
1037
991244
43
307168
1080
692832
32
2 'J
299034
1036
991218
43
307815
1078
692185
31
3!)
31
299655
9.300276
1034
991193
9.991167
43
43
308463
9.309109
1077
691537
30
29
1032
1075,
10.690891
32
300895
1031
991141
43
309754
1074
690246
28
33
301514
1029
991115
43
310398
1073
689602
27
34
302132
1028
991090
43
811042
1071
688958
26
35
302748
1026
991064
43
311685
1070
688315
25
36
303364
1025
991038
43
312327
1068
687673
24
37
303979
1023
991012
43
312967.
1067
687033
23
38
304593
1022
990986
43
313608
1065
686392
22
39
305207
1020
990960
43
314247
1064
685753
21
40
305319
1019
990934
44
314885
1062
685115
20
41
9.306430
1017
9.990908
44
9.315523
1061
10.684477
19
42
307041
1016
990882
44
316159
1060
683841
18
43
307650
1014
990855
44
316795
1058
683205
17
44
308259
1013
990829
44
317430
1057
682570
16
45
308887
1011
990803
44
318064
1055
681936
15
46
309474
1010
990777
44
318697
1054
681303
14
47
310080
1008
990750
44
319329
1053
680671
13
48
310685
1007
990724
44
319961
1051
680033
12
49
311289
1005
990697
44
320592
1050
679408
11
50
311893
1004
990671
44
321222
1048
678778
10
51
9.312495
1003
9.990644
44
9.321851
1047
10.678149
9
52
313097
1001
990618
44
322479
1045
677521
8
53
313698
1000
990591
44
323106
1044
676894
7
54
314297
998
990565 44
323733
1043
676267
6
55
314897
997
990538 44
324358
1041
675642
5
56
315495
996
990511 45
324983
1040
675017
4
57
316092
994
J90485 45
325607
1039
674393 3
58
316689
993
990458 45
326231
1037
673769 2
59
317284
991
990431 45
326853
1036
673117 1
€0
317879
990
990404 45
327475
10.35
672525
I Cosine
• 8iiie 1
) (Jolaiu.
Tan-, j
78 Degrees
30
(12 Degrees.) a
TABLE OP I.OGARITHMIC
M.
1 Si.ie
! i>.
1 Cosine | D.
1 Timti.
1 D.
1 Cofaii;;. i 1
"IT
9.317879
990
9.&9U404
45
9.327474
1035
10.672526
60
1
318473
988
990378
45
328095
1033
671905
59
2
319066
987
990351
45
328715
1032
671285
58
3
319658
986
990324
45
329334
1030
670G66
57
4
320249
984
990297
45
329953
1029
670047
56
5
6
320840
983
990270
45
330570
1028
669430
55
"3til475tr
~ 982
990243
45
331187
1026
668813
54
7
322019
980
990215
45
331803
1025
668197
53
8
322607
979
990188
45
332418
1024
667582
52
9
323194
977
990161
45
333033
1023
066967
51
10
11
323780
9.324366
976
990134
9.990107
45
46
333646
1021
666354
10.66.5741
50
49
975
9.334259
1020
12
324950
973
990079
46
334871
1019
665129
48
13
325534
972
990052
46
335482
1017
664518
47
14
326117
970
990025
46
336093
1016
663907
46
15
326700
969
98'.997
46
336702
1015
663298
45
16
327281
968
989970
46
337311
1013
662689
44
17
327862
966
989942
46
337919
1012
662081
43
18
328442
965
989915
46
338527
1011
661473
42
19
329021
964
989887
46
339133
1010
660867
41
20
21
329599
962
989860
46
46
339739
1008
1007
660261
10.659656
40
39
9.330176
961
9.989832
9.340344
22
330753
960
989804
46
340948
1006
6590.52
38
23
331.329
958
989777
46
341552
1004
658448
37
24
331903
957
989749
47
342155
1003
657845
36
25
332478
956
989721
47
342757
1002
657243
35
26
333051
954
989693
47
343358
1000
656642
34
27
333624
953
989665
47
343958
999
656042
33
28
334195
952
989637
47
344558
998
655442
32
29
334706
950
989609
47
345157
997
654843
31
30
31
335337
949
989582
9.989553
47
47
345755
9.346353
996
994
654245
10.653647
30
29
9.335906
948
32
336475
946
989525
47
346949
993
653051
28
33
337043
945
989497
47
347545
992
652455
27
34
337610
944
989469
47
.348141
991
651859
26
35
338176
943
989441
47
348735
990
651265
25
36
338742
941
989413
47
349329
988
650671
24
37
339306
940
989384
47
349922
987
650078
23
38
. 339871
939
989356
47
350514
986
649480
22
39
340434
937
989328
47
351106
985
64P894
21
40
41
340996
936
989300
9.989271
47
47
351697
983
982
648303
20
19
9.341558
935
9.352287
10.647713
42
342119
934
989243
47
352876
981
647124
18
43
342679
932
989214
47
353465
980
646535
17
44
343239
931
989186
47
354053
979
645947
16
45'
343797
930
9891.57
47
354640
977
645360
16
46
344355
929
989128
48
355227
976
644773
14
47
344912
927
989100
48
355813
975
644187
13
48
345469
926
989071
48
356398
974
643602
12
49
346024
925
989042 48 |
356982
973
643018
11
50
346579
924
989014
48
357566
971
642434
10
51
9.347134
922
9.988985
48
9.358149
970
10.641851
9
62
347687
921
988956
48
358731
969
641269
8
53
348240
920
988927
48
359313
968
640687
7
54
348792
919
988898
48
359893
967
^ 640107
6
55
349343
917
988869
48
360474
966
639526
5
66
349893
916
988840
48
361053
965
638947
4
57
350443
915
988811
49
361632
963
638368
3
58
350992
914
988782
49
362210
962
637790
2
69
351540
913
9887531 49 1
r62787
961
637213
1
60
352088
911
9887241 49 '
363364
960
6.36636
Li
Cosine j
Sine 1 1
CuUMg. 1
1
Tang 1 M. |
7T Ufigrces.
SINES AND TANGENTS. ^^13 Degrees.)
31
I Cosine I I). I T.iMg. | I). \ Ci
I 9.3-V20*o!8
' 3.3263.=
.373933
374452
374970
375487
376003
376519
377035
377549
378003
378577
.379089
379601
380 113
380624
381134
381643
382152
3S2661
383 16S
383675
898
897
896
895
893
092
891
890
889
888
887
885
884
8S3
882
881
880
879
87r
876
875
874
873
872
871
870
869
867
866
865
864
883
862
861
860
859
858
857
856
854
853
852
851
850
849
848
847
846
845
844
9.i
.988103
988073
988043
988013
987983
987953
987922
987892
987862
987832
.987496
987465
987434
987403
987372
987341
997310
987279
987248
987217
49
9.363304
960
49
363940
959
49
3645 1&
958
49
365090
957
49
365664
955
49
368237
954
49
366810
953
49
367382
952
49
367953
951
49
368524
950
49
49
369094
949
9.369663
948
49
370232
946
49
370799
945
50
371367
944
50
371933
943
50
372499
942
50
373064
941
50
373629
940
50
374193
939
50
50
374756
938
9.375319
937
50
375881
935
50
376442
934
50
377003
933
50
377563
932
50
378122
931
50
378681
930
50
379239
929
50
379797
928
51
380354
927
51
9.380910
926
51
381466
925
51
382020
924
51
382575
923
51
383129
922
51
383682
921
51
384234
920
51
384786
919
51
385337
918
51
51
385888
91V
9.386438
915
51
386987
914
51
387536
913
52
388084
912
52
388631
911
52
389178
910
52
389724
909
52
390270
908
52
390815
907
52
52
391360
906
9.391903
905
52
392447
904
52
392989
903
52
393531
902
62
394073
901
52
394614
900
52
395154
899
52
395694
898
{,2
396233
897
52
396771
896
10.6386361 »J0
6360801 ?Q
635485' :.<<
6349101 57
6343361 50
6b3763' 55
633190 54
632618
632047
631476
630908
10^30337
629768
62920 1
628633
623067
627501
626936
626371
625807
625244
53
52
51
50
49
48
47
46
45
44
43
42
41
40
3&
38
37
36
35
34
33
32
31
30
29
28
27
26
25
24
23
22
21
20
107613562 19
613013 18
612464
611918
611369
610822
610276
609730
609185
608640
10.624881
624119
623558
622997
622437
621878
621319
620781
620203
619848
1.0.619090
618534
617980
617425
616871
616318
615768
615214
614683
614112
ov/
10.608097
607553
607011
606489
605927
605386
604846
604306
6037671
6033291
I C oine I
I I
r.-iiip. I M.
EE*
76 Degrees.
sj
(14 Degrees.; a
TABLE OF LOGARITHailC
M.
Sine
1).
Cosine | 1).
1 Tan-.
D.
Cotanj!. i 1
U
9.383675
844
9. 9^6904
52
9.396771
896
10.6032^9
60
1
384182
843
986873
53
397309
896
602691
59
2
384687
842
986841
53
397846
895
602154
58
3
385192
841
986809
53
398.383
894
601617
5~
4
385697
840
986778
53
398919
893
601081
5d
5
386201
^39
986746
53
399455
892
600545
55
6
386704
838
986714
53
399990
891
600010
54
7
387207
837
986683
53
400524
890
599476
53
8
387709
836
986651
53
401058
889
598942
52
9
388210
835
986619
53
401591
888
598409
51
10
11
388711
834
986587
53
53
402124
887
886
597876
10..597'344
50
49
9.38^11
833
9.986555
9.402656
12
389711
832
986523
53
403187
885
.596813
48
13
390210
831
986491
53
403718
884
696282
47
14
390708
830
986459
53
404249
883
59.5751
46
15
391206
828
986427
53
404778
882
,595222
45
16
391703
827
986395
53
405308
881
594692
44
17
392199
826
986363
54
405836
880
594164
43
18
392695
825
986331
54
406364
879
593636
42
19
393191
824
986299
54
406892
878
593108
41
20
21
393685
9.394179
823
822
986266
9.986234
54
54
407419
877
592581
40
39
9.407945
876
10.592055
22
394673
821
986202
54
408471
875
591529
38
23
395166
820
986169
54
408997
874
591003
37
21
395658
819
986137
54
409521
874
590479
36
25
396150
818
986104
54
410045
873
589955
35
26
396641
817
986072
54
410569
872
589431
34
27
397132
817
986039
54
411092
871
588908
33
23
397621
816
986007
54
411615
870
58.8385
32
29
398111
815
935974
54
412137
869
587863
31
30
3i
398600
814
985942
54
55
4126.58
868
587342
10.. 586821
30
29
9.399088
813
9.985909
9.413179
867
32
399575
812
985876
55
41.3699
866
586301
28
33
400062
811
985843
55
414219
865
58578 I
27
34
400549
810
985811
55
414738
864
585202
26
35
401035
809
985778
55
41.5257
864
584743
25
3G
401520
808
985745
55
415775
863
584225
24
37
402005
807
985712
55
416293
862
583707
23
38
402489
806
985679
55
416810
861
683190
22
33
402972
805
985646
55
417326
860
582674
21
40
41
403455
804
985613
9.985580
55
55
417842
9.418358
859
.582158
20
19
9.403938
803
858
10.581642
42
404420
802
985547
55
418873
857
581127
18
43
404901
801
985514
55
419387
S5G
580613
17
44
' 405382
800
985480
55
419901
855
580099
16
45
405862
799
985447
55
420415
855
579.585
15
48
406341
798
985414
56
420927
854
579073
14
47
406820
797
985380
56
421440
853
578560
13
48
407299
796
985347
56
421952
852
678048
12
49
407777
795
985314
56
422463
851
577537
11
50
51
408254
794
985280
9.985247
56
56
422974
850
577026
10.576516
10
9
9.408731
794
9.423484
849
52
409207
793
985213
56
423993
848
576007
8
53
409682
792
985180
56
424503
848
575497
7
54
410157
7!)1
985146
56
42.5011
847
574989
6
55
410632
790
985113
56
425519
846
674481
5
5fi
411106
789
985079
56
426027
845
573973
4
57
411579
788
985045
56
4265.34
844
573466
3
58
412055i
787
985011
56
427041
843
572959
2
59
412524
786
984978
58
427547
843
572453
1
fiO
412996
785
994944
58
428052
842
.57194^
1
Oiisiiie
riinn |
1 C.tu,. 1
1 Tang 1 M. |
75 Degrees.
SINES AND TANGENTS. (15 Degrees.)
33
M.
1 Sine
1 0.
1 Cosine 1 D.
1 Tan!.'.
1 1>.
! Cotaiii:. j
"U
9.412996
785
9.984944
' .57
9.428052
842
10.571948 60
1
413467
784
984910
57
428557
841
571443 b^
2
413938
783
984876
57
429062
840
570938 !'^
3
414408
783
984842
57
429666
839
570434 ■ 57
4
414878
782
984808
57
430070
838
569930 : 56
5
415347
781
984774
57
430573
838
569427
55
6
415815
780
984740
57
431075
837
568925
54
7
416283
779
9S4706
57
431577
836
568423
53
8
416751
778
984672! 57
432079
835
667921
62
9
417217
777
984637 57
432580
834
567420
51
10
417684
770
9846031 57
4330S0
833
566920
50
11
9.418150
775
9.984569! 67
'9.433580
832
10.566420
49
12
418615
774
984535
57
434080
832
565920
48
13
419079
773
984500
57
434579
831
66.5421
47
14
419544
773
984466
57
435078
830
564922
46
15
420007
772
984432
58
435676
829
564424
45
16
420470
771
984397
58
436073
828
563927
44
17
420933
770
984363
58
436670
828
563430
43
18
421395
769
984328
58
437067
827
662933
42
19
421857
768
984294
58
437563
826
562437
41
20
422318
767
984259
58
438059
'82.'>
561941
40
21
9 422778
767
9.984224
58
9.438554
824
I0r561446
.39
22
423238
766
984190
58
439048
823
560952
38
23
423697
765
984155
58
439543
823
560457
37
24
424156
764
984120
58
440036
822
569964
36
25
424615
763
984085
58
440529
821
5.59471
35
26
425073
762
984050
68
441C22
820
658978
34
27
425530
761
984015
58
441514
819
558486
33
28
425987
760
983981 58
442006
819
557994
32
29
426443
760
983946 58
442497
818
657503
31
30
31
426899
759
083911 .58
9.983875158
442988
817
5.57012
30
29
9.427354
758
9.'W3479
816
10.556521
32
427809
757
983840 59
443968
816
556032
28
33
428263
756
983805 59
983770 59
444458
815
555542
27
34
428717
755
444947
814
565053
26
35
429170
754
9»3735 59
445435
813
654565
25
36
429623
753
9837001 59
445923
812
654077
24
37
430075
752
983664
59
446411
812
553n89
23
38
430527
752
983629
59
446898
811
553102
22
39
430978
751
983594
59
447384
810
552616
21
40
4f
431429
750
983558
59
59
447870
809
552130
20
19
9.431879
749
9.983523
9.448.356
809
10.. 55 1644
42
432329
749
98.3487
59
448841
808
6511.59
18
43
432778
748
983452
59
449326
807
660674
17
44
433226
747
9834161 59 I
449810
806
550190
16
45
433675
746
983381
59
450294
806
.549706
16
46
434122
745
983345
59
460777
805
649223
14
47
434569
744
983309
59
451260
804
.548740
13
48
435016
744
9832731 60
451743
803
548257
•12
49
435462
743
983238 60
462225
802
547775
11
50
51
435908
9.436353
742
983202
9.983166
60
60
452706
802
647294
10.546813
10
9
741
9.4.53187
801
52
436798
740
983130
60
453668
SOO
546332
8
53
437242
740
983094
60
464148
799
645852
7
54
437686
739
983058
60
454628
799
545372
6
55
438129
738
983022
60
455107
798
641893
6
56
438572
737
982986 60
455586
797
614414
4
57
439014
736
982950 60
466064
• 796
543936
3
58
439456
736
982914 60
456542
796
543458
2
59
439897
735
9828781 60
4.57019
795
542981
60
440338
734
9828421 60
457496
794
542504
Cosine |
1
Sine 1 1
Colling. 1
Taxa. 1 M. j
74 Degrees.
34
(16 Degrees.) a
TAHLE OF LOGARITHMIC
M.
Sine
D.
VoifUic
I).
Tang.
D.
Cot:ir</ j 1
^
9.440338
734
9. 98:..' 842
60
9.457496
794
"107542504
lo
1
440778
733
982805
60
457973
793
542027
59
2
441218
732
982769
61
458449
793
541551
58
3
441658
731
982733
61
458925
792
541075
57
4
442090
731
982696
61
459400
791
540600
56
5
442535
730
982660
61
459875
790
540125
55
6
442973
729
982624
61
460349
790
639651
54
7
443410
728
982587
61
460323
789
539177
53
8
443847
727
982551
61
461297
788
538703
52
9
444284
727
982514
61
461770
788
533230
51
10
11
444720
726
982477
61
61
462242
787
537758
50
49
9.445155
725
9.982441
9.462714
786
10.537286
12
445590
724
982404
61
463186
785
530814
48
I'd
446025
723
9823G7 61
463658
785
536342
47
14
446459
723
982301
61
464129
784
.535871
46
15
446893
722
982294
61
464599
783
5J5401
45
16
447326
721
982257
61
46506:
783
rm^jn
44
17
447759
720
982220
62
465539
782
534461
43
18
448191
720
982183
62
466008
-781
533992
42
19
448623
719
982146
62
466476
780
533524
41
20
21
449054
9.449485
718
717
982109
9.982072
62
62
466945
780
533055
10.532587
40
39
9.467413
779
22
449915
716
982035
62
467880
778
532120
38
23
450345
716
981998
62
468347
778
53?653
37
24
450775
715
981961
62
468814
777
531186
36
25
451204
714
981924
62
469280
776
530720
35
26
451632
713
981886
62
469746
775
530254
34
27
452060
713
981849
62
470211
775
529789
33
28
452488
712
981812
62
470676
774
529324
32
29
452915
711
981774
62
471141
773
528859
31
30
453342
710
981737
62
471605
773
5283iJ5
30
31
9.453768
710
9.981699
63
9.472008
772
10.527932
¥j
32
454194
709
981662
63
472532
771
527468
28
33
454619
708
981625
63
472995
771
527005
27
34
455044
707
981587
63
473457
770
526543
26
35
455469
707
981549
63
473919
769
.526081
25
36
455893
706
981512
63
474381
769
52.5619
24
37
456316
705
981474
63
474842
768
5251.58
23
38
456739
704
981436
63
475303
767
524697
22
39
457162
704
981399
63
475763
767
524237
21
40
457584
703
981361
63
476223
766
523777
20
41
9.458006
702
9.981323
63
9.476683
765
10.523317
19
42
458427
701
981285
63
477142
765
522858
18
43
458848
70.1
981247
63
477601
764
522399
17
44
459268
700
981209
63
478059
763
521941
16
45
459688
699
981171
63
478517
763
521483
15
46
460108
698
981133
64
478975
762
.521025
14
47
460527
698
981095
64
479432
761
52056S
13
48
460946
697
981057
64
479889
761
520111
12
49
461364
696
981019
64
480345
760
519655
11
50
51
461782
695
980Q81
9.980942
64
64
480801
759
519199
10.518743
10
9
9.462199
695
9.481257
759
52
462616
694
980904
64
481712
758
518288
8
53
463032
693
980866
64
482167
757
^ 517833
7
54
463448
693
980^27
64
482621
757
617379
6
55
46386^.
692
980789
64
483075
756
516925
5
56
464279
691
980750! 64
483529
755
616471
4
57
464694
690
• 9807 12
64
483982
755
616018
3
58
465108
69a
980673
64
484435
754
615565
2
59
465522
68»
980635
64
484887
7i=a
615113
1
60
465935
688
98059b i 64
485339
753
5T4f>fi1
^
Coi-iiie
1 Sine 1
Coluiig.
1 Tang. |M.|
73- l)<igitesk
SINES AND TANGENTS.
(l7 Degrees
35
M^
Si IK! 1
n !
Cosine | !).
Ttuig.
U.
Ootan^. . 1 1
9.46G935
688
9.980596
64
9.485339
755
10.514661 i 60 1
1
466348
688
980558
64
485791
752
614209
C9
2
466761
087
980519
65
486242
751
5137i38
58
3
467173
686
980480
65
486693
751
613307
57
4
467585
685
980442
65
487143
750
5128.57
56
5
467996
685
980403
65
487593
749
612407
55
6
468407
684
980364
65
488043
749
611957
54
7
468817
683
980325
65
488492
748
511.508
53
8
469227
683
980286
65
488941
747
6110.59
52
9
469637
682
980247
65
489390
747
510610
51
10
470046
681
980208
65
489838
746
610162
50
11
9.470455
6^0
9.980169
65
9.490286
746
10 .009714
49
12
470863
680
980130
65
490733
745
509267
48
13
471271
679
980091
65
491180
744
608820
47
14
471^9
678
980052
65
491627
744
608373
46
15
472086
678
980012
65
492073
743
607927
45
16
472492
677
979973
65
492519
743
507481
44
17
47289S
676
979934
66
492965
742
6070.35
43
18
473304
676
979895
66
49.3410
741
606590
42
19
473710
675
979855
66
493854
740
606148
41
20
474115
674
979816
66
494299
740
505701
40
21
9.474519
674
9.979776
66
9.494743
740
10.50.5257
39
22
474923
673
979737
66
496186
739
604814
88
23
475327
672
979697
66
495630
738
504370,371
24
475730
672
979668
66
496073
737
503927
36
25
476133
671
979618
66
496615
737
503485
35
26
476536
670
979.'i79
66
496957
736
503043
34
:i7
476938
669
979539
66
497399
736
602601
33
28
477340
669
979499
66
497841
735
• 502159
32
29
177741
668
979459
66
498282
734
501718
31
30
478142
667
979420
66
498722
734
601278
30
31
9.478542
667
9.979380
66
9.499163
733
10.500837
29
32
478942
666
979340
66
499603
733
600397
28
33
479342
665
979300
67
500042
732
499958
27
34
479741
665
979260
67
500481
731
499519
26
35
480140
664
979220
67
500920
731
499080
25
36
480539
6 53
979180
67
601359
730
498641
24
37
480937
663
979140
67
601797
730
498203
23
38
481334
662
979100
67
602235
729
497765 ! 22 |
39
481731
661
979059
67
602672
728
497328 j 21 !
40
482128
66i
979019
67
603109
728
496891 i20|
41
9.482525
600
9.978979
67
9.603546
727
10.496454
10
42
482921
659
978939
67
503982
727
496018
18
43
483316
659
978898
67
604418
726
495582
17
44
483712
658
978858
67
504854
725
495146
16
45
484107
657
978817
67
506289
725
494711
15
46
484501
657
978777
67
605724
724
494276
14
47
484895
656
978736
67
.6061.59
724
493841
13
48
485289
656
978696
68
506593
723
493407
12
49
485682
655
978655
68
507027
722
492973
.11
50
486075
654
978615
68
607460
722
492.540
10
51
9.486467
653
9.978.574
68
9.507893
721
10.492107
9
52
486860
653
978.533
68
508326
721
491674
8
53
487251
652
978493
68
508759
720
491241
7
54
487643
651
978452
68
509191
719
490809
6
55
488034
651
978411
68
509622
719
490378
5
56
488424
050
978370
68
5100.54
718
489940
4
57
488814
650
978329
68
510485
"18
489515 3
58
489204
649
978288
68
610916
717
489084 2
59
489593
648
978247
68
511346
710
48R6,54 1
60
489982
648
978206
68
611776
716
488224'
~
Ciit-iiie
Sine j
Coiani'.
1 ■! ann j M.
18
71 Uegiees.
36
ri 8 Degrees.) a
TAWiB OF LOGARITHMIC
"m"
1 Sine
1 I>.
1 Cosine | 1).
1 'I'aiig.
1 D
1 Cutaiiu.
9.489982
648
9.978206168
9.511776
512206
1 716
10.488224 1 60
1
490371
648
978165
68
716
487794
59
2
490759
647
978124
68
512635
715
487365
58
3
491147
646
978083
69
513064
714
486936
57
4
491535
646
978042
69
513493
714
486507
56
5
491922
645
978001
69
613921
713
486079
55
6
492308
644
977959
69
614349
713
485651
54
7
492695
644
977918
69
614777
712
485223
53
8
493081
643
977877
69
615204
712
484796
52
9
493466
642
977835
69
615631
711
484369
51
10
11
493851
642
977794
9.977752
69
69
516057
9.616484
710
483943
10.483516
50
49
9.494236
641
710
12
494621
641
9V7711
69
516910
709
483090
48
13
495005
640
977669
69
517335
709
482665
47
14
495388
639
977628
69
517761
708
482239
46
15
495772
639
977586
69
618185
708
481815
45
16
496154
638
977544
70
618610
707
481390
44
17
496537
637
977503
70
519034
706
480966
43
18
496919
637
977461
70
519458
706
480542
42
19
497301
636
977419
70
519882
705
480118
41
20
21
497682
636
977377
70
70
620305
705
479695
10.479272
40
39
9.498064
635
9.977.335
9.520728
704
22
498444
634
977293
70
621151
703
478849
38
23
498825
634
977251
70
621573
703
478427
37
24
499204
633
977209
70
621995
703
478005
36
25
499584
632
977167
70
622417
702
477583
35
26
499963
632
977125
70
522838
702
477162
34
27
500342
631
977083
70
523259
701
476741
33
28
600721
631
977041
70
523680
701
476320
32
29
501099
630
976999
70
524100
700
475900
31
30
31
501476
629
976957
70
70
524520
699
475480
30
29
9.501854
629
9.976914
9.524939
699
10.475061
32
502231
628
976872
71
525359
698
474641
28
33
502607
628
9768.30
71
525778
698
474222
27
34
502984
627
976787
71
526197
697
473803
26
35
503360
026
976745
71
526615
697
473385
25
36
503735
626
976702
71
527033
696
472967
24
37
504110
625
976660
71
527451
696
472549
23
38
504485
625
976617
71
527868
695
472132
22
39
504860
624
976574
71
628285
695
471715
21
40
41
505234
623
976532
9.976489
71
71
628702
694
4712118
0.470881
20
19
9.505608
623
9.529119
693
42
505981
622
976446
71
529535
693
470465
18
43
606354
622
976404
71
629950
693
470050
17
44
506727
621
976361
71
630366
692
469634
16
45
507099
620
976318
71
530781
691
469219
15
46
507471
620
976275
71
531196
691
468804
14
47
507843
619
976232
72
631611
690
468389
13
48
508214
619
976189
72
532025
690
467975
12
49
508585
618
976146
72
532439
689
467561
11
50
51
508956
618
976103
9.976060
72
72
532853
689
467147
10
9
9.509326
617
9.533266
688
10.466734
52
509696
6]6
976017
72
533679
688
466321
8
53
510065
616
975974
72
534092
687
465908
7
54
510434
615
975930
72
534504
687
, 465496
8
55
510803
615
975887
72
634916
686
465084
5
56
511172
614
975844
72
535328
686
464672
4
57
511540
613
975800
72
535739
685
464361
3
C'8
611907
613
975757
72
536150
685
463850
2
59
612275
612
97.5714
72
536561
684
403439
1
60
512642
G12
975670
72
536972
684
463028
J.
Cosine
1
Sine 1
(J(Kan<r.
'J'aiifr. 1 M. 1
71 De;!j;ree8.
SINES AND TANGENTS. (19 Degrees.)
S7
M.
Sine
D.
Cosine 1 D.
'I'itng.
n.
Coiatif. !
"IT
9.512642
612
9.975670
73
9.536972
684
10.463028 60
1
513009
611
975627
73
537382
683
462618 .59
2
513375
611
975583
73
537792
683
462208
68
3
513741
610
975539
73
538202
682
461798
67
4
514107
609
975496
73
538611
682
461389
56
5
514472
609
976452
73
639020
681
460980
55
6
514837
608
9/5408
73
639429
681
460671
54
7
515202
608
975366
73
639837
680
460163
53
8
515566
607
976321
73
.540246
680
459756
62
9
615930
607
976277
73
640653
679
469347
51
}0
'll
516294
606
605
975233
73
73
.541061
679
4.58939
10.4.586.32
60
49
9.516657
9.975189
9.641468
678
12
517020
605
975145
73
641875
678
458125
48
13
517382
604
975101
73
.542281
677
457719
47
14
517745
604
975057
73
542688
677
457312
46
15
518107
603
97.5013
73
643094
676
456906
45
16
518468
603
974969
74
643499
676
456601
44
17
518829
602
974925
74
543905
675
456095
43
18
519190
601
974880
74
544310
675
455690
42
19
519551
601
974836
74
544715
674
■ 456285
41
20
21
519911
600
974792
74
74
546119
674
673
4,54881
10.4.544V6
40
39
9.520271
600
9.974748
9.. 546524
22
520631
599
974703
74
545928
673
454072
38
23
520990
599
974659
74
546331
672
463669
37
24
521349
598.
974614
74
546736
672
453265
36
25
.521707
598
974570
74
647138
671
452862
35
26
522066
597
974525
74
647540
671
452460
34
27
522424
596
974481
74
647943
670
452067
33
28
522781
596
974436
74
648346
670
461655
32
29
.523138
595
974391
74
548747
669
451253
31
30
523495
595
974347
75
649149
669
4.50861
30
31
9.. 523852
594
9.974302
75
9.549550
668
10.450450
29
32
524208
594
974267
76
649951
668
460049
28
33
524564
593
974212
76
550352
667
449648
27
34
524920
593
974167
76
5507.52
667
449248
26
35
525275
592
974122
75
651152
666
448848
25
36
525630
591
974077
75
651552
666
448448
24
37
525984
591
974032
75
551952
666
448048
23
38
526339
590
973987
75
562351
665
447649
22
39
526693
590
973942
75
652750
666
447250
21
40
41
527046
589
973897
9.973852
75
75
553149
9.5.53548
664
664
446851
20
19
9.527400
589
10.446452
42
527753
588
973807
75
563946
663
446054
18
43
.528105
588
973761
76
554344
663
445656
17
44
528458
587
973716
76
554741
662
445259
16
45
.528810
587
973671
76
655139
662
444861
16
46
529161
586
973625
76
556536
661
444464
14
47
629513
586
973580
76
555933
661
444067
13
48
529864
585
973535
76
556329
660
443671
12
49
.^^302 15
585
973489
76
556725
660
443275
11
50
51
b30565
1/530915
584
973444
76
76
557121
659
442879
10.442483
10
9
584
9.973398
9.667517
659
62
531265
583
973352
76
557913
659
442087
8
53
531614
582
973307
76
658308
658
441692
7
54
531963
682
973261
76
568702
658
441298
6
55
532312
581
973215
76
559097
657
440903
5
56
532661
581
973169
76
569491
667
440509
4
57
633009
580
973124
76
659885
656
440115
3
58
533357
680
973078
76
560279
656
439721
3
59
533704
679
973032 77
500673
655
439327
1
il
534052
578
972986 77
561066
655
438934
^
Cosine
1
Sine 1
Cotang.
1
Tang. 1 M. |
70 Degrees
38
\^20 Degrees.} a
TABLE OF LOGARITHMIC
1 Sine
1 D.
1 <;osine | U
TaiiR.
} »
1 C.iju.a. 1 1
~0
9.534052
1 578
9.9729S6
l77
"9.561066
655
10.43.Si^34
60
1
534399
577
972940
77
561459
6.64
438541
59
2
534745
577
972894
77
561851
654
438149
68
3
535092
577
972848
77
562244
663
437756
o7
4
535438
676
972802
77
562636
663
437364
56
s
535783
676
972766
77
663028
653
436972
55
6
536129
576
972709
77
66.3419
652
436681
54
7
536474
574
972663
77
563811
652
436189
53
8
536818
574
972617
77
564202
661
435798
52
9
537163
673
972570
77
564592
661
436408
51
10
537507
573
972624
77
564983
660
435017
50
11
9.537851
572
9:972478
77
9.. 5653713
650
10.434627
49
12
538194
672
972431
78
565763
649
434237
48
13
538538
671
972386
78
666153
649
433847
47
14
538880
671
972338
78
566542
649
433468
46
15
539223
670
972291
78
566932
648
433068
45
16
539565
670
972246
78
567320
648
432680
44
17
539907
669
972198
78
667709
647
432291
43
18
540249
569
972161
78
668098
647
431902
42
19
540590
568
972105
79
668486
646
431514
41
20
21
540931
9.541272
568
972068
9.972011
78
78
568873
9.669261
646
431127
10.430739
40
39
567
645
22
541613
567
971964
78
669648
645
430352
38
23
541953
666
971917
78
570035
645
429965
37
24
542293
566
971870
78
570422
644
429578
36
25
542632
505
971823
78
570809
644
429191
35
26
542971
565
971776
78
571195
643
428805
34
27
543310
564
971729
79
571681
643
428419
33
28
.543649
564
971682
79
671967
642
428033
32
29
543987
563
971635
79
572352
642
427648
31
30
541325
.663
971588
79
572738
642
427262
30
31
9.544663
662
9.971540
79
9.573123
641
10.426877
29
32
545000
562
971493
79
673507
641
426493
28
33
545338
561
971446
79
573892
640
426108
27
34
545674
561
971398
79
574276
640
425724
26
35
646011
560
971351
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52
669703
398
946471
111
723232
509
276768
8
53
669942
398
946404
111
723538
609
270462
7
54
670181
397
946337
111
723844
609
2761.56
6
55
670419
397
946270
112
724149
509
27.5851
5
56
670658
397
946203
112
7244.54
509
275546
4
57
670896
397
946136
112
724759
508
275241 3|
58
671134
396
946069
112
725065
608
274935
2
59
671372
396
946002
112
725309
608
274631
1
=
671609
396
945935
112
725674
508
274326
C.isiue
Sine 1
Cotaii?.
1 Tawti. |M. j
62 Degrees.
46 (28 Defrrces.j a tablr of LonfAKiTiiMTC
M
1 Sl:.-
1 1».
' «Ji)i«in(!
1 I)
•; I'ai;-.
1 ".
1 »•..;,„.
1
1)
, 9.67lfi0<
;}96"
9.915935 m
r"9.72567l
H 50^
10. 27 13;; j
|-6(r
1
67 1847
395
915SW 11-^
7259/91 508
274021
59
f!
6720S'i
395
9458001 ir*
726S8^
507
27371(1
58
672321
395
945733! Hi
72658fc
507
273412
57
672558
395
945666 112
72639-^
507
273108
56
5
,672795
394
945598 112
72719?
507
2728G3
55
6
"a73032
394
945531 :i2
727501
507
272499
54
7
673268
3.94
945464 113
-727805
506
272195
53
8
673505
394
945396 113
728109
506
271891
52
9
673741
393
945328 113
728412
506
271588
51
10
673977
1 393
9452(31] 113
728716
506
2712841 50
U
9.674213
i 393
9.945193
1 113
9.729020
50ii
10.270980
49
12
674448
392
945125
i 113
729323
505
270677
48
13
674684
392
945058
1113
729626
505
270374
47
14
674919
392
9449901 113
729929
505
270071
46
15
675155
392
944922! 113
730233
505
269767
45
16
675390
391
944854 143
730535
505
269465
44
17
675624
391
944786! 113
730838
504
269162
43
18
675859
391
944718
113
731141
504
268859
42
19
676094
391
944650
113
731444
504
268556
41
20
676328
390
944582
114
731746
504
268254
40
21
9.676562
3J0
9.944514
.114
9.732048
504
10.'2'67952
39
22
676793
390
944446
114
732351
503
267649
38
23
677030
390
9443771 114
732653
503
267317
37
24
677264
389
944309 ; 14
732955
503
267045
36
25
677498
3!>!«
944241 114
733257
503
206713! 35
2fi
677731
389
944172 114
733558
503
266412! 34
27
677964
383
944104 114
733860
502
2661401 33
2S
678 197
388
944036 114
734162
502
2658381 32
29
678430
388
943967 114
734463
502
265537
31
30
678668
388
943899] 114
734764
502
265236
30
31
9 673395
387
9.943330! 114
9.735066
602
10.264931
29
32
679128
387
943761
114
735367
502
264633
28
33
679360
387
943693
115
735 368
501
264332
27
34
6'79r.92
337
943624
1 15
735969
501
264031
26
35
679824
386
943555
115
736269
.501
2637311 251
36
680056
386-
943486
115
736570
.501
263430 241
37
680288
336
943417
115
736871
501
263129
23
38
630519
385
943348
115
737171
500
262829
22
39
680750
385
943279
115
737471
.500
262529
21
40
680982
385
943210
115
737771
500
262229 20 |
41'
9.681213
385
9.943141 115
9.738071
500
10.261929
19
42
681443
384
943072 115
738371
503
261629
18
43
681674
384
943003 115
738671
499
281329
17
44
681905
384
942934 115
7389 71
499
261029
16
45
632135
384
942864 115
739271
499
260729
15
46
"682365
383
942795 116
739570
499
2S0430
14
47
682595
383
942726 116
739870
499
260130
13
48
682825
383
942656
116
740169
499
259831
12
49
683055
383
942587
116
740468
498
239532
11
50
683284
332
942517
116
740767
493
259233 10 1
51
9.683514
382
9.942448
116
9.741066
493
I0.25.s934
9
c)2
683743
332
942378
116
741365
498
258635
8
53
6S3972
382
942308
116
741664
498
258336
7
54
684201
331
0422391 116
741962
49?
^58038
6
55
684430
381
9421691 116
742261
497
257739 5
56
684658
381
9420991 116
742559
497
257441 4
57
684887
380
9420291 116
742858
497
257142 3
5S
685115
380
941959! 116
743156
497
25684-4 2
59
r/}5343
380
9418891 117
743454
497 1
256546 1
60
6s5n71
380
9418191 117
743752
496 1
256248 i
_J
C<isiii« j
1
»\,u^. 1 1
Cdiaii!:.
i
'J'iiML'. 1 M.
Ql Dogfeas.
SINES AND TANGENTS
. (29 I
)egrecs.;
47
_M^
Sine
D.
(Cosine 1 n.
Tuuu.
D.
C(.ia!.s. 1 1
"(T
y.6S5571
380
y.94i8iy! 1171
9.7437.-)2
496
10.256218
60
1
685799
379
941749
117
744050
496
2,55y50
59
2
686027
379
941679
117
744348
496
255652
58
3
68C254
379
941609^
117
744645
496
2.5.5355
57
4
686482
379
941539
117
744943
496
256057
56
5
686709
378
941469
117
745240
496
254730
55
G
686936
378
941398
117
745S38
495
2.54432
54
7
687163
378
941328
117
745835
495
2.54 lf)5
53
8
6S7389
878
941258
117
746132
495
2.538 ('.8
52
9
687616
377
941187
117
746429
495
253571
51
10
687843
377
941117
117
746726
495
253274
50
11
9.^88069
377
9.941046
118
9.747023
494
10.2529;7
49
12
688295
377
940975
118
747319
494
252681
48
13
688521
376
940905
118
747616
494
252384
47
14
688747
376
940834
118
747913
494
25208V
40
15
688972
376
940763
118
748209
494
251791
45
16
689198
376
940693
118
748S05
493
251495
4^1
17
689423
375
940622
118
748801
493
251199
43
18
68964.S
375
940.')51
118
749097
493
250903
42
19
689873
375
940480
118
749393
493
250607
41
20
690098
375
940409
118
749689
493
2.50311
40
21
9.690323
374
9.940.T38
118
9.749985
493
10.250015
39
22
690548
374
940267
118
750281
492
249719
38
23
690772
374
940196
118
750576
492
249424
37
24
690996
374
940125
119
750872
492
249128
36
25
691220
37.3
940054
119
751167
492
248833
35
26
691444
373
939982
119
751462
492
248538
34
27
691668
373
939911
119
7517.57
492
248243
33
28
691892
373
939840
119
752052
491
247948
32
29
692115
372
939768
939697
119
752347
491
247653
31
30
692339
372
113
752642
491
247358
30
31
9.692562
372
9.939625
119
9.7.52937
491
10.247063
29
32
692785
371
939554
119
753231
491
246769
28
33
693008
371
939482
119
753526
491
246474
27
34
693231
371
939410
119
753820
490
246180
26
35
693453
371
939339
119
754115
490
245885
25
36
693676
370
939267
120
7.54409
490
24.5.59 J
24
37
693898
370
939195
120
754703
490
245297
23
38
694120
370
939123
120
754997
490
245003
22
39
694342
370
939052
120
75.5291
490
244709
2)
40
41
694564
9.694786
369
369
93S980
9.938908
120
120
755585
9.755878
489
244415
10.244122
20
19
489
42
695007
369
938836
120
756172
489
243828
18
43
695229
369
938763
120
756465
489
243535
17
44
695450
368
938691
120
756759
489
243241
16
45
695671
368
938619
120
757052
489
242948
15
46
695892
368
938547
120
757345
488
242655
14
47
696113
368
938475
120
■ 767638
488
242362
13
48
696334
367
938402
121
757931
488
242069
12
49
6965.'^,4
367
938330
121
758224
488
241776
11
50
51
696775
9.696995
367
367
9382581 121
9.9.38185! 121
7.58517
488
488
241483
10.241190
10
Q
9.758810
52
697215
366
938113
121
7.59102
487
240898
8
53
697435
366
038040
121
759395
487
240605
7
54
697654
366
937967
121
759687
487
240313
6
55
697874
366
937895
121
759979
487
240021
5
56
698094
365
937822
121
760272
487
239728
4
57
698313
365
937749: 121
760564
487
239436
3
58
C98532
365
937676 121
760856
486
239144
2
59
698751
365
937604, 121
761148
486
238852
1
60
1 698970
364
9375311 121
761439
486
238561
1 Cnsilie
i
Sine 1
j Coiaiin:.
1
1 Tuup. 1 M.
19
GG
Degi
cea.
48 (39 Degrees.) a takle of looaiuth.hic
M
Si..« 1
■»•■ 1
('..sin. 1 1.. 1
T.-iiiu. 1
l>
(•..:m... 1
U
9 . 698970
364
9.9375L'l! 121
9.7614391
486
10.2385(511 60
1
699 189
36.1
937458 122
761731
486
238269, .59
2
699407
364
• 937385 122
76:^023
480
237!t77! .58
:i
699626
364
937312 122
762314
486
237686 57
4
6998441
363
937238 122
762606
485
237394 56
5
700062
363
937165 122
762897
485
237103 55
6
70;}280
363
9370921 122
763188
485
236812
54
7
700498
363
937019
122
763479
485
236.521
53
8
700716
363
936946
122
763770
485
236230
52
9
700933
362
936872
122
76406 1
485
235939
51
10
701 iSl
362
936799 122]
764352
484
235648
50
11
9.701368
362
9.936725
122
9.764643
48^1
10.235357
49
12
701585
302
936652
123
764933
484
235067
48
13
701802
361
936578
123
765224
765514
484
234776
47
14
702019
361
936505
123
484
234186
46
15
702236
.S61
936431
123
765M05
484
234195
45
16
702452
361
936357
123
766095
484
233905
44
17
702669
360
936284
123
766385
483
233615
43
18
702885
360
936210
123
766675
483
233325
42
19
703101
360
936136
123
766965
483
233035 41
20
703317
360
93«i062
123
767255
483
232745 40
21
9.703533
359
9 . 935988
123
9.767545
483
10.232455 39
22
703749
359
935914
123
767834
4!^3
232 J 66 38
23
703964
359
935840
123
768124
482
231876
37
24
704179
359
935766
124
768413
482
231.587
35
25
704395
359
935692
124
768703
482
231297
35
26
704610
358
9350 18
124
768992
482
231008
34
27
704825
358
935543
124
76928 1
482
230719
33
2S
705040
358
935469
124
769570
482
230430
32
29
705254
358
935395
124
769860
481
230l40|3ll
30
705469
357
935320
124
770148
481
229852
30
31
9 705683
357
9.935246
124
9.770437
481
10.229563
29
32
705898
357
935171
124
770726
481
229274
28
33
706112
357
935097
124
771015
481
228985
27
34
706326
358
935022
124
771303
481
228697
26
35
706539
356
934948
124
771.592
481
228408
25
36
706753
356
934873
124
771880
4.80
228120
24
37
706967
356
934798
125
772168
480
227832
23
38
707180
355
934723
125
772457
480
227543
22
39
707393
355
934649
125
772745
480
227255
21
40
707600
355
934574
125
773033
480
226967
20
41
9.707819
355
9.934499
125
9.773321
480
10.226679
.19
42
708032
354
934424
125
773608
479
226392
18
43
708245
354
934349
125
773896
479
226104
17
•14
708458
354
934274
125
774181
479
225816
16
45
708670
354
934199
125
774471
479
225529
15
46
' 708882
353
934123
125
774 75W
479
225241
14
47
709094
.353
934048
125
775046
479
224954
13
48
709306
353 •
933973
125
775333
479
224667
12
49
709518
353
933898
126
775621
478
224379
11
50
709730
353
933822
126
775908
478
224092
10
51
9 709'94l
352
9.933747
126
9.776195
478
10.22.3805
9
52
710153
352
933671 126
776482
478
'223518
8
53
710364
352
933596 126
776769
478
223231
7
.',4
710575
352
933520 126
777055
478
,22294:
6
55
710786
351
933445 126
777342
478
222658
5
50
710997
.351
933369 120
777628
477
222372
4
57
711208
351
933293 126
777915
477
222085
3
58
711419
351
93.3217 126
778201
477
221799
2
59
711629
350
933141 126
778487
477
221512
1
60
711839
350
933066 126
778774
477
2212261
1 .,., >u^
!
1 ^o: !
1 »...,....
1
j .....:^ 1 ..
.
>9 4.>c>
!,ir-i»
S1INK3 AND TA^-GK^TS
{31 Docrrces
V
49
M.
, fiiU,'.
I I*-
1 ^..^i|.e 1 1).
1 T.....^
.).
! (.n.lM!!'.'. I
"iT
"y 71 1839
350
9.933060
126
y. 778774
477
10.221226 60
I
712050
350
932990
127
779060
477
220940 59
2
7122G0
350
932914
127
779346
176
220654 58
3
7124H9
349
932838
127
779632
476
22(»368 .57
4
712079
349
932762
127
779918
476
220082
56
5
712889
349
932685
127
780203
476
219797
55
6
713098
349
9326091 127
780489
780775
476
219511
54
7
713308
349
932533
127
476
219225
53
8
713517
348
932457
127
781060
476
218940
52
9
713726
348
932380
127
781346
475
218654
51
10
11
713935
348
932304
9.932228
127
127
781631
9.781916
475
475
218369
10.218084
50
49
9.714144
348
12
714352
.347
932151
127
782201
475
217799
48
13
714561
347
932075
128
782486
475
217514
47
14
714769
347
93 1998
128
782771
475
217229
46
15
714978
347
931921
128
783056
475
216944
45
Hi
715186
347
931845
128
783341
475
216659
44
17
715394
346
931768
128
783626
474
216374
43
18
715602
346
931691
128
783910
474
216090
42
19
715809
346
931614
128
■ 784195
474
21.5805
41
20
716017
346
931.537
128
784479
474
21.5521
40
21
9.71(5224
345
9.9314^0
128
9.784764
785048
474
10.21.5236
39'
22
716432
345
931383
128
474
214952
38
23
716639
345
931306
128
785332
473
214668
37
24
716846
345
931229
129
78.5616
473
214384
36
25
717053
345
931152
129
785900
473
214100
35
26
717259
344
931075
129
- 786184
473
213S10
34
,27
717466
344
930998, 129
786468
473
213532
33
28
717673
344
930921
129
786753
473
213248
32
29
717879
344
930843
129
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9.798877
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10.201123
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9.726426
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9.927549
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799157
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200563
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14
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927310
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799717
467
200283
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15
727228
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927231
133
799997
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200003
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16
727428
333
927151
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800277
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199 723
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727628
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927071
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10.198325
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19,5813
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193863
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19.3307
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19.3029
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046325
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04.5817
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9.871073
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10.045563, 60
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04.5309! 59
2
825791
233
870816
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0'i5055
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825931
233
870732
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'044800
57
4
826071
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423
044.546
56
5
826211
233
870504
190
955707
423
044293
55
6
826351
233
870390
190
955961
423
044039
54
7
826491
£33
870276
190
956215
423
043785
53
8
826631
233
870161
190
956469
423
04.3531
52
9
826770
232
870047
191
956723
423
043277
51
10
11
826910
9.827049
232
232
869933
9.8')9S1>'^
191
Wl
956977
423
043023
10.042769
50
49
i). 95/231
423
12
827189
232
869704
191
957485
423
042515
48
13
827328
232
869589
191
957739
423
042261
47
14
827467
232
869474
191
957933
423
042007
46
15
827606
232
869360
191
958246
423
0417.54
45
16
827745
232
869245
191
958500
423
0415001 44
17
827884
231
8691.30
191
9587.54
423
041246J43
18
82^023
231
869015
192
959008
423
040992 42
19
828162
231
868900
192
959262
423
040738141
20
828301
231
868785
192
959516
423
040484
40
21
9.828439
231
9.868670
192
9.9.59769
423
10.040231
39
22
828578
231
8685.55
192
960023
423
039977
38
23
828716
231
868440
192
960277
423
039723
37
24
8.28855
8^28993
230
868324
192
960531
423
039469
36
25
230
868209
192
960784
423
039216
35
20
829131
230
868093
192
961033
423
038962
34
27
829269
230
867978
193
961291
423
038709
33
28
629407
230
867862
193
961.545
423
038455
32
29
829545
230
' 867747
193
961799
423
038201
31
30
8296S3
230
867631
193
962052
423
037948
30
31
9.829821
229'
9.867515
193
9.962306
423
10.037694
29
32
829959
229
867399
193
962560
423
037440
28
33
830097
229
867283
193
.962813
423
0371871271
34
830234
229
867167
193
963067
423
0.36933 261
35
830372
229
867051
193
963320
423
036680
25
3H
830509
229
866935
194
963574
423
0.36426
24
37
830646
229
866819
194
963827
423
036173
23
38
830784
229
866703
194
964081
423
035919
*'2
39
830921
228
8665861 194
964335
423
035665
21
40
831058
228
866470
194
964.588
- 422
035412
20
4!
9.831195
228
9.866353
194
9.964S42
422
10.0351.58
19
42
831332
228
866237
194
965095
422
034905
18
43
831460
228
866120
194
965349
422
034651
17
44
831606
228 .
866004
195
965602
422
034398
16
45
831742
228
8658H7
195
96.5855
422
034145
15
4(5
831879
228
865770
195
966109
422
0.33891
14
47
832015
227
8656.53
195
966362
422
03363S
13
4«S
832152
227
865536
195
966616
422
033384
12
•19
832288
227
86.5419
195
96686J
422
0.33131
11
50
832425
227
865302
195
967123
422
032877
10
5i
97832561
227
9. 865 1 85
195
9.967376
422
10.032624
9
£')
832697
227
865068
195
967629
422
032371
8
53
832833
227
864950
19.)
967883
422
032117
7
54
832969
226
864833
196
968136
422
,031864
6
5;
833105
226
864716 196
968389
422
03 1611
5
5f5
83324 1
226
864598
1 96
968643
422
031357 4|
57
833377
226
864481
196
968896
422
031104
3
5:S
833512
226
864363
196
969149
422
030851
2
5:)
833648
226
864245
196
969403
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030597
1
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8337.S3
226
864127 196
969656
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030344
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iT i).
SINES AND TANGftKTS. (43 DpfrrceS )
O'l
I r..siii(
n.l
I ».
I
2
3
4
r>
6
7
.?
if
12
13
14
ir>
If)
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
30
37
3f<
39
40
41
42
43
44
45
4G
4?
48
49
50
51
52
53
54
55
5fi
57
58
59
60
0.833783
226
9.864127
1961
9.969656
422
10.030344, 60
833919
225
864010
196
969909
422
U3009 1 1 59
831054
225
863892
197
970162
422
029838; .58
834189
225
863774
197
970416
422
029584 57
834325
225
863656
197
970669
422
029331
50
834460
225
863538
197
970922
422
029078
55
834595
225
863419
197
971175
422
028825
54
834730
225
863301
197
971429
422
028571 .53
834865
225
863183
197
971682
422
028318 52
834999
224
863064
197
971935
422
028005! 51
835134
224
862946
198
972188
422
0278121
50
9.835269
224
9.862827
198
9.972441
422
10.027559'
49
815403
224
862709
19S
972694
422
027306!
48
8355.38
224
862.590
198
972948
422
0270.52
47
8.35672
224
862471
198
973201
422
026799
46
83580?
224
862353
198
973454
423
026.546
45
83594 1
224
862234
198
973707
422
026293
44
836075
223
802 1 15
198
973960
422
026040
43
836209
223
861996
198
974213
422
025787
42
836343
223
861877
198
974466
422
02.5534
41
836477
233
8617.58
199
974719
4%2
02.5281
40
9.836611
223
9.861638
199
9.974973
423
10.025027
39
836745
223
861519
199
97.5226
422
024774
38
836878
223
861400
199
97.5479
422
024.521
37
837012
222
861280
199
975732
422
024268
36
837146
222
861161
199
975985
422
024015
35
837279
222
861041
199
976238
422
023762
34
837412
222
860922
199
976491
422
023509
33
837546
222
860802
199
976744
422
023256
32
837679
222
860682
200
97699?
422
023003
31
837812
• 222
860562
200
977250
422
022750
30
9.837945
222
9.860442
200
9.977503
422
10.022497
29
838078
221
860322
200
977756
422
022244
28
8382 1 1
221
860202
200
978009
422
021991
27
838344
221
860082
200
978262
422
021738
26
838477
221
859962
200
978515
422
021485
25
. 838610
221
859842
200
978768
422
0212.32
24
833742
221
8.59721
201
979021
422
020979
23
838875
221
859601
201
979274
422
020726
22
839007
221
859480
201
979527
422
L 020473
21
8.39140
220
859360
201
979780
422
020220
20
9.839272
220
9.859239
201
9.980033
422
10.019967
19
839404
220
859119
201
980286
422
019714
18
839536
220
8.58998
201
980538
422
019462
17
839668
220
858877
201
980791
421
019209
10
839800
220
858756
202
981044
421
018956
15
83:)932
220
8.58635
202
981297
421
018703
14
840064
219
858514
202
981550
421
018450
13
840196
219
858393
202
98 1803
421
018J97
12
840328
219
858272
202
982056
421
017944
11
840459
219
858151
202
982309
421
017691
10
9.840591
219
9.858029
202
9.982562
421
10.017438
9
840722
219
857908
202
982814
421
017186
8
8408.54
219
857786
202
983067
421
016933
7
840985
219
857665
203
983320
421
016680
6
841116
218
857543
203
983573
421
016427
5
841247
218
857422
203
983826
421
016174
4
841378
218
857300
203
984079
421
015921
3
841509
218
857178
203
9>^4331
421
01.5669
2
811640
218
857056
203
984584
421
015416
1
841771
218
856931
203
1 984837
421
01 51 63
I C(..-ii..!
Col arm.
I nu:,. |M.
46 Degrees.
62
(14 Degrees^ a
TABLE OF LOGARITHMIC
>J 1
Sim..'
'»• 1
t'osiiie 1 ji
T.-.,... I I) ,
Cul.liiy..
9.841771
218
9.856934
203
9.9318371 421
10.015163
6U
I
841^02
218
856312
203
935090 421
014910
59
2
8420:13
218
85G690
204
935343
421
014657
53
3
842163
217
856568
204
935^96
421
014404
57
4
842294
217
856446
204
985343
421
0l415ii
56
5
842424
217
856323
204
986101
421
013S99
55
6
842555
217
85320 1
204
986354
421
013646
54
?
842685
217
856078
204
986607
421
013393
53
8
842S15
217
855956
204
986860
421
013140
52
9
842946
217
855333
204
937112
421
012833
51
10
843076
217
855711
2^
937365
421
012635
50
11
9.813206
2i6
9.855538
205
9.987618
421
10.012332
49
12
843336
216
855465
205
987871
421
012129
43
13
843466
216
855342
205
988123
421
011877
47
14
843595
216
855219
205
938376
421
011624
46
15
843725
216
855096
205
988629
421
011371
45
16
843S55
216
854973
205
988882
421
011118
44
17
843934
216
854850
205
989134
421
010366
43
18
844114
215
854727
206
939387
421
010613
42
19
844243
215
854603
206
989640
421
010360
41
20
844372
215
85+180
206
939393
421
010107
40
21
9.844502
2j5
9.854356
206
9. 99 J 145
421
10.009355
3 J
22
844631
215
854233
206
990393
421
009602
33
23
841760
215
854109
2(r,
990651
421
009349
37
24
8448S9
215
8.53936
206
990903
421
009007
36
25
845018
215
853862
206
991156
421
008344
35
26
845147
215
853738
206
991409
421
00859 1
34
27
845276
214
853614
207
991662
421
003333
33
28
845405
214
853490
207
991914
421
003036
32
23
845533
214
853366
207
992167
421
007833
31
30
845662
214
853242
207
992420
421
007530
30
31
9.845790
214
9.8.53118
207
9.992672
421
10 007323
29
32
845919
214
852994
207
992925
421
007075
23
33
846047
214
852869
207
993178
421
006S2->.
27
34
846175
214
852745
207
993430
421
006570
26
35
846304
214
852620
207
993633
421
006317
25
36
846432
213
852496
203
993938
421
006064
24
37
846560
213
852371
203
994189
421
005311
23
33
8466S8
213
852247
203
994441
42 i
005559
22
39
840316
213
852122
203
994694
421
0053061 21 1
40
846944
213
851997
203
994947
421
005053
20
41
9.847071
213
9.851872
203
9.9951991 421
10.004301
19
42
847199
213
851747
203
995452
421
004548
18
43
847327
213
851622
208
995705
421
004295
17
44
84X}5;i
212
851497
209
995957
421
O04043
16
45
8 7583
212
851372
209
996210
421
003790
15
46
• 770}
212
851246
209
996463
421
003537
14
47
847836
212
851121
209
995715
4-^1
0032S5
13
48
847964
212
850996
209
996963
421
003032
12
49
818091
212
850870
209
997221
421
002779
IL
50
848218
212
850745
209
997473
421
002527
10
51
9.848345
212
9.850519
209
9.997726 421
10.002274
9
52
848472
211
85049!J
210
997979 421
002021
8
53
848599
211
850368
210
998231 421
001769
7
54
848726
211
850242
210
998484 421
O0i5l6
6
55
848852
211
850116
210
9987371 421
001263
5
56
848979
211
849990
210
998989 421
OOlOll
4
57
849106
211
849364
210
999242 421
0007 5S
3
58
84)232
211
849738
210
9994951 421
0005(15
2
59
849359
211
849611
210
999743 1 421
000253
1
60
819185
211
8494S51210
10. 000000 1 421
000000
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(Jo:illie
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^ (
4
THIS BOOK IS DUE ON THE LAST DATE
STAMPED BELOW
AN INITIAL FINE OF 25 CENTS
WILL BE ASSESSED FOR FAILURE TO RETURN
THIS BOOK ON THE DATE DUE. THE PENALTY
WILL INCREASE TO SO CENTS ON THE FOURTH
DAY AND TO $1.00 ON THE SEVENTH DAY
OVERDUE.
\ APH y J 936
.. :'50^6
fi'JG '^Q 1070 m
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^"^ u . ij IJ/y ^
^ M^^- AUG 12 m
3
1
LD 21-100m-7,'33
■'meen^-
nvm
//
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