Lawrence S. (Lawrence Sluter) Benson.

# Geometry : the elements of Euclid and Legendre simplified and arranged to exclude from geomtrical reasoning the reductio ad absurdum : with the elements of plane and spherical trigonometry, and exercises in elementary geometry and trigonometry / /c By Lawrence S. Benson online

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Font size dividing the third and fourth terms li^'the result by AC ; as by
this means we get the analogy li X sin CAD : sin D x sin BCA
: : DC : AB. Hence, to find the logarithm of AB, to the loga-
rithm of DC add the logarithmic sines of D and BCA, and from

Prob. Ill, â€” To find the distance oftxno objects A a7idB on a
horizontal plane.

This may be effected in different ways according to circum-
stances.

1. A base AC may be taken, terminated at one of the ob-
jects. The angles A and C, and the side AC
are then measured ; and the required distance
AB is found by the analogy, sinB : sin C : :
AC : AB.

2. This method fails if the objects A and B
be not visible from one another, as then the
angle A can not be measured. In this case, a
station C may be taken as before, from which both A and B
are visible. Then, having measured the angle C, and the sides
AC, BC, we compute the distance AB by means of the second
case of trigonometry.

PLANE TKIGONOMETKY.

203

3. When from inequalities in the ground, or other causes, the
preceding methods are inapplicable, the solution may be effect-
ed in the following manner: Measure a base CD, such that A
and B are both visible from each of its extremities ; measure,
also, the two angles at C, and the two at D. Then, by the
first case in trigonometry, -we compute AC in the triangle
ACD, and BC in the triangle BCD ; from which, and from the
contained angle ACB, AB is computed by means of the second
case. The operation may be verified by computing AD, BD,
by the first case, and thence AB by the second case.

Prob. IV. â€” Let AFB be a great circle of the earth, supposed
to be a sphere; E a point in the diameter B A produced, EF a
straight line touching the circle in F, and ED a straight line in
its plane, per2Jendicular to AB; it is required to compute the
angle DEF, and the straight line EF.

Draw the radius CF. Then, since (III. 12) CFE is a right
angle, we have (hyp. and I. 20, cor. 3)
DEC=CEF+ECF." Take away CEF,
and there remains DEF=:ECF. Now
(III. 21) EF- = BE.EA. Hence EF will
be found by adding AE to AB, multiply-
ing the sum by EA, and extracting the
square root. To find CE, add AE to the

radius AC. Then (Trig. 1) CE : EF : :

R : sin ECF, or sin DEF ; or CE : CF : : ~

Pv : cos ECF, or cos DEF.

8cho. These examples have been selected from the " Ele-
ments of Plane Trigonometry," by Prof Thomson, of the Uni-
versity of Glasgow, since they exhibit in the simplest manner
the elementary principles of trigonometrical computation.

4*

THE ELEMENTS OF SPHERICAL TRIGONOMETRY.

DEFINITIONS.

1. Spherical Trigonometry explains the processes of cal-
culating the unknown parts of a spherical triangle when any
three parts are given; and certain formulae derived from Plane
Trigonometry are employed to express the relations between
the six parts of a spherical triangle.

2. A spherical triangle is that portion of the surface of a
sphere bounded or contained by the arcs of three great circles
intersecting each other ; the spherical triangle being formed by
three planes passing through the sphere, and intersecting each
other, each angle of the triangle (VI. 24) is contained by
the tangents of the sides at their point of intersection, and is
measured by the arcs of great circles described from the ver-
tices as poles, and limited by the sides of the triangle produced
if necessary. Also, the angles of a spherical triangle vary
(VL 24, cor. 2) between two and six right angles.

3. The spherical angle being contained by the tangents of
the sides at their point of intersection, the properties of the
spherical triangle are explained by means of Plane Trigonome-
try, and its analogies are applied to imaginary rectilineal tri-
angles, the sides of the spherical triangle being regarded
functions of rectilineal angles having the sides of the spherical
triangle as arcs measuring (I. def 19) them. Spherical Trig-
onometry treats of the angles at the apex of a triangular pyra-
mid; but Plane Trigonometry treats oi plane angles ^ therefore
Spherical Trigonometry treats of solid angles.

4. Let ABC be a spherical triangle, and H the center of the
sphere ; the angles of the tnangle are equal to the angles in-
cluded by the planes IIAB, HAC, and IIBC (VI 24),"which
are the angles formed by the planes at the apex of a triangular
pyramid, and the arcs AB, BC, and CA measure the angles on
the planes at the apex of the pyramid AHB, BHC, AHC re-

SPHERICAL TRIGONOMETRY.

205

spcctively. And we can represent the side opposite the angle
A by a, the side opposite the
angle B by b, and the side op-
po>ite the angle C by c. On
the line HA take any point as
L, and draw perpendiculars as
FL, LG to HA. Then, GLF
will be equal to the angle A,
LEG equal to B, and FGL
equal to C ; and the sides CB,
BA, and AC of the spherical triangle ABC will measure the
angles CHB, AHB, and AHC respectively ; hence these an-
gles are denoted by a, c, b.

5. If FG be joined in the triangles FHG and FLG, we will
have (Plane Trig., 6 cor.),

HF'+HG'-FG*

cos BHC=cos a=

cos FLG = cos A=

2HFxHG
LF=+LG=-FG*

2LGxLF â€˘

Reducing and subti-acting second from the first, we will
have,

2 [cos a (HFxHG)â€” cosA (LGxLF)]=.2HLl

Dividing both members by 2 (HFxHG), we get.

cos aâ€” cos A

LGxLF HL HL

HFxHG"

HF^HG"

Since regular and similar polygons have their perimeters
proportionate to their apothems, and circles have their circum-
ferences proportionate to their diameters (V. 14, cor. 3), the
sine of an angle is the ratio of the radius, or the hypothenuse
of a liglit-angled triangle to the perpendicular from the vertex
of the right angle to the hypothenuse.

â€ž LG , , LF . HL HL

Hence we get ^- = sin 5, g^ = sm e, g-, = cos c, jj^ = cos b.

Substituting and ti-ansposing, we derive the formula,

cos a=cos b cos c+sin b sin c cos A, )
cos b = cos. a cos c+sin a sin c cos B, /â€˘ (1)
cos c=cos a cos 6+ sin a sin b cos C. )

206

THE ELEMENTS OF

Or, The cosine of either side of a spherical triangle is equal
to the product of the cosiyies of the other two sides increased by
the product of their sines into the cosine of the an^le included
hy them.

The three formulae show the relations between the six parts
of a spherical triangle such, that if any three of them be given,
the other three can be determined.

6. Then (VI. 24, cor. 1), if we denote the angles and sides of
the spherical triangle polar to ABC, respectively, by A', B'', C^,
a\ h\ c', we will have,

Â«'=180Â°â€” A, d'=180Â°â€” B, c'=180Oâ€” C,
A'==180Â°â€” a, B'r=l80Â°â€” 6, C"=180Oâ€” c.
Since any of the formulae (l) is applicable to polar spherical
triangles, we have, after substituting the respective values
and changing the signs of the terms, other formulae :

cos A=sin B sin C cos a â€” cos B cos C, \
cos B=sin A sin C cos h â€” cos A cos C, \ (2)
cos C=sin A pin B cos c â€” cos A cosB. ;

Or, The cosine of either angle of a spherical triangle is equal
to the product of the sines of the two other angles into the co-
sine of their included side, diminished by the product of the
cosines of those angles.

V. Transposing the first and second formulae (l) we get,
cos a â€” cos h cos c=sin h sin c cos A,^
cos b â€” cos a cos c=:sin a sin c cos B ;

respectively adding and subtracting, we get,

cos a+ cos 6â€” cos c (cos a+ cos 5)=sin c (sin 5 cos A+ sin a

cos B),
cos a â€” cos &+COS c (cos aâ€” cos b) =m\ c (sin b cos Aâ€” sin a

cos B) ;

which can be put in the forms

(1 â€” cose) (cos a -|- cos J) = sin c (sin b cos A -|- sin a cos B),
(1 +COS c) (cos aâ€” cos J) = sin c (sin b cos A â€” sin a cos B) ;

multiplying these equations together, substituting for (1 â€” cos')
its value (sin'), and for (cos') its value (1 â€” sin'), and dividing by
Bill* c, we have,

SPHERICAL TRIGONOMETRY.

207

COS* aâ€” cos' 5=sin'' 5â€” sin' h sin" Aâ€” sin' a+sin* a sin' B;
then, since (1â€” sin'a)â€” (1â€” sin= h) =sin' h â€” sin' Â«, we have,

cos' a â€” cos* 6=sin' h â€” sin' a, and we get,
sin" h sin' Am sin' a sin' B ;

extracting tlie sqiiare root, sin 5 sin A = sin a sin B, or '
sin A sin a

sin B sin b'
sin A

derive

And from first and third formulae (l) we
From the second and third formu-

sin C sin c

sin C sin c

\i?)

lae we derive â€”. â€” t^_ . ,.
sin Â±> sin

Or, In every spherical triangle^ the sines of the angles are
to each other as the sines of their opposite sides.

8. Taking the third formula (l), and substituting for (cos h) its
value as expressed in the second, and for (cos' a) its value
(1â€” sin' a), and dividing by sin a, we will have,

cos c sin a=sin c cos a cos B + sin b cos C.

But (Spher. Trig. 7) we have sin b~ ._ ^, â€” ; substituting for

sin C

cose

sin b its value, and dividing by sin c, we get, -. sin a=C08 a

cosB4

sin B cos C

sin C

_, cos

But -r- =COt,

sin

Hence we can derive, by similar processes,

cot a sin b=cot A sin C + cos b cos C,
cot a sin c=cot A sin B+cos c cosB,
cot b sin a=cot B sin C + cos a cosC,
cot b sin c=cot B sin A+cos c cos A,
cot c sin a=:cot C sin B+cos a cos B,
cot c sin b=cot C sin A+cos b cos A.

} (4)

The formulae (1) are the fundamental analogies of Spherical
Trigonometry, from which all the others are derived, which
others are more adapted for logarithmic computations.

208 THE ELEMENTS OF

THE SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES BY

LOGARITHMS,

The following equations give the unknown parts of a right-
angled spherical triangle when C is the right angle and any
two other parts are known. There are six cases.

Let C be the right angle, and c be the hypothenuse.

Case 1. Given a and b to find c, A, and B ;

, . tan a _, tnn b
cos e=cos a cos o; tan A= r; tan l3= .

â– ^ tan o tan a

Case 2. Given c and a side b to find a, A, and B ;

cos c . tan b . â€ž sin &

cos a= r; cosA=: â€” â€” : sin 13=^ .

cos b tan c sin c

Case 3. Given aside a and opposite angle A to find others;

. , tan a . â€ž sin a . _ cos A

sm 0= â€” r ; sin C=-. â€” -; sin B= -.

tan A sm A cos a

Both acute or both not acute.

Case 4. Given a side a and adjacent angle B to find others;

tan 6=sin a tan B ; cot c=:cot a cos B ; cos A = cos a sin B.

Case 5. Given the hypothenuse c and an angle A to find
others ;
sin Â« = sine sin A ; tan 5i=tan c cos A; cot B = cose tan A.

Case 6. Given the oblique angles A and B to find others ;

cos A - cos B . _

cos a=â€”. â€” ^rr; cos o=â€” â€” - ; cos c = cot A cot B.
sin B sin A

Napier's circtdar parts are much the simplest method of re-
solving right-angled spherical triangles; they are the two
sides about the right angle, the complements of the oblique
angles, and the complement of the hyjiothenuse. Hence there
are five circular parts; the right angle not being a circular
part, is supposed not to separate the two sides adjacent to the
right angle ; therefore these sides are regarded adjacent to
each other, so that when any two parts are given, tlieir corre-
sponding circular ])aits are also known, and these with the re-
quired part constitute the three parts under consideration ;
therefore these three paias will lie together, or one of tliem

SPHERICAL TRIGONOMETRY.

209

â– will be separated from both tlie others. Hence one part is
known as the middle part; and when three parts are undei- con-
sideration, the parts separated by the middle part are caUcd
t\\G adjacent parts ; and the parts separated liuni tlie middle
parts are called the opposite parts. IS'ow, assume any part in

B

the diagram for the middle part, and using the formulae (1)
â€˘when the other parts are the ojiposite parts, and we get, 27ie
sine of the middle part is equal to the product of the cosines of
the opposite parts. Then, assume again any part for the mid-
dle, and use the formulae (2) when the other parts are the ad-
jacent parts, and we get, The sine of the middle pa. t is equal
to the pyroduct of the tangents of the adjacent parts. Hence
"We derive the five following equations :

sin a=tan h tan (90"â€” B)=cos (90Â° â€” A) cos (90Â° â€” c;,
sin 6=tan a tan (90Â°â€” A)=:cos (90Â°â€” B) cos (90Â°â€” c),
sin (90Â°â€” A)=tan b tan (90Â°â€” c)=cos (90Â°â€” c) cos c/,
sin (90Â° â€” c)=tan (90Â°â€” A) tan (90Â°â€” B)==cos a cos 6,.
sin (90Â°â€” B)â€” -tana tan (90Â°â€” c)=cos 6 cos (90Â° â€” A).

The a7za7o^/es of Xapier are derived from the formulce (l) by
eliminating the cosines of any of the sides, reducing and chang-
ing to linear sines and cosines (Plane Trig. p. 193), and we
bave,

cos I {a-\-h) : cos | (a â€” b) : : cot | C : tan | (A + B),
sin \ (a+b) : sin -^ (aâ€”b) : : cot -^ C : tan ^ (Aâ€” B),
cos -^ (a-f c) : cos | {aâ€”c) : : cot ^ B : tan ^ (A+C),
sin ^ {a+c) : sin -^ (aâ€”c) : : cot | B : tan ^ (A â€” C),
cos ^ {b-hc) : cos -^ (bâ€”c) : : cot ^ A : tan ^ (B + C),
sin ^ {b+c) : sin -^ {bâ€”c) : : cot | A : tan | (B â€” C),
14

210 THI. ELEMENTS OF

The same proportions applied to tlic triangle polar to ABC,
with accents omitted, we have,

cos I (A + B) : cos i (Aâ€” B) : : tan | c : tan i {a+b),
sin i (A+B) : sin i (Aâ€” B) : : tan -J c : tan i (a-b),
cos i (A+C) : cos i (Aâ€” C) : : cot ^ & : tan ^ (a+c),
sin i (A+C) : sin | (Aâ€” C) : : cot i 6 : tan ^ (aâ€” c),
cos i (B + C) : cos ^ (Bâ€” C) : : cot i a : tan i (6+c),
sin i (B + C) : sin i (Bâ€” C) : : cot i a : tan | (*â€” c).

The same ambiguity that there is between plane triangles
(I 3, Fourth Case) exists also between spherical triangles,
which may be avoided by remembering that every angle and
gide of a spherical triangle are each less than two right angles,
and that the greater angle is opposite to the greater side, and
the least angle is opposite to the least side; and conversely.

Quadrantal spherical triangles are such which have one side
equal to ninety degrees ; hence they can very easily be solved
by formulae for right-angled spherical triangles.

SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES BT

LOGARITHMS.

Case 1. Given the three sides to find the angles.

Ain {sâ€”fi) sin (5â€” 5) sin (Â»â€” c)
Find s=i (a+6+c) ;let M=|/ â–  ^-'^ ^ 5

then, tan iA = ^-^^^^3:^; tan i B=^^^^^-; tan i C=

]\I

ein (.Â« â€” c)'

Case 2. Given two sides a and b and the included angle C,

to find others.

Tan^(A+B)=^^^^i^4NotiC; tan^A-B^'-i^^-^
â– ^^"^^^^^^ cosi(a+6) ' ' Biui(u+6)

cot \ C.

But A=i (A+B)+i (A-B) ; B=| (A+B)-i (A-B) ;

sin c=sin a^!^=sin b ^^, or find an angle cot 9=tan o
sin A sin 15

cos a sin (J+(j))

cos C ; cos 0= -. â€˘

' sin (J)

BPOEKICAL TKIGONOMETKY.

211

Case 3. Given the sides a and b and an angle opposite to
one of them, to find others.

Find cot (p=tan b cos A, and tanj^^cos b tan A; then, sin

, , . cos a sin (p . ^ . . sin b . , , . .

{c+(p)=i â€” ; sm JL> = sm A -; â€” : sin (c +>-)=: cot a tan

^ ' cos 6 ' sin a ^ ^'

fiinx-

Case 4. Given the angles A and B and the inchiaed side, to

find others.

Find tan (p=cos c tan A, and tan p^=cos c tan B ;

, tan c sin (p , tan c sin y

then, tan aâ€” . .,, , â€” -; tan 6 = - . , . , t;

p_cos A cos (B + (p) _ cosB cos (A+y)
cos 9 cos -/^

Case 5. Given A and B and a side opposite one of them, to
find others.

Find tan (p=:tan a cos B ; cot ;^=:C08 a tan B;

ein irzsin a

sin B
sin A â–

sin (<3â€” (p)=cot A tan B sin 9,

â€˘ //-I \ cos A sin Y

sm (C â€” y) = â€” ^.

^ ^' cosB

Case 6. Given the three angles A, B, C, to find the sides.
Take â€˘

S=|(A+B + C);andN=,4/ â€” - ""'*''' ^

' cos 10 â€” J

cos (Sâ€” A) cos (Sâ€” B) cos (tSâ€” C) ;

then, tan \ aâ€”1^ cos (S â€” A) ;
tan| &=N' cos (S â€” B);
tan \ c=N cos (S â€” C).

* THE SURFACE OP A SPHERICAL TRIANGLE.

Let ABC be a spherical triangle, AC=:
DF, BC=FE, ABC=DEF.

S=surface of ACB, 5=surface of hemis-
phereâ€” BHDCâ€”AGECâ€”DCE=0 R'â€”
(lune AHD â€” S) â€” (lune BGEâ€” S)â€” (lune
CDFE-S)

=eR'(,-j

80

180""" W + ^^-

212 BPHKRICAL TRIGONOMETKT.

180*^, or equivalent to the sum of the three angles above 180Â°;
hence, its spherical excess is sometimes taken as the measure
of the ti'iangle.

Or in terms of its sides, formula given by L'Huiller,

tan ^ E= V [tan ^ s tan ^ {s â€” a) tan ^ (s â€” b) tan ^ {s â€” c)].

EXERCISES IN

ELEMENTARY GEOMETRY,

AND m

PLANE AND SPIIEKICAL TKIGONOMETRY.*

DEFI^nXIOXS.

1. Lines^ avgUs^ and spaces are said to be given in magni-
tude, when they are either exhibited, or when the method of
finding them is known.

2. Points, lines, and spaces are said to be given in position,
â– which have always the same situation, and which are actually
exhibited or can be found.

3. A circle is said to be given in magnitude when its radius
is oiven ; and \w position, when its center is given.

Mao-nitudes, instead of being said to be given in magnitudey
or given in position, are often said simply to be given, when no
ambiguity arises from the omission.

* For these Exercises I am indebted to Thomson's Euclid (Belfast), they
being judiciously selected by that eminent writer, and their presentation
here is a valuable acquisition to an American school text-book. I
would gladly acknowledge my obligations for many propositions in
this volume, but they being culled for more than two thousand years
from the best writers on Geometry, and being so much modified by each
succeeding age, that it is impossible at this day to attribute tliem to their
rightful authors, and many of them being more or less contained in
every work on the subject, they have become public property. What I
have introduced myself will be well recognized by every student of
Geometry, and my only apolugy is, the desire to advance the cause of
Truth.

214 EXEBCISK8.

4. A ratio is said to be giveti when it is the same as that of
two cciven maGrnitudes.

5. A rectilineal figure is said to be given in species, when its
several anarles and the ratios of its sides are iriven.

6. When a series of unequal magnitudes, unlimited in num-
ber, agree in certain relations, the greatest of them is called a
ma/x'mum ^ the least, a minimum.

Thus, of chords in a given circle, the diameter is the maxi-
mum ; and of straiiiht lines drawn to a given straight line,
from a given point without it, the minimum line is the perpen-
dicular.

7. A line which is such that any point whatever in it fulfills
certain conditions, is called the locus of that point.

8cho. 1. Several instances of loci have already occurred in
the preceding books.

1. Tlius, it was stated in the fifth corollary to the fifteenth
proposition of the first book, that all triangles on tlie same
base, aiid between the same pai'allels, are equivalent in area;
and hence, if only the base and area of a triangle be given, its
vertex may be at any point in a straight line parallel to the
base, and at a distance from it which may be determined by
ap])lying (11. 5, scho.) to half the base a parallelogram equiva-
lent to the given area ; and therefore the parallel is the locus
of the vertex. Here the conditions fulfilled are, that straight
lines drawn from any point in the parallel to the extremities of
the given line, form with it a triangle having a given area.

2. It was stated in the first corollary to the eighteenth proposi-
tion of the third book, that all anHes in the same segment of a
cinOe are equal ; and hence, if only the base and vertical angle
of a triangle be given, the vertex may be at any point of the
arc of a segment described on the base, in the manner pointed
out in the nineteenth pi'oposition of the third book; that arc,
therefore, is the locus of the vei-tex.

3. It will be seen in the sixth proposition of these Exercises that
straight lines drawn from any point whatever in the circumfer-
ence of the circle ABGC to the points E, F, have the same
ratio â€” that of EA to AF. Hence, therefore, if the base of a
triangle, and the ratio of the sides be given, the locus of the
â– vertex is the circumference of the circle described in the man-

EXEUCISES. 215

ner pointed out in the corollary to this proposition ; unless
the ratio be that of equality, in which case the locus is evi-
dently a perpendicular bisecting the straight line joining the
points.

4, It follows likewise, from the fifth corollary to the twcnty-
fomth proposition of the first book, that when the base of a tri-
angle and the difference of the squares of the sides are given,
if the point D be found (II. 12, scho.) in the base BC, or its
continuation, such that the difference of the squares of BD, CD
is equivalent to the difference of the squares of the sides; and
if through D a perpendicular be drawn to ]3C, straight lines
drawn from any point of the perpendicular to B, C will have
the difference of their squares equivalent to the given difference;
and hence the perpendicular is the locus of the vertex, when the
base and the difference of the squares of the sides are given.

5. It will appear in a similar manner from the corollary to the
twelfth proposition of the second book, that if BC the base of a
triangle, and the sum of the squai'es of the other sides AB, AG
be given, the locus of the vertex is the circumference of a circle
described from D, the middle point of the base as center, and
with the radius T>A. To find DA, take the diagonal of the
square of BD as one leg of a right-angled triangle, and for its
bypothenuse take the side of a square equivalent to the given
sum of the squares of AB, AC ; then the diagonal of the square
described on half the remaininoj le<x of that rigcht-aniiled trian-
gle will be the radius of DA. The proof of this is easy, de-
pending on the third and fourth coi'ollaries to the twenty-fourth
jDroposition of the first book, and on the corollai-y to the twelfth
proposition of the second book.

Scho. 2. In discovering loci, as well as in other investigations
in geometry, the stuflent is assisted by what is termed geomet-
rical analysis ; of the nature of which it may be proper here
to give some explanation.

Take this proposition : If a chord of a given circle have one
extremity given in positior), and if a segment terminated at
that extremity he taken on the ch rd, produced if necessary^
such that the rectangle under the segment and chord may be
equivalent to a given space ; the locus of the point of section is
a straight I i?ie given inp)osition.

216 EXERCISES.

Let AB be the diameter of the circle and AC a chord of the
circle.

If, in the proposition, instead of being informed that the locns
is a strai!j;ht line, we were required to find what tlie locus is,
we might proceed in the following manner: Lft D be any point
in the required line, so that the rectangle ACAD is equivalent
to the given space ; and having drawn the diameter AB, find
E, so that the rectangle AB.AE may be equal to ACAD, and
therefore E a point in the required line ; and join DE, BC
Then (V. 10, cor.) AB : AC : : AD : AE. Hence (V. 6) tho
triangles DAE, BAC, having the angle A common, are equi-
angular; and thereiore AED is equal to ACB, which is a right
angle. The point D is therefore in a perpendicular passing
through E; and in the same manner it would be t^hown, that
any other point in the required line is in the perpendicular ;
that is, the perpendicular is its locus.

The investigation just given is called the analysis of the
projjosition, while the solutions hitherto given are called the
synthesis or compositio7i. In analysis we commence by sup-
posing that to be effected which is to be done, or that to be
true which is to be proved ; and, by a regular succession of con-
sequences founded on that supposition, and on one another,
we arrive at something which is known to be true, or which
we know the means of effecting. Thus, in the second corollary
to the seventeenth proposition of the sixth book, the conclusion
obtained for the area of the circle is shown by the third corol-
lary to be consistent with the proportion established by Ar-
chimedes between the cone, sphere, and cylinder, and also
consistent with the geometrical truth in regard to the sur-
faces of the sphere and cylinder. Plence, analysis takes into
consideration this consistence, and confirms the second corol-
lary from its agreement with established truths of geometr3^

Again: in the corollary to the twenty-fourth proposition of
the fifth book, since circles are in propoition to the squares of