Jeremiah Day.

A practical application of the principles of geometry to the mensuration of superficies and solids. Being the third part of a course of mathematics, adapted to the method of instruction in the American colleges online

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1. Required the square root of 648.3.

Numbers. Logarithms. .

Power 648.3 2)2.81176 *

Root 25.46 1.40589

2. Required the cube root of 897.1.

Power 897.1 3)2.95284

Root 9.645 0.98428

In the first of these examples, the logarithm of the given
number is divided by 2 ; in the other, by 3..

3. Required the 10th root of 6948.

Power 6948 10)3.84186
Root 2.422 0.38418

4. Required the 100th root of 983.

Power 983 100)2.99255

Boot 1.071 042992

The division is performed here, as in other cases of decim-
als, by removing die decimal point to the left. .

5. What is the ten thousandth root of 49680000 ?
Power 49680000 10000)7.69618

Root 1.Q0179 0.00077

We have, here, an example of the great rapidity with which,
arithmetical operations are performed by logarithms.


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48. If the index of the logarithm is negative, and is nof
divisible by the given divisor, without a remainder, a difficulty
will occur, unless the index he altered.

Suppose the cube root of 0.0000692 is required. The

logarithm of this is 546036. If we divide the index by 3,
die quotient wfll bfe — 1, with —2 remainder. This remain-
der, if it were' positive, might, as in other cases of division*,
be prefixed to the next figure. But the remainder is nega-
tive, while the decimal part of the logarithm is positive; so
that, when the former is prefixed to toe latter, it will make
neither +2.9 nor —2.9, but — 2+.9. This embarrassing in-
termixture of positives and' negatives may be avoided, by
adding to the index another negative number, to make it ex-
actly divisible by* the divisor. Thus, if to the index —5
there be added — 1, the sum — 6 will be divisible by 3. But
this addition of a negative number must be compensated, by
the addition of an equal positive number, which may be pre-
fixed to the decimal part of the logarithm. The division
may then be continued, without difficulty, through the whole.
Thus, if the logarithm 7.95036 be altered to "5+ 1.95036

it may be divided by 3, and the quotient will be ~2.65012.
We have then this rule,

49. Add to the index, if necessary, such a negative number
as will make it exactly divisible by the divisor, and prefix) an
equal positive number to the decimal part of the logarithm.

1. Required the 5th root of 0.009642

Power 0.009642 jog. &98417

or 5 + £98417

Root 0.3952 L59683

2. Required the 7th root of 0.0004935

Power 0.0004935 Jog. 4.69329

or 7)7 + 3.69329

Root , 0.337 1.52761

50. If, for the sake of performing the division conveniently,.
the negative index be rendered positive, it will be expedient
to borrow as many tens, as there are units in the number de-
noting* the root*


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*Wh*l is the fourth root of 0.03608 ?

Power 0.03698 4)2756797 or 4)38.56797
Root 0.4385 1.64199 9.-64190

Here the index, by borrowing, is made 40 too great, that
4s, +38 instead of —2. When, therefore, it is divided by 4,
it is stiH 10 too great, +9 instead of — 1.

What is the 5th root of 0*008926 ?
Power 0.008926 5)^95066 or 5)47.95066

Root X).389*16 1.59013 9.59013

51. A power of a root may be found by first multiplying
the logarithm of <the given quantity into the index of the

Eower, (Art. 45.) and then dividing the product by the num-
er expressing the root. (Art. 47.)

1. What is the value of (53)% that is, the 6th power of
the 7th root of 53?

Given number 53 log. 1 .72428

.Multiplying by 6 .

Dividing by 7)10.34568

Tower required 30.06 1.47795

J2L What is the 8th power of the 9th root of 654 ? *<

Proportion bt Logarithms.

*52. In a proportion, when three terms are given, the fourth .
is found, in common arithmetic, by multiplying together the
second and third, and dividing by the first. But, when log-
arithms are ufeed, addition takes the place of multiplication,
and subtraction, of division.

To find then, by logarithms, the fourth terra in a propor-
tion, Add the logarithms of the second and third terms, and
from the sum subtract the logarithm of the first term. The
"remainder will be the logarithm of the term required.

Ex. L Find a fourth, proportional to 7964, 378, and 27960.


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Second term
Third term

Numbers. •




First tcna


Fourth term


2. Find a 4th proportional to 768, 381 and 9780.

Second term 381 2.58092

Third term 9780 3.99034

First term
Fourth term





Arithmetical Complement.

53* When one number is to be subtracted from another,
it is oftep convenient, first to subtract it from 10, then to add
the difference to the other number, and afterwards to reject
the 10.

Thus, instead of <*— i, we may put 10— b+a—lQ. .

In the first of these expressions, h is subtracted from a. In
the other, b is subtracted from 10, the difference is added to
a, and 10 is afterwards taken from the sum. The two ex-
pressions are equivalent, because Jhey consist of the same
terms, with the addition, in one of them, of 10—10=0. The
alteration is, in fact, nothing more than borrowing 10, for the
sake of convenience, and then rejecting it in the result.

Instead of 10, yre may borrow, a* occasion requires, 100,
1000, &c.

Thus a-&r*lOO^i+a-lQ0=lOOO-4+€r- 1000, &c.

54. The difference between a given number and 10, or
100, or 1000, fyc. is ceiled the arithmetical complement
of that number.

The arithmetical complement of a number consisting of
one integral figure, either with or without decimals, is found,
by subtracting the number from 10. If there are two inte-


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gral figures, they are subtracted from 100 ; if three, from
1000, be-
Thus the arithmetical compPt of 3.46 is 10—3.46=6.54

of 34.6 is 100-34.6=65.4
of 346. is 1000- 346^=654. &c.

According to the rule for subtraction in arithmetic, any
number is subtracted from 10, 100, 1000, be. by beginning
on the right hand, and taking each figure from 10, after in-
creasing aU except the first, by carrying 1.

Thus, if from 10.00000

We subtract 7.63125

The difference, or arithl comp't is 2.36875, which is ob-
tained, by taking 5 from 10, 3 from 10, 2 from 10, 4 from 10,
7 from 10, and 8 from 10. But, instead of taking each fig-
ure, increased by 1, from 10; we may take it without being,
increased, from 9.

Thus 2 from 9 is the same as 3 from 10,

3 from 9, the same as 4 from 10, &c. Hence,

55. To obtain the arithmetical complement of a num-
ber, subtract the right hand significant figure from 10, and each
of the other figures from 9. if, however, there are ciphers
on the right hand of all the significant figures, they are to be
set down without alteration.

In taking the arithmetical complement of a logarithm, if
the index is negative, it must be added to 9 ; for adding a
negative quantity is the same as subtracting a positive one.
(Alg. 81.) The difference between —3 and +9, is not 6,
but 12.

The arithmetical complement

of 6.24897 is 3.75103 of 2".70649 is 1 1.29351

of 2.98643 7.01357 of 3.64200 6.35800

of 0.62430 9.37570 of 9.35001 0.64999.

56. The principal use of the arithmetical complement, is
in working proportions by logarithms ; where some of the
terms are to be added, and one or more to be subtracted. In
the. Ride of Three or simple proportion, two terms are to be
added, and from the sum, the first term is to be subtracted.


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But if, instead of the logarithm of the first term, we suhptir
tute its arithmetical complement, this may be added to the
" sum of the other two, or more simply, all three may be ad-
ded together, by one operation. After the index is dimin-
ished by 10, the result will be the same as by the common
method. For subtracting a number is the same, as adding
its arithmetical complement, and then rejecting 10, 100, or
1000, from the sum. (Art. 53.)

In the following proportion, .the calculation is made in
both ways.

If* the profit on 2625 dollars employed in trade, is 831 dol-
lars ; what is the profit on 536 dollars ?

By the common method.
-fid term 831 2.91960
3d term 536 2.72916

By the Arith'I Compjemejit
2d term 2.91060

5d term 2.72916

1st term 2625 3.41913

1st term a. c. 6.58089

Ath term 169.68 2.22963

4th -term


The second method here, after rejecting 10, gives the same
result as the first. But it is unnecessary, first to add two of
flie terms, and then the arithmetical complement of the oth-
er. The three may be added at once ; and it will generalir
be expedient, to place the terms in the same order, in which
,*hey are arranged in the statement of the proportion.

Thus, As 2625$ stock,
Is to 831 profit
So is 536 stock

a. c. 6.58087

To 169.68 profit


2. As 6273 o.
Is to 769.4
So is 37.61

c. 6.20252

As 253 a. c.
Is to 672,5
So is 497


'To 4.613 0.66397

To 1321.1 3.12083


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4. As 46.34 a. c. 8.33404
Is to 892.1 2.95041
So is 7.636 0.88298

To 147


5. As 9.85 a. c. 9.00656
Is to 643 2.80821

So is 76.3 1.86252

To 4981


Compound Pboportion.

57. In compound, as in single proportion, the term requir-
ed-may be found by logarithms, if we substitute addition for
multiplication, and subtraction for division-
Ex. 1. If the interest of $365, for 3 yfws and 9 months, be
$82.13 ; what will be the interest of $9940, for 2 yean and
6 months ?

In common arithmetic, the statement of the question is
made in this manner,

3.75 years > v £ 2.5 years >

And the method of calculation is, to divide the product of
the third, fourth, and fifth terms, by the product of the two
first.* This, if logarithms are used, will be to subtract the
sum of the logarithms of the two first terms, from the sum of
the logarithms of the other three.

Two first terms $ *** ,d * * 5 * 229
xwo nrsi terms j Mfi 0#57403

Sum of the logarithms 3.13632

Third term 82.13


Fourth and fifth terms <



Sum of the logs, of the 3d, 4th, and 5th 6.26378
Pp. 1st and 2d 3.13632

Term required



* Set Webber's Arithmetic

&': )


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58. The calculation will be more simple, if, instead of
subtracting the logarithms of the two first terms, we add their
arithmetical complements. But it must be observed, that each
arithmetical complement increases the index of the logarithm
by 10. If the arithmetical complement be introduced into
two of the terms, the index of the sum of the logarithms
will be 20 too great 5 if it be in three terms, the index will be
30 too great, &c.

Two first terms \ 365 * c ' 1M111
lwo nrst terms £ 3 ?5 Q c 9.43537

Third term 82.13 1.91450

■n .a. ,i<uvw $8940 3.95134

Fourth and fifth terms < ^5 0.39794

Term required 1341 23.12746

The result is {the same as before, except that the index of
the logarithm is 20 too great.

Ex. 2. If the wages of 53 men for 42 days be 2200 dol-
lars ; what will be the wages of 87 men for 34 days?

53 ™ en Laarn. J 87 men)


42 days 5 : <" w:: { 34 days J

t,™ ««♦ f-^ fl S 53 *• «• 8.27572

Two first terms j 42 fl c MWW

Third term 2200 3.34242

Fourth and fifth terms | *J }"SJS

Term required 2923.5 3.46589

59. In the same manner, if the product of any number of
quantities, is to be divided, by the product of several others ;
we may add together the logarithms of the quantities to be
divided, and the arithmetical complements of the logarithms
of the divisors.

Ex. If 29.67 x 346.2 be divided by 69J24 X 7.862 X 497 5
what will be the quotient ?


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Climbers to be divided y t

•-fftiltf* . f ...






a. c. 8 J 5964


a,c. 9,10443


«*. c. 7130864,

Quotient 0.03797 8.57949

In thie-way, the calculations in Conjoined Proportion may-
be expeditiously perfdntoed. • .

Compound Interest.

60. In calculating compound l inte re st, tbe araotiirt* for tht
Xrstyear, is made 4he principal for the second year; the
amount for the second year, the principal for the tnird year ?
&c. Now t&e im'otfrtf &t the end of each year, mast be pro-
portioned to the principal at the beginning of the .year. If

the principal for the first year be 1 dollar, and if the amount
of 1 dollar for i year t=«; then, {AJg. 377.)

C a :ct* =the am't for fte 2d y'r, ortheprin. for the 3d;

Jr. a:: < a 2 : a* =the arii't Tot the 3d y'f, or the prin. for the 4th;

(a* : a* =the im't for'the 4th y'r, or the prln. for the 5th,

That is, the amdunt of 1 dolkr'for any number of years is
■ obtained, by finding the amount for 1 year, and involving thi^
*to a power whose index is equal ,tp the number of years.
And the amount of any other principal, for the given timo,
is found, by multiplying the amount of 1 dollar, into the
number of dollars, or the fractional par^ of a dollar*

If logarithms are used, the multiplication required here may
be performed by addition; and the involution, by Multiplica-
tion. (Art. 4 ,

61. To cj He amount of
I dtffttfi for he number of
yeati ; ( anS\ he principal.
Tlfe'&ini wi >r the given
time. From the amount subtract the principal, and the re-
mainder will he the interest. , . t% { .

If tBe interest becomes due half yearly or quarterly; fina
♦the amount of one dollar, for the half-year or quarter, and


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34 EXPOirfEIrtlAL

multiply the logarithm, by the number of balf-jre&rs or quar-
ters in tie given time.

Ex* 1. What is the amount of 2Q dollars, aj^fc per. fen t
compound interest, for 100 years?

L Amount of 1 dollar for 1 year 1.06 log. 0.0253059
M^Wpiyaigby WO ,

. . . 3153059 ., :
Gto» piincipd . 20 4.301.03, ,

y • ■ I "

Amount required '$6186 3.83162 j

' :ft!j . * What &&a aipount of 1. cent, at 6. per cent compound

jlmoiint of 1 dollar for 1 year LOS " log. 0.02*3060
c Multiplying by 500 ?

Ourea principal 0.01 -2.00000

Amount $44,973,000,000 10.65295

. More exact answers may be obtained, by using^ftgariihms
-of. * jreater number of decimal places.

3. What is the amount of 1000 doQais, at 6 per cent com-
ptund, interest, for 10 years? Ans. 1790.80.

Exponential Equations- l

4& An Exponential equation is one in which the letter
4*bttessing the unknown quantity is an exponent.

Thus a* s=rA, and a? =6c, are exponential equations.* These
-are most easily solved by logarithms. As the two members
' of an equation are equal, their logarithms must also be equal.
If the logarithm of each side be taken, the equation may
-then' be reduced, by the rules given in algebra.
* • * •* '

Ek. What is the value of ^ in the equation 3* =243?


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* ■

. Tatyog the typritjuna of M* sides log. (3*)^log, 243
But tbe logarithm of a power is equal to the lo^anmtti of
the root, multiplied into the index of the power. (Art. 4ft)'

Therefore (log. 3) xisslog. 243 ; and dividing by log; 3,
log. 243 2.38561

63. The preceding is an exponeirtiai equation of the sim-
plest form. Other cases, after the logarithm of each side is
taken} may be solved by Trial and Errour An the same man-
ner ait affected equations. (Alg. 503.) For this pwrppse,
make two suppositions of the value of the unknown quanti-
ty, and find tneir errours ; then say, *

As the difference of the errours, to the dif-
- m ference of the assumed numbers ;

So is the least errour, to the e*rrecti*a required
in the corresponding assumed tiamer.
Ek. 1, Find th£ value of « in the equation x* =r25Q.
Taking the logarithms of both sides (log. x) X a? =log.366
Let x be -supposed equal to 3.5, or 3.6.

By the second supposition.
2=3.6, atad log,#*4MIJ630
Multiplying by 3.6

in<i ■
(log. x) xx =2.00268
log. 256=2.40824

IJy fbe first supposition.
* a 3.5, and log. x =0.54407
Multiplying by 3.5

^log. x) x*s= 1.90424
tyg, 256=2.40824

Errour -0.50400 Errour' ML40W6

"DHfiteiice tif the errourr <fcQB844

Then 0.09844 : 0.1 : : 0.40556 : 0.41 19, the Correction. Wis
added to 3.6, the second assumed number, makes the value
of a?— 4.0119. . r

To correct this farther, suppose a; =4.011, or 4.012.

By the first supposition, By the seflond sOjfpojMon,

a?=^.Oll,andtog-r=0.6O325 «=4;Oia,andlagur=O^)0336

Mufcifrtyfogby 4,011

M ' (Tog. *Tx»'=2.419lB3
' -' * teg. &56*»2.4082ft

\_ .*. tn ii Win

k . Errour " +0.01139
Difference of the errours

Multiplying by 4.012

■■ ■ i m i ■

log. af«=fiUMI824




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ar exponehtiai; equations;

Tim 040165 : (fcOOl : : QM 139 : 0.01 1 ray aeaiiy.

Subtracting this correction from the first assumed number
4*011, we have the value of a?=4, which satisfies the condi-
tions of the proposed equation ^ fcr 4* =^256.

2. Reduce the equation 4r* = I00a? 8 . Ans-a?=5.-

3, Reduce the equation x x =9^,


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A 71 ^MGONOMETRY treats, of the relation* <sf the
. w<fe» and ang-Je* of triangles. Its first object
is, to determine the length of tbe sides, and the quantity oT
the arises* In addition to this, from its principles are deriv-
ed risany interesting methods of investigation in the higher
branches of analysis, particularly in physical astronomy..
Scarcely any department of mathematics is more important^
or more extensive in its applications. By trigonometry, the
mariner traces his path on the ocean ; the geographer deter-
mines the latitude- and longitude of places, the dimension*
and positions of countries, the altitude of mountains, the
exxtrses of rivers, &c. and' the astronomer calculates the dis-
tances and magnitudes of the heavenly bodies, predicts the
eclipses of the sun and moon, and measures the progress of
light from the stars. . ,

72; Trigonometry is either plant or spherical. The for-
mer treats of triangles bounded by right Hues; the latter, of
triangles bounded by arcs of cirdes.

Divisions of the Circle*

73. In a triangle there are two classes of quantities whicl*
are the subjects of inquiry, the sides and the angles. For
thepurpose of measuring the latter, a circle is introduced. .

The periphery of every circle, whether great or small, is
supposed to- be divided into 360 equal parts called degrees,
each degree into 60 minutes^ each minute into 60 secondly


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each f second kto HW f Ai'rdEr', -fee* marked with tfc* chngnrftra
•, ', ", "', be. Thus 32° flt' IS" SBf" ifl 32 degrees 24 nun*
utes 13 second £2 thirds*'

A degree, then, is not a magnitude of a given length; bat
a certain portion of the whofc circumference of any circle.
It is evident, that the 386th part of a large circle is greats,
than the same part of a small one. On the other bandy the
number of degrees, in a small circle, is' the same as in a large

The fourth part of a circle is called a quadrant, and con-
tains 90 cl^rees.

74. Tb measure an angle, a circle is so described that its
centre shall be the angular point, and its periphery shall cut
the two hn^s^fciph include tfctt apgfo Toe axe betwe^the
tw$ liqts is considered a measxart^f the angle, because, by
Euc. 33. 6, angles at the centre of a given circle, have the
sarfie ratio to each other, as the *rcs on *bich th«j! stand.
Vkm the arc AB t (Fijp. fc) is & mei&tk ©f th* rtgle ACB.

It is immaterial what is the size of the circle, provided it
'eutrtbe lines whieh instate : the. angle. < Thus the angle
r AGD (Tig.4.) .is measured! by eitiwr of the axes > AG* «S-
iWACD tatoACH^»(A&to,AH,oraft4F^to«^. (Eao.
«&."&>• ■■■ i.- ■> ..•••• .' '. « . . . . •. . -i ..; . . -.. .
' vW* In the. circle ADGH (Kg.-3t) let ttaiwotfhMBftfM
AG and DH be perpendfrwdarjto each other. . The angles
AGO, DCO, GCH, andrUQAifwiH/beiiight an^bs > add the
periphery of the circle wiU jbedmitajk into ioar equal p^rts,
earth containing 00 degrees. < As* tight, angle «& subtended
^jT an. are of 80*, the angie itself iaaaid 4n contain; ^(W.
i*eiiae T hfr tore right angles, these am lBQ^.infoer right aft-
"gle*3GQ* ; sad in any other sagte, a* jmanyi Agrees, as in
the arc by which it is subtended. . .. i : . * .«

76. The sum of the three angles of any triangle being

3|ual to two right asgfos, {Em. 82.. li) k eqwi to 160*.
ence, there can never he more than one obtuse angle in a
triangle. For the sum of two obtuse angles is more 'than
WO< ^ ■ . i . ■ . . . •. .-a

'' 77. The complement of an are or an angle, is the differ-
ence between the atb wrangle 0flrf •§&*£/■«*.•' J " " /

The complement of the are A» V F%. «.) 4s D8j arid die
^ortiplement of thean^te ACB b£K)B/ 'Hb<? co***pteih*ht
•f the arc BDG is also DB. -» • .

"« » «« .» »'i -f '•«: -• >-. . I .\t ,1 I: '/- r* i. i


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- * ' of 9V h 70% of 120 i«30% *

of 50* is 40% ■«. of !Tfc is 80\«ee,

!*Hence,aii acute angle and its complement are qlwayi
equal to 99*. ^Tbe angles ACB and DCB are, tosethqr
fequal to a right 7 angle* , The two acute angles of a rij^it 'atf-
<gjed tri&igle are eqqal to 90° : therefore each is the comple-
ment of the other.

-f 78, The s wpleveht Kff an arc or an qnghis thfi $ffer«
ehce between the arc or angle and 180 degrees.

The supplement of the. spc BDG (Fig, 2.J is AB j . and the
supplement of the angle BCG is BCA. , .

y: 'lbe supplement of HPfc 170* of 120° is 60% *

V f - of 90«ift'100», of ISO* is 3&% &6.'

• • • 1

1 Hence ^Ln angte and its -luppiemeiit are always equal .to

ISO*. The angles BCA »d BGG are together equal to tttd

*i£te angles. ■ - • - ■* ■«• • •'

.- 79. Ger< -.44 the tbre«.ai^e^0f>a,plaiie'im^le ( atoteqtiel

to< MA right angle*, tbatds, ta WW (Euc. 3&1.) thtf-shln

ofaoy teaof the^iediafi^ptetnontiof ihe othen )Sotbtft

the third angle may be found, by subtracting the sum- of tltfe

uethertwo Aom 180°. Or the sum of any two may he found,

by.SBbtracting the third from ISO . . ...i .- .-*

-<^&L A straight lice drawn/ fk)*i the eentre ef fttt&dto/lQ

. my . part of the periphery, is, called a radix* \ of the* «jnri#.

Jn many calculations, k «i& convenient to oonttdenthe radius,

whatever he kg length, as a unit. (Algi 610;) To. thi* mvrft

he referred the numbers expressing the lengfchsef etfier lioti.

Think 20 mil' be twenty times < the radius, a«d 0.75 v tfa1«e

fourths of the radius. v ,, - .......♦

. • Definitions t>f^Sints t Tangemtz, Secants, $c»t .*■»,.

81. Tp facilitate the calculations in trigonometry, liw*
are drawn, within and about the circle, a number of straight
lines, Oifed Sine*, Tangent*, Secant, $c< With, these; the
learner should make, himself perfectly familiar. • • . s
>■ $& Wp f>**c of wane is a straight line dwym frwone
md\<^^ nrc^perpendic^mt^ adiemtterivhickpcwmthrti^
the other end. ti , . ; . .,

Thus BG (Fig. 3.) is the sine of the arc AG. For BG is
a line drawn from the end G of the arc, perpendicular to the


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diataetflr AM which passes through the other end A oTIbe


*Ooi». • *Tbe sine is hedf the chord of double the arc. r fhe
«ine BG is half PG, which is the chord of the arc PAG,
dotibte the arc AG. * ' '

83; The versed smc of an are is that part of the diameter
tafcicfc is between' the sine and the arc.

Thus BA is the versed sine of the arc AG.
•"S4>. The tangent of an arc, is a straight line drawn per-
pttodicutar from the extremity of the diameter which passes
Jhrtugh one end of the arc, and extended till it meets a link
drawn from the centre through the other end.

Thus AD (Fig. 3.) is the tangent of the arc AG.

85. The Secant, of an orchis a straight line drawn from the

^centre, through one end of the arc y and extended to the tangent

jwhich is drawn from the other end.

* Thus OD (iHg. 3:) iff the secant of the arc AG: " '

Online LibraryJeremiah DayA practical application of the principles of geometry to the mensuration of superficies and solids. Being the third part of a course of mathematics, adapted to the method of instruction in the American colleges → online text (page 3 of 26)