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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 traced up in the right hand column to the same horizontal line. This being apparent, the reason is manifest, why the columns designated sine, cosine, tang., and cotang., when the degrees are pointed out at the top of the page, and the minutes counted 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.

If 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

COS.

For, sec.= ; or, taking the logarithms, log. sec.-2 log. R-log. cos.-20-log. cos. ; that is, the logarithmic secant is found by substracting the logarithmic cosine from 20. And R2 or log. cosec.=2 log. R―log. sine 20—log.

9

cosec.

sine

sine; that is, the logarithmic cosecant is found by subtracting the logarithmic sine from 20.

It has been shown that R2-tang. x cotang.; therefore, 2 log. R=log. tang.+log. cotang.; or 20=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 D, omitting the decimal point. Now, supposing the increase of the logarithmic sine to be proportional 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 applicable 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

to either of these columns; since of the same arc, the log. tang. + log. cotang 20. Therefore, having two arcs, a and b, log. tang blog. cotang blog. tang a+log. cotang a; or, log. tang blog. tang a=log. cotang a-log. cotang b.

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'

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Tabular difference 247
Number of seconds = 28

Cosine 3° 40'

Tabular difference
Number of seconds

Product 69.16, to be added =

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 cotangent, it must be remembered, that they increase while the arcs decrease, and decrease while the arcs are increased, consequently, the proportional numbers found for the seconds must be subtracted, not added.

Ex. To find the cosine 3° 40′ 40′′.

9.811952

CASE II.

69.16

9.999110

13

40

Product 5.20, which being subtracted = 5.20

Gives for the cosine of 3° 40′ 40' 9.999104.80

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 number be found; the degrees are shown either at the top or bottom 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

given logarithm, annex two ciphers, and then divide the remainder 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
Sine 49° 20′, next less in the table,

Tab. Diff. 181)9100(50" Hence the arc 49° 20′ 50′′ corresponds to the given sine A.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.

THEOREM I.

9.880054 9.879963

PRINCIPLES FOR THE SOLUTION OF RECTILINEAL TRI ANGLES.

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.

E

Let ABC be the proposed triangle, 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 the 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 proportion,

CE EF:: CB: BA : hence
Rsin C: BC: : BA.

FD

B

A

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But we also have,

CE: CF :: CB: CA: hence
R: cos C: CB: CA.

Cor. If the radius R=1, 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 the base; the radius being equal to unity.

THEOREM II.

In every right angled triangle, radius is to the tangent of either of the acute angles, as the side adjacent to the side opposite.

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THEOREM III.

D

B

A

Cor. 1. If the radius R=1,
AB CA tang C.

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 substituted in the proportion for tang C, which will give

R: cot B:: CA: AB.

In every rectilineal triangle, the sines of the angles are to each other as the opposite sides.

Let ABC be the proposed triangle; AD the perpendicular, let fall from the vertex A on the opposite side BC: there may be two

cases.

First. If the perpendicular falls within B 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 B::AB: AC.

Secondly. If the perpendicular falls A without the triangle ABC, the rightangled triangles ABD, ACD, will still give the proportions,

R: sin ABD::AB: AD,
R: sin C :: AC: AD;

from which we derive

D B

THEOREM IV.

hence BD=

ABD, we have

sin C: sin ABD::AB: AC.

But the angle ABD is the supplement of ABC, or B; hence sin ABD sin B; hence we still have

sin C: sin B::AB: AC.

Let ABC be a triangle: then will

cos B-R

First. If the perpendicular falls within
the triangle, we shall have AC-AB2+
BC2-2BCX BD (Book IV. Prop. XII.); B
AB2+BC-AC

2BC

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.

D

AB2+BC2—AC2

2AB × BC.

D

R: cos B::AB: BD;

But in the right-angled triangle

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