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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 9.880054 Sine 49° 20′, next less in the table, 9.879963

Tab. Diff. 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.

PRINCIPLES FOR THE SOLUTION OF RECTILINEAL TRIANGLES.

THEOREM I.

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 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

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sine. The triangles CBA, CEF, are similar, and give the pro

portion,

CE EF: CB: BA: hence

R sin C:: BC: BA.

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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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.

THEOREM III.

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.

D

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;

C

D B

from which we derive

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.

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.

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First. If the perpendicular falls within
the triangle, we shall have AC2¬AB2+
BC2-2BCX BD (Book IV. Prop. XII.); B
AB2+BC2-AC2

=

D

But in the right-angled triangle

hence BD:

2BC

ABD, we have

R: cos B::AB: BD;

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hence by substituting the value of BD, we shall again have

cos B-Rx

AB2+BC2AC2

2ABX BC

Scholium. Let A, B, C, be the three angles of any triangle ; a, b, c, the sides respectively opposite them: by the theorem,

we shall have cos B=Rx

a2 + c2_b2
2ac

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And the same principle,

when applied to each of the other two angles, will, in like man

b2+c2-a2

ner give cos A=R×

9 2bc

and cos C-R×·

a2+b2--c2

2ab

Either of these formulas may readily be reduced to one in which the computation can be made by logarithms.

Recurring to the formula R2-R cos A-2 sin2 A (Art. XXIII.), or 2sinA-R2-RcosA, and substituting for cosA, we shall have

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4bc

sin ¿A=R √ ((a+b~~c) (a+c—b)).

For the sake of brevity, put

(a+b+c)=p, or a+b+c=2p; we have a+b—c=2p—2c, a+c-b=2p-2b; hence

sın ‡A=R √ ((P—b) (p—o
—c)).

bc

THEOREM V.

In every rectilineal triangle, the sum of two sides is to their difference 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 (Theorem III.). Hence, AB+BC: AB-BC :: sin C+sin A: sin C-sin A.

But

C+A

B

sinC+sin A: sin C-sin A:: tang

2

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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 be 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 multiplication of their corresponding numbers, and their subtraction to the division of their numbers; 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. Thus 10-9.274687-0.725313: hence, 0.725313 is the arithmetical complement of 9.274687.

Q

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