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To check and verify operations like these, the proportions should be changed at certain stages. Thus,

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The coincidence of the results thus obtained, with the analogous results in the preceding operations, will manifestly establish the correctness of both.

The sines and cosines of the degrees and minutes up to 30°, being determined by these or other processes, they may be continued thus:

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And these being continued to 60°, the cosines also become known to 60°; because

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The sines and cosines from 60° to 90°, are deduced from those between 0° and 30°.

For

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The sines and cosines being found, the versed sines are determined by subtracting the cosines from radius in arcs less than 90°, and by adding the cosines to radius in arcs greater than

90°.

The tangents may be found from the sines and cosines.

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*Take the first formula of Art. XXIII., make p=30o 1', q=29° 59′, recollecting that sine 30°-R, or ; R being equal to 1. Then let p=30° 2′, q=29° 58', &c.

RECTILINEAL TRIGONOMETRY.

277

Above 45° the process may be considerably simplified by the theorem for the tangents of the sums and differences of For, when the radius is unity, the tangent of 45° is also unity, and tan (A+B) will be denoted thus:

arcs.

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And this, again, may be still further simplified in practice. The secants may readily be found from the tangents by addition. For sec A=tan ▲+tan 1⁄2 comp s. Or, for the odd minutes of the quadrant the secants may be found from the

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Other methods for all the trigonometrical lines are deduced from the expressions for the sines, tangents, &c. of multiple arcs; but this is not the place to explain them, even if it were requisite to introduce them at large into a cursory outline.

2

PRINCIPLES FOR THE SOLUTION OF RECTILINEAL TRIANGLES.

XXXI. In all right-angled triangles, the radius is to the sine of one of the acute angles, as the hypotenuse is to the side opposite this angle.

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

B

E

ED

A

will be the sine of the angle C. The triangles CBA, CEF are similar, and give the proportion CE: EF:: CB: BA ; hence

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XXXII. In all right-angled triangles, radius is to the tangent of one of the acute angles as the side lying adjacent to this angle is to the side lying opposite.

Having described the arc DE (see the last figure), as in the preceding Article, draw DG perpendicular to CD; it will be the tangent of the angle C. From the similar triangles CDG, CAB, we shall have the proportion CD: DG: CA: AB; hence

R: tang C:: CA: AB.

XXXIII. In any 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 the triangle ABC, the right-angled trian- B gles ABD, ACD (Art. 31.) will give

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D

In these two propositions, the extremes are equal; hence with the means we shall have

sin C sin B :: AB: AC.

Secondly. If the perpendicular falls without the triangle ABC, (see the fig. in the next page) the right-angled triangles ABD, ACD will still give the proportions

R: sin ABD :: AB: AD,

R: sin C :: AC: AD;

But

from which we derive sin C: sin ABD : : AB : AC. the angle ABD is the supplement of ABC, or B; hence sin ABD=sin B; hence we again have

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XXXIV. In all rectilineal triangles, the cosine of one angle is to radius, as the sum of the squares of the sides which contain it, minus the square of the third side, is to twice the rectangle of the two former sides; in other words, we have (last fig.)

cos B R

AB2+BC2—AC2: 2AB. BC, or
AB+BC-AC

cos BRX

2AB. BC

From the vertex A, let AD be again drawn perpendicular

to the side BC.

First. If this perpendicular falls within the triangle (see the preceding figure) we shall have AC ́=AB+BC2—2BC × BD

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angled triangle ABD, we have R : sin BAD : : AB : BD ; also the angle BAD being the complement of B, we have

sin BAD=cos B; hence cos B=

RX BD

AB

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or by substituting

the value of BD,

AB+BC2-AC2

cos B=RX

2ABX BC

Secondly. If the perpendicular falls A

without the triangle, we shall have AC AB2+ BC2+2BCX BD (Art.

=

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2BC

But in the right-angled triangle BAD, D B

we still have sin BAD or cos ABD: =

RX BD

AB; and the angle

ABD being supplemental to ABC, or B, we have cos B=

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

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

manner give cos A=R×

b2+c2-a2
2bc

a2 + b2-c2 and cos C Rx

2ab

These three formulas are of themselves sufficient for solving all the problems of rectilineal trigonometry; because, when three of the six quantities A, B, C, a, b, c, are given, we have by these formulas the equations necessary for determining the other three. The principles already explained, and whatever others may be added to them, can, therefore, only be consequences of these three principal formulas.

Accordingly the value of cos B gives

4a2 c2—(a2 + c2—b2)2

sin' B-R-cos2 B-R2.

=

4a2 c2

R2

4a2c (2 a2 b2+2 a2 c2 +2 b2 c2-a'—b'-c') : hence

sin B R

b

2abc √(2 a2 b2+2 a2 c2+2 b2 c2—a—b'—c').

The second member being a function of a, b, c, in which these three letters all occur under the very same form, we may evidently change two of these letters at will, and thus have sin B sin A sin C ; which is the principle of Art. 33. And

b

a

с

from this, the principles of Art. 30, and 32, are easily deducible.

XXXVI. In any rectilineal triangle, the sum of two sides is to their difference, as the tangent of half the sum of the angles opposite those sides, is to the tangent of half the difference of those same angles.

From the proportion AB : AC :: sin C : sin B (see the figures in Art. 33, 34.); we derive AC+ AB: AC-AB:: sin B+sin C: sin B-sin C. But, according to the formulas of Art. 24, we have

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which is the property we had to demonstrate.

With this small number of principles, we are enabled to solve all the cases of rectilineal trigonometry.

SOLUTION OF RIGHT-ANGLED TRIANGLES.

XXXVII. Let A be the right angle of the proposed rightangled triangle, B and C the other two angles; let a be the hypotenuse, 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 as

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