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Draw one of the given sides. From one end of it lay off the given angle, and draw the other given side, making the required angle with the first side. Then connect the extremities of the two sides, and there will be formed the triangle required. Ex. 1. Given the angle A, 37° 25', the side AC, 675, and the side AB, 417, to construct the triangle, and find the other parts.

Ex. 2. Given the angle A, 75°, the side AC, 543, and the side AB, 721, to construct the triangle.

IV. Given the three sides, to find the angles.

Draw one of the sides as a base; and from one extremity of the base, with a radius equal to the second side, describe an arc of a circle. From the other end of the base, with a radius equal to the third side, describe a second arc intersecting the former; the point of intersection will be the third angle of the triangle.

Ex. 1. Given AB, 678, AC, 598, and BC, 435, to find the angles.

Ex. 2. Given the three sides 476, 287, and 354, to find the angles.

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I

Values of the Sines, Cosines, &c., of certain Angles (64.) We propose now to examine the changes which the Lines, cosines, &c., undergo in the dif ferent quadrants of a circle. Draw two diameters, AB, DE, perpendicular to each other, and suppose one of them to occupy a horizontal position, B the other a vertical. The angle ACD is called the first quadrant, the angle DCB the second quadrant, the angle BCE the third quadrant, and the an

C

E

H

gle ECA the fourth quadrant; that is, the first quadrant is above the horizontal diameter, and on the right of the vertical diameter; the second quadrant is above the horizontal diameter, and on the left of the vertical, and so on.

Suppose one extremity of the arc remains fixed in A, while the other extremity, marked F, runs round the entire circumference in the direction ADBE.

When the point F is at A, or when the arc AF is zero, the sine is zero. As the point F advances toward D, the sine increases; and when the arc AF becomes 45°, the triangle CFG being isosceles, we have FG : CF ::1: √2 (Geom., Prop. 11, Cor. 3, B. IV.); or sin. 45°: R: 1:√2.

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The sine of 30° is equal to half radius (Art. 22). Also, since sin. A= √ R2—cos. 'A, the sine of 60°, which is equal to the cosine of 30°, = √R2—4R2= √¿R2={R√3.

2

4

2

The arc AF continuing to increase, the sine also increases till F arrives at D, at which point the sine is equal to the radius; that is, the sine of 90°=R.

As the point F advances from D toward B, the sines diminish, and become zero at B; that is, the sine of 180°=0.

In the third quadrant, the sine increases again, becomes equal to radius at E, and is reduced to zero at A.

(65.) When the point F is at A, the cosine is equal to radius. As the point F advances toward D, the cosine decreases, and the cosine of 45°-sine 45° R√2. The arc continuing to increase, the cosine diminishes till F arrives at D, at which point the cosine becomes equal to zero. The cosine in the second quadrant increases, and becomes equal to radius at B; in the third quadrant it decreases, and becomes zero at E; in the fourth quadrant it increases again, and becomes equal to radius at A.

(66.) The tangent begins with zero at A, increases with the arc, and at 45° becomes equal to radius. As the point F approaches D, the tangent increases very rapidly; and when the difference between the arc and 90° is less than any assignable quantity, the tangent is greater than any assignable quantity. Hence the tangent of 90° is said to be infinite.

In the second quadrant the tangent is at first infinitely great, and rapidly diminishes till at B it is reduced to zero. In the third quadrant it increases again, becomes infinite at E, and is reduced to zero at A.

The cotangent is equal to zero at D and E, and is infinite at A and B.

(67.) The secant begins with radius at A, increases through

D

D

the first quadrant, and becomes infinite at D; diminishes in the second quadrant, till at B it is equal to the radius; increases again in the third quadrant, and becomes infinite at E; decreases in the fourth quadrant, and becomes equal to the B radius at A.

A

C

G

H

E

The cosecant is equal to radius at D and E, and is infinite at A and B. (68.) Let us now consider the algebraic signs by which these lines are to be distinguished. In the first and second quadrants, the sines fall above the diameter AB, while in the third and fourth quadrants they fall below. This opposition of directions ought to be distinguished by the algebraic signs; and if one of these directions is regarded as positive, the other ought to be considered as negative. It is generally agreed to consider those sines which fall above the horizontal diameter as positive; consequently, those which fall below must be regarded as negative. That is, the sines are positive in the first and second quadrants, and negative in the third and fourth.

In the first quadrant the cosine falls on the right of DE, but in the second quadrant it falls on the left. These two lines should obviously have opposite signs, and it is generally agreed to consider those which fall to the right of the vertical diameter as positive; consequently, those which fall to the left must be considered negative. That is, the cosines are positive in the first and fourth quadrants, and negative in the second and third.

(69.) The signs of the tangents are derived from those of R. sin. the sines and cosines. For tang.

COS.

(Art. 28). Hence, when the sine and cosine have like algebraic signs, the tangent will be positive; when unlike, negative. That is, the tangent is positive in the first and third quadrants, and negative in the second and fourth.

Also, cotangent

R2 tang.

(Art. 28); hence the tangent and

cotangent have always the same sign.

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R2 sin.

Also, cosec. = ; hence the cosecant must have the same

sign as the sine.

(70.) The preceding results are exhibited in the following tables, which should be made perfectly familiar :

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(71.) In Astronomy we frequently have occasion to consider arcs greater than 360°. But if an entire circumference, or any number of circumferences, be added to any arc, it will terminate in the same point as before. Hence, if C represent an entire circumference, or 360°, and A any arc whatever, we shall have

sin. A sin. (C+A)=sin. (2C+A)=sin. (3C+A)=, &c. The same is true of the cosine, tangent, &c.

We generally consider those arcs as positive which are estimated from A in the direction ADBE. If, then, an arc were estimated in the direction AEBD, it should be considered as negative; that is, if the arc AF be considered positive, AH must be considered negative. But the latter belongs to the fourth quadrant; hence its sine is negative. Therefore, sin. (-A)=—sin. A.

The cosine CG is the same for both the arcs AF and AH. cos. (-A)=cos. A.

Hence,

Also,

And

tang. (-A)=-tang. A.
(—A)=

cot. (-A)=-cot. A.

TRIGONOMETRICAL FORMULÆ.

(72.) Expressions for the sine and cosine of the sum and difference of two arcs.

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D.

F

E

L

Let AB and BD represent any two given arcs; take BE equal to BD: it is required to find an expression for the sine of AD, the sum, and of AE, the difference of these arcs. Put AB=a, and BD=b; then AD= a+b, and AE-a-b. Draw the chord DE, and the radius CB, which may be represented by R. Since DB is by construction equal to BE, DF is equal to FE, and therefore DE is perpendic

C

K

TH

GA

ular to CB. Let fall the perpendiculars EG, BH, FI, and DK

upon AC, and draw EL, FM parallel to AC.

Because the triangles BCH, FCI are similar, we have

CB: CF :: BH: FI; or R: cos. b:: sin. a : FI.

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Also, CB: CF :: CH: CI; or R: cos. b:: cos. a : CI. .

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The triangles DFM, CBH, having their sides perpendicular each to each, are similar, and give the proportions

CB : DF :: CH: DM; or R : sin. b :: cos. a : DM.

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Also, CB: DF :: BH: FM; or R sin. b:: sin, a: FM.

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