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Ε L Ε Μ Ε Ν Τ S

OF

PLAIN AND SPHERICAL

TRIGONOME TRY.

PLAIN TRIGONOMETRY.

The

THE business of trigonometry is to find the angles when

the fides are given, and the sides, or ratio of the sides, when the angles are given ; and to find sides and angles, when fides and angles are given. For which, it is necessary, that, not only the periphery of the circle, but likewise certain right lines in it, be supposed divided into fome determinate number of parts. The ancient geometers have supposed the periphery divided into 360 parts or degrees, and every degree into 60 minutes, and every minute into 60 seconds, &c.; and every angle is said to be of such a number of degrees and minutes as there are in that part of the periphery measuring the angle.

I. An arch is any part of the periphery or circumference, and is the measure of the angle at the center which it subtends.

II. The quadrant of a circle is one fourth part of the circumference;

the difference of an arch from a quadrant or go degrees, is called the complement of that arch.

III. A chord or subtense, is a right line drawn from one part of an arch to another.

a 3. 3.

IV.
The right sine, or fine of any arch, is a right line drawn from

the vertex of an arch perpendicular to the diameter of the

circle, and is equal to half the chord of double that arch a. If the arch DB, (fig. for the def.) is an arch of 30 deg. DE is the fine of 30 deg. and twice :) E, equal DO, is the subtense of 60

bis, 4o deg. The fine of 30 deg. is equal one half radius b.

V.
Every fine, as DE, divides the radius into two parts, that part

betwixt the center and fine, as CE, is called the cosine ; and
the part betwixt the fine and arch, as EB, is the versed fine
of the arch DB. For the same reason, AE is the versed sine
of the arch AD; therefore, the versed fine may be equal,
greater, or less than the radius.

VI.
The arch HD is the complement of BD to a quadrant; and FD,
equal CE, is the fine of that arch or cosine of BD.

VII.
If a right line, BG, is drawn from the point B, at right angles

to the diameter, and meeting the right line CG, palling thro'
the point D; then BG is the tangent of the arch BD, and CG

is the secant of that arch. The right line HI, drawn from the point H, at right angles to

CH, and meeting CG produced in I, is the tangent of the arch HD, or cotangent of BD; and CI is the fecant of HD,

or cosecant of BD. The fine totus, or greatest fine, is the radius of the circle, which

is the fine of 90 deg. Characters used. + Addition. - Subtraction. Multipli

cation. = Equality. :: Proportion. V Extraction of the square-root.

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When a line is drawn over any number of quantities, these quantities are to be considered as one quantity. The marks

, put overany numbers, are to be read degr. min. second third minutes, &c. as 23°, 17, 18", 25", &c. Likewife, R. fignifies Rad. S. Sine, Cof Cofine, T. Tang. Cot. Cotangent, Sec. Secant, and Cofec. Cofecant.

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Because the triangles CED, CBG, (fig. for the definitions,) are similar, CE: EV :: CB : BG, by alter. CE : CB :: ED: BG, i.e. Cof. : R :: Sine: Tangent.

Again,

d 34. 1

Again, CE:CD :: CB: CG, i. e Cof. :R. ::R.: Secant. And, because the triangles CDF, CED, CBG, and CHI, are similar, CE :ED :: CF : FD; but CE-FD ; therefore ED=CF d, therefore CE is the line of the angle CDE=DCF.

Again, EC :D :: CB : BG; altern. EC: CB:ED:BG; there!ore, if EC be nearly equai to CB, ED will be nearly equal to EG ; therefore, if the arch DB be a very small arch, the fine and tangent are nearly to one another in the ratio of equality:

II. Because the chord of any arch, and its supplement to a circle, is the same, so the fine, tangent, or fecant, of any arch, and its supplement to a femicircle, is likewise the fame.

III. If two sides of a right angled triangle be given, the other can be found by 47. El. 1.

PRO P. I.

IN N a right angled triangle, if the hypothenuse be made radius,

then the sides are the fines of their opposite angles; and, if either of the sides about the right angle be made radius, the other side is the tangent of its opposite angle, and the hypothenuse is the secant of that angle.

& def. 1.

For, in the triangle ABC, if, with the center A, and distance AC, a circle be described, and AB produced till it cut that circle in D, (No 1.); then CB is the fine of the arch CD, or of the angle Aa, and AB is the cofine of CD, or fine of the angle C. Again, (No 2.) if AB is made radius, and the arch BD drawn, then BC is the tangent of the arch BD, or of the angle A ; and AC is the secant of that angle. Wherefore, &c.

COR. Hence, as AC, Rad. taken in any given measure, is to BC, taken in the same measure, are so any parts into which the radius is supposed to be divided, viz. 10.000000, to a number expressing the parts in proportion to the length of the fine of the angle, that is,

AC being radius, AC:BC ::R:SA
And

AC:BA::R:S,C
AB Rad.

AB: BC ::R:T,A
And

AB: AC ::R: Sec. A.
BC Rad.

BC:BA :: R:TC
And

BC: AC :: R: Sec. C.

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THE sides of plain triangles are to one another as the fines of

their oppojite angles.

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Let ABC be the triangle, about which defcribe a circle ABC"; from the center D let fall perpendiculars upon each of a 5. 4. the sides AB, BC, AC, which will be bisccted in the points E, F, and Gb; but the angle BDE ° is equal to the angie b 3. 3. CDE, and BE is the fine of the angle BDE d, or of the angle BAC ®;. For the same reason, BF is the one of the angle ACB, e 21. 1. and GC the fine of the angle ABC: Theretore, BE is to BF as twice BE is to twice DF; that is, BC is to BA as the fine of the angle A is to the fine of the angle C.

2d. If the triangle is right angled, then BD, the Rad. is the fine of the right angle; the other two angles as pefore.

3d. If the triangle is obtufe angle, then, if a triangle is formed upon the same base, in the opposite Pznent in the point I, then that angle will be acute; and BE is the line of the angle BIC, or BACf. Wherefore, &c,

f schol. 1,

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I Nany right lined triangle, the sum of any two sides, is to their

difference, as the tangent of half the sum of the angles at the base, is to the tangent of half their difference.

Let ABC be the triangle, the sum of any two of its sides, as AB, BC, is to the difference of these fides, as the tangent of half the sum of the angles BAC, ACB, at the base, is to the tangent of half their difference.

For, let AB be produced to H; make BH equal to BC ; and cut off BI equal to BA; then AH is equal to the sum of the fides, and HI to the difference of the fides, and the angle HBC equal to the sum of the angles at the base“, viz. the angles BAC, a 32. I. ACB. Join HC ; and, from B, let BE fall perpendicular upon HC b; then, because HB is equal to BC, the angle BHC is equal b 12. I. to BCH“, and the angles BEC, BEH, are equal b; therefore the c 5. I. angles HBE, EBC, are likewise equal d; and, it BE be made ra- d 32. I. dius, then EC is the tangent of half the sum of the angles at the base. Draw BD parallel to AC, then the angle DBC is equal to the angle ACB. Take HF equal to DC, and join FB;€ 29. s!

then

f 2. 6.

8 14. 5. h 4. 6.

then FBD is the difference of the angles, and EBD half their difference: Through I draw IG parallel to BD or AC; then IB is to BA as GD is to DCF; but IB is equal to BA ; therefore GD is equal to DC 8; but AH is to HC as HI is to HG b; and, by altern. AH is to HI as HC is to HG, the consequents being halved, as HA is to HI, so is HC to HG; but HF, GD, are each equal to DC, and therefore equal to one another. Add, or take away, GF to or from both, then HG is equal FD; but half FD'is ED; therefore AH is to HI as EC is to ED. Wherefore, &c.

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IN any triangle, the rectangle under half the sum of the sides, and

excess of the same, above any of the sides, taken as the base, is to the reclangle contained by the right lines, by which the half of the sum of the sides exceeds the other two fides, as the square of the rad. is to the square of the tangent of half the angle opposite to the base.

a 4. 4

Let ABC be the triangle, BC the base ; in the triangle ABC let a circle be inscribed a, of which let G be the center, and let fall GD, GE, GF, perpendiculars to the sides AB, BC, AC; then AD is equal to AE, BD to BF, and CE to CF; and the angles at A and B bisected by the right lines - AG, BG ; produce AB, AC, to H, L; make BH equal to FC, and CL to BF; at the points H, L, raise the perpendiculars HK, LK, meeting the right line AG, produced in K; from the point K let fall KM perpendicular to BC; and join BK, KC; then the rectangle HAD will be to the rectangle BFC as the square of AD to the square of DG.

For, the triangles ADG, AEG, are equiangular ?, and the angle ADG equal to the angle AHK, for each are right ones ; therefore DG is parallel to HKb; and, fince the angles AHK, HAK, are equal to the angles KAL, ALK, for AK bifects them, AH is equal to AL; but the angles ABC, CBH, are equal to two right angles d; and DGF, DBF, equal to two right angles"; for the angles at D and F are right ones. Take the angle DĚF from both, and there remains DGF equal to HBM ; but HBM, HKM, are equal to two right angles, for the angles at H and M are right ones; therefore the quadrilateral figures BDGF, BHKM, are equiangular, and BG bisects the angles DBF, DGF ; therefore BK will likewise bisect the angles

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