be 90 of thofe Parts: And if it contains 100 Parts, a Quadrant will be 25 of thefe Parts.

The Complement of an Arc is the Difference thereof from a Quadrant.

A Chord, or Subtenfe, is a Right Line drawn from one End of the Arc to the other.

The Right Sine of any Arc, which alfo is commonly called only a Sine, is a Perpendicular falling from one End of an Arc, to the Radius drawn through the other End of the faid Arc. And is therefore the Semifubtenfe of double the Arc, viz. DE➡ DO, and the Arc DO is double of the Arc DB. Hence, the Sine of an Arc of 30 Degrees, is equal to the one half of the Radius. For (by 15. El. 4.) the Side of an Hexagon inferib'd in a Circle, that is, the Subtenfe of 60 Degrees is equal to the Radius. A Sine divides the Radius into two Segments CE, EB; one of which, CE, which is intercepted between the Center and the Right Sine, is the Sine Complement of the Arc DB to a Quadrant, (for CE FD which is the Sine of the Arc DH,) and is called the Cofine. The other Segment BE, which is intercepted between the Right Sine and the Periphery, is called a verfed Sine, and fometimes a Sagitta.

And if the Right Line CG be produced from the Center C, thro' one End D of the Arc, until it meets the Right Line BG, which is perpendicular to the Diameter drawn thro' the other End B of the Arc, then CG is called the Secant, and BG the Tangent of the Arc DB.

The Cofecant and Cotagent of an Arc is the Secant or Tangent of that Arc, which is the Complement of the former Arc to a Quadrant. Note, As the Chord of an Arc, and of its Complement to a Circle, is the fame; fo likewife is the Sine, Tangent, and Secant of an Are the fame as the Sine, Tangent, and Secant of its Complement to a Semicircle.

The Sinus Totus is the greatest Sine, or, the Sine of 90 Degrees, which is equal to the Radius of the Circle.

A Trigonometrical Canon is a Table, which, be ginning from one Minute, orderly expreffes the Lengths that every Sine, Tangent, and Secant, have in respect of the Radius, which is fuppofed Unity; and is conceived to be divided 10,000,000, or more decimal Parts. And

fo

fo the Sine, Tangent, or Secant of any Arc, may be bad by Help of this Table; and contrarywife, a Sine, Tangent, or Secant, being given, we may find the Arc it expreffes. Take Notice, That in in the following Tract, R fignifies the Radius, S a Sine, Col. a Cofine, Ta Tangent, and Cot, a Cotangent.

The CONSTRUCTION of the Trigonometrical Canon.

PROPOSITION I

THEOREM.

The two Sides of any Right-angled Triangle being given, the other Side is also given.

FOR (by 47. of the firft Element) ACq=ABq +BCq and ACq-BCq=ABq, and interchangeably ACq-ABq=BCq. Whence, by the Extraction of the fquare Root, there is given AC=✔ABq+BCq and AB=✔ACq—BCq. And BCVACq-ABq.

The Sine DE of the Arc DB, and the Radius CP, being given, to find the Cofine DF.

TH

HE Radius CD and the Sine DE, being given in the Right-angled Triangle CDE, there will be given (by the laft Prop.) ✔CDq-DEq=DF.

PROPOSITION INI.

PROBLEM.

The Sine DE of any Arc DB being given, to find DM or B M the Sine of half the Arc.

DE being given, CE (by the laft Prop.) will be given, and accordingly EB which is the Difference between the Cofine and Radius. Therefore DE, EB, being given in the Right-angled Triangle DBE, there will be given DB, whofe half DM is the Sine of the Arc DL the Arc BD.

PROPOSITION IV.
PROBLEM.

The Sine BM of the Arc BL being given, to find the Sine of double that Arc.

HE Sine BM being given, there will be given (by Prop. 2.) the Cofine CM. But the Triangles CBM, DBE, are equiangular, because the Angles at E and M are Right Angles, and the Angle at B common. Wherefore (by 4. 6.) we have CB: CM :: BD, or 2 BM : DE. Whence, fince the three firft Terms of this Analogy are given, the fourth also, which is the Sine of the Arc DB, will be known.

Coroll. Hence, CB: 2 CM:: BD: 2 DE, that is, the Radius is to double the Cofine of one half of the Arc DB, as the Subtenfe of the Arc D B, is to the Subtenfe of double that Arc. Alfo CB: 2 CM (2 BM: 2 DE::) BM: DE:: CB CM. Wherefore the Sine of any Arc, and the Sine of its Double being given, the Cofine of the Arc it felf is given.

PRO.

PROPOSITION V.

The Sines of two Arcs BD, FD, being given, to find FI the Sine of the Sum, as likewife EL, the Sine of their Difference.

LET the Radius CD be drawn, and then CO is

the Cofine of the Arc FD, which accordingly is given, and draw OP thro' O parallel to DK. Alfo let OM, GE, be drawn parallel to CB. Then because the Triangles CDK, COP, CHI, FOH, FOM, are equiangular. In the firft Place, CD: DK CO OP, which confequently is known, Also we have CD: CK:: FO: FM, and fo likewife this fhall be known. But becaufe FO= EO, then will FM MGON. And fo OP +FM FI Sine of the Sum of the Arcs: And OP FM, that is, OP-ON EL Sine of the Difference of the Arcs. W. W. D.

Corall. Because the Differences of the Arcs BE, BD, BF, are equal, the Arc BD fhall be an Arithmetical Mean between the Arcs B E, BF.

PROPOSITION VI.

The fame Things being fuppofed, Radius is to double the Cofine of the mean Arc as the Sine of the Difference, to the Difference of the Sines of the Extremes.

FOR

OR we have CD: CK:: FO: FM, whence by doubling the Confequents CD: 2 CK :: FO: 2 FM, or to FG; which is the Difference of the Sines EL, FI. W.W.D.

Coroll. If the Arc BD be 60 Degrees, the Difference of the Sines FI, EL, fhall be equal to the Sine, FO; of the Distance, For in this Cafe, CK is the Sine of 30 Degrees, Double thereof being equal to Radius; and fo fince CD 2 CK we fhall have FOFG. And confequently, if the two Arcs BE, BF, are Equidiftant from the Arc of 60

T 3

De.

Degrees, the Difference of the Sines fhall be equal to the Sine of the Distance FD.

Coroll. 2.

Hence, if the Sines of all Arcs be given diftant from one another by a given Interval, from the Beginning of a Quadrant to 60 Degrees, the other Sines may be found by one Addition only. For the Sine of 61 Degrees Sine of 59 Degrees Sine of 1 Degree. And the Sine of 62 Degrees Sine of 58 Degrees + Sine of 2 Degrees. Alfo the Sine of 63 Degrees Sine of 57 Degrees +

Sine of 3 Degrees, and fo on.

Coroll. 3. If the Sines of all Arcs, from the Beginning of a Quadrant to any Part of the Quadrant, diftant from each other, by a given Interval be. given, thence we may find the Sines of all Arcs to the Double of that Part. For Example, Let all the Sines to 15 Degrees be given; then, by the precedent Analogy, all the Sines to 30 Degrees, may be found. For Radius is to double the Cofine of 15 Degrees, as the Sine of 1 Degree, is to the Difference of the Sines of 14 Degrees, and 16 Degrees; fo alfo is the Sine of 3 Degrees, to the Difference between the Sines of 12 and 18 Degrees; and fo on continually until you come to the Sine of 30 Degrees.

After the fame Manner, as Radius is to double the Cofine of 30 Degrees, or to double the Sine of 60 Degrees, fo is the Sine of 1 Degree to the Difference of the Sines of 29 and 31 Degrees: Sine 2 Degrees, to the Difference of the Sines of 28 and 32 Degrees: Sine 3 Degrees, to the Difference of the Sines of 27 and 33 Degrees, But in this Cafe, Radius is to double the Cofine of 30 Degrees, as 1 to 3.. And accordingly, if the

FIG. for the DEFINITIONS.

Let B D be an Arch of 30es
Rad. Fan Co-fine Sine

Then as CB: BG:

VCB

FD: DE. DOCB ergo DE = 1⁄2, DECE=√ √ CB:CE: 1:√ 2.E.D.

=√3.

CD: 2 CE :1:2 = √ = = √3.