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will be known. Then becaufe of the fimilar Triangles CFE, CAT, and CDH, it will be (by 5.4.)

1. CF. FE :: CA: AT; whence the Tangent is known,

2. CF: CE (CA) :: CA: CT; whence the Secant is known.

3. EF: CF:: CD: DH; whence the Co-tangent is known.

4. EF: EC (CD):: CD: CH; whence the Cofecant is known.

Hence it appears,

1. That the Tangent is a Fourth-proportional to the Co-fine, the Sine, and Radius.

2. That the Secant is a Third-proportional to the Co-fine and Radius.

3. That the Co-tangent is a Fourth-proportional to the Sine, Co-fine, and Radius.

4. And that the Co-fecant is a Third-proportional to the Sine and Radius.

5. It appears moreover (because AT: AC:: CD (AC): DH), that the Rectangle of the Tangent and Co-tangent is equal to the Square of the Radius (by 3.4.): Whence it likewife follows, that the Tangent of Half a Right-angle is equal to the Radius; and that the Co-tangents of any two different Arches (represented by P and Q are to one another, inversely as the Tangents of the fame Arches: For, fince Tang. Px Co-tang. P = fqu. Rad. = Tang. QxCo-tang. Q; therefore will Co

tang.

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tang. P: Co-tang. Q:: Tang. Q: Tang. P; or as Co-tang. P: Tang. Q:: Co-tang. Q: Tang. P (by 3.4.)

PROP. II.

If there be three equidifferent Arches AB, AC, AD, it will be, as Radius is to the Co-fine of their common Difference BC, or CD, fo is the Sine C F, of the Mean, to Half the Sum of the Sines BE + DG, of the two Extremes: And, as Radius to the Sine of the common Difference, fo is the Co-fine FO of the Mean, to Half the Difference of the Sines of the two Ex

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meeting DG in H and v.

For let BD be drawn, interfecting the Radius OC in m; alfo draw mn parallel to CF, meeting AO in ʼn ; and BH and m v5. parallel to AO,

Then, the Arches BC and CD being equal to each other (by Hypothefis), OC is not only perpendicular to the Chord BD, but also bifects it (by 1. 3.) and therefore Bm (or Dm) will be the Sine of BC (or DC), and Om its Co-fine: Moreover mn, being an arithmetical Mean between the Sines B E, DG of the two Extremes (because Bm Dm) is therefore equal to Half their Sum, and Dv equal to Half their Difference. But, because of the fimilar Triangles OCF, Omn and Dum,

It will be SOC: Om :: CF : mn
OC Dm :: FO: Dv

} Q. E.D.

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30°

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COROLLARY II.

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Hence, if the mean Arch AC be fuppofed that of 60°; then OF being the Co-fine of 60°, = Sine Chord of 60° OC, it is manifeft that DG-BE BE will, in this Cafe, be barely = Dm; and confequently DG=Dm+BE. From whence, and the preceding Corollary, we have these two ufeful Theorems.

1. If the Sine of the Mean, of three equidifferent Arches (fuppofing Radius Unity) be multiplied by twice the Co-fine of the common Difference, and the Sine of either Extreme be fubtracted from the Product, the Remainder will be the Sine of the other Extreme.

2. The Sine of any Arch, above 60 Degrees, is equal to the Sine of another Arch, as much below 60°, together with the Sine of its Excefs above 60°.

• Note,

Om x CF

OC

taken in a geometrical Senfe, de

notes a Fourth-proportional to OC, Om and CF; but, arithmetically, it fignifies the Quantity arifing by dividing the Product of the Measures of Om and CF by that of OC. Understand the like of others.

PROP.

PROP. III.

To find the Sine of a very small Arch; fuppofe that of 15'.

It is found, in p. 139. of the Elements, that the Length of the Chord of 4 of the Semi-periphery is expreffed by ,00818121 (Radius being Unity); therefore, as the Chords of very fmall Arches are to each other nearly as the Arches themselves (vid. p. 140.) we fhall have, as 34:35::,00818121: ,008726624, the Chord of, or Half a Degree; whofe Half, or ,004363312, is therefore the Sine of 15', very nearly.

From whence the Sine of any inferior Arch may be found by bare Proportion. Thus, if the Sine of 1' be required, it will be, 15: 1'::,004363312: ,000290888, the Sine of the Arch of one Minute, nearly.

But if you would have the Sine of 1' more exactly determined (from which the Sines of other Arches may be derived with the fame Degree of Exactnefs); then let the Opperations, in p. 139, be continued to II Bifections, and a greater Number of Decimals be taken; by which Means you will get the Chord of 4 Part of the Semi-periphery to what Accuracy you please: Then, by proceeding as above (for finding the Sine of 15′), the Sine of 1 Minute will also be obtained to a very great Degree of Exactness.

PROP. IV.

To fhew the Manner of constructing the Trigonometrical Canon.

First, find the Sine of an Arch of one Minute, by the preceding Prop. and then its Co-fine, by Prop.

Prop. 1. which let be denoted by C; then (by Theor. 1. p. 13.) we shall have

2C x Sine 1'-Sine o'

Sine 2'.

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And thus are the Sines of 6', 7, 8', &c. fucceffively derived from each other.

The Sines of every Degree and Minute, up to 60°, being thus found; thofe of above 60° will be had by Addition only (from Theor. 2. p. 13.) then, the Sines being all known, the Tangents and Secants will likewise become known, by Prop. 1.

Note, If the Sine of every 5th Minute, only, be computed according to the foregoing Method, the Sines of all the intermediate Arches may be had from thence, by barely taking the proportional Parts of the Differences, and that so near as to give the firft 6 Places true in each Number; which is fufficiently exact for all common Purposes.

Scholium.

Although what has been hitherto laid down for conftructing the Trigonometrical-Canon, is abundantly fufficient for that Purpose, and is also very eafily demonftrated; yet, as the first Sine, from whence the reft are all derived, must be carried on to a great Number of Places, to render the numerous Deductions from it but tolerably exact (because in every Operation the Error is multiply'd). I fhall here fubjoin a different Method, which will be found to have the Advantage, not only in that, but in many other Respects.

First, then, from the Co-fine of 15°, which is given (by p. 138 and 139 of the Elements)=√2+ √3 =,965925826, &c. (the Supplement Chord of 30°) and the Sine of 18°, which is√=

,309017,

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