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will be known. Then because of the similar Triangles CFE, CAT, and CDH, it will be (by

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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 Co

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

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

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

5. It appears moreover (because AT; AC: : CD (AC): DH), that the Rećtangle of the Tangent and Co-tangent is equal to the Square of the Radius (by 3. 4.): Whence it likewise 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 same Arches: For, since Tang. Px Co-tang. P = squ. Rad. = Tang, Q_x Co-tang, Q; therefore will Cotang.

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If there be three equidifferent Arches AB, AC, AD, it will be, as Radius is to the Cosine of their common Difference BC, or CD, so is the Sine C F, of the Man, to Half the Sum of the Sines B E + DG, of the two Extremes: And, as Radius to the Sine of the common Difference, so is the Co-sine FO of the Mean, to Half the Difference of the Sines of the two Extroots. -

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AE En G. O parallel to AO,

meeting DG in H and v. Then, the Arches BC and CD being equal to each other (by Hypothesis), OC is not only perpendicular to the Chord BD, but also bisects it (by I. 3.) and therefore B m (or D m) will be the Sine of BC (or DC), and Om its Co fine: Moreover mn, being an arithmetical Mean between the Sines BE, 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 similar Triangles OCF, Omn and Dvn,

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1. If the Sine of the Mean, of three equidifferent Arches (supposing Radius Unity) be multiplied by twice the Cosine of the common Difference, and the Sine of either Extreme be subtraffed from the Produći, 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 Exces above 60°.

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It is found, in p. 139. of the Elements, that the Length of the Chord of ris of the Semi-periphery is expressed by ,oo818121 (Radius being Unity) ; therefore, as the Chords of very small Arches are to each other nearly as the Arches themselves (vid. p. 140.) we shall have, as #: 3+8 ::,00818121 :

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whose Half, or soo43633 12, 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 :: ,oo43633 12 : ,ooo29.0888, the Sine of the Arch of one Minute, nearly. But if you would have the Sine of 1 more exaćtly determined (from which the Sines of other Arches may be derived with the same Degree of Exačtness); then let the Opperations, in p. 139, be continued to 11 Bise&tions, and a greater Number of Decimals be taken; by which Means you will get the Chord of zir; 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 Exačtness.

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First, find the Sine of an Arch of one Minute,

by the preceding Prop. and then its Co-sine, by Prop.

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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 first 6 Places true in each Number; which is sufficiently exačt for all common Purposes.

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