PROPOSITION. II.

PROBLEM.

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

The 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 III.

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, whole half DM is the Sine of the Arc DL the Arc BD.

PROPOSITION IV.

PROBLEM.

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

THE

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

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 first Terms of this Analogy are given, the fourth alfo, 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

the Arc DB, as the Subtenfe of the Arc DB, 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.

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.

LE

ET 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 becaufe the Triangles CDK, COP, CHI, FOH, FOM, are equiangular. In the first Place, CD: DK CO: OP, which Confequently is known. Alfo we have CD: CK; FO: FM, and so likewife this fhall be known. But because FO EO, then will FM MGON, And fo OPFMFI Sine of the Sum of the Arcs And OP-FM, that is, OPON= EL Sine of the Difference of the Arcs. W.W. D.

=

Coroll. Because the Differences of the Arcs BE, BD, BF, are equal, the Arc BD fhall be an Arithme tical Mean between the Arcs BE, BF.

PROPOSITION VI.

The fame Things being fuppofed, Radius is to double the Cofine of the mean Arc as the Sine of the Diffe rence, to the Difference of the Sines of the Ex

treams.

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

Coroll

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 CD2 CK, we fhall have FOFG. And confeqently, if the two Arcs BE, BF, are Equidiftant from the Arc of 60 Degrees, the Difference of the Sines shall 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 Quardrant 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 Degree. Alfo the Sine of 63 Degrees Sine of 57 Degrees + Sine of 3 Degrees, and so 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 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 Dif ference 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

Sines of the Distances from the Arc of 30 Degrees, be multiply'd by 3, the Differences of the Sines will be had.

So likewife may the Sines of the Minutes in the Begining the Quadrant be found, by having the Sines and Cofines of one and two Minutes given. For as the Radius is to double the Cofine of 2':: Sine i': Difference of the Sines of 1' and 3':: Sine 2': Difference of the Sines of o' and 4', that is, to the Sine of 4. And fo the Sines of the four firft Minutes being given; we may thereby find the Sines of the others to 8', and from thence to 16, and fo

on.

PROPOSITION VII.

THEOREM.

In Small Arces, the Sines and Tangents of the fame Arcs are nearly to one another, in a Ratio of Equality.

FOR because the Triangles CED, CBG, are equiangular, CE: CB:: ED: BG. But as the Point E approaches B, EB will vanish in Refpect of the Arc BD: Whence CE will become nearly equal to CB. Aud fo ED will be (alfo near

ly equal to BG. If EB be lefs than the

I

10,000,000

Part of the Radius, then the Difference between the

Sine and the Tangent will be alfo lefs than the

Part of the Tangent.

I

10,000,000

Coroll. Since any Arc is lefs than the Tangent, and greater than its Sine, and the Sine and Tangent of avery small Arc, are nearly equal; it follows that the Arc fhall be nearly equal to its Sine; and fo in very final Arcs it shall be, as Arc is to Arc, fo is Sine to Sine..

PRO

PROPOSITION VIII.

To find the Sine of the Arc of one Minute.

THE

HE Side of a Hexagon infcribed in a Circle, that is, the Subtenfe of 60 Degrees, is equal to the Radius, (by 15th of the 4th,) and fo the half of the Radius fhall be the Sine of the Arc 30 Degrees. Wherefore the Sine of the Arc of 30 Degrees being given, the Sine of the Arc of 15 Degrees may be found, (by Prop. 3.) Alfo the Sine of the Arc of 15 Degrees being given, (by the fame Prop.) we may have the Sine of 7 Degrees 30 Minutes: So likewife can we find the Sine of the half of this, viz. 3 Degrees 451; and so on, until twelve Bifections being inade, we come to an Arc of 522, 443, 03*, 45, whole Cofine is nearly equal to the Radius, in which Cafe (as is manifeft from Prop. 7.) Arcs are propotional to their Sines: And fo, as the Arc of 52, 443, 3*, 453, is to an Arc of one Minute, fo fhall the Sine before found, be to the Sine of an Arc of one Minute, which therefore will be given. And when the Sine of one Minute is found, then (by Prop. 2. and 4.) the Sine and Cofine of two Minutes will be had.

PROPOSITION IX.

THEOREM

If the Angle BAC, being in the Periphery of a Circle, be bifected by the Right Line AD, and if AC be produced until DEAD meets it in E: then ball CE=AB.

IN

N the quadrilateral Figure A B D C (by 13. 1.) the Angles Band ACD are equal to two Right Angles DCE+DCA (by 22. 3.) Whence the the Angle BDCE. But likewife the Angle E= DAC (by. 1.) DAB and DC=DB. Wherefore the Triangles BAD and CED are congruous, and CE is equal to AB. W. W. D.

PRO