From hence, if the Sines of the Arcs diftant one Mrnute from each other be given, the Sines of all the Arcs that are in the fame Progreffion may be found by an exceeding eafy Operation.

In the first and fecond Series, if Ao; then shall a=o, and b its Cofne, will became Radius, or I. And hence, if the Terms wherein a is, are taken away, and I to be put instead of b, the Series will become the Newtonian. In the third and fourth Series, if A be 90 Degrees, we shall have bo, and a 1. Whence again, taking away all the Terms wherein b is, and putting instead of a, we fhall have the Newtonian Series arife.

Note, All the faid Series eafily flow from the New tonian ones. By the fifth Propofition.

PROPOSITION XI.

In a Right-angled Triangle, if the Hypothenufe be made the Radius, then are the Sides the Sines of their oppofite Angles; and if either of the Legs be made the Radius, then the other Leg is the Tangent of its oppofite Angle, and the Hypothenuje is the Secant of that Angle.

IT

T is manifeft that CB is the Sine of the Arc CD, and AB the Cofine thereof; but the Arc CD is the Measure of the Angle A, and the Complement of the Measure of the Angle C. Moreover, if AB, in the Figure to this Propofition, be fuppofed Radius, then BC is the Tangent, and AC the Secant of the Arc BD, which is the Measure of the Angle A. So alfo if BC be made the Radius, then is BA the Tangent, and AC the Secant of the Arc BE, or Angle C. W. W. D. Therefore as AC being taken as fome given Meafure, is to BC taken in the fame Meafure; fo fhall the Number 10,000,000 Parts in which the Radius is fuppofed to be divided, be to a Number expreffing in the fame Parts the Length of the Sine of the Angle that is, it will be

A;

as

And fo if any three of thefe Proportionals be given, the fourth may be found by the Rule of Three.

PROPOSITION XII.

The Sides of Plain Triangles are as the Sines of their oppofite Angles.

[F the Sides of a Triangle, infcribed in a Circle, be

Sides be the Sines of the Angles at the Periphery; for the Angle BDC at the Center, is double of the Angle BAC at the Periphery; (by 20 El. lib. 3.) and fo the half of every of them, viz. BDE BAC, and BE is the Sine thereof. For the fame Reason, BF fhall be the Sine of the Angle BCA, and AG the Sine of the Angle ABC.

In a Right-angled Triangle we have BD2 BC Radius (by 31. Eucl. 3.) but Radius is the Sine of a Right Angle: Whence BC is the Sine of the Angle A.

In an Obtufe-angled Triangle, let BI, CI, be drawn, and then the Angle L fhall be the Complement of the Angle A to two Right Angles, (by 22 El. 3.) and fo they fhall both have the fame Sine. But the Angle BDE, (whofe Sine is BE) Angle L. Therefore BE fhall be the Sine of the Angle BAC. And fo in every Triangle, the Halves of the Sides are the Sines of the oppofite Angles; but it is manifeft that the Sides are to one another as their Halves. W.W.D.

PRO

PROPOSITION XIII.

In a plain Triangle, the Sum of the Legs, the Dif ference of the Legs, the Tangent of the half Sum of the Angles at the Bafe, and the Tangent of one half their Difference, are proportional.

ET there be a Triangle ABC, whofe Legs are AB, BC, and Bafe AC. Produce AB to H, fo that BHBC, then fhall AH be the Sum of the Legs; and if you make BIBA, then IH will be the Difference of the Legs. Alfo the Angle HBC

Angles A+ ACB, (by 32. El. 1.) and fo EBC the half thereof= half the Sum of the Angles A and ACB, and its Tangent (putting the Radius = EB) is EC. Again, let BD be drawn parallel to AC, and make HFCD. Then fince HBCB, we shall have (by 4. El. 1.) the Angle HBF CBD-BCA. (by 29. El. 1.) Alfo the Angle HBD= Angle A; whence F BD fhall be the Difference of the Angles A and ACB; and EBD, whofe Tangent is ED, half their Difference. Let IG be drawn thro' I pa rallel to AC or BD, and then (by 2. El. 6.) AB: BI:: CD: DG, but AB=BI; whence we shall have CD=DG, but CD=HF, and fo HF=DG, and confequently, HG DF, and HG= DF

DE; and because the Triangles AHC, IHG, are equiangular, it fhall be as AH: IH :: HC: HG:: HC HG :: EC: ED. That is, AH the Sum of the Legs, to IH the Difference of the Legs, fhall be as EC the Tangent of one half the Sum of the Angles at the Base, to ED the Tangent of one half their Difference. W. W.D.

PROPOSITION XIV.

In a plain Triangle, the Bafe, the Sum of the Sides, the Difference of the Sides, and the Difference of the Segments of the Bafe, are proportional. ET DC be the Base of the Triangle BCD Circle be described. Produce DB to G, and from B

let

let fall BE perpendicular to the Bafe; then shall DG=DB+BC=Sum of the Sides, and DH= Difference of the Sides; and DE, CE, are the Segments of the Bafe whofe Difference is DF; because (by Cor. Prop. 38. El. 3.) the Rectangle under DC and DF, is equal to the Rectangle under DG, DN, it fhall be (by 16. El. 6.) as DC: DG :; DH : DF.

6

PROBLEM.

The Sum and Difference of any two Quantities being given, to find the Quantities themselves.

F one half of the Sum be added to one half of the Difference, the Aggregate fhall be equal to the greater of the Quantities; and if from one half of the Sum be taken one half of the Difference, the Refidue fhall be equal to the leffer of the Quantities. For let there be two Quantities AB, BC, and let there be taken AD=BC; then DB will be their Difference, and AC their Sum; which, bifected in E, gives AE or EC the half Sum, and DE or EB the half Difference. Hence ABAE+EB= the half Sum the half Difference, and BC=CEEB the half Sum-the half Difference.

=

In any plain Triangle if two Angles be given, the third Angle is alfo given, because it is their Complement to two Right Angles.

If one of the acute Angles of a Right-angled Triangle be given, the other acute Angle will be given, because it is the Complement of the given Angle to a Right Angle.

And if two Sides of a Right-angled Triangle be given, the other Side may be found by the firft Propofition without a Canon.

The Trigonometrical Solutions of a Rightangled Triangle, may be as follow. Vid.

Fig. A.

The Trigonometrical Solutions of Oblique angled Triangles. Vid. Fig. to Prop. 12.

Given Sought
The The
Angles Sides
A, B, C, BC and
I and the AC.
Side AB.

Make as

S, C: SA :: AB:BC. Also S, C: S, B:: AB: AC: But when two Angles are given, the third is alfo given; whence the Cafe wherein two Angles and a Side are given, to find the reft, falls into this Cafe.