as AC: BC:R: S, A. by the fame Reason, as AC: BA:R:S, C. also as AB: BC:R: T, A. and BC: BA:R:T, C. And so if any three of these 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 opposite Angles. F the Sides of a Triangle, inscribed in a Circle, be bifected by perpendicular Radi, then thall the half Sides be the Sines of the Angles at the Periphery; for the Angle B DC at the Center, is double of the Angle BAC at the Periphery; (by 20+El. lib. 3.) and so the half of every of them, viz. B DE=BAC, and BE is the Sine thereof. For the fame Reason, BF shall be the Sine of the Angle BCA, and AG the Sine of the Angle ABC. In a Right-angled Triangle we have BD=;B°C = 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 Obtuse-angled Triangle, let BI, CI, be drawn, and then the Angle L shall be the Complement of the Angle A to two Right Angles, (by 22 El. 3.) and so they fhall both have the fame Sine. But the Angle B DE, (whose Sine is B E) = Angle L. Therefore BE shall be the Sine of the Angle BAC. And so in every Triangle, the Halves of the Sides are the Sines of the opposite Angles; but it is manifest 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 Base, and the Tangent of one half their Difference, are proportional. ET there be a Triangle ABC, whose Legs are AB, BC, and Base AC. Produce AB to H, so that BH=BC, then shall AH be the Sum of the Legs; and if you make BI=BA, then IH will be the Difference of the Legs. Also the Angle HBC = Angles A + ACB, (by 32. El. 1.) and so EBCthe half thereof = half the Sum of the Angles A and ACB, and its Tangent (putting the Radius = EB) is EC. Again, let B D be drawn parallel to AC, and make HF=CD. Then fince HB=CB, we shall have (by 4. El. 1.) the Angle HBF=CBD=BCA. (by 29. El. 1.) Also the Angle H BD=Angle A; } whence F BD shall be the Difference of the Angles A and ACB; and EBD, whose Tangent is ED, half their Difference. Let IG be drawn thro' I parallel to AC or BD, and then (by 2. El. 6.) AB: BI :: CD:DG, but AB=BI; whence we hall have CD=DG, but CD=HF, and so HF=DG, and consequently, HG=DF, and HG=DF =DE; and because the Triangles AHC, IHG, are equiangular, it shall 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, shall be as E C 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 Base, the Sum of the Sides, the Difference of the Sides, and the Difference of the Segments of the Base, are proportional. ET DC be the Base of the Triangle BCD about the Center B, with the Radius B C, let a Circle be described. Produce DB to G, and from B Ict let fall BE perpendicular to the Base; then shall DG=DB+BC=Sum of the Sides, and DH= Difference of the Sides; and DE, CE, are the Segments of the Base whose. Difference is DF; because (by Cor. Prop. 38. El. 3.) the Rectangle under DC and DF, is equal to the Rectangle under DG, DM, it shall be (by 16. El. 6.) as DC: DG :: DH:DF. PROBLEM. The Sum and Difference of any two Quantities be ing given, to find the Quantities tbemselves. 7 3 F one half of the Sum be added to one half of the 1 Difference, the Aggregate dhall be equal to the 17 24. 41 z . 20 3 Ž 24 48 greater of the Quantities; and if from one half of the Sum be taken one half of the Difference, the Refidue shall be equal to the leffer of the Quantities. For let there be two Quantities A B, BC, and let there be taken AD=BC, then DB will be their Difference, and AC their Sum; which, bifected in E, gives A E or EC the half Sum, and D E or E B the half Difference. Hence AB= AE +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 also 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 first Proposition without a Canon. The Trigonometrical Solutions of a Right angled Triangle, may be as follow. Vid. Fig. A. 2 Given Sought Make as ment is the Angle C. AC: AB :: R:S, C Angle A. ROT, A:: AB : BC. S, C:R :: AB : A C. and the Side BC, 3 Angle A. and the Hypo- AC AB. Leg AB. AC, and Make as The Trigonometrical Solutions of Oblique angled Triangles. Vid. Fig. to Prop. 12. Given Sought FH The The S, C:SA :: AB:BC. Allo Angles Sides SC:S, B:: AB: AC: But A,B,C; BC and when two Angles are given, the I and the AC third is also given; whence the Side AB. Case wherein two Angles and a Side are given, to find the rest, falls into this Case. Given Make as 2 The two and C, Given Sought portions of the Sides may be first known. The AB:BC::S,C:S, A; which Sides Angles therefore may be found. When AB,BC, A and B. AB the Side opposite to C, the given Angle is longer than BC the Angle the Side opposite to the fought opposite Angle, the sought Angle is less 3 to one of than a right one. But when it them, is shorter, because the Sine of an Angle, and that of its Complement to two Right Angles, is the same, the Species of the AnIgle A must be first known, or the Solution will be ambiguous. The two The Vid. Fig. to Prop.13.BC+AB: Sides Angles A + A-C T. Whence is known the Difference of the Angles A and C, whose cent An Sum is given; and so (by the Anoles themselves will be given. Sides AB, Angles. cular be drawn from the Vertex BC, AC. to the Base, and find the Seg ments of the Base by Prop. 14. viz. Make as BC:AC FAB:: 5 AC AB:DC-DB. And so BD, DC, are given from this Analogy; and thence the Angles ABD, ADC, will be given by the Resolution of Right-angled Triangles. 2 2 4 interja Igle B. 7 |