In any plane Triangle, it will be, as the Sum of any two Sides is to their Difference, so is the Tangent of Half the Sum of the two opposite Angles, to the Tangent of Half their Difference. Sides AC and AB: Join F, B and B, D, and draw DE parallel to FB, meeting BC in E. == C+ ABC (by Cor. 3. to Io. 1.) it is plain that ADB is equal to Half the Sum of the Angles op * Co Hence, in two Triangles ABC and AbC, having two Sides equal, each to each, it will be (by Equa lity), as Tang. Acrae : I ang. Abc —AC, Case our Proportion will become, Radius : Tang. A bC—ACb = AbC — 45°):: Tang. ABC+ACB, Tang ABC-ACB, which 2 gives the following Theorem, for finding the An- known. to the greater (AC), so is Radius to the Tangent of an Angle (Ab C, see Theor. 2.) And as Radius to the * This Theorem, tho' it requires two Proportions, is commonly us’d by Affronomers in determining the Elongation and Parallaxes s Two Sides The Angles and one Side the other Two Sides the other AC, AB and Angles C the included and ABC TwofidesACThe other Sought e Side BC The other|Let the Angle ABC be foundby AB, BC and Side AC Solution As Sine C. AB :: Sin. A : BC (by Theor. 3.) (by Theor. 3.) which added to C, and theSum subtracted from 180 ives the other Angle ABC. than the given Side BC, adjacent to it (except the Angle found is exactly a Right one): For then another Right-line Ba, equal to BA, may be drawn from B to a Point in the Base, somewhere between C and the Perpendicular BD, and therefore the Angle found by the Proportion, A B (aB): Sin. C :: BC: Sin. A (or of Ca B,) may, it is evident, be either the acute Angle A, or the obtuse one C a B (which is its Supplement), the Sines of both being exačtly the same. , Having laid down the Method of resolving the different Cases of plane Triangles, by a Table of Sines and Tangents; I shall here shew the Manner of constructing such a Table (as the Foundation upon which the whole Doctrine is grounded); in order to which, it will be requisite to premise the following Propositions. |