1 In any right-angled plane triangle ABC, it will be at the hypothenufe is to the perpendicular, fo is the radius of the table) to the fine of the angle at the bafe. For, let AE or AF be the radius to which the table of fines, &c. is adapted, and ED the fine of the angle A or arch EF (Vid. Def. 3. and 6.); then, BD F because of the fimi lar triangles ACB and AED, it will be AC: BC:: AE: ED (by 14.4.) 2. E. D.... Thus, if AC,75, and BC,45; then it will be, 75,45 (radius): the fine of A6; which, in the table, anfwers to 36° 52', the measure, or value of A.. THEOREM II. In any right-angled plane triangle ABC, it will be, as the bafe AB is to the perpendicular BC, fo is the radius of the table) to the tangent of the angle at the bafe. A For, let AE or AF be the radius of the table, or canon (fee the preceding figure), and FG the tangent of the angle A, or arch: EF (Vid. Def. 3. and 9.); then, by reafon of the fimilarity of the triangles ABC, AFG, it will be, AB BC:: AF: FG. Q E. D. you is made Note, In the quotations where meet with two numbers (as 14 4) without any mention of Prop Theor. &c. referenc to the Elements of Geometry published by the fame author; to which this little tract is defigned as an Appendix. Thus Thus let AB =,8, and BC=,5; then we shall have,8,5 I (radius) tangent A =,625; whence A itself is found, by the canon, to be 32° 00'. THEOREM III. In every plane triangle ABC, it will be, as any one fide is to the fine of its oppofite angle, fo is any other fide to the fine of its oppofite angle. Now the triangles C CBD,' CFE being fimilar, we have CB: BD (fin. A):: CF (AB): FE (fin. C). Q. E. D. THEOREM IV. As the bafe of any plane triangle ABC, is to the fum of the two fides, fo is the difference of the fides to twice the distance DE of the perpendicular from the middle of the bafe. d THEOREM V. In any plane triangle, it will be, as the fum of any two fides is to their difference, fo is the tangent of half the fum of the two oppofite angles, to the tangent of half their difference. For, let ABC be the triangle, and AB and AC the two propofed fides; and from the center A, with the radius AB, let a circle be defcribed, in terfecting CA produced, in D and F; fo that CF may ex prefs the fum, and CD the difference, of the fides AC and AB join F, B and B, D, and draw DE parallel to FB, meeting BC in E. Then, because 2ADB = ADB + ABD (by 12. 1.) = C + ABC (by 9. 1.) it is plain that ADB is equal to half the fum of the angles oppofite to the fides propofed. Moreover, fince ABC ABD (ADB) + DBC, and C ADB DBC (by 9. 1.) it is plain that ABC - C is = 2DBC; or that DBC is equal to half the difference of the fame angles. Now, because of the parallel lines BF and ED, it will be CF: CD:: BF: DE, but BF and DE, because DBF and BDE are right-angles (by 13.3. and 7. 1.) will be tangents of the forefaid angles FDB (ADB) and DBE (DBC) to the radius BD, 2. E. D. COROL COROLLARY. Hence, in two triangles ABC and A/C, having two fides equal, each to each, it will be (by equa АЪС+АСЬ АБС-АСЬ 2 lity), as tang. : tang. 2 2 if CAb be fuppofed a right-angle, then will AbC+ AC alfo a right angle (by Cor. 3. to 10. 1.) and the tangent of 1 gives the following Theorem, for finding the angles oppofite to any two propofed fides; the in-. cluded angle, and the fides themfelves, being known. As the leffer of the propofed fides (Ab or AB) is to the greater (AC), fo is radius to the tangent of an angle (AbC, fee Theor. 2.) And as radius to the tangent of the excess of this angle above 45°, fo is the tangent of half the fum of the required angles to the tangent of half their difference. This Theorem, though it requires tavo proportions, is commonly ufed by Aftronomers in determining the elongation and parallaxes of the planets (being beft adapted to logarithms); for which reason is is here given. 1 The The folution of the cafes of right-angled plane triangles. The hyp. One leg IAC and BC the angles 2 AC and one legBC The hyp. The other 3 AC and leg AB one legBC Solution. As radius is to the fine of A, fo is the hyp. AC to the leg BC (by Theor. I.) As AC: BC :: radius : fin. A (Theor. I.) whofe complement is the angle C. Let the angles be found, by Cafe 2. and then the required leg AB. by Cafe 1. The an- The hyp. As fine A: radius : : the 4 gles and one legBC AC leg BC: to the hyp. AC (Theor. I.) The an- The other As fine A: BC:: fine C gles and leg AB 5 one leg BC The two The an 6 legs AB gles and BC AB (by Theor. III.) Or, as radius tang. C:: BC AB (by Theor. II.) As AB: BC :: radius: tang. A (by Theorem I.) whofe complement is the angle C.. The two The hyp. Let the angles be found, 7 legs AB AC by Cafe 6. and then the and BC hyp. AC, by Cafe 4. |