Note, The logarithms found according to this method, in numbers between 10000 and 20000, are true to 8 or 9. places of figures: those of numa bers between 20000 and 50000 érr only in the gek or 10 place; and those of above 50000 are true to 10 places, at least. B Having explained the manner of constructing a table of logarithms, and that by various methods, I now come to shew the use of such a table in the business of trigonometry. First, in the rightangled plane triangle ABC, let there be given the hypothenuse AC = 17910 feet, and the angle A = 35° 20'; to find the perpendicular BC and the base AB. A Here, because radius : line 35° 20' :: 17910:BC (by Theor. 2.p.6.) we have BC –sine 35° 20'x17910 radius therefore, because the addition and fubtraction of logarithms answers to the multiplication and division of the natural numbers (fee. p. 38, 39.) we have log. BC=log. fine 35° 20' + log. 17910 log. radius. But, by the tables of artificial, or logarithmic, fines *, the log. sine of 35° 20' will appear to be 9,7621775; to which add 4,2530956, the log. of 17910, and from the sum (14,0152731) take 10, the log. of radius, and there results 4,0152731 = the log. of BC; which, in the tables, answers to 10358, the length of BC required. A table of artificial fines is nothing more than a table of the Ingarithms of the numbers expressing the natural fines, to the radius 10000000000; wbofe logarithm-is 10. E 2 Again, 1 Again, for AB, it will be, as radius : fine of C (54° 40'): ; AC (17910). : AB (by Theorem 2.-). Whence, by adding the logarithms of the second and third terms together and subtracting that of the first (as above), we have AB = 14611. See the operation, Log, sine C (54° 40').9,9115844 14611 - 4,1646800 in Which 14° added to 66'; the half sum of the angles-Card B, gives the greater C = 80°; and subtracted theretrom, leaves the leffer B = 52°. 4 Lastly, 1 Lastly, in the rightangled spherical triangle ABC, let there be given the hypothenuse AC 60°, and the angle A = 23° 29'; to find the base B В. and perpendicular. Then (by Theor. 1. p. 25.) the operation will be as follows: Having exhibited the manner of resolving all the common cases of plane and spherical triangles, both by logarithms and otherwise ; I shall here subjoin a few propositions for the solution of the more difficult cases which sometimes occur ; when, instead of the sides and angles themselves, their fums, or differences, &c. are given. PROPOSITION I. The fine, co-fine, or versed fire of an arch being given; to find the fine and co-line, &c. of balf that arch. ex D From the two tremes of the diameter let the radius CQ bifect с F A AE, perpendicularly, in D (Vid. 1. 3.); then will AD be the fine, and CD the co-fine, of the angle ACD, or (ACE. But 4AD? = AE? (by Cor. 1. to 6. 2.) АВх AF (by Cor. to 19. 4.) =2ACxAF; whence AD* ={AC ~ AF : also 4C!? = BE = AB x BF = 2AC * BF; whence -CD = JAC x BF. From which it appears, that the square of the fine of half any arch, or angle, is equal to a rectangle under half the radius and the versed fine of the whole; and that the square of its co-fine is equal to a re&tengle under half the radius and the verjed fine of the supplement of the whole arch, or angle. PROP. II. The fines and co-fines of two arches being given, to find the fines, and the co-fines, of the fum and difference of these arches. D Let AC and CD (=BC) be the two proposed arches; let CF and OF be BA Н. the fine and cofine of the greater AC, and let mD AE Fn G 0 (Bm) and Om, be thofe of the lefser CD (or BC): moreover, let DG and OG be the fine and co-fine of the sum AD; and BE and OE, those of the difference AB. Draw min parallel to CF, meeting AO in n; also draw mv whence and BH parallel to AO, meeting GD in v and H: OC:Om::CF:mn mn X OC-Omx CF. mu x OC=DmX CF. Now, by adding the two first of these equations together, we have mn + Dux OC (DG XOC) = Om X CF + Dm x OF; whence DG is known. Moreover, by taking the latter from the former, we get mn - Dvx OC (BE OC)= Om CF Dm' x OF; whence BE is known, In like manner, by adding the third and fourth equations together, we have On + mu x OC (OEX OC) = Om x OF + Dm x CF; and, by subtracting the latter from the former, we have On-my x OC (OG XOC)=Om x OF-Dm x CF; whence QE and OG are also known. 2. E. I. -COROLLARY I. Hence, if the lines of two arches bc denoted by S and s; their co-fines by C and c; and radius by R; then will the fine of their fum = Sc + sC R the line of their difference = SSC R the co-fine of their fum = Çc - $s the co-fine of their difference = Cc+S4: R EA COROL |