Then the loga A being 4,252853031 B rithm of equal 4,255272505 с 4,257678575 we shall have 2 log: B log. A - log. C to 10000 ,00000000134 (See Rule 2.) which multiplied by 49,5, and the product added to log. B – log. A gives ,00002426107 for the excess of the logarithm of A + 1 above that of A (by Rule 3.) From whence the work, being continued according to Rule 4 and 5, will stand as follows: 25705 3d rem. 25571 4th rem. ,000024 26107 excess. 4,25 2853031 log. 17900 134 2426107 excess. 25973 It rem. 287729207 log. 17901 134 2425973 25839 2° rem. 290155180 log. 17902 134 2425839 292581019 log. 17903 134 2425705 295006724 log. 17904 134 2425571 25437 5th rem. 297432295 log. 17905 134 2425437 299857732 log. 17906 134 2425303 25169 7th rem. 302283035 log. 17907 134 2425169 25035 gth rem. 304708204 log. 17908 134 2425035 24901 9th rem. 307133239 log. 17909 134 2424901 24767 Toth rem. 309558140 log. 17910 &c. &c. &c. &c. Notes 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 numbers between 20000 and 50000 err only in the 9th or 10 place; and those of above 50000 are true to 10 places, at least. 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 А B Here, because radius : sine 35° 20':: 17910:BC (by Theor. 2. p. 6.) we have BC= fine 35° 20' x 17910 radius therefore, because the addition and subtraction of logarithms answers to the multiplication and division of the natural numbers (see p. 38, 39.) we have log. BC= log. fine 35° 20' + log. 17910 — log. radius. But, by the tables of artificial, or logarithmic, fines *, che 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 chere 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 logarithms of the numbers expressing the natural fines, to the-radius 10000000000; whose logarithm is 10. Again, E 2 Again, for AB, it will be, as radius: sine 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. radius · 10, Which 14° added to 66°, the half sum of the angles C and B, gives the greater C = 80°; and fubtracted therefrom, leaves the lefser B = 52o. Lastly, с Lastly, in the rightangled spherical triangle ABC, let there be given the hypothenuse AC = 60°, and the angle A = 23° 29'; to find the bafe B. and perpendicular. Then (by Tbeor. 1.p.25.) the operation will be as follows: Having exhibited the manner of refolving all the common cases of plane and spherical triangles, both by logarithms and otherwise ; I shall here subjoin a few propositions for the folution of the more difficult cases which sometimes occur ; when, instead of the sides and angles themselves, their sums, or differences, &c. are given. PROPOSITION I. The fine, co-fine, or versed fine of an arch being given, to find the fine and co-fine, &c. of half that arch. D A From the two ex- tremes of the diameter let the radius CQ bisect D (Vid. 1. 3.); then will AD be the fine, and CD the co-fine, of the angle ACD, or ACE. But 4AD2 = AE? (by Cor. 1. to 6. 2.) = AB x AF (by Cor. to 19. 4.) = 2AÇ AF; whence AD2 = ACX AF : alfo 4CD? = BE = AB x BF = 2 AC ~ BF; whence CD2 = {AC BF. From which it appears, that the square of the fine of balf 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-line is equal to a reEtangle under half the radius and the versed 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 sum and difference of these arches. D Let AC and CD (=BC) be the two proposed arches; let CF and OF be the fine and co- AC, and let mD (Bm) and Om, be those of the lesser 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 mn parallel to CF, meeting AO in n; also draw mv and |