,00000000134 (See Rule 2.) which multiplied by equal 4,255272505 - log. C 49,5, and the product added to log. B log. A 100 gives ,00002426107 for the excess of the logarithm of A+ above that of A (by Rule 3.) From whence the work, being continued according to Rule 4 and 5, will ftand as follows: Note, The logarithms found according to this method, in numbers between 10000 and 20000, are true to 8 or 9 places of figures: thofe of numbers between 20000 and 50000 err only in the 9th or 10th 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 fhew the use of fuch a table in the bufinefs of trigonometry. First, in the rightangled plane triangle ABC, let there be given the hypothenufe AC= 17910 feet, and the angle A35° 20′; to find the perpendicular BC and the bafe AB. Here, because radius: fine 35° 20′:: 17910: BC (by Theor. 2. p. 6.) we have BC-fine 35° 20′ x 17910 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. 17910log. radius. But, by the tables of artificial, or logarithmic, fines*, the log fine of 35° 20' will appear to be 9,7621775; to which add 4,2530956, the log. of 17910, and from the fum (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 logarithms of the numbers expreffing the natural fines, to the radius 10000000000; whofe logarithm is 10. 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 fecond and third terms together and fubtracting that of the first (as above), we have AB 14611. See the operation. Log. fine C (54° 40′) 9,9115844 Which 14° added to 66°, the half fum of the angles C and B, gives the greater C = 80°; and fubtracted therefrom, leaves the leffer B = 52°. Lastly, Laftly, in the rightangled fpherical triangle ABC, let there be given the hypothenuse AC = 60°, and the angle A = 23° 29'; to find the bafe and perpendicular. Then C B (by Theor. 1. p. 25.) the operation will be as follows: Having exhibited the manner of refolving all the common cafes of plane and fpherical triangles, both by logarithms and otherwife; I fhall here fubjoin a few propofitions for the folution of the more difficult cafes which fometimes occur; when, inftead of the fides and angles themfelves, their fums, or differences, &c. are given. PROPOSITION I. The fine, co-fine, or verfed fine of an arch being given, to find the fine and co-fine, &c. of half that arch. will AD be the fine, and angle ACD, or ACE. From the two extremes of the diameter AB, let the chords AE and BE be drawn, and let the radius CQ bifect AE, perpendicularly, in D (Vid. 1. 3.); then CD the co-fine, of the But 4AD2 = AE2 (by Cor. 1. to 6. 2.) = AB × AF (by Cor. to 19.4.) = 2AÇ × AF; whence AD2 =AC × AF: alfo 4CD2 BE2 = AB × BF = 2AC × BF; whence CD2 = AC × 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 verfed fine of the whole; and that the Square of its co-fine is equal to a rectangle under half the radius and the verfed fine of the fupplement 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. Let AC and CD (=BC) be the two propofed arches ; let CF and OF be the fine and cofine of the greater AC, and let mD (Bm) and Om, be those of the leffer CD (or BC): moreover, let DG and OG be the fine and co-fine of the fum AD; and BE and OE, thofe of the difference AB. Draw mn parallel to CF, meeting AO in n; alfo draw mu and |