less logarithm. Divide the difference of these logarithms, with two ciphers annexed, by the tabular difference taken from column D, the quotient will give seconds, which seconds are to be added to the degrees and minutes already found, in the case of a sine or tangent, but to be subtracted in the case of a cosine or cotangent. EXAMPLES. 1. Find the arc whose logarithmic sine is 9-365365. 9-365365 = given logarithmic sine. and 34900883 = 39, the number of seconds. Hence, the arc 13° 24' 39" corresponds with the logarithmic sine of 9-365365. 2. Find the arc whose cosine is 9·142857. 9.142857 = given logarithmic cosine. = 9.142655 cos. 82° 1' subtract 13" 202001500 = 13". 82° 0′ 47′′ = arc required. 3. Find the arc whose tangent is 10.528017. 10-528017 = given logarithmic tangent. tang. 73° 29' =10.527931 $38. The secants and cosecants are omitted in this table, since they are so readily found by the aid of the cosines and sines. By equation B, § 8, we have Taking the logarithms, and observing to add 10 to each logarithm, we have log. sec. A+log. cos. A=20; log. cosec. A+ log. sin. A=20. Hence, A20-log. cos. A, log. sec. log. cosec. A = 20-log. sin. A. From this we see that The logarithmic secant is found by subtracting the logarithmic cosine from 20; and the logarithmic cosecant is found by subtracting the logarithmic sine from 20. EXAMPLES. 1. Find the logarithmic secant and cosecant of 41° 41'. 20. sin. 41° 41' 20. 9.822830; cos. 41° 41'= 9.873223. cosec. 41° 41'=10-177170; sec. 41° 41'=10∙126777. 2. What is the secant and cosecant of 15° 15′ 15′′? = 20. = 20. sin. 15° 15′ 15′′ 9-420123; cos. 15° 15′ 15′′ 9.984423. cosec. 15° 15′ 15"=10-579877; sec. 15° 15′ 15′′=10-015577. § 39. Since we have (equation B, §8) tang. A = it follows that and tang. A x cotan. A = 1, log. tang. A + log. cotan. A = 20. 1 cotan. A' Hence, the logarithmic cotangent may be found by subtracting the logarithmic tangent from 20; and, conversely, the logarithmic tangent may be found by subtracting the logarithmic cotangent from 20. EXAMPLES. 1. The logarithmic tangent of an arc is 9-545454; what is the logarithmic cotangent of the same arc? Ans. 10-454546. 2. The logarithmic cotangent is 9-333444; what is that of the tangent of the same angle? Ans. 10.666556. § 40. In any plane triangle there are three sides and three angles, making in all six parts to be considered. Any three of these six parts, provided one at least is a side, being given or known, the other three parts can be found. Thus, in Book I., Geometry, we have seen that when two triangles had three parts of the one respectively equal to the three corresponding parts of the other, provided the three parts compared were not all angles, the two triangles were identical, or equal in all respects. The case in which three angles of a triangle are respectively equal to three angles of a second triangle, does not lead of necessity to an equality of these two triangles, but simply to their similarity. Hence, when in a triangle three parts, provided one at least is a side, are given, the remaining parts can be found. §41. In the case of right triangles, one angle, that is, the right angle, is always given; consequently two parts in addition to the right angle must be given, one of which must be a side, in order that we may find the remaining parts. The geometrical or graphic method of solving a triangle when a sufficient number of parts are known, has already been exhibited in the Problems at the end of Book Second of Geometry. The method of calculating the numerical values of these parts which we now proceed to explain-belongs to Trigonometry. As an example, suppose the hypotenuse to be 125, and the angle at A to be 37° 30'. Required the other parts. h = 125; A=37° 30'. We find = 90°-37° 30' = 52° 30'. BY NATURAL NUMBERS. (TABLE III.) sin. 37° 30′ = 0.60876; cos. 37° 30′ = 0·79335 log. sin. 37° 30′ = 9.784447; log. cos. 37° 30′ =9-899467 log.p=1.881357 Hence, p=76-095, and 699-169. log. b=1.996377 NOTE. In adding the log. sin. and log. cos. respectively to the log. 125, we rejected 10 from the sum to correct for the 10 which had been added to all logarithms of the sines, cosines, &c. (§ 34). For example, suppose the perpendicular to be 75, and the angle at A to be 39° 15'. Required the other parts. p=75; A39° 15'. We find C = 90°-39° 15′ = 50° 45'. |