The tangent of an arc where there are seconds is found by the same method as has been employed for finding the sine. In finding the values of the cosine and cotangent, we must bear in mind that the cosines and cotangents decrease as the arcs increase, consequently the corrections found by aid of columns D must be subtracted instead of being added. Find the cosine of 23° 6′ 47′′. cos. 23° 6' = 9.963704 subtract 42.30% of 90 (D.) cos. 23° 6' 47" 9.963662 Find tangent and cotangent of 30° 30′ 30′′. tang. 30° 30′ = 9.770148 add 144,3% of 481 (D.) 30 100 tang. 30° 30′ 30′′= 9.770292 cotan. 30° 30' =10·229852 subtract 144% of 481 (D.) cotan. 30° 30′ 30′′—10-229708 §37. To find the arc expressed in degrees, minutes, and seconds, when its logarithmic sine, cosine, tangent, or cotangent is given. Seek for the given value in the column having the proper designation of sine, cosine, tangent, or cotangent, as the case may require, keeping in mind (§ 34) that the columns designated by sine and cosine at the top of the page are marked cosine and sine respectively at the bottom of the page. Also, those marked tangent and cotangent at the top are respectively cotangent and tangent at the bottom. If the exact value is found in the table, the arc will be accurately given in degrees and minutes. The degrees are to be taken from the top of the page, and the minutes from the lefthand column, when the proper precept of the column is found at the top. But when the precept is found at the bottom, the degrees are to be taken from the bottom of the page and the minutes from the right-hand column. If the number cannot be exactly found in the table, then find, as above, the degrees and minutes corresponding to the nearest 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. sin. 13° 24′ = 9.365016 = 349 difference, and 3490088339, the number of seconds. Hence, the arc 13° 24′ 39′′ corresponds with the logarithmic sine of 9-365365. 3. Find the arc whose tangent is 10-528017. 10.528017 = given logarithmic tangent. =10.527931 tang. 73° 29' §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 Clearing of fractions, we have sec. A × cos. A = 1; cosec. A × sin. A = 1. 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, A=20-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'. sin. 41° 41' 9.822830; cos. 41° 41'= 9.873223, 2. What is the secant and cosecant of 15° 15′ 15′′ ? 20. sin. 15° 15′ 15′′= 9.420123; cos. 15° 15′ 15′′ 20. 9.984423. cosec. 15° 15′ 15′′-10-579877; sec. 15° 15′ 15′′=10·015577. 1 § 39. Since we have (equation B, §8) tang. A= it follows that cotan. A' and tang. A × cotan. A =1, log. tang. A + log. cotan. A = 20. 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. |