prefixed from the column headed 0%; to this add the propor. tional part for the odd seconds at the bottom of the page. Required the logarithmic sine of 24° 27' 34". The logarithmic sine of 24° 27' 30" is 9.617033 18 Logarithmic sine of 24° 27' 34" is 9.617051. Required the logarithmic tangent of 73° 35' 43". The logarithmic tangent 73° 35' 40" is 10.531031 23 Logarithmic tangent of 73° 35' 43'' is 10.531054. When a cosine is required, the degrees and seconds must be sought at the bottom of the page, and the minutes on the right, and the correction for the odd seconds must be subtracted from the number in the table. Required the logarithmic cosine of 59° 33' 47". The logarithmic cosine of 59° 33' 40" is 9.704682 25 Logarithmic cosine of 59° 33' 47" is 9.704657. So, also, the logarithmic cotangent of 37° 27' 14" is found to be 10.115744. It will be observed that for the cosines and cotangents, the seconds are numbered from 10" to 60', so that if it is required to find the cosine of 25° 25' 0" we must look for 25° 24' 60"; and so, also, for the cotangents. (37.) The proportional parts given at the bottom of each page correspond to the degrees at the top of the page, increased by 30', and are not strictly applicable to any other number of minutes; nevertheless, the differences of the sines change so slowly, except near the commencement of the quadrant, that the error resulting from using these numbers for every part of the page will seldom exceed a unit in the sixth decimal place. For the first two degrees, the differences change so rapidly that the proportional part for 1" is given for each minute in the right-hand column of the page. The correction for any number of seconds less than ten will be found by multiplying the proportional part for 1" by the given num. ber of seconds. Required the logarithmic sine of 1° 17' 33". The logarithmic sine of 1° 17' 30" is 8.352991. The correction for 3'' is found by multiplying 93.4 by 3, which gives 280. Adding this to the above tabular number, we obtain for the sine of 1° 17' 33', 8.353271. A similar method may be employed for several of the first degrees of the quadrant, if the proportional parts at the bottom of the page are not thought sufficiently precise. This correction may, however, be obtained pretty nearly by inspection, from comparing the proportional parts for two successive degrees. Thus, on page 26, the correction for 1", corresponding to the sine of 2° 30', is 48; the correction for 1", corrosponding to the sine of 3° 30', is 34. Hence the correction for 1", corresponding to the sine of 3° 0', must be about 41; and, in the same manner, we may proceed for any other part of the table. (38.) Near the close of the quadrant, the tangents vary so rapidly that the same arrangement of the table is adopted as for the commencement of the quadrant. For the last, as well as the first two degrees of the quadrant, the proportional part to 1" is given for each minute separately. These proportional parts are computed for the minutes placed opposite to them, increased by 30", and are not strictly applicable to any other number of seconds; nevertheless, the differences for the most part change so slowly, that the error resulting from using these numbers for every part of the same horizontal line is quite small. When great accuracy is required, the table on page 114 may be employed for arcs near the limits of the quadrant. This table furnishes the differences between the logarithmic sines and the logarithms of the aros expressed in seconds. Thus the logarithmic sine of 0° 5', from page 22, is 7.162696 the logarithm of 300" (=5') is 2.477121 the difference is 4.685575. This is the number found on page 114, under the heading log. sine A-log. A", opposite to 5 min.; and, in a similar manner, the other numbers in the same column are obtained. These numbers vary quite slowly for two degrees; and hence, to find the logarithmic sine of an arc less than two degrees, we have but to add the logarithm of the arc expressed in seconds to the Tabular number from page 114, 4.685575 Logarithmic sine of 0° 7' 22" is 7.330997. 4.685584 The logarithm of 1656'' is 3.219060 Logarithmic tangent of 0° 27' 36" is 7.904644. The column headed log.cot. A+log. A", is found by adding the logarithmic cotangent to the logarithm of the arc expressed in seconds. Hence, to find the logarithmic cotangent of an aro less than two degrees, we must subtract from the tabular number the logarithm of the arc in seconds. Required the logarithmic cotangent of 0° 27' 36". Tabular number from page 114, 15.314416 3.219060 Logarithmic cotangent of 0° 27' 36" is 12.095356. The same method will, of course, furnish cosines and cotan gents of arcs near 90°. (39.) The secants and cosecants are omitted in this table, since they are easily derived from the cosines and sines. We R? have found, Art. 28, secant = ; or, taking the logarithms, cosine log. secant =2. log. R-log. cosine =20-log. cosine. R? Also, cosecant sine' or log. cosecant =20-log. sine. That is, 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. Thus we have found the logarithmic sine of 24° 27' 34" to be 9.617051. Hence the logarithmic cosecant of 24° 27' 34" is 10.382949 The logarithmic cosine of 54° 12' 40" is 9.767008. Hence the logarithmic secant of 54° 12' 40" is 10.232992. (40.) To find the arc corresponding to a given logarithmic sine or tangent. If the given number is found exactly in the table, the corresponding degrees and seconds will be found at the top of the page, and the minutes on the left. But when the given number is not found exactly in the table, look for the sine or tangent which is next less than the proposed one, and take out the corresponding degrees, minutes, and seconds. Find, also, the difference between this tabular number and the number proposed, and corresponding to this difference, at the bottom of the page, will be found a certain number of seconds which is to be added to the arc before found. Required the arc corresponding to the logarithmic sine 9.750000. The next less sine in the table is 9.749987. The difference between its sine and the one proposed is 13, corresponding to which, at the bottom of the page, we find 4" nearly. Hence the required arc is 34° 13' 4". In the same manner, we find the arc corresponding to logarithmic tangent 10.250000 to be 60° 38' 57". When the arc falls within the first two degrees of the quadrant, the odd seconds may be found by dividing the difference between the tabular number and the one proposed, by the proportional part for 1". We thus find the arc corresponding to logarithmic sine 8.400000 to be 1° 26' 22" nearly. We may employ the same method for the last two degrees of the quadrant when a tangent is given; but near the limits of the quadrant it is better to employ the auxiliary table on page 114. The tabular number on page 114 is equal to log. sin. A-log. A". Hence log. sin. A- tabular number=log. A"; that is, if we subtract the corresponding tabular number on page 114, from the given logarithmic sine, the remainder will be the logarithm of the arc expressed in seconds. Required the arc corresponding to logarithmic sine 7.000000. We see, from page 22, that the arc must be nearly 3'; the corresponding tabular number on page 114 is 4.685575 The difference is 2.314425, which is the logarithm of 206.265. Hence the required arc is 3' 26."265. We see from page 22, that the arc is about 34'. The corresponding tabular number from page 114 is 4.685568, which, subtracted from 8.000000, leaves 3.314432, which is the logarithm of 2062."68. Hence the required arc is 34' 22."68. In the same manner, we find the arc corresponding to logarithmic tangent 8.184608 to be 0° 52' 35". SOLUTIONS OF RIGHT-ANGLED TRIANGLES. THEOREM I. (41.) In any right-angled triangle, radius is to the hypothenuse as the sine of either acute angle is to the opposite side, or the cosine of either acute angle to the adjacent side. Let the triangle CAB be right angled B at A, then will R: CB :: sin. C: BA :: cos. C: CA. Α. radius equal to the radius of the tables, c describe the arc DE, and on AC let fall the perpendicular EF. Then EF will be the sine, and CF the cosine of the angle C. Because the triangles CAB, CFE are similar, we have CE : CB :: EF: BA, R: CB :: sin. C : BA. CE : CB :: CF : CA, FD or or THEOREM II. (42.) In any right-angled triangle, radius is to either side as the tangent of the adjacent acute angle is to the opposite side, or the secant of the same angle to the hypothenuse. Let the triangle CAB be right angled at A, then will R:CA :: tang. C : AB :: sec. C:CB. C A radius equal to the radius of the tables, B F D |