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Example 2.

Required the arch corresponding to the given log. co-secant 10.562114?

Solution. The next greater co-secant, in the Table, is 10.562129, corresponding to which is 15:45.25"; the difference between this log. Co-secant and that given, is 15; answering to which, in the column of proportional parts abreast of the tabular log., is 2"; which, being added to the above-found arch, gives 15:45:27" for that required.

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Remark.-The log. secant of any arch is expressed by the difference between twice the radius and the log. co-sine of that arch; and the cosecant of an arch, by the difference between twice the radius and the log. sine of such arch. Hence, to find the log. secant of 50:40..—The log. co-sine of 50:40? is 9,801974, which, being taken from twice the radius, viz., 20.000000, leaves 10. 198026 for the log. secant: from this, the manner of computing the co-secant will be obvious,

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TABLE XXXVI.

Logarithmic Sines. Of all the Logarithmic Tables in this work, this is, by far, the most generally useful, particularly in the sciences of Navigation and Nautical Astronomy; and, therefore, much pains have been taken in reducing it to that state of simplicity which appears to be best adapted to its direct application to the many other purposes for which it is intended, besides those above-mentioned.

In this Table, the log. sines of the two first degrees of the quadrant are given to every

second. The next eight degrees, víz,, from 2 to 10, have their corresponding log. sines to every fifth second, with proportional parts answering to the intermediate seconds in the adjacent right-hand column ; and because the log.sines increase rapidly in those degrees, two pages are allotted to a degree; every page being divided into two parts, and each part containing 15 minutes of a degree : so that no portion whatever of the proportional parts might be lost, and that the whole might have room to be fully inserted. In the following seventy degrees, that is, from 10 to 80, the log. sines are also given to every fifth second, with proportional parts corresponding to the intermediate seconds in the right-hand column of each page. In this part of the Table, each page contains a degree; and, for want of sufficient room, the indices are only prefixed to the logs. expressed in the first column.

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From 80 to 90 degrees, the log. sines are only given to every tenth second, because of the small increments by which the sines increase towards the end of the first quadrant; the proportional parts for the intermediate seconds are given in the right-hand column of each page, as in the preceding part of the Table.

The Table is so arranged, as to be extended to every second in the semicircle, or from 0 to 180 degrees; as thus : the arcs corresponding to the log. sines are given in regular succession at top, from 0 to 90 degrees, and then continued, at bottom, reckoning towards the left hand, from 90 to 180 degrees. The arcs corresponding to the co-sines are given at bottom of the Table, and ranged in numerical order towards the left hand, from 0 to 90 degrees, (according to the order of the sines between 90 and 180 degrees,) and then continued at top, from 90 degrees to the end of the semicircle, agreeably to the order of the sines in the first quadrant. This mode of arrangement does away with the necessity of finding the supplement of an arch when it exceeds 90 degrees, and possesses the peculiar advantage of enabling the navigator to take out the log. sine or co-sine of any arch, and conversely, at sight, as will appear obvious by the following Problems.

Note.---The log. co-sine of a given degree is found in the same page with the next less degree in the column marked 0” at top, it being the first number in that column; and the co-sine of a given degree and minute is found on the same line with the next less minute in the column marked 60' at bottom of the page.

PROBLEM I.

To find the Logarithmic Sine, and Co-sine of any given Arch, expressed

in Degrees, Minutes, and Seconds.

RULE.

If the given arch be comprised within the limits of the two first degrees of the quadrant, the Table will directly exhibit its corresponding log. sine or co-sine; but when it exceeds those limits, then find the log. sine, or cosine, in the angle of meeting made by the given degree and next less fifth or tenth second at top, and the minutes in the left-hand column; to which add the proportional part corresponding to the odd seconds in the righthand column abreast of the angle of meeting, if a sine be wanted, or a co-sine above 90 degrees; but subtract that part when a co-sine is required, or a sine above 90 degrees : and the sum, or difference, will be the log sine, or co-sine, answering to the given arch.

Example 1.
Required the log. sine, and co-sine, corresponding to 23:14:23". ?

To find the Log. Sine :

23:14:20", ans, to which is . . . 9.596119. Odd seconds 3 propor. part to which is +

15

Given arch = 23:14:23" Corresponding log. sine

9.596134

To find the Log. Co-sine :

Odd seconds

23:14:20", ans. to which is ... 9.963253
3 propor. part to which is

3

Given arch = 23:14:23" Corresponding log.co-sine 9. 963250

Example 2. Required the log. sine, and co-sine, corresponding to 113:2347". ?

To find the Log. Sine :

Odd seconds

113:23:45", ans. to which is .. 9.962740
2 propor. part to which is

2

Given arch = 113:23:47". Corresponding log. sine 9.962738

To find the Log. Co-sine :

113:23:45", ans. to which is 9,598879 Odd seconds

2 propor. part to which is + 10

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Given arch = 113:23:47". Corresponding log. co-s. 9.598889

PROBLEM II.

To find the Arch corresponding to a given Logarithmic Sine, or Co-sine,

RULE. If the given log. sine, or co-sine, be less than the sine of 2 degrees, viz., 8.542819, its corresponding arch will be found at first sight in the Table; but if it exceeds that number, find the arch answering to the next less sine, or next greater co-sine; the difference between which and that given, being found in the column of proportional parts abreast of the tabular log., will give a certain number of seconds, which, being added to the above-found arch, will give that required.

Note. Since the arcs corresponding to the sines between 90 and 180 degrees are found at the bottom of the Table, and those corresponding to the co-sines between the same limits at its top; if, therefore, it be required to find the arch above 90 degrees answering to a given log. sine, or co-sine, the first term is to be taken out as if it were a co-sine under 90 degrees, and the other term as if it were a sine under 90 degrees.

Example 1.

Required the arch corresponding to the given log. sine 9.437886 ?

Solution. The next less log. sine in the Table is 9. 437871, corresponding to which is 15:54'25"; the difference between this and that given, is 15; answering to' which, in the column of proportional parts, abreast of the tabular log., is 2" ; which, being added to the above-found arch, gives 15:54:27" for that required.

Example 2. Required the arch corresponding to the given log. co-sine 9.764579?

Solution. The next greater co-sine in the Table is 9.764588, corresponding to which is 54:26:25; the difference between this and that given, is 9; answering to which, in the column of proportional parts, abreast of the tabular log., is 3"; which, being added to the above-found arch, gives 54:26:28" for that required.

Remark. --The log. sines are deduced directly from the natural sines ; as thus :

Multiply the natural sine by 10000000000; find the common log. of the product, and it will be the log. sine.

Example 1. Required the log.sine of 39:30! ?

Solution. The natural sine of 39:30! is . 636078, which, being multiplied by 10000000000, gives 6360780000.000000, the common log. of which is 9. 803511; which, therefore, is the log. sine of 39 degrees and 30 minutes, as required.

Example 2. Required the log. co-sine of 68 degrees ?

Solution.-The natural co-sine of 68 degrees is .374607, which, being multiplied by 10000000000, gives 3746070000.000000, the common log. of which is 9,573575; which, therefore, is the log. co-sine of 68 degrees, as required.

TABLE XXXVII.

Logarithmic Tangents. This Table is arranged in a manner so very nearly similar to that of the log. sines, that it is not deemed necessary to enter into its description any farther than by observing, that it is computed to every second in the two first and two last degrees of the quadrant, or semicircle, and to every fifth second in the intermediate degrees. The log. tangent, or co-tangent, of a given arch, and conversely, is to be found by the rules for the log. sines in pages 94 and 95.

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Example 1.
Required the log, tangent, and co-tangent, corresponding to 31:10:47"?

To find the Log. Tangent :-
31:10:45., ans. to which is ... 9.781845
2 propor. part to which is +

10

Odd seconds

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Given arch = 31:10:47" Corresponding log. tang.=9.781855

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Given arch = 31:10:47". Corres. log. co-tang. = 10.218145

Example 2.
Required the log, tangent, and co-tangent, corresponding to13911:53"?

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To find the Log. Tangent :-
139:11:50", ans. to which is .: 9.936142
3 propor. part to which is

13

Odd seconds

Given arch = 139:11:53" Corres. log. tang. =

9.936129

To find the Log. Co-tangent:-
139:11:50", ans. to which is 10.063858

3
propor, part to which is +

13

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Given arch = 139:11:53" Corres. log. co-tang. = 10.063871

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