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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.

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Example 1.

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

To find the Log. Sine :

Odd seconds

23:14:20", ans, to which is . . 9.596119.

3

propor. part to which is + 15

9.596134

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

To find the Log. Co-sine :

23:14:20", ans. to which is ...

propor. part to which is

9. 963253

3

Odd seconds

Given 'arch = 23:14:23" Corresponding log. co-sine 9.963250

Example 2. Required the log. sine, and co-sine, corresponding to 113:23:47"?

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

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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.

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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.

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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 loz. 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,

.

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

To find the Log. Tangent :

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Odd seconds

31:10:45", ans. to which is

9.781845 2 propor. part to which is + 10

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

To find the Log. Co-tangent:

.

31:10:45', ans. to which is

2 propor. part to which is

10. 219155

10

Odd seconds

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

Example 2. Required the log. tangent, and co-tangent, corresponding to 139:11:53"?

To find the Log. Tangent :

Odd seconds

139:11:50", ans. to which is . . 9.936142
3 propor, part to which is

13

Given arch = 139:11:53". Corres. log. tang. = .9.936129

. =

To find the Log. Co-tangent:

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Odd seconds

139:11:50", ans. to which is 10.063858

3 propor. part to which is + 13

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

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Example 3.

Required the arch corresponding to the given log. tang. 10. 155436 ?

Solution. The next less log. tangent in the Table is 10. 155423, corresponding to which is 55:2:25; the difference between this log. tangent and that given, is 13; answering to which, in the column of proportional parts abreast of the tabular log., is 3" ; which, being added to the abovefound arch, gives 55.2 28. for that required.

Example 4. Required the arch corresponding to the given log. co-tang. 9. 792048 ?

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Solution. The next greater log. co-tangent in the Table is 9.792057, corresponding to which is 58:13:15"; the difference between this log. cotangent and that given, is 9; 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 58:13:17" for that required.

Remark.

The arch corresponding to a given log. tangent may be found by means of a Table of log.sines, in the following manner; viz., .

Find the natural number corresponding to twice the given log. tangent, rejecting the index, to which add the radius, and find the common log. of the sum ; now, half this log. will be the log. secant, less radius, of the required arch; and which, being subtracted from the given log. tangent, will leave the log. sine corresponding to that arch.

Example. Let the given log. tangent be 10.084153; required the-arch corresponding thereto by a table of log. sines? Given log. tang. .084153 x 2 = . 168306, Nat. num. = 1.473349

to which add the radius = 1.000000

Sum =

2.473349, common log. of which is 0.393286; the half of this is 0. 196643,

secant, less radius of the required arch. Given log. tangent =

10.084153

the the

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Corresponding log. sine =

9.887510, answering to which is 50:31: ; and which, therefore, is the

;
required arch corresponding to the given log. tangent.

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