If the given logarithm was 2.32858; I find that 32838 stands in the column marked 0 at the top or bottom, directly opposite to 213 which is the number sought, because the index being 2, the number must consist of 8 places of figures.

If the number corresponding to the logarithm 2.57345 was required; I look in the column 0, and find in it, against the number 374, the logarithm 57287, and guiding my eye along that line, I find the given logarithm 57345 in the column marked 5; therefore the mixed number sought is 3745, and since the index is 2, the integer part must consist of 3 places, therefore the number sought is 374,5. If the index had been 1, the number would have been 37,45; and if the index had been 0, the number would have been 3,745. If the index had been 8 corresponding to a number less than unity, the answer would have been 0,03745, &c.

Again, if the number corresponding to the logarithm 5,57811 was required; I look in the column 0, and find in it against 378, and under 5, the logarithm 57807, the difference between this and the next greater logarithm 57818 being 11, and the difference between 57807 and the given number 57811 being 4, to this 4 I affix two ciphers, which make 400, and divide it by 11, the quotient is 36 nearly; this number connected with the former four figures make 378536, which is the number required, since the index being 5 the number must consist of six places of figures.

MULTIPLICATION BY LOGARITHMS.

RULE. Add the logarithms of the two numbers to be multiplied, and the sum will be the logarithm of their product.

EXAMPLE I.

EXAMPLE II.

In the last example the sum of the two indices is 16, but since 10 was borrowed in each number, I have neglected 10 in the sum, and the remainder 6 being less than the other 10, is evidently the index of the logarithm of a fraction less than unity.

DIVISION BY LOGARITHMS.

RULE. From the logarithm of the dividend subtract the logarithm of the divisor, the remainder will be the logarithm of the quotient.

Quotient 3,26 log.

8.47712

0,51322 Quotient 0,03 log.

In Example III. both the divisor and dividend are fractions less than unity, and the divisor is the least, consequently the quotient is greater than unity. In Example IV. both fractions are less than unity, and since the divisor is the greatest, its logarithm is greater than that of the dividend; for that reason it was necessary to borrow 10 in the index previous to making the subtraction, hence the quotient is less than unity.

INVOLUTION BY LOGARITHMS.

RULE. Multiply the logarithm of the number given, by the index of the power to which the quantity is to be raised, the product will be the logarithm of the power sought. But in raising the powers of any decimal fraction it must be observed, that the first significant figure of the power must be put as many places below the place of units as the index of its logarithm wants of 10 multiplied by the index of the power.

Answer 40,96 log.

In the last example the index 28 wants 2 of 30 (the product of 10 by the power 3) therefore the first significant figure of the answer, viz. 1, is placed two figures distant from the place of units.

EVOLUTION BY LOGARITHMS.

RULE. Divide the logarithm of the number by the index of the power, the quotient will be the logarithm of the root sought. But if the power whose root is to be extracted is a decimal fraction less than unity, prefix to the index of its logarithm a figure less by one than the index of the power,' and divide the whole by the index of the power, the quotient will be the logarithm of the root sought.

TO WORK THE RULE OF THREE BY LOGARITHMS. When three numbers are given to find a fourth proportional in arithmetic, we make a statement and say, as the first number is to the second so is the third to the fourth; and by multiplying the second and third together, and dividing the product by the first, we obtain the fourth number sought. To obtain the same result by logarithms, we must add the logarithms of the second and third numbers together, and from the sum subtract the logarithm of the first number, the remainder will be the logarithm of the sought fourth number.

EXAMPLE I.

EXAMPLE II.

If 6 yards of cloth cost 5 dollars, what If a ship sails 20 miles in 7 hours, how much will she sail in 21 hours at the same

will 20 yards cost?

As 6

Is to 5

In this rule it is supposed that 10 was borrowed in finding the index of the decimal according to the rule page 30.

To 60

log. 1.77815

The answer is

60 miles.

To calculate COMPOUND INTEREST by Logarithms.

To 100 dollars add its interest for one year; find the logarithm of this sum and reject 2 in the index, then multiply it by the number of years and parts of a year, for which the interest is to be calculated; to the product add the logarithm of the sum put at interest; the sum of these two logarithms will be the logarithm of the amount of the given sum for the given time.

EXAMPLE.

Required the amount of the principal and interest of 355 dollars, let at 6 per cent. compound interest for 7 years?

Adding 6 to 100 gives 106, whose logarithm, rejecting 2 in the index, is

0.02531

Therefore the amount of principal and interest is 533 dollars and 83 cents. To find the Logarithm-sine, Tangent, or Secant, corresponding to any number of Degrees and Minutes, by Table XXVII.

The given number of degrees must be found at the bottom of the page when between 450 and 1350, otherwise at the top, the minutes being found in the column marked M, which stands on the side of the page on which the degrees are marked; thus, if the degrees are less than 45, the minutes are to be found in the left hand column, &c. and it must be noted that if the degrees are found at the top, the names of hour, sine, co-sine, tangent, &c. must also be found at the top; and if the degrees are found at the bottom, the names sine, co-sine, &c. must also be found at the bottom. Then opposite to the number of the minutes will be found the log-sine, log-secant, &c. in the column marked sine, secant, &c. respectively.

EXAMPLE I.

Required the log. sine of 28° 37' ?

EXAMPLE II.

Required the log. secant of 126° 20′ ?

Find 28 at the top of the page, directly Find 126 at the bottom of the page, dibelow which, in the left hand column, find rectly above which, in the left hand column, 37'; against which in the column marked find 20'; against which, in the column marksine is 9.68029, the log. sine of the given ed secant, is 10.22732 required. number of degrees; and in the same man

ner the tangents, &c. are found.

To find the Logarithm-sine, Co-sine, &c. for Degrees, Minutes, and Seconds, by Table XXVII.

Find the numbers corresponding to the even minutes next above and below the given degrees and minutes, and take their difference; then say, as 60′′ is to the number of seconds in the proposed number, so is that difference to a correction to be applied to the number corresponding to the least number of degrees and minutes; additive if it is the least of the two numbers taken from the table, otherwise subtractive.

Then, as 60": 48′′ :: 29: 23, which, added Then as 60′′ : 16′′ :: 46: 12, which, subtracted to the number corresponding to 24° 16', from the number corresponding to 105- 20′, gives 9.61405 the log. sine of 24° 16′ 43′′. gives 10.57756, the log. sec. of 105° 20′ 16′′.

If the given seconds be §, 1, 4, 1, or, or any other even parts of a minute, the like parts may be taken of the difference of the logarithms, and added or subtracted as above, which may be frequently done by inspection.

G

To find the Degrees, Minutes, and Seconds, corresponding to any given Logarithm-sine, Co-sine, &c. by Table XXVII.

Find the two nearest numbers to the given logarithm-sine, co-sine, &c. in the column marked sine, co-sine, &c. respectively, one being greater and the other less, and take their difference; take also the difference between the given logarithm and the logarithm corresponding to the least number of degrees and minutes: then say, as the first found difference is to the second found difference, so is Go" to a number of seconds to be annexed to the smallest number of degrees and minutes before found.

EXAMPLE I.

Find the degrees, minutes, and seconds (less than 90) corresponding to the log. sine
9.61405.
Next less log. 24° 16'
Greater

24 17

9.61382 Log. of least numb. 24° 16' is
9.61411 Given log.

9.61382

9.61405

Then say, as 29: 23 :: 60"; 48′′ which annexed to 24° 16' give 24° 16' 48" answering to log. sine 9.61405. Subtracting 24° 16' 48" from 180°, there remains 155° 43′ 12′′, the log. sine of which is also 9.61405.

EXAMPLE II.

Find the degrees, minutes, and seconds (above 90°) corresponding to the log. secant 10.56703.

Log. of the least numb. 105° 43′ 10.56722 Given log. 10.56703

Then as 45 is to 19, so is 60′′ to 25", which annexed to 105° 43', gives 105° 43′ 25′′, the degrees, minutes, and seconds required.

To find the Arithmetical Complement of any Logarithm.

The arithmetical complement of any logarithm, is what it wants of 10,00000 and is used to avoid subtraction. For when working any proportion by logarithms, you may add the arithmetical complement of the logarithm of the first term, instead of subtracting the logarithm itself, observing to neglect 10 in the index of the sum of the logarithms. The arithmetical complement of any logarithm is thus found:-Begin at the index, and write down what each figure wants of 9, except the last significant figure, which take from 10*.

EXAMPLE.

Required the arithmetical complement of 9.62595?

For the first figure 9, write 0; for 6, 3; for 2,7; for 5, 4; for 9, 0; and for the last figure 5, write 5; thus the arithmetical complement is 0,37405. In the same manner the arithmetical complement of 1.86563 is 8.13437, the ar. co. of 10.33133 is 9.66867, and the ar. co. of 1.22800 is 8.77200. To illustrate the method of using the arithmetical complement of any logarithm, I shall here calculate the examples as given in page 32.

When the index of the given logarithm is greater than 10, as in some of the numbers of Table XXVII. the left hand figure of it must be neglected; and when there are any ciphers to the right hand of the last significant figure, you may place the same number of ciphers to the right hand of the other figures of the arithmetical complement.

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PLANE TRIGONOMETRY.

PLANE TRIGONOMETRY is the science which shows how to find the measures of the sides and angles of plane triangles, some of them being already known. It is divided into two parts, right-angled and oblique-angled: in the former case, one of the angles is a right angle or 90°; in the latter they are all oblique.

In every plane triangle there are six parts, viz. three sides and three angles; any three of which being given (except the three angles) the other three may be found by various methods, viz. by Gunter's Scale, by the sliding rule, by the sector, by geometrical construction, or by arithmetical calculation. We shall explain each of these methods;* but the latter is by far the most accurate; it is performed by the help of a few theorems, and a trigonometrical canon, exhibiting the natural or the logarithmic sines, tangents, and secants, to every degree and minute of the quadrant. The theorems alluded to are the following:

THEOREM I.

In any right angled triangle, if the hypotenuse be made radius, one side will be the sine of the opposite angle, and the other its co-sine; but if either of the legs be made radius, the other leg will be the tangent of the opposite angle, and the hypotenuse will be the secant of the same angle.

1st. If in the right-angled plane triangle ACB (fig. 1) we make the hypotenuse AB radius, and upon the centre A describe the arch BE, to meet AC produced in E; then it is evident that BC is the sine of the arch BE (or the sine of the angle BAC) and that AC is the co-sine of the same angle: and if the arch AD be described about the centre B, AC will be the sine of the angle ABC, and BC its co-sine.

2dly. If the leg AC (fig. 2) be made radius, and the arch CD be described about the centre A; CB will be the tangent of that arch, or the tangent of the angle CAB; and AB will be its secant.

Sdly. If the leg BC (fig. 3.) be made radius, and the arch CD be described about the centre B; CA will be the tangent of that arch, or the tangent of the angle B; and AB will be its secant.

Now it has been already demonstrated (in art. 55. Geom.) that the sine, tang. sec. &c. of any arch in one circle, is to the sine, tang. sec. &c. of a similar arch in another circle as the radius of the former circle to the radius of the latter. And since in any right-angled triangle there are given either two sides, or the angles and one side, to find the rest; we may, if we wish to find

• It will not be necessary to add any further description of the uses of the sector or sliding rule, for what we have already given will be sufficient for any one, tolerably well versed in the use of Gunter's Scale. + See Tables XXIV. and XXVIL