A System of Geometry and Trigonometry

When the given logarithmic sine or tangent is not found exactly, or very nearly, then, for the seconds,

RULE 2. From the given logarithm subtract the next less in the table, annex two cyphers to the remainder, and divide it, thus augmented, by the tabular difference D, and the quotient will be the number of seconds to be added to the degrees and minutes of the tabular logarithm, for sines and tangents, but which must be subtracted for co-sines and co-tangents.

EXAMPLE. Required the degrees, minutes, and seconds corresponding to the logarithmic sine of 9.759567.

Given logarithm

Next less in the table=

Divide by tab. diff.

Angle required,=

9.759567
9.759492=

35° 5'

MULTIPLICATION BY LOGARITHMS.

RULE. Take from the table the logarithms of all numbers to be multipliplied, add them together, and their sum will be the logarithm of the product. Then by means of the table, take out the natural number, corresponding to this sum, for the product sought.

Observe, that whatever is to be carried from the decimal part of the logarithm, is always positive, and must be added to the positive, or subtracted from the negative index or indices.

3. Multiply the following numbers together: 3.902, 597.16, and .0314728.

Logarithms.

0.591287

2.776091

2.497935

1.865313

Here, the 2 cancels the 2, and the 1 to carry, from the decimal part of the logarithms is set down,

4. Multiply 3.586, and 2.1046, and 0.8372, and 0.0294 all together.

Logarithms.

Numbers.

3.586

2.1046

0.8372

0.0294

Pro. 0.105762

Here the 2 to carry cancels the 2, and there remains the ī to set down,

In practice, however, it is usual to make all the indices positive. This is done by adding 10 to each negative index; observing, to reject an equal number from the final result.

Thus, for negative --1 we may put down positive 9.

for negative-2, we may put down positive 8.

for negative-3, we may put down positive 7, &c. Because, minus 1, plus 10, equals 9.

minus 2, plus 10, equals 8.
minus 3, plus 10, equals 7, &c.

5. Repeating the third example we have

Numbers.

3.902

597.160

.0314728

Prod. 73.3333

Logarithms.

Logarithms.
0.591287

2.776091

Here the sum of the indices is 11: from which reject 10, and the result is the same as before.

COMPOUND INTEREST BY LOGARITHMS.

RULE. Find the amount of 1 dollar for 1 year; multiply its logarithm by the number of years; and to the product, add the logarithm of the principal. The sum will be the logarithm of the amount for the given time. From the amount subtract the principal, and the remainder will be the in

terest.

EXAMPLE. The last example in Daboll's arithmetic, under Compound Interest, is as follows:-" What will 50 dollars amount to in 20 years at 6 per cent."

This question wrought out in the most expeditious manner by common arithmetic, would take the student scarcely less than three hours, and the final result would be, if done correctly,

$160.3567736106422365941496492288974572748800. But, by logarithms, this sum may be done in as many minutes, and the correct answer, in dollars and cents, obtained with very few figures. Thus : Amount of 1 dollar for 1 year is 1.06 logarithm

Multiply by the time

0.0253059

From the foregoing general principles of the nature and application of logarithms, are derived an infinite number of specific rules, adapted to particular cases.

OF THE TABLE OF NATURAL SINES AND TANGENTS.

To those who are unacquainted with logarithms, it will be interesting to know, that all the cases in right and oblique angled trigonometry, contained

help of natural sines and tangents, be solved exactly in the same way, and with the same facility as he would solve a simple question in the Rule of Three. Natural sines are merely decimals, bearing the same proportion to unity, or 1, that the corresponding number of degrees and minutes bears to radius, or 90°. Natural tangents bear the same proportion to unity or 1, that the corresponding number of degrees and minutes bears to 45°, because it is a well known principle, that the sine of 90° and the tangent of 45° are each equal to radius. That is, 1 is assumed as the natural sine of 90° in the table of natural sines, and as the tangent of 45° in the table of tangents, and every other number in each of these tables is calculated accordingly.

GENERAL RULE. 1. State the question in every case, as already taught: 2. Multiply the second and third terms together, and divide their product by the first.

The manner of taking natural sines and tangents from the tables, is the same as for logarithmic sines and tangents; only that there is in the tables, no column of differences as in the latter, for the more readily finding the odd seconds, when required. But these may be found by making a proportion for the aliquot parts.

There are some problems to which natural tangents afford a much more simple and ready solution, than any process by logarithms. The following one, in heights and distances, will illustrate this.

EXAMPLE. The altitude of an inaccessible object taken at an unknown distance from its base, is 55° 54'; and when taken again at the distance of 93 feet from the place of the first observation in a direct line with it, the altitude is 33° 20': Required the height of the object.

RULE. Divide the difference of the natural co-tangents of the angles of elevation, by the distance between the stations. Thus-

Co-tangent of 33° 20' is 1.52043

of 55° 54' is .67705

Divide by the diff. =

feet.

.84338)93.0000(110.27 Ans.

NOTE. This is the shortest solution possible, and perfectly easy.

Again: Given the latitude and departure, in transverse sailing or surveying, to find the course.

RULE. Divide the departure by the latitude, the quotient will be the natural tangent of the course: or divide the latitude by the departure, and the quotient will be the co-tangent of the course. Universally, If in any right angled triangle, the perpendicular be divided by the base, the quotient will be the tangent of the angle at the base; and if the base be divided by the perpendicular, the quotient will be the tangent of the angle at the vertex of the perpendicular.

OF THE TRAVERSE TABLE, OR TABLE OF LATITUDE AND DEPARTURE.

This is calculated for degrees and quarters of degrees, and for any distance up to 100 rods, chains, &c.; by which the northings and southings,

PROBLEM XII.-To find the latitude and departure, or northing, &c. for any course and distance.

If the course be less than 45°, look for it at the top, but if more than 45°, at the bottom of the page, and look for the distance in the right or left hand column; against the distance, and directly under or over the course, stand the northing, &c. in whole numbers and decimals.

If the course be less than 45°, the northing or southing will be greater than the easting or westing; but if more than 45°, the easting or westing will be the greatest.

When the distance exceeds 100, take any two or more numbers, which, added together, will equal the distance, and find the latitude and departure for each of these numbers; add the several latitudes together, and the sum will be the whole latitude; and so for the departure. And when the distance is in chains and links, or whole numbers and decimals, find the latitude, &c. for the chains or whole numbers, and then for the links and decimals, remembering to remove the decimal point in the table further to the left, according to the given decimal.

1. Required the latitude and departure for 45 rods, on a course N. 15° 15′ W.

Under 15° 15' and against 45 is 43.42 for the northing, and 11,84 for the westing.

2. Required the latitude and departure for 120 rods, on a course S. 580 30' E.

Take one third of 120, which is 40; against this number, over 58° 30', i 20.90 for the latitude, and 34.11 for the departure. These multiplied by 3 give 62.70 for the southing, and 102.33 for the easting.

3. Required the latitude and departure for 37.36 rods, or 37 chains and 36 links, on a course N. 26° 45′ E.

NOTE. When the minutes are not 15, 30, or 45, the northings, &c. may be had by proportion, or they may be calculated by natural sines, or by trigonometry.

PROBLEM XIII.-To calculate the Northing or Southing, &e. for any course and distance, by natural sines.

Find the nat. sine and co-sine of the course, and into each of these multiply the distance; the products will be the latitude and departure required. Required the latitude atd departure for 6 chains and 22 links on a course N. 38° 27', W.

Nat sine of 38° 27', 0.62183

Nat co-sine, 0.78315

1-24

3.14159265359

0.13089969390

-1.1169387

in minutes,
in seconds,

1-4
of do. or area of cir. to diam. 1 0.78539816340
365th root of $1.05, or amount of $1. for 1 day, 1.00013368072
365th root of $1.06, or amount of $1. for 1 day, 1.00015965359
12th root of $1.05, or amount of $1. for 1 mo. 1.00407412
12th root of $1.06, or amount of $1. for 1 mo. 1.0048675505
360 degrees expressed in seconds,
Arc, equal to radius, in degrees,

Length of an arc of 1"=sine of 1′′

of 2"-sine of 2"
of 3"-sine of 3"
of l' sine of 1'

1296000 57.295780

0.017452406 --2.2418553

-1.8950899

0.00005805

0.00006933

0.00176577

0.00210882

6.1126050

1.7581226

3437.74677