not, however, computed to this radius, on account of the inconvenience which would attend the continual use of negative indices in all the sines and cosines; but they are computed to a radius of 1010. Hence, in all formulas of trigonometry intended for logarithmic computation, the radius R must always be introduced, so as to make the terms homogeneous; and, although in the formulas which will be hereafter given, we shall but seldom encumber the expressions by actually inserting in them R, and its powers, yet the computist must not fail to take account of them in the logarithmic process.

Introducing R into the foregoing equations we may write them thus:

and all these equations may be comprehended in a single rule, expressed as below

As the tabular radius

: the radius in the figure,

:: any tabular line

: the corresponding line in the figure;

and from this it immediately follows that

Any tabular line

: corresponding line in the figure,

:: any other tabular line

: corresponding line in the figure;

which proportion, obviously, comprehends the former.

It appears from this rule that when we want to find a side, we must begin the proportion with a given tabular line, that is, either with the tabular radius, of which the logarithm is 10, or else with the tabular sine, cosine, &c. of a given angle; but when we want to find an angle, then we must invert this proportion, beginning with a given side, which must be made the geometrical radius, as no other tabular line but the radius will be given, seeing that angles are in this case unknown.

(14.) In operating with logarithms, the logarithm of the first term of

the proportion must be subtracted from the sum of the logs. of the other two, to obtain the logarithm of the sought fourth term; and thus the logarithmic process will consist of five lines or rows of figures. If, however, the first term, or that to be subtracted were 10, we might save a line, by adding the two other logs. together, and rejecting 10 in the index; when the first term is not 10 we may still save a line by the following artifice, viz. instead of putting down for the first term the log. given by the table, put down its deficiency from the number 10, which may be done with as much readiness as transcribing the number itself, provided we begin at the left-hand figure and subtract each in succession from 9, till we come to the last significant figure, which must be taken from 10; we shall thus have instead of the logarithm, what is called its arithmetical complement, which, being added in with the other two terms, rejecting 10 from the index, must give the same result as if we had subtracted the log. of the first term from the sum of the other two. An example or two will fully illustrate what has now been said.

EXAMPLES.

(15.) 1. Given the angles and the base to find the perpendicular and hypotenuse, viz. A=53° 8', AB = 288.

BC might have been found by making any other side radius, although not quite so easily, as we should then have had to seek out in the table the tabular line for the first term, corresponding to the known line AB; thus if AC had been made radius, then the tabular line we should have commenced with, would have been that corresponding to AB, viz. the cosine of the angle A. If CB had been made radius we should have commenced with the cotangent of A, that is, the tangent of C, for such would be the tabular line corresponding to BA.

II. To find the Hypotenuse AC.

Preserving the same radius we have,

If we had made AC radius, the proportion would have been cos. A: AB :: rad. : AC. By way of showing the use of the arithmetical complement, let us determine AC by this proportion

2. Given the two perpendicular sides to find the hypotenuse and angles, viz. AB=472, BC=765, (see last fig.)

1. To find the Angle A.

We must here, agreeably to the rule, begin with a given side, say AB, which we shall make radius.

Here we must begin with a tabular line; we shall choose the radius.

Or without employing the angle A we may determine AC by the formula.

For the method of determining the angle corresponding to any tabular number to seconds, see the introductory explanation prefixed to the tables.

ADC, AC 288, ACD=30° 6′ ... A=90° - 39° 6′ = 50° 54′;

=

hence, to find AD, we have by making AC radius,

4. Given the base AB=53 · 42, and the perpendicular BC=75·18, to find the hypotenuse and angles?

A=54° 36′ 14′′, C=35° 23′ 46′′, AC=72.23.

5. Given the hypotenuse AC=643 · 7, and the base AB=473-8, to find the other parts?

A = 42° 36′ 12′′, C= 47° 23′ 48′′, BC=35·87. 6. Given the angle A = 37° 2′ 43′′, and the hypotenuse AC=173 · 2 to find the other parts?

C=52° 57′ 17′′, AB = 138.24, BC=104·34.

(16.) We shall now proceed to investigate rules and formulus for the solution of triangles in general.

Oblique-Angled Triangles.

Let ABC be any plane triangle, and let us denote the angles by the capital letters, A, B, C, at their vertices, and the sides opposite to them by the small letters a, b, c.

From either vertex, as C, draw the perpendicular CD to the opposite side.

C

B. A

B

Then the sine of A to the radius b will, obviously, be the line CD, and the A value of this sine in terms of the trigonometrical sine of the same angle to radius 1 is (art. 5,) CD = b sin. A. In like manner the sine of B to the radius a, is the same line CD, whose value, therefore, in term of the trigonometrical sine, is CD = a sin. B; consequently, by