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dius, we have tables of sines, tangents, &c. to every arc of the quadrant, called natural sines, tangents, &c. and the logarithms of these give us tables of logarithmic sines, tangents, &c. and such are usually bound up with logarithms of numbers.

In which you may observe, that each page is divided into 3 columns, the first and last of which are minutes, and the intermediate ones contain the sines, tangents, and secants, the upper and lower columns contain degrees, the column of the minutes on the left hand of each page, answers to the degrees in the top column; and the sines, tangents, and secants belonging to those degrees and minutes, are in the columns marked at the top with the words sine, tangent, and secant; . the column of minutes on the right hand of each page, answers to the degrees in the bottom of the page; and the sines, tangents, and secants, answering to those degrees and minutes, are in the columns, marked at the bottom with the words sine, tangent secant; the degrees in the top column beginning at 0, proceed to 44, where they end; and those at the bottom of the page begin at 89, and proceed to 45 in a decreasing series; the degrees in the different columns being the complement of each other. From what has been said, we may easily find the sine, tangent, or secant of any arc, from the tables, by looking for the given number of degrees at the head or foot of the page, according as they are less or greater than 45, and in the proper side column for the odd minutes, if there be any; then below or above the word sine, tangent or secant, and on the same line with the minutes, we shall have that which was required.

EXAMPLE I.

Required, the sine of 36 degrees 40 minutes.

Look at the head of the page for 36 degrees, and in the side column on the left hand, for 40 minutes; then below the word sine, on the same line with 40, we find 9.77609; which is that required.

EXAMPLE II.

Required the tangent of 54 degrees 30 minutes.

Look at the foot of the page (because the proposed degrees are more than 45.) for 54 degrees, and in the right hand column for 30 minutes ; then in the column marked tangent at its bottom, and on the same line with 30 minutes, in the side column, we find 10.14673, which is the log-tangent required.

The reverse of this, viz. The logarithm of a sine, tangent, or secant, being given, to find the arc belonging to it, is performed by only looking in the proper column, for the nearest logarithm to that proposed, and the degrees and minutes answering thereto, are those required.

We will now shew how any sign, tangent, or secant may be had, tho' the figures in the tables were defaced, mis-printed, or obliterated.

PROBLEM I.

To find the tangent which is defaced, by the sine and co-sine.

The co-sine taken from the sine added to 90, or radius, which is 10.00000, the remainder is the tangent. (By part 1. theo. 24.)

EXAMPLE.

1. Suppose the tangent of 41o. 20', was defaced, but the sine and co-sine of it visible.

From the sine of 41o. 20' + 10.00000, or radius,

Take the co-sine of 41° 20'

19.81983 9.87557

The rem. is thetan. of 41o. 20req. viz. 9.94426

2. To find a sine which is mis-printed, by help of the co-sine and tangent.

From the sum of the tangent and co-sine, take 10.00000, or radius, (which is the same thing) cut off the first figure in the index, the remainder is the sine required (by part 2. theo. 24.)

EXAMPLE.

Suppose the sine of 46°. 50' was defaced, but the tangent and co-sine visible.

To the tangent of 46o. 50
Add the co-sine of 46° 50'

10.02781 9.83513

Their sum is the sine of 46°. 50' req. viz. 19.86294

The co-tangent and co-sine of any arc, may be had by the same method; the complement of any degree, being only its residue from 90, or a

quadrant; as before observed, by theo. 24, part 3

and 4.)

3. To find a tangent by the help of a co-tangent only.

From twice the radius, which is 20.00000, take the co-tangent, the remainder is the co-tangent, (by theo. 24. part. 5.)

EXAMPLE.

Required, the tangent of 29o. 50' being defaced, as also the sine and co-sine defaced, by the co-tangent only.

From twice the radius,
Take the co-tangent of 29o. 50

20.00000 10.24148

The rem. is the tang. of 29o. 50' req.

9.75852

4. To find the secant by the help of a co-sine; which may be found of great use when a table of sines and tangents can only be had.

From twice the radius, which is 20.00000, take the co-sine, and the remainder will be the secant, (by theo. 24. part 6.)

EXAMPLE.

Required, the secant of 57°. 20' by the help of the co-sine only.

From the double radius,
Take the co-sine of 570. 20

20.00000 9.73219

The rem. is the secant of 57° 20' req. 10.26781

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5. To find a secant by the help of the sine and tangent.

From the tangent added to radius, take the sine, the remainder will be the secant, (by theo, 24. part 7.)

EXAMPLE

Required, the secant of 57o. 20by help of the sine and tangent.

From the tan. of 57o. 20' + 10.00000 the radius,

20.19303 Take the sine of 57°. 20'

9.92522

The rem. is the secant of 57o. 20 req. 10.26781

The secants in these tables might have been omitted, because all proportions in which they are concerned, may be wrought by sines and tangents only, as shall be shewn in the several cases of plane trigonometry; and are here only inserted that all the various methods of resolving triangles may be shewn.

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