metry. It will therefore be proper to begin with the mode of constructing it, which may be done in the following manner:

PROBLEM I.

To find the sine and cosine of a given arc.

This problem is resolved after various ways. One of these is as follows, viz. by means of the ratio between the diameter and circumference of a circle, together with the known series for the sine and cosine, hereafter demonstrated. Thus, the semicircumference of the circle, whose radius is 1, being 3.141592653589793, &c., the proportion will therefore be,

as the number of degrees or minutes in the semicircle,

is to the degrees or minutes in the proposed arc,

so is 3.14159265, &c. to the length of the said arc.

This length of the arc being denoted by the letter a; also its sine and cosine by s and c; then will these two be expressed by the two following series, viz.

EXAMPLE I.-If it be required to find the sine and cosine of one minute. Then, the number of minutes in 180° being 10800, it will be first, as 108001 :: 3.14159265, &c.: 000290888208665 = the length of an arc of one minute. Therefore, in this case,

Here, as 180° 5° :: 3·14159265, &c. :·08726646 = a the length of 5 degrees.

After the same manner, the sine and cosine of any other arc may be computed. But the greater the arc is, the slower the series will converge, in which case a greater number of terms must be taken to bring out the conclusion to the same degree of exactness.

Or, having found the sine, the cosine will be found from it, by the property of the right-angled triangle CBF, viz. the cosine CF = √CB2 — BF2, c = VT — s2.

or

There are also other methods of constructing the canon of sines and cosines, which, for brevity's sake, are here omitted.

PROBLEM II.

To compute the tangents and secants.

The sines and cosines being known, or found by the foregoing problem; the tangents and secants will be easily found, from the principle of similar triangles, in the following manner :—

In the first figure, where, of the arc AB, BF is the sine, CF or BK the cosine, AH the tangent, CH the secant, DL the cotangent, and CL the cosecant, the radius being CA, or CB, or CD; the three similar triangles CFB, CAH, CDL, give the following proportions:

1st, CF: FB:: CA: AH; whence the tangent is known, being a fourth proportional to the cosine, sine, and radius.

2d, CF: CB:: CA: CH; whence the secant is known, being a third proportional to the cosine and radius.

3d, BF: FC:: CD: DL; whence the cotangent is known, being a fourth proportional to the sine, cosine, and radius.

4th, BF: BC::CD: CL; whence the cosecant is known, being a third proportional to the sine and radius.

Having given an idea of the calculation of sines, tangents, and secants, we may now proceed to resolve the several cases of Trigonometry; previous to which, however, it may be proper to add a few preparatory notes and observations, as below.

Note 1.-There are usually three methods of resolving triangles, or the cases of trigonometry; namely, Geometrical Construction, Arithmetical Computation, and Instrumental Operation.

In the First Method. The triangle is constructed by making the parts of the given magnitudes, namely, the sides from a scale of equal parts, and the angles from a scale of chords, or by some other instrument. Then, measuring the unknown parts, by the same scales or instruments, the solution will be obtained near the truth.

In the Second Method. Having stated the terms of the proportion according to the proper rule or theorem, resolve it like any other proportion, in which a fourth term is to be found from three given terms, by multiplying the second and third together, and dividing the product by the first, in working with the natural numbers; or, in working with the logarithms, add the logs. of the second and third terms together, and from the sum take the log. of the first term; then the natural number answering to the remainder is the fourth term sought. In the Third Method.-Or Instrumentally, as suppose by the log. lines on one side of the common two foot scales; Extend the compasses from the first term, to the second or third, which happens to be of the same kind with it; then that extent will reach from the other term to the fourth term, as required, taking both extents towards the same end of the scale.

Note 2.-In every triangle, or case in trigonometry, there must be given three parts, to find the other three. And, of the three parts that are given, one of

them at least must be a side; because the same angles are common to an infinite number of triangles.

Note 3.-All the cases in trigonometry may be comprised in three varieties only; viz.

1st, When a side and its opposite angle are given.

2d, When two sides and the contained angle are given.

3d, When the three sides are given.

For there cannot possibly be more than these three varieties of cases; for each of which it will therefore be proper to give a separate theorem, as follows:

THEOREM I.

When a side and its opposite angle are two of the given parts.

Then the sides of the triangle have the same proportion to each other, as the sines of their opposite angles have.

That is, As any one side,

Is to the sine of its opposite angle;

So is any other side,

To the sine of its opposite angle.

Demonstr.-For, let ABC be the proposed triangle, having AB the greatest side, and BC the least. Take AD BC, considering it as a radius; and let fall the perpendiculars DE, CF, which will evidently be the sines of the angles A and B, to the radius AD or BC. But the triangles ADE, ACF, are equiangular, and therefore AC: CF :: AD or BC: DE; that is, AC is to the sine of its opposite angle B, as BC to the sine of its opposite angle A.

E

F

B

Note 1.-In practice, to find an angle, begin the proportion with a side opposite a given angle. And to find a side, begin with an angle opposite a given side.

Note 2.-An angle found by this rule is ambiguous, or uncertain whether it be acute or obtuse, unless it be a right angle, or unless its magnitude be such as to prevent the ambiguity; because the sine answers to two angles, which are supplements to each other; and accordingly the geometrical construction forms two triangles with the same parts that are given, as in thè example below; and when there is no restriction or limitation included in the question, either of them may be taken. The degrees in the table, answering to the sine, is the acute angle; but if the angle be obtuse, subtract those degrees from 180o, and the remainder will be the obtuse angle. When a given angle is obtuse, or a right one, there can be no ambiguity; for then neither of the other angles can be obtuse, and the geometrical construction will form only one triangle.

1. Geometrically.

Draw an indefinite line, upon which set off AB = 345, from some convenient scale of equal parts.-Make the angle A = 37°3.—With a radius of 232, taken from the same scale of equal parts, and centre B, cross AC in the two points C, C.-Lastly, join BC, BC, and the figure is constructed, which gives two triangles, showing that the case is ambiguous.

Then, the sides AC measured by the scale of equal parts, and the angles B and C measured by the line of chords, or other instrument, will be found to be nearly as below; viz.

angle B 27°

AC 174

or 3744

angle C 115°
64

In the first proportion.-Extend the compasses from 232 to 345 upon the line of numbers; then that extent will reach, on the sines, from 37° to 64°), the angle C.

In the second proportion.-Extend the compasses from 37° to 27° or 78°1, on the sines; then that extent will reach, on the line of numbers, from 232 to 174 or 374, the two values of the side AC.

When two sides and their contained angle are given.

Then it will be,

As the sum of those two sides,

Is to the difference of the same sides;

So is the tang. of half the sum of their opposite angles,
To the tang. of half the difference of the same angles.

Hence, because it has been shown under Algebra, that the half sum of any two quantities increased by their half difference, gives the greater, and diminished by it gives the less, if the half difference of the angles, so found, be added to their half sum, it will give the greater angle, and subtracting it will leave the less angle.

Then, all the angles being now known, the unknown side will be found by the former theorem.

Demonstr.-Let ABC be the proposed triangle, having the two given sides AC, BC, including the given angle C. With the centre C, and radius CA, the less of these two sides, describe a semicircle, meeting the other side BC produced in D and E. Join AE, AD, and draw DF parallel to AE.

E

D

B

Then, BE is the sum, and BD the difference of the two given sides CB, CA. Also, the sum of the two angles CAB, CBA, is equal to the sum of the two CAD, CDA, these sums being each the supplement of the vertical angle C to two right angles: but the two latter CAD, CDA, are equal to each other, being opposite to the two equal sides CA, CD: hence, either of them, as CDA, is equal to half the sum of the two unknown angles CAB, CBA. Again, the exterior angle CDA is equal to the two interior angles B and DAB; therefore the angle DAB is equal to the difference between CDA and B, or between CAD and B; consequently the same angle DAB is equal to half the difference of the unknown angles B and CAB; of which it has been shown that CDA is the half sum.

Now the angle DAE, in a semicircle, is a right angle, or AE is perpendicular to AD; and DF, parallel to AE, is also perpendicular to AD: consequently, AE is the tangent of CDA the half sum, and DF the tangent of DAB the half difference of the angles, to the same radius AD, by the definition of a tangent. But, the tangents AE, DF, being parallel, it will be as BE: BD :: AE : DF; that is, as the sum of the sides is to the difference of the sides, so is the tangent of half the sum of the opposite angles, to the tangent of half their difference. Note. The sum of the unknown angles is found, by taking the given angle from 180°.