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 CFCB2 — BF2, or c = √ ] − s2.

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.

As for the log. sines, tangents, and secants, in the tables, they are only the logarithms of the natural sines, tangents, and secants, calculated as above.

HAVING given an idea of the calculation and use 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

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. Every triangle has six parts, viz. three sides and three angles. And in every triangle, or case in trigonometry, there must be given three of these parts, to find the other three. Also, of the three parts that are given, one of them at least must be a side; because, with the same angles, the sides may be greater or less in any proportion.

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 varie ties of cases; for each of which it will therefore be give a separate theorem, as follows:

THEOREM 1.

proper to

When a Side and its Opposite Angle are two of the Given Parts.

THEN the unknown parts will be found by this theorem ; viz. 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.

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. Now the triangles ADE, ACF, are equiangular; they therefore have their like sides proportional, namely, AC: CF :: AD or BC DE; that is, the side AC is to the sine of its opposite angle B, as the side BC is to the sine of its opposite angle A.

Note 1. In practice, to find an angle, begin the proportion with a side opposite to a given angle. And to find a side, begin with an angle opposite to 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 the example below; and when there is no restriction or limitation included in the question, either of them may be taken. The number of degrees in the table, answering to the sine, is the acute angle; but if the angle be obtuse, subtract those degrees from 180°, 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 construc tion will form only one triangle.

EXAMPLE I.

In the plane triangle ABC,

AB 345 yards

Given BC 232 yards
LA 37° 20′

Required the other parts.

1. Geometrically.

Draw an indefinite line; on which set off AB = 345, from some convenient scale of equal parts.-Make the angle A = 37°.—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 con

structed,

structed, which gives two triangles, and showing that the case is ambiguous.

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

In the first proportion.-Extend the compasses from 232 to 345 on the line of numbers; then that extent reaches, on the sines, from 370 to 640, the angle c.

In the second proportion.-Extend the compasses from 37° to 27° or 78°4, on the sines; then that extent reaches, 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. FIRST add the two given sides together, to get their sum, and subtract them, to get their difference. Next subtract the given angle from 180°, or two right angles, and the remainder will be the sum of the two other angles; then divide that by 2, which will give the half sum of the said unknown angles. Then say,

As the sum of the two given sides,

Is to the difference of the same sides;

So is the tang. of half the sum of their op. angles, To the tang. of half the diff. of the same angles. Then add the half difference of the angles, so found, to their half sum, and it will give the greater angle, and subtracting the same will leave the less angle: because the half sum of any two quantities, increased by their half difference, gives the greater, and diminished by it gives the less.

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

Note. Instead of the tangent of the half sum of the unknown angles, in the third term of the proportion, may be used the cotangent of half the given angle, which is the same thing.

Demonst. 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, E, and the unknown side AE in A, G. Join AE, AD, CG, and draw DF parallel to AE.

Then BE is the sum of the two given sides AC, CB, or of IC, CB; and BD is the difference of the same two given sides

AC,