Note.-The 2d and 3d cases are ambiguous, or admit of two different answers each, when the side AB opposite the given angle C (see fig. 8.) is less than the given side BC, adjacent to it (except the angle found be exactly a right one): for then another right line Ba, equal to BA, may be drawn from B to a point in the base, somewhere between C and the perpendicular BD, and therefore the angle found by the proportion AB (aB): sin. C:: BC: sin. A (or of CaB), may, it is evident, be either the acute angle A, or the obtuse one CaB, which is its supplement, the sines of both being exactly the same.

Having laid down the method of resolving the different cases of plane triangles, by a table of signs and tangents; I shall here show the manner of constructing such a table, as the foundation upon which the whole doctrine is grounded; in order to which, it will be requisite to premise the follow ing propositions.

PROPOSITION I.

The sine of an arch being given, to find its co-sine, versed sine, tangent, co-tangent, secant, and co-secant.

Let AE (fig. 9.) be the proposed arch, EF its sine, CF its co-sine, AF its versed sine, AT its tangent, CT its secant, DH its co-tangent, and CH its co-secant.

Then (by 47. 1.) we have CF the square root of CE* EF2; whence, not only the co-sine CF, but also the versed sine AF, will be known. Then, because of the similar triangles CFE, CAT, and CDH, it will be (by 4. 6.)

1. CF FE:: CA: AT; whence the tangent is known.

2. CF: CE (CA) :: CA: CT; whence the secant is known.

3. EF: CF CD: DH; whence the co-tangent is known.

4. EF: EC (CD) :: CD: CH; whence the co-secant is known.

Hence it appears,

1. That the tangent is a fourth proportional to the co-sine, the sine, and radius.

2. That the secant is a third proportional to the co-sine and radius.

3. That the co-tangent is a fourth proportional to the sine, co-sine, and radius.

4. And that the co-secant is a third proportional to the sine and radius.

5. It appears, moreover (because AT: AC:: CD (AC) ; DH), that the rectangle of the tangent and co-tangent is equal to the square of the radius (by 17. 6.): whence it likewise follows, that the tangent of half a right angle is equal to the radius; and that the co-tangents of any two different arches, represented by P and Q, are to one another, inversely as the tangents of the same arches: for, since tang. Px co-tang. P =squ. rad. tang. Qx co

1

tang. Q; therefore will co-tang. P: co-tang. Q: : tang. Q : tang. P; or as co-tang. P: tang. Q: co-tang. Q: tang. P (by 16. 6.).

PROP. II.

If there be three equidifferent arches AB, AC, AD (fig. 10.), it will be, as radius is to the co-sine of their common difference BC or CD, so is the sine CF of the mean, to half the sum of the sines BE + DG, of the two extremes: and, as radius to the sine of the common difference, so is the co-sine FO of the mean to half the difference of the sines of the two extremes.

For let BD be drawn, intersecting the radius OC in m; also draw mn parallel to CF, meeting AO in n; and BH and mv, parallel to AO, meeting DG in H and v.

Then, the arches BC and CD being equal to each other (by hypothesis), OC is not only perpendicular to the chord BD, but also bisects it (by 3. 3.), and therefore Bm (or Dm) will be the sine of BC (or DC), and Om its co-sine: moreover mn, being an arithmetical mean between the sines BE, DG of the two extremes (because Bm Dm) is therefore equal to half their sum, and Do equal to half their difference. But, because of the similar triangles OCF, Omn, and Dom,

It will be

SOC: Om:: CF : mn
OC: Dm:: FO: Dv

2. E. D.

=

Hence, if the mean arch AC be supposed that of 60°; then, OF being the co-sine of 60°, sine 30° chord of 60° OC, it is manifest that DGBE will, in this case, be barely Dm; and consequently DG Dm+ BE. From whence, and the preceding corollary, we have these two useful theorems.

1. If the sine of the mean of three equidifferent arches (supposing radius unity) be multiplied by twice the cosine of the common difference, and the sine of either extreme be subtracted from the product, the remainder will be the sine of the other extreme.

* Note.

Om X CF

ос

taken in a geometrical sense, denotes a fourth pro.

portional to OC, Om, and CF; but, arithmetically, it signifies the quantity arising by dividing the product of the measures of Om and CF by that of OC. Understand the like of others.

C

2. The sine of any arch, above 60 degrees, is equal to the sine of another arch, as much below 60°, together with the sine of its excess above 60°.

PROP. III.

15'.

To find the sine of a very small arch; suppose that of

4

It is found, in p. 181 of the Elements*, that the length of the chord of of the semi-periphery is expressed by ,00818121 (radius being unity); therefore, as the chords of very small arches are to each other nearly as the arches themselves (vid. p. 181), we shall have, as 1 : 380 :: ,00818121,008726624, the chord of, or half a degree; whose half, or ,004363312, is therefore the sine of 15', very nearly.

From whence the sine of any inferior arch may be found by bare proportion. Thus, if the sine of 1' be required, it will be 15' 1'::,004363312,000290888, the sine of the arch of one minute, nearly.

But if you would have the sine of 1' more exactly determined (from which the sines of other arches may be derived with the same degree of exactness); then let the operations in p. 181 be continued to 11 bisections, and a greater number of decimals be taken; by which means you will get the

* The Elements of Geometry pubk shed by the same author are here referred to. The same conclusion is easily derived from the scholium to Prop. 11, book 1, Supplement to Play fair's Geometry.