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8. The co-fine of an arch is the part of the diameter intercepted between the center and fine ; and is equal to the line of the complement of that arch. Thas CF is the co-fine of the arch AB, and is equal to BI, the fine of its complement HB.
9. The tangent of an arch is a right-line touching the circle in one extremity of that arch, produced from thence till it meets a right-line paffing through the center and the other extremity. Thus AG is the tangent of the arch AB.
1o. The secant of an arch is a right-line teaching, without the circle, from the center to the extremity of the tangent: Thus CG is the fecant of AB.
11. The co-tangent, and co-fecant, of an arch are the tangent, and secant, of the complement of that arch. Thus HK and CK are the cotangent and co-secant of AB.
12. A trigonometrical cahon is a table exhibiting the length of the fine, tangent, and secant, to every degree and minute of the quadrant, with respect to the radius ; which is supposed unity, and conceived to be divided into 10000, or more, decimal parts. By the help of this table, and the doctrine of similar triangles, the whole business of trigonometry is performed ; which I shall now proceed to shew. But, first of all, it will be proper to observe, that the fine of any arch Ab greater than 90°, is equal to the fine of another arch AB as much below 90°; and that, its co-fine Cf, tangent Ag, and secant Cg, are also respectively equal to the co-line, tangent, and secant of its supplement AB ; but only are negative, or fall on contrary fides of the points C and A, from whence they have their origin. All which is manifest from the definitions.
For, let AE or AF
be the radius to which
Def. 3. and 6.); then,
BDF becaufe of the fimi
lar triangles ACB and AED, it will be AC : BC :: AE : ED (by 14. 4.) 2. E. D.
Thus, if AC = ,75, and BC = ,45; then it will be, ,75 : ,45 :: 1 (radius) : the fine of A=6; which, in the table, answers to 36° 52', the measure, or value of A.
THEOREM II. In any right-angled plane triangle ABC, it will be, es the base AB is to tbe perpendicular BC, fo is the radius (of the table) to the tangent of the angle at the base.
For, let AE or AF be the radius of the table, or canon (see the preceding figure), and FG the tangent of the angle A, or arch EF (Vid. Def. 3. and 9.); then, by reason of the similarity of the triangles ABC, AFG, it will be, AB : BC :: AF : FG. 2. E. D.
Note, In the quotations where you meet with two numbers (as 14. 4.) without any mention of Prop. Theor. &c. reference is made to the second edition of the Elements of Geometry published by the fame author ; to which this little tract is defigned as an Appendix. I
Thus let AB = ,8, and BC =,5; then we shall have ,8 : ,5 :: 1 (radius) : tangent A = ,625 ; whence A itself is found, by the canon; to be
TÁ LOREM III. In every plane triangle ABC, it will be, as any one fide is to the fine of its opposite angle, fo is any other fide to the fine of its opposite angle.
For take CF = В. AB, and upon AC let fall the perpen
F diculars BDand FE; which will be the sines of the angles A and C to the equal radii AB and CF. A D D Now the triangles CBD, CFE being similar, we have CB : BD Uin. A) :: CF (AB) : FE (fin. C). 2. E. Da
THE OREN IV As the base of any plane triangle ABC, is to the Jurn of the two fides, jo is the difference of the fides to twice the distance DE of the perpendicular from the middle of the base. For (by Cor. t09. 2.)
B AB+BC AB-BC = AC 2DE, whence AC : AB + BC :: AB BC : 2DE (by 10. 4.) 2. E. D.
In any plane triangle, it will be, as the sum of any two fides is to their difference, so is the tangent of balf the sum of the two opposite angles, to the tangent of half their difference.
For, let ABC be the triangle, and AB and AC the
two proposed fides; F
and from the cen-
ter A, with the ra3
tersecting CA pro-
duced, in D and F;
so that CF may express the fum, and CD the difference, of the sides AC and AB : join F, B and B, D, and draw DE parallel to FB, meeting BC in E.
Then, because 2ADB = ADB + ABD (by 12. 1.) = C + ABC (by 9. 1.) it is plain that ADB is equal to half the fum of the angles opposite to the sides proposed. Moreover, since ABC = ABD (ADB) + DBC, and C= ADB DBC (by 9. 1.) it is plain that ABC — C is = 2DBC ; or that DBC is equal to half the difference of the fame angles.
Now, because of the parallel lines BF and ED, it will be CF : CD :: BÈ: DE; but BF and DE, because DBF and BDE are right-angles (by 13. 3. and 7. 1.) will be tangents of the foresaid angles FDB (ADB) and DBE (DBC) to the radius BD. 2. E. D.
Hence, in two triangles ABC and A C, having two fides equal, each to each, it will be (by equa
AC + ACb ABC-ACE lity), as tang
: tang ABC + ACB ABC ACB tang : tang
But, if CAb be supposed a right-angle, then will ABC to ACb also = a right-angle (by Cor. 3. to 10. 1.) and
A6C + ACE the tangent of
= radius. There
2 fore in this case our proportion will become,
AbC АСЬ Radius : tang
(= ABC - 45°):: ABC + ACB ABC - ACB tang. : tang
Which gives the following Theorem, for finding the angles opposite to any two proposed sides; the included angle, and the sides themselves, being known.
As the lesser of the proposed fides (Ab or AB) is to the greater (AC), so is radius to the tangent of an. angle (ABC, see Theor. 2.) And ás radius to the tangent of the excess of this angle above 45", so is the tangent of balf the sum of the required angles. to the tangent of half their difference
* This Theorem, though it requires two proportions, is commonly used by Astronomers in determining the elongation and parallaxes of the planets (being beft adapted to logarithms); for which reason it is bere given.