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4. The difference of any arch from 90° (or a quadrant)is called its complement ; and its difference from 180° (oria a semicircle) its supplement.
5. A chord, or subtense, is a right line drawn from one extremity of an arch to the other : thus the right line BE is the chord, or subtense, of the arch BAE or BDE.
6. The sine, or right sine, of an arch is a right line drawn from one extremity of the arch, perpendicular to the diameter passing through the other extremity. Thus BF is the sine of the arch AB or DB.
7. The versed sine of an arch is the part of the diameter intercepted between the sine and the periphery. Thus AF is the versed sine of AB; and DF of DB.
8. The co-sine of an arch is the part of the diameter intercepted between the centre and sine, and is equal to the sine of the complement of that arch. Thus CF is the cosine of the arch AB, and is equal to BI, the sine 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 passing through the centre and the other extremity. Thus AG is the tangent of the arch AB.
10. The secant of an arch is a right line reaching, without the circle, from the centre to the extremity of the tangent. Thus CG is the secant of AB.
11. The co-tangent and co-secant of an arch are the tangent and secant of the complement of that arch. Thus HK and CK are the co-tangent and co-secant of AB.
12. A trigonometrical canon is a table exhibiting the length of the sine, 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 10,000 or more decimal parts. By the help of this table, and the doctrine of similar triangles, the whole business of trigono." metry is performed; which I shall now proceed to show. But, first of all, it will be proper to observe, that the sine of any arch Ab greater than 90° is equal to the sine of another arch AB as much below 90°; and that its co-sine Cf, tangent Ag, and secant Cg, are also respectively equal to the co-sine, tangent, and secant of its supplement AB; but only are negative, or fall on contrary sides of the points C and A, from whence they have their origin. All which is manifest from the definitions.
In any right-angled plane triangle ABC (fig. 2.), it will be as the hypothenuse is to the perpendicular, so is the radius (of the table) to the sine of the angle at the base.
For, let AE or AF be the radius to which the table of sines, &c. is adapted, and ED the sine of the angle A or arch' EF (Vid. Def. 3. and 6.); then, because of the similar triangles ACB and AED, it will be AC; BC :: AE ;ED (by 4. 6.). 2. E. D.
Thus, if AC=,75, and BC = ,45; then it will be, ,75 : 245 :: 1 (radius) : the sine of A=6; which, in the table, answers to 36° 52', the measure or value of A.
In any right-angled plane triangle ABC (fig. 2.), it will be, as the buse AB is to the perpendicular BC, so 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, 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. 9. E. D.
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 32° 00'.
In every plane triangle ABC (fig. 3.), it will be, as any one side is to the sine of its opposite angle, so is any other side to the sine of its opposite angle.
For, take CF = AB, and upon AC let fall the perpendiculars BD and FE; which will be the sines of the angles A and C to the equal radii AB and CF. Now the triangles CBD, CFE being similar, we have (B:BD (sin. A) :: CF (AR): FE (sin. C). 2. E. D
As the base of any plane triangle ABC (fig. 4.) is to the sum of the two sides, so is the difference of the sides to twice, the distance DE of the perpendicular from the middle of the base.
From the vertex C, with the greater side AC, describe a circle, and produce AB and CB to the circumference; then it is evident that BF is the sum, and BH the difference of the sides; also, since AB + BG = 2AD (3. 3.) = 2AE +2ED = AB + 2ED; BG= 2ED. Join AF, HG; then the angle AFB = BGH; FAB = BHG (21. 3.), and ABF = HBG; therefore AB : BF (AC + BC) : : BH (AC - BC): BG (2ED), by 4. 6. Q: E. D.
In any plane triangle, it will be, as the sum of any two sides is to their difference, so is the tangent of half the sum of the two opposite angles, to the tangent of half their difference.
For, let ABC (fig. 5.) be the triangle, and AB, AC une two proposed sides; and from the centre A, with the radius AB, let a circle be described, intersecting CA, produced, in D and F; so that CF may express the sum, 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 5. 1.) = C + ABC (by 32. 1.) it is plain that ADB is equal to half-the sum of the angles opposite to the sides proposed. Moreover, since ABC = ABD (ADB) + DBC, and C = ADB - DBC (by 32. 1.) it is plain that ABC — C is = 2DBC; or that DBC is equal to half the difference of the same angles.
Now, because of the parallel lines BF and ED, it will be CF: CD:: BF: DE ; but BF and DE, because DBF and
: ; BDE are right angles (by 31. 3. and 29. 1.), will be tangents of the foresaid anglés FDB (ADB) and DBE (DBC) to the radius BD. 2. E. D.
Hence, in two triangles ABC and AbC, having two sides equal, each to each, it will be (by equality), as tang. AC + ACb АБС AC
ABC + ACB : tang
: : tang
ABC ACB tang.
But, if CAl be supposed a right an2 gle, then will ABC + ACb also = a right angle (by 32. 1.),
A6C + AC6 and the tangent of
= radius*. Therefore in
2 this case our proportion will become,
AbC АСЬ Radius : tang:
45) :: 2 ABC + ACB ABC ACB tang: : tang:
Which gives 2
2 the following theorem, for finding the angles opposite to any
* See Prop. 1. Cor. 5.