Trigonometria Britannica

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## Chapter Eleven

The ratio of other Chords should be found with no less certainty, for this the following Lemma serves.

LEMMA
.

If the line AC shall bisect the Angle in the Periphery BAD; and AC, CE shall be equal; DE, BA shall be equal. For the Angles BAC, CAD, CED shall be equal from the thesis, and from Prop. 5, Book 1. And [angles] ABC, CDE also [are] equal, by Prop. 22, Book 3 and Prop. 13, Book 1 . Therefore CDE, CBA are similar triangles; And with CA, CE being equal from the construction, BA, DE will be equal.

Let AB, BC, CD, DE, EF be inscribed, and equal: in like manner AC, CO; AD, DV; AE, EY; AF, FX; AG, GR are equal; AB, DO; AC, EV; AD, FY; AE, GX; AF, HR will be equal by the preceding lemma.

Let the first AB, which being written

AD will be the third, for which DO, equal to the first, will have been taken.

With all of these Triangles the ratio is the same of the equal legs to their own base1.

If AE, let it be 1 - 2: AY will be 1 - 2, from which AD or FY should be taken away: 1 - 1 and there will remain AF, 1 - 3 + 1.

If any lines shall be in continued proportion, of which the first shall be the Chord subtended by any Arc, the second truly shall be the Chord of double the Arc; the Third, if the first is taken away, will be the Chord of the triple Arc: the Fourth, with double the second taken away, will be the Chord of the quadruple Arc. The Fifth with the first added on, with three of the third taken away, will be the Chord of the Quintuple of the Arc: the Sixth with three of the second added on, and with four of the fourth taken away, will be the Chord of six times the Arc. The Seventh with six of the third added on and five of the fifth taken away with the first, shall be the Chord of seven times the Arc, etc., as in the following table you can see.

Column A gives the lines in continued proportion, for Columns B,D,F,H, taken away; and C,E,G with the proportions added.
For the Chord being subtended by the Arc times ten, it is the tenth proportional, increased by 21 of the sixth and 5 of the second, from which ought to be taken away 8 of the eighth and twenty of the fourth.

[Note: Tables 11-3 and 11-4 have been badly affected by errors, that presumably have originated by the letters E and D occupying each other's positions in the original Figure 11-3: Accordingly, E and D have been interchanged in the diagram reproduced here from those in the original diagram, to give agreement with the original tables, and to make mathematical sense.]

If the multiple Arc shall be left with a single circle or three, the contrary sign being assumed; as in the examples A and B. For they were the signs in the Table (from section three of this chapter) 1 - 1 and 1 - 2; but contrary to this 1 - 1 and 2 - 1. If however there should remain two or four, the signs being serves as in examples C and D.

Notes On Chapter Eleven

1 Following Briggs' usual procedure with sets of similar isosceles triangles, if (a, a, pa) are the lengths of the sides of the first triangle ABC in Figure 11-2, where a is the original length of the equal side, and p is the common ratio used; the sides of the second triangle ACO as constructed will be (pa, pa, p2a), the third (p2a, p2a, p3a), and so on. The length of the first chord AB = a = 1 ; the second chord AC = pa = 1; while the length of the third chord AD = AO - DO = AO - AB = p2a - a = 1 - 1 . The lengths of the chords Ln can then be found from the iterative scheme: Ln = p Ln-1 - Ln-2
L1 = a, L2 = pa, L3 = (p2 - 1)a , L4 = p L3 - L2 = p(p2- 1)a -pa = p((p2- 1) - 1)a = p3a -2pa,
L5 = p L4 - L3 = p2((p2 - 1) - 1)a - (p2 - 1)a = (p4 - 3p2 + 1)a,
L6 = p L5 - L4 =p( p2((p2 - 1) - 1)a - (p2 - 1)a ) - p((p2 - 1) - 1)a = (p5 - 4p3 + 3p)a, ... , etc

The length of the side a of a regular polygon with n sides inscribed in a circle with unit radius is given by L1 = a = 2sin(p/n); the next chord is subtended by double the angle, and hence L2 = pa = 2 sin(2p/n): hence, p = sin(2p/n)/sin(p/n) = 2 cos(p/n). The third chord has length L3 = p L2 - L1 = 2 cos(p/n).2 sin(2p/n) - 2 sin(p/n)
= 8 cos2(p/n).sin(p/n) - 2 sin(p/n) = (p2 - 1)a.

For a 20 sided figure, a = 2 sin(p/20), and p = 2 cos(p/20).

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Ian Bruce January 2003