Trigonometria Britannica

Previous page
(Chapter Two)
Contents Next page
(Chapter Four)

Chapter Three

Concerning [a General Method of] Triplication

And these are the principal [propositions] being had from the writings of Ptolemy and the Ancient Mathematicians; Let us now come to these which by the favour of God have been found a few years ago [i.e. by Briggs himself], by which the whole Table of Subtended Chords will be able to be constructed most easily and accurately.

To Trisect a Given Arc.


If the straight line AC drawn through the Centre should cut the Base of the section EB [ the given minor arc] in the point G, and the Arc in the point C, being equidistant from the angle [i.e. vertex] B; it will cut the Arc in the duplicate ratio; and the Arc EDC will be double the Arc CB.1 [Figure 3 - 1.]
For if BC, BG being equal; the Angles BCG, BGC, CBA will be equal, by Prop. 4, Book 1; and CBG, CAB, also equal, by Prop. 32, Book 1: and therefore CDE, the base of the angle CBE in the Arc, will be the Double of the Arc CB, the base of the Angle CAB in the Centre, by Prop.20, Book 3.


FIRST CONSEQUENCE.

If the Line CH being drawn,
[Figure 3-2, for the major arc case], Parallel to the line DA, the Triangles ABC, BCG, GCH will be similar: and the Lines AB, BC, CG, GH [will be] continued proportionals. And if the Radius AB shall be unity; BC will be the side; CG the square [of the side]; GH the cube [of the side]. If by the square and the cube, we should know the second and the third continued proportionals from unity: indeed from the radius AB, which we place to be unity.


SECOND CONSEQUENCE.

BE ([composed of] the three equal subtending chords, BC, CD, DE) being inscribed together with the cube GH will be equal to these three Subtended Chords taken together: for ED, EF; BC, BG; and DC, FH being equal from the construction2, and Prop. 33, Book 1.

THIRD CONSEQUENCE.

Given any Subtended Chord whatsoever BC; it will be permitted to find BE the Subtended Chord of the triple Arc, and conversely.
Being sought of a given Subtended Chord the square and the Cube. The cube being taken from the triple of the Subtended Chord shall leave the Subtended Chord of triple the Arc. For let these be3:

With these be exceedingly careful, lest the [triple of the] Subtended Chord or the Cube should be taken from the other place [i.e. the signs have reversed in the defining equation at the start of the chapter], and not from its own place; it is allowed of course that the same noted [numbers] may be kept, of these however the value [of the resultant chord] for the different place being changed the most, and the result by the subtraction agrees with the remaining Subtended Chord, if the place of the Cube truly being served: otherwise all will become frustration4.

Because if the Arc of which the Subtended Chord shall be given, the triplicate shall be greater than the whole Circle, the triplicate of the given Subtended Chord being taken from the Cube of the same; the remainder shall be the departure of the Subtended Chord of the given Arc triplicated over the whole Circle, for:




This will be shown in this way. Let BC, CF, FG be inscribed equal; BG, CF will be Parallel: and BG, being continued in either direction, will make BC, BD; CF, GH; CH, HE equal: ABC, BCD, CDE will be similar Triangles; and AB, BC, CD, DE, continued proportionals [1 : p : p2 : p3 ]; and by subtracting, DB,GH, HE, (of which each being equal to the given Subtended Chord BC) will leave BG, the Subtended Chord the departure above of the whole Circle.

And according to this method which we will be able to call Triplationem [a general method of Triplication], most easily being found the Lines being subtended for any Arc whatsoever, if the Subtended Chord of a third of the Arc shall be given.




Notes On Chapter Three

1 For AB = 1, and BC = BG = p; then from the similar triangles ABC, BCG, and CGH in Figures 3-1:
p/1 = CG/p = GH/CG; hence, CG = p2 = CH, and GH = p. CG = p3 . Thus, the three triangles have sides: (1, 1, p); (p, p, p2); (p2, p2, p3). It follows that the chord for the triple arc BE has length BG + EF + GF = 2p + p - p3 ; hence, BE = 3p - p3.
We may isolate the method used by Briggs' of constructing nested similar isosceles triangles, as in Figure 3-4 (i), (ii), and (iii); the construction can obviously be extended indefinitely.

Note: Although the sub-heading is for trisection, or cutting into three, the theorem is concerned with the triplication of a chord of length p, to find the length of the chord that corresponds to triple the original arc. Obviously, if the angle exceeds 360degrees, the defining equation has to be modified, the cause of Brigg's concern in this chapter.

Briggs has distinguished three cases: the first where the triple arc is less than 1800, or the initial arc 0 < theta < 600, as in Figure 3-1; the second where the triple arc lies between 180degrees and 360degrees, Figure 3-2, corresponding to 60degrees < theta < 120degrees; and the third where the triple arc is greater than 360degrees, as in Figure 3-3, and theta > 120degrees.


2 The analysis of Figure 3-2 is similar to 1: BE + GH = (EF + FB) + GF + FH = ED + (FB + GF) + CD = ED + BG + CD, as required. Or, BE + p3 = 3p, where BE is the subtended chord for the triple arc.
We recognise this equation to be a form of the sine triplication identity:
2 sin(3theta/2) + (2 sin(theta/2))3 = 3.(2 sin(theta/2)); where we identify BE = 2 sin(3theta/2), and
p = 2 sin(theta/2). Examples of this are shown in Table 3-1. Briggs has chosen an angle of 117degrees34' in Table 3-2 to show that the initial scheme still works, as theta < 120degrees. However, if theta > 120degrees In the following Table 3-4, the angle chosen is 147degrees, which gives a triple arc of 441degrees, or 81degrees beyond the whole circle.

3 Subtended chord = 2Rsin (theta/2) = 2 cross 1010 cross sin 8degrees = 2783462019, etc

4 For 2 sin(3theta/2) = 3.(2 sin(theta/2)) - (2 sin(theta/2))3 = 3p - p3 = p(3 - p2) < 0 if p2 > 3,
or theta/2 > 60degrees. In which case the signs are reversed on the right hand side.
From Figure 3-3, according to Note 1: AB, BC, CD, DE, are the continued proportionals 1 : p : p2 : p3. From the construction, DE = p3 = (BD + GH + HE) + GB = 3p + GB : hence GB = p3 - 3p, as required.


Previous page
(Chapter Two)
Contents Next page
(Chapter Four)

Ian Bruce January 2003