Trigonometria Britannica

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## Chapter Seven

For Septisection

If the Arc BCDEFGHI shall be cut in these seven equal parts with the lines BE, BG, BI being drawn, OD will be 1 , and BO, 2 - 1. And the Subtended Chord [BI] of seven times the Arc [BC] will be, as here concluded, equal to 7 - 14 + 7 - 1 .

With being given therefore eight lines in continued proportion of which the first is the Radius, the second whatever Chord you wish: 7 of the Chord being increased by adding 7, will be equal to 14 + 1 and [i.e. added to] the Subtended Arc of seven times [the angle].

And by this equation: with any given Subtended [Chord], we will be able to find the Subtended Arc of seven times; or if seven of the given Arc shall be in excess of one, two or three circles, we will be able to find the Subtended Chord excess above the whole number of circles.

It should be noted with all the equations of the section by this method, if the number of the entire Circles shall be less than one: the same signs of the equation will be used, because the Periphery together come to less than 360 Degrees; if on the contrary it will have been more than a whole circle, the opposite signs should be placed. Let the given Subtended Chord [angle] be 140 Degrees. Seven times 140 has the value 980. That is 260 beyond two whole circles.2

For the Subtended 140 . . . . . . . . . . . . . . . . . 18793852415718167610822, the equation 7 - 14 + 7 - 1 .

Let the Subtended Chord be 157:2':40": 196000039119128.
Seven times 157:2':40" has the value 1099:18':40" or 19: 18':40" beyond three whole circles.

Because the number of Circles is more than a whole circle the equation will be
14 - 7 - 7 + 1 . [See Note 2]

Let the given Subtended Chord of 10 Degrees be 17431148549. The Chord of 70:0' being sought:

2. We are able also by the same equation: For any given Subtended Arc, to find the Subtended Chord of the seventh part of the same Arc, or the total being composed from the given Arc, and with one, two, or three circles.

The way of working will be scarcely different from that which was expounded above, for the Subtended Chords of the Third and Fifth parts. But the working for the multiplication of the terms is not without a little more labour.

And by this easiest of methods the equations are being found and are able to be demonstrated, if any periphery you wish being cut into equal parts however big, by the method shall be the number deficient [determined]: which all will give the Subtended Chords of multiplied Arcs, by the same preceding method, and the parts also, but not with the same facility. For where there are many Arcs of equal segments, with these there are equations with more terms, and to any term being added a larger number: As with all operations the other will be the source of more difficulty.

Notes On Chapter Seven

1 This beautiful piece of elementary geometry is based on the set of similar isosceles that Briggs considered previously; see the Notes 1 of Chapters Three and Five. The constructions 'work' because the angle subtended at the centre is double the angle subtended at the circumference by the equal angles GBI, GBE, and EBC. As AH bisects the arc GI, etc, it follows that the angles at the centre such as HAI are equal to the angles GBI, etc. The rest follows from the set of similar isosceles discussed previously. Thus, for the first set of proportionalities, [Table 7-1] becomes:

Hence, the total length of the subtending chord IB = 2.BQ + QR = 7p -14p3 +7p5 - p7.

2 For sin 7/2 = 7 sin /2 - 56 sin3 /2 + 112 sin5 /2 - 64 sin7 /2, and
2 sin 7/2 = 7(2 sin /2) -14(2 sin /2)3 + 7(2 sin /2)5 - (2 sin /2)7.

If p = 2 sin 7/2 < 0, then this has the effect of changing the signs in the equation for the chord; otherwise the equation is used as it stands.

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Ian Bruce January 2003