But BO, BP are equal because PBO, OED are similar triangles: therefore BP, not less than BO, will be 2 
, and PO will be 2
- 1
; 
And therefore OD, 1 , the total PD will be 3 - 1 which taken from the Radius AD, 1, leaves AP, 1 - 3 + 1 , and by proportionality PQ will be 1
- 3 + 1
. But BP is 2 - 1, and with this being equal to GQ: the total BPQG will be 5 - 5 + 1.
2. With the total Arc BCDEFG, if it shall be the larger to the semicircle2, nevertheless BG will be equal to 5 - 5 + 1. For DO is 1 as before; and BO, 2 - 1, and therefore OP, 2
- 1; to which being added DO, 1, the total DP will be 3 - 1: from which if AD, 1. being taken away, AP will be 3 - 1 - 1. And PQ will be 3 - 1 - 1. As either BO, BP, or GQ is 2 - 1: therefore BP and GQ have the value 4 - 2; from which if PQ, 3 - 1 - 1, being taken away there will remain 5 - 5 + 1 being equal to the line BG.
3. Being given a given Subtended Chord of any Arc, we will be able to find the Quintuple Arc subtending chord; and conversely.
Let the given Subtended Chord of 10 Degrees be 17431148549.
Should be found by Multiplication of the given Subtending Chord into itself, and in products by itself with itself, the Square of the same, together with the Cube, the Biquadratic [Fourth power], and the whole [Fifth power]. If from five of the Subtended Chord [i.e. times] and with one of the whole, should be taken away five of the Cube; what has been left should be the Subtended Chord of 50 Degrees.
Let the given Subtended Chord of 50 Degrees be 845236523481.
4. For if the five-fold of the given Arc should exceed the whole Circle: the signs of the whole equation should be changed, and for 5 - 5 + 1 being placed 5 - 5 - 1 , which should be the equal of the Subtending Chord of five times the given Arc: or rather of the excess of five times the subtending chord over the whole Circle.
Let the Arc with the given Subtending Chord be 140 Degrees: 0': if 5 + 1 should be taken from 5, the Subtended Chord of 20:0' or 340 Degrees shall remain.
With these continued proportionals, if from the three nearest being sought the fourth; with the triple of the second being added to the first; the total will be the fourth. For: With the same Subtended Chord of 20 Degrees will be found by the same way, if the Subtended Chord of 76 Degrees will have been given: of which the quintuple Arc will exceed the Circle by 20 Degrees.
5. For if the quintuple of the given Arc should exceed two whole circles; the equation resumes that which was demonstrated from the first situation, and will be 5 - 5 + 1 the equal of the subtending Chord over two Circles. As:
Let the given Subtended Chord of 157:2':40" be 19600039119128
And this method will be called 'Quintuplication' [Latin: Quintuplatio, times five], as with any Subtended Chord, the Subtended Chord of the quintuple being sought.
1 This beautiful piece of elementary geometry is based on the set of similar isosceles that Briggs has considered previously; see Chapter Three, Note 1. Thus, for the first set of proportionalities, [Table 5-1] becomes:
Hence, the total length of the subtending chord BG = 2.BP + PQ = 5p - 5p3 + p5.
2 This derivation is almost the same as Note 1. We will, however, set out the working for this case. Again, it is based on a number of similar isosceles triangles:
OE = BV = p, AO = 1 - p2, AO/OV = 1/p gives OV = p - p3. Then
BO = PV + VO = 2p - p3 = BP, PO = 2p2 - p4, and AP = AO - PO = 1 - 3p2 + p4 . Again, AP/1 = PQ/p gives PQ = p - 3p3 + p5 , and finally: BG = 2BP + PQ = 5p - 5p3 + p5 as before.
This may be an appropriate place to comment on Briggs' 5th power equation for finding the subtended chord of the 5-fold arc given the single arc, and the 5th part of a given subtended chord. The first is dealt with by Briggs in a straightforward manner: however, the second requires a few remarks. We note initially the customary relation:
sin 5
= 5sin - 20 sin3 + 16 sin5. Now for unit radius, the subtended chord has length A = 2 sin5(/2) ; hence, A = 2 sin5(/2) = 10 sin(/2) - 40 sin3(/2) + 32 sin5(/2) = 5(2 sin(/2)) - 5(2 sin(/2))3 +(2 sin(/2))5. This is the equation solved by Briggs in Chapter 6, where the variable is (2 sin(/2)), leading to very accurate values of the sine of the 5th part of the original angle. Four of the 5 real roots are determined by Briggs, by adding on multiples of 360, taking supplements, etc. He does not seem to know that the sum of the roots is zero, or in fact that the equation has 5 roots: his attention is focused on finding the sines of useful angles for his tables.
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