LEMMA. Trigonometriae Britannicae Chapter Eleven

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Nathaniel Caldwell
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1 Trigonometriae Britannicae -. hapter leven The ratio of other chords can be found with no less certainty, and for this purpose the following Lemma is useful. LMM. B If the line bisects the angle in the periphery B; and, are equal; then, B are equal also. For the angles B,, are equal by hypothesis, and from Prop. 5, Book. nd the angles B, also are equal, by Prop. 22, Book 3 and Prop. 3, Book. Therefore, B are similar triangles; nd with, equal from the construction, B, will be equal. [Figure -] Let B, B,,, F be equal chords inscribed in a circle : in like manner, O;, V;, Y; F, FX; G, GR are equal; then B, O;, V;, FY;, GX; F, HR are equal by the preceding lemma. H B Let the first B, which is written thus: second 2 O third cont. proport. G F O will be the third, for which O, equal to the first, is taken. R X Y V [Figure -2]

2 - 2 Trigonometriae Britannicae B 2 O B V Y 2 3 F 2 3 F X G G R H F H [Table -] With all of these triangles the ratio of the equal legs to their own base is the same. Proportionals B 2 V 4 2 If, let it be : Y will be 3 taken away:, 2, from which or FY is there remains F,. If any lines are in continued proportion, of which the first is the chord subtended by some arc, the second truly is the chord of double the arc; the third, if the first is taken away, is the chord of the triple arc: the fourth, with double the second taken away, is the chord of the quadruple arc. The fifth with the first added on, with three of the third taken away, is 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, is 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, is the chord of seven times the arc, etc., as you can see in the following table.

3 Trigonometriae Britannicae - 3 First Second Third Fourth 2 2 Fifth Sixth Seventh ighth Ninth Tenth leventh Twelfth Thirteenth Fourteenth Fifteenth Sixteenth B F G H [Table -2] olumn gives the lines in continued proportion, with columns B,,F,H, taken away; and,,g with the proportions added. For the chord subtended by the arc times ten, it is the tenth proportional, increased by 2 of the sixth and 5 of the second, from which is taken 8 of the eighth and twenty of the fourth. B Sides, subtending 2 angles Sides, subtending 4 angles Numbers in continued proportion B 44 egrees hord egrees hord [=] 360 & 72 ifference of 3 rd & st [=] 360 & 26 iff. 4 th & two of 2 nd [=] 360 & 360 iff. 5 th st, & 3 of 3 rd [Table -3] B [Figure -3]

4 - 4 Trigonometriae Britannicae Second Fourth Fifth First Sum Three of the third. By taking 0 remains [Table -4] [Note: Tables -3 and -4 have been badly affected by errors, that presumably have originated by the letters and occupying each other's positions in the original Figure -3: ccordingly, and have been interchanged in the diagram reproduced here from those in the original diagram, to give agreement with the tables, and to make mathematical sense.] hord 62 eg hord 324 eg Numbers in continued proportion [Table -5] [Figure -4] 2

5 Trigonometriae Britannicae - 5 Numbers in continued proportion hord of 62 egrees hord of 486, 26 above a single circle nd Subtending 324 egrees th B hord of 648, above a single circle rd st.5 th Sum hord 80, 90 above two circles th nd th Sum.6 th hord: 972, First hord 62:0': 6803 Second hord 324:0': or 36:0': [Table -6] If the multiple arc is left with a single circle or three circles, the contrary signs are assumed; as in the examples and B. For they were the signs in the Table (from section three of this chapter) and ; but contrary to this 2 and 2 4. If however there should remain two or four circles, then the signs are as in examples and.

6 - 6 Trigonometriae Britannicae. Notes on hapter leven. Following Briggs' usual procedure with sets of similar isosceles triangles, if (a, a, pa) are the lengths of the sides of the first triangle B in Figure -2, where a is the original length of the equal side, and p is the common ratio used; the sides of the second triangle O constructed will be (pa, pa, p 2 a), the third (p 2 a, p 2 a, p 3 a), and so on. The length of the first chord B = a = ; the second chord = pa = 2 ; while the length of the third chord = O - O = O - B = p 2 a - a =. The lengths of the chords L n can then be found from the iterative scheme: L n = pl n- - L n-2 L = a; L 2 = pa; L 3 = (p 2 - )a; L 4 = pl 3 -L 2 = p(p 2 - )a -pa = p((p 2 - ) - )a = p 3 a -2pa; L 5 = pl 4 -L 3 = p 2 ((p 2 - ) - )a - (p 2 - )a = (p 4-3p 2 )a; L 6 = pl 5 -L 4 =p( p 2 ((p 2 - ) - )a - (p 2 - )a ) - p((p 2 - ) - )a = (p 5-4p 3 3p)a,, etc Let us examine how the various regular figures fare according to this scheme: Im(z) B e ι π/2 I. The equilateral triangle: p =, as L = a; L 2 = pa = L ; II. The square: L = a; L 2 = pa; L 3 = p 2 a - a = L = a; Hence p = 2. III. The pentagon: L 4 = p 3 a -2pa = L = a; and L 2 = pa = L 3 = p 2 a - a; hence, p 2 = p : this has solution τ = ( 5)/2, (and -/τ). IV. The hexagon: L 5 = (p 4-3p 2 )a = L = a; and L 2 = L 4, being symmetric about the central chord L 3. Hence, pa = p 3 a -2pa, or p = 3. e ι π/2 e ι 2π/5 e ι π/2 e ι 4π/5 V. The heptagon: L = L 6 ; L 2 = L 5 ; and L 3 = L 4. From the last equality, p 2 a - a = p 3 a -2pa, or p 3 - p 2-2p = 0 O ι 2π/5 e ι π/2 e Re(z) ι 4π/5 e ι π/2 e

PTOLEMY DAY 7 BUILDING A TABLE OF CHORDS AND ARCS: PART 1

PTOLEMY DAY 7 BUILDING A TABLE OF CHORDS AND ARCS: PART 1 PTOLEMY Y 7 UILING TLE OF HORS N RS: PRT 1 (1) WHY 360 N 120? Ptolemy divides his standard circle s diameter, R, into 120 parts, in keeping with his sexagesimal ways this way the radius of the circle is