Trigonometria Britannica

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## Chapter Two

`Concerning How the Subtended Chords Are Sought in the Writings of the Ancients.`

1. . Prop. Ptol.
Ptolemy1 showed First, following Euclid, the way the [lengths of the] sides of the particular figure to the inscribed circle ought to be found, namely of the equilateral triangle by Prop. 12. Book 13. Of the square by Prop. 6. Book 4. Of the Pentagon, Hexagon, and Decagon by Prop 9. and 10. Book 13.

2. Prop. Ptol.
Secondly, by Prop. 30, Book 3: from Euclid, he teaches, the Square of the Given Subtended Chord, taken from the Square of the given Diameter, to leave the Square of the Complement of the Subtended [Chord] to the Semi-circle. And according to this method the Subtended Chords of 60, 90,120, 36, 72 and of the Complements 144 and 108 degrees have been found.

3. Thirdly, this Lemma is proposed according to the need in subsequent proofs. If the quadrilateral [lit. quadrangle] being inscribed in the Circle, the Rectangle comprising the Diagonals being equal to the [the sum of the] Rectangles comprising the opposite sides.

As for the Circle ABCD, let the Diagonals be DB, 25; AC, 20. The Rectangle from the Diagonals 500 being equal to [the sum of] the Rectangles AB, DC, 360. And AD, BC, 140. For the line BE being drawn such that the angles ABE, DBC should be equal. The Triangles DBC, ABE are similar; because the Angles CDB, CAB being equal, since they shall be from the same section, by Prop. 21, Book 3. and CBD, EBA shall be equal from the construction; the remaining therefore being equal by Prop. 31, Book 1.

BD, DC : BA, AE are therefore in proportion: and the Rectangles BD, AE; DC, BA equal from Prop.16, Book 6. Likewise, the Triangles BDA, BCE are similar, because the Angles BCE, BDA are on the same section, and therefore equal. And as ABE, CBD shall be equal from the construction, with the common part DBE taken away, the remaining [angles] CBE, DBA being equal; and therefore CB, CE : BD, DA, are proportionals, and the rectangles CB, DA; CE, BD being equal; but [the sum of] the rectangles DB, EA; DB, CE, was equal to the rectangle DB, CA by Prop. 1, Book 2, Euclid: and therefore the Rectangle with the Diagonals DB, CA being equal to [the sum of] the Rectangles DC, BA, and BC, DA, which had to be shown2.

4. Fourthly, as permitted from the above Lemma, from two given arcs with unequal Subtended Chords, the Subtended Chord of the difference being found.

Let the given Diameter [Figure 2-2] be AB, 20.

Of the Inscribed [Chords] AD, 12. AC 53/5. CD being sought. Firstly, DB and CB should be found by the 2nd Prop. Since the angles ADB, ACB shall be in the semi-Circle, they shall be right, by Prop. 30, Book 3. and therefore if the Square AD, 144, being taken from the Square AB, 400, there will remain the Square DB, 256. By Prop. 47, Book 1. DB will be therefore 16. By the same method, by taking the Square AC, 319/25 from the Square AB, 400, there will remain the square CB, 36816/25, or 36864. And CB will be 192. Therefore being given AB, 20; AD, 12; AC, 56 ; and by finding DB, 16. CB, 192. CD being sought by the preceding Lemma. The rectangle3 AC, DB, 896, being taken from the rectangle Ad, CB, from the diagonals, 2304 , will leave the rectangle 1408 , being taken from the diameter AB, 20 CD. Therefore with 1408 being divided by 20, the Quotient 704 will be the length of the straight line CD, being sought.

By the same Lemma: From two given arcs with Subtended Chords, the Subtended Chord of the sum being found4.

Let the given diameter [Figure 2-3] be AB, 20. AC, 12. AD, 56. DC being sought. CB, DB especially being sought, as before5: and CB will be 16, DB, 192 We have therefore AB, AC, AD, being given; and CB, DB being found. The rectangles AD, CB, 896 , and AC, BD, 2304 being taken, of which the sum will be 320.0, to which the rectangle being comprised of the Diameter and DC, by the preceding Lemma, Prop. 3. Therefore given the Diameter AB, 20. Let it divide 320.0 , the rectangle being taken from the Diameter and DC, the Quotient 16. will be the Subtended Chord DC being sought6.

` Notes On Chapter Two`

1 The First Lemma looks at the problem of finding the lengths of the sides of some regular figures inscribed in a circle of unit radius. The Second Lemma applies the Theorem of Pythagoras to the sides of the appropriate right angled triangle to determine the ratios for the specified angles.

2 In modern terminology, we have AE.BD = AB.CD, and CE.BD = CB.DA; hence,
(AE + EC).BD = AB.CD + CB.DA, or AC.BD = AB.CD + CB.DA, as required. See the CRC Handbook of Modern Mathematics, p.84, for the 'whole story' on this theorem from a modern perspective. This theorem is the main 'workhorse' used by Ptolemy in the construction of his Table of Sines alluded to in the first chapter.

3 These are referred as 'oblongs' in the text always.

4 This should be called the 5th Lemma, or the converse of Lemma 4.

5 That is, we find the remaining sides of the cyclic quadrilateral first.

6 This is a rather perfunctory and incomplete look at Ptolemy's method. Ptolemy uses these results to find the lengths of half chords in terms of the known chords of larger angles: in this way he subdivides 12 formed from the difference of 72 from the regular pentagon and 60 from the regular hexagon to find the half chords corresponding to 60, 30, 11/20, 3/40 and eventually for 1/20, which he uses as the unit to build up his table. See his Almagest I for details. Briggs, however, wishes to move on to his own methods...

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Ian Bruce January 2003