Trigonometria Britannica

 Previous page (Chapter Nine) Contents Next page (Chapter Eleven)

## Chapter Ten

We are able therefore in accordance with the rules propounded further up: Given the Arc of any Chord, to find the Chord both for any multiples of the Arc as for fractions. But any Chords you wish found by the above methods, will give the Squares of the Chords of the other Arcs. And the Squares will give the Chords also, as will be evident from the following propositions.

1. If the Chord of any Arc whatsoever should be added to the Diameter, the sum will be the square of the Chord of the same Arc increased by half the complement to the Semicircle. As the Chord of 36 Degrees of which the complement 144, and of this half 72, the sum of the Arcs 108. Or the Arc of the square of the Chord composed of the Quadrant and from half the given Arc. As for the Chord 36:0': 061803398874989484824.

The Arc composed from the Quadrant 90:0': and from half the Arc given 18:0': The Square of the Chord of 108:0': 2618033988.

Let AC be the Diameter of the Circle, FE the Chord of 36:0': to which CG shall be equal; I assert AG to be the Square of the line AF
of the subtending Arc ADF, 108:0': For the triangles BAF, FAG are equiangular. Thus because EF, CG shall be equal and parallel from the construction and the proposition, the Angles ECA, FGA equal to the Angle BAO, and between themselves. Therefore AB, AF, AG are continued proportionals. And as the radius shall be from the preposition unity, AB shall be the side and AG the square of the same side.
[i.e. AF/AG = 1/AF].

2. If the Chord of any Arc whatsoever should be taken away from the Diameter, the remainder will be the Square of the Chord of half the complement to the Semicircle.

Let DE, DF be equal, and from the Arc EB bisected in C being drawn the line CD: CE, CF will be equal; and the triangles ACB, CBF with equal angles and equal legs; and the Angles ACB, CBF, BFC equal. AB, BC, BF are therefore continued proportionals, and AB unity, BC the side, and BF the Square.
[i.e. BD - DE = BF = BC2; and 1/BC = BC/BF].

Let DE or DF be the Chord of 117:0': 1705280328708184, of which the complement to the Semicircle 63:0': of which the half 31:30': Of which the square of the Chord is 0294719671291816. And the Chord of 31:30': itself 0542880899730149. [This approach resembles the work in Chapter Three].

3. With the difference of the Diameter and the Square of the Chord of the Arc given, the Chord is in excess or deficient to the double of the Arc given to the Semicircle. For if the double of the Arc shall be greater than the Semicircle , AF the Arc 108: the double 216. The excess over the Semicircle 36. AG the Square of the Chord AF, CG being equal to the Chord EF, 36:0': And if the double Arc shall be smaller than the Semicircle , BC 31:30': BE the double Arc 63:0': BF the Square of the Chord BC. DF being equal to the Chord DE the Arc 117:0': of the complement to the Semicircle.

 Previous page (Chapter Nine) Contents Next page (Chapter Eleven)

Ian Bruce January 2003