If the Quadrant should be cut in any number of equal parts, and Tangents being prepared for half of the same of these from the start or the end of the Quadrant: the Tangents of the rest of the parts will be able to be found by Addition or Subtraction:
And the Secants also of the other parts; as it will be shown by the following Propositions
1. Half the sum of the Tangents of any Arc and its Complement: is the secant of the Difference of the same arcs.
2. Half the difference of the Tangents of any arc and its complement: is the Tangent of the Difference of the same Arcs.
Let CAB be 19:0', CAE or BAD 35:30'. BAE or CAD 45:30'. EAD 90:0'. CAE,CEA, VAO equal to the angle BAD 35:30' BDA, BAE, CAD equal 54:30' CA, CE therefore are equal; in the same manner CA, CD are equal by Prop. 5, Book 1, Euclid. And CA half the sum of the Tangents EB 54:30' and BD, 35:30' . The Secant of the angle CAB 19:0' is the Difference of the angles EAB 54:30' and BAD 35:30'.1
3. The sum of the Secant and of the Tangent of the same Arc, being equal to the Tangent of the same Arc increased by half the Complement2.
4. The Difference of the Secant and the Tangent of the same Arc, being equal to the Tangent of half the Complement.
Let CAB be 33:0'. The Complement CAO 57:0'. CASE, BAD 28:30'. EAB 61:30'.
5. The Tangents of any Arc you please and of half the Complement, being equal to the Secant of the same Arc. [See Figure 17-2].
Let CAB 47:0'. The Complement CAO 43:0'. Half the Complement 21:30' BAN.
6. The Tangent being doubled of any Arc together with the Tangent of half the Complement, being equal to the Tangent of the same Arc increased by half the Complement3.
This being proved by Props. 5 and 3. For by 5, the Tangents of the Arc and of half the Complement being equal to the Secant of the same. To which if the tangent of the given Arc being added, the sum by 3 will be equal to the tangent of the same arc increases by half the Complement.
Let BED be 39:0'. The Complement BEO 51:0'. Half of the Complement DEP 25:36'.
7. The tangent of any Arc you wish by being taken from the tangent of the Complement; there remains the double of the Tangent of the difference4.
Let the arc PED be 2737 . The Complement DEF 6263. The difference DEB 3526 .
8. Double of the difference of the Tangent of any arc you wish and of the Complement being taken from the Tangent of the larger arc, there will remain the Tangent of the smaller arc.
PED 3975 , of which the Complement DEF 5025, the Difference DEB 1050 .
9. The Radius is the mean proportional between the Tangents of any Arc you please and of the Complement.
CB, to BA : as AE, to ED. Because the Triangles CBA, AED are equal angled.[Similar].
10. Any Sine SQ, or OV, is to the Sine of the Complement QV: as the Radius OR, to the Tangent of the same Complement RP.
11. The Radius is the mean proportional between any Sine and the Secant of the Complement.
SQ to QO; as OR to OP. Because the Triangles SQO, ROP are equiangular.
12. The Sine is to the Tangent of its own Arc; as the radius to the Secant of the same.
For EOF, EHG, and in like manner, EON, EYX are similar Triangles.
13. The Sine is to the Tangent of its own Arc, as the Tangent of the Complement to the Secant of the same Complement.
Because the Radius is the mean proportional between the second and the third, by Prop. 9, and between the first and the fourth by Prop. 11. And therefore the Rectangle of the means being equal to the Rectangle of the extremes. And by Prop. 16, Book 6, Euclid, the given lines are proportional.
14. The Radius is to the Secant of any arc; as the Tangent of the Complement to the secant of the same Complement.
For the first to the second; as OF to HG by Prop.12. And the third to the fourth as OF to HG by Prop. 13.
15. The Diameter of the Circumscribed Circle is to any side of inscribed triangle; as the other side of the same to the perpendicular drawn from the angle being comprised of the said sides, to the third side.
Let the inscribed triangle be ABC, the Diameter of the Circle AD, the Perpendiculars AE in BC, and CF in BA continued. I say AD, AC: AB, AE to be proportionals: For the Triangles DAC, BAE are similar, because the angles at D and B are in the same section; and the angles ACD, AEB are right. Therefore DA, AC: BA, AE are proportionals. Also DAC, BCF are equal angled Triangles: and therefore DA, AC: BC, CF are proportionals: and therefore DA, AC: BC, CF are proportionals5.
The radius is to the Sine of the angle, as the sine of the other angle to half the perpendicular from the same parts drawn from the third angle to the opposite side. For these lines are the halves of these above which were said to be in proportion. As XA the Radius, AM the Sine of the angle AXM or ACB equal to this , by Prop. 20, Book 3, Euclid. And AN the Sine of the angle AXN or ABC.
16. If twice four lines shall be proportional; the Rectangles from similar terms being taken together are proportional.
For two ratios being given, the products from the homologous terms, have the ratio from the given ratios being composed. Therefore since ratios being composed from the same of these shall be the same [also], it is necessary [for] these rectangles (or the products expressing the sizes of the rectangles) to be in proportion. For let
4, 3: 8, 6 be proportionals. In the same way:
8, 7: 16, 14 are proportionals. The proportional products will be:
32, 21: 128, 84.
2. The Quadrant being cut in any number of equal parts; suppose, for example, eighteen. If the Tangents will have been given for the first nine parts: the Tangents will be able to be found for the remainder of the nine parts by Addition only.
The given Tangents being duplicated, and being added to the tangents of half the Complements: the sums will be (by Proposition 6) the Tangents of the Arcs being composed from the given arcs and from half of the Complements. For
If however the Tangents will have been given for the nine last parts of the Quadrant; the remainder will be able to be found by Subtraction.
If double the Tangent of the difference of the larger arc above the Complement being taken away from the Tangent of the Larger arc: the Tangent of the Complement will remain, by Proposition 6.
3. With the same Tangents of the Semiquadrant; the Secants will be able to be found for the other parts.
In the first place the Tangents should be found for the separate parts of the whole Quadrant: Then by Prop. 5 the Tangents being added of any arc you wish and with half the Complements ; the sum will give the secant of the same arc.
4. And by this method the Tangents and secants will be able to be prepared: the Logarithms of the same truly will be able to be computed, either as the Sines by Chapter 14, Arith. Logar.; or preferably with the minimum bother and with none by the Logarithms of Sines first found, by Propositions 10 and 11.
But with the Logarithms for the Tangents of the semiquadrant being given; the Logarithms being computed of the half left by Subtraction alone from double the logarithm of the Radius. And the Logarithm of the Tangent of this will be the Arithmetical Complement of the Complement of the Tangent. And for this reason the Logarithms from 45:0' as far as the end of the Quadrant, without much loss will be able to be omitted. However, it is better to add these, as the place that they occupy being left empty.
4. As by Prop. 11 the Radius shall be the mean proportional between any Sine and the Secant of the Complement: If the Logarithm of the Sine of any of these being taken from the Logarithm of the Radius doubled; the remainder will be the Logarithm of the Complement of the Secant. As
Because of this, the Logarithms of the Secants being found so much easier by the Logarithms of the Sines; I have considered these should be omitted altogether: the lest of these studious things, for all that I desire aiding by calculation, the exacting labour would hinder me to much.
To be noted, if beforehand these numbers shall have been found, the remainder by Quinquisection being added on most conveniently. That was being shown with Sines before6.
1 These results are similar to those considered in Chapter 15, where the same construction has been employed. Thus, two tangents are known, corresponding to BE and BD. Half their sum is the secant AC of , while half their difference is the tangent BC of .
2 For AC + CB = BE. For the complement of is 2, and half the complement is : BE is the tangent for the arc + . Similarly, AC - CB = BD , which is the tangent of , half the complement.
3 For 2BC + BD = CD + BC = BE.
4 For BF - DP = BP - DP = BD, and DF = BF + DB ; hence, DF - DP = BD + (BF - DP) = 2BD.
5 In modern terms, we set AD = 2R;
AC = b, BC = a, and CF = a sin B. Then:
Prop. 15 becomes 2 R/b = a/(a sin B), or
sin B/b = 2R. As similar results follow for the other sides, the Sine Rule is established. This result is apparent directly from proposition 15.
6 Thus, in this rather abrupt manner, the last words of Briggs are given in this work. One may presume his health was failing during this time: as we see in the preface by Gellibrand, no attempt had been made to start Book II. Here, too, our translation comes to an end: for Gellibrand's contribution has long since been rendered into English, (being the practical or business end of the tables), and of course it lacks the wonderful ab initio constructs, both geometrical and numerical, used by Briggs to produce this work, his last sadly forgotten masterpiece.