CHAP. I.
Geometricall Definitions.

OF things Mathematicall there are two principall kindes, Number and Magnitude; and each of these hath his proper Science. The Science of Number is A­rithmetick, and the Science of Magnitude is commonly called Geometry, but may more properly be termed Megethelogia, as com­prehending all Magnitudes whatsoever, whereas Geometry, by the very Etymologie of the word, doth seeme to confine this Sci­ence to Land-measuring only.

Of this Megethelogia Geometry, or Sci­ence [Page 2] of Magnitudes, we will set down such grounds and principles as are necessary to be known, for the better understanding of that which follows, presuming that the rea­der hereof hath already gotten some com­petent knowledge in Arithmetick.

Concerning then this Science of Magni­tude, two things are to be considered: First, the severall heads to which all Magnitudes may be referred: And then secondly, the terms and limits of those Magnitudes.

All Magnitudes are either Lines, Plains, or Solids, and do participate of Length, Breadth, or Thicknesse.

1. A Line is a supposed length, or a thing extending it self in length, without breadth or thickness, whether it be a right line or a crooked; and may be divided into parts in respect of his length, but admitteth no o­ther division: as the line AB.

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2. The ends or limits of a line are points, as having his beginning from a point, and ending in a point, and therefore a Point [Page 3] hath neither part nor quantity, it is only the term or end of quantity, as the points A and B are the ends of the aforesaid line AB, and no parts thereof.

3. A Plain or Superficies is the second kind of magnitude, to which belongeth two di­mensions, length, and breadth, but not thickness.

4. As the ends, limits or bounds of a line are points confining the line, so are lines the limits, bounds and ends inclosing a Su­perficies; as in the figure you may see the plain or Superficies here inclosed with four lines, which are the extreams or limits thereof.

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5. A Body or Solid is the third kinde of magnitude, and hath three dimensions be­longing [Page 4] to it, length, breadth, and thick­ness. And as a point is the limit or term of a line, and a line the limit or term of a Su­perficies, so likewise a Superficies is the end or limit of a Body or Solid, and represent­eth to the eye the shape or figure thereof.

6. A Figure is that which is contained under one or many limits, Under one bound or limit is comprehended a Circle, and all other figures under many.

7. A Circle is a plain figure contained under one round line, which is called a cir­cumference, as in the Figure following, the Ring CBDE is called the circumference of that Circle.

8. The Center of a Circle is that point which is in the midst thereof, from which point, all right lines drawn to the circum­ference are equal the one to the other; as in the following figure, the lines AB, AC, AD, and AE, are equal.

9. The Diameter of a Circle, is a right line drawn through the center thereof, and ending at the circumference on either side, dividing the Circle into two equal parts, as the lines CAD and BAE, are either of them the diameter of the Circle BCDE, because that either of them doth passe through the center A, and divideth the [Page 5] whole Circle into two equal parts.

10. The Semidiameter of a Circle is half the Diameter, and is contained betwixt the center and one side of the Circle; as the lines AB, AC, AD, and AE, are either of them the Semidiameters of the Circle CBDE.

11. A Semicircle is the one halfe of a Circle drawn upon his Diameter, and is contained by the half circumference and the Diameter; as the Semicircle CBD is halfe the Circle CBDE, and contained a­bove the Diameter CAD.

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[Page 6] 12. A Quadrant is the fourth part of a Circle, and is contained betwixt the Semi­diameter of the Circle, and a line drawn perpendicular unto the Diameter of the same Circle, from the center thereof, divi­ding the Semicircle into two equal parts, of the which parts the one is the Quadrant or fourth part of the same Circle. Thus, the Diameter of the Circle BDEC is the line CAD, dividing the Circle into two equal parts, then from the center A raise the per­pendicular AB, dividing the Semicircle likewise into two equal parts, so is ABD, or ABC, the Quadrant or fourth part of the Circle.

13. A Segment or portion of a Circle is a figure contained under a right line and a part of the circumference of a Circle, either greater or lesser than the Semicircle; as in the former figure, FBGH is a segment or part of the Circle CBDE, contained un­der the right line FHG lesse than the Dia­meter CAD.

14. By the application of several lines [...]terms of a Superficies one to another, are made Parallels, Angles, and many sided Figures.

15. A Parallel line is a line drawn by the side of another line, in such sort that they [Page 7] may be equidistant in all places, and of such there are two sorts, the right lined pa­rallel, and the circular parallel.

Right lined Parallels are two right lines equidistant in all places one from the other, which being drawn to an infinite length would never meet or concur; as may be seen by these two lines, AB and CD.

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A Circular Parallel is a Circle drawn within or without another Circle, upon the same center, as you may plainly see by the two Circles BCDE, and FGHI, these Circles are both of them drawn upon the same center A, and therefore are parallel one to the other.

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16. An Angle is the meeting of two lines in any sort, so as they both make not one line; as the two lines AB and AC incline the one to the other, and touch one another in the point A, in which point is made the angle BAC. And if the lines which con­tain the angle be right lines, then it is cal­led a right lined angle; as the angle BAC. A crooked lined angle is that which is con­tained of crooked lines; as the angle DEF: and a mixt angle is that which is contained both of a right and crooked line; as the an­gle GHI: where note that an angle is (for [Page 9] the most part) described by three letters, of which the second or middle letter represent­eth the angular point; as in the angle BAC, A representeth the angular point.

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17. All Angles are either Right, Acute, or Obtuse.

18. When a right line standeth upon a right line, making the angles on either side equal, either of those angles is a right an­gle, and the right line which standeth erect­ed, is a perpendicular line to that upon which it standeth. As the line AB (in the following figure) falling upon the line CBD perpendicularly, doth make the an­gles on both sides equal, that is, the angle [Page 10] ABC is equal to the angle ABD, and ei­ther of those angles is therefore a right an­gle.

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19. An acute angle is that which is lesse than a right angle; as the angle ABE is an acute angle, because it is less than the right angle ABD, in the former figure.

20. An Obtuse Angle is that which is greater than a right angle; CBE in the former figure is greater than the angle ABC by the angle ABE, and therefore it is an obtuse angle.

21. The measure of every angle is the arch of a Circle described on the angular point, as in the following figure, the arch CD is the measure of the right angle CED. The arch BC is the measure of the acute angle BEC. And the arch BCD is the measure of the obtuse angle BED. But of their measure there can be no certain know­ledge, unlesse the quantity of those arches be exprest in numbers.

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22. Every Circle therefore is supposed to be divided into 360 equall parts, called Degrees, and every Degree into 60 Minutes, every Minute into 60 Seconds, and so for­ward. This division of the Circle into 360 parts we shall retain, but every Degree we will suppose to be divided into 100 parts or Minutes, & every Minute into 100 Seconds: and thus all Calculations will be much ea­sier, and no lesse certain.

23. A Semicircle is the halfe of a whole circle containing 180 degrees. A Quadrant or fourth part of a circle is 90 degrees. And thus the measure of the right angle CED is the arch CD 90 degrees. The mea­sure of the acute angle BEC is the arch BC 30 degrees. And the measure of the ob­tuse angle BED is the arch BD 120 degrees.

24. The complement of an angle to a Quadrant is so much as the angle wanteth [Page 12] of 90 degrees, as the complement of the angle AEB 60 degrees is the angle BEC 30 degrees; for 30 and 60 do make a Qua­drant or 90 degrees.

25. The complement of an angle to a Semicircle is so much as the said angle wanteth of 180 degrees, as the comple­ment of the angle BED 120 degrees, is the angle AEB, 60 degrees; for 60 and 120 do make 180 degrees.

26. Many sided figures are such as are made of three, four, or more lines, though for distinction sake, those only are so called which are contained under five lines or terms at the least.

27. Four sided figures are such as are contained under four lines or terms, and are of divers sorts.

• 1. There is the Quadrat or Square whose sides are equall and his angles right.
• 2. The Long Square whose angles are right, but the sides unequal.
• 3. The Rhombus or Diamond, having e­quall sides but not equal angles.
• 4. The Rhomboides, having neither equal sides nor equal angles, and yet the oppo­site sides and angles equal.

All other figures of four sides are called Trapezia or Tables. The dimension whereof [Page 13] as also of all figures whatsoever, depend­eth upon the knowledge of three sided fi­gures, or Triangles, of which in the Chap­ter following.

CHAP. II.
Of the nature and quality of Triangles.

1. A Triangle is a figure consisting of three sides and three angles.

2. Every of the two sides of any Triangle are the sides of the angle compre­hended by them, the third side is the Base, as in the figure following, the sides AC and BC and sides of the angle BCA, and AB is the Base of the said angle.

3. Every side is said to subtend the angle that is opposite to that side; as the side AB subtendeth the angle ACB, the side AC subtendeth the angle ABC, and the side BC subtendeth the angle BAC: the great­er sides subtend the greater angles, the les­ser sides lesser angles, and equal sides equal angles.

[Page 14]

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4. Of Triangles there are diverse sorts; as,

1. There are Equilateral Triangles, ha­ving three equal sides.

2. There is an Isoscheles, which is a Tri­angle that hath two equal sides.

3. Scalenum, which is a Triangle whose sides are all unequal.

4. An Orthigonium, or a right angled Triangle, having one right angle.

5. An Ambligonium, or an obtuse angled Triangle, having in it one obtuse angle.

6. An Oxigonium, or an acute angled Triangle, having all his angles acute.

7. All these Triangles are either Plain or Spherical.

8. The sides of Plain Triangles in Trigo­nometria are right lines only, concerning [Page 15] which we have added these Theorems fol­lowing.

9. Theorem. If one right line cut through two parallel right lines, then are the an­gles opposite one against another equal.

In the following Scheme the two lines WX and YZ are parallel, and therefore the an­gles XIC, and ICY are equal.

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Demonstration.

The two angles XIC and WIC are e­qual to two right angles, as also ICY and ICZ, because on the parallel lines at the points I and C there may be drawn two Se­micircles, each of which are the measures of two right angles. If then the angle XIC be lesse than ICY, the angle WIC must as much exceed the angle ICZ, and the angles XIC and ZCI would be lesse than [Page 16] two right angles, and consequently the lines WX and YZ may be extended on the side X and Z til at length they shall concur together, and then the lines WX and YZ are not parallels, as is supposed here, and therefore the angles XIC and ICY are equal.

10. Theor, If four right lines be propor­tionall, the right angled figure made of the two means, is equal to the right ang­led figure made of the two extreams.

Let the four proportional lines be AB two foot, EF three foot, FG six foot, and BC nine foot: I say then that the right angled figure made of the two means EF and FG, that is, the right angled figure EF GH, is equal to the right angled figure made of the extreams AB and BC, that is, to the right angled figure ABCD; for as twice 9 is 18, so likewise three times is 18.

[Page 17]

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11. Theor. If three right lines be propor­tionall, the Square made of the mean is equal to the rightangled figure made of the extreams.

[Page 18] Demonstration.

The Demonstration of this proposition is all one in effect with the former, the dif­ference is, that here is spoken of three lines, there of four, and therefore if we take the mean twice, of which the square is made, the work will be the same with that in the former proposition. As if the length of the first line were two foot, the second four, and the third eight; it is evident, that as four times four is 16, so two times eight is 16, and therefore what hath been said of four pro­portionals, is to be understood of three pro­portionals also.

12. Theor. If a right line being divided into two equal parts, shall be continued at pleasure, then is the right angled fi­gure made of the line continued, and the line of continuation, with the square of one of the bi-segments, equal to a square made of one of the bi-segments and the line of continuation.

The line PQ is divided into two equal parts, the midst is C, to the same is added a right line, as QN; and of the whole line PQ, and the added line QN is made [Page 19] PN as one line, and of this line PN, and the added line QN is inclosed the right angled figure [...]M, and upon the halfe line CQ and the line of continuation QN is made the square CF. Now if you draw the line QG parallel to NF, and equal to the same, then is the right angled figure [...]M with the square of CQ that is, the square IG, equal to the square of CN, that is, the square CF.

[figure]

Demonstration.

Forasmuch as CQ is equal unto MF, the which is also equal unto IO or IL, it fol­loweth that IG is a Square, which with the right angled figure PM is equal to the square CF, because the right angled figure GM is equal to CO, which is also equal to PI.

[Page 20]13. Theor. To divide a right line in two parts, so that the right, angled figure made of the whole line and one part shall be equal to the square of the other part.

The right line given is AB, upon the same line AB, make a square, as ABCD; and divide the side AD in two equal parts, the midst is M, from M draw a line to B, and produce AD to H, so that MH be e­qual to MB; and upon AH make a square, as AHGF. Then extend GF to E, and then is the right angled figure FC, being made of the whole line FE (which is e­qual to AB) and the part BF, equall to the square of the other part AF, that is, to the square AHGF.

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Forasmuch as by the last aforegoing, the [Page 21] right angled figure comprehended of HD and HA, or the right angled figure of HD and HG, as the figure GHED, with the square of AM, are together equal to the square of HM, being equall to BM: it followeth, that if we take away the square [...]f AM, common to both, that the square [...]f AB, that is, the square ABCD is equal [...]o the right angled figure HGED, and the [...]ommon right angled figure AE being taken from them both, there shall remain the right angled figure FC, equal to the square [...]HFG, which was to be proved.

14 Theor. To divide a right line given by extream and mean proportion.

A right line is said to be divided by an extream and mean proportion, when the whole is to the greater part, as the greater is to the lesse. And thus a right line being divided, as the right line AB is divided in the preceding Diagram in the point F, it is divided in extream and mean proportion; that is, As AB, is to AF: so is AF, to BF.

Demonstration.

Forasmuch as the right lined figure in­cluded with AB and FB, as the figure [Page 22] FBCE is equal to the square of AF, that is, to the square AFGH; it followeth, by the eleventh Theorem of this Chapter, that the line AB is divided in extream and mean proportion; that is, As AB, is to AF: So is AF, to FB.

15 Theor. In all plain Triangles, a line drawn parallel to any of the sides, cut­teth the other two sides proportionally.

As in the plain Triangle ABC, KL be­ing parallel to the base BC, it cutteth off from the side AC one fourth, and also it cutteth off from the side AB one third part: the reason is, because the right line EH cutteth off one third part from the whole space DGFB, & therefore it cutteth off one third part from all the lines that are drawn quite through that space.

And hereupon parallel lines bounded with parallels are equal; as the parallels ED and GH being bounded with the pa­rallels DG and HE are equal, for since the whole lines DB and GF are equall, DE and GH being one fourth part thereof, must needs be equal also.

[Page 23]

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16 Theor. Equiangled Triangles have their sides about the equall angles pro­portionall, and contrarily.

Let ABC and ADE be two plain equi­angled Triangles, so as the angles at B and D, at A and A, and also at C and E be e­qual one to the other; I say, their sides a­bout the equal angles are proportionall; that is,

• 1 As AB, is to BC: So is AD, to ED.
• 2 As AB, is to AC: So is AD, to AE.
• 3 As AC, is to CB: So is AE, to ED.

[Page 24] Demonstration.

Because the angles BAC and DAE are equal by the Proposition; therefore if A + B be applied to AD, AC shall fall in AE; and by such application is this figure made. In which, because that AB and AD do meet together, and also that the angles at B and D are equall, by the Proposition; therefore the other sides BC and DE are parallel; and, by the last aforegoing, BC cutteth the sides AD and AE proportional­ly: and therefore,

As AB, to AD: So is AC, to AE.

[figure]

Moreover, by the point B, let there be drawn the right line BF parallel to the [Page 25] base AE, and it shall cut the other two sides proportionally in the points B and F, and therefore,

1. As AB to AD: so is EF to ED,

Or thus.

As AB to AD: so is CB to ED: because that FE and BC are equal, by the last a­foregoing.

1. Theor. In all right angled plain Tri­angles, the sides including the right an­gle are equal to the the third side.

In the right angled plain triangle ABC, right angled at B, the sides AB and BC are equal in power to the third side AC; that is the squares of the sides AB and BC, to wit, the squares ALMB and BEDC ad­ded together, are equal to the square of the side AC, that is to the square ACKI.

Demonstration.

Let ABC be a triangle, right angled at B, and let the side BC be 3 foot, the side AB 4 foot, and the side AC 5 foot. Let e­very side be squared severally, so shall you finde the square of the side AC to contein as much as the squares of the sides AB and [Page 26]

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BC added together. For, the square of the side AB is 16, the square of BC is 9, which added together make 25, which is equall to the square of the side AC, which was to be demonstrated.

18. Theor. The three angles of a right li­ned Triangle are equal to two right an­gles.

As in the following plain Triangle ABC the three angles ABC, ACB, and CAB are equal to two right angles. Let the side [Page 27] AB be extended to D, and let there be a se­micircle drawn upon the point B, and let there be also dawn a line parallel unto AC, from B unto G.

[figure]

Demonstration.

I say that the angle GBD is equal to the angle BAC, by the 9 th hereof, and the angle CBG is equal to the angle ACB by the same reason, and the angles CBG and GBD, are together equal to the angle CBD, which is also equal to the angle ABC, by the 18 th. of the first: and there­fore; the three angles of a right lined Tri­angle are equal to two right angles, which was to be proved.

19. Theor. If a plain Triangle be inscri­bed in a Circle, the angles opposite to the circumference are halfe as much as that part of the Circumference which is opposite to the angles.

[Page 28] As if in the circle ABC the circumference BC be 120 degrees, then the angle BAC which is opposite to that circumference shall be 60 degrees. The reason is, because the whole circle ABC is 360 degrees, and the three angles of a plain triangle cannot ex­ceed 180 degrees, or two right angles, by the last aforegoing, therefore, as every arch is the one third of 360, so every angle oppo­site to that arch is the one third of 180, that is 60 degrees.

Or thus, From the angle ABC, let there be drawn the diameter BED, and from the center E to the circumference, let there be drawn the two Radii or semidiameters AE and AC, I say then that the divided angles ABD and DBC are the one halfe of the angles AED and DEC: for the an­gles ABE and BAE are equall, because their Radii AE and EB are equall, and also the angle AED is equal to the angles ABE and BAE added together, for if you draw the line EF parallel to AB, the an­gle FED shall be equal to the angle ABE by the 9 th. hereof; and by the like reason the angle AEF is also equal to the angle BAE. and therefore the angle AED is e­qual to the angles ABE, and BAE: or, [Page 29] which is all one, the angle AED is double to the angle ABD.

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In like manner, the angles EBC and ECB are equal, and the angle DEC is e­qual to them both: therefore the angle DEC is double to the angle DBC. Then because the parts of the angle AEC are double to the parts of the angle ABC; therefore also the whole angle AEC is double to the whose angle ABC; and thereupon the angle ABC is half the angle AEC; and consequently, half the arch ADC; is the measure of the angle ABC, as was to be proved. Hence it followeth,

[Page 30] 1. If the side of a plain Triangele in­scribed in a circle be the diameter, the an­gle opposite to that side is a right angle, that is, 90 degrees; for that it is opposite to a semicircle, which is 180 degrees.

2. If divers right lined triangles be in­scribed in the same segment of a circle up­on one base; the angles, in the circumfe­rence are equal. As the two triangles ABD and ACD being inscribed in the same seg­meut of the circle ABCD, upon the same base AD are equiangled in the points B and C, falling in the circumference. For the same arch AD is opposite to both those angles; that is, to the angle ACD, and also to the angle ABD.

20 Theor. If two plaine Triangles inscri­bed in the same segment of a circle, up­on the same base, be so joyned together in the top, (or in the angles falling in the circumference) that thereof is made a four-sided figure, intersected with Di­agonals, the right angled figure made of the Diagonals, is equall to the right an­gled figures made of the opposite sides added together.

Let ABD and ACD be two triangles, [Page 31] inscribed in the same segment of the circle ABCD upon the same base AD so joyned in the top by the right line BC, that there­upon is made the four sided figure ABCD. I say, that the right angled figures made of the opposit sides AB and DC, and also of the sides BC and AD added together are equall to the right angled figure made of the Diagonals AC and BD.

[figure]

Demonstration.

If at the point B you make the angle ABE equall to the angle DBC, and so cut the Diagonall AC into two parts by the right line EB at the point E, then shall the angles ABD and EBC be equall, because [Page 32] the angles ABE and DBC are equal by the proposition, and the angle EBD com­mon to both, and the angles ADB & ECB are equal, because the arch AB is the double measure of them both by the last aforegoing, and therefore the triangles ABD & EBC are equiangled and there sides proportional by the 18 th and 16 th Theoremes of this chapter, that is, as BD to DA, so is BC to CE, and therefore also the rectangles of BD in CE is equal to the rectangle of DA in BC by the 10 th hereof.

And because the angles DBC and ABE are equal by the proposition, and the angles BDC and EAB equall because the arch BC is the double measure of them both by the last aforegoing, the triangles BDC, & EAB are equiangled and there sides propor­tional by the 18 th and 16 Theoremes of this chapter, that is as BD to DC so is AB to AE; and therefore also the rectangle of BD in AE is equal to the rectangle of DC in AB.

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[Page 33] And because the rectangled figure [...] of AD and DF is equall to the two rectan­gled figures of AD in DC and BC in CF, therefore also the rectangled figure of BD in AC is equal to the rectangled figures of BD in AE and BD in EC. From hen [...] and the two former proportions the proposi­tion is thus Demonstrated.

And by Composition.

BD in CE more by BD in AE is equal to DA in BC more by DC in AB, now then because BD in AC is equal to BD in [...] EC more by BD in EA, therefore also BD in AC is equall to DA in BC more by DC in AB which was to be demonstrated.

21. Theor. If two right lines inscribed in a circle cut each other within the circle, the rectangle under the segments of the one, is equall to the rectangle under the segment of the other.

Let the two lines be FD and BC, inter­secting each other in the point A; I say, the triangles ABF and ADC are like, [Page 34] because of their equal angles AFB and ACD, which are equal, because the arch BD is the double measure of them both, and because of their equal angles BAF and DAC, which are equal by the ninth hereof, and where two are equal, the third is eqaul by the 18 aforegoing; therefore AD in AF is equal to AC in AB, which was to be proved.

[figure]

Theor. 22. In a plain right angled Trian­gle, a perpendicular let fall from the right angle upon the Hypothenuse, di­vides the triangle into two triangles, both like to the whole, and to one ano­ther.

[Page 35] The triangle ABC is right angled at B, the hypotenuse or side subtending the right angle is AC, upon which from the point B is drawn the perpendicular BD which divideth the triangle ABC into two triangles, ADB and BDC, each of them like to the whole triangle ABC, and each like to one another also, that is equiangled one to another.

[figure]

Demonstration.

In the triangle ABD, the angles ABD and ADB are equal to the angles ACB and ABC, because of their common angle at A, and their right angles at B and D, and in the triangle CDB, the angle. CBD and BDC are equal to the angles ABC and CAB, because of their common angle at C, and their right angles at B and D; these triangles are therefore each of them like to the whole triangle ABC, and by consequence like to one another.

[Page 36]23. Theor. If two sides of one triangle be e­qual to two sides of another, & the angle comprehended by the equal sides equal, the third side or base of the one, shall be equal to the base of the other, and the re­maining angles of the one equal to the remaining angles of the other.

Of these two triangles CBH and DEF, the sides CB & BH in the one are equal, to DE & EF in the other and the angle CBH equal to the angle DEF, therefore CH in one is equal to DF in the other, for if the base CH be greater then the base CF from CH let be taken.

[figure]

CG equal to DF and let there be drawne the right line BG, now if BC and BG be equal to DE and EF, yet the angle CBG cannot be equal to the angle DEF by the [Page 37] angle GBH which is contrary to the Pro­position, and therefore CH must be equal to DF, and consequently the angle BCH equal to EDF, and CHB equal to DFE which was to be proved.

24. Theor. An Isocles or triangle of two equal sides, hath his angles at the base equal the one to the other, and contrarily.

Let the sides AB and AC in the triangle ABC be equal and produced at pleasure, so that AD may be equal to AE then draw the lines CD and BE, forasmuch as the two sides AD and AC in the triangle DAC are equal to the two sides AE and AB in

[figure]

the triangle ABE, and the angle at A com­mon to both, the base BE shall be equal to the base CD, and the angle at D to the an­gle at E, and the angle ABE to the angle ACD by the last aforegoing, therfore also [Page 38] the angles DCB and EBC are equal, now if you take these equal angles from the equal angles ABE and ACD the angles remaining ABC and ACB must needs be equal, which was to be proved.

25. Theor. If the Radius of a circle be di­vided, in extreame and mean proportion, the greater segment shall be the side of a Decangle, in the same circle.

In the semicircle AGC let AG be the side of a Decangle DG or DA the Radius, then because the arch AG is the tenth part of a circle it is also the fift part of a semicir­cle, and the arch CG, which is four times as much as the arch AG is the double measure of the angle DAG, and because the sides AD and DG are equal, therefore the an­gles AGD and DAG are equal by the last aforegoing, therefore either of the angles

[figure]

AGD or GAD is double to the angle ADG, now then if you divide the angle AGD in­to two equal parts by the right line EG the [Page 39] angles EGD or EGA shall either of them be equal to the angle ADG and therefore ED & EG are equal by the last aforegoing, & the triangles AGD & AEG are equian­gled because of their common angle DAG and there equal angles AGE and ADG, as before, and EG which is equal to DE is equal to AG, therefore as AD to AG (or ED) so is AG to AE, and the Radius AD is divided in extreame and meane propor­tion by the 14 th hereof, and ED the greater segment is the side of a decangle.

These foundations being laid we will proceed to the making of the tables, where­by any triangle may be measured.

Chap. III
of Trigonometria, or the mea­suring of all Triangles.

THe dimension of triangles, is per­formed by the Golden Rule of Arith­metick, which teacheth of four num­bers proportional one to another, any three of them being given, to finde out a fourth.

Therefore for the measuring of all tri­angles [Page 40] there must be certain proportions of all the parts of a triangle one to another, and these proportions must be explained in numbers.

2. And the proportions of all the parts of a triangle one to another cannot be cer­tain unlesse the arches of circles (by which the angles of all triangles, and of Spheri­cal triangles, also the sides are measured) be first reduced into right lines, because the proportions of arches one to another, or of an arch to a right line, is not as yet found out.

3. The arches of every circle are after a sort reduced to right lines, by defining the quantity, which the right lines to them applied have, in respect of Radius, or the Semidiameter of the circle.

4. The arches of a circle thus reduced to right lines are either Chords, Sines, Tangents, or Secants,

5. A Chord or Subtense is a right line in­scribed in a circle, dividing the whole circle into two segments, and in like man­ner subtending both the segments.

6. A chord or subtense is either the great­est or not the greatest.

7. The greatest Subtense is that which divideth the whole circle into two equal [Page 41] segments, as the right line GD, and is also commonly called a diameter.

8. A subtense not the greatest, is that which divideth the whole circle into two unequal segments: and so on the one side subtendeth an arch less then a semicircle; and on the other side subtendeth an arch more then a semicircle, as the right line CK on the one side subtendeth the arch CDK, lesse then a semicircle; and on the other side subtendeth the arch CGK more then a semicircle.

9. A sine is either right or versed.

10. A right sine is the one half of the subtense of the double arch, as the right sine of the arches CD and CG is the right line AC, be [...] half the chord or subtense of the double arches of CD and CG, that is, half of the right line CAK, which sub­tendeth the arches CDK and CGK; whence it is manifest, that the right sine of an arch lesse then a Quadrant, is also the right sine of an arch greater th [...]n a Qua­drant. For as the arch CD is lesse then a Quadrant by the arch CE; so the arch CG doth as much exceed a Quadrant, the right line AC being the right sine unto them both. And hence instead of the ob­tuse angle GBC, which exceeds 90 degrees, [Page 42] we take the acute angle CBA, the comple­ment thereof to 180: and so our Canon of sines doth never exceed a quadrant or 90 d.

[figure]

11. Againe, a right [...] is either Sinus totus, that is the Radius or whole sine, as in the triangle ABC, AC is the Radius, semidiameter, or whole sine, Or else the right sine is the Sinus simpliciter, that is, the first sine, as CA or BA, the one where­of is alwayes the complement of the other to 90 degrees; we usualy call them sine and co-sine.

12. The versed sine of an arch is that part of the diameter, which lieth between the right sine of that arch and the circum­ference. Thus AD is the versed sine of [Page 43] the arch CD, and AG the versed Sine of the arch CEG; therefore of versed Sines some are greater, and some are lesse.

13. A greater versed Sine is the versed Sine of an arch greater then a Quadrant, as AG is the versed Sine of the arch CEG greater then a Quadrant.

14. A lesser versed sine is the versed Sine of an arch lesse then a Quadrant, as AD is the versed sine of the arch CD less then a Quadrant.

15. A tangent of an arch or angle is a right line drawn perpendicular to the Ra­dius or semidiameter of the circle of the triangle, so as that it toucheth the outside of the circumference, And thus the right line FD is the tangent of the arch DC.

16. A secant is a right line proceeding from the center of the circle, and extended through the circumference to the end of the tangent; and thus BF is the Secant of the arch DC.

17. The definition of the quantity which right lines applyed to a circle have, is the making of the Tables of Sines, tan­gents and secants; that is to say, of right Sines and not of versed; for the versed Sines are found by the right without any labour.

18. The lesser versed sine with the sine [Page 44] of the complement is equal to the Rad [...] as the lesser versed sine AD with the right sine of the complement AB is equal to the Radius BD; therefore if you substract the right sine of the complement AB from the Radius BD, the remainder is the ver­sed sine AD.

19. The greater versed sine is equal to the Radius added to the right sine of the excesse of an arch more then a Quadrant, as the greater versed sine AG is equal to the Radius BG with the sine of the excess AC: therefore if you adde the right sine of the excess AB to the Radins BG, you shal have the versed sine of the arch CEG, & so there is no need of the table of versed sines, the right sines may thus be made.

20. The Tables of Sines, Tangents, and Secants may be made to minutes, but may, by the like reason, be made to se­conds, thirds, fourths, or more, if any please to take that paines: for the making whereof the Radius must first be taken of a certain number of parts, and of what-parts soever the Radius be taken, the Sines, Tangents, and Secants are for the most part irrational [...] i [...], that is, they are in­explicable in any true whole numbers or fractions precisely, because there are but [Page 45] few proportional parts to any Radius, [...], whose square root multiplied in it self will produce the number from whence it was taken, with­out some fraction still remaining to it, and therefore the Tables of Sines, Tangents, and Secants cannot be exactly made by any meanes; and yet such may and ought to be made, wherein no number is different from the truth by an integer of those parts, where­of the Radius is taken, as if the Radius be taken of ten Millions, no number of these Tables ought to be different from the truth by one of ten Millions.

That you may attain to this exactnesse, either you must use the fractions, or else take the Radius for the making of the Ta­bles much greater then the true Radius, but to work with whole numbers and fractions is in the calculation very tedious; be­sides here no fractions almost are exqui­sitely true: therefore the Radius for the making of rhese Tables is to be taken so much the more, that there may be no er­rour, in so many of the figures towards the left hand as you would have placed in the Tables; and as for the numbers superflu­ous, they are to be cut off from the right hand towards the left after the ending of [Page 46] the supputation, Thus, to finde the num­bers answering to each degree and minute of the Quadrant to the Radius of 10000000 or ten millions, I adde eight ciphers more, and then my Radius doth consist of sixteen places.

This done, you must next finde out the right Sines of all the arches lesse then a Quadant, in the same parts as the Radi­us is taken of, whatsoever bignesse it be, and from those right Sines the Tangents and secants must be found out.

21. The right Sines in making of the Tables are either primary or secondary. The primarie Sines are those, by which the rest are found, And thus the Radius or whole Sine is the first primary Sine, the which how great or little soever is equall to the side of a six-angled figure inscribed in a circle, that is, to the subtense of 60 de­grees, the which is thus demonstrated.

Let BC be the side of a six angled fi­gure inscribed in a circle, then because the arch BC is the sixt part of a circle, and that every circle is suppoosed to be divided into 360 parts the side BC must needs be 60 parts, because six times 60 makes 360, and the angle BAC is 60 parts also, by the [...] of the first. And the angles ABC and [Page 47]

[figure]

ACB are 120, by the 18 of the second, and are also equal, because the sides AC and AB which are opposite unto them are equal, for they are two Radii, by the work, and therefore either of the angles are 60 parts, and consequently the whole triangle is equiangled, and the whole tri­angle being equiangled, and the sides AB and BC being Radii, the side [...]C must be Radius also. Therefore the Radius or whole Sine is equal to the side of a six an­gled figure inscribed in a circle, as was to be proved.

Out of the Radius or subtense of 60 de­grees the sine of 30 degrees is easily found, [Page 48] the halfe of the subtense being the measure of an angle at the circumference opposite thereunto by the 19 of the second; if therefore your Radius consists of 16 places being 1000.0000.0000.0000. The sine of 30 degrees will be the one half thereof, to wit, 500.0000.0000.0000.

22. The other primary sines are the sines of 60, 45, 36, and of 18 degrees, being the halfe of the subtenses of 120, 90, 72, and of 36 degrees.

23. The subtense of 120 degrees is the side of an equilateral triangle inscribed in a circle, and may thus be found.

The Rule.

Substract the Square of the subtense of 60 degrees, from the Square of the diame­ter, the Square root of what remaineth is the side of an equilateral triangle inscribed in a circle [...] or the subtense of 120 degrees.

The reason of the Rule.

The subtense of an arch with the subtense of the complement thereof to 180 with the diameter, make in the meeting of the two subtenses a right angled triangle. As the subtense AB 60 degrees, with the subtense AC 120 degrees, and the diameter CB, make the right angled triangle ABC, right angled at A, by the 19 of the second. [Page 49] And therfore the sides including the right angle are equal in power to the third side, by the [...] of the second. Therefore the square of AB being taken from the square of CB, there remaineth the square of AC, whose squar root is the subtense of [...] de­grees or the side of an equilateral triangle inscribed in a circle,

[figure]

Example.

Let the diameter CB be 2000.0000. 0000.0000. the square thereof is 400000. 00000.00000.00000.00000.00000. The subtense of AB is 100000.00000.00000. The square thereof is 100000.00000.00000. 00000.00000.00000, which being substract­ed [Page 50] from the square of CB, the remainder is 300000.00000.00000.00000.00000.00000, whose square root 173205.08075.68877. the subtense of 120 degrees.

CONSECTARY.

Hence it followeth, that the subtense of an arch lesse then a Semicircle being given, the subtense of the complement of that arch to a Semicirc [...]e is also given.

24. The Subtense of 90 degrees is the side of a square inscribed in a circle, and may thus be found.

[figure]

The Rule.

Multiply the diameter in it self, and the [Page 51] square root of half the product is the sub­tense of 90 degrees, or the side of a square inscribed in a circle.

The reason of this Rule.

The diagonal lines of a square inscribed in a circle are two diameters, and the right angled figure made of the diagonals is e­qual to the right angled figures made of the opposite sides, by the 20 th. of the second, now because the diagonal lines AB and CD are equal, it is all one, whether I mul­tiply AC by it self, or by the other diagonal CD, the p [...]oduct will be still the same, then because the sides AB, AC, and BC do make a right angled triangle, right angled at C, by the [...] of the second, & that the [...] AC and [...]B are equal by the work, the half of the square of AB must needs be the square of AC or CB, by the 17 th. of the second, whose square rootes the sub­tense of CB, the side of a square or 90 de­gree.

Example.

Let the diameter AB be 200000.00000. 00000, the square thereof is 400000.00000. 00000.00000.00000.00000, the half where­of is 200000.00000.00000.00000.00000. 00000. whose square root 14142 [...].356 [...]3. 73095. is the subtense of 90 degrees, or the [Page 52] side of a square inscribed in a Circle.

25. The subtense of 36 degrees is the side of a decangle, and may thus be found.

The Rule.

Divide the Radius by two, then multiply the Radius by it self, and the half there­of by it self, and from the square root of the summe of these two products substract the half of Radius, what remaineth is the side of a decangle, or the subtense of 36 degrees.

The reason of the rule.

In the following Diagram, let EB repre­sent the Radius of a circle on which draw

[figure]

the square EG, then is GB equal to EB, [Page 53] which being bisected in the point H draw the line HE, then continue the segment HB to K, making HK equal to HE and upon the line KB make the square BD, then the Ra­dius EB is divided into extreame and meane proportion by the 14 th of the second, and the greater segment MB is the side of a decangle by the 25 of the second, and KB is equal thereunto; now then because the Radius EB and the half Radius HB with the right line HE, do make the right angled triangle EBH right angled at B, by the 21 th. of the first, and therefore the squares of EB and BH are together equal to the square of HE or HK, by the 17 th. of the second, now if from the square root of the square of HE, that is from the side HE or HK you deduct the side HB, the remainer is KB the side of a decangle.

For example.

Let the Radius EB be 100000.00000.00000. then is BH, or the half thereof 500000. 00000.00000. the square of EB is 100000 00000.00000.00000.00000.00000. and the square of BH 250000.00000.00000.00000. 00000.00000.00000. The summe of these two squares, viz 125000.00000.00000. 00000, 00000. 00000, is the square of HE [Page 54] or HK, whose square root is 1118033 [...] 88|7 [...]9895, from which deduct the halfe Radi­us BH 500000000000000, and there re­maineth 618033988749895, the right line KB, which is the side of a decangle, or the subtense of 36 degrees.

26 The subtense of 72 degrees is the side of a Pentagon inscribed in a circle, and may thus be sound.

The Rule.

Substract the side of a decangle from the diameter, the remainer multiplied by the Radius, shall be the square of one side of a Pentagon, whose square root shall be the side it self, or subtense of 72 degrees.

The Reason of the Rule.

In the following Diagram let AC be the side of a decangle, equal to CX in the di­ameter, and let the rest of the semicircle be bisected in the point E, then shall either of the right lines AE or EB represent the side of an equilateral pentagon, for AC the side of a decangle subtends an arch of 36 degrees the tenth part of a circle, and therefore AEB the remaining arch of a se­micircle is 144 degrees, the half whereof AE or EB is 72 degrees, the fift part of a circle, or side of an equilateral pentagon, [Page 55] the square whereof is equal to the oblong made of DB and BX.

[figure]

Demonstration.

Draw the right lines EX, ED, and EC, then will the sides of the angles ACE and ECX be equal, because CX is made equal to AC, and EC common to both; and the angles themselves are equal, because they are in equal segments of the same circle by the 19 of the second; and their bases AE and EX are equal by the 23 of the second; and because EX is equal to AE, it is also equal to EB, and so the triangle EXB is equicrural, and so is the triangle EDB, be­cause the sides ED and DB are Radii, and the angles at their bases X and B, E and B, by the 24th. of the second, and because the angles at B is common to both, therefore [Page 56] the two triangles, EXB and EDB are e­quiangled, and their sides proportional, by the 18 th. and 16 th. Theoremes of the se­cond Chapter, that is as DB to EB; so is EB to BX, and the rectangle of DB in BX is equal to the square of EB, whose square root is the side EB, or subtense of 72 de­grees,

[figure]

Example.

Let AC, the side of a decangle or the subtense of 36 degrees, be as before: 618033988749895, which being substract­ed from the diameter BC 200000.00000, 00000. the remainer is XB, 1381966011|151105, which being multiplied by the Ra­dius DB, the product 1381966011251105 00000.00000.0000, shall be the square of EB whose square root 1175570504584946 [Page 57] is the right line EB, the side of a Pentagon or subtense of 72 degrees.

CONSECTARY.

Hence it followes, that the subtense of an arch lesse then a semicircle being given, the subtense of half the complement to a semicircle is given also,

Thus much of the primarie Sines, the secondary Sines or all the Sines remaining may be found by these and the Propositions following

27. The subtenses of any two arches together lesse then a semicircle being given, to finde the subtense of both those arches.

The Rule.

Finde the subtense of their complements to a semicircle, by the 23 hereof; then mul­tiply each subtense given by the subtense of the complement of the other subtense given, the sum of both the products being divided by the diameter, shall be the subtense of both the arches given.

[Page 58] The reason of the Rule.

Let the subtenses of the given arches be the right lines AE and AI, and let the sub­tense of both those arches be the right line EI, let the diameter AO be drawn to the

[figure]

very point in which the subtenses of the given arches do concurre, to wit, in the point A. Then draw the right lines EO and IO, which with the diameter and the subtenses given, do make the two right an­gled Triangles AEO and AIO, right an­gled at E and I (by the 19 th. of the second.) [Page 59] And therefore the sides EO and IO are given by the 23 hereof, and consequently the right angled figures made of AE and IO, AI and EO, to which the right angled figure made of the diagonals EI and AO is equal by the 20 th. of the second, and therefore the summe of the right angled fi­gures made of AE and IO, and also of AI and EO, being divided by the diameter AO, the quotient is EI, the subtense of both the arches given.

Example.

Let AI, the side of a square or subtense of 90 degrees be 141421.35623.73059. And EO, the side of a triangle, or subtense of 120 degrees, 173205.08075.68877, the pro­duct of these two will be 244948974278|3▪ 77659465844164315. Let AE, the side of a sixangled figure, or the subtense of 60 degrees be 100000 00000.00000. And IO, the side of a square, or subtense of 90 de­grees 141421.35623.73059 the product of these two will be 141421.35623.73059. 00000.00000.00000. the summe of these two products 38637033051562726594658|44164315. And this summe divided by the diameter AO, 200000.00000.00000. lea­veth [Page 60] in the quotient for the side EI, or sub­tense of 150 degrees, 1931851652578136. the half whereof 965925826289068, is the Sine of 75 degrees.

28 The subtenses of any two arches lesse then a Semicircle being given, to finde the subtense of the difference of those arch­es,

The Rule.

Finde the subtenses of their complements to a semicircle, by the 23 hereof, as before, then multiply each subtense given, by the subtense of the complement of the other subtense given; the lesser product being substracted from the greater, and their dif­ference divided by the diameter, shall be the subtense of the difference of the arch­es given.

The Reason of the Rule.

Let the subtenses of the given arches be AE and EI, and let the subtense sought be the right line EI; then because the right angled figure made of the diagonals. AI and EO is equal to the right angled figures [Page 61] made of their opposite sides, by the 20 of the second; therefore if I subtract the right angled figure made of AE and IO, from the right angled figure made of AI and EO the remainer will be the right angled figure of AO and EI, which being divided by the diameter AO, leaveth in the quo­tient EI.

Example,

Let the right angled figure AI and

[figure]

EO be the same with the former, viz, 2449489742783177659465844164315. And the right angled figure of AE and IO 1414213562373059. 00000.00000.00000. Their difference shall be 10352761804100|82659465844164315, which divided by the diameter AO, leaveth in the quotient [Page 62] 517638090205041, for the subtense of the difference of the arches of 60 and 90, that is, for the subtense of 30 degrees. The half whereof, viz. 258819045102520, is the sine of 15 degrees

29. The sine of an arch lesse then a Quadrant being given, together with the sine of half his complement, to finde the sine of an arch equal to the commplement of the arch given, and the half comple­ment added together.

The Rule.

Multiply the double of the sine given, by the sine of half his complement, the product divided by the Radius, will leave in the quotient, a number, which being ad­ded to the sine of the half complement shall be the sine of the arch sought.

The reason of the Rule.

Let EAI be a quadrant, and in that let the arches IS, SV, VE be equal, then let the last arch VE be bisected in Y, and let the Quadrant be made into a Semicir­cle, and the arches OC, CL, LE, equall [Page 63] to the former: then shall the right lines LV and CS be parallels to the diameter OAI, and bisected by the Radius AE, and because YV is half of the arch EV, it is also the half of the arch VS or SI, and equal to the arches IB, CG, or GO, Then let there be drawn the right lines EV, LS, CI, and GB, perpendicular to the Radius AY, and bisected by it. I say, then that the

[figure]

right angled triangles IAM, CPM, and SPN are equiangled, for the arches CO, CL, and SI are equal, by the work, and, the double measures of the angles AIM, PCM, and PSN, and the angles AMI CMP, and PNS are equall, that is, right angles, because the right line AY doth fall perpendicularly upon the parallel right [Page 64] lines LS and CI, and now where two an­gles are equal, there the third is equall, by the 18 th. of the second; and consequently the sides of the triangles IAM, CPM, and SPN are proportional.

[figure]

That is, as AI, is to AM: so is CP, to PM; and so is PS, to PN, and then by composition, as AI, AM: so is CS, to MN. Now then let ES be the arch given, and SI the complement thereof to a Qua­drant, then is CG or IB, being equal to EY, the half of the said complement SI, and AM is the Sine thereof, and the Sine of ES is the right line HS, and the double CS, MN is the difference between AM, the Sine of CG or IB, and AN the Sine of SB, and AI is the Radius, and it is al­ready [Page 65] proved, that AI is in proportion to AM, as CS, is to MN, therefore if you multiply AM by SC, and divide the pro­duct by AI, the quotient will be NM, which being added to AM, doth make AN, the Sine or the arch sought.

Example.

Let ES, the arch given, be 84 degrees, and the Sine thereof 9945219, which doubled is 19890438, the Sine of 3 de­grees, the halfe complement is 523360, by which the double Sine of 84 degrees be­ing multiplied, the product will be 104098| [...]9.631680, which divided by the Radius, the quotient will be 10409859, from which also cutting off the last figure, because the Sine of 3 degrees was at first taken too little, and adding the remainer to the Sine of 3 degrees, the aggregate 1564345 is the Sine of 6 degrees, the complement of 84, and of 3 degrees, the halfe comple­ment added together, that is, it is the sine of 9 degrees.

30. The subtense of an arch being gi­ven, to find the subtense of the triple arch.

[Page 66] The Rule.

Multiply the subtense given by thrice Ra­dius square, and from the product substract the cube of the subtense given, what remain­eth shall be the subtense of the triple arch.

The reason of the Rule.

If in a circumference you distinguish three equall parts from O the end of the Diame­ter, with the letters ABC and draw the sub­tenses as in the scheame, making MX equal to MB [...], drawing also AX and AB and the diameter NRA, then shall the triangles BMX and ARO be equicrurall all because RA and RO are two Radii, and MB and MX are equal by the worke, and the angles BMX and ARO are equal by the 19 th of the second; and therefore the triangles BMX and ARO are equiangled by the 23 of the second, and because the sides MB and MX are equal and AM common to both the triangles AMB and AMX, therefore AX is equal to AB by the 24 th of the second, and AB is equal to AO by the work, and there­fore AX is also equal to A [...] ▪ and the angles AXO and AOX are equal by the 24 th of [Page 67] the second, and the riangles AOX and ARO are like, because the angle AOR is common to both & ther [...]o [...]e as AR to AO, so is AO to OX, that is [...] O square divided by Radius, is equal to OX and OS is equal to AO and OX to AS, because the tr [...]angles AOX and AOS are equiangled, the an­gles SAO and AOX are equal because they are the same with two of the angles in the equiangled triangle ARO; and the an­gles AOS and XAO are equal, because

[figure]

they are measured by equal arches, for AC the double of AO, is the double mea­sure of the angle AOS, by the nineteenth of the second, and AO is the measure of [Page 68] ARO equal to XAO, because the trian­gles ARO and AOX are like. And then because AS is equal to OX, SN must needs be equal to MX or MB, and the right ang­led figure made of OS and SC, is equal to the right angled figure made of AS and SN, by the 21 th. of the second, that is, as OS, to NS, so is SA to SC.

[figure]

Now then we have already proved, that the square of AO divided by Radius, is e­qual to OX, and also that OX is equal to SA, and therefore SN is less then twice Radius by the right line AS; or thus, NS is twice Radius less by AO square divided [Page 69] by Radius: and NS multiplied by SA is the same with twice Radius lesse by AO square divided by Radius, multiplied into AO square divided by Radius, and NS mul­tiplied by SA is equal to SC multiplied by OS; and therefore twice Radius less AO square divided by Rad. multiplied by AO square divided by Radius, is equal to SC, multiplied by SO: or thus, 2 Radius less AO square divided by Radius, multiplied into AO square divided by Radius, and di­vided by AO or SO is equal to SC. All the parts of the first side of this Equation are fractions, except AO and the two Ra­dii, as will plainly appear, by setting it down according to the form of Symbolical or spe­cious Arithmetick; thus. [...]. Which being reduced into an im­proper fraction, by multiplying 2 Radius by Radius, the Equation will run thus: [...]

And then these two fractions having one common denominator, they may be redu­ced [Page 70] into one after the manner of vulgar fractions, that is, by multiplying the nume­rators, the product will be a new numera­tor, and by multiplying the denominators the product will be a new denominator; thus multiplying the numerators, 2 Rad. aa − AO aa by the numerator AO aa, the product is 2 Rad. square into AO square, less AO square square, as doth appear by the operation; [...]

And then the denominators being multi­plied by the other, that is, Radius being multiplied by Radius, the product will be Radius aa for a new denominator; and then the Equation will run thus; [...]: but before this fraction can be divided by AO, AO being a whole number, must be reduced into an improper fraction, by subscribing an Unite, and then the Equation will be; [Page 71] [...]. Now as in vulgar fractions, if you multiply the numerator of the dividend by the denominators of the divisor, the pro­duct shall be a new numerator; again, if you multiply the denominator of the divi­dend by the numerator of the divisor, their product shall be a new denominator, and this new fraction is the Quotient sought in this example, the numerator will be still the same, and the denominator will be Ra­dius square multiplied in AO, and the fra­ction will be [...]. And in its least termes it is [...]. In words thu [...]: Twice Radius square multiplied in AO, lesse by the cube of AO divided by Radius square is equal to SC. And by ad­ding AO to both sides of the Equation, it will be, twice Radius square in AO, lesse AO cube divided by Radius square, more AO, is equal to SC more AO, that is, to [Page 72] OC. Here again AO, the last part of the first side of this Equation is a whole num­ber, and must be reduced into an improper fraction, by being multiplied by Radius square, the denominator of the fraction; and then it will be Radius square in AO di­vided by Radius square, which being added to twice Radius square in AO, divided by Radius, the summe will be 3 Radius square in AO divided by Radius square, and the whole Equation [...], the sub­tense of the triple arch.

For Example.

Let AO or AB, 17431. 14854. 95316. the subtense of 10 degrees be the subtense gi­ven, and let the subtense of 30 degrees be required; the Radius of this subtense given consists of 16 places, that is, of a unite and 15 ciphers, and therefore thrice Rad. square is 3, and 30 ciphers thereunto annexed, by which if you multiply the subtense given, the product will be 52293. 44564. 85948. 00000. 00000. 00000. 00000. 00000. 00000. the square of this subtense given is 30384|49397.55837.60253.85793.9856, and the cube 529.63662.80907.48519.77452.00270. 23994. 54977. 14496, which being substracted [Page 73] from the former product, there will remain 51763.80902.05040.51480.22547.99729. 76005.45022.85504. this remainer divi­ded by the square of Radius, will leave in the quotient, 51763.80902.05040. for the subtense of 30 degrees.

31. The subtense of an arch being given, to finde the subtense of the third part of the arch given.

The Rule.

Multiply the subtense given by Radius square, and divide the product by thrice Radius square, substracting in every ope­ration the cube of the figure placed in the quotient from the triple thereof; so shall the quotient in this division be the subtense of the third part of the arch given.

The reason of the Rule.

The reason of the rule is the same with the triple arch, but the manner of working is more troublesome, the which I shall en­deovour to explain by example.

Let there be given the subtense of 30 de­grees, 517638090205040, and let the sub­tense of 10 degrees be required: First, I multiply the subtense given by the square [Page 74] of Radius, that is, I adde 30 ciphers there­unto, and for the better proceeding in the work, I distinguish the subtense given thus inlarged by multiplication into little cubes, setting a point between every third figure or cipher, beginning with the last first, and then the subtense given will stand thus: 517.638.090.205.040.000.000.000.000.000 000.000.000.000.000. And so many points as in this manner are interposed, of so ma­ny places the quotient wil consist, the which in this example is 15, and because here are too many figures to be placed in so narrow a page, we will take so many of them one­ly as will be necessary for our present pur­pose; as namely, the 15 first figures, which being ordered, according to the rules of de­cimal Arithmetick, may be divided into lit­tle cubes, beginning with the first figure, but then you must consider whether the number given to be thus divided be a whole number or a fraction, if it be a whole num­ber, you must set your point after or over the head of the first figure, if it be a fracti­on, place as many ciphers before the fra­ction given, as will make it consist of equal places with the denominator of the Fra­ction given; thus the subtense given being a fraction, part of the supposed Radius [Page 75] of a circle, the which, as hath been said doth consist of 16, and the subtense given but of fifteen, I set a cipher before it, and distinguish that cipher from the subtense given by a point or line, and every third fi­gure after, so will the subtense given be di­stinguished into little cubes, as before. This done, I place my divisor thrice Radius square, that is, 3 with ciphers (or at least supposing ciphers to be thereunto annexed) as in common division under the first fi­gure of the subtense given, that is, as we have now ordered it under the cipher, and ask how often 3 in nought, which being not once, I put a cipher in the margine, and move my divisor a place forwarder, setting it under 5, and ask how often 3 in 5, which being but once, I place one in the quotient, and the triple thereof being 3, I place un­der 3 my divisor, and the cube of the figure placed in the quotient, which in this case is the same with the quotient it self, I set un­der the last figure of the first cube, and sup­posing ciphers to be annexed to the triple root, I substract this cube from it, and there doth remain 299, which is my divisor corrected; with this therefore I see whe­ther I have rightly wrought or not, by ask­ing, how often 299 is contained in the [Page 76] first cube of the subtense given, 517, which being but once, as before, the former work must stand, & this divisor corrected must be subtracted from the first cube in the subtense given, and there will rest 218, and so have I wrote once. To this remainer of the first cube 218, I draw down 638, the figures of the next cube & moving my divisor a place forwar­der, I ask, how often 3 in 21, which being 7 times, I put 7 in the quotient, and under the first figure of this second cube, that is, under 6 I set the triple square of the first figure in the quotient, that is, 3, for the quotient being but one, the square is no more, and the triple thereof is 3; under the second figure of this second cube I set the triple quotient, the which in this exam­ple is likewise 3, and both these added to­gether, do make 33, which being substra­cted from my divisor 3000, there will re­main 2967, for the divisor corrected, and by this also I finde the quotient to be 7, and yet I know not whether my work be right or not, I must therefore proceed, and set the triple of the figure last placed in the Quotient under the first figure of the re­mainer of the first cube, that is, I must set 21, the triple of 7 under 2, the first figure of 218, and now having two figures in the [Page 77] quotient, for distinction sake I call the first a, and the second e, that so the method of the work may the better be seen in the margine, and I set 3 aae, that is, 3, the square of the first figure noted with the let­ter a, viz. 1. multiplied by the second fi­gure, noted with the letter ( e) to wit, 7, under the first figure of the next cube, now the square of ( a) that is, of one is one, and the triple of this square is 3, and 3 times 7 is 21, which is (3 aae) or thrice ( a) square in e, the last figure whereof, to wit, one, I place under 6, the first figure of the next cube 638: next I set (3 aee) that is, three times one multiplied by the square of 7, that is, 3 multiplied by 49, which is 1 [...] under the 2 figure of the cube 638: and last­ly, I set ( eee, that is) the cube of e, that is, the cube of 7, viz. 343, under the last figure of the cube 638, and these 3 sums added toge­ther do make 3 [...]3, which being substracted from the triple root, that is, from 21, suppo­sing ciphers to be thereunto annexed, as be­fore, there will remain [...], and because this may be subtracted from the 2d. cube, & the remainer of the first, I finde that 7 is the true figure to be placed in the quotient, and such a subtraction being made, the remain­er will be 12511, and so have I wrote twice. [Page 78] The work following must be done in all things, as this second, save onely in this particular, that both the figures in the quo­tient are reckoned but as one, which for distinction sake I called a, and the figure to be found by division I called e, and therefore in this third work 3 aa, or thrice a square is the square of 17, that is 289, 3 a or thrice a is 3 times 17, that is, 51, and so of the rest, in the fourth work the three first figures must be called a, in the fifth work the four first figures found, and so forward, till you have finished your di­vision, and therefore this second manner of working being well observed, there can be no difficulty in that which followes.

[Page 79]

[Page 80] [Page 81]

[Page 82] 32. The subtense of an arch being given to finde the subtense of the arch quintuple, or of an arch five times as much.

The Rule.

From the product of the subtense given, multiplied by 5 times Radius square square, subtract the cube of the subtense given mul­tiplied by 5 times Radius square, the squa­red cube of the subtense given being first ad­ded thereunto, the remainer divided by Radius square square, shall leave in the quotient the subtense of the arch quintu­ple, or the arch 5 times as much.

The reason of the Rule.

In the annexed Diagram, twice ET more CB is equall to OE, because OE is the subtense of five equall arches, by the work, and by letting fall the perpendicular CT, the right line OT doth answer to three e­quall arches, AO, AB, and BC; and therefore ET doth answer to the other two: now if you deduct the right line CB from the right line OT, the remainer must be e­quall to ET, and so it followes, that 2 ET + CB = OE. And the triangles OBM and OCT are equiangled, because of their equall angles CTO and MBO, which are [Page 83] both right, and the angles BMO & COT are equal, because they are measured by equal arches; and therefore, as MO is to MB: so is OC, to TO: that is, as hath been shewed in the triplication of an an­gle. As twice Radius, is to twice Radi­us, lesse by the square AO divided by Ra­dius: so is thrice Radius square in AO, less by the cube of AO divided by Radius square, to a fourth number represented by the right line OT, what that number is by the rule of proportion may be thus found:

[figure]

Multiply the numerator of the fractions in the second place by the numerator of [Page 84] the fraction in the third, and their product will be a new numerator, the numerator of the fraction in the second term is 2 Rad. − AO aa And in the third, 3 Rad. aa × AO − AO aaa

That one of these termes may be the better multiplied by the other, the first of the second term, 2 Rad. must be reduced into an improper fraction, by the multipli­cation thereof by Radius, the denomina­tor of that fraction, and then the 2d. term will be 2 Rad. aa − AO aa, and because this second term is the lesse, we will multi­ply the third thereby, the work stands thus: [...]

Thrice Radius square in AO multiplied by twice Radius square, doth make 6 Rad. square squares, and AO cube multiplied by 2 Radius square is 2 Rad. square in AO cube, and because it hath the signe lesse, therefore the first product is 6 Rad. aaaa × AO − 2 Rad. aa × AO aaa. Again, 3 R. aa in AO, multiplied by AO aa, doth make 3 R. aa in AO aaa, & AO aa multiplied [Page 85] by AO aaa, doth make AO aaaaa, & because it hath the sign less, therefore the 2. product is 3 R. square × AO aaa + AO aaaaa, and so both the products will be 6 Rad. square of squares multiplied by AO lesse by 5 Rad. square in AO cube more by AO square cube. And if you multiply Rad. square, the denominator of the third term by Rad. the denominator of the second, the product will be Rad. cube, and the whole product will stand thus, [...]

To divide this product by twice Radius, twice Radius being a whole number must be first reduced into an improper fraction, by subscribing an unite thus, [...]. then if you multiply the numerator of the product by one, the denominator of this fraction, the product will be still the same, and if you multiply the denominator of the product Rad. aaa by 2 Radius, the nume­rator of this improper fraction, the product will be 2 Rad. square square for a new de­nominator, and the Quotient will be [...] [Page 86] the quantity of the right line OT, the double whereof is [...] which is the quantity of the right line OE more by CB, and therefore CB or AO be­ing deducted, the remainer will be the right line OE, which is the quintuplation of an angle, and to this end AO must be reduced into an improper fraction of the same denomination, that is, by multiply­ing thereof by 2 Rad. aaaa, and then the fraction will be [...] and this being deducted from [...] the remainer will be [...]. And this reduced into its least terms, will be [...], which was to be proved.

[Page 87] For example.

Let AO or AB 349048, the subtense of 2 degrees be given, and let the subtense of 10 degrees be demanded, 5 times Radius square square is 50000000.000000.00000 00.0000000. by which if you multiply the subtense given, the product will be 1745240 0000000.0000000.0000000.0000000. The Cube of the subtense given multiplied by 5 times Radius square is 21263045378199|2960.0000000.0000000. the squared cube of the subtense given is 51846392428249|21385360723968, the which being added to the product of 5 Rad.aa in AO, that is, to 212630453781992960.0000000.0000000 the summe will be 212682300174421209|21385360723968. And this being subtract­ed from the product of the subtense given multiplied by 5 times Radius square square, the remainer will be 17431141769982557|879078614639276032, and this remainer divided by Radius square square, that is, cutting off 28 figures, their quotient will be 1743114, the subtense of 10 degrees.

33. The subtense of an arch being given to finde the subtense of the fift part of the arch given.

The Rule.

Divide the subtense given by five roots, [Page 88] lesse 5 cubes, more one Quadrato cube, the quotient shall be the fift part of the arch given.

The reason of the rule depends upon the foregoing Probleme, in which we have pro­ved, that the subtense of five equall arch­es is equall to 5 roots, lesse 5 cubes, more by one quadrato cube, of which 5 roots one of them is the subtense of the fift part of the arch given. And consequently, if I shall divide the subtense of five equall ar­ches by 5 roots, lesse 5 cubes, more one quadrato cube, the quotient shall be the subtense of the fift part of the arch.

The manner of the work is thus: First, consider whether the subtense given to be divided doth consist of equal, or of fewer places then the Radius thereof, if it con­sist of equal places, set a point over the head of the first figure of the subtense gi­ven, if of fewer places, make it equal, by prefixing as many ciphers before the sub­tense given as it wanteth of the number of places of the Radius thereof.

For example.

Let the subtense of 10 degrees be given, viz. 0.17431.14854.95316.34711. This is lesse then the Rad. by one place, and there­fore [Page 89] I have set one cipher before, and have distinguished it from the subtense given by a point set between, the which is all one, as if it had been put over the head thereof: next you must distinguish the subtense gi­ven into little cubes, & into quadratocubes, which may be conveniently done thus; having found the place of the first point, which is alwayes the place of the Radius, the subtense given must be distinguished in­to little cubes, by putting a point under e­very third figure, as in the trisection of an angle: thus in this example the first cubick point will fall under the figure 4, and the subtense given must be distinguished into quadrato cubes, by setting a point over the head, or else between every fift figure from the place of the Radius: thus in this ex­ample the first quadrato cubick point must be set over the head, or after the figure of 1, the second after 4, as here you see.

After this preparation made, you must place your two divisors, 5 roots and 5 cubes in this manner, the first as in ordinary di­vision under the first figure of the subtense given, the other 5 under the first cubick point, and they will stand as in the work you see; then ask how often 5 in one, which being not once, I put a cipher in the [Page 90] quotient, and remove my first divisor a place forwarder, as in ordinary division, but the other 5 I remove to the next cubick point, then, as before, I ask how often 5 in 17, which being 3 times, I set 3 in the quotient, and of this quotient I seek the quadrato cube, and finde it to be 243, the last figure whereof, namely, 3, I set under the last figure of the second quadrato cu­bick point (because there are but 3 figures between my divisor 5 and the first cubick point, whereas there must be alwayes four at the least) then I multiply the figure 3 placed in the quotient by my divisor 5, and the product thereof is 15, the first figure whereof I place under my said divisor 5, to which having annexed ciphers, or at least supposing them to be annexed, (as to the triple root in the trisection) I draw the qua­drato cube of the figure in the quotient, and these 5 roots or 5 quotients into one summe, the which is 1500000243, under this summe I draw a line, so have we five roots more one quadrato cube, from which I must subtract 5 cubes, I therefore seek the cube of 3, the figure placed in the quotient, and finde it to be 27, which multiplied by 5, the product will be 135, the last figure of these five cubes, viz. 5, I set under my se­cond [Page 91] 5 or cubick divisor, and substracting these 5 cubes from the 5 roots more one quadrato cube, the remainer will be [...], which remainer being also substra­cted from the figures of the subtense given standing over the head thereof, the remain­er of the subtense given will be 244464611, and so have I wrought once.

To this remainer of the two first quadra­to cubes, I draw down 95316, the figures of the next quadrato cube, and setting my first divisor a place forwarder, I ask how often 5 in 24, which being four times, I set 4 in the quotient, not knowing yet whether this be the true quotient or not, but with this I pro­ceed to correct my divisor, and first I seek the quadrato quadrat of 3, the first quoti­ent, and finde it to be 81, this multiplied by 5, will make 405, this product I set un­der my divisor, and 5 the last figure thereof I set under 9, the first figure of the 3 qua­drato quadrate; next I seek the cube of 3, & finde it to be 27, which being multiplied by 10, the product will be 270, and this I set a place forwarder under the former product: thirdly, I seek the square of 3 which is 9, and this multiplied by 10 is 90, which I set a place forwarder under the se­cond product 270. Lastly, I multiply 3, [Page 92] the figure in the quotient by 5 my divisor, this product which is 15, I set a place for­warder under 90, the third product, and now these 4 products together with my di­visor and ciphers thereunto annexed, being gathered into one summe, will be 500000|432915, under which I draw a line. And thrice the square of 3, multiplied by 5, which is 135, I set under this summe, the last figure thereof 5, under the first figure of the third cubick point, that is, under 4, and the triple of 3 multiplied by 5, which is 45, I set under the former summe 135, a place forwarder, and my cubick divisor 5 under the last summe a place forwarder, that is, under the third cubick point, these drawn into one summe will be 13955, and being substracted from the former summe 500000432915, the remainer 498.60493. + 2915 is my divisor corrected, and yet I know not whether I have a true quotient or not; under this remainer therefore I draw a line, and work with 4, which I sup­pose to be the true quotient in manner fol­lowing; and that the manner of the work may be the more perspicuous, (as in the trisection of an angle, so here) 3 the first figure found I call ( a) and 4 the second figure I call ( e) the square of three I note [Page 93] with aa, the cube with aaa, the quadrate quadrat with aaaa, the quadrato cube with aaaaa, so likewise the square of 4 the second figure I note with ee, the cube with eee, the quadrato quadrate with eeee, the quadrato cube with eeeee; my first divisor I note with ffff, because this Equation is qua­drato quadratick, and 5 my second divisor, I note with cc, because the divisor it self is cubick: these things premised, I proceed thus: First, I multiply 405, which is 5 aaaa or 5 times the quadrato quadrate of 3 by e, that is, by 4, and the product thereof 1620, I set under my divisor corrected, so as the last figure thereof may stand under the first figure of the third quadrato cubick num­ber, and against this number I put in the margine 5 aaaae, that is, five times the qua­drato quadrate of 3 multiplied by 4: next 270, ten times the cube of 3, by 16 the square of 4, and this product 4320, I set un­der the former a place forwarder, and 90, which is 10 times the square of 3, I multi­ply by 64, the cube of 4, & this product 5760 I set under the last a place forwarder then that, and 15, which is 5 times 3, I multiply by 256, the quadrato quadrate of 4, & the product thereof 3840, I set under the third product a place forwarder, and 1024, [Page 94] the quadrato cube of four under that: lastly, I multiply four, the last figure placed in the quotient by 5 my divisor, and the last figure of this product I set under 5 my divisor, and supposing ciphers to be there­unto annexed, I collect these several pro­ducts into one summe, and their aggreagate 20000021135424, is five roots more one quadrato quadrate, under which I draw a line, and seek the five cubes to be substract­ed, thus.

First, I multiply 135 (which is thrice the square of three multiplied by five my cu­bick divisor) by four, the figure last pla­ced in the quotient, and the product there­of 540 I set under the last summe, so as the last figure thereof may be under the first fi­gure of the third cube; next I multiply 45 that is, five times the triple of three, by 16 the square of four, and this product 720 I set under the former a place forwarder, and under that 320, which is five times the cube of 4, a place forwarder too, these pro­ducts drawn into one summe do make 61520, the five cubes to the substracted from the five roots more one quadrato qua­drate before found, which being done, the remainer will be 19938501135424, and this remainer being substracted from the [Page 95] figures of the subtense given over the head thereof, the remainer will be 450.79600. 59892, and because such a substraction may be conveniently made, I conclude, that I have found the true quotient, and so have I wrought twice.

The work following must be done in all things like as this second, onely remem­ber that as in the trisection of an angle, both the figures in the quotient are termed a in the third operation, the three figures found are a in the fourth work; and so forward till your division be finished.

[Page 96] [Page 97] [Page 98] [Page 99] [Page 100]

[Page 101] 34 The Sines of two arches equally di­stant on both sides from 60 degrees, being given, to finde the Sine of the distance.

The Rule.

Take the difference of the Sines given, and that difference shall be the Sine of the arch sought.

The reason of the Rule.

Let CN and PN be the two arches gi­ven, and equally distant from 60 deg. MN, that is equally distant on both sides from the point M. And let the right lines CK and PL be the Sines of those arches, being drawn perpendicular to the right line AN, and thereupon parallel to one another.

Moreover, let the right line PT be drawn perpendicular upon the right line CK, and so parallel to the right line KL, then this right line TP cutteth from the right line CK another line TK, equal unto PL, by the 15 of the second, and leaveth the right line TC for the difference of the Sines CK and PL. Lastly, the Sines of the distance of either of them from 60 de­grees let be the right line CD or DP, I say, that the right line TC is equal to the right line CD or DP.

Demonstration.

Because in the triangle GCP, that the [Page 102] perpendicular GD doth bisect the base CP, by the proposition: therefore the sides GC and GP are equall, and the angles GCP and GPC are equal, because equal sides subtend equal angles: and lastly, the angles CGD and DGP are also equal, by the same reason; but the angle CGD

[figure]

is 30 degrees, for that it is equal to the an­gle BAM, because a [...] right line drawn through two parallel right lines maketh the angles opposite to one another equall. And therefo [...]e the angle CGP is 60 de­grees, because it is double to the angle CGD. And because the angle CGP is [...]0 degrees, therefore the other two angles GCP and GPC are 120, by the 18 th. of [Page 103] the second, and these two angles are de­monstrated to be equall; and therefore e­very of them is 60 degrees. And the angle CGP is also 60 degrees, and therefore the triangle CGP is equiangled, but because the triangle CGP is equiangled, therefore also it is equilateral. Moreover, because the triangle CGP is equilateral, therefore the perpendicular PT bisecteth the base CG into two equall parts, or else it could not be perpendicular. Then the sides CP and CG are equall, and therefore also their bi-segments CT and CD are equal: which was to be demonstrated. The Sines therefore or whatsoever 60 degrees being given, you may finde the Sines of the other 30 degrees, by Addition or Substraction onely.

Example.

Let the arches CN be 70 degrees, PN 50, CM or PM 10 degrees; for so many degrees are the arches of 70 degrees; and 50 degrees distant from the arch of 60 de­grees on both sides. And let first the Sines of 70 degrees and 10 degrees be given, and let the Sine of 50 degrees be demanded.

[Page 104]

Then let the Sine of 70 degrees and 50 degrees be given, and let the Sine of ten degrees be demanded.

Lastly, let the Sines of 50 degrees and 10 degrees be given, and let the Sine of 70 degrees be demanded.

And thus far of the making of the Ta­bles of right Sines, the Tables of versed Sines are not necessary, as hath been said

CHAP. IV.
By the Tables of Sines to make the Tables of Tangents and Secants.

1. AS the Sine of the complement, Is to the Sine of an arch: so is the Radius, to the tangent of that arch.

2. As the Sine of the complement, is to the Radius; so is the Radius, to the secant of that arch. For, by the 16 th. of the se­cond:

• 1. As the Sine of the complement AB, is to the Sine CA: so is the Radius BD or BC, to DF the tangent.
• 2. As the sine of the complement AB, is to the Radius BD or BC: so is the Ra­dius BC, to the secant BF.

Example.

Let the tangent and secant of the arch CD 30 degrees be sought for. The sine AC 30 degrees is 5000000, the sine of the [Page 106] complement AB 60 degrees is 8660254. Now then if you multiply the sine AC 5000000, by the Radius CB 10000000, the product wil be 50000000000000, which divided by the sine of the complement AB 8660254: the quotient will be 5773503, the right line FD or the tangent of the arch of 30 degrees.

[figure]

2. As the sine of the complement AB 8660254, Is to the Radius DB 10000000: so is the Radius BC 10000000, to FB, the secant of the arch of 30 degrees: and so for any other: but with more ease by the help of these Theorems following.

CHAP. V.
Of the nature and construction of Logarithmes.

LOgarithmes are borrowed numbers, which differ amongst themselves by Arithmetical proportion, as the num­bers that borrow them differ by Geometrical [Page 114] [...] [Page 115] [...] [Page 116] proportion: So in the first column of the ensuing Table the numbers Geometrically proportional being 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, &c. you may assigne unto them for bo [...]rowed numbers or Loga­rithmes, the numbers subscribed under the letters A, B, C, D, or any other at plea­sure; provided, that the Logarithmes so assigned still differ amongst themselves by Arithmetical proportion, as the numbers of the first column differ by Geometrical pro­portion: For example. In the column C, if you will appoint 5 to be the Logarithme of one, 8 the Logarithme of 2, and 11 the Logarithme of 4, 14 must needs be the Logarithme of 8, the next proportional, because the numbers 5, 8, 11, and 14 differ amongst themselves by Arithmetical pro­portion, as 1, 2, 4, and 8 (the proportional numbers unto which they are respectively assigned) differ by Geometrical proportion, that is, as the numbers 5, 8, 11, and 14 have equal differences: so the numbers 1, 2, 4, and 8 have their differences of the same kinde: for as the difference between 5 and 8, 8 and 11, 11 and 14, is 3: so in the other numbers, as 1 is half 2, so 2 is half 4, 4 half 8, &c. The same observation may be made of the Logarithmes placed in [Page 117] the columns, A, B, and D, or of any other numbers which you shall assigne as Loga­rithmes unto any rank of numbers, which are Geometrically proportional, and these Logarithmes or borrowed numbers you may propound to increase, and to be conti­nued upwards, as those of the columnes A, B, C, or otherwise to decrease, and to be continued downwards, as those of the column D.

[Page 118] The numbers continually proportional, which Mr. Briggs (after a conference had with the Lord Nepeir) hath proposed to himself in the Calculation of his C [...]ili [...]des, are 1, 10, 100, 1000, &c. to which numbers he hath assigned for Logarithmes 000, &c. 1000, and 2000, and 3000, that is to say, to 1, the Logarithme 0.000, and to 10, the Logarithme 1,000, and to 100 the Loga­rithme 2.000, as in the table following you may perceive. In the column marked by the letter A, there is a rank of numbers continually proportional from 1, and over against each number his respective Loga­rithme in the other column, signed by the letter B.

Having thus assigned the Logarithme to the proportional numbers of 1, 10, 100, 1000, &c. in the next place, it is requisite [Page 119] to finde the Logarithmes of the mean num­bers situate amongst those proportionals of the same table, viz. of 2, 3, 4, &c. which are numbers scituate betwixt 1 and 10, of 11, 12, 13, &c. which are placed betwixt 10 and 100; and so consequently of the rest: wherefore how this also may be done we intend to explain by that which followeth.

1. §. Make choice of one of the propo­tional numbers in the Table AB, and by a continued extraction of the square root create a rank of continuall meanes betwixt that number and 1, in such sort, that the continuall mean which cometh nearest 1 may be a mixt number, lesse then 2, and so near 1, that it may have as many ciphers before the significant figures of the nume­rator, as you intend that the Logarithmes of your Table shall consist of places.

Example.

In the premised Table AB, I take 10, the second proportional of that Table, then annexing unto it a compent company of ci­phers, as twenty and four, thirty and six, fourty and eight, or any other number at pleasure; onely observe, that the more ci­phers you annex unto the number given, the more just and exact the operation will prove; to make the Logarithmes of a [Page 120] Table to seven places 28 ciphers will be sufficient, they being therefore added to 10, I extract the square root thereof, and finde it to be 3.16227766016837; again, annex­ing unto this root thus found 14 ciphers more, and working by that entire number so ordered, as if it were a whole number, I extract the root thereof, which I finde to be 1.77827941003892: and so proceeding successively by a continued extraction, I produce 27 square roots, or continual means betwixt 10 and 1, and write them down in the first column of the Table hereunto an­nexed, in which you may observe, that the three last numbers marked by the letters G, H, and L, viz.

• 1.00000006862238
• 1.00000003431119
• 1.00000001715559

are each of them mixt numbers lesse then 2, and greater then 1, and likewise to have seven ciphers placed bef [...]re the significant figures of their numerators, according to the true meaning and intention of this pre­sent rule.

2. §. Having thus produced a great company of continual meanes, annex unto them their proper Logarithmes, by half­ing first the Logarithme of the number ta­ken, [Page 121] and then successively the Logarithme of the rest.

For example.

1.000000000000000 being assigned the Logarithme of 10, the number taken 0.500000, &c. marked by the letter D, in the second column of the following Table, which is the half of 1.0000, &c. is the Logarithme of the number A, the square root of 10: in like manner 0.25000, &c. being half 0.5000, and is the Logarithme of the number B, and 0.125000, &c. is the Logarithme of the number C, and so of the rest in their order. So that at last, as you have in the first column of the following Table 27 continuall meanes, betwixt 10 and 1, as aforesaid: So in the other co­lumn you have to each of those continuall meanes, his respective Logarihme.

3. §. When a number which being lesse then 2, and greater then 1, comes so neer to 1, that it hath seven ciphers placed be­fore the significant figures of the numera­tor, the first seven significant figures of the numerator of such a number, and the first seven significant figures of the numerator of his square root lessen themselves like their Logarithmes, that is, by halfes.

This is proved by the Table following; [Page 122] for there in the second column thereof, the number N being the Logarithme of the number G, I say, as the Logarithme K is half the Logarithme N, so 3431119, the first seven figures of the numerator of the number H, are half 6862238, the first seven significant figures of the numerator of the number G. Any two numbers of this kinde therefore being given, their Logarithmes and the significant figures of their nume­rators are proportional.

Example. The numerators G and H be­ing given, I say, as 6862238, the significant figures of the numerator of the number G; are to 3431119, the significant figures of the numerator of the number H; so is 29802322, the Logarithme of the number G, to 140901161, the Logarithme of the number H. In like manner, G and L being given, as 6862238, is to 1715559, so is 29802322, the Logarithme of the number G, to 7450580, the Logarithme of the number L. This holdeth also true in any other number of this kinde, though it be not one of the continual means betwixt 10 and 1, for the significant figures of the nu­merator of any such number bear the same proportion to his proper Logarithme, that the significant figures of any of the numbers marked by the letters G, H, or L bear to his.

[Page 123]

[Page 124] 4. §. These things being thus cleared, it is manifest, that a number of this kinde being given, the Logarithme thereof may be found by the Rule of three direct. For as the significant figures of the numerator of any one of the numbers (signed in the first column of the last Table by the letters G, H, or L) are to his respective Loga­rithme: so are the significant figure of the numerator of the number given, to the Logarithme of the same number.

Example. The number 1.00000001021301 being given, I demand the Logarithme thereof: I say then,

As 6862238, the significant figures of the numerator of the number G, are to 2980|2322, the logarithme of the same number G: so are 1021301, the significant figures of the numerator of the number given, to 4357281, the Logarithme sought; before which if you prefix 9 ciphers, to the intent it may have as many places as the Loga­rithme in the last premised Table, ( viz. 16) the true and entire Logarithme of 1.00000|001021301, the number given is 0.000000|004357281, as before. And to every Lo­garithme thus found, you must prefix as many ciphers as will make the said Loga­rithme to have as many places as the other [Page 125] Logarithmes in the same table: for though you make your Table of Logarithmes to consist of as many places as you please, yet when you are once resolved of how many places the Logarithmes of your Ta­ble shall consist, you must not alter your first resolution, as to make the Logarithme of 2 to consist of six places, and the Loga­rithme of 16 to have seven, but if the sig­nificant figures of the numerator of the Logarithme of 2 have not so many places as the significant figures of the Logarithme of 16, you must prefix a cipher or ciphers to make them equal; because (as hath been said, the Logarithmes of this kinde ought all to consist of equal places in the same Table.

5. §. Now then to finde the Logarithme of any number whatsoever, you are first to search out so many continual means be­twixt the same number and 1, till the con­tinual mean that cometh neerest 1 hath as many ciphers placed before the significant figures of his numerator, as you intend the Logarithmes of your Table shall consist of places; Again, this being done, you are to finde the Logarithme of that conti­nual mean: And lastly, by often doubling and redoubling of that Logarithme so [Page 126] found (according to the number of the continual meanes produced) in conclusion you shall fall upon the Logarithme of the number given.

Example. the number 2 being given, I demand the Logarithme thereof to seven places: Here first in imitation of that which is before taught in the first rule of this Chapter, I produce so many continual meanes between 2 and 1, till that which cometh nearest 1 hath seven ciphers before the significant figures of the numerator, which after three and twenty continued extractions, I finde to be 1.00000008262958 This continual mean being thus found (by the direction of the last rule aforegoing) I finde the Logarithme thereof to be 0.000000035885571. for,

• As 6862238, is to 29802322:
• So 8262958, is to 35885571.

This Logarithme being doubled will produce the Logarithme of the continual mean next above 1.00000008262958, and so by doubling successively the Logarithme of each continual mean one after another, according to the number of the extractions ( viz. three and twenty times in all) at last you shall happen upon the Logarithme 0.301029987975168, which is the Loga­rithme [Page 127] of 2 the number propounded: The whole frame of the work is plainly set down in the table following; for in the first column thereof you have 23 continu­al meanes betwixt 2 and 1, and in the o­ther column their respective Logarithmes, found by a continual doubling and redou­bling of 0.000000035885571, the Loga­rithme of the last continual mean in the table.

[Page 128]

[Page 129] But now because the Logarithme of the number propounded was to con [...]ist onely of seven places; therefore of the Logarithme so found I take onely the first seven figures rejecting the rest as superfluous, and then at the last the proper Logarithme of 2, the number given will be found to be 0.301029, and because the eighth figure being 9, doth almost carry the value of an unit to the same seventh figure, I adde one thereto, and then the precise Logarithme of 2 will be 0.301030. And thus as the Logarithme of 2 is made, so may you likewise make the Logarithme of any other number what­soever: Howbeit, the Logarithmes of some few of the prime numbers being thus discovered, the Logarithmes of ma­ny other derivative numbers may be found out afterwards without the trouble of so many continued extractions of the square root, as shall appear by that which fol­lowes.

6. §. When of four numbers given, the second exceeds the first as much as the fourth exceeds the third; the summe of the first and fourth is equal to the summe of the second and third; and contrarily.

As 8, 5: 6, 3. here 8 exceeds 5, as much as 6 exceeds 3: therefore the summe of [Page 130] the first and fourth, namely, of 8 and 3 is equall to the summe of the second and third; namely of 5 and 6: from whence necessarily followes this Corollary; ‘When four numbers are proportionall, the summe of the Logarithmes of the mean numbers is equal to the summe of the Logarithmes of the extreams.’

Example.

Let the four proportional numbers be those exprest in the first column of the first Table in this Chapter, viz. 4, 16, 32, 128, in which Table the Logarithme of 4 under the letter A is 3, the Logarithme of 16, 5, the Logarithme of 32, 6; and the Loga­rithme of 128 is 8. Now as the summe of 5 and 6, the Logarithmes of the mean numbers do make 11, so the summe of 3 and 8, the Logarithmes of the extreames, do make 11 also.

7. §. When four numbers be proportio­nal, the Logarithme of the first substracted from the summe of the Logarithmes of the second and third, leaveth the Logarithme of the fourth.

Example.

Let the proportion be, as 128, to 32; so is 16, to a fourth number: here adding 5 and 6, the Logarithmes of the second and [Page 131] third, the sum is 11, from which substract­ing 8, the Logarithme of 128, the first pro­portional, the remainer is 3, the Logarithm of 4, the fourth proportional.

8. §. If instead of substracting the a­foresaid Logarithme of the first, we adde his complement arithmetical to any num­ber, the totall abating that number, is as much as the remainer would have been.

The complement arithmetical of one number to another, (as here we take it) is that, which makes that first number equall to the other; thus the complement arith­metical of 8 to 10 is 2, because 8 and 2 are 10. Now then whereas in the example of the last Proposition, substracting 8 from 11, there remained 3, if instead of substracting 8, we adde his complement arithmeticall to 10, which is 2, the totall is 13, from which abating 10, there remains 3, as be­fore: both the operations stand thus:

from which abate 10, there remaines 3, and the like is to be understood of any other.

[Page 132] The reason is manifest, for whereas we should have abated 8 out of 11, we did not onely not abate it, but added moreover his complement to 10, which is 2, wherefore the total is more then if should be by 8 & 2, that is by 10; wherefore abating 10 from it, we have the Logarithme desired; which rule, although it be generall, yet we shall seldome have occasion to use any other complements, then such as are the comple­ments of the Logarithmes given either to 10,000000, or to 20,000000, the [...] comple­ment arithmetical of any Logarithme to either of these numbers, is that which makes the Logarithme given equal to either of them. Thus the complement arithmetical of the Logarithme of 2 viz. 0301030, is 9698970, because these two numbers added together, do make 10.000000, and thus the complement thereof to 20.000000is 1969|8970: if therefore 0301030 be substracted from 10.000000, the remainer is his com­plement arithmetical.

But to finde it readily, you may instead of substracting the Logarithme given from 10.000000, write the complement of every figure thereof unto 9, beginning with the first figure toward the left hand, and so on, till you come to the last figure towards the [Page 133] right hand, and thereof set down the resi­due unto 10. Thus for the complement a­rithmetical of the aforesaid Logarithme, 0301030; I write for 0, 9: for 3, 6: for 0, 9: for 1, 8: for 0, 9: for 3 again I should write 6: but because the last place of the Logarithme is a cipher, and that I must write the complement thereof to 10, instead of 6 I write 7, and for 0, 0: and so have I this number, 9698970, which is the complement arithmetical of 0301030, as before.

9. § Every Logarithme hath his proper Characteristick, and the Character or Cha­racteristicall root of every Logarithme is the first figure or figures towards the left hand, distinguished from the rest by a point or comma. Thus the Character of the Lo­garithmes of every number lesse then 10 is 0, but the Character of the Logarithme of 10 is 1; and so of all other numbers to 100, but the Character of the Logarithme of 100 is 2; and so of the rest to 1000; and the Character of the Logarithme of 1000 is 3; and so of the rest to 10000: in brief, the Characteristick of any Loga­rithme must consist of a unite lesse then the given number consisteth of digits or pla­ces, And therefore by the Character of a [Page 134] Logarithme you may know of how many places the absolute number answering to that Logarithme doth consist.

10. §. If one number multiply another, the summe of their Logarithme is equal to the Logarithme of the product.

As let the two numbers multiplied toge­ther be 2, and 2 the products is 4, I say then that the summe of the Logarithmes of 2 and 2, or the Logarithme of 2 doubled is equal to the Logarithme of 4, as here you may see.

Again, let the two numbers multiplied together be 2, and 4, the product is 8, I say then that the summe of the Logarithmes of 2 and 4 is equall to the Logarithme of 8, as here you may also see,

And so for any other.

The reason is, for that (by the ground of multiplication) as unit is in proportion to the multiplier: so is the multiplicand, to [Page 135] the product: therefore (by the sixth of this Chapter) the sum of the Logarithmes of a unit, and of the product is equall to the summe of the Logarithmes of the mul­tiplier and multiplicand, but the Loga­rithme of a unit is 0, therefore the Loga­rithme of the product alone is equal to the summe of the Logarithmes of the multi­plier and multiplicand.

And by the like reason, it three or more numbers be multiplied together, the summe of all their Logarithmes is equall to the Logarithme of the product of them all.

11. §. If one number divide another, the Logarithme of the Divisor being sub­stracted from the Logari [...]hme of the Di­vidend, leaveth the Logarithme of the Quotient.

As let 10 be divided by 2, the quotient is 5. I say then, if the Logarithme of 2 be substracted from the Logarithme of 10, there will remain the Logarithme of 5, as here is to be seen.

For seeing that the quotient multiplied by the divisor produceth the dividend, [Page 136] therefore, by the last proposition, the sum of the Logarithmes of the quotient and of the divisor is equal to the Logarithme of the divi [...] if therefore the Logarithme of the divid [...]ol, be substracted from the Logarithme of the divi [...] there remaines the Logarithme of the quotient.

12. §. In any continued rank of num­bers Geometrically proportionall from 1, the Logarithme of any one of them being divided by the denomination of the pow­er which it challengeth in the same rank, the quotient will give you the Logarithme of the root. In the rank of the proporti­onal numbers of the Table ABCD, 2 be­ing the root, or first power; 4 the square or second power, 8 the cube, or third power, 16 the bi-quadrate or fourth, 32 the fifth power, 64▪ the sixth power, &c. I say, the Logarithme of 4, 8, 16, 32, 64, or of any of the other subsequent proportionals in that rank, being divided by the demonina­tion of the power that the same proportio­nal claimeth in the same rank, you shall finde in the quotient the Logarithme of 2 the root.

For example.

In the same Table the Logarithme of 4. the square or second power, viz. 3. being [Page 137] given, I demand the Logarithme of 2, the root: here the denomination of the power that the proportional 4 challengeth in that rank (being the square or second power) is 2, wherefore if 3, the Logarithme of 4 be divided by 2, the quotient will be 1, and there will remain 1 for a fraction; so that you see it cometh very near in the Loga­rithmes of but one figure, but if you take it to seven places, as in this table is intend­ed, you shall finde it exactly: for then the Logarithme of 4 will be 0.602060, and this being divided by 2, the quotient will be 0.301030, the Logarithme of 2 the root. So likewise 0.903090, the Logarithme of 8 the third power, being divided by 3, leaves 0.301030 in the quotient, as before, and so of any other.

13. §. In any rank of numbers Geome­trically proportionall from 1, the Loga­rithme of the root being multiplied by the denomination of any of the powers, the product is the Logarithme of the same power.

This Rule is the inverse of the last.

For example.

In the rank produced in the last rule 0.301030, (the Logarithme of 2 the root) being doubled, or multiplied by 2, produ­ceth [Page 138] 0.602060, the Logarithme of 4, the square or second power, and the same Lo­garithme of 0.301030, being trebled or multiplied by 3, produceth 0.903090, the Logarithme of 8, the cube or third power, and so of the rest.

The truth of these two last rules may thus be proved. In arithmeticall proportion, when the first term is the common diffe­rence of the terms, the last term being di­vided by the number of the terms, the quotient will give you the first term of the rank: again, in this case, the first term multiplied by the number of the termes produceth the last term. So this rank 3, 6, 9, 12, 15, 18, 21 being propounded, wherein three is both the first term and also the common difference of the terms: I say, 21, the last term being divided by 7, the number of the termes, the quotient is 3, the first term. Contrariwise, 3 the first term multiplied by 7, produceth 21, the last term; and by the like reason, 0.301030 being the first term, and also the common difference of the termes, that is, of the Logarithmes of 4, 8, 16, 32 and 64, the Logarithme of 2 the first term, being multiplied by 6, the number of the terms, produceth the Logarithme of 64, the last term, and the [Page 139] garithme of 64, the last term, being divi­ded by 6, leaveth in the quotient the Loga­rithme of 2 the root.

Hence it also followes, that if you adde the Logarithme of 2, the common diffe­rence of the termes, to the Logarithme of any term, their aggregate shall be the Lo­garithme of the next term. Thus if I adde 0.301030, the Logarithme of 2 the root or first term, to 0.903090, the Logarithme of 8, the third term, their aggregate is 1.204|120, the Logarithme of 16, the fourth term; and so of the rest.

14. §. Thus having shewed the constru­ction of the Logarithmetical Tables, the converting of the Table of natural Sines, Tangents, and Secants into artificiall can­not be difficult, the artificiall Sines and Tangents being nothing but the Loga­rithmes of the naturall.

15. §. In the conversion whereof Mr. Briggs in his Trigonometria Britannica, thought fit to make the Radius of his na­tural Canon to consist of 16 places, and to confine his artificiall to the Radius of ele­ven, whose Characteristick is 10, but the Characteristick of the rest of the Sines till you come to the sine of 5 degrees and 73 centesmes is 9, and from thence to 57 [Page 140] centesmes, the Characteristick is 8, and from thence 7, till you come to 5 cen­tensmes, and from thence but 6, to the be­ginning of the Canon. The Characteri­stick still decreasing in the same proportion with the naturall numbers, and the number of the places in the naturall Ca­non, do therefore exceed the Characteri­stick in the artificiall, that so the artifi­ciall numbers might be the more ex­act.

16. §. In the Canon herewith printed, the Characteristick in the artificiall num­bers doth exceed the number of places in the naturall, which is not done so much out of necessity as conveniency, for the artifi­ciall numbers in this Canon might in all respects have been made answerable to the natural, and so the Characteristick of the Radius, or whole Sine would have been seven, the Characterick of the first minute 3, but thus the subduction of the Radius would not have been so ready as now it is, nor yet the Canon it self altogether so exact, and therefore as Master Briggs confined the Radius of his artificiall Ca­non to eleven places for conveniency sake, though he made the Logarithmes to [Page 141] the Radius of sixteen: so here for conve­niency and exactnesse both, the same Characterick is here continued, though the naturall numbers do not require it, if any think this a defect, I answer, that it could not well be avoided here, but may be supplied by Master Briggs his Canon, of which this is an abbreviation: and yet even here there is so small a difference between the Logarithmes of these natu­rall numbers, and the Logarithmes in the Canon, that any one may well per­ceive the one to be nothing else but the Logarithme of the other, if they do but change the Characteristick.

17. §. The Sine of an arch and half the Radius are mean proportionals be­tween the Sine of halfe that arch, and the Sine complement of the same half.

In the annexed Diagram, let DE be [Page 142] the sine of 56 degrees, BC the sine of 28, AC the sine complement thereof, that is, of 62. DB the subtense of 56. CF per­pendicular to the Radius, then are ABC and ACF like triangles, by the 22 of the second, and their sides proportional that is,

[figure]

And therefore the oblongs of BC×AC, AO × DE, and AB × CF are equal, and the sides of equal rectangled figures reci­procally proportional, that is, as BC, AO ∷ DE, AC. or as AO, BC ∷ AC, DE.

If therefore you multiply AO, the half [Page 143] Radius, by DE, the sine of the arch given, and divide the product by BC, the sine of half the arch given, the quotient shall be AC, the sine complement of half the gi­ven arch.

Or if you multiply BC, the sine of an arch by AC, the sine complement of the same arch, and divide the product by AO, the half Radius, the quotient shall be DE, the sine of the double arch. And therefore the sines of 45 degrees being given, or the Logarithmes of those sines, the rest may be found by the rule of proportion. For il­lustration sake we will adde an example in naturall and artificiall numbers.

Naturall,

Logarith.

18. §. The composition of the naturall Tangents and Secants, by the first and se­cond [Page 144] of the fourth are thus to be made.

1. As the sine of the complement, is to the sine of an arch: So is the Radius, to the tangent of that arch.

2. As the sine of the complement, is to the Radius: so is the Radius, to the Se­cant of that arch; and by the same rules may be also made the artificiall; but with more ease, as by example it will appear.

Let the tangent of 30 degrees be sought.

And thus having made the artificiall Tangents of 45 degrees, the other 45 are but the arithmeticall complements of the former, taken as hath been shewed in the eighth rule of the fifth Chapter.

Again, let the secant of 30 degrees be sought. [Page 145]

And thus the Radius being added to the arithmetical complement of the sine of an arch, their aggregate is the secant of the complement of that arch. And this is suf­ficient for the construction of the naturall and artificiall Canon. How to finde the Sine, Tangent or Secant of any arch given in the Canon herewith printed, shall be shewen in the Preface thereunto: here followeth the use of the naturall and arti­ficiall numbers both; first, in the resolving any Triangle, and then in Astronomy, Dialling, and Navigation.

CHAP. VI.
The use of the Tables of natural and artificial Sines, and Tan­gents, and the Table of Logarithmes.
In the Dimension

CHAP. VII.
Of Sphericall Triangles.

A Sphericall Triangle is a figure de­scribed upon a Sphericall or round superficies, consisting of three arch­es of the greatest circles that can be descri­bed upon it, every one being lesse then a Semicircle.

2. The greatest circles of a round or Spherical superficies are those which divide the whole Sphere equally into two Hemi­spheres, and are every where distant from [Page 176] their own centers by a Quadrant, or fourth part of a great circle.

3. A great circle of the Sphere passing through the poles or centers of another great circle, cut one another at right an­gles.

4. A spherical angle is measured by the arch of a great circle described from the angular point betwixt the sides of the tri­angle, those sides being continued to qua­drants.

5. The sides of a Spherical triangle may be turned into angles, and the angles into sides, the complements of the greatest side or greatest angle to a Semicircle, being ta­ken in each conversion.

It will be necessary to demonstrate this, which is of so frequent use in Trigonometry. In the annexed Diagram let ABC be a sphericall triangle, obtuse angled at B, let DE be the measure of the angle at A. Let FG be the measure of the acute angle at B, (which is the complement of the obtuse angle B, being the greatest angle in the gi­ven triangle) and let HI be the measure of the angle at C, KL is equal to the arch DE, because KD and LE are Quadrants, and their common complement is LD. LM is equall to the arch FG, because LG and [Page 177] FM are Quadrants, and their common complement is LF. KM is equal to the arch HI, because KI and MH are Qua­drants, and their common complement is KH. Therefore the sides of the triangle KLM are equal to the angles of the trian­gle ABC, taking for the greatest angle ABC, the complement thereof FBG.

[figure]

And by the like reason it may be demon­strated, that the sides of the triangle ABC are equal to the angles of the triangle KLM. For the side AC is equall to the arch DI, being the measure of the angle DKI, which is the complement of the ob­tuse angle MKL. The side AB is equall to [Page 178] the arch OP, being the measure of the an­gle MLK. And lastly, the side BC is equal to the arch FH, being the measure of the angle LMK, for AD and CI are Qua­drants: so are AP and OB, BF and CH. And CD, AO, and CF are the common complements of two of those arches. Therefore the sides of a spherical triangle may be changed into angles, and the an­gles into sides, which was to be demonstra­ted.

6. The three sides of any spherical tri­angle are lesse then two Semicircles.

7. The three angles of a spherical tri­angle are greater then two right angles, and therefore two angles being known, the third is not known by consequence, as in plain triangles.

8. If a spherical triangle have one or more right angles, it is called a right an­gled spherical triangle.

9. If a spherical triangle have one or more of his sides quadrants, it is called a quadrantal triangle.

10. If it have neither right angle, nor any side a quadrant, it is called an oblique spherical triangle.

11. Two oblique angles of a spherical triangle are either of them of the same [Page 179] kinde of which their opposite sides are.

12. If any angle of a triangle be neerer to a quadrant then his opposite side: two sides of that triangle shall be of one kinde, and the third lesse then a quadrant.

13. But if any side of a triangle be near­er to a quadrant then his opposite angle, two angles of that triangle shall be of one kinde, and the third greater then a qua­drant.

14. If a spherical triangle be both right angled and quadrantal, the sides thereof are equall to the opposite angles.

For if it have three right angles, the three sides are quadrants, if it have two right angles, the two sides subtending them are quadrants; if it have one right angle, and one side a quadrant, it hath two right angles and two quadrantal sides, as is evi­dent by the third Proposition. But if two sides be quadrants, the third measureth their contained angle, by the fourth pro­position. Therefore for the solution of these kindes of triangles, there needs no further rule: But for the solution of right angled, quadrantall, and oblique sphe­rical triangles there are other affections proper to them, which are necessary to be known as well as these general affections [Page 180] common to all spherical triangles. The affections proper to right angled and qua­drantal triangles we will speak of first.

CHAP. VIII.
Of the affections of right angled Sphericall Triangles.

IN all spherical rectangled Triangles, having the same acute angle at the base: The sines of the hypothenusals are pro­portional to the sines of their perpēdiculars. As in the annexed diagram, let ADB re­present a spherical triangle, right angled at B: so that AD is the sine of the hypo­thenusal, AB the sine of the base, and DB is the perpendicular. Then is DAB the angle at the base, and IH the sine, and LM the tangent thereof: Also DF is the sine of the perpendicular DB, and KB is the tangent thereof: I say then, As AD, is to FD: So is AI, to IH, by the 16 th. The­oreme of the second Chapter.

And because it is all one, whether of the mean proportionals be put in the second [Page 181] place; therefore I may say: As AD, the sine of the hypothenusal, is in proportion to AI Radius: So is FD, the sine of the perpendicular, to IH the sine of the angle at the base.

[figure]

2. In all rectangled spherical triangles, having the same acute angle at the base. The sines of the bases, and the tangents of the perpendiculars are proportional.

For as AB, to KB; so is AM, to ML, by the 16 th. Theorem of the second Cha­pter: or which is all one; As AB, the sine of the base, is in proportion to AM Ra­dius: so is BK, the tangent of the perpen­dicular, to ML, the tangent of the angle at the base.

[Page 182] 3. If [...] circles of the Sphere be so ordered, that the first intersect the second, the second the third, the third the fourth, the fourth the fift, and the fift the fift at right angles: the right angled triangles made by their in­tersections do all consist of the same circu­lar parts.

As in this Scheme, let IGAB be the first circle, BLF the second, FEC the third, GAD the fourth, HLEI the fift.

[figure]

Then do these five circles retain the con­ditions required. The first intersecting the second in B, the second the third in F, the third the fourth in C, the fourth the [Page 185] fift in H, the fift the angle, we mark or note intersections at B, F, [...] to a quadrant. As angles; therefore I say [...]nt as the comple­triangles made by the inte [...]or AD we write circles; namely, ABD, D [...] write compl. EGI, and GCA do all co [...]d AB be­same circular parts; for the circu [...] [...] in every of these triangles are, as h [...]d by peareth.

Where you may observe, that the side AB in the first triangle is equal to compl. HLD in the second, or compl. ELF in the third, or IG in the fourth, or com. GA in the fift; and so of the rest.

To expresse this more plainly, AB in the first triangle is the complement of the angle HLD in the second, or the comple­ment of the angle ELF in the third, or the side IG in the fourth, or the complement of the hypothenusal GA in the fift. And from these premises is deduced this univer­sall proposition.

[Page 184] 4. The sine of the middle part and Radius are reciprocally proportional, with the tangents of the extreams conjunct, and with the co-sines of the ex­treams disjunct.

Namely; As the Radius, to the tangent of one of the extreames conjoyned: so is tangent of the other extream conjoyned, to the sine of the middle part.

And also; As the Radius, to the co-sine of one of the extreams dis-joyned: so the co-sine of the other extream dis-joyned, to the sine of the middle part.

Therefore if the middle part be sought, the Radius must be in the first place, if ei­ther of the extreams; the other extream must be in the first place.

For the better Demonstration hereof, it is first to be understood, that a right an­gled Spherical Triangle hath five parts be­sides the right angle. As the triangle ABD in the former Diagram, right angled at B, hath first, the side AB: secondly, the an­gle at A: thirdly, the hypothenusal AD: fourthly, the angle ADB: fifthly, the side DB. Three of these parts which are far­thest [Page 185] from the right angle, we mark or no [...]e by their complements to a quadrant. As the angle BAD we account as the comple­ment to the same angle. For AD we write comp. AD, and for ADB we write compl. ADB. But the two sides DB and AB be­ing next to the right angle, [...] are not noted by their complements. Of these five parts, two are alwayes given to finde a third, and of these three one is in the middle, and the other two are extreams either adjacent to that middle one, or opposite to it. If the parts given and required are all conjoyned together, the middle is the middle part con­junct, and the extreams the extream parts conjunct. If again any of the parts given or required be dis-joyned, that which stands by it self is the middle part dis-joyned, and the extreames are extream parts dis-joyned. Thus, if there were given in the triangle ABD, the side AB, the angle at A, to finde the hypothenusal AD, there the angle at A is in the middle, and the sides AD and AB are adjacent to it; and there­fore the middle part is called the middle conjunct, and the extreames are the ex­treames conjunct; but if there were given the side AB, the hypothenusal AD, to finde [Page 186] the angle at D, here AB is the middle part dis-junct, because it is dis-joyned from the side AD by the angle at A, and from the angle at D by the side DB, for the right angle is not reckoned among the circular parts, and here the extreams are extreams dis-junct.

These things premised, we come now to demonstrate the proposition it self, consist­ing of two parts: first, we will prove, that the sine of the middle part and Radius are proportional with the tangents of the ex­treams conjunct.

The middle part is either one of the sides, or one of the oblique angles, or the hypothenusal.

CASE 1.

Let the middle part be a side, as in the right angled spherical triangle ABD of the last diagram, let the perpendicular AB be the middle part, the base DB and comp. A the extreame conjunct, then I say, that the rectangle of the sine of AB and Radius is equal to the rectangle of the tangent of DB, and the tangent of the complement of DAB: for, by the second proposition of this Chapter, As the sine of AB, is in proportion to Radius: so is the tangent of [Page 187] DB, to the tangent of the angle at A. Therefore if you put the third term in the second place, it will be, as the sine of AB, to the tangent of DB: so is the Radius, to the tangent of the angle at A. But Ra­dius is a mean proportional between the tangent of an arch, and the tangent of the complement of the same arch, by the Co­rollary of the first reason of the second Axiome of plain Triangles: and therefore as Radius, is to the tangent of the angle at A; so is the tangent complement of the same angle at A unto Radius: Therefore as the sine of AB is in proportion to the tangent of DB; so is the co-tangent of the angle at A, to Radius: and therefore the rectangle of AB▪ Radius, is equall to the rectangle of the tangent of DB, and the co-tangent of the angle at A.

CASE 2.

Let the middle part be an angle, as in the triangle DHL of the former Diagram, and let compl. HLD be the middle part, HL and compl. LD the extreames conjunct; then I say, that the rectangle made of the co-sine of HLD and Radius, is equal to the rectangle of the tangent of HL and the co [...]tangent of LD. For▪ by the third [Page 188] proposition of this Chapter, compl. HLD is equal to AB, and compl. LD to DB, and HL to compl. DAB; and here we have proved before, that the rectangle of the sine of AB and Radius, is equal to the rectangle of the tangent of DB, and the co-tangent of the angle at A; therefore also the rectangle of the co-sine of HLD and Radius, is equal to the rectangle of the co-tangent of LD, and the [...] tangent of HL.

CASE 3.

Let the middle part be the hypothenusal, as in the triangle GCA, let compl. AG be the middle part, compl. AGC, and compl. CAG the extreams conjunct; then I say, that the rectangle of the co-sine of▪ AG and Radius, is equal to the rectangle of the co-tangent of AGC, and the co-tangent of CAG: for we have proved before, that the rectangle of the sine of AB and Radius is equal to the rectangle of the tan­gent of DB and the co-tangent of DAB, but, by the third proposition of this Cha­pter, compl. AG is equal to AB, compl. AGC to DB, and compl. CAG to compl, DAB; therefore also the rectangle of the co-sine of AG and Radius, is equal to the [Page 189] rectangle of the co-tangent of AGC and the co-tangent of CAG, which was to be proved.

It is further to be proved, that the sine of the middle part and Radius are propor­tional with the co-sines of the extreams dis-junct. Here also the middle part is ei­ther one of the sides, or the hypothenusal, or one of the oblique angles.

CASE 1.

Let the middle part be a side: as in the triangle ABD, let DB be the middle part, compl. AD and compl. A the opposite ex­treams: then I say, that the rectangle of the sine of BD and Radius is equal to the rectangle of the sine of AD, and the sine of the angle at A; for, by the first propo­sition of this Chapter, as the sine of AD, is to Radius; so is the sine of DB, to the sine of the angle at A. Therefore, the rectangle of the sine of DB and Radius, is equal to the rectangle of the sine of AD and the sine of the angle at A.

CASE 2.

Let the hypothenusal be the middle part; as in the triangle DHL, let compl. LD [Page 190] be the middle part, DH and HL the ex­treams dis-junct. Then I say, that the rectangle of the co-sine of LD and Ra­dius is equal to the rectangle of the co-sine of DH and the co-sine of HL: for compl. LD is equal to DB, and DH is equall to compl. AD, and HL to compl. DAB, by the third proposition of this Chapter: therefore the rectangle of the co-sine of LD and Radius, is equal to the rectangle of the co-sine of DH and the co-sine of HL.

CASE 3.

Let one of the oblique angles be the middle part, as in the triangle IEG, let compl. IGE be the middle part: then I say, that the rectangle of the co-sine of IGE and Radius is equal to the rectangle of the sine of GEI and the co-sine of IE: for compl. IGE is equall to DB, and GEI is equal to AD, and EI to compl. DAB.

5. In any Spherical triangle, the sines of the sides are proportional to the sines of their opposite angles.

Let ABC be a spherical triangle, right angled at C, then let the sides AB, AC, and CB be continued to make the qua­drants AE, AF, and CD, and from the [Page 191] pole of the quadrant AF, to wit, from the point D, let be drawn down the other qua­drants DF and DH; so there is made three new triangles BDE, GDE, and the obliquangled triangle BDG. I say, in the right angled triangle ABC, that the sine of the side AB is in proportion to the sine of his opposite angle ACB: as the sine of the side AC, is to his opposite angle ABC; or as BC, to BAC: likewise in the obli­quangled spherical triangle BDG, I say, that as BG, is to BDG: so is BD, to BGD; or so is DG, to DBG.

[figure]

For first, in the right angled triangle ABC, the angle ACB and the arch AE are of the same quantity, to wit, quadrants, so likewise the angle BAC and the arch [Page 192] EF are of the same quantity, it being the measure of the said angle. Now then as AB, to AE; so is BC, to EF, by the first proposition of this Chapter: therefore al­so as AB, to ACB; so is BC, to BAC. Then in the obliquangled Triangle BDG, because, by the demonstration of right an­gled triangles, they are as DB, to DEB; so is DE, to DBE: and as DG, to DEG▪ so is DE, to DGE, or to DGB. Therefore changing of the proportional termes, it shall be, as DG, to DB: so is DBE, or DBG, to DGB, which was to be demon­strated.

These foundations being thus laid, the businesse of right angled spherical trian­gles is easily dispatcht. And the proporti­ons to be used in every case may be disco­vered either by the first, second and fift pro­positions; or by the fourth proposition on­ly. The severall cases in a right angled sphericall triangle are sixteen in number, whereof six may be resolved by the first pro­position: seven by the second, and three by the fift; an example in each will suffice.

In the triangle ABC, let there be given the hypothenusal AB, and the perpendicu­lar BC, to finde the base AC; then by the first proposition, the Analogie is,

[Page 193] As the co-sine of the perpendicular, is to Radius: so is the co-sine of the hypothe­nusal, to the co-sine of the base.

2. Let there be given the base AC, and the angle at the base BAC, to finde the perpendicular BC, by the second proposi­tion, the analogie is:

As Radius, to the sine of the base; so is the tangent of the angle at the base, to the tangent of the perpendicular.

3. Let there be given the hypothenusal AB, the angle at the base BAC, to finde the perpendicular BC, by the fifth propo­sition, the analogie is:

As Radius, to the sine of the hypothenu­sal: so is the sine of the angle at the base, to the sine of the perpendicular: and so of the rest.

By the fourth or universall Proposition, the proportions for right angled sphericall triangles may be found two wayes:

First, by the equality of the Sines and Tangents of the circular parts of a trian­gle, that is, of the Logarithmes of the na­tural, thus by the universal proposition in the aforesaid triangle ABC, the hypo­thenusal AB, and the angles at AM and B being noted by their complements, I say.

• 1. The sine of AC added to Radius; is [Page 194] equal to the sine of AB added to the sine of the angle at [...].
• 2. The cosine of A added to Radius is equal to the co-sine of BC added to the sine of the angle at B.
• 3. The co-sine of AB added to Radius, is equall to the co-sine of AC added to to the co-sine of BC.
• 4. The co-sine of AB added to Radius is equal to the co-tangent of A, added to the co-tangent of the angle at B.
• 5. The cosine of the angle at B added to Radius is equal to the tangent of BC, ad­ded to the cotangent of AB.
• 6. The sine of BC added to Radius is equal to the co-tangent of the angle at B added to the tangent of AC.

And thus he that listeth may set down the equality of the sines and tangents of the other sides and angles, and so there will be ten in all; but these may here suf­fice: for to these may the sixteen cases of a right angled spherical triangle be reduced; namely, three to the first, three to the se­cond, two to the third, two to the fourth, three to the fift, and three to the sixt.

As admit there were given the hypothe­nusal BA, and the angle at B, to finde the base AC; then, by the first, seeing that [Page 195] the sine of AB added to the sine of the an­gle at B, is equal to the sine of AC added to Radius. Therefore, if working by na­tural numbers I multiply the sine of AB by the sine of B, and divide the product by Radius, the remainer will be the sine of AC: and working by Logarithmes, if from the summe of the sines of AB and B I substract Radius, the rest is the sine of AC.

Secondly, admit there were given AB and AC, to finde B, then seeing that the sine of AC added to Radius is equal to the sines of AB and B. Therefore, if working by naturall numbers I multiply the sine of AC by Radius, and divide the product by AB, the remainer is the sine of B. Or working by Logarithmes, if from the sum of the sines of AC and Radius, I substract the sine of AB, the remainer will be the sine of B.

Or thirdly, if there were given AC and the angle at B, to finde AB: then foras­much as AC and Radius is equal to the sines of AB and B, therefore if working by natural numbers I multiply AC by the Radius, and divide the product by the sine of B, the remainer is the sine of AB. Or working by Logarithmes, if from the sine [Page 196] of AC and Radius, I substract the sine of B the remainer is the sine of AB: an so of the rest.

Which that you may the better perceive, I have here added in expresse words, the Canons or rules of the proportions of the things given and required in every of the sixteen cases of a right angled sphericall triangle, as they are collected from the Catholick Proposition. And here the side subtending the right angle we call the hy­pothenusal, the other two containing the right angle we may call the sides; but for further distinction, we call one of these containing sides (it matters not which) the base, and the other the perpendicular.

The base an angle at the base given, to finde

• 1. The Perpendicular.] As Radius, to the sine of the base; so is the tangent of the angle at the base, to the tangent of the perpendicular.
• 2. Angle at the perpendicular.] As Radi­us, to the co-sine of the base; so the sine of the angle at the base, to the co-sine of the angle at the perpendicular.
• 3. Hypothenusal.] As Radius, to the co-sine of the angle at the base: so the co­tangent [Page 197] of the base, to the co-tangent of the hypothenusal.

The perpendicular and angle at the base given, to finde

• 4. Angle at perpend.] As the co-sine of the perpendicular, to Radius; so the co-sine of the angle at the base, to the sine of the angle at the perpendicular.
• 5. Hypothenusal.] As the sine of the an­gle at the base, to Radius; so the sine of the perpendicular, to the sine of the hypo­thenusal.
• 6. The Base.] As Radius, to the co-tangent of the angle at the base; so is the tangent of the perpendicular, to the sine of the base.

The hypothenusal and angle at the base given, to finde

• 7. The base.] As Radius, to the co-sine of the angle at the base; so the tangent of the hypothenusal, to the tangent of the base.
• 8. Perpendicular.] As Radius, to the sine of the hypothenusal, so the sine of the angle at the base, to the sine of the perpen­dicular.
• 9. Angle at perpend.] As Radius, to the [Page 198] co-sine of the hypothenasal; so the tan­gent of the angle at the base, to the co-tan­gent of the angle at the perpendicular.

The base and perpendicular given, to finde

• 10. Hypothenusal.] As Radius, to the co-sine of the perpendicular: so the co-sine of the base, to the co-sine of the hypo­thenusal.
• 11. Angle at the base] As Radius, to the sine of the base: so is the co-tangent of the perpendicular, to the co-tangent of the an­gle at the base.

The base and hypothenusal given, to finde the

• 12. Perpendicular.] As the co-sine of the base, to Radius; so the co-sine of the hypothenusal, to the co-sine of the perpen­dicular.
• 13. Angle at the base.] As Radius, to the tangent of the base; so the co-tangent of the hypothenusal, to the co-sine of the angle at the base.
• 14. Angle at the perpend.] As the sine of the hypothenusal, to Radius; so the sine of the base, to the sine of the angle at the per­pendicular.

[Page 199] The angles at the base and perpendicular given, to finde

• 15. The perpendicular.] As the sine of the angle at the perpendicular, is to Ra­dius: so the co-sine of the angle at the base, to the co-sine of the perpendicular.
• 16. The hypothenusal.] As Radius, to co-tangent of the angle at the perpendicular; so the co-tangent of the angle at the base, to the co-sine of the hypothenusal.

Secondly, the proportions of all the ca­ses of a right angled spherical triangle, may by the aforesaid Catholick Proposition be known thus: If the middle part be sought, put the Radius in the first place; if either of the extreams, the other extream put in the first place.

And note, that when a complement in the proposition doth chance to concur with a complement in the circular parts, you must take the sine it self, or the tangent it self, because the co-sine of the co-sine is the sine, and the co-tangent of the co-tan­gent is the tangent.

As in the following triangle ABC, let there be given the base AB, and the angle at C, to finde the hypothenusal BC. Here [Page 200] AB is the middle part, BC and C are the opposite extreams, or the extreams dis­junct. Now because the extream BC is sought, therefore I must put the other ex­tream, that is, the angle at C, in the first place; and because that angle, as also the side sought are noted by their complements, therefore I must not say: As the co-sine of the angle at C, is to Radius: so is the sine of the base AB, to the co-sine of the hy­pothenusal BC: but thus;

As the sine of the angle at the perpendi­cular ACB, is to Radius; so is the sine of the base AB, to the sine of the hypothenu­sal BC. The like is to be understood of the rest.

Thus much concerning right angled sphe­rical triangles: as for Quadrantal there needs not much be said, because the circu­lar parts of a quadrantal triangle, are the same with the circular parts of a right an­gled triangle adjoyning.

As let ABC be a triangle, right angled at A, and let one of the sides thereof; namely, AC be extended, till it become a quadrant, that is to D; then draw an arch from D to B; then is DBC a quadrantal triangle, to which there is a right angled triangle adjoyning, as ABC. I say there­fore [Page 201] that the circular parts of the qua­drantal triangle BCD are the same with the circular parts of the right angled tri­angle ABC: for the circular parts of ei­ther of them are as here appeareth.

[figure]

The five circular parts of the triangle.

• ABC are AC AB co ABC cō B [...] cō BCA
• BCD are com CD CDB DBC cō BC cō BCD

Where it is evident, that AD and DB being quadrants, DBA is a right angle, and BA is the measure of the angle at D, [Page 202] so that the side AC in the one is equall to compl. CD in the other: and the side AB in the one is equal to the angle BDC in the other: and compl. ABC in the one is equal to DBC in the other, and compl. BC in the one is the same with BC in the other: and lastly, compl. BCA in the one is the same with compl. DCB in the other; for the compl. of the acute angle A [...] unto a quadrant is also the comple­ment of the obtuse angle BCD, and the circular parts of both triangles being the same, it followes, that that which is here proved touching right angled triangles is also true of quadrantal. And all the six­teen cases thereof may also be resolved by the aforesaid Catholick Proposition.

As let there be given the side DC, and the angle at C, to finde the angle at D, then is the side DC the middle, and the angles at D and C are extreams adjacent; now because the angle at D, one of the ex­treams is sought, we must put the other ex­tream, to wit, the angle at C in the first place, and that is noted by its complement: and therefore the Analogie is▪

As the co-tangent of the angle at C, to Radius; so the co-sine of DC, to the tan­gent of the angle at D: and so of the rest; [Page 203] and what is said of the addition of the ar­tificial numbers is to be understood of the rectangles of the natural.

CHAP. IX.
Of Oblique angled Sphericall Triangles.

IN an obliquangled spherical triangle, there are twelve Cases; two whereof, that is, those wherein the things given and required are opposite, may be resolved by the fift proposition of the last Chapter.

CHAP. I.
Of the Tables of the Suns motion, and of the equation of time for the difference of Me­ridians.

WHereas it is requisite that the Rea­der should be acquainted with the Sphere, before he enter upon the practise of Spherical Trigonome­tri, the which is fully explained in Blunde­viles Exercises, or Ch [...]lmades translation of Hues on the Globes, to whom I refer those that are not yet acquainted therewith: that which I here intend is to shew the use of Trigonometrie in the actuall resolution of so me known Triangles of the Sphere.

[Page 250] And because the Suns place or distance from the next Equinoctial point is usually one of the three terms given in Astronomi­cal Questions, I will first shew how to com­pute that by Tables calculated in Decimal numbers according to the Hypothesis of Bullialdus, and for the Meridian of London, whose Longitude reckoned from the Cana­rie or Fortunate Islands is 21 deg. and the Latitude, North, 51 deg. 57 parts (min.) or centesms of a degree.

Nor are these Tables so confined to this Meridian, but that they may be reduced to any other: If the place be East of London, adde to the time given, but if it be West make substraction, according to the diffe­rence of Longitude, allowing 15 deg. for an houre, and 6 minutes or centesms of an houre to one degree, so will the sum or dif­ference be the time aequated to the Meridi­an of London, and for the more speedy ef­fecting of the said Reduction, I have added a Catalogue of many of the chiefest Towns and Cities in diverse Regions, with their Latitudes and difference of Meridians from London in time, together with the notes of Addition and Substraction, the use whereof is thus.

[Page 251] Suppose the time of the Suns enterance into Taurus were at London Aprill the 10th. 1654, at 11 of the clock and 16 centesms before noon, and it be required to reduce the same to the Meridian of Vraniburge, I therefore seeke Vraniburge in the Cata­logue of Cities and Places, against which I finde 83 with the letter A annexed, there­fore I conclude, that the Sun did that day at Uraniburge enter into Taurus at 11 of the clock and 99 min. or centesms before noon, and so of any other.

CHAP. II.
THE ART OF SHADOWS: Commonly called DIALLING. Plainly shewing out of the Sphere, the true ground and reason of making all kinde of Dials that any plain is capable of.