A TREATISE OF Angular Sections By JOHN WALLIS D. D. Professor of Geometry in the University of Oxford and a Member of the Royal Society LONDON LONDON Printed by Iohn Playford for Richard Davis Bookseller in the University of OXFORD 1684. A TREATISE OF Angular Sections CHAP. 1. Of the Duplication and Bisection of an ARCH or ANGLE I. LET the Chord or Subtense of an Arch proposed be called A or E of the Double B of the Treble C of the Quadruple D of the Quintruple F c. The Radius R the Diameter 2R But sometimes we shall give the name of the Subtense A E c. to the Arch whose Subtense it is yet with that care as not to be liable to a mistake II. Where the Subtense of an Arch is A let the Versed sine be V where that is E let this be U. Which drawn into or Multiplied by the remainder of the Diameter 2 R − V makes 2 R VVq the Square of the Right-sine this Sine being a Mean-proportional between the Segments of the Diameter on which it stands erect by 13 ◠6. That is Q ½ B the Square of the Right-sine or half the Subtense of the double Arch That is 2 R V − Vq = Q ½ B = ¼ Bq. III. If to this we add Vq the Square of the Versed-sine it makes 2 RV = ¼ Bq + Vq = Aq. And by the same reason 2 R U = Eq. That is IV. The Subtense of an Arch is a Mean Proportional between the Diameter and the Versed-sine V. Again because 2 R V = Aq therefore dividing both by 2 R And the Square thereof Which subtracted from Aq leaves the Square of the Right-sine And in like manner and and That is VI. If from the Square of the Subtense we take its Biquadrate divided by the Square of the Diameter the Remainder is equal to the Square of the Right-sine And the Square-root of that Remainder to the Sine it self And the double of this to the Subtense of the double Arch. VII Accordingly because therefore its Quadruple and And in like manner That is VIII If from Four-times the Square of a Subtense are taken its Biquadrate divided by the Square of the Radius the Remainder is the Square of the Subtense of the double Arch And the Quadratick Root of that Remainder is the Subtense it self IX But That is X. The Rect-angle of the Subtense of an Arch and of its Remainder to a Semicircle divided by the Radius is equal to the Subtense of the double Arch. XI Because therefore AE = RB = 2R × ½B And R. A B. B And therefore because A E contain a Right-angle as being an Angle in a Semicircle XII In a Right-angled Triangle the Rect-angle of the two Legs containing the Right-angle is equal to that of the Hypothenuse and the Perpendicular from the Right-angle thereupon And XIII As the Radius to the Subtense of an Arch so the Subtense of its Remainder to a Semicircle is to that of the double Arch. XIV Because B the Subtense of a double Arch doth indifferently subtend the two Segments which compleat the whole Circumference and consequently the half of either may be the single Arch of this double It is therefore necessary that this Equation have two Affirmative Roots the greater of which we will call A and the lesser E And therefore That is XV. Any Arch and its Remainder to a Semicircumference as also its excess above a Semicircumference and either of them increased by one or more Semicircumferences will have the same Subtense of the double Arch. For in all these Cases the Subtense of the single Arch will be either A or E. XVI Because And therefore and 4Aq Rq − Aqq = Bq Rq = 4Eq Rq − Eqq Therefore by Transposition 4Aq Rq − 4Eq Rq = Aqq − Eqq and dividing both by That is XVII The Square of the Diameter is equal to the difference of the Biquadrates of the Subtenses of two Arches which together complete a Semicircumference divided by the difference of their Squares And this also equal to the sum of the Squares of those Subtenses That is because A E contain a Right-angle XVIII In a Right-angled Triangle the Square of the Hypothenuse 4Rq is equal to the Squares of the sides containing the Right-angle Aq+Eq XIX Or thus Because B is the common Subtense to two Segments which together complete the whole Circumference and therefore the half of both complete the Semicircumference If therefore in a Circle according to Ptolemy's Lemma a Trapezium be inscribed whose opposite sides are A A and E E The Diagonals will be Diameters that is 2R And consequently 4Rq = Aq + Eq as before XX. Hence therefore The Radius R with the Subtense of an Arch A or E being given we have thence the Subtense of the double Arch B which is the Duplication of an Arch or Angle For R A being given we have or R E being given we have A = 4Rq − Eq And having R A E we have by § 9. XXI The Radius R with B the Subtense of the double Arch being given we have thence the Subtense of the single Arch A or E. which is the Bisection of an Arch or Angle For by § 14 And therefore 4Rq Aq − Aqq = Rq Bq = 4Rq Eq − Eqq. And the Roots of this Equation or Eq. And the Quadratick Root of this is A or E. XXII Hence also we have an easie Method for a Geometrical Construction for the Resolution of such Biquadratick Equations or Quadratick Equations of a Plain Root wherein the Highest Power is Negative Understand it in Mr. Oughtred's Language Who puts the Absolute Quantity Affirmative and by it self and the rest of the Equation all on the other side Suppose Rq Bq = 4Rq Aq − Aqq or putting P = ½ B 4Rq Pq = 4Rq Aq − Aqq. For dividing the Absolute term Rq Bq or 4Rq Pq by the Co-efficient of the middle term 4Rq the Result is ¼Bq or Pq and its Root ½ B or P. Which being set Perpendicular on a Diameter equal to 2 R the Square Root of that Co-efficient a streight Line from the top of it Parallel to that Diameter will if the Equation be not impossible cut the Circle or at least touch it From which Point of Section or Contact two streight-streight-lines drawn to the ends of the Diameter are A and E the two Roots of that ambiguous Biquadratick Equation or if we call it a Quadratick of a Plain-root the Root of the Plain-root of such Quadratick Equation XXIII And this Construction is the same with the Resolution of this Problem In a Right-angled Triangle the Hypothenuse being given and a Perpendicular from the Right-angle thereupon to find the other sides and if need be the Angles the Segments of the Hypothenuse and the Area of the Triangle ½ R B or P R. XXIV Or thus Having R and B as at § 22. with the Radius R describe
a Circle and therein inscribe the Chord B and another on the middle hereof at Right-angles which will therefore bisect that and be a Diameter And from both ends of this to either end of B draw the Lines A E as before And this Construction is better than the former because of the uncertainty of the precise Point of Contact or Section in case the Section be somewhat Oblique XXV Now if it be desired in like manner to give a like Construction in Case of such Biquadratick Equations or Quadraticks of a Plain-root where the highest Power is Affirmative though that be here a Digression as in all the rest that follow to § 35. It is thus Suppose the Equations Aqq − VqAq = VqEq = Pqq + VqPq Whose Affirmative Roots are Aq and Pq and therefore Vq VqEq and consequently Eq are known Quantities Therefore by Transposition Aqq − VqAq + VqPq and dividing by And therefore Aq − Vq = Pq and Pq + Vq = Aq And by Multiplication Aqq − VqAq = AqPq = Pqq + VqPq = VqEq XXVI The Equation therefore proposed dividing all by Vq comes to this That is Whose Roots are and Namely And And these Multiplied into V a known Quantity make Aq and Pq Namely And And consequently A is a mean Proportional between V and And P a mean Proportional between V and Therefore XXVII And Equation being proposed in one of these Forms Aqq − VqAq = VqEq = Pqq + VqPq The absolute term VqEq being divided by the Co-efficient of the middle term Vq the quantity resulting is Eq whose Square-root E set Perpendicular on the end of a streight Line equal to V the Square-root of the Co-efficient which we may suppose the Diameter of a Circle to which that Perpendicular is a Tangent On the same Center with this Circle and on the same Diameter continued by the Top of that Perpendicular draw a second Circle The Diameter of this second Circle is by that Perpendicular E cut into two Segments which are the Roots of these Equations That is and XXVIII Or without drawing that second Circle from the Top of that Perpendicular in a streight Line through the Center of the first which will cut the Circumference in two Points to the first Section is to the second XXIX These two Roots Multiplied one into the other become equal to the Absolute quantity And Multiplied into V become Aq Pq Or thus P is a mean Proportional between V and U and A between V and V+U Or thus P is a mean Proportional between V and and because by § 25. Aq = Pq + Vq A is the Hypothenuse in a Right-angled Triangle to the Legs P V. And this is no contemptible Method For the resolving Quadratick Equations of a Plain-root wherein the highest term is Affirmative The whole Geometrick Construction is clear enough from the Figures adjoined where yet the Circles for the most part serve rather for the Demonstration than the Construction XXX Again by the same § 25. And therefore A and E are also the Legs of a Right-angled Triangle whose Hypothenuse is V + U Which by P a Perpendicular on it from the Right-angle is cut into those two Segments XXXI From the same Construction therefore we have also the Geometrical Construction of this Problem In a Right-angled Triangle having one of the Legs E with the farther Segment of the Hypothenuse V to find the other Segment and so the whole V + U and the Perpendicular P and the other Leg A and the whose Triangle XXXII We have thence also this Analogy And Or thus And XXXIII If therefore we make V the Radius of a Circle then is A the Secant P the Tangent E a Parallel to the Right-sine in contrary position from the end of the Secant to the Diameter produced If we make A the Radius then is P the Right-sine and E the Tangent of the same Arch and V the Sine of the Complement or Difference between the Radius and Versed Sine From hence therefore XXXIV The Tangent E and Sine of the Complement V being given we have the Right-sine P and the Radius A. But § 25 and all hitherto is a Digression XXXV If in a Semicircle on the Diameter 2 R we inscribe B the Subtense of a double Arch A Perpendicular on the middle Point hereof will cut the Arch of that Semicircle into two Segments whose Subtenses are A E either of which is a single Arch to the double whereof B is a Subtense This as to E is evident from 4 è 1 and 28 è 3 And as to A from § 15 of this XXXVI But also by the same reason the Arch β the difference of the Arches A E and B the double of either will if doubled have the same Subtense of their double Arch. That is The double of the double of either and the double of their difference will have the same Subtense XXXVII If an Arch to be doubled be just a third part of the Circumference the Subtense of the double is equal to that of the single Arch. For the same Subtense which on one side subtends two Trients doth on the other side subtend but one That is by § 7 And therefore by Transposition and 3Rq = Aq. That is XXXVIII The Square of the Subtense to a Trient of the Circumference or of the side of an Equilater Triangle inscribed is equal to three Squares of the Radius XXXIX Again the same being the Subtense of the double Trient and of the double Sextant for a Trient and a Sextant compleat the half ⅓ + ⅙ = ½ the Square of the Subtense of a Sextant Eq is the difference of the Squares of that of the Trient and the Diameter or that of the Semicircumference That is 4Rq − Aq = Eq that is by § preced 4Rq − 3Rq = Rq = Eq And E = R. That is XL. The Subtense of a Sextant or side of the inscribed Equilater Hexagon is equal to the Radius CHAP. II. Of the Triplication and Trisection of an ARCH or ANGLE I. IF in a Circle be inscribed a Quadrilater whose three sides are A A A Subtenses of a single Arch and the fourth C the Subtense of the Triple Arch the Diagonals are B B the Subtense of the double as is evident But it is evident also that in this Case A is less than a Trient of the whole Circumference II. And therefore the Rect-angle of the Diagonals being equal to the two Rect-angles of the opposite sides Bq = Aq+AC and therefore Bq − Aq = AC and That is III. The Square of the Subtense of the double Arch is equal to the Square of the Subtense of the Single Arch less than a Trient of the Circumference and the Rectangle of the Subtenses of the Sngle and Treble Arch. And therefore IV. The Square of the Subtense of the double Arch wanting the Square of the Subtense of the single Arch being
or E less than a Quadrant then A B and A D will be opposite sides and B C Diagonals And therefore CB − AB = AD. And consequently into equal to AD. That is And as before And for the same reason And LXXXVIII If the Subtense of the single Arch be P or S greater than a Quadrant and even greater than a Trient but less than two Trients Then B C and B P or B S will be opposite sides and D P or D S Diagonals And therefore BC+BP = PD or BC+BS = SD And consequently into equal to PD That is And As before And by the same reason BC+BS = SD if S also be greater than a Trient and LXXXIX But if the single Arch be that of S greater than a Quadrant but less than a Trient or P greater than two Trients but less than three Quadrants then B C and D S are opposite sides and B S Diagonals And therefore BS − BC = DS. And consequently into That is And As before And in like manner BP − BC = PD if the Arch of P be greater than two Trients which is the same as if less than one and XC From all which ariseth this General Theorem The Rect-angle of the Subtenses of the single and of the Quadruple Arch is equal to the Subtense of the double Multiplied into the Excess of the Subtense of the Triple above that of the single in case this be less than a Quadrant or more than three Quadrants or into the Excess of the Subtense of the single above that of the Triple in case the single be more than a Quadrant but less than a Trient or more than two Trients but less than three Quadrants or lastly into the Sum of the Subtenses of the Triple and single in case this be more than a Trient but less than two Trients That is AD = B into C − A if the Arch of A be less than a Quadrant or greater than three Quadrants A − C if it be greater than a Quadrant but less than a Trient or greater than two Trients but less than three Quadrants A+C if it be greater than a Trient but less than two Trients XCI And universally That is if the Difference of 2RqA and A c whereof that is the greater if the single Arch be less than a Quadrant or greater than three Quadrants but this if contrarywise divided by Rc be Multiplied into Product is equal to D. XCII And therefore That is XCIII As the Cube of the Radius to the Solid of the Subtense of the single Arch into the Difference of the Square of it self and of the double Square of the Radius So is the Subtense of the Difference of that single Arch from a Semicircumference to the Subtense of the Quadruple Arch. XCIV Now what was before said at § 15 Chap. 29. That the Subtense of an Arch with that of its Remainder to a Semicircumference or of its Excess above a Semicircumference will require the same Subtense of the double Arch is the same as to say that From any Point of Circumference two Subtenses drawn to the two ends of any inscribed Diameter as A E will require the same Subtense B of the double Arch. XCV And what is said at § 12 26 Chap. preced That the Subtense of an Arch less than a Trient and of its Residue to a Trient as A E and of a Trient increased by either of those as Z will have the same Subtense of the Triple Arch is the same in effect with this that From any Point of the Circumference three subtenses drawn to the three Angles of any inscribed Regular Trigone as A E Z will have the same Subtense C of the Triple Arch. XCVI And what is said here at § 18 20. That the Subtense of an Arch less than a Quadrant and of its Residue to a Quadrant as A E and of a Quadrant increased by either of these as P S will have the same Subtense of the Quadruple Arch Is the same with this that From any Point of the Circumference Four Subtenses drawn to the four Angles of any inscribed Regular Tetragone as A E P S will have the same Subtense D of the Quadruple Arch. XCVII But the same holds respectively in other Multiplications of Arches as five Subtenses from the same Point to the five Angles of an inscribed Regular Pentagon and six to the six Angles of an Hexagon c. Will have the same Subtense of the Arches Quintuple Sextuple c. For they all depend on the same common Principle That a Semicircumference Doubled a Trient Tripled a Quadrant Quadrupled a Quintant Quintupled a Sextant Sextupled c. Make one entire Revolution which as to this business is the same as nothing And therefore universally XCVIII From any Point of the Circumference two three four five six or more subtenses drawn to so many ends of the Diameter or Angles of a Regular Polygone of so many Angles however inscribed will have the same Subtense of the Arch Multiplied by the number of such ends or Angles And therefore CXIX An Equation belonging to such Multiplication or Section of an Arch or Angle must have so many Roots Affirmative or Negative as is the Exponent of such Multiplication or Section As two for the Bisection three for the Trisection four for the Quadrisection five for the Quinquisection And so forth C. And consequently Such Equations may accordingly be resolved by such Section of an Angle As was before noted at § 61 Chap. preced of the Trisection of an Angle CHAP. IV. Of the Quintuplation and Quinquisection of an ARCH or ANGLE I. IF in a Circle be inscribed a Quadrilater whose sides A F the Subtenses of the single Arch and the Quintuple be Parallel B B subtenses of the double opposite The Diagonals will be C C the subtenses of the Triple as is evident from the Figure But it is evident also that in this case the single Arch must be less than a Quintant or fifth part of the whole Circumference II. And therefore the Rect-angle of the Diagonals being equal to the two Rect-angles of the opposite sides Cq − Bq = AF. And by the same reasons cq − bq = EF. That is III. The Square of the Subtense of the Triple Arch wanting the Square of the Subtense of the double Arch is equal to the Rect-angle of the Subtenses of the single and of the Quintuple the single Arch being less than a fifth part of the whole Circumference IV. And therefore if it be divided by one of them it gives the other That is and And in like manner and V. But C+B into C − B is equal to Cq − Bq. And therefore That is VI. As the Subtense of the single Arch less than a fifth part of the whole Circumference to the Aggregate of the subtenses of the Triple and double so is the Excess of the Subtense of the Triple above that of the double to that of