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
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
less than a Trient is equal to the Rect-angle of the Subtenses of the single and Treble Arch. And consequently V. If the Square of the Subtense of the double Arch wanting the Square of the Subtense of the single Arch less than a Trient be divided by the Subtense of the single Arch the Result is the Subtense of the Triple Arch. VI. Because that by § 2. and that B+A into B − A is equal to Bq − Aq as will appear by Multiplication Therefore That is VII As the Subtense of a single Arch less than a Trient to the sum of the Subtenses of the single and double Arch so is the Excess of that of the double above that of the single to the Subtense of the Triple VIII Again because by § 7 of the precedent Chapter Therefore And therefore That is IX The Triple of the Subtense of an Arch less than a Trient wanting the Cube thereof divided by the Square of the Radius is equal to the Subtense of the Triple Arch. X. But because the same Subtense C subtends also to another Segment of the same Circle the Subtense of whose Trient we shall call E Therefore XI And because the three Arches A A A and the three Arches E E E complete the whole Circumference as is evident Therefore once A and once E complete a Trient or third part thereof Therefore XII An Arch less than the Trient of a Circumference and the Residue of that Trient A and E have the same Subtense of their Triple Arch. XIII Again because as is shewed already and therefore 3RqA − Ac = 3RqE − Ec and 3RqA − 3RqE = Ac − Ec Therefore dividing both by A − E As will appear upon dividing Ac − Ec by A − E or Multiplying A − E into Aq+AE+Eq XIV But by § 37 38 Chap. preced 3Rq is the Square of the Subtense of a Trient that is by § 11 of this of the sum of the Arches A and E. Therefore XV. The Square of the Subtense of the Trient of the Circumference of a Circle or three Squares of the Radius is equal to the Squares of the Subtenses of any two Arches completing that Trient and the Rect-angle of them That is putting T for the Subtense of a Trient Tq = 3Rq = Aq+AE+Aq XVI But the Angle which AE contain as being an Angle in the Trient of a Circle or insisting on two Trients is an Angle of 120 Degrees And therefore by § 15. XVII In a Right-lined Triangle one of whose Angles is 120 Degrees the Square of the Subtense to that Angle is equal to the two Squares of the sides containing it and a Rect-angle of those sides For if such Triangle be inscribed in a Circle the Base of that Triangle will be the subtendent of a Trient in such Circle or Rq. XVIII If a Quadrilater be inscribed in a Circle three of whose sides are A E A or E A E and the fourth Z Each of the Diagonals by § 11. is T the Subtense of a Trient And therefore by § 13 14 15 ZE+Aq = ZA+Eq = Tq = 3Rq = Aq+AE+Eq And consequently ZE = AE+Eq and ZA = Aq+AE and therefore Z = A+E And therefore XIX If to the Aggregate of two Arches A E completing a Trient be added a third equal to either of them Z the Subtense of the Aggregate of all the three is equal to the sum of the Subtenses of those two That is Z = A+E XX. But the same Chord Z doth subtend on the one side to a Trient increased by the Arch A and on the other side to a Trient increased by the Arch E as is evident That is to an Arch which doth as much exceed a Trient or want of two Trients as the Arch A or E wants of a Trient Therefore XXI The Aggregate of the Subtenses of two Arches which together make up a Trient is equal to the Subtense of another Arch which doth as much exceed a Trient or want of two Trients as either of those two wants of a Trient XXII The same will in like manner be inferred if we inscribe a Quadrilater whose opposite sides are A T and E T and the Diagonals TZ For then TA+TE = TZ and therefore A+E = Z as before XXIII But if either of the Arches to which Z subtendeth greater than a Trient and less than two Trients be Tripled the Subtense of this Triple is the same with that of the Triple of A or E. For the Triple of an Arch greater than a Trient is equal to one whole Circumference with the Triple of that Excess For the Triple of â…“+A is 1+3A Now because when we have once gone round the whole Circumference we are just there where at first we began this therefore as to this Point is as nothing and the whole distance to be acquired is but the Triple of such Excess and just the same as if onely this Excess had been thrice taken XXIV As for Example If the Arch subtended by Z be β γ δ that is a Trient increased by the Arch E and to this we add a second equal to it δ ζ θ the Aggregate β γ δ ζ θ is the double Arch and the Subtense thereof is B or β θ which is also the Subtense of the Difference of the Arches A E and if to these two we add a third equal to either of them θ γ χ then is β γ δ ζ θ γ χ the Triple of the Arch first proposed and the Subtense hereof that is the streight-Streight-line which joyns the beginning and the end of this Triple Arch is β χ = C the very same which subtends the Triple of E. XXV And just the same would come to pass if for the first Arch we take β θ ζ δ that is a Trient increased by A to which Z is a subtendent likewise For taking a second equal to it δ χ γ β θ the Aggregate β θ ζ δ χ γ β θ more than one entire Circumference is the double Arch and the Subtense thereof B as before And if to these two we add a third equal to either θ ζ δ χ the Triple Arch is β θ ζ δ χ γ β θ ζ δ χ and the Subtense hereof as before β χ or C the same with the subtendent of the Triple of A. And therefore XXVI The Triple of an Arch greater than a Trient hath the same Subtense with the Triple of its Excess above a Trient And the same for the same reason holds in Arches greater than 2 3 or more Trients XXVII But note here that in this case That is if the Arch to be Tripled be greater than a Trient but less than two Trients for if more than two Trients but less than the whole Circumference it is the same as if it were less than a Triâ—â—â—â—â— the Subtense of the double is less than that of the single For in such case the Arch will differ from that of a Semicircle either
which is ever equal to χ+δ±μ however these parts be intermingled Which where it is +μ is commonly more obvious to the Eye but where it is − μ is more perplex and will need more consideration to discern but it is equally true in both cases The Square of the Base of an Angle of 135 Degrees is equal to the Squares of the Legs with a Rectangle of them Multiplied into VI. If A be 45 Degrees It will in like manner be shewed that because of B = χ+δ − μ. into χ+δ − μ = B = Bq. That is The Square of the Base of an Angle of 45 Degrees is equal to the Squares of the Legs wanting a Rectangle of them Multiplied into VII And universally what ever be the Angle A it will by like process be shewed That That is The Square of the Base whatever be the Angle at the Vertex is equal to the Squares of the Legs together with if it be greater than a Right-angle or wanting if less than such a Plain which shall be to the Rect-angle of the Legs as a Portion in the Base-line intercepted between two Lines from the Vertex making at the Base a like Angle with that of the Vertex to one of those two Lines so drawn VIII Of this we are to give great variety of Examples in the following Chapter where this General Theorem is applied to particular cases And which is further improved by these two ensuing Propositions IX The Radius of a Circle with the subtenses of two Arches being given the Subtense of their Aggregate is also given For supposing the subtenses of the given Arches to be A E The subtenses of their Remainders to a Semicircle are also had Suppose And And therefore inscribing a Quadrilater whose opposite sides are A ε and E α one of the Diagonals is the Diameter = 2R the other the Subtense of the Sum or Aggregate of those Arches suppose X. The same being given the Subtense of the Difference of those Arches is also given For having as before A α E ε 2R we have by a Quadrilater duly inscribed the Subtense of the Difference XI It is manifest also from what is before delivered that the same Triangle GΓμ doth indifferently serve for the Angle of 120 Degrees and of 60 Degrees And in like manner for 135 and 45 And so for any two Arches whereof one doth as much exceed as the other wants of a Quadrant For the Angle V is in both the same and the Angles at the Base differ only in this That in one the External Angle in the other the Internal which is the others Complement to two Right-angles is equal to the Angle of CD at the Vertex XII Hence it follows That Of two Angles where the Legs of the one are respectively equal to those of the other the one as much exceeding a Right-angle as the other wants of it The Square of the Base in the one doth as much exceed the two Squares of the Legs as in the other it wants thereof XIII And consequently In any Right-lined Triangle however inclined the Squares of the Axis or Diameter and of the half Bases twice taken are equal to the Squares of the Legs For supposing C C the two halfs of the Base and B the Diameter or Axis of the Triangle meaning thereby a streight Line from the Vertex to the middle of the Base and B β the two Legs It is manifest that of the two Angles at the Base which are each others Complement to two Right-angles the one doth as much exceed as the other wants of a Right-angle And therefore the Square of one of the Legs as Bq doth as much exceed as the other βq doth come short of Dq+Cq And therefore both together Bq+βq = 2Dq+2Cq XIV And therefore The Base and Axis or Diameter of a Triangle remaining the same however differently inclined the Aggregate of the Squares of the two Legs remains the same XV. And the same is to be understood of the Squares of Tangents of a Parabola Hyperbola Elipsis or other Curve Line having Diameter and Ordinates from the two ends of an Inscribed Ordinate to the Point of the Diameter produced if need be wherein those Tangents meet XVI The same may be likewise accommodated to the Segments of such Legs or Tangents Cut off by Lines Parallel to the Base Namely The Squares of such Segments intercepted by those Parallels together taken the Axe of such Trapezium remaining the same are the same Whether such Trapezium be Erect or however inclined For such Segments are still Proportional to their Wholes CHAP. VII Application thereof to particular cases I. IF A be a Right-angle or of 90 Degrees GΓ are Co-incident and μ = 0. and therefore And consequently by § 7 Chap. preced II. If A = 120 Degrees then is V that is the Angle contained of GΓ = 60 Degrees as being always the Difference of 2 A from two Right-angles And consequently GΓμ an Equilater Triangle for such also are the Angles at the Base each of which is the Complement of A to two Right-angles And therefore μ = G and Bq = Cq+Dq+CD III. If A = 60 Degrees Then also is V = 60 Degrees and μ = G as before And therefore Bq = Cq+Dq − CD IV. If A = 135. Then V = 90 And therefore by § 1. μq = Gq+Γq that is because G = Γ μq = 2Gq and And therefore V. If A = 45. Then also V = 90 And therefore as before and consequently VI. If A = 150 Then V = 120. And therefore by § 2. μq = Gq+Γq+GΓ that is because G = Γ μq = 3Gq and And VII If A = 30 Then V = 120. And therefore by § 2. μq = Gq+Γq+GΓ that is because G = Γ μq = 3Gq and And VIII If A = 157½ Then V = 135. And by § 4. And therefore IX If A = 22½ Then V = 135. And by § 4. And therefore X. If A = 112½ Then V = 45. And by § 5. And therefore XI If A = 6 − ½ Then V = 45. And by § 5. And therefore XII If A = 165 Then V = 150. And by § 6. And therefore XIII If A = 15 Then V = 150. And by § 6. And therefore XIV If A = 105 Then V = 30. And by § 7. And therefore XV. If A = 75 Then V = 30. And by § 7. And therefore XVI If A = 172½ Then V = 165. And by § 12. And therefore XVII If A = 7½ Then V = 165. And by § 12. And therefore XVIII If A = 97½ Then V = 15. And by § 13. And XIX If A = 82½ Then V = 15. And by § 13. And XX. If A = 142½ Then V = 105. And by § 14. And XXI If A = 37½ Then V = 105. And by § 14. And XXII If A = 127½ Then V = 75. And by § 15. And XXIII If A = 52½ Then V = 75. And by § 15. And And in