Mathesis enucleata, or, The elements of the mathematicks by J. Christ. Sturmius ; made English by J.R. and R.S.S.

Proposition XXXIV.

IN Equiangular Triangles (ACB and abc, Fig. 62.) the Sides ••¦bout the equal Angles are Proportional, viz. as AB to BC, so ab to bc, and as BC to CA so is bc to ca. &c. (β)

Demonstration.

For having described Circles thro' the Vertex of each Trian∣gle, according to Consect. 6. Definit. 8. by reason of the supp••¦sed equality of the Angles A and a, B and b, C and c, th•• Arches also AB and ab, &c. will necessarily agree in the num¦ber of Degrees and Minutes, by the foregoing 33 Prop, a•••• so also the Chords AB and ab, BC and bc, &c. will agree in th•••• number of Parts of the Radius or whole Sine ZA and za, 〈◊〉〈◊〉 Consect. 2. Definit. 10. Wherefore as many such Parts as A•••• has, whereof az has also 10000000, so many such also will a•• have, whereof az has also 10000000, &c. Therefore AC•• to CB as ac to cb, &c. Q. E. D.

CONSECTAYS.

I. WHerefore by the same necessity the Bases of such T••••¦angles AB and ab, will be proportional to their A••¦titudes CD and cd, as being Right Sines of the like Arches (〈◊〉〈◊〉 and cb, or rather CE and c e; and so for similar or like Tria••¦gles (and consequently also Parallelograms) we may rightly sup¦pose that their Bases are as a to ea, and their Heighths as b 〈◊〉〈◊〉 eb; tho we must not immediately conclude on the contrary, tha•• because their Bases and Altitudes are so, therefore they are S••¦milar.

II. As also in Similar Parallelepipeds it will be manifest 〈◊〉〈◊〉 any attentive Person, that the Bases are in a duplicate Propo••¦tion of the Altitudes. For since the Planes of Similar Solids a•••• equal in number, and Similar each to the other, if for A•• (Fig. 63.) we put a, and for BC b, AB will = ea and BC=eb; and so that Basis will be to this as ab to eeab. Moreover having let fall the Perpendiculars EH and EH, the Triangle

••BH and EBH are similar, and by putting c for BE, BE will be ec, putting also d for EH, EH will consequently be ed. But the Reason of the Base ab to the Base eeab, is du∣plicate of the Reason of d to e d, by Def. 34. Wherefore in Similar Parallelepipeds we may rightly suppose, that their Bases are as a b to eeab, or as a to eea, and their Altitudes as d to ed.

SCHOLIUM I.

FRrom this Proposition flows first of all the chiefest part of Trigonometry for the Resolution of Right Angled ▵▵: For since in any Right Angled Triangle, if one side, e. g. AB (Fig. 64.) be put for the whole Sine, the other BC will be the Tangent of the opposite Angle at A (and in like manner if CB be the whole Sine, BA will be the Tangent of the Angle C;) but if the Hypothenuse AC be made Radius or whole Sine, then the Side BC will be the Right Sine of the Angle A, or the Arch CD described from the Center A, and AB the Right Sine of the Angle C, or the Arch AE, described from the Center C, (we will omit mentioning the Secants, because the business may be done without them) which all follow from Def. 10. Wherefore you may find,

• I. The Angles.
  • 1. From the Sides by inferring As one leg to the other, so the whole Sine to the Tangent of the Angle opposite to the other Leg.
  • 2. From the Hy∣poth. & one side, by inferring As the Hyp. to the W.S. (whole sine) so the given leg to the S. of the opp. angle
• II. The Sides.
  • 1. From the Hy∣poth. and Angles: As the W. S. to the Hypoth. so the Sine of the Angle, opposite to the Leg sought, to the Leg it self.
  • 2. From one Leg and the Angles: As the W. S. to the given Leg, so the Tan∣gent of the Angle adjacent to it, to the Leg sought.
  • 3. From the Hypoth. and one of the Sides: Having first found the Angles, it's done by the 2, 1. or by the Pythagorick Theorem.

• III. The Hypothenuse.
  • 1. From the An∣gles and one of the Legs. As the S. of the Angle, opposite to the gi¦ven Leg, to that Leg, so the W. S. to the Hypoth.
  • 2. From the Legs given; Having first found the Angles its done by the 1. or by the Pythagorick Theorem.

III. Inversly also, if two Triangles ABC and ABC (〈◊〉〈◊〉 the Figure of the present Proposition) have one Angle of o•••• equal to one Angle of the other (e. g. B and B) and the Sid•••• that contain these equal Angles proportional (viz. as AB to B•• so AB to BC) then the other Angles (A and A, C and C will be also equal, and the Triangles similar(α) 1.1 for to 〈◊〉〈◊〉 like Chords AB and AB, BC and BC, there answer by t•••• Hypoth. like or similar Arches, i. e. equal in the number 〈◊〉〈◊〉 Degrees and Minutes; and to these also there answer equal A••¦gles both at the Periphery and Center.

IV. (Fig. 65. N^o 1.) If(β) 1.2 the Sides of the Angle BA•• are cut by a Line DE, parallel to the Base BC, the Segments 〈◊〉〈◊〉 those, Sides will be proportional, viz. AE to EC as AD to BD•• for by reason of the Parallelism of the Lines BE and BC, th•• Triangles ADE and ABC are Equiangular: Therefore as th•• whole BA to the whole AC, so the part AD to the part A•• and consequently also the remainder EC to the remainder D•• as the part EA to the part AD, by Prop. 26. and alternative•••• by Prop. 24. EC will be to EA as BD to AD.

SCHOLIUM II.

THere are several useful Geometrical Practices depend 〈◊〉〈◊〉 this Consectary and its Proposition. 1. That(γ) 1.3 where•• we are taught to cut off any part required, e. •• ⅓ from a given Line AB, and so generally to 〈◊〉〈◊〉 or divide any given Line AC, in the same pr••¦portion as any other given Line, is supposed 〈◊〉〈◊〉 be divided in D, (and consequently into as ma•••• equal parts as you please;) viz. if in the fi•••• Case, having drawn any Line AF, you take AD•• 1, and make DB 2, and having joined CB, dra••••

the Parallel LE: for as AD to DB so is AE to AC; that is, as 1 to 2, by this 4th Consect. therefore AE is one third of the whole AC, &c.

II. A Rule(α) 1.4 to find a third Proportional to the 2 Right Lines given AB and BC (N^o 2. Fig. 65.) (or a fourth to three given;) if, viz. having drawn AF at pleasure, you make AD equal to BC, and Joining DB, draw the Parallel EC: For as AB to BC, so AD i. e. BC) to DE. Now if AD be not equal to BC but to another (viz. a) third Proportional, then by the same Reason DE will be a fourth Proportional.

III. Another Rule(β) 1.5 to find a mean Proportional between two Right Lines given AC and CB; which is done by join∣••ng both the Lines together, and from the middle of the whole AB describing a Semicircle, and from C erecting the Perpendi∣cular CD: For since the Angle ADB is a Right one, by Con∣sect. 1. of the preceding Proposition, and the two Angles at C are Right ones, and those at A and B common to the whole Triangle ADB, and to the two partial ones ACD and BCD, ••hese two will be Equiangular and Similar to the great one, and consequently to one another: Therefore by the present Proposition, as AC to CD, so CD to CB, Q. E. D. and also as AB to BD so BD to BC, and as AB to AD so AD to AC, &c.

IV. The Analytical Praxis of multiplying and dividing Lines ••y Lines, so that the Product or Quotient may be a Line; and ••lso the way of Extracting Roots out of Lines: Which Des Cartes, gives us, p. 2. of his Geom. viz. assuming a certain Line ••or Unity, e. g. AB (in Fig. 65. N^o 2.) if AC is to be multi∣••lied by AD, having joined BD, and drawn the Parallel CE, ••he Product will be AE; for it will be as 1 to ••he Multiplier AD, so the Multiplicand AC to ••he Product AE; or if AE is to be divided ••y AC, having joined EC and drawn the Pa∣••allel BD, the Quotient will be AD; (for AC ••he Divisor, will be to AE the Dividend, as an ••nit AB to the Quotient AD;) all which are e∣••ident from the Nature of Multiplication and Division, and the

Precedent Praxes. As also taking CB (in the same Fig. N^o 3.) for Unity, if the Square Root is to be extracted out of any o∣ther Line AC, this being joined to your Unity in one Line AB▪ and having described thereon a Semicircle, the Perpendicular CD will be the Root sought, as being a Mean Proportional be∣tween the two Extremes CB and AC, according to Prop. 17.

V. A Right Line AG which divides(α) 1.6 any given Angle A into two equal Parts (Fig. 66.) being prolonged, divides the Base BC proportionally to the Legs of the Angle AB and AC For having prolonged CA to E, so that AE shall be = to AB•• the Angles ABE and AEB will be equal, by Consect. 2. Def. 13. and consequently also equal to each of the halves of the exter∣nal Angle CAB, by Consect. 3. of the antecedent Proposition Therefore the lines AG and EB will be parallel, by Cons. 1. Def. 11. Therefore as AC to AE, i. e. to AB, so GC to GB, by Co••¦sect. 3. of this Proposition. Q.E.D.

VI. Hence also there follows further, by conversion of th•• last inference, as AC+AB to AC, so GC+GB (i. e. BC) 〈◊〉〈◊〉 GC; and inversly GC to BC as AC to AC+AB; and lastly alternatively, GC to AC as BC to AC+AB.

N. B. This last Inference follows also immediately from the preceding Consectary. For by reason of the Similitude of the ▵▵ ACG and ECB, as GC to AC so BC to CE, i. e. to AC+AB.

SCHOLIUM III.

FRom these two last Consectarys there an•• these or two or three Practical Rules, t•••• first whereof shews, how having the two Legs A•• and AC given, and also the Base BC, to find the Se••∣ments GC and GB, made by the Bisection of the Intercrural Angle•• (viz. by this inference, according to Consect. 6: As the Sum 〈◊〉〈◊〉 the Sides to one Side (e. g.) AC;) so the Sum of the Segmen•• of the Base, i. e. the whole Base to one of the Segments, vi•• that next the said Side GC. 2. It shews on the contrary, how having the Base and one of its Segments given, and moreover the S•••• of the Sides, to find separately the Side AC next the known Segment

by inferring as the Sum of the Segments, or the Base BC to the Sum of the Sides, so the given Segment GC to the sought AC: or also, 3dly, Having only the Base and Sum of the Sides given, but not the Segment GC, yet to express its Proportion to the next side AC, — viz. in the Quantities of the given Terms, by putting (by Consect. 6.) for GC the value of the Base BC, and for AC the value of the Sum AB+AC; the great use of which last Rule will appear hereafter in the Cy∣clometry (or Quadrature of the Circle) of Archimedes.

VII. In any Triangle ABC (Fig. of the present Proposition) the Sides are to one another as the Sines of their opposite An∣gles: For they are as the Chords of the double Angles at the Center, by Prop. 33. therefore they are also one to another as half those Chords, i. e. by Definit. 10. as the Sines of the half Angles.

SCHOLIUM IV.

HEnce flow two new Rules of Plane Trigonometry, for Oblique-angled Triangles to find, viz.

• 1. The other Angles:
  • From 2 gi∣ven Sides, & an Angle op∣posite to one of them: by inferring As the Side opposite to the given Angle to the other Side, so is the Sine of the given Angle to the sine of the angle opposite to the other Side; which being given, the third is easily found.
• II. The other Sides:
  • From one side and the angles given, As the Sine of the Angle opposite to the given side, to that side; so is the Sine of the Angle op∣posite to the side sought to the side sought.

So that this way we have reduced all the Cases excepting one of Plane Trigonometry, and consequently all Euthymetry to their original Foundations (for in that Case of having two Sides, and the included Angle given, we may find the rest by

the Resolution of the Obliqueangled Triangle into two Right Angled ones; and so it's done by the Rules we have deduc'd in (Schol. 1.) I say, excepting one, in which from the three sides of an Obliqueangled Triangle given, you are required to find the Angles: the Rule to resolve which we will hereafter deduce in the 2d Consect. of Prop. 45. from that Theorem which Euclid gives us, lib. 2. Prop. 13.

VIII. Because in in the Right Angled ▵ BAC (Fig. 67) BC is to CA as CA to CD, by N^o 3. of the 2d Schol. of this Prop. the □ of CA will be = ▭ CE, by Prop. 17. In like manner because as CB to BA so is BA to BD; the □ of BA will be = to ▭ BE: Wherefore the two Rectangles BE and CE taken together, that is, the □ of the Hypothenuse BC, will be = to the two □'s BA and CA taken together: Which is the very Theorem of Pythagoras demonstrated two other ways in Schol. of Definit. 13.

SCHOLIUM V.

THis Theorem of Pythagoras as it furnishes us with Rules of adding Squares into one Sum, or subtracting one Square from another; so likewise it helps us to some Foundations where∣on, among the rest, the structure of the Tables of Sines relies, &c. Whose use we have already partly shewn in Schol. 1 and 4. 1. If several Squares are to be collected into one Sum, having joined the Sides of two of them so as to form a Right Angle, e. g. AB and BC (Fig. 68. N^o 1.) the Hypothenuse AC being drawn, is the Side of a Square equal to them both; and if this Hypothenuse AC be removed from B to D, and the Side of the third Square from B to E, the new Hypothenuse DE will be the Side of a Square equal to the three former taken together. 2. If the Square of the side MN (N^o 2.) is to be subtracted from the Square of the side LM. Having described a Semicircle upon LM, and placed the other MN within that Semicircle, then draw the Line LN and that will be the Side of the remain∣ing Square. 3. Having the Right Sine EG of any Arch ED gi∣ven (but how to find the Primary Sines we will shew in another place), you may obtain the Sine Complement CG or EF, by the preceding

Numb. viz. by subtracting the □ of the given Sine from the □ of the Radius; and moreover the versed Sine GD by subtract∣ing the Sine Complement CG from the Radius CD. 4. The Squares of the versed Sine GD, and of the Right sine EG being added together, give the □ of the Chord ED of the same Arch, (which all are evident from the Pythagorick Theorem) and half of that EH gives the Right Sine of half that Arch. 5. From the Right Sine EG you have the Tangent of that Arch, if you make, as the Sine Complement CG to the Right Sine GE, so the whole Sine CD to the Tangent G I. 6. Lastly, From these Data you may also have the Secants (if required) thus, as the Sine Complement CG to the W. S. CE, so the W. S. CD to the Secant CI; or as the Right Sine EG to the W. S. E.C. so the Tangent ID to the Secant IC; both which are e∣vident by our 34th Proposition.

Consect. 9. If the Quadrant of a Circle (CBEG, Fig. 70.) be inclined to another Quadrant (CADG) and two other Per∣pendicular Quadrants cut both of them, viz. FBAG and FEDG, and the latter do so in the extremities of them both) having let fall Perpendiculars from the common Sections E and B, thro' the Planes of the Perpendicular Quadrants, and the inclined Quadrant, (viz. on the one side EG and BH, as Right Sines of the Segments EC and BC; on the other EI and BK, as Right Sines of the Segments ED and BA) you'l have 2 Tri∣angles EIG and BKH Right Angled at I and K, Equiangular at G and H (by reason of the same inclination of the Plane CBEGC) and consequently similar, by our 34th Proposition; wherefore as the Sine EG to the Sine EI, so the Sine BH to the sine BK, or as EG to BH so EI to BK, and contrariwise.

SCHOLIUM VI.

HEnce you have several Rules of Spherical Trigonometry for resolving Right Angled ▵▵(α) 1.7 1. Having given in the Rightangled ▵ ABC the Hypothenuse BC and the Oblique Angle ACB, for the Leg AB op∣posite to this Angle, make: as the sine T (EG) to the sine of the Hypoth. (BH) so the sine of the given Angle (EI) to the sine of the Leg sought (BK). 2. Having given the Hypothe∣nuse

BC and the Leg AB for the opposite Angle ACB make as the sine Hypoth. (BH) to S. T. (EC) so the sine of the given Leg (BK) to the sine of the Angle sought (EI) 3. Having given the side AB and the Angle opposite to it ACB, for the Hypothenuse BC (supposing you know whether it be greater or less than a Quadrant) make as the sine of the given Angle EI to the sine T. (EG) so the sine of the given Leg (BK) to the sine of the Hypoth. BH). 4. Having given in the Right Angled ▵ EBF (which we take instead of ABC that so we may not be obliged to change the Figure) one Leg EB and the Hypothenuse BF for the other Leg EF, you may find its complement, if you make as the sine Complement o•• the given side (BH) to S. T. (EG) so the sine Compleme•••• of the Hypothenuse (BK) to the sine Compl. of the side sought (EI) 5. Having both Legs EB and EF given, for the Hy∣pothenuse BF its Compl. BA may be found thus: as S. T. (EG) is to the sine Cpmpl. (BH) of one side EB, so the sine Compl. (EI) of the other side (EF) to (BK) the sine Compl. of the Hypothenuse.

6. Having given in the same Right Angled Triangle EBF one Leg EF, and the Angle adjacent to it ETB, first prolong into whole Quadrants BA to f, that A f may = BF Hypoth. & BC to e that Ce may = EB, and AC to d that Cd may = D•• the measure of the given Angle EFB: secondly from d thro' e and f let fall a Quadrant thro' the extremities of the Quadrant Bf and Be, that so the ▵ C de may be Right Angled, in which there are given the Hypoth. Cd = to the given Angle, and the Angles C = to the Compl. of the given Leg (viz. to the Arch ED) and so, thirdly, there is sought the side de, as the Complement of the Arch ef, or of the Angle sought ABC, or EBF; viz. by the first case of this, by inferring, as S. T. to the sine Hypoth. cd (i. e. of the given Angle EFB;) so the An∣gle dce (i. e. DE the Compl. of the given Leg RF) to the sine de (as the Compl. of the Angle fBe or EBF.

7. Having given, in the same Triangle, the side EF and the opposite Angle EBF (i. e. the Arch ef) for the other Angle EFB (that is the Hypoth. cd in the ▵ cde) make by the third of this:

As the Sine of the Angle dce (i. e. the sine Compl. of the gi∣ven Leg DE) to the S. T. so the sine of the Leg de (i. e. the

fine Compl. of the Angle EBF) to the Hypoth. cd (i. e. the sine of the Arch DA or Angle EFB.).

8. Having the Oblique Angles given to find either of the sides, viz. EF; which may be done thus by the second of this:

As the sine of the Hypoth. cd (i. e. the sine of the Angle at F) to the W. S. so the sine de (i. e. the sine Compl. of the Angle at B) to the sine of the Angle dce (i. e. the sine Compl. of the side sought EF.)

Consect. 10. The same being given as in Consect. 7. if instead of the Right Sines EI and BK, you erect Perpendicularly DL and AM(Fig. 71) because of the similitude of the Triangles DGL and AHM, you'l have, as DG sine T. to DL the Tangent of the Arch DE, so AH the Right Sine of the Arch AC to AM the Tangent of the Arch AB; or as DG to AH, so DL to AM, and contrariwise.

SCHOLIUM VII.

HEnce flow the other Rules of Spherical Trigonometry for Re∣solving Right Angled Triangles, viz. 9. Having given the side AC in the ▵ ABC, and the adjacent Angle ACB, for the other side AB, make as the W. S. (DG) to the sine of the given side (AH) so the Tangent of the given Angle (ACB) to the Tangent of the Angle sought (AB.) 10. Having given the side (AB) and the opposite Angle (at C) for the other side (AC, so you know whether it be greater or less than a Quadrant) make as the Tangent of the given Angle (DL) to the Tangent of the given Leg (AB) so the whole S. (DG) to the sine of the Leg sought (viz. at AH.) 11. Both sides being given, for the Angles, make, as the sine of one Leg (AH) to the W. S. (DG) so the T. of the other Leg (AM) to the Tangent of the Angle opposite to the same (at C.) 12. Having given moreover in the Right Angled Triangle EBF the Hypothenuse (BF) and the Angle (EFB) for the adjacent side EF, make, as the sine Compl. of the given Angle (AH) to the W. S. so the Tangent Compl. of the Hypoth. (AM) to the Tang. Compl. of the Leg sought (DL) 13. Having given the side (EF) and the adjacent Angle F for the Hypoth. BF make; as the W. S. to the sine Compl. of the given Angle (AH) so the Tangent Complement of the given Leg (DL)

to the Tangent Compl. of the Hypoth. (AM.) 14. Having given the Hypoth. (BF) and one side EF for the adjacent An∣gle (F) make as the Tang. Compl. of the given Leg (D. L.) to the W. S. so the Tang. Compl. of the Hypoth. (AM) to the Sine Compl. of the Angle sought (AH). 15. Having gi∣ven the Hypoth. (BF, i. e. the arch Af, or the angle at d) and either of the oblique angles (at F) for the other angle (EBF) make by help of the new Triangle cde, by the 12th of this.

As the Sine Compl. of the angle cde (i. e. the Sine Compl. of the Hypoth. (AH) to the W. S. so the Tang. Compl. of the Hypoth. cd (i. e. Tang. Compl. of the given angle) to the Tang. Compl. of the Side de (i. e. to the Tang of the angle sought ABC or EBF.)

16. Having given the oblique Angles to find the Hypoth. (BF, or the arch Af, or the angle cde) it is done by the 14 of this Schol.

As the Tang. Compl. of the Leg de (i. e. the Tangent of the angle ABC or EBF) to the W. S. so the Tangent Com∣plement of the Hypoth. cd (i. e. the Tangent Complement of the other angle EFB) to the Sine Compl. of the angle cde (i. e. the Sine Compl. of the Hypoth. BC sought.)

So that now we have with Lansbergius (but much more com∣pendiously) Scientifically Resolved all the Cases of Rigt-angled Triangles; the Resolution of Oblique-angled ones only now remaining.

Consect. 11. In Oblique-angled Spherical Triangles, as well as Right-angled ones, the Sines of the angles are directly proportio∣nal to the Sines of the opposite Sines. 1. Of the Right-angled ones this is evident from N^o 3. Schol. 6. and from the 9th Consect. For as the Sine of the angle A (Fig. 72.) to the Sine of BD, so the W. S. (i. e. of the angle D) to the Sine of AB. 2. The same is immediately evident of an Oblique-angled Tri∣angle ABC, resolved into 2 Right-angled ones. For,

The Sine of the angle C is to the Sine of BD as the sine of the gle D to the Sine of AB; and also,

The Sine of the angle C to the Sine of BD as the Sine of the angle D to the sine of BC, by the 1.

In each Proportionality the means are the Sines of BD and D; therefore the Rectangles of the Extremes of the Sines of AB in∣to the Sine of A, and the Sine of BC into the Sine of C, will be equal among themselves, since the Rectangles of the same Means are equal, by Prop. 18. therefore by Prop. 19. as the Sine of A to the Sine of BC, so the Sine of C to the Sine of AB, Q. E. D.

SCHOLIUM VIII.

THe latter may appear of Oblique-angled Triangles after this way also; since the Sine of the angle A is to the Sine of BD as the Sine of the angle D to the sine of AB, call the first a, the second ea, the third b, the fourth eb; and because the Sine of the angle C (which we call c) is likewise to the Sine of BD (i. e. to ea) as the Sine of D (i. e. b) to the Sine of BC (which will consequently be 〈 math 〉〈 math 〉) it will be manifest, that the Sine of the angle

A is to the sine of BC as the sine of the angle C to the sine of AB, i. e. as a to .. 〈 math 〉〈 math 〉, i. e... c to .. eb. by multiplying the Means and Extremes, whose Rectangles are on both sides eab. Therefore as by the present and precedent Consectary 7, it is universally true, That in any Triangle whether Right Lined or Spherical, Right-Angled or Oblique-angled, the Sides or their Sines, are to one another, as the Sines of their opposite Angles (which therefore is commonly called a Common Theorem:) so also hence flow 2 new Rules of Spherical Trigonometry for Oblique-angled Triangles, like those we found in Schol. 4.

To find

• I. The other Angles.
  • From 2 sides given of an an∣gle opposite to one of them, by inferring As the sine of the side opposite to the gi∣ven angle to the sine of the other side, so the sine of the given angle to the sine of the angle sought.

• II. The other Sides.
  • From one side and the angles given, by inferring As the sine of the angle opposite to the given side to the sine of that side, so the sine of the angle opposite to the side sought to the sine of the side sought.

And thus we have reduced all the Cases and Rules of Sphe∣rical Trigonometry to their original Fountains (for from 2 Sides given and the interjacent Angle, or 2 Angles and their adjacent side, we may find the rest in Oblique-angled Triangles, by resolving them into 2 Right-angled ones; and so by the Rules we have deduc'd in Schol. 6 and 7) excepting two Cases, viz. when from 3 sides given, the Angles, or from 3 Angles the Sides are sought; to resolve which, we are supplied with Rules from the following

Cons. 12. In the given Oblique angled Spherical Triangle ABC (Fig. 73.) whose Sides are unequal and each less than a Quadrant, having produced the sides AB and AC to the Quadrant AD and AE, and effected besides what the Figure directs, then will

The Arch DE be the Mea∣sure of the angle A, AF=AC, and so FB the difference of the Sides AB and AC.

BC=BG, and so GF the difference of the third side, and the differences of the rest FB.

But now, 1. As EH or DH to CM or FM so will PH be to NM (by reason of the Equi∣angular Triangles EPH and CNM;) therefore by Prop. 26. so will also DP be to FN. Make therefore DH=a FM=ea, DP=b FN=eb.

AI the R. Sine of the side AB. GM the Sine of the Side AC GL the Sine of GB, or of the side BC.

FK the R. Sine of the Arch FB. BI the versed Sine of AB.

BL the vers. Sine of GB or BC. BK the versed Sine of FB. KL or NO the difference of the versed Sines we have now men∣tioned.

EP the Right Sine and DP the versed Sine of the arch DE.

CN the R. Sine and FN the ver∣sed Sine of the arch FC.

[illustration]

Pag. 108.

73

74

75

76 n. 1.

76 n. 2.

77 n. 1.

77 n. 2.

2. By reason of the Equiangular ▵▵ FNO and HAI (for FNO is Equiangular to the ▵ FKQ, and that to the ▵ HQM, by reason of the Vertical Angles at Q; and that also to the ▵ HAI by reason of the Common Angle at H) and you have also,

As HA,

Or DH to AI, so FN to NO.

〈 math 〉〈 math 〉.

Wherefore now, 3. you'l have evidently, the □ DH to FM into AI as DP to NO.

〈 math 〉〈 math 〉. DH NO DP.

And Inversly as oea to a so oeb to b.

SCHOLIUM IX.

SInce therefore the Radius DH or a is known, and also NO the Difference of the versed Sines BL and BK, it is evident, that DP the versed Sine of the angle A will be known also; supposing that the first Quantity oea is likewise known. But this may be had by another Antecedent Inference, if you make, as AH to FM so AI to a fourth oea.

a ea oa

Hence therefore arises, 1. the Rule: Having given the 3 Sides of an Oblique-angled Triangle, to find any one of the Angles, viz. by inferring,

1. As the Sine of T to the Sine of R, one of the sides com∣prehending AC; so the sine of the other side AB to a fourth, 〈 math 〉〈 math 〉

2. As this fourth to the sine of T, so the difference of the versed sines of the third side, BC, and the differences of the o∣thers to the versed sine of the Angle sought, viz.

〈 math 〉〈 math 〉.

But since the sides of a Spherical Triangle may be changed

into Angles, and contrariwise the sides being continued [as 〈◊〉〈◊〉 the side AB of the given Triangle ABC (Fig. 47.) be con••¦nued a Circle, the rest into Semicircles from the Poles b and 〈◊〉〈◊〉 and likewise the Semicircle HI from the Pole A, and the Sem••¦circle FG from the Pole B, and the Semicircle EA from th•• Pole C, you'l have a new Triangle a, b, c, the 3 angles o•• which will be equal to the 3 sides of the former ABC; as th•• angle a or its measure IG, is equal to the side AB, by reaso•• each makes a Quadrant joined with the third arch AG; b•• the measure of the angle b, is the side AC (viz. in this cas•• wherein the side AC is a Quadrant, in the other wherein 〈◊〉〈◊〉 would be greater or less than a Quadrant, it would be th•• measure of the angle of the Compl. for then the Semicircle H•••• described from the Pole A, would not pass thro' C but beyon•• or on one side of C. See Pitisc. lib. 1. Prop. 61. p. m. 25.) — the angle c or its measure KL, is equal to the 〈◊〉〈◊〉 BC, because with the third KC they make the Quadrants BK and CL] Therefore, 2. Having given the three Angles 〈◊〉〈◊〉 the Oblique-angled Triangle abc, you may find any side, e. g. ac, if there be sought the Angle ABC, or rather its Comple∣ment KBF, or its measure FK=ac, — from th•• 3 sides given of the ▵ ABC, by the preceding Rule, by infer∣ring, viz. 1. As S. T. to the sine R, of one side comprehending the angle of one side AB (i. e. of one angle a adjacent to the side sought) so the sine of the other side BC (i. e. of the other an∣gle C) to a fourth.

2. As the fourth to the S. T. so the difference of the ver∣sed Sines of the third side AC, and the differences of the others (i. e. of the 3d angle b, and the differences of the rest) to the versed Sine of the comprehended angle, or Complement to a Semicircle (i. e. of the side sought ac.)

Notes

• (α) 1.1

  Eucl prop. 6. lib. 6.
• (β) 1.2

  Eucl. 2. lib. 6.
• (γ) 1.3

  Eucl. l. 9 & 10. l. 6.
• (α) 1.4

  Eucl. 11 & 12, l. 6.
• (β) 1.5

  Eucl. 13. lib. 6 & and Eucl. 8. lib. 6.
• (α) 1.6

  Eucl. 3. lib. 6.
• (α) 1.7

  Lansberg. Geom. Triang. lib. 4. Prop. 12.