OF things Mathematical there are two principal kinds Number, and Magnitude, and each of these hath his proper science.

2 The science of Number is A­rithmetick, and the science of Magni­tude is commonly called Geometry; but may more properly be termed Magethelogia as comprehending all Magnitudes whatsoever, whereas Geometry, by the very Etymologie of the word doth seeme to confine, this science to Land measuring onely.

3 Of this Magethelogia, Geome­try or science of Magnitudes, we will set down such grounds and principles, as are necessary to be known, for the better understanding of that which followeth, presuming that the Reader hereof hath already gotten some com­petent knowledge in Arithmetick.

[Page 2] 4 Concerning then this science of Magnitudes, two things are to be con­sidered. 1 The several heads to which all Magnitudes may be referred. 2 The terms and limits of those Mag­nitudes.

5 All magnitudes are either lines planes, or solids, and doe participate of one or more of these dimensions, length, breadth, and thickness.

6 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 oft [...] length, but admitteth no other division as the line AB.

7 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 hath neither part nor quantity. As the points at A & B [Page 3] are the ends of the afore-saidline AB, and no parts thereof.

8 A plain or superficies is the se­cond kind of magnitude, to which be­longeth two dimensions length, and breadth, but not thickness.

As the end, limits, or bounds of a line are points confining the line: So lines are the limits bounds and ends in­closing a superficies, as in the figure A BCD, the plain or superficies thereof is inclosed with the four lines AB, BD, DC, CA, which are the extreams or limits thereof.

9 A body or solid is the third kind of magnitude, and hath three dimen­sions, length, breadth, and thickness. [Page 4] 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 and limit of a body or solid, and representeth to the eye the shape or figure thereof.

10 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.

11 A Circle is a figure contained under one round line, which is the cir­cumference thereof: Thus the round line CBDE is called the circumfe­rence of that Circle.

12 The center of a Circle is the point which is in the middest thereof, from which point all right lines drawn to the circumference are equal to one another: As in the following figure, the lines AB, AC, and AD, are equal.

13 The Diameter of a Circle, is a right line drawn through the center thereof, and ending at the circumfe­rence on the other side, dividing the [Page 5] circle into two equal parts.

As the lines CAD and BAE, are either of them the Diameter of the Circle CBDE, because that either of them doth pass through the Center A, and di­videth the Circle into two equal parts.

14 The Semidiameter of a circle is halfe the Diameter, and is contained between the center, and one side of the Circle. As the lines AB, AC, AD, and AE, are either of them the Semi­diameter of the Circle CBDE.

15 A Semicircle is the one half of [Page 6] a circle drawn upon his Diameter, & is contained by the half circumference and the Diameter. As the Semicircle CBD is half the Circle CBDE and drawn upon the Diameter CAD.

16 A quadrant is the fourth part of a Circle, and is contained between 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 circle. Thus from the center A the perpendicular AB being raised perpendicularly upon the Diameter CAD, divideth the Semîcircle CBD into the two equal parts CFB & FGD each of which is a Quadrant or fourth part of the circle CBDE.

17 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 then a Semicircle. As in the former figure FBGH is a Segment or part of the [Page 7] circle CBDE, contained under the right line FHG, less then the Semicircle CBD.

And by the application of the se­veral lines or terms of a superficies one to another, are made parallels, angles, and many sided figures.

18 A parallel line, is a line drawn by the side of another line, in such sort that they may be equi-distant in all places, and of such there are two sorts, the right lined parallel, and the circular parallel.

19 Right lined parallels are two

[Page 8] right lines equidistant in all places one from another, which being drawn to an infinite length would never meet or concur. As the right lines FHG & CAD in the annexed Diagram.

20 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 CFGD, and ABHE which are both drawn upon the same center K, & are there­fore parallel to one another.

21 An angle is the meeting of two lines in any sort, so as they both make not one line: as the two lines AB

[Page 9] and AC incline the one to the other and touch one another in the point A, in which point is made the angle BAC. Where note, that an angle is for the most part described by three letters of which the second or middle letter repre­senteth the angular point. Thus in the angle BAC the letter A representeth the angular point.

If the lines which containeth the angle be right lines, it is called a right lined angle: As the angle BAC.

A crooked line 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 a croo­ked line, as the angle GHI.

22 All angles are either right or oblique.

23 A right angle is an angle con­tained between two right lines, drawn perpendicular to one another. Thus the angle. ABC is a right angle, be­cause the right line AB is perpendicu­lar to the right line CD and the contrary.

[Page 10] 24 An oblique angle, is an angle contained between to right lines not perpendicular to one another: as the angle CBE or EBD, the one whereof

[viz.] EBD is acute or less then a right angle, and the other [viz.] CBE is ob­tuse or more then a right, & are the complements of each other to a Se­micircle.

25 The measure of an angle is the arch of a circle described on the an­gular point, and intercepted between the two sides of the angle.

Thus in the annex­ed Dia­gram, the arch AB is the measure of the an­gle [Page 11] AEB, & that the quantity there­of may be the better known.

26 Every circle is supposed to be divided into 360 parts or deg. every degr. into 60 min. or 100 parts, &c. Therefore a Semicircle as the arch ACD is 180 deg. a quadrant or fourth part of a Circle, as the arch ABC is 90 deg.

27 Complements of arches are either in reference to a quadrant or a Semicircle, the complement of an arch or angle to a quadrant is so much as the arch given wanteth of a quadrant or 90 deg. as if the arch AB be 60 deg. the complement thereof to a quadrant is the arch BC 30.

In like manner, the complement of an arch or angle to a Semicircle, is so much as the arch or angle given wan­teth of a Semicircle; as if the arch BED be 120 degrees, the complement thereof is the arch AB 60 deg.

28 When a right line falling upon two right lines doth make on one and the same side, and two inward angles. [Page 12] less the two right, the right lines being produced will at length concur in that part in which the two inward angles are less then two right. As in the figure, the right line AB falling upon the right lines CD and EF maketh the angles CGH and GHE together less

then two right angles, if therefore the lines CD and EF be produced from C and E they will at last concur.

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

[Page 13] In this Treatise we have to do with such onely as are contained under three lines or sides, & these are there­fore called Triangles, for the better understanding whereof we will here set down some necessary and funda­mental Propositions of Geometry.

1. Prop. If two sides of one Triangle be e­qual to two sides of another, and 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.

Demonst. In the Triangles CBH & FED make CB = FE and BH = ED [Page 14] and ang. CB = FED, then shall CH = FD, For if CG = FD, the angle CBG = FED, but CBH = FED by construction, therefore CH = FD, ang. C = F, and the angle H = D as was to be proved.

2 Prop. If a Triangle have two equal sides the angles at the base are also equal to one an­other, and the contrary.

Demonst. In the Triangle ABC let AB = AC, and let AB be extended to D and AE = AD, and draw the lines BE and DC, now then because AD = AE and AB = AC and the angle A common to the Triangle ADC and ABE, the base BE = DC the angle D = E and ABE = ang. ACD by the former prop. and there­fore [Page 15] ang. DCB = EBC and being taken from the equal angles ABE & ACD, there shall remain the angle ABC = ACB as was to be pro­ved.

3 Prop. If two right lines doe cut through one another, the angles opposite to one another are equal.

Demonst. Ang. ACB + BCD = 2 right ang. 24 of the first, ang. ACB + ACE = 2 right, and ang. ABC common, therefore ang. BCD = ACE as was to be proved.

4 If the side of a Triangle be extended the outward angle shall be greater then any one of the inward opposite angles.

Demonst. In the Triangle ABC [Page 16]

let AC be bisected in F and FG= FB, then is ang. AFG = BFC by the 3 prop. and BC = AG by the first hereof, as also, ang. GAF = FCB and therefore ang. EAC greater then ACB as was to be proved.

5 Prop. In all Triangles, the greatest sides sub­tend the greatest angles, and the lesser sides, the lesser angles.

Demonst. In the Triangle ABC [...] BD = BC and draw DC, then [...] BDC = BCD, & ang. BDC [...]eater then DAC by the 4th hereof. [...]refore ang. ACB greater then BAC [...] to be proved.

[Page 17] 6 If a right line drawn through two other right lines, doe make the alternate angles equal, then are those two right lines parallel.

Demonst. Let ang. AGH = GHD and then if AB and CD be not pa­rallel they will at length concurre

(suppose at K) by the 28 of the first, & the angle GHD greater then AGH by the fourth hereof, the which is con­trary to the proposition, if therefore the alternate angles be equal, the lines being extended shall not concur, and are therefore parallel, by the 19th of the first, as was to be proved.

7 If a right line drawn through two other right lines, shall make the outward angle of the one, equal to the inward angle of the other on the same side, or both the inward angles equal to two right, then are those two right lines pa­rallel.

Demonst. Ang. AGE = BGH by [Page 18] the third hereof, and ang. AGE = GHC by the proposition, therefore BGH = GHC and AB parallel to CD by the last afore-going.

Again, ang. AGE+AGH= 2 R. ang. by 24 first, and AGH+ GHC = 2 R. ang. by the proposi­tion, therefore AGE = GHC & AB parallel to CD as before.

18 If one right line cut through two parallel right lines, the angl [...]s opposite to one another are equal.

Demonst. Ang. AGH+BGH= 2 R. ang. and ang. CHG+GHD = 2 R. ang. by the 24 of the first, therefore, ang. CHG = BGH, for if BGH be greates CHG ang. CHG+ AGH − 2 R. ang. and AB not pa­rallel to CD by the 28 of the first, which is contrary to the proposition, and therefore AB being parallel to CD, ang. AGH = GHD as was to be proved.

[Page 19] 19 If the side of a Triangle be continued, the outward angle made by the continuation, is equal to the inward opposite angles, and three inward angles are together equal to two right.

Demonst. Let CD be parallel to AB then shall ang. BAC = ACD and ang. ABC = DCE by the last afore-going, and therefore ang. ACE =ABC+BAC.

Again, ang. ACB+ACE = 2 R. by the 24th of the first, and the angle ACE = ABC+BAC, therefore ang. ACB+BAC+ABC = 2 R. ang. as was to be proved.

10. If two equal parallel lines be joyned to­gether with two other right lines, those other lines are equal and also parallel.

Demonst. Ang. VXW = XWY and VX parallel WY by the pro­position, and XW common to both, [Page 20] therefore VW = XY by the first hereof, and also parallel by the 7th hereof, as was to be proved.

11 In all Paralellograms, the angles and sides that are opposite to one another are equal, and the Diamet [...]r or Diagonal divideth the same into two equal parts.

Demonst. In the preceeding Dia­gram ang. VWX = YXW, and angle XWY = VXW by the 8th hereof, and XW common to both the Triangles VXW and YXW, therefore the other ang. and sides are also equal by the first hereof, and the paralellogram VY is divided into two equal parts by the diagonal XW as was to be proved.

12 All paralellograms and Triangles stand­ing upon the same or equal base, and between two parallel lines (that is having the same common height) are equal.

Demonst. The two paralellograms [Page 21] NL and NO stand upon the same base KN, and between the two paral­lel lines KR and LP. The side KL= NM and KO = NP by the 11th hereof: and the Triangle KOL= NPM. From these two take the

common Trian. MIO & there will re­main the quadrilateral figure LKIM =NIOP to which if you adde the common Triangle KIN the paralel­logram LN = NO.

Again upon the equal bases KN & QR do stand the paralellograms LN & PQ which let be connected by the lines KO and NP being pa­rallel by the 10th hereof, and include the paralellogram NO = NL as was proved before, and may in like man­ner be proved to be equal to QP▪ [Page 22] therefore NL = QP, and what is said of Paralellograms is also true of Triangles they being the halves of pa­ralellograms.

THat the Proportions which the parts of a Triangle have one to another may be certain, the arches of circles (by which the an­gles of all Triangles, and of Spherical Triangles the sides are also measured) must be first reduced into right lines, by defining the quantity of right lines, as they are applyed to the arches of a circle.

2 Right lines are applyed to the arches of a circle three wayes, viz. ei­ther as they are drawn within the circle, without the circle, or as they are drawn through it.

3 Right lines within the circle are Chords and sines.

4 A Chord or subtense is a right line inscribed in a ci [...]cle, dividing the whole circle into two segments: and in like manner subtending both the [Page 34] segments: as the right line CK divi­deth the circle GEDK into the two segments CEGK and CDK, and subtendeth both the segments, that is, the right line CK is the chord of the arch CGK, and also the chord of the arch CDK.

5 A Sine is a right line in a semi­circle falling perpendicular from the term of an arch.

6 A Sine is either right or versed.

7 A right Sine is a right line in a Semicircle, which from the term of an arch is perpendicular to the diameter, dividing the Semicircle into two seg­ments, and in like manner referred to both: Thus the right line CA is the sine of the arch CD less then a qua­drant, and also the sine of the arch CEG greater then a quadrant, and hence instead of the obtuse angle GBC, we take the acute angle CBA the complement thereof to a Semicir­cle, and so our Canon of Triangles doth never exceed 90 deg.

[Page 35] 8 A right sine is either Sinus totus, that is, the Radius or whole Sine, as the right line EB: or Sinus simplici­ter the first sine, or a sine less then Radius, as AC or AB, the one where­of is alwayes the complement of the other to 90 degrees, we usually call them sine and co-sine.

9 A Versed sine is a right line in a Semi-circle perpendicular to the right

Sine of the same arch; Thus A D is the versed sine of the arch CD and GBA is the versed sine of the arch [Page 36] GEC the complement of CD.

10 Right lines without the Circle whose quantity we are to define, are such as touch the circle, and are cal­led Tangents.

11 A Tangent is a right line which touching the circle without, is perpen­dicular from the end of the diameter to the Radius, continued through the term of that arch of which it is the Tangent: Thus the right line FD is the Tangent of the arch CD.

12 Right lines drawn through the circle, whose quantity we are to define are such as cut the circle and are cal­led Secants.

13 The Secant of an arch, is a right line drawn through the term of an arch, to the Tangent line of the same arch: and thus the right line BF is the Secant of the arch CD: as also of the arch CEG the complement thereof to a Semicircle.

14 A Canon of Triangles then is that which conteineth the Sines, Tan­gents, and Secants of all degrees & [Page 37] parts of degrees in a quadrant, accor­ding to a certain diameter, or measure of a circle assumed: The constructi­on whereof followeth, and first of the Sines.

15 The right Sines as they are to be considered in order to their con­struction are either Primary or Se­condary.

16 The Primary Sines are those by which the rest are found: And thus the Radius or whole sine is the first pri­mary sine, and is equal to the side of a six-angled figure inscribed in a Circle.

Demonst. AC = AB by the work,

[Page 38] and ang. C = B by the second of the second hereof, and BC the measure of the angle at A is 60 degr. and the angles A, B, C, are together equal to two right angles, and therefore ang. A = C or B and BC = AC or AB as was to be proved.

A Plain Triangle is contained un­der three right lines, and is ei­ther Right-angled, or Obli­que.

2 In all plain Triangles, two an­gles being given the third is also gi­ven: and one angle being given the sum of the other two is given: because the three angles together are equal to two right by the 9th of the second.

Therefore in a plain right angled triang. one of the acute angles is the complement of the other.

3 In the resolution of plain Trian­gles, the angles onely being given, the sides cannot be found, but onely the the reason of the sides: It is therefore necessary, that one of the sides be known.

[Page 53] 4 In a Right-angled Triangle two termes (besides the right-angle) will serve to find the third; so the one of them be a side.

5 In Oblique angled Triangles there must be three things given to find a fourth.

6 In right-angled plain Triangles there are seven cases, and five in Ob­lique, for the solution of which the four Axiomes following are suffi­cient.

HAving done with plain triangles, we come next to speak of Sphe­rical.

1 A Spherical triangle is that which is described on the sur-face of the Sphere.

2 The sides of a Spherical Triangle are the arches of three great circles of the Sphere, mutually intersecting each other.

[Page 68] 3 The measures of Spherical an­gles are the arches of great circles de­scribed from the angular point be­tween the sides of the angles, those sides being continued to quadrants.

4 Those are said to be great circles which bi-sect the Sphere.

5 Those circles which cut each o­ther at right angles, the one of them passeth through the poles of the other and the contrary.

6 In every Spherical triangle, each side is less then a semicircle, for if in the triangle ABC, you produce the

sides AB, BC, till they meet in the point D the arches BAD, and BCD are each of them a semicircle because they intersect each other in the points B and D, and therefore the arch BA [Page 69] or BC is less then a semicircle. In like manner if the sides AB and AC be produced, the side AC may be also proved to be less then a Semicircle.

7 In every Spherical Triang. any two sides are together greater then the third, for otherwise they cannot possi­bly make a triangle.

8 The sum of the sides of a Sphe­rical Triangle are less then two se­micircles. For if any two sides be pro­duced as suppose AB, BC till they concur in the point D, the arches BAD, BCD shall be each of them a semicircle, but in the train. ADC, the sides AD and CD are together grea­ter then AC by the last aforegoing, therefore the three sides AB, BC, AC are together less then the two semi­circles BAD, BCD,

9 If two sides of a Spherical tri­angle be equal to a semicircle, the two angles at the base shall be equal to two right. If they be less then a se­micircle, the two angles shall be less [Page 70] then two right: but if greater then a semicircle, the two angles shall be greater then two right. As in the sphe­rical Triangle ABC let the sides AB AC be equal to a Semicircle, ther [...] are the angles at B & C equal to two right, for the arch BC [...] produced to D, the an­gle ACB shall be equal to B, & seeing that the two angles at C are equal to two right, the two

angles ACD and B shall be also equal to two right.

Again. Let the sides AB and AC be less then a semicircle, seeing that the two angles at C are equal to two right, and the angle B less then the angle ACD, the angles ACB and B are together less then two right.

Lastly, Let the sides AB and AC be more then a semicircle, the angles at C being equal to a semicircle, and the angle at B greater then the an­gle [Page 71] ACD, the angles ACB & B shall be greater then two right.

10 The sum of the three angles of a Spherical triangle are greater then two right angles and less then six.

Demonst. In the triangle ABC let the side BC be produced to D, then shall the angle ACD be either more or less or e­qual to the angle ABC; first, sup­pose them equal, then the arches AB and AC

shall be equal to a semicircle, by the last aforegoing, and the angles ABC and ACB are equal to two right, and therefore the three angles A.B.C are more then two right.

Again. Let the angle ACD be less then the angle B, then the sum of the arches AB & AC shall be more then a Semicircle, and therefore the angles ABC and ACB greater also then two [Page 72] right, and therefore much more are the three angles A.B.C. greater then two right.

Lastly. Let the angle ACD be greater then the angle ABC, and make the angle DCE equal thereto, and the side AB being produced to E that the arch BE and CE may meet, and let the arch CA be produced to F, then shall the arches EB and EC be together equal to a semicircle, and therefore AE and EC are together less then a semicircle, and the angle EAF or BAC is greater then the angle ACE by the ninth hereof, but the angles ACE, ACB and B are equal to two right, therefore the an­gle ACB, ABC & BAC are grea­ter then two right.

And because every angle of a Sphe­rical Triangle is less then two right, the three angles together must needs be less then six, as was to be proved. Therefore,

11 Two angles of any Spherical Triangle are greater then the diffe­rence [Page 73] between the third angle and a semicircle also.

12 Any side being continued, the exteriour angle is less then the two interiour opposite ones.

13 In any Spherical Triangle, the difference of the sum of two angles, and a whole circle is greater then the difference of the third angle, and a semicircle.

14 In any Spherical Triangle, one side being produced, if the other two sides be equal to a semicircle, the out­ward angle shall be equal to the in­ward opposite angle upon the side produced: if they be less then a se­micircle, the outward angle shall be greater then the inward opposite an­gle: if greater then a semicircle, the outward angle shall be less then the inward opposite angle.

In the Spherical Triangle ABC let the side BC be continued to D, and let the sides AB and AC be together equal to a semicircle. I say then that the outward angle ACD is equal to [Page 74] the inward opposite angle at B, be­cause

the angle B = D and the angle ACD = D and therefore angle ACD = B.

Again, Let the sides AB and AC be less then the semicircle BAD if the common arch AB be taken away, there shall remain the arch AC less then the arch AD and therefore the ang. ACD shall be greater then the angle D, therefore also more then B.

Lastly, If the sides AB and AC be together more then a semicircle, taking away the common arch AB, the remaining arch AC shall be grea­ter then AD, and the angle ACD lesser then D, and therefore also lesser then B as was to be proved.

[Page 75] 15 A Spherical Triangle is either right or angled or oblique.

16 A right angled Spherical Tri­angle is that which hath on right an­gle at the least.

17 The legs of a right angled sphe­rical Triangle are of the same affection with their opposite angles.

In the spherical Triangle ABC right angled at A let the side AB be a quadrant, I say then that the an­gle ACB shall be a quadrant also, be­cause B is then the pole of the arch AC, and the arch BC perpendicu­lar

thereunto, there­fore also in the Triangle ACD, the side AD, being more then a quadrant, the an­gle ACD shall be also more then a quadrant, it being more then the right angle ACB, and in the right angles Spherical Triangle AEC the side AE being less then a qua­drant, [Page 76] the angle ACE shall be also less then a quadrant, it being less then the right angle ACB.

18 In a right angled spherical Tri­angle, if either leg be a quadrant, the Hypotenusa shall be also a quadrant; but if both the legs shall be of the same affection, the Hypotenuse shall be less then a quadrant; if of different, then greater, and the contrary.

In the right angled triangle ABC right angled at A let the side AB be a quadrant, I say then that the Hypo­tenuse BC is also a quadrant, because the angle ACB is right, by the last a­fore-going, and the arches AB and BC which are perpendicular to the arch AC doe meet in the pole B.

Again, Let the sides AB and AC be continued to their opposite pole at F, then shall the Triangle FBC be equal to the triangle▪ ABC, but the arch GH being drawn by the points G and H, the base GH will be common to the right angled triangle [Page 77] AHG, whose legs AG and AH are greater then the quadrants AB & AC and also to the other right angled tri­angle FGH whose legs FG and FH are less then the quadrants FB & FC, and GH is less then the quadrant BC, which is the measure of the right an­gles

at F and A, if it be not less, it must be either more or equal to it, it cannot be more because the triangle ABC having all the angles right, can have no side greater then a qua­drant, and it cannot be equal, because neither of the legs are a quadrant.

Lastly, In the triangle DAH right angled at A, the leg AD is less then the quadrant AB, and the leg HA [Page 78] is greater then the quadrant AC, therefore the Hypotenusa DH is also greater then a quadrant, for AC and DC are each of them quadrants by the work, if therefore upon the pole D you describe the arch CI it will cut the Hypotenuse DH in the point I, and therefore DI is a qua­drant and DH more then a quadrant, as was to be proved.

19 In a right angled spherical tri­angle, if either of the angles at the Hypotenusa be a right angle, the hy­potenusa shall be also a quadrant, but if both shall be of the same affection it shall be less, if of different then greater and the contrary.

In the triangle ABC right angled at C if either of the angles at A or B be right, the side opposite thereto shal be also right by the 17 hereof, and the Hypotenusa AB shall be a quadrant by the last aforegoing, but if the an­gles at A aud B be both acute or ob­tuse the sides AC and CB shall be both acute or obtuse also by the 17th [Page 79] hereof, and the Hypotenuse AB less then the quadrant by the last afore­going: but if either of the angles at A & B be acute, and the other obtuse, one of the legs shall be less, the other more then a quadrant, by the 17th hereof, and the Hypotenuse AB more then a quadrant by the last afore­going, as was to be proved. Therefore

20 In a right angled spherical Tri­angle either of the Oblique angles is greater then the complement of the other, but less then the difference of the same complement to a semicircle.

21 An Oblique angled spherical Triangle, is either acute or obtuse.

22 An acute angled spherical Tri­angle hath all its angles acute.

23 An obtuse angled spherical Triangle hath all its angles either ob­tuse or mixt, viz. acute and obtuse.

24 In any Spherical Triangle, whose angles are all acute, each side is less then a quadrant.

In the Triangle ABC let all the angles be acute, and A the greatest [Page 80] angle, then is BC greater then either of the other arches AB, AC, and the arch A E being drawn at right angles to the arch AB and made equal to AC, the arch BE is less then a quadrant, because the legs of the right angled Triangle ABE, viz. AE

and AB are each of them less then a quadrant, and therefore the arch BC is much more less then a quadrant, and if BC the greatest arch be less then a quadrant, the sides AB & AC which are each of them less then the arch BC, must needs be less then qua­drants also.

25 In any oblique angled spheri­cal Triangle, if the angles at the base be of the same affection, the perpen­dicular [Page 81] drawn from the vertical an­gle shall fall within, if of different without.

In the oblique angled Triangle ABC whose angles at B & C are acute, the perpendicular AD shall fall within the triangle, for if it fall not within, it must be the same with one of the sides, or els fall without the Triangle if it be the same with either side, the angle at B or C must be right, which is contrary to the proposition, if it fall without the triangle, suppose at E the angle AEB shall be right,

but the angle ABE is obtuse, viz: the complement of the acute angle ABC [Page 82] and therefore the side AE is greater then a quadrant, by the 17th thereof, and the angle ACE being acute AE shall be also less then a quadrant by the same Theorem, but that the same side should be both more and less then a quadrant is absurd in this case there­fore the perpendicular shall fall with­in the triangle.

But in the Triangle AEB, obtuse-angled at B, acute at E, the perpendi­cular AD shall fall without the tri­angle, upon the side EB continued, or if otherwise, it must be the same with one of the sides, or fall within the triangle, it cannot be the same with either of the sides, for then the angle at B or E should be a right angle, and cannot fall within the triangle be­cause then the angles at B and E must either be both acute or both obtuse as hath been already proved, if therefore the angles at the base be of different affection, the perpendicular shall fall without, as was to be proved.

IN Spherical Triangles there are 28 varieties or cases, 16 in rectangular and 12 in oblique angular, whereof all the rectangular, and 10 of the ob­lique angular may be resolved by these two Axiomes following.

IN Oblique angled Spherical trian­gles, there are 12 Cases, 10 where­of may be resolved by the Catholick proposition, if the Spherical Trian­gle propounded be first converted in­to two Right, by letting fall of a per­pendicular, [Page 102] sometimes within, some­times without the triangle, if the an­gles at the base be both acute, or both obtuse, it falleth within the triangle, but if one of the angles at the base be acute, and the other obtuse, it falleth without, by the 25 of the 5th. chapter: however it falleth it must be alwayes opposite to a known angle, for your better direction, take this general Rule.

From the end of a side given, being adjacent to an angle given, let fall the perpendicular,

As in the triangle ABC, if there were given the side AB and the angle at A: by this rule the perpendicular must fall from B upon the side AC.

But if there were given the side AC, and the angle at A, the perpendicular must fall from C upon the side AB continued if need req [...]ire.

And to know whether the side upon which the perpendicular shall fall, must be continued or not, that is to know whether the perpendicular [Page 103] shall fall within or without the triang. If the former direction be not suffici­ent, the calculation will determine it, for if the arch found at the first ope­ration (whether side or angle) be more then the arch given, the per­pendicular shall fall without, if less, within the triangle, as shall plainly appear by our explanation of the ensu­ing Problemes.

HItherto we have explained the Catholick or Universal Propo­sition, and shewed how service­able it is, or may be made in Obli­que angled spherical Triangles, as well as right; and now for varietie sake we will shew how all the case in an Oblique angled spherical Triangle may be resolved without letting fall a perpendicular, and that by help of these Propositons following.