GEOMETRICAL TRIGONOMETRY. OR The explanation of such GEOMETRICAL PROBLEMS As are most useful & necessary, either for the construction of the CANONS of TRIANGLES. Or for the solution of them. Together with The Proportions themselves suitable unto every Case both in plain and Spherical Triangles, those in Spherical being deduced from the Lord Nepeirs Catholic or Universal proposition. By J. NEWTON M. A.  
LONDON, Printed for George Hurlock, at Magnus' Church-corner, and Thomas Pierrepont, at the Sun in Paul's churchyard. 1659.   

TO THE READER  
MAthematical Sciences in their full latitude do comprehend all Arts and Sciences whatsoever, but in a more strict sense, those only are esteemed such, which do belong to the Doctrine of Quantity, & amongst these this of Triangles is not of least concernment, of which though many have written and that excellently, yet still there may be something added, at least by way of explanation, if not some matter absolutely new. What we have here published is intended for beginners only, in which we have endeavoured after brevity & perspicuity both, in the construction of the Canon of natural sines, Tangents and Secants first, & then the Axioms, & Problems for the solution of all Triangles, both plain & Spherical, either by those natural numbers, or by the Logarith. of them, & in the solution of plain Triangles by logarithmetical numbers, sometimes the logarithmes of a Decimal fraction is required, for the finding whereof having given no direction in our Preface to those Tables, we have (though somewhat out of place) inserted it her.  
A fraction whether Vulgar or Decimal being given to find the logarithm thereof.  
That the way & method here propounded may be the better conceived, the proper charecteristique to the logar of any integer or whole number must be considered, and the characteristic of the log arithmes of all numbers, under 10, is 0, of all numbers hetweens 10 and 100 is 1, between 100 & 1000 is 2, & so forward, nor will there, as I couceive, be any inconvenience, if 10 be the characteristic of the logarithm of any digit, & 11 the characteristic of all numbers between 10 & 100, & 12 the characteristic of all numbers between 100 & 1000, & so forward; & yet if this shall be supposed, the characteristic of a number that is but one place beneath unity shall be 9, if it be two places beneath unity it shall be, 8 &c.  
Thus the Log. of 5 will be 10.698970  
The Log. of 5 tenths will be 9.698970  
The Log. of 05 hundreds will be 8.698970  
And upon this ground to find the Log. of a vulgar fraction, you must subtract the logarithm of the denominator from the logarithm of the numerator, and what remaineth shall be the logarithm of the vulgar fraction given.  
Example, let 6/12 be the vulgar fraction given, 

Logar. 6 the Nume●ator is 0.778151  
Logar. 12 the Denominator is 1.079181  
There difference 9.698970   is the log. of 6/12 the fraction propounded.  
And thus the logarithm of any fraction, or the decimal fraction answering to any logarithm, may be as easily found, as by any directions hitherto given, and in their use are much more ready, as will be manifest by the Example following.  
If 6 tenths of an ounce of Gold cost .95 hundreds of a pound sterling, what shall .07 hundreds of an ounce of the same gold cost. Answer is 1108. 

Logar. 6 tenths is 9.778151  
Logar. 95 hundreds is 9.977724  
Logar. 07 hundreds is 8.845098  
1.8822822  
Adswer 1108 9●044671    
To resolve this question, you must add the logarithms of the second & third terms together, & from their sum subtract the first, what remaineth will be the logar.. of the fourth proportional required, the characteristic of which fourth proportional doth by inspection show, whether it be the logarithm of a fraction or not, there is but one difficulty in this way of working which is to subtract a greater number from a less, in which if you will but borrow from a supposed figure as we do from a real, and set down the difference that difficulty is also av●ided, as in the first Example. Though I can subtract 1 out of 1 yet I cannot subtract 8 out of 5, but borrowing one from the next figure, I can subtract 8 out of 15 and there will remain 7, and so proceeding till you come the characteristic the remainder will be found in the ordinary way, but the characteristigue of the upper number being a cipher, and that of the lower an unite I cannot subtract 1 from 0, but supposing an unite to be beyond the cipher I can subtract 1 from 10 and there will remainder 9, the like is to be done in other cases, which I must leave to the consideration of the Ingenious Reader, here being no room for to explaim myself.  
JOHN NEWTON.   

An explication of the Symbols.  

= Equal to  
+ More  
− Less  
× Drawn into  
∷ Proportionality  
Z The Sum  
X The Difference  
s Sine  
cs cousin  
To Tangent  
ct cotangent  
R. ang. Right angle  
2 R. ang. Two right angles  
q. Square.    

CHAP. I.  
Definitions Geometrical.  
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 arithmetic, 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 etymology of the word doth seem to confine, this science to Land measuring only.  
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 arithmetic.  
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 do participate of one or more of these dimensions, length, breadth, and thickness.  
6 A line is a supposed length, or a thing extending itself 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 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 enclosing a superficies, as in the figure A BCD, the plain or superficies thereof is enclosed with the four lines AB, BD, DC, CA, which are the extremes or limits thereof.  
9 A body or solid is the third kind of magnitude, and hath three dimensions, length, breadth, and thickness. And as a point is the limit or term of a line, and a line the limit or term of a superficies, So likewise a superficies is the end 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 centre of a Circle is the point which is in the midst 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 centre thereof, and ending at the circumference on the other side, dividing the circle into two equal parts.  
As the lines god and BAE, are either of them the Diameter of the Circle CBDE, because that either of them doth pass through the centre A, and divideth the Circle into two equal parts.  
14 The Semidiameter of a circle is half the Diameter, and is contained between the centre, 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 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 god.  
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 centre 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 centre A the perpendicular AB being raised perpendicularly upon the Diameter god, 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 than a Semicircle. As in the former figure FBGH is a Segment or part of the circle CBDE, contained under the right line FHG, less than 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.  
20 A Circular parallel is a circle drawn within or without another circle, upon the same centre, as you may plainly see by the two circles CFGD, and ABHE which are both drawn upon the same centre K, & are therefore parallel to one another.  
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 mixed 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.  
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 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 compliments of arches are either in reference to a quadrant or a Semicircle, the compliment 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 compliment thereof to a quadrant is the arch BC 30.  
In like manner, the compliment 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 compliment thereof is the arch AB 60 deg.  
29 Many sided figures are such as are made of three, four, or more lines, though for distinction sake those only are so called which are contained under five lines or terms at the leas▪.  
In this Treatise we have to do with such only 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.   

CHAP. II.  
Propositions Geometrical.  
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 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 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 do 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 than any one of the inward opposite angles.  
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, than 〈◊〉 BDC= BCD, & ang. BDC ●●eater than DAC by the 4th hereof. ●●●refore ang. ACB greater than BAC 〈◊〉 to be proved.  
6 If a right line drawn through two other right lines, do make the alternate angles equal, then are those two right lines parallel.  
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 the third hereof, and ang. AGE= GHC by the proposition, therefore BGH= GHC and AB parallel to CD by the last aforegoing.  
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.  
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 aforegoing, 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 joined together with two other right lines, those other lines are equal and also parallel.  
Demonst. Ang. VXW= XWY and VX parallel why by the proposition, and XW common to both, 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 preceding 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 parallelogram 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.  
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 parallelogram NO= NL as was proved before, and may in like manner be proved to be equal to QP▪ therefore NL= QP, and what is said of Paralellograms is also true of Triangles they being the halves of paralellograms.  

COROLLARY.  
In a rectangular parallelogram, the Area is found by multiplying the unequal sides together, and any Obliquangular parallelogram is equal to a Rectangular one made upon the same base and altitude, the Area thereof is therefore found by multiplying the base by its altitude which is the other side of the Rectangular parallelogram, the same is to be understood of there halves being Triangles.  
13. Paralellograms and Triangles having the same common height are in proportion as there bases.  
Demonst. Let the paralellograms AM and GW stand between the parallel lines BW and AC, and have for there equal and common height the perpendicular PC and by the corollary of the 12th hereof CP × AE= AM an PC × GC= GW, therefore PC × AE, PC × GC ∷ AE, GC, and also AM, GW, ∷ AE, GC, as was to be proved.  
14. In all plane Triangles, if a line be drawn parallel to any of the sides, the intersegment of the other sides are proportional.  
15 Equiangled Triangles have there sides which contain or subtend equal Angles proportional, and the contrary.  
AB. BF (= CE) ∷ AC, CD, and also AC, CD ∷ EF, DE.  
And DC, CA ∷ BF, AB as was to be proved.  
16. In all paralellograms the supplements or fillings which stand upon the Diameter are eequal.  
17 If four right lines be proportional, the Rectangled figure made of the two means, is equal to the Rectangled figure made of the two extremes.  
Demonst. In the preceding Diagram, the Triangles GFD and BHF are like, because the ang. BFH= FDG and H= G by the 7th hereof, and therefore BH. HF ∷ FG. GD by the 15th hereof, & the Rectangle FC= BH × GD and the rectangle HG= HF × FG by the last aforegoing as was to be proved. The like may be said of three proportionals taking the mean twice.  
18 In a plain rightangled Triangle a perpendicular let fall from the right angle upon the side subtending the same, divideth the Triangle into two triangles both like to the whole, and to one another.  
Demonst. The Triangle ABD is like to the Triangle ABC, because B and D are right, and A common to both: and the Triangles ABC and BDC are like, because B and D are right & C common to both, and therefore the Triangles BCD and BAD are like, as was to be proved.  
19 In all right angled plain Triangles the sides including the right angle are equal in power to the third side.  
Demonst. In the precding Diagram the Triangles ABC, ABD, and BCD are like, by the last aforegoing, therefore AC, AB ∷ AB, AD, and AC, CB, ∷ CB, BD, and AC × CD+ AC × AD= ACq.  
And ABq+ CBq= AC × CD+ AC × AD, therefore ABq+ CBq= ACq, as was to be proved.  
20 If a right line being divided into two equal parts shall be continued at pleasure, the rightangled figure made of the line continued, and the line of continuation, with the square of one of the bi-segments, is equal to a square made of one of the bi-segments, and the line of continuation.  
Demonst. Let PQ be bisected in C and continued to N and let QC= OI= LI than the square OL+ MP= NCq for foe= OC or IP and MC common to both, therefore NP × MN+ QCq= NCq as was to be proved.  
21 To divide a right line given by extreme and mean proportion, that is, that the right angled figures made of the whole line and one part shall be equal to the square of the other part.  
Demonst. If the right line QP= QB be bisected in C and a line drawn from C to B, and the line NC= CB and MN= NQ, the rectangled figure made of BQ × boy= OQq for by the last aforegoing MP+ QCq= NCq, or BCq. And BQq+ QCq= BCq by the 19 hereof.  
22 If a plain Triangle be inscrihed in a circle the angles opposite to the circumference, are half as much as that part of the circumference which is opposite to the angles.  
Demonst. In the Triangle ebbed, ang. EDB= ebbed by the second hereof, and ang. AEB equal to both by the 9th hereof, the arch AB is the measure of the angle AEB, by the 25th of the first, therefore the arch AB is the double measure of the angle ADB as was to be proved.   

1 Consectary.  
If the side of a plain Triangle inscribed in a Circle be the Diameter, the angle opposite to that side is a right angle. As the angle ABD opposite to the diameter AD.  
  

2 Consectary.  
If divers right lined Triangles, be inscribed in the same segment of a circle upon one base, the angles in the circumference are equal. As the Triang. ABD & ACD being inscribed in the same segment of the circle ABCD and upon the same base AD, have their angles at B and D falling in the circumference equal.  
23 If a quadrilateral figure be inscribed in a Circle, the angles thereof which are opposite to one another, are together equal to two right angles.  
Demonst. Ang. CDB= CAB & BDA= BCA by the last aforegoing, therefore ang. CDA= BCA+ BAC and ABC+ BAC+ BCA= 2 R. ang. by the 9th hereof, & therefore ang. ABC+ ADC= 2 R. ang. as was to be proved.  
24 If in a quadrilateral figure inscribed in a circle, there be drawn two Diagonal lines, the rectangle under the Diagonals, is equal to the two rectangles under the opposite sides:  
Demonst. Let ang. DAE= CAB by construction, then shall ang. DAC= EAB and ang. ACD= ABE because the arch A D is the double measure to them both and therefore the triangles. ADC & AEB are like.  
Again ang. ADB= ACB because the arch AB is the double measure to them both and ang. DAE= CAB by construction, & the Triang. AED and ABC like, therefore, AC, CB ∷ AD, DE.  
And AC, CD ∷ AB, BE.  
Therefore AC × DE= CB × AD.  
And also AC × BE= CD × AB  
And AC × DE+ AC × BE= AC × DB. Therefore AC × DB= CB × AD+ CD × AB as was to be proved.    

CHAP. III.  
Of the Construction of the Canon of Triangles.  
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 applied to the arches of a circle.  
2 Right lines are applied to the arches of a circle three ways, 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 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 than a quadrant, and also the sine of the arch CEG greater than a quadrant, and hence instead of the obtuse angle GBC, we take the acute angle CBA the compliment thereof to a Semicircle, and so our Canon of Triangles doth never exceed 90 deg.  
8 A right sine is either Sinus totus, that is, the Radius or whole Sine, as the right line ebb: or Sinus simpliciter the first sine, or a sine less than Radius, as AC or AB, the one whereof is always the compliment of the other to 90 degrees, we usually call them sine and cousin.  
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 compliment thereof to a Semicircle.  
14 A Canon of Triangles than is that which containeth the Sines, Tangents, and Secants of all degrees & 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.  

Consectary.  
The Radius of a circle being given, the sine of 30 deg. is also given, for by this proposition, the Radius of a circle is the subtense of 60 deg. and the half thereof is the sine of 30, and therefore the Radius AB, or BC being 1000.0000 the sine of 30 deg. is 500.0000.  
17 The other primary sines are the sines of 60.18 and 12 deg. being the half of the subtenses of 120. 36 & 24 degr. and may be found by the problems following.  
18 The right sine of an arch & the right sine of its compliment, are in power equal to Radius.  
Demonst. In the first diagram of this chapter, AC is the sine of CD, and AB the sine of CE the compliment thereof, which with the Radius BC make the rightangled Triangle ABC, therefore ABq+ ACq= BCq by the 19 of the second, as was to be proved.  
And hence the sine of 60 deg. may thus be found, let the sine of 30 deg. AC be 500.0000 the square whereof 250.00000 being subtracted from the square of BC Radius, the remainder is 750.00000 the square of AB, whose square root is 8660254 the sine of 60 deg.  
19 The subtense of 36 deg. is the side of a Dec-angle inscribed in a circle, or the greater segment of a Hexagon divided into extreme and mean proportion.   

Corsectary.  
The side of a Hexagon being given, the side of a Dec-angle or subtense of 36 degr. is also given, for by the second of the second the Semi-radius being deducted from the square root of the squares of Radius, and the half Radius added together, the remainder is 6180339 the subtense of 36 deg.  
20 The subtense of 24 deg. is the side of a Quin-decangle, or the difference between the subtenses of 60 and 36 deg. and may be found by the 24 of the second.  
21 Having thus found the primary sines, the secondary sines as the sines of 6 d●g 3 deg. 1 d. 50 cent. 0 deg. 75 centesm●s, and 0 d. 01 centesme may be found from them, and all the other sines, to every degree, and part of a degree in the quadrant.  
22 The right sine and versed sine of an arch are together equal in power to the subtense of the same arch.   

Consectary.  
The right sine and versed sine of an arch being given, the sine of half that arch is also given, for EH the ½ EC is the sine of ED the half of the arch EDC.  
23 The right sine of an arch is a mean proprotional between the Semi-radius, and the versed sine of the double arch.  
Demonst. In the preceding diagram the Triangles EHA and ECF are like, because ang. F= EHA and E common to both, therefore  
AE, EC ∷ EH, EF or AE. EC ∷ ½ AE: ½ EC that is, ½ AE, ½ EC= EH ∷ EH, EF as was to be proved.   

1 Consectary.  
Therefore by this or the former proposition, the fine and sine compliment of an arch being given, the sine of half that arch is also given, I say the fine compliment, because the versed sine is found, by deducting it from Radius; Thus FA the sine of BC being deducted from AE Radius the remainder is FE the versed sine of EDC and ½ AE × EF= EHq whose root is EH or the sine of ED.  
24 The rectangle of the sine and sine compliment of an arch is equal to the rect●ngle of half the Radius, and the sine of the double arch.  
Demonst. In the preceding diagram, EH is the sine of ED and HA the cousin thereof, and CF is the sine of EDC the double arch, the Triangle ha' and EFC are like, as befo●e.  
And AE, EC ∷ HA, CF And ½ AE, EH= ½ EC ∷ HA CE And the rectangle of AH × HE= ½ AE × CF as was to be proved.  
25 By these Propositions the sine of 12 deg. being given the sines and sin●s complem●nts of these arches 6 deg. 3 deg. 1 deg. 50 cent. 0 deg. 75 cent. &c. were found to be as followeth. Deg. parts Sines Cosines. 6.00 10452. 846●2 99452.18953 3.00 5233.59562 99862.95347 1.50 2617. 691●3 99965.73249 0.75 1308.95955 99991. 43●75 0.375 654.49379 99997.85816 0.1875 327.24865 99999.46453  
And from the sine of 0 deg. 1875 parts, the sine of 10 centesmes may be found, in this manner.  

As, 0 deg. 1875  
Is to the sine thereof 327.14865  
So is 0 deg. 10 centesmes.  
To the sine thereof 174.53261  
Therefore the sine of 0d. 5 cent. 87.26930  
The sine of 0d. 01 cent. is 17.43326  
And the cousin thereof is 99999.99847   
Which being given the rest of the Table of sines may be ea●●ly made by this proposition following.  
26 Three equi-different arches being propounded, the rectangle made of the cousin of the common difference and the double sine of the mean arch, is equal to the R●ctangle made of the Radius, and the sum of the sines of the two extreme arches.   

Consectary.  
The sines of ED, and EC with the cousin of CD or BC being given, the sine of ebb is also given, for if from BH the sum of BM and DK you deduct MH= DK the remainder is BM the sine of BE, for illustration sake we have added these examples following.  

1 Example.  
ED 5 cent. and EC 10 centesmes being given to find BM the sine of ebb 15 centesmes.  

As the Radius AB 10000.0000  
To AN the cos. BC 5′ viz. 9999.9962  
So BF the doub.. of LC 10′ 349.0650  
To BH= BM+ DK 349.0659  
From which subt.. MH= DK 5′ cent. 87.2663  
Resteth the sine of 0d 15 cent. BM 261.7987    

2 Example.  
ED 0 deg. 10 cent.  
EC 0 deg. 15 cent. given, to find ebb 0 deg. 20  

As the Radius AB 10000.0000  
To cos. BC 5 cent. viz. AN 9999.9962  
So the double of LC 15 cen. viz BF. 523.5974  
To BH= BM+ DK 523.5976  
Subt. MH= DK 10 cent. 174.5326  
Rest sine 0 deg. 20 cent. BM. 349.0650    

3 Example.  
Let ebb 0 d. 25 cent. be inquired. There being given ED 0 deg. 15 cent. EC 0 deg. 20 cent.  

As the Radius AB 90 10000.0000  
To Cos. BC viz. s AN 89d 95 9999●9962  
So the doub.. of LCs 0d 20 cent. 698.1300  
To BH= BM+ DK 698.1297  
Sub. MH= DK 0 d. 15 cent. ●61. 798●  
Rest BM the sine of 0d. 25 cent. 436. 330●    

4 Example.  
Let ebb 0 d. 30 cent. be inquired. There being given ED 0 d. 20 cent. EC 0 d. 25 cent.  

As the Radius 90 1000.0000  
To Cos. BC viz. s AN 89.95 9999.9962  
So twice L●= BF viz. 25 cent. 872.6618  
To BH= BM+ DK 872.6613  
Subt. DK= MH 0d. 20 cent. 149.0950  
Rest▪ MB the sine of 0 d. 30 cent. 523.5963   
And in this manner may the whole Canon of sines be completed, or the sines of the first or last 60 deg. being thus made, or the sines of 45 deg. bein● made by this rule and the sines of 15 d more by the 24th thereof, the other 30 d. may be more easily made by the following problem.  
26 The sum of the Sines of any two arches equally distant from 30 deg. is equal to the cousin of the distance.  
Demonst. In the preceding diagram we have already proved.  
AB. AN ∷ BF. BH therefore also ½ AB. AN ∷ ½ BF. BH, now than if ½ BF= ½ AB sine of 30 deg. AN the cousin of the distance must be also equal to BH, the sum of the sines, as was to be proved.   

Example.  
Let DK the sine of DE 27 d. be 45399.04997  
And BM the sine of ebb 33 d. be 54463.90350  
Their sum is BH= AN sine of 87d. 99862.95347  
27 The Canon of sines being thus made, a Table of Tangents and Secants may be easily deduced from them, by the following problems.  
28 As the cousin of an arch, is to the sine thereof, so is Radius, to the Tangent of that arch.  
Demonst. In the annexed diagram, the Triangles AEF and AHG are like, because of their right angles at F and G, & their common angle at A. Therefore, AF. FE ∷ AG. GH.  
29 The Radius is a mean proportional, between the tangent and the tangent compliment of an arch.  
Dem. HG is the tangent of an arch, CK the cotang thereof & LH= AG and the triangles, ALH and ACK are like, because of their right angles at L and C, and their common angle at A. Therefore all= HG. LH ∷ AC. CK.  
30 The Radius is a mean proportional, between the right sine of an arch, and the secant of its compliment.  
Demonst. In the preceding diagram the triangles AEF & AHG are like, therefore, AF. AE ∷ AG. AH.  
31 As the sine of an arch or angle is to Rad. so is the tangent of the same arch, to the secant thereof.  
Demonst. In the preceding diagram the triangles AEF & AGH are like, therefore EF. AE ∷ HG. AH.  
32 As Radius, is to the secant of an arch, so is the cotangent of the same arch, to the cosecant thereof.  
Demonst. In the preceding diagram, the triangles ALH and ACK are like, therefore LH, AH ∷ CK. AK.  
Other more easy and expeditious ways of making the Tangents and Secants, you may see in the first Chap. of my Trigonometria Britannica, but the Canons being now already made, these Rules we deem sufficient.  
The construction of the Artificial Sines and Tangents, we have purposely omitted, they being nothing else but the logarithms of the Natural, of which logarithms we have showed the construction in a former Institution, by the extraction of roots, and in my Trigonometria Britannica by multiplication: and therefore shall now proceed to the use of the Canon of Sines, Tangents and Secants, in the solution of all Triangles, whether plain or Spherical.     

CHAP. IV.  
Of the Calculation of plain Triangles.  
A Plain Triangle is contained under three right lines, and is either rightangled, 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 compliment of the other.  
3 In the resolution of plain Triangles, the angles only being given, the sides cannot be found, but only the the reason of the sides: It is therefore necessary, that one of the sides be known.  
4 In a rightangled Triangle two terms (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 rightangled plain Triangles there are seven cases, and five in Oblique, for the solution of which the four axioms following are sufficient.  

1 axiom.  
In a rightangled plain Triangle: The rectangle made of Radius & one of the sides containing the right-angle, is equal to the rect-angle made of the other containing side, and the tangent of the angle thereunto adjacent.  
Dem. In the right angled plain triang. BED draw the periphery FE, then is BE Radius & DE the Tangent of the angle at B, make CA parallel to DE, then are the Triangles ABC and ebbed like, because of their rightangles at A and E and their common angle at B, therefore BA, BE ∷ AC. ED. and BA × ED= BE × AC. that is BA × tB= Rad. × AC. as was to be proved.   

2 axiom.  
In all plain Triangles: The sides are proportional, to the sines of their opposite angles.  
Demonst. In the ●plain triangle BCD extend BC to F making BF= DC, and draw the arches FG & CH, then are the perpendiculars FE & CA the sines of the angles at B & D by the 7th of the third, and the triangles BEF and BAC are like, because of their right angles at E and A, and their common angle at B. Therefore C. C A ∷ BF. FE. that is BC. sine D ∷ DC= BF. sine B. as was to be proved.   

3 axiom.  
In all plain Triangles; As the half sum of the sides, is to their half difference: so is the tangent of the half sum of their opposite angles, to the tangent if their half difference.  
Demonst. In the triangle BCD let the sides be CB and CD, and CG= CB. wherefore ½ Z crur.= EG. & ½ × crur.= EC, draw CH bi-secting BG at right angles, and make the angle GCI= D, then will the angle GCH= ½ Z angle B and D whose tangent is HG, and the angle ICH= ½ × ang. B and D whose tangent is HI.  
But EG. EC ∷ HG. HI. that is. ½ Z crur. ½ × crur. ∷ t ½ Z ang t ½ X ang.   

4 axiom.  
In all plain triangles: As the base, is to the sum of the other sides, so is the difference of those sides, to the difference of the segments of the base.  
DB. BF ∷ HB. GA. That is, DB. Z crur. ∷ × crur. × seg. base.  
These things premised, we will now set down the several Problems or cases in all plain triangles rightangled and oblique, with the proportions by which they may be solved, & manner of solving them both by natural and Artificial numbers.   

Of right angled Plain Triangles.  
IN right angled plain triangles, the sides comprehending the right-angle we call the Legs, and the side subtending the right angle, we call the Hypotenuse.  
1 Probl. The legs given to find an Angle.  
The given legs 

AB. 230  
AC. 143.72    
AC. Rad ∷ AB. tA CB by the 1 axiom.  
That the quantity of this angle, or any other term required may be expressed in numbers, if the solution be to be made in natural numbers, multiply the second term given by the third, and their product divide by the first, the Quotient is the fourth proportional sought.  
But if the solution be to be made in artificial numbers, from the sum of the logarithms of the second and third terms given, subtract the logarithm of the first, the remainder shall be the logarithm of the fourth proportional required.  

Illustration by natural numbers.  

As the Leg AC 143.72  
Is to the Radius AC 10000000  
So is the Leg. AB 235  
To the tang. of ACB grad. 58.55   
The product of the second and third terms is 2350000000, which being divided by 143.72 the first term given, the quotient is 16351238 the tangent of the angle ACB which being sought in the Table the nearest less is 16350527, and the arch answering thereto is Grad. 58. 55 parts.   

Illustration by logarithms.  
 Logarithms. As the leg AC 143.72 2.157517 Is to the rad.. IC 10.0000 So is the leg AB 235 2.371068 To the tang. of. C. gr. 58.55 10.213551  
2 Probl. The angles and one leg given, to find the other leg.  
In the right angled plain Triangle ABC, the leg AC is inquired:  
The given 

Angle ABC  
Leg. AB   Rad. AB▪ ∷ t ABC. AC. by the first axiom.  
3 Prob. The Hypotenuse and a leg given to find an angle.  
In the right angled plain Triangle ABC, the angle ACB is inquired.  
The given 

Hypoth. BC.  
Leg AB.    
BC. Rad ∷ AB. s. ACB. by 2 Ax.  
4 Probl. The Hypotenuse and angles given, to find either leg.  
In the right angled plain Triangle ABC, the leg. AB is inquired:  
The given 

Hypot. BC.  
Angle ACB.    
Rad. BC ∷ s. ACB. AB. by 2 Ax.  
5 Probl. The angles and a leg given, to find the Hypotenuse.  
In the rightangled plain Triangle ABC, the Hypotenuse BC is inquired;  
The given 

Angle ABC.  
The given Leg AC.    
s. ABC. AC ∷ Rad. BC. by the second axiom.  
6 Probl. The Hypotenuse and leg given, to find the other leg.  
In the rightangled plain Triangle ABC, the leg AB is inquired,  
The given 

Hypot. BC  
Leg. AC    
1. BC. Rad ∷ AC. s. ABC, by the 3 Problem.  
2. t. ABC. AC ∷ Rad. AB. by the 2 Probl.  
7 Probl. The legs given, to find the Hypotenuse.  
In the rightangled plain Triangle ABC the Hypotenuse BC is inquired.  
The given legs 

AB  
AC    
1 AB. Rad ∷ AC. t. ABC by the 1 Problem.  
2 s. ABC. AC ∷ Rad. BC. by the 5 Problem.    

Of Oblique angled plain Triangles.  
1 Probl. Two sides and an angle opposite to one of them given, to find the angle opposite to the other side.  
In the Oblique angled plain Triangle DCB the angle CBD is inquired.  
The given Sides 

DC 865  
CB 632    
The given Angle CDB 26. 37  
But here it must be known whether the angle sought, be acute or obtuse. CB. s CDB ∷ CD, s. CBD. by the second axiom.  

Illustration by natural Numbers.  

As the side CB 632  
Is to the sine of CDB 26.37 4441661  
So is the side CD 865  
To the sine of CBD 37.43 6079172   
For if you multiply 4441661 by 865, the product will be 3842036765, which product being divided by the first term 632, the quotient 6079172 is the sine of 37 deg. 43 parts.   

Illustration by Artificial numbers.  

As the side CB 632 2.800717  
Is to the sine of D 26.37 9.647545  
So is the sides CD 865 2.937016  
12.584561  
To the sine of B 37.43 9.783844   
2 Probl. Two sides with the angle comprehended by them given, to find either of the other angles.  
In the Oblique angled plain Triangle BDC.  
The angle DBC is inquired.  
The given Sides 

DC  
BC    
The given Angle DCB  
Subtract the angle given, out of 180d. the remainder is the sum of the other angles, and ½ Z c●ur. DC and BC. ½ X crur ∷ t ½ Z ang. t ½ X ang. by the 3 Ax. To the half sum of the other angles, add the half difference found, and you will have the greater angle, subduct it & you will have the lesser, in our example ½ Z ang.+ ½ Xang. is= DBC sought, because that is the greater ang.  
3 Probl. The angles and a side given, to find either of the other sides.  
In the Oblique angled plain triangle CBD,  
The side DB is inquired.  
The given Angles 

DCB  
CDB    
The given Side CB  
s. CDB, CB ∷ s. DCB. DB. by the second axiom.  
4 Probl. Two sides with the angle comprehended by them given, to find the third side.  
In the oblique angled plain triangle DCB  
The side CB is inquired.  
The given Sides 

DB  
CB    
The given Angle CBD  
First, find the angle BCD by the third axiom.  
Secondly, find the side CB by the second axiom.  
5 Probl. The three sides given to find an angle.  
In the Oblique angled plain triangle CBD.  
The angle CDB is inquired.  
The three sides being given. 

DC  
CB  
DB    
The resolution of this Problem doth require two operations, first, for the segment of the base GB.  
DB. Z crur. DC & CB ∷ X crur. GB. ●y the fourth axiom.  
And DB − GB= DG, and ½ DG= DA or AG.  
And DC. Rad ∷ AD. s ACD, by the second axiom whose compliment ●s ADC.  
And AG+ GB= AB, therefore ang. ABC may be also found in like manner.     

CHAP. V.  
Of the affections of Spherical Triangles.  
HAving done with plain triangles, we come next to speak of Spherical.  
1 A Spherical triangle is that which is described on the surface of the Sphere.  
2 The sides of a Spherical Triangle are the arches of three great circles of the Sphere, mutually intersecting each other.  
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 bisected 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.  
7 In every Spherical Triang. any two sides are together greater than the third, for otherwise they cannot possibly make a triangle.  
8 The sum of the sides of a Spherical Triangle are less than 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 than AC by the last aforegoing, therefore the three sides AB, BC, AC are together less than the two semicircles BAD, BCD,  
Again. Let the sides AB and AC be less than a semicircle, seeing that the two angles at C are equal to two right, and the angle B less than the angle ACD, the angles ACB and B are together less than two right.  
Lastly, Let the sides AB and AC be more than a semicircle, the angles at C being equal to a semicircle, and the angle at B greater than the angle ACD, the angles ACB & B shall be greater than two right.  
10 The sum of the three angles of a Spherical triangle are greater than two right angles and less than six.  
Again. Let the angle ACD be less than the angle B, then the sum of the arches AB & AC shall be more than a Semicircle, and therefore the angles ABC and ACB greater also then two right, and therefore much more are the three angles A.B.C. greater than two right.  
Lastly. Let the angle ACD be greater than 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 ebb and EC be together equal to a semicircle, and therefore AE and EC are together less than a semicircle, and the angle EAF or BAC is greater than 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 than two right.  
And because every angle of a Spherical Triangle is less than two right, the three angles together must needs be less than six, as was to be proved. Therefore,  
11 Two angles of any Spherical Triangle are greater than the difference between the third angle and a semicircle also.  
12 Any side being continued, the exterior angle is less than the two interior opposite ones.  
13 In any Spherical Triangle, the difference of the sum of two angles, and a whole circle is greater than 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 than a semicircle, the outward angle shall be greater than the inward opposite angle: if greater than a semicircle, the outward angle shall be less than the inward opposite angle.  
Again, Let the sides AB and AC be less than the semicircle BAD if the common arch AB be taken away, there shall remain the arch AC less than the arch AD and therefore the ang. ACD shall be greater than the angle D, therefore also more than B.  
Lastly, If the sides AB and AC be together more than a semicircle, taking away the common arch AB, the remaining arch AC shall be greater than AD, and the angle ACD lesser than D, and therefore also lesser then B as was to be proved.  
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.  
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 than a quadrant; if of different, than 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 aforegoing, and the arches AB and BC which are perpendicular to the arch AC do meet in the pole B.  
Lastly, In the triangle DAH right angled at A, the leg AD is less than the quadrant AB, and the leg HA is greater than the quadrant AC, therefore the Hypotenusa DH is also greater than 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 than 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 than 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 shall 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 and B be both acute or obtuse the sides AC and CB shall be both acute or obtuse also by the 17th hereof, and the Hypotenuse AB less than 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 than a quadrant, by the 17th hereof, and the Hypotenuse AB more than 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 than the compliment of the other, but less than the difference of the same compliment 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 mixed, viz. acute and obtuse.  
24 In any Spherical Triangle, whose angles are all acute, each side is less than a quadrant.  
25 In any oblique angled spherical Triangle, if the angles at the base be of the same affection, the perpendicular drawn from the vertical angle shall fall within, if of different without.  
But in the Triangle AEB, obtuse-angled at B, acute at E, the perpendicular AD shall fall without the triangle, upon the side ebb 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.   

CHAP. VI.  
Of the Calculation of Spherical Triangles.  
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 axioms following.  

1 axiom.  
In all Spherical rectangled Triangles, having the same acute angle at the base: The sines of the Hypotenusaes' are proportional to the sines of their perpendicular.   

2 axiom.  
In all Spherical rectangled Triangles, having the same acute angle at the base: The sines of the bases, and the tangents of the perpendiculars are proportional.  
Demonst. Let ADB and AIM represent two Spherical Triangles having the same angle at A, then is IH the sine of IM and FD the sine of DB. But  
s. AD. AI Rad ∷ s DF s. IH.  
Again. KB is the tang. of DB & LM the tang. of IM. And  
s. AB. t. KB ∷ s. AM. t. LM. as was to be proved.  
That all the case of a rightangled Spherical Triangle may be resolved, by these two axioms, the several parts of the Spherical triangle, proposed must sometimes be continued to quadrants, that so the angles may be turned into sides, the hypotenusaes into bases and perpendiculars, and the contrary, by which means, the proportions as to the parts of the Triangle given, instead of sines do sometimes fall in cosines, and sometimes in cotangents instead of tangents, which the L. Nepeir observing, those parts of the right-ang. Spherical triangle which in such conversion do for the most part change their proportion, he noteth with their compliments, viz. the hypotenuse and both the acute angles, but the sides containing the right angle are not so noted: and these five he calleth the Circular parts of the triangle, amongst which the right angled is not reckoned, and therefore the two sides which do contain it, are supposed to be joined together.  
Of these five Circular parts, one is always in the middle, and two of the five are adjacent to that middle part, and the other two are disjunct, the parts adjacent, are called extremes adjacent, and the parts disjoined, are called extremes disjunct.  
Each of these five circular parts, may by supposition be made the middle part, and then the two circular parts which are next to that which is by supposition made the middle, are the extremes conjunct, the other two remote from the middle part assumed are the extreme parts disjoined.  
As in the Triangle ABC, if comp. AC be made the middle part, comp. A, and comp. C are the extremes conjunct, and the sides AB and BC are the extremes disjunct, and so of the rest as in the Table following.  
Mid. part Extr. conj. Extr. disj. Leg. AB. Comp. A Leg BC Comp. AC Com. C Comp. A Comp. AC Leg. AB Com. C Leg. BC Comp. AC Comp. A Comp. C Leg. AB Leg. BC Comp. C Comp. AC Leg. BC Comp. AB Leg. AB Leg. BC Comp. C Leg. AB Comp. A Comp. AC  
The parts of a right angled Spherical Triangle, being thus distinguished into five circular parts, (besides the right angle) the three parts remote from the right angle, being noted by their compliments, and of these five one accounted the middle, the other either extremes adjacent or disjunct, the L. Nepeir as a consectary from the two preceding axioms, for the help of memory, and therefore the more easy resolving of all spherical Triangles hath composed this Catholic and Universal Proposition.  
The Sine of the Middle part and Radius are reciprocally proportional, with the tangents of the extremes conjunct, and the cosines of the extremes disjunct.  
That is, As Radius, to the tangent of one of the extremes conjoined: So is the tangent of the other extreme conjunct to the Sine of the Middle part.  
And also, As Radius, to the cousin of one of the extremes disjunct: So is the cousin of the other extreme disjunct, to the sine of the Middle part.  
Therefore, if the Middle part be sought, the radius must be in the first place, If either of the extremes, the other extreme must be in the first place.  
only note, That if the Middle part, or either of the extremes conjunct, be noted with its compliment in the circular parts of the Triangle, instead of the sine or tangent, you must use the cousin or cotangent of such circular part or parts.  
If either of the extremes disjunct, be noted by its compliment in the circular parts of the triangle, instead of the cousin, you must use the sine of such extreme disjunct.  
That these directions may be the better conceived, we have in the table following set down the circular parts of a Triangle under their respective titles, whether they be taken for the Middle part, or for the extremes, whether conjunct or disjunct, & unto those parts we have perfixed the sine or cousin, the tangent or cotangent as it ought to be by the former Rules.  
Middle part Extr. conjunct Extr. disjunct. Sine AB cotang A Sine AC Tang. BC Sine C Cousin A cotang AC Sine C Tang. AB Cousin BC Cousin AC cotang A Cousin AB cotang C Cousin BC Cousin C cotang AC Sine A Tang. BC Cousin AB Sine BC cotang C Sine A Tang. AB Sine AC  
These things premised, we will now set down the several cases with the Analogies by which they may be solved, according to this Catholic proposition of right angled Spherical triangles first, and then of oblique, and in every case in which we deem it necessary, we will Demonstrate the truth of the Proposition by the first & second axioms of this chapter, as where the middle part, is either one of the legs containing the right angle, or one of the oblique angles or the Hypotenusal.   

Of right angled Spherical Triangles, Problem. 1.  
A leg with an angle opposite thereunto being given, to find the other leg. If it be known whether the hypotewse or the other angle be more or less than a quadrant.  
In the right angled Spherical Triangle ABC.  
The leg AB or middle part is inquired  
The given 

Leg BC  
ang. come. A   are extreme conjunct.  
Analogy Rad. cot. ang. A ∷ tang. BC. sine AB.  
Demonst. The Triangles AGD and ACB have the same acute angle at the base.  
Therefore ●an. DG. AD ∷ tang. BC ●ine AB by the second Ax. And by the 29th of the third.  
Rad. AD. cotang DG tang. DG. AD— therefore, Rad. cot. DG ∷ tang. BC. sine AB and DG is the measure of the angle, at A therefore, as Rad. &c.  
Illustration by Natural numbers.  

As the Radius. 10000000  
To cotang A grad. 30. 17320508  
So tang. BC. grand. 22.89 4222108  
To sine AB. 46 d. 994 7312904   
Illustration by Artificial numbers.  

As the Radius 90 d. 10.000000  
To cotang A 30 10.238561  
So tang. BC 22.89 9.625529  
To sine AB. 46.994 9.864090    

Problem 2.  
A leg and an angle coonterminate therewith being given, to find the other leg  
In the R. angled spherical trian. ABC.  
The leg BC one of the extremes conjunct is inquired.  
The given Angl. Comp. A the other extr. conj. Leg AB the middle part.  
Cot. A. Rad ∷ sine AB. tang. BC by the analogy in the first Problem.   

Problem 3.  
The legs given to find an angle.  
In the right angled Spherical Triangle ABC.  
The angle comp. A one of the extremes conjunct is inquired.  
The given legs 

AB the middle part.  
BC the other extrea. conjunct.    
analogy Tang. BC. sine AB ∷ Rad. cot. A. The inverse of that analogy in the ●st problem.   

Problem 4.  
The Hypotenuse and a leg given, to find the ●●gle contained by them.  
In the right angled Spherical Tri●ngle, ABC.  
The angle, comp. C, or middle part ●s inquired.  
The given 

Hypot. comp. AC  
Leg BC   the adjac. extr.  
analogy, rad.. cotang. AC ∷ tang. BC. cousin C.  
Demonst. The triangles EFH and EIG have the same acute angle at the base, therefore, ●IG= AC. Rad. IE ∷ t. FH= BC.— sine FE. by the 2 Ax.  
Rad. cotang IG ∷ tang. IG. Rad. by the 29 of the third. Therefore  
Rad. cot. IG ∷ t. FH sine FE, or Rad. cot. AC ∷ t. BC. cousin C, because FI the compliment of FE is the measure of ACB.   

Problem 5.  
A leg and an angle conterminate with it given to find the Hypotenuse.  
In the right angled Spherical Triangle ABC.  
The Hypot. comp. AC. one of the extremes conjunct is inquired.  
The given Leg BC. the other extr. conjunct. Angle comp. C, the middle part.  
analogy. tang. BC. cousin C ∷ rad.. cot. AC, it being the inverse of the analogy in the former Problem.   

Problem 6  
The Hypotenuse and angle given, to find the leg conterminate with the given angle.  
In the right angled Spherical Triangle, ABC.  
The leg BC one of the extremes adjacent is inquired.  
The given Hypot. comp. AC the other extr. conj. Angle comp. C. the middle part.  
analogy, cotang. AC. Rad ∷ cos. C. tang. BC.  
This also is the inverse of the 3 problem, and therefore needs no further demonstration.   

Problem 7.  
The Oblique angles given, to find the Hypotenuse.  
In the right angled Spherical Triangle ABC.  
The Hypot. comp. AC, the middle part is inquired.  
The given ang. 

Comp. A.  
Comp. C   The extremes conjunct.  
analogy, Rad cotang. C ∷ cotang. A cousin AC:  
Demonst. The triangles CGH & CIF have the same acute angle at the base, CG is the compliment of AC, HG the compliment of GD the measure of A, and FI the measure of C, but  
Tang. FI. CI ∷ tang. GH, sine GC by 2 axiom. therefore,  
Tang. C. Rad. ∷ cot. A. cos. AC. And as Rad. cot. C ∷ tang. C. Rad. Therefore Rad. cot. C ∷ cot. A. cousin AC.   

Problem 8.  
The Hypotenuse and an angle given, to find the other angle.  
In the right angled Spherical Trigle ABC.  
The angle comp. C, one of the extremes conjunct is inquired.  
The given Hypot. comp. AC, the middle part. Ang, comp. A the other extr. conj. Cotang. A. cousin AC ∷ Rad. cot. C, by the analogy in the preceding Probl.   

Problem 9  
The Hypotenuse, and an oblique angle given, to find the leg opposite to the given angle.  
In the right angled Spherical Triangle ABC.  
The leg BC the middle part is inquired.  
The given Hyp. comp. AC Angle comp. A extreme disjunct.  
analogy, Rad. sine A ∷ sine AC sine BC.  
Demonst. The Triangles ACB & AGD have the same acute angle at the base, therefore by the first axiom, AG. s. GD ∷ sine AC. sine BC.   

Problem 10.  
A leg and an angle opposite thereunto, being given, to find the Hypotenuse. If it be known whether it, or the other leg or angle be acute or obtuse.  
In the right angled Spherical Triangle ABC.  
The Hypotenuse comp. AC one of the extremes disjunct is inquired.  
The given Ang. come. A the other extr. d. Leg. BC the middle part.  
Anal. Sine A. R. ∷ sine BC. sine C, by the analogy in the preceding Probl.   

Problem 11.  
The Hypotenuse and a leg given, to find the angle opposite to the given leg.  
In the right angled Spherical Triangle ABC.  
The angle comp. A one of the extremes disjunct is inquired.  
The given Hyp. comp. AC, the other extr. disj. Leg. BC the middle part.  
Anal. Sine AC, sine BC ∷ R. sine A by the analogy in the 10 Probl.   

Problem 12.  
An angle and leg conterminate with it given, to find the other angle.  
In the right angled Spherical Triangle ABC.  
The angle comp. A the middle part is inquired.  


The given Angle comp. C  
The given Leg. BC   extremes disjunct.  
Anal. Rad. sine C ∷ cs BC. cos. A.  
Demonst. The triangles CFI and CHG have the same acute angle at the base, therefore by the first axiom● Rad. FC. sine FI ∷ sine HC. sine HG, and the compl. of HC is the leg BC, & GD the measure of the angle at A is the compliment of HG.  
Therefore,  
R. sine C ∷ cousin BC. cousin A.   

Problem 13.  
An angle and leg opposite thereunto being given, to find the other angle. If it be known whether it, the Hypotenuse or the other leg be acute or obtuse.  
In the right angled Spherical Triangle ABC.  
The ang. comp. C one of the extremes disjunct is inquired.  
The given Leg BC the other extreme disjunct. Angle, comp. A the middle part.  
Anal. cs BC. cs A ∷ Rad. sine C. by the analogy in the preceding problem.   

Problem 14.  
The oblique angles given, to find either leg.  
In the right angled Spherical Triangle ABC.  
The leg BC one of the extremes disjunct is inquired.  
The given Angle, comp. C the other extr. disj. Angle, comp. A, the middle part.  
Anal. Sine C Rad ∷ cs A. cs BC. the inverse of that analogy in the last Problem.   

Problem 15.  
The legs given, to find the Hypotenuse.  
In the right angled Spherical Tri●gle ABC.  
The Hypotenuse comp. AC the middle part is inquired.  
The given legs 

AB  
BC   extreme disjunct.  
Anal. Rad. cs AB ∷ cs BC cs AC.  
Demonst. Tsie triangles HBD and HCG have the same acute angle at the base, therefore  
Rad. HB. sine BD ∷ fine CH. s CG, But BA is the compliment of BC, and BC the compliment of CH, and CG the compliment of AC.  
Therefore, Rad. cs AB ∷ cs BC. cs AC.   

Problem 16.  
The Hypotenuse and a leg given, to find the other leg.  
In the right angled Spherical Triangle ABC.  
The leg AB one of the extremes disjunct is inquired.  
The given Leg BC the other extreme dijunct. Hypot. comp. AC the middle part.  
Anal. cs BC. cs AC ∷ R. cs AB, by the analogy in the 15th problem.    

CHAP. VII.  
Of Oblique angled Spherical Triangles.  
IN Oblique angled Spherical triangles, there are 12 Cases, 10 whereof may be resolved by the Catholic proposition, if the Spherical Triangle propounded be first converted into two Right, by letting fall of a perpendicular, 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 always 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 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 than the arch given, the perpendicular shall fall without, if less, within the triangle, as shall plainly appear by our explanation of the ensuing problems.  

Problem 1.  
Two sides with an angle opposite to one of them being given, to find the angle opposite to the other. If it be known whether the angle sought be acute or obtuse.  
In the oblique angled Spherical triangle ABC.  
The angle BAC is inquired.  
The given Sides 

AB 42.15  
BC 29.83    
The given Angle ACB 36. 14  
The Anal. s. BA. s. BC ∷ s. C. s A.  
For by the Catholic proposition.  

1 R. s AB ∷ s A. s BD.  
2 R. s BC ∷ s C. s BD.  
Therefore, s AB. s BC ∷ s C. s A.   

Illustration by natural numbers.  

As the sine of AB 42.15 6710738  
To the sine of BC 29.83 4974282  
So is the fine of ACB 36.14 5897603  
To the sine of BAC 25.92 4371552    

Illustration by artificial numbers.  

As the sine of AB 42.15 9.826770  
To the sine of BC 29.83 9.696730  
So the sine of ABC 36.14 9.770675  
19.467405  
To the sine of BAC 25.92 9.640635     

Problem 2  
Two angles with a side opposite to one of them being given, to find the side opposite to the other. If it be known whether the side sought be more or less than a quadrant.  
In the Oblique angled Spherical Triangle ABC.  
The side AB is inquired.  
The given Angles 

BAC  
ACB    
The given Side BC  
The analogy is s A. s C ∷ s BC. s AB, by the last aforegoing.   

Problem 3.  
Two sides and their contained angle given, to find the third side.  
In the Oblique angled Spherical Triangle ACD, the side DC is inquired.  
The given Sides 

AD  
AC    
The given Angle DAC  
In this case the perpendicular may fall from the extremity of either side, but opposite to the angle given.  
The terms of proportion.  

1 ct AC. R ∷ cs DAC. To AB  
And AD − AB= BD in the 1 Triang.  
But AD+ AB= BD in the 2 Triang.  
2 cs AB. cs AC ∷ R. cs BC  
3 R cs BC ∷ cs BD. cs DC   
Therefore  
cs AB. cs AC ∷ cs BD. cs DC.   

Problem 4.  
Two sides with an angle opposite to one of them, to find the third side. If it be known whether the side sought, or the angle opposite to the other given side, be acute or obtuse.  
In the Oblique angled Spherical Triangle ACD the side AD is inquired.  
The given Sides 

AC  
DC    
The given Angle ADC  
In this case let the perpendicular fall from the concourse of the given sides on the side inquired, continued if need be.  
The terms of proportion.  

1 ct CD. R ∷ cs ADC. t. BD.  
2 cs BD. cs CD ∷ Res CB  
3 cs CB. R ∷ ct AC. cs AB   
Therefore  

cs CD. cs BD ∷ cs AC. cs AB  
And BD+ AB= AD in the 1 Triang.  
But BD − AB= AD in the 2 Triang.    

Problem 5.  
Two sides and their contained angle given to find one of the other angles.  
In the Oblique angled Spherical Triangle ACD, the angle ADC is inquired.  
The given Sides 

AC  
AD    
The given Angle god  
In this case the perpendicular may fall from the extremity of either side given opposite to the angle given.  
The terms of proportion.  

1. ct AC. R ∷ cs DAC. To AB.  
And AD − AB= BD in the 1 Triang.  
But AD+ AB= BD in the 2 Traing.  
ct god. s AB ∷ R. t BC.  
To BC. R ∷ s BD. ct ADC   
Therefore,  
s AB. ct god ∷ s BD. ct ADC.   

Problem 6  
Two angles and a side opposite to one of them being given, to find the side between then. If it be known whether the side sought or the side opposite to the other given angle be acute or obtuse.  
In the Oblique angled Spherical Triangle ACD, the side AD is inquired.  
The given Ang. 

ADC  
God    
The given Side AC  
In this case let the perpendicular fall from the extremity of the given side on the side inquired, continued if need be.  
The terms of proportion.  

1 ct AC. R ∷ cs DAC. To AB.  
2 ct god. s AB ∷ R t BC  
3 R. t BC ∷ ct ADC. s BD.   
Therefore,  

ct god. s AB ∷ ct ADC. s BD.  
And AB+ BD= AD in the 1 Triang.  
But DB − AB= AD in the 2 Triang.    

Problem 7.  
Two angles and a side opposite to one of them being given, to find the third angle. If it be known whether the angle inquired or the side opposite to the other given angle be acute or obtuse.  
In the Oblique angled Spherical Triangle ACD, angle ACD is inquired.  
The given Angles 

DAC  
ADC    
The given Side AC  
In this case let the perpendicular fal● from the angle inquired.  
The terms of proportion.  

1 ct god. R ∷ cs AC. ct ACB  
2 s ACB. cs CAB ∷ R. cs BC  
3 cs BC. R ∷ cs BDC. s BCD   
Therefore.  
cs CAB. s ACB ∷ cs BDC. s BCD. And ACB+ BCD= ACD in the first Triangle. But BCD − ACB= ACD in the second Triangle.   

Problem 8.  
Two angles and the side between them given, to find the third angle.  
In the Oblique angled Spherical Triangle ACD, the angle ADC is inquired.  
The given Angle 

ACD  
God    
The given Side AC  
In this case, let the perpendicular fall from the extremity of the given side, and opposite to the angle inquired.  
The terms of proportion.  
1 ct CAB. R ∷ cs AC. ct ACB. And ACD − ACB= BCD in the first Triangle.  
But ACD+ ACB= BCD in the second Triangle.  
2 s ACB. cs CAB ∷ R. cs BC  
3 R. cs BC ∷ s BCD. cs CDB  
Therefore  
s ACB. cs CAB ∷ s BCD. cs CDB.   

Problem 9  
Two sides with an angle opposite to one of them being given, to find their contained angle. If it be known whether the angle inquired or the angle opposite to the other given side be acute or obtuse.  
In the Oblique angled Spherical Triangle ACD, the angle ACD is inquired.  
The given Sides 

AC  
CD    
The given Angle ADC  
In this case, let the perpendicular fall from the angle inquired.  
The terms of proportion.  

1 ct CDB. R ∷ cs CD. ct BCD.  
2 ct CD. cs BCD ∷ R. t BC  
3 R. t BC ∷ ct AC. cs ACB   
Therefore  
ct CD. cs BCD ∷ ct AC. cs ACB. And BCD+ ACB= ACD in the first Triangle.  
But BCD − ACB= ACD in the second Triangle.   

Problem 10.  
Two angles and the side between them given, to find either of the other sides.  
In the Oblique angled Spherical Triangle ACD, the side DC is inquired.  
The given Angles 

DAC  
ACD    
The given Side AC  
In this case, let the perpendicular fall from the concourse of the side given and sought, on the third side continued if need be.  
The terms of proportion.  
1 ct CAB. R ∷ cs AC. ct ACB. And ACD − ACB= BCD in the first Triangle.  
But ACD+ ACB= BCD in the second Triangle.  
2 ct AC. cs ACB ∷ R. t BC  
3 t B C. R ∷ cs BCD. ct CD  
Therefore  
cs BCA. ct AC ∷ cs BCD. ct CD.   

Problem 11.  
Three sides being given to find an angle.  
To resolve this Problem there must be some preparation made, for that the Catholic proposition is not of itself sufficient either for this or for the ensuing Problem. And therefore the Lord Nepier to bring these Problems within some compass of the Catholic Proposition, first, findeth the difference of the segments of that side, which being made the base of the Triangle is divided into two parts by letting fall a perpendicular, and that by help of this analogy, which we have demonstrated in the second Book of our Trigonemotria Britannica, chap. 2.  
As the tangent of half the base.  
Is to the tangent of half the sum of the other sides.  
So is the tangent of half the difference of those sides.  
To the tangent of half the difference of the segments of the base.  
In the Oblique angled spherical Triangle ACD, the angle god is inquired.  
The given Sides 

AC  
CD    
The given Base AD  
The terms of proportion.  
To ½ AD. To ½ Z cru. AC. CD ∷ t ½ X cr. t ½ AE  
And ½ AD+ ½ AE= AB ½ AD − ½ AE= BD  
In the first Triangle.  
½ AE+ ½ AD= AB ½ AE − ½ AD= BE or BD.  
In the second Triangle.  
Hence to find the angle at A.  
R. ct AC ∷ t AB. cs CAB.   

Problem 12.  
The three angles being given, to find a side.  
This Problem is the converse of the last aforegoing, and to be resolved after the same manner, if so be we convert the angles into sides: for the two less angles are always equal unto two sides of a Triangle comprehended by the arches of great circles drawn from there Poles, and the compliment of the greatest angle to a semicircle must be taken for the third side. If in this triangle therefore thus connected, you shall by the preceding Problem find an angle, that angle so found shall be one of the three sides inquired.  
As the Triangle ACD the poles of those arches are H. R and Q, which connected make the Triangle HRQ, the sides of the former Triangle being equal to the angles of the latter, taking for one of them, the compliment of the greater angle to a Semicircle: as AD is equal to the angle at H, or the arch EN.  
DC is equal to the angle at Q, or the arch MF.  
And AC is equal to the compliment of the angle HRQ, or the arch GL. Therefore if the angles, A, D, C, be given, the sides QR. QH. RH are likewise given.  
If therefore we resolve the triangle HRQ by the directions of the last Problem, the angle so found, shall be the side inquired.    

CHAP. VIII.  
Of the Solution of Oblique angled Spherical Triangles without letting fall a perpendicular.  
HItherto we have explained the Catholic or Universal Proposition, and showed how serviceable it is, or may be made in Oblique angled spherical Triangles, as well as right; and now for variety sake we will show 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.  

Propos. 1.  
In any spherical Triangle: the sines of the angles are proportional to the sines of their opposite sides, and the contrary.  
This we have demonstrated from the Catholic Proposition, in the first Problem of the last chapter.   

Propos. 2.  
In all Oblique angled Triangles whose sides together are less than a semicircle.  
As the sine of half the sum of the angles at the base,  
Is to the sine of the half difference of those angles,  
So is the tangent of half the base,  
To the tangent of half the difference of the sides.  
And also:  
As the cousin cf ½ the sum of the angles at the base  
Is to the cousin of the half difference of the same angles,  
So is the tangent of half the base:  
To the tangent of the half sum of the sides.  
This you may see demonstrated in the 2 chapter of Spherical triangles in our Trigonometria Britannica.   

Propos. 3.  
In all spherical Triangles: As the difference of the versed sines of the sum and difference of any two sides (including an angle) is to the Diameter: So is the difference between the versed sine of the third side, & the versed fine of the difference of the other two sides, to the versed sine of the angle comprehended of the said two sides.  
MH is the difference between PH the versed sine of PS, and PM the versed sine of PC the difference of the sides, QV is the diameter and OV the versed sine of PAS the angle sought: and the right lines NC. KL. and RG being parallel by the work, their intersegments MB and RC, MH and SC are proportional by the 14 of the second chapter. 

MB. MH ∷ RC. SC  
RC. SC ∷ QV. OV    
Therefore, MB. MH ∷ QV. OV. as was to be proved.  
Or thus, MB. MH ∷ ½ QV. ½ OV.   

Propos. 4.  
In all spherical Triangles: As the Rectangle of the sines of the sides containing the angle inquired, is to the square of Radius: So is the difference between the versed sine of the base and the versed sine of the difference of the other two sides, to the versed sine of the angle sought.  
Or rejecting the comm.. altitude OK, it will be LO × NE. DC × EC ∷ OB. DX▪ the versed sine of the angle sought as was to be proved.   

Propos. 5.  
In all Spherical Triangles: As the Rectangle of the sines of the sides containing the angle inquired.  
Is to the square of Radius,  
So is the Rectangle of the sines of the half sum and half difference of the base, and difference of the legs.  
To the Rectangle made of Rad. and half the versed sine of the angle inquired.  
We have already proved, by the last aforegoing.  
LO × NE. DC × AC ∷ OB. DX.  
And therefore also  
LO × NE. Rad. q ∷ ½ OB. ½ DX.  
And in the following Diag. OEGH the sum of ES and OE is the double measure of the angle BSO, the arch OS is the difference between the base ES and EO, the difference of the ●ides AK and AE, and R. ½ OS ∷ OH. ½ OB, ½ OS and ½ OH= OB × R.  
Therefore,  
LO × NE. Rq ∷ ½ OB × R. ½ DX × R  
And also,  
LO × NE. Rq ∷ ½ OS × ½ OH ½ DX × R which was to be proved.   

Propos. 6  
In all Spherical Triangles: As the Rectangle of the sines of the sides containing the angle inquired.  
Is to the square of Radius,  
So is the Rectangle of the sines of the half sum and half difference of the base, and difference of the legs.  
To the square of the sine of half the angle inquired.  
We have already proved, by the last aforegoing.  
LO × NE. Rq ∷ ½ OS × ½ OH. ½ DX × R.  
But the Rectangle made of Rad. & half the versed sine of an arch, is equal to the square of the sine of half the arch, as in the former figure, let the arch given be DT, then is DX the versed sine of that arch, and DF the right fine of half the arch, and the Triangles DFR and DTX are like.  
Therefore,  
DR. DF ∷ ½ DT (= DF.) ½ DX and DR × ½ DX= DFq  
Therefore,  
LO × NA. Rq ∷ ½ OS × ½ OH. D Fq as was to be proved.   

Problem 1.  
Two sides with an angle opposite to one of them being given, to find the angle opposite to the other.  
In the Oblique angled Spherical Triangle ABC, the angle BAC is inquired,  
The given Sides 

AB  
BC    
The given Angle ACB  
The Anal. s BA. s BC ∷ s C. s A. by the first Prop.   

Problem 2.  
Two angles with a side opposite to one of them being given, to find the side opposite to the other.  
In the Oblique angled Spherical Triangle ABC, the side AB is inquired.  
The given Angles 

BAC  
ACB    
The given Side BC  
The Anal s A. s C ∷ s BC. s AB by the first Prop.   

Problem 3.  
Two sides and their contained angle being given to find the other angles.  
In the Oblique angled Spherical Triangle ABC.  
The angles 

BAC  
ACB   are inquired.  
The given Sides 

AB  
BC    
The given Angle ABC  

1 Operation.  
s ½ Z crur. s ½ X crur. ∷ ct ½ B. t ½ X ang. A and C.   

2 Operation.  
cs ½ Z crur. cs ½ X crur. ∷ ct ½ B t ½ Z ang. A and C, by the second Proposition of this chapter. ½ Z ang.+ ½ X ang.= ACB, ½ Z ang. − ½ X ang.= BAC.    

Problem 4.  
Two angles and their contained side being given to find the other sides.  
In the Oblique angled Spherical Triangle ABC.  
The sides 

AB  
BC   are inquired.  
The given Angles 

BAC  
ACB    
The given Side AC  

1 Operation.  
s ½ Z ang. s ½ X ang. t ½ AC. To ½ X crur. AB and BC.   

2 Operation.  
cs ½ Z ang. cs ½ X ang. t ½ AC. To ½ Z cr. AB and BC. by the second Proposition of this chapter. ½ Z crur.+ ½ X crur.= AB. ½ Z crur. − ½ X crur.= BC.    

Problem 5  
Two sides with an angle opposite to one of them to find the third side.  
In the Oblique angled Spherical Triangle ABC, the side AC is inquired.  
The given Sides 

AB  
BC    
The given Angle ACB  

1 Operation.  
s AB. s BC ∷ s C. s A. by the first Proposition.   

2 Operation.  
s ½ X ang. A and C. s ½ Z ang. t ½ X crur. t ½ AC.  
By the first operation of the last probl. And ½ AC being doubled, giveth the side AC inquired.    

Problem 6.  
Two angles with a side opposite to one of them being given, to find the third angle.  
In the Oblique angled Spherical triangle ABC, the angle ABC is inquired.  
The given Angles 

BAC  
ACB    
The given Side AB  

1 Operation.  
s C. s A ∷ s AB. s BC. by the first Proposition.   

2 Operation.  
s ½ X crur. s ½ Z crur. t ½ X ang. A and C. ct ½ ABC. by the 1 oper. of the third probl.  
And the double of ½ ABC is the angle ABC inquired.    

Problem 7.  
Two angles and a side opposite to one of them being given, to find the side between them.  
In the oblique angled Spherical Triangle ABC, the side AC is inquired.  
The given Angles 

BAC  
ACB    
The given Side AB  

1 Operation.  
s C. s A ∷ s AB. s BC. by the first proposition.   

2 Operation.  
s ½ X ang. s ½ Z ang. t ½ X crur. t ½ AC. By the 1 oper. of the 4 Problem.  
And the double of ½ AC, is the side AC inquired.    

Problem 8.  
Two sides and an angle opposite to one of them being given, to find their contained angle.  
In the Oblique angled Spherical Triangle ABC, the angle CBA is inquired.  
The given Sides 

AB  
BC    
The given Angle ACB  

1 Operation.  
s AB. s BC ∷ s C. s A. by the 1 Prop.   

2 Operation.  


s ½ X crur. s ½ Z crur.  
To ½ X ang. ct ½ ABC,   by the 1 oper. of the 3 Probl.  
And the double of ½ ABC, is the angle ABC inquired.    

Problem 9  
Two sides and their contained angle given, to find the third side.  
In the Oblique angled Spherical Triangle ABC, the side AC is inquired.  
The given Sides 

AB  
BC    
The given Angle ABC  
First, find the other angles by the third Probl. Then you may find the third side by the 1 Prop.  
Or thus▪  
Radq. s AB × s BC s q ½ ABC. X us AC & us X crur. by the 4 & 6 Prop of this chap.  
And the versed sine X cr.+ X vs AC and us X cr.= vs AC.   

Problem 10.  
Two angles and the side between them given. to find the third angle.  
This problem is the converse of the last, and to be resolved in the same manner, either at 3 operations, finding first the other two sides by the 4th Probl. and then the third angle, by the 1 Propos.  
Or, else by the 4 & 6 Propositions of this chap. the angles being first converted into sides, and their contained side into an angle as hath been showed in the 12 Probl. of the last chap.   

Problem 11.  
Three sies being given, to find an angle.  
In the Oblique angled Spherical Triangle ABC, the angle CBA is inquired.  
The sides AB. BC. & AC being given.  
analogy, s AB × s BC. Radq. s ½ ZAC and X cru. × s ½ XAC and X cru. sq ½ B. by the 6 Propos.   

Problem 12.  
Three angles being given, to find a side.  
This is the converse of the last, and to be resolved after the same manner, the angles being first converted into sides, (as hath been already showed in the 12 Probl: of the last chapt.) for then the angle so found, shall be the side inquired.  
FINIS.