And here note I. That the Centres of all Azimuth Circles fall in the Horizon H A D being extended where need is The Centres of all the Hour-Circles fall in the Aequinoctial Line AE A ae being extended The Centres of the Tropicks and Parallels of Declination fall in the Axis of the World P A S extended And the Centres of the Circles of Altitude fall in the Prime Verticall Circle Z A N. Note II. That if the middle Point of any Hour-Circle do not fall just in the Aequinoctial or any Azimuth Circle just in the Horizon but on either side of them then you may find the Centres by the Geometricall Propositions at the beginning of this Book though there be other waies to find the Centres upon the Projection it self which I omit for that I would not cumber the Scheme with unnecessary Lines Seventhly Every Circle in the Projection hath its proper Pole as was before intimated Now for the finding of them you are to note that the Pole of every great Circle is 90 degr or a Quadrant of a Circle distant from the Circle it self upon that Line which cutteth the Circle at right Angles Thus the Poles of all the Hour-Circles are upon the Aequinoctial and the Poles of all the Azimuths upon the Horizon Now if you would find the Pole of the Hour-Circle P D S lay a Ruler upon P and D and it will cut the Meridian Circle in e then take 90 degr of your Line of Chords and set them from e to f a Ruler laid from P to f will cut the Aequinoctial in Y so is Y the Pole of the Hour-Circle P D S. Lastly The finding of the Poles of the Azimuth Circles is the same with the Hour-Circles So if you would find the Pole of the Azimuth Circle Z G N lay a Ruler upon Z and G it will cut the Meridian Circle in g then set 90 degr of your Chord from g to d so a Ruler laid from Z to d will cut the Horizon H A O in the Point ☉ which Point ☉ is the Pole of the Azimuth Circle Z G N. And thus have you found the Poles of one of the Hour and one of the Azimuth Circles And by the same manner of Work you may find the Poles of all the rest As The Pole of the Hour-Circle P D S will be found at Y The Pole of the Hour-Circle P C S will be found at V The Pole of the Hour-Circle P A S will be found at AE or ae The Pole of the Hour-Circle P B S will be found at X The Pole of the Azimuth Circle Z G N will be found at ☉ The Pole of the Azimuth Circle Z A N will be found at H or O The Pole of the Azimuth Circle Z F N will be found at T The Pole of the Azimuth Circle Z ☉ N will be found at G The Poles of the Horizon H A O are Z and N the Zenith and Nadir The Poles of the Aequinoctial AE A ae are P and S the Poles of the World The Poles of the Ecliptick ♋ A ♑ are Q and R. Thus have I given you at large a plain and easie method how to project the Sphere upon the Plain of the Meridian Circle by help of the Line of Chords onely Vpon which Projection by the intersection or crossing of the severall Circles thereof are constituted divers Sphericall Triangles some Right-angled and others Oblique-angled By the resolving of which Triangles variety of Questions appertaining to Astronomie Geographie and Navigation may with speed and exactness be resolved But before I come to shew the manner of working particular Questions of any kind it will be expedient that I shew you 1. how to measure or find the quantity of the Sides and Angles of a Sphericall Triangle as they are here projected and 2. how to project or lay down an Angle or Side of any quantity that shall be required I. A Sphericall Triangle being projected how to find the quantity of any Angle thereof LAY a Ruler to the angular Point and the extremity of the Sides containing the Angle they being continued to Quadrants and note where the Ruler cuts the Meridian or outward Circle at both which places make marks upon the Meridian the distance between those two marks being measured upon your Line of Chords shall give you the quantity of the Angle required Example I. IN the Triangle P ☉ O in the Projection let it be required to find the quantity of the Angle ☉ P O. First lay a Ruler upon the angular Point P and to the extreme ends of the Sides P ☉ and P O they being extended to Quadrants which is to that Circle which measures that Angle as the Aequinoctial measures all the Angles at P the Pole of the World the Horizon all the Angles at Z the Zenith c. So the Ruler laid from P to ae will cut the Meridian in ae and being laid from P to B it will cut the Meridian in the Point b. The distance b ae being taken in your Compasses and measured upon your Line of Chords will be found to contain 62 degr 46 min. which is the quantity of the Angle ☉ P O. But if upon the Point P you were to project an Angle to contain 62 degr 46 min. then take 90 degr of your Chords and set them from P to ae and through the Centre A draw the Line AE A ae then take 62 degr 46 m. out of your Line of Chords and set them from ae to b and laying a Ruler from P to b it will cut AE A ae in the Point B the Circle P B S being drawn the Angle at P will contain 62 degr 46 min. Example II. LET it be required to find the quantity of the Angle Z E P. Lay a Ruler to ☉ the Pole of the Circle Z E N and the Point E it will cut the Meridian Circle in M from M set 90 degr to z a Ruler laid from ☉ to z will cut the Circle Z E N it being extended beyond the Zenith Z at the Point δ. Again Lay a Ruler upon Y the Pole of the Circle P E S and it will cut the Meridian Circle in v set 90 degr from v to x upon the Meridian a Ruler laid from Y to x will cut the Circle P E S in y. This done lay a Ruler from E to δ and it will cut the Meridian in θ also lay the Ruler from E to y it will cut the Meridian in λ the distance θ λ being taken in your Compasses and applied to your Line of Chords will be found to contain 21 degr 45 min. And such is the quantity of the Angle Z E P. These two sorts of Angles are the most troublesome to find their quantities and therefore I have instanced in them There are other Angles in the Projection which render their measures to the eye without farther Instructions for finding their quantities II. A Sphericall Triangle being projected to find the quantity of
also the manner how to make all such Instruments as will be in use to him in the Practice of these Exercises As of his Line of Chords which is the chief and also of a Sea-Chart according to Mercator's Projection without the help of a Table of Meridionall Parts or a Meridian Line but easily and speedily by a Line of Chords onely and that either General from the Aequinoctial towards either Pole or Particular for any designed Navigation Moreover in the Working of the severall Questions by Mercator's Chart I have at the end of every Problem given an account of the Difference arising between Working by that and the Plain Sea-Chart whereby the Errours of the one are clearly detected and the Verity of the other discovered I might have proceeded farther to the performance of other things by the Line of Chords onely as to shew the manner of Sailing by the Arch of a great Circle but that with some other things tending to Navigation I shall shortly make publick in a Treatise formerly written by my self and Mr. Vincent Wing now in my hands and almost ready for the Press and wholly designed for the Use of Sea-men And for farther Use of this Line I have in a Treatise now in the Press taught the manner how to delineate all manner of Sun-Dials by projecting of the Sphere in Plano whereby not onely the Making but the Reason also of Dials is discovered To which Treatise if my present occasions will permit I shall adde both an Arithmeticall and an Instrumentall way of Dialling also Thus friendly Reader at present I present thee with This and in a few Months thou shalt God willing participate of those above mentioned All which I hope will be as kindly accepted as they are freely tendered by Will. Leybourn Arts and Sciences MATHEMATICALL Professed and Taught by the Authour Arithmetick In Whole Numbers and Fractions In Decimals and by Logarithms Instrumentally by Decimal Scales Napiers Bones and to extract the Square and Cube Roots by Inspection Geometrie The Principles thereof with the Practice and Demonstration Astronomie The Description of the Circles of the Sphere The Use of the Globes Celestiall and Terrestriall To project the Sphere in Plane upon any Circle Right or Oblique And upon these Foundations the following Superstructures The Use of Geometricall Instruments in the Practice of Longimetria or the Mensuration of Heights as of Trees Towers c. Depths as of Mines Wells Descents c. Distances as of Churches Towers c. Planometria or the Mensuration of Board Glass Pavement Tiling c. Or any other Superficies Stereometria or the Mensuration of Timber growing or squared Stone regular or irregular Cask commonly called Gageing Geodaesia or the Measuring of Land divers waies and by severall Instruments to draw the Plot of a whole Mannor or Lordship to cast up the Content thereof and to beautifie the same with all necessary Ornaments thereunto belonging Trigonometria Or the Mensuration of Triangles both Plain and Sphericall The Application thereof in the solution of Problems in Geometry Astronomie Geographie Navigation Fortification Dialling c. Navigation The Principles thereof and the manner of Sailing by The Plain Sea-Chart Mercator's Chart. The Arch of a great Circle Horologiographia Or Dialling Arithmetically by the Tables of Sines Tangents Logarithms Geometrically by Scale and Compasses Instrumentally by the Sector Quadrants Scales and other Instruments accommodated with Lines for that purpose You may hear of him at Mr. Hayes's at the Cross-daggers in Moor-fields ADVERTISEMENT AND here I thought good to give notice That if any Gentlemen studious in the Mathematicks have or shall have occasion for Instruments thereunto belonging or Books to shew the Vse of them they may be furnished with all sorts usefull both for Sea or Land either in Silver Brass or Wood by Walter Hayes at the Cross-daggers in Moor-fields next door to the Pope's head Tavern where they may have all sorts of Maps Globes Sea-plats Carpenters Rules Post and Pocket-Dials for any Latitude This Scheme having relation to the Fourth and Fifth Cases of Sphericall Triangles in Page 53 and 54 being casually omitted is here inserted GEOMETRICALL PROPOSITIONS and THEOREMS Necessary to be known and practised for the more easie understanding of the subsequent EXERCISES The First EXERCISE THE ARGUMENT THE following Propositions Theorems and Problems are such as will come in continuall use in the Practice of the subsequent EXERCISES and therefore I have here promiscuously inserted such as do relate to the following Treatises and ought throughly to be understood and often practised by which means nothing in the ensuing Cases Propositions and Problems will be difficult to be understood but that you may gradually proceed in your Practice from Exercise to Exercise without being referred to any other Book or Books GEOMETRICALL PROPOSITIONS PROP. I. A right Line being given to divide the same into two equal parts at right Angles LET the right Line to be divided be A B. First open your Compasses to any distance greater then the length of half the given Line then setting one foot in A with the other foot describe the Arches C C both above and below the Line A B. Secondly the Compasses still resting at the same distance set one foot in B and with the other describe the Arches D D cutting the former Arches C C in the Point E above and F below the given Line A B. Lastly draw the Line E F which will divide the given Line A B in two equal parts in the Point G and at right Angles PROP. II. Vpon a right Line given to erect a Perpendicular upon any part thereof LET the given Line be H K and from the Point L let it be required to erect a Perpendicular First open your Compasses to any convenient distance and setting one foot in the given Point L with the other make a Mark or Point at pleasure as M. Then keeping the Compass-point in M with the other describe the Arch N N above the Point L and also another Arch at O cutting the given Line in O. Lastly lay a Ruler from O to M which will cut the Arch N N before drawn in the Point P. So a Line drawn from P to L shall be a Perpendicular to the given Line H K and from the Point L. PROP. III. From a Point above to let fall a Perpendicular upon a right Line given LET the Point given be Q and the Line upon which the Perpendicular is to fall be R S. First opening the Compasses to any distance greater then Q X set one foot in Q and with the other cross the given Line R S in the Points T and V. Secondly the Compasses unaltered set one foot in T and with the other describe the Arch w w below the given Line R S. Thirdly remove the Compasses to V being still at the same distance and cross the Arch w w in the Point Z Lastly a Ruler laid from Q to Z will cut the Line R S in the
Point X. So a Line drawn from Q to X shall be perpendicular to the given Line R S. PROP. IV. A right Line being given to draw another right Line which shall be parallel thereto at any distance required LET A B be a Line given and let it be required to draw another right Line which shall be parallel thereunto and at the distance of the length of the Line C. First take the length of the Line C in your Compasses and setting one foot towards one end of the given Line as at D describe the Arch E. Secondly set one foot of the Compasses towards the other end of the given Line as at F and describe the Arch G. Lastly lay a Ruler to the Arches E and G so that the Ruler onely touch the Arches and not cut or cross them in any part So a Line drawn thereby shall be parallel to the given Line A B and at the distance of the Line C. PROP. V. A right Line being given to draw another right Line parallel thereunto which shall pass through a given Point LET the given Line be H K to which let it be required to draw a parallel Line which shall pass through the Point L. First take in your Compasses the distance K L and setting one foot of the Compasses in H with the other describe the Arch M M. Secondly take the Line H K in your Compasses and setting one foot in the given Point L with the other cross the Arch M M in the Point O. So a Line drawn from L to O shall be parallel to the given Line H K and shall pass through the Point L. PROP. VI. Three right Lines being given to make a Triangle whose three Sides shall be equal to the three given Lines LET the three Lines given be N P Q. First take the Line N in your Compasses and lay that down from R to S. Secondly take the Line P in your Compasses and setting one foot in S with the other describe the Arch V V. Thirdly take the Line Q in your Compasses and setting one foot in R with the other cross the Arch V V in the Point T. Lastly draw the Lines T R and T S. So shall you have constituted the Triangle T R S whose three Sides are equal to the three given Lines N P Q. PROP. VII Three Points which lie not in a straight Line being given to finde the Centre of a Circle which being described shall pass through the three given Points LET the three given Points be A B C. First open your Compasses to any distance greater then half the distance between A and B and setting one foot in B with the other describe the Arch G D. Then remove the Compasses set one foot in A and with the other cross the former Arch in the Points D and F and draw the Line D F. Secondly the Compasses still continuing at the same distance set one foot in the Point C and with the other cross the Arch before drawn in the Points E and G and draw the Line E G crossing the other Line D F in the Point H. So shall H be the Centre of the Circle which being described shall pass directly through the three given Points A B C. PROP. VIII Two Points within any Circle being given how to describe the Arch of another great Circle which shall pass through those two given Points and also divide the Circumference of the given Circle into two equal parts LET the two Points given be E and F within the Circle A B C D. First through either of them as through E draw the right Line E D passing through the Centre of the Circle at K. Secondly draw the Line A C at right Angles to B D so shall the Circle be divided into four equal parts or Quadrants by the Lines A C and B D. Thirdly draw the Line E A and upon the Point A by the II. Prop. erect the Perpendicular A G cutting the Line B D it being extended in the Point G so have you three Points E F and G through which by the last Prop. ou may draw a Circle to pass whose Centre will be at H upon which Point if you describe an Arch of a Circle at the distance H E or H F it will pass through the two given Points E and F and divide the Circle A B C D into two equal parts in the Points L and M which was required And that this Arch thus drawn doth divide the Circle into two equal parts is evident for a Line drawn from L to M will pass directly through the Centre K. These are such Geometricall Propositions as are absolutely necessary for the working of the severall Conclusions in the following Exercises More might have been added but these well understood and practised will be sufficient to carry you through this Work And now I will describe unto you the making of the slight Instrument by which all contained in this Book is performed namely The Line of Chords and shew you the general Vse thereof in the protracting or laying down of Angles of any quantity Or if any Angle be already laid down to finde thereby the quantity thereof Then will I give you some few usefull and necessary Theorems chiefly appertaining to Trigonometry or the Solution of Triangles and so conclude this first EXERCISE How to make a Line of Chords ACcording to the largeness of your Line of Chords you intend to make draw a right Line as A B and upon the Point A by the II. Prop. erect the Perpendicular A C and upon A as a Centre describe the Quadrant B D E C which you must divide into 90 equal parts or Degrees Which that you may readily doe your Compasses being opened to the distance A B set one foot in B and the other will reach to E also set one foot in C and the other will reach to D so is your Quadrant divided into three equal parts each part containing 30 degr This done divide each of these three parts into three more so shall you have divided your Quadrant into 9 equal parts each containing 10 degr and each of these 9 parts being divided into halves will contain 5 degr and if you make your Line large enough you must divide those into 5 equal parts which you may very well doe if the Line A B be but two inches long as all the Schemes and Figures in these Exercises are drawn by a Line of Chords of that length Your Quadrant being thus divided into 90 degr draw the Line B C and parallel thereto two other Lines one pretty close to B C to contain the small Divisions and the other at a larger distance to set the Figures in Now it is the Line B C which is called the Line of Chords possibly for this Reason the Arch or Ark B D E C representing Arcus a Bow and B C the String or Chord thereof the divisions whereof are to be transferred from the degrees of the Quadrant
B D E C in this manner First setting one foot of your Compasses in B extend the other to 80 degr in the Quadrant and from the division of 80 degr in the Quadrant draw the Arch 80 80 which will cut the Chord-Line in 80 doe so with 70 60 50 c. and the like with every fifth degree as you see in the Figure And if your Line be very large you may doe so to every single degree and part of a degree And by this means have you reduced the degrees of the Quadrant B D E C to the straight Line B C more commodious to be set upon a Ruler then the crooked Arch B D E C. The Uses of the Line of Chords THE Vses of this Line are principally two The one is To protract or lay down upon Paper an Angle of any quantity that is of any number of degrees required The other Vse is If an Angle be already protracted or laid down to finde how many degrees and parts of a degree it containeth In both which I would have the Reader very perfect because very much contained in this Book hath dependence thereupon And here it will be necessary that I give you the Definition of an Angle Know therefore that an Angle is the Inclination or bowing of two right Lines the one to the other As the two right Lines C A and B A incline the one to the other and touch or meet each other in the Point A in which Point by reason of the inclination of the said Lines is made the Angle C A B. And here note that an Angle is commonly signed by three Letters the middlemost whereof signifies the angular Point As in this Figure when we say the Angle C A B you are to understand the very Point at A. I. How to protract or lay down upon Paper an Angle containing any number of Degrees and Minutes by the Line of Chords DRaw a right Line at pleasure as A B and from the Point A let it be required to protract or lay down an Angle containing 40 degrees First open your Compasses alwaies to 60 degr of your Line of Chords which is equal to the Line A C of the Quadrant and with this distance setting one foot of the Compasses upon the Point A with the other foot describe the Arch B C. Secondly take in your Compasses 40 degr which is the quantity of the Angle to be laid down out of the Line of Chords from the beginning thereof and setting one foot in B the other will reach to C upon the Arch wherefore through the Point C draw the Line C A. So shall the Angle at A contain 40 degr as was required II. An Angle that is already protracted how to finde the quantity of Degrees it containeth SUppose C A B were an Angle already protracted and it were required to finde the quantity thereof First open your Compasses to 60 degr of your Line of Chords and setting one foot in A the angular Point with the other describe the Arch B C. Secondly take in your Compasses the distance between B and C which distance apply to your Line of Chords by setting one foot in the beginning thereof and you shall finde the other to fall upon 40 degr which is the quantity of the Angle at A. Thus have you the Vses of your Line of Chords in protracting and finding the quantities of Angles And now it will not be impertinent if in this place I shew you how Angles may be protracted and laid down and also their quantities found by an Instrument which I shall make use of towards the end of this Book which I call a Protracting Quadrant It s Description IT is no other then a Quadrant made upon a piece of very thin Brass and divided into 90 degr the Brass being cut away close to the divisions of the degrees on the out-side and also the hollow within so that there remains nothing but the Limb and the two Sides as you may discern by the Figure In the protracting or laying down of Angles and in finding of the quantity of Angles already laid down this is Its Vse SUppose you were to finde the quantity of the Angle C A B. Hold a Pin or Needle upon the angular Point at A to which bring the Centre of your Quadrant noted also with A and there turn it about till the Meridian Line thereof A B lie upon the Line A B of the Angle then see under what degrees of the Quadrant the Line A C lieth which you shall find to lie just under 40 degr And such is the quantity of the Angle C A B. And if upon the Line B A you were to protract such an Angle of 40 degr Lay the Meridian Line of the Quadrant upon the Line A B the Centre of the Quadrant upon the Point A and with your Needle make a prick or point just against 40 degr of the Quadrant's limb So a Line drawn from A through this Point shall make an Angle of 40 degr Trigonometricall Theorems 1. A Triangle is a Figure consisting of three Sides and as many Angles as is the Figure C A B. 2. Any two Sides of a Triangle are called the Sides of the Angle contained by them as the Sides C B and A B are the Sides containing the Angle C B A. 3. The measure of an Angle is the quantity of the Arch of a Circle described upon the angular Point and cutting both the Sides containing the Angle As in the Triangle A C B the Arch f g is the measure of the Angle at C the Arch d e is the measure of the Angle at B and the Angle at A is a right Angle containing 90 degr 4. A Degree is the 360. part of any Circle Therefore 5. A Semicircle contains 180 degr And 6. A Quadrant or right Angle contains 90 degr 7. The Complement of an Angle less then 90 degr is so much as that Angle wanteth of 90 deg As the Angle A C B of the Triangle containeth 53 degr 7. min. the Complement whereof is 36 degr 53 min. which is so much as 53 degr 7 min. wanteth of 90 degr For if you subtract 53 degr 37 min. from 90 degr or from 89 degr 60 min. for ease in subtracting the remainder will be 36 degr 53 min. 8. The Complement of an Angle to a Semicircle is so much as that Angle wanteth of 180 degr So the Angle C being 53 degr 7 min. take 53 degr 7 min. from 180 degr or from 179 degr 60 min. and the remainder will be 126 degr 53 min. which is the Complement of the Angle C to 180 degr 9. An Angle is either right acute or obtuse 10. A Right Angle is that whose measure is 90 degr or a Quadrant 11. An Acute Angle is less then a right Angle and alwaies contains less then 90 degr 12. An Obtuse Angle is greater then a right Angle and alwaies contains more then 90 degr 13. A Triangle is either right-angled
length of the Perpendicular C A. The Analogie or Proportion is As the Sine of the Angle at C is to the Log. of A B So is the Sine of the Angle B to the Logar of C A. Or As the Radius is to the Logar of A B So is the Tangent of B to the Logar of C A. CASE IV. The Hypotenuse C B 225 the Angle C 53 degr 7 min. and the Angle at B 36 degr 53 min. given to finde the Base B A and the Perpendicular C A. DRaw a right Line C B containing 225 of your Line of equal parts then taking 60 deg out of your Line of Chords set one foot of the Compasses in B and with the other describe the Arch e d also the Compasses continuing at the same distance place one foot in C and with the other describe the Arch g f. Then from the Point B and through the Point d draw a right Line also from the Point C and through the Point f draw another right Line these two Lines will intersect or cross each other in the Point A forming the Triangle C A B. Lastly take the Line A B in your Compasses and applying it to your Scale of equal parts you shall finde it to contain 180 and that is the length of the Base A B. Likewise A C being taken in the Compasses and measured upon the Line of equal parts will be found to contain 135 which is the length of the Perpendicular C A. The Analogie or Proportion is As the Radius is to the Logarithm of C B So is the Sine of C to the Logarithm of A B And the Sine of B to the Logarithm of C A. CASE V. The Hypotenuse C B 225 and the Base A B 180 being given to finde the Perpendicular C A. DRaw a right Line A B containing 180 of your Scale of equal parts and upon the end A erect a Perpendicular A C. Then take out of your Scale of equal parts 225 the length of your Hypotenuse given and setting one foot of the Compasses in B with the other describe the Arch h k cutting the Perpendicular A C in C then draw the Line C B so have you constituted the Triangle C A B. Lastly take in your Compasses the length of the Line A C and apply it to your Line of equal parts where you shall finde that it will contain 135 and that is the length of the Perpendicular C A. The Analogie or Proportion is 1. Operation As the Logarithm of C B is to the Radius So is the Logarithm of A B to the Sine of C. 2. Operation As the Radius is to the Logarithm of C B So is the Sine of B the Complement of C to the Log. of C A. CASE VI. The Base A B 180 the Angle C 53 degr 7 min. and the Angle B 36 degr 53 min. being given to finde the Hypotenuse C B. DRaw a right Line A B containing 180 parts of your Scale for the Base of your Triangle and on the end A erect a Perpendicular A C. Then take 60 degr out of your Line of Chords and upon the Point B with that distance describe the Arch d e and because the Angle at B is 36 degr 53 min. take 36 degr 53 min. from your Line of Chords and set it upon the Arch from d to e. Then from B through the Point e draw a right Line till it meet with the Perpendicular before drawn which it will do in the Point C. And thus have you protracted your Triangle C A B. Lastly take in your Compasses the length of the Hypotenuse C B and measure it upon your Scale of equal parts and you shall finde it to contain 225. The Analogie or Proportion is As the Sine of C is to the Logarithm of A B So is the Radius to the Logarithm of C B. CASE VII The Base A B 180 and the Perpendicular C A 135 being given to finde the Hypotenuse C A. DRaw a right Line A B containing 180 equal parts and upon the end A erect the Perpendicular A C and out of your Scale of equal parts take the length thereof 135 which set from A to C and draw the Line C B which constitutes the Triangle C A B. Lastly take the length of the Hypotenuse C B in your Compasses and measuring it upon your Line of equal parts you shall finde it to contain 225. The Analogie or Proportion is 1. Operation As the Logarithm of A B is to the Logarithm of C A So is the Radius to the Tangent of B. 2. Operation As the Sine of B is to the Logarithm of C A So is the Radius to the Logarithm of C B. These are the severall Varieties or Cases that can at any time fall out in the Solution of Right-angled plain Triangles wherefore we will now proceed to the Solution of Oblique plain Triangles II. Of Oblique-angled plain Triangles THE Triangle which I shall make use of in the Solution of the severall Cases appertaining to an Oblique-angled plain Triangle shall be this following C D B in which   parts  D B the Base contains 335  C B the longer Side 271  D C the shorter Side 100  And  deg m. C the obtuse Angle contain 122 00 D the 2 acute Angles 43 20 B the 2 acute Angles 14 40 CASE I. Two Sides as the Base D B 335 and the Side C B 271 and the Angle D 43 degr 20 min. opposite to C B to finde the Angle at C opposite to the Base D B. DRaw a right Line D B representing the Base of your Triangle which by help of your Scale of equal parts make to contain 335. Then upon the Point D with the distance of 60 degr of your Line of Chords describe the Arch k l and from your Chords take 43 degr 20 min. the quantity of the Angle at D and set it upon the Arch-line from l to k drawing the Line C D. And because your other given Side B C contains 271 parts take 271 out of your Line of equal parts and setting one foot in B with the other describe the Arch m n crossing the former Arch k l in the Point C then draw the Line C B. So shall you have constituted the Triangle C D B. Lastly because it is the Angle at C that is required take 60 degr of your Chords and upon C describe the Arch g h and taking the distance between g and h apply it to your Line of Chords and you shall finde it to reach from the beginning thereof beyond the end of the Line wherefore take 90 degr the whole Line and set that distance from g to o then take the remainder of the Arch o h and measure that upon your Chord and you shall finde it to contain 32 degr which added to 90 degr make 122 degr and that is the quantity of the Angle at C which was required The Analogie or Proportion is As
and K H both parallel to the Line B A. Then taking in your Compasses the distance A 6 set one foot of them in C and with the other describe the Arch Z and draw the Line A X so that it onely touch the Arch Z. Then taking in your Compasses the distance A H set one foot upon the Line A C moving it along till the other foot being turned about will onely touch the Line A Z and where the point of the Compasses resteth upon the Line A C which it will doe at e through that Point draw the Line o e parallel to B A. So shall B o being measured upon your Line of Chords give you 69 degr 22 min. the quantity of the enquired Angle at B. CASE VIII The Base A C 27 degr 54 min. and the Hypotenuse A B 30 degr being given to finde the Perpendicular B C. The Analogie or Proportion is As the Co-sine of the Base A C 62 degr 6 min. is to the Radius So is the Co-sine of the Hypotenuse 60 degr to the Co-sine of the Perpendicular B C. HAving drawn your Quadrant A B C take out of your Line of Chords 62 degr 6 min. the Co-sine of the Base A C and set them from B to S. Also take from the Chords 60 degr the Co-sine of the Hypotenuse and set them from B to m and draw the Lines S T and m n both parallel to B A. Then taking the distance A T in your Compasses set one foot in C and with the other describe the Arch V and draw the Line A W so that it may onely touch the Arch V. Then taking A n in your Compasses move one foot thereof gently along the Line C A till the other being turned about doth onely touch the Line A W and where the Point resteth upon the Line C A which you will finde to be at c there make a mark and draw the Line c a parallel to B A. Lastly take the distance from B to a and measure it upon your Line of Chords where you shall finde it to contain 78 degr 30 min. the Complement of the Perpendicular or C A measured upon the Chord will give you 11 d. 30 min. the Perpendicular it self These Five last are all the Cases in a Right-angled Sphericall Triangle that are resolvable by Sines alone Those which follow are to be resolved by Sines and Tangents joyntly and so will require another manner of Operation then the former CASE IX The Hypotenuse A B 30 degr and the Angle at the Base A 23 d. 30 min. being given to finde the Angle at the Perpendicular B. The Analogie or Proportion is As the Radius is to the Co-sine of the Hypotenuse A B 60 deg So is the Tangent of the Angle at the Base A 23 degr 30 min. to the Co-tangent of the Angle at the Perpendicular B. FIrst draw a right Line as C H and upon one end thereof as at C erect the Perpendicular C A and with the distance of 60 degr of your Line of Chords upon the Centre C describe the Quadrant C A D. Also upon the Point A with 60 degr of your Chord describe the Quadrant A B C. Being thus prepared First take 60 deg the Co-sine of the Hypotenuse A B and set them from A to O also take 23 d. 30 m. the quantity of the Angle at A and set them from C to r and draw the Line A M F and the Line O G parallel to A C. Then take in your Compasses the distance between C and G and setting one foot in D with the other describe the Arch K and draw the Line A L so that it onely touch the Arch K. Then placing one foot of the Compasses in F take the least distance you can to the Line C L which set from C to E. Lastly draw the Line A E cutting the Quadrant C B in N. So the distance C N measured upon your Chords shall give you 20 degr 38 min. the Complement of the Angle at B which was required or the distance B N will give you 69 degr 22 min. the Angle it self CASE X. The Hypotenuse A B 30 degr and the Angle at the Base A 23 d. 30 min. being given to finde the Base A C. The Analogie or Proportion is As the Radius is to the Co-sine of the Angle at A 66 d. 30 m. So is the Tangent of A B the Hypotenuse 30 degr to the Tangent of the Base A C. HAving prepared your Quadrants out of your Line of Chords take 66 degr 30 min. the Complement of the Angle at A and set them from A to a and draw the Line a b parallel to A C also take 30 degr the Hypotenuse from your Chord and set them from C to c and draw the Line A c prolonging it to d. This done take in your Compasses the distance from C to d and setting one foot in D with the other describe the Arch P and draw the Line C Q so that it may onely touch the Arch P. Then setting one foot of the Compasses in b take the least distance between b and the Line C Q which distance will reach from C to e. Lastly draw the Line A e cutting the Quadrant A C B in the Point M. So shall the distance C M being taken in the Compasses and measured upon the Line of Chords contain 27 degr 54 m. which is the quantity of the Base required CASE XI The Base A C 27 degr 54 min. and the Angle at the Base A 23 degr 30 min. being given to finde the Perpendicular B C. The Analogie or Proportion is As the Radius is to the Sine of the Base A C 27 degr 54 min. So is the Tangent of the Angle at A 23 degr 30 min. to the Tangent of the Perpendicular B C. YOur Quadrants being prepared out of your Line of Chords take 27 d. 54 m. the quantity of the given Base A C and set them from A to S drawing the Line S R parallel to C D. Also take out of your Chords 23 degr 30 minutes the quantity of the given Angle at A and set them from C to r and draw the Line A r prolonging it to F. This done take in your Compasses the distance A R and setting one foot in D with the other describe the Arch T and by the side thereof draw the Line C V onely to touch it Then set one foot of your Compasses in F and with the other take the nearest distance to the Line C V which distance will reach from C to X. Lastly draw the Line A X which will cut the Quadrant A B C in the Point Z. So shall C Z being measured upon the Line of Chords contain 11 degr 30 min. which is the quantity of the Perpendicular B C which was required CASE XII The Base A C 27 degr 54 min. and the Perpendicular B C 11 d. 30 min. being given to finde the Angle at the Base A. The
any Side thereof A Ruler laid upon the Pole of the Circle which is to be measured and to the extreme ends of the Side of the Triangle note where the Ruler so laid cuts the Meridian at both ends of the Side that distance taken in your Compasses and measured upon the Line of Chords will give you the quantity of the Side of the Triangle Example I. LET it be required to find the Side E Z of the Triangle Z E P. Lay a Ruler to ☉ the Pole of the Circle Z E N and the angular Point E it will cut the Meridian in M and a Ruler laid to Z will cut the Meridian in Z. So the distance M Z taken in the Compasses and measured upon the Line of Chords will be found to contain 78 degr And such is the quantity of the Side Z E. Example II. LET it be required to find the Side ☉ B of the Sphericall Triangle A ☉ B. Lay a Ruler upon X the Pole of the Circle P B S and the Point B it will cut the Meridian Circle in ae Also lay a Ruler from X to ☉ it will cut the Meridian in the Point ♌ The distance between ae and ♌ being taken and measured on the Line of Chords will contain 20 d. And such is the quantity of the Side ☉ B. I could instance in divers other Examples concerning the Measuring of the Sides and Angles of Triangles upon the Projection but I here omit them because in the resolving of the following Propositions they will come in practice and the Manner of the performance is there plainly expressed onely I deemed it convenient here to give some taste thereof as a Preparative to that which followeth But before I come to shew the Manner of resolving of particular Questions in Astronomie Geography c. I will declare the Variety of Sphericall Problems that will naturally arise out of every Sphericall Triangle being projected THE VARIETY OF SPHERICALL PROBLEMS Naturally arising out of every Sphericall Triangle both Right and Oblique-angled and that are resolvable thereby described as they are perspicuous to the Eye in the Projection The Fifth EXERCISE IN the foregoing Part of this Book you have the Doctrine of Plain and Sphericall Triangles Geometrically performed And in the Solution of Right-angled Sphericall Triangles there were 16 Cases and in Oblique-angled there were 12 Cases but the 16 Cases of Right-angled Triangles will by this projective way be reduced to 5 Cases and the 12 of Oblique-angled will be reduced to 6 so that in both there will be but 11 Cases whereas before there were 28. That this may appear plain to the Reader I will make use of two Triangles in the Projection one whereof shall be Right-angled as the Triangle P O ☉ Right-angled at O and the other shall be the Oblique-angled Triangle Z E P. The Right-angled Triangle is constituted by the Intersection of three great Circles of the Sphere namely of P O an Arch of the Meridian ☉ O an Arch of the Horizon and P ☉ the Arch of an Hour-Circle The Oblique-angled Triangle Z E P is constituted also of three Arches of great Circles of the Sphere as all Sphericall Triangles whatsoever are namely of Z P an Arch of the Meridian P E an Arch of an Hour-Circle and Z E an Arch of an Azimuth Circle In the Right-angled Sphericall Triangle P ☉ O The Side P O is the Latitude of the Place The Side ☉ O is the Sun's Amplitude from the North. The Side P ☉ is the Sun's distance from the Pole or the Complement of his Declination The Angle ☉ P O is the Hour from Midnight or from the North part of the Meridian The Angle P ☉ O is the Angle of the Sun's Position at the time of the Question The Angle P O ☉ is the Right Angle The Parts of the Triangle being declared and of what Circles of the Sphere the Sides do consist I will now come to the Cases which as I said before are 5 in every Right-angled Triangle So that any two parts of the Triangle besides the Right Angle being given I will shew in every of the 5 Cases what parts may be found I. In a Right-angled Sphericall Triangle CASE I. The Base and Perpendicular being given to finde the other parts of the Triangle IN the Triangle P ☉ O here is given P O the Latitude and ☉ O the Amplitude from the North part of the Meridian by which you may find 3. 1. P ☉ the Complement of the Sun's Declination 2. ☉ P O the Hour from Midnight 3. P ☉ O the Angle of the Sun's Position CASE II. The Hypotenuse and Perpendicular being given to find the other parts IN the Triangle P ☉ O here is given P O the Latitude and P ☉ the Complement of the Sun's Declination by which may be found 6. 1. ☉ O the Amplitude from the North. 2. O P ☉ the Hour from Midnight 3. P ☉ O the Angle of the Sun's Position And if in stead of the Perpendicular there had been given the Base ☉ O you might then find 4. P O the Latitude 5. ☉ P O the Hour from Midnight 6. P ☉ O the Angle of the Sun's Position CASE III. The Hypotenuse and an Angle being given to find the other Parts HERE is given in the Triangle P ☉ O the Complement of the Sun's Declination P ☉ and ☉ P O the Hour from Midnight by which may be found 6. 1. ☉ O the Amplitude from the North. 2. P O the Latitude 3. C ☉ O the Angle of the Sun's Position And if in lieu of the Angle at P the Angle at ☉ P had been taken then you might have found 4. ☉ O the Amplitude from the North. 5. P O the Latitude 6. ☉ P A the Hour from Midnight CASE IV. The Perpendicular or Base and either of the Angles given to find the other Parts IN the Triangle P ☉ O let there be given the Amplitude ☉ O and the Angle of the Sun's Position P ☉ O by which you may find 12. 1. P O the Latitude 2. ☉ P O the Hour from Midnight 3. P ☉ the distance of the Sun from the Pole But if the Side given had been ☉ O and the Angle given ☉ P O then you might have found 4. P ☉ the Complement of the Sun's Declination 5. P O the Latitude 6. P ☉ O the Angle of the Sun's Position But again if ☉ O and ☉ P O had been given then might be found 7. ☉ P the Complement of the Sun's Declination 8. ☉ O the Amplitude from the North. 9. P ☉ O the Angle of the Sun's Position And again if P O and O ☉ P had been given we might then have also found 10. ☉ P the distance of the Sun from the Pole 11. ☉ O the Amplitude from the North. 12. ☉ P O the Hour from Midnight CASE V. The Angles being given to find the other Parts IF the two Angles P ☉ O and O P ☉