The description and use of the universall quadrat.: By which is performed, with great expedition, the whole doctrine of triangles, both plain and sphericall, two severall wayes with ease and exactness. Also the resolution of such propositions as are most usefull in astronomie, navigation, and dialling. By which is also performed the proportioning of lines and superficies: the measuring of all manner of land, board, glasse; timber, stone. &c. / By Thomas Stirrup, Philomathemat.

The Second Book. (Book 2)

Shewing the most plentifull, easie, and speedy use of the Ʋniversal Quadrat, in the Resolution of the whole Do∣ctrine Trigonometrical, as well plain as sphericall, and that two severall wayes upon the Instrument, with sur∣passing facility, and with the least in∣tricacie that may be. (Book 2)

CHAP. I. To finde the chord of any arch, the radius being gi∣ven, not exceeding the side of the quadrat.

THe radius of a circle, is the semidiameter of the same circle, or the totall sine or sine of 90 degrees; so AE is the semidiame∣ter or radius upon which the divided arch EG was dramn; and AB is the ra∣dius upon which the arch BFC was drawne.

And a Chord is a right line subtending an arch; so BC is the chord of the arch BFC, for the finding of which chord for any arch required, place the radius given upon the side of the quadrat from the center, and note the point where the same endeth, and there place the bead; then opening the threed to the arch given, counted in the quadrant, the distance between the point noted in the side of the quadrat, and the head shall be the chord re∣quired.

[illustration] diagram of the measurement of an arch

As for example, let AB be the radius of a circle placed upon the side of the quadrat, and let it be required to finde a chord of 40 degrees. First I place the bead at the point B where the given radius ended, and opening the thred to 40 degrees, in the quadrant, I take the dist∣ance betwixt the point B, and the bead; which is the chord of 40 degrees, viz. the chord of the arch BFC, drawn upon the given radius AB, as was required, and thus may you finde the chord of any arch as well as upon the sector.

CHAP. II. To finde the right sine of any arch given, the radius being put 1000.

HAving shewed the use of these parallel lines as they stand in their own proper signification; it resteth now to shew the use of them, as they signifie right fines and tangents: and first to finde the right fine of any arch, seeke the degree given upon the quadrant counted from any side of the quadrant; but here let it be from the be∣ginning of the degrees at E, so shall the right parallel running through the degree given in the quadrant; give the number of the fine required; but if you would have the fine in a right line, take the length of the contrary parallel from the side of the quadrat AE, to the degree given in the quadrat; and that shall be your desired sine; whereof the whole side is radius.

As for example, let it be required to finde the right sine of 53 degrees 8 minutes I look 53 degrees 8 minutes

in the quadrant, counted from the begining of the degree at E, unto O, through which point O runeth the right parallel of 800, this 800 is the right sine of 53 degrees 8 minutes, 100 being radius.

But if you would haue this sine in a right line, take the length of the contrary paaallel from the point O unto N, for the right sine of 53 degrees 8 minutes which was re∣quired.

CHAP. III. To finde the arch of any sine given, the radius being put 1000.

IF your sine be given in numbers, seeke the right paral∣lel of the number given, for where that parallel cut∣teth the quadrant, there is the arch required, counted from the beginning of the degrees at E as before.

But if your sine be given in a right line, then from the side of your quadrat AE, extend your given sine upon any of the contrary parallels; and that right parallel that shall cut through the extream point thereof, shall cut the quadrant in the required arch.

As for example, let 800 be a sine given, unto which it is required to finde the arch belonging thereto; I seeke 800 in the right parallels, and finding it, I follow it to the quadrant, and there I see it cut the arch of 53 degrees 8 minutes, counted from E, which 53 degrees 8 minutes is the arch belonging to the sine 800, the radius being 1000.

But if the sine be given in a right line, take it betwixt

your compasses, and setting one foot in the side of the quadrat AE, extend the other upon any of the contrary parallels, as here I set one foot in the point N, and ex∣tend the other unto O, through which point the right parallel of 800 runneth, and cutteth the quadrant in 53 degrees 8 minutes counted from E as before.

CHAP. IV. Any radius not exceeding the side of the quadrat being given, to finde the right sine of any arch or angle thereunto belonging.

FIrst, take the given radius and place it upon the con∣trary parallel of 100, which is the total sine; and where it shall end there place the threed, which being so placed; see where it cutteth the contrary parallel that passeth through the given arch, counted in the quadrant from G: for the segment so cut, which lieth betweene the threed and the side AE, is the fine required.

[illustration] diagram of the measurement of an arch (lines A and B)

Let the line A be the radius given, and let it be requi∣red to finde the sine of 36 degrees 53 minutes.

First, I take the line A, and place it upon the contrary pa∣rallel of 100, as from E to D, upon which point D, I place

the threed; this done I count the given arch viz. 36 de∣grees 33 minutes from the end of the quadrant at G unto O, and see what contrary parallel there I finde, which is 60 intersecting the threed at M; so shall the distance NM be the line B the sine of 36 degrees 53 minutes, as was re∣quired.

CHAP. V. The right sine of any arch being given, to finde the radius.

FIrst, take the given sine, and place it from the side AE of your quadrat, along upon that contrary paral∣lel which cutteth the degrees of the sine given, counted in the quadrant from G, and where the sine endeth there place the threed; so shall it shew upon the contrary paral∣lel of 100, the radius required.

Let the line B in the foregoing Chapter, be a given sine of 36 degrees 53 minutes. First, place it from N to M, upon that parallel which runneth through the degrees of 36 and 53 minutes as at O, then upon the point M, I place the threed; so doth it cut the contrary parallel of 100, in the point D; making DE the radius, which is the line A in the foregoing Chapter, as was required.

CHAP. VI. The radius being given, with a streight line resembling a sine, to finde the quantitie of that unknown sine.

FIrst, place the given radius upon the contrary parallel of 100, and thereto apply the threed; then take the right sine given, and setting one foot of your compasses in the side of the quadrat, carry it parallel to the former, till the other foot cut the threed, and there stay; for the parallel where the compasses so resteth, shall cut the qua∣drant in the degrees answering to that unknown sine gi∣ven.

As if the line A in the fourth Chapter, were the radius given, and B the streight line resembling a sine: First, I place the line A from E to D, upon the contrary parallel of 100, and apply the threed thereto; the threed lying in this position, I take the line B, and fixing one foot in the side of the quadrat AE, I carry it parallel to the former, until the other foot touch the threed, so will one foot rest in the point N, and the other in M, upon the parallel of 60, which parallel cutteth the quadrat in the point O at 36 degrees 53 minutes, counted from G, this 36 degrees 53 minutes, is the arch, of which the given line B is the sine, the line A being radius.

CHAP. VII. The use of these parallels, as they signifienaturall sines.

I Told you in the fourth Chapter, of the foregoing Book, that these lines did somtimes signifie themselves alone, somtimes sines, and somtimes tangents, and now for (distinction sake) when I call them parallels, I would have you to understand them as they are in their own pro∣per signification; and when I call them Sines, I would have them understood as the Sines of those arches through which they run; and when I call them Tangents, I would have them understood as the Tangent of those degrees, against which they butt in the sides of the quadrant.

And further for distincton, whereas in their own sig∣nification I call them right and contrary parallels, so now I call them right and contrary sines, and right and contra∣ry tangents.

These lines as they signifie sines, hath like use in find∣ing a fourth proportional sine, as the ordinary Canon of Natural sines: and the manner of finding it, is alwayes such as in this example.

• As the sine of 90 degrees unto the sine of 30 degree:
• So is the sine of 23 degrees 30 minutes unto a fourth sine.

Wherefore place the threed at the intersection of the contrary Sine of 90 degrees, with the right Sine of 30 de∣grees, and it will cut the contrary Sine of 23 degres 30 minutes, at the right Sine of 11 degrees 30 minutes, and such is the fourth proportional Sine required.

Or you may place the threed at the intersection of the contrary Sine of 90 degrees with the right Sine of 23 de∣grees 30 minutes, and it shall cut the contrary Sine of 30 degrees, at the right sine of 11 degrees 30 minutes, as before.

And thus may all the rest of the sinical propositions be wrought both wayes.

CHAP. VIII. The use of these parallels, as they signifie both Sines and Tangents.

HEre the work is in a manner the same as before, as will appear by this example.

• As the Sine of 90 degrees, to the Sine of 51 degr. 30 min.
• So is the Tangent of 30 degrees, to a fourth Tangent.

Thus placeing the threed upon the intersection of the contrary Sine of 90 degrees, with the right Sine of 51 degrees 30 minutes, it will cut the contrary Tangent of 30 degrees, at the right Tangent of 24 degrees 20 mi∣nutes.

Or if you place the threed at the intersection of the contrary Sine of 90 degrees, with the right Tangent of 30 degrees, it will cut the contrary Sine of 51 degrees 30 minutes, at the right Tangent of 24 degrees 20 minutes, which is the fourth proportional Tangent required.

And this manner of work will hold, until the Tan∣gents

be greater then 45 degrees the side of the quadrat. But when the given tangent or tangents exceed 45 de∣grees the side of the quadrat, if the two first given num∣bers be Sines and the third a Tangent; set the second Sine in the place of the first, and the first Sine in the place of the second; and in steed of the given Tangent, take his complement unto a quadrant; so shall the fourth propor∣tional, be the complement of the Tangent required. As for example, in this proportion.

• As the Sine of 90 degrees,
• To the Sine of 20 degrees;
• So the Tangent of 70 degrees,
• To a fourth Tangent.

And because the Tangent of 70 degrees is greater then 45 degrees, the side of the quadrat, therefore I turn the proportion thus.

• As the Sine of 20 degrees,
• To the Sine of 90 degrees;
• So the Co-tangent of 70 degrees,
• To the Co-tangent of the sum required.

Thus placing the threed upon the intersection of the contrary sine of 20 degrees, with the right Tangent of 20 degrees, (the complement of 70 degrees,) it should cut the contrary Sine of 90 degrees, at the right Tangent of the complement of the arch required, but the threed intersecteth not the said contrary Sine of 90 degrees up∣on the quadrat, but besides it; which never happeneth but

when the one Tangent is greater then 45 degrees, and the other lesse. Wherefore you are to note, that at any such time, though the threed cutteth not the contrary Sine of 90 degrees upon the quadrat, yet is the threed (by the former position) placed at the true angle of the Tangent required; which is 46 degrees 47 minutes, the comple∣ment whereof 43 degrees 13 minutes, is the arch requi∣red. And if the first given numbers be Tangents, and the third a Sine; take the complements of the two Tangents and exchange their places, as in the examples following. But if one be more and the others lesse, as in this propor∣tion.

• As the Tangent of 70 degrees,
• To the Tangent of 43 degrees 13 minutes;
• So the Sine of 90 degrees:
• To a fourth Sine.

Herein regard the Tangent of 70 degrees is more then 45 degrees, and the Tangent of 43 degrees lesse then the same, if you open the threed to the greater Tangents an∣gle, viz. to the angle of 70 degrees, it shall cut the right Tangent of the lesser, viz. the right Tangent of 43 de∣grees 13 minutes at the contrary Sine of 20 degrees, and such is the fourth proportional Sine required.

I have beene the larger in this, to prevent mistakes in the rest.

CHAP. IX. The use of these parallels, as they are in their owne signification, joyned with their signification of Sines and Tangents.

THis manner of work is the same with the former, only you must know when the lines signifie them∣selves alone, and when Sines and Tangents; the which you may know by the proportions given, as here in this example.

• As the Sine of 53 degrees 8 minutes, unto 80:
• So is the Sine of 36 deg. 52 min. unto a fourth number.

Here if you place the threed at the intersection of the contrary Sine of 53 degrees 8 minutes with the right pa∣rallel of 80; it shall cut the contrary Sine of 36 degrees 52 minutes at the right paralel of 60, and such is the fourth proportionall number required.

Of the resolution of right lined Triangles

CHAP. X. Having three angles and one side to finde the other two sides.

IN all Triangles there being six parts viz. three Angles and three Sides, any three whereof being known in a sphericall Triangle, the other three may be found by the universal quadrat; but in a right lined Triangle, one of the three given termes must be a Side in the finding of the o∣ther parts.

Thus having three angles and one Side, you may finde the other two Sides by this proportion.

• As the Sine of the angle opposite to the given Side,
• To the number belonging to the same Side:
• So is the Sine of the angle opposite to the Side required,
• To the number belonging to the Side required.

Let the following Triangle ABC be given, whereof the three angles and the Side AB is known; viz. the an∣gle ACB 53 degrees 8 minutes, the angle CAB 36 de∣grees 52 minutes, and the angle ABC 90 degrees, the

given side AB being 80, it is required to finde the other two sides viz. AC and BC.

Knowing the angle ACB opposite to the given side AB to be 53 degrees 8 minutes, and the angle opposite to the required side BC to be 36 degrees 52 minutes, I place the threed upon the intersection of the contrary Sine of 53 degrees 8 minutes with the right parallel of 80 his opposite side; the threed lying in this position I see it cut the contrary Sine of 36 degrees 52 minutes, at the right parallel of 60 his opposite side; therefore I conclude the side BC to be 60 such parts as AB is 80.

The threed lying still in the former position see where it cutteth the contrary Sine of the third angle, for there is the right parallel of the third side: so here the threed cutteth the contrary Sine of 90 degrees, (which is the third angle) at the right parallel of 100, which is the side AC, as was required.

So here you see the threed once placed, giveth both the sides required, and that right speedily in all Triangles; as well in oblique angled as right angled triangles.

CHAP. XI. Having two sides given, and one angle opposite to either of them; to finde the other two angles, and the third side.

• AS the side opposite to the angle given,
• Is to the Sine of the angle given:
• So is the other side given,
• To the sine of that angle to which it is opposite.

[illustration] diagram of the measurement of a triangle (ABC)

Thus in the Triangle ADC, having the two sides AD 35, and CD 75, with the angle ACD 16 degrees 16 minutes, being opposite to the side AD; I may finde the angle CAD which is opposite to the other side CD, for if I place the threed upon the intersection of the con∣trary parallel of 35, with the right Sine of 16 degrees 16 minutes, it shall cut the contrary parallel of 75, at the right sine of 36 degrees 52 minutes, his opposite angle, the angle CAD.

Then keeping the threed in the former position, adde these two known angles together, and they will give you the outward angle ADC 53 degrees 8 minutes, now where the right Sine of 53 degrees 8 minutes, cutteth the threed, there is the contrary parallel of 100, which is the third side AC, as was required.

CHAP. XII. Having two sides and the angle between them, to finde the two other angles and the third side.

IF the angle contained betweene the two sides be a right angle, the other two angles will be found readily by this proportion

• As the greater side given,
• Is to the lesser side;
• So is the Tangent of 45 degrees,
• To the Tangent of the lesser angle.

So in the rectangle Triangle ABC, knowing the side AB to be 80, and the side BC to be 60. If I place the threed upon the intersection of the contrary parallel of 80, with the right parallel of 60; it shall cut the contrary Tangent of 45 degrees at the right Tangent of 36 de∣grees 52 minutes, and such is the lesser angle CAB; the complement whereof unto a quadrant is the greater angle ACB 53 degrees 8 minutes, the angles being known, the third side AC may be found by the 10 Chapter.

But if it be an oblique angle that is betweene the two sides given; the Triangle may be reduced into two rect∣angles and then resolved as before.

As in the Triangle ADC, where the side AC is 100,

and the side AD 35, and the angle CAD 36 degrees 52 minutes, if you let down the perpendicular DE, up∣on the side AC, you shall have two rectangle Triangles, AED, DEC; and in the rectangle AED, the angle at A being 36 degrees 52 minutes, the other angle ADE will be 53 degrees 8 minutes, by complement: and with these angles and the side AD, you may finde both AE and DE, by the tenth Chapter.

Then taking AE out of AC, there remaines EC for the side of the rectangle DEC; and therefore with this side EC and the other DE, you may finde both the an∣gle at C, and the third side CD, by the former part of this Chapter.

Or you may finde the angles required, without letting down any perpendicular.

For

• As the sum of the sides,
• Is to the difference of the sides:
• So the Tangent of the halfe sum of the opposite angles,
• To the Tangent of halfe the difference betweene those angles,

As in the former Triangle ADC, the sum of the sides AC and AD, is 135, and the difference betweene them 65; the angle contained 36 degrees 52 minutes; and therefore the sum of the two opposite angles 143 degrees 8 minutes, and the halfe sum 71 degrees 34 minutes, and because the halfe sum 71 degrees 34 minutes, is greater then the side of the quadrat 45 degrees, therefore I turne the proportion after this manner.

• As the difference of the sides 65,
• Is to the sum of the sides 135:
• So is the co-tangent of the halfe sum of the opposite angles 71 degrees 34 minutes.
• To the co-tangent of half the difference betweene those angles.

Therefore if you place the threed upon the intersecti∣on of the contrary parallel of 65, with the right parallel of 135, you shall see it cut the contrary Tangent of 18 degrees 26 minutes, (which is the complement of 71 de∣grees 34 minutes,) at the right Tangent of 34 degrees 42 minutes, the complement whereof is 55 degrees 18 minutes, and such is the halfe difference betweene the op∣posite angles at B and D, this halfe difference being ad∣ded to the half sum, giveth 126 degrees 52 minutes, for the greater angle ADC, and being substracted, leaveth 16 degrees 16 minutes for the lesser angle ACD, the three angles being thus found, you may finde the third side CD by the tenth Chapter.

CHAP. XIII. Having the three sides of a right lined Triangle, to finde the three angles.

LEt one of the three sides given be the base, but rather the greater side, that the perpendicular may fall with∣in the Triangle; then gather the sum and difference of the two other sides, which being done the proportion will be.

• As the base of the Triangle,
• Is to the sum of the sides;
• So the difference of the sides
• To the alternate base.

This alternate base being taken out of the true base, if we let down the perpendicular from the opposite angle, it shall fall upon the middle of the remainder.

As in the former Triangle ADC, where the base AC is 100, the sum of the sides AD and CD is 110, and the difference of them 40.

Therefore if you place the threed upon the intersecti∣on of the contrary parallel of 100, with the right paral∣lel of 110, it will cut the contrary parallel of 40, at the right parallel of 44, for the alternate base; this alternate base take out of 100 the true base, and the remainder will be 56, the halfe whereof is 28, and sheweth the distance from A unto E, where the perpendicular shall fall, from the angle D, upon the base AC, dividing the former Triangle ACD into two right angled Triangles, viz. AED and DEC, in which the angles may be found by the 11th. Chapter.

Thus you may see how by this universal quadrat, we have resolved the four propositions of Mr. Gunters upon his Crosse-staffe; whereby you may perceive the agree∣ment betwixt this quadrat and his staffe.

And now I will shew how they may be otherwise re∣solved upon this uniuersal quadrat, agreeing neerer with the Sector; but more spendily performed, then by either Sector or Cross-staffe.

CHAP. XIIII. In a rectangle Triangle having the angles and one of the sides given, to finde the other side and the base.

OPen the threed to the angle opposite to the side gi∣ven, so shall it intersect the right parallel of the side given, at the contrary parallel of the side required; and if to the intersection you place the bead, and apply the threed to the side of the quadrat, it shall there shew you the length of the base.

As in the right angled Triangle ABC in the 11 Chapter, the Side BC being given, opposite to the an∣gle BAC 36 degrees 52 minutes, it is required to finde the other Side AB, and the base AC.

First, open the threed to the angle of 36 degrees 52 minutes, so shall it cut the right parallel of 60, at the con∣trary parallel of 80, and such is the side AB, which was required.

The threed lying in the same position, if you place the bead to this intersection, and then apply the threed to the side of the quadrat, it shall there give you 100, for the base AC as was required.

CHAP. XV. Having both sides of a rectangle Triangle, to finde the two angles and the base.

AT the intersection of the contrary parallel of the greater side, with the right parallel of the lesser side, there place the threed; which being so placed giveth the lesser angle required. And if to the intersection you place the bead and then apply the threed to the side of the quadrat, it shall there shew the base required.

As in the rectangled Triangle ABC in the 11 Chapter, where the two given sides is AB 80, and BC 60: Wherefore I place the threed at the intersection of the contrary parallel of 80, with the right parallel of 60, so doth the threed give me the lesser angle at A, which is 36 degrees 52 minutes, which being known, the angle ACB will be found by complement, to be 53 degrees 8 minutes, thus have I both the angles required.

The threed lying in the former position, I place the bead to the intersection of the two sides; and then apply∣ing the threed to the side of the quadrat, it giveth me 100 for the base AC, which was required.

CHAP. XVI To finde a side and both the angles, by having the base and the other side given.

FIrst, lay the threed upon the side of the quadrat, and apply the bead to the length of the base; then open the threed untill the bead fall directly upon the contrary parallel of the side given, and it shall likewise lie up∣on the right parallel of the side required, and the threed shall also shew the angle opposite to the required side.

As in the former rectangle Triangle ABC in the 11th. Chapter, having the base AC 100, and the side AB 80; if you apply the threed to the side of the qua∣drat, and place the bead to 100, and then open the threed until the bead fall directly upon the contrary pa∣rallel of 80, so shall the same bead so posited, cut the right parallel of 60 the side required. And further, the threed lying in this constitution, giveth the angle opposite to the required side to be 36 degrees 52 minutes, which being known, the other is soon found out by complement unto 90 degrees.

CHAP. XVII. To finde the sides by having the base, and the angles given.

FIrst, place the bead to the base given on the side of the quadrat, then open the threed to any of the two an∣gles, and the bead shall fall upon the right parallel of that side which is opposite to the angle, unto which the threed was opened; and also upon the contrary parallel of the o∣ther side.

As for example, in the right angled Triangle ABC in the 11th. Chapter, the angles at A and C are given, viz. the angle ACB 53 degrees 8 minutes, and the an∣gle BAC 36 degrees 52 minutes, with the base AC 100, and it is required to finde the two sides AB, and BC.

First, therefore I place the bead unto 100 upon the side of the quadrat, and opening the threed to the angle of 36 degrees 52 minutes, I finde the bead to fall upon the right parallel of 60, which is the side CB subtending the said angle; through which intersection doth passe the contrary parallel of 80, which is the other side AB sub∣tending the other angle ACB, which was required.

CHAP. XVIII. In any right lined Triangle whatsoever, to finde a side by knowing the other two sides, and the angle contained by them.

FIrst, place the bead to the shorter side given, and o∣pen the threed to the given angle, then setting one foot of your Compasses in the other side given up∣on the side of the quadrat, with the other foot extended to the bead, you shall have the third side betwixt your compasses.

Thus in the triangle ADC in the 11th. Chapter, having the sides AC 100, and BC 75, and the angle be∣tweene them ACD 16 degrees 16 minutes if you place the bead to 75, and open the threed to the angle of 16 degrees 16 minutes, then setting one foot of your com∣passes in 100 and extending the other unto the bead, you shall have 35 betwixt your compasses for the third side AD, which was required.

CHAP. XIX. To finde an angle by knowing the three sides.

FIrst, place the bead to the shortest of the sides contain∣ing the angle required; then take the side subtending the same angle betwixt your compasses, and setting one foot thereof in the third side counted in the side of the

quadrat, with the other foot open or shut the threed, un∣til the bead and the movable foot of the compasses meet both in one point, so shall the threed shew the angle required.

Thus having the three sides of the Triangle ADC in the 11th. Chapter. First, I place the bead to 75 the side CD, then I take 35 betwixt my compasses for the side AD, (which is opposite to the required angle ACD) and setting one foot in 100 (upon the side of the qua∣drat) for the side AC, I turne the other foot towards the threed, moving it to and fro until the compass point fall just upon the bead, so shall the threed cut 16 degrees 16 minutes for the angle ACD, as was required.

And this may suffice for right lined Triangles, where∣by you may perceive the agreement of this Universal quadrat, both with Sector & Cros-staff. I will set down only two or three Chapters more, for the ready redu∣cing of hypothenusal to horizontal lines in the art of Sur∣veying. And for the speedy searching out of all perpen∣dicular altitudes.

CHAP. XX. Of the ready reducing of hypothenusal to horizontal lines.

FIrst, measure the hypothenusal line with your chain, then place your bead to the length thereof counted on the side of your quadrat; this being done, observe by your instrument the angle either ascending or descending,

the threed and plummet having free liberty to play; So shall the bead fall upon the contrary parallel of the hori∣zontal line required.

Suppose ABCD be a hill or mountain to be pro∣tracted, and laid down in your plot amongst your other grounds.

[illustration] diagram of the measurement of a triangle (ABC)

It is apparent by the figure, that the hypothenusal lines AB, and C cannot be laid down exactly in a right line, between the other grounds which bounder on this hill at the point A and C: wherefore we are to finde the true level and horizontal distance betweene A and C, which is a right line extending overthwart the ground whereon the hill standeth, which to do first measure the hypothenusal line BC, which in this example will be found to be 60 pole, unto which 60 place the bead; then planting your instrument at C, and observing the ascent from C to B, (by lifting up your Instrument towards B)

you shall finde the threed to fall upon 36 degrees 52 mi∣nutes, for the angle of the ascent BCD; and the bead to fall upon the contrary parallel of 48, which is the hori∣zontal distance CD, which was required.

So likewise, the threed hanging at the former angle the bead as before fixed shall fall upon the right parallel of 36, which is the length of the perpendicular BD, be∣ing let down from the top of the hill at B through the, same, unto the horizontal line AC whereon the hill standeth. And what is said here of the ascent CB, the same may be understood of the descent BA.

CHAP. XXI. To finde the height of an object accessible at one obsen∣vation. or the distance betwixt the top of the object and that point thereof which is level with your eye.

FIrst, lift up the quadrat looking through both the sights to the top of the object, going neerer or farther from it, till the threed fall directly on the opposite angle of the quadrat, which is at the intersection of the right and contrary parallels of 100, or at 45 degrees, in the quadrant; then letting down the quadrat still looking through the sights, until the threed fall directly upon the side of the quadrat; then what point soever you see in the object through the sights, is level with your eye; then measure the distance betwixt your eye and the level

point in the object, for that is equal to the height requi∣red, above the point level with your eye, this needeth no example.

But more readily thus, take a station at any convenient distance, which distance measure and count the number thereof amongst the contrary parallels; then lifting up your quadrant looking through the sights to the top of the obiect, the threed shall cut the contrary parallel of the measured distance, at the right parallel of the height required.

Let BC be the height of some Steeple or Tower whose height is required; first, I measure out any conve∣nient distance; as here I measure from the base of the ob∣ject at B, 80 paces unto A, and at AI take my station; and lifting up my quadrat until I can see the very top of the obiect at C, through both the sights, I finde the threed to cut the contrary parallel of 80 at the right pa∣rallel of 60, which is the number of places contained in the height BC, as was required. And this of all others I hold the speediest way for attaining of all altitudes ac∣cessible: and if to this intersection you place the bead, it shall give you the length of the hypothenusal or scaling ladder AC.

And you may also speedily search out any perpendi∣cular heights by the rules of proportion after this man∣ner.

[illustration] diagram of the measurement of a triangle (ABC)

Having made choice of a convenient station as at A, observe the angle at A made betweene the hypothenusal line AC, and the level line with your eye AB, the com∣plement whereof is the angle at the top of the object at C, made between the hypothenusal line AC, and the line CB, the height required. These being known, with the distance AB: Say,

• As the sine of the angle ACB,
• To the measured distance AB:
• So the sine of the angle BAC,
• To the required height BC.

Thus having found the distance AB to be 80 yards or

paces, and the angle at A to be 36 degrees 52 minutes, the complement whereof is 53 degrees 8 minutes, for the angle at C; I place the threed at the intersection of the contrary Sine of 53 degrees 8 minutes with the right pa∣rallel of 80; so doth the threed cut the contrary Sine of 36 degrees 52 minutes, at the right parallel of 60, wherefore I conclude the height BC, to be 60 such parts as AB is 80, which is the thing required.

CHAP. XXII. To finde the height of an object inaccessible, at two observations.

FIrst, as far off your object as conveniently you may, make choice of your first station; and observe the an∣gle of your eye, made between the visual line to the top of the object, and the level line with your eye, this being done, measure forward in a right line towards the object, approaching as neere thereunto as you may, and there make choice of your second station where observe the an∣gle at your eye as before: now take the complement of this second angle, out of the complement of the first an∣gle, and note the difference, then, say,

• As the sine of the difference of the complements,
• To the measured distance;
• So the Sine of the first angle observed,
• To the second hypothenusa.

Then againe,

• As the Sine of 90 degrees,
• To the second hypothenusa:
• So is the Sine of the second observed angle,
• To the height required

As for example, let BC in the former Chapter, be the height of some Tower or Steeple to be measured, unto which I may not approach, by reason of some River or o∣ther obstacle betwixt the object and me; and yet it is re∣quired to finde the altitude BC, and also the breadth of the river BD, with the length of the scaling ladder CD.

First, therefore I make choice of a station with the best convenience I can, as at A, where I make observation, and finde the angle at my eye to be 36 degrees 52 mi∣nutes, the complement whereof is 53 degrees 8 minutes, then measuring in a right line towards the object, I ap∣proach as neere as I may, which is at the point D, finding AD to be 35 paces; and now at D, I make a second ob∣servation, and finde the angle at my eye to be 53 degrees 8 minutes, the complement whereof is 36 degrees 52 mi∣nutes, this 36 degrees 52 minutes I take out of the for∣mer complement 53 degrees 8 minutes, and the differ∣ence is 16 degrees 16 minutes, the angle ACD, where∣fore I place the threed at the intersection of the contrary Sine of 16 degrees 16 minutes, with the right parallel of 35, and I finde the threed to cut the contrary Sine of 36 degrees 52 minutes (which is the angle at A) at the right

Parallel of 75, and such is the scalling ladder, or hypo∣thenusal line CD.

Then again, I place the threed at the intersection of the contrary Sine of 90 degrees, with the right parallel of 75, and it cutteth the contrary Sine of 53 degrees 8 minutes, (which is the angle at D) at the right parallel of 60, and such is the line BC, the height required. And the threed in the same position, cutteth the contrary Sine of the complement of the second angle observed, viz. 36 de∣grees 52 minutes, at the right parallel of 45, and such is the breadth of the River or other obstacle BD.

Or having found the length of the second hypothenu∣sal line to be 75, if you place the bead thereto, and then open the threed to the second observed angle, viz. 53 degrees 8 minutes, the bead shall fall upon the right pa∣rallel of 60, which is the height BC; and also upon the contrary parallel of 45, which is the breadth BD, as be∣fore, and as was required.

Or more speedily thus, at two observations and one opperation.

First, make choice of your stations as before, then make choice of some one of your right parallels, that the num∣ber thereof may signifie the height required; then at your first station make observation as before, and note the parts cut by the threed upon the said right parallel; this being done, measure your stationary distance, as before; and at the second station, make observation, as you did at the first station; noting likewise the parts cut by the threed upon the said right parallel; now taking these latter parts out of the former, or the lesser out of the greater, the dif∣ference shall signifie the stationary distance: wherefore,

• As the distance upon the parallel,
• To the measured distance:
• So the number of the said parallel,
• To the height required.

Thus in the figure of the foregoing Chapter, I take my first station at A, and making choice of some one of the right parallels, (as here of the right parallel of 45) to signifie the height required, I make observation to∣wards C, and I finde the threed to cut the said parallel at 60: then measuring the stationary distance AD, I finde it to be 35, and at D, I make a second observation, and I finde the threed to cut the foresaid parallel at about 33¾ which being taken out of 60, leaveth 26¼ for to signifie the stationary distance; wherefore I place the threed at the intersection of the contrary paral∣lel of 26¼, with the right parallel of 35, & it will cut the contrary parallel of 45, at the right parallel of 60, which is the height BC, as was required.

Many propositions to this purpose might be framed, but these may suffice for a tast of the use of this universal quadrat.

And if an index with sights were fitted to turne upon the center it would serve then by the same reason, for the findeing of all other distances.

Of the resolution of spherical Trian∣gles; and first of those which be right angled.

CHAP. XXIII. To finde a side by knowing the base, and the angle opposite to the required side.

A Spherical Triangle, is a figure included by three arches of great circles; the angles whereof are mea∣sured by the arch of a great circle subtending the angle, and intercepted betweene the containing sides continued to quadrants.

Now in every Triangle there being six parts, viz. three angles and three sides, any three of them being gi∣ven in a spherical Triangle, the other three may be found by the Uniuersal quadrat, and that two severall wayes, and as it were by two several Instruments; the former part agreeing nearest with the sector, the latter part (per∣formed by the Planisphere on the back of the quadrat) a∣greeing neerest to the Mathematical Jewel of Master Blagraves; but the working hereby will appeare more cleare then through the thick rete of the Jewel.

First, to finde a side by knowing the base, and the angle opposite to the required side.

• As the radius, is to the sine of the base:
• So is the sine of the opposite angle,
• To the sine of the side required.

[illustration] diagram of the measurement of a spherical triangle (ABC)

Let the Triangle ABC be given, wherein let A stand for the point of East or West; AB an arch of the Hori∣zon representing the azimuth of the Sun from the East or West points; BC an arch of an azimuth or vertical cir∣cle, making right angles with the Horizon AB, in the point B, representing the Suns altitude; and AC an arch of the Equator representing the houre from Sun rising or setting when he hath no declination. and further, let the angle at A be the complement of the latitude, and the angle at C the vertical angle, made betweene the equator and the vertical circle passing by the Sun at C, perpendi∣cularly

to the horizon AB, wherefore knowing the base AC to be 65 degrees 23 minutes, and the angle at A to be 37 degrees 30 minutes, I may finde the side BC, for if I place the threed to the intersection of the contrary Sine of 90 degrees with the right Sine of 65 degrees 22 minutes it will cut the contrary Sine of 37 degrees 30 mi∣nutes, at the right Sine of 33 degrees 36 minutes, and such is the side BC, which was required.

CHAP. XXIV. To finde a side by knowing the base, and the other side.

• AS the Sine of the complement of the side given,
• To the sine of the complement of the base:
• So is the radius,
• To the Sine of the complement of the side required.

Thus in the rectangle ABC, having AC 65 degrees 23 minutes, and AB, 60 degrees, let it be required to finde the side BC, place the threed at the intersection of the contrary Sine of 30 degrees, (the complement of AB) with the right Sine of 24 degrees 37 minutes, (the complement of AC) and it shall cut the contrary Sine of 90 degrees, at the right sine of 56 degrees 24 mi∣nutes, the complement whereof is 33 degrees 36 minutes, the side BC, which was required.

CHAP. XXV. To finde a side, by knowing the two oblique angles.

• AS the Sine of either angle,
• To the cosine of the other angle:
• So is the sine of 90 degrees,
• To the cosine of the side opposite to the second angle.

So in the rectangle ABC, having ACB for the first angle 72 degrees 16 minutes, and BAC for the second 37 degrees 30 minutes, it is required to finde the side BC, wherefore I place the the threed at the intersection of the contrary Sine of 72 degrees 16 minutes, with the right Sine of 52 degrees 30 minutes, (the complement of the angle at A) and I finde it cut the contrary Sine of 90 degrees, at the right Sine of 56 degrees 24 minutes, whose complement is 33 degrees 36 minutes, the side BC, as was required.

CHAP. XXVI To finde the base, by knowing both the sides.

• AS the radius, to the cosine of the one side:
• So the cosine of the other side,
• To the cosine of the base.

Thus in the rectangle ABC, knowing the latitude of the place, with the altitude of the Sun having no decli∣nation, we may finde the houre of the day; as having BC the altitude of the Sun 33 degrees 36 minutes and the an∣gle BAC the complement of the latitude 37 degrees 30 minutes, the base AC which is the houre of the day counted from Sun rising or setting, will be found by the quadrat. For if you apply the threed to the intersection of the contrary Sine of 37 degrees 30 minutes with the right Sine of 33 degrees 36 minutes, it will cut the con∣trary Sine of 90 degrees, at the right Sine of 65 degrees 23 minutes and such is the base AC, which being redu∣ced into time (by allowing 15 degrees to an houre, and for every 4 minutes a degree) giveth 4 houres 22 minutes for the time after Sun rise, or before Sun set: so that if it be in the forenoon it is 10 a clock and 22 minutes, the Sun then rising at 6; and if it be in the afternoon it is 38 mi∣nutes past one.

CHAP. XXVIII. To finde an angle, by knowing the other oblique angle, and the side opposite to the inquired angle.

• A the radius,
• To the sine of the angle given:
• So the cosine of the side,
• To the cosine of the angle required.

Thus in the Rectangle ABC, having the angle BAC 37 degrees 30 minutes, with the side AB 60 de∣grees, to finde the angle ACB: I place the threed at the intersection of the contrary Sine of 90 degrees with the right Sine of 37 degrees 30 minutes, so will it cut the contrary sine of 30 degrees, (the complement of the side AB) at the right Sine of 17 degrees 44 minutes, the complement whereof is 72 degrees 16 minutes, the angle ACB, which was required.

CHAP. XXIX. To finde an angle, by knowing the other oblique angle, and the side opposite to the angle given.

• As the cosine of the side,
• To the cosine of the angle given:
• So is the radius,
• To the Sine of the angle required.

So in the rectangle ABC, having the angle BAC 37 degrees 30 minutes, and the side BC 33 degrees 36 minutes, to finde the angle ACB: place the threed at the intersection of the contrary Sine of 56 degrees 24 minutes, (the complement of BC) with the right sine of 52 degrees 30 minutes (the complement of the angle BAC) and it will cut the contrary sine of 90 degrees, at the right sine of 72 degrees 16 minutes, the angle ACB, which was required.

CHAP. XXX. To finde an angle, hy knowing the base, and the side opposite to the inquired angle.

• AS the sine of the base,
• To the sine of 90 degrees:
• So the sine of the side given,
• To the sine of the angle required.

Thus in the rectangle ABC, having the base AC 65 degrees 23 minutes, and the side BC 33 degrees 36 mi∣nutes, to finde the angle BAC: I place the threed to the intersection of the contrary sine of 65 degrees 23 mi∣nutes, with the right sine of 90 degrees, and I finde it cut the contrary sine of 33 degrees 36 minutes, at the right sine of 37 degrees 30 minutes, the angle BAC which was required.

Thus far by sines alone; but now we must require joynt help of Tangents.

And forasmuch as the Tangents extend but to 45 de∣grees, which is equal to the side of the quadrat. I shall set down the two ways for the resolution of each propo∣sition, delivered by Master Gunter; and if one will not hold, the other will; observing the rules given to this purpose in the 8 Chapter.

CHAP. XXXI. To finde a side, by knowing the other side, and the angle opposite to the inquired side.

• AS the radius,
• To the sine of the side given:
• So is the Tangent of the angle,
• To the Tangent of the side required.

Or

• As the sine of the side given,
• Is to the radius:
• So the cotangent of the angle,
• To the cotangent of the side required.

Thus in the rectangle ABC, having the side AB 60 degrees, and the angle BAC 37 degrees 30 minutes, to finde the side BC; I place the threed at the intersection of the contrary sine of 90 degrees, with the right sine of 60 degrees, and it doth cut the contrary Tangent of 37 degrees 30 minutes, at the right Tangent of 33 degrees 36 minutes, the side BC, which was required.

CHAP. XXXII. To finde a side, by knowing the other side, and the angle next to the inquired side.

• AS the Tangent of the angle given,
• To the Tangent of the side given:
• So is the radius,
• To the sine of the side required.

Or,

• As the cotangent of the side given,
• To the cotangent of the angle given;
• So is the radius,
• To the Sine of the side required.

Thus in the rectangle ABC having the side AB 60 degrees, and the angle ACB 72 degrees 16 minutes, it is required to finde the side BC. Now here in regard the given Tangents exceed the side of the quadrat; I let passe the first proportion, and make use of the second: and opening the threed to the angle of 30 degrees, or which is all one, if I place the threed at the intersection of the contrary Sine of 90 degrees, with the right Tan∣gent of 30 degrees, (which is the complement of the side AB) it will cut the right Tangent of 17 degrees 44 mi∣nutes, (the complement of the angle ACB) at the con∣trary 〈…〉〈…〉 33 degrees 36 minutes, the side BC, or

placing the threed at the intersection of the contrary Tan∣gent of 30 degrees, (the complement of the side AB) with the right Sine of 90 degrees, it will cut the contrary Tangent of 17 degrees 44 minutes, (the complement of the angle ACB) at the right sine of 33 degrees 36 mi∣nutes, the side BC. Or if the threed be placed at the in∣tersection of the contrary Tangent of 30 degrees, with the right Tangent of 17 degrees 44 minutes, it will cut the contrary Sine of 90 degrees, at the right Sine of 33 degrees 36 minutes, for the side BC, as before, which was required.

CHAP. XXXIII. To finde a side by knowing the base, and the angle next to the inquired side.

• AS the radius,
• To the cosine of the angle given:
• So is the Tangent of the base,
• To the Tangent of the side required.

Or,

• As the cosine of the angle given,
• Is to the radius:
• So is the cotangent of the base,
• To the cotangent of the side required.

So in the rectangle ABC, having the base AC 65

degrees 23 minutes, for the houre from Sun-rise; (he ha∣ving no declination) and the angle BAC 37 degrees 30 minutes, for the complement of the latitude; we may finde the Azimuth from the East, which is the side AB: for if the threed be applied to the intersection of the con∣trary Sine of 52 degrees 30 minutes, (which is the com∣plement of the angle BAC,) with the right Tangent of 24 degrees 37 minutes, (the complement of the base AC) it shall cut the contrary Sine of 90 degrees at the right Tangent of 30 degrees, the complement whereof is 60 degrees, the side AB, or the Azimuth from the East, which was required.

CHAP. XXXIV. To finde the base by knowing the oblique angles.

• As the Tangent of the one angle,
• To the cotangent of the other angle:
• So is the radius,
• To the cosine of the base.

Thus in the rectangle ACB, having the angle BAC 37. degrees 30 minutes, and the angle ACB 72 degrees 16 minutes, to finde the base AC: I place the threed at the intersection of the contrary Tangent of 37 degrees 30 minutes with the right Sine of 90 degrees, and it will cut the contrary Tangent of 17 degrees 44 minutes, (the complement of the angle ACB) at the right Sine of 24

degrees 37 minutes, the complement whereof is 65 de∣grees 23 minutes, the base AC, as was required.

CHAP. XXXV. To finde the base, by knowing one of the sides, and the angle next the same side.

• AS the radius,
• To the cosine of the angle given:
• So is the cotangent of the side given,
• To the cotangent of the base.

Or

• As the cosine of the angle given,
• Is to the radius:
• So the tangent of the side given,
• To the tangent of the base required.

Thus in the rectangle ABC, having AB 60 degrees, and the angle BAC 37 degrees 30 minutes, to finde the base AC; I place the threed at the intersection of the contrary Sine of 90 degrees, with the right tangent of 30 degrees, (the complement of the side AB) and it cut∣teth the contrary Sine of 52 degrees 30 minutes, (the complement of the angle BAC) at the right tangent of 24 degrees 37 minutes, the complement whereof 56 de∣grees 23 minutes, is the base AC, which was required.

CHAP. XXXVI. To finde an angle by knowing both the sides.

• As the radius,
• To the Sine of the side next the inquired angle:
• So is the cotangent of the opposite side,
• To the cotangent of the angle required.

Or,

• As the sine of the side next the inquired angle,
• Is to the radius:
• So is the tangent of the opposite side,
• To the tangent of the angle required.

Thus in the rectangle ABC, having the side BC 33 degrees 36 minutes, and the side AB 60 degrees, to finde the angle BAC; place the threed upon the inter∣section of the contrary Sine of 60 degrees, with the right tangent of 33 degrees 36 minutes, and it shall cut the con∣trary Sine of 90 degrees, at the right tangent of 37 de∣grees 30 minutes, and such is the angle BAC, as was re∣quired.

CHAP. XXXVII. To finde an angle, by knowing the base, and the side next to the inquired angle.

• AS the cotangent of the side given,
• To the cotangent of the base:
• So is the radius,
• To the cosine of the angle required.

Or

• As the tangent of the base,
• To the tangent of the side:
• So is the radius,
• To the cosine of the angle required.

Thus in the rectangle ABC, having the base AC 65 degrees 23 minutes, and the side BC 33 degrees 36 minutes, it is required to finde the angle ACB.

Now in regard the tangent of the base is greater then the side of the quadrat, and the side BC is lesser then the same; I open the threed to an angle equal with the base, viz. to an angle of 65 degrees 23 minutes, and it will cut the right tangent of 33 degrees 36 minutes, at the contrary Sine of 17 degrees 44 minutes, the complement whereof being 72 degrees 16 minutes, is the angle ACB which was required.

CHAP. XXXVIII. To finde an angle, by knowing the other oblique angle, and the base.

• AS the radius,
• To the cosine of the base:
• So the Tangent of the angle given,
• To the cotangent of the angle required.

Or,

• As the cosine of the base,
• Is to the radius:
• So is the cotangent of the angle given,
• To the tangent of the angle required.

Thus in the rectangle ABC, having the base AC 65 degrees 23 minutes and the angle BAC 37 degrees 30 minutes, to finde the angle ACB; place the threed at the intersection of the contrary Sine of 90 degrees, with the right Sine of 24 degrees 37 minutes, (the comple∣ment of the base) and it will cut the contrary tangent of 37 degrees 30 minutes, at the right tangent of 17 de∣grees 44 minutes, the complement whereof being 72 de∣grees 16 minutes, is the angle ACB, which was required.

Thus have we by the Universal quadrat, resolved all the 16 cases of a rectangle spherical triangle. Now fol∣loweth the resolution of all manner of spherical triangles whatsoever.

In any Spherical Triangle whatsoever.

CHAP. XLI. To finde a side opposite to an angle given, by knowing one side, and two angles, the one opposite to the given side, the other to the side required.

• AS the sine of the opposite angle to the side given,
• Is to the sine of that side given:
• So is the sine of the angle opposite to the side required,
• To the sine of the side required.

Thus in the following triangle PZS, having the side PZ 37 degrees 30 minutes, and the angle ZPS 53 de∣grees 40 minutes with the angle ZSP 34 degrees 47 mi∣nutes, we may finde the side ZS; for if you apply the threed to the intersection of the contrary Sine of 34 de∣grees 47 minutes, with the right Sine of 37 degrees 30 minutes, it shall cut the contrary Sine of 53 degrees 40 minutes, at the right Sine of 59 degrees 15 minutes, and such is the side ZS, which was required.

CHAP. XL. To finde an angle opposite to a side given, by knowing one angle and two sides, the one opposite to the given angle, the other to the angle required.

• AS the sine of the side opposite to the angle given,
• Is to the sine of that angle given:
• So is the sine of the side opposite to the angle required,
• To the sine of the angle required.

Thus in the triangle PZS, having the angle PZS 113 degrees 17 minutes, the Azimuth from the North; and the side PS 78 degrees 30 minutes, the complement of the Suns declination; with the side ZS 59 degrees 15 minutes, the complement of the Suns altitude; we may finde the angle ZPS which is the houre from the meri∣dian: for if we place the threed at the intersection of the contrary sine of 78 degrees 30 minutes, with the right sine of 66 degrees 43 minutes the outward angle at Z, (because the inward angle exceedeth a quadrant) it will cut the contrary sine of 59 degrees 15 minutes, at the right sine of 53 degrees 40 minutes, which is the hour from the meridian as was required.

[illustration] diagram of the measurement of a spherical triangle (PZS)

CHAP. XLI. To finde an angle by knowing the three sides.

IN respect of the three sides of a not right angled sphe∣rical triangle, let that which subtendeth the angle re∣quired be called the base; and the other two the contain∣ing sides.

Then, first, take the sum and difference, of the lesser side and the complement or excesse of the greater side; and adde the sines of the said sum and difference together, the halfe whereof shall be the first found sine: then out of the sine of the sum take this first found sine, and note the remainder.

Now if every of the three sides be lesse then a qua∣drant, take the difference betweene the remainder noted, and the cosine of the base: but if any of the three sides

be greater then a quadrant, adde the remainder noted to the cosine or excesse of the base; so shall the sum or dif∣ference be the second found sine. Then say,

• As the first found sine,
• To the second sound sine:
• So is the radius,
• To a fourth fine.

Now when every side of the triangle is lesse then a quadrant; if the remainder noted be lesse then the cosine of the base, the arch of this fourth sine must be taken out of 90 degrees but if the remainder noted, be greater then the cosine of the base, then adde the arch of this fourth sine to 90 degrees, so shall the sum or difference be the angle required.

But when one side is greater then a quadrant, and that side be made the base; then alwayes add this fourth arch to 90 degrees, to make up your angle required.

And if the side exceeding a quadrant be one of the containing sides, then alwayes take this fourth arch out of 90 degeees, and the remainder shall be the angle re∣quired.

As in the triangle ZPS, having the side ZP, the com∣plement of the latitude 37 degrees 30 minutes, and the side PS the complement of the declination 78 degrees 30 minutes with the base ZS the complement of the Suns altitude 59 degrees 15 minutes, we may finde the angle ZPS, the houre of the day.

For the side ZP, the complement of the latitude being 37 degrees 30 minutes, and the complement of the side

PS, the declination of the Sun, being 11 degrees 30 mi∣nutes, their sum is 49 degrees 00 minutes, and their dif∣ference 26 degrees 00 minutes, and if we put the side of the quadrat 1000, the sine of the sum 49 degrees 00 mi∣nutes, is 754, and the sine of the difference 26 degrees 00 minutes, is 438, these two sines joyned together is 1192, the halfe whereof is 596 for the first found sine, which taken out of the sine of the sum leaveth 158, which re∣mainder I note, & because either of these sides is lesse then a quadrant, I take the difference betweene this remain∣der noted and the sine of the complement of the base ZS being 30 degrees 45 minutes, the sine whereof is 511, out of which I take the remainder noted, and the remainder is 353, for the second found sine; then placing the threed at the intersection of the contrary parallel of 596, with the right parallel of 353, it will cut the contrary sine of 90 degrees, at the right sine of 36 degrees 20 minutes, and in regard the sides are all lesse then quadrants, and the remainder noted, lesse then the cosine of the base, I take this 36 degrees 20 minutes, out of 90 degrees, and there resteth 53 degrees 40 minutes, for the houre from the meridian which was required. The like may you doe with any of the other angles, as to finde the Azimuth, or the angle of the Suns position in regard of Pole and Zenith.

But you shall have a more speedy way to perform this probleme, by the Planisphere on the back of the quadrat; for by knowing the three sides of any spherical Triangle, you shall finde an angle as spedily as if you had the other two angles also.

CHAP. XLII. To finde a side by knowing the three angles.

IN any spherical Triangle the sides may be changed in∣to angles, and the angles into sides: by taking first for any one angle and his subtending side, the remainder to a semicircle: then shall the operation be the same, as in the former Chapter.

Thus in the Triangle ZPS, having the angle ZPS 53 degrees 40 minutes, and ZSP 34 degrees 47 minutes, with SZP 113 degrees 17 minutes, if you take the greater angle of 113 degrees 17 minutes, out of 180 degrees, there will remain 66 degrees 43 mi∣nutes, then as if you had a Triangle of three known sides, one of 53 degrees 40 minutes, another of 34 degrees 47 minutes, and a third of 66 degrees 43 minutes, you may finde an angle opposite to any one of these sides by the last Chapter. So shall the angle thus found, be the side required, if it be not the angle opposite to that side which was made by substraction from 180 degrees, if it shall be that angle, then take it likewise from 180 degrees, and the remainder shall be the side required.

CHAP. XLIII. To finde a side, by knowing the other two sides, and the angle contained by them.

HEre, first, let down a perpendicular, the which you may performe by the 23 Chapter, so shall the Tri∣angle

angle given be reduced into two right angled triangles Thus in the Triangle ZPS, having ZP the complemen of the latitude, and PS the complement of declination with ZPS the angle of the houre from the meridian, we may finde SZ the complement of the Suns altitude.

For having let down the perpendicular Z X by the 23 Chapter: we have two Triangles ZXP, and ZXS, both right angled at X, then may we finde the side PX, either by the 24, 32, or 33, Chapters, which taken out of PS, leaveth the side XS: now with this side XS, and the perpendicular ZX, we may finde the base ZS by the 26 Chapter.

Or if you let down the perpendicular SA, you shall have two right angled Triangles, viz. SAZ, and SAP, then having the base PS, and the perpendicular SA you may by the 24, or 32 or 33 Chapters finde the side AP, from which if you take ZP, there will remain ZA: then with the sides AZ, and AS, you may by the 26 Chap∣ter, finde the base ZS.

CHAP. XLIV. To finde a side by knowing the other two sides, and one angle next the inquired side.

THus in the Triangle ZPS, having the side ZP the complement of the latitude, and PS the complement of the declination, with PZS the angle of the Azimuth, we may finde the side ZS the complement of the Suns altitude.

For having ZP, and the angle at Z, we may to SZ pro∣duced, let down the perpendicular PB by the 23 Chap∣ter: then have we two right angled Triangles, PBZ, and PBS, then by the 24 Chapter, having the base PS of the Triangle PBS, and the perpendicular PB, we may finde the side BS; and by the base PZ, of the Triangle PBZ, and the same perpendicular PB, we may finde the side BZ which being taken out of BS leaveth ZS for the side required.

CHAP. XLV. To finde a side, by knowing one side, and the two angles next the inquired side.

THus in the Triangle PZS, having PS the declina∣tion of the Sun, and PZS, the angle of the Azi∣muth, and ZPS the angle of the houre from the meridi∣an, we may finde the side ZP the complement of the la∣titude.

For if by the 23 Chapter, we let down the perpendi∣cular SA, we shall have two right angled Triangles, SAZ, and SAP, then by the 24 Chapter, having the base PS, of the Triangle SAP, with the perpendicular SA, we may finde the side AP; and by the angle SZA, of the triangle SAZ, and the same perpendicular SA, we may finde the side AZ by the 32 Chapter, which be∣ing taken out of AP leaveth ZP the side required.

CHAP. XLVI. To finde a side by knowing two angles, and the side inclosed by them.

THus in the triangle ZPS, having the angles at Z and P, with the side intercepted ZP, we may finde the

side PS, for having by the 23 Chap. let down the per∣pendicular PB, we have two right angles PBZ, and PBS, then may we by the 29, or 37, or 38, Chap. finde the angle BPZ, which added to the angle ZPS giveth the angle BPS, then with this angle BPS, and the perpen∣dicular BP, we may finde the base PS, according to the 35 Chapter.

CHAP. XLVII. To finde an angle by knowing the other two angles, and the side inclosed by them.

THus in the triangle ZPS, having the angles at Z and P with the side intercepted ZP, we may finde the o∣ther angle ZSP, for having by the 23 Chapter, let down the perpendicular ZX, we have two right angled triangles ZXP, and ZXS; then may we finde the angle PZX, by the 37 or 38 Chapter, and that taken out of the angle PZS leaveth the angle SZX: then with the angle SZX, and the perpendicular ZX, we may finde the angle re∣quired ZSX, according to the 28 Chapter.

CHAP. XLVIII. To finde an angle, by knowing the other two angles, and one of the sides next the inquired angle.

THus in the triangle ZPS, having the angles at P, and S, with the side ZP, we may finde the angle PZS, for by the 23 Chapter having let down the per∣pendicular ZX we have two right angled triangles, PXZ, and SXZ, then may we by the 37 or 38 Chapter, finde the angles PZX, and SXZ, which being both ad∣ded together maketh the angle PZS which was required.

CHAP. XLIX. To finde an angle, by knowing two sides, and the angle conteined by them.

THus having the sides ZP, and P S, in the triangle ZPS, with the angle comprehended ZPS, we may finde the angle PZS, for having let down the perpendicular AS, by the 23 Chapter, we have two right angled trian∣gles SAZ, and SAP; then may we by the 14, or 32, or 33 Chapters, finde the side AP, out of which if we take ZP, there will remain AZ: with which side AZ, and the perpendicular SA, we may finde the angle AZS, by the 36 Chapter, this angle AZS, taken out of 180 degrees, leaveth the angle PZS, which was required.

CHAP. L. To finde an angle, by knowing the two sides next it, and one of ahe other angles.

THus in the triangle ZPS, having the sides ZP, and PS, with the angle PZS, we may finde the angle ZPS for having let down the perpendicular PB by the 23 Chapter, we have two right angled triangles PBZ, and PBS; then may we by the 37, or 38 Chapter finde the angles BPZ, and BPS; and taking BPZ, out of BPS, there remaineth the angle ZPS, which was re∣quired.

Thus you may see Master Gunters 28 cases of a sphe∣rical

triangles, which he applyed to his sector, all resolved by the foreside of ray Universal quadrat. And now you sh•…•… see them more speedily performed, by the moveable planisphere on the back of the Quadrat; comming som∣thing neere to that of Mr. Blagraves Iewel. But where∣as he for the making up of his triangles, useth his dark and intricate reete, (which is no small let to to the eye, for beholding of the true intersections upon the Iewel,) I shall only use a fine threed, which may always hang in the center on the foreside, to serve redily for the afore∣said work, and being of a resonable length, will be as use∣ful for this side too.

CHAP. LI. In a right angled triangle, having the base and one of the oblique angles, to finde the two sides, and the other angle.

LEt the triangle ABC be given again, being the same right angled triangle as was given in the 23 Chapter, wherein knowing the base AC to be 65 deg. 23 min. and the angle BAC, to be 37 deg. 30 min. let it be requir∣ed to finde the sides AB, and BC, with the angle ACB.

First move the sphere about, until the horizon cut the deg. of the angle given upon the limb, counted from the equinoctial point, viz. 37 deg. 30 min. which resting in this position, count the base AC 65 deg. 23 min. from the center upon the horizon; and where this sum endeth, there the horizon cutteth the parallel of 33 deg. 36 min. which is the side BC subtending the given angle BAC,

and farther, at this intersection of 65 deg. 23 min. on the Horizon, with the parallel of 33 deg. 36 min. doth meet the sixtieth meridian, which is the side AC next to the given angle at A, and here you may see your whole trian∣gle upon your Instrument, as plainly as on the sphere it selfe, included by these three arches, viz. 65 deg. 23 min. on the Horizon, and 60 deg. in the Equinoctial, and 33 deg. 36 min. upon the said sixtieth meridian, and for the angles, that at the intersection of the meridians with the Equator must needs be right, the given angle is pla∣ced id the center, and denominated by the Horizon upon the limb; the other resteth to be found. And thus of the six parts of a right angled triangle, five is in open vew.

[illustration] diagram of the measurement of a spherical triangle (ABC)

But now to finde the other angle subtended by the side AB, exchange the places of the sides, and let that which was counted upon the Equator amongst the meridians, be counted amongst the parallels; then moving the plani∣sphere until the base 65 deg. 23 min. counted on the Ho∣rizon, meet with the parallel of 60 deg. the side AB sub∣gnidnet

rending the angle required: so shall the Horizon cut the limb at the angle required, counted from the Equator. And here is your triangle made up as before, viz. The base 65 deg. 23 min. on the Horizon, the side BC 33 deg. 36 min. on the Equator, and 60 deg. the side AB amongst the parallels.

CHAP. LII. By knowing the base and one of the sides, to finde the other side, with both the oblique angles.

THus in the right angled triangle ABC, having the base AC 65 deg. 23 min. and the side BC 33 deg-36 min. we may finde the other side AB, and the two oblique angles at A and C.

For if you count the base AC 65 deg. 23 min. upon the Horizon, and the side BC 33 deg. 36 min. among the parallels; and then move the planisphere to and fro, until the 65 deg. 23 min. do cut the parallel of 33 deg. 36 min. counted from the Equinoctial, and there shall meet you the 60 meridian to make up your triangle; which 60 degrees is the required side AB, and now at this very instant you may see the Horizon cut the limbe at 37 deg. 30 min. which is one of the angles required, viz. the angle BAC. And now for to finde the other angle you must shift the sides as in the last Chap. for the side subtending the angle required, must be counted a∣mong the parallels.

Thus moving about the planisphere til the intersection of the sixtieth parallel and meridian of 33 deg. 36 min. do cut the Horizon, which will be at 65 deg. 23 min. the base AC as before, and now doth the horizon cut the limbe at 72 deg. 16 min. which is the angle ACB as was required.

CHAP. LIII. Having one side, and one of the oblique angles of a right an∣gled triangle, to finde the base, the other side, and the other oblique angle

AS in the right angled triangle ABC, having the an∣gle BAC 37 deg. 30 min. and the side AB 60 deg. we may finde the base AC, and the side BC, with the angle ACB.

For if you move about the planisphere, until the Ho∣rizon cut upon the limbe, the degrees of the angle given, viz. 37 deg. 30 min. and then reckon the side given viz. 60 deg. upon the Equator among the meridians; you shall upon the same 60 meridian, finde 33 deg. 36 min. intercepted between the equator and horizon; which 33 deg. 36 min. is the side BC subtending the given angle BAC.

Now for to finde the base, mark where the said sixti∣eth meridian cutteth the horizon, which is at 65 deg. 23 min. and such is the base AC. As was required. And for to finde the other oblique angle, you may shift the sides, as was shewed in the two last Chapters.

CHAP. LIV. Having the two sides of a right angled triangle, to finde the base, and the two oblique angles

AS in the right angled triangle ABC, having the two sides AB 60 deg. and BC 33 deg. 36 min. we

may finde the base AC, and the two oblique angles A and C.

For if we count the two sides one among the meridi∣ans, and the other amongst the parallels; as here 60 deg. among the meridians, and 33 deg. 36 min. amongst the parallels; and then turn the planisphere about, until their intersection cut the Horizon, we shall have thereon cut 65 deg. 23 min. for the base AC. and the Horizon also cut∣ting 37 deg 30 min. on the limb which is the angle BAC, subtended by the side BC, which was counted amongst the parallels.

Thus have you the base, and one of the angles required, and the other is soon found by shifting the sides as in the 51 and 52 Chapters.

CHAP. LV. Having the two oblique angles, of a right angled Triangle, to finde a side opposite to either of them.

LEt the right angled Triangle ABC be given, where∣in the two oblique angles A and C are known, viz. the angle BAC 37 degrees 30 minutes, and the an∣gle ACB 72 degrees 16 minutes, and let it be required to finde the two sides AB and BC.

First, place the planisphere so that the two poles there∣of may both lye in the Horizon, so shall the Equator be in the zenith. This done, count the angle next the side required, amongst the parallels to the Equator, from a∣ny of the poles inward, and the angle subtended by the side required, among the almicanters, counted from the zenithward downward on both sides, upon which points

place the threed, which by this means lyeth parallel to the Horizon. Now where the threed crosseth the for∣mer parallel, there passeth by the meridian which giveth you the angle required, counted from the limb.

As for example, both poles being in the Horizon, I count the angle ACB 72 deg. 16 min. among the paral∣lels, and the angle BAC 37 deg. 30 min. among the Almicanters, being both counted from the limb, then ap∣plying the threed to the said Almicanter, I finde it cut the aforesaid parallel, at the meridian of 33 deg. 36 min. which is the side BC which was required.

Now for to finde the other side, let the poles remain in the Horizon, and apply the threed to the Almicanter of 72 deg. 16 min. counted from the zenith, and you shall see it cut the parallel of 37 deg. 30 min. counted from the poles, as the sixtieth meridian counted from the limb, and such is the side AB as was required.

CHAP. LVI. Having the three sides of any spherical Triangle, to finde an angle opposite to any of them.

LEt the Triangle ZPS be given, whereof the three sides are all known, and the three angles all unknown. First, therefore let the planisphere be so placed, that one of the sides next the angle required, be contained be∣tweene the pole and zenith. Or place the sphere so that the Horizon may cut upon the limb, the degrees of one of the sides containing the angle required, which is all one thing, this being done count the other containing

side from the pole among the parallels; and the side sub∣tending the angle required, always from the zenith a∣mongst the almicanters, and thereto place the threed now where the threed cutteth the parallel of the second con∣taining side, there passeth by the meridian which giveth you the angle required, counted from the limb.

Thus in the Triangle ZPS, having ZP the comple∣ment of the latitude, and PS the complement of the Suns declination, with ZS the complement of the Suns alti∣tude, we may finde the angle of the hour from the meri∣dian ZPS.

For having placed the Horizon to the complement of the latitude 37 deg. 30 min. the side ZP, and counted the complement of declination 78 deg. 30 min. from the pole among the parallels of declination, or the declina∣tion it selfe from the Equator 11 deg. 30 min. and also 59 deg. 15 min. amongst the almicanters from the zenith which is the side ZS, or the complement of the Suns al∣titude from the Horizon, and, thereto applyed the threed parallel to the Horizon; it shall cut the parallel of decli∣nation at the meridian or hour circle of 53 deg. 40 min. from the limb or general meridian, and such is the angle ZPS, which was required.

In like manner may we finde the angle PZS which is the Azimuth from the North, for letting the sphere rest as before, with the side ZP at the Horizon; if we count the side ZS 59 deg. 15 min. among the parallels count∣ed from the pole, and the side PS 78 deg. 30 min. from the zenith amongst the almicanters, and thereto apply the threed parallel to the Horizon, it shall cut the said parallel of 59 deg. 15 min. at the meridian of 113 deg. 17 min. for in this manner of work, the angles are all counted from

the South part of the limb. This 113 deg. 17 min. is the angle PZS, which was required.

Then for to finde the third angle, you may shift the sides, placing one of the other sides at the Horizon, and then proceed as before.

Thus you may see this proposition which is most use∣ful, but withal most difficult of all others, (as in Arith∣metique, so by the Sector and Crosse-staffe, and also by the fore-side of this Universal Quadrat) speedily perfor∣med by the planisphere on the back of this quadrat.

[illustration] diagram of the measurement of a spherical triangle (ZPS)

CHAP. LVII. To finde a side, by knowing the other two sides, and the angle comprehended.

FIrst, set the Horizon to one of the given sides, counted on the limb from the Equator, and count the other side from the pole among the parallels, then count the given angle amongst the meridians from the limb; and at the point of intersection between that meridian and the parallel, there place the threed parallel to the Horizon; which you may do by help of the almicanters, and note what almicanter the threed so cutteth, for that is the third side counted from the zenith.

As in the Triangle ZPS, having the sides ZP 37 deg. 30 min. ZS 59 deg. 15 min. with the angle PZS 113 degrees 17 min. we may finde the side PS, for first I place the Horizon at 37 deg. 30 min. and count the other side

ZS 59 deg. 15 min. from the pole among the parallels, and the angle PZS 113 deg. 17 min. among the meridi∣ans from the south part of the limb, and where this meri∣dian of 113 deg. 17 min. cutteth the parallel of 59 deg. 15 min. there I place the threed parallel to the Horizon, and I finde it to cut the Almicanters at 78 deg. 30 min. from the zenith, and such is the side SP, which was re∣quired.

Thus having all three sides, you may finde any of the other two angles by the last Chapter.

CHAP. LVIII. To finde a side, by knowing the other two sides, and one of the angles next the inquired side.

FIrst, set the Horizon to that side given, which is next the given angle, counting on the limb from the Equa∣tor; then place the threed (parallel to the Horizon) upon the other side counted from the zenith amongst the Almi∣canters; now where the threed intersecteth the angle gi∣ven counted among the meridians from the limb, there passeth by the parallel of the side required counted from the pole.

As in the Triangle ZPS, having the side ZP the com∣plement of the latitude 37 deg. 30 min. and the side PS the complement of the declination 78 deg. 30 min. with the angle PZS, the azimuth from the North 113 deg. 17 min. we may finde the side ZS the complement of latitude.

For if we place the Horizon to the side ZP 37 deg. 30 min. from the Equator in the limb, and then place the threed to the almicanter of 78 deg. 30 min. counted from the zenith, it shall cut the meridian of 113 deg. 17 min. from the limb, at the parallel of 59 deg. 15 min. counted from the pole, and such is the side ZS, which was required.

CHAP. LIX. Having two sides and one angle, to finde any of the other two angles.

FIrst, finde the third side by one of the two last Chap. then having all three sides you may by the 56 Chap. finde any angle required.

Here needeth no example more then what those Chap∣ters affords.

CHAP. LX. Having either all three angles to finde a side, or having two angles and one side, to finde any of the other two sides.

TO finde a side by knowing the three angles, you must first convert the angles into sides by the 42 Chap. which being done, you may by the 56 Chap. finde an angle opposite to any one of these sides; so shall the angle which is thus found, be the side which is here required. As in the said 42 Chap. so likewise having two angles & one side, you may by the said 42 Chap. convert these angles into sides, and the sides into angles; so shall you have two sides and one angle, by which and the 57 or 58 Chapters, you may finde a third side, then having all three sides, you may by the 56 Chap. finde an angle opposite to any of them, which angle thus found, shall be the side required

The end of the second Book.