The geometrical square, with the use thereof in plain and spherical trigonometrie chiefly intended for the more easie finding of the hour and azimuth / by Samuel Foster ...

Of the use of the Square in General for the Solution of Spherical Triangles.

In any Spherical Triangle whatsoever.

¶By having the Legs and Base, to find the Vertical Angle.

THe Angle given or sought is the Vertical Angle, The sides comprehending it are the legs. The side subtend∣ing it is the Base.

From the top of the Square, count the sum of the legs upon one side, the difference of them on the other side, To this sum and difference apply the threed, Then from the same top of the Square count the base also, And mark where it cuts the threed, for the line passing through the intersection, and standing Square to the top, (if it be numbred from that side of the Square whereon the difference of the legs was counted) gives the Vertical Angle required.

This is the general manner of work for this Proposition, which may be illustrated by these particulars.

FIRST,

Having the Latitude of the place, the Declination and Altitude of the Sun, to find the Hour of the day.

BY the declination of the Sun, may be had his distance from the elevated Pole, By subtracting it from 90 gr. when the Declination is of the same denomination with the said

Pole; Or by adding the Declination to 90 gr. when the De∣clination and elevated Pole are of several denominations.

In this case, we have the three sides of a Spherical Triangle given, and an Angle sought.

The two legs are The complement of the Latitude, and The Suns distance from the Pole. The base is, The complement of the Suns Altitude: The Angle is the Hour required, which must be accounted from the Coast of a contrary name, to the elevated Pole.

According then to the former general prescript, and this par∣ticular declaration, For the hour take the sum and difference of the complement of the Latitude, and of the Suns distance from the Pole, and from the top of the Square, upon one side, count the difference, the sum on the other, to these terms apply the threed; Then from the top of the Square also, count the com∣plement of the Suns Altitude, and where it cuts the threed, the line that crosseth it Square in the same point (being reckoned from that side whereon the difference of the legs was counted) gives the hour from the Meridian or noon.

To make it plain by an Example.

In a North Latitude of 52 gr. 30 min. the Sun declining 20 gr. to the North, the Altitude of the Sun being by obser∣vation 43 gr. I would know the Hour of the day. The legs of this Triangle are the complements of the latitude and de∣clination, that is 37 gr. 30 min. and 70 gr. 0 min. The sum of them is 107 gr. 30 min. their difference is 32 gr. 30 min. Then from the top of the Square at D upon the side DG, I reckon this difference 32 gr. 30 min. downward to k. And on the other side of the Square from the top at B, I also count the sum of the legs 107 gr. 30 min. downward to l. To k and l. I apply the threed. Which done from the top of the Square, again, I count the base 47 gr. the complement of 43 gr. the altitude observed, downward also to 0, and the line that there meets me, I follow till it cut the threed, which is at n, and the line that there ariseth Square to it is n r. I say now that nr, if it be counted from the side DC whereon the difference of the legs was counted, shall give 44gr. 8 min. which turned into hours and minutes of an hour, (allowing 15 gr. to an hour; and 15 min. of a degree to one minute of an hour) will make two hours ••nd 56½ min. from

the Meridian or South, And such is the Hour for that Latitude, Altitude, and Declination.

So also, If in the same Latitude and distance of the Sun from the Pole, but in the altitude of 10 gr. I would know The boure of the day. Here because the legs, that is, the com∣plement of the latitude and the distance from the Pole, are the same, therefore the same position of the threed remaines still, I therefore onely reckon the base (as before) which here is 80 gr. from D to p, then I follow the line p, till it cuts the threed at m, and the line there arising is m s, which counted from DC, whereon the difference of the legs was reckoned, shall give 99 gr. 50 min. that is 6 hours 39⅔ min. of an hour from the Meridian or South.

Another Example.

In the same latitude of 52 gr. 30 min. let the declination of the Sun be 20 gr. to the South, where his distance from the elevated Pole is 110 gr. and let the altitude of the Sun be by observation 10 gr. I require the Hour. The legs are 37 gr. 30 min. the complement of the latitude, And 110 gr. the Suns distance from the Pole. The sum of them is 147 gr. 30 min. The difference 72 gr. 30 min. which I count upon the sides of the Square down to u and t; and the base which is 80 gr. the complement of 10 gr. I count also from D to p, then I follow the line p, till it cut the threed at x, and the line there arising is x y, which counted from DC, whereon the difference of the legs was reckoned, shall give 38 grad. 56 min. that is, two hours and almost 36 min. of an houre from the Meridian or South.

Note, That the threed in this situation, shewes on the dia∣meter of the Square (which in this case represents the Hori∣zon) the Semidiurnal and Seminocturnai Arks, for where the threed crosseth the middle line, the line there arising, (counted from that side of the Square, whereon the difference was num∣bred) shewes the Semidiurnal ark, and counted from the other side, shewes the Seminocturnal ark.

Observe also, If you would known the Crepusculum or Twi∣light, the threed is to be placed as before, according to the sum and difference of the legs, and if you allow 18 gr. for the

Crepusculin line (as they usually doe) the base will alway be 108 gr. which in the two first Examples will not touch the threed at all, and therefore in that latitude and parallel of the Sun, the twilight continues all night. But in the last Example you shall find the Crepusculin line to cut the threed, 6 hours and 15 min. from the Meridian, which shewes that the twi∣light begins at 5¾ a clock in the morning, and ends at 6¼ in the evening, and the rest of the time is dark night which is 11½ hours.

If the sum of the the legs be more then 180 gr. that is, if it would reach beyond the bottom of the Square, you must when you have reckoned to the bottom, count upward back again till you have ended the whole sum.

SECONDLY,

Having the Latitude of the place, the Declination and Altitude of the Sun, To find the Azimuth of the Sun.

HEre also the 3 sides are given, the same with the former, and an Angle sought. The two legs are the Comple∣ments of the latitude, and Suns altitude, The base is the Suns distance from the Pole which is elevated above the Horizon. The angle sought is the Suns Azimuth, from that part of the Meridian, which is of the same denomination with the elevated Pole.

So then according to the former general prescript, and this particular declaration, for the Azimuth, doe thus.

Take the sum and difference of the Complements of the lati∣tude and Suns altitude, and count from the top of the Square, the one upon one side, the other on the other side, and to these terms apply the threed; Then from the top of the Square also, count the Suns distance from the Pole, and where it doth crosse the threed, the line that there ariseth Square to the former, being reckoned from that side of the Square whereon the difference of the legs was counted, gives the Azimuth from that part of the Meridian which is of the same denomination with the elevated Pole, and counted from the other side, gives the Azimuth from the other coast.

To Illustrate it by an Example.

In a North latitude of 52 gr. 30 min. let the altitude of the Sun be 22 gr. and the declination 10 gr. Northerly, By these given I would know the Suns Azimuth, the two legs of the Triangle are the complements of the latitude and Suns altitude, that is 37 gr. 30 min. and 68 gr. the sum of them is 105 gr. 30 min. the difference is 30 gr. 30 min. The sum of them I count on the side BA, from the top at B down to k, The difference I count on the other side from D down to h, and to these points k and h, I apply the threed k h, And last∣ly, because the de∣clination is 10 gr. North-ward in a North latitude, ther∣fore his distance from the elevated Pole is 80 gr. which I count from the top D, down to l, and follow the line at l, till it meet with the threed at n, where I find the line m n, to crosse it also, which numbred from the side DC, whereon the difference of the legs was numbred, gives 102 gr 38 m. the Azimuth from the North: And so also if it be account∣ed from the side BA, it gives the Azimuth from the South 77 gr. 22 min. the residue of the former, or the complement of it to 180 gr.

[illustration]

Another Example, In the same latitude and the same al∣titude, and therefore also the same situation, of the threed, let the declination be Northerly 23½ gr. therefore the distance from the Pole will be 66½ gr. which I count from D to f, and following the line f till it meet with the threed at i, I find the line g i, to crosse there also, which being counted from the side DC, whereon the difference of the legs was counted,

shewes 79 gr. 38 min. the Azimuth from the North, Or count∣ed from the other side, gives the residue of the former, 100 gr. 22 min. The Azimuth from the South.

A third Example. In the same latitude and altitude, and therefore also in the same situation of the threed, let the de∣clination of the Sun be 10 gr. to the South, then shall his di∣stance from the elevated North Pole be 100 gr. and because this 100 gr. is the base, I therefore count it from the top D, down to o, and following the line o, I find it to cut the threed at r, and the line r p there crossing, shewes me from DC, (the side whereon the difference of the legs was counted) 146 gr. 32 min. for the Azimuth from the North, or if the same line be numbred from the side BA, it shewes 33 gr. 28 min. the residue of the former, for the Azimuth from the South.

These Examples may suffice for this kind, and according to these patternes, all others are to be framed.

In any Spherical Triangle whatsoever.

¶By having the legs and Vertical Angle, to find the Base.

From the top of the Square, count the sum of the legs upon one side, the difference of them on the other side. To this sum and difference apply the threed: Then from that side of the Square whereon the difference of the legs was numbred, count the Vertical Angle given, and where it cuts the threed, mark the line that passeth there-through parallel to the top of the Square, for that line, counted from the top, gives the base required.

This is general for all works of this kind, which may be illustrated in particular: thus,

Having the latitude of the place, the declination of the Sun, and the hour of the day, to find the altitude of the Sun, for that latitude, declination, and hour.

HEre have we the two legs of the Triangle, with the in∣tercepted Vertical Angle, given, and the base sought. The legs are the complement of the latitude, and the Suns distance, from the Pole; The angle intercepted is the

hour whose altitude we seek. And the base is the complement of the altitude sought for.

Wherefore by the former general prescipt, and this parti∣cular explication, we may attain to the thing required thus, as in the former practice, so here; Apply the threed to the sum and difference of the complement of the latitude, and of the Suns distance from the Pole, Then reckon the hour given from the Meridian, from the side whereon the difference of the legs was counted, and where it crosseth the threed observe the line that passeth there-through parallel to the top of the Square, for that line reckoned from the top, shewes the base, that is, The com∣plement of the altitude, or reckoned from the middle line of the Square, it gives the altitude it self of the Sun, for that parallel and Hour; And so the threed (which now represents the Suns parallel:) lying still, you may count the altitudes for all the rest of the hours for that parallel.

For Example.

In a North latitude of 52 gr. 30 min. let the Sun decline 20 gr. to the North, so that his distance from the elevated North-pole will be 70 gr. which is one of the legs given, and the complement of the latitude, 37 gr. 30 min. is the other, The sum of them is 107 gr. 30 min. The difference is 32 gr. 30 min. This difference is the complement of the Suns Meri∣dian altitude; and I count it from D the top of the Square, to k: (in the first figure) thereto applying one end of the threed And on the other side from B, I count the sum, 107 gr. 30 min. down to l, thereto applying the other end of the threed. The threed thus laid, resembles the Suns parallel, for that decli∣nation, Now from the side D k, whereon the difference was numbred, I count the Vertical Angle, As first 15 gr. for the first hour from the Meridian, either 11 in the morning, or one in the afternoon, and where it cuts the threed, I observe the other line there crossing also, which counted from the top, gives for the base, 34 gr. 32 min. the complement of the al∣titude required.

Or rather. Count it from the middle line, which in this case represents the Horizon, and then you shall have 55 gr. 28 m. the altitude it self, for that hour and parallel, So the second

hour from the Meridian (10 or 2) gives for the altitude 50 gr. 4 min. The third hour (9 or 3) gives 42 gr. 31 m. The fourth hour from the Meridian (8 or 4) gives 33 gr. 53 min. The fifth (7 or 5) gives 24 gr. 48 min. The sixth (6 in the morn∣ing and evening) gives 15 gr. 45 m. The seventh (5 in the morn∣ing, or 7 in the evening) gives 7 gr. 6 m. And so farrethe Sun is above the Horizon in that parallel, and then begins to go down.

And observe further, That the threed thus placed taken in that part below the Horizon, gives the altitudes for the hours in the declination which is equal to this, but to a contrary coast; so that the threed in this situation, gives the altitudes for the declination of 20 grad. towards the South, for that part of the threed that is under the Horizon or middle line, is the Semi-nocturnal ark for the parallel lying 20 gr. from the Equinoctial Northward, and is therefore equal to the Semi-diurnal ark that belongs to the parallel which lies 20 gr. from the Equinoctial Southward, and is of like si∣tuation below the Horizon that the other is above, wherefore the depressions belonging to the hours in this, are the same with the altitudes of the same hours in the other. To go on then where we left, The next hour counted from the Meridian of the Winter parallel is the fourth, that is, either 8 in the morning, or 4 in the afternoon, and his depression is 0 gr. 50 min. The next hour the third from the Meridian (either 9 or 3) is de∣pressed 7 gr. 39 m. The second hour, (10 or 2) is 12 gr. 57 m. The first hour (11 or 1,) is depressed in this North parallel 16 gr. 20 min. that is, it is elevated so much in the like South parallel. Thus of each two opposite parallels of declination may the altitudes be had at one and the same situation of the threed. But if the other way seeme plainer, do as before. Let the Sun in the same latitude decline 20 gr. to the South, his distance from the North elevated Pole, is then 110 gr. the sum of the complement of the latitude 37 gr. 30 min. and this distance is 147 gr. 30 min. The difference is 72 gr. 30 m. This difference I count from D to t, (in the first figure,) The sum I also account as before, from B to u, And to t u, I placed the threed, Now from the side D t, I count the hours as I did before, and find the altitude of 11 and 1, 16 gr. 20 min. of 10 and 2, 12 gr. 57 min. of 9 and 3, 7 gr. 39 min. of 8 and 4, 0 gr. 50 min. All the same that the former depressions were;

And if now you take the depressions of the hours upon the threed in this situation, you shall find them all the same that the altitudes in the former parallel of 20 gr. North declination were; So that ever, one side of the threed will afford the al∣titudes for the hours in any two opposite parallels.

The Meridian altitude is the complement of the difference of the legs, And in the opposite parallel it is the excesse of the sum of the legs above 90 gr. And as you have done for the Altitudes of the whole hours, so may you doe for their halves and quarters. Thus much for this also.

In any Spherical Triangle, whatsoever.

¶If the Proportions be in right Sines alone, they are resolved in this manner.

Count the first sine given (upon one of the sides of the lesser Square EIDG,) from the Center E, and upon the line there arising count the second sine, whereto apply the threed, Then upon the same side with the first, count the third, and observe the line there arising, for from it doth the threed cut off the fourth sine required.

This general may be illustrated in particular thus.

Having the greatest Declination of the Sun, and his distance from the next Equinoctial point▪ to find the Declination of the Sun for that distance.

THis particular belongs to the solution of a Rect-angled Spherical Triangle, yet the manner of the work in this is the same with the work belonging to the solution of the Obliquangled ones. The proportion stands thus; As the radius, is to the sine of the greatest declination; So the sine of the Suns distance from the next Equinoctial point, to the sine of the declination of that point.

For an Example.

Let the distance from the Equinoctial be 30 gr. The great∣est declination 23 gr. 30 min. I would know the declination for that distance of 30 gr. The Proportion is, As the radius,

is to the sine of 23 gr. 30 min. So the sine of 30 degrees, to what sine? Count EG for the radius, and upon GD the line there arising, reckon 23 deg. and 30 min. up to a, and thereto apply the threed, Then again upon EG, count e b, the sine of 30 gr. and follow the line there arising which is b c, till it cut the threed at c, and the line c d, there crossing also (being counted from EG) gives for the declination required, 11 gr. 30 min. So that the sine of 11 gr. 30 min. is the fourth Pro∣portional Sine to the former three.

By the Hour of the Day given, with the Suns distance from the elevated Pole, and the complement of his altitude above the Horizon, to find his Azimuth.

The Azimuth thus gotten is counted in North latitudes from the North, in South latitudes from the South. Let the hour be 3 from the meridian. The Suns distance from the Pole 66 gr. 30 min. The complement of the altitude, 44 gr. 20 m. By these things known, I would find the Azimuth. The pro∣portion whereby it is wrought is this. As the Sine of the hour, is to the Sine of the complement of the altitude; So is the sine of the distance, to the sine of the Azimuth. Wherefore upon the side of the Square EG, I count the sine of 45 gr. to h, and upon the line there arising, I count the sine of 44 gr. 20 min. up to i, and thereto apply the threed. Then upon the same side with the first I reckon the sine of the Suns distance 66 gr. 30 min. from E to k, and following the line there arising till it meet with the threed at n, I find the line nl, to crosse there also, which counted from EG, gives the sine of 65 gr. for the fourth proportional, so that 65 gr. shewes the residue of the Angle required, that is to say, In our North latitude it shewes me the Azimuth from the South; because the Angle of the A∣zimuth from the North is an obtuse Angle, namely 115 gr. and the same sine serves both to it and to 65 gr. his residue.

[illustration]

¶And here also is to be noted, That when any Proportion in right sines alone is offered, and the radius is the first leader

in the Proportion, that then I say, it may be resolved by the former kind of work, by the sum and difference, counting the complement of the arks of the two Sines given, for the legs of the Triangle, and the ark of the radius or 90 gr. for the Ver∣tical Angle, and the base found out to be the complement of the ark required. As in the first Example,

The two middle arks were 30 gr. and 23 gr. 30 min. their complements are 60 gr. and 66 g. 30 min. The sum of these is 126 gr. 30 min. there difference 6 gr. 30 min. to this sum and difference, I apply the threed, as in the former Examples, and then count the Vertical Angle 90 gr. which fals in the middle line and where the threed cuts it, there is the quantity of the Declination, 11 gr. 30 min. as before: And these degrees are counted from the Center of the Square at E.

And thus may all others of this nature, having the radius in the first place, be absolved. And not onely these of sines alone, but with sines intermingled with Tangents also, If it so fall out that these Tangents be lesse then the radius, And if instead of their proper arks be taken the complements of the arks of sines equal unto those Tan∣gents.

And thus much for Exemplifying in this kind also. Those that follow are appropriate to rectangled Spherical Triangles only.

In any Rectangled Spherical Triangle whatsoever.

¶If the Proportion stand between right sines (whereof the Radius is alway one) and Tangents, they are to be re∣solved in this manner.

Vpon one of the sides of the lesser square EIGD. Count the first term, and upon the line there arising count the second, whereto apply the threed. Then upon the same side whereon the first was reckoned, count the third, and follow it till it crosse the threed, for the quantity of it comprehended between the third term and the threed, gives the fourth proportional term required, alway remembring that every term be taken on his proper Scale.

Here because the proportions are divers, we shall need more explication then in all the rest. Yet the variety herein, may

be reduced to three wayes according as one of these three, either the Radius; Sine, or Tangent, doth lead in the Proportion, the three wayes are these:

• 1 As the radius, is to a tangent, So is a sine to a Tangent.
• 2 As a sine, is to a tangent, So the Radius is to another tangent.
• 3 As a tangent, is to the radius, So another tangent, is to a sine.

But this variety is not all, for each of these three wayes is subject to variation, and that upon this occasion.—Vpon the square we have no tangent greater then the radius, or tangent of 45 degrees. Wherefore the proportion must be so contrived, as that no tangent greater then of 45 gr. be ingredient into it. To that purpose serves this general direction, namely,—If the tangent which is co-partner, in the proportion with the sine, be greater then of 45 gr. (alway provided that the two tangents doe never stand immediately together, which if they doe, may be brought into frame by transposition or alteration of the middle term.) Then, In the two first wayes the radius and sine must change places; and for the two tangents must be taken the tan∣gents of their complements; In the third way, the co-tangents of the third and first terms must remove into the first and third places.

To shew this more particularly in the 3 former wayes.

In the first,

If the tangent required in the fourth place prove greater then of 45 gr. (which how to discover is shewed hereafter) then by the former direction this alteration must be made.

As the sine in the third. place, is to the co-tangent in the se∣second,

So is the radius in the first place, to the co-tangent of the fourth.

In the Second,

If the tangent in the second place, be greater then of 45 gr. then by the former direction this proportion must be thus changed.

As the radius in the third place, is to the co-tangent of the second.

So is the sine in the first place, to the co-tangent of the fourth.

In the third

If the tangent in the third place, be greater then of 45 gr. then according to the former prescript this proportion must thus be varied.

As the co-tangent of the third place, is to the radius in the second;

So the co-tangent of the first place, is to the sine required in the fourth place.

Because in the first proportion it is unknown whether the tan∣gent required in the fourth place be greater or lesser then of 45 gr. and yet is necessary it should be known before it can be found out, you shall therefore in practice discover it thus.

If the line whereon it is to be accounted doth not meet with the threed rightly situated upon the Square, then is it greater then of 45 gr. and then the proportion must be altered as be∣fore, but if it doe meet with the threed, then is it lesse then of 45 gr. as it should be.

Observ•• that the tangents are actually in the limb onely, yet may be understood to be all over the plain, for some line or other standing even against them in the plain will supply them as well as if they were actually there drawn.

And note that if such a Proportion as this do at any time happen, namely, As a sine is to a tangent, So another sine, is to another tangent. And that these tangents, be discovered to be one of them greater then of 45 gr. the other lesse, That then the radius is to be brought into the Proportion, by saying, As the first sine, is to the first, So is the radius to another Tan∣gent; Then leaving out the first sine and tangent, and using for them the radius and this later tangent, say, As the radius is to the last found tangent, So is the sine in the third place to the tangent in the fourth; Which Proportion suites with those going before. But if both the tangents be either greater or less••r then of 45 gr. then may the solution be made with∣out the help of the radius.

According to the former Rules generally delivered are these following Examples framed, and will fully illustrate every Case.

[ I] For the first of the three several wayes there are three cases, For either both the tangents are lesse then 45 gr. or both grea∣ter, or no lesse, the other greater.

1 As the radius is to the tangent of 40 gr. So the sine of 50 gr. to the tangent of what?

Vpon the Square EIDG, I count EG for the radius, and upon the end of it in the Scales of tangents I reckon G a, 40 gr. thereto applying threed. Then upon EG, I count the sine of 50 gr. from E to b, and follow the line there arising till it cut the threed E a, at c, So that b c is the fourth term requi∣red, which because it is a tangent, must be numbred in the Scale of tangents, and therefore by help of the line c e, I transferre it thither, and find that the line lies even against 32 gr. 44 min. in the Scale of Tangents, and this is the ark of the fourth proportional term required.

2 As the radius is to the Tangent of 60 gr. So the sine of 50 gr. to the Tangents of what?

[illustration]

Upon the Square I count (as be∣fore) from G up to D and so for∣ward to n, 60 gr. among the Tan∣gents, and thereto apply the threed. Then I count the third term. E b the sine of 50 gr. and I follow the line there arising to the top of the Square, and yet it meets notwith the threed, but beyond the Square I see that it would concurre with it at r, which shewes that the fourth Tangent re∣quired, is greater then the Tangent of 45 gr. and therefore standing as it doth, cannot be expressed upon the Scale of Tangents, wherefore I alter the Proportion, by transposing the first and third terms into one anothers places, and for the Tangents themselves I take their complement thus.

As the sine of 50, is to the Co-tangent of 60, So is the Radius, to a fourth Tangent, which will be the complement of that which should be produced by the former Proportion. By this alteration it comes to passe that the sine leads in the Proportion, and so this Example now fals under the Examples of the second General way, and therefore shall be resolved there.

3 As the radius is to the Tangent of 50 gr. so the sine of 50 gr. to the Tangent of what?

Upon the Square I take EG for the radius, and at the end of it I reckon up to D, which is 45 gr. and so forward on the other side to g, that is to 50 gr. Then upon EG, I count e b the sine of 50 gr. and follow the line there arising till it cut the threed at i, so that b i is the fourth term, and because it is a Tangent, therefore by help of the line passing through i, that is, by the line i m, I transferre it to the Scale of Tan∣gents, and find that lies even against 42 gr. 24 min. which is the fourth ark required.

For the second of the three general wayes, there are two [ II] Cases; For the Tangent that is Copartner with the sine in the Proportion, may be either lesser or greater then of 45 gr. for the lesser, take the Example which before was preferred hither, namely,

1 As the sine of 50 gr. is to the Tangent of 30 gr. So the radius is to the Tangent of what?

First, upon the side EG, I count the sine of 50 gr. and to the line there arising, I transferre Gt the Tangent of 30 gr. by help of the line t s, and to s, I apply the threed, which threed cuts the limb in u, so that G u I find to be the Tangent of 37 gr. and this is the fourth term required in this Propor∣tion; But in the second Example going before, whereof this is also the solution, this 37 gr. is the complement of the fourth ark there required, so that the fourth ark there, should be 53 gr. which because it is greater then 45 gr. is therefore absolved this way, and not the other.

2 As the sine of 50 gr. is to the tangent of 50 gr. So the radius is to the Tangent of what?

Here because the Tangent of 50 (being Co-partner in the Proportion with the sine) is greater then the Tangent of 45 gr. and so cannot be expressed upon the square, therefore the Proportion must be altered by changing the places of the first and third terms, and by taking the complement of the second and fourth, after this manner.

As the radius is to the Tangent of 40 gr. So the sine of 50 gr. to the Tangent of 32 gr. 44 min. the complement whereof 57 gr. 16 min. answers to the question in the former Proportion, and this last Proportion fals under the first general

way where the radius leads, and was resolved before in the first practice upon the Square, As EG, to G a, So E b, to b c, or G e the Tangent of 32 gr. 44 min.

[ III] In the third of the three general wayes, there are two cases, according as the Tangent of the third place, which is Co-partner in the proportion with the sine, is lesser or greater then the Tangent of 45 degrees.

1 As t••e tangent of 40 gr. is to the Radius, so the Tangent of 32 gr. 44 min. to what sine?

Upon the side GD, I count G a, the Tangent of 40 gr. and thereto apply the threed, then upon the same side GD, I reckon also the Tangent of 32 gr. 44 min. from G to e, and follow the line meeting at e, till it cut the threed at c, and the line there crossing also is c b, which counted from e, the Center of the Square, gives the sine of 50 gr. which is the fourth term required.

2 As the Tangent of 60 gr. is to the radius, So the Tan∣gent of 53 gr. to what sine?

Here because the Tangent of 53 gr. being Co-partner in Proportion with the sine, is greater then of 45 gr. therefore the first and third terms must change places, and their com∣plements are also to be taken, thus, As the co-tangent of 53 gr. or Tangent of 37 gr. is to the Radius, So is the co-tan∣gent of 60 gr. or Tangent of 30 gr. to the fourth sine re∣quired.

Upon GD the side of the Square, I count G u the Tangent of 37 gr. thereto applying the threed. Then on the same side of the Square, I also count G t, the Tanegnt of 30 gr. and follow the line at t, till it cut the threed at s, and the line s b, there crossing being counted from e, the Center of the Square, gives me the Sine of 50 gr. the fourth term required.

These Examples are sufficient to give light to the rest, For no Proportion can fall out in these kinds, whereunto these Proportions and their Examples are not suitable.

And so much of Spherical Triangles.