Fig. the Parallels; namely, the Sides themselves, from the Pole,. XLIX. or their Complements and Excesses, from the Diameter.

Note, That of a Triangle, the Conjunct Angle, or the Outer
Angle, is all one.

4. Two Sides of an Oblique-angled Spherical Triangle, where-
of one is BD, with the Angle intercepted, being given : To
find out, at one Work, the Third Side; and the other Angle.
at B D.

5. The Rule. If the Angles D and B, be reckoned in the Upper Semicircle of the Planisphere, at the Pole N, and the Angle D be Obtufe, or B Acute; the Point C shall be taken in the former Semicircle towards: But, if the Angle D be Acute, and B Obtuse, the Point C shall be taken in the latter Semicircle towards w; and contrariwise, for the lower Pole S. -Then, upon the Diameter, count the Angle given, Dor. B,. either by it felf, or by its Conjunct, according as was taught in Set. 3. And from the end of that reckoning, imagine an Arch of a Great Circle aptly traced, until it meet with the Parallel of the Complement of the respective Side given. DCor BC, for that Point of Concurrence shall be the Point C, for the given Side, either DC or BC. To this Point C, thus found, apply the Index noted with the proper Letter of the Angle given, Dor B; and mark the Distance of it from the Centre for this Distance being exactly taken on the other Index, shall in the Planisphere give the true Place of the Point fought: Which being aptly estimated, will, upon the Diameter, shew the Angle fought; either B, or the Conjunct D, from; or else, the Conjunct of B, or the Angle D, from w. And, it will also shew upon the Limb, the Complement of the Side fought.

By what hath been before shewed, (that the Point C, both of the Sides DC and BC, are evermore equally diftant from the Centre,) there is opened a Way, whereby Any Two of Five Circular Parts; together with the Side DB, being given, the other Two may be found at one Operation.

:

In the Upper Semicircle of the Planisphere, if the Angle D be Obtufe: Or if the Side DC be much Leffer than B C, the Point C shall be taken in the former Quadrant towards B: But, if D'be Acute, or DC much Greater than BC; the Point C shall be taken in the latter Quadrant towards vs. And, contrariwife, in the Under Semicircle. -And in both, if the Angles D and Babe Acute, or if the Sides DC and BC, differ but little in length, the true Point fought, shall fall near the Axis, towards w, within a Space, equal to the Index opened at the Wideness of B D.

Set, therefore, the Index thus opened, so that the Two Points C, may fall within it. And among the Great Circles and Parallels, or both (according as Reason shall direct) reckon the... Two Circular Parts given, tracing them proportionally, with a Pin in each Hand, until they shall both exactly meet with their proper Legs of the Index, at an equal Distance from the Centre, and as near to it as poffibly may be: So have you upon the Planisphere, the true Places of both the Points C; and thereby, the Two Circular Parts fought, respectively, to each Point C; both of DC, and B С.

And note farther, That if any Point C, falleth, either among the Parallels, near the Diameter; or among the Great Circles, near the Limb; where they be almost equidistant, there may be some Uncertainty in finding the true Points exactly: The remedying whereof requireth the more Diligence in the Computer; but this may be remedied, by reducing the Oblique Triangle into Two Right-angled.

Other Inconveniencies may arise in the Use of this, as in all other Instrumental Operations; for the remedying whereof,.. Jus optimus Magifter.

Fig. XLXI.

The

Fig. The Solution of Spherical Triangles, by the Orthographi

XLIX.

1.THE

cal Projection, or Analemma.

I. Of Right-angled Spherical Triangles, HE Hypotenuse is, always, represented upon the Index. 2. The Right Angle, at the Section of any Meridian, and the Æquinoctial.

3. One Leg in the Meridian, and the other in the Æquino

Fial.

4. One Angle only entering the Question, is represented at the Centre, between the Æquinodial and the Index, and is numbered in the Limb.

5. But the Two Angles entering the Question, you must turn the Cafe into an Angle, and its opposite Leg. For in the Triangle ABC, whose Right Angle is A; the other Two Angles ABC,

Fig. L. and ACB, both entering the Question, you must lengthen the Hypotenuse to a full Quadrant to Es as also, the Legs adjacent to that Extremity of the Hypotenuse, from which it was lengthned. Thus in the Scheme, the Leg A C, adjacent to C, from which Extremity the Hypotenuse was lengthened: And fo, in the Triangle CDE, Right-angled at E: The Angle DCE is equal the Angle BCA; and the Leg DE, equal to he Complement of the other Angle CBA..

II. Of Oblique-angled Spherical Triangles.

1. Three Sides being given; To find an Angle opposite to any of them.

Reckon the Greatest Leg from the Pole upon the Limb, and where it endeth, apply the Index; upon the Index reckon, from the Limb, the Base; (that is, the Side opposite to the Angle fought:) And to the Point of the Index, where this Base endeth; apply (or bring to) the Curfor: Then look where the Curfor cutteth the Parallel of the Leffer Leg, to be reckoned from the Pole, and obferve what Meridian passeth by this Section of the Curfor and Parallel; for, where this Meridian cutteth the Equinotial, there you have the Measure of the Angle fought; to be accounted from thence to the Limb.

2. Three Angles given, to find a Side. Turn the Angles into Sides, and deal with them, as with the

Sides.

3. The

3. The Parts Given and Sought, being altogether Opposite.

Reckon the Greater of the Two first Terms upon the Index, and where this Number endeth upon the Index; apply that Point to the leffer of the faid Two first Terms; which Parallel is to be reckoned from the Equinoctial; then order the Terms, reckoning the First and Third both upon the Index; or both upon the Parallel; and fo likewife do with the Second and Fourth Terms.

:

4. Two Sides, with an Angle between them Given: To find the Third Side.

Reckon the Greater Side given, from the Pole upon the Limb; and to the end of it, apply the Index. Reckon the other Side Given upon the Parallels, from the Pole: And the Angle given, upon the Equinoctial, from the Limb; and the Meridian it comes to, pursue till it comes to the Parallel of the Leffer Leg: And to this Settion of the faid Meridian and Prallel, apply the Curfor, so it may justly lye on the Index; and mark what Point of the Index it pointeth out: For, this Point of the Index, counted from the Limb; is the Third Side required. And after the Third Side is found, you may find either of the Two unknown Angles, by the Rute for Oppofite Parts.

5. Two Sides, and an Angle opposite to the Lesser of them, Given: To find the Third Side.

Reckon the Difference of the given Sides, and the Sum of them (one and the same Way) from the Pole, and mark the Points where both of them do end. Count also, the Angle given, upon the Æquinoctial from the Limb, and mark what Meridian it cometh to. Then extending a strait Line (or applying a strait Ruler) between the Two Points marked in the Limb; it will cut the faid Meridian in Two Places': So you are to observe the Parallels, where the strait Line (or Ruler) cutteth the faid Meridian; for, these being reckoned from the Pole, will give you the Quantity of the Third Side required: For, this same Third Side, may be of a Twofold Quantity: The Leffer it is, when the Angle oppofite to the Greater given Side is Obtufe, and the Greater it is, when the faid Angle (opposite to the Greater given Side) is Acute

1

Fig. L. Fig. L.

6. Two Sides, and an Angle opposite to the Greater of them Given; To find the third Side.

Reckon the Difference of the given Sides from the Pole, one way; and the Sum of the same, the other way. Count also, the given Angle upon the Æquinodial: And extend the strait Line (or lay a Ruler) cutting the Meridian, as in the last; for now it will cut but once, and so the Third Side will admit only of one fingle Answer.

The End of the Second Part.

POSTSCRIPT.

MAny are the ways by

which Plain and Spherical Triangles may be both Geometrically and Instrumentally performed: As by Scales of Natural Sines, Tangents, Secants, and Equal Parts, by Protraction: Also, by Scales of Artificial Numbers, Sines, Tangents, and Verjed Sines, as Mr. Gunter long time fince contrived them, to be used with Compasses: And I have now lately contrived an Instrument, which I call TRISSOTETRAS, which printed on a large Sheet of Paper, and pafted upon a Board, all Triangles, both Plain and Spherical, may be. Resolved by Inspection, without Pen, Compasses, opening Joints, or other Moveable; fave only the Extension of a Thread, or thin Streight Ruler, upon the Instrument; the Description and Use whereof I may hereafter publish by it self. But, the best and most absolute Way of Resolving Triangles; and to what Ufes foever they be applied, is by the Canons of Artificial Sines, Tangents and Logarithms; both Decimal and Sexaginary: And such a Canon was intended to be joined to this Book, at this time; but must be referred till farther Opportunity, which may be shortly.

ANCILLA