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The Base B A, is 47.8
The half Bafe

Fig. XL.




The Perpendicular DC 17.33-
The Logarithm of 414.2

Which is the Area of the Triangle.

After the fame manner, if the Base of a Triangle were 42, and the Perpendicular 16, the Area will be found to be 336.


There is an Oblique-angled Plain Triangle A B C, one of whofe Angles at the Bafe BC is 53.48 d. And the Sum of the Two Sides oppofite to the Baje B C, viz. A B and A C, is 129, and the Bafe CB, is longer than the longer fide A C, by 16.8. The reSpective Sides and Angles of the Triangle are required.


Raw a Right Line BC, at Liberty, for the Bafe; and upon one end thereof, at B, make an Angle ABC, to contain 53.84 d. Then the Sum of the Two Sides A B and A C being together 120, break it into any Two equal or unequal Parts, as into 43.2 and 76.8, then (by the Propofition) muft the Bafe B €, be 93.6, which is greater than A C by 16.8.-So in the Obliqueangled Triangle ABC, you have given, (1.) The Side AB 43.2. (2.) The Side A C 76.8. (3) The Angle oppofite thereto, ABC 53 d. 50 m. by which you may find the Angle A C B. For,

As AC 76.8 Is.tos ABC 53.84 d.
So is AB 43.2: To s A CB 27.02 d.


Of the Menfuration of the Area of a Spherical Triangle.

Lemma I. The Lunary Superficies of the Hemifphericks, are as the
Angles of the fame Superficies.

HE Proof of this Lemma (amongst many other Ways) may

Fig. XLI. T be this:

Let the Meridian Semicircle A ED be imagined to be equally moved over the Longitude of the Equator BE C, upon the Poles A and D. Therefore the Angles on the other fide of A and D


(to wit, F and G) fhall be as the Times. The Superficies alfo, Fig. XLI, K and M, fhall be as the Times; therefore F fhall be to G, as the Superficies K, to the Superficies M. And G unto M, as the Superficies M, to the Superficies K, &c. howfoever they can be taken. For,

Thofe Things that agree to a Third,
Agree among themselves.


Therefore, by Compofition,

As FGG:: KM: M,
Or, As FG: FRM: K.
That is,

As the Two Right Angles are unto G,
So is half the Spherical Superficies, to M..
And, As Two Right Angles, are to F;
So is half the Spherical Superficies, to K.
Lemma II. The Triangle G, is
the Angles and Sides of one,
of the other: To wit, A
L= 0, MP, N=Q.
and Equal.

equal to the Triangle H, because Fig.XLII.
are equal to the Angles and Sides
D, BE, C= F. Also,
Therefore they are Congruous


The Excess of the Three Angles, over and above Two Right An gles, divided by 720, fhews what the Area of the Triangle, is, in refpect of the whole Spherick.

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(by the Second Lemma.)



Therefore, As 180 ABC: half the Spherick to

3 GRIST, (by 24 El. 5 Ev.)

As 180: A+B+C 180::.

:: So half the Spherick: 3 GRST-half the Sph.. But GRIST is equal to half the Spherick;


Fig. Therefore, 3 GRIST - half the Sph.
XLII. So that, As 180 ABC-180::

:: So is half the Spherick: to 2 G.

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And the Antecedent Terms being Quadropled, it fhall be,
As 720.: A+B+C-180:: 2 Sphericks: 2 G.
And fo the Spherick to G. Therefore,

A B C 180

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720 whole Spherick.

180 }

fhews what part the Triangle is of the

Thefe Things are likewife true in all Spherical Polygons, of what Ordinate Figure foever they be; or in-ordinate, fo all the Angles be given. And the Reafon is, becaufe all Polygons may be refolved into Triangles. Therefore, this Rule fhall hold in thefe Multangles alfo.


Multiply 180 d. by the Number of the Angles; subdu&t the Product out of the Aggregate of all the Angles increased by 360d. The Refidue divided by 720 d. gives the Area of the Polygon.

Of the Completion of a Solid Body.

Rom the foregoing Menfuration of the Area of a Spherical Triangle, this Fruit arifeth.

If the Radius of the Sphere be 100000.00, the Side of an infcribed Icofaedrum fhall be 105146.22, equal to the Subtense of 63 deg. 26 min. 10 fec. Therefore the Plain Equilateral Triangle FEO (in the Icofaedrum) answers to the Equilateral Spherical Triangle in the Sphere; whofe Three Spherical Triangles are con nected in the Plain Angles, in the fame Points, F, E, O.

And the Sides of this Spherical Triangle are feparately_taken 63 deg. 26 min. 10 fec. to wit, because their Subtenfes FE, EO, OF, in the Plain Triangle, are equal to one another.

Fig. Let fall now the Perpendicular E P, the Spherical Triangle XLIII. E PO fhall be Rectangled; where, over and above the Right Angle at P, are given, E O and PO, equal to half EO; wherefore the Vertical Angle PEO fhall be 36 deg. juft, and the whole Angle at.E, 72 deg. And the Sum of the Three equal Angles, E, F, O, fhall be 216 deg. from whence taking Two Right



Angles, equal to 180 Deg. there remains 36 Deg. Therefore, the Triangle E FO is, of the whole Spherick, that is Part: And this moft truly, for 20 Pylamides FE OC, fill the XLIII. Solid Place of the Icofaedrum. And fo 20 Spherical Bafes, (covered over with 20 Triangular Plain Bafes,) compleat the whole Spherick.

FEOC is One of the 20 Pyramids in the Icofaedrum: The Plain Triangle B, is One of the Hedra or Bafes C is the Center of the Body, or Sphere, that circumfcribes it.

4.77 Icofaedres 4.24 Dodecaedres 9.244 Octaedres 8.000 Cubes

Fill a Solid Place, as will appear out of Precedent Practice by Triangles, and Spherical Polygons.

That is to fay, None of the Five Regular Bodies fill a Solid
Place, the Cube only excepted.

Contrary to what Potamon, and from him Ramus, and all
that have followed Ramus; to wit, Snellius, and others,
have delivered.

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Fig. XLIV.



Spherical Trigonometry, Geometrically


S the Sides of Plain Triangles are Three Right Lines, interfecting each other upon a Plain: And a Right Line is, The Shortest Extenfion between any Two Points upon a Plain Su perficies: So the Sides of a Spherical Triangle are Three Arches, of Three Great Circles of the Sphere, interfetting each other upon the Globe. And an Arch of a Great Circle, paffing through any Two Points upon a Spherical Superficies, is the neareft Distance between thofe Two Points..

In Pursuance of the Work in this Chapter intended, I fuppose the Reader to be acquainted with the Circles of the Sphere; that is, to know their Names and Situations upon the Globe, to what Use each of them ferveth; and alfo, how to project any of them upon a Plain, anfwerable to any Pofition of the Globe: For to fuch as do not, this Section will be but of little Ufe; and therefore I would advife my Reader, before he enter upon this Geometrical Way of refolving Spherical Triangles, to perufe the Beginnings of the Second and Third Sedions of the Third Part hereof; which treat of the Circles of the Sphere, and their feveral Pofitions and Affections; in which he may receive very much Satisfa&tion concerning thofe Particulars. So that it fhall fuffice, in this Place, that I declare,

1. What a Great Circle is.

2. How to Project fuch a Circle of the Sphere upon a 'Plain, fuitable to any Day and Time of the Day, at any time of the. Year, and in any Latitude.

3. To difcover the Triangle, which is made by the Interfection. of thofe Great Circles fo projected.


To find the Poles of thofe Great Circles. And,

To Meafure the Sides and Angles of the Triangle fo laid down: Which to do, is ufually called, The Dourine of the Dimenfion of Triangles.

I.. What

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