(to wit, F and G) shall be as the Times. The Superficies also, Fig, XLI. K and M, thall be as the Times; therefore F shall be to G, as the Superficies K, to the Superficies M. And Gunto M, as the Superficies M, to the Superficies K, &c. howsoever they can be taken. For,

Lemma II. The Triangle G, is equal to the Triangle H, because Fig.XLII.. the Angles and Sides of one, are equal to the Angles and Sides of the other: To wit, A = D, B = E, C= F. Also, L = 0, M = P, N= Q. Therefore they are Congruous and Equal.

THEOREM.

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

For by Lemma I.

{

180: A:: Sph.:G+R

2

180 : B::Sph.: G+S

180:C::Sph.: G+T= H+ T.

(by the Second Lemma.)

Therefore, As 180:A+B+C:: half the Spherick : to

3G+R+S+T, (by 24 El. 5 Ευ.)

As 180: A+B+C-180::

:: So half the Spherick: 3G+R+S+T--half the Sph... But G+R+S+T is equal to half the Spherick;

Therefore,

Fig. Therefore, 3G+R+S+T- half the Sph. = 2 G.
XLII. So that, As 180: A+B+C-180::

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

And the Antecedent Terms being Quadropled, it shall be,

As 720.: A+B+C-180:: 2 Sphericks: 2G.

And fo the Spherick to G. Therefore,

A+B+C-180

These Things are likewise true in all Spherical Polygons, of what Ordinate Figure soever they be; or In-ordinate, so all the Angles be given. And the Reason is, because all Polygons may be resolved into Triangles. Therefore, this Rule shall hold in these Multangles also.

RULE.

Multiply 180 d. by the Number of the Angles; fubduct 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.

F

Of the Completion of a Solid Body.

Rom the foregoing Mensuration of the Area of a Spherical Tri. angle, this Fruit arifeth.

If the Radius of the Sphere be 100000.00, the Side of an inscribed Icosaedrum shall be 105146.22, equal to the Subtense of 63 deg. 26 min. 10 sec. Therefore the Plain Equilateral Triangle FEO (in the Icosaedrum) answers to the Equilateral Spherical Triangle in the Sphere; whose Three Spherical Triangles are connected in the Plain Angles, in the same Points, F, E, O.

And the Sides of this Spherical Triangle are separately taken 63 deg. 26 min. 10 sec. to wit, because their Subtenses FE, EO, OF, in the Plain Triangle, are equal to one another.

Fig. Let fall now the Perpendicular E P, the Spherical Triangle XLIII. EPO shall 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 shall be 36 deg. just, and the whole Angle at. E, 72 deg. And the Sum of the Three equal Angles, E, F, O, shall be 216 deg. from whence taking Two Right Angles,

Fig.

Angles, equal to 180 Deg. there remains 36 Deg. Therefore, the Triangle EFO is, of the whole Spherick, that is Part: And this most truly, for 20 Pylamides FEOC, fill the XLIII. Solid Place of the Icosaedrum. And so 20 Spherical Bafes, (covered over with 20 Triangular Plain Bafes,) compleat the whole Spherick.

FEOC is One of the 20 Pyramids in the Icosaedrum : The Plain Triangle B, is One of the Hedre or Bases: C is the Center of the Body, or Sphere, that circumscribes it.

4.77 Icosaedres

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 say, 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.

Fig.

XLIV.

JANUA MATHEMATICA.

SECTION II.

Spherical Trigonometry, Geometrically
Performed.

S the Sides of Plain Triangles are Three Right Lines, ina Plain: And a Right Line is, The Shortest Extension between any Two Points upon a Plain Superficies: So the Sides of a Spherical Triangle are Three Arches, of Three Great Circles of the Sphere, intersetting 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 those Two Points..

A terfecting each other upon

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 Nomes and Situations upon the Globe, to what Use each of them serveth; and also, how to project any of them upon a Plain, anfwerable to any Position of the Globe: For to fuch as do not, this Se&ion will be but of little Ufe; and therefore I would advise my Reader, before he enter upon this Geometrical Way of refolving Spherical Triangles, to peruse the Beginnings of the Second and Third Sedions of the Third Part hereof; which treat of the Circles of the Sphere, and their several Pofitions and Affections; in which he may receive very much Satisfa&ion concerning those Particulars. So that it shall suffice, in this Place, that I declare,

1. What a Great Circle is.

2. How to Project such 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 discover the Triangle, which is made by the Interfection. of those Great Circles so projected.

4. To find the Poles of those Great Circles. And,

5. To Measure the Sides and Angles of the Triangle so laid down: Which to do, is usually called, The Doctrine of the

Dimension of Triangles.

I... What

T

Fig. XLIV.