Thofe Things which agree to a Third, agree to each other.

:

Whence by Compofition{FG: G :: K+M; M, or

That is in Words at length,

As two Right Angles are to G,

F:: K+M : K.

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.

In the Scheme adjoined, the Triangle G is equal

to the Triangle H.

For the Angle A =D, and the Angle E, and the An

B

The Excess of the three Angles of a Spherical Triangle over and above two Right Angles, divided by 720, fhews what Part of the whole Superficies of the Sphere or Globe, the Area of the faid Triangle is. VOL. II. Demon

E e

Therefore, As 180: A+B+C:: Spheric : 3G+ R+S+T, (by Eucl. Lib. 5, Prop. 24.) As 180: A+B+C 180 Spheric : 3G+R+S+T— Spheric; but G+R+S+T=Spheric; therefore 3G + R + S + T― Spheric =2G.

:

В

Hence 180 A+B+C-180:: Spheric : 2G. Quadruple the Antecedent Terms, and it will be, As 720: A+B+C—180 :: 2 Spherics: 2G A+B+C-180 1 Spheric G. Therefore

720

: :

fhews what

Part of the whole Superficies of the Sphere, the Area

of the given Triangle is.

Wherefore the Area of fuch a Triangle as was used in the last Chapter is 0,00893 of the whole Superficies of the Sphere or Globe.

Now suppose the Diameter of any Sphere or Globe be 16 Inches, then the Superficies of fuch a Sphere will contain 804 Square Inches nearly. Then 8044 X 0,00893 7.1819 Inches, the Area of the Triangle on that Sphere, whofe Angles are fuch as are above fpecified.

Because (by Theorem 16,) the Sum of all the An-gles of any Spherical Triangle is greater than two Right Angles, and lefs than fix; and all the internal and external Angles of a Triangle are together but fix Right Angles; therefore, if to the Sum of the Angles of any Triangle you add four Right Angles, and from that Sum deduft fix Right Angles, the Remainder divided by 720 will fhew what Part of the Sphere's Surface the Area of the Triangle is.

Hence because all Spherical Polygons, or Multangular Figures of every kind, may be reduced to Spherical Triangles, therefore we have this general Rule for finding the Area of any fuch Figure alfo; viz.

The Rule.

Multiply 180 by the Number of the Angles; deduct this Product from the Aggregate of the Angles increased by 360°; the Remainder divided by 720, gives the Area of the Polygon.

The Completion of a Solid Body.

From the foregoing Menfuration of the Area of a Spherical Triangle is derived the following Invention.

If the Radius of the Sphere be 100000, the Side of an Infcribed Icofihedron fhall be 105146.22, equal to the Subtenfe of 63° 26' 10". Therefore the Plane equilateral Triangle FEO (in the Icofihedron) anVOL. II. Ec 2

fwers

E

fwers to the Equilateral Spherical Triangle in the Sphere; whofe three Spherical Angles, are connected in the Plain Angles, in the fame Points F, E, O. And the Sides of this Spherical Triangle are each equal to 63° 26' 10", because their Subtenfes FE, EO, OF, are equal to one another.

Let fall the Perpendicular F E P, then will the Spherical Triangle EPO be right an

=

=

'26′

10

P 31°43'5

gled at P, where there is given (befides the Right Angle) EO 63° 26' 10", and PO 31° 43' 5"; whence the Vertical Angle PEO will be found precisely 36°, and fo the whole Angle E=72°; hence the Sum of the three equal Angles, E+F+0=216°; from which deduct two Right Angles, or 180, and there will remain 36°; therefore the Triangle E FO is of the whole Spherical Superficies, that is Part; and this moft truly; for 20 Pyramids FEOC fill the folid Place of the Icofiedron; and fo 20 Spherical Bafes (covered over with 20 Triangular Plain Bases) compleat the whole Sphere or Globe.

C

20

36

720