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of the Meridian of the Place, BDF an Arch of the Horizon, FEC an Arch of the Circle described about the Sun, CAH an Arch of the Meridian of the Sun, HLI an Arch of the Equinoctial. Then do thefe five Arches retain the Conditions required.

The first interfecting the fecond in B; the fecond the third in F; the third the fourth in C; the fourth the fifth in H; and the fifth the first in I; and these Interfections at B, F, C, H, I, are at right Angles : there I fay the right angled Triangles made by the Interfections of the fe Circles, namely ABD, DHL, LFE, EIG, and GCA, do confift of the fame circular Parts; as here appeareth.

Triangle

The five circular Parts in the

ABD, are AB,BD,com.BDA, com. AD,com.DAB. DHL, are com.HLD,com.LD,com. LDH,DH,HL. LFE, are com. ELF, LF, FE, com. FEL, com. EL. EIG, are IG, com. IGE, com. GE, com. GEI, EI. (GCA, are com. GA,com. AGC, GC,CA,com.CAG.

Where you may obferve, that to the Side AB in the first Triangle, is equal the Complement HID in the fecond, or the Complement of ELF in the third, or IG in the Fourth, or the Complement of GA in the fifth, &c. The fame uniformity of the circular Parts is alfo apparent in quadrantal Triangles.

As in the fame Scheme G from D, D from E, E from A, A from L, and L from G, are diftant by Arches each equal to a Quadrant. Here are therefore five quadrantal Triangles, as appears in the Scheme.

of

Of the five circular Parts in a spherical Triangle, right angled, or quadrantal.

HE Sine of the middle Part with Radius, is

Tequal to the Tangents of the Extremes adjacent,

or to the Co-fines of the Extremes disjunct.

PARTI. Touching the first part of this Axiom, in right angled Triangles; the middle Part is either one of the Sides, or one of the oblique Angles, or the Hypothenufe.

CASE 1. Let the middle Part be a Side; as in the Triangle ABD, let AB be the middle Part, and BD, and the Complement of A, the Extremes adjacent: Then I fay, that the Sine of AB the middle Part, with Radius, is equal to the Tangent of DB, with the Tangent of the Complement of DAB.

For (by Axiom 2. Page 70, 71.) as the Sine of AB is to Radius, fo is the Tangent of DB to the Tangent of the Angle at A: Therefore alfo alternately, As the Sine of AB, to the Tangent of DB; fo is Radius to the Tangent of A.

But Radius is a mean proportional between the Tangent of an Arch and the Tangent Complement of the Arch. So then it is, As Radius is to the Tangent of A, fo is the Tangent Complement of A to Radius. Then it will be, As the Sine of AB to the Tangent of DB, fo is the Tangent Complement of A to Radius.

CASE 2. Let the middle Part be an Angle, as in the Triangle DHL, let the Complement of HLD be the middle Part, and HL and Complement of LD the Extremes adjacent: Then I fay, that the Sine Complement of HLD with Radius, is equal to the Tan

G 4

gent

Part I. gent of HL, with the Tangent of the Complement

of LD.

For the Complement of HLD is equal to AB, and the Complement of LD is DB; and HL is the Complement of DAB; and we proved before, that the Sine of AB with Radius, is equal to the Tangent of DB with the Co-tangent A: Therefore alfo the Co-fine of HLD with Radius, is equal to the Co-tangent of LD with the Tangent of HL.

CASE 3. Let the middle Part be the Hypothenufe, as in the Triangle GCA, let the Complement of AG be the middle Part, and the Complement of AGC, and Complement of CAG, the Extremes adjacent.

Then alfo I fay, that the Co-fine of AG with Radius is equal to the Co-tangent of AGC, with the Cotangent of CAG.

For we have before proved, that the Sine of AB with Radius is equal to the Tangent of DB with the Co-tangent of A; but the Complement of AG is equal to AB, and the Complement of AGC is equal to DB, and the Complement of CAG equal to the Complement of DAB. Therefore alfo the Co-fine of AG with Radius, is equal to the Co-tangent of AGC, with the Co-tangent of CAG.

Therefore in a right angled Triangle, the Sine of the middle Part with Radius, is equal to the Tangents of the Extremes adjacent.

I fay further, that,

PART II. The Sine of the middle part with Radius is equal to the Sines Complement of the oppofite Extremes.

For here alfo the middle Part is either one of the Sides, or the Hypothenufe, or one of the oblique Angles.

CASE

CASE 1. Let the middle part be a Side, as in the Triangle ABD, let DB be the middle part, and the Complement of AD, and Complement of A, the Extremes Disjunct. Then I fay, that the Sine of BD with Radius, is equal to the Sine of AD with the Sine A.

For (by Axiom 1. Page 68.) As the Sine of AD to Radius, fo is the Sine of DB to the Sine of A; therefore, the Sine of DB with Radius, is equal to the Sine of AD with the Sine of A.

CASE 2. Let the Hypothenufe be the middle Part, as in the Triangle DHL; let the Complement LD be the middle Part, and DH and HL, the Extremes Disjunct: Then I fay, that the Co-fine of LD with Radius, is equal to the Co-fine of DH with the Co-fine of HL.

For the Complement of LD is equal to DB, and DH equal to the Complement of AD, and HL equal to the Complement of DAB: And it was proved before, that the Sine of DB with Radius, is equal to the Sine of AD with the Sine of DAB.

CASE 3. Let one of the oblique Angles be the middle Part, as in the Triangle EIG; let the Complement of IGE be the middle Part: then I say, that the Co-fine of IGE with Radius, is equal to the Sine of GEI with the Co-fine of EI. For the Complement of IGE is equal to DB; and GEI is equal to AD; and El is equal to the Complement of DAB.

Therefore in a right angled Triangle, the Sines of the middle Part with Radius, are equal to the Sines Complement of the Extremes disjunct. Which was to be demonstrated.

But here Note, That when a Complement in any Proportion does chance to concur with a Complement

in

in the circular Parts, you must always take the Sine it felf, or the Tangent it felf, inftead of the Co-fine or Co-tangent in the circular Parts; because the Cofine of the Co-fine, is the Sine; or the Co-tangent of the Co-tangent is the Tangent it felf. As in Cafe the first, where DB is the middle Part, and the Complement of AD, and Complement of A, the Extremes Disjunct; here because the two Extremes fall upon Complements in the circular Parts, therefore it must be the Sine of AD, and the Sine of A, because the Co-fine of the Complement is the Sine.

Having explained and demonftrated the feveral Axioms, I fhall next proceed to the Solution of the Sixteen Cafes of right angled Spherical Triangles, both by Lord Neper's Univerfal Propofition, and alfo by the three first Axioms, as followetb.

CHAP. VIII.

The Solution of the Sixteen Cafes of right angled Spherical Triangles.

CASE I.

The Hypothenuse AC, and the Angle A given, to find the oppofite fide BC, the middle Part.

יע

D

B

A

Ss. AG: s. AC :: s. DG: s. BC. (by Ax. 1.) That is, As Radius: s. AC: s. A: s. BC.

Which is agreeable to the univerfal Propofition; for BC being the middle Part, the Product of the Extremes Disjunct,

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