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: BF (BE) the Value of x in the first Cafe, where x2 + ax = b2. Again, Radius: Co-tang. BAF (Tang. F): BF (BE): AB (by Theor. 2.); and BE :AB::AB: BD (by 17. 3. and 3. 4.); whence, by Equality, Radius: Co-tang. BAF:: AB(b): BD; which is the Value of x in the fecond Cafe; where x2- ax = b2. Q. E.D.

PROP. XVII.

In any plane Triangle ABC, it will be, as the Line CE bifecting the vertical Angle, is to the Bafe AB, fo is the Secant of half the vertical Angle ACB, to the Tangent of an Angle; and, as the Tangent of half this Angle is to Radius, fo is the Sine of half the vertical Angle, to the Sine of either Angle, which the bifecting Line makes with the Bafe.

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Let ACBD be a Circle defcribed about the Triangle, and let CE be produced to meet the Periphery thereof in D; moreover let AD B and BD be drawn, and likewife DF, perpendicular to the Base AB; which will, alfo, bifect it, because (BCD being=ACD) the Subtenfes BD and AD are equal (by 10. 3.) Moreover, fince the Angle DBE ACD (by Cor. to 9. 3.) = DCB, the Triangles DEB and DBC (having D common) are equiangular, and therefore DE x DC = DB2 (by 18. 3.) or, which is the fame, DE + DEX CE = DB2. Therefore (by Prop. 16.) CE: DB:: Radius: Tangent of an Angle (which we will call Q); and, as Radius : Tang Q: DB : DE.

But DB: BF (AB) :: Secant FBD (BCE): Radius; therefore, by compounding this, with the

firft

firft Proportion, we have, CE x DB: AB x DB 1444 :: Radius x Secant BCE: Radius x Tang. Q(by 9. 4.) and confequently CE: AB:: Secant BCE: Tang. Q(by 1. 4.) Again, DE: DB:: Sine DBE (BCE): Sine of DEB (or of CEB); whence, by Equality, Tang. Q: Radius :: Sine BCE: Sine CEB. 2. E. D.

PROP. XVIII.

In any plane Triangle ABC, it will be, as the Perpendicular is to the Sum of the two Sides, fo is the Tangent of half the Angle at the Vertex, to the Tangent of an Angle; and, as Radius is to the Tangent of half this Angle, fo is the Sum of the two Sides, to the Bafe of the Triangle.

Let DP, perpendicular to A B, be the Diameter of a Circle defcribed about the Triangle; let CF be perpendicular to DP, and DG to AC, and let DA, DC and DB be drawn, and IA alfo FI, parallel to BD,

meeting B A, produced,

in I.

P

F

E

B

It is manifeft, from Prop. 13. that CG is equal to Half the Sum of the Sides AC and BC: It alfo appears, from 8. 1. and Cor. to 9. 3. that the Triangles DCG, ADE, BDE and IFE are all equiangular. Therefore, it will be, CG2: BE:: DC. (DF × DP) : BD3 (DE × DP) : : DF (DE + EF): DE :: BE + EI: EB :: BE+EI × BE: EB2; and, confequently, CG2 (= BE+EI × BE) = BE2+ EIX BE. But, by Prop. 16. it will be, EI : CG :: Radius: Tangent of an Angle (which we will call Q); and as Radius: Tang. Q:: CG:BE 44

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(:: 2CG, or AC + BC, to AB). But (by Theor. 2.) EF: EI:: Tang. I (ACD): Radius; therefore, by compounding this Proportion with the laft but one, we fhall have, EI × EF: EI × CG :: Tang. ACD x Radius Tang. Qx Radius (by 9. 4.) and confequently EF: 2 CG (AC + BC):: Tang. ACD: Tang. Q. Whence the Truth of the Propofition is manifeft.

PROP. XIX.

In any plane Triangle ABC, it will be, as the Perpendicular is to the Difference of the two Sides, fo is the Co-tangent of half the vertical Angle, to the Tangent of an Angle; and, as Radius is to the Co-tangent of half this Angle, fo is the Difference of the Sides to the Bafe of the Triangle.

P

E

F

Let DP, DG, CF &c. be as in the preceding Propofition; alfo let PA and PC be drawn, and FI, parallel to PA, meeting AB in I.

The right-angled Triangles ADG and DPC, having DAG = DPC (by Cor. to 9. 3.) are fimilar; and therefore, AG: PC2 :: AD: DP (by 5. and 9. of 4.) : : AE2 : AP2 (by 11. 4.); whence, alternately, AG2: AE1 :: PC2 (PF × PD) : AF2 (PE × PD) :: PF: PE :: AE : AI (by 2. 4.) :: AE2 : AI x AE; and, confequently, AG AI x AE= AE-EIXAE. Therefore, by Prop. 16. EI: AG :: Radius: Tangent of an Angle (Q); and as Radius Co-tang. Q:: AG: AE. But (by Theor. 2.) EF: EI:: Co-tang. EFI (ACD): Radius; which Proportion being compounded with the laft

3

but

but one, &c. we fhall have, EF: 2AG (AC-BC, fee Prop. 13.):: Co-tang. EFI: Tang. Q; and as Radius: Co-tang. Q:: AG: AE:: 2AG (ACBC): 2AE (AB). 2. E. D.

PROP. XX.

The Hypothenufe AC, and the Sum, or Difference, of the Legs AB, BC, of a right-angled Spherical Triangle ABC, being given, to determine the Triangle.

Let A E be the Sum, and AF the Difference of the two Legs. Becaufe, Radius : Co-f. AB :: Co-f.

BC: Co-f. AC (by

B

E

Theor. 2.) therefore Co-f. AB x Co-f. BC = Rad. x Co-f. AC; but the former of these is =

Rad. x Co-f. AE+Co-f. AF (by Corol. 3. to Prop. 2.); therefore 2 x Co-f. AC = Cof. AE + Co-f. AF. Whence it appears, that, if from twice the Co-fine of the Hypothenufe, the Co-fine of the given Sum, or Difference, of the Legs, be fubtracted, the Remainder will be the Co fine of an Arch, which added to the faid Sum, or Difference, gives the Double of the greater Leg required.

COROLLAR Y.

Hence, if the two Legs be fuppofed equal to each other (or the given Difference = 0), then will the Co-fine of the Double of each, be equal to twice the Co-fine of the Hypothenufe minus the Radius.

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PROP. XXI.

One Leg BC and the Sum, or Difference, of the Hypothenufe and the other Leg A B being given, to determine the Hypothenuse ( fee the laft Fig.)

Since Rad. Co-f. BC:: Co-f. A B: Co-f. A C (by Theor. 2.), it will be (by Comp. and Div.) Radius + Co-f. BC: Rad. Co-f. BC :: Co-f. AB+ Co-f. AC: Co-f. AB-Co-f. AC. But the Radius may be confidered as the Sine of an Arch of 90°, or the Co-fine of o: And, therefore, fince (by the Lemma in p. 28.) Co-fine o + Co-fine BC: Co-f.o-Co-f.

BC :: Co-tang.

BC+o.

2

Tang.

BC-o

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; and, Co-f.

2

Co-f. AC:: Co-
AC-AB, it follows, by

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2

BC

: Tang.

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AC-AB

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BC

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; that is, As the Co

tang. of half the given Leg, is to its Tangent; fo is the Co-tang. of half the Sum of the Hypothenufe and the other Leg, to the Tangent of half their Difference,

PROP. XXII.

The Angle at the Bafe and the Sum, or Difference, of the Hypothenufe and Bafe, of a right-angled fpherical Triangle being given, to determine the Triangle,

First,

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