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For, AC + BC: AQ-BQ:: AB: AC-BC

(by 8. 2. and 3. 4.) :: Co-fine of

ACB
2

: Sine of

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As the Tangent of the vertical Angle C of a plane Triangle ABC, is to Radius, fo is half the Bafe AB to a Fourth-proportional; and as half the Bafe is to the Excess of the Perpendicular above the faid Fourthproportional, fo is the Sine of the vertical Angle, to the Co-fine of the Difference of the Angles at the Bafe.

Let ABCD be a Circle defcribed about the Triangle, and from O, the Center thereof, let OB and OC be drawn; moreover, draw CD parallel to B A, meeting the Periphery in D, and EOF, perpendicular

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to AB, meeting DC in F. Then it is evident, that EF will be equal to the perpendicular Height of the Triangle, EOB equal to the vertical Angle ACB, and FOC (= DAC) equal to the Difference of the Angles (ABC and BAC) at the Base.

But (by Theor. 2.) as Tang. EOB (ACB) : Radius:: EB (AB): EO; moreover as EB: OF (EF-EO) :: Sine EOB (ACB): Sine OCF, or Co-fine of FOC. Q. E. D.

PROP. XII.

As the Tangent of the vertical Angle, of a plane Triangle ABC, is to Radius, fo is the Bafe A B to a

I

Fourth

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Fourth-proportional; and, as the faid Fourth-proportional, is to the Sum of the Semi-bafe and the Line CD bifecting the Bafe, fo is the Difference of these two, to the perpendicular Height of the Triangle.

F

D

C

Let a Circle be defcribed about the Triangle,and from O, the Center thereof, let OA, OC and OD be drawn ; alfo let CF, parallel to AB, be drawn, meeting DO, produced (if need be) in F. It is evident that DF will be perpendicular to AB, and equal to the Height of the Triangle. But DC2OC2 + OD2 + 20Dx OF (by 11.2.) = OA2 + OD2+2OD× DF-OD = 0 A2 — OD2 + 2OD × DF = AD2 + 2QD × DF (by 7.2.); whence, by taking away A D2 from the first and laft of thefe equal Quantities, we have DC‘—AD2 = 20D × DF; or DC+AD×DC¬AD =20D x DF (by 6. 2.) and therefore 2OD: DC + AD:: DC-AD: DF; but (by Theor. 2.) as the Tang. AOD (= ACB, by 9. 3.) is to Radius (:: AD: OD) :: AB: 20D. Q, E. D.

PROP. XIII.

As twice the Rectangle under the Bafe and Perpendicular of a plane Triangle ABC, is to the Rectangle under the Sum, and Difference, of the Bafe and Sum of the two Sides; fo is Radius, to the Co-tangent of half the vertical Angle.

Let

Let HG, perpendicul ar to A B, be the Diameter of a Circle defcribed about the Triangle, and let HD and Hd be perpendicular to the two Sides of the Triangle; alfo let CF be parallel to A B, and let HA, HB and HC be drawn.

=

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=

B

Since the Diameter HG is perpendicular to AB, therefore is AE = BE (by 1. 3.) AH BH, and the Angle ACH BCH (by 10. 3.); whence, alfo, CD=Cd, and HD Hd (by 20. 1.) Therefore, the right angled Triangles HAD and HBd, having AH HB and HD Hd, have, likewise, AD = Bd (by 7. 1.) From whence it is manifeft, that CD will be equal to half the Sum, and AD equal to half the Difference, of the two Sides of the Triangle. Moreover, because of the fimilar Triangles AEH and HCD, it will be, AE: CD:: HA* (= HE × HG, by Cor. to 11. 4.): HC2 (HF × HG) ::HE: HF (by 1.4.) Whence, by Divifion, &c. AE2: CD-AE':: HE: EF:: HEX AE: EF x AE; therefore, by Inverfion and Alternation, EF x AE : CD2 — AE' (CD + AE x CD—AE): : HE × AE AE. HE: AE :: Radius: Co-tang. EAH (ACH), by Theor. 2.): Whence the Truth of the Propofition is manifeft.

PROP. XIV.

As twice the Rettangle of the Bafe and Perpendicular of a plane Triangle ABC, is to the Rectangle under the Sum, and Difference, of the Bafe and the Difference of the two Sides; fo is Radius, to the Tangent of half the vertical Angle.

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Let the preceding Figure and Conftruction be retained, and let AG and C G be drawn. The Triangles AHD and GHC (being right-angled at D and C, and having HAD = HGC, by Cor. to 9. 3.) are equiangular; and fo AD: GC:: AH: HG:: AE: AG (by 11.4.); whence, alternately, AD: AE:: GC AG, and AD: A E:: GC (GFx HG): AG' (GEx HG) :: GF: GE; therefore, by Divifion, &c. AE-AD: AE2:: EF: GE :: EF x AE: GE x AE; whence, again, by Alternation, &c. EF x AE: AE2- AD (AE+AD x AE-AD) :: GE x AE: AE::GE: AE:: Radius: Tang. AGE (by Theor. 2.); from which the Truth of the Propofition is manifeft.

PROP. XV.

If the Relation of three Right-lines a, b and x, be fuch that ax-x2 = b2; then it will be, asa: b :: Radius, to the Sine of an Angle; and, as Radius, to the Tangent (or Co-tangent) of half this Angle, fo is

b: x.

A

C

D

B

Make A B equal to a, upon which let a Semicircle ADB be described; alfo let CD, equal to b, be perpendicular to A B, and meet the Periphery in D (for it cannot exceed the Radius of the Circle when the Propofition is poffible): Moreover, let AD, BD, and the Radius OD, be drawn. Because ACXCB=CD2=b2 (by Cor. to 11. 4.) it is plain that AC xa-AC, or BC xa-BC is alfo = b2; and, therefore, xx a-x being be, it is manifest that x may be equal, either, to AC, or to BC. Now

(by

(by Theor. 1.) OD (ža): CD (b) :: Radius: Sine DOC; whofe Half is equal to A, or BDC (by 9. 3.) But, as Radius: Tang. BDC :: DC (b): BC; or, as Radius: Co-tang, BDC (Tang. CDA): : DC (b): AC. 2. E. D.

PROP. XVI.

If the Relation of three Lines a, b and x be fuch that x'ax= b2; then it will be, as a): b:: Radius to the Tangent of an Angle; and as Radius is to the Tangent, or Co-tangent, of half this Angle, (ac cording as the Sign of axis pofitive or negative) :: b: x.

Make AB b, and AC, perpendicular to AB, equal to a; about the latter of which, as a Diameter, let a Circle be defcribed; and, thro' O, the Center thereof, let BD be drawn, meeting the Periphery in E and D; alfo let A, E and C, E bet

D

F

E

A

joined, and draw BF parallel to AC, meeting AE, produced, in F. Then, fince (by 17.3.) BEX BD (= BEX BE+a = BD x BD — a) = A B2 (b2 } and xxx a = b2, by Supposition, it is manifeft, that BE will be = x, when x x x + a = b2; and BDx, when xxx-a b2.

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Furthermore, because the Angle F = OAE (by 8.1.) OEA (by 6.1.) = BEF (by 5. 1.) it is evident that BF BE (by 22. 1.) and that the Angles BAF and C, (being the Complements of the equal Angles F and OAE) are likewise equal.

Now (by Theor. 2.) AO (a): AB (b) :: Radius: Tang. AOB; whofe Half is equal to C, or BAF (by 9.3.) But, as Radius: Tang. BAF :: AB (b)

F

: BF

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