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1D In AC produced, take

CD = BC, and let BD be
drawn : then (by Tbeor. 5.
p. 8.) it will be, AD +
AB : AD - AB::tang.
ABD + D 180° - A

2
= 90° LA) : tang.
B

ABD-D / ABD-CBD

2

A

2

2

ABC

that is, AC + BC + AB : AC + BC – AB:: CO-tang. A : tang ABC. 2. E. D.

ABC)

PROP. VI.

In any plane triangle ABC, it will be, as the base plus the difference of the two sides, is to the base minus the same difference, so is the tangent of balf the greater angle at the base, to the tangent of balf

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In the lefser side CA, produced, take CD = CB, so that AD may be

the difference of the two A

fides; and let BD be B drawn : then it is mani

fest that the angle CBD

will be equal to D: but (by Theor. 5. p. 8.) AB + AD : AB - AD ::

D + DBA CAB tangent

(by 9. 1.) : tangent D - DBA CBD - DBA CBA

2. E. D. 2

2

2

2

2

PROP

PROP. VII.

As the base of any plane triangle ABC, is to the fum of the two fides, so is the fine of half the vertical angle, to the co-fine of balf the difference of the angles at the base.

In AC, produced, take CD = CB; join B, D, and draw CE

E parallel to AB, and CF perpendicular to BD.

Since CD = CB, therefore is the angle A

B D = DBC (by 12. 1.) and, consequently, half the vertical angle ACB =

D + CBD

(by 9. 1.) = D.

Moreover, seeing DCB is = the sum of the angles A and CBA, at the base (by 9. 1.) it is evident that BCF (or DCF) is equal to half that sum ; and, therefore, as ECF is the excess of the greater ABC = BCE (by 7. 1.) above the half sum (BCF), it must, manifestly, be equal to half the difference of the same angles A and СВА: :

But (by Theor. 3.) AB : AD (AC + BC):: sine D(ACB): fine ABD = fine CED (by Cor. 1. to 7. 1.) = sine FEC = co-fine ECF. 2. E. D.

2

PROP. VIII.

As the base of any plane triangle ABC, is to the difference of the two sides, so is the co-fine of half the vertical angle, to the fine of half the difference of the angles at the base.

с

E

In the greater fide CA
let there be taken CD

CB, and let BD bedrawn,
D

and likewise CE, per

pendicular to BD. It is B manifest, because CD=

CB, that CDB and CBD are equal to one another, and that each of them is also equal to half the sum of the angles CBA and A at the base (by Cor. 2. to 10. 1.); therefore ABD, being the excess of the greater CBA above the half fum, must consequently be equal to half the difference of the fame angles.

But (by Theor. 3.) AB : AD (AC -- BC):: sine D (co-fine DCE, or C): sine ABD. 2. E. D.

PROP. IX. As the difference of the two sides AC, BC, of a plane triangle, is to the difference of the segments of the base AQ, BQ (made by letting fall a perpendicular from the vertex), so is the fine of half the vertical angle, to the co-fine of half the difference of the angles at the base.

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PROP. X. As the sum of the two sides of a plane triangle, is to the difference of the segment of the base (see the preceding figure), so is the co-fine of half the vertical angle, to the fine of half the difference of the angles at the base. 3

For,

For, AC + BC: AQ-BQ:: AB: AC – BC

ACB (by 9. 2. and 10. 4.) :: co-line of : fine of

2

BA (by Prop. 8.) 2. 2. D.

PROP. XI. As the tangent of the vertical angle C of a plane triangle ABC, is to radius, so is half the base AB to a fourth proportional; and as half the base is to the excess of the perpendicular above the said fourthproportional, so is the fine of the vertical angle, to the co-fine of the difference of the angles at the base.

D

C

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Let ABCD be a circle described about the triangle, and from 0, the center thereof, let OB and OC be drawn; moreover, draw CD parallel to BA, meeting the periphery in D, and EOF, perpendicular to AB, meeting DC in E. 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 GAB): EO; moreover, as EB : OF (EF - EO) :: sine EOB (ACB) : sine OCF, or co-line of FOC. 2. E. D.

PROP,

PROP. XII. As the tangent of the vertical angle, of a plane triangle ABC, is to radius, so is the base AB to a fourth-proportional; and, as the said fourth-proportional, is to the sum of the semi-base and the line CD biseating the base, so is the difference of these two, to the perpendicular beight of the triangle.

Let a circle be described about the triangle, and from O, the center thereof, let OA,OC and OD be drawn;

also let CF, parallel to AB, D

be drawn, meeting DO, proB

duced (if need be) in F. It

is evident that DF will be perpendicular to AB, and equal to the height of the triangle. But DC= OC + OD' + 2OD X OF (by 11. 2.) = OA' + OD' + 2OD DF-OD = OA’ — OD: + 2OD X DF = AD + 2OD X DF (by 7. 2.); whence, by taking away AD’ from the first and last of these equal quantities, we have DC'-AD = 2OD RDF; or DU+AD xDC-AD = 2OD X DF (by 7. 2.) and therefore 2OD : DC +AD :: DC --- AD : DF; but (by Theor. 2.) as the tang. AOD = ACB (by 10. 3.): to radius (: AD : OD) :: AB : 20D. 2. E. D.

PROP. XIII. As twice the rectangle under the base and perpendicular of a plane triangle ABC, is to the rectangle under the fum, and difference, of the base and sum of the two sides ; so is radius, to the co-tangent of half the vertical angle.

Let

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