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THEOREM V.

In any plane Triangle, it will be, as the Sum of any two Sides is to their Difference, fo is the Tangent of Half the Sum of the two oppofite Angles, to the Tangent of Half their Difference.

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For, let ABC be the Triangle, and AB and A C the two propofed Sides; and from the Center A, with the Radius AB, let a Circle be defcribed, interfecting ÇA produced,inD and F; fo that CF

may exprefs the Sum,and CD the Difference, of the Sides AC and AB: Join F, B and B, D, and draw DE parallel to FB, meeting BC in E.

Then, because 2ADB = ADB + ABD (by 6. 1.) C+ABC (by Cor. 3. to 10. 1.) it is plain that ADB is equal to Half the Sum of the Angles oppofite to the Sides propos'd. Moreover, fince ABC = ABD (ADB) + DBC, and C = ADBDBC (by 10. 1.) it is plain that A B C C is = 2 DBC; or that DBC is equal to Half the Difference of the fame Angles.

Now, because of the parallel Lines BF and ED, it will be CF CD:: BF: DE; but BF and DE, because DBF and BDE are Right-angles (by 11.3. and 8. 1.) will be Tangents of the forefaid Angles FDB (ADB) and DBE (DBC) to the Radius BD. 2. E. D.

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COROLLARY.

Hence, in two Triangles ABC and AbC, having two Sides equal, each to each, it will be (by Equa

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CAb be fuppofed a Right-angle, then will AbC+
ACb alfo a Right-angle (by 10. 1.) and the Tan-
AbC+Aсb

gent of

2

Cafe our Proportion will become,

Radius. Therefore in this

(= AbC — 45°)::

2

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gives the following Theorem,. for finding the Angles oppofite to any two propos'd Sides; the included Angle, and the Sides themselves, being known.

As the leffer of the propofed Sides (Ab or AB) is to the greater (AC), fo is Radius to the Tangent of an Angle (AbC, fee Theor. 2.) And as Radius to the Tangent of the Excefs of this Angle above 45°, fo is the Tangent of Half the Sum of the required Angles to the Tangent of Half their Difference*.

*This Theorem, tho' it requires two Proportions, is commonly us'd by Aftronomers in determining the Elongation and Parallaxes of the Planets (being beft adapted to Logarithms); for which Reafon it is here given.

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B

Given

The Solutions of the Cafes of right-angled plane Triangles.

Cafe

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The Hyp. The other Let the Angles be found,

AC and Leg AB

3

one Leg

BC

by Cafe 2. and then the re

quired Leg AB, by Cafe 1.

The An- The Hyp. As Sine A: Radius :: the

gles and

AC

4

one Leg

Leg BC to the Hyp.
AC (Theor. I.)

BC

The An The other As Sine A: BC :: Sine C

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The two The Ar- As AB: BC:: Radius:

6

Legs AB

and BC

gles

Tang. A (by Theor. 2.)

whofe Complement is the

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Angle C.

The two The Hyp. Let the Angles be found,

7 Legs A E and BC

AC

by Cafe 6. and then the Hyp. AC, by Cafe 4.

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Cafe

The Solution of the Cafes of oblique plane Triangles.

Given
The Angles

1 and one Side
AB

3

4

5

6

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Two Sides The other As AB: Sin. C:: BC: Sin. A

AB, BC and Angles A an Ang. Cop. and ABC

to one of 'em

Two Sides The other AB, BC and Side AC an opp. Angle

C

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(by Theor. 3.) which added to C, and theSum fubtracted from 180 gives the other Angle ABC.

Let the Angle ABC be found,by the preceding Cafe, and then it will be, Sine C: AB :: Sin.ABC AC (by Theor. 3.)

Two Sides The other As Sum of AB and AC: their Dif. AC, AB and Angles C:: Tan. of half the Sum of ABC the included and ABC and C: Tang. of half their Diff. Angle A (by Theor. 5.) which added to, and fubtracted from, the Sum gives the 2 Angles.

Two fides AC The other Let the Angles be found by the AB and the Side BC laft Cafe, and then BC, by Cafe 1. incl. Angle A

All the three An Angle, Let fall a Perp. BD opp. to the fuppofe A req. Angle: Then, (by Theor.4.)

Sides

as AC: Sum of AB and BC:: their Dif.: Dift. DG of the Perp. from the Middle of the Bafe; whence, AD being also known, the Angle A will be found by Cafe 2. of Right-angles.

Note, The 2d and 3d Cafes are ambiguous, or admit of two different Answers each, when the Side A B oppofite the given Angle C (see Fig. 2.) is lefs than the given Side BC, adjacent to it (except the Angle found is exactly a Right one): For then another Right-line Ba, equal to BA, may be drawn from B to a Point in the Bafe, fomewhere between C and the Perpendicular BD, and therefore the Angle found by the Proportion, AB (aB): Sin. C :: BC: Sin. A (or of Ca B,) may, it is evident, be either the acute Angle A, or the obtufe one C a B (which is its Supplement), the Sines of both being exactly the fame.

Having laid down the Method of refolving the different Cafes of plane Triangles, by a Table of Sines and Tangents; I fhall here fhew the Manner of constructing fuch a Table (as the Foundation upon which the whole Doctrine is grounded); in order to which, it will be requifite to premise the following Propofitions.

PROPOSITION I.

The Sine of an Arch being given, to find its Cofine, Verfed Sine, Tangent, Co-tangent, Secant, and Co-fecant.

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