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PROP. XII. THEOR.

In any triangle, the rectangle under the semiperimeter and its excess above the base, is to the rectangle under its excesses above the two sides, as the square of the radius, to the square of the tangent of half the contained angle.

In the triangle ABC, the perimeter being denoted by P, |P(AP–AC : (AP—AB) (AP–BC): : R* : tank B".

For, employing the construction of Prop. 29., Book VI. of the Elements; since the triangles BIE and BGD

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PROP. XIII. THEOR.

In any triangle, the rectangle under two sides, is to the rectangle under the semiperimeter, and its excess above the base, as the square of the radius, to the square of the cosine of half the contained angle. - - -

In the triangle ABC, the perimeter being denoted by P, AB.BC : P(, P-AC) : : R* : cos, B2. For, the same construction remaining ; in the rightangled triangles BIE and BGD, r BE : BI: : R : sin BEI, or cos B, and BD: BG : : R : sin BDG, or cos' B; whence B.E.BD : BI.BG : : R* : cos; B2. But the quadrilateral figure EADC, being right angled at A and C, is (III. 17. El.) contained in a circle, and consequently (III. 16. El.) the angle AED or AFB is equal to ACD or to DCB; wherefore, since by construction the angle ABE is equal to DBC, the triangles BAE and BDC are similar, and BE : AB :: BC : BD, or BE.BD = AB, BC. Hence AB.BC : BI.BG : : R* : cos, B*. Now BI is the semiperimeter, and BG its excess above IG or AC; wherefore the proposition is demonstrated.

PROP. XIV. THEOR,

In any triangle, as the rectangle under two sides is to the rectangle under the excesses of the semiperimeter above those sides, so is the square of the radius, to the square of the sine of half their contained angle.

In the triangle ABC, the perimeter being still denoted by P, AB.BC : (AP–AB) (AP–BC): : R* : sin, B'. For, the same construction being retained, in the rightangled triangles BIE and BGD, BE : IE : : R : sink B, - and BD ; GD :: R : sin}B; whence B.E.BD: IE.GD :: R* : sing Bo. But it has been proved that BE.BD=AB.BC, or therectangleunderthecontaining sides of the triangle; and IE.GD= AI.AG, or the rectangle under the excesses of the semiperimeter above the sides AB and BC. Wherefore the proposition is established.

Scholium. The three last propositions are demonstrated here by an independent process; but they are only modifications of the same principle, and might consequently be derived from a comparison with the first of the train.

The eight preceding theorems contain the grounds of trigonometrical calculation. A triangle has only five va

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riable parts—the three sides and two angles, the remaining angle being merely supplemental. Now, it is a general principle, that, three of those parts being given, the rest may be thence determined. But the right-angled triangle has necessarily one known angle; and, in consequence of this, the opposite side is deducible from the containing sides. In right-angled triangles, therefore, the number of parts is reduced to four, any two of which being the assigned, the others may be found.

FROP. XV. PROB.

Two variable parts of a right-angled triangle being given, to find the rest.

This problem divides itself into four distinct cases, according to the different combination of the data.

1. When the hypotenuse and a side are given. 2. When the two sides containing the right angle are given, 3. When the hypotenuse and an angle are given. 4. When either of the sides and an angle are given, The first and third cases are solved by the application of Proposition 7, and the second and fourth cases receive their solution from Proposition 8. It may be proper,

however, to exhibit the several analogies in a tabular form.

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In the first and second cases, BC or AC might also be deduced, by the mere application of Prop. 11. Book II. of the Elements:

For AC*=AB* +BC”, or AC= w (AB*-i-BC*) and BC*=AC*-AB*=(AC+AB) (AC-AB), or BC= v((AC+AB) (AC–AB).

Cor. Hence the first case admits of a simple approximation. For, by the scholium to Proposition 6, it appears, that, AC being made the radius, 2AC+ AB is to 3AC, as the side BC is to the arc which measures its opposite angle

CAB, or alternately 2AC+ AB is to BC, as 3AC to the

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