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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 ra

dius, to the square of the cosine of half the contained angle. > ... In the triangle ABC, the perimeter being denoted by P, AB.BC : A P(; Pi—AC); : R* : cosy B4.

For, the same construction remaining; in the rightangled triangles BIE and BGD,

. BE : BI: ; R : sin BEI, or cos) B, and BD: BG : : R : sin BDG, or cos; B;

- whence BE.BD : BI.BG : : R* : cos, B*.

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 AEB 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.


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 : (#P-AB) (AP–BC): : R* : sin; B'. For, the same construction being retained, in the rightangled triangles BIE and BGD, BE : IE : : R : sin; B, and BD ; GD :: R : sin; B; whence BE.BD: IE.GD :: R* : sing Bo. But it has been proved that BE.BD=AB.BC, or therectangle under thecontaining sides of the triangle; and IE.GD=ALAG, or the rectangle under the excesses of the semiperimeter above the sides AB . and BC. Wherefore the proposition is establish


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.

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


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.


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^+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-HAB is to BC, as 3AC to the arc corresponding to BC. But the radius is equal to an

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the arc which corresponds to BC, as 3.x 57, or 172°, to the number of degrees contained in the angle CAB, and consequently 2AC+AB: BC: ; 172°: the expression of the angle at A, or AC+}AB : BC :: 86°: number of degrees in the angle at A. o This approximation will be the more correct, when the side opposite to the required angle becomes small in comparison with the hypotenuse; but the quantity of error can never amount to 4 minutes,


Three variable parts of an oblique angled triangle being given, to find the other two.

This general problem includes three distinct cases, one of which again is branched into two subordinate divisions.

1. When all the three sides are given.

2. When two sides and an angle are given; which angle may either (1.) be contained by these sides, or (2) subtended

by one of them.

3. When a side and two of the angles are given.

The first case admits of four different solutions, derived from Propositions 11, 12, 13, and 14, and which have their several advantages. The second case, consisting of

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