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But fince 4EF2BK, 4EF.ED=2BK.ED2ED.BK DB.BK HB.BG; therefore, 4AC.AD: DB.BK:: AD2: ED2. Now AD: ED :: R fin: EAC➡ fin1⁄2 BAC (1. Trig.) and AD2: ED2:: R2: (fin BAC)2; therefore, ex æquo, 4AC.AD: HB.BC:: R2: (fin BAC), or fince AB= AD, 4AC.AB: HB.BC :: R2: (fin BAC)2. Now 4AC.AB is four times the rectangle contained by the fides of the triangle; HB.BC is that contained by BC + (AB-AC) and BC — (AB—AC.). Therefore, &c. Q. E. D.

COR. Hence 2AC.AD: √/HB.BC :: R : fin ¦ BAC.

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PROP. VIII.

OUR times the rectangle contained by any two fides of a triangle, is to the rectangle contained by two ftraight lines, of which one is the fum of those fides increased by the bafe of the triangle, and the other the fum of the fame fides diminished by the bafe, as the square of the radius to the fquare of the co-fine of half the angle included between the two fides of the triangle.

Let ABC be a triangle, of which BC is the bafe, and AB the greater of the other two fides, 4AB.AC: (AB + AC + BC)(AB + AC — BC) : R2 : (cof. 1⁄2 BAC)2.

From the centre C, with the radius CB, defcribe the circle BLM, meeting AC, produced, in L and M. Produce AL to N, fo that AN AB; let AD AB; draw AE perpendicular to BD; join BN, and let it meet the circle again in P; let CO be perpendicular to BN, and let it meet AE in R. It is evident that MN AB+ AC + BC; and that LN AB+ AC-BC. Now, because BD is bifected in E: (3.3.) and DN in L, BN is parallel to AE, and is therefore perpendicular to BD, and the triangles DAE, DNB are equiangular; wherefore, fince DN2AD, BN2AE, and BP 2RE; alfo PN 2AR.

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(1.) .cof(BAC): Wherefore 4AC.AD MN.NL:: R2: (cof1⁄2 BAC)2.

Now 4AC.AD is four times the rectangle under the fides AC and AD, (for AD AB), and MN.NL is the rectangle under the fum of the fides increased by the bafe, and the fum of the fides diminished by the base. Therefore, &c. Q E. D.

COR. 1. Hence 24/AC,AB:: /MN.NL:: R: cof BAC. COR. 2. Since by Prop. 7. 4AC.AB: (BC+(AB-AC)) (BC (AB-BC)) : : R2: (fin BAC); and as has been now proved, 4AC.AB; (AB+ AC + BC) (AB + AC — BC) :: R2: (cof BAC); therefore ex æquo, (AB + AC + BC) (AB+ AC — BC): (BC + (AB — AC)) (BC (AB AC)) (cof BAC): (fin. BAC). But the co-fine of 1-4 any arch is to the fine, as the radius to the tangent of the fame arch; therefore, (AB + AC + BC) (AB+AČ - BC: (BC+ AB — AC) ((BC —(AB—AC)) :: R2: (tan 1⁄2 BAC)2; and √√(AB+ AC + BC) (AB + AC — BC):

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(BC + AB- AC) (BC — (AB — AC)): : R: tan BAC.

LEMMA

I

LEMM A II.

there be two unequal magnitudes, half their difference added to half their fum is equal to the greater; and half their difference taken from half their fum is equal to the lefs.

Let AB and BC be two unequal magnitudes, of which AB is the greater; fuppofe AC bifected in D, and AE equal to BC. It is manifeft, that

AC is the fum, and EB the

difference of the magnitudes. A E D B And because AC is bifected

in D, AD is equal to DC; but AE is alfo equal to BC, therefore DE is equal to DB, and DE or DB is half the difference of the magnitudes. But AB is equal to BD and DA, that is to half the difference added to half the fum; and BC is equal to the excefs of DC, half the fum, above DB, half the difference. Therefore, &c. Q. E. D.

COR. Hence, if the fum and the difference of two magnitudes be given, the magnitudes themfelves may be found; for to half the fum add half the difference, and it will give greater; from half the fum fubtract half the difference, and it will give the less.

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SECTION II.

OF THE RULES OF TRIGONOMETRICAL CAL

CULATION.

THE

HE GENERAL PROBLEM which Trigonometry propofes to refolve is: In any plane triangle, of the three fides and the three angles, any three being given, and one of thefe three being a fide, to find any of the other three.

The things here faid to be given are understood to be expreffed by numbers, or by their numerical values; the angles, in degrees, minutes, &c.; and the fides in feet, or any other known measure.

The reason of the restriction in this problem to those cacafes in which at least one fide is given, is evident from this, that by the angles alone being given, the magnitudes of the fides are not determined. Innumerable triangles, equiangular to one another, may exist, without the fides of any one of them being equal to thofe of any other; though the ratios of their fides to one another will be the fame in them all, (4.6.) If, therefore, only the three angles are given, nothing can be determined of the triangle but the ratios of the fides, which may be found by trigonometry, as being the fame with the ratios of the fines of the oppofite angles.

For the conveniency of calculation, it is ufual to divide the general problem into two; according as the triangle has, or has not, one of its angles a right angle.

PROB. I..

N a right angled triangle, of the three fides and three angles, any two being given, befides the

IN

right

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right angle, and one of those two being a fide, it is required to find the other three.

It is evident, that when one of the acute angles of a right angled triangle is given, the other is given, being the complement of the former to a right angle; it is also evident that the fine of any of the acute angles is the co-fine of the other.

This problem admits of several cafes, and the folutions, or rules for calculation, which all depend on the firft Propofition, may be conveniently exhibited in the form of a Table; where the first column contains the things given; the second the things required; and the third the rules or proportions by which they are found.

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