But since 4EF=2BK, 4EF.ED=2BK.ED=2ED.BK DB.BK =HB.BG; therefore, 4AC.AD: DB.BK :: AD?: ED?. Now AD:ED:: Rsin: EAC=fin BAC (1. Trig.) and AD: ED:: R? :(sin BAC) ; therefore, ex æquo, 4AC.AD: HB.BC ::R?: (sin BAC), or since AB = AD, 4AC.AB : HB.BC :: R?:(fin BAC)?. Now 4AC.AB is four times the rectangle contained by the sides of the triangle; HB.BC is that contained by BC +(AB - AC) and BC—(AB - AC.). Therefore, &c. Q. E. D.

Cor. Hence 2N AC.AD: WHB.BG :: R: sin BAC.

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PRO P. VIII. POUR times the rectangle contained by any two

sides of a triangle, is to the rectangle contained by two straight lines, of which one is the sum of those fides increased by the base of the triangle, and the other the sum of the same fides diminished by the base, as the square of the radius to the square 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 base, and AB the greater of the other two sides, 4AB.AC: (AB + AC + BC)(AB + AC - BC): R2: (cos. BAC).

From the centre C, with the radius CB, describe the circle BLM, meeting AC, produced, in L and M. Produce AL to N, so 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 LNAB + AC - BC. Now, because BD is bisected 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 DN 32AD, BN = 2AE, and BP2B0=2RE ; alfo PN = 2AR. X 2



But because the tri

N angles ARC and AED are equiangular, AC: AD :: AR: AE, and because rectangles of the same altitude are as their bases, (1. 6.), AC.AD : AD2 : : AR.AE : AE, and alternately AC.AD : PI AR.AE:: AD: AE?, & 4AC.AD: HAR.AE

IR : : AD : AE. But 4 AR.AE = 2 AR X

D 2 AE — NP.NB


MN.NL; therefore
AD2: AE?. But AD:
AE ::R: cos DAE

M (1.) cos (BAC): Whereforę 4ĄC.AD : MN.NL :: R? :(cof_BAC).

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

Cor. 1. Hence 2N AC,AB :: NMN.NL::R: cof1 BAC.

Cor. 2. Since by Prop. 7. 4AC.AB:(BC+(AB— AC)) (BC-(AB-BC)):: R?: (sin BAC) ; and as has been now proved, 4AC.AB; (AB + AC + BC) (AB + AC -- BC) :: R? :(cos { BAC)”; therefore ex æquo,, (AB + AC + BC) (AB + AC – BC): (BC + (AB - AC)) (BC - (AB – AC))::(cof BAC)?: (sin. BAC). But the co-fine of any

arch is to the fine, as the radius to the tangent of the same arch ; therefore, (AB + AC + BC) (AB + AC -- BC:(BC+ AB - AC) ((BC-(AB-AC)) :: R’:(tan BĄC); and N(AB,+ AC + BC) (AB + AC-BC): NBC + AB-AC) (BC- (AB - AC)) ::R: tan , BAC.


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F there be two unequal magnitudes, half their dif

ference added to half their sum is equal to the greater; and half their difference taken from half their sum is equal to the less.

Let AB and BC be two unequal magnitudes, of which AB is the greater; suppose AC bifected in D, and AE equal to BC. It is manifeft, that AC is the sum, and EB the difference of the magnitudes. A E D B C And because AC is bifected in D, AD is equal to DC; but AE is also equal to BC, therefore DE is equal to DB, and DE or DB is half the dif. ference of the magnitudes. But AB is equal to BD and DA, that is to half the difference added to half the sum ; and BC is equal to the excess of DC, half the sum, above DB, half the difference. Therefore, &c. Q. E. D.

Cor. Hence, if the sum and the difference of two magni. tudes be given, the magnitudes themselves may be found ; for to half the fum add half the difference, and it will give the greater ; from half the sum subtract half the difference, and it will give the less.

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HE GENERAL PROBLEM which Trigonometry proposes

to resolve is : In any plane triangle, of the three sides and the three angles, any three being given, and one of these three being a fide, to find any of the other three.

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

The reason of the restri&tion in this problem to those cacases 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 sides of any one of them being equal to those of any other; though the ratios of their fides to one another will be the same 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 sides, which may be found by trigonometry, as being the same with the ratios of the fines of the opposite angles.

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

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N a right angled triangle, of the three sides and three angles, any two being given, besides the


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 cases, and the folutions, or rules for calculation, which all depend on the first 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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