PROP. V. THEOR. If a perpendicular be drawn from any angle of a triangle to the opposite side, or base; the sum of the segments of the base is to the sum of the other two sides of the triangle as the difference of those sides to the difference of the segments of the base. For (K. 6.), the rectangle under the sum and difference of the segments of the base is equal to the rectangle under the sum and difference of the sides, and therefore (16. 6.) the sum of the segments of the base is to the sum of the sides as the difference of the sides to the difference of the segments of the base. PROP. VI. THEOR. In any triangle, twice the rectangle contained by any two sides is to the difference between the sum of the squares of those sides, and the square of the "base, as the radius to the cosine of the angle included by the two sides. Let ABC be any triangle, 2AB.BC is to the difference between AB2+BC2 and AC2 as radius to cos. B. From A draw AD perpendicular to BC, and (11. and 12. 2.) the difference between the sum of the squares of AB and BC, and the square on AC is equal to 2BC.BD. A B D C But BC.BA: BC.BD :: BA : BD:: R: cos. B, therefore also 2BC.BA: 2BC. BDR cos. B. Now 2BC.BD is the difference between AB2+BC" and AC2, therefore twice the rectangle AB.BC is to the difference between AB2+BC2, and AC2 as radius to the cosine of B. = COR. If the radius 1, BD=BA Xcos. B, (1.), and 2BC.BAX cos. B =2BC.BD, and therefore when B is acute, 2BC.BAX cos. BBC2+BA2 -AC2, and adding AC2 to both; AC2 +2 cos. B x BC.BA = BC2+ BA2; and taking 2 cos. Bx BC.BA from both, AC2=BC2-2 cos. BX BC.BA +BA2. Wherefore AC=√(BC2—2 cos. B× BC.BA+BA2). If B is an obtuse angle, it is shewn in the same way that AC= √(BC2+2 cos. BxBC.BA+BA2), PROP. VII. THEOR. Four 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 base or third side of the triangle increased by the difference of the two sides, and the other the base diminished by the difference of the same sides, as the square of the radius to the square of the sine of half the angle included between the two sides of the triangle. Let ABC be a triangle of which BC is the base, and AB the greater of the two sides; 4AB.AC: (BC+(AB—AC)) × (BC—(AB—AC)): : R2 : (sin.BAC)2. Produce the side AC to D, so that AD=AB; join BD, and draw AË, CF at right angles to it; from the centre C with the radius CD describe the semicircle GDH, cutting BD in K, BC in G, and meeting BC produced in H. It is plain that CD is the difference of the sides, and therefore that BH is the base increased, and BG the base diminished by the difference of the sides; it is also evident, because the triangle BAD is isosceles, that DE is the half of BD, and DF is the half of DK, wherefore DE-DF-the half of BD-DK (6. 5.), that is, EF BK. And because AE is drawn parallel to CF, a side of the triangle CFD, AC: AD:: EF: ED, (2. 6.); and rectangles of the same altitude being as their bases AC.AD: AD2:: EF.ED: ED2 (1. 6.), and therefore 4AC.AD: AD2 :: 4EF.ED: ED2, or alternately, 4AC.AD: 4EF.ED :: AD2: ED2. But since 4EF=2BK, 4EF.ED=2BK.ED=2ED.BK=DB.BK= 'HB.BG; therefore 4AC.AD: DB.BK:: AD2: ED2. Now AD: ED :: R: sin. EAC sin. BAC (1. Trig.) and AD2: ED2 :: R2: (sin. BAC)2: therefore, (11. 5.) 4AC.AD: HB.BG :: R2: (sin. BAC)2, or since AB =AD, 4AC.AB HB.BG:: R2: (sin. BAC)2. "Now 4AC.AB is four times the rectangle contained by the sides of the triangle; HB.BG is that contained by BC+(AB-AC) and BC-(AB-AC). COR. Hence 2 VAC.AD: HB.BG :: R: sin+ BAC. PROP. VIII. THEOR. Four 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 sides increased by the base of the triangle, and the other the sum of the same sides diminished by the base, as the square of the radius to the square of the cosine of half the angle included between the two sides 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. RAC)2. 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 LN-AB+ACBC. Now, because BD is bisected in E, and DN in A, BN is parallel to AE, and is therefore perpendicular to BD, and the triangles DAE, DNB are equiangular; wherefore, since DN=2AD.BN=2AE, and BP=2BO 2RE; also PN=2AR. But because the triangles ARC and AED are equiangular, AC: AD :: AR: AE, and because rectangles of the same altitude are as their bases N (1. 6.), AC.AD: AD2 :: AR.AE: AE2, and alternately AC.AD: AR.AE :: AD2: AE2, and 4AC.AD: 4AR.AE:: AD2: AE2. But 4AR.AE= 2ARX2AE=NP.NB=MN.NL; therefore 4AC.AD: MN.NL:: AD2: AE2. But AD: AE:: R: cos. DAE (1) cos. (BAC): Wherefore 4AC.AD: MN.NL:: R2: (cos.BAC). Now 4AC.AD is four times the rectangle under the sides AC and AB, (for AD=AB), and MN.NL is the rectangle under the sum of the sides increased by the base, and the sum of the sides diminished by the base. COR. 1. Hence 2 VAC.AB: √MN.NL:: R: cos. BAC. COR. 2. Since by Prop. 7. 4AC.AB: (BC+(AB-AC)) (BC-(AB -BC)) :: R2: (sin. BAC)2; and as has been now proved 4AC.AB: (AB+AC+BC) (AB+AC-BC): R2: (cos.BAC)2; therefore, ex æquo, (AB+ AC+ BC) (AB+AC-BC): (BC+(AB—AC)) (BC— (AB-AC)): (cos. BAC)2: (sin. BAC)2. But the cosine of any arc is to the sine, as the radius to the tangent of the same arc; therefore, (AB +AC+BC) (AB+AC-BC): (BČ+(AB−AC)) BC—(AB—AC)):: R2: (tan. BAC)2; and √(AB+AC+BC) (AB+AC-BC: √(BC+AB-AC) (BC-(AB-AC)) :: R: tan. BAC. LEMMA II. If there be two unequal magnitudes, half their difference added to half their sum is equal to the greater; and half their difference taken from half their sum is equal to the less. A ED B C Let AB and BC be two unequal magnitudes, of which AB is the greater; suppose AC bisected in D, and AE equal to BC. It is manifest that AC is the sum, and EB the difference of the magnitudes. And because AC is bisected 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 difference 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. COR. Hence, if the sum and the difference of two magnitudes be given, the magnitudes themselves may be found; for to half the sum add half the difference, and it will give the greater: from half the sum subtract half the difference, and it will give the less. SCHOLIUM. This property is evident from the algebraical sum and difference of the two quantities a and b, of which a is the greater; let their sum be denoted by s, and their difference by d: then, SECTION II. OF THE RULES OF TRIGONOMETRICAL CALCULATION. THE 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 side, to find any of the other three. The things here said to be given are understood to be expressed by their numerical values: the angles, in degrees, minutes, &c.; and the sides in feet, or any other known measure. The reason of the restriction in this problem to those cases in which at least one side is given, is evident from this, that by the angles alone being given, the magnitudes of the sides are not determined. Innumerable tri angles, 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 sides 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 sines 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 the angles a right angle. PROBLEM I. In 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 side, 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 sine of any of the acute angles is the cosine of the other. This problem admits of several cases, and the solutions, or rules for calculation, which all depend on the first Proposition, 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 propositions by which they are found. |