About the centre A with the radius AB, the greater of the two sides, describe a circle meeting BC produced in D, and AC produced in E and F. Join DA, EB, FB; and draw FG parallel to CB, meeting EB in G. Because the exterior angle EAB is equal to the two interior ABC, ACB, (32. 1.): and the angle EFB, at the circumference is equal to half the angle EAB at the centre (20. 3.); therefore EFB is half the sum of the angles opposite to the sides AB and AC. Again, the exterior angle ACB is equal to the two interior CAD, ADC, and therefore CAD is the difference of the angles ACB, ADC, that is of ACB, ABC, for ABC is equal to ADC. Wherefore also DBF, which is the half of CAD, or BFG, which is equal to DBF, is half the difference of the angles opposite to the sides AB, AC Now because the angle FBE in a semicircle is a right angle, BE is the tangent of the angle EFB, and BG the tangent of the angle BFG to the radius FB; and BE is therefore to BG as the tangent of half the sum of the angles ACB, ABC to the tangent of half their difference. Also CE is the sum of the sides of the triangle ABC, and CF their difference; and because BC is parallel to FG, CE: CF :: BE: BG, (2.6.) that is, the sum of the two sides of the triangle ABC is to their difference as the tangent of half the sum of the angles opposite to those sides to the tangent of half their difference. Q. E. D. PROP. V. 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 seg ments 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. Q. E. D. PROP. VI. 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 AB+BC2 and AC as radius to cos. B. A From A draw AD perpendicular to BC, and (12. and 13.2.) the difference between the sum of the squares of AB and BC, and the square on AC is equal to 2BC.BD. B D C But BC.BA: BC.BD :: BA: BD :: R: cos. B, therefore also 2BC.BA : 2BC.BD :: R:cos. B. Now 2BC.BD is the difference between AB +BC and AC2, therefore twice the rectangle AB.BC is to the dif ference between AB2 + BC2, and AC2 as radius to the cosine of B. A Wherefore, &c. Q. E. D. Cor. If the radius=1, BD=BA Xcos. B, (1.), and 2BC.BAXcos. B=2BC.BD, and therefore when B is acute, 2BC.BAXcos. B=BC2 +BA-AC2, and adding AC to B both; AC2+2 cos. B×BC.BA = BC2+BA2; and taking 2 cos. BBC.BA from both, AC2=BC3_2 cos. BXBC.BA+BA2. Wherefore AC = (BC2 -2 cos. BXBC. BA+BA2). CD If B is an obtuse angle, it is shewn in the same way that AC = ✓ (BC2+2 cos. B×BC.BA+BA3). PROP. VII. 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 AE, 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): therefore, (11.5.) 4AC.AD: HB.BG :: R2: (sin. BAC), or since AB=AD, 4AC.AB: HB.BG :: R2: (sin BAC). 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). Therefore, &c. Q. E. D. COR. Hence 2✓AC.AD: ✓ HB.BG :: R: sin. BAG.. PROP. VIII. Four times the rectangle contained by any two sides of a triangle, is ta 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. BAC)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+AC -BC. Now, because BD is bisected in E, (3. 3.) 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, (1. 6.), AC.AD: AD2:: AR.AE: AE, and alternately AC.AD: AR.AE :: AD2: AE2, and 4AC.AD : 4AR.AE :: AD2: AE. But 4AR.AE = 2AR X 2AE = 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)2. 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. Therefore, &c. Q. E. D. COR. 1. Hence 2 AC.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): (sin. BAC)2. But the cosine of any arch is to the sine, as the radius to the tangent of the same arch; therefore, (AB+AC+BC) (AB+AC-BC): (BC+ (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. 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, A E D B that AC is the sum, and EB the differ C ence 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. Therefore, &c. Q. E. D. 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, SECT. II. OF THE RULES OF TRIGONOMETRICAL 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, Gg |