Remark. Though the algebraic investigations commonly lead to results which are apparently simple, yet they are often, especially in polygons of many sides, inferior in practice to the methods suggested by subdividing the figures. The following examples are added for the purpose of explaining those methods: the operations however are merely indicated; the detail being omitted to save room. EXAMPLES. Er. 1. In a hexagon ABCDEF, all the sides except AF, and all the angles except a and F, are known. Required the unknown parts. Suppose we have Ext. ang. CD = 2400 C = 48° E = 66° Then, by cor. 3 th. 2, tan BAF = BC. sin B+ CD. Sin (B+C)+ DE . Sin (B+C+D) + EF. Sin (B+C+D+E) AB + BC. COS B + CD. COS (B+C) +DE. COS (B+C+D) + EF. COS (B+ C + D + E ) sin 80° + DE sin 132+ EF. sin 1980 = BC. sin 32° + CD Whence BAF is found 106°31′38′′; and the other angle AFE= 91°28′22′′. So that the exterior angles A and F are 73°28′22′′, and 88°31′38′′ respectively: all the exterior angles making 4 right angles, as they ought to do. Then, all the angles being known, the side AF is found by th. 1 = 4621-5. If one of the angles had been a re-entering one, it would have made no other difference in the computation than what would arise from its being considered as subtractive. Ex. 2. In a hexagon ABCDEF, all the sides except AF, and all the angles except c and D, are known: viz, The second of these will give for c, a re-entering angle; the second will give exterior angle c = 33° 23′26′′, and then will D 14° 36′ 34". Lastly, AF Er. 3. In AB. cos 54° +BC. COS 64° + CD. cos 30°36′34′′ = 3885-905. +DE. COS 44° - EF. cos 72° hexagon ABCDEF, are known, all the sides except AF, and all the angles except в and E; to find the rest. Given AB 1200 Exterior angles A = 64° BC 1500 Suppose the diagonal BE drawn, dividing the figure into two trapeziums. Then, in the trapezium BCDE, the sides except BE, and the angles except в and E, will be known; and these may be determined as in exam. 1. Again, in the trapezium ABEF, there will be known the sides except AF, and the angles except the adjacent ones в and E. Hence, first for BCDE: (cor. 3 th. 2), tan CBE = CD sin C+ DE . sin (C + D) BC + CD .cos C+ DE .COs (c + D) CD. sin 720 + DB. sin 1470 CD. sin 72+ DE. sin 33 BC+CD.COS 72+ DE. cos 147° BC+ CD. cos 72°-DE. cos 33°* Whence CBE = 79° 2′1′′; and therefore DEB = 67° 57′59′′. Then EB= { BC. cos 79° 2′ 1′′ +CD. cos 7° 2′ 1′′ +DE. cos 67°57′59′′ Secondly, in the trapezium ABEF, 2548 581. AB . sin a † BE. sin (A + B) EF. sin F: whence BE { 4' S 20°55'54", sin (A+B) = = sin 159° 4′ 6′′. Taking the lower of these, to avoid re-entering angles, we have B (exterior ang.) = 95°4′6′′; ABE=84°55′54′′; FEB= 63°4′6′′: therefore ABC 163°57′55′′; and FEd=131°2′5′′ : and consequently the exterior angles at B and E are 16°2′5′′ and 48° 57'55" respectively. = Lastly, AF=- AB.COS A BE.COS (A+B) - EF .COS F—— AB. Cos 64+ BE. cos 20° 55′ 54′′- EF. cos 84° 1645·292. Note. The preceding three examples comprehend all the varieties which can occur in Polygonometry, when all the sides except one, and all the angles but two, are known. The unknown angles may be about the unknown side; or they may be be adjacent to each other, though distant from the unknown side; and they may be remote from each other, as well as from the unknown side. Ex. 4. In a hexagon ABCDEF, are known all the angles, and all the sides except AF and CD: to find those sides. Given AB 2200 Ext. Ang. A = 96° Here, reasoning from the principle of cor. th. 2, we have Ex. 5. In the nonagon ABCDEFGHI, all the sides are known, and all the angles except A, D, G: it is required to find those angles. Given AB 2400 FG = 3800 Ext. Ang. B = 40° BC 2700 GH 4000 CD 2800 HI 4200 DE 3200 IA = 4500. C= 32 E = 36° F = 45° H = 48° D H B A BC. sin 40° + CD. sin 72° AB+ BC. Cos 40°+ CD.cos72°* 32° 29′18′′. =6913.292. 2dly. In the quadrilateral DEFG, where DG and the angles about it are unknown; we have tan EDG EF. Sin E+FG. Sin (E + F) =3 DE + EF, COS E + FG, COS (E + F EF. Sin 36°+ FG . sin 81° DEEF. COS 360+ FG.cos 81°° Whence Whence EDG = 41° 14′ 53′′, FGD = 39° 45′ 7′′. 3dly. In the trapezium GHIA, an exactly similar process gives HGA = 50° 46′ 53′′, IAG = 47° 13′ 7′′, and AG = 9780-591. 4thly. In the triangle ADG, the three sides are now known, to find the angles: viz, DAG 60°53′ 26", AGD=43° 15′ 54′′, ADG = 75° 50′ 40′′. Hence there results, lastly, = = IAB = 47° 13' 7"+60° 53′26′′ +39° 30′42′′ 147° 37′ 15, CDE = 32° 29′ 18" +70° 50' 40"+41° 14'53" =149° 34′51", FGH=39° 45′ 7′′ +43° 15′ 54′′ +50° 46′53′′ = 133° 47′54′′. Consequently, the required exterior angles are A=32°22′45′′, D= 30° 25′ 9′′, G = 46° 12′ 6′′. Ex. 6. Required the area of the hexagon in ex. 1. Ans. 16530191. Ex. 7. In a quadrilateral ABCD, are given AB=24, BC=30, CD 34; angle ABC = 92°18', BCD = 97°23'. Required the side AD, and the area. Ex. 8. In prob. 1, suppose PQ2538 links, and the angles as below; what is the area of the field ABCDQP? APQ 89°14', BPQ=68°11′, crQ=36°24′, DPQ= 19°57′; AQP 25°18', BQP=69°24', cap=94° 6', DQP=121°18', CHAPTER VII. PROBLEMS RELATIVE TO THE DIVISION OF FIELDS OR OTHER SURFACES. PROBLEM I. To Divide a Triangle into Two Parts having a Given Ratio, m: n. 1st. By a line drawn from one angle of the triangle. Make AD AB::m: m +n; draw cò. Here, evidently, AD =; m m+ n n AB, DB= AB. m+ n 2dly. By a line parallel to one of the sides of the triangle, Let ABC be the given triangle, to be divided into two parts, in the ratio of m ton, by a line parallel to the base AB. Make CE to EB as m to n: erect ED perpendicularly to CB, till it meet the semicircle described on CB, as a diameter, in D. Make CF CD: and draw through F, GF || AB. So shall GF divide the triangle ABC in the given ratio. CD2 CE For, CE: CB== :: CD2(=CF2): CB. But CE: EB :: M: N, or CE CB; m : m+n, by the construction: therefore, CF2: CB2 :: M: m +n. And since ▲ CGF: A CAB:: CF2: CB2; it follows that CGF: CAB:: m : m+n, as required. Computation. Since CB2: CF :: m+n: m, therefore, (m + n)cF2 = m. CB2; whence CF(m + n) = CB/m, or m CFCB In like manner, CG = CA m+ n Sdly. By a line parallel to a given line. Let HI be the line parallel to which a line is to be drawn, so as to divide the triangle ABC in the ratio of m to n. By case 2d draw GF parallel to AB, so as to divide ABC in the given ratio. Through F draw FE parallel to HI. On CE as a diameter describe a semicircle; draw GD perp. to AC, to cut the semicircle in D. Make CP = CD: through P, parallel to EF, draw pa, the line required. H B The demonstration of this follows at once from case 2; because it is only to divide FCE, by a line parallel to FE, into two triangles having the ratio of FCE to FCG, that is, of CE to Co. Computation. CG and CF being computed, as in case 1, the distances CH, CI being given, and CP being to co as CH to CI; the triangles CGF, GPQ, also having a common vertical angle, are to each other, as CG.CF to ca.CP. These products therefore are equal; and since the factors of the former are known, the latter product is known. We have hence given the ratio of the two lines CP(=x) to ca(=y) as CH to CI; say, as p to g; and their product CF. CG, say ab: to find x and y. b. That is, Here we find x = CF.CG. CH ab, y = CPV ; CQ # CF CG CI CH N. B. If the line of division were to be perpendicular to one of the sides, as to CA, the construction would be similar: VOL. III. M CP |