POLYGONOMETRY. Thus AB and BC are determined. Like expressions will serve for the determination of any other two sides, whether contiguous or not: the co-efficients of x and y being designated by different letters for that express purpose; which would have been otherwise unnecessary in the solution of the individual case proposed. CD DE EF 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 rooin. EXAMPLES. Ex. 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 10 Whence AB=1284 80° 2400 C B+C 2700 D 529 2860 E 66° 1629. Then, by cor. 3 th. 2, tan BAF = BC. sin B+ CD. sin (BC)+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.CÓS (B+C+D+E) BC. sin 32o + CD. sin 80° DE. sin 132°EF. sin 1980 + AB+BC. COS 32°+CD.cos 80o+DE. cos 132°+EF.cos198? BC .sin 32o+CD. sin 80°+DE. sin 48° EF sin 18° AB+BC.COS 32°+CD.cos 80o Whence BAF is found 106° 31' 38"; AB=2400 Ex. Ang. DE 3500 DE.Cos 48° -EF.cos18° and the other angle AFE 91° 28′ 22′′. So that the exterior angles A and ƒ 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 B, are known: viz. We shall have, by th. 2 cor. 1, }={ AB sin A B=62° +BC. sin (A+B) E=64° +CD. sin (A+B+C) 15 105 DE.Sin(E+F) +EF. Sin F There 149° 23′ 26′′, Or, 116°+c={ +33° 36′ 34′′. The second of these will give for c, a re-entering angle; the first will give exterior angle c=33° 23′ 26′, and then will D=14° 36′ 34′′. Lastly, "AB. Cos 54° +BC. COS 64° +CD. cos 30°36′34′′ =3885.905. +DE COS 44° EF COS 72° • Ex. 3. In a hexagon ABCDEF, are known, all the sides ex- CD = 1600 AF -AB. sin 54° -BC. sin 116o +DE. Sin 136° +EF. Sin 7.29. C = 729 D 759 F = 84°. Suppose the diagonal BE drawn, dividing the figure into two trapeziums. Then, in the trapezium BCDE the sides, except BE, and the angles except B and E, will be known; and these may be determined as in exam. 1. Again, in a trapezium ABEF, there will be known the sides except AF, and the angles except the adjacent ones B 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) cos CD. sin 72°+DE . sin 147o sin 72°DE . Sin 33° BC+CD.COS 72°DE.COS 147° Whence CBE CD. BC+CD . cos 72o — de . cos 33o 79° 2′ 1"; and therefore DEB = 67° 57′59′′. BC. COS 79° 2' 1" +CD. cos 7° 2′ 1′′ +DE. cos 67°57′59′′ Secondly, in the trapezium ABEF, AB. Sin A+BE. Sin (A+B): = EF. sin F : whence Then EB 2548.581. EF . Sin F-AB. sin B = sin { BE sin (A+B) 20/55/54", 159° 4′ 6′′, Taking the lower of these to avoid re-entering angles, we have в (exterior ang.) = 95° 4′ 6′′: ABE = = 95° 4′6′′ : Abe = 84° 55′54′′; FEB= 63° 4'6": therefore ABC 163° 57′ 55′′; and FED=131°25" and " 1 POLYGONOMETRY. 107 and consequently the exterior angles at в and E are 16o 2 5′ Lastly, AF AB. COS A-BE. COS (A+B)—EF COS F —— Ex. 4. In a hexagon ABCDEF, are known all the angles, and all the sides except AF and GD: to find those sides. Whence CD Here, reasoning from the principle of cor. th. 2, we have, DE.Sin1660 or AB.sin 84o DE.Sin14°.cosec 10°-AB.sin84°.cosec100 +CD.Sin1700+EF.sin 1480 Given AB = 2200 Ext. Ang. A ≈ 96° BC= 2400 B 54° And AF DE =4800 EF = 5200 +EF.sin32°.cosec10°-Bc.sin30°.cosec 10°) DE.Sin24°.cosec10°-CB.sin20° =14874-98. +EF.Sin42o.cosec10o-BA.sin 74° 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. c = 20° D = 24° E 18° F =148° Suppose diagonals drawn to join the unknown angles, and dividing the poly gon into three trapeziums and a triangle; as in the marginal figure. Then, Given AB = 2400 FG = 3800 Ext. ang. в — BC 32° C CD 2700 GH = 4000 2800 HI = 4200 3200 IA = 4500 EF = 3500 36° 45° DE F = H = 48° 1 = 50°. F 1st. In the trapezium ABCD, where AD, and the angles about it are unknown we have (cor. 3. th. 2). DE. Sin14° +EF. Sin32° BC. Sin BCD. sin (B+c) tan BAD = AB+BC.COS B+CD.COS. (B+C) Whence BAD = 39° 30′ 42′′, CDA Ab . cos 39° 30′42′′ +BC. COS 0 29 18 +CD. cos 32 29 18 And AD= 6913.292. 2 2dly. In the quadrilateral DEFG, where DG and the angles about it are unknown; we have EF. Sin EFG. sin (EF). tan EDG And DG 3dly. In the trapézium GHIA, an exactly similar process gives 50° 46′ 53′′, IAG 47° 13′ 7', and AG = 9780.591. HGA D BC. sin 409 CD sin 70° ABBC. COS 40° + CD.cos 72o 32° 29′ 18′′ EF. Sin 36°+ FG sin 81° DEEF. Cos 36°+ FG. cos 81? { DE. COS 41° 14′ 53′′ 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° 39° 45′ 7"+43° 15′ 54′′ +50° 46′53′′ = 133° 47′ 54′′. Consequently, the required exterior angles are A=32°22′45′′, 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° 28'. Required the side AD, and the area. Ex. 8. In prob. 1, suppose rq-2538 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′, CQP—94° 6′, dqp=121° 18′. OF : OF MOTION, FORCES, &c. [ 109 ] Art. 1. BODY is the mass, or quantity of matter, in any material substance; and it is always proportional to its weight or gravity, whatever its figure may be. DEFINITIONS. 2. Body is either Hard, Soft, or Elastic. A Hard Body is that whose parts do not yield to any stroke or percussion, but retains its figure unaltered. A Soft Body is that whose parts yield to any stroke or impression, without restoring themselves again; the figure of the body remaining altered. And an Elastic Body is that whose parts yield to any stroke, but which presently restore themselves again, and the body regains the same figure as before the stroke. We know of no bodies that are absolutely, or perfectly, either hard, soft, or elastic; but all partaking these properties, more or less, in some intermediate degree. 3. Bodies are also either Solid or Fluid. A Solid Body, is that whose parts are not easily moved among one another, and which retains any figure given to it. But a Fluid Body is that whose parts yield to the slightest impression, being easily moved among one another; and its surface, when left to itself, is always observed to settle in a smooth plane at the top. 4. Density is the proportional weight or quantity of matter in any body. So, in two spheres, or cubes, &c. of equal size or magnitude; if the one weigh only one pound, but the other two pounds; then the density of the latter is double the density of the former; if it weigh 3 pounds, its density is triple; and so on. 5. Motion is a continual and successive change of place.If the body move equally, or pass over equal spaces in equal times, it is called Equable or Uniform Motion. But if it increase or decrease, it is Variable Motion; and it is called Accelerated Motion in the former case, and Retarded Motion in the latter. Also, when the moving body is considered with respect |