From the bearings of the lines are found the angle ADE=63° 30', AED=80° 15', = AFG=97° 00′ and AGF 46° 45' 2. The bearings and distances of a triangular piece of land ABC are, AB, S. 69° E. 21.40 ch. BC, N. 31° E. 18.66 ch. and CA, S. 74°1 W. 30.85 ch., and it is required to divide it by a line FG running due north, so that the part FBCG may be to the part AFG as 5 to 4; what will be the distance AF? Ans. 17.40 ch. PROBLEM XVI. The bearings and distances of the sides AB, BC, CD, DA, Fig. 103, of a trapezoidal tract of land being given, to divide it into two parts having a given ratio, by a line EF running parallel to the parallel sides AB, CD. RULE. and in the triangle GDC having given all the angles and side DC, find the side GD; from GD subtract AD and the remainder will be GA. Then, As the sum of the numbers expressing the ratio of the parts, Is to the greater or less of them, according as the So is the rectangle of AD and sum of GD and GA, Subtract this fourth term from the square of GD; the square root of the remainder will give GE, and GE less GA will be the distance AE*. EXAMPLES. 1. Let the bearing of AB be N. 14° E. dist. 10 ch. BC, N. 55°1 E. 18.67 ch. CD, S. 14° W. 20.98 ch. and DA, DEMONSTRATION. Let m to n be the ratio of the part AEFB to the part EDCF; then (18.5.) m+n: n :: ADCB: EDCF :: (19.6. and division) GD3—. GA2:GD3—GE3; hence because (cor.5,2.) GD2—GA3—GD + GA. × AD, the truth of the rule is evident. Construction. Produce DA, CB, Fig. 104, to meet in G and on GD describe the semicircle GLD; join AC and parallel to it draw BH meeting GD in H; make HK to KD in the given ratio of AEFB to EDCF, and draw KL perpendicular to GD, meeting the semicircle in L; take GE equal to GL and parallel to AB or DC, draw EF which will divide the trapezoid as required. For join KC, HC, then (37.1.) the triangle AHC is equal to the triangle ABC; to each of these add the triangle ADC, then it is plain that the triangle HDC is equal to the trapezoid ADCB; but (1.6.) the line KC divides the triangle HDC in the given ratio: and from the demonstration to the construction of the last problem, it is manifest that the triangles GKC and GEF are equal; consequently KDC is equal to EDCF. Whence the truth of the construc tion is evident. Cor. If HM be drawn perpendicular to GD, and DM, DL be joined, we shall have DGa: DM2 :: DEFC: DABC. For GD: GA :: GC : GB :: GA : GH; hence GA'=DGXGH=GM3, and GD3—GM2 (GA3)—DM2; also D West, 12.70 ch.; it is required to divide the trapezoid into two parts by a line EF parallel to AB or DC, so that the part AEFB may be to the part EDCF as 3 to 2. 2. The boundaries of a trapezoidal field ABCD are given as follow: viz. AB, N. 80° W. 60 per. BC, N. 39°1 W. 45.5 per. CD, S. 80° E. 89.4 per., and DA, South, 30 per., and it is required to divide it into two equal parts by a line EF parallel to AB or CD; what PROBLEM XVII. The bearings and distances of the sides AB, BC, CD, DA, Fig. 105, of any quadrilateral tract of land being given to divide it into two parts having a given ratio, by a line EF running parallel to one of the sides as AB. RULE. Suppose the sides DA, CB produced to meet in G, and as in the preceding problem, find GD and GA. Let HI parallel to AB be a line making the trapezoid AHIB equal to the given trapezium ABCD. Then, ད As the rectangle of the sines of the angles H and C, Is to the rectangle of the sines of the angles D and I; So is the square of GD, To the square of GH. The square root of which gives GH; and GH less GA gives AH; then as in the preceding problem it will be, As the sum of the numbers expressing the ratio of the parts, Is to the greater or less of them, according as the So is the rectangle of AH and sum of GH and GA, square root of the remainder will give GE, and GE less GA will be the distance AE*. Note. If the division line is to run parallel to CD, then HI must also be supposed to be drawn parallel to CD, and the manner of working varied accordingly. EXAMPLES. 1. Let the bearing of AB be North, dist. 12 ch. BC, N. 56°1 E. 20.78 ch. CD, S. 33°1 E. 22.21 ch. and DA, S. 80°1 W. 30 ch. ; it is required to divide the tract into two parts by a line EF parallel to AB, so that the part AEFB may be to the part EDCF as 3 to 5. *The truth of this rule is evident from the demonstrations to the 11th and preceding problems. Construction. Draw CM, Fig. 106, parallel to AB, and on GM describe the semicircle GLM; then proceed in every other respect as in the construction to the preceding problem. Note. When the division line is to run parallel to CD, the construction is exactly as in the preceding problem. |