Let the bearing of EA, Fig 94, be East, AB, N. 10° E. dist. 15 ch, and BF, S 58° 30′ W, then the angles G, GAB, GBA, the distance GA, and area GAB, being the same as in Fig. 93, it is required to cut off the same area by a line DC parallel to AB. From GA2 462.3 Subtract the fourth term 291.105 171.195 13.08 ch.-GD 8.42 ch.=AD Fig. 94. 2. Given the bearings of three adjacent sides of a tract of land and the length of the middle one as follow; Ist. N. 20° W. 2nd. N. 60° 30' E., dist. 6 ch. 3rd. S. 61° 30' E., to cut off a lot containing 24 acres, by a line parallel to the 2nd side; required the distance on the first side. Ans. 3.45 ch. 3. Given as follow; 1st side N. 31° 15′ W. 2nd. N. 58° 45' E. dist. 13.50 ch. 3rd. S. 14° 45' E. to cut off 8 acres by a line running parallel to the 2nd side; the distance on the 1st side is required. Ans. 6.58 ch. PROBLEM XI. The bearings of three adjacent sides, EA, AB, BF, Fig. 96, of a tract of land, and the length of the middle side AB, being given, to cut off a trapezium ABHI containing a given area by a line HI running a given course. RULE. and let DC parallel to AB, be a line cutting off the given area. From the given bearings, find the angles A, G, B, H, and I, and by the preceding problem find GA and GD. Then, As the rectangle of the sines of the opposite angles Is to the rectangle of the sines of the opposite an- So is the square of the distance GD, To the square of the distance GI. The square root of which gives GI, and the difference between GI and GA will be the distance AI*. EXAMPLES. 1. Let the bearing of EA, Fig. 96, be N. 80° 30′ W. AB, North, dist. 12 ch., and BF, N. 58° E.; it is re DEMONSTRATION. Draw CM, Fig. 97, parallel to HI; then by trig. As sin. GCD (B) : sin. GDC (A) :: GD: GC. sin. GMC (I): sin. GCM (H) :: CC : GM. Therefore (C.6.) sin. B×sin. I : sin. A×sin. H :: GDXGC : GM × GC :: (cor. 1.6.) GD: GM ; but (19.6.) GI2: GM3 :: GHI (GDC): GMC :: (1.6.) GD: GM :: (cor. 1.6.) GD3: GMXGD; therefore (11.16.5.) GIa: GD3: GM1 : GMXGD :: GM: GD, or GD: GM :: GD2: GI2; consequently (11.5.) sin. BX sin. I sin. A X sin. H :: GD2: G12. Construction. By the construction to the preceding problem, draw DC, Fig. 97, parallel to AB, cutting off the trapezoid ABCD containing the given area. From C, draw CM according to the given bearing of HI, on GM describe the semicircle GLM, and from D draw DL perpendicular to GM, meeting the semicircle in L; join GL, make GI equal to it, and draw IH parallel to MC, then will the trapezium ABHI, be equal to the trapezoid ABCD, and consequently contain the given area. Since (31.3, and cor. 8.6.) GD: GL (GI) :: GL (GI) :: GM, or (20,6. cor. 2.) GD: GM :: GD3: GIa, the trath of quired to cut off 10 acres by a line HI, running S. 14° 30' E. AG, N 80°30'W GE, S 80°30′E BG, S 58°00'W :: twice the area to be cut off, 200 sq. ch. 2.30103 : a fourth term 259.53 2.41419 GA2=235.87 2. Given the bearings of three adjacent sides of a tract of land and the length of the middle one as follow: 1st. N. 31° 15′ W. 2nd. N. 58° 45' E. dist. 13.50 ch. 3rd. S. 14° 45 'E. to cut off 8 acres by a line running N. 87° 30' W; required the distance on the first side. Ans. 2.67 ch. 3. Given as follow: 1st side, N. 74° 45′ W. 2nd. N. 37° E. dist. 17.24 ch. 3rd. N. 84° E., to cut off a field containing 20 acres by a line running S. 20° W.; the distance on the 1st side is required. Ans. 14.01 ch. PROBLEM XII. The bearings of several adjacent sides, LA, AB, BC, CD, DE, EF, Fig. 98. of a tract of land, and the distance of each, except the first and last, being given, to cut off a given area by a line IH running a given course from a point somewhere in AL, and falling on EF. RULE. Suppose a line drawn from A to E, and calculate the Y Subtract the area of ABCDEA from the area to be cut off, the remainder will be the area of AEHI. Then having the bearings of LA, AE, EF, HI, the distance AE and the area of AEHI, find AI by the preceding problem*. . EXAMPLES. 1. Let the bearing of LA, Fig. 98, be N. 48° 30′ W. AB, S. 78° 00′ W. dist. 8 ch. BC, N. 26° 30' W. dist. 11.08 ch. CD, N. 38° 30′ E, dist. 12.82 ch. DE, S. 64° 00′ E. dist. 10.86 ch., and EF, S. 86° 00′ E.; it is required to cut off 30 acres by a line HI running S. 32° 15′ W. To find the area of ABCDEA. |