Now, to find the distances SI and IF, find the bearing With the difference of latitude and departure, SF is found to bear S. 48° 31', E. 4.72 chains; then in the triangle SFI, are given the angles by the bearing of the lines, viz. FSI 14° 29′, SFI 53° 46', SIF 111° 45', and the side SF 4.72 chains, to find the sides SI and IF; and by Case 1, Oblique Angled Trigonometry, they are found to be SI 4.10 chains, IF 1.27 chains; then to find the area of ARSIFA, are the following field notes. From the point E, let fall a perpendicular on the line GI; then in the right angled triangle VEI, are given the hypothenuse s chains, 73 links; and by the bearings of the lines, the angle VEI 21° 45', VIE 68° 15', to find thebase VI, 1 chain, 38 links, and perpendicular VE, 3 chains 46 links GO. From the point H, let fail a perpendicular on the line OC; then in the right angled triangle HNC, are given the hypothenuse 4 chains, 16 links, and by the bearings of the lines, the angle NCH 54°, and NHC 36o, to find the base NC, 2 chains, 44 links, and perpendicular NH = LO, 3 chains, 37 links. x +7.21 = 2A. 3R. 14P To find the area of ELHDE, the following notes arise. Suppose a survey to be made as follows, viz. AB east 20. chains, BC north 18. chains, CD north 30° west, so far as will inclose 1201 acres, the north boundary DE, to be due west 17. chains from the north 30° west line. Required the distance on this line; also the bearing and distance of the closing line EA, to include 120 acres. Note. The propriety of this example being under the head of Division, may, perhaps, be doubted by some; but the calculations are so nearly allied thereto, that I thought it admissible. From the point C, draw Cd due west 17 chains; then by problem 9, page 110, the area of ABCdA, is found to be 33.3 acres; continue AB, and from the point C, reverse CD, to meet AB in the point L; then in the right angled triangle BCL, are given the perpendicular BC, 18 chains, and the angles, by Position 4, page 12, and Po |