2. The boundaries of a field ABCD are given as follow; viz. AB, S. 10° W. 7.20 ch. BC, S. 67° W. 12.47 ch. CD, N. 23° W. 13.33 ch. and DA, S. 89° E. 18 ch. and it is required to divide it into two parts by a line EF parallel to the side AB, so that the part AEFB may be to the part EDCF as 3 to 4; what will be the distance PROBLEM XVIII. The boundaries of a tract of land ABCDEFGHIA, Fig. 107, being given, to divide it into two equal parts by a line IN running from the corner I and falling on the opposite side CD. RULE. Suppose lines drawn from I to C and D, and calculate the area of the whole tract. Take the latitudes and departures of IA, AB, and BC, and by balancing find the latitude and departure of CI; also calculate the area of the part IABCI; from half the area of the whole tract, subtract the area of the part IABCI, the remainder will be the area of the triangle ICNI. Take the latitudes and departures of IC and CD, and by balancing, find the latitude and departure of DI, and calculate the area of the triangle ICDI. Then, Also, As the area of the triangle ICDI, Is to the area of the triangle ICNI;. To the latitude of CN. As the area of the triangle ICDI, Now take the latitudes and departures of IC and CN, and by balancing, find the latitude and departure of the division line NI, with which find its bearing and distance*. Note. It is the corrected latitudes and departures that are to be used throughout the calculation. EXAMPLES. 1. Let the bearing of AB be N. 19° E. dist. 27 ch. BC, S. 77° E. 22.75 ch. CD, S. 27° E. 28.75 ch. DE, S. 52° W. 14.50 ch. EF, S. 15°1 E. 19 ch. FG, West, 17.72 ch. GH, N. 56° W. 11.75 ch. HI, North, 16.07 ch., and IA, N. 62° W. 14.88 ch.; it is required to divide the tract into two equal parts by a line IN running from the corner I and falling on the opposite side CD. First, calculate the whole area, thus: This rule needs no demonstration. |