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°; E. 19 ch. FG, West, 17.72 ch. GH, N. 36° 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. A a Sta. Courses. Dist. N. AB N. 19°E. 27.00 25.53 BC S. 77 E. 22.75 CD S. 27 E. 28.75 DE S. 52 W. 14.50 EF S. 15 E. 19.00 FG West, 17.72 GHN. 36 W. 11.75 HI North, IA N. 62 W. 14.88 (172.42 16.07 9.51 16.07 6.98 58.09 .11 Er. S. 0 | 4 57.98 49.09 49.19 49.09 .02 11.42.01 01.01 .02 .02 .01 3.01 .01 .01 .0.1 .01 .01 .01 .10 Er. E. ..01 .10 58.04 58.04 49.15 49.15 Area of ABCDEFGHIA (sq. ch.) Half do. (sq. ch.) 110.4583 136.7145 1587.1160 568.2264 1048.6368 .4462 400.8022 3451.5982 400.8022 2)3050.7960 1525.3980 762.6990 To find the latitude and departure of CI, and area of the part IABCI. As area of ICDI : area of ICNI :: latitude of CD : latitude of CN As area of ICDI 27.35 27.35 30.92 Area of IABCI (sq. ch.) Area of ICNI (sq. ch.) To find the area of ICDI. E. W. M. Dist. N.area. 0.00 17.85 E (30.92) 30.92 | 407.57 215.46 M. Dist. N.area. 0.00 13.13 W 4.33 W 4.47 E 25.64 S. 13.55 S. 407.57 215.46 26.65 E 48.83 E 13.07 E. 30.98 E 13.13 E 30.92 E 43 99 E 13.07 E 17 85 W S. area. 110.4583 136.7145 847.3030 2)1094.4758 547.2379 762.6990 215.4611 2)815.1385 Area of ICDI (sq. ch.) 407.56925 S. area. 792.7888 22.3497 Ar. Co. 7.38980 2.33337 1.40892 1.13209 Ar. Co. 7.38980 2.33337 1.11628 To find the latitude and departure of NI. Sta N. IC 27.35 CN NI E. 13.55 6.91 As diff. of lat. of NI, 13.80 S. do. 24.76 W. 17.85 (13.80) 27.35 27.35 24.76-24.76 sec. bearing of NI, 60° 52′ :: diff. lat. do. 13.80 W. To find the bearing and distance of NI. : dep. :: rad. : tang. bearing of NI, S. 60° 52′ W. As rad. 24.76 Ar. Co. 0.00000 10.31261 1.13988 1.45249 ; dist. NI, 28.35 ch. Hence IN bears, N. 60° 52′ E. dist. 28.35 ch. 2. Given the boundaries of a tract of land as follow; viz. 1st. S. 35°4 W. 11.20 ch.. 2nd. N. 45° W. 24.36 ch. 3rd. N. 15° E. 10.80 ch. 4th. S. 77 E. 16 ch. 5th. N. 87° E. 21.50 ch. 6th. S. 60° E. 14.80 ch. 7th. South, 10.91 ch. 8th. N. 85° W, 29.28 ch. to the place of beginning; to divide the tract into two equal parts by 1 a line running from the first station and falling on one of the opposite sides: the bearing and distance of the division line are required. Ans. N. 7° 18′ E. 15.28 ch. VARIATION OF THE COMPASS, A meridian pointed out by the magnetic needle is not in general a true one; for the needle does not point truly to the north point of the horizon, but varies from it, in some places to the eastward, and in others to the westward. The angle contained between the true meridian and that pointed out by the needle, is called the variation of the compass. The variation is named east or west, according as the needle points to the eastward or westward of the true north. As the variation is different in different places, so also in the same place it does not remain the same, but differs sensibly in the course of a few years. Hence in running a lìne that was run a number of years previously, the bearing will be found different from what it was at that time; this, together with the difference in compasses, causes many difficulties, and frequently inaccuracies, in |