The difference of latitude (54.6) and departure (99.5) of the line NI, in the third table are found by balancing those of IC and CN; and as they are the base and perpendicular of a right angled triangle, of which the line NI is the hypothenuse, and the angle opposite to the departure, the bearing, we have the answer by two trigonometrical statings, as above; and thus may any tract be accurately divided, or any proposed quantity readily cut off or inclosed. Now, the student or practitioner may calculate the content of the part ABCNIA (the bearing and distance, or the diff. lat. and dep. of CN and of NI being known) and if it be found equal to the intended quantity, it proves the truth of the operation. Plate XII. fig. 3. EXAMPLE III. It is proposed to cut off 38A. 161P. to the south end of this tract, by a line running from E due West 40 perches to a well at O, and from thence a right line to a point M in the boundary HI; the place of M, and the bearing and length of the line OM are required; the field-notes being as in example 2d. Answer, M from H, north, 43.23 In this example we find As HI happens to be a meridian, the area of HOMH divided by half OV (19.1) quotes HM (43.33) without finding the area of HOIH, as we did of ICDI in example 2d. and HM-HV = VM8.03 diff. lat. of OM, which with its dep. VO 38.2 gives the bearing and distance as before. 1 Plate XII. fig. 4. EXAMPLE IV. A trapezoidal field ABCD, bounded as under specified, is to be divided into two equal parts by a right line EF parallel to AB or CD; required AF or BF? In the triangle CBG are given BC and all the angles (known by the bearings) to find BG, and thence the area by prob. 9. sect. 4. which + half the area of ABCD-area of EFG; then as the area of CBG to that of EFG, so is the square of BG to the square of FG, and FG-BG=BF. Operation at large. Angle G 39° 30', log. S. Co. Ar. 0.19649 Side BC 60 per. log. 1.77815 add Angle C 40° 30', sine Side BG 61.26 per. 9.81254 1.78718) Side BC 60 per. 1.77815 add Angle B 100° 0' sine 9.99335 By the application of this method a tract of land may be divided accurately, in any proportion, by a line running in any assigned direction. Note. When the practitioner would wish to be very accurate, it will be much better to work by four-pole chains and links than by perches and tenthis; one tenth of a perch square being equal to 61 square links. 276 EXAMPLE V. The following Field-Notes (from A. Burns) are of a piece of land, which is proposed, as an example to be divided into three equal parts by two right-lines running from the sixth and seventh stations; and proved, by calculating the content of the middle part. Plate VIII. fig. 5 EXAMPLE VI. The plot ABCDEFGHA is proposed to be divided, geometrically, in the proportion of 2 to 3, by a right line from a given point in any boundary or angle thereof, suppose the point D. Reduce the plot to the triangle cDe, as already taught; divide the base ce in the point N, so that eN be to Nc in the ratio of two to three, by prob. 14, page 53; draw DN, and it is done. Plate XII. fig. 3. EXAMPLE VII. Example 2d may likewise be performed geometrically. Produce CD both ways for a base, and reduce the whole to a triangle, making I the vertical point; then bisect the base in N, and draw IN. But, Notwithstanding this geometrical method is demonstrably true in theory, it is not as safe, on practical occasions requiring accuracy, as the calculation, even when performed with the greatest care; for which reason we will not enlarge on it here. EXAMPLE VIII. Suppose 864 acres to be laid out in form of a rightangled parallelogram, of which the sides shall be in proportion as 5 to 3, required their dimensions? For the greater side, multiply the area by the greater number of the given proportion, and divide |