since sin (180° +6° 23′) is negative, the value of Z is 180° +6° 23′ 186° 23' The angle Z' is equal to the back azimuth of A, A., or (180° +186° 23′) - 360° 6° 23′. For the length of A, A., we have = It is not necessary to actually calculate /, as will be seen from the transformations given below. The angle K is equal to the back azimuth of A, A,, that is (325° +180°) — 360° = 35° = M 145°. Also, L = = = 325° 180° - K = 180° 145° = 51. Inaccessible Intersections.-The intersection P, Fig. 17, of two lines being inaccessible, and the conditions being such that the methods explained in Arts. 38 and 39 cannot be applied, the distances BP and DP and the angle / are determined by running a traverse line B A, A, A, D between two points on the given lines. For convenience, one of the given lines (AB in this case) is taken as a meridian. After measuring the azimuths and lengths of la, la, la, la, the transit is set at D, on CD, oriented on DA,, and the azimuth of DC is measured. The angle I is equal to this azimuth, since AB, or A E, is the meridian. In the polygon BA, A, A, DI, all azimuths and lengths are known, except the lengths of the two sides DP and PB. These sides can be supplied by the methods already explained. It should be carefully borne in mind that the azimuth of DP is equal to the back azimuth of DC, or 180° + I, and that the azimuth of PB is zero. EXAMPLE.-To determine D P and PB from the following measure For the azimuth Z of the closing line DB we have 1. The azimuths of four courses of a deflected traverse that has been run to pass an obstacle in the line of survey are 273° 12′, 49° 56', 312° 15', and 42° 45', respectively. The lengths of the first three courses are 250 feet, 150 feet, and 200 feet, respectively. Determine: (a) the length of the fourth course; (b) the distance between the two points. Ans. (551.1 ft. (a) 416.6 ft. 2. In order to determine the distances of two points B and D, Fig. 17, on the lines A B and C D, from their point of intersection P, a traverse of five lines was run from B to D. The lengths of these lines and their azimuths, taking A B as the meridian, were as follows: = 200 feet Ꮓ = = 102° 45' Azimuth of DC = 113° 37' Determine the dista aces BI and DI and the angle of intersection I. PROBLEMS ON AREAS GENERAL REMARKS ON THE DETERMINATION OF AREAS AREA BOUNDED BY STRAIGHT LINES 52. Division Into Triangles.-When the lengths and bearings (or azimuths) of all the courses of a closed field are known, the best method for calculating the area is by double longitudes, as explained in Compass Surveying, Part 2. If the angles at the different corners have been measured, instead of the bearings or azimuths, any of the sides may be assumed as a meridian; the azimuths or bearings of all the sides with reference to that meridian can be easily determined, the ranges calculated, and the method of double longitudes applied. Another way of computing the area is by dividing the field into triangles. This method is especially adapted to a field surveyed by the chain only, as explained in Chain Surveying. As there stated, the division into triangles should, if possible, be so made that the angles will be neither too acute nor too obtuse. In this respect, the surveyor must use his judgment and exercise some ingenuity. Fig. 18 shows a tract divided into triangles by the diagonals EC, EB, etc. If the chain alone is used, these diagonals must be measured in the field. If a transit or compass is used, it is not necessary to measure the diagonals CE, BE, etc., but, the angles C1, B1, B,, etc. having been measured, the areas T., T1, etc. are computed by the formula given in Plane Trigonometry, Part 2, for the area of a triangle of which one side and the angles are known. When two of the sides of a triangle T, are parts of the boundary and the angle between them has been measured, the area is equal to one-half the product of the two sides and the sine of the included angle. If there is a point inside the field visible from all the corners, the B method illustrated in Fig. 19 may be used. Here O is a point inside the field visible from all the corners. The lengths of the sides AB, B C, etc. having been measured, as well as the C D D angles A,, A,, B1, B,, etc., the areas of the triangles T1, T2, T., etc. are computed by the formula of trigonometry just referred to. For the conditions shown in Fig. 19, a method that is in some cases very expeditious, when it is not necessary to measure the boundary lines of the tract, is to set up the transit at and measure the angles AOB, BOC, COD, etc. and also measure the radial lines OA, OB, O C, etc. The area of each triangle is equal to one-half the product of the two sides and the sine of the included angle. 53. When not very accurate results are required, the area of a field, or of any other figure bounded by straight |