Figure 15-29.-Bearing and distances of a quadrilateral.Thus we have, by computation of two routes, valuesThe following paragraphs explain how to determine thefor BC of 433.322 feet and 433.315 feet. There is adiscrepancy here of 0.007 feet For third-order work thiswould usually be considered within tolerable limits; andthe computed value of BC would be taken to be theaverage between the two, or (to the nearest 0.01 foot)433.32 feet.Suppose, now, that the precision requirements forthe base line check are 1/5,000. This means that the ratiobetween the difference in lengths of the measured andcomputed base line must not exceed 1/5,000. Youmeasure the base line BC and discover that it measures433.25 feet. For a ratio of error of 1/5,000, the maximumallowable error (discrepancy between computed andmeasured value of base line) is 433.25/5,000, or 0.08feet. The error here is (433.32 – 433.25), or 0.07 foot,which is within the allowable limit.LOCATIONS OF POINTSThe end result desired in a triangulation survey isthe horizontal locations of the points in the system, bybearing and distance. Methods of converting deflectionangles to bearings and converting bearings to exterioror interior angles are described in the EA3 TRAMAN.bearings of lines of a quadrilateral.Bearing and DistanceFigure 15-29 shows the quadrilateral we have beenworking on, with the computed values of the sidesinscribed. Take station D as the starting point. Supposethat, by an appropriate method, you have determined thebearing of DA to be N15°00´00´´E, as shown. To have agood picture of how you proceed to compute for thebearing of the next line, AB, you must superimpose themeridian line through the starting point, laying offapproximately the known bearing; in this case,N15°00´00´´E. Now draw your meridian through pointA. From figure 15-29 you can see that the line AB bearssoutheast, and you can find its bearing by subtracting15°00´00´´ from angle A. Angle A is the sum of angles 1and 2 (38°44´08´´ + 23°44´35´´), or 62°28´43´´, as youshould recall from figure 15-28. The bearing angle ofAB, then, is 62°28´43´´ – 15°00´00´´, or 47°28´43´´.Therefore, the complete bearing of line AB isS47°28´43´´E.15-39