By Eq. 4, 7602 =^1 / 2 GC(265.5 sin 108°59.6'); solving gives GC = 60.6 ft (18.47 m). By
Eq. 5, tan EGC = 265.5 sin 108°59.6V(60.6 - 265.5 cos 108°59.6'); EGC = 59°38.8'. By
the law of sines, £G/sin GCE = EC/sin EGO, EG = 291.0 ft (88.70 m); CEG = 180° -
(108°59.6' + 59°38.8')=11°21.6'.
- Find the bearing of course EG
Thus, OLEG = OLEC + CEG = 11°55.9' + 11°21.6' = 23°17.5' bearing of EG = S23°17.5'E.
The surveyor requires the length and bearing of EG to establish this line of demarca-
tion. She or he is able to check the accuracy of both the fieldwork and the office calcula-
tions by ascertaining that the point G established in the field falls on BC and that the
measured length of GC agrees with the computed value.
AREA OF TRACT WITH MEANDERING
BOUNDARY: OFFSETS AT
IRREGULAR INTERVALS
The offsets below were taken from stations on a traverse line to a meandering stream, all
data being in feet. What is the encompassed area?
Station 0 + 00 0 + 25 0 + 60 0 + 75 1+010
Offset 29.8 64.6 93.2 58.1 28.5Calculation Procedure:
- Assume a rectilinear boundary between successive offsets;
develop area equations
Refer to Fig. 7. When a tract has a meandering boundary, this boundary is approximated
by measuring the perpendicular offsets of the boundary from a straight line AB. Let dr de-
note the distance along the traverse line between the first and the rth offset, and let /Z 1 , h 2 ,
..., hn denote the offsets.
Developing the area equations yields
Area = V^d 2 (H 1 - A 3 ) + djfa - A 4 ) + - - • + dn^(hn_ 2 - hn) + dn(hn^ + /*„)] (6)Or,
Area = Y 2 [^d 2 + h 2 d 3 + h 3 (d 4 - d 2 ) + h 4 (d 5 - J 3 ) + • - - + hn(dn - ^ 1 )] (7)- Determine the area,
using Eq. 6
Thus, area = l/ 2 [25(29.% - 93.2) +
60(64.6 - 58.1) + 75(93.2 - 28.5) +
110(58.1 + 28.5)] = 6590 ft^2 (612.2 m^2 ). - Determine the area,
using Eq. 7
Thus, area = ^[29.S x 25 + 64.6 x 60 +
93.2(75 - 25) + 58.1(110 - 60) + FIGURE 7. Tract with irregular boundary.
Traverse lineBoundaryOffset