FIGURES FIGURE 6. Partition of tract.
DMDEC = 143.6 + 143.6 + 54.9 = 342.1 ft (104.27 m); DMDCD = 342.1 + 54.9 - 198.5 =
198.5 ft (60.50 m). This is acceptable.
- Determine angle GCE by finding the bearings of courses
EC and BC
Thus tan OLEC = 54.9/259.8; bearing of EC = S11°55.9'E; tan OLBC = 77.5/9.6; bearing of
5C = N82°56.3'W; angle GCE= 180°-(82°56.3' - 11°55.9') = 108°59.6'. - Determine the area of triangle GCE
Calculate the area of triangle DEC; then find the area of triangle GCE by subtraction.
Thus
Course Latitude x DMD = 2 x Projection area
CD +97.9 198.5 +19,433
DE +161.9 143.6 +23,249
EC -259.8 342.1 -88,878
Total -46,196So the area of DEC= '/2(46,196) = 23,098 ft^2 (2145.8 m^2 ); area of GCE = 30,700 - 23,098
= 7602 ft^2 (706.22 m^2 ).
- Solve triangle GCE completely
Apply Eqs. 4 and 5. To ensure correct substitution, identify the corresponding elements,
making A the known angle GCE and c the known side EC. Thus
Fig. 5 Fig. 6 Known values Calculated values
A GCE 108°59.6'
B CEG 11°21.6'
C EGG 59°38.8'
a EG 291.0 ft (88.7Om)
b GC 60.6 ft (18.47m)
c EC 265.5 ft (80.92m)