In the preceding examples the bearings and distances of all the boundaries are given ; but when the field-work is accurately performed, the area may be calculated, if any two of the bearings or distances, or one bearing and distance be omitted. The method of doing this in the cases most likely to occur in practice, is exemplified in the three following examples. Either of the preceding rules may be used in the calculation.

EXAMPLE 1.

In taking a survey of a tract of land bounded by six straight sides, I was prevented going directly from the 3rd to the 4th corner by a pond of water. I therefore set up two stakes near the edge of the pond, and took the bearing and distance from the 3rd corner to the first stake, from the first stake to the second, and from the se. cond to the 4th corner, and noted them in my field-book as all belonging to the 3rd station of the survey. The field-notes being as follow, the bearing and distance of the 3rd side, and the area of the survey are required.

Ch. 1. North,

7.81 2. S. 76°1 E. 18.15

S. 52 W. 10.70 3, s. 7; W. 13.92

S. 331 E. 9.00 4. N. 841 W. 27.12 5. N. 41 W. 22.00 6. East,

16.58 To find the bearing and distance of the 3rd side, Fig. 80.

Find the difference of latitude and departure for each of the devious courses, EA, AB, and BC. Then the difference between the sums of the north and south latitudes, and the difference between the sums of the east

2

and departure corresponding to the 3rd side, and of the same name with the less sums respectively.

Dep. E. 5.32
Draw CD parallel to NS, and on it let fall the perpen-
dicular ED; then will CD be the difference of latitude,
and ED the departure corresponding to the 3rd side, and
the angle DCE will be the bearing, which will be between
the north and east in going from C to E. Therefore, by
trigonometry,

As diff. of lat. CD = 27.92 N. 1.44592
Is to the dep. ED= 5.32 E. 0.72591
So is rad.

10.00000

10.72591
* To the tang. of DCE, or bearing of
CE, N. 10° 47' E.

9.27999
Consequently the bearing of EC is S. 10° 47' W.
As rad.

10.00000 Is to sec. of DCE 10° 47'

10.00774 So is the diff. of lat. CD = 27.92 1.44592

11.45366

To the dist. EC 28.42

1.45366 The bearing and distance of the 3rd side is therefore S. 10° 47' W. 28.42 Ch.

R

4N 841 W27.12 2.72

26.981

27.02 2.56 W

5)N

41 W 22.00 21.93

1.73

1.76131.34 W

.05 .04) 2.67

6.8352

29.58 W ,04 .03|| 21.89

686.0326

133.10 W ,03 .03

.03| 16.55

16.55 W 0.4965

0.00 .221 .18|| 32,36) 32.361 34.151 34.15

10.4965 1603.66001

.4965 2)1603.1635

801.58715 Ch. Ans. Area. 80 A. O R. 25 P.

130

EXAMPLE 2.

In a survey of which the following are the field-notes, the bearing and distance of the last side were not taken on account of obstacles in the way; but depending on the accuracy of the others, it is required to find them and the area

of the survey

Ch. 1. N. 60 W. 9.72 2. N. 171 E. 7.65 3. N. 15. W. 9.40 4. N. 63: E. 10.43 5. S. 49 E. 8.12 6. S. 131 E. 8.45 7. S. 16. E. 6.44 8.

To find the area. With the given bearings and distances find their corresponding latitudes and departures, and what they want of balancing will be the difference of latitude and departure of the closing line. The area may then be found as in the preceding examples.

By Rule 3.

With the difference of latitude and departure of the closing line, its bearing and distance may be found as in the preceding example. Thus,

The bearing and distance of the last side is therefore S. 60° 8' W. 12.27 ch.