2. Given the bearings and distances of the sides of a tract of land, as follow: 1st. N. 15 W. 9.40 ch.; 2d N. 633 E. 10.43 ch. ; 3d. S. 49° E. 8.12 ch.; 4th. S. 13 E. 8.45 ch.; 5th. S. 163 E. 6.44 ch.; 6th. Unknown; 7th. N. 60° W. 9.72 ch.; 8th. N. 174 E. 7.65 ch.; required the bearing and distance of the 6th. side.

Ans. S. 60° 8' W. 12.27 ch.

3. One side of a tract of land of which a survey is to be taken, passes through a pond. Two stations are therefore taken on one side of the pond as represented in Fig. 80. The bearings and distances from the first end of the side to the first station, from that to the second, and thence to the other end of the side are; 1st. S. 52° W. 10.70 ch.; 2d. S. 7° W. 13.92 ch.; and 3d. S. 34 E. 9 ch. Required the bearing and distance of the side. Ans. S. 10° 33′ W. 28.31 ch.

4. Given the bearings and distances of an old road, running between two places, as follow; 1st. S. 10° E. 92.20 ch.; 2d. S. 15° W. 120.50 ch.; 3d. S. 18 W. 205. ch.; 4th. S. 71 E. 68 ch. Required the bearing and distance of a straight road, that shall connect the two places. Ans. S. 2° 8' W. 423.47 ch.

PROBLEM II.

Given all the bearings and distances of the sides of a survey, except the distances of two sides, to find these.

RULE.

By prob. 9, of the preceding chapter, change all the given bearings, in a corresponding manner, so that one of the sides whose bearings only are given, may become a meridian. With the changed bearings and given distances find the corresponding differences of latitude, and the departures. Add up the eastings and westings,

departure of that unknown side, which is not made a meridian. With this departure and the changed bearing, find by prob. 10, of the preceding chapter, the distance and difference of latitude of this side, which place in their proper columns. Now add up the northings and southings, and take their difference, which will be the distance of the side made a meridian.*

EXAMPLES.

Given the following bearings and distances of the sides of a survey; 1st. S. 45 W. 15.16 ch.; 2d. N. 50° W. 22.10 ch.; 3d. North 18.83 ch.; 4th. N. 85° E. 35.65 ch.; 5th. S. 47° E. dist. unknown; 6th. S. 20 W. dist. unknown; 7th. N. 514 W. 26.47 ch. to the place of beginning. Required the unknown distances.

no departure, the difference of the sums of the departures, must be the departure of the other unknown side. And when the difference of latitude of this side has been found and placed in its proper situation, the difference of the sums of the latitudes must evidently be the difference of latitude of the

0.95945

: cotang, chang. bearing 671

: diff. lat. 6th. side . - 9.11

Ans. 5th. side 29.02 ch. and 6th. side 23.80 ch.

2. Given the bearings and distances of a tract of land as follow: 1st. S. 40 E. 31.80 ch.; 2d. N. 54° E. dist. unknown; 3d. N. 294 E. 2.21 ch.; 4th. N. 284 E. 35.35 ch.; 5th. N. 57° W. dist. unknown; 6th. S. 47° W.31.30 ch.; to the place of beginning. Required the distances of the 2d. and 5th. sides.

Ans. 2d. side. 2.08 ch. and 5th. side 20.90 ch.

PROBLEM III.

Given the bearings and distances of all the sides of a sur vey except two; one of which has only its bearing given, and the other, the distance and the points of the compass between which it runs; to find the unknown bearing and distance.

RULE.

As in the last problem, change all the given bearings, so that the side whose bearing only is given, may become a meridian. Find the differences of latitude and the departures, corresponding to the changed bearings and the given distances. Take the difference of the

parture of the side whose bearing is not given. With the given distance and this departure, find by chap. 1. prob. 10. the changed bearing and difference of latitude, and place them in their proper columns. From the changed bearing, the true bearing may be readily found by note to prob. 9. chap. 1. Lastly, take the difference of the sums of the northings and southings, and it will be the distance of the side, changed to a meridian.

Note. The changed bearing as found by the rule, must be reckoned from the north, or the south point of the compass, according as the one, or the other, will render the true bearing when found from it, conformable to the given points. The point from which the changed bearing must be reckoned determines also the column in which the difference of latitude must be placed. Sometimes the changed bearing when reckoned from either north or south, will render the true bearing conformable to the given points. In such cases, there are two different bearings and distances that will answer the conditions of the problem; and we can only know which of them is the right one by previously knowing the required bearing nearly.

EXAMPLES.

1. Given the bearings and distances of a survey as follow: 1st. S. unknown W. 15.16 ch.; 2d. N. 50° W. 22.10 ch.; 3d. N. 18.83 ch.; 4th. N. 85° E. 35.65 ch. ; 5th. S. 47° E. 29.02 ch.; 6th. S. 20 W. dist. unknown, 7th. N. 51° W. 26.47 ch. Required the unknown bearing and distance.

Ans. 1st. S. 45° 36′ W.; 6th, 23.81 ch.

2. Given the following bearings and distances of a survey: 1st. S. 40° E. 31.80 ch.; 2d. N. 54° E. dist.

unknown 3rd. N. 29° E. 2.21 ch.; 35.35 ch.; 5th. N. 57° W. 20.90 31.30 ch.; to place of beginning.

4th. N. unknown E. ch.; 6th. S. 47° W. Required the bearing

of the 4th. side and distance of the 2d. side.

Ans. Bearing of 4th. side N. 28° E.; dist. of 2d