quired to cut off 10 acres by a line HI, running S. 14 30 E.

AG, N 80°30'W GE, S 80°30'E BG, S 58°00′W
AB, N 0 00 W GF, N 58 00 E BA, S 000W

:: twice the area to be cut off, 200 sq. ch.

: a fourth term 259.53

GA2=235.87

2.30103

2.11119

2. Given the bearings of three adjacent sides of a tract of land and the length of the middle one as follow; 1st. N. 31° 15′ W. 2nd. N. 58° 45' E. dist. 13.50 ch. 3rd. S. 14° 45′ E. to cut off 8 acres by a line running N. 87° 30′ W.; required the distance on the first side. Ans. 2.67 ch.

3. Given as follow; 37° E. dist. 17.24 ch.

1st side, N. 74° 45′ W. 2nd. N. 3rd. N. 84° E., to cut off a field containing 20 acres by a line running S. 20° W.; the distance on the 1st side is required. Ans. 14.01 ch.

PROBLEM XII.

The bearings of several adjacent sides, LA, AB, BC, CD, DE, EF, Fig. 98, of a tract of land, and the distance of each, except the first and last, being given, to cut off a given area by a line IH running a given course from a point somewhere in AL, and falling on EF.

RULE.

Suppose a line drawn from A to E, and calculate the

Subtract the area of ABCDEA from the area to be cut off, the remainder will be the area of AEHI.

Then having the bearings of LA, AE, EF, HI, the distance AE and the area of AEHI, find AI by the preceding problem.*

EXAMPLES.

1. Let the bearing of LA, Fig. 98, be N. 48° 30′ W. AB, S. 78° 00′ W. dist. 8 ch. BC, N. 26° 30′ W. dist. 11.08 ch. CD, N. 38° 30′ E. 64° 00′ E. dist. 10.86 ch., and required to cut off 30 acres by a line HI running S. 32°

15' W.

dist. 12.82 ch. DE, S. EF, S. 86° 00′ E.; it is

tang. of bearing of EA, S. 20° 9' W.

9.56450

Now we have given the bearing of LA, N. 48° 30′ W. AE, N. 20° 9′ E. dist. 14.40 ch., and EF, S. 86° 00′ E. to cut off a trapezium AEHI containing 82.3 sq. ch. by a line HL running S. 32° 15′ W. Hence by the preceding problem we find the distance AI=3.51 ch.

2. Given as follow; 1st side, N. 62° 15′ W. 2nd. N. 19° 00′ E. dist. 18 ch. 3rd. S. 77° 00′ E. dist. 15.25 ch. 4th. S. 27° 00′ E., to cut off 35 acres by a line running S. 82° 30′ W., from a point somewhere in the last side and falling on the first; required the distance on the first side. Ans. 5.14 ch.

PROBLEM XIII.

The bearings of several adjacent sides, AB, BC, CD, DE, Fig. 99, of a tract of land, and the distance of each, except the last, being given, to cut off a given area by a line AH running from the angle A and falling on the side DE.

RULE 1.

Suppose a line drawn from A to D, and calculate the

subtract the area of ABCDA from the area to be cut off, the remainder will be the area of ADHA. Also from the bearings of DA and DE, find the angle ADE.

Then having the angle ADH, distance AD and area ADHA, find DH by problem 5th.

With AD, DH and the included angle ADH, find, by case 3, trig. the angle DAH and distance AH; the an

gle DAH, applied to the bearing of AD, will give the bearing of AH.*

EXAMPLES.

1. Let the bearing of AB, be N. 62° 15′ W. dist. 14.75 ch. BC, N. 19° E. dist. 27 ch. CD, S. 77° E. dist. 22.75 ch., and DE, S. 27° E.; it is required to cut off 70 acres by a line AH, running from, the angle A and falling on the side DE.