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and 17 chains 97 links from B to b, the line ab will cut off 100 acres and be parallel to AB.

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ld A f

B Put AB=a; Co. tang. of A=s ; Co. tang. of B=r, and the perpendicular af=x; then as 1: x:: :: sx=Af; and as 1 : x::r:rx=Bd; therefore, fd, or its equal ab is=2sx+rx and į AB +į aba— sx+1 rx put n=smr; and R for the area to be cut off,

2a Then ax-nx=R; hence we have x?-- and

2R

X

n

n

by completing the square x=a—a?_2nR which gives X=17.92; hence Aa=20. 11, and Bb=17.97, the same as before.

Note. If the angles at A and B were both obtuse,

then would n=8+r, and x =a*+2nR -a but if one be

1 obtuse, and the other acute, and the Co.tang. of the ob

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tuse angle the greater, then will n=8 Sr, and *=

3 a' +2nR-las before ; but if both be acute, n-8+r

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and x=a-a-2nR and these theorems will afford

the most expeditious solution that the above useful problem can admit of.

Note 2. The quantity aʼ+2nR or a?—2nR expresses the length of the division line ab.

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To divide off 50 acres by a right line, parallel to DC in the same survey with the preceding examples.

Here the angle D=64 45' its Co. tang. =-1.471630

C=138° 00' its Co. tang. =-1.110606

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EXAMPLE III.

It is required to cut off 80 acres adjoining and parallel to BC, in the same survey.

Here the angle B is 94° 15', Co. tang.

angle C is 138°00', Co. tang.

.07431 1.11061

therefore n = 1.18492

a=39.20; and a2 +2nR=ab = 58.582

2 a! hence

Max is found=16.36

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To divide off 120 acres, adjoining and parallel to AD.

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As S. < D:X:: Rad. : Db= 18.15
As S. LA:x:: Rad.: Aa= 18.43

If it were required to cut off from a given farm any number of acres by a line, having a given bearing, the problem might be solved from the precepts already given; for by drawing a line from any convenient angle in the direction required, find its length, by the foregoing problem, and then the content it incloses ; which done, the last problem will give the position of the line required.

EXAMPLE

Let it be required to cut off 100 acres from the N. end of the foregoing survey, by a line running north

From C, draw CE, North East; then since DC is 8. 37° 45' E. the angle DCE must be 7° 15' and the angle D is 64° 45', therefore angle E is 108° 00'.

As S. E.: DC :: S. D: CE = 39.94 = a
hence the triangle DCE = 105.85 chs.

area to be cut off = 1000.
difference

891.15 ch=R
Co. tang. E 72° is =.32492
Co. tang. C 430 43' =.86466
Therefore n = .53674
2 n R = 959.85214

a = 1595.2036
a’+2nR* =50.55 nearly
a+2nRX-a=x=19.761

n
hence Ed = 20.78

cb = 26.08

39

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