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: tang. changed bear. of HA, S. 79° 34′ W. 1073458

Subtract 27 00

16998

As rad.

: sec. changed bearing of HA, 79° 34' :: diff. lat. 5.22

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1

Hence AH, bears N. 52° 34′ É. dist. 28.83 ch.

2. Given as follow: 1st. side S. 78° W. 8 ch.; 2d. N. 261° W. 11.08 ch.; 3d. N. 38° E. 12.82 ch.; 4th. S. 64° E. 10.86 ch.; 5th. S. 231°E.; to cut off 25 acres by a line running from the place of beginning, and falling on the 5th. side; required its bearing and distance.

Ans. N. 45° 1' E. dist. 10.67 ch.

PROBLEM XIV.

The sides AB, BC, CA, Fig. 100, of a triangular piece of ground being given, to divide it into two parts having a given ratio, by a line FE, running parallel to one of the sides as BC.

RULE.

As the sum of the numbers expressing the ratio of the parts,

Is to that number of the ratio which corresponds to the part to be adjacent to A;

So is the square BC,

To the square of FE.

Then, As BC: AB :: FE: AF.*

* DEMONSTRATION. Let m to n be the ratio of the part AFE to the par^ FECB; then (18.5) m+n:m:: ABC: ADE: (19.6) BC2 : FE2.

Construction. On AB describe the semicircle AMB, and by Prob. 17, Page 34, divide AB in K, so that AK may be to KB in the given ratio of the

EXAMPLES.

1. Let AB be 21.26 ch. ; BC, 12.76 ch.; and AC, 19.30 ch. ; it is required to divide the triangle by the line FE, parallel to BC, so that the part AFE may be to the part FECB as 2 to 3.

As 52
FE

As 12.76

12.762 : FE-65.12704. 65.12704=8.07

21.26: 8.07: AF-13.45.

2. The three sides of a triangular piece of land, taken in order, measure 15, 10, and 13 chains respectively; it is required to divide it into two equal parts by a line parallel to the second side. What will be the length of the division line and its distance from the place of beginning, measured on the first side?

Ans. Division line 7.07 ch.; dist. on 1st. side 10.61 ch.

PROBLEM XV.

The bearings and distances of the sides AB, BC, CA, Fig. 101, of a triangular piece of ground being given, to divide it into two parts having a given ratio, by a line FE, running a given course.

RULE.

As the product of the sines of F and E,
Is to the product of the sines of B and C ;

micircle in M, and with the radius AM and centre A, describe the arc MF. From F, draw the division line FE parallel to BC. Since, (35.3, and cor. 8.6) AB: AM (AF) :: AM (AF): AK, we have (20.6 cor. 2) AB : AK:: AB

1

So is the square of BC,

To a fourth term.

Multiply this fourth term by that number of the ratio which corresponds to the part to be adjacent to the angle A, and divide the product by the sum of the numbers expressing the ratio. The square root of the result will be FE. Then,

As sin. A sin. E:: FE: AF.*

:

EXAMPLES.

1. Let the bearing of AB, be S. 82° E. dist. 14.17 ch. ; BC, N. 183° W. 8.51 ch.; and CA, S. 614° W. dist. 12.87 ch.; it is required to divide the triangle by the line FE, running N. 143° E. so that the part AFE may be to the part FECB in the ratio of 2: 3.

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* DEMONSTRATION. Draw CG and BR, Fig. 102, each parallel to EF; and let VW, also parallel to EF, make the triangle AVW equal to ABC. Then (15.6) AB: AV:: AW: AC; but (4.6) AB: AV:: BR: VW, and AW : AC:: VW: GC. Therefore BR: VW:: VW: GC; and hence (17.6) BRX GC But by trigonometry,

VW2.

As sin. R (sin. E): sin. C:: BC: BR,

sin. G. (sin. F): sin B:: BC: GC.

Hence (23.6) sin. Ex sin. F: sin. Bx sin. C:: BC2: BR× GC :: BC2 VW2.

Also, m+n: m :: ABC: AFE :: AVW: AFE :: (19.6) VW2 : FE2.

Construction. From C, Fig. 101, draw CG according to the reverse bearing of FE, and on AG describe the semicircle AMG. By prob. 17, page 34, divide AB in K, so that AK may be to KB in the given ratio of the part AFE to the part FECB. Draw KM perpendicular to AB, and with the radius AM and centre A, describe the arc MF. From F, and parallel to GC, draw FE the required division line. For join KC, Fig. 102; then (1.6) KC divides the triangle in the given ratio. Now AC: AE:: AG: AF (AM) .: (cor. 9.6) AM (AF): AK; therefore (11.5) AC : AE :: AF: AK, and hence (15.6) the triangle AFE is equal to AKC. Consequently FE divides the triangle in

Angle A=364°, B=634°, C=801°, E=463° and F=83°. As sin. Ex sin. F, SE, 46° 45′

F, 83 00

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Ar. Co. 0.13765

0.00325

- 9.95179

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2. The bearings and distances of a triangular piece of land ABC are, AB, S. 69° E. 21.40 ch.; BC, N. 311⁄2 E. 18.66 ch.; and CA, S. 744 W. 30.85 ch.; and it is required to divide it by a line FE, running due north, so that the part AEF, may be to the part FECB, as 4 to 5. What will be the length of the division line FE, and the distance AF? Ans. FE 10.74, and AF 17.40.

PROBLEM XVI.

The bearings and distances of the sides AB, BC, CD, DA, fig. 103, of a trapezoidal tract of land being given, to divide it into two parts having a given ratio, by a line FE, running parallel to the parallel sides AB, CD.

RULE.

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