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As the product of the sines of the angles adjacent to

the parallel side, Is to the product of the sines of the angles adjacent to

the opposite side : So is the square of the opposite side, To a fourth term.

Multiply this fourth term by that number of the ratio which corresponds to the part to be adjacent to the parallel side, and to the product add the product of the square of the parallel side by the other number of the ratio; and divide the sum by the sum of the numbers expressing the ratio. The square root of the quotient will be the length of the division line FE. Then,

As the sine of P,
Is to the sine of E;
So is the difference between FE and the parallel side,
To the distance of F from the adjacent end of the

parallel side.

* DEMONSTRATION. Produce DA and CB, Fig. 106, to meet in P. Draw DR and CG each parallel to AB; and let VW, also parallel to AB, make the triangle PVW equal to PCD. Then (15.6) PD: PV :: PW : PC. But (4.6) PD: PV :: DR : VW, and PW: PC :: VW: CG. Therefore (11.5) DR: VW:: VW:CG; and hence (17.6) DR X CG=VW?. But by trigonomctry,

As sin. CRD (sin. B) : sin. C :: CD : DR.

sin. CGD (sin. A): sin D :: CD : CG. Hence (23.6.) sin. Axsin. B : sin. Cx sin. D :: CD2 : DRX CG.

Or, sin. AX sin. B : sin Cx sin. D :: DC2 . VW2

But by the demonstration to the rule in the last problem, we have

m+n. FE?=m. VW2+n. AB2. Hence the truth of the rule is evident.

Construction. From C, Fig. 105, draw CG parallel to AB, and on it de

EXAMPLES.

1 Let the bearing of AB be North, 12 ch.; BC, N. 561° E. 20.78 ch.; CD, S. 331° E. 22.21 ch.; and DA, S. 80° W. 30 ch.; it is required to divide the tract into two parts by a line FE, parallel to AB, so that the part ABEF may be to the part FECD as 3 to 5. Angle A=801°, B=1231°,C=90°, D=66°, and P=24°. . A ,

Ar. Co. 0.00600

0.07889 : sin. Cx sin. D, C, 90 00

10.00000 D. 66.00

9.96073 CD, 22.21

1.34655 :: CD CD, 22.21

1.34655

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Join CA, and draw BH parallel to it meeting DA produced in H. Divide HD in K, so that HK may be to KD in the given ratio of the part ABEF to FECD. Draw KL parallel to BC and LM perpendicular to CG. With the radius CM and centre C, describe the arc MN. Draw NF parallel to KL and FE parallel to AB. Then will FE be the division line. When the division line is to be parallel to CD, the semicircle must be described on CD; and the line CG need not be drawn. The demonstration of the construction is the same as for the last problem.

2. The boundaries of a field ABCD are given as follow : viz. AB, S. 10° W. 7.20 ch.; BC, S. 67° W. 12.47 ch. ; CD, N. 23° W.13.33 ch.; and DA, S. 89° E. 18 ch.; and it is to be divided into two parts by a line FE, parallel to the side AB, so that the part ABEF may be to the part FECD, as 3 to 4. Required the length of the division line FE, and the distance AF.

Ans. FE 10.69 ch,; and AF 7.15 ch. 3. Given the boundaries of a field the same as in the preceding example, to divide it into two parts by a line FE, parallel to the side CD, so that the part ABEF may be to the part FECD, as 3 to 4. Required FE and AF.

Ans. FE 10.14, and AF 10.16.

PROBLEM XVIII.

The bearings and distances of the sides AB, BC, CD, DA,

Fig. 107, of any quadrilateral tract of land, being given, to divide it into two parts having a given ratio by a line FE, running a given course from some point in AD to another in BC.

RULE.

From A, draw AS parallel to BC, meeting FE in S. f'rom the bearings find the angles A, B, C, D, E, and F.* Take the difference between the sum of the angles A and B, and 180° and call it P. Then, As the product of the sines of the angles E and F,

the product of the sines of the angles A and B, : : the square of AB,

a fourth term. Also, As the product of the sines of the angles E and F,

the product of the sines of the angles C and D,

:

:

:

* It is iminaterial whether it is the angle BEF or CEF, that is found, als whether, AFE or DFE.

:

: : the square of CD,

to a fourth term. Multiply this latter fourth term by that number of the ratio which corresponds to the part to be adjacent to AB, and the other fourth term by the other number of the ratio; add the two products together, and divide the sum by the sum of the numbers expressing the ratio. The square root of the quotient will give the length of the division line FE. Then,

As sin. E. : sin. B :: AB : ES. The difference between FE and ES, gives FS. Then,

As sin. P. : sin. E:: FS: AF.*

EXAMPLES,

1. Let the bearing of AB be North, dist. 12 ch.; BC N. 561° E. 20.78 ch.; CD, S. 331° E. 22.21 ch.; and DA, S. 80° W. 30 ch. ; it is required to divide the tract into two parts by a line FE, running N. 20° W. so that the part ABEF may be to the part FECD as 3 to 5.

Angle A=801°, B=1231°, C=90°, D=66°, E=763",
F=79}° and P=24°.
As sin. Ex sin. F,

E, 76° 30 Ar. Co. 0,01217
F, 79 30

0.00733 . sin. A x sin. B, A, 80 30

- 9.99400 B, 123 30

- 9.92111 AB, 12

1.07918 :: AB AB, 12

1.07918

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: fourth term

• 2.09297

123.87

5

619.35

The truth of this rule is evident from the demonstration to the rule in the last problem.

Construction. Draw CG according to the reverse bearing of FE, and then proceed with the construction exactly as in the last problem.

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1413.99
619.35

8)2033.34

FE=V 254.17=15.94 As sin. E, 76° 30 : sin. B, 123 30 : : AB, 12

Ar. Co. 0.01217

9.92111 1.07918

: ES, 10.29

1.01246

As sin. P, 24° 00' : sin. E, 76 30 :: FS, 5.65

Ar. Co. 0.39069

9.98783 0.75205

: AF, 13.51

1.13057

2. The boundaries of a field ABCD are given as follow : viz. AB, S. 103° W.7.20 ch.; BC, S. 67° W. 12.47 ch.: CD, N. 23° W. 13.33 ch.; and DA, S. 89° E. 18 ch. ; and it is to be divided into two parts ABEF and FECD, in the ratio of 3 to 4, by a line FE, running due South. Required the length of the division line FE and the

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