CASE 3.

397. When the starting point is in the line AD.

Given as before to run the line from a point G in AD at 4.97 chains from A.

Produce DA and BC (Fig. 186) to meet in P. Calculate AP: let the given ratio be represented by r: then, As sin. P (36° 10′): sin. ABC (81° 55′) :: AB (24.10): AP = 40.432.

Put

and

Lay off GD = A±√‡ A2 + M2.382 + 5.900 = 6.282, (the lower sign being used when G is between A and P.) Then GH parallel to DB will be the division line.

DEMONSTRATION.-Since GD = } A + √} A' + M3,

we have GD } A = √} A® + M3, and GD'

A. GD =

M3,

or GD (GD — A) = A . PG; whence PG: DG :: DG — A : A,

and composition, PD: DG :: DG: A

whence

:: AP. DG: r. AGa;

r. PD. AG" = AP. DG3,

and r. AG3: DG3 :: AP : PD :: PC: PB :: PF: PA :: AF: AD,

or, r. AE2 : EB2 :: AF: AD. As this agrees with (A) in the demonstration to Case 1, the truth of the work is clear.

Having found AD, the bearing of DB, which is parallel to GH, may be found by calculating the angle ADB; thus: As (AB + AD) 35.352 : (AB AD) 12.848: tan. ADB + ABD

2

30° 57′ tan.
:

ADB · ABD

2

= 12° 17' 55".

Whence the angle ADB is 43° 15′ 25′′, and the bearing of DB or GH is S. 89° 59' 35" E.

The whole of the preceding construction might be made geometrically, but some of the lines required would be so small that no dependence could be had on the work; the method is therefore omitted.

If the given point were not on one of the lines, the problem becomes very complicated. It may, however, be solved by running "guess-lines."

SECTION II.

DIVISION OF LAND.

Problem 1.-To divide a triangle into two parts having a given ratio.

CASE 1.

398. By a line through one of the corners.

Divide the base into two parts having the same ratio as the parts into which the triangle is to be divided, and draw a line from the point of section to the opposite angle, (1.6).

EXAMPLES.

Ex. 1. A triangular field ABC contains 10 acres, the base AB being 22.50 chains. It is required to cut off 4 acres towards the point A by a line CD from the angle C. What is the distance AD?

Calculation.

As 10: 41: AB (22.50): AD = 10.125 chains.

Ex. 2. The area of a triangle ABC is 7 acres, the side AC being 15 chains. To determine the distance AD to a point in AC, so that the triangle ABD may contain 3 acres. Ans. AD: = 6.43 chains.

CASE 2.

399. By a line through a given point in one of the sides.

Say, As the whole area is to the area of the part to be cut off, so is the rectangle of the sides about the angle towards which the required part is to lie, to a fourth term.

This fourth term divided by the given distance will give the distance on the other side.

DEMONSTRATION.-Let ABC (Fig. 187) be the given triangle, and ADE the part cut off. Then we shall have (Art. 357) rad. sin. A: AB. AC: 2 ABC, and rad. : sin. A :: AD. AE: 2 ADE; wherefore 2 ABC: 2 ADE :: AB. AC: AD. AE, or ABC: ADE :: AB. AC: AD AE.

EXAMPLES.

=

Ex. 1. Given the side AB = 25 chains, AC 20 chains, and the distance AD 12 chains, to find a point E in AB, such that the triangle cut off by DE may be to the whole triangle as 2 is to 5.

whence

Calculation.

As 5:2:: AB. AC (500): AD. AE (200);

Ex. 2. Given AB = 12.25 chains, AC = 10.42 chains, and the area of ABC = 5 A. 3 R. 8 P., to cut off 3 acres towards the angle A by a line running through a point E in AB 8.50 chains from the point A. Required the distance on AC. Ans. 7.77 chains.

CASE 3.

400. By a line parallel to one of the sides.

Since the part cut off will be similar to the whole, say, As the whole area is to the area to be cut off, so is the square of one of the sides to the square of the corresponding side of the part.

The problem may be constructed thus: Let ABC (Fig. 188) be the given triangle. Divide AB in F, so that AF may be to FB in the ratio of the parts into which the triangle is to be divided. Take AD

A

Fig. 188.

E

F D

C

B

a mean proportional between AF and AB. Then, DE parallel to BC will divide the triangle as required.

For AFC FCB:: AF: FB, and (lemma) ADE = AFC; therefore ADE: DECB:: AF: FB.

=

EXAMPLES.

Ex. 1. The three sides of a triangle are AB = 25 chains, AC 20 chains, and BC = 17 chains, to divide it into two parts ADE and DECD, having the ratio of 4 to 3, by a line parallel to BC.

Say, As 7:4:: AB2 (625): AD2 = 357.1428;

whence

=

Ex. 2. The three sides of a triangle are AB = 25 chains, AC 20 chains, and BC = 15 chains, to divide it into two parts ADE and DECB, which shall be to each other as 2 to 3, by a line parallel to BC. What is the distance on AC to the division line? Ans. 12.65 chains.

CASE 4.

401. By a line running a given course.

Construction.-Divide AB in G, (Fig. 189,) so that AG may be to GB in the ratio of the parts of the triangle. Run CF according to the given course. Take AD a mean proportional between AF and AG. Then DE parallel to CF is the division line.

For ACG CGB:: AG: GB, and, by the lemma, ADE = ACG.

..

ADE: DECB:: AG: GB.

Calculation.

In ACF find AF. Then AD = ✔AG. AF; or say, As the rectangle of the sines of D and E is to the rectangle of the sines of B and C, so is the square of BC to a fourth

term.

Then, if the ratio of the parts is to be as m to n, m corresponding to the triangular portion, multiply this fourth term by m, and divide by m +n: the quotient will be the square of DE. Whence AD is readily found.

DEMONSTRATION.-Draw zy parallel to CF, making Axy ABC, and draw BR parallel to xy. Then, as was shown in Art. 385, sin. D. sin. E: sin. B . sin. C:: BC2: xy3, and (Cor. 2, 20.6) Axy : ADE or m + n : m :: xya : DEa

EXAMPLES.

Ex. 1. The bearings and distances of the sides of a triangular plat of ground are AB N. 71° E. 17.49 chains, BC S. 15° W. 12.66 chains, and CA N. 63° W. 14.78 chains, to divide it into two parts ADE and DECB, in the ratio of 2 to 3, by a line running due north. The distance AD is