As diff. lat of HA. 5.22 S.

: dep. do. 28.33 W. :: rad.

Ar. Co. 9.28233

1.45225 10.00000

: tang. changed bearing of HA. S. 79° 34' W. 10.73458

Subtract 27 00

Proper bearing of HA, S. 52 34 W. As rad.

Ar. Co. 0.00000 : sec. changed bearing of HA, 79° 34' 10.74210 :: diff. lat. 522

0.71767

: dist. AH. 28.83

1.45977

Hence AH bears N. 52° 34' E. dist. 28.83 ch. the same as found by the preceding rule very nearly.

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2. Given as follow : 1st side. S. 780 W. 8 ch. 2nd. N. 26°1 W. 11.08 ch. 3rd. N. 38; E. 12.82 ch. 4th. S. 64o 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 the bearing and distance of the division line. 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 DE, running parallel to one of the sides as BC.

RULE.

As the sum of the numbers expressing the ratio of the

parts, Is to the greater or less of them, according as the greater

So is the square of AB,
To the square of the distance AD*.

Note. In like manner the square of the distance AE may be found by taking AC for the third term instead of AB,

EXAMPLES.

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

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

beginning, measured on the first side ? Ans. 10.61 ch.

/

* DEMONSTRATION. Let m to n be the ratio of the part DBCE to the part ADE;

then (18.5.) m+n: n :: ABC : ADE :: (19.5.) ABP : AD%. Construction. On AB describe the semicircle AMB and divide AB in L, so that BL may be to AL in the given ratio of the part DBCE to the part ADE; draw LM perpendicular to AB, meeting the semicircle in M, and make AD =AM; parallel to BC, draw DE which will divide the triangle in the given ratio. For (31,3, and cor. 8.6.) AB : AM (AD) :: AM (AD): AL, or (206, cor. 2.) AB : AL :: ABS : AD: :: (19.6.) ABC:ADE ; therefore (17,5.) BL: AL :: DBCE: ADE.

:

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 EF running a given course.

Let DG, parallel to BC, be a line dividing the triangle in the given ratio and by the preceding problem find the square

of the distance AD. Then,
As the rectangle of the sines of the angles E and G,
Is to the rectangle of the sines of the angles D

and F;
So is the square of AD,
To the square of AF*.

EXAMPLES.

1. Let the bearing of AB, be S. 82°1 E. dist. 14.17 ch. BC, N. 18°1 W. 8.51 ch., and CA, S. 61°4 W. 12.87 ch. ; it is required to divide the triangle by the line FG, running N. 14° E., so that the part FBCG may be to the part AFG as 3 to 2.

• The demonstration of this rule is the same as of that in prob. XI.

Construction. Divide AB, Fig. 102, in K, so that AK may be to KB in the given ratio of the part AEF to the part EBCF; from C, draw CI according to the given bearing of the division line, on AI describe the semicircle ALI, make KL perpendicular to AB, meeting the semicircle in L, and take AE= AL ; then parallel to IC, draw EF which will divide the triangle as required. For, join KC, AL and LI, then it is evident (1.6.) that KC divides the triangle in the given ratio ; therefore it will only be necessary to prove that the triangle AEF is equal to the triangle AKC. Now (31.3,and cor. 8.6) AI: AL (AE):: AL (AE): AK, or (20.6. cor.2.) AI: AK :: AIP: AER :: (19.6.) AIC : AEF; but (1.6.) AI: AK :: AIC : AKC; therefore AIC : AEF :: AIC : AKC, and consequently AEF=A&C.

7

From the bearings of the lines are found the angle ADE=63° 30', AED=80° 15', AFG=97° 00' and AGF=46° 45'

2. The bearings and distances of a triangular piece of land ABC are, AB, S. 69° E. 21.40 ch. BC, N. 31°į E. 18.66 ch. and CA, S. 74°; W. 30.85 ch., and it is required to divide it by a line FG running due north, so that the part FBCG may be to the part AFG as 5 to 4; what will be the distance AF? Ans. 17.40 ch.

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 EF running parallel to the parallel sides AB, CD.

RULE.

Suppose the sides DA, CB, produced to meet in G,

and in the triangle GDC having given all the angles and side DC, find the side GD; from GD subtract AD and the remainder will be GA. Then,

As the sum of the numbers expressing the ratio of

the parts,

Is to the greater or less of them, according as the

greater or less part is to be adjacent to DC; So is the rectangle of AD and sum of GD and GA, To a fourth term.

Subtract this fourth term from the square of GD; the square root of the remainder will give GE, and GE less GA will be the distance AE*.

EXAMPLES.

1. Let the bearing of AB be N. 14° E. dist. 10 ch. BC, N. 5501 E. 18.67 ch. CD, S. 14° W. 20.98 ch. and DA,

# DEMONSTRATION. Let m to n be the ratio of the part AEFB to the part EDCF ; then (18.5.) m+n:n:: ADCB: EDCF :: (19.6. and division) GDE. GAR:GDS–GER; hence because (cor.5.2.) GDS-GAS=GD + GA. X AD, the truth of the rule is evident.

.

Construction. Produce DA, CB, Fig. 104, to meet in G and on GD describe the semicircle GLD; join AC and parallel to it draw BH meeting GD in H ; make HK to KD in the given ratio of AEFB to EDCF, and draw KL perpendicular to GD, meeting the semicircle in L; take GE equal to GL and paral. lel to AB or DC, draw EF which will divide the trapezoid as required. For join KC, HC, then (37.1.) the triangle AHC is equal to the triangle ABC ; to each of these add the triangle ADC, then it is plain that the triangle HDC is equal to the trapezoid ADCB ; but (1.6.) the line KC divides the triangle HDC in the given ratio : and from the demonstration to the construction of the last problem, it is manifest that the triangles GKC and GEF are equal ; consequently KDC is equal to EDCF. Whence the truth of the construc. tion is evident.

:

Cor. If HM be drawn perpendicular to GD, and DM, DL be joined, we shall have DGS: DMR :: DEFC: DABC. For GD: GA :: GC : GB :: GA : GH; hence GA'=DGXGH=GM”, and GDP-GMP (GA)=DM2; also D