ducts in each of those columns, subtract the less sum from the greater, and take the difference between the remainder and double the area to be cut off; divide this difference by the upper meridian distance corresponding to the side DH, on which the division line is to fall, the quotient will be the latitude of the side DH, which place against it in the column of north or south latitude according as its bearing is north or south; add together the numbers in each latitude column, and take the difference of their sums, which will be the latitude of the division line AH, and of the same name with the less sum; then with the latitude and departure of AH, find, by trigonometry, its bearing and distance. Change the bearing thus found by applying to it the angle expressing the bearing of the line, made a meridian, in a manner contrary to that in which it was applied in changing the original bearings, and it will give the proper bearing of the division line*. EXAMPLES. 1. Let the bearing of AB be N. 62°1 W. 14.75 ch. BC, N. 19° E. 27 ch. CD, S. 77° E. 22.75 ch. and DE, S. 27° E.; it is required to cut off 70 acres by a line AH, running from the angle A and falling on the side DE. * The truth of this rule is too evident to need a demonstration. : tang. changed bearing of HA. S. 79° 34' W. 10.73458 Hence AH bears N. 52° 34' E. dist. 28.83 ch. the same as found by the preceding rule very nearly. 2. Given as follow: 1st side. S. 78° W. 8 ch. 2nd. N. 26°1 W. 11.08 ch. 3rd. N. 381 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 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 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.) AB3 : 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 (20 6, . 2.) AB : AL :: AB2 : AD2 :: (19.6.) ABC: ADE; therefore (17.5.) BL: cor. 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°3 W. 8.51 ch., and CA, S. 61°1 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 :: AI2 : AE3 :: (19.6.) AIC : AEF; but (1.6.) AI: AK :: AIC: AKC; therefore AIC: AEF:: AIC : AKC, and consequently AEF=A«C. 7 |