BC. BC. NM. 2 2 is the area, therefore BC. CA. is the area. Ex. Required the area of a triangle whose sides are, 45, and 62 12 chains, and the contained angle, 52°. 10". Ans. 110a 1r 22P. Prob. 7. To find the area of a triangle when one side and the two adjacent angles are given. Rule. Multiply the square of the side, by the rectangle of the two natural sines of said angles divided by 2 sine of the third angle, for the area. Demonstration.-Look to the last figure. Now, CA. =the area. And sine A: BC: :sine B: CA. (16. 6.) sine A. CA.=BC. sine B., and BC. area. MN. 2 S.A.CA. Take b this value of BC. into the first equation and it will give, AC2.S.A.E PA. =BC.X 2 2 S b = S Ex. What is the area of a lot of land in the form of a triangle, one side of which is found to be 44 chains, and the adjacent angles 50° 10' and 70°. Ans. 187a 1r 8P. Prob. 8. To find the content of a trapezium, having the diagonals and the angle of intersection given. Rule.-Multiply the product of the two diagonals by half the natural sine of the angle of intersection, for the area. Demonstration.-Let ABCD. fig. 3, be the trapezium; DB, and AC the diagonals, and O the angle of intersection. From problem the 6 we get AO. OB.+BO. OC. natural sine of the angle O. to give the area of the triangle ABC. For the same reason, AC. OD. the natural sine of the angle O, gives the area of the triangle ACD,::(1. 2.) AC. DB. natural sine of the angle O, gives the area of the whole figure ABCD, as required. 2 Ex. What is the content of a field, whose diagonals are 20, and 30 chains, angle of intersection 40° 10'. Ans. 19a 1r 16p. SECTION II. SURVEYING BY THE CHAIN ONLY. In Surveying there are only two distinct methods by which the area can be found. First, Arithmetically; Secondly, Geometrically. The necessary rules and theorems for the former we have already explained. There are also but two methods by which the dimensions of any estate or farm can be taken. First, by the chain only; Secondly, by the chain accompanied with angular instruments. Each of these methods I will fully explain. I have before observed, that all slant or inclined surface are /measured horizontally, and not on the surface of the hill. To measure any of the regular figures passed by, requires no more explanation. Prob. 9. To measure any irregular figure. Divide the whole into triangles and trapeziums. Then measure the side required for the area. Prob. 10. To survey an estate, or any large tract of land by the chain. On commencing the survey, you should first make yourself acquainted with the form of the estate. This must, of course, materially assist in the selection of a base-line, and this line should run through the greatest extent of the property to be surveyed. As nearly through centre as possibly. Upon this base-line let principal triangles be formed as nearly equilateral as possible, and let the vertices of these triangles be connected by other measured lines. Upon these connecting lines let as many other triangles be formed as may be found sufficient to enclose the whole property. As you proceed with the mea surement of the base-line, as well as with all the other principal lines, let marks be made in the ground at the crossing of all the fences, roads, rivers, &c. And enter the distance from the beginning of the line to each of these marks in the field-book, for it may be necessary to run other lines to or from these points. Also, it will help to give a correct map of the whole. Now, before you proceed to the second principal triangle, it is better you should finish each triangle as you go on—that is, run such lines as will truly give the position of the fences. Having gone through the whole in this manner, the survey is complete. = Prob. 11. Required to divide a field or farm into any proposed number of parts. Let ABCD, fig. 4, be the pro, posed farm, to be divided suppose into two parts, by a line in the direction of RQ. First run the division line QP, as near the truth as possible. Then survey each lot if the area is not before truly given. Now suppose the parts QPCD, is short 10 perches of the required number of perches. Reduce this into square links, by multiplying the 10 by 625, and divide the product by the number of links in QP; then you get the perpendicular for a rectangle on QP, to contain 10 perches. Or double of this perpendicular will do for a triangle, on the same base, Erect this perpendicular, (and as near the bounds as possible, if not on the bounds of the field itself,) for either the rectangle or triangle, as OX, or YZ: and through the end of said perpendicular, draw a line parallel to the base as NXM, or ZR then for the triangle join QR, and for the rectangle NM, either of which will give the position of the bounds to cut off the required lot. But if you use the rectangle, if NM differs from QP, take half their sum and divide it into the number of square links in the 10 perches, which will give a perpendicular nearer the truth than the last. After this manner you may cut off any required part of. any figure. Prob. 12. Required to divide 100 acres among A, B, and C, whose claims upon it are in proportion as the numbers, 1, 3, 6. Ans. 10: 100::1: 10a, A's share. 10:100: 3:30, B's share. 10: 100;6:60, C's share. Now, you may go on the land, and cut of each lot as before directed. Prob. 13. Required to divide any quantity of land among any number of persons, in proportion to their several estates, and the value of the land that falls to each man's share. Rule. Divide the yearly value of each persons estate by the value per acre of the land that falls for his share, then say as the sum of the quotients is to the whole quantity of land, so is each particular quotient to its proportional share of land. Ex. Let 100 acres of land be divided among A, B, and C, whose estates are 100, 150, and 300 dollars per year; and the value of the land allotted to each, 5, 6, and 10, shillings per acre. Then 10°=20, 15°=25, 300=30; and the sum 75. 75: 100::25:33, B's share. Now having found each person's share it may be laid out as required. Prob. 14. To find the number of degrees in an angle by the chain. Let ACB fig.. 5, be the proposed angle, Set off 50 links from C on both sides, as CB and CA, and measure the chord AB. Now look upon the links in the chord as a decimal, and find in the tables of natural sines 10 said number: the degrees answering this number must be the degrees on half the angle. Proff-Allow AD=AC, and CM=Radius. Let fall the perpendiculars CO, DP, and MN. Now per similar triangles MN=AB÷100. In this case it is plain AB must always be a decimal, and MN the natural sine of half the proposed angle. Prob. 15. To take an inaccessible distance with the chain only. Let DC fig. 6, be the horizontal distance required, which is inaccessible by reason of the lake. Measure a line at right angles to AD as DO; and another at right angles to DO as PO. Set up a pole in the middle of DO as at T and another at C. Now go along OP until you come in a straight line with T and C, so will PO be=DC. (26 1.) OZ measure OB, and per 47. 1, BO2-OD2=DC2. OR produce PO until a perpendicular falling from PO produced would fall on any required point as, C, N, or B: then will OQ or OM be=DN, or DB, &c. Or take any point R, and measure RD and RB:. then take RX and RY, the same parts of RD and RB, and measure XY which must be the same part of DB that RX is of RD. Or if you take the DOB by the last problem, then you may find DB. Note. There are various other methods to find the length of an inaccessible distance. SECTION III. OF THE CIRCUMFERENTOR. Before we proceed to consider the principles involved in surveying with the compass, it will be necessary to describe the above instrument, and its use in the field. This instrument consists of a compass-box, a magnetic needle, a brass plate, two sights, sometimes a telescope, &c. |