15, 20. 2. It is required to divide 300 acres of land : B, C, D, E, F, G, and H, whose claims upor spectively in proportion as the numbers 1. w lines to the oppor The sum of these proportional nu my number of parts, 300 being divided, the quotient is estion to each other ing multiplied by each of the nu have for the several shares as fiumber of parts, and in Ао la; then from the several AS to the proposed angle, and B= as required. * 3 C, "ус the 2+3+5+9=20, and 1600 20 = 80; ring multiplied by each of the proportional num have , 400 then, by drawing the lines Aa, Ab, Ac, Ad, the *** Jus C180, ab 160, bc 240, cd=400, and dB after he is divided as required. PROBLEM * DEMONSTRATION. For the several parts are triangles of altitude, and which therefore are as their bases, which Diss are taken in the assigned proportion. the same PROBLEM IV, side of a given triangle, to draw lines des, dividing the triangle into any number shall be in any assigned ratio. B I K H the given point D DB to the angle opusite the side AC, in which the point is taken ; then divide the same side E F D G AC into as many parts AE, EF, TG, GC, and in the same proportion with the required parts of the triangle, as in the last Problem ; and ; from the points of division draw lines EK, FI, GH, parallel to the line BD, and meeting the other sides of the triangle in K, I, H ; lastly, draw KD, ID, HD; so shall ADK, KDI, IDHB, HDC, be the parts required.* An exainple of this is performed like that of the last Problem. LEVELLING. * DEMONSTRATION. The triangles ADK, KDI, IDB, being of the same height, are as their bases. AK, KI, IB ; which, by means of the parallels EK, FI, DB, are as AE, EF, FD); in like manner, the triangles CDH, HDB are to cach other as CG, GD : but the two triangles IDB, BDH, having the same base BD, are to each oiher as the distances of I. and H from BD, or as FD . DG ; consequently the parts DAK, DKI, DIDH, DHC, are to each other as AF, EF, FG, GC. : From an angle, in a given triangle, to draw lines to the oppos site side, dividing the triangle into any number of parts, which shall be in any assigned proportion to each other. Divide the base into the same ríumber of parts, and in the same proportion, by Prob. I. ; then from the several points of division draw lines to the proposed angle, and they will divide the triangle as required.* Here 1 +2+3+5+9=20, and 1600 20 80 ; which being multiplied by each of the proportional numbers, we have 80, 160, 240, 400, and 720. Therefore make Cas80, ab=160, be=240, cd=400, and dB =720 ; then, by drawing the lines Aa, Ab, Ac, Ad, the triangle is divided as required. PROBLEM * DEMONSTRATION. For the several parts are triangles, of the same altitude, and which therefore are as their bases, which bases are taken in the assigned proportion. PROBLEM IV. From any point, in one side of a given triangle, to draw lines to the other two sides, dividing the triangle into any number of parts, which shall be in any assigned ratio. B I G From the given point D draw DB to the angle opposite the side AC, in which the point is taken ; then divide the same side А E F D AC into as many parts AE, EF, FG, GC, and in the same proportion with the required parts of the triangle, as in the last Problem ; and from the points of division draw lines EK, FI, GH, parallel to the line BD, and meeting the other sides of the triangle in K, I, H ; lastly, draw KD, ID, HD; so shall ADK, KDI, IDHB, HDC, be the parts required.* An example of this is performed like that of the last Problem. LEVELLING. DEMONSTRATION. The triangles ADK, KDI, IDB, being of the same height, are as their bases. AK, KI, IB ; which, by means of the parallels EK, FI, DB, are as AE, ÈF, FD); in like manner, the triangles CDH, HDB are to each other as CG, GD : but the two triangles IDB, BDH, haring the same base BD, are to each other as the distances of I. and H from BD, or as FD to DG ; consequently the parts DAK, DKI, DILH, DHC, are to each other as AE, EF, FG, GC. LEVELLING. BY LEVELLING' We find how much one point on the surface of the earth is higher or lower than another ; or the diffcrence in their distances from the centre of the earth. PROBLEM I. To find the difference between the true and apparent levels at a given distance. • Let O be the centre of the earth, B OA, OD, any two radii, AB a tan 70 gent to the surface ADE at A. Now, AB, which is the line of sight in the level when properly adjusted, is called the line of apparent level. But a line, as AD, equally distant from the centre of the earth in all its points is called the line of true level. Hence DB is the height of the apparent above the true level at the distance AD or AB, being the difference of these heights above the centre of the earth. It is therefore evident, that the difference of the true and apparent height is always cqual to the difference of the radius of the earth and the secant of the arc of distance. By a property of the circle, we have+ 20D+DBX DB=AB'; or because the diameter of the earth is so great with respect to the line BD at all distances, to which an operation of levelling commonly extends, that 200 may safely be taken for 20D+DB, without any sensible AB? AD: error ; 20D X DB = AD', or DB = 20D 20D nearly That . |