CHAPTER VI. PROBLEMS AND FORMULÆ OF UTILITY IN LAND AND ENGINEERING SURVEYING. The areas of triangles, &c., may be expeditiously found by the following Problems, without mapping them for the purpose of finding perpendiculars, &c. PROBLEM I. To find the area of a triangle, when two of its sides and their included angle are given. Let a and b be the two sides of the triangle, a their included angle, and A the area; then A = ab sine a This Problem may be solved by logarithms, by which the multiplication of large numbers is avoided, and by which the area is found in acres and decimals; thus log. A = log. a + log. b + log. sine a — 15.30103. EXAMPLE. The two sides of a triangle are 1920 and 1152 links, and their included angle 53° 8'; required the area. log. 1920.........3.28330 log. 1152.........3 06145 log. sine 53° 8'.........9·90311 16.24786 15.30103 log. A = 8.848.........0-94683 PROBLEM II. To find the area of a triangle, when two of its angles and their included side are given. Let a and B be the two angles, and a their included side; then By logarithms for the area in acres. log. A = 2 log. a + log. sine a + log. sine ß — log. sine (a+B) — 15.30103 EXAMPLE. A side of a triangle is 2000 links, and its adjacent angles 60° 14′ and 59° 46'; required the area by logarithms. PROBLEM III Ans. 17 a. 1r. 12p To find the area of a triangle when the three sides are given. Let a, b and c be the sides, and s their half sum, then A = √s (sa) (s—b) (s—c). By logarithms, for the areas as above, log. A=log. s+log. (s—a)+log. (s—b)+log. (s—c) ÷ 2 — 5. EXAMPLE. The three sides of a triangle are 3050, 2520, and 2040 links, required the area by logarithms. Ans. 25a. 2r. 4p. PROBLEM IV. To find the area of a trapezium, when its four sides are given, two of its opposite angles being together = 180°. Let a, b, c, and d be the four sides, and s = half their sum; A = √(sa) (s—b) (s—c) (s—d). then, The method of solving this Problem by logarithms is sufficiently obvious from Problem III.. PROBLEM V. To find the area of a trapezium when its two diagonals and the angle of their intersection are given. Let D and be the two diagonals, and a the angle of their intersection; then The method of solving this Problem by logarithms is the same as that given in Problem I. NOTE. The investigations of the Formulæ used in these Problems are given in various works on Analytical Trigonometry. PROBLEM VI. A B C is a triangle, in which the base A B and a point D therein are given, DC is a quality line, making a given angle with AB; it is required to determine C D so that the triangle A B C may contain land of a given value. Put A Da, DB=b, CD= x, l = square links in an acre, sine D=σ, the values of the land adjacent to a and b respectively, m and n per acre, and V the given value; then the areas of the triangles A CD, B C D are respectively ασπ box ασηπ bon x whence -and and their values are -and 21 21 21 21 2 VI = = CD; whence the position of the point C becomes known. COR. When CD is perpendicular to A B, σ = 1, whence the above formula becomes NOTE. This Problem would be available in solving Case IV, Problem V, page 121; however, it is there proposed to be solved by assumed or "Guess lines," conformable to the old practice of surveyors. PROBLEM VII. In making an extensive survey, the fundamental lines A B, B C, CD, DA were measured, and the distance CP noted; but the distance AP was accidentally omitted. It is required to lay down the lines independent of this distance; and give a solution when AD=DP. B P Put A Ba, BC=b, CP=c, PD=d, DA=e, and AP; then PB = a- -x; and by trigonometry cos. APD x2 + d2 — e2 ___ (a — x)2 + c2 — b3 2 d x X c+d =0, a cubic equation, from which the value of x may be found, which determines the distance AP. = When A D DP, that is, when de, the above equation becomes a quadratic equation, from which the value of x = AP may be readily found. Investigation of formula. As the formula for dividing land, at pages 119 and 121, Cases II. and III., Problem V., are given without investigation, to obviate all doubts of their accuracy, the processes of deducing them are here given. (1.) By referring to the definition of the symbols, and to the diagram at page 119, Case II., it will be seen that the right angled triangles W Ss, WUu, W Fb, are similar, therefore, a2 x the area in acres of W Ss=WS • Ss= 2 st (2 a+b) b x 2 sl (2 a+2b+c) cx of Ss Uu (Ss+Uu) SU= 2 sl Then these areas, being multiplied by their respective values per acre, and their sum equated to the required value V', give a2m x (2 a + b) b n x 2 sl + + (2a + 2 b + c) co x _V', 2 sl 21s 2ls V' = Fb. Q.E.I whence x = a2m+(2a+b)bn+(2a+2b+c)co COR. I. If there be only two qualities of land to be thus laid ont, c must be made to vanish in the above formula, whence it becomes COR. II. If there be four different qualities of land, let the breadth of the additional quality be d, and its value p; then the formula becomes x= 2 ls V' a2m+(2a+b) b n+(2a+2b+c)co+(2 a+2 b+2 c + d) dp the law of extension being sufficiently clear. = Fb (2.) By referring to the diagram at page 120, Case III., Problem V., and to the symbols on the same and following page, it will be readily perceived, that by drawing a perpendicular from a on NB, (which is not shewn in the diagram) that a q = a − 2 x a, whence the area in acres of the trapezoid (a—a x) x, in a similar way three areas by their respective values, and putting their sum equal to the required value V', their results (a − a x) m x + (c − В x) o x + b n x =V'; whence (am + Bo) x2 (am + b n + c o) x + 1 V′ = 0; and, by solving this quadratic for the reciprocal of x, there results =pg or rs; am + bn + co + √ (am + bn + co)2−4(am+Bo) l V' NOTE The method of adapting this formula to a greater or less number of qualities of land is sufficiently clear, from the different modifications of the preceding formula in Cor. I. and II. PROBLEMS, ORIGINAL AND SELECT, FOR THE EXERCISE OF STUDENTS. I. The side of an equilateral triangle is = a, and its area = A; prove that A = 1 a2 √3. II. The base of an isosceles triangle is = a, one of its equal sides = b, and its area A; prove that A = a √(2b + a) (2 b − a). III. In a triangle are given the perpendicular = p, the angles opposite the perpendicular = a, and B, and consequently the third angle = 7; prove that p2 sine y A = IV. ABCD is a trapezium, in which the angles at A and C are equal, and AB = a, BC = b, CD = c, DA = d, the half sum of all the sides = s, and the area = A; then prove that V. In a trapezium A B C D, are given two opposite sides AB = a, CD = b, the angles at A, B, and C respectively = a, B, and y, and consequently the fourth angle D = 8; then prove that A a2 sine a sine B b2 sine y sine & 2 sine (a+B) VI. In a trapezium A B C D, are given the three sides |