BDE are given DE and all the angles to find DB. Lastly, in the triangle ADB are known AD, DB, the included angle BDA = BDc – Adc, to find AB = 345.5 yards. - FXAMPLE XI. In order to determine the distance between two inac cessible objects E and w, on a horizontal plane, we mea. sured a convenient base AB of 536 yards, and at the extremities A and B took the following angles, viz. BAw = 40° 16', wa E = 57° 40', ABE = 42°22', EBw = 71°7', Required the distance Ew. First, in the triangle ABE are given E W all the angles, and the side AB, to find BE. So again, in the triangle ABw, are given all the angles and AB to find Bw, Lastly, in the triangle BEw are given the two sides EB, Bw, and the included angle EBw, to find Ew = 939:52 yards. A. B Remark. In like manner the distances taken two and two, between any number of remote objects posited around a convenient station line, may be ascertained. ExAMPLE XII. ... Suppose that in carrying on an extensive survey, the distance between two spires A and B has been found equal to 6594 yards, and that c and D are two eminences conveniently situated for extending the A b triangles, but not admitting of the de- B termination of their distance by actual admeasurement: to ascertain it, there- * fore, we took at c and D the following angles, viz. . | ACB = 8.5° 46' ADC = 31°48' * }. = 23° 56' }. = 68° 2 ° Required CD from these data. In order to solve this problem, construct a similar quadrilateral Acab, assuming cd equal to 1, 10, or any gled at B, and having the angle BAE other convenient number: compute Ab from the given angles, according to the method of the preceding example. Then, since the quadrilaterals Acdb, AcDB, are similar, it will be, as Ab : cal:: AB : CD; and CD is found = 4694 yards. ExAMPLE XIII. If the height of the mountain called the Pike of Teneriffe be 3 miles, and the angle formed at its top between the vertical or plumb line and a right line conceived to touch the earth in the horizon, or at the far thest visible point, be 87° 46' 33”; it is required from hence to determine the diameter of the earth, supposing it to be a perfect sphere. Let c, in the marginal diagram, be the centre of the earth, the circle BTG a vertical section passing through the centre, AB the height of the Pike of Teneriffe, and At the tangential line drawn to the visible horizon : let, also, BE, a tangent to the earth’s surface at B, meet the other tangent AT in E. Then, in the triangle ABE, right an = 87° 46' 33”, it is, as rad: A B :: tan A : BE, and :: sec A : AE. But it is evident, from Euc. iii. 37, that BE = ET; therefore, AE + EB = AE + ET = AT. In the triangle Atc right angled at T, we have, as rad : AT:: tan A : Tc, the radius of the earth. The operation performed as above described occupies but small compass; it may, however, be shortened by means of ch. ii. prop. 4. For, since tan A + sec A = tan (A + # comp. A), we shall, by incorporating the roportions from which AE, BE, and cT are deduced, ave CT. rad? = AB tan . + # comp. A) tan A. Or, log cT = log AB + log tan(A + 3 comp. A) + log tan A — 20, in the index. E: The double of this, or 7958.3 miles, is the diameter of the earth. If AT be required, we have only to take radius (10) from the sum of the first two lines, the remainder 2:1890522, is the log of 154'54 the distance sought. Note 1. The log tangents are found in this example by the method taught at p. 153 of the Introduction to Dr. Hutton's Logarithm Tables. Note 2. This method of determining the earth's radius, though elegant in theory, is rendered useless in practice, by the extreme irregularity of the horizontal refractions. ExAMPLE XIV. Given the angles of elevation of any distant object, taken at three places in a horizontal right line, which does not pass through the point directly below the object; and the respective distances between the stations; to find the height of the object, and its distance from either station. Let AEc be the horizontal plane, F FE the perpendicular height of the object above that plane, A, B, c, the * three places of observation, FAE, IE E FBE, FCE, the angles of elevation, C and AB, Bc, the given distances. Then, since the triangles AEF, BEF, CEF, are all right angled at E, the distances AE, BE, CE, will manifestly be as the cotangents of the angles of elevation at A, B, and c. Put AB = D, Bc = d, EF = or, and then express algebraically the following theorem, demonstrated at p. 128, Simpson’s Select Exercises: viz. AE”. BC + CE*. AB = BE". AC + AC. AB . BC, the line EB being drawn from the vertex E of the triangle AcE, to any point B in the base. The equation thence resulting is, da” cot “A + Dz” cot *c = (D + d) to cot *B + (D + d) Dd. Hence, transposing all the unknown terms to one side of the equation, dividing by the sum of the coefficients, and extracting the square root, we shall have Thus EF becoming known, the distances AE, BE, CE, are found, by multiplying the cotangents of A, B, and c, respectively, by EF. Cor. When D = d, or D + d = 2D = 2d, the expression becomes * = d -- v (; cot “A + 3 cot ‘c — cot *B), which is pretty well suited for logarithmic computation. The rule may, in that case, be thus expressed.— Double the log. cotangents of the angles of elevation of the extreme stations, find the natural numbers answering thereto, and take half their sum; from which subtract the natural number answering to twice the log. cotangent of the middle angle of elevation: then half the log. of this remainder subtracted from the log. of the measured distance between the 1st and 2d, or the 2d and 3d stations, will be the log. of the height of the object.” To illustrate these soners methods, two particular examples are subjoined. ExAMPLE XV. * For geometrical constructions of this general problem the student may consult Hutton's Course, vol. iii. p. 128, Leybourn's Ladies' Diaries, vol. i. p. 897, or the Ladies' Diary for 1748. and 14°, and the two equal measured distances 84 feet. From a convenient station P, where could be seen three objects A, B, and C, whose distances from each other were known (viz. AB = 800, Ac = 600, Bc = 400 yards), I took the horizontal angles Apc = 33° 45', BPC = 22° 30'. It is hence required to determine the respective distances of my station from each object. Here it will be necessary, as preparatory to the computation, to describe the manner of Construction. Draw the given triangle ABC from any convenient scale. From the point A draw a line AD to make with AB an angle equal to 22° 30' and from B a line BD to make an angle DBA = 33°45'. Let a circle be described to pass through their intersection D, and through the points A and B. Through c and D draw a right line to meet the circle again in P: so shall P be the point required. For, drawing PA, PB, the angle APD is evidently = ABD, since it stands on the same arc AD : and for a like reason BPD = B.A.D. So that P is the point where the angles have the assigned value. . . . Manner of computation. In the triangle ABC where the sides are known, find the angles. In the triangle ABD, where all the angles are known, and the side AB, |