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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 scught.

Note 1. The log tangents are found in this example by the method taught at p. 153 of the Introduction to 1)r. 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, FE the perpendicular height of the object above that plane, A, B, C, the three places of observation, FAE, FBE, FCE, the angles of elevation, 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 = a, and then express algebraically the following theorem, demonstrated at p. 128, Simpson’s Select Exercises; viz.

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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, dr” cot “A + procot *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

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Thus EF becoming known, the distances AE, BE, CE, are found, by multiplying the cotangents of A, B, and c, respectively, by E.F. or. When D = d, or D + d = 2D = 2d, the expression becomes * = d -- v(; cot “A + , 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 general methods, two particular examples are subjoined.

ExAMPLE XV.
Let the three angles of elevation be 36° 50', 21°24',

* 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. - E 2

and 14°, and the two equal measured distances 84 feet. Required the height of the object. Ans. 53.964 feet.

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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 = BA.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,

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find one of the other sides AD. Take BAD from BAc, the remainder, DAC is the angle included between two known sides AD, Ac; from which the angles ADc and AcD may be found, by chap. iii. case 2. The angle cAP = 180°– (APC + AcD). Also, BCP = Bca — AcD; and PBC = ABC + PBA = ABC + sup. ADC. Hence, the three required distances are found by these proportions. As sin Apc : Ac:: sin PAc: PC, and :: sin PCA: PA; and lastly, as sin BPC : Bc :: sin Bcp: BP. The results of the computation-are, PA = 709:33, Pc = 1042-66, PB = 934 yards.

*** The computation of problems of this kind, however, may be a little shortened by means of the following

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Hence, a being thus determined, we get y from the equation y = R — w, and ce from either of the expressions above given.

Note. It will be a useful exercise for the student, to work out the computation by both these methods. The comparison of the results will serve to give him confidence in the deductions from the analytical investigations.

ExAMPLE XVIII.

It is required to find the distances from Edystone light-house to Plymouth, Start Point, and the Lizard, respectively, from the following data: . Plymouth to Lizard.... 60 The distance from #: to Start Point .. 70 % miles. Start Point to Plymouth 20 y

Plymouth.... North.

Lizard ...... } bears from Edystone rock };

Start Point... - E. by N. Ans. From Edystone to Lizard . . . . . 53-04 miles

to Plymouth ... 14.333 ditto to Start Point... 17-36 ditto. Remark. The general problem of which the last two examples are particular cases, was originally proposed by Richard Townley, Esq. and solved by Mr. John Collins, in the Philosophical Transactions, N°69, A. D. 1671. “See New Abridgement, vol. i. p. 563,

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