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Given the interval between two stations, and the directions of two remote objects viewed from them in the same plane, to find the mutual distance, and relative position of those objects.

Let the points A, B represent the two objects, and C, D the two stations from which these are observed ; the interval or base CD being measured, and also the angles CDA, CDB at the first station, and DCA, DCB at the second; it is thence required to determine the transverse distance AB, and its direction.

It is obvious that each of the points A and B would be assigned geometrically by the intersection of two straight lines, and consequently that the position of the objects will not be determined, unless each of them appears in a different direction at the successive stations.

1. Suppose one of the stations C to lie in the direction of the two objects A and B,

At C observe the angle BCD, and at
D the angles CDA and BDC. Then
by Prop. 9..sincAD: sincDA : : CD :
CA, and sincBD : sincDB : : CD :
CB; the difference or sum of CA and
CB is AB, the distance sought. c

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2. When neither station lies in the direction of the two objects, and the base CD has a transverse position.


Find by Prop. 19. the distances AC and BC of both objects from one of the stations C; then the contained angle ACB, or the excess of DCA above DCB, being likewise given, the angles at the base AB of the triangle BCA, and the base itself, may be calculated,

from the analogies exhibited for the solution of the second branch of Case second. For AC BC : AC–BC : : cots ACB : cot(#ACB+CAB), and thus the angle CAB is found. Or more conveniently by two successive operations, AC : BC : : R : tan b, and R : tan(45°–b) : : cot;ACB : cot(#ACB+CAB. Now, sincAB : sinACB:: BC : AB, or AB = W(AC*-i-BC*—2AC.BC cosACB). The inclination of AB to CD in the first case is given by observation, and in the second case it is evidently the supplement of the interior angles CAB and DCA. A parallel to AB may hence be drawn from either station.

Cor. Hence the converse of this problem is readily solved. Suppose two remote objects A and B, of which the mutual distance is already known, are observed from the stations C and D, and it were thence required to determine the interval CD. Assume unit to denote CD, and calculate AB according to the same scale of measures; the actual distance AB being then divided by that result, will give CD: For the several triangles which combine to form the quadrilateral figure CABD, are evidently given in species.



Given the directions of two inaccessible objects viewed in the same plane from two given stations, to trace the extension of the straight line connecting them.

Let the angles ACD, BCD be observed at C, and * ADC, BDC at D, with the base CD; to find a point E in the straight line ABF produced through A and B.

By the last proposition, find AD and the angle DAB, and * assume any angle ADE. In the triangle DAE, the angles at the base AD, and consequently the vertical angle AED, being known, it fol. “ ” lows, by Prop. 9., that sin AED : sinEAD :: AD : DE. Wherefore, measure out DE on the ground, and its ex

, tremity E will mark the extension of AB.


Given on the same plane the direction of two remote objects separately seen from two stations and their direction as viewed at once from an intermediate station, with the distances of those stations, from the middle station,--to find the mutual distance of the objects.


Let object A be visible from the station D, and B from E, and both of them be seen at once from the station C; the compound base DC, CE being measured, and the angle DCA, ACB and BCE, with ADC and BEC, observed,—to determine AB.

In the triangles DAC, CBE, the sides AC and BC are found by Prop. 19., and in the triangle o ACB, the base AB is thence found "TTTTTT E. by the application of Prop. 20.

It is evident that the mode of investigation will not be altered, if the three stations D, C and E should lie in the

same straight line.

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Given four stations, with the direction of a remote object viewed from the first and second stations, and the direction of another remote object viewed from the third and fourth stations, all in the same plane,—to find the distance between

the objects.

Let the bases EC, CD, and DF be given, with the angles ECD and CDF, and suppose that at the stations E and C the angles CEA and ECA are observed, and the angles BDF and BFD at D and F; to find the transverse

distance AB.


In the triangles EAC and A DBF, find by Prop. 19. the sides AC and BD; and in the triangle CAD, the sides AC, CD, with their contained angle ACD, being given, the base DA and the angle CDA are found by Case II. But the distances DA, DB being now given, with their contained angle ADB, the base AB is found by Prop. 20.


The mutual distances of three remote objects being given, with the angles which they subtend at a station in the same plane, to find the relative place of that station.

Let the three points A, B, and C, and the angles ADB and BDC which they form at a fourth point D, be given; to determine the position of that point. ..

1. Suppose the station D to be situate in the direction of two of the objects A, C.

All the sides AB, AC and BC of the triangle ABC being given, the angle BAC is B found by Case I. ; and in the ~\ triangle ABD, the side AB with A Is) I3 & the angles at A and D being

given, the side AD is found by ~

Case III. and consequently the * * C position of the point D is determined,

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