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Given the distances of two objects from any station and the angle which they subtend, to find their mutual distance,

Let AC, BC, and the angle ACB be

given, to determine AB.
In the triangle ABC, since two B

sides and their contained angle are gi-
ven, therefore, by corollary to Propo-
sition 10. AC + BC : AC – BC : :
cotáC : cot(A++C), then sin A : sinc : :
BC : AB; or (from the cor. to Prop. 11.) C
AB= W(AC*-i-BC*—2AC.BC cosc.)

Cor. By combining this with the preceding proposition, the distance of an object may be found from two stations, between which the communication is interrupted. Thus let A be visible from B and C, though the straight line BC cannot be traced. Assume a third station D, from which B and C are both seen. Measure DB and DC, and observe the angles BDC, ABC and ACB. #-----------. In the triangle BDC, the base BC is found as above ; and thence, by the preceding proposition, the sides AB and AC of the triangle ABC are determined

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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 dis. tance, 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 : sincI)A : : CD :
CA, and sincBD : sinCDB : : CD :
CB; the difference or sum of CA and
CB is AB, the distance sought. C

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 A 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 : : cot;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 : cos(ACB+CAB. Now, sincAB : sinACB:: BC : AB, or AB = W(AC*-ī- BC*—2AC.BC cosACB). The inclination of AB to CDs 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.

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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.


PROP. XXII. PROB. 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- C () o lows, by Prop. 9., that sin AED : sinEAD :: AD : DE. Wherefore, measure out DE on the ground, and its extremity 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 ACB, the base AB is thence found by the application of Prop. 20.

It is evident that the mode of investigation will not b altered, if the three stations D, C and E should lie in the same straight line.

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mote 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 F3 to find the transverse

distance AB,

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