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In the right-angled triangle DBC, to find DC.

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REMARK I. It might, at first, appear that the solution which we have given, requires that the points B and A should be in the same horizontal plane; but it is entirely independent of such a supposition.

For, the horizontal distance, which is represented by BA, is the same, whether the station A is on the same level with B, above it, or below it. The horizontal angles CAB and CBA are also the same, so long as the point C is in the vertical line DC. Therefore, if the horizontal line through A should cut the vertical line DC, at any point as E, above or below C, AB would still be the horizontal distance between B and A, and AE which is equal to AC, would be the horizontal distance between A and C.

If at A, we measure the angle of elevation of the point D, we shall know in the right-angled triangle DAE, the base AE, and the angle at the base; from which the perpendicular DE can be determined.

Applications.

Let us suppose that we had measured the angle of elevation DAE, and found it equal to 20° 15'.

First: In the triangle BAC, to find AC or its equal AE.

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In the right-angled triangle DAE, to find DE.

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Now, since DC is less than DE, it follows that the station B is above the station A. That is,

DE DC= 283.66 218.64 65.02 = EC, which expresses the vertical distance that the station B is above the station A.

REMARK II. It should be remembered, that the vertical distance which is obtained by the calculation, is estimated from a horizontal line passing through the eye at the time of observation. Hence, the height of the instrument is to be added, in order to obtain the true result.

SECOND METHOD.

When the nature of the ground will admit of it, measure a base line AB in the direction of the object D.

Then mea

sure with the instrument the angles of elevation at A and B. Then, since the outward angle DBC is equal to the sum

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vation at A and B. Hence, we can find all the parts of the triangle ADB. Having found DB, and knowing the angle DBC, we can find the altitude DC.

This method supposes that the stations A and B are on the same horizontal plane; and therefore can only be used when the line AB is nearly horizontal.

Let us suppose that we have measured the base line, and the two angles of elevation, and

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First: ADB = DBC- A = 27° 29' - 15° 36' 11° 53'.

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Applications.

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PROBLEM III.

To determine the perpendicular distance of an object below a given horizontal plane.

Suppose to be directly over

the given object, and A the point through which the horizontal plane is supposed to pass.

Measure a horizontal base line AB, and at the stations A and B conceive the two horizontal lines A

AC, BC, to be drawn. The oblique

lines from A and B to the object will be the hypothenuses of two right-angled triangles, of which AC, BC, are the bases. The perpendiculars of these triangles will be the distances from the horizontal lines AC, BC, to the object. If we turn the triangles about their bases AC, BC, until they become horizontal, the object, in the first case, will fall at C', and in the second at C".

Measure the horizontal angles CAB, CBA, and also the angles of depression C'AC, C"BC.

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First In the triangle ABC, the horizontal angle ACB =

1800

(A + B)

= 180°

111° 49' 68° 11'.

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