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 (Art. 74). The horizontal angles CAB and CBA are also the same, so long as the point C is in the vertical line DC (Art. 75). 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 DAE, the base AE, and the angle at the base; from which the perpendicular DE can be determined.

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.

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

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

SECOND METHOD.

90. When the nature of the ground will admit of it, measure a base

line AB in the direction of the object D. To

do this, it will be well to A

C

place the theodolite at A, and range the chain staves by means of the upper telescope. Having measured the base, measure with the instrument the angles of elevation at A and B.

Then, since the outward angle DBC is equal to the sum of the angles A and ADB, it follows, that the angle ADB is equal to the difference of the angles of elevation 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

AB=975 yards

found A=15° 36'

DBC 27° 29';

required the altitude DC.

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

In the triangle ADB, to find BD.

PROBLEM III.

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

91. Suppose C 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 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.

Let us suppose that we have

AB=672 yards

BAC 72° 29'

found ABC-39° 20'

C'AC 27° 49'

C"BC 19° 10'

First In the triangle ABC, the horizontal angle
ACB=180°-(A+B)=180°-111° 49′-68° 11'.

To find the horizontal distance AC.

Hence also, CC'—CC"=242.06-239.93=2.13 yards; which is the height of the station A above station B.

REMARK. In measuring a base line, if great accuracy is required, the theodolite should be placed at one extremity, and the telescope directed to the other, and the alignment of the staves made by means of the intersection of the spider's lines. If the highest degree of accuracy is necessary, the base line should be measured with rods, which admit of being adjusted to a horizontal position by means of a spirit level.

APPLICATIONS.

1. Wanting to know the distance between two inaccessible objects, which lie in a direct line from the bottom of a tower. of 120 feet in height, the angles of depression are measured, and are found to be, of the nearest 57°, of the most remote 25° 30' required the distance between them.

Ans. 173.656 feet.

2. In order to find the distance between two trees A and B, which could not be directly measured because of a pool which occupied the intermediate space, the distances of a third point C from each of them were measured, and also the included

CB=672 yards
CA=588 yards

ACB 55° 40';

required the distance AB.

Ans. 592.967 yards.

3. Being on a horizontal plane, and wanting to ascertain the height of a tower, standing on the top of an inaccessible hill, there were measured, the angle of elevation of the top of the hill 40°, and of the top of the tower 51°; then measuring in a direct line 180 feet farther from the hill, the angle of elevation of the top of the tower was 33° 45'; required the height of the tower.

4. Wanting to know the horizontal distance between two inaccessible objects E and W, the following measurements were made,

Ans. 83.998 feet.

other, equal to 200 yards; from the former of these points A could be seen, and from the latter B, and at each of the points C and D a staff was set up. From C a distance CF was measured, not in the direction DC, equal to 200 yards, and from D a distance DE equal to 200 yards, and the following angles taken,

Ans. AB=345.467 yards.