angles to the river; and AB and CD can then, if necessary, be made perpendicular to it.

EXAMPLE 1. Took a line, 6 chains, alongside a river, and having had a flag placed on the other side, in the direction I was desirous of going, measured in a range with it from one end of the line, 4.50 chains, then took a second line parallel to the former, 8 chains, to such a point that I covered the flag and the second station of the first line. What is the width of the river in the required direction? Answer 13 chains 50 links. EXAMPLE 2. Wanted the perpendicular distance across a river: placed a flag on the other side; and from the water's edge on this side, measured 3 chs. back from the river at right angles to it, and in line with the flag: then measured along the side of the river, 8·20 chs. and having set a flag there; ranged from that point, back in a line with the flag beyond the river, until the line from here to the 3 chs. was parallel to the line by the river's side. This line was 10.50 chs. long. Answer 10.69 chains.

PROBLEM IV.

To continue the measurement and direction of a given line, when any obstacle stands in the way, which cannot be crossed but can be avoided by going to the right or the left.

At any point (A), on the given line AC, take an angle with the given line, of 120°, if you would turn to the left, or 240°, if to the right, as in this case, and proceed measuring to B, till an angle of

60° made with this line towards the first line AC, will carry you clear of the obstacle.

Take this angle, ABC, 60 degrees, and measure BC the same distance as AB; the point C shall be in the given line and AC shall be equal to AB or BC. By taking an angle of 240 degrees with the line BC, the range of the line can be continued.

PROBLEM V.

To ascertain the height of an object, when the base
is accessible.

Measure any distance from the base of the object, as nearly equal as possible to the height, and take the angle of elevation by the theodolite.

Let BA be the object; measure BC, B

and take the angle BCA, then BA is determinable by the case of right-angled triangles.

EXAMPLE 1. What is the height of a tower, whose top, at the distance of 5 chains, 75 links, subtends an angle of 33° 17/? Answer 249 feet. EXAMPLE 2. What is the height of a church spire, whose angle of elevation is 6o at the distance of § of a mile? Answer 185 feet.

EXAMPLE 3. Given the angle of elevation 7° 45', at the distance of a mile; required the height of the object? Answer 359 feet.

A

If the object be inaccessible, so that DB cannot be measured, range DB on to C, and taking BC, nearly equal to DB: measure BC, D

B

and at B and C take the angles of elevation. The angle DBA, being the outward angle, is equal to the angle at C, the angle BAC, and BC is measured; therefore, the triangle ABC, in the vertical plane, comes under the first case; and AB being thus determined, and the angle ABD, at the base, being known, the triangle ABD is determinable by the case of right-angled triangles, and AD becomes known.

EXAMPLE 1. Wanted the height of a tower, and the width of a moat around it, when the angle subtended by the top of it, at the edge of the moat was 64° 20′; and at 4 chains 50 links off, was 25° 54′.

Ans. Height of tower 188 ft. And width of moat 90·4 ft.

EXAMPLE 2. When the angle, at the moat was 45°, and at the distance of 5 chs. was 15°; what was the height? Answer. 120.8 feet.

Where DB cannot, from any cause, be produced towards C, another plan must be adopted.

In the last case, because neither AD nor DB could be measured, it was requisite that AB should be made the connecting link between some new triangle, that could be determined; and the given one AB was therefore made a common line to the triangles ABC, and ABD; and being determined by the former, it was the means of determining the latter. These triangles are both in the same vertical

plane. This, however, cannot be done in the case before us, as DB cannot be produced; we can, therefore, no longer make AB the connecting line. By supposing, however, the horizontal distance DB to be the common intersection of the vertical and horizontal planes, we may be enabled to determine its distance in the horizontal plane, and employ that distance to determine the vertical height AD.

DB comes under the case of Problem II., and AB under that of Problem V., and must be determined in the same way accordingly.

PROBLEM VI.

It was required to find the height above C, of a high tower at D which was inaccessible.

BD

To do so, it was requisite that CD should be obtained, which is the line of intersection of the horizontal and vertical planes.

The base line was first measured, noting the station C. The horizontal angles at A, C, and B, were taken; and at C, the angle of elevation DCD was also found; the height of the instrument was 5 feet.

AB being given, and the angles at A and B, the sides AD, BD are determinable; and AC being given, AC, AD, and the included angle at A, will give the distance CD,* which thus becomes a base for the right-angled triangle, the angle at whose base is the angle of elevation 6° 17', whence the height can be determined by the case of right-angled triangles.

The base CD was 1206 feet, and the height 133 feet above the instrument; adding, therefore, 5 feet, the height of the instrument, we have 138 feet as the height of the tower, above the point where the observation was taken.

Note.—The distance being under a of a mile, the correction for curvature and refraction would be inappreciable.