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. 38. When the nature of the ground will admit of it, measure a base line AB in the direction of the object D. Then measure with the instrument the angles of elevation at A and B. Then, since the exterior angle DBC is equal to the sum of the angles A and ADB, it follows that the an B gle ADB is equal to the difference of the angles of eleva tion 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 it 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 First: ADB DBC - A=27° 29' -15° 36′ = 11° 53'. III. To determine the perpendicular distance of an object below a given horizontal plane. 39. 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 are the hy pothenuses of two right-angled triangles, of which AC, BC, are the bases. The perpendiculars of these triangles are 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. First: in the triangle ABC, the horizontal angle ACB=180° - (A+B) = 180° - 111° 49' 68° 11'. = Hence also, CC-CC" 242.06-239.932.13 yards, = which is the height of the station A above station B. PROBLEMS. 1. Wanting to know the distance between two inacces sible objects, which lie in a direct level line from the bot tom of a tower of 120 feet in height, the angles of depres sion are measured from the top of the tower, and are found to be, of the nearer 57°, of the more remote 25° 30′: required the distance between the objects. 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 angle ACB: it was found that, A. B 3. Being on a horizontal plane, and wanting to ascer tain 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. 5. Wanting to know the horizontal distance between and B, and not finding any station from which both of them could be seen, two points C and D, were chosen D at a distance from each 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 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. find the three distances PA, PC, and PB. 7. This problem is much used in maritime survey ing, for the purpose of locating buoys and sounding boats. The trigonometrical solution is somewhat tedious, but it may be solved geometrically by the following easy construction. |