First: ADB = DBC - A= 27° 29' -15° 36' 11° 53'. In the triangle ADB, to find BD. = 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 hypothenuses 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 PROBLEMS. 1. Wanting to know the distance between two inaccessible objects, which lie in a direct level line from the bottom 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. 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. 5. Wanting to know the horizontal distance between two inacessible objects A and B, and not finding any station from which both of them could be seen, two points C and D, were chosen F. A B 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. PA=710.193 yards. Ans. PC1042.522 PB934.291. 7. This problem is much used in maritime surveying, 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 con as a centre, and OA or OC as a radius, describe the circumference of a circle: then, any angle inscribed in the segment APC, will be equal to 33° 45'. Subtract, in like manner, twice CPB = 45°, from 180°, and lay off half the remainder = 67° 30', at B and C, determining the centre of a second circle, upon the circumference of which the point P will be found. The required point P will be at the intersection of these two circumferences. If the point P fall on the circumference described through the three points A, B, and C, the two auxiliary circles will coincide, and the problem will be indeterminate. |