NAVIGATION AND SURVEYING. 6.00 .00 .02 5.98 26.80 59.58 .08.18 66.68 26.80 0.00 26.80 1787.0240 110.85 110.55 79.85 79.06 .30 .79 110.72 110.72 79.31 79.31 79.06 .30 N. .79 E. 2500.7637, 11037.9905 2500.7637 4 40 'IA •H5] Ans. The Area = 426 A. 3 R. 18 r. Areas of irregular field. 2. Given the lengths and bearings of the sides of a field, as in the following table; to find its area. 64. Problem. To find the area of a field bounded by sides, irregularly curved. Solution. Let ABCEFHIKL (fig. 22) be the field to be measured, the boundary ABCEFHIKL being irregularly curved. Take any points C and F, so that by joining AC, CF, and FL, the field ACFL, bounded by straight lines, may not differ much from the given field. Find the area of ACFL, by either of the preceding methods, and then measure the parts included between the curved and the straight sides by the following method of offsetts. bc, Take the points a, b, c, d, so that the lines A a, a b, cd, d C may be sensibly straight. Let fall on AC the perpendiculars a a', b b', cc', dd'. Measure these perpendiculars, and also the distances A a', a'b', b'c', c'd', d'C. Areas of irregular field. The triangles A a a', C dd, and the trapezoids ab a' b', b c b'c', co c' d' are then easily calculated, and their sum is the area of ABC. с In the same way may the areas of CEF, FHI, and IKL be calculated; and then the required area is found by the equation ABCEFHIKL = ACFL ABC + CEF+ FHI - IKL. EXAMPLE. Given (fig. 22) A a' = 5 ch., a' b' = 2 ch., b'd' = 6 ch., = 6 c' d' = 1 ch., d'C = 4 ch. ; also a a' = :3 ch., b b': = 2 ch., cd' = 2.5 ch., dd 1 ch.; to find the area of ABC. Ans. Required area = 2 A. 3 R. 36 r. Horizon. Bearing. CHAPTER VII. HEIGHTS AND DISTANCES, 65. The plane of the sensible horizon at any place, is the tangent plane to the earth's surface at that place. [B. p. 48.] The horizontal plane coincides with that of the surface of tranquil waters, when this surface is so small that its curvature may be neglected; and it is perpendicular to the plumb line. 66. The angle of elevation of an object is the angle which the line drawn to it makes with the horizontal plane, when the object is above the horizon ; the angle of depression is the same angle when the object is below the horizon. 67. The bearing of one object from another is the angle included by the two lines which are drawn from the observer to these two objects. 68. Problem. To determine the height of a vertical tower, situated on a horizontal plane. [B. p. 94.] Solution. Observation. Let AB (fig. 23) be the tower, whose height is to be determined. Measure off the distance BC on the horizontal plane of any convenient length. At the point C observe the angle of elevation ACB. Height of vertical tower. Calculation. We have, then, given in the right triangle ACB the angle C and the base BC, as in problem, § 34 of Pl. Trig., and the leg AB is found by (26). EXAMPLE. At the distance of 95 feet from a tower, the angle of elevation of the tower is found to be 48° 19'. Required the height of the tower. Ans. 106.69 feet. 69. Problem. To find the height of a vertical tower situated on an inclined plane. Solution. Observation. Let AB (fig. 24) be the tower situated on the inclined plane BC. Observe the angle B, which the tower makes with the plane. Measure off the distance BC of any convenient length. Observe the angle C, made by a line drawn to the top of the tower with BC. Calculation. In the oblique triangle ABC, there are given the side BC and the two adjacent angles B and C, as in 73 of Pl. Trig EXAMPLE. Given (fig. 24) BC = 89 feet, B = 113° 12', C = 23° 27'; to find AB. Ans. AB = 51.595 feet. 70. Problem. To find the distance of an inaccessible object. [B. p. 89 and 95.] Solution. Observation. Let B (fig. 2) be the point, the distance of which is to be determined, and A the place of the |