PROBLEM III. Coming from sea, at the point D I observed twɔ headlands, A and B, and inland, at C, a steeple, which appeared between the headlands. I found, from a map, that the headlands were 5.35 miles apart; that the distance from A to the steeple was 2.8 miles, and from B to the steeple 3.47 miles; and I found, with a sextant, that the angle ADC was 12° 15, and the angle BDC, 15° 30'. Required my distance from each of the headlands, and from the steeple. CONSTRUCTION. The angle between the two headlands is the sum of 15° 30′ and 12° 15', or 27° 45'. Take double this sum, 55° 30'. Conceive AB to be the chord of a circle, and the arc on one side of it to be 55° 30'; and, of course, the other will be 304° 30'. The point D will be somewhere in the circumference of A E this circle. Consider that point as determined, and draw CD. In the triangle ABC, we have all the sides, and, of course, we can find all the angles; and if the angle ACB is less than 180°. 27° 45′ = 152° 15', then the circle cuts the line CD in a point E, and C is without the circle. Draw AE, BE, AD, and BD. AEBD is a quadrilateral in a circle, and ADB = 180°. = AEB + The ADE= = ABE, because both are measured by one half the arc AE. Also, | EDB LEAB, for a similar reason. Now, in the triangle AEB, its side AB, and all its angles, are known; and from thence AE can be computed. Then, having the two sides, AC and AE, of the triangle AEC, and the included angle CAE, we can find the angle AEC, and, of course, its supplement, AED. Then, in the triangle AED, we have the side AE, and the two angles AED and ADE, from which we can find AD The computation, at length, is as follows: 70° 47′50, supplement 109° 12' 10", angle CAD The elevation of a spire at one station was 23° 50' 17', d the horizontal angle at this station, between the spire and another station, was 93° 4′ 20′′. The horizontal angle at the latter station, between the spire and the first station, was 54° 28' 36", and the distance between the two stations was 416 feet. Required the height of the spire. Let CD be the spire, A the first station, and B the second; then the vertical angle CAD is 23° 50′ 17′′; and as the horizontal angles, CAB and CBA, are 93° 4′ 20′′ and 54° 28′ 36′′, respectively, the angle ACB, the supplement of their sum, is 32° 27′ 4′′. To find AC. B m M B For example, let M be a prominent tree or rock near the top of a mountain, and by observations taken at A, we can determine the perpendicular Mn. By like observations taken at B, we can determine the perpendicular Mm. The difference between these two perpendiculars is nm, or the difference in the elevation between the two points A and B. If the distances between A and n, or B and m, are considerable, or more than two or three miles, corrections must be made for the convexity of the earth; but for less distances such corrections are not necessary. PRACTICAL PROBLEMS. 1. Required the height of a wall whose augle of elevation, at the distance of 463 feet, is observed to be 16° 21'. Ans. 135.8 feet. 2. The angle of elevation of a hill is, near its bottom, 31° 18', and 214 yards further off, 26° 18'. Required the perpendicular height of the hill, and the distance of the perpendicular from the first station. Ans. { The height of the hill is 565.2 yards, and the distance of the perpendicular from the first station is 929.6 yards. 3. The wall of a tower which is 149.5 feet in height, makes, with a line drawn from the top of it to a distant object on the horizontal plane, an angle of 57° 21′. What is the distance of the object from the bottom of the tower? Ans. 233.3 feet. 4. From the top of a tower, which is 138 feet in height, I took the angle of depression of two objects standing in a direct line from the bottom of the tower, and upon the same horizontal plane with it. The depression of the nearer object was found to be 48° 10', and that of the further, 18° 52'. What was the distance of each from the bottom of the tower? Ans. Distance of the nearer, 123.5 feet; and of the further, 403.8 feet. 5. Being on the side of a river, and wishing to know the distance of a house on the opposite side, I measured 312 yards in a right line by the side of the river, and then found that the two angles, one at each end of this line, subtended by the other end and the house, were 31° 15′ and 86° 27'. What vas the distance between each end of the line and the house? Ans. 351.7, and 182.8 yards. 6. Having measured a base of 260 yards in a straight line, on one bank of a river, I found that the two angles, one at each end of the line, subtended by the |