Plate V. Fig. 20. PROBLEM II. To find the height of a perpendicular object, on an horizontal plane; by having the length of the shadow given. Provide a rod, or staff, whose length is given, let that be set perpendicular, by the help of a quadrant, thus; apply the side of the quadrant AC, to the rod, or staff; and when the thread cuts 90° it is then perpendicular; the same may be done by a carpenter's, or mason's plumb. Having thus set the rod or staff perpendicular; measure the length of its shadow, when the sun shines, as well as the length of the shadow of the object, whose height is required; and you have the proper requisites, given. Thus, ab, the length of the shadow of the staff, 15 feet, bc, the length of the staff, 10 feet. AB, the length of the shadow of the steeple, or object, 135 feet. Required BC, the height of the object. The triangles abc, ABC, are similar, thus: the angle b-B, being both right; the lines ac, AC are parallel, being rays, or a ray of the sun; whence the angle a=A (by part 3. theo. 3. sect. 1.) and consequently c-C. The triangles being therefore mutually equiangular, are similar (by theo. 16. sect. 1.) it will be, ab: bc:: AB: BC. 15 10 135 90. the steeple's height, required. The foregoing method is most to be depended on; however, this is mentioned for variety's sake. PROBLEM III. To take the altitude of a perpendicular object, at the foot of a hill, from the hill's side. Turn the centre A of the quadrant, next your eye, and look along the side AC, or 90 side, to the top and bottom of the object; and noting down the angles, measure the distance from the place of observation to the foot of the object. Thus, Given, Angle to the foot of the object, 55° 1⁄2 Angle to the top of it, 31° or 31° 15′ Required the height of the object. Geometrically. Draw an infinite blank line AD, at any point in which A makes the angle EAB of 55° 15', and EAC of 31° 15'; lay 250 from A to B; from B, draw the perpendicular BE (by prob. 7. of geometry) crossing AC in C; so will BC be the height of the object required. Plate V. fig. 21. In the triangle ABC there is given, ABE the complement of EAB to 90° which is 34° 45'. ACB the difference of the given angle 24° 00', The sides AB, 250. Required, BC. This is performed as case 2. of oblique angular trigonometry. Thus, 180-the sum of ABE 34°. 45', and CAB 24° 00' ACB 121°. 15'. Then, S. ACB: AB:: S. CAB: BC. 121°. 15′ 250 24°.00 119, the height required. PROBLEM IV. To take the altitude of a perpendicular object, on the top of a hill, at one station; when the top and bottom of it can be seen from the foot of the hill. Plate V. Fig. 22. As in prob. I. take an angle to the top, and another to the bottom of the object; and measure from the place of observation to the foot of the object, and you have all the given requisites. Thus, Given, A Tower on a hill. Angle to the bottom, 48°. 30. Angle to the top, 67°. 00'. Dist. to the foot of the object, 136 feet. Required, the height of the object. Plate V. fig. 22. Geometrically. Make the angle DAB=48° 30', and lay 136 feet from A to B; from B, let fall the perpendicular BD; and that will be the height of the hill: produce BD upwards by a blank line: again, at A, make the angle DAC-67° 00′ by a blank line, and from C where that crosses the perpendicular produced, draw the line CB, and that will be the height of the object required. Let AC be drawn. In the triangle ABC, there is given, The angle ACD the complement of DAC23° 00'. CAB the difference between the two given angles-18° 30'. And the side AB 136. To find BC. S.C: AB:: S.CAB: BC. 23° 136 18°.30' 110. If BD were wanted, it is easily obtained, by the first case of right-angled plane trigonometry. THEOREM II. To take an inaccessible perpendicular altitude, on an horizontal plane. Plate V. fig. 23. This is done at two stations, thus: Plate V. fig. 23. Let DC be a tower which cannot be approached, by means of a moat or ditch, nearer than B; at B, take an angle of altitude, to C: measure any convenient distance backward to A, which note down; at A, take another angle to C; so have you the given requisites, thus ; First angle, 55°. 00' Given, Stationary distance, 87 feet. The height of the tower CD is required. Geometrically. Upon an infinite blank line, lay off the stationary distance 87, from A to B; from B, set off your first; and from A, your second angle; from C, the point of intersection of the lines which form these angles, let fall the perpendicular CD; and that will be the height of the object required. The external angle CBD, of the triangle ABC; is equal to the two internal opposite ones, A, and ACB: (by theo. 4. sect. 1.) wherefore if one of the internal opposite angles be taken from the external angle; the remainder will be the other internal opposite one, thus; CBD 55°--A 37°-AÇB 18° Therefore in the triangle ABC we have the angles A, and ACB, with the side AB given to find BC. |