The adding and subtracting these of angles, are omitted to try the judgment of the learner. PROBLEM IV. To take the Height of an inaccessible Object, on a plane, at two Stations. Angle at the nearest station to the top 55°, Given Stationary distance 87 feet backwards, Angle at the farthest station to the top 37°. RULE. By Case 1, Oblique Angled Trigonometry, find the distance from either station, to the top of the object; from the nearest is 169.4 feet; from the farthest is 230.6 feet; then, by Right Angled Trigonometry, the height of the object will be found to be 138.8 feet. Let BC be a pole 100 feet high, and broken off at D, so that the part broken off, viz. DC, will reach from the top of the stump to A, on a plane 34 feet from the bottom or foot of the pole. Required the length of the part broken off?. RULE. In the right angled triangle ABC, the base and perpendicular are given, to find the angles; and by Case 6 and 7, Right Angled Trigonometry, the angle ACB will be found to be 18° 47', and 90°. -18° 47'71° 13′ CAB; then by position 8, page 15, intersect BC, in the broken place at D; then, by position 11, page 16, AD and CD being equal, their opposite angles must be equal; therefore, in the right angled triangle ABD, the angles and base are given to find the hypothenuse AD 55.77 feet, the part required. To take the Height of a perpendicular object, on a Hill, at two Stations, from a Plane beneath it. Given Farthest station, angle to the bottom 21°, angle to the top 35°, Stationary distance 104 feet directly forwards, RULE. By Case 1, Oblique Angled Trigonometry, find the distance from the farthest station to the top of the object, viz. 333.6 feet; then by the same, the height of the object will be found to be 86.76 feet. Note. The addition and subtraction of the angles, are omitted to exercise the judgment of the learner. PROBLEM VII. To find the Length of an Object which stands obliquely on a Hill, at two stations, on a Plane beneath it. Nearest station, angle to the bottom 36° 30', angle to the top 44° 30', Given Stationary distance, 104 feet backward, Farthest station, angle to the bottom 24° 30', angle to the top 32°. RULE. By Case 1, Oblique Angled Trigonometry, find the distance from either station to the top and bottom of the object, as from the nearest to the top 254.7 feet, and to the bottom 207.4 feet; then by Case 4 and 5, the length of the object may be found to be 57.15 feet. ACB 12° 30', to find BC 254.7 feet, Angles ADB 12°, to find BD 207.4 feet, CBD 8°, to find the length of the object 57.15 feet. |