The Theodolite is a semi-circle divided into 180°, with an index which turns about on its centre, and retains any fituation given it, on which are two lights, called the moveable fights; there are also two other fights fixed on the diameter of the theodolite, which are called the fixed fights. Fig. 2. plate 4. Sights are small pieces of wood or brass, having small holes or slits in them, to view the object through ;---They are fixed perpendicular to the plane of the theodolite, but parallel to the plane of the quadrant. The geometrical square may be made of brass, wood, or any solid body, having equal sides and angles; from one of the angles, a thread is suspended, with a small weight at the end, fo so as to point always to the centre. The two sides opposite to the centre of fuspension, are divided each of them into 100 equal parts; there is also an index, which, (when occasion ferves), may be fixed to the centre of suspension, and is made so as to turn round, and retain any situation ; on this index, are two fights. See fig. 3. plate 4. Heights and distances are of two kinds, viz. accesible and inaccessible : accessible objects are houses, growing trees, &c. inaccesible ones are all mountains, celestial bodies, also houses and trees, in certain situations. Let AB be a horizontal plane and BC a tower, whose height a , is required : From B, the foot of the tower, measure any convenient distance, 80 feet upon the horizontal plane AB. Suppose the tower to subtend an angle of 39° 49' from A. What is its height? Ag 11.70950 To the height of the tower 66.69 = = 1.82408 EXAMPLE II. A tower, surrounded by a ditch 40 feet broad: from the other side of the ditch, the tower subtends an angle of 53° 13' Required the height of the tower, also the length of a ladder sufficient to scale the tower. See fig. 58. plate 4. . 10.0000 To find the height of the tower. Tofind the length of the ladder. 90.10000 As radius 90° is to the breadth of is to the br. of ditch 40 1.60206 the ditch 40 1.60206 So is fec. elev. 53° 13' 10.22256 So is tan. el. 53° 13' 10.1 2631 To ladder 66.78 = 1.82462 To the height of the tower 53:5 =1.72837 EXAMPLE III. Plate 4. fig. 59. From the top of a ship-mast 100 feet above the level of the water, I took an angle of depression of another ship's hull, 74° 15'; required the distance of the other fhip. PROBLEM II. To measure inaccessible heights and difrances. 1 EXAMPLE I. Plate 4. fig. 60. At the foot of a hill, I took an angle of elevation of its top, and found it to be 50° 42'. I then measured back 120 yards on the horizontal plane, and observed the angle to be 40° 12'. Required the perpendicular height of the hill. N. B. When any side AB of the triangle ADB is produced, the exterior angle DBC is equal to both the interior and opposite angles DAB, ADB; therefore the angle ADB will be 10° 300 10.00000 O To find BD. To find DC the height. 2.07918 2.62839 So is sine an. A 40°12' 9.80987 So is fine DBC 50°42' 9.88865 11.88905 To the height 328.9 2.5:704 • 2.62842 TO BD 425 EXAMPLE II. Plate 4. fig. 67. I observed an object on the other side of a river, on a level with the place where I stood ; behind me was a regular declivity, which I might reckon a straight line. I marked my station by the side of the river, and measured back 170 yards, when I observed I was higher than the object. I took the angle of depression of the mark by the river side 42° 18', of the H bottom bottom of the object 72° 8', and of its top 78° 20'. Required the height and distance of the object. Here, because the angle ABC is 42° 18' the angle BAC is 47° 42; consequently, its supplement, the angle BAD will be 132° 18. And since all the angles of a triangle are equal to two right angles, and that the angle DBA is 29° 50', the remaining angle BDA will be 17° 52' Again, the angle CDE is a right angle, of which the angle BDC is a part; therefore, the angle BDE is 72° 8', and the angle at E 101° 40'; also the angle DBE will be 6° 12'. To find the dist. of the object. To find BD. As fine ADB 17°52' 9.48686 | As fine BDA 17° 52' 9.48686 is to AB 170 - 2.23045 is to AB 170 2.23045 So is fine ABD 29° 50' 9.69677 | fois sineBAD=132°18'9.86902 Being on a horizontal plane, I took the angle of elevation of the summit of a hill, and of the top of a tower built upon it, and found them to be 48° 20' and 61° 25'. I then measured back 150 yards, and found the angle subtended by the height of the tower above the plane to be 38° 19'. Required the height of the tower. The The exterior angle CBD, is equal to both the interior and opposite angles, CAB, ACB; but CAB is 38° 19' ; therefore, ACB will be 23° 6': and since all the angles of a triangle are equal to two right angles, angle ABC will be 118° 35'. Or it is the supplement of the angle CBD; also angle BCD is 28° 35', and CEB will be 138° 20'. ' To find BC. To find the tower's height. As fine an. ACB 23°6' 9.59366 | As fine CEB 138° 20' 9.82269 is to AB 150 2.17609 is to BC 237 2.37475 So is finean. A 38°19' 9.79240 Sois fine CBE 13° 5' 9.35481 TO BC 237 11.96849 11.72956 2.37483 To the height of the tower 80.7 1190687 EXAMPLE IV. Plate 5. fig -2 a From a window on a level with the bottom of a steeple, I took the angle of elvation of the top of the steeple 50° ; from another window, 20 feet perpendicular above the former, I took another angle of the top of the steeple 45° 15' Required the height and distance of the steeple. Because the angle ACD is a right angle, of which the angle SCD=50° is a part, the angle SCA will be 40°, consequently, the alternate angle CSD will also be 40°. And fince the angle SAB is 45° 15', and the angle BAD a right angle : therefore, the whole angle SAC 135° 15', and the angle ASC 4° 45'. a To find CS. To find the height of the steeple. As fine ASC 4° 45' 8.91807 | Assec. ang. SCD 50° 10.19193 is to AC 20 1.30103 is to SC 170 2.23045 So is fineSAC 135° 15'9.84758 So is tan. SCD 50° 10.07619 To CS 170 11.14861 130.2 feet. 12.30664 2.11471 |