As cosine ang. elev. 39° 49' 9.88542 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 fubtends an angle of 53° 13' Required the height of the tower, also the length of a ladder fufficient to scale the tower. See fig. 58. plate 4. To find the height of the tower. Tofind the length of the ladder. 90.10000 10.0000 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 ico 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 diffances. 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° 30': To find BD. To find DC the height. Assine ADB=10°30' 9.26063 | As rad. 90 10.00000 is to AB 120 2.07918 is to BD 425 2.62839 So is fine an. A 40°12' 9.80987 So is fine DBC 50°42' 9.88865 To BD 425 11.88905 To the height 328.9 2.5:704 2.62842 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 fide 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° 78', 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 dift. of the object. . To find BD. As fine ADB 17° 52' 9.48686 | As sine 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 fineBAD=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 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'. To find CS. To find the height of the steeplé. As sine 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 sineSAC 135° 15'9.84758 So is tan. SCD 50° 10.07619 12.30664 To CS 170 11.14861 130.2 fcet. } 2.11471 EXAMPLE V. Plate 5. fig. 3. From the top of a tree 70 feet high, I took the angle of depression of two other trees, lying directly in a straight line from me, and on the same horizontal plane with the tree on which I then stood, viz. that of the nearer 36°, and of the other, 55° 30'. Required their distance from the tree from which the. observation was taken, and from one another. To find the dist. of the nearer. To find the dift of the other. 1 0.00000 10.00000 to height of tree 70 1.84510 is to height of tree 70 1.84510 So is tan. dep. 36° 9.86120 | Soistan.2.depr.55°30'10.16287 To the dist. 50.86 1.70636 To the dift. 101.9 2.00797 The distance of the farthest lor 9 feet. 50.86 feet. 51.04 feet. Their distances from one another EXAMPLE VI. Plate 5. fig. 4. Wanting to know the distance between a house and a tree, the tree being on the other side of a river; I took my first station at the house, and marked my second at B; the angle subtended by the distance between my second station, and the tree is 60°. I then measured the distance between my first and fecond ftations, 380 yards, and found the angle subtended by the house and tree to be 43°. Required the distance between the house and the tree. As |