To find the difference of the bases. As the sum of the base BA and CA 105 2.02119 1.54407 2.13033 3.67440 2.02119 To the diff. of the bases BA and CA 1) 45 1.65321 22.5 To half the sum, 52.5, add half the difference, 22.5, the sum is 75, the greater base BA, and subtracted, leaves 30, the less base CA. By right angled Trigonometry, to find the Angles. By position 3, page 11, 18053° 8' + 28° 5′, = 98° 47′ = ang. BDC, the several angles required. BY GUNTER. The extent from 105 to 135, will reach from 35 to 45, on the line of numbers. The extent from 85 to 75, on the line of numbers, will reach from radius to 61° 55', [BDA, on the line of aes. The extent from 50 to 30, on the line of numbers, will reach from radius to ang, ADC 36° 59′, on the line of sines. Examples for Practice. 1. Given the angle BDC 100°, and the angle DCB 54°, the leg BD 220 perches, to find the other two legs. Answer; BC 267.8, DC 119.2. Tables vary in this sum. 2. Given the side BC 365, and the side AB 640, the angle BAC 26°, to find the other side and angles. Answer; the side AC 808.7, ang. at C 50° 14', B 103° 46'. 3. Given the side BC 110, AC 80, and their contained angle 102° 30', to find the other angles and side. Answer; the greater angle 45° 58', the less angle 31° 32', and the third side 149.3. 4. Given the side BA 88, BC 54, AC 108, to find the angles. Answer; the least angle 29° 49', next greater 54°, 07′, and the greatest 96° 04'. N 4 OF HEIGHTS. Note. The several figures in Heights and Distances, are drawn by the principles of the preceding Geometrical Problems. PROBLEM 1. To find the Height of a perpendicular Object, at one Station, on a Plane. Given A STEEPLE. The angle of altitude 53 degrees, Distance to the foot of the object, 85 feet, RULE. By Case 1, or 2, in Right Angled Trigonometry, find the perpendicular; to which add 5 feet, the height of the observer's eye, and it makes 117.8 feet, the height of the steeple required. The work in this, and the following problems, is omitted, to exercise the judgment of the learner in the To find the Height of a perpendicular Object, on the top of a Hill, from the Hill's foot. Angle to the bottom 48° 30′, Given Angle to the top 67°, Distance to the bottom, 136 feet RULE. By Case 1, Oblique Trigonometry, the angles and one side being given; thus, the angle to the top, 67°, its complement is 23°, the angle at C; then the difference between Angles the two altitudes, is 18° 30'; and of course the angle ABC 138° 30′, by position 3; hence the height of the object will be found to be 110.5 feet. And by Right Angled Trigonometry, the height of the hill may be found to be 101.8 feet and depth to the perpendicular distance of the object 90.12 feet. From the top of a Hill to find the Height of a perpendicular Object, at the foot thereof. Angle to the foot of the object 55°15′, Given Angle to the top 31° 15', Distance to the foot of the object 250 feet. Note. By the same Case, as the last Problem, the height of the object will be found to be 119 feet; the horizontal distance AE, 142.5 feet, and heigh of the hill 205.4 feet; from the height of the hill, take the height |