The angle CBE-A-ACB, as in the last prob. In the triangle ABC, find AC thus ; S. ACB: AB:: S. ACB (or sup. of CBE): AC 13. 30' 104: 131.30' The angle CAE-DAE-CAD. : 333.6 The angle ACD=AED×EAD, by theo. 4. In the triangle CAD, find CD thus, S. ADC: AC :: S. CAD: DC 111. 333.6 :: 14: 86.46 required. CE, BE, or DE, may be found by other various statings, as set forth in the first and second cases of right-angled trigonometry, PROB. VIII. PL. 5. fig. 26. To find the length of an object, that stands obliquely on the top of a hill, from a plane beneath. Let CD be a tree whose length is required. This is done at two stations. Make a station at B, from whence take an angle to the foot, and another to the top of the tree; measure any convenient distance backward to A, from whence also let an angle be taken to the foot, and another to the top; and you have the requisites given. Thus, First station. Angle to the foot EBD=36°. 30′. Angle to the top EBC=44°. 30′. Stationary distance AB 104 feet. Second station. Angle to the foot EAD=24°. 30. Angle to the top EAC=32°. 00ʻ. Let DC and DE be required. The geometrical constructions of this and the next problem are omitted; as what has been already said, and the figures, are looked upon as sufficient helps. EBC-A-ACB, or 44°. 30'-32°.-12°. 30". as before. In the triangle ABC, find BC. Thus, 1. S. ACB: AB:: S. A: BC. 12°. 30' 104 32° 254.7. EBD-EAD=ADB,or 36°. 30′-24°. 30′=12° 00′ In the triangle ADB, find DB, thus ; 2. S. ADB : AB : : S. DAB : DB. CBE-DBE-CBD,or 44°. 30′-36°. 30′ =8°00′ In the triangle CBD there is given, CB 254.7, DB 207.4, and the angle CBD 8°00′; to find DC. This is performed as case 3. of oblique angled trigonometry, thus; 3. BC × BD; BC~BD: : T. oft BDC+ BCD: 86° 00′+55° 40′=141°. 40′=BDC. 4. S. BCD: BD :: S. CBD : DC. 30°. 20' 207.4 the tree. 8°.00' 86°.00' 57.15 length of To find DE in the triangle DBE. Say R.: BD :: S. DBE : DE, 90°. 207.4 36. 30' 123.4 height of the hill. PROB. IX. To find the height of an inaccessible object CD, on a hill BC, from ground that is not horizontal. PL. 6. fig. 1. From any two points, as G and A, whose distance GA, is measured, and therefore given; let the angles HGD, BAD, BAC, and EAG, be taken; because GH is parallel to EA (by part 2. theo. 3. geom.) the angle HGA-EAG; therefore EAG× HGD=AGD: and (by cor. 1. thee. 1. geom) 180-the sum of EAG and BAD-GAD; and, (by cor. 1. theo. 5. geom.( 180-the sum of the angles AGD and GÅD=GDA: thus we have the angles of the triangle AGD, and the side AG given; thence (by case 2. of obl. ang. trig.) AD may be easily found. The angle DAB-CAB DAC, and 90°-BAD-ADC; and 180°-the sum of DAC and ADC-ACD: so have we the = several angles of the triangle ACD given, and the side AD; whence (by case 2. of obl. trig.) CD may be easily found. We may also find AC, which with the angle BAC, will give CB the height of the hill. The solutions of the several problems in heights and distances, by Gunter's scale, are omitted; because every particular stating has been already shewn by it, in trigonometry. 2d. OF DISTANCES. THE principal instruments used in surveying, will give the angles or bearings of lines; which has been particularly shewn, when we treated of them. PROB. I. PL. 6. fig. 2. Let A and B be two houses on one side of a river, whose distance asunder is 293 perches: there is a tower at C on the other side of the river, that makes an angle at A, with the line AB of 53° 20′; and another at B, with the line BA of 66°20'; required the distance of the tower from each house, viz. AC and BC. This is performed as case 2. of oblique angled trigonometry, thus; 1. S. C: AB : : S. A: BC. 2. S. C: AB:: S. B: AC. PROB. II PL. 6. fig. 11, Let B and C, be two houses whose direct distance asunder, BC, is inaccessible: however it is |