Plate V. fig. 26. 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° SO'~-24° 30'=12° 00' In the triangle ADB, find DB, thus; 2. S. ADB : AB::S. DAB: DB. 12° 0 0 104 249.30. 207.4. CBE-DBE=CBD, or 44° 30-36° 30'=80 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 angular trigonometry, thus; 3. BC+BD: BC-BD::T. of BDC+BCD; , 462.1. 47.3 860.00 T. of į BDC-BCD. 55°. 40. 4. S. BCD : BD :: S. CBD : DC. 30°. 20' 207.4 8° 00' 57.15 length of the tree. To find DE in the triangle DBE. Say R. : BD:: S. DBE : DE, 90°. 207.4 56°. 30 123.4 height of the hill. PROBLEM IX. To find the height of an inaccessible object CD, on a hill BC, from ground that is not horizontal. Plate VI. 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. sect. 1.) the angle HGA=EAG; therefore EAG +HGD=AGD: and (by cor. 1. theo. 1. sect. 1.) 180—the sum of EAG and BAD=GAD: and, (by cor. 1. theo. 5. sect. 1.) 180—the sum of the angles AGD and GAD=GDA: thus we have the angles of the triangle AGD, and the side AG given ; thence (by case 2, of obl. trig.) AD may be easily found. The angle DAB-CABEDAC, 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 the rectangular and oblique angular trigonometry R OF DISTANCES. A NY of the instruments used in surveying, will give you the angles or bearings of lines; which will be particularly shewn, when we come to treat of them. PROBLEM I. Plate VI. 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 angular trigonometry thus 1. S. C: AB::S. A : BC. 2. S. C:AB::S. B : AC. PROBLEM II. Plate VI. fig. 11. Let B and C, be two houses whose direct distance asunder, BC, is inaccessible: however it is known that a house at A is 252 perches from B, and 230 from C; and that the angle BAC, is found to be 70°. What is the distance BC, between the two houses ? This is performed as case 3. of oblique angular trigonometry, thus ; 1. AB + AC : AB-AC : : T. of C+B: 482 22 55°. 00' T. of įC-B 3° 44' 55° +3°. 44' = 58°. 44'= C 556_3°. 44' = 51'. 16-B. 2. S. C.: AB::S. A: BC. 58° 44' 252 70° 277. PROBLEM III. Plate VI. fig. 3. Suppose ABC a triangular piece of ground, which by an old survey we find to be thus : AB 260, AC 160, BC 150 perches the mearing lines AC and BC, are destroyed or plowed down, and the line AB, only remaining. What angles must be set off at A and B, to run new mearings by exactly where the old ones were ? This is performed as in case 4. of oblique angled trigonometry, thus; 1 AB : AC + BC : : AC-BC : AD-DB. 260 310 10 11.92 130 +5.96=135.96%AD. 2. AD: R:: AC : Sec. A. 136 90° 160 31°. 47'. 3. BC: S. A:: AC : S. B. 150 31°. 47' 160 34° 10'. PROBLEM IV. Plate VI. fig. 4. Let D and C, be two trees in a bog, to which you can have no nearer access than at A and B : there is given DAB 100°, CAB 36o. 30, CBA 121°. DBA 49°, and the line AB 113 perches. Required, the distances of the trees DC. 180°—the sum of DBA and DAB-ADB=31'. 180°—the sum of CAB and CBA-ACB=22.30. In the triangle ABD, find DB, thus 1. S. ADB : AB :: S. DAB: DB. 31° 113 100 216. |