it subtended at a station to be 55° 40′, and measured from the station to the trees 588 and 672 yards. tance. Required their dis- PROB. XIII. To find the distance between two places, both of them inaccessible. 1. To find the distance of two places A and B, on the opposite side of a river, I took two stations, C and D, distant 1267 links from one another, and such, that from each of them the other station and the places A and B were seen. At C I took the angles BCA 53° 38', and BCD 34° 50′, and at D the angles ADC D 43° 44', and ADB 58° 38'. Required the distance between A and B. In the triangle ADC, the angle ACD is 88° 28', and CAD 47° 48'; hence sin. A: sin. C :: CD: DA= 1709.69. In the triangle BCD, the angle CDB is 102° 22′, and CBD 42° 48′; hence sin. B: sin. C:: CD: DB = 1065·14. In the triangle ADB are given AD and DB, and the angle ADB; therefore (AD+DB) 2774.83 : (AD — DB) 644-55 :: tan. (A+B) 60° 41' tan. (A — B) =22° 28'; whence ABD is 83° 9', and sin. ABD : sin. ADB :: DA : AB = 1470-3 links. 2. To find the distance between two steeples A and B, I took two stations C and D, distant 428 yards from one another; and at C took the angles ACB 54° 30′, and BCD 42° 26'; and at D took the angles CDA 40° 44′, and ADB 57° 42'. Required the distance of the steeples. Ans. 546-7 yards. 3. To find the distance between two places M and P, I took two stations A and B, distant from one another 908-36 feet; and at A took the angles PAM 14° 34', and MAB 46° 16′; and at B took the angles ABP 96° 44′, and PBM 18° 39′. Required the distance between M and P. Ans. 674-6375 ft. NOTE. If the distance between the objects be known, and the distance between the stations be required, assume 1 or 1000 for the distance between the stations, and with it find the distance between the objects. Then, as the distance found is to the given distance, so is 1000 to the true distance between the stations. 4. Suppose the distance AB 700 feet, and at the station C let the angles ACB be 42° 45′, and BCD 54° 12′, and let the angles at D be ADB 50° 19′, and ADC 57° 33′. Required the distance CD. Ans. 330-04 feet. 5. To find the distance between two lighthouses A and B, I measured the distance between two stations M and R, 3370 yards, and at M took the angles AMB 37° 52′, and BMR 91° 27', and at R the angles ARM 29° 56', and ARB 40° 27'. Required the distance AB. Ans. 7063-36 yards. 6. At a station C, I took the angle ACB, subtending a line AB 3291 yards, and found it 4° 35', and the angle BCD between B and another station D 86° 52′; and at D took the angles ADB 8° 24′, and ADC 70° 23'. Required the distance of the stations from one another. Ans. 3370-248 yards. PROB. XIV. Given the distances of three places, A, B, C, from one another, viz. AB 317, AC 308, and BC 478 feet, and the angles which these distances subtend at a station D in the same plane with them, viz. ADB 24° 50', and ADC 27° 44′; to find the distance of the station D from each of the places. Having drawn the triangle ABC, make at the point C, on the side of BC, opposite to that on which the station D lies, the angle BCd 24° 50′, and at B the angle CBd 27° 44', and about the triangle BCd describe a circle, and join Ad, meeting the circle again in D, and join BD and DC. The three sides of the triangle ABC are given to find the angle ABC-39° 25′ 14-6′′; = B C then ABd ABC±dBC= 67° 9′ 14′′, when A and d are on different sides of BC, or = 11° 41′ 14·6", when, as here, A and d are on the same side of BC. Also, the angles of the triangle BCd are given, with the side BC, to find Bd=252.7 feet. Again, in the triangle ABd are given the sides AB and Bd, and the included angle ABd, to find the angles AdB = 131° 53′ 53′′, and BAd=36° 24′ 53′′. Then in the triangle ABD are given the angles and the side AB, to find BD = 448.066, and AD = 661.738. And in the triangle DBC are given the angles and BC, to find DC = 591.563 feet. 2. If A be the place nearest to D, the angle BAd is 46° 47′ 32.2"; then BD is 550-153, AD 282.25, and CD 528-4 feet. NOTE 1. If the given station be within the triangle, as at d, make the angles BCD and CBD equal to the supplements of BdA and AdC. NOTE 2. If two of the given places, A and B, be in a straight line with the station D, the distances BC and CA subtend the same angle BDC. After finding the angle at B, work the triangle DBC. NOTE 3. If the three places A, B, C, be in a straight line, the first operation will not be required. The rest are the same as before. 3. The three sides of the triangle ABC are AB 280, BC 314, and AC 326 yards; and from the station D without the triangle, the angle ADB was 25° 52′, and ADC 23o 6', the point C being the nearest to D. Required their distances from D. Ans. AD 586·163, BD 413·4114, CD 308·1078 yds. 4. Suppose AB 267 feet, BC 209, and AC 346, and at the point D, within the triangle, the angle ADC is 128° 40′, and ADB 91° 20′. Required the distances of D from the angles. Ans. AD 195.357, BD 85-98, and CD 188.5074 feet. NOTE. When D is in one of the sides, describe a segment on BC containing the given angle. 5. Suppose AB 122, BC 74, and AC 82 chains, and at D in AB, produced beyond B, the angle ADC is 22° 45'. Required the distance of D from the angles. Ans. AD 181.79, BD 59.79, and CD 125-434 chains. 6. Suppose AB 1234, BC 873, and AC 632 yards, and at D in AB the angle ADC is 120°. Required its distance from the angles. Ans. AD 226-117, BD 1007.883, and CD 487.84 yards. 7. Suppose AB 138, BC 224, and AC 326, and at D the angles are ADB 7° 22′, and ADC 19° 58′. Required the distance of D from the angles. Ans. AD 510-9635, BD 385-2876, and DC 204-875. PROB. XV. Given the angles of elevation of a tower PS, taken at three stations A, B, and C, on a level plane, no two of which are in the same vertical plane with the tower, viz. PAS 20° 10′, PBS 18° 50′, and PCS 34° 30′, and also the distances between the stations AB 324, BC 568, and AC 672 yards; to find the height of the tower. Make the triangle ABC, of which AB is 324, BC 568, and AC 672 yards; make BE = BC, and BD BA. Join ED, and upon it make the triangle EDF on either side of DE, so that BE: EF:: cot. PBS : cot. PAS and BD: DF:: cot. PBS : cot. PCS; or make EF = 527.495, DF160-792, and join BF, and make the angle BAP = BFE. Then erect PS perpendicular to the plane ABP, and in the plane passing through AP and PS make the angle PAS = 20° 10, and PS will be the tower required. Join PC, CS, BS, the triangles APB, FBE, being similar, AP: PB:: FE: EB:: cot. SAP: cot. SBP, therefore SBP is 18° 50′; also PB: BE = BC :: BA = BD: BF, there fore the triangles PBC and FBD are similar; and BP: PC :: BD:DF :: cot. PBS: cot. PCS, therefore PCS is 34° 30′. In each of the triangles EBD, EFD, are given the three sides, to find the angles BED 28° 45′ 31′′, and FED 6° 47′ 24"; then their difference 21° 58' 7", or their sum 35° 32' 55", is the angle BEF, from which, with the sides BE and EF, the angle BFE or BAP is found in the first case to be 89° 48′ 3"-7, and in the other 78° 48′ 18"-2. Therefore AP is 804-313 or 507·692, and PS is 295·3986 or 186·4592. 2. Let AB be 326, BC 584, and AC 683, and the angles of elevation SAP 30°, SBP 26°, and SCP 23°; to find PS. Ans. PS is 952.161 or 168.645. 3. Let AB be 80, BC 119, and AC 140 feet, and the elevation at A 50°, at B 60°, and at C 55°. Required the height of the object D. Ans. 305-431 or 97.3602 feet. 4. Let AB be 60, BC 72, and AC 132 feet, and the elevations of S at A 30° 48', at B 40° 33', and at C 50° 23′. Re*quired the height of S. Ans. 94-8328 feet. 5. Let AB and BC be each 84 feet, and the points A, B, C, in a straight line, and the elevation at A 36° 50', at B 21° 24', and at Č 14°. Required the height of the object. C OF LEVELLING. Ans. 53·9606 feet. When the altitudes of the several parts of an irregular ascent are to be determined, the surveyor should be provided with a SPIRIT. LEVEL, with telescopic sights, and one or two square poles, which slide out to the length of 20 or 25 feet, divided into feet and hundredth parts of a foot. On each pole is fitted a moveable vane, with a strong black line drawn horizontally between two white ones. A small level is also fixed upon the top of the under-part of the pole, to assist in holding it perpendicular during the observations. PROB XVI. To find the height of g above a. Place an assistant with the pole ab at a, and another with the pole cd at c, and having fixed the level nearly midway between them, turn the telescope towards a, and direct the assistant to move the vane upwards and downwards b a upon the pole till the black line on it coincide with the horizontal hair in the telescope, and then let him fix the vane. The feet and hundredth parts of a foot, cut by the under-part of the vane upon the pole, are then carefully read off, and entered into the surveyor's book. The telescope is then turned towards the pole at c, and the assistant is directed, the height read off, and entered as at the first station. The pole at a is now removed and placed at e, whilst that at c still remains; the level is again placed in the middle between them, and the observations made and registered as before; and so on till the whole is finished. The difference between the sums of the heights of the back-observations, or those taken with the telescope directed towards a, and that of the fore-observations, or those taken towards g, will show the height of g above a.* To find the height of any point c in a regular ascent: The distance ag is to ac as the height of g above a to the height which c ought to have above a. It is not necessary to place the poles in the same direction with ab and gh, but it is necessary to erect them perpendicular, or nearly so. NOTE. When the distance between the poles ab and cd is very great, the line bm will differ a little from the true level; for bm is a tangent to a great circle of the earth, passing through the centre of the instrument, and the true level is the arc of that circle between the poles ab and od. The correction may be neglected when the distance between the stations does not exceed 300 or 400 yards, and the instrument is placed in the middle between them: for a mile it is 7.96 or 8 inches; and for other distances from the instrument, the correction varies as the square of the distance. 1. To determine the height of an eminence, the following observations were taken : Two poles are not necessary, for, after taking the back-observation upon the pole at a, it may be removed to c, and the fore-observation taken; then, removing the level into the second position, another back-observation is taken, and the pole-removed to e for another fore-observation, and so on. |