that is, 726576: 200 :: 576: 768 links = BA, 4. Should impediments prevent the lines BC, HI being measured perpendicular to HBA, measure BC to make any angle with HA, and at a convenient distance BH measure HI parallel to BC, which may be done by erecting two equal perpendiculars to BC, the point I being in a right line with A and C as before. H B The rule for finding BA will be the same as in the last example; the calculation in both cases may be readily performed on the ground. 5. Lastly, if the nature of the ground be such that none of the four preceding methods can be adopted, neither to the right nor to the left of the base-line, measure BC in the most convenient direction, and take the angles ABC, ACB with the theodolite; then by taking the sum of these angles from 180°, the angle A will become known, whence, by Case I. of oblique-angled triangles : As sin. A: BC :: sin. C : BA. This problem may also be solved by construction. Examples for practice. 1. BC= 308 links (fig. to first method) is measured perpendicular to BA and CD is ranged perpendicular to CA, meeting DA in D, the distance BD being 400 links, and the flags at B and a being at the water's edge; required the distance BA. Ans. By the second method BA is found = 237 links nearly. 400 (fig. to first method), HI = = 2. Given BC links, to find BA. 180 520, and BH = = 600 links. Ans. By the third method BA is found 3. Given BC= 8 chains, and the angles at B and C respectively 30° and 45°; required the distance BA, and the perpendicular width AD of the river. Ans. BA=585 links, and AD = 292 links. * This rule is derived from similar triangles; for HI: BC:: HA: BA, and HIBC HA-BA :: BO : BA, i.e., HI - BC: HB:: BC: BA. PROBLEM II. To continue the direction and measurement of a given line when buildings or other obstructions stand in the way, which can be avoided by going to right or left. A C H 1. At any convenient point A in the given line aACC, which is obstructed by the buildings H, take an angle aAB = 120° (thus making the angle CAB = 60°); range and measure the line AB, till, by taking an angle B = 60°, the line BC will clear the obstructions at H; then measure BC = AB, and take the angle BCC = 120°. Then Cc is in the range of the given line aAcc, and AC=AB or BC, the triangle ABC being thus made equilateral; hence the measurement of the line aACC may proceed from C, after having added the distance thus found to the measure at A. B 2. If the nature of the ground be such as not to admit line AB to make with as an angle aAB= 120°, take any other angle, and suppose it to be 132°, which taken from 180° leaves the angle CAB = 48°; measure AB, till by taking an angle at B = 84° twice the complement of the angle CAB, the line BC may clear the buildings; then BA, taking the angle CCB also measure BC = = = 132°, and continue the line in the direction cc. The triangle ABC is thus made isosceles, and a perpendicular let fall from B to H on AC will divide ABC into two equal right-angled triangles. Hence as rad. or sin. 90° side AB :: sin. B, or 42° : AH = AC, whence AC becomes known. 3. If it be convenient to take the angle aAB = 135°, the angle B = 90°, AB = BC = (suppose) 800 links, and lastly the angle CCB = 135°, thus making each of the angles CAB, ACB = 45°, the triangle being right-angled at B, as well as isosceles. Hence, by Euc. I. 47, NOTE.-By any of these methods the distance AC between any two given points A and c may be found, when sight is not obstructed, as when H is a lake or projection of a sea-coast, intervening between A and c, that cannot be measured across with the chain. PROBLEM III. To find the distances of three accessible and visible objects, when obstructions prevent the measurement of these distances in direct lines. B Let A, B, and D be the three objects, the distances of which, AB, AD, and BD, cannot be measured directly, on account of the obstruction shown in the figure. Set up a flag at C in the direction AB, take the angle ACD, measure the distance CD, and take the angles ADC and ADB. Then in the triangle ACD are given the angles ACD and ADC, and their included side CD, to find AD. In the triangle ADB are given the angles ADB and DAB = ACD + ADC, and their included side AD, to find AB and BD. D Ex. Let CD1962 links, and the angle c=57°, CDA = 14°, and 41° 30′; required the distances AD, DB, and AB. ADB = As sin. 109°: CD=1962 :: sin. 57° : AD = 1740 links. The angle B 180° (71° 41° 30′) = 67° 30′. As sin. 67° 30′ : AD = 1740 :: sin. 41° 30′ : AB = 1248 links, = : :: sin. 71° : BD = 1781 links. By construction. Draw the indefinite line CB, make the angle C = 57; draw CD, which make 1962 links, lay off the angles CDA = 14°, CDB = 14° + 41° 30' == 55° 30′, and draw the lines DA and DB meeting CB in A and B, then the distances AD, AB, and BD will be found as above. PROBLEM IV. To find the height of a tower or any other object, standing on an horizontal plane, the base of the tower, &c., being accessible. C Let CB be a tower, standing on an horizontal plane. Measure any convenient distance, AB, in a direct line from the tower; fix the theodolite at A, and take the vertical angle Cab, the line ab being in the level of centre of the vertical arc of the instrument. Then in the right-angled triangle acb are given ab = AB, and the angle cab, to find cb; to which add AaBb = height of the centre of the instrument, and the sum will be the height of the tower. Ex. Let AB = 200 feet, the angle Cab 46° 30′, and the height of the centre of the instrument 5 feet; required the height of the tower. As rad. sin. 90°: tan. Cab = 46° 30′ :: 200= ab: cb= 210.76 = AaBb= 5 = 215.76 feet. The height of the tower BC PROBLEM V. To find the height of a tower or any other object, standing on an horizontal plane, the base of the tower, &c., being inaccessible. Ce Let CB be a spire or steeple inaccessible on every side, on account of trees, a portion AD of the ground line AB being horizontal, and passing through the bottom of the Take the angle CDB = 51° measure the distance AD 75 feet, and take the angle A = spire. 30', = 26° 30'. Required the height of the spire and the distance of the first station from the centre of its base. A D B Subtract the angle A from the angle CDB, the remainder 25° = angle ACD. Then in the oblique-angled triangle ADC, AD and all the angles are given to find DC=79-18; and in the right-angled triangle DBC, the hypotenuse DC and angle BDC are given to find BC 61.97, and DB equal 49-29 feet. Or BC may be found by a single operation by the formula BC = AD × sin. A x sin. D that is, add together the logarithms of AD, rad. × sin. (D– A)’ and of the sines of the angles A and D; from the sum subtract the log. sin. (DA) with its index increased by 10, and the remainder is the log. of the height BC. Thus, To this result the height of the instrument should be added. Examples for practice on the two last problems. 1. At the distance of 45 feet from the bottom of a steeple, on an horizontal plane, the angle of elevation is 48° 12', and the height of the instrument 5 feet. Required the height of the steeple. Ans. 55 feet 4 inches. 2. The angles taken at two stations A and D (see last fig.) on an horizontal plane passing through the base of an inaccessible tower CB, are 28° 34′ = CAB, 50° 9' = CDB, the distance of stations being 30 yards = AD; required the height of the tower, and its distance from the station D, the height of the instrument being 51 feet. Ans. Height of tower 31-69 yards; distance from D = 24.99. PROBLEM VI. To find the height of a tower standing on the side or top of an Let CE be a tower on the inaccessible hill E, ADB the horizontal plane, the point B being in the prolongation of the line CE; adb the level of the centre of the instrument, with which the angles a, cdb, Edb were taken, and found to be respectively 44°, 67° 50', = = 230.36 In the triangle CdE all the angles are given, viz. cdE 67° 50' 51° = 16° 50′, dcE · 90° - 67° 50′ = 22° 10′. Hence the angle cEd = 141°, and CE = 106 yards, the height of the tower. = In the right-angled triangle cdb, the angle cdb and DC are given to find cb 213.34 yards: hence be bc which add the height of the instrument = = CE 107.34 yards, to = 5 feet = 1.66 yards, and there results EB 109-01 yards, the height of the hill. = |