the mearing lines 4C and BC are destroyed or ploughed 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. 10 11.92. 130+5.96=135.96=AD. 130—5.96=124.04=DB. 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. Pl. 6. 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 36° 30', CBA 121°, DBA 49°, and the line AB 113 perches. Required the distance 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. And in the triangle ABC, find BC thus, 2. S. ACB: AB::S.CAB: BC. 22° 30' 113 36° 30' 175.6. In the triangle DBC, you have DBC=ABC-ABD=72°, likewise the sides BD, BC as before found, given, to find DC. 3. BD+BC: BD-BC :: T. of }(DCB+CDB) : T. 391.6 40.4 54° of }(DCB-CDB). 8° 05', 54° +8° 05'=62° 05'=DCB. 54°-8° 05'=45° 55'=CDB. 4. S.CDB : BC :: S.DBC: DC. 45° 55' 175.6 72° 232.5. ! LEMMA. Pl. 6. fig. 10. f from a point of a triangle ABC, inscribed in a circle, there be a perpendicular CD let fall upon the opposite side AB, that perpendicular is to one of the sides, including the angle, as the other side, including the angle, is to the diameter of the circle, i. e. DC: AC :: CB: CE. Let the diameter CE be drawn, and join EB; it is plain, the angle CEB=CAB (by cor. 2, theo. 7, geom.), and CBE is a right angle (by cor. 5, theo. 7, geom.), and ADC: whence ECB=ACD. The triangles CEB, CAD are therefore mutually equiangular, and (by theo. 16, geom.) DC : AC :: CB: CE, or DC : CB::AC: CE. Q. E. D. PROBLEM V. Pl. 6. fig. 5. Let three gentlemen's seats A, B, C be situate in a triangular form : there is given, AB 2.5 miles, AC 2.3, and BC 2. It is required to build a church at E, that shall be equidistant from the seats A, B, C. What distance must it be from each seat, and by what angle may the place of it be found ? By Construction. By prob. 15, geom., find the centre of a circle that will pass through the points A, B, C, and that will be the place of the church; the measure of which, to any of these points, is the answer for the distance : draw a line from any of the three points to the centre, and the angle it makes with either of the sides that contain the angle it was drawn to; that angle laid off by the direction of an instrument, on the ground, and the distance before found, being ranged thereon, will give the place of the church required. By Calculation. .3 .516. 1.25+.258=1.508=AD. By cor. 2, theo. 14, geom., the square root of the difference of the squares of the hypothenuse AC, and given leg AD, will give DC. That is, 5.29—2.274064=3.015936. Its square root is 1.736=CD. 2. CD: AC :: CB: the diameter. 2.65. the half of which, viz. 1.325, is the semidiameter, or distance of the church from each seat, that is, AE, CE, BE. From the centre E let fall a perpendicular upon any of the sides as EF, and it will bisect it in F (by theo. 8, geom.). Wherefore, AF=CF={AC=1.15. In the right-angled triangle AFE you have AF 1.15, and AE the radius 1.325 given, to find FÅE. Thus : 3. AF: R:: AE: sec. FAE. 1.15 90° 1.325 29° 47'. Wherefore, directing an instrument to make an angle of 29° 47' with the line AC, and measuring 1.325 on that line of direction, will give the place of the church, or the centre of a circle that will pass through A, B, and C. The above angle FAE may be had without a secant, as before. Thus : AE:R:: AF: S.AEF. 1.325 90° 1.15 60° 13'. Its complement 29° 47' will give FAE, as before. PRACTICAL QUESTIONS. Ex. 1. At 170 feet distance from the bottom of a tower the angle of its elevation was found to be 52° 30'. Required the altitude of the tower. Ans. 221.55 feet. Ex. 2. From the top of a tower, by the seaside, of 14.3 feet high, it was observed that the angle of depression of a ship’s bottom, then at anchor, measured 35°. What then was the ship's distance from the bottom of the wall ? Ans. 204.22 feet. Ex. 3. From a window near the bottom of a house which seemed to be on a level with the bottom of a steeple, I took the angle of elevation of the top of the steeple, equal 40°; then from another window, 18 feet directly above the former, the like angle was 37° 30'. What then is the height and distance of the steeple ? 57.26. Ans. height distance 150.50. Ex. 4. Wanting to know the height of an inaccessible tower, at the least distance from it, on the same horizontal plane, I { took its angle of elevation, equal to 58°; then going 300 feet directly from it, found the angle there to be only 32°. Required its height and my distance from it at the first station. S height 307.53. Ans. distance 192.15. Ex. 5. Being on the side of a river, and wanting to know the distance to a house which was seen on the other side, I measured out for a base 400 yards in a right line by the side of the river, and found that the two angles, one at each end of this line, subtended by the other end and the house, were 68° 2' and 73° 15'. What then was the distance between each station and the house? 593.08 Ans. { Ex. 6. Wanting to know the breadth of a river, I measured a base of 500 yards in a straight line close by one side of it, and at each end of this line I found the angles subtended by the other end and a tree close to the bank on the other side of the river to be 53° and 73° 15'. What then was the perpendicular breadth of the river ? Ans. 529.48 yards. Ex. 7. Two ships of war, intending to cannonade a fort, are, by the shallowness of the water, kept so far from it that they suspect their guns cannot reach it with effect. In order therefore to measure the distance, they separate from each other a quarter of a mile, or 440 yards ; then each ship observes and measures the angle which the other ship and the fort subtend, which angles are 83° 45' and 85° 15'. What then is the distance between each ship and the fort ? S 612.38 } yards. 2298.05 } yards. Ex. 8. A point of land was observed by a ship at sea to bear east by south ; and after sailing north-east 12 miles, it was found to bear south-east by east. It is required to determine the place of that headland, and the ship's distance from it at the last observation. ns. 26.0728 miles. Ex. 9. If the height of the mountain called the Peak of Teneriffe be 23 miles, as it is very nearly, and the angle taken at the top of it, as formed between a plumb-line and a line conceived to touch the earth in the horizon, or farthest visible point, be 88° 2'; it is required from these measures to determine the magnitude of the whole earth, and the utmost distance that can be seen on its surface from the top of the mountain, supposing the form of the earth to be perfectly globular. distance 135.943 Ans. { diameter 7916} miles. Ex. 10. Wanting to know the extent of a piece of water, or distance between two headlands, I measured from each of them to a certain point inland, and found the two distances to be 735 yards and 840 yards; also the horizontal angle subtended between these two lines was 55° 40'. What then was the distance required ? Ans. 741.2 yards. Ex. 11. Wanting to know the distance between a house and a mill which were seen at a distance on the other side of a river, I measured a base line along the side where I was of 600 yards, and at each end of it took the angles subtended by the other end and the house and mill, which were as follows, viz. at one end the angles were 58° 20' and 95° 20', and at the other end the like angles were 53° 30' and 98° 45'. What then was the distance between the house and mill? Ans. 962.5866 yards. Ex. 12.* Wanting to know my distance from an inaccessible object 0 on the other side of a river, and having no instrument for taking angles, but only a chain or cord for measuring distances; from each of two stations, A and B, which were taken at 500 yards asunder, I measured, in a direct line from the object 0, 100 yards, viz. AC and BD, each equal 100 yards ; also the diagonal AD measured 550 yards, and the diagonal BC 560. What then was the distance of the object O from each station A and B? AO 526.81. . . 3 SECTION III. MENSURATION OF AREAS, OR THE VARIOUS METHODS OF CALCULATING THE SUPERFICIAL CONTENTS OF ANY FIELD, Definition. The area or contents of any plane surface in perches is the number of square perches which that surface contains. * These practical examples are taken from Hutton's Mathematics, vol. ii. seventh London edition. |