Let us suppose that we have AB = 672 yards BAC = 72° 29' found ABC = 39° 20' C AC= 27° 49 CBC=19° 10'. First: in the triangle ABC, the horizontal angle ACB=180° — (A + B)=180° — 111° 49' = 68° 11'. To find the horizontal distance AC. sino 68° 11' ar. comp. sin B 39° 20' AB 672 : 0.032275 9.801973 2.827369 . AC 458.79 2.661617. To find the horizontal distance BC. AB 672 : 0.032275 9.979380 2.827369 . . . BC 690.28 2.839024. In the triangle CAC", to find CC". ar. comp. : tan C'AC 27° 49' AC 458.79 0.000000 9.722315 2.661617 CC 242.06 2.383932 In the triangle CBC", to find CC". ar. comp. BC 690.28 Hence also, CC" – CC"=242.06 - 239.93= 2.13 yards, which is the height of the station A above station B. PROBLEMS. 1. Wanting to know the distance between two inaccessible objects, which lie in a direct level line from the bottom of a tower of 120 feet in height, the angles of depression are measured from the top of the tower, and are found to be, of the nearer 57°, of the more remote 25° 30': required the distance between the objects. Ans. 173.656 feet. 2. In order to find the distance B between two trees, A and B, which could not be directly measured because of a pool which occupied the intermediate space, the distances CB = 672 yards, ACB=55° 40'; required the distance AB. Ans. 592.967 yards. 3. Being on a horizontal plane, and wanting to ascertain the height of a tower, standing on the top of an inaccessible hill, there were measured, the angle of elevation of the top of the hill 40°, and of the top of the tower 51°; then measuring in a direct line 180 feet farther from the hill, the angle of elevation of the top of the tower was 33° 45'; required the height of the tower. Ans. 83.998. 4. Wanting to know the hori. zontal distance between two inaccessible objects E and W, the fol. lowing measurements were made. AB =536 yards BAW=40° 16' viz. WAL=57° 40' ABE =:42° 22' EBW=71° 07'; required the distance EW. Ans. 939.527 yards. 5. Wanting to know the horizontal distance between B two inacessible objects A F and B, and not finding any station from which both of them could be seen, two D points C and D, were chosen E at a distance from each other, equal to 200 yards ; from the former of these points A could be seen, and from the latter B, and at each of the points C and D a staff was set up. From C a distance CH was measured, not in the direction DC, equal to 200 yards, and from D a distance DE equal to 200 yards, and the following angles taken, AFC = 83° 00', BDE=54° 30', viz. { ACD=53° 30', BDC = 156° 25', ACF = 54° 31', BED=88° 30'. Ans. AB= 345.467 yards. 6. From a station P there can be seen three objects, A, B and C, whose distances from A. each other are known: viz., AB=800, AC=600, and BC = 400 yards. Now, there are measured the horizontal angles. APC=33° 45' and BPC P = 22° 30': it is required to find the three distances PA, PC, and PB. PA= 710.193 yards. Ans. PC=1042.522 PB= 934.291. 7. This problem is much used in maritime survey ing, for the purpose of locating buoys and sounding boats. The trigonometrical solution is somewhat tedious, but it may be solved geometrically by the following easy construction. Let A, B, and C be the three fixed points on shore, and P the position of the boat from which the angles B APC=33° 45', CPB=22° 30', and APB=56° 15', have been measured. Subtract twice APC=67° 30' from 180°, and lay off at A and C two angles, CAO, ACO, each equal to half the remainder = 56° 15'. With P the point 0, thus determined, as a centre, and 0A or OC as a radius, describe the ciroumference of a circle: then, any angle inscribed in the segment APC, will be equal to 33° 45'. Subtract, in like manner, twice CPB=45°, from 180°, and lay off half the remainder = 67° 30', at B and C, determining the centre of a second circle, upon the circumference of which the point P will be found. The required point P will be at the intersection of these two circumferences. If the point P fall on the circumference described through the three points A, B, and C, the two auxiliary circles will coincide, and the problem will be in: determinate. ANALYTICAL PLANE TRIGONOMETRY. 40. WE have seen (Art. 2) that Plane Trigonometry explains the methods of computing the unknown parts of a plane triangle, when a sufficient number of the six parts is given. To aid us in these computations, certain lines were employed, called sines, cosines, tangents, cotangents, &c., and a certain connection and dependence were found to exist between each of these lines and the arc to which it belonged. All these lines exist and may be computed for every conceivable arc, and each will experience a change of value where the arc passes from one stage of magnitude to another. Hence, they are called functions of the arc; a term which implies such a connection between two varying quantities, that the value of the one shall always change with that of the other. In computing the parts of triangles, the terms, sine, cosine, tangent, &c., are, for the sake of brevity, applied to angles, but have in fact, reference to the arcs which measure the angles. The terms when applied to angles, without reference to the measuring arcs, designate mere ratios, as is shown in Art. 88. 41. In Plane Trigonometry, the numerical values of these functions were alone considered (Art. 13), and the arcs from which they were deduced were all less than 180 degrees. Analytical Plane Trigonometry, explains all the processes for computing the unknown parts of rectilineal triangles, and also, the nature and properties of the circular functions, together with the methods of deducing all the formulas which express relations between them. |