Consequently, in the triangle ABC, we have all the angles and one side AB given; hence, by Case 1. At the top of a castle, which stood on a hill near the sea-shore, the angle of depression HTS, of a ship at anchor, was observed to be 4° 52′; at the bottom of the castle the angle of depression OBS was 4° 2′. Required the horizontal distance AS of the vessel, and the height of the hill above the level of the sea, the height of the castle being 60 feet. As TH, BO, are parallel to AS, we have TSA=4° 52′, and BSA =4°2′. Bearing this in mind we have E In ▲ BSA, AS=SB cos. BSA, AB=SB sin. BSA The distances of three objects A, B, C, from each other, are as follow, viz. AB=462 yards, AC: = 328 yards, and BC 297 yards; a person at D, wishing to know his distance from each object, takes the angle ADB, and finds it to be 34° 16′ 21′′; it is required to determine DA, DC, and DB. B As the three sides of the triangle ABC are given we may find the angle CAB, and, consequently, the supplemental angle DAB, so that we shall have in the triangle DAB the two angles D, A and the side AB to find the rest. The computation will, therefore, be as follows. Hence the three distances are DA= 301.01, DC = 629·101, DB= 719 522. PROBLEM V. Suppose that from the top of a mountain, three miles high, the angle of depression of the remotest visible point of the earth's surface is taken and found to be 2° 13′ 27′′; it is required thence to determine the diameter of the earth, supposing it to be a perfect sphere. Let O be the centre of the earth, BA the mountain, AC the visual the earth's surface in C. ray or line touching Draw the tangent then the angle of de BD, and join OD, OC; BOD=DOC= comp. A=1° 6′ 49′′, and, therefore, BDO=88° 53′ 16′′. Now in the right-angled triangle ABD we have PROBLEM VI. Given the distances between three objects A, B, C, and the angles subtended by these distances at a point D in the same plane with them; to determine the distance of D from each object. Let a circle be described about the triangle ADB, and join AE, EB, then will the angles ABE, BAE, be respectively equal to the given angles ADE, BDE, (Geom. p. 50;) thus all the angles of the triangle AEB are known, as also the side AB; we may find, therefore, the remaining sides AE, EB. Again, the sides of the triangle ABC being known, we may find the angle BAC; hence the angle CAE becomes known, so that in the triangle CAE we shall have the two sides AE, AC, and the included angle given, from which we may find the angle AEC in fig. 1, or the angle ACE in fig. 2, and thence its supplement AED or ACD; this with the given side AE and angle ADE, in the first figure, or with the given side AC and B angle ADC in the second, will enable us to find AD, one of the required lines, and thence DC and DB the other two. Or the solution may be conducted more analytically as follows. Ꮖ Put a for the angle DAC, and x' for the angle DBC; also call the given angles ADC, BDC, a and a', then a, b, c, representing as usual the sides opposite to A, B, C, we have This is one equation between the unknown quantities x, x'. Another is easily obtained; for since the four angles of the quadrilateral ADCB make up four right angles or 360°, we have x + x' + a + a + ACD |