EXERCISES.-OBLIQUE-ANGLED PLANE TRIANGLES. If ABC be any triangle, complete the solution in the following cases : 1. a = 1056, b = 359, c = 1267. 2. a = 364, b = 217, c = 494. 3. a = 96, b = 80, c = 63. = 37° 54′, B = 74° 25′, a = 104.6. 5. B 14° 20′, C = 101° 13', b = 296. 6. A = 7. a = 16. 97° 34', C = 50° 16', c = 8. a = 267, c = 341, B 9. b 61, c = = 68° 46'. 49, A = 97° 18'. 865, b = 742, A = 52° 26'. = 12. a = 1065, c = 897, A = 49° 36'. = 21° 23'. 15. a = 600, b = 500, c = 1000. 16. a = 50, b = 60, c = 70. 17. A = 18° 20′, B = 150°, b = 10,000. 18. A 78° 40', C = 39° 55', c = : 135. 20. c = 21. a = 200, b = 300, A = 35° 15'. 24. The angular elevation of a tower at a place A due south of it is 45°, and at a place B, due west of A, the elevation is 30°; AB = 300 ft. Find height of tower. 25. AB is a line 250 ft. long, in the same horizontal plane as the foot D of a tower CD; the angles DAB and DBA are respectively 61° 23′ and 47° 14′; the angle of elevation CAD is 34° 50'. Find height of tower. MISCELLANEOUS PROBLEMS. 1. The difference between the base and perpendicular of a right-angled triangle is one-fifth of their sum. Find the angles. 2. P and Q are two stations 1000 yards apart on a straight stretch of seashore which bears east and west. At P a rock bears S. 42° W., at Q it bears S. 35° E. Find distance of rock from shore. 3. A person on a ship, which is sailing on a straight course observes two lighthouses, one of which, L, is due north of the other, M. When M is due east, L bears 45° to the east of north, and when L is due east, M bears 60° to the east of south. Find direction of ship's course. 4. AB is a vertical pole 50 feet high, A being above B; BC is inclined upward at an angle of 20° to the horizontal line, BD, so that ABC is an angle of 70°; the shadow of AB on BC is 50 feet long. If the shadow fell on BD, what would be its length? 5. From the top of a vertical cliff 100 ft. high, forming one bank of a river, the angles of depression of the top and bottom of a vertical cliff, which forms the opposite bank of the river, are 28° 40′ and 64° 30′ respectively. Find height of opposite cliff and width of river. 6. The summit of a spire is vertically over the middle point of a horizontal square enclosure, side = 1000 ft. When the shadow of the spire just reaches a corner of the square the altitude of the sun is 27° 29′ 50′′. Find height of spire. 7. A tower near the coast is 197 feet high. From a rock the angle of elevation of the top of the tower is 8° 15'. A ship steams so as to clear the rock by 1.5 cables. What is the angle of elevation of tower from ship? 8. At two boats 600 yds. apart, the angles of elevation of a hydroplane flying directly between the boats are observed to be 57° and 65° 30′. Find the height of the hydroplane. 9. Two ships, A and B, leave harbour together. A sails S. 12° W. at a rate of 10.5 knots, and B S. 67° E. at 8 knots. How far apart will they be in 24 hours, and what will be the bearing of A from B? 10. From a ship the bearings of two objects, A and B, are N. 30° E., and N. 20° W. After the ship has sailed 5 miles due N., the bearings are N. 60° E. and N. 40° W. Find distance and bearing of A from B. 11. The foot, Q, of a tower and two stations, A and B, are in the same horizontal plane. The angular elevation of the tower at A is 60°, and at B is 45°, the distance from A to B is 100 feet, and the angle AQB is 60°. Find height of tower. 12. From P the bearing and distance of A is NE. 5 miles, and of B, S. 70° E. From Q, due east of P, the bearing of A is N. 25° E. If B is due south of A, what is the bearing of B from Q? 13. The hypotenuse of a right-angled triangle is 1000 feet long, and the difference between the other two sides is 240 feet: calculate the other sides and angles. 14. The height of a tower is 67.375 yds. At P, which is 5'35 yds. above the level of its base, the angle subtended by the tower is 74° 40'. Find the angle of elevation of the top of the tower as seen from P, and horizontal distance of P from tower. CHAPTER V. SOLUTION OF SPHERICAL TRIANGLES. ART. 17.-A spherical triangle is a portion of the surface of a sphere bounded by arcs of three great circles, and, like a plane triangle, has three sides and three angles. The sides, being arcs of circles, are measured like angles, in degrees, minutes, and seconds (°'"). When one of the angles is 90°, the triangle is right-angled; when one side is 90°, it is quadrantal; and when no side or angle is 90°, it is an oblique-angled triangle. Unlike a plane triangle, a spherical triangle may have two or three right angles, or one, two, or three sides of 90°, but no angle or side can be as much as 180°. The sum of any two sides is greater than the third side. The sum of the three angles of a spherical triangle is always more than 180°, but less than 540°, and the sum of the three sides is less than 360°. The greater angle is opposite to the greater side, as in plane triangles. The angles at the base of an isosceles triangle are equal, as in plane triangles. ART. 18. Solution of Right-angled Spherical Triangles.-Here one of the given parts is the right angle; any other two parts being given, the triangle can be solved. In order to simplify the solution, instead of using the five remaining parts of the triangle, Napier devised rules which refer only to his "Circular Parts." Napier's Circular Parts.-These are five in number, and are the two sides which contain the right angle, the complement of the hypotenuse, and the complements of the two angles. Thus in triangle ABC, right-angled at A, the circular parts are― B FIG. 56. AB; AC; Co. C; Co. BC; Co. B These are split into three groups, consisting of1 middle, 2 adjacents, 2 opposites. Any part may be taken as middle. Thus Middle. Opposites. AC Co. BC Napier's Rules. Co. B; co. BC AB; AC I. Sine mid part = tangents of adjacents multiplied. Each of these rules connects three parts. When used it must be arranged that two are given parts; also that the given parts are the only ones used to find the others. To do this, always start by considering whether the given parts are adjacents or opposites; then take as middle each given part in succession, and use the Rule which will connect the given parts. Hence Again sin AC =tan AB. tan co. C =tan AB. cot C and sin co. BC = cos AB. cos AC or cos BC= cos AB. cos AC ART 19.-Hints on construction of Spherical Triangles. Draw a circle as a primitive, with radius of sphere, we shall use one inch; but the larger the radius the better the result will be; with the Gunter rule it would be two inches. Let C be the centre, then any diameter drawn will represent a great circle at right angles to the primitive: thus ACB (Fig. 57) is a right circle. Draw DCE at right angles to AB. Then D and E are the Poles of ACB. The centres of all oblique circles through A and B lie on the line DCE. Thus with any centre F and radius FA describe an arc of a circle, then AGB is an oblique circle. To find its pole, join BG and produce to b, make be equal to EA (90°), join eB cutting DE in P: then P is the pole of AGB. FIG. 57. N.B. Great circles at right angles to AGB must pass through its pole P. Hence great circles can be drawn through a point on or within the primitive, or through two points, one on, the other inside the primitive. Thus in Fig. 58 let A and B be two points, A on, and B inside. Draw diameter ACD, and a line through C, as CO, at right angles. Join AB, and bisect it by a line cutting CO in X. Then X is centre of great circle through A and B. If an oblique circle is to be drawn through any two points inside the primitive as through A and B, Fig. 59, proceed B B FIG. 58. FIG. 59. thus-Draw diameter AC, and CD through C at right angles to AC. Join AD, make ADE a right angle, cutting AC in E: join AB, BE, bisect these by lines at right angles intersecting in F. F will be the centre for describing oblique circle required. To draw a great circle at right angles to a given oblique circle through a given point, first find the pole of the given oblique circle; then by the preceding examples draw a great circle through the given point and this pole; this will be at right angles to the given oblique circle, the given point may be either on the primitive, or on the oblique circle. A great circle is drawn making a given angle with the primitive (say 35°) through a given point as follows: From centre with radius = tan 35° describe a small arc, and with given point as centre and radius = sec 35°: draw another arc, this intersects the first arc at the centre from which the great circle is to be drawn with sec 35° as radius. N.B. With a radius of one inch, the tan and sec of any angle may be obtained from the traverse table thus-Take angle (say 35°) as course and radius (100) in lat. col., then sec (1.22) is found in dist. col. and tan (0·7) in dep. col., the semi-tangent tangent of half the angle = 0.31. In Fig. 60, A is given point, and o is centre of required great circle. = = If oC be drawn as a diameter, the arc ab measures the angle: Ca = 0.52; in traverse table find 1 in lat. 0.52 in dep.; angle 27°, hence Ca on scale of semi-tangents = 55°, or ab = 35°. A small circle parallel to the primitive and at any distance. from it is drawn from the centre of primitive, with radius = semi-tangent of complement of required distance. |