From a table of sines we find that the angle whose sine is '6 is 36° 52′ very nearly; 81. A and B, two points on the opposite banks of a river, are a feet apart. Between them is PN, the mast of a ship, and the elevations of P 1. A tower subtends an angle of 25° 10' to an observer standing at 200 yards distance from its base. Find its height, given L tan25°10' = 9.671963, log9 397972993. 2. The angular elevation of the top of a rock overlooking a perfectly level plain is measured from two points, in a line with the rock, one a hundred feet behind the other. The angles are observed to be 35° and 55°. Find the height of the rock, given tan35° = '7. 3. ABC is a triangle with a right angle at C. CB is 30 feet long, and BAC is 20°. If CB is produced to P so that PAC 55°, find the length of PC, given 4. a and b are the heights of two towers. At a point between them the elevations of the tops are a and ẞ respectively, and at another point between them the elevations are a' and B' respectively. Show that α cotẞ' - cotß 5. AB is a tower, height a, surmounted by a flagstaff BC, height h. P is a point on a level with A the base of the tower, and distant b from the tower. Show that Calculate the angle BPC to the nearest minute in the case when a 80 feet, h=20 feet, b=300 feet; given = tan3°30′ = '06116, tan3°31′ = '06145. 6. A straight street ismile long and has an ascent of 1 in 20. At the top of the street is a church spire subtending at the bottom an angle of 5°. Find the height of the spire to the nearest foot; given sin2°52′ = '05, cos2°52′ = 999, tan7°52′ = 138. CHAPTER VIII. SECTION I. GENERAL PROPERTIES OF 82. In this Section we shall give some general properties of triangles, and in the next Section we shall apply these properties to the solution of oblique-angled triangles. 83. The three angles of any triangle are together equal to two right angles. If A, B, C denote the angles of any triangle, therefore A+B+C=180°; A = 180° - (B+C), 3 which cosA = + 5' , taking the positive sign, since A is acute; 4 3 64 225 289 289' 15 8 15 = 17 17 8 from which sinB and 84. In any triangle the sides are proportional to the sines of the opposite angles. Let ABC be any triangle, and let a perpendicular AD be drawn from the angular point A to the side BC. In the first figure, in which the angles are all acute, also therefore AD AB sinABC; AD AC sin ACB; AB sin ABC= AC sinACB. In the second figure, in which the angle C is obtuse, Hence in every triangle, we have, in the usual notation, or, in any triangle the sines of the angles are proportional to the opposite sides. 85. Formulæ for the Cosines of the Angles in terms of the Sides. In the first figure of last Article, BC= BD + DC=AB cosABC + AC cosACB. In the second figure, BC BD-CD = AB cosABC-AC cosACD = AB cos ABC – AC cos(180° – ACB) = AB cosABC + AC cosACB. Hence in every triangle, BC AB cos ABC+AC cosACB, = or, in usual notation, (Art. 43). |