CHECK ON THE CORRECTNESS OF THE WORK a2 = c2 — b2 = (c + b)(c—b) Remark. The results obtained in the solution of the preceding exercise without logarithms are less accurate than those obtained in the solution by the use of logarithms; the cause of this is that fourplace tables have been used in the former method, five place in the latter. EXERCISES 28. (1.) In a right triangle b = 96.42, c=114.81; find a and A. (2.) The hypotenuse of a right triangle is 28.453, a side is 18.197; find the remaining parts. (3.) Given the hypotenuse of a right triangle = 747.24, an acute angle =23° 45′ ; find the remaining parts. (4.) Given a side of a right triangle 37.234, the angle opposite = 54° 27′; find the remaining parts and the area. (5.) Given a side of a right triangle = 1.1293, the angle adjacent =74° 13′ 27′′; find the remaining parts and the area. (6.) In a right triangle A = 15° 22′ 11′′, c=.01793; find b. (7.) In a right triangle B=71° 34′ 53′′, b = 896.33; find a. The angle of elevation or depression of an object is the angle a line from the point of observation to the object makes with the horizontal. Thus angle x (Fig. 1) is the angle of elevation of P if O is the point of observation; angle y (Fig. 2) is the angle of depression of P if O is the point of observation. (11.) At a horizontal distance of 253 ft. from the base of a tower the angle of elevation of the top is 60° 20′; find the height of the tower. (12.) From the top of a vertical cliff 85 ft. high the angle of depression of a buoy is 24° 31′ 22′′; find the distance of the buoy from the foot of the cliff. (13.) A vertical pole 31 ft. high casts a horizontal shadow 45 ft. long; find the angle of elevation of the sun above the horizon. (14.) From the top of a tower 115 ft. high the angle of depression of an object on a level road leading away from the tower is 22° 13′ 44′′; find the distance of the object from the top of the tower. (15.) A rope 324 ft. long is attached to the top of a building, and the inclination of the rope to the horizontal, when taut, is observed to be 47° 21′ 17′′; find the height of the building. (16.) A light-house is 150 ft. high. How far is an object on the surface of the water visible from the top? [Take the radius of the earth as 3960 miles.] (17.) Three buoys are at the vertices of a right triangle; one side of the triangle is 17,894 ft., the angle adjacent to it is 57° 23′ 46′′. Find the length of a course around the three buoys. (18.) The angle of elevation of the top of a tower observed from a point at a horizontal distance of 897.3 ft. from the base is 10° 27′ 42′′; find the height of the tower. (19.) A ladder 424 ft. long leans against the side of a building; its foot is 25 ft. from the building. What angle does it make with the ground? (20.) Two buildings are on opposite sides of a street 120 ft. broad. The height of the first is 55 ft.; the angle of elevation of the top of the second, observed from the edge of the roof of the first, is 26° 37'. Find the height of the second building. (21.) A mark on a flag-pole is known to be 53 ft. 7 in. above the ground. This mark is observed from a certain point, and its angle of elevation is found to be 25° 34'. The angle of elevation of the top of the pole is then measured, and found to be 34° 17'. Find the height of the pole. (22.) The equal sides of an isosceles triangle are each 7 in. long; the base is 9 in. long. Find the angles of the triangle. Hint.-Draw the perpendicular BD. BD bisects the base, and also the angle ABC. In the right triangle ABD, AB=7 in., AD=4 in., hence ABD can be solved. Angle C angle A, angle ABC=2 angle ABD. (23.) Given the equal sides of an isosceles triangle each 13.44 in., and the equal angles are each 63° 21' 42"; find the remaining parts and the area. (24.) The equal sides of an isosceles triangle are each 377.22 in., the angle between them is 19° 55' 32". Find the base and the area of the triangle. (25.) If a chord of a circle is 18 ft. long, and it subtends at the centre an angle of 45° 31' 10", find the radius of the circle. (26.) The base of a wedge is 3.92 in., and its sides are each 13.25 in. long; find the angle at its vertex. (27.) The angle between the legs of a pair of dividers is 64° 45', the legs are 5 in. long; find the distance between the points. (28.) A field is in the form of an isosceles triangle, the base of the triangle is 1793.2 ft.; the angles adjacent to the base are each 53° 27' 49". Find the area of the field. (29.) A house has a gable roof. The width of the house is 30 ft., the height to the eaves 25 ft., the height to the ridge-pole 33 ft. Find the length of the rafters and the area of an end of the house. (30.) The length of one side of a regular pentagon is 29.25 in.; find the radius, the apothem, and the area of the pentagon. Hint.-The pentagon is divided into 5 equal isosceles triangles by its radii. Let AOB be one of these triangles. AB=29.25 in.; angle AOB={} of 360° 72°. Find, by the methods previously given, OA, OD, and the area of the triangle 40B. These are the radius of the pentagon, the apothem of the pentagon, and the area of the pentagon respectively. (31.) The apothem of a regular dodecagon is 2; find the perimeter. (32.) A tower is octagonal; the perimeter of the octagon is 153.7 ft. Find the area of the base of the tower. (33.) A fence extends about a field which is in the form of a regular polygon of 7 sides; the radius of the polygon is 6283.4 ft. Find the length of the fence. (34.) The length of a side of a regular hexagon inscribed in a circle is 3.27 ft.; find the perimeter of a regular decagon inscribed in the same circle. (35.) The area of a field in the form of a regular polygon of 9 sides is 483930 sq. ft.; find the length of the fence about it. SOLUTION OF OBLIQUE TRIANGLES BY THE AID OF RIGHT TRIANGLES 29. Oblique triangles can always be solved by the aid of right triangles without the use of special formulas; the method is frequently, however, quite awkward; hence, in a later chapter, formulas are deduced which render the solution more simple. The following exercises illustrate the solution by means of right triangles: (1.) In an oblique triangle a=3.72, B=47° 52′, C= 109° 10′; find the remaining parts. The given parts are a side and two angles. Solve the right triangle BCD. Having thus found CD, solve the right triangle ACD. (2.) In an oblique triangle a = 89.7, c = 125.3, B= 39° 8'; find the remaining parts. The given parts are two sides and the included angle. |