MISCELLANEOUS PROBLEMS. 1. The angle of elevation of a vertical tower is observed to be 30°, at the end of a horizontal base line of 100 yards, measured from its foot. Required the height of the tower. 2. A rope-dancer wishes to ascend a steeple 100 feet high, by means of a rope 196 feet long. If he can do so, find at what inclination he must be able to walk up the rope. 3. From the summit of a pier which rises 100 feet above the margin of a river, the angle of depression of the opposite margin was found to be 33° 16'. Required the width of the river. 4. If the distance of the moon from the earth be taken at 238500 miles, and the angle subtended by the semidiameter of the moon be 15' 33".5 at that distance, what is the moon's diameter? Ans. 2158 miles. 5. A point of land was observed by a ship at sea to bear east by south, that is, 11° 15' S. of E.; and after sailing northeast 12 miles, it was found to bear southeast by east, that is, 33° 45′ S. of E. Required the distance of the headland from the ship at the last observation. Ans. 26.07 miles. 6. From the top of Mont Blanc, 3 miles high, the angle of depression of the remotest visible point of the earth's surface is 2° 13' 27". Required the diameter of the earth, supposing it to be a perfect sphere; and, also, the utmost distance from which the mountain is visible. Ans. Diameter, 7958 miles; distance, 154.5 miles. 7. A side of the base of a square pyramid is 200 feet, and each edge is 150 feet; required the slope of each face. Ans. 26° 34', nearly. 8. I have a meadow in the form of a parallelogram, whose two adjacent sides are 20 rods and 18 rods, including an angle of 78° 9'; the same has been divided into two equal lots by a fence running diagonally. Required the area of each lot. Ans. 176.16 square rods. 9. A traveler wishing to know the distance and height of a mountain-top over which he had to pass, took the angle of its elevation at two stations, in a direct line towards it, the one 3 miles, or 5280 yards, nearer the mountain than the other, and found the angles to be 2° 45′ and 3° 20'. Required the horizontal distance of the mountain-top from the nearer station, and its height. Ans. Distance, 24840 yards; height, 1447 yards. 10. From the top of a light-house the angle of depression of a ship at anchor was observed to be 4° 52', from the bottom of the light-house the angle was 4° 2. Required the horizontal distance of the vessel, and the height of the hill on which the light-house is placed, the height of the light-house being 60 feet. Ans. Horizontal distance, 4100.4 feet; height, 289.12 feet. 11. When a tower 150 feet high throws a shadow 75 feet long upon the horizontal plane on which the tower stands, what is the sun's altitude (Art. 189)? Ans. 63° 26' 6". 12. The sides of a triangle are equal to 3 and 12, respectively, and the included angle is 30°; find the hypothenuse of an equal right-angled isosceles triangle. Ans. 6. 13. From a window near the bottom of a house, which seemed to be on a level with the bottom of a steeple, I took the angle of elevation of the top of the steeple, equal to 40°; then from another window, 18 feet directly above the former, the like angle was 37° 30'. Required the height and distance of the steeple. Ans. Height, 210.4 feet; distance, 250.8 feet. 14. Two pulleys, whose diameters are 6 inches and 4 feet 3 inches, respectively, are placed at a distance of 3 feet 6 inches from centre to centre. What must be the length of a belt which shall connect them, by passing around their circumferences, without crossing? Ans. 15 feet 5.9 inches. 15. A tower is surrounded by a circular moat. At noon on a certain day, the shadow of the top of the flag-staff is observed to project 45 feet beyond the edge of the moat. When the sun is due west, on the same day, the shadow projects 120 feet beyond the moat. The distance between the extremities of the shadows is 375 feet. The angle of elevation of the top of the flag-staff from any point of the edge of the moat is 60°. Find the height of the tower, and the altitude of the sun at noon. Ans. 311.77 feet; 54° 10′ 57′′. BOOK V. SPHERICAL TRIGONOMETRY. DEFINITIONS. 145. SPHERICAL TRIGONOMETRY treats of methods of computing spherical triangles. 146. A SPHERICAL TRIANGLE is a portion of the surface of a sphere bounded by three arcs of a great circle, each of which is less than a semi-circumference. The three planes in which the arcs lie form a polyedral angle at the centre of the sphere. The ANGLES of a spherical triangle are the diedral angles made by the plane faces which form the polyedral angle. 147. The sides and angles of spherical triangles are usually both expressed in degrees, minutes, &c. The circumference, however, is sometimes supposed to be divided into 24 equal parts, called hours; each hour into 60 equal parts, called minutes of time; each minute into 60 equal parts, called seconds of time. Then a side is expressed by the number of hours, minutes, seconds, and decimal parts of a second, which it contains. Hours, minutes, and seconds are denoted by h., m., and s. Thus, 3h. 35m. 5.8s. RELATIONS BETWEEN THE SIDES AND ANGLES OF SPHERICAL TRIANGLES. 148. In any spherical triangle, the sines of the sides are proportional to the sines of the opposite angles. DA', DC, perpendicular to OA, O C, respectively; join B'A', B' C'. BCO is a right angle (Geom., Prop. VI. Bk. VII.); therefore, and B COB' sin B' O C' = 0 B' sin a, BD BC' sin B C D B C' sin COB' sin a sin C. = In like manner, = and, by the two preceding equations, O B' sin a sin C = 0 B′ sin c sin A, The figure supposes a, c, B, C, &c. each less than 90°, but the relation stated may be shown to hold when the figure is modified to meet any case whatever. For instance, if C alone is greater than 90°, the point D will fall beyond O C, instead of between OC and OA; then, BCD will be the supplement of C, and thus, since the sine of an angle and its supplement are the same, the sine of BCD is still equal to the sine of C. 149. In any spherical triangle, the cosine of any side is equal to the product of the cosines of the other two sides, plus the product of the sines of those two sides into the cosine of their included angle. Let ABC be any spherical triangle, O the centre of the sphere. Draw the plane B'A'C' perpendicular to OA. Then the angle B'A'C' is equal to the angle A, the angle B'OC measures the side a, and in the triangles A'B'C', O B'C' we have, by Art. 113, 2 2 Α' B B a C A BCA B+A' C-2 A' B' X 4' C cos A, B' C = 0 B' + 0 C — 2 0 B' × O C' cos a. Subtracting the first equation from the second, observing that 2 2 2 O B - A' B' and OC. 2 2 C ·A'C' are each equal to O A', since the triangles OA' B', O A'C' are right-angled at A', we have 0 = 2 0 A2+2 A' B' A' C cos A-2 O B × O C cos a; therefore, cos α = Ο Α' ΧΟ Α' + A' B'X A' C' cos A. Substituting the functions derived from the triangles O A' B, The preceding construction supposes the sides b and c, which contain the angle A, to be both less than 90°, but the formule obtained may be shown to be applicable in all cases. |