CHAPTER XVI ON THE MEASUREMENT OF HEIGHTS AND DISTANCES 161. By the aid of the solution of triangles we can find the distance between points which are inaccessible; we can calculate the magnitude of angles which cannot be practically observed; we can find the relative heights of distant and inaccessible points. The method on which the trigonometrical survey of a country is conducted affords the following illustration: 162. To find the distance between two distant objects. Two convenient positions A and B, on a level plain as far apart as possible, having been selected, the distance between A and B is P A measured with the greatest possible care. This line AB is called the base line. Next, the two distant objects, P and Q (church spires, for instance), visible from A and B, are chosen. Then by Case II. Ch. The angles PAB, PBA are observed. Again, the angles QAB, QBA are observed; and by Case II. the lengths of QA and QB are calculated. Thus the lengths of PA and QA are found. The angle PAQ is observed; and then by Case III. the length of PQ is calculated. Thus we are able to find not only the length PQ, but the angle which PQ makes with any line in the figure. The points P and Q are not necessarily accessible, the only condition being that P and Q must be visible from both A and B. 163. In practice, the points P and Q will generally be accessible, and then the line PQ, whose length has been calculated, may be used as a new base to find other distances. 164. To find the height of a distant object above the point of observation. Let B be the point of observation; P the distant object. From B measure a base line BA of any convenient length, in any convenient direction; observe the angles PAB, PBA, and by Case II. Ch. XV. calculate the length of BP. Next observe at B the 'angle of elevation' of P; that is, the angle which the line BP makes with the horizontal line BM, M being the point in which the vertical line through P cuts the horizontal plane through 3. Then PM, which is the vertical height of P above B, can be calculated, for PM = BP • sin MBP. EXAMPLE 1. The distance between a church spire A and a milestone B is known to be 1764.3 feet; C is a distant spire. The angle CAB is 94° 54', and the angle CBA is 66° 39'; find the distance of C from A. ABC is a triangle, and we know one side, c, and two angles (A and B), and therefore it can be solved by Case II. Ch. XV. The angle ACB180° - 94° 54' - 66° 39' 18° 27'. = Therefore the triangle is the same as that solved in Art. 148. Therefore AC 5118.2 feet. = EXAMPLE 2. If the spire C, in the last example, stands on a hill, and the angle of elevation of its highest point is observed at A to be 4°19'; find how The required height x = AC sin 4° 19' and AC is 5118.2 feet; (EXAMPLES VII. CONSIST OF EASY EXAMPLES ON THIS SUBJECT.) 1. Two straight roads, inclined to one another at an angle of 60°, lead from a town A to two villages B and C; B on one road distant 30 miles from A, and C on the other road distant 15 miles from A. Find the distance from B to C. Ans. 25.98 m. 2. Two ships leave harbor together, one sailing N.E. at the rate of 71⁄2 miles an hour, and the other sailing north at the rate of 10 miles an hour. Prove that the distance between the ships after an hour and a half is 10.6 miles. 3. A and B are two consecutive milestones on a straight road, and C is a distant spire. The angles ABC and BAC are observed to be 120° and 45° respectively. Show that the distance of the spire from A is 3.346 miles. 4. If the spire C in the last question stands on a hill, and its angle of elevation at A is 15°, show that it is .896 of a mile higher than A. 5. If in Question (3) there is another spire D such that the angles DBA and DAB are 45° and 90° respectively and the angle DAC is 45°, prove that the distance from C to D is 23 miles very nearly. 6. A and B are two consecutive milestones on a straight road, and C is the chimney of a house visible from both A and B. The angles CAB and CBA are observed to be 36° 18′ and 120° 27', respectively. Show that C' is 2639.5 yards from B. 7. A and B are two points on opposite sides of a mountain, and C is a place visible from both A and B. It is ascertained that C is distant 1794 feet and 3140 feet from A and B, respectively, and the angle ACB is 58° 17'. Show that the angle which the line pointing from A to B makes with AC is 86° 55′ 49′′. 8. A and B are two hill-tops 34920 feet apart, and C is the top of a distant hill. The angles CAB and CBA are observed to be 61° 53′ and 76° 49', respectively. Prove that the distance from A to C is 51515 feet. log 34920 4.54307; = log 515154.71193; log sin 76° 49′ = 9.98840; log cosec 41° 18′ = 10.18045. 9. From two stations A and B on shore, 3742 yards apart, a ship C is observed at sea. The angles BAC, ABC are simultaneously observed to be 72° 34' and 81° 41', respectively. Prove that the distance from A to the ship is 8522.7 yards. log 3742 = 3.57310; log 8522.7 3.90057; log sin 81° 41′ = 9.99540; log cosec 25° 45' = 10.36206. 10. The distance between two mountain peaks is known to be 4970 yards, and the angle of elevation of one of them when seen from the other is 9° 14'. How much higher is the first than the second? Sin 9° 14' = .16045. 35 miles apart. Ans. 797.5 yards. 11. Two straight railways intersect at an angle of 60°. From their point of intersection two trains start, one on each line, one at the rate of 40 miles an hour. Find the rate of the second train that at the end of an hour they may be Ans. Either 25 or 15 miles an hour. (Art. 153.) 12. A and B are two positions on opposite sides of a mountain; C is a point visible from A and B; AC and BC are 10 miles and 8 miles, respectively, and the angle BCA is 60°. Prove that the distance between A and B is 9.165 miles. 13. A and B are consecutive milestones on a straight road; C is the top of a distant mountain. At A the angle CAB is observed to be 38° 19'; at B the angle CBA is observed to be 132° 42', and the angle of elevation of C at B is 10° 15'; show that the top of the mountain is 1243.5 yards higher than B. log sin 38° 19' = 9.79239; log cosec 8° 59' = 10.80646; log sin 10° 15' = 9.25028. = log 1760 3.24551; log 1243.5 = 3.09465; 14. A base line AB, 1000 feet long, is measured along the straight bank of a river; C is an object on the opposite bank; the angles BAC and CBA are observed to be 65° 37' and 53° 4' respectively; prove that the perpendicular breadth of the river at C is 829.87 feet. 15. A is the foot of a vertical pole, B and C are due east of A, and D is due south of C. The elevation of the pole at B is double that at C, and the angle subtended by AB at D is tan-1. Also BC= 20 feet, CD = 30; find the height of the pole. - Hobson's Trig. 16. Two towers, one 200 feet high, the other 150 feet high, standing on a horizontal plane, subtend, at a point in the plane, angles of 30° and 60° respectively. The horizontal angle that their bases subtend at the same point is 120°; how far are the two towers apart? 17. The diagonals of a parallelogram are in length d1 and d2, the angle between them is ; show that the area of a parallelogram is d1d2 sin . 18. A man walking along a straight road at the rate of three miles an hour sees in front of him at an elevation of 60° a balloon which is travelling horizontally in the same direction at the rate of six miles an hour; ten minutes after he observes that the elevation is 30°; prove that the height of the balloon above the road is 440 √3 yards. 19. A person standing at a point A, due south of a tower built on a horizontal plain, observes the altitude of the tower to be 60°. He then walks to a point B due west from A and observes the altitude to be 45°, and then at the point C in AB produced he observes the altitude to be 30°; prove that AB = BC. 20. The angle of elevation of a balloon, which is ascending uniformly and vertically, when it is one mile high is observed to be 35° 20′; 20 minutes later the elevation is observed to be 55° 40'. How fast is the balloon moving? 21. A tower stands at the foot of an inclined plane whose inclination to the horizon is 9°; a line is measured up the incline from the foot of the tower of 100 feet in length. At the upper extremity of this line the tower subtends an angle of 54°; find the height of the tower. Ans. 114.4 feet. 22. The altitude of a certain rock is observed to be 47°, and after walking 1000 feet towards the rock, up a slope inclined at an angle of 32° to the horizon the observer finds that the altitude is 77°; prove that the vertical height of the rock above the first point of observation is 1034 feet. Sin 47° = .73135. 23. At the top of a chimney 150 feet high standing at one corner of a triangular yard, the angle subtended by the adjacent sides of the yard are 30° and 45° respectively; while that subtended by the opposite side is 30°; show that the lengths of the sides are 150 feet, 86.6 feet, and 106 feet respectively. 24. A flagstaff h feet stands on the top of a tower. From a point in the plane on which the tower stands, the angles of elevation of the top and bottom of the flagstaff are observed to be a and 8 respectively; prove that the height h sin B cos a of the tower is feet. sin (a - B) 25. The angular elevation of the top of a steeple at a place due south of it is 45°, and at another place due west of the former station and distant a feet from it the elevation is 15°; show that the height of the steeple is (33) feet. |