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 B 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. 163 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 B. 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 ACB 180° — 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 7 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 349204.54307; log 51515 4.71193; = 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. 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 B. 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 ACB 180° - 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 (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 7 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 349204.54307; log 51515 = 4.71193; 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. |