the observer stands on the former his eye is on a level with the latter. With the sextant the angles PEF and PFE are taken, and found to be 83° 20' and 74° 45'; and the angle PER, 22° 6'. The latter is the angle of elevation at the point E, since the tree is upright and the points E and R on the same level: what is the height of the tree? * Ans. 113 feet, nearly. 33. In an experiment similar to the preceding, the base EF was level and of 2 chains in length. taken with the theodolite, were as follows: 46′, ▲ REF = 78° 18′, RFE - 79° 56'. What was the height of the tree in this instance? The angles, PER = 19° Ans. 126 feet. 34. In order to ascertain the height and distance of a mountain top, P, a base, EF, of 34 E P F chains, was measured, on a level plain, across the direction in which the mountain was seen, and the angle of elevation at E was taken and found to be 3° 18'. When the instrument was placed at E, and the telescope directed to P, the index of the horizontal circle pointed to 182° 34'; and when the telescope was turned round till it pointed to F, the index then marked 272° 14'. The difference of these two numbers, viz. 89° 40', gives the horizontal angle at E. In like manner the horizontal angle at F was found to be 87° 33'. To explain this, let us refer to the annexed diagram.‡ Let PR be a perpendicular dropt on the same horizontal plane with E and F; and let ER and FR be E joined: ERP and FRP are both right angles; and FER and EFR are the horizontal angles taken by the theodolite at E and F, while PER is the angle of elevation at E. The height of P above the F R *The example of a tree is taken, not as the best, but that the learner may distinguish more readily, on the diagram, the vertical from the horizontal and oblique lines. The exercises which follow should be illustrated with models. † It is usual and proper to take the angles of elevation both at E and at F, the one serving as a check upon the other, since the height may be found by either; but both are not necessary. This diagram is intended to represent the same thing as the preceding, though in a different way. But both may fail in conveying the proper ideas to the learner. He may not easily see that the angles ERP and FRP can both be right angles. The difficulty arises from our being obliged to delineate both horizontal and vertical lines on the same sheet of paper. But if the teacher will show him the same lines in a model-that is, with the point P raised above the plane of E and F, the whole will be clear to him. The only construction required for this purpose will be a straight wire placed upright on a board (to telescope of the theodolite at E and F is required, in feet, taking the surface of the earth as a plane (that is, the line PR); and also the horizontal distance of the mountain top from the station E, in miles (that is, the line ER). Ans. ER 8.744 miles, and PR=2,662 feet, nearly. 36. With a view to determine the distance and height of a rock seen rising from the sea, remote from the shore, two stations are taken along the shore, both on the high water mark. The direct distance between the two stations, being measured with a chain, is 2,527 links; and the two horizontal angles taken at the two stations (as described in Ex. 34), are 89° 15′ and 86° 21'. The angle of elevation of the highest point of the rock, taken at the first station, being 1° 48'. What is the distance of the top of the rock from the first station, in miles, and its height, in feet, above the high water level, taking 5 feet as the height of the instrument, and regarding the earth as a plane? Ans. Distance, 4:11+ miles; height, 687-feet. 41. A meteorologist was desirous of ascertaining, as nearly as he could, the general height of the clouds on a particular occasion. As the wind blew gently and steadily, he first obtained their velocity from that of their shadows, and found that they travelled at about the rate of 15 miles per hour. He then marked the exact time at which the margin of a cloud was right over head; and precisely 8 minutes after, observed that its angle of elevation was about 31°. What conclusion could he draw from that observation, no allowance being made for the Earth's curvature, which would make no difference worth estimating? Ans. The height of the cloud was about 1 mile. 43. On another occasion the height of the clouds was tried by a different method, their velocity not being regarded as uniform. A station was taken on an eminence overlooking an extensive plain. The observer waited till he perceived a small cloud, or rather part of a cloud, vertically over the station. He watched the same cloud, keeping the telescope of the theodolite directed to it, till its shadow touched a suitable point on the plain, a friend keeping his eye on the shadow while he observed the cloud. He then took its angular elevation, which was 65°. He still kept the telescope upon it till its shadow touched represent PR), with the horizontal lines drawn on the surface of the board, and with threads for the oblique lines; or simply a piece of wood cut into the form EPRF. (The latter will be found in the set of Models prepared for the illustration of this work, and sold by the publishers, Messrs Sutherland and Knox.) another convenient point on the plain, and then found its elevation 42°. He next measured the distance between the two points on the plain, which proved to be 923 yards. What was the height of the cloud above the station? Ans. 1,433 yards, or 814 of a mile. 44. Taking the Earth as a sphere of 7,912 miles in diameter, what will be the dip of the sea-horizon as seen from a mountain 3 miles in height, making no allowance for terrestrial refraction? Ans. 2° 14'. NOTE. By the dip of the sea-horizon is meant the angle of depression at which the extreme surface of the sea appears below a horizontal line or plane at the point of observation. 51. The distances of three stations, P, Q, R, from each other, are as follows: PQ=4, QR = 5, PR= 6, chains. A point E is in the same line with PR and at such a distance from P that the angle QER is of 36°. What is the distance of E from P, Q, and R? E P R Ans. 2.30+, 5.63, and 8.30+, chains. CHAPTER X. UNRESOLVED EXERCISES. EXERCISES IN CHAPTER II, PROBLEM VI. 1. What are the logarithms of the numbers 26,583; 8,425.9; 164.51; and 98.506? 2. Of 4.387,28; 6-3; 305-472; and 9,845.54? EXERCISES IN CHAPTER II, PROBLEM IX. 1. Find the natural numbers corresponding to the logarithms 2.000000; 3.220108; 5.831166; and 1.597146. 2. Of 4.820136; and 2.271377. 3. Of 1.371070; 2·472440; 0.837958; and 3.450100. EXERCISES IN CHAPTER II, PROBLEM X. Find the following products by logarithms : 1. 7 x 6 x 5. 2. 258-7 × 6·844 3. 5.48 × 6·72 × 3·81 × 2.99. 4. 825 × 351 × 26.2 × 3·08, completing with ciphers. 5. 208.91 x 34.228. (See Example 2.) 6. 5.3 × 2.8. 7. 48.574 × 5·8347 × 12.1499. 8. 786.6 × 0428. (See Note 1.) 9. 004,99 × ⚫658. 10. 67·433×5·8628 ×·38 × ·006. EXERCISES IN CHAPTER II, PROBLEM XI. 1. Divide 3,312 by 92. 3. Divide 92.61 by 44.1. 5. Divide 16.56 by 46. (See Note 1.) 7. Divide 4.83 by 0.882. (See Note 2.) 9. Divide 0.882 by 4.833. (See Note 3.) 10. Divide .0084 by 0·6076. EXERCISES IN CHAPTER II, PROBLEM XII. Compute the fourth term of each of the following proportions, by logarithms : 1. As 279: 651 :: 657 : 2. As 94.25 : 486·3 :: 8.828 : 5. As 0656 : 0835 :: 2.24 : (See Note 2.) : (See Note 3.) (See Note 4.) 079,63 : (See Note 5.) EXERCISES IN CHAPTER II, PROBLEM XIII. 1. Square the numbers 25; 38-3; 5.4962; and 48.38. 2. Cube the numbers 15; 6.3; and 18.499. 3. Square 384-5, and cube 198, completing the results with ciphers. 4. Square 0.852,63; 0179; and '008. (See the Note.) 5. Cube 0.2835 and 0899. EXERCISES IN CHAPTER II, PROBLEM XV. 1. Extract the square roots of 84,100; 88.36; 285,168; and 10. 2. Extract the cube roots of 6,859; 3,375,000; 857·4; and 600. 3. Find the square roots of 0081; 0·658; 00055; and 04. (See Note.) 4. Calculate the cube roots of 000,585; 0.5; 0688; and 0.8. EXERCISES IN CHAPTER III. 1. An ancient Roman mile was equivalent to 1612 yards of British Imperial Measure. Convert 556 Roman miles into Imperial miles. 2. If 8.546 acres of land are valued at £564, 10s., what should be the value of 2.868 acres, at the same rate? 3. What is the interest of £457, 8s., for 7 months, at 44 per cent. per annum ? 4. Compute the interest of £64: 10: 9, for 288 days at 33 per cent. per annum. 5. Calculate the amount of £10, for fifty years, at 4 per cent. per annum, compound interest. 6. What sum will one penny amount to in 500 years, if accumulating, all the while, at 5 per cent. per annum, compound interest? 7. What is the compound interest of £383 for 5 years at 3 per cent. per annum? 8. Find the diameter of a circle, the circumference of which is 408.2. 9. What is the area, in acres, of a circle whose radius is 2.85 chains? |