3. In the right angled triangle DBE, we have the angle BDE = 51°, and the side DE = 138.1 yards, to find BE = 107.3 yards, the height of the hill. EXAMPLE 10. Fig. 64. An obelisk AD standing on the top of a declivity, I measured from its bottom a distance AB = 40 feet, and then took the angle ABD = 41°; going on in the same direction 60 feet farther to C, I took the angle ACD – 23° 45'; what was the height of the obelisk ? Calculation. 1. In the triangle BCD, we have given the angle BCD - 23° 45', the angle BDC = ABD - BCD = 17° 15', and side BC 60, to find BD = 81.19. 2. In the triangle ABD are given the side AB = 40, BD = 81.49, and the angle ABD = 41', to find AD = 57.63 feet, the height of the obelisk. EXAMPLE 11. Fig. 65. Wanting to know the height of an object on the other side of a river, but which appeared to be on a level with the place where I stood, close by the side of the river; and not having room to go backward on the same plane, on account of the immediate rise of the bank, I placed a mark where I stood, and measured in a direct line from the object, up the hill, whose ascent was so regular that I might account it a right line, to the distance of 132 yards, where I perceived that I was above the level of the top of the object; I there took the angle of depression of the object equal 27°, and of its top equal 19°: required the height of the object. Calculation. 1. In the triangle ACD, are given the angle CAD = EDA = 27°, ACD = 180° - CDE (FCD) = 138o and the side CD = 132, to find AD = 194.55 yards. 2. In the triangle ABD, we have given ADB = ADE BDE = 8°, ABD = BED + BDE - 109° and AD = 194.55, to find AB = 28.64 yards the required height of the object. EXAMPLE 12. Fig. 66. A May-pole whose height was 100 feet, standing on a horizontal plane, was broken by a blast of wind, and the extremity of the top part struck the ground at the distance of 34 feet from the bottom of the pole : required the length of each part. Construction. Draw AB = 34, and perpendicular to it, make BC = 100; join AC and bisect it in D, and draw DE perpendicular to AC, meeting BC in E; then AE = CE = the part broken off.* * DEMONSTRATION. In the triangles AED, DEC, the angle ADE = CDE, the side AD = CD, and DE is common to the two triangles, therefore (4.1) AE = CE. Note. This question may be neatly solved in the following manner without finding either of the angles. Thus, draw DF perpenCalculation. 1. In the right angled triangle ABC, we have AB = 34 and BC = 100, to find the angle. C = 18° 47'. 2. In the right angled triangle ABE, we have AEB = ACE + CAE = 2 ACE - 37° 34' and AB = 34, to 2 find AE = 55.77 feet, one of the parts; and 100 - 55.77 = 44.23 feet, the other part. PRACTICAL QUESTIONS. 1. At 85 feet distance from the bottom of a tower, the angle of its elevation was found to be 52° 30': required the altitude of the tower. Ans. 110.8 feet. 2. To find the distance of an inaccessible object, I measured a line of 73 yards, and at each end of it took the angle of position of the object and the other end, and found the one to be 90°, and the other 61° 45'; required the distance of the object from each station. Ans. 135.9 yards from one, and 154.2 from the other. 3. Wishing to know the distance between two trees C and D, standing in a bog, I measured a base line AB = 339 feet; at A the angle BAD was 100° and BAC 36° 30°; at B the angle ABC was 121° and ABD 49° : required the distance between the trees. Ans. 232; feet. FC DC? AC? AB + BC2 sequently CE = ; and FC 4 4 AB + BC2 342 + 1002 therefore CE 2BC 200 1156 + 10000 11156 ABC ; . 4. Observing three steeples A, B and C, in a town at a distance, whose distance asunder are known to be as · follows, viz. AB = 213, AC = 104, and BC= 262 yards, , = I took their angles of position from the place D where I stood, which was nearest the steeple B, and found the angle ADB = 13° 30', and the angle BDC = 29° 50': Required my distance from each of the three steeples: Ans. AD = 571 yards, BD = 389 yards, and CD = 514 yards. 5. A may-pole, whose top was broken off by a blast of wind, struck the ground at 15 feet distance from the foot of the pole : what was the height of the whole maypole, supposing the length of the broken piece to be 39 feet? Ans. 75 feet. 6. At a certain place the angle of elevation of an inaccessible tower was 26° 30'; but measuring 75 feet in a direct line towards it, the angle was then found to be 51° 30': required the height of the tower and its distance from the last station. Ans. Height 62 feet, distance 49. 17. From the top of a tower by the sea side, of 143 feet high, I observed that the angle of depression of a ship's bottom, then at anchor, was 35o; what was its distance from the bottom of the wall? Ans. 204.2 feet. 1 8. There are two columns left standing upright in the ruins of Persepolis; the one is 64 feet above the plane, s and the other 50 : In a right line between these stands .an ancient statue, the head of which is 97 feet from the summit of the higher, and 86 from that of the lower co. lumn; and the distance between the lower column and the centre of the statue's base is 76 feet: required the SURVEYING. Surveying is the art of measuring, laying out, and dividing land. MEASURING LAND. Preliminary Definitions, Observations, &c. The Instrument used for measuring the sides of fields, or plantations, is a GUNTER's Chain, which is 4 poles or 66 feet in length, and is divided into 100 equal parts or links; consequently the length of each link is 7.92 inches : also 1 square chain is equal to 16 square perches, and 10 square chains make an acre. When the land is uneven or hilly, a four-pole chain is too long to be convenient, and the measures cannot be taken with it as accurately as with one that is shorter. Surveyors therefore generally make use of a chain that is two poles in length and divided into 50 links. The measures thus taken are, for the sake of ease in the calculation, reduced either to four-pole chains or to perches. The following rules sbew the method of making these, |