3. In the right angled triangle DBE, we have the an = = 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 = 81.49. 2. In the triangle ABD are given the side AB = 40, BD81.49, 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 1 180° CDE (FCD) = 138° 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. Draw AB = Construction. 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) AECE. Note. This question may be neatly solved in the following manner without finding either of the angles. Thus, draw DF perpen Calculation. 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 CAE2 ACE 37° 34′ and AB = 34, to 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. = 55.78, the same as before nearly. = 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 404, 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. 7. 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 35°; what was its distance from the bottom of the wall? Ans. 204.2 feet. 8. There are two columns left standing upright in the ruins of Persepolis; the one is 64 feet above the plane, 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 column; 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 shew the method of making these, |