Calculation. 1. In the triangle ADC, we have the angle DAC = 44°, the angle ACD = BDC — DAC = 23° 50', and the side AD, to find DC 230.4. = 2. In the triangle DEC all the angles are given, viz. CDE BDC = - BDE = 16° 50', DCE 90° — BDC 22° 10′, DEC = 180°- the sum of the angles CDE and DCE, = 141°, and CD = 230.4, to find CE = 106 yards, the height of the tower. 3. In the right angled triangle DBC, we have the angle BDC = 67° 50', and the side DC 230.4, to find BC= 213.4; then BE BE=BC-CE=213.4-106= 107.4 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, BD = 81.49, and the angle ABD = 41o, to find AD = 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 mark by the river's side equal 42°, of the bottom 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) = 138° and the side CD 132, to find AD = 194.55 yards. 2. In the triangle ABD, we have given ADB = ADE — BDE = 8o, 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 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 AECE = the part broken off.* 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 + CAE =2 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 inaccessable object, I measured a line of 73 yards, and at each end of it took * 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 perpendicular to BC, then (31.3 and cor. 8.6) FC: DC: DC: CE; consequently AC2 AB2+ BC2 CE= DC2 but DC2 and FC = 115610000 200 BC; there 11156 200 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. 697 feet. 4. Observing three steeples A, B and C, in a town at a distance, whose distances 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 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 distance between the top of the columns. Ans. 157 feet. |