angle therefore in the right-angled triangle DAC, we have the angle ADC=ABC=60° 16′, and the side AC to find CD=52.98 perches. The half of CD is=26.49 perches CE the distance of the well from each corner. 3. The angle ACD = 90o — ADC 29° 44'; but because AEC is an isosceles triangle, the angle CAE ACE =29° 44' the angle required. EXAMPLE 7. Fig. 61. Wishing to know the height of a steeple situated on a horizontal plane, I measured 100 feet in a right line from its base, and then took the angle of elevation* of the top, which I found to be 47° 30', the centre of the quadrant being 5 feet above the ground: required the height of the steeple. Calculation. In the right-angled triangle DEC, we have the angle CDE 47° 30' and the base DE AB 100 feet to = * Angles of elevation, or of depression are usually taken with an instrument called a quadrant, the arc of which is divided into 90 equal parts or degrees, and those when the instrument is sufficiently large may be subdivided into halves, quarters, &c. From the centre a plummet is suspended by a fine silk. thread. Fig. 60 is a representation of this instrument. To take an angle of elevation, hold the quadrant in a vertical position, and, the degrees being numbered from B towards C, with the eye at C, look along the side CA, moving the quadrant till the top of the object is seen in a range with this side; then the angle BAD made by the plummet with the side BA, will be the angle of elevation required. Angles of depression are taken in the same manner, except that then the eye is applied to the centre of the quadrant. Note. In finding the height of an object, it is best so to contrive it that the observed angle of altitude may be about 45°; for when the observed angle is 45°, a small error committed in taking it, makes the least error in the com find CE 109.13 feet; to CE add EB = DA = 5 feet the height of the quadrant, and it will give BC= 114.13 feet, the required height of the steeple. EXAMPLE 8. Fig. 62. Wishing to know the height of a tree situated in a bog, at a station D which appeared to be on a level with the bottom of the tree, I took the angle of elevation BDC 51° 50'; I then measured DA 75 feet in a direct line from the tree, and at A, took the angle of elevation BAC = 26° 30'. Required the height of the tree. = Calculation. 1. Because the exterior angle of a triangle is equal to the sum of the two interior and opposite ones, the angle BDC=DAC + ACD; therefore ACD=BDC -- DAC =25°: now in the triangle ADC we have DAC = 26° 30', ACD = 25°, and AD = 75 to find DC=79.18. 2. In the right-angled triangle DBC, are given DC = 79.18, and the angle BDC=51° 30′ to find BC=61.97 feet, the required height of the tree. EXAMPLE 9. Fig. 63. Wanting to know the height of a tower EC, which stood upon a hill, at A, I took the angle of elevation CAB = 44°; I then measured AD 134 yards, on level ground, in a straight line towards the tower; at D the angle CDB was 67° 50' and EDB 51°. Required the 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 BC213.4; then 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. ABD-- BCD = 17o 15', 1. In the triangle BCD, we have given the angle BCD 23° 45', the angle BDC = ABD 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 = 41°, 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 42o, of the bottom of the object equal 27°, and of its top equal 19o: 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 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 AE CE = the part broken off.* 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 44.23 feet, the other part. 55.77 PRACTICAL QUESTIONS. 1. At 85 feet distance from the bottom of a tower, the angle of its elevation was found to be 52o 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 * 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 perpendicular to BC, then (31.3 and cor. 8. 6) FC ; DC : : DC : CE; consequently AC2 AB2+ BC2 DC2 CE= fore CE and FC=BC; there ; but DC2= |