3. In the triangle CAE we have given the side AC = 8, AE= 5.624, and the angle CAE = BAC - EAB = 10° 35', to find the angle ACE = 22° 41'. 4. In the triangle DAC, all the angles are given, viz. ADC = 19°, ACD = 22° 41' and CAD= 180° - the sum of the angles ADC and ACD, = 138° 19', and the side AC = 8, to find AD= 9.47 miles and CD= 16.34 5. In the triangle ABD, we have the angle ADB = ADC + BDC = 44°, the angle BAD = CAD- BAC = 102° 44', and the side AB 12, to find BD = 16.85 miles. EXAMPLE 6. Fig. 59. A person having a triangular field, the sides of which measure AB = .50 perches, AC = 46 perches and BC = 40 perches, wishes to have a well dug in it, that shall be equally distant from the corners A, B and C. What must be its distance from each corner, and by what angle from the corner A, may its place be found. Construction. With the given sides construct the triangle ABC, and (by Prob. 10.) describe a circle that shall pass through the points A, B and C; then the centre E of this circle is the required place of the well. * Calculation. 1. In the triangle ABC, all the sides are given, to find the angle ABC = 60° 16'. 2. Join CE and produce it to meet the circumference in D; also join AE and AD; then the angles ADC, * The demonstration of this is plain (1. 3. cor.) K ABC being angles in the same segment are equal; also the angle DAC being an angle in a semicircle, is a right 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 = 90° 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. * a 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 Calculation. In the right-angled triangle DEC, we have the angle CDE = 47° 30' and the base DE = AB= 100 feet to 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° 30'; 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 51o. Required the height of the tower and also of the hill. 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° — ' = 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=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= 41o; 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 = 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 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= 180o -- CDE (FCD) = 1389 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. |