apart I knew to be, AB = 3 miles, AC2, and BC = 1.8, I took a station D, in a straight line with the objects A and C, being nearer the former, and found the angle ADB = 17° 47'. Required my distance from each of the objects. Construction. With the three given distances, describe the triangle ABC; from B, draw BE parallel to CA, and draw BD making the angle EBD = 17° 47' (the given angle ADB) and meeting CA produced, in D: then AD, CD and BD will be the distances required.* Calculation. 1. In the triangle ABC we have all the sides given, to find the angle C 104° 8'. = 2. Subtract the sum of the angles D and C from 180°, the remainder 58° 5' will be the angle DBC; then in the triangle BCD we know all the angles and the side BC to find DC5.003 and BD = 5.715; therefore DA = DC - AC 3.003. EXAMPLE 5. Fig. 58. From a station at D, I perceived three objects, A, B and C, whose distances from each other I knew to be as follow: AB = 12 miles, BC= 7.2 miles and AC=8 miles; at D, I took the angle CDB 25° and ADC = 19o. Hence it is required to find my distance from each of the objects. = * DEMONSTRATION. By construction, the distances AB, BC and AC are equal to the given distances; also the angle (29.1) BDC = the Construction. With the given distances describe the triangle ABC; at B, make the angle EBA = 19° the given angle ADC, and at A, make the angle EAB 25° the given angle BDC; draw AE, and BE meeting in E, and (by prob. 10.) describe a circle that shall pass through the points A, E and B: join CE and produce it to meet the circle in D, and join AD, BD, then will AD, CD and BD be the distances required.* Calculation. 1. In the triangle ABC, all the sides are given, to find the angle BAC = 35° 35'.* 2. In the triangle AEB are given all the angles, viz. EAB = 25o, EBA=19° and AEB = 136°, and the side AB 12, to find AE = 5.624. › * DEMONSTRATION. The angle ADC standing on the same arc with the angle ABE is equal to it (21.3.) For the same reason the angle BDC is equal to the angle BAE; but by construction the angles ABE and BAE are equal to the given angles; therefore the angles ADC and BDC are equal to the given angles. Note. When the given angles ADC, BDC are respectively equal to the angles ABC, BAC, the point E will fall on the point C, the circle will pass through the points A, C, and B, and the point D may be any where in the arc ADB; consequently, in this case, the situation of the point D, or its distance from each of the objects A, B, C, cannot be determined. It may not be improper also to observe that even when the angle ADB, which is the sum of the given angles, is equal to the sum of the angles ABG, BAC, or which is the same thing, is the supplement of the angle ACB, the circle passes through the points A, C, B; but then the angle ADC, BDC, unless they have been erroneously 1 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 an. gle 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 = 60o 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. * 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 BAC26° 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 |