Ex. 1. Having measured AB equal to 100 feet from the bottom of a tower on a horizontal CDE, of the top to be 47° 30', the What is the C E B Ex. 2. From the edge of a ditch 18 feet wide, surrounding a fort, the angle of elevation of the wall was found to be 62° 40'. Required the height of the wall, and the length of a ladder necessary to reach from my station to the top of it. Ans. The height is 34.82 feet. Length of ladder, 39.20 feet. PROBLEM II. (144.) To find the distance of a vertical object whose height is known. Measure the angle of elevation, and we shall have given the angles and perpendicular of a right-angled triangle to find the base (Art. 46). C Ex. 1. The angle of elevation of the top of a tower whose height was known to be 143 feet, was found to be 35°. What was its distance? Here we have given the angles of the triangle ABC, and the side CB, to find AB. Ans., 204.22 feet. If the observer were stationed at the A D B top of the tower BC, he might find the length of the base AB by measuring the angle of depression DCA, which is equal to BAC. Ex. 2. From the top of a ship's mast, which was 80 feet above the water, the angle of depression of another ship's hull was found to be 20°. What was its distance? Ans., 219.80 feet. X PROBLEM III. (145.) To find the height of a vertical object standing on an inclined plane. Measure the distance from the object to any convenient station, and observe the angles which the base-line makes with lines drawn from its two ends to the top of the object. If we measure the base-line AB, and the two angles ABC, BAC, then, in the triangle ABC, we shall have given one side and the angles to find BC. Ex. 1. Wanting to know the height of a tower standing on an inclined plane, BD, I measured from the bottom of the tower a distance, AB, equal to 165 feet; also the angle ABC, equal to 107° 18', and the angle BAC, D equal to 33° 35'. Required the height of the object. B A sin. ACB AB :: sin. BAC : BC=144.66 feet. : The height, BC, may also be found by measuring the distances BA, AD, and taking the angles BAC, BDC. The dif ference between the angles BAC and BDC will be the angle ACD. There will then be given, in the triangle DAC, one side and all the angles to find AC; after which we shall have, in the triangle ABC, two sides and the included angle to find BC. Ex. 2. A tower standing on the top of a declivity, I measured 75 feet from its base, and then took the angle BAC, 47° 50'; going on in the same direction 40 feet further, I took the angle, BDC, 38° 30'. What was the height of the tower? Ans., 117.21 feet. PROBLEM IV. (146.) To find the distance of an inaccessible object, Measure a horizontal base-line, and also the angles between this line and lines drawn from each station to the object. Let C be the object inaccessible from A and B. Then, if the dis tance between the stations A and B be measured, as also the angles at A and B, there will be given, in the triangle ABC, the side AB and the angles, to find AC and BC, the distances of the object from the two stations. Ex. 1. Being on the side of a river, and wanting to know the distance to a house which stood on the other side, I measured A B 400 yards in a right line by the side of the river, and found that the two angles at the ends of this line, formed by the other end and the house, were 73° 15′ and 68° 2′. the distance between each station and the house? The angle C is found to be 38° 43'. Then What was Ex. 2. Two ships of war, wishing to ascertain their distance from a fort, sail from each other a distance of half a mile, when they find that the angles formed between a line from one to the other, and from each to the fort, are 85° 15′ and 83° 45'. What are the respective distances from the fort? Ans., 4584.52 and 4596.10 yards. PROBLEM V. (147.) To find the distance between two objects separated by an impassable barrier. Measure the distance from any convenient station to each of the objects, and the angle included between those lines. If we wish to know the distance between the places C and B, both of which are accessible, but sep arated from each other by water, we may c measure the lines AC and AB, and also the angle A. We shall then have given two sides of a triangle and the included angle to find the third side. A B Ex. 1. The passage between the two objects C and B being obstructed, I measured from A to C 735 rods, and from A to B 840 rods; also, the angle A, equal to 55° 40'. What is the distance of the places C and B? Ans., 741.21 rods Ex. 2. In order to find the distance between two objects, C d B, which could not be directly measured, I measured 1 ɔm C to A 652 yards, and from B to A 756 yards; also, the angle A equal to 142° 25'. What is the distance between the objects C and B? Ans. PROBLEM VI. (148.) To find the height of an inaccessible object above a horizontal plane. First Method.-Take two stations in a vertical plane passing through the top of the object; measure the distance between the stations and the angle of elevation at each. If we measure the base AB, and the angles DAC, DBC, then, since CBA is the sup A B plement of DBC, we shall have, in the triangle ABC, one side and all the angles to find BC. Then, in the right-angled triangle DBC, we shall have the hypothenuse and the angles to find DC. D Ex. 1. What is the perpendicular height of a hill whose angle of elevation, taken at the bottom of it, was 46°; and 100 yards farther off, on a level with the bottom of it, the angle was 31°? Ans., 143.14 yards. Ex. 2. The angle of elevation of a spire I found to be 58°, and going 100 yards directly from it, found the angle to be only 32°. What is the height of the spire, supposing the instrument to have been five feet above the ground at each observation? Anş., 104.18 yards. (149.) Second-Method.-Measure any convenient base-line, also the angles between this base and lines drawn from each of its extremities to the foot of the object, and the angle of elevation at one of the stations. Let DC be the given object. If we measure the horizontal base-line AB, and the angles CAB, CBA, we can compute the distance BC. Also, if we observe the angle of elevation CBD, we shall have given, in the right-angled triangle BCD, the base and angles to find the perpen dicular. Ex. 1. Being on one side of a river, and wanting to know the height of a spire on the other side, I measured 500 yards, AB, along the side of the river, and found the angle ABC=74° 14', and BAC=49° 23'; also, the an A B D 悦悦 gle of elevation CBD=11° 15'. Required the height of the spire. Ans., 271.97 feet. Ex. 2. To find the height of an inaccessible castle, 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'; also, the elevation of the castle from the latter station, 10° 35'. Required the height of the castle. Ans., 86.45 feet. PROBLEM VII. (150.) To find the distance between two inaccessible objects. Measure any convenient base-line, and the angles between this base and lines drawn from each of its extremities to each of the objects. D Let C and D be the two inaccessible objects. If we measure a base-line, AB, and the angles DAB, DBA, CAB, CBA, then, in the triangle DAB, we shall have given the side AB and all the angles to find BD; also, in the triangle ABC, we shall have one side and all the angles to find BC; and then, in the triangle BCD, we shall have two sides, BD, BC, with the included angle, to find DC. ' A B Ex. 1. Wanting to know the distance between a house and a mill, which were separated from me by a river, I measured a base-line, AB, 300 yards, and found the angle CAB=58° 20', CAD=37°, ABD=53° 30', DBC-45° 15'. What is the distance of the house from the mill? Ans., 479.80 yards. |