Ex. 2. Given the logarithms of BC and AC 3.964217 and 3.729415 respectively, and the angle C = 63° 17′ 24′′, to find AB. Ans. 8317. Ex. 3. Given the logarithms of AB and BC 1.963425 and 2.416347, and the angle B = 129° 42', to find AC. CASE 3. Ans. 327.27. 142. Given the three sides, to find the angles. RULE 1. Call the longest side the base, and on it let fall a perpendicular from the opposite angle. Then, as the base is to the sum of the other sides, so is the difference of those sides to the difference of the segments of the base. Half this difference added to half the base will give the greater segment, and subtracted will give the less segment. Having the segments of the base, and the adjacent sides, the angles may be found by Rule 2, Art. 137. Ex. 1. Given the three sides of a triangle,-viz.: AB= 467, AC = 413, and BC = 394, to find the angles. Whence C=180−(A+B) = 70° 40′ 18′′. Ex. 2. Given the three sides of a triangle, BC 167, AB 214, and AC 195 yards, respectively, to find the angles. Ans. A= 47° 55′ 13′′, B = 60° 4′ 19, C = 72° 0′ 28′′. = Ex. 3. Given AB 51.67, AC = 43.95, and BC= 27.16, to find the angles. Ans. A 31° 42′ 42′′, B = 58° 16' 34", C=90° 0'44". = RULE 2. 143. As the rectangle of two sides is to the rectangle of the half sum of the three sides and the excess thereof above the third side, so is the square of radius to the square of the cosine of half the angle contained by the first mentioned sides. DEMONSTRATION.-Let ABC (Fig. 50) = K be a triangle, of which AB is greater EGD = BGH and EDG: = GBH, (26.1,) DG = GB and EG = GH. On EH describe a circle, and it will pass through F. Now, 2 AK=2AG+2 GK AC + AD+2 DG+2 GK=AC+AB+BC; or and AI-AK-KI = (AC+ AB+BC)—BC. But, (Rule 2, Art. 137,) As AD: AE r cos. DAE (cos. BAC), and whence (23.6) AB AF: cos. BAC; AB. AD: AE. AF::: cos." BAC. But (36.3, Cor.) AE. AF AK. AI (AC+AB+ BC). (AC+ AB+ = = ≥ † BC) - BC; whence AB.AC: (AC+AB+BC) · (} (AC+AB+BC)—BC): : ra: cos.31⁄2 BAC. EXAMPLES. Ex. 1. Given AB = 467, AC = 413, and BC= 394, to find the angle C. Here, put s= half sum of the sides: we have s=637 and s-AB=170; whence In the above calculation the R' and its logarithm might have been omitted, since we have to deduct 20 in consequence of having taken two arithmetical complements. The sum of the logarithms is divided by 2, to extract the square root, (Art. 16.) The rule may be expressed thus:— Add together the arithmetical complements of the logarithms of the two sides containing the required angle, the logarithm of the half sum of the three sides, and the logarithm of the excess of the half sum above the side opposite to the required angle: the half sum of these four logarithms will be the logarithmic cosine of half that angle. Ex. 2. Given AB=167, AC=214, and BC=195, to find the angles. Ans. A 60° 4′ 22′′, B = 72° 0′ 28′′, C= 47° 55′ 16′′. = Ex. 3. Given AB = 51.67, AC = 43.95, and BC= 27.16, to find the angles. Ans. A 31° 42' 40", B = 58° 16' 28", C= 90° 0′52′′. = SECTION V. INSTRUMENTS AND FIELD OPERATIONS. 144. The Chain, GUNTER'S CHAIN is the instrument most commonly employed for measuring distances on the ground. For surveying purposes, it is made 66 feet or 4 perches long, and is formed of one hundred links, each of which is therefore .66 feet or 7.92 inches long. The links are generally connected by two or three elliptic rings, to make the chain more flexible. A swivel link should be inserted in the middle, that the chain may turn without twisting. In order to facilitate the counting of the links, every tenth link is marked by a piece of brass, having one, two, three, or four points, according to the number of tens, reckoned from the nearest end of the chain. Sometimes the number of links is stamped on the brass. The middle link is also indicated by a round piece of brass. The advantage of having a chain of this particular length is, that ten square chains make an acre. The calculations are therefore readily reduced to acres by simply shifting the decimal point. There being one hundred links to the chain, all measures are expressed decimally, which renders the calculations much more convenient. Eighty chains make one mile. In railroad surveying, a chain of one hundred feet long is preferred, the dimensions being thus at once given in feet. When the measurements are required to be made with great accuracy, rods of wood or metal, which have been made of precisely the length intended, are used. In the surveys of the American Coast Survey, the unit of length employed is the French metre, equal to the 10000000th part of the quadrant of the meridian. The metre is 39.37079 inches 3.280899 feet 1.093633 yards long. = = It were much to be desired that the metre, or some other unit founded on the magnitude of the earth, or on some other natural length, such as that of a pendulum beating seconds at a given latitude, were universally adopted as the unit. The metre will probably gradually come into general use. To reduce chains and links to feet, express the links decimally and multiply by 66. Thus, 7 chains 57 links = 7.57 chains are equal to 7.57 × 66 = 499.62 feet = 499 feet 7.4 inches. To reduce feet and inches to chains, divide by 66, or by 6 and 11. The inches must first be reduced to a decimal of a foot. Thus, 563 feet 8 inches = 563.67 feet= 8.54 chains. Instead of a chain of 66 feet, one of 33 feet, divided into fifty links, is sometimes used. This is really a half chain, and should be so recorded in the notes. The half chain is more convenient when the ground to be measured is uneven. 145. The chain is liable to become incorrect by use; its connecting rings may be pulled open, and thus the chain become too long, or its links may be bent, which will |