this kind of exercise wholly to the student, remarking that in our general solutions we have given those methods which, under all circumstances, seemed to be the safest and best. $83. EXAMPLES FOR PRACTICE. 1. Given, in a right spherical triangle, A = 91° 11′ ; B=111° 11', to find the remaining parts. 2. Given, in a right spheric triangle, a=35° 44′; A=37° 28′, to find the other parts. 3. Given, in a spheric oblique triangle, a=138°; A= 95°; C=104°, to find the other parts. 4. Given a = 81° 17′; b=114° 3′; c= 59° 12′, to find the angles. A: 62° 39′ 43′′. Ans. B = 124° 50′ 50′′. C 50° 31′ 43′′. 5. Given, in a right spheric triangle, a=118° 54′; B=12° 19′, to 6. Given, in a right spheric triangle, A = 110°; B=100°, find the other part. (This is the same as first example under Case VI. of Oblique Spherical Trigonometry.) CHAPTER VIII. MENSURATION OF SURFACES. $84. THE area of a surface is found by comparing it with some known and fixed surface, which is taken as the unit surface, which unit surface is usually in the form of a square; as, a square inch, a square foot, a square yard, &c. The superficial unit receives its name from that of the linear unit, which measures its side. Thus, a square whose side is one inch is called a square inch; one whose side is one foot is called a square foot; one whose side is one yard is called a square yard. There are some superficial units which aave no corresponding linear unit, such as the acre, in land measure, and its fourth part, called a rood. There is no such linear measure as an acre or a rood. It would be absurd to speak of an acre long or a rood long. The usual units of square measure are given by the following 4014489600 — 27878400 = 3097600 = 102400 = 6400 = 640 640 1 § 85. In measuring land, Gunter's Chain is most commonly employed. Its length is 4 rods, or 66 feet, and is divided into 100 links, so that we have as follows: 7,92 inches, or 7.92 inches 100 80 1 link. links 792 inches=66 feet 22 yards=4 rods 1 chain. 1 mile. 1 sq. chain. chains 10000 sq. links 100000 sq. links 16 sq. rods 10 sq. chains 160 sq. rods 1 acre. It cannot be given The unit of measure for land is the acre. in an exact square, since it is made to consist of 160 square rods, and 160 is not a square number, which proves that the side of a square containing 160 sq. rods cannot be accurately expressed in rods; and if not in rods, then not in any multiple or submultiple of rods. One-fourth of an acre is called a rood, and the square rod in land measure is frequently called a perch. If links be multiplied by links, the product will be square links, which may be reduced to acres by dividing by 100000, or by simply pointing off five decimal places. If chains be multiplied by chains, the product will be square chains, which may be reduced to acres by dividing by 10, or pointing off one decimal figure. If chains are multiplied by links, the product will be converted into acres by pointing off three decimals. we We will also add, that to convert square feet into acres, we must divide by 43560. To convert square yards into acres, must divide by 4840; and to convert square rods into acres, we must divide by 160. PROBLEM 1. To find the area of a square or rectangular piece of land. The obvious rule in this case will be as follows: Take the product of two adjacent sides for the area. [Geom., B. III., T. XXI., Scholium.] EXAMPLES. 1. How many acres in a piece of ground in the form of a square, each of whose sides is 13 chains, 25 links, or 13.25 chains ? 13·25 × 13·25 = 175·5625 Or, since 13.25 chains = 1325 links, we have 1325 × 1325 = 1755625 square links, Converting this decimal of an acre into roods and perches, 2. How many acres in a rectangular piece of land whose adjacent sides are 30.25 chains and 25.21 chains? 2521 × 3025 = 7626025 sq. links = 76A. 1R. 1,84P. 3. How many rods in a rectangular piece of ground 282 feet wide and 325 feet long? Ans. 336-639 sq. rods. and area ABCD = DE × AB = AD × AB × sin. A. (1.) Using logarithms, log. area = log. AD + log. AB + log. sin. A—10. We subtract 10 to correct for the 10 added to log. sin. of A. Hence we have this (2.) SECOND RULE. Multiply the product of two adjacent sides by the natural sine of the included angle. OR BY LOGARITHMS. To the sum of the logarithms of two adjacent sides add the logarithmic sine of the included angle, and subtract 10, and the result will be the logarithm of the area. EXAMPLES. 1. How many acres in a field in the form of a parallelogram, having a base of 13 chains 14 links, and an altitude of 10 chains 37 links? 1314×1037=1362618 sq. links-13A. 2R. 20P. 2. How many acres in a field in the form of a parallelogram, the adjacent sides being 25.17 chains and 30-25 chains, and having 70° 30' for the included angle? BY LOGARITHMS. log. 25.17 = 1·400883 log. area = 2.855955 = Hence, area 717-72 sq. chains, which is Ans. 71A. 3R. 31P. 3. A field, in the form of a parallelogram, has 10-21 chains, 12.12 chains for the adjacent sides, and 30° 45′ for the included angle. How many acres does it contain? Ans. 63.27 sq. chains. PROBLEM IIL To find the area of a triangle. FIRST RULE. When the altitude is given, multiply its half by the base. [Geom., B. III. T. XXIII.] Or multiply the altitude by half the base, or take half the product of the altitude and base. Since the triangle ABD is obviously just one-half of the par A D E B |