2 and each of the remainders =351-234-117; hence the area 3×1 17 × 117 × 1 17 × 117 = 117x117 √3 = 13689 x√3. But the square root of 3=1·73205, which being multiplied by 13689, gives 23710 square links, which is equal to •23710 acres. 4 Ans. Sq. links 23710=4.374932 Log. area. Ans. 30 yds. 270 feet =2·431364 Log. area. Or the sides expressed in yards are 13, 12, and 5, hence half their sum is 15, and the several remainders are 2, 3, and 10; therefore the area in square yards is √15×2×3×10 =√√30 × 30=30 Ans. (3.) Here the sides expressed in chains are 5, 5, and 6, therefore their half sum is 8, and the remainders are 3, 3, and 2; hence the area in square chains is 8×3×3×2 √144=12; and since 10 square chains make an acre, the area is 1.2 acres =1 acre 32 poles. = (4.) Here (74+82+90)=123, and the three remainders are 49, 41, and 33; hence the area will be = = √123 × 49 × 41 × 33 815431, the square root of which is equal to 2855-61 the answer. (1.) Here the sum of the angles 68° 24′ and 71° 13' is 139° 37', the cosecant of which is the secant of its excess above 90°, that is of 49° 37'. Hence the Ans.. 7 acres, 2 roods, 39.84 poles. (3) In this exercise there are given in each of the triangles ABD and DBC two sides and the contained angle; hence we must use Rule II. Problem II., to find the area. poles. DBC=172876 ..▲ABD+DBC=3,51560= 3 acres, 2 roods, 2.496 9.993725 = 5.237735 4 2,06240 40 2,49600 4. Here in the triangles ABD and DBC there are given the three sides of each, the diagonal BD being common to both; hence their areas must be found by Rule III., Problem II. Hence the sum of the areas is 777·084 +675⋅6=1452.684, which being divided by 9, gives 161.409 sq. yards. (5.) By Rule II. of this Problem we have Log. 760 Log. 810 Log. sin. 79° 16' Subtract 3 ac. 3.864 poles =3,02415 = 2.880814 = 2.908485 = 9.992335 15.781634 10.301030 = 5.480604 4 .9660 40 3.86400 (6.) In the ADC there are given the two sides AD= 320 and DC=360, and the included LADC=80° 30′; from which, by Trig., Art. 40, the diagonal AC can be found, and then there are given the three sides of each of the triangles ADC and ABC, from which the area can be found, by Rule III., Problem II.; thus, |