For this case we employ the two first of Napier's Analogies. cos (a+b): cos (a—b): : cot C: tang (A+B) sin(a+b) sin (a-b): cot C: tang (A-B). Having found the half sum and the half difference of the angles A and B, the angles themselves become known; for, the greater angle is equal to the half sum plus the half difference, and the lesser is equal to the half sum minus the half diffe rence. The greater angle is then to be placed opposite the greater side. The remaining side of the triangle can then be found by Case II. Ex. 1. In a spherical triangle ABC, there are given a=68° 46′ 2′′, b=37° 10′, and C-39° 23'; to find the remaining parts (a+b)=52° 58′ 1′′, (a—b)=15° 48′ 1′′, C=19° 41′ 30′′. (a+b) 52° 58′ 1" log. ar.-comp. C 19° 41′ 30′′ As cos Is to cos So is cot As sin Is to sin 0.220210 9.983271 10.446254 (a+b) 52° 58' 1" log. ar.-comp. So is cot C 19° 41' 30" Totang (A-B) 43° 37′ 21′′ 9.435016 10.446254 9.979110 Hence, A=77° 22′ 25′′ +43° 37′ 21′′"=120° 59′ 46′′ B=77° 22′ 25′′-43° 37′ 21′′: 33° 45′ 04′′ Ex. 2. In a spherical triangle ABC, there are given b=83" 19' 42", c=23° 27′ 46", the contained angle A=20° 39′ 48′′; to find the remaining parts. In a spherical triangle, having given two angles and the included side to find the remaining parts. For this case we employ the second of Napier's Analogies. (A—B) :: tang c: tang (a+b) (A-B) :: tang c: tang (a-b). cos (A+B): cos sin (A+B) sin From which a and b are found as in the last case. maining angle can then be found by Case I. The re Ex. 1. In a spherical triangle ABC, there are given A=81° 38′ 20′′, B=70° 9′ 38′′, c=59° 16′ 23"; to find the remaining parts. (A+B)=75° 53′ 59′′, (A-B)=5° 44′ 21′′, c=29° 38′ 11′′. (A+B) 75° 53' 59" log. ar.-comp. 0.613287 (A-B) 5° 44′ 21′′ As cos To cos 9.997818 So is tang C 29° 38′ 11′′ 9.755051 To tang (a+b) 66° 42′ 52′′ 10.366156 As sin To sin (A+B) 75° 53′ 59′′ log. ar.-comp. 0.013286 (A-B) 5° 14′ 21′′ 9.000000 So is tang c 29° 38′ 11′′ 9.755051 To tang (a-b) 3° 21′ 25′′ 8.768337 Hence a=66° 42′ 52′′+3° 21′ 25′′=70° 04′ 17′′ b=66° 42′ 52′′-3° 21′ 25′′=63° 21′ 27′′ Ex. 2. In a spherical triangle ABC, there are given A=34° 15′ 3′′, B=42° 15′ 13′′, and c=' =76° 35′ 36′′; to find the remaining parts. Ans. a = -40° 0' 10" b=50° 10′ 30′′ C=58° 23′ 41′′. S 274 MENSURATION OF SURFACES. The area, or content of a surface, is determined by finding how many times it contains some other surface which is assumed as the unit of measure. Thus, when we say that a square yard contains 9 square feet, we should understand that one square foot is taken for the unit of measure, and that this unit is contained 9 times in the square yard. The most convenient unit of measure for a surface, is a square whose side is the linear unit in which the linear dimensions of the figure are estimated. Thus, if the linear dimensions are feet, it will be most convenient to express the area in square feet; if the linear dimensions are yards, it will be most convenient to express the area in square yards, &c. We have already seen (Book IV. Prop. IV. Sch.), that the term, rectangle or product of two lines, designates the rectangle constructed on the lines as sides; and that the numerical value of this product expresses the number of times which the rectangle contains its unit of measure. PROBLEM I. To find the area of a square, a rectangle, or a parallelogram. RULE.-Multiply the base by the altitude, and the product will be the area (Book IV. Prop. V.). 1. To find the area of a parallelogram, the base being 12.25 and the altitude 8.5. Ans. 104.125. 2. What is the area of a square whose side is 204.3 feet? Ans. 41738.49 sq. ft. 3. What is the content, in square yards, of a rectangle whose base is 66.3 feet, and altitude 33.3 feet? Ans. 245.31. 4. To find the area of a rectangular board, whose length is 12 feet, and breadth 9 inches. Ans. 93 sq. ft. 5. To find the number of square yards of painting in a parallelogram, whose base is 37 feet, and altitude 5 feet 3 inches. Ans. 217. When the base and altitude are given. RULE.-Multiply the base by the altitude, and take half the product. Or, multiply one of these dimensions by half the other (Book IV. Prop. VI.). 1. To find the area of a triangle, whose base is 625 and altitude 520 feet. Ans. 162500 sq. ft. 2. To find the number of square yards in a triangle, whose base is 40 and altitude 30 feet. Ans. 663. 3. To find the number of square yards in a triangle, whose base is 49 and altitude 25 feet. Ans. 68.7361. CASE II. When two sides and their included angle are given. RULE. Add together the logarithms of the two sides and the logarithmic sine of their included angle; from this sum subtract the logarithm of the radius, which is 10, and the remainder will be the logarithm of double the area of the triangle. Find, from the table, the number answering to this logarithm, and divide it by 2; the quotient will be the required area. Let BAC be a triangle, in which there are given BA, BC, and the included angle B. From the vertex A draw AD, perpendicular to the base BC, and represent the area of the triangle by Q. Then, B R : sin B :: BA: AD (Trig. Th. I.): A D BAX sin B R BCX AD But, Q = (Book IV. Prop. VI.); 2 hence, by substituting for AD its value, we have Taking the logarithms of both numbers, we have which proves the rule as enunciated. 1. What is the area of a triangle whose sides are, BC= 125.81, BA=57.65, and the included angle B=57° 25'? Then, log. 2Q= + log. BC +log. sin B 57° 25′ 2.099715 1.760799 and 2Q=6111.4, or Q=3055.7, the required area. 2. What is the area of a triangle whose sides are 28 and 40 and their included angle 28° 57′ ? Ans. 290.427. 3. What is the number of square yards in a triangle of which the sides are 25 feet and 21.25 feet, and their included angle 45°? Αης. 20.8694. CASE III. When the three sides are known. RULE.-1. Add the three sides together, and take half their sum. 2. From this half-sum subtract each side separately. 3. Multiply together the half-sum and each of the three remainders, and the product will be the square of the area of the triangle. Then, extract the square root of this product, for the required area. Or, After having obtained the three remainders, add together the logarithm of the half-sum and the logarithms of the respective remainders, and divide their sum by 2: the quotient will be the logarithm of the area. D B Let ABC be the given triangle. Take CD equal to the side CB, and draw DB; draw AE parallel to DB, meeting CB produced, in E: then CE will be equal to CA. Draw CFG perpendicular to AE and DB, and it will bisect them at the points G and F. Draw FHI parallel to AB, meeting CA in H, and EA produced, in I. Lastly, with the centre H and radius HF, describe the circumference of a circle, meeting CA produced in K: this circumference will pass through I, because AI-FB-FD, therefore, HF=HI; and it will also pass through the point G, because FGI is a right angle. K ¡A E Now, since HA=HD, CH is equal to half the sum of the sides CA, CB; that is, CH=CA+CB; and since HK is equal to IFAB, it follows that CK=¦AC+CB+1AB=¦S, by representing the sum of the sides by S. Again, HK=HI=IF=AB, or KL=AB. Hence, CL=CK-KL-1S-AB, and AK-CK-CA-1S-CA, and AL=DK=CK-CD-4S--CB. the area of the triangle ACE, the area of the triangle ABE; the area of the triangle ACB. Now, AGX CG and X = |