CASE III. Having given the three sides of a spherical triangle, to find the angles. 21. For this case we use equations (4). cos A = Rsin s sin (}s — a) sin b sin c Ex. 1. In an oblique-angled spherical triangle, there are given a 56° 40', b = 83° 13', and c = 114° 30': requir ed the angles. log sin b - log sin 1 (a + b + c) = s = 127° 11' 30", a) 70° 31' 30". = 127° 11' 30". 83° 13' 114° 30' angle A 48° 31' 18". ar. comp. ar. comp. - log sin с Sum Half sum = log cos A 24° 15′ 39′′ Hence, The addition of twice the logarithm of radius, or 20, to the numerator of the quantity under the radical, just cancels the 20 which is to be subtracted on account of the arithmetical complements, so that the 20, in both cases, may be omitted. Applying the same formulas to the angles B and C, we find, Ans. = 62° 55′ 46" B = 9.901250 9.974413 0.003051 0.040977 19.919691 9.959845 Ex. 2. In a spherical triangle there are given a = 40° 18′ 29′′, b = 67° 14′ 28′′, and c = 89° 47′ 06′′ required the three angles. Having given the three angles of a spherical triangle, to find the three sides. 22. For this case we employ equations (9). cos (SB) cos (S-C) Ꭱ cos a = R√ Hence, Ex. 1. In a spherical triangle ABC there are given A = 48° 30', B = 125° 20', and C sides. = = 62° 54'; required the CASE IV. = (A + B + C)= } S= - log sin - log sin Sum Half sum = log cosa log cos (SB) 6° 58' log cos (SC') 55° 28' B 125° 20' ar. comp. C 62° 54' ar. comp. 118° 22' 69° 52' 6° 58' 55° 28' 28° 19' 48" side a = 56° 39′ 36′′. In a similar manner we find, 9.996782 9.753495 0.088415 0.050506 19.889198 9.944599 114° 29′ 58" b = Ex. 2. In a spherical triangle ABC, there are given A 109° 55′ 42′′, B= 116° 38′ 33′′, and C=120° 43′ 37′′; required the three sides. α= 98° 21' 40" Ans. b 109° 50′ 22′′ c = 115° 13′ 26′′ CASE V. Having given in a spherical triangle, two sides and their included angle, to find the remaining parts. 23. For this case we employ the two first of Napier's Analogies. cos (a + b): cos sin(a + b) sin (a - b) :: cot C: tang sin sin :: cot 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 difference. The greater angle is then to be placed opposite the greater side. The remaining side of the triangle can be found by Case II. : tang Exc. 1. In a spherical triangle ABC, there are given a = 68° 46′ 02′′, b = 37° 10′, and C 39° 23'; to find the remaining parts. = Hence, +(a+b)=52° 58′ 1′′, 1(a — b)= 15°, 48′ 01′′, C=19° 41′ 30′′. (a+b) 52° 58′ 01′′ log. ar. comp. cos : cos (a - b) 15° 48′ 01′′ :: cot 19° 41' 30" C : tang (A+B) 77° 22′ 25′′ (A + B), (A - B). (a+b) 52° 58′ 01′′ log. ar. comp. (a - b) 15° 48′ 01′′ C 19° 41' 30" (AB) 43° 37′ 21′′ A 77° 22′ 25′′ + 43° 37′ 21′′ = 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. = Ans. 0.220205 9.983272 10.446253 10.649730 0.097840 9.435023 10.446253 9.979116 = 120° 59′ 47" 33° 45' 03" = 43° 37' 37" B = 156° 30 16" C= 9° 11' 48" α= 61° 32′ 12′′ CASE VI. In a spherical triangle, having given two angles and the included side, to find the remaining parts. 24. For this case, we employ the second of Napier's Analogies. cos (A+B): cos 1⁄2 (A − B) :: tang c: tang 1(a+b), sin (A + B) : sin }(4 – B) :: tang c: tang (a — b). From which a and b are found as in the last case. The remaining angle can then be found by Case I. Ex. 1. In a spherical triangle ABC, there are given A = 81° 38′ 20′′, B = 70° 09′ 38′′, c = 59° 16′ 23′′: to 'find the remaining parts. }(A+B)=75° 53′ 59′′, (A—B)=5° 44′ 21′′, c=29° 38′ 11′′. COS (A+B) 75° 53′ 59′′ log. 0.613287 cos (AB) 5° 44′ 21′′ 9.997818 ar. comp. : COS :: tang c 29° 38′ 11′′ 9.755051 tang 66° 42′ 52" 10.366156 (a + b) 75° 53′ 59′′ log. ar. comp. 0.013286 5° 14′ 21′′ 9.000000 29° 38′ 11′′ 9.755051 3° 21' 25" 8.768337 Hence, a = 66° 42′ 52" + 3° 21′ 25′′ 70° 04′ 17′′ Ex. 2. In a spherical triangle A = 34° 15′ 03′′, B = 42° 15 to find the remaining parts. ABC, there are given 13", and c = 76° 35′ 36′′: α = 40° 00' 10" Ans. b 50° 10′ 30′′ = C= 121° 36' 19" MENSURATION OF 1. WE determine the area, or contents of a surface, by finding how many times the given surface 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. 2. 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. 3. We have already seen (B. IV., P. 4, s. 2), 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. 4. To find the area of a square, a rectangle, or a parallel ogram. Multiply the base by the altitude, and the product will be the area (B. IV., P. 5). Ex. 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 Ans. 41738.49 sq. ft. feet? 3. What are the contents, in square yards, of a rectangle whose base is 66.3 feet, and altitude 33.3 feet? Ans. 245.31. |