CASE III. : angles. 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. ila + b + c) = js = 127° 11' 30", i (b +c - a) = (i's – a) = 70° 31' 30". log sin is 127° 11' 30". . . 9.901250 log sin (1s – a) 70° 31' 30". . . 9.974413 – log sin b 83° 13' ar. comp. 0.003051 – log sin c 114° 30' ar. comp. 0.040977 Sum . . . . . . . . 19.919691 Half sum = log cos į A 24° 15' 39" ... 9.959845 Hence, angle A = 48° 31' 18". 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, B= 62° 55' 46" C = 125° 19' 02" 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. (A = 34° 22' 16" Ans. B= 53° 35' 16'' IC= 119° 13' 32'' CASE IV. Having given the three angles of a spherical triangle, to find the three sides. 22. For this case we employ equations (9). cos ta = R Vcos (S – B) cos (+ S-C). sin B sin C Ex. 1. In a spherical triangle ABC there are given A = 48° 30', B = 125° 20', and C = 62° 54'; required the sides. *(A + B + C)=S= 118° 22' (IS – C). . = 55° 28' log cos (1S - C) 55° 28' . . 9.753495 - log sin B 125° 20' ar. comp. 0.088415 – log sin C 62° 54' ar. comp. 0.050506 Sum. . . . . 19.889198 Half sum = log cos I 28° 19'48" . 9.944599 side a = 56° 39' 36". In a similar manner we find, b = 114° 29' 58" c= 83° 12' 06" 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. ( a = 98° 21 40” Ans. { b = 109° 50' 22'' lc= 115° 13' 26" Hence, CASE V. living given in a spherical triangle, two sides and their in. cluded angle, to find the remaining parts. · 23. For this case we employ the two first of Napier's Analogies. cos f(a + b) : cost (a - b) :: cot 1C : tang I(A + B), sin } (a + b) : sin f(a - b) :: cot 1C : tang 3 (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 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. 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. f(a+b)=52° 58' 1", }(a - b)=15°, 48' 01", IC=19° 41' 30". cos ž (a + b) 52° 58' 01" log. ar. comp. 0.220205 : cos }(a - b) 15° 48' 01" 9.983272 :: cot 1C 19° 41' 30" 10.446253 : tang {(A + B) 77° 22' 25" 10.649730 sin }(a + b) 52° 58' 01" log. ar. comp. 0.097840 i sin }(a - b) 15° 48' 01" 9.435023 :: cot 1C 19° 41' 30" . 10.446253 : tang }(A - B) 43° 37' 21" . . . 9.979116 Hence, A = 77° 22' 25" + 43° 37' 21" = 120° 59' 47" B= 77° 22' 25" – 43° 37' 21" = 33° 45'03" side c . . . . . = 43° 37' 37" 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. (B= 156° 30 16" Ans. C= 9° 11' 48' la= 61° 32' 12" CASE VI. In a spherical triangle, having given two angles and the in. cluded side, to find the remaining parts. 24. For this case, we employ the second of Napier's Analogies. cos ž (A + B) : cos ž (A – B) :: tang ic : tang i (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. ar. comp. 0.613287 : cos }(A – B) 5° 44' 21" . . . 9.997818 :: tang įc 29° 38' 11" . 9.755051 : tang 1 (a + b) 66° 42' 52" . . . 10.366156 sin }(A + B) 75° 53' 59" log. ar. comp. 0.013286 : sin }(A - B) 5° 14' 21" . 9.000000 :: tang lc 29° 38' 11" . . . 9.755051 : tang ila – 1) 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" angle C . . . . = 64° 46' 33" Ex. 2. In a spherical triangle ABC, there are given A = 34° 15' 03", B = 42° 15 13", and c= 76° 35' 36" : to find the remaining parts. (a= 40° 00' 10" Ans. 0= 50° 10' 30" (C= 121° 36' 19" MENSURATION OF SURFACES. 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 feet? Ans. 41738.49 sq. ft. 3. What are the contents, in square yards, of a rectan. gle whose base is 66.3 feet, and altitude 33.3 feet ? Ans. 245.31. |