. Mathematics, compiled from the best authors and intended to be the text-book of the course of private lectures on these sciences in the University at Cambridge [microform]. e. This rule is preferable, when the required angleis between 90° and 180° ; and the other, when it is less than90°. Half the sum of j ^ 3g, ^ the sine 9-9203841the three sides JSide opposite Z A 110 DifferenceCo-arc sineCo-arc sine Cos. 13 36 37, the sine 9*372694553 0*0715795 79 17 14 00076359 2)19*3722940 60 57 28 9*6861470 2 Angle A = 121 54 56 CASE 6. Example. Given the three angles, namely, A=121c54 56, B=50°, and C
. Mathematics, compiled from the best authors and intended to be the text-book of the course of private lectures on these sciences in the University at Cambridge [microform]. e. This rule is preferable, when the required angleis between 90° and 180° ; and the other, when it is less than90°. Half the sum of j ^ 3g, ^ the sine 9-9203841the three sides JSide opposite Z A 110 DifferenceCo-arc sineCo-arc sine Cos. 13 36 37, the sine 9*372694553 0*0715795 79 17 14 00076359 2)19*3722940 60 57 28 9*6861470 2 Angle A = 121 54 56 CASE 6. Example. Given the three angles, namely, A=121c54 56, B=50°, and C=62° 34 6 ; to find the rest. „ , , .r3 X sin. is—a x sin. 4 s Cos. I angle = J :—f : 2— sin. b x sin. c Whence, logarithms being used, Cos.£ angle = sin. is—a -f sin.^s + co-ar. sin. b -\- which, expressed by words, is the rule-Vol II. Iii 450 MATHEMATICS. PROJECTION OF THE TRIANGLE. > r r Draw the oblique circleABd, making the angleBAe = 58° 5 4, the sup-plement of the given Z Aset the angle C =62° 34 6from d to b ; reduce b toa, and describe the quad-rant ac ; about the pole pof ABd describe the smallparallel circle fgh, and the ^K. Sl^S^&l^ intersection f of fgh and the quadrant ac will be the pole ofeBC ; describe eBC about f, as its pole, and through e, ABC is the triangle required. CALCULATION. To find the side BC. 1. From half the sum of the three angles subtract each ofthe angles next to the required side. 2. Add together the cosines of these two remainders andthe co-arcs of the sines ol each of the adjacent angles. 3. Half the sum of these four logarithms will give the co-sine of half the required side.* * Let s be put for the sum of the three angles, a for the angleopposite to the required side, b and c for the adjacent angles;then the Theorem may be expressed thus : „ , ., r5 x cos i —Ay cos. i-v—r Cos. i side = s/ .-^- : - ; ? sm. b x sin. c Whence, logarithms being used, Cos.^ side = cos iy—b -J-cos. it—c 4-co-ar.
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