SG4 - 2 53.7 2-2(37.2(53.7)cos111.5° V=75.71 8.6(32 X 3 1=0 therefore X 3 =-1 r=-1 and =270° x 3 =-1=0-1=1(cos270°-sin270° since r 3(cos3

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8.1 (28) sinA/a = sinB/b Therefore sin40°/35 = sinB°/30 B = 33.43° Another possibility = 180°-33.43° = 146.57° Two possible triangles 8.2 (28) AB= 298m, AC=421m and BC=324m Let AB=c, AC=b and BC=a C 2 =a 2 +b 2 -2abcosC therefore cosC = 324 2 +421 2 -298 2 /(2*324*421) Cos C = .70897 C=44.85° Sin B/b =Sin C/c therefore Sin B/421 = Sin 44.85°/298 sinB=.9963 B=85.10° A=180°-44.85° - 85.10 = 50.05° 8.3 (52) Ơ = 180 – 68.5° = 111.5° V 2 = 37.2
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Unformatted text preview: 2 +53.7 2-2(37.2)(53.7)cos111.5° V=75.71 8.6 (32) X 3 +1=0 therefore X 3 =-1 r=-1 and ơ =270° x 3 =-1=0-1=1(cos270°-sin270°) since r 3 (cos3 ơ-sin3 ơ )=1cos(270°-sin270°) ơ =(270°+360°*K)/3 = 90°+120°*k if K = 0, ơ =90° if K= 1, ơ =210° if K=2, ơ =333° cos90°-sin90°=-1, cos210°-sin210° = -1/2-√3/2 cos330°-sin330°=1/2+√3/2 solution set (-1,-1/2-√3/2,1/2+√3/2)...
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This note was uploaded on 01/18/2012 for the course MATH 129 taught by Professor Nilanjanpaul during the Spring '11 term at Thomas Edison State.

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