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Problem 3.3,3.4.ppt

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- Virginia
- George Mason University
- Engineering
- Engineering 107
- Hassan
- Problem 3.3,3.4.ppt

anas m.

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26 Feb 07 Engr 107 Prob 3.1 Watson James 1 3 Problem 3.1 Given one side and two angle of an oblique triangle. Theory Solution A B C a c b C = 40.0˚ a = 7.5 m A = 110.0 ˚ Using the law of sines and the sum of angles, determine angle B and sides b and c of the triangle. The sum of the three internal angles of a triangle equal 180˚ . a = b = c = 2R sin A sin B sin C A + B + C = 180˚ = 110.0 ˚ + B + 40.0 ˚ B = 180 ˚ - 110.0 ˚ - 40 ˚ and B = 30.0˚ b = a sin B/Sin A = 7.50 (sin 30.0˚)/ (sin 110.0 ˚) = 3.99m b = 3.99 m c = a sin C/Sin A = 7.50 (sin 40.0˚)/ (sin 110.0 ˚) = 5.13m c = 5.13 m 26 Feb 07 Engr 107 Prob 3.1, 3.3 Watson James 1 3 Problem 3.3 Given the sides of an oblique triangle a = 11.1 inches b = 9.35 inches C= 6.79 inches Determine the angles A, B, and C to the nearest tenth of a degree using the law of cosines and the sum of angles. Theory The sum of the three internal angles of a triangle equal 180˚ . Law of cosines: c2 = a2 + b2 – 2ab cos C Solution c2 = a2 + b2 – 2ab cos C 2ab cos C= +a2 + b2 - c2 cos C = +(a2 + b2 - c2) / 2ab cos C = (11.1)2 +(9.35)2-(6.79)2 / 2(11.1)(9.35) cos C =.7926 & C = cos-1.7926 = 37.5º 26 Feb 07 Engr 107 Prob 3.1, 3.3 Watson James 1 3 Problem 3.3 a2 = c2 + b2 – 2bc cos A 2bc cos A= +c2 + b2 - a2 cos A = +(c2 + b2 - a2) / 2bc cos A = (6.79)2 +(9.35)2-(11.1)2 / 2(6.79)(9.35) cos A =0.08125 & A = cos-10.08125 = 85.4º B = 180º - 37.5º -85.4º = 57.1º