Maths Q 快! 40分－航空百大

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Maths Q 快! 40分

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Q 1 圖片參考：http://imgcld.yimg.com/8/n/HA06048535/o/701108150015613873442970.jpg Please help me! Thanks!

最佳解答:

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(a) Consider triangle PAB AB^2 = 248^2 + 200^2 – (2)(248)(200) cos 70 deg AB^2 = 61504 + 40000 – (2)(248)(200)(0.342) AB = 259.95 m (b) Consider triangle ABC Using Heron’s formula Semi-perimeter s = (a + b + c)/2 = (200+300 +259.95)/2 = 379.95 Area of triangle = [s(s-a)(s-b)(s-c)]^0.5 = [379.95 (379.95-200)( 379.95- 300379.95– 259.95)]^0.5 = 5748.365 m^2 (c) Consider triangle PAB , Sin angle PBA/200 m = Sin angle 70 deg/259.95 m Sin angle PBA = 200 m x Sin angle 70/259.95 m = 0.723 angle PBA = 46.3 deg Consider triangle ABC b^2 = a^2 + c^2 – 2ac cos B 300^2 = 200^2 + 259.95^2 – (2)(200)(259.95) cos B cos B = 0.1690 B = 80.27 deg (angle ABC) Angle CBQ = 180- 46.3- 80.27= 53.43 deg Angle BQC = 32 + 58 = 90 deg Cos angle CBQ = BQ/BC Cos 53.43 deg = BQ/200 BQ = 200 Cos 53.43 deg = 119.16 m PQ = PB + BQ = 248 m + 119.6 m = 367.16 m Consider triangle PRQ Cos P = PQ/PR Cos 70 deg = 367.16 m /PR PR = 367.16 m/0.342= 1073 m The distance he walked is 1073 m

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