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急! F5 Trigonometry 11q11
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請詳細步驟教我計以下二條 : 不要網址回答: 圖片參考：http://imgcld.yimg.com/8/n/HA05788109/o/20130128003143.jpg
14. (a) (i) PQ = 12 x 3 = 36 km (a) (ii) PR2 = PQ2 + QR2 - 2 PQ QR cosQ PR2 = 362 + 512 - 2 x 36 x 51 x cos40° km2 PR = 32.9 km (b) Average speed of ship M = 32.9/2 km/h = 16.5 km/h (c) sinR / PQ = sinQ / PR sinR / 36 = sin40° / 32.9 R = 44.7° Bearing of R from P = S 44.7° E ===== 15. (a) The bearing of P from aeroplane B = S [180° - (285° - 180°)] E = S 75° E (b) In DLBP : ∠PLB = 230° - 155° = 75° ∠BPL = 180° + 155° - 285° = 50° ∠PBL = 180° - (75° + 50°) = 55° sin∠BPL / BL = sinPLB / BP (sine law) sin50° / BL = sin 75° / (60 km) BL = 60 sin50° / sin75° km = 47.6 km (c) Denote H as the hangar. ∠LBH = 230° - 180° = 50° In DBLH : cos∠LBH = BH / BL cos50° = BH / (47.6 km) Distance that the aeroplane have to move BH = 47.6 cos50° km = 30.6 km
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急! F5 Trigonometry 11q11
發問:
請詳細步驟教我計以下二條 : 不要網址回答: 圖片參考：http://imgcld.yimg.com/8/n/HA05788109/o/20130128003143.jpg
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最佳解答:14. (a) (i) PQ = 12 x 3 = 36 km (a) (ii) PR2 = PQ2 + QR2 - 2 PQ QR cosQ PR2 = 362 + 512 - 2 x 36 x 51 x cos40° km2 PR = 32.9 km (b) Average speed of ship M = 32.9/2 km/h = 16.5 km/h (c) sinR / PQ = sinQ / PR sinR / 36 = sin40° / 32.9 R = 44.7° Bearing of R from P = S 44.7° E ===== 15. (a) The bearing of P from aeroplane B = S [180° - (285° - 180°)] E = S 75° E (b) In DLBP : ∠PLB = 230° - 155° = 75° ∠BPL = 180° + 155° - 285° = 50° ∠PBL = 180° - (75° + 50°) = 55° sin∠BPL / BL = sinPLB / BP (sine law) sin50° / BL = sin 75° / (60 km) BL = 60 sin50° / sin75° km = 47.6 km (c) Denote H as the hangar. ∠LBH = 230° - 180° = 50° In DBLH : cos∠LBH = BH / BL cos50° = BH / (47.6 km) Distance that the aeroplane have to move BH = 47.6 cos50° km = 30.6 km
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