標題:
maths ~-00~~唔係好多題
發問:
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最佳解答:
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1. Let x m be the distance between the building and Danny, l m be the vertical distance between Sam and Danny, then (l - 20)m be the vertical distance between Cherry and Danny. l / x = tan36° l = xtan36° ───── (1) (l - 20) / x = tan22° ───── (2) Sub (1) into (2), we have (xtan36° - 20) / x = tan22° xtan36° - 20 = xtan22° x(tan36° - tan22°) = 20 x = 62 (cor. to the nearest m) So, the distance between Danny and the building is 62 m. 2. a. ∠PQR = (90° - 38°) + (90° - 52°) = 90° So, ΔPQR is a right-angled triangle. PQ = speed X time PQ = 2 X (2 X 60) = 240 m QR = speed X time QR = 1 X (5 X 60) = 300 m So PR = √(PQ2 + QR2) PR = √[(240)2 + (300)2] PR = √147 600 PR = 384 m (cor. to the nearest m) b. tan∠PRQ = PQ / QR tan∠PRQ = 240 / 300 = 4 / 5 ∠PRQ = 38.6598° So, the bearing of P from R = N[90° - 38.6598° - (90° - 52°)]E = N13°E (cor. to the nearest degree)
其他解答:
1.)19 2a.)384 2b.)唔明問咩
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