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(f5) trigonometry 2題
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最佳解答:
1. In ΔABD : ∠ABD = 25° ∠BAD = 180° - 60° =120° (adj.∠s on a st. line) ∠ADB = 180° - (25° -120°) = 35° (∠sum of Δ) AD / sin∠ABD= AB / sin∠ADB(sine law) AD / sin25° = (500 m) / sin35° AD = 500 sin25° /sin 35° m AD = 368.4 m In ΔABC : ∠ABC = 80° ∠BAC = 180° - 120° =60° (adj. ∠son a st. line) ∠ACB = 180° - (80° +60°) = 40° (∠sum of Δ) AC / sin∠ABC= AB / sin∠ACB(sine law) AC / sin80° = (500 m) / sin40° AC = 500 sin80° / sin40° m AC = 766.0 m In ΔADC : CD2 = AD2 + AC2 - 2?AD?AC?cos∠DAC(cosine law) CD2 = 368.42 + 766.02 - 2 x 368.4 x 766.0 x cos(120° - 60) m2 CD = 663.5 m (b) In ΔADC : AD / sin∠ACD= CD / sin∠DAC 368.4 / sin∠ACD= 663.5 / sin(120° - 60°) sin∠ACD = 368.4 sin60° /663.5 ∠ACD = 28.7° The bearing of D from C = 180° + 80° + 40° + 28.7° = 328.7° 2. (a) In ΔPBR : tanθ = h / PR PR = h / tanθ In ΔPRQ : PQ / sin∠PRQ= PR / sin∠PQR(sine law) c / sin[180°-(α+β)] = (h / tanθ) / sinβ c / sin(α+β) = h / tanθ sinβ Hence, h = c tanθ sinβ / sin(α+β) (b)(i) h = c tanθ sinβ / sin(α+β) h = c tan40° sin46° / sin(54°+46°) h = c tan40° sin46° / sin100° h = 0.6129c h/c = 0.6129 (b)(ii) In ΔQBR : tan∠BQR = h / QR QR = h / tan∠BQR In ΔPQR : PQ / sin∠PRQ= QR / sinα (sine law) c / sin[180°-(α+β)] = (h / tan∠BQR)/ sinα tan∠BQR = (h/c) sin(α+β)/ sinα tan∠BQR = 0.6129 sin100°/ sin54° tan∠BQR = 36.7° Angle of elevation of B from Q = 36.7° (c) From (a) : PR = h / tan40° PR = 0.6129c / tan40° PR = 0.7304c In ΔPRM : RM2 = PR2 + PM2 - 2?PR?PM?cosα (cosine law) RM2 = (0.7304c)2 + (0.5c)2 - 2 x 0.7304c x 0.5c x cos54° m2 RM = 0.5951c m In ΔBRM : tan∠BMR = BR / RM tan∠BMR = h / 0.5951c tan∠BMR = 0.6129 / 0.5951 ∠BMR = 45.8° Angle of elevation of B from M = 45.8° In ΔPRM : RM / sinα = PR / sin∠PMR 0.5851c / sin54° = 0.7304c / sin∠PMR sin∠PMR = 0.7304 sin54° /0.5951 ∠PMR =83.2° Bearing of B from M = N 83.2° E
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