Maths Problem_1－rll33xb99t的部落格

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Maths Problem

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4)(a)Factorize 4x^2+6x-4(b)By using the substitution x= tan A, solve 4 sin^2A+6sinAcosA-4cos^2A=0 where 0
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4. (a) 4x2 + 6x - 4 = 2(2x2 + 3x - 2) = 2(2x - 1)(x + 2) (b) Let x = tanA 4sin2A + 6sinAcosA - 4cos2A =0 (4sin2A + 6sinAcosA - 4cos2A)/cos2A = 0 4tan2A + 6tanA - 4 = 0 4x2 + 6x - 4 = 0 2(2x - 1)(x + 2) = 0 x = 1/2 or x = -2 tanA = 1/2 or tanA = -2 (rejected) A = 26.6° ===== 5. In the question, OC = OD = r and arc CD = π (a) Length of arc CD : 2 x π x OC x (A/360) = π π x r x (A/180) = π rA/180 = 1 rA = 180 Length of arc AB : 2 x π x (OC + CB) x (A/360) = 4π π x (r + 6) x (A/360) = 4π (r + 6)A/180 = 4 (r + 6)A = 720 (b) rA = 180 ...... [1] (r + 6)A = 720 ...... [2] From [2] : rA + 6A = 720 ...... [3] [3] - [1] : 6A = 540 A = 90 Put A = 90 into [1] : r(90) = 180 r = 2 (c) Total surface area = 2 x (Area of ABCD) + 2 x (Area BCGF) + (Area CDHG) + (Area ABFE) = 2 x [π x 82 x (90/360) - π x 22 x(90/360)] + 2 x (6 x7) + (π x 7) + (4π x 7) = 30π + 84 + 7π + 28π = 65π + 84 ≈ 288.2 (to 4 sig. fig.) ===== 6 (a) In ΔABD : AB2 = BD2 +AD2 (Pythagorean theorem) AB2 = (x - 2)2 +(2x + 6)2 AB2 = x2 -4x + 4 + 4x2 + 24x + 36 AB2 = 5x2 +20x + 40 AB2 = 5(x2 +4x + 8) (b) In ΔACD : AC2 = DC2 + AD2 AC2 = (2x)2 + (2x+ 6)2 AC2 = 4x2 + 4x2 + 24x + 36 AC2 = 8x2+24x + 36 (c) From (a): AB2 = 5x2 + 20x + 40 Hence, AB = √[5(x2 + 4x + 8)] From (b): AC2 = 8x2+ 24x + 36 Hence, AC = √[4(2x2+ 6x + 9)] AB : AC = 5 : 6 √[5(x2 + 4x + 8)] : √[4(2x2+ 6x + 9)] = 5 : 6 {√[5(x2 + 4x + 8)]}2 : {√[4(2x2+ 6x + 9)]}2 = 52 :62 5(x2 + 4x + 8) : 4(2x2 + 6x + 9) = 25 : 36 (x2 + 4x + 8) : (2x2 + 6x + 9) = 5 : 9 9(x2 + 4x + 8) = 5(2x2 + 6x + 9) 9x2 + 36x + 72 = 10x2 +30x + 45 x2 - 6x - 27 = 0 (x + 3)(x - 9) = 0 x = -3 (rejected) or x = 9

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