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標題:
F.4 數學...(3D)
發問:
http://photo.xanga.com/freestylenba2k9/bab7e177385410/photo.html 第20題
最佳解答:
a) By sine law, PR / sin β = c / sin(180 - (α + β) PR = c sin β / (sinα + β) tan θ = h / PR tan θ = h sin (α + β) / c sin β h = c tanθ sin β / sin (α + β) b)(i) h = c tanθ sin β / sin (α + β) h/c = tan 40°sin 46° / sin (54° + 46°) h/c = 0.613 (ii) By sine law, QR / sin α = c / sin(α + β) QR = c sin 54° / sin 100° tan ∠BQR = h / QR = h/c ( sin 100° / sin 54°) ∠BQR = 36.7° (iii) By cosine law, cos α = (PR^2 + PM^2 - MR^2) / (2)(PR)(PM) cos 54° = ((c sin 46° / sin 100°)^2 + (c / 2)^2 - MR^2) / 2(c sin 46° / sin 100°)(c / 2) MR = 0.595c tan ∠BMR = h/ 0.595c = (h/c)(1 / 0.595) ∠BMR = 45.8° PR = 0.7304c , MR = 0.595c , PM = 0.5c By cosine law, cos ∠PRM = PR^2 + MR^2 - PM^2 / 2(PR)(PM) ∠PRM = 42.8° ∠PRM = 180° - 54° - 42.8° = 83.2° The bearing is N 83.2° E
其他解答:
F.4 數學...(3D)
發問:
http://photo.xanga.com/freestylenba2k9/bab7e177385410/photo.html 第20題
最佳解答:
a) By sine law, PR / sin β = c / sin(180 - (α + β) PR = c sin β / (sinα + β) tan θ = h / PR tan θ = h sin (α + β) / c sin β h = c tanθ sin β / sin (α + β) b)(i) h = c tanθ sin β / sin (α + β) h/c = tan 40°sin 46° / sin (54° + 46°) h/c = 0.613 (ii) By sine law, QR / sin α = c / sin(α + β) QR = c sin 54° / sin 100° tan ∠BQR = h / QR = h/c ( sin 100° / sin 54°) ∠BQR = 36.7° (iii) By cosine law, cos α = (PR^2 + PM^2 - MR^2) / (2)(PR)(PM) cos 54° = ((c sin 46° / sin 100°)^2 + (c / 2)^2 - MR^2) / 2(c sin 46° / sin 100°)(c / 2) MR = 0.595c tan ∠BMR = h/ 0.595c = (h/c)(1 / 0.595) ∠BMR = 45.8° PR = 0.7304c , MR = 0.595c , PM = 0.5c By cosine law, cos ∠PRM = PR^2 + MR^2 - PM^2 / 2(PR)(PM) ∠PRM = 42.8° ∠PRM = 180° - 54° - 42.8° = 83.2° The bearing is N 83.2° E
其他解答:
此文章來自奇摩知識+如有不便請留言告知
(a) angle PQR = 180 - α - β By sine law PR / sin β = c / sin (180 - α - β) PR = c sin β / sin (180 - α - β) PR = h / tan θ h / tan θ = c sin β / sin (α + β) h = c tanθ sin β / sin (α + β) (b)(i) α=54, β= 46,θ =40 from result (a) h/c = tanθ sin β / sin (α + β) = tan 40 sin 46 / sin (46+54) = 0.613 (ii) QR / sinα = c / sin (α + β) h/QR = tan(angle BQR) h/ (c sinα / sin(α + β) = tan (angle BQR) tan (angle BQR) = 0.613 sin (46 + 54) / sin 40 = 0.939 angle BQR = arc tan 0.939 = 43.2 (iii) PR = h / tanθ, PM = c/2 = h/(2 x 0.613) = 0.816 h By cosine law: RM^2 = PR^2 + PM^2 - 2(PM)(PR)cos α = (h/ tan40)^2 + (0.816h)^2 - 2(h/tan40)(0.816h)cos54 = 1.42h^2 +0.665h^2 - 1.1429h^2 = 0.943h^2 RM = 0.971 h tan (angle BMR) = h / RM = h / (0.971h) = 1.03 angle BMR = arc tan (1.03) = 45.8 2008-03-09 23:33:10 補充： Further to question (3) MQ = PM =0.816h By sine law MQ / sin (angle MRQ) = MR / sin β 0.816h / sin (angle MRQ) = 0.971h / sin 46° angle MRQ = 37.2° Bearing from B to Q is S 46°W Bearing from B to M is S46°W + 37.2° = S83.2°W
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