標題:
Trigonometry Maths
發問:
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最佳解答:
a) Let angle OAC be a AC2 + AO2 - 2(AO)(AC)(cos a) = OC2 62 + 32 - 2(6)(3)(cos a) = 42 -2(6)(3)(cos a) = -29 cos a = (-29) / (-36) a = 36.336° b) i) tan40° = BC / OC BC = 4(tan40°) tan30° = BC / CD CD = 4(tan40°) / tan30° CD = 5.813m ii) Let angle CAD be A AC2 + AD2 - 2(AC)(AD)(cos A) = CD2 62 + 82 - 2(6)(8)(cos A) = 5.8132 A = 46.395° iii) from above a) and bii) Angle EAD = Angle CAD – Angle OAC = A - a = 46.395° - 36.336° = 10.059° Angle AED = 180° - θ ED / sin(A-a) = 8 / sin(180° - θ) (note: sin(180° - θ) = sinθ) ED = 8(sin10.059°) / sinθ CE / (sin a) = 6 / sinθ CE = 6(sin36.336°) / sinθ CD = CE + ED CD = [6(sin36.336°) / sinθ] + [8(sin10.059°) / sinθ] 5.813 = [6(sin36.336°) + 8(sin10.059°)] / sinθ sinθ = [6(sin36.336°) + 8(sin10.059°)] / 5.813 θ = 58.425°
其他解答:
55621
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