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急! F5 Trigonometry 11q2

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12. Let O be the center of equilateral triangle PQR Let M be the mid-point of PQ Height h = TO i.e. ∠TOP = 90° and ∠OMQ = 90° ∠OPM = 60°/2 = 30° PM/PO = cos30° = 3^(1/2)/2 PO = 2PM/3^(1/2) = PQ/3^(1/2) = 30/3^(1/2) cm PO^2 + TO^2 = TP^2 [30/3^(1/2)]^2 + h^2 = 25^2 300 + h^2 = 625 h^2 = 325 h = 5(13)^(1/2) cm i.e. height = 5(13)^(1/2) cm. 17. (a) YF = XC and BF < BC i.e. YF/BF > XC/BC tan∠YBF > tan∠XBC In tanθ, greater tanθ forms greater θ when 0 < θ < 180° Therefore, ∠YBF > ∠XBC (b) ∠XBC = 56° XC/BC = tan56° YF/BC = tan56° YF/(FB/cos24°) = tan56° YF/FB = tan56°/cos24° tan∠YBF = tan56°/cos24° = 1.62 ∠YBF = 58.36°

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