急! F5 Trigonometry 11q2

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°