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In \(\Delta\) BPY, \(\tan 35^\circ = \frac{h}{x}\) \(h = x \tan 35 = 0.7 x ... (1)\) In \(\Delta\) BPX, \(\frac{h}{x + 4000} = \tan 15\) \(\frac{h}{x + 4000} = 0.3\) \(h = 0.3 (x + 4000) = 0.3x + 1200 ... (2)\) Equating (1) and (2) as the values of h, we have \(0.7x = 0.3x + 1200 \implies 0.7x - 0.3x = 1200\) \(0.4x = 1200 \times x = \frac{1200}{0.4} = 3000m\) \(\therefore h = 0.7x = 0.7 (3000) = 2100m\) \(\text{The total height = } 5200m + 2100m = 7300m\)

(a)(i) \(\log_{10} 45 = \log_{10} (3 \times 3 \times 5)\) = \(\log_{10} (3^{2} \times 5)\) = \(\log_{10} 3^{2} + \log_{10} 5\) = \(2 \log_{10} 3 + \log_{10} 5\) = \(2(0.477) + 0.699\) = \(0.954 + 0.699 = 1.653\) (ii) \(x^{0.8265} = 45\) Taking the log of both sides, \(\log_{10} x^{0.8265} = \log_{10} 45\) \(0.8265 \log_{10} x = \log_{10} 45\) \(\log_{10} x = \frac{1.653}{0.8265}\) \(\log_{10} x = 2\) \(x = 10^{2} = 100\) (b) \(\sqrt{\frac{2.067}{0.0348 \times 0.538}}\) No Log 2.067 \(0.0348\) = 0.3513 - 0.0348 \(\bar{2}.5416 +\) 0.538 \(\bar{1}.7308\) \(\bar{2}.2724\) = \(\bar{2}.2724\) = \(2.0789 \div 2 = 1.0395\) Antilog - 10.95 \(\therefore \sqrt{\frac{2.067}{0.0348 \times 0.538}} = 10.95\)

(a) Angular difference = 33^\circ + 23^\circ = 56^\circ Distance PQ along parallel of latitude = \(\frac{56}{360} \times 2 \pi r\) where r = R cos 55^\circ \(\therefore |PQ| = \frac{56}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 55\) = \(3589.46 km \approxeq 3589 km\) (to the nearest degree) (b) \(|PO|^{2} = r^{2} + r^{2} - 2r^{2} \cos 56^\circ = R^{2} + R^{2} - 2R^{2} \cos \theta \) = \(2r^{2} - 2r^{2} \cos 56^\circ = 2R^{2} - 2R^{2} \cos \theta\) But \(r = R cos 55^\circ = 6400 \times 0.5736 = 3671\) \(\therefore 2r^{2} (1 - \cos 56^\circ) = 2R^{2} (1 - \cos \theta)\) \(2 \times 3671^{2} ( 1 - 0.5592) = 2 \times 6400^{2} (1 - \cos \theta)\) \(1 - \cos \theta = \frac{2 \times 3671^{2} \times 0.4408}{2 \times 6400^{2}}\) \(1 - \cos \theta = 0.1259\) \(\cos \theta = 1 - 0.1259 = 0.8741\) \(\theta = \cos^{-1} (0.8741) = 29.06^\circ\) \(\therefore\) Distance PQ along great circle = \(\frac{29.06}{360} \times 2 \times \frac{22}{7} \times 6400\) = \(3247.34 km \approxeq 3247km\)

Using sinØ = \(\frac{6\sqrt3}{12}\) → \(\frac{\sqrt3}{2}\) sinØ = \(\frac{\sqrt3}{2}\) or 60^\circ The bearing of Z from X = [270 - 60]^\circ → 210^\circ

H will represent \(\frac{108}{630}\) x \(\frac{360^o}{1}\) ≈ 62^\circ

Area of \(\Delta OXY\) = \(12\frac{1}{4} cm^2\) Area of sector OXY = \(\frac{30}{360} \times \frac{22}{7} \times 7 \times 7\) = \(\frac{77}{6} = 12\frac{5}{6} cm^2\) Area of the shaded portion = \(12\frac{5}{6} - 12\frac{1}{4}\) = \(\frac{7}{12} cm^2\)

∠RQT = 180^\circ - (95^\circ + 48^\circ) = 37^\circ ∠OQR = 90^\circ - 37^\circ = 53^\circ ∠QOR = 180^\circ - (53^\circ + 53^\circ) = 74^\circ QSR = 74^\circ/2 = 37^\circ

Reflex < POR = 360^\circ - 126^\circ = 234^\circ \(\therefore\) < PQR = \(\frac{1}{2} \times 234^\circ\) = 117^\circ

< QOS = 2 < QPS (angle subtended at the centre) \(\therefore\) < QOS = 2 x 36^\circ = 72^\circ

Construction: Join PQ. Then < RSP = 37^\circ = < RQP (angles on the same segment) But < RPQ = 90^\circ (angle in a semi-circle) \(\therefore\) < PRQ = 180^\circ - (90^\circ + 37^\circ) = 53^\circ

Draw line ANB// ML < MNA = 40^\circ (alternate angle AB/ ML) < ANP = 55^\circ - 40^\circ = 15^\circ < PQR = 15^\circ = < ANP (corresponding angles, PN // RQ)

Sum of exterior angles = 360^\circ 2x + 3x + 2x + 3x + 5x = 360^\circ 15x = 360^\circ x = 360^\circ/15 = 24^\circ

360/18 = 20

In the second quadrant, \(\sin 120 = \sin (180 - 120)\) = \(\sin 60\)

No expanded working has been attached for this question yet. The verified answer is Option D.