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Rice = 75g, Margarine = 40g, Meat = 105g, Bread = 20g, Total = 240 Angle of sector represented by meat = \(\frac{105}{240}\) x \(\frac{360^\circ}{1}\) = 157.5^\circ

The gradient of any curve is equal to zero at the turning points. i.e. maximum or minimum points. The points in the above curve are b, e, g, j, m

QMN = 60º MRQ = 60º(angle in the alternate segment are equal) MQN = 80º(angle sum of a \(\bigtriangleup\) = 180º) 60 + x = 80º(exterior angle = sum of opposite interior angles) x = 80º - 60º = 20º RMQ = 20º

Since QS = QR then, angle SQR = angle SRQ 2 SQR = 180 - 56, SQR = \(\frac{124}{2}\) = 62o QTP = 62o QTP = 62o, corresponding angle 3y + 56 + 62 = 180 = 3y = 180 - 118 3y = 62 = 180 3y = 180 - 118 3y = 62 y = \(\frac{62}{3}\) = 20\(\frac{3}{2}\)

(a) \(6 \log (x + 4) = \log 64\) \(\implies (x + 4)^{6} = 64\) \((x + 4)^{6} = 2^{6}\) \(\implies x + 4 = 2 ; x = 2 - 4 = -2\) \(\therefore x = -2\) (b) \(U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}\) \(X = {1, 2, 4, 6, 7, 8, 9}; Y = {1, 2, 3, 4, 7, 9} ; Z = {2, 3, 4, 7,9}\) \(X \cap Y = {1, 2, 4, 7, 9}\) \({1, 2, 4, 7, 9} \cap Z' = {1, 2, 4, 7, 9} \cap {1, 5, 6, 8, 10}\) = {1}

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

(a) (i) Given \(\cos \alpha = 0.6717\) \(\alpha = \cos^{-1} 0.6717 = 47.8^\circ\) (From cosine of angle table) (ii) \(\sin 47.8^\circ = 0.7408\) (b) From the diagram, \(< BPT = < PTR = 23.6^\circ\) \(\therefore |BT| = |BP| \tan 23.6^\circ\) = \(50 \tan 23.6^\circ = 50 \times 0.4369\) = \(21.8469 m\) \(\therefore\) The height of the building = 22m (to the nearest metre).

Table of values of \(d = 60t - 10t^{2}\) t 1 2 3 4 5 \(60t\) 60 120 180 240 300 \(- 10t^{2}\) -10 -40 -90 -160 -250 d 50 80 90 80 50 (a)(i) It takes 1.6 seconds to reach a height of 70 metres. (ii) Height of stone at 5 seconds was 50 metres but after 5 seconds, d < 50 metres. (iii) It reaches its maximum height after 2.9 seconds. (b) From the graph, the gradient of the slope = \(\frac{AB}{BC}\) = \(\frac{56}{2.8} = 20 m/s\) But it is a negative slope, hence the slope = -20 m/s.

(a) From the diagram and by Pythagoras theorem, \(|PR|^{2} = |PQ|^{2} + |QR|^{2}\) \(|PR|^{2} = (3a)^{2} + (4a)^{2}\) \(|PR|^{2} = 9a^{2} + 16a^{2} = 25a^{2}\) \(|PR| = \sqrt{25a^{2}} = 5a\) (b) (ii) From the figure, Bearing of Y from X = 060^\circ Using cosine rule, \(\cos XYZ = \frac{24^{2} + 18^{2} - 30^{2}}{2 \times 24 \times 18}\) \(\cos XYZ = 0 \implies < XYZ = 90^\circ\) \(< RYZ = 90^\circ - 60^\circ = 30^\circ\) \(< N_{2} YZ = 180^\circ - 30^\circ = 150^\circ\) (angles on a straight line) The bearing of Z from Y is 150^\circ. (iii) \(\cos XZY = \frac{30^{2} + 18^{2} - 24^{2}}{2 \times 30 \times 18}\) = \(\frac{900 + 324 - 576}{1080}\) \(\cos XZY = \frac{648}{1080} = 0.6\) \(< XZY = \cos^{-1} (0.6) = 53.13^\circ\) \(\therefore\) The bearing of X from Y = 360^\circ - (53.13^\circ + 30^\circ) = 276.87^\circ.

0.06 x 0.09 = 0.0054 = 5.4 x 10-3

cos 60^\circ = 1/2, cos(180^\circ - 60^\circ) = -1/2 cos120^\circ = -1/2

cos 64^\circ = x/9 x = 9 cos 64^\circ =3.94 ≅ 4.0m

Using Pythagoras' theorem: Hyp\(^2\) = Adj\(^2\) + Opp\(^2\) → 24\(^2\) + 7\(^2\) Hyp = \(\sqrt{625}\) → 25 \(\tan P = \frac{7}{24}\) tan P = 0.2917 P = tan\(^{-1}\) (0.2917) = 16.26^\circ Bearing of Q from P = 180^\circ - 16.26^\circ = 163.74^\circ \(\approxeq\) 164^\circ

sinθ = 3/5 tanθ = 3/4

vertically opposite angles are equal, hence 3x + 6x + 3x = 180^\circ 12x = 180^\circ x = 15^\circ