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o (Corresponding angles) 245 + o (o

Given \(\sin x = 0.6 = \frac{3}{5}\) \(\therefore Opp = 3; Hyp = 5\) Using Pythagoras theorem, \(5^{2} = 3^{2} + Adj^{2}\) \(Adj = \sqrt{5^{2} - 3^{2}}\) \(Adj = \sqrt{25 - 9} = \sqrt{16} = 4\) \(\therefore \cos x = \frac{4}{5} = 0.8\) \(2\cos x + 3\sin x = 2(\frac{4}{5}) + 3(\frac{3}{5})\) = \(\frac{8}{5} + \frac{9}{5}\) = \(\frac{17}{5}\). (b) Join O to X and O to Y. Considering either \(\Delta OXC\) or \(\Delta OYC\), \(\sin 65 = \frac{r}{30}\) \(r = 30 \sin 65\) \(r = 30 \times 0.9063\) \(r = 27.189 cm \approxeq 27.2 cm\) (to one decimal place) (b) Area of shaded portion = Area of triangle ABC - Area of semi-circle WXYZ. = \(\frac{1}{2} bh - \frac{1}{2} \pi r^{2}\) = \((\frac{1}{2} \times 2c \times 30) - (\frac{1}{2} \times \frac{22}{7} \times (27.189)^{2})\) From \(\Delta COA, \tan 65 = \frac{c}{30}\) \(\implies c = 30 \tan 65 \approxeq 64.34 cm\) \(\therefore\) Area of shaded portion = \(30(64.34) - \frac{8131.659}{7}\) = \(1930.056 - 1161.666\) = \(768.39 cm^{2}\)

(c) |XP| = 3.8 cm.

Score 1 2 3 4 5 6 Frequency 2 5 x 11 9 10 P(3) = 0.26 \(\frac{x}{37 + x} = 0.26\) \(x = 0.26(37 + x) \implies x = 9.62 + 0.26x\) \(x - 0.26x = 9.62 \implies 0.74x = 9.62\) \(x = \frac{9.62}{0.74} = 13\) Total toss frequency = 37 + x = 37 + 13 = 50. (a) Median = \(\frac{n + 1}{2}\) = \(\frac{50 + 1}{2}\) = 25.5 From the table, 25.5 occurs at 4 which is the median. (b) Standard deviation = \(\sqrt{\frac{\sum fd^{2}}{\sum f}}\) \(x\) \(f\) \(\fx\) \(d = |x - \bar{x}|\) \(d^{2}\) \(fd^{2}\) 1 2 2 -3 9 18 2 5 10 -2 4 20 3 13 39 -1 1 13 4 11 44 0 0 0 5 9 45 1 1 9 6 10 60 2 4 40 \(\sum\) 50 200 100 \(Mean(\bar{x}) = \frac{\sum fx}{\sum f}\) = \(\frac{200}{50} = 4 S.D = \(\sqrt{\frac{100}{50}}\) = \(\sqrt{2}\) = 1.4142

(a) \(\theta = \frac{60^\circ}{2} = 30^\circ\) Area of trapezium = \(\frac{a + b}{2} h\) \(60 = \frac{15 + 25}{2} h \implies 60 = 20h\) \(h = \frac{60}{20} = 3 cm\) (ii) Considering \(\Delta PSR\), \(\sin \theta = \frac{Opp}{Hyp}\) \(\sin 30 = \frac{3}{|PS|}\) \(|PS| = \frac{3}{\sin 30}\) = \(\frac{3}{0.5} = 6 cm\) (b) Let the total amount to be shared = k. Ade's share = \(\frac{3}{5} k \) Remainder = \(k - \frac{3}{5}k \) = \(\frac{2}{5}k\) Nelly's share : \(\frac{1}{3} \times \frac{2}{5}k\) = \(\frac{2}{15}k\) Austin's share : \(\frac{2}{5}k - \frac{2}{15}k\) = \(\frac{4}{15}k\) \(\frac{4}{15}k - \frac{2}{15}k = \frac{2}{15}k\) \(\frac{2}{15}k = N3,000\) \(\implies k = \frac{3,000 \times 15}{2}\) = \(N22,500\) Ade's share = \(\frac{3}{5}k\) = \(\frac{3}{5} \times N22,500 = N13,500\).

(a) \(P \propto Q\) and \(P \propto \frac{1}{R^{2}}\). \(P \propto \frac{Q}{R^{2}} \implies P = \frac{kQ}{R^{2}}\) When P = 1, Q = 8, R = 2 \(1 = \frac{8k}{2^{2}} \implies 8k = 4\) \(k = \frac{4}{8} = 0.5\) \(P = \frac{Q}{2R^{2}}\) When P = 3, R = 5, Q = ? \(3 = \frac{Q}{2(5^{2})}\) \(3 = \frac{Q}{50} \implies Q = 150\). (b) Angular difference : 60^\circ - 20^\circ = 40^\circ \(\frac{\theta}{360} \times 2\pi r \) \(r = R\cos \theta\) \(\frac{40}{360} \times 2 \times 3.142 \times 6400 \cos 20 = \frac{1511687.278}{360}\) = \(4199.13 km\) \(Speed = \frac{Distance}{Time}\) = \(\frac{4199.13}{6}\) = \(699.85 km/hr\) \(\approxeq 700 km/hr\) (to 3 significant figures) (ii) \(D = \frac{\theta}{360} \times 2\pi r\) \(420 = \frac{\theta}{360} \times 2 \times 3.142 \times 6400\) \(\theta = \frac{420 \times 360}{2 \times 3.142 \times 6400}\) \(\theta = \frac{151200}{40217.6} = 3.76^\circ\) Latitude of C : 20^\circ + 3.76^\circ = 23.76^\circ Note that due North implies same longitude but different latitude.

Median = \(\frac{\text{N}}{2}\) th item = \(\frac{150}{2}\) th ≈ 36.5 from the graph.

∠RTS = 90o (∠ in semi circle) ∠UTS = 25o (∠ in alternate segment) X + 90 + 25 = 180(sum of ∠s on a str line) X + 115 = 180 X = 180 - 115 X = 65o

cot\(\theta\) = \(\frac{1}{\cos \theta}\) = \(\frac{8}{15}\)(given) tan\(\theta\) = \(\frac{15}{8}\) By Pythagoras' theorem, x\(^2\) = 15\(^2\) + 8\(^2\) x\(^2\) = 225 + 64 = 289 x = \(\sqrt{289}\) = 17 Hence sin\(\theta\) = \(\frac{15}{x}\) = \(\frac{15}{17}\)

0.21 x 0.34 = 0.0714 = 7.14 x 10-2

Area of a sector = \(\frac{\theta}{360}\) x \(\pi\)r\(^2\) = \(\frac{50}{360} \times \frac{22}{7}\) x 36 = \(\frac{110}{7}\)

Area of a triangle = \(\frac{1}{2}\) ab Sin\theta 12\(\sqrt{3}\) = \(\frac{1}{2}\) x 8 x q sin 60 12\(\sqrt{3}\) = 4 x q x \(\frac{\sqrt{3}}{2}\) 12\(\sqrt{3}\) = 2q\(\sqrt{3}\) q = \(\frac{12\sqrt{3}}{2\sqrt{3}}\) q = 6cm.

∫(sin χ + 2)dχ = (∫sin χ + ∫2)dx = -cos χ + 2χ + K

Using simple interest = \(\frac{P \times T \times R}{100}\), where: P denoted principal = N400 T denotes time = 3 years R denotes interest rate = ? 24 = \(\frac{400 \times 3 \times R}{100}\) 24 x 100 = 400 x 3 x R R = \(\frac{24 \times 100}{400 \times3}\) = 2%

(x + 15)º + (2x - 45)º + (x + 10)º + (x - 30)º = (2n - 4)90º when n = 4 x + 15º + 2x - 45º + x - 30º + x + 10º = (2 x 4 - 4) 90º 5x - 50º = (8 - 4)90º 5x - 50º = 4 x 90º = 360º 5x = 360º + 50º 5x = 410º x = \(\frac{410^o}{5}\) = 82º Hence, the value of the least interior angle is (x - 30)º = (82 - 30) º = 52º