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x o\(^o\) 15\(^o\) 30\(^o\) 45\(^o\) 60\(^o\) 75\(^o\) 90\(^o\) 105\(^o\) 120\(^o\) 135\(^o\) \(^o\) y 0.0 1.5 2.6 2.1 2.6 1.5 0 - 1.5 - 2.6 - 3 -2.6 y = 35m\(^2\)x when x = 15\(^o\) y = 3 sin = (15\(^o\)) y = sin 30\(^o\) y = 3 x \(\frac{1}{2}\) = 1.5 When x = 30\(^o\) y = 3 sin 2(30\(^o\)) y = 3 sin 60\(^o\) = 2.6 When x = 45\(^o\) y = 3sin45\(^o\) = 2.1 When x = 60\(^o\) y = 3 sin 2(60\(^o\)) = 2 sin 120\(^o\) = 2.6 When x = 75\(^o\) y = 3 sin 2(75\(^o\)) = 3 sin 150\(^o\) = 1.5 When x = 90\(^o\) y = 3 sin 2(90\(^o\)) = 3 sin 180\(^o\) = 0 When x = 105\(^o\) y = 3 sin 2(105\(^o\)) = sin 210\(^o\) = -1.5 When x = 120\(^o\) y = 3 sin 2(120\(^o\)) = 3 sin 240\(^o\) = -2.6 When x = 135\(^o\) y = 3 sin 2(135\(^o\)) y = 3 sin 270\(^o\) y = -3 (b SEE THE GRAPH ABOVE... (c)(i) 113 (ii) 3\(^o\) and 82\(^o\)

(a) diagram above (b)(i) m\(^2\) = 5\(^2\) + 9\(^2\) m\(^2\) = 25 + 81 m\(^2\) = 106 m = \(\sqrt{106}\) = 10.296 km/hr Speed = \(\frac{distance}{times}\) distance = speed x time = 10.296 x 2 = 21km (b)(ii) tan\(\theta\) = \(\frac{9km/hr}{5km/hr}\) tan\(\theta\) = 1.8 \(\theta\) = tan\(^{-1}\) \(\theta\) = 60.9\(^o\) the bearing of cyclist x from y 90\(^o\) + 19.9\(^o\) = 109.9\(^o\) = 110\(^o\) (c) The average speed = \(\frac{distance }{time}\) = \(\frac{21}{4}\)km = 5km/her

Marks Upper Class Boundary Frequency (F) Cumulative Frequency (F) (CF) 0 - 9 10 - 19 20 - 29 30 - 39 40 - 49 50 - 59 60 - 69 70 - 79 80 - 89 90 - 99 9.5 19.5 29.5 39.5 49.5 59.5 69.5 79.5 89.5 99.5 7 11 17 20 29 34 30 25 21 6 7 18 35 55 84 118 148 173 194 200 (b) (c) (I) Median Mark = 54.5 (ii) Lowest mark, if 5% of the students passed = 14

< AOB = 108\(^o\) - (90\(^o\) + 40\(^o\)) = 180\(^o\) - 130\(^o\) = 50\(^o\) < DCO = 40\(^o\) (Alternate angle) < ODE = \(\frac{180^o - 80^o}{2}\) = \(\frac{100^o}{2}\) = 50\(^o\) (b)(i) (b)(ii) Area of the parallelogram = ab sin \(\theta\) = 5cm x 7cm sin 125\(^3\) = 35 sin 125 = 28.7cm\(^2\) (c) x = \(\frac{1}{2}(1 - \sqrt{2})\) evaluate (2x\(^2\) - 2x) = 2[\(\frac{1}{2}(1 - \sqrt{2})^2\)] - 2[\(\frac{1}{2} (1 - \sqrt{2}\))] = 2[\(\frac{1}{4}(1 - 2\sqrt{2} + 2)\)] - 1 + \(\sqrt{2}\) = 2[\(\frac{1}{4}(3 - 2\sqrt{2})\) - 1 + \(\sqrt{2}\)] 2[\(\frac{3}{4} - \frac{1}{2} \sqrt{2} - 1 + \sqrt{2}\)] = \(\frac{3}{2} - \sqrt{2} - 2 + 2\sqrt{2}\) = \(\frac{1}{2} + \sqrt{2}\)

x = 180\(^o\) - (52\(^o\) + 70\(^o\)) = 180\(o\) - 122\(^o\) = 58\(^o\)

E\(G\)R = 50\(^o\) (Alternative angles) n + 55\(^o\) + 50\(^o\) = 180\(^o\) (Sum of angles on a straight line) n = 180\(^o\) - 105\(^o\) n = 75\(^o\)

\(\tan 78 = \frac{h}{60}\) \(h = 60 \tan 78\) \(h = 60 \times 4.705 = 282.27m\) \(\approxeq\) 282m to the nearest whole number.

Total number of pupils : 4 + 13 + 30 + 44 + 9 = 100 The number of 8 - year olds = 13 The angle represented by the 8-year olds on the pie chart = \(\frac{13}{100} \times 360^\circ\) = 46.8^\circ

Length of chord = \(2r \sin (\frac{\theta}{2})\) = \(2(3) \sin (\frac{68}{2})\) = \(6 \sin 34\) = \(6 \times 0.559\) = 3.354 cm \(\approxeq\) 3.4 cm

\(\sin \theta = \cos (90 - \theta)\) \(\sin \theta = \cos (90 - 58)\) = \(\cos 32\)

The locus of a points equidistant from two intersecting straight lines is a pair of bisectors that bisect the angles formed by the two intersecting lines.

The sum of two opposite angles of a cyclic quadrilateral = 180^\circ \(\therefore\) (2x + 18)^\circ + 84^\circ = 180^\circ 2x + 102^\circ = 180^\circ \(\implies\) 2x = 78^\circ x = 39^\circ

\(4\sin^2 x - 3 = 0\) \(4 \sin^2 x = 3 \implies \sin^2 x = \frac{3}{4}\) \(\sin x = \frac{\sqrt{3}}{2}\) \(\therefore x = \sin^{-1} (\frac{\sqrt{3}}{2})\) x = 60^\circ

Q(m, n) and R(n, -4) Midpoint : P(2, m) \(\implies (\frac{m + n}{2}, \frac{n - 4}{2}) = (2, m)\) \(m + n = 2 \times 2 \implies m + n = 4 ... (i)\) \(n - 4 = 2 \times m \implies n - 4 = 2m ... (ii)\) Solving (i) and (ii) simultaneously, m = 0 and n = 4.

Since there are 4 angles given, the polygon is a quadrilateral. Sum of angle in a quadrilateral = 360^\circ \(\therefore\) 2x + 5x + x + 4x = 360^\circ 12x = 360^\circ x = 30^\circ