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The sum of angles in a triangle = 180º (180 - 108)º = 72º Isosceles triangle has both two equal sides and two equal angles. ∠OSN = ∠SON = \(\frac{72}{2}\) = 36º

The sum of angles in a triangle = 180º (180 - 108)º = 72º Isosceles triangle has both two equal sides and two equal angles. ∠OSN = \(\frac{72}{2}\) = 36º

The exterior angle theorem states that the measure of each exterior angle of a triangle is equal to the sum of the opposite and non-adjacent interior angles. ∠M = ∠R + ∠N → 141 = 41 + ∠N ∠N = 141 - 41 = 100º ∠QNR = 180 - 100 → 80º

where x=-4 and y = 16, 16 = -4m - 4 4m = -16 - 4 4m = -20 M = -5

Using pythagoras formula: Hyp\(^2\) = adj\(^2\) + opp\(^2\) 5\(^2\) = opp\(^2\) + 3\(^2\) 5\(^2\) - 3\(^2\) = adj\(^2\) 4 = adj length of the chord = 2 * 4 = 8cm

\(\frac{28 * 360}{100}\) = 100.8º

(a) 2x + 40 + 2y + 10 = 180^\circ...(1) 2x + 2y = 130 x + y = 65º 5x - 35 = 2(2y+ 10) ... (2) 5x - 35 = 4y +20 5x - 4y = 55 5x - 4(5-x) = 55 5x - 260 + 4x =55 9x = 315 x = 35^\circ y= 65 - 35 y=30^\circ ∠ABC + [2 x º35 + 40] + [5(35)-350)] + 40^\circ =360º ∠ABC + 110 + 140^\circ+ 40º = 360 ∠ABC + 290^\circ = 360^\circ ∠ABC = 70^\circ

a) m = tan 30º = \(\frac{\sqrt{3}, n = tan 45º = 1 \(\frac{ m - n}{mn}\), = \(\frac{\frac{\sqrt{3}}{3} - 1}{\frac{\sqrt{3}}{3}(1)}\) = \(\frac{\frac{\sqrt{3} - 3}{3}}{\frac{\sqrt{3}}{3}}\) = \(\frac{\sqrt{3} - 3}{3}\) X \(\frac{3}{\sqrt{3}}\) = \(\frac{\sqrt{3} - 3}{\sqrt{3}}\) = \(\frac{\sqrt{3} - 3}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\) = \(\frac{3 - 3\sqrt{3}}{3}\) = 1 - \(\sqrt{3}\) b) see the diagram above.. Let the number of women wearing both wristwatches and glasses be x 15 - x + x + 10 - x = 20 25 - x = 20 x = 5 Therefore, the number of women wearing wrist watches only = 10 - x = 10 - 5 = 5 and number of women wearing glasses only = 15 - x = 15 - 5 = 10 Pr(both) = \(\frac{5}{20}\) = \(\frac{1}{4}\) = 0.25

(a) x-axis: 2 cm to 2 units or 1 cm to 1 unit y-axis: 2 cm to 10 units or 1 cm to 5 units (b) x = -2, x = 4 (x+2)(x-4) = 0 x\(^2\) -4x + 2x -8 = 0 x\(^2\) -2x - 8 = 0 y = 8 + 2x - x(curve has a max. point) m = -1,n = 2, r =8 (c) P(-5,-27),Q(3,5) Gradient = \(\frac{5+27}{3+5}\) = \(\frac{32}{8}\) = 4 (d) Range: {x:-2 < x < 4)

xº 0 20 40 60 80 100 120 140 160 180 y 7.0 7.6 7.3 6.1 4.2 1.7 -0.9 -3.4 -5.6 7.0 (b) (C)(i) From the graph, When x is 150^\circ y = -4.6 \pm 0.2 (ii) Range of value of x for which y > is x: 0º \le x \le 113

Mean = \(\frac{\sum fx}{ \sum f}\) = (3x2) +(4x6) + (5x5) + (6x) + (7x6) + (8x9) + (9x8) + (10x5) \div (2 + 6 + 5 + x + 6 + 9 + 8 + 5) 7 = (291 + 6x) \div (41+x) 7(41+x) = 291 + 6x 287 + 7x = 291 + 6x x = 4 (b) Age(x in yrs) No.of child Fx\(^2\) 3 2 18 4 6 96 5 5 125 6 4 144 7 6 294 8 9 576 9 8 648 10 5 500 Σf = 45 Σfx\(^2\) = 2401 Standard deviation = \(\sqrt \frac{196}{45} = \sqrt 4.3555\) = 2.087

(a) ∠XYZ = \(\frac{1}{2}\) (146) = 73º ∠OZX = 34º ∠OZY = 73 - 34 ∠OYZ = ∠OZY = 39 (b) x = y - 7 42 + 38 + 57 + x + x + y + 2x - 15 + 3x - y = 360 122^\circ+7x = 360^\circ 7x = 360º - 122 = 238^\circ 7x = 238º x = 34 : 34 = y - 7 34 + 7 = y 41 = y

209 = 2 * \(\frac{22}{7}\) * \((\frac{7}{2})^2\) + 2 * \(\frac{22}{7}\) * \(\frac{7}{2}\) * h 209 = 77 + 22h 132 = 22h h = 6cm (b) tan 43º = \(\frac{h}{y}\) 0.9125y = h ...[i] tan 38º = \(\frac{h}{y+19}\) 0.7813(y+19) = h ...[ii] 0.7813(y+19) = 0.9125y 0.7813y + 14.8447 = 0.93257 14.8447 = 0.1512y y= 98.1792 = 98.2m h = 0.9325 x 98.1792 = 91.55 =91.6m

PQ\(^2\) = (x2 - x1)\(^2\) + (y2 - y1)\(^2\) = 12\(^2\) + 5\(^2\) = 144 + 25 = 169 PQ = \sqrt169 = 13 But PQ = diameter = 2r, r = PQ/2 = 6.5 units

Vertically opp angles are equal ∠CDE = 70^\circ Sum of interior angles in a quadilateral = 360^\circ (90 + 90 + 70 + t)^\circ = 360^\circ t^\circ = 360^\circ - 250^\circ t^\circ = 110^\circ