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\(\angle\)KGL = 87º (exterior angle of a cyclic quadrilateral) \(\angle\)KGL + \(\angle\)GLK + \(\angle\)LKG = 180º 87º + x + 38º = 180º x = 180 - 87 - 38 = 55º

Vol of cylinder = \(\pi\)r\(^2\)H = vol of cone = \(\frac{1}{3}\)\(\pi\)r\(^2\)h H = height of cylinder and h = height of cone Let y = radius of cylinder = y, then radius of cone = 2y \(\pi\)y\(^2\)H = \(\frac{1}{3}\)\(\pi\)(2y)\(^2\)h y\(^2\)H = \(\frac{1}{3}\)4y\(^2\)h ( \(\pi\) cancels out and cross multiplying) The ratio of the height of the cylinder to that of the cone = \(\frac{\text{H}}{\text{h}}\) = \(\frac{4y^2}{3y^2}\) = \(\frac{4}{3}\) = 4: 3 (y\(^2\) cancels out )

From the diagram on the right-hand side above, x\(^2\) = 20\(^2\) - 19\(^2\) = 39 ( from Pythagoras theorem) x = \(\sqrt{39}\) = 6.245m As can be seen from the diagram on the right-hand side above, h = x + 17 = 6.245 + 17 = 23.245 m ≈ 23 m(to the nearest whole number)

P(2, -8) and Q(1, y) Gradient of slope(m) = \(\frac{y_2 - y_1}{x_2 - x_1}\) - 4 = \(\frac{y - (-8))}{1 - 2}\) -4 = \(\frac{y + 8}{- 1}\) cross multiplying 4 = y + 8 y = 4 - 8 = -4

\(\angle\)XWY = \(\angle\)ZYW = 44º (alternate \(\angle\)s) Considering \(\triangle\)WTX \(\angle\)W + \(\angle\)T + \(\angle\)X = 180º 44 + \(\angle\)T + 50 = 180 \(\angle\)T = 180 - 44 - 50 = 86º

Let the length of the garden be x, then, the width = x - 7 perimeter of rectangle = 2(L + B) = 90 m = 2(x + x - 7) = 90 = 2x - 7 = 45 ( after dividing through by 2) = 2x = 45 + 7 = 52 x = \(\frac{52}{2}\) = 26 m Thus, the length of the rectangular garden = 26 m.

Hexagon = 6 sides Sum of interior angles of a regular hexagon = (n - 2)180 = (6 - 2)180 = 4 x 180 = 720º But 4 of these angles sum up to 420º and the remaining ones are equal. Let the remaining two that are equal = 2x SuM of the angles = 420 + 2x = 720º 2x = 720 - 420 = 300 x = \(\frac{300}{2}\) = 150º Thus, the value of each of the remaining angles = 150º

\(\angle\)N + \(\angle\)P + \(\angle\)M = 180 ( sum of angles in \(\triangle\) NPM) 65º + 55º + \(\angle\)M = 180 \(\angle\)M = 180 - 65 - 55 = 60º x + \(\angle\)M = 180(opp interior angles of parallel lines are supplementary) x = 180 - 60 = 120º

Mean(\(\overline{x}\)) = \(\frac{\sum{x}}{n}\) Mean(\(\overline{x}\)) = \(\frac{ 59 + 44 + 53 + 57 + 49 + 40 + 48 + 50}{8}\) Mean(\(\overline{x}\)) = \(\frac{397}{8}\) ≈ 50 kg

Variance(s\(^2\)) = \(\frac{\sum{(x - \overline{x})^2}}{n}\) But mean = 50 (from question 39 same information) x x - \(\overline{x}\) (x - \(\overline{x})^2\) 59 9 81 44 - 6 36 53 3 9 57 7 49 49 - 1 1 40 -10 100 48 - 2 4 50 0 0 \(\sum{(x - \overline{x})^2}\) = 280 Variance(s\(^2\)) = \(\frac{\sum{(x - \overline{x})^2}}{n}\) = \(\frac{280}{8}\) = 35

Opposite sides of a rectangle are equal, then (5x + 3) = (2x + 9) 5x - 2x = 9 - 3 3x = 6 x = \(\frac{6}{3}\) = 2. Put x = 2 in either (5x + 3) or (2x + 9) = 13m Adjacent side (6x - 7) (given) also put x = 2 into it, 6(2) - 7 = 5 Area of rectangle = length x breadth(width) = 13 x 5 = 65 m\(^2\)

Sample space = { 1, 2, 3, 4, 5, 6} Prime numbers = ( 2, 3, and 5) Pr(prime Nos) = \(\frac{\text{number of required outcome}}{\text{total possible outcome}} = \(\frac{3}{6}\) = \(\frac{1}{2}\)

Area of a sector = \(\frac{\theta}{360}\) x \(\pi\) r\(^2\) 51.3 = \(\frac{\theta}{360}\) x \(\frac{22}{7}\) x 7 x 7 \(\theta\) = \(\frac{51.3 \times 360}{22 \times 7}\) = 119.9 ≈ 120º to the nearest whole number.

From the figure above, using SOH CAH TOA Tan\(\alpha\) = \(\frac{\text{opposite}}{\text{adjacent}}\) Tan\(\alpha\) = \(\frac{87}{22}\) = 3.9545 \(\alpha\) = Tan\(^{-1}\)(3.9545) = 76º

\(\angle\)SPT = 35º (alternate segment) From \(\triangle\)TSP 44 + \(\angle\)SPT + \(\angle\)PST = 180º (sum of angles in a triangle) \(\angle\)PST = 180 - 44 - 35 = 101º