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In the diagram above, \(\alpha\) = 54o(alternate angles; KL||MN) o 35o + o = 180o(sum of angles of \(\bigtriangleup\)) o = 180o o - 143o = 37o

In the figure above, qo = 30o (vertically opposite angles) (P + 2q)o + 30o = 180o(angles on a straight line) p + 2 x 30o + 30o = 180o p + 60o + 30o = 180o p + 90o = 180o p = 180o - 90o = 90o

In the figure above, \(\frac{x}{\sin 60^o} = \frac{10}{\sin 30^o}\) (Sine rule) x = \(\frac{10 \sin 60^o}{\sin 30^o}\) = 10 x \(\frac{\sqrt{3}}{2} \times \frac{1}{2}\) = 10 x \(\frac{\sqrt{3}}{2} \times \frac{2}{1}\) = 10\(\sqrt{3}\)cm

Angle of sector subtended by yam = 360o - (70 + 80 + 50)o = 360o - 200o = 160o But \(\frac{80^o}{360^o}\) x T = 8000 T = \(\frac{8000 \times 360^o}{80^o}\) = N36,000 Hence the amount spent on yam = \(\frac{160^o}{260} \times N36,000\) = N16,000

p(4, 3) Q(2 - 1) distance = \(\sqrt{(x_2 - x_1)^2 + (Y_2 - y_1)^2}\) = \(\sqrt{(2 - 4)^2 + (-1 - 3)^2}\) = \(\sqrt{(-2)^2 = (-4)^2}\) = \(\sqrt{4 + 16}\) = \(\sqrt{20}\) = \(\sqrt{4 \times 5}\) = 2\(\sqrt{5}\)

(2, -3)(2, 5) gradient = \(\frac{y_ = y_1}{x_2 = x_1} = \frac{5 - (-3)}{2 - 2}\) = \(\frac{5 + 3}{0}\) = \(\frac{8}{0}\) = undefined

let the chord be AB = 7cm radius OA = 3.7cm distance of the = OM using Pythagoras theorem OA2 = AM2 + OM2 3.72 = 3.52 + OM2 13.69 = 12.25 + OM2 13.69 - 12.25 = OM2 1.44 = OM2 OM = \(\sqrt{1.44}\) OM = 1.2cm distance of chord = 1.2cm

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

cos x\(\frac{12}{13}\) 132 = 122 + a2 169 = 144 + a2 a2 = 169 - 144 a2 = 25 a \(\sqrt{25}\) a = 5 tan x = \(\frac{5}{12}\) \(\frac{1 - \tan x}{\tan x} = \frac{1 - \frac{5}{12}}{\frac{5}{12}}\) \(\frac{\frac{1 - \frac{5}{12}}{12 - 5}}{12} = \frac{\frac{7}{12}}{\frac{5}{12}}\) = \(\frac{7}{2} \div \frac{5}{12}\) = \(\frac{7}{12} \times \frac{12}{5} = \frac{7}{5}\)

tan \(\theta = \frac{opp}{adj} = \frac{6.6}{13}\) tan \(\theta = 0.5077\) \(\theta\) = tan-1 0.5077 \(\theta = 27^o\)

perimeter of minor sector 2r + \(\frac{\theta}{360} \times 2 \pi r\) = 2 x 3.5 + \(\frac{120^o}{360^o} \times 2 \frac{22}{7} \times 3.5\) = 7 + \(\frac{154}{21}\) = 7 + 7.33 = 14.33 = 14\(\frac{1}{3}\)cm

number of sides = \(\frac{360^o}{\theta} = \frac{360^o}{306o}\) n = 12o Sum of interior angle = (n - 2) 180o (12 - 2) 180v = 10 x 180o = 1800o

y + z = 180o z = 180o - y x + k = 180o k = 180o - x the sum of angles in the \(\bigtriangleup\) are: n + z + k = 180 but n + 180o - y + 180o - x = 180o n + 180o - y - x = 0 but we want to find n n + 180o = x + y but x + y = 220o n + 180o = 220o n = 220o - 180o n = 40o

From the diagram SR//UW given that o and o o(corresponding angle) xo = 45

Construction: draw a line from Q to point P and another line from S to point P. o