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\(\tan \theta = \frac{9}{20} = 0.45\) \(\theta = \tan^{-1} (0.45) \) = 24.23^\circ \(\therefore\) The bearing of Y from X = 180^\circ - 24.23^\circ = 155.77^\circ = 156^\circ (to the nearest degree)

< BAC = \(\frac{130}{2}\) (angle subtended at the centre) < BAC = 65^\circ Also, x = 26^\circ (theorem) y = 65^\circ - 26^\circ = 39^\circ < AOC = 180^\circ - (39^\circ + 39^\circ) = 102^\circ

Since each interior angle = 140^\circ; Each exterior angle = 180^\circ - 140^\circ = 40^\circ Number of sides of the polygon = \(\frac{360^\circ}{40^\circ}\) = 9 Sum of angles in the polygon = 140^\circ x 9 = 1260^\circ

Let the number of days worked by the assistant = t \(\therefore\) The bricklayer worked (t + 10) days. 1500(t + 10) + 500(t) = N 95,000 1500t + 15,000 + 500t = N 95,000 2000t = N 95,000 - N 15,000 2000t = N 80,000 t = 40 days \(\therefore\) The assistant worked for 40 days and received N (500 x 40) = N 20,000

\(x^2 - 7x + 10 \leq 0\) Solve for \(x^2 - 7x + 10 = 0\) We have, (x - 5)(x - 2) \(\leq\) 0. Conditions: Case 1: (x - 5) \(\leq\) 0, (x - 2) \(\geq\) 0. \(\implies\) x \(\leq\) 5; x \(\geq\) 2. 2 \(\leq\) x \(\leq\) 5. Choosing x = 3, 3\(^2\) - 7(3) + 10 = 9 - 21 + 10 = -2 \(\leq\) 0. \(\therefore\) 2 \(\leq\) x \(\leq\) 5.

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

= \(\sqrt{(5 - 2)^2 + (6 - 2)^2}\) = \(\sqrt{3^2 + 4^2}\) = \(\sqrt{9 + 16}\) = \(\sqrt{25}\) = 5 units

Length of arc = \(\frac{\theta}{360^\circ} \times 2 \pi r\) = \(\frac{57}{360} \times 2 \times \frac{22}{7} \times 7\) = 6.97 cm

\(\sin x = \frac{opp}{Hyp} = \frac{4}{5}\) 5\(^2\) = 4\(^2\) + adj\(^2\) adj\(^2\) = 25 - 16 = 9 adj = \(\sqrt{9}\) = 3 \(\tan x = \frac{4}{3}\) \(\cot x = \frac{1}{\frac{4}{3}} = \frac{3}{4}\) \(\cot^2 x = (\frac{3}{4})^2 = \frac{9}{16}\) \(\csc x = \frac{1}{\sin x}\) = \(\frac{1}{\frac{4}{5}} = \frac{5}{4}\) \(\csc^2 x = (\frac{5}{4})^2 = \frac{25}{16}\) \(\therefore \frac{1 + \cot^2 x}{\csc^2 x - 1} = \frac{1 + \frac{9}{16}}{\frac{25}{16} - 1}\) = \(\frac{25}{16} \div \frac{9}{16}\) = \(\frac{25}{9}\)

(a) finding a = 10\(^{1.3010}\) and b = 10\(^{1.4771}\) Then, ab = 10\(^{1.3010}\) x 10\(^{1.4771}\) = 10\(^{2.4771}\) = 599.9 (b)(i) they were expected to find < ABC = 180^\circ - 62^\circ = 118^\circ Exterior < AOC a 2 x 118^\circ - 236^\circ (< at centre is twice angle at the circumference) Interior < AOC = 360^\circ - 236^\circ = 124^\circ(angle at a point) as required. (b)(ii) < BAC in 180^\circ - 39^\circ - 118^\circ = 23^\circ (sum of angles of a triangle)

Let C be the cost and n the number of tourists. then C = x + ny. Where x and y are constants 740 = x + 20y...........(1) 960 = x + 30y.............(2) Solving (1) and (2) simultaneously will yield x = 300 and y = 22. Then, C = 300 + 22n. If we substitute 15 for n this implies that C = 300 + (22 x 15)= $630.00

(a) Using length of chord = 2r sin (\(\frac{\theta}{2}\)) = 22.42 and simplifying sin(\(\frac{\theta}{2}\)) = \(\frac{22.42}{33.18}\) = 0.6757 Taking the sine inverse of 0.6757 to obtain \(\frac{\theta}{2}\) = 42.51\(^o\) and multiplying through to get \(\theta\) = 58.02\(^o\) = 885\(^o\) (b(i), the expected diagram is (b)(ii) since T is equidistant from P and Q, then ITPI = ITQI. Using sum of the angles in a triangle. 70^\circ + x + x = 180^\circ and solving for x. x = 55^\circ. Therefore. the bearing of Q from P = 270^\circ — 30^\circ — 55^\circ = 185^\circ.

(a) The range; 20 - 15 = 5. (b) They drew the frequency distribution table as \(\sum\) f = 11\(\sum\) fx = 191 \(\sum\)f\(^2\) = 3337 Age(x) Tally Freq (f) Com. Freq fx fx\(^2\) 15 16 17 18 19 20 I II III III I I 1 2 3 3 1 1 1 3 6 9 10 11 15 32 51 54 19 20 225 512 867 972 361 400 (c) Median age as 17. (d)(i) Mean = \(\frac{\sum fx}{\sum f}\) = \(\frac{191}{11}\) = 17.3636 (d)(ii). The standard deviation was \(\sqrt{\frac{2227}{11} - (\frac{191}{11})^2}\) = \(\sqrt{1.8677}\) = 1.37(correct to two decimal places)

(a) The required diagram was (b)(i) we have sin 75\(^o\) = \(\frac{x}{11}\) and this will give x = 10.625m. Since the ladder has moved up by 0.2 , the new point where the ladder touches the wall is 0.2 + 10.625 = 10.825m. Then sin\(\theta\) = \(\frac{10.825}{11}\) Sin\(\theta\) = 0.9841 and \(\theta\) = sin\(^{-1}\)(0.9841) = 79.77\(^o\) = 80\(^o\) (b)(ii) Cos 75\(^o\) = \(\frac{y}{11}\) so that y = 2.2847m Similarity, cos 79.77 = \(\frac{q}{11}\) and q = 1.9536m. Therefore, the distance the ladder was dragged = 2.8470 - 1.9536 = 0.8934m = 1m

(a). < SOT = 2 < SOT = 2 x 68^\circ =136. Observe that < OSR = < OTR = 90\(^o\) So that < OSR + x + < OTR + < SOT = 360\(^o\). Next is to substitute and simplified to get x + 316\(o\) = 360\(o\) so that x = 44\(o\) (b) If we let capacity of tank A = a and capacity of tank B = b, then a = 600 + b......(1) a - 100 = 3(b - 100) a = 3b - 200........(2) Substituting (2) into (1) yielded 3b - 200 = 600 + b and solving for b, b = 400. Therefore. Capacity of tank B = 400litres Substituting b in (1) yielded A = 600 + 400 = 1000 Capacity of A =1000.