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(a) Given \(\sin \alpha = 0.3907\) From the sine table, \(\alpha = 23^\circ\) (i) \(\tan \alpha = \tan 23^\circ = 0.4245\) (ii) \(\cos \alpha = \cos 23^\circ = 0.9205\) (b) Let T be the top of the ladder and B, the bottom of the vertical wall. From the figure, above \(\sin 50^\circ = \frac{BT}{4.5m}\) \(BT = 4.5 \sin 50^\circ = 4.5 \times 0.7660\) \(BT = 3.447m\) Distance between the top of the ladder and bottom of the window = \(4 - 3.447 = 0.553m\) = \( 0.553m \times 100\) = \(55.3 cm\)

Subject Period Angle of sector in pie chart English 10 \(\frac{10}{36} \times 360^\circ = 100^\circ\) Mathematics 7 \(\frac{7}{36} \times 360^\circ = 70^\circ\) Biology 3 \(\frac{3}{36} \times 360^\circ = 30^\circ\) Statistics 4 \(\frac{4}{36} \times 360^\circ = 40^\circ\) Igbo 3 \(\frac{3}{36} \times 360^\circ = 30^\circ\) Others 9 \(\frac{9}{36} \times 360^\circ = 90^\circ\) Total 36 360^\circ

(a) Length of the arc XY = Circumference of the base. \(\frac{\theta}{360^\circ} \times 2\pi r\) where \(\theta\) = 270^\circ ; r = 7cm. Length of arc XY = \(\frac{270^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 7 = 33cm\) Let the radius of the base of the cone = x \(2 \pi r = 33cm\) \(2 \times \frac{22}{7} \times x = 33\) \(\frac{44}{7} \times x = 33 \implies x = \frac{33 \times 7}{44}\) \(x = 5.25cm\) (b) Area of the base of the cone = \(\pi r^{2}\) = \(\frac{22}{7} \times \frac{21}{4} \times \frac{21}{4}\) = \(\frac{693}{8} cm^{2} = 86.625 cm^{2}\) \(\approxeq 87 cm^{2}\)

(a) A = {m, a, p, e}; B = {a, e, i, o, u} ; C = {l, m, n, o, p, q, r, s, t, u} (i) \(A \cup B = {a, e, i, m, o, p, u}\) (ii) \(A \cup C = {a, e, l, m, n, o, p, q, r, s, t, u}\) (iii) \(A \cup (B \cap C)\) \(B \cap C = {o, u}\) \(A \cup (B \cap C) = {m, a, p, e, o, u}\) Biology students = n(B) = 190 Chemistry students = n(A) = 300 Let \(n(A \cap B) = a\) (i) From the Venn diagram, \((300 - a) + a + (190 - a) + 70 = 400\) \(560 - a = 400 \implies a = 160\) \(\therefore\) 160 students offer both Biology and Chemistry. (ii) Students offering at least one of Biology and Chemistry = \(n(A) + n(B) + n(A \cap B)\) = \((300 - 160) + (190 - 160) + 160\) = \(140 + 30 + 160\) = 330 students 330 students offer at least one of Biology and Chemistry.

Draw the table of value such that \(-x + 5y \leq 10 \implies -x + 5y = 10\) \(\implies 5y = 10 + x \) \(y = 2 + \frac{x}{5}\) and \(3x - 4y \leq 8 \implies 3x - 4y = 8\) \(y = -2 + \frac{3x}{4}\) Table of value for \(-x + 5y = 10\) x -4 0 4 y 1.2 2 2.8 Table of value for \(3x - 4y = 8\) x -4 0 4 y -5 -2 1

(a) (b) The greatest value of y = 2.25 and this occurs at x = 0.5. (c) Table of values of \(y = 1 - x\) x -2 2 y 3 -1 (d) \(1 + 2x - x^{2} = 0\) \(1 + 1 + 2x - x - x^{2} = 1 - x\) \(2 + x - x^{2} = 1 - x\) The solution of \(1 + 2x - x^{2} = 0\) occurs at the point of intersection of \(y = 2 + x - x^{2}\) and \(y = 1 - x\). This is at points A (x = -0.8) and B (x = 2.8).

(i) The length of the parallel of latitude 46^\circN = \(2 \pi R \cos 46^\circ\) = \(2 \times 3.142 \times 6400 \times 0.6947\) = \(27,939.167 km \approxeq 28000 km\) (to the nearest 100km) (ii) Longitudinal difference between K and Q = 130^\circ - 103^\circ = 27^\circ Distance between K and Q = length of arc KQ = \(\frac{27}{360} \times 2 \times 3.142 \times 6400 \times \cos 46^\circ\) = \(2,095.31 km \approxeq 2100km\) (to the nearest 100km) (iii) Longitudinal difference between Q and P = 103^\circ + 23^\circ = 126^\circ Distance between Q and P = \(\frac{126}{360} \times 2 \times 3.142 \times 6400 \times \cos 46^\circ\) = \(9778.12 km \approxeq 9780 km\) (to the nearest 10km)

(a) Given any \(\Delta\) ABC, Required: That < A + < B + < C = 180^\circ. Construction : Draw CP // BA and produce BC to Q. Proof: Using the above figure, \(a_{1} = a_{2}\) (alternate angles) \(b_{1} = b_{2}\) (corresponding angles) \(a_{1} + b_{1} + c = 180^\circ\) (angles on a straight line) \(\therefore a_{2} + b_{2} + c = 180^\circ\) \(< A + < B + < C = 180^\circ\) (proved) (b) < LNP = 46^\circ + 80^\circ = 126^\circ (exterior angles of triangle LMN) \(\theta = \frac{126^\circ}{2} = 63^\circ\) (bisector divides into 2 equal halves) \(< MLN = 180^\circ - 80^\circ = 100^\circ\) (angles on a straight line) \(< LQN = 180^\circ - (63^\circ + 100^\circ) = 17^\circ\)

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

(a)(i) Marks (%) Freq Cum Freq 0 - 9 10 10 10 - 19 40 50 20 - 29 80 130 30 - 39 140 270 40 - 49 170 440 50 - 59 130 570 60 - 69 100 670 70 - 79 70 740 80 - 89 40 780 90 - 99 20 800 (ii) (iii) From the Ogive, the 50th percentile is 70 marks. (b) Those who scored below 25% (24% and below) = 85 students Those who scored 75% and below = 710 \(\therefore\) Those who scored above 75% = 800 - 710 = 90 students Those to be retained without award = 800 - (85 + 90) = 800 - 175 = 625 students.

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

The underlined word "intake" refers to the number of students admitted into the university each year. The opposite of intake would be rejection, which means refusing admission to students.

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

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

Densely means closely packed; having many people living in a small area. The opposite of densely populated is sparsely populated, meaning having very few people spread over a wide area.