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To find the cosine of the complement of \(30^\circ\), we first calculate the complement: \(90^\circ - 30^\circ = 60^\circ\) Next, we need to find \( \cos 60^\circ \). The known trigonometric value is: \(\cos 60^\circ = 0.5\) Therefore, the cosine of the complement of \(30^\circ\) is: 0.5

Let the two parallel chords be AB = 46 cm and CD = 40 cm. Radius of the circle, r = 25 cm. The centre O lies between the two chords. Draw perpendiculars from O to the chords (these perpendiculars bisect the chords). - Half of AB = 23 cm - Half of CD = 20 cm Distance from centre to each chord: For chord AB (46 cm): d\(_1 = \sqrt{r^2 - 23^2} = \sqrt{625 - 529} = \sqrt{96} = 4\sqrt{6} \approx 9.80 \text{ cm}\) For chord CD (40 cm): d\(_2 = \sqrt{r^2 - 20^2} = \sqrt{625 - 400} = \sqrt{225} = 15 \text{ cm}\) Since the centre lies between the two chords, the distance between the chords is: d\(_1 + d_2 = 4\sqrt{6} + 15 \text{ cm}\) Approximate value: ≈ 24.80 cm

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

When a shirt is sold for ₦12,000 at a 20% loss, it means ₦12,000 is 80% of the cost price. Find the Cost Price (CP) CP = \(\frac{12,000}{80\%} = \frac{12,000}{0.8}\) = ₦15,000 Find the selling price for 20% gain To gain 20%, the new selling price should be 120% of the cost price. SP = 120\(\% \text{ of } ₦15,000 = 1.2 \times\) 15,000 = ₦18,000.

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

Given: L = 8cm, B = 7\(\frac{1}{2}\)cm = 7.5cm, H = 6\(\frac{1}{2}\)cm = 6.5cm Volume of cuboid = LBH = 8 x 7.5 x 6.5 = 390cm\(^3\) but in m\(^3\) = \(\frac{390}{1000000}\) = 0.00039m\(^3\) 0.00039m\(^3\) = 3.9 x 10\(^{-4}\)m\(^3\)

(a) \((2a + b)^{2} - (b - 2a)^{2}\) Using the method of difference of two squares, \((2a + b)^{2} - (b - 2a)^{2} = ((2a + b) + (b - 2a))((2a + b) - (b - 2a))\) = \((2b)(4a)\) = \(8ab\). (b)(i) \(S = K\sqrt{m^{2} + n^{2}}\) \(S^{2} = K^{2}[m^{2} + n^{2}]\) \(\frac{S^{2}}{K^{2}} = m^{2} + n^{2}\) \(m^{2} = \frac{S^{2}}{K^{2}} - n^{2}\) \(m = \sqrt{\frac{S^{2}}{K^{2}} - n^{2}}\) (ii) When S = 12.2, K = 0.02, n = 1.1 \(m = \sqrt{\frac{12.2^{2}}{0.02^{2}} - (1.1^{2})}\) \(m = \sqrt{\frac{148.84}{0.0004} - 1.21}\) \(m = \sqrt{372,098.79}\) = \(609.999 \approxeq 610\)

(a) \(\frac{3}{\sqrt{3}}(\frac{2}{\sqrt{3}} - \frac{\sqrt{12}}{6})\) = \(\frac{6}{\sqrt{9}} - \frac{3\sqrt{12}}{6\sqrt{3}}\) = \(2 - \frac{2\sqrt{3}}{2\sqrt{3}}\) = \(2 - 1 = 1\) (b)(i) \(< ABC = 180^\circ - 68^\circ = 112^\circ\) \(< AOC = 2 \times 112^\circ = 224^\circ\) \(\hat{O} = < AOC (obtuse)\) = \(360^\circ - 224^\circ = 136^\circ\) (ii) In \(\Delta AOC\), OA = OC (radii) \(\therefore \Delta AOC = isosceles\) \(\hat{A} = < OAC ; \hat{C} = < ACO ; \hat{A} = \hat{C} = x\) \(\hat{O} + \hat{A} + \hat{C} = 180^\circ\) \(136^\circ + x + x = 180^\circ \implies 2x = 44^\circ\) \(x = 22^\circ\) In \(\Delta BOC\), OB = OC (radii) \(\therefore \Delta BOC = isosceles\) \(< OCB = 49^\circ + 22^\circ= 71^\circ\) \(< OBC = < OCB = 71^\circ\) \(< OBC + < OCB + < BOC = 180^\circ\) \(71^\circ + 71^\circ + < BOC = 180^\circ\) \(< BOC = 180^\circ - 142^\circ = 38^\circ\)

(a) Width = x m; Length = 2x m; Height = (x - 4) m. Total surface area = 2Lh + WL + 2Wh = \(2(2x)(x - 4) + x(2x) + 2(x)(x - 4) = 1440 m^{2}\) = \(4x^{2} - 16x + 2x^{2} + 2x^{2} - 8x = 1440\) = \(8x^{2} - 24x = 1440\) = \(x^{2} - 3x - 180 = 0\) = \((x - 15)(x + 12) = 0\) \(x = 15m\) Length = 2(15m) = 30m. (b) \(I = \frac{PRT}{100}\) \(I_{P} + I_{Q} = I_{T}\) \(\frac{3000 \times x \times 1}{100} + \frac{2000 \times y \times 1}{100} = 640\) \(30x + 20y = 640 \implies 3x + 2y = 64 .... (1)\) \(\frac{2000 \times x \times 1}{100} + \frac{3000 \times y \times 1}{100} = 660\) \(20x + 30y = 660 \implies 2x + 3y = 66 .... (2)\) To eliminate x , multiply (1) by 2 and (2) by 3. \(6x + 4y = 128 ... (1)\) \(6x + 9y = 198 .... (2)\) (2) - (1) : \(9y - 4y = 198 - 128 \implies 5y = 70\) \(y = 14\) \(2x + 3y = 66 \implies 2x + 3(14) = 66\) \(2x + 42 = 66 \implies 2x = 24\) \(x = 12\) \(\therefore x = 12 ; y = 14\).

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

(a)(i) < PQR = (180^\circ - 130^\circ) + 50^\circ = 100^\circ \(PR^{2} = PQ^{2} + QR^{2} - 2(PQ)(QR) \cos < PQR\) = \(8^{2} + 13^{2} - 2(8)(13) \cos 100\) = \(233 + 36.118\) \(PR^{2} = 269.118\) \(PR = 16.405 m \approxeq 16.4 m\) (ii) \(\frac{QR}{\sin < QPR} = \frac{PR}{\sin < PQR}\) \(\frac{13}{\sin < QPR} = \frac{16.4}{\sin 100}\) \(\sin < QPR = \frac{13 \times \sin 100}{16.4}\) \(\sin < QPR = 0.7806\) \(< QPR = 51.32^\circ \approxeq 51.3^\circ\) The bearing of R from P = 50^\circ + 51.3^\circ = 101.3^\circ \(\approxeq\) 101^\circ. (b) In \(\Delta PQD\), \(\frac{h}{8} = \sin 51.3\) \(h = 8 \sin 51.3\) = \(6.243 m \) \(\approxeq 6.24 m\) Area of \(\Delta PQR = \frac{1}{2} \times PR \times h\) \(\frac{1}{2} \times 6.24 \times 16.4 = 51.166 m^{2}\) \(\approxeq 51.2 m^{2}\)

(a) Class Interval Freq (f) Tally Mid-point (x) \(fx\) 45 - 49 1 | 47 47 50 - 54 3 ||| 52 156 55 - 59 6 |||| | 57 342 60 - 64 7 |||| || 62 434 65 - 69 6 |||| | 67 402 70 - 74 4 |||| 72 288 75 - 79 2 || 77 154 80 - 84 1 | 82 82 \(\sum\) 30 1905 (b) (c) Mean \(\bar{x} = \frac{\sum fx}{\sum f}\) = \(\frac{1905}{30}\) = \(63.5\). (d) P(iron rod in modal class) = \(\frac{7}{30}\)

Hint: Make a sketch of an isosceles triangle with two of its sides and angles angles. : PQ (r) = PR (q) = 8cm : R^\circ = Q^\circ = 30^\circ Sum of angles in a triangle = 180^\circ P^\circ + Q^\circ + R^\circ = 180^\circ P^\circ + 30^\circ + 30^\circ = 180^\circ P^\circ = 180^\circ - 60^\circ p^\circ = 120^\circ. : PQ = r, PR = q, QR = p Using sine rule: \(\frac{p}{sinP}\) = \(\frac{q}{sinq}\) \(\frac{p}{sin120^\circ}\) = \(\frac{8}{sin30^\circ}\) cross multiply p = \(\frac{8 X sin120^\circ}{sin30^\circ}\) p = \(\frac{8 X \sqrt3/2 }{1/2}\) p = 8\sqrt3

The line joining (P, 4) and (6, -2). Gradient: \(\frac{-2 - 4}{6 - P} = \frac{-6}{6 - P}\) The line joining (2, P) and (-1, 3) Gradient: \(\frac{3 - P}{-1 - 2} = \frac{3 - P}{-3}\) For perpendicular lines, the product of their gradient = -1. \((\frac{-6}{6 - P})(\frac{3 - P}{-3}) = -1\) \(\frac{6 - 2P}{6 - P} = -1 \implies 6 - 2P = P - 6\) \(6 + 6 = P + 2P \implies P = \frac{12}{3} = 4\)

Let the ext. angle = x Thus int. angle = 2x But sum of int + ext = 180 (angle of a straight line). 2x + x = 180 3x = 180 x = 180/3 = 60 Each ext angle = 360/n => 60 = 360/n n = 360/60 = 6