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\(\bigtriangleup\)PUS is right angled \(\frac{US}{5}\) = sin60o US = 5 x \(\frac{\sqrt{3}}{2}\) = 2.5\(\sqrt{3}\) = 4.33cm

Mean \(\bar{x}\) = \(\frac{156}{10}\) = 15.6 Median = \(\bar{y}\) = \(\frac{15 + 16}{2}\) \(\frac{31}{2}\) = 15.5 \(\frac{x}{y}\) = \(\frac{15.6}{15.5}\) = 1.0065 1.0(1 d.p)

The total sample space when two dice are thrown together is 6 x 6 = 36 \(\begin{array}{c|c} & 1 & 2 & 3 & 4 & 5 & 6\\ 1. & 1.1 & 1.2 & 1.3 & 1.4 & 1.5 & 1.6\\ 2 & 2.1 & 2.2 & 2.3 & 2.4 & 2.5 & 2.6\\3 & 3.1 & 3.2 & 3.3 & 3.4 & 3.5 & 3.6\\ 4 & 4.1 & 4.2 & 4.3 & 4.4 & 4.5 & 4.6\\ 5 & 5.1 & 5.2 & 5.3 & 5.4 & 5.5 & 5.6\\6 & 6.1 & 6.2 & 6.3 & 6.4 & 6.5 & 6.6 \end{array}\) At least 10 means 10 and above P(at least 10) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

SRT is a straight line, where QRT = 120 SRQ = 180o - 120o = 60o - (angle on a straight line) also angle QRS = 180o - 100o (angle on a straight line) . In angles where QR = SR and angle SRQ = 60o x = 100 - 60 = 40o

From the figure, 3x + x - 40o = 180o 4x = 180o + 40o 4x = 220o x = \(\frac{220}{4}\) = 55o

From the figure, PS = QR = YT = 7cm Area of parallelogram PQRS = 56cm 56 = base x height, where base = 7 7 x h = 56cm, h = \(\frac{56}{7}\) = 8cm Area of trapezium \(\frac{1}{2}\) (sum of two sides)x height where two sides are QT and PS but QT = QR + RY + YT = 7 +9 + 7 = 23cm Area of trapezium PQTS = \(\frac{1}{2}\)(23 + 7) x 8 \(\frac{1}{2}\) x 30 x 8 = 120cmsq

From similar triangle, \(\frac{QS}{QP} = \frac{TQ}{QR} = \frac{5}{12} = \frac{6}{QR}\) QT = \(\frac{6 \times 12}{5} = \frac{72}{5} = SR = QR - QS\) = \(\frac{72}{5} - 5 = \frac{72 - 25}{5}\) = \(\frac{47}{5}\)

In the figure PS = RS = QS, they will have equal base QR = RP In angle SQR, angle S = 50º In angle QRP, 65 + 65 = 130º Since RQP = angle RPQ = \(\frac{180 - 130}{2}\) = \(\frac{50}{2} = 25^o\) QPR = 25^\circ

Given: ∠ZXR = 45º (tangent XR), ∠YZX = 55º (inscribed angle). ∠YZX subtends arc YX, so arc YX = 2 \times 55^\circ = 110º. Major arc YZX = 360º - 110º = 250º. By the alternate segment theorem, ∠ZXR subtends arc XZ, so arc XZ = 2 \times 45º = 90º. Arc YZ = 250º - 90º = 160º. Major arc YZ = 360º - 160º = 200º. ∠ZYQ (tangent YQ) = \(\frac{1}{2}\) \times major arc YZ = \(\frac{1}{2}\) \times 200º = 100º (by alternate segment theorem, larger arc for obtuse angle).

The volume of the solid = vol. of cone + vol. of hemisphere volume of cone = \(\frac{1}{3} \pi r^2 h\) = \(\frac{1 \pi}{3} \times (3)^2 x 6 = 18 \pi cm^2\) vol. of hemisphere = \(\frac{4 \pi r^3}{6} = \frac{2 \pi r^3}{3}\) = \(\frac{2 \pi}{3} \times (3)^3 = 18\pi cm^3\) vol. of solid = 18\(\pi\) + 18\(\pi\) = 36\(\pi\)cm3

(i) (ii) Average production in the period = \(\frac{4100 + 2500 + 1500 + 1800 + 9200}{5}\) = \(\frac{19100}{5}\) = \(3820\)

(a) Let \(< SQR = \theta ; < QRS = 90^\circ - \theta ; < QPS = \theta\) In \(\Delta\) PQS, \(\tan \theta = \frac{x}{9} ... (1)\) In \(\Delta\) QRS, \(\tan \theta = \frac{16}{x} ... (2)\) From (1) and (2), \(\frac{x}{9} = \frac{16}{x}\) \(x^{2} = 16 \times 9 \implies x = \sqrt{16 \times 9} = 12 cm\) (b) (i) From \(\Delta\) QRS, \(\tan \theta = \frac{16}{12} = 1.333\) \(\theta = \tan^{-1} (1.333) = 53.1^\circ\) \(\therefore < QRS = 90^\circ - 53.1^\circ = 36.9^\circ\) \(\approxeq 37^\circ\) (to the nearest degree) (ii) In \(\Delta\) PQS, \(|PQ|^{2} = 12^{2} + 9^{2}\) = \(144 + 81 = 225\) \(|PQ| = \sqrt{225} = 15 cm\)

(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 = \(\pi r^{2}\) = \(\frac{22}{7} \times \frac{21}{4} \times \frac{21}{4}\) = \(\frac{693}{8} cm^{2} = 86.625 cm^{2}\) \(\approxeq 87 cm^{2}\)