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Let < POQ = \(\theta\) \(\implies\) SR = \(4\theta\). \(< PSQ = \frac{\theta}{2} ; < SPR = 2\theta\) In \(\Delta PSK\), \(2\theta + \frac{\theta}{2} = 90^\circ\) \(\frac{5\theta}{2} = 90^\circ \implies 5\theta = 180^\circ\) \(\theta = \frac{180^\circ}{5} = 36^\circ\) \(\therefore < SOR = 4\theta = 4 \times 36^\circ = 144^\circ\) Length of arc = \(\frac{\theta}{360^\circ} \times 2\pi r\) \(\frac{144}{360} \times 2 \times \frac{22}{7} \times 7 \) = \(17.6 cm\).

(a) \(\frac{\frac{1}{3}c^{2} - \frac{2}{3}cd}{\frac{1}{2}d^{2} - \frac{1}{4}cd}\) = \(\frac{\frac{1}{3}c(c - 2d)}{\frac{1}{4}d(2d - c)}\) = \(\frac{\frac{1}{3}c (c - 2d)}{-\frac{1}{4}d (c - 2d)}\) = \(-\frac{4c}{3d}\) (b)(i) In \(\Delta XYP\), \(\frac{7}{XP} = \sin 44^\circ\) \(XP = \frac{7}{\sin 44}\) = \(10.08 cm\) < XPM + 44^\circ + 46^\circ = 180^\circ < XPM = 90^\circ. In \(\Delta XPM\), \(XM^{2} = XP^{2} + PM^{2}\) \(XM^{2} = (10.08)^{2} + 9^{2}\) = \(101.6064 + 81 = 182.6064\) \(XM = \sqrt{182.6064}\) = \(15.513 cm \approxeq 13.5 cm\) In \(\Delta XYP\), \(\frac{7}{YP} = \tan 44\) \(YP = \frac{7}{\tan 44}\) \(YP = 7.249 cm \approxeq 7.2 cm\) In \(\Delta MPF\), \(\frac{PF}{9} = \cos 46\) \(PF = 9 \cos 46 = 6.2519 cm\) \(YF = YP + PF = 7.249 + 6.2519\) = \(13.5009 cm \approxeq 13.5 cm\) (ii) \(\frac{10.1}{9} = \tan \theta\) \(\tan \theta = 1.122\) \(\theta = \tan^{-1} (1.122)\) = 48.296^\circ

(a) Class Boundaries Tally Frequency 59.5 - 64.5 || 2 64.5 - 69.5 ||| 3 69.5 - 74.5 |||| | 6 74.5 - 79.5 |||| |||| | 11 79.5 - 84.5 |||| ||| 8 84.5 - 89.5 |||| || 7 89.5 - 94.5 || 2 94.5 - 99.5 | 1 (b) An histogram to represent the distribution of 40 students in an examination. (c) Modal mark : \(\frac{74.5 + 79.5}{2} = \frac{154}{2} = 77\) (d) Number of students who obtained marks greater than 79 : 8 + 7 + 2 + 1 = 18. Total number of students = 40 \(\therefore\) P(mark greater than 79) = \(\frac{18}{40} = \frac{9}{20}\)

(a) < QRT = < RQS (alternate angles) < RPS = < RQS (angles in the same segment) \(\therefore\) < RPS = < QRT (ii) < QRT = < PQS (corresponding angles) < PQS = < PRS (angles in the same segment) \(\therefore\) < QTR = < PRS. (b) \(< ABC = \hat{B} = 180^\circ - 132^\circ = 48^\circ\) \(< ACB = \hat{C} = 180^\circ - 96^\circ = 84^\circ\) \(< BAC = \hat{A} = 180^\circ - (48^\circ + 84^\circ) = 48^\circ\) \(< BAC = < ABC = 48^\circ\) \(\therefore \Delta ABC = isosceles\).

(a) \(\frac{r}{R} = \cos 50\) \(r = 6400 \cos 50 = 4,113.84 km\) \(\approxeq 4,114km\) The distance PQ along the parallel of latitude = speed (km/h) \(\times\) time (h) = 600km/h x 3h = 1800km \(\frac{\theta}{360} \times 2 \times \frac{22}{7} \times r = 1800\) \(\theta = \frac{1800 \times 7 \times 360}{44 \times 4113.84}\) \(\theta = 25.059^\circ \approxeq 25.1^\circ\) \(\therefore\) The longitude of Q = \(60 + \theta\) = \(60 + 25.1 = 85.1^\circW\) (b) The distance (Q) along the meridian \(\frac{\theta}{360} \times 2\pi R\) \(500 = \frac{\theta}{360} \times 2 \times \frac{22}{7} \times 6400\) \(\theta = \frac{500 \times 360 \times 7}{44 \times 6400}\) \(\theta = 4.47^\circ\) \(\therefore \text{The latitude of Y} = 50^\circ - 4.47^\circ = 45.53^\circ\) \(\approxeq 45.5^\circN\)

(d) /PM/ = 9.7 cm.

POM is a diameter, so arc MP (lower semicircle via Q) = 180^\circ. ∠MNQ = 42^\circ is an inscribed angle subtending arc MQ. Therefore, arc MQ = 2 \times 42^\circ = 84^\circ Arc MQ + arc QP = lower semicircle: arc QP = 180^\circ − 84^\circ = 96^\circ - ∠QMP is an inscribed angle subtending arc QP. Therefore, ∠QMP = \(\frac{1}{2}\) \times 96^\circ = 48^\circ. (Alternatively: in right triangle PQM, ∠PQM = 90^\circ (angle in a semicircle), so the angles at M and P sum to 90^\circ; the arc method directly gives the result.)

To find the maximum of \(f(\theta) = \frac{4}{6 + 2\cos \theta}\) for \(\frac{\pi}{2} \leq \theta \leq 2\pi\), simplify to \(f(\theta) = \frac{2}{3 + \cos \theta}\). The maximum occurs when \(3 + \cos \theta\) is minimized. In the interval, \(\cos \theta\) ranges from \(-1\) to \(1\), with minimum \(-1\) at \(\theta = \pi\). Thus, minimum denominator is \(3 - 1 = 2\), so maximum \(f(\theta) = \frac{2}{2} = 1\).

\((343)^{\frac{1}{3}} \times (0.14)^{-1} \times (25)^{-\frac{1}{2}}\) = \(\sqrt[3]{343} \times (\frac{14}{100})^{-1} \times (\sqrt{25})^{-1}\) = \(7 \times \frac{100}{14} \times \frac{1}{5}\) = 10.

\(y = \sin^{2} (5x)\) Let u = sin 5x \(\frac{\mathrm d u}{\mathrm d x} = 5 \cos 5x\) \(\therefore y = u^{2}\) \(\frac{\mathrm d y}{\mathrm d u} = 2u\) \(\frac{\mathrm d y}{\mathrm d x} = 2u . 5 \cos 5x\) = \(10u \cos 5x\) = \(10 \sin 5x \cos 5x\)

y = 3x\(^2\) - 2x + 5 Slope = \(\frac{\mathrm d y}{\mathrm d x} = 6x - 2\) At x = 1, Slope : 6(1) - 2 = 4.

\(\int \sin 3x \mathrm d x = - \frac{\cos 3x}{3} + c\) = \(-\frac{1}{3} (\cos 3x) + c\)

Area of the circle (A) = \(\pi r^{2}\) \(\frac{\mathrm d A}{\mathrm d r} = 2\pi r\) \(r = 5cm\); \(\frac{\mathrm d r}{\mathrm d t} = 0.2\) \(\frac{\mathrm d A}{\mathrm d t} = \frac{\mathrm d A}{\mathrm d r} . \frac{\mathrm d r}{\mathrm d t}\) \(\frac{\mathrm d A}{\mathrm d t} = 2\pi \times 5 \times 0.2 = 2\pi\)

Diameter = 4\(\sqrt{3}\) cm radius = 2\(\sqrt{3}\) cm Area of major sector = \(\frac{\theta}{360} \times \pi r^{2}\) \(\theta = 360 - 120 = 240^\circ\) = \(\frac{240}{360} \times \pi \times 12\) = \(8\pi cm^{2}\)

\(\tan \theta = \frac{opposite}{adjacent} = \frac{4}{3}\) Hyp\(^2\) = 4\(^2\) + 3\(^2\) Hyp = 5. \(\sin \theta = \frac{4}{5}; \cos \theta = \frac{3}{5}\) \(\sin^{2} \theta - \cos^{2} \theta = \frac{16}{25} - \frac{9}{25}\) = \(\frac{7}{25}\)