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Since it is divided into 3 sectors then \(8x^\circ = 360^\circ\) \(x^\circ = 45^\circ\)The 3rd sector = 3 times the size of the other two sectors put together = \(3(45 + 45) = 3(90) = 270^\circ\) Perimeter of a sector = \(2r + l\) where r = radius of the circle ; l = length of arc = \(\frac{\theta}{360^\circ} \times 2\pi r\) Area of circle = \(\frac{22}{7} \times r^{2} = 154 cm^{2}\) \(r^{2} = \frac{154 \times 7}{22} = 49\) \(r = 7 cm\) \(l = \frac{270}{360} \times 2 \times \frac{22}{7} \times 7 = 33 cm\) Perimeter of sector = \(2(7) + 33 = 14 + 33 = 47 cm\).

(a) (b) \(\tan \theta = \frac{4.2}{6 + x} ; \tan \theta = \frac{1.2}{x}\) \(\tan \theta = \tan \theta\) \(\implies \frac{4.2}{6 + x} = \frac{1.2}{x}\) \(4.2 x = 1.2(6 + x) \) \(4.2 x = 7.2 + 1.2 x \implies 4.2x - 1.2x = 7.2\) \(3x = 7.2 \implies x = 2.4 m\) (length of the shadow cast by the lamp) (ii) \(\tan \theta = \frac{1.2}{2.4} = 0.5\) \(\theta = \tan^{-1} (0.5) \) = \(27^\circ\) (angle of elevation)

(a) diagram above. (b) From the figure above, (i) |WZ| = 5.25 cm ; (ii) |ZN| = 4.86 cm.

(a) Length of the chord = 14 cm Radius = \(\frac{3}{2} \times 14 cm = 21 cm\) The chord AB = \(2r \sin \frac{\theta}{2}\) \(14 = 2(21) \sin \frac{\theta}{2}\) \(\sin \frac{\theta}{2} = \frac{14}{42} = 0.333\) \(\frac{\theta}{2} = \sin^{-1} (0.333)\) \(\frac{\theta}{2} = 19.469^\circ\) \(\theta = 19.469^\circ \times 2 = 38.938^\circ\) Length of the arc = \(\frac{\theta}{360^\circ} \times 2\pi r\) \(\frac{38.938}{360} \times 2 \times \frac{22}{7} \times 21\) = \(14.277 cm\) Perimeter of the remaining portion = 22 + 12 + 12 + 5 + 3 + 14.277 = 68.277 cm \(\approxeq\) 68.28 cm (2 decimal places). (b) \(\frac{3}{\sqrt{3}}(\frac{2}{\sqrt{3}} - \frac{\sqrt{12}}{6}\) = \(\frac{3}{\sqrt{3}}(\frac{12 - \sqrt{36}}{6\sqrt{3}}\) = \(\frac{3}{\sqrt{3}}(\frac{12 - 6}{6\sqrt{3}}\) = \(\frac{3}{\sqrt{3}}(\frac{6}{6\sqrt{3}}\) = \(\frac{18}{6\sqrt{9}}\) = \(\frac{18}{18}\) = 1.

(a) see diagram above. (b)(i) \(Speed = \frac{distance}{time}\) \(950 = \frac{d}{4} \implies d = 950 \times 4\) \(d = 3800km\) \(d = \frac{\theta}{360} \times 2\pi r\) \(3800 = \frac{\theta}{360} \times 2 \times \frac{22}{7} \times 6400\) \(3800 = \frac{\theta \times 281600}{2520}\) \(\theta = \frac{2520 \times 3800}{281600}\) \(\theta = 34.01^\circ\) \(\theta = 34^\circN\) (to the nearest degree) (ii) Distance between P and Q, correct to four significant figures. Longitude difference = 65^\circ - 15^\circ = 50^\circ Using, \(d = \frac{\theta}{360} \times 2 \pi r\) where \(r = R \cos \theta\) \(d = \frac{\theta}{360} \times 2 \pi R \cos \theta\) = \(\frac{50}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 34.01^\circ\) = \(4631.53 km\) \(\approxeq 4632km\) (to 4 significant figures).

(a) see diagram above (b) From the above diagram, \(\sin 37^\circ = \frac{5}{|LM|}\) \(|LM| = \frac{5}{\sin 37}\) \(|LM| = \frac{5}{0.6018}\) = \(8.308 m\) \(\approxeq 8.31 m\) \(\cos 37 = \frac{|AM|}{|LM|}\) \(|AM| = 8.308 \times \cos 37\) \(|AM| = 8.308 \times 0.7986\) \(|AM| = 6.6348 m\) (c) The new |AM| = 6.6348 m - 2 = 4.6348 m \(\cos y = \frac{4.6348}{8.308}\) \(\cos y = 0.5579\) \(y = \cos^{-1} (0.5579)\) \(y = 56.089^\circ\) \(y = 56^\circ\) (to the nearest degree).

Sum of angles in a quadrilateral = 360^\circ \(\therefore\) (3y + 10) + (2y + 30) + (y + 20) + 4y = 360 10y + 60 = 360 \(\implies\) 10y = 300 y = 30^\circ

Length of the tile = 30 cm = 0.3m Area of the tile = 0.3 \(\times\) 0,3 = 0.09 m\(^2\) Area of the room = (7.2 \(\times\) 4.2)m\(^2\) Number of tiles = \(\frac{7.2 \times 4.2}{0.09}\) = 336

Length of chord = \(2r \sin (\frac{\theta}{2})\) = \(2 \times 8 \times \sin (\frac{90}{2})\) = \(16 \times \frac{\sqrt{2}}{2}\) = \(8\sqrt{2} cm\)

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

If you take any point that is equidistant from the line PQ, it lies at a constant distance above and below the line. Therefore, the correct answer is: a pair of parallel lines to PQ.

Using the cosine rule, we have \(p^{2} = q^{2} + r^{2} - 2qr \cos P\) \(p^{2} = 8^{2} + 6^{2} - 2(8)(6)(\frac{1}{12})\) = \(64 + 36 - 8\) \(p^{2} = 92 \therefore p = \sqrt{92} cm\)

\(\tan \theta = \frac{opp}{adj} = \frac{3}{4}\) \(hyp^{2} = opp^{2} + adj^{2}\) \(hyp = \sqrt{3^{2} + 4^{2}}\) = 5 \(\sin \theta = \frac{3}{5}; \cos \theta = \frac{4}{5}\) \(\sin \theta + \cos \theta = \frac{3}{5} + \frac{4}{5}\) = \(\frac{7}{5} = 1\frac{2}{5}\)

\(\int_{0} ^{\frac{\pi}{2}} \sin x \mathrm d x\) = \(- \cos x |_{0} ^{\frac{\pi}{2}\) = \(- \cos (\frac{\pi}{2}) - (- \cos 0)\) = \(0 + 1\) = 1

Since there are 5 angles given, the polygon is a pentagon. Sum of interior angles of a pentagon = (2(5) - 4) x 90^\circ = 540^\circ \(\therefore\) x + 2x + 3x + 4x + 5x = 15x 15x = 540^\circ \(x = \frac{540}{15} = 36^\circ\)