Question Library

XTY = 50o = TXY = \(\frac{130^o}{2}\) = 65o

\(\frac{1}{2}\)(PQ x 10 sin 30) = 35 PQ = \(\frac{35 \times 2}{10 \sin 30}\) = 14cm

(a) \(y = k + \frac{c}{x^{2}}\) where c and k are constants. (b) When x = 1, y = 11 \(11 = k + \frac{c}{1^{2}} \implies 11 = k + c ... (1)\) When x = 2, y = 5 \(5 = k + \frac{c}{2^{2}} \implies 5 = k + \frac{c}{4}\) \(\equiv 20 = 4k + c ... (2)\) (2) - (1) : \(4k - k = 20 - 11 \implies 3k = 9\) \(k = 3\) Put k = 3 in (1), we have \(11 = 3 + c \implies c = 11 - 3 = 8\) \(\therefore y = 3 + \frac{8}{x^{2}}\) When x = 4, \(y = 3 + \frac{8}{4^{2}} = 3 + \frac{1}{2}\) = \(3\frac{1}{2}\)

(a) \(/QY/ = 2cm; /RS/ = 3cm\) \(\therefore /YZ/ = /RS/ = /PQ/ = 3cm\) (opp sides of a parallelogram) But \(/PZ/ = /PQ/ + /QY/ + /YZ/\) \(\therefore /PZ/ = 3cm + 2cm + 3cm = 8cm\) (b) Difference in longitude = 95^\circ - 25^\circ = 70^\circ Distance PQ = \(\frac{70}{360} \times 2\pi R \cos \theta\) = \(\frac{70}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 56\) = \(\frac{7}{36} \times 2 \times \frac{22}{7} \times 6400 \times 0.5592\) = \(4374.19 km \approxeq 4374 km\) (to the nearest km)

(a) Total no of cards = 52; No of fives = 4; No of red nines = 2. Prob(of 5) = \(\frac{4}{52} = \frac{1}{13}\) Prob(red nine) = \(\frac{2}{52} = \frac{1}{26}\) \(\therefore\) Prob (of 5 or red nine) = \(\frac{1}{13} + \frac{1}{26} = \frac{3}{26}\) (b) \(\tan < RPQ = \frac{3}{5} = 0.6\) \(< RPQ = 30.96^\circ\) \(\therefore\) Bearing of R from P = 150^\circ - 30.96^\circ = 119.04^\circ \(\approxeq\) 119^\circ (to the nearest degree).

(a - d)

(a) < XZY = 180^\circ - (30^\circ + 45^\circ) = 105^\circ \(\frac{20}{\sin 105^\circ} = \frac{YZ}{\sin 30^\circ}\) \(YZ = \frac{20 \sin 30^\circ}{\sin 105^\circ} = 10.35km\) \(\approxeq 10.4km \) (to 3 sig. figs) (ii) \(\frac{20}{\sin 75} = \frac{XZ}{\sin 45}\) \(XZ = \frac{20 \sin 45}{\sin 75} = 14.64km\) \(\approxeq 14.6km\) ( to 3 sig. figs) (b) Let P and Q be two airfields. Difference in latitude = 36^\circ + 36^\circ = 72^\circ (i) Distance between P and Q = \(\frac{72}{360} \times 2 \times \frac{22}{7} \times 6400\) = \(\frac{22}{7} \times 2560\) = \(8,045.71km \approxeq 8,050 km\) (3 sig. figs) (ii) Time taken = \(\frac{8,045.71}{800} = 10.06 hours\) \(\approxeq 10hours\) (to the nearest hour)

\(\sin \theta = \frac{k}{1}\) \(\implies 1^2 = k^2 + adj^2\) \(adj = \sqrt{1 - k^2}\) \(\therefore \tan \theta = \frac{k}{\sqrt{1 - k^2}}\)

\(\frac{8.75}{0.025}\) = \(\frac{8750}{25}\) = \(350\) = \(3.5 \times 10^2\)

-1-(-3)/-2-(-0.5) = 2/-1.5 = 20/-15 = -113

\(\frac{r}{6} = \sin 60 \) \(r = 6 \sin 60\) = \(6 \times \frac{\sqrt{3}}{2}\) = \(3\sqrt{3}\) Chord = 2r = \( 2 \times 3\sqrt{3}\) = \(6\sqrt{3}\)

Length of arc = \(\frac{\theta}{360} \times 2\pi r\) = \(\frac{75}{360} \times 2 \times \frac{22}{7} \times 3.5\) = \(4.583 cm\) \(\approxeq\) 4.6 cm (to 2 sig. figs)

A = θ 360 x πr2 = 135 360 x π x x2 = (3πx2)/8

Join PR to a pentagon 90 + 90 + 140 + 55 + x = 540o x = 165o

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