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\(|TQ|^{2} = 4^{2} + 6^{2} - 2 \times 4 \times 6 \times \cos 30^\circ\) = \(16 + 36 - 48 \times 0.8660\) = \(52 - 41.568 = 10.432\) \(\therefore |TQ| = \sqrt{10.432} \approxeq 3.23 cm\) (a) By the rules of similar triangles, \(\frac{|PT|}{|TQ|} = \frac{|PS|}{|SR|}\) \(\frac{4}{3.23} = \frac{10}{|SR|}\) \(|SR| = \frac{3.23 \times 10}{4}\) = \(8.075 cm \approxeq 8 cm\) (b) Using sine rule, \(\frac{|PQ|}{\sin \alpha} = \frac{|TQ|}{\sin 30}\) \(\frac{6}{\sin \alpha} = \frac{3.23}{\sin 30}\) \(\sin \alpha = \frac{6 \sin 30}{3.23}\) sin \(\alpha\) = 0.9288 From the diagram, \(\alpha\) = \(\beta\) \(\sin \alpha = \sin \beta = 0.9288\) \(\sin \beta = \frac{h}{6}\) h = \(6 \sin \beta\) Hence, area of quadrilateral TQRS = \(\frac{1}{2} (TQ + SR) \times h\) = \(\frac{1}{2} (3.23 + 8.075) \times 6\sin \beta\) = \(\frac{1}{2} (11.305) (6 \times 0.9288)\) = 31.501 cm\(^2\) \(\approxeq\) 32 cm\(^2\) (to the nearest whole number)

(a) In triangle SPQ, \(|SQ|^{2} = 5^{2} + 12^{2}\) (Pythagoras theorem) = \(25 + 144 = 169\) \(|SQ| = \sqrt{169} = 13 cm\) Angle b is common to triangles SPQ and PRS are similar. Using \(\sin b = \frac{12}{|SQ|} = \frac{|PR|}{5}\) \(\sin b = \frac{12}{13} = \frac{|PR|}{5}\) \(|PR| = \frac{12 \times 5}{13} \approxeq 4.62 cm\) (to 3 s.f) (b)(i) \(h^{2} = 10^{2} - 8^{2} = 36\) \(h = \sqrt{36} = 6 cm\) (ii) In the smaller triangle, \(\cos x = \frac{8}{10} = 0.8\) \(\cos^{-1} (0.8) = 36.87^\circ\) Since these are corresponding angles, x = x in the bigger triangle. \(\sin x = \frac{y}{12}\) \(y = 12 \sin x = 12 \sin 36.87\) = \(12 \times 0.6\) = 7.20 m The ladder M touches the wall at a height 7.2 m above the horizontal ground.

(a) x -5 -4 -3 -2 -1 0 1 2 3 4 y 35 18 5 -4 -9 -10 -7 0 11 26 (b) (c)(i) 2x\(^{2}\) + x = 10 \(\equiv\) 2x\(^{2}\) + x - 10 = 0 Thus, the solution of 2x\(^{2}\) + x - 10 = 0 are the values of x at which the curve cuts the x- axis. x = -2.5 or x = 2. (ii) 2x\(^{2}\) + x - 10 = 2x To solve the given equation, we first draw the graph of y = 2x: when x = -2, y = 2(-2) = -4; when x = 3, y = 2(3) = 6. The values of x at which y = 2x\(^{2}\) + x - 10 and y = 2x intersect give the solution of 2x\(^{2}\) + x - 10 = 2x. These are x = -2 or x = 2.5.

(a) px + qy = z \(p \begin{pmatrix} 2 \\ 3 \end{pmatrix} + q \begin{pmatrix} 5 \\ -2 \end{pmatrix} = \begin{pmatrix} -4 \\ 13 \end{pmatrix}\) \(\begin{pmatrix} 2p + 5q \\ 3p - 2q \end{pmatrix} = \begin{pmatrix} -4 \\ 13 \end{pmatrix}\) Thus, \(2p + 5q = -4 .....(1)\) \(3p - 2q = 13 ......(2)\) Solving the equations, we have p = 3 and q = -2. (b)(i) (ii) See attached graph for the quadrilateral W(2, 3), X(4, -1), Y(-3, -4) and Z(-3,2). (iii) See attached for the image for the quadrilateral WXYZ under an anticlockwise rotation of 90^\circ is \(W_{1}(-3, 2), X_{1}(1, 4), Y_{1}(4, -3)\) and \(Z_{1}(-2, -3)\).

(a) Marks(x) Frequency (f) fx 10 1 10 20 1 20 30 x 30x 40 5 200 50 y 50y 60 1 60 70 4 280 80 3 240 90 1 90 Total 16 + x + y 900 + 30x + 50y \(\sum f = 16 + x + y = 30\) \(\implies x + y = 14 ... (1)\) \(\bar{x} = \frac{\sum fx}{\sum f}\) \(52 = \frac{900 + 30x + 50y}{30}\) \(1560 = 900 + 30x + 50y \implies 30x + 50y = 660\) \(3x + 5y = 66 .... (2)\) Solving equation (1) and (2), From (1), x = 14 - y \(3(14 - y) + 5y = 42 - 3y + 5y = 66\) \(2y = 24 \implies y = 12\) \(x = 14 - y = 14 - 12 = 2\) (x, y) = (2, 12). (b) Class interval Frequency Upper class boundary 1 - 10 1 10.5 11 - 20 1 20.5 21 - 30 2 30.5 31 - 40 5 40.5 41 - 50 12 50.5 51 - 60 1 60.5 61 - 70 4 70.5 71 - 80 3 80.5 81 - 90 1 90.5 (c) (d) From the histogram, Mode = 44.

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