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(a) < ABC = 100^\circ \(\therefore b^{2} = 8^{2} + 13^{2} - 2(13)(8) \cos 100^\circ\) = \(64 + 169 - (208 \times - 0.1736)\) = \(233 + 36.12\) \(b^{2} = 269.12 \implies b = \sqrt{269.12} = 16.405 km\) \(\approxeq 16.4 km\) (3 significant figures) (b) \(\frac{\sin B}{b} = \frac{\sin A}{a}\) \(\frac{\sin 100}{16.4} = \frac{\sin A}{13}\) \(\sin A = \frac{13 \times \sin 100}{16.4}\) \(\sin A = \frac{12.803}{16.4}\) \(\sin A = 0.7806\) \(A = \sin^{-1} (0.7806) = 51.32^\circ\) \(\therefore\) Bearing of C from A = 180^\circ - (50^\circ + 51.32^\circ) = 180^\circ - 101.32^\circ = 78.68^\circ \(\approxeq\) 78.7^\circ (3 significant figure). (c) \(\cos 40 = \frac{BD}{13}\) \(BD = 13 \cos 40\) = \(13 \times 0.7660\) = 9.959 km \(\approxeq\) 9.96 km east of B.

X\(^2\) + x\(^2\) = 2\(^2\) 2x\(^2\) = 4 x\(^2\) = 4/2 x\(^2\) = 2 x = +/-\sqrt2 Area of \Delta = 1/2bh = 1/2 x \sqrt2 x \sqrt2 = 2/2 = 1 cm\(^2\)

x\(^2\) + 5\(^2\) = 13\(^2\) x\(^2\) + 25 = 169 x\(^2\) = 144 x = \sqrt 144 = 12 Length of the chord AB = 2x 2\(\times\)12= 24cm

C.P = N34000 + N2000 = N36000 Gain = 100 + 15 = 115% S.P = \(\frac{115}{100}\times \frac{N3600}{1}\) = N41400

If S \(\subset\) T \(\subset\) W, S \(\cap\) T \(\cap\) W = S is true since S \(\cap\) T = S and S \(\cap\) W = S. S \(\cup\) T \(\cup\) W = W is also true. S \(\cup\) T = T and T \(\cup\) W = W. However, to say that T \(\cap\) W = S is not very true mathematically. Instead, it is safe to say S \(\subset\) (T \(\cap\) W).

y = x, Gradient (m1) = 1 y = √3x gradient (m2) = √3 \(tan\theta = \frac{m_2 - m_1}{1+m_2 m_1}\\ tan\theta = \frac{\sqrt{3}-1}{1+\sqrt{3}\times 1}\\ =\frac{1.73 -1 }{1+1.73}= \frac{0.73}{2.73}=0.268\) θ = Tan-1 (0.268) = 15o

Sum of interior angles of a polygon = each interior ∠ x n (n-2)180 = 150 x n 180n - 360 = 150n 180n-150n = 360 30n = 360 n = 360/30 n = 12 sides

Locus of a particle which moves in the first quadrant so that it is equidistant from the lines X = 0 and Y = 0 are the x and y axes as the lines bisecting the angle between x and y axes.

x\(^2\) + 3\(^3\) = 5\(^2\) x\(^2\) + 9 = 25 x\(^2\) = 25 – 9 x\(^2\) = 16 x = \sqrt 16 = 4 cm ∴ The locus is a circle of radius 4 cm with the center O

Gradient = \(\frac{y_2 - y_1}{x_2 - x_1}\) \(\frac{2}{3}\) = \(\frac{4 - p}{p - (-1)}\) \(\frac{2}{3}\) = \(\frac{4 - p}{p + 1}\) 3(4-p) = 2(p+1) 12 - 3p = 2p + 2 12 – 2 = 2p + 3p 10 = 5p p= \(\frac{10}{5}\) = 2

A(4,2) and B(1,r), AB = 3 units 3 = \sqrt (x2- x1)2 + (y2-y1)2 3 = \sqrt (4-1)2 + (2-r)2 3 = \sqrt 32 + (2-r)2 3 = \sqrt 9 + 4 – 4r + r2 3 = \sqrt r2 - 4r + 13 9 = r2 - 4r + 13 By squaring both sides r2 - 4r + 4 = 0 (r-2)(r-2) = 0 r = 2

y = 3 cos 4x Applying product rule: \(\frac{dy}{dx}\) = u\(\frac{dv}{dx}\) + v\(\frac{du}{dx}\) Where u = 3 and v = cos 4x \(\frac{du}{dx}\) of 3 = 0 and \(\frac{dv}{dx}\) of cos 4x = -4sin 4x \(\frac{dy}{dx}\) = 3 \(\times\) -4 sin 4x + cos 4x \(\times\) 0 \(\frac{dy}{dx}\) = -12 sin 4x

\(Sin 45 - cos 30\\ \frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\\ =\frac{\sqrt{2}-\sqrt{3}}{2}\)

Tan 60^\circ = \(\frac{300}{\text{x}}\) x = \(\frac{300}{tan 60}\) x = \(\frac{300}{\sqrt{3}}\) x = \(\frac{300}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\) x = \(\frac{300\sqrt{3}}{3}\) x = 100\(\sqrt{3}\)m

S = t\(^3\) - t\(^2\) - t + 5 ds/dt = 3t\(^2\) - 2t - 1 As ds = 0 3t\(^2\) - 2t - 1 = 0 (3t+1)(t-1) = 0 ∴ t = 1 or -1/3 At min pt t = 1 S = t\(^3\) - t\(^2\) - t + 5 put t = 1 = 1\(^3\) - \1(^2\) - 1 + 5 = 1 - 1 - 1 + 5 = 4