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(a) \(< ADB = < ACB = 20^\circ\) (Angle in the same segment). In \(\Delta ACY\), \(\hat{C} = < ACB\) (i) \(\hat{C} + \hat{A} + \hat{Y} = 180^\circ\) \(20^\circ + \hat{A} + 60^\circ = 180^\circ\) \(\hat{A} = 180^\circ - 80^\circ = 100^\circ\) \(< CAY = \hat{A} = 100^\circ\) \(< CAX = 180^\circ - 100^\circ = 80^\circ\) (ii) In \(\Delta ACB\), \(\hat{B} = \hat{C} = \frac{180^\circ - 20^\circ}{2} \) = \(\frac{160^\circ}{2}\) = \(80^\circ\) In \(\Delta ABY\), \(< ABY = \hat{B} = 180^\circ - 80^\circ\) = \(100^\circ\) (b) \((m + 1)(m - 3) \equiv m^{2} - km + c\) \(m^{2} - 3m + m - 3 = m^{2} - 2m - 3\) \(m^{2} - 2m - 3 \equiv m^{2} - km + c\) \(\implies k = 2 ; c = -3\)

(a) (i) Let the exterior angle be x Let the interior angle be x + 120^\circ. Exterior angle + Interior angle = 180^\circ \(\therefore x + x + 120 = 180 \implies 2x + 120 = 180\) \(2x = 60 \implies x = 30^\circ\) Number of sides = \(\frac{360}{\text{exterior angle}}\) = \(\frac{360}{30} = 12 sides\) (ii) Sum of interior angle = \((2n - 4) \times 90^\circ\) = \((2(12) - 4) \times 90^\circ\) = \(20 \times 90\) = \(1800^\circ\) (b) Note : < PQR = 65^\circ + 34^\circ = 99^\circ\) \(|PR|^{2} = |PQ|^{2} + |QR|^{2} - 2(|QR|)(|PQ|) \cos 99\) \(|PR|^{2} = 6^{2} + 13^{2} - 2(6)(13) \cos 99\) \(|PR|^{2} = 205 - 156 \cos 99\) \(|PR|^{2} = 205 + 24.40 = 229.40\) \(|PR| = \sqrt{229.40}\) = \(15.146 km\) \(\approxeq 15 km\)

(a) Longitude difference = 85^\circ - 25^\circ = 60^\circ. Distance AB along the parallel of latitude = \(\frac{\theta}{360^\circ} \times 2\pi R \cos \alpha\) \(AB = \frac{60}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 53\) = \(\frac{1}{6} \times \frac{44}{7} \times 3851.62\) = \(4,035.03 km\) \(\approxeq 4035 km\) (b) \(Speed = \frac{Distance}{Time}\) \(\therefore Time = \frac{Distance}{Speed}\) = \(\frac{4035}{400}\) \(\approxeq 10 hours\). (c) Distance BC measured along the meridian \(BC = \frac{\theta}{360} \times 2 \pi R\) \(2000 = \frac{\theta}{360} \times 2 \times \frac{22}{7} \times 6400\) \(\theta = 17.898^\circ \approxeq 18^\circ\) \(\theta\) = Latitude difference Let the latitude of C = x. \(\theta = 53 - x\) \(18 = 53 - x\) \(x = 35^\circ\)

(a) \(|AB| = 85m = \frac{85}{10}\) = \(8.5cm\) \(|TC| = 6cm \equiv 60 m\).

Length of an arc = \(\frac{\theta}{360}\) x 2\(\pi\)r 22 = \(\frac{3x}{360}\) x \(\frac{22}{7}\) x 7 (since D = 2r) 22 x 7 x 360 = 3x X 22 X 7 x = \(\frac{22 \times 7 \times 360}{3 \times 22 \times 7\times 2}\) = 60º

PQ = \(\sqrt{(\beta - 1)^{2} + (1 - \alpha)^{2}}\) 3 = \(\sqrt{(\beta^{2} -2\beta^{2} + 1 + 1 - 2\alpha + \alpha^{2})}\) 3 = \(\sqrt{(\alpha^{2} + \beta^{2} - 2\alpha + 2\beta + 2)}\) 3 = \(\sqrt{(\alpha^{2} + \beta^{2} - 2(\alpha + \beta) + 2)}\) 3 = \(\sqrt{(\alpha^{2} + \beta^{2} - 2 * 2 + 2)}\) 3 = \(\sqrt{(\alpha^{2} + \beta^{2} - 2)}\) 9 = \(\alpha^{2} + \beta^{2} - 2)\) \(\alpha^{2} + \beta^{2}\) = 9 + 2 \(\alpha^{2} + \beta^{2}\) = 11

6x + 6y = (n - 2) 180 6x + 6y= (5 - 2) 180 6(x + y) = 3 * 180 x + y = (3 * 180)/6 x + y = 90o y = 90 - x

The diagonal QS bisects the angle formed by PQ and QR ∴ [A]

\(\frac{X}{\sin 45^\circ} = \frac{15}{\sin 60^\circ}\) \(X = \frac{15 \sin 45^\circ}{\sin 60^\circ}\) \(X = \frac{15 \times \left( \frac{1}{\sqrt{2}} \right)}{\sqrt{\frac{3}{2}}}\) \(X = \frac{15}{\sqrt{2}} \times \frac{2}{\sqrt{3}}\) \(X = \frac{30}{\sqrt{6}}\) \(X = \frac{30}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{30\sqrt{6}}{6} = 5\sqrt{6}\)

The locus of a point which is 5cm from the line LM is a pair of lines on opposite sides of LM and parallel to it, each distance 5cm from LM.

y = 3cos(\(\frac{x}{3}\)) \(\frac{dy}{dx}\) = 3x(\(\frac{1}{3}\))x - sin (\(\frac{x}{3}\)) = - sin \(\frac{x}{3}\) But x = \(\frac{3\pi}{2}\) ∴ -sin(\(\frac{x}{3}\)) = -sin(\(\frac{3\pi}{6}\)) = -sin \(\frac{(3 \times 180)}{6}\) = - sin 90 = -1

Sectorial angle of beans = 360 - (150 + 90 + 60) = 360 - 300 = 60º If 150º represents 3000 tonnes 1º = \(\frac{3000}{150}\) 60º will represents (\(\frac{3000}{150}\)) x (\(\frac{60}{1}\)) = 1200 tonnes

A histogram is described by the following; Rectangular bars of equal width II. The height of each rectangular bar is proportional to the frequency of the corresponding class interval. III. Rectangular bars have common sides with no gaps in between.

The sum of angles at point = 360o (x+10)o + (4x+50)o + 20o + 3xo + 2xo = 360o 10x + 80o = 360o 10x = 360o - 80o = 280o \(\frac{280}{10}=28^{\circ}\)

QOP = 180 - 100(straight line angle) → 80^\circ OPQ and OQP are isosceles of 40^\circ each. ∠QRS = ∠RPQ (theorem: angle in the alternate segment are equal) ∠RPQ = 40º ∠QRS = 40º