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The sum of the exterior angles of a regular polygon = 360^\circ. \(\therefore x + (x + 5) + (x + 10) + (x + 15) + (x + 20) = 360^\circ\) \(5x + 50 = 360^\circ \implies 5x = 360^\circ - 50^\circ = 310^\circ\) \(x = \frac{310^\circ}{5} = 62^\circ\)

\(\frac{BC}{60}\) = \(\frac{tan 62}{1}\) BC = 60 tan 62 \(\frac{AC}{60}\) = \(\frac{tan 62}{1}\) AC = 60 tan 64 AB = AC - BC = 60(tan 64o - tan 62o)

\(\cos^{2} x (\sec^{2} x + \sec^{2} x \tan^{2} x)\) = \(\cos^{2} x \sec^{2} x + \cos^{2} x \sec^{2} x \tan^{2} x\) = \(1 + \tan^{2} x\) = \(1 + \frac{\sin^{2} x}{\cos^{2} x}\) = \(\frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x}\) = \(\frac{1}{\cos^{2} x} = \sec^{2} x\)

cosx = \(\sqrt{\frac{a}{b}}\) y2 + \(\sqrt{(a)^2}\) = \(\sqrt{(b)^2}\) by pythagoras y2 = b - a ∴ y = b - a cosec x = \(\frac{1}{sin x}\) = \(\frac{1}{y}\) \(\frac{b}{y}\) = \(\frac{\sqrt{b}}{\sqrt{b - a}}\) = \(\sqrt{\frac{b}{b - a}}\)

tan \(\theta\) = \(\frac{60\sqrt{3}}{60}\) = \(\sqrt{3}\) \(\theta\) = tan \(\frac{1}{3}\) = 60o Bearing of y from x = 060o

The total surface area of a cylinder = \(2\pi r (r + h)\) = \(2 \pi (4) (4 + 5)\) = \(8 \pi (9) = 72 \pi cm^{2}\)

Education = 11% = \(\frac{11}{100} \times 360^\circ = 39.6^\circ\) Food = 59% = \(\frac{59}{100} \times 360^\circ = 212.4^\circ\)

No of defective(x) No of boxes (f) \(fx\) 4 2 8 5 7 35 6 17 102 7 10 70 8 8 64 9 6 54 \(\sum\) 50 333 \(Mean = \frac{\sum fx}{\sum f}\) = \(\frac{333}{50} = 6.66 \approxeq 6.7\) The median is the average of the 25th and 26th position = 6.

% of boxes containing at least 5 defective bolts each = \(\frac{7 + 17 + 10 + 8 + 6}{50}\) = \(\frac{48}{50}\) x \(\frac{100}{1}\) = 96%

Coca-cola = 10 bottles, Fanta = 8 bottles Sprite = 6 bottles, Total = 24 P(cola-cola) = \(\frac{10}{24}\) P(not coca-cola) = 1 - \(\frac{10}{24}\) \(\frac{24 - 10}{24}\) = \(\frac{14}{24}\) = \(\frac{7}{12}\)

From the diagram above x = 60 + 70 = 130º (sum of angles on a straight line was used to obtain 60º and 70º, and alternate angles are equal to obtain the value of x)

\(\frac{PQ}{PN} = \frac{PM}{PR} = \frac{QM}{NR}\) \(\frac{4.8}{12} = \frac{5}{PR}\) PR = \(\frac{5 \times 12}{4.8} = \frac{50}{4}\) = 12.5 \(\frac{PQ}{PN} = \frac{PM}{PT} = \frac{TM}{NT}\) \(\frac{PT}{12} = \frac{5}{PR}\) PT2 = 60 PT = \(\sqrt{60}\) = 7.746 = 7.7

PQM = 110o(Ext. o - (110 + 55) = 180o - 165o = 15o

ADB = 180o - 152o = 28o 28o + y + 30o = 180o y = 180o - 58o = 122o

Since PQRS is quadrilateral 2y + 2x = QPS = 360o i.e. 2(y + x) + QPS = 360o QPS = 360o - 2(y + x) But x + y + 68o = 180o x + y = 180o - 68o = 180o x + y = 180o - 68o = 112o QPS = 360o - 2(112o) 360o - 224o = 136o