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(a) \(\tan 50^\circ = \frac{|TF|}{20} \implies |TF| = 20 \tan 50\) = \(20 \times 1.1918\) = \(23.836 m\) \(\tan 30 = \frac{|PF|}{20} \implies |PF| = 20 \tan 30\) = \(20 \times 0.5774\) = \(11.548 m\) \(\therefore |PT| = 23.836 m - 11.548m\) = \(12.288 m \approxeq 12.3 m\) (b) \(|TH| = \frac{20}{\cos 50}\) = \(\frac{20}{0.6428}\) = \(31.11 m\) (c) \(|HF| = \frac{20 \tan 50^\circ}{\tan 40^\circ}\) = \(\frac{20 \times 1.1918}{0.8391}\) = \(28.41 m\)

(b) Two possible ways (c) The distances of the locations from town X = 27.6 km and 12.4 km. (d) 12.4 km

Class Interval Freq (f) Mid-point (x) \(fx\) \(x^{2}\) \(fx^{2}\) 60 - 64 2 62 124 3844 7688 65 - 69 3 67 201 4489 13487 70 - 74 6 72 432 5184 31104 75 - 79 11 77 847 5929 65219 80 - 84 8 82 656 6724 53792 85 - 89 7 87 609 7569 52983 90 - 94 2 92 184 8464 16928 95 - 99 1 97 97 9409 9409 \(\sum\) 40 3150 250590 (a) Mean \(\bar{x} = \frac{\sum fx}{\sum f}\) = \(\frac{3150}{40}\) = 78.75 (b) \(SD = \sqrt{\frac{\sum fx^{2}}{\sum f} - (\frac{\sum fx}{\sum f})^{2}}\) = \(\sqrt{\frac{250590}{40} - (\frac{3150}{40})^{2}}\) = \(\sqrt{6264.75 - (78.75)^{2}}\) = \(\sqrt{6264.75 - 6201.5625}\) = \(\sqrt{63.1875}\) = 7.95.

Area of sector = \(\frac{\theta}{360^\circ} \times \pi r^{2}\) = \(\frac{150}{360} \times \pi \times 12^{2}\) = 60\(\pi\) cm\(^{2}\)

If \(y = \cos x\) \(\frac{\mathrm d y}{\mathrm d x}\) =—\(\sin x\)

(a)(i) < PTS = 38^\circ (alternate angle) 42^\circ + x + 38^\circ = 180^\circ (angles on a straight line) 80^\circ + x = 180^\circ x = 180^\circ - 80^\circ = 100^\circ (ii) < SQT = 42^\circ (alternate angle) < SQR = 60^\circ (sum of angles in a triangle) y + 42^\circ + 60^\circ = 180^\circ (angle on a straight line) y + 102^\circ = 180^\circ y = 180^\circ - 102^\circ = 78^\circ (b) \(< QSP = \frac{150}{2}\) (angle at the centre is twicw angle at the circumference) \(< QSP = 75^\circ\) \(\hat{R} + \hat{P} = 180^\circ\) (opposite angles in a cyclic quad are supplementary) \(\hat{P} + 75^\circ + 45^\circ = 180^\circ\) (sum of angle in a triangle) \(\hat{P} = 180^\circ - 120^\circ\) \(\hat{P} = 60^\circ\) \(\hat{R} + \hat{P} = 180^\circ\) \(\hat{R} = 180^\circ - 60^\circ\) \(\hat{R} = 120^\circ\) \(< RQS + < QRS + < QSR = 180^\circ\) (sum of angle in a triangle) \(< RQS = 180^\circ - (40^\circ + 120^\circ)\) = \(180^\circ - 160^\circ\) \(< RQS = 20^\circ\)

volume of rectangular container = L x B x H 1 litre = 1000cm3 96 litres = \(\frac{1000cm^3}{1 litres}\) x 96 litres volume = 96000cm3 96000 = 220 x 40 x 11 96--- = 8800H H = \(\frac{96000}{8800}\) = 10.97cm = 11cm

Interior angle = (n - 2)180 but, n = 12 = (12 -2)180 = 10 x 180 = 1800 let each interior angle = x x = \(\frac{(n - 2)180}{n}\) x = \(\frac{1800}{12}\) = 150o

From Pythagoras theorem |OA|2 = |AN|2 + |ON|2 72 = |AN|2 + (5)2 49 = |AN|2 + 25 |AN|2 = 49 - 25 = 24 |AN| = \(\sqrt {24}\) = \(\sqrt {4 \times 6}\) = 2\sqrt6 cm |AN| = |NB| (A line drawn from the centre of a circle to a chord, divides the chord into two equal parts) |AN| + |NB| = |AB| 2\sqrt6 + 2\sqrt6 = |AB| |AB| = 4\sqrt6 cm

Volume of cube = L3 33 = 27cm3 volume of rectangular tank = L x B X h = 3 x 4 x 5 = 60cm3 volume of H2O the tank can now hold = volume of rectangular tank - volume of cube = 60 - 27 = 33cm3

tan\(\theta\) = \(\frac{3}{4}\) from Pythagoras tippet, the hypotenus is T i.e. 3, 4, 5. then sin \(\theta\) = \(\frac{3}{5}\) and cos\(\theta\) = \(\frac{4}{5}\) cos\(\theta\) - sin\(\theta\) \(\frac{4}{5}\) - \(\frac{3}{5}\) = \(\frac{1}{5}\)

tan\(\theta\) = \(\frac{100}{100}\) = 1 \(\theta\) = tan-1(1) = 45o The bearing of x from z is N45oE or 135o

\(\int\) cos4 x dx let u = 4x \(\frac{dy}{dx}\) = 4 dx = \(\frac{dy}{4}\) \(\int\)cos u. \(\frac{dy}{4}\) = \(\frac{1}{4}\)\(\int\)cos u du = \(\frac{1}{4}\) sin u + k = \(\frac{1}{4}\) sin4x + k

\(\begin{array}{c|c}Class Interval & 3 - 3 & 6 - 8 & 9 - 11 \\ x & 4 & 7 & 10 \\ f & 2 & 2 & 2 \\ f - x & 8 & 14 & 20 \\ |x - \bar{x}|^2 & 9 & 0 & 9 \\ |x - \bar{x}|^2 & 18 0 & 18 \end{array}\) \(\bar{x}\) = \(\frac {\sum fx}{\sum f}\) = \(\frac {8 + 14 + 20}{2 + 2 + 2}\) = \(\frac{42}{6}\) \(\bar{x}\) = 7 S.D = \(\sqrt\frac{\sum f(x - \bar{x})^2}{\sum f}\) = \(\sqrt\frac{18 + 0 + 18}{6}\) = \(\sqrt\frac{36}{6}\) = \(\sqrt {6}\)

ELATION Since there are 7 letters. The first letter can be arranged in 7 ways, , the second letter in 6 ways, the third letter in 5 ways, the 4th letter in four ways, the 3rd letter in three ways, the 2nd letter in 2 ways and the last in one way. therefore, 7 x 6 x 5 x 4 x 3 x 2 x 1 = 7! ways