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(a) From \(\Delta\) AMC, \(\sin 60^\circ = \frac{\sqrt{3}}{2}\) \(\cos 30^\circ = \frac{\sqrt{3}}{2}\) \(\tan 60^\circ = \frac{\sqrt{3}}{1} = \sqrt{3}\) Hence, \(\frac{2 \tan 60^\circ + \cos 30^\circ}{\sin 60} = \frac{2 \times \sqrt{3} + \frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}}\) = \(\frac{4\sqrt{3} + \sqrt{3}}{2} \div \frac{\sqrt{3}}{2}\) = \(\frac{5\sqrt{3}}{2} \times \frac{2}{\sqrt{3}} = 5\) (b) In \(\Delta\) POT, \(\tan 36^\circ = \frac{|TO|}{1050}\) \(|OT| = 1050 \tan 36^\circ = 1050 \times 0.7265\) = 762.87m In \(\Delta\) POB, \(\tan 41^\circ = \frac{|OB|}{1050}\) \(|OB| = 1050 \tan 41^\circ = 1050 \times 0.8693\) = 912.751m \(|TB| = 912.751m - 762.87m = 149.88m\) Hence, the height of the control tower = 150 m (to the nearest meter). (ii) In \(\Delta\) POB, \(\cos 41^\circ = \frac{1050}{|PB|}\) \(|PB| = \frac{1050}{\cos 41^\circ} = \frac{1050}{0.7547}\) = 1,391.28m The shortest distance between the aeroplane and the base of the control tower = 1,391m (to the nearest metre).

(a) \(h = \frac{mt}{d(m + p)}\) \(dh(m + p) = mt\) \(dhm + dhp = mt \implies dhp = mt - dhm\) \(dhp = m(t - dh) \implies m = \frac{dhp}{t - dh}\) (b) In the diagram above, < WXM = 90^\circ (angle in a semicircle) < WMX = 180^\circ - (90^\circ + 48^\circ) = 42^\circ < XMZ = 180^\circ - 42^\circ (angles on a straight line) = 138^\circ < WYZ = 180^\circ - 138^\circ (opp. angles of a cyclic quadrilateral) = 42^\circ (c) \(\star\) 1 3 5 6 1 4 6 1 2 3 6 1 3 4 5 1 3 5 6 6 2 4 6 0 (ii) From the table, the truth set of : (I) \(3 \star n = 3; n = {5}\) (II) \(n \star n = 3; n = { }\)

(a) (i) Using volume of reservoir = volume of hemisphere + volume of cone. \(\implies 333\frac{1}{3}\pi = \frac{2}{3}\pi r^{3} + \frac{1}{3}\pi r^{2} h\) \(\implies \frac{1000}{3}\pi = \frac{1}{3}\pi (2r^{3} + r^{2}h)\) Let the radius of the hemisphere be x m; Then \(\frac{1000}{3} = \frac{1}{3}(2x^{3} + x^{2}(6x))\) \(1000 = 2x^{3} + 6x^{3}\) \(1000 = 8x^{3} \implies x^{3} = \frac{1000}{8} = 125\) \(x = \sqrt[3]{125} = 5 m\) Hence, the volume of the hemisphere = \(\frac{2}{3}\pi r^{3}\) (where r = x = 5 m) = \(\frac{2}{3} \times \frac{22}{7} \times 5^{3} m^{3}\) = \(\frac{5500}{21} m^{3} \approxeq 261.905 m^{3}\) = \(262 m^{3} \) (nearest whole number) (ii) Total surface area of reservoir = surface area of hemisphere + surface area of cone = \(2\pi r^{2} + \pi rl\) From the diagram above, \(l^{2} = 30^{2} + 5^{2}\) \(l^{2} = 900 + 25 = 925\) \(l = \sqrt{925} = 30.41 m\) Hence, the total surface area of the reservoir = \(2 \times \frac{22}{7} \times 5 \times 5 + \frac{22}{7} \times 5 \times 30.41\) = \(157.143 + 477.871\) = \(635.104 m^{2}\) \(\approxeq 635 m^{2}\) (to the nearest whole number).

Marks (%) frequency (f) Cumulative frequency Upper class boundaries 0 - 9 7 7 9.5 10 - 19 11 18 19.5 20 - 29 17 35 29.5 30 - 39 20 55 39.5 40 - 49 29 84 49.5 50 - 59 34 118 59.5 60 - 69 30 148 69.5 70 - 79 25 173 79.5 80 - 89 21 194 89.5 90 - 99 6 200 99.5 (b) (i) If 5% of the candidates passed the examination, then (100 - 5)% = 95% passed with a mark \(\leq\) the lowest mark for distinction. \(\text{95% of 200} = \frac{95}{100} \times 200 \) = 190 candidates. From the ogive, 190 corresponds to 79.5 + 8 = 87.5 marks (to one decimal place) (ii) From the ogive, the number of candidates who scored at most 45% is 69. Hence, the probability of selecting a candidate who scored at most 45% = \(\frac{69}{200} = 0.345 \approxeq 0.3\) (1 d.p)

Using the general gas laws: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) P\(_1\) = 800mmHg, V\(_1\) = 76cm\(^3\), T\(_1\) = 273 + 27 = 300k P\(_2\) = 760mmHg, V\(_2\) = ? T\(_2\) = 273 + 0 = 273 (@ STP) \(\frac{800 \times 76}{300} = \frac{760 \times V_2}{273}\) V\(_2 = \frac{800 \times 76 \times 273}{300 \times 760}\) = 72.8cm\(^3\) Therefore, the volume of gas at S.T.P = 72.8 cm\(^3\)

Boyle‘s law stats that the volume of a given mass of a gas, at a constant temperature is inversely proportional to us pressure. Mathematically V ∝ l/p, v = k/p (where k is constant) K = VP Let k = k1 Then, if k = k, V and P = V1P1 V1P1 = V2P2

Ray – a beam of light or radiation

XL = WL = 2πFL (w = 2πf) XL = reactance inductor (S.I unit Ω) L = inductance (L) in Henry XL = 1Ω, f = 50/π, L = ? 1 = 2× π × 50/π × L 1 = 100L i.e. 100L = 1 L = 1/100 0.01H ∴ Inductance (L) of a inductor = 0.01H

Stroboscope is a scientific instrument used to make a wave appears stationary when used

In an inelastic collision, linear momentum is conserved and energy is decreased Momentum and kinetic energy are conserved in elastic collision

Note:The principle of floatation states that a floating body displaces its own weight of the fluid or which its float. For object to sink weight of the object (W) > upthrust of the fluid (v) i.e. w > u (i.e. weight of the body (or object) mass of the fluid)

Formula for general gas law V × 1/T × P N [(PV) ÷ T] = K is constant

The appearance of pool of water on road surfaces is called MIRAGE His observation show that light from the sun ray refracted through a layer of hot air(layer) which is closed to the ground, passes from a hot air layer to the cold layer

Sun are self luminous source Moon is a non – luminous

Note: moment of a couple = force × length (fL) The tangential force is actingin an opposite direction alongside its length i.e f1 L1 = f2L2 f2L2 − f1 L1 = 0(i.e Zero)