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y = 6 sin (2x - 25)º \(\frac{dy}{dx}\) = 12 cos(2x - 25)º \(\frac{dy}{dx}\) = 0 12 cos x (2x - 25º) = 0 = cos(2x - 25º) = 0 = 2x - 25º = cos\(^{-1}\)(0) = 2x - 25º = 90º x = \(\frac{115}{2}\)

o) = 90o (angle substended by a semi-circle) ∴ o o - 30o = 60o o - 35o = 55o o + 55o) = 180 - 115o = 65o o(angles in the same segment) o o + 35o = 180o o o - 155o = 25o

A\(\bigtriangleup\) = \(\frac{1}{2}\) x 8 x h = 20 = \(\frac{1}{2}\) x 8 x h = 4h h = \(\frac{20}{4}\) = 5cm A\(\bigtriangleup\)(PQTS) = L x H A\(\bigtriangleup\)PQRT = A\(\bigtriangleup\)QSR + A\(\bigtriangleup\)PQTS 20 + 50 = 70cm\(^2\) ALTERNATIVE METHOD A\(\bigtriangleup\)PQRT = \(\frac{1}{2}\) x 5 x 28 = 70cm\(^2\)

< TXR = < QTR = 40o (alternate segment) < TWR = < TXR = 40o(Angles in the same segments) < YXR = 40o + 32o = 72o < YXR + < YTR = 180o(Supplementary) 72o + < YTR = 180o < YTR = 180o - 72o = 108o

\(\frac{10}{\sin 30^o} = \frac{15}{\sin x} = \frac{10}{0.5} = \frac{15}{\sin x}\) \(\frac{15}{20} = \sin x\) sin x = \(\frac{15}{20} = \frac{3}{4}\) N.B x = < PTR

Length of chord = \(2r \times \sin(\frac{\theta}{2})\) = \( 2 \times \sqrt{3} \times \sin(\frac{60}{2})\) = \(2 \times \sqrt{3} \times \frac{1}{2}\) = \(\sqrt{3}\) cm.

Let the monthly expenditure on school fees be x let the monthly expenditure on housing be 2x; angle of housing and school fee in a pie chart = 360o - (120o + 90o) = 360o - 210o = 150o Angle of housing in a pie-chart = \(\frac{2}{3}\) x 150 = 100 \(\frac{100}{360}\) x 7200 = N2,000

Let one of the sides be x metres. The other is (x + 1.5) metres. (a) \(Area = Length \times breadth\) \(x(x + 1.5) = 13.5\) \(x^{2} + 1.5x = 13.5 \implies x^{2} + 1.5x - 13.5 = 0\) \(x^{2} + 4.5x - 3x - 13.5 = 0 \implies x(x + 4.5) - 3(x + 4.5) = 0\) \((x - 3)(x + 4.5) = 0 \implies x = 3 ; x = -4.5\) Since x cannot be negative, therefore x = 3m. The longer side = (3 + 1.5) = 4.5m (b) \(N250 = 1m^{2}\) \(N2000 = \frac{2000}{250} = 8m^{2}\) \(\therefore \text{The area of the floor that will be covered by the carpet} = 8m^{2}\)

(a)(i) < ASO = < AMO = 90^\circ (radius to a tangent) < SOM = 360^\circ - (3 x 90^\circ) (sum of angles of a rectangle) = 90^\circ. \(\therefore\) obtuse < SOT = 360^\circ - (90^\circ + 60^\circ + 60^\circ) = 150^\circ. (ii) Arc SRT = \(\frac{150^\circ}{360^\circ} \times 2\pi r\) = \(\frac{150}{360} \times 2 \times \frac{22}{7} \times 10 = \frac{550}{21}\) = \(26.19cm\) (iii) \(\tan 60^\circ = \frac{TB}{10}\) \(TB = 10\tan 60^\circ = 17.32cm\) (b) Length of the string ASRTB = AS + arc SRT + TB = 10cm + 26.19cm + 17.32cm = 53.51cm \(\approxeq\) 54cm.

XM = \sqrt132 - 52 = 12cm MZ = 152 - 122 = 9cm YZ = 9 + 5 = 14cm

P(67^\circW, 0^\circN) Q(67^\circW 90^\circN), R = 6400km Distance PQ = 90/360 x 2 x 6400 \pi = 3200\pikm

Area = 1/2(6 + 10) x 5 40cm2

44 = 200/360 x 2 x 22/7 x r r = 360 x 7 x 44/200 x 2 x 22 = 12.6cm

Area = 1/2 x y x yz sinθ 23.5 = 1/2 x 10 x 8 sinθ sinθ = 23.5 x 2/10 x 8 = 0.5875 θ = sin-1 0.5875 = 36o

No expanded working has been attached for this question yet. The verified answer is Option C.