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

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

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

The sum of exterior angles = 360^\circ \(60^\circ + 2(90 - m)^\circ + 2(30 + 2m)^\circ = 360^\circ\) \(60^\circ + 180^\circ - 2m^\circ + 60^\circ + 4m^\circ = 360^\circ\) \(300^\circ + 2m = 360^\circ\) \(\implies 2m = 360^\circ - 300^\circ = 60^\circ\) \(m = 30^\circ\) (b) In \(\Delta QRS, \cos Q = \frac{\sqrt{3}}{2}\) \(\cos Q = 0.8660 \implies Q = 30^\circ\) \(\therefore < SQR = 30^\circ^\) \(< PQS = 180^\circ - 30^\circ \) (angles on a straight line = 180^\circ) \(< PQS = 150^\circ\)

Class Interval Tally Classmark (x) Freq (f) \(fx\) 21 - 25 IIII || 23 7 161 26 - 30 |||| |||| | 28 11 308 31 - 35 |||| |||| 33 9 297 36 - 40 |||| |||| 38 9 342 41 - 45 |||| | 43 6 258 46 - 50 |||| ||| 48 8 384 \(\sum\) 50 1750 (b) (c) Mode : \(L_{1} + (\frac{f_{0} - f_{1}}{2f_{0} - f_{1} - f_{2}})t\) Where \(L_{1}\) = lower class boundary of modal class = 25.5 \(f_{0}\) = frequency of modal class = 11 \(f_{1}\) = frequency of pre-modal class = 7 \(f_{2}\) = frequency of post modal class = 9 \(t\) = interval mark = 5. Mode : \(25.5 + (\frac{11 - 7}{22 - 7 - 9})\times 5 = 25.5 + 3.3\) = 28.8. (d) Mean : \(\frac{\sum fx}{\sum f} = \frac{1750}{50}\) = 35.

(a) (i) \(\cos C = \frac{a^{2} + b^{2} - c^{2}}{2ab}\) \(\cos C = \frac{12^{2} + 10^{2} - 17^{2}}{2(12)(10)}\) \(\cos C = \frac{-45}{240} = -0.1875\) \(C = \cos^{-1} (-0.1875)\) = \(100.81^\circ\) (ii) Area of \(\Delta ABC = \frac{1}{2} ab \sin C\) = \(\frac{1}{2} \times 12 \times 10 \times \sin 100.81\) = \(60 \times 0.9822\) = \(58.936 m^{2}\) (b) \(\tan 63 = \frac{38}{x}\) \(x = \frac{38}{\tan 63}\) \(x = 19.362 m\) \(\approxeq 19 m\)

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

(a) 105^\circ + 26^\circ + x = 180^\circ 131^\circ + x = 180^\circ x = 180^\circ - 131^\circ = 49^\circ. 131^\circ + x + 131^\circ + 360 - y = 360^\circ 131^\circ + 49^\circ + 131^\circ + 360 - y = 360^\circ 671^\circ - y = 360^\circ y = 671^\circ - 360^\circ = 311^\circ. (b)(i) \(\stackrel\frown{ABC} = \frac{1}{2} \stackrel\frown{AOC} reflex\) = \(\frac{1}{2} \times 190^\circ = 95^\circ\) (ii) \(\stackrel\frown{ADC} = \frac{1}{2} \stackrel\frown{AOC}\) = \(\frac{1}{2} (360^\circ - 190^\circ)\) = \(\frac{1}{2} \times 170^\circ\) = 85^\circ (c) (i) \(\stackrel\frown{ABC} = \frac{360 - 190}{360} \times 2\pi r\) = \(\frac{170}{360} \times 2 \times 3.142 \times 7\) = \(20.77 cm\) \(\approxeq 20.8 cm\) (3 significant figures) (ii) \(AD = 2 \times 7\cos 35\) = \(14 \times 0.8192\) = \(11.468 cm \approxeq 11.5 cm\) (3 significant figures)

y = x cos x dy/dx = 1. cos x + x (-sin x) = cos x - x sin x

\(\int_0 ^{\frac{\pi}{2}}(-2cos x)dx = [-2sin x + c]_0 ^{\frac{\pi}{2}}\) =(-2sin\(\frac{\pi}{2}+c+2sin0-c)\) =-2sin90+c+2sin0-c =-2(1)+2(0) =-2

Length of an arc = \(\frac{\theta}{360}\) x 2\pir \(\frac{90}{360}\) x 2 x \pi x 7 = \(\frac{7\pi}{2}\)cm\(^2\)

If the line 3y = 4x – 1 is parallel t[ line qy = x + 3 Implies gradient of 3y = 4x – 1 Y = 4/3x - 1/3 ∴Gradient = 4/3 Gradient of line qy = x + 3 Y = 1/qx + 3/q ∴ Gradient = 1/q 4/3 = 1/q 4q = 3 Q = 3/4

x = 30o (alternate ∠s b/c PQ//Sr) Y = 50o (vertical opposite ∠s) r = 30o (alternate ∠s b/c PQ//SR) But = y + q + t = 180o (∠s on a straight line) 50 + q + 30 = 180 q + 80 = 180 q = 180 -80 q = 100o

When a particle P moves such that the angle ∠SPT is constant while ST remains fixed, it traces a circular path. Specifically, the locus of P is a semicircle because points on the circle subtended by the line segment ST maintain that angle.

Tan 30 = \(\frac{h}{40}\) \(\frac{1}{\sqrt{3}}=\frac{h}{40}\) h\(\sqrt{3}\) = 40 h = \(\frac{40}{\sqrt{3}}\) h = \(\frac{40}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\) h = \(\frac{40\sqrt{3}}{3}\)