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for a regular polygon to be possible, it must have all sides angles equal. \(\frac{360}{18}\) = 20 sides and \(\frac{360}{15}\) = 24 sides (ii) and (iii) are right

\(\cot \theta = \frac{x}{y}\) \(\implies \tan \theta = \frac{y}{x}\) \(opp = y; adj = x\) Using Pythagoras theorem, \(Hyp^{2} = Opp^{2} + Adj^{2}\) \(Hyp^{2} = y^{2} + x^{2}\) \(Hyp = \sqrt{y^{2} + x^{2}}\) \(\sin \theta = \frac{y}{\sqrt{y^{2} + x^{2}}}\) \(\therefore \csc \theta = \frac{\sqrt{y^{2} + x^{2}}}{y}\) = \(\frac{1}{y}(\sqrt{y^{2} + x^{2}})\)

PR2 = PQ2 + QR2 - 2(QR)(PQ) COS 120o PR2 = 12 + 22 - 2(1)(2) x - cos 60o = 5 - 2(1)(2) x -\(\frac{1}{2}\) = 5 + 2 = 7 PR = \(\sqrt{7}\)cm

\(\bigtriangleup\)PUS is right angled \(\frac{US}{5}\) = sin60o US = 5 x \(\frac{\sqrt{3}}{2}\) = 2.5\(\sqrt{3}\) = 4.33cm

SRT is a straight line, where QRT = 120 SRQ = 180o - 120o = 60o - (angle on a straight line) also angle QRS = 180o - 100o (angle on a straight line) . In angles where QR = SR and angle SRQ = 60o x = 100 - 60 = 40o

From similar triangle, \(\frac{QS}{QP} = \frac{TQ}{QR} = \frac{5}{12} = \frac{6}{QR}\) QT = \(\frac{6 \times 12}{5} = \frac{72}{5} = SR = QR - QS\) = \(\frac{72}{5} - 5 = \frac{72 - 25}{5}\) = \(\frac{47}{5}\)

Given: ∠ZXR = 45º (tangent XR), ∠YZX = 55º (inscribed angle). ∠YZX subtends arc YX, so arc YX = 2 \times 55^\circ = 110º. Major arc YZX = 360º - 110º = 250º. By the alternate segment theorem, ∠ZXR subtends arc XZ, so arc XZ = 2 \times 45º = 90º. Arc YZ = 250º - 90º = 160º. Major arc YZ = 360º - 160º = 200º. ∠ZYQ (tangent YQ) = \(\frac{1}{2}\) \times major arc YZ = \(\frac{1}{2}\) \times 200º = 100º (by alternate segment theorem, larger arc for obtuse angle).

(a) Given \(\sin \alpha = 0.3907\) From the sine table, \(\alpha = 23^\circ\) (i) \(\tan \alpha = \tan 23^\circ = 0.4245\) (ii) \(\cos \alpha = \cos 23^\circ = 0.9205\) (b) Let T be the top of the ladder and B, the bottom of the vertical wall. From the figure, above \(\sin 50^\circ = \frac{BT}{4.5m}\) \(BT = 4.5 \sin 50^\circ = 4.5 \times 0.7660\) \(BT = 3.447m\) Distance between the top of the ladder and bottom of the window = \(4 - 3.447 = 0.553m\) = \( 0.553m \times 100\) = \(55.3 cm\)

Subject Period Angle of sector in pie chart English 10 \(\frac{10}{36} \times 360^\circ = 100^\circ\) Mathematics 7 \(\frac{7}{36} \times 360^\circ = 70^\circ\) Biology 3 \(\frac{3}{36} \times 360^\circ = 30^\circ\) Statistics 4 \(\frac{4}{36} \times 360^\circ = 40^\circ\) Igbo 3 \(\frac{3}{36} \times 360^\circ = 30^\circ\) Others 9 \(\frac{9}{36} \times 360^\circ = 90^\circ\) Total 36 360^\circ

(i) The length of the parallel of latitude 46^\circN = \(2 \pi R \cos 46^\circ\) = \(2 \times 3.142 \times 6400 \times 0.6947\) = \(27,939.167 km \approxeq 28000 km\) (to the nearest 100km) (ii) Longitudinal difference between K and Q = 130^\circ - 103^\circ = 27^\circ Distance between K and Q = length of arc KQ = \(\frac{27}{360} \times 2 \times 3.142 \times 6400 \times \cos 46^\circ\) = \(2,095.31 km \approxeq 2100km\) (to the nearest 100km) (iii) Longitudinal difference between Q and P = 103^\circ + 23^\circ = 126^\circ Distance between Q and P = \(\frac{126}{360} \times 2 \times 3.142 \times 6400 \times \cos 46^\circ\) = \(9778.12 km \approxeq 9780 km\) (to the nearest 10km)

(a) Given any \(\Delta\) ABC, Required: That < A + < B + < C = 180^\circ. Construction : Draw CP // BA and produce BC to Q. Proof: Using the above figure, \(a_{1} = a_{2}\) (alternate angles) \(b_{1} = b_{2}\) (corresponding angles) \(a_{1} + b_{1} + c = 180^\circ\) (angles on a straight line) \(\therefore a_{2} + b_{2} + c = 180^\circ\) \(< A + < B + < C = 180^\circ\) (proved) (b) < LNP = 46^\circ + 80^\circ = 126^\circ (exterior angles of triangle LMN) \(\theta = \frac{126^\circ}{2} = 63^\circ\) (bisector divides into 2 equal halves) \(< MLN = 180^\circ - 80^\circ = 100^\circ\) (angles on a straight line) \(< LQN = 180^\circ - (63^\circ + 100^\circ) = 17^\circ\)

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

1m = 1.1 yard, length(L)= 50m = (50 x 1.1)yards = 55 yards Area(A) = length(L) y width (W) 1815 = 55y width (W) width (w) = \(\frac{1815}{55}\) = 33 yards But 33 yards = 30 meters

Let the distance from P to Q = d km (We don't need the actual value of d — it will cancel out.) Time taken from P to Q = \(\frac{\text{d}}{90}\) hours Time taken from Q to P = \(\frac{\text{d}}{45}\) hours Total distance travelled = d + d = 2d km Total time taken = \(\frac{\text{d}}{90}\) + \(\frac{\text{d}}{45}\) = \(\frac{\text{d}}{90}\) + \(\frac{2d}{90}\) = \(\frac{3d}{90}\) = \(\frac{\text{d}}{30}\) hours Average speed = \(\frac{\text{total distance}}{\text{total time}}\) = 2d \div (\(\frac{\text{d}}{30}\)) = 2d \times (\(\frac{30}{\text{d}}\)) = 60 km/h

(0.008) -\(\frac{1}{3}\) x (0.16) - \(\frac{3}{2}\) = (8 x 10-3)10-3 x (16 x 10-2)-\(\frac{3}{2}\) = \(\frac{(2^3)}{10^3}\) - 3 x \(\frac{(2^4)}{10}\) - \(\frac{-3}{2}\) = \(\frac{625}{8}\)