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Volume of pyramid with square base = \(\frac{\text{base area} \times height}{3}\) \(80 = \frac{\text{base area} \times 15}{3}\) \(80 = 5 \times \text{base area}\) \(\text{Base area} = 16 cm^2\)

Total volume = (5 x 5) x 5 + (5 x 5) x 2 = 125 + 50 = 175 cm\(^3\)

< T = < S = 50^\circ (OS = OT) < SOT = 180^\circ - 2(50^\circ) = 80^\circ < ROP = 80^\circ (vertically opposite angle) \(\therefore\) < OPQ = 180^\circ - 80^\circ = 100^\circ (adjacent angles)

Sum of interior angle in a hexagon = (6 - 2) x 180^\circ = 720^\circ \(\therefore\) 107^\circ + 2x^\circ + 150^\circ + 95^\circ + (2x - 15)^\circ + 123^\circ = 720^\circ 460 + 4x = 720 \(\implies\) 4x = 720 - 460 4x = 260^\circ \(\implies\) x = 65^\circ

(a) \(\cos x = 0.7431\) \(x = \cos^{-1} (0.7431)\) \(x = 42^\circ\) (i) \(2 \sin x = 2 \sin 42\) = \(2 \times 0.6692\) = \(1.3384\) (ii) \(\tan \frac{x}{2} = \tan \frac{42}{2}\) = \(\tan 21^\circ\) = \(0.3839\) (b) Sum of the interior angles of a polygon = \((2n - 4) \times 90^\circ\) For a pentagon, n = 5 \((2(5) - 4) \times 90^\circ = 6 \times 90^\circ\) = \(540^\circ\) Ratio of sides = 2:3:4:4:5 Total = 2 + 3 + 4 + 4 + 5 = 18 Largest angle = \(\frac{5}{18} \times 540^\circ = 150^\circ\)

circumference of the bicycle wheel \(= 2\pi r = 2 \times \frac{22}{7} \times 42 = 264cm = 2.64m\\\) Number of revolution \(=\frac{66}{2.64} = 25\)

Length of arc = \(\frac{\theta}{360} \times 2\pi r\) \(50 = \frac{75}{360} \times 2 \times \frac{22}{7} \times r\) \(r = \frac{50 \times 360 \times 7}{75 \times 2 \times 22}\) \(r = \frac{420}{11}\) = 38.18 cm \(\approxeq\) 38.2 cm (3 sig. figs)

From the figure above, and by measurement < RMQ = 60^\circ.

In triangle ABC, \(\frac{BC}{AC} = \cos \theta\) \(\implies \cos \theta = \frac{5}{10} = 0.5\) \(\theta = \cos^{-1} (0.5) = 60^\circ\)

(a) see figure a above \(\frac{h}{100} = \sin 24\) \(h = 100 \sin 24\) = \(100 \times 0.4067\) = \(40.67m\) (b) see figure b above \(\sin \alpha = \frac{4}{5} = 0.8\) \(\alpha = \sin^{-1} (0.8) = 53.13^\circ\) \(< BAC = 2 \theta = 2(53.13^\circ)\) = \(106.26^\circ\) \(\approxeq 106^\circ\) (to the nearest degree). (c) See Figure C above In \(\Delta PSQ\), \(\frac{h}{PQ} = \tan 71.6\) \(PQ = \frac{h}{\tan 71.6}\) In \(\Delta QSR\), \(\frac{h}{QR} = \tan 45\) \(QR = \frac{h}{\tan 45}\) \(PR = PQ + QR\) \(\frac{h}{3} + h = 70 \implies \frac{4}{3}h = 70\) \(h = \frac{70 \times 3}{4} = 52.5m\)

\(x^2 = 8^2 + 8^2\) \(x^2 = 64 + 64 = 128\) \(x = \sqrt{128}\) = \(8\sqrt{2}\) km

The bearing of Kofi from the building = 180^\circ + 45^\circ = 225^\circ

S50^\circW = 180^\circ + 50^\circ = 230^\circ

The bearing of the ship from its original position is S90ºE

A = \(\int^{x = 4}_{x = 0}\) ydx = \(\int^{4}_{0}\)16 - x\(^2\)dx = [16 - \(\frac{x^3}{3} + c]^{4}_{0}\) = \(\frac{128}{3}\) square units