(a) \(5 \cos (x + 8.5)^\circ - 1 = 0\) \(5 \cos (x + 8.5)^\circ = 1\) \(\cos (x + 8.5)^\circ = \frac{1}{5} = 0.2\) \((x + 8.5)^\circ = \cos^{-1}(0.2) = 78.463^\circ\) \(x = 78.463^\circ - 8.5 = 69.963^\circ\) \(\approxeq 70^\circ\) (to the nearest degree). (b)(i) (ii) By the cosine rule, \(|QR|^{2} = 32^{2} + 24^{2} - 2 \times 32 \times 24 \times \cos 45\) \(|QR|^{2} = 1024 + 576 - 1536 \cos 45\) = \(1600 - 1086.1056\) \(|QR|^{2} = 513.8944\) \(|QR| = \sqrt{513.8944} = 22.669 km\) \(\approxeq 22.67 km\) ( 2 decimal place) (iii) By the sine rule, \(\frac{32}{\sin \alpha} = \frac{22.67}{\sin 45}\) \(\sin \alpha = \frac{32 \times \sin 45}{22.67}\) = \(0.9981\) \(\alpha = \sin^{-1} (0.9981) = 86.4787^\circ\) The diagram below shows all the angles at Q; The bearing of R from Q is given by the reflex angle NQR. Thus reflex < NQR = 360^\circ - (86.47^\circ + 30^\circ) = 360^\circ - 116.47^\circ = 243.53^\circ Hence, the bearing of R from Q = 244^\circ (to the nearest degree).