(a) Note that triangles AXC and BXD are similar. \(\therefore \frac{6}{|XD|} = \frac{3}{4}\) \(|XD| = \frac{6 \times 4}{3}\) = \(8 cm\) (b)(i) From \(\Delta ZYX\), \(\stackrel\frown{YZX} = 90^\circ - 22^\circ = 68^\circ\) \(\stackrel\frown{YXZ} = 312^\circ - 270^\circ = 42^\circ\) \(\stackrel\frown{ZYX} = 180^\circ - (68^\circ + 42^\circ)\) = \(180^\circ - 110^\circ\) = \(70^\circ\) \(a^{2} = b^{2} + c^{2} - 2bc \cos A\) \(|XZ|^{2} = 50^{2} + 32^{2} - 2(50)(32) \cos 70\) = \(2500 +1024 - 3200 \times 0.342\) = \(3524 - 1094.46\) \(|XZ|^{2} = 2429.54\) \(|XZ| = \sqrt{2429.54} = 49.29 km\) \(\approxeq 49.3 km\) (to 1 d.p) (ii) \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\) \(\frac{50}{\sin \theta} = \frac{49.3}{\sin 70}\) \(\sin \theta = \frac{50 \times \sin 70}{49.3}\) \(\sin \theta = 0.9530\) \(\theta = \sin^{-1} (0.9530)\) = \(72.37^\circ\) \(\therefore\) The bearing of Z from X = \(312^\circ - 72.37^\circ\) = \(239.63^\circ \approxeq 239.6^\circ\)