(a) < ABC = 100^\circ \(\therefore b^{2} = 8^{2} + 13^{2} - 2(13)(8) \cos 100^\circ\) = \(64 + 169 - (208 \times - 0.1736)\) = \(233 + 36.12\) \(b^{2} = 269.12 \implies b = \sqrt{269.12} = 16.405 km\) \(\approxeq 16.4 km\) (3 significant figures) (b) \(\frac{\sin B}{b} = \frac{\sin A}{a}\) \(\frac{\sin 100}{16.4} = \frac{\sin A}{13}\) \(\sin A = \frac{13 \times \sin 100}{16.4}\) \(\sin A = \frac{12.803}{16.4}\) \(\sin A = 0.7806\) \(A = \sin^{-1} (0.7806) = 51.32^\circ\) \(\therefore\) Bearing of C from A = 180^\circ - (50^\circ + 51.32^\circ) = 180^\circ - 101.32^\circ = 78.68^\circ \(\approxeq\) 78.7^\circ (3 significant figure). (c) \(\cos 40 = \frac{BD}{13}\) \(BD = 13 \cos 40\) = \(13 \times 0.7660\) = 9.959 km \(\approxeq\) 9.96 km east of B.