In the diagram, /AB/ = 8 km, /BC/ = 13 km, the bearing of A from B is 310° and the bearing of B from C is 230°. Calculate, correct to 3 significant figures,
(a) the distance AC ;
(b) the bearing of C from A ;
(c) how far east of B, C is.
Explanation
(a)
< ABC = 100°
(therefore b^{2} = 8^{2} + 13^{2} – 2(13)(8) cos 100°)
= (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°)
(therefore) Bearing of C from A = 180° – (50° + 51.32°)
= 180° – 101.32°
= 78.68° (approxeq) 78.7° (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.