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.
(a) the distance AC ;
(b) the bearing of C from A ;
(c) how far east of B, C is.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option n
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.
(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.