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2010 WAEC Mathematics Theory  In the diagram, /AB/ = 8 km, /BC/ = 13 km, the bearing of A...

Mathematics
WAEC 2010

 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.

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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.


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