Question on: WAEC Mathematics - 1999
The diagram shows the position of three ships A, B and C at sea. B is due north of C such that |AB| = |BC| and the bearing of B from A = 040°. What is the bearing of A from C
A
040o
B
070o
C
110o
D
290o
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Correct Option: D
< ABC = 40° (alternate angles)
\(\therefore\) < ACB = \(\frac{180° - 40°}{2}\)
= 70°
\(\therefore\) Bearing of A from C = 360° - 70°
= 290°
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