The diagram shows the position of three ships A, B and C ...
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
040o
070o
110o
290o
Correct answer is D
< ABC = 40° (alternate angles)
∴ < ACB = \frac{180° - 40°}{2}
= 70°
\therefore Bearing of A from C = 360° - 70°
= 290°
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