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.
040^{o} 
B.
070^{o} 
C.
110^{o} 
D.
290^{o}
Correct Answer: Option D
Explanation
< ABC = 40° (alternate angles)
(therefore) < ACB = (frac{180° – 40°}{2})
= 70°
(therefore) Bearing of A from C = 360° – 70°
= 290°