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
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°