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 is D
< ABC = 40° (alternate angles)
\(\therefore\) < ACB = \(\frac{180° - 40°}{2}\)
= 70°
\(\therefore\) Bearing of A from C = 360° - 70°
= 290°