WAEC - Mathematics (1999 - No. 22) | ExamPlay

WAEC - Mathematics (1999 - No. 22)

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

040^o

070^o

110^o

290^o

Explanation

< ABC = 40° (alternate angles)

\(\therefore\) < ACB = \(\frac{180° - 40°}{2}\)

= 70°

\(\therefore\) Bearing of A from C = 360° - 70°

= 290°

Comments (0)

Advertisement