Question

A ship sailed from port A to port B on the bearing N65°E a distance of 20km sailed from port B to port C on S25°E, a distance of 240km.
a) Calculate the distance between A and C
b) Find i) Bearing of C from A ii)Bearing of A from C

Answer 1

As always, draw the diagram. You will see that you have a right triangle with legs of 20 and 240, so you can easily figure
(a) the hypotenuse.
(b) the angle at C has tanC = 20/240, so you can easily figure the direction of the hypotenuse. The bearings are 180° apart.

Answer 2

All angles are measured CW from +y-axis.
AC = AB + BC = 20km[65o] + 240km[155o],
X = 20*sin65 + 240*sin155 = 119.6 km.
Y = 20*Cos65 + 240*Cos155 = -209.1 km.

a. AC = sqrt(X^2+Y^2) = 241 km.

b. Tan A = X/Y,
A = -29.8o = 29.8o E. of S. = 150.2o CW = bearing of C from A.
150.2 + 180 = __Degrees = bearing of A from C.