Question

A ship sails 95 km on a bearing of 140 degree,then a further 102 km on a bearing of 260 degree and then returns directly to its starting point.find the length and bearing of the return journey.

Answer 1

The word bearing has been misused, abused and overused like no tomorrow, in navigation. There are many definitions and reference points. But for the sake of our problem and simplicity, we can assume that we are talking about true bearing. This means, that the angle is in reference to geometric north pole.

So, the direction of motion makes an angle given by the bearing with respect to the geometric north south line. If the ship is pointing north, it will be 0 degrees and increasing clockwise until 359.99.. degrees. Hope that gives a basic idea.

For our problem, let's take the north to be the Y axis, and east to be X-axis. And assume that the ship starts at O (0,0) for the sake of simplicity.

So, the first line makes 140 degrees wrt Y-axis, or makes 50 degress wrt X-axis and is in the fourth quadrant. Now, you can draw the first line, say OA.
Co-ordinates of A:
xa = 95 sin(140) = 95 sin(40) = 61.06km
ya = 95cos(140) = - 95cos(40) = -72.77km

Now, the next leg goes from A to B, having angle of 260 wrt Y axis, or having 80 degrees wrt -Y axis:
Co-ordinates of B:
xb = xa +102*sin(260) = xa - 102*sin(80) = 61.06 - 100.45= - 39.39km
yb = ya + 102*cos(260) = ya - 102*cos(80) =-72.77 -17.71 = -90.48km
So, B is in the third quadrant.
Return journey is from B to O, BO will have bearing in the first quadrant (If you draw a line from B to O and extend it beyond the origin, you will note the angle lies in the first quadrant)

Length of journey L= sqrt(xb ^2 + yb^2) = 98.68 km <== Ans

Let T(theta) be the bearing, yb = L cos , xb = L sinT
tan T = xb/yb = 39.39/90.48
T = 23.52 degrees. <== Answer

NOTE: The other answer is beautiful, except he made a mistake in the final part, since bearing is measured wrt Y axis, the y = L cos beta, and x = L sin beta :)