On sailing ships, the bearing is measured from the North axis, clockwise.
So a bearing of 240 degrees is a ray in the 3rd quadrant that is 240 degrees clockwise from the North axis.
It is actually, S 60deg W.
It is actually at 210 degrees by normal, counterclockwise, measuring of angles.
If you can draw correctly the figure, you'd have 3 ships A, B, C on a straight line that is 30deg above the x-axis or East axis.
With A on the (0,0), the buoy is located at 2.5km, N 25deg E.
Ship B is somewhere that is N 60deg E of A.
Ship C is a lot fartheraway, that is also at N 60deg E of A.
Let point P be the location of the buoy.
In triangle APB,
angle PAB = 90 -25 -30 = 35deg ....say, angle A.
AP = 2.5km ------given.
BP = 2km -------given.
By Law Of Sines,
2.5/sinB = 2/sinA
2sinB = 2.5sinA
sinB = (2.5sinA)/2 = 2.5sin(35deg) / 2 = 0.71697 .....positive sine value.
That means angle B is in the 1st or 2nd quadrant.
If angle B is in the1st quadrant,
angle B = arcsin(0.71697) = 45.8 deg or 46deg
So, angle PBA = 46deg
But that is not possible according to the figure.
So we use the other angle B that is in the 2nd quadrant,
angle B = arcsin(0.71697) = 180 -46 = 134deg --------------**
Then, the bearing of point P from point B:
240 +134 = 374deg
That is more than 360....more by 14 degrees.
Hence, the bearing of the buoy from the ship B is 014 degrees. -----answer.
For the bearing of the buoy from ship C.
Per givens, triangle APC is like triangle APB, in that
angle A = 35deg
AP = 2.5km
CP = 2km
By Law of Sines,
2.5/sinC = 2/sin(35deg)
sinC = 2.5sin(35deg) / 2 = 0.71697 .......positive, same as above.
In the 1st quadrant,
angle C = 46deg --------okay, according to the figure.
So, bearing of point P from point C,
240 +46 = 286 degrees
Hence, the bearing of the buoy from ship C is 286 degrees. ----answer.