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

Cross multiply,

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.

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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.