A labellled diagram of the triangle formed is shown below
The bearing of ship C from ship B is the angle x shown in the labelled diagram.
Looking at triangle ABC, we would find BC by applying the law of cosines which is expressed as
a^2 = b^2 + c^2 - 2bcCosA
In this case, we have
BC^2 = AB^2 + AC^2 - 2 * AB * AC * CosA
BC^2 = 150^2 + 275^2 - 2 * 150 * 275 * Cos120
BC^2 = 98125 - 82500Cos120 = 98125 + 41250 = 139375
BC = square root of 139375 = 373.33
We would find angle ABC by applying the sine rule which is expressed as
a/SinA = b/SinB = c/SinC
a = 373.33, b = 275, A = 120, B = angle ABC
Thus, we have
373.33/Sin120 = 275/SinABC
By crossmultiplying, we have
373.33 SinABC = 275Sin120
SinABC = 275Sin120/373.33 = 0.638
We would find angle ABC by finding the sine inverse of 0.638
angle ABC = Sin^-1(0.638)
angle ABC = 39.64
Angle x and angle ABC are linear pairs. This means that their sum is 180 degrees. Thus,
angle x = 180 - 39.64 = 140.36
To the nearest degree, the bearing of ship C from ship B is 140 degrees