Consider X-axis as West - East.

Consider Y-axis as North - South.

The plane starts from O (0, 0).

First it goes 200 miles due West to A (- 200, 0).

Then it goes to B, say (p, q).

So, AB is 150 miles 60° North of West.

So, p - (- 200) = - 150 * cos 60°

Or, (p + 200) = - 150 * 1 / 2 = - 75

Or, p = (- 200 - 75)

Or, p = - 275

And, q - 0 = - 150 * sin 60°

Or, q = - 150 * (√3) / 2 = - 75√3

Or, q = - 75√3

Thus, final location of the aircraft = B (- 275, - 75√3).

Therefore, magnitude of resultant displacement = |√[{(- 275)^2} + {(- 75√3)^2}]| miles = √(75625 + 16875) miles = √92500 miles ≈ 304.138 miles.

And, direction of resultant displacement = arc tan {(- 75√3) / (- 275)} = arc tan {(3√3) / 11} = 25.284° North of West.

i copied this answer from a forum in quora, dont get me wrong. but the end results varies