Answer
$(40, 41.5)$
Work Step by Step
Step 1. Put axes to the figure given by the Exercise: assume y-axis is along B-A and x-axis is horizontal with the origin at the midpoint between A and B. Assume the ship is at point $C$
Step 2. We can identify the following coordinates: $A(0, 150), B(0, -150), C(40, y)$ where $y$ is an unknown.
Step 3. Calculate the distances: $CB=\sqrt {(40)^2+(y+150)^2}$ and $CA=\sqrt {(40)^2+(y-150)^2}$
Step 4. We know that $CB-CA=80$, thus $\sqrt {(40)^2+(y+150)^2}-\sqrt {(40)^2+(y-150)^2}=80$
Step 5. Rearrange the above the equation as $\sqrt {(40)^2+(y+150)^2}=\sqrt {(40)^2+(y-150)^2}+80$
Step 6. Take the square on both sides to get $(40)^2+(y+150)^2=(40)^2+(y-150)^2+80^2+160\sqrt {(40)^2+(y-150)^2}$
Step 7. Simplify the above equation to get $4\sqrt {(40)^2+(y-150)^2}=15y-160$. Again, take the square on both sides to get $209y^2=16(150^2+40^2)-160^2=360000$
Step 8. We solve for $y$ to get $y\approx41.5$ mi.
Step 9. We conclude that the ship is at a location $(40, 41.5)$ with the origin at the midpoint between A and B.
