A Diving Dilemma!

In the above diagram $A$ represents my ship, $B$ represents the pirate ship, and $C$ represents the treasure.

We were given a precise angle of $53.13010235°$ , and considering it took him $10$ minutes to travel to the treasure at $30$ km/h, then the distance must have been 5 km. So now we know

$\frac{A}{(5)} = \cos 53.13010235° = 0.6$ , $A = 3$ km, and $\frac{O}{5} = \sin 53.13010235° = 0.8$ , $O = 4$ km.

Now, it says that facing in the direction of where I came, I turned $65°$ upwards from the ocean floor in the direction I came from and ascended. Since we know the vertical distance is $3$ km, then we know $\frac{(3)}{A_{2}} = \tan 65°$ , and therefore $A_{2} = \frac{3}{\tan 65°} \approx 1.4$ . Therefore, the distance from one ship to the other is $4 - 1.4 = \boxed{2.6}$ .