Boat $\mathrm{A}$ is $20 \mathrm{~km}$ from port on a bearing of $025^{\circ}$ and boat $\mathrm{B}$ is $25 \mathrm{~km}$ from port on a bearing of $070^{\circ}$. Boat B is in distress. What bearing should boat A travel on to reach boat B?

Step 1 ：Given that boat A is 20 km from port on a bearing of \(025^{\circ}\) and boat B is 25 km from port on a bearing of \(070^{\circ}\).

Step 2 ：We can form a triangle with the two boats and the port. The sides of the triangle are 20 km, 25 km, and the distance between the two boats.

Step 3 ：We can use the law of cosines to find the angle between the two boats. The law of cosines is \(c^2 = a^2 + b^2 - 2ab\cos(C)\), where a, b, and c are the sides of the triangle and C is the angle we want to find.

Step 4 ：Substituting the given values into the law of cosines, we get \(c^2 = 20^2 + 25^2 - 2*20*25\cos(45)\). Solving for c, we get \(c \approx 17.83\) km.

Step 5 ：We can then use the law of cosines again to find the angle A. The law of cosines is \(\cos(A) = (b^2 + c^2 - a^2) / (2bc)\).

Step 6 ：Substituting the given values into the law of cosines, we get \(\cos(A) = (25^2 + 17.83^2 - 20^2) / (2*25*17.83)\). Solving for A, we get \(A \approx 52.48^{\circ}\).

Step 7 ：The bearing that boat A should travel on to reach boat B is the sum of the bearing of boat A and the angle A. So, the bearing is \(025^{\circ} + 52.48^{\circ} = 77.48^{\circ}\).

Step 8 ：Final Answer: The bearing that boat A should travel on to reach boat B is approximately \(\boxed{77.48^{\circ}}\).