A ship leaves its port and sails on a bearing of $\mathrm{N} 40^{\circ} 30^{\prime} \mathrm{E}$, at speed $27.6 \mathrm{mph}$. Another ship leaves the same port at the same time and sails on a bearing of $S 49^{\circ} 30^{\prime} \mathrm{E}$, at speed $23.6 \mathrm{mph}$.
How far apart are the ships after 3 hours? miles (Round to the nearest integer as needed.)

Step 1 ：Convert the bearings of the two ships into degrees. The bearing of the first ship is N 40° 30' E, which is equivalent to 40.5°. The bearing of the second ship is S 49° 30' E, which is equivalent to 180° - 49.5° = 130.5°. The angle between the two ships is therefore 130.5° - 40.5° = 90°.

Step 2 ：Calculate the distance each ship has traveled after 3 hours. The distance is equal to the speed times the time, so the first ship has traveled 27.6 mph * 3 hours = 82.8 miles and the second ship has traveled 23.6 mph * 3 hours = 70.8 miles.

Step 3 ：Use the law of cosines to calculate the distance between the two ships. The formula is \(c² = a² + b² - 2ab \cos(C)\), where a and b are the lengths of the two sides of the triangle (the distances the ships have traveled), C is the angle between the two sides (the difference in their bearings), and c is the length of the third side of the triangle (the distance between the two ships).

Step 4 ：Final Answer: The ships are \(\boxed{109}\) miles apart after 3 hours.