Two ships leave a port at 9 a.m. One travels at a bearing of N 53° W at 13 miles per hour, and the other travels at a bearing of S 67° W at 17 miles per hour. Approximate how far apart they are at noon that day. (Round your answer to one decimal place.)

To solve this problem, we need to use the law of cosines. Here are the steps:

Step 1: Determine the time each ship has been traveling. Since they left at 9 a.m. and we're finding the distance at noon, each ship has been traveling for 3 hours.

Step 2: Determine the distance each ship has traveled. The first ship has traveled 13 miles/hour * 3 hours = 39 miles. The second ship has traveled 17 miles/hour * 3 hours = 51 miles.

Step 3: Determine the angle between the paths of the two ships. The first ship is traveling at a bearing of N 53° W, and the second ship is traveling at a bearing of S 67° W. The angle between North and West is 90°, so the first ship is 90° - 53° = 37° from West. The second ship is 67° from West. So, the angle between the two ships is 37° + 67° = 104°.

Step 4: Use the law of cosines to find the distance between the two ships. The law of cosines is c² = a² + b² - 2ab*cos(C), where a and b are the sides of a triangle (the paths of the two ships), C is the angle between them, and c is the length of the third side of the triangle (the distance between the ships). Plugging in the values we have:

c² = 39² + 51² - 23951cos(104°) c² = 1521 + 2601 - 23951(-0.242) c² = 4122 + 953.778 c = sqrt(5075.778)

So, the two ships are approximately 71.2 miles apart at noon.