Two ships leave at harbor at the same time. When ship travels on a bearing South 13° west at 15 mph. The other ship travels on a bearing north 75° east at 11 mph. How far apart with the ships be after two hours

44.76 miles

Step-by-step explanation:

The First Ship travels at 15 mph from A to C

Distance covered in 2 hours =15 X 2 =30 Miles

The Second Ship travels at 11 mph from A to B

Distance covered in 2 hours =11 X 2 =22 Miles

The path of the two ships forms a triangle ABC where:

AB=22 Miles

AC=30 miles

We want to determine the distance BC between the two ships.

Using Law of Cosines

BC= 44.76 miles

After two hours, the two ships are 44.76 miles apart.

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step-by-step explanation:

if i'm not wrong, i would say that the third angle has (72) because the three sides of all triangles have 180 degrees and if we subtract 180 - 62 - 40 = (72) / but i'm not sure about 40 yd.