Question

two ships leave a harbor at the same time. one ship travels on a bearing s11°w at 18 miles per hour. the other ship travels on a bearing n75°e at 9 miles per hour. how far apart will the ships be after 2 hours? the distance is approximately 40.0 miles. (round to the nearest tenth as needed.)

Explanation:

Step1: Calculate the distances traveled by each ship

The first ship travels at a speed of \(18\) miles per hour for \(2\) hours. Using the formula \(d = vt\) (distance = speed×time), its distance \(a=18\times2 = 36\) miles.
The second ship travels at a speed of \(9\) miles per hour for \(2\) hours. Using the formula \(d = vt\), its distance \(b = 9\times2=18\) miles.

Step2: Find the included angle

The angle between their paths: \(180^{\circ}-11^{\circ}-75^{\circ}=94^{\circ}\)

Step3: Use the Law of Cosines

The Law of Cosines is \(c^{2}=a^{2}+b^{2}-2ab\cos C\), where \(a = 36\), \(b = 18\), and \(C = 94^{\circ}\), \(\cos(94^{\circ})\approx - 0.06976\)

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Answer:

The ships will be approximately \(41.4\) miles apart.