Question

two ships leave a harbor at the same time. one ship travels on a bearing s14°w at 14 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 □ miles. (round to the nearest tenth as needed.)

Explanation:

Step1: Calculate the distances each ship travels

The first ship travels at a speed of \(14\) miles per hour for \(2\) hours. Using the formula \(d = vt\) (distance = speed×time), the distance \(a=14\times2 = 28\) miles.
The second ship travels at a speed of \(9\) miles per hour for \(2\) hours. So the distance \(b = 9\times2=18\) miles.

Step2: Find the included angle

The angle between the two paths:
The first - ship's bearing is \(S14^{\circ}W\) and the second - ship's bearing is \(N75^{\circ}E\).
The angle \(\theta=180^{\circ}-14^{\circ}-75^{\circ}=91^{\circ}\)

Step3: Apply the Law of Cosines

The Law of Cosines formula is \(c^{2}=a^{2}+b^{2}-2ab\cos\theta\).
Substitute \(a = 28\), \(b = 18\), and \(\theta = 91^{\circ}\) (\(\cos91^{\circ}\approx - 0.01745\)) into the formula:

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Answer:

\(33.6\)