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 33.6 miles. (round to the nearest tenth as needed.)

Explanation:

Step1: Calculate the distances each ship travels

• For the first ship:
• Speed \(v_1 = 14\) miles per hour, time \(t = 2\) hours.
• Using the formula \(d=v\times t\), the distance \(a=14\times2 = 28\) miles.
• For the second ship:
• Speed \(v_2=9\) miles per hour, time \(t = 2\) hours.
• Using the formula \(d = v\times t\), the distance \(b = 9\times2=18\) miles.

Step2: Find the included angle \(\theta\)

• The first - ship's bearing is \(S14^{\circ}W\) and the second - ship's bearing is \(N75^{\circ}E\).
• The included 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:
• \(c^{2}=28^{2}+18^{2}-2\times28\times18\times\cos91^{\circ}\).
• \(c^{2}=784 + 324-1008\times(-0.01745)\).
• \(c^{2}=784+324 + 17.69\).
• \(c^{2}=1125.69\).
• \(c=\sqrt{1125.69}\approx33.6\) miles.

Answer:

The distance is approximately \(33.6\) miles.