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Question
a ferry transports passengers to three different ports. from port a, the ferry travels east 3.5 miles to port b. the ship then turns 75° south and travels for 6 miles to reach port c. what is the distance d from port c to port a? approximate to the nearest tenth of a mile
Response
- First, identify the information given:
- In triangle \(ABC\), \(AB = 3.5\) miles, \(BC = 6\) miles, and \(\angle ABC=180 - 75=105^{\circ}\).
- Then, use the Law - of - Cosines:
- The Law of Cosines states that for a triangle with sides \(a\), \(b\), \(c\) and the angle \(\theta\) opposite to side \(c\), \(c^{2}=a^{2}+b^{2}-2ab\cos\theta\).
- In triangle \(ABC\), if we want to find the length of \(AC = d\), where \(a = 3.5\), \(b = 6\), and \(\theta = 105^{\circ}\), and \(\cos105^{\circ}=\cos(60^{\circ}+45^{\circ})=\cos60^{\circ}\cos45^{\circ}-\sin60^{\circ}\sin45^{\circ}=\frac{1}{2}\times\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}=\frac{\sqrt{2}-\sqrt{6}}{4}\approx - 0.259\).
- According to the Law of Cosines, \(d^{2}=3.5^{2}+6^{2}-2\times3.5\times6\times\cos105^{\circ}\).
- Calculate each part:
- \(3.5^{2}=12.25\), \(6^{2}=36\).
- \(2\times3.5\times6 = 42\).
- \(d^{2}=12.25 + 36-42\times(-0.259)\).
- \(d^{2}=12.25 + 36+10.878\).
- \(d^{2}=59.128\).
- Take the square - root of both sides:
- \(d=\sqrt{59.128}\approx7.7\) miles.
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