Question

a life preserver is seen floating down a river. the life preserver is first spotted 60 feet away. a few moments later the life preserver is 55 feet away, making a 40° angle between the two sightings. how far did the life preserver travel? round to the nearest tenth.

Explanation:

Step1: Apply the Law of Cosines

The Law of Cosines formula is \(c^{2}=a^{2}+b^{2}-2ab\cos C\). Here, \(a = 60\), \(b = 55\), and \(C=40^{\circ}\).
Substitute the values into the formula: \(c^{2}=60^{2}+55^{2}-2\times60\times55\times\cos(40^{\circ})\).
First, calculate \(60^{2}=3600\), \(55^{2}=3025\), and \(2\times60\times55 = 6600\).
\(\cos(40^{\circ})\approx0.7660\).
So, \(c^{2}=3600 + 3025-6600\times0.7660\).

Step2: Simplify the expression

Calculate \(6600\times0.7660 = 5055.6\).
Then \(c^{2}=3600+3025 - 5055.6\).
\(3600+3025=6625\).
\(c^{2}=6625-5055.6=1569.4\).

Step3: Find \(c\)

Take the square root of \(c^{2}\). \(c=\sqrt{1569.4}\approx39.6\).

Answer:

\(39.6\) feet.