Question

two airplanes leave an airport at the same time. an hour later, the planes are 278 km apart. if one plane has traveled 211 km and the other has traveled 254 km during the hour, find the angle θ between their flight paths. see the figure below.
carry your intermediate computations to at least four decimal places.
round your answer to the nearest tenth of a degree.

Explanation:

Step1: Apply the Law of Cosines

The Law of Cosines formula is \( c^{2}=a^{2}+b^{2}-2ab\cos C\). In this triangle, let \(a = 211\), \(b = 254\), \(c = 278\), and \(C=\theta\). Then \(278^{2}=211^{2}+254^{2}-2\times211\times254\times\cos\theta\).
First, calculate \(278^{2}=77284\), \(211^{2}=44521\), \(254^{2}=64516\).
So, \(77284=44521 + 64516-2\times211\times254\times\cos\theta\).

Step2: Simplify the equation

Simplify the right - hand side: \(44521+64516 = 109037\), and \(2\times211\times254=106988\).
The equation becomes \(77284=109037-106988\cos\theta\).
Rearrange it to get \(106988\cos\theta=109037 - 77284\).
Calculate \(109037-77284 = 31753\). So, \(\cos\theta=\frac{31753}{106988}\approx0.2968\).

Step3: Find the angle \(\theta\)

Take the inverse cosine of \(0.2968\), \(\theta=\cos^{-1}(0.2968)\approx72.7^{\circ}\).

Answer:

\(72.7^{\circ}\)