Question

select the correct answer.

two airplanes leave an airport at the same time. after one hour, airplane a is 315 kilometers away from the airport, airplane b is 405 kilometers away from the airport, and the airplanes are 285 kilometers apart. what is the approximate angle between the flight paths of the airplanes?

\\(30.1^{\circ}\\)
\\(44.5^{\circ}\\)
\\(50.8^{\circ}\\)
\\(84.8^{\circ}\\)

Explanation:

Identify the triangle parameters

We model the airport as point \(P\), airplane A as point \(A\), and airplane B as point \(B\).
The distances from the airport are:

• \(b = PA = 315\text{ km}\)
• \(a = PB = 405\text{ km}\)
• \(c = AB = 285\text{ km}\)

We need to find the angle \(\theta\) between the flight paths, which is \(\angle APB\).

Apply the Law of Cosines

Using the Law of Cosines to solve for the angle \(\theta\) opposite side \(c\):

$$ c^2 = a^2 + b^2 - 2ab \cos(\theta) $$

Rearranging the formula to solve for \(\cos(\theta)\):

$$ \cos(\theta) = \frac{a^2 + b^2 - c^2}{2ab} $$

Calculate the cosine value

Substitute the given values into the equation:

$$ \cos(\theta) = \frac{405^2 + 315^2 - 285^2}{2 \cdot 405 \cdot 315} $$

$$ \cos(\theta) = \frac{164025 + 99225 - 81225}{255150} $$

$$ \cos(\theta) = \frac{182025}{255150} \approx 0.7134 $$

Find the angle \(\theta\)

Calculate the inverse cosine of the value:

$$ \theta = \arccos(0.7134) \approx 44.48^\circ $$

Rounding to the nearest tenth gives:

$$ \theta \approx 44.5^\circ $$

Answer:

• (A) \(30.1^\circ\)
• (B) \(44.5^\circ\) (Correct answer)
• (C) \(50.8^\circ\)
• (D) \(84.8^\circ\)