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Two airplanes leave an airport at the same time, traveling on courses that have an angle of 125° between them. If they each travel 500 miles, how far apart are they to the nearest mile?

1 Answer

4 votes

Answer:

887 mi

Explanation:

You want the distance between two planes that have each traveled 500 miles on courses that are 125° apart.

Law of cosines

The law of cosines can help you find the side c in a triangle when sides a and b are given, along with angle C between them.

c² = a² +b² -2ab·cos(C)

Here, this means ...

c² = 500² +500² -2·500·500·cos(125°) ≈ 786788.2

c = √786788.2 ≈ 887.01

The airplanes are about 887 miles apart.

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Additional comment

The second attachment shows an alternate solution. The distance halfway between the planes will be 500 miles times the sine of half the angle between them. That half-distance is ...

500·sin(62.5°) ≈ 443.5 miles

so the distance between the planes is 887 miles.

Two airplanes leave an airport at the same time, traveling on courses that have an-example-1
Two airplanes leave an airport at the same time, traveling on courses that have an-example-2
User Astropringles
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