Question

An airplane is at a location 800 miles due west of city X. Another airplane is at a distance of 1,200 miles southwest of city X. The angle at city X created by the paths of the two planes moving away from city X measures 60°. What is the distance between the two airplanes to the nearest mile? Assume that the planes are flying at the same altitude

Answer 1

Use Law of cosines

d = √(800^2 + 1200^2 - 2·800·1200·COS(60°)) = 1058 miles

Answer 2

The distance between the two airplanes (to the nearest mile) is 1058 miles.

Explanation:

An airplane A is at a location 800 miles due west of city X. So AX = 800 miles.

Another airplane is at a distance of 1,200 miles southwest of city X. So BX = 1200 miles.

The angle at city X created by the paths of the two planes moving away from city X measures 60°. So angle ∠AXB = 60°.

In triangle ΔAXB, AX = 800 miles, BX = 1200 miles, ∠AXB = 60°.

Using law of cosines:-

AB² = AX² + BX² - 2 * AX * BX * cos(∠AXB).

AB² = 800² + 1200² - 2 * 800 * 1200 * cos(60°).

AB² = 640000 + 1440000 - 2 * 960000 * 1/2

AB² = 2080000 - 960000

AB² = 1120000

AB = √(1120000) = 1058.300524

Hence, the distance between the two airplanes (to the nearest mile) is 1058 miles.