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2 years ago

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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.

1 answer:

zaharov [31]2 years ago

6 0

Answer:

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

Step-by-step 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.

Read more on Brainly.com - brainly.com/question/12103659#readmore

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