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.
