Can someone explain to me how to get the answer for this problem?
At 9am, two airplanes leave the airport. Airplane A travels at a bearing of N 53° W at 12 mph. Airplane B travels at a bearing of S 67° W at 16 mph. How far apart will the planes be at noon?
At 9am, two airplanes leave the airport. Airplane A travels at a bearing of N 53° W at 12 mph. Airplane B travels at a bearing of S 67° W at 16 mph. How far apart will the planes be at noon?
-
Angle between directions of planes = 120°
After 3 hours (at noon)
- distance traveled by A = 12 * 3 = 36 miles
- distance traveled by B = 16 * 3 = 48 miles
If distance between two planes at noon is D
Cos 120° = (36² + 48² - D²) / (2 * 36 * 48)
-1/2 = (3600 - D²) / 3456
D² = 5328
D = √5328 = 72.99 miles
After 3 hours (at noon)
- distance traveled by A = 12 * 3 = 36 miles
- distance traveled by B = 16 * 3 = 48 miles
If distance between two planes at noon is D
Cos 120° = (36² + 48² - D²) / (2 * 36 * 48)
-1/2 = (3600 - D²) / 3456
D² = 5328
D = √5328 = 72.99 miles